JI. 3. 3JlbCfOJibU .li.l1<1><1>EPEHUHAJlbH biE YPABHEHH5I l1 BAPHAUHOHHOE HC4HC.JIEHHE H3.li.ATEJlbCTBO cHAYKA• MOCKBA ---------------L. ELSGOLTS Differential equations and the calculus of variations TRANSLATED FROM THE RUSSIAN BY GEORGE YANKOVSKY MIR PUBLISHERS • MOSCOW First published 1970 Second printing 1973 Third printing 1977 TO THE READER Mir Publishers welcome your comments on the content, translation and design of this book, We would also be pleased to receive an} proposals you care to make about our future publications. Our address is: USSR, 129820, Moscow 1-110, GSP Pervy Rizhsky Pereulok, 2 MIR PUBLISHERS Ha aHuuacKOM 1l3bUCe © English translation, Mir Pn hlishers. 1977 The subject of this book is the theory of differential equations and the calculus of variations. It is based on a course of lectures which the author delivered for a number of years at the Physics Department of the Lomonosov State University of Moscow. Contents Introduction • • • • • • PART ONE DIFFERENTIAL EOUATIONS Chapter 1. First-Order Dillerential Equations 1. First-Order Differential Equations Solved for the Derivative 2. Separable Equations . . . . . . . . . . . . . . • . . . • 3. Equations That Lead to Separable Equations • • ; • • • • 4. Linear Equations of the ·First Order . . . . • • • • • • • 5. Exact Differential Equations . . . . . . . . . . . . . . .·. . • . 6. Theorems of the Existence and Unioueness of Solution of the Equa- dy tton dx =I (.t, y) • • • • . . . . • • . • . • . . • • •. • • 7. Approximate Me~hods of Integrating First-Order· Equations. . . • • 8. Elementary Types of Equations Not Solved for the Derivative .. 9. The Existence and Uniqueness Theorem for Differential Equations Not Solved for the Derivative. Singular Solutions • • • • Problems . • • • . . . . . • • . . • • . . . • • • • • Chapter 2. Differential Equations of the Second Order and Higher 1. The Existence and Uniqueness Theorern for an ntn Order Differential Equation . • . . . . . . . . . . . . . . . . • • • • . . • 2. The Most Elementary Cases of Reducing the Order ..•. ·..• 3. Linear Differential Equations of the nth Order . . . . . . . . . 4. Homogeneous Linear Equations with Constant Coefficients and Euler's E.quations . . . . . . . . . . . . . . . . . . . . . . . . . . 5. Nonhomogeneous Linear Equations . . . . . . . . . . . . . . . 6. Nonhomogeneous Linear Equations with Constant Coefficients and Euler's Equations . . . . . . . . . . . . . . . . . . . . . . . 7. Integration of Differential Equations by Means of Series . . . . . 8. The Small Parameter Method and lts Application in the Theory of Quasilinear Oscillations • . . . . . . . . . • . • • . • . . . . 9. Boundary-Value Problems. Essentials Problems • . • • • . • • • . • . • • • • Chapter 3. Systems of Differential Equations 13 91 19 23 29 32 37 44 66 73 81 88 91 91 93 98 112 119 130 143 153 165 172 176 I. Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . 176 2. Integrating a System of Differential Equations by Reducing It to a Single Equation of Higher Order . . • . . • • • . • • • • • • 179 8 CONTENTS 3. Finding Integrable Combinations . . . . . . . . . . . . . . . . 186 4. Systems of Linear Differential Equations . . . . . . . . . . . . 189 5. Systems of Linear Differential Equations with Constant Coefficients 200 6. Approximate Methods of Integrating Systems of Differential Equa· tions and Equations of Order n . . • . . . . . . . . . . . . 206 Problems . . . . . . . 209 Chapter 4. Theory of Stability 211 211 214 223 229 I. Fundamentals . . . . . . . . . 2. Elementary Types of Rest Points . . . . . . . 3. Lyapunov's Second Method . . . . . . . . . . 4. Test for Stability Based on First Approximation . . . . . ·5. Criteria of Negativity of the Real Parts of All Roots of a Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . 6. The Case of a Small Coefficient of a Higher-Order Derivative 7. Stability Under Constantly Operating Perturbations Problems . . . . . . . . . . . . . . . . Chapter 5. First-Order Partial Differential Equations I. Fundamental! . . . . . . . . . . . . . . .......... 2. Linear and Quasillnear First-Order Partial 3. Pfaffian Equations . . . . . . . . • . 4. First-Order Nonlinear Equations . • . . Problems . . . . . . . . . . . • . . . Differential Equations PART TWO THE CALCULUS OF VARIATIONS Introduction • • • • • 236 238 244 247 251 251 253 265 271 288 293 Chapter 6. The Method of Variations in Problems with Fixed Boundaries 297 I. Variatlon and Its Properties . 2. Euler's Equation . . . . . . x, 297 304 S. Functionals of the Form ~ F (x, y1, y,, ... , Yno y~. y~• ..., y~) dx 3!8 "•4. Functionals Dependent on Higher-Order Derivatives . . . . . . . 321 5. Functionals Dependent on the Functions of Several Independent Variables . . . . . . . . . . . . . . . 325 6. Variational Problems in Parametric Form 330 7. Some Applications 333 Problems • • . • • . . . . • . . . . • 338 Chapter 7: Variational Problems with Moving Boundaries and Certain Other Problems . . . • . . . . . . . . 341 1. An Efernentary Problem with Moving Boundaries . 2. The Moving-Boundary Problem for a Functional of the Form 341 x, ~ F (x, y, z, y', z') dx • • • • • • • • • • • • • • •••• 347 .a:, CONTENTS 9 3. Extremals with Corners 352 4. One-Sided Variations . 360 Problems . . . . . . . 363 Chapter 8. Sufficient Conditions for an Extremum 365 I. Field of Extremals . . . . . . . . . . . • • • • • • 365 2. The Function E (x, g, p, g') . . . . . . . . . . . . . 371 3. Transforming the Euler Equations to the Canonical Form 383 Problems . . . . . . . . . . . . . . . . ~ . . . . • . M1 Chapter 9. Variational Problems Involving a Conditional Extremum 389 1. Constraints of the Form q> {x, g1, y1 , •• • , Yn)=O . • • • • . • • • 389 2. Constraints of the Form q> (x, Y~t y,, ..., Yn• y;, g~, ••• , y,.)=O 396 3. lsoperlmetrlc Problems • • . • . 399 Problems . . . . . . . • . • • • 407 Chapter 10. Direct Methods In Variational Problems 408 1. Direct Methods . . . . 2. Euler's Finite-Difference S. The Ritz Method . . 4. Kantorovich's Method Problems .• Answers to Problems . • Recommended literature Index • • • • • • • • • Method ..... ...... 408 409 411 420 427 429 436 431 PART ONE Differentia1 equations Introduction In the study of physical phenomena one is frequently unable to find directly the laws relating the quantities that characterize a phenomenon, whereas a relationship between the quantities and their derivatives or differentials can readily be established. Oue then obtains equations containing the unknown functions or vector functions under the sign of the derivative or differential. Equations in which the unknown function or the vector function appears under the sign of the derivative or the differential are called differential equations. The following are some examples of differential equations: (I) ~~ = -kx is the equation of radioactive disintegration (k is the disintegration constant, x is the quantity of undisintegrated substance at time t, and ~ is the rate of decay proportional to the quantity of disintegrating substance). (2) m :ts:= F ( t, r, :;) is the equation of motion of a particle of mass m under the influence of a force F dependent on the time. the position of the particle (which is determined by the radius vector r), and its velocity : The force is·equal to the product of the mass by the acceleration. (3) ~~+:;~+~~ = 4np (x, y, z) is Poisson's equation, which for example is satisfied by the potential .u (x, y, z) of an electrostatir field, p(x, y, z) is the charge density. The relation between the sought-for quantities will be found if methods are indicated for finding the unknown functions which are defined by differential equations. The finding of unknown functions defined by differential equations is the principal task of the theory of differential equations. If in a differential equation the unknown functions or the vector functions are functions of one variable, then the differential equation is called ordinary (for example, Eqs. I and 2 above). But if the unknown function appearing in the differential equation is a function of two or more independent variables, the differential equation is called a partial diOerential equation (Eq. 3 is an instance). 14 I. DIFFERENTIAL EQUATIONS The order of a differential equation is the highest order of the derivative (or differential) of the unknown function. A solution of a differential equation is a function which, when substituted into the differential equation, reduces it to an iOentity. To illustrate, the equation of radioactive disintegration dx dt= -kx (I. I) has the solution where c 'is an arbitrary constant. It is obvious that the differential equation (1.1) does not yet fully determine the law of disintegration x = x (t). For a full determination, one must know the quantity of disintegrating substance X0 at some initial instant of time t0 • If x0 is known, then, taking into account the condition x(t0)=X0 from (1.1 1), we find the law of radioactive disintegration: x = xoe-k. The procedure of finding the solutions of a differential equation is called integration of the differential equation. In the above case, it was easy to find an exact solution, but in more complicated cases it is very often necessary to apply approximate methods of integrating differential equations. Just recently these approximate methods still led to arduous calculations. Today, however, highspeed computers are able to accomplish such work at the rate of several hundreds of thousands of operations per second. Let us now investigate more closely the above-mentioned more complicated problem of finding the law of motion r = r (t) of a particle of mass m under the action of a specified force F (t, r, r). By Newton's law, mr=F (t, r, r). (1.2) Consequently, the problem reduces to integrating this differential equation. Quite obviously, the law of motion is not yet fully defined by specifying the mass m and the force F; one has also to know the initial position of the particle and the initial velocity r (to)= ro r(to)= ro. We shall indicate an extremely natural approximate method for solving equation (1.2) with Initial conditions (1.21) and (1.22); the INTRODUCTION 15 idea of this method can also serve to prove the existence of a solution of the probl~m at hand. We take the interval of time t0 ::::; t::::; T over which it is required to find a solution of the equation (1.2) that will satisfy the initial conditions (1.21) and (1.22) and divide it into n equal parts of length h= T-to: n (t0 , 11], {tl> 12], ••• , [tn-l• T], where (k= I, 2, ..., n-1). For large values of n, within the limits of each one of these small intervals of time, the force F (t, r, r) changes but slightly (the vector function F is assumed to be continuous); therefore it may be taken, approximately, to be constant over every subinterval [tk_ 1, tk], for instance, equal to the value it has at the left-hand boundary point of each subinterval. More exactly, on the subinterval (t0 , 11] the force F (t, r, r) is considered constant ·and equal to F (10 , r0 , r0 ). On this assumption, it is easy, from (1.2) and the initial conditions (1.21) and (1.2~), to determine the law of motion rn(t) on the subinterval [t9, t1 ] (the motion will be uniformly variable) and, hence, in particular, one knows the values of rn (t1) and rn (t1). By the same method, we approximate the law of motion rn (t) on the subinterval (t1, t2] considering the force F as constant on this subinterval and as equal to F (t1, rn (t1), rn (/1)). Continuing this process, we get an approximate solution rn (t) to the posed problem with initial conditions for equation (1.2) over' the whole interval [t 0 , T]. It is intuitively clear that as n tends to infinity, the approximate solution rn (t) should approach the exact solution. Note that the second-order vector equation (1.2) may be replaced by an equivalent system of two first-order vector equations if we regard the velocity v as the second unknown vector function: dr dv at= v, dt = F (t, r, v). (1.3) Every vector equation in three-dimensional space may be replaced by three scalar equations by projecting onto the coordinate axes. Thus, equation (1.2) is equivalent to a system of three scalar Pquations of the second order, and system (1.3) is equivalent to a system of six scalar equations of the first order. Finally, it is possible to replace one second-order vector equation (I.~) in three-dimensiona I space by one vector equation of the first ·•rdt·r in six-dimensional space, the coordinates here being rJI;, r1 , r:& 16 I. DIFFERENTIAL EQUATIONS of the radius vector r (t) and vx, vy, v, of the velocity vector v. Phase space is the term physicists use for this space. The radius vector R(t) in this space has the coordinates (rx, ry, r,, vx, vy, v,). In this notation, (1.3) has the form _ dR di=O y'>O Fig. I·" Fig. 1-6 Let us now det.~rrriine the signs of the second derivative in various regions of the plane: y"=xy'+y or y"=x(l+xy)+y=x+(x'+l)y. The curve x +(x2 +1) y = 0 or X y=- l+xz (1.1) (Fig. 1.7) partitions tht> plane into two parts, in one of which y" < 0, and, hence, the integral curves are convex upwards, and y Fig. 1-7 in the other y" > 0, and thus the integral curves are concave upwards. When passing through the curve (1.1) the integral curves pass from convexity to concavity, and, consequently, it is on this curve that the points of intlection of the integral curves are located. I. FIRST ORDER O:FFERENTIAL EQUATIONS 23 As a result of this investigation we now know the regions of increase and decrease of the integral curves, the position of the points of maximum and minimum, the regions of convexity and concavity and the location of the inflection points, and also the isocline k =I. This information is quite sufficient for us to sketch the locations of the integral curves (Fig. I.8), but we could dra\\< a few more isoclines, and this would enable us to specify more accurately the location of the y integral curves. In many problems, for instance in almost all problems of a geometrical nature, the variables x and y are absolutely equivalent. It is therefore natural in such problems, if they reduce to solving a differential equation z dy Idx = (x, y), (1.2) to consider, alongside equation (1.2), also the equation dx 1 dy = f (x, y) · (1.3) If both of these equations Fig. t-8 are meaningful, then they are equivalent, because if the function y = y (x) is a solution of the equation (1.2), then the inverse function x = x (y) is a solution of (1.3), and hence, (1.2) and (1.3) have common integral curves. But if at certain points one of the equations, (I .2) or (1.3), becomes meaningless, then it is natural at such points to replace it by the other equation. For instance, ~! = ~ becomes meaningless at x = 0. Replacing it by the equation ~= = ; , the right side of which is already meaningful at x = 0, we find another integral curve x = 0 of this equation in addition to the earlier found solufions y =ex (see page 20). 2. Separable Equations Differential equations of the form {, (y) dy = {1 (x) dx are called equations with separated variables. and {1 (y) will be considered continuous. (1.4) The functions 11 (x) 24 I. DIFFERENTIAL EQUATIONS Assume that y (x) is a solution of this equation; then by substituting y(x) into (1.4) we get an identity, which when integrated yields (1.5) where c is an arbitrary constant. We obtained a finite equation (1.5) which is satisfied by all the solutions of (1.4); note that every solution of (1.5) is a solution of (1.4), because if some function y(x) when substituted reduces (1.5) to an identity, then by differen~iating this identity we· find that y (x) also satisfies the equation (1.4). The finite equation ci> (x, y) = 0, which defines the solution y (x) of the differential equation as an implicit function of x, is called the integral of the differential equation under study. If the finite equation defines all solutions of a given differential equation without exception, then it is called the complete (general) integral of that differential equation. Thus, equation (1.5) is the complete integral of equation (1.4). For (1.5) to define y as an implicit function of x, it is sufficient to require that {, (y) =1= 0. It is quite possible that in certain problems the indefinite integrals ~ {1 (x) dx and ~ {, (y) dy will not be expressible in terms of elementary functions; nevertheless, in this case as well we shall consider the problem of integrating the differential equation (1.4) as completed in the sense that we have reduced it to a simpler problem, one already studied in the course of integral calculus: the computation of indefinite integrals (quadratures). * If it is required to isolate a particular solution that satisfies the condition y (x0) = y0 , it will obviously be determined from the equation II X ~ !, (y) dy = ~ {1 (x) dx, Yo Xo which we obtain from y X ~ {,(y)dy= ~ {, (x)dx+c, Yo Xo taking advantage of the initial condition y (x0 ) =Yo· • Since the term 'integral' in the theory of ditTerential equations is often used in the meaning of the integral of a differential equation, the term •quadrature' is ordinarily used to avoid confusion when dealing with integrals of the functions ~ f (x) dx. I. FIRST-ORDER DIFFERENTIAL EQUATIONS 25 Example J; xdx+ydy=O. The vadables are separated since the coefficient of dx is a function of x alone, whereas the coefficient of dy is a function of y alone. Integrating, we obtain ~ x dx + ~ y dy = c or x2 + y2 = ct, which is a family of circles with centre at the coordinate origin (compare with Example 2 on page 20). Example 2. Integrating, we get erdx= ldy • ny SeX. dx = S~~YY +c. The integrals SeX2 dx and S1~YY are not expressible in terms of elementary functions; nevertheless, the initial equation is considered integrated because the problem has been reduced to quad- ratures. Equations of the type 2 (X) "¢2 (y) dy in which the coefficients of the differentials break up into factors depending solely on x and solely on y are called differential equations with variables separable, since by division by '¢1 (y) q>2 (x) they may be reduced to an equation .with separated variables: (jl1 (xl dx = '1'2 (y) dy (jl2 (x) ¢, (y) · Note that division by "¢1 (y) q>2 (x) may lead to loss of particular solutions that make the product "¢1 (y) ·q>2 (x) vanish, and if the functions "¢1 (y) and q>2 (x) can be discontinuous, then extraneous solutions converting the factor '1'1 (y) (jl2 (x) to zero may appear. 26 I. DIFFERENTIAL EQUATIONS Example 3. ~ = ~ (compare with Example 1, page 20). Separate the variables and integrate: dy =dx 5dy= 5dx y X ' y X ' In Iy I= In (x) +Inc, c > 0. Taking antilogarithms, we get IYI=clx/. If we speak only of smooth solutions, the equation Iy 1= c x , where c > 0, is equivalent to the equation y =±ex or y = c1x, where c1 can take on either positive or negative values, but c1 =1= 0. If we bear in mind, however, that in dividing by y we lost the solution y= 0, we can take it that in the solution y = c1x the constant c1 also assumes the value c1 = 0, in which way we obtain the solution y = 0 that was lost earlier. Note. If in Example 3 we consider the variables x and y to be equivalent, then equation ~ = ~ , which is meaningless at x = 0, must be supplemented by the equation : = ; (see page 23), which obviously also has the solution x = 0 not contained in the solution y = c1x found above. Example 4. x(l +y')dx-y(l +x1 )dy=0. Separate the variables and integrate: ydy xdx 5ydy 5xdx l+y2 =l+x2 ; l+y1 = l+x•+c; In (I+ y1) = ln{l +x1)+ Inc1; I+ y• = c1 (I +x1). Example 5. dx 1 r.:Tt= 4t r x. Find the solution x(t) that satisfies the condition x{l) =I. Separating variables and integrating, we have % t f 2 ~x-=S2tdt. Vx=t1 , x=t·. I I Example 6. As was mentioned in the Introduction, it has been established that the rate of radioactive decay is proportional to the quantity x of substance that has not yet decayed. Find x as I. FIRST-ORDER DIFFERENTIAL EQUATIONS 27 a function of the time t if at the initiaJ instant t = 10 we have X=X0 . The constant of proportionality k, called the decay constant, is assumed known. The differential equation of the process will be of the form dx dt=-kx (1.6) (the minus sign indicates a decrease in x as t increases, k > 0). Separating the variables and integrating, we get dx d-x=-k t; In lxl-lnlxoI=- k(t-10) and then Let us also determine the half-life 't (that is, the time during which x0 /2 decays). Assuming t- 10 = 't, we get x0 /2 = x0e-kt, ln2 whence 't = T . Not only radioactive disintegration, but also any other monomolecular reaction is described on the basis of the mass action law, by the equation : =- kx. where x is the quantity of substance that has not yet reacted. The equation dx Tt=kx, k> 0, (1.7) which differs from (1.6) only in the sign of the right side, describes many processes of multiplication, like the multiplication of neutrons in nuclear chain reactions or the reproduction of bacteria on the assumption of an extremely favourable environment, in which case the rate of reproduction will be proportional to the number of bacteria present. The solution of (1.7) that satisfies the condition x{t0)=x0 is of the form x = x0e1nt -tol and, unlike the solutions of (1.6), x (t) does not diminish but increases exponentially as t increases. Example 7. dp dqJ = p (p-2) (p-4). Draw the integral curves without integrating the equation; p and 'P are polar coordinates. The equation has the obvious solutions p = 0, p = 2, and p = 4. For 0 < p < 2, dp >0; for 2 < p < 4, . ddp < 0 and for p> 4, aqJ qJ ~~ >0. 28 l. DIFFERENTIAL EQUATIONS Consequently, the integral curves are the circles p = 2 and p = 4 and the spirals that wind around the circle p = 2 as

-z. Note that the right side of the homogeneous equation is a homogeneous function of the variables x and y (zero degree of homogeneity) and so an equation of the form M (x, y)dx+N (x, y)dy=O I. FIRST-ORDER DIFFERENTIAL EQUATIONS 31 ------------------------------------------------- will be homogeneous if M (x, y) and N (x. y) are homogeneous functions of x and y of the same degree, because in this case dy M (x, y) f( yx ) • dx=- N(x, y) Example 3. dy = .!L + tan .!L . dx X X Putting y = xz, :~ = x ~ +z and substituting into the initial equation, we have dz cos z dz dx x ilx + z = z +tan z, SfilZ = x , In Isin z I= In IxI+ Inc, sin z =ex, sin .!L =ex.X Example 4. (x +y) dx-(y-x) dy = 0. Putting y = xz, dy = xdz + zdx, we get (x +xz) dx-(xz-x) (x dz + z dx)·= 0, (1 +2z-z'.a)dx+x(l-z)dz=0, (I - z) dz dx I z I 1+ 22 _ 22 +-;=0, 2lnl1+2z-z l+lnlxi= 2 Jnc, x2 (1 +2z-z2 )=c, x2 +2xy-y2 =c. Equations of the form dy = f(a1x+bty+ct) (1.8) dx a2x-t-baY+c~ are converted into homogeneous equations by translating the origin of coordinates to the point of intersection (x1, y1 ) of the straight lines a1x+b1y+c1 =0 and a2x+b2y+c2 =0. Indeed, the constant term in the equations of these lines in the new coordinates X= x-x1, Y = y-!h will be zero, the coefficients r h.l dy dY o the running coordinates remain unchanged, w 1 e dx = dX The equation (1.8) is transformed to ~y =f (a1X +b1Y)dX a2 X+b2Y or dY (a'+b, ~) ( y ) dX =f Y =q> X a2-t-b2}[ and is now a homogeneous equation. 32 I. 0 IFFERENTIAL EQUATIONS This method cannot be used only when the lines a1x +b1y + + c1 = 0 and a2x + b2y +C2 = 0 are parallel. But in this case the coefficients of the running coordinates are proportional; ~ = ~ = k a1 .b1 and (1.8) may be written as dy_f( a1x+b1y+c1 )-F(ax+b) dx- k(a1x+b1y)+c2 - 1 . 1Y' and consequently, as indicated on page 29, the change of variables z = a1x+b1y transforms the equation under consideration into an equation with variables separable. Example 5. dy x-y+l dx= x+y-3 · Solving the system of equations' x-y+ I =0, x+ y-3=0, we get x1 = I, y1 = 2. Putting x =X+ I, y = Y +2, we will have dY X-Y dX = X+Y. y The change of variables z=X or Y = zX leads to · the separable equation z+ X ~ = 1-z (I +z) dz = dX dX I+ z' l-7z-z2 X ' -; In II-2z-z~ I= In IX 1- ~ Inc, (l-2z-z:l) X2 =c, X2 -2XY-Y2 =c, x2 -2xy-y2 +2x+6y=c1 • 4. Linear Equations of the First Order A first-order linear differential equation is an equation that is linear in the unknown function and its derivative. A linear equation has the form :~ + p (x) y = f(x), (1.9) where p (x) and f(x) wi II henceforward be considered continuous functions of x in the domain in which it is required to integrate equation (1.9). If f (x) ==0, then the equation (L9) is called homogeneous linear. The variables are separable in a homogeneous linear equation: dy dy iiX +p (x) y = 0, whence Y = -p (x) dx, I. FIRST·ORDER DIFFERENTIAL EQUATIONS 33 and, integrating, we get In Iy I=- ~ p (x) dx +In c1, c1 > 0, -Jp lXI dx ....L. Oy=ce , C-r- , (1.10) ln dividing by y we lost the solution y ==0, however it can be included in the set of solutions (1.10) if we assume that c can take the value 0 as well. The nonhomogeneous linear equation ~ ~~ + p(x)y=f(x) (1.9) may be integrated by the so-called method of variation of parameters. ln applying this method, one first integrates the appropriate (having the same left-hand member, that is) homogeneous equation dy dx +p (x) y = 0, the general solution ol which, as already indicated, is of the form -J' p (%) dx y=ce . Given a constant c, the function ce-f P 1"1d> is a solution of the homogeneous equation. Let us now try to satisfy the nonhomogeneous equation considering c as a function oi .-:, that is actually perfor· ming the change of variables () -fp(X)d% y=cxe • where c (x) is a new unknown function of x. Computing the derivative dy t.JL - J0 IX) dx ( ) ( ) - f P (X) dx dx=die -cxpxe and substituting it into the original nonhomogeneous equation (I.9), we get de -j'ocx)d.t () () -foox)dJ<+ () () -j'p(x)dJ< /() die -cxpxe pxcxe = x or de-l<) J/)t.t)dJ< dx- x e , whence, integrating we find S J/) 1%1 dJ< c(x) = I (x) e dx +c1; 34 I. DIFFERENTL<\L EQUATION and consequently _ {) -Jp (x) dx _ - Jp (X) dx + -Jp (X) dx 5f{ ) JP (X) dxd ( 1 11 )y- c x e - c1e e x e x. . To summarize: the general solution of a nonhomogeneous linear equation is the sum of the general solution of the corresponding homogeneous equation -Jp (X) dx c1e and of the particular solution of the nonhomogeneous equation e-JP dx sf(x) efP (x> dxdx obtained from (1.11) for C1 =0. Note that in specific cases it is not advis~ble to use the cumbersome and involved formu Ia (1.11 ). It is much easier to repeat each time all the calculations given above. Example I. dy y 2 ---=X dx x · Integrate the corresponding homogeneous equation dy_.JL=o '!1!=~ lnlul=lnlxl+lnc, u=cx.dx x y x' Consider c a function of x, then y = c(x) x, fx = ~ x + c(x) and, substituting into the original equation and simplifying, we &et de x1 dxx=r or dc=xdx, c(x)= 2 +c1 • Hence, the general solution is Example 2. ddy-ycot x = 2x sin x. X . Integrate the corresponding homogeneous equation dy- y cot x = 0 dy = c~s x dx dx ' y Stn X ' In Iy I= In Isin xI+ Inc, y = c sin x. We vary the constant y = c(x) sin x, y' = c' (x) sin x +c(x) cos x. I. FIRST-ORDER DIFFERENTIAL EQUATIONS 35 ------- Substituting into the original equation, we get c' (x) sin x + c (x) cos x- c(x) cos x = 2x sin x, c' (x) = 2x, c(x)= x2 +c1, y=x2 sinx+c1 sinx. Example 3. In an electric circuit with self-inductance, there occurs a process of establishing alternating electric current. The voltage V is a given function of the time V = V (t); the resistance R and the self-inductance L are constant; the initial current is given: I (0) =I0 • Find the dependence of the current I= I (t) on the time. Using Ohm's law for a circuit with self-inductance, we get dl V-Ld1=RI. The solution of this linear equation that satisfies the initial condition I (0) =I0 has by (1.11) the form l~e-:'[1,+ qV(f)e~ 1dtJ. (1.12) For a constant voltage V = V0 , we get R 1=\;;+(10-~)e-r'. An interesting case is presented by a sinusoidally varying voltage V=Asinffit. Here, by (1.12), we get _!it( As1 !it )I =e L I 0 +I 0 eL sin ffii dt . The integral on the right side is readily evaluated. Numerous differential equations can be reduced to linear equations by means of a change of variables. For example, Bernoulli's equation, of the form ~~ +p (x) y = f (x) yn, n =F 1 or y-n :~ +P (x) yl-n = f (x), (1.13) is reduced to a linear equation by the change of variables y1 -n = z. Indeed, differentiating y1 -n·= z, we find (I-n) y-n ~~=:~and, sub· :r·· 36 I. DIFFERENTIAL EQUAl IONS stituting. into (1.13), we get the Iinear equation I dz f-1 -d- +p (X) Z= (X).-n x Example 4. and further as in Example 1 on page 34. Equation : +p (x) y +q (x} y2 = f (x) is calied Riccati's equation and in the general form is not integrable ~ by quadratures, but may be transformed into Bernoulli's equation by a change of variable if a single particular solution y1 (x) of this equation is known. Indeed, assuming y = y1 +z, we get y; +z' + p (x) (y1 +z) +q (x) (y1 +z)2 = f (x) or, since y; + p (x) y1 + 4 (x) y; == f (x), we will have the Bernoulli equation Example 5. Z1 + [p (x) + 2q (x} y,J z +q(x) z2 = 0. dy 2 2 dx=Y- xa· In this example it is easy to choose a particular solution y1 =+. p tt' + I t , 1 I 1 I ( + I )a :2u mg y= z x , we ge y = z - x2 , z - x2 = z x - x2 or z' = z2 +2 ~, which is Bernoulli's equation.X z' 2 I du z' -=-+ 1 u=-z, -= --z2 xz ' dx z2 ' du = - 2u- 1 du = - 2dx In Iu I= -2 In IxI + In c,dx x 'u x c c (x) U=x2 U=~ X3 C1 X c(x)=-T+cl, u=x2-'?f' I c1 X I ::!x2 --1 = x2 -3' u=-x+ c,-x8 • y--x I. FIRST-ORDER DIFFERENTIAL EQUATIONS 37 5. Exact Difierential Equations It may happen that the left-hand side of the differential equation M(x, y)dx+N(x, y)dy=O (1.14) is the total differential of some function u (x, y): du (x, y) = M (x, y) dx+N (x, y)dy and hence equation (I .14) takes the form du(x, y)=O. lf the function y(x) is a solution of (1.14), then du (x, y (x)) =0 and consequently u (x, y (x)) =c. (I. 15) where c is a constant, and conversely, if some function y (x) reduces the finite equation (1.15) to an identity, then, by differentiating the ic'...ntity, we get du (x, y (x)) = 0, and hence u (x, y) = c, where c is an arbitrary constant, is the complete integral of the original equation. If the initial values y (x0 ) =Yo are given, then the constant c is determined from (1.15), c=u(x0 , y0), and u (x, y) = u (x0 , y0) (1.151) is tl ~ desired particular integral. If ~~ = N (x, y) =1= 0 at the point (x0 , y0), then equation (1.151) defines y as an implicit function of x. For the left-hand side of (1.14) M(x, y)dx+N(x, y)dy to be the total differential of some function u (x, y), it is necessary and sufficient, as we know, that oM (x, y) _oN (x, y) (l.IB) oy = ox If this condition, first pointed out by Euler, is fulfilled, then (1.14) is readily integrable. Indeed, du = Mdx +Ndy. On the other hand, d ou ouU=-dx+-dy.ox oy Consequently, ou ou ax= M (x, y); oy = N (x, y), 38 I. DIFFERENTIAL EQUATIONS whence u (x, y) = ~ M (x, y) dx +c (y). When calculating the integral ~ M (X, y) dx, the quantity y is regarded as a constant, and so c(y) is an arbitrary function of y. To determine the function c (y) we differentiate the function u (x, y) with respect to y and, since ~~ = N (x, y), we have ~(S M(x, y)dx)+c'(y)=N(x, y). From this equation we determine c' (y) and, integrating, find c (y). !/ (:c,g} !/ (:JJo,g} (X.!/) ~ I(.z'g'!lo) (.z:flu } .27 f.ZOo·fluJ :c fl fl Fig. 1- 10 As is known from the course of mathematical analysis, it is still simpler to determine the function u (x, y) from its total differential du = M (x, y) dx +.N (x, y) dy, taking the line integral from M (x, y) dx +N (x, y) dy. between some fixed point (x0 , y0) and a point with variable coordinates (x, y) over any path: (X, y) u(x, y)= ~ M (x, y)dx+ N(x, y)dy. (Xo, Yol In most cases, it is convenient to take for the path of integration a polygonal line consisting of two line segments parallel to the coordinate axes .(Fig. 1.10). In this case (X, Yl (X, Yol (X, Y) ~ M dx +N dy = ~ M dx + ~ N dy (Xo, Yol (Xo. Yol (X, Yol or (X, y) P'o. yl \X, Yl ~ Mdx+Ndy= ~ Ndy+ ~ Mdx. tx•• Yo) (Xu, Yul (xe. Y) I. FIRST-ORDER DIFFERENTIAL EQUATIONS 39 Example 1. (x+u+ l)dx+(x-y2 +3) dy=O. The left-hand member of the equation is the total differential of some function u (x, y), since a(x+y+ I)_ a(x-yt+3) ay = ax au x2 ax=x+u+ I, U= 2 +xu+x+c(y), ~==x+c' (y), x+c' (y) =x-y2 +3, c'(y)= -y2 +3, c(y)= _u; +3y+c11 xs y3 U=2+xy+x- a+3y+c1• Hence, the complete integral is of the form 3x2 +6xy+6x-2y8 + 18y=c2 • (1.17) A different method may also be used to determine the function u(x, y)1 tx. II) u (x, y) = ~ (x+ y+ 1)dx+(x-y2 +3)dy. (Xo, Yo) For the initial point (x0 , y0 ) we choose, for instance, the origin of II Fig. 1-11 coordinates, and we take the path of integration as shown m Fig. 1.11 (polygonal line). Then (x. 0) (x. y) u(x, y)= 5(x+l)dx+ 5(x-y2 +3)dy=i+x+xy-~ +3y (0, 0) (x. 0) 40 -- I. DIFFERENTIAL EQUATIONS and the complete integral is of the form xz ys 2 +x+xy-3 +3y=c or as in (1.17). In certain cases when the left-hand side of the equation M(x, y)dx+N(x, y)dy=O (1.14) is not the total differential, it is easy to choose a function J.t (x, y) such that after multiplying by it the left side of (1.14) is transformed to the total differential du = J.~.M dx + J.LN dy. Such a function J.t is called an integrating factor. Observe that multiplication by the integrating factor J.t (x, y) can lead to the appearance of extraneous particular solutions that reduce this factor to zero. Example 2. xdx+ ydy+ (x2 + y2 )x2 dx = 0. It is obvious that multiplying by the factor J.t = 2 +1 2 makesX g left-hand member a total differential. Indeed, multiplying I J.t= 2 + 2 , we getX g xdx+ydy + 2 d =O x2+g2 X X the by or, integrating, ~ ln(x2 +y2 ) +x; =In c1• Multiplying by 2 and then taking antilogarithms, we will have .!_ x• (x2+y2)e3 =c. Of course it is not always so easy to find the integrating factor. In the general case, to find the integrating factor it is necessary to choose at least one particular solution (not identically zero) of the partial differential equation iJ!J.M O!J.N ay-=ax-· or in expanded form :: M + J.t aa~ = :: N +~~ J.t, which, when it is divided by J.l. and certain terms are transposed, yields ~~M-iJln!J. N-aN _aM iJy ax - iJx ay • (1.18) I. FIRST-ORDER DIFFERENTIAL EQUATIONS 41 In the general case, integrating this partial differential equation is by no means an easier task than integrating the original equation, though in some cases a particular solution to equation (l.J8) may easily be found. Besides, if we consider that the integrating factor is a function solely of one argument (for example, only of x+ y or x2 +y1, or a function of x alone, or of y only, and so forth), we can then easily integrate the E'quation (1.18) and indicate the conditions under which an integrating factor of the form under consideration exists. In this way, classes of equations are isolated for which an integrating factor is readily found. For example, let us find the conditions for which the ttquation M dx +N dy = 0 has an integrating factor dependent solely on x, 11 = J.t (x). Equation (1.18) is then simplified to _dIn 1.1. N =oN_ oM dx ox oy ' oM oN whence, taking Ty~ax to be a continuous function of x, we get oM oN r oy- ox In!'= J N dx+ Inc, iJM iJN r iJy- iJx dx J.I.=CeJ N • (l.J9) We can take c=l, since it is sufficient to have only one integrating factor. oM oN If Ty:Tx is a function of x alone, then there exists an integrating factor dPpendent solely on x and equal to (1.19), otherwise there does not exist an integrating factor of the form J.t (x). The condition for the existence of an integrating factor dependent solely on x is, for example, fulfilled for the linear equation ~!+p (x) y =I(x) or [p (x) y- I (x)] dx + dy = 0. oM oN Indeed, au:Tx = p (x) and hence 11 =efp dx Quite analogously we can find the conditions for the existence of integrating factors of the form 11 (y), 11 (x ± y), 11 (x2 ± y2), J.t (x ·y), 11 ( ~), and so forth. 42 I. DIFFERENTIAL EQUATIONS Example 3. Does the equation xdx +ydy+xdy-ydx = 0 (1.20) have an integrating factor of the form 1.1. = 1.1. (r +y2)? Put x2 +y2 = z. Equation (1.18), for 1.1. = 1.1. (x1 +y2) = J.' (z), takes the form and from this or where 2 (My-Nx) dIn J& =aN_aM az ax ay lnlJ.LI=! Sq>(z)dz+lnc _21 scp (z) dz J.L=ce ' aN aM ax -ay q> (z) = M N •y- X (1.21) For the existence of an integrating factor of the given form, it Is necessary and, on the assumption of the continuity of q> (z), suffi. aN aM cient that ! -: be a function of x2 +y2 alone. In that casey- X ~N aM -ax-ay 2 My-Nx =-x2+y2 and hence the integrating factor 1.1. = 1.1. (x2+y2) exists and is equal to (1.21). For c=l we have -s~ I I J.L=e z =-=--z x2+g2. Multiplying (1.20) by J.lo = x2~Y2 reduces it to the form xdx+ydy +xdy-ydx =O xz+y2 x2+y2 or I. FIRST-ORDER DIFFERENTIAL EQUATIONS Integrating, we get In Vx2 +y2 = -arctan .!L -t IncX and after taking antilogarithms we will have V-2- -2 arctan.!!... x+y=ce x. 43 or in polar coordinates p =ce-Q), i. e. a family of logarithmic spirals. Example 4. Find the shape of a mirror that reflects, parallel to a given direction, all the rays emanating from a given point. Locate the origin at a given point and direct the axis of abscissas parallel to the direction given by hypothesis. Let a ray fall on the mirror at the point M (x, y): Consider (Fig. 1.12) the section of the z FIJ. 1-12 mirror (cut by the xy-plane) that passes through the axis of abscissas and the point M. Draw a tangent MN to the section of the surface of the mirror at the point M (x, y). Since the angle of incidence of the ray is equal to the angle of reflection, the triangle MNO is an isosceles triangle. Therefore, t ' y an IJl =y = x+ yxa+yt • The homogeneous equation thus obtained is readily integrable by the change of variables X y=Z, but a still easier way is to rationalize the denominator and write the equation as xdx+ydy= v·x'+y2 dx. The equation has the obvious integrating factor J.L 1 x dx +y dy _ dx Vxa +y' =X +~. = Jlxa+ya ' yxa+ya - ' y1 =2cx +c' (a family of parabolas). l4 I. DIFFERENTIAL EQUATIONS Note. This problem is still more easily solved in the coordinates " and p, where p = Vx2 +y2 ; here the equation of the section of desired surfaces takes the form dx=dp, p=x+c. One can prove the existence of an integrating factor, or, what is the same thing, the existence of a nonzero solution of the partial differential equation (1.18) (see page 40) in a certain domain if the functions M and N have continuous derivatives and if at least one of these functions does not vanish. Thus, the integrating-factor method may be regarded as a general method of integrating equations of the form M (x, y)dx+ N (x, y) dy=O; however, because of the difficulty of finding the integrating factor this method is for the most part used only when the integrating factor is obvious. 6. Theorems of the Existence and Uniqueness of Solution of the Equation ~=/(x, y) The class of differential equations that are integrable by quadratures is extremely narrow; for this reason, since Euler's day, approximate methods have become very important in the theory of differential equations. At the present time, in view of the rapid development of computational technology, approximate methods are of incomparably greater importance. It is now frequently advisable to apply approximate methods even when an equation may be integrated by quadratures. What is more, even if the solution may be simply expressed in terms of elementary functions, it will often be found that using tables of these functions is more cumbersome than an approximate integration of the equation by computer. However, in order to apply one or another method of .aproximate integration of a differential equation, it is first necessary to be sure of the existence of the desired solution and also of the uniqueness of the solution, because in the absence of uniqueness it will not be clear what solution is to be determined. In most cases, the proof of the theorem of the existence of a solution yields at the same time a method for finding an exact or approximate solution. This elevates still more the significance of existence theorems. For example, Theorem 1.1, which is proved below, substantiates Euler's method of approximate integration of I. FIRST-ORDER DIFFERENTIAL EQUATIONS 45 differential equations, which consists in the fact that the desired . integral curve of the differential equation ~~ = f (x, y) that passes through the point (x0 , Yo) is replaced by a polygonal line consisting of straight lines (Fig. 1.13), each segment of which is tangent to the integral curve at one of its boundary points. When applying this method for the approximate calculation of the value of the desired solution of y (x) at the point x = b, the segment x0 ~ x ~ b (if b >x0 ) is subdivided into n equal parts by the points x0 , X1, X2 , ••• , Xn-t• xn, where Xn =b. The length of each subdivision x1+1 -x1=h is called the i.nterval of calculation, or step. Denote by y1 the approximate values of the desired solution at the points x1• To compute y1 , on the interval h (J Fig. 1-13 x0 ~ x ~ x1 replace the desired integral curve by a segment of its tangent at the point (x0 , y0). Hence, Y1 =Yo +hy~ where y~ = f (X0 , y0) (see Fig. 1.13). In similar fashion we calculate: Ys = y1 +hy~, where y; = f (x1, y1); Ya = Ys +hy~. where y~ = f (X2 , y2); Yn=Yn-t+hY~-1• where Y~-t=f(xn-1• Yn-1>· If b X0 , then at x > x1 the solution may no longer be defined (if x1 < x0 , then the solution may not be defined for x l ~ N IY1-Yzl may be replaced by a somewhat cruder, yet a more easily verifiable condition of the existence of a partial derivative f~ (x, y) bounded in absolute value in the region D. Indeed, if in the rectangle D If~ (x, y) I~ N' we get, by the mean-value theorem, If (x, Yl)- f (x, Yz) I=~~ (x, s) IYl-Y21. where 6 is a value intermediate between y1 and y,. Hence, the point (x, ~) lies in D and for this reason If~ (x, i,) I~ N and If(x, Y1)- f (x, Yz) I~ N IY1-Ys lIt is easy to give examples of functions f (x, y) (say, f (x,y) ==I y 1 in the neighbourhood of the points (x, 0)) for which the Lipschitz condition holds, but the derivative ~~ does not exist at certain points and hence the condition I:~ I~ N is weaker than the Lipschitz condition. Proof of the existence and uniqueness theorem. Replace the differential equation ~~= f(x, y) (1.22) having the initial condition (1.23) 48 I. DIFFERENTIAL EQUATIONS by the equivalent integral equation X Y=Yo+ ~ f(x, y)dx. (1.24) Xo Indeed, if a certain function y = y(x), when substituted, turns equation (1.22) into an identity and satisfies the condition (1 23), then, integrating the identity (1.22) and taking into account the condition (1.23), we find that y=y(x) reduces equation (1.24) to an identity as well But if some function y = y(x), when substituted, reduces (1.24) to an identity, it will obviously also satisfy the condition (1.23); differentiating the identity (1.24), we will find that u='iJ(x) also reduces equation (1.22) to an identity. Construct Euler's polygonal line y = y,. (x) emanating from the point (x0 , y0) with calculation interval h,. = H on the segmentn x0 ~ x ~ x0 +H, where n is a positive integer. (in exactly the same way we prove the existence of a solution on the interval x0 - H ~x~0) Euler's polygonal line that passes through the point (x0 , y0) cannot leave the rt!gion D for X0 ~ x ~ X0 +H ~or X0 - H ~ x ~x0), since the slope of each segment of the polygonal line is tess than M in absolute value. We break up the subsequent proof of the theorem into three parts: (1) The sequence y= y,. (X) converges uniformly. (2) The function y (x) = lim y,. (x) is a solution of the integral eq>taliQn (1.24). (3) The solution y(x) of the equation (1.24) is unique. Proof. (1) By the definition of Euler's poly:;'>nal line, y:, (x)=f(x,, y,) for x,~x~x,.+t• k=O, 1, ... , n-1 (ti)e right-hand derivative is taken at the corner point x11), or we denote y~ (x) = f (x, Yn (x)) + [f (x,, y,)- f (x, Yn (x))] (1.25) as f (X11, y,)- f (x, Yn (x)) = T),. (x). By virtue of the uniform continuity of the function f (x, y) in D we have IT),. (x) I= If (x,, y,.)- f (X, Yn (x)) I N (e,.), where e,.- 0 as n- oo, since Ix-x.l ~h,. and ly,-y,. ).) I< Mh,. and h,. = ~ -0 as n- oo. I FIRST·ORDER DIFFERENTIAL EQUATIONS 49 Integrating (1.25) with respect to x from x0 to x and taking into account that Yn (x0 ) = y0 , we get X X Yn(X)=Yo+ ~ f(t, Yn(t))dt+ ~ fln(t)dt. (1.27) Here n may take on any positive integer value and so for integer m>O X X Yn+m(X)=Yo+ ~ f(t, Yn+m(t)) dt+ ~ fln+m(t)dt. (1.28) Subtracting (1.27) from (1.28) termwise and taking the absolute value of the difference, we get IYn+m lX) -Yn (x)j =If[f (t, Yn+m {/)) -f(t, Yn(t)] dt+ + ffin+,.(/) dt- ffin (I) dt I~X., a;, %' ~~if(t, Yn+m(t))-{(1, Yn(t))ldt+ "•X X +~I 'ln+m (/)I dt + ~ ITin (I) Idt .Co l'o for x0 ~ x ~ x0 +H or, taking into consideration (1.26) and the Lipschitz condition: X IYn+m (X) -Yn (X) I~ N ~I Yn+m (t)-Yn (I) Idt +(8nu +8n)·H. X. H~nce, max IYn+m (X)- Yn (X) I~ .co<.c<;;.c,+H .c ~ N max~ IYn+m (t)-Yn (t) ldt + (8n+m+sn) H, .c. whence I () ()1 .-(En+m+En)H< max Yn+m X -Yn X :::::::: l-NH 8 x.,<;;.c<;;.c,+H for any e > 0 given sufficiently large n > N 1 (B). And so max IYn+m(X)-YnlX)I Nde), that ·is, the sequence of continuous functions Yn (x) converges uniformIy for X0 ~ x ~ X0 +H: Yn (x) =tY(x), where y(x) is a continuous function. (2) In equation (1.27) let us pass to the limit as n--+ oo: Jt % lim Yn (x) =Yo+ lim ~ I (x, Yn (x)) dx + lim S'ln (x) dx fi,..CD II ... CDX. II ... CIDz0 or % % y(x) =Yo+ lim ~I(x, Yn (x)) dx+ lim ~ 'ln (x) dx. (1.29) n-m~ n-e~ By virtue of the uniform convergence of Yn (x) to y(x) and the uniform continuity of the function I (x, y) in D, the sequence I (x, Yn (x)) =tI (x, y(x)). Indeed, II (x, y(x))- I (x, Yn (x)) I< e, where e>O if ly(x)-Yn(x)l<6(e), but ly(x)-Yn(x)l<6(e), if n>N1 (6(e)) for all x of the interval X0 ~x~x0 +H. And so ll(x, y(x))-l(x, Yn(x))l N.(6(e)), where N1 is not dependent on x. By virtue of the uniform convergence of the sequence I(x, Yn (x)) to I (x, y(x)), in (1.29) a passage to the Iimit is possible under the integral sign. Besides, taking into account that I'ln (e) I< eno where en--+0 as n-+oo, in (1.29) we finally get % y(x) =Yo+ SI (x, y(x)) dx. %o Thus, y(x) satisfies equation (1.24). (3) Assume the existence of two noncoincident solutions y1 (x) and y, (x) of equation (1.24); thus, max IY1 (x)- y, (x) I:::/= 0. t 0 ..;;x..;;x0 +H A termwise subtraction from the identity % y, (x) ===Yo+ SI (x, Y1 (x)) dx Xo of the identity )( Ya (x) =Yo+ ~ I (x, y, (x)) dx, x. I. FIRST-ORDER DIFFERENTIAL EQUATIONS 5! yields X y1 (x)-y,(x)== ~ (f(x, y.(x))-f(x, y,(x))]dx. .to Hence max Iy1 (x)- y, (x) I= .t0 .;;;x.;;;x0 +H = max ~~ [f(x, y1 (x))-f(x, y,(x))]dxl~x0 .;;;x<;x0 +H x, ~ max I~ If (x, Y1 (x))-f (x, y, (x)) Idx 1-x.<.:x<.:x,+H ·x0 Taking advantage of the Lipschitz condition, we will have max jy1(x)-Y2(x)I~N max ~~IY1 (x)-y,(x)ldxl.,;;;;x0 0 there ili a number N (e) such that for n ?- N (e) the distance p (y,. Yn+m> < e for any interval m > 0. I. DIFFERENTIAL EQUATIONS If we now apply the triangle rule m-1 times and use the inequality (1.31), we get P(Ym Yn+m) ~ P (yn, Yn+I) +P(Yn+l• Yn+t) + ••• ···+P (Yn+m-t• Yn+m) ~(an+ an+t + ···+ an+m-t] P (Yu Yo)= "n_"n+m "n = 1_" P(Yt• Yo) < 1-" P (yt, Yo) < 8 for sufficiently large n. Consequently, the sequence y0 , y1 , y1 , ••• , Yn• ... is fundamental and, by virtue of the completeness of the M space, it converges to a certain element of this space: lim Yn = y, yc M. n-+CD · We will now prove that y is a fixed point. Let A (yl = y. Applying the triangle rule twice, we get P('Y, Y) ~ P (y, Yn) +P 0, we can choose N (8) such that for n ~ N (8) - 8 (I) p (y, Yn) < 3, since y = lim Yn; n ... "' (2) p and hence IY;- Y;o I~ b;. It remains to verify fulfillment of the condition (2) of the contraction-mapping principle: p (A (Y]), A (Z]) = =i~l max II[{;(X, Y1 • Ys• .•. , Yn)-{;(X, Z1 , Z2, ••• , Zn)]dxl~ ~.± maxlf /f;(X, y1 , y2 , ••• , Yn)-/;(X, Z1 , Z2, ... , Zn)/dxl~I= I Xo ~N ,t, max li1~1 IY;-z;ldxl~ ~N 1~1 max IY;-Z; Ii~lmax Iidx I=Nnh,)p (Y, l). Consequently, if one chooses h0 ~ n~ where 0 0, a 6 (e, b)> 0 may be found such that from the inequalities lx0 -Xol <6(e,b) and Jy0 -y0 J<6(e, b) there will follow the inequality JIJ (X, X0 , Yo)- Y (x, X0 , Yo) I< 8 (1.36) for X0 ~x~l (Fig. 1.18). Generally speaking, the number 6 (e, b) decreases with increasing b, and as b-. oo it can approach zero. For this reason, it is by 0 (:Cg-Zo)< ~(E, b) (!fo-ffoJ<~(t, b) Fig. 1-18 b far not always possible to choose a number 6 (e)> 0 for which the inequality (1.36) would be satisfied for all x > x0 , that is, it does not always happen that solutions which have close-lying initial values remain so for arbitrarily large values of the argument. A solution that changes but slightly for an arbitrary but sufficiently small variation of the initial values, given arbitrarily large values of the argument, is called stable. Stable solutions will be examined in more detail in Chapter 4. Theorem 1.3 (on the analytical dependence of a solution on a parameter, Poincare's theorem). The solution x(t, fl) of the differential equation X= f(t, x, fl) which satisfies the condition x (t0 ) = x0 , depends analytically on the parameter fl in the neig.'.bourhood of the value fl = fl0 , if the function f in the given range of t and x and in some neighbourhood of the point flo is continuous with respect to I and is analytically dependent on fl and x. An analogous assertion also holds for the system of equation~ xi(t)=f;(t, X1 , x2 , ••• , Xn,fl) (i= 1, 2, ... , n), I. FIRST-ORDER DIFFERENTIAL EQUATIONS 61 and in this case it is assumed that the functions f; are continuous with respect to the first argument and are analytically dependent on all the other arguments. We do not give a detailed proof of this theorem (and the same goes for other theorems requiring application of the theory of analytic functions) and refer the reader to a paper by A. Tikhonov [4] which contains the simp!est proof of the theorem on the analytic dependence of a solution on a parameter. The underlying idea of Tikhonov's proof is: assuming that J-L can take on complex values as well, the existence is proved of the limit lim 6~'-:(t, J.l) = ~x , which thus signifies analytic dependence of the Al.l- o J.1 J.1 solution on J-L. The existenc~ of this limit follows from the fact that 6 X the ratio X satisfies the linear differential equation "'..:!._ t\"'x =f(t, x(t, J.L+L\J.L), J.L+t\!J.)-/(1, x(t, J.L), J.L+L\J.L) t\1'-x(t, J.l) + dt t\J.L l\"'x (t, J.L) 6J.1 + f(t,x(I,J.L), !l+l\J.L)-f (t,x(I,J.L),!J) l\"'x I =0 t\J.L ' 6J.1 l=to ' the solution of which is unique and, as the increment dJ-L tends to zero by any law whatsoever, approaches the unique solution of the equation dz of of dt =ax z + OJ.l. • z Uo> = 0. Theorem 1.4 (on the differentiability of solutions). If in the neighbourhood of a point (x0 , y0 ) a function f (x, y) has continuous derivatives to the order k inclusive, the solution y (x) of the equation dy fdx = (X, y), (1 37) that satisfies the initial condition y (x0 ) = y0 , has continuous derivatives to order (k +1) inclusive in some neighbourhood of the patnt (Xo, Yo)Proof. Substituting y (x) into equation (1.37), we get the identity ~ === f (x, y (x)), (1.371) and hence the solution y (x) has a continuous derivative f (x, y (x)) in some neighbourhood of the point under consideration. Then, by virtue of the existence of continuous derivatives of the function f, there will exist a continuous second derivative of the solution d2y of of dy d/ of dx3 =ox+ oydx =ox+ ay f (x, Y (x)). 62 I. DIFFERENTIAL EQUATIONS If k·> I, then by virtue of the existence of continuous second-order derivatives of the function f, it is possible, by differentiating the identity (1.371) once again, to detect the existence also of a third derivative of the solution: tJ.Sy=iJSf+ 2 iJSf f+iPf f+of(of+o't) dx8 ox' OX og ogl og ox og . Repeating this argument k times, we prove the theorem. Now consider the points (x0 , y0 ) in the neighbourhood of which there is no solution of the equation : = f (x, y) that satisfies the condition y (x0 ) = y0 , or the solution exists but is not unique. Such points are called singular points. A curve that consists entirely of singular points is called singular. If the graph of a certain solution consists entirely of singular points, then the solution is called singular. To find singular points orsingular curves it is first of all necessary to find a set of points in which the conditions of the existence and uniqueness theorem are violated, since only such points can include singular points. Of course, not every point at which the conditions of the theorem of existence and uniqueness of solution are violated is a singular point, since the conditions of this theorem are sufficient for the existence and uniqueness of the solution but are not necessary. The first condition of the existence and uniqueness theorem (see page 45) is violated at the discontinuity points of the function f (x, y); note that if the function f (x, y) increases without bound in absolute value upon approaching (by any path) some isolated point of discontinuity (x0 , y0), then in those problems in which the variables x and y are equivalent (as we have already agreed) the equation : = f (x, y) must be replaced by the equation ~; = f (:. y), for which the right-hand side is now continuous at the point (x0 , y0) if it is taken that f ( 1 ) 0.Xo, Yo Hence, in problems in which the variables x and y are equivalent the first condition of the existence and uniqueness theorem is violated at those points at which both the function f (x, y) and f (:. y) are discontinuous. One particularly often has to consider equations of the form dy M (x, y) dx = N (x, y) ' (1.38) where the functions M (x, y) and N (x, y) are continuou~. In this I FIRST-ORDER DIFFERENTIAL EQUATIONS 63 case the functions M (x, y) and N (x, Y) will at the same time be N (x, y) M (x, y) discontinuous only at those points (x0 , y0) at which M (x0 , Yo) = = N (x0 , Yo)= 0 and the limits and do not exist. lim M (x, y) %-+Xo N (x, y) IJ-+1/o lim N (x, y) X-+Xo M (X, y) IJ-+1/o Let us consider several typical singular points of the equation (1.38). Example 3. dy 2y dx=x The right sides of this equation and of the equation :; = ;Y are discontinuous at the point x = 0, y = 0. Integrating the equation, we get y = cx1, which is a family of parabolas (Fig. 1.19), and Fig. 1-1~. Fig. 1-20 x = 0. The singular point at the coordinate origin is called a nodal point. Example 4. dy y dX=--x 64 I DIFFERENTIAL EQl1ATIONS The right sides of this equation and of equation d~ =-!.... are. y y discontinuous at the point x = 0, y = 0. Integrating the equation, we get y =!..., which is a family of hyperbolas (Fig. 1.20), and X the straight line x = 0. The singular point at the origin is called a saddle point. Example 5. dy x+y dx = x-y · The right sides of this equation and of equation d~ =x+y arey X y discontinuous at the point x = 0, y = 0. Integrating the homogeneous y Fig. 1-21 Fig. 1-22 equation (compare with Example 3 on page 42), we have V-1- -1 arcran ...!L X +y =Ce -¥ or in polar coordinates p=ce'f, which represents logarithmic spirals (Fig. 1.21). The singular point of this kind is called a focal point. Example 6. dy X dx=--y The right sides of this equation and of the equation d~ = - !ly X are discontinuous at the point ~ = 0, y =0. Integrating the equation, we get x3 +y2 = c'. which is a family of circles with centre at the. coordinate origin (Fig. 1.22). The singular point of this type, i.e. the singular point whose neighbourhood is filled with a family of closed h•tegral curves, is called a centre. In this example there is no solution that satisfies the condition y (0) == 0. In Chapter 4 we shall return to the problem of singular points and their classification from a somewhat different point of view. I. FIRST-ORDER DIFFERENTIAL EQUATIONS 65 The second condition of Theorem 1.1 on the existence and uniqueness of solution-the Lipschitz condition, or a cruder condition requiring the existence of a bounded partial derivative : is most often violated at poiuts, upon approaching which :~ increases without bound, i.e. at points at which a~ = 0. ag Generally speaking, the equation a~ = 0 defines a certain curve ag at points of which uniqueness may be violated. If at points of this curve uniqueness is violated, then the curve will be singular; if, besides, the curve turns out to be integral, then we have a singular integral curve. The curve ;, = 0 may have several branches, then with respect ag to each branch one has to decide whether that branch will be a singular curve and whether it will be an integral curve. Example 7. Has the equation ~~=!I +x2 a singular solution? The conditions of the existence and uniqueness theorem are fulfilled in the neighbourhood of any point; hence there is no singular solution. Example 8. Has the equation Z= V 0. Thus, equation F(x, y, y')=O (1.51) may be integrated by solving for y' and by integration of the thus obtained equations y' = {; (x, y) (i = 1, 2, ... ) that have already been solved for the· derivative. ----tl Fig. 1-25 However, equation (1.51) is not always so readily solved for y', and even more infrequently are the equations y' = f1(x, y) obtained after solving for y' easily integrated. Therefore, one often has to integrate equations of the form (1.51) by other methods. Let us consider a number of cases. 1. Equation (1.51) is of the form F(y') = 0, (1.55) and there exists at least one real root y' = k1 of this equation. ---~~-~1'-=,:;..o__;..zo Since (1.55) does not contain x and y, k1 is a constant. Consequently, integrating the equation y' =k1, we get y=k1x+c orFig. 1-26 I. FIRST-ORDER DIFFERENTIAL EQUATIONS 75 k; = Y x c , but k; is a root of the equation (1.55); consequently, F ( Y x c ) = 0 is an integral of the equation in question. Example I. (y'f -(y')' + y' +3 = 0. The integral of the equation is ( y X c r-(y / r+ y X c +3 = 0. 2. Equation (l.5I) is of the form F(x, y')=O. (I .56) If this equation is hard to solve for y', it is advisable to introduce the parameter t and replace (1.56) by two equations: x =

(t) and y' = 'ljl (t). Since , dy rp' (I) dt 5rp' (I) dt dy = y dx, then dx =IT= "'(t) , whence x = "'(t) +c. Thus, in parametric form the desired integral curves are defined by the equations 5rp' (f) dt x= 1jl(t) +c and y=q>(t). As a particular case, if equation (1.61) is readily solvable for y, it is usually convenient to take y' as the parameter. Indeed, if y = q> (y'), then, putting y' = t, we get = (t) d = ~ = rp' (t) dt yq>,X y' t ' -5rp' (t) dt +X- t C. Example 4. y = (y')' + (y')3 + y' + 5. Put y' = t; then y=t'+t3 +t+5. (1.62) dx = .!!_y_ = (.514 +312 +l).dt = (5ta+3t +_!_) dt y' t t • Stt 312 x= 4 + 2 +InJtl+c. (1.63) Equations (1.62) and (1.63) are parametric equations of a family of integral curves. Example 5. - y-=1 VI +y'z Put y' = sinht; then y=cosht dx = dY_ = sinh t dt = dt, y' sinh t x=t+c ;1.64) (1.65) or, eliminating the parameter t from (1.64) and (1.65), we get y=cosh (x-c). I. FIRST·Of{DER DIFFERENTIAL EQUATIONS 77 Now consider the general case: the left-hand side of equation F(X, y, y') = 0 (1.51) depends on all three arguments x, y. y'. Replace (1.51) by its parametric representation: x = q> (u, v), !J = 'IJ1 (u, v), y' = x(u v). Taking advantage of the dependence dy = y'dx, we will have ~: du+!atdv=x(u, v) [~~ du+: dvj; from this, solving for the derivative : , we get Oq> a"'dv X(U, v) au--audu = o\j) Otp (1.66 -ar; -x (u, v) av We have thus obtained a first-order equation that is already sulved for the derivative, and we have thus reduced the problem to one that has been considered in earlier sections. However, the resulting equation (1.66) is of course by far not always integrable by quad- ratures. If the equation F (x, y, y')=O is readily solvable for y, it is often convenient to take x and y' for the parameters u and v. Indeed, if equation (1.51) is reduced to the form !J =I (x, y'), (1.67) then, considering x and y' = p as parameters, we obtain at at!J=I(x, p), dy= ax dx+apdp or dy at at dp dx = ox +op dx • - ~ ~!!.E._ P- iJx + iJp dx · (1.68) Integrating (1.68) (of course, it is by far not always integrable by quadratures), we get t1> (x, p, c)= 0. The collection of equations <1> (x, p, c)= 0 and y =I (x, p), where p is a parameter, defines a family of integral curves. Note that (1.68) may be obtained by differentiating (1.67) with respect to x. Indeed, differentiating (1.67) with respect to x and 78 I. DIFFERENTIAL EQUATIONS assuming t/ = p, we get p = ;~ + ;~ ~~, which coincides with ( 1.68). For this reason, this method is often called integration of differential equations by means of differentiation. In quite analogous fashion, the equation F(x, y, y')=O, can frequently be integrated if it is readily solvable for x: x=f(y, y'). (1.69) In this case, taking y and y' = p for the parameters and using the relation dy=y'dx, we get dy=p l;~ dy+ ;~ dp] or (1.70) Integrating equation {I.70), we get <1> (y, p, c)= 0. This equation together with x = f (y, p) defines the integral curves of the original equation. Equation (1.70) may be obtained from (1.69) by differentiation with respect to y. To illustrate the use of this method, let us consider the following equation, which is linear in x and y: y = Xq> (y') +'iJ (y'), it is called Lagrange's equation. Differentiating with respect to x and putting y' = p, we have p=q>(p)+xq>'(p) :~ +'i'' (p) ::, (1.71) or [ dx p- q> (p)] dp = xq>' (p) +'I>' (p). (1.72) This equation is linear in x and :; and, hence, is readily integrable, for example, by the method of variation of parameters. After obtaining the integral Ill (x, p, c)= 0 of equation (l.72) and adjoining to it y = xcp (p) +'I> (p), we get equations that define the desired integral curves. When passing from (1.71) to (1.72) we had to divide by 'Jx · But in the process we lose solutions (if they exist) for which p is constant, and hence ~~ = 0. Taking p constant, we note that equation (1.71) is satisfied only if pis a root of the equation p-q>(p)=O. I. FIRST-ORDER DIFFERENTIAL EQUATIONS 79 Thus, if the equation p-ep (p) = 0 has real roots p = p1, then to the solutions of the Lagrange equation that were found above we have to add y=xcp(p)+'IJ1(p), p=p1 or, eliminatingp, y=xcp(p1)+ + 'lj1 (p1) are straight lines. We have to consider separately the case when p-cp(p)=O and hence, when dividing by ~~ , we lose the solution p = c, where c is an arbitrary constant. In this case, cp (y') =y' and the equation y = xcp (y') + 'lj1 (y') takes the form y = xy' + 'lj1 (y'), which is Claimut's equation. Putting y' = p, we have y=xp+'l? (p). Differentiating with respect to x, we will have p=p+x :~+'I'' (p) ~~ or (x+'lj)'(p)) :: =0, whence either :: =0, and hence p=c, or x+'lj)' (p)=O. In the first case, eliminating p, we get y=cx+'IJ1 (c), (1.73) which is a one-parameter family of integral curves. ln the second case, the solution is defined by the equations y=xp+'f(p) and x+'lj)'(p)=O. (1.74) lt can readily be verif,ed that the integral curve defined by the equations (1.74) is the envtlope of the family of integral curves (1.73). Indeed, the envelope of .:! family (x. y, c)= 0 is defined by the equations OCD (x, y, c) = 0 and 7F = 0, (1.75) which for the family y =ex+ 'lj1 (c) are of the form y =ex+'¢ (c), x +'lj)' (c) = 0 and differ from equations (1.74) (Fig. 1.27) only in the designation of the parameter. Note. As we know, besides the envelope, the equations (1.75) can define the loci of multiple points and sometimes other curves as well; however, if even one of the derivatives ~ and OCD is diffeox oy rent from zero and both are bounded at points satisfying the equations (1.75), then these equations define only the envelope. In the 80 I DIFFERENTIAL E(JU A TJONS given case, th~se conditions are fulfilled: : =-c. : =I. Consequently, equations (l.75) define an envelope that can degen~rate into a point if the family (I.73) is a pencil of Jines. Example 6. y = xy'- y'2 is Clairaut's equation. A one-parameter family of integral straight lines has the form y :c [J Fig. 1-27 Fig. 1-28 y=cx-c2• Besides, the envelope of this family, defined by the equations y=cx-c' and x-2c=0, is an integral curve. Elimtnating c, we get y = ~~ (Fig. J.28). Example 7. y= 2xy' -y'" is Lagrange's equation. y'=p, y=2xp-p1 • Differentiating, we get p=2p+2x dp -3p1 dp dx dx and, after dividing by ~~ , we arrive at the eqa.:afion dx 2 3 I pdp=- x+ p. (J.76) (I.77) lntegrat ing this linear equation, we obtain x= ;! +! p1 • Hence, the integral curves are defined by the equations y= 2xp-p1, c 3p2 X= -p~ +-4-. As mentioned above, in dividing by :~ we lose the solutions p = P;. where the p,. are roots of the equation p-ff! (p) = 0. Here, we I f'IRST·ORDER DIFFERENTIAL EQUATIONS 81 lose the solution p=O of equation (1.77), to which corresponds, by virtue of equation (1.76), the solution of the original equation y=O. 9. The Existence and Uniqueness Theorem for Differential Equations Not Solved lor the Derivative. Singular Solutions In Sec. 6 we proved the theorem of existence and uniqueness of the solution y (x) of the equation :~ = f (x, y) that satisfies the con· dition y (x0 ) =Yo· A similar question arises for equations of the form F (x, y, y') = 0. It is obvious that for such equations, generally speaking, not one but several integral curves pass through some point (x0 , y0 ), since, as a rule, when solving the equation F (x, y, y') = 0 for y' we get several (not one) real values y' = f; (x, y) (i = I 2, ... ), and if each of the equations y' = f; (x, y) in the neighbourhood of the point (x0 , y0) satisfies the conditions of the existence and uniqueness theorem of Sec. 6, then for each one of these equations there will be a unique solution satisfying the condition y (x0 ) = y0 . Therefore the property of uniqueness of solution of the equation F (x, y, y')=O, which satisfies the condition y(x0 )=y0 , is usually understood in the sense that not more than one integral curve of the equation f(x, y, y') = 0 passes through a given point (X0 , Yo) in a given direction. For example, for the solutions of the equation (:~) 2 -I= 0, the property of uniqueness is everywhere fulfilled, since through every point (x0 , y0 ) there pass two integral curves, but in different directions. Indeed, dy ddx =+I, y=x+c an y= -x+c. For equation (y')2 - (x +y) y' +xy = 0 consi<:tered on page 73. the property of uniqueness is violated at points of the straight line y = x, since the integral curves of the l quations y' = x and y' = y pass through points of this line in the same direction (Fig. 1.29). Theorem I .5. There exists a unique solution y = y (x), ..0 -hr::;;;;;, ::;;;;;, x::;;;;;, X0 +h0 (where h0 is sufficiently small) of the equation F(x,y,y')=O, (1.78) that satisfies the condition y (x0 ) =Yo for which y' (x0 ) = y~. where y;, is one of the real roots of the equation F (x0 , y0 , y') = 0, if in a closed neighbourhood of the point (x0 , y0 , y~) tfoe function F (x, y, y'] satis/ie~> the conditions; G • for then it will be possible to assert that the equation y' = f (x, y) (1.79) satisfies the conditions of the existence and uniqueness theorem (see Sec. 6, page 46) and, consequently, that there exists a unique solution of the equation (1.79) that satisfies the condition y (X0 ) = y0 , I. FIRST-ORDER I. DIFFERENTIAL EQUATIONS 83 and there also exists a unique integral curve of the equation (L78) that passes through the point (x0 , y0) and has, at that point, the slope of the tangent y~. In accordance with the familiar theorem on implicit functions, it may be asserted that if conditions (1), (2) and (3) are fulfilled, the derivative :~ exists and may be computed by the rule of differentiating implicit functions. Differentiating the identity F (x, y, y') = 0 with respect to !I and taking into account that y' = f (x, y), we get dF +oF of _ 0 iii ou' ou- • oF of ou ou =- oF • oy' whence, taking into account the conditions (2) and (3), it follows that I:~ I~ N in a closed neighbourhood of the point (x0 , y0 ). The set of points (x, y) at which the uniqueness of solutions of the equation F (x, y, y') = 0, (1.78) is violated is called a singular set. At the points of a singular set at least one of the conditions of Theorem 1.5 must be violated. In differential equations encountered in applied problems, conditions (1) and (3) are usually fulfilled but condition (2), ::. =1= 0, is frequently violated. If conditions (1) and (3) are fulfilled, then at the points of a singular set the equations F(x, y, y')=O and ;:, =0 (1.80) must be satisfied simultaneously. Eliminating y' from these equations, we get the equation (x, y)=O, (L8l) which must be satisfied by the points of the singular set. However, the uniqueness of solution of the equation (1.78) is not necessarily violated at every point that satisfies the equation (L81), because the conditions of Theorem L5 are only sufficient for uniqueness of solution, but are not necessary, and hence violation of a condition of the theorem does not of necessity imply violation of uniqueness. 6''' 84 I. DIFFERENTIAL EQUATIONS Thus, only among points of the curve (x, y) = 0, called the p-discriminant curve (since the equation (1.80) is most frequently written in the form F (x, y, p) = 0 and ~: =OJ, can there be points of the singular set. If some kind of branch y = cp (x) of the curve (x, y) = 0 belongs to the singular set and at the same time is an integral curve, it is called a singular integral curve, and the function y = cp (x) is called a singular solution. Thus, in order to find the singular solution of the equation F(x, y, y')=O (1.78) it is necessary to find the p-discriminant curve defined by the equations aF F(x, y, p)=O, ap-=0, to find out [by direct substitution into equation (1.78)] whether there are integral curves among the branches of the p-discriminant curve and, if there are such curves, to verify whether uniqueness is violated at the points of these curves or not. If uniqueness is violated, then such a branch of the p-discriminant curve is a singular integral curve. Example 1. Does the Lagrange equation y = 2xy'- (y')2 have a singular solution? Conditions (1) and (3) of the existence and uniqueness theorem are fulfilled. The p-discriminant curve is defined by the equations y=2xp-p2 , 2x-2p=0 or, eliminating p, y=x2• The parabola y = x2 is not an integral curve since the function y = x2 does not satisfy the original equation. There is no singular solution. Example 2. Find a singular solution of the Lagrange equation 4 (y' )' 8 ( ')3 (1 82)X-Y=g -27 y . . Conditions (1) and (3) of the existence and uniqueness theorem are fulfilled. The p-discriminant curve is defined by the equations 4 2 8 3 8( 2 0 x-y= 9 P -27 P • 9 p-p) = · From the second equation we find p=O or p=l; substituting into the first equation, we obtain 4 y=x or y=x- 27 . Only the second of these functions is a solution of the original equation. I. FIRST-ORDER DIFFERENTIAL EQUATIONS 85 To find out whether the solution y =x-2~ is singular, we have to integrate the equation (I .82) and find out whether other integral curves pass through points of the straight line y = x- 2~ in the direction of this line. Integrating tht! Lagrange equation (1.82), we get (y-c)2 = (x-c)3 • (1.83) From (1.83) and Fig. 1.30 it is seen that the straight line y = x-2~ is the envelope of a family of semicubical parabolas (y-c)2 = = (x -c)3 and, hence, uniqueness is violated at every point of the straight line y = x- 2~: there are two integral curves in one and the same direction- the straight I. 4 d . Fig. 1-30 me y = x - 27 an a semicubical parabola that is tangent to this strai~ht line at the point under consideration. Thus, y = x- 2~ is a singular solution. In this example, the envelope of the family of integral curves is a singular solution. If the envelope of the family (x, y, c)=O (1.84) is a curve which is tangent at each of its points to some curve of the family (1.84), and to each segment of which are tangent an infinite set of curves of this family, then th~ envelop~ of a family of integral curves of some equation F (x, y, y') = 0 will always be a singular integral curve. Indeed, at points of the envelope the values x, y and y' coincide with the values x, y and y' for the integral curve tangent to the envelope at the point (x, y), and hence at every point of the envelope the values x, y and y' satisfy the equation F (x, !J, y')= 0; that is, the envelope is an integral curve (Fig. 1.31 ). Uniqueness is violated at every point of the envelope, since at least two integral curves pass through the points of the envelope in the same direction: the envelope and the integral curve of the family (1.84) tangent 86 I. DIFFERENTIAL EQUATIONS to it at the point under consideration. Consequently, the envelope is a singular integral curve. Knowing the family of integral curves (x, y, c)= 0 of some differential equation F(x, y, y') = 0, it is possible to determine its singular solutions by finding the envelope. As we know from the course of differential geometry or mathematical analysis, the envelope is contained in the c-discriminant curve defined by the equa- tions (}$ (x, y, c)= 0 and ac = 0, however, the c-discriminant curve Fig. 1-31 can include, besides the envelope, other sets as well, for instance, the set of multiple points of curves of the family under study in which ~~ =: = 0. For some branch of the c-discriminant curve definitely to be an envelope, it is sufficient that on it: (I) there exist the following partial derivatives bounded in absolute value: /aa~ I~ N1, laa~ I~ N2; (2) aa~*o or ~~ =:~=o. Note that these conditions are only sufficient, and so curves involving a violation of one of the conditions (1) or (2) can also be enve- lopes. Example 3. Given a family of integral curves (y- c)2 = (x-c)3 of some differential equation (see Example 2 on page 84). Find a singular solution of this equation. Find the c-discriminant curve: (y-c)2 = (x-c)3 and 2 (y-c) = 3 (x-c)'. Eliminating the parameter c, we get 4 y=x and x-y-27 =0. The straight line y=x- 2~ is an envelope since on it are fulfilled all the conditions of the envelope theorem. The function y = x does not satisfy the differential equation. The straight line y = x I. FI~ST·O~OE~ OIFFE~ENTIAL EQUATIONS 87 is a cusp locus (see Fig. 1.30). The second condition of the envelope theorem is violated at the points of this straight line. Example 4. Given a family of integral curves (1.85) of some differential equation of the first order. Find a singular solution of this equation. The problem reduces to finding the t>nvelope of the desired family. If one applies directly the above-indicated method for finding the envelope, one gets the contradictory equation 1= 0, Fig. 1-32 Fig. 1·33 whence it would seem natural to draw the conclusion that the family (1.85) does not have an envelope. However, in this ca~. , the derivative of the left side of (1.85) with respect to y, ~ = i 4 = !y-5 becomes infinite when y= 0 and hence there is a possibility that y = 0 is the envelope of the family (1.85) that could not be found by the general method since the conditions of the envelope theorem were violated on the straight line y = 0. One has to transform equation (1.85) so that the conditions of the envelope theorem are fulfilled for the transformed equation, which is equivalent to the original equation. For example, write (1.85) in the form y-(x-c)' = 0. The conditions of the envelope theorem are now fulfilled and, using the general method, we get y = (x-c)', 5 (x-c)' =0 or, eliminating c, we will have the equation of the envelope y = =0 (Fig. I:32). Example 5. Given the family of integral curves (1.86) 88 I. DIFFERENTIAL EQUATIONS of some first-order differential equation. Find a singular solution of this equation. The c-discriminant curve is defined by the equations y1 -(x-c)3 =0 and x-c=O or, eliminating c, we get y=O. On the straight line y=O, both partial derivatives C:: and ~:of the left-hand side of equation (1.86) vanish, hence y =0 is the locus of multiple points of curves of the family (1.86), in the given case, the cusp locus. However, the cusp locus in this example is at the same time an envelope. Fig. 1.33 depicts the semicubical parabolas (1.86) and their envelope y = 0. PROHLEMS ON CHAPTER I I. tanydx--cotxdy=O. 2. (12x +5y-9)dx+ (5x+2y-3)dy=0. 3. x~= y+ Vx2 +Y2 4. x:~+Y=X3• 5. ydx-xdy = x2 ydy. 6 dx 3 _ 21 "dt+x-e. 7. y sin x +y' cos x = I. 8. y'=e"-Y. 9. :; =X+ sin/. 10. x(lnx-lny)dy-ydx=O. 11. xy(y')2 -(X2 +l)y' t xy=O. 12 (j/)2 =9!/. dx .! x 13. dt = et +T • 14. x1 +(y')1 = 1. 15. y = xy' +_I_ •y 16 X=(y')3 -y'+2. dy y 17. dx = x+ya 18. y= (y')'-(y')3 -2. I. FIRST-ORDER DIFFERENTIAL EQUATIONS 89 19. Find the orthogonal trajectories of the family xy =c; i.e., find lines that orthogonally intersect the curves of the indicated family. 20. Find the curve whose subtangent is twice the abscissa of the point of tangency. 21. Find the curve whose y-intercept cut by a tangent is equal to the abscissa of the point of tangency. 22. Find the orthogonal trajectories of the family x2 -l-y2 =2ax. 23. Considering that the rate at which a body cools in the air is proportional to the difference between the temperature of the body and that of the air, solve the following problem: if the air temperature is 20°C and the body cools from 100° to 60°C in 20 min, how long will it take the body to reach 30° C? 24. A motor boat is in motion in calm water with a velocity of 10 km/hr. The motor is cut out and in t = 20 sec the velocity falls to v, = 6 km/hr. Determine the speed of the boat 2 minutes after the motor was cut out (assume that the resistance of the water is proportional to the speed of the boat). 25. Find the shape of a mirror that reflects, parallel to a given direction, all the rays emanating from a specified point. 26. y'2 + y2 = 4. 27. Find a curve whose tangent segment lying between the coordinate axes is divided into equal parts at the point of tangency. 28 dy 2y-x,--4 29 dy Y 2 O • dx=2x-y-j-5" · dx-1-j-x+Y = · 30. Integrate the following equation nume1 ically: ~~~~X -j- y2 , y (0) = 0. Determine y(0.5) to within 0.01. 31. Integrate numerically the equation dy 3 2 (0. 0dx = xy +x , y )= . Determine y (0.6) to within 0.01. 32. y' = 1.31x-0.2y2 , y (0) = 2. Form a table of fifteen values of y with interval of computation h=0.02. 33. y=2xy'-y'2• 34. ~=cos(x--y). 35. Using the method of isoclines (see page 21), sketch a family of integral curves of the equation dy ~ 2 dx=X -y · 90 I. DIFFERENTIAL EQUATIONS 36. (2x+2y--l)dx+ (x+y-2)dy=0. 37. y'8 -y'e3"·= 0. 38. Find the orthogonal trajectories of the parabolas y1 + 2ax=a'. 39. Does the differential equation y = 5xy' -(y')1 have a singular solution? 40. Integrate in approximate fashion the equation ~=x-y1, y(l)=O by the metho4 of successive approximations (determine y1 and y,). " 41. y·=x'+ 5fdx. I 42. Has the equation y' = Vx 5y+ 2 a singular solution? 43. (x-y)ydx-x2 dy=0. 44. Find the orthogonal trajectories of the family y2 = cx3• 45. x+ 5x = IOt + 2 for t = I, x = 2. • x x2 46. x = T + 13 for t = 2, x = 4 47. y=xy' +y'2 for x=2, y= -I. 48. y = xy' +y'2 for x = I, y = - 1. 49 dy _ 3x-4y-2 · dx- 3x-4y-3 · 50. x-xcot t = 4 sin t. 51. y=x'+2y'x+Y~2 • 52. y'- 3Y +x3y2 =0.X 53. y(l +y'2)=a. 54. (x3 -y)dx+(x2y1 +x)dy=0. 55. Find the integrating factor of the equation' (3y2 -x) dx +2y (y2 -3x) dy = 0 having the form 1-L = J1 (x + y2). 56. (x-y)ydx-x2 dy=0. , x+y-3 57.y=l +"-X y 58. xy' -y2 1nx+ !I= 0. 59. (x2 -l) y' +2xy-cosx= 0. 60. (4f+2x+3)y'-2y-x-l =0. 61. (y -x)y'-y+x2 =0. 62. (y2 -x2 ) y' +2xy = 0. 63. 3xy2y' +y3 -2x= 0. 64. (y')2 +(x+a)y'-y=0, where a is constant. 65. (y')2 -2xy' +y=O. 66. (y')3 +2yy' cot x- y2 = 0. CHAPTER 2 Differential equations of the second order and higher 1. The Existence and Uniqueness Theorem for an nth Order Differential Equation Differential equations of the tHh order are of the form y'n»=f(x, y, y', ... , yln-1l) (2.1) or, if they are not solved for the highest derivative, F(x, y, y', ... , y1m) =0. The theorem of existence and uniqueness for an nth order equation may readily be derived by reducing it to a system of equations for which the existence and uniqueness theorem has already been proved (see page 56). Indeed, if in the equation y1n1 = f (x, y, y', ... , y1n- 1l) we consider as unknown functions not only y but also y' = y1 , y" = y,, .•• . . ., y1n- 11 =yn_1, then equation (2.1) is replaced by the system y'=y.. ly~ =y2, .· ~.. }Yn-2- Yn-l• J Y~-• = f (x, y, Yt• · · ·• Yn-t). (2.2) and we can then apply the theorem of the existence and uniqueness of solution of a system of equations (see page 56), according to which if the right sides of all equations of the system (2.2) are continuous in the region under consideration and if they satisfy the Lipschitz condition with respect to all arguments, except x, then there exists a unique solution of the system (2.2) that satisfies the conditions Y (Xo) = Yo• Yt (xo) = Yw · · ·• Yn-t (xo) = Yn-1, o· The right sides of the first n-1 equations (2.2) are continuous and satisfy not only the Lipschitz condition but even the cruder condition of the existence of bounded derivatives with respect to y, Y1, y2 , •••• Yn- 1· Hence, the conditions of the existence and uniqueness theorem will be fulfilled if the right side of the last 92 I. DIFFERENTIAL EQUATIONS equation Y~- 1 = f (x, y, y1, ••• , Yn- 1 ) is continuous in the neighbourhood of the initial values and satisfies the Lipschitz condition with respect to all arguments, from the second onwards, or satisfies the cruder condition of the existence of bounded partial derivatives with respect to all arguments from the second onwards. Thus, reverting to the earlier variables x and y, we finally get the following existence and uniqueness theorem. Theorem 2./. There exists a unique solution of an nth order differential equation yJD OtlDER AND HIGHER 93 unique law of motion x = x (t) if, of course, the function f satisfies the conditions of the existence and uniqueness theorem. The theorem (considered on page 58) on the continuous dependence of the solution on parameters and on the initial values can be extended. without altering the method of proof, to systems of differential equations and hence to equations of the nth order. 2. The Most Elementary Cases of Reducing the Order In certain cases the order of a differential equation may be reduced. This ordinarily simplifies its integration. The following are some frequently encountered classes of equations that permit reducing the order. I. The equation does not contain the desired function and its derivatives up to order k-l inclusive: (2.3) In this case the order of the equation may be reduced to n-k by changing the variables: y(kl = p. Indeed after changing the variables the equation (2.3) becomes F (x, p, p', .•. ' p(ra-kl) =0. From this equation we find p = p (x, c1, c2 , ••• , c,_k) and y is found from y(kl = p (x, c1 , c2 , ••• , c,_k) by k-fold integration. In particular, if a second-order equation does not contain y, the change of variables y' = p produces an equation of the first order. Example 1. d6y - _.!_ d•y = 0 dx& x dx• · Assuming dd•~ = p, we get ddp _ _!__ p = 0; separating variables and X X X integrating we have: In Ip) =In Ix) +Inc, or p =ex, :~=ex, whence y = c1x5 + C2X3 + C3X2 +C4X + c5. Example 2. Find the law of motion of a body falling in the air without an initial velocity. Take the air resistance to be proportional to the square of the velocity. The equation of motion is of the form d2s (ds )2m dt2 = mg-k ,di , where s is the distance covered, m the mass of the body, and t the ds time. For t = 0, we have s = 0 and. dt = 0. 94 I. DIFFERENTIAL EQUATIONS The equation does not explicitly contain the unknown function s, and so the order of the equation may be reduced by taking : = v. Then the equation of motion will be of the form dv 1w• mdt=mg- · Separating the variables and integrating, we get 0 m dv dt t s dv I t h-1 kv mg-kv" = ; = m mg-ku• = k Vi an Vi; 0 from this we have v = ~i tanh (k Vg t); multiplying by dt 'and integrating again. we find the law of motion: s = :. In cosh (k Vg t). 2. The equation does not contain an independEnt variable: f (y, y', y•, , , ., yCnl) = 0. Here, the order of the equation may be reduced by unity by the substitution y' = p, p being regarded as a new unknown function of y, p = p (y), and, hence, all derivatives ~! have to be expressed in terms of the derivatives of the new unknown function p (y) with respect to y: dy=p dx • dig _dp _dpdy _dp dx3 - dx- dydx- dyP' z=!(:;p) = :~~ ( :: p) ~= ~; p' +(::rp and similarly for derivatives of higher order. It is obvious here that the derivatives :! are expressed in terms of derivatives of order not higher than k-1 of p with respect to y, which is what reduces the order by one. In particular, if a second-order equation does not contain an independent variable, the indicated change of variable leads to a first-order equation. Example 3. dly (dy)ly dxl- dx = 0· 2. DiffERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 95 Putting 2= p, Z= p :: , we get the separable equation yp :: - p2 = 0, the general solution of which is p = c1y or 2= CJJ. Again separating variables and integrating we get In Iy I=c1x+Inc, or y = c,ec•Jf. Example 4. Integrate the equation of a simple pendulum x+a' sin x = 0 for initial conditions x (0) = x0 , x(0) = 0. We reduce the order by putting · ·· dux=u, x=vdx' vdu=-a'sinxd.r, ul 2 =a2 (cos x-cos x0), tl::a: ±a V2 (cos x-cos x0), " 1=±-·-s dx a Y2 Ycos x- cos x, •~ d.r dl =±aY2 (cos x-cos x0), The integral on the right side is not expressible by elementary functions but is readily reducible to elliptic functions. 3. The left-hand side of the equation. F(X, y, Y1 I y", ... ,y'"1) = 0 (2.4) is a derivative of some diOeren.tlal expression. of the (n.-l)th de ..,. (x 1 '"-'')or r 'V , y, y , .•. , y . Here we readily find a so-called first integral, i.e. a differential equation of the (n- I)th order containing one arbitrary constant. This equation is equivalent to the given equation of the n.th order, so we have reduced the order of the equation by one. Indeed, (2.4) may be rewritten in the form !!. (x, y, Y1 , ••• , y"'-1') is identically zero. Consequently, the function (x, y, yl, .•• , y"'-11) is equal to a constant, and we get the first integral «<>(x, y, y~, .•., y"'-1')=c. Example 5. yy" +(yl)l = 0. This equation may be written as d (yy1 ) = 0, whence yy' = c, or y dy =c1 dx. Thus, y1 = c1x +c1 is the complete integral. Sometimes the left-hand side of the-equation F(x, y, Y1 , ••• , !1'"1) = 0 becomes the derivative of the (n.-1 )th order differential expression (x, y, y1 , ••• , y'"-11) only after multiplication by some factor J.L(X, y, y'• ••. , y'"-1'). 96 I. DIFFERENTIAL EQUATIONS Example 6. yy" -(y')' = 0. 1 " ( ')~Multiplying by the factor J.L = 2 , we get Y ~ Y = 0 ory y d ( ') • ddx i =0, whence i=C1 or dxlnlul=c1 . Hence,lnlyl=c1x+lnc2, c, > 0, whence y = c,e:•x, c, =1= 0, as in Example 3 of this section. Note. When multiplying by the factor J.L (x, y, y', ... , ycn-1') extraneous solutions may be introduced that make this factor vanish. If the factor J.L is discontinuous, a loss of solutions is possible. ln Example 6, when multiplying by J.L = ~, the solution y = 0 was lost. However, this solution can be included in the solution obtained y= c,e:•x if it is taken that c, can assume the value 0. 4. The equation F (x, y, y', ... , y141) = 0 is homogeneous in the arguments y, y', ... , y1111• The order of the following equation (which is homogeneous in y, y', ..• , ylnl) F(X, y, y'' ... ' !/41) = 0, (2.5) that is, an equation for which the identity F(x, ky, ky', ... , ky1" 1)=kPF(x, y, y', ••• , ylnl) holds, may be reduced by unity by the substitution y=efzdx, where z is a new unknown function. Indeed, differentiating, we get y' =efz dxz, y" = efz dx (z' +z'), y'" =efzax (Z3 +3zz' +z"), (that this equality is true can be demonstrated by the method of induction). By substitution into (2.5) and by noting that by virtue of homog!!neity the factor t! fz dx may be takeu outside the sign of the function F, we get t! fz dx I (x, Z, z'' ••. ' zln- II)= 0 or, cancelling eP Iz dx, we have f (x, z, z'• ••. • z,,._,,) = 0. 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORL:lER AND HIGHER 97 Example 7. yy"- (y')1 = 6xy1• Putting y=efzd~.weget z'=6x, z=3x'+cv y=ef1ax'+c,Jdx or y = c,elx'+c,x). Especially frequently encountered in applications are second-order differential equations that allow for reducing the order. (1) F (x, y")=O. (2.6) Here, we can lowPr the order by the substitution y' = p and we can reduce it to the equation F ( x, ::) = 0 considered on page 75. Equation (2.6) can be solved )or the second argument y" = f(x) and integrated twice, or we can introduce a parameter and replace (2.6) by its parametric representation whence (2) tPy dx' = cp (t), X= 'I> (t), dy' = y" dx = cp (t) '!>' (t) dt, y' = )

' (t) dt +c1, dy = y' dx, Y= ~ Ucp (/) '!>' (I) dt +C1] '!>' (t) dt +C2• F(y', y")=O. (2.7) Putting y' =p, transform (2.7) to equation (1.61), page 76, or represent (2.7) parametrically: y:= cp (t), y;Jt = w(t), whence dy' q>' (t) dt sq>' (t) dt dx=7=~· X= ~+c1 , y is then determined by a quadrature: I d q>' (/) sq> (/) q>' (f) d dy=y x=cp(t)w(t) dt, y= 'lj!(t) t+c,. (3) F(y, y")=O. (2.8) The order can be reduced by putting dy d2y - dp dy - dp dx = p, dx2 - dy dx- P dy • If equation (2.8) is readily solv8ble for the second argument y" = f (y), then by multiplying this equation termwise by 2y' dx = 2dy, we get .l7K 98 1. DIFFERENTIAL EQUATIONS d (y' )2 = 2f (y) dy, whence 2=+ V2 ~ f (y)dy+cl' Equation (2.8) may be replaced by its parametric representation y = cp (t), y" ='If' (t); then from dy' = y" dx and dy = y' dx we get y' dy' = y" dy or ~ d (y')1 = "' (t) cp' (t) dt, (y')1 = 2 ~ 1j) (t) cp' (t) dt +C1,· y' = ± y2 ) 1j) (t) cp' (t) dt +c1, then from dy = y' dx we find dx and then x: dx=dy= cp'(t)dt !i' ± l/2) '¢ (t) cp' (t) dt+c1 ' ± r cp'(t)dt +c X= J v2)¢(t)cp'(f)dt+cl I' <2·9) lt is equation (2.9) and y = cp (t) that define in parametric form a family of integral curves. Example 8. y" = 2!/, y (0) = 1, y' (0) = 1. Multiplying both sides of the equation by 2y' dx, we get d (y')1 = 4!/ dy, whence (y')' = y' +c1• Taking the initial conditions into account, we find that c1 = 0 and y' = y1 • Hence, d;= dx, _..!._ = x +c2, c2 =y y I =-1, Y=-1-.-% 3. Linear Differential Equations of the nth Order An nth order linear differential equation is an equation linear in the unknown function and its derivatives and, hence, of the form ao (X) y1n1+al (X) Jfn-1) + ... +an-l (X) y' +a, (X) Y= cp (X). (2.10) 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 99 If the right side q> (x) ==0, then the equation is called homogeneous linear since it is homogeneous in the unknown function y and its derivatives. If the coefficient a0 (x) is not zero at any point of some interval a =s:;;:; x =s:;;:; b, then by dividing by a0 (x) we can reduce the homogeneous linear equation, for x varying on this interval, to the form y +p1 (x) y-- ~Pi (x)y (t), where q> (t) is an arbitrary n times differentiable function, the derivative of which, on the interval of variation of t under consideration, i~ q>' (t) =1= 0. Indeed, A derivative of any order ~~ is a homogeneous linear function of th d . t' dg dig dky d tl h b t'e enva tves dt , dt' , ••• , dtk an , consequen y, w en su s t· tuted into equation (2.11) retains its linearity and homogeneity. Linearity and homogeneity are also retained in a homogeneous linear transformation of the unknown function y (x) =a(x) z(x). 7* 100 I. DIFFERENTIAL EQUATIONS Indeed, by the formula for differentiating a product, you = a (x) z +Pt (x) y~n-u + ··· ···+Pn (x) Y2J· A corollary of properties (1) and (2) is L [ ~ C;Y;] = 'i:C;L [Y;]' I= I I= I where c1 are constants. Proceeding from the properties of the linear operator L, we shall prove a number of theorems on the solutions of a homogeneous linear equation. 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 101 Theorem 2.2. If y1 is a solution of a homogeneous linear equation L [y] = 0, then cy1 , where cis an arbitrary constant, is likewise a solution of that equation. Proof. Given L [y1] =0. It is required to prove that L [cy1l =0. Taking advantage of property (I) of the operator L, we gel L [cy1] ==cL [y1] ==0. Theorem 2.3. The sum y1 +y, of solutions y1 andy, of a homogeneous linear equation L [y] = 0 is a solution of that equation. Proof. Given L [y1] ==0 and L [y,] == 0. It is required to prove that L [y1 +y,] ==0. Taking advantage of Property (2) of the operator L, we get L [y1 +y,] ==L [y1] +L [y,] ==0. Corollary to Theorems 2.2 and 2.3. A linear combination with m arbitrary constant coefficients ~C;Yt of solutions y1, y,, ... , Ym i= I of a homogeneous linear equation L [yJ = 0 is a solution of that equation. Theorem 2.4. If a homogeneous linear equation L [yJ = 0 with real coefficients P; (x) has a complex solution y (x) = u (x) + iv (x), then the real part of this solution u (x) and its imaginary part v (x) are separately solutions of that homogeneous equation. Proof. Given L [u (x) +iv (x)] ==0. It is required to prove that L[u]==O and L[v]=O. Taking advantage of Properties (I) and (2) of the operator L, we get L [u +iv] = L [u] +iL [v] ==0, whence L[u]==O and L[v]=O, since the complex function of a real variable vanishes identically when, and only when, its real and imaginary parts are identically equal to zero. Note. We applied Properties (I) and (2) of the operator L to the complex function u (x) +iv (x) of a real variable, which is obviously admissible since in proving Properties (I) and (2) use was made only of the following properties of the derivatives: (cy)' = cy', where c is a constant, and (y1 +y,)' = y~ +y;, which hold true for complex functions of a real variable as well. The functions y1 (x), y, (x), ... , Yn (x) are called linearly dependent over a certain interval of variation of x, a~ x ~ b, if there exist constant quantities et1, a.2, ••• , a.n such that on this interval CX.1Y1 +CX.2Y2 + ••·+CX.nYn = 0, (2.12) and at least one a.; =1= 0. Now if the identity (2.12) holds true only 102 I. DIFFERENTIAL EQUATIONS for a1 = a2 = ... =an= 0, then the functions y1 , y2, ••• , Yn are called linearly independent functions over the interval a~ x ~b. Example l. The functions 1, x, xS, ... , xn are linearly independent on any interval a~ x ~ b, since the identity a1 +a,x+a8X2 + . •• +an+lxn=O (2.13) is only possible if all the a1= 0. If even one a1=fo 0, the left member of the identity (2.13) would be a polynomial of degree not higher than n that can have no more than n distinct roots and, hence, vanishes at no more than n points of the interval under consideration. Example 2. The functions ek,x, ek•x, ••. , ek..x, where k1=P k1 for i =P j, are linearly independent on any interval a~ x ~b. Suppose that the functions under consideration are linearly dependent. Then a1ek•x+a2ek.x+ .•. +aneknx =0, (2.14) where at least one of the a19'= 0, for instance, for the sake of definiteness, an=t=O. Dividing the identity (2.14) by ek,x and differentiating, we get a 2 (k2 -k1)e x and differentiating, we obtain a linear relation between n-2 exponential functions with distinct exponents. Continuing this process n-1 times, we get an (kn-kl) (kn-k,) ••. (kn-kn-1) e x = 0. (2.17) The degrees of the polynomials Q1 and P1 (i = 2, 3, ... , p) coincide since on differentiating the product P,(x) eP~, p =1= 0, we get [P1(x)p +PI (x)] eP~, that is, the coefficient of the highest-degree term of the polynomial P,(x), after differentiating the product P,(x)eP~, acquires only the nonzero factor p. In particular, the degrees of the polynomials PP. (x) and QP (x) coincide, and hence the polynomial QP (x) is not iaentically zero. Dividing (2.17) by e x and differentiating n1 + 1 times, we get a linear relation with a still smaller number of functions. Continuing this process p-1 times, we obtain Rp(x)e (xo) = 0 and so that not all the a; are zero. Such a choice is possible since the determinant of tht' homogeneous linear system (2.21) of n equations in n unknowns a.; is zero, W (x0 ) = 0, and, hence, there exist nontrivial solutions of this system. In such a choice of a;, the linear combination Y= a.1Y1 (x) +a.zYz (x) + · · · +a.nYn (x) will be a solution of the homogeneous linear equation (2.20) that satisfies, by virtue of the equations of the system (2.21), the zero initial conditions y (x0) = 0, y' (x0) = 0, ... , yln- 11 (x0) = 0. (2.22) Obviously, such initial conditions are satisfied by the trivial solution y === 0 of the equation (2.20) and, by the uniqueness theorem, the initial conditions (2.22) are satisfied only by this solution. Consequently, a.1y1 (x)+a2y2(x) + ... +a.nYn (x) =0 and the solutions y" Yz, ... , Yn• despite the hypothesis of the theorem, are linearly dependent. Note. I. From the Theorems 2.5 and 2.6 it follows that the solutions Y~t y2 , ••• , Yn of equation (2.20), that are linearly independent on the interval a~ x ~ b, are also linearly independent on any interval a1 ~ x ~ b1 located on the interval a~ x ~b. 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 105 Note. 2. In Theorem 2.6, in contrast to Theorem 2.5, it was assumed that the functions Y1 , y2, ••• , Yn are solutions of the homogeneous linear equation (2.20) with continuous coefficients. It is not possible to reject this demand and to consider the functions Y1• Yz• ... , Yn arbitrary n- 1 times continuously differentiable functions. It is easy to give examples of linearly independent functions that are of course not solutions of the equation (2.20) with continuous coefficients for which the Wronskian not only vanishes at separate points but is even identically zero. For instance, let there be defined two functions y1 (x) and y2 (x) on the interval 0:::;;;; x:::;;;; 2: 0 YdX)=(x-1)2 for o:::;;;;x:::;;;; 1 and Yl (x) = 0 for 1 + P1 (x) y + • • • + Pn (x) Y= 0, (2.20) it is possible, by the substitution y = y1 ) u dx, to reduce the order of the equation and retain its linearity and homogeneity. Indeed, the substitution y = y1 ) u dx may be replaced by two substitutions: y = y1z and z' = u. The homogeneous linear transfor- mation y = y1z (2.23) preserves the linearity and homogeneity of the equation(see pages 99-100) consequently, (2.20) is thus transformed to a0 (x)z+a1 (x) z + a1 (x) z+ ... + an_1(x) z' = 0, and the substitution z' = u reduces the order by one: a0 (x) u + a1(x) u+ ... + an_.(x) u = 0. Note that the same substitution y = y1 ) u dx, where y1 is a solution of the equation L [y] = 0, also reduces by unity the order of the nonhomogeneous linear equation L [y] = f (x), since this substitution does not affect the right-hand side of the equation. Knowing k linearly independent (on the interval a~x~b) solutions y1, y2 , ••• , y, of a homogeneous linear equation, it is possible 108 I. DIFFERENTIAL EQUATIONS to reduce the order of the equation to n-k on the same interval a~x~b. Indeed, reducing the order of the equation L [y] = 0, (2.20) by unity by the substitution y = y11 ) u dx, we again get a homogeneous linear equation a0 (x) u ycn-Ht ... Yhn-1> y(ll-1) y~n> y~"' ... y<"'fl y(ll) or, expanding the elements of the last column, Yt Y2 Yn y~ y~ y~ W [yl, Y2• • • •• Yn] Y(ll)_ y2 Yh"' (2.31) The equation obtained, (2.31), is the desired homogeneous linear equation having the specified fundamental system of solutions y1 , y2, ... , Yn (since for y = ydi = 1, 2, ... , n) W [y1, y2, ... , Yn• y] ==0). Dividing both sides of (2.31) by the nonzero coefficient W [y1, y2, ••• •. •, Yn] of the highest derivative, we reduce it to the form (2.28). From this it follows, in particular, that Yt y' 1 ··· Yn ... y~ y~n-2l y~n-2) .. , yhn-2> y , , , y~n-2> y~n-Il y~n-ll , , . Yhn-n is equal to the sum over i from I to n determinants differing from the Wronskian in that in them the elements of the ith row have been differentiated while the remaining rows of the Wronskian are preserved without change. In this sum, only the last determinant, for i = n, which coincides with the determinant (2.32), can be different from zero. The other determinants are zero, since their ith and i +1st rows coincide. Consequently, p1 (x) = - ~ . Whence, by multiplying by dx and integrating, we get or In IW I= - ~ P1 (x) dx +Inc, W = ce- I Pa " X -I p, (x) dx W =Ce Xo For X=x0 we get c= W(x0 ), whence X -I p, (x) dx W (x) = W (x0 ) e x. (2.33) (2.34) Formulas (2.33) or (2.34), which were first derived by M. Ostrogradsky and, independently, by Liouville, are called OstrogradskyLiouville formulas. The Ostrogradsky-Liouville formula (2.34) may be employed for integrating a second-order homogeneous linear equation y" +P1 (x) y' +P2 (x) Y= 0, (2.35) if a single nontrivial solution of this equation, y1 , is known. According to the Ostrogradsky-Liouville formula (2.34), any solution 112 I DIFFERENTIAL EQUATIONS ------ of (2.35) must also be <:1 solution of the l'quation or This linear equation of the first order is most easily integrated by the integrating factor method. Multiplying by f.1. = ~2 , we get yl ..!!___ (L) = ~e-J p 1 tx)dx dx Yt y~ • wh{·nce or 4. Homogeneous Linear Equations with Constant Coefficients and Euler's Equations I. Ifomogeneous linear equations with constant coefficients. If in a homogeneous linear equation (2.36) all the coefficients a; are constant, then its particular solutions may be found in the form y=ekx, where k is a constant. Indeed, putting into (2.36) y = ekx and yx and etm-!li>x that correspond to the pair of complex conjugate roots k1 =a+~i and k, =a-~i, can be replaced by two real solutions: the real and imaginary parts (see page IOI) of one of the solutions elm+flilx = emx (COS ~X+ i sin ~X), or etm-flilx = emx (COS~X-i sin ~X). Thus, to the pair of complex conjugate roots k1 , 2 =a±fH there correspond two real solutions: emx cos ~x and e-x sin ~x. Example 3. y" +4y' +5y = 0. The characteristic equation is of the form k2 -t 4k +5 = 0; its roots are k1 • 1 = -2 ± i. The general solution is y = e-sx (c1 cosx +c, sin x). Example 4. y"+a2y=0. The characteristic equation k2 +a1 = 0 has the roots k1 , 1 = ± ai. The general solution is y=c1 cosax+c, sin ax. 114 I. DIFFERENTIAL EQUATIONS If there are multiple roots among the roots of the characteristic equation, then the number of different solutions of the form ekx is less than n and, hence, the lacking linearly independent solutions have to be sought in a different form. We shall prove that if a characteristic equation has a root k; of multiplicity a.;. then the solutions of the original equation will be not on!y efl;x, but also xek;x, x2ek;x, ... , xa•- 1ek;x. First suppose that the characteristic equation has a root k; = 0 of multiplicity a.i. Hence, the left-hand side of the characteristic equation (2.37) has a common factor ka.; in this case, i.e. the coefficients an= an- 1 = ... =an-a.;+ 1 = 0 and the characteristic equation is of the form aok"+atkn-l + ... +an-a;ka;=O. The corresponding homogeneous linear differential equation aoy +a1y+ ... +an-a1y =0 obviously has the particular solutions 1, x, x2 , ••• , ;ca.;- 1 since the equation does not have derivatives of order lower than a.;. Thus, to the multiple root k; = 0 of multiplicity a.; there correspond a.; linearly independent (see page 102, Example 1) solutions 1, x, x2, ... , xa•-1. If the characteristic equation has a root k; =1= 0 of multiplicity a.i, then a change of variables y=ek•xz (2.38) reduces the problem to the already considered case of a zero multiple root. . Indeed, as was pointed out on pages 99-IOO,a homogeneous linear transformation of the unknown function (2.38) preserves linearity and homogeneity of the equation. In the change of variables (2.38) the coefficients are also held constant, since y

= (zektx) +pzk; + P(P2~ 1 ) zkr+ ... +zk~) , and after substitution into equation (2.36) and cancelling of ektx only constant coefficients remain in z, z', •• , , z. And so the transformed equation will be a homogeneous linear equation of the nth order with constant coefficients boz x = ePx (cos qx +i sin qx) .~ 116 J. DIFFERENTIAL EQUATIONS and, separating the real and imaginary parts, one can obtain 2a. real solutions: ePxcosqx, xePxcosqx, x2ePxcosqx, ... , xa.- 1ePxcosqx, } (2.43 ePx sin qx, xePx sin qx, x2ePx sin qx, •.. , xa.-1ePx sin qx. ) Taking the real parts and the imaginary parts of the solutions corresponding to the conjugate root p-qi of the characteristic equation, we will not get any new linearly independent solutions. Thus, to the pair of complex conjugate roots p ± qi of multiplicity a there correspond 2a. linearly independent real solutions (2.43). Example 6. yiV +2y" +y=O. The characteristic equation k' + 2k1 + I = 0 or (k2 +I)1 = 0 has double roots ±i. Hence, the general solution is of the form y= (c1 +c2X) cosx+(c3 +c,x) sin x. 2. l!uler's equations. Equations of the form aoX"y+atxy+ ... +an-txy' +any=O, (2.44) where all the a1 are constants, are called Euler's equations. An Euler equation can be transformed, by changing the independent variable x = e1 •, into a homogeneous linear equation with constant coefficients. Indeed, as was shown on page 99, the linearity and homogeneity of an equation are preserved during transformation of an independent variable, and the coefficients become constant because dy dy -t dx= dte ' d2y -2t ( d2y dy) dt2 = e dt2 -eli • (2.45) where all ~~ are constants; upon substitution into equation (2.44) the factors e-llt and x 11 = e11t cancel. The validity of (2.45) can readily be proved by the method of induction. Indeed, assuming that (2.45) is true and differentiating it once again with respect to x, we prove the truth of equality * Or x = -et if x < 0; for the sake of definiteness, we will henceforth consider x > 0. 2. DIFFERENTIAL EQUATIO,"IIS OF THE SECO:-.JD ORDER AND HIGHER 117 dk+ly (2.45) for -k- as well: dx +1 dk+ly- - t(~ dy+A d2y+···+~ dky)= 1 dt pz dt2 k dt" -e-<.t+l>t ( dy d2y + + d"+~y) - "h dt +i'z dt2 • • • i'.t+l dtk+J ' where all the '\'; are constants. Thus, the validity of the formula (2.45) is proved and, consequently, the products with constant coefficients k dky A dy d2y dky X dx" = t'l dt +~~ d/2 + ···+~k dtk that linearly enter into the Euler equation n dk ~ a "xk _1!. = 0~ n- dxk k=o (2.44') are linearly (with constant coefficients) expressed in terms of the derivatives of the function y with respect to the new independent variable t. From this it follows that the transformed equation will be a homogeneous linear equation with constant coefficients: (2.46) Instead of transforming the Euler equation into a linear equation with constant coefficients, the particular solutions of which are of the form y=e"t, it is possible immediately to seek the solution of the original equation in the form y=x", since e"t =x". The resulting equation (after cancelling x") a0 k (k-1) .. .(k-n + 1) +a1k (k-1) .. .(k-n +2) + ... . . . +an= 0 (2.47) for a determination of k should coincide with the characteristic equation of the transformed equation (2.46). Hence, to the roots k; of (2.47) of multiplicity a.; there correspond the solutions e"•t, tek1t, tzek'', ••• , ta;-teJt1t of the transformed equation or xk1, xk; In x, xk; ln2 x, .•• , xk• Jna•- 1x 118 I. DIFFERENTIAL EQUATIONS of the original equation, and to the complex conjugate roots p ± qi of (2.47) of multiplicity a there correspond the solutions ePI COS qf, fePI COS qf, ... , ta.-tept COS qf, eP1 sin qt, teP1 sin qt, ... , ta.-teP1 sin qt of the transformed equation or xP cos (q In x), xP In x cos (q In x), ... , xP lna.-t x cos (q In x), xP sin (q In x), xP In x sin (q In x), ... , xP lna.- 1 x sin (q In x) of the original Euler equation. Example 7. x2y" + ~ xy' -y=O. We seek a solution in the form y=x"; k(k-I)+~k-1=0, whence k1 = ~ , k1 = -2. Hence, the general solution for x > 0 is of the form Example 8. x•y"-xy' +y=O. We seek a solution in the form y=x"; k(k-1)-k+l =0, or (k-1)1 = 0, k1 ,, =I. Hence, the general solution for x > 0 will be y= (c1 +c.lnx)x. Example 9. x2y" +xy' +y=O. We seek a solution in the form y=x"; k(k-l)+k+ I =0, whence k1, • = ±i. Hence, the general solution for x > 0 is of the form y = c1 cos In x +c1 sin In x. Equations of the form ao (ax+b)n y are any real numbers and x0 is any point of the interval a< x (x0) = y~k> (k = 0, 1, 2, ... , n-1 ), (2.52) where a~x0 ~b. By requiring that the solution (2.51) should satisfy the initial conditions (2.52), we arrive at the following system of equations " ~ C;Y; (Xo) +Y(X0 ) = Yo• 1=1 " ~ c,y; (X0 ) +y' (Xo) = y~, 1=1 (2.53)" 1~ C;Yi (X0) +y" (Xo) = y;, l ~;,y<,~-:,(~o;;yl~--1>(~o;_·y~~-:,. 1=1 This system of n equations, which is linear with respect to the constants c1, in n unknowns with arbitrary right-hand sides admits of a unique solution for the c; (i = 1, 2, ... , n), since the determinant of the system (2.53), being the Wronskian W [Yt> y1 , ••• , Yn] for the linearly independent system of solutions of the corresponding homogeneous equation, is different from zero for any values of x on the interval a~x~b and, in particular, for x=x0 • Hence, integration of a nonhomogeneous linear equation reduces to finding one particular solution of the equation and to integrating the corresponding homogeneous linear equation. Example t. y" +u=x. A particular solution of this equation, y = x, which is obviously the general solution of the corresponding homogeneous equation, is of the form y = c1 cos x +c1 sin x (see page 113, Example 4). Thus, the general solution of the original nonhomogeneous equation is y = c1 cos x+c2 sin x +x. 122 I. DIFFERENTIAL EQUATIONS If choice of a particular solution of the nonhomogeneous equation is difficult, but the general solution of the corresponding homoge- n neous equation y = ~ c1y1 is found, then it is possible to integrate l=l the nonhomogeneous linear equation by the method of variatwn of parameters. In applying this method, we seek the solution of the nonhomo- n geneous equation in the form y= ~ct(x)y1; that is, in place of I l=l the unknown function y we actually introduce n unknown functions c1 (x). Since the choice of functions c1 (x) (i = I, 2, ... , n) has to satisfy only one equation y +Pl (x) yln-1> +•.•+Pn (x) y = f (x), (2.49) we can demand that these n functions c1 (x) should satisfy some other n-1 equations, which we cpoose so that the derivatives of n the function y = ~ ct(x) y1(x) should be, as far as possible, of the l=l form that they have in the case of constant c1• Choose the c; (X) so that the second sum on the right of n n y' = ~ c1 (x) yl (x) + ~c; (x) y1 (x) l=l l=l should be equal to zero, n ~ci (x) y1(x) =0, 1=1 and, consequently, n y' = ~ c1 (x) yj (x), l=t that is, y' Is of the same form as in the case of constant c1• In the same fashion, we demand that the second sum in the second derivative n " y" = ~ c1(x) y• + ~ c; (x) Yi I= 1 l= I vanish and we thus subject c1 (x) to the second condition: n ~cl (x) yi = 0. I= I n· Continuing to evaluate the derivatives of the function y = ~ c1 (x) y1 l=l 2. DIFFERENTIAL EQUATIONS OF THE.SECOND ORDER AND HIGHER 123 up to order n-1 inclusive and demanding each time that the sum II ~ c~ (x) y]k' (x) vanish, 1=1 we get II ~ ci (x) y}k' (x) = 0 (k=O, 1, 2, •.. , n-2), 1=1 II y= ~c,.(x)y1, 1=1 II y' = ~c1 (x)y{, 1=1 n y" = ~c1 (x) yi, 1=1 n yen-I> = ~ C; (x) y~11-u, i= 1 n 11 ytnl = ~ c;(x) y}11 >+ ~ci (x) y]11-u. 1=1 1=1 J n (2.54) (2.55) In the last equation, we cannot demand that ~c/y~n-u = 0 since l=t the functions c1(x) are already subject to the n-1 conditions (2.54), and we still have to satisfy the original equation (2.49). Substituting y, y', ... , y1n1 of (2.55) into the equation ytn> +Pt (x) ytn-1) +... +Pn (x) y = f (x), (2.49) we get the equation we need to determine c1 (x) (i = 1, 2, ... , n). And it is obvious here that in the left-hand member of (2.49) only n the sum ~ c/ (x) y}11-u will remain, since all the other terms are of I= 1 the same form as in the case of constant c1, and when the c1 are n constant the function y = ~ c1y1 satisfies the corresponding homo- l=t geneous equation. We can also convince ourselves of this by means of direct cal- culation: n n n 11 ~ c{yJII-u + ~ c,y~·n' +Pt (x) ~ ciy}n-u + P2 (x) ~ CtY111- 11 +·••i:1 i=l 1=1 1=1 n •·· + Pn (x) ~CiYI = f (X) 1=1 124 I. DIFFERENTIAL EQUATIONS or II II ~cjy}11-u + ~ci [Y1111 +Pi (x) Y~11-11 + ... +P11 (x) Y1J = f (x). (2.56) 1=1 1=1 All the y1 are particular solutions of the corresponding homogeneous equation, consequently, y}111 +p1(x)y}11- 11 + ... +P11 (x) y,. ==0 (i = 1, 2, ... , n) and equation (2.56) takes the form II ~ cjy~11-11 = f (x). 1=1 To summarize, then, the functions cdx) (i = 1, 2, ... , n) are determined from the system of n linear equations II ) ~ci (x) Y1 = 0, 1=1 II ~ci (x) y; = 0, 1=1 II ~cj (x) yj = 0, I= 1 II ~ cj (x) y~ll-21 = 0, 1=1 II ~ cj (x) y}11- 11 = f (x) 1=1 (2.57) with a nonzero determinant of the 1'ystem, since this determinant is the Wronskian for linearly independent solutions of the corresponding homogeneous equation. Having detennined all the cj (x) = q>1(x) from (2.57), we find, using quadratures, c1 (x) = ~ q>1 (x) dx +~Example 2. , + I Y Y =cos x' 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 125 The general solution of the corresponding homogeneous equation is of the form y = c1cos x+c1 sin x. We vary c1 and c2: y= c1 (x) cos x+c, (x) sin x. c1 (x) and c, (x) are found from the system of equations (2.57) c; (x)cosx+c~ (x) sinx=O, -c~ (x) sin x+c~ (x)cosx=-1-, COS X whence • sin x c (x)- - -1 - COS X' c; (x) = 1, cl (x) =In Icos X I+cl; C2 (x) = x+c2• The general solution of the original equation is y_=c1 cosx +c2 sin x +cosx In Icosx I+x sin x. Example 3. x+a'x=f(t). The general solution of the corresponding homogeneous equation is of the form x = c1 cos at +c2 sin at. Varying the constants x = = c1 (t) cos at+ c2 (t) sin at, we obtain c~ (t) cos at +c~ (t) sin at= 0, -ac~ (t) sin at+ ac~ (t) cos at= f (t), and from this t c; (t)' = -! f (t) sin at, C1 (t) = --!J5f (u) sin au du +Cu 0 t c~(t)=..!..f(t)cosat, c2(t)=..!..5f(u)cosaudu + c,,a a 0 t t cos at 5 . sin at 5fx(t)= --a- f (u) sm audu+-a- (u)cosaudu + 0 0 +clcos at+c,sin at, or t X (t) = !sf (u)[cos au sin at -sin au COS atj du +C1 COS at +c2 sin at, 0 126 I. DIFFERENTIAL EQUATIONS whence we finally get t x(t) =-;}-sf(u) sin a (t-u)du +clcos at +c, sin at. 0 Note that the first summand on the right is a particular solution of the original equation that satisfies the initial conditions x (0) = 0, x(O)=O. Thus, a knowledge of n linearly independent particular solutions of the corresponding homogeneous equation permits us, by usiQg the method of variation of parameters, to integrate the nonhomogeneous equation L [y]=f(x). Now if only k, where k < n, linearly independent solutions y10 y,, ... , Y~c of the corresponding homogeneous equation are known, then, as pointed out on pages 107-108, a change of variables permits reducing the order of the equation to n-k while retaining its linearity. Observe that if k = n- 1, then the order of the equation is reduced to the first, and a first-order linear equation can always be integrated by quadratures. In similar fashion we can utilize the k solutions y1 , y,, ...,Y~t of the nonhomogeneous equation, since their differences are already solutions of the corresponding homogeneous equation. Indeed, L [y.] == f (x), consequently L [y,-yp] == L rYJ]-L [Yp] =f (x)-f (x) ==0. If the particular solutions of the corresponding homogeneous equation (yl-YII>• . ..., (2.58) are linearly independent, then the order of the equation L(y) = f (x) may be reduced to n-(k-1). Obviously, the other differences y1- Yp are linear' combinations of the solutions (2.58) u1-Yp = - and consequently cannot be employed for further reduction of the order. There is also the so-called Cauchy method for finding a particular solution of a nonhomogeneous linear equation L [y (x)] = f (x). (2.59) In this method, it is assumed that we know the solution K(x, s) 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 127 (dependent on a single parameter) of the corresponding homogeneous equation L [y (x)] = 0, which solution satisfies the conditions K(s, s)=K'(s, s)= ... =K1n-21 (s, s)=0; K'n-1) (s, s) =I. It is easy to verify that in this case X y (x) = ~ K (x, s) f (s) ds (2.60) (2.61) (2.62) will be a particular solution of the equation (2.59), which solution satisfies the zero initial conditions Y(Xo} = y' (X0) = ... = y (x, s)f (s) ds +f (x) Xo Putting (2.62) and (2.63) into (2.59), we get }f. ~ L [K (x, s)] f (s) ds +f (x) = f (x), x., (2.63) since K (x, s) is a solution of the corresponding homogeneous equation and L[K(x, s)] =0. The solution K(x, s) may be isolated from the general solution n y = ~ c1y1(x) of the homogeneous equation if the arbitrary constants l=I c1 are chosen so as to satisfy the conditions (2.60) and (2.61). Example 4. The general solution of the equation y" +a2y = f (x) (2.64) 128 I. DIFFERENTIAL EQUATIONS is y = c1 cos ax+ c, sin ax; the conditions (2.60) and (2.61) lead to the following equations: Hence, c1 cos as+ c, sin as =0, -ac1 sin as+ ac, cos as= 1. sin as cos as c.=--a-' c,=-aand the sought·for solution K (x, s) is of the form K(x, s) = .! sin a (x-s). According to (2.62), the solution of l =1uation (2.64) that satisfies zero initial conditions is representable as "I 5 .y(x) =a sin a (x-s) f (s) ds. X, For x0 =0, this solution coincides with the one obtained earli'er (see pages 125-126) by a different method. We can give a phy~ical interpretation to the function K (x, s) and to the solution of the linear equation with right-hand side in the form (2.62). It will be more convenient here to denote the independent variable by t. In many problems the solution y (t) of the equation !/"1+Pt (t) Y'"-11 + ... +p,. (t) Y= f (t) (2.65) describes the displacement of some system, while the function f (t) describes ~ force acting on the system, and t is the time. First suppose that when t s + e, we get s+e y. (t) = K (t, s+ e*) ~ f. ('r:) dr. = K(t, s +e•), where 0 < e* < e; hence, limy, (t) = K(t, s). lt is therefore natural to call the function K (t, s) the influence function of instantaneous momentum at time t = s. Partitioning the interval (t0 , t) by points s; (i = 0, l, ... , m) into m equal parts of length !ls = 1 mto, we represent the function f (t) in (2.65) as a sum of the functions f1(t), where fdt) is different from zero only on the ith interval s1_ 1 < t < s;, on which it coincides with the function f (t): m f (t) = ~ It (t). I= I By virtue of the superposition principle (page 120) the solution of the equation (2.65) is of the form m y (t) = ~ yt(t), l=l where y1 (t) are solutions of the equations ylni+Pt (t)y!n-1)+ • • • +p,.(t)y=f.-(t) with zero initial values. If m is sufficiently great, the solution Y;(t) may be regarded as the influence function of instantaneous momentum of intensity f; (s1) !ls. Consequently, m y (t)- ~ K(t, s;) f (s1) !ls. I= I Passing to the limit as m -+oo, we get the solution of the equation (2.65) with zero initial conditions in the form I y = ) K (t, s) f (s) ds, In which indicates that the effect of a constantly acting force may be regarded as the superposition of the influences of iustantaneous momenta. 130 I. DIFFERENTIAL EQUATIONS 6. Nonhomogeneous Linear Equations with Constant Coefficients and Euler's Equations In many cases, when solving nonhomogeneous linear equations with constant coefficients, it is possible to select without difficulty certain particular solutions and thus to reduce the problem to integration of the appropriate homogeneous equation. For example, let the right-hand side be a polynomial of degree s and hence the equation will be of the form aoy• (2.69) where all the a1 and A1 are constants. As has been shown above (see page 114), the change of variables y = eP~z transforms equation (2.69) to the form eP:. [b0 Z1",+b1Z1"_ 1,+ ... +b,.z] =eP~ (A0x"+A1XS- 1 + ... +As) or b0zrnl +b1Z1"_ 1, + ... + b,.z = A0x' +A1x"' -t + ... +A,, (2.70) where all the b1 are constants. 9"' I. DIFFERENTIAL EQlJATIONS A particular solution of (2.70), if b,. =1= 0, is of the form z= B0X5 +B,xs-l + ... +B,, and, hence, a particular solution of (2.69) is y=eP>< (B0x' +B1x5 - 1 + ... +B~)The condition b,. =1= 0 means that k= 0 is not a root of the characteristic equation b0k"+b1k"-1 + .. , +bn=O, (2.71) and hence k = p is not a root of the characteristic equation a0k"+a1k"-1 + ... +a,.=O, (2.72) since the roots of these characteristic equations are connected by the relationship k=k+p (see page 115). Now if k= 0 is a root of multiplicity a of the characteristic equation (2.71), in other words, if k = p is a root of the same multiplicity a of the characteristic equation (2.72), then the particular solutions of the equations (2. 70) and (2.69) are, respectively, of the form z= x• (B0XS + B,xs-l + ... +B.,), y= x•eP>< (B0xs +B1xs-I + ... +B,). To summarize, then: if the right-hand member of a linear differential equation with constant coefficients is of the form ePX (Aox'· +A,xs-l + ... -1- As), then, if p is not a root of the characleristic equation, a particular solution is to be sought in the same form: y=ePX(Boxs + Blxs-1+ ... +Bs). But if p is a root of multiplicity a of the characteristic equation (this case is called singular or resonance), then a particular solution has to be sought in the form y=x•ePx (B0X5 + B1xs-1 +... +B,). Example 3. y" +9y = e~x. A particular solution has to be sought in the form .iJ= Be5x Example 4. y" +y = e3·\ {x-2). 2. Dlf'FERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 133 A particular solution ha~ to be sought in tt~ form y= e3" (B0x +B1). Example 5. y" -y=e" (x2 -l). A particular solution· has to be sought in the form y=xeX (B0X2 +B1x+ B2 ), since k = 1 is a simple root of the characteristic equation. Example 6. y"' +3y" +3y' +y =e-x (x-5). A particular solution has to be sought in the form y= x~e-x (B0x +B1), since k = -1 is a triple root of the characteristic equation. Observe that our arguments hold true for a complex p as well, therefore if the right-hand member of a linear differential equation is of the form eP" [Ps (x) cosqx +Q8 (x) sin qxJ, (2.73) where one of the polynomials Ps(x) or Qs (x) is of degree s, and the other is of degree not higher than s, then, transforming the trigonometric functions by Euler's formulas to the exponential form, we obtain on the right e kr 2 D2 v + ... + DPv] = n =ekx ~ an-p(D+ k)Pv=e"X F(D+k)v(x). p=O The sum of the operators F1 (D) and F2 (D) is the operator [F1 (D)+ F2 (D)], whose operation on a certain function f (x) is determined by the equality [F1 (D)+ F2 (D)] f (x) = F1 (D) f (x) + F2 (D) f (x). From this definition it follows that since the operation of the left and right members of this equality on a certain n times difYerentiable function f (x) leads to one and the same result, that is, the rule of adding operator polynomials does not differ from the rule of adding ordinary (nonoperator) polynomials. The product of two operators F1 (D)· F2 (D) is an operator whose operation on a certain function f (x) differentiable a sufficiently large number of times is determined by the equality that is, the function f (x) is first operated on by the right-hand factor and then the result of the operation of the right-hand factor on the function f (x) is operated on by the left-hand factor. On the .basis of this definition, it is easy to see that the rule for multiplication of operator polynomials docs not difYer from that of ordinary (nonoperator) polynomials. Indeed, " m 11 m '' DP '\.'I Dq- ~ ,., b Dp+q ,.;;_, (I II-'' • ""' ) tn- Q - ,.;;_, ,.;;_, Q n -· p Ill- iJ 0 p=O q=O p=O q=O (2.77, 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHE!~ 137 since " "'~ a"_PDP ~ bm-qDq f (x) = p=O q=O n ["' ] n m= ~a DP ~ b f1q' (x) = ~ ~a b {1p+ql (x)n-p ~ m-q ~ ~ n-p m-q • pa q•O p=O q= 0 which coincides with the result of operating on f (x) with the operator From (2.77), in particular, it follows that multiplication of operators is commutative: F1 (D) F2 (D)= F2 (D) F1 (D). The validity of the distributive law F (D) [F1 (D)+ F2 (D)]= F (D) F1 (D)+ F (D) F, (D) follows directly from the rule of differentiating a sum. Hence, the operations of addition and multiplication of operator polynomials do not differ from the same operations involving ordinary (nonoperator) polynomials. I Now let us define the operator F (D) . The result of the operation of operator F (lD) on a certain continuous function f (x) is the solution y = F :D)f (x) of the equation F (D) y = f (x), (2.78) Consequently, F (D) [F :D)f (x)] =f (x). (2.79) It might be considered that F (1D) f (x) is the solution of equation (2.78) defined by some specific, say zero, initial conditions; however, for our purposes it is more convenient to consider that F :D) f (x) is one of the solutions (which one is immaterial) of equation (2.78) and, hence, the operation of the operator F to) on a certain function f (x) is defined only up to a summand equal to the solution of the corresponding homogeneous equation. 138 I. DIFFERENTIAL EQUATIONS In that meaning of the operation of the operator F :D) the equation I F (D) [ F (D) f (x)] = f (x) (2.80) will be valid, since f (x) is obviously a solution of the equation F(D) y = F(D) f (x). The product of the operators F, (D) f (x), that is, Y= F,\D) [F,\D)I (x)] (2.81) is a solution of the equation F, (D) F, (D) y =I (x). (2.82) Indeed, substituting (2.81) into (2.82), we get F, (D) F, (D) F, 1(i5) [F, \D)I (x)] == F, (D) F, ~D) f (x) = f(x). Some examples of finding particular solutions of nonhomogeneous linear equations with constant coefficients by the operator method are given below: (I) y"+4y=ex, or(D'+4)y=eX, whence I ex y= [)2+4 eX=s. (2) ylV +y = 2 cos 3x, or (D4 +I) y = 2 cos 3x, 1 2 cos :lx I y= D•+ 1 2cos3x=(- 9)2 + 1= 41 cos3x. (3) y" +9y= 5 sin x, (D2 +9) y=5 sin x. I 5 . 5 sin x 5 . y= b2 +9 sm x = _ 1+9 = 8 sin x. 140 I. DIFFERENTIAl. EQUATIONS ~------------------- ----------------------- _ I 2.\" 2 _ 2x I 2 _ 2x X~ Y- (D- 2)2-e x -e l:i2 x -c 12 . (5) y'" -3y" +3y' -y=e'\ (D-OS y =e", I ¥ y= (D-1)3 e4. F (k) = 0, and so in place of the second formula we use formula (5) (page 139), regarding ex as the product ex ·1: I I x3 y= e"·1 =e"-1 =e"-6 .(D-1)3 Ds (6) y'"-y =sin x, (D9 -1)y=sinx. (2.83) y = D3 1 1sin x. Since the operator contains odd degrees of D, formula (4) cannot be employed. Therefore in place of the original equation we consider the equation (D3 - 1) y = eix or (D3 -l) y =COS X+ i sin X. (2.84) The imaginary part of the solution of (2.84) will be the solution of ihL' original equation (see page 120): I . eix -eix (-l+i)(cosx+isinx) Y=va-lelx=is-1= 1+i = 2 = I( . ) i( . )=- 2 cosx+smx +2 cosx-smx. The imaginary part of the solution cosx-;sinx of equation (2.83) is a solution of the equation (2.83). (7) y" +y =cos X, (D2 + l) y =cos :1:, I y= D2 +l cosx. Formula (3) (page 138) cannot be applied since F (- a2 ) = 0; and so once again in place of the given equation we consider the equation y" +y = eix or y" +y = cos x +i sin x and take the real part of its solution _ I ix -· I I ix _ Y- D2+ I e -(D-i)(D+i)e -- 1 eix eix I eixx x(cosx+isinx} = D-i 21 = 2i 15. I = 2i = ·--2"'"'i-----" Taking the real part of the thus found solution of the auxiliary 2. DIFFERENT!AL EQUATIONS OF THE SECOND ORDER AND HIGHER 141 t. x sin x bt .e(fua wn - 2- we o am the solution of the original equation, I (8) yiV_y=ex, (D'-l)y=ex, £;= 0,_ 11f'= 1 1 1 ex I I xex = D-1 (D+ I)(D2 +1) ex= D-1 4 =-.rex v 1=-:r · I Now let us find out how the operator F (D) operates on the po- lynomial Pp(X)=A0xP+A1xr1 + ... +Ar We formally divide l by the polynomial F(D) =an +a,_1 D + ... -j-a0 Dn, an =P 0, arranged in increasing powers of D, by the rule of division of ordinary (nonoperator) polynomials. We stop the process of division when the quotient is an operator polynomial of degree p: b0 -j-b1D-j- ... -f-bpDP=Qp(D). Then the remainder will be the polynomial R(D) =cp+1DP+ 1 +cP+2DP+2 + ... +cp+nDp+n, which contains the operator D to powers not lower than p + l. By virtue of the relationship between the dividend, divisor, quotient and remainder, we get F(D) QP (D)+ R(D)== l. (2.85) This identity holds true for ordinary (nonoperator) polynomials, but since the rules of addition and multiplication of operator polynomials do not differ from the rules of addition and multiplication of ordinary polynomials, the identity also holds true for operator polynomials. Operating with the right side and left side of the identity (2085) on the polynomial A0xP +A1xr 1 + 0. 0+AP, we get (F(D) Qp(D) +R (D)] (A0xP + A1xr1 -f- ... + Ap) == == A0xP + A1xP-1 + 0 . 0 + AP or, taking into account that R(D) (A0 xP +A1xr• + o • • +Ap) = 0, since R(D) contains D to powers not lower than p +I, we will have F(D)[QP (D) (A 0xP +A1xr• + .o. + Ap)] == = A0xP +A,xr• +. 0. +AP, 142 I. DIFFERENTIAL EQUATIONS that is, QP (D) (A0xP +A1xr 1+ ... + Ap) is a solution of the equation F(D)y=A0xP+A,xr1+ ... +Ar And so F (~) (A 0xP + A1xr1+ ... + Ap) = Qp(D) (A0xP +A1xP- 1 + ... +Ap). For example: (9) y" +y=x2 -x+2, (D2 + l)y=x2 -x+2, Y=o2~1 (x~-x+2). Dividing 1 by I +D2 , we get Q2 (D)= I-D2 • Hence, y= (I-D2) (x2-x+ 2)=x2 -x. (10) y"+2y'+2y=x2e-x, (D2 +2D+2)y=x2e-x, y = D2+~D+ 2x2e-x =e-x 02~ I xs =e-x (I-D2) x2 =e-x (x2-2). (II) y'"+y=xcosx, (D2 +1)y=xcosx. Let us pass over to the equation (D2 + I) y = xtx and then take the real part of the solution _ I ix _ ix I _ lx _!_(_!_+E..) _Y-02+1 xe -e D(D+2i) x-e D 2i 4 Xix I ( X + I ) tx ( x2 + X ) ( • + . . )(x2 + X )= e 0 2i . 4 = e 4i 4 = cos x t sm x 4i 4 . Taking the real part ~2 sin x + ; cos x, we get the desired solution. Note. The last example indicates how one should operate on the polynomial with the operator F (~) if an= 0. Representing F (D) in the form Ds (D), where the absolute term of the polynomial (D) is no longer zero, we operate on the polynomial first with the operator Ill :D) -and then with the operator ~s . The nonhomogeneous Euler equations aoxny'n'+a~xn-1yln-11+ . •• +any=f(x) (2.86) or a0 (ax+ b)n y'nl +a, (ax+ b)n- 1 y1n-11 + ... +anY= f (x) (2.87) may be integrated by solving the corresponding homogeneous equations (see page 120) and by choosing one particular solution of the nonhomogeneous equation, or by applying the method of variation of parameters. However, it is ordinarily simpler at first 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 143 to integrate the homogeneous equation and, for choice of a particular solution, to transform the Euler equation (2.86) by the change of variable x = ± e1 [for equation (2.87) ax+ b= ± e1] to an equation with constant coefficients for which methods of finding particular solutions have been thoroughly developed. Example 11. x2y• (x) -xy' (x) +y (x) = x ln3 x. (2.88) We seek the solution of the corresponding homogeneous equation in the form y = xk: (2.89) k1, 2 = 1; hence, the general solution of the homogeneous equation is of the form y=(c1 +c2 lnx)x. The change of variables x=e' transforms equation (2.88) to an equation with constant coefficients y(t)- 2y(t) +y = t3e1 [the left-hand side of this equation can straightway be written in accordance with the characteristic equation (2.89)]. Using the operator method, it is easy to find a particular solution of the transformed equation: Y =_I_ e'ta = e' _!_ ts =efta (D-1)2 02 20 ' x ln6 x Y=2(). Consequently, the general solution of equation (2.88) is of the form y = (c1 +c,lnx +~~~x) x. 7. Integration of Differential Equations by Means of Series The problem of integrating homogeneous linear equations of the nth order Po (X) y'n' +p, (x) y'n-1'+ .. • +Pn (X) Y = 0 (2.90) reduces to choosing n or at least n-1 linearly independent particular solutions. However, particular solutions are readily selected only in exceptional cases. In more involved cases, particular solu"'tions are sought in the form of a sum of a certain series ~ cti 1) and a zero of order not lower than s-2 of the coefficient of p2 (x) (if s > 2), then there exists at least one nontrivial solution of the equation (2.91) in the form of a sum of the generalized pawer series y=a0 (X-X0 ),.+a1 (X-X0 )k+'+ . .. +an(X-X0 )1<+"+ .•• , (2.92) where k is some real number that may be either integral or fractional, either positive or negative. The second solution, linearly independent with respect to (2.92), is also as a rule in the form of a sum of a generalized power series, but sometimes may also contain the product of the generalized power series by ln (x-x0). However, in specific examples one can dispense with the two theorems just formulated, all the more so since these theorems (as they are stated) do not establish the domains of convergence of the series under consideration. In concrete problems the most used procedure is to choose a power series or a gent!ralized power st!ries that formally satisfies the differential equation; that is, such that if substituted turns the equation (2.90) of order n into an identity if one assumes convergence of the series and the possibility of n-fold termwise differentiation. Having formally obtained a solution in the form of a series, the next step is to investigate the convergence and the possibility of n-fold termwise differentiation. In the region where the series converges and admits an n-fold termwise differentiation, it not only formally satisfies the equation, but its sum is indeed the desired solution. Example 1. y"-xy=O. We seek the solution in the form of a power series: "'~ II y == ~ a11X • ,..o (2.93) 2. DIFFERENTIAL EQUI\ TJONS OF THE SECOND ORDER AND HIGHER 145 Proceeding from Theorem 2.9 or formally differentiating this series termwise twice and substituting into (2.93), we get "' ... ~ ann(n-l)x"-3 -x ~ anx"=O. na2 n=O Comparing the coefficients of identical powers of x in the left-hand and right-hand members of the identity, we get: a~= 0, 3· 2a3 -a0 = 0, whence a3 = 2~~; 4·3a.-a1 =0, whence a4 =;.~; 5·4a5 -a2 =0, h a2 ( 1) O h an-a w ence afi= 4.5 , .•• , n n- a,.-a,._ 3 = , w encea,.=(n-l)n'"'' Consequently a = a. Bn+l 3·4·6·7 ... 3n (3n + 1) (n= 1, 2, ... ), a0 and a1 remain arbitrary. Thus, [ xs xB x3n ] Y =ao 1+ 2·3 + 2·3·5·6 + · · · + 2·3·5·6 ... (3n--l) :3n + • · · + [ x4 x7 x:w ' ' l+at x+3.4 +3·4·6·7+ · · · + 3-4-6·7 ... 3n \3n; IJ + ··· · (2-94) The radius of convergence of this power series is equal to infinity. Therefore, the sum of the series (2.94) is, for any values of x, a solution of the equation under consideration. Example 2. (2.95) This equation is called Bessel's equation of order n, although it first appeared in the works of Euler and Bernoulli. Many problems of mathematical physics reduce to the Bessel equation, and so we shall investigate it in somewhat more detail. By Theorem 2.I0, at least one nontrivial solution of the Bessel equation can be found in the form of the sum of the generalized power series liD y= ~ apx"+P. p=O Differentiating this series twice term-by-term and substituting into equation (2.95), we get 00 x2 ~ ap(k+p)(k+p-1)xk+r2 + P=O "' .s> +x ~ ap(k+p)xk+r1 +(x2 -n2 ) ~a xk+p=O. p=O p=O p 10 :m; 146 I. DIFFERENTIAL EQUA TJONS Comparing the coefficients of identical powers of x in the left and right members of the equation, we obtain a0 (k2 -n2] = 0, a1 [(k+ 1)2 -n2 ] =0, [(k+ 2J2 -n2] a2 +a0 = 0, ((k +3)2 -n2] a3 +a1 = 0, [(k +p)2 -n2 ] aP +ap-s = 0. Since the coefficient a0 of the lowest power of x may be considered nonzero, the first equation reduces to k'-n' =0, whence k= ± n. For definiteness we will meanwhile regard k = n ~ 0; then from the second equation a1 [(n + l)2 -n2] =0 we get a. =0 and, hence, all the a2p+t = 0, a =- ao ao 2 (n+2)2-n2 2' 0, r (p +1) = pf (p). 0 ( x)2p+n00 (-1)P 2 y = Epi f(n+p+ 1) p=O (2.96) . DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 147 This solution is ordinarily written as Jn (x) and is called Bessel's function of the first kind of order n. For k = -n, and for a0 chosen as a0 = 2-nr (~ n+ I), we similarly get Bessel's function of the first kind of order -n: "' (-J)P ( !._ rp-n J_n(X)= ~plf(-~!P+I)" p=O (2.97) The series (2.96) and (2.97) converge for any values of x [in (2.97) x =1= 0] and admit two-fold termwise differentiation; hence, Jn(x), J:n(x) are solutions of Bessel's equation (2.95). For nonintegral n the solutions Jn (x) and J -n (x) are obviously linearly independent, since their expansions in series begin with different powers of x and, consequently, the linear combination cx.Jn (x) +cx.2J -n (x) can be identically zero only when cx.1 =a~= 0. Now if n is some integer, then, since for integral negative values of p and for p = 0 the function r (p) becomes infinite, expansions in series of the functions Jn (x) and J -n (x) begin with the same powers of x and, as is readily verifiable, the functions Jn (x) and J-n (x) will exhibit the following linear relation: J -n (x) = (- l)nJn(x). Hence, when n is integer, one must seek, in place of J-n (x), another solution that would be linearly independent of Jn (x). Such a solution may be obtained by various methods; for instance, it is possible, knowing a single particular solution of Jn (x), to reduce the order of the equation (2.95) by the substitution indicated on page 107, or to seek straightway a solution in the form of the sum of a generalized power series and the product of a generalized power series into In x. The solution [linearly independent of Jn (x)] obtained by any one of these procedures in the case vf a completely definite choice of the arbitrary constant factor is called Bessel's function of the second kind and is denoted as Yn (x). However, Yn (x) is most often defined as follows: taking n nonintegral, consider the solution Yn (x) of the Bessel equation, which solution is a linear combination of the solutions Jn (x) and J-n (x): y (X)= Jn(x)cos_nn-J_n(X). n stn nn ' then, p!issing to the limit for n approaching an integer, we get a particular solution [which is linearly independent of Jn (x)] of the Bessel equation Yn (x), which is now defined for integral values of n as well. l!F 148 I. DIFFERENTIAL EQUATIONS --------- ·--·-·--···---··------ - - - Thus, the general solution of Bessel's equation for n a noninteger is of the form y = c1Jn (x) +cJ -n (x) and for n an integer, y=cJ,. (x) +c2 Yn (x), where C1 and C2 are arbitrary constants. Bessel's functions of the first and second kind have been studied in great detail; in particular, detailed tables of their values have been compiled. For this reason, if a problem has been reduced to Bessel's functions, then it may be considered solved to the same extent that we consider as solved a problem in which the answer is given for example in trigonometric functions. The following equation is frequently encountered in applications: x2 y" +xy' +(m2x2 -n2) y = 0. (2.98) This equation may be reduced to Bessel's equation by a change of variables x1 = mx. Indeed, given such a change of variables, dy dy dx1 dy d2y d2y 3 -=-·-=-m -=---mdx dx1 dx dx1 ' dx2 dxr and equation (2.98) turns into the Bessel equation: 2 d2y ' dy +( 2 2) - 0Xt-2 -,·xl-d- x~-n Y- . dx1 x, Thus, the general solution of (2.98) for n a noninteger is of the form y = clJn (mx) +c2J -ll (mx), and for n an integer, y = C1J n (mx) +C2 Y" (mx). Example 3. x2y" +xy' + (4x2 - :5) y=O. The general solution of the equation is of the form y = c1J..::_ (2x) +cJ_J!.. (2x). ; 6 Example 4. x2y" +xy' +(3x2 -4) y = 0. The general solution is Y = clJ 2 (xV3) +c2Y2 (x J-r3). 2. DIFFEf~ENTIAL EQUATIONS OF THE SECOND OIWER AND HIGHER 149 Example 5. Integrate the equation x2y'' +xy' +(4x2 -_!_) y=O \ 9 provided that the solution must be continuous at the point x = 0 and y (0.3) = 2. The general solution has the form y = c1J 2.. (2x) +C2./ _...!... (2x). 3 3 The function J _2_ (2x) is discontinuous at x = 0 since the series 3 (2.97) begins with negative powers of x. Hence, the solution y is continuous at the point x = 0 only for c2 = 0: y = cJ2.. (2x). a Satisfying tl1e second condition, y (0.3) = 2, we get 2 cl = J 1 (0.6) 3 Tables of Bessel's functions yield J .!....(0.6) = 0.700; thus, C1 ~ 2.857 and 3 y ~ 2.857J 1 (2x). 3 Applied problems frequently demand finding periodic solutions of some differential equation. In this case it is ordinarily advisable to seek the solution in the form of the sum of some Fourier series: Observe that if the equation xln) = F (t, X, X, ... ' x(n-l)) (2.99) ~· has a periodic solution x0 (t) with period T, then the right side of (2.99) along the integral curve under consideration is a periodic function of the period T with respect to the first argument. Indeed, substituting into equation (2.99) the periodic solution x=x0 (t), we get the identity x~m(t)=F(t, X0 (t), X0 (t), ... , x~n-u(t)). If in this identity we replace t by t +T, we will not - by virtue of the periodicity of the function x0 (t) and its derivatives- alter the left-hand member of the equation and will not !50 I. DIFFERENTIAL EQUATIONS change the arguments of the right-hand member beginning with the second; hence · F (t, X0 (/), X0 (/), ••• , x1~-tl (/)) == == F (I+ T, X0 (t), X0 (/), · 1n-11 (/)) • •• ' Xo ' that is, the function F along the integral curve x = x0 (/) has a period T with respect to the explicitly appearing argument t. Therefore, if the right side of equation (2.99) is not a periodic function [for any choice of X0 (t)) with respect to the first argument, then no periodic solutions exist either. If the function F is not explicitly dependent on t, that is, it is constant with respect to the argument t, then F may be regarded as a periodic function (with respect to t) of any period and therefore the existence of periodic solutions of any period whatscever is not excluded. For example, Jet it be required to find the periodic solutions of the equation x+a'x=l(t). (2.100) For a periodic solution to exist we have to assume that I is a periodic function. Without any essential Joss of generality it may be taken that I (t) is a periodic function with period 2n, since if the function I (t) had a period T, then after transformation of the independent variable /1 = 2; t the right side would become a function with period 2n with respect to the new independent variable t1' Further suppose that the function f (t) is continuous and can be expanded in a Fourier series: .. . f(t)=a.; +~ (a11 coskt+b11 sinkt). k=l (2.101) We seek the periodic solution in the form "' x(t) = ~0 +~ (A11 cos kt + B, sin kt). k=l (2.102) FormaMy differentiating the series (2.102) termwise twice and ~ubstituting into equation (2.100), we get ao -£.k' (A11 cos kt +811 sin kt) + k=l +a2 [~o+t.(A, cos kt +B11 sin kt)] == ... = ~ + L.(a11 cos kt +b11 sin kt). k=l 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 151 wht!nce, if a is not an integer, we determine the coefficients of the series (2.102): a2Ao ao A Oo I-2- 2· o = 0 z • ( a3 -k2) Ak = ak, A ak (2.103)k = aZ-kZ • J(a2 -k2 ) Bk = bk, B bk k = ""2Ji2.aConsequently, the equation (2.100) is formally satisfied by the series 00 .E!__ +~ ak cos kt +bk sin kt 2a2 ~ 0z_Ra • k=l (2.104) Obviously, series (2.104) converges and admits of twofold termwise differentiation since the series (2.10 I) converges uniformly by virtue of the continuity of f (t), and the coef_ficients of the series (2.105) made up of the second derivatives of the term3 of the series (2.1 04), differ from the coefficients ak and bk of the series (2.101) solely in the factor - 02 k2 k2 , which does not depend on t and monotonically approaches 1 as k-+ oo (this proof is not sufficitmtly rigorous). Consequently, the series (2.105) converges uniformly and, hence, the series (2.104) may be differentiated termwise twice. And so the series (2.104) not only formally satisfies the equation (2.100), but its sum x (t) exists and is a periodic solution of the equation (2.100). If a differs but slightly fro111 an integer n and an =;6 0 or bn =;6 0, then resonance sets in; this consists in a sharp rise, as a approaches n, of at least ont> of the coefficients But if a= n and at least one of the coefficients an or bn is not zero, then no periodic solutions exist since to the resonance terms an cos nt +bn sin nt in the right-hand member of equation (2.100), as indicated on page 114, there corresponds, in accordance with the principle of superposition, a nonperiodic term of the form t (An cos nt + Bn sin nt), 152 I. DIFFERENTIAL EQUATIONS in the general solution of the equation (2.1 00), whereas the othl'r summands in the general solution of the equation will be periodic functions. Consequently, for a= n, a periodic solution of the equation (2.100) exists only if there are no resonance terms a,. cos nt + +b,. sin nt in the right-hand member, that is in the case of 2n 2n a,.=!~ f(t)cosntdt=O, 1 (' b,. = -nJ I (t) sin nt dt = o. 0 (2. I06) In the latter case, i.e. when a= n, a,.= b,. = 0, a periodic solution of the equation (2.100) exists, and for k =1= n, the coefficients are determined from the formulas (2.103), while the coefficients An and B,. remain arbitrary, since A,. co:> nt +Bn sin nt is, for arbitrary A,. and B,., a solution of the corresponding homogeneous equation. Example 6. Determine the periodic solution of the equation We seek the solution in the form of a series, "' x(t)=~0 + L (Akcoskt+Bksinkt) k=l and, determining the coefficients Ak and Bk from formulas (2.103), we get "''\:..., sin kt x(t)= ~ k'(2-k~)· k=l Example 7. Determine the periodic solution of the equation x+4x=sin2 t. Since the conditions of the existence of the periodic solution (2.106) are not satisfied, 2:11 ~ sin2 t sin 2t dt = 0, 0 but 2n ~ sin2 t cos 2t dt =1= 0, 0 no periodic solution exists. 2. DIFFE!{ENTII\L EQUATIONS OF THE SECOND ORDER AND HIGHER !53 Example 8. Determine the periodic solution of the equation •• ~ cos kt x+x= ~fir· k=2 The resonance terms a1 cos t +b1 sin t are absent in the right-hand member. Therefore, a periodic solution exists and is determined by the formulas (2.1 03): where c1 and c2 are arbitrary constants. 8. The Small Parameter Method and Its Application in the Theory of Quasilinear Oscillations In the preceding section we indicated a method for finding periodic solutions of linear equations of the torm x+a2X = f (t). In many practical problems it is necessary to find the periodic solution of an analogous equation, but having a small nonlinear term in the right-hand member: x+a2x=f(t>+v.F(t, X, X, v->. (2.107) where v- is a small parameter. If we discard the term v.F(t, x, x, v.). i.e. if we consider v-=0 in the equation (2.107), then we have a linear equation x+a2x=f(t), which is known as the generating equation for (2.107). One of the most effective methods of finding periodic solutions of an equation of nonlinear oscillations with a small nonlinearity (2.1 07) is that devised by Poincare and Lyapunov- a method of expanding the solution in a series of powers of a small parameter v-. which is widely used at present in solving a great diversity of problems. Proceeding from the theorem on the analytic dependence of a solution on a parameter (see page 60), which theorem is readily grneralized to equations of the second and higher orders, it may be asserted that the solutions x (t, v.) of equation (2.107) will be analytic functions of the parameter v- for sufficiently small absolute values of J.l., if the function f (t) is continuous and the function 154 I. DIFFERENTIAL EQUATIONS F(t, x, x, J.L), continuous with respect to t, is analytically dependent on the remaining arguments: on x and xin the region in which these variables will in future continue to vary, and on J.L for sufficiently small absolute values of J.L· Assuming that these conditions are fulfilled, we seek the periodic solution x (t, J.L) in the form of the sum of the series X(t, J.L) = X0 (I)+ J.LX1 (t) + J.L2X2 (t) + ... + J.LnXn (I)+ .... We differentiate this series twice term by term: X(t, J.L) = X0 (t) + llXt (t} + ••• + llnXn (I)+ ... , K(t, J.L)=x0 (t)+J.Lx1 (t)+ ... +J.Lnxn(t>+ ... , and substitute into equation (2.107), in which the function F (t, x, x, J.L) has first been expanded in a series of powers of x-xo, x-xo and J.L: i+ a'x ~f (I)+~ [ F (1, x,, x,. 0)+ (:~)~~r (x -x,)+ +(a~ ) (~-Xn) +(aF) J.L + .. ·l·ax ~ = ~· a"' ~ = ~· X=Xo X=X0 11=0 11=0 J (2.108) Comparing the coefficients of identical powers of J.L in the left and right members of (2.108), we get X0 +a2X0 = f (t), x.+a2x.=F(t, Xn, Xo, 0), x,+a2x,=(aF) x1 +(a~) x1 +(aF) .(2.109) ax~=~· ax~=~· a"'-!=~· X = X't JC = X" X = Xo 11=0 11=0 11=0 The first of these linear equations coincides with the generating equation. Integrating it and substituting the obtained solution x0 (t) into the second equation, we again get a linear equation for determining x1 (t), and so forth. For the determination of xn (t) we also get a linear equation, since the right-hand member of this equation will contain only xj and xj with indices less than n because, due to the presence of the factor J.L in F, the terms containing xn and xn on the right, 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIOHER 155 and all the more so xk and xk with large indices, will have the factor 1.1. to a power of at least n +1. In this section we consider only the problem of finding periodic solutions, and so it is natural to impose on the right-hand side of the equation x+a2x=f(t)+t-tF(t, X, X, f.l.), in accordance with the note on pages 149-150, yet another restriction; we require that the right-hand side be a periodic function with respect to the explicitly· occurring argument t. Without any essential Joss of generality, it may be taken that the smallest period of the right side, if the right side is explicitly dependent on t, is equal to 2n; then if f (t) is not equal to a constant quantity, the periodic solutions of the equation (2.107), if they exist, can only have periods equal to 2n or that are multiples of 2n, given sufficiently small 1.1. (see page 150). To find the periodic solution of equation (2.108) in the form x(t, f.l.)=X0 (t)+J.txdt)+ ... +!.l.nxn(t)+... (2.110) it is necessary to determine the periodic solutions x, (t) of equations (2.109). Indeed, if the solution x(t, 1.1.) has a constant period 2n (or 2mn, where m is an integer) for any sufficiently small absolute value of f.l., then X0 {t)+f.l.X1 (t)+ ... +f.l.nxn(t)+ ... ==x0 (t+2n)+ + f.l.X1 (t+2n)+ ... + f.l.nXn (t+2n)+... . (2.111) Hence, the coefficients of identical powers of 1.1. in the right and left members of the identity (2.111) must be equal, that is, Xn (t) == Xn (t +2n), and this signifies the periodicity of the functions xn (t) (n = 0, 1, 2, ... ). That the coefficients of identical powers of 1.1. in the left and right members of the identity (2.110) coincide may be seen, for example, by differentiating the identity (2.110) n times with respect to f.l.; then, assuming 1.1. = 0, we get Xn (2n + t) = Xn (t) (n = 0, 1, 2, ... ). Thus we have to find the periodic solutions of the equations (2.109). Here it is advisable to consider the following cases sepa- rately. l. Nonresonance case: a is not an integer. If a is a noninteger, the first of the equations (2.109) has a unique periodic solution X0 = cp0 (t), which is found by the method of the _preceding section (see page 150). Then find x1 (t), x2 (t), etc. by the same method. !56 I. DIFFERENTIAL EQUATIONS If, using this method, we found the general term of the series (2.110), and if we established the convergence of the series and the validity of its twofold term-by-term differentiation, then the sum of the series (2.110) would be the desired periodic solution with period 2n. However, finding the general term of the series (2.110) is usually an exceedingly complicated problem, and so one has to confine oneself to computing only the first few terms of the series, which would be· sufficient for an approximation of the periodic solution if we were confident that the series converges and its sum is a periodic solution. Of great importance in this connection are the theorems of Poincare on the existence of periodic solutions. In particular, these theorems permit finding the conditions under which there definitely exists a unique periodic solution of the equation (2.107) that approaches the periodic solution of the generating equation as !l -+ 0. If the conditions of Poincare's theorem are fulfilled and, hence, there exists a unique periodic solution of equation (2.107) that approaches a periodic solution of the generating equation as !l- 0, then the sum of the unique series with periodic coefficients (2.110), which series formally satisfies the equation (2.107), must exist and must coincide with the sought-for periodic solution. It is not necessary then to seek the general term of the series (2.110) for investigating the series for convergence and it is possible, after finding the first few terms of the series (2.110), to state that, given a small !l, their sum is approximately equal to the desired periodic solution. Poincare's theorems are based on information from the theory of analytic functions and are rather involved. We therefore give only the most elementary of these theorems at the end of this section, but even so it will permit us to assert that in the nonresonance case under consideration equation (2.107) always has a unique periodic solution for sufficiently small ll· Example t. Determine approximately the periodic solution of the equation x+2X= sin t +!!X~, where !l is a small parameter [determine two terms of the series (2.110)]. We seek the solution in the form X (t, !l) = X0 (i) +!!X1 (t) + ... +!lnxn (t) + .... We find the periodic solution of the generating equation X0 +2x0 =sin t, x0 (t) =sin t. The periodic solution of the equation .. • 2 .. 2 I - cos 2t x1 +2x1 = sm t or X1 + xl = 2 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER !57 is of the form 1 +cos 2t Xl=4 -4-, Hence, the periodic solution x (t, f.t) ~sin t + i- (I +cos 2t) ll· 2. Resonance case. The small-parameter method may also be employed in the resonance case, i.e. in the case when in equation (2.107) a is equal to an integer n or tends to an integer n as fl--7 0. If in the equation (2.107) a differs but slightly from an integer n, more precisely, the difference a2 - n2 is of an order not lower than fl, where a1 is bounded as 11---+ 0, then the equation x+a2x=f(t)+!1F(t, x,x, !l) may be rewritten in the form x+n2X=f(t)+(n2 -a2 )X+f1F(t, X, X, f.t), whence, by virtue of (2.112), x+n•x = f (t) +flF1 (t, x, x, p.), (2.112) where the function F1 satisfies the same conditions which by assumption are satisfied by the function F. Hence, in the resonance case we can henceforth consider a equal to an integer: x+n1X= f (t) +!1F (t, X, X, p.). Applying the small-parameter method, we seek the periodic solution in the form of a series X (t, J.L) = X0 (t) +J.LX1 (t) + ... +!J.kXk (t) +... , To determine the functions xk (t) we again get equations (2.109), in which a2 = n2, but in the given case the generating equation X0 +n2X0 = f (t) (2.113) has a periodic solution only if there are no resonance terms in the 158 I. DIFFERENTIAL EQUATIONS right-hand member, that is, if the conditions (see page 152) rf(t)cosntdt _;_0, l 2n } ~ f (/) sin nt dt = 0 1 0 ) are fulfilled. (2.106) If these conditions are fulfilled, then all solutions of the equation (2.113) will be periodic with a period 2n (see page 152): X0 (t) = C10 cos nt +C20 sin nt +cp0 (/). The function x1 (t) is determined from the equation •• 2 X1 +n X1 = F (t, X0 , X0 , 0). (2.. 114) This equation also has periodic solutions only if resonance terms in the right-hand member are absent, i.e. if the following conditions are fulfilled: rf(t, Xo, Xo, O)cosntdt=O, l ~ J~ F (t, X0 , X0 , 0) sin nt dt = 0. (2.115) Equations (2.115) contain C10 and c20 which, generally speaking, are determined from this system. Let c10 and C20 satisfy the system (2.115); then all solutions of (2.114) have the period 2n: X1 (t) = c11 cos nt +Cu sin nt +cp1 (f), (2.116) and c11 and c21 are again determined from the two conditions of the absence of resonance terms in the following equation of (2.109): x,+n2Xs = ( iJF) x1 +(0~) x, +(iJF) , ox ~=~· OX ~=~· OJI. ~=~· and so forth. Consequently, not to every periodic solution X0 = C10 cos nt +C20 sin nt +Yo (t) of the generating equation but only to some [the values C10 and C20 !>f which satisfy the equations (2.115)] do there correspond periodic solutions of equation (2.107) for small J.L. Of course, in the resonance 2 DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 159 case as well, in order to be sure [without finding the general term of the series (2.110)] that a periodic solution will be found by the indicated process it is first necessary to prove the existence theorem of periodic solutions. This observation also refers to the cases described in Items 3 and 4. 3. Resonance of the nth kind. Occasionally in systems described by the equation (2.107) which satisfies the conditions given above, intensive oscillations are observed when the proper frequency differs but slightly from 1/n, where n is an integer. This phenomenon became known as resonance of the nth kind. From the mathematical viewpoint, this means that for a differing only slightly from 1/n, where n is an integer greater than unity, equation (2.107) may have periodic solutions with a period 2nn, which are not periodic solutions with a period 2n. Let .. I x+ -a x=f(t)+f.LF(t, x, x, f.L)n (2.117) [if a differs only slightly from l;n, more precisely a1 -~ = f.La1, n where a1 remains bounded as f.L - 0, then, transposing the term (a2 - :~ )x to the right side and including it in f.LF(t, x, x, f.L), we get an equation of the form of (2.117)). We. seek the periodic solution of equation (2.117) with a period 2nn in the form of a series (2.110) Putting (2.110) into (2.117) and comparing coefficients of identical powers of f.L, we get the equations (2.109) in which a=*. To determine x0 (t) we obtain the generating equation .• 1 X0 +-2 X0 = f (t), (2.] 18)n which has a periodic solution with a period 2nn only in the absence of resonance terms in the right-hand member, i.e. for 2nn 2nn Sf (I) cos ~ dt = 0 andSf (t) sin ~ dt = 0. 0 0 If these conditions are fulfilled, then all solutions of the equation 160 I. DIFFERENTIAL EQUATIONS (2.118) have the period 2nn: X0 = C10 cos_!___+ C20 sin_!_+ q>0 (t), n n where c10 and C20 are arbitrary constants. The equation which determines x1, " I . x1 +/ii""x1 = F (t, x0 , x0 , f..t), (2.119) will have periodic solutions with a period 2nn only in the absence of resonance terms in the right members, i.e. when the conditions 2nn lJF(t, X0 , X0 , f..t)COS! dt=O, linn S . . t 0 F(t, Xo, Xg, f..t)Sm ndt=O J (2.120) are fulfilled. Generally speaking, c10 and c20 are determined from these conditions. If conditions (2.120) are satisfied, then all solutions of equation (2.119) have the period 2nn: t t X1 =C11 cos--+cu sin -+cp1 (t). n n To determine the arbitrary constants c11 and c21 , we make use of the two conditions of the absence of resonance terms in the following equation of (2.109): .. 1 ( aF ) (aF) (aF)X +-x = - X + - X + -I n2 2 · ax ~=7o l ax ~=Xo l aJ.Io ~=~o ' and so forth. X=Xo 1.1=0 4. Autonomous case. Let us assume that the right-hand side of equation (2.107) is not explicitly dependent on t, and the equation is of the form (2.121) where the function F satisfies the conditions posed above. At first glance it would seem that an investigation of (2.121) should be simpler than an investigation of equation (2.107), in which the right side depends on the argument t, however the absence of t in the right member of the equation actually complicates the problem. If the right side is explicitly dependent on t, then, as has already been pointed out, the possible periods of the solutions are known, 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 161 since the periods of the solutions can be only equal to, or multiples of, the period of the right-hand member along the solutions with respect to the argument t that occurs explicitly. Now if the right-hand side does not contain t, it may be regarded as a periodic function of an arbitrary period, and, hence, the possibility remains of the existence of solutions of any period; the period of solutions, generally speaking, will be a function of the parameter J..l.. In view of the fact that the period of the solution x(f, J..l.) is, generally speaking, a function of J..l., it would not be advisable to seek the solution in the form of the series x(l, J.L)=X0 (t)+J.Lxdf)+ ... +J.LnXn(t)+ ••• , (2.110) since each one of the functions xn (t) taken separately does not necessarily have to be a periodic function and, hence, the functions xn (t) could not be found by the methods given above. It is therefore necessary to transform equation (2.121) to a new independent variable so that the equation now has a constant period with respect to the new variable, and then seek the solution· in the form of the series (2.110). First, for purposes of simplification, we transform equation (2.121) by a change of the independent variable t1 =at to the form d2x -~ +x= J.LF1 (x, x, J.L). (2.122) dt, Each solution of the generating equation x0 (t1)=c1 cos(t1 -t0 ) will have a period 2n and the periodic solutions of the equation (2.122), when J.L =¥= 0, if they exist, will have the period 2n +a (J.L), and it may be proved that a (J.L) is an analytic function of f..1. for sufficiently small J.L. Expand a (J.L) in a series of powers of J.L; then 2n+a(J.L)=2n(l +h1J.L+h1J.L1 + ... +hnJ.Ln+ •. •), (2.123) where h1 are certain constant quantities that we do not yet know. Transform the variables so that the periodic solutiQn x (t, J.L) of the equation (2.122) has a constant period 2n and not the period 2n+a (J.L). This is attained by the change of variables 11 =11 (1 +h1J..1.+h2 J.L1 + .•. +hnJ.Ln+ •. •), (2.124) since, by virtue of the relationship (2.123), the new variable /1 varies from 0 to 2n when t1 varies from 0 to 2n+a(J.L). In the process. equation (2.122) is transformed to Xt,t, +(1 +h1J.L+ ... +hnJ.Ln+, . .)2 X= = J.L (1 +h1J.L+ ... +hnJ.Ln+ .. . )1 F1 (x, (1 +h1J.L+ ... ••• +hnJ.Ln + .•.)-1 Xt01 J.L). (2 125) 11-37~ 162 I. DIFFERENTIAL EQUATIONS We seek the periodic solution of this equation in the form X (11 , ~) = X0 (/1) +~xd/2) +... +~nxn (!2) + ... , (2.126) where xn (t,) are periodic functions of the argument 12 of period 2n. Putting (2.126) into equation (2.125) and comparing the coefficients of identical powers of ~ in the left and right members of the equality, we get or x0 +xo = 0, whence X0 =ccos (12 -10 ), x1 +x1 = -2h1x0 +F1 (X0 , X0 , 0) X1 + X1 = -2h1 C cos (12 - / 0 ) + F1 (c cos (/2 -10 ), -c sin (12 -t0 ), 0) (2.127) For equation (2.127) to have periodic solutions, it is necessary and sufficient that resonance terms (see (2.106)] be absent in its right-hand member, that is, that 2n \ ~ F1 (c cos (t, -- !0 ), -c sin (t,- 10 ), 0) sin (!2 -10 ) dt2 = 0, I -2h.c+ ~ rF.(ccos(t,-to). -csin(t2-to), O)x IX ot:OS (t2 - t0) d/2 = 0. (2.128) The first of these equations permits finding the values of c, and the second, those . of h1 ; having determined them, we find those solutions of the generating equation x0 = c cos {i2 - t0) in the neighbourhood of which, for small ~. there appear periodic solutions of the equation (2.122); and we approximately determine the period of the desired solution 2n +a(~)~ 2n (1 + h1~). Knowing c and h1 , it is possible to determine x1 (t2) and, if necessary, to compute by the same method x2 (t2 ), x3 (t2), and so forth. Example 2. (2.129) Determine the solutions of the generating equation to which the periodic solutions of the equation (2.129) approach as ~ -.. 0. The solutions of the generating equation are of the form x = ., ccos (t- /0). To determine the desired values of c, we take ad- 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 163 vantage of the first equation of (2.128): 2n Sc (9-c2 cos2 (t- t0)) sin2 (t- t0) dt = 0 0 or nc(9- ~) =0, whence c1 =0, c2 • 8 = ± 6. For c1 = 0 we get the trivial solution x ==0 of the generating equation, which remains a solution of the equation (2.129) for any J.L· For C2 , 3 =±6, we get x=±6cos(t-t0 ). Let us prove the most elementary of Poincare's theorems on the existence and uniqueness of a periodic solution tending to a periodic solution of the generating equation as J.L- 0, as applied to an equation of the form X=f(t, X, X, J.L), (2.130) where the function f satisfies the conditions of the theorem on the analytic dependence of the solution upon the parameter J.L for sufficiently small absolute values of J.L. Besides, let us assume that the function f is explicitly dependent on t and has a period 2n with respect to t. Also assume that the generating equation x= f (t, x. .x, 0) has a unique periodic solution x = from the initial values o 1 (J.L, ~0, ~1), respectively, we write the system (2-131) in the form m"X = t. 41. yV'-3yV +3ylV -y"' =X. 42. x'v +2x" +x =cos t. 43. (I +x)2 y" +(I +x) y' +y = 2 cos In (I +x). 44. Determine the periodic solution of the equation .., ·· "" sin nt x +2x +2x = .t...,. ---rl'. n=l 45. Find the periodic solution of the equation x+a.x+ a2x = f(t), where a1 and a2 are constants and f(t) is a continuous periodic function with period 2n that can be expanded in a Fourier series, a1 =I= 0 and a2 =1= 0. 2. DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER 175 46. x'+ 3x =cost+ f.l.X2 ; f.l is a small parameter. Give an approximation of the periodic solution. 47. x3y"-xy'+y=0; integrate the equation if y1 =x is a particular solution. 48. Find the homogeneous linear equation with the following I fundamental system of solutions: Y1 = x, Ys = x . 49. x1V +x= t3 • 50. X=(y")3 +y" +1: 51. x+ 10x+25x=2' +te··it. 52. xyy"- x (y')2 - yy' = 0. 53. yVI_y=e2x. 54. yVl -f- 2y1V -f- y" =X -f-eX. 55. 6y"yiV-5 (y"')1 = 0. 56. xy" = y' In JL .X 57. y" +y = sin 3x cos x. 58. y"=2y3 , y(l)=l, y'(l)=L 59. yy"- (y')2 = y'. CHAPTER 3 Systems of differential equations t. Fundamentals The equation of motion of a particle of mass m under the action of a force F (t, r, r) d'~r . m dt2 = F (t, r, r) can be replaced by a system of three scalar equations of second order by projection on the coordinate axes: d2x m dt~ =X (t, x, y, z, x, y, i), d2y . . m dtz = Y (t, x, y, z, x, y, z), m :;: =l (t, x, y, z, x, y, i), or a system of six equations of the first order, if for the unknown functions we take not only the coordinates x, y, z of the moving particle, but also the projections x, y, i of its velocity :~ : X=U, y=v, Z=W; mit= X (t, x, y, z, u, v, w), mv = Y (t, X, y, Z, U, V, W), mW= l (t, X, y, Z, U, V, W). It is then usual to specify the unitial position of the point X (t0 ) = X0 , y (t0) = y0 , z (t0) = Z0 and the initial velocity u (t0 ) = u0 , V Uo) = Vo, W (to)= Wo· This basic problem with initial values has alr~::ady been considered in Sec. 6, Chapter l (page 56). There, proof was given of the theorem of existence and uniquenl;!ss of the solution of a system 3. SYSTEMS OF DIFFERENTIAL EQUATIONS of differential equations ~/ = ft (t, X1, X2• ••• , X,.), d~~ = f2 (t, x1 , x2 , ••• , x,.), d:t= f,. (t, xl, Xs, ••• ' x,.), which satisfies the initial conditions 177 (3.1) X; (t0 ) = X;0 (i = 1, 2, ... , n). (3.2) We recall that the sufficient conditions for the existence and uniqueness of solution of a system (3.1 ), given initial conditions (3.2), are: (1) continuity of all functions f1 in the neighbourhood of the initial values; (2) fulfillment of the Lipschitz condition for all functions f1 with respect to all arguments, beginning with the second one in the same neighbourhood. Condition (2) may be replaced by a cruder condition by requiring the existence of partial derivatives b0unded in absolute value: iJf; iJXj (i, j = I, 2, ... , n). The solution of the system of differential equations x1 (t), x2 (t), ... , x,. (t) is an n-dimensional vector function which we will briefly designate as X (t). In this notation, system (3.1) may be written as dX Tt=F(t, X), where F is a vector function with coordinates (f1, f,, ... , /,.) and the initial conditions are in the form X {10) = X0 , where X0 is an n-dimensional vector with coordinates (xlO' X10 , ••• , x,.0 ). The solutions of the system of equations X1 =X1 (t), X2 =X1 (t), ... , x,.=x,. (t) or, briefly, X= X (t) defines in Euclidean space with coordinates t, X1 , X2 , ••• , x,. a certain curve called the integral curve. Upon fulfillment of the conditions (1) and (2) of the theorem of existence and uniqueness, a unique integral curve passes through every point of this space and the assemblage of such curves forms an n-parameter family. As parameters of this family, one can, for example, take the initial values X10 , x~0, ••• , Xno. 178 I. DIFFERENTIAL EQUATIONS A different interpretation of the solutions X1 = X1 (t), X~= X2 (t), ... , Xn = Xn (/), or, briefly X= X (t), is possible; it is particularly convenient if the right sides of (3.1) do not depend explicitly on t. In Euclidean space with rectangular coordinates xt> X2 , ••• , xn the solution X1 =X1 (t), X2 =X2 (t), ... , xn=Xn(t) defines a law of motion of some trajectory depending on the variation of the parameter t, which in this interpretation will be called the time. In such an interpretation, the derivative ~~ will be the velocity of t. f · t d dx1 dx2 dxn "II b th d" tmo wn o a pom , an Tt , Tt , ... , dt wt e e coor ma es of the velocity of that point. Given this interpretation, which is extremely convenient and natural in many physical and mechanical problems, the system dx· IJ}=f;(t, X1, X2, ••• , xn) (i= 1, 2, ... , n) (3.1) or dX df=F (t, X) is ordinarily called dynamical, the space with coordinates x1, X2 , ••• , xn is called the phase space, and the curve X= X (t) is called the phase trajectory. At a specified instant of time t, the dynamical system (3.1) defines a field of velocities in the space X1 , x2 , ••• , xn. If the vector function F is explicitly dependent on t, then the field of velocities varies with time and the phase trajectories can intersect. But if the vector function F or, what is the same thing, all the functions h are riot dependent explicitly on t, then the field of velocities is stationary, that is to say, it does not vary with time, and the motion will be steady. In the latter case, if the conditions of the theorem of existence and uniqueness are fulfilled, then only one trajectory will pass through each point of the phase space (x1, x2, ••• , xn>· Indeed, in this case an infinite number of different motions X= X (t +c), where c is an arbitrary constant, occur along each trajectory X= X (t); this is easy to see if we make a change of variables t, = t +c after which the dynamical system does not change form: dX =F(X) dt, and consequently X= X {t1) will be its solution, or, in the old variables, X= X (t +c). 3. SYSTEMS OF DIFFERENTIAL EQUATIONS 179 If In the case at hand two trajectories passed through a certain point X0 of the phase space, X= X1 (t) and X= X2 (t), X1 {t0 ) = X2 (t0 ) = X0 , then, taking on each of them that motion for which the point X0 is reached at time t = t0 , i.e., considering the solutions X= X1 (t-t0 +fo) and X= X2 (t-to +t0), we obtain a contradiction with the existence and uniqueness th-:?rem, since two different solutions X1 (t-t0 +t0 ) and X2 (t-t0+t0 ) satisfy one and the same initial condition X (t0) = X0 • Y Example. The system of equations dx dy dt = y, dt =-X (3.3) has the following family of solutions (as may readily be verified by direct substitution): X=C1 cos(t-c1); y = -c1 sin (t -c2). Regarding t as a parameter, we get a family of circles on the phase plane x, y with centre at the origin of coordi- Fig. 3-l nates (Fig. 3.1). The right member of :c (3.3) is not dependent on t and satisfies the conditions of the existence and uniqueness theorem, and so the trajectories do not intersect. Fixing c1 , we get a definite trajectory, and to different c2 there will correspond different motions along this trajectory. The equation of the trajectory x2 +y2 = c~ does not depend on c2 so that all the motions for fixed c1 are executed along one and the same trajectory. When c1 = 0 the phase trajectory consists of a single point called in this case the rest point of the system (3.3). 2. Integrating a System of Differential Equations by Reducing It to a Single Equation of Higher Order One of the main methods of integrating a system of differential equations consists in the following: all unknown functions (except one) are eliminated from the equations of the system (3.1) and from the equations obtained by differentiation of the equations that make up the system; to determine this one function, a single differential equation of higher order is obtained. Integrating the equation of 12* 180 I. DIFFERENTIAL EQUATIONS higher order, we find one of the unknown functions; the other unknown functions are determined (if possible without integrations) from the original equations and from the equations obtained as a result of their differentiation. The following examples will serve as an illustration. Example t. dx dy dt=y, dt=X. Differentiate one of these equations, for instance the first, ~;: = ~~ ; eliminating ':ft by means of the second equation, we get~~ -x=O, whence x = c1e1 + c,e-t. Utilizing the first equation, we get dx t -t y = dt =cte -c2e . We determined y without integrations by means of the first equation. If we had determined y from the second equation dy t + -t t -t +dt=x=c1e C2e , y=c1e -c2e c3 , we would have introduced extraneous solutions, since direct substitution into the original system of equations shows that the system fs satisfied by the functions x=c1e1+c2e-1, y=c1e1-c2e-1+ca not ior an arbitrary c3 but only for C3 = 0. Example 2. dx -=3x-2ydt , dy dt=2x-y. Differentiate the second equation: d2y dx dy dt2 = 2 dt- dt • From (3.42) and (3.5) we determine x and ~: X=~(~~+y), dx I (d2y dy) dt = 2 dt1 +dt . Substituting into (3.41) we get d2y dy dt2 - 2 dt +y=O. (3.5) (3.6) Integrate the resulting homogeneous linear equation with constant 3. SYSTEMS OF DIFFERENTIAL EQUATIONS coefficients y=e1 (c1 +c2t); substituting into (3.6), we find x(t): x =}e1 (2c1 +C2 +2c2t). Example 3. d2x d2y dt 2 =y, dt2 =X. 181 Differentiating the first equation, we get :;; = :;: and, substituting into the second equation, we have :;: = x. Integrating this homogeneous linear equation with constant coefficients, we obtain X= c1et +c2e-t +c3 cost +c, sin t and, substituting into the first equation, we find y=c1et +c2e-1 -c8 cos t-c, sin t. Now we shall describe more exactly the process of eliminating all unknown functions except one from the system of equations. First we will show that one of the unknown functions, say, x1 (t), which is a component of the solution x1 (t), X2 (t), ... , xn (t) of the system of differential equations: dJt1 = { 1 (t, X1, X1, • • ·, Xn), ~2 ={2 (t, X1 , X2 , ••• , Xn), (3.1) satisfies a certain nth order equation; here we assume that all the functions f1 have continuous partial derivatives up to order n-1 inclusive with respect to all arguments. Substituting some solution x1 (t), x2 (t), ... , xn (t) into the system (3.1), we reduce all the equations of the system to identities. In particular, the first equation of the system, will be reduced to an identity. Differentiate this identity with respect to t: 182 I. DIFFERENTIAL EQUAliONS or n tPxl =oft+""" iJft f (3 7a) dt2 iJt ~OX· I• • 2 i= 1 l and designating the right side of the latter identity by F2 (t, x1 , X2 , ••• , xn), we get Again differentiating this identity: or n d3xt = iJF2 +""" iJF2 f. dt3 at ~ iJx· 1' i= 1 I (3.7~) (3.7~) and denotin~ the right side of the latter identity by F3 (t, x1 , X2 , ••• , Xn), we get (3.7~) We again differentiate this identity; continuing the process n-2 times, we finally get the identity which~ when differentiated once again and when the identities (3.1) are taken advantage of, yields dnx dt"t=Fn(t, xl, x•• ... , xn)· We thus have n-1 identities ~1 = fdt, X1, x,, ..., Xn), (3.71) (3.7) and one more identity (3.8) 3. SYSTEMS OF DIFFERENTIAL EQUATIONS 183 Suppose that in the given range of the variables the determinant !!._(/1~f2,_!'a. · · ·• fn-t) ::;e 0 D (x2, x8, x,, ... , X11) • Then the system (3.7) may be solved for X2 , x3 , ••• , X11 by expres. th · t f th · bl t dx1 dn-Ixt p tsmg em m erms o e varta es , xto dt , ... , dtn- 1 • u · ting into the last equation (3.8) the variables x2 , x3 , ••• , X11 found from the system (3.7), we get an equation of the nth order: ~~ ( d~ ~-~) dtn = t, x1, dt ' ... ' dtn-1 ' (3.81) which is satisfied by the function x1 (t), which by hypothesis was a function X1 (t) of the solution x1 (t), X2 (t), ... , X11 (t) of the system (3.1). Now let us prove that if we take any solution X1 (t) of this nth order equation (3.81), put it into the system (3.7) and determine from this system x1 (t), x8 (t), ... , X11 (t), then the system of functions X1 (t), x, (t), ... , X11 (t) (3.9) will be the solution of the system (3.1). We put this system of functions (3.9) into (3.7) and thus reduce all the equations of the system to identities; in particular we obtain the identity ~1 =- ft (t' x1, x,, ... ' xn>· Differentiating this identity with respect to t, we will have 11 d2x1 = at1 +L. at1 dxi (3.10) dt2 at i=l ax; dt . In this identity it is not yet possible to replace ~~ with the functions{;, since we have not yet proved that the functions x1, X2 , ••• , X11 obtained by the above-mentioned method from the equation (3.8) and the system (3.7) satisfy the system (3.1); what is more, it is precisely this assertion that is the aim of our proof. Subtracting identity (3.72), taken in the expanded form (3. 7~). termwise from the identity (3.10), we get or, by virtue of (3.71), n ""aft (dx,-_t·) =0 ~ax· dt 1 - • 1=2 I 184 I. DIFFERENTIAL EQUATIONS In analogous fashion, differentiating identity (3.72) and subtracting (3.7~). and then differentiating identities (3.73 ) and subtracting (3.7~). and so on, we obtain n n ~ ofz (~i-f·) =0 ~ox· dt 1 ' 1=2 I ~ oFn- 1 (dxt_f·) = O. ~ OX· dt I 1=2 I Since the determinant of the homogeneous linear system of equa- tions (3.11) .i;Fn--~·(;xt~-~ ). .o. ·1 ~ OXt dt I ' J1=2 consisting of (n-1) equations in n-1 unknowns (~1 -11) li = 2, 3, ... , n) coincides with the nonzero functional determinant D :;i=O D (x2, x3 , ••• , Xn) ' the system (3.11) has only trivial solutions at each point of the region under consideration: dx·d; -/1=0 (i=2, 3, ... , n). Taking into account also (3.71), we find that the n functions X1, x1 , ••• , xn are the solution of the system of equations (i=1,2, ... ,n). Note. 1. This process of eliminating all functions except one presupposes that D(ft, f2, ... • Fn-t> :;e:O (3.12) D (x1, x3, ••• , Xn) • If this condition is not fulfilled, then the same process may be employed, but in place of the function x1 take some other one of 3. SYSTEMS OF DIFFERENTIAL EQUATIONS 185 the functions x2 , x3 , ••• , xn that make up the solution of the system (3.1). Now if the condition (3.12) is not fulfilled for any choice of some function X2 , X3 , ••• , xn in place of X1 , then various exceptional cases are possible. We illustrate them in the following examples. Example 4. d;t' = fdt, xl), d~2 = fs (t, X2), ~~3 =fs(t, Xa)· The system has disintegrated into quite independent equations, each of which has to be integrated separately. Example 5. ~; = '1(t, xl), ~2 = '2 (t, x2, ) df2 0 Xa' a=/= 'Xa ~t=fa(t, x2, Xa)· The latter two equations may be reduced to 0:1e equation of the second order in the manner indicated above, but the first equation, which contains the unknown function X1 that does not appear in the other equation, has to be integrated separately. Note. 2. If we apply the above-indicated process of eliminating all unknown functions except one to the system n d;: =I.ail (t) x1 (i = 1, 2, ... , n), i=I called a homogeneous linear system, then, as is readily verifiable, the nth order equation dnx1 (f) (t dx1 dn-lx) dtn = ' X1, dt' ''' ' dtn-/ (3.81) will also be homogeneous linear, and if all the coefficients a11 were constant, then the equation (3.81) as well will be a homogeneous linear equation with constant coefficients. A similar remark holds true for the nonhomogeneous linear system n d;: = "2:ail (t) x1+fdt) (i = 1, 2, ... , n), i= I 186 I. DIFFERENTIAL EQUATIONS for which the equation (3.81) will be a nonhomogeneous linear equation of the nth order. 3. Finding Integrable Combinations Integration of the system of differential equations dxl • Tt=fdt, X1, X2 , ••• , xn) (t=l, 2, ... , n) (3.1) is often accomplished by choosing so-called integrable combinations. An integrable combination is a differential equation which is a consequence of the equations (3.1) but one which is readily integrable, for example an equation of the type del> (t, X1, XJ, ••• , Xn) = 0 or an equation that, by a change of variables, reduces to some kind of integrable type of equation in one unknown function. Example t. dx dy dt =y, dt =X. Adding the equations term by term, we find a single integrable combination d(x+u> + or d(x+v> =dt, dt X Y x+y whence Inlx+yl=t+ lnc10 x+y=c1e'. Subtracting termwise the second equation of the system from the first, we get a second integrable combination d~-0 ( ) d~-0 dt--'--::-:-'""'"=- x-y or =dt x-y ' In jx-yl =-t+ lnc1 , x-y=c.e-t. We have thus found two finite equations x+y=c1e1 and x-y=c,e-t, from which the solution of the original system can be determined, or x=c1e1 +c,e-', y=~e1 -c2e-t. One integrable combination permits obtaining one finite equation <1>1 (t, xl, x~, ... , Xn) = cl, 3. SYSTEMS OF DIFFERENTIAL EQUATIONS 187' which relates the unknown functions and the independent variable; such a finite equation is called a first integral of the system (3.1 ). Thus, the first integral (t, X1 , X2 , ••• , xn)=c (3.13) of the system of equations (3.1) is a finite equation that is converted into an identity for some value c if the solution of the system (3.1) is substituted in place of xi (t) (i = 1, 2, ... , n). The left-hand member (t, X1 X2 , ••• , xn) of equation (3.13) is often also called a first integral, and then the first integral is defined as a function not identically equal to a constant, but retaining a constant value along the integral curves of the system (1). Geometrically, the first integral <1> (t, x1 , X2 , ••• , xn) = c for fixed c may be interpreted as an n-dimensional surface in (n +1)-dimensional space with coordinates t, x1 , X2 , ••• , xn, the surface possessing the property that each integral curve having a common point with the surface lies entirely within the surface. For variable c, we get a family of nonintersecting surfaces possessing the same property, that is consisting of the points of some (n-1 )-parameter family of integral curves of the system (3.1). If k integrable combinations have been found, we get k first integrals: <1>1 (t, X1 , X2 , ••• , xn) _ C1 , } ~z(~, .x~: ~2·. ·:·: ~n~~c~, (3.14) k(t, Xl, Xz, ... 'Xn)=Ck. If all these integrals are independent, i.e., if at least one determinant D ($1, $z, · · · ,$k) =/= 0 D (xit, x1,, ..• , xfk) ' where x1,, x1,, ••• , x1k are certain k functions made up of x1 , x2 , ••• , xn, then it is possible to express the k unknown functions of the system (3.14) in terms of the others; substituting into (3.1), we reduce the problem to integration of a system of equations with a smaller number of unknowns. If k = n and all the integrals are independent, then all the unknown functions are determined from the system (3.14). Example 2. dx dy dz dt = y-z, dt =z-x, dt =x-y. Adding the equations of this system term by term, we get dx +dy +dz = O d ( + + ) 0dt dt dt or dt x Y z = . 188 I. DIFFERENTIAL EQUATIONS and from this we have x+y+z=c1 • The first integral that has been found permits expressing one of the unknown functions in terms of the remaining variables and this enables us to reduce the problem to integration of a system of two equations in two unknown functions. However, in the given case it is easy to find one more first integral. Multiply the first equation termwise by x, the second by y, the third by z and add: dx dy dz X dt +y dt +z dt = O, or, multiplying by 2, we get :t(x2 +y2 +z2) = 0, whence x2 +Y2+z~ =C~. Using the two first integrals thus found it is possible to express the two unknown functions in terms of the other variables and thus to reduce the problem to integration of one equation in one unknown function. Example 3. dp dq C A Cdr AAdf=(B-C)qr, Bdf=( - )rp, df=( -B)pq, where A, Band C are constants (this system is encountered in the theory of motion of rigid bodies). Multiplying the first equation by p, the second by q, the third by r and adding, we get dp dq dr Ap df+Bqdf+Cr df=O, w!lence we find the first integral Ap2 +Bq' +Cr2 = c1• Multiplying the first equation by Ap, the second by Bq, the third by Cr and adding, we have A2 dp 8 2 dq cz dr O Pdf+ qTt+ 'tit= • and then integrating we get yet another first integral, Azp'+82q2 +C2r2 = Cz· If we exclude the case of A= B = C, in which the system is integrated directly, the first integrals that have been found are independent and, hence, it is possible to eliminate the two unknown functions by taking advantage of these first integrals; to determine the third function we get one equation with variables separable. · 3. SYSTEMS OF DIFFERENTIAL EQUATIONS 189 When finding integrable combinations, it is often convenient to use the so-called symmetric form of writing the system of equations (3.1 ): -q>2 (t, Xt, x2, ••• , Xn) dxn dt IPn (t, x1 , X2, ••• , Xn) !po (t, x1, x2, ••• , Xn) ' (3.15) where f (t ) q>; (t, Xt, X2, ••• , Xn) i ' x,, X2, , •.' Xn = (t ) IPo ' Xt, x2, .•• ' Xn (i = I, 2, ... , n). The variables are involved equivalently in a system given in symmetric form, and this sometimes simplifies finding integrable com- binations. Example 4. dx dy dz xt-y2-z2 = 2xy = 2xz. Integrating the equation dy · dz 2xy=2xz' (3.16) we find J!..=c1 • Multiplying numerators and denominators of the first z of the relations of the system (3.16) by x, the second by y, the third by z and forming a derived proportion, we get and from this or xt+y2+z2 _.:.....::...y-'-- =Cz· The independent first integrals thus found, .!L =c andz 1 determine the desired integral curves. 4. Systems of Linear Differential Equations A system of differential equations is called linear if it is linear in all unknown functions and their derivatives. A system of n linear equations of the first order written in normal form looks like n ~; = L.a;1 (t)x1+{;(t), (i= I, 2, ... , n), /=I (3.17) 190 or, in vector form, I. DIFFERENTIAL EQUATIONS dX dT=AX+F. (3.18) where X is an n-dimensional vector with coordinates x1 (t), x, (/), ... , xn (/), F is an n-dimensional vector with coordinates f.(/), fs (/), ... , fn (/), which it will be convenient to regard in future as one-column matrices: x. f. x, f, X= F= Xn fn dx1 Tt au au ... aln dx2 (if A= an an ... a,n dX . . dt= anl an2 ... ann I dxn (if According to the rule of matrix multiplication, the rows of the first factor must be multiplied by the column of the second; thus, n n ~ a11x1 ~ alfx1+ f. J=I /=1 n n AX= ~ a21x1I= I AX+F= i~l a2/xl+fs n· ~ an,xj J=l The equality of any matrices implies the equality of all ments, hence the one matrix equation (3.18) or dxl I "Tt ~ a11x1+/1 dx2 i=l dt n ~ atfx1+f2 J= I n ~( 1~1 anfxl+ fn is equivalent to the system (3.17). their ele- 3. SYSTEMS OF DIFFERENTIAL EQUATIONS 191 .If all the functions ail. (t) and /; (t) in (3.17) are continuous on the interval a~ t ~ b, then in a sufficiently small neighbourhood of every point (t0 , X10, x~0, ••• , Xn0 ), where a~ t0 ~ b, the conditions of the existence and uniqueness theorem are fulfilled (see page 177) and, hence, a unique integral curve of the system (3.17) passes through every such point.. Indeed, in the case at hand the right-hand members of the system (3.17) are continuous and their partial derivatives with respect to any x1 are bounded, since these partial derivatives are equal to the coefficients ail (t) which are continuous on the interval a~ t ~b. We define the linear operator L by the equality dX L [X]=--AXdt ' then equation (3.18) may be written still more concisely as L [X]= F. (3.19) If all the f;(t)==O (i= 1, 2, ... , n) or, what is the same thing, the matrix F = 0, then the system (3.17) is called homogeneous linear. In abridged notation, a homogeneous linear system is of the form L [X] =0. The operator L has the following two properties: (1) L (eX) =eL [X), where e is an arbitrary constant. (2) L[X1 +X2)=L[X1]+L(X2). Indeed, d~tX>_,~ (eX)= e [~~-AX] , d(X~d;X2) A (XI +X2) = (d;~-AXl) + (d;2-AX2) . A consequence of Properties (1) and (2) is L[ ~ eixi] = ~ e;L [X;), l= I l= I where e1 are arbitrary constants. (3.20) Theorem 3.1. If X is a solution of a homogeneous linear system L[X]=O, then eX, where cis an arbitrary constant, is a solution of the same system. Proof. Given L [X] == 0; it is required to prove that L [eX]== 0. Taking advantage of Property (l) of the operator L, we have L [eX] =cL [X] =0. 192 I. DIFFERENTIAL EQUATIONS Theorem. 3.'!. The sum x.+x2 of two solutions XI and x2 of a homogeneous linear system of equations is a solution of that system. m Proof. Given L [X1) =0 and L [X2) =0. It is required to prove that L [X1 +X2] =0. Using Property (2) of the operator L, we obtain L [X1 +X2) ==L [X1] +L [X2) =0. Corollary to Theorems 3.1 an.d 3.2. A linear combination ~ c1X1with arbitrary constant coefficients of the solutions X1, X 2 , ••• , l= I X,. ofahomogeneous linear system L [X]= 0is asolution of that system. Theorem 3.3. If the homogeneous linear system (3.20) with real coefficients ail (t) has a complex solution X= U+iV, then the real and imaginary parts ut u~ V= and V= Un Vn separately are solutions of that system. Proof. Given L [U+iV] = 0. It is required to prove that L [U] ==0 and L [V] =0. Using Properties (1) and (2) of the operator L, we get L (U +iV] =L [U] +iL [V] ==0. Hence, L [U] == 0 and L [V] =0. The vectors X 1, X., ... , Xn, where X,= x.,(t) X1; (t) Xnl(t) are calted linearly dependent on the interval a~ t ~ b if there exist constants a1, ~. • •• , an such that a.X1 +a,X, +... +anXn =0 (3.21) for a~ t ~ b; and at least one a1::;6 0. But if the identity (3.21) holds true only for a1 =a,= ... =an= 0, then the vectors X10 X,, ••. , Xn are termed linearly independent. 3. SYSTEMS OF DIFFERENTIAL EQUA r IONS 193 We note that the single vector identity (3.21) is equivalent to n identities: " 1~ a;Xti (t) = 0, I= I " I~ a 1x 11 (t) ==0, ~ (3021.)I= I I . . I" ~ a 1x,1(t) ==00 J'=I If the vectors Xi (i = 1, 2, ... , n) are linearly dependent and thus there exists a nontrivial system a1 (i.e., not all a1 are zero) that satisfies the system of n homogeneous (with respect to a1) linear equations (3.211), then the determinant of the system (3.211) Xu Xu ... x," W= Xu Xn ... x,n Xnt Xnz ... Xnn must be zero for all values of t of the interval a~ t::::;; b. This determinant of the system is called the Wronskian determinant of the system of vectors X1 , X,. 0 • • , Xno Theorem 3.4. If the Wronskian W of the solutions X1 , X2 , •• 0, X" of the homogeneous linear system of equations (3.20) with coefficients ail (t) continuous on the interval a~ t ~ b is zero at least in one point t =to of the interval a~ t~ b, then the solutions x•. x•. .0 • • xn are linearly dependent on that interval, and, hence, W;::;: 0 on that interval. Proof. Since the coefficients a/J (t) (l, j = 1, 2, ... , n) are continuous, the system (3.20) satisfies the conditions of the existence and uniqueness theorem. Hence, the initial value X (t0 ) = 0 [or 10 more detail, x1 (t0) = 0, x, (t0) = 0, ... , xn (t0) = 0] determines the unique solution of this system, and this solution is obviously the trivial solution of the system (3.20) X (t) ==0 [or, in more detail, x1 (t)=O, x,(t)=O, ... , x"(t)=O]o The determinant W(t0 )=0. Hence, there exists a nontrivial system c1, c,, ..•, c" that satis· fies the equation c1Xdt0 ) +c,X,(t0 ) + ... + c"X" (t0 ) =0, since this single vector equation is equivalent to a system of n homogeneous (with respect to c1) linear equations with the zero 13--37H 19<1 I. DIFFERENTIAL EQUATIONS determinant "~ C;Xti (10 ) = 0, I= I n ~ C;X2; (10 ) = 0, I= I n ~ C;Xn; (10 ) = 0. l= I n The solution of the equation (3.20) X (t) = ~ c,.X,. (t) that corre- 1= I sponds to this nontrivial system c1, c,, ... , en satisfies the zero initial conditions X (t.,) =0 and, consequently, coincides with the trivial solution of the system (3.20): n ~ C;X; (t) ==0, I= I i.e., the X; are linearly dependent Note. As the most elementary examples show, this theorem does not extend to the arbitrary vectors X1, X,, ... , Xn which are not solutions of the system (3.20) with continuous coefficients. F.xample I. The system of vectors X• =II:II and X2 =II::II is linearly independent, since from or cx1 X1 + cx2 X 2 =0 { cx1t +cx2t2 = 0, cx1t +cx2L2 = 0 it fallows that cx1 = cx2 = 0 (see page 102, Example I). At the same time the Wronskian I~ ::j is identically zero. Hence, the vectors X1 and X2 cannot be solutions of one and the same homogeneous linear system (3.20) with continuous coefficients a,.j (t) (i. ; = 1. 2) n Theorem 3.5. The linear combination ~ c;X; of n linearly I= I independent solutions Xh X2 , ••• , Xn of the homogeneous linear system (3.20) with coefficients ail (t) continuous on the interval a~ t ~ b is the general solution of the system (3.20) on that interval. Proof. Since the coefficients ail (t) are continuous on the interval a~ t ~ b, the system satisfies ttie conditions of the existence and 3. SYSTEMS OF DIFFERENTIAL EQUATIONS 195 uniqueness theorem and, hence, to prove the theorem it is sufficient to notice that by proper choice of the constants C; in the solution n ~ c;X; it is possible to satisfy arbitrarily chosen initial conditions i= I X (10) =Xu, where t0 is one of the values of t on the interval a,.;;; t,.;;; b, i.e., it is possible to satisfy the single vector equation n ~ C;X{ (t(l) =XII l= I · or the equivalent system of n scalar equations: n ~ C;Xli {10 ) = X10, •= I n ~ C;Xz; (to)= X2o• I= I n ~ C;Xni Uo) = Xno· i= I This system is solvable for c1 for any X;0 , since the determinant of the system is the Wronskian determinant for a linearly independent system of solutions X,, X2 , ••• , Xn and, hence, does not vanish at any point of the interval a,.;;; t ::;;;;; b. Example 2. dx }dt=y, dy dt=-X. (3.22) It may readily be the solutions verifled that the system (3.22) is satisfied by x, =cost, y, =-sin t and x2 =sin t, y, =cost. These solutions are linearly independent since the Wronskian / cost -sintl sin t cost = 1 196 I. DIFFERENTIAL EQUATIONS -------------------------------------------------- is not zero. And so the general solution is of the form x=c1 cost+c,sin t, y=- C1 sin t +c, cost, where c1 and c, are arbitrary constants. Theorem 3.6. If X is a solution of the nonhomogeneous linear system L [X] =F, (3.19) and X1 is a solution of the corresponding homogeneous system L [X]= 0 then the sum X,+ X is also a solution of the nonhomogeneous system L [X]_=F. Proof. Given L [X]==- F and L [X1] ==- 0. Prove that L [X1 +X]==- F. Using Property (2) of the operator L, we get L [X1 +X]== L [X1] + L [X]== F. Theorem 3.7. The general solution, on the interval a~ t ~ b, of the nonhomogeneous system (3.19) with coefficients a11 (t) continuous on that interval and with right sides f1 (t) is equal to the sum of n the general solution ~ c1X1 of the correspondinghomogeneous system 1=1 and the particular solution X of the nonhomogeneous system under consideration. Proof. Since the conditions of the existence and uniqueness theorem are fulfilled (see page 191 ), to prove the theorem it will suffice to notice that by means of a proper choice of arbitrary constants c1 II in the solution X= ~ c1X1+X it is possible to satisfy arbitrary I= I specified initial conditions i.e., we have to prove that the one matrix equation n ~ c1X1 (t0) +X (/0) = X0 l= I 3. SYSTEMS OF DIFFERENTIAL EQUA TJONS 197 or an equivalent system of equations ±e;x1; (to) +x1 (to)= xto• l 1=1 l~1elx,, (to)+ Xz (/0) = X~o· f t elXnl (to)+ Xn (to)= Xno 1=1 (3.23) always has the solution e1, e,, ... , em no matter what the righthand members. But in this form this assertion is obvious, since the determinant of the system (3.23) is the Wronskian at the point t = t0 for the linearly independent solutions Xl> X2, ••• , Xn of the corresponding homogeneous system and, by Theorem 3.4, is not zero. Hence, the system (3.23) has the solution el> Cu .•• , en for any right-hand members. · Theorem 3.8 (the principle of superposition). The solution of the system of linear equations m L[X]=~f1, F1= i=1 m is the sum ~ X1 of the solutions Xi of the equations 1=1 L[X;]=Fi (i=l, 2, ... , m). Proof. Given L [X;]== Fi (i = 1, 2, ... , m). Prove that L [i: xi]==i:F1• 1=1 1=1 Taking advantage of Property (2) of the operator L, we obtain L [i:xi]==i L [X;]== iF;. 1=1 1=1 1=1 Note. Theorem 3.8 can, without any change in the proof, be "'obviously extended to the case when m-+ oo, if the series ~ X1 1=1 converges and admits term-by-term differentiation. 198 I. DIFFERENTIAL EQUATIONS Theorem 3.9. If the system of Linear equations L(X] = U+iV. where V= Un lvn with real functions a11 (t), u1(t), v1(t) (i, j= l, 2, ... , n) has the solution - -ul ~1 X=U+fV, 0= u2 v v2 - -un Vn then the real part of the solution [J and its imaginary part V are respectively solutions of the equations L[X]=V and L[X]=V. Proof. Given L [U +iV) ~ U +iV; prove that L [U] ~ U, L [V] ~v. Using Properties (1) and (2) of the operator L, we get L [0+iV] ~ L[U] +iL [V] == U +iV. Hence, L (U] = U and L (V] = V. If the general solution of the corresponding homogeneous system L (X)= 0 is known, and one cannot choose a particular solution of the nonhomogeneous system L (X)= F and, consequently, one cannot take advantage of Theorem 3.7, then the method of variation of parameters may be applied. n Let X=~ c1X1 be the general solution of the corresponding hoI= I mogeneous system dX --AX=Odt for arbitrary constants c1, and, hence, X1 (i = 1, 2, ... , n) are linearly independent particular solutions of the same homogeneous 3. SYSTEMS OF DIFFERENTIAL EQUATIONS system. We seek the solution of the nonhomogeneous system dX df-AX=F in the form II X=~ C; (t)Xl, I= I 199 where the c1 (t) are new unknown functions. Substitution into the nonhomogeneous equation yields 11 II dX· II L.c;(t)X1 +Lc1(t)df=A L.c1(t)X1+F, 1=1 t=l 1=1 . dX;_AX hor. smce (if= 1, we ave II ~ c; (t) X1=F. 1=1 This vector equation is equivalent to a system of n equations: II ) ~c{(t)x11 =f,(t) II= I II ~ ci (t) x2; = {, (t). l ~.~~. . . . . . . If ! ci (t) X11; = fn (t). I= I J (3.24) All the ci (t) are determined from this system of n equations in n unknowns ci (t) (i = 1, 2, ... , n) with the determinant of the system W coinciding with the Wronskian for the linearly independent solutions X1, X2 , ••• , Xn and, consequently, not zero: c/ (t) = cp1 (t) (i = 1, 2, ... , n), whence, by integrating, we find the unknown functions c1 (I): c1 (t) = ~ cp1(t) dt +C; (i = 1, 2, ... , n). Example 3. dx dy I 'iii= y, dt = -x+ cost · The general solution of the corresponding homogeneous system dx dy (jj=Y, dt =-X 200 I. DIFFERENTIAL EQUATIONS is of the form x= c1 cost +c2 sin t, y=- c1 sin t+c, cost (see page 195, Example 2). We vary the constants x=c1 (t) cost +cz (t) sin t, y=- c1 (t) sin t +c, (t) cost. c~ (t) and c~ (t) are determined from the system (3.24), which, in this case, is of the form whence Therefore, c~ (t) cost +c~ (t) sin t = 0, -c; (t) sin t +c~ (t) cost= _!._t ,cos c; (t) =- ~~~tt , c~ (t) = 1. c1 (t) =In Icost I+~. C1 (f)=t+~ and we finally get x=c1 cost +c, sin t +cost ln I cost 1+t sin t, y=-~ sin t +~cost-sint In Icost I+ t cost. 5. Systems of Linear Differential Equations with Constant Coefficients The linear system of equations dx· n lif= I.al/x1+f;(t) (i = 1, 2, ..• , n) i== I is a linear system with constant coefficients. In the vector form it is dX Tt=AX+F. in which all the coefficients a11 are constant or, what is the same thing, the matrix A is constant. A system of homogeneous or nonhomogeneous linear equations with constant coefficients is most simply integrated by reducing it to a single equation of higher order. As was noted on page 185, the resulting equation of higher order will be linear with constant coefficients. However, it is possible directly to find the fundamental system of solutions of a homogeneous linear system with constant coeffici- ents. 3. SYSTEMS OF DIFFERENTIAL EQUATIONS ------ We will seek the solutions of the system d~1 =auxi+aizXs+ · · · +alnxno ldx2 dt = auxl + aux• + ... + a,nxn, ~~~·=·a~1X~ ~a~~~~~.·...+.an.nx~,) where all the a;1 are constant, in the form xl = a.Iekt' x2 = a.~ekt, ... ' Xn = a."ekt' 201 (3.25) with constants a.1 (j = 1, 2, ... , n). Substituting into the system (3.25), cancelling ek1 and transposing all terms to one side, we get (a11-k)a.1 + a12a.2+ ... +a111fX11 = 0,}a2la.l + (aaz-k) a..+ ... +a.,a.n = 0, (3.26). . . . . . . . . . . . . . . . anlal +an,a,+ ... +(an,-k) an= 0. For this system of n homogeneous linear equations in n unknowns a.1 (j = 1, 2, ... , n) to have a nontrivial solution, it is necessary and sufficient that the determinant of the system (3.26) be zero: =0. (3.27) . . . . . . . . . . anl ani ... ann- k From this equation of degree n we determine the values of k at which the system (3.26) has nontrivial solutions a.1 (j = 1, 2, ... , n). Equation (3.27) is called the characteristic equation. If all the roots k; (i = 1, 2, ... , n) of the characteristic equation are distinct, then, by putting them into the system (3.26) in succession, we determine the corresponding nontrivial values a.J1' (i, j = 1, 2, ... , n) and, consequently, we find the n solutions of the original system (3.25) in the form X't>- ,..m..k,t ••d>- ,..,,,..k,, ..d> - ,.,,,,..k,t (; - 1 2 n) (3 28)1 - '""'I ~· , Art - ~2 c:-- ' • • • ' AriJ - """n ~- • - ' ' • • • , , • where the upper index indicates the number of the solution and the lower index, the number of the unknown function. Using vector notation, we get the same result more compactly: dX Tt =AX; (3.251) 202 I. DIFFERENTIAL EQUATIONS we seek the ·solution in the form X= Aek1, where A= or (A -kE) A=0, where E is a unit matrix: 1 0 0 0 E= 0 1 0 0 0 0 0 (3.29) So that the equation (3.29) is satisfied by the nontrivial matrix A 0 0 0 it is necessary and sufficient that the matrix A-kE should be singular, i.e., that its determinant be zero: IA-kE I=0. For each root k1 of this characteristic equation IA-kE I=0 we determine, from (3.29), the nonzero matrix A and, if all roots k1 of the characteristic equation are distinct, we get n solutions: Xl=.rflt'ek•', X,=Al2>ek,t, .•. , X,.=Al"'eknt, where lt is easy to show that these solutions are linearly independent. 3. SYSTEMS OF DIFFERENTIAL EQUATIONS 203 lndeed, if there were a linear dependence n ~~; A'o~,t =0 i-J or, in expanded form, (3.30) then, by virtue of the linear independence of the functions eki1 (i= 1, 2, ... , n)(see pages 101-102) it would follow from (3.30) that ~;a.\1> = 0, } p,a.(il = 0 1 l ' (i= 1, 2, ... , n). (3.31) . . . . . R,..lll-0 t';-n - . But since for every i, at least one of the a.\;1, a.~1 , ••• , a.:/1 (i = 1, 2, ... , n) is different from zero, it follows from (3.31) that ~; = 0 (i = 1, 2, ... , n). And so the solutions A'11ek,, (i = 1. 2, ... , n) are linearly independent and the general solution of the system (3.25) is of the fort11 n X= ~c;Aiilek;t 1=1 or n x1= ~c;aY'ek;t (j = 1, 2, ... , n), 1=1 where the C; are arbitrary constants. The constants a.t (j = 1, 2, ... , n) are ambiguously determined from the system (3.26) for k = k;, since the determinant of the system is zero and, hence, at least one equation is a consequence of the others. The ambiguity in the determination of a.~11 is due to the fact that a solution of the system of homogeneous linear equations remains a solution of the same system when it is multiplied by an arbitrary constant factor. 204 I. DIFFERENTIAL EQUATIONS To the complex root k1 = p +qi of the characteristic equation (3.27) there corresponds the solution (3.32) which, if all the coefficients a;1 are real, can be replaced by two real solutions: the real part and imaginary part of the solution (3.32) (see page 192). A complex conjugate root k1+1 = p-qi of the characteristic equation will not yield any new linearly independent real solutions. If the characteristic equation has a multiple root k3 of multiplicity y, then, taking into account that the system of equations (3.25) can be reduced (by a process indicated on pages 181-183) to a single homogeneous linear equation with constant coefficients of order n or less (see Note 2 on page 185), it is possible to assert that the solution of the system (3.25) is of the form where A-(S){ - ,..CS} ""'nt (3.33) aw are constants. It should be noted that even in cases when the system of n equations (3.25) is reduced to an equation of order lower than n (see Note 1 on page 184), the characteristic equation of the latter necessarily has roots that coincide with the roots of the equation (3.27) [since the equation to which the system was reduced has to have solutions of the form ek-t where the k3 are the roots of the equation (3.27)). But it may be that the multiplicities of these roots, if the order of the equation obtained is less than n, will be lower than the multiplicities of the roots of the equation (3.27) and, hence, it may be that in the solution (3.33) the degree of the first factor will be lower than y-1, i.e., if we seek the solution in the form of (3.33), it may turn out that some of the coefficients AJ5', including the coefficient of the highest-degree term, vanish. Thus, we have to seek a solution of the system (3.25), which solution corresponds to a multiple root of the characteristic equation, in the form of (3.33). Putting (3.33) into (3.251) and demanrling that it become an identity, we define the matrices A)5'; some of them, including A~~ 1 as well, may turn out equal to zero. 3 SYSTEMS OF DIFFERENTIAL EQUATIONS 205 Note. There is a more precise way of indicating the type of solution of the system (3.25) that corresponds to a multiple root of the characteristic equation (3.27). Transforming the system (3.25) by means of a nonsingular linear transformation to a system in which the matrix II A-kE II has the Jordan normal form and then integrating the resulting readily integrable system of equations. we find that the solution which corresponds to the multiple root k8 of multiplicity y of the characteristic equation (3.27) is of the form X (t) = (A~1 + A\51/ + ... +A~~ ,tr>- 1) l.t, where ~ is the highest degree of the elementary divisor of the matrix 11 A-kE II corresponding to the root k8 • Example t. dx dy dt =x+2y, dt =4x+3y. The characteristic equation l14k 3_:kI=0 or k2-4k-5 =0 has roots k1 =5, k1 =-l. Hence, we seek the solution in the form xt = a.\l>e5t, x, = a.j2>e-t' Yt = aiuest, 11, = a.~2>e-t. (3.34) Putting (3.34) in the original system, we get - 4et\11 + 2a.~u = 0, whence a.~11 = 2a.\11; a.:v remains arbitrary. Consequently, xt = c1e6t, Yt = 2c~e~t, cl = a.:u. For determining the coefficients a.\21 and a.~21 we get the equation 2a~~~ +2a.~21 = 0, whence a~21 =- a.\21 ; the coefficient a\2 ' remains arbitrary. Hence, x2 = c,e-1, y, =- c,e-1, Cs =a.~''· The general solution Example 2. x = c1e51 +c2e-1, y = 2c1e51 -c,e-t. dx -=x-5ydt • dy dt=2x-y. The characteristic equation II2k -l-5k I= 0 or k~ +9 = 0 206 I. DIFFERENTIAL EQUATIONS has roots k, .~ =±3i, x, =a.1e3a, y1 = a.2e3' 1, (I -3i) a.1 -5a.2=0.This equation is satisfied, for example, by a.,= 5, a.2 = 1-3i. Therefore, x1 = 5e3i1 = 5 (cos 3t +i sin 3/), y, = (1- 3i) e3it = (1- 3i) (cos 3t +i sin 3/). The real part and the imaginary part of this solution are Iikewise solutions of the system at hand, and their linear combination with arbitrary constant coefficients is the general solution: x =,5c1 cos3t +5c2 sin 3t, y =c1 (cos 3t +3 sin 3t) +c, (sin 3t -3 cos 3/). Example 3. dx }dt=x-y, ~~=x+3y. (3.35) The characteristic equation j 1}k 3=!1=0 or k2 -4k+4=0 has a multiple root k1 , 2 = 2. Hence, the solution is to be sought in the form x=(a..+Btt)e2'.} Y = (a.2 +~2t) e'' Putting (3.36) into (3.35), we get whence ~2=-~t· a.2 =-a.,-~,. (3.36) a.. and ~~ remain arbitrary. Denoting these arbitrary constants by c, and C2 , respectively, we get the general solution in the form x = (c1 +c2t) e21 , y =- (c1 +c2 +c2t) e21 • G. Approximate Methods of Integrating Systems of Dift'erential Equations and Equations of Order n . All the methods (given in Sec. 7, Ch. I) of approximate integrati?n of differential equations of the first order may be carried over, Without essential changes, to systems of first-order equations and also to equations of order two and higher, which are reduced in 3. SYSTEM~ OF DIFFERENTIAL EQUATIONS 207 the usual manner to systems of first-order equations (see page 91). I. Successive approximation. As was pointed out on page 56, the method of successive approximations is applicable to systems of equations (i=1, 2, ... , n) (3.37) with initial conditions Y; (x0 ) = Y;o (i = 1, 2, ... , n) if the functions f; are continuous in all arguments and satisfy the Lipschitz conditions with respect to all arguments, from the second onwards. The zero approximation Y;o (X) (i = 1, 2, ... , n) may be chosen arbitrarily as long as the initial conditions are satisfied, and further approximations are computed from the formula .. Y1, HI (x) = Y;o + ~ /; (x, Y11~.• Y~k· •••• Ynk) dx (i = 1, 2, ... , n). JC. Just as for one equation of the first order, this method is rarely applied in practical calculations due to the relatively slow convergence of the approximations and the complexity and diversity of the compu- ~ tations. 2. Euler's method. An integral curve of the system of differential equations dy· d; =/;(X, y1, !/2, ••• , Yn) (i = 1, 2, ... , n), defined by the initial conditions Y; (x0 ) = =Y;0 (i= 1, 2.... , n) is replaced by a polygonal line that is tangent, at one of the boundary points of each segment, to the integral curve passing through the Fig. 3-2 same point (Fig 3.2 depicts Euler's polygonal line and its projection only on the xy1-plane). The interval x0 ~ x ~ b on which the solution has to be computed is partitioned into segments of length h. and the computation is performed using the formu.las (i = l, 2, ... , n). The convergence of Euler's polygonal lines to the integral curve as h--. 0 is proved in the same way as for a single first-order equation (see page 48). Iteration may be employed to increase the precision. 208 '· DIFFERENTIAL EQUATIONS 3. Expansion by means of Taylor's formula. Assuming that the right-hand members of the system of equations (3.37) are differentiable k times (in order to ensure differentiability of the solutions k + l times), we replace the desired solutions by the first few terms of their Taylor expansions: , ) • ) (X-X0)2 Y; (x) ~ Yt (Xo) +Yi (Xo) (X-Xo +Yt (Xo 21 + ··· tkl( )(X-Xo)k ("-} 2 )... +Yt Xo kl t - • • ••• • n . The error may be estimated by estimating the remainder term in Taylor's formula R;,. = y;k+t> [xo + e(X-Xo)] (x(,;~~~;l. where 0 < e< 1. This method yields good results only in a small neighbourhood of the point x0 . 4. Stormer's method. The interval x0 ::;;;;, x::;;;;, b is partitioned into subintervals of length h, and the computation of the solution of the system (3.37) is performed on the basis of one of the following formulas: where (i=l,2, ...' n), Ytk = y. _..,,.). Q;k = Yi (x,.) h, !J.q;, k-1 =qo.-Qt, k-1• !J.2qt, ,._, = !J.q;, k-1-!J.q;, ,._,, lJ.3Q;, 11.-s = !J.2Q;, 11.-2- !J.2qi, 11.-a· (3.38) (3.39) The formulas (3.38), (3.39), and (3.40) may be obtained in exactly the sC1me way as for one equation of the first order (see page 68). When using these formulas the order of error remains the same as for one equation. · To start computations by means of Stormer's formula, one has to know the first few values y1 (xk) which may be found by a Taylor expansion or by Euler's method with a diminished interval; just as in the case of one equation, the precision may be enhanced by applying iteration (see pages 67-68) or by Runge's method. 3. SYSTEMS OF DIFFERENTIAL EQUATIONS 5. Runge's method. The following numbers are computed m,l = {; (xk• Y1k• Yak• · · ·, Ynk), f( h hm11 m,2 = ; xk+ 2, Ylk+-2- • +hm21 +hmn1) Yak -2- • • • • • Ynk -2- • f ( h hm11 m;a= ; xk+2• Y1k+-2-, +hm22 hmn2 lY2k -2-' · · ·• Ynk + -2-; • m1, = {1 (xk +h. Ytk +hm18, Ytk+hm2a• · · ·• Ynk+hmna>· Knowing these numbers, we find y1, k+t from the formula h Yi,k+I=Yik+6(m11 +2m,,+2m18 +m1,) (i=l, 2, ... , n). The order of the error is the same as for one equation. 209 Depending on the required precision of the result, the interval (step) h is roughly chosen with account taken of the order of errors in the formulas used and is improved by means of trial computations with step h and : . The most reliable approach is to perform the computations with h and : of all required values of y1(xk), and if the results all coincide within the limits of the given precision, then h is considered to ensure the given accuracy of computations, otherwise the step is again reduced and the computations are performed with step ~ and : , etc. With a properly chosen step h, the differences t1q1k, t12q1k, • • • should vary smoothly, and the last differences in Stormer's formulas should affect only reserve decimals. PROBLEMS ON CHAPTER 3 dx dy 1. dt = y, dt =-X, X(0) = 0, y (0) = l. d2x1 d2x2 • 2. dtl = Xa, (jj2 = x., X1 (0) = 2, X1 (0) = 2, x, = 2. .t2 = 2. 3 dx dy 2 . dt+5x+y=e1, di-x-3y=e t. dx dy dz 4· di=y, dt=z, dt=x. dx dy ya 5· dt=y, dt=-x· dx dy dx ay 6· dt+ dt=-X+Y+ 3• dt-di=x+Y- 3• 14-378 210 I. DIFFERENTIAL EQUATIONS dy z dz 7· dx=x• dx=-xy. 8. ~ = _!!1!._ = _!1!__, • z-y x-z y-x dx dy dz 9. dt=-x+y+z, di=x-y+z. di=x+y-z. dx O dy 10. t(jj+Y= , t dt+x=O. dx dy I II (ii=Y+ I, dt=-x+sint · 12 dx __Y_ dy = _x_ · dt- x-y' dt x-y 13. x+ y=COS t, Y+ X= sin t 14. x+3x-y=0, y-8x+ y=O. x(O)= t. y(0)=4. d28 . n diJ 15. dt2 +sm9=0 for 1=0, 0= 36 , dt=O. Determine a(1) to within 0.001. 16. x(t)=ax-y, y(t)=x+ay; a is a constant. 17. x+3x+4y=0, y+2x+5y=0. 18. x=-5x-2y, y=x-7y . . 19. x=y-z, y=x+y. z=x+z 20 x-y+z=O, y-x-y= t, z-x-z=t. 21 dx _ dy _ dz x (y-z)- y (z-x)- z (x-y) · 22 ~ ~ • . x(y2-z2) 1J(z2- x2) z(x2- y2). 23. X=AX, where X=ll~:ll and A=ll~ =JII· CHAPTER 4 Theory of stability 1. Fundamentals Jn order to make a real phenomenon amenable to mathematical description, one unavoidably has to simplify and idealize it, isolating and taking into account only the more essential of the influencing factors and rejecting all other less essential factors. Inescapably, one is confronted by the question of how well the simplifying suppositions have been chosen. It may very well be that ignored factors produce a strong effect on the phenomenon under consideration, substantially altering its quantitative and even qualitative characteristics. The question is finally resolved by practicethe correspondence between the conclusions obtained and experimental findings-yet in many cases it is possible to indicate the conditions under which certain simplifications are definitely impossible. If a phenomenon is described by the following system of differential equations dy{ "" )Tt=~;(t, Y1• Y2• .•• , Yn (i=l, 2, ... , n) (4.1) with initial conditions Y; (t0 ) = Y;o (i = l, 2, ... , n), which are ordinarily the results of measurements and, hence, are inevitably obtained with a certain error, the question naturally arises as to the effect of small changes in the initial values on the desired solution. If it turns out that arbitrarily small changes in the initial data are capable of producing a substantial change in the solution, then the solution defined by the chosen inaccurate initial data is ordinarily devoid of any practical meaning and cannot describe the given phenomenon even approximately. This brings us to the important question of finding the conditions under which a sufficiently small change in the initial values brings about an arbitrarily small change in the solution. If t varies on a finite interval /0 :::;;;; t:::;;;; T, the answer to this question is given by the theorem on the continuous dependence of solutions on the initial values (see pages 58-59). But if t can take on arbitrarily large values, then the problem is dealt with by the theory of stability. 14 * 212 I. DIFFERENTIAL EQUATIONS The solution cp1 (t)(i=l, 2, ... , n) of the system (4.1) is called stable, or, more precisely, Lyapunov stable, if for any e > 0 we can choose all (e)> 0 such that for any solution !/i (t) (i = 1, 2, ... , n) of that system the initial values of which satisfy the inequalities 1Ydi0)-cp;(f0)l 0, the inequalities (4.2) are not fulfilled for at least one solution ydt) (i = 1, 2, ... , n), then the solution cp1 (t) is called unstable. Unstable solutions are rarely of interest in practical problems. If a solution cp1 (t) (i = I, 2, ... , n) is not only stable but, in addition, satisfies the condition lim ly1(t) -q>1 (t) I=0, t-.., (4.3) if lydt0)-c:p1(t0)l < 61 , 61 > 0, then the solution c:p;(t)(i= 1, 2, ... , n) is called asymptotically stable. Note that the stability of a solution !Jl; (t) (i = 1, 2, ... , n) does not yet follow from the single condition (4.3). Example 1. Test. for stability the solution of the differential equation :~ =- a2 y, a=FO, defined by the initial condition y (t0 ) =Yo· The solution y = y0e-a' (t-to) is asymptotically stable since IYoe-a' (t- to)-Yoe-a' (t- to) I= e-a• (t- to) IYo- Yo I< 8 for t;;;;;: t0 if IYo- y:I..(: se-a•t. and lim e-a• (t-to) IYo- Yo I=0. t ... ,., Example 2. Test for stability the solution of the equation dy 2 0 fdt =a y, a=F , de ined by the condition y (t0) =Yo· 4. THEORY OF STABILITY 213 The solution y = y0ea• 0 so small that there should follow from the inequality IY.,-y0 I< 6 (e) or for all t ;;z; t0 • Investigating the stability of some solution Yi = y;(t) (i = I. 2, ... , n) of the system of equations dyi (lf=Cf>t(t, Yt· Ys• · • ., Yn) (i=I, 2, ... , n) (4.1) may be reduced to investigating the stability of a trivial solution: a rest point located at the coordinate origin. Indeed, transform the system of equations (4.1) to new variables, putting (i= I. 2, ... , n). (4.4) The new unknown functions xi are the deviations y1-y1 (t) of the earlier unknown functions from the functions "it (t) that define the solution being tested for stability. By virtue of (4.4), the system (4.1) in new variables takes the form dxi dit - (lf= -dt +Cf>t(t, xt+ Ydt), x,+Ys (t), .. ·• Xn+Yn (t)) (i=I, 2, ... , n). (4.5) It is obvious that to the solution (being tested for stability) y1 =y~t(i=I, 2, ... , n) of the system (4.1) there corresponds the trivial solution xi_:_O (i=I, 2, .•. , n) of the system (4.5) by virtue of the dependence x1 = y1- ih (t); investigating the stability of the solution Yi=yt(t)(i= I. 2, ... , n) of the system (4.I) may be replaced by the stability testing of the trivial solution of the system (4.5). Therefore, henceforward we can take it, without loss of generality, that we test for stability the trivial solution or, what is the same thing, the rest point of the system of equations located at the coordinate origin. Let us formulate the conditions of stability as applied to the rest point xi== 0 (i = I, 2, ... , n). The rest point xi== 0 (i =I. 2, ... , n) of the system (4.5) is stable in the sense of Lyapunov if for each e > 0 it is possible to 214 I. DIFFERENTIAL EQUAliONS choose a ·6 (e)> 0 such that from the inequality there follows IX;(t0 )1<6(e) (i=l. 2, •.• , n) lx;(t)j 0 it is possible to choose a (\(e)> 0 such that from the inequality there follows n ~ x] (t) 0, k2 > 0. This case passes into the preceding one when t is replaced by -t. Hence, the trajectories have the same shape as in the preceding case, but the point moves along the trajectories in the opposite direction (Fig. 4.2). It is obvious that as t increases, points that are arbitrarily close to the origin recede from the e-neighbourhood of the origin- the rest point is unstable in the sense of Lyapunov. This type of rest point is called an unstable nodal point. 216 I. DIFFERENTIAL EQUATlONS (3) If k1 > 0, k2 < 0, then the rest point is also unstable, since a point moving along the trajectory (4.9) for arbitrarily small values of c1 goes out of the e-neighbourhood of the origin as t increases. Note that in the case under consideration there are motions which approach the origin, namely: X= C2~1ekat' y = c,~2ekat. Given different values of c2 , we get different motions along one and the same straight line y = {-; x. As t increases, the points on this straight line move in the direction of the origin (Fig. 4.3). Note Fig. 4-3 further that the points of the trajectory (4.9) move, as t increases, along the straight line y = ~ x receding from the origin. But if c1 =1= 0 and c2 =1= 0, then both as t-oo and as t--oo, the. trajectory leaves· the neighbourhood of the rest point. A rest point of this type i.s called a saddle point (Fig. 4.3) because the arrangement of trajectories in the neighbourhood of such a point resembles the arrangement of level lines in the neighbourhood of a saddle point of some surface z=f(x, y). (b) The roots of the characteristic equation are comple.x.. k1 , 2 =p±qi, q=foO. The general solution of this system may be represented in the form (see page 204) x = ePt (c1 cos qt +c, sin qt), } y = ePt (c~ cos qt +c; sin qt), (4.10) where c1 and c2 are arbitrary constants and c~ and c; are certain linear combinations of these constants. 4. THEORY OF STABILITY 217 -------------------------------------------------- The following cases art: then possible: (l) k,.,,=p±qi, p 0, q=/=0. This case passes into the preceding one when t is replaced by -t. Consequently, the trajectories do not differ from the trajectories of the preceding case, but motion along them occurs in the opposite direction as t increases (Fig. 4.6). Because of the presence of the increasing factor eP1, points which at the initial instant were arbitrarily close to the origin leave the e-neighbourhood of the origin as t incrt:ases; the rest point is uustable. It is termed an unstable focal point. (3) kl,l = ± qi, q =I= 0. As has already been mentioned, due to the periodicity of the solutions, the trajectories are closed curves containing within them 218 I. DIFFERENTIAL EQUATIONS the rest point (Fig. 4.4), which in this case is called a centre. The centre is a stable rest point, since for a given 8 > 0 it is possible to choose a 6 > 0 such that closed trajectories whose initial points lie in the 6-neighbourhood of the origin do not go out of the 8-neighbourhood of the origin or, what is the same thing, it is possible to choose such small c1 and c2 that the solutions will satisfy the X=C1 cosqt +c2 sin qt, } y = c~ cos qt +c; sin qt inequality x2 (t) +ys (t) < 8a. (4.11) Note, however, that there is no asymptotic stability in this case, since x(t) and y(t) in (4.11) do not tend to zero as t-oo. Fig. 4-6 (c) Roots that are multiples of k1 = k7 • (I) k.=k2<0. The general solution is of the form x (t) = (c1a 1 +c2~ 1 t) ek,t, y (t) = (c1a, +c2~ 2t) ek,t; Fig. 4-7 the possibility of ~~ = ~ 2 = 0 is not t!Xcluded, but then a 1 and a1 will be arbitrary constants. Due to the presence of the factor ek,t rapidly tending to zero as t -+ oo, the product (c1a; +c2~;t) ek,t (i = 1, 2) tends to zero as t----. oo, and for a sufficiently large t all the points of any 6-neighbourhood of the origin enter the given 8-neighbourhood of the origin, and, hence, the rest point is asymptotically stable. Fig. 4.7 depicts this type of rest point, which [like in Case (a) of (I)) is called a stable nodal point. This nodal point occupies an intermediate position 4. THEORY OF STABILITY 219 between the nodal point (a) (1) and the focal point (b) (1), since for an arbitrarily small change in the real coefficients aw a120 a21 , a22 it can turn either into a stable focal point or into a stable nodal point of type (a) (1) because in the case of an arbitrarily small change in the coefficients the multiple root can pass either into a pair of complex conjugate roots or into a pair of real distinct roots. If ~~ = ~2 = 0, we again get a stable nodal point (the so-called proper node) depicted in Fig. 4.8. X Fig. 4-8 Fig. 4-9 (2) If k1 =k2 > 0, then changing t to-t leads to the preceding case. Hence, the trajectories do not differ from the trajectories of the preceding case shown in Figs. 4.7 and 4.8, but motion along them occurs in the opposite direction. In this case the rest point is called an unstable nodal point, like in case (a) (2). Thus all the possibilitit!$ have been exhausted, since the case k1 = 0 (or k2 = 0) is excluded by the condition Note 1. If Ia11 a12 j=F 0. a21 an then la11 au the characteristic equation Iau - k a12 j= 0 au a22-k has a zero root k1 = 0. Assume that k1 = 0, but k2 =F 0. Then the general solution of the system (4.6) will be of tht: form X= clccl +c2~lekzt, !I= clcc2 +c2B~ek•t. 220 I. DIFFERENTIAL EQUATIONS Eliminating t, we get a family of parallel straight lines ~~ (y-c1ct2) = ~2 (x-c1ct1). When C2 = 0 we get a one-parameter family of rest points located on the straight line a1y = a2x. If k2 < 0, then as t --... oo the points on every trajectory approach the rest point x=c1a1 , y=c1ct2 lying on this trajectory (Fig. 4.9). The rest point x ==0, y _ 0 is stable, but there is no asymptotic stability. But if k2 > 0, then the trajectories are arranged in the same way, but points on the trajectories move in the opposite direction, and the rest point x ==0, y = 0 is unstable. But if k1 = k2 = 0, then two cases are possible: 1. The general solution of the system (4.6) is of the form x = c1 , y = c2 - all the points are rest points and all the solutions are stable. 2. The general solution is of the form X=C1 +c2t, y=c7 +c;t, where c7 and c; are linear combinations of arbitrary constants c1 and c2 . The rest point x = 0, y ==0 is unstable. Note 2. The classification of rest points is closely associated with the classification of singular points (see pages 62-64). Indeed, in the case under consideration the system (4.6) where lau ~~~:;60, ~I a2Z may be reduced, by eliminating t, to the equation dy a21x+a22Y (4.12) ax= aux+auy , the integral curves of which coincide with the trajectories of motion of the system (4.6). Here, the rest pointx=O, y=O of the system (4.6) is a singular point of the equation (4.12). It will be noted that if both roots of the characteristic equation have a negative real part [cases (a) (I); (b) (1); (c) (1)], then the rest point is asymptotically stable. But if at least one root of the characteristic equation has a positive real part [cases (a) (2); (a) (3); (b) (2); (c) (2)], then the rest point is unstable. Similar assertions hold true also for a system of n homogeneous linear equations with constant coefficients: (i = 1, 2, ... , n). (4.13) 4. THEORY OF STABILITY 221 If the real parts of all roots of the characteristic equation of the system (4.13) are negative, then the trivial solution x1 ==0 (i = 1, 2, ... , n) is asymptotically stable. Indeed, the particular solutions corresponding to a certain root k5 of the characteristic equation are of the form (pages 201 and 203) (i= 1, 2, ... , n) if k8 are real, x 1=eP•1 (~1 cosqi+y1 sin q5t) if k5 =Ps+qi and, finally, in the case of multiple roots, the solutions are of the same kind, but multiplied by certain polynomials P1(t). It is obvious that all solutions of this kind, if the real parts of the roots are negative (Ps < 0, or if ks is real, then k5 < 0) tend to zero as t-+ oo not more slowly than ce-m1, where c is a constant factor and -m < 0 and greater than the greatest real part of the roots of the characteristic equation. Consequently, given sufficiently large t, the trajectory points whose initial values lie in any 6-neighbourhood of the origin enter an arbitrarily small e-neighbourhood of the origin and, as t-+ oo, approach the origin without bound: the rest point X;== 0 (i = 1, 2, ... , n) is asymptotically stable. But if the real part of at least one root of the characteristic equation is positive, Re k1= p1 > 0, then a solution of the form x1= ca1.ekt1 corresponding to this root, or in the case of a complex k;, its real (or imaginary) part cePit (~1 cos q1t +y1 sin q;t) (j = 1, 2, ... , n), no matter how small the absolute values of c, increases in absolute value without bound as t increases; and, consequently, the points located at the initial moment on these trajectories in an arbitrarily small 6-neighbourhood of the origin leave any specified e-neighbourhood of the origin as t increases. Hence, if the real part of at least one root of the characteristic equation is positive, then the rest point x1 ==0(j=1,2, ... ,n) of the system (4.13) is unstable. Example 1. What typP of rest point does the following system of equations have? dx dt =X-y, dy dt =2x+ 3y. The characteristic equation ll;k 3=~1=0 or 222 I. DIFFERENTIAL EQUATIONS has· the roots k1 , 1 = 2 ± i, hence, the rest point x = 0, y = 0 is an unstable focal point. Example 2. x= -a2x-2bx is the equation of elastic oscillations with account taker. of !:-:dian or resistance of the medium (forb> 0). Going over to an equivalent system of equations, we have x=y, y= -a2x-2by. The characteristic equation is of the form 1 -k • I 2 = 0 or k2 +2bk +a2 = 0, -a -2b-k whence k1 .~= -b±Vb2 -a2 . Consider the following cases: (l) b=O, i.e., the resistance of the medium is ignored. All motions are periodic. The rest point at the origin is a centre. (2) b2 -a2 < 0, b > 0. The rest point is a stable focal point. The oscillations die out. (3) b2 -a2 ~ 0, b > 0. The rest point is a stable nodal point. All solutions are damped and nonoscillating. This case sets in if the resistance of the medium is great (b ~a). (4) b < 0 (the case of negative friction), b2 -a2 < 0. The rest point is an unstable focal point. (5) b < 0, b2 -a2 ~ 0 (the case of large negative friction). The rest point is an uristable nodal point. Example 3. Test for stabil.ity the rest point of the system of equations dx -=2y-zdt ' dy -=3x-2zdt ' :: =5x-4y. The characteristic equation is of the form -k 2 -1 or 3 -k -2 =0 5 -4 -k k8 -9k+8=0. In the general case it is rather difficult to determine the roots of a cubic equation; however, in the given case one root k1 = 1 is readily found, and since this root has a positive real part, we may assert that the rest point x = 0, y = 0, z = 0 is unstable. 4. THEORY OF STABILITY 223 3. Lyapunov's Second Method At the end of last century, the celebrated Russian mathematician Aleksandr Mikhailovich Lyapunov elaborated an extremely general method for investigating the solutions of a system of differential equations for stability: (i=l, 2, ... , n). (4.14) It became known as Lyapunov's second method Theorem 4.1 (Lyapunofi'S stability theorem). If there exists a differentiable function v (x1 , x,, ... , xn), called Lyapunov's function. that satisfies the following conditions in the neighbourhood of the ~oordinate origin: (1) v(x., x,, .. ., xn)~O and v=O only for X;=O (i= 1, 2, .•• • . . , n), i.e., the function v has a strict minimum at the origin; n dv ~ ~ f(2) dt=.t....ax- f;(t. x1, ••• , xn)::::;;;;O or t~t0, I= I 1 then the rest point x;==O (i= I. 2, ... , n) is stable. In condition (2), the derivative ~~ is taken along the integral curve, i.e. it is computed on the assumption that the arguments X; (i = 1, 2, ... , n) of the function v (x1, x,, ... , xn) are replaced by the solution xdt) (i = 1, 2, ... , n) of the system of differential equations (4.14) d n ~ d . d Indeed. on this assumption ij =Lox· ~~ or, replacing ~; by I= I I the right sides of the sys~em (4.14), we finally get n dv ~ ~ dt = .t.... OX· fdt' xl, x,, •••• xn>· 1=.1 I Proof of Lyapunov's stability theorem. In the neighbourhood of the origin, as also in the neighbourhood of any point of a strict minimum (Fig. 4 10), the level surfaces v (X1 , x,, ... , xn) = c of the function v(x1 , x,, ... , xn) are closed surfaces; inside of which is located a minimum point, the coordinate origin. Given e > 0 For a sufficiently small c> 0 the surface of the level v=c lies completely in thee-neighbourhood of the origin,* but does not pass through the coordinate origin; hence, a 6 > 0 may be chosen such that the 6-neighbourhood of the origin completely lies inside the • More preci'lely, at least one closed component of the surface level v = c lies in the £-neighbourhood of the origin. 224 I. DIFFERENTIAL EQUATIONS surface v = c; and in this neighbourhood v 10 the point of the trajectory defined by these initial conditions cannot go beyond the limits of the e-neighbourhood of the origin and even beyond the limits of the z Fig. 4·10 Fig. 4·11 level surface v= c, since, by virtue of condition (2) of the theorem, the function v does not increase along the trajectory and, hence, for t ;;;a. t0 V (x1 (t), X1 (t), ... , Xn (t)) ~ C1 x,, ..., x,.) that satisfies the conditions: (I) v (x1, x., ... , x,.) has a strict minimum at the origin: v(0, 0, ... , 0) = 0; (2) the derivative of the function V1 computed along the integral curves of the system (4.14) n dv=L::.f;(t, x1 , x,, ... ,x,.)~O, dt I= I 1 and outside an arbitrarily small neighbourhood of the origin, that II is, for L xj ~ 6~ > 0, t ~ T0 ~ t0 , the dertvative :; ~-~ <0, I=I where ~ is a constant, then the rest point x1 =0 (i = 1, 2, ••. , nJ of the system (4.14) is asymptotically stable. Proof. Since the conditrons of the stability theorem are fulfilled, it follows that for every e > 0 a 6 (e) > 0 may. be chosen such that 15-378 226 I. DIFFERENTIAL EQUATIONS the trajectory, the initial point of which lies in the 6-neighbourhood of the origin, does not, for t ·~ t0 , leave the e-neighbourhood of the origin. Hence, in particular, condition (2) is fulfilled along such a trajectory for t > T0 , and for this reason, the function v monotonically decreases along the trajectory as t increases, and along the trajectory there is a limit to the function v as t--+ oo: lim v (t, x, (t), X2 (t), ... , Xn (t)) =a~ 0. t .... r:I:J We have to prove that a= 0, since if a= 0, then from condition (I) it follows that lim x;(t)=O (i=l, 2, ... , n), i.e. the rest t-+ r:I:J point X;= 0 (i = I, 2, ... , n} is asymptotically stable. Suppose a > 0; then the trajectory, for t > t0 , lies in the region v ~a, and therefore outside a certain ~.-neighbourhood of the origin; that is, where we have, by condition (2), ~; ~-~ < 0 for t ~ T0 • Multiplying inequality ~~ ~ -~ by dt and integrating along the trajectory from T0 to t, we get or v (X1 (t), X2 (t), ... , xn {t)) -v (X1 (T0}, X2 (T0 ), ••• , Xn (T0 )) ~ ~-~U-To) v(x1 (f), X2 (f), .•. , xn (t)) ~ ~V(X1 (T0), X2 (T0 ), •••• ""n(T0))-~(f-T0). Given a sufficiently large t, the right side is negative, and hence also v(x1 (t), x,(t), ... , xn(t)) <0, which contradicts condition (1). Note. The theorem on asymptotic stability is generalized to the case of the function v dependent on t, x1 , x2 , ••• , xn if the first condition, as in the preceding theort!m, is replaced by the following: where the function w has a strict minimum at the origin and, besides, if we ask that the function v (t, X1 , X1 , ••• , xn) uniformly n approach zero in t as L x~ --+ 0. l=l Theorem 4.3 (Chetaye 0) in which v > 0, and v = 0 on a part 4. THEORY OF STABILITY 227 of the boundary of the region (v > 0) lying in U; (2) in the region (v > 0) the derivative and in the rest stable. n dv ~ av fdt =~ax· i (t, XI' X2, • •• ' Xn) > 0, I= I I the region (v ~ ct), ct > 0, the derivative ~~ ~ ~ > 0, then point x;=O (i= l, 2, ... , n) of the system (4.14) is unProof. We take the initial point X1 (t0 ), X2 (t0 ), ••• , xn (t0 ) in an arbitrarily small neighbourhood of the coordinate origin in the region (v > 0), v (xl Uo), x2 Uo), · · · ••• , Xn {t0 )) = ct > 0 (Fig. 4.13). Since :~ ~ 0 along the trajectory, the function vdoes not diminish along the trajectory and hence as long as the trajectory lies inside the h-neighbourhood of the origin we are interested in (where the conditions of the theorem are fulfilled), the trajectory must lie in the region (v ~ ct). Assume that the trajectory does not leave the h-neighbourhood of the origin. Then, by virtue of condition (2), the Fig. 4-13 derivative :~~ ~ > 0 along the trajE>r.tory for t ~ t0 • Multiplying this inequality by dt and integrating, we get V (X1 (t), .X2 (t), ... , Xn (t))-v (X1 (/0 ), x, (t0 ), ••• , Xn {t0 )) ~ ~ (t -t0 ), whence it follows that as t -. oo the function v increases without bound -along the trajectory; but this runs counter to the assumption that the trajectory does not leave the closed h-neighbourhood of the origin, since the continuous function v is bounded in this h-neigh- bourhood. Note. N. G. Chetayev proved the instability theorem on the assumption that v can also depend on t; then the hypothesis is somewhat modified; in particular, we have to demand that the function v be bounded in the region (v ~ 0) in the h-neighbourhood under consideration of the origin. 15"" Example I. Test for stability the trivial solution of the system: dx 3 dy 3 dt = -y-x • iii =x-y · 228 I. DIFFERENTIAL EQUATIONS The function v (x, y) = x~ +y~ satisfies the conditions of Lyapunov's theorem on asymptotic stability: (I) v(x, y)~O, v(O, 0)=0; (2) ~~ = 2x(- y-x3 ) +2y (x-y3) =-2 (x'+ y') ::;;;;o. Outside the neighbourhood of the origin, ~~::;;;;-~<0. Consequently, the solution x == 0, y =0 is asymptotically stable. Example 2. Test for stability the trivial solution x = 0, y ==0 of the system dx , dy , dt= -xy; ([I=YX. The function v (x, y) = x' + y' sati~fies the conditions of the Lyapunov stability theorem: (I) v(x, y)=x'+y'~O. v(O, 0)=0; (2) ~~ = -4x'y' +4x'y' ==0. Therefore, the trivial solution x == 0, y ==0 is stable. Example 3. Test for stability the rest point x =0, y =0 of the system of equations dx=ya'x' dt I ' ~~ =xa+y'. The function theorem: v = x'-y' satisfies the (1) v>O for lxl>ryl; conditions of Chetayev's (2) : = 4x3 (y3 +x6)-4y3 (x8 +y6) = 4 (x" -y8) > 0 dv for IxI > Iy I; and for v ~a > 0, Tt ~ ~ > 0. Hence, the rest point x = 0, y ==0 is unstable. Example 4. Test for stability the trivial solution xi== =0 (i = I, 2, ... , n) of the system of equations dxi iJu (x1, x,, ... , x,.) (. I 2 ) dt = iJxi ' = , ' ..., n if it is given that the function u (x1, x,, ... , x,.) has a strict maximum at the coordinate origin. For the Lyapunov function we take the difference v (x1, Xz, ••• , x,.) = u (0, 0, .•. , 0)-u lX1, Xa, ••• , X11), 4. THEORY OF STABILITY 229 which obviously vanishes when x1= 0 (i = 1, 2, ... , n), has a strict minimum at the origin, and, hence, satisfies condition (1) of Lyapunov's stability theorem. The derivative along the integral curves n n dv = - ~ au dx; = - ~ ( au )' ~ 0. ~ ~~- ~ ~ ~- -I= I 1 I= I 1 Thus, the conditions of Lyapunov's stability theorem are fulfilled and so the trivial solution is stable. Example 5. Test for stability the trivial solution x1 == = 0 (i = 1, 2, ... , n) of the system of equations II ~~~ = L.a11 (t) x1, where a11 (t) = -a11 (t) for i :::/= j /=I and all au (t) ~ 0. n The trivial solution is stable since the function v = ~ x] satis' =I fies the conditions of Lyapunov's stability theorem: (1) v~ 0 and v(0, 0, . . . , 0) = 0; n n 11 n . (2) :~ = 2Lx1 ~~ = 2I.I.a11 (t) x1xr- 2 I.a1;(t) xl ~ 0. i=l i=li=l <=I 4. Test for Stability Based on First Approximation When testing for stability the rest point x1 = 0 (i = 1, 2, ... , n) of the system of differential equations (i = 1, 2, ... , n), (4.14) where f1 are functions differentiable in the neighbourhood of the coordinate origin, freq~ent use is made of the following method: taking advantage of the differentiability of the functions f1(t, x1 , x,, ... , x,.), we represent the system (4.14) in the neighbourhood of the origin x1= 0 (i = 1, 2, ... , n) in the form n d:: = L.a11 (t)x1+Rt 0 (i.e., if the R; do not depend on t, then their order is greater than first with respect to ~) and all the roots of the characteristic equation a11 -k au aln au a22-k a2n =0 (4.17) anl an2 ... ann-k have negative real parts, then the trivial solutions X;= 0 (i = 1, 2, .... n) of the system of equations (4.15) and the system of equations (4.16) are asymptotically stable; hence in this case a test for stabiliiy based on a first approximation is permissible. Theorem 4.5. If the system of equations (4.15) is stationary to a first approximation, all the functions R; satisfy the conditions of the preceding theorem, and at least one root of the characteristic 4. THEORY OF STABILITY 231 equation (4.17) has a positive real part, then the rest points x;=O(i= 1, 2, ... , n) of the system (4.15) and the system (4.16) are unstable; consequently, this case too permits testing for stability on the basis of a first approximation. Theorems 4.4 and 4.5 fail to embrace only the so-called critical case with respect to restrictions imposed on the roots of the characteristic equation: all real parts of the roots of the characteristic equation are nonpositive, and the real part of at least one root is zero. In the critical case, the nonlinear terms R1 begin to influence the stability of the trivial solution of the system (4.15) and, generally speaking, it is impossible to test for stability on the basis of a first approximation. Theorems 4.4 and 4.5 are proved in Malkin's book (2]. To give the reader an idea of the methods of proof of such theorems, we prove Theorem 4.4 on the assumption that all the roots k; of the characteristic equation are real and distinct k; 0, t ~T. \i=l Relative to the system (4.18), the Lyapunov function that satisfies the conditions of the asymptotic stability theorem is Indeed, (1) v(yt, y, ... , .fln)::::.:::O,' ~ v(0, 0, ... ' 0) = 0; n n n n (2) ~; = 2LYi ~: = 2 L.k,y] + 2 L.k,y,R, ~ Lk,yj ~ 0 i=l 1=1 1=1 1=1 for sufficiently small y1, since all k1 <0, and the double sum II 2 1~ k1y1R1 may, for sufficiently small y1, be made less, in absolute II value, than the sum ~~ k1yf. Finally, outside the neighbourhood of the coordinate origin r • Exampl~~;1. Test for stability the rest point x = 0, y = 0 of the system i·~ :: =x-y+x'+y'sint, } dy I dt=x+y-y. (4.19) 4. THEORY OF STABILITY 233 The nonlinear terms satisfy the conditions of Theorems 4.4 and 4.5. Test for stability the rest point x=O, y=O of the first-approximation system :; =x-y,} dy dt=X+Y· (4.20) The characteristic equation IIlk 1 1k I=0 has the roots k1, 1 = = 1 ± i, and so, by virtue of Theorem 4.5, the rest point of the systems (4.19) and (4.20) is unstable. Example 2. Test for stability the rest point x = 0, y = 0 of the system : =2x+8siny, } it =2-e"-3y-cosy. (4.21) Expanding sin y, e" and cosy by Taylor's formula, we represent the system as dx dy df=2x+By+R1, df=-x-3y+R,, where R1 and R, satisfy the conditions of the Theorems 4.4 and 4.5. The characteristic equr.tion 12_ / _ 3 ~k /= 0 for the firstapproximation system dx dy df = 2x + By, Tt = -x-3y (4.22) has roots with negative real parts. Consequently, the rest point x=O, y=O of the systems (4.21) and (4.22) is asymptotically stable. Example 3. Test for stability the rest point x=O, y=O of the system dx = -4y-x' }dt ' :~ =3x-y•. (4.23) The characteristic equation ~-~ =!I= 0 for the first-approximation system has pure imaginary roots, which is the critical case. 234 I. DIFFEI~ENTIAL EQUATIONS A first-approximation investigation is impossible. In this case it is easy to choose a Lyapunov function v=3x2 +4y2• (1) v(x, y)~O. v(O, 0)=0: (2) ~~ = 6x (-4y-x3 ) +By (3x-y3 ) = -(6x' +By') ~0; note that outside a certain neighbourhood of the origin ~ ~- ~ < 0, hence the rest point x = 0, y = 0 is asymptotically stable by the theorem of the preceding section. Let us examine this example in somewhat more detail. The first-approximation system of equations dx dy Tt= -4y, df=3x (4.24) had the centre at the origin. The nonlinear terms in the system (4.23) converted this centre into a stable focal point. The general case too exhibits a similar but somewhat more complicated geometrical pattern. Let the first-approximation system of the system d:/ = a11x1+a12X2+R1 (xl, x2), } d:r = auxl +a22x2 +R2 (xlt x2) (4.25) have a rest point of the centre type at the origin. As on page 229, assume that the nonlinear terms R1 (x1 , X2) and R2 (x1, x2) are of order higher than first with respect to Vxi+ xi. These nonlinear terms are small, in a sufficiently small neighbourhood of the origin, in comparison with linear terms, but still they somewhat distort the direction field defined by the first-approximation linear system. For this reason, a trajectory emanating from some point (x0 , y0) is slightly displaced (after a circuit of the origin) from the linear-system trajectory passing through the same point, and, generally speaking, does not arrive at the point (x0 , y0). The trajectory is not closed. If after such a circuit of the origin all the trajectories approach the origin, then a stable focal point arises at the origin; but if the trajectories recede from the origin, an unstable focal point develops. An exceptional case is possible in which all the trajectories of the nonlinear system which are located in the neighbourhood of the origin remain closed; however, the most typical case is that in which only certain closed curves (or, possibly, none) remain closed, while the others are converted into spirals. 4. THEORY OF STABILITY 235 Such closed trajectories in the neighbourhood of which all trajectories are spirals are called limit cycles. If the trajectories close to the limit cycle are spirals that approach the limit cycle as t -+oo, then the limit cycle is called stable (Fig. 4.14). If the trajectories close to the limit cycle are spirals receding from the limit cycle as t-oo, then the limit cycle is called unstable. And if the spirals approach the limit cycle from one side as t-oo, and recede from it on the other side (Fig. 4.15), then the limit cycle is called half-stable. !I Fig. 4·14 Fig. 4-15 Thus, transition from the first-approximation system (4.16) to the system (4.25), generally speaking, leads to a transformation of the centre into a focal point surrounded by p (the case p = 0 is not excluded) limit cycles. On pages 160-161, when studying the periodic solutions of the autonomous quasi-linear system x+a2x = J.Lf (x, x, J.L), (4.26) we encountered a similar instance. Indeed, replacing (4.26) by an equivalent system, we get x=y, } y=- a2x+ J.Lf (x, y, J.L). (4.27) The corresponding linear system: • • 2 x=y, y= -ax has a rest point of the centre type at the origin; the addition of small (for small J.L) nonlinear terms converts the centre, generally speaking, into a focal point surrounded by several limit cycles whose radii were found from the equation (2.128), page 162. The only difference between the cases (4.25) and (4.27) consists in the fact that the terms R1 and R2 are small only in a sufficiently 236 I. DIFFERENTIAL EQUATIONS small neighbourhood of the origin, whereas in the ·case (4.27) the term JJ.f (x, y, Jl) can be made small, for a sufficiently small Jl, not only in a sufficiently small neighbourhood of the coordinate origin. In Example 2 (page 162), a limit cycle appears, for small Jl, in the neighbourhood of a circle of radius 6 with centre at the coordinate origin, which circle is the trajectory of the generating equation. In applications, stable limit cycles are usually found to correspond to auto-oscillatory processes, i.e. periodic processes in which small perturbations practically do not alter the amplitude and fre· quency of the oc;cillations. 5. Criteria of Negativity of the Real Parts of All Roots of a Polynomial In the preceding section, the problem of the stability of a trivial solution of a broad class of systems of differential equations was reduced to investigating the signs of the real parts of the roots of the characteristic equation. If the characteristic equation has a high degree, then its solution is complicated; for this reason, very important are methods which permit establishing (without solving the equation) whether all its roots have a negative real part or not. Theorem 4.6 (Hurwltz,s theorem)*. A necessary and suffi· cient condition for the negativity of the real parts of all the roots of the polynomial z"+a,z"-1 + ... +a,_,z+a, with real coefficients i8 the positivity of all the principal diagonals of the minors of the Hurwitz matrix a, as a2 a, a. a7 aft 0 0 0 a, as a, 0 0 1 a, a. 0 0 0 The principal diagonal of the Hurwitz matrix exhibits the coeffi. cients of the polynomial under consideration in the order of their numbers from a, to a,. The columns alternately consist of coeffici· • The proof of the Hurwitz theorem may be found in courses of higher algebra (for example, A. Kurosh Course of Higher Algebra). .f. THEORY OF STABILITY 237 ents with odd only or even only indices, including the coefficient a0 = 1; hence the matrix element b1, = a,1_,. All missing coefficients (that is, coefficients with indices greater than n or less than 0) are replaced by zeros. Denote the principal diagonal minors of the Hurwitz matrix: Ia1 1I at 0 Ill= Ial I, ll, = a. at ' lla= aa a, al ' ... a, a, aa al 0 0 aa a, at •.•• ll,. = a& a. aa 0 0 0 ... a,. Observe that since ll,. = ll,._ 1a,., the last of the Hurwitz conditions ll1 > 0, ll, > 0, ... , ll,. > 0 may be replaced by the demand that a,.> 0*. Let us apply the Hurwitz al theorem to polynomials of second, third, and fourth degree. (a) z1 +a1z +a,. Uz a, Fig. 4-16 Fig. 4-17 The Hurwitz conditions reduce to a1 > 0, a2 > 0. These inequa· lities d~fine the first quadrant in the space of the coefficients a1 and a, (Fig. 4.16). Fig. 4-16 depicts the region of asymptotic :stability of a trivial solution of some system of differential equations • Note that from the Hurwitz conditions it follows that all the a; > 0; however, the positivity of all coefficients is not enough for the real parts of all roots to be negative. 238 l. DIFFERENTIAL EQUATIONS that satisfies the couditions of Theorem 4.1, provided that Z2 -1- a1 z+ +a2 is its characteristic polynomial. (b) z3 +a1z2 +a2 z +a~. The Hurwitz conditions reduce to a1 > 0, a1a2 -a3 > 0, a3 > 0. The region defined by this inequality in the coefficient space is depicted in Fig. 4-17. (c) z' +a1z3 +a2z2 +a3z +a4 • The Hurwitz conditions reduce to a1 > 0, a1a2 -a8 > 0, (a1a2 -a3) a3 -aia4 > 0, a4 > 0. The Hurwitz conditions are very convenient and readily verifiable for the polynomials we have just considered. But the Hurwitz conditions rapidly become complicated as the degree of the polynomial increases, and it is often more convenient to apply other criteria for the negativity of the real parts of the roots of a poly- nomial. Example. For what values of the parameter a is the trivial solution X1 = 0, X2 = 0, x3 = 0 of the system of differential equations dx1 dx2 3 dxs , 2dt=Xs, dt=- xl, dt=axl, x2-Xa asymptotically stable? The characteristic equation is of the form -k 0 1 -3 -k 0 =0 or k3 +k2 -ak+6=0. a 2 -1-k By the Hurwitz criterion, a1 > 0, a1a2 -a3 > 0, a3 > 0 will be the conditions of asymptotic stability. In the given case, these conditions reduce to -a-6 > 0, whence a< -6. 6. The Case of a Small Coefficient of a Higher-Order Derivative The theorem on the continuous dependence of a solution upon a parameter (see pages 58-59) asserts that the solution of the differential equation x(t) = f (t, x (t), !l) is continuously dependent on the parameter fl if in the closed range of t, x and !l under consideration, the function f is continuous with respect to the collection of arguments and satisfies the Lipschitz condition with respect to x: lf(t, x, ~J.)-f(t x, ~J.))~N/x-x), where N does not depend on t, x and fl· 4. THEORY OF STABILITY 239 The conditions of this theorem are ordinarily fulfilled in problems of physics and mechanics, but one case of the discontinuous dependence of the right side on a parameter is comparatively often encountered in applications. This section is devoted to a study of the given case. Consider the equation dx p. dt =I(t, x), (4.28) where p. is a small parameter. The problem is to find out whether for small values of IJ.LI it is possible to ignore the term p. :; , i.e. whether it is possible to replace approximately a solution of the equation p. :;=I(t, x) by a solution of the so-called degenerate equation l(t, x)=O. (4.29) We cannot take advantage here of the theorem on the continuous dependence of a solution on a parameter, since the right side of the equation dx I -=-l(t x) dt J1 ' is discontinuous when p. = 0. For the time being, let us assume for the sake of simplicity that the degenerate equation (4.29) has only one solution x = cp (t); also assume for definiteness that p. > 0. As the parameter p. tends to zero, the derivative ~~ of the solutions of the equation :; =_!_I (t, x) . J1 will, at every point at which f(t, x):;60, increase without bound in absolute value, having a sign that coincides with the sign of the function f (t, x). Consequently, at all points at which f (t, x) =F Q, tangents to the integral curves tend to a direction parallel to the x-axis as p.-+ 0; and if f (t, x) > 0, then the solution x (t, p.) of the equation (4.281) increases with increasing t, since :; > 0, and if f(t, x) < 0, then the solution x(t, p.) diminishes with increasing t, . dx t0 and a sufficiently small ~. we can, approximately, replace the solution x (t, ~) of the equation (4.28) by the solution of the degenerate equation. In the case at hand, the solution x =

0, then the solution x = cp (t) is unstable, since in the first case the function f decreases with increasing x in the neighbourhood of the curve x=cp(t) and, hence, changes sign from+ to-, whereas in the second case it increases with increasing x and, hence, the function f changes sign from - to + when crossing the graph of the solution x = cp (t). If the degenerate equation has several solutions x = cp; (t), then each one of them has to be investigated for stability; depending on the choice of the initial values, the integral curves of the initial equation may behave differently as J.1- 0. For example, in the case depicted in Fig. 4.21 of three solutions x = fP; (t) (i = 1, 2, 3) of the degenerate equation, the graphs of which do not intersect, th~ 16-378 242 I. DIFFERENTIAL EQUATIONS - - - - solutions x = x (t, J.L), J.L > 0, of the original equation defined by the initial points lying above the graph of the function X= CJ12 (t) tend to a stable solution of the degenerate equation x = t0 , while the solutions x = x (t, J.L) defined by the initial points lying below the graph of the function x = /0 (Fig. 4.21). Fig. 4-21 Example t. Find out whether the solution x = x (t, J.L) of the equation J.L ~ = x~ t, J.L > 0 satisfying the initial conditions x (t0 )=X0 tends to the solution of the degenerate equation x- t = 0 as J.L --+ 0 for t > 10 • The solution x=x(t, J.L) does not tend to the solution of the degenerate equation x = t, since the solution of the degenerate equation is unstable because 0 (xi;;t) = l > 0 (Fig. 4.22). Example 2. The same with respect to the equation dx . 2 t 3nXJldf =Stn - c;;··. The solution of the degenerate equation x = 2 In Isin t l-In 3 is b . o(sln2 t-3eX) O . sta le smce ox = -3ex < . Hence, the solut10n of the original equation x = x (t, f.L) tends to the solution of the degenerate equation as J.L --+ 0 for t > t0 • 4. THEORY OF STABILITY 243 Example 3. The same with respect to the solution of the equation ll ~~ =xW-x+ 1), ll > 0, X(t0 )=X0 • Of the two solutions x= 0 and x= t2 +1 of the degenerate equation xW-x+1)=0, the first is unstable, since ax(t 2 axx+l) L=o= = t2 + l > 0 and the second is stable, since ax (t2 ;x+ I) I =X x=t1 +1 .=-t2 -1<0. If the initial point (t0 , x0 ) lies in the upper half-plane x > 0, then the integral curve of the original equation as r..t ...... 0 approaches the graph of the solution x = t2 + I of the degenerate equation (Fig. 4.23) and remains in its neighbourhood. Fig. 4-22 Fig. 4-23 But if the initial point lies in the lower half-plane, x <0, the lim x (t, r..t) = - oo for t > t0 (Fig 4.23). f.l-0 The problem of the dependence of a solution on a small coefficient r..t of the highest derivative also arises in regard to nth-order equations r..tX(n) (t) = f (t, X, X, X, ..• ' x(n-l>), and systems of differential equations. An equation of the nth order may, in the usual manner (see page 91 ), be reduced to a system of first-order equations, and hence the principal problem is to investigate the system of first-order equations with one or several small coefficients of the derivatives. This problem has been studied in detail by A. Tikhonov (4) and A. Vasilieva. Iii'' 244 I. DIFFERENTIAL EQUATIONS 7. Stability Under Constantly Operating Perturbations If the system of equations being investigated dx· d/ =!; (t, xt, x2, ... ' Xn), (4.30) (i=1, 2, ... , n) is subjected t~ small brief perturbations, then, for a small range of t, t 0 ~ t ·!£:;;,}0 , the system (4.30) should be replaced by a perturbed system: d:: =f;(t,_x1'_X=, _:_ .. , Xn~~R;(t, Xt, X2, • .. , Xn), ) (4.31) X; (/0 ) -X; (t0) (t- 1, 2, ... , n), where all _!he R; (t, X1 , X21 ••• , xn) are small in absolute value; when t ~ ro the perturbations cease and we again rever~ to ('!:30), but w~th somewhat altered initial values at the point t"o. X; (fa) = =X;(t0)+1\(i= 1, 2, ... , n), where x;(t) (i= 1, 2, ... , n) is .r the solution under investigation of the system (4.30), and all the fJ; are small in absolute value for small IR; I by virtue of the theorem on the continuous dependence of a solution upon a parameter (Fig. 4.24). Fig. 4·24 Consequent)y, the operation of short-time perturbations ultimately reduces to perturbations of the initial values, and the problem of stability with respect to such short-time or, as they are often called, instantaneous perturbations reduces to the above-considered problem of stability in the sense of Lyapunov. But if perturbations operate constantly, then the system (4.30) must be replaced by the system (4.31) for all t ~ 10 and a completely new problem of stability arises under constantly acting perturbations. This problem has been investigated by I. Malkin and G. Duboshin. As in the investigation of stability in the sense of Lyapunov, it is possible, by a change of variables X;= Y;- fP; (t) (i = 1, 2, ... , n), to transform the solution under investigation Y; = fP; (t) (i = 1, 2, ... , n) of the system~; =Cl>;(t, y1 , y2 , ••• , Yn) (i=l, 2, ... , n) into the trivial solution X;= 0 (i = 1, 2, ... , n) of the transformed 4. THEORY OF STABILITY 245 system. Therefore, from now on it may be assumed that, given constantly operating perturbations, we test for stability the trivial solution x,. ==0 (i = 1, 2, ... , n) of the system of equations (4.30). The trivial solution of the system (4.30) is called stable with respect to constantly operating perturbations if for every e > 0 it is possible to choose <\ > 0 and 62 > 0 such that from the inequa. n n lities ~ Ri < 6: for t ~ t0 and ~ x[0 < 6~ it follows that j'-J i c- I n ~ x l (t) < £2 for t ~ 10 , l=i where xdf) (i= 1, 2, ... , n) is the solution of the system (4.31) defined by the initial conditions x,. (t0) = x,-0 (i = 1, 2, ... , n). Theorem 4.7 (Malkin's theorem). If for the system of equations (4.30) there exists a differentiable Lyapunov function v (t, x1 , X2, • • • x,) which satisfies the following conditions in the neighbourhood of the coordinate origin for t ~ t0 : (1) v(t, x1, X2 , ••• , x,)~wdx1 , X2 , ... , x,)~O, v(t, 0, 0, ... , 0) = 0, where w, is a continuous function that only vanishes at the origin; (2) the derivatives :' (s = 1, 2, ... , n) are bounded in absolute ~~ > 0) and for t ~ t0 , by virtue of conditions (2) l=l and (3), the derivative n n dv iJv ~ iJv ~ iiv Tt=ar+ kA iJx· h+~ iJx· R;~-k <0 {=! I {=! I for sufficiently small absolute values of R; (i = 1, 2, ... , n). Let us specify e >0 and choose some level surface (or one of its components) w1 = l, l >0, lying wholly in the e-neighbourhood of the coordinate origin. By virtue of condition (1 ), the moving (given a variable t ~ /0 ) level surface v (t, x1, x~, ... , x,) = l lies inside the level surface w1 = l, whereas, by virtue of the fact that the function v tends "to zero uniformly in t as ~ xf-+ 0, it lies outside a certain ~2- i=l neighbourhood of the origin in which v< l and, hence, on the level surface v(t, x1, X2, ... , x,) = l, given any t ~ t0 , the derivative dv iJv " iJv " iJv Tt=at+ Lax;{;+I. iJx; R;~-k 0, where c51 is sufficiently small. The trajectory l= I defined by the initial point X; (t0 ) = X;0 (i = 1, 2, ... , n) lying in the above-indicated t\-neighbourhood of the origin cannot, given t ~ /0 , go beyond the e-neighbourhood of the origin, since, by virtue of the choice of c52 , v (t0 , x10 , X20 , ••• , x,0) < l and, hence, if for t ~ t0 the trajectory went beyond the e-neighbourhood of the origin or at least beyond the level surface W1 = l, then it should, at some value t = T, cross the level surface v (t, x11 X2 , ••• , x,) = l for the first time; and the function v should, in the neighbourhood of the point of intersection, increase along the trajectory but this contradicts the condition ~~ ~-k < 0 along the trajectory at points of the level surface v (t, X1 , X2 , ••• , x,) = l. Comparing the conditions of the Malkin theorem with those of the Lyapunov theorem on asymptotic stability (see note on page 226), we see that they nearly coincide; the additional feature of Malkin's theorem is the demand of boundedness of the derivatives iJiJv (s= I, 2, ... , n) so that asymptotic stability and stability Xs with respect to constant perturbatiors are extremely close properties, though they do not coincide. 4. THEORY OF STABILITY 247 Example I. Is the trivial solution x = 0, y = 0 of the system of equations dx 2 a Tt=ay-x, !l1L = -b2x- ..adt y, where a and b are constants, stable with respect to constantly operating perturbations? The Lyapunov function that satisfies all the conditions of the Malkin theorem is v=b'x2 +a2y2 • Thus the rest point x = 0, y = 0 is stable with respect to constantly acting perturbations. Example 2. Do we have a stable rest point X;== 0 (i = 1, 2, ... , n) of the system (i = 1, 2, ... , n) (4.32) with respect to constantly operating perturbations if all the ail are constant and the R; satisfy the conditions of the Lyapunov theorem, page 230, that is IR; I~N (~/~}~-+a., a> 0, N is constant, and all the roots of the characteristic equation of the first-approximation system are distinct and negative? On page 231, after a change of variables that reduced the linear parts of equation (4.32) to the canonical form, a Lyapunov function n v= ~ yf was indicated that satisfied all the conditions of Malkin's 1=1 theorem; hence, the rest point X;= 0 (i = 1, 2, ... , n) is stable with respect to constantly operating perturbations. The same result may be obtained on the assumption that the real parts of all the roots of the characteristic equation (multiple ones may also occur among them) are negative; only in this case the choice of the Lyapunov function is substantially more involved. PROBLEMS ON CHAPTER 4 1. Test for stability the rest point x = 0, y = 0 of the system dx ..o Tt= -2x-3y+A;, dy 3 Tt=x+y-y. 248 I. DIFFERENTIAL EQUATIONS 2. Test for stability the rest point x=O, y=O, z=O of the system dx dy dz di"=x-y-z, di"=x+y-3z Tt=x-5y-3z. 3. For what values of a is the rest point x = 0, y = 0, z = 0 of dx dy dz the system Tt=ax-y, Tt=ay-z, Tt=a.z-x stable? 4. For what values of a does the system dx .... Tt=y+ax-x., :~ = -x-!1have a stable rest point x = 0, y = 0? 5. To what limit does the solution of the differential equation 1.1. :; =(x2 +t1 -4)(x1 +t2 -9), x(l)= I tend a$ 1.1. - 0, for 1.1. > 0, t > 1? 6. To what limit does the solution of the differential equation d I 1.1. d; =x-t+5. x(2)=5 tend as 1.1.-0, for J.L>O. t>2? 7. Test for stability the rest point x = 0, y = 0 of the system of equations dx Tt=x+eY-cosy, dy . Tt= x-y-smy. 8. Is the solution x = 0, y = 0 of the following system of equations stable with respect to constantly operating perturbations: dx .... Tt=-2y-x.·, dy !I-=5x- ·?dt 9. Is the solution x ==0 of the equation stable? 10. Is the solution x;;;;;; 0 of the equation x'+ 5x+6x+x =0 stable? 4. THEORY OF STABILIIY 249 11. What type of rest point x = 0, y = 0 does the system of equa- tions have? dy =5x-y dt 12. Determine the periodic solution of the equation x+ 2X + 2x =sin t and test it for stability. 13. x+2x+ 5x+ 3x= cost. Is the periodic solution of this equation stable? 14. Test for stability the rest point x == 0, y == 0 of the system x=y3 +x•. y=x3 +!/. 15. Test for stability the solutions of the system of equations x=3y-2x+e1, y=5x-4y+2. 16. Test for stability the trivial solution of the equation x·+ 2x+ 3x+ 7 sinhx=O. 17. Test for stability the trivial solution of the equation x+(ex-1) x+ (4 -ex1) x=O, where ex is a parameter. 18. Is the solution x== 0, y ==0 of the system x=3y-x3, y= -4x-3y1 stable for constantly operating perturbations? 19. Is the trivial solution stable of the system X(t) =AX (t), where X (t) is a vector in three-dimensional space, and (1 2 0\ A=\0 1 1)? \1 3 1 20. Test the solutions of the equation x+ 4x+5x=t for stability. 250 I. DIFFERENTIAL EQUATIONS 21. Test the solutions of the equation x+9x= sin t for stability. 22. x +x =cost. Find the periodic solution and test it for sta- bility. 23. Find the region of stability of x+ai+(l-a)x=O. 24. x+x+a2x+5ax=0. Find the region of stability. CHAPTER 5 First-order partial differential equations I. Fundamentals As was pointed out in the introduction (page 13), partial differential equations are differential equations in which the unknown functions are functions of more than one independent variable. Very many physical phenomena are described by partial differential equations. The equation ( iJu ) 2 ( iJu ) 2 ( iJu ) z Tx + ay + \Tz = n (x, y, z) describes the propagation of light rays in an inhomogeneous medium with refractive index n (x, y, z); the equation iJu 2 iJ2u Tt=a ax~ describes the temperature variation of a rod; the equation iJ2u 2 iJ2u iJtz =a iJx2 is the equation of the vibration of a string; the Laplace equation iJ2u iJ2u iJ2u iJx2 + iJy2 + iJz2 = 0 is satisfied by a field potential in regions devoid of charges, and so forth. In this chapter we will deal briefly only with methods of integrating first-order partial differential equations, the theory of which is closely associated with the integration of certain systems of ordinary equations. Higher-order partial differential equations, which are integrated by quite different methods, are taken up in another book of this series. We consider a few elementary examples. Example 1. iJz (x, y) iJx =y+x. 252 I. DIFFERENTIAL EQUATIONS Integrating with respect to x, we get x2 z(x, y)=xy+ 2 + (c11 c,) = 0 between the parameters c1 and c2 • Eliminating the parameters c1 and c, from the system '1'1 (x, y, z) = C1 , 'iJ2 (x, y, z) = c2 , (c1 , c.) = 0, we get the desired equation of vector surfaces: is an arbitrary function. We have thus found the integral of the quasilinear equation (5.3) that depends on an arbitrary function. If it is required to find not an arbitrary vector surface of a field F = P (x, y, z) i +Q (x, y, z) j +R (x, y, z) k, but a surface passing through a given line defined by the equations <1>1 (x, y, z) = 0 and in (5.6) will no longer be arbitrary but will be determined by eliminating the variables x, y, z from the system of equations <1>1 (x, y, z) = 0, <1>2 (x, y, z) = 0. '1'1 (x, y, z) = C1 , '1'2 (x, y, z) = C2, which must simultaneously be satisfied at the points of the given line 2 = 0, through which we draw the characteristics defined by the equations 'iJ1 (x, y, z) = C1 , '1'2 (x, y, z) = C2 • Note that the problem becomes indeterminate if the given line <1>1 (x, y, z) = 0, 2 (x, y, z) = 0, '1'1 (x, y, z) = C1, '~'~ (x, y, z) = c!, as a result of which we obtain the equation cD (c1 , C2) = 0, and the desired integral will be cD ('lll1 (x, y, z), '1'2 (x, y, z)) = 0. Example I. Determine the integral, dependent on an arbitrary function, of the equation iJz +iJz = 1 ax iJy . Form an auxiliary system of equations: dx=dy=dz. Its first integrals will be of the form x-y=c11 z-x=c2 . The integral of the original equation cD (x- y, z-x) = 0, where (\jl1 , \jl1, ... , 'IJ',._1)=c, where is an arbitrary function, is a first integral of the system (5.9), since all the functions 'IJ'1, 'i'u ... , 'i'n-l are turned into constants along the integral curve of the system (5.9); consequently, ('IJ'1, \j)1, ••. . . . , \jl,._1) too becomes a constant along the integral curve of the system (5.9). Hence, z = ('IJ'1, 'IJ'1, ••• , 'IJ'11 _ 1), where (}) is an arbitrary differentiable function, is a solution of the homogeneous linear equation (5.8). We will prove that z = ('1'1 (x1, ••• , x,.), '1'1 (x1, • o o , x,.), 'IJ',._1 (x1, ••• , x,.)) Is the general solution of the equation (5.8). Theorem 5.2. z= ('IJ'1, \jl1, ••• , 1j1,,_1), where ls an arbitrary function, is the general solution of the equation (5.8) that is, a solution containing all the solutions of the equation without exception. 17* 260 I. DIFFERENTIAL EQUATIONS Proof. Assume that z = 'IJl (x1, x.. . .. , X11) is some solution of tht: equation (5.8) and prove that there is a function <1> such that 'IJl = , iJ"IJ's ax. ax2 ••• axn a¢n-1UI!Jn-1 UI!Jn.-1 iJxl ax, . . • axn Is identically zero in the interval under consideration. However, the fact that the Jacobian of the functions 'IJl, '1Jl1, '1'1••.• , '1Jl,_1 vanishes identically shows that there is a functional relation bet· ween these functions: F('iJ', lj71, '1Jl1, ... , 'tJ111_1)=0. (5.13) By virtue of the independence of the first integral~ '1Jl1(x1, X2 , ••• ... , x,)=c1 (i=l, 2, ... , n-l)ofthesystem(5.9), at least one of the minors of the (n- 1) th order of the Jacobian D (¢, 11'1• "IJ'z, · · · • 'l>n-1) D (x1, x1, x1, ••• , X11) 5. FIRST·ORDER PARTIAL DIFFERENTIAL EQUATIONS 261 of the form D ('i't• 'i'z• ··· • 'i'n) D (xa, • xa.• . . . • Xan-t) is nonzero. Hence, equation (5.13) may be given in the form 'iJ = (11'1• 'I'a• • • • • 'l'n-1). Example 5. Integrate the equation n ~ iJz ,1,... X; iJx· = 0. I= l I (5.14) The system of equations defining the characteristics is of the form dx1 _ dx2 _ _ dxn x;- - 'X;- - · · ' - Xn • The independent first integrals of this system are Xt Xs Xn-l -=c., -=c,, ... '--=Cn-t· Xn Xn Xn The general solution of the original equation = ~ (Xl X a Xn-1)z"", , ... ,Xn Xn Xn is an arbitrary homogeneous function of degree zero. Euler's theorem on homogeneous functions asserts that homogeneous functions of degree zero satisfy the given equation (5.14); we have now proved that only homogeneous functions of degree zero possess this property. A nonhomogeneous linear equation of the first order n LX;(X1 , X1 , ••• , Xn, z) :z = Z (X1 , X2 , ••• , xn, z), (5.15) I= I X{ where all the Xi and Z are continuously differentiable functions that do not vanish simultaneously in the given range of the variables x1, x1 , ••• , Xn, z, is integrated by reducing it to a homogeneous linear equation. For this purpose, as in the case of three variables, it suffices to seek the solution z of the equation (5.15) in implicit form: iJu where iJz =1=0. u (x1• X1 , ••• , Xn, z) = 0, (5.16). Indeed, assuming that the function z = z (x1, x,, ... , xn) is determined from the equation (5.16), and differentiating the identity u (x1 , x,, ... , x", z (x1, x1, ••• , xn)) = 0 262 I. DIFFERENTIAL EQUATIONS with respect to X;, we get ~+au~=O ax, oz ox; ' whence au iJz iJx1 ax,= -a;;. oz S b t·t !' oz . t (5 15) It' I . b au d t .u s 1 uLtng ax-: m o . , mu 1p ymg y - az an ran~posmg I all terms to the left-hand side of the equation, we get the homogeneous linear equation n ~ au . au ..t,.. X1 (x1, x,, ... , x,., z) ax1 +Z (x.. x,, ... , x,., z) az =0, (5.17) i -) which the function u must satisfy; however, this only on the assumption that z is a function of x1 , X2 , ••• , x,. defined by the equation u (x1 , x,, ... , x,., z) = 0. Thus, we have to find the functions u that reduce the homogeneous linear equation (5.17) to an identity by virtue of the equation u (x10 x2 , ••• , x,., z) = 0. First find the functions u that reduce the equation (5.17) to an identity, given independently varying x1, X2, ••• , x,., z. All such functions !" are solutions of the homogeneous equation (5.17) and may be found by a method we already know: we form a system of equations that defines the characteristics dx1 · dx2 X1 (x1 , x2 , ••• , x,., z) = X1 (x1, x2 , .••• x,., z)- · · · dx,. dz · · • =X,. (x1, x1, .•• , x,., z) = Z (x1, x2, ••• , x,., z) ; we find n independent first integrals of this system: '11'1 (x1, x,, ... , x,., z) = c1, '11'1 (x1, x,, •.• , x,., z) = c2 , 'II',. (x1, x,, •.. , x,., z) = c,.; then the general solution of (5.17) is of the form U = ('1'1, 'i'z• · •· • 'i'n), where <11 is an arbitrary function. (5.18) 5. FIRST·ORDER PARTIAL DIFFERENTIAL EQUATIONS 263 The solution z of equation (5.15), which depends on an arbitrary function, is determined from the equation U (X1, X2 , ••• , Xn, Z) = 0 or ("IJ>I> "IJ>2, ••• , "IJ>n) = 0. However, besides the solutions found by this method there may be solutions z which are determined from the equations u (x1, x2 , ••• . . . , xn, z) = 0, where the function u is not a solution of the equation (5.17), but reduces this equation to an identity only by virtue of the equation u (x1, x2 , ••• , xn, z) = 0. Such solutions are called special. In a certain sense, there are not very many special solutions; they cannot even form one-parameter families. Indeed, if the special solutions formed a one-parameter family and were defined by the equation u(x1• X2 , ••• , Xn, z)=c, (5.19) where c is a parameter, c0 ~ c ~ c1 , then the equation (5.17) should reduce to an identity by virtue of the equation (5.19) for any c. But since (5.17) does not contain c, it cannot reduce to an identity by virtue of (5.19), which contains c, and, hence, must be an identity with respec.t to all the variables x1 , x2, ••• , xn, z which vary independently. · The last statement admits of a simple geometrical interpretation. When we say that the equation (5.17) reduces to an identity by virtue cf the equation u (x1 , X2 , ••• , xn, z) = 0, we assert that (5.17) reduces to an identity at points of the surface u = 0, but cannot reduce to an identity at other points of the space x1, x2, ••• , xn, z But if equation (5.17), which does not contain c, reduces to an identity by virtue of the equation u = c, where c is a continuously varying parameter, then this means that the equation (5.17) reduces to an identity on all surfaces u = c, c0 ~ c ~ c, that do not intersect and that fill a certain part D of the space x" x2 , ••• , xn, z; and, hence, (5.17) reduces to an identity in the domain D for independently varying X1 , x,, ... , xn, z. In concrete problems it is ordinarily required to find the solution of the equation (5.15) that satisfies some other initial conditions as well, and since there are comparatively few special solutions in the above-indicated meaning, only in quite exceptional cases will they satisfy the initial conditions; for this reason it is only in rare cases that they need to be allowed for. Example 6. Integrate the equation n ~ iJz ~X; iJx = pz, 1=1 I (5.20) where p is a constant. 264 I. DIFFERENTIAL EQUATIONS The system of equatwns dx1 dx2 dxn dz x1 ="X;= · · · = x;; = pz has the following independent integrals: x•=cX l• n Hence, solution z of the original equation is determined from the equation (!> (X1 Xs Xn-1 .!._) _O Xn • Xn , • • • ' Xn ' X g - ' whence _ p ,., (XI X a X n-l)Z- Xn 't' -, - , ••• , - • Xn Xn Xn And so the solution is an arbitrary homogeneous function of degree p. It may be proved that the equation (5.20) does not have special integrals and, hence, Euler's theorem on homogeneous functiolls is invertible, that is, only homogeneous functions of degree p satisfy the equation (5.20). The concept of a characteristic can be extended to systems of quasilinear equations of the following special type: au au .. P (x, y, u, v) ax+ Q (x, y, u, v) ay = R1 (x, y, u, v), ~ ~ (D) P (x, y, u, v) ax+ Q (x, y, u, v) 011 = R, (x, y, u, v). The characteristics of this system are the vector Jines of a vector field in the four-dimensional space F = P(x, y, u, v) i +Q(x, y, u, v) j +R1(x, y, u, v) k1 +R,(x, y, u,v) k,, where i, j, k,, k2 are unit vectors along the respective axes (x, y u and v). The characterisctics are defined by the system of equations dx _ dy _ du dv (E) P (x, y, u, v) - Q(x, y, u, v)- R1 (x, y, u, v) R1 (x, y, u, u) · The system of equations (D) is written as follows in vector notation: (F · N1) = 0 and (F · N1) = 0, where N1 and N, are vectors with the coordinates ( ~~, :; , - I,0) and (:, :, 0, -1) and directed normally to the desired threedimensional cylindrical surfaces u = u (x, y) and v= v(x, y), res- pectively. 5. FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 265 Hence, from the geometrical point of view. integration of the system (D) reduces to finding two three-dimensional cylindrical surfaces u = u (x, y) and v = v (x, y), the normals to which are orthogonal to the vector lines at the points of intersection of the surfaces. It is obvious that this condition will be fulfilled if the twodimensional surface S, on which, generally speaking, the threedimensional cylindrical surfaces u = u (x, y) and v = v (x, y) intersect, consists of vector lines, since these vector lineswilllie simultaneously on the surfaces u = u (x, y) and v = v (x, y) and, consequently, will be orthogonal to the vectors N1 and N2• If we take any two- independent of u and v-first integrals (p, q) = 0 between the numbers p and q which define the direction of the normal N(p, q, -1) to the desired integral surfaces z = z (x,y) of equation (5.28). Thus, the direction of the normal to the desired integral surfaces at a certain point (x, y, z) is not defined exactly; there is only isolated a one-parameter family of pos sible directions of normals-a .zo certain cone of admissible directions of the normals ~------------~-y Fig. 5-3 N (p, q, -I), where p and q satisfy the equation qJ (p, q) = 0 (Fig. 5.3). Thus, the problem of integrating the equation (5.28) reduces to finding the surfaces z=z(x, y), the normals to which would, at every point. be directed along one of the permissible directions of the cone of normals at that point. Proceeding from this geometric interpretation, we shall indicate a method of finding the integral of the equation (5.28) which depends on an arbitrary function, if its integral (x, y, z, a, b) =0 that depends on two parameters a and b is known. The integral {x, y, z, a, b)= 0 of (5.28), which depends on two essential arbitrary constants a and b, is called the complete integral. 272 I. DIFFERENTIAL EQUATIONS Since the original differential equation (5.28) imposes restrictions solely on the direction of the normals to the desired integral surfaces, every surface, the normals to which coincide with the normals to the integral surfaces at the same points, will be an integral surface. Consequently, the envelopes of a two-parameter or one-parameter family of integral surfaces will be integral surfaces, since the normal to the envelope coincides with the normal to one of the integral surfaces of the family passing through the same point. The envelope of a two-parameter family of integral surfaces, on the assumption of the existence of bounded partial derivatives : , : , 0 0~ not vanishing simultaneously, and the existence of the derivatives a~ and : , is defined by the equations acD acD (x, y, z, a, b)= 0, in arbitrary fashion, a one-parameter family (to do this, we take b as an arbitrary differentiable function of the parameter a); and in finding the envelope of the one-parameter family (x, y, z, a, b (a))= 0, we also obtain an integral surface. Assuming that bounded derivatives of the function exist with respect to all arguments and that the derivatives : , : , : do not vanish simultaneously, the envelope of this one-parameter family is given by the equation acD (x, y, z, a, b (a))= 0 and aa { (x, y, ? a, b (a)}= 0 or (x, y, z, a, b(a))= 0 and : +: b' (a)= 0. (5.30) These two equations define a set of integral surfaces that depends on the choice of an arbitrary function b= b (a). Of course, the presence in equation (5 30) of an arbitrary function does not permit us to assert that the equations (5.30) define the set of all integral surfaces of the original equation (5.28) without exception. For example, this set. generally speaking, does not contain the integral surface defined by the equations (5.29), but still the presence of an arbitrary function in (5.30) is usually sufficient for one to isolate an integral surface that satisfies the given initial Cauchy conditions (see page 252). Thus, knowing the complete integral, it is possible to construct an integral that depends on an arbitrary function. 5 FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 273 In many cases, it is not at all difficult to find the complete integral, for instance: (I) If equation (5.28) is of the form F (p, q) = 0 or p = q> (q), then by putting q =a, where a is an arbitrary constant, we get p= q> (a), dz = pdx +qdy= q>(a)dx+ady, whence Z=q>(a)x+ay+b is the complete integral. (2) If equation (5.28) can be reduced to the form q>1 (x, p) = = q>z (y, q), then, putting q>1 (x, p) = q>1 (y, q) =a, where a is an arbitrary constant, and solving (if this is possible) for p and q, we get p = 'IJ>1 (x, a), q = 'IJ>, (y, a), dz=pdx+qdy='¢dx. a)dx+'IJ>,(y, a)dy, Z= ~ '¢1 (X, a)dx+ ~ 'IJ>,(y, a)dy+b is the complete integral. (3) If the equation (5.28) is of the form F (z, p, q) = 0, then by putting z = z (u), where u =ax+ y, we obtain F ( z, a, ~ , :~ ) = 0. Integrating this ordinary equation, we get z = (u, a, b), where b is an arbitrary constant, or z=(ax+y. a, b) is the complete integral. (4) If the equation (5.28) is of a form resembling the Clairaut equation: z = px +qy +q> (p, q), then, as may be readily verified by direct substitution, the complete integral is Example 1. Find the complete integral of the equation p = 3q1 • q=a, p= 3a8 , dz=3a8 dx+ady, z= 3a8x+ay+b. Example 2. Find the complete integral of the equation pq = 2xy. 18--.17/l p 2y -=-=aX q ' 2y p=ax, q=-a· dz =axdx+ ~dy, 274 I. DIFFERENTIAL EQUATIONS ::..:....:._______ ------------------ Example 3. Find the complete integral of the equation z3 = pq'. dz dz z= z(u), where u=ax+ y, p=adii, q= du , 1 z3 =a (::r or :: =a.z, where a. =a-. lnlzl=a1u+lnb, z=iJet'•", a, ( :a +11) Z=be 1 • Example 4. Find the complete integral of the equation z=px+qy+p'+q'. The complete integral is z =ax+by+ a'+ b'. In more complicated cases, the complete integral of the equation F (x, y, z, p, q) = 0 is found by one of the general methods. The simplest idea is that underlying the me~hod of Lagrange and Charpit. In this method, an equation U (x, y, z, p, q) =a (5.31) is chosen for the equation F(x, y, z, p, q)=O 0>.28) so that the functions p = p (x, y, z, a) and q = q (x, y, z, a), which are determined from the system of equations (5.28) and (5.31 ), should lead to the Pfaffian equation that is integrable by a single relation az = p (x, y, z, a) dx +q(x, y, z, a) dy. (5.32) Then the integral of the Pfaffian equation (x, y, z, a, b)= 0, where b is an arbitrary constant appearing during integration of (5.32), will be the complete integral of the equation (5.28). The function U is determined from the integrability condition of the equation (5.32) by single relation: (f·rotf)=O, where F=p(x, y, z, a)i+q(x, y, z, a)j-k, that is, in expanded form, from the equation oq op op iJq P(ji-q iJz- iJy + iJx =0. (5.33) 5. FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 275 The derivatives ~!, Z,: ,~: are computed by differentiating the identities F(x, y, z, p, q) = 0, } (5.34) U (x, y, z, p, q)=a, in which p and q are regarded as functions of x, y and z, which are defined by tht.- system (5.34). Differentiating with respect to x, we get whence iJF +oF iJp +iJF ~ = O ax op ax oq ax • au+ au op +au~ =O OX iJp ox iJq OX • D (F, U) iJq D (p, x) iJx =- D (F, U) • D (p, q) Similarly, defferentiating (5.34) with respect to y iJp iJy, we have Op iJy = D (F, U) D (y, q) D(F,U)' D (p, q) and determining Differentiating (5.34) with respect to z and solving for ~: , ~, we will have D (F, U) iJp D (z, q) Tz = D(F, U) ' D (p, q) D (F, U) iJq D (p, z) oz= D(F,U)' D (p. q) Substituting the computed derivatives into the integrability condition (5.33) and multiplying by the determinant ~r· Ul. whtch p, q) we assume to be different from zero, we get ( aF au aF au) (oF au dF au) p iJz ap - ap az +q iJz iJq - aq az + + fiiF au_ aF iJU) +(iJF iiU _ iiF iiU) -O \au oq iJq dy iJx ap iJp iJx - Ill' 276 I. DIFFERENTIAL EQUATIONS ---------------- or oF au oF au ( oF oF) au ap ax +aq ay + P ap +qaq Tz_ (aF oF) au_ (oF oF) au= 0 ax +p az iJp iJy +qiJz iJq . (5 35) To determine the function U we obtained a homogeneous linear equation (5.35), which may be integrated by the method indicated in Sec. 2 of this chapter: an equation of the characteristics is formed: dx dy dz dp dq (5 36) iJF = iJF = JF iJF iJF iJF = iJF iJF ' . ap aq Pa,;+qaq ax+Paz- ay+q CiZ then at least one first integral of the system (5.36) is found, U1 (x, y, z, p, q)=a, and if the functions F and U1 are independent of p and q, i.e. ~(~.uq1i =1= 0, then the first integral U1 (x, y, z, p, q) will be the desired solution of the equation (5.35). Thus, by determining p = p (x, y, z, a) and q = q (x, y, z, a) from the system of equations and substituting into F (x, y, z, p, q) = 0, U1 (x, y, z, p, q)=a dz=p(x, y, z, a)dx+q(x, y, z, a)dy, we get the Pfaffian equation integrable by a single relation, which, when solved, yields the complete integral of the original equation: <1> (x, y, z, a, b)= 0. Example 5. Find the complete integral of the equation yzp%-q=O. (5.37) The system (5.36) is of the form dx dz dp dp 2pyz= -dy=2p2yz-q = -yp•= -,P'+yp'q' Taking advantage of the original equation, we simplify the denominator of the third ratio and obtain the integrable combination _..!!!_ =- dp whence p'yz ply ' a p=-z . (5.38) 5. FIRST·ORDER PARTIAL DIFFERENTIAL EQUATIO!\:S 277 a u2y From equations (5.37) and (5.38) we find p = z, q = -z-, whence dz = .!!_ dx +a 2 y dy. Multiplying by 2z and integrating, we find the z z complete integral of the original equation z' = 2ax +a2.t/ +b. Knowing the complete integral a> (x, y, z, a, b)= 0 of the equation F(x, y, z, p, q)=O, it is, generally speaking, possible to solve the basic initial problem (see p. 252) or even the more general problem of determining the integral surface that passes through a given curve, X= X (t), y = y (t), Z = Z (t). (5.39) Define the function b = b (a) so that the envelope of the one-parameter family (x, y, z, a, b (a))= 0, defined by the equations (5.40) and iJ (x (t), y (t), z (t), a, b (a))= 0 (5.42) and iJ= 0, and hence also to the desired envelope at the appropriate points. The condition (5.44) is geometrically obvious, since the desireJ surface must 278 I. DIFFERENTIAL EQUATIONS pass through the given curve and, consequently, the tangent to this curve must lie. in the plane tangential to the desired surface. Example 6. Find the integral surface of the equation z = px+ +qy + P: that passes through the curve y =0, z =x'. The complete integral of this equation (see case (4) on page 273) . ab is of the form z =ax +by+ 4 . The equation of the given curve may be written in parametric form x = t, y = 0, z = /2 • To determine the function b = b (a), we form a system of equations (5.42) and (5.44), which in the given case are of ab a2 the form t2 =at+ 4 and 2t=a, whence b=-a, z=a(x-y)-4 . The envelope of this family is determined by the equations a'z=a(x-y)- 4 and a x-y-2 =0. Eliminating a, we get Z=(x-y)'. If the system (5.36) (page 276) is easy to integrate, then the method of characteristics (Cauchy's method-see below) is very convenient for solving the generalized Cauchy problem that has been posed. The integral surface z = z (x, y) of the equation F (x, y, z, p, q) = 0 that passes through the given curve Xo = Xo (s), Yo= Yo (s), Zo = Zn (s) may [as in the case of the quasilinear equation {see page 256)] be pictured as consisting of points lying on a certain one-parameter family of curves x=x(t, s), y=y(t, s), z=z(t, s), where s is the parameter of the .family, called characteristics. First we find the family of characteristics that depends on several parameters, and then, drawing the characteristics through points of the curve X0 =X0 (s), Yo= Yo (s), Z0 = Z0 (s) and satisfying certain other conditions as well, we isolate the one- 5. FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 279 parameter family of curves in which s may be taken as the para· meter: X=X(t, s), y=y (t, s), z=z(t, s) (Fig. 5.4). The set of points lying on these curves is what forms the desired integral surface. That in brief is the underlying idea of Cauchy's method. Let z= z (x, y) be the in· tegral surface of the equation F(x, y, z, p, q) = 0. (5.45) Then, by differentiating the identity (5.45) with respect to x and with respect to y, we get ap aq Fx+ pFz+FPox +Fqax =0, ap aq Fy+qFz+ FPiJy +Fq.ay=O, . iJq iJp or, smce ax = ay , we have Fig. 5_4 F F iJp iJp }x+ zP +FPax+Fq ay =0, F aq aq (5.46) Fv+ zq+FPax+Fq ay =0. The equations of the characteristics for the system of equations (5.46), which is quasilinear in p ami q, and z is considered a known function of x and y, are of the form (see p. 264) ~= dy =- dp =- dq =dt Fp Fq Fx+PFz F1 +qFz . Since z is connected with p and q by the equation dz=pdx+qdy, it follows that, along the characteristic, dz dx dy Tt=P Tt+qTt=pFP+qFq or (5.47) (5.48) dz =dt (5 49) pF"+qFq ' . which enables us to supplement the system (5.47) w!th yet another equation (5.49). 280 I. DIFFERENTIAL EQUATIONS Thus, assuming that z = z (x, y) is the solution of equation (5.45) we arrive at the system dx =~= dz =- dp =- dq =dt (5.50) Fp Fq pFp+QFq Fx+PF, F1 +qFz . From (5.!10) it is possible, without knowing the solution z = z (x, y) of the equation (5.45), to find the functions x = x (t), y = y (t), z = z (t), p = p (f), q = q (t); that is, we can find the curves X=X(t), y= y (t), Z= Z(t), called characteristics, and in each point of a characteristic we can find the numbers p = p (t) and q = q (t) that determine the direction of the plane Z-z=p(X-x)+q(Y -y). (5.51) The characteristic, together with the plane (5.51) referred to each of its points, is called a characteristic strip. We shall show that it is possible, from characteristics, to form the desired integral surface of the equation F (x, y, z, p, q) = 0. First of all note that the function F retains a constant value along the integral curve of the system (5.50): F(x, y, z, p, q)=c, in other words, the function F (x, y, z, p, q) is a first integral of the system (5.50). lndt'ed, along the integral curve of the system (5.50). d ~ ~ ~ ~ ~ TtF(x, y, z, p, q)=Fxdf+F1 Tt+F,df+FPdt+Fq([[= =FxFp+ FyFq+ F~ (pFP+ qFq)-Fp(Fx+PF,)-Fq (Fy+qF,)=O; consequently, along the integral curve of the system (5.50), F (x, y, z, p, q) = c, where c= F (X0 , Yo• Z0 , Po• Q0 ). ln order that the equation F (x, y, z, p, q) = 0 should be satisfied along the integral curves of the system (5.50), the initial values X0 (s), Yo (s), Z0 (s), Po (s), q0 (s) must be chosen so that they will satisfy the equation F (Xo, Yo• Zo, Po• Qo) = 0. Integrating the system (5.50) for initial values X0 = X0 (s), Yo= =Yo (s), Z0 = Z0 (s), Po= Po (s), Q0 = q0 (s) that satisfy the equation F(x0 , y0 , Z0 , p•• q0)=0, we get x=x(t, s), y=y(t, s), z=z(t, s), p=p(t, s), q=q(t, s). For a fixed s we will have one of the characteristics x=x(t, s}, y=y(t, s), z=z(t, s); 5. FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 281 varying s, we get a certain surface, in each point of which, for p = p (t. s), q = q (t, s), the equation F (x, y, z, p, q) = 0 is satisfied, but it is also necessary to find out whether, in the process, p = ~; and q= ~: or, what is the same thing, whether dz = pdx+ +qdy or dz = p ( ~= ds + ~: dt) +q ( ~; ds + ~~ dt) = ~: ds+~: dt, , which is equivalent to the two conditions (5.52) (5.53) The latter of these equations obviously reduces to an identity, since we have already required, when forming the system (5.50), that dz=pdx+qdy along the characteristic. Incidentally, this is quite evident from direct inspection, if one takes into account that by virtue of the system (5.50). ax ay az ar=Fp, ar=Fq, at=pFP+qFq ( . . f 0% oy oz dx dy dz m (5.50) m place o at, at, Tt we wrote dt, (if, (if, since we considered s fixed). · In order that the equation (5.52) may be satisfied, it is necessary to impose certain other restrictions on the choice of the initial values X0 (s), Yo (s), Z0 (s), Po (s), q0 (s). Indeed, put P _ax +q oy -~=U (5.54) OS OS OS and prove that U ==0 if the initial value U It=o = 0, whence it will follow that if the initial functions X0 (s), Yo (s). Z0 (s), Po (s), q0 (s) are so chosen that Po (s) x~ (s) +q0 (S)·Y~ (s)- z~ (s) = 0, then U === 0 for all t. Differentiating (5.54) with respect to t, we get au op ax cJ2x oq oy a2y if'z at = at7fS = p ot i)s +Tt'" ds +q iJt os - ot as 282 I. DIFFERENTIAL EQUATIONS and, taking into account the result of differentiating the identity (5.53) with respect to s: iJp iJx iJ2x iJq iJy iJ2y iJ2z a;ar+P asat +a;ar+q asat-asat=O, we will have au = iJp ax + !!.9.... iJy- iJp iJx _!!.9._ ay iJt iJt as at iJs as iJt iJs iJt or, by virtue of the equations (5.50), au ax ay op aq Tt=-(Fx+PFz) as -(Fv+qFz) 7fi-FP7JS-Fq7fi= =- (F !:_+F ~+F ~+F op +F !!.9....)-x iJs Y iJs z iJs P iJs q iJs ( iJx ay az ) a } -Fz P7fi+qa;-a; =-a;{F -FzU=-FzU• . a since F ==0, and hence the total partial derivative Ts {F} = 0. From the equation '- f F2 dt au -=-FUiJt z (5.55) we find U=U0e0 • Hence if U0 =0, then U=O, which incidentally follows also from the uniqueness of the solution U ==0 of the linear equation (5.55), which satisfies the condition U It=o = 0. Thus, when integrating the equation F (x, y, z, p, q) = 0 (5.45) with initial conditions X0 = X0 (s), Yo= Yo (s), z~ = Z0 (s), use the Cauchy method to determine the functions Po=- Po('>) and q0 = q0 (s) from the equations F(X0 (s), Yo (s). Z0 (S), Po (s), q0 (s)) = 0 and Po (s)x; (s) +qo (s)y~ (s)-z~ (s) =0 and then integrate the system of equations dx dy dz dp dq = dt (5.50) -p;= Fq = pFp+qFq Fx+PFz =- F1 +qFz with the initial conditions: for t =0 X= X0 (s), y =Yo (s), z = Z0 (s), P=Po (s), q = q0 (s). The three functions X=X(t, S), y=y(t, S), Z=Z(t, S) 5. FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 283 of the solution of the system (5.50) yield in parametric form the equation of the desired integral surface of the equation (5.45). The foregoing is readily generalized to nonlinear partial differential equations with an arbitrary number of independent variables F (x1, X1 , ••• , x,., z, p1 , p1 , ••• , p,.) = 0, (5.56) where the · P;= :;; (i= 1, 2, ... , n). It is required to determine the integral n-dimensional surface z = z (x., x2 , ••• , x,.) of the equation (5.56) that passes through a given (n-1)-dimensional surface: X;0 = x10 (s1, s,, ... , s,._1) (i = 1, 2, .•. , n), (5.57) 20 = 20 (s1 , s1 , •.. , s,._ 1). For the time being suppose that we know the initial values of the functions P;o=P;0 (s., 52 , ••• , s,._ 1) (i= l, 2, ... , n); (5.58) then, integrating the auxiliary system of equations dpl =- dp,. dt Fx, +Ptfz- '' ' Fx,. +PnFz = (5.59) with initial conditions (5.57) and (5.58), we get x1=x1 (t, S1, s,, ... , s,._ 1) , } z=z(t,s1,s,, ... ,s,._1), (i=l,2, ... ,n). p1=p,(t, s1 , s,, ... , s,._ 1). (5.60) For fixed s10 s,, ... , s,._ 1, the equations (5.60) define, in a space with coordinates x1, x,, ... , x,., z, curves called characteristics, to each point of which are also referred the numbers p1= p1(t, s1 , s,, ... , s,._1) that define the direction of certain planes II Z-z= ~pdX1-x1). (5.61) i=t The charact~ristics together with the planes (5.61) form so-called characteristic strips. Changing the parameters s1 , s,, ... , s,._1 yields an (n-1)-parameter family of characteristics x1=x1 (t, s1, .•• , s,._ 1), z=z(t, s1, .••• s,. 1) that pass through the given (n-l)-dimension11l surface (5 57). 284 I. DIFFERENTIAL EQUATIONS We shall show that for a definite choice of the functions Pio=Pio(SI, s2, ... , Sn-1) (i= 1, 2, ... , n) points lying on the characteristics of the family (5.60) form the desired n-dimensional integral surface. Hence we have to prove that for a definite choice of the functions p10 (s1, s,, ... , s,._ 1): (1) F (x1 (t, s 1, ... , sn_ 1), ••• , x,. (t, s 1, ••• , s,._ 1), z (t, sl, ... 'Sn-1), PI (t, sl, •.• ' Sn-1), ••. 'Pn (t, sl, ... 'sn-1)) ==0, (2) p1 = !~~ (i = 1, 2, ... , n) or, what is the same thing, n dz= ~p1 dx1• I== I lt can be readily verified that the function F (x1 , x1 , ••• , xn, z, p1 , p2 , ••• , Pn) is a first integral of the system of equations (5 59). Indeed, along the integral curves of the system (5.59), d dt F (xi, x,, · · •• Xn, Z, Pt• p,, · · •• Pn) == ".., dx· dz n dp· = L FXj d/ +FII dt +~ FPiTt== i==1 1==1 n n n == ~ Fx;fp, + F11 ~PtFP,- ~ Fp, (Fx, +P;F11) == 0 1=1 1==1 1==1 and, hence, along the integral curves of the system (5.59) F(x1 , X2 , ••• , Xn, z, p1, p1 , ••• , Pn)=c, where c is a constant equal to F (X10 , X20 , ••• , Xno• Z0 , P10 , P2o• ..., P,.0). In order that .the functions (5.60) should satisfy the equation (5 56) along the integral curves of the system (5.59), one has to choose the initial values p10 (s1, s 1, ••• , sn_ 1) so that F (XIO (Sp .•• ' Sn-I), ••• ' Xno (s1, • • •' Sn-1), Z (sp ... ' Sn-1), P1 (si, · · · • sn-1), • • • • Pn (sl, · · · • Sn-1)) = 0. "It remains to verify that dz = ~ p1dx1 or 1=1 iJz . n-1 iJz n 'iJx· n-1 iJx· ) Ft dt +~ os; ds1 == ~ p1 ( a/ dt + ~ as; ds1 • i=l i==l (==1 This identity is equivalent to the following: "iJz ~ ox1 -O at- .k. P; ar = (5.62) I= I 5. FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 286 and az II ax; --~p.-=0 as/ ~ I ()s.i= I '} (j= l, 2, ... , n-1). (5.63) The validity of identity (5.62) becomes obvious if one takes into account that by virtue of the system (5.59) az ~ ax· at=}: p,.Fp, and ai = Fp1 (i=l,2, ... ,n) I= I (in place of ~: and ~j we write the partial derivatives since all the s1 in the system (5.59) were assumed to be fixed). To prove the indentities (5.63), which are true only for a definite choice of the initial values P;o (s1 • S2 , ••• , S11 _ 1 ) we put u - az ~ ox; I- as/-~ P; as/ i=l (j = l, 2, ... , n-I) and, differentiating U1 with respect to t, we get auI a2z II a2x; II ap,. ax{ ar= at as}-:EP; at as}-L ar as/. l=n t=o (5 64) Taking into account the result of differentiating the identity (6.62) with respect to s1 we can rewrite the equation (5.64) as auj = ~ ap1ax1 _ ~ ap1ax1 at ~as/ at ~ iJJ as/ . t=o =o Taking advantage of the system (5.59), we haVP 286 I. DIFFERENTIAL EQUATIONS The total partial derivative aa {F} = 0, since F==0 and, hence,Sj the functions U1 are solutions of the homogeneous linear equations a~,= -FzU1, which have the unique solution V1 ==0 if V1 !t..o=0. Consequently, if the initial values p10 (s1; S1, ••• , sn_ 1) (i =- I, ( az ~ ax;)2, ..., n) are chosen so that U11t:o = 0 or as1 - ~ Pt as, = 0 I= I t=o (j =I, 2, ... , n-1), then (/=I, 2, ... , n-1) n az and, hence, on the surface (5.60) dz = ~ P; dx1, i.e., p1= ax1 I: I (i = I. 2, ... , n). Thus, to find the integral surface of the equation F (x1 , x,, ... , Xn, z, PJt P1 , ••• , Pn)=O passing through the (n-1)-dimensional surface X;0 = X;0 (s1 , S1 , ••• , Sn_ 1) Z0 = Z0 (s1 , S1 , ••• Sn_ 1), (l = I, 2, ... , n), it is necessary to determine the initial values P;o (s1 , s1, ••• , s,._ 1) from the equations F (x,o, :•o• · · · • Xno• Z• Pto• Pto• · · •• Pno) == 0, } iJzo ""' ax;o 0 (" 1 2 I) (5.65)rs-~Pio7fS= J=, •... ,n-; I 1";'1 I then, integrating the system (5.59) (page 283) with initial conditions x10 == X;0 (s1 , S3, ••• , Sn_ 1), } Z0 =Z0 (S1 , S1, ••. , Sn_1), (i-1, 2, ..., n), P1o- P1o (s,, s., ·•••Sn-1) we get XI=X;(t, S1 , S1 , ••• , Sn_ 1) (i= I, 2, ..., n), Z=Z(t, S1 , S1, ••• , S11_ 1), (6.66) Pt-P;(t, S1 , s,, ..• , sn_ 1) (i= I, "2, ••• , n). (5.67) The equations (5.66) and (5.67) are the parametric equations of the desired integral surface. Note. We assumed that the system of equations (5.65) was solvable for P;o and also that the system (5.59) satisfies the condi· tions of the existence and uniqueness theorem. 5. FIRST ORDER PARTIAL DIFFERENTIAL EQUATIONS 287 Example 1. Find the integral surface of the equation z = pq passing through the straight line x = 1, z= y. Write the equation of the straight line x = 1, z= y in parametric form, X0 = l, Yo= s, Z0 = s. Find Po (s) and q0 (s) from the equations (5.65): s= p0q0 , l-q0 = 0, whence Po= s, q0 = l. Integrate the system (5.59): Taking into account that for t = 0 X= l, y=s, Z=S, p=S, q= 1, we have p = se', q = e1, x = e1, y = se1, z= se'1• Consequently, the desired integral surface is x=e1, y=se1, z=se21 or z=xy. Example 2. Integrate the equation (:;r+(:;r= 2 provided that for x=O, z=y, or in parametric form X0 =0, y0 =S, Z0 =s. Determine Po (s) and q0 (s): p~+q~=2, 1-q0 =0, whence q0 = 1, Po= ± 1 Integrate the system of equations (5.59): dx _ ay _ dz _ dp _ dq _ dt 2P-2q--4- o- o- · p =Cu q=c2 , x= 2c1t +c3 , y= 2cit +c,, z = 4t +c,; using the initial conditions Po=± 1, q0 = 1, X0 = 0, Yo= s, Z0 = s, we obtain p = ± 1, q = 1, x= ± 2t, y= 2t +s. Z= 4t +s. The last three equations are the parametric equations of the desired integral surface. Eliminating the parameters t and s, we get z = y ± x. In problems of mechanics one often has to solve the Cauchy problem for the equation ihJ ift +H (t, xl, x,, ... ' Xn, Pl• p•• •.. ' Pn> = 0, (5.68) where the P; = ::. , which is a special case of the equation (5.56) I (p. 283). The Cauchy method, which, as applied to equation (5.68), 288 I. DIFFERENTIAL EQUATIONS ~--------------------- ------------------------ is frequently called Jacobi's first method, yields the system of equations whence and or dt - ax. - dxz - iJH - iJH iJpl iJp, dp, =- iJH = dp; an Tt = -ax1 ' (i = 1, 2, .... n) (5.69) (5.70) The system of 2n equations (5.69) does not contain v and may be integrated independently of equation (5.70); then the function v is found from (5.70) by a quadrature. Therein lies the specificity of applying Cauchy's method to equation (5.68). Besides, in this case there is no necessity of introducing an auxiliary parameter into the system (5.50), since that role can be successfully played by the independent variable t. PROBLEMS ON CHAPTER 5 1 az_az_0 . ax ag- . az Oz 2 ax+av=2z. Oz 3 Xag=Z. oz az 4. zax-Yav=O. dz 5. Yax=z for x=2, z=y. oz dz 6. x iJx-y011 = z for y = 1, z= 3x. 5. FIRST-ORDER PARTIAL DIFFERENTIAL EQUATIONS 289 az az Of 0 a 7. yz ax+ ag = or %= ' z= y . 8. Find surfaces that are orthogonal to the surfaces of the family z= axy. 9. Find surfaces that are orthogonal to the surfaces of the fa· mily xyz=a. 10 ~ az_J!... az = z-5 · 3ax sag · 11 au+au+du=O ·ax ag az 0 au au au 12. x ax+ 2y ag +3za;-= 4u. 13 Otz (x, g) -O . axz - . az az Of 1 •14. ax-2xag= or%= . z=y. 15. Can the equation (J/ +z•-x1) dx +xz dy +xydz = 0 be integrated by one relation? 16. Integrate by one relation the equation (y+3z1 )th+(x+y)dy+6xzdz=0. 17. Find the complete integral of the equation pq=x•y•. 18. Find the complete integral of the equation z= px +qy+p1q1 • 19. Find the complete integral of the equation pq=9z'. 20. Find the complete integral of the equation p=sinq. 21. Find the surfaces that are orthogonal to the vector lines of the vector field F = (2xy-3yz) I+(x2 -3xz)j -3xu k. 19 :m.; 290 I. DIFFE~ENTIAL EQUATIONS 22. Find a family of surfaces that are orthogonal to the vector lines of the vector field F = (2x-y) i +(3y-z) j +(x-2y) k. 23. Find the vector Jines, the vector surfaces and surfaces that are orthogonal to the vector Jines of the field F =xi+yj-zk. 24. z=pq+ 1 for y=2, Z=2x+ 1. 25. 2z=pq-3xy for x=5, Z= 15y. 26. 4z = p2 +q2 for x= 0, z = y'. PART TWO The calculus of variations Ill Introduction Besides problems in which it is necessary to determine the maximal and minimal values of a certain function z= f (x), in physics there are often encountered problems w·here one has to find the maximal and minimal values of special quantities cailed functionals. Functionals are variable quantities whose values are determined by the choice of one or several functions. y y B :r: 0 0 Fig. A Fig. B For example, the arc length I of a plane (or space) curve connecting two given points A (x0 , y0 ) and 8 (x1 , y1) (see Fig. A) is a functional. The quantity l may be computed if the equation of the curve y = y (x) is given; then x, l(y (x)] = ~ V1+(y')2 dx. Xo The area S of a surface is also a functional, since it is determined by the choice of surface, i.e., by the choice of the function z (x, y) that enters into the equation of the surface z= z(x, y). As is known, where D is a projection of the surface on the xy-plane. 294 II. THE CALCULUS OF VAlUATIONS Moments of inertia, static moments, the coordinates of the centre of gravity of a homogeneous curve or surface are also functionals, since their values are determined by the choice of the curve or surface, i. e. by the choice of functions that enter into the equation of the curve or surface. In all these examples we have a relationship that is characteristic of functionals: to a function (or vector function) there corresponds a number, whereas when we specify a function z = f (x), to a number there corresponds a number. The calculus of variations investigates methods that permit finding maximal and minimal values of functionals. Problems in which it is required to investigate a function for a maximum or a minimum are called variational problems. Numerous laws of mechanics and physics reduce to the statement that a certain functioual in a given process has to reach a minimum or a maximum. Thus stated, such laws are termed variational principles of mechan:cs or physics. The following are some variational principles or elementary consequences of them: the principle of least action, the law of conservation of energy, the law of conservation of momentum, the law of conservation of angular momentum, various variational principles of classical and relativistic field theory, Fermat's principle in optics, the principle of Castigliano in the theory of elasticity, and so forth. The calculus of variations began to develop in 1696 and became an independent mathematical discipline with its own methods of investigation after the fundamental works of Euler (1707-1783), whom we may justifiably consider the founder of the calculus of variations. Three problems exerted a considerable influence on the development of the calculus of variations: The problem of the brachistochrone. In 1696 Johann Bernoulli published a letter in which he advanced the problem or the line of quickest descent (brachistochrone). In this problem it is required to find the line connecting two specified points A and B that do not lie on a vertical line and possessing the property that a moving particle slides down this line from A to B in ti1P. shortest time (Fig. B). It is easy to see that the line of quickest descent will not be the straight line connecting A and B, though. that is the shortest distance between the two points, because the velocity of motion in a straight line will build up comparatively slowly; whereas if we take a curve that is steeper near A, even though the path becomes longer, a considerable portion of the distance will be covered at a greater speed. The problem of the brachistochrone was solved by Johann Bernoulli, Jacob Bernoulli, Leibnitz, Newton, L'Hospital. INTRODUCTION 295 It turned out that the line of quickest descent is a cycloid (see pages 316-317). The problem of geodesics. It is required to determine the line ol minimum length connecting two given points on a surface

(x). This compels us to introduce the following definitions of proximity of the curves y = y (x) and y = y1 (x). The curves y = y (x) andy= y1 (x) are close in the sense of zero-order proximity if the absolute value of thediOerence y(x)-y1 (x) is small. The curves y = y (x) and y = y1 (x) are close in the sense of firstorder proximity if the absolute values of the diOerences y (x)-y1 (x) and y' (x)-y; (x) are small. The curves y = y (x) and y = Y1 (x) are close in the sense of kth order proximity if the absolute values of the differences are small. y (x)-y1 (x), y' (x)-y~ (x), y (x) 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 299 Fig. 6.1 exhibits curves close in the sense of zero-order proximity but not close in the sense of first-order proximity, since the ordinates are close but the directions of the tangents are not. In Fig. 6.2. are depicted curves close in the sense of first-order proximity. y ~B II y ./B 0 :r 0 Fig. 6-1 Fig. 6-2 From these definitions it follows that if the curves are close in the sense of kth order proximity, then they are definitely close in the sense of any lesser order of proximity. We can now refine the concept of continuity of a functional. 3'. A function I (x) is continuous at X=X0 if for any positive e there is a IS > 0 such that II (x)-l(x0) I< e for Ix-x0 I 0 such that 1v [y (x) ]- v [Yo (x)] I< e for IY (x)-Yo (x) I< 6, IY' (x)-y~ (x) I< 6, Iy (x)-y~k) (x) I< 6. It is assumed here that the function y (x) is taken from a class of functions on which the functional v [y(x)] is defined. One might also define the notion of distance p (y1, y2) between the curves y = Y1 (x) and y = Y2 (x) (x0 ..;:;;; x ~ x1) and then close-lying curves would be curves with small separation. ::JOO II. THE CALCULUS OF VARIATIONS If we assume that P(yl, Yz) = max IY1 (x)-Yz (x) 1.X0 ...,; X...,; X 1 that is if we introduce the space metric C0 (see pages 54-55), we have the concept of zero-order proximity. If it is taken that k p (y1, y,) = ~ max Iy~P> (x)-yiP> (x) I P = I X0 <: X < x, (it is assumed that y1 and y., have continuous derivatives up to order k inclusive), then the proximity of the curves is understood in the sense of kth order proximi ty. 4. A linear function is a function l (x) that satisfies the following conditions: l (ex)= cl (x), where c is an arbitrary constant, and l (X1 +X2) = l (X1) +l (X2). A linear function of one variable is of the form l (x) = kx, where k is constant. 5. If the increment of a func- tion M= f (x+ tlx)-f (x) may be represented in the form M=A (x) tlx +~ (x, tlx) ·tlx, where A (X) does not depend on tlx, and ~ (x, tlx) -+ 0 as tlx .... 0, then the function is called diOerentiable, while the part of the increment that is linear with respect to tlx-A (x) tlx-is called the diOerential of the function and is denoted by df. Dividing 4. A linear functional is a functional L [y (x)] that satisfies the following conditions L [cy (x)] = cL [y (x)], where c is an arbitrary constant and L [Yt (x) +Yz (x) ] = = L [Y1 (x)] +L [Yz (x) ]. The following is an instance of a linear functional: x, L [y (x)] = ~ (p (x) y +q (x) y') dx. Xo 5. If the increment of a func- tional tlv = v [y (x) +6y] -v [y (x)) may be represented in the form tlv = L [y (x), ~y] + +~ (y (x), 6y) max 16y 1. where L [y (x), 6y] is a functional linear with respect to 6y, and max 16y I is the maximum value of l6y I and ~ (y (x), 6y)-+ 0 as max 16y I-+ 0, then the part of the increment of the functional that is linear with respect to 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 301 by .1x and passing to the limit as .1x ----0, we find that A (x) = f' (x) and, hence. df = f' (x) .1x. 6y, i.e. L (y (x), l'Jy]. is called the variation of the functional and is denoted by 6v. Thus the variation of a functional is the principal part of the increment of the functional, which part is linear in l'Jy. In the examination of functionills, the variation plays the same role as the differential does in investigations of functions. Another, almost equivalent definition of the differential of a function and the variation of a functional may be given. Consider the value of the function f (x +a..1x) for fixed x and .1x and varying values of the parameter a.. For a.= 1 we get an increased value of the function f (x +.1x), for a.= 0 we get the initial value of the function f (x). It may be readily veri lied that the derivative of f (x +a..1x) with resp~t to a. for a.= 0 is equal to the differential of the function f (x) at the point x. Indeed, by the rule for differentiating a composite function :af (x +a. .1x) /a= o= f' (x +a..1x) /:1. =o= f' (x) .1x = df (x), In the same way, for a function of several variables Z = f (x1, X1 , ••• , Xn) one can obtain the differential by differentiating f (x1 +a..1x1 , X2 +a..1x1 , ••• , Xn +a..1xn) with respect to a., assuming a.=O. Indeed, " !_f (x1 +a. .1x1, X2 +a..1x,, ... , xn +a..1xn) Ia= o= L ~~ .1x1 = df. l= I i Likewise, for functionals of the form v [y (x)] or more complicated ones depending on several unknown functions or on functions of several variables, it is possible to determine the variation as a derivative of the functional v [y (x) +a.6y] with respect to a. for a.= 0. Indeed, if the functional has a variation in the sense of the principal linear part of the increment, then its increment will be of the form .1v=v [y(x) +a.l'Jy] -v[y(x)] =L(y, a.l'Jy) +~ (y, a.6y) Ia. Imax j6y J. The derivative of v [y +a.6y] with respect to a. at a.= 0 is lim 11v=lim 11v=lim L(y, a6y)+~(y(x), a6y])a)maxi6YI = <\a-+0 11a a-+0 a <7.-+0 a I. L (y, a 6y) +1. ~ (y (x), a 6y) I a I max )6y I L (y Jl.y)= tm tm = , u 01~0 a 01~0 a 302 II. THE CALCULUS OF VARIATIONS since by virtue of linearity L (y, a. fJy) = aL (y, fJy) and lim ~(y(x), a6ytlalmaxl6yl=lim ~[y(x), a.t'ly]maxl6yl=O a-+0 a a-+0 because ~ [y (x), a. fJy]- 0 as a.- 0. Thus, if there exists a variation in the sense of the principal linear part of the increment of the functional,then there also exists a variation in the sense of the derivative with respect to the parameter for the initial value of the parameter, and both of these definitions are equivalent. The latter definition of a variation is somewhat broader than the former, since there are instances of functionals, from the increments of which it is impossible to isolate the principal linear part, but the variation exists in the meaning of the second definition. 6. The diOerential of a function 6. The variation of a functiof (x) is equal to nal v [y (x)] is equal to a a 00 / (x+a. ~X)Ia =0· aa V [y (X)+ a c'\y] la=O· Definition. A functional v [y (x)] reaches a maximum on a curve y =Yo (x) if the values of the functional v [y (x)] on any curve close toy= Yo (x) do not exceed v [Yo (x)); that is ~v = v [y (x)]- v [Yo (x)) ~ 0. If ~v ~ 0, and ~v = 0 only for y (x) =Yo (x), then it is said that a strict maximum is reached on the curve y =Yo (x). The curve y =Yo (x), on which a minimum is achieved, is defined in similar fashion. In this case, ~v ~ 0 for all curves close to the curve y =Yo (x). 7. Theorem. If a diOerentiable function f (x) achieVes a maximum or a minimum at an interior point x = x0 of the domain of definition of the function, then at this point df=O. 7. Theorem. If a functional v [y (x)J having a variation achieves a maximum or a minimum at y =Yo (x), where y (x) is an interior point of the domain of definition of the functional, then at y =Yo (x), t'lv=O. Proof of the theorem for functionals. For fixed Yo (x) and fJy v [Yo (x) +a t'lyJ = cp (a) is a function of a, which for a= 0, by hypothesis, reaches a maximum or a minimum; hence, the derivative cp' (0) = 0*, and aiJ v [Yo (x) +a t'ly] I = 0, a a=O • a can take on either positive or negative values in the neighbourhood ol the point a=O, since Yo (x) is an interior point of the domain of definition of the functional. 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 303 i. e. fJv = 0. Thus, the variation of a functional is zero on curves on which an extremum of the functional is achieved. The concept of the extremum of a functional must be made more specific. When speaking of a maximum or a minimum, more precisely, of a relative maximum or minimum, we had in view the largest or smallest value of the functional only relative to values of the functional on close-lying curves. But, as has already been pointed out, the proximity of curves may be understood in different ways, and for this reason it is necessary, in the definition of a maximum or minimum, to indicate the order of proximity. If a functional v [y (x)] reaches a maximum or a minimum on a curve y=y0 (x) with. respect to all curves for which the absolute value of the difference y (x)-Yo (x) is small, i.e. with respect to curves close to y =Yo (x) in the sense of zero-order proximity, then the maximum of minimum is called strong. However, if a functional v [y (x)] attains, on the curve y =Yo (x), a maximum or minimum only with respect to curves y = y (x) close to y=y0 (x) in the sense of first-order proximity, i.e. with respect to curves close to y =Yo (x) not only as regards ordinates but alsJ as regards the tangent directions, then the maximum or the minimum is termed weak. Quite·obviously, if a strong maximum (or minimum) is attained on a curve y=y0 (x), then most definitely a weak one has been attained as well, since if the curve is close to y =Yo (x) in the sense of first-order proximity, then it is also close in the sense of zero-order proximity. It is possible, however, that on the curve y =Yo (x) a weak maximum (minimum) has been attained, yet a strong maximum (rnnimum) is not achieved; in other words, among the curves y = y (x; close to y = y1 (x) both as to ordinates and as to the tangent diredions, there may not be any curves for which v [y (x)] > v lYo (x)j (in the case of a minimum v [y (x)] < v[y0(x)J), and among the curves y = y (x) that are close as regards ordinates but not close as regards the tangent directions there may be those for which v [y (x)] > v [Yo (x)] (in the case of a minimum v [y (x)] < J, linear in 6z, or as a derivative with respect to the parameters for the initial value of the parameter !, v [z (x, y) +a 6z) Ia= 0 , and if for z= z(x, y) the functional v attains an extremum, then for z = z (x, y) the variation 6v = 0, since v [z (x, y) +a6z] is a function of a, which for a= 0, by hypothesis, attains an extremum and, hence, the derivative of this function with respect to a for a=O vanishes, !,v [z (x, y)+a6z) la=o =0 or 6v=0. 2. Euler's Equation Let us investigate the functional x, v (y (x)] = ~ F (x, y (x), y' (x)) dx (6.1) Xo • It is assumed that a can take on any values close to a=O and 00 lg (x, a))l exists. da. a=O 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 305 for an t:xtreme value, the boundary points of the admissible curves being fixed: y (x0 ) =Yo and y (x1) = y1 (Fig. 6.3). We will consider the function F (x, y, y') three times differentiable. We already know that a necessary condition for ttn extremum is that the variation of the functional vanish. We will now show how this basic theorem is applied to the functional under consideration, and we will repeat the earlier argument as applied to the functional (6.1). Assume that the extremum is attained on a twicedifferentiable curve y = y (x) (by only requiring that admissible y !I B n !It A !/a r :r 0 Xq :r, 0 Fig. 6-3 Fig. 6-4 curves have first-order derivatives, we can prove by a different method that the curve which achieves the extremum has a second derivative as well). Take some admissible curve y = y(x) close to y = y (x) and include the curves y = y (x) and y = y(x) in a one-parameter family of curves y (x, a) = y (x) + a(Y (x)- y (x)); for a= 0 we get the curve y = y (x), for a= 1 we have y = y(x) (Fig. 6.4). As we already know, the difference y (x)-y (x) is called the variation of the function y (x) and is symbolized as 6y. In variational problems, the variation 6y plays a role similar to that of the increment of the independent variable !lx in problems involving investigating functions f (x) for extreme values. The variation 6y = y(x)- y (x) of the function is a function of x. This function may be differentiated once or several times; (6y)' = y' (x)- y' (x) = 6y', that is, the derivative of the variation is equal to the variation of the ':lerivative, and similarly ({)yr = y" (X)- y" (X)= l5y", ............. (l5y)l"' = ifk• tx) _ ylkl (x) = l5y (a)= ~ F (x, y (x, a), y~ (x, a)) dx, x. it follows that i, cp' (a)= S[F, :cty (x, a.)+ FY' !y' (x, a)Jdx, where or since and X. Fv = :YF (x, y (x, a), y' (x, a)), F11• = 0~, F (x, y (x, a), y' (x, a)), a aoct y (x, a)= oct [y (x) +a&y] = &y 0°ct y' (x, a) = 0°ct [y'(x) +a6y') = &y'. 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 307 we get x, ' (0) is called the variation of the functional and is denoted by 6v. A necessary condition for the extremum of a functional v is that its variation vanish: .Sv = 0. For the functional x, v[y(x)]=~F(x, y, y')dx this condition has the form x, ~ [Fv6y+ F"' 6y'] dx=O. X, We integrate the second term by parts and, taking into account that t3y' = (6y)', we get x, 6v = [Fv' 6y];~+S(FY -!Fy') 6y dx. "•But t3ylx=x.= y(x0)-y(x0)=0 and oylx=x, =y(x1)-y(xt)=0, because all admissible curves in the elementary problem under consideration pass through fixed boundary points and, hence, x, t3v= S(Fy- !Po') 6y dx. ~n Thus, the necessary condition for an extremum takes the form x, s(FY-:xfv,) 13ydx=0, (6.2) Xo the first factor Fy-!FY' on the extremizing curve y = y (x) i:> a given continuous function, while the second factor lly, because of the arbitrary choice of the comparison curve y = y (x), is an arbitrary function that satisfies only certain very general conditions, :20* 308 II. THE CALCULUS OF VARIATIONS namely: at the boundary points x = x0 and x = x1 the function ~y vanishes, it is continuous and differentiable once or several times; lJy or 6y and ~y' are small in absolute value. To simplify the condition obtained, (6.2), let us take advantage of the following lemma. The fundamental lemma of the calculus of 'Variations. If for every continuous function TJ (x) "'• ~ $ (x) TJ (x) dx = 0, %,, where the function (x) is continuous on the interval [x0 , x1], /hen (x) ==0 on that interval. Note. The statl•ment of the lemma and its proof do not change if the following restrictions are imposed on the functions: TJ (x0 ) = = 'l (x1) = 0; 'l (x) has continuous derivatives to order p, l'l (x) l< (x) =F 0, we arrive at a contradiction. Indeed, from the continuity of the function (x) it follows that if (x) maintains its sign in a certain neighbourhood (x0 ~ x ~ ~xl) of the point x; but then, having chosen a function T] (x), which also maintains its sign in this neighbourhood and is equal to zero outside this neighbourhood (Fig. 6.5), we get X1 Xa ~ $ (x) TJ (x) dx = i (x) TJ (x) does not change sign on the interval (x0 ~ x~ xJ and vanishes outside this interval. We have thus arrived at a contradiction; hence, (x) ==0. The function TJ (x) may for example be chosen thus: TJ (x) ==0 outs!de the interval (X.,~ X~ X1); TJ (X)= k(X-Xo)2n (x-X.)2n On the interval,(Xo ~X~ X1), where n is a positive integer and k is a constant factor. It is obvious that the function TJ (x) satisfies the above conditions: it is continuous, has continuous derivatives up to order 2n-l, vanishes at the points X0 and X1 and may z be made arbitrarily small in absolute value together with its derivatives by reduCing the absolute value of the constant k. Note. Repeating this argument word for word, one can prove that if the function region D on the plane (x, y) 8 (x, y) is continuous in the Q) and~~ (x, y) TJ (x, y) dx dy=O .r for a~ arbitrary choice of the Fig. 6-6 function TJ (x, y) satisfying only certain general conditions (continuity, differentiability once or several times, and vanishing at the boundaries of the region D, ITJ I< e, ITJ~ I< e, I~ I (x, y) ==0 in the region D. When proving the fundamental lemma, the function TJ (x, y) may be chosen, for example, as follows: TJ (x, y) ==0 outside a circular neighbourhood of sufficiently small radius e1 of the point {x, y) in which (x, y) =1= 0, and in this neighbourhood of the point (i, y) the function TJ (x, y) = k [(x-x)' + +(y-Y}'-eWn (Fig. 6.6). An analogous lemma holds true for n-fold multiple integrals. Now let us use the fundamental lemma to simplify the aboveobtained condition (6.2) for the extremum of the elementary functional (6.1) (6.2) All conditions of the lemma are fulfilled: on the extremizing curve the factor ( FY-! F11,) is a continuous function, and the variation 6y is an arbitrary function on which only restrictions of a general nature that are provided for by the fundamental lemma have 310 II. THE CALCULUS OF VARIATIONS been imposed; hence, Fy-! F11,_0 on the curve y=y(x) which extremizes the functional under consideration, i.e. y = y (x) is a solution of the second-order differential equation d F --F11·=0Y dx ' or in expanded form Fv- Fxy'- Fuu'Y'- Fu'u'Y" = 0. This equation is called Euler's equation (it was first published in 1744). The integral curves of Euler's equation y=y(x, C1, C2) are called extremals. It is only on extrernals that the functional x, v [y (x)] = ~ F (x, y, y') dx Xo can be extremized. To find the curve that extremizes the functional (6.1), integrate the Euler equation and determine both arbitrary constants that enter into the general solution of this equation, proceeding from the conditions on the boundary y(x0) = = y0 , y (x1) = y1 . Only on extremals that satisfy these conditions can the functional be extremized. However, in order to establish whether indeed an extremum (and whether it is a maximum or a minimum) is achieved on them, one has to take advantage of the sufficient conditions for an extremum given in Chapter 8. Recall that the boundary-value problem d F -- f 11,=0 Y dx ' does not always have a solution and if the solution exists, it may not be unique (see page 166). Note that in many variational problems the existence of a solution is obvious from the physical or geometrical meaning of the problem and if the solution of Euler's equation satisfying the boundary conditions is unique, then this unique extremal will be the solution of the given variational problem. Example 1. On what curves can the functional 11 2 v(y(x)]= Sr 0. .... This lower bound of values of the Fig. 6-9 functional is attained on the discontinuous function (Fig. 6.9) y(xo)=Yo• y (x) = 0 for x0 < x < x, y(x.)=y•. (2) The function F is linearly dependent on y': F(x, y, y') = M (x, y) +N (x, y) y'; v [y (x)] = S[M (x, y) +N (x, y) ~~] dx. Xo The Euler equation is of the form iJM iJN I d iJg +iJy y - d:c N (x, y) = 0, or or 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 313 aM_ aN =O· ay ax ' but again, as in the preceding case, this is a finite, not a differential, equation. Generally speaking, the curve ~:- C::: = 0 does not satisfy the boundary conditions; consequently, the variational problem does not, as a rule, have any solutions in the class of continuous functions. But if aa~- a;: ==0, then the expression M dx +N dy is an exact differential and x, x, v= S(M+N~) dx= S(Mdx+Ndy) x. x. is independent of the integration path, the value of the functional v is constant on admissible curves. The variational problem becomes meaningless. Example 4. I v[y(x)]= ~ (y2 +x2y')dx; y(O)=O, y(l)=a. 0 Euler's equation is of the form ~: - ~~ = 0 or y- x = 0. The first boundary condition y (0) == 0 is satisfied, but the second boundary condition is satisfied only for a= I. But if a=1= I, then there is no extremal that satisfies the boundary conditions. Example 5. ~ ~ v[y(x)) = ~ (y+xy')dx or v[y(x)] = ~ (ydx+xdy); x, x. Y (Xo) = Yo• Y (x,) = Y1• Euler's equation reduces to the identity 1== 1. The integrand is an exact differential and the integral does not depend on the path of integration: x, v [y (x)) = ~ d (xy) = x,y,-X0y0 , no matter which curve we integrate along. The variational problem is meaningless. 314 II THE CALCULUS OF VARIATIONS (3) F is dependent solely on y': F= F (y'). Euler's equation is of the form Fll v·Y" = 0, since Fy = Fxy' = Fvii= 0. Whence y" = 0 or Fv·v = 0. If y" = 0, then y= C1x+C2 is a twoparameter family of straight lines. But if the equation Fv'v' (y') = 0 has one or several real roots y' = k1, then y = kix +C and we get a one-parameter family of straight lines contained in the aboveobtained two-parameter family y = C1x +C2• Thus, in the case of F= F(y'), the extremals are all possible straight lines y = C1x +C2• Example 6. The arc length of the curve x, l [y (x)J = ) V 1+y'2 dx x. has for extremals the Straight Jines y = C1X +C2• Example 7. The time t [y (x)J spent on displacement along a certain curve y = y (x) from the point A (x0 , Yo) to the point B (x1 , y1) if the velocity ~; = v(y') is dependent solely on y', is a functional of the form t [y (x)J ·= xs, Vi+? dx tl (y') x. ( ds dt = _!!!__ = V1+y'2 dx • dt = v (y'); tl (y') tl (y') ' x,~~) Jf 1+y'' t = S v (y') dx . Xo Hence the extremals of this functional are straight lines. (4) F is dependent solely on x and y': F=F (x, y'). Euler's equation takes the form :xFv'(x, y') = 0 and, hence, has a first integral Fy' (x, y') = C1; and since the first-order equation Fv (x, y') = C1 thus obtained does not contain y, the equation may be integrated either by direct solution for y' and integration, or by means of introducing a properly chosen parameter (see p~ 75). Example 8. The functional t [y (x)J = sv~ dx "· (t is the time spent on translation along the curve y =- y (x) from 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 315 one point to another if the rate of motion v = x, since if :~ = x, ds xS, V1+y'2 ) then dt = x and t = x dx . The first integral of the Fuler Xo equation fu·=C1 is of the form Vy' =C1 . This equation is X !+y'2 most readily integrated if one introduces a parameter putting y' =tan t; then I y' 1 . X=- =-smt C1 VI +u'~ cl . - I or x=C1 smt, where C1 =c;,; ~=tan t; dy=tan tdx=tan t. C1 costdt=C1 sintdt; integrating, we get y= - C1 cost+ C2. Thus, X= cl sin t, y-C2 = -Cl cost or, eliminating t, we have x2 +(y-C2)2 =C~. which is a family of circles with centres on the axis of ordinates. (5) F is dependent on y and y' alone: F=F(y, y'). Euler's equation is of the form: Fy- Fuu·Y'- Fu·u'Y'' = 0, since Fxy' = 0. If we multi ply this equation termwise by y', then, as is readily verifiable, the left-hand side becomes an exact derivative :X(F- y'Fy-}. Indeed, fx (F -y'Fy·) = Fuy' +Fu·Y"-y"Fu· -F11y·y'" -Fvy.y'y'' = =y' (Fu- Fyy' y'-Fu·u·Y"). Hence, Euler's equation has the first integral F-y'Fu·=C1 , and since this first-order equation does not contain x explicitly, it may be integrated by solving for y' and separating the variables, or by introducing a parameter. Example 9. The minimum-surface-of-revolution problem: find a curve with specified boundary points whose rotation about the axis of abscissas generates a surface of minimum area (Fig. 6.10). As we know, the area of a surface of revolution is x, S [y (x)] = 2n ~ y V1+y'•dx. x. 316 II. THE CALCULUS OF VARIATIONS The integrand is dependent solely on y and y' and, henct!, a first integral of Euler's equation will have the form F-y'Fu·=C1 or in the given case yVt+y'z_ uu'z =Cl. YI+u'2 y 0 :r Fig. 6-10 After simpIification we have V Y C1• The simplest way to I +u'~ integrate this equation is by the substitution y' =sinh t, then y=C1 cosh t, and d = dy = cl sinh t dt = c dt· = c t +cX . y' sinh I 1 ' X 1 z· And so the desired surface is formed by revolution of a line, the equation of which, in parametric form, is x=C1t+C,, y=C1 cosht. EJiminating the parameter t, we get y = C1 cosh x c~' . a family of catenaries, the revolution of which forms surfaces called catenoids. The constants C1 and C, are found from the condition of the passage of the desired line through given boundary points (depending on the position of the points A and B, there may be one, two or zero solutions). Example tO. The problem of the brachistochrone (see page 294): find the curve connecting given points A and B which is traversed 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 317 by a particle sliding from A to B in the shortest time (friction and resistance of the medium are ignored). Put the coordinate origin at A, make the x-axis horizontal and the y-axis vertical. The speed of the particle is ~ = V2gy, whence we find the time spent in moving from A (0, 0) to B (x1, y1): x·v--1 S 1+u'•t [y (x)] = .r- y dx; ,. 2g 0 y y (0)=0, Although in this case the integral is not proper, it is easy to establish that here as wen we can take advantage of the preceding theory. Since this functional also belongs to the most elementary type and its integrand does not contain x explicitly, the Euler equation has a first integral F- y'FII'= C or in the given case Vt+u'' u'1 YY -;-;Vr=-=y=(l=+=Y,=:=1) = C, whence, after simplification, we get V 1 Cory (1 +y'1)=C1 • u(t +u'1) Introduce the parameter t, putting y' =cot t; then we have Y= 1+~~12 1 =c. sin1 t =~1 (l-co:S2t); d.x _dy _2C1 sintcostdt _ 2C . 2 tdt-C (l 2t)dt·--.- t 1 - 1 sm - 1 -cos ,y co C ( sin 2t) C . X= 1 t--2 - +C, =-f (2t-sm 2t) +C,. Consequently, in parametric form the equation of the desired line is x-C, = c; (2t-sin 2t), y = c; (1-cos2t). If the parameter is transformed by the substitution 2t = t1 and if we take into account that C, = 0, since x = 0 for y = 0, then we get the equation of a family of cycloids in the ordinary form: x = c; (t1 - sin t1), cy=-f(l-cos t1), where c2 is the radius of a rolling circle, which is found from the condition of passage of the cycloid through the point 8 (x1, y1). Thus, the brachistochrone is a cycloid. 318 II. THE CALCULUS OF VAR!AliONS 3. Functionals of the Form x, SF(x, Yt• y •• ... ,Yn• y;, y;, •••. y~) dx Ln order to obtain the necessary conditions for the extremum of a functional v of a more general type x, v [Yt• y,, · · • ,Yn] = ~ F (x, Yt• Yz, · · ·, Yno y~. y~, · · · , Y~) dx for the given boundary conditions of all functions Y1 (xo) = Yto• Ya (xo) = Y2o• • • ••Y,. (Xo) = Yno• Y1 (x,) = Yw y, (xl) = Yn• ••·, Yn (xl) = Ynt• we shall vary only one of the functions y1 (x) (/ = 1, 2, ... , n), holding the other functions unchanged. Then the functional v [y1 , y2 , ••• , YnJ will reduce to a functional dependent only on a single varied function, for example, on y1 (x). v[y,, Yz, · •·, Yn) = v[Yt] of the form considered in Sec. 2, and, hence, the extremizing function must satisfy Euler's equation d fv.-d-F ·=0.1 X V{ Since this argument is applicable to any function y1(i = 1, 2, ... , n), we get a system of second-order differential equations d fv -d-F ·=0 (i= 1, 2, ...,n),I X Vt which, generally speaking, define a 2n-parameter family of integral curves in the space x, y1, y2, ••• , Yn-which is the family of extremal!' of the given variational problem. If, for example, the functional depends only on two functions y (x) and z(x): X, v [y (x), z(x)] = ~ F (x, y, z, y', z') dx; x. y (Xo) =Yo· z (Xo) = Zo, y (xl) = Yto z(xl) = zl, that b to say, it is defined by the choice of space curve y = y (x), z= z (x) (Fig. 6.11 ), then by varying y (x) alone and holding z(x) 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 319 constant we can change our curve so that its projection on the xz-plane does not· change, i.e. the curve all the time remains on the projecting cylinder z= z(x) (Fig. 6.12). z Fig. 6-ll z y·y(.z}_ B z•z .x .c 0 g .X • Fig. 6-12 Similarly, by fixing y(x) and varying z(x), we vary the curve so that all the time it lies on the projecting cylinder y = y (x). We then obtain a system of two Euler's equations: d d F11 - dx F11·=0 and F6 - dxFr =0. 320 II. THE CALCULUS OF VARIATIONS E:umple 1. Find the extremals of the functional n 2 v [y (x), z (x)] = S[y'1 +z'1 +2yz] dx, y(O)=O, 0 z(0) =0, z(~)=-1. The system of Euler's differential equations is of the form y"-z =0, z"-y=O. Eliminating one of the unknown functions, say z, we get y1v- y = 0. Integrating this linear equation with constant coefficients, we obtain y=C1~ +C,e-~ +Cs cosx+C, sinx; z=y"; z=C1~+C,e-~-C3 cosx-C,sinx. Using the boundary conditions, we find C1 =0, C,=O, C8 =0, C4 =1; hence, y =sin x, z=- sin x. Example 2. Find the extremals of the functional ... v[y (x), z (x)] = ~ F (y', z') dx. "• The system of Euler's equations is of the form FII'ill+ F1/'z-z" = 0; F11·z-Y" +Fz'z'z" = 0, whence, assuming Fy11 Fz-z·-(F11z·)'=I=O, we get y"=O and z"=Oor y=C1x+C,, z=C,x+C, are a family of straight lines in space. Example 3. Find the differential equations of the lines of propa· gation of light in an optically nonhomogeneous medium in which the speed of light is v(x, y, ~~According to Fermat's principle, light is propagated from one point A (x0 , y0) to another B (x1 , y1) along a curve for which the time T of passage of light will be least. If the equation of the desired curve y = y (x) and z = z (x), then T = JVI +u''+z'' dx t1 (x, y, z) · "'- 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 321 For this functional, the system of Euler's equations av V 1+y''+z'' +~ y' =O, ay v2 dx v v1+Y''+z'2 av V1-t-y'2+z'' d ,. - +- =0 az v2 dx v v1+y''+z't will be a system that defines the lines of light propagation. 4. Functionals Dependent on Higher-Order Derivatives Let us investigate the extreme value of the functional x, v [y (x)] = ~ F (x, y (x), y' (x), ... , ylnl (x)) dx, x. where we consider the function F differentiable n +2 times with respect to all arguments and we assume that the boundary conditions are of the form Y (Xo) = Yo• y' (xo) = y~, . • . ' yln-11 (Xo) = y~n-u; y(x1)=y1, y'(x.)=y~, ... , yln-ll(x.)=y~n-p, i.e. at the boundary points are given the values not only of the function but also of its derivatives up to the order n-l inclusive. Suppose that an extremum is ~ttained on the curve y = y (x), which is 2n times differentiable, and let y = y(x) be the equation of some comparison curve, which is also 2n times differentiable. ·Consider the one-parameter family of functions y(x, a)=y(x)+a[y(x)-y(x)] or y(x, a)=y(x)+afly. For a=O, y(x, a)=y(x) and for a=l, y(x, a)=y(x). If one considers the value of the functional v[y(x)] only on curves of the family y = y (x, a), then the functional reduces to a function of the parameter a, which isextremized fora= 0; hence, d~ v(y(x, a)] Ia=O =0. According to Sec. 1, this derivative is called the variation of the functional v and is symbolized by 6v: 6v- [ :. IF(x, y (x, a), y' (x, a), •.• , !I"' (x, a)) dx].=•- x, = ~ (F11 fly+F116y' +F11w6y" +... +F111n16y1" 1)dx. ~ 322 II. THE CALCULUS OF \'ARIA 1101\;S Integrate the second summand on the right once ter111-by-term x, x. ~ F'" 6y' dx = [F11• oy];;- ~ :x F11- 6y dx, ~ ~ the third summand twice: ~ ~ S FII"6y"dx=[F11.6y']x,_ [-dd F11.6y]x'+S dd2 2 Fll"6ydx,· Xo X Xo X ~ ~ and so forth; the last summand, n times: x, SFyl"l 6y1" 1 dx = [F11,.., 6yw-ll]~~-l1x F111n16yln-ZIt:+ .. , x. Taking into account the boundary conditions, by virtue of which for x = x0 and for x = X1 , the variations 6y = 6y' = 6y" = ... = = 6y1" -ll = 0, we finally get Xo Since on the extremizing curve we have xs· ( d d2 d" ) bv= F11 -TxF11+ dx2 Fr~·+ ... +(-1)" dx" f,,.., 6ydx=0 for an arbitrary choice of the function 6y and since the first factor under the integral sign is a continuous function of x on the same curve y = y (x), it follows that by virtue of the fundamental lemma the first factor is identically zero: d d2 d" Fw- dx F11·+ dx2 F!l'+ ... +(-1)" dx" F111n,=O. Thus, the function y = y (x), which extremizes the functional x, v[y(x)]= ~ F(x, y, y', y", ••• t Xo must be a solution of the equation d d2 d" Fv- dx fv· + dx' Fll' + ... +{-1)" dx" Fvl"l = 0. 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 323 This differential equation of order 2n is called the Euler-Poisson equation, and its integral curves are termed extremals of the variational problem under consideration. The general solution of this equation contains 2n arbitrary constants, which, g~nerally speaking, may be determined from the 2n boundary conditions: Y(Xo) =Yo· y' (Xo) = y~, ... ' /ln-I) (Xo) = y~n-u; y(x1)=y., y'(x1)=y;, ... , y) dx. Varying some one function y1(x) and holding the others fixed, we get the basic necessary condition for an extremum in the form d d11• F11,--d F ,+ ... +(-l)"i-11 F 111 .,=0 (i=l, 2, ... , m). X ~ dxl ~· 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 325 5. Functlonals Dependent on the Functions of Several Independent Variables Let us investigate the following functional for an extremum: v [z (x, y)] = SSF ( x, y, z, ~~ , ~~) dx dy; D the values of the function z (x, y) are given on the boundary C of domain D, that is, a spatial path (or contour) C is given, through z Z•l(Z,!/} ho~------------~~Y Flig. 6-13 which all permissible surfaces have to pass (Fig. 6.13). To abbreviate notation, put ~; = p, ~~ = q. We will consider the function F as three times differentiable. We assume the extremizing surface z = z (x, y) to be twice differentiable. Let us consider a one-parameter family of surfaces z = z (x, y, a)= = z (x, y) +a6x, where 6z = z(x, y)- z (x, y), including for a= 0 the surface z = z (x, y) on which the extremum is achieved, and for a= I, a certain permissible surface z = z (x, y). On functions of the family z (x, y, a), the functional v reduces to the function a, which has to have an extremum for a= 0; consequently, !v[z(x, y, a)]lm=o=O. If, in accordance with Sec. 1, we call the derivative of v [z (x, y, a)] with respect to a, for a=O, the variation of the functional and symbolize it by 6v, we will have 6v={:XSS F(x, y, z(x, y, a), p(x, y, a), q(x, y, a))dxdy},.t=O= D = ss[F.,6z +Fp6p+Fq6q]dxdy, n 326 II. THE CALCULUS OF V.'\IUATIONS where z (x. y, a.)= z (x, y) +a.6z, 1 ) 0Z (x, y, ct.) ( ) I f: P\X, y, a.= ox =p X, y -r-a.up, ( ) iJz (x, y, a.) ( ) + ·"q X, y, a. = iJy = q X. y a. uq. Since it follows that 5S(Fp6p+F9 6q)dxdy = D =55[! {Fp6z}+ :Y {Fq6z}ldxdy- D -5}[! {Fp} + :Y {Fq}] 6zdxdy, where :x {Fp} is the so-called total partial derivative with respect to x. When calculating it, y is assumed to be fixed, but the dependence of z, p and q upon x is taken into account: :x {Fp} = FP"+Fpz ~: +FPP ~: +Fpq~ and similarly 0 {F } F F iJz iJp iJq Ty q = qy + qz iJx +Fqp iJy +Fqq iJg • Using the familiar Green's function 55(~~+a::) dx dy= 5(Ndy-M dx) D C we get 55[! {FP6z} + :U {Fq 6z}Jdxdy= 5(FPdy-F9 dx) 6z =0. D C The last integral is equal to zero, since on the contour C the variation 6z = 0 because all permissible surfaces pass through one and 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 327 the same spatial contour C. Consequently, ss(Fp6p+Fq6q)dxdy=- SS[:x {Fp}+ :y {Fq}]6zdxdy, D D and the necessary condition for an extremum, H(Fz6z+FP6p+Fq6q)dxdy=0, D takes the form SS(Fz-:x {Fp}-:u {Fq})6zdxdy=0. D Since the variation 6z is arbitrary (only restrictions of a general nature are imposed on 6z that have to do with continuity and differentiability, vanishing on the contour C, etc.) and the first factor is continuous, it follows from the fundamental lemma (page 308) that on the extremizing surface z = z (x, y) a aFz- ax {Fp} -ay- {Fq} ==0. Consequently, z (x, y) is a solution of the equation iJ iJ Fz-CJX {Fp}-0y {Fq} =0. This second-order partial differential equation that must be satisfied by the extremizing function z (x, y) is called the Ostrogradsky equation after the celebrated Russian mathematician M. Ostrogradsky who in 1834 first obtained the equation (for rectangular domains D it had already appeared in the works of Euler). Example I. v[z(x, y))=ss [(~;y +(~;rJdxdy, D the values of the function z are given on the boundary C of the domain D: z = f (x, y). Here the Ostrogradsky equation is of the form iJ2z iJ2z ax2 +a;T=O or, in abbreviated notation, l!lz=O, which is the familiar Laplace equation; we have to find a solution, continuous in D, of this equation that takes on specified values on the boundary of the domain D. This is one of the basic problems of mathematical physics, called the Dirichlet problem. 328 II. THE CALCULUS OF VARIATIONS Example 2. v[z(x. y)] = ss[(~; r+(~; r+2zf(x, y)] dxdy. D a function z is given on the boundary of a domain D. Here, the Ostrogradsky equation is of the form iJ2z iJ2.: axz + iJy2 = f (x, y), or, in abbreviated notation, dz = f (x, y). This equation is called Poisson's equation and is also frequently encountered in problems of mathematical physics. Example 3. The problem of finding a surface of minimal area stretched over a given contour C reduces to investigating the func- tional s [z (x, y)] = J.fJ/1 + (:; r+(~;-rdxdy D for a minimum. Here the Ostrogradsky equation has the form fx {Jlt +:2+q2 }+ :U { Vt +:2+q2 }= 0 or ~z [t + (~)2]-2~~ ~z +~ [t ' (.?.~ \2] -0 iJx2 iJy ox iJy ox iJy iJy2 1 dx ) - ' i.e. at every point. the mean curvature is zero. It is known that soap bubbles stretched on a given contour C are a physical realization of minimal surfaces. For the functional v[z(x1, X2 , ••• , Xn)] = = ~ ~ ... ~ F (X1 , X2 , ••• , Xn, Z, p1 , p2 , ••• , Pn) dx, dx2 ••• dxn, D where P; = ::. , in quite the same fashion we obtain, from the I basic necessary condition for an extremum, 6v = 0, the following Ostrogradsky equation: n Fz- L a~- {Fp;} =0, i= I I 6. f\\ETHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 329 which the function z = Z (x1, X2 , • • • Xn) extremizing the functional v must satisfy. For example, for the functional V= 555[(~~r +( ~=r+ (~~ YJdxdydz D the Ostrogradsky equation is of the form d2u azu a2u axz + ayz + azz =0. If the integrand of the functional v depends on derivatives of higher order, then, by applying several times the transformations used in deriving the Ostrogradsky equation, we find, as the necessary condition for an extremum, that the extremizing function has to satisfy an equation similar to the Euler-Poisson equation (pages 322-333). For example, for the functional v [z (x y)] - 5SF (x y z j!_ az Qlz azz d2z_)dx dy' - ' ' ' ax ' 7fij ' iJxZ ' axay ' i)yZ D we get the equation a a az d2 az Fz-ax{Fp}-ay{Fq}+ axz {f,} +axay{Fs} + ayz {ft}=O, where ilz az iJ2z iJ2z iJ2z p = ax , q = 7iii , r = axz • s = ax ay , t = iJy2 . This fourth-order partial differential equation must be satisfied by the function extremizing the functional tl. For example, for the functional V= 55[(g:~r+(:U!r+2 (a:;uY]dxdy D the extremizing function z must satisfy the so-called biharmonic equation which is ordinarily written briefly as L\L\z=O. For the functional V= 55 [(:;:r+ (g;!-r+2 (a~~yr -2zf(x, y)]dxdy D 330 II. THE CALCULUS OF VARIATIONS the extremizing function z (x, y) must satisfy the equation !J. !J.z = f(x, y). The biharmonic equation is also involved in extremum problems of the functional V= ss(~:! +~;:rdxdy D or the more general functional v= SS{(:!+ ~;~)2 - 2 (1- ~) [ ~:~ ~;~ - (a~2~Y)2 J}dxdy•D where ~ is the parameter. 6. Variational Problems in Parametric Form In many variational problems the solution is more conveniently sought in parametric form. For example, in the isoperimetric problem (see page 295) of finding a closed curve of given length l bounding a maximum area S, it is incon!/ venient to seek the solution in --~----------------~X0 the form y = y (x), since by the very meaning of the problem the function y (x) is ambiguous (Fig. 6.14). Therefore, in this problem it is advisable to seek the solution in parametric form: x = x (t), y = y (t). Hence, in the given case we have to seek the extremum of the functionalFig. 6-14 T ) n . . S fx(t), y(t)] = 2 \ (xy-yx)dt • 0 T provided that l = ~ vX2 +il dt. where l is a constant. 0 • In the investigation of a certain functional x, v[y(x)]=~F(x, y, y')dx Xo for an extremum let it be more advisable to seek the solution in the parametric form x = x (t), y = y (t); then the functional will be reduced to the following form: t~(x(t), y(t)]=SF(x(t), y(t), t~!~)x(t)dt. to 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 331 Note that after transformation of the variables, the integrand F(x(t), y (t), ~U>) x(t) x(l) does not contain t explicitly and, with respect to the variables xand y, is a homogeneous function of the first degree.· Thus, the functional v [x (t), y (t)] is not an arbitrary functional of the form I, ~ (t, x(t), y(t), x(t), y(t)) dt to that depends on two functions x (t) and y (t), but only an extremely particular case of such a functional, since its integrand does not contain t explicitly and is a homogeneous function of the first degree iu the variables xand y. If we were to go over to any other parametric representation of the desired curve x = x (-r), y = y (-r), then the functional v [x, y) would be reduced to the form sF(X, y, t:)x~d't. Hence, the in- 'to tegrand of the functional v does not change its form when the parametric representation of the curve is changed. Thus, the functional v depends on the type of curve and not on its parametric represen- tation. It is easy to see the truth of the following assertion: if the integrand of the functional t, v [x (t), y (t)] = ~ (t, x (t), y (t), x(I), y(t)) dt '·does not contain t e·~plicitly and is a homogeneous function of the first degree in xand y, then the functional v [x (t), y (t)] depends solely on the kind of curve x = x (t), y = y (t), and not on its parametric representation. Indeed, let t, v[x(t), y(t))= ~ (x(t), y(t), x(t), y(t)) dt, t. where <1>( . ") ( . ")X, y, kx, ky =k X, y, X, y. Let us pass to a new parametric representation putting 't=q>(/) (q>(/):;o6:0), X=X(T), y=y(T). 332 II. THE CALCULUS OF VARIATIONS Then '· '• '(J> (x (I), y (t). X(I), y (t)) dt = s(J>(x(-r), y('t), x,(-r) is a homogeneous function of the first degree in xand y, we have whence t, ~ S(J> (x, y, Xt, y,) dt = S is a homogeneous function of the first degree in xand y, and also for functionals with an arbitrary integrand function "-dt 0. 6. METHOD OF VA~IATIONS IN P~OBLEMS WITH FIXED BOUNDA~IES 333 of a solution of a system of differential equations. This is an indication that for functionals of the form t, v[x(t), y(t)]=~u] dxdt, 10 o and the equation of forced vibrations of the string is :1 (pui)-:x ( ku~) - pf (t, x) = 0, or, if the string is homogeneous, a2u k aju atj -P axj = f(t, x). We can similarly obtain the equation of a vibrating membrane. Example 3. Let us derive the equation of vibrations of a rectilinear bar. Direct the axis of abscissas along the axis of the bar in the equilibrium position. A deviation from the equilibrium position u (x, t) will be a function of x and of the time t, the kinetic energy of a bar of length l is I T=; Spu?dx. 0 We a~sume the bar to be inextensible. The potential energy of an elastic bar with constant curvature is proportional to the square of the curvature. Consequently, the differential dU of the potential energy of the bar is I { ~ }a dU=2k [t+(:~rr~~ . and the potential energy of the whole bar, the curvature of the axis of which, generally speaking, is variable, will be u-! t{[1+ ~rr.fdx. Suppose that the deviations of the bar from the equilibrium position are small and the term ( ~; r in the denominator may be neglected; then I 1 5 (aau )'u=2 k ax2 dx. 0 6. METHOD OF VARIAfiONS IN PROBLEMS WITH FIXED BOUNDARIES 337 --- The Ostrogradsky-Harnilton integral is of the form t, ' SS [~ pu?- { ku~i] dxdt. i 0 0 Consequently, in the case of the free vibrations of an elastic bar we will have the following equation of motion: ! (pui) +:;2(ku~x) = 0. If the bar is homogt>neous, then p and k are constants, and the equation of vibrations of the bar is transformed to iJ2u iJ'u p atz + k ax' =0. If the bar is acted upon by an external force f (t, x), then we also have to take into account the potential of this force (see preceding example). The principle of least action may be applied in deriving field equations. Consider the scalar, vector or tensor field w=w(x, y, t, z, t). The integral ~ (T- U) dt here will, generally speaking, be t, equal to the quadruple integral taken over the space coordinates x, y, z and the time t of some function L, called the density of the Lagrange function or the Lagrangian. Ow iJw d~ Ordinarily, the Lagrangian is a function of w, ax, Ty, Tz, aw ar= ( iJw iJw aw iJw ) L = L w, 7fX, Ty, Tz , Tt , and, therefore, the action is of the form SSSSL (w, ~~ , ~~ , Z, Z)dxdydzdt. (6.3) D According to the principle of least action, the field equatiou is the Ostrogradsky equation for the functional (6.3): a a d a Lw- ax {Lp,} - dy {Lp.}- az {LpJ- Tt {Lp.} =0, where 338 II. THE CALCULUS OF VARIATIONS PROBLEM~ ON CHAPTER ti 1. Find the extremals of the functional v[y(x))= sv~ dx. ~. 2. Test for an extremum the functional v [!J (x)) = ~ (y~ +2xyy') dx; y (X0 ) =Yo; y (x,) = y1• 3. Test for an extremum the functional I v [!J (x)) = ~ (xy+ y1 -2y'y')dx; y(O) = 1; y( 1) =2. 0 4. Find the extremals of the functional "'• v [y (x)) = ~ y' (1 +x1y') dx. ro 5. Find the extremals of the functional .. v[y(x)]= ~ (y''+2yy'-l6y1 )dx. •• 6. Find the extremals of the functional r, v[y(x)]= ~ (xy' +y'')dx. ~. 7. Find the extremals of the functional x, Sl+!fv [y (x)) = -,2- dx.y ro 8. Find the extremals of the functional x, v [y (x)] = ~ (y2 +y''-2y sin x) dx. Xo 9. Find the extremals of the functional •• v[y(x)) = ~ (16l-y"2 +x2 )dx. r. 6. METHOD OF VARIATIONS IN PROBLEMS WITH FIXED BOUNDARIES 3:l9 10. Find the extremals of the functional •• v[y(x)] = ~ (2xy+y'"2 )dx. "• II. Find the extremals of the functional <, v[y(x), z(x)]= ~ (2yz-2y2 +y'2 -z'2 )dx. "• 12. Write the Ostrogradsky equation for the functional v[z(x, y)J=S S[(~;Y-(~;Y]dxdy. D 13. Write the Ostrogradsky equation for the functional •(u(x, y, z)]=sss [( :~ r+( ~~ r+( ~~ r+2uf(x, y, z)] dxdydz. D 14. Find the extremals of the functional r '2 v[y (x)) = J ~3 dx. ... 15. Find the extremals of the functional •• v [y (x)] = ~ (y2 +y'' +2yex) dx. "• 16. Find the extremals of the functional <, v [y (x)) = ~ (y1 -y'2 -2y sin x) dx. r equation contains two arbitrary constants, for the determination of which we need two conditions. In the problem with fixed boundary points, these conditions were y (x0 ) =Yo and y (X1) = y1• In the moving-boundary problem, one or both of these conditions are absent and the missing conditions for a determination of the arbitrary constants of the genera! solution of Euler's equation have to be obtained from the basic necessary condition for an extremum which is .that the variation 6v be equal to zero. Since in the moving-boundary problem an extremum is attained only on solutions y = y (x, C1, C,) of the Euler equation, from now on we can consider the value of the functional only on functions 342 II. THE CALCULUS OF VARIATIONS of this family Then the functional v [y (x, CI, C2)] is reduced to a function of the parameters CI and c2 and of the limits of integration x0 and xi, while the variation of the functional coincides with the differential of this function. For the purpose of simplifi· cation, we shall assume that one of the boundary points, say (x0 , y0), is fixed, and the other (xi, y1) can be moved and passes to point (X1 +~xi> y1 +~y1) or, as ordinarily denoted in the calculus of ~,+G.r,,g,+ay,J , . - - •MI IJ !I 0 Fig. 7-1 variations, (xi+ 6x1, Y1 +6yi). We will call the permissi· ble curves y = y (x) and y = = y (x) +f>y neighbouring if the absolute values of the variations 6y and 6y' are small, and the absolute values of the increments 6xi and 6y1 are also small (the increments 6xi and 6yi are ordinarily called variations of the limit values X1 and y1). The extremals passing through the point (x0 , y0) form a pencil of extremals y = y (x, C1). The functional v [y (x, CI)] on the curves of this pencil is reduced to a function of C1 and X1 If the curves of the pencil y = y (x, CI) in the neighbourhood of the extremal under consideration do not intersect, then v [y (x, C1)] may be regarded as a one-valued function of X1 and y1 since specification of xi and y1 determines tht! extremal of the pencil (Fig. 7.1) and thus determines the value of the functional. Let us compute the variation of the functional v [y (x, C1)] on the extremaIs of the pencil y = y (x, C1) when the boundary point is displaced from the position (xi, y1) to the position (x1 +6x1 , y1 +{)y1). Since the functional v on the curves of the pencil is reduced to a function of xi and Y~> its variation coincides with the differential of this function. From the increment ~v. remove the principal part that is linear in 6x1 and {)y1 : x1 +0-'- x, ~v= ~ F(x, y+{)y, y'+{jy')dx-~ F(x, y, y')dx= x,+Ox, = ~ F(x, y+{)y, y' +f>y')dx + x, x, + ~ [F(x, y+{)y, y'+fJy')-F(x, y, y')]dx (7.1) ... 7. VARIATION,\L PROBLEMS WITH MOVING BOUNDARIES 343 Using the mean-value theorem, transform the first term on the right: x1+0x1 ~ F(x, y+fly, y'+fly')dx=Fix=x,+60x,6x1 , where 0<8< 1, x, by virtue of the continuity of the function F we will have Fix=x,+60x, = F(x, y, y') lx=x, +e1, where e1 __... 0 as 6x1 __... 0 and 6y1 ---. 0. Thus ~,+llx, ~ F(x, y+ 6y, y' +6y') dx = F (x, y, y') lx=x, 6x1 +e, 6x1• x, We transform the second term on the right-hand side of (7.1) by means of Taylor's expansion of the integrand: x, ~ [F(x, y+6y, y'+6y')-F(x, y, y')]dx= Xo x, = ~ [Fy(x, y, y')6y+F11·(x, y, y')6y']dx+R1• Xo where R1 is an infinitesimal of higher order than 6y or 6y'. In turn, the linear part x, ~ (Fy6y +F11·6y') dx Xo may be transformed, by integrating by parts the second summand of the integrand, to the form x, [F11·6y]~~+ 5(fy- :xf11,J 6ydx. Xo The values of the functional are only taken on extremals, he11ce Fy-! F11• =0. Since the boundary point (x0 , y0 ) is fixed, it follows that 6y lx=x. = 0. Hence x, ~ (FY 6y+ F11• 6y') dx = rf 11• 6y] ix=x,. Xo Note that 6y lx=x, is not equal to 6y1, the increment of y., since 6y1 is the increment of y, when the boundary point is displ~ced 344 II. THE CALCl)LlJS OF VARIATIONS to the position (x1 +Bx1 , y1 +6y1), and By lx;x, is the increment of the ordinate at the point x1 when going from the extremal passing through the points (x," y0 ) and (x1 , y,) to the extremal passing through the points (x0 , Yo) and (x1 +Bx1 , Y1 +By1) (Fig. 7.2). From the figure it is clear that BD =By lx=x,; FC = By1; EC:::::::; y' (x1 ) 6x1; BD = FC- EC or Here, the approximate equality holds true to within infinitesimals of higher order. cfx,...&:r,,y,...s!l:l And so we finally have [ x1 +6x, ~x, x, R ifF(x, y+f>y, y'+6y')- Xo -F (x, y, y'}] dx:::::::; 0 .r, .r,- .r, :::::::; fll, lx=x1 ·(f>yl-y' (xt)llxt), Fig. 7-2 where the approximate equalities likewise hold true to within terms higher than the first in 6x1 and 6y1 • Hence, from (7.1) we get or llv= F lx=x, 6x1 +F11• lx=x, (6y1 - y' (X1) 6x1 ) = = (F-y' f",) lx=x, 6x1 +Fy· lx=x, 6y1, dv (x1, y1) = ( F- y' F11·) lx=x, dx1 +Fy' lx=x, dy1 , where v(x1, y1) is a function into which the functional v was transformed on the extremals y = y(x, C1), and dx1 = ~x1 = 6x1 , dy1 = ~y1 = 6y1 are increments of the coordinates of the boundary point. The basic necessary condition for an extremum 6v = 0 takes the form (f-y' f 11·) lx=x, 6xl +f y' lx=x, l\yl = 0. (7.2) If the variations llx1 and 6y1 are independent, then it follows that (f- y'F'") lx=x, = 0 and F11• lx=x, = 0. However, more often one has to consider the case when the variations 6x1 and f% are dependent. For instance, allow the right boundary point (x1, y1) to move along a certain curve 7. VARIATIONAL PROBLEMS WITH MOVING BOUNDARIES 345 Then 6y1 ~q>'(x1)6x1 and, consequently, the condition (7.2) takes the form (F+(q>'-y')Fy•J6x1 =0 or, since 6x1 varies arbitrarily, it follows that [F +(q>' -y') Fy•)x=x, = 0. This condition establishes a relationship between the slopes of q>' and y' at the boundary point. It is called the transuersality condition. The transversality condition together with the condition y1 = q> (x1) generally speaking enables us to determine one or several extremals of the pencil y = y (x, C1) on which an extremum may be achieved. If the boundary point (x0 , Yo) can move along some curve Yo= 'ljJ (x0), then in the very same way we will find that the transversality condition [F +(ljJ'- y') Fy•)x=x• = 0 must be satisfied also at the point (x0 , y0). Example t. Find the transversality condition for functionals of the form x, v = ) A (x, yq(1+y'1 dx. Xo The transversality condition F+Fy,('y'1 = 0; Y I +y'1 VI+ y'1 assuming that A (x, y) =1= 0 at the boundary point, we get l +y' in this case reduced to the orthogonality condition. Example 2. Test for an extremum the functional xf Vi+"?" dx ~ y 0 given that y(O) =0 and y1 =X1 -5 (Fig. 7.3). The integral curves of the Euler equation (Problem 1, page 338) are the circles (x-C1)'+y2 =q. The first boundary condition yields C1 =C2 • Since for· the given functional the transversality condition reduces to the orthogonality condition (see preceding example), it follows that the straight line y1 = x1 --5 must be a diameter of the circle, and, hence, the centre of the desired circle lies at the point (0, 5), where the straight line y1 = x1 - 5 meets the axis of abscissas. Consequently, (x-5)2 +y2 = 25 or y = ± Y10x-x2• And so, an extremum can be achieved only on arcs of the circle y = VTox -x• andy=- Yl0x-x2 • 346 II. THE CALCULUS OF VARIATIONS If the boundary point (x1 , y1) can move only along a vertical straight line (Fig. 7.4) and hence 6x1 = 0, then the condition (7.2) passes into Fy' lx=x, = 0. For example, in the problem of the brachistochrone (see pages 316-317) let the left boundary point be fixed and the right allowed to !I y II :r: 0 :r=:r:, Fig. 7-3. Fig. 7-4 move along a vertical straight line. The extremals of the functional V= rvrdx are cycloids, the equations of which, if the con- 0 IJ dition y (0) = 0 is taken into consideration, will be of the form x=C1 (t-sin t), y= cl (1-cos t). To determine C1, use the condition Fy·!x=x, = 0, which in the given case is of the form y· I =0 VIJ ~ x=x1 ' whence y' (x1) = 0; that is, the desired cycloid must intersect the straight line x=x1 at right angles and, hence, the point x=x1, y = y1 must be a cusp of the cycloid (Fig. 7.5). Since to the cusp there corresponds the value t = n, it follows that x1 = C1n, C1 = ~~ . Hence, an extremum can be achieved only on the cycloid X=~ (t-sin t); y=~(l-cost). a 7. VARIATIONAL PROBLEMS WITH MOVING BOUNDARIES :l47 If the boundary point (x~' y1) in the problem of the extrt-mum .r, of the functional v = ~ F (x, y, y') dx is permitted to move along Xo 0 .r B !I :r·:r, Fig. 7-5 the horizontal straight line y = y1, then {% = 0 and the condition (7.2), or the transversality condition, takes the form [F-y'Fy'}x=x, =0. 2. The Moving-Boundary Problem for a Functional .r, of the Form ~ F(x, y, z, y', z') dx If in investigating the functional x, v= ~ F(x, y, z. y', z')dx Xo for an extremum, one of the boundary points, say B (x1 , y1 , z1) is moved, and the other, A (x0 , y0 , Z0), is fixed (or both boundary points are movable), then it is obvious that an extremum may be achieved only on the integral curves of th~ system of Euler's equations d d F,- dxp!l, =0; Fz- dxFz' =0. Indeed, if on a certain curve C an extremum is achieved in the moving-boundary problem, i.e. a maximum or minimum value of v is obtained compared with the values of v on all close-lying permissible curves, which include both those curves that have common boundary points with the extremizing curve C and also those curves whose boundary points do not coincide with the boundary points of C, then an extremum is surely achieved on the curve 348 II. THE CALCULUS OF VARIATIONS C with respect to a narrower class of neighbouring curves having common boundary points with the curve C. Hence, the necessary conditions for an extremum in the problem with fixed boundary points must be satisfied on the curve C, and in particular, the curve C must be an integral curve of a system of Euler's equations. The general solution of the system of Euler's eq,uations contains four arbitrary constants. Knowing the coordinates of the boundary point A (x0, Yo• Z0), which we consider to be fixed, it is possible, generally speaking, to eliminate two arbitrary constants. To determine the other two arbitrary constants we have to have another two equations, which will be obtained from the condition l>t.• = 0; note that in computing the variation we will now assume that the functional is specified only on solutions of the system of Euler's equations, for an extremum is attainable only on them. Then the functional v is reduced to the function $ (x1, y1 , Z1) of the coordinates x1 , y1 , Z1 of the point B (x1 , y1 , Z1), and the variation of the functional is reduced to the differential of this function. • The calculation of the variation of v may be performed exactly as indicated on pages 342-344: x1 +6x, ~v= ~ F(x, y+fJy, z+Bz, y'+fJy', z'+fJz')dx- "'• x, +llx, = sx, x, "'• - ~ F(x, y, z, y', z')dx= "'• F (x, y+fJy, z+fJz, y' + 6y', z' + 6z') dx+ + ~ [F(x, y+fJy, z+6z, y'+fJy', z'+6z')- x. -F (X, q, Z, y', z')] dx. Apply the mean-value theorem to the first integral and take advantage of the continuity of the function F; in the second integral isolate the principal linear part by means of Taylor's formula. These transformations yield x, l>v=Fix=x, 6x1 + ~ [FyfJy+Fz6z-+f11,6y'+f?,fJz'Jdx -"• • The function Cll will be single·valued if the extremals of the pencil centred at A do not intersect, for then the point 8 (x1, y1, z1) uniquely defines an extremal. 7. VARI!\TIONAL PROBLEMS WITH MOVING BOUNDARIES 349 Integrating by parts the last two terms under the integral s1gn yields 6v= F k=x, 6x1 + [F11• 6y}x=x, +[fz, 6z}x=x, + "• +s[(Fy-:x F11·) 6y +(Fz-(J~Fz·) 6z'] dx. ... Since the values of v are calculated only on extremals, it follows that and hence, 6v = F lx=x, 6x1 + [FII' 6y)x=x, + [Fz' 6z)x=x,• Arguing in the same way as on page 344, we get 6y lx=x, ~ 6y,- y' (x1) 6x1 and 6z lx=x, ~ 6z1 - z' (x1) 6x1, and, consequently, 6v = [F -y'Fy'- z'Fz·]x=x, 6xl +FU' lx=x, 6yl +fz· lx=x, 6zl = 0. If the variations 6x1 , 6y1 , 6z1 are independent, then from the condition 6v=0 we have [F-y'F11·-z'Fz•]x=x, =0; F11•lx=x, =0 and fz· lx=x, =0. If the boundary point B (x1 , y1 , z1) can move along some curve y1 =

'- y') F11• +('ljl' -z') Fz• )x=x, = 0. This condition is called the transversality condition in the problem of investigating the functional x, v= ~ F (x, y, z, y', z')dx for an extremum. Together with the equations y1 = (x1, y1), then flz1 = q>x,flx1 +q>y,fly1 and the variations flx1 and l5y1 are arbitrary. Consequently, the condition flv= 0 or, in expanded form, [F -y1 Fll' -Z1 F z']x=x, flx1 + ft/ Ix=x, f>y1 + fz• Ix=x, flz1 = 0 is transformed to the condition [f-y~FI( -Z1 Fr +q>;Fz]x=x, <'>X1 + [Fv· + Fz·q>~}x=x, fly1 = 0. Whence, since 6x1 and fly1 are independent, we get [f-y1 F11 +(q>~ -z') FrJx=x, = 0, [Fv + Fz'{jl~]x=x, = 0. These two conditions, together with the equation Z1 = q> (x1 , y1), generally speaking, enable one to determine two arbitrary constants in the general solution of the system of Euler's equations If the boundary point A (x0 , y0 , Z11 ) is moving, then by the same method at this point we get similar conditions. If we consider the functional "'• V= ~ F (X, Y1 , y,, ••. , Yn• y;, y;, ••• , y~) dx, then, without changing the proof procedure, we find that in the case of the moving point B (x1• y111 Yw ... , y,1) at this point ( f- ~ y;F v·.)l flx1 + t F ,/, 6y;1 =0. l=l t x=x, l=1 ' xa:x, Example t. Find the transversality condition for the functional x, v = ~ A (x, y, z) V1+y'2 +Z11 dx if Z1 = q> (X1 , y1). Xo The transversalitv conditions [F -Y1 F11• +(QJ~ -z') F z•)x•z, = 0 and [Fv + Fz'QJ~]z=z, = 0 in this case are of the form 1+QJ~z' = 0 and y' +q>~z~ = 0 for x = x1 I I I or., =J.=~ for x=x1 , that is, they are the condition for 'Px cpy parallelism of the vector of the tangent t (1, yl, z') to the desired extremal at the point (x1, y1, z1) and the vector of the normal N(q>~. q>~, - 1) to the surface z = q> (x, y) at the same point. Hence, in this case, the transversality condition becomes an orthogonality condition of the extremal to the surface Z=QJ(X, y). 7. VARIATIONAL PROBLEMS WITH MOVING BOUNDARIES 351 Example 2. Find the extremal distance between two surfaces z=

(x0 , y0 ) and the coordinates of the other boundary point (x1, y1, z1) satisfy the equation Z1 = ljl (x1, y1). Since the integrand depends solely on y' and z', the extremaIs ate straight lines (Example 2, page 320). Since the functional x, ~ V1+ y'• + z'1 dx is a particular case of the functiong_l Xo x, ~ A (x, y, z) Jt-, + y'1 + z'1 dx considered in the preceding example, Xo the transversality conditions both at the point (x0 , y0 , Z0 ) and at the point (x1 , y1 , z1 ) pass into orthogonality conditions. Hence, an extremum can be achieved only on straight lines that are orthogonal both to the surface z=

x1 = 0 and 6y1 and 6z1 are arbitrary. In the example under consideration, F11 = 2y'; Fz· = 2z', hence y' (x1) = 0 and z' (X1 ) = 0 352 II. THE CALCULUS OF VARIATIONS or If cosx1 :;6 0, then C, =c.= 0 and an extremum may be achieved only on the straight line y=O; z =0; but if cosx1 =0, i.e., x1 = ~ +nn, where n is an integer. then C, = 0, C4 is an arbitrary constant, y =C4 sin x, z =-c. sin x. It is easy to verify that in the last case the functional v= 0 for any C4• 3. Extremals with Corners Up till now we have considered variational problems in which the desired function y = y (x) was assumed continuous with a continuous derivative. However, in ma11y problems the latter requireFig. 7-7 ment is not natural; what is more, in certain classes of variatiog·u{zJ nal problems the solution is, as a :1· rule, attained on extremals having corner points. Such problems include for instance problems involving the reflection and refraction of extremals, which are a generalization of the corresponding problems involving the reflection and refraction of light. The reflection-of-extremals problem. Find the curve that extremizes the functional v = rF (X, y, y') dx and passes through the given ·.r. points A (x0 , y0) and 8 (x,, y,); the curve must arrive at B only after being reflected from a given line y = cp (x) (Fig. 7.7). It is natural to assume that at the point of reflection C (x1 , y1) there can be a corner point of the desired extremal and, consequently, at this point the left-hand derivative y' (x1 -0) and the right-hand derivative y' (x1 +0) are, generally speaking, distinct. It is therefore more convenient to represent the functional v [y (x)] in the form X1 Xt v [y (x)] = ~ F (x, y, y') dx + ~ F (x, y, y') dx; J'o :r. here, on each of the intervals x0 (x), and the boundary points A and B are located on different sides of the discontinuity line (Fig. 7.10). 356 II. THE CALCULUS OF VARIATIONS Represent the functional v in the form x, X:.: v= ~ F1 (x, y; y')dx+ ~ F,(x, y, y')dx, x, where F1 (x, y, y') = F (x, y, y') is on one side of the discontinuity line, and F, (x, y, y') = F (x, y, y') is on the other side. Suppose that f 1 and F1 are three times differentiable. It is natural to expect a corner point at the point C of the intersection of the desired curve with the discontinuity line. Arcs AC and CB are obviously extremals (this again follows from the fact that by fixing one of these arcs and varying the other alone we get a problem with fixed boundary points). For this reason, for the comparison curves we can take only polygonal lines consisting of two arcs of extremals, and then the variation, because of the movable nature of the boundary point C (x1, y1) that is translated along the curve y = cp (x), wi II take the following form (see page 345): .l'a Xa 6v= 6 ~ f 1 (x, y, y') dx+ 6 ~ F, (x, y, y') dx = x0 -x1 = [fl +(cp' -y') F tv']x=x,-o 6X1- [f, +(cp' -y') F :y]x=x,+o 6x1> and the basic necessary condition for an extremum, 6v = 0, reduces to the equality [fl +(cp'- y') Ftv'] x=x1 - o = [f2+(cp'- y') Fa.v'] x=x1 +o· Since only y' can be discontinuous at the point of refraction, this refraction condition may be written as follows: Fttx1, Y1. y' (x~-0)) + +(cp' (X1) - y' (x1 -0)) F IV (x1, y1 , y' (X1 -0}) = = F, (x1, y1, y' (x1 +0)) + +(cp' (X1) - y' (X1 +0)) F sg' (X1, y1, y' (x1 +0)). The refraction condition, together with the equation y1 = cp (x1), makes it possible to determine the coordinates of the point C. If, in particular, the functional v is equal to x. ~A (x, y) V1+"y'1 dx= Xo x, x. = ~ A1 (x, y) V 1+y'• dx + ~ A, (x, y) V 1+ y'1 dx, ~. x, 7. VARIATIONAL PROBLEMS WITH MOVING BOUNDARIES 357 then the refraction condition takes the form A1 (x, y) q> 11 =A, (x, y) q> II ,1+ .. I •+ .. IV1 +11'1 x=x,-o V1 +11'1 x=x,+o or, retaining the notation of pages 354-355, y' (x1 - 0) =tan fi1, y' (x1 +0) =tan~~· «p' (x1) =tan a; after simplifying and multiplying by cos a we get sin [..::_-(a-~1)]cos (a- ~1) A2 (x1, g1) 2 A1 (x1, g1) cos (a- ~1) = A1 (x1, g1) or . [ n ( A >] = A1 (x1, g1)' sm T- a-.,1 which is a generalization of the familiar law of the refraction of light: the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities 1 I v1 (x, y) =A ( ) and v11 (x, y) =A < ) (compare page 354) 1 X, 1/ I X, 1/ in media, on the boundary between which the refraction occurs. One should not think that extremals with corners only occur in problems of refraction or reflection of extremals. An extremum may be achieved on extremals with corner points even in extremum x, problems of the functional v= SF (x, y, y') dx, where the function Xu F is three times differentiable, and the permissible curves must pass through the boundary points A and B without any supplementary conditions whatsoever. Let us investigate, say, the functional I v=Su''(l-y')'dx, y(O)=O; y(2)=1. Since the integrand is positive, v ~ 0 and hence if the functional v = 0 on some curve, it follows that on this curve there will definitely be achieved the absolute minimum of the functional v, that is, the least value of the functional on the permissible curves. It will readily be seen that on the polygonal line y = x for O~x~ l and y= 1 for 1 lx=x,+O 6xl -F,,. lx=x,+o ()yl = 0, whence (F -y'Fy•) lx=x,-u <'IX1 +Fy' lx=x,-o 6y1 = = (F- y' Fy') lx=x,+o 6xl +Fy' lx=x,+O 6y1, or, since 6x1 and 6y1 are independent, we have (F -y'Fv) lx=x,-o = (F-y' fv) lx=x1 +o and Flf !x=x,-o = Fv lx=x,+o· These conditions, together with the continuity conditions of the desired extremal, permit determining the coordinates of the corner point. Example t. Find the broken-line extremals (if they exist) of the a functional v =) (y'3 -y2) dx. Write the second of the conditions 0 that must be fulfilled at the point of inflection, Fy· lx=x,-o = = fu·lx=x,+o or, in the given case, 2y'(x1 -0)=2y'(x1 +0), whence y' (x1 -0) = y' (x1 +0); that is, the derivative y' at the point X1 is continuous and there is no point of inflection. Hence, in the problem at hand an extremum may be reached only on smooth curves, Example 2. Find the broken-line extremals of the functional x. v = ~ y'2 ( 1-y')2 dx. Since the integrand depends only on y', the ... extremals are the straight lines y=Cx+C (see page 314). In this case the conditions at the point of deflection take the form -y'2 (1-y') (1-3y') lx=x,-o = -y'1 (1-y') (l-3y') lx=x,+o and 2y' (1-y') (1-2y') lx=x,-o = 2y' (1-y') (1-2y') lx=41 +o• These conditions, if one disregards the trivial possibility y' (x1 -0) = y' (x1 +0), are satisfied for y' (X1 -0) =0 and y' (X1 +0) = 1 360 II. THE CALCULUS OF VARIATIONS or and y' (x1 -0) =I y' (x1 +0)=0. Fig. 7-13 Consequently, broken-line extremals may consist solely of segments of straight Jines that belong to the families y = C1 and y =X+C1 (Fig. 7.13). 4. One-Sided Variations In certain variational problems involving an extremum of a functional v [y (x)], a restriction may be imposed on the class of permissible curves that prohibits them from passing through points of !I A 0 Fig. 7-14 a certain region R bounded by the curve (x, y) = 0 (Fig 7.14). In these problems the extremizing curve C either passes completely outside the boundaries of the region R, and then it must be an extremal, since in this case the presence of the prohibited region R does not in the least affect the properties of the fuuctional and 7. VARIATIONAL PROBLEMS WITH MOVING BOUNDARIES 361 its variations in the neighbourhood of the curve C, and the arguments in Chapter 6 hold true, or C consists of arcs lying outside the boundary of R and also consists of parts of the boundary of the region R. In this latter instance, a new situation arises; only one-sided variations of the curve C are possiblt~ on parts of the boundary of the region R, since permissible curves cannot enter the region. Parts of the curve C that lie outside the boundary of R must, as before, be extremals, since if we vary the curve C only on such a segment that permits two-sided variations, the presence of the region R will not affect the variations of y, and th~ conclusions of Chanter 6 continue to hold true. !I Fig. 7-15 Thus, in the problem under consideration, an extremum can be reached only on curves consisting of arcs of extremals and parts d the boundary of the region R, and hence, in order to construct the desired extremizing curvE we have to obtain conditions, at the points of transition of the extremal to the boundary of the region R. which permit determining these points. In the case depicted .in Fig. 7.15, it is necessary to obtain conditions at the points M, N, P and Q. Let us, for instance, obtain a condition at the point M. In quite analogous fashion one could obtain conditions also at other points of transition of the extremal to the boundary of the region. When calculating the variation 6v of the functional x1 x x, v = ~ F (x, y, y') dx = ~ F(x, y, y') dx +~ F(x, y, y') dx Xo .Co x we can consider that the variation is caused solely by the displacement of the point M (x, y) on the curve «!) (x, y) = 0, i.e. it may be taken that for any position of the point M on the curve 362 II. THE CALCULUS OF VARIATIONS (x, y) = 0, the arc AM is already an extremal, and the segment MNPQB does not va~y. The functional -· X .. V1 = ~ F(x, y, y')dx has a boundary point moving along the boundary of the region R, whose equation is (x, y) = 0, or in the form solved for y in the neighbourhood of the point M: y = cp (x). Thus, according to Sec. 1 (page 344) <'>v1 = [ F -t- (cp'-y') Fy·]x == x 6X. x, The functional v2 = ~ F (x, y, y') dx also has a moving boundary X point (x, y). However, in the neighbourhood of this point the curve on which an extremum y = cp (x) can be achieved does not vary. Consequently, the variation of the functional v2 in the translation of point (x, y) to the position (t +6x, y+6y) only reduces to a change in the lower limit of integration and x. x, ~v2 = ~ F(x,y,y')dx-SF(x,y,y')dx= x+~x % x+~x x+o5x = - l F (x, y, y') dx = - l F (x, cp (x), cp' (x)) dx, X X since y= cp (x) on _the interval (x, x+6x). Applying the mean-value theorem and taking advantage of the continuity of the function F, we get ~v2 =- F (x, q> (x), cp' (x)) lx = x<'>x +~ ·<'>x, where 6-0 as <'>x-0. Consequently, <'>vz = -F (x, cp (x), cp' (x)) lx=x 6x, f>v=f>v1 +f>v2 = [F(x, y, y')+ +(cp'- y') Fu' (x, y, y')]x=x<'>x- F (x, y, rp') lx == x6x = =[F(x, y, y')-F(x, y, q>')-(y'-cp')Fu·(x, y, y')lx=x<'>x, since y (x) = cp (X}. Due to the arbitrary nature of 6x, the necessary condition for an extremum, f>v = 0, takes the form [F(x, y, y')-F(x, y, cp')-(y'-cp')Fu'(x, y, y')]x="i=O. 7. VARIATIONAL PROBLEMS WITH MOVING BOUNDARIES 363 Applying the mean·value theorem, we obtain (y'- q>')[F !f (x, y, q)- F !f (x, y, y')]x= x= 0, where q is a value intermediate between q>' (X) and y' (x). Again applying tht! mean-value theorem, we get (y' -q/) (q-y') F!f!f (x, y, q) lx =x= 0, where q is a value intermediate between q and y' (x). Suppose Fv'v (x, y, g)=1= 0. This supposition is natural for many variational problems ·(see Chapter 8). In this case the condition at the point M is of the form y' (x) = rp'(x) (q = y' only when y' (x) = rp' (x), since q is a value intermediate between y' (X) and q>' >Hence, at the point M the extremal AM and the boundary curve MN have a common tangent (the left tangent for the curve y = y (x) and the right tangent for the curve y = q> (x)). Thus, the extremal is tangent to the boundary of the region R at the point M. PROBLEMS ON CHAPTER 7 1. Find a solution with one corner point in the minimum problem of the functional 4 v[y(x)]=) (y'-l)2 (y'+l)2 dx; y(O)=O; y(4)=2. 0 2. Do solutions with corner points exist in the extremum problem of the functional x, v [y (x)] =) (y'2 +2xy-y2 )dx; Y (X0 ) =Yo; Y (xl) = Y1? X 3. Are there any solutions with corner points in the extremum problem of the functional x, v[y(x)]=) (y''-6y' 1 )dx; y(O)=O; y(x1)=y1? 0 4. Find the transversality condition for the functional x. - · - v [y (x)J= ) A (x, y) earctan !I~ I +y'2 dx, A (x, y) =I= 0. Xo 364 II. THE'CALCULUS OF VARIATIONS 5. Using the basic necessary condition for an extremum, llv = 0, find the function on which the functional can be extremized 1 v fy (x)] = ~ (y""- 2xy) dx; y (0) = y' (0) = 0, 0 y(l) = 1~; y' (I) is not given. 6. Find curves on which an extremum of the functional 10 v[y(x)]= ~ y'"dx; y(O)=O; y(IO)=O 0 can be achieved, provided that the permissible curves cannot pass inside a circle bounded by the circumference (x-5)2 +y2 = 9 7. Find a function on which an extremum of the functional !! 4 'Jly(x)]= ~ (y2 -y'1)dx; y(O)=O 0 can be achieved if the other boundary point is permitted to slide along the straight line x = ~ . 8. Using only the basic necessary condition llv = 0, find the curvP. on which an extremum of the functional x,v-.-,.v[y(x)]= ~ :u dx; y(O)=O 0 can be achieved if the second boundary point (x1, y1) can move along the circumfereuce (x- 9)2 +y2 = 9. CHAPTER 8 Sufficient conditions for an extremum t. Field of Extremals If on an·xy-plane, one and only one curve of the family y=y(x, C) passes through every point of a certain region D, then we say that this family of curves forms a field, more precisely, a proper field, in the region D. The slope of the tangent line p (x, y) to the curve of the family y = y (x, C) that passes through the point (x, y) is called the slope of the field at the point (x, y). For instance, inside the circle x1 +y1 ~ 1 the parallel straight linesy=x +C forma field (Fig. 8.1), the slope of which is p (x, y)= 1. On the contrary, the family of parabolas y=(x-a)1 -1 (Fig. 8.2) inside the same circle does not form a field since the parabolas of this family intersect inside the circle. If all the curves of the family y = y (x, C) pass through a certain point (x0 , y0), i.e. if they form a pencil of curves, then they defi- !I fig. 8·1 nitely do not form a proper field in the region D, if the centre of the pencil belongs to D. However, if the curves of the pencil cover the entire region D and do not intersect anywhere in this region, with the exception of the centre of the pencil, i.e. the requirements imposed on the field are fulfilled at all points other than the centre of the pencil, then we say that the family y = y (x, C) also fJrms a field, but in contrast to the proper field it is called a central field (Fig. 8.3). For example, the pencil of sinusoids y = C sin x for 0 ~ x ~a, a< :rt forms a central field (Fig 8.4). The very same pencil of sinusoids forms a proper field in a sufficiently small neighbourhoJd of the segment of the axis of abscissas 6 ~ x ~a, where 6 > 0, a<:rt (Fig. 8.4). The very same pencil of sinusoids does not form a field in the neighbourhood of the segment of the axis of abscissas O::;:;;;x~a 1 , a1 >:rt (Fig. 8.4). 366 II. THE CALCULUS OF VARIATIONS --------------------------------- If a proper field or a central fieid is formed by a family of extremals of a certain variational problem, then it is called an extremal field. \ \ \ \ '\ \ \ \ \ '\ \ \ \ \ \ \ \ \ !I Fig. 8-:? I I I I I I I I I I lr The field concept is extended almost without any changes to the case of a space of any number of dimensions. The family y1 = = y1 (x, C1 , ••• , Cn) (i = 1, 2, ... , n) forms a field in the region D of the space x, y1, ••• , Yn if through every point of D there passes !/ .r ;r: 0 Fig. 8-3 Fig. a-4 one and only one curve of the family y1 = y1 (x, Cu ... , Cn)· The partial derivatives of the functions y1(x, C1 , C2, ••• , Cn) with respect to x calculated at the point (x, Yu Yz• ... , Yn) are called functions of the slope of the field p1 (x, y1 , y2 , ••• , Yn) (i= 1, 2, •.., n); 8. SUFFICIENT CONDITIONS FOR AN EXTREMUM 367 hence, to obtam PdX, y1 , y2 , ••• , Yn) one has to take :Xydx, C17 C2 , •••• C") and replace C1 , C2 , ••• , Cn by their expressions given in terms of the coordinates x, y1, y2 , ••• , Yn· The central field is defined in similar fashion. Let the curve y = y (x) be an extremal of a variational problem involving the extremum of an elementary functional x, v [y (x)] = ~ F (x, y, y') dx, X. and let the boundary points A (x0 , y0) and B (x1, y1) be fixed. We say that the extremal y = y (x) is included in the extremal field if !/ lj .~ B :r: 0 0 Fig. 8-5 Fig. 8-6 a family of extremaIs y = y (x, C) has been found that forms a field containing the extremal y = y (x) for some value C = C0 , and the extremal y = y (x) does not lie on the boundary of the region D in which the family y = y (x, C) forms the field (Fig. 8.5). If a pencil of extremals centred at the point A (x0 , y0) forms a field in the neighbourhood of the extrernal y = y (x) that passes through this point, then the central field including the given extremal y = y (x) has thus been found. In the given case, for the parameter of the family we can take the slope of the tangent line to the curves of the pencil at the point A (x0 , y0) (Fig. 8.6). Example 1. Given the functional it is required to include the arc of the extremal y = 0 that counects the points (0, 0) and (a, 0), where 0 (X, y) = 0, then .•~~~~~~~~~~· close-lying curves of the ~ family y=y(x, C) will intersect near the curve ct> (x, y) = 0 and, in particular, the curves of this .r, Fig. 8-8 family that are close to the extremal under consideration y = y (x) that passes through the points A (x0 , Y0) and B (x1, y1), will intersect at points close to the points of tangency (or intersection) of the curve y = y (x} with the· C-discriminant curve (see Fig. 8.7, where the C-discrim1nant curve is depicted by a heavy line). If the arc AB of the extremal y = y (x) does not have common points (different from A) with the C-discriminant curve of the pencil of extremals that includes the given extremal, then extremals of the pencil sufficiently close to the arc AB do not intersect; that is to say, they form, in the neighbourhood of the arc AB, a central field that includes this arc (Fig. 8.8). 8. SUFFICIENT CONDITIONS FOR AN EXTREMUM 369 If the arc AB of the extremal y= y(x) has a point A* (different from A) in common with the C-discriminant curve of the pencil y=y(x, C), then curves of the pencil close to y=y(x) can intersect among themselves and with the curve y=y(x) near the point A*; generally speaking, they do not form a field (Fig. 8.7). The point A* is called a point that is conjugate to the point A. The result obtained may be formulated as follows: to construct a central field of extremais with centre at the point A. which field contains the arc AB of the extremal, it is sufficient for the point A*, conjugate to A, not to lie on the arc AB. This condition of the possibility of constructing a field of extremals including a given extremal is called the Jacobi condition. This condition may be readily stated analytically as \veil. Let y=y(x, C) be the equation of a pencil of extremals with centre at the point A, the parameter C being, for dt>finiteness, considered as coinciding with the slope y' of the extremals of the pencil at the point A. The C-discriminant curve is defined by the equations Y- y (x C)· cJy (x, C) = 0 - · • ac · Along every fixed curve of the family, the derivative ay ~c C) is a function of x alone. Denote this function briefly by u: u = iJy ~c C) , where C is given; whence u~ = iJ'Io~i:xaxC). The functions y = y (x, C) are solutions of the Euler equation. Therefore fv(x, y(x, C), y~(x, C))-fxF11 (x, y(x, C), y~(x, C))=O. Differentiating this identity with rt>spect to C and putting ay~c C)= u, we get or (Fyy-fxF11,/) u -fx(F,/y'u')=O. Here, FYY (x, y, y'), FYY' (x, y, y'), F!lu' (x, y, y') are known functions of x, since the second argument y is equal to a solution of Euler's equation y=y(x, C) taken for the value C=C0 that corresponds to the extremal AB. This homogeneous linear equation of the second order in u is called Jacobi's equation. If the solution of this equation u = ay ~[- C) , which vanishes at the centre of the pencil for x = X0 (the centre of the pencil always 370 II. THE CALCULUS OF VARIATIONS belongs to the C~discriminant curve) also vanishes at some point of the interval X0 < x < x1, then the point conjugate to A defined by the equations ( C ) d ay (x, C) 0 0 y = y x, 0 an ac = or u = lies on the arc AB of the extremal.• But if there exists a solution of the Jacobi equation that vanishes for x = x0 and that does not further vanish .at any point of the interval X0 ~ x ~ xl> then there are no points conjugate to A on the arc AB: the Jacobi condition is fulfilled, and the arc AB of the extremal may be included in the central field of extremals centred at the point A. Note. It may be proved that the Jacobi condition is necessary for achieving an extremum; i. e. for the extremizing curve AB the point conjugate to A cannot lie in the interval X0 < x < x1• Example 2. Is the Jacobi condition fulfilled for the extremal of IJ the functional v = ~ (y'1 - y2) dx that passes through the points 0 A (0, 0) and B (a, 0)? The Jacobi equation has the form -2u-!(2u')=0 or u•+u=O, whence u = C1 sin (x-C,). Since u (0) = 0, it follows that c.=0; u = clsin X. The function u vanishes at the points x = kn, where k is an integer, and, hence, if 0 0, b > 0. fhe straight lines y = C1x+C, are extremals. An extremum may be achieved only on the straight line y = !x. The pencil of straight lines y = C1x centred at the point (0, 0) forms a central field b !I that includes the extremal y=- xa (Fig. 8.10). The function E(x, y, p, y')=y'8 -p8 -~p• (y' -p) = (y' -p)' (y' +2p). b On the extremal y = - x, the slope of --~~----------~~z a o the field p= ~ > 0, and if y' assumes b Fi~. 8-10 values close to p=a, then E ~ 0, and, hence, all the conditions that are sutficient for achieving a weak minimum are fulfilled. Thus, a week minimum is achieved on the extremal y=!!.. x. But ifa y' takes on arbitrary values, then (y' +2p) may have any sign and, hence, the function E changes sign, and the conditions sufficient for achieving a strong minimum are hot fulfilled. If we take'into account the note on pages 373-374, it will be possible to assert that a strong minimum is not achieved on the straight line y =!!..x.a Example 2. Test for an extremum the functional a ~(6y' 1 -y''+yy')dx; y(O)=O; y(a)=b; a>O and b>O 0 in the class of continuous functions with continuous first derivative. The extremals are the straight lines y = C1x+C,. The boundary conditions are satisfied by the straight line y = : x, which is inclu- 8. SUFFICIENT CONDITIONS FOR AN EXTREMUM 375 ded in the pencil of extremals y = C1x that form the central field. The function E(x, y, p, y')=6y'1 -y''+yy'-6p'+p'-yp-(y'-p) x X(l2p-4p3 +Y) = -(y'-p)2 [y'1 +2py'-(6-3p2)]. The sign of the function E is opposite that of the last factor y''+2py'-(6-3p'). This factor vanishes and can change sign only when y' passes through the value y'=-p±V6-2p'. For 6-2p'~O. or p;;:::V3 for any y' we have [y'1 + 2py' -(6-3p2)];;:::: 0 but if 6-2p' > 0 or p 1 and its negative sign for p < l. Consequently, for p=.!!.. < 1ora b 1orb >a Fig. 8-11 we have a weak maximum. For p= !_;;:::: V3 we have a strong maximum since E ~ 0 for any values ofa y'. For p =.!!.. < V3, on the basis of the note on pages 373-374, therea is neither a strong minimum nor a strong maximum (Fig. 8.11). Even in the above very simple examples, testing the sign of the function E involved certain difficulties and for this reason it is advisable to replace the condition of retaining the sign by the fun~-· tion E by a more readily verifiable condition. Suppose that the function F (x, y, y') is three times differentiable with respect to the argument y' By Taylor's formula we get F(x, y, y')= I (JI'-p)l =F(x, y, p)+(y -p)Fp(x, y, p)+ 21 Fy•g•(x, y, q), where q lies between p and y'. The function E(x, y, p, y')=F(x, y, y')-F(x, y, p)-(y'-p)Fp(x, y, p), 376 II. THE CALCULUS OF VARIATIONS ------ aftt>r replacement of the function F(x, y, y') by its Taylor expansion, takes the form E ' (y'-p)~ (x, y, p, y) = - 2-1 - F11•11• (x, y, q). From this we see that the function E does not change sign if F11•11• (x, y, q) doesn't. When investigating for a weak extremum, the function F11•11• (x, y, q) must retain sign for values of x and y at points close to the points of the extremal under study, and for values of q close to p. If F11•11• (x, y, y') =t= 0 at points of the extremal C, then by virtue of continuity this second derivative maintains sign both at points close to the curve C and for values of y' close to the values of y' on the curve C. Thus, when testing for a weak minimum, the condition E ~ 0 may be replaced by the condition F11•11 • <0 on the t'Xtremal C, and when testing for a weak maximum the condition E ~ 0 may be replaced by the condition F11•11• < 0 on the curve C. The condition F11 •11• > 0 (or F11•11• < 0) is called the T..egendre condition. • When testing for a strong minimum the condition E ~ 0 may be rt>plzced by the requirement F11•11• (x, y, q) ~ 0 at points (x, y) close to points of the curve C for arbitrary values of q. Her~. of course, it is assumed that the Taylor expansion F(x, y, y')= = F(x, y, p) +(y'- p) FP (x, y, p) +(y';t>2 F11•11• (x, y, q) holds true for any y'. When testing for a strong maximum, we get the condition F11•11• (x, y, q) ~ 0, for the very same assumptions regarding the range of the arguments and the expansibility of the function F (x, y, y'\ in Taylor's series. Example 3. Test fbr an extremum the functional a vly(x)]= ~(y'2 -y2 )dx, a>O; y(O)=O, y(a)=O. The Euler equation has the form y" +y = 0, its general solution is y = C1 cosx +C, sin x. Using the boundary conditions, we get C1 = 0 and C, = 0, if a =t= kn, where k is an integer. Thus, for a =t= kn ~n extremum may be achieved only on the straight line y= 0. If a< n, then the pencil of extremaIs y = C1 sin x with centre at the point (0, 0) forms a central field. For a> n, the Jacobi condition is not fulfilled (see page 365). • The condition f 11•11• > 0 (or fy•y• < 0) is often called the strong Legendre 'ondition, while the Legendre condition is the inequality f y•y•;;;.. 0 (or fy•y•.:;;;, 0). Summary of Sufftcient Conditions for a Minimum of the Elementary functional * z, v (JI (x))= ~ F (x, Jl, Jl') dx; Jl (Xo)=Jio• Jl (xl)=J11 Jto Weak minimum I Strong m'inimum I Weak minimum I Strong minimum I Weak minimum I Strong minimum d II. Fy- :xFg•=O kFy-:X F11·=0 II. Fy-:XFv·=O II. Fy-i:Fy·=O II. Fy- :XFy·=O1. Fy--Fu•=O dx · 2. Jacobi condition 12. Jacobi condition ~- Jacobi condition 12. Jacobi condition t2: An extremal field 12. exists that includes the given extremal An extremal field exists that includes the given extremal 3. Fy•y• > 0 on the extremal under study 3. Fy•y• (x, y, y') ~ 0 E (x, y, p, y'}~ 3. E (x, y, p, y')~O E (x, y, p, y') ~ 0 3. E (x, y, p, y');30 for points (x, y) for points (x, y) for points (x, y) for points (x, y) for points (x, y) close to points on close to points close. to points close to the points close to the the extremal under on the extremal on the extremal on the extremal un- points on the study and for arbit- under study and under study and der study and for extremal under rary values of y'. for y' close to for arbitra y y'. y' close to p (x, y) study, and for It is here assumed p (x, y) arbitrary y' that the function F (x, y, y') is three times differentiable with respect to y' for any values of y' • To obtain the sufli.,ient conditions for a maximum. take the Inequalities, given here. in the oppo:;: ;e sense. 378 II. THE CALCULUS OF VARIATIONS Since the integrand is three times differentiable with respect to y' for any values of y' and F11,11,= 2 > 0 for any values of y', it follows that on the straight line y = 0 a strong minimum is achieved for a < n. If we take into account the note on page 370, it may be asserted that for a> n a minimum is not achieved on the straight line y=O. Example 4. Test for an extremum the functional x, v[y(x)] = J~ dx, y(O)=O, y(x1)=y1 0 (see the problem of the brachistochrone, pages 316-317). The extremals .are the cycloids x=C1 (t-sin t)+C1 , y=C1 (I-eos t). The pencil of cycloids x=C1 (t-sint), y=C.(1-cost) with centre !I Fig. 8-12 in the point (0, 0) forms a central field including the extremal x=a(t-sint), y=a(1-cost), where a is determined from the condition of the passage of a cycloid through the second boundary point B (x,, y,), if x1 < 2na (Fig. 8.12). We have Fy,= !I Jig Vt+y'1 I F/1'11'= y ( ')'I > 0y I+y' I for any y'. Hence, for x, < 2na, a strong minimum is achieved on the cycloid x=a(t-sint), y=a(l-cost). 8. SUFFICIENT CONDITIONS FOR AN EXTREMUM 379 Example 5. T~st for an extremum the functional 4 v [y (x)] = ~ y'3 dx; y (0) = 0, y (a)= b, a> 0, b > 0. 0 This example was solved on page 374, but we can now simplify the investigation with respect to a weak extremum. The extremals are straight lines. The pencil y = Cx forms a cent- b rat field that includes the extremal y =- x. On the extremala y = !x, the second derivative FY'Y' = 6y' = 6 !> 0. Hence, the straight line y = !!_ x achieves a weak minimum. For arbitrary y',a the second derivative FY'll' = 6y' changes sign; thus the above-indicated sufficient conditions for achieving a strong minimum are not fulfilled. However, one cannot conclude from this that a strong extremum is not achieved. Example 6. Test for an extremum the functional 4 v [y (x)] = 5;.2 dx; y (0) = I, y (a)= b, a> 0, 0 < b < 1. 0 The first integral of the Euler equation (see case (5), page 315) is of the form ..!L+y' 2Y =C or y'2 =4C-wy'2 y'B L:1' extracting the root, separating the variables and integrating, we get y=(C1x+C,)1 , which is a family of parabolas. From the condition y (0) = 1 we find C, = 1. The pencil of parabolas y = (C1x +I)1 with centre in the point A (0, 1) has a Ccdiscriminant curve y = 0 (Fig. 8.13). Two parabolas of this pencil pass through the point B (a, b). On the arc AB of one of them (L1) lies the point A*, which is conjugate to the point A, on the other (L,) there is no conjugate point and, hence, the Jacobi condition is fulfilled on the arc L'l, and an extremum can be achieved on this arc of the parabola. In the neighbourhood of the extremal under study Fv'v' = :~ > 0 for arbitrary y'; however, on this basis we cannot assert that a strong minimum is achieved on the arc L,, since the function F (x, y, y') = ~ cannot be represented in the form y' , , (y'-p)' F(x, y, y)=F(x, y, p)+(y -p)Fp(x,y,p)+-2-1-Fv·v·(x,y, q) for arbitrary values of y' due to the presence of a discontinuity of 380 II. THE CALCULUS OF VARIATIONS the function F(x, y, y') when y' = 0. One can only assert that a weak minimum is achieved on L2 , since for values of y' close to the slope of the field on the curve L2 we have an expansion of the function F (x, y, y') by Taylor's formula. A full investigation of this function for an extremum involves considering the function E (x, y, p, y'): E (x y p y') = .!!_ _jl_ +2y (y' -p) = y (y' -p)2 (2y' +P). ' ' ' y'' p2 p3 y''p3 Since the factor (2y' +p) changes sign for arbitrary y', on the basis of the note on pages 373-374 we can assert that a strong minimum is not achieved on the arc L2 • y Fig. 8-13 The foregoing theorem can, without substantial modifications. b£ extended to functionals of the form x-, v[y., y,• .•• , Yn)=) f(x, Yt• Y2• •• ·• Yn• Y~· Y~. ·· ·• Y~)dx; JC Y; (xo) = Yio• Y; (xt) = Yit (i = l, 2, · ·., n). The function E takes the form n E=F(x, y1 , y,, ...• Yn· y;, y;, ···• Y~)- F (x, Y~" Y2• • •·, Yn· Pt• P2• · · · · Pn)-~ (yi-P;)F/],(x, y,, Y2• ... , Yn• Pt• P2• ... , Pn), I: I where the P; are functions of the slope of the field, on which certain 8 SUFFICIENT CONDITIONS FOR AN EXTREMUM 381 restrictions are imposed (under these restrictions it IS called a spec1al field). The Legendre condiiion £11•11• ~ 0 is replaced by the followmg conditions: F 11• 11• ••• F ·.I· 2 1/1 lin F11 • .. • ••• F · .2 '2 liz Yn 0 . . . . . . ;;;;::: . F . • F . · ... F ..1/11 1/1 lln/12 lin Yn Both in the elementary problem and in more complicated problems, the sufficient conditions for a weak minimum may be obtained by a different method based on a study of the sign of the second variation. Using the Taylor formula, transform the increment of the function in the elementary problem to the following form: x, llv= s[F(x, y+6y, y'+6y')-F(x, y, y')]dx= ~. ~ ~ = s (F,6y+F,.6y')dx+ ~ s[Fv,6y2 +2F,,.6y6y'+F,.,.6y'1 Jdx+R. ~ ~ where R is of order higher than second in 6y and 6y'. When investigating for a weak extremum, 6y and 6y' are sufficiently small. and in this case the sign of the increment !lv is determined by the sign of -the term on the right containing the lowest degrees of 6y and 6y'. On the extremal, the first variation r, ~ (fy 6y +F,.6y') dx = 0 r. and, hence, the sign of the increment !lv, generally speaking. coin· cides with the sign of the second variation ~. 62v = ~ (f vv 6y2 + 2F,,. 6y 6y' +F,.,. 6y'1) dx. X.. The Legendre condition and. the Jacobi condition together are the conditions that ensure constancy of sign of the second variation, and thus also constancy of sign of the increment !lv in the problem involving a weak extremum. 382 II. THE CALCULUS Of VARIATIONS Indeed, consider the integral z, ~ [ro' (x) fJy' +2ro (x) fJy fJy'] dx, (8.2) where ro (x) is an arbitrary differentiable function. This integral is equal to zero: ~ ~ ~ [(ro' (x) fJy' + 2ro (x) fJy fJy'] dx = ~ d (ro fJy') dx = [ro (x) fJy'];: = 0 ~ ~ (because fJy lx = fJy lx = 0). Adding the integral (8.2) to the second variation, we get x, fJ'v = ~ [(FYY +ro') fJy1 +2 (Fyy• +ro) fJy fJy' +Fy•y• fJy''] dx.. " 0) or maximum (f,,.11.< 0). 3. Transforming the Euler Equations to the Canonical Form A system of n Euler equations (see page 318) d F11.--F ·=0• dx lit (i = I, 2, ... , n) (8.3) may be replaced by a system of 2n first-order equations. Putting in (8.3) (k = I, 2, ••. , n), (8.4) we obtain dq11 iJF ([X= oy, (k = 1, 2, ••• , n). (8.5) Solve the system of equations (8.4) for y', (to make such a solution possible, assume that D ( F11~ , F11~, ••• , F 11~) ( • • ') =I= 0), D Y1 Y•· ••·• Yn (8.6) where and substitute (8.6) into (8.5). We then get a system of 2n firstorder equations in the normal form1 dy, ) } ::k ro{,0~,} ~s• q$ , dx iJy, (8.7) Here and henceforward the braces signify that in place of yj, they contain rok (x, Ys• qs)· With fhe aid of the function n H (X, Ys• Q5 ) = ~ (i = 1. 2, ... , n) (8.13) W~ hetV~ thus obtained a 2n-parHmeter family of extrema)s. ft may be proved that (8.13) is the general s:>lution of the system of Euler's equations. and the functions and (i = I, 2. . .. , n) are tile general solution of the system (8-t<) Example. Find the equation of geodesics on a surface on which the element of length of the curve is of the form ds1 = (q>1 (x) +q>1 (y)] (dx1 + dt/), that is, find the extremals of the functional Since S = ~ VlQl, (x) +~x>+IJ>.2(Y) =Vq>, (x)+q>2(y)·V"l-qi, I+II' q = V y' ' H2 +q• = q>, (X) + qlz (y), I +Y't it follows that the Hamilton-Jacobi equation has the form (:r+ (~r =q>, (x)+q>t(Y) or (:)' -cp, (X) =cp,(y)-(:y. For equations of this type (equations with separated variables) cD, (x. ~:)-<~>.(u. :) the first integral is easily found. Putting (:)'-q>,(X)=cx and cp1 (y)-(:)=cx 8. SUFFICIENT CONDITIONS FOR AN EXTREMUM 387 or Ov vax= t (x)+cx J V!J>z (y)-cx Note. The Hamilton-Jacobi equation can be approached by different reasoning as well. Consider a central field of extremals with centre in the point A (x0 , Yo) for the functional x, v[y (x)) = ~ F (x, y, y') dx. Xo On extremaIs of the field, the functional v [y (x)] is transformed into the function v(x, y) of the coordinates of the second boundary point B (x, y). As was pointed out on page 385, av avax=-H(x.y.ql. ay=q. Eliminatinli, q, we get : =-H (x, y, ~~). And so the function v(x, y) is a solution of the HamiltonJacobi equation. Quite analogous arguments also hold true for the functional x, ~ F (x, Yt• Y2• · · ·, Yn• y;, Y~. ••·, Y~) dx. to PROBLEMS ON CHAPTER 8 Test the following functionals for extrema: t- 1. v [y (x)] = ~ (xy' +y'1 ) dx; y (0) = 1; y (2) = 0. 0 25* 388 II. THE CALCULUS OF VARIATIONS a 2. v [y (x)] = ~ (y'2 +2yy'- 16y2 ) dx; a> 0; y (0) = 0; y (a)= 0. 0 2 3. v [y (x)l = ~ y' (I +x2y') dx; y (-1) = 1; y (2) = 4. -I 2 4. v(y(x)]=~y'(l+x2y')dx; y(l)=3; y(2)=5. I 2 5. v[y(x)]= ~y'(l+x1y')dx; y(-l)=y(2)= 1. -I n 4 6. v[y(x)1= s(4y'-y'1 +8y)dx; y(0)=-1; u( ~ )=0. 0 2 7. f' [y (X)]=~ (x2y'1 +12y1) dx; y (1) = 1; y(2) = 8. I I 8. v[y(x)]= J(y''+y1 +2ye1x)dx; y(O)=+; y(1)=+e2 • 0 n 4 9. v lY (x)l = S(y'-11'1 + 6.u sin 2x) dx; y (0) = 0; y ( ~ ) = 1. 0 xs, dx 10. v lY (x)) = -, ; y (0) = 0; y (x1) = y1; X1 > 0; y1 > 0. II 0 .t, dx 11. v[y(x)J= S11,,; y(O)=O; y(x1)=y1; x1 > 0; y1 >0. 0 2xl 12. v[y(x))= Sy.adx; y(l)= 1; y(2)=4. I s 13. v[y(x)l=~(l2xy+y'1)dx; y(I)=O; y(3)=26. I 2 14. v[y (x)) = ~ [y' +(y')' -2xy) dx; y (0) = 0; y (2) = 3. 0 CHAPTER 9 Variational problems involving a conditional extremum 1. Constraints of the Form CJ> (X, y 1, y,, ..., y,.) == 0 Variational problems involving a conditional extremum are problems in which it is required to find an extremum of a functional v; certain constraints are imposed on the functions on which the functional v is dependE:nt. For example, it is required to investigate for an extremum the functional x, v ly., Y2• ... , Y,.)= ~ F(x, Yt• Ys• · · ., Yn• Y~. y;, . · ., Y~)dx ""• given the conditions for an unconditional extremum. This approach can also be used, of course, to solve the above variational problem. Solving the system 'Pi (x, y1, y2, ••• , Yn) = 0 390 11. THE CALCULUS OF VAlUATIONS (i = 1, 2, ... , m) for y1 , y2 , ••• , Ym (01 any other m functions Y;) and substituting their expressions into v [y1, y2 , ••• , Ynl, we get the functional W lYmr•· Ym+Z• ... , Yn) which depends only on n-m arguments that are a ready independent, and, hence, the methods given in Sec. 3, Chapter 6, can now be applied to the functional W. However, both for functions and for functionals a different and more convenient method is commonly employed, that of undeter· mined coefficients, which retains complete equivalence of all variables. As we know, when investigating a function z= f (x1, X2 , ••• , xn) for an extremum, given the constraints IJl; (x1, X2, ••• , xn) = 0 (i = 1, 2, ... , m), this method consists in constructing a new auxiliary function m z* = f + ~ A.;IJl;. I= I where the A; are certain constant factors and the function z* is now investigated for an unconditional extremum; that is, we form a system of equations ~; = 0 (j = 1, 2, ... , n) supplemented by the J constraint equations 1 = 0 can also be considered Euler's equations for the functional v * if we consider as arguments of the functional not only the functions y1 , y1 , ••• , Yn but also A1 (x), A., (x), ... , A.,. (x). The equations q>1 (x, y1, y,, ... , y,) = 0 (i = I, 2, ... , m) are assumed independent, i.e. one of the Jacobians of order m is different from zero, for instance, D\!h,cp2, .. ,cp,.)=FO D (y,, ~~~· ... y,.) . Proof In the given case, the basic condition of an extremum, 6v = 0, takes the form .c, n ~ ~(F111 6y1 +F,,;6u/)dx=O. .c. I= I Integrating by parts the second terms in each parenthesis and noting that we get Since the functions y1, y1 , ••• , y, are subject to m independent constraints 1(11 (x, y1, y,, ••• , y,) = 0 {i = 1, 2, ..• , m), it follows that the variations 6y1 are not arbitrary and the fundamental lemma cannot as yet be applied. The variations 6y1 must satisfy the following conditions obtained by means of varymg the constraint equations cp1 = 0: • More exactly, applying Taylor's formula to the difference cp/(.t, y1 +6y1, ..• , y,+lly,)-cpj{x, y1, ... , y,) of the left-hand sides of the equations 'Pi (x, y1 +6y1, ... , y,+6y,)=0 and 'Pi (x, lito ... , y,) = 0, we should write n '\." OIJ't.k. iJy 6y1+R;=O, i= I I where the R1 are of order higher than first in 6y1(j =I, 2, ... , n). However, as may readily be verified, the terms R; do not exert any appreciable influem:e on subsequent reasoning, since when calculating the variation of the functional we are only interested in frrst-order terms in 6y U= 1, 2, ... , n). 9. VARIATIONAL-i'ROBLEMS WITH A CONDITIONAL EXTREMUM 39.3 and, hence, only n-m of the variations 6y1 may be cousidered arbitrary, for example 6ym+l• 6ym+~• ... , 6yn, while the resr are determined from the equations that have been obtained. Multiplying each of these equations term by term by 'A.1 (X) dx and integrating from x0 to xl> we get Jts n A; (x) L.:cp16y1 dx = 0 p.., 1 Yt (l= I, 2, ••• ' m). x. Adding termwise all these m equations, which are permissible variations 6y1, with the equation we will st.(Fy1-! Ff!i) 6y1 dx= 0. x, I= I or, if we introduce the notation we get x, n sL (F;,- :x F~;) 6y; dx = 0. .t'n i= f satisfied by the Here too it is impossible as yet to employ the fundamt-ntal lemma due to th~:: fact that the variations 6y1 are not arbitrary Choose m factors 'A1 (x), A2 (x), ••• , A.. (x) so that they should satisfy the m equations (/=1, 2, •••• m), or m iJF ~ iJcp1 d iJF iJy + ~ At (X) oy - dx f}y' == 0 I 1=1 1 I (/ = I, 2, ••• , m). These equations form a system that is linear in 'A.1, with a nonzero determinant D (!p., 'P2• ••. , !J!m) =I= 0· D (Yt· II~. · . • lim) ' hence this system has the solution Al \X), A.. (X), • • • • A.IJI (X). 394 II. THE CALCUL,US OF VARIATIONS Given this choice of 1..1 (x), A., (x), o o o , J..m (x), the basic necessary condition for an extremum takes the form Xs II S L (F;1- :x F;;) fJy1dx= 0. X0 /=m+l Since, for the extremizing functions y1, y,, .•• , y,. of the functional v, this functional equation reduces to an identity already for an arbitrary choice of 6y1 (j=m+ 1, m+2, ... , n), it follows that the fundamental lemma is now applicable. Putting all the 6y1 equal to zero in turn, except one, and applying the lemma, we obtain F;1- :x F;; =0 (/ =m+ 1, m+2, ... , n). Taking into account the above obtained equations F11 _..!!_p·. = 0 (J' -1 2 m) 1 dx llf - ' ' • ' ' ' ' we finally find that the functions which achieve a conditional extremum of the functional v, and the factors 1..1 (x) must satisfy the system of equations . • d • F111 -dxF11;=0 (}=1,2, .. o,n), yj = (6y1)' and (6y1)x=x, = (6y1)x=x, = 0, we will have x, II SL. [f.., :';-!(f..; :0]f>y1dx = oo (9.3) x, /= 1 0 From the basic necessary condition for an extremum, 6v = 0, w~> have x, II sL (Fyj-#xF,i) 6y,dx=0, x0 /=1 (9.4) since ~ II ~ II f>v= sL (Fy1f>y1+FYj f>yj) dx = sL.(F111 - :xFYj) f>yidxo ~/=1 ~/=1 • Here too (as on page 392), summands containing terms of erder higher than first in {jyi and l'Jy{' (j = I. 2, 0 o 0, n) should be included in the left-hand sides of the equations: i is now cpnsiderably more difficult to take into account the effect of these nonlinear terms. 398 II. THE CALCULUS OF VARIATIONS Adding termwise all the equations (9.3) and equation (9.4) and m introducing the notation P = F + ~ A.1(x) ~;. we will have h•l S±.(FZ1 - !Fz;) 6y1dx=o. (9.5) Xo/=1 Since the variations 6y1 (j = 1, 2, ... , n) are not arbitrary, we cannot yet use the funi:lamental lemma. Choose m factors A.1(x), A., (x), ... , A.,. (x) so that they satisfy the equations rv,-!FZj = 0 (/ = 1t 2, ... ' m), When written in expanded form, these equations form a system of linear differential equations in A.1 (x) and {1=1, 2, ... , m). which, given the assumptions we have, has the solution A.1 (x), A.1 (x), ... , A.111 (x), which depends on m arbitrary constants. With this choice of A.1 (x), A.1 (x), ••. , A.,. (x) the equation (9.5) is reduced to the form where the variations 6y1 (j=m+ 1, m+2, ... , n) are n001 arbitrary, and hence, assuming all variations lly; = 0, except some one 6y;. and applying the fundamental lemma, we obtain F* d F*· 0lit- dx llj = (J=m+ 1, m+2, ..., n). Thus, the functions y1 (x), y. (x), ... , y,. (x) that render the functional v a conditional extremum, and the factors A.1 (x), A.1 (x), ... , A.,. (x) must satisfy the system of n+m equations: (/= 1, 2, ... , n) and ~,=0 (i = 1, 2, ..., m), that is they must satisfy the Euler equations of the auxiliary functional v*, which is regarded as a functional dependent on the n+m functions lit• y.. . . .• /J,., A.l, A... •••• A.•• 9. VARIATIONAL PROBLEMS WITH A CONDITIONAL EXTREMUM 399 3. lsoperimetrlc Problems In the strict sense of the word, isoperimetric problems are problems in which one has to find a geometric figure of maximum area for a given perimeter. Among such extremum problems, which were even studied in ancient Greece, were also variational problems like the one on page 295 (to find a closed curve, without self-intersection, of a given length bounding a maximum area).• Representing the curve in -parametric form x=x(t), y=y(t), we can formulate the problem as follows: maximize the functional t, t, S = ~ xy dt or S = ~ S(xy - yx) dt lo lo provided that the functional t, ~ v.ta+!i~ dt lo maintains a constant value: It ~ y x•+y2 dt = z. '·We thus have a variational problem involving a conditional extre- t, mum with a peculiar condition: the integral Syx2 +y' dt main· lo tains a constant value. At the present time, isoperimetric problems embrace a much more general class of problems, namely: all variational problems in which it is required to determine an extremum of the functional -t, tl= SF(x, y1, y,, .• ., Yn• y~, y;, ••. , y~)dx given the so-called isoperimetric conditions -"• ~ Ft(X, y1, y,, ... , Yn• y~. y;, .•. , y~)dx=lt -to (i= l, 2, ..., m), where the 11 are constants, m may be greater than, less than or equal to n; and also analogous problems for more complicated functionals. • Though the solution of this problem was known in ancient Greece, its peculiar variational nature was understood only at the end of the seventeenth century. 4.00 II. THE CALCULUS OP VARIATIONS lsoperimetric problems can be reduced to conditional-extremum problems considered in the preceding section by introducing new unknown functions. Denote Jt ~ F;dX=Z;(X) (i=1, 2, ... , m), x, whence Z; (x0)..:: 0 and from the condition ~ F; dx =I; we have Z; (x1) = 1,.. Differentiating z,. with respect to x, we get zi (x) = F;(x, Y1• y,, . · ·, Yn• y~. Y~· • • •• Y~) (i= 1, 2, .•. , m). .... In this way, the integral isoperimetric constraints ~ F,-dx =I; are replaced by the differential constraints: zi = Fdx. Y1• y,, · · ·• Yn• y;, y;, ••·• Y~) (i= 1, 2, ... , m) and hence the problem is reduced to the problem considered In the preceding section. Applying the factor rule, it is possible, given the constraints F;-zi=O (i= 1, 2, ... , m), to replace an investigation of the "• functional v= ~ F dx for a conditional extremum by an investiga"•tion or 'he functional tf 1[F+ f.t.,.(x)(F1-zi)] dx= rF-dx x., I= 1 Xo for an unconditional extremum; here m F-== F + ~A.dx)(F,.-z~) i=l The Euler equations for the functional rt are of the form or FZ1-~F:i =0 F* _!!.p•. = 0 z, dx ', (j = 1, 2, •.. , n), (i= 1, 2, ••• , m), F +~.... I.-F1 - - F +, .... t..f1 • =0m d ( m ) "' .4-1. 1 IIJ dx Yl .4-1. 1 111 i=l i=l 9. VARIATIONAL PROBLEMS WITH A CONDITIONAL EXTREMUM 401 (/=I, 2, ... , n), d d./•dx) = 0 (i = 1, 2, ... , m). From the last m equations we find that all the ')..1 are constant and the first n equations coincide with Euler's equations for the functional rr=J(f+t).1F1) dx. We thus get the following rule: to obtain the basic necessary condition in an isoperimetric problem involving finding an extre~ ~ mum of a functional V= ~ F d'C, given the constraints ~ F1dx = 11 ~ ~ (i = 1, 2, ... , m), it is necessary to form the auxiliary functional tr= i(f+t).1f 1)dx, where the A; are constants, and write the Euler equations for it. The arbitrary constants C,, C,, ... , C,n in the general solution of a system of Euler's equations and the cons-tants ).1, ).1, ••• , Am are determined from the boundary conditions YJ(Xo)=YJo• Y,(x1)=YJ1 (i= I,~•... , n) and from the isoperimetric conditions (i = 1, 2, ... , m). The system of Euler's equations for the functional v•• does not vary if v** is multiplied by some constant factor ~o and, hence, is given in the form where the notations F0 =F, ~1 =')..1~0, /=1, ... , m have been introduced. Now all the functions F1 enter symmetrically, and therefore the extremals in the original variational problem and in the x, problem involving finding an extremum of the functional ~ Fsdx, Xo 402 II. THE CALCULUS OF VARIATIONS given the isoperimetric conditions x, ~F,.dx=l; (i=O, 1, 2, ... , s-l,s+l, ... , m) Xo coincide with any choice of s (s = 0, 1, ... , n). This property is called the reciprocity principle. For example, !J B II the problem of a maximum area bounded by a closed curve of given length, and the problem of the minimum length of a closed curve bounded by a given area are reciprocal and have common extremals. Example 1. Find the curve y=y(x) of given length l, for which :r the area S of the curvilinear tra-0::+----';c:-------JJ~-- pezoid CABD depicted in Fig. 9.1 is a maximum. Fig. 9-1 x, Investigate for an extremum the functional S= ~ ydx, y(x0 )=Yo• Xo y(x1) = y1 , given the isoperimetric condition x, ~ V l +y'2 dx = l. x. First form the auxiliary functional x, s•• = ~ (u+AV 1+y'1 ) dx. x. Since the integrand does not contain x, Euler's equation for S** has a first integral F-y'Fy·=C1 or, in the given case, V-- A ,z y-t-A 1-t-y'z__y =Cu Vt +u'2 whence 9. VARIATIONAL PROBLEMS WITH A CONDITIONAL EXTREMU.'•I 403 Introduce the parameter t, putting y' =tan t; this yields y-C1 = -A. cost; dy dy A. sin t dt ax= tan t, whence dx =tan/= tan t =A. cost dt; x= A. sin t +C2 • Thus, the equation of the extremals in parametric form is x-C2 =A. sin t, y-C1 =-A. cost. or, eliminating t, we get (x-C2} 3 + + (y-C1 ) 2 = A.2 or a family of circles. The constants C10 C2 and').. are determined from the conditions Y(Xo)=Yo• x, Y (xi)= Y1 and ~ V1+ y'2 dx = l. ... !I 0 II Example 2. Find a curve AB of given length I bounding, together with a given curve y = f (x), the maximum area cross-hatched · F' g 2 Fig. 9-2m tg... J It is required to determine an extremum of the functional x, S= ~ (y-f(x))dx; Xo given the condition Y(xo) = !/0 • !J (xl) = !/1 x, ~ V 1+y'2 dx =I. x. Form the auxiliary functional x, S** =o ~ (y- f (x) -i- ')..VI + y'2) dx. x. The Euler equation for this functional does not differ from the Euler equation of the preceding problem and so in the given problem the maximum may be achieved only on arcs of circles. Example 3. Find the form of an absolutely flexible, nonextens:ble homogeneous rope of length l suspended at the points A aud B (Fig. 9.3). 404 II. THE CALCULUS OF VARIATIONS Since in the equilibrium position, the centre of gravity must occupy the lowest position, the problem reduces to finding the minimum of the static moment P about the x-axis, which is assumed to be horizontal. Investigate for an extremum the functional ~ ~ P = j y V 1+ y'• dx provided that j Vl +y'• dx = l. Form the auxi~ ~ liary functional x, p•• = ~ (y+ A.) Vl +y'• dx, Xo for which Euler's equation has a first integral F-y'Fy·=C -~---------~;r or, in the given case, 0 (y +A.) V 1+y'1 - (Yf A.) y'• =C1, VI+ y'•Fig. 9-3 whence y+ A.= C1 V1+y'•. Introduce a parameter putting y' = sinht, whence V1+y'2 =cosh t and y+ A.= C1 cosh t; ~~=sinh t; dy c 0 dx=-,--h1= 1dt. x=C1t+C2 , or, eliminating t, we getSin y +A.= C1 cosh x;1 c2 , which is a family of catenaries. The foregoing rule for solving isoperimetric problems can also be extended to more complicated functionals. We shall mention one more problem involving a conditional extremum- the problem of optimal control. Consider the di fferential equation dx fdt = (t, X (t), U (t)) (9.6) with the initial condition x (t.,) = X0 • Besides the unknown function (or vector function) x (t), this equation also contains a so-called control function (or vector function) u (t). The control function u (t} has to be chosen so that the given functional '·ti=~F(x(t}, u(t)}dt is extremized. t, 9. VAlUATIONAL PROBLEMS WITH A CONDITIONAL EXTREMUM 405 The function y (t) which yields the solution of this problem is called the optimal function or optimal control. This problem may be regarded as a problem involving a conditional extremum of a functional v with differential constraints (9.6). However, in practical problems the optimal functions freq!lently lie on the boundary of a set of admissible control functions (for example, if the control function is the engine power to be switched on, then obviously this power is bounded by the maximum power output of the engines; and in the solutions of optimum problems it is often necessary to run the engines at peak-power output, at least over certain portions). Now if the optimal function lies on the boundary of a set of admissible control functions, then the foregoing theory of problems involving a conditional extremum and presuming the possibility oi two-sided variations is not applicable. For this reason, other methods worked out by L. Pontry1.1gin (see [8]) and R. Bellman (see t9]) are ordinarily applied in solving problems of optimal control. Example. In the system of differential equations ~~ = v, ~ = u (t is the time), (9. 7) which describe the motion, in a plane, of a particle with coordinates x, v, determine the control function u (t) so that the point A (x0 , v0 ) moves to the point B (0, 0) in a least interval of time; IuI~ I (since u = ~:~, it follows that u may be considered a force acting on a particle of unit mass). The control function u (t) is piecewise continuous. To simplify our reasoning, let us assume that it does not have more than one point of discontinuity, though the final result holds true even without this assumption. It is almost obvious that on optimal trajectories u= ± I, since for these values I~~ j and j~~I attain maximum values and, hence, the particle moves with maximum speed. Putting u = 1 in (9.7) we get or v' == 2 (x-C) and similarly for u == -1: C 12 t v=-t+ 1, x=-2 +C1t+C,, v =-2(x-C). 406 II. THE CALCULUS OF VARIATIONS Figures 9.4 and 9.fi d~pict these families of parabolas, tht• arrows indicating the direction of motion as t increases. If the point A (x0 , V0 ) lies on arcs of the parabolas v=- Vi or v= V-x (9.8) (Fig. 9.6) passing through the coordinate origin, then the optimal trajectory is an arc of one of these parabolas connecting the point u u .r Fh!. 9-4 Fig. 9-5 A with the point B. But if A does not lie on these parabolas, then the optimal trajectory is the arc AC of the parabola passing through A, and the arc CB of one of the parabolas (9.8) (see Fig. 9-6 Fig. 9.6 which indicates two possible positions of the points A and C). In this problem, the time T of translation of the point from position A to position B is a functional defined by the first of the equations (9.7); the second equation of (9.7) may be regarded as a constraint equation. It would, however, be difficult to apply to this problem the earlier described classical methods of solution, since optimal control lies on the boundary of the region of admissible controls Iu j.;;;;; 1 and two-sided variations are impossible here; moreover, the solution is sought in the class of piecewise continuous controls. Both of ~hese circumstances are extremely characteristic of most practical problems involving optimal control. 9. VARIATIONAL PROBLEMS WITH A CONDITIONAL EXTREMUM 407 PROBLEMS ON CHAPTER 9 1. Find the extremaIs of the isoperimetric problem v [y (x)] = I I =~(y'2 +x2 )dx given that ~y1 dx=2; y(O)=O; y(l)=O. 0 0 2. Find the geodesics of a circular cylinder r = R. Hint. It is convenient to seek the solution in cylindrical coordinates r, d the I~ itz method, in the development of which a very substantial contribution has been made by the Soviet mathematicians N. Krylov, N. Bogolyubov, and others, finds wide application in the solution of various variational problems. A third direct method, proposed by L. Kantorovich, is applicable to functionals that depend on the functions of several independent variables and is finding ever broader uses in areas in which the Ritz method is employed. We shall examine only these three basic direct methods (the proofs of many of the assertions will not be given). The reader who wishes to study mort! closely the direct methods now in use is referred to L. Kantorovich and V. Krylov (10) and S. Mikhlin [11). 2. Euler's Finite-Difference Method The underlying idea of the method of finite differences is that the values of a functional v (y(x)), for example, .. ~ F(x, y, y')dx, y(x.)=a, y(x1 )=b, •• are considered not on arbitrary curves that are admissible in the given variational problem, but only on polygonal curves made up of a given numbt•r n of straight-line segments with specified abscissas of the vertices: X11 +L\x, X0 +21\x, ... , X0 +(n- 1) Ax, where Ax=~) -xu (Fig. 10.1 ). n On such polygonal curves, the functional v [y (x)] is transformed into a function q>(y1, y2 , ••• ,y,_ 1) of thcordinatcsy1,!J2 , ••• ,y,_1 elf the vertices of the polygonal curve, since the curve is completely defined by these ordinates. 410 II. THE CALCULUS OF VARIATIONS We choose the ordinates y,, Yz· ... , y,._, so that the function q> (y,, y,, ... , y,._ 1) is extremized, that is we determine y,, y1, ••• , y,._1 from the system of equations cJcp -0 cJcp -0 ~=0 iJy, - ' iJyz- ' • • •' iJy,._, • and then pass to the limit as n -~ oo. Given certain restrictions imposed on the function F, we obtain, in the limit, the solution of the variational problem. 9 !fn !f(Z,J :z 0 z;tn-ljtJz Fig. 10-1 tiowever, it is more convenient to calculate approximately the value of the functional v [y (x)] on the above-indicated polygonal curves; for instance in the most elementary problem it is best to replace the integral x, n=l x0 +(11+J).ax SF(x, y, y') dx ~ L. S F(x, y, Y~~+~;YII) dx "• 11=0 t,+k .ax by the integral sum t F (x;. Y;. ~~)Ax.I= I By way of an illustration, let us derive Euler's equation for the functional x, v [y (x)] = ~ F (x, y, y') dx. "• In this case, on the polygonal curves under investigation n-1 v[y(x)j ~q>(y,, Ya• ••• , y,._,j=(;.; F (x;, Y;. Yi+~;Y;)Ax. 10. DIRECT METHODS IN VARIATIONAL PROBLEMS 411 Since only two terms of this sum, the ith and the (i -1) th, depend on the y,.: F ( Y;+t-Yi) F ( Yi-Yi-t)X;, y,., -liX tu and X;_ 1 , y,._1 , £\x Ax, it follows that the equations :,. =0 (i = 1, 2, ... , n-1) take the form F ( - Y;+t-Yi) A F ( Yi+t-Yi' ( I ) AY X;, y,., L\x x+ y' X;, y,., £\x } - £\x X+ +fu•(x,._1 , Y;- 1, Yi~~i-t) ;xAx=O (i= 1, 2, ... , (n-1)), or F ( £\y;) F ( £\Yi-t) 1 tly;) _ u• Xfo Yi· .ix - y' Xi-t• Yi-t• -x;-, FY \X;, y,., .ix .ix 0, or Passing to the limit as n--. oo, we get Euler's equation d Fy- dxfv=O which must be satisfied by the desired extremizing function y(x). In similar fashion it is possible to obtain the basic necessary extremum condition in other variational problems. If we do not pass to the limit, then from the system of equations ~,. = 0 (i = 1, 2, ... , n-1) it is possible to determine the desired ordinates y1 , y,, ... , y,._1 and thus obtain a polygonal curve which is an approximate solution of the variational problem. 3. The Ritz Method The underlying idea of the Ritz method is that the values of a functional v [y(x)] are considered not on arbitrary admissible curves of a given variational problem but only on all possible linear com- ,. binations y,. =2: a,.W,. (x) with constant coefficients composed of n to= I first functions of some chosen sequence of functions W1 (x), W2 (x).... , W,.(x), ... n The functions y,. = ~ a,.W; (x) must be admissible in the problem at '""' 412 II. THE CALCULUS OF VARIATIONS hand; this imposes certain restrictions on the choice of sequence of fuoctions W; (x). On such linear combinations, the functional o[y(x)] is transformed into a function If' (a10 a2 , ••• , a,) of the coefficients eta, ct2 , ••• , a,. These coefficients ct1 , tt2 , ••• , a, are chosen so th<1t the function If' (eta, ct2 , ••• , a,) is extremized; hence, a 1 , a~, ... , a11 must be determined from the system of equations a A minimizing sequenct may not tend to the extremizing function in the class of admissible functions. 10. DIRECT METHODS IN VARIATIONAL PROBLEMS 415 Indeed, the functional x, v [Yn (x)] = ) F (x, Yn (x), y~ (x)) dx x. may differ but slightly from x, v [y (x)] = ) F (x, y (x), y' (x)) dx x. not only when throughout the interval of integration y,. (x) is close in the sense of first-order proximity to y(x), but also when, over sufficiently small (JOrtions of !I the interval (x0 , x,), the functions y,. (x) and y(x) or their derivatives differ radically, though remain close on the rest of the interval (x0 , X1) (Fig. 10.2). For this reason, the minimizing sequence y" y2 , ••• , Yn may not even have -+---'--------'--~r a limit in the class of admis- 0 Zo z, sible functions, though the functions y1, y2 , ••• , y,. will themselves be admissible. Fig. 10-2 The conditions of convergence of the sequence Yn• obtained by the Ritz method, to a solution of the variational problem and the evaluation of the speed of convergence for concrete, frequently encountered fuuctionals have been worked out by N. Krylov and N. Bogolyubov. For instance, for functionals of the type 1 v · S[p (x) y'2 +q(x) y2 +f (x) y] dx; y (0) = y (1) = 0, 0 where p (x) > 0; q (x) ~ 0, which are often met with in applications, not only has the convergence been proved of approximations (obtained by the Ritz method) to the function y (x) that minimizes the functional, given the coordinate functions Wk (x) = V2 sin kn x (k = I, 2, •.. ), but extremely precise error estimations Iy (x)- Yn (x) I have been given. 416 II. THE CALCULUS OF VARIATIO"iS We give one of these estimations of the maximum jy(x)-y,.(x)j on the interval (0, 1): 1 • I r12 dx I I [ max q (x) ] 2 V J maxjy-y,. ~n-+-1 maxp(x)+(n-t-l)~a~ o s X n2 Y2[minp(x)J7 x [max Ip' (x)l +! max q (x) +:rt min p (x)] *. Even in this comparatively simple case, the estimation of error is very complicated. For this reason, to estimate the accuracy of results obtained by the Ritz method or by other direct methods, one ordinarily uses the following procedure, which is of course theoretically imperfect but sufficiently reliable in a practical way: after calculating y,. (x) and y,. t-l (x), a comparison is made between 0 them at several points of the interval [x0 , x1]. If their values coincide within the limits of accuracy required, then it is taken that to within the required Fig. 10-3 accuracy the solution of the variational problem at hand is y,. (x). But if the values of y,. (x) and Yn+t (x) do not coincide even at some of the chosen points within' the limits of the given at:curacy, then Yn+2 (x) is calculated and the values of y,.+l (x) and y,.+ 2 (x) are compared. This process is continued until the values of Yn+k (x) and y,. 1-k+1 (x) coincide within the limits of the given accuracy. Example I. In studying the vibrations of a fixed wedge of constant thiclmess (Fig. 10.3), one has to test for an extremum the functionaI 1 v = S(ax3yn2 -bxy2 ) dx; y (l) = y' (1) = 0, 0 where a and b are positive constants. For the coordinate functions satisfying the boundary conditions we can take (x-1)2 , (x-1)2 x, (x-1)2x2 , ••• , (x-l)2x"-\ ..• , hence, n ,, ( 1)2 k - 1y,.= ~ ak x- x . k~l * St't' 1\.anlorovich and Krylov JIOJ. Conli11i11;_; oursel\"l•s o1il\' to the lirst hw1 knn-;, \H' !-:d ,, ( r I ,.~ Ir;'' . r'- \1. then l'~ ct'[!le)• ~ (ar"jfir!"<-: 2r-.t1 -·'frl."}~-/,\(.\- ll11',:, r!.,XI~jtf.r " HNe, the Jl1'('('"8ary <.:ontlitions for .111 exlrunum, take the forrn ( b \ / •) ,, \ :... a -- .;-ll·l- .· -:,, : - i -:_ u- . . ".t"\ :') IIV• 0 ·and tla., / •l !' I I c) h ! ~- a - 1~1:) \ "J., -t ~. ; II - :0:~0' r;_ . -0. t 17 To obtain ~(·lutions di!Terent from the ~olution ".t,, "!... - n. ,,·hich corresp011fls tu the <1 hst·nce of vibrations of the \n•df,!e. it i:- llcn·ssarv that tlw determinant of this homo~encous linear ~vstem of eqtin tions be zero: · Ia- 1~- 1 :30 I '2 h jy, a- lfl:'i nr h'/2 h\ !2 h\~ a - - . ~ a--- 1- . - u - 11.1,:;: .·'· =· 0.,\!1 \ .. :!'10 . :1 ,, This equatiot• i-; calll:'d the frc£Jlll'ncy equation. It detines the frequency b of nJtur;.~l \·ibrations of the we1=SS[(~:r+(~~r+2zt] dxdy D (see page 328). The boundary condition is maintained: z = 0 on the boundary of the domain D. Let us investigate this functional for an extremum by the Ritz method. For a system of coordinate functions take . nx. ny( 12smma smnb m, n= , , ... ). Each of these functions and their linear combinations satisfy the boundary condition z= 0 on the boundary of D. These functions also possess the property of completeness. Taking we will have v[z,."') = sS[(a~:m )2 +(a~~m)2 + 0 0 ~., --\.., . nx . ny] ·+2z,.m ~ ~ ~pqsm Pa smqb dxdy= po=l qo=l n m n 111 n2ab ~ ~..., ( p2 q2 ) 2 ab ~.., '\.' =-4- ~ .L. a2 +b2 apq+2 ~ ~ apq~pq· po=l qo=l po=l qo=l This· result is readily obtained if we take into account that the coordinate functions sin p :x sin q ~Y (p, q= 1, 2, ...} form iiJ.D an orthogonal system, i.e. ,// 27 , SJsin p :x sin q ~~sinp, ';; sin'17 420 II. T!IE CALCCLUS OF VARIATIONS for any positive integral p, q, p., q,, with the exception of the case p=p1, q=q,. for p=p1 and q=q,, we get f(' .., ax . :wd ·' "b JJ srn- p a- srn- q -,; Xu!J= T. /) Therefore, of all the terms under the sign of the double integral, equal to v [z,,.), only those are taken into account that contain f th f t. . ;"IX • :~y . ;;tX ;ry d squares o e unc 1ons sm p a srn q -6- . sm p -£1 cos q T an :IX • :ny 0'· . I [ I . f t. (cosp -a- Sill q -;;-. uvrous y, v z,., IS a unc 1on \·ariatiunal problem is sou~ht :1! in the form z,, = ~ a..wk (x1, x~, .•. , x"), whl're the codticients all k=l ilfl• constants. The 1\untoro;JidL metllt'll also requires choosin~ a coordinatt> system of iunctiow. \\'1 ,1x,, x~, ... , x,,), W~(x.. x~, ... , x,), ... , Wm(x.,x", ... , x,J, ... and an approximate solution is also sought in the form '"- ~· w .z,, -- ._. ct.k (x;) k (x., x.,, ••. , x,). k=l ltowevtr, the cocflici<.>nts ct.dx;) are not constants hut are unknown , functions of one of the inII z,, = ~ rJ.1,1X1i \f'., (x, x,, . _. , x,) 1.: =- i it is possible to lind fu11ctious that a!Jpruxi:uatl' belte1 the solution II. THE CALCULUS OF VARIATIONS of the variational problem than among functions of the form m ~ a~W~ (x1, X2 , ••• , xn) where the a11 are constant. .t=l For example, let it be required to investigate for an extremum the functional JC !l'z (XI V= ssF (x, y, Z, ~;, ~;) dxdy, extended over the y = «p2 (x) and two ICe ll'a (X) D bounded by the curves y = «p1 (x), lines x = x0 and x = x1 (Fig. 10.4). Values of the functions z (x, y) are given on the boundary of the domain D. Choose a sequence of coordinate functions: Wt:i and on the hound:.!f\ of tlw ink;.:ration dllmain z =' 0. Pr11ceedinL' b\ tile K:mturovich method. we :,hdl! "e~k the thl' it.;flil . I ., :' .,... :~ \ ~ II ;r-: -- X • . ll (.d. l .. \ .J • ! ,\i!cr i:lk:_r;:ti(JI1 with n·..;p~·d to ''· tht" functional t' [z,J take-s tht· ic.rrn ,, . '-' [z 1' = ::_y) ,· (2.r''u'~ _: IOx'uu' --: .10x'u~ ·; 15x"u)dx 1 '-~\ '·' '" Eulcr ·._ equation for this fundi]wt{x)dx=0 (i=l, 2, ... , n). (10.3) It is natural to expect that Yn tends to the exact solution - ... y = ~ al.wi (x), I= I as n- oo, since if the series obtained converges and admits two times termwise differentiation, then the function L (y)- f (x) is orthogonal on the interval [x0 , x1] of each function w; (x) of the system (10.2), and since the system (10.2) is complete, it follows that L (y)- f (x) ==0, and this signifies that yis a solution of the equation (10.1). Obviously, yalso satisfies the boundary conditions y(x0)=y(x1)=0 [since all the w1 (x0 )=w1(x1)=0]. Only very rarely is it possible to determine all the a; from the system (10.3) that is linear in them and to pass to the limit as n- oo; for this reason, one ordinarily confines oneself to a finite (very small) number n (n = 2, 3, 4, 5, and sometimes even n = 1). Here, of course, one has to choose only n functions w1 (x), and so the condition of completeness is discarded and one has only to choose them linearly independent and satisfying the boundary con- ditions W; (X0) = W; (X1) = 0. Very often, for these so-called coordinate functions we take the polynomials (x-x0 ) (X-X1), (x-x0 ) 2 (x-x1), (X-X0 ) 3 (X-X1 ), ••• . . . , (X-X0 )n (x-x,), . . . (10.4) [it is convenient here ·to transfer the coordinate origin to the point x0 , and then in (I 0 4) x0 =OJ or the trigonometric functions . nn (x-x0) Stn-'---= Xt-Xo (n = 1, 2, ... ). 10. DIRECT METHODS IN VARIATIONAL PROBLEMS 427 This method is applicable to equations of any order n, to systems of equations, and to partial differential equations. PROBLEMS ON CHAPTER 10 l. Find an approximate solution of the equation ~z .- l inside the square -a~ x ~a, -a~ y ~a, which vanishes on the boundary of the square. Hint. The problem reduces to investigating for an extremum the functional 55 [(::r+ (:;r-2z] dxdy. D An approximate solution may be sought in the form Zo =a (x'-ax) (y~ - az). 2. Find an approximate solution of the problem of the extremum of the functional I v LY (x)] = ~ (x3y"2 + l00xy1 -20xy) dx; y (l) = y' (1) = 0 0 Hint The solution may be sought in the form Yn (x) = (x-1)2 (ct0 +ct1x+ ... +ctnXn); carry out the calculations for n = l. 3. Find an approximate solution of the problem of the minimum of the functional I v [y (x)] = ~ (y'1 -y2 -2xy) dx; y (0) = y (l) = 0, 0 and compare it with the exact solution. Hint. The approximate solution may be sought in the form Yn =X (I -X) (ct0 +ct1X + ... + anxn); carry out the calculations for n = 0 and n = 1. 4. Find an approximate solution of the problem of the extremum of the functional 2 v [y(x)] = 5(xy'2 - xz x 1y2 -2x2.1/ Jdx; y (1) = y (2) = 0, I and compare it with the exact solutwn. 2H* 428 II. THE CALCULUS OF VARIATIONS Hint. The solution may be sought in the form y=a.(x-l)(x-2). 5. Using the Ritz method, find an approximate solution of the problem of the minimum of the functional 2 v [y(x)] = ~ (y'' +y2 +2xy)dx; y(O) = y(2) =0, 0 and compare it with the exact solution. Hint. See Problem 3. 6. Using the Ritz mdhod, find an approximate solution of the differential equation y" +x2y = x; y (0) = y ( 1) = 0. Determine y2 (x) and Ya (x) and compare their values at the points x = 0.25, x = 0.5, and x =0.75. Answers to problems CHAPTER I 1. sin y cos x=c. 2. 6x2+5xy+y2-9x- 3y=c. c xs y2y I 4. u=-x+T· 5. 2+-x=C. 6. X=ce- 3'+se21 • 1. y=ccosx+sinx. 8. e"-eY=c. 9. x=ce1-! (cost+sint). tO. Homogeneous equation: x= I -!... =yecY-H. II. y=cx and y2-x2 =c. 12. y2 =(3x+c)2 . 13. lnltl=c-e t. f X= sin I, 14. A parameter may be introduced, putting y' =cost\ t +sin 21 + t 11=2 -4- c. J X=p3 -p+2, 15. y=cx+_!_; singular solution y2 =4x. 16. \ 3 4 p2 11. Equac Y=4P -2+c. dx y3 {X=~ p3 -~ p2 +c tion is linear in x anddy' x=cu+ 2 . 18. 3 2 ' g=p4-p3-2. 19 Hyperbolas x2 -y2 =c. 20. The differential equation of the required curves is !fx= y'. Ans. y2= 2cx. 21. The differential equation of the required curves is y-xy' =X. Ans. y=cx-x In IxI· 22. x2 +y2 -2cy=0. The problem is solved very simply in polar coordinates. 23. The differential equation of the problem is ~~ = k (T -20). Ans. In one hour. 24. The differential equation of the problem is : = kv, where u is the velocity. Ans. u:::::0.466 kmjhr. 25. If the origin is put in the given point and the x-axis directed parallel to the direction given in the problem, then the differential equation of the curves, the rotation -x ± Vx2+y2 of which forms the desired surface, is of the form y' = (or dx- II -dp = 0, where p= VxLJ-112). Ans. The axial section of the desired surface is defined by the equation y2=2cx+c2, the surface is a paraboloid of revolution. 26. y= 2 sin (x-c). 27. The differential equation of the desired curves is 11' = _JL. Ans. Hyperbolas xy=c. 28. (x+y+ 1)3 =c (x-y +3). 29. y= X 2 (1 +x) =c+ 2x+x2 • 30. y(0.5):::::0.13. 31. y(0.6)~0.07. 32. y(0.02):::::1.984; 11 (0.04) ~ 1.970; y (0.06) ~ 1.955; y (0.08) ::::: 1.942; y (0.10) ~ 1.930; y (0.12) ~ :::::1.917; y(O.l4) ~ 1.907; y(0.16)::::::1.896; y(O.J8)::::::1.886; 11(0.20)::::: 1.877; 430 ANSWERS TO PROBLEMS y (0.22) ;:::: 1.869; y (0.24) ;:::: 1.861; y (0.26) ;:::: 1.854; y (0.28) ;:::: 1.849; y (0.30)::::: { x=~+~e.. x-y . ::::: 1.841. 33. p2 3 and y=O. 34. x+cot - 2-=c. 36. (x+y+ y=2px-p2 +1)3 =ce2X+Y. 37. y=c; y=ex+c; y=-ex+c. 38. y2=2cx+cz. 39. No. r x2 -I x2-l 2 1 x3 x' 40. y1 =-2-; y2 =-2-+yg- 4 x+ 6 - 20 . 41. y=2x2-x. 42. No. X 43. X=CeY. 44. xZ+ Jr=C2. 45. X=2t . .46. X=tz. 47. y=-x+l and 48. A real solution does not exist. 49. 3x-4y+l=cex-Y. cz-xz 50. x=(4t+c) sin t. 51. y=cx+-2- and the singular solution y= -xz. ~ a a 52. y=x7 +c' y=O. 53. x-c= 2 (2t-sin2t), Y='2(1-cos2t)isafamily of cycloids. A singular solution is y=a. Hint: it is convenient. to introduce a parameter t, putting y'=cot t. 54. 3 (x2+u>+xy3 =cx. 55. Jl.= (yz~x)a. ~ 1 56. x=ceY. 57. x2+2xy-y2 -6x-2y=c. 58. y= 1+cx+lnx and y=O. 59. (x2-l) y-sin x=c. 60. By+4x+5=ce'X-8Y-•. 61. y3 +x&-3xy=c. 62. y=c (x2+y2). 63. ya~x+.E... 64. y=c (x+a)+c2 and a singular solution X . (x +a)2 2 c 3 ts y = ---4- . 65. x = '3 t +ji' , y =2xt- t2 and y = 0, y =4 xz 66. y= c I ±cos X. CHAPTER 2 ANSWERS TO PROBLEMS 431 18. Y= ( -}x+ I)'. 19. y=c1 cos x+c2 sin x+ I+x cos x-sin x In 1sin xI· 't . d2r k 20. u=,-+c2• 21. The differential equation of the problem Is dt' =?i dv k or vd- = 2 , where r is the distance from the centre of the earth to the body,, , u is the velocity, and k=-64002 g. Ans. u::::: II kmjs. 22. The differential d2x (dx)' 752 gequation of motion is (fji"= -g+k dt . Ans. x=-g In cosh 75 t. 23. The . d's d2s g differential equation of motion 1s dt' =k (s+ I) or tfii"=s (s +I). . /6 .r- 3 .r- F-a Ans. t= V g ln(6+ r 35). 24. t= Vi ln(9+ r 80). 25. s=-b- t_(F-a)p ( 1_ -~g t) ./g ,/g b2g e . 28. x=Acos V at. 27. x=acos V ~t. 28. The differential equation of motion is x+k1x-k2x=0, kt > 0. ( _!!_+ y k~ +k,) t ( _!!__ y ki +k,) t Ans. x=c1e 2 ' +c1e 2 • • 29. x =~ (- 1)11 sin nt = 12 kA (n2 _ 2)ns . 30. Y2 =c1 (x'+x Vl+x2 +lnlx+ Jfl +x21}+c1• n=l 31. y=c2ec,x+c1, y=c 4 x· 32. x"=c1 cos3t+c2 sin3t-/2 t2 cos3t+~sln3t. ( I ) I -..!...x( y333. y=e-z c1 +c2x-4x' +ge~. 34. y=c1e~+e 2 c1 cos-2-x+ . y3 ) I xe~ cos x x1e~ sin x + c3 sm-2-x + 3 xe~. 35. y=e~(c1 cosx+c,slnx)+ 4 + 4 • 36 y = c1 (x - r) + c1 [4 - 6x2 + 3 (r - x)J In I;+! IJ - !. 37. u = tion of motion is mx= mg-ki. Ans. ( k. ) mg m'g -- t X=- t-- 1-e m k k2 • t-10 = ., r du . 2 x3 x' =m J f(u), where u=x. 41. y=c1+c2x+c3x +e~(c4 +c&x+c1x2)- 2 - 24 • "• 432 ANSWERS TO PROBLEMS 42. x=(c1 +c2t)cost+(c3 +c4 t)sint-}t2 cost. 43. y=c1 cosln(l·-t-x)+ Qll . . V (2-n2)sinnl-2ncusnl +c2 sm In (I +x>+ ln(l +x) sm In (I +x). 44. X=~ [(2-n2) 2 + 4n2) n4 • n=l 45 -~+1J [-(n2 -Gz)an-atn~n t+atn(ln.=._0!-=.a2)~n. t l. x- 2a ( 2 ) 2 + • 2 cos n ( 2 ) 2 + 2 2 sm n ,1 n -a2 ain n -a2 a1n n=l where a 0, an, ~n are Fourier's coefficients of the function f (t). 46. x =co; 1 + +~(I +3 cos 2t). 47. y=c,x+ c2xe " . 48. xzy• + xy'-y = 0. 49. 56. y=e'+c,x(~--1-)+cz. C1 c:. sin 2x sin 4x I c x I 57.y=c1 cosx+c2 smx--6- - -30 .58. Y= ---2 .59. y=c2e' +-. x- c1 CHAPTER 3 1. x=sint, y=cost. 2. x1 =2e1, x2 =2e1• 3 x=c1e(-t+V\6)t+ + ce<- t- Vli) t +.! e1+ _!_ e21• we find y from the first equation:2 II 6 ' 1 dx 5 _..!._ t ( J"J JfJ )y=e -iii- x. 4 x=c1e'+e 2 c2 coc; - 2- t +c3 sin- 2- t ; y and z . dx d2x t are found from the equations Y= dt, z= dt 2 • 5. x=c1ec, y=c1c2ec,t. 6. X=c1 cost+c2 sint+3; y=-c1 sinl+c2 cost. 7. y=c1J 0 (x)+c2Y0 (x); z=x [c1J~(x)+c2Y~(x)]. 8. x+y+z=c~o x2 +y2 +z2 =c~. 9. x=c1el+c2e-21, y=c1e1 +c3e-21; z=c1e1-(c2 +c3)e-21• 10. x=c1t+ ~2 ; y=-c1t+ ~2 • II x=c1 cost+c2 sint-lcost-j-sinllnlsintl; y is determined from the equation Y=~ -I. 12. x2 -y2 =c1, y-x-t=c2. 13. x=c1e1-t-c2e-1+sin t; ANSWERS TO PIWBLEMS 433 -------·-·-·----------------------- y. c -- c1e1·i c~e- 1 • 14. x ___ e1; y= 4e1• 15. IJ (I) ~ 0.047. 16. x=~ ea' (c1cost -1-c~ sin t_), u-~e"1 (c,sint-c2cost). 17. X=2c,e- 1+c2e- 11, Y=-c,e- 1 +c~-·1• 18. x=e- 61 (2c1 cost+2ctsin t), y=e- 61 l(c1 -c2)cost+(c.-J-c2)sin t). 19. x=c1e1+c2, y=(c1t+c3 )e1 -t-l-c2, z=y-c1e1• 20. x+y+z=c.. • 2 2-. - 22 x-llclet+c2e-' IIxyz=c~. 21. x·+y +z -cf. xyZ-Cz. · - c1et+ 3c~-t · CHAPTER 4 1. The rest point is asymptotically stable. 2. The rest point is unstable. 3. For a.<- ~ , the rest point is asymptotically stable, for a.=- ~_stable, and for a. >- ~ unstable. 4. For a.,.;;;; 0, the rest point is asymiJtotically stable, for a.> 0 unstable. 5. For I< t < 2x(t, Jl) _,. Y4-t2 ; for 2 < t < < 3x(t, Jl) _,._ Y9-t'1 ; for t > 3x(t, Jl)-+ oo. 6. x(t, Jl) _,.co. 7. The rest point is unstable. 8. The rest point is stable. 9. The rest point is unstable. 10. The rest point is stable. 11. Saddle point. 12. The periodic solution x=+sint-; cost is asymptotically stable. 13. All solutions, including pe· riodic solutions, are asymptotically stable. 14. The rest point is unstable. The function U=x'-11' satisfies the conditions of the Chetayev theorem. 15. All solutions are unstable. 16. The solution x = 0 is unstable. 17. For I 2 and for a.< I the solution x=='O is unstable. 18. The solution x =0, y =0 is stable for constantly acting perturbations. The function u=4x2 +3y2 satisfies the conditions of Malkin's theorem. 19. The solution x (t) = 0 is unstable. 20. All solutions are stable, but there is no asymptotic stability. 21. All solutions are stable, but there is no asymptotic stability. 22. The periodic solution x= cost-:; sin t is unstable. 23. The region of stability is 0.;;;;;; a..;;;;;; I, the region of asymptotic stability is 0 < a. < I. 24. The region of stability is a.;;:;. 5, the region of asymptotic stability is a. > 5. CHAPTER 5 I. z=lll(x+y). 2. z=e2X$(x-y). 3. z=e~lll(x). 4. CD(z, ye:)=o. ID (x3y5) ( y z ) 5. z=5+ y' . 6. u=(x-y, y-z). 7. U=x' ( ~ ) • The equation of surfaces orthogonal to vector lines is x2+y2-z2=c. 24. z=xy+l. 25. z=3xy. 26. z=x2+y2. CJIAPTER 6 t. The extremals are the circles (x- C1)2 + y2 = C~. 2. The integraI is independent of the path of integration. The variation problem is meaningless. 3. An extremum is not achieved in the class of continuous functions. 4. The extremals are cl x2 the hyperbolas y= x-+C2. 5. y=C1 sin (4x-C2 ). 6. Y= - 4 +C1x+C2• ·,. !J= ~.inh (C1x+C2). 8. y=C1ex+C~-x+ ~ sin x. 9. y=C1e2X+C2e-2x + x7 -1- C3 cos 2x-!-C4 sin 2x. 10. Y=71+ C1x1 -!-C2x4 + C3x3 -!-C4x2-!-C&x-!-C6• 11. y=(C1x-!-C2)cosx-!-(C3x+C4 )sinx, z=2y-!-y", whence z is readily deter· . iJ2z iJ2z iJ2u iJ2u iJ2u mmed. 12. iJx2 -iJy2 =0 13. Dx2 +oy2 + iJz2 =/(x, y, z). 14. y=C1x'-!-C2• I X cos X • 15. u=2 xex+C1ex+C2e-x. 16 y=--2-+ C1 cosx+C2smx. 17. y=C1 coshx-!-C2 sinhx+xsinhx-coshxlncoshx. 18. y=C1x+~+ ~ xlnlxj. . x2 sinx 19. y= (C1 +C2x) cosx+(C3 +C4x) s1n x--4--. 20. y=C1ex+c,e-x + 4( Y3 . }·~'3 ) - ; ( c y3 c . ¥3 ) 3 +e C3 cos-2-x+C4 sm-2-x +e 1 cos-2-x+ 8 sm-2-x +x. CHAPTER 7 I. y=-x for Oc;;;;x...;;;l; y=x-2 for I 4 , then there is no minimum. ANSWERS TO PROBLEMS 4.15 3. An extremum is not achieved on continuous curves. 4. There is a strong mini· mum for y = 7- i. . 5. There is a strong minimum for y = I. 6. A strong maxi· X mum is achieved for y= sin2x-l. 7. A strong minimum is achieved for y=x'. 8. A strong minimum is achieved for y= ~ esx. 9. A strong maximum is achieved for y =sin 2x. 10. A weak minimum is achieved on the straight line y = 1!1 x. xl It. A weak minimum is achieved on the straight line y = !!J x. 12. A weak mini- xl mum is achieved for y = x2. 13. A strong maximum is achieved for y = xa -1. 4 A t . . . h. d f sinh x +1 . s rong mmtmuril IS ac 1eve or y =sinh 2 x. CHAPTER 9 t. y=±2sinnnx, where n is an integer. 2. q>=C1 +C2z; r=R. 3. y=l..x2+C1x+C2, where· C1, C2 and A are determined from the boundary conditions and from the isoperimetric condition. 4. :x (p (x) y') + (Ar (x)-q (x)J y = 0; y (0) = 0; y (x1) = 0. The trivial solution y = 0 does not satisfy the isoperimetric condition, and nontrivial solutions, as is known, exist only for certain values of A called eigenvalues. Hence, A must be an eigenvalue. One arbitrary constant of the general solution of Euler's equation i.~ determined from the condition y (0)=0, 5 7 the other, from the isoperimetric condition. 5. y= - 2 x2+ 2 x; z=x. CHAPTER 10 5 1. z1 = 16a2 (x2-a2)(y'-b2). may be sought in the 2. y1 =(x-1)2 (0.124+0.2l8x). If greatt"r accuracy is needed, then the solution form z2 =(x2-a2) (y2-b2) [a0 +a1 (x2+y2)): 3. The exact solution is y = s~n xl-x. 4. The sm solution of Euler's equation is y=3.6072J1 (x)+0.75195Y1 (x)-x, where J 1 and Y 1 are Bessel functions. 5. The exact solution is y= 2 ~inhh2x -x. 6. If the solu- sm tion is saught in the form: Yt=X (x-1) (al +ctzx), Ya=X (x-1) (al +aax+aax2), then y2 =x(x-l) (0.1708+0.17436x), y3 =x (x-1) (0.1705+0.1760x-0.0018x2). The values of y2 and y3 coincide at the specified points to within 0.0001. Recommended Literature PART I I Petrovskii, J.. Leklsii po leorii obyknooennykh diflerenlsia/'nykh uravnenii (Lectures on Ordinary Differential Equations). 5th ed., Nauka, 1964 2. Malkin, I. Teoriya usloychivosli dvizheniya (Theory of Stability of Motion). Gostekhizdat, 1952 (to Ch. 4) 3. Malkin, I. Nekoloryie zadachi leorii nelineinykh kolebanii (Certain Problems in the Theory of Non-Linear Vibrations). Gostekhizdat, 1956 (to Sec. 8, Ch. 2) 4. Tikhonov, A. 0 zavisimosli reshenii di/ferenlsia/'nykh uravnenii ol malogo paramelra (On the Dependence of Solutions of Differential Equations on a Small Parameter). Matematicheskii sbornik, Vol. 22 (64):2 (1948) and Vol. 31 (72):3 (1952) (to Sec. 6, Ch. 4) 5. Stepanov, V. Kurs diOerentsia/'nykh uravnenii (A Course of Differential Equations), 8th ed., Fizmatgiz, 1959 6. Krylov, A. Lektsii o priblizhennykh vychisleniyakh (Lectures on Approximate Calculations). 5th ed., Gostekhizdat, 1950 (to Sec. 7, Ch. 1 and Sec. 6, Ch. 3) I. Berezin, 1., Zhidkov, N. Melody vychislenii (Methods of Calculations). Vol. 11. Fizmatgiz, 1960 (to Sec. 7, Ch. 1 and Sec. 6, Ch. 3) PART II I. Gelfand, 1., Fcmin, S. Variatsionnoye ischisleniye (The Calculus of Variations). Fizmatgiz, 1961 2. Lavrentiev, M., Lyusternik, L. Kurs varialsionnogo ischisleniya (A Course of the Calculus of Variations). 2nd ed. Gostekhizdat, 1950 3. Smirnov, V., Krylov, V., Kantorovich, L. Variatsionnoye ischisleniye (The Calculus of Variations). KUBUCH, 1933 4. Smirnov, V. Kurs vysshey malematiki (A Course of Higher Mathematics). Vol. 4, 4th ed. Fizmatgiz, 1958 5. Gunther, N. Kurs variatsionnogo ischisleniya (A Course of the Calculus of Variations). Gostekhizdat, 1941 6. Akhiezer, N. Lektsii po variatsionnomu ischisleniyu (lectures on the Calculus of Variations). Gostekhizdat, 1955 7. Lavrentiev, M., Lyusternik, L. Osnooy varialsionnogo ischisleniya (Fundamentals of the Calculus of Variations). Parts I and 2. Gostekhizdat, 1935 8. Pontryagin, L., Boltyanskii, V., Gamkrelidze, R., Mishchenko, E. Matemalicheskaya teoriya optimal'nykh protsessoo (Mathematical Theory of Optimal Processes). Fizmatgiz, 1961 9. Bellman, R. Dynamic Programming, Princeton University Press, 1957 10. Kantorovich, L., Krylov, V. Priblizhennyie melody vysshego analiza {Approximate Methods of Higher Analysis). 5th ed. Fizmatgiz, 1962 I1. Mikhlin, S. Pryamyie melody v matemalicheskoy fizike (Direct Methods in Mathematical Physics). Gostekhizdat, 1950 Asymptotically stable solution Bellman, R. 405 Berezin, I. S. 72 Bernoulli, Jacob 294, 295 Johann 294 Bernoulli equation 35 et seq. Bessel equation of order n 145 function of the first kind 147, 148 function of the second kind 147, 148 Bogolyubov, N. 409, 415 Boundary-value problems 17, 165 et seq. for second-order Iinear equations 167 Brachistochrone 317 problem of 294, 316, 346 Broken-line extremals 360 Calculus of variations 291 et seq. Castigliano, principle of 294 Catenary 404 Catenoids 316 Cauchy method 126, 278, 279, 287, 288 problem 17 C-discriminant curve 86, 368 Central field 365 Centre 64, 218 Charasteristic equation 112, 201 strips 283 Characteristics 278, 279, 283 of an equation 255, 258 Charpit 274 Chetayev, N. G. 227, 230 Chetayev instability theorem 226 Clairaut equation 79, 273 Comparison curve 306 Complete integral 271 Constraints, holonomir 396 nonholonomic 396 Continuous function 297 functional 297 Contraction·mapping principle 53 et seq. Control function 404 Coordinate functions 426 Decay constant 27 Degenerate equation 239 Index Density of Lagrange function 337 Dlnerentiable function 300 DifTerential 300 DifTerential equation 13 complete (general) integral of ?4 exact 37 et seq. existence and uniqueness theorem for 91 et seq. general solution of 19 integral of 24 integration of 14 order of 14 ordinary 13 partial 13 solution of 14 systems of 146 et seq. with variables separable 25 Direct method of finite differences 408 methods in variational problem 408 et seq. Direction field 19 Dirichlet rroblem 327 Disintegra ion constant 13 Distance 53 Duboshin, G. 244 Dynamical system 178 Equation of radioactive disintegration 13, 14 Equations, independent 389 with variables separable 23 et seq., 29 Euler 145, 294, 295 equations 116, 117, 118, 304 et seq., 310, 369 general solution of 341 in theory of differential equations 424 transformin,g to canonical form 383 et seq. Euler-Poisson equation 323 Euler finite-difference method 409 et seq 438 INDEX gamma function 146 method 207 method of integrating linear equations 113 method with iteration 67 polygonal curve 16 theorem on homogeneous functions 261, 264 . Exact differential equations 37 et seq. Existence and· uniqueness theorem for nth order differential equation 91 et seq. Extremals 310 broken-line 360 field of 365 et seq. of a variational problem 323 with corners 352 et seq. Extremum, conditional 295 connected 295 of a functional 303 sufficient conditions for 365 et seq. Fermat principle 294, 320 Field of extremals 365 et seq. Finite equation 187 Finite-difference method, direct 408 Euler's 409 et seq. First integral 95, 187 Focal point 64 Frequency equation 417 Function E (x, y, p, y') 371 et seq. Functional 295. 299 Functionals dependent on functions of several independent variables 325 et szq. on higher-order derivatives 321 et seq. Fundamental lemma of the calculus of variations 308 system of solutions of homogeneous lmear equations 106 Galerkin, B. 425 method 425 Geodesics 295 Green's function 168 et seq., 326 Half-stable solution 240 Hamilton 333 Hamilton-Jacobi t>quation 385 et seq. Holonomic constraints 396 Homogeneous differential equations of the first order 30 Homogeneous linear equations 32, 99, et seq., 253 fundamental system of solutions of 106 with constant coefficients 112 et seq. Hurwitz conditions 237 et seq. matrix 236 theorem 236 Influence function 168 Initial-value problem 17, 165 Integrable combinations 186, 189 Integral curves 20, 177 Integrating factor 44 Integrating-factor method 44 Integration of differential equations by differentiation 78 system of differential equal ions, approximate methods of et seq. 206 Interval of calculation 45 Isoclines 21 Isoperimetric conditions 399 problems 295, 330, 399 et seq. Jacobi condition 369 et seq., 373, 381, 383 equation 369, 370, 382 Jacobian 163, 260 Jacobi first method 288 Jordan normal form 205 Kantorovich, L. 409, 414, 416 Kantorovich method 420 et seq. Kovalevskaya, S. 252 Kovalevskaya theorem 252 Krylov, A. N. 72 Krylov, N. 409, 415 Krylov, V. 409, 414, 416 Kurosh, A. 236 Lagrange 295 equation 78 function, density of 337 Lagrangian 337 Laplace equation 327 Law of conservation of energy 384 Legendre condition 376, 381, 383 strong 376 Leibnitz 294 L'Hospital 294 INDEX 4111 Limit cycle 28, 235 half-stable 235 stable 235 unstable 235 Line of quickest descent (brachistoch· rone) 294 Linear differential equations, first-order 32 nth order 99 systems of 189 et seq. with constant coefficients 200 et seq. Linear difTerential operator 100 Linear equation 32 Linearly dependent functions 101 vectors 192 Linearly independent functions 102 vectors 192 Liouville Ill Lipschitz condition 46 et seq., 59; 91 et seq., 177, 238 Lyapunov, A. M. 153, 223 '-yapunov function 223, 247 second method 223 et seq. stability theorem 223 proof of 223 theorem 246 theorem on asymptotic stability 225 Lyusternik, L. 409 ~alkin, l. 230, 244 Malkin theorem 245, 246, 247 Method of characteristics 278 of double computation 97 of fixed points 53 of Lagrange and Charpit 2'i4 of successive approximation~ 57 of variation of parameters 33, 126 Metric space 53 complete 53 Mikhlin, S. 409, 414 Miln 67 Minimizing sequence 414 Minimum of elementary functional, sufficient conditions for 377 Moving-boundary prodlem 341 et seq., 347 et seq. Negativity of real parts of all roots of a polynomial, criteria of 236 et seq. Newton 294 interpolation formula 69 Nonholonomic constraints 396 Nonhomogeneous linear e4uation 33, 119, et seq .. :!53 Nonlinear equations, first-order :.!71 tl seq One-sided variations 361 et seq. Operator polynomial 135 Optimal control 405, 406 function 405 problem of 404 et seq. Ordinary (nonoperator) polynomials 136 et seq. division of 141 Orthogonality condition 20, 350 Ostrogradsky, M. Ill, 327, 333 Ostrogradsky equation 327. 328, 337, 414, 419 Ostrogradsky-Hamilton integral 395 principle 336, 337, 395 Ostrogradsky-Liouville formulas Ill Partial differential equations 13 first order 251 et seq. integration of 251 et seq. P discriminant curve 84 Periodic solutions 153 autonnmous case 160 nonresonance case 155 resonance case 157 resonance of nth kind 159 !Jeriodicity, conditions of 165 Perron, 0. 230 Persidsky, K. 230 Perturbations, instantaneous 244 Petrovsky, I. 409 Pfaffian equations 265 et seq. 271, 276 Poincare, A. 153 Poincare theorem 60 theorems on existence of periodic solutions 156, 163, 164 Point conjugate to another point 369 Poisson equation 13, 328. 413 Pontrya!;in, L. 405 Principle of least action 333, 337 Proper field 365 node 219 Proximity (of curves) 298 first-order 298 kth order 299 zero·order 298 Quadrature 24 Quasi! int>ar first order differential cquu· tion 253 440 INDEX Reciprocity principle 402 Reflection-of-extremals method 352 Refraction of extremals 355 et seq. Resonance case of characteristic equation 134 Rest points 179 elementary types of 214 et seq. Riccati equation 36 Ritz method 409, 411 et seq. Runge 67 Runge method 69, 72, 209 Saddle point 64, 215 Separable equations 23 et seq., 29 .Singular curve 62 integral curve 84 point 62 set83 solution 62, 84 Small-parameter method 153 et seq. Solution, half-stable 240 asymptotically -;table 212 stable 240, 245 unstable 240 Space of uniform convergence 55 Special field 381 Stable focal point 217 nodal point 215 o;olution 240, 245 Stability in the sense of Lyapunov 244 theory of 211 et seq. Step 45 Stokes theorem 268 Stormer 67 method 68, 72, ·208 Strict maximum 302 Strong maximum 303 minimum 303 Successive approximations 207 Superposition principle 120, 129, 139, 151, 197 Symmetric form ol a system ol equations 189 System of equations ol first dpproximation lor a system 230 Taylor expansion~ 208 series expansion method 71 Test for stability based on first approximation 229 et seq. Theorem on analytical dependence of solution on parameter 60 continuous dependence of solution on parameter and initial values 58 differentiability of solutions 61 Tikhonov, A. N. 61, 243 Total partial derivative 326 Transversality condition 345, 349, 350 Triangle rule 53 Unstable focal point 217 in the sense of Lyapunov 215 nodal point 215, 219 solution 240 Variation 297 of a function 305 of a functional 301, 307, 321 of limit values 342 of parameters, method of 33 Variational principles of physics 294 problems 294 direct methods in 408 et seq. extremals of 323 in parametric form 330 et seq. involving conditional extremum 389 et seq. Vasilieva, A, 243 Vector function 404 lines 253 surface 254 Weierstrass condition 373 function 373 Wronski, G. 103 Wronskian 103 et seq., 193, 195 Zhidkov, N. P. 72