Non-linear waves and solitons
Lecture notes1
Linus Wulff
Spring 2020
1
Thanks to M. Bartova for correcting several typos in these notes.
2
Chapter 1
Wave equations
The existence of waves is one of the most universal phenomena in physics. From mechanical
vibrations and sound to water waves to light and electromagnetic waves. Even at the most
fundamental level string theory suggests that our world is governed by waves on tiny strings.
In these lectures we will confine ourselves to waves in one spatial dimension, which we will
see is enough to see some very interesting and rich phenomena.
The waves you have encountered so far are solutions of what is often referred to simply
as the wave equation. In one dimension it takes the form
∂2u
∂t2
− v2 ∂2u
∂x2
= 0 . (1.1)
Here u(x, t) describes the profile of the wave as a function of space (x) and time (t) and v
is a constant – the velocity of the wave. This is the first wave equation to be systematically
studied, starting with d’Alembert in 1746. The general solution takes the form of the sum of
a right-moving and a left-moving wave1
u(x, t) = f(x − vt) + g(x + vt) , (1.2)
where f and g are arbitrary functions of one variable specifying the profile of the right-moving
and left-moving wave respectively at t = 0. Note that the sum (or superposition) of any
two such solutions gives a new solution, this is due to the fact that the wave equation (1.1)
is a linear equation, i.e. u enters linearly. Note also that the right-moving and left-moving
wave retain their shape as a function of time and do not interact with each other. This is
again a consequence of the linearity of the wave equation.
1.1 Dispersion, dissipation and non-linearity
In many situations is physics one encounters waves that do not quite satisfy this simple
idealized wave equation. They may for example display the phenomena of dispersion, i.e.
1
To see this notice that the wave equation factorizes as
0 =
∂
∂t
− v
∂
∂x
∂
∂t
+ v
∂
∂x
u = −v2 ∂
∂x−
∂
∂x+
u ,
where we have defined x±
= x ± vt. This implies that ∂u
∂x+ = g (x+
) and therefore u(x−
, x+
) = f(x−
) + g(x+
)
for some functions f and g.
3
4 CHAPTER 1. WAVE EQUATIONS
the wave tend to spread out with time, or dissipation, i.e. the amplitude decays with time
and the wave loses energy to the surroundings. These two phenomena can also be modeled
with simple linear wave equations. To see this we focus on the right-moving solution of the
wave equation (1.1), u(x, t) = f(x − vt). A wave equation which describes this is simply
˙u + vu = 0 , (1.3)
where we have defined the short-hand notation ˙u = ∂u
∂t and u = ∂u
∂x . If we look for a solution
of the form u = Aei(ωt−kx) (we can always take the real part at the end) we find that the
frequency ω and wave-number k are related to the velocity v as
v =
ω
k
. (1.4)
Now consider instead the wave equation
˙u + au + bu = 0 , (1.5)
with a, b constants. Taking again the same ansatz for u we now find
ω = ak − bk3
. (1.6)
The relation between ω and k is known as the dispersion relation. The phase velocity of
the wave is given by ω/k and we see that for non-zero b it depends on the wave number k.2 This
means that if we superpose waves with different wave number (note that this is allowed since
the equation is still linear) these will move with different velocity and spread out or disperse.
Therefore adding the term bu has introduced dispersion. Adding instead a term cu one
would find that ω gets an imaginary piece and therefore the solution u = Aei(ωt−kx) will have
an exponentially decaying (or increasing) amplitude and therefore the wave is loosing energy
to its surroundings – there is dissipation. A slight generalization of this argument shows that
terms with an odd number of spatial derivatives introduce dispersion while terms with an
even number of derivatives introduce dissipation.
So far we have considered linear wave equations which satisfy the superposition principle.
A more realistic wave equation may also contain non-linearities, for example our simplest
wave equation considered above (with v = 1 for simplicity) could be an approximation, valid
for small wave amplitude, to the non-linear wave equation
˙u + (1 + u)u = 0 . (1.7)
Clearly the quadratic term in u means that the sum of two solutions will in general not be a
solution – there is no superposition principle in this case. Such non-linear wave equations are
much harder to solve, in fact in general they can only be solved numerically.3 However it turns
out that the equation above is an approximation to a more interesting equation which can
be solved exactly. Let us first note that shifting u → u − 1 the equation becomes ˙u + uu = 0
and the coefficient of the second term can be changed to any non-zero number by rescaling
2
The group velocity, the speed at which physical information in the wave travels, is given instead by
∂ω/∂k but this does not affect the discussion here.
3
However, in this simple case it is not hard to see that the general solution is u(x, t) = f(x − (1 + u)t) for
some function f and one can then in principle solve for u given the initial profile u(x, 0) = f(x).
1.2. THE SOLITARY WAVE 5
u. The equation we will spend a large fraction of this course studying is this equation with
the incorporation of dispersion4
KdV equation: ˙u − 6uu + u = 0 (1.8)
and it is known as the Korteweg-de Vries (or KdV) equation. It is the simplest wave
equation incorporating both non-linearity and dispersion. Note that the coefficients in front
of each term are purely conventional and can be set to any non-zero value by rescaling u,
x and t. This equation has the very special property of being ’exactly integrable’ which
means that we will be able to solve it (in a sense that will become clear later on). Later on
in this course we will meet a few other examples of exactly integrable equations.
1.2 The solitary wave
The story of the KdV equation goes back all the way to 1834. John Scott Russell, a Scottish
engineer and naval architect, was performing experiments in water flow in channels to
determine the most efficient design for canal boats when he witnessed a very interesting
phenomenon. In his ’Report on Waves’ from 1844 he recalls the event as follows:
I was observing the motion of a boat which was rapidly drawn along a narrow channel
by a pair of horses, when the boat suddenly stopped - not so the mass of water in the
channel which it had put in motion; it accumulated round the prow of the vessel in a state
of violent agitation, then suddenly leaving it behind, rolled forward with great velocity,
assuming the form of a large solitary elevation, a rounded, smooth and well-defined
heap of water, which continued its course along the channel apparently without change
of form or diminution of speed. I followed it on horseback, and overtook it still rolling
on at a rate of some eight or nine miles an hour, preserving its original figure some
thirty feet long and a foot to a foot and a half in height. Its height gradually diminished,
and after a chase of one or two miles I lost it in the windings of the channel.
He continued to study the properties of these waves, which eventually became known as
solitary waves, in water tank experiments showing for example that their velocity is related
to the (undisturbed) water depth h and amplitude a by the equation
v2
= g(h + a) , (1.9)
where g is the acceleration of gravity. However a theoretical understanding had to await the
works of Boussinesq (1871) and Rayleigh (1876) who were able to derive Russell’s empirical
velocity formula. Finally Korteweg and de Vries (1895) derived the KdV equation5 (1.8) for
water waves in a shallow channel and showed that it had a solution with the properties of
Russell’s solitary wave.
Let us look for a solution of the KdV equation in the form of a traveling wave. To do this
we take the ansatz u(x, t) = f(x − vt), with f and function of one variable and v a constant
to be determined. The KdV equation then reduces to
−vf − 6ff + f = 0 , (1.10)
4
The corresponding equation with dissipation, i.e. ˙u − 6uu + u = 0, is known as the Burgers equation
and can also be solved exactly.
5
In fact this equation had already been written down by Boussinesq 20 years earlier.
6 CHAPTER 1. WAVE EQUATIONS
which integrates to −vf − 3f2 + f = a for some constant a and multiplying by f and
integrating again we find
−
v
2
f2
− f3
+
1
2
f 2
= af + b , (1.11)
with a, b integration constants. Requiring f, f , f → 0 as |x| → ∞ sets a = b = 0 and we
have
f 2
= f2
(2f + v) , (1.12)
and therefore a real solution exists only if 2f + v ≥ 0. Integrating this equation gives
df
f
√
2f + v
= ± dx−
, (1.13)
and the substitution f = −1
2v sech2
y (recall that sech x = 1/ cosh x) gives y = 1
2v1/2(x− −
x−
0 ) = 1
2v1/2(x − vt − x0) (the sign of y is clearly irrelevant) and therefore
u(x, t) = f(x − vt) = −1
2v sech2 1
2v1/2
(x − vt − x0) . (1.14)
This is the solitary wave solution. The minus sign reflects our choice of signs in (1.8) and
we see that for u to describe the amplitude of a water wave we should send u → −u. Note
that the greater the amplitude of the wave the narrower it is and the faster it travels. Note
that the amplitude (and position x0 at t = 0) uniquely specifies the solution. Notice also that
from the dispersive term in the KdV equation one might have expected a wave like this to
spread out in time and lose its character, but the non-linearity of the equation is perfectly
balancing this tendency for the wave to disperse.
1.3 Solitons
After the paper of Korteweg and de Vries there was not much work on the subject until the
1960’s. In 1965 Zabrusky and Kruskal did some computer simulations of the KdV equation
and made some very interesting observations. They found that there are solutions which
consist of several separated sech2
waves, which by itself is not so surprising, but since these
waves travel at different velocity they will typically collide with each other. The remarkable
thing they observed is that after such a collision the waves would continue undeformed. Since
these waves, while interacting strongly with each other, nevertheless retained their individual
identity in collisions they called them solitons (cf. electron, proton, etc.) to emphasize their
similarity to particles.
The fact that these sech2
waves, or KdV solitons, can pass through each other undeformed
sounds like there is a principle of superposition at work. However, it is easy to see that this
not quite what is happening. While the waves retain their shape when they pass through
each other they do interact. Amazingly the only effect of the interaction is to displace the
waves relative to where they would have been if they had passed through each other without
interacting.
Such solitons have since been found in other (typically one-dimensional) systems and are
often associated with the system being integrable (solvable). We will meet a few examples
later in this course.
1.4. APPLICATIONS OF THE KDV EQUATION 7
1.4 Applications of the KdV equation
We have seen that the KdV equation is the simplest equation one can write which incorporates
both dispersion and non-linearity. This also suggests that it should have many applications.
Indeed, for a linear wave with dispersion in one dimension the dispersion relation takes the
form
ω(k) = kc(k2
) , (1.15)
since only odd derivatives of u are allowed. In the long wavelength limit, k 1, we find
ω
k
= c0 − c1k2
. (1.16)
This approximate dispersion relation is obtained from the wave-equation
˙u + c0u + c1u = 0 . (1.17)
If the wave is propagating in a continuous material then ˙u is replaced by the ’material derivative’
˙u + uu where we assume the underlying fluid velocity to be proportional to u itself. In
that case the equation becomes
˙u + c0u + auu + c1u = 0 , (1.18)
which is transformed to the KdV equation (1.8) by the change of variables x → x − c0t,
t → at. This shows that the KdV equation can be expected to describe the propagation of
long wavelength waves in various circumstances.
8 CHAPTER 1. WAVE EQUATIONS
Chapter 2
Elementary solutions of the KdV
equation
In the last chapter we derived the solitary wave solution of the KdV equation (1.8). Assuming
a solution of the form of a right-moving wave of a fixed form
u(x, t) = f(ξ) , ξ = x − vt . (2.1)
With this ansatz we found that we could integrate the KdV equation twice to obtain (1.11)
−
v
2
f2
− f3
+
1
2
f 2
= af + b , (2.2)
where a, b are integration constants. We then assumed a solution which is constant far away,
f, f , f → 0 as ξ ± ∞, which sets a = b = 0 and leads to the solitary wave solution (1.14).
2.1 General wave solutions of fixed shape
We will now derive the most general wave solutions of fixed shape. We therefore want to solve
the equation
−
v
2
f2
− f3
+
1
2
f 2
= af + b , (2.3)
without assuming anything about the constants a, b. We first rearrange this equation to read
1
2
f 2
= F(f) , (2.4)
where we have introduced the function
F(f) = f3
+
v
2
f2
+ af + b . (2.5)
We will be interested only in real bounded solutions of this equation. Clearly real solutions
exist only for F > 0. We see from the equation that f changes monotonically until f vanishes,
which happens at a zero of F. To begin with we therefore analyze what happens near a zero
of F(f). Since F(f) is a cubic polynomial it can have a simple, double or triple zero.
9
10 CHAPTER 2. ELEMENTARY SOLUTIONS OF THE KDV EQUATION
(i) Simple zero: Taylor expanding around the point f = f1 where F vanishes we find
(f )2
= 2(f − f1)F (f1) + O((f − f1)2
) , (2.6)
with solution
f = f1 +
1
2
(ξ − ξ1)2
F (f1) + O((ξ − ξ1)3
) . (2.7)
We see that f has a local minimum(maximum) at ξ1 for F (f1) > 0 (F (f1) < 0).
(ii) Double zero: Taylor expanding around this point we find
(f )2
= (f − f1)2
F (f1) + O((f − f1)3
) . (2.8)
We see that we must have F (f1) ≥ 0 and the solution is
f − f1 = ce±ξ
√
F (f1)
, as ξ → ∞ , (2.9)
with c some constant, for the solution to be bounded. The solution extends from −∞
to +∞, where f → f1, with a bump somewhere in between.
(iii) Triple zero: Here F(f) = (f − f1)3 so that
1
2
(f )2
= (f − f1)3
, (2.10)
with solution
f = f1 +
2
(ξ ± c)2
, (2.11)
for some constant c. This solution is however unbounded at ξ = c and we will therefore
discard it.
We conclude that there are only two possibilities: Either f changes sign across f1 (since that
point is a local minimum or maximum) or f → 0 as ξ → ±∞.
Consider first the situation where F crosses zero from below at a simple zero and remains
positive to the right of the crossing (so the solution exists there). If f (ξ0) > 0 for some ξ0
then, since f is increasing, F(f) > 0 for all ξ > ξ0. But that means that f → +∞ as ξ → +∞
and the solution is unbounded. If instead f (ξ) < 0 f will decrease until it reaches f1. This
is a simple zero so f changes sign and once again f → +∞ as ξ → +∞ and the solution is
unbounded.
This leaves only two possible cases to analyze. Either F has a simple zero at f1, where
it crosses from negative to positive, and a double zero at f2 where it touches the line F = 0
and then is positive again. Or F has a simple zero at f1, where it crosses from negative to
positive, a simple zero at f2, where it crosses from positive to negative, and finally a simple
zero at f3 where it again crosses from negative to positive. In the first case the solution has
a minimum at f = f3 since F(f3) > 0 and approaches f = f1 as ξ ± ∞. This is the solitary
wave solution we found previously.
It therefore remains only to analyze the case with three simple zeros. The solution can
exist between f = f2 and f = f3 since F is positive there. It has a local maximum at f2
(F (f2) < 0) and a local minimum at f3 (F (f3) > 0). At these points f changes sign.
2.1. GENERAL WAVE SOLUTIONS OF FIXED SHAPE 11
Therefore the solution oscillates between f2 and f3. Half the period of the solution is given
by the distance from the minimum to the maximum
ξ2
ξ3
dξ =
f2
f3
df
f
=
f2
f3
df
2F(f)
. (2.12)
The solution f itself is implicitly given by the integral
ξ = ξ3 ±
f
f3
dg
2F(g)
, (2.13)
with the ± according to whether f > 0 or f < 0. Since F(g) is cubic with simple roots are
f1, f2, f3 we can write
F(g) = (g − f1)(g − f2)(g − f3) , f1 > f2 > f3 . (2.14)
And substituting
g = f3 + (f2 − f3) sin2
θ (2.15)
in the integral we find
ξ = ξ3 ±
2
f1 − f3
φ
0
dθ
1 − m sin2
θ
, m =
f2 − f3
f1 − f3
≤ 1 , (2.16)
with
f = f3 + (f2 − f3) sin2
φ . (2.17)
Consider the function
v =
φ
0
dθ
1 − m sin2
θ
, 0 ≤ m ≤ 1 . (2.18)
Let us compare it to the function
w =
ψ
0
dt
√
1 − t2
. (2.19)
Making the substitution t = sin θ we find that w arcsin ψ or sin w = ψ.
In analogy with this Jacobi defined two new inverse functions as
sn v = sin φ , cn v = cos φ . (2.20)
They are two of the Jacobi elliptic functions. They are often written
sn(v|m) , cn(v|m) , (2.21)
denoting also the dependence on the parameter m. In the special cases m = 0 and m = 1
these elliptic functions reduce to more familiar functions. Setting m = 0 we find v = φ so
that
sn(v|0) = sin φ = sin v , cn(v|0) = cos φ = cos v . (2.22)
In this sense the functions sn and cn are a generalization of sin and cos, as suggested by their
names. Setting m = 1 we also find an integral we can do and one finds for example
cn(v|1) = cos φ =
1
cosh v
= sech v , (2.23)
12 CHAPTER 2. ELEMENTARY SOLUTIONS OF THE KDV EQUATION
precisely the function that appeared in the solitary wave solution. Note that cn and sn are
periodic functions except for at m = 1 where the periodicity is lost.
The solution we found above to the KdV equation (2.16) and (2.17) can now be expressed
using these elliptic functions. Since
f = f3 + (f2 − f3) sin2
φ = f2 − (f2 − f3) cos2
φ , (2.24)
we find that the solution takes the form
f = f2 − (f2 − f3) cn2 f1 − f3
2
(ξ − ξ3) m , m =
f2 − f3
f1 − f3
. (2.25)
This solution was found by Korteweg and de Vries in their 1895 paper and they dubbed the
solution the cnoidal wave. The solution depends on three parameters f1 ≥ f2 ≥ f3, the
roots F(f) defined in (2.5). From this equation we see that the speed of the wave is given by
v = −2(f1 + f2 + f3).
2.2 Limits of the cnoidal wave
The cn function has two limits were it reduces to more familiar function, when the parameter
takes the limiting values m = 0 and m = 1. Correspondingly there are two limits in which
the cnoidal wave solution simplifies. Consider first the limit where the amplitude of the wave,
defined as a = 1
2(f2 −f3), tends to zero. Since m = 2a/(f1 −f3) we find that m → 0 as a → 0
so that cn(v|m) → cos v and cnoidal wave solution (2.25) becomes
f ∼ f2 − 2a cos2 f1 − f3
2
(ξ − ξ3) as a → 0 . (2.26)
Setting k = 2(f1 − f3) and ˆf2 = f2 − a this can be written as
f = ˆf2 − a cos (k(x − vt − x0)) + O(a2
) . (2.27)
This is a linear wave of amplitude a oscillating about the level f = ˆf2. Furthermore we have
ω = kv = −2k(f1 + 2 ˆf2) = −k(k2
+ 6 ˆf2) + O(a) . (2.28)
When a → 0 this reduces to the dispersion relation for the linear equation
˙u − 6 ˆf2u + u = 0 , (2.29)
obtained by linearizing the KdV equation setting u → u + ˆf2 and letting |u| 1. Therefore
the limit of small amplitude of the cnoidal wave gives a linear wave with the correct dispersion
relation.
The solitary wave limit requires the two roots f1, f2 to become a double root. We therefore
take the limit f2 → f−
1 with f3 fixed. This implies
m =
f2 − f3
f1 − f3
→ 1−
(2.30)
2.3. OTHER SOLUTIONS OF THE KDV EQUATION 13
so that cn(v|m) → sech v and the cnoidal wave solution (2.25) reduces to
f → f1 − (f1 − f3) sech2 f1 − f3
2
(x − vt − x0) . (2.31)
Setting f1 − f3 = 1
2a this becomes
f → f1 −
a
2
sech2
√
a
2
(x + 6f1t − at − x0) . (2.32)
This is the original solitary wave transformed by u → u + f1 and (x, t) → (x + 6f1t, t).
Alternatively we may set f1 and note that this coindices with (1.14) (setting v = a).
2.3 Other solutions of the KdV equation
The KdV equation has many more solutions than the ones we’ve met so far. Examples of
simple type are so-called similarity solutions and rational solutions. Similarity solutions are
for examples solutions of the form
u(x, t) = tm
f(η) , η = xtn
, (2.33)
with the integers m, n chosen so that f satisfies an ordinary differential equation. For example
the non-linear equation
˙u + uu = 0 , (2.34)
has a solution
u(x, t) = tm
f(xtn
) (2.35)
if m + n = −1 and
mf − (1 + m)ηf + ff = 0 . (2.36)
Setting m = 0 we find that either f = 0, so that f =constant, or f = η. Thus u(x, t) = x
t is
a similarity solution of (2.34).
Similarly the dispersive equation
˙u + u = 0 , (2.37)
has a solution u(x, t) = f(xt−1/3) with
−
1
3
ηf + f = 0 , (2.38)
which can be solved in terms of Airy functions.
Combining these two examples we have the KdV equation
˙u − 6uu + u = 0 , (2.39)
which can be solved by setting
u(x, t) = −(3t)−2/3
f(η) , η = x(3t)−1/3
(2.40)
14 CHAPTER 2. ELEMENTARY SOLUTIONS OF THE KDV EQUATION
with
f + (6f − η)f − 2f = 0 . (2.41)
This equation can be reduced to a so-called Painlev´e equation, with solution describing a wave
profile that decays as η → +∞ and oscillates as η → −∞. The appearance of an equation of
Painlev´e type is directly related to the existence of soliton solutions.
Finally we have the solutions of rational type, i.e. solutions which are rational functions
of the variables. We have already met one example, the solution u(x, t) = x
t of (2.34) which
is also a similarity solution. The KdV equation
˙u − 6uu + u = 0 , (2.42)
also has a simple rational solution. Let’s assume that
u(x, t) = u(x) , u, u , u → 0 as |x| → ∞ . (2.43)
The the equation reduces to
−6uu + u = 0 , (2.44)
which is not hard to solve yielding u = 2/x2 (picking the pole to be at x = 0). This
is essentially the solution corresponding to F(f) in (2.5) having a triple zero encountered
before.
In fact there exists a whole hierarchy of rational solutions of the KdV equation. The next
one is
u(x, t) =
6x(x3 − 24t)
(x3 + 12t)2
. (2.45)
In the KdV case these solutions are all singular, but there are other integrable equations for
which this is not the case.
Chapter 3
The scattering and inverse
scattering problems
So far we have discussed some special solutions of the KdV equation
˙u − 6uu + u = 0 , (3.1)
such as traveling waves of fixed form. Our next task is to understand how to solve the general
initial value problem for (3.1). That is to find u(x, t) for all t > 0 given the initial condition
u(x, 0). This problem turns out to be related to the scattering problem of quantum mechanics.
We therefore need to first recall how that works.
The scattering problem in quantum mechanics involves solving the time independent
Schr¨odinger equation (in 1 dimension)
ψ + (λ − u)ψ = 0 , (3.2)
where u(x) is the potential and λ is the eigenvalue, i.e. the energy. The scattering problem is
to determine the eigenvalues λ given a potential u(x). Mathematically this is called a SturmLiouville
problem. Less familiar is the fact that the scattering data, i.e. the form of ψ(x; λ)
as x → ±∞, can in fact determine uniquely the potential u(x) which gave rise to these data.
This is the inverse scattering problem.
The fact that we call the potential u by the same letter as a solution to the KdV equation
is not an accident. The relation between u(x) and the KdV solution u(x, t) will become clear
later.
3.1 The scattering problem
For appropriate solutions of
ψ + (λ − u)ψ = 0 , (3.3)
to exist we will require that u(x) decays fast enough at infinity. In fact we need that (Faddeev
1958)
∞
−∞
(1 + |x|)|u(x)|dx < ∞ . (3.4)
The reasons for this condition are technical and won’t be discussed here. We will also typically
assume that u(x) is infinitely differentiable, although this condition is not necessary. A
15
16 CHAPTER 3. THE SCATTERING AND INVERSE SCATTERING PROBLEMS
familiar fact about this eigenvalue problem from quantum mechanics is that there are two
types of eigenvalues: those with λ > 0 (positive energy) and those with λ < 0 (negative
energy). Since u → 0 as x → ±∞ we have
ψ ∼ −λψ . (3.5)
So for λ > 0 the solution is
λ > 0 : ψ ∼ aei
√
λ x
+ be−i
√
λ x
as x → ±∞ . (3.6)
While for λ < 0 we have
λ < 0 : ψ ∼ ce−
√
−λ x
+ de
√
−λ x
as x → ±∞ . (3.7)
This solution is unbounded as x → +∞ unless we set d = 0. This condition leads to a
discrete spectrum of allowed eigenvalues λ (energies). Physically these solutions correspond
to bound states since the wave function decays at infinity. Whereas for λ > 0 we need no
extra condition and we have a continuous spectrum. These solutions are called scattering
solutions since they oscillate at infinity rather than decay. To summarize we have
λ < 0 Neg. Energy Bound state solutions Decay at infinity Discrete spectrum
λ > 0 Pos. Energy Scattering solutions Oscillate at infinity Continuous spectrum
For some u(x) there may not exist a discrete spectrum at all (e.g. u(x) ≥ 0).
Integrating the Scr¨odinger equation twice one finds that ψ is continuous. So we consider
continuous, bounded, and usually at least once differentiable eigenfunctions ψ.
For the discrete eigenvalues the eigenfunctions decay exponentially at x → ±∞ so they
can be integrated
∞
−∞
|ψ|dx < ∞ and
∞
−∞
|ψ|2
dx =
∞
−∞
ψ2
dx < ∞ . (3.8)
The last equality says that ψ is square integrable (recall that ψ is real for the discrete eigenvalues).
This is not true for the continuous eigenfunctions (in that case
∞
−∞ ψ2dx = ∞).
We introduce a convenient representation for the eigenvalues and eigenfunctions as follows.
For the discrete spectrum we write
κn = −λn n = 1, 2, 3, . . . , N , (3.9)
and we order them according to κ1 < κ2 < · · · < κN (we won’t consider cases with degenerate
eigenvalues). Then
ψn(x) ∼ cne−κnx
as x → +∞ , (3.10)
where the constant cn is fixed by normalizing so that
∞
−∞ ψ2dx = 1.
For the continous spectrum we write
k =
√
λ (3.11)
3.1. THE SCATTERING PROBLEM 17
and we define the solution with the following oscillatory behavior at infinity
ˆψ(x; k) =
e−ikx + beikx as x → +∞
ae−ikx as x → −∞
(3.12)
Physically this corresponds to an incident wave of unit amplitude coming in from x = +∞,
a transmitted wave of amplitude a going out to x = −∞ and a reflected wave of amplitude b
going back out to x = +∞. The complex constants a(k) and b(k) can be determined uniquely
for a given u(x).
Consider two different discrete eigenfunctions
ψn + (κ2
n − u)ψn = 0 ψm + (κ2
m − u)ψm = 0 . (3.13)
We find
(κ2
n − κ2
m)ψnψm = ψnψm − ψmψn = (W(ψn, ψm)) , (3.14)
where the Wronskian is defined as W(α, β) = αβ − βα . Integrating we find
(κ2
n − κ2
m)
∞
−∞
ψnψmdx = [W(ψn, ψm)]∞
−∞ . (3.15)
Since ψ → 0 as x → ±∞ the RHS vanishes as we find
∞
−∞
ψnψmdx = 0 for m = n , (3.16)
i.e. the discrete eigenfunctions are orthogonal. The continous eigenfunctions ˆψ are also
orthogonal to all the discrete eigenfunctions. (In fact the discrete and continuous eigenfunctions
form a complete set, so that any square integrable function can be represented as a
linear combination of ψns and an integral of ˆψ over k.)
If θ, φ are two solutions with the same eigenvalue we have
(W(θ, φ)) = 0 ⇒ W(θ, φ) = const. (3.17)
If φ is proportional to θ then the Wronskian vanishes. For the continuous eigenfunction ˆψ it is
easy to see that its complex conjugate ˆψ∗ has the same eigenvalue. We can find the constant
Wronskian by evaluating it at x = ±∞ giving
W( ˆψ, ˆψ∗
) = 2ik|a|2
(3.18)
and
W( ˆψ, ˆψ∗
) = 2ik(1 − |b|2
) (3.19)
respectively. These expressions must be equal and we find the condition
|a|2
+ |b|2
= 1 . (3.20)
In the scattering problem a, b are transmission and reflection coefficients and this condition
follows from energy conservation.
To learn more about the solution of the scattering problem we have to consider specific
forms of the potential u(x).
18 CHAPTER 3. THE SCATTERING AND INVERSE SCATTERING PROBLEMS
Example 1: delta function
We take
u(x) = −U0δ(x) (3.21)
with U0 a constant and δ(x) the Dirac delta function. Integrating the Schr¨odinger equation
from − to gives
[ψ ]− +
−
(λ + U0δ(x))ψ(x)dx = 0 (3.22)
so that the discontinuity of the derivative is given by
disc(ψ ) = lim
→0
(ψ ( ) − ψ (− )) = −U0ψ(0) . (3.23)
We see that ψ is continuous but not differentiable at x = 0. The discrete eigenfunctions are
ψn(x) =
αne−κnx x > 0
βneκnx x < 0
(3.24)
since u = 0 for x = 0. For ψn to be continuous we need αn = βn and normalizing we find
1 =
∞
−∞
ψ2
dx =
0
−∞
α2
ne2κnx
dx +
∞
0
α2
ne−2κnx
dx =
α2
n
κn
, (3.25)
so that αn =
√
κn. Finally the discontinuity of ψ requires
disc(ψn) = −2κnαn = −U0ψn(0) = −U0αn , (3.26)
so that κn = 1
2U0. There is only one eigenvalue λ1 = −1
4U2
0 (κ1 = 1
2U0) and only if U0 > 0
since αn =
√
κn = U0/2 must be real.
Next we consider the continuous eigenfunctions. We have
ˆψ(x; k) =
e−ikx + beikx as x > 0
ae−ikx as x < 0
. (3.27)
Continuity at x = 0 requires a = 1 + b. The discontinuity of ψ at x = 0 gives
disc( ˆψ ) = 2ikb = −U0(1 + b) , (3.28)
or
b(k) = −
U0
U0 + 2ik
. (3.29)
We see that in this example there is always a continuous spectrum but only a discrete spectrum
for U0 > 0 (negative potential) with only one bound state.
Note that the pole in b(k), in the upper half complex plane, at k = − i
2U0 corresponds
to λ = k2 = −1
4U0 which is the discrete eigenvalue! This is a general result that allows the
discrete eigenvalues to be determined from the continuous eigenfunctions.
Next we consider another example of a potential which we will need when we discuss the
KdV equation.
3.1. THE SCATTERING PROBLEM 19
Example 2: sech2
function
We take the potential to be
u(x) = −U0 sech2
x , (3.30)
with U0 a constant. The Schr¨odinger equation becomes
ψ + (λ + U0 sech2
x)ψ = 0 . (3.31)
It is convenient to introduce a new variable T = tanh x (−1 < T < 1 for , −∞ < x < ∞).
We have
d
dx
= sech2
x
d
dT
= (1 − T2
)
d
dT
(3.32)
and the Schr¨odinger equation takes the form
(1 − T2
)
d
dT
(1 − T2
)
dψ
dT
+ λ + U0(1 − T2
) ψ = 0 , (3.33)
or equivalently
d
dT
(1 − T2
)
dψ
dT
+ U0 +
λ
1 − T2
ψ = 0 . (3.34)
This is in fact the associated Legendre equation. This is the equation you solve to find the
spherical harmonics, for example when finding the wave functions for the hydrogen atom. To
use our knowledge from that case we will assume, to start with, that
U0 = ( + 1) , (3.35)
where is an integer (the angular momentum quantum number in the case of the hydrogen
atom). Consider the discrete eigenvalues
λ = −κ2
< 0 . (3.36)
Bounded solutions exist only for κm = m (m = 1, 2, . . . , ), as is familiar from the case of the
hydrogen atom (where m is the quantum number associated to the angular momentum in the
z-direction). The eigenfunctions are (up to normalization) the associated Legendre functions
(in the spherical harmonics case we have T = cos θ)
Pm
(T) = (−1)m
(1 − T2
)m/2 dm
dTm
P (T) , (3.37)
where
P (T) =
1
2 !
d
dT
(T2
− 1) (3.38)
is the Legendre polynomial of degree .
For example for = 2 (U0 = 6) we have
ψ1 ∝ P1
2 (tanh x) = −3 tanh x sech x , ψ2 ∝ P2
2 (tanh x) = 3 sech2
x . (3.39)
Normalizing we find
ψ1 =
3
2
tanh x sech x , ψ2 =
√
3
2
sech2
x , (3.40)
20 CHAPTER 3. THE SCATTERING AND INVERSE SCATTERING PROBLEMS
corresponding to eigenvalues λ1 = −1 and λ2 = −4, respectively.
Now we turn to the continuous spectrum, λ = k2, for which the analysis is bit more
involved. We will keep U0 arbitrary in the following. Recall that we defined the eigenfunctions
behaving asymptotically as
ˆψ(x; k) =
e−ikx + beikx as x → +∞
ae−ikx as x → −∞
(3.41)
The solution of the associated Legendre equation with this behavior involves a hypergeometric
function. The hypergeometric function F(a, b; c; z) (often 2F1(a, b; c; z)) is the solution of the
second order ODE
z(1 − z)F + (c − (a + b + 1)z)F − abF = 0 , (3.42)
where a, b, c are constants and z is the variable. Relating this equation to the associated
Legendre equation gives the solution with the correct behavior at infinity
ˆψ(x; k) = a(k)2ik
(sech x)−ik
F(c+, c−; 1 − ik; 1
2(1 + T)) , c± = 1
2 − ik ± U0 + 1
4 . (3.43)
As x → −∞ we have 1
2(1 + T) → 0 and
ˆψ(x; k) ∼ a(k)e−ikx
(3.44)
while x → +∞ corresponds to 1
2(1 + T) → 1 giving
ˆψ(x; k) ∼ a(k)
Γ(1 − ik)Γ(ik)
Γ(c+)Γ(c−)
e−ikx
+ a(k)
Γ(1 − ik)Γ(−ik)
Γ(1 − ik − c+)Γ(1 − ik − c−)
eikx
. (3.45)
Comparing to (3.41) we read off
a(k) =
Γ(c+)Γ(c−)
Γ(1 − ik)Γ(ik)
, b(k) = a(k)
Γ(1 − ik)Γ(−ik)
Γ(1 − ik − c+)Γ(1 − ik − c−)
. (3.46)
The expression for b(k) can be simplified using the identity
Γ(1
2 + z)Γ(1
2 − z) =
π
cos(πz)
. (3.47)
Using this we have
Γ(1 − ik − c+)Γ(1 − ik − c−) = Γ(1
2 − U0 + 1
4)Γ(1
2 + U0 + 1
4) =
π
cos π U0 + 1
4
, (3.48)
so that
b(k) =
a(k)
π
cos π U0 + 1
4 Γ(1 − ik)Γ(−ik) . (3.49)
But now we notice something curious: for certain U0 b(k) vanishes for all k. Namely if
U0 + 1
4 = + 1
2 ⇒ U0 = ( + 1
2) − 1
4 = ( + 1) , (3.50)
the same values we considered before when discussing the discrete eigenvalues. Recall that
b(k) is the amplitude of the reflected wave in the scattering problem. Potential for which this
3.2. THE INVERSE SCATTERING PROBLEM 21
vanishes are called reflectionless potentials. The case U0 = ( + 1), with discrete eigenfunctions
given by the associated Legendre functions, is therefore an example of a reflectionless
potential.
Recall from last time that the poles in a(k), b(k) in the upper half-plane correspond to
the discrete eigenvalues. We can therefore use our expressions for a(k), b(k) to learn about
the discrete spectrum for general U0.
The poles in a(k), b(k) arise from poles in the Gamma functions and Γ(x) has poles at
x = −m for m = 0, 1, 2, . . .. Poles in the upper half-plane can come only from c− and we find
−m = c− = 1
2 − ik − U0 + 1
4 (3.51)
or
k = i U0 + 1
4 − (m + 1
2) . (3.52)
If U0 > 0 there are a finite number of poles in the upper half-plane, i.e. a finite number
of discrete eigenvalues. The eqigenfunctions are a generalization of the associated Legendre
functions Pm to non-integer , m.
In summary we have learned that for U0 > 0 there are a finite number of discrete eigenvalues,
while for U0 < 0 there is only a continuous spectrum. We have also seen that for some
special values of U0 (U0 = ( + 1)) the potential is reflectionless, i.e. b(k) = 0 for all k.
This completes our study of the scattering problem. We now turn to the inverse problem.
3.2 The inverse scattering problem
Eigenvalue problems such as the scattering problem were fairly well understood by about
1850, but the inverse problem was not solved until 1951. Physically the problem is like the
problem of finding the shape of a drum from the sounds it makes. In our terms, for example,
given b(k) can we find u(x)? This is much less straightforward than the scattering problem.
This section will therefore be a bit more technical.
To motivate the first part of the calculation we start with the classical linear wave equation
¨φ − φ = 0 . (3.53)
One way to solve it is to use the Fourier transform. Writing (we will set t = z for later
convenience)
φ(x, z) =
1
2π
∞
−∞
ψ(x; k)e−ikz
dk (3.54)
then
ψ(x; k) =
∞
−∞
φ(x, z)eikz
dz (3.55)
is the Fourier transform of φ(x, z) with respect to z. The wave equation becomes
ψ + k2
ψ = 0 . (3.56)
Suppose we are interested in solutions with
ψ ∼ eikx
as x → +∞ . (3.57)
22 CHAPTER 3. THE SCATTERING AND INVERSE SCATTERING PROBLEMS
This can be done by taking
φ(x, z) = δ(x − z) + K(x, z) , (3.58)
with K(x, z) = 0 for z < x and satisfying ¨K − K = 0, since then
ψ(x; k) = eikx
+
∞
x
K(x, z)eikz
dz , (3.59)
which has the required behavior as x → +∞.
The equation we are really interested in is
ψ + (k2
− u)ψ = 0 , (3.60)
the Schr¨odinger equation with λ = k2, corresponding to the continuous spectrum. If differs
from (3.56) only by the uψ-term. We look again for solutions with
ψ ∼ eikx
as x → +∞ . (3.61)
We may suppose that ψ is again of the form
ψ+(x; k) = eikx
+
∞
x
K(x, z)eikz
dz , (3.62)
The subscript + denotes a solution with boundary condition at x → +∞. The original
problem is now traded for the problem of finding K(x, z). To find what the Schr¨odinger
equation implies for K we first calculate the derivatives of ψ+. We find
ψ+ = ikeikx
− Keikx
+
∞
x
K (x, z)eikz
dz (3.63)
where we introduced K(x) = K(x, x). The second derivative becomes
ψ+ = eikx
−k2
− K − ikK − K |z=x +
∞
x
K eikz
dz . (3.64)
Note that K = (K + ˙K)|z=x. It will be convenient to rewrite ψ+ itself in a similar way.
Integrating by parts twice and assuming that K, ˙K → 0 as z → +∞ we have
ψ+ = eikx
+
1
ik
[Keikz
]∞
x −
1
ik
∞
x
˙Keikz
dz = eikx
1 −
K
ik
−
1
k2
˙K|z=x −
1
k2
∞
x
¨Keikz
dz .
(3.65)
Using these facts the Schr¨odinger equation becomes
0 =ψ+ + (k2
− u)ψ+ = eikx
−K − K |z=x − ˙K|z=x +
∞
x
(K − ¨K)eikz
dz − uψ+
= − eikx
u + 2K +
∞
x
(K − ¨K − u(x)K)eikz
dz , (3.66)
which implies
u = −2K , and K − ¨K − u(x)K = 0 (z > x) . (3.67)
3.2. THE INVERSE SCATTERING PROBLEM 23
Together with the conditions K, ˙K → 0 as z → +∞ this defines the problem for K(x, z).
Given u(x) K(x, z) is a real valued function satisfying a wave equation (with z playing
the role of time) with initial data at z = x (u = −2K = −2(K + ˙K)|z=x) and at z = +∞.
It is well known that the solution exists and is unique.
So far we have just reformulated the scattering problem as the problem of finding K(x, z)
given u(x). However, we can now try to invert the expression
ψ+(x; k) = eikx
+
∞
x
K(x, z)eikz
dz , (3.68)
to obtain K from ψ and then u(x) from u = −2K . This is possible because we have just
argued that a unique K exists.
Before we attack this problem we have to relate the solution ψ+ to the continuous eigenfunctions
ˆψ introduced before. We can construct ˆψ from ψ+ as
ˆψ = ψ∗
+ + b(k)ψ+ . (3.69)
Indeed, we then find
ˆψ ∼ e−ikx
+ b(k)eikx
x → +∞ , (3.70)
as required. Using the expression for ψ+ we obtain
ˆψ = e−ikx
+ b(k)eikx
+
∞
−∞
K(x, z)e−ikz
dz + b(k)
∞
−∞
K(x, z)eikz
dz , (3.71)
since K(x, z) = 0 for z < x. Writing this as
∞
−∞
K(x, z)e−ikz
dz = ˆψ − e−ikx
− b(k)eikx
− b(k)
∞
−∞
K(x, z)eikz
dz (3.72)
we can invert it by performing the inverse Fourier transform yielding
K(x, z) =
1
2π
∞
−∞
ˆψ − e−ikx
− b(k)eikx
− b(k)
∞
−∞
K(x, y)eiky
dy eikz
dk . (3.73)
Let F(x) be the Fourier transform of b(k),
F(x) =
∞
−∞
b(k)eikx
dk . (3.74)
We find
K(x, z) =
1
2π
∞
−∞
ˆψ(x; k) − e−ikx
eikz
dk − F(x + z) −
∞
−∞
K(x, y)F(y + z)dy . (3.75)
To evaluate the first integral we note that since we are assuming that there are no discrete
eigenvalues ˆψ has no poles in the upper half complex k-plane. Furthermore the integrand
ˆψ(x; k)eikx
− 1 eik(z−x)
(3.76)
decays exponentially at large |k| with (k) > 0 since z > x. We can therefore close the
integration contour with a semicircle of radius R in the upper half plane. Since the semicircle
24 CHAPTER 3. THE SCATTERING AND INVERSE SCATTERING PROBLEMS
part of the contour does not contribute as R → ∞ and since there are no poles inside the
contour we find, by Cauchy’s theorem,
∞
−∞
ˆψ(x; k) − e−ikx
eikz
dk = 0 . (3.77)
We therefore obtain
K(x, z) + F(x + z) +
∞
x
K(x, y)F(y + z)dy = 0 z > x . (3.78)
This is known as the Marchenko equation for K(x, z) given F(x). Once this integral equation
is solved, we will discuss how shortly, the potential is obtained from u(x) = −2K (x).
But first, to complete the analysis, we must take into account the discrete spectrum.
The analysis leading to (3.75) is unchanged but now the evaluation of the integral involving
ˆψ changes since now there are poles in the upper half plane corresponding to the discrete
eigenvalues. Using the same contour as before (with R large enough to enclose all poles)
Cauchy’s residue theorem gives
∞
−∞
ˆψ(x; k) − e−ikx
eikz
dk = 2πi
N
n=1
Rn , (3.79)
where Rn = Resk→iκn ( ˆψeikz) is the residue on the nth pole. To find these residues we need
to introduce the other solution with boundary condition at x → −∞
ψ−(x; k) = e−ikx
+
x
−∞
L(x, z)e−ikz
dz , (3.80)
where L(x, z) plays the same role for this solution that K(x, z) does for ψ+. The relation
between ˆψ and ψ− is fixed by the behavior at −∞ and we have
ˆψ = a(k)ψ− = ψ∗
+ + b(k)ψ+ , (3.81)
so that
ψ− = a−1
ψ∗
+ + a−1
bψ+ . (3.82)
These equations imply that ψ+ and ψ− are well defined at the poles k = iκn and from their
asymptotic behavior we have
ψ+(x; iκn) ∼ e−κnx
x → +∞ ψ−(x; iκn) ∼ eκnx
x → −∞ (3.83)
But this is precisely the behavior of the discrete eigenfunction ψn so we may write
ψn(x) = cnψ+(x; iκn) = dnψ−(x; iκn) , (3.84)
for some constants cn, dn. Next we will use these relations to derive some identities involving
the Wronskian which will allow us to find the residues we are after.
Recall that the Wronskian W(ψ−, ψ+) is a constant (since ψ± are solutions to the Schr¨oding
equation with the same eigenvalue) so we may evaluate it at x → +∞ and using (3.82) we
find
W(ψ−, ψ+) = ψ−ψ+ − ψ+ψ− = 2ika−1
. (3.85)
3.2. THE INVERSE SCATTERING PROBLEM 25
Taking a derivative of this equation with respect to k we find
W(
d
dk
ψ−, ψ+) + W(ψ−,
d
dk
ψ+) = 2ia−1
− 2ika a−2
. (3.86)
This equation can be used to show that a has only simple poles (see the problems). If we
take instead the k-derivative of the original Schr¨odinger equation we find
d
dk
ψ + 2kψ + (k2
− u)
d
dk
ψ = 0 . (3.87)
Multiplying with ψ and using the Schr¨odinger equation again we find
ψ
d
dk
ψ + 2kψ2
− ψ
d
dk
ψ = 0 , (3.88)
or
[W(
d
dk
ψ, ψ)] = 2kψ2
. (3.89)
Taking ψ = ψ+, k = iκn and integrating from −∞ to +∞ we get
W+
n (
d
dk
ψ+, ψ+) − W−
n (
d
dk
ψ+, ψ+) = 2iκn
+∞
−∞
ψ2
+dx =
2iκn
c2
n
, (3.90)
where W±
n denotes the Wronskian evaluated at x = ±∞ and k = iκn. On the RHS we used
the fact that ψ+ = ψn/cn and the normalization of the discrete eigenfunctions ψn.
Evaluating instead the identity (3.86) at x = −∞ and k = iκn we find
W−
n (
d
dk
ψ−, ψ+) + W−
n (ψ−,
d
dk
ψ+) = −2ik
a
a2
n
, (3.91)
or
W−
n (
d
dk
ψ−,
dn
cn
ψ−) + W−
n (
cn
dn
ψ+,
d
dk
ψ+) = −2ik
a
a2
n
, (3.92)
where the superscript n on the RHS denotes evaluation at k = iκn. Using (3.90) to eliminate
the second term we find
W+
n (
d
dk
ψ+, ψ+) −
d2
n
c2
n
W−
n (
d
dk
ψ−, ψ−) =
2iκn
c2
n
− 2κn
dn
cn
a
a2
n
. (3.93)
From the expression for ψ+ in (3.62) we see that d
dk ψ+ → ikeikx as x → ∞, which means
that the first term on the LHS vanishes. Similarly the second term on the LHS vanishes and
we find
a
a2
n
=
i
cndn
. (3.94)
We can now easily evaluate the residues of ˆψeikz. We have
ˆψ(x; k)eikz
= a(k)ψ−(x; k)eikz
=
cn
dn
a(k)ψ+(x; k)eikz
. (3.95)
Now only a(k) has poles so the residue at k = iκn is
Rn =
cn
dn
ψ+(x; iκn)e−κnz
Resk→iκn a(k) . (3.96)
26 CHAPTER 3. THE SCATTERING AND INVERSE SCATTERING PROBLEMS
Since a(k) has only simple poles the residue is equal to
Res a = Res
1
a−1
=
1
(a−1) n
= −
a2
a n
= icndn (3.97)
and we find
Rn = ic2
nψ+(x; iκn)e−κnz
. (3.98)
Therefore we have
∞
−∞
ˆψ(x; k) − e−ikx
eikz
dk = −2π
N
n=1
c2
ne−κnz
e−κnx
+
∞
x
K(x, y)e−κny
dy . (3.99)
Using this in (3.75) we find again the Marchenko equation
K(x, z) + F(x + z) +
∞
x
K(x, y)F(y + z)dy = 0 z > x , (3.100)
but with F now given by
F(x) =
N
n=1
c2
ne−κnx
+
1
2π
∞
−∞
b(k)eikx
dk . (3.101)
This agrees with our previous result if there is no discrete spectrum. These two equations
complete our formulation of the inverse scattering problem. Given the discrete spectrum,
normalization constants cn and reflection coefficient b(k) the potential u(x) can be determined
from K(x, z). For the final step we need to solve the Marchenko equation, which is what we
now turn to.
3.3 The solution of the Marchenko equation
The Marchenko equation is a Fredholm integral equation, meaning that it can be written
with constants as integration limits,
K(x, z) + F(x + z) +
∞
−∞
K(x, y)F(y + z)dy = 0 , (3.102)
with K(x, z) = 0 for z < x. Note that x just plays the role of a parameter here. A direct way
of solving this equation is by iteration. Define
K1(x, z) =
−F(x + z) z > x
0 z < x
(3.103)
and
K2 = − F(x + z) −
∞
−∞
K1(x, y)F(y + z)dy
K3 = − F(x + z) −
∞
−∞
K2(x, y)F(y + z)dy
3.3. THE SOLUTION OF THE MARCHENKO EQUATION 27
and so on. If Kn(x, z) → K(x, z) pointwise as n → ∞ the existence of the solution is
established. The infinite expansion so obtained is called the Neumann series.
This gives a formal solution to the Marchenko equation, but it is unlikely to lead to simple
closed form expressions. If we are lucky and F is sufficiently simple it may be possible to
instead find the solution directly. In the case relevant to the KdV equation it turns out
that we are lucky and F takes a form which reduces the Marchenko equation to a standard
problem.
Suppose that F(x + z) is a separable function, i.e. that
F(x + z) =
N
n=1
Xn(x)Zn(z) , (3.104)
where N is finite. Then the Marchenko equation becomes
K(x, z) +
N
n=1
Xn(x)Zn(z) +
N
n=1
Zn(z)
∞
−∞
K(x, y)Xn(y)dy = 0 . (3.105)
It is clear from this expression that the solution is of the form
K(x, z) =
N
n=1
Ln(x)Zn(z) , (3.106)
for some functions Ln(x) that we need to find. Substituting this into the Marchenko equation
we find a sum of terms each of which should vanish giving
Ln(x) + Xn(x) +
N
m=1
Lm(x)
∞
x
Zm(y)Xn(y)dy = 0 n = 1, 2, . . . , N . (3.107)
Therefore the Marchenko equation is reduced to solving N algebraic equations for N unknowns,
Ln(x). The solution is now straightforward. We will come back to it in the next
chapter when we discuss the case relevant to the KdV equation. Here we will instead just
analyze two simple examples to illustrate in detail how the inverse scattering problem is
solved.
Example 1: Reflection coefficient with one pole
We assume the reflection coefficient has the form
b(k) = −
β
β + ik
(3.108)
for some constant β > 0. There is only one discrete eigenvalue corresponding to the pole in
the upper half plane at k = iβ, i.e. κ1 = β. The remaining piece of scattering data we need is
the normalization coefficient c1. It can be found as follows. We have seen that we can write
ˆψ = ψ∗
+ + b(k)ψ+ , (3.109)
while we have also seen that (3.98)
Resk→iκn ( ˆψeikz
) = ic2
nψ+(x; iκn)e−κnz
. (3.110)
28 CHAPTER 3. THE SCATTERING AND INVERSE SCATTERING PROBLEMS
Putting these two facts together and noting that ψ+ has no poles we find
iβ = Resk→iκn b(k) = ic2
1 , (3.111)
so that c1 =
√
β and the discrete eigenfunction goes as
ψ1 ∼ βe−βx
x → +∞ . (3.112)
From the definition (3.101) we now find
F(x) = βe−βx
−
β
2π
∞
−∞
eikx
β + ik
dk . (3.113)
To evaluate the integral imagine closing the contour with a semicircle in the upper half plane.
The semicircular part does not contribute provided that x > 0, since then the integrand
decays exponentially. Cauchy’s residue theorem then gives
∞
−∞
eikx
β + ik
dk = 2πiResk→iβ
eikx
β + ik
= 2πe−βx
x > 0 . (3.114)
If x < 0 we instead close the contour in the lower half plane and since there are no poles there
we find
∞
−∞
eikx
β + ik
dk = 0 x < 0 . (3.115)
Therefore we find
F(x) = βe−βx
H(−x) , (3.116)
where the Heaviside step function is defined as
H(x) =
1 x > 0
0 x < 0
. (3.117)
The Marchenko equation now gives K(x, z) = 0 for x + z > 0 since F vanishes there, while
for x + z < 0 we get
K(x, z) + βe−β(x+z)
+ β
−z
x
K(x, y)e−β(y+z)
dy = 0 , (3.118)
since F(y + z) vanishes for y + z > 0. One way to find the solution is to integrate by parts
giving
K(x, z)+βe−β(x+z)
−K(x, −z)+K(x, x)e−β(x+z)
+
−z
x
∂
∂y
K(x, y)e−β(y+z)
dy = 0 . (3.119)
It is clear from this expression that one solution is K(x, z) = −β. By uniqueness this must
be the required solution. We then have
K(x, z) = −βH(−x − z) . (3.120)
The potential becomes
u(x) = −2[K(x, x)] = 2β[H(−2x)] = 2β[H(−x)] = −2βδ(x) . (3.121)
This recovers our first example of the solution of the scattering problem, setting β = U0/2.
3.3. THE SOLUTION OF THE MARCHENKO EQUATION 29
Example 2: Zero reflection coefficient
Since b(k) = 0 only the discrete eigenvalues contribute to F. We suppose that we have two
of them with
ψ1 ∼ c1e−κ1x
, ψ2 ∼ c2e−κ2x
, x → +∞ . (3.122)
We then obtain
F(x) = c2
1e−κ1x
+ c2
2e−κ2x
. (3.123)
We see that
F(x + z) =
2
n=1
c2
ne−κnx
e−κnz
(3.124)
i.e. F(x+z) is separable with Xn(x) = c2
ne−κnx and Zn(z) = e−κnz. The Marchenko equation
gives (3.107)
Ln(x) + Xn(x) +
2
m=1
Lm(x)
∞
x
Zm(y)Xn(y)dy = 0 n = 1, 2 . (3.125)
In particular
0 = L1+c2
1e−κ1x
+c2
1L1
∞
x
e−2κ1y
dy+c2
1L2
∞
x
e−(κ1+κ2)y
dy = L1+c2
1e−κ1x
+c2
1
2
m=1
Lm
κ1 + κm
e−(κ1+κm)x
(3.126)
and similarly for L2 giving
Ln + c2
ne−κnx
+ c2
n
2
m=1
Lme−(κm+κn)x
κm + κn
= 0 . (3.127)
It is convenient to write this in matrix notation as
AL + X = 0 (3.128)
with
L =
L1
L2
, X =
c2
1e−κ1x
c2
2e−κ2x , (3.129)
and
Amn = δmn + c2
m
e−(κm+κn)x
κm + κn
. (3.130)
This system of equations can be easily generalized to the case of more than two eigenvalues.
The solution is
L = −A−1
X (3.131)
and then
K(x, x) = ZT
L , Z =
e−κ1x
e−κ2x . (3.132)
The important simplifying observation is that
d
dx
Amn = −c2
me−(κm+κn)x
= −XmZn (3.133)
30 CHAPTER 3. THE SCATTERING AND INVERSE SCATTERING PROBLEMS
so that
K(x, x) = ZmLm = −ZmA−1
mnXn = A−1
mn
d
dx
Anm = tr A−1 d
dx
A =
d
dx
tr log A =
d
dx
log det A .
(3.134)
Therefore we find
u(x) = −2[K(x, x)] = −2
d2
dx2
log det A . (3.135)
In our example A is a 2 × 2 matrix and
det A = 1 +
c2
1
2κ1
e−2κ1
1 +
c2
2
2κ2
e−2κ2
−
c2
1c2
2
(κ1 + κ2)2
e−2(κ1+κ2)x
(3.136)
which gives the required potential.
As a check of this result let us retain just one eigenvalue by setting c2 = 0. Then we find
u(x) = −2
d2
dx2
log 1 +
c2
1
2κ1
e−2κ1
= −2κ2
1 sech2
(κ1x + x0) , (3.137)
where we defined c2
1/2κ1 = e−2x0 . By shifting and rescaling x this can be brought to the
form u(x) = −2 sech2
(x) and since 2 = ( + 1) with = 1 this is precisely one of the
class of reflectionless potentials we encountered when discussing the corresponding scattering
problem.
Chapter 4
Initial value problem for the KdV
equation
We are now ready to apply the techniques from the inverse scattering problem to the solution
of the initial value problem for the KdV equation. Let us first summarize the solution to the
inverse scattering problem, i.e. the problem of finding the potential u(x) in the Schr¨odinger
equation
ψ + (λ − u)ψ = 0 , (4.1)
given the scattering data, the asymptotic behavior of ψ. The data needed is summarized as
Continuous spectrum: λ > 0, k =
√
λ ˆψ ∼
e−ikx + b(k)eikx x → +∞
a(k)e−ikx x → −∞
Discrete spectrum: λ < 0, κn =
√
−λn ψn ∼ cne−κnx x → +∞ (n = 1, . . . , N)
We have seen in the last chapter that the solution to the inverse scattering problem is given
by
u(x) = −2
d
dx
K(x, x) , (4.2)
where K(x, z) is the solution to the Marchenko equation
K(x, z) + F(x + z) +
∞
x
K(x, y)F(y + z)dy = 0 , (4.3)
where F encodes the scattering data as
F(x) =
N
n=1
c2
ne−κnx
+
1
2π
∞
−∞
b(k)eikx
dk . (4.4)
4.1 Relation to the KdV equation
To connect to the KdV equation
˙u − 6uu + u = 0 , (4.5)
31
32 CHAPTER 4. INITIAL VALUE PROBLEM FOR THE KDV EQUATION
consider writing
u =
ψ
ψ
(4.6)
for some ψ(x; t) = 0, where t just enters as a parameter. We can write this as
ψ − uψ = 0 . (4.7)
This is almost the Schr¨odinger equation we want. In fact, noting that the KdV equation is
invariant under the Galilean transformation
u(x, t) → λ + u(x + 6λt, t) (4.8)
for constant λ and that the x-dependence is unchanged we can simply replace u by u − λ
giving
ψ + (λ − u)ψ = 0 , (4.9)
the Schr¨odinger equation with eigenvalue λ. Therefore, if we could solve this equation for ψ
we would find u from (4.6). This is however diffcult since u appears already in the equation
for ψ. The way out is to interpret this as an inverse scattering problem.
Let u(x, t) be the solution of
˙u − 6uu + u = 0 (4.10)
with u(x, 0) = f(x) given. This is the initial value problem for the KdV equation. Now let
ψ(x; t) be the solution of
ψ + (λ − u)ψ = 0 , (4.11)
with the potential given by this u(x, t) and λ = λ(t). This is the Schr¨odinger equation but
with everything depending on the extra parameter t. The solution of the KdV equation can
be described in three steps:
1. At t = 0 we are given u(x, 0) = f(x). We can solve the usual scattering problem for
this potential to find the scattering data S(0) = {b(k), κn, cn}.
2. If we can determine the time-evolution of these scattering data we will know the scattering
data for all t > 0, S(0) = {b(k; t), κn(t), cn(t)}
3. With this information we can solve the inverse scattering problem to find u(x, t) for all
t > 0.
We summarize the procedure in the following diagram:
u(x, 0) -scattering
S(0)
?
time-
evolution
S(t)
inverse scattering
u(x, t)
?
KdV
4.2. TIME EVOLUTION OF THE SCATTERING DATA 33
It remains of course to see how the time-evolution of the scattering data can be determined.
This is the question we now turn to.
4.2 Time evolution of the scattering data
We start with the Schr¨odinger equation depending on the extra parameter t
ψ + (λ − u)ψ = 0 , (4.12)
for ψ(x; t), λ(t) and u(x, t) a solution to the KdV equation. Taking the x- and t-derivative of
this equation we get
ψ − u ψ + +(λ − u)ψ = 0 , (4.13)
˙ψ + ( ˙λ − ˙u)ψ + (λ − u) ˙ψ = 0 , (4.14)
where u(x, t) satisfies the Kdv equation
˙u − 6uu + u = 0 . (4.15)
It is convenient to define
R(x, t) = ˙ψ + u ψ − 2(u + 2λ)ψ . (4.16)
We then find
(ψ R − ψR ) = ψ R − ψR = −ψ[(λ − u)R − R ] , (4.17)
where we used the Schr¨odinger equation. Using the definition of R, the Schr¨odinger equation
and its derivatives derived above we find
(ψ R − ψR ) = −ψ2
[ ˙u − 6uu + u − ˙λ] = ˙λψ2
, (4.18)
where we used the KdV equation in the last step. This equation determines the time-evolution
of the spectrum and scattering data as we will now show.
The discrete spectrum
Taking λ = −κ2
n and ψ = ψn in (4.18) and integrating over all x we find
[ψnRn − ψnRn]∞
−∞ = −
d
dt
(κ2
n)
∞
−∞
ψ2
ndx = −
d
dt
(κ2
n) , (4.19)
where Rn denotes R with the discrete eigenfunction ψn substituted and we used the fact that
these are normalized in the last step. Because ψn and therefore also Rn decays exponentially
as |x| → ∞ we find
d
dt
(κ2
n) = 0 ⇒ κn = constant . (4.20)
The discrete eigenvalues are constants of motion, once they are determined from the initial
potential (profile) u(x, 0) they remain fixed for all t!
We also need to determine the time-evolution of the normalization constants cn(t). Since
κn is constant (4.18) says that
(ψnRn − ψnRn) = 0 (4.21)
34 CHAPTER 4. INITIAL VALUE PROBLEM FOR THE KDV EQUATION
which we can integrate to give
ψnRn − ψnRn = gn(t) , (4.22)
for some arbitrary functions gn(t). But since ψn, Rn → 0 as |x| → ∞ we find gn(t) = 0. Thus
ψnRn − ψnRn = 0 , (4.23)
or dividing by ψ2
n,
Rn
ψn
= 0 ⇒
Rn
ψn
= hn(t) . (4.24)
Multiplying by ψ2
n and integrating over all x gives
hn(t) =
∞
−∞
ψnRndx =
∞
−∞
ψn( ˙ψn + u ψn − 2(u − 2κ2
n)ψn)dx
=
1
2
d
dt
∞
−∞
ψ2
ndx +
∞
−∞
(uψ2
n − 2ψ 2
n + 4κ2
nψ2
n) dx = 0 (4.25)
where we used the normalization of ψn the Schr¨odinger equation and the fact that ψn → 0 as
|x| → ∞. Therefore we have
0 = Rn = ˙ψn + u ψn − 2(u − 2κ2
n)ψn , (4.26)
which determines the time-evolution of the discrete eigenfunctions ψn. Using the fact that
u, u → 0 and
ψn ∼ cne−κnx
(4.27)
as x → +∞ we finally find
˙cn − 4κ3
ncn = 0 ⇒ cn(t) = cn(0)e4κ3
nt
, (4.28)
where cn(0) (n = 1, . . . , N) are the normalization constants at t = 0.
The continuous spectrum
We can apply the same procedure to this case, for which λ = k2. Since k can take any real
value we are allowed to consider the time-evolution with k fixed. With this choice, taking
ψ = ˆψ (4.18) gives
( ˆψ ˆR − ˆψ ˆR ) = 0 ⇒ ˆψ ˆR − ˆψ ˆR = g(t; k) , (4.29)
for some function g(t; k). Since
ˆψ ∼ ae−ikx
x → −∞ (4.30)
we find from the definition of R that
ˆR ∼
˙ˆψ − 4k2 ˆψ ∼ (˙a + 4ik3
a)e−ikx
x → −∞ , (4.31)
and it follows that g(t; k) = 0. Therefore
ˆψ ˆR − ˆψ ˆR = 0 ⇒
ˆR
ˆψ
= h(t; k) . (4.32)
4.3. SUMMARY OF THE SOLUTION 35
Looking at the behavior as x → −∞ we find
˙a + 4ik3
a = h(t; k)a , (4.33)
while as x → +∞ we have
ˆψ ∼ e−ikx
+ beikx
(4.34)
and
ˆR ∼ ˙beikx
+ 4ik3
(e−ikx
− beikx
) (4.35)
and we get the condition
ˆψ ∼ e−ikx
+ beikx
(4.36)
and
˙beikx
+ 4ik3
(e−ikx
− beikx
) = h(e−ikx
+ beikx
) . (4.37)
The coefficients of both exponentials have to match and we find
h = 4ik3
⇒ ˙a = 0 (4.38)
and
˙b − 8ik3
b = 0 ⇒ b(k; t) = b(k; 0)e8ik3t
. (4.39)
This completes the derivation of the time-evolution of the scattering data. The important
ones for the inverse scattering problem are
κn = constant , cn(t) = cn(0)e4κ3
nt
(n = 1, . . . , N) (4.40)
b(k; t) = b(k; 0)e8ik3t
. (4.41)
4.3 Summary of the solution
Before analyzing examples of solutions we will summarize here the solution to the initial value
problem for the KdV equation via what is usually called the inverse scattering transform. We
want to solve the equation
˙u − 6uu + u = 0 , (4.42)
with the initial profile u(x, 0) = f(x) given. We assume that f is sufficiently well-behaved to
ensure the existence of a solution of the KdV equation and also of the Schr¨odinger (SturmLiouville)
equation
ψ + (λ − u)ψ = 0 . (4.43)
We first solve this equation for u(x, 0) = f(x) to determine the scattering data, in particular
the discrete spectrum −κ2
n (n = 1, . . . , N), normalization constants cn(0) and reflection
coefficient b(k; 0). The time-evolution of these data are then given by
κn = constant , cn(t) = cn(0)e4κ3
nt
, b(k; t) = b(k; 0)e8ik3t
. (4.44)
The function entering the Marchenko equation is
F(x; t) =
N
n=1
c2
n(0)e8κ3
nt−κnx
+
1
2π
∞
−∞
b(k; 0)e8ik3t+ikx
dk , (4.45)
36 CHAPTER 4. INITIAL VALUE PROBLEM FOR THE KDV EQUATION
which now depends also on the parameter t. The Marchenko equation is
K(x, z; t) + F(x + z; t) +
∞
x
K(x, y; t)F(y + z; t)dy = 0 . (4.46)
Finally the solution to the KdV equation is expressed in terms of the solution to this equation
as
u(x, t) = −2
∂
∂x
K(x, x; t) . (4.47)
Note that the solution outlined here involves two potentially difficult steps: solving the
scattering problem for the initial profile u(x, 0) = f(x) and solving the Marchenko equation.
Nevertheless we have managed to reduce solving a non-linear PDE to solving two linear
problems, a second order ODE and an ordinary integral equation. We will now consider some
simple examples of solutions to illustrate how the method works.
4.4 Reflectionless potentials
To illustrate how the inverse scattering transform works we will take the initial profile u(x, 0)
to be a sech2 function. In fact we will take the coefficient in front to correspond to a reflectionless
potential, i.e. one that leads to b(k) = 0. We showed in the previous chapter when
we analyzed the scattering problem that this is the case for
u(x, 0) = −N(N + 1) sech2
x . (4.48)
We will see that this choice of initial profile describes the N-soliton solution.
N = 1: The solitary wave
In this case the initial profile is
u(x, 0) = −2 sech2
x . (4.49)
The scattering problem is
ψ + (λ + 2 sech2
x)ψ = 0 . (4.50)
When analyzing the scattering problem for a sech2 potential we showed that introducing
T = tanh x this becomes the associated Legendre equation
d
dT
(1 − T2
)
dψ
dT
+ 2 +
λ
1 − T2
ψ = 0 . (4.51)
This is the equation we solve to determine spherical harmonics (for = 1). For λ = −κ2
bounded solutions exist only for κ = m with m ≤ an integer. Setting = N = 1 we find that
there is only one discrete eigenvalue, κ1 = 1. The corresponding eigenfunction is proportional
to the associated Legendre function
ψ1(x) ∝ P1
1 (tanh x) = − sech x (4.52)
and since ∞
−∞
sech2
xdx = 2 (4.53)
4.4. REFLECTIONLESS POTENTIALS 37
the normalized eigenfunction is (the sign is irrelevant)
ψ1(x) =
1
√
2
sech x . (4.54)
Its asymptotic behavior is
ψ1(x) ∼
√
2e−x
as x → +∞ , (4.55)
and we read off that c1(0) =
√
2. Therefore
c1(t) =
√
2e4t
. (4.56)
This is all we need since we already know from our previous analysis that b(k) = 0. We find
F(x; t) = c2
1(t)e−κ1x
= 2e8t−x
. (4.57)
The Marchenko equation becomes
K(x, z; t) + 2e8t−(x+z)
+ 2
∞
x
K(x, y; t)e8t−(y+z)
dy = 0 . (4.58)
Looking at the z-dependence we see that the solution is of the form
K(x, z; t) = L(x, t)e−z
(4.59)
and plugging this in we find
0 = L(x, t) + 2e8t−x
+ 2L(x, t)e8t
∞
x
e−2y
dy = L(x, t) + 2e8t−x
+ e8t−2x
L(x, t) , (4.60)
so that
L(x, t) =
−2e8t−x
1 + e8t−2x
. (4.61)
Finally we get
u(x, t) = −2
∂
∂x
−2e8t−x
1 + e8t−2x
e−x
= 4
∂
∂x
1
1 + e2x−8t
= −2 sech2
(x − 4t) . (4.62)
This is the solitary wave solution of amplitude −2 and speed of propagation 4. We have
already found this solution by different methods, looking for a traveling wave of fixed form.
However, the following solutions would be very difficult to find by other methods.
N = 2: two-soliton solution
In this case the initial profile is
u(x, 0) = −6 sech2
x (4.63)
and we need to solve the scattering problem
ψ + (λ + 6 sech2
x)ψ = 0 . (4.64)
38 CHAPTER 4. INITIAL VALUE PROBLEM FOR THE KDV EQUATION
As before we transform this to the associated Legendre equation by introducing the variable
T = tanh x. Now there are two discrete eigenvalues with κ1 = 1 and κ2 = 2. The normalized
eigenfunctions are
ψ1(x) =
3
2
tanh x sech x , ψ2(x) =
√
3
2
sech2
x . (4.65)
Their asymptotic behavior is
ψ1(x) ∼
√
6e−x
, , ψ2(x) ∼ 2
√
3e−2x
as x → +∞ , (4.66)
and we read off c1(0) =
√
6 and c2(0) = 2
√
3. Their time-dependence is then given by
c1(t) =
√
6e4t
, c2(t) = 2
√
3e32t
. (4.67)
As before our choice of profile ensures that b(k; t) = 0. The function F becomes
F(x; t) = 6e8t−x
+ 12e64t−2x
(4.68)
and the Marchenko equation is
K(x, z; t) + 6e8t−(x+z)
+ 12e64t−2(x+z)
+
∞
x
K(x, y; t) 6e8t−(y+z)
+ 12e64t−2(y+z)
dy = 0 .
(4.69)
It is clear that the solution has the form
K(x, z; t) = L1(x, t)e−z
+ L2(x, t)e−2z
, (4.70)
since F is a separable function. Collecting the coefficients of e−z and e−2z gives the pair of
equations
L1 + 6e8t−x
+ 6e8t
L1
∞
x
e−2y
dy + L2
∞
x
e−3y
dy = 0 , (4.71)
L2 + 12e64t−2x
+ 12e64t
L1
∞
x
e−3y
dy + L2
∞
x
e−4y
dy = 0 . (4.72)
Evaluating the integrals they become
L1 + 6e8t−x
+ 3L1e8t−2x
+ 2L2e8t−3x
= 0 , (4.73)
L2 + 12e64t−2x
+ 4L1e64t−3x
+ 3L2e64t−4x
= 0 , (4.74)
with solution
L1 = 6
e72t−5x − e8t−x
D
, L1 = −12
e72t−4x + e64t−2x
D
, (4.75)
where
D = 1 + 3e8t−2x
+ 3e64t−4x
+ e72t−6x
. (4.76)
Finally the solution to the KdV equation becomes
u(x, t) = −2
∂
∂x
(L1e−x
+ L2e−2x
) = 12
∂
∂x
e8t−2x + e72t−6x − 2e64t−4x
D
(4.77)
4.4. REFLECTIONLESS POTENTIALS 39
and after a bit of algebra one finds
u(x, t) = −12
3 + 4 cosh(2x − 8t) + cosh(4x − 64t)
(3 cosh(x − 28t) + cosh(3x − 36t))2
. (4.78)
We will now analyze this solution a bit to see why it is called the two-soliton solution. The
first thing to note is that the solution is valid for all t so we can consider how it evolves into
the initial profile at t = 0 and one from there.
The solution is shown at different times in figure 4.1. For large negative t is describes
two (almost) solitary waves. The taller one is to the left and since it moves faster it catches
up with the shorter one to form a single wave – our initial profile at t = 0. Finally the
taller wave overtakes the shorter one. Superficially this looks like a linear process but careful
examination shows that the taller wave has moved forward and the shorter wave backward
relative to where they would be if they just passed each other.
Each solitary wave appearing at t → ±∞ and interacting in this way, via a phase shift, is
called a soliton. The solution we have described is therefore called the two-soliton solution.
We can gain a better understanding of the solution by looking at its asymptotics as
t → ±∞. Writing ξ = x − 16t the solution takes the form
u(x, t) = −12
3 + 4 cosh(2ξ + 24t) + cosh(4ξ)
(3 cosh(ξ − 12t) + cosh(3ξ + 12t))2
. (4.79)
We can now take t → ±∞ with ξ held fixed, which means that we are following the wave
with speed 16. We get
u(x, t) ∼ − 96
e±(2ξ+24t)
(3e (ξ−12t) + e±(3ξ+12t))2
= −96
1
(3e 2ξ + e±2ξ)2
= − 32
1
(e 2ξ+ln 3/2 + e±2ξ−ln 3/2)2
= −8 sech2
2ξ 1
2 ln 3 as t → ±∞ .
(4.80)
Doing the same for the wave of speed 4 we can add the two together to get the solution (the
error is exponentially small since the waves are far apart at very early/late times)
u(x, t) ∼ −8 sech2
2ξ 1
2 ln 3 − 2 sech2
η ± 1
2 ln 3 as t → ±∞ , (4.81)
where ξ = x − 16t and η = x − 4t. Here we see that the taller wave of amplitude −8 is shifted
to the right by 1
2 ln 3 while the shorter one of amplitude −2 is shifted to the left by ln 3 by
the interaction.
N-soliton solution
We now want to analyze the case of general N. The initial profile is
u(x, 0) = −N(N + 1) sech2
x (4.82)
There are now N discrete eigenvalues and no contribution from the continuous spectrum
since b(k) = 0. The discrete eigenvalues are λ = −κ2
n with κn = n and n = 1, 2, . . . , N. The
normalized eigenfunctions go as
ψn(x) ∼ cne−nx
as x → +∞ . (4.83)
40 CHAPTER 4. INITIAL VALUE PROBLEM FOR THE KDV EQUATION
They are proportional to the associated Legendre functions
ψn(x) ∝ Pn
N (tanh x) (4.84)
and cn(0) is determined from the normalization condition. Then
cn(t) = cn(0)e4n3t
. (4.85)
The function F becomes
F(x; t) =
N
n=1
c2
n(0)e8n3t−nx
(4.86)
and the Marchenko equation becomes
K(x, z; t) +
N
n=1
c2
n(0)e8n3t−n(x+z)
+
∞
x
K(x, y; t)
N
n=1
c2
n(0)e8n3t−n(y+z)
dy = 0 . (4.87)
The solution must take the form
K(x, z; t) =
N
n=1
Ln(x, t)e−nz
. (4.88)
In section 3.3 we analyzed the solution of the Marchenko equation for a general separable F
(i.e. F(x+z) = N
n=1 Xn(x)Zn(z)). We can apply what we learned there to the present case.
The Marchenko equation reduces to the algebraic system
AL + X = 0 (4.89)
with
L =





L1
L2
...
LN





, X =





c2
1(0)e8t−x
c2
2(0)e64t−2x
...
c2
N (0)e8N3t−Nx





, (4.90)
The N × N matrix A has elements
Amn = δmn +
c2
m(0)
m + n
e8m3t−(m+n)x
(4.91)
and we showed that the solution to the KdV equation takes the form
u(x, t) = −2
∂2
∂x2
log det A . (4.92)
As in the 2-soliton example we can determine the asymptotic form of the solution. Setting
ξn = x − 4κ2
nt = x − 4n2t and taking t → ±∞ with ξn held fixed the behavior of u is
u(x, t) ∼ −2n2
sech2
(n(x − 4n2
t) xn) as t → ±∞ , (4.93)
where the phase, xn, is given by
e2xn
=
N
m=1, m=n
n − m
n + m
sgn(n−m)
n = 1, 2, . . . , N . (4.94)
4.5. SOLUTIONS WITH B(K) = 0 41
The N-soliton form of the solution is evident from the full asymptotic solution
u(x, t) ∼ −2
N
n=1
n2
sech2
(n(x − 4n2
t) xn) as t → ±∞ . (4.95)
The asymptotic solution consists of N separate solitons ordered according to their speeds: as
t → +∞ the tallest (fastest) is at the front followed by progressively shorter waves. All N
solitons interact to form a single sech2 pulse at t = 0 – the initial profile u = −N(N+1) sech2
x.
4.5 Solutions with b(k) = 0
So far we have considered reflectionless initial profiles. It is clear that a more general choice
of u(x, 0) will however lead to to b(k) = 0. Unfortunately when b(k) = 0 it is not possible
to solve the Marchenko equation in closed form. Instead we have to resort to numerical and
asymptotic analysis of the solution. We will now describe the features of the solutions for
some simple initial profiles.
Example 1: delta-function initial profile
We take the initial profile to be
u(x, 0) = −U0δ(x) , (4.96)
with U0 > 0 and constant. We know from our analysis of the scattering problem that there
is a single discrete eigenvalue λ = −κ2
1 with κ1 = 1
2U0 and eigenfunction
ψ1(x) =
√
κ1e κ1x
x 0 . (4.97)
The reflection coefficient is given by
b(k) = −
U0
U0 + 2ik
. (4.98)
The time-evolution of these scattering data is
c1(t) =
√
κ1e4κ3
1t
, b(k; t) = −
U0e8ik3t
U0 + 2ik
. (4.99)
The function F becomes
F(x; t) = κ1e8κ3
1t−κ1x
−
U0
2π
∞
−∞
e8ik3t+ikx
U0 + 2ik
dk , (4.100)
but now the Marchenko equation cannot be solved completely. However, it is clear that the
solution should incorporate the single soliton associated with the discrete eigenvalue κ1 = 1
2U0,
i.e.
u(x, t) ∼ −1
2U2
0 sech2 1
2U0(x − U2
0 t − x1) , x1 = −
ln 2
U0
. (4.101)
This solution will be valid where the integral term in F is zero (or very small). To see the role
of the integral term in F we take t → +∞ and consider the region x < 0 where the soliton
solution is exponenially small. Writing k = l x/t the integral becomes
−
U0
2π
∞
−∞
ei
√
x3/t(8l3+l)
U0 t/x + 2il
dl = −
U0
2π
∞
−∞
eiy3/2(8l3+l)
U0t1/3y−1/2 + 2il
dl (4.102)
42 CHAPTER 4. INITIAL VALUE PROBLEM FOR THE KDV EQUATION
where we introduced y = xt−1/3. Taking now t → +∞ with y held fixed gives an oscillatory
dispersive wave propagating to the left with amplitude decaying as t−1/3 as t → +∞. The
solution is depicted in figure 4.2.
If instead we would take U0 < 0 there is no discrete eigenvalue and the solution has no
soliton component, only a dispersive wave train for t > 0.
Example 2: Negative sech2
initial profile
Consider the initial profile
u(x, 0) = −4 sech2
x . (4.103)
The number of discrete eigenvalues is
U0 +
1
4
−
1
2
+ 1 = 2 , (4.104)
for U0 = 4, where [· · · ] denotes the integer part, so we will get a solution with two solitons.
But since 4 cannot be written as N(N + 1) the solution will include a dispersive component.
The solution is depicted in figure 4.3.
Example 3: Positive sech2
initial profile
Whenever u(x, 0) > 0 there are no discrete eigenvalues (there are no bound states for a
positive potential) and the solution will develop without the emergence of a soliton. The
initial profile collapses into a wave train which disperses to the left. The solution for the
profile
u(x, 0) = sech2
x (4.105)
is depicted in figure 4.4. The solution can be seen to approach the (inverted) Airy function.
Here we will leave the discussion of explicit solutions of the KdV equation. Instead we
want to now address the question whether there are other equations that can also be solve in
a similar way. This will lead us to develop some new and powerful tools.
4.5. SOLUTIONS WITH B(K) = 0 43
Figure 4.1: The two-soliton solution at various times: (a) t = −0.5 (b) t = −0.1 (c) t = 0 (d)
t = 0.1 (e) t = 0.5 [Figure taken from Drazin & Johnson.]
44 CHAPTER 4. INITIAL VALUE PROBLEM FOR THE KDV EQUATION
Figure 4.2: The delta-function initial profile: (a) inital profile (b) solution at a later time.
[Figure taken from Drazin & Johnson.]
4.5. SOLUTIONS WITH B(K) = 0 45
Figure 4.3: Solution with two solitons and dispersive wave, inital profile u(x, 0) = −4 sech2
x:
(a) t = 0 (b) t = 0.4 (c) t = 1 [Figure taken from Drazin & Johnson.]
46 CHAPTER 4. INITIAL VALUE PROBLEM FOR THE KDV EQUATION
Figure 4.4: Solution with dispersive wave only, initial profile u(x, 0) = sech2
x: (a) t = 0 (b)
t = 0.1 (c) t = 0.5 [Figure taken from Drazin & Johnson.]
Chapter 5
Lax formulation and more general
inverse methods
So far we have described the solution of the KdV equation via the inverse scattering transform.
To understand if there are other problems that can be solved in a similar way we need to
develop some more general tools. A very important tool was introduced by Lax in 1968.
5.1 Lax formulation
Suppose we are interested in an evolution equation of the form
˙u = N(u) , (5.1)
where N is some non-linear operator this is independent of t but may involve x and partial
derivatives with respect to x. For example, in the KdV case
N(u) = 6uu − u . (5.2)
Now suppose that the evolution equation (5.1) can be expressed in the form
˙L = [M, L] = ML − LM , (5.3)
where L and M are some operators linear in x, which can depend on u(x, t) and which operate
on some auxiliary Hilbert space H. Note that
˙L =
∂
∂t
L (5.4)
means the derivative with respect to t that appear explicitly in L, e.g. for
L = −
∂2
∂x2
+ u(x, t) we have ˙L = ˙u . (5.5)
The Hilbert space has an inner product (φ, ψ) and we assume that L is self-adjoint (Hermitian),
i.e. (Lφ, ψ) = (φ, L†ψ) = (φ, Lψ) ∀φ, ψ ∈ H (cf. the Hamiltonian in quantum
mechanics).
47
48 CHAPTER 5. LAX FORMULATION AND MORE GENERAL INVERSE METHODS
Consider now the spectral problem for L, i.e. the eigenvalue equation
Lψ = λψ , (5.6)
where λ = λ(t). Taking a t-derivative we find
˙Lψ + L ˙ψ = ˙λψ + λ ˙ψ (5.7)
and using ˙L = [M, L] this becomes
˙λψ = (L − λ) ˙ψ + [M, L]ψ = (L − λ)( ˙ψ − Mψ) , (5.8)
where we used again the fact that Lψ = λψ. Taking the inner product with ψ we find
˙λ(ψ, ψ) = (ψ, (L − λ)[ ˙ψ − Mψ]) = ((L − λ)ψ, ˙ψ − Mψ) = 0 , (5.9)
since L is self-adjoint. Therefore we conclude that the eigenvalues of the operator L are
constant. Equation (5.8) now gives
(L − λ)( ˙ψ − Mψ) = 0 , (5.10)
so that ˙ψ − Mψ is also an eigenfunction of L with eigenvalue λ. Therefore we must have
(assuming a non-degenerate case)
˙ψ − Mψ ∝ ψ . (5.11)
But we can always shift M by a function of t times the identity operator since [M, L] remains
invariant, so we may take
˙ψ = Mψ , (5.12)
which gives the time-evolution of the eigenfunction ψ.
We have shown that if the evolution equation
˙u = N(u) (5.13)
can be expressed as the Lax equation
˙L + [L, M] = 0 , (5.14)
where L, M are referred to as the Lax pair, the spectrum of L
Lψ = λψ (5.15)
evolves according to
λ = constant , ˙ψ = Mψ . (5.16)
There is a converse to this statement. The two equations
Lψ = λψ , ˙ψ = Mψ , (5.17)
with λ a constant, imply, taking the time-derivative of the first and using the equations, that
0 = ˙Lψ + L ˙ψ − λ ˙ψ = ˙Lψ + LMψ − λMψ = ( ˙L + [L, M])ψ . (5.18)
This means that the Lax equation is the compatibility condition for the system of equations
(5.17). This is often a useful way of thinking about the Lax equation.
5.2. LAX FORMULATION OF THE KDV EQUATION 49
5.2 Lax formulation of the KdV equation
For the KdV equation it is natural to assume that we should take
L = −
∂2
∂x2
+ u , (5.19)
so that Lψ = λψ becomes the Schr¨odinger (Sturm-Liouville) equation. We still need to find
M, if it exists. Since L is hermitian (self-adjoint), L† = L, we have
[M, L] = ˙L = ˙L†
= [M, L]†
= LM†
− M†
L = −[M†
, L] , (5.20)
and we see that we should take M to be anti-hermitian, M† = −M. Therefore we should
construct M from a combination of odd derivatives in x. (For example
(
∂n
∂xn
φ, ψ) =
∞
−∞
∂nφ
∂xn
ψ = (−1)n
∞
−∞
φ
∂nψ
∂xn
= (−1)n
(φ,
∂n
∂xn
ψ) , (5.21)
where we partially integrated n times using the fact that φ, φ , . . . , ψ, ψ , . . . → 0 as x → ±∞,
i.e. ∂n
∂xn is anti-hermitian for n odd.)
The simplest possibility is to take a single derivative
M = c
∂
∂x
. (5.22)
Taking c to be a constant we find
[L, M] = −
∂2
∂x2
+ u, c
∂
∂x
= −cu . (5.23)
Therefore the Lax equation becomes
0 = ˙L + [L, M] = ˙u − cu , (5.24)
which is the simplest linear wave equation. This is not very useful since we can solve this
equation directly. The next possibility is to take M to be a third order operator
M = −α
∂3
∂x3
+ U
∂
∂x
+
∂
∂x
U + A , (5.25)
where α is a constant, U = U(x, t) and we wrote the ∂
∂x -terms in a form that will be convenient
later. In this case we find
[L, M] = αu − U − A − 2u U + 3αu − 4U − 2A
∂
∂x
+ 3αu − 4U
∂2
∂x2
. (5.26)
Since this should equal ˙L = ˙u which is a multiplicative operator the last two terms must
vanish which happens if
U =
3
4
αu and A = A(t) . (5.27)
In this case the Lax equation becomes
0 = ˙L + [L, M] = ˙u −
3
2
αuu +
1
4
αu , (5.28)
50 CHAPTER 5. LAX FORMULATION AND MORE GENERAL INVERSE METHODS
which for α = 4 is the KdV equation. In this case the operator M becomes
M = −4
∂3
∂x3
+ 3u
∂
∂x
+ 3
∂
∂x
u + A(t) , (5.29)
so the evolution of ψ is
˙ψ = Mψ = −4ψ + 6uψ + 3u ψ + Aψ . (5.30)
Using the Schr¨odinger equation in the first term this becomes
˙ψ = 4[(λ − u)ψ] + 6uψ + 3u ψ + Aψ = 2(2λ + u)ψ − u ψ + Aψ . (5.31)
With A = 0 we recover the time-evolution of the discrete eigenfunctions (4.26) and with
A = 4ik3 we recover that of the continuous ones.
5.2.1 The KdV hierarchy
The KdV equation appeared as our second example in the Lax formulation with L the
Schr¨odinger operator. The procedure can be extended to higher-order non-linear equations.
A little work shows that the appropriate choice of M is
M = −α
∂2n+1
∂x2n+1
+
n
m=1
Um
∂2m−1
∂x2m−1
+
∂2m−1
∂x2m−1
Um + A , (5.32)
where α is a constant, Um = Um(x, t) and A = A(t) (A does not enter the evolution equation).
The condition that [L, M] be a multiplicative operator imposes n conditions on the n unknown
functions Um, namely that the coefficient of ∂2m−1
∂x2m−1 should vanish for m = 1, 2, . . . , n. For
n = 1 we recover the KdV equation while for n = 2 we get the equation
˙u + 30u2
u − 20u u − 10uu + u = 0 , (5.33)
a fifth order KdV equation.
We have now presented the first three evolution equations in the KdV hierarchy, each of
which can be solved with the inverse scattering transform. This already gives us infinitely
many non-linear integrable equations. The Lax formulation has provided this for free, so we
see that it is indeed quite powerful. In fact it can me extended in various ways which makes
it even more powerful. We will now consider another remarkable fact which follows from a
slightly different version of the Lax formulation.
5.2.2 Infinitely many conservation laws
A remarkable property of integrable systems like the KdV equation is that they possess
infinitely many conserved quantities. A nice way to see this uses a slightly different version
of the Lax formulation.
Suppose that an evolution equation can be formulated as the zero-curvature condition
˙L − M + [L, M] = 0 , (5.34)
5.3. THE ZAKHAROV-SHABAT SCHEME 51
where now L and M are not operators but just matrices. However, they are required to
depend non-trivially on an extra auxiliary spectral parameter λ, i.e. L = L(x, t; λ) and
M = M(x, t; λ). Consider for example the choice
L =
0 −1
λ − u 0
, M =
−u −2(2λ + u)
2(λ − u)(2λ + u) + u u
. (5.35)
We then get the equation
0 = ˙L − M + [L, M] = −
0 0
˙u − 6uu + u 0
, (5.36)
which is equivalent to the KdV equation.
The zero-curvature condition for L, M can again be thought of as a compatibility condition,
this time for the two linear equations
V = LV , ˙V = MV . (5.37)
Indeed, taking the t-derivative of the first equation and subtracting the x-derivative of the
second we find
0 =
∂
∂t
(LV ) −
∂
∂x
(MV ) = ˙LV + L ˙V − M V − MV = ( ˙L − M + [L, M])V . (5.38)
For the case of the KdV equation these equations read
V1
V2
=
−V2
(λ − u)V1
,
˙V1
˙V2
=
−u V1 − 2(2λ + u)V2
[2(λ − u)(2λ + u) + u ]V1 + u V2
. (5.39)
Consider the equation for ˙V2. It is
˙V2 = 2(λ − u)(2λ + u)V1 + u V1 + u V2 = [2(2λ + u)V2 + u V1] , (5.40)
as is easily verified using the other equations. Writing f = V2 and g = 2(2λ + u)V2 + u V1
this equation reads
˙f = g . (5.41)
This is a local conservation equation. It says that the integral of f over all x is a constant
of the motion. But since L, M depend also on the auxiliary spectral parameter λ the same
must be true for V1, V2 and therefore in particular f = f(t, x; λ). This means that Taylor
expanding f in powers of the spectral parameter each term is a conserved quantity. Therefore
we have found infinitely many conservation laws for the KdV equation.
The Lax formulation is indeed powerful since it has given us both the KdV hierarchy and
an infinite set of conserved quantities. We will now see how it can be exploited to construct
more general methods for solving integrable equations.
5.3 The Zakharov-Shabat scheme
So far we have discussed the KdV equation and its solution via the inverse scattering transform
in some detail. We are now finally ready to see how these methods can be generalized to other
equations. In the early 70’s two groups generalized the inverse scattering transform so that it
could be applied to several problems. Zakharov and Shabat (ZS) generalized the Lax method
while Ablowitz, Kaup, Newell and Segur (AKNS) generalized the Sturm-Liouville scattering
problem. We will describe only the ZS scheme, which tends to be more useful, here.
52 CHAPTER 5. LAX FORMULATION AND MORE GENERAL INVERSE METHODS
5.3.1 Integral operators
The approach of ZS starts by defining three integral operators. Let F(x, z) and K±(x, z) be
N × N-matrices with
K+(x, z) = 0 if z < x
K−(x, z) = 0 if z > x
(5.42)
We define the integral operators JF and J± by
JF ψ =
∞
−∞
F(x, z)ψ(z)dz , J±ψ =
∞
−∞
K±(x, z)ψ(z)dz , (5.43)
where ψ is an N-component vector. We further demand that these operators satisfy the
identity
(I + J+)(I + JF ) = I + J− (5.44)
where I is the N × N unit matrix. We also require that I + J+ is invertible so that
I + JF = (I + J+)−1
(I + J−) . (5.45)
The identity (5.44) gives
(J+ + JF )ψ + J+JF ψ = J−ψ (5.46)
or
∞
−∞
(K+ + F)(x, z)ψ(z)dz +
∞
−∞
K+(x, z)
∞
−∞
F(z, y)ψ(y)dy dz =
∞
−∞
K−(x, z)ψ(z)dz .
(5.47)
Let’s suppose that ψ(z) = 0 for z < x. Then the RHS is zero and the double integral can be
written (relabeling the integration variables)
∞
−∞
∞
x
K+(x, y)F(y, z)ψ(z)dydz . (5.48)
Therefore we find
∞
−∞
K+(x, z) + F(x, z) +
∞
x
K+(x, y)F(y, z)dy ψ(z)dz = 0 . (5.49)
Since this has to hold for all ψ such that ψ(z) = 0 for z < x we must have
K+(x, z) + F(x, z) +
∞
x
K+(x, y)F(y, z)dy = 0 for z > x . (5.50)
This is the matrix Marchenko equation. Similarly one can show by taking z < x that
K−(x, z) = F(x, z) +
∞
x
K+(x, y)F(y, z)dy , (5.51)
which determines K− in terms of K+ and F.
5.3. THE ZAKHAROV-SHABAT SCHEME 53
5.3.2 Differential operators
We now let K± and F depend on two auxiliary variables, e.g. t, y. The evolution of K± and
F in t, y will be determined by appropriate linear differential operators. We define first the
N × N matrix differential operator ∆0 acting on ψ(x; t, y), which has constant coefficients
and commutes with JF
[∆0, JF ] = 0 . (5.52)
We also introduce an associated operator ∆ defined by
∆(I + J+) = (I + J+)∆0 . (5.53)
(The same holds with J+ replaced by J−.) Recall that I + J+ is invertible so this equation
indeed defines ∆. ∆0 is sometimes referred to as ’undressed’ and ∆ as ’dressed’ operator.
To see how this works consider the simple example, which will be useful later,
∆0 = I α
∂
∂t
−
∂2
∂x2
, (5.54)
where I is the N × N unit matrix and α is a constant. We now get
0 =[∆0, JF ]ψ = ∆0JF ψ − JF ∆0ψ (5.55)
= α
∂
∂t
−
∂2
∂x2
∞
−∞
F(x, z; t)ψ(z; t)dz −
∞
−∞
F(x, z; t) α
∂
∂t
−
∂2
∂z2
ψ(z; t)dz .
Assuming ψ, ∂
∂z ψ → 0 as z → ±∞, we may integrate by parts to get
∞
−∞
α ˙F − F +
∂2
∂z2
F ψdz = 0 . (5.56)
Since this holds for all ψ we find the linear wave equation for F
α ˙F − F +
∂2
∂z2
F = 0 . (5.57)
The associated operator ∆ is obtained from the definition in (5.53). Acting on ψ we get
∆ ψ(x; t) +
∞
x
K+(x, z; t)ψ(z; t)dz = α ˙ψ(x; t)−ψ (x; t)+
∞
x
K+(x, z; t) α
∂
∂t
−
∂2
∂z2
ψ(z; t)dz .
(5.58)
Partially integrating we have
∞
x
K+
∂2
∂z2
ψdz = −K+ψ +
∂
∂z
K+
z=x
ψ +
∞
x
∂2
∂z2
K+ψdz (5.59)
where K+(x; t) = K+(x, x; t). Setting ∆ = ∆0 + ∆1 we therefore get
0 = ∆1 ψ +
∞
x
K+ψdz +∆0
∞
x
K+ψdz−α
∞
x
K+
˙ψdz−K+ψ +
∂
∂z
K+
z=x
ψ+
∞
x
∂2
∂z2
K+ψdz .
(5.60)
54 CHAPTER 5. LAX FORMULATION AND MORE GENERAL INVERSE METHODS
Now the second term is
∆0
∞
x
K+ψdz = α
∞
x
˙K+ψdz +α
∞
x
K+
˙ψdz +
d
dx
K+ψ+K+ψ + K+ z=x
ψ−
∞
x
K+ψdz
(5.61)
and using this we get
0 = ∆1 ψ +
∞
x
K+ψdz +
∞
x
α ˙K+ − K+ +
∂2
∂z2
K+ ψdz+
d
dx
K+ψ+ K+
∂
∂z
K+
z=x
ψ
(5.62)
which simplifies to
0 = ∆1 + 2
d
dx
K+ ψ + ∆1
∞
x
K+ψdz +
∞
x
α ˙K+ − K+ +
∂2
∂z2
K+ ψdz . (5.63)
This must be true for all ψ which implies that
∆1 = −2
d
dx
K+ = −2 K+
∂
∂z
K+
z=x
, (5.64)
so that ∆1 is just a multiplicative operator and K+ satisfies
α ˙K+ − K+ +
∂2
∂z2
K+ + ∆1K+ = 0 . (5.65)
These equations are similar to ones we found when discussing the KdV equation.
Returning now to the main development of the ZS scheme, the next step is to introduce
a pair of operators ∆0 and ∆. We take
∆
(1)
0 = Iα
∂
∂t
− M0 ∆
(2)
0 = Iβ
∂
∂y
+ L0
∆(1)
= Iα
∂
∂t
− M ∆(2)
= Iβ
∂
∂y
+ L (5.66)
where α, β are constants and L0, M0, L, M are differential operators in x only. L0, M0 have
constant coefficients and we require
0 = [∆
(1)
0 , ∆
(2)
0 ] = [L0, M0] (5.67)
as well as
[∆
(i)
0 , JF ] = 0 i = 1, 2 . (5.68)
As before ∆(i) is defined by
∆(i)
(I + J+) = (I + J+)∆
(i)
0 i = 1, 2 . (5.69)
Using this we find
[∆(1)
, ∆(2)
](I + J+) = ∆(1)
∆(2)
(I + J+) − (1 ↔ 2) = (I + J+)[∆
(1)
0 , ∆
(2)
0 ] = 0 (5.70)
and since I + J+ is invertible this implies
[∆(1)
, ∆(2)
] = 0 . (5.71)
5.3. THE ZAKHAROV-SHABAT SCHEME 55
Using the expressions for these operators in (5.66) this becomes
0 = Iα
∂
∂t
− M, Iβ
∂
∂y
+ L = α
∂
∂t
L + β
∂
∂y
M + [L, M] . (5.72)
This is a generalization of the Lax equation (5.14) to two auxiliary variables. Taking α = 1,
β = 0 recovers the original Lax equation.
Systems of non-linear evolution equations that can be formulated as
α
∂
∂t
L + β
∂
∂y
M + [L, M] = 0 (5.73)
can be solved by the ZS scheme. The procedure for solving them is as follows. The variable
coefficients which arise in the ’dressed’ operators L, M constitute the functions which satisfy
the system of evolution equations. They are found from K+, c.f. our example (5.64), where
K+ is a solution to the matrix Marchenko equation. This equation requires F and F is given
by the solution to the pair of equations (5.68), c.f. our example (5.57). Note that in the ZS
scheme the eigenvalue does not appear explicitly anywhere.
Before looking at how to carry out the solution of a system in detail, we will first look
at some examples of equations that can be cast in the ZS form and therefore solve in this
scheme.
5.3.3 Example 1: The KdV equation
In this case the operators involved are scalars. We set
∆
(1)
0 =
∂
∂t
+ 4
∂3
∂x3
∆
(2)
0 = −
∂2
∂x2
. (5.74)
(i.e. α = 1, β = 0, M0 = − ∂3
∂x3 and L0 = − ∂2
∂x2 in (5.66).) We also write
∆(2)
= L = −
∂2
∂x2
+ u(x, t) . (5.75)
Since ∆
(2)
0 coincides with (5.54) for α = 0 we conclude from (5.64) with ∆1 = u that
u = −2
d
dx
K+ (5.76)
and from (5.57) and (5.65) that
∂2
∂z2
F − F = 0 ,
∂2
∂z2
K+ − K+ + uK+ = 0 . (5.77)
Similarly the condition [∆
(1)
0 , JF ] = 0 turns out to give
˙F + 4
∂3
∂z3
F + 4F = 0 . (5.78)
Finally we have
∆(1)
=
∂
∂t
− M , M = M0 + M1 = −4
∂3
∂x3
+ M1 . (5.79)
56 CHAPTER 5. LAX FORMULATION AND MORE GENERAL INVERSE METHODS
The condition ∆(1)(I + J+) = (I + J+)∆
(1)
0 becomes
−M1ψ +
∂
∂t
+ 4
∂3
∂x3
− M1
∞
x
K+ψdz =
∞
x
K+
∂
∂t
+ 4
∂3
∂z3
ψdz (5.80)
or
−M1ψ +
∞
x
˙K+ψdz + 4
∂3
∂x3
∞
x
K+ψdz − M1
∞
x
K+ψdz = 4
∞
x
K+
∂3
∂z3
ψdz . (5.81)
Expanding the ∂3
∂x3 -term and partially integrating the ∂3
∂z3 -term we get
− M1(ψ +
∞
x
K+ψdz) +
∞
x
˙K+ψdz − 4
∂2
∂x2
(K+ψ) − 4
∂
∂x
(K+|z=xψ) − 4K+|z=xψ
+ 4
∞
x
K+ ψdz = −4K+ψ + 4
∂
∂z
K+
z=x
ψ − 4
∂2
∂z2
K+
z=x
ψ − 4
∞
x
∂3
∂z3
K+ψdz
(5.82)
Now suppose that
M1 = 4A(x, t)
∂
∂x
+ 4B(x, t) . (5.83)
The we get
− 4
d2
dx2
K+ +
d
dx
(K+|z=x) + K+|z=x −
∂2
∂z2
K+
z=x
− AK+ + B ψ
− 4 3
d
dx
K+ + A ψ +
∞
x
˙K+ + 4
∂3
∂z3
K+ + 4K+ + 4AK+ − 4BK+ ψdz = 0 (5.84)
To cancel the ψ -term we take A = −3 d
dxK+ = 3
2u and to cancel the first term
B =
∂2
∂z2
K+ − K+
z=x
− (K+ +
∂
∂z
K+)|z=x +
3
2
uK+ −
d2
dx2
K+ (5.85)
which can be written as
B =
3
2
∂2
∂z2
K+ − K+ + uK+
z=x
−
1
2
(K+ + 2
∂
∂z
K+ +
∂2
∂z2
K+)|z=x −
d2
dx2
K+ , (5.86)
The first term vanishes by (5.77) and the second and third are the same so we get
B = −
3
2
d2
dx2
K+ =
3
4
u . (5.87)
And so we have
∆(1)
=
∂
∂t
+ 4
∂3
∂x3
− 6u
∂
∂x
− 3u . (5.88)
The evolution equation
α
∂
∂t
L + β
∂
∂y
M + [L, M] = 0 (5.89)
becomes, setting α = 1, β = 0, L = − ∂2
∂x2 + u and M = −4 ∂3
∂x3 + 6u ∂
∂x + 3u ,
0 = ˙u + [L, M] = ˙u − [
∂2
∂x2
, 6u
∂
∂x
+ 3u ] − [−4
∂3
∂x3
+ 6u
∂
∂x
, u] = ˙u − 6uu + u , (5.90)
recovering the KdV equation. This shows how the ZS scheme can be applied to the KdV
equation.
5.3. THE ZAKHAROV-SHABAT SCHEME 57
5.3.4 Example 2: The 2d KdV equation
The KdV case involved only one of the auxiliary variables, t. A straightforward generalization
is to include also dependence on the other variable, y. We take
∆
(1)
0 =
∂
∂t
+ 4
∂3
∂x3
∆
(2)
0 =
∂
∂y
−
∂2
∂x2
, (5.91)
and write the dressed operators as
∆(2)
= ∆
(2)
0 + u(x, t, y) , ∆(1)
= ∆
(1)
0 − 6u
∂
∂x
− 3u + w(x, t, y) . (5.92)
Following the same steps as in the previous section we find the equations for F
∂
∂y
F +
∂2
∂z2
F − F = 0 , ˙F + 4
∂3
∂z3
F + 4F = 0 . (5.93)
And in the end the equations for u and w become
˙u − 6uu + u −
∂
∂y
w = 0 , w = −3
∂
∂y
u . (5.94)
Taking the x-derivative of the first equation and eliminating w using the second equation we
find the equation for u
( ˙u − 6uu + u ) + 3
∂2
∂y2
u = 0 . (5.95)
This equation is known as the 2d KdV equation or the Kadomtsev-Petviashvili equation. It
can also be solved by the ZS scheme.
So far we took the operators ∆
(i)
0 to be scalars. It should be clear that much more freedom
exists if we take them to be matrices. We will now consider such a case.
5.3.5 Example 3: The non-linear Schr¨odinger equation
Let us take
∆
(1)
0 = I iα
∂
∂t
−
∂2
∂x2
∆
(2)
0 =
l 0
0 m
∂
∂x
, (5.96)
where α, l, m are real constants and I is the 2 × 2 identity matrix. We have already seen that
for an operator of the form of ∆
(1)
0 above we have, see (5.54) and (5.64)
∆(1)
= ∆
(1)
0 + U(x, t) (5.97)
where
U(x, t) = −2
d
dx
K+(x, t) . (5.98)
It remains to determine ∆(2).
∆(2)
= ∆
(2)
0 + V (x, t) (5.99)
the condition
∆(2)
(I + J+) = (I + J+)∆
(2)
0 (5.100)
58 CHAPTER 5. LAX FORMULATION AND MORE GENERAL INVERSE METHODS
gives
∆
(2)
0 J+ψ + V (ψ + J+ψ) = J+∆
(2)
0 ψ (5.101)
or
l 0
0 m
∂
∂x
∞
x
K+ψdz + V ψ +
∞
x
K+ψdz =
∞
x
K+
l 0
0 m
∂
∂z
ψdz . (5.102)
Partially integrating on the RHS and expanding the ∂
∂x -term we get
−
l 0
0 m
K+ψ + K+
l 0
0 m
ψ + V ψ
+
∞
x
l 0
0 m
K+ +
∂
∂z
K+
l 0
0 m
+ V K+ ψdz = 0 . (5.103)
The unintegrated terms must vanish separately so that
V =
l 0
0 m
, K+ . (5.104)
Writing
K+ =
A B
C D
, (5.105)
this becomes
V = (l − m)
0 B
−C 0
. (5.106)
From the vanishing of the integral terms we get the equation for K+
l 0
0 m
K+ +
∂
∂z
K+
l 0
0 m
+ V K+ = 0 . (5.107)
Evaluating this at z = x we find for example
lA = −(l − m)BC , mD = (l − m)BC . (5.108)
The conditions [∆
(i)
0 , JF ] = 0 give the following equations for F
l 0
0 m
F +
∂
∂z
F
l 0
0 m
= 0 , iα ˙F +
∂2
∂z2
F − F = 0 . (5.109)
Now we let B = u and C = ±u∗, the complex conjugate of u. Then we have
∆(1)
= I iα
∂
∂t
−
∂2
∂x2
− 2
d
dx
K+(x, t) , (5.110)
where
−2
d
dx
K+(x, t) = −2
A B
C D
= −2
l−m
l |u|2 u
±u ∗ ±l−m
m |u|2 . (5.111)
5.3. THE ZAKHAROV-SHABAT SCHEME 59
While
∆(2)
=
l 0
0 m
∂
∂x
+ (l − m)
0 u
u∗ 0
. (5.112)
The Lax equation now becomes
0 =iα(l − m)
0 ˙u
˙u∗ 0
+
l 0
0 m
∂
∂x
+ (l − m)
0 u
u∗ 0
,
∂2
∂x2
+ 2
l−m
l |u|2 u
±u ∗ ±l−m
m |u|2 (5.113)
The commutator becomes
2
l 0
0 m
l−m
l (uu∗) u
±u ∗ ±l−m
m (uu∗)
+ 2
l 0
0 m
,
l−m
l |u|2 u
±u ∗ ±l−m
m |u|2
∂
∂x
− 2(l − m)
0 u
u ∗ 0
∂
∂x
− (l − m)
0 u
u ∗ 0
+ 2(l − m)
0 u
u∗ 0
,
l−m
l |u|2 u
±u ∗ ±l−m
m |u|2 . (5.114)
It is easy to see that the ∂
∂x -terms cancel and we are left with
2
(l − m)(uu∗) lu
±mu ∗ ±(l − m)(uu∗)
− (l − m)
0 u
u ∗ 0
+ 2(l − m)
±(uu ∗ + u∗u ) ±l2−m2
lm u|u|2
l2−m2
lm u∗|u|2 (u∗u + u ∗u)
. (5.115)
The diagonal terms cancel and the Lax equation reduces to the equation
iα(l − m) ˙u + (l + m)u ± 2(l − m)
l2 − m2
lm
u|u|2
= 0 , (5.116)
together with its complex conjugate. By rescaling x, t and u it can be brought to the form
i ˙u + u + u|u|2
= 0 , (5.117)
which is known as the non-linear Schr¨odinger (NLS) equation. This equation has applications
to water waves, plasma waves and to the propagation of light in non-linear optical fibers. Since
it can be cast in the Lax form it can be solved by the ZS scheme.
It is also possible to describe another important equation, the Sine-Gordon equation
u − ¨u = sin u (5.118)
in the ZS scheme but we will not have to to do it here. Instead we will look at how solutions
to the NLS equation can be constructed using the ZS scheme.
60 CHAPTER 5. LAX FORMULATION AND MORE GENERAL INVERSE METHODS
5.4 Solving the non-linear Schr¨odinger equation
We will see how to find the simplest solutions to the NLS equation
i ˙u + u + u|u|2
= 0 . (5.119)
The first step is to solve the equations for F in (5.109). Writing
F =
0 r
s 0
(5.120)
the first equation becomes
0 lr
ms 0
+
0 m ∂
∂z r
l ∂
∂z s 0
= 0 , (5.121)
or
lr + m
∂
∂z
r = 0 , ms + l
∂
∂z
s = 0 , (5.122)
which implies
r = r(mx − lz; t) , s = s(lx − mz; t) . (5.123)
Using this the second equation for F becomes
iα
0 ˙r
˙s 0
+
0 (l2 − m2)r
(m2 − l2)s 0
= 0 , (5.124)
where the prime denotes the derivative with respect to the first variable. It is clear that there
exists an exponential solution
r(x, z; t) = r0eρ(mx−lz)+iρ2(l2−m2)t/α
, s(x, z; t) = s0eσ(lx−mz)+iσ2(m2−l2)t/α
, (5.125)
where r0, s0, ρ, σ are arbitrary constants.
The next step is to plug the solution for F into the Marchenko equation for K+
K+(x, z, ; t) + F(x, z; t) +
∞
x
K+(x, y; t)F(y, z; t)dy = 0 . (5.126)
Writing
K+ =
a b
c d
(5.127)
this becomes, suppressing the arguments,
a b
c d
+
0 r
s 0
+
∞
x
a b
c d
0 r
s 0
dy = 0 , (5.128)
so that
a(x, z; t) +
∞
x
b(x, y; t)s(y, z; t)dy = 0 (5.129)
b(x, z; t) + r(x, z; t) +
∞
x
a(x, y; t)r(y, z; t)dy = 0 (5.130)
5.4. SOLVING THE NON-LINEAR SCHR ¨ODINGER EQUATION 61
and two similar equations for c and d. Looking at the z-dependence we find that
a(x, z; t) = e−σmz
L(x, t) , b(x, z; t) = e−ρlz
M(x, t) . (5.131)
Using this the equations reduce to
L + s0M
∞
x
e(σ−ρ)ly+iσ2(m2−l2)t/α
dy = 0 (5.132)
M + r0eρmx+iρ2(l2−m2)t/α
+ r0L
∞
x
e(ρ−σ)my+iρ2(l2−m2)t/α
dy = 0 . (5.133)
For the integrals to be well defined we need
((σ − ρ)l) < 0 and ((ρ − σ)m) < 0 , (5.134)
so that l and m must have opposite sign. Let us take
l = 2 , m = −1 and α =
1
3
. (5.135)
The NLS equation (5.116) for the lower sign becomes
i ˙u + u + 9u|u|2
= 0 , (5.136)
so at the end we will need to take u → u/3 to get a solution to the equation in the standard
form (5.119). Provided that (σ − ρ) < 0 we get the following equation for M (solving the
first equation for L and plugging into the second)
M 1 − r0s0e9i(ρ2−σ2)t
∞
x
e2(σ−ρ)y
dy
∞
x
e(σ−ρ)y
dy = −r0e−ρx+9iρ2t
. (5.137)
Doing the integrals this reduces to
M 1 −
r0s0
2(ρ − σ)2
e3(σ−ρ)x+9i(ρ2−σ2)t
= −r0e−ρx+9iρ2t
. (5.138)
Setting
ρ = k + iλ , σ = −k + iλ ,
r0s0
8k2
= −1 , (5.139)
with k > 0 we get
M = −r0
e−(k+iλ)x+9i(k+iλ)2t
e−6kx−36kλt + 1
. (5.140)
Therefore
b = e−2ρz
M = −r0
e−(k+iλ)(x+2z)+9i(k+iλ)2t
e−6kx−36kλt + 1
. (5.141)
But u(x, t) = B(x, t) = b(x, x; t), see (5.105), so we get
u = − r0
e−3(k+iλ)x+9i(k+iλ)2t
e−6kx−36kλt + 1
= −r0e−3iλx+9i(k2−λ2)t e−3kx−18kλt
e−6kx−36kλt + 1
= −
r0
2
e−3iλx+9i(k2−λ2)t
sech (3kx + 18kλt) . (5.142)
62 CHAPTER 5. LAX FORMULATION AND MORE GENERAL INVERSE METHODS
Finally the equations for c and d and the requirement that C(x; t) = c(x, x; t) = −u∗ implies
s0 = −r0 so that r0s0 = −8k2 gives r0 = ±2
√
2k. To get a solution to our original equation
(5.119) we also need to send u → u/3 and doing this we find the solitary wave solution to the
NLS equation
u(x, t) = ±aei[c
2
(x−ct)+nt] sech
a
√
2
(x − ct) , (5.143)
where
a = 3
√
2k , c = −6λ and n =
1
2
a2
+
1
4
c2
. (5.144)
The generalization to the N-soliton solution is now straight-forward. Instead of constructing
r, s in F from a single exponential as we did one now takes a sum of N of them.
Unfortunately we have to leave the subject of the ZS scheme here, although there is much
more to be said. Hopefully this course has served as an invitation to the vast and fascinating
subject of integrable models which will inspire you to learn more.