Nonlinear waves and solitons (F7780)
Lecture notes
Linus Wulﬀ
Spring 2024
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 get 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.
1
the wave tends to spread out with time, or dissipation, i.e. the amplitude decays with time
and the wave loses energy to its 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, which is interpreted as the
wave losing 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).
2
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 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 on 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 a 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.
3
which integrates to −vf −3f2 +f = a for some constant a. 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 that f, f , f → 0 as |x| → ∞ gives a = b = 0 and
we have
f 2
= f2
(2f + v) . (1.12)
Note that a real solution can exist only if 2f + v ≥ 0. Integrating this equation gives
df
f
√
2f + v
= ± dx−
. (1.13)
To evaluate the LHS we make the substitution f = −1
2v sech2
y (recall that sech x = 1/ cosh x),
which leads to the solution y = 1
2v1/2(x− − x−
0 ) = 1
2v1/2(x − vt − x0) (the sign of y is clearly
irrelevant). The solution is therefore
u(x, t) = f(x − vt) = −
v
2
sech2
√
v
2
(x − vt − x0) . (1.14)
This is the solitary wave solution. The minus sign in front 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 also that the amplitude and position x0 at t = 0 uniquely speciﬁes the solution. 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 rather remarkably the non-linearity of the equation is
countering this tendency for the wave to disperse, leading to a wave that retains its shape.
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 Zabusky 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 other examples
later in this course.
4
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 into the KdV equation (1.8) by the change of variables x → x − c0t,
followed by rescaling t and x. This shows that the KdV equation can be expected to describe
the propagation of long wavelength waves in various circumstances.
5
Chapter 2
Elementary solutions of the KdV
equation
In the last chapter we derived the solitary wave solution of the KdV equation (1.8) by 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 and b are integration constants. We then assumed a solution which goes to zero 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.
6
(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 f3, where
it crosses from negative to positive, and a double zero at f1 where it touches the line F = 0
and then is positive again. Or F has a simple zero at f3, 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 f1 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 (we will show this explicitly below).
It therefore remains only to analyze the case with three simple zeros at f3 < f2 < f1. The
solution can exist only between f = f3 and f = f2 since F is positive there. It has a local
minimum at f3 (F (f3) > 0) and a local maximum at f2 (F (f2) < 0). At these points f
7
changes sign. Therefore the solution oscillates between the values f3 and f2. 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 at
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)
8
precisely the function that appeared in the solitary wave solution. Note that cn and sn are
periodic functions except when 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 it
the cnoidal wave. The solution depends on three parameters f1 ≥ f2 ≥ f3, the roots of
F(f) deﬁned in (2.5). From this equation we see that the speed of the wave, appearing in
ξ = x − vt, 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 functions, 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)
9
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 = 0 and note that the solution then coincides 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 example 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)
10
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 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 a triple root of F(f) in (2.5) 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.
11
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). Surprisingly, 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
12
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 0 < 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 continuous spectrum we write
k =
√
λ (3.11)
13
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)
Multiplying the ﬁrst equation by ψm and the second by ψn and taking the diﬀerence we ﬁnd
(κ2
m − κ2
n)ψnψm = ψnψm − ψmψn = (W(ψn, ψm)) , (3.14)
where the Wronskian of two functions is deﬁned as W(α, β) = αβ − βα . Integrating this we
ﬁnd
(κ2
m − κ2
n)
∞
−∞
ψnψmdx = [W(ψn, ψm)]∞
−∞ . (3.15)
Since ψ → 0 as x → ±∞ the RHS vanishes and we learn that
∞
−∞
ψnψmdx = 0 for m = n , (3.16)
i.e. the discrete eigenfunctions are orthogonal. The continuous 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 ﬁnd instead
(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).
14
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 = 2
∞
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 κn > 0 by construction.
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.
It is interesting to 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.
15
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)
16
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(c− + ik)Γ(c+ + ik)
. (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(c− + ik)Γ(c+ + ik) = 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. Potentials for which this
17
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 corresponding eigenfunctions 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)
18
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. It 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)
19
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
20
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¨odinger
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)
21
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 subscript 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)
22
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 Solving 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
23
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)
24
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.
25
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)
26
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.
27
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)
28
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(t) = {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
29
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[(u − λ)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)
30
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)
31
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
˙beikx
+ 4ik3
(e−ikx
− beikx
) = h(e−ikx
+ beikx
) . (4.36)
The coeﬃcients of both exponentials have to match and we ﬁnd
h = 4ik3
⇒ ˙a = 0 (4.37)
and
˙b − 8ik3
b = 0 ⇒ b(k; t) = b(k; 0)e8ik3t
. (4.38)
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) (4.39)
b(k; t) = b(k; 0)e8ik3t
. (4.40)
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 , (4.41)
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 . (4.42)
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
. (4.43)
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 , (4.44)
32
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 . (4.45)
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) . (4.46)
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 . (4.47)
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 . (4.48)
The scattering problem is
ψ + (λ + 2 sech2
x)ψ = 0 . (4.49)
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 . (4.50)
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 (4.51)
and since ∞
−∞
sech2
xdx = 2 (4.52)
33
the normalized eigenfunction is (the sign is irrelevant)
ψ1(x) =
1
√
2
sech x . (4.53)
Its asymptotic behavior is
ψ1(x) ∼
√
2e−x
as x → +∞ , (4.54)
and we read oﬀ that c1(0) =
√
2. Therefore
c1(t) =
√
2e4t
. (4.55)
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
. (4.56)
The Marchenko equation becomes
K(x, z; t) + 2e8t−(x+z)
+ 2
∞
x
K(x, y; t)e8t−(y+z)
dy = 0 . (4.57)
Looking at the z-dependence we see that the solution is of the form
K(x, z; t) = L(x, t)e−z
(4.58)
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) , (4.59)
so that
L(x, t) =
−2e8t−x
1 + e8t−2x
. (4.60)
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) . (4.61)
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 (4.62)
and we need to solve the scattering problem
ψ + (λ + 6 sech2
x)ψ = 0 . (4.63)
34
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 . (4.64)
Their asymptotic behavior is
ψ1(x) ∼
√
6e−x
, , ψ2(x) ∼ 2
√
3e−2x
as x → +∞ , (4.65)
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
. (4.66)
As before our choice of proﬁle ensures that b(k; t) = 0. The function F becomes
F(x; t) = 6e8t−x
+ 12e64t−2x
(4.67)
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 .
(4.68)
It is clear that the solution has the form
K(x, z; t) = L1(x, t)e−z
+ L2(x, t)e−2z
, (4.69)
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 , (4.70)
L2 + 12e64t−2x
+ 12e64t
L1
∞
x
e−3y
dy + L2
∞
x
e−4y
dy = 0 . (4.71)
Evaluating the integrals they become
L1 + 6e8t−x
+ 3L1e8t−2x
+ 2L2e8t−3x
= 0 , (4.72)
L2 + 12e64t−2x
+ 4L1e64t−3x
+ 3L2e64t−4x
= 0 , (4.73)
with solution
L1 = 6
e72t−5x − e8t−x
D
, L1 = −12
e72t−4x + e64t−2x
D
, (4.74)
where
D = 1 + 3e8t−2x
+ 3e64t−4x
+ e72t−6x
. (4.75)
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
(4.76)
35
and after a bit of algebra1 one ﬁnds
u(x, t) = −12
3 + 4 cosh(2x − 8t) + cosh(4x − 64t)
(3 cosh(x − 28t) + cosh(3x − 36t))2
. (4.77)
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 on 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
. (4.78)
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 → ±∞ .
(4.79)
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 → ±∞ , (4.80)
where ξ = x − 16t and η = x − 4t. Here we see that the taller wave of amplitude −8 is shifted
to the right by ln 3 while the shorter one of amplitude −2 is shifted to the left by ln 3 by the
interaction.
1
First take out the square root of the last term in D so that
D = 2e36t−3x
(3 cosh(28t − x) + cosh(36t − 3x))
and
u(x, t) = −6
∂
∂x
3 sinh(28t − x) + cosh(28t − x) − cosh(36t − 3x) − sinh(36t − 3x)
3 cosh(28t − x) + cosh(36t − 3x)
and the rest is easy.
36
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 (4.81)
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 → +∞ . (4.82)
They are proportional to the associated Legendre functions
ψn(x) ∝ Pn
N (tanh x) (4.83)
and cn(0) is determined from the normalization condition. Then
cn(t) = cn(0)e4n3t
. (4.84)
The function F becomes
F(x; t) =
N
n=1
c2
n(0)e8n3t−nx
(4.85)
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 . (4.86)
The solution must take the form
K(x, z; t) =
N
n=1
Ln(x, t)e−nz
. (4.87)
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 (4.88)
with
L =





L1
L2
...
LN





, X =





c2
1(0)e8t−x
c2
2(0)e64t−2x
...
c2
N (0)e8N3t−Nx





, (4.89)
The N × N matrix A has elements
Amn = δmn +
c2
m(0)
m + n
e8m3t−(m+n)x
(4.90)
37
and we showed that the solution to the KdV equation takes the form
u(x, t) = −2
∂2
∂x2
ln det A . (4.91)
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 → ±∞ , (4.92)
where the phase, xn, is given by
e2xn
=
N
m=1, m=n
n − m
n + m
sgn(n−m)
n = 1, 2, . . . , N . (4.93)
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 → ±∞ . (4.94)
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) , (4.95)
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 . (4.96)
The reﬂection coeﬃcient is given by
b(k) = −
U0
U0 + 2ik
. (4.97)
The time-evolution of these scattering data is
c1(t) =
√
κ1e4κ3
1t
, b(k; t) = −
U0e8ik3t
U0 + 2ik
. (4.98)
38
The function F becomes
F(x; t) = κ1e8κ3
1t−κ1x
−
U0
2π
∞
−∞
e8ik3t+ikx
U0 + 2ik
dk , (4.99)
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
. (4.100)
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 exponentially 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 (4.101)
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 . (4.102)
The number of discrete eigenvalues is
U0 +
1
4
−
1
2
+ 1 = 2 , (4.103)
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 (4.104)
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 solved
in a similar way. This will lead us to develop some new and powerful tools.
39
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.]
40
Figure 4.2: The delta-function initial proﬁle: (a) inital proﬁle (b) solution at a later time.
[Figure taken from Drazin & Johnson.]
41
Figure 4.3: Solution with two solitons and dispersive wave, initial proﬁle u(x, 0) = −4 sech2
x:
(a) t = 0 (b) t = 0.4 (c) t = 1 [Figure taken from Drazin & Johnson.]
42
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.]
43
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 nonlinear operator that 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 linear operators (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).
44
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.
45
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)
46
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 be 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)
47
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.
48
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.
49
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 ∆. It is sometimes referred to as the “dressed” version of the operator ∆0.
To see how this works consider the following 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. In this case we ﬁnd
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)
Integrating by parts twice 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
∆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)
50
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)
the last two terms add up and we ﬁnd
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+ , (5.64)
so that ∆1 is just a multiplicative operator and K+ satisﬁes the wave equation
α ˙K+ − K+ +
∂2
∂z2
K+ + ∆1K+ = 0 . (5.65)
Note that these equations are very 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 ∆
(i)
0 and ∆(i) with i = 1, 2. 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, i.e. they
don’t contain derivatives with respect to the auxiliary variable t, y. L0, M0 have constant
coeﬃcients and we require the two operators to commute
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)
51
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, which in turn
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 solved 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 = −4 ∂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)
52
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 takes the form
M1 = 4A(x, t)
∂
∂x
+ 4B(x, t) . (5.83)
Then 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.
53
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)
54
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)
55
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 do it here. Instead we will look at how solutions to the NLS
equation can be constructed using the ZS scheme.
56
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)
57
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)
58
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.
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