3Inverse scattering transform
II Integrable Systems
3.2 Inverse scattering problem
We might be interested in the inverse problem. Given scattering data
S =
{χ
n
, c
n
}
N
n=1
, R(k), T (k)
,
can we reconstruct the potential u = u(x) such that
L = −
∂
2
∂x
2
+ u(x)
has scattering data
S
? The answer is yes! Moreover, it turns out that
T
(
k
) is
not needed.
We shall write down a rather explicit formula for the inverse scattering
problem, but we will not justify it.
Theorem
(GLM inverse scattering theorem)
.
A potential
u
=
u
(
x
) that decays
rapidly to 0 as |x| → ∞ is completely determined by its scattering data
S =
{χ
n
, c
n
}
N
n=1
, R(k)
.
Given such a scattering data, if we set
F (x) =
N
X
n=1
c
2
n
e
−χ
n
x
+
1
2π
Z
∞
−∞
e
ikx
R(k) dk,
and define k(x, y) to be the unique solution to
k(x, y) + F (x + y) +
Z
∞
x
k(x, z)f(z + y) dz = 0,
then
u(x) = −2
d
dx
k(x, x).
Proof. Too hard.
Note that this equation
k(x, y) + F (x + y) +
Z
∞
x
k(x, z)f(z + y) dz = 0
is not too hard to solve. We can view it as a linear equation of the form
x + b + Ax = 0
for some linear operator
A
, then use our familiar linear algebra techniques to
guess a solution. Afterwards, we can then verify that it works. We will see an
explicit example later on when we actually use this to solve problems.
Now that we’ve got this result, we understand how scattering problems work.
We know how to go forwards and backwards.
This is all old theory, and not too exciting. The real exciting thing is how
we are going to use this to solve PDE’s. Given the KdV equation
u
t
+ u
xxx
− 6uu
x
= 0,
we can think of this as a potential evolving over time, with a starting potential
u
(
x,
0) =
u
0
(
x
). We then compute the initial scattering data
T
,
R
,
χ
and
c
.
Afterwards, we obtain the corresponding equations of evolution of the scattering
data form the KdV equation. It turns out this is really simple — the
χ
n
are
always fixed, and the others evolve as
R(k, t) = e
8ik
3
t
R(k, 0)
T (k, t) = T (k, 0)
c
n
(t) = e
4χ
3
n
t
c
n
(0).
Then we use this GLM formula to reconstruct the potential u at all times!