3Inverse scattering transform
II Integrable Systems
3.4 Evolution of scattering data
Now we do the clever bit: we allow the potential
u
=
u
(
x, t
) to evolve via KdV
u
t
+ u
xxx
− 6uu
x
= 0.
We see how the scattering data for
L
=
−∂
2
x
+
u
(
x, t
) evolves. Again, we will
assume that u has compact support. Note that this implies that we have
A = 4∂
3
x
as |x| → ∞.
3.4.1 Continuous spectrum (λ = k
2
> 0)
As in Section 3.1.1, for each
t
, we can construct a solution
ϕ
to
Lϕ
=
k
2
ϕ
such
that
ϕ(x, t) =
(
e
−ikx
x → −∞
a(k, t)e
−ikx
+ b(k, t)e
ikx
x → ∞
.
This time, we know that for any
u
, we can find a solution for any
k
. So we can
assume that k is fixed in the equation
Lϕ = k
2
ϕ.
We assume that
u
is a solution to the KdV equation, so that (
L, A
) is a Lax
pair. As in the proof of the isospectral flow theorem, we differentiate this to get
0 = (L − k
2
)(ϕ
t
+ A
ϕ
).
This tells us that
˜ϕ = ϕ
t
+ Aϕ
solves
L ˜ϕ = k
2
˜ϕ.
We can try to figure out what
˜ϕ
is for large
|x|
. We recall that for large
|x|
, we
simply have A = 4∂
3
x
. Then we can write
˜ϕ(x, t) =
(
4ik
3
e
−ikx
x → −∞
(a
t
+ 4ik
3
a)e
−ikx
+ (b
t
− 4ik
3
b)e
ikx
x → ∞
We now consider the function
θ = 4ik
3
ϕ − ˜ϕ.
By linearity of L, we have
Lθ = k
2
θ.
Note that by construction, we have
θ
(
x, t
)
→
0 as
x → −∞
. We recall that the
solution to Lf = k
2
f for f = f
0
as x → −∞ is just
f = (I − K)
−1
f
0
= (I + K + K
2
+ ···)f
0
So we obtain
θ = (1 + K + K
2
+ ···)0 = 0.
So we must have
˜ϕ = 4ik
3
ϕ.
Looking at the x → +∞ behaviour, we figure that
a
t
+ 4ik
3
a = 4ik
3
a
b
t
− 4ik
3
b = 4ik
3
b
Of course, these are equations we can solve. We have
a(k, t) = a(k, 0)
b(k, t) = b(k, 0)e
8ik
3
t
.
In terms of the reflection and transmission coefficients, we have
R(k, t) = R(k, 0)e
8ik
3
t
T (k, t) = T (k, 0).
Thus, we have shown that if we assume
u
evolves according to the really compli-
cated KdV equation, then the scattering data must evolve in this simple way!
This is AMAZING.
3.4.2 Discrete spectrum (λ = −κ
2
< 0)
The discrete part is similar. By the isospectral flow theorem, we know the
χ
n
are constant in time. For each
t
, we can construct bound states
{ψ
n
(
x, t
)
}
N
n=1
such that
Lψ
n
= −χ
2
n
ψ
n
, kψ
n
k = 1.
Moreover, we have
ψ
n
(x, t) = c
n
(t)e
−χ
n
|x|
as x → +∞.
From the isospectral theorem, we know the function
˜
ψ
n
= ∂
t
ψ
n
+ Aψ
n
also satisfies
L
˜
ψ
n
= −χ
2
n
˜
ψ
n
It is an exercise to show that these solutions must actually be proportional to
one another. Looking at Wronskians, we can show that
˜
ψ
n
∝ ψ
n
. Also, we have
hψ
n
,
˜
ψ
n
i = hψ
n
, ∂
t
ψ
n
i + hψ
n
, Aψ
n
i
=
1
2
∂
∂t
hψ
n
, ψ
n
i + hψ
n
, Aψ
n
i
= 0,
using the fact that
A
is antisymmetric and
kψ
n
k
is constant. We thus deduce
that
˜
ψ
n
= 0.
Looking at large x-behaviour, we have
˜
ψ
n
(x, t) = ( ˙c
n
− 4χ
3
n
c
n
)e
−χ
n
x
as x → +∞. Since
˜
ψ
n
≡ 0, we must have
˙c
n
− 4χ
3
n
c
n
= 0.
So we have
c
n
(t) = c
n
(0)e
4χ
3
n
t
.
This is again AMAZING.
3.4.3 Summary of inverse scattering transform
So in summary, suppose we are given that
u
=
u
(
x, t
) evolves according to KdV,
namely
u
t
+ u
xxx
− 6uu
x
= 0.
If we have an initial condition
u
0
(
x
) =
u
(
x,
0), then we can compute its scattering
data
S(0) =
{χ
n
, c
n
(0)}
N
n=1
, R(k, 0)
.
Then for arbitrary time, the scattering data for L = −∂
2
x
+ u is
S(t) =
n
{χ
n
, c
n
(0)e
4χ
3
n
t
}
N
n=1
, R(k, 0)e
8ik
3
t
o
.
We then apply GLM to obtain u(x, t) for all time t.
u
0
(x) S(0) =
{χ
n
, c
n
(0)}
N
n=1
, R(k, 0)
u(x, t) S(t) =
n
{χ
n
, c
n
(0)e
4χ
3
n
t
}
N
n=1
, R(k, 0)e
8ik
3
t
o
Construct scattering data
L=−∂
2
x
+u
0
(x)
KdV
equation
Evolve
scattering
data
L
t
=[L,A]
Solve GLM equation
The key thing that makes this work is that
u
t
+
u
xxx
−
6
uu
x
holds if and only if
L
t
= [L, A].
For comparison, this is what we would do if we had to solve
u
t
+
u
xxx
= 0
by a Fourier transform:
u
0
(x) ˆu
0
(k)
u(x, t) ˆu(u, t) = ˆu
0
(k)e
ik
3
t
Fourier transform
u
t
+u
xxx
=0
ˆu
t
−ik
3
ˆu=0
Inverse Fourier
Transform
It is just the same steps, but with a simpler transform!