3Inverse scattering transform
II Integrable Systems
3.5 Reflectionless potentials
We are now going to actually solve the KdV equation for a special kind of
potential — reflectionless potentials.
Definition
(Reflectionless potential)
.
A reflectionless potential is a potential
u(x, 0) satisfying R(k, 0) = 0.
Now if u evolves according to the KdV equation, then
R(k, t) = R(k, 0)e
8ik
3
t
= 0.
So if a potential starts off reflectionless, then it remains reflectionless.
We now want to solve the GLM equation in this case. Using the notation
when we wrote down the GLM equation, we simply have
F (x) =
N
X
n=1
c
2
n
e
−χ
n
x
.
We will mostly not write out the
t
when we do this, and only put it back in at
the very end. We now guess that the GLM equation has a solution of the form
K(x, y) =
N
X
m=1
K
m
(x)e
−χ
m
y
for some unknown functions
{K
m
}
(in the second example sheet, we show that
it must have this form). We substitute this into the GLM equation and find that
N
X
n=1
"
c
2
n
e
−χ
n
x
+ K
n
(x) +
N
X
m=1
c
2
n
K
m
(x)
Z
∞
x
e
−(χ
n
+χ
m
)z
dz
#
e
−χ
n
y
= 0.
Now notice that the
e
−χ
n
y
for
n
= 1
, ··· , N
are linearly independent. So we
actually have N equations, one for each n. So we know that
c
2
n
e
−χ
n
x
+ K
n
(x) +
N
X
m=1
c
2
n
K
m
(x)
χ
n
+ χ
m
e
−(χ
n
+χ
m
)x
= 0 (∗)
for all
n
= 1
, ··· , N
. Now if our goal is to solve the
K
n
(
x
), then this is just a
linear equation for each x! We set
c = (c
2
1
e
−χ
1
x
, ··· , c
2
N
e
−χ
N
x
)
T
K = (K
1
(x), ··· , K
N
(x))
T
A
nm
= δ
nm
+
c
2
n
e
−(χ
n
−χ
m
)x
χ
n
+ χ
m
.
Then (∗) becomes
AK = −c.
This really is a linear algebra problem. But we don’t really have to solve this.
The thing we really want to know is
K(x, x) =
N
X
m=1
K
m
(x)e
−χ
m
x
=
N
X
m=1
N
X
m=1
(A
−1
)
mn
(−c)
n
e
−χ
m
x
Now note that
d
dx
A
nm
(x) = A
0
nm
(x) = −c
2
n
e
−χ
n
x
e
−χ
m
x
= (−c)
n
e
−χ
m
x
.
So we can replace the above thing by
K(x, x) =
N
X
m=1
N
X
n=1
(A
−1
)
mn
A
0
nm
= tr(A
−1
A
0
).
It is an exercise on the second example sheet to show that this is equal to
K(x, x) =
1
det A
d
dx
(det A) =
d
dx
log(det A).
So we have
u(x) = −2
d
2
dx
2
log(det A).
We now put back the
t
-dependence we didn’t bother to write all along. Then we
have
u(x, t) = −2
∂
2
∂x
2
log(det A(x, t)),
where
A
nm
(x, t) = δ
nm
+
c
n
(0)
2
e
8χ
3
n
t
e
−(χ
n
+χ
m
)x
χ
n
+ χ
m
.
It turns out these are soliton solutions, and the number of discrete eigenstates
N is just the number of solitons!