Introduction
The inverse scattering transformation gives a way to solve the KdV equation
exactly. You can think about is as being an analogous transformation to the
Fourier transformation, except it works for a non linear equation. We want to
be able to solve
[math]\displaystyle{ \begin{align}
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\
u(x,0) & =f\left( x\right)
\end{align} }[/math]
with [math]\displaystyle{ \left\vert u\right\vert \rightarrow0 }[/math] as [math]\displaystyle{ x\rightarrow\pm\infty. }[/math]
The Miura transformation is given by
[math]\displaystyle{
u=v^{2}+v_{x} \,
}[/math]
and if [math]\displaystyle{ v }[/math] satisfies the mKdV
[math]\displaystyle{
\partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0
}[/math]
then [math]\displaystyle{ u }[/math] satisfies the KdV (but not vice versa). We can think about the Miura
transformation as being a nonlinear ODE solving for [math]\displaystyle{ v }[/math] given [math]\displaystyle{ u. }[/math] This
nonlinear ODE is also known as the Riccati equation and there is a well know
transformation which linearises this equation. It we write
[math]\displaystyle{
v=\frac{\left( \partial_{x}w\right) }{w}
}[/math]
then we obtain the equation
[math]\displaystyle{
\partial_{x}^{2}w+uw=0
}[/math]
The KdV is invariant under the transformation [math]\displaystyle{ x\rightarrow x+6\lambda t, }[/math]
[math]\displaystyle{ u\rightarrow u+\lambda. }[/math] Therefore we consider the associated eigenvalue
problem
[math]\displaystyle{
\partial_{x}^{2}w+uw=-\lambda w
}[/math]
The eigenfunctions and eigenvalues of this scattering problem play a key role
in the inverse scattering transformation. Note that this is Schrodinger's equation.
Properties of the eigenfunctions
The equation
[math]\displaystyle{
\partial_{x}^{2}w+uw=-\lambda w
}[/math]
has two kinds of solutions for [math]\displaystyle{ u\rightarrow0 }[/math] as [math]\displaystyle{ x\rightarrow\pm\infty. }[/math] The
first are waves and the second are bound solutions. It is well known that
there are at most a finite number of bound solutions (provided [math]\displaystyle{ u\rightarrow0 }[/math]
as [math]\displaystyle{ x\pm\infty }[/math] sufficiently rapidly) and a continum of solutions for the
incident waves.
Example: Scattering by a Well
The properties of the eigenfunction is prehaps seem most easily through the
following example
[math]\displaystyle{
u\left( x\right) =\left\{
\begin{matrix}
0, & x\notin\left[ -1,1\right] \\
b, & x\in\left[ -1,1\right]
\end{matrix}
\right.
}[/math]
where [math]\displaystyle{ b\gt 0. }[/math]
Case when [math]\displaystyle{ \lambda\lt 0 }[/math]
If we solve this equation for the case when [math]\displaystyle{ \lambda\lt 0, }[/math] [math]\displaystyle{ \lambda=-k^{2} }[/math] we
get
[math]\displaystyle{
w\left( x\right) =\left\{
\begin{matrix}
a_{1}\mathrm{e}^{kx}, & x\lt -1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1\lt x\lt 1\\
a_{2}\mathrm{e}^{-kx} & x\gt 1
\end{matrix}
\right.
}[/math]
where [math]\displaystyle{ \kappa=\sqrt{b-k^{2}} }[/math] where we have assumed that [math]\displaystyle{ b\gt k^{2} }[/math] (there is
no solution for [math]\displaystyle{ b\lt k^{2}). }[/math] We then match [math]\displaystyle{ w }[/math] and its derivative at [math]\displaystyle{ x=\pm1 }[/math]
to solve for [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math]. This leads to two system of equations, one for the
even ([math]\displaystyle{ a_{1}=a_{2} }[/math] and [math]\displaystyle{ b_{2}=0 }[/math] ) and one for the odd solutions
([math]\displaystyle{ a_{1}=-a_{2} }[/math] and [math]\displaystyle{ b_{1}=0) }[/math]. The solution for the even solutions is
[math]\displaystyle{
\left(
\begin{matrix}
\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & \sin\kappa
\end{matrix}
\right) \left(
\begin{matrix}
a_{1}\\
b_{1}
\end{matrix}
\right) =\left(
\begin{matrix}
0\\
0
\end{matrix}
\right)
}[/math]
This has non trivial solutions when
[math]\displaystyle{
\det\left(
\begin{matrix}
\mathrm{e}^{-kx} & -\cos\kappa\\
k\mathrm{e}^{-kx} & - \kappa \sin\kappa
\end{matrix}
\right) =0
}[/math]
which gives us the equation
[math]\displaystyle{
- \kappa \sin\kappa \mathrm{e}^{-kx}+\left( \cos\kappa\right) k\mathrm{e}^{-kx}=0
}[/math]
or
[math]\displaystyle{
\kappa \tan\kappa=k=\sqrt{b-\kappa^{2}}
}[/math]
We know that [math]\displaystyle{ 0\lt \kappa\lt \sqrt{b} }[/math] and if we plot this we see that we obtain a
finite number of solutions.
Similarly we repeat the above process for the odd solutions.
Case when [math]\displaystyle{ \lambda\gt 0 }[/math]
When [math]\displaystyle{ \lambda\gt 0 }[/math] we write [math]\displaystyle{ \lambda=k^{2} }[/math] and we obtain solution
[math]\displaystyle{
w\left( x\right) =\left\{
\begin{matrix}
\mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x\lt -1\\
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1\lt x\lt 1\\
a\mathrm{e}^{-\mathrm{i}kx} & x\gt 1
\end{matrix}
\right.
}[/math]
where [math]\displaystyle{ \kappa=\sqrt{b+k^{2}}. }[/math] Matching [math]\displaystyle{ w }[/math] and its derivaties at [math]\displaystyle{ x=\pm1 }[/math] we
obtain
[math]\displaystyle{
\left(
\begin{matrix}
-\mathrm{e}^{-ik} & \cos\kappa & -\sin\kappa & 0\\
ik\mathrm{e}^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\
0 & \cos\kappa & \sin\kappa & -\mathrm{e}^{-ik}\\
0 & -\kappa\sin\kappa & \kappa\cos\kappa & ik\mathrm{e}^{-ik}
\end{matrix}
\right) \left(
\begin{matrix}
r\\
b_{1}\\
b_{2}\\
a
\end{matrix}
\right) =\left(
\begin{matrix}
\mathrm{e}^{ik}\\
ik\mathrm{e}^{-ik}\\
0\\
0
\end{matrix}
\right)
}[/math]
Connection with the KdV
If we substitute the relationship
[math]\displaystyle{
\partial_{x}^{2}w+uw=-\lambda w
}[/math]
into the KdV after some manipulation we obtain
[math]\displaystyle{
\partial_{t}\lambda w^{2}+\partial_{x}\left( w\partial_{x}Q-\partial
_{x}wQ\right) =0
}[/math]
where [math]\displaystyle{ Q=\partial_{t}w+\partial_{x}^{3}w-3\left( \lambda-u\right)
\partial_{x}w. }[/math] If we integrate this equation then we obtain the result that
[math]\displaystyle{
\partial_{t}\lambda=0
}[/math]
provided that the eigenfunction [math]\displaystyle{ w }[/math] is bounded (which is true for the bound
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
and [math]\displaystyle{ u\left( x,t\right) }[/math] evolves according to the KdV.
Scattering Data
For the discrete spectrum the eigenfunctions behave like
[math]\displaystyle{
w_{n}\left( x\right) =c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}
}[/math]
as [math]\displaystyle{ x\rightarrow\infty }[/math] with
[math]\displaystyle{
\int_{-\infty}^{\infty}\left( w_{n}\left( x\right) \right) ^{2}dx=1
}[/math]
The continuous spectrum looks like
[math]\displaystyle{
v\left( x,t\right) \approx \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
,\ \ \ x\rightarrow-\infty
}[/math]
[math]\displaystyle{
v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ x\rightarrow
\infty
}[/math]
where [math]\displaystyle{ r }[/math] is the reflection coefficient and [math]\displaystyle{ a }[/math] is the transmission
coefficient. This gives us the scattering data at [math]\displaystyle{ t=0 }[/math]
[math]\displaystyle{
S\left( \lambda,0\right) =\left( \left\{ k_{n},c_{n}\left( 0\right)
\right\} _{n=1}^{N},r\left( k,0\right) ,a\left( k,0\right) \right)
}[/math]
The scattering data evolves as
[math]\displaystyle{
k_{n}=k_{n}
}[/math]
[math]\displaystyle{
c_{n}\left( t\right) =c_{n}\left( 0\right) \mathrm{e}^{4k_{n}^{3}t}
}[/math]
[math]\displaystyle{
r\left( k,t\right) =r\left( k,0\right) \mathrm{e}^{8ik^{3}t}
}[/math]
[math]\displaystyle{
a\left( k,t\right) =a\left( k,0\right)
}[/math]
We can recover [math]\displaystyle{ u }[/math] from scattering data. We write
[math]\displaystyle{
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}
x}+\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
}[/math]
Then solve
[math]\displaystyle{
K\left( x,y;t\right) +F\left( x+y;t\right) +\int_{x}^{\infty}K\left(
x,z;t\right) F\left( z+y;t\right) \mathrm{d}z=0
}[/math]
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
}equation. We then find [math]\displaystyle{ u }[/math] from
[math]\displaystyle{
u\left( x,t\right) =2\partial_{x}K\left( x,x,t\right)
}[/math]
Reflectionless Potential
In general the IST is difficult to solve. However, there is a simplification
we can make when we have a reflectionless potential (which we will see gives
rise to the soliton solutions). The reflectionless potential is the case when
[math]\displaystyle{ r\left( k,0\right) =0 }[/math] for all values of [math]\displaystyle{ k }[/math] for some [math]\displaystyle{ u. }[/math] In this case
[math]\displaystyle{
F\left( x,t\right) =\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}x}
}[/math]
then
[math]\displaystyle{
K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right)
\mathrm{e}^{-k_{n}\left( x+y\right) }+\int_{x}^{\infty}K\left( x,z,t\right)
\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( y+z\right) }dz=0
}[/math]
From the equation we can see that
[math]\displaystyle{
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left(
x\right) \mathrm{e}^{-k_{m}y}
}[/math]
If we substitute this into the equation
[math]\displaystyle{
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right)
\mathrm{e}^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left(
y+z\right) }dz=0
}[/math]
which leads to
[math]\displaystyle{
-\sum_{n=1}^{N}c_{n}\left( t\right) v_{n}\left( x\right) \mathrm{e}^{-k_{n}y}
+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) \mathrm{e}^{-k_{n}\left( x+y\right) }
-\sum_{n=1}^{N}\sum_{m=1}^{N}\frac{c_{m}\left( t\right) c_{n}^{2}\left(
t\right) }{k_{n}+k_{m}}v_{m}\left( x\right) \mathrm{e}^{-k_{m}x}\mathrm{e}^{-k_{n}\left(
y+x\right) }=0
}[/math]
and we can eliminate the sum over [math]\displaystyle{ n }[/math] , the [math]\displaystyle{ c_{n}\left( t\right) , }[/math] and the
[math]\displaystyle{ \mathrm{e}^{-k_{n}y} }[/math] to obtain
[math]\displaystyle{
-v_{n}\left( x\right) +c_{n}\left( t\right) \mathrm{e}^{-k_{n}x}-\sum_{m=1}
^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right) }{k_{n}+k_{m}}
v_{m}\left( x\right) \mathrm{e}^{-\left( k_{m}+k_{n}\right) x}=0
}[/math]
which is an algebraic (finite dimensional system)\ for the unknows [math]\displaystyle{ v_{n}. }[/math] We
can write this as
[math]\displaystyle{
\left( \mathbf{I}+\mathbf{C}\right) \vec{v}=\vec{f}
}[/math]
where [math]\displaystyle{ f_{m}=c_{m}\left( t\right) \mathrm{e}^{-k_{m}x} }[/math] and
[math]\displaystyle{
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left( t\right) c_{m}\left( t\right)
}{k_{n}+k_{m}}\mathrm{e}^{-\left( k_{m}+k_{n}\right) x}
}[/math]
[math]\displaystyle{
K\left( x,y,t\right) =-\sum_{m=1}^{N}c_{m}\left( t\right) \left(
\mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}\mathrm{e}^{-k_{m}y}
}[/math]
This leads to
[math]\displaystyle{
u\left( x,t\right) =2\partial_{x}^{2}\log\left[ \det\left( \mathbf{I}
+\mathbf{C}\right) \right]
}[/math]
Lets consider some simple examples. First of all if [math]\displaystyle{ n=1 }[/math] (the single soliton
solution) we get
[math]\displaystyle{ \begin{matrix}
K\left( x,x,t\right) & =-\frac{c_{1}\left( t\right) c_{1}\left(
t\right) \mathrm{e}^{-k_{1}x}\mathrm{e}^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left(
t\right) }{k_{1}+k_{1}}\mathrm{e}^{-\left( k_{1}+k_{1}\right) x}}\\
& =\frac{-1}{1+\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}
\end{matrix} }[/math]
where [math]\displaystyle{ \mathrm{e}^{-\alpha}=2c_{0}^{2}\left( 0\right) . }[/math] Therefore
[math]\displaystyle{ \begin{matrix}
u\left( x,t\right) & =2\partial_{x}K\left( x,x,t\right) \\
& =\frac{4k_{1}\mathrm{e}^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+\mathrm{e}^{2k_{1}
x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\
& =\frac{-8k_{1}^{2}}{\left( \sqrt{2k_{1}}\mathrm{e}^{\theta}+\mathrm{e}^{-\theta}
/\sqrt{2k_{1}}\right) ^{2}}\\
& =2k^{2}\sec^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1}^{3}t\right\}
\end{matrix} }[/math]
where [math]\displaystyle{ \theta=k_{1}x-4k^{3}t-\alpha/2 }[/math] and [math]\displaystyle{ \sqrt{2k}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
} }[/math]. This is of course the single soliton solution.
