Difference between revisions of "Introduction to the Inverse Scattering Transform"

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(Created page with '{{nonlinear waves course | chapter title = Introduction to the Inverse Scattering Transform | next chapter = Reaction-Diffusion Systems | previous chapter = [[Conservation…')
 
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  | previous chapter = [[Conservation Laws for the KdV]]
 
  | previous chapter = [[Conservation Laws for the KdV]]
 
}}
 
}}
 +
 +
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
 +
<center><math>\begin{matrix}
 +
\partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u  &  =0\\
 +
u(x,0)  &  =f\left(  x\right)
 +
\end{matrix}</math></center>
 +
with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math>
 +
 +
The Miura transformation is given by
 +
<center><math>
 +
u=v^{2}+v_{x}
 +
</math></center>
 +
and if <math>v</math> satisfies the mKdV
 +
<center><math>
 +
\partial_{t}v-6v^{2}\partial_{x}v+\partial_{x}^{3}v=0
 +
</math></center>
 +
then <math>u</math> satisfies the KdV (but not vice versa). We can think about the Miura
 +
transformation as being a nonlinear ODE solving for <math>v</math> given <math>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
 +
<center><math>
 +
v=\frac{\left(  \partial_{x}w\right)  }{w}
 +
</math></center>
 +
then we obtain the equation
 +
<center><math>
 +
\partial_{x}^{2}w+uw=0
 +
</math></center>
 +
The KdV is invariant under the transformation <math>x\rightarrow x+6\lambda t,</math>
 +
<math>u\rightarrow u+\lambda.</math> Therefore we consider the associated eigenvalue
 +
problem
 +
<center><math>
 +
\partial_{x}^{2}w+uw=-\lambda w
 +
</math></center>
 +
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
 +
<center><math>
 +
\partial_{x}^{2}w+uw=-\lambda w
 +
</math></center>
 +
has two kinds of solutions for <math>u\rightarrow0</math> as <math>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>u\rightarrow0</math>
 +
as <math>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
 +
<center><math>
 +
u\left(  x\right)  =\left\{
 +
\begin{matrix}
 +
[c]
 +
0 & x\notin\left[  -1,1\right] \\
 +
b & x\in\left[  -1,1\right]
 +
\end{matrix}
 +
\right.
 +
</math></center>
 +
where <math>b>0.</math>
 +
 +
\paragraph{Case when <math>\lambda>0</math>}
 +
 +
If we solve this equation for the case when <math>\lambda<0,</math> <math>\lambda=-k^{2}</math> we
 +
get
 +
<center><math>
 +
w\left(  x\right)  =\left\{
 +
\begin{matrix}
 +
[c]
 +
a_{1}e^{kx}, & x<-1\\
 +
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
 +
a_{2}e^{-kx} & x>1
 +
\end{matrix}
 +
\right.
 +
</math></center>
 +
where <math>\kappa=\sqrt{b-k^{2}}</math> where we have assumed that <math>b>k^{2}</math> (there is
 +
no solution for <math>b<k^{2}).</math> We then match <math>w</math> and its derivative at <math>x=\pm1</math>
 +
to solve for <math>a</math> and <math>b</math>. This leads to two system of equation, one for the
 +
even (<math>a_{1}=a_{2}</math> and <math>b_{2}=0</math> ) and one for the odd solutions
 +
(<math>a_{1}=-a_{2}</math> and <math>b_{1}=0)</math>. The solution for the even solutions is
 +
<center><math>
 +
\left(
 +
\begin{matrix}
 +
[c]
 +
e^{-kx} & -\cos\kappa\\
 +
ke^{-kx} & \sin\kappa
 +
\end{matrix}
 +
\right)  \left(
 +
\begin{matrix}
 +
[c]
 +
a_{1}\\
 +
b_{1}
 +
\end{matrix}
 +
\right)  =\left(
 +
\begin{matrix}
 +
[c]
 +
0\\
 +
0
 +
\end{matrix}
 +
\right)
 +
</math></center>
 +
This can non trivial solutions when
 +
<center><math>
 +
\det\left(
 +
\begin{matrix}
 +
[c]
 +
e^{-kx} & -\cos\kappa\\
 +
ke^{-kx} & \sin\kappa
 +
\end{matrix}
 +
\right)  =0
 +
</math></center>
 +
which gives us the equation
 +
<center><math>
 +
e^{-kx}\sin\kappa+\left(  \cos\kappa\right)  ke^{-kx}=0
 +
</math></center>
 +
or
 +
<center><math>
 +
\tan\kappa=-k=-\sqrt{b-\kappa^{2}}
 +
</math></center>
 +
We know that <math>0<\kappa<\sqrt{b}</math> and if we plot this we see that we obtain a
 +
finite number of solutions.
 +
 +
\paragraph{Case when <math>\lambda>0</math>}
 +
 +
When <math>\lambda>0</math> we write <math>\lambda=k^{2}</math> and we obtain solution
 +
<center><math>
 +
w\left(  x\right)  =\left\{
 +
\begin{matrix}
 +
[c]
 +
e^{-ikx}+re^{ikx}, & x<-1\\
 +
b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
 +
ae^{-ikx} & x>1
 +
\end{matrix}
 +
\right.
 +
</math></center>
 +
where <math>\kappa=\sqrt{b+k^{2}}.</math> Matching <math>w</math> and its derivaties at <math>x=\pm1</math> we
 +
obtain
 +
<center><math>
 +
\left(
 +
\begin{matrix}
 +
[c]
 +
-e^{-ik} & \cos\kappa & -\sin\kappa & 0\\
 +
ike^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\
 +
0 & \cos\kappa & \sin\kappa & -e^{-ik}\\
 +
0 & -\kappa\sin\kappa & \kappa\cos\kappa & ike^{-ik}
 +
\end{matrix}
 +
\right)  \left(
 +
\begin{matrix}
 +
[c]
 +
r\\
 +
b_{1}\\
 +
b_{2}\\
 +
a
 +
\end{matrix}
 +
\right)  =\left(
 +
\begin{matrix}
 +
[c]
 +
e^{ik}\\
 +
ike^{-ik}\\
 +
0\\
 +
0
 +
\end{matrix}
 +
\right)
 +
</math></center>
 +
 +
 +
==Connection with the KdV==
 +
 +
If we substitute the relationship
 +
<center><math>
 +
\partial_{x}^{2}w+uw=-\lambda w
 +
</math></center>
 +
into the KdV after some manipulation we obtain
 +
<center><math>
 +
\partial_{t}\lambda w^{2}+\partial_{x}\left(  w\partial_{x}Q-\partial
 +
_{x}wQ\right)  =0
 +
</math></center>
 +
where <math>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
 +
<center><math>
 +
\partial_{t}\lambda=0
 +
</math></center>
 +
provided that the eigenfunction <math>w</math> is bounded (which is true for the bound
 +
state eigenfunctions). This shows that the discrete eigenvalues are unchanged
 +
and <math>u\left(  x,t\right)  </math> evolves according to the KdV.
 +
 +
==Scattering Data==
 +
 +
For the discrete spectrum the eigenfunctions behave like
 +
<center><math>
 +
w_{n}\left(  x\right)  =c_{n}\left(  t\right)  e^{-k_{n}x}
 +
</math></center>
 +
as <math>x\rightarrow\infty</math> with
 +
<center><math>
 +
\int_{-\infty}^{\infty}\left(  w_{n}\left(  x\right)  \right)  ^{2}dx=1
 +
</math></center>
 +
The continuous spectrum looks like
 +
<center><math>
 +
v\left(  x,t\right)  \approx e^{-ikx}+r\left(  k,t\right)  e^{ikx}
 +
,\ \ \ x\rightarrow-\infty
 +
</math></center>
 +
<center><math>
 +
v\left(  x,t\right)  \approx a\left(  k,t\right)  e^{-ikx},\ \ \ x\rightarrow
 +
\infty
 +
</math></center>
 +
where <math>r</math> is the reflection coefficient and <math>a</math> is the transmission
 +
coefficient. This gives us the scattering data at <math>t=0</math>
 +
<center><math>
 +
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></center>
 +
The scattering data evolves as
 +
<center><math>
 +
k_{n}=k_{n}
 +
</math></center>
 +
<center><math>
 +
c_{n}\left(  t\right)  =c_{n}\left(  0\right)  e^{4k_{n}^{3}t}
 +
</math></center>
 +
<center><math>
 +
r\left(  k,t\right)  =r\left(  k,0\right)  e^{8ik^{3}t}
 +
</math></center>
 +
<center><math>
 +
a\left(  k,t\right)  =a\left(  k,0\right)
 +
</math></center>
 +
We can recover <math>u</math> from scattering data. We write
 +
<center><math>
 +
F\left(  x,t\right)  =\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}
 +
x}+\int_{-\infty}^{\infty}r\left(  k,t\right)  e^{ikx}dk
 +
</math></center>
 +
Then solve
 +
<center><math>
 +
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)  dz=0
 +
</math></center>
 +
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko
 +
}equation. We then find <math>u</math> from
 +
<center><math>
 +
u\left(  x,t\right)  =2\partial_{x}K\left(  x,x,t\right)
 +
</math></center>
 +
 +
 +
==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>r\left(  k,0\right)  =0</math> for all values of <math>k</math> for some <math>u.</math> In this case
 +
<center><math>
 +
F\left(  x,t\right)  =\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}x}
 +
</math></center>
 +
then
 +
<center><math>
 +
K\left(  x,y,t\right)  +\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)
 +
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)  e^{-k_{n}\left(  y+z\right)  }dz=0
 +
</math></center>
 +
From the equation we can see that
 +
<center><math>
 +
K\left(  x,y,t\right)  =-\sum_{m=1}^{N}c_{m}\left(  t\right)  v_{m}\left(
 +
x\right)  e^{-k_{m}y}
 +
</math></center>
 +
If we substitute this into the equation
 +
<center><math>
 +
-\sum_{n=1}^{N}c_{n}\left(  t\right)  v_{n}\left(  x\right)  e^{-k_{n}y}
 +
+\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}\left(  x+y\right)  }
 +
+\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left(  t\right)  v_{m}\left(  x\right)
 +
e^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  e^{-k_{n}\left(
 +
y+z\right)  }dz=0
 +
</math></center>
 +
which leads to
 +
<center><math>
 +
-\sum_{n=1}^{N}c_{n}\left(  t\right)  v_{n}\left(  x\right)  e^{-k_{n}y}
 +
+\sum_{n=1}^{N}c_{n}^{2}\left(  t\right)  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)  e^{-k_{m}x}e^{-k_{n}\left(
 +
y+x\right)  }=0
 +
</math></center>
 +
and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left(  t\right)  ,</math> and the
 +
<math>e^{-k_{n}y}</math> to obtain
 +
<center><math>
 +
-v_{n}\left(  x\right)  +c_{n}\left(  t\right)  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)  e^{-\left(  k_{m}+k_{n}\right)  x}=0
 +
</math></center>
 +
which is an algebraic (finite dimensional system)\ for the unknows <math>v_{n}.</math> We
 +
can write this as
 +
<center><math>
 +
\left(  \mathbf{I}+\mathbf{C}\right)  \vec{v}=\vec{f}
 +
</math></center>
 +
where <math>f_{m}=c_{m}\left(  t\right)  e^{-k_{m}x}</math> and
 +
<center><math>
 +
c_{mn}=\sum_{m=1}^{N}\frac{c_{n}\left(  t\right)  c_{m}\left(  t\right)
 +
}{k_{n}+k_{m}}e^{-\left(  k_{m}+k_{n}\right)  x}
 +
</math></center>
 +
<center><math>
 +
K\left(  x,y,t\right)  =-\sum_{m=1}^{N}c_{m}\left(  t\right)  \left(
 +
\mathbf{I}+\mathbf{C}\right)  ^{-1}\vec{f}e^{-k_{m}y}
 +
</math></center>
 +
This leads to
 +
<center><math>
 +
u\left(  x,t\right)  =2\partial_{x}^{2}\log\left[  \det\left(  \mathbf{I}
 +
+\mathbf{C}\right)  \right]
 +
</math></center>
 +
Lets consider some simple examples. First of all if <math>n=1</math> (the single soliton
 +
solution) we get
 +
<center><math>\begin{matrix}
 +
K\left(  x,x,t\right)  &  =-\frac{c_{1}\left(  t\right)  c_{1}\left(
 +
t\right)  e^{-k_{1}x}e^{-k_{1}x}}{1+\frac{c_{1}\left(  t\right)  c_{1}\left(
 +
t\right)  }{k_{1}+k_{1}}e^{-\left(  k_{1}+k_{1}\right)  x}}\\
 +
&  =\frac{-1}{1+e^{2k_{1}x-8k_{1}^{3}t-\alpha}}
 +
\end{matrix}</math></center>
 +
where <math>e^{-\alpha}=2c_{0}^{2}\left(  0\right)  .</math> Therefore
 +
<center><math>\begin{matrix}
 +
u\left(  x,t\right)  &  =2\partial_{x}K\left(  x,x,t\right)  \\
 +
&  =\frac{4k_{1}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left(  1+e^{2k_{1}
 +
x-8k_{1}^{3}t-\alpha}\right)  ^{2}}\\
 +
&  =\frac{-8k_{1}^{2}}{\left(  \sqrt{2k_{1}}e^{\theta}+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></center>
 +
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0}
 +
}</math>. This is of course the single soliton solution.

Revision as of 22:24, 15 September 2010

Nonlinear PDE's Course
Current Topic Introduction to the Inverse Scattering Transform
Next Topic Reaction-Diffusion Systems
Previous Topic Conservation Laws for the KdV


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{matrix} \partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\ u(x,0) & =f\left( x\right) \end{matrix} }[/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} [c] 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]

\paragraph{Case when [math]\displaystyle{ \lambda\gt 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} [c] a_{1}e^{kx}, & x\lt -1\\ b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1\lt x\lt 1\\ a_{2}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 equation, 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} [c] e^{-kx} & -\cos\kappa\\ ke^{-kx} & \sin\kappa \end{matrix} \right) \left( \begin{matrix} [c] a_{1}\\ b_{1} \end{matrix} \right) =\left( \begin{matrix} [c] 0\\ 0 \end{matrix} \right) }[/math]

This can non trivial solutions when

[math]\displaystyle{ \det\left( \begin{matrix} [c] e^{-kx} & -\cos\kappa\\ ke^{-kx} & \sin\kappa \end{matrix} \right) =0 }[/math]

which gives us the equation

[math]\displaystyle{ e^{-kx}\sin\kappa+\left( \cos\kappa\right) ke^{-kx}=0 }[/math]

or

[math]\displaystyle{ \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.

\paragraph{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} [c] e^{-ikx}+re^{ikx}, & x\lt -1\\ b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1\lt x\lt 1\\ ae^{-ikx} & 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} [c] -e^{-ik} & \cos\kappa & -\sin\kappa & 0\\ ike^{-ik} & \kappa\sin\kappa & \kappa\cos\kappa & 0\\ 0 & \cos\kappa & \sin\kappa & -e^{-ik}\\ 0 & -\kappa\sin\kappa & \kappa\cos\kappa & ike^{-ik} \end{matrix} \right) \left( \begin{matrix} [c] r\\ b_{1}\\ b_{2}\\ a \end{matrix} \right) =\left( \begin{matrix} [c] e^{ik}\\ ike^{-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) 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 e^{-ikx}+r\left( k,t\right) e^{ikx} ,\ \ \ x\rightarrow-\infty }[/math]
[math]\displaystyle{ v\left( x,t\right) \approx a\left( k,t\right) e^{-ikx},\ \ \ 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) e^{4k_{n}^{3}t} }[/math]
[math]\displaystyle{ r\left( k,t\right) =r\left( k,0\right) 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) e^{-k_{n} x}+\int_{-\infty}^{\infty}r\left( k,t\right) e^{ikx}dk }[/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) dz=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) e^{-k_{n}x} }[/math]

then

[math]\displaystyle{ K\left( x,y,t\right) +\sum_{n=1}^{N}c_{n}^{2}\left( t\right) 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) 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) 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) e^{-k_{n}y} +\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) } +\int_{x}^{\infty}-\sum_{m=1}^{N}c_{m}\left( t\right) v_{m}\left( x\right) e^{-k_{m}y}\sum_{n=1}^{N}c_{n}^{2}\left( t\right) 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) e^{-k_{n}y} +\sum_{n=1}^{N}c_{n}^{2}\left( t\right) 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) e^{-k_{m}x}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{ e^{-k_{n}y} }[/math] to obtain

[math]\displaystyle{ -v_{n}\left( x\right) +c_{n}\left( t\right) 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) 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) 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}}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}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) e^{-k_{1}x}e^{-k_{1}x}}{1+\frac{c_{1}\left( t\right) c_{1}\left( t\right) }{k_{1}+k_{1}}e^{-\left( k_{1}+k_{1}\right) x}}\\ & =\frac{-1}{1+e^{2k_{1}x-8k_{1}^{3}t-\alpha}} \end{matrix} }[/math]

where [math]\displaystyle{ 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}e^{2k_{1}x-8k_{1}^{3}t-\alpha}}{\left( 1+e^{2k_{1} x-8k_{1}^{3}t-\alpha}\right) ^{2}}\\ & =\frac{-8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+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}e^{-\alpha/2}=e^{-kx_{0} } }[/math]. This is of course the single soliton solution.