Introduction to the Inverse Scattering Transform
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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
with [math]\displaystyle{ \left\vert u\right\vert \rightarrow0 }[/math] as [math]\displaystyle{ x\rightarrow\pm\infty. }[/math]
The Miura transformation is given by
and if [math]\displaystyle{ v }[/math] satisfies the mKdV
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
then we obtain the equation
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
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
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
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
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
This can non trivial solutions when
which gives us the equation
or
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
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
Connection with the KdV
If we substitute the relationship
into the KdV after some manipulation we obtain
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
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
as [math]\displaystyle{ x\rightarrow\infty }[/math] with
The continuous spectrum looks like
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]
The scattering data evolves as
We can recover [math]\displaystyle{ u }[/math] from scattering data. We write
Then solve
This is a linear integral equation called the \emph{Gelfand-Levitan-Marchenko }equation. We then find [math]\displaystyle{ u }[/math] from
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
then
From the equation we can see that
If we substitute this into the equation
which leads to
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
which is an algebraic (finite dimensional system)\ for the unknows [math]\displaystyle{ v_{n}. }[/math] We can write this as
where [math]\displaystyle{ f_{m}=c_{m}\left( t\right) e^{-k_{m}x} }[/math] and
This leads to
Lets consider some simple examples. First of all if [math]\displaystyle{ n=1 }[/math] (the single soliton solution) we get
where [math]\displaystyle{ e^{-\alpha}=2c_{0}^{2}\left( 0\right) . }[/math] Therefore
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.