|
|
| Line 1: |
Line 1: |
| {{nonlinear waves course | | {{nonlinear waves course |
| | chapter title = Introduction to the Inverse Scattering Transform | | | chapter title = Introduction to the Inverse Scattering Transform |
| | next chapter = [[Reaction-Diffusion Systems]] | | | next chapter = [[Properties of the Linear Schrodinger Equation]] |
| | previous chapter = [[Conservation Laws for the KdV]] | | | previous chapter = [[Conservation Laws for the KdV]] |
| }} | | }} |
|
| |
|
|
| |
| == Introduction ==
| |
|
| |
|
| The inverse scattering transformation gives a way to solve the KdV equation | | The inverse scattering transformation gives a way to solve the KdV equation |
| Line 12: |
Line 10: |
| Fourier transformation, except it works for a non linear equation. We want to | | Fourier transformation, except it works for a non linear equation. We want to |
| be able to solve | | be able to solve |
| <center><math>\begin{align} | | <center><math>\begin{matrix} |
| \partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\ | | \partial_{t}u+6u\partial_{x}u+\partial_{x}^{3}u & =0\\ |
| u(x,0) & =f\left( x\right) | | u(x,0) & =f\left( x\right) |
| \end{align}</math></center> | | \end{matrix}</math></center> |
| with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> | | with <math>\left\vert u\right\vert \rightarrow0</math> as <math>x\rightarrow\pm\infty.</math> |
|
| |
|
| The Miura transformation is given by | | The Miura transformation is given by |
| <center><math> | | <center><math> |
| u=v^{2}+v_{x} \, | | u=v^{2}+v_{x} |
| </math></center> | | </math></center> |
| and if <math>v</math> satisfies the mKdV | | and if <math>v</math> satisfies the mKdV |
| Line 45: |
Line 43: |
| The eigenfunctions and eigenvalues of this scattering problem play a key role | | The eigenfunctions and eigenvalues of this scattering problem play a key role |
| in the inverse scattering transformation. Note that this is Schrodinger's equation. | | 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}
| |
|
| |
| 0, & x\notin\left[ -1,1\right] \\
| |
| b, & x\in\left[ -1,1\right]
| |
| \end{matrix}
| |
| \right.
| |
| </math></center>
| |
| where <math>b>0.</math>
| |
|
| |
| ==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}
| |
|
| |
| a_{1}\mathrm{e}^{kx}, & x<-1\\
| |
| b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
| |
| a_{2}\mathrm{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 equations, 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}
| |
|
| |
| \mathrm{e}^{-k} & -\cos\kappa\\
| |
| k\mathrm{e}^{-k} & -\kappa \sin\kappa
| |
| \end{matrix}
| |
| \right) \left(
| |
| \begin{matrix}
| |
|
| |
| a_{1}\\
| |
| b_{1}
| |
| \end{matrix}
| |
| \right) =\left(
| |
| \begin{matrix}
| |
|
| |
| 0\\
| |
| 0
| |
| \end{matrix}
| |
| \right)
| |
| </math></center>
| |
| This has non trivial solutions when
| |
| <center><math>
| |
| \det\left(
| |
| \begin{matrix}
| |
|
| |
| \mathrm{e}^{-k} & -\cos\kappa\\
| |
| k\mathrm{e}^{-k} & -\kappa \sin\kappa
| |
| \end{matrix}
| |
| \right) =0
| |
| </math></center>
| |
| which gives us the equation
| |
| <center><math>
| |
| - \kappa \sin\kappa \mathrm{e}^{-k}+\left( \cos\kappa\right) k\mathrm{e}^{-kx}=0
| |
| </math></center>
| |
| or
| |
| <center><math>
| |
| \kappa \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.
| |
|
| |
| In other words, we solve the final equation above for <math>k</math> to obtain our eigenvalues corresponding to even solutions.
| |
| Similarly we repeat the above process for the odd solutions.
| |
|
| |
| ==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}
| |
|
| |
| \mathrm{e}^{-\mathrm{i}kx}+r\mathrm{e}^{\mathrm{i}kx}, & x<-1\\
| |
| b_{1}\cos\kappa x+b_{2}\sin\kappa x & -1<x<1\\
| |
| a\mathrm{e}^{-\mathrm{i}kx} & 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}
| |
|
| |
| -\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></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) \mathrm{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 \mathrm{e}^{-\mathrm{i}kx}+r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}
| |
| ,\ \ \ x\rightarrow-\infty
| |
| </math></center>
| |
| <center><math>
| |
| v\left( x,t\right) \approx a\left( k,t\right) \mathrm{e}^{-\mathrm{i}kx},\ \ \ 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) \mathrm{e}^{4k_{n}^{3}t}
| |
| </math></center>
| |
| <center><math>
| |
| r\left( k,t\right) =r\left( k,0\right) \mathrm{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) \mathrm{e}^{-k_{n}
| |
| x}+ \frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}\mathrm{d}k
| |
| </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) \mathrm{d}z=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) \mathrm{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)
| |
| \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></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) \mathrm{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) \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></center>
| |
| which leads to
| |
| <center><math>
| |
| -\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></center>
| |
| and we can eliminate the sum over <math>n</math> , the <math>c_{n}\left( t\right) ,</math> and the
| |
| <math>\mathrm{e}^{-k_{n}y}</math> to obtain
| |
| <center><math>
| |
| -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></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) \mathrm{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}}\mathrm{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}\mathrm{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) \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></center>
| |
| where <math>\mathrm{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}\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}\mathrm{sech}^{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}\mathrm{e}^{-\alpha/2}=\mathrm{e}^{-kx_{0}
| |
| }</math>. This is of course the single soliton solution.
| |
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.
