Connection betwen KdV and the Schrodinger Equation

Nonlinear PDE's Course
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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. Many other properties can be found

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 \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) 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}+\frac{1}{2\pi}\int_{-\infty}^{\infty}r\left( k,t\right) \mathrm{e}^{\mathrm{i}kx}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}v_{m}\left( x,t\right) e^{-k_{m}y} }[/math]

If we substitute this into the equation,

[math]\displaystyle{ -\sum_{n=1}^{N}v_{n}\left( x,t\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}v_{m}\left( x,t\right) e^{-k_{m}z}\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} v_{n}\left( x,t\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_{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] and the [math]\displaystyle{ e^{-k_{n}y} }[/math] to obtain

[math]\displaystyle{ -v_{n}\left( x,t\right) +c_{n}^2\left( t\right) e^{-k_{n}x}-\sum_{m=1} ^{N}\frac{c_{n}^2\left( t\right) }{k_{n}+k_{m}} v_{m}\left( x,t\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}^2\left( t\right) e^{-k_{m}x} }[/math] and

[math]\displaystyle{ c_{mn}=\sum_{m=1}^{N}\frac{c_{n}^2\left( t\right)} {k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x} }[/math]

This gives us

[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]

Single Soliton Example

If [math]\displaystyle{ n=1 }[/math] (a single soliton solution) we get

[math]\displaystyle{ \begin{matrix} K\left( x,x,t\right) & =-\frac{c_{1}^2\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{align} 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_{1}^{2}\mbox{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1} ^{3}t\right\} \end{align} }[/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.