Connection betwen KdV and the Schrodinger Equation

Nonlinear PDE's Course
Current Topic	Connection betwen KdV and the Schrodinger Equation
Next Topic	Example Calculations for the KdV and IST
Previous Topic	Properties of the Linear Schrodinger Equation

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 we obtain

[math]\displaystyle{ \int^\infty_{-\infty}\partial_{t}\lambda w^{2}\,\mathrm{d}x+\left( w\partial_{x}Q-\partial _{x}wQ\right)\Bigg|^\infty_{-\infty}=0 }[/math]

and, since [math]\displaystyle{ w^2\gt 0 }[/math] except for some isolated points,

[math]\displaystyle{ \partial_{t}\lambda=0 }[/math]

provided that the eigenfunction [math]\displaystyle{ w }[/math] vanishes at [math]\displaystyle{ x=\pm\infty }[/math] (which is true for the bound state eigenfunctions, i.e. those from the discrete spectrum). 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.

Some more details on the content of this lecture can be found in Dunajski, Maciej (2009). Solitons, Instantons, and Twistors (Chapters 2.2 and 2.3).

Scattering Data

For the discrete spectrum the eigenfunctions behave like

[math]\displaystyle{ w_{n}\left( x,t\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,t\right) \right) ^{2}\mathrm{d}x=1 }[/math]

The continuous spectrum looks like

[math]\displaystyle{ w\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{ w\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}(t)=k_{n}(0) = 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}\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 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) }\mathrm{d}z=0 }[/math]

From the equation we can see that

[math]\displaystyle{ K\left( x,y,t\right) =-\sum_{n=1}^{N}v_{n}\left( x,t\right) e^{-k_{n}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) }\mathrm{d}z=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 the elements of [math]\displaystyle{ \mathbf{C} }[/math] are given by

[math]\displaystyle{ c_{mn} = \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} \left( \mathbf{I}+\mathbf{C}\right) ^{-1}\vec{f}e^{-k_{m}y} }[/math]

We then find [math]\displaystyle{ u(x,t) }[/math] from [math]\displaystyle{ K }[/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) e^{-2k_{1}x}}{1+\frac{c_{1}^2\left( t\right) } {2k_{1}}e^{-2 k_{1} x}}\\ & =\frac{-1}{e^{2k_{1}x - 8k_{1}^{3}t-\alpha} + 1/2k_1} \end{matrix} }[/math]

where [math]\displaystyle{ e^{-\alpha}=1/c_{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( e^{2k_{1} x-8k_{1}^{3}t-\alpha} + 1/2k_1\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.

Lecture Videos

Part 1

Load video

YouTube

YouTube might collect personal data. Privacy Policy

ContinueDismiss

Part 2

Load video

YouTube

YouTube might collect personal data. Privacy Policy

ContinueDismiss

Part 3

Load video

YouTube

YouTube might collect personal data. Privacy Policy

ContinueDismiss

Part 4

Load video

YouTube

YouTube might collect personal data. Privacy Policy

ContinueDismiss