Connection betwen KdV and the Schrodinger Equation

Nonlinear PDE's Course
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If we substitute the relationship

${\partial }_x^{2}w+uw=-\lambda w$

into the KdV after some manipulation we obtain

${\partial }_t\lambda w^2+{\partial }_x(w{\partial }_{x}Q-{\partial }_{x}wQ)=0$

where $Q={\partial }_{t}w+{\partial }_x^{3}w-3(\lambda -u){\partial }_{x}w.$ If we integrate this equation then we obtain the result that

${\partial }_t\lambda =0$

provided that the eigenfunction $w$ is bounded (which is true for the bound state eigenfunctions). This shows that the discrete eigenvalues are unchanged and $u(x,t)$ evolves according to the KdV. Many other properties can be found

Scattering Data

For the discrete spectrum the eigenfunctions behave like

$w_n(x,t)=c_n\left(t\right)e^{-k_{n}x}$

as $x\to \mathrm{\infty }$ with

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\left(w_n(x,t)\right)}^2\mathrm{d}x=1$

The continuous spectrum looks like

$w(x,t)\approx {\mathrm{e}}^{-\mathrm{i}kx}+r(k,t){\mathrm{e}}^{\mathrm{i}kx},x\to -\mathrm{\infty }$ $w(x,t)\approx a(k,t){\mathrm{e}}^{-\mathrm{i}kx},x\to \mathrm{\infty }$

where $r$ is the reflection coefficient and $a$ is the transmission coefficient. This gives us the scattering data at $t=0$

$S(\lambda ,0)=({\{k_n,c_n\left(0\right)\}}_{n=1}^N,r(k,0),a(k,0))$

The scattering data evolves as

$k_n(t)=k_n(0)=k_n$ $c_n\left(t\right)=c_n\left(0\right)e^{4k_n^{3}t}$ $r(k,t)=r(k,0)e^{8i{k}^{3}t}$ $a(k,t)=a(k,0)$

We can recover $u$ from scattering data. We write

$F(x,t)=\sum _{n=1}^N{c}_n^2\left(t\right)e^{-k_{n}x}+\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}r(k,t){\mathrm{e}}^{\mathrm{i}kx}\mathrm{d}k$

Then solve

$K(x,y;t)+F(x+y;t)+{\int }_x^{\mathrm{\infty }}K(x,z;t)F(z+y;t)\mathrm{d}z=0$

This is a linear integral equation called the Gelfand-Levitan-Marchenko equation. We then find $u$ from

$u(x,t)=2{\partial }_{x}K(x,x,t)$

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 $r(k,0)=0$ for all values of $k$ for some $u.$ In this case

$F(x,t)=\sum _{n=1}^N{c}_n^2\left(t\right)e^{-k_{n}x}$

then

$K(x,y,t)+\sum _{n=1}^N{c}_n^2\left(t\right)e^{-k_n(x+y)}+{\int }_x^{\mathrm{\infty }}K(x,z,t)\sum _{n=1}^N{c}_n^2\left(t\right)e^{-k_n(y+z)}\mathrm{d}z=0$

From the equation we can see that

$K(x,y,t)=-\sum _{n=1}^N{v}_n(x,t)e^{-k_{n}y}$

If we substitute this into the equation,

$-\sum _{n=1}^N{v}_n(x,t)e^{-k_{n}y}+\sum _{n=1}^N{c}_n^2\left(t\right)e^{-k_n(x+y)}+{\int }_x^{\mathrm{\infty }}-\sum _{m=1}^N{v}_m(x,t)e^{-k_{m}z}\sum _{n=1}^N{c}_n^2\left(t\right)e^{-k_n(y+z)}\mathrm{d}z=0$

which leads to

$-\sum _{n=1}^N{v}_n(x,t)e^{-k_{n}y}+\sum _{n=1}^N{c}_n^2\left(t\right)e^{-k_n(x+y)}-\sum _{n=1}^N\sum _{m=1}^N\frac{c_n^2\left(t\right)}{k_n+k_m}{v}_m\left(x\right)e^{-k_{m}x}{e}^{-k_n(y+x)}=0$

and we can eliminate the sum over $n$ and the $e^{-k_{n}y}$ to obtain

$-v_n(x,t)+c_n^2\left(t\right)e^{-k_{n}x}-\sum _{m=1}^N\frac{c_n^2\left(t\right)}{k_n+k_m}{v}_m(x,t)e^{-(k_m+k_n)x}=0$

which is an algebraic (finite dimensional system) for the unknows $v_n.$.

We can write this as

$(\mathbf{I}+\mathbf{C})\overrightarrow{v}=\overrightarrow{f}$

where $f_m=c_m^2\left(t\right)e^{-k_{m}x}$ and the elements of $\mathbf{C}$ are given by

$c_{mn}=\frac{c_n^2\left(t\right)}{k_n+k_m}{e}^{-(k_m+k_n)x}$

This gives us

$K(x,y,t)=-\sum _{m=1}^N{(\mathbf{I}+\mathbf{C})}^{-1}\overrightarrow{f}{e}^{-k_{m}y}$

We then find $u(x,t)$ from $K$.

Single Soliton Example

If $n=1$ (a single soliton solution) we get

$\begin{array}{cc}K(x,x,t)& =-\frac{c_1^2\left(t\right)e^{-2k_{1}x}}{1+\frac{c_1^2\left(t\right)}{2k_1}{e}^{-2k_{1}x}}\\ & =\frac{-1}{e^{2k_{1}x-8k_1^{3}t-\alpha }+1/2k_1}\end{array}$

where $e^{-\alpha }=1/c_0^2\left(0\right).$ Therefore

$\begin{array}{rl}u(x,t)& =2{\partial }_{x}K(x,x,t)\\ & =\frac{4k_1{e}^{2k_{1}x-8k_1^{3}t-\alpha }}{{(e^{2k_{1}x-8k_1^{3}t-\alpha }+1/2k_1)}^2}\\ & =\frac{8k_1^2}{{(\sqrt{2k_1}{e}^{\theta }+e^{-\theta }/\sqrt{2k_1})}^2}\\ & =2k_1^2{\text{sech}}^2\{k_1(x-x_0)-4k_1^{3}t\}\end{array}$

where $\theta =k_{1}x-4k^{3}t-\alpha /2$ and $\sqrt{2k}{e}^{-\alpha /2}=e^{-k{x}_0}$. This is of course the single soliton solution.