Connection betwen KdV and the Schrodinger Equation: Difference between revisions
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\partial_{t}\lambda=0 | \partial_{t}\lambda=0 | ||
</math></center> | </math></center> | ||
provided that the eigenfunction <math>w</math> | provided that the eigenfunction <math>w</math> vanishes at <math>x=\pm\infty</math> (which is true for the bound | ||
state eigenfunctions). This shows that the discrete eigenvalues are unchanged | state eigenfunctions, i.e. those from the discrete spectrum). This shows that the discrete eigenvalues are unchanged | ||
and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other | and <math>u\left( x,t\right) </math> evolves according to the KdV. Many other | ||
properties can be found. | properties can be found. | ||
Some more details on the content of this lecture can be found in [https://www.rexresearch1.com/SpinorsLibrary/SolitonsInstantonsTwistorsDunajski.pdf Dunajski, Maciej (2009). Solitons, Instantons, and Twistors] (Chapters 2.2 and 2.3). | |||
==Scattering Data== | ==Scattering Data== | ||
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+\sum_{n=1}^{N}c_{n}^{2}\left( t\right) e^{-k_{n}\left( x+y\right) } | +\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( | -\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( | t\right) }{k_{n}+k_{m}}v_{m}\left( x, t\right) e^{-k_{m}x}e^{-k_{n}\left( | ||
y+x\right) }=0 | y+x\right) }=0 | ||
</math></center> | </math></center> | ||
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v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0 | v_{m}\left( x,t\right) e^{-\left( k_{m}+k_{n}\right) x}=0 | ||
</math></center> | </math></center> | ||
which is an algebraic (finite dimensional system) for the | which is an algebraic (finite dimensional system) for the unknowns <math>v_{n}.</math>. | ||
We can write this as | We can write this as | ||
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the elements of <math>\mathbf{C}</math> are given by | the elements of <math>\mathbf{C}</math> are given by | ||
<center><math> | <center><math> | ||
c_{ | c_{nm} = \frac{c_{n}^2\left( t\right)} | ||
{k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x} | {k_{n}+k_{m}}e^{-\left( k_{m}+k_{n}\right) x} | ||
</math></center> | </math></center> | ||
This gives us | This means | ||
<center><math> | |||
v_n(x, t)=\sum_{m=1}^{N}\left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_m | |||
</math></center> | |||
Since <math>\partial_x\left(\delta_{nm}+c_{nm}\right)=-e^{-k_mx}f_n</math>, this gives us | |||
<center><math>\begin{align} | |||
K\left( x,x,t\right) =-&\sum_{n=1}^{N} v_n(x, t)e^{-k_{n}x} \\ | |||
=-&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}f_me^{-k_{n}x} \\ | |||
=&\sum_{m=1}^{N}\sum_{n=1}^{N} \left(\left(\mathbf{I}+\mathbf{C}\right) ^{-1}\right)_{nm}\partial_x\left(\left(\mathbf{I}+\mathbf{C}\right)_{mn}\right) \\ | |||
=\,&\operatorname{Tr}\left(\left(\mathbf{I}+\mathbf{C}\right)^{-1}\partial_x\left(\mathbf{I}+\mathbf{C}\right)\right) \\ | |||
=\,&\dfrac{\partial_x\left(\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right)}{\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)} \\ | |||
=\,&\partial_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right) | |||
\end{align} | |||
</math></center> | |||
And therefore | |||
<center><math> | <center><math> | ||
u\left( x,t\right) =2\partial_{x}K\left(x,x,t\right)=2\partial^2_x\left(\log\operatorname{det}\left(\mathbf{I}+\mathbf{C}\right)\right) | |||
\mathbf{I}+\mathbf{C}\right) | |||
</math></center> | </math></center> | ||
=== Single Soliton Example === | === Single Soliton Example === | ||
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& =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt | & =\frac{8k_{1}^{2}}{\left( \sqrt{2k_{1}}e^{\theta}+e^{-\theta}/\sqrt | ||
{2k_{1}}\right) ^{2}}\\ | {2k_{1}}\right) ^{2}}\\ | ||
& =2k_{1}^{2}\ | & =2k_{1}^{2}\operatorname{sech}^{2}\left\{ k_{1}\left( x-x_{0}\right) -4k_{1} | ||
^{3}t\right\} | ^{3}t\right\} | ||
\end{align}</math></center> | \end{align}</math></center> | ||
where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0} | where <math>\theta=k_{1}x-4k^{3}t-\alpha/2</math> and <math>\sqrt{2k}e^{-\alpha/2}=e^{-kx_{0} | ||
}</math>. This is of course the single soliton solution. | }</math>. This is of course the single soliton solution. | ||
== Lecture Videos == | == Lecture Videos == | ||
Latest revision as of 06:50, 30 September 2025
| 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
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 we obtain
and, since [math]\displaystyle{ w^2\gt 0 }[/math] except for some isolated points,
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
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 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] and the [math]\displaystyle{ e^{-k_{n}y} }[/math] to obtain
which is an algebraic (finite dimensional system) for the unknowns [math]\displaystyle{ v_{n}. }[/math].
We can write this as
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
This means
Since [math]\displaystyle{ \partial_x\left(\delta_{nm}+c_{nm}\right)=-e^{-k_mx}f_n }[/math], this gives us
And therefore
Single Soliton Example
If [math]\displaystyle{ n=1 }[/math] (a single soliton solution) we get
where [math]\displaystyle{ e^{-\alpha}=1/c_{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.
