Introduction to the Inverse Scattering Transform: Difference between revisions

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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.
== Lax Pair ==
A powerful way to study the KdV equation is via the idea of a '''Lax pair'''.
=== The Lax Pair ===
A Lax pair consists of two linear operators <math> L </math> and <math> P </math> such that the KdV equation is equivalent to the so-called '''Lax equation''':
:<math> \frac{dL}{dt} = [P, L] = PL - LP </math>
For the KdV equation, a classical Lax pair is:
:<math> L = -\partial_x^2 + u(x,t) </math>
:<math> P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u </math>
These are differential operators acting on a function <math> \psi(x,t) </math>. The idea is that the evolution of <math> u(x,t) </math> is such that the operator <math> L </math> undergoes an isospectral deformation: its spectrum does not change in time.
If <math> L\psi = \lambda \psi </math> and <math> \psi_t = P\psi </math>, then consistency requires:
:<math> \frac{d}{dt}(L\psi) = L_t \psi + L\psi_t = PL\psi </math>
so
:<math> L_t = [P, L] </math>
This is the Lax equation.
=== Derivation of the KdV Equation from the Lax Pair ===
Let us now compute <math> [P, L] = PL - LP </math> explicitly, using:
:<math> L = -\partial_x^2 + u </math>
:<math> P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u </math>
We compute the commutator acting on a test function <math> \psi </math>:
'''Step 1:''' Compute <math> PL\psi </math>.
<math>
\begin{align}
PL\psi &= P(-\psi_{xx} + u\psi) \\
&= -4\partial_x^3(-\psi_{xx} + u\psi) + 3u\partial_x(-\psi_{xx} + u\psi) + 3\partial_x u (-\psi_{xx} + u\psi)
\end{align}
</math>
We compute each term:
* <math> -4\partial_x^3(-\psi_{xx}) = -4\partial_x^3(-\psi_{xx}) = 4\psi_{xxxxx} </math>
* <math> -4\partial_x^3(u\psi) = -4[(u\psi)_{xxx}] </math>
* <math> 3u\partial_x(-\psi_{xx}) = -3u\psi_{xxx} </math>
* <math> 3u\partial_x(u\psi) = 3u(u\psi)_x = 3u^2\psi_x + 3uu_x\psi </math>
* <math> 3\partial_x u (-\psi_{xx}) = -3u_x\psi_{xx} </math>
* <math> 3\partial_x u (u\psi) = 3u_x u \psi + 3u u_x \psi = 6u u_x \psi </math>
Add all the terms together:
<math>
\begin{align}
PL\psi &= 4\psi_{xxxxx} -4(u\psi)_{xxx} -3u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi -3u_x\psi_{xx} + 6uu_x\psi
\end{align}
</math>
'''Step 2:''' Compute <math> LP\psi </math>.
<math>
\begin{align}
LP\psi &= (-\partial_x^2 + u)P\psi = -\partial_x^2(P\psi) + u P\psi
\end{align}
</math>
We focus on the leading terms in <math> P\psi </math>:
<math>
P\psi = -4\psi_{xxx} + 3u\psi_x + 3u_x\psi
</math>
Then:
<math>
\begin{align}
LP\psi &= -\partial_x^2(-4\psi_{xxx} + 3u\psi_x + 3u_x\psi) + u(-4\psi_{xxx} + 3u\psi_x + 3u_x\psi) \\
&= 4\psi_{xxxxx} -3\partial_x^2(u\psi_x) -3\partial_x^2(u_x\psi) -4u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi
\end{align}
</math>
Compute:
* <math> \partial_x^2(u\psi_x) = u_{xx}\psi_x + 2u_x\psi_{xx} + u\psi_{xxx} </math>
* <math> \partial_x^2(u_x\psi) = u_{xxx}\psi + 2u_{xx}\psi_x + u_x\psi_{xx} </math>
So:
<math>
\begin{align}
LP\psi &= 4\psi_{xxxxx} -3(u_{xx}\psi_x + 2u_x\psi_{xx} + u\psi_{xxx}) -3(u_{xxx}\psi + 2u_{xx}\psi_x + u_x\psi_{xx}) \\
&\quad -4u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi
\end{align}
</math>
Simplify:
<math>
\begin{align}
LP\psi &= 4\psi_{xxxxx} -3u_{xx}\psi_x -6u_x\psi_{xx} -3u\psi_{xxx} -3u_{xxx}\psi -6u_{xx}\psi_x -3u_x\psi_{xx} \\
&\quad -4u\psi_{xxx} + 3u^2\psi_x + 3uu_x\psi
\end{align}
</math>
Group like terms:
* <math> -3u\psi_{xxx} -4u\psi_{xxx} = -7u\psi_{xxx} </math>
* <math> -6u_x\psi_{xx} -3u_x\psi_{xx} = -9u_x\psi_{xx} </math>
* <math> -3u_{xx}\psi_x -6u_{xx}\psi_x = -9u_{xx}\psi_x </math>
* <math> -3u_{xxx}\psi </math>
So:
<math>
\begin{align}
LP\psi &= 4\psi_{xxxxx} -7u\psi_{xxx} -9u_x\psi_{xx} -9u_{xx}\psi_x -3u_{xxx}\psi + 3u^2\psi_x + 3uu_x\psi
\end{align}
</math>
'''Step 3:''' Compute <math> [P, L]\psi = PL\psi - LP\psi </math>
Subtract term by term (all <math> \psi </math> terms):
* The <math> 4\psi_{xxxxx} </math> terms cancel.
* Remaining terms:
<math>
\begin{align}
[P, L]\psi &= (-4(u\psi)_{xxx} + 3u\psi_{xxx}) + 3u_x\psi_{xx} + 9u_{xx}\psi_x + 3u_{xxx}\psi + 6uu_x\psi
\end{align}
</math>
Use:
<math> (u\psi)_{xxx} = u_{xxx}\psi + 3u_{xx}\psi_x + 3u_x\psi_{xx} + u\psi_{xxx} </math>
So:
<math>
\begin{align}
-4(u\psi)_{xxx} &= -4(u_{xxx}\psi + 3u_{xx}\psi_x + 3u_x\psi_{xx} + u\psi_{xxx}) \\
&= -4u_{xxx}\psi -12u_{xx}\psi_x -12u_x\psi_{xx} -4u\psi_{xxx}
\end{align}
</math>
Now add everything:
<math>
\begin{align}
[P,L]\psi &= (-4u_{xxx}\psi -12u_{xx}\psi_x -12u_x\psi_{xx} -4u\psi_{xxx}) + 3u\psi_{xxx} + 3u_x\psi_{xx} + 9u_{xx}\psi_x + 3u_{xxx}\psi + 6uu_x\psi \\
&= (-u_{xxx}\psi -3u_{xx}\psi_x -9u_x\psi_{xx} -u\psi_{xxx}) + 6uu_x\psi
\end{align}
</math>
So the final result is:
<math>
[P,L]\psi = ( -u_{xxx} + 6uu_x ) \psi
</math>
Thus, we have:
:<math> \frac{dL}{dt} = [P, L] \quad \Rightarrow \quad \frac{du}{dt} = -u_{xxx} + 6uu_x </math>
Which is exactly the KdV equation.
=== Summary ===
The KdV equation:
:<math> u_t + 6uu_x + u_{xxx} = 0 </math>
can be written as the Lax equation:
:<math> \frac{dL}{dt} = [P, L] </math>
where
:<math> L = -\partial_x^2 + u, \qquad P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u </math>
This reformulation reveals the integrable structure of the KdV equation and allows powerful solution methods such as the inverse scattering transform.


== Lecture Videos ==  
== Lecture Videos ==  

Latest revision as of 03:23, 10 September 2025

Nonlinear PDE's Course
Current Topic Introduction to the Inverse Scattering Transform
Next Topic Properties of the Linear Schrodinger Equation
Previous Topic Conservation Laws for the KdV



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}-\partial_x v\, }[/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 known 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.

Lecture Videos

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