Introduction to the Inverse Scattering Transform

Nonlinear PDE's Course
Current Topic	Introduction to the Inverse Scattering Transform
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Miura transform

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.

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]\displaystyle{ L }[/math] and [math]\displaystyle{ P }[/math] such that the KdV equation is equivalent to the so-called Lax equation:

[math]\displaystyle{ \frac{dL}{dt} = [P, L] = PL - LP }[/math]

For the KdV equation, a classical Lax pair is:

[math]\displaystyle{ L = -\partial_x^2 + u(x,t) }[/math]

[math]\displaystyle{ P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u }[/math]

These are differential operators acting on a function [math]\displaystyle{ \psi(x,t) }[/math]. The idea is that the evolution of [math]\displaystyle{ u(x,t) }[/math] is such that the operator [math]\displaystyle{ L }[/math] undergoes an isospectral deformation: its spectrum does not change in time.

If [math]\displaystyle{ L\psi = \lambda \psi }[/math] and [math]\displaystyle{ \psi_t = P\psi }[/math], then consistency requires:

[math]\displaystyle{ \frac{d}{dt}(L\psi) = L_t \psi + L\psi_t = PL\psi }[/math]

so

[math]\displaystyle{ L_t = [P, L] }[/math]

This is the Lax equation.

Derivation of the KdV Equation from the Lax Pair

Let us now compute [math]\displaystyle{ [P, L] = PL - LP }[/math] explicitly, using:

[math]\displaystyle{ L = -\partial_x^2 + u }[/math]

[math]\displaystyle{ P = -4\partial_x^3 + 3u\partial_x + 3\partial_x u }[/math]

We compute the commutator acting on a test function [math]\displaystyle{ \psi }[/math]:

Step 1: Compute [math]\displaystyle{ PL\psi }[/math].

[math]\displaystyle{ \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]\displaystyle{ -4\partial_x^3(-\psi_{xx}) = -4\partial_x^3(-\psi_{xx}) = 4\psi_{xxxxx} }[/math]
• [math]\displaystyle{ -4\partial_x^3(u\psi) = -4[(u\psi)_{xxx}] }[/math]
• [math]\displaystyle{ 3u\partial_x(-\psi_{xx}) = -3u\psi_{xxx} }[/math]
• [math]\displaystyle{ 3u\partial_x(u\psi) = 3u(u\psi)_x = 3u^2\psi_x + 3uu_x\psi }[/math]
• [math]\displaystyle{ 3\partial_x u (-\psi_{xx}) = -3u_x\psi_{xx} }[/math]
• [math]\displaystyle{ 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]\displaystyle{ \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]\displaystyle{ LP\psi }[/math].

[math]\displaystyle{ \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]\displaystyle{ P\psi }[/math]:

[math]\displaystyle{ P\psi = -4\psi_{xxx} + 3u\psi_x + 3u_x\psi }[/math]

Then:

[math]\displaystyle{ \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]\displaystyle{ \partial_x^2(u\psi_x) = u_{xx}\psi_x + 2u_x\psi_{xx} + u\psi_{xxx} }[/math]
• [math]\displaystyle{ \partial_x^2(u_x\psi) = u_{xxx}\psi + 2u_{xx}\psi_x + u_x\psi_{xx} }[/math]

So:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ -3u\psi_{xxx} -4u\psi_{xxx} = -7u\psi_{xxx} }[/math]
• [math]\displaystyle{ -6u_x\psi_{xx} -3u_x\psi_{xx} = -9u_x\psi_{xx} }[/math]
• [math]\displaystyle{ -3u_{xx}\psi_x -6u_{xx}\psi_x = -9u_{xx}\psi_x }[/math]
• [math]\displaystyle{ -3u_{xxx}\psi }[/math]

So:

[math]\displaystyle{ \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]\displaystyle{ [P, L]\psi = PL\psi - LP\psi }[/math]

Subtract term by term (all [math]\displaystyle{ \psi }[/math] terms):

• The [math]\displaystyle{ 4\psi_{xxxxx} }[/math] terms cancel.
• Remaining terms:

[math]\displaystyle{ \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]\displaystyle{ (u\psi)_{xxx} = u_{xxx}\psi + 3u_{xx}\psi_x + 3u_x\psi_{xx} + u\psi_{xxx} }[/math]

So:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ [P,L]\psi = ( -u_{xxx} + 6uu_x ) \psi }[/math]

Thus, we have:

[math]\displaystyle{ \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]\displaystyle{ u_t + 6uu_x + u_{xxx} = 0 }[/math]

can be written as the Lax equation:

[math]\displaystyle{ \frac{dL}{dt} = [P, L] }[/math]

where

[math]\displaystyle{ 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.

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