Inverse scattering transform

Definition

Consider the Korteweg–De Vries equation $u_t - 6 u u_x + u_{xxx} = 0$ for $u(x, t)$ real, smooth and decaying fast as $\abs{x} \to \infty$. For each $t$, let $L_t = -\partial_x^2 + u(\cdot, t)$ be the one-dimensional Schrödinger operator with potential $u(\cdot, t)$. Its scattering data $S(t)$ consist of:

1. the finitely many eigenvalues $-\kappa_1^2, \dots, -\kappa_N^2$ of $L_t$, with $\kappa_n > 0$;
2. the norming constants $c_n(t) > 0$, where the normalised eigenfunction $\psi_n$ satisfies $\psi_n(x) \sim c_n e^{-\kappa_n x}$ as $x \to +\infty$;
3. the reflection coefficient $b(k, t)$ for $k \in \R$, where a solution of $L_t \psi = k^2 \psi$ satisfies $\psi(x) \sim e^{-ikx} + b(k, t) e^{ikx}$ as $x \to +\infty$ and $\psi(x) \sim a(k, t) e^{-ikx}$ as $x \to -\infty$.

The inverse scattering transform solves the initial value problem with data $u(x, 0)$ in three steps.

1. Direct scattering: compute $S(0)$ from $u(\cdot, 0)$.
2. Time evolution: $\kappa_n$ stays constant, $c_n(t) = c_n(0) e^{4 \kappa_n^3 t}$ and $b(k, t) = b(k, 0) e^{8 i k^3 t}$.
3. Inverse scattering: set $F(x) = \sum_{n} c_n(t)^2 e^{-\kappa_n x} + \frac{1}{2\pi} \int_{\R} b(k, t) e^{ikx} \, dk$, solve the Gelfand–Levitan–Marchenko equation $$K(x, y) + F(x + y) + \int_x^\infty K(x, z) F(z + y) \, dz = 0 \quad (y > x)$$ for $K$, and put $u(x, t) = -2 \frac{d}{dx} K(x, x)$.