[proofplan] The constancy of $\kappa_n$ follows immediately from the [Isospectral Flow Theorem](/theorems/???). For the normalisation constants, we apply the Isospectral Flow Theorem to obtain $\tilde{\psi}_n = \psi_{n,t} + A\psi_n$ as an eigenfunction at the same eigenvalue $-\kappa_n^2$. Since the eigenvalue is simple, $\tilde{\psi}_n = \tau\psi_n$ for some scalar $\tau(t)$. The antisymmetry $A^* = -A$ forces $(\psi_n, \tilde{\psi}_n)_{L^2} = 0$, so $\tau = 0$. Reading off the $x \to +\infty$ asymptotics of $0 = \psi_{n,t} + A\psi_n$ gives the ODE $\dot{c}_n = 4\kappa_n^3 c_n$, whose solution is $c_n(t) = c_n(0)e^{4\kappa_n^3 t}$. [/proofplan]

[step:Establish constancy of $\kappa_n$ from the Isospectral Flow Theorem] The KdV equation admits the Lax pair $(L, A)$ with $L = -\partial_x^2 + u$ and $A = 4\partial_x^3 - 3(u\partial_x + \partial_x u)$. Since $u(x,t)$ satisfies KdV, we have $L_t = [L, A]$. By the [Isospectral Flow Theorem](/theorems/???), all eigenvalues of $L(t)$ are time-independent. In particular, the discrete eigenvalues $-\kappa_n^2$ are constant, so \begin{align*} \dot{\kappa}_n = 0, \qquad n = 1, \ldots, N. \end{align*} [/step]

[step:Show that $\tilde{\psi}_n = \psi_{n,t} + A\psi_n$ is proportional to $\psi_n$]Let $\psi_n(x, t)$ be the $n$-th normalised bound state: $L\psi_n = -\kappa_n^2\psi_n$ with $\|\psi_n\| = 1$. By the second part of the [Isospectral Flow Theorem](/theorems/???), the function \begin{align*} \tilde{\psi}_n := \partial_t\psi_n + A\psi_n \end{align*} also satisfies $L\tilde{\psi}_n = -\kappa_n^2\tilde{\psi}_n$. Since the eigenvalue $-\kappa_n^2$ is simple (each bound state eigenspace of $L$ is one-dimensional for a Schrodinger operator with compactly supported potential), any eigenfunction at this eigenvalue is a scalar multiple of $\psi_n$. Therefore there exists $\tau = \tau(t) \in \mathbb{R}$ such that \begin{align*} \tilde{\psi}_n = \tau\, \psi_n. \end{align*}[/step]

[guided]Why is $\tilde{\psi}_n$ an eigenfunction at $-\kappa_n^2$ rather than at some other eigenvalue? The Isospectral Flow Theorem states that if $L\psi = \lambda\psi$, then $L(\psi_t + A\psi) = \lambda(\psi_t + A\psi)$. We apply this with $\lambda = -\kappa_n^2$. The simplicity of each eigenvalue $-\kappa_n^2$ is a standard result for one-dimensional Schrodinger operators. Geometrically, a second-order ODE on $\mathbb{R}$ with the boundary condition $\psi \to 0$ as $x \to +\infty$ constrains the solution to a one-dimensional subspace of the two-dimensional solution space at each energy, and the condition $\psi \to 0$ as $x \to -\infty$ imposes a second constraint. The intersection of these two constraints is generically zero-dimensional (no eigenvalue) or one-dimensional (simple eigenvalue).[/guided]

[step:Use the antisymmetry of $A$ to show $\tau = 0$] We compute the $L^2$ inner product $(\psi_n, \tilde{\psi}_n)_{L^2}$ in two ways. **First computation.** Since $\tilde{\psi}_n = \tau\psi_n$: \begin{align*} (\psi_n, \tilde{\psi}_n)_{L^2} = \tau\, (\psi_n, \psi_n)_{L^2} = \tau\, \|\psi_n\|^2 = \tau. \end{align*} **Second computation.** Expanding $\tilde{\psi}_n = \partial_t\psi_n + A\psi_n$: \begin{align*} (\psi_n, \tilde{\psi}_n)_{L^2} = (\psi_n, \partial_t\psi_n)_{L^2} + (\psi_n, A\psi_n)_{L^2}. \end{align*} For the first term, since $\|\psi_n\|^2 = 1$ for all $t$, differentiating in $t$ gives \begin{align*} 0 = \frac{d}{dt}\|\psi_n\|^2 = 2\,(\psi_n, \partial_t\psi_n)_{L^2}, \end{align*} so $(\psi_n, \partial_t\psi_n)_{L^2} = 0$. For the second term, the operator $A$ is antisymmetric: $A^* = -A$ in the $L^2$ inner product. This is verified from the explicit form $A = 4\partial_x^3 - 3(u\partial_x + \partial_x u)$, since $\partial_x$ is antisymmetric and $u\partial_x + \partial_x u$ is symmetric (being $2u\partial_x + u_x$, whose symmetric part is $u_x$ and whose antisymmetric part is $2u\partial_x$; actually $u\partial_x + \partial_x u = 2u\partial_x + u_x$, and the full expression $-3(u\partial_x + \partial_x u)$ ensures $A^* = -A$). Therefore \begin{align*} (\psi_n, A\psi_n)_{L^2} = -(A\psi_n, \psi_n)_{L^2} = -\overline{(\psi_n, A\psi_n)_{L^2}}. \end{align*} Since $\psi_n$ is real-valued, $(\psi_n, A\psi_n)_{L^2}$ is real, and the above forces $(\psi_n, A\psi_n)_{L^2} = 0$. Combining: $\tau = 0 + 0 = 0$, so $\tilde{\psi}_n = 0$. [/step]

[step:Extract the ODE for $c_n(t)$ from the $x \to +\infty$ asymptotics of $\tilde{\psi}_n = 0$]The equation $\tilde{\psi}_n = \partial_t\psi_n + A\psi_n = 0$ holds for all $x$. We evaluate it in the asymptotic region $x \to +\infty$, where $u$ vanishes and the bound state has the form \begin{align*} \psi_n(x, t) \sim c_n(t)\, e^{-\kappa_n x}. \end{align*} In this region $A = 4\partial_x^3$ (since $u = 0$), so \begin{align*} \partial_t\psi_n &\sim \dot{c}_n(t)\, e^{-\kappa_n x}, \\ A\psi_n &\sim 4\partial_x^3\bigl(c_n(t)\, e^{-\kappa_n x}\bigr) = 4c_n(t)(-\kappa_n)^3 e^{-\kappa_n x} = -4\kappa_n^3 c_n(t)\, e^{-\kappa_n x}. \end{align*} Substituting into $\tilde{\psi}_n = 0$: \begin{align*} 0 = \bigl(\dot{c}_n - 4\kappa_n^3 c_n\bigr)\, e^{-\kappa_n x}. \end{align*} Since $e^{-\kappa_n x} \neq 0$, we obtain the first-order linear ODE \begin{align*} \dot{c}_n = 4\kappa_n^3\, c_n. \end{align*}[/step]

[guided]The equation $\tilde{\psi}_n = 0$ is an identity in $x$, valid for all $x$. We evaluate it in the tail $x \to +\infty$ where the expressions simplify. The bound state decays as $c_n(t) e^{-\kappa_n x}$, so time differentiation produces $\dot{c}_n(t) e^{-\kappa_n x}$ (the spatial part $e^{-\kappa_n x}$ does not depend on $t$). For the $A\psi_n$ term, we use $A = 4\partial_x^3$ in the asymptotic region. Applying $\partial_x^3$ to $e^{-\kappa_n x}$ gives $(-\kappa_n)^3 e^{-\kappa_n x} = -\kappa_n^3 e^{-\kappa_n x}$, so $A\psi_n \sim -4\kappa_n^3 c_n e^{-\kappa_n x}$. The sign here is crucial: $(-\kappa_n)^3 = -\kappa_n^3$ because $\kappa_n > 0$. This gives $\dot{c}_n - 4\kappa_n^3 c_n = 0$, a growth equation rather than a decay equation. The positive sign of $4\kappa_n^3$ means that the normalisation constants grow exponentially in time.[/guided]

[step:Solve the ODE to obtain $c_n(t) = c_n(0)\,e^{4\kappa_n^3 t}$] The ODE $\dot{c}_n = 4\kappa_n^3 c_n$ is a first-order linear ODE with constant coefficient $4\kappa_n^3$. Separating variables: \begin{align*} \frac{dc_n}{c_n} = 4\kappa_n^3\, dt. \end{align*} Integrating from $0$ to $t$: \begin{align*} \ln c_n(t) - \ln c_n(0) = 4\kappa_n^3 t, \end{align*} giving \begin{align*} c_n(t) = c_n(0)\, e^{4\kappa_n^3 t}. \end{align*} Combined with the constancy $\dot{\kappa}_n = 0$ established in the first step, this completes the proof: the scattering data of the discrete spectrum evolves as $\kappa_n(t) = \kappa_n(0)$ and $c_n(t) = c_n(0)\, e^{4\kappa_n^3 t}$. [/step]