[proofplan] We use the Lax pair $(L, A)$ for KdV with $L = -\partial_x^2 + u$ and $A = 4\partial_x^3 - 3(u\partial_x + \partial_x u)$. The Isospectral Flow Theorem guarantees that $\tilde{\varphi} = \varphi_t + A\varphi$ satisfies $L\tilde{\varphi} = k^2\tilde{\varphi}$ whenever $L\varphi = k^2\varphi$. Since $u$ has compact support, $A$ reduces to $4\partial_x^3$ for large $|x|$, allowing explicit computation of the asymptotics of $\tilde{\varphi}$. A Wronskian argument then shows $\tilde{\varphi} = 4ik^3\varphi$, and matching the $x \to +\infty$ asymptotics yields the ODEs $\dot{a} = 0$ and $\dot{b} = 8ik^3 b$, from which the stated evolution formulas follow. [/proofplan] [step:Apply the Isospectral Flow Theorem to the scattering solution] For each fixed $t$, the scattering solution $\varphi(x, t)$ satisfies $L\varphi = k^2\varphi$ with boundary conditions \begin{align*} \varphi(x, t) &\sim e^{-ikx} \quad \text{as } x \to -\infty, \\ \varphi(x, t) &\sim a(k, t)\, e^{-ikx} + b(k, t)\, e^{ikx} \quad \text{as } x \to +\infty. \end{align*} Since $u(x, t)$ satisfies KdV, the pair $(L, A)$ with $L = -\partial_x^2 + u$ and $A = 4\partial_x^3 - 3(u\partial_x + \partial_x u)$ is a Lax pair: $L_t = [L, A]$. By the [Isospectral Flow Theorem](/theorems/???), the function \begin{align*} \tilde{\varphi} := \varphi_t + A\varphi \end{align*} satisfies $(L - k^2)\tilde{\varphi} = 0$, i.e., $\tilde{\varphi}$ is also a solution of the same Schrodinger equation $L\psi = k^2\psi$. [/step] [step:Compute the asymptotics of $\tilde{\varphi}$ for large $|x|$ using $A = 4\partial_x^3$] Since $u$ has compact support, for $|x|$ sufficiently large the potential vanishes and the operator $A = 4\partial_x^3 - 3(u\partial_x + \partial_x u)$ simplifies to $A = 4\partial_x^3$. **Asymptotics as $x \to -\infty$.** Here $\varphi \sim e^{-ikx}$, so $\varphi_t \sim 0$ (the boundary condition $e^{-ikx}$ is independent of $t$) and \begin{align*} A\varphi \sim 4\partial_x^3(e^{-ikx}) = 4(-ik)^3 e^{-ikx} = 4ik^3 e^{-ikx}. \end{align*} Therefore \begin{align*} \tilde{\varphi} \sim 4ik^3 e^{-ikx} \quad \text{as } x \to -\infty. \end{align*} **Asymptotics as $x \to +\infty$.** Here $\varphi \sim a\, e^{-ikx} + b\, e^{ikx}$, so \begin{align*} \varphi_t &\sim \dot{a}\, e^{-ikx} + \dot{b}\, e^{ikx}, \\ A\varphi &\sim 4\partial_x^3\bigl(a\, e^{-ikx} + b\, e^{ikx}\bigr) = 4a(-ik)^3 e^{-ikx} + 4b(ik)^3 e^{ikx} \\ &= 4ik^3 a\, e^{-ikx} - 4ik^3 b\, e^{ikx}. \end{align*} Adding these: \begin{align*} \tilde{\varphi} \sim (\dot{a} + 4ik^3 a)\, e^{-ikx} + (\dot{b} - 4ik^3 b)\, e^{ikx} \quad \text{as } x \to +\infty. \end{align*} [guided] The computation of asymptotics is the heart of the proof. The operator $A = 4\partial_x^3 - 3(u\partial_x + \partial_x u)$ has two types of terms: the pure derivative $4\partial_x^3$ and the terms involving $u$. Since $u$ vanishes for large $|x|$, only the $4\partial_x^3$ term survives asymptotically. This is the critical simplification that makes the entire method work: the nonlinear part of $A$ drops out in the asymptotic region, leaving a constant-coefficient operator whose action on exponentials is immediate. As $x \to -\infty$, the scattering solution has the fixed form $\varphi \sim e^{-ikx}$. This normalization is part of the definition of $\varphi$ and does not depend on $t$, so $\varphi_t \sim 0$ in this region. The operator $4\partial_x^3$ applied to $e^{-ikx}$ yields $4(-ik)^3 e^{-ikx} = 4ik^3 e^{-ikx}$ (using $(-i)^3 = i$). As $x \to +\infty$, the scattering solution has two components: $a(k,t) e^{-ikx}$ and $b(k,t) e^{ikx}$, where $a$ and $b$ depend on $t$. Differentiating in $t$ gives $\dot{a} e^{-ikx} + \dot{b} e^{ikx}$. Applying $4\partial_x^3$ gives $4(-ik)^3 a\, e^{-ikx} + 4(ik)^3 b\, e^{ikx} = 4ik^3 a\, e^{-ikx} - 4ik^3 b\, e^{ikx}$. Note the sign difference: for $e^{-ikx}$ we get $+4ik^3$, for $e^{ikx}$ we get $-4ik^3$, because $(-ik)^3 = ik^3$ while $(ik)^3 = -ik^3$. [/guided] [/step] [step:Use a Wronskian argument to show $\tilde{\varphi} = 4ik^3\varphi$] Both $\varphi$ and $\tilde{\varphi}$ satisfy the same second-order ODE $L\psi = k^2\psi$, which for fixed $t$ has a two-dimensional solution space. We show they are linearly dependent by examining their Wronskian. Define $W(x) := \varphi(x)\tilde{\varphi}'(x) - \varphi'(x)\tilde{\varphi}(x)$, where primes denote $\partial_x$. Since both $\varphi$ and $\tilde{\varphi}$ satisfy $-\psi'' + u\psi = k^2\psi$, Abel's theorem gives $W'(x) = 0$, so $W$ is constant in $x$. Evaluate $W$ as $x \to -\infty$. In this limit, $\varphi \sim e^{-ikx}$ and $\tilde{\varphi} \sim 4ik^3 e^{-ikx}$. Therefore \begin{align*} W &= e^{-ikx} \cdot 4ik^3(-ik)e^{-ikx} - (-ik)e^{-ikx} \cdot 4ik^3 e^{-ikx} \\ &= 4ik^3(-ik)e^{-2ikx} - (-ik)(4ik^3)e^{-2ikx} \\ &= 0. \end{align*} Since $W$ is constant and vanishes as $x \to -\infty$, we have $W(x) = 0$ for all $x$. A vanishing Wronskian implies that $\varphi$ and $\tilde{\varphi}$ are linearly dependent: there exists a scalar $\mu$ such that $\tilde{\varphi} = \mu\varphi$. Matching the leading behaviour as $x \to -\infty$: $\tilde{\varphi} \sim 4ik^3 e^{-ikx} = \mu\, e^{-ikx}$, giving $\mu = 4ik^3$. Therefore \begin{align*} \tilde{\varphi} = 4ik^3\, \varphi. \end{align*} [guided] This is the central step of the proof. We have two solutions of the same second-order ODE, and we need to show they are proportional. The Wronskian is the standard tool: for any two solutions $f, g$ of $-\psi'' + u\psi = k^2\psi$, the Wronskian $W[f, g] = fg' - f'g$ satisfies $W' = 0$ (since the ODE has no first-derivative term). So $W$ is constant. We evaluate this constant by computing $W$ in the asymptotic region $x \to -\infty$, where both $\varphi$ and $\tilde{\varphi}$ are proportional to $e^{-ikx}$. Since both are proportional to the same exponential (with ratio $4ik^3$), the Wronskian of two proportional functions vanishes: if $g = cf$ then $W[f, g] = f(cf') - f'(cf) = 0$. More explicitly: $\varphi \sim e^{-ikx}$ and $\tilde{\varphi} \sim 4ik^3 e^{-ikx}$. Then $\varphi' \sim -ike^{-ikx}$ and $\tilde{\varphi}' \sim 4ik^3 \cdot (-ik) e^{-ikx} = 4k^4 e^{-ikx}$. Computing: \begin{align*} W = e^{-ikx} \cdot 4k^4 e^{-ikx} - (-ik)e^{-ikx} \cdot 4ik^3 e^{-ikx} = 4k^4 e^{-2ikx} + 4k^4 e^{-2ikx} \cdot i^2 \cdot (-1) \ldots \end{align*} Actually, let us be more careful. We have $\tilde{\varphi} = 4ik^3 \varphi + (\text{terms that vanish as } x \to -\infty)$. But in the asymptotic region both are exactly proportional to $e^{-ikx}$, so $W[{\varphi}, \tilde{\varphi}] = W[\varphi, 4ik^3\varphi] = 4ik^3 W[\varphi, \varphi] = 0$. Since $W$ is constant in $x$ and equals zero as $x \to -\infty$, it is zero everywhere. A vanishing Wronskian for two solutions of a second-order linear ODE implies linear dependence (at least on any interval where one of the solutions is nonzero; for the scattering solution $\varphi$, this holds generically for $k \in \mathbb{R}$). So $\tilde{\varphi} = \mu\varphi$ for some constant $\mu$, and matching at $x \to -\infty$ gives $\mu = 4ik^3$. [/guided] [/step] [step:Read off the evolution equations for $a(k,t)$ and $b(k,t)$] From $\tilde{\varphi} = 4ik^3\varphi$, we compare the $x \to +\infty$ asymptotics. On the left: \begin{align*} \tilde{\varphi} \sim (\dot{a} + 4ik^3 a)\, e^{-ikx} + (\dot{b} - 4ik^3 b)\, e^{ikx}. \end{align*} On the right: \begin{align*} 4ik^3\varphi \sim 4ik^3 a\, e^{-ikx} + 4ik^3 b\, e^{ikx}. \end{align*} Equating the coefficients of $e^{-ikx}$ and $e^{ikx}$ separately (since these are linearly independent functions): \begin{align*} \dot{a} + 4ik^3 a &= 4ik^3 a, \\ \dot{b} - 4ik^3 b &= 4ik^3 b. \end{align*} The first equation gives $\dot{a} = 0$, so \begin{align*} a(k, t) = a(k, 0). \end{align*} The second gives $\dot{b} = 8ik^3 b$, which is a first-order linear ODE with solution \begin{align*} b(k, t) = b(k, 0)\, e^{8ik^3 t}. \end{align*} [/step] [step:Derive the evolution of $R(k,t)$ and $T(k,t)$] Recalling $R(k, t) = b(k, t)/a(k, t)$ and $T(k, t) = 1/a(k, t)$: \begin{align*} R(k, t) = \frac{b(k, 0)\, e^{8ik^3 t}}{a(k, 0)} = R(k, 0)\, e^{8ik^3 t}, \end{align*} and \begin{align*} T(k, t) = \frac{1}{a(k, 0)} = T(k, 0). \end{align*} The transmission coefficient is constant in time, while the reflection coefficient acquires a purely oscillatory phase factor $e^{8ik^3 t}$. In particular, $|R(k,t)| = |R(k,0)|$ for all $t$, confirming that the amplitude of the reflection coefficient is conserved. [/step]