[problem:Reflection Of A Travelling Wave | 1] Let $u$ solve the one-dimensional wave equation on the half-line $x > 0$ with Dirichlet boundary condition $u(0, t) = 0$ for all $t \geq 0$, and initial data $g \in C^2([0, \infty); \mathbb{R})$ with $g(0) = 0$ and $h \in C^1([0, \infty); \mathbb{R})$ with $h(0) = 0$. Derive the solution formula by extending $g$ and $h$ to all of $\mathbb{R}$ as odd functions and applying d'Alembert's formula. [/problem]

[solution] **Step 1: Construct the odd extensions.** Define \begin{align*} g_{\text{odd}}(x) := \begin{cases} g(x) & x \geq 0, \\ -g(-x) & x < 0, \end{cases} \qquad h_{\text{odd}}(x) := \begin{cases} h(x) & x \geq 0, \\ -h(-x) & x < 0. \end{cases} \end{align*} The conditions $g(0) = 0$, $h(0) = 0$ ensure $g_{\text{odd}} \in C^2(\mathbb{R})$ and $h_{\text{odd}} \in C^1(\mathbb{R})$ (the odd extension is smooth at the origin when the function vanishes there with compatible derivatives). **Step 2: Apply d'Alembert's formula on all of $\mathbb{R}$.** By [D'Alembert's Formula](/theorems/665), the unique solution on $\mathbb{R} \times [0, \infty)$ with data $(g_{\text{odd}}, h_{\text{odd}})$ is \begin{align*} \tilde{u}(x, t) = \frac{1}{2}\bigl(g_{\text{odd}}(x + t) + g_{\text{odd}}(x - t)\bigr) + \frac{1}{2}\int_{x - t}^{x + t} h_{\text{odd}}(y) \, d\mathcal{L}^1(y). \end{align*} **Step 3: Verify the boundary condition.** At $x = 0$: \begin{align*} \tilde{u}(0, t) = \frac{1}{2}\bigl(g_{\text{odd}}(t) + g_{\text{odd}}(-t)\bigr) + \frac{1}{2}\int_{-t}^{t} h_{\text{odd}}(y) \, d\mathcal{L}^1(y) = 0, \end{align*} since $g_{\text{odd}}$ is odd (so $g_{\text{odd}}(t) + g_{\text{odd}}(-t) = 0$) and $h_{\text{odd}}$ is odd (so its integral over the symmetric interval $[-t, t]$ vanishes). **Step 4: Write the formula explicitly for $x > 0$.** For $x > 0$ and $t < x$, both $x + t > 0$ and $x - t > 0$, so $g_{\text{odd}} = g$ and the formula reduces to the standard d'Alembert expression. For $t > x$ (when the backward characteristic $x - t < 0$ crosses the boundary), the odd extension introduces a sign change: $g_{\text{odd}}(x - t) = -g(t - x)$. The solution for $x > 0$, $t > x$ becomes \begin{align*} u(x, t) = \frac{1}{2}\bigl(g(x + t) - g(t - x)\bigr) + \frac{1}{2}\int_{t - x}^{x + t} h(y) \, d\mathcal{L}^1(y), \end{align*} where the integral limits follow from splitting $\int_{x-t}^{x+t} h_{\text{odd}} \, d\mathcal{L}^1$ at $y = 0$ and using the substitution $y \mapsto -y$ on the negative part. The term $-g(t - x)$ represents the **reflected wave**: the left-travelling component $G(x - t)$ hits the boundary at $x = 0$, reverses sign (to satisfy the Dirichlet condition), and returns as a right-travelling wave with inverted amplitude. The physical interpretation is that a wave pulse approaching the fixed endpoint reflects with inverted phase. [/solution]

[problem:Energy Equipartition For The Wave Equation | 2] Let $n = 1$ and let $u$ solve the wave equation with compactly supported initial data $g \in C_c^2(\mathbb{R}; \mathbb{R})$ and $h \in C_c^1(\mathbb{R}; \mathbb{R})$. Define the kinetic energy $K(t) := \frac{1}{2}\int_{\mathbb{R}} (\partial_t u)^2 \, d\mathcal{L}^1$ and the potential energy $P(t) := \frac{1}{2}\int_{\mathbb{R}} (\partial_x u)^2 \, d\mathcal{L}^1$. Show that $K(t) = P(t)$ for all sufficiently large $t$. [/problem]

[solution] **Step 1: Compute $\partial_t u$ and $\partial_x u$ from d'Alembert's formula.** By [D'Alembert's Formula](/theorems/665): \begin{align*} u(x, t) = \frac{1}{2}\bigl(g(x + t) + g(x - t)\bigr) + \frac{1}{2}\int_{x - t}^{x + t} h(y) \, d\mathcal{L}^1(y). \end{align*} Differentiating: \begin{align*} \partial_t u &= \frac{1}{2}\bigl(g'(x + t) - g'(x - t)\bigr) + \frac{1}{2}\bigl(h(x + t) + h(x - t)\bigr), \\ \partial_x u &= \frac{1}{2}\bigl(g'(x + t) + g'(x - t)\bigr) + \frac{1}{2}\bigl(h(x + t) - h(x - t)\bigr). \end{align*} **Step 2: Factor the difference $(\partial_t u)^2 - (\partial_x u)^2$.** Observe that \begin{align*} \partial_t u + \partial_x u &= g'(x + t) + h(x + t), \\ \partial_t u - \partial_x u &= -g'(x - t) - h(x - t). \end{align*} This follows by direct addition and subtraction of the expressions in Step 1. The identity $a^2 - b^2 = (a+b)(a-b)$ gives: \begin{align*} (\partial_t u)^2 - (\partial_x u)^2 = -\bigl(g'(x + t) + h(x + t)\bigr)\bigl(g'(x - t) + h(x - t)\bigr). \end{align*} **Step 3: Identify the separated regime.** Since $g$ and $h$ are compactly supported, there exists $R > 0$ with $\mathrm{supp}(g'), \mathrm{supp}(h) \subseteq [-R, R]$. The factor $g'(x + t) + h(x + t)$ is supported where $x + t \in [-R, R]$, i.e., $x \in [-R - t, R - t]$. The factor $g'(x - t) + h(x - t)$ is supported where $x - t \in [-R, R]$, i.e., $x \in [t - R, t + R]$. For $t > R$, these two intervals are disjoint since $R - t < 0 < t - R$. **Step 4: Conclude.** For $t > R$, the product in Step 2 vanishes identically for all $x \in \mathbb{R}$, since the two factors have disjoint supports. Therefore: \begin{align*} K(t) - P(t) = \frac{1}{2}\int_{\mathbb{R}} \bigl((\partial_t u)^2 - (\partial_x u)^2\bigr) \, d\mathcal{L}^1 = 0 \quad \text{for all } t > R. \end{align*} The kinetic and potential energies are exactly equal once the two travelling pulses have fully separated. Before separation ($t < R$), the cross-term is generally nonzero and the two energies oscillate. The equipartition result is specific to one dimension; in higher dimensions, a weaker asymptotic equipartition holds as $t \to \infty$, but exact equality for large finite $t$ is not guaranteed. [/solution]

[problem:Domain Of Dependence Via Kirchhoff | 3] Let $n = 3$ and let $u$ solve the wave equation with initial data $g \in C^3(\mathbb{R}^3; \mathbb{R})$ and $h \in C^2(\mathbb{R}^3; \mathbb{R})$ satisfying $g = 0$ and $h = 0$ on $\overline{B}(0, 1)$. Use Kirchhoff's formula to show directly that $u(0, t) = 0$ for all $0 \leq t \leq 1$. [/problem]

[solution] **Step 1: Write the Kirchhoff representation at $x = 0$.** By [Kirchhoff's Formula](/theorems/666): \begin{align*} u(0, t) = \partial_t\bigl[t \, (M_t g)(0)\bigr] + t \, (M_t h)(0), \end{align*} where $(M_t g)(0) = \frac{1}{4\pi t^2}\int_{\partial B(0, t)} g \, d\mathcal{H}^2$. **Step 2: Use the vanishing hypothesis on the displacement term.** For $0 \leq t \leq 1$, the sphere $\partial B(0, t) \subseteq \overline{B}(0, 1)$, so $g = 0$ on $\partial B(0, t)$. Therefore: \begin{align*} (M_t g)(0) = \frac{1}{4\pi t^2}\int_{\partial B(0, t)} 0 \, d\mathcal{H}^2 = 0 \quad \text{for all } 0 \leq t \leq 1. \end{align*} Since $t \mapsto t(M_t g)(0)$ is identically zero on $[0, 1]$, its $t$-derivative vanishes: \begin{align*} \partial_t\bigl[t(M_t g)(0)\bigr] = 0 \quad \text{for } 0 \leq t \leq 1. \end{align*} **Step 3: Handle the velocity term.** Similarly, $h = 0$ on $\partial B(0, t) \subseteq \overline{B}(0, 1)$ for $0 \leq t \leq 1$, so: \begin{align*} t \, (M_t h)(0) = \frac{t}{4\pi t^2}\int_{\partial B(0, t)} 0 \, d\mathcal{H}^2 = 0. \end{align*} **Step 4: Conclude.** Combining: \begin{align*} u(0, t) = 0 + 0 = 0 \quad \text{for all } 0 \leq t \leq 1. \end{align*} This is the same conclusion as the [Finite Speed Of Propagation For The Wave Equation](/theorems/670), obtained here by direct inspection of the representation formula rather than by an energy argument. The Kirchhoff-based proof is more transparent in $n = 3$: the formula *shows* that only the data on $\partial B(0, t) \subseteq \overline{B}(0, 1)$ appear. The energy proof is more general — it works in all dimensions and does not require an explicit formula — but gives less geometric insight into why the result holds. [/solution]