**Proof plan.** By linearity, it suffices to show that if $u$ solves the [wave equation](/page/Wave%20Equation) with zero data and zero forcing, then $u \equiv 0$. We define the energy $e(t) := \frac{1}{2}\int_{\mathbb{R}^n}((\partial_t u)^2 + |\nabla u|^2) \, d\mathcal{L}^n$ and show $e'(t) = 0$ by differentiating under the [integral](/page/Integral) and integrating by parts. Since $e(0) = 0$ (both initial data vanish), we conclude $e(t) = 0$ for all $t$, which forces $\partial_t u = 0$ and $\nabla u = 0$, hence $u \equiv 0$. **Step 1: Reduce to the zero-data problem.** Let $u_1, u_2$ be two $C^2$ solutions of the full initial-value problem (same $f$, $g$, $h$). Set $w := u_1 - u_2$. Then $w$ solves \begin{align*} \begin{cases} \partial_t^2 w - \Delta w = 0 & \text{in } \mathbb{R}^n \times (0, T), \\ w = 0, \quad \partial_t w = 0 & \text{on } \mathbb{R}^n \times \{t = 0\}. \end{cases} \end{align*} It suffices to show $w \equiv 0$. **Step 2: Define the energy functional.** Set \begin{align*} e(t) := \frac{1}{2}\int_{\mathbb{R}^n} \bigl((\partial_t w(x, t))^2 + |\nabla w(x, t)|^2\bigr) \, d\mathcal{L}^n(x). \end{align*} The integral is well-defined and finite because $w$ has bounded support in $x$ for each fixed $t$ (this follows from the finite propagation speed, which can be established independently by the domain of dependence argument, or assumed here as a consequence of the explicit representation formulas). **Step 3: Compute $e'(t)$.** Differentiating under the integral sign (justified by the smoothness of $w$ and compact spatial support): \begin{align*} e'(t) = \int_{\mathbb{R}^n} \bigl(\partial_t w \cdot \partial_t^2 w + \nabla w \cdot \nabla \partial_t w\bigr) \, d\mathcal{L}^n. \end{align*} [Integration by parts](/theorems/210) in the second term (with no [boundary](/page/Boundary) contribution, since $w$ has compact support in $x$): \begin{align*} \int_{\mathbb{R}^n} \nabla w \cdot \nabla \partial_t w \, d\mathcal{L}^n = -\int_{\mathbb{R}^n} \Delta w \cdot \partial_t w \, d\mathcal{L}^n. \end{align*} Therefore: \begin{align*} e'(t) = \int_{\mathbb{R}^n} \partial_t w \bigl(\partial_t^2 w - \Delta w\bigr) \, d\mathcal{L}^n = 0, \end{align*} since $w$ solves the homogeneous wave equation. **Step 4: Conclude.** Since $e'(t) = 0$ for all $t \in (0, T)$, $e$ is constant. At $t = 0$, $w = 0$ and $\partial_t w = 0$, so $e(0) = 0$. Therefore $e(t) = 0$ for all $t$, which implies $\partial_t w = 0$ and $\nabla w = 0$ pointwise. Combined with $w(\cdot, 0) = 0$, this gives $w \equiv 0$ on $\mathbb{R}^n \times [0, T]$.