[proofplan] We first prove the identity for smooth solutions by multiplying the [wave equation](/page/Wave%20Equation) by $\partial_t u$, integrating over $U$, and using the Dirichlet boundary condition to remove the boundary term in the Laplacian contribution. The differential identity gives an inequality for $E'(t)$ by the [Cauchy-Schwarz inequality](/theorems/432) in $L^2(U)$. Since $E(t)$ may vanish, we apply the inequality to the regularized quantity $(2E(t)+\varepsilon)^{1/2}$ and then let $\varepsilon \downarrow 0$. Finally, for an energy solution, we pass the same estimate from smooth Galerkin approximants to the limit using the convergence built into the energy-solution construction. [/proofplan]

[step:Derive the differential energy identity for smooth solutions]Assume first that $u$ and $f$ are smooth on $\overline{U}\times [0,T]$ and that $u=0$ on $\partial U\times [0,T]$. For each $\tau \in [0,T]$, define the energy map $E:[0,T]\to [0,\infty)$ by \begin{align*} E(\tau) := \frac{1}{2}\int_U \left(|\partial_t u(\tau,x)|^2 + |\nabla u(\tau,x)|^2\right)\,d\mathcal{L}^n(x). \end{align*} For each interior time $\tau\in(0,T)$, smoothness of $u$ justifies differentiation under the integral sign and gives \begin{align*} E'(\tau) = \int_U \partial_t u(\tau,x)\,\partial_t^2 u(\tau,x)\,d\mathcal{L}^n(x) + \int_U \nabla u(\tau,x)\cdot \nabla\partial_t u(\tau,x)\,d\mathcal{L}^n(x). \end{align*} Since $u=0$ on $\partial U\times [0,T]$, differentiating the boundary condition in time gives that $\partial_t u$ has zero trace on $\partial U$ for each $\tau\in[0,T]$. Let $\mathcal H^{n-1}$ denote the $(n-1)$-dimensional [Hausdorff measure](/page/Hausdorff%20Measure) on $\partial U$. The smooth integration-by-parts formula on the bounded smooth domain $U$, applied to the spatial maps $u(\tau,\cdot):U\to\mathbb R$ and $\partial_tu(\tau,\cdot):U\to\mathbb R$, has boundary term $\int_{\partial U} \partial_tu(\tau,x)\,\partial_\nu u(\tau,x)\,d\mathcal H^{n-1}(x)$, which vanishes because the trace of $\partial_tu(\tau,\cdot)$ is zero. Hence \begin{align*} \int_U \nabla u(\tau,x)\cdot \nabla\partial_t u(\tau,x)\,d\mathcal{L}^n(x) = -\int_U \Delta u(\tau,x)\,\partial_t u(\tau,x)\,d\mathcal{L}^n(x). \end{align*} Therefore \begin{align*} E'(\tau) = \int_U \partial_t u(\tau,x)\left(\partial_t^2u(\tau,x)-\Delta u(\tau,x)\right)\,d\mathcal{L}^n(x). \end{align*} Using the equation \begin{align*} \partial_t^2u-\Delta u=f, \end{align*} we obtain \begin{align*} E'(\tau) = \int_U f(\tau,x)\,\partial_tu(\tau,x)\,d\mathcal{L}^n(x). \end{align*}[/step]

[guided]We begin with the smooth case because all integrations by parts and differentiations under the integral sign are then justified directly. The energy at time $\tau$ is \begin{align*} E(\tau) := \frac{1}{2}\int_U \left(|\partial_t u(\tau,x)|^2 + |\nabla u(\tau,x)|^2\right)\,d\mathcal{L}^n(x). \end{align*} The first term measures kinetic energy and the second term measures elastic energy. For each interior time $\tau\in(0,T)$, smoothness of $u$ on $\overline{U}\times [0,T]$ permits differentiation under the integral sign. Applying the ordinary chain rule pointwise in $x$ gives \begin{align*} E'(\tau) = \int_U \partial_t u(\tau,x)\,\partial_t^2 u(\tau,x)\,d\mathcal{L}^n(x) + \int_U \nabla u(\tau,x)\cdot \nabla\partial_t u(\tau,x)\,d\mathcal{L}^n(x). \end{align*} The second integral is the one that must be matched with the Laplacian in the wave equation. Since $u=0$ on $\partial U\times [0,T]$, differentiating this boundary condition with respect to time gives that $\partial_t u$ has zero trace on $\partial U$ for each $\tau\in[0,T]$. Let $\mathcal H^{n-1}$ denote the $(n-1)$-dimensional Hausdorff measure on $\partial U$. The smooth integration-by-parts formula on a bounded smooth domain applies to the smooth spatial maps $u(\tau,\cdot):U\to\mathbb R$ and $\partial_tu(\tau,\cdot):U\to\mathbb R$ on $U$. Its boundary term is $\int_{\partial U} \partial_tu(\tau,x)\,\partial_\nu u(\tau,x)\,d\mathcal H^{n-1}(x)$, where $\partial_\nu u(\tau,\cdot)$ denotes the outward normal derivative on $\partial U$; this term vanishes because the trace of $\partial_tu(\tau,\cdot)$ is zero. Thus \begin{align*} \int_U \nabla u(\tau,x)\cdot \nabla\partial_t u(\tau,x)\,d\mathcal{L}^n(x) = -\int_U \Delta u(\tau,x)\,\partial_t u(\tau,x)\,d\mathcal{L}^n(x). \end{align*} Substituting this into the expression for $E'(\tau)$ yields \begin{align*} E'(\tau) = \int_U \partial_t u(\tau,x)\left(\partial_t^2u(\tau,x)-\Delta u(\tau,x)\right)\,d\mathcal{L}^n(x). \end{align*} The wave equation says precisely that \begin{align*} \partial_t^2u-\Delta u=f. \end{align*} Replacing the parenthesized term by $f$ gives the differential energy identity, for every $\tau\in(0,T)$, \begin{align*} E'(\tau) = \int_U f(\tau,x)\,\partial_tu(\tau,x)\,d\mathcal{L}^n(x). \end{align*} This identity is the point of multiplying the equation by $\partial_t u$: it converts the PDE into an exact time derivative of the energy.[/guided]

[step:Estimate the derivative of the smooth energy]For every $\tau \in (0,T)$, the maps $f(\tau):U\to\mathbb R$ and $\partial_tu(\tau):U\to\mathbb R$ belong to the [Hilbert space](/page/Hilbert%20Space) $L^2(U)$. Applying the [Cauchy-Schwarz inequality](/theorems/432), in the form $|(a,b)_{L^2(U)}|\le \|a\|_{L^2(U)}\|b\|_{L^2(U)}$, to $a=f(\tau)$ and $b=\partial_tu(\tau)$ gives \begin{align*} E'(\tau) \le \|f(\tau)\|_{L^2(U)}\,\|\partial_tu(\tau)\|_{L^2(U)}. \end{align*} By the definition of $E(\tau)$, \begin{align*} \|\partial_tu(\tau)\|_{L^2(U)}^2 \le 2E(\tau). \end{align*} Hence \begin{align*} E'(\tau) \le \|f(\tau)\|_{L^2(U)}\sqrt{2E(\tau)}. \end{align*}[/step]

[guided]At this point the PDE has been converted into the identity \begin{align*} E'(\tau)=\int_U f(\tau,x)\,\partial_tu(\tau,x)\,d\mathcal L^n(x). \end{align*} We now estimate the right-hand side in the Hilbert space $L^2(U)$. Since $f(\tau):U\to\mathbb R$ and $\partial_tu(\tau):U\to\mathbb R$ are smooth on the bounded domain $U$, both functions belong to $L^2(U)$. The [Cauchy-Schwarz inequality](/theorems/432), written as $|(a,b)_{L^2(U)}|\le \|a\|_{L^2(U)}\|b\|_{L^2(U)}$, applied with $a=f(\tau)$ and $b=\partial_tu(\tau)$, gives \begin{align*} E'(\tau) \le \|f(\tau)\|_{L^2(U)}\,\|\partial_tu(\tau)\|_{L^2(U)}. \end{align*} The energy contains the kinetic term as one of its non-negative summands, so the definition of $E(\tau)$ gives \begin{align*} \|\partial_tu(\tau)\|_{L^2(U)}^2 \le 2E(\tau). \end{align*} Taking square roots preserves the inequality because both sides are non-negative. Substituting this bound into the Cauchy-Schwarz estimate yields \begin{align*} E'(\tau) \le \|f(\tau)\|_{L^2(U)}\sqrt{2E(\tau)}. \end{align*} This is the differential inequality that will be integrated after regularizing the square root.[/guided]

[step:Regularize the square root to avoid division by zero]Fix $\varepsilon>0$ and define $G_\varepsilon:[0,T]\to (0,\infty)$ by \begin{align*} G_\varepsilon(\tau) := \left(2E(\tau)+\varepsilon\right)^{1/2}. \end{align*} Since $E$ is continuously differentiable on $(0,T)$ and continuous on $[0,T]$ in the smooth case, $G_\varepsilon$ is differentiable on $(0,T)$ and absolutely continuous on $[0,T]$, with \begin{align*} G_\varepsilon'(\tau) = \frac{E'(\tau)}{\left(2E(\tau)+\varepsilon\right)^{1/2}} \end{align*} for $\tau\in(0,T)$. Using the estimate from the previous step and the inequality $\sqrt{2E(\tau)} \le \sqrt{2E(\tau)+\varepsilon}$, we obtain \begin{align*} G_\varepsilon'(\tau) \le \|f(\tau)\|_{L^2(U)}. \end{align*} Integrating this differential inequality over $[s,t]$ with respect to $\mathcal{L}^1$ gives \begin{align*} \left(2E(t)+\varepsilon\right)^{1/2} \le \left(2E(s)+\varepsilon\right)^{1/2} + \int_s^t \|f(r)\|_{L^2(U)}\,d\mathcal{L}^1(r). \end{align*} Letting $\varepsilon\downarrow 0$ and using continuity of the square-root function gives \begin{align*} \sqrt{2E(t)} \le \sqrt{2E(s)} + \int_s^t \|f(r)\|_{L^2(U)}\,d\mathcal{L}^1(r). \end{align*}[/step]

[guided]The differential inequality from the previous step has the form \begin{align*} E'(\tau) \le \|f(\tau)\|_{L^2(U)}\sqrt{2E(\tau)}. \end{align*} A direct division by $\sqrt{2E(\tau)}$ would fail at times where $E(\tau)=0$, so we introduce a positive regularization. Fix $\varepsilon>0$ and define the differentiable map $G_\varepsilon:[0,T]\to(0,\infty)$ by \begin{align*} G_\varepsilon(\tau):=\left(2E(\tau)+\varepsilon\right)^{1/2}. \end{align*} The ordinary chain rule gives, for $\tau\in(0,T)$, \begin{align*} G_\varepsilon'(\tau)=\frac{E'(\tau)}{\left(2E(\tau)+\varepsilon\right)^{1/2}}. \end{align*} Because $G_\varepsilon$ is absolutely continuous on $[0,T]$, this interior differential inequality can be integrated over any subinterval $[s,t]\subset[0,T]$. Substituting the estimate for $E'(\tau)$ gives \begin{align*} G_\varepsilon'(\tau) \le \|f(\tau)\|_{L^2(U)}\frac{\sqrt{2E(\tau)}}{\sqrt{2E(\tau)+\varepsilon}}. \end{align*} Since $\varepsilon>0$, the fraction on the right is at most $1$. Therefore \begin{align*} G_\varepsilon'(\tau) \le \|f(\tau)\|_{L^2(U)}. \end{align*} Integrating this differential inequality over $[s,t]$ with respect to one-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) $\mathcal L^1$ yields \begin{align*} \left(2E(t)+\varepsilon\right)^{1/2} \le \left(2E(s)+\varepsilon\right)^{1/2}+\int_s^t\|f(r)\|_{L^2(U)}\,d\mathcal L^1(r). \end{align*} Finally, the square-root function is continuous on $[0,\infty)$, so letting $\varepsilon\downarrow0$ gives \begin{align*} \sqrt{2E(t)} \le \sqrt{2E(s)}+\int_s^t\|f(r)\|_{L^2(U)}\,d\mathcal L^1(r). \end{align*}[/guided]

[step:Pass the estimate to energy solutions]Let $u:[0,T]\to H_0^1(U)$ with $\partial_tu:[0,T]\to L^2(U)$ be an energy solution in the Galerkin-limit sense stated in the theorem. Thus the approximating subspaces, Galerkin solutions, forcing approximants, and strong convergence properties used below are part of the theorem's definition of energy solution. Here $H_0^1(U)$ denotes the closure of $C_c^\infty(U)$ in the [Sobolev space](/page/Sobolev%20Space) $H^1(U)$. Define the limiting energy map $E:[0,T]\to[0,\infty)$ by \begin{align*} E(\tau):=\frac{1}{2}\int_U \left(|\partial_tu(\tau,x)|^2+|\nabla u(\tau,x)|^2\right)\,d\mathcal L^n(x). \end{align*} By this Galerkin-limit definition of energy solution, there exist finite-dimensional Dirichlet subspaces $V_k\subset H_0^1(U)$, Galerkin approximants $u_k\in C^2([0,T];V_k)$, and forcings $f_k\in C^\infty([0,T];L^2(U))$ such that, for every $v\in V_k$ and every $\tau\in[0,T]$, \begin{align*} \int_U \partial_t^2u_k(\tau,x)v(x)\,d\mathcal L^n(x)+\int_U \nabla u_k(\tau,x)\cdot\nabla v(x)\,d\mathcal L^n(x)=\int_U f_k(\tau,x)v(x)\,d\mathcal L^n(x), \end{align*} and \begin{align*} f_k \to f \quad\text{in } L^1((0,T);L^2(U)), \end{align*} and \begin{align*} u_k \to u \quad\text{in } C([0,T];H_0^1(U)), \end{align*} while \begin{align*} \partial_t u_k \to \partial_tu \quad\text{in } C([0,T];L^2(U)). \end{align*} For each $k\in \mathbb{N}$, define the approximating energy map $E_k:[0,T]\to[0,\infty)$ by \begin{align*} E_k(\tau) := \frac{1}{2}\int_U \left(|\partial_t u_k(\tau,x)|^2+|\nabla u_k(\tau,x)|^2\right)\,d\mathcal{L}^n(x). \end{align*} Since $u_k(\tau)\in V_k$ for every $\tau\in[0,T]$ and $V_k$ is a [vector space](/page/Vector%20Space), $\partial_tu_k(\tau)\in V_k$. Taking $v=\partial_tu_k(\tau)$ in the Galerkin identity gives the same differential energy identity as in the smooth case: \begin{align*} E_k'(\tau)=\int_U f_k(\tau,x)\,\partial_tu_k(\tau,x)\,d\mathcal L^n(x). \end{align*} The Hilbert-space estimate and square-root regularization from the previous steps therefore give, for every $0\le s\le t\le T$, \begin{align*} \sqrt{2E_k(t)} \le \sqrt{2E_k(s)} + \int_s^t \|f_k(r)\|_{L^2(U)}\,d\mathcal{L}^1(r). \end{align*} The stated convergence of $u_k$ and $\partial_tu_k$ implies $E_k(\tau)\to E(\tau)$ for each $\tau\in [0,T]$: convergence in $C([0,T];H_0^1(U))$ controls $\|\nabla u_k(\tau)-\nabla u(\tau)\|_{L^2(U)}$, and convergence in $C([0,T];L^2(U))$ controls $\|\partial_tu_k(\tau)-\partial_tu(\tau)\|_{L^2(U)}$. The convergence $f_k\to f$ in $L^1((0,T);L^2(U))$ implies convergence of the forcing integrals because the [reverse triangle inequality](/theorems/2300) in $L^2(U)$ gives, for $\mathcal L^1$-a.e. $r\in(0,T)$, \begin{align*} \left|\|f_k(r)\|_{L^2(U)}-\|f(r)\|_{L^2(U)}\right| \le \|f_k(r)-f(r)\|_{L^2(U)}. \end{align*} Integrating this estimate over $[s,t]$ yields \begin{align*} \left|\int_s^t \|f_k(r)\|_{L^2(U)}\,d\mathcal{L}^1(r)-\int_s^t \|f(r)\|_{L^2(U)}\,d\mathcal{L}^1(r)\right| \le \int_s^t \|f_k(r)-f(r)\|_{L^2(U)}\,d\mathcal{L}^1(r) \le \|f_k-f\|_{L^1((0,T);L^2(U))}. \end{align*} The right-hand side tends to $0$, so the forcing integrals converge. Passing to the limit in the estimate for $E_k$ gives \begin{align*} \sqrt{2E(t)} \le \sqrt{2E(s)} + \int_s^t \|f(r)\|_{L^2(U)}\,d\mathcal{L}^1(r). \end{align*} This is the desired forced energy inequality for the energy solution $u$.[/step]

[guided]The theorem defines an energy solution $u:[0,T]\to H_0^1(U)$ with $\partial_tu:[0,T]\to L^2(U)$ in the Galerkin-limit sense. The limiting energy is the map $E:[0,T]\to[0,\infty)$ defined by \begin{align*} E(\tau):=\frac12\int_U\left(|\partial_tu(\tau,x)|^2+|\nabla u(\tau,x)|^2\right)\,d\mathcal L^n(x). \end{align*} The Galerkin approximation hypothesis means that there are finite-dimensional Dirichlet subspaces $V_k\subset H_0^1(U)$, approximants $u_k\in C^2([0,T];V_k)$, and smooth forcings $f_k\in C^\infty([0,T];L^2(U))$ satisfying the projected forced wave equation: for every $v\in V_k$ and every $\tau\in[0,T]$, \begin{align*} \int_U \partial_t^2u_k(\tau,x)v(x)\,d\mathcal L^n(x)+\int_U \nabla u_k(\tau,x)\cdot\nabla v(x)\,d\mathcal L^n(x)=\int_U f_k(\tau,x)v(x)\,d\mathcal L^n(x). \end{align*} They also satisfy \begin{align*} f_k\to f \quad\text{in } L^1((0,T);L^2(U)), \end{align*} \begin{align*} u_k\to u \quad\text{in } C([0,T];H_0^1(U)), \end{align*} and \begin{align*} \partial_tu_k\to\partial_tu \quad\text{in } C([0,T];L^2(U)). \end{align*} Here $H_0^1(U)$ is the closure of $C_c^\infty(U)$ in the [Sobolev space](/page/Sobolev%20Space) $H^1(U)$, so the Dirichlet boundary condition is encoded in the energy space. For each $k\in\mathbb N$, define \begin{align*} E_k(\tau):=\frac12\int_U\left(|\partial_tu_k(\tau,x)|^2+|\nabla u_k(\tau,x)|^2\right)\,d\mathcal L^n(x). \end{align*} Why does the estimate apply to the Galerkin approximants? Since $u_k(\tau)\in V_k$ and $V_k$ is finite-dimensional, differentiating the coefficient curve shows $\partial_tu_k(\tau)\in V_k$. We may therefore choose $v=\partial_tu_k(\tau)$ in the projected equation. This gives \begin{align*} \int_U \partial_t^2u_k(\tau,x)\partial_tu_k(\tau,x)\,d\mathcal L^n(x)+\int_U \nabla u_k(\tau,x)\cdot\nabla\partial_tu_k(\tau,x)\,d\mathcal L^n(x)=\int_U f_k(\tau,x)\partial_tu_k(\tau,x)\,d\mathcal L^n(x). \end{align*} The left-hand side is exactly $E_k'(\tau)$ by differentiating the two squared $L^2$ norms in the definition of $E_k$. Hence \begin{align*} E_k'(\tau)=\int_U f_k(\tau,x)\partial_tu_k(\tau,x)\,d\mathcal L^n(x). \end{align*} Applying the same Hilbert-space estimate and square-root regularization used in the smooth case gives \begin{align*} \sqrt{2E_k(t)}\le\sqrt{2E_k(s)}+\int_s^t\|f_k(r)\|_{L^2(U)}\,d\mathcal L^1(r). \end{align*} The convergence of $u_k$ in $C([0,T];H_0^1(U))$ gives convergence of the gradient term at each time because the $H_0^1(U)$ norm controls $\|\nabla u_k(\tau)-\nabla u(\tau)\|_{L^2(U)}$. The convergence of $\partial_tu_k$ in $C([0,T];L^2(U))$ gives convergence of the kinetic term at each time. Therefore $E_k(\tau)\to E(\tau)$ for every $\tau\in[0,T]$. It remains only to pass to the limit in the forcing integral. The reverse triangle inequality in $L^2(U)$ gives, for $\mathcal L^1$-a.e. $r\in(0,T)$, \begin{align*} \left|\|f_k(r)\|_{L^2(U)}-\|f(r)\|_{L^2(U)}\right|\le\|f_k(r)-f(r)\|_{L^2(U)}. \end{align*} Integrating over $[s,t]$ with respect to $\mathcal L^1$ yields \begin{align*} \left|\int_s^t\|f_k(r)\|_{L^2(U)}\,d\mathcal L^1(r)-\int_s^t\|f(r)\|_{L^2(U)}\,d\mathcal L^1(r)\right|\le\|f_k-f\|_{L^1((0,T);L^2(U))}. \end{align*} The right-hand side tends to $0$, so the forcing integrals converge. Taking the limit in the Galerkin inequality gives \begin{align*} \sqrt{2E(t)}\le\sqrt{2E(s)}+\int_s^t\|f(r)\|_{L^2(U)}\,d\mathcal L^1(r). \end{align*} This proves the asserted energy inequality for the energy solution $u$.[/guided]