[proofplan] We construct the solution by expanding the initial data in the Dirichlet eigenfunction basis of $-\Delta$ on $U$ and solving the resulting scalar harmonic oscillator equations. The spectral expansion gives convergence in the energy topology, so the series defines a function $u \in C([0,T];H^1_0(U))$ with $\partial_t u \in C([0,T];L^2(U))$ and $\partial_t^2u \in C([0,T];H^{-1}(U))$. The weak equation and initial conditions follow by testing first against finite eigenfunction combinations and then passing to the limit. Energy conservation is obtained from exact conservation in each scalar oscillator and summing the convergent series, while uniqueness follows by projecting any zero-data solution onto every eigenfunction. [/proofplan]

[step:Choose the Dirichlet eigenfunction basis and expand the initial data]Let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$, and let $\mathcal{L}^1$ denote one-dimensional Lebesgue measure on $\mathbb{R}$. The space $H^1_0(U)$ is the zero-trace [Sobolev space](/page/Sobolev%20Space), and $H^{-1}(U)$ denotes its [topological dual](/page/Topological%20Dual). Since $U \subset \mathbb{R}^n$ is bounded with smooth boundary, the Dirichlet Laplacian $A = -\Delta$ with domain $H^2(U) \cap H^1_0(U)$ is a positive self-adjoint operator on $L^2(U)$ with compact resolvent. The [Spectral Theorem for the Dirichlet Laplacian with Compact Resolvent](/theorems/6447) therefore gives a sequence of real-valued functions \begin{align*} e_k: U \to \mathbb{R}, \qquad k \in \mathbb{N}, \end{align*} and a sequence of positive eigenvalues $(\lambda_k)_{k \in \mathbb{N}} \subset (0,\infty)$ with $\lambda_k \to \infty$, such that $(e_k)_{k \in \mathbb{N}}$ is an [orthonormal basis](/page/Orthonormal%20Basis) of $L^2(U)$ and \begin{align*} -\Delta e_k = \lambda_k e_k \quad \text{in } U, \qquad e_k \in H^1_0(U) \cap C^\infty(U). \end{align*} For $v \in H^1_0(U)$ with Fourier coefficients $v_k := \int_U v(x)e_k(x)\,d\mathcal{L}^n(x)$, The [Parseval Identity](/theorems/248) and the Dirichlet form identity from the same spectral theorem give \begin{align*} \|v\|_{L^2(U)}^2 = \sum_{k=1}^{\infty} |v_k|^2 \end{align*} and \begin{align*} \|\nabla v\|_{L^2(U)}^2 = \sum_{k=1}^{\infty} \lambda_k |v_k|^2. \end{align*} Define the initial Fourier coefficients $a_k \in \mathbb{R}$ and $b_k \in \mathbb{R}$ by \begin{align*} a_k := \int_U u_0(x)e_k(x)\,d\mathcal{L}^n(x), \qquad b_k := \int_U u_1(x)e_k(x)\,d\mathcal{L}^n(x). \end{align*} Since $u_0 \in H^1_0(U)$ and $u_1 \in L^2(U)$, \begin{align*} \sum_{k=1}^{\infty} \lambda_k |a_k|^2 = \|\nabla u_0\|_{L^2(U)}^2 < \infty \end{align*} and \begin{align*} \sum_{k=1}^{\infty} |b_k|^2 = \|u_1\|_{L^2(U)}^2 < \infty. \end{align*}[/step]

[guided]The boundary condition is homogeneous Dirichlet, so the correct spatial basis is the eigenfunction basis of the Dirichlet Laplacian rather than the full Fourier basis on a box. We use $\mathcal{L}^n$ for $n$-dimensional Lebesgue measure and $\mathcal{L}^1$ for one-dimensional Lebesgue measure. The [Spectral Theorem for the Dirichlet Laplacian with Compact Resolvent](/theorems/6447) gives functions \begin{align*} e_k: U \to \mathbb{R}, \qquad k \in \mathbb{N}, \end{align*} and positive eigenvalues $\lambda_k > 0$ such that $(e_k)_{k \in \mathbb{N}}$ is an orthonormal basis of $L^2(U)$ and \begin{align*} -\Delta e_k = \lambda_k e_k, \qquad e_k \in H^1_0(U) \cap C^\infty(U). \end{align*} The condition $e_k \in H^1_0(U)$ is what builds the boundary condition into the expansion. We define the Fourier coefficients of the initial data by \begin{align*} a_k := \int_U u_0(x)e_k(x)\,d\mathcal{L}^n(x), \qquad b_k := \int_U u_1(x)e_k(x)\,d\mathcal{L}^n(x). \end{align*} Because $(e_k)$ is an orthonormal basis of $L^2(U)$, the [Parseval Identity](/theorems/248) gives \begin{align*} \sum_{k=1}^{\infty} |b_k|^2 = \|u_1\|_{L^2(U)}^2. \end{align*} For the displacement $u_0$, the energy norm is controlled by the Dirichlet form of the Laplacian. The spectral theorem identifies this Dirichlet form with the weighted coefficient sum: \begin{align*} \sum_{k=1}^{\infty} \lambda_k |a_k|^2 = \|\nabla u_0\|_{L^2(U)}^2. \end{align*} This is the key reason the construction works in the energy space: $H^1_0(U) \times L^2(U)$ is exactly the space in which the two sums needed for the wave energy are finite.[/guided]

[step:Solve the scalar oscillator equations for each Fourier mode] For each $k \in \mathbb{N}$, define \begin{align*} q_k: [0,T] \to \mathbb{R} \end{align*} by \begin{align*} q_k(t) := a_k \cos(c\sqrt{\lambda_k}t) + \frac{b_k}{c\sqrt{\lambda_k}}\sin(c\sqrt{\lambda_k}t). \end{align*} Then $q_k \in C^\infty([0,T])$ and direct differentiation gives \begin{align*} q_k''(t) + c^2\lambda_k q_k(t) = 0 \end{align*} for every $t \in [0,T]$, with \begin{align*} q_k(0) = a_k, \qquad q_k'(0) = b_k. \end{align*} Moreover, the scalar energy \begin{align*} E_k(t) := |q_k'(t)|^2 + c^2\lambda_k |q_k(t)|^2 \end{align*} is independent of $t$. Indeed, \begin{align*} E_k'(t) = 2q_k'(t)q_k''(t) + 2c^2\lambda_k q_k(t)q_k'(t) = 2q_k'(t)\bigl(q_k''(t) + c^2\lambda_k q_k(t)\bigr) = 0. \end{align*} Therefore \begin{align*} |q_k'(t)|^2 + c^2\lambda_k |q_k(t)|^2 = |b_k|^2 + c^2\lambda_k |a_k|^2 \end{align*} for every $t \in [0,T]$. [/step]

[step:Define the solution by the convergent energy series]For $N \in \mathbb{N}$, define the finite-mode function \begin{align*} u_N: [0,T] \to H^1_0(U) \end{align*} by \begin{align*} u_N(t) := \sum_{k=1}^{N} q_k(t)e_k. \end{align*} For $M > N$ and every $t \in [0,T]$, the scalar energy identity applied to the coefficient differences gives \begin{align*} \|\partial_t u_M(t)-\partial_t u_N(t)\|_{L^2(U)}^2 + c^2\|\nabla u_M(t)-\nabla u_N(t)\|_{L^2(U)}^2 = \sum_{k=N+1}^{M} |b_k|^2 + c^2\sum_{k=N+1}^{M}\lambda_k |a_k|^2. \end{align*} The right-hand side tends to $0$ as $N,M \to \infty$, uniformly in $t \in [0,T]$. Hence $(u_N)$ is Cauchy in $C([0,T];H^1_0(U))$, and $(\partial_t u_N)$ is Cauchy in $C([0,T];L^2(U))$. Let \begin{align*} u: [0,T] \to H^1_0(U) \end{align*} be the first limit, and let \begin{align*} v: [0,T] \to L^2(U) \end{align*} be the second limit. For every $N \in \mathbb{N}$ and every $s,t \in [0,T]$, the Banach-valued version of the [Fundamental Theorem of Calculus](/theorems/632) in $H^1_0(U)$ gives \begin{align*} u_N(t)-u_N(s)=\int_s^t \partial_tu_N(\tau)\,d\mathcal{L}^1(\tau) \end{align*} as an identity in $H^1_0(U)$. Passing to the limit in $L^2(U)$, using $H^1_0(U) \hookrightarrow L^2(U)$ continuously and the [uniform convergence](/page/Uniform%20Convergence) of $\partial_tu_N$ to $v$ in $L^2(U)$, gives \begin{align*} u(t)-u(s)=\int_s^t v(\tau)\,d\mathcal{L}^1(\tau) \end{align*} in $L^2(U)$. Therefore $u$ is continuously differentiable as an $L^2(U)$-valued map and $\partial_tu=v$. Thus \begin{align*} u \in C([0,T];H^1_0(U)), \qquad \partial_t u \in C([0,T];L^2(U)). \end{align*} For $v_0 \in H^1_0(U)$, define the weak Dirichlet Laplacian $\Delta v_0 \in H^{-1}(U)$ by \begin{align*} (\Delta v_0)(\phi):=-\int_U \nabla v_0(x)\cdot\nabla\phi(x)\,d\mathcal{L}^n(x), \qquad \phi \in H^1_0(U). \end{align*} With the standard $H^1_0(U)$ norm, this definition gives the bound \begin{align*} \|\Delta v_0\|_{H^{-1}(U)}\le \|\nabla v_0\|_{L^2(U)}. \end{align*} Therefore the convergence of $u_N$ in $C([0,T];H^1_0(U))$ implies that $c^2\Delta u_N$ converges in $C([0,T];H^{-1}(U))$ to $c^2\Delta u$. Since $U$ is bounded, the embedding $J: L^2(U) \to H^{-1}(U)$ defined by $J(f)(\phi)=\int_U f(x)\phi(x)\,d\mathcal{L}^n(x)$ is continuous by the [Cauchy-Schwarz inequality](/theorems/432) and the Poincare inequality on $H^1_0(U)$. Hence $\partial_tu_N \to \partial_tu$ in $C([0,T];H^{-1}(U))$. Since $\partial_t^2u_N=c^2\Delta u_N$, the same Banach-valued integral identity applied to $\partial_tu_N$ shows that $\partial_tu$ is continuously differentiable as an $H^{-1}(U)$-valued map and \begin{align*} \partial_t^2u=c^2\Delta u \quad \text{in } C([0,T];H^{-1}(U)). \end{align*} In particular, \begin{align*} \partial_t^2 u \in C([0,T];H^{-1}(U)). \end{align*}[/step]

[guided]The convergence estimate is strong enough to construct both the solution and its time derivatives, but we must identify the derivative of the limit rather than only the limit of the derivatives. For $N \in \mathbb{N}$, the finite-mode function is \begin{align*} u_N: [0,T] \to H^1_0(U) \end{align*} defined by \begin{align*} u_N(t) := \sum_{k=1}^{N} q_k(t)e_k. \end{align*} For $M>N$, the difference $u_M-u_N$ contains exactly the modes $N+1,\dots,M$. Applying the scalar energy identity to those modes gives, for every $t \in [0,T]$, \begin{align*} \|\partial_t u_M(t)-\partial_t u_N(t)\|_{L^2(U)}^2 + c^2\|\nabla u_M(t)-\nabla u_N(t)\|_{L^2(U)}^2 = \sum_{k=N+1}^{M} |b_k|^2 + c^2\sum_{k=N+1}^{M}\lambda_k |a_k|^2. \end{align*} The two tails on the right tend to $0$ because the initial data satisfy \begin{align*} \sum_{k=1}^{\infty}|b_k|^2<\infty \end{align*} and \begin{align*} \sum_{k=1}^{\infty}\lambda_k|a_k|^2<\infty. \end{align*} The estimate is uniform in $t$, so $u_N$ converges in $C([0,T];H^1_0(U))$ to a map $u: [0,T]\to H^1_0(U)$, and $\partial_tu_N$ converges in $C([0,T];L^2(U))$ to a map $v: [0,T]\to L^2(U)$. Why is $v$ really $\partial_tu$? For each finite $N$, the Banach-valued version of the [Fundamental Theorem of Calculus](/theorems/632) gives \begin{align*} u_N(t)-u_N(s)=\int_s^t \partial_tu_N(\tau)\,d\mathcal{L}^1(\tau) \end{align*} in $H^1_0(U)$. Passing to the limit in $L^2(U)$ is justified because $H^1_0(U)$ embeds continuously into $L^2(U)$ and $\partial_tu_N\to v$ uniformly in $L^2(U)$. Thus \begin{align*} u(t)-u(s)=\int_s^t v(\tau)\,d\mathcal{L}^1(\tau) \end{align*} in $L^2(U)$. This integral identity proves that $u$ is continuously differentiable as an $L^2(U)$-valued map and that $\partial_tu=v$. It remains to identify the second time derivative. For $v_0 \in H^1_0(U)$, define the weak Dirichlet Laplacian $\Delta v_0 \in H^{-1}(U)$ as the distributional operator induced by the Dirichlet form by \begin{align*} (\Delta v_0)(\phi):=-\int_U \nabla v_0(x)\cdot\nabla\phi(x)\,d\mathcal{L}^n(x), \qquad \phi \in H^1_0(U). \end{align*} The Cauchy-Schwarz inequality on $L^2(U;\mathbb{R}^n)$ gives \begin{align*} |(\Delta v_0)(\phi)|\le \|\nabla v_0\|_{L^2(U)}\|\nabla\phi\|_{L^2(U)}. \end{align*} Hence \begin{align*} \|\Delta v_0\|_{H^{-1}(U)}\le \|\nabla v_0\|_{L^2(U)}. \end{align*} This bound shows that $c^2\Delta u_N\to c^2\Delta u$ in $C([0,T];H^{-1}(U))$. We also need the velocities to converge in $H^{-1}(U)$. Since $U$ is bounded, the map $J: L^2(U)\to H^{-1}(U)$ given by $J(f)(\phi)=\int_U f(x)\phi(x)\,d\mathcal{L}^n(x)$ is continuous: the Cauchy-Schwarz inequality gives $|J(f)(\phi)|\le \|f\|_{L^2(U)}\|\phi\|_{L^2(U)}$, and the Poincare inequality on $H^1_0(U)$ bounds $\|\phi\|_{L^2(U)}$ by a constant times $\|\phi\|_{H^1_0(U)}$. Therefore $\partial_tu_N\to \partial_tu$ in $C([0,T];H^{-1}(U))$. Since $\partial_t^2u_N=c^2\Delta u_N$ for each $N$, applying the same integral-identity argument to $\partial_tu_N$ proves that $\partial_tu$ is continuously differentiable as an $H^{-1}(U)$-valued map and \begin{align*} \partial_t^2u=c^2\Delta u \quad \text{in } C([0,T];H^{-1}(U)). \end{align*}[/guided]

[step:Pass the finite-mode equation to the weak limit]For every $N \in \mathbb{N}$ and every finite linear combination \begin{align*} v: U \to \mathbb{R} \end{align*} of eigenfunctions $e_1,\dots,e_N$, the scalar oscillator equations give \begin{align*} (\partial_t^2 u_N(t))(v) + c^2\int_U \nabla u_N(t,x)\cdot \nabla v(x)\,d\mathcal{L}^n(x) = 0. \end{align*} Here $(\partial_t^2 u_N(t))(v)$ denotes the action of the element $\partial_t^2u_N(t) \in H^{-1}(U)$ on $v \in H^1_0(U)$. The [Spectral Theorem for the Dirichlet Laplacian with Compact Resolvent](/theorems/6447) also gives density of finite eigenfunction spans in $H^1_0(U)$ with respect to the $H^1_0(U)$ norm. Fix $K\in\mathbb{N}$ and let $v\in H^1_0(U)$ be a finite linear combination of $e_1,\dots,e_K$. For every $N\ge K$, the finite-mode identity above holds with this $v$. Since \begin{align*} u_N \to u \quad \text{in } C([0,T];H^1_0(U)) \end{align*} and \begin{align*} \partial_t^2u_N \to \partial_t^2u \quad \text{in } C([0,T];H^{-1}(U)), \end{align*} the continuity of the $H^{-1}(U)$ action on $H^1_0(U)$ gives $(\partial_t^2u_N(t))(v)\to(\partial_t^2u(t))(v)$, and the continuity of the Dirichlet [inner product](/page/Inner%20Product) gives $\int_U \nabla u_N(t,x)\cdot\nabla v(x)\,d\mathcal{L}^n(x)\to\int_U \nabla u(t,x)\cdot\nabla v(x)\,d\mathcal{L}^n(x)$. Thus the weak identity holds for every finite eigenfunction combination. For arbitrary $v\in H^1_0(U)$, choose finite eigenfunction combinations $v_j$ with $v_j\to v$ in $H^1_0(U)$; continuity of both terms in $v$ then passes the identity from $v_j$ to $v$. Hence \begin{align*} (\partial_t^2 u(t))(v) + c^2\int_U \nabla u(t,x)\cdot \nabla v(x)\,d\mathcal{L}^n(x) = 0 \end{align*} for every $v \in H^1_0(U)$ and every $t \in [0,T]$. This is precisely \begin{align*} \partial_t^2u - c^2\Delta u = 0 \quad \text{in } C([0,T];H^{-1}(U)). \end{align*}[/step]

[guided]We must be careful about what is being passed to the limit. Fix $K\in\mathbb{N}$ and let $v:U\to\mathbb{R}$ be a finite linear combination of $e_1,\dots,e_K$. For every $N\ge K$, the scalar oscillator equations imply \begin{align*} (\partial_t^2 u_N(t))(v) + c^2\int_U \nabla u_N(t,x)\cdot \nabla v(x)\,d\mathcal{L}^n(x) = 0. \end{align*} The convergence $\partial_t^2u_N\to\partial_t^2u$ in $C([0,T];H^{-1}(U))$ implies convergence of the first term because $v\in H^1_0(U)$ is fixed. The convergence $u_N\to u$ in $C([0,T];H^1_0(U))$ implies convergence of the second term because $w\mapsto\int_U\nabla w(x)\cdot\nabla v(x)\,d\mathcal{L}^n(x)$ is a continuous linear functional on $H^1_0(U)$. Therefore the weak identity holds for every finite eigenfunction combination. To extend the identity to an arbitrary $v\in H^1_0(U)$, use the density of finite eigenfunction spans in $H^1_0(U)$ supplied by the [Spectral Theorem for the Dirichlet Laplacian with Compact Resolvent](/theorems/6447). Choose $v_j$ in the finite span with $v_j\to v$ in $H^1_0(U)$. The $H^{-1}(U)$ action is continuous in the [test function](/page/Test%20Function), and the Dirichlet inner product is continuous in the test function, so passing $j\to\infty$ gives \begin{align*} (\partial_t^2 u(t))(v) + c^2\int_U \nabla u(t,x)\cdot \nabla v(x)\,d\mathcal{L}^n(x) = 0. \end{align*} This is exactly the weak formulation of $\partial_t^2u-c^2\Delta u=0$ in $C([0,T];H^{-1}(U))$.[/guided]

[step:Identify the initial data]At $t=0$, \begin{align*} u_N(0) = \sum_{k=1}^{N} a_k e_k \end{align*} and \begin{align*} \partial_tu_N(0) = \sum_{k=1}^{N} b_k e_k. \end{align*} By the spectral expansions of $u_0$ in $H^1_0(U)$ and $u_1$ in $L^2(U)$, \begin{align*} u_N(0) \to u_0 \quad \text{in } H^1_0(U) \end{align*} and \begin{align*} \partial_tu_N(0) \to u_1 \quad \text{in } L^2(U). \end{align*} Since $u_N \to u$ in $C([0,T];H^1_0(U))$ and $\partial_tu_N \to \partial_tu$ in $C([0,T];L^2(U))$, evaluating at $t=0$ gives \begin{align*} u(0)=u_0 \quad \text{in } H^1_0(U) \end{align*} and \begin{align*} \partial_tu(0)=u_1 \quad \text{in } L^2(U). \end{align*}[/step]

[guided]At time $t=0$, the scalar coefficients satisfy $q_k(0)=a_k$ and $q_k'(0)=b_k$. Therefore the finite-mode approximations have initial values \begin{align*} u_N(0)=\sum_{k=1}^{N}a_ke_k \end{align*} and \begin{align*} \partial_tu_N(0)=\sum_{k=1}^{N}b_ke_k. \end{align*} The [Parseval Identity](/theorems/248) gives convergence of the velocity expansion in $L^2(U)$, and the Dirichlet form identity from the [Spectral Theorem for the Dirichlet Laplacian with Compact Resolvent](/theorems/6447) gives convergence of the displacement expansion in $H^1_0(U)$. Hence \begin{align*} u_N(0)\to u_0 \quad \text{in } H^1_0(U) \end{align*} and \begin{align*} \partial_tu_N(0)\to u_1 \quad \text{in } L^2(U). \end{align*} On the other hand, the construction already proved $u_N\to u$ in $C([0,T];H^1_0(U))$ and $\partial_tu_N\to\partial_tu$ in $C([0,T];L^2(U))$. Evaluation at $t=0$ is continuous in these spaces, so the two limits must be \begin{align*} u(0)=u_0 \quad \text{in } H^1_0(U) \end{align*} and \begin{align*} \partial_tu(0)=u_1 \quad \text{in } L^2(U). \end{align*}[/guided]

[step:Sum the scalar energy identities to obtain conservation of energy]For each $N \in \mathbb{N}$ and each $t \in [0,T]$, summing the scalar energy identities for $k=1,\dots,N$ gives \begin{align*} \|\partial_t u_N(t)\|_{L^2(U)}^2 + c^2\|\nabla u_N(t)\|_{L^2(U)}^2 = \sum_{k=1}^{N}|b_k|^2 + c^2\sum_{k=1}^{N}\lambda_k |a_k|^2. \end{align*} The convergence $\partial_tu_N(t)\to \partial_tu(t)$ in $L^2(U)$ and $u_N(t)\to u(t)$ in $H^1_0(U)$ implies \begin{align*} \|\partial_t u_N(t)\|_{L^2(U)}^2 \to \|\partial_t u(t)\|_{L^2(U)}^2 \end{align*} and \begin{align*} \|\nabla u_N(t)\|_{L^2(U)}^2 \to \|\nabla u(t)\|_{L^2(U)}^2. \end{align*} The coefficient sums converge to the norms of the initial data: \begin{align*} \sum_{k=1}^{\infty}|b_k|^2 = \|u_1\|_{L^2(U)}^2 \end{align*} and \begin{align*} \sum_{k=1}^{\infty}\lambda_k |a_k|^2 = \|\nabla u_0\|_{L^2(U)}^2. \end{align*} Letting $N \to \infty$ yields \begin{align*} \|\partial_t u(t)\|_{L^2(U)}^2 + c^2\|\nabla u(t)\|_{L^2(U)}^2 = \|u_1\|_{L^2(U)}^2 + c^2\|\nabla u_0\|_{L^2(U)}^2. \end{align*}[/step]

[guided]The finite-dimensional energy identity is exact because every Fourier mode is an independent harmonic oscillator. For the $N$-mode approximation, \begin{align*} u_N(t)=\sum_{k=1}^{N} q_k(t)e_k, \end{align*} orthonormality in $L^2(U)$ gives \begin{align*} \|\partial_t u_N(t)\|_{L^2(U)}^2 = \sum_{k=1}^{N}|q_k'(t)|^2, \end{align*} and the Dirichlet form identity for eigenfunctions gives \begin{align*} \|\nabla u_N(t)\|_{L^2(U)}^2 = \sum_{k=1}^{N}\lambda_k |q_k(t)|^2. \end{align*} Therefore \begin{align*} \|\partial_t u_N(t)\|_{L^2(U)}^2 + c^2\|\nabla u_N(t)\|_{L^2(U)}^2 = \sum_{k=1}^{N}\bigl(|q_k'(t)|^2 + c^2\lambda_k |q_k(t)|^2\bigr). \end{align*} Each summand is conserved, so \begin{align*} \sum_{k=1}^{N}\bigl(|q_k'(t)|^2 + c^2\lambda_k |q_k(t)|^2\bigr) = \sum_{k=1}^{N}\bigl(|b_k|^2 + c^2\lambda_k |a_k|^2\bigr). \end{align*} Now we pass from finite sums to the full solution. This is legitimate because the convergence is strong in the exact spaces where the energy is measured: \begin{align*} \partial_tu_N(t) \to \partial_tu(t) \quad \text{in } L^2(U), \end{align*} and \begin{align*} u_N(t) \to u(t) \quad \text{in } H^1_0(U). \end{align*} Strong convergence in $L^2(U)$ gives convergence of the velocity norms, and strong convergence in $H^1_0(U)$ gives convergence of the gradient norms. On the initial side, [Parseval's identity](/theorems/434) and the Dirichlet form identity give \begin{align*} \sum_{k=1}^{\infty}|b_k|^2 = \|u_1\|_{L^2(U)}^2 \end{align*} and \begin{align*} \sum_{k=1}^{\infty}\lambda_k |a_k|^2 = \|\nabla u_0\|_{L^2(U)}^2. \end{align*} Taking the limit in the finite-dimensional identity therefore gives the full energy equality \begin{align*} \|\partial_t u(t)\|_{L^2(U)}^2 + c^2\|\nabla u(t)\|_{L^2(U)}^2 = \|u_1\|_{L^2(U)}^2 + c^2\|\nabla u_0\|_{L^2(U)}^2. \end{align*} The important point is that no lower semicontinuity loss occurs here: the spectral construction gives strong convergence in the energy topology.[/guided]

[step:Prove uniqueness by projecting onto every eigenfunction]Let \begin{align*} w: [0,T] \to H^1_0(U) \end{align*} be another weak solution with the same initial data, and define $z := u-w$. Then \begin{align*} z \in C([0,T];H^1_0(U)), \qquad \partial_tz \in C([0,T];L^2(U)), \qquad \partial_t^2z \in C([0,T];H^{-1}(U)), \end{align*} and \begin{align*} \partial_t^2z - c^2\Delta z = 0 \end{align*} with \begin{align*} z(0)=0, \qquad \partial_tz(0)=0. \end{align*} For each $k \in \mathbb{N}$, define \begin{align*} z_k: [0,T] \to \mathbb{R} \end{align*} by \begin{align*} z_k(t) := \int_U z(t,x)e_k(x)\,d\mathcal{L}^n(x). \end{align*} Testing the weak equation for $z$ against $e_k$ gives \begin{align*} z_k''(t) + c^2\lambda_k z_k(t)=0 \end{align*} in the sense of distributions on $(0,T)$. Since $z \in C([0,T];H^1_0(U))$ and $\partial_tz \in C([0,T];L^2(U))$, the coefficient $z_k$ belongs to $C^1([0,T])$, and the differential equation implies $z_k \in C^2([0,T])$. The initial conditions give \begin{align*} z_k(0)=0, \qquad z_k'(0)=0. \end{align*} The [Picard-Lindelöf Existence and Uniqueness Theorem](/theorems/2774), applied to the equivalent first-order system $Y_1'=Y_2$ and $Y_2'=-c^2\lambda_kY_1$ with initial data $Y_1(0)=0$ and $Y_2(0)=0$, gives $z_k(t)=0$ for every $t \in [0,T]$. Thus every $L^2(U)$ Fourier coefficient of $z(t)$ is zero for every $t \in [0,T]$. Since $(e_k)_{k \in \mathbb{N}}$ is an orthonormal basis of $L^2(U)$, it follows that \begin{align*} z(t)=0 \quad \text{in } L^2(U) \end{align*} for every $t \in [0,T]$. Because $z(t)\in H^1_0(U)$, this also identifies the same function as zero in the energy-space representative. Hence $u=w$, and the weak solution is unique.[/step]

[guided]Let $w:[0,T]\to H^1_0(U)$ be another weak solution with the same initial data, and set $z:=u-w$. Linearity of the [wave equation](/page/Wave%20Equation) gives \begin{align*} \partial_t^2z-c^2\Delta z=0 \end{align*} with \begin{align*} z(0)=0, \qquad \partial_tz(0)=0. \end{align*} For each $k\in\mathbb{N}$, define the Fourier coefficient \begin{align*} z_k:[0,T]\to\mathbb{R}, \qquad z_k(t):=\int_U z(t,x)e_k(x)\,d\mathcal{L}^n(x). \end{align*} Testing the weak equation against $e_k$ gives \begin{align*} z_k''(t)+c^2\lambda_kz_k(t)=0 \end{align*} in the sense of distributions on $(0,T)$. Since $z\in C([0,T];H^1_0(U))$ and $\partial_tz\in C([0,T];L^2(U))$, the coefficient $z_k$ lies in $C^1([0,T])$; the differential equation then implies $z_k\in C^2([0,T])$. The initial conditions give \begin{align*} z_k(0)=0, \qquad z_k'(0)=0. \end{align*} By the [Picard-Lindelöf Existence and Uniqueness Theorem](/theorems/2774), applied to the first-order system $Y_1'=Y_2$ and $Y_2'=-c^2\lambda_kY_1$, the only solution with these initial data is $Y_1=0$ and $Y_2=0$. Thus $z_k(t)=0$ for every $t\in[0,T]$ and every $k\in\mathbb{N}$. For each fixed $t$, all $L^2(U)$ Fourier coefficients of $z(t)$ vanish. Since $(e_k)_{k\in\mathbb{N}}$ is an orthonormal basis of $L^2(U)$, the [Parseval Identity](/theorems/248) gives $\|z(t)\|_{L^2(U)}=0$. Hence $z(t)=0$ in $L^2(U)$ for every $t\in[0,T]$, and because $z(t)\in H^1_0(U)$ this is the same energy-space representative. Therefore $u=w$, proving uniqueness.[/guided]