[proofplan] Multiply the damped [wave equation](/page/Wave%20Equation) by $u_t$ and integrate over $U$. The Laplacian term is converted to the time derivative of the Dirichlet energy by [integration by parts](/theorems/210), using the homogeneous boundary condition to remove the boundary contribution. The damping term remains with a negative sign and gives the dissipation identity. Integrating the differential identity in time gives the stated balance law. [/proofplan] [step:Differentiate the kinetic and potential energy terms] By the assumed regularity, $E$ is differentiable and \begin{align*} \frac{d}{dt}\frac12\|u_t(t)\|_{L^2(U)}^2=\int_U u_{tt}(t,x)u_t(t,x)\,d\mathcal L^n(x). \end{align*} Similarly, \begin{align*} \frac{d}{dt}\frac12\|\nabla u(t)\|_{L^2(U)}^2=\int_U \nabla u(t,x)\cdot\nabla u_t(t,x)\,d\mathcal L^n(x). \end{align*} Since $u(t,\cdot)$ has zero trace on $\partial U$, differentiating the boundary condition in time gives $u_t(t,\cdot)=0$ on $\partial U$ in the trace sense. Green's identity therefore gives \begin{align*} \int_U \nabla u(t,x)\cdot\nabla u_t(t,x)\,d\mathcal L^n(x)=-\int_U \Delta u(t,x)u_t(t,x)\,d\mathcal L^n(x). \end{align*} Thus \begin{align*} E'(t)=\int_U \left(u_{tt}(t,x)-\Delta u(t,x)\right)u_t(t,x)\,d\mathcal L^n(x). \end{align*} [guided] The energy has two parts. The kinetic part differentiates by the chain rule in the [Hilbert space](/page/Hilbert%20Space) $L^2(U)$: \begin{align*} \frac{d}{dt}\frac12\|u_t(t)\|_{L^2(U)}^2=\int_U u_{tt}(t,x)u_t(t,x)\,d\mathcal L^n(x). \end{align*} The potential part is the Dirichlet energy. Its derivative is \begin{align*} \frac{d}{dt}\frac12\|\nabla u(t)\|_{L^2(U)}^2=\int_U \nabla u(t,x)\cdot\nabla u_t(t,x)\,d\mathcal L^n(x). \end{align*} To compare this with the equation, move the derivative from $u_t$ onto $u$ by [integration by parts](/theorems/2098). The boundary term is zero because the boundary condition $u=0$ on $\partial U$ implies $u_t=0$ there. Hence \begin{align*} \int_U \nabla u(t,x)\cdot\nabla u_t(t,x)\,d\mathcal L^n(x)=-\int_U \Delta u(t,x)u_t(t,x)\,d\mathcal L^n(x). \end{align*} Adding the kinetic and potential derivatives gives \begin{align*} E'(t)=\int_U \left(u_{tt}(t,x)-\Delta u(t,x)\right)u_t(t,x)\,d\mathcal L^n(x). \end{align*} [/guided] [/step] [step:Use the equation to identify the dissipation] The differential equation gives \begin{align*} u_{tt}(t,x)-\Delta u(t,x)=-a u_t(t,x) \end{align*} for $(t,x)\in(0,\infty)\times U$. Substituting this into the expression for $E'(t)$ gives \begin{align*} E'(t)=-a\int_U |u_t(t,x)|^2\,d\mathcal L^n(x). \end{align*} By definition of the $L^2(U)$ norm, \begin{align*} \int_U |u_t(t,x)|^2\,d\mathcal L^n(x)=\|u_t(t)\|_{L^2(U)}^2. \end{align*} Therefore \begin{align*} E'(t)=-a\|u_t(t)\|_{L^2(U)}^2. \end{align*} Because $a>0$, the right-hand side is non-positive, so $E$ is non-increasing. [/step] [step:Integrate the differential identity] Integrating the identity for $E'$ over $(0,t)$ with respect to $\mathcal L^1$ gives \begin{align*} E(t)-E(0)=-a\int_0^t\|u_s(s)\|_{L^2(U)}^2\,d\mathcal L^1(s). \end{align*} Rearranging yields \begin{align*} E(t)+a\int_0^t\|u_s(s)\|_{L^2(U)}^2\,d\mathcal L^1(s)=E(0). \end{align*} This is the asserted energy balance. [/step]