[proofplan] The plain energy is decreasing, but its derivative controls only $\|u_t\|_{L^2}^2$. Add a small multiple of the cross term $(u,u_t)_{L^2}$ and a matching multiple of $\|u\|_{L^2}^2$ to build a Lyapunov functional equivalent to the energy. The derivative of this modified energy controls both $\|u_t\|_{L^2}^2$ and $\|\nabla u\|_{L^2}^2$. Equivalence and Gronwall's inequality then give exponential decay of $E$. [/proofplan] [step:Record the basic energy dissipation identity] As in the standard damped-wave energy computation, testing the equation with $u_t(t,\cdot)$ and integrating over $U$ gives \begin{align*} E'(t)=-a\|u_t(t)\|_{L^2(U)}^2. \end{align*} Here the [integration by parts](/theorems/210) has no boundary contribution because $u(t,\cdot)\in H_0^1(U)$ and the boundary condition is homogeneous. [/step] [step:Define a modified energy equivalent to $E$] For a parameter $\delta>0$ to be chosen, define $F:[0,\infty)\to\mathbb R$ by \begin{align*} F(t):=E(t)+\delta (u(t),u_t(t))_{L^2(U)}+\frac{\delta a}{2}\|u(t)\|_{L^2(U)}^2. \end{align*} The Poincare inequality and [Young's inequality](/theorems/244) give \begin{align*} |(u(t),u_t(t))_{L^2(U)}|\le \|u(t)\|_{L^2(U)}\|u_t(t)\|_{L^2(U)}\le C_P\|\nabla u(t)\|_{L^2(U)}\|u_t(t)\|_{L^2(U)}. \end{align*} Applying Young's inequality to the last product shows that there is a constant $K>0$, depending only on $a$ and $C_P$, such that \begin{align*} \left|(u(t),u_t(t))_{L^2(U)}+\frac{a}{2}\|u(t)\|_{L^2(U)}^2\right|\le K E(t). \end{align*} Choose $\delta>0$ so small that $\delta K\le \frac12$. Then \begin{align*} \frac12 E(t)\le F(t)\le \frac32 E(t) \end{align*} for every $t\ge0$. [guided] The extra term must be small enough not to change the size of the energy. Poincare's inequality converts $\|u\|_{L^2(U)}$ into $\|\nabla u\|_{L^2(U)}$, and Young's inequality converts the product $\|\nabla u\|_{L^2(U)}\|u_t\|_{L^2(U)}$ into a sum of squares. Thus \begin{align*} |(u(t),u_t(t))_{L^2(U)}|\le C_P\|\nabla u(t)\|_{L^2(U)}\|u_t(t)\|_{L^2(U)}\le K_1 E(t) \end{align*} for a constant $K_1>0$. Also \begin{align*} \frac{a}{2}\|u(t)\|_{L^2(U)}^2\le \frac{aC_P^2}{2}\|\nabla u(t)\|_{L^2(U)}^2\le K_2E(t). \end{align*} With $K:=K_1+K_2$, the perturbation has absolute value at most $\delta K E(t)$. Choosing $\delta K\le 1/2$ gives \begin{align*} \frac12E(t)\le F(t)\le \frac32E(t). \end{align*} So proving exponential decay for $F$ is equivalent to proving it for $E$. [/guided] [/step] [step:Show the modified energy has a coercive negative derivative] Differentiate the cross term: \begin{align*} \frac{d}{dt}(u(t),u_t(t))_{L^2(U)}=\|u_t(t)\|_{L^2(U)}^2+(u(t),u_{tt}(t))_{L^2(U)}. \end{align*} Using $u_{tt}=\Delta u-a u_t$ and [integration by parts](/theorems/2098), \begin{align*} (u(t),u_{tt}(t))_{L^2(U)}=-\|\nabla u(t)\|_{L^2(U)}^2-a(u(t),u_t(t))_{L^2(U)}. \end{align*} Therefore \begin{align*} \frac{d}{dt}\left((u(t),u_t(t))_{L^2(U)}+\frac{a}{2}\|u(t)\|_{L^2(U)}^2\right)=\|u_t(t)\|_{L^2(U)}^2-\|\nabla u(t)\|_{L^2(U)}^2. \end{align*} Combining this identity with $E'(t)=-a\|u_t(t)\|_{L^2(U)}^2$ gives \begin{align*} F'(t)=-(a-\delta)\|u_t(t)\|_{L^2(U)}^2-\delta\|\nabla u(t)\|_{L^2(U)}^2. \end{align*} Choose $\delta<a$. Then, with \begin{align*} \mu:=\min\{a-\delta,\delta\}>0, \end{align*} we have \begin{align*} F'(t)\le -\mu\left(\|u_t(t)\|_{L^2(U)}^2+\|\nabla u(t)\|_{L^2(U)}^2\right)=-2\mu E(t). \end{align*} Since $F(t)\le \frac32E(t)$, equivalently $E(t)\ge \frac23F(t)$, this gives \begin{align*} F'(t)\le -\frac{4\mu}{3}F(t). \end{align*} Set $\gamma:=\frac{4\mu}{3}$. [/step] [step:Apply Gronwall's inequality] The differential inequality for $F$ implies \begin{align*} F(t)\le e^{-\gamma t}F(0) \end{align*} for every $t\ge0$. Using the equivalence between $F$ and $E$, \begin{align*} E(t)\le 2F(t)\le 2e^{-\gamma t}F(0)\le 3e^{-\gamma t}E(0). \end{align*} Thus the result holds with $C=3$ and with the positive constant $\gamma$ chosen above. [/step]