[proofplan] We first prove the estimate for initial data in $H^1_0(U)$ by differentiating the $L^2$ energy along the Dirichlet heat flow. The boundary condition removes the boundary term, and the variational characterization of $\lambda_1$ converts the gradient dissipation into a multiple of the $L^2$ energy. Gronwall's inequality gives exponential decay for regular data, and density plus strong continuity of the Dirichlet heat semigroup extends the estimate to arbitrary $L^2$ data. [/proofplan]

[step:Use the Dirichlet energy identity for $H^1_0(U)$ initial data]Assume first that $u_0 \in H^1_0(U)$. Let $\mathcal{L}^n$ denote $n$-dimensional [Lebesgue measure](/page/Lebesgue%20Measure) on $\mathbb{R}^n$, and let $H^{-1}(U) := (H^1_0(U))^*$ denote the dual [Sobolev space](/page/Sobolev%20Space), whose elements act on $H^1_0(U)$ by functional evaluation. Let $u: [0,\infty) \to L^2(U)$ be the map defined by \begin{align*} u(t) = e^{t\Delta_D}u_0. \end{align*} This is the corresponding Dirichlet heat flow. For each $t>0$, the function $u(t)$ belongs to $H^1_0(U)$ and satisfies the homogeneous [heat equation](/page/Heat%20Equation) \begin{align*} \partial_t u(t) = \Delta_D u(t) \end{align*} in the usual weak sense. Define the energy function $E: [0,\infty) \to [0,\infty)$ by \begin{align*} E(t) = \|u(t)\|_{L^2(U)}^2. \end{align*} The Dirichlet energy identity gives, for almost every $t>0$, \begin{align*} \frac{1}{2}E'(t) + \int_U |\nabla u(t,x)|^2\,d\mathcal{L}^n(x) = 0. \end{align*} Equivalently, \begin{align*} E'(t) = -2\|\nabla u(t)\|_{L^2(U)}^2. \end{align*}[/step]

[guided]We begin with the case where the initial datum has one [weak derivative](/page/Weak%20Derivative) and zero trace, because then the standard energy computation is justified directly. Let $\mathcal{L}^n$ denote $n$-dimensional Lebesgue measure on $\mathbb{R}^n$, and let $H^{-1}(U) := (H^1_0(U))^*$ denote the dual Sobolev space, whose elements act on $H^1_0(U)$ by functional evaluation. Let $u_0 \in H^1_0(U)$, and define $u: [0,\infty) \to L^2(U)$ by \begin{align*} u(t) = e^{t\Delta_D}u_0. \end{align*} The operator $-\Delta_D$ is the non-negative self-adjoint operator associated with the closed Dirichlet form \begin{align*} a(v,w)=\int_U \nabla v(x)\cdot \nabla w(x)\,d\mathcal{L}^n(x) \end{align*} on the form domain $H^1_0(U)$. Since $u_0 \in H^1_0(U)$, the semigroup form identity gives $u(t) \in H^1_0(U)$ for $t>0$ and permits testing the weak heat equation against $u(t)$ for almost every $t>0$. This is the form-domain justification for the energy computation, replacing any need to assume classical smoothness. The zero Dirichlet trace is encoded in the form domain $H^1_0(U)$, so the boundary contribution in the weak Green identity vanishes. Define $E: [0,\infty) \to [0,\infty)$ by \begin{align*} E(t) = \|u(t)\|_{L^2(U)}^2. \end{align*} For an energy solution generated by the closed Dirichlet form, the map $E$ is absolutely continuous on compact subintervals of $(0,\infty)$, and for almost every $t>0$ its derivative is obtained by testing the weak equation against $u(t)$. Thus \begin{align*} \frac{1}{2}E'(t) = \int_U u(t,x)\,\partial_t u(t,x)\,d\mathcal{L}^n(x). \end{align*} Since the weak heat equation is $\partial_t u(t)=\Delta_D u(t)$ in $H^{-1}(U)$, the preceding identity becomes \begin{align*} \frac{1}{2}E'(t) = \Delta_D u(t)(u(t)). \end{align*} Here $\Delta_D u(t)$ is being evaluated as an element of $H^{-1}(U)$ on the [test function](/page/Test%20Function) $u(t) \in H^1_0(U)$. By the definition of the weak Dirichlet Laplacian associated with the Dirichlet form, \begin{align*} \Delta_D u(t)(u(t)) = -\int_U |\nabla u(t,x)|^2\,d\mathcal{L}^n(x). \end{align*} Therefore, for almost every $t>0$, \begin{align*} \frac{1}{2}E'(t) + \int_U |\nabla u(t,x)|^2\,d\mathcal{L}^n(x) = 0. \end{align*} This is the analytic heart of the argument: the heat equation dissipates exactly the Dirichlet energy.[/guided]

[step:Convert the energy identity into a scalar differential inequality]By the variational characterization of $\lambda_1$, every $v \in H^1_0(U)$ satisfies \begin{align*} \int_U |\nabla v(x)|^2\,d\mathcal{L}^n(x) \geq \lambda_1\int_U |v(x)|^2\,d\mathcal{L}^n(x). \end{align*} Applying this with $v = u(t)$, which belongs to $H^1_0(U)$ for $t>0$, gives \begin{align*} \|\nabla u(t)\|_{L^2(U)}^2 \geq \lambda_1\|u(t)\|_{L^2(U)}^2 = \lambda_1 E(t). \end{align*} Substituting this bound into the energy identity yields, for almost every $t>0$, \begin{align*} E'(t) \leq -2\lambda_1 E(t). \end{align*}[/step]

[guided]The next task is to convert the energy identity into an inequality involving only the scalar energy $E(t)$. The bridge is the variational characterization of the first Dirichlet eigenvalue: every $v \in H^1_0(U)$ satisfies \begin{align*} \int_U |\nabla v(x)|^2\,d\mathcal{L}^n(x) \geq \lambda_1\int_U |v(x)|^2\,d\mathcal{L}^n(x). \end{align*} This characterization applies to $v=u(t)$ because the semigroup solution satisfies $u(t) \in H^1_0(U)$ for every $t>0$ in the form-domain argument above. Therefore \begin{align*} \|\nabla u(t)\|_{L^2(U)}^2 \geq \lambda_1\|u(t)\|_{L^2(U)}^2 = \lambda_1 E(t). \end{align*} Substituting this lower bound for the gradient dissipation into \begin{align*} E'(t) = -2\|\nabla u(t)\|_{L^2(U)}^2 \end{align*} gives, for almost every $t>0$, \begin{align*} E'(t) \leq -2\lambda_1 E(t). \end{align*} This is the point where the spectral gap $\lambda_1>0$ is used; it turns dissipation of gradient energy into decay of the $L^2$ norm itself.[/guided]

[step:Apply Gronwall to obtain exponential decay for $H^1_0(U)$ data]Define $F: [0,\infty) \to [0,\infty)$ by \begin{align*} F(t) = e^{2\lambda_1 t}E(t). \end{align*} Since $E'(t) \leq -2\lambda_1 E(t)$ for almost every $t>0$, we have \begin{align*} F'(t) = e^{2\lambda_1 t}\bigl(E'(t)+2\lambda_1 E(t)\bigr) \leq 0 \end{align*} for almost every $t>0$. Hence $F$ is non-increasing on $(0,\infty)$. Since the semigroup $e^{t\Delta_D}$ is strongly continuous on $L^2(U)$ at $t=0$, we have $E(t) \to E(0)$ as $t \downarrow 0$, and therefore $F(t) \to F(0)$ as $t \downarrow 0$. It follows that $F(t) \leq F(0)$ for every $t \geq 0$. Therefore \begin{align*} e^{2\lambda_1 t}\|u(t)\|_{L^2(U)}^2 \leq \|u_0\|_{L^2(U)}^2. \end{align*} Taking square roots gives \begin{align*} \|u(t)\|_{L^2(U)} \leq e^{-\lambda_1 t}\|u_0\|_{L^2(U)} \end{align*} for every $t \geq 0$ whenever $u_0 \in H^1_0(U)$.[/step]

[guided]The scalar inequality from the previous step has the form \begin{align*} E'(t) \leq -2\lambda_1 E(t) \end{align*} for almost every $t>0$. To remove the negative zeroth-order term, define $F: [0,\infty) \to [0,\infty)$ by \begin{align*} F(t) = e^{2\lambda_1 t}E(t). \end{align*} Because $E$ is absolutely continuous on compact subintervals of $(0,\infty)$, the product rule holds for almost every $t>0$, and gives \begin{align*} F'(t) = e^{2\lambda_1 t}\bigl(E'(t)+2\lambda_1 E(t)\bigr) \leq 0. \end{align*} Thus $F$ is non-increasing on $(0,\infty)$. The endpoint $t=0$ requires a limiting argument rather than a direct appeal to the almost-everywhere derivative statement. Strong continuity of the Dirichlet heat semigroup on $L^2(U)$ gives $u(t) \to u_0$ in $L^2(U)$ as $t \downarrow 0$, hence $E(t) \to E(0)$ and $F(t) \to F(0)$. Therefore $F(t) \leq F(0)$ for every $t \geq 0$, which means \begin{align*} e^{2\lambda_1 t}\|u(t)\|_{L^2(U)}^2 \leq \|u_0\|_{L^2(U)}^2. \end{align*} Taking square roots is legitimate because both sides are non-negative, and yields \begin{align*} \|u(t)\|_{L^2(U)} \leq e^{-\lambda_1 t}\|u_0\|_{L^2(U)}. \end{align*} This proves the theorem for initial data $u_0 \in H^1_0(U)$.[/guided]

[step:Extend the estimate from $H^1_0(U)$ to all of $L^2(U)$]Let $u_0 \in L^2(U)$. Let $C_c^\infty(U)$ denote the space of smooth real-valued functions on $U$ with compact support contained in $U$. Since $U$ is open, $C_c^\infty(U)$ is dense in $L^2(U)$, and since $C_c^\infty(U) \subset H^1_0(U)$, there exists a sequence $(u_{0,k})_{k=1}^{\infty}$ in $H^1_0(U)$ such that \begin{align*} \lim_{k \to \infty}\|u_{0,k}-u_0\|_{L^2(U)} = 0. \end{align*} For each $k \in \mathbb{N}$, define $u_k: [0,\infty) \to L^2(U)$ by \begin{align*} u_k(t) = e^{t\Delta_D}u_{0,k}. \end{align*} The estimate already proved gives, for every $t \geq 0$, \begin{align*} \|u_k(t)\|_{L^2(U)} \leq e^{-\lambda_1 t}\|u_{0,k}\|_{L^2(U)}. \end{align*} For each fixed $t \geq 0$, the operator $e^{t\Delta_D}:L^2(U)\to L^2(U)$ is bounded and linear, so convergence of $u_{0,k}$ to $u_0$ in $L^2(U)$ gives \begin{align*} \lim_{k \to \infty}\|e^{t\Delta_D}u_{0,k}-e^{t\Delta_D}u_0\|_{L^2(U)} = 0. \end{align*} Passing to the limit in the preceding inequality gives \begin{align*} \|e^{t\Delta_D}u_0\|_{L^2(U)} \leq e^{-\lambda_1 t}\|u_0\|_{L^2(U)}. \end{align*} Since $u(t)=e^{t\Delta_D}u_0$, this is precisely \begin{align*} \|u(t)\|_{L^2(U)} \leq e^{-\lambda_1 t}\|u_0\|_{L^2(U)}. \end{align*} This completes the proof for all $u_0 \in L^2(U)$.[/step]

[guided]The estimate has been proved for initial data in $H^1_0(U)$. To pass to arbitrary $L^2(U)$ data, we approximate by better initial data and use continuity of the semigroup. Let $u_0 \in L^2(U)$. Let $C_c^\infty(U)$ denote the space of smooth real-valued functions on $U$ with compact support contained in $U$. Since $U$ is open, $C_c^\infty(U)$ is dense in $L^2(U)$, and because $C_c^\infty(U) \subset H^1_0(U)$, there exists a sequence $(u_{0,k})_{k=1}^{\infty}$ in $H^1_0(U)$ such that \begin{align*} \lim_{k \to \infty}\|u_{0,k}-u_0\|_{L^2(U)} = 0. \end{align*} For each $k \in \mathbb{N}$, define $u_k: [0,\infty) \to L^2(U)$ by \begin{align*} u_k(t) = e^{t\Delta_D}u_{0,k}. \end{align*} The $H^1_0(U)$ case gives \begin{align*} \|u_k(t)\|_{L^2(U)} \leq e^{-\lambda_1 t}\|u_{0,k}\|_{L^2(U)} \end{align*} for every $t \geq 0$. Fix $t \geq 0$. The semigroup operator $e^{t\Delta_D}:L^2(U)\to L^2(U)$ is bounded and linear, so convergence $u_{0,k} \to u_0$ in $L^2(U)$ implies \begin{align*} \lim_{k \to \infty}\|e^{t\Delta_D}u_{0,k}-e^{t\Delta_D}u_0\|_{L^2(U)} = 0. \end{align*} Also $\|u_{0,k}\|_{L^2(U)} \to \|u_0\|_{L^2(U)}$ by the [reverse triangle inequality](/theorems/2300). Passing to the limit in the inequality for $u_k(t)$ gives \begin{align*} \|e^{t\Delta_D}u_0\|_{L^2(U)} \leq e^{-\lambda_1 t}\|u_0\|_{L^2(U)}. \end{align*} Since $u(t)=e^{t\Delta_D}u_0$, this is exactly \begin{align*} \|u(t)\|_{L^2(U)} \leq e^{-\lambda_1 t}\|u_0\|_{L^2(U)}. \end{align*} Thus the decay estimate holds for every $u_0 \in L^2(U)$.[/guided]