[proofplan] We apply the [Lumer-Phillips Theorem](/theorems/3140) to both $A$ and $-A$. Skew-adjointness of $A$ on $X$ — that is, $(A\Phi, \Phi)_X + (\Phi, A\Phi)_X = 0$ for all $\Phi \in D(A)$ — yields dissipativity of both $A$ and $-A$ via Green's identity, in fact equality with zero real part. Range surjectivity reduces, after eliminating $v$, to solving the elliptic problem $u - \Delta u = h \in L^2(U)$ in $H^2 \cap H^1_0$, which holds by Lax-Milgram and elliptic regularity. The two contraction semigroups $\{T_+(t)\}_{t \ge 0}$ generated by $A$ and $\{T_-(t)\}_{t \ge 0}$ generated by $-A$ paste into a group $T(t)$ on $\mathbb{R}$, and the energy identity forces $\|T(t)\Phi\|_X = \|\Phi\|_X$, i.e. each $T(t)$ is an isometry. [/proofplan] [step:Verify density of $D(A)$ in $X$] The space $C_c^\infty(U) \times C_c^\infty(U)$ is dense in $X = H^1_0(U) \times L^2(U)$: this is the definition of $H^1_0(U)$ (closure of $C_c^\infty(U)$ in the $H^1$ norm) and the standard density of test functions in $L^2(U)$. Since $C_c^\infty(U) \subset H^2(U) \cap H^1_0(U)$ and $C_c^\infty(U) \subset H^1_0(U)$, we have $C_c^\infty(U) \times C_c^\infty(U) \subset D(A)$. Therefore $D(A)$ contains a dense subspace, hence is dense in $X$. [/step] [step:Compute $\operatorname{Re}(A\Phi, \Phi)_X = 0$ for all $\Phi \in D(A)$ via Green's identity] Let $\Phi = (u, v) \in D(A)$, so $u \in H^2(U) \cap H^1_0(U)$ and $v \in H^1_0(U)$. By definition of $A$ and the energy inner product: \begin{align*} (A \Phi, \Phi)_X &= \left( \begin{pmatrix} v \\ \Delta u \end{pmatrix}, \begin{pmatrix} u \\ v \end{pmatrix} \right)_X = \int_U \nabla v \cdot \overline{\nabla u}\, d\mathcal{L}^n + \int_U \Delta u \cdot \bar{v}\, d\mathcal{L}^n. \end{align*} For the second integral, we apply Green's First Identity on $U$. Since $u \in H^2(U)$ and $v \in H^1_0(U)$, the boundary integral vanishes (the trace of $v$ is zero on $\partial U$): \begin{align*} \int_U \Delta u \cdot \bar{v}\, d\mathcal{L}^n &= -\int_U \nabla u \cdot \overline{\nabla v}\, d\mathcal{L}^n + \int_{\partial U} \bar{v}\, \partial_\nu u\, d\mathcal{H}^{n-1} = -\int_U \nabla u \cdot \overline{\nabla v}\, d\mathcal{L}^n. \end{align*} Therefore \begin{align*} (A \Phi, \Phi)_X &= \int_U \nabla v \cdot \overline{\nabla u}\, d\mathcal{L}^n - \int_U \nabla u \cdot \overline{\nabla v}\, d\mathcal{L}^n. \end{align*} Writing $\int_U \nabla v \cdot \overline{\nabla u}\, d\mathcal{L}^n = \overline{\int_U \nabla u \cdot \overline{\nabla v}\, d\mathcal{L}^n}$, the right-hand side is of the form $\bar{z} - z = -2 i \operatorname{Im}(z)$, which is purely imaginary. Hence \begin{align*} \operatorname{Re}(A \Phi, \Phi)_X &= 0. \end{align*} The same identity with $-A$ gives $\operatorname{Re}(-A \Phi, \Phi)_X = 0$. In particular, both $A$ and $-A$ are dissipative on $X$. [/step] [step:Show $I - A: D(A) \to X$ and $I + A: D(A) \to X$ are surjective] Fix $(f, h) \in X = H^1_0(U) \times L^2(U)$. We seek $(u, v) \in D(A)$ with $(I - A)(u, v) = (f, h)$, i.e. \begin{align*} u - v &= f, & v - \Delta u &= h. \end{align*} Eliminating $v = u - f$ and substituting into the second equation: \begin{align*} u - \Delta u &= h + f =: w \in L^2(U). \end{align*} Since $f \in H^1_0(U) \subset L^2(U)$ and $h \in L^2(U)$, we have $w \in L^2(U)$. The bilinear form \begin{align*} B: H^1_0(U) \times H^1_0(U) &\to \mathbb{C}, & B(u, \varphi) &= \int_U u\, \bar{\varphi}\, d\mathcal{L}^n + \int_U \nabla u \cdot \overline{\nabla \varphi}\, d\mathcal{L}^n, \end{align*} is continuous (Cauchy-Schwarz) and coercive ($B(u, u) = \|u\|_{H^1}^2$). The map $\varphi \mapsto \int_U w \bar\varphi\, d\mathcal{L}^n$ is a bounded conjugate-linear functional on $H^1_0(U)$. By the Lax-Milgram Theorem, there exists a unique $u \in H^1_0(U)$ such that $B(u, \varphi) = \int_U w \bar\varphi\, d\mathcal{L}^n$ for all $\varphi \in H^1_0(U)$. This $u$ is the weak solution of $u - \Delta u = w$ with zero boundary data. By interior and boundary elliptic regularity for the Laplacian on a smooth bounded domain, $w \in L^2(U)$ implies $u \in H^2(U) \cap H^1_0(U)$, with the equation $u - \Delta u = w$ holding in $L^2(U)$. Setting $v := u - f$, we have $v \in H^1_0(U)$ (since both $u, f \in H^1_0(U)$) and the second equation $v - \Delta u = h$ is satisfied. Therefore $(u, v) \in D(A)$ and $(I - A)(u, v) = (f, h)$. This proves $I - A$ is surjective. The argument for $I + A$ is identical with the sign of $v$ flipped: $u + v = f$, $v + \Delta u = h$ leads to the same elliptic equation $u - \Delta u = f - h \in L^2(U)$ (from $v = f - u$ substituted into $v + \Delta u = h$, which gives $f - u + \Delta u = h$, i.e. $u - \Delta u = f - h$), with $w := f - h \in L^2(U)$. The same Lax-Milgram and elliptic regularity argument gives $u \in H^2(U) \cap H^1_0(U)$, then $v := f - u \in H^1_0(U)$, so $(u, v) \in D(A)$ and $(I + A)(u, v) = (f, h)$. [/step] [step:Apply Lumer-Phillips to $A$ and to $-A$] We have shown $A$ is densely defined on the Hilbert space $X$, $\operatorname{Re}(A \Phi, \Phi)_X \le 0$ for all $\Phi \in D(A)$ (in fact equals zero), and $\lambda I - A$ is surjective for $\lambda = 1 > 0$. By the [Lumer-Phillips Theorem](/theorems/3140), $A$ generates a $C_0$-semigroup of contractions $\{T_+(t)\}_{t \ge 0}$ on $X$: \begin{align*} \|T_+(t)\|_{\mathcal{L}(X)} &\le 1 \quad \text{for all } t \ge 0. \end{align*} The same theorem applied to $-A$ — which is also densely defined, dissipative ($\operatorname{Re}(-A\Phi, \Phi)_X = 0 \le 0$), and has $I - (-A) = I + A$ surjective — yields a $C_0$-semigroup of contractions $\{T_-(t)\}_{t \ge 0}$ on $X$ generated by $-A$: \begin{align*} \|T_-(t)\|_{\mathcal{L}(X)} &\le 1 \quad \text{for all } t \ge 0. \end{align*} [/step] [step:Paste $T_+$ and $T_-$ into a $C_0$-group $\{T(t)\}_{t \in \mathbb{R}}$] [claim:For every $t \ge 0$, the operators $T_+(t)$ and $T_-(t)$ are mutually inverse: $T_+(t) T_-(t) = T_-(t) T_+(t) = I$.] [/claim] [proof] Fix $\Phi \in D(A)$ and $t \ge 0$. By the differentiation formula from the [Closure and Density of the $C_0$-Semigroup Generator](/theorems/3144) applied to both semigroups (noting $D(A) = D(-A)$, since $-A$ has the same domain), the function \begin{align*} \psi: [0, t] &\to X, & s &\mapsto T_+(s) T_-(s) \Phi \end{align*} is differentiable with \begin{align*} \psi'(s) &= T_+(s)\, A\, T_-(s) \Phi + T_+(s)\, (-A)\, T_-(s) \Phi = 0. \end{align*} Hence $\psi$ is constant, and $\psi(t) = \psi(0) = \Phi$, so $T_+(t) T_-(t) \Phi = \Phi$ for all $\Phi \in D(A)$. Since $D(A)$ is dense in $X$ and $T_+(t) T_-(t) \in \mathcal{L}(X)$ is bounded, the identity extends to all of $X$: $T_+(t) T_-(t) = I$. By the same argument with the order swapped, $T_-(t) T_+(t) = I$. [/proof] Define \begin{align*} T: \mathbb{R} &\to \mathcal{L}(X), & T(t) &:= \begin{cases} T_+(t) & t \ge 0, \\ T_-(-t) & t < 0. \end{cases} \end{align*} We verify the group properties. - $T(0) = T_+(0) = I$. - For $s, t \ge 0$: $T(s) T(t) = T_+(s) T_+(t) = T_+(s + t) = T(s + t)$ by the semigroup law for $T_+$. - For $s, t \le 0$: same argument with $T_-$. - For $s \ge 0$, $t < 0$ with $s + t \ge 0$: $T(s) T(t) = T_+(s) T_-(-t)$. Using $T_+(s) = T_+(s + t) T_+(-t)$ (semigroup law, valid since $-t \ge 0$) and the previous claim $T_+(-t) T_-(-t) = I$: \begin{align*} T(s) T(t) &= T_+(s + t)\, T_+(-t)\, T_-(-t) = T_+(s + t)\, I = T(s + t). \end{align*} - The remaining sign cases are symmetric. Hence $T(s + t) = T(s) T(t)$ for all $s, t \in \mathbb{R}$. Strong continuity on $[0, \infty)$ holds for $T_+$ and on $(-\infty, 0]$ for $t \mapsto T_-(-t)$; continuity at $0$ from both sides matches at $T(0) = I$. So $\{T(t)\}_{t \in \mathbb{R}}$ is a strongly continuous group. [/step] [step:Conclude unitarity (energy preservation)] Each $T(t)$ is contractive ($\|T(t)\|_{\mathcal{L}(X)} \le 1$) and invertible with contractive inverse $T(-t)$: \begin{align*} \|T(t)\|_{\mathcal{L}(X)} &\le 1, & \|T(-t)\|_{\mathcal{L}(X)} &\le 1. \end{align*} For any $\Phi \in X$, \begin{align*} \|\Phi\|_X &= \|T(-t) T(t) \Phi\|_X \le \|T(-t)\|_{\mathcal{L}(X)}\, \|T(t) \Phi\|_X \le \|T(t) \Phi\|_X \le \|T(t)\|_{\mathcal{L}(X)}\, \|\Phi\|_X \le \|\Phi\|_X. \end{align*} All inequalities are equalities, so $\|T(t) \Phi\|_X = \|\Phi\|_X$ for every $\Phi \in X$ and every $t \in \mathbb{R}$. In particular, \begin{align*} \|T(t)\|_{\mathcal{L}(X)} &= 1 \quad \text{for all } t \in \mathbb{R}. \end{align*} Each $T(t)$ is a surjective isometry on the Hilbert space $X$, hence unitary. The energy norm $\|(u, v)\|_X^2 = \int_U |\nabla u|^2 \, d\mathcal{L}^n + \int_U |v|^2 \, d\mathcal{L}^n$ is conserved along solutions, completing the proof. [/step]