 Solution

The Lumer-Phillips theorem says that a densely defined operator on a Hilbert space generates a contraction $C_0$-semigroup exactly when it is maximal dissipative:and $\mathrm{Ran}(\lambda I-A)$ is the whole space for some, equivalently every, $\lambda >0$.

For $Z=(u,v)\in D(A)$, self-adjointness of $B$ givesThus both $A$ and $-A$ are dissipative. To check maximality, solveThe equations giveOn Fourier mode $n$, the last operator has multiplier $1+n^2+{\lambda }^2>0$, so it gives a unique $u\in H^2$ and then $v\in H^1$ whenever $(f,g)\in \mathcal{H}$. Hence $\lambda I-A$ is onto; the same calculation applies to $-A$.

The two contraction semigroups generated by $A$ and $-A$ are inverses. They form a strongly continuous unitary group $U(t)$ on the complexification of $\mathcal{H}$, or an orthogonal group on the real space, andfor every $t\in \mathbb{R}$.