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Under the Fourier transform, the operator $A(t)={\partial }_x^3+\varphi (t){\partial }_x$ is the Fourier multiplier operatorIts symbol is purely imaginary because $\varphi$ is real. Consequently $A(t)$ is skew-adjoint on $L^2(\mathbb{R})$ with common domain $H^3(\mathbb{R})$ and generates the strongly continuous unitary groupThe Plancherel theorem gives $\Vert e^{rA(t)}u{\Vert }_2=\Vert u{\Vert }_2$, so every $A(t)\in \mathcal{G}(1,0)$. Products of the frozen groups are also unitary; hence this is a stable family of semigroup generators with constants $1,0$.

For $u\in H^3(\mathbb{R})$,Since $\varphi \in C^1(\mathbb{R})$, the map $t\mapsto A(t)$ is continuously differentiable from $H^3$ to $L^2$. All hypotheses from part f are satisfied, so an evolution family exists on every finite interval $[0,T]$.

In this commuting Fourier multiplier operator example the solution operator can also be written explicitly:Its multiplier has absolute value one, directly confirming strong continuity, the evolution family law, preservation of $H^3$, and the required derivatives. The equation combines the dispersive Airy equation with a time-dependent linear transport equation.

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 $t\in \mathbb{R}$.