= Solution
Set
$$
Z=\binom{u}{u_t},
\qquad
Z_0=\binom{u_0}{u_1},
\qquad
F(t)=\binom0{f(t,\cdot)}.
$$
Then the forced equation is the <abstract Cauchy problem>
$$
\dot Z=AZ+F,
\qquad
Z(0)=Z_0.
$$
For $Z_0\in\mathcal H$ and $F\in C([0,T];\mathcal H)$, a <mild solution of an abstract Cauchy problem> is a function $Z\in C([0,T];\mathcal H)$ satisfying the <variation-of-constants formula>
$$
\boxed{
Z(t)=U(t)Z_0+int_0^tU(t-s)F(s)\,ds}.
$$
Suppose $F(s)\in D(A)$ and both $F$ and $AF$ are continuous. If also $Z_0\in D(A)$, then the closedness of $A$ permits differentiation under the <Bochner integral>:
$$
\frac d{dt}\int_0^tU(t-s)F(s)\,ds
=F(t)+\int_0^tU(t-s)AF(s)\,ds.
$$
Thus $Z\in C^1([0,\infty);\mathcal H)$ and $Z'=AZ+F$. The assumption $Z_0\in D(A)$ is necessary here: a unitary group has no smoothing, so the conditions on $F$ alone cannot make $U(t)Z_0$ differentiable for arbitrary $Z_0\in\mathcal H$.
For the resulting strong solution, skew symmetry of $A$ gives the <energy estimate>
$$
\frac d{dt}\|Z(t)\|_{\mathcal H}^2
=2\operatorname{Re}(Z(t),F(t))_{\mathcal H}.
$$
Integration yields
$$
\boxed{
\|Z(t)\|_{\mathcal H}^2
\leq\|Z(0)\|_{\mathcal H}^2
+2\int_0^t|(Z(s),F(s))_{\mathcal H}|\,ds}.
$$
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