\beta \right\rangle^2
\int_t^{s}e^{\lambda(s-r)} G(r)  {\mathrm{d}}r.$$

Similarly if $\left\langle \alpha,x\right\rangle > 0$, taking the control $\left\langle \beta,\bar\alpha(t) \right\rangle$ the associated state trajectory is $$x^{op}(s)= e^{(s-t)A}x + \int_t^{s}e^{(s-r)A} R(\left\langle
\beta,\bar\alpha(r) \right\rangle) {\mathrm{d}}r$$ and it easy to check that it satisfies $\left\langle
\alpha,x^{op}(s) \right\rangle > 0 $ for every $s\ge t$.

We now apply Theorem \[thdeterministicverification\] taking $q(s)=\partial_t V(s, x^{op}(s))$, $$p_1(s)=\left\lbrace
\begin{array}{ll}
\bar \alpha(s) & if \; \left\langle \alpha,x^{op}(s)\right\rangle \le 0\\
-\bar \alpha(s) & if \; \left\langle \alpha,x^{op}(s)\right\rangle
>0
\end{array}
\right.$$ and $p_2(s)=0$. It is easy to see that $(q(s), p_1(s), p_2(s)) \in
E^{1,+}V(s,x^{op}(s))$. The argument for $\bar u^{op}$ is completely analogous.

We continue by giving a specific example of the Hilbert space $\mathcal{H}$, the operator $A$, and the data $\alpha$ and $\beta$. This example is related to the vintage capital problem in economics, see e.g. [@BarucciGozzi01; @BarucciGozzi98]. Let $\mathcal{H}=L^2(0,1)$. Let $\{e^{tA}; \; t \ge 0\}$ be the semigroup that, if we identify the points $0$ and $1$ of the interval $[0,1]$, “rotates” the function: $$e^{tA}f(s) = f(t+s - [t+s])$$ where $[\cdot]$ is the greatest natural number $n$ such that $n\leq
t+s$. The domain of $A$ will be $$D(A)= \left\lbrace f\in W^{1,2}(0,1) \; : \; f(0)=f(1) \right\rbrace$$ and for all $f$ in $D(A)$ 