the one for the optimal control problem with the objective functional $\int_t^T
\left\langle \alpha,x(s)\right\rangle + \frac{1}{2}u(s)^2 {\mathrm{d}}s$ (respectively $\int_t^T -\left\langle \alpha,x(s)\right\rangle +
\frac{1}{2}u(s)^2 {\mathrm{d}}s$) and it is known in the literature (see [@FaggianGozzi] Theorem 5.5) that its solution is $$v_1(t,x)= \left\langle \bar\alpha(t),x \right\rangle - \int_t^T
\frac{1}{2} \left( R^*\bar\alpha(s)\right)^2 {\mathrm{d}}s$$ (respectively $$v_2(t,x)= -\left\langle \bar\alpha(t),x
\right\rangle - \int_t^T \frac{1}{2} \left(
R^*\bar\alpha(s)\right)^2 {\mathrm{d}}s).$$ Note that on the separating hyperplane $\left\langle
\alpha,x\right\rangle=0$ the two functions assume the same values. Indeed, since $\alpha$ an eigenvector for $A^*$, $$\bar\alpha(t) = G(t) \alpha$$ where $$G(t)= \int_t^T e^{\lambda(s-t)}{\mathrm{d}}s$$ So, if $\left\langle \alpha,x\right\rangle=0$, $${\left\langle}\bar \alpha (t) ,x {\right\rangle}=0\;\;\;\; \text{ for all $t\in [0,T]$}.$$ Therefore we can glue $v_1$ and $v_2$ writing $$W(t,x)=\left\{ \begin{array}{ll} v_1(t,x) & \hbox{if } \left\langle
\alpha,x\right\rangle \le 0 \\
 v_2(t,x) & \hbox{if } \left\langle
\alpha,x\right\rangle>0
 \end{array}\right.$$ It is easy to see that $W$ is continuous and concave in $x$. We claim that $W$ is a viscosity solution of (\[example1HJB\]). For $\left\langle \alpha,x\right\rangle<0$ and $\left\langle
\alpha,x\right\rangle>0$ it follows from the fact that $v_1$ and $v_2$ are explicit regular solutions of the corresponding HJB equations.

For the points $x$ where $\left\langle \alpha,x\right\rangle=0$ it is not difficult to see that $$\left \{
\begin{array}{l}
D^{1,+} W(t,x)= \left \{\left(\frac{1}{2}\left(
R^*\bar\alpha(t)\right)^2,
\gamma 