Pontryagin principle for optimal control problem with a terminal cost involving the control

Let $$T >0$$ and consider the problem of minimizing $$P(v(.)) \triangleq \int_0^T l(x_t,v_t) d t + h(x_T)$$ over a broad class of control $$v(.)$$ where $$\begin{equation} \dot x_t = f(x_t,v_t), \quad x_{t=0} = \xi \in \mathbb{R}. \end{equation}$$ For the sake of simplicity, here $$x(.)$$ and $$v(.)$$ are real valued functions and $$f,l: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$$ and $$h: \mathbb{R} \to \mathbb{R}$$ are smooth functions. The problem can be studied via the so-called Pontryagin principle that states a necessary condition for optimality as follows: $$\begin{equation} \begin{cases} & \dot x_t = f(x_t,v^\star[x_t,p_t])\\ & \dot p_t = -H_x(x_t,v^\star[x_t,p_t],p_t)\\ & H(x,v^\star[x,p],p) = \min \limits_{v \in \mathbb{R}} H(x,v,p) \end{cases} \end{equation}$$ with $$x_{t=0} = \xi \quad \mbox{and} \quad p_{t=T} = h_x(x_T)$$ where $$\begin{equation} H(x,v,p) \triangleq l(x,v) + p \cdot f(x,v). \end{equation}$$

Comment The functional $$P$$ above does not involve the value of the control at the terminal time $$T$$. In all the references I have found it always appears in this way, but it would be quite reasonable to have a terminal cost in the form $$h(x_T,v_T)$$.

Question What becomes the terminal condition at $$t = T$$ for the variable $$p$$ when $$P(v(.)) \triangleq \int_0^T l(x_t,v_t) d t + h(x_T, \color{red}{v_T})?$$ I had a glimpse of the design of the variable $$p(.)$$, page 112 in https://math.berkeley.edu/~evans/control.course.pdf (see the variation of the functional $$P(.)$$ around its optimal control) and it seems to me that the terminal condition in $$p$$ remains the same when the terminal is of the form $$h(x,v)$$.

• Can't you just define $h(x_T) = \min_{v_T} h(x_T,v_T)$ with $v^*[x_t,p_t]$ being potentially discontinuous at $t=T$. As long as this value for $v_T$ is finite (no Dirac delta function) then the contribution of it to $x_T$ should be zero. – fibonatic Nov 27 '18 at 9:58
• @fibonatic: thank you for your comment. I agree. This is certainly the most pragmatic approach. – megaproba Nov 28 '18 at 0:37