We consider a zero sum differnetial game.
Let $x \in (0, M] \subset \mathbb R_{++}$ denote the state and $(u,v) \in [0,x]$ the control of player 1 and 2 respectively with $u + v \leq x$.
Denote the lower $\underline V(x)$ and upper $\overline V(x)$ value by a Hamilton-Jacobi-Bellman-Isaacs equation
\begin{align}
r\underline V(x) = \max_u \min_v \{\ln(u) - \underline V'(x)(u + v)\}
\end{align}
Consider the first order condition of player 2 for the lower value
\begin{align}
- \underline V'(x) < 0 \Rightarrow v^* = x
\end{align}
If $v^* = x$, then $u^* = 0$ such that $\underline V(x) = -\infty$.
For the upper value I'm not too sure. \begin{align} r\overline V(x) = \min_v \max_u \{\ln(u) - \overline V'(x)(u + v)\} \end{align} Consider the first order condition of player 1 for the upper value \begin{align} \frac{1}{u^*} - \underline V'(x) = 0 \Leftrightarrow \underline V'(x) = \frac{1}{u^*} \end{align} The maximized HJBI becomes \begin{align} r\overline V(x) = \min_v \left\{\ln(u^*) - 1 - \frac{v}{u^*}\right\} \end{align} The first order conditions of the second player then reads \begin{align} - \frac{1}{u^*} < 0 \Rightarrow v^* = x. \end{align} Again $v^* = x$, then $u^* = 0$ such that $\overline V(x) = -\infty$.
Is the reasoning correct? I was wondering if we may argue that for the upper value player 1 chooses $u^*$ first and then $v^* = x - u^*$. Then \begin{align} r\overline V(x) = \ln(u^*) - \frac{x}{u^*} \end{align} Now guess linear control $u^* = Ax$ \begin{align} &r\overline V(x) = \ln(Ax) - \frac{1}{A}\\ &r\overline V'(x) = \frac{r}{Ax} = \frac{1}{x} \Leftrightarrow A = r \end{align} The upper value is then rather given by $\overline V(x) = (\ln(rx) - \frac{1}{r})/r$.
Edit: I tend to believe that $\underline V(x) = \overline V(x) = V(x)$ and that we cannot impose an implcit timing of moves but, must consider the first order conditions simultaneously. So we either have \begin{align} \frac{1}{u^*} - V'(x) = 0 &\Leftrightarrow V'(x) = \frac{1}{u^*}\\ - \underline V'(x) < 0 &\Rightarrow v^* = x - u^* \end{align} Such that \begin{align} V(x) = \frac{1}{r}\left(\ln(rx) - \frac{1}{r}\right) \end{align} Can someone confirm that?
Or simply $V(x) = -\infty$ for $v^* = x$. What's correct? The latter seems to make more sense. Because player 2 only cares about minimizing such that he consumes the entire state.
