I want to find the function $\gamma(t)$ that makes the following stationary.
$S(t) = \int_0^t \gamma(t'') \Big(\beta\int_0^{t''}(1-\gamma(t'))dt' + e_0\Big) dt''$
Constraints:
$\beta\gt0$,
$e_0\ge0$,
$1\ge\gamma(t)\ge0$ for all $t$.
Can this be solved using the calculus of variations? I don't know how to deal with the inner integral.
Solutions obtained by numerical methods suggest that the $\gamma(t)$ that makes $S(t)$ stationary is a step function, transitioning from $0$ to $1$ at some point. If I assume it's a step function I get the solution
$ \gamma(t) = \begin{cases} 0, & \text{if } t\lt \frac{t}{2}-\frac{e_0}{2\beta}, \\ 1, & \text{otherwise}. \end{cases} $
By Fubini's theorem: $$ \int_0^t \int_0^{t''} \gamma(t'') \gamma(t') \, d t' \, d t''= \frac12 \int_0^t \int_0^t \gamma(t'') \gamma(t') \, d t' \, d t''. $$ Maybe this helps?