I would like to solve a maximization problem below using optimal control and/or calculus of variations. I'm slightly more familiar with the former but still a beginner who learned from economics books that take a slightly handwavy approach rather than what I'd guess is a more rigorous [normed function space]-based approach that mathematicians would learn. Here goes.
Problem
Solve the following problem which can be interpreted as an insurance design problem. You can ignore these interpretations if you want, but if you care: $\pi$ is the premium and $m(s)$ is expenditure in state $s$, $H(s,m(s))$ is health which goes in the second argument of utility, and the first argument of $U$ is monetary component of utility. $w$ is initial wealth.
\begin{align*} \max_{m(s), \pi} \quad & \int_{t_0}^{t_1} U \left(w - \pi, H\left(s, m(s)\right) \right) f(s)ds \\ \text{s.t} \quad & \pi = \int_{t_0}^{t_1} m(s) f(s)ds \end{align*}
Assume all functions are smooth and so on.
Conjectured Answer
Based on the discrete analogue, I think the necessary condition should come out to be
$$\int_{t_0}^{t_1}U_1\left(w-\pi, H\left(s,m(s)\right)\right)f(s)ds = U_2\left( w-\pi , H\left(s,m(s)\right) \right) H_2(s,m(s)),$$
where subscripts indicate partial derivatives, but I don't quite see how to show it.
Attempt to Derive Conjectured Answer
The regular FOC wrt $\pi$ (without any fancy CoV or Control stuff) yields
$$\int_{t_0}^{t_1}U_1\left(w-\pi, H\left(s,m(s)\right)\right)f(s)ds = \lambda, $$ where $\lambda$ is the multiplier for the constraint $\pi = \int_{t_0}^{t_1} m(s) f(s)ds$.
To deal with FOC's regarding $m(s)$, I try to take a control approach. I treat $m(s)$ as a control variable. I then introduce a new state variable $x(s)$ which will play the role of $\int_{t_0}^{s} m(\tau) f(\tau)d\tau$. I thus impose the constraints $x(t_0) = 0$, $x(t_1) = \pi$, and $x'(s) = m(s)$.
Then the Hamiltonian is $$U\left(w-\pi, H\left(s,m(s)\right)\right) + \mu(s) m(s),$$ and the optimality conditions wrt $m(s)$ and $x(s)$ are respectively
$$U_2\left( w-\pi , H\left(s,m(s)\right) \right) H_2(s,m(s)) = \mu(s),$$
and $\dot{\mu}(s) = 0.$
The latter implies $\mu(s)=\mu$ is constant. In order to get the desired result, I think I need to show $\mu = \lambda$.
I'm not seeing how to do that. Presumably I need to use the initial conditions $x(t_0) = 0$, $x(t_1) = \pi$ but not sure how.
If someone has a totally different approach to solving the problem I'm interested. Thanks!
We can eliminate the variable $\pi$ to end up with the maximization of $$ F(m) = \int_{t_0}^{t_1} U\left( w- \int_{t_0}^{t_1}m(s)f(s)ds, \ H(t,m(t)) \right) f(t)dt. $$ If $m$ is a maximum then the first derivative of $F$ in direction $h$ is zero, which amounts to $$ 0 = F'(m)h =\int_{t_0}^{t_1} U_1\left( w- \int_{t_0}^{t_1}m(s)f(s)ds, \ H(t,m(t)) \right) f(t) \cdot\left(- \int_{t_0}^{t_1}h(s)f(s)ds\right) dt\\+ \int_{t_0}^{t_1} U_2\left( w- \int_{t_0}^{t_1}m(s)f(s)ds, \ H(t,m(t)) \right) H_2(t,m(t))h(t)f(t)dt. $$ Let me set $$ \alpha:=\int_{t_0}^{t_1} U_1\left( w- \int_{t_0}^{t_1}m(s)f(s)ds, \ H(t,m(t)) \right) f(t) dt. $$ Then we get $$ 0 = -\alpha \int_{t_0}^{t_1}h(t)f(t)dt + \int_{t_0}^{t_1} U_2\left( w- \int_{t_0}^{t_1}m(s)f(s)ds, \ H(t,m(t)) \right) H_2(t,m(t))h(t)f(t)dt. $$ Now $h$ was arbitrary, hence we get $$ \alpha f(t) = U_2\left( w- \int_{t_0}^{t_1}m(s)f(s)ds, \ H(t,m(t)) \right) H_2(t,m(t)), $$ which implies with $\pi=\int_{t_0}^{t_1}m(s)f(s)ds$, $$ \int_{t_0}^{t_1} U_1\left( w- \pi, \ H(s,m(s))\right) f(s) ds \cdot f(t) = U_2\left( w- \pi, \ H(t,m(t)) \right) H_2(t,m(t)) , $$ which is your equation. This is not completely rigorous but should give you the general idea.