Constrained linear optimization problem on $C^1$

Question

I am dealing with a problem of the form ($a<b$) $$ \displaystyle \max_{v \in C^1([a, b])} \int_a^b v(x)~\mathrm{d}x, \quad \mathrm{s.t.} \int^b_a \big(-o'(x)v(x)-v'(x)o(x)\big)f(x)~\mathrm{d}x \leq 0,~ \frac{o(b)}{o(a)} \leq v \leq 1 $$ where $o \in C^\infty(\mathbb{R})$ with $o>0$ and $o'< 0$ on $(a, b)$ and $f \in C([a, b])$ with $f>0$ on $(a, b)$. I would already be very happy if someone could provide me with input for the case $f(x)=1$.

Problem is, I can find little to no publications on such problems. Moreover, the usual optimality conditions are useless because we lose $v$ in the derivative of both objective and constraint. My suspicion is that $$ v(x) = \frac{o(b)}{o(x)} $$ for all $x \in [a, b]$ is optimal, since it solves the ODE $-o'v-v'o$ and is $1$ on the boundary $b$.

Furthermore, using Gronwall, I was able to show that if some optimal $\tilde{v}$ exists and $$ -o'(x)\tilde{v}(x) -\tilde{v}'(x)o(x) \leq 0 $$ on some interval $(c, d) \subseteq (a, b)$ holds, then we get the upper bound $$ \tilde{v}(x) \leq \frac{o(d)}{o(x)} $$ on $\tilde{v}$ for all $x \in (c, d)$, which is at least somewhat consistent with my hypothesis. Any input would really be appreciated.

Answer 1

As you suggest, let me consider the case $f \equiv 1$. Without loss of generality, assume also that $a = 0$ and $b = 1$. Let $\sigma := o(1)/o(0) \in (0,1)$. The problem becomes $$ \sup_{v \in C^1([0,1])} \int_0^1 v(x) \mathrm{d}x \quad \text{s.t.} \quad v(0)\leq \sigma v(1) \quad \text{and} \quad \sigma \leq v(x) \leq 1. $$ First, note that I changed the maximum to a supremum, since it is not clear a priori that the supremum will be achieved (it will turn out not to be the case). Second, note that the function $o(x)$ has disappeared from the problem, only remaining through the single real value $\sigma \in (0,1)$.

Since $v(x) \leq 1$ on $[0,1]$, one has $\int_0^1 v \leq 1$ for any admissible $v \in C^1([0,1])$. Let us check that $1$ is the value of the supremum, but that it is not achieved.

• If $v \in C^1([0,1])$ is such that $v(x) \leq 1$ on $[0,1]$ and $\int_0^1 v = 1$, then $v \equiv 1$. But the constraint $v(0) \leq \sigma v(1)$ becomes $1 \leq \sigma$, which fails because $\sigma \in (0,1)$. So the supremum cannot be achieved.
• For $n \in \mathbb{N}^*$, let $v_n(x) := 1$ for $x \in [1/n,1]$ and $v_n(x) := 1 + (\sigma-1)n^2(x-1/n)^2$ for $x \in [0,1/n]$. Then $v_n \in C^1([0,1])$ because the junction at $x=1/n$ is $C^1$, $\sigma \leq v_n \leq 1$ on $[0,1]$, and $\sigma = v(0) \leq \sigma v(1) = \sigma$ because $v(1) = 1$. Moreover, one checks that $\int_0^1 v_n \geq 1 - 1/n$. So the supremum is indeed equal to $1$.