Maximize a functional

$$F\{a(s)\} = \int\limits_0^t \left( g(a(s)) - \alpha\, v(s)^2 \right) ds, \ a(s) \in \left[0, \infty\right)$$

where $g(x) = x e^{-x}$ , $\alpha=\mathrm{const}$, and $\displaystyle v(t) = \int\limits_0^t a(s) ds$

• @MhenniBenghorbal It was the first thing I've tried. I don't undestand how to deal with $v(t)^2$. – centuri0n Jul 15 '13 at 11:47
• Could you please clarify the definition of your functional? I think that the integral is from $0$ to $\infty$. Also, waht is the precise definition of $v(t)$. Is $v(t)=\int_0^t a(s)ds$? – Tomás Jul 15 '13 at 13:37
• @centuri0n, so you mean that $t$ is fixed and $F$ sends $a$ to some real number. Right? – Norbert Jul 15 '13 at 17:21
I don't understand how to deal with $v^2$
Reformulate the problem in terms of $v$. That is, you seek the maximum of $$F\{v\} = \int\limits_0^t \left( g(v'(s)) - \alpha\, v(s)^2 \right) ds, \ a(s) \in \left[0, \infty\right) \tag1$$ over increasing differentiable functions $v$ with $v(0)=0$. The Euler-Lagrange equation for (2) is easy to state: $$-\frac{d}{ds}(g'(v'(s))) -2\alpha v(s)\equiv 0 \tag2$$ Since $g''(x)=(x-2)e^{-x}$, the equation (2) becomes $$(2-v'(s))e^{-v'(s)}v''(s) -2\alpha v(s)\equiv 0 \tag3$$ I wouldn't expect an explicit solution of (3), but a numeric solution should not be hard to obtain.