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Please help me how to deal with maximization of functional like this:

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

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

Thank you for reading.

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    $\begingroup$ Try to use calculus of variation. Check Euler–Lagrange equation. $\endgroup$ – Mhenni Benghorbal Jul 15 '13 at 11:37
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    $\begingroup$ @MhenniBenghorbal It was the first thing I've tried. I don't undestand how to deal with $v(t)^2$. $\endgroup$ – centuri0n Jul 15 '13 at 11:47
  • $\begingroup$ 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 $v(t)=\int_0^t a(s)ds$? $\endgroup$ – Tomás Jul 15 '13 at 13:37
  • $\begingroup$ Yes, I updated the question. I hope it's more clear now. $\endgroup$ – centuri0n Jul 15 '13 at 15:34
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    $\begingroup$ @centuri0n, so you mean that $t$ is fixed and $F$ sends $a$ to some real number. Right? $\endgroup$ – Norbert Jul 15 '13 at 17:21
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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.

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