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Let $\phi \in \mathcal ( [0,1]^2)$ symetric , can we find a solution to the following minimisation problem?

$$ \inf \left\{ F[\nu] : \nu \in L^2 , \nu \geq 0, \int _0 ^1 \nu=1\right\}$$

with $$ F[\nu] := \frac{1}{2} \int_0^1 \nu^2(t) ~dt + \frac{1}{2} \int_0^1\int_0^1 \phi(s,t) ~\nu(s) ~\nu(t) ~ds~dt$$

And if it exist such a $\nu$ what optimality conditions it satisfies ?

If I made no mistakes by making the Frechet derivative equals to zero we should have for any $h \in L^2$, $\int _0^1 h=0$ :

$$ \int_0^1 \nu(t)~h(t) ~dt = - \frac{1}{2} \int_0^1\int_0^1 \phi(s,t) ~[\nu(s)~h(t)+\nu(t)~h(s)] ~ds~dt$$

After this point I can't see how to conclude exploring probably the symmetry of $\phi$. Could someone give me some advice or the solution if you see how to solve it easily?

Many thanks.

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Besides symmetry, do you know anything else about $\phi$? Nonnegative, bounded, continuous...? –  Thursday Jun 30 at 17:30
    
@Thisismuchhealthier. There is no additional conditions about $\phi$. Please see my idea for approach the optimality condition satisfied by $nu$. –  Paul Jun 30 at 20:05

1 Answer 1

Here one idea of solution. Any corrections, additional steps, alternative solutions and suggestion are welcome.

Admitting that there is a solution for the optimisation problem, fact that still must be proved, we have:

$$ \int_0^1 \nu(t)~h(t) ~dt = - \frac{1}{2} \int_0^1\int_0^1 \phi(s,t) ~[\nu(s)~h(t)+\nu(t)~h(s)] ~ds~dt$$

where $\phi$ is symmetric so it implies that

$$ \int_0^1 \nu(t)~h(t) ~dt = - \int_0^1\int_0^1 \phi(s,t) ~\nu(s)~h(t) ~ds~dt$$

therefore $\forall h \in L^2, ~ \int h =0$

$$ \int_0^1 \left( \nu(t) +\int_0^1 \phi(s,t) ~\nu(s) ~ds \right) ~h(t) ~dt = 0$$

then

$$ \nu(t) +\int_0^1 \phi(s,t) ~\nu(s) ~ds =0$$

so

$$ \int_0^1 \Phi (s) ~\nu(s) ~ds = 1$$

where $\Phi(s) := -\int_0^1 \phi(s,t) ~dt$.

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