problem of calculus of variations

**ltrapani** · February 22nd 2011, 06:09 AM

Hello everyone

I have the following problem. I need to find a function, say $f(x)$ , that maximizes
$\int_{a}^{b}u(x)f(x)dx$ , where $u(x)$ is the square of the cumulative distribution function of the standard normal minus one half, i.e. $u(x)=(\phi(x)-1/2)^2$ .
The maximisation problem does not have prespecified boundary values for $a$ and $b$ , and is subject to the constraint $\int_{a}^{b}f(x)dx = 1$

any help is really precious

thanks

Lorenzo

**Ackbeet** · February 22nd 2011, 07:04 AM

Hmm. Typically, a problem with an integral constraint is dealt with using the Lagrange multiplier method. That is, instead of setting

$L=u(x)\,f(x),$ and solving the DE

$\displaystyle\frac{d}{dx}\,\frac{\partial L}{\partial f'}-\frac{\partial L}{\partial f}=0,$

you do the following. Set $L=u(x)\,f(x)+\lambda\,f(x),$ and set

$\displaystyle\frac{d}{dx}\,\frac{\partial L}{\partial f'}-\frac{\partial L}{\partial f}=0.$

The problem is that in all these cases,

$\dfrac{\partial L}{\partial f'}=0,$

which makes all the resulting equations much too constrained. Even if you let $\lambda=\lambda(x),$ you still have problems. You end up with $\lambda=-u(x),$ or $\lambda(x)=-u(x),$ an absurdity.

There's something I'm missing here, or something is really whacked out. Can you please double-check your problem statement, and make sure it's correct? In particular, I'd expect to see an $f'(x)$ floating around somewhere, at least if the calculus of variations is going to be the right tool for the job.

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