"Roger Stafford" wrote in message <im5eb8$rih$1@fred.mathworks.com>...
> "toms Seidel" wrote in message <im4us8$7kc$1@fred.mathworks.com>...
> > Hello!
> >
> > I am just about to solve the following minimization problem with equality and inequality constrains:
> >
> > min int_{\inf}^{\inf} log(1+ a * f(x)) dx
> >
> > s.t. f(x) >= 0, int_{\inf}^{\inf} f(x) dx = C, int_{\inf}^{\inf} f(x) * b(x) >= d *C
> >
> > f(x) and b(x), respectively, are scalar function of x, a, C and d can be considered as constants. The optimization parameter is f(x) What's the best way to tackle such a problem? I would have started using fmincon, but I am not sure whether this can be used because of the infinite integrals.
> >
> > What integral solver should be employed for this purpose? According to the Matlab reference quadgk seems to be the right choice ...
> >
> > Any hints appreciated!
>         
> ...... That makes it a problem in the calculus of variations which leads to differential equations of the Euler Lagrange type. .......
After thinking over your problem for a while I retract my statement that it is a problem in the calculus of variations. I contend that, strictly speaking, it is a problem without a solution, but it is trivial in any case. If the problem is to be meaningful at all there will exist functions f(x) which bring the integral in question arbitrarily close to zero, but no single f(x) can ever achieve that minimum, which means there are no solutions.
To begin with, I claim that a and c must both be positive quantities to make the problem meaningful. Next I claim that, assuming b(x) is a continuous function, there must be intervals in x for which b(x) >= d. Otherwise the last constraint could not be satisfied.
Suppose for a small interval of width 'delta' for which b(x) >= d we define a (discontinuous) f(x) = c/delta and f(x) = 0 elsewhere. This satisfies all the constraints. It also means our integral will equal delta*log(1+a*c/delta). But this quantity can be made arbitrarily close to zero by making delta sufficiently small. Moreover continuous functions f(x) can approach such discontinuous f(x) arbitrarily close. And yet it is clear that any function f(x) satisfying the constraints must always yield a positive value to this integral since 1+a*f(x) must be greater than one over some set of x values of positive measure. Thus there can be no solution.
If 'a' were to have a negative value, one could use similar arguments to the above to show that 1+a*f(x) could be brought arbitrarily close to zero over a finite interval and therefore give a minimum integral of minus infinity.
Roger Stafford
