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  Minimizing a functional definite integral

+ 3 like - 0 dislike
1270 views

I have a definite integral defined by

$$T\left(G\left(g\right)\right)=\int_{g_{1}}^{g_{2}}G(g)\mathrm{d}g$$

where $G$ is a continuous function of a variable $g$, and $g_{1}$ and $g_{2}$ are known numbers. I want to minimize $T\left(G\left(g\right)\right)$, that is I want to find a continuous function $G=f\left(g\right)$ that makes $T\left(G\left(g\right)\right) $ minimum. Ideally I would differentiate it and equate to zero, but because $T\left(G\left(g\right)\right)$ is too complicated to be obtained and then differentiated analytically, I would like to know if there is a numeric technique or any other technique by which this problem can be solved.

This post imported from StackExchange Mathematics at 2014-06-02 20:31 (UCT), posted by SE-user James White
asked Dec 19, 2012 in Mathematics by James White (15 points) [ no revision ]
I don't completely understand your problem. Your functional doesn't look "complicated" at all - it maps a function $G$ to it's integral $\int_1^2 G$. However, the 'minimisation' isn't well-defined. If you take $G(g) \equiv c \equiv \text{constant}$, you can make $T(G)$ any value you want, so what would be the minimum? Maybe I'm completely misreading the question, but in that case many others are probably having the same problem.

This post imported from StackExchange Mathematics at 2014-06-02 20:31 (UCT), posted by SE-user Vibert
Look up calculus of variations (en.wikipedia.org/wiki/Calculus_of_variations), to see how these type of problems are normally dealt with. In your case, as Vibert points out, either $f(g)=0$ or $f(g) = -\inf$, depending on how you define minimisation, is the solution to your problem.

This post imported from StackExchange Mathematics at 2014-06-02 20:31 (UCT), posted by SE-user Jaime
Also note that $g$ of the right hand side is swallowed, and this $T$ doesn't depend on $g$ (but on $G$ and $g_1,g_2$).

This post imported from StackExchange Mathematics at 2014-06-02 20:31 (UCT), posted by SE-user Berci

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