# A variational problem with the constraint that the function be decreasing

1. May 15, 2012

### Lennart Stern

The problem is in the attached document. Thanks for any suggestions!
Lennart

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2. May 16, 2012

### Lennart Stern

The mathematical problem:

$\theta$ is a constant that equals 0.8.

We consider the set

'$S=\left\{(F,h):F is a decreasing function from R^{+} to R^{+}, h\in R, 0=1- \frac{\theta+1)}{\theta} \frac {(\int^{h}_{y=0} F(y) dy)}{/F(0)} \frac{F(0)-\frac{1}{2}}{F(0)-F(h)} \right\}$'
The function L is defined on S by
$L(F,h)= \frac{\int^{h}_{x=0} \int^{h}_{y=x} F(y) dy dx} {\int^{h}_{x=0} \int^{h}_{y=0} F(y) dy dx} h$
We want to find the maximal value of L.

A further problem:
Denote by $L(\theta)$ the maximal value of L. What is $max_{\theta \in (0,1)} \frac {L(\theta)}{\theta}$ ?

Last edited: May 16, 2012
3. May 16, 2012

### Ray Vickson

If you want LaTeX to be readable in this forum, you need to remove the \$ delimiters and enclose your commands between "[tex ]" and "[/tex ]" delimiters (no space after the word tex--I used a space so as to not confuse the system). For example, here is one of your formulas (justa copied and pasted between the delimiters I:
$$L(F,h)= \frac{\int^{h}_{x=0} \int^{h}_{y=x} F(y) dy dx} {\int^{h}_{x=0} \int^{h}_{y=0} F(y) dy dx} h$$

RGV

4. May 16, 2012

### Lennart Stern

$$\theta$$ is a positive constant.

We consider the set

$$S=\left\{(F,h):F is a decreasing function from R^{+} to R^{+}, h\in R, 0=1- \frac{\theta+1}{\theta} \frac {\int^{h}_{y=0} F(y) dy}{F(0)} \frac{F(0)-\frac{1}{2}F(h)}{F(0)-F(h)} \right\}$$
The function L is defined on S by
$$L(F,h)= \frac{\int^{h}_{x=0} \int^{h}_{y=x} F(y) dy dx} {\int^{h}_{x=0} \int^{h}_{y=0} F(y) dy dx} h$$
We want to find the maximal value of L, denoted by $$L(\theta)$$.

A special question:
What is L(0.8) ?

A general question:
What is $$max_{\theta \in (0,1)} \frac {L(\theta)}{\theta}$$ ?