Optimizing the ratio of two integrals

In summary: Therefore, in summary, for optimization problems involving maximizing the ratio of two integrals, there is a general approach using variational calculus. However, for these types of problems, a necessary condition for finding the maximum is that the derivative of the functional at \epsilon=0 must equal 0, which leads to a differential equation that the function must satisfy. To solve this differential equation, we use a set of comparison functions and solve for the parameters d and n. In simpler cases, the solution may be an algebraic equation in the function itself, but in more complex cases, it may require solving a system of algebraic equations.
  • #1
aetherane
2
0
Frequently, I have encountered optimization problems which involve finding the optimal function to maximize the ratio of two integrals (bounds from 0 to infinity). e.g: Maximize [tex]\frac{\int_0^\infty (g(x))^3 g'(x) dx}{\int_0^\infty (g(x))^3(1-g(x)) dx}[/tex]

I am aware that variational calculus works on a single integral, but is there a general approach that might work for these types of problems?
 
Mathematics news on Phys.org
  • #2
Yes, variational calculus!


Define:
[tex]F(\epsilon)=\frac{\int_{0}^{\infty}(g(x)+\epsilon\gamma(x))^{3}(g^{,}(x)+\epsilon\gamma^{,}(x))dx}{\int_{0}^{\infty}(g(x)+\epsilon\gamma(x))^{3}(1-(g(x)+\epsilon\gamma(x)))dx}[/tex]
where [itex]\gamma(x)[/itex] is an arbitrary function that vanishes at the boundaries.

Now, a necessary condition in order to let [itex]\epsilon=0[/itex] be the maximum of F would be that [itex]\frac{dF}{d\epsilon}_{\epsilon=0}=0[/itex]

This condition will yield the differential equation g must satisfy.
 
Last edited:
  • #3
To treat this problem in a somewhat general manner, let us assume that y's boundary values are fixed, say
[tex]y(a)=y_{a},y(b)=y_{b}[/tex]
We look then at the set of comparison functions:
[tex]Y(x,\epsilon)=y(x)+\epsilon\gamma(x),\gamma(a)=\gamma(b)=0[/tex]
That is, the [itex]\gamma[/itex]-function is arbitrary except for vanishing at the boundaries.

We have a functional,
[tex]F(\epsilon)=\frac{\int_{a}^{b}N(Y,Y^{,},x)dx}{\int_{a}^{b}D(Y,Y^{,},x)dx}[/tex]
and also define the quantities:
[tex]n=\int_{a}^{b}N(y,y^{,},x)dx, d=\int_{a}^{b}D(y,y^{,},x)dx (*)[/tex]

Now, differentiating F with respect to [itex]\epsilon[/itex] and then setting thederivative of F equal to 0 at [itex]\epsilon=0[/itex] yields, with some rearrangement:
[tex]\frac{\int_{a}^{b}(d(\frac{\partial{N}}{\partial{y}}-\frac{d}{dx}\frac{\partial{N}}{\partial{y^{,}}})-n(\frac{\partial{D}}{\partial{y}}-\frac{d}{dx}\frac{\partial{D}}{\partial{y^{,}}}))\gamma(x)dx}{d^{2}}=0[/tex]

Thus, we get the following diff.eq problem to solve:
[tex]d(\frac{\partial{N}}{\partial{y}}-\frac{d}{dx}\frac{\partial{N}}{\partial{y^{,}}})-n(\frac{\partial{D}}{\partial{y}}-\frac{d}{dx}\frac{\partial{D}}{\partial{y^{,}}})=0, y(a)=y_{a},y(b)=y_{b}[/tex]

The solution of this diff.eq problem will typically be a function of the two parameters d and n, (in addition of course, of being a function in x)!

In order to determine d and n, (*) represents a system of algebraic equations in d and n, so we solve this system to complete our solution.
 
Last edited:
  • #4
In your case, the differential equation becomes exceedingly simple:
[tex]3g^{2}-4g^{3}=0[/tex]
where d and n vanish as determining parameters of the equation, and we retain an algebraic equation in g.

Thus, the only acceptable solution for a stationary point for the functional is [tex]g(x)=\frac{3}{4}[/tex]

However, since this yields a divergent integral in the denominator, that particular ratio cannot be said to have a maximizing (or minimizing) function.
 

1. What is the purpose of optimizing the ratio of two integrals?

The purpose of optimizing the ratio of two integrals is to find the best possible values for the variables in the integrals that will result in the maximum or minimum value of the ratio. This can be useful in various scientific and mathematical applications, such as in optimization problems or in finding the most efficient solution to a problem.

2. How is the ratio of two integrals calculated?

The ratio of two integrals is calculated by dividing the value of one integral by the value of the other integral. This ratio can then be simplified or manipulated to optimize its value.

3. What are the key factors to consider when optimizing the ratio of two integrals?

The key factors to consider when optimizing the ratio of two integrals include the variables in the integrals, the boundaries of integration, and any constraints or limitations on the variables. It is also important to consider the context and purpose of the problem and to determine what the optimal ratio would signify.

4. Are there any specific techniques or methods for optimizing the ratio of two integrals?

Yes, there are various techniques and methods for optimizing the ratio of two integrals, such as using calculus and mathematical analysis to find the critical points of the ratio. Other methods may involve using numerical methods or computer algorithms to approximate the optimal value of the ratio.

5. Can the ratio of two integrals always be optimized?

In theory, yes, the ratio of two integrals can always be optimized. However, in practice, it may not always be possible to find the exact optimal value due to limitations in the variables, boundaries, or available techniques. In some cases, a close approximation may be used instead.

Similar threads

  • Calculus and Beyond Homework Help
Replies
2
Views
134
Replies
3
Views
1K
Replies
1
Views
907
Replies
6
Views
3K
Replies
4
Views
384
  • Calculus
Replies
6
Views
1K
  • General Math
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
2
Views
830
  • Calculus and Beyond Homework Help
Replies
1
Views
663
Replies
6
Views
1K
Back
Top