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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?

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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?

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arildno

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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.

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.

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arildno

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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.

[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.

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arildno

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[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.

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