A functional depending upon x and y'(x)

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SUMMARY

The discussion centers on solving the functional equation S[y] = ∫21 dx ln(1 + xny') with boundary conditions y(1) = 1 and y(2) = 21-n, where n is a constant integer greater than 1. Participants confirmed that the term O(ε) vanishes when y' satisfies the differential equation dy/dx = 1/c - 1/xn, leading to the stationary path y(x) = (n-2)(2n-1 -1)(1-x)/2n-1(n-1) + 1/(n-1)xn-1 + n-2/n-1. The solution involves substituting the derived y(x) into the ε term to validate the vanishing condition.

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Homework Statement



S[y] = [itex]\int[/itex]21dx ln(1 + xny'), y(1) = 1, y(2) = 21-n

where n > 1 is a constant integer, and y is a continuously differentiable function
for 1 ≤ x ≤ 2. Let h be a continuously differentiable function for 1 ≤ x ≤ 2 and ε
a constant. Let ∆ = S[y + εh] − S[y].

Show that if h(1) = h(2) = 0, the term O(ε) in this expansion vanishes if
y′(x) satisfies the equation

dy/dx = 1/c -1/xn


where c is a constant.
Solve this equation to show that the stationary path is

y(x) = (n-2)(2n-1 -1)(1-x)/2n-1(n-1) + 1/(n-1)xn-1 + n-2/n-1

Homework Equations





The Attempt at a Solution




Right, I've found S[y+ εh] -s[y] = [itex]\int[/itex]21dx (xnh(x)'/(1 + xny(x)')ε - ε/2 [itex]\int[/itex]21dx x2n(h(x)')2/(1+ xny(x)')2 + O(ε3)

I'm aslo pretty sure I've proved the O(ε) vanishes. I did this substituting y(x) = (n-2)(2n-1 -1)(1-x)/2n-1(n-1) + 1/(n-1)xn-1 + n-2/n-1 into ε term and got it to equal 0. I don't know how to find y(x) though. Through integration i got y(x) = 1/c - (1-n+1)/(-n+1)

I don't know how to progress from here or if I'm doing it correctly. Some guidance would be great! :)





 
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