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I am stuck on a possibly trivial question. I have the Greens function for the equation


with the boundary value $y(\alpha)=y(\beta)=0$ and I need to solve the equation


with the boundary conditions $y(\alpha)=0, y(\beta) = A \not =0$

I know that if the boundary conditions of the second was the same as the first, then I could do

$$y(x)=\int_{\alpha}^{\beta} G(x,t) r(t) dt$$

But what does one do with the different boundary conditions?

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That's rather an unusual boundary condition. It doesn't fix a solution but excludes certain solutions. Since the space of solutions is the space of solutions for $y(\alpha)=0$ minus the space of solutions for $y(\alpha)=y(\beta)=0$, you need to find those two spaces. The second you can find using your Green's function. – joriki Apr 21 '11 at 5:18
@joriki, I meant to say that it is a constant, but one that is not equal to 0. – picakhu Apr 21 '11 at 5:36
I don't think you'll be able to solve this with that Green's function. You could calculate $B=\int_\alpha^\beta r(t)\mathrm dt$ and then look for a Green's function with $G(\alpha,t)=0$ and $G(\beta,t)=A/B$. – joriki Apr 21 '11 at 5:52
The FAQ explicitly says you can answer your own question. I think you should do that so that the question doesn't appear unanswered. – joriki Apr 21 '11 at 6:54
up vote 3 down vote accepted

To those interested, I got the solution. It uses the following trivial theorem:

Given that $y_1$ is a solution to the problem


with BC $y(\alpha)=A, y(\beta) = B$

and that $y_2$ is a solution to the problem


with BC $y(\alpha)=0, y(\beta) = 0$


$y=y_1+y_2$ is a solution to


with boundary conditions $y(\alpha)=A, y(\beta) = B$

So the original problem is easily solved.

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