I have the question:

Solve Laplace's equation in the square

$\partial^2u/\partial x^2 +\partial^2u/\partial y^2 =0$

$0<x, y<\pi$

With the boundary conditions



This is a standard seperation of variables problem, and breaking u into functions of x and y, one gets:



Then one uses the superposition principle and has:

$u(x,y) = \sum_{n=1}^{\infty} b_n \sinh(ny) \sin(nx)$

Then applying the boundary condition $u(x,\pi)=1$

$1 = u(x,\pi) = \sum_{n=1}^{\infty} b_n \sinh(\pi y) \sin(nx)$

This all makes sense to me. What does not make sense to me is the following step in the solution to this problem:

"From the orthogonality of the eigenfunctions

$b_n sinh(n \pi) = 2/\pi \int_{0}^{\pi} \sin (nx) dx$"

I get that this is a sturm-liouville problem. I get that the two functions $\sinh(nx)$ and $\sin(nx)$ are eigenfunctions, meaning that $\sinh(n\pi)$ and $\sin(nx)$ are also eigenfuntions and are thereby orthogonal. The thing I do not understand is how this orthogonality leads to the above integral for the coefficient of the sum.

If anyone could explain this to be I would be very much appreciatied

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    $\begingroup$ I missed in my edit, in your expression after applying the boundary condition, you should have $\sinh(n\pi)$ not $\sinh(\pi y)$ $\endgroup$ – Tyberius Feb 25 '18 at 4:33
  • $\begingroup$ The only eigenfunction problem is in the variable $x$ where you have endpoint conditions at both endpoints of an interval. The resulting eigenfunctions in $x$ are orthogonal. And that's what is being used. $\endgroup$ – DisintegratingByParts Feb 25 '18 at 16:45

Well, the orthogonality is not referring to $\sin(nx)$ and $\sinh(nx)$, but to $\{\phi_n(x,y) = \sin(nx)\sinh(ny)\}$.

Also, perhaps you wrote the equation incorrectly: You have $u(x,y) = \sum_{n \geq 1}b_n\phi_n(x,y)$ and so the boundary conditions give:

$$1 = u(x,\pi) = \sum_{n \geq 0}b_n\phi_n(x,\pi) = \sum_{n\geq 1}[b_n\sinh(n\pi)]\sin(nx) = \sum_{n\geq 1}\tilde{b}_n\sin(nx).$$ Now, we get a Fourier sine series and we know that they are orthogonal over $[0,\pi]$ (which has nothing to do with the orthogonality of "the eigenfunctions" of this problem, but orthogonality of "the eigenfunctions" from which sines are a solution which is also a Sturm-Louiville problem: $v'' + v = 0$ on $[0,\pi]$), so multiplying both sides by $\sin(mx)$ and integrating gives $$\int_0^\pi\sin(mx)dx = \int_0^\pi \left(\sum_{n\geq 1}\tilde{b}_n\sin(nx)\right)\sin(mx) ``=" \sum_{n\geq 1}\left(\int_0^\pi \tilde{b}_n\sin(nx)\sin(mx)dx\right),$$ where the equal in quotes is allowing an interchange of integral and sum, which has not been justified (but allowed for what we are doing, I suppose). Then we see by orthogonality of the sines that it equals $$\int_0^\pi \tilde{b}_n\sin(nx)\sin(nx) dx= \tilde{b}_n\int_0^\pi\sin^2(nx)dx = \frac{\pi}{2}b_n\sinh(n\pi),$$ which gives the result that you were looking for. So, I suppose that this is what was intended.

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    $\begingroup$ The first sentence of my answer is incorrect: Please interpret it to mean that "The orthogonality of this Sturm-Louiville problem would not be referring to $\sin(nx)$ and $\sinh(nx)$, but to \{\phi_n(x,y)\}". It turns out the "the orthogonality" seems to not have actually been referring to that at all, but orthogonality of a different problem: the orthogonality of sines. $\endgroup$ – user357980 Feb 25 '18 at 4:48

When you have a regular Sturm-Liouville problem with homogeneous endpoint conditions, then the eigenvalues are a discrete set with a lower bound and no finite point of accumulation. The corresponding eigenfunctions for different eigenvalues are automatically orthogonal with respect to the weight of the problem. That's what is at work here.

The relevant equation is $$ X''(x)=\lambda X,\;\; X(0)=X(\pi)=0. $$ The corresponding eigenfunctions are $\{ \sin(nx) \}_{n=1}^{\infty}$ and the weight function is $1$. So $$ \int_{0}^{\pi}\sin(nx)\sin(mx)dx = 0,\;\;\; n\ne m. $$ You can then use orthogonality to isolate every coefficient in your expansion $$ 1 = \sum_{n=1}^{\infty}b_n\sinh(n\pi)\sin(nx). $$ What you do is multiply both sides by $\sin(kx)$ and integrate over $[0,\pi]$, and use this orthogonality to conclude that $$ \int_{0}^{\pi}\sin(kx)dx = b_k \sinh(k\pi)\int_{0}^{\pi}\sin^2(kx)dx \\ \implies b_k = \frac{\int_{0}^{\pi}\sin(kx)dx}{\sinh(k\pi)\int_{0}^{\pi}\sin^2(kx)dx}. $$


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