Sturm Liouville applied to a Laplace equation

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

$u(0,y)=u(\pi,y)=u(x,0)=0$

$u(x,\pi)=1$

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

$f(x)=\sin(nx)$

$g(y)=\sinh(ny)$

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

• I missed in my edit, in your expression after applying the boundary condition, you should have $\sinh(n\pi)$ not $\sinh(\pi y)$ – Tyberius Feb 25 '18 at 4:33
• 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. – 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.
• 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. – user357980 Feb 25 '18 at 4:48
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 $$\int_{0}^{\pi}\sin(kx)dx = b_k \sinh(k\pi)\int_{0}^{\pi}\sin^2(kx)dx \\ \implies b_k = \frac{\int_{0}^{\pi}\sin(kx)dx}{\sinh(k\pi)\int_{0}^{\pi}\sin^2(kx)dx}.$$