Sturm Liouville applied to a Laplace equation

Question

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

Answer 1

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.

Answer 2

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 $$ \int_{0}^{\pi}\sin(kx)dx = b_k \sinh(k\pi)\int_{0}^{\pi}\sin^2(kx)dx \\ \implies b_k = \frac{\int_{0}^{\pi}\sin(kx)dx}{\sinh(k\pi)\int_{0}^{\pi}\sin^2(kx)dx}. $$