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I am trying to solve this problem in my analysis book in a chapter on Fourier series:

Solve the differential equation $$(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}) u(x,y) = 0$$ In the rectangle $0 \le x \le a, 0 \le y \le b$ with boundary conditions that $u(x,0) = f(x)$ on the bottom side and $u = 0$ on the other three sides by expanding $u(x,y)$ in a Fourier sine series in x, for each fixed y.

I'm not really sure how to tackle this problem and so far all I can come up with is the trivial solution $u(x,y) = 0$. I'd very much appreciate any help. Thanks!

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  • $\begingroup$ Let's start with the basics. Why are we expanding in a sine series? $\endgroup$ – Ron Gordon Apr 8 '15 at 17:58
  • $\begingroup$ We would want to use a sine series to represent an odd function, but I don't think we know that about $u$ so I'm not sure $\endgroup$ – Sliingshot Apr 8 '15 at 18:23
  • $\begingroup$ Is $u$ continuous on the boundary? What does that imply for $f$? $\endgroup$ – Lutz Lehmann Apr 8 '15 at 18:57
  • $\begingroup$ is this a homework problem? $\endgroup$ – abel Apr 8 '15 at 19:01
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I'm assuming you've been through Fourier Series before, so I've skipped some minor details. From your question, we get the BCs

$$\begin{align} u(x,0) &= f(x) \\ u(x,b) &= 0 \\ u(0,y) &= 0 \\ u(a,y) &= 0 \\ \end{align}$$

Assuming $u(x,t) = X(x)Y(y)$ and using separation of variables (I'm using $- \lambda$ for my separation constant), we find

$$\begin{align} X'' &= - \lambda X \ \ \ (1) \\ Y'' &= \lambda Y \ \ \ (2) \\ \end{align}$$

and from our BCs we get

$$X(0) = X(a) = 0 \ \ \ (3)$$ $$Y(b) = 0 \ \ \ (4)$$

So solving $(1)$ we get

$$X = A\cos(\sqrt{\lambda} x) + B\sin(\sqrt{\lambda} x)$$

and applying $(3)$ we find

$$\begin{align} X(0) &= 0 \\ &= A\cos(0) + B\sin(0) \\ \implies A &= 0 \\ \implies X(x) &= B\sin(\sqrt{\lambda} x) \ \ \text{(this is where your sin comes from)}\\\\ X(a) &= 0 \\ \implies B\sin(\sqrt{\lambda} a) &= 0 \\ \implies \sqrt{\lambda} a &= n \pi \ \ \text{(B $\ne 0$, otherwise trivial solution)} \\ \implies \lambda &= \bigg( \frac{n \pi}{a} \bigg)^{2} \\\\ \implies X(x) &= B_{n}\sin \bigg( \frac{n \pi x}{a} \bigg), \ \ n \ge 1 \\ \end{align}$$

Solving $(2)$, we get

$$\begin{align} Y(y) &= D\cosh( \sqrt{\lambda} y) + E\sinh( \sqrt{\lambda} y) \\ &= D\cosh \bigg( \frac{n \pi y}{a} \bigg) + E\sinh \bigg( \frac{n \pi y}{a} \bigg) \\ \end{align}$$

and applying $(4)$ we find

$$\begin{align} Y(b) &= 0 \\ &= D\cosh \bigg( \frac{n \pi b}{a} \bigg) + E\sinh \bigg( \frac{n \pi b}{a} \bigg) \\ \implies D &= -E\tanh \bigg( \frac{n \pi b}{a} \bigg) \ \ \ \text{(as $\cosh$ can't equal $0$)} \\\\ \implies Y(y) &= -E\tanh \bigg( \frac{n \pi b}{a} \bigg) \cosh \bigg( \frac{n \pi y}{a} \bigg) + E\sinh \bigg( \frac{n \pi y}{a} \bigg) \\ &= \frac{E}{\cosh \bigg( \frac{n \pi b}{a} \bigg)} \sinh \bigg( \frac{n \pi}{a} (y - b) \bigg) \\ \end{align}$$

Hence, our solution is given by

$$\begin{align} u(x,y) &= \sum_{n = 1}^{\infty} B_{n}\sin \bigg( \frac{n \pi x}{a} \bigg) \frac{1}{\cosh \bigg( \frac{n \pi b}{a} \bigg)} \sinh \bigg( \frac{n \pi}{a} (y - b) \bigg) \\ &= \sum_{n = 1}^{\infty} B_{n} \text{sech} \bigg( \frac{n \pi b}{a} \bigg ) \sin \bigg( \frac{n \pi x}{a} \bigg) \sinh \bigg( \frac{n \pi}{a} (y - b) \bigg) \\ \end{align}$$

Solving for our other BC, we find

$$\begin{align} u(x,0) &= \sum_{n = 1}^{\infty} B_{n} \text{sech} \bigg( \frac{n \pi b}{a} \bigg ) \sin \bigg( \frac{n \pi x}{a} \bigg) \sinh \bigg( \frac{n \pi b}{a} \bigg) \\ &= \sum_{n = 1}^{\infty} B_{n} \tanh \bigg( \frac{n \pi b}{a} \bigg) \sin \bigg( \frac{n \pi x}{a} \bigg) \\ &= f(x) \\ \end{align}$$

Now you just need to solve for your $B_{n}$ using orthogonality. If you need anymore help, just comment below.

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