Today at our math exam we got a very weird Laplace equation which to me seemed to only have solution $u(x,y)=0$. I also tried putting it in Mathematica to no avail. Here is the problem, is there any other solution than $0$ possible with the separation of variables?

Use the separation of variables to solve the Laplace equation $u_{xx}+u_{yy}=0$ for a function $u=u(x,y)$ in the area $0 \le x \le \pi, 0 \le y \le \pi$ with the following boundary conditions:

  1. $u(x,0)=0;u(0.y)=0;u(\pi,y)=0;u(\pi,y)=\sin y \cos y$
  2. $u(x,0)=0;u(0.y)=0;u(x,\pi)=\sin x \cos x;u(\pi,y)=0$
  3. $u(x,0)=0;u(0.y)=0;u(x,\pi)=\sin x \cos x;u(\pi,y)=\sin y \cos y$

For the first boundary conditions I got $u(x,y)=\sum_{n=1}^{\infty}c_{n}(e^{nx}-e^{-nx})\sin ny$. Using a fourier series I tried to determine the coefficient $c_{n}=\frac{2}{\pi(e^{n \pi}-e^{-n \pi})} \int_{0}^{\pi} \cos x \sin x \sin nx dx$. Which is $0$ for all $n$. Where did I make a mistake, if anywhere?

  • $\begingroup$ $\cos(x)\sin(x) = 0.5\sin(2x)$, so for $n =2$, the integral isn't 0, is it? $\endgroup$ – Andy Walls Dec 21 '17 at 2:17
  • $\begingroup$ ahh yes now I see it, thanks a lot! $\endgroup$ – stijn delnoij Dec 21 '17 at 9:42
  • $\begingroup$ $u=0$ is not the solution of the problem because the BC's are not zero. Perhaps you may take a look at the solution. $\endgroup$ – daulomb Dec 21 '17 at 21:22

The function $u=0$ cannot be a solution of the problem because of the boundary conditions (which are not all zero on the boundary). In the first boundary conditions the first $u(\pi, y)=0$ must be replaced by $u(x,\pi)=0$ in order to have zero boundary conditions for $y$. After coorecting this and applying the separation of variables we get: $$u(x,y)=\sum_{n=1}^{\infty}\big(c_{n}\sinh(nx)+d_n\cosh(nx)\big)\sin(ny).$$ Now using the othogonality of the functions $\sin(ny)$ and BC's for $x$ we will determine $c_n$ and $d_n$. $$u(0, y)=0\Longrightarrow d_n=0.$$

$$u(\pi, y)=\sin y\cos y=\frac{1}{2}\sin(2y)=\sum_{n=1}^{\infty}c_n\sinh(n\pi)\sin(ny).$$ Orthogonality of $\sin(ny)$ implies that $$\int_0^\pi\sin(2y)\sin(ny)dy=\pi c_n\sinh(n\pi).$$ In this case we should be more careful, because the integral on the LHS is not zero for all $n$. It is zero (by othogonality) for all $n$ except for $n=2$. When $n=2$, $$\int_0^\pi\sin^2(2y)dy=\frac{\pi}{2}$$ and hence $c_2=\displaystyle\frac{1}{2\sinh(2\pi)}$. Therefore, $$u(x,y)=\frac{\sinh(2x)\sin (2y)}{2\sinh(2\pi)}.$$


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