Solution of boundary value problem using Fourier series

I want to solve the following PDE using Fourier series:

• $$u(x,y): \Omega \to \mathbb{R}$$,
• $$\Omega=(0,\pi)\times (0,2\pi)$$
• $$u-3u_{xx}-u_{yy}= 3\sin(2x)-\sin(5x)$$

$$u_{xx}$$ and $$u_{yy}$$ are second derivatives with respect to $$x$$ and $$y$$ respectively. The boundary conditions of the problem are:

• Dirichlet boundary condition in $$x$$ direction: $$u(x=0,y)=u(x=\pi,y)=0$$
• Neumann boundary condition in $$y$$ direction: $$u_y(x,y=0)=u_y(x,y=2\pi)=0$$

I chose my base functions for $$u$$ such that they satisfy boundary conditions: $$\sin(nx)\cos(my/2)$$. Therefore, the left-hand side of the equation would be: $$\sum a(n,m)\left[1+3n^2+\frac{m^2}{4}\right]\sin(nx)\cos(my/2)$$ My problem is that I should be able to write the right-hand side of the equation in form of $$\sum b(n,m)\sin(nx)\cos(my/2)$$, so that I can obtain $$a(n,m)$$. Any suggestions?

• Please edit the question, instead of using * to bold the text, use \$text \$. For example: \$\pi\$ renders as $\pi$. – fosho Jan 30 '16 at 13:55
• See this page for info on how to typeset math formulas using mathjax/latex. It's really easy to learn. – Winther Jan 30 '16 at 14:07
• Thanks for the tips and also the edit. – Eman Jan 30 '16 at 14:13
• $m=0$ is allowed, which means $\sin(2x),\sin(5x) \in \{ \sin(nx)\cos(my) : 1 \le n < \infty, \;\; 0 \le m <\infty \}$ – DisintegratingByParts Jan 31 '16 at 0:11

You have to project the left-hand and right-hand sides of the PDE on the basis $$\sin(nx)\cos(my/2)$$, that is you have to compute $$\int_0^\pi\int_0^{2\pi} \Bigl(\sum_{n,m}a(n,m)[1+3n^2+m^2/4]\sin(nx)\cos(my/2)-3\sin(2x)+\sin(5x)\Bigr)\sin(nx)\cos(my/2)\, \mathrm{d}y\,\mathrm{d}x,\quad \forall n,m$$ It will give you the missing terms so that you can solve in $$a(n,m)$$.