2
$\begingroup$

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?

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

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 $a(n,m)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.