# Thread: Fourier Series Wave Equation

1. ## Fourier Series Wave Equation

It's been forever since I've done a course in PDE's, so I'd really appreciate a step by step process for solving this

$\displaystyle u_{tt}-\alpha^2 u_{xx}$ $\displaystyle 0<x<L, t>0$
$\displaystyle u(0,t)=u(L,t)=0$ $\displaystyle t \geq 0$
$\displaystyle u(x,0)=f(x)$ $\displaystyle 0 \leq x \leq L$
$\displaystyle u_t(x,0)=g(x)$ $\displaystyle 0 \leq x \leq L$

leads to a solution of the form
$\displaystyle u(x,t) = \sum_{n=1}^\infty \left[a_n \cos\left(\frac{n\pi \alpha}{L}t\right)+b_n \sin\left(\frac{n\pi \alpha}{L}t\right)\right]\sin\left(\frac{n\pi x}{L}\right)$

If $\displaystyle \alpha=3$,$\displaystyle L=\pi$, $\displaystyle f(x)=6\sin(2x)+2\sin(6x)$ and $\displaystyle g(x)=11\sin(9x)-14\sin(15x)$.

I want to say the $\displaystyle a_n=0$ due to the boundary condition, but I'm not sure.

I'm also not sure what the integral's should look like when I'm trying to find the fourier coefficents of u,f and g.

2. ## Re: Fourier Series Wave Equation

u(x,0) = sum(n,1,Infinity, (a_n cos(n pi alpha 0/L) + b_n sin(n pi alpha 0/L))sin(n pi x/L)

u(x,0) = sum(n,1,Infinity, a_n sin(n pi x/L)) = f(x)

so it looks to me like your a_n are a sort of Fourier coefficients of f(x)

3. ## Re: Fourier Series Wave Equation

Nevermind, just figured it out. Thanks anyways.