Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

So my current solution to the $1D$ wave equation is (with my given boundary and initial conditions): $$y(x,t) = \sum_{n=1}^\infty C_n\cdot \sin\frac{n \pi x}{2 l}\cdot\cos\frac{n \pi c t}{2 l}$$

However there is one final initial condition that is piecewise, I'm unsure of how to apply this and the solutions expected are of a infinite series (fourier series).

The last initial condition is:

$$y(x,0)= \begin{cases} R\cdot\frac{x}{l} & \quad 0\le x\le 1 \\ R\cdot\left(2 - \frac{x}{l}\right) & \quad l\le x\le 2 l \end{cases}$$

Any help or advice on solving is much appreciated.

share|improve this question
    
Please, check if I add correct LaTeX for your formula. –  m0nhawk Nov 29 '12 at 17:56
    
I'm going to assume that because there is an $n$ in $y(x,t)$ it isn't just some single fixed $n$ and you should actually sum over them all...also should there really be a $ct$ in the $\sin$ part? I think it should just be $x$. –  Matt Nov 29 '12 at 18:14
    
The $x$ was erroneously replaced by $ct$ by an earlier formatting edit. @Sam, you can avoid these sorts of problems by formatting the post yourself. We have a basic tutorial and quick reference for that. –  joriki Nov 29 '12 at 18:28
add comment

1 Answer 1

Assuming all of our changes are correct, you can get all the $C_n$ by using the formula of the Fourier sine series coefficients:

The solution tells you that $y(x,0)=\sum_{n=1}^\infty C_n\sin\left(\frac{n\pi x}{2l}\right)$, and the condition tells you that this series must equal that piecewise function.

Thus $$C_n=\frac{1}{l}\int_0^{2l}y(x,0)\sin\left(\frac{n\pi x}{2l}\right)dx.$$

Since the function is given piecewise you have to break it up into two integrals:

$$C_n=\frac{1}{l}\int_0^l R\cdot\frac{x}{l}\sin\left(\frac{n\pi x}{2l}\right)dx+\frac{1}{l}\int_l^{2l}R\left(2-\frac{x}{l}\right)\sin\left(\frac{n\pi x}{2l}\right)dx$$

It is kind of tedious to make sure you don't lose any of these constants while integrating, but you should be able to solve it from here (look up $\int x\sin(ax)dx$ in a table to make it go quicker if you want).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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