I am really struggling with the problem:
Consider a vibrating string occupying the interval $0\leq x \leq l$. Suppose the string is strucked in the middle so that its initial displacement is zero but its initial velocity $u_{t}(x,t)$ is $1$ for $|x-\frac{1}{2}l|<\delta$ and $0$ elsewhere. Find $u(x,t)$ for $t>0$.
I am having the most trouble with the fact that the velocity is only $1$ on $(\frac{1}{2}l-\delta, \frac{1}{2}l+\delta)$. I don't know how to find a Fourier series that satisfies this condition. If I ignore this and assume that the velocity is $1$ on $[0,l]$, then I get something like
$$u(x,t) = \sum_{n=1}^{\infty} \frac{4l}{(2n-1)^{2}\pi^{2}c} \sin \frac{(2n-1)\pi x}{l} \sin \frac{(2n-1) \pi ct}{l}$$
The correct answer should be $$u(x,t) = \sum_{n=1}^{\infty} \frac{4l (-1)^{n+1}}{(2n-1)^{2}\pi^{2}c} \sin \frac{(2n-1) \pi \delta}{l} \sin \frac{(2n-1)\pi x}{l} \sin \frac{(2n-1) \pi ct}{l}$$
I am thinking that, because I assumed the velocity was $1$ on the whole interval $[0,l]$, so if I multiply the expression by the Fourier series for $\begin{cases}1 & |x-\frac{1}{2}l|<\delta \\ 0 & |x-\frac{1}{2}l|\geq\delta\end{cases}$, then I should get the correct answer? And comparing the correct answer to what I got, it seems like that Fourier series would be $\sum_{n=1}^{\infty} (-1)^{n+1} \sin \frac{(2n-1)\pi \delta}{l}$. I have no clue how to find this series.
Any help is appreciated!
p.s. We have learned that for a vibrating string, the conditions are $$ u_{tt} = c^{2} u_{xx}, \quad u(x,0) = f(x), \quad u_{t}(x,0) = g(x), \quad u(0,t) = u(l,t) = 0. $$
where $f(x)$ and $g(x)$ are defined on $[0,l]$ and their Fourier series are $$f(x) = \sum_{n=1}^{\infty} b_{n} \sin \frac{n \pi x}{l}, \quad g(x) = \sum_{n=1}^{\infty} B_{n} \sin \frac{n \pi x}{l}$$
And the general solution is $$ u(x,t) = \sum_{n=1}^{\infty} \sin \frac{n\pi x}{l} \left(b_{n} \cos \frac{n \pi c t}{l} + \frac{lB_{n}}{n\pi c} \sin \frac{n \pi c t}{l} \right) $$
In this problem, since the initial displacement is $0$, I took $f (x)$ to be $0$. Then I think $u(x,t)$ will be in the form $$ u(x,t) = \sum_{n=1}^{\infty} \sin \frac{n\pi x}{l} \left(\frac{lB_{n}}{n\pi c} \sin \frac{n \pi c t}{l} \right) $$