I know how to do (a).
I know the sine expansion of $\phi(x)$ on $(0,l)$: $\phi(x)=\sum_{n=1}^\infty B_n \sin \frac{n\pi x}{l}$, but could not get the desired form.
Through the formula I mentioned above, we can write $\tilde{\phi}(x)=\phi(2l-x)=-\sum_{n=1}^\infty B_n \sin \frac{n\pi x}{l}$ for $x\in (l,2l)$. I guess there might be something wrong here.
If (b) is solved, the rest should not be too hard.
Thank you!
Best Answer
Here's a hint: Forget about the sine expansion of $\phi$ on $(0,l)$; that's not going to get you anywhere. Consider instead (as the text suggests) the sine expansion of $\tilde{\phi}$ on $(0,2l)$. You will get an equality of the form "$\tilde{\phi}(x) = \sum\dots$ for $0 < x < 2l$", which when restricted to $0 < x < l$ gives you the desired result.