I have the given problem,
\begin{equation}
\alpha u_{xx}=u_{tt}
\end{equation}
with conditions:
\begin{equation}
\begin{cases}
u(0,t)=u(L,t)=0\\
u(x,0)=x \\
u_t(x,0)=0
\end{cases}
\end{equation}
Separation of variables gives
\begin{equation}
u(x,t)=\begin{cases}
A\sin\frac{kx}{\alpha}\sin\frac{\omega t}{\alpha}\\
B\cos(\frac{kx}{\alpha})\sin\frac{\omega t}{\alpha}\\
B\cos(\frac{kx}{\alpha})\cos\frac{\omega t}{\alpha}\\
B\sin(\frac{kx}{\alpha})\cos\frac{\omega t}{\alpha}\\
\end{cases}
\end{equation}
Having the given IC., we can disregard from the cosine terms for the position function, X(x), which leaves
\begin{equation}
u(x,t)=\begin{cases}
A\sin\frac{kx}{\alpha}\sin\frac{\omega t}{\alpha}\\
B\sin(\frac{kx}{\alpha})\cos\frac{\omega t}{\alpha}\\
\end{cases}
\end{equation}
We find, using X(x) that:
\begin{equation}
X(x)= A\sin\frac{kL}{\alpha}=n\pi
\end{equation}
which gives that $k=\frac{n\pi\alpha}{L}$
This gives
\begin{equation}
u(x,t)=\begin{cases}
A\sin\frac{n\pi x}{L}\sin\frac{\omega t}{\alpha}\\
B\sin\frac{n\pi x}{L}\cos\frac{\omega t}{\alpha}\\
\end{cases}
\end{equation}
Then we discard the term with sine of $t$, since the derivative of the time-function T(t) must also be zero at t=0. So we are left with:
\begin{equation}
u(x,t)= B\sin\frac{n\pi x}{L}\cos\frac{\omega t}{\alpha}
\end{equation}
expanding to a sum:
\begin{equation}
u(x,t)= \sum_{n=1}^\infty B\sin\frac{n\pi x}{L}\cos\frac{\omega t}{\alpha}
\end{equation}
But how do I find B?
I tried to use the I.C. $ u(x,0)=x$ but that didn't work.
Best Answer
In the Fourier series you need a different constant $B_n$ in each term, like this: $$ u(x,t) = \sum_{n=1}^\infty B_n \sin\frac{n\pi x}{L} \cos\frac{\omega t}{\alpha} . $$ Then you want to find the constants $B_1$, $B_2$, $B_3$, etc., so that you achieve $u(x,0) = x$, i.e., $$ \sum_{n=1}^\infty B_n \sin\frac{n\pi x}{L} = x , $$ for $0 < x < L$. And you do this through the usual Fourier trick of multiplying both sides by by $\sin\frac{m \pi x}{L}$ and integrating from $0$ to $L$, using the orthogonality property $\int_0^L \sin\frac{n\pi x}{L} \sin\frac{m \pi x}{L} dx = 0$ for $m \neq n$. Then only the $m$th term on the left-hand side survives, and you get $$ B_m \int_0^L \sin^2 \frac{m\pi x}{L} dx = \int_0^L x \sin\frac{m\pi x}{L} dx , $$ where you can compute the integrals and solve for $B_m$.