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We have $u_{xx} = u_{tt}$ on $0<x<L, t\geq 0$

with boundary conditions $u(0, t) = 0$ and $u(L, t)=0$

The string is released from rest with initial displacement $u(x,0) = \left\{\begin{matrix} \ 1 &\text{ for } 0 <x\leq \frac{L}{2}\\ 0 &\text{for} \frac{L}{2}<x<L \end{matrix}\right.$

What are the steps to solve this by separation of variables?

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Recovering an old question by providing my answer. Hopefully this will be of instructional value.

Find u(x,t) using separation of variables. Assume $u(x,t)=X(x)T(t)$ and $\lambda$ is the separation constant.

Step 1) $\frac{X''}{X}=\frac{T''}{T}=-\lambda$ $\rightarrow T''+\lambda T = 0,t\geq0$ Notice that $X$ satisfies the boundary conditions $X(0)=X(L)=0$ and $T$ satisfies the initial condition $T'(0)=0$, (this is from $u(x,t)$).

Step 2) $\lambda >0$ must be true to have a non-trivial soplution for X. Assum $\lambda >0$, then $X(x)=Acos\sqrt{\lambda x} + Bsin\sqrt{\lambda x}$. Now the boundary conditions are $Acos(0)+Bsin(0)=Acos\sqrt{\lambda L} + B sin \sqrt{\lambda L}=0$

Step 3) For $A=0$ and nontrivial solution, $sin\sqrt{\lambda L}=0$ is a requirement. Now we can find the eigenvalues which are $\lambda_n$=($\frac{n\pi}{L})^2$, $n= 1, 2, ...$ with associated eigenfunction $X_n(x)=B_n sin \frac{n\pi x}{L}$.

Step 4) With these values of $\lambda,$ the general solution of the equation for $T$ is $T_n(t)=C_n cos\frac{n\pi t}{L} + D_Nsin\frac{n\pi t}{L},$ $n= 1, 2, ...$

To satisfy the initial condition $T'(0)=0$, we take $D_n=0$ so $T_n(t)=C_n cos\frac{n\pi t}{L},$ $n=1, 2, ...$

We now have a collection of functions call it $u_n(x,t)=X_n(x)T_n(t)=B_n sin\frac{n\pi x}{L}cos\frac{n\pi t}{L}$

Step 5) Take a linear combination of $u_n(x,t)$ we have $u(x,t)=\sum\limits_{n=1}^\infty B_n sin\frac{n\pi x}{L} cos\frac{n\pi t }{L}$ Find $B_n$ such that $u(x,0)=\sum\limits_{n=1}^\infty B_n sin\frac{n\pi x}{L}=f(x)$, where $f(x)= \begin{cases} 1, & \mbox{for } 0<x\leq L \\ 0, & \mbox{for } L/2<x\leq L \end{cases}$

The $B_n$ are the coefficients of the sine series $f(x).$ We extend $f(x)$ to an odd function with period $2L$

$g(x) = \begin{cases} f(x), & \mbox{for } 0<x\leq L \\ 0, & \mbox{for } x=0 \\-f(x), &\mbox{for }L\leq x <0\\ g(x-2L), &\mbox{everywhere}\end{cases}$

Now we can take an integral $B_n=\frac{1}{L}\int_{-L}^{L}sin\frac{n\pi x}{L} dx =-\frac{1}{L}\int_{-L/2}^{0}sin\frac{n\pi x}{L}dx+\frac{1}{L}\int_{0}^{L/2}sin\frac{n\pi x}{L}dx=$ $\frac{2(1-cos(\frac{\pi n}{2}))}{\pi n}$

The solution is $u(x,t)=\sum\limits_{n=1}^\infty \frac{2(1-cos(\frac{\pi n}{2}))}{\pi n}sin\frac{n\pi x}{L}cos\frac{n\pi t}{L}$

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