Help to find the general solution for a separable PDE in the form: $$u_{tt} = c^2 u_{xx}$$ Conditions: $$u(0,t) = u(L,t) = 0 $$ $$u(x,0) = f(x)$$ Initial derrivative: $$u_t(x,0) = g(x)$$
I was given the following solution: $$G_n(x)T_n(t) = u(x,t)$$
where $$G_{n}(x) = \sin (\sqrt k x)$$ $$k=(n\pi)^2$$
and $$T_n(t)=B_n \cos(cn\pi t)+B_n^* \cos(cn\pi t)$$ but I don't know how to find $$B_n$$ and $$B_n^*$$
Please help I need this for my final!
$$u_{tt}=c^2u_{xx}$$ Search for particular solutions on the form $u=X(x)T(t)$
$XT''=c^2X''T\quad;\quad \frac{T''}{T}=c^2\frac{X''}{X}=-\nu^2=$constant.
Sign $-$ in order to get sinusoidal functions and comply to the conditions $u(0,t)=u(L,t)=0$.
$X(x)=\sin(\nu x)=\sin(n\pi\frac{x}{L})\quad$ with $\quad\nu=n\pi\frac{1}{L}\quad$ which satisfies the condition.
$T(t)=\sin(\frac{\nu}{c}t)\quad$ and $\quad T(t)=\cos(\frac{\nu}{c}t)$
General solution complying with the condition $u(0,t)=u(L,t)=0$ : $$u(x,t)=\sum_{\forall n} \sin\left(n\pi\frac{x}{L}\right)\left(a_n\sin(\frac{n\pi}{cL}t)+b_n\cos(\frac{n\pi}{cL}t)\right) \tag 1$$
Condition $u(x,0)=f(x)$ :
$$u(x,0)=\sum_{\forall n} b_n\sin\left(n\pi\frac{x}{L}\right)=f(x)$$ The Fourier series of $f(x)$ on $0<x<L$ gives the values of $b_n$. $$b_n=\frac{2}{L}\int_0^L f(x) \sin\left(n\pi\frac{x}{L}\right)dx$$
Condition $u_t(x,0)=g(x)$ :
$$u(x,t)=\sum_{\forall n} \frac{n\pi}{cL}a_n\sin\left(n\pi\frac{x}{L}\right) =g(x)$$
The Fourier series of $g(x)$ on $0<x<L$ gives the values of $a_n$. $$\frac{n\pi}{cL}a_n=\frac{2}{L}\int_0^L g(x) \sin\left(n\pi\frac{x}{L}\right)dx$$
With the $a_n$ and $b_n$ found, the above equation $(1)$ is solution of the PDE and satisfies all the specified conditions.