Damped wave equation BVP solution with Fourier series Consider the problem where PDE given with boundary and initial conditions: 
$u_{tt}+2\alpha u_t-u_{xx} = 0, \ 0<x<\pi, \ t>0 \\ u(x,0)=f(x), \ u_t(x,0)=0,\ 0<x<\pi \\ u(0,t)=u(\pi,t)=0 \  t>0 \\$
I used method of separable variables and stuck at the point assumption of alpha, how I have to proceed? 
$u(x,t)=X(x)T(t)\\ XT''+2\alpha XT'-X''T=0 \\ \frac{X''}{X}=\frac{T''+2\alpha T'}{T}=\lambda\\$
To avoid trivial solutions we set $X(0)=X(\pi)=0$ due to boundary conditions, then we solve the cases of $\lambda$ for X(x) and found out that we get solution when $\lambda$ is negative.  $\lambda= -\mu^2 <0 $ where $\mu=n,\ n=1,2,3..$.
then solution for X is $X_n=sin(nx)$ for $n=1,2,3...$
then I plugged the value of $\lambda$ for equation of T and I got $T''+2\alpha T'+n^2 T=0$
when I am solving this equation I got $r^2+2\alpha r+n^2=0$ what kind of assumption I need take in consideration? thanks a lot. After founding out with roots I will get series of cosine and sine then I need to find coefficients of Fourier series by giving initial conditions above.
 A: The separation of variables $u(x,t) = X(x) T(t)$ leads to
$$
T'' + 2 \alpha T' + \lambda T = 0
\qquad\text{and}\qquad
X'' + \lambda X = 0 \, ,
$$
where $\lambda$ is the separation constant. Solutions are of the form
\begin{aligned}
T(t) &= e^{-\alpha t}\big( Ae^{-\text i t \sqrt{\lambda - \alpha^2}} + B e^{\text i t \sqrt{\lambda - \alpha^2}}\big) \\
X(x) &= C \cos(x\sqrt{\lambda}) + D \sin(x\sqrt{\lambda})
\end{aligned}
Imposing the boundary conditions $X(0) = X(\pi) = 0$, non-trivial solutions are obtained provided that $C = 0$ and $\sqrt{\lambda} = n$ with $n \in \Bbb N^*$. Hence, due to the superposition principle, the solution might be sought as
$$
u(x,t) = e^{-\alpha t} \sum_{n = 1}^{+\infty} \left(A_n \cos \beta_n t + B_n \sin \beta_n t \right) \sin nx \, ,
$$
where $\beta_n = \sqrt{n^2 - \alpha^2}$ is real and $0\leq \alpha < 1$ (weak damping is assumed).
The initial conditions are satisfied if
$$
f(x) = \sum_{n = 1}^{+\infty} A_n \sin nx
\qquad\text{and}\qquad
0 = \sum_{n = 1}^{+\infty} ( \beta_n B_n - \alpha A_n) \sin nx \, .
$$
By using Fourier series, we obtain
$A_n = \frac{2}{\pi}\int_0^{\pi} f(x) \sin nx\,\text dx$ and $B_n = \alpha A_n/\beta_n $.

Note: This is a subcase of Exercise 4.4.3 p. 142 of (1).
Note: this equation is also known as telegraphers' equation or simply telegraph equation.
(1) R. Habermann, Applied Partial Differential Equations; with Fourier Series and Boundary Value Problems, 5th ed., Pearson Education Inc., 2013.
