Solving wave equation I want to solve the following PDE:
$$\begin{align}
u_{tt}&=c^2u_{xx}-\gamma u_x, \quad 0<x<1, \quad t>0,\\ \\
u(0,t)&=u(1,t)=0, \\
u(x,t=0)&=x(1-x),\\
u_t(x,t=0)&=0. 
\end{align}$$
Though I see the literature has examples with the damped wave equation, I don't have an explanation for the $-\gamma u_x$ term. How would I solve it?
 A: One can, as I mentioned in the comment above, do the problem by Fourier series. 
One can however to a funny change of variables to convert the equation to one that is perhaps better studied. 
We observe the following fact:
$$ (c \partial_x - \frac{\gamma}{2c})(c\partial_x - \frac{\gamma}{2c}) = c^2 \partial^2_{xx} - \gamma \partial_x + \frac{\gamma}{4c^2} $$
which means that your equation can be re-written as
$$ u_{tt} = (c \partial_x - \frac{\gamma}{2c})(c\partial_x - \frac{\gamma}{2c})u - \frac{\gamma}{4c^2}u $$
Now, we have that
$$ c \partial_x u - \frac{\gamma}{2c} u = c e^{\gamma x / (2c^2)} \partial_x \left( e^{- \gamma x/(2c^2)} u\right) $$
which rewrites the equation as
$$u_{tt} = e^{\gamma x / (2c^2)} c^2 \partial^2_{xx}  \left(e^{-\gamma x /(2c^2)}u\right) - \frac{\gamma}{4c^2}u $$

So we make the substitution $v(x,t) = e^{-\gamma x / (2c^2)} u(x,t)$. The equation for $v$ is simply the following Klein-Gordon equation
$$ v_{tt} = c^2 v_{xx} - \frac{\gamma}{4c^2} v $$
with the boundary conditions
$$ v(0,t) = v(1,t) = 0 $$
and the initial values
$$ \begin{align} v_t(x,0) &= 0 \\ v(x,0) &= e^{-\gamma x / (2c^2)} x(1-x) \end{align}$$
A: If $\gamma \neq \gamma(u)$, or even simpler, $\gamma$ is a constant, then your problem can be solved using separation of varaiables since the PDE is linear. Furthermore, the boundary conditions are homogenous, as well as the PDE.
Set $u(t,x) = P(x)Q(t)$, with $P\neq0, Q\neq 0$. Plug it into the equation to obtain:
$$PQ'' = c^2 P''Q - \gamma P' Q,$$ which can be rewritten as:
$$\frac{\frac{c^2}{\gamma} P'' -  P'}{P} = \frac{Q'}{\gamma Q} = \lambda, $$ where $\lambda$ is some real constant, which in general can be zero, positive or negative. The equation written above leads you to two problems, one for $P(x)$ and another for $Q(t)$. Let's solve the problem for $P$ (you may solve the problem for $Q$ as well, but it's not relevant). 
The boundary conditions for $P$ are given by the original boundary conditions for $u$, indeed: $P(0) = P(1) = 0$; so the problem for $P$ becomes:
$$P''- \beta P'- \beta \lambda P = 0, \quad 0 < x < 1, \quad P(0) = P(1) = 0,$$ where $\beta = \frac{\gamma}{c^2}$. The solution of this equation depends on the value of $\lambda$ (in general it'll be: $P(x) = Ae^{s_1x} + Be^{s_2 x}$, where $s_i$ are the two -different- roots of $s^2 - \beta s - \beta \lambda$.), whether it's positive, zero or negative. 
Solve for the so called eigenfunctions, $P_n(x)$, and expand the solution in terms of them as follows:
$$u(t,x) = \sum^\infty_n P_n(x) Q_n(t),$$ 
plug this into the original PDE and solve for $Q_n(t)$ by applying the prescribed initial conditions.
Can you take it from here?
Cheers!
