Solve the damped quasilinear wave equation $u_t+uu_x+u=0$ with $u(x,0)=f(x)$. 
Solve the damped quasilinear wave equation $u_t+uu_x+u=0$ with $u(x,0)=f(x)$.  Determine if the solution breaks when $f$ satisfies the condition $f^\prime(x)>-1$ for all $x\in\mathbb{R}$.

I'm stumped at how to begin. Please help.
 A: Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dt}{ds}=1$ , letting $t(0)=0$ , we have $t=s$
$\dfrac{du}{ds}=-u$ , letting $u(0)=u_0$ , we have $u=u_0e^{-s}=u_0e^{-t}$
$\dfrac{dx}{ds}=u=u_0e^{-s}$ , letting $x(0)=g(u_0)$ , we have $x=g(u_0)+u_0(1-e^{-s})=g(ue^t)+u(e^t-1)$ , i.e. $u=e^{-t}G(x+u(1-e^t))$
$u(x,0)=f(x)$ :
$G(x)=f(x)$
$\therefore u=e^{-t}f(x+u(1-e^t))$
A: As shown in this post and in the accepted answer, the method of characteristics gives
$$
u = f\big(x - (e^{-t}-1) u\big) \, e^{-t} .
$$
The linked post presents several methods to analyze the existence of shocks (breakdown of classical solution). Let us consider the dependence to initial data by differentiating the expression $x = f(x_0)(1-e^{-t}) + x_0$ of the characteristic curves w.r.t. $x_0$:
$$
\frac{\text d x}{\text d x_0} = f'(x_0)(1-e^{-t}) + 1 \, .
$$
If $f'(x_0) > -1$, then $\frac{\text d x}{\text d x_0} > 0$ for all positive times. Therefore, since the derivative $\frac{\text d x}{\text d x_0}$ does not vanish, no shock forms. The solution does not break.
