= Solution
For zero viscosity, the <method of characteristics> gives $u=u_0(a)e^{-t}$ and $x=a+u_0(a)(1-e^{-t})$. On the sloping part of the initial profile, $u_0(a)=-2a$, so $x=a(2e^{-t}-1)$. While this map is increasing, inversion gives
$$
\boxed{u(x,t)=\begin{cases}
2e^{-t},&x\leq1-2e^{-t},\\
-\dfrac{2e^{-t}x}{2e^{-t}-1},&1-2e^{-t}<x\leq0,\\
0,&x>0.
\end{cases}}
$$
The sloping interval shrinks from width one to zero while its height decreases from two to one. Its <spatial derivative> is $-2e^{-t}/(2e^{-t}-1)$ and diverges as $2e^{-t}-1\downarrow0$. Thus the first <shock formation by characteristic intersection> occurs at
$$
\boxed{t_* =\log2,\qquad x_*=0.}
$$
For $0\leq t<t_*$ the profile consists of a left plateau, a straight descending ramp ending at zero, and a right zero plateau. At $t_*$ the ramp becomes a vertical jump. This is a <damped Burgers equation>; viscosity smooths that jump into a thin transition.
\Image[/past-exam-of-the-differential-equations-and-applications-course-of-the-university-of-oxford/2011-damped-burgers-shock.png]
{title=Decay and steepening of the inviscid ramp, its characteristic collapse, and the entropy shock after formation}
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