 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 givesThe 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 $2e^{-t}-1\downarrow 0$. Thus the first shock formation by characteristic intersection occurs atFor $0\le t<t_{\ast }$ the profile consists of a left plateau, a straight descending ramp ending at zero, and a right zero plateau. At $t_{\ast }$ the ramp becomes a vertical jump. This is a damped Burgers equation; viscosity smooths that jump into a thin transition.