Consider the Burgers equation $u_t+uu_x=0$ with the initial condition $$u_0(x) =
\begin{cases}
u_l,x\leq0\\
u_r,x>0
\end{cases}$$
My attempt to solve this:
Using the method of characteristics we parametrize:
$$\frac{dx}{ds}=u, x(0,r)=x_0(r)=r$$
$$\frac{dt}{ds}=1,\ t(0,r)=t_0(r)=0 \implies t=s$$
$$\frac{du}{ds}=0,\ u(0,r)=u_0(r)=\begin{cases} u_l,r\leq0\\ u_r,r>0 \end{cases} \implies u=u_0(r)$$
$$\frac{dx}{ds}=u,x(0,r)=r \implies x=us+r=u_0(r)t+r$$ so solutions are constant along $x=u_0(r)t+r$.
For $r\leq 0, u_0=u_l \implies \text{slope}=\frac{1}{u_l}$ on the $x-t$ plane
For $r>0, u_0=u_r\implies \text{slope}=\frac{1}{u_r}$ on the $x-t$ plane
So why and where does a shock form at $t = 0$?
Best Answer
The initial condition is discontinuous, so the shock is there from the very beginning, and if $u_l>u_r$ it remains.
I will also note that it's not true that the border between area where $u = u_l$ and area where $u=u_r$ is given by equation $x-t=0$. If $u_l>u_r$ the equation of the discontinuity line is given by $x = \frac{u_l+u_r}{2}t $ (which can be derived from Rankine-Hugoniot condition), and the solution is $$ u(x,t) =\left\{\begin{array}{ll} u_l & \text{for } x< \frac{u_l+u_r}{2}t\\ u_r & \text{for } x >\frac{u_l+u_r}{2}t\end{array}\right.$$ If $u_l<u_r$, then there's no shockwave, but rarefaction appears instead: $$ u(x,t) =\left\{\begin{array}{ll} u_l & \text{for } x \le u_l t\\ \frac{x}{t} & \text{for } u_lt \le x \le u_rt \\u_r & \text{for } x \ge u_r t\end{array}\right.$$