Since the question is old, I'll give the complete solution, not just the $x_s$ part.
Before $T_c$
Draw the $xt$ coordinates, with $t$-axis pointing up. From the given $u(x,0)$, we find three families of characteristic lines:
- vertical lines $x=x_0$ with $x_0\le 0$
- slope $1/2$ lines $t=\frac12(x-x_0)$ with $0<x_0<1$
- vertical lines $x=x_0$ with $x_0\ge 1$
There is a gap between the families 1 and 2: it's a triangular region bounded by $x=0$ and $t=\frac12 x$.
This is the rarefaction wave: within this region, all characteristic lines go through $(0,0)$, the origin of the wave.
Therefore, the line through $(x,t)$ has slope $t/x$, which yields
$$
u(x,t)=x/t ,\qquad 0<x< 2t
$$
This is family 4 of characteristics.
The families 2 and 3 appear to overlap. This means they are separated by shock wave, which originates at $x=1$ at time $t=0$ and
moves to the right with velocity $\frac12(2+0)=1$ (the mean of velocities in front and behind the shock, as the jump condition says). Its trajectory is $x=1+t$.
The front edge of rarefaction wave $x=2t$ meets the shock wave $x=1+t$ when $t=1$. Thus, $T_c=1$. The meeting point is $(2,1)$.
After $T_c$
The characteristics of family 2 are no more; the shock is between families 3 and 4. The speed of shock wave as it passes through $(x,t)$ is the mean
of velocities in front and behind the shock:
$$
\frac{1}{2}\left( \frac{x}{t} +0 \right) = \frac{x}{2t}
$$
Therefore, the trajectory of shock wave is described by the ODE
$$
\frac{dx}{dt}=\frac{x}{2t}
$$
Solve this separable ODE with the initial condition $x(1)=2$ to get
$$
x_s(t) = 2 \sqrt{t}
$$
This is the trajectory of shock for $t>1$.
Second method, conservation law
Introduce the quantity $P(t)=\int_{-\infty}^\infty u(x,t)\,dx$. It is actually independent of $t$ because
$$\frac{dP}{dt} = \int_{-\infty}^\infty u_t \,dx = - \int_{-\infty}^\infty (u^2/2) _x \,dx = (u^2/2) \bigg|_{-\infty}^\infty =0$$
Since $P =2$ at $t=0$, it stays at $2$ for all times. At time $t>T_c$ the function $u$ is equal to $x/t$ for $0< x <x_s$, and is zero otherwise. Thus, its integral is
$$
\frac{1}{t} \frac{x_s^2}{2}
$$
Equating the above to $2$, we once again get
$$
x_s(t) = 2 \sqrt{t}
$$
If needed, here is a sketch of the characteristic curves in $x$-$t$ plane, which may help to see better what happens:
The solution obtained by the method of characteristics up to the breaking time satisfies $u = \phi (x-ut)$ for $t<1$, i.e.
$$
u(x,t) = \left\lbrace\begin{aligned}
&1 & &\text{if}\quad x \leq t \\
&\tfrac{1-x}{1-t} & &\text{if}\quad t \leq x\leq 1\\
&0 & &\text{if}\quad x\geq 1
\end{aligned}\right.
$$
At the breaking time, the shock speed $s$ is given by the Rankine-Hugoniot condition $s = \tfrac12 (1 + 0)$.
Therefore, after the breaking time $t\geq 1$,
$$
u(x,t) = \left\lbrace\begin{aligned}
&1 & &\text{if}\quad x < 1 +\tfrac12 (t-1) \\
&0 & &\text{if}\quad x > 1 +\tfrac12 (t-1)
\end{aligned}\right.
$$
To your questions:
- The Rankine-Hugoniot condition $[\![\frac{1}{2}u^2]\!] = s [\![u]\!]$ is a condition satisfied by discontinuous weak solutions of the Burgers' equation. A weak solution which satisfies the Rankine-Hugoniot condition is not necessarily a solution to the proposed initial-value problem $u(x,0) = \phi(x)$. To be a solution, it must be in agreement with the characteristics equation until they cross, i.e. the fact that $u$ is constant along the curves $u = \phi(x - ut)$. Moreover, since weak solutions are not unique, it must satisfy an additional entropy condition. In the case of Burgers' equation, the Lax entropy condition amounts to the following statement: "if characteristics cross, then a shock-wave arises; but if they separate, a rarefaction wave occurs".
- The location of a discontinuity is deduced from the characteristics equation, whereas its speed is deduced from the Rankine-Hugoniot condition. In the present case, a shock occurs at the time $t=1$ (characteristics intersect). Its speed deduced from the Rankine-Hugoniot condition is $s=\frac{1}{2}$.
Best Answer
The situation is very similar to this post and related ones, where piecewise constant initial data is considered (i.e, a series of Riemann problems).
To link with the Burgers equation, one may consider the change of variable $v=8(1-u)$ such that $$ v_t + (\tfrac12 v^2)_x = 0\, , $$ $$ v(x,0) = \begin{cases} 0, & x\leq 1\\ 4, & 1<x\leq 3\\ -4, & x>3 \end{cases} $$