Burgers' equation after rarefaction wave catches up with the shock Burgers' equation 
$$u_t+\Bigl(\frac{u^2}{2}\Bigr)_x=0$$
with initial conditions as below:
$$\left\{\begin{aligned}&u(x,0)=2&&\text{if $0<x<1$}\\ 
&u(x,0)=0&&\text{otherwise}
\end{aligned}\right.$$
So on the right hand side of $1$, it would have a shock wave of $x-1=t$.
On the left hand side of $0$, it will have a rarefaction wave of $x=2t$.
At some $T_c$, the rarefaction wave would catch up with the shock wave.
But the question asks what's the location of $x_s(t)$ after the two waves meet?  And it asks to solve it in two ways:


*

*Solve an ODE using Rankine-Hugoniot jump condition

*For $t>T_c$, the exact solution is triangular-shaped. Use conservation law(?) to determine $x_s(t)$ based on the area of this triangle?


I am very lost on this problem...  When $t<T_c$, I know the shock location is $(1+t,t)$, does it change after they meet? 
 A: 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 
$$
x_s(t) = 2 \sqrt{t} 
$$
