Entropy solution to inviscid Burgers with triangular initial data 
Find the entropy solution of
  $$\begin{cases}
u_t + \left( \frac{u^2}{2} \right)_x = 0 & \text{ in } \mathbb{R}\times(0,\infty)
\\ u = g & \text{ on } \mathbb{R}\times\{0\},
\end{cases}$$
  where
  $$g(x) = \begin{cases}
0&\text{ if } x\leq -1 \\
1+x&\text{ if } -1\leq x\leq 0 \\
1-x&\text{ if } 0\leq x\leq 1 \\
0&\text{ if }x\geq 1.
\end{cases}$$

This is what I have so far. To get the characteristics we have $x=g(x_0)t+x_0$ which gives us
$$\begin{cases}
x_0&\text{ if } x_0<-1 \\
(1+x_0)t+x_0&\text{ if } -1<x_0<0 \\
(1-x_0)t+x_0&\text{ if } 0<x_0<1 \\
x_0&\text{ if } x_0>1
\end{cases}$$
After this step I get a bit confused. I believe the next step is finding the equations for the shocks at the discontinuous points, in this case $(-1,0)$, $(0,0)$, and $(1,0)$. Here is my attempt at calculating the shocks:
$$ \frac{dx}{dt} = \frac{0+(1+x)}{2} = \frac{1+x}{2} ~~~~~\Rightarrow~~~~~ \int_x^{-1}\frac{dy}{1+y} = \int_0^t \frac{ds}{2} ~~~~~\Rightarrow~~~~~ \boxed{x=e^{-t/2}-1}$$
$$\frac{dx}{dt} = \frac{(1+x)+(1-x)}{2} = \frac{2}{2} = 1 ~~~~~\Rightarrow~~~~~ \int_0^x dy = \int_0^t ds ~~~~~\Rightarrow~~~~~ \boxed{x=t}$$
$$\frac{dx}{dt} = \frac{(1-x)+0}{2} = \frac{1-x}{2} ~~~~~\Rightarrow~~~~~ \int_1^x \frac{dy}{1-y} = \int_0^t \frac{ds}{2} ~~~~~\Rightarrow~~~~~ \boxed{x=1-e^{-t/2}}$$
Assuming I've done everything right so far, I'm lost after this point. How do I get my entropy solution from this? Also, are there other shocks I need to look at? For example, where my current shocks intersect do new shocks get created?
Any help, guidance, and feedback is greatly appreciated.
 A: Following the steps in this post, the solution $u = g(x-ut)$ obtained with the method of characteristics reads
$$
u(x,t) = \left\lbrace
\begin{aligned}
&0 &&\text{for}\; x < -1\\
&\tfrac{1+x}{1+t} &&\text{for}\; {-1}\leqslant x \leqslant t\\
&\tfrac{1-x}{1-t} &&\text{for}\; t\leqslant x \leqslant 1\\
&0 &&\text{for}\; x > 1
\end{aligned}
\right.
$$
which is valid for times $0\leqslant t <1$. At the breaking time $t=1$, the base characteristics intersect in the $x$-$t$ plane:

Starting from the breaking time, the entropy solution includes a shock wave, which abscissa $x_s(t)$ satisfies the Rankine-Hugoniot condition. Here, the value on the left of the shock is $\tfrac{1+x_s}{1+t}$, while the value on the right is zero. Thus, the shock trajectory satisfies
$$
\frac{\text d x_s}{\text d t} = \frac{1}{2}\left(\frac{1+x_s}{1+t} + 0\right)
$$
with $x_s(1)=1$. Therefore,
$x_s(t) = \sqrt{2(1+t)} - 1$,
and the entropy solution for $t>1$ reads
$$
u(x,t) = \left\lbrace
\begin{aligned}
&0 &&\text{for}\; x < -1\\
&\tfrac{1+x}{1+t} &&\text{for}\; {-1}\leqslant x < x_s(t)\\
&0 &&\text{for}\; x > x_s(t)
\end{aligned}
\right.
$$
The solution is maximal at the left of the shock, and the supremum $u|_{x=x_s^-} = \sqrt{2/(1+t)}$ goes to zero as $t$ goes to infinity.
