Entropy Solution of the Burgers' Equation

I am working on the following problem, which gives the Burgers' equation $$u_t + uu_x=0$$ with the initial data $$g(x)= \begin{cases}1, & x < 0, \\ 2, & 0 < x < 1,\\ 0, & x > 1.\end{cases}$$ It then asks to find the entropy solution of $u(x,t)$ for all $t>0$.

This type of problem, with shock and rarefaction waves, is discussed in Evans, but I don't really know how to apply it to solving this problem. Any help with this would be greatly appreciated. Thank you.

So, let's start looking at the first jump in the initial conditions. Here we left $1$ on the left side and $2$ on the right. As we are looking for an entropy solution and these can only jump down across shocks, we have a rarefaction wave here. At $x=1$ we have a shock (as $2 > 0$) with speed given by Rankine-Hugeniot as $\frac 12 (2+0) = 1$. So for small $t$ we have $$ u(x,t) = \begin{cases} 1 & x \le t\\ \frac xt & t < x < 2t\\ 2 & 2t \le x < 1+t \\ 0 & x \ge 1+t\end{cases} \quad \quad (0 \le t \le 1) $$ The next time something interesting happends is when the charateristic with speed 2 starting at 0, hits the shock. That is when $t+1 = 2t$, i. e. $t=1$. Now the shock is built from characteristics of the rarefaction fan, let's denote the shock curve by $s$, we have $s(1) = 2$, and by the jump condition $$ s'(t) = \frac 12 \cdot \frac{s(t)}t $$ The solution of this ode is $s(t) = 2\sqrt t$, that is we have for the next part $$ u(x,t) = \begin{cases} 1 & x \le t \\ \frac xt & t < x < 2\sqrt t \\ 0 & x \ge 2\sqrt t \end{cases} \quad\quad (1 \le t \le 4) $$ Then next interesting time is when the first speed 1 characteristic hits the speed 0 characteristics, that is when $2\sqrt t = t \iff t = 4$ (as $t\ge 0$). After that the shock travels with speed $\frac 12(1+ 0) = \frac 12$, that is we have $$ u(x,t) = \begin{cases} 1 & x \le 2 +\frac 12t \\ 0 & x> 2 + \frac 12t \end{cases} \quad \quad (t \ge 4). $$