Find the weak solution of the conservation law 
The picture above is the question I have been asked to do.
For part ii) a) what I have done is that I have said that $g'(v) = v^{\frac{1}{2}}=u$ therefore it does satisfy the scalar conservation laws and it does satisfy Burgers' equation.
Now for part ii) b) I have an answer but I don't think I am correct.
What I have said is that 
$$v_0(x) =
\begin{cases}
u_l^2,  & x<0\\[2ex]
u_r^2, & x>0
\end{cases}$$
This would give me that $$\dot\sigma = \frac{\frac{2}{3}(u_L^3 - u_R^3)}{u_L^2-u_R^2}$$
which comes from the Rankine Hugionot condition. This means, along with initial condition that $\sigma(0)=0$, that the shock comes at 
$$\sigma = \frac{\frac{2}{3}(u_L^3 - u_R^3)}{u_L^2-u_R^2}t$$
which gives me weak solution 
$$v(x,t) =
\begin{cases}
u_l^2,  & x<\frac{\frac{2}{3}(u_L^3 - u_R^3)}{u_L^2-u_R^2}t\\[2ex]
u_R^2, & x>\frac{\frac{2}{3}(u_L^3 - u_R^3)}{u_L^2-u_R^2}t
\end{cases}$$
Which means that $v\neq u^2$
What I am looking for here is for someone to tell me whether my solution is correct or not. And if I am wrong, what have I do wrong and how do I rectify it.
 A: Your solution is correct. Written differently:
(a) If $u$ is smooth, positive and $v=u^2$, then
\begin{aligned}
v_t &= 2 u u_t \\
&= -2 u^2 u_x \\
&= - 2v ({v}^{1/2})_x \\
&= -v^{1/2} v_x \\
&= -\left(\tfrac{2}{3}v^{3/2}\right)_x \, .
\end{aligned}
Therefore, $v$ satisfies the conservation law $v_t + g(v)_x = 0$ with $g(v) = \tfrac{2}{3}v^{3/2}$.
(b) We have to solve the Cauchy problem for $v_t + g(v)_x = 0$ with initial data $v(x,0) = v_0(x)$ (Riemann problem), where
$$
v_0(x) =
\left\lbrace
\begin{aligned}
&v_l &&\text{if } x<0 \\
&v_r &&\text{if } x>0
\end{aligned}
\right.
\qquad \text{and}\qquad v_l>v_r.
$$
The Rankine-Hugoniot condition gives the shock speed
\begin{aligned}
s &= \frac{2}{3} \frac{(v_r)^{3/2} - (v_l)^{3/2}}{v_r-v_l}\\
&= \frac{2}{3} \frac{(u_r)^{3} - (u_l)^{3}}{(u_r)^2-(u_l)^2} \\
&= \frac{2}{3}\left( u_l+u_r - \frac{u_r u_l}{u_r+u_l} \right) ,
\end{aligned}
so that
$$
v(x,t) = \left\lbrace
\begin{aligned}
&v_l = (u_l)^2 &&\text{if } x<s t \\
&v_r = (u_r)^2 &&\text{if } x>s t 
\end{aligned}\right.
$$
and $v\neq u^2$.
Note that $u$ is not smooth here, which invalidates the derivation in (a).
