Shock trajectory deduced from Rankine-Hugoniot in Burgers' equation Question:

Find the entropy solution to Burgers' equation:
  $$\partial_tu+u\partial_x u=0, t>0, x \in \mathbb{R}$$
  with the initial data
  $$u(x,0)= \left\{\begin{array} .x & x<1 \\ 0 & x\geq 1 \end{array} \right.$$

My attempt:
I have solved similar problems to this but only when my initial conditions were constants. I am having trouble when I reach the part I need to apply the Rankine-Hugoniot condition.
I have found that my characteristic curves are given by 
$$x= \left\{\begin{array} .rt+r & r<1 \\ r & r\geq 1 \end{array} \right.$$
I will have a shock due to the characteristics crossing which will be at the discontinuity of the initial data, i.e the shock will start from $t=0$ at $x=1$.
Therefore, the entropy solution to Brugers' equation is given by
$$u(x,t)= \left\{\begin{array} .\frac{x}{1+t} & x<\sigma(t) \\ 0 & x>\sigma(t) \end{array} \right.$$
How exactly can I apply the RH condition in this case. Can someone walk me trough the steps I need follow to find $\sigma(t)$?
 A: The classical solution $u=g(x-ut)$ where $g = u(\cdot,0)$ denotes the initial data is correct. One notes that the characteristic curve starting at $x_0=1^-$ intersects the characteristic curve starting at $x_0=1^+$  already at time $t=0$ (see representation of the base characteristics in the $x$-$t$ plane below). Hence, since the classical solution is no longer well-defined in the vicinity of $x=1$, we need to look for weak solutions.

Because the characteristics intersect at $(x,t)=(1,0)$, a shock wave is generated. On the left of the shock located at the position $x=\sigma(t)$, we have $u^-=\frac{x}{1+t}$. On the right of the shock, we have $u^+=0$. The shock trajectory $\sigma(t)$ is given by the Rankine-Hugoniot condition
$$
\sigma'(t) = \frac{u^-+u^+}2 =\frac12\left(\frac{\sigma(t)}{1+t} + 0\right)
$$
with the initial condition $\sigma(0)=1$. Hence, $\sigma(t)=\sqrt{1+t}$, and
$$
u(x,t) =\left\lbrace
\begin{aligned}
&\tfrac{x}{1+t}, && x< \sqrt{1+t}\\
&0, && x> \sqrt{1+t}
\end{aligned}\right.
$$
