The Rankine-Hugoniot condition In this notes, example 5, page 5. It gives an example on how to solve a PDE where characteristics might cross using the Rankine-Hugoniot condition. The example is in an initial-value problem for Burgers' equation $u_t + u u_x = 0$, with piecewise linear initial data
$$
u(x,0) = \phi(x) = \left\lbrace\begin{aligned}
&1 &&\text{if}\quad x<0\\
&1-x &&\text{if}\quad 0<x<1\\
&0 &&\text{if}\quad 1<x
\end{aligned}\right.
$$
I have two questions about it:
1). In my opinion, Rankine-Hugoniot condition is just a necessary condition for the discontinuous points of a solution. How does it guarantee the thing we get is indeed a solution?
2).Using the Rankine-Hugoniot condition we can only calculate the speed of the curve of discontinuities. How do we locate its exact formula?(i.e. how to choose one-point on it). Can we choose any arbitrary point where the characteristics cross?
 A: If needed, here is a sketch of the characteristic curves in $x$-$t$ plane, which may help to see better what happens:

The solution obtained by the method of characteristics up to the breaking time satisfies $u = \phi (x-ut)$ for $t<1$, i.e.
$$
u(x,t) = \left\lbrace\begin{aligned}
&1 & &\text{if}\quad x \leq t \\
&\tfrac{1-x}{1-t} & &\text{if}\quad t \leq x\leq 1\\
&0 & &\text{if}\quad x\geq 1
\end{aligned}\right.
$$
At the breaking time, the shock speed $s$ is given by the Rankine-Hugoniot condition $s = \tfrac12 (1 + 0)$.
Therefore, after the breaking time $t\geq 1$,
$$
u(x,t) = \left\lbrace\begin{aligned}
&1 & &\text{if}\quad x < 1 +\tfrac12 (t-1) \\
&0 & &\text{if}\quad x > 1 +\tfrac12 (t-1)
\end{aligned}\right.
$$
To your questions:

*

*The Rankine-Hugoniot condition $[\![\frac{1}{2}u^2]\!] = s [\![u]\!]$ is a condition satisfied by discontinuous weak solutions of the Burgers' equation. A weak solution which satisfies the Rankine-Hugoniot condition is not necessarily a solution to the proposed initial-value problem $u(x,0) = \phi(x)$. To be a solution, it must be in agreement with the characteristics equation until they cross, i.e. the fact that $u$ is constant along the curves $u = \phi(x - ut)$. Moreover, since weak solutions are not unique, it must satisfy an additional entropy condition. In the case of Burgers' equation, the Lax entropy condition amounts to the following statement: "if characteristics cross, then a shock-wave arises; but if they separate, a rarefaction wave occurs".

*The location of a discontinuity is deduced from the characteristics equation, whereas its speed is deduced from the Rankine-Hugoniot condition. In the present case, a shock occurs at the time $t=1$ (characteristics intersect). Its speed deduced from the Rankine-Hugoniot condition is $s=\frac{1}{2}$.

