ut + uux = 0,
for u+ < u_
Consider Burger's equation
-? < x < ?,
t > 0,
u(x, 0) = \begin{cases} u_{-}, & x \le 0, \\ u_{+}, & x > 0. \end{cases}
Apply the transform u(x, t) = exp(v(x, t)), derive a conservation law equation for v. From this
conservation law, deduce that the shock at x = 0 has velocity,
s = \frac{u_{+} - u_{-}}{\log u_{+} - \log u_{-}}.
calculate the shock speed after first transforming with u = v^2.