# Solve the ivp for a scalar conservation law

Solve the ivp for a scalar conservation law, show that it is a rarefaction and find it $\frac{\partial u}{\partial t} + \frac{\partial }{\partial x}F(u) = 0$ where $F(u) = \frac{u^{4}}{4}$, $u = u(x,t)$ and $u(x,0) = \begin{cases} u_0^- & x <0 \\ u_0^+ & x >0 \end{cases}$

where $u_0^- < u_0^+$.

$F'(u) = u^3$ using characteristics, $$u(s,0) = \begin{cases} u_0^- & s <0 \\ u_0^+ & s >0 \end{cases}$$ $$x = F'(u_0(s))t + s \ \ \ \ and \ \ \ \ z=u(x,t)=u_0(s)$$ for s < 0 $x = {u_{0}^{-}}^{3}t + s < 0 \implies s = x - {u_{0}^{-}}^{3}t < 0 \implies x < {u_{0}^{-}}^{3}t$

for s > 0 $x = {u_{0}^{+}}^{3}t + s > 0 \implies s = x - {u_{0}^{+}}^{3}t > 0 \implies x > {u_{0}^{+}}^{3}t$

since $u_0^- < u_0^+$. $u(x,t) = \begin{cases} u_0^- & x <u_0^-t \\ u_0^+ & x >u_0^+t \end{cases}$

no solution from $u_0^-$ to $u_0^+$

so assume at $s = 0$ we will get $x \in [u_0^-, u_0^+]$ $s = 0 \implies x = u_0(s)^3t \implies u(x,t)^3 = x/t \implies u(x,t) = \sqrt[3]{\frac{x}{t}}$

so is the rarefraction wave? $u(x,t) = \begin{cases} u_0^- & x <u_0^-t \\ \sqrt[3]{\frac{x}{t}} &u_0^-t<x< u_0^+t\\ u_0^+ & x >u_0^+t \end{cases}$

In facts, the proposed solution is almost correct. There is a mistake on the bounds of the rarefaction fan, which should be u(x,t) = \left\lbrace\begin{aligned} &u_0^- &&\text{if}\quad x < F'(u_0^-) t\, ,\\ &\sqrt[3]{x/t} &&\text{if}\quad F'(u_0^-) t \leq x\leq F'(u_0^+) t\, ,\\ &u_0^+ &&\text{if}\quad F'(u_0^+) t < x \, . \end{aligned}\right. Since $F'(u) = u^3$, one can verify that this solution is continuous.