1

Consider $$u_t+(|u|^\alpha)_x=0, \quad\alpha>1$$ Given the initial condition $$u(x,0)=\begin{cases} 0, x<0\\1,x>0\end{cases}$$

a) Find a solution for $u(x, t)$ that is continuous for all $t > 0$ and satisfies the initial condition.

b) Find a solution for $u(x, t)$ that includes a shock obeying the appropriate jump condition.

(c) Show that one of these solutions does not satisfy the entropy condition.

My attempt:

We can rewrite the equation $$u_t+(|u|^\alpha)_x=u_t+\alpha u |u|^{\alpha-2}u_x$$ then we parametrize \begin{align} x(0,r) &= r \\ t(0,r) &= 1 \\ u(0,r) &= u_0 \end{align}

The characteristics satisfying IVP are

\begin{align} x_s &= \alpha u |u|^{\alpha-2} \\ \implies x &= \alpha u |u|^{\alpha-2}t+r \\ t_s &= 1 \\ \implies t &= s \\ u_s &=0 \\ \implies u &= u_0 \end{align}

The projection on $(x,t)$-plane is given by $$t = \frac{x-r}{\alpha u |u|^{\alpha-2}}$$ So if $x<0$, the denominator is $0$. How do I find the solution?

EditPiAf
  • 20,898
dxdydz
  • 1,371

1 Answers1

0

The characteristics are the curves $x=x_0+\alpha u|u|^{\alpha-2}t$ along which $u=u(x_0,0)=F(x_0)$ is constant. With the given initial data, we have $$ u(x,t) = \left\lbrace \begin{aligned} &0 &&\text{if}\quad x <0\\ &U (x/t) &&\text{if}\quad 0 \leq x \leq \alpha t\\ &1 &&\text{if}\quad \alpha t <x \end{aligned} \right. $$ where $U (x/t)$ is a rarefaction wave solution. The latter is obtained from assuming a smooth solution of the form $u (x,t) = U (\xi)$ with $\xi=x/t$. Using this self-similarity Ansatz and the PDE, one obtains $$ U (x/t) = \left.\left (\xi\to\alpha \xi|\xi|^{\alpha-2}\right)^{-1}\right|_{\xi=x/t} = \left(\frac{x}{\alpha t}\right)^\frac {1}{\alpha-1} . $$ Concerning the shock wave solution $$ u(x,t) = \left\lbrace \begin{aligned} &0 &&\text{if}\quad x <st\\ &1 &&\text{if}\quad st <x \end{aligned} \right. $$ with $$ s = \frac{|1|^\alpha - |0|^\alpha}{1 - 0} = 1 $$ following from the Rankine-Hugoniot condition, one shows that this solution does not satisfy the Lax entropy condition since $$ \alpha\, 0\, |0|^{\alpha-2} < s < \alpha\, 1\, |1|^{\alpha-2} . $$ Indeed, the Lax entropy condition requires opposite inequalities. Finally, the entropy solution is the rarefaction wave.

EditPiAf
  • 20,898
  • How can we find s, the jump condition gives us $s=\frac{|u^-|^\alpha-|u^+|^\alpha}{u^- - u^+}$, and how can we show that the shockwave solution is a weak solution? I know the formula I do not see how that condition is easy to check – dxdydz Dec 09 '18 at 16:26
  • thnx, got it, how can we show that the shock wave solution is a weak solution? – dxdydz Dec 09 '18 at 18:14
  • @dxdydz to prove that the shock is a weak solution, we can follow the steps in this post – EditPiAf Dec 09 '18 at 19:42