Question Consider the equation $$u_t + (u^4 /4)_x =0. $$ Find the admissible solution for each of the following initial data: $$u(x,0) = \begin{cases} 1 \quad & x<0 \\ 0 & x>0\end{cases} \quad \text{and} \quad u(x,0)=\begin{cases} 0 \quad & x<0 \\ 1 & x>0\end{cases} .$$ Attempt For the initial data $$u(x,0) = \begin{cases} 1 \quad & x<0 \\ 0 & x>0\end{cases}$$ the speed of the shock is $$ \frac{dx}{dt} = \frac{1/4 -0}{1-0} = \frac{1}{4} $$ therefore the solution is $$u(x,t) = \begin{cases} 1 \quad & x<t/4 \\ 0 & x>t/4\end{cases} .$$ How can I find the solution for $u(x,0) = \begin{cases} 0 \quad & x< 1\\ 1 & x>0\end{cases}$ ? Thank you kindly in advance.
Asked
Active
Viewed 78 times
1
-
1The max entropy solution would require a rarefaction wave – Ninad Munshi Jan 01 '23 at 21:34
-
@NinadMunshi Thank you. How can I find the rarefaction wave? – Philomath Jan 01 '23 at 22:35
1 Answers
1
We apply the method described in this post to the present initial value problem. For the dropping unit discontinuity given by $u(x<0,0)=1$ and $u(x>0,0)=0$, the admissible solution is a shock wave with Rankine-Hugoniot speed $\frac14$. Hence the solution proposed in OP is correct. For the positive initial jump $$u(x,0) = \begin{cases} 0 ,\quad & x<0, \\ 1, & x>0,\end{cases} $$ the admissible solution is a rarefaction wave: $$u(x,t) = \begin{cases} 0 ,\quad & x\leq 0, \\ \sqrt[3]{x/t}, & 0\leq x\leq t, \\ 1, & x\geq t.\end{cases} $$
EditPiAf
- 21,328