$u_t+[u(1-u)]_x=0$ with initial conditions - Need help with rarefaction wave portion

I need help solving $$u_t+[u(1-u)]_x=0$$ with initial conditions $$u(x,0) = \begin{cases} 0.75 & |x|<0.5\\ 0 & else \end{cases}$$ I am trying to use the method of characteristics but I am having trouble with the rarefaction wave part.

First, we can rewrite the PDE as $$u_t+(1-2u)u_x=0$$ So, we have $$\frac{dx}{dt}=1-2u$$ and we know that characteristics are constant because $$\frac{d u(x(t),t)}{dt}=\frac{\partial u}{\partial t} + \frac{\partial u}{\partial x}\frac{dx}{dl t} = u_t+u_x(1-2u)=0$$

So, we have $$\frac{d}{dt}\frac{dx}{d t}=1-2u(x,t)=1-2u(\xi,0)$$ and solving for this gives $$x=(1-2u)t+\xi$$

So, I have that there are characteristics that have slope $$1-2u(\xi,0)$$ so the slope is 1 for $$x\leq-0.5$$ and $$x\geq -0.5$$ and we have slope -2 for $$-0.5.

So, there is a shock at $$x=-0.5$$. The propagation can be found as $$s'=\frac{0(1-0)-0.75(1-0.75)}{0-0.75}=1/4$$ So the characteristic propagates as $$\frac{dx}{dt}=1/4$$ $$x(0)=-0.5$$ and we get $$x(t)=-0.5+t/4$$.

Then there is a rarefaction fan at $$x=0.5$$. I think that it starts at $$x=-t/2 +0.5$$ and ends at $$x=t+0.5$$. I am having a hard time figuring out what the equation is for the fan though.

So, my solution is $$u(x,t) = \begin{cases} 0 & x<-0.5+t/4\\ 0.75 & -0.5+t/4\leq x\leq -t/2+0.5\\ rarefaction & -t/2+0.5 I would really appreciate help on how to define the rarefaction fan part. I have tried variations of $$\frac{x-0.5}{t}$$ and $$\frac{t+x-0.5}{t-0.5}$$ but they don't seem to work right.

Thank you.

• You've written $u_x + (u(1 - u))_x = 0$; shouldn't there be a $t$-deriviative in there somewhere; otherwise, it's an ODE . . . – Robert Lewis Mar 26 at 4:37
• Is it supposed to be $u_t + (1-2u)u_x = 0$? – Dylan Mar 26 at 4:39
• @Dylan Sorry about that. You are right. I will edit it. – MathIsHard Mar 26 at 4:42
• @RobertLewis you are definitely correct. There is a t derivative. Sorry about the typo there. I fixed it in the post – MathIsHard Mar 26 at 4:43

I think I figured this out after working on this since yesterday afternoon almost nonstop. We need that the rarefaction satisfies $$0.75$$ at $$x=-0.5t+0.5$$ and we need that it also satisfies $$0$$ at $$x=t+0.5$$ since these are the initial densities to the left and right of the fan.
So with a bit of algebra, one can find that the rarefaction wave should behave like $$-0.5\bigg(\frac{x-0.5}{t}-1\bigg)$$ since then,
$$-0.5\bigg(\frac{(-0.5t+0.5)-0.5}{t}-1\bigg)=0.75$$ and $$-0.5\bigg(\frac{(t+0.5)-0.5}{t}-1\bigg)=0.$$