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I want to consider the following equation $$u_t+\mathrm{sgn}(x)u_x=0,\,\,u(0,x)=u_0(x)$$ Now if $x>0$ or $x<0$ I can use the method of characteristics to obtain $u(t,x)=u_0(x-t)$ if $x>t$ and $u(t,x)=u_0(x+t)$ if $x<-t$. What about the region $-t<x<t$?

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  • $\begingroup$ Maybe using a smooth approximation of the signum function $\endgroup$ – max Oct 1 '14 at 15:13
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Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dx}{ds}=\text{sgn}(x)$ , letting $x(0)=0$ , we have $|x|=s$

$\dfrac{dt}{ds}=1$ , letting $t(0)=t_0$ , we have $t=s+t_0=|x|+t_0$

$\dfrac{du}{ds}=0$ , letting $u(0)=f(t_0)$ , we have $u(t,x)=f(t_0)=f(t-|x|)=F(|x|-t)$

$u(0,x)=u_0(x)$ :

$F(|x|)=u_0(x)$

$F(x)=u_0(\pm x)$

$\therefore u(t,x)=u_0(\pm|x|\mp t)$

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