Exercise on discontinuous coefficients in first order pde

Question

I'm trying to solve the following exercise on first order pde with discontinuous coefficients, which I've found online. It consists in giving an unambiguous meaning to the following equation $$u_t+\textrm{sign}(x)u_x=0,\,\,\,\,u(0,x)=u_0(x)$$ The hint consists in approximating the problem with the following $$u_t+f_\epsilon(x)u_x=0$$ with $f_\epsilon (x)$ a smooth non decreasing function, such that $f_\epsilon(x)=\textrm{sign(x)}$ if $\vert x\vert>\epsilon$ and then in passing to the limit of $\epsilon\to 0$. I've solved the regularized problem with the method of characteristics getting as characteristic curves $x(t)=t+c$ in $x>\epsilon$, $x(t)=-t+c$ in $x<-\epsilon$ and in the remaining region I draw curves in a qualitative way. But now, what about the limit? I have problems in the region $\vert x\vert<t$ in crossing the axis $x=0$. Any suggestions for the limit?

Answer 1

Take $f_\epsilon$ to be an odd function (which is a natural thing to do) and observe that the characteristics are even as functions of $x$: that is, writing a characteristic as $t=t(x)$ gives $t(x)=t(-x)$. With the initial condition $u(0,x)=u_0(x)$, there is no solution to regularized problem unless $u_0$ is an even function. And if $u_0$ is even, the solution satisfies $u(x,t)=u_0(|x|-t)$ for $|x|>t+\epsilon$. In the limit $\epsilon\to0$ this determines the solution to original problem for $|x|\ge t$. The region $|x|\le t$ remains undetermined; the initial value problem does not tell us what $u$ does there.