On the propagation of singularities in PDE

Question

This question might be a little generic, but i wanted to get some idea on the concept of propagation of singularities in PDE. Searching the internet i only found very complicated things about the subject, so if someone could give an idea, more or less precise of this and it is related to hypoellipticity i would be grateful! Also providing some examples for classical heat, wave, schrodinger equations. Thanks for any help.

Answer 1

There is no propagation of singularities for the heat equation, because the solution instantly becomes smooth: the heat kernel is a mollifier.

The simplest equation for which singularities propagate is the 1d wave equation $u_{tt}=c^2u_{xx}$: see the nice exposition by Jiří Lebl with some plots of solution.

I also think that Searching the internet is not a viable way to learn PDE theory in depth. You should be reading books such as Hörmander's Analysis of Linear Partial Differential Operators II. (Others might suggest more accessible sources, but Hörmander's volumes are generally readable).

Answer 2

The propagation of singularity means that the singularity of the initial data propagates left and right at a speed c as time goes on. More specifically, if at $t=0$ the solution has a singularity at $x=x_1$ then at $t=t_1$ the solution will have a singularity at $x=x_1\pm ct_1$. The following example is taken from the book Differential Equations for Engineers written by Jiří Lebl. The solution of the wave equation $y_{tt}=y_{xx}$, $y(0, t)=y(2, t)=0$, $y(x, 0)=f(x)$ and $y_t(x, 0)=0$. enter image description here The initial solution(that is, $y(x, 0)=f(x)$) $f(x)$ has a singularity at $x=1$ (the upper left picture). At time $t=0.4$ the solution has a singularity at $x=1\pm 0.4$, that is at $x=0.6$ and $x=1.4$(the upper right picture). At time $t=0.8$ the solution has a singularity at $x=1\pm 0.8$, that is at $x=0.2$ and $x=1.8$ (the lower left picture). Finally, at time $t=1.2$ the solution has a singularity at $x=1\pm 1.2$, that is at $x=2.2$ and $x=-0.2$. But since the speed of the propagation is one, if time $t=1$ the wave hits the wall and reflects and as we are confined in $0\leq x\leq 2$ we subtract $2$ from $x=2.2$ this gives us $0.2$ and we add $2$ to $-0.2$ this gives us $1.8$. Thus the singularity occurs at $x=0.2$ and $x=1.8$ (figure lower right).