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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.

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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).

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Thank you. But what do mean by 'propagation of singularities'? I guess it's the fact that if the data has singularities this are "carried on" by the solution? So the heat operator regolarizes (no propagation) while the wave operator leaves the same regularity of the data so if these have singularities the "continue" in the solution. Is this fairly right? –  balestrav Jun 27 '12 at 21:09
    
@balestrav Yes, as the plots in Lebl's book show, the solution that was nonsmooth at $x=0, t=0$ will be nonsmooth at $x=\pm ct$, $t=0$: so, the singularity propagates left and right at speed $c$. In this example it's just the lack of derivative, but a jump discontinuity (wave front) would behave in the same way. –  user31373 Jun 27 '12 at 21:15
    
And what about the hypoellipticity? Can we say that an hypoelliptic operator doesn't propagate singularities? For What I know this means the if the data is smooth then the solution is smooth, but is it true in general that if the data is singular the solution is smooth (like in the case of the heat operator) or this was just a particular case? –  balestrav Jun 27 '12 at 21:20
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