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I am curious about weak solutions of the parabolic problem $ u_t - \Delta u + a(x,t) \cdot \nabla u = f(x,t)$ in $ \Omega \times (0,\infty)$ with $ u=0$ on $ \partial \Omega$. Here $a(x,t)$ is some divergence free vector field with minimal smoothness properties. We assume that $ u(x,0)$ is smooth.

My question is: what minimal properties can I assume of $a(x,t)$ which will imply that $u(x,t)$ is locally bounded in time?

In the elliptic case one essentially doesn't need any restrictions on $a$ besides the diverge free property. I suspect this is the case for the parabolic versions also but I am curious.
Lately these questions are a fairly hot topic but there people want to put minimal conditions on $a$ so as to have the solutions Holder continuous.

thanks Craig

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Doesn't this follow directly from the maximum principle? –  Deane Yang Mar 5 '13 at 2:36
    
I am sure you are right and I am missing something. The problem (the way i see it) is that I want $a(x,t)$ to basically have no regularity assumptions and wouldn't i need to assume some regularity of $u$ before I can apply the maximum principle? it? I assume the proof (and any conditons one would need to impose on $a$) should come directly off the $L^\infty$ estimate in the parabolic versions of the De Giorgi (which i am not familiar with). At least in the elliptic case this is the case. Thanks for your interest. –  Craig Mar 5 '13 at 18:34
    
There is a parabolic maximum principle you can use directly. If you can prove existence and uniqueness for the initial value problem, then all you need to do is apply the maximum principle to the equation with a smooth approximation to $a$. Then if you take a limit, the solution to the smoothed equation has to converge to the solution of the original equation. The bound obtained by the maximum principle does not depend on $a$ at all, so it still holds in the limit. –  Deane Yang Mar 7 '13 at 2:42
    
Thanks Deane, I was attempting this method but was having some difficultly smoothing out $a$ and keeping it divergence free (in particular I was having some issues near the boundary of $ \Omega$ that were causing problems). –  Craig Mar 7 '13 at 15:48

2 Answers 2

up vote 4 down vote accepted

The $L^\infty$ a priori estimate for u does not depend on $a$ at all provided that the initial data is either bounded (you would apply maximum principle) or $L^1$ (you would apply Nash bound on the heat kernel).

The only subtlety (if any) would be at the definition of weak solution. If $a$ is too rough, weak solutions may not be unique and then everything breaks. There is a series of papers by Pierre Louis Lions and Claude Le Bris discussing the issue of uniqueness.

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Thanks Luis, I will pull up the references and take a look. –  Craig Mar 7 '13 at 15:48

Do you know the sign of f(x,t)? I would say it should be work fine for nonpositive functions f. Could you please write the reference for the elliptic case?

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