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I am told that for a non-linear PDE $$u_t = f(x,t,u,u_x,u_{xx})$$ is parabolic if, writing $f = f(x,t,z,p,q)$, $$\frac{\partial f}{\partial q} > 0.$$

Is this the usual definition of parabolic in the field?

Also, suppose I have $$u_t = \frac{1}{u_x}$$ or $$u_t = u_x$$ then do I need $u_x$ to be bounded for parabolicity? Bounded means what exactly? I saw it somewhere but it was not explained to me.

Thanks.

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  • $\begingroup$ Your two questions are, as best as I can tell, independent of each other. In these situations it is best to ask two separate questions instead of just one. $\endgroup$ Jul 3, 2012 at 12:59
  • $\begingroup$ @timur Yes, sorry, thought I already had. $\endgroup$
    – blahb
    Jul 6, 2012 at 7:45

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There exist various definitions, or rather, different shades of parabolicity. The one (that $\partial f/\partial q$ does not change sign) bartgol is referring to is typically given when classifying second order equations. In some books (e.g., Qing Han's new book, if I remember correctly) this notion is called degeneracy, in order to avoid confusion with the "time-irreversible" notion of parabolicity (which seems to be used in your source). The latter is the idea that parabolic equations should generalize the heat equation, not the backward heat equation. This general class roughly includes all equations that are even order in space when written in a first order form in time, and are well-posed. For linear equations, the most general form of parabolicity that I know of is called Shilov parabolicity. Slightly more restrictive class is that of Petrowsky parabolicity, which arises when you want to have a classification in terms of the principal part in space. For nonlinear equations you linearize around a solution to transfer those notions.

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  • $\begingroup$ Thanks. Do you have any idea as to why in the PDE $u_t = \frac{(1-u)}{\sqrt{(1-u)^2 + u_x^2}}f(u_{xx})$ where $f$ is smooth, we need $u_x$ to be bounded? I know the authors say this is necessary for it to be parabolic and you say there are lots of definitions of parabolicity, but maybe there is an obvious reason why they say this. $\endgroup$
    – blahb
    Jul 3, 2012 at 20:14
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    $\begingroup$ @blahb: If $u_x$ were not bounded, your equation fails to be uniformly parabolic. To get useful estimates on the forward in time solution, generally one uses some given control $0 < \lambda < F_q < \Lambda$ for constants $\lambda,\Lambda$, where the equation is $u_t = F(x,t,u, u_x, u_{xx})$ as in your OP. If $F_q \to 0$ the behaviour there more closely resembles hyperbolic equations. $\endgroup$ Jul 4, 2012 at 8:33
  • $\begingroup$ @WillieWong Thank you! $\endgroup$
    – blahb
    Jul 4, 2012 at 17:21
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You don't need

$$\frac{\partial f}{\partial q} > 0.$$

Indeed, the two equations

$$u_t = u_{xx}\qquad u_t=-u_{xx}$$

(which correspond to a forward and backward heat equation respectively) are both parabolic. The important thing is that the sign of $f_q$ does not change. Its sign will then determine only if the equation is forward or backward in time (that is if you need to give an initial or a final condition).

I don't understand the second question. "Do I need $u_x$ to be bounded..." in order for what? What property do you want your solution/equation to enjoy?

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  • $\begingroup$ Thanks for replying. Strange then that I read that it needs to be positive to be parabolic (assuming $f$ is smooth). I meant do I need $u_x$ to be bounded in order for the equation to be parabolic? That is what I have read. I was wondering why. The PDE in that context was $u_t = \frac{(1-u)}{\sqrt{(1-u)^2 + u_x^2}}$, and I am told that I need to $1-u > 0$ and $u_x$ needs to be bounded for it to be parabolic. $\endgroup$
    – blahb
    Jul 3, 2012 at 14:48
  • $\begingroup$ Your PDE that you just wrote down is not parabolic. It has no dependence on $u_{xx}$. What @bartgol is referring to is the fact that by changing the time parameter $t \leftrightarrow -t$ you make $u_t \leftrightarrow -u_t$. Basically for $f_q > 0$ you expect to be able to solve forward in time, while for $f_q < 0$ you expect to be able to solve backwards in time. $\endgroup$ Jul 3, 2012 at 15:08
  • $\begingroup$ Thanks @WillieWong. I wrote it down wrong (trying to simply the PDE I had in mind..). There should be on the right hand side a factor of $f(u_{xx})$ where $f$ is a smooth function. $\endgroup$
    – blahb
    Jul 3, 2012 at 20:10

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