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I went to a talk today and the speaker mentioned when you add a Laplacian term to a PDE, the Laplacian will dominate (in what sense?), which I don't quite understand. I know this question is a bit vague but I appreciate if someone could clarify the question more.

Update: The speaker didn't actually mention specific pdes. He said it's more like a general "rule" or principle. Since he said it like a convention, I didn't have gut to ask him this question. I understand there are exceptions and I appreciate if there are common pdes that follow his "principle".

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    $\begingroup$ It would definitely depend on the context. Can you at the very least specify which equation you are talking about? For example, you can think of Navier-Stokes as "adding a Laplacian" to Euler's equations. That the regularity of Navier-Stokes is a Clay Problem is precisely because you cannot always say that the Laplacian will dominate! $\endgroup$
    – Willie Wong
    Commented Apr 18, 2017 at 2:38
  • $\begingroup$ @WillieWong Thanks! I just updated the question. $\endgroup$
    – qie wen
    Commented Apr 18, 2017 at 3:19
  • $\begingroup$ @qiewen I am interested in your question. Some how I encountered this concept in the comment conversation of this post: mathoverflow.net/questions/182415/… $\endgroup$
    – Ali Taghavi
    Commented Apr 18, 2017 at 9:12
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    $\begingroup$ If you're looking for any example showing that adding a Laplacian term (even with a small coefficient) may change the game, see Burgers' equation. Googling "vanishing viscosity" might also be helpful. $\endgroup$
    – Michał Miśkiewicz
    Commented Apr 18, 2017 at 18:35
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    $\begingroup$ Similarly, if you take a linear transport equation, it has hyperbolic nature. If you add a diffusive term, say $\epsilon\Delta$, then no matter how small $\epsilon>0$ is, the corresponing PDE will have parabolic nature. Understanding how the solutions to this regularized PDE converge as $\epsilon\to 0$ is a subtle issue, and all the more so if boundary conditios play a role (think of the number of necessary boundary conditions). On the other hand, if you perturb a bi-Laplacian $\Delta^2$ by a Laplacian, you surely won't get anything essentially different. $\endgroup$
    – Delio Mugnolo
    Commented Apr 20, 2017 at 16:08

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