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There are, of course, plenty of approaches on how to solve a non-linear PDE. Two of them are the following:

  1. Solve (easier) approximate problems, show some form of compactness for the approximate solutions to get a limit, then identify the limit as a solution to the original problem.

  2. Show local well-posedness for the problem (that is, for short times and without changing the initial value too much), establish a priori bounds for the solution, conclude global well-posedness by a blowup criterion (that is, local well-posedness contradicts a finite blow-up time in virtue of the a priori bounds).

I'm interested in heuristically comparing both approaches and identifying advantageous of one over the other. A first advantage of the blowup approach over the compactness approach is that it applies to problems that lack compactness properties, for instance when working in the full Euclidean space instead of a bounded domain. What are other general observations when comparing both approaches?

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  • $\begingroup$ The two things are not disjoint. Proving local well-posedness often involves some compactness method. This usually happens for quasilinear equations, e.g. Euler's equation. $\endgroup$ Commented Feb 12 at 14:33
  • $\begingroup$ Also, compactness methods can also work on the Eucledian space, not just on bounded domains. The main difference between the two approaches is that compactness methods alone do not give you the uniqueness of the solution. So, even if you can prove that solutions exist for all times, the solution could still "blow up" at some time, meaning that it ceases to be unique. And this can also happen when you can prove l.w.p.: for example, nobody knows whether Leray solutions of 3D Navier-Stokes (global weak solutions found via compactness) are unique or not, even though local well-posedness is known. $\endgroup$ Commented Feb 12 at 14:41
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    $\begingroup$ @LorenzoPompili I'm aware that the question is not really sharp. Your comment highlights an advantage of the compactness approach: it can yield existence in situations in which there is not uniqueness (you don't get a full global well-posedness result then, of course). $\endgroup$ Commented Feb 13 at 12:17

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One well-known example is Navier-Stokes' equation. It can be proved local well-posedness in any dimension $d\geq 2$ in suitable Lebesgue or Sobolev spaces (we call the solutions coming from the well-posedness theorems strong solutions), using Banach fixed-point theorem. Except in dimension $2$, it is still unknown whether strong solutions are global in time, or blow-up in finite time. Nevertheless, it is possible to construct global weak solutions using the energy balance of the equation, which are called Leray solutions. These solutions are constructed with a compactness method, and it is not known whether they are unique (although it is known that they are unique as long as the strong solution does not blow up).

Let me mention that often, one still uses compactness methods for the well-posedness of some equations, because the fixed-point scheme does not work. This is the case for quasilinear equations, like for instance Euler's equation, for which uniqueness of solutions can be proved under suitable assumptions, but one still needs to use compactness methods for the existence.

An excellent reference for the above topics is of course the lecture notes of Prof. Tao on incompressible fluids, in his blog. See Notes 1,2 and 3.

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    $\begingroup$ "it is not known whether they are unique": annals.math.princeton.edu/2019/189-1/p03 $\endgroup$
    – YangMills
    Commented Feb 13 at 14:06
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    $\begingroup$ @YangMills Nice remark. I just want to make clear that I was talking specifically about Leray solutions. The authors of the linked article write "We note that while the weak solutions of Theorem 1.2 may attain any smooth energy profile, at the moment we do not prove that they are Leray-Hopf weak solutions, i.e., they do not obey the energy inequality or have $L^2_t {\dot Ḣ}_x^1$ integrability.". But yes, it is known that "very weak" solutions are not unique. $\endgroup$ Commented Feb 13 at 14:40

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