Timeline for Different types of a test functions in weak solutions of a PDEs
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| Sep 3, 2018 at 12:19 | comment | added | Mark | @HaraldHanche-Olsen: $x$ is not multidimensional. I have $t$ in the equation. I've seen a few times (for example in the papers of above mentioned E.Feireisl) that when he defines solution of a (stochastic) pde he just uses $\varphi (x)$ and doesn't do $dx$ integration (think of the second definition given above just without $\phi (t)$ and without integrals $dx$). There, it is some form of weak solution. But it isn't look like a traditional weak solution that we used to. Also I agree too that if you only use $\varphi(x)$ you are not involving $t$ at all. | |
| Sep 3, 2018 at 10:44 | comment | added | Harald Hanche-Olsen | I am not sure I understand your second question. But I'd say if you only use $\varphi(x)$ then you are not involving the $t$ variable at all, so how can it be a reasonable definition of weak solution? And if you don't have a $t$ variable, as you allude to in your followup comment, isn't your PDE then an ODE? Unless, of course, $x$ is multidimensional, in which case notions of weak solutions certainly do exist. | |
| Sep 2, 2018 at 11:33 | comment | added | Mark | @WillKwon: Thank you for your comment and for the lecture note. Faedo-Galerkin with Aubin-Lions is a nice example (also E. Feireisl use something similar in a few of his papers). I am not working on Navier-Stokes currently but this maybe could help me with something else I was working on. But somehow I think there is more to it with using different types of test functions. Hope I'll figure it out. | |
| Sep 2, 2018 at 9:31 | comment | added | Mark | @HaraldHanche-Olsen: Thanks for your comment. I agree that the second approach is a little bit easier to check. In the problem I have I do not use Kruzkov's doubling of variables (but it is a nice reminder). What do you think of question two? I am thinking of a PDE problem where I do not have a problem with a variable t (so I wouldn't use $\phi(t)$). I would use just $\varphi(x)$. Could I think then of a solution of a PDE as a weak solution too? Also sorry for a late answer (I was on two weaks vacation). | |
| Aug 22, 2018 at 2:06 | comment | added | Will Kwon | Actually, I'm not familiar with the one-dimensional case. So my answer may not appropriate to your question. So I explain the one I know currently. The second definition of a weak solution has an advantage when we proceed with the Galerkin approximation with compactness argument such as Aubin-Lions lemma. In the case of the Navier-Stokes equation, see Lemma 2.1 and the proof of Theorem 2.6(in particular Step 3) in the following lecture note goo.gl/gLmEZp | |
| Aug 19, 2018 at 8:26 | comment | added | Harald Hanche-Olsen | The second definition is just like the first one, but with test functions restricted to the form $\psi(x,t)=\varphi(x)\phi(t)$. I'm pretty sure (though this is new to me, and I haven't checked it) that linear combinations of test functions on this form are dense in the space of all test functions, so the two approaches should be equivalent. The advantage of the second approach would be that it might be easier to check. Note also that Kružkov's famous doubling of variables proof only relies on test functions on product form. | |
| Aug 19, 2018 at 4:08 | history | edited | Martin Sleziak | CC BY-SA 4.0 |
added a top-level tag; https://meta.mathoverflow.net/questions/1457/why-are-mo-tags-formatted-as-they-are
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| Aug 18, 2018 at 16:51 | history | asked | Mark | CC BY-SA 4.0 |