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Jun 17, 2022 at 14:17 answer added Robert Bryant timeline score: 3
Jun 17, 2022 at 9:30 comment added mlainz @WillieWong this is an interesting suggestion, but I think that it restricts too much the classes of solution that I can consider. Locally this is equivalent than considering that the rotational vanishes. The obvious generalization would be to set $δu = g$..
Jun 16, 2022 at 20:10 answer added MyShepherd timeline score: 4
Jun 16, 2022 at 19:36 comment added Willie Wong The other way to use the Hodge idea is to fix $u$ to purely a gradient. Is that possible in your case? The scalar equation $\Delta u = f(x, \nabla u)$ certainly has been previously studied.
Jun 16, 2022 at 18:54 history edited mlainz CC BY-SA 4.0
deleted 375 characters in body
Jun 16, 2022 at 18:50 comment added mlainz I also though about using the Hodge decomposition and fix the codifferential of $u$ and try to use boundary conditions to determine the harmonic part, but I do not think that this approach is possible if you allow $g$ to depend on $u$
Jun 16, 2022 at 18:45 comment added mlainz Also, you are right that $f$ is not a top form, but a fiber bundle morphism from $E \to M$ to $\Lambda^k(M) \to M$, so $f(\cdot, u(\cdot))$ is the actual volume form. I only look for locally defined solution, so the manifold is completely irrelevant.
Jun 16, 2022 at 18:37 comment added mlainz @WillieWong You are right about the EDIT, I will remove it.
Jun 16, 2022 at 17:08 comment added Willie Wong The first displayed equation, contrary to asserted, is not linear with constant coefficients.
Jun 16, 2022 at 16:18 history edited mlainz CC BY-SA 4.0
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S Jun 16, 2022 at 15:34 review First questions
Jun 16, 2022 at 17:48
S Jun 16, 2022 at 15:34 history asked mlainz CC BY-SA 4.0