Timeline for Neumann boundary conditions in NDSolve over nontrivial region
Current License: CC BY-SA 3.0
19 events
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| Dec 11, 2018 at 3:01 | history | tweeted | twitter.com/StackMma/status/1072325363236253696 | ||
| Dec 3, 2018 at 11:10 | history | edited | user21 |
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| Nov 14, 2014 at 2:12 | comment | added | Greg P | @Secchi, which statement are you referring to? It has been a while since I've looked at all this. | |
| Nov 13, 2014 at 0:59 | comment | added | user18607 | From this last statement, it seems that one can not use NeumannValue for solving the Poisson equation, i.e. Laplacian u=0 in some 3D volume. Is this true? | |
| Nov 11, 2014 at 23:36 | answer | added | JEP | timeline score: 2 | |
| Sep 5, 2014 at 15:46 | comment | added | user21 | @GregP, thanks! | |
| Sep 5, 2014 at 14:37 | comment | added | Greg P | @user21, sure thing, go ahead and use it! | |
| Sep 5, 2014 at 9:48 | comment | added | user21 | @GregP, I wanted to ping you again and ask if it were OK to use some variation of this in the FEM documentation of NDSolve? That would be great. Thanks. | |
| Sep 2, 2014 at 6:48 | comment | added | user21 |
@GregP, the coefficients in the PDE and the ones addressed through NeumannValue are the same - that's why you specify a Neumann *Value*. There are actually three sign changes, those appear during integration by parts in the derivation of the FEM.
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| Sep 1, 2014 at 23:15 | comment | added | Greg P | @user21 That seems consistent with the "Classical Partial Differential Equations" section (see "The Coefficient Form...") of the FEM documentation, although that suggests that the natural conditions have $+\gamma$ rather than $-\gamma$. | |
| Sep 1, 2014 at 23:06 | comment | added | Greg P | My assumption that default (natural) boundary conditions represented zero flux came from reading several parts of the documentation, @user21. For example, the NeumannValue documentation says that if no bc's are specified, then $\vec{n}\cdot (c \nabla u + \alpha u - \gamma) = 0$ is used. It is not explained how these coefficients are related to the input, but I assumed that the PDE was of the form $\nabla \cdot$current density vector$=0$. | |
| Aug 29, 2014 at 20:28 | comment | added | user21 | @george2079, the natural Neumann bc depends on the equation. In this case the equation evaluated to something where the natural bc is $\nabla u \cdot n = 0$. If you do find that claim natural==zero flux in the docs, I'd appreciate if you could let me know. | |
| Aug 29, 2014 at 19:39 | vote | accept | Greg P | ||
| Aug 29, 2014 at 18:54 | comment | added | george2079 | I suspect the natural Neumann b.c. is just what you have observed ( zero gradient, which in this case is not zero flux ). That is to say the result is correct and the issue lies in the docs if they claim natural==zero flux. (My guess they only make that claim in the context of examples w/o the potential term ). | |
| Aug 29, 2014 at 16:22 | answer | added | user21 | timeline score: 16 | |
| Aug 29, 2014 at 14:57 | comment | added | Greg P |
Hi user21. When I do that, I get equal fluxes into and out of the box, as desired. But then I'm also solving a different PDE, since your op differs from my op by a term - Div[Grad[v[x, y], {x, y}], {x, y}] u[x, y].
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| Aug 29, 2014 at 9:41 | comment | added | user21 |
Or alternatively, this equivalent: op = 2*E^(-x^2 - y^2)*y*Derivative[0, 1][u][x, y] + Derivative[0, 2][u][x, y] + 2*E^(-x^2 - y^2)*x*Derivative[1, 0][u][x, y] + Derivative[2, 0][u][x, y]
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| Aug 29, 2014 at 9:36 | comment | added | user21 |
I think I have an idea what the issue might be, could you try this op = Div[Grad[u[x, y], {x, y}] + Grad[v[x, y], {x, y}] u[x, y], {x, y}] - Div[Grad[v[x, y], {x, y}], {x, y}] u[x, y] and see if this solves your problem. (I know the equation looks strange),if this is the case, I have an idea what is going on.
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| Aug 29, 2014 at 3:45 | history | asked | Greg P | CC BY-SA 3.0 |