Timeline for How to apply different equations to different parts of a geometry in PDE?
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2019 at 16:28 | comment | added | Alex Trounev | @MichaelE2 I compared the solution according to your model and the model of Henrik Schumacher (see my answer). | |
| Mar 25, 2019 at 9:44 | review | Suggested edits | |||
| Mar 25, 2019 at 10:59 | |||||
| Mar 25, 2019 at 9:39 | comment | added | MOON | In Latex equations (now numbered) equation 1 governs the evolution of variable v inside the disk. Equation 2 governs the evolution of the m on the boundary and conversion of v to m on the boundary with a rate proportional to the value of v on the boundary. The affix of on the boundary in equation 2 tries to tell that m exist only on the boundary and it also depends on the value of v on the boundary. I will add the physical meaning of the system to the question. | |
| Mar 24, 2019 at 22:51 | comment | added | Michael E2 | @MOON I should admit that I answered without understanding what you could mean by v is converted to m: (1) In your TeX equations, there's no PDE governing the evolution of m except on the boundary. (2) The PDE in your code has m depending on both m and v, so both would have to coexist (even though I understood you to say you don't want them to). (3) It sounds to me that you might want a single solution that behaves piecewise, differently inside the small disk than outside it. The points (1) - (3) seem to be inconsistent, so I'm hoping you can clarify. | |
| Mar 24, 2019 at 22:39 | comment | added | MOON | I tried to use DirichletCondition[m[x, y, t] == 0, x^2 + y^2 < 1] to keep m inside the disk zero while it should be non-zero at the boundary. The second part of your answer is closer to what I need. However, there is still a small problem. While m is zero in the disk with radius 1/2 and non-zero from r = 1/2 to r = 1, in the region r=1/2 to r=1 both m and v coexist. v is converted to m with a rate proportional to the value of v on the boundary (r = 1). If m exist for r<1 then m coverts from v with a rate that depends on the value of v for r<1. | |
| Mar 24, 2019 at 22:03 | comment | added | Michael E2 | @MOON I adapted what you indicated: "Adding` DirichletCondition[m[x, y, t] == 0, x^2 + y^2 < 1]` to enforce the value of m inside the geometry (here the disk) gives this error:" I don't know what you meant, but you included t and I thought you meant for all time. | |
| Mar 24, 2019 at 22:02 | history | edited | Michael E2 | CC BY-SA 4.0 | added 1415 characters in body |
| Mar 24, 2019 at 22:01 | comment | added | MOON | Thank you for your answer. The solution msol obtained by your answer is zero on the boundary for all the time. I think it should be non-zero because the variable v converts to variable m on the boundary and there is no decay term for m. Even if somehow m at the end should be zero at least initially it should be non-zero because the first time that v reaches the boundary it converts to m. | |
| Mar 24, 2019 at 21:47 | history | answered | Michael E2 | CC BY-SA 4.0 |