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Boundary conditions for a ring PDE

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Joshua Chang
Joshua Chang 2020 年 1 月 21 日
回答済み: Josh Meyer 2020 年 1 月 30 日
I'm trying to model a Fitzhugh-Nagumo like equation set along a ring geometry. The equation is set in 1-D, but the boundary conditions are described such that the u(0,t) = u(1, t) and du/dt(0, t) = du/dt(1, t). I know how to format the PDE itself into the required form for pdepe, but I'm not sure how to formulate the boundary conditions such that this ring effect is correct. Is pdepe the right way to be approaching this problem? Thanks!

回答 (1 件)

Josh Meyer
Josh Meyer 2020 年 1 月 30 日
pdepe does not allow for time derivatives in the boundary conditions. The boundary conditions are expressed in terms of the flux term f(x,t,u,dudx) in the PDE equation, which only includes the spatial derivative.
If you instead had du/dx(0, t) = du/dx(1, t) on the ring, then you could use m = 1 for a spherical geometry. Whenever m > 0 pdepe automatically imposes its own left boundary condition to account for the singularity at the origin, so then you would just need to code the right boundary condition.

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