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with pdepe, how to solve coupled ode-pde with ode citing pde's BC

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Rui
Rui el 4 de Oct. de 2013
Comentada: Rui el 2 de Nov. de 2013
Hi everyone,
I have coupled pde-ode equations, where pde's one boundary condition is ode's input parameter. I want to write both equations as PDE so that I can use pdepe, but cannot figure out how to get boundary condition of pde and pass that to ode in pdefun. Could anyone tell me how to code it in order to use pdepe? Thanks!
for example, on x from 0 to 1, i have
dC1/dt = d^2 C1 / d x^2
dC2/dt = C2 - C1(x = 1)
pde's BC : dC1/dx = C1(x=1) - C2
best, Rui
  2 comentarios
Bill Greene
Bill Greene el 5 de Oct. de 2013
Hi,
Let me ask a couple of questions first. In the line: pde's BC : dC1/dx = C1(x=1) - C2 is this the BC at x=1? That is, are dC1/dx and C2 both values at x=1? Since, the equation for C1 is second-order, you also need a BC at x=0. What is that?
Bill
Rui
Rui el 2 de Nov. de 2013
Bill, sorry for the slow feedback. The system did not send me the notification.
at X = 0, dC1/dt = 0, no flux BC.
C1 diffuses to the surface at x=1, where some of it goes across interface between C2/C1.

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