How to solve a system of pdes coupled in their boundary condition
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Initially i had the following system of pdes to solve:
with the following initial and boundary conditions
at
at and
Both domains are coupled by a first order reversible reaction such as
at
where are constants.
I have successfuly solved the following system using an implicit euler scheme for the x-space derivative and second order central difference for the y-space second derivative for the internal nodes. Points at and were discretized using second order backward discritization.
Then to link between both domains the interface conditions was discritized also using backward discritization.
Now i want to developp more my model by considering the following system of 4 pdes:
the boundary conditions are the same as the previous system except that this time the interface condition is a non linear condition of the following form
at
and now i am confused on how to solve numerically such a system i have read about the segregated approach to solve coupled pdes. My understanding is that we fix all dependant variable but one where we solve for it and do the same for each variable. Then we iterate until convergence. However, i am not really sure how to implement it.
If anyone can point me in the right direction or useful literature, i would be grateful. Note that if it is feasable, i want to stick with my previously developed scheme and then elaborate it to take into account the coupling.
Thanks in advance for any help
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Accepted Answer
Torsten
on 25 Apr 2022
Ok, then the discretization at y = alpha should be for the nonlinear case
(A(n+1)-A(n-1))/(2*dy)=-Bi_A * A(n)*A1(n)^2*B(1)
(A1(n+1)-A1(n-1))/(2*dy) = -2*Bi_A1 * A(n)*A1(n)^2*B(1)
(B(2)-B(0))/(2*dy) = 2*Bi_B * A(n)*A1(n)^2*B(1)
(B1(2)-B1(0))/(2*dy) = -Bi_B1 * A(n)*A1(n)^2*B(1)
These are four equations for A and A1 in the artificial point yA(n+1) and for B and B1 in the artificial point yB(0).
Assuming that your PDE equations also hold at the interface point, you can use the usual central difference discretization for the 2nd derivatives through the points yA(n-1), yA(n), yA(n+1) and yB(0),yB(1) and yB(2) to get A(n), A1(n), B(1) and B1(1).
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