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March 3, 2005, 04:28 
Periodic BC + FVM

#1 
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Hi all,
I am trying to implement, Periodic BC to the Unsteady finite volume code, that I have, which solves the solution of equations by line by line TDMA. So, no longer I can use this Line by line TDMA, for periodic BC, as I will be having some more enteries in the Matrix that I am solving, so I need to go into Cyclic Tridiagonal systems, where in, there are two additional enteries that are positoned at the end of the First row and at the begining of the last row. I am not sure how these values at the end of the first row, and the begining of the last row are generated, as per my reading, the last entry in the first row is the East coefficient of the last node (considering 1D) and the first entry in the last row, is West coefficient of the first node in the grid, but these coefficients are ZERO, which means that should I consider the coefficeints at the 2nd and the (n1) th node????????? Any help is HIGHLY appreciated kilda 

March 4, 2005, 08:44 
Re: Periodic BC + FVM

#2 
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Is you "cyclic tridiagonal system" a function/subroutine you have got from somewhere and do not understand the arguments or do you not understand how to solve a periodic tridiagonal system using code for a nonperiodic system?
The coefficients in the corners are the coefficients linking your first point to the last point and the last point to the first point. If they are zero then you do not have a periodic problem. 

March 4, 2005, 20:17 
Re: Periodic BC + FVM

#3 
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Hi Andy,
Thanks for the reply, I am trying to figure out what the coefficients will be at the end of first row, and the begining of the N th row, in the cyclic system????? As per my understanding so far, (you have to correct me), the last entry (alpha) in the first row will be the west coefficient of the Nth node(last node), and the first entry in the last row(beta), will be the east coefficient of the first Node (starting node). The matrix will luk something like this b1 c1 0 0 0 alpha a2 b2 c2 0 0 0 0 a3 b3 c3 0 0 0 0 a4 b4 c4 0 0 0 0 a5 b5 c5 beta 0 0 0 a6 b6 thanks, kilda 

March 6, 2005, 09:39 
Re: Periodic BC + FVM

#4 
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Assuming west is i1 and east i+1 then alpha is the west coefficient of the first point and beta is the east coefficient of the last point. This seems an odd question and I suspect it may not be where your problem/misunderstanding lies.


March 9, 2005, 05:02 
Re: Periodic BC + FVM

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Hi andy,
Thanks for the reply. The fact that I am applying PBC, is that there will be no i1 for the 1st point(boundary point,1st row in the matrix) and no i+1 to the Nth (last boundary point, last row in Matrix). Where as for the interior nodes there will be no alpha and beta...(which u will know!!!) Kilda. 

March 15, 2005, 21:28 
Re: Periodic BC + FVM

#6 
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The point is, since you are imposing periodic bc, that means the first node value u(1) = the point west to the last node value u(N+1), so that (because derivatives at 1 and N are also periodic) u(2) = u(N+2), u(0) = u(N),,,
Hence the west point of your first node u(1) is u(0), and you should substitute u(0) by u(N);; Also, the left point of your last node u(N) is u(N+1), and you should substitute it by u(1), Then, you will have a cyclic equation. sat your discretization for any point i=1,,,N is: a*u(i1) +b*u(i) + c*u(i+1), i=1,,,N then you have(N=6): c b c 0 0 0 a u(1) a b c 0 0 0 u(2) 0 a b c 0 0 * u(3) = (R1,R2,...,RN)^T 0 0 a b c 0 u(4) 0 0 0 a b c .. c 0 0 0 a b u(N) c Google MorisonSherman formula for solving the upper equation. Wen 

March 15, 2005, 21:30 
Re: Periodic BC + FVM

#7 
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The point is, since you are imposing periodic bc, that means the first node value u(1) = the point east to the last node value u(N+1), so that (because derivatives at 1 and N are also periodic) u(2) = u(N+2), u(0) = u(N),,,
Hence the west point of your first node u(1) is u(0), and you should substitute u(0) by u(N);; Also, the East point of your last node u(N) is u(N+1), and you should substitute it by u(1), Then, you will have a cyclic equation. say your discretization for any point i=1,,,N is: a*u(i1) +b*u(i) + c*u(i+1), i=1,,,N then you have(N=6): c b c 0 0 0 a u(1) a b c 0 0 0 u(2) 0 a b c 0 0 * u(3) = (R1,R2,...,RN)^T 0 0 a b c 0 u(4) 0 0 0 a b c .. c 0 0 0 a b u(N) c Google MorisonSherman formula for solving the upper equation. Wen 

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