Recursive method for solving the block Tridiagonal Matrix Equation Appendix A

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Appendix A

Recursive method for solving the block Tridiagonal Matrix Equation

To solve the equation (3.32)

A  N y N1    B N y N C  N y N1F N , 2NL1 A1 for

^y ^N ^p   ^N ,^nN , ^N ^T , 2NL1

assuming that equation A1 is transformed into the following equation involving unknown vectors at two points instead of three:

B^'   N y N C^'  N y N1F^' N A2 this is written explicitly for ^y ^N to yield

y N B^' N ^1F^' N B^' N ^1C^'  N y N1 A3 Division point N is decreased by one, and result is substituted in equation A1to yield





^B'      ^{N B N AN B' N1}^1C'^N1

^C'   ^{N C N} A5 ^F'      ^{N F N AN B' N1}^1F'^N1 3NL1

hence for N=2, direct comparison of equations A2and A1 yields ^y ^10

^B'   ^{2 B2 , C'}   ^{2 C2 , F'} ^{2 F} ² . A6 Thus, starting with equation A6, ^B'     ^{N ,C' N ,F' N} in equation A5 are determined for N=3, 4, …, L-1 where the inverse matrix for is directly obtained through the Sylvester’s formula.

Namely, let

 ^'

' bij

B  and B^1 ij^' with i,j1,2,3.

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then the interrelation between these two matrices is given by,

' '23 '32 '33 '22

11 detB

b b b b 

 

' '33 '12 13' '32

12 detB

b b b b 

 

' '22 '12 '13 '12

13 detB

b b b b 

 

' '33 '21 '23 '31

21 detB

b b b b 

 

' '13 '31 '33 '11

22 detB

b b b b 

 

' '33 '11 '13 '21

23 detB

b b b b 

  A7

' '32 '31 '32 '21

31 detB

b b b b 

 

' '32 '11 '12 '31

32 detB

b b b b 

 

' '12 '21 '22 11'

33 detB

b b b b 

 

The final step is to calculate ^y ^{N L1,L2,....2}by equation A.3, where ^y ^L0is used as the starting condition.