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2. Two-Point Boundary Value Problems
52
Then, if m j = β j /δ j−1 times (2.20) is subtracted from this equation, we
get
where
δ j = α j − m j γ j−1 ,
= b j − m j c j−1 .
After k − 1 iterations of this procedure we obtain a system of the form
δ 1 v 1 + γ 1 v 2 c j δ j v j + γ j v j+1 = c j , = c 1 ,
. . . . . . . . .
+ = c k ,
δ k v k γ k v k+1
= b k+1 ,
β k+1 v k + α k+1 v k+1 + γ k+1 v k+2
. . . . . .
β n−1 v n−2 + α n−1 v n−1 + γ n−1 v n = b n−1 ,
β n v n−1 + α n v n = b n .
(2.21)
Here the variables δ j and c j are defined from the given coefficients α j ,β j ,γ j ,
and b j of (2.19) by the recurrence relations
δ 1 = α 1 ,c 1 = b 1 ,

β j
m j = , (2.22)
δ j−1
δ j = α j − m j γ j−1 , j =2, 3,... ,k ,
c j = b j − m j c j−1 .

Note that in the derivation of the system (2.21) from the original system
(2.19) we have implicitly assumed that the variables δ 1 ,δ 2 ,... ,δ k−1 will
be nonzero. Furthermore, if δ 1 ,δ 2 ,... ,δ k−1 are nonzero, the two systems
(2.19) and (2.21) are equivalent in the sense that they have the same solu-
tions.
If the computed values of the δ k s are always nonzero, we can continue
to derive a system of the form (2.21) until k = n. Hence, in this case we
obtain a system of the form

= c 1 ,
δ 1 v 1 + γ 1 v 2
δ 2 v 2 + γ 2 v 3 = c 2 ,
. . . . . . (2.23)
δ n−1 v n−1 + γ n−1 v n = c n−1 ,
δ n v n = c n .