Solution of Strictly Diagonal Dominant Tridiagonal Systems on Vector Computers

24 restriction. As and , the minimum of δ (l) is achieved on the boundary of R. It is not difficult to see that this minimum is in x = y = , independently of the values of the greek letters. With this, we can see that: Now, if , we have: which allows to finish the proof: x y , () R ∈ 0 δ l 1 − () 1 − , [ ] 0 δ l 1 − () 1 − , [ ] × = x ∂ ∂δ l () 0 ≠ y ∂ ∂δ l () 0 ≠ δ l 1 − () 1 − δ l () δ l 1 − () a i 2 − a i 1 c i 2 − δ l 1 − () 1 − − c i 2 + c i 1 a i 2 + δ l 1 − () 1 − − + ≥ ∇ () = c i 2 ± a i 2 ± + δ l 1 − () 1 − = a i 2 − 1 c i 2 − δ l 1 − () 1 − − δ l 1 − () 1 − ≤ c i 2 + 1 a i 2 + δ l 1 − () 1 − − δ l 1 − () 1 − ≤ , ∇ () δ l 1 − δ l 1 − 1 − a i δ l 1 − 1 − c i + δ l 1 − 2 a i c i + δ l 1 − 3 ≥ ≥ ≥ 23 Also, the values of d' i-1 , d' i and d' i+1 can be found easily by applying transformations to the corresponding equations. With this, the diagonal dominance of equation i after the transformations of step l is exactly: Now, we want to bound δ (l) in terms of δ (l-1). We do that by finding the minimum for δ (l). This minimum bounds the diagonal dominance of the system at step l with respect to the equations at a given step l-1. Note that the worst case for δ (l) (δ (l) minimum) is when. All other cases do not cause so many cancellations as this case. Taken this into account, we know that in the worst case and. So, defining and we have where greek letters are defined …