Improving Results on Convergence of AOR Method

Definition 2.1 (see [2].)A matrix A ∈ C^n×n is called a strictly diagonally dominant matrix (SD) if

A matrix A ∈ C^n×n is called a strictly doubly diagonally dominant matrix (DD) if

Definition 2.2 (see [3].)A matrix A ∈ C^n×n is called a strictly α diagonally dominant matrix (D(α)) if there exits α ∈ [0,1], such that

Definition 2.3 (see [4].)Let A = (a_ij) ∈ C^n×n, if there exits α ∈ [0,1] such that

then A is called a strictly doubly α diagonally dominant matrix (DD(α)).

Theorem 2.4 (see [3].)Let A ∈ D(α), then AOR method converges for

Theorem 2.5 (see [6].)Let A ∈ DD, then AOR method converges for

where

Lemma 3.1 (see [4].)If A ∈ DD(α), then A is a nonsingular H-matrix.

Theorem 3.2. Let A ∈ DD(α), if 1 − σ²[αR_i(L)R_j(L) + (1 − α)S_i(L)S_j(L)] > 0, for  all  j ∈ N,   i ≠ j, then

where

Proof. Let λ be an eigenvalue of M_σ,ω such that

that is, If λ(I − σL) − ((1 − ω)I + (ω − σ)L + ωU) ∈ DD(α), then by Lemma 3.1, λ(I − σL) − ((1 − ω)I + (ω − σ)L + ωU) is nonsingular and λ is not an eigenvalue of iterative matrix M_σ,ω, that is, if then λ is not an eigenvalue of M_σ,ω. Especially, if then λ is not an eigenvalue of M_σ,ω.

If λ is an eigenvalue of M_σ,ω, then there exits at least a couple of i, j (i ≠ j), such that

that is, Since A₁ = 1 − σ²[αR_i(L)R_j(L) + (1 − α)S_i(L)S_j(L)] > 0, and the discriminant Δ of the quadratic in |λ| satisfies Δ ≥ 0, then the solution of (3.8) satisfies So

Theorem 4.1. Let A ∈ DD(α), then AOR method converges if ω, σ satisfy either

where

Proof. It is easy to verify that for each σ, which satisfies one of the conditions (I)–(III), we have

Firstly, we consider case (I). Since A be a α diagonally dominant matrix, then by Lemma 3.1, we know that A is a nonsingular H-matrix; therefore, M(A) is a nonsingular M-matrix, and it follows that from paper [7], ρ(M_σ,σ) < 1 holds for 0 ≤ σ < s and for σ ≠ 0,

If 0 < ω < max {2σ/(1 + ρ(M_σ,σ)), t} = 2σ/(1 + ρ(M_σ,σ)), then by extrapolation theorem [6], we have ρ(M_σ,ω) < 1.

If 0 < ω < max {2σ/(1 + ρ(M_σ,σ)), t} = t, then it remains to analyze the case

Since when ρ(M_σ,σ) < 1, then 0 ≤ σ < ω. From we have $$.

• (1)

  When ω ≤ 1, it easy to verify that (4.8) holds.

• (2)

  When ω > 1, since

  $$ ()

then by A₂ − A₃ < A₁ and 1 − σ²[αR_i(L)R_j(L) + (1 − α)S_i(L)S_j(L)] > 0, for  all  i, j ∈ N,   i ≠ j, we have It is easy to verify that the discriminant Δ of the quadratic in ω satisfies Δ > 0, and so there holds or For ω₁, we have A₂ > 2, it is in contradiction with ((4.8)b). Therefore, ω₁ should be deleted.

Secondly, we prove (II).

• (1)

  When 0 < ω ≤ 1,  σ < 0,

  $$ ()
  By A₂ − A₃ < A₁, we have
  $$ ()
  It is easy to verify that the discriminant Δ of the quadratic in σ satisfies Δ > 0, and so there holds
  $$ ()
  By σ < 0 and 1 − σ²[αR_i(L)R_j(L) + (1 − α)S_i(L)S_j(L)] > 0, for  all  i, j ∈ N, i ≠ j, we obtain
  $$ ()

• (2)

  When 1 < ω < t, σ < 0,

  $$ ()
  By A₂ − A₃ < A₁, we have
  $$ ()
  It is easy to verify that the discriminant Δ of the quadratic in σ satisfies Δ > 0, and so there holds
  $$ ()
  By σ < 0 and 1 − σ²[αR_i(L)R_j(L) + (1 − α)S_i(L)S_j(L)] > 0, for  all  i, j ∈ N,  i ≠ j, we obtain
  $$ ()
  Therefore, by (4.16) and (4.20), we get
  $$ ()

Finally, we prove (III).

• (1)

  When 0 < ω ≤ 1, σ ≥ t,

  $$ ()
  By A₂ − A₃ < A₁, we have
  $$ ()
  It is easy to verify that the discriminant Δ of the quadratic in σ satisfies Δ > 0, and so there holds
  $$ ()
  By σ ≥ t, we obtain
  $$ ()

• (2)

  When 1 < ω < t,  σ ≥ t,

  $$ ()
  By A₂ − A₃ < A₁, we have
  $$ ()
  It is easy to verify that the discriminant Δ of the quadratic in σ satisfies Δ > 0, and so there holds
  $$ ()
  By σ ≥ t, we obtain
  $$ ()
  Therefore, by (4.25) and (4.29), we obtain
  $$ ()

Theorem 4.2. Let A ∈ DD(α). If R_i(L + U)R_j(L + U) − S_i(L + U)S_j(L + U) > 0, when 0 < ω ≤ 1, the following conditions hold:

or when 1 < ω < t, the following conditions hold: then the area of convergence of AOR method obtained by Theorem 4.1 is larger than that obtained by Theorem 2.5.

Theorem 4.3. Let A ∈ DD(α). If R_i(L + U)R_j(L + U) − S_i(L + U)S_j(L + U) < 0, when 0 < ω ≤ 1, the following conditions hold:

or when 1 < ω < t, the following conditions hold: then the area of convergence of AOR method obtained by Theorem 4.1 is larger than that obtained by Theorem 2.5.

Example 5.1 (see [6].)Let

where Obviously, A ∉ SD, but A ∈ DD(1/2).

Example 5.2. Let

Obviously, A ∈ DD(1/3), A ∉ D(α), A ∉ DD. So we cannot use Theorems 2.4 and 2.5. By Theorem 4.1, we have the following area of convergence: