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SOR on a positive-definite system with relaxation parameter in (0, 2)

Example

Take

A=(2−1−12),b=0,x(0)=(10),ω=32.

The matrix is positive definite and ω∈(0,2). For the fixed order 1,2, the SOR iteration matrix is

Gω=(−1/23/4−3/81/16),

with eigenvalues (−7±i207)/32, both of modulus 1/2. Thus the error converges to zero for every initial vector.

Facts & Assumptions

Given: The displayed matrix, right-hand side, initial vector, and relaxation parameter.

[F1]

SOR uses Mω=ω−1D−L and Nω=(ω−1−1)D+L∗ (Finite-budget successive over-relaxation stationary iteration).

[L1]

For Hermitian A with positive diagonal, SOR converges for every start if and only if A is positive definite and 0<ω<2 (SOR converges exactly for positive-definite A and relaxation parameter in (0, 2)).

[L2]

Hermitian positive definite means A=A∗ and z∗Az>0 for every nonzero z (Hermitian positive-definite matrices and Cholesky factorisation A = LL* with positive diagonal).

Verification

1.1givenL2algebra

For z=(z1,z2)T, z∗Az=∣z1∣2+∣z2∣2+∣z1−z2∣2, so [L2] shows that A is positive definite; its diagonal is positive and 0<3/2<2.

1.2F1algebra

Here D=2I, L21=1, and U=L∗. By [F1], Mω=(4/30−14/3) and Nω=(−2/310−2/3), hence Gω=Mω−1Nω=(−1/23/4−3/81/16). Its characteristic polynomial is λ2+(7/16)λ+1/4, whose roots are (−7±i207)/32 and have modulus 1/2.

2.1L1step 1.1step 1.2∎

The hypotheses of [L1] hold by step 1.1, so SOR converges for every start; in particular the stated start converges to the solution 0.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources