Alphabeta Math
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An explicit SDC sweep diverges on a stiff scalar mode

Example

Assume Countable Choice for the Lebesgue interval-measure formula. Analyze the formal one-node Gauss SDC stage-iteration recurrence for y′=λy with λ=−6, h=1, and initial value u=1. The test is stiff in the sense that h∣λ∣=6; the fully solved midpoint stage remains nonsingular and its endpoint is stable. This example studies the recurrence without invoking a residual-acceptance test.

Facts & Assumptions

Given: The one-node Gauss tableau, the explicit-left-Euler SDC preconditioner and the scalar test equation in the Example.

[F1]

One-node Gauss data are midpoint data: Q11=1/2 and b1=1.

[F3]

SDC iteration matrix for the scalar linear test equation: stage errors obey Ek+1=B(z)Ek, with B(z)=(I−zQΔ)−1z(Q−QΔ).

[F4]

Collocation tableau from distinct quadrature nodes: the midpoint stage solves U=1+(z/2)U and its endpoint is 1+zU.

[A1]

Countable Choice, required to instantiate the one-node Gauss rule for Lebesgue measure on [0,1].

Verification

1.1A1F1F2F3F4inductionalgebra

Here z=hλ=−6, so [F3] and [F1] under [A1] give B(−6)=(−6)(1/2)=−3. The collocation stage is U∗=1−3U∗, hence U∗=1/4. The constant predictor is U0=1, so E0=3/4≠0 and Ek=(−3)k(3/4) by induction; its magnitude grows by a factor 3 at every sweep.

2.1F1F3F4algebra∎

The exactly solved endpoint is 1−6(1/4)=−1/2, which is also the midpoint stability factor (1+z/2)/(1−z/2)=−1/2 multiplying the initial value. Its magnitude is below one, while the SDC contraction ratio from the scalar matrix is ∣z∣/2=3>1. Thus stability of the fully solved midpoint step does not imply convergence of the explicit SDC iteration, and the contraction theorem correctly makes no claim here.

Depends on

Used by

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