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SDC iteration matrix for the scalar linear test equation

Statement

For the scalar test equation y′=λy and z=hλ, let U∗ be a collocation stage vector and Uk the SDC iterates using the explicit-left-Euler stage block QΔ from Finite explicit-left-Euler spectral deferred-correction sweep. Then Ek+1=B(z)Ek,B(z):=(I−zQΔ)−1z(Q−QΔ),Ek:=Uk−U∗. The inverse exists for every real z because QΔ is strictly lower triangular. In the induced infinity norm from Collocation tableau from distinct quadrature nodes, ∥Ek∥∞,2≤∥B(z)∥∞k∥E0∥∞,2. This identity makes no convergence claim when the powers of B(z) do not tend to zero; stability of the fully solved collocation step alone is not an SDC iteration guarantee.

Facts & Assumptions

Given: A fixed node set, its collocation matrices Q,QΔ, a scalar λ∈R, step h>0, and the scalar ODE y′=λy.

[F1]

Finite explicit-left-Euler spectral deferred-correction sweep: the stage update is Uk+1=1u+hQΔF(Uk+1)+h(Q−QΔ)F(Uk).

[F2]

Fixed points of the SDC sweep are collocation stages: the SDC fixed point is the collocation stage vector.

[F3]

Finite explicit-left-Euler spectral deferred-correction sweep makes QΔ strictly lower triangular; Collocation tableau from distinct quadrature nodes defines ∥A∥∞=max⁡i∑j∣aij∣.

Proof

1.1givenF1F2F3algebra

On this equation, F(U)=λU. Subtract the fixed-point instance of [F1] from its kth sweep to obtain (I−zQΔ)Ek+1=z(Q−QΔ)Ek. The diagonal of I−zQΔ consists of ones and it is lower triangular by [F3], so forward substitution uniquely solves this system for every z∈R.

2.1step 1.1F3algebra∎

Solving step 1.1 gives Ek+1=B(z)Ek with the displayed B(z); induction yields Ek=B(z)kE0. For any vector v, the row-wise triangle inequality gives ∥B(z)v∥∞,2≤∥B(z)∥∞∥v∥∞,2, so applying this bound k times gives the stated power estimate. No bound less than one is asserted for general z or node sets.

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