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Derivative of the exact-branch FAS two-grid cycle

Statement

Consider one exact-branch two-grid cycle from Finite-budget nonlinear FAS two-grid cycle. Let Fh:Uh→Rnh and FH:UH→RnH be C1 on open sets, let I,R,P be linear, and suppose Fh(u∗)=fh and Iu∗∈UH. Assume the exact coarse solver selects the local root branch through Iu∗ and DFH(Iu∗) is invertible. Suppose the pre/post smoothing maps are C1 and fix u∗. The selected coarse root v(u) is locally C1, and the derivative of the actual FAS cycle E at u∗ is DE(u∗)=DSpost(u∗)[Inh−P DFH(Iu∗)−1R DFh(u∗)]DSpre(u∗).

No identity between R and I, or between the coarse operator and a rediscretized fine operator, is assumed in this derivative formula.

Facts & Assumptions

Given: The finite-dimensional maps, linear transfers, fine root, selected exact coarse branch, invertible coarse Jacobian, and C1 pre/post smoothers in the statement.

[F1]

Finite-budget nonlinear FAS two-grid cycle: after pre-smoothing at u, the coarse target is gH(u)=FH(Iu)+R(fh−Fh(u)), and an exact coarse solve returns v(u) before the correction u+P(v(u)−Iu).

[F2]

The Euclidean implicit function theorem with derivative formula: for a C1 map G(u,v) with G(u∗,v∗)=0 and invertible partial derivative DvG(u∗,v∗), there is a local C1 root branch v(u) and Dv(u∗)=−DvG(u∗,v∗)−1DuG(u∗,v∗).

[F3]

The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a): for totally differentiable Euclidean maps, D(g∘f)(a)=Dg(f(a))∘Df(a).

Proof

1.1givenF1constructalgebra

Define G(u,v):=FH(v)−FH(Iu)−R(fh−Fh(u)) on the open product domain where its terms are defined. The given fine root gives G(u∗,Iu∗)=0. Since I and R are linear and Fh,FH are C1, G is C1; its partial derivatives at (u∗,Iu∗) are DvG=DFH(Iu∗) and DuG=−DFH(Iu∗)I+RDFh(u∗). The first is invertible by hypothesis.

2.1F2step 1.1algebra

Apply [F2] to G. Its selected local branch satisfies Dv(u∗)=−DFH(Iu∗)−1(−DFH(Iu∗)I+RDFh(u∗))=I−DFH(Iu∗)−1RDFh(u∗).

3.1F1step 2.1algebra

The coarse correction map is C(u)=u+P(v(u)−Iu) by [F1]. Differentiating and using step 2.1 gives DC(u∗)=Inh+P(Dv(u∗)−I)=Inh−P DFH(Iu∗)−1RDFh(u∗).

4.1givenF3step 3.1algebra∎

The whole cycle is E=Spost∘C∘Spre. The pre-smoother fixes u∗, and the stated smoother maps are C1, so [F3] and step 3.1 give the displayed formula for DE(u∗).

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