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Fortin stability and quasi-optimality for practical DPG

Statement

Assume Countable Choice and the real form and spaces of Ultraweak Poisson DPG spaces and plan-consistent bilinear form on a finite conforming triangular mesh. For every admissible local input to Exact local finite Riesz solves for practical DPG tests, the exact factorization terminates after O(NK3) scalar arithmetic operations once its Gram entries and right-hand side are formed; it returns singular-test-matrix exactly for a dependent input family, and on an independent family it returns the unique finite Riesz representative. Let Uh⊂U consist of volume fields σh∣K∈P1(K)2, uh∣K∈P1(K), continuous scalar traces u^h∣∂K∈P~2(∂K), and conservative normal traces σ^n,h∣∂K∈P1(∂K). This is finite-dimensional. Let Vr=∏K(P3(K)2×P3(K)), presented cellwise by fixed ordered bases, and let Tr:Uh→Vr be the assembled finite Riesz map. On each fixed basis the unique local Riesz representatives assemble to a bounded linear map satisfying ∥Trwh∥V≤∥b∥∥wh∥U. Now assume the continuous primal and adjoint graph inf-sup constants have minimum β>0. Let Π:V→Vr be the bounded Fortin map of A uniformly bounded Fortin map for the practical P1/R3 DPG family, with b(Uh,(I−Π)V)=0 and ∥Π∥≤CΠ. Then for every bounded load F∈V′, the practical DPG system tested with Tr(Uh) has a unique solution xh, and ∥x−xh∥U≤(1+∥b∥CΠ/β)inf⁡zh∈Uh∥x−zh∥U, where x is the exact continuous solution.

Facts & Assumptions

Given: Countable Choice, the real DPG form and spaces on a finite triangular mesh, admissible local finite-Riesz inputs, the stated P1/R3 trial and test spaces, the two continuous inf-sup hypotheses, and the specified Fortin map.

[A1]

Countable Choice is used by the mixed BNB solve. (The Axiom of Countable Choice (ACω))

[F1]

The DPG spaces and bilinear form are those of the ultraweak definition. (Ultraweak Poisson DPG spaces and plan-consistent bilinear form)

[F2]

The continuous form is bounded. (The ultraweak Poisson DPG bilinear form is bounded)

[F3]

The practical Fortin map is bounded and preserves the form on Uh. (A uniformly bounded Fortin map for the practical P1/R3 DPG family)

[F4]

The BNB theorem gives a unique mixed solution when the constraint inf-sup holds and the first form is coercive on its kernel. (Babuska-Brezzi solvability by kernel coercivity and inf-sup)

[F5]

Riesz represents the bounded form and datum on the real Hilbert trial and test spaces; the adjoint operator is the Hilbert adjoint. (Riesz representation for Hilbert spaces)

[F6]

The practical test map Tr is the finite Riesz representative in the broken polynomial test space. (Exact local finite Riesz solves for practical DPG tests)

[F7]

The trial and test Hilbert factors are complete: real L2 and H1 are Hilbert, H(div) is Hilbert, and the trace quotient factors are Hilbert. Finite products preserve completeness. (L2 with the integral pairing is a Hilbert space, The Sobolev space H1 is a Hilbert space, H(div) is Hilbert and H0(div) is closed, Ultraweak Poisson skeleton spaces are Hilbert trace quotients)

Proof

1.1F2F5F6givenalgebra

On an arbitrary local input family, its Gram matrix is positive semidefinite. As long as the preceding vectors are independent, the next unpivoted LDLT pivot is the squared norm of that vector's residual after orthogonal projection onto their span; this follows by solving the preceding Gram equations. Hence every pivot is positive exactly while the vectors remain independent, and the first dependent vector gives a zero pivot, so the algorithm returns singular-test-matrix exactly for a dependent family. If the family is independent, the finite loop and triangular solves terminate, the resulting coefficients solve Gc=g, and the returned vector is therefore the unique finite Riesz representative. The factorization and solves use O(NK3) exact scalar operations after entry formation. For each cell this representative satisfies (Tr,Kwh,vK)V(K)=bK(wh,vK); summing over cells gives (Trwh,vr)V=b(wh,vr) for every vr∈Vr. For the fixed bases of Vr, local uniqueness and linearity of b show that the assembled Tr is linear. Its Riesz identity and [F2] give ∥Trwh∥V≤∥b∥∥wh∥U.

1.2F1F2F5F7givenalgebra

By [F7], U,V are real Hilbert spaces; boundedness [F2] and Riesz [F5] define A:U→V by (Aw,v)V=b(w,v). The primal inf-sup gives ∥Aw∥V≥β∥w∥U, so A is injective and its range is closed: if Awn converges, the lower bound makes wn Cauchy, and its limit maps to the range limit. The adjoint inf-sup gives ∥A∗v∥U≥β∥v∥V, so ker⁡A∗={0}. By the defining adjoint identity, v⊥ran⁡(A) iff (Aw,v)V=0 for every w, iff (w,A∗v)U=0 for every w, iff A∗v=0. Hence ran⁡(A)⊥=ker⁡A∗, the closed range is dense, and therefore all of V. Riesz represents F∈V′, so it has a unique preimage x∈U.

1.3F3givenalgebra

For wh∈Uh, the continuous primal bound gives β∥wh∥U≤sup⁡v∈Vb(wh,v)/∥v∥V. Fortin orthogonality in [F3] gives b(wh,v)=b(wh,Πv) and ∥Πv∥V≤CΠ∥v∥V, so the restricted inf-sup on Vr is at least γ=β/CΠ.

2.1A1F2F4F6step 1.3

On Vr×Uh, put a(r,v)=(r,v)V and c(v,wh)=b(wh,v). The first form is coercive on all of Vr, and [F4]’s constraint inf-sup is γ by step 1.3. Applying [F4] with datum F∣Vr and zero constraint gives a unique pair (rh,xh) with (rh,v)+b(xh,v)=F(v) and b(wh,rh)=0. The first equation makes rh the Riesz representative of the restricted residual; the second is orthogonality to Tr(Uh), exactly the practical DPG equations by [F6].

3.1F2F5step 1.2step 1.3step 2.1algebra∎

Give Vr′ the inner product transported from Vr by its Riesz map. The second mixed equation makes F−Brxh orthogonal to every Brwh in this inner product. For any zh, F−Brzh=(F−Brxh)+Br(xh−zh), so Pythagoras gives ∥Br(xh−zh)∥Vr′≤∥F−Brzh∥Vr′=∥Br(x−zh)∥Vr′≤∥b∥∥x−zh∥U. Apply the discrete inf-sup γ to xh−zh, then the triangle inequality and infimize over zh to obtain the displayed estimate.

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Cited to discharge well-definedness by Exact local finite Riesz solves for practical DPG tests.

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