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Weakly Imposed Symmetry and Robust Preconditioners for Biot’s Consolidation Model

Trygve Bærland ORCID logo, Jeonghun J. Lee ORCID logo, Kent-Andre Mardal ORCID logo and Ragnar Winther ORCID logo

Abstract

We discuss the construction of robust preconditioners for finite element approximations of Biot’s consolidation model in poroelasticity. More precisely, we study finite element methods based on generalizations of the Hellinger–Reissner principle of linear elasticity, where the stress tensor is one of the unknowns. The Biot model has a number of applications in science, medicine, and engineering. A challenge in many of these applications is that the model parameters range over several orders of magnitude. Therefore, discretization procedures which are well behaved with respect to such variations are needed. The focus of the present paper will be on the construction of preconditioners, such that the preconditioned discrete systems are well-conditioned with respect to variations of the model parameters as well as refinements of the discretization. As a byproduct, we also obtain preconditioners for linear elasticity that are robust in the incompressible limit.

MSC 2010: 65N30; 65N55; 74S05

Funding statement: The research leading to these results has received funding the European Research Council under the European Union’s Seventh Framework Programme (FP7/2007-2013)/ERC grant agreement no. 339643. The work of Kent-Andre Mardal has also been supported by the Research Council of Norway through grant no. 179578/F50.

A Appendix: Proofs of (3.3) and (3.5)

Here we provide proofs of inequalities (3.3) and (3.5).

Proof of (3.3).

Fix τΣ and recall that |Γt|>0 and τν^=0 on Γt. By the pointwise decomposition

τ=PDτ+(I-PD)τ,

and the fact that (I-PD)τ=1ntrτ𝕀, it suffices to show that

trτ02C((12μPDτ,PDτ)+𝐝𝐢𝐯τ02)

for some constant C independent of τ. To prove this, we use a well-known result for the right inverse of the divergence operator: There exists ϕHΓd1(Ω;𝕍):={φH1(Ω;𝕍):φ|Γd=0} such that

(A.1)

divϕ=trτ,ϕ1Ctrτ0

with C>0 independent of τ, cf. Appendix B. We then have that

trτ02=(trτ,divϕ)=(trτ𝕀,𝐠𝐫𝐚𝐝ϕ).

Since trτ𝕀=n(τ-PDτ), we get

trτ02=n(τ,𝐠𝐫𝐚𝐝ϕ)-n(PDτ,𝐠𝐫𝐚𝐝ϕ)
=-n(𝐝𝐢𝐯τ,ϕ)-n(PDτ,𝐠𝐫𝐚𝐝ϕ),

where the first term of the final form is a result of integration by parts. Next, we may use the Cauchy–Schwarz inequality, which results in

trτ02n(𝐝𝐢𝐯τ0ϕ0+PDτ0gradϕ0)
n(𝐝𝐢𝐯τ02+PDτ02)12ϕ1
C(𝐝𝐢𝐯τ02+PDτ02)12trτ0,

and so the result follows after dividing by trτ0. ∎

When |Γt|=0, i.e., Γd=Ω, (A.1) can only hold if trτ has mean value zero. However, with this constraint, we can prove (3.5) with almost the same argument as above.

Proof of (3.5).

Fix any τΣ. From the decomposition P0τ=PDτ+(P0-PD)τ, it suffices to prove the estimate for (P0-PD)τ component. Denoting the mean value of the trace by

trτ¯:=1|Ω|Ωtrτdx,

we have that (I-PD)P0τ=(P0-PD)τ=1n(trτ-trτ¯)𝕀, and so it is sufficient to show that

trτ-trτ¯02C((12μPDτ,PDτ)+𝐝𝐢𝐯τ02).

Since trτ-trτ¯ is mean-value zero, there exists ϕ{φH1(Ω;𝕍):φ|Ω=0} such that

divϕ=trτ-trτ¯,ϕ1Ctrτ-trτ¯0

with C>0 independent of τ (cf. [14, Theorem 5.1]). The rest of the proof is completely analogous to the proof of (3.3) above. ∎

B Appendix: Right Inverse of Divergence Operator

A result for the right inverse of the divergence operator, as expressed by (A.1), is closely related to the inf-sup condition for the Stokes problem, and therefore well-known. However, we are not aware of a proper reference for the case when |Ω|>|Γt|>0, i.e., for the case when |Γd|>0, but Γd is not all of Ω. Therefore, for completeness, we include a proof here.

Lemma B.1.

Assume that |Γt|>0 and set HΓd1(Ω;V)={ϕH1(Ω;V):ϕ|Γd=0}. Then there is a constant C>0 so that for every fL2(Ω) there is a ϕHΓd1(Ω;V) so that

divϕ=f,ϕ1Cf0.

Proof.

Take any fL2(Ω). We first decompose f into its mean value zero- and mean value part as f=f0+fc, where f0L02(Ω) and fc=af1Ω for af. Further, we can decompose HΓd1(Ω;𝕍)=H01(Ω;𝕍)V1, where

V1:={ϕHΓd1(Ω;𝕍):(𝐠𝐫𝐚𝐝ϕ,𝐠𝐫𝐚𝐝ψ)=0 for all ψH01(Ω;𝕍)}.

Consider then the problem of finding ζV1 so that

(B.1)(𝐠𝐫𝐚𝐝ζ,𝐠𝐫𝐚𝐝ψ)=(𝕀,𝐠𝐫𝐚𝐝ψ)for all ψV1.

By the Lax–Milgram lemma (cf. e.g. [8, Theorem 4.1.6]), problem (B.1) has a unique solution ζ and ζ1C1 for some constant C1>0 depending on Ω. Taking ψ=ζ in (B.1), we obtain

Ωdivζdx=𝐠𝐫𝐚𝐝ζ02.

Therefore, if we set

ω=af𝐠𝐫𝐚𝐝ζ02ζ,

we have

Ωdivωdx=af

and ω1Cfc0 for some constant C depending on ζ. It follows that f-divωL02(Ω), i.e., f-divω has mean value zero. From the theory of Stokes equation, we can thus find a ω0H01(Ω;𝕍) so that

(B.2)divω0=f-divω,ω01C2f-divω0,

where the constant C2 is independent of f-divω (cf. [14, Theorem 5.1]). We set ϕ=ω0+ω, and it follows from (B.2) that divϕ=f. Using the triangle inequality, (B.2) and the properties of ω, we estimate ϕ1 as

ϕ1ω01+ω1C(f-divω0+fc0)C(f0+ω1)Cf0,

which completes the proof. ∎

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Received: 2017-3-15
Revised: 2017-5-15
Accepted: 2017-5-16
Published Online: 2017-6-17
Published in Print: 2017-7-1

© 2017 Walter de Gruyter GmbH, Berlin/Boston

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