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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
From the journal Computational Methods in Applied Mathematics
https://doi.org/10.1515/cmam-2017-0016

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.

Keywords: Poroelasticity; Finite Elements; Preconditioning
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 τ∈Σ and recall that |Γt|>0 and τ⋅ν^=0 on Γt. By the pointwise decomposition

τ=PD⁢τ+(I-PD)⁢τ,

and the fact that (I-PD)⁢τ=1n⁢tr⁡τ⁢𝕀, it suffices to show that

∥tr⁡τ∥02≤C⁢((12⁢μ⁢PD⁢τ,PD⁢τ)+∥𝐝𝐢𝐯⁡τ∥02)

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 ϕ∈HΓd1⁢(Ω;𝕍):={φ∈H1⁢(Ω;𝕍):φ|Γd=0} such that

(A.1)

div⁡ϕ=tr⁡τ,∥ϕ∥1≤C⁢∥tr⁡τ∥0

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

∥tr⁡τ∥02=(tr⁡τ,div⁡ϕ)=(tr⁡τ⁢𝕀,𝐠𝐫𝐚𝐝⁡ϕ).

Since tr⁡τ⁢𝕀=n⁢(τ-PD⁢τ), we get

∥tr⁡τ∥02=n⁢(τ,𝐠𝐫𝐚𝐝⁡ϕ)-n⁢(PD⁢τ,𝐠𝐫𝐚𝐝⁡ϕ)
=-n⁢(𝐝𝐢𝐯⁡τ,ϕ)-n⁢(PD⁢τ,𝐠𝐫𝐚𝐝⁡ϕ),

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

∥tr⁡τ∥02≤n⁢(∥𝐝𝐢𝐯⁡τ∥0⁢∥ϕ∥0+∥PD⁢τ∥0⁢∥grad⁡ϕ∥0)
≤n⁢(∥𝐝𝐢𝐯⁡τ∥02+∥PD⁢τ∥02)12⁢∥ϕ∥1
≤C⁢(∥𝐝𝐢𝐯⁡τ∥02+∥PD⁢τ∥02)12⁢∥tr⁡τ∥0,

and so the result follows after dividing by ∥tr⁡τ∥0. ∎

When |Γt|=0, i.e., Γd=∂⁡Ω, (A.1) can only hold if tr⁡τ 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 τ∈Σ. From the decomposition P0⁢τ=PD⁢τ+(P0-PD)⁢τ, it suffices to prove the estimate for (P0-PD)⁢τ component. Denoting the mean value of the trace by

tr⁡τ¯:=1|Ω|⁢∫Ωtr⁡τ⁢d⁢x,

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

∥tr⁡τ-tr⁡τ¯∥02≤C⁢((12⁢μ⁢PD⁢τ,PD⁢τ)+∥𝐝𝐢𝐯⁡τ∥02).

Since tr⁡τ-tr⁡τ¯ is mean-value zero, there exists ϕ∈{φ∈H1⁢(Ω;𝕍):φ|∂⁡Ω=0} such that

div⁡ϕ=tr⁡τ-tr⁡τ¯,∥ϕ∥1≤C⁢∥tr⁡τ-tr⁡τ¯∥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)={ϕ∈H1⁢(Ω;V):ϕ|Γd=0}. Then there is a constant C>0 so that for every f∈L2⁢(Ω) there is a ϕ∈HΓd1⁢(Ω;V) so that

div⁡ϕ=f,∥ϕ∥1≤C⁢∥f∥0.

Proof.

Take any f∈L2⁢(Ω). We first decompose f into its mean value zero- and mean value part as f=f0+fc, where f0∈L02⁢(Ω) and fc=af⁢1Ω for af∈ℝ. Further, we can decompose HΓd1⁢(Ω;𝕍)=H01⁢(Ω;𝕍)⊕V1, where

V1:={ϕ∈HΓd1⁢(Ω;𝕍):(𝐠𝐫𝐚𝐝⁡ϕ,𝐠𝐫𝐚𝐝⁡ψ)=0⁢ for all ⁢ψ∈H01⁢(Ω;𝕍)}.

Consider then the problem of finding ζ∈V1 so that

(B.1)(𝐠𝐫𝐚𝐝⁡ζ,𝐠𝐫𝐚𝐝⁡ψ)=(𝕀,𝐠𝐫𝐚𝐝⁡ψ) for all ⁢ψ∈V1.

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

∫Ωdiv⁡ζ⁢d⁢x=∥𝐠𝐫𝐚𝐝⁡ζ∥02.

Therefore, if we set

ω=af∥𝐠𝐫𝐚𝐝⁡ζ∥02⁢ζ,

we have

∫Ωdiv⁡ω⁢d⁢x=af

and ∥ω∥1≤C⁢∥fc∥0 for some constant C depending on ζ. It follows that f-div⁡ω∈L02⁢(Ω), i.e., f-div⁡ω has mean value zero. From the theory of Stokes equation, we can thus find a ω0∈H01⁢(Ω;𝕍) so that

(B.2)div⁡ω0=f-div⁡ω,∥ω0∥1≤C2⁢∥f-div⁡ω∥0,

where the constant C2 is independent of f-div⁡ω (cf. [14, Theorem 5.1]). We set ϕ=ω0+ω, and it follows from (B.2) that div⁡ϕ=f. Using the triangle inequality, (B.2) and the properties of ω, we estimate ∥ϕ∥1 as

∥ϕ∥1≤∥ω0∥1+∥ω∥1≤C⁢(∥f-div⁡ω∥0+∥fc∥0)≤C⁢(∥f∥0+∥ω∥1)≤C⁢∥f∥0,

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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