32.1 Weak mixed formulation

Transcription

1 Part VII, Chapter 32 arcy s equations arcy s equations consist of the following PEs in the domain : 1 σ + p = f, a.e. in, (32.1a) σ = g, a.e. in. (32.1b) Theunknownsaretheprimalvariablepandthedualvariableσ;pisalsocalled the potential and σ the flux. The data are, f, and g; is a second-order tensor if the material is anisotropic and it may depend on x if the material is non-homogeneous. The PEs (32.1), together with appropriate boundary conditions, are often used to model porous media flows, e.g., fluid flows in aquifers and petroleum reservoirs. In this context, σ is the seepage velocity, p the pressure, and the material permeability; (32.1a) is called arcy s law, and (32.1b) expresses mass conservation. In this chapter, we derive well-posed weak formulations for(32.1) with various boundary conditions. Then we study mixed finite element approximations using H(div)-conforming spaces for the dual variable Weak mixed formulation We assume that f L 2 (), g L 2 (), and that takes symmetric values and the eigenvalues of are bounded from below and from above uniformly in by λ and λ, We consider irichlet, Neumann, and mixed irichlet Neumann conditions for (32.1). We will see that contrary to the primal formulation studied in Chapter 21, irichlet conditions on p are enforced weakly, and Neumann conditions on σ are enforced strongly irichlet boundary condition In this section we consider the irichlet condition p = a d on. Let us first proceed formally by assuming that all the functions are smooth enough.

2 422 Chapter 32. arcy s equations Multiplying (32.1a) by a smooth R d -valued test function τ, integrating over, integrating by parts the pressure term, and using the irichlet boundary condition, we infer that (τ ( 1 σ) p τ)dx = τ f dx (τ n)a d ds. Furthermore, multiplying (32.1b) by a smooth scalar-valued test function q and integrating over gives q σdx = qgdx. Since f L 2 () and g L 2 (), the volume integrals make sense if we assume that σ,τ H(div;) = {ς L 2 () ς L 2 ()} and that p,q L 2 ().Togivearigorousmeaningtotheboundaryintegral,letusrecall that the normal component of fields in H(div;) can be defined through the map γ n : H(div;) H 1 2( ), see (B.71). Assuming that a d H 2( ), 1 the boundary integral is understood as γ n (τ),a d, where, denotes the duality pairing between H 1 2( ) and H 2( ). 1 We can now define the linear forms F d (τ) := τ f dx γ n(τ),a d and G d (q) := qgdx and the bilinear forms (recall that the tensor is symmetric) a(ς,τ) := 1 τ ς dx, b(ς,q) := q ς dx, (32.2) leading to the following problem: Find σ V := H(div;) and p Q := L 2 () such that a(σ,τ)+b(τ,p) = F d (τ), τ V, b(σ,q) = G d (q), q Q. (32.3) The above assumptions imply that F d V, G d Q Q, a B(V,V ), and b B(V,Q). Proposition 32.1 (Well-posedness). Assume f L 2 (), g L 2 (), and a d H 1 2( ). Then (32.3) is well-posed. Moreover, the pair (σ,p) satisfies the PEs (32.1) a.e. in, and p H 1 () satisfies the boundary condition γ 0 (p) = a d a.e. on where γ 0 : H 1 () H 1 2( ) is the trace map. Proof. (1) We apply Theorem Set B = : V Q so that ker(b) = {v V v = 0}. We have already seen that the bilinear forms a and b are continuous. Moreover, the inequality a(ς,ς) λ 1 ς 2 L 2 () implies that the bilinear form a is coercive in ker(b) since ς L2 () = ς H(div;) if ς ker(b); as a result, the two conditions in (30.5) hold. Moreover, (30.6) follows from Lemma 32.2 below. Hence, all the required assumptions hold. (2) Let us recover the PEs (32.1) and the boundary condition. Upon testing

3 Part VII. PEs in Mixed Form 423 (32.3)withanarbitraryfunctionτ C0 ()andq = 0,weinferthat 1 σ+ p,τ = τ f dx, where, is the duality product in the distribution sense. Since C0 () is dense in L 2 (), this proves that 1 σ + p = f L 2 (). Since σ,f L 2 (), this implies that p L 2 (), so that p H 1 (). A similar distribution argument shows that σ = g in L 2 (). Since p H 1 (), we can use the integration by parts formula (p τ +τ p)dx = γ n (τ),γ 0 (p). We then infer that γ n (τ),γ 0 (p) a d = 0 for all τ H(div;). The surjectivity of γ n from Theorem B.113 implies that φ,γ 0 (p) a d = 0 for all φ H 1 2( ), which proves that γ 0 (p) = a d in H 2( ). 1 Lemma 32.2 (Surjectivity of divergence). Let be a Lipschitz domain. The operator : H(div;) L 2 () is surjective: there is β > 0 such that q τ dx inf sup β. (32.4) q L 2 () q L2 () τ H(div;) τ H(div;) Proof. Let q L 2 (). Let φ H0() 1 be such that ( φ, ψ) L2 () = (q,ψ) L2 () for all ψ H0(). 1 Setting ς q = φ, we have ς q H(div;), ς q = q, and ς q H(div;) c q L 2 (). Setting β = c 1, (32.4) follows from q τ dx q ςq dx sup τ H(div;) τ H(div;) ς q H(div;) β q L 2 (). Remark 32.3 (Regularity of ς q ). It is known from elliptic regularity that the field ς q from the above proof is in H s () with s ( 1 2,1] (see 21.4). The embedding argument from Theorem B.101 then implies that ς q L t () where 1 t 1 2 s d, i.e., t 2 d d 2s (e.g., t [1, ) for s = 1 and d = 2) Neumann boundary condition We now consider the Neumann boundary condition σ n = a n on. We still look for σ H(div;), we assume a n H 1 2( ) and we interpret the boundary condition as γ n (σ) = a n. Since γ n (σ),1 = σdx, we infer that a n and g must satisfy the compatibility condition a n,1 = gdx. (32.5) It is convenient to lift the Neumann boundary condition, by considering a field σ n H(div;) so that γ n (σ n ) = a n (recall that γ n is surjective onto H 1 2( )),andtomakethechangeofvariableσ 0 = σ σ n sothatσ 0 satisfies a homogeneous Neumann boundary condition. Let us set H 0 (div;) := {ς H(div;) γ n (ς) = 0} = ker(γ n ); this is a Hilbert space equipped with the natural norm. Moreover, only the gradient of the primal variable p now

4 424 Chapter 32. arcy s equations appears in the problem, so that we conventionally enforce pdx = 0 and we set L 2 0() := {q L 2 () qdx = 0}. The weak formulation is as follows: Find σ 0 V := H 0 (div;) and p Q := L 2 0() such that a(σ 0,τ)+b(τ,p) = F n (τ), τ V, (32.6) b(σ 0,q) = G n (q), q Q, with linear forms F n (τ) := τ (f 1 σ n )dx and G n (q) := q(g σ n )dx; the bilinear forms a and b are the same as in (32.3) for irichlet conditions. Proposition 32.4 (Well-posedness). Assume that f L 2 (), g L 2 (), a n H 1 2( ) and that the compatibility condition (32.5) holds. Then (32.6) is well-posed. Moreover, the pair (σ := σ 0 +σ n,p) satisfies the PEs (32.1) a.e. in, and the boundary condition γ n (σ) = a n holds in H 1 2( ). Proof. See Exercise Remark 32.5 (Choice of σ n ). One possibility to define σ n H(div;) is to set σ n = φ where φ H () 1 = {q H 1 () qdx = 0} solves the pure Neumann problem φ rdx = grdx+ a n,γ 0 (r) for all r H () 1 (this problem is well-posed owing to (32.5)). The compatibility condition (32.5) implies that any test function r H 1 () can be considered. Taking first r C0 () yields σ n = φ = g, whence γ n (σ n ) a n,γ 0 (r) = 0 for all r H 1 (). The trace map γ 0 being surjective onto H 1 2( ), we conclude that γ n (σ n ) = a n. Note that this construction of the lifting σ n yields G n = Mixed irichlet Neumann boundary conditions ( ) Let d n be a partition of the boundary with d 0 and n 0. We want to enforce the mixed irichlet Neumann conditions p = a d on d and σ n = a n on n. A rigorous mathematical setting for these conditions entails some subtleties. We have seen in that a Neumann condition on is enforced using the normal trace operator γ n which maps onto the dual space H 1 2( ). Here, the Neumann condition is to be enforced on n only, and this leads us to consider the restriction to n of a linear form in H 1 2( ). Let H ( n ) be composed of the functions θ defined on n whose zero-extension to, say θ, is in H 1 2( ). Then, for all ς H(div;), the restriction γ n (ς) n is defined in H ( n ) by using the duality product γ n (ς) n,θ n := γ n (ς), θ. (32.7) The restriction γ n (ς) d is defined in H ( d ) similarly.

5 Part VII. PEs in Mixed Form 425 We assume that a d H ( d ) and that a n H ( n ). We lift the Neumann condition by proceeding as in Remark We set σ n = φ where φ Hd 1() := {q H1 () γ 0 (q) d = 0} solves the mixed irichlet Neumann problem φ rdx = a n,γ 0 (r) n n for all r Hd 1 (). Observing that σ n = 0 and that for all θ H ( n ) there is r θ Hd 1() such that γ 0 (r θ ) = θ (the zero-extension of θ to ), we infer that γ n (σ n ) n a n,θ n = γ n (σ n ), θ φ r θ = 0, i.e., γ n (σ n ) n = a n. We make the change of variable σ 0 = σ σ n. Let us set H n (div;) := {ς H(div;) γ n (ς) n = 0}; this is a Hilbert space equipped with the natural norm. The weak formulation is as follows: Find σ 0 V := H n (div;) and p Q := L 2 () such that a(σ 0,τ)+b(τ,p) = F dn (τ), τ V, (32.8) b(σ 0,q) = G n (q), q Q, with linear form F dn (τ) := τ (f 1 σ n )dx γ n (τ) d,a d d, and G n, a, b are defined as above. Proposition 32.6 (Well-posedness). Assume f L 2 (), g L 2 (), a d H ( d ), and a n H ( n ). Then (32.8) is well-posed. Moreover, the pair (σ := σ 0 +σ n,p) satisfies the PEs (32.1) a.e. in, the irichlet condition γ 0 (p) d = a d hold a.e. on d, and the Neumann condition γ n (σ) n = a n holds in H ( n). Proof. We just sketch the differences with respect to the above proofs. The surjectivityofthedivergencefollowsbydefiningφ H 1 ()suchthat φ = r, φ d = 0, n φ n = 0 and setting ς r = φ (for mixed conditions, the regularity result holds only for s (0, 1 2 ) (see Jochmann [289]), which still verify regularity implies ς r L t () for all t [1,2 d d d 2s ], note that 2d 2s > 2). To recover the irichlet boundary condition, we first infer that γ n (τ),γ 0 (p) = γ n (τ) d,a d d, τ H n (div;). Let ψ C0 ( d ) and let ψ be the zero-extension of ψ to ; there is τ ψ H n (div;) such that γ n (τ) = ψ. This implies that d ψ(γ 0 (p) a d )ds = 0, which in turn gives γ 0 (p) d = a d since ψ is arbitrary Primal, dual, and dual mixed formulations Recall that p is called the primal variable and σ the dual variable. In this section, we consider alternative formulations where either p or σ is eliminated.

6 426 Chapter 32. arcy s equations We focus on homogeneous irichlet boundary conditions for simplicity. The material extends to other types of (non-homogeneous) boundary conditions. The primal formulation of the PEs (32.1) with the boundary condition p = 0 on is obtained by eliminating σ using that σ = (f p). This leads to the following formulation: { Find p H 1 0 () such that a (p,r) := p rdx = F (r), r H0(), 1 (32.9) withlinearformf (r) := ( r f+rg)dx.thisproblemhasbeenanalyzed in Chapter 21 (notice that F H 1 (), see Remark 21.7). In particular, p is the unique minimizer of the energy functional E (q) := 1 2 a (q,q) F (q) over H0(), 1 see Remark The dual formulation is obtained by eliminating p using divergence-free test functions in arcy s law (observe that p τ dx = 0 if τ is divergencefree since p = 0), and enforcing the mass conservation equation explicitly. This leads to the following formulation: { Find σ H(div;) with σ = g such that a (σ,τ) := σ 1 τ (32.10) dx = F (τ), τ H(div = 0;), with the space H(div = 0;) := {ς H(div;) ς = 0} and the linear form F (τ) := τ f dx. The well-posedness of (32.10) can be established by lifting the divergence constraint using σ g H(div;) such that σ g = g, making the change of variable σ 0 = σ σ g H(div = 0;), and observing that the bilinear form a is coercive in H(div = 0;) equipped with the natural norm. Moreover, defining the complementary energy functional E (ς) := 1 2 (ς f) 1 (ς f)dx and observing that (32.10) amounts to E (σ)(τ) = 0 for all τ H(div = 0;), we infer that σ solving (32.10) is the unique maximizer of E on the affine subspace of H(div;) with divergence equal to g, The dual formulation is seldom used for approximation purposes since it requires to manipulate divergence-free vector fields; an interesting application is presented in We can now relate the primal and dual formulations (32.9) and (32.10) to the mixed formulation (32.3), which in the present context is called the dual mixed formulation. Proposition 32.7 (Equivalence, energy identity). The primal formulation (32.9), the dual formulation (32.10), and the dual mixed formulation (32.3) are equivalent in the sense that the solutions p from (32.9) and (32.3) coincide, the solutions σ from (32.10) and (32.3) coincide, and 1 σ + p = f holds. Moreover, the following energy identity holds: min E (q) = E (p) = E (σ) = max E (ς). (32.11) q H0 1() ς H(div;), ς Proof. See Exercise 32.3.

7 Part VII. PEs in Mixed Form 427 Remark 32.8 (Lagrangian). Proposition implies that the pair (σ, p) solving the dual mixed formulation (32.3) is the unique saddle point of the Lagrangian L(ς,q) = (1 2 ς 1 ς q ς)dx F d (ς) G d (q). Since L(ς,q) = E (ς) 1 2 f f dx + q(g ς)dx, we infer that E (p) = E (σ) = L(σ,p) 1 2 f f dx. Remark 32.9 (Elasticity). The above formalism can be applied to the linear elasticity equations, see Chapter 27. enoting by u the displacement and the stress tensor, arcy s law is replaced by the constitutive equation S: = 1 2 ( u+( u)t ), where S is the (fourth-order) compliance tensor, and the mass conservation equation is replaced by the equilibrium equation = g. Weak mixed formulations are discussed in Gatica [232, 2.4.3] Approximation of the mixed formulation In this section we analyze an H(div)-conforming approximation of the weak formulation (32.3) focusing on irichlet boundary conditions for simplicity iscrete problem and well-posedness Let(T h ) h>0 beashape-regularsequenceofaffinesimplicialmeshes,letp d k (T h) be the H(div)-conforming Raviart Thomas finite element space from of order k 0, and let P b k (T h) be the broken finite element space built using P k polynomials; recall that Pk(T d h ) = {ς h H(div;) ψk(ς d h K ) RT k, K T h }, P b k(t h ) = {q h L 2 () ψ g K (q h K) P k, K T h }, (32.12a) (32.12b) where ψk d is the contravariant Piola transformation and ψg K the pullback by the geometric map T K. Note that ς h K RT k and q h K P k since the mesh is affine. It is possible to work with rectangular meshes using Cartesian Raviart Thomas elements for the dual variable and Q k polynomials for the primal variable. Moreover, other boundary conditions can be treated; for mixed irichlet Neumann conditions, boundary faces are assumed to be located in either d or n, and an H n (div;)-conforming subspace is built by taking fields in P d (T h ) with zero normal component on n. The discrete counterpart of (32.3) is Find σ h V h := Pk d(t h) and p h Q h := Pk b(t h) such that a(σ h,τ h )+b(τ h,p h ) = F(τ h ), τ h V h, (32.13) b(σ h,q h ) = G(q h ), q h Q h. Let ker(b h ) = {ς h V h b(ς h,q h ) = 0, q h Q h } and recall that ker(b) = {ς H(div;) ς = 0}.

8 428 Chapter 32. arcy s equations Lemma (iscrete divergence). ker(b h ) ker(b), and there is a uniform constant β = (β 2 Jh d 2 L(L 2 ;L 2 ) + 1) 1 2 > 0, where Jh d is the L2 - stable commuting projection from 15.3, such that inf sup q h ς h dx β. (32.14) q h Q h ς h V h q h L 2 () ς h H(div;) Proof. Thatker(B h ) ker(b)followsfromlemma13.11which,inparticular, shows that : V h = Pk d(t h) Pk b(t h) = Q h. Let us now prove (32.14). Let q h Q h. Using Lemma 32.2 we know that there is ς qh H(div;) such that ς qh = q h and β ς qh H(div;) q h L2 (). Let us set ςh := J h d(ς q h ). Theorem 15.4 implies that ςh = J h b( ς q h ) = Jh b(q h) = q h (since Jh b leaves Pk b(t h) pointwise invariant). Since ςh L 2 () Jh d L(L 2 ;L 2 ) ς qh L 2 (), we infer that β ςh H(div;) q h L2 (), whence sup ς h V h q h ς h dx ς h H(div;) q h ςh dx ςh H(div;) = q h 2 L 2 () ςh β q h L 2 (). H(div;) Corollary (Well-posedness). (32.13) is well-posed. Proof. We apply Proposition Condition (31.4a) follows from the coercivity of a on ker(b) and ker(b h ) ker(b); condition (31.4b) is just (32.14). Remark (Variant of (32.14)). Adapting the last step of the proof of Lemma using the L 2 -norm for ς h instead of the H(div)-norm, we infer that inf qh Q h sup q h ς h dx ςh V h q h L 2 () ς h β L 2 () L := β J d 2 h 1 L(L 2 ;L 2 ) > β. Remark (Fortin operator). Since : H(div;) L 2 () is surjective, it admits a bounded right inverse which we denote ( ) 1. Then Π h = Jh d ( ) 1 Ih b ( ) is a Fortin operator, see Exercise Error analysis One can derive an error estimate from the general setting of Lemma 31.3, see Exercise However, we can exploit the particular structure of arcy s equations to derive a more specific estimate which distinguishes the approximation of the dual variable and of its divergence, both with constants independent of the inf-sup constant β in (32.14) (this is not the case in Exercise 32.7). Let I b h denote the L2 -orthogonal projector onto P b k (T h), see Theorem (Error estimate). Let (σ,p) and (σ h,p h ) solve (32.3) and (32.13), respectively. Let V h,g := {ς h V h B h ς h = Ih b (g)}. Then, σ σ h L 2 () c 1 inf σ ς h L ς h V 2 (), h,g (σ σ h ) L 2 () = inf σ φ h L 2 (), Φ h Pk b(t h) (32.15a) (32.15b) p p h L 2 () c 3 inf σ ς h L ς h V 2 () +2 inf p q h L h,g q h Q 2 (), (32.15c) h

9 Part VII. PEs in Mixed Form 429 with c 1 = λ λ and c 3 = c1 λ β with β L 2 L := β J d 2 h 1 L(L 2 ;L 2 ). Moreover, if σ H r (), σ H r (), and p H r () with r (0,k +1], then σ σ h L 2 () ch r σ H r (), p p h L2 () ch r ( σ Hr () + p Hr ()). (σ σ h ) L 2 () ch r σ H r (), Proof. (1) We first observe that the following Galerkin orthogonality holds: a(σ σ h,τ h )+b(τ h,p p h ) = 0, τ h V h, (32.16a) b(σ σ h,q h ) = 0, q h Q h. (32.16b) Since B h σ h I b h (g) Q h and (B h σ h I b h (g),q h) L2 () = b(σ h σ,q h ) = 0 for all q h Q h owing to (32.16b), we infer that σ h V h,g. (2) Let ς h V h,g, then σ h ς h ker(b h ) ker(b). Since (32.16a) implies that a(σ σ h,τ h ) = 0 for all τ h ker(b), we infer that λ 1 σ σ h 2 L 2 () a(σ σ h,σ σ h ) = a(σ σ h,σ ς h )+a(σ σ h,ς h σ h ) = a(σ σ h,σ ς h ) λ 1 σ σ h L2 () σ ς h L2 (). Thus, σ σ h L 2 () c 1 σ ς h L 2 (), and (32.15a) holds. Since σ h = B h σ h = I b h (g) and Ib h is an L2 -orthogonal projection, (32.15b) also holds. (3) Let q h Q h. Owing to Remark 32.12, we infer that there is τ h V h such that τ h = p h q h and β L 2 τ h L 2 () p h q h L 2 (). We infer that p h q h L 2 () = b(τ h,p h q h ) = b(τ h,p h p)+b(τ h,p q h ) = a(σ σ h,τ h )+b(τ h,p q h ) λ 1 σ σ h L2 () τ h L2 () + τ h L2 () p q h L2 (), owing to (32.16a). The above properties of τ h combined with (32.15a) and the triangle inequality imply that (32.15c) holds. (4) Finally, if σ H r () with r > 0, then σ L p () with p > 2 owing to the Embedding Theorem B.101. Since σ = g L 2 (), we infer using (10.1b) that σ is in the domain of the Raviart Thomas interpolation operator I d h. Taking ς h = I d h (σ), which is in V h,g since I d h (σ) = Ib h ( σ) = Ib h (g), the convergence rate on σ σ h L 2 () results from Corollary The other rates follow from this one and the approximations properties of Ih b. Remark (Proof variant). If g = 0, we can also take ς h = Jh d(σ) since Jh d(σ) = J h b( σ) = J h b(g) = 0 so that ς h V h,0. This way, there is no need to use the operator Ih d and invoke Theorem B.101. We now use duality techniques to derive an improved-order error estimate on the primal variable. One difference with respect to the primal formulation analyzed in is that we now bound the (discrete) error I b h (p) p h,

10 430 Chapter 32. arcy s equations instead of the full error p p h. As in , we assume that the following smoothing property holds: There is s (0,1] and a uniform constant c sp such that the (adjoint) solution z H 1 0() of the PE ( z) = φ satisfies the a priori bound z H 1+s () c sp φ L 2 (). Sufficient conditions for this smoothing property to hold follow from elliptic regularity, see We also make the (so-called) multiplier assumption on the field stating that the map H s () ξ ξ H s () is bounded, see (21.29). Theorem (Potential supercloseness). Under the above smoothing property and multiplier assumption, the following holds: I b h(p) p h L2 () ch s σ σ h H(div;). (32.17) Proof. Let z H 1+s () H 1 0() be the adjoint solution with data φ = I b h (p) p h and set ξ = z. We observe that I b h(p) p h 2 L 2 () = (p p h,i b h(p) p h ) L2 () = (p p h, ξ) L2 () = (p p h, J d h(ξ)) L 2 () = ( 1 (σ σ h ),J d h(ξ)) L 2 () = ( 1 (σ σ h ),J d h(ξ) ξ) L2 () +( 1 (σ σ h ),ξ) L2 (), where we have used (p I b h (p),q h) L2 () = 0 for all q h P b k (T h) and the definition of the adjoint solution in the first line, and J d h (ξ) = J b h ( ξ) = ξ since ξ P b k (T h) and (32.16a) in the second line. The first term on the right-hand side, say T 1, is bounded as T 1 λ 1 σ σ h L 2 () J d h(ξ) ξ L 2 () c σ σ h L 2 ()h s ξ H s (), and ξ H s () c z H 1+s () owing to the multiplier assumption. For the second term, say T 2, (32.16b) implies that (σ σ h, z) L2 () = ( (σ σ h ),z) L2 () = ( (σ σ h ),z I b h(z)) L2 (). This leads to the bound T 2 c (σ σ h ) L 2 ()h z H 1 (), and we conclude using the smoothing property z H 1+s () c sp I b h (p) p h L 2 (). Remark (L -bounds). L -bounds on the dual and primal errors are derived, e.g., by Gastaldi and Nochetto [231] Hybridization ( ) Hybridization was introduced by Fraejis de Veubeke in 1965 (see [225] for a reprint) as a computationally effective technique that allows one to transform the original (symmetric indefinite) linear system on the primal and dual variables into a symmetric positive definite system on an auxiliary variable that

11 Part VII. PEs in Mixed Form 431 can be interpreted as the Lagrange multiplier associated with the continuity constraint on the normal component of the dual variable. Let us focus on simplicial Raviart Thomas elements so that the discrete spaces are V h = Pk d(t h) and Q h = Pk b(t h). Let Λ h := Pk b(f h ) be spanned by piecewise polynomials of order at most k on each mesh interface. Recall the broken Raviart Thomas space V hy h := P d,b k (T h ) spanned by piecewise RT k polynomials in each mesh cell. The hybridized version of (32.13) is Find σ h V hy h, p h Q h, and λ h Λ h such that a(σ h,τ h )+b(τ h,p h )+c(τ h,λ h ) = F(τ h ), τ h V hy h, (32.18) b(σ h,q h ) = G(q h ), q h Q h, c(σ h,µ h ) = 0, µ h Λ h, with bilinear form c(τ h,µ h ) := F Fh F [[τ h]] n F µ h ds. The third equation in (32.18) implies that σ h has continuous normal component, i.e., σ h V h. Restricting test functions in the first equation to V h then shows that the pair (σ h,p h ) coincides with the unique solution of (32.13). The crucial advantage of (32.18) is that (σ h,p h ) can be eliminated locally. This operation is called static condensation. To state our result, we need some notation. For all µ h Λ h and all K T h, we define µ K = (µ h F ) F FK, where µ h is extended by zero on boundary faces; thus, µ K is a piecewise polynomial of order at most k on the boundary of K. Furthermore, we write ψ = (ψ σ,ψ p ) V k := RT k P k, and we define the local bilinear form c K (ψ,ζ) := ( 1 ψ σ,ζ σ ) L2 (K) (ψ p, ζ σ ) L2 (K) (ζ p, ψ σ ) L2 (K). Let Φ K : L 2 (K) L 2 (K) V k be such that Φ K (f,g) V k solves c K (Φ K (f,g),ψ) = (f,ψ σ ) L 2 (K) (g,ψ p ) L 2 (K) for all ψ V k. Let Θ K : L 2 ( K) V k be such that Θ K (µ) V k solves c K (Θ K (µ),ψ) = (µ,ψ σ n K ) L2 ( K) for all ψ V k. Proposition (Static condensation). The pair (σ h,p h ) solves (32.13) if and only if (σ h K,p h K ) = Φ K (f K,g K )+Θ K (λ K ) where λ h Λ h (extended by zero on boundary faces) is the unique solution of 1 ΘK(λ σ K ),ΘK(µ σ K )) L 2 (K) = l(µ h ), µ h Λ h, (32.19) K T h ( with l(µ h ) := K T h (f,θk σ(µ K)) L2 (K) (g,θ p K (µ K)) L2 (K). The algebraic realization of this problem leads to a symmetric positive definite matrix. Plugging this formula into the third equation of (32.18) leads to a linear problem in the Lagrange multipliers whose algebraic realization is a symmetric positive definite matrix; more details can be found in Boffi et al. [60, ] and Cockburn and Gopalakrishnan [150]. Remark (One unknown per cell). Implementation of lowest-order Raviart Thomas methods with one unknown per mesh cell and relations to other methods (including Finite Volumes and Mimetic Finite ifferences) are discussed in Younes et al. [469], Vohralík and Wohlmuth [454].

12 432 Chapter 32. arcy s equations (i) Show that Φ K and Θ K are defined by solving well-posed (local) irichlet problems in mixed form. Exercises Exercise 32.1 (Neumann condition). Prove Proposition (Hint: for the surjectivity of the divergence, pose a pure Neumann problem.) Exercise 32.2 (Integration by parts). Let Hd 1() and H n(div;) be defined in Prove that ( q ς+q ς)dx = 0 for all q H1 d () and all ς H n (div;). (Hint: observe that γ 0 (q) n H ( n ).) Exercise 32.3 (Primal, dual formulations). Prove Proposition Exercise 32.4 (Primal mixed formulation). Consider the problem of finding p H 1 () such that p = f and γ 0 (p) = g with f L 2 () and g H 1 2( ). erive a mixed formulation of this problem with unknowns (p,λ) H 1 () H 1 2( ) and show that it is well-posed; this method is introduced in Babuška [30]. (Hint: set b(v,µ) = µ,γ 0 (v) and observe that B = γ 0 : H 1 () H 1 2( ).) Recover the PE and the boundary condition. Exercise 32.5 (Fortin operator). Justify Remark (Hint: use arguments similar to those of the proof of Lemma ) Exercise 32.6 (Inf-sup condition). The goal is to prove the inf-sup condition (32.14) using the canonical Raviart Thomas interpolation operator. (i) o this by using elliptic regularity. (Hint: pose a irichlet problem.) (ii) o this by using the surjectivity of : H 1 () L 2 (). Exercise 32.7 (Error estimate). Prove that σ σ h H(div;) c 1 inf ς h V h σ ς h H(div;), p p h L2 () c 3 inf ς h V h σ ς h H(div;) +2 inf q h Q h p q h L2 (), no 1+ with c 1 = (1+ λ λ )(1+ 1 β ) and c 3 = c 1 λ β. Assuming that σ H r (), σ L H r (), and p H r () with r (0,k + 2 1], prove that σ σ h H(div;) ch r ( σ Hr ()+ σ Hr ()) and p p h L2 () ch r ( σ Hr ()+ σ Hr ()+ p H r ()). (Hint: use the commuting projection Jh d.)

13 Part VII. PEs in Mixed Form 433 Solution to exercises Exercise 32.1 (Neumann condition). There are only two differences with the proof of Proposition The first one concerns the divergence operator which now maps from H 0(div;) to L 2 0 () (owing to the divergence theorem). To prove that this operator is surjective, let q L2 0 () and let us solve the pure Neumann problem φ H 1 () such that φ = q and φdx = 0. Then ς q = φ has all the desired properties. The second difference concerns the recovery of the boundary condition which is now a simple consequence of γ n(σ) = γ n(σ 0) + γ n(σ n) = a n. Exercise 32.2 (Integration by parts). Observe that ( q ς+q ς)dx = γ n(ς),γ0(q). 1 Now γ 0(q) n H 2 00 ( n) since its zero-extension to is γ0(q) (since q H1 d ()) which is in H 1 2( ). This implies that γ n(ς),γ 0(q) = γ n(ς) n,γ 0(q) n n, and this last quantity vanishes since ς H n(div;). Exercise 32.3 (Primal, dual formulations). Let p H 1 0 () solve (32.9) and define σ = 1 (f p). Since ( p τ + p τ)dx = 0 for all τ H(div;), the first equation of (32.3) holds; moreover, we infer that σ qdx = gqdx for all q H1 0 (), implying that σ = g so that the second equation of (32.3) holds. Let now (σ,p) H(div;) L 2 () solve (32.3). Then σ = g and taking a divergence-free test function τ in the first equation of (32.3) shows that (32.10) holds. Finally, let σ H(div;) with σ = g solve (32.10); let p L 2 () be defined such that p τ dx = ( 1 σ f) τ dx for all τinh(div;) (note that the righthand side vanishes if τ is divergence-free). Since : H(div;) L 2 () is surjective owing to Lemma 32.2, p is well-defined, and its definition implies that the first equation of (32.3) holds; the second one follows from σ = g. Finally, since p = f 1 σ, the energy identity results from E (p) = 1 p pdx = 1 (ς 1 (ς f) f) = E (σ). 2 2 Exercise 32.4 (Primal mixed formulation). The weak mixed formulation is Find p V := H 1 () and λ Q := H 1 2( ) such that a(p,q) + b(q,λ) = f(q), q V, b(p,µ) = g(µ), µ Q, with a(p,q) = p qdx, b(v,µ) = µ,γ0(v), f(q) = fqdx, and g(µ) = µ,g. All these forms are bounded. Moreover, by reflexivity of H 1 2( ), B = γ 0 so that ker(b) = ker(γ 0) = H 1 0 () owing to Theorem B.106, whence we infer that the bilinear form a is coercive in ker(b). Furthermore, B is surjective still by Theorem B.106, so that well-posedness follows from Theorem Finally, the second equation in the weak mixed formulation implies that µ,γ 0(p) g = 0 for all µ H 1 2( ), so that we recover the boundary condition γ 0(p) = g a.e. on, and the PE follows by taking q arbitrary in H 1 0 () in the first equation. Exercise 32.5 (Fortin operator). Let us verify the two properties stated in Lemma 18.9(i) with W = H(div;), W h = P d k (T h), V = L 2 (), and V h = P b k (T h). The operator Π h maps from H(div;) to P d k (T h) as required. Moreover, we infer that, for all q h P b k (T h) and all v H(div;), q h Π h (v)dx = q h (J d h (( ) 1 (I b h ( v))))dx = q h J i h (Ib h ( v))dx = q h I b h ( v)dx = q h vdx, since J d h = J i h ( ), J i h leaves Pb k (T h) pointwise invariant, and q h is in P b k (T h). Moreover, using the stability of all the operators, one can see that Π h (v) H(div;) c v L 2 (). Exercise 32.6 (Inf-sup condition). (i) Let q h Q h. Let φ H 1 0 () solve φ = q h and set ς qh = φ. Elliptic regularity implies that ς qh H s () with s > 1 2, i.e., ςq h ˇV d (), see (13.14c) with p = 2. This means that ς qh is in the domain of canonical interpolation operator I d h. Moreover, Id K (ς q h K) L 2 (K) c( ς qh K L 2 (K) +h s K ς q h K H s (K) ) for all K T h. Summing over mesh cells and since h K is bounded by the diameter of, we infer that I d h (ςq h ) L 2 () c ςq h H s () c q L 2 (). We can now conclude as in the proof of Lemma 32.10, but this time we take ς h = Id h (ςq h ).

14 434 Chapter 32. arcy s equations (ii) Let again q h Q h. We use the hint to infer that there is ς qh H 1 () such that ς qh = q h and ς qh H 1 () c q L 2 (). Since I d K (ς q h K) L 2 (K) c( ς qh K L 2 (K) + h 1 K ς q h K H 1 (K) ) for all K T h, we can now conclude as in Step (i). Exercise 32.7 (Error estimate). The error bound on the dual variable follows from Lemma 31.3 since a = λ, α = λ, and b = 1. For the primal variable, we proceed as in the proof of Theorem and use the bound on the dual variable. To bound the best-approximation error on the dual variable, we use the commuting projection J d h and Theorem 15.4 to infer that σ J d h (σ) H(div;) σ J d h (σ) L 2 () + σ J b h ( σ) L 2 () c inf σ ς h ς h P k d L 2 () + inf σ φ h (T h ) φ h P k b L 2 (), (T h ) and we conclude invoking Corollary 14.8 (with p = 2 and x = d,b).