Chapter 2 Finite Element Spaces for Linear Saddle Point Problems

Transcription

1 Chapter 2 Finite Element Spaces for Linear Saddle Point Problems Remark 2.1. Motivation. This chapter deals with the first difficulty inherent to the incompressible Navier Stokes equations, see Remark 1.19, namely the coupling of velocity and pressure. The characteristic feature of this coupling is the absence of a pressure contribution in the continuity equation. In fact, the continuity equation can be considered as a constraint for the velocity and the pressure in the momentum equation as a Lagrangian multiplier. This kind of coupling is called saddle point problem. Appropriate finite element spaces for velocity and pressure have to satisfy the so-called discrete inf- condition. This condition is derived on the basis of the theory for an abstract linear saddle point problem. Several techniques for proving the discrete inf- condition will be presented and applied for concrete pairs of finite element spaces for velocity and pressure. All special cases of models for incompressible flow problems given in Remark 1.22 possess the same coupling of velocity and pressure, in particular the linear models of the Stokes and the Oseen equations. Linear problems are also of interest in the numerical simulation of the Navier Stokes equations. After having discretized these equations implicitly in time, a nonlinear saddle point problem has to be solved in each discrete time. The solution of this problem is performed iteratively, requiring in each iteration step the solution of a linear saddle point problem for velocity and pressure. These linear saddle point problems will be discretized with finite element spaces. The existence and uniqueness of a solution of these discrete linear problems is crucial for performing the iteration. Altogether, the theory of linear saddle problems plays an essential role for the theory of all models for incompressible flows from Chapter 1. A comprehensive presentation of the theory of linear saddle point problems can be found in the monograph Boffi et al. (2013). 21

2 22 2 Finite Element Spaces for Linear Saddle Point Problems 2.1 Existence and Uniqueness of a Solution of an Abstract Linear Saddle Point Problem Remark 2.2. Contents. This section presents an abstract framework for studying the existence and uniqueness of solutions of those types of linear saddle point problems which are of interest for incompressible flow problems. The presentation follows (Girault & Raviart, 1986, Chapter I, 4). Remark 2.3. Abstract linear saddle point problem. Let V and Q be two real Hilbert spaces with inner products (, ) V and (, ) Q and with induced norms V and Q, respectively. Their corresponding dual spaces are given by V and Q, with the dual pairing denoted by, V,V and, Q,Q. The norms of the dual spaces are defined in the usual way by φ,v V φ V :=,V ψ,q Q, ψ v V,v 0 v Q :=,Q. (2.1) V q Q,q 0 q Q Two continuous bilinear forms are considered a(, ) : V V R, b(, ) : V Q R, (2.2) with the usual definition of their norms a = v,w V,v,w 0 a(v, w) v V w V, b = v V,q Q,v,q 0 b(v, q) v V q Q. (2.3) The following problem is studied: Find (u,p) V Q such that for given (f,r) V Q a(u,v)+b(v,p) = f,v V,V v V, b(u,q) = r,q Q,Q q Q. (2.4) System (2.4) is called linear saddle point problem. Concrete choices of the spaces and bilinear forms for incompressible flow problems are discussed in Section 2.2. Remark 2.4. Operator form of the linear saddle point problem. Problem (2.4) can be transformed into an equivalent form using operators instead of bilinear forms. Linear operators can be defined which are associated with the bilinear forms given in (2.2): A L(V,V ) defined by Au,v V,V = a(u,v) u,v V, B L(V,Q ) defined by Bu,q Q,Q = b(u,q) u V, q Q. Using the definition of the norms of the dual spaces (2.1), the norms of the operators are given by

3 2.1 Solution of an Abstract Linear Saddle Point Problem 23 Av,w V,V Av V = = w V,w 0 w V A L(V,V ) = Av V a(v, w) = = a, v V,v 0 v V v,w V,v,w 0 v V w V and analogously B L(V,Q ) = b. Let B L(Q,V ) be the adjoint (dual) operator of B defined by B q,v V,V = Bv,q Q,Q = b(v,q) v V, q Q. With these operators, Problem (2.4) can be written in the equivalent form: Find (u,p) V Q such that Au +B p = f in V, Bu = r in Q. Definition 2.5. Well-posedness of Problem (2.5). Let (2.5) Φ L(V Q,V Q ) : Φ(v,q) = (Av +B q,bv) be a linear operator, where (, ) denotes a vector with two components. Problem (2.5) is said to be well-posed if Φ(, ) is an isomorphism from V Q onto V Q. Remark 2.6. On Definition 2.5. Definition 2.5 means that Problem (2.5) possesses for all possible right-hand sides a unique solution. The purpose of the following studies consists in deriving necessary and sufficient conditions for (2.5) to be well-posed. Remark 2.7. The finite-dimensional case. Consider for the moment that V and Q are finite-dimensional spaces of dimension n V and n Q, respectively. Then, the operators in (2.5) can be represented with matrices, with B = B T, and the functions with vectors. The well-posedness of (2.5) means that the linear system of equations ( )( A B T u = B 0 p) ( f r ), ( ) A B T R (nv +nq) (nv +nq), (2.6) B 0 has a unique solution for all right-hand sides or, equivalently, that the system matrix is non-singular. Here, conditions will be derived such that this property is given. These considerations should provide an idea of the kind of conditions to be expected in the general case. Separate consideration of velocity and pressure. A possible way to solve (2.6) starts by solving the first equation of (2.6) for u

4 24 2 Finite Element Spaces for Linear Saddle Point Problems u = A 1( f B T p ). (2.7) Inserting this expression in the second equation gives ( BA 1 B T) p = BA 1 f r. (2.8) If (2.8) possesses a unique solution p, this solution can be inserted in (2.7) and a unique solution u is obtained, too. This way to compute a unique solution works if A : V V is an isomorphism, i.e., A is non-singular, BA 1 B T : Q Q is an isomorphism, i.e., BA 1 B T is non-singular. Let p be a solution of (2.8). Then, also p+ p with p ker ( B T) is a solution of (2.8). Thus, for BA 1 B T to be non-singular, it is necessary that ker ( B T) = {0} or equivalently that B T : Q V = V is injective. With a similar argument, one finds that B must be injective on the range of A 1 B T, i.e., ker(b) range ( A 1 B T) = {0}. Joint consideration of velocity and pressure. One can also consider the system matrix (2.6) as a whole. A first necessary condition for the matrix to be non-singular is n Q n V, since the last rows of the system matrix span a space of dimension at most n V (only the first n V entries of these rows might be non-zero). Assume that A is non-singular, then the system matrix is nonsingular if and only if B has full rank, i.e., rank(b) = n Q. It will be shown now that rank(b) = n Q if and only if inf q R n Q,q 0v R n V,v 0 v T B T q v 2 q 2 β > 0. (2.9) Let (2.9) be satisfied and let rank(b) < n Q. Then, there is a q R nq, q 0, such that q ker ( B T), i.e., B T q = 0. For this vector, it is v T B T q = 0 for all v R nv such that the remum of (2.9) is zero and (2.9) cannot be satisfied. This result is a contradiction and hence rank(b) = n Q. On the other hand, let rank(b) = n Q. Then, for each q R nq, q 0, one has that B T q 0 with B T q R nv. Choosing v = B T q gives inf q R n Q,q 0v R n V,v 0 v T B T q v 2 q 2 inf q R n Q,q 0 = inf q R n Q,q 0 B T q 2 2 BT q q 2 2 B T q 2 q 2. (2.10) It is B T q 2 q = qt BB T q q T. q This expression is a Rayleigh quotient and it is known that

5 2.1 Solution of an Abstract Linear Saddle Point Problem 25 q T BB T q ( inf q R n Q,q 0 q T = λ min BB T ), q ( where λ ) min BB T is the smallest eigenvalue of BB ( T, see Lemma A.19. Since B was assumed to have full rank, one has λ ) min BB T > 0 and hence with (2.10) v T B T q inf q R n Q,q 0 v 2 q λ 1/2 min( BB T ) > 0. 2 v R n V,v 0 Altogether, under the assumption that A is non-singular, i.e., A : V V is an isomorphism, (2.9) is satisfied, the system matrix (2.6) is non-singular. The result presented here just states that the given problem has a unique solution because (2.9) is satisfied. In the finite element theory it turns out that there is another important aspect to study, namely the dependency of β on the dimension of the finite element spaces. To obtain optimal orders of convergence, β has to be independent of the dimension, e.g., compare Remark This aspect can also be taken into account in the matrix-vector formulation of linear saddle point problems, see Section Then, one has to solve a generalized eigenvalue problem, see (2.150). It turns out that one gets similar conditions in the general case, see Lemma 2.12 and Theorem Whether or not these conditions are satisfied depends finally on the spaces V and Q. Remark 2.8. A manifold and a subspace in V. A manifold of V will be defined that contains all elements which fulfill the second equation of (2.5) V(r) = {v V : Bv = r}, V 0 := V(0) = ker(b). The manifold V 0 is even a subspace of V. From Hilbert space theory, it follows that there is an orthogonal decomposition, with respect to the inner product of V, V = V0 V 0, where V 0 is the orthogonal complement of V 0. Lemma 2.9. Properties of V 0 and V 0. The spaces V 0 and V 0 are closed subspaces of V. Proof. First, the closeness of V 0 will be proved. Let {v n} n=1 be an arbitrary Cauchy sequence with v n V 0 for all n. Since V is complete, there exists a v V with lim n v n = v. One has to show that v V 0. By the continuity of the linear operator B, it follows that ( ) Bv = B lim n vn = lim (Bvn) = lim 0 = 0. n n Hence v V 0 and V 0 is closed.

6 26 2 Finite Element Spaces for Linear Saddle Point Problems The closeness of V0 follows from the fact that the orthogonal complement of every subspace is closed, see Lemma A.17. Remark Functionals vanishing on V 0. A subset of V is defined for the following analysis: Ṽ = {φ V : φ,v V,V = 0 v V 0} V. (2.11) This subset, which is even a closed subspace of V, contains all linear functionals on V that vanish for all v V 0 = ker(b). Remark Reduction of the system to a single equation in a subspace. In the next step, the following problem is associated with Problems (2.4) and (2.5): Find u V(r) such that a(u,v) = f,v V,V v V 0. (2.12) Clearly, if (u,p) V Q is a solution of (2.4) or (2.5), then u V(r). In addition, one obtains B p,v V,V = Bv,p Q,Q = b(v,p) = 0 v V 0. (2.13) Since the first equation of (2.4) holds for all v V, it holds in particular for all v V 0. With (2.13) it follows that u is a solution of (2.12). The aim of the analysis consists now in finding conditions to ensure that the converse of this statement holds: if u V(r) is a solution of (2.12), one can find a unique p Q such that (u,p) is the unique solution of (2.4) or (2.5), respectively. Lemma The inf- condition. The three following properties are equivalent: i) There exists a constant β is > 0 such that inf q Q,q 0v V,v 0 b(v, q) v V q Q β is. (2.14) ii) The operator B is an isomorphism from Q onto Ṽ and B q V β is q Q q Q. (2.15) iii) The operator B is an isomorphism from V 0 onto Q and Bv Q β is v V v V 0. (2.16) Proof. The proof follows Girault & Raviart (1986). i) and ii) are equivalent. ii) = i). From the definition of the norm of a linear functional and the definition of the adjoint operator, it follows that

7 2.1 Solution of an Abstract Linear Saddle Point Problem 27 B q B q,v V V =,V v V,v 0 v V Hence, (2.15) gives Bv,q Q =,Q b(v, q) =. (2.17) v V,v 0 v V v V,v 0 v V b(v, q) β is q v V,v 0 v Q q Q. V Dividing by q Q and taking the infimum with respect to q on both sides of this inequality shows that ii) implies i). i) = ii). The inequality (2.15) follows from (2.14) and (2.17). It remains to prove that B is an isomorphism from Q onto Ṽ. First, it will be shown that B is an isomorphism from Q onto range(b ) with a continuous inverse, see also Theorem A.70. By the definition of the range, B is surjective on range(b ). If it would be not injective then there would be q 1 q 2 Q such that B q 1 = B q 2 or equivalently B (q 1 q 2 ) = 0. In this case, it follows that B (q 1 q 2 ),v V,V = b(v,q 1 q 2 ) = 0 for all v V and q 1 q 2 Q > 0. Then, (2.14) cannot hold, in contrast to the assumption. Hence, B is an isomorphism from Q onto range(b ). Moreover, the inverse operator is continuous (or bounded) if and only if range(b ) is a closed subspace of V. By assumption, it is B L(Q,V ), hence B is bounded. Thus, one can apply the Closed Range Theorem of Banach, see Theorem A.71 iv), to B and one obtains immediately range ( B ) = Ṽ. Since Ṽ is a closed subspace of V, see Remark 2.10, the inverse operator is continuous. Alternatively, the closeness of range(b ) follows from Theorem A.71 iii). Altogether, B is an isomorphism from Q onto Ṽ. ii) and iii) are equivalent. A standard property of an operator B and its adjoint operator B is that range(b ) = (domain(b)). Hence, one has to show that the range Ṽ of B can be identified with the dual space ( V 0 ) of the domain of B. ) will be constructed. The space V An isomorphism between Ṽ and ( V0 0 is defined in Remark 2.8. Let v V and denote the projection of v onto V0 by v. Then, with φ ( ) V0, one associates the element ψ V defined by ψ,v V,V = φ,v V,V v V. (2.18) In particular, one obtains for v V 0 that v = 0 and ψ,v V,V = φ,v V,V = φ,0 V,V = 0 v V 0. Hence, it is ψ Ṽ. Thelinearcontinuousmapping ( ) V0 Ṽ,φ ψ isinjectivesinceforφ 0,φ 1 ( ) V0 with φ 0 φ 1 and ψ,v V,V = φ 0,v V,V = φ 1,v V,V v V, it follows that φ0 φ 1,v V,V = 0 v V. Since all elements of V0 are arguments of the linear functional φ 0 φ 1, it follows that φ 0 φ 1 is the null element of ( ) V0. The mapping is also surjective. Consider an arbitrary element ψ Ṽ and an arbitrary element v V. Using the decomposition v = v 0 +v, v 0 V 0, v V0, gives

8 28 2 Finite Element Spaces for Linear Saddle Point Problems ψ,v V,V = ψ,v 0 V,V + ψ,v V,V = ψ,v V,V v V. Comparing this equation with (2.18) shows that the restriction of ψ from V to V0 defines the inverse image of ψ. Thus, surjectivity is proved. Altogether, the mapping is bijective and the spaces ( ) V0 and Ṽ can be identified. In essence, each functional from Ṽ is mapped to the functional that corresponds to its restriction (of its domain) from V on V0 and there are no two functionals that both vanish on V 0 and give the same result in V0. The equivalence of the range of B and the dual space of the domain of B proves the statements concerning the isomorphisms. For proving the equivalence of the inequalities (2.15) and (2.16), first Bv,q Q,Q Bv,q Q,Q = (2.19) v V,v 0 v V v V0,v 0 v V will be shown. That the left-hand side is larger or equal than the right-hand side follows from the fact that the remum is taken in a larger set. Using the orthogonal decomposition v = v 0 +v, v 0 V 0, v V0, yields Bv,q Q,Q v V,v 0 v V = v V,v 0 = v V,v 0 v V,v 0 Bv0 +Bv,q Q,Q ( v 0 2 V + ) v 2 1/2 V Bv,q Q,Q ( v 0 2 V + ) v 2 1/2 V Bv,q Q,Q v V = v V 0,v 0 Bv,q Q,Q v V, since reducing the denominator can at most increase the quotient. Combining both inequalities proves (2.19). One gets from (2.15), the definition of the norm in V, the definition of the adjoint operator, (2.19), and the definition of the norm in Q β 1 is q Q,q 0 q Q B q V q Q v V = q Q,q 0v V,v 0 B q,v V,V q Q v V q Q v V = = q Q,q 0v V,v 0 Bv,q Q,Q q Q,q 0v V0,v 0 Bv,q Q,Q = v V0,v 0 q Q,q 0 q Q v V Bv,q Q,Q v = V. v V0,v 0 Bv Q From this estimate, (2.16) follows. Analogously, one can derive (2.15) from (2.16). Thus, both estimates are equivalent. Theorem Inf- condition for the complete bilinear form, Babuška s inf- condition. Let V 1 and V 2 be two Hilbert spaces with inner products (, ) V1 and (, ) V2. Let A(, ) be a bilinear form on V 1 V 2 such that

9 2.1 Solution of an Abstract Linear Saddle Point Problem 29 A(v 1,v 2 ) C 1 v 1 V1 v 2 V2, v 1 V 1,v 2 V 2, A(v 1,v 2 ) C 2 v 2 v 1 V 1, v 1 V1 0 v 1 V2 v 2 V 2, V1 A(v 1,v 2 ) C 3 v 1 v 2 V 2, v 2 V2 0 v 2 V1 v 1 V 1, V2 with C 1 < and C 2,C 3 > 0. Let f V 2, then there exists exactly one element u V 1 such that A(u,v 2 ) = f,v 2 V 2,V 2 v 2 V 2 and u V1 1 C 3 f V 2. Proof. For the proof, it is referred to Babuška (1971). Remark Inf- conditions, Babuška Brezzi condition. Condition (2.14) is called inf- condition. It was introduced in this form in Brezzi (1974). In Babuška (1971), the inf- condition presented in Theorem 2.13 was proved. Choosing in this theorem V 1 = V 2 = V Q and setting the bilinear form to be A(v 1,v 2 ) = A((v +w),(q +r)) = a(v,w)+b(w,q)+b(v,r) leads to the inf- condition a(v,w)+b(w,q)+b(v,r) ( ) β is,bab w (v,q) V Q v V + q V + r Q Q v V + q Q 0 for all (w,r) V Q. If a(, ) is symmetric, then both inf- conditions in Theorem 2.13 are identical. The inf- condition from Babuška (1971) was applied to the analysis of finite element problems with Lagrangian multipliers in Babuška (1973). Note that the pressure in the Navier Stokes equations can be interpreted as a Lagrangian multiplier associated with the imposition of the divergence-free constraint for the velocity. Nevertheless, it became common to call (2.14) Babuška Brezzi condition. The relation between the inf- conditions from Babuška (1971, 1973) and Brezzi (1974) in the context of finite element methods is discussed briefly in Remark Sometimes, the inf- condition (2.14) is even called Ladyzhenskaya Babuška Brezzi condition or LBB condition. It is mentioned in the literature, e.g., in Gunzburger (2002), that in the book Ladyzhenskaya (1969) the property (2.16) is proved. The proof of the uniqueness of the pressure in (Ladyzhenskaya, 1969, Chapter 2.1) uses an argument that is based on the uniqueness of the Helmholtz decomposition of vector fields in L 2 (Ω), see Section 2.7.

10 30 2 Finite Element Spaces for Linear Saddle Point Problems Remark On the inf- condition and the coercivity. For a bounded bilinear form c : V V R, the theorem of Lax Milgram, see Theorem B.4, states that the coercivity of c(, ) is a sufficient condition for the existence of a unique solution of a corresponding problem. The coercivity can be written in the form that there is a constant m > 0 m w 2 V c(w,w) w V = m w V c(w,w) w V w V \{0}. From the last formulation, it follows that also c(v, w) m w V w V, (2.20) v V,v 0 v V holds. Observe that in this more general condition, the spaces in the bilinear form are allowed to be different. Assume now that u V(r) is the unique solution of (2.12). Then, the first equation of (2.4) can be written in the form b(v,p) = f,v V,V a(u,v) v V. The concept of coercivity cannot be applied for the analysis of this equation. But note that the inf- condition (2.14) can be written in the form b(v, q) β is q Q q Q, v V,v 0 v V which is the same form as (2.20). Altogether, the inf- condition can be viewed as a generalization of the coercivity condition to bilinear forms where the arguments are from different spaces. Remark The space V(r) is not empty. From the inf- condition, one gets that V(r) is not empty. Let r Q, then it follows from Lemma 2.12 iii) that there is a v V0 such that Bv = r. Remark An imbedding operator. Before stating the results concerning the well-posedness of Problem (2.5), see Definition 2.5, the linear continuous operator E 0 L(V,V 0) given by E 0 φ,v V,V = φ,v V,V φ V, v V 0, (2.21) is introduced. This operator is an imbedding operator. Since V 0 V, it is V 0 V and E 0 φ is the restriction of the functional φ from V onto V 0. The subspace V 0 is equipped with the same norm as V, hence one obtains

11 2.1 Solution of an Abstract Linear Saddle Point Problem 31 E 0 φ,v V,V E 0 φ V = 0 v V 0,v 0 v V φ,v V,V = v V 0,v 0 v V φ,v V,V = φ v V,v 0 v V. (2.22) V Theorem Well-posedness of Problem (2.5). Problem (2.5) is wellposed if and only if the following two conditions hold: i) The operator E 0 A is an isomorphism from V 0 onto V 0. ii) The bilinear form b(, ) satisfies the inf- condition (2.14). Proof. The conditions i) and ii) are sufficient. Assume that conditions i) and ii) are satisfied. In the first step, it will be proved that Problem (2.12) possesses a unique solution u V(r). Then, the second step proves the existence of a unique solution (u,p) V Q of Problem (2.5). Unique solution u V(r) of Problem (2.12). It follows from Lemma 2.12 iii) that there is a unique u 0 V 0 such that Bu 0 = r, u 0 V 1 β is r Q. (2.23) Therefore, Problem (2.12) can be stated equivalently in the following way: Find w = u u 0 V 0 such that a(w,v) = f,v V,V a(u 0,v) v V 0. Now, the ansatz and the test space are the same. Writing the equation in the form Aw,v V,V = f Au 0,v V,V v V 0, (2.24) then it follows from (2.21) that w satisfies the operator equation (E 0 A)w = E 0 (f Au 0 ) in V 0. (2.25) By assumption i), E 0 A is an isomorphism from V 0 onto V 0, such there is a unique w satisfying this equation. Hence, Problem (2.12) has the unique solution u = u 0 +w V(r). Unique solution (u,p) V Q of Problem (2.5). From the boundedness of A and E 0, it follows that E 0 A is bounded. Therefore, the inverse (E 0 A) 1 is a continuous bounded operator, see Theorem A.70. From (2.25), this property, and (2.22), it follows that w V = (E0 A) 1 E 0 (f Au 0 ) V C E 0 (f Au 0 ) V 0 C f Au 0 V. The application of the triangle inequality, the boundedness of the operator A, and twice (2.23) yields u V u 0 V + w V 1 r β Q +C f Au 0 V is ) C ( r Q + f V + u 0 V C ( r Q + f V ). (2.26) Using the decomposition of u and (2.24) yields f Au = f Au 0 Aw Ṽ. Thus, according to Lemma 2.12 ii) there exists a unique p Q such that

12 32 2 Finite Element Spaces for Linear Saddle Point Problems B p = f Au. Using Lemma 2.12 ii), the triangle inequality, the boundedness of A, and (2.26), one obtains p Q 1 f Au β V 1 ( ) f V + Au is β V is 1 β is ( f V +C u V ) C ( f V + r Q ). (2.27) Hence, Problem (2.5) has a unique solution (u,p) and the norm of the solution is bounded bythenormofthedataoftheproblem,whichisstability.orinotherwords,from(2.26)and (2.27), it follows that the mapping from the right-hand side to the solution (f,r) (u,p) is bounded. Thus, this mapping is continuous from V Q onto V Q. This property means, together with the already proved uniqueness of u and p, that Φ is an isomorphism from V Q onto V Q. The conditions i) and ii) are necessary. Assume now that Φ is an isomorphism from V Q onto V Q. It will be proved in the first step that then the inf- condition (2.14) holds, i.e., condition ii) of the theorem. The second step proves that also condition i) of the theorem is satisfied. Condition ii) of the theorem holds. Consider the restriction of B to V0 and denote it by B. Every u V can be decomposed uniquely into u = u 0 + ũ, u 0 V 0, ũ V0. Then, one obtains, using the definition of V 0, Bu = Bu 0 +Bũ = Bũ = B ũ. (2.28) Since Φ is an isomorphism, it is range(b) = Q and because of (2.28), one gets range ( B ) = Q. Hence, B is surjective. Next, one has to show that B is injective. Let u 0,u 1 V0 with B u 0 = B u 1 = ψ. Hence B (u 0 u 1 ) = 0 from what follows that u 0 u 1 V 0. Because u 0,u 1 V0 also each linear combination is element of V0, in particular u 0 u 1 V0. The only element which is in V 0 and V0 is the zero element such that u 0 = u 1 follows and B is injective. Altogether, B is an isomorphism between V0 and Q. Consequently, the inverse map (B ) 1 is also an isomorphism. Theorem A.70 gives that it is a bounded operator, i.e., for all ψ Q there is a constant C such that (B ) 1 ψ V C ψ Q. Choosing ψ = Bv, one observes that this inequality is equivalent to the existence of a constant β is such that for all v V 0 β is v V Bv Q. From Lemma 2.12, it follows that then condition (2.14) holds. Condition i) of the theorem holds. Let φ V 0 be arbitrary. By the Hahn Banach Theorem, Theorem A.72, there exists at least one element f V such that φ = E 0 f. Setting (u,p) = Φ 1 (f,0), then one has u V 0, since Bu = 0, and Au+B p = f in V. (2.29) By the definition of E 0, it follows that for all v V 0 E0 B p,v V,V = B p,v V,V = Bv,p Q,Q = 0. That means E 0 B p = 0 in V 0 and one gets with (2.29) E 0 Au = E 0 f = φ in V 0.

13 2.1 Solution of an Abstract Linear Saddle Point Problem 33 Hence, for each φ V 0 there is an element u V 0 with E 0 Au = φ, such that E 0 A is surjective. Now, the injectivity of E 0 A will be proved. Let v 1 v 2 V 0 with Let q Q be arbitrary. Then E 0 Av 1 = E 0 Av 2 = φ = E 0 A(v 1 v 2 ) = 0. (2.30) Av 1 +B q = f 1, Av 2 +B q = f 2 V, with f 1 f 2 since Φ is an isomorphism and Bv 1 = Bv 2 = 0. It follows that A(v 1 v 2 ) = f 1 f 2 V Φ(v 1 v 2,0) = (f 1 f 2,0). (2.31) By the definition (2.21) of E 0, one obtains with (2.30) and (2.31) 0 = E 0 A(v 1 v 2 ),v V,V = A(v 1 v 2 ),v V,V = f 1 f 2,v V,V v V 0. With the definition of Ṽ, it follows that f 1 f 2 Ṽ. It was already proved that the inf- condition holds. Thus, it follows from Lemma 2.12 ii) that there is a unique q Q such that B q = f 1 f 2. Hence, Φ(0,q) = (f 1 f 2,0). Since Φ is an isomorphism, it follows with (2.31) that (v 1 v 2,0) = (0,q), and in particular that v 1 = v 2. Thus, E 0 A is a one-to-one linear continuous mapping from V 0 onto V 0 and therefore an isomorphism from V 0 onto V 0. Lemma Sufficient condition on a(, ) for the well-posedness of (2.5). Assume that the bilinear form a(, ) is V 0 -elliptic, i.e., there is a constant α > 0 such that a(v,v) α v 2 V v V 0. Then, Problem (2.5) is well-posed if and only if the bilinear form b(, ) satisfies the inf- condition (2.14). Proof. V 0 -ellipticity and inf- condition = well-posed problem. It will be shown that the V 0 -ellipticity implies the first condition of Theorem Let f V 0 be arbitrary. Since a(, ) is V 0 -elliptic, the Lax Milgram theorem, see Theorem B.4, gives that there is a unique u V 0 such that a(u,v) = f,v V,V v V 0, or equivalently, that there is a unique u V 0 such that E 0 Au = f. Since f was chosen to be arbitrary, E 0 A is surjective. Consider now the injectivity of E 0 A. Let u V 0 with E 0 Au = 0. By the definition (2.21) of E 0, it is and in particular 0 = E 0 Au,v V,V = Au,v V,V v V 0, 0 = Au,u V,V = a(u,u). From the V 0 -ellipticity, it follows that u = 0, which implies injectivity.

14 34 2 Finite Element Spaces for Linear Saddle Point Problems Altogether, E 0 A is an isomorphism from V 0 onto V 0 and Theorem 2.18 gives the well-posedness of Problem 2.5. V 0 -ellipticity and well-posed problem = inf- condition. The satisfaction of the inf- condition in case Problem 2.5 is well-posed follows directly from Theorem The V 0 -ellipticity of the bilinear form a(, ) is a special case of the first condition of Theorem Remark Formulation as an optimization problem, saddle point problem. Problems (2.5) and (2.12) can formulated as optimization problems under certain conditions. Let J 0 : V R and J 1 : V Q R be two quadratic functionals defined by J 0 (v) = 1 2 a(v,v) f,v V,V, J 1(v,q) = J 0 (v)+b(v,q) r,q Q,Q. The functional J 0 is called energy functional associated with Problem (2.12) and J 1 is the Lagrangian functional associated with Problem (2.5). Consider the following problem: Find a saddle point (u,p) V Q of the Lagrangian functional J 1 over V Q, i.e., find a pair (u,p) V Q such that J 1 (u,q) J 1 (u,p) J 1 (v,p) v V, q Q. (2.32) This form is the classical formulation of a saddle point problem. The characterization (2.32) inspired the notation saddle point problem also for Problem (2.5). Theorem Existence and uniqueness of a solution of (2.32). Assume the conditions i) and ii) of Theorem Assume in addition that the bilinear form a(, ) is symmetric and positive semi-definite on V, i.e., a(v,v) 0 v V. Then, Problem (2.32) has a unique solution (u,p) V Q that is precisely the solution of Problem (2.5). Proof. It is referred to (Girault & Raviart, 1986, p. 62) for the proof. Remark Some generalizations. Generalizations of the saddle point problem considered in this section have been studied in the literature. In Ciarlet et al. (2003), solvability and stability conditions for a saddle point problem of the form a(u,v)+b 1 (v,p) = f,v V,V v V, b 2 (u,q) c(p,q) = r,q Q,Q q Q, are established. Abstract saddle point problems of form (2.4) can be also studied in Banach spaces, see (Ern & Guermond, 2004, Chapter 2.4) and the references therein.

15 2.1 Solution of an Abstract Linear Saddle Point Problem 35 The course covered the complete subsection until Rem. 2.20