A Posteriori Error Estimates and Domain Decompositionwith non Matching Grids.J. POUSIN, T. SASSI,Mathematical modelling and scientic computing Laboratory, CNRS UMR5585,National Institute of Applied Sciences in Lyon,20 av. Einstein, F-69621 Villeurbanne Cedex FranceKeywords Numerical Analysis for P.D.E; nonlinear elliptic Problems; A poste-riori error estimates; Domain Decomposition.MSC(91) 65J10, 65N55, 65M60.AbstractLet F be a non linear mapping dened from an Hilbert space X into its dual X 0, andlet x be in X the solution of F (x) = 0. Assume that a priori, the zone where thegradient of the function x varies very much is known. The aim of this article is toprove a posteriori error estimates for the problem F (x) = 0 when it is approximatedwith a Petrov-Galerkin nite element method combined with a domain decompositionmethod with non matching grids. An estimator of the error based on residual isproposed and an algorithm for computing the approximated solution is proved to beconvergent for a semi linear problem.1 IntroductionAdaptive nite element techniques have been introduced at the beginningof the eighties for linear elliptic problems and nowadays are commonly used inindustrial codes. For a general presentation the reader is refereed to [2]. Veryfew non linear elliptic problems have been considered, except recently, wherea general framework has been proposed for adaptive nite element techniquesbased on residual estimator [10]; [13]; [17]. The cornerstone of the adaptivenite element technique based on residual estimator consists in the two followingproperties:1) the equivalence of the error and of the residual;2) the location of the residual by means of an estimator.Please remark that even if the residual is proved to be related to the error,algorithmic diculties have to be considered such as minimising the estimator byadapting the mesh with a procedure consuming not too much time. Practically,it is easy to check that the mesh renement propagates outside of the zones wherethe error is large. This is a consequence of some regularity requirements of themesh for the nite element method. When the zones of large variations of thegradient of x solution to F (x) = 0 are a priori known, the domain decompositionmethod with non matching grids is a way for controlling the spread of the meshrenement. For a general presentation of the domain decomposition methodwith non matching grids, the reader is referred to [4]. Combustion problems, orchemical problems with localised reaction zones, or non smooth problems-such as problems stated in non convex domains- can be quoted as problemsfor which the zone of large variations of the gradient of the solution is known a1

priori. In such a context, from a practical point of view, one can easily verifythat during the mesh adapting process, far from the reaction zones or far fromthe singular zones, the mesh does not change a lot accordingly with the localregularity properties ( [12]; [11]).The Heuristic proposed strategy is to adapt the mesh, if necessary, only in sin-gular zones which constitute the sub domains. In order to make such a strategypossible, we have to be sure that the residual for multi domain in non matchinggrids setting can be related to the error for a non linear elliptic problem. Theorganisation of this paper is as follows: the end of the introduction is dedicatedto the notations, the functional spaces, the topologic properties of the sub do-mains, and the meshes of sub domains are specied. In section two the abstractresults are given. In section 3, a simple semi linear problem is considered, andnally, in section 4, a residual estimator is derived and an algorithm for solvingthe approximated problem is proposed.For the sake of clarity, we are going to consider the case of only two sub domains,the extension to the case of more than two is straightforward. Let be aparallelepiped bounded open domain of R3 which is split into two parallelepipedssub domains 1, 2 such that = 1 [ 2, and such that = 1 \ 2 is arectangle (see Figure 1).

AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA

Γ

Ω2Ω1

corner nodes

Figure 1.Let Th1 , Th2 be regular triangulations of 1, 2 (see [5] p. 124 for a denition)which do not match each other on . We denote by hi the maximal diameterof parallelepipeds constituting Thi, and we set h = max(h1; h2). It is assumedthat the corner nodes of sub domains 1 and 2 belonging to belong to Th1and to Th2 and are corner nodes of . For i = 1; 2 let Trhi denotes the trace oftriangulation Thi on . The following hypothesis H1) is assumed to be satised:H1) The triangulation Th of is coarser than the traces of the induced triangu-lations Trhi 1 i 2.In this work, we are dealing with Th, the coarsest trace of the induced triangu-lations Trhi . Please remark this choice for the nite dimensional space in which2

the weak matching between 1 and 2 will be ensured is not the only possible inthe framework of mortar nite element technique (see [4]). Now let us introducethe functional spaces which will be used in the following. We denote by H10 ()the classical Sobolev space of functions vanishing on the boundary @ of , thenorm of which is the H1 semi-norm denoted by j j1; (see [7]). For i = 1; 2 thespaces 0H1(i) are dened by:0H1(i) = f'i for ' 2 H10 ()g; (1)and are equipped with the semi-norm j j1;i which is a norm on 0H1(i) thanksto Poincare inequality. Let us recall the well known trace results (see [7]).Lemma 1. For i = 1; 2 there exists a linear surjective continuous operator ifrom H1(i) into H 12 (@i) which to ' associates its trace i(') on @i, in ageneralised sense.If for ' 2 L2(), e' stands for its extension by zero to @i, then the space eH 12 ()is dened byeH 12 () = f' 2 L2() such as e' 2 H 12 (@i); 1 i 2g; (2)the norm of which is given byk'k212@ = k'k20; + Z j'(s) '(t)j2js tj2 ds dt: (3)If B stands for a Banach space, the norm of which is denoted by kkB, B0 standsfor its dual and L(B;B0) denotes the Banach space of linear continuous mappingfrom B into B0, the norm of which is denoted by kkB;B0 . We set Xi =0 H1(i)for i = 1; 2; M = eH 12 ()0; X = X1 X2 equipped with the product norm,and X =0 H1(i)0 H1(i) eH 12 ()0 = X M: (4)Now let us introduce the linear continuous operator K, used for ensuring thematching conditions on for the functions of X1 and X2, which is dened byK : X = X1 X2 ! eH 12 () =M 0V = (v1; v2) 7! 2Xi=1 (1)i i(vi): (5)The space X0 and X0 are dened byX0 = KerK; and X0 = X0 M: (6)Finally, let us state the problem (P) and its formulation in domain decom-position setting, the Problem (P), which will be under our consideration. LetF : H10 () ! H1() be a C1 mapping given, Problem (P) is dened by:(P) nd u 2 H10 () verifying F (u) = 0: (7)3

If for i = 1; 2 Fi denotes the restriction of F to Xi; Fi : Xi ! X 0i, then the C1mapping F : X ! X 0 is dened by8(; ) 2 X ; < F(; ); (V; q) >X 0;X= 2Xi=1 < Fi('i); vi >X0i;Xi + (8)< KV; >M 0;M + < K; q >M 0;M ; 8(V; q) 2 X :The Problem (P) is dened by:(P) nd U = (U; p) 2 X verifying F(U) = 0: (9)Please note that in the formulation (9), extra terms dealing with functions inM are consequences of the existence of a Lagrange multiplier related to thematching conditions on .2 Abstract resultsThe matching conditions on have been expressed with a mixed formulation,so it is possible to reformulate Problem (P) in the setting introduced in [13] inorder to get a posteriori error estimates. Essentially, this consists in expressingthe approximated problem in the space X itself, by using a projector in theapproximated space built with the derivative of the function F thanks to anInf-Sup condition (see H5)). A xed point theorem then yields existence for theapproximated solution and a priori and a posteriori error estimates.Let nXh = Xh1 Xh2 Mh = Xh Mho0<h1 be a family of nite dimensionalsubspaces of X . An approximation of Problem (P) by using a Petrov-Galerkinand domain decomposition mixed formulation consists in seeking Uh 2 Xh veri-fying < F(Uh);Vh >X 0;X= 0; 8Vh 2 Xh: (10)In what follows if G is a C1 mapping, then DG denotes its Frechet derivative.Now, let us specify the hypotheses, under which, it will be proved that Problem(10) is a suitable approximation of Problem (P).H2) There exists U 2 X such that F(U) = 0;H3) DF(U) is an isomorphism from X into X 0,H4) DF(U) is Lipchitzian at U ,H5) 9 > 0; such that 8h; 0 < h (11)minYh 2 XhkYhkX = 1 maxVh 2 XhkVhkX = 1< DF(U)Yh;Vh >X 0;X ; (12)H6) limh!0 kYh YkX = 0; 8Y 2 X :In order to make possible the estimate of the distance between U solution toProblem (P) and Uh solution to Problem (10), we have to reformulate Problem4

(10) as an equivalent problem stated in X . To do so, we need to introduce aprojector Xh : X ! Xh by using DF(U). Let b(; ) : X X ! R be thecontinuous bilinear form dened by:b(Y;V) =< DF(U)Y;V >X 0;X ; 8Y; V 2 X : (13)For all h 0 the projector Xh is dened by:b((I Xh)Y;Vh) = 0; 8Vh 2 Xh; 8Y 2 X : (14)Lemma 2. Assume Hypotheses H5), H6) hold true, then the operator Xh iswell dened by (14). Moreover, there exists C > 0 independent of h such that:kXhYkX ;X CkYk and (15)limh!0 k(I Xh)YkX = 0; 8Y 2 X : (16)Proof. Thanks to Hypothesis H5), Theorem 5.2.1 p.112 of [2] applies, thus theoperator Xh is well dened. Moreover the following estimate is veried:kXhkX ;X 1 supV;Y 2 XkVkX = kYkX = 1 b(Y;V) (17)By considering the inequalityk(I Xh)YkX kY YhkX + kYh XhYkX 8Yh 2 Xh; (18)the result is k(I Xh)YkX (1 + kXhkX ;X ) infYh2Xh kY YhkX : (19)Thus Estimate (17) combined with Hypothesis H6) yields (16).Lemma 2 is proved.Now we are in position for giving the main result of this section.Theorem 1. Assume that hypotheses H2) to H6) are satised. Then there existthree positive constants h0, 0 and C0 such as for all h h0 there exists a uniqueUh 2 Xh solution to Problem (10) and verifyingkU UhkX 0. Furthermore, the following error estimates are valid:kU UhkX C0 minYh2Xh kU YhkX ; (20)kU UhkX 2kDF(Uh)1kX 0;XkF(Uh)kX 0; (21)kF(Uh)kX 0 2kDF(Uh)kX ;X 0kU UhkX : (22)Theorem 1 is a corollary of a more general xed point theorem, the proof ofwhich is given in [13] theorems 2 and 3 p.217, and which reads:5

Theorem 2. Let G : X ! X 0 be a C1 mapping satisfying the following hy-potheses: there exists U0 2 X such as G(U0) = 0; (23) DG(U0)is an isomorphism from X onto X 0; (24) DG is Lipchitzian at (U0): (25)Let fGhg0<h<1 be a family of C1 mappings from X into X 0 satisfying:DGh is uniformly Lipchitzian at (U0) ; (26) limh!0kGh(U0)kX 0 = 0; (27)DGh(U0)is an isomorphism from X onto X 0 withuniformly bounded inverse with respect to h: (28)Then there exist 0 < h0, 0 < 0 such as for all h h0 there is a unique Uh 2 Xverifying: Gh(Uh) = 0 and kU UhkX < 0: (29)Moreover, the following estimates hold true:kU0 UhkX 2kDGh(U0)1kX 0;XkGh(U0)kX 0; (30)kU0 UhkX 2kDG(Uh)1kX 0;XkG(Uh)kX 0: (31)Before proving Theorem 1, let us establish intermediate results useful for derivingTheorem 1 from Theorem 2. Set G = F and U = U0. A family of mappingsGh : X ! X 0 remains to be build such asFinding Y 2 X ; such that < Gh(Y);V >X 0;X= 0; 8V 2 X ; (32)is equivalent toFinding Y 2 Xh; such that < F(Y);Vh >X 0;X= 0; 8Vh 2 Xh: (33)Set < Gh(Y);V >X 0;X=< F(Y);XhV >X 0;X+ b(Y; (I Xh)V); 8V 2 X ; 8Y 2 X : (34)Lemma 3. If Gh is given by (34), Xh by (14), then the equivalence of items(32) and (33) holds true.Proof. If Y 2 Xh veries (33), then accounting for the denitions of Gh and Xhthe result is < Gh(Y);V >X 0;X=< F(Y);XhV >X 0;X= (35)b(Y; (I Xh)V) = b((I Xh)Y;V) = 0; 8V 2 X :6

So it is checked that Y veries (32). Conversely, if Y 2 X veries (32), rst byputting V =W XhW in (32) we getb(Y; (I Xh)W) = 0; 8W 2 X : (36)Accounting for the denition of Xh (14) we getb((I Xh)Y;W) = 0; 8W 2 X ; (37)which with Hypothesis H3) provides Y XhY = 0, and thus (32) implies (33).Lemma 3 is proved.Proof of Theorem 1.let us check that hypotheses (26) to (28) are satised by Gh given by (34).Accounting for the denition (13) of b(; ) , the denition (14) of Xh we have< (DGh(U0)DGh(Z))Y;V >X 0;X=< (DF(U0)DF(Z))Y;XhV >X 0;X :Thus (26) is a consequence of Hypothesis H4) combined with Item (15) of Lemma2.Hypothesis H3) with DF(U0) = DGh(U0) imply that DGh(U0) is an isomor-phism, the inverse of which is uniformly bounded with respect to h. So Ghsatises (28).In order to establish (27), kGh(U0)kX 0 has to be estimated.kGh(U0)kX 0 CkDF(U0)kX ;X 0k(I Xh)U0kX : (38)By combining Hypothesis H2) and Item (16) of Lemma 2, we get (27). It is asimple matter to check that the hypotheses (23) to (25) are satised. Theorem 2applies , so thanks to Lemma 3 we get existence of Xh 2 X solution to Problem(10) and uniqueness in the ball B(U0; 0) with the a priori and a posterioriestimates (20), (21). For the Inequality (22), rst note that a Taylor expansionleads to F(Uh) = nI + 1Z0 F(tUh + (1 t)U)DF(Uh)DF(Uh)1 dtoDF(Uh)(Uh U):Then thanks to (20) we deduce that the integral term goes to zero with h, whichimplies (22). Theorem 1 is proved. Remark. Hypothesis H5) can be weakened by introducingXh0 = fYh 2 Xhj < KYh; qh >M 0;M= 0 for all qh 2Mhgand by assuming that discrete Inf-Sup condition holds true only in Xh0 = Xh0 Mh (see for example [16]).The results presented can be extented to the case where the space H1() isreplaced by W 1;p() (see for example [15]). 7

3 A semi linear elliptic model ProblemIn this section, the abstract results of the previous section are specied with thefollowing semi linear elliptic problem: for f 2 L1() non negative given, nd usatisfying: u+ u3 = f in ;u = 0 on @: (39)First, it is recalled that this problem has a unique solution, and that the mappingwhich to f associates u is regular. Then it is checked that the hypotheses H3)to H6) are satised in order that Theorem 1 applies to Problem (39).The function F : H10 () ! H1() is dened by< F (); w >H1();H10 ()= Z rrw dx+ Z (3 f)w dx 8; w 2 H10 ():(40)The existence and uniqueness of u 2 H10 ()\L1() verifying (39) is well known(see for example [9]). Moreover, F 2 C1(H10 ();H1()) and< DF (u); w >H1();H10 ()= Z rrw dx + Z 3u2w dx 8; w 2 H10 ():(41)Accordingly with notations introduced in section 1, the mixed formulation withdomain decomposition setting of Problem (39) leads to the following denitionof F . 8(; ) 2 X ;< F(; ); (V; q) >X 0;X= 2Xi=1 Zi r'irvi dx+ Zi ('3i f)vi dx + (42)< KV; >M 0;M + < K; q >M 0;M ; 8(V; q) 2 X :We have F 2 C1(X ;X 0) and 8Y = (Y; ) 2 X ,8(; ) 2 X ;< DF(Y; )(; ); (V; q) >X 0;X= 2Xi=1 Zi r'irvi dx+ Zi 3y2i'ivi dx+ (43)< KV; >M 0;M + < K; q >M 0;M ; 8(V; q) 2 X :Now we are dealing with Hypotheses H2) to H6). Let U = (U; P ) 2 X be asolution to F(U) = 0 (U = (U; P ) with ui = uji , p = @n1u1 + @n2u2 = 0). Thebilinear symmetric form a : X X ! R is dened by:a(; V ) = 2Xi=1 Zi r'irvi dx+ Zi 3u2i'ivi dx; 8V; 2 X: (44)So Hypothesis H2) is satised. Before dening the nite dimensional subspaces,let us check that Hypotheses H3) and H4) are satised. It is easily seen that8

Hypotheses H4) is satised by using the factorisation 3(u2i y2i ) = 3(ui+yi)(uiyi) and by using the Sobolev embedding of H1() into L4(). Now we turn toprove that DF(U) is an isomorphism. For all = (; ), V = (V; q) belongingto X , the bilinear form b(; ) is dened by:b(;V) =< DF(U);V >X 0;X= a(; V ) +< KV; >M 0;M + < K; q >M 0;M : (45)Lemma 4. For the bilinear continuous form b(; ), the following Inf-Sup condi-tions hold true: there exists > 0 such asinf 2 XkkX = 1 supV 2 XkVkX = 1b(;V) ;sup 2 XkkX = 1b(;V) > 0; 8V 6= 0: (46)Proof. Introduce the bilinear continuous formk : X M ! R(V; q) 7! k(V; q) =< KV; q >M 0;M=< V;Kq >X;X0 : (47)First recall that X0 = kerK. Since a(; ) is a bilinear symmetric form X-coercitive, it induces a scalar product onX. Thus, X can be split in the followingway: X = X0 X>a , i. e.8Y 2 X; Y = Y0 + Y >a with a(Y >a ; V0) = 0; 8V0 2 X0: (48)In what follows, it is proved that K 2 Isom(M ; (X>a )0). Let 6= 0 be given, let 2M 0 be such that k kM 0 = 1, and such that< ; >M 0;M= kkM 0: (49)By using the trace operator 1 (see Lemma 1), w1 is dened by w1 = 11 ( ),then set W = (w1; w2) with w2 = 0. DeneV0 2 X0 by a(V0; Y0) = a(W;Y0); 8Y0 2 X0 (50)and V >a 2 X>a by V >a = W V0: (51)Thanks to Lemma 1 we havekV >a kX kWkX C1k kM 0; (52)and thus kkM 0C1 supV >a 2 X>akV >a kX = 1 k(V >a ; ) = supV 2 XkV kX = 1 k(V; ): (53)9

Now, let V >a 2 X>a be such that V >a 6= 0, then KV >a 6= 0, so choose q 2 M , withkqkM = 1 be such that < KV >a ; q >M 0;M= kKV >a kM 0: (54)The result is 0 < kKV >a kM 0 supq 2MkqkM = 1 k(V >a ; q): (55)Theorem 5.1.2 of [2] applies and we get K 2 Isom(M ; (X>a )0), and thus K 2Isom(X>a ;M 0).Since the bilinear form a(; ) is X0-coercitive and continuous, the induced linearoperator A 2 L(X0; (X0)0) dened bya(0; V0) =< A0; V0 >X0;X 80; V0 2 X0 (56)veries A 2 Isom(X0; (X0)0). The Bezzi-Babuska Theorem applies (see [16]Theorem 10.1 p. 111) and for f 2 X 0 given, the following system has a uniquesolution (; ) 2 X M8<: a(>a ; V >a ) + k(V >a ; ) =< f; V >a >X0;X 8V >a 2 X>a ;a(0; V0) =< f; V0 >X0;X 8V0 2 X0;k(>a ; q) = 0 8q 2M: (57)This linear system, to which is associated the following matrix of operators0@ A> 0 K0 A 0K 0 0 1A (58)has a unique solution as an operator acting from X into X 0. We can concludethat Inf-Sup conditions (46) hold true (see [16]). Lemma 4 is proved.Now the nite dimensional spaces Xhi and Mh are constructed. The space Mhconsidered in what is following allowed to treat the case of many sub domainsfor which a corner node of a sub domain may be a corner node of many othersub domains. In such a context, the matching in value for such a node is weaklyimposed (see [4]) since it is not judicious to write many times the same equation.For the presented case of just two sub domains a more simple subspace Mh couldbe considered (Mh = Wh the denition of which is given below).For i = 1; 2 the space Xhi Xi is dened by:Xhi = fvi 2 C0(i); vijT 2 Q1(T ) 8T 2 Thi; vij@ = 0g: (59)The denition of Mh is a little bit more involved.Mh f' 2 C0(); 'jA 2 Q1(A) 8A 2 Th; such that A \ @ = ;g: (60)10

Now let us specify the restriction of functions of Mh to A for polygons A 2 such that A \ @ 6= ;. Let lj or lk be the edges of A belonging to @ for

j; k 2 f1; 2; 3; 4g (see gure 2).A

A

A

1

2

Ω

3

4

k

j

jΓ

boundary of

Figure 2.For i = j or i = k, di 2 Q1(A) denotes the function verifying:di(x) = 0 8x 2 li;1 8x 2 lp; for p = i+ 2 mod (4) (61)In the case of only one edge li of A belonging to @ then ' 2 Mh is such that'jA = qdi where q 2 Q1(A) is such that qjli = 0. If li and lk are edges of Abelonging to @ then ' 2 Mh is such that 'jA = qdkdi where q 2 Q1(A) is suchthat qjlj = 0 for j = i or j = k.The nite element space Xh is dened byXh = Xh1 Xh2 Mh = Xh Mh: (62)Finally, let j 2 f1; 2g be such that the trace of Trhj on is Th (the coarsest traceof the grids on ). If j denotes the trace operator from Xj into M 0, then Wh isdened by: Wh = jXhj : (63)Now let us give some technical results concerning the existence and stability ofprojector operators which will be useful in the following. Let h be the projectordened by h : L2() !Whq 7! hq verifyingR h(q hq) dx = 0 8h 2Mh: (64)Concerning the operator h the result is:Lemma 5. There exists 0 < C3 independent of h such thatkhqk0; C3kqk0;; 8q 2 L2(): (65)11

A proof of Lemma 5 can be found in ([3]). For the convenience of the reader, ashort proof of this result for the nite element case is given in Appendix A.By using interpolation technique between Hilbert spaces and Lemma 5, the fol-lowing stability and convergence results in H 12 norm for the projector h areestablished.Lemma 6. Assume that Th is regular and satises an inverse assumption (see[5] p.140), then there exists 0 < C4 independent of h such that:khqkM 0 C4kqkM 0; 8q 2M 0 = eH 12 (): (66)Furthermore we have 8q 2M 0; limh!0 k(I h)qkM 0 = 0: (67)Proof. Let rh be the Clement interpolation operator, then the following estimateshold true (see [6]):8q 2 H1(); krhqk1; C5kqk1;; k(I rh)qk0; C6hkqk1;: (68)The L2 continuity of the bilinear form c(; ) which to (h; qh) 2 Mh Wh asso-ciates c(h; qh) = Z hqh dx (69)and the Inf-Sup condition onMhWh for the bilinear form c(; ) (see AppendixA) provide the estimatek(I h)qk0; ( 1CA + 1)k(I rh)qk0;; 8q 2 L2(): (70)In order to prove (66), we proceed by interpolation between the Hilbert spacesL2() and H10 (). So let us give the stability result in H1-norm for the projectorh. Let q 2 H10 () be given, by splitting h in the following wayhq = hq rhq + rhq (71)and by using an inverse inequality (see [5] p. 140), the following inequalities holdtrue. khqk1; krhqk1; + k(h rh)qk1; krhqk1; + C7h1k(h rh)qk0; krhqk1; + C8h1nk(h I)qk0; + k(I rh)qk0;o krhqk1; + C9h1k(I rh)qk0;: (72)Inequalities (68) then provideskhqk1; C10kqk1;: (73)12

Interpolation technique yields Inequality (66).We proceed analogously as before for the expression (I h)q, thus (70) com-bined with an inverse inequality between L2 and H1 providek(I h)qkM 0 C11k(I rh)qkM 0: (74)Then arguing by density and accounting fork(I rh)qkM 0 C12h 12 jqj1; (75)gives Item (67). Lemma 6 is proved.We end this technical part with a corollary of Lemma 6 concerning the projectorh which is dened by h 2 L(M ;Mh) with< qh; (I h) >M 0;M= 0 8qh 2 Wh; 8 2M: (76)Lemma 7. Assume the hypothesis of Lemma 6 holds true, thenkhqkM C4kqkM ; 8q 2M = ( eH 12 ())0: (77)Moreover we have 8 2 M; limh!0 k(I h)kM = 0: (78)Now, we are in position for checking that the hypotheses H5) and H6) are sat-ised.Lemma 8. Assume Hypothesis H1) is satised, then there exists 0 < suchthat for all 0 < h, minYh 2 XhkXhkX = 1 maxVh 2 XhkVhkX = 1< DF(U)Yh;Vh >X 0;X : (79)Proof. We argue in the same way as in the proof of Lemma 4. Keeping the samenotations for the bilinear forms a(; ) and k(; ) (see (44), (47)) the subspaceXh0 is dened by:Xh0 = fVh 2 Xh; k(Vh; qh) = 0; 8qh 2Mhg: (80)Please remark that Xh0 * X0. The space Xh is expressed as the following sumXh = Xh0 X>ha, i. e.8Yh 2 Xh; Yh = Yh0 + Y >ha with a(Y >ha; Vh0) = 0; 8Vh0 2 Xh0 : (81)Since the bilinear form a(; ) is X-coercitive, it is sucient to check that thefollowing discrete Inf-Sup condition holds true, that is to say, there exists 0 < 1such that for all 0 < h1 minh 2MhkhkM = 1 maxVh 2 XhkVhkX = 1k(Vh; h): (82)13

Let 2 M 0 be such that k kM 0 = 1, and such that< ; h >M 0;M= khkM 0: (83)By denition of h (see 64), we have:< ; h >M 0;M=< h ; h >M 0;M : (84)Thanks to Hypothesis H1), for i = 1; 2, the extension operator 1i is denedfrom Wh into Xhi and is continuous. So, by settingwh1 = 11 (h ) 2 Xh1; wh = (wh1; 0); (85)we prove that the following inequalities are veried with constants independentof h. kWhkX C11kh kM 0 C12k kM 0 = C12: (86)Then the result iskhkM 0C12 khkM 0kwhkX k(wh; h)kwhkX supVh 2 XhkVhkX = 1k(Vh; h); (87)which proves that K>h : Mh ! X>ha dened by< Vh; K>h h >M 0;M= k(Vh; h); (88)is an isomorphism. We proceed in the same way as in the proof of Lemma 4, forestablishing the the Inf-Sup condition (79). Lemma 8 is proved.Finally, the last hypothesis H6) is a consequence of Lemma 5 and of the wellknown interpolation results for linear Lagrange nite element. Theorem 1 appliesand we have the following result:Theorem 3. There exist three positive constants h0, 0 and C0 such as for allh h0 there exists a unique Uh = (Uh; ph) 2 Xh solution to< F(Uh);Vh >X 0;X= 0 8Vh 2 Xh (89)and verifying kU UhkX 0. Furthermore, the following error estimates arevalid: kU UhkX C13 minYh2Xh kU YhkX ; (90)kU UhkX 2kDF(Uh)1kX 0;XkF(Uh)kX 0; (91)kF(Uh)kX 0 2kDF(Uh)kX ;X 0kU UhkX : (92)14

4 Error estimatorIn Theorem 3 we proved that the error is related to the residual. In this sectionwe dene the estimator which allows to localised the residual. Let us mentionthat the Verfurt localisation technique could be used, but this technique is moreinvolved than the following rough estimator we are going to derive.Let us call iin the union of all interior faces of the triangulation Thi . For eachface ` iin let us choose an arbitrary normal direction n, for any vhi 2 Xhi , let[@vhi=@n]` denote the jump of @vhi=@n a cross the face `.For i = 1; 2, let rhi be the Clement interpolation operators dened from L1(i)!Xhi, from which the operator Rh = (rh1 ; rh2) X ! Xh is derived.For V 2 X , Vh 2 Xh by using (89) the result is:< F(Uh);V >X 0;X=< F(Uh);V Vh >X 0;X :Choose Vh = (RhV; 0) 2 Xh and by integrating by parts we have< F (Uh);V >X 0;X = 2Xi=1 n XTi2Thi nZTi (uh + u3h f)(vi rhivi) dx+ Z@Ti @uhi@n (vi rhivi) dso+ X@Ti2 Z@Ti ((1)iph(vi rhivi)) dso+ < KUh; q >M 0;M ;Let us denote by Th2 the nest grid, and introduce Q a solution to Q = 0 in2; @Q@n = q on ; Q = 0 on @2 n . If h2 stands for an extension by 0 to Xh2of KUh then we have< KUh; q >M 0;M= XT 2 Th2T \ 6= ; ZT rh2rQ dx: (93)Now, accounting for the estimates between a function and its Clement interpolate(see [6]), we have:kv rhivk0;T Chi PT 02ST kvk1;T 0; ST 0 = f[T =T \ T 0 6= ;g;kv rhivk0;@Tl C1h 12@TlPT 02ST kvk1;T 0; for l = 1; 2; 3; (94)where, here the constants are independent of the parallelepiped T . It followsthat15

< F (Uh);V >X 0;X C2n 2Xi=1 nn XTi2Thi nh2Tik(uhi + u3hi fi)k20;Tio 12+ n X@Ti =2 h@Tik[@uhi@n ]k20;@Tio 12o + n X@Ti2h@Tik(1)iph + @uhi@n k20;@Tio 12o+ n XT 2 Th2; T \ 6= ; kh2k20;To 12okVkX : (95)Remark. The rst and second terms of the above inequality are classical, whenwe solve the semi linear problem (39) (see [10]). The other terms are due to thenon matching grids Remark. The right-hand side of (95) can be used as an a posteriori error esti-mator, but it requires the knowledge of KUh at the interface , which preventto compute the estimator in parallel on each sub domains. This part of theestimator can be neglected in practice [14].For practical computations, the last term of the right-hand side of (95) is nottaking into account since it is negligible compared to the other terms. Theresidual estimator can be written independently on each sub domains i. To dothis, we dene for any parallepiped Ti 2 ThiTif = 1jTij ZTi f dx;the L2(T ) projection of f onto the piecewise constant functions and for each `of the triangulation J i = 8><>:[@uhi@n ]` if ` iin;2((1)i ph + @uhi@n j`) if ` ;0 if ` @: (96)Then the estimator on parallelepiped Ti of each sub domain writesi(Ti) = jTij2(Ti(f uhi3))2 + 12 X`2ETi j`j2 J i12 ; (97)where jTij is the measure of the parallelepiped Ti, ETi stands for the set of itsfour faces and j`j is the surface of `. Finally we set = 2Xi=1 XTi2Thi i(Ti)2 12 : (98)Let us end this section with some considerations concerning the way in whichthe approximated problem can be rephrased as an interface problem. Theorem16

1 states that for all h h0 the following approximated problem has a uniquesolution Uh = (Uh; ph) 2 Xh verifying:< F(Uh; ph); (Vh; qh) >X 0;X = P2i=1 Ri ruhirvhi + u3hivhi dx Ri fvhi dx+< KVh; ph >M 0;M + < KUh; qh >M 0;M ; 8(Vh; qh) 2 Xh: (99)Now a domain decomposition method for solving the discrete problem (99), asan interface problem is introduced. Let ' 2 Mh, fi = fji be given, introducethe following problems on each sub-domain: for 1 i 2 nd uhi('; f) 2 Xhi,hi 2Mh verifying Ri ruhi('; f)rvhi + u3hi('; f)vhi dx = < hi; ivhi >M;M 0 + < fi; vhi >X0i ;Xi;< iuhi('; f) '; qh >M;M 0 = 0 ; 8 qh 2Mh ; 8 vhi 2 Xhi: (100)The discrete Steklov-Poincare operator (cf [1]) which to ' associates hi thegeneralised normal derivative of uhi('; f)on is dened as follows:Sih : iXhi ! Mh ;uhij 7! hi ;where (uhi('; f); hi) is the solution of (100).The result isLemma 9. Assume h h0, if ' is a solution of the following interface problem2Xi=1 < Sih'; ivhi >M;M 0 0; 8 vh 2 Xh0; (101)then the original discrete problem (99) and the interface Problem (101) are equiv-alent.Proof. Let ' 2Mh be a solution of (101). For 1 i 2, Tr1ih h stands for anyelement vhi 2 Xhi the trace of which on is equal to h 2 Mh. Please remarkthat Vh = (vh1; vh2) 2 Xh0 (that is to say k(Vh; qh) = 0 8qh 2 Mh). Then fromthe denitions of Sih it follows that 2Xi=1 < Sih' ; ivhi >M ;M 0 =+ 2Xi=1 8<:Zi ruhi('; f) rTr1ih ivhi + u3hi('; f)Tr1ih ivhi dx Zi fi Tr1ih ivhi dx9=;Since Problem (99) has a unique solution, then we deduce that(uh1('; f); uh2('; f)) = Uh:from which it is easy to obtain the equivalence between (99) and (101).Remark. The resulting non linear interface problem (101) can be solved with aNeumann preconditionned Conjugate Gradient operator in the same way as forthe linear case ( see [8]). For 2-D domains numerical results have been carriedout and will be presented in a forthcoming article. 17

5 Appendix ALemma 10. There exists 0 < C3 independent of h such thatkhqk0; C3kqk0;; 8q 2 L2(): (102)Proof. The projector h will be well dened and (102) will be established if thefollowing holds true (see [2] Theorem 5.2.1 p. 112): there exists 0 < C such that80 < h and 8qh 2 Yh Ckqhk0; maxh 2 Mhh 6= 0 R hqh dxkhk0; ; (103)dimMh = dimYh: (104)It is easily seen that Item (104) is true, so let prove Item (103). Let qh 2 Yh bexed, Yh eH 12 () \ C0(), thus qh@ = 0. SethA = 8><>: qhjA if A 2 Th is such that A \ @ = ;;qhjAdj if @A \ @ = lj;qhjAdjdk if @A \ @ = lj [ lk (105)h 2Mh and we haveZ hqh dx = XA 2 ThA \@ = ; kqhk20;A + XA 2 ThA \@ 6= ; ZA hqh dx: (106)Please note that 1 1dj and 1 1djdk a. e. in A; (107)so kqhk20; Z hqh dx: (108)Now, let us estimate the L2 norm of h in term of kqhk0;.khk20; = XA 2 ThA \@ = ; kqhk20;A + XA 2 ThA \@ 6= ; khk20;A: (109)The second part of the right hand side of (109) is split in two partsXA 2 ThA \@ 6= ; khk20;A = XA 2 ThA \@ = lj khk20;A + XA 2 ThA \@ = lj [ lk khk20;A = I1 + I2:(110)18

Let A be xed, with the reference element technique, FA denotes the anemapping from A, the reference element, onto A. We set qh(x) = qh FA(x). Wehave ZA 2h dx = jdetF1A j ZA (h)2 dx: (111)If A contributes to I1, we have h = qhdj : (112)Please remark that dj(x) = dj(xi) where i = 1 or 2, becausedj(x) = 0 if x 2 lj1 if x 2 lk with k = j + 2 mod. (4): (113)The expression for the L2 norm iskhk20;A = jdetF1A j ZA qh(xdj(xi)2 dx: (114)By using the Hardy inequality for non negative function ' 2 L2(0; 1) with (y) =R y0 '(s) ds 1Z0 (y)y 2 dy 12 2 1Z0 '(y)2 dy12 ; (115)since Q1(A) = P 1(0; 1)NP 1(0; 1) the result isZA qh(xdj(xi)2 dx 4 ZA @xi qh(x2 dx 4 ZA jrxi qh(x)j2 dx: (116)The equivalence of norms in nite dimensional vector space yields the existenceof C independent of h such that:ZA qh(xdj(xi)2 dx 4C ZA qh(x2 dx: (117)By using the ane change of variables for expressing the integral in (117) as anintegral on A, combined with Identity (114) provideskhk20;A = 4CjdetF1A jjdetFAjkqhk20;A = 4Ckqhk20;A: (118)For the case where A \ @ = lj [ lk, we argue in the same way as before,using the following Hardy inequality: for '; 2 L2(0; 1) non negative, then for(y) = R y0 (s) ds, (y) = R y0 '(s) ds1Z0 1Z0 (y)y (s)s 2 dyds 16 1Z0 1Z0 '(y)2 (s)2 dyds: (119)19

Finally, Identity (109) leads tokhk20; = 4Ckqhk20;: (120)Gathering the estimates (108) and (120) gives Estimate (103) with C3 = 12pC .Lemma 10 is proved.References[1] V.I. Aghoskov, Poincare-Steklov's operators and domain decompositionmethods in nite dimensional spaces. In Proceedings of the rst Inter-national Symposium on Domain Decomposition Methods, Paris, France,January (1987), SIAM, Philadelphia (1988).[2] I. Babuska, Feedback , Adaptivity and A Posteriori estimates in nite ele-ment Aims, Theory, and Experience,John Willey sons LTD 1986.[3] F. Ben Belgacem, The mortar Finite Element Method with Lagrange Mul-tipliers, Rapport interne, MIP, RI 94.1,[4] C. Bernardi, Y. Maday, T. Patera, A new non conforming approach to do-main decomposition ; the mortar element method, in Nonlinear PartialDierential Equations and their Applications (H. Brezis and J.L Lions edi-tors), College de France Seminar, Vol. XI, Pitman (1993)[5] P. Ciarlet, The Finite Element Method for Elliptic Problems, North-Holland, Amesterdam (1978).[6] P. Clement, Approximating by nite element function using local regulari-sation RAIRO Anal. Numer., 9, 77-84(1974)[7] P. Grisvard, Elliptic problems in non smooth domains, Pitman Boston 1985.[8] P. Le Tallec, T. Sassi, M. Vidrascu, Three-dimensional Domain Decompo-sition Methods with Nonmatching Grids and Unstructured Coarse SolversContemporary Mathematics, Volume 180, 1994, p. 61-73.[9] P.L. Lions, On the existence of positive solutions of semi-linear elliptic equa-tions SIAM Review, 24, 1982, p. 441-467.[10] J. Pousin, J. Rappaz, Consistance, stabilite, erreurs a priori et a posterioripour des problemes non lineaires CRAS, Paris, t. 312, Serie I 312 (1991)699-703.[11] M. Picasso, Simulation numerique de traitements de surface par laser TheseEPFL, no. 1011 (1992) 20

[12] L.Pouly, J. Pousin, Adaptive nite element for semi-linear convection-dision problem Advances in Computational Mathematics 7(1997), p. 135-259.[13] J. Pousin, J. Rappaz, Consistency, stability, apriori and a posteriori errorsfor Petrov-Galerkin methods applied to nonlinear problems, Numer. Math.69 (1994) 213-232.[14] J. Pousin, T. Sassi, Adaptive nite element and domain decomposition withnonmatching grids. ECCOMAS 96 J. Willey Sons p. 820-829; 1996[15] J. Pousin, T. Sassi, Domain decomposition with nonmatching grids andmixed formulation in spaces W 1;p0 , W 1;q0 . ZAMM vol. 77 issue 9 p. 639-644;1997[16] P. Robert, J.M. Thomas, Mixed Formulation in Handbook of NumericalAnalysis, vol II, North Holland, Amesterdam 1991. Hybrid and mixed[17] R. Verfurth, a posteriori errors for nonlinear problems. Finite element dis-cretizations of elliptic equations. Math. of Comp., 62 (206), p. 445-475;1994

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