Chacón Rebollo, Tomás

Chacón Vera, Eliseo

idUS. Depósito de Investigación Universidad de Sevilla

Analysis of a Non-overlapping Domain Decomposition Methodfor Stokes EquationsToma´s Chaco´n Rebollo, Eliseo Chaco´n VeraDepto de Ecuaciones Diferenciales y Analisis Numerico. Universidad de Sevillachacon@numer.us.es, eliseo@numer.us.esMonograf´ıas del Semin. Matem. Garc´ıa de Galdeano. 27: 201–208, (2003).AbstractIn this note we extend the analysis for elliptic problems performed in [1] to saddlepoint problems like the Stokes equations. We use a non overlapping domain decom-position and the introduction of a penalty term. In a simply connected boundeddomain  ⊂ R2 with Lipschitz boundary, we decompose  into two non-overlappingLipschitz subdomains 1 and 2 with 1 ∩2 = ∅, and suppose that ∂i =  i ∪  where  i is the common boundary with ,  i = ∂∩∂i and   is the interface withj,   = ∂1 ∩ ∂2. The Stokes equations on  are solved via the following parallelprocess: For n = 0, 1, 2, ..., given uni , pni we compute un+1i and pn+1i (i = 1, 2) suchthat −un+1i +∇pn+1i = f in i∇ · un+1i = 0 in iun+1i = 0 on  i∂un+1i∂nij− pn+1i nij = −1(un+1i − unj ) on  where nij is the outward normal vector on   pointing from i into j,  > 0 is aparameter that tends to cero and inforce the transmision conditions on the interface  and we stress that the pressures pi do not longer have cero mean average. Wepresent the convergence analysis of this technique and some numerical tests. Anampliation of this work will appear in [2].Keywords: Stokes Problem, Parallel technique, Non OverlappingAMS CLASSIFICATION: 65L50, 65L60, 65L701 IntroductionIn a simply connected bounded domain Ω ⊂ Rd (d = 2, 3) with a Lipschitz boundary∂Ω and with f ∈ [L2(Ω)]d, we search for a velocity field u ∈ [H10 (Ω)]d and a pressure201p ∈ L20(Ω) such that {−∆u +∇p = f, ∇ · u = 0 in Ω,u = 0 on ∂Ω.In the classical mixed formulation of this problem we look for (u, p) ∈ [H10 (Ω)]d × L20(Ω)with(∇u,∇v)Ω − (p,∇ · v)Ω − (∇ · u, q)Ω = (f,v)Ω, (1)for all (v, q) ∈ [H10 (Ω)]d×L20(Ω). Now we decompose Ω into two non-overlapping Lipschitzsubdomains Ω1 and Ω2 with Ω1 ∩ Ω2 = ∅ (this choice is made to ease the exposition ofthe main ideas, but these can be extended to more than two subdomains). Suppose that∂Ωi = Γi ∪ Γ where Γi is the common boundary with Ω, Γi = ∂Ω ∩ ∂Ωi and Γ is theinterface with Ωj, Γ = ∂Ω1∩∂Ω2, all of these boundaries are Lipschitz (d−1)-dimensionalmanifolds. Next, we consider the Sobolev spacesXi =[H10 (Ωi; Γi)]d= {v ∈ [H1(Ωi)]d s.t. v|Γi = 0}normed by |v|21,Ωi = (∇v,∇v)Ωi and the Hilbert spaces Mi = L2(Ωi) normed as usual.Now for  > 0 we consider the problem (P):Find (ui, pi) ∈ Xi ×Mi with(P)(∇u1,∇v1)Ω1 − (p1,∇ · v1)Ω1 − (q1,∇ · u1)Ω1 +1(u1 − u2,v1)0,Γ = (f,v1)Ω1,(∇u2,∇v2)Ω2 − (p2,∇ · v2)Ω2 − (q2,∇ · u2)Ω2 +1(u2 − u1,v2)0,Γ = (f,v2)Ω2,for all (vi, qi) ∈ Xi × Mi, i = 1, 2. This problem is the variational formulation of thefollowing coupled partial differential equations−∆u1 +∇p1 = f in Ω1∇ · u1 = 0 in Ω1u1 = 0 on Γ1∂u1∂n12− p1n12 = −1(u1 − u2) on Γ−∆u2 +∇p2 = f in Ω2∇ · u2 = 0 in Ω2u2 = 0 on Γ2∂u2∂n21− p2n21 = −1(u2 − u1) on Γwhere nij is the outward normal vector on Γ pointing from Ωi into Ωj and we stress thatthe pressures pi do not longer have cero mean average. The apprpriated transmissionconditions are enforced when  −→ 0 because we show that ‖u1 − u− 2‖0,Γ = O().The iteration process that we proposse is the following: For n = 0, 1, 2, ..., given un1 , un2202we compute un+11 , un+12 and pn+11 , pn+12 such that the following problems are satisfied−∆un+11 +∇pn+11 = f in Ω1,∇ · un+11 = 0 in Ω1,un+11 = 0 on Γ1,∂un+11∂n12− pn+11 n12 = −1(un+11 − un2 ) on Γ,−∆un+12 +∇pn+12 = f in Ω2,∇ · un+12 = 0 in Ω2,un+12 = 0 on Γ2,∂un+12∂n21− pn+12 n21 = −1(un+12 − un1 ) on Γ.We remark that our method may be viewed as a variation of the Robin2 Analysis of problem (P)Let us introduce the product spaces X = X1 ×X2, M = M1 ×M2 and denote by capitalletters the elements (pairs) of X and M. Then we norm M with ‖P‖2M=∑2i=1 ‖pi‖20,Ωiand X via ((U,V)) =∑2i=1(∇ui,∇vi)Ωi+ 1 (u1−u2,v1−v2)0,Γ i.e., the norm in X is givenby ‖U‖ = ((U,U)). Next we define the forms b(P,V) = −∑2i=1(pi,∇ · vi)Ωi, F (V) =∑2i=1(f,vi)Ωi and write problem (P) in terms of the variational problem:{Find (U,P) ∈ X×M such that((U,V)) + b(P,V) + b(Q,U) = F (V), ∀ (V,Q) ∈ X×M.Now we consider the following symmetric and continuous, according to ‖ · ‖ and ‖ · ‖M,bilinear form on X×M given byB(U,P;V,Q) = ((U,V)) + b(P,V) + b(Q,U)for all pairs (U,P), (V,Q) ∈ X×M. We haveLemma 1 There exists a positive constant γ independent of  > 0 such that for all(U,P) ∈ X×MS = sup(V,Q)∈ XxM|B(U,P;V,Q)|‖V‖ + ‖Q‖M≥  γ (‖U‖ + ‖P‖M).As a consequence, given f ∈ [L2(Ω)]d and for each  > 0 problem (P) has a uniquesolution (U,P) ∈ X×M.We introduce next the consistency error of problem (P) as an approximation of the Stokesequations in variational form. This error is the result of plugging the solution of (1) into(P).203Lemma 2 Let (u, p) be the solution of the Stokes problem and U = (u|Ω1 ,u|Ω2 ), P =(p|Ω1 , p|Ω2 ). Then, we consider the consistency error of problem (P) viaG(V) = ((U,V)) + b(P,V)− F (V)=2∑i=1(∇u,∇vi)Ωi −2∑i=1(p,∇ · vi)Ωi −2∑i=1(f,vi)Ωifor all V = (v1,v2) ∈ X. Then, assuming u ∈ [H2(Ω) ∩H10 (Ω)]d and p ∈ H1(Ω), letn1,2 = n, we haveG(V) =∫Γ(∂nu− pn) · (v1 − v2) dσand therefore|G(V)| ≤ ‖∂nu− pn‖0,Γ‖v1 − v2‖0,Γ.Now we can estimate the error in approximating the variational formulation of the StokesEquations with problem (P)Lemma 3 Suppose that u ∈ [H2(Ω) ∩H10 (Ω)]d and p ∈ H1(Ω) is the solution to theStokes problem. For each  > 0 let (U,P) ∈ X ×M be the unique solution of problem(P), with U = (u1,u2) and P = (p1, p2). Let c(u, p) = ‖∂nu− pn‖0,Γ, U = (u|Ω1 ,u|Ω2 )and constructpi = p1χΩ1 + p2χΩ2 −1|Ω|(∫Ω1p1 +∫Ω2p2).Then‖U−U‖ ≤ c(u, p)√ and ‖p− pi‖0,Ω ≤ c(u, p)√.As a consequence we have2∑i=1|u− ui |1,Ωi ≤ c(u, p)√ and ‖u1 − u2‖0,Γ ≤ c(u, p) .3 Discrete problem and error estimatesWe suppose that the domain Ω is polygonal and take for h > 0 an admissible and regulartriangulation Th of Ω formed by polygons (d = 2) or polyhedra (d = 3) elements such thatΓ is formed by faces or sides of elements K in Th. Then we use T ih = Th ∩Ωi, for i = 1, 2.These triangulations of Ωi are compatible on Γ, i.e., they share the same edges on Γ. Forthe triangulation Th we consider finite element subspaces (Vh, Ph) of ([H10 (Ω)]d , L20(Ω))satisfying the discrete inf-sup condition of Ladyzhenskya-Brezzi-Babusˇka on Ω. Now weconsider the discrete solution (uh, ph) ∈ Vh × Ph of the discrete version of the Stokes204problem posed on Vh × Ph and assume that, when the solution (u, p) to the continuousStokes problem in Ω satisfies u ∈ [Hk+1(Ω) ∩H10 (Ω)]d and p ∈ Hk(Ω) (k ≥ 1), then|uh − u|1,Ω + ‖ph − p‖0,Ω ≤ C0 hk (2)for some constant C0 = C0(u, p). Now, based on T ih , use finite element subspaces of(Xi, Mi), denoted by (Xi,h, Mi,h), such that each pair (Yi,h, Ni,h), where Yi,h = Xi,h ∩[H10 (Ωi)]dand Ni,h = Mi,h ∩L20(Ωi) also satisfies the discrete inf-sup condition on Ωi. Forinstance we could use the restriction of the spaces Vh and Ph to each of the Ωi. Set nowXh = X1,h ×X2,h and Mh = M1,h ×M2,h and pose the discrete version of (P), that wedenote by (P,h):{Find (Uh,Ph) ∈ Xh ×Mh such that((Uh,Vh)) + b(Ph,Vh) + b(Qh,Uh) = F (Vh), ∀(Vh,Qh) ∈ Xh ×Mh.The existence and uniqueness of solution for (P,h) is carried out as for (P) and we havethe estimatesTheorem 4 Let u ∈ [Hk+1(Ω) ∩H10 (Ω)]d and p ∈ Hk(Ω) (k ≥ 1) be the solution to theStokes problem in Ω and for each h > 0 let (uh, ph) ∈ Vh × Ph solve the discrete Stokesproblem on Vh × Ph. Now consider Uh = (uh|Ω1 ,uh|Ω2 ) ∈ Xh, Ph = (ph|Ω1 , ph|Ω2 ) ∈ Mh.For each  > 0 let (Uh,Ph) ∈ Xh × Mh solve (P,h) and write Uh = (u1,h,u2,h) andPh = (p1,h, p2,h). Now constructpih = p1,hχΩ1 + p2,hχΩ2 −1|Ω|(∫Ω1p1,h +∫Ω2p2,h) (3)then, the following error estimate hold‖Uh −Uh‖ ≤ C (hk +√) (4)‖ph − pih‖0,Ω ≤ C (hk +√) (5)where C = C(u, p) is a positive constant just depending on (u, p). As a consequence of(4) we have2∑i=1|uh − ui,h|1,Ωi ≤ C (hk +√) and ‖u1,h − u2,h‖0,Γ ≤ C (√ hk + ).Via the triangular inequality, we give the main result of this sectionTheorem 5 Let u ∈ [Hk+1(Ω) ∩H10 (Ω)]d and p ∈ Hk(Ω), (k ≥ 1) be the solution to theStokes problem in Ω. For each h > 0 and  > 0 let (Uh,Ph) ∈ Xh ×Mh solve (P,h) withfinite dimensional spaces of accuracy k ≥ 1 and write Ph = (p1,h, p2,h). Then constructpih as in (3). The following bounds hold2∑i=1|u− ui,h|1,Ωi +1√‖u1,h − u2,h‖0,Γ ≤ C (hk +√) (6)‖p− pih‖0,Ω ≤ C (hk +√) (7)205where C = C(u, p, f) is a positive constant just depending on the data. When  = O(h2k)we have2∑i=1|u− ui,h|1,Ωi + ‖p− pih‖0,Ω ≤ C hk and ‖u1,h − u2,h‖0,Γ ≤ C h2 k.4 Iteration processWe search for the solution of (P,h) via the following parallelizable technique: For n =0, 1, 2, ..., given un1 = u,n1,h and un2 = u,n2,h we compute un+11 ∈ X1,h, un+12 ∈ X2,h andpn+11 ∈ M1,h, pn+12 ∈ M2,h such that the following problem (P n,h) is satisfied(∇un+11 ,∇v1)Ω1 − (pn+11 ,∇ · v1)Ω1 − (q1,∇ · un+11 )Ω1 +1(un+11 − un2 ,v1)0,Γ = (f,v1)Ω1 ,(∇un+12 ,∇v2)Ω2 − (pn+12 ,∇ · v2)Ω2 − (q2,∇ · un+12 )Ω2 +1(un+12 − un1 ,v2)0,Γ = (f,v2)Ω2for all (v1,v2) ∈ Xh (we drop the indices  and h when not needed). We obtain thefollowing geometric rate of convergenceTheorem 6 Let Uh = (u1,h,u2,h) ∈ Xh and Ph = (p1,h, p2,h) ∈ Mh be the solution of(P,h) and U,nh = (u,n1,h,u,n2,h) ∈ Xh, P,nh = (p,n1,h, p,n2,h) ∈ Mh be the solution of (P n,h). Letus define pih and pi,nh as in (3). Then, starting off the iterative process, for instance, withu0,i,h = 0, there exists a positive constant C0 such that for each , h > 0 and all n ≥ 02∑i=1|u,n+1i,h − ui,h|1,Ωi ≤P ‖f‖0,Ω√ (1 + 2 C0 )n/2,‖pi,n+1h − pih‖0,Ω ≤P ‖f‖0,Ω (1 + 2 C0 )n/2for some constant P proportional to the constant in Poincare’s Inequality.Via the triangular inequality we obtain the final boundTheorem 7 Let u ∈ [Hk+1(Ω) ∩H10 (Ω)]d and p ∈ Hk(Ω), for k ≥ 1, be the solution tothe Stokes problem in Ω. For each h > 0 and  > 0 let (U,nh ,P,nh ) ∈ Xh ×Mh (n ≥ 1)solve the iteration problem (P n,h) starting off the iteration with U,0h = 0, and using finiteelement spaces of accuracy k ≥ 1. Then the following bounds hold for all n ≥ 02∑i=1|u− un+1,i,h |1,Ωi ≤ C (hk +√ +1√ (1 + 2 C0 )n/2) (8)‖p− pin+1,h ‖0,Ω ≤ C (hk +√ +1 (1 + 2 C0 )n/2) (9)206where C = C(u, p) is a positive constant just depending on (u, p). When  = O(h2k) andn large enough we obtain error bounds O(hk) for velocity and pressure2∑i=1|u− un,i,h |1,Ωi + ‖p− pin,h ‖0,Ω ≤ C hkwhere C = C(u, p) is a positive constant just depending on (u, p).5 Numerical experimentsWe use a known solution of the incompressible Stokes equations to compute the errorbetween the exact solution and the numerical approximation in the case k = 1. In thistest Ω = (0, 1) × (0, 1) and the boundary condition is u = 0 on the boundary ∂Ω of Ω.The exact solution isu(x, y) = − cos(2pix) sin(2piy) + sin(2piy)v(x, y) = sin(2pix) cos(2piy)− sin(2pix)p(x, y) = 2pi(− cos(2pix) + cos(2piy))and we take viscosity ν = 1. We consider the interface Γ as the line y = 0.5 and thenΩ1 = (0, 1) × (0, 0.5) and Ω2 = (0, 1) × (0.5, 1). Next, we consider a uniform triangularmesh of mesh size h = hx = hy, take  = h2 and use P1 finite elements with the Brezzi-Pitkaranka stabilization technique for computing the solutions ui,h and pi,h on each Ωi.Then we construct the approximated velocity field uh and pressure pih ∈ L20(Ω) viauh = ui,h, in Ωiuh = (u1,h + u2,h)/2, on ∂Γ,pih = p1,h χΩ1 + p2,h χΩ2 −1|Ω|(∫Ω1p1,h +∫Ω2p2,h), in Ωwhere |Ω| = 1. Finally we compute the errors eu(h) = (∑2i=1 ∫Ωi |∇(uh−un,ih )|2 dx)1/2 andep(h) = ‖p− ph‖0,Ω. The following table shows the values obtained for these measures.wesh 16× 16 (h = 1/16) 32× 32 (h = 1/32) 64× 64 (h = 1/64)eu(h) 0.4600 0.13413 0.0412ep(h) 0.5773 0.1942 0.066Indeed, an order of convergence slightly larger that 1 is obtained on this example.References[1] Chaco´n Rebollo, T., Chaco´n Vera, E., Dom´ınguez Delgado, A. Analysisof a Non-overlapping Domain Decomposition Method for Elliptic Equations XVIICEDYA/VII CMA, Salamanca (Spain) 2001.207[2] Chaco´n Rebollo, T., Chaco´n Vera, E.A non-overlapping Domain Decomposi-tion Method for the Stokes equations via a penalty term on the interface C.R. Acad.Sci. Paris, t. 334, Serie I, p. 2002.[3] J.L. Lions, Pironneau, O., Overlapping domain decomposition of evolution oper-ators. C.R. Acad. Sci. Paris, t. 330, Serie I, p. 937-942, 2000.[4] P.L. Lions On the Schwarz alternating method III: a variant for non-overlappingsubdomains Third International Symposium on Domain Decomposition Methods forPartial Differential Equations, T.F. Chan et al. eds., SIAM, Philadelphia, pp. 202-231,1990.208

Analysis of a non overlapping domain decomposition method for stokes equations

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Analysis of a non overlapping domain decomposition method for stokes equations

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