Analysis of a non overlapping domain decomposition method for stokes equations

Analysis of a Non-overlapping Domain Decomposition Method for Stokes Equations Tom´as Chac´on Rebollo, Eliseo Chac´on Vera Depto de Ecuaciones Diferenciales y An´alisis Num´erico. Universidad de Sevilla [email protected], [email protected] Monograf´ıas del Semin. Matem. Garc´ıa de Galdeano. 27: 201–208, (2003). Abstract In this note we extend the analysis for elliptic problems performed in [1] to saddle point problems like the Stokes equations. We use a non overlapping domain decomposition and the introduction of a penalty term. In a simply connected bounded domain Ω ⊂R2with Lipschitz boundary, we decompose Ω into two non-overlapping Lipschitz subdomains Ω1and Ω2with Ω1∩Ω2=∅, and suppose that ∂Ωi= Γi∪Γ where Γiis the common boundary with Ω, Γi=∂Ω∩∂Ωiand Γ is the interface with Ωj, Γ = ∂Ω1∩∂Ω2. The Stokes equations on Ω are solved via the following parallel process: For n= 0,1,2, ..., given un i,pn iwe compute un+1 iand pn+1 i(i= 1,2) such that                −∆un+1 i+∇pn+1 i=fin Ωi ∇ · un+1 i= 0 in Ωi un+1 i= 0 on Γi ∂un+1 i ∂nij −pn+1 inij =−1 (un+1 i−un j) on Γ where nij is the outward normal vector on Γ pointing from Ωiinto Ωj, > 0 is a parameter that tends to cero and inforce the transmision conditions on the interface Γ and we stress that the pressures pido not longer have cero mean average. We present the convergence analysis of this technique and some numerical tests. An ampliation of this work will appear in [2]. Keywords: Stokes Problem, Parallel technique, Non Overlapping AMS CLASSIFICATION: 65L50, 65L60, 65L70 1 Introduction In 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∈[H1 0(Ω)]dand a pressure 201 p∈L2 0(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)∈[H1 0(Ω)]d×L2 0(Ω) with (∇u,∇v)Ω−(p, ∇·v)Ω−(∇·u, q)Ω= (f,v)Ω,(1) for all (v, q)∈[H1 0(Ω)]d×L2 0(Ω). Now we decompose Ω into two non-overlapping Lipschitz subdomains Ω1and Ω2with Ω1∩Ω2=∅(this choice is made to ease the exposition of the main ideas, but these can be extended to more than two subdomains). Suppose that ∂Ωi= Γi∪Γ where Γiis the common boundary with Ω, Γi=∂Ω∩∂Ωiand Γ is the interface with Ωj, Γ = ∂Ω1∩∂Ω2, all of these boundaries are Lipschitz (d−1)-dimensional manifolds. Next, we consider the Sobolev spaces Xi=H1 0(Ωi; Γi)d={v∈[H1(Ωi)]ds.t. v|Γi= 0} normed by |v|2 1,Ωi= (∇v,∇v)Ωiand the Hilbert spaces Mi=L2(Ωi) normed as usual. Now for  > 0 we consider the problem (P): Find (ui, pi)∈Xi×Miwith (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 the following coupled partial differential equations                −∆u1+∇p1=fin Ω1 ∇·u1= 0 in Ω1 u1= 0 on Γ1 ∂u1 ∂n12 −p1n12 =−1 (u1−u2) on Γ                −∆u2+∇p2=fin Ω2 ∇·u2= 0 in Ω2 u2= 0 on Γ2 ∂u2 ∂n21 −p2n21 =−1 (u2−u1) on Γ where nij is the outward normal vector on Γ pointing from Ωiinto Ωjand we stress that the pressures pido not longer have cero mean average. The apprpriated transmission conditions are enforced when −→ 0 because we show that ku1−u−2k0,Γ=O(). The iteration process that we proposse is the following: For n= 0,1,2, ..., given un 1,un 2 202 we compute un+1 1,un+1 2and pn+1 1, pn+1 2such that the following problems are satisfied                −∆un+1 1+∇pn+1 1=fin Ω1, ∇·un+1 1= 0 in Ω1, un+1 1= 0 on Γ1, ∂un+1 1 ∂n12 −pn+1 1n12 =−1 (un+1 1−un 2) on Γ,                −∆un+1 2+∇pn+1 2=fin Ω2, ∇·un+1 2= 0 in Ω2, un+1 2= 0 on Γ2, ∂un+1 2 ∂n21 −pn+1 2n21 =−1 (un+1 2−un 1) on Γ. We remark that our method may be viewed as a variation of the Robin 2 Analysis of problem (P) Let us introduce the product spaces X=X1×X2,M=M1×M2and denote by capital letters the elements (pairs) of Xand M. Then we norm Mwith kPk2 M=P2 i=1 kpik2 0,Ωi and Xvia ((U,V))=P2 i=1(∇ui,∇vi)Ωi+1 (u1−u2,v1−v2)0,Γi.e., the norm in Xis given by kUk= ((U,U)). Next we define the forms b(P,V) = −P2 i=1(pi,∇·vi)Ωi, F(V) = P2 i=1(f,vi)Ωiand write problem (P) in terms of the variational problem: (Find (U,P)∈X×Msuch 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 k·kand k·kM, bilinear form on X×Mgiven by B(U,P;V,Q) = ((U,V))+b(P,V) + b(Q,U) for all pairs (U,P),(V,Q)∈X×M. We have Lemma 1 There exists a positive constant γindependent of  > 0such that for all (U,P)∈X×M S= sup (V,Q)∈XxM |B(U,P;V,Q)| kVk+kQkM≥ γ (kUk+kPkM). As a consequence, given f∈[L2(Ω)]dand for each  > 0problem (P) has a unique solution (U,P)∈X×M. We introduce next the consistency error of problem (P) as an approximation of the Stokes equations in variational form. This error is the result of plugging the solution of (1) into (P). 203 Lemma 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) via G(V) = ((U,V))+b(P,V)−F(V) = 2 X i=1 (∇u,∇vi)Ωi− 2 X i=1 (p, ∇·vi)Ωi− 2 X i=1 (f,vi)Ωi for all V= (v1,v2)∈X. Then, assuming u∈[H2(Ω) ∩H1 0(Ω)]dand p∈H1(Ω), let n1,2=n, we have G(V) = ZΓ (∂nu−pn)·(v1−v2)dσ and therefore |G(V)| ≤ k∂nu−pnk0,Γkv1−v2k0,Γ. Now we can estimate the error in approximating the variational formulation of the Stokes Equations with problem (P) Lemma 3 Suppose that u∈[H2(Ω) ∩H1 0(Ω)]dand p∈H1(Ω) is the solution to the Stokes problem. For each  > 0let (U,P)∈X×Mbe the unique solution of problem (P), with U= (u 1,u 2)and P= (p 1, p 2). Let c(u, p) = k∂nu−pnk0,Γ,U= (u|Ω1,u|Ω2) and construct π=p 1χΩ1+p 2χΩ2−1 |Ω|(ZΩ1 p 1+ZΩ2 p 2). Then kU−Uk≤c(u, p)√and kp−πk0,Ω≤c(u, p)√. As a consequence we have 2 X i=1 |u−u i|1,Ωi≤c(u, p)√and ku 1−u 2k0,Γ≤c(u, p). 3 Discrete problem and error estimates We suppose that the domain Ω is polygonal and take for h > 0 an admissible and regular triangulation Thof Ω formed by polygons (d= 2) or polyhedra (d= 3) elements such that Γ is formed by faces or sides of elements Kin Th. Then we use Ti h=Th∩Ωi, for i= 1,2. These triangulations of Ωiare compatible on Γ, i.e., they share the same edges on Γ. For the triangulation Thwe consider finite element subspaces (Vh, Ph) of ([H1 0(Ω)]d, L2 0(Ω)) satisfying the discrete inf-sup condition of Ladyzhenskya-Brezzi-Babuˇska on Ω. Now we consider the discrete solution (uh, ph)∈Vh×Phof the discrete version of the Stokes 204 problem posed on Vh×Phand assume that, when the solution (u, p) to the continuous Stokes problem in Ω satisfies u∈Hk+1(Ω) ∩H1 0(Ω)dand p∈Hk(Ω) (k≥1), then |uh−u|1,Ω+kph−pk0,Ω≤C0hk(2) for some constant C0=C0(u, p). Now, based on Ti h, 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 ∩ [H1 0(Ωi)]dand Ni,h =Mi,h ∩L2 0(Ωi) also satisfies the discrete inf-sup condition on Ωi. For instance we could use the restriction of the spaces Vhand Phto each of the Ωi. Set now Xh=X1,h ×X2,h and Mh=M1,h ×M2,h and pose the discrete version of (P), that we denote by (P,h): (Find (U h,P h)∈Xh×Mhsuch 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 have the estimates Theorem 4 Let u∈Hk+1(Ω) ∩H1 0(Ω)dand p∈Hk(Ω) (k≥1) be the solution to the Stokes problem in Ωand for each h > 0let (uh, ph)∈Vh×Phsolve the discrete Stokes problem on Vh×Ph. Now consider Uh= (uh|Ω1,uh|Ω2)∈Xh,Ph= (ph|Ω1, ph|Ω2)∈Mh. For each  > 0let (U h,P h)∈Xh×Mhsolve (P,h)and write U h= (u 1,h,u 2,h)and P h= (p 1,h, p 2,h). Now construct π h=p 1,hχΩ1+p 2,hχΩ2−1 |Ω|(ZΩ1 p 1,h +ZΩ2 p 2,h) (3) then, the following error estimate hold kUh−U hk≤C(hk+√) (4) kph−π hk0,Ω≤C(hk+√) (5) where C=C(u, p)is a positive constant just depending on (u, p). As a consequence of (4) we have 2 X i=1 |uh−u i,h|1,Ωi≤C(hk+√)and ku 1,h −u 2,hk0,Γ≤C(√ hk+). Via the triangular inequality, we give the main result of this section Theorem 5 Let u∈Hk+1(Ω) ∩H1 0(Ω)dand p∈Hk(Ω),(k≥1) be the solution to the Stokes problem in Ω. For each h > 0and  > 0let (U h,P h)∈Xh×Mhsolve (P,h)with finite dimensional spaces of accuracy k≥1and write P h= (p 1,h, p 2,h). Then construct π has in (3). The following bounds hold 2 X i=1 |u−u i,h|1,Ωi+1 √ku 1,h −u 2,hk0,Γ≤C(hk+√) (6) kp−π hk0,Ω≤C(hk+√) (7) 205 where C=C(u, p, f)is a positive constant just depending on the data. When =O(h2k) we have 2 X i=1 |u−u i,h|1,Ωi+kp−π hk0,Ω≤C hkand ku 1,h −u 2,hk0,Γ≤C h2k. 4 Iteration process We search for the solution of (P,h) via the following parallelizable technique: For n= 0,1,2, ..., given un 1=u,n 1,h and un 2=u,n 2,h we compute un+1 1∈X1,h,un+1 2∈X2,h and pn+1 1∈M1,h, pn+1 2∈M2,h such that the following problem (Pn ,h) is satisfied            (∇un+1 1,∇v1)Ω1−(pn+1 1,∇·v1)Ω1−(q1,∇·un+1 1)Ω1+1 (un+1 1−un 2,v1)0,Γ= (f,v1)Ω1, (∇un+1 2,∇v2)Ω2−(pn+1 2,∇·v2)Ω2−(q2,∇·un+1 2)Ω2+1 (un+1 2−un 1,v2)0,Γ= (f,v2)Ω2 for all (v1,v2)∈Xh(we drop the indices and hwhen not needed). We obtain the following geometric rate of convergence Theorem 6 Let U h= (u 1,h,u 2,h)∈Xhand P h= (p 1,h, p 2,h)∈Mhbe the solution of (P,h)and U,n h= (u,n 1,h,u,n 2,h)∈Xh,P,n h= (p,n 1,h, p,n 2,h)∈Mhbe the solution of (Pn ,h). Let us define π hand π,n has in (3). Then, starting off the iterative process, for instance, with u0, i,h = 0, there exists a positive constant C0such that for each , h > 0and all n≥0 2 X i=1 |u,n+1 i,h −u i,h|1,Ωi≤P kfk0,Ω √(1 + 2 C0)n/2, kπ,n+1 h−π hk0,Ω≤Pkfk0,Ω (1 + 2 C0)n/2 for some constant Pproportional to the constant in Poincare’s Inequality. Via the triangular inequality we obtain the final bound Theorem 7 Let u∈Hk+1(Ω) ∩H1 0(Ω)dand p∈Hk(Ω), for k≥1, be the solution to the Stokes problem in Ω. For each h > 0and  > 0let (U,n h,P,n h)∈Xh×Mh(n≥1) solve the iteration problem (Pn ,h)starting off the iteration with U,0 h= 0, and using finite element spaces of accuracy k≥1. Then the following bounds hold for all n≥0 2 X i=1 |u−un+1, i,h |1,Ωi≤C(hk+√+1 √(1 + 2 C0)n/2) (8) kp−πn+1, hk0,Ω≤C(hk+√+1 (1 + 2 C0)n/2) (9) 206 where C=C(u, p)is a positive constant just depending on (u, p). When =O(h2k)and nlarge enough we obtain error bounds O(hk)for velocity and pressure 2 X i=1 |u−un, i,h |1,Ωi+kp−πn, hk0,Ω≤C hk where C=C(u, p)is a positive constant just depending on (u, p). 5 Numerical experiments We use a known solution of the incompressible Stokes equations to compute the error between the exact solution and the numerical approximation in the case k= 1. In this test Ω = (0,1) ×(0,1) and the boundary condition is u= 0 on the boundary ∂Ω of Ω. The exact solution is u(x, y) = −cos(2πx) sin(2πy) + sin(2πy) v(x, y) = sin(2πx) cos(2πy)−sin(2πx) p(x, y) = 2π(−cos(2πx) + cos(2πy)) and we take viscosity ν= 1. We consider the interface Γ as the line y= 0.5 and then Ω1= (0,1) ×(0,0.5) and Ω2= (0,1) ×(0.5,1). Next, we consider a uniform triangular mesh of mesh size h=hx=hy, take =h2and use P1finite elements with the BrezziPitkaranka stabilization technique for computing the solutions ui,h and pi,h on each Ωi. Then we construct the approximated velocity field uhand pressure πh∈L2 0(Ω) via          uh=ui,h,in Ωi uh= (u1,h +u2,h)/2,on ∂Γ, πh=p1,h χΩ1+p2,h χΩ2−1 |Ω|(ZΩ1 p1,h +ZΩ2 p2,h),in Ω where |Ω|= 1. Finally we compute the errors eu(h) = (P2 i=1 RΩi|∇(uh−un, ih )|2dx)1/2and ep(h) = kp−phk0,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 0.0412 ep(h) 0.5773 0.1942 0.066 Indeed, an order of convergence slightly larger that 1 is obtained on this example. References [1] Chac´ on Rebollo, T., Chac´ on Vera, E., Dom´ ınguez Delgado, A. Analysis of a Non-overlapping Domain Decomposition Method for Elliptic Equations XVII CEDYA/VII CMA, Salamanca (Spain) 2001. 207 [2] Chac´ on Rebollo, T., Chac´ on Vera, E.A non-overlapping Domain Decomposition 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 operators. 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-overlapping subdomains Third International Symposium on Domain Decomposition Methods for Partial Differential Equations, T.F. Chan et al. eds., SIAM, Philadelphia, pp. 202-231, 1990. 208