Analysis of time-dependent Navier-Stokes flow coupled with Darcy

Transcription

1 Analysis of time-dependent Navier-Stokes flow coupled wit Darcy flow Ayçıl Çeşmelioğlu and Béatrice Rivière Abstract Tis paper formulates and analyzes a weak solution to te coupling of time-dependent Navier-Stokes flow wit Darcy flow under certain boundary conditions, one of tem being te Beaver-Josep-Saffman law on te interface. Existence and a priori estimates for te weak solution are sown under additional regularity assumptions. We introduce a fully discrete sceme wit te unknowns being te Navier-Stokes velocity, pressure and te Darcy pressure. Te sceme we propose is based on a finite element metod in space and a Crank-Nicolson discretization in time were we obtain te solution at te first time step using a first order backward Euler metod. Convergence of te sceme is obtained and optimal error estimates wit respect to te mes size are derived. Keywords: time-dependent, Navier-Stokes, Darcy, Beaver-Josep-Saffman s condition, Crank-Nicolson, backward Euler 1 Introduction Tis work follows a series of papers on te coupling of surface flow wit subsurface flow. Te domain is divided into two subdomains: in te surface region, flow is caracterized by te time-dependent Navier- Stokes equations and in te subsurface region, flow is caracterized by te Darcy equations. Te coupling of te two types of flow is accomplised troug interface conditions. In tis work, we define a weak solution and sow its existence and uniqueness. We propose a numerical sceme tat is second order in time and optimal in space. Te underlying space discretization is te classical finite element metod. Te weak problem of a similar coupling is analyzed in [4], in wic an interface problem wit Steklov-Poincaré operators is formulated. In [9, 14], we analyze te steady-state problem of Navier-Stokes coupled wit Darcy. We sow well-posedness of te weak problem and convergence of te numerical algoritms. If te nonlinearity is removed from te Navier-Stokes equations, we obtain te coupling of Stokes and Darcy. Tis problem as been extensively studied in te literature. Te reader can refer to [17, 1] for te analysis of te weak solution and to [1, 6, 11, 1,, 16,, 19] for a variety of numerical scemes. We denote by Ω R a bounded domain decomposed into two disjoint domains Ω 1 and Ω. Te fluid velocity and pressure in Ω 1 are denoted by u and p 1 respectively. Te deformation tensor is D(u) = 1 ( u + ( u)t ). Te flow in Ω 1 over te time interval (, T ) is caracterized by te time-dependent Navier-Stokes equations: u t (µd(u) p 1I) + u u = f 1 inω 1 (, T ), (1.1) Department of Matematics, University of Pittsburg, 31 Tackeray Hall, Pittsburg, PA, 156. Te autors acknowledge te support of NSF troug te grant DMS

2 u = in Ω 1 (, T ). (1.) Te fluid pressure in Ω is denoted by p. Te flow in Ω over te time interval (, T ) is caracterized by te Darcy equation: K p = f in Ω (, T ). (1.3) Te coefficients in te equations are µ > te fluid viscosity, f 1 a body force acting on Ω 1 [, T ], K a positive definite symmetric matrix corresponding to te permeability of Ω and f a body force acting on Ω [, T ]. Te system of equations is completed by an initial condition u = u at time t =, and a set of boundary conditions. Let Ω i denote te boundary of Ω i wit exterior unit normal n Ωi, let Γ 1 = Ω 1 Ω and let Γ i = Ω i \Γ 1 for i = 1,. We decompose te boundary Γ = Γ D Γ N and we assume tat Γ D >. u = on Γ 1 (, T ), (1.4) p = on Γ D (, T ), (1.5) K p n Ω = on Γ N (, T ). (1.6) Let n 1 be equal to n Ω1 on Γ 1 and let τ 1 be te tangential unit vector to Γ 1. We assume continuity of te normal component of velocity across te interface: u n 1 = K p n 1. (1.7) We assume tat te Beaver-Josep-Saffman law olds [5, ] wit a positive constant G > (usually obtained from experimental data.) u τ 1 = µg(d(u)n 1 ) τ 1. (1.8) Finally, we write te balance of forces across te interface by writing ((µd(u) + p 1 I)n 1 ) n (u u) = p. (1.9) Te balance of forces includes te inertial forces. In [14], tis new interface condition was considered in te steady-state coupling of Navier-Stokes wit Darcy. Weak Formulation We define te following Sobolev spaces using te notation in [1] X = {v H 1 (Ω 1 ) : v = on Γ 1 }, M 1 = L (Ω 1 ), M = {q H 1 (Ω ) : q = on Γ D }. In general, if Z is a Banac space, ten te space L (, T ; Z) denotes te space of square-integrable functions from [, T ] into Z. It is a Banac space wit te norm ( Z dt)1/. For any domain O, we denote by (v, w) O te L inner-product of two functions v, w defined on O. We now define a form γ tat takes into account te interface conditions as follows: u X, p M, γ(u, p; v, q) = (p 1 (u u), v n 1) Γ1 + 1 G (u τ 1, v τ 1 ) Γ1 (u n 1, q) Γ1. Consequently, we observe tat γ(v, q; v, q) = ( 1 (v v), v n 1) Γ1 + 1 G v τ 1 L (Γ 1). (.1)

3 We propose te following weak formulation: Find (u, p 1, p ) (L (, T ; X) H 1 (, T ; L (Ω 1 ) )) L (, T ; M 1 ) L (, T ; M ) suc tat (v, q) X M, ( u + µ(d(u), D(v)) Ω1 + (u u, v) Ω1 (p 1, v) Ω1 + (K p, q) Ω +γ(u, p (Q) ; v, q) = (f 1, v) Ω1 + (f, q) Ω, q M 1, ( u, q) Ω1 =, v X, (u(), v) Ω1 = (u, v) Ω1. Lemma.1. Assume tat f 1 L (, T ; L (Ω 1 ) ), f L (, T ; L (Ω )), K L (Ω ), (.) and K is uniformly bounded and positive definite in Ω : tere exist λ min, λ max > suc tat λ min x Kx x λ max x a.e x Ω. (.3) In addition, let u L (Ω 1 ). Ten any solution (u, p 1, p ) (L (, T ; X) H 1 (, T ; L (Ω 1 )) ) L (, T ; M 1 ) L (, T ; M ) of (1.1)-(1.9) is also a solution to (Q). Conversely any solution to (Q) satisfies (1.1)-(1.9). Proof. First, we prove tat if (u, p 1, p ) (L (, T ; X) H 1 (, T ; L (Ω 1 ) )) L (, T ; M 1 ) L (, T ; M ) is a solution to (1.1)-(1.9), ten it satisfies problem (Q). Indeed, let v X. Ten taking te scalar product of (1.1) wit v X over Ω 1 yields ( u ( (µd(u) p 1 I), v) Ω1 + (u u, v) Ω1 = (f 1, v) Ω1. Applying Green s formula to te second term and using te duality pairing,, we obtain ( u + (µd(u), v) Ω1 (p 1, v) Ω1 + (µd(u) + p 1 I)n Ω1, v Ω1 + (u u, v) Ω1 = (f 1, v) Ω1. Since D(u) is a symmetric tensor, we ave (D(u), v) Ω1 = (D(u), D(v)) Ω1. (.4) Tis and te assumption tat v = on Γ 1 gives ( u + (µd(u), v) Ω1 (p 1, v) Ω1 + (µd(u) + p 1 I)n 1, v Γ1 + (u u, v) Ω1 = (f 1, v) Ω1. (.5) Now let q M. Taking te scalar product of (1.3) wit q over Ω yields ( K p, q) Ω = (f, q) Ω. After applying Green s formula wit te boundary condition (1.6) and te fact tat n Ω = n 1, we get Adding (.5) and (.6) yields (K p, q) Ω + (K p ) n 1, q Γ1 = (f, q) Ω. (.6) ( u + (µd(u), D(v)) Ω1 + (u u, v) Ω1 + (K p, q) Ω (p 1, v) Ω1 + (µd(u) + p 1 I)n 1, v Γ1 + (K p ) n 1, q Γ1 = (f 1, v) Ω1 + (f, q) Ω. (.7) We write v as a sum of its normal and tangential components, i.e., v = (v n 1 )n 1 + (v τ 1 )τ 1. 3

4 From [14], we ave ((µd(u) p 1 I)n 1 ) n 1 L (Γ 1 ), µ(d(u)n 1 ) τ 1 L 4 (Γ 1 ). So we can write (µd(u) + p 1 I)n 1, v Γ1 = (((µd(u) + p 1 I)n 1 ) n 1, v n 1 ) Γ1 + (((µd(u))n 1 ) τ 1, v τ 1 ) Γ1. Ten by (1.8) and (1.9), (µd(u) + p 1 I)n 1, v Γ1 = (p 1 (u u), v n 1) Γ1 + 1 G (u τ 1, v τ 1 ) Γ1. By (1.7), we also ave So (.7) becomes, (K p ) n 1, q Γ1 = (u n 1, q) Γ1. ( u +(µd(u), D(v)) Ω1 +(u u, v) Ω1 +(K p, q) Ω (p 1, v) Ω1 +γ(u, p ; v, q) = (f 1, v) Ω1 +(f, q) Ω. Now let q M 1 and multiply (1.) by q and integrate over Ω 1 to get ( u, q) Ω1 = wic completes te weak formulation (Q). Conversely, assume tat (u, p 1, p ) (L (, T ; X) H 1 (, T ; L (Ω 1 )) L (, T ; M 1 ) L (, T ; M ) is a solution of (Q). As u(t) X and p (t) M, by definition of te spaces, equations (1.4) and (1.5) are satisfied immediately. Te assumption ( u, q) Ω1 = for all q M 1 gives (1.). To get (1.1), let v D(Ω 1 ) and q =. We recall tat D(Ω 1 ) is te space of smoot functions wit compact support. Wit tis coice of (v, q), te first equation in (Q) becomes Terefore in te sense of distributions on Ω 1 ( u t µ D(u) + u u + p 1, v) Ω1 = (f 1, v) Ω1. wic gives (1.1). Next, letting v = and q D(Ω ) in (Q) yields So in te distributional sense on Ω, we ave u t µ D(u) + u u + p 1 = f 1. (.8) ( K p, q) Ω = (f, q) Ω. wic gives (1.3). Taking te scalar product of (.8) wit v X yields By Green s formula, we ave K p = f. (.9) ( u (µ D(u), v) Ω1 + (u u, v) Ω1 + ( p 1, v) Ω1 = (f 1, v) Ω1. ( u +(µd(u), v) Ω1 +(u u, v) Ω1 (p 1, v) Ω1 + (µd(u)+p 1 I)n Ω1, v Ω1 = (f 1, v) Ω1. (.1) Multiplying (.9) by q M and integrating over Ω gives ( K p, q) Ω = (f, q) Ω. 4

5 As q H 1 (Ω ) by Green s formula, we ave Adding (.1) and (.11) and using (.4) gives (K p, q) Ω (K p ) n Ω, q Ω = (f, q) Ω. (.11) ( u + (µd(u), D(v)) Ω1 + (u u, v) Ω1 (p 1, v) Ω1 + (K p, q) Ω + (µd(u) + p 1 I)n Ω1, v Ω1 + (K p ) n Ω, q Ω = (f 1, v) Ω1 + (f, q) Ω. Comparing tis wit (Q), we end up wit (v, q) X M, (p 1 (u u), v n 1) Γ1 + 1 G (u τ 1, v τ 1 ) Γ1 (u n 1, q) Γ1 = (µd(u) + p 1 I)n Ω1, v Ω1 + (K p ) n Ω, q Ω. (.1) Letting v = in (.1) (u n 1, q) Γ1 = K p n Ω, q Ω. (.13) Coosing q = on Γ 1 and as q = on Γ D K p n Ω, q ΓN =. wic implies (1.6), i.e., K p n Ω = on Γ N. Hence, since n Ω = n 1 on Γ 1 and q = on Γ D, equation (.13) becomes (u n 1, q) Γ1 = K p n 1, q Γ1 q M. Terefore we obtain (1.7). Next, we take q = in (.1): v X, ((p 1 (u u))n G (u τ 1)τ 1, v) Γ1 = (µd(u) + p 1 I)n 1, v Γ1. (.14) Tus we ave (µd(u) + p 1 I)n 1 = (p 1 (u u))n G (u τ 1)τ 1. (.15) in te sense of distributions on Γ 1. We obtain immediately: and i.e., (1.8) and (1.9). ((µd(u) + p 1 I)n 1 ) τ 1 = p 1 G (u τ 1), ((µd(u) + p 1 I)n 1 ) n 1 = p 1 (u u). 3 Existence and uniqueness of weak solution We start tis section by recalling Poincaré, Sobolev and trace inequalities. We use te notation v H 1 (Ω 1) = v L (Ω 1), wic is a norm for X. Tere exist constants P 1, C, C 1, C 4, C 4 wic only depend on Ω 1 suc tat for all v X, v L (Ω 1) P 1 v H 1 (Ω 1), v L 4 (Ω 1) C 4 v H 1 (Ω 1), v H 1 (Ω 1) C 1 D(v) L (Ω 1), (3.1) 5

6 v L (Γ 1) C v H 1 (Ω 1), v L 4 (Γ 1) C 4 v H 1 (Ω 1). (3.) Tere exist constants P and C tat only depend on Ω suc tat for all q M In addition, from te assumption (.3), we ave q L (Ω ) P q H 1 (Ω ), q L (Γ 1) C q H 1 (Ω ). (3.3) 1 K 1/ 1 q L (Ω ) q H 1 (Ω ) K 1/ q L (Ω λmax λmin ). (3.4) Now denote by Y te product space Y = X M equipped wit te norm and te associated scalar product (v, q) Y, (v, q) Y = (µ D(v) L (Ω 1) + K1/ q L (Ω ) )1/. (v, q), (w, r) Y, ((v, q), (w, r)) Y = µ(d(v), D(w)) Ω1 + (K q, r) Ω. Because of (3.1) and (3.4) te norm (, ) Y is equivalent to te following product norm (v, q) Y, (v, q) = ( v H 1 (Ω 1) + q H 1 (Ω ) )1/. So (Y, (, ) Y ) is a Hilbert space. Define te space of divergence free functions by V = {v X : v = in Ω 1 }, and te associated subspace W of Y by W = V M. Te space W is also a Hilbert space wit te norm and scalar product of Y. Restricting te test functions v to V in (Q), we obtain a second variational formulation: Find (u, p ) (L (, T ; V ) H 1 (, T ; L (Ω 1 ) ) L (, T ; M ) suc tat (v, q) W, (P ) v V, (u(), v) Ω1 = (u, v) Ω1. ( u + µ(d(u), D(v)) Ω1 + (u u, v) Ω1 + (K p, q) Ω + γ(u, p ; v, q) = (f 1, v) Ω1 + (f, q) Ω, Clearly if (u, p 1, p ) is a solution to (Q), ten (u, p ) is a solution to (P ). We will now sow existence of a solution to problem (P ) using te Galerkin metod. Te spaces V and M are separable Hilbert spaces as tey are closed subspaces of separable Hilbert spaces H 1 (Ω 1 ) and H 1 (Ω ). So we can find a basis {w i, r i } i 1 of W suc tat w i V H (Ω 1 ) and r i M H (Ω ). Fix m N and let W m = span{(w i, r i ), i = 1,..., m}. Denote by π m te ortogonal projection of V onto span{w i, i = 1,..., m}. Ten a Galerkin approximation to problem (P ) is te finite-dimensional problem (P m ) defined as: Find (u m, p m ) L (, T ; W m ) wit u m H 1 (, T ; L (Ω 1 ) ) suc tat 1 i m, (P m ) ( um t, w i) Ω1 + µ(d(u m ), D(w i )) Ω1 + (u m u m, w i ) Ω1 + (K p m, r i ) Ω +γ(u m, p m ; w i, r i ) = (f 1, w i ) Ω1 + (f, r i ) Ω, 1 i m, (u m (), w i ) Ω1 = (π m u, w i ) Ω1. We want to sow te existence of a unique solution to (P m ) and also a uniform bound for te solution. We look for a solution (u m, p m ) of te form m m u m (t)(x) = u m (x, t) = a m j (t)w j(x), p m (t)(x) = p m (x, t) = b m j (t)r j(x). j=1 were we wis to select a m j and b m j so tat (P m ) is satisfied. Wit tese u m and p m, problem (P m ) becomes j=1 1 i m, m d m m m dt am j (w j, w i ) Ω1 + µ a m j (D(w j), D(w i )) Ω1 + a m j am k (w j w k, w i ) Ω1 j=1 j=1 j=1 k=1 6

7 m m + b m j (K r j, r i ) Ω + b m j (r j, w i n 1 ) Γ1 1 m m a m j a m k (w j w k, w i n 1 ) Γ1 j=1 j=1 j=1 j=1 j=1 k=1 + 1 m m a m j G (w j τ 1, w i τ 1 ) Γ1 a m j (w j n 1, r i ) Γ1 = (f 1, w i ) Ω1 + (f, r i ) Ω, m 1 i m, a m j ()(w j, w i ) Ω1 = (π m u, w i ) Ω1. j=1 We rewrite te system in matrix form and define te following mass and stiffness matrices: M = ((w j, w i ) Ω1 ) 1 i,j m, A 1 = (µ(d(w j ), D(w i )) Ω1 ) 1 i,j m, A = ((K r j, r i ) Ω ) 1 i,j m, B = ((r j, w i n 1 ) Γ1 ) 1 i,j m, C = 1 G ((w j τ 1, w i τ 1 ) Γ1 ) 1 i,j m, 1 i m, N i = (α ijk ) 1 j,k m, wit α ijk = (w j w k, w i ) Ω1 1 (w j w k, w i n 1 ) Γ1. We tus obtain a first order nonomogeneous nonlinear system of ordinary differential equations M da dt + (A 1 + C)a + Bb = g 1 (a), A b B T a = g, Ma() = g 3. (3.5) were a = a m 1., b = b m 1., and te rigt and side vectors are (f 1, w 1 ) Ω1 N 1 a a g 1 (a) =. (f 1, w m ) Ω1 N m a a a m m, g = b m m (f, r 1 ) Ω. (f, r m ) Ω, g 3 = (π m u, w 1 ) Ω. (π m u, w m ) Ω. As te w i s are linearly independent, te Gram matrix M is invertible and positive definite. Te matrix A is also invertible as te r i s are linearly independent. Tus we can solve for b in (3.5) as b = A 1 (BT a + g ), substitute tis expression in te first equation and multiply by te inverse of M: { da dt + M1 (A 1 + C + BA 1 BT )a = M 1 (g 1 (a) BA 1 g ), a() = M 1 (3.6) g 3. By Carateodory s teorem [8], tis system as a maximal solution a defined on some interval [, t m ]. We will sow a priori bounds on te solution. Tis will imply tat t m = T. Once te solution a is obtained, we ave a unique solution b = A 1 (BT a + g ). Coosing w i = u m and r i = p m in (P m ) yields, ( u m t, u m) Ω1 + µ(d(u m ), D(u m )) Ω1 + (u m u m, u m ) Ω1 + (K p m, p m ) Ω + γ(u m, p m ; u m, p m ) = (f 1, u m ) Ω1 + (f, p m ) Ω. (3.7) Observe tat (u m u m ) = u m u m + u m u m = u m u m. By Green s teorem ( u m, u m u m ) Ω1 = (u m, (u m u m )) Ω1 + (u m n Ω1, u m u m ) Ω1 7

8 = (u m, u m u m ) Ω1 + (u m n Ω1, u m u m ) Ω1. for all u m V. Terefore as u m = and u m = on Γ 1, From (.1) and (3.7), we obtain (u m, u m u m ) Ω1 = 1 (u m n 1, u m u m ) Γ1. 1 d dt u m L (Ω + µ D(u 1) m) L (Ω + 1) K1/ p m L (Ω + 1 ) G u m τ 1 L (Γ 1) = (f 1, u m ) Ω1 + (f, p m ) Ω. Te terms on te rigt-and side are bounded using Caucy-Scwarz s inequality and te inequalities (3.1)- (3.4) (f 1, u m ) Ω1 + (f, p m ) Ω f 1 L (Ω 1)P 1 u m H1 (Ω 1) + f L (Ω )P p m H1 (Ω ) f 1 L (Ω 1)P 1 C 1 D(u m ) L (Ω 1) + f L (Ω )P 1 λmin K 1/ p m L (Ω ) Terefore, we obtain 1 4µ P 1 C1 f 1 L (Ω + µ D(u 1) m) L (Ω + 1 1) λ min f L (Ω + 1 ) K1/ p m L (Ω. ) 1 d dt u m L (Ω + µ D(u 1) m) L (Ω + 1 1) K1/ p m L (Ω + 1 ) G u m τ 1 L (Γ 1) 1 4µ P 1 C 1 f 1 L (Ω + 1 P 1) f L λ (Ω. (3.8) ) min Multiplying (3.8) by and integrating from to t, we conclude tat wit P u m (t) L (Ω 1) C e, (3.9) C e = ( u L (Ω 1) + 1 µ P 1 C 1 f 1 L (,T ;L (Ω 1)) + P λ min f L (,T ;L (Ω )) )1/. (3.1) Again multiplying (3.8) by and integrating tis time from to T we obtain (u m, p m ) L (,T ;Y ) C e. Tis a priori bound implies existence of a solution to (3.6) on te interval (, T ). We summarize wat we ave so far by te following teorem: Teorem 3.1. Under te assumptions of Lemma.1 tere exists a solution (u m, p m ) W m to te problem (P m ) satisfying sup u m (t) L (Ω 1) + (u m, p m ) L (,T ;Y ) C e, (3.11) t [,T ] were C e is te constant independent of m defined explicitly by (3.1). We now pass to te limit to obtain a solution for te problem (P ). Te sequence {(u m, p m )} m N is bounded in L (, T, W ). Since W is a Hilbert space, it is reflexive and so is L (, T, W ). Hence we can find a subsequence still denoted by {(u m, p m )} m N and a pair (u, p ) L (, T ; W ) suc tat u m u weakly in L (, T ; V ), (3.1) p m p weakly in L (, T ; M ). (3.13) Also by Banac-Alaoglu Teorem [18] since {u m } m N is bounded in L (, T ; L (Ω 1 ) ), tere exists a furter subsequence, still denoted by {u m } m N suc tat for some u L (, T ; L (Ω 1 ) ) u m u in weak topology of L (, T ; L (Ω 1 ) ), (3.14) 8

9 i.e., By (3.1), we ave wic implies (u m (t) u (t), v(t)) Ω1 dt, v L 1 (, T ; L (Ω 1 ) ) L (, T ; L (Ω 1 ) ). (3.15) u m (t) u(t), v(t) Ω1 dt, v L (, T ; V ) L (, T ; L (Ω 1 ) ) (3.16) Terefore comparing (3.15) and (3.17) So (u m (t) u(t), v(t)) Ω1 dt, v L (, T ; L (Ω 1 ) ) (3.17) v L (, T ; L (Ω 1 ) ), (u(t) u (t), v(t)) Ω1. u = u L (, T ; V ) L (, T ; L (Ω 1 ) ). (3.18) To pass to te limit in (P m ) wit te subsequence we extracted consider Ψ : [, T ] R suc tat Ψ(T ) = and Ψ C 1 [, T ]. Multiply te first term in te first equation in (P m ) by Ψ(t) and integrate from to T. Apply integration by parts (u m (t), w j) Ω1 Ψ(t)dt = = (u m (t), w j ) Ω1 Ψ (t)dt + (u m (t), w j ) Ω1 Ψ(t) T (u m (t), w j ) Ω1 Ψ (t)dt (u m (), w j ) Ω1 Ψ(). So te first equation in (P m ) becomes (as u m () = π m u ) + (u m (t), Ψ (t)w j ) Ω1 dt (π m u, w j ) Ω1 Ψ() + µ + (u m (t) u m (t), Ψ(t)w i ) Ω1 dt + (p m (t) 1 (u m(t) u m (t)), Ψ(t)w j n 1 ) Γ1 dt + 1 G (u m (t) n 1, Ψ(t)r i ) Γ1 dt = (D(u m ), Ψ(t)D(w j )) Ω1 dt (K p m (t), Ψ(t) r i ) Ω1 dt (f 1 (t), Ψ(t)w i ) Ω1 dt + (u m (t) τ 1, Ψ(t)w i τ 1 ) Γ1 dt (f (t), Ψ(t)r i )Ψ(t)dt. By (3.14), (3.17), (3.18) and as u m () = π m u u strongly in L (Ω 1 ), letting m, for all j {1,..., m} we can replace u m and p m wit u and p in te linear terms and π m () wit u. For te nonlinear terms and te interface terms observe tat by Sobolev imbeddings for any 1 s <, we can extract anoter subsequence (u m, p m ) suc tat for any 1 s < Observe also tat for any u V and any v, w X we ave u m u strongly in L (, T ; L s (Ω 1 ) ), (3.19) (u v, w) = (u w, v) 9

10 Indeed, (u v, w) = u i v j,i w j dx = u i,i v j w j dx u i v j w j,i dx + u i n i v j w j dx Ω 1 Ω 1 Ω 1 Ω 1 = u i v j w j,i dx + u i n i v j w j dx = u i w j,i v j dx = (u w, v) Ω 1 Γ 1 Ω 1 Hence by (3.1) and (3.19) we ave (u m (t) u m (t), Ψ(t)w i ) Ω1 dt = (u(t) Ψ(t) w i, u(t)) Ω1 dt = (u m (t) Ψ(t) w i, u m (t)) Ω1 dt (u(t) u(t), Ψ(t)w i ) Ω1 dt By te continuity of te trace operator from H 1 (Ω i ) to H 1/ ( Ω i ) in te weak topology we ave u m Ω1 u Ω1 weakly in L (, T ; H 1/ ( Ω 1 )), (3.) p m Ω p Ω weakly in L (, T ; H 1/ ( Ω )). (3.1) Hence again by Sobolev imbeddings after extracting anoter subsequence wic will take care of te interface terms. Finally we ave + (u(t), w j ) Ω1 Ψ (t)dt + (u, w j ) Ω1 Ψ() + µ (K p (t), r j ) Ω1 Ψ(t)dt + u m Ω1 u Ω1 strongly in L (, T ; L 4 ( Ω 1 )), (3.) (u(t) n 1, r j ) Γ1 Ψ(t)dt = (D(u), D(w j )) Ω1 Ψ(t)dt + (p (t) 1 (u(t) u(t)), w j n 1 ) Γ1 Ψ(t)dt + 1 G (f 1 (t), w j ) Ω1 Ψ(t)dt + (u(t) u(t), w i ) Ω1 Ψ(t)dt (u(t) τ 1, w j τ 1 ) Γ1 Ψ(t)dt (f (t), r j )Ψ(t)dt. (3.3) Te second equation in P m is true for u and u as π m u u strongly in L (Ω 1 ), i.e., letting m in (u m (), w j ) = (π m u, w j ) we obtain (u(), w j ) = (u, w j ), j {1,..., m}. (3.3) olds for any v span{(w i, r i )} m. We ave cosen {(w i, r i )} i N to be total in W. So any (v, q) W can be approximated by elements of W m s. Terefore, for any (v, q) W, + (u(t), v) Ω1 Ψ (t)dt + (u, v) Ω1 Ψ() + µ (K p (t), q) Ω1 Ψ(t)dt+ (D(u), D(v)) Ω1 Ψ(t)dt + (p (t) 1 (u(t) u(t)), v n 1) Γ1 Ψ(t)dt+ 1 G (u(t) n 1, q) Γ1 Ψ(t)dt = (f 1 (t), v) Ω1 Ψ(t)dt + (u(t) u(t), v) Ω1 Ψ(t)dt (u(t) τ 1, v τ 1 ) Γ1 Ψ(t)dt (f (t), q) Ω Ψ(t)dt. (3.4) As D(, T ) C 1 [, T ] contains functions wic vanis at bot and T, restricting Ψ to D(, T ) to get rid of te term wit Ψ(), we get (u(t), v) Ω1 Ψ (t)dt + µ (D(u), D(v)) Ω1 Ψ(t)dt + 1 (u(t) u(t), v) Ω1 Ψ(t)dt

11 + (K p (t), q) Ω1 Ψ(t)dt+ (u(t) n 1, q) Γ1 Ψ(t)dt = (p (t) 1 (u(t) u(t)), v n 1) Γ1 Ψ(t)dt+ 1 G (f 1 (t), v) Ω1 Ψ(t)dt + By te definition of weak derivatives, So, for any Ψ D(, T ), (u (t), v) Ω1 Ψ(t)dt + µ (u(t), v) Ω1 Ψ (t)dt = (u (t), v) Ω1 Ψ(t)dt. + (K p (t), q) Ω1 Ψ(t)dt+ Hence for all (v, q) W, (u(t) n 1, q) Γ1 Ψ(t)dt = (D(u), D(v)) Ω1 Ψ(t)dt + (p (t) 1 (u(t) u(t)), v n 1) Γ1 Ψ(t)dt+ 1 G (f 1 (t), v) Ω1 Ψ(t)dt + (u(t) τ 1, v τ 1 ) Γ1 Ψ(t)dt (f (t), q) Ω Ψ(t)dt. (u(t) u(t), v) Ω1 Ψ(t)dt (u(t) τ 1, v τ 1 ) Γ1 Ψ(t)dt (f (t), q) Ω Ψ(t)dt. (u (t), v) Ω1 + µ(d(u), D(v)) Ω1 + (u(t) u(t), v) Ω1 + (K p (t), q) Ω1 + (p (t) 1 (u(t) u(t)), v n 1) Γ1 + 1 G (u(t) τ 1, v τ 1 ) Γ1 (u(t) n 1, q) Γ1 = (f 1 (t), v) Ω1 + (f (t), q) Ω (3.5) in te distributional sense. To see u = u() we multiply tis wit Ψ C 1 [, T ] suc tat Ψ(T ) =. Ten integration by parts yields So, (u (t), v) Ω1 Ψ(t)dt = (u(t), v) Ω1 Ψ (t)dt (u(), v) Ω1 Ψ(). (u(t), v) Ω1 Ψ (t)dt (u(), v) Ω1 Ψ() + µ + (K p (t), q) Ω1 Ψ(t)dt+ (u(t) n 1, q) Γ1 Ψ(t)dt = (D(u), D(v)) Ω1 Ψ(t)dt + (p (t) 1 (u(t) u(t)), v n 1) Γ1 Ψ(t)dt+ 1 G (f 1 (t), v) Ω1 Ψ(t)dt + (u(t) u(t), v) Ω1 Ψ(t)dt (u(t) τ 1, v τ 1 ) Γ1 Ψ(t)dt (f (t), q) Ω Ψ(t)dt. Comparing tis wit (3.4) yields (u, v) Ω1 Ψ() = (u(), v) Ω1 Ψ(). Coosing Ψ() we get (u u(), v) Ω1 =, v V. Terefore letting v = u u() we finally get u = u(). Te following a priori estimate follows trivially; Corollary 3.. Under te same assumptions as in Lemma.1 every solution (u, p ) of (P) satisfies (u, p ) L (,T ;Y ) C e (3.6) were C e is defined by (3.1). 11

12 Now we will sow tat te solution for (P ) is unique. For tat purpose assume tat (u, p ) and (ũ, p ) are two solutions. Let w = u ũ and r = p p. Ten (w, q) L (, T ; W ) satisfies ( w + µ(d(w), D(v)) Ω1 + (w u, v) Ω1 + (ũ w, v) Ω1 + (K r, q) Ω + (r, v n 1 ) Γ1 + 1 G (w τ 1, v τ 1 ) (w n 1, q) Γ1 1 (w u, v n 1) Γ1 1 (ũ w, v n 1) Γ1 = Coose v = w and q = r, 1 d dt w L (Ω + 1) µ D(w) L (Ω + 1) K1/ r L (Ω + 1 ) G w n 1 L (Γ + (w u, w) 1) Ω 1 Observing +(ũ w, w) Ω1 1 (w u, w n 1) Γ1 1 (ũ w, w n 1) Γ1 (ũ w, w) Ω1 = ( ũ, w w) Ω1 (ũ w, w) Ω1 + (ũ n Ω1, w w) Ω1 we ave = (ũ w, w) Ω1 + (ũ n 1, w w) Γ1 (ũ w, w) Ω1 = 1 (ũ n 1, w w) Γ1 So te equation becomes 1 d dt w L (Ω 1) +µ D(w) L (Ω 1) + K1/ r L (Ω (w u, w) ) Ω 1 1 ((w w, ũ n 1) Γ1 (w (u+ũ), w n 1 ) Γ1 ) Te rigt and side can be bounded by te virtue of (3.1), (3.13), (3.17) and (3.6) by Tus we ave w L 4 (Ω 1) u L (Ω 1) + 1 w L 4 (Γ 1) ( u L (Γ 1) + ũ L (Γ 1)) C1 D(w) 3 ( L (Ω C 1) 4 D(u) L (Ω 1) + 1 C 4(C D(u) L (Ω 1) + C D(ũ) ) L (Ω 1)) C e C1 3 ( C µ C C4 ) D(w) L (Ω 1) 1 d dt w L (Ω + (µ C e ( 1) C3 1 C µ C C4 )) D(w) L (Ω + 1) K1/ r L (Ω. ) Since w() = multiplying by and taking te integral from to T we get w(t ) L (Ω 1) + (µ C3 1 C e µ ( C C C 4 )) D(w) L (,T ;L (Ω 1)) + K1/ r L (,T ;L (Ω )). So under te condition (µ) 3/ > C 3 1C e ( C C C 4), we ave (w, r) = (, ). Now we will sow te existence of te pressure p 1 in te distributional sense. We follow te argument in [3] and define U(t) = t u(s)ds, F 1 (t) = t f 1 (s)ds, β(t) = t u(s) u(s)ds. 1

13 Ten U, F 1, β C(, T ; V ). Integrating (P ) between and t, coosing v V wit v = on Γ 1 and q = yields t (, T ), µ(d(u(t)), D(v)) Ω1 = (u() u(t) β(t) + F 1 (t), v) Ω1. So for all t [, T ] tere exists a P 1 (t) L (Ω 1 ) suc tat t (, T ), u(t) u() µ D(U(t)) + β(t) + P 1 (t) = F 1 (t). (3.7) Since te gradient operator is an isomorpism from L (Ω 1 ) \ R into H 1 (Ω 1 ), we conclude tat P 1 belongs to C([, T ]; H 1 (Ω 1 )) and tus P 1 C([, T ]; L (Ω 1 )). We now differentiate (3.7) in te distributional sense in Ω 1 (, T ) and we obtain u t µ D(u) + u u + p 1 = f 1 wit p 1 = P 1 t. Wat we acieved in tis section can be stated as follows; Teorem 3.3. Let u V and suppose tat te assumptions of Lemma.1 olds. If in addition we assume tat (µ) 3/ > C 3 1C e ( C C C 4), ten te problem (P) as a unique solution (u, p ) (L (, T ; V ) H 1 (, T ; L (Ω 1 ) ) L (, T ; M ) suc tat (u, p ) L (,T ;Y ) C e, (3.8) wit te constant defined in Teorem 3.1. Moreover, tere exists p 1 L (, T ; L (Ω 1 )) suc tat (u, p 1, p ) is a solution to te problem (Q). 4 Numerical Sceme We discretize te coupled problem by a finite element metod in space and a Crank-Nicolson sceme in time. Let X X, M 1 M 1 and M M be finite element spaces to be specified later. We regroup all te linear terms involving u and p by defining a bilinear form B B([u, p ]; [v, q]) = µ(d(u), D(v)) Ω1 +(K p, q) Ω +(p, v n 1 ) Γ1 (u n 1, q) Γ1 + 1 G (u τ 1, v τ 1 ) Γ1. Clearly B is bilinear and so it is bounded since we are in finite dimension. We also ave (4.1) B([v, q]; [v, q]) = µ D(v) L (Ω + 1) K1/ q L (Ω + 1 ) G v τ 1 L (Γ 1). (4.) Te nonlinear reaction term u u and te nonlinear term in γ are discretized using te form N N(u; w, v) = 1 (u w, v) Ω 1 1 (u v, w) Ω (u v, w n 1) Γ1 1 (u w, v n 1) Γ1. Ten, te form N is linear wit respect to all tree arguments, and N satisfies te following property: N(u; v, v) =. (4.3) 13

14 Lemma 4.1. wit u, w, v X, N(u; w, v) C N u L (Ω 1) v L (Ω 1) w L (Ω 1), (4.4) C N = C 4 + C 4 C. Proof. Using Hölder s inequality, we ave: N(u; w, v) 1 u L 4 (Ω 1) w L (Ω 1) v L 4 (Ω 1) + 1 u L 4 (Ω 1) v L (Ω 1) w L 4 (Ω 1) + 1 u L 4 (Γ 1) w L4 (Γ 1) v n 1 L (Γ 1) + 1 u L 4 (Γ 1) v L4 (Γ 1) w n 1 L (Γ 1). By te bounds (3.1), (3.) we obtain N(u; w, v) 1 C 4 u L (Ω 1) w L (Ω 1) v L (Ω 1) + 1 C 4 u L (Ω 1) v L (Ω 1) w L (Ω 1) + 1 C 4 C u L (Ω 1) w L (Ω 1) v L (Ω 1) + 1 C 4 C u L (Ω 1) v L (Ω 1) w L (Ω 1) C N u L (Ω 1) v L (Ω 1) w L (Ω 1). Let N T > be te number of time steps, let t 1 be te first timestep and let = (T t 1 )/(N T 1) and let t i = t 1 + (i 1) for i. We use te standard notation φ i+1/ = φi+1 + φ i, for a sequence {φ i } or a function φ i = φ(t i ). We propose te following sceme: Find {u i } i in X, {p i 1 } i 1 M 1 and {p i } i 1 in M suc tat v X, (u, v) = (u(), v), (4.5) v X, q M, ( u1 u t 1, v) Ω1 + B([u 1, p1 ]; [v, q]) (p1 1, v) Ω 1 +N(u 1 ; u1, v) = (f 1 1, v) Ω 1 + (f 1, q) Ω, (4.6) i 1, v X, q M, ( ui+1 +N(u i+1/ u i, v) Ω1 + B([u i+1/, p i+1/ ]; [v, q]) (p i+1/ 1, v) Ω1 ; u i+1/, v) = (f i+1/ 1, v) Ω1 + (f i+1/, q) Ω, (4.7) i, q M 1, ( u i+1, q) Ω 1 =. (4.8) Equation (4.5) represents te initial condition wereas equation (4.6) computes te solution at te first time step using a first order backward Euler sceme. We will coose t 1 small enoug so tat te resulting sceme is of second order. Equation (4.7) defines te Crank-Nicolson sceme. Finally te incompressibility condition is enforced discretely at eac time step by equation (4.8). Let us now prove existence of te numerical solution. As in te continuous problem, we restrict te discrete problem (4.5)-(4.8) to te space of discretely divergent-free velocities: V = {v X : q M 1, (q, v) Ω1 = }. 14

15 We sow existence of {u i } i V, {p i } i 1 M satisfying (4.5) and v V, q M, ( u1 u t 1, v) Ω1 + B([u 1, p1 ]; [v, q]) + N(u1 ; u1, v) = (f 1 1, v) Ω 1 + (f 1, q) Ω, (4.9) i 1, v V, q M, ( ui+1 u i, v) Ω1 + B([u i+1/, p i+1/ ]; [v, q]) + N(u i+1/ ; u i+1/, v) = (f i+1/ 1, v) Ω1 + (f i+1/, q) Ω. (4.1) Clearly u is uniquely defined. We first give te proof for existence of (ui, pi ) i. Te proof of existence of (u 1, p1 ) is simpler and outlined at te end. We assume tat ui and pi are given for some i 1. We sow tat te solution (u i+1, p i+1 ) satisfying (4.1) exists using a corollary of Brouwer s fixed point teorem. We can modify te argument to sow existence of (u 1, p1 ). We introduce a mapping F i : V M V M defined by (v, q) V M, (F i (z, t), (v, q)) Y = ( (z ui ), v) Ω1 + B([z, t]; [v, q]) + N(z; z, v) (f i+1/ 1, v) Ω1 (f i+1/, q) Ω. (4.11) So F i is a well-defined map from V M into itself by te Riesz representation teorem. Te mapping F i is also continuous. Furtermore if (z, t ) is a zero of F i, ten (z u i, t p i ) is a solution to (4.1). Compute (F i (z, t), (z, t)) Y = ( (z ui ), z) Ω1 + µ D(z) L (Ω + 1) K1/ t L (Ω + 1 ) G z τ 1 L (Γ 1) (f i+1/ 1, z) Ω1 (f i+1/, t) Ω 1 z L (Ω 1 1) ui L (Ω + 1) µ D(z) L (Ω + 1) K1/ t L (Ω ) Using te bound (3), we ave (f i+1/ 1, z) Ω1 (f i+1/, t) Ω. (F i (z, t), (z, t)) Y 1 (z, t) Y 1 4µ P 1 C1 f i+1/ 1 L (Ω 1 1) P f i+1/ L λ (Ω 1 ) min ui L (Ω. 1) We conclude tat (F i (z, t), (z, t)) Y for all (z, t) suc tat (z, t) Y = R i, wit te radius R i defined by R i = ( 1 µ P 1 C i+1/ 1 f 1 L (Ω + P 1) f i+1/ L λ (Ω + 1 1/. ) min ui L (Ω 1)) Tis implies tat tere is a zero of F i denoted by (u i+1, pi+1 ). Tis zero is a solution to (4.1). To sow tat (u 1, p1 ) exists, we follow a similar argument wit te mapping F 1 : V M V M defined by (v, q) V M, (F 1 (z, t), (v, q)) Y = ( z u t 1, v) Ω1 + B([z, t]; [v, q]) + N(z; z, v) Tis yields a solution in te ball of radius R 1. (f 1 1, v) Ω 1 (f 1, q) Ω. (4.1) (u 1, p 1 ) Y R 1, 15

16 wit R 1 = ( 1 µ P 1 C1 f 1 1 L (Ω + P 1) f 1 L λ (Ω + 1 ) 1/. ) min t 1 u() L (Ω 1) (4.13) Coosing (v, q) = (u 1, p1 ) in (4.9) and (v, q) = (ui+1/, p i+1/ ) in (4.1) yields a priori bounds on te solution. We skip te proof as it follows te argument above. Lemma 4.. If {(u i, pi )} i 1 is a solution of (4.9)-(4.1), it satisfies u 1 L (Ω 1) + t1 (u 1, p1 ) Y u() L (Ω 1) + 1 µ P 1 C 1 t1 f 1 1 L (Ω 1) + P λ min t 1 f 1 L (Ω ), (4.14) m N T, + 1 NT 1 µ P 1 C 1 u m L (Ω + 1) f i+1/ 1 L (Ω + P N T 1 1) λ min (u i+1/, p i+1/ ) Y u 1 L (Ω 1) f i+1/ L (Ω ). (4.15) Te next result states uniqueness of te solution under some condition on te data and on te time step. Lemma 4.3. Let R 1 be defined by (4.13). Under te following conditions wit R = ( t 1 R N µ P T 1 1 C1 (µ) 3/ > C 3 1C N max(r 1, R), f i+1/ 1 L (Ω + P N T 1 1) λ min tere exists a unique solution {(u i, pi )} i 1 satisfying (4.9)-(4.1). f i+1/ L (Ω )) 1/. (4.16) Proof. First, we sow uniqueness of (u 1, p1 ). Assume tat tere are two solutions say ({u1 }, {p1 }) and ({ũ 1 }, { p 1 }). Let w1 = u 1 ũ1 and r 1 = p 1 p1. From (4.9), we ave Equivalently, v X, q M, ( w1 t 1, v) Ω 1 + B([w 1, r 1 ]; [v, q]) + N(u 1 ; u 1, v) N(ũ 1 ; ũ 1, v) =. v X, q M, ( w1 t 1, v) Ω 1 + B([w 1, r 1 ]; [v, q]) + N(w 1 ; u 1, v) + N(ũ 1 ; w 1, v) =. Coosing v = w 1 and q = r 1 yields 1 t 1 w1 L (Ω 1) + (w1, r 1 ) Y N(w 1 ; u 1, w 1 ) C 3 1 C N D(w 1 ) L (Ω 1) D(u1 ) L (Ω 1). From (4.14) and te definition (4.13), we ave Terefore, we obtain D(u 1 ) L (Ω 1) R 1 (µ) 1/. 1 t 1 w1 L (Ω 1) + ( µ C3 1 C NR 1 (µ) 1/ ) D(w 1 ) L (Ω 1) + K1/ r 1 L (Ω ). 16

17 Tis means tat w 1 and r 1 are zero if te following condition is satisfied (µ) 3/ > C 3 1C N R 1. Next, we fix i 1 and sow uniqueness of (u i+1, pi+1 ). We assume tat (ui, pi ) exists and is unique. As above, we take te difference of two solutions w i+1 = u i+1 ũ i+1 and r i+1 = p i+1 pi+1. Ten from (4.1) we ave v V, q M, ( wi+1, v) + B([wi+1/, r i+1/ ]; [v, q]) + N(u i+1/ Adding and subtracting te term N(u i+1/ ; ũ i+1/, v) yields: ; u i+1/, v) N(ũ i+1/ ; ũ i+1/, v) =. 1 (wi+1, v) + B([w i+1/, r i+1/ ]; [v, q]) + N(u i+1/ ; w i+1/, v) + N(w i+1/ ; ũ i+1/, v) =. Letting v = w i+1/ and q = r i+1/ and using te fact tat N(u i+1/ ; w i+1/, w i+1/ ) vanises, we are left wit From (4.15) we ave 1 wi+1 + µ D(w i+1/ ) L (Ω + 1) K1/ r i+1/ L (Ω ) C N w i+1/ L (Ω 1) ũi+1/ L (Ω 1) C N C1 3 D(wi+1/ ) L (Ω 1) D(ũi+1/ ) L (Ω 1). D(u i+1/ ) L (Ω 1) R (µ) 1/, wit R defined by (4.16). Terefore if we assume tat (µ) 3/ C N C 3 1 R >, te functions wi+1/ and r i+1/ vanis. Since in fact w i+1/ = w i+1 / and r i+1/ = r i+1 /, we can conclude. We assume tat te spaces X and M 1 are conforming, i.e. tey satisfy an inf-sup condition wit β > independent of. ( v, q) Ω1 inf sup β. (4.17) q M 1 D(v) L (Ω 1) q L (Ω 1) v V A straigtforward consequence is te existence and uniqueness of te Navier-Stokes pressure p i 1 for all i 1. 5 Error analysis Before we prove some error estimates, we sow tat te proposed sceme is consistent. Lemma 5.1. Te weak solution (u, p 1, p ) of (Q) also satisfies v X, q M, ( u + B([u, p ]; [v, q]) (p 1, v) Ω1 + N(u; u, v) = (f 1, v) Ω1 + (f, q) Ω. Proof. It suffices to ceck tat But N(u; u, v) = (u u, v) Ω1 1 (u u, v n 1) Γ1. N(u; u, v) = 1 (u u, v) Ω 1 1 (u v, u) Ω (u v, u n 1) Γ1 1 (u u, v n 1) Γ1. Tis is enoug to conclude since te second and tird term are equal to te first term by using integration by parts and te fact tat u =. (5.1) 17

18 We decompose te errors into approximation and numerical errors. For any t, let ũ(t) X be an approximation of u satisfying ( (u(t) u(t)), q) Ω1 = for any q in M 1. Existence of suc an approximation can be found for instance in [13]. Also let p 1 M 1 and p M be approximations of p 1 and p, respectively. We take p 1 to be te L -projection of p 1, tat is, (p 1 p 1, q) Ω1 = for all q M 1 and we take p to be te Lagrange interpolant. We furter assume tat te approximation errors are optimal, i.e., for any t and for some positive integers k 1, k D(u(t)) D(ũ(t)) L (Ω 1) C k1 u(t) H k 1 +1 (Ω 1), u L (, T ; H k1+1 (Ω 1 ) ) L (, T ; X), (5.) i =, 1, i p 1 (t) i p 1 (t) L (Ω 1) C k1i p 1 (t) H k 1 (Ω1), p 1 L (, T ; H k1 (Ω 1 )) L (, T ; M 1 ), (5.3) i =, 1, i p (t) i p (t) L (Ω 1) C k+1i p (t) H k +1 (Ω ), p L (, T ; H k+1 (Ω 1 )) L (, T ; M ). (5.4) By te virtue of triangle inequality and (5.) we ave te following stability condition D(ũ(t)) L (Ω 1) D(u(t)) D(ũ(t)) L (Ω 1) + D(u(t)) L (Ω 1) C a u(t) H1 (Ω 1), t. (5.5) were C a > is a constant independent of. Let us give examples of conforming spaces tat satisfy te above assumptions [3, 15]. Let T = T 1 T be te union of regular triangulations of te subdomains Ω 1 and Ω suc tat te meses matc at te interface Γ 1. For any element T in T, let P k (T ) denote te space of polynomials of degree less tan or equal to k and defined on T. Te space M is cosen to be te usual continuous finite element space of piecewise polynomials of degree k on eac mes element. We give below two examples of Navier-Stokes velocity and pressure spaces. Example 5.. P P 1 Taylor-Hood spaces wit continuous piecewise quadratic functions in velocity space and continuous piecewise linear functions in pressure space, i.e., X = {v C (Ω 1 ) : v T P (T ) T T 1 } X, M 1 = {q C (Ω 1 ) : q T P 1 (T ) T T 1 } M 1. In tis case, te estimates (5.) and (5.3) are satisfied wit k 1 =. Example 5.3. MINI element wit continuous piecewise linears wit bubbles for velocity and continuous piecewise linears for pressure space. Let B 1 (T ) = span{λ 1 λ λ 3 } were λ i P 1 (T ) wit λ i (x j ) = δ ij for eac vertex x j of T T. X = {v C (Ω 1 ) : v T (P 1 (T ) B 1 (T )) T T 1 } X, M 1 = {q C (Ω 1 ) : q T P 1 (T ) T T 1 } M 1. In tat case, te estimates (5.) and (5.3) are satisfied wit k 1 = 1. Next we write u u = χ η, χ = u ũ, η = u ũ, p p = ξ ζ, ξ = p p, ζ = p p. Te following teorem states error bounds of te quantities χ and ξ. Teorem 5.4. Let u ttt L (, T ; X). Assume tat te following condition olds µ 3/ C NC1 3 max(r, R 1 ). 18

19 Tere exists a constant C independent of, t 1, and µ suc tat and for any m 1 χ 1 L (Ω 1) + µt1 D(χ 1 ) L (Ω 1) + t1 K 1/ ξ 1 L (Ω ) χ L (Ω 1) + C(1 + µ + µ1 + R 1 + C e µ )t 1 D(η 1 ) L (Ω 1) + C(1 + µ1 )t 1 ζ 1 H 1 (Ω ) +Cµ 1 t 1 p 1 1 p 1 1 L (Ω ) + Cµ1 t 3 u tt ( t ) L (Ω 1) + Cµ1 1 t 1 η1 η L (Ω 1), χ m L (Ω + µ 1) D(χ i+1/ ) L (Ω + 1) K 1/ ξ i+1/ L (Ω ) χ 1 L (Ω + C(1 + µ + 1) µ1 + R + Ce µ ) D(η i+1/ ) L (Ω + C(1 + 1) µ1 ) ζ i+1/ H 1 (Ω ) +Cµ 1 p i+1/ 1 p i+1/ 1 L (Ω + ) Cµ1 5 ( u t ( t i ) 4 L (Ω + u 1) ttt(ť i ) L (Ω ) 1) +Cµ 1 1 η i+1 η i L (Ω 1). Proof. First, we take te average of (5.1) at times t = t i and t = t i+1 : v X, q M, (u i+1/ t, v) Ω1 + B([u i+1/, p i+1/ ]; [v, q]) + 1 (N(ui+1 ; u i+1, v) + N(u i ; u i, v)) From (5.6) and (4.7), we ave for any i 1: ( χi+1 χ i (p i+1/ 1, v) = (f i+1/ 1, v) Ω1 + (f i+1/, q) Ω. (5.6) +N(u i+1/ ) Ω1 + B([χ i+1/, ξ i+1/ ], [v, q]) (p i+1/ 1, v) Ω1, u i+1/, v) = (u i+1/ t (ũi+1 ũ i, v) Ω1, v) Ω1 +B([η i+1/, ζ i+1/ ], [v, q]) (p i+1/ 1, v) Ω1 + 1 N(ui+1, u i+1, v) + 1 N(ui, u i, v). (5.7) Coose v = χ i+1/ and q = ξ i+1/ in (5.7). Ten 1 ( χi+1 L (Ω 1) χi L (Ω 1) ) + µ D(χi+1/ ) L (Ω 1) + K1/ ξ i+1/ L (Ω ) + 1 G χi+1/ τ 1 L (Γ 1) N(u i+1/, u i+1/, χ i+1/ ) 1 N(ui+1, u i+1, χ i+1/ ) 1 N(ui, u i, χ i+1/ ) + (u i+1/ t Let us first consider te nonlinear term, χ i+1/ (ũi+1 ũ i ) Ω1, χ i+1/ ) Ω1 + B([η i+1/, ζ i+1/ ], [χ i+1/, ξ i+1/ ]) + (p i+1/ 1 p i+1/ 1, χ i+1/ ) Ω1. (5.8) N = N(u i+1/, u i+1/, χ i+1/ ) 1 N(ui+1, u i+1, χ i+1/ ) 1 N(ui, u i, χ i+1/ ). 19

20 After adding and subtracting N(u i+1/ ; u i+1/, χ i+1/ ), we obtain N = N((u u) i+1/ ; u i+1/, χ i+1/ )+N(u i+1/ ; u i+1/, χ i+1/ ) 1 N(ui+1 ; u i+1, χ i+1/ ) 1 N(ui ; u i, χ i+1/ ). Next, separating te approximation error and discrete error and adding and subtracting N(u i+1/, u i+1/, χ i+1/ ) we ave N = N(χ i+1/ ; u i+1/, χ i+1/ ) N(η i+1/ ; u i+1/, χ i+1/ ) + N(u i+1/ ; u i+1/ u i+1/, χ i+1/ ) +N(u i+1/, u i+1/, χ i+1/ ) 1 N(ui+1 ; u i+1, χ i+1/ ) 1 N(ui ; u i, χ i+1/ ). Again, using te decomposition of u u in te tird term gives: N = N(χ i+1/ ; u i+1/, χ i+1/ ) N(η i+1/ ; u i+1/, χ i+1/ ) + N(u i+1/ ; χ i+1/, χ i+1/ ) N(u i+1/ ; η i+1/, χ i+1/ ) +N(u i+1/, u i+1/, χ i+1/ ) 1 N(ui+1 ; u i+1, χ i+1/ ) 1 N(ui ; u i, χ i+1/ ). From (4.3), te term N(u i+1/ ; χ i+1/, χ 1+1/ ) vanises. We next rewrite te last tree terms. N(u i+1/, u i+1/, χ i+1/ ) 1 N(ui+1 ; u i+1, χ i+1/ ) 1 N(ui ; u i, χ i+1/ ) = 1 4 N(ui ; u i+1 u i, χ i+1/ ) 1 4 N(ui+1 ; u i+1 u i, χ i+1/ ) = 1 4 N(ui+1 u i ; u i+1 u i, χ i+1/ ). Terefore we ave N = N(χ i+1/ ; u i+1/, χ i+1/ ) N(η i+1/ ; u i+1/, χ i+1/ ) N(u i+1/ ; η i+1/, χ i+1/ ) 1 4 N(ui+1 u i ; u i+1 u i, χ i+1/ ). Using te bounds (4.4), (3.1) and (4.15) wit te definition (4.16), we ave and N(χ i+1/ ; u i+1/, χ i+1/ ) C N C1 3 D(χi+1/ ) L (Ω 1) D(ui+1/ ) L (Ω 1) R (µ) C NC 3 1/ 1 D(χi+1/ ) L (Ω, 1) N(η i+1/ ; u i+1/, χ i+1/ ) R (µ) 1/ C NC 1 D(χ i+1/ ) L (Ω 1) η i+1/ L (Ω 1). Similarly using (4.4), (3.1) and (3.8), we obtain N(u i+1/ ; η i+1/, χ i+1/ ) Finally we ave for some t i (t i, t i+1 ): C e (µ) 1/ C NC 1 D(χ i+1/ ) L (Ω 1) η i+1/ L (Ω 1). 1 4 N(ui+1 u i ; u i+1 u i, χ i+1/ ) 1 4 C NC 1 D(χ i+1/ ) L (Ω 1) (u i+1 u i ) L (Ω 1) 1 4 C NC 1 D(χ i+1/ ) L (Ω 1) u t ( t i ) L (Ω 1). Terefore by Young s inequality, we obtain for any δ > : N RC NC 3 1 (µ) 1/ D(χi+1/ ) L (Ω 1) + µδ D(χ i+1/ ) L (Ω 1) + C µ δ (R + C e ) ηi+1/ L (Ω 1) + C µδ 4 u t ( t i ) 4 L (Ω 1).

21 Next, we consider te terms D = (u i+1/ t, χ i+1/ ) Ω1 (ũi+1 ũ i, χ i+1/ ) Ω1 = (u i+1/ t From a Taylor expansion, we ave for some t i 1, ti (ti, t i+1 ) u i+1/ t ui+1 u i Ten by (3.1) and Young s inequality, for any δ 1 >, we obtain D µδ 1 D(χ i+1/ ) L (Ω 1) + C µδ 1 ( 4 ui+1 u i = u ttt (t i 1) 8 u ttt(t i ) 4. θ=1, χ i+1/ ) Ω1 + ( ηi+1 η i, χ i+1/ ) Ω1. u ttt (t i θ ) L (Ω 1) + 1 ηi+1 η i L (Ω 1) ). Using (3.1)-(3.4), we bound te linear term B(, ) for any positive numbers δ, δ 3 B([η i+1/, ζ i+1/ ], [χ i+1/, ξ i+1/ ]) µ D(η i+1/ ) L (Ω 1) D(χ i+1/ ) L (Ω 1) + K 1/ ζ i+1/ L (Ω ) K 1/ ξ i+1/ L (Ω ) + C C C 1 D(χ i+1/ ) L (Ω 1) ζ i+1/ H1 (Ω ) + C C K 1/ ξ i+1/ L (Ω λmin ) η i+1/ H 1 (Ω 1) + 1 G ηi+1/ τ 1 L (Γ 1) χ i+1/ τ 1 L (Γ 1) δ (µ) D(χ i+1/ ) L (Ω 1) + δ 3 K 1/ ξ i+1/ L (Ω ) + 1 G χi+1/ τ 1 L (Γ 1) +C(1 + 1 δ 3 + µ δ ) D(η i+1/ ) L (Ω 1) + C( 1 µδ + 1 δ 3 ) ζ i+1/ H 1 (Ω ). Since ( (u i+1/ u i+1/ ), q) Ω1 = for all q M 1, we can rewrite te pressure term as (p i+1/ 1 p i+1/ 1, χ i+1/ ) Ω1 = (p i+1/ 1 p i+1/ 1, χ i+1/ ) Ω1 ( p i+1/ 1 p i+1/ 1, χ i+1/ ) Ω1. Te second term vanises because of te property of te interpolant. Te first term is bounded by Young s inequality for any δ 4 > : (p i+1/ 1 p i+1/ 1, χ i+1/ ) Ω1 µδ 4 D(χ i+1/ ) L (Ω 1) + C µδ 4 p i+1/ 1 p i+1/ 1 L (Ω ). We combine te bounds above wit (5.8) and coose δ = δ 1 = δ = δ 4 = 3/16 and δ 3 = 1/. We multiply te resulting inequality by and sum from i = 1 to i = m 1 for some m 1. If te following condition is satisfied µ 3/ C NC 3 1 R ten we obtain: χ m L (Ω + µ 1) D(χ i+1/ ) L (Ω + 1) K 1/ ξ i+1/ L (Ω ) χ 1 L (Ω + C(1 + µ + 1) µ1 + R + Ce µ ) D(η i+1/ ) L (Ω + C(1 + 1) µ1 ) ζ i+1/ H 1 (Ω ) +Cµ 1 p i+1/ 1 p i+1/ 1 L (Ω + ) Cµ1 5 ( u t ( t i ) 4 L (Ω + u 1) ttt(ť i ) L (Ω ) 1) +Cµ 1 1 η i+1 η i L (Ω 1). 1

22 It remains to find a bound for χ 1 L (Ω. 1) For tis, we consider te equation (4.6) and follow a similar derivation as above. We skip te details. Assume tat Ten, we can prove µ 3/ C NC 3 1 R 1. χ 1 L (Ω 1) + µt1 D(χ 1 ) L (Ω 1) + t1 K 1/ ξ 1 L (Ω ) χ L (Ω 1) + C(1 + µ + µ1 + R 1 + C e µ )t 1 D(η 1 ) L (Ω 1) + C(1 + µ1 )t 1 ζ 1 H 1 (Ω ) +Cµ 1 t 1 p 1 1 p 1 1 L (Ω ) + Cµ1 t 3 u tt ( t ) L (Ω 1) + Cµ1 1 t 1 η1 η L (Ω 1). A straigtforward corollary is te following result. Teorem 5.5. In addition to te assumptions of Teorem 5.4 assume tat t 1. Ten tere exists a constant C independent of, t 1 and but dependent on µ suc tat for any m χ 1 L (Ω + 1) χm L (Ω + 1) µt1 D(χ 1 ) L (Ω + µ 1) +t 1 K 1/ ξ 1 L (Ω + ) D(χ i+1/ ) L (Ω 1) K 1/ ξ i+1/ L (Ω ) C(k1 + k + 4 ). Remark 5.6. From te analysis above, it is easy to derive te error estimates for a backward Euler time discretization at eac time step. Te resulting metod is ten first order in time. Anoter extension of tis work is to consider non-omogeneous boundary conditions for te Darcy pressure as in [7]. For instance, assume tat p = g D on Γ D wit g D H 1/ (Γ D). It suffices to consider a lift of te g D inside Ω 1, say p D and te weak solution becomes (u, p 1, ϕ ) wit ϕ = p + p D and wit (u, p 1, p ) satifying te problem (Q). 6 Conclusions We formulate a weak problem of te coupling between time-dependent Navier-Stokes and Darcy equations and proved its well-posedness. We approximate te weak solution by a continuous finite element solution. Uniqueness of te solution is obtained under a condition on te data. We sow tat te sceme is optimal in space and second order in time. References [1] R. Adams. Sobolev Spaces. Academic Press, New-York, [] T. Arbogast and D. Brunson. A computational metod for approximating a Darcy-Stokes system governing a vuggy porous medium. Computational Geosciences, 11(3):7 18, 7. [3] D.N. Arnold, F. Brezzi, and M. Fortin. A stable finite element for te Stokes equations. Calcolo, 1: , 1984.

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