An explicit nite element method for convection-dominated compressible viscous Stokes system with inow boundary

Journal of Computational and Applied Mathematics 156 (2003) 319 343 www.elsevier.com/locate/cam An explicit nite element method for convection-dominated compressible viscous Stokes system with inow boundary Jae Ryong weon a;1, Philsu im b; ;2 a Department of Mathematics, Pohang University of Science and Technology, Pohang 790-784, South orea b Major in Mathematics, Dong-A University, 840, Hadan-2 Dong, Saha-u, Pusan 604-714, South orea Received 28 May 2002; received in revised form 30 October 2002 Abstract A linearized steady-state compressible viscous Stokes system with inow boundary is considered on a plane domain. An explicit nite element method for the system is presented with convection-dominance and O(h) viscous numbers where h is a given mesh size. With small viscous numbers it is a degenerate hyperbolic problem and also the Dirichlet boundary condition may generate a layer near the outowboundary. The method is applied over a triangulation of the domain in an explicit fashion from triangle to triangle and gives a continuous nth degree piecewise polynomial approximation. We show a local stability and a global one for the method and derive error estimates for each variable of velocity, pressure and their derivatives. It is observed that the compressibility number := = is an essential ingredient in showing our stability results. c 2003 Elsevier B.V. All rights reserved. eywords: Convection-dominance; Compressible viscous Stokes ow 1. Introduction and results In this paper we consider a steady-state compressible viscous Stokes system which can be obtained by linearizing the stationary compressible viscous Navier Stokes equations around an ambient ow [5,6]. The system contains not only convection terms in the momentum equation and the continuity one, respectively but also the viscous terms in the momentum one. With convection dominance and small viscous numbers we are interested in solving the corresponding discrete problem explicitly. Corresponding author. E-mail addresses: kweon@postech.ac.kr (J.R. weon), kimps@donga.ac.kr (P. im). 1 Partially supported by Com 2 MaC-OSEF. 2 This paper was supported by orea Research Foundation grand RF-2002-003-C00010. 0377-0427/03/$ - see front matter c 2003 Elsevier B.V. All rights reserved. doi:10.1016/s0377-0427(02)00919-6

320 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 The considered equations in this paper are u div u + U u + p = f in ; div u + U p = g in ; u = u 0 on ; p = p 0 on ; (1.1) where is a domain in R 2 with boundary := 9; u =(u 1 ;u 2 ) and p are the unknown velocity and pressure variables, respectively; = (P) is a strictly increasing positive function that provides density as a C 1 function of pressure; = (P) := (P)=(P), =d=dp; U := (U; V ) and P are given functions that describe the ambient ow; u 0 and p 0 are given boundary data; f and g are given functions; and are the viscous constants with 0 and + 0. For simplicity we let := U and denote by w = w. Throughout this paper it is assumed that is a unit vector function and, are positive constant functions, for simplicity. The inow and outowboundaries, and + are dened by = {(x; y) : n 0}; + = {(x; y) : n 0}; (1.2) where n =(n 1 ;n 2 ) denotes the unit outward normal to. The nonzero ambient velocity eld appearing in the continuity equation of (1.1) gives the characteristic direction for the equation (see (1.7)), and values of the pressure are specied on those portions of the boundary where the ambient velocity vector points into the region. As an illustration a jet of liquid owing into a region can be such a phenomenon [1]. Also it is seen that the system (1.1) is an elliptic hyperbolic system. When the viscous numbers and are zero, it is a linearized compressible Euler system while it is regarded a degenerate hyperbolic system when the viscous numbers become small. For the existence of solution of (1.1) on smooth domains we refer to [5, Theorem 2.1] and [6, Lemma 3.2 3.3]. The uniqueness and existence results for (1.1) are given there, and its regularity results are shown under condition that guarantee that the ambient ow is close to a constant ow. Recently in [7] an analysis for a very simple form of the system (1.1) was given on (concave or convex) polygonal domains. So far the explicit nite element method with respect to convection has been applied to several types of PDEs. Especially, Falk and Richter [3] applied the method to the rst-order scalar hyperbolic equation, and Richter [8] considered it for convection-diusion equations. In addition, Falk and Richter [4] applied to the linear symmetric and hyperbolic system in the time and space variable the method by a generalization of the discontinuous Galerkin method, and showed that the method is explicit in time variable. More specically it is seen in [3,8] that the nite element method is applied over a triangulation of the domain and can be developed in an explicit fashion from triangle to triangle and gives a continuous piecewise polynomial approximation. In doing so, the following method was applied to the convection-diusion equation with the Dirichlet boundary condition [8]: ( u h + u h ;v h ) =(f; v h ) ; v h P n &() (); (1.3)

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 321 where ( ; ) denotes the L 2 inner product on, &() is the number of inowsides that has, P n () is the space of polynomials of degree 6 n on. There the method was shown to be stable under the condition h 1 q 0 (q 0 0) and the assumption that the triangles can be ordered explicitly with respect to the convection term, and then the following estimates were derived: u u h L 2 ( ) 6 Ch n u H n+1 ( ); (u u h ) L 2 ( ) 6 Ch n 1=2 u H n+1 ( ); where is any sub-triangulation for which ( ) (). One may wonder if a similar stability result for (1.1) can be obtained with the method like (1.3). So the purpose of this paper is to present an explicit nite element method for (1.1) and showa unique existence of the discrete problem and derive error estimates for the velocity and pressure variables and their tangential derivatives over sub-domains staying away from the outow boundary. Let T h be a quasi-uniform triangulation for, which was constructed in such a way that no triangle has a side parallel to the characteristic direction at any point. For a triangle T h, let P n ()=P n () P n (), Mh n = { C(): P n ()} and Vh n = M h n M h n. We dene the following bilinear forms on each triangle T h as follows: a (u; v)= (u; v) ( div u; v) + ( u; v) ; b (v;)=( ; v) ; c (p; )=( p; ) ; d (u;) = (div u;) : (1.4) Note that the bilinear form a can be regarded a bilinear form for the convection-diusion equation and the form c a bilinear form for the pure hyperbolic equation. With an interpolation [u h ;p h ]of the initial data [u 0 ;p 0 ]on, we seek a nite element solution [u h ;p h ] Vh n M h n such that a (u h ; v)+b (v;p h )=(f; v) ; v P n &() (); c (p h ;)+d (u h ;)=(g; ) ; P n &() (): (1.5) The discrete solution [u h ;p h ] starts as an interpolation of [u 0 ;p 0 ]on and the triangles are ordered explicitly with respect to the convection term. We see that the approximate solution [u h ;p h ]of(1.5) has a total of 3% n degrees of freedom in each triangle where % n =(n + 1)(n +2)=2. In an one-inowside triangle, there are 3(n + 1) degrees of freedom in [u h ;p h ] along the inow, leaving a total of 3% n 1 to be determined from (1.5) with &() = 1. In a two-inow side triangle, there are 3(2n + 1) degrees of freedom in [u h ;p h ] along the inow, leaving 3% n 2 to be determined from (1.5) with &() = 2. Thus the number of equations in (1.5) equals the number of unknowns. It is assumed in this paper that n 2 even though the case n = 1 is of some interest. The latter is a degenerate case in which (1.5) with &() = 2 is vacuous and (1.5) with &() = 1 completely determines the approximate solution.

322 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 As is the case given in [8], we show that the method (1.5) is stable provided a condition of the form 1 := 1 h 6 0 (1.6) is satised where 1 = +. In addition, we require the triangle sides to be bounded away from the characteristic direction, i.e., n c 0: (1.7) Note that conditions (1.6) (1.7) are needed because the convection-dominated case is considered. In showing our stability we need to impose a condition on the compressibility number: := 1 ; (1.8) where is a positive constant to be given later. Let D be a domain in R 2 and a line segment. Dene, respectively ( 1=2 ( 1=2 f D = f dx) 2 and f = f ds) 2 : D Denote by l;d and l; the corresponding norms for the Sobolev spaces H l (D) and H l ( ), respectively. For simplicity, we dene the following notations for the norms: [ av; b] 2 := a v 2 + b 2, [ av; b] 2 D := a v 2 D + b 2 D; [ av; b] 2 := a v 2 + b 2 ; (1.9) where a and b are positive numbers. Denote by P k f the L 2 projection over D into P k (D). We here give a main result in this paper, which is shown in Section 4. Let [u;p] and [u h ;p h ]be the solutions of (1.1) and (1.5), respectively. Assume that the solution [u;p] satises the a priori estimate M := h u n+1; + p n; 6 C. Let be any subtriangulation of such that ( ) (). If (1.6) (1.8) hold, we then have [u u h ;p p h ] 6 Ch n 1 M; [ (u u h ) ; 0 (p p h ) ] 6 Ch n 1 M; { [ (u u h ); 1=2 0 (p p h )] 2 n ds} 6 Ch n 1 M; +( ) [ (u u h ); (p p h )] 6 Ch n 3=2 M; { [ (u u h ) t ; 1=2 0 (p p h ) t ] 2 n ds} 1 6 Ch n 3=2 M; +( ) where 0 := with a constant = C(; c ; 0 ) with a minimum angle 0 in the triangulation T h, v t := t v is the tangential derivative along and n is the unit normal vector on and C is a generic constant not depending on h.

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 323 Note that the a priori estimate for (1.1) can be obtained with a large condition on, not assuming a large condition on (see [5, Theorem 2.1]). Note that the constant is a positive number depending on only ; c and 0 (see (3.10), (3.28), (4.6), etc.) and 0 can be positive under the condition (1.8). With condition (1.6), system (1.1) can be viewed a degenerate hyperbolic system. For small viscous numbers and the boundary condition on + () may generate a boundary layer which has width O( 1 ) with 1 := +. The layer cannot be resolved in the regime 1 h. Consider a reduced problem: system (1.1) with = = 0 and u not specied on +. In fact, the solution of the reduced problem does not have a boundary layer. However, if the values attained by the solution u of the reduced problem on the outowboundary + do not coincide with the boundary value specied in the full problem, the solution of the full problem may have a boundary layer at +. Let be any triangle in a given triangulation T h. For i =1; 2; 3, we denote by 9 i the sides of numbered counterclockwise, by a i the vertices of opposite 9 i,byn i the unit outward normals to 9 i, and by t i the unit tangents along 9 i taken in a counterclockwise direction (see Fig. 1). We always take 9 3 to be the inow(outow) side of a type I (type II) triangle. In addition, the layers consisting of the triangles in the triangulation T h are dened as follows S 1 = { T h 9 ()}; ( S i+1 = { T h 9 )} S k ; k6i i=1; 2;::: : (1.10) With this partition of T h, the approximate solution may be obtained in an explicit fashion, rst in S 1, then in S 2, etc. Within each layer, the approximate solution can be obtained in parallel since the solution in any of triangles within a layer does not depend on the solution in other triangles in that layer. In the triangulation in Fig. 1, the number 1 indicates the triangles for the layer S 1 and 2 the ones for the layer S 2, and so on. Also, the dark area of Fig. 1 indicates a layer near the outowboundary. In our analysis, it is assumed that T h is a given triangulation of satisfying the following conditions [2,3]: (i) all angles of triangles satises a minimum angle condition: 0 0 and (ii) the triangles in T h can be partitioned into O(h 1 ) layers. Fig. 1.

324 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 The paper is organized as follows: In Section 2 a local stability for (1.5) is established and in Section 3 a global stability is shown and nally in Section 4 error estimates are derived. In our proofs, C denote a generic constant, depending on certain quantities. We shall make this dependence explicitly, for example, writing C() if C depends only on (for example, in the Sobolev inequalities) or C(c ; 0 ;)ifc depends both on c ; 0, and, and so on. 2. Local stability This section shows that the solution [u h ;p h ]of(1.5) is well dened and satises local estimates (see Theorem 2.1 below). To do this we rst introduce a reference triangle ˆ with vertices â 1 =(1; 0), â 2 =(0; 1) and â 3 =(0; 0). Let x =(x; y) and ˆx =(ˆx; ŷ) ˆ. The reference triangle ˆ can be mapped into the triangle by the ane transformation x = B ˆx + a 3 ; B=( 9 1 t 1 ; 9 2 t 2 ); which is mapped into the triangle, where 9 i denotes the length of 9 i. Let û(ˆx; ŷ) = u(x(ˆx; ŷ);y(ˆx; ŷ)) and h = max Th h where h is the diameter of the triangle. Since the ane mapping is invertible, we may also have u(x; y) =û(ˆx(x; y); ŷ(x; y)). Let J =(9(x; y)=9(ˆx; ŷ)) T be the transpose of the Jacobian and ˆ =(9=9 ˆx; 9=9ŷ) T. Then the gradient, divergence and Laplacian operators are transformed over the reference triangle ˆ as follows: u = J 1 ˆ û := h 1 û; div v = ˆv ˆx ˆx + ˆvŷ ŷ := h 1 div ˆv; u = ˆx T x Ĥ 1û ˆx x + ˆx T yĥ 1û ˆx y := h 2 û; where ˆv ˆx =(û ˆx ; ˆv ˆx ), ˆx x =(ˆx x ; ŷ x ) T and Ĥ 1 = 9 2 9 ˆx 2 9 2 9 ˆx9ŷ 9 2 9 ˆx9ŷ 9 2 9ŷ 2 : Also one can see that div v=h 2 div ˆv. Hence problem (1.5) can be transformed over the reference triangle ˆ as follows: ( ûh div û h + û h + ˆp h ) ˆv h d ˆx ˆ ˆ =h ˆf ˆv h d ˆx; ˆ ( divû h + ˆp h )ˆ h d ˆx = h ˆv h P n &( ˆ ) ( ˆ); (2.1) ˆ g ˆ h d ˆx; ˆ h P n &( ˆ ) ( ˆ); (2.2) where = =h and = =h. In ˆ we derive a linear algebraic system for (2.1) (2.2) by expressing the approximate solution in a linear combination of the Lagrange basis functions. The functions û h

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 325 and ˆp h can be expressed in the form % n % n % n û h =[û h ; ˆv h ]= û j j ; ˆv j j and ˆp h = j=1 j=1 j=1 ˆp j j ; where û j := [û j ; ˆv j ]=[û h (P j ); ˆv h (P j )], ˆp j = ˆp h (P j ), P j are the usual (equispaced) nodes for nth degree interpolation in ˆ and j (x) are the corresponding Lagrange basis functions for P n ( ˆ) and % n =(n + 1)(n +2)=2. Recalling that (2.1) and (2.2) are true for each basis function i for P n &( ˆ ) ( ˆ) and letting u j = û h (P j ), p j =ˆp h (P j ) for P j ( ˆ), they can be written in the following equations: for each i =1;:::;% n &( ˆ ) and on ˆ, % n j=1 % n j=1 {A ji û j + B ji ˆp j } = f i := h( ˆf; i ) ˆ ; (2.3) {C ji ˆp j + D ji û j } = g i := h(ĝ; i ) ˆ ; (2.4) û j ; ˆp j are given for P j ( ˆ); (2.5) where i =( i ; 0) is chosen for the rst equation of (2.3), i =(0; i ) for the second equation of (2.3) and A ji = A ;ji + A 1;ji with A ;ji û j = (û j j ) i d ˆx 1 div(û j j ) i d ˆx ˆ and A 1;ji = ( j ; i ) ˆ, C ji = ( j ; i ) ˆ, and B ji and D ji are given by B ji ˆp j =ˆp j j i d ˆx and D ji û j = div(û j j ) i d ˆx: ˆ ˆ ˆ Denoting by û =(û 1 ;:::;û %n ) t, ˆp =(ˆp 1 ;:::; ˆp %n ) t, f =(f 1 ;:::;f %n ), g =(g 1 ;:::;g %n ), and ẑ =(û; ˆp) t, ẑ =(û ; ˆp ) t, F =[f;g] t,(2.3) (2.5) can be written in a matrix form: (S + T )ẑ = M; (2.6) where ( ) ( ) A1 B A 0 S = ; T = D C 0 0 ( A ) B and M = F + D C ẑ : We recall that A is a block matrix generated by A ;ji and similarly A 1, B, C and D are such matrices. Hence (2.6) is a linear algebraic system with as many equations as unknowns, and S, T are uniformly bounded and also M is bounded (see Theorem 2.1 below).

326 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 If the matrix S is invertible, a unique existence of the solution of (2.6) easily follows under a condition of the form 1 := + 6 0 ; h where 0 0 is to be chosen later. For this, let Q denote the centroid of and 0 = (Q) with (x) 0 6 Ch 1;. So the matrix S can be split into S = S 0 + hs 1, where ( ) ( ) A0 B h 1 (A 1 A 0 ) 0 S 0 = ; S 1 = D C 0 0 h 1 (C C 0 ) with the block matrices A 0 and C 0 generated by A 0;ji = ( 0 ˆ j ; i ) ˆ, C 0;ji = ( 0 ˆ j ; i ) ˆ, respectively. We observe that the invertibility of S 0 is equivalent to showthat the only solution to (1.5) with replaced by 0 and =0=, u h 9 =0, p h 9 =0, f = 0 and g =0 is u h = 0 and p h =0: in other words, the following problem has only trivial solution: nd [u;p] P n () P n () such that a 1 (u; v)+b (v;p)=0; c (p; )+d (u;)=0; v P n &() (); P n &() (); u =0; p=0 on 9 ; (2.7) where a 1 (u; v)=(u 0 ; v) and c (p; )=(p 0 ;). For solving (2.7), we construct a function, which will be used later, having zero value on the inowsides of T h. Dene a function : R by (x)= 1 ( 1 n 1 + 2 n 2 ) (x a 3 ) n 1 t 2 1 2 (n 1 (x a 3 ))(n 2 (x a 3 )) (n 1 t 2 ) 2 1 2 (type I triangle); (type II triangle); where n i and t i are the unit normal and tangential vectors to i = 9 i, respectively and i is the length of i. Then the function has the following properties: (2.8) Lemma 2.1. Let 0 = (Q) with Q the centroid of. The function (x) dened in (2.8) satises the following properties: (a) (x) P &() () where &() is the number of inow sides of, and (b) (x) 0 on, and (c) (x)=0 on 9, and (d) 0 (x) 0 on if h is small enough. Proof. (a) (c) easily followfrom the denition of (x). Next the gradient of (x) is given by 1 ( ) n 1 n 1 t 2 2 + n2 (type I triangle); 1 (x)= (2.9) n 1 n 2 (x a 3 )+n 2 n 1 (x a 3 ) (type II triangle): (n 1 t 2 ) 2 1 2

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 327 For type I triangle, we have n 1 t 2 0 and n i 0 0(i =1; 2). For type II triangle, we have 0 n i 0 and n i (x a 3 ) 0(i =1; 2). Hence (d) follows. Using Lemma 2.1 and a large condition on the coecient (see (1.8)), we obtain Lemma 2.2. Let 0 be given in Lemma 2.1. Assume that (1.6) (1.8) hold. Then if 2=c 2 and the mesh size h is small enough (2.7) has only trivial solution. Proof. From the boundary conditions in (2.7), the solutions u P n () and p P n () can be factored into u = w and p = q, where is dened in (2.8) and (w;q) P n &() () P n &() (). Now taking v = w, = q in (2.9), and letting A 1 = a 1 (w; w)+c (q; q), A 2 = b (w;q)+d (w;q), we have A 1 + A 2 = 0. Since (w) 0 = 0 w + w 0, the integration by parts and Lemma 2.1 yields A 1 = [ w; q] 2 0 dx + w 0 w dx + qq 0 dx = 1 ( [ w; q] 2 0 dx + [ w; ) q] 2 2 0 n ds 9 + w q 0 dx + w q 0 n ds; (2.10) 9 + where = = was used in the above inequality. Again using Lemma 2.1 and the integration by parts, we have A 2 = q w dx + qw n ds 9 + q w dx q w ds: (2.11) 9 + Noting that 0 n c with h small, the above two inequalities gives 0=A 1 + A 2 ( q w 0 ) ( dx + w q 0 n 1 ) ds: (2.12) 0 9 + c From (2.9), we see that for type I triangle, 2 2 ( 0 ) = i=1 ( 1 n 1 i + 2 n 2 i ) 2 2 ( 1 0 n 1 + 2 0 n 2 ) 6 2 (2.13) 2 c 2 and for type II triangle, 2 2 ( ) ( 0 ) = n1 e i n 2 (x a 3 )+n 2 e i n 1 (x a 3 2 ) 2 0 n 1 n 2 (x a 3 )+ 0 n 2 n 1 (x a 3 ) = i=1 2 i=1 ( ) n1 e i + n 2 e i 2 ; (2.14) 0 n 1 + 0 n 2

328 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 where e 1 =(1; 0), e 2 =(0; 1) and := (n 1 (x a 3 )=n 2 (x a 3 )) 0on. Recalling that ( 0 n 1 ) ( 0 n 2 ) 0 and h() := n1 e i + n 2 e i 0 n 1 + 0 n 2 is a function over (0; ), we have { } n 1 e i + n 2 e i 0 n 1 + 0 n 2 6 max n 1 e i 0 n 1 ; n 2 e i 0 n 2 6 1 : c Indeed, the absolute value of the above function h() can be rephrased by h() := a + b c + d = b c=d d 1+a=b + c=d ; where a; b; c and d are real numbers with cd 0. We then have two cases: rst, if a=b c=d 0, then h() is decreasing on (0; ) and so h() 6 h(0) = a, and second, if a=b c=d 0, then c h() is increasing on (0; ) and h() 6 h( ) = b. Hence the required inequality follows. d Hence, for type II triangle, we have 2 ( 0 ) 6 2 : (2.15) 2 c 2 Combining (2.12) (2.15), we get ( ) 2 0 q w 0 c ( dx + q w 0 n 1 ) ds 0; 9 + c provided that 2c 2. Then w q =0 in 9 +. Using (2.11), we have A 2 0, and 0 = A 1 + A 2 A 1. Thus it follows from (2.10) that w and q are identically zero on. Note that condition (1.8) is essential in showing Lemma 2.2. Since the uniqueness for a matrix problem of a nite order is equivalent to its invertibility, it follows from Lemma 2.2 that if 2c 2 is positive, the matrix S 0 dened above has a bounded inverse S 1 0 with S 1 0 6 C, where is a matrix norm and C is a constant independent of h. SoS = S 0 + hs 1 has a bounded inverse if h is small enough. If 0 is suciently small such that ( + )=h 6 0, then S 1 T 1 with = =h, so problem (2.6) is solved uniquely, with its coecient matrix S + T having a bounded inverse (S + T ) 1 6 S 1 1 S 1 T 6 C 1; (2.16) where C 1 is a constant not depending on the mesh size h. From the above arguments we see that the solution [u h ;p h ]of(2.6) is well dened with small conditions on 0 and h. We next showthat the solution and its derivative are bounded by the data f;g, and their values on the inowsides of the triangle. Theorem 2.1. Let T h be any triangle. Let u h and p h be given on 9. Assume that S 1 T 1 where = =h and that (1.6) (1.8) hold. Then problem (1.5) has a unique solution

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 329 [u h ;p h ] P n () P n (). Furthermore there is a constant C such that the inequalities hold: [u h ;p h ] 6 C(h 1=2 [u h ;p h ] 9 + h [f;g] ); (2.17) [ u h ; p h ] 6 C(h 1=2 [u ht ;p ht ] 9 + [f;g] ); (2.18) where v t = t v with the tangential vector t on 9, and the norms [ ; ] and [ ; ] 9 are dened in (1.9). Proof. If is small so that S 1 T 1, then S + T has a uniformly bounded inverse and using (2.6), z =(S + T ) 1 M =(S + T ) 1 F + z. Hence max i z i 6 C 1 max F i + max z i ; (2.19) i i where C 1 is the constant given in (2.16) and F i =[f i ;g i ]. Since F i 6 Ch( f 0; + g 0; ) and z i 6 C [û h ; ˆp h ] 9 ˆ, we have, using (2.19) and transforming back ˆ to, [u h ;p h ] 6 C(h 1=2 [u h ;p h ] 9 + h [f;g] ): (2.20) For showing (2.18), for P 0 9 let u h = u h (P 0 )+v h and p h = p h (P 0 )+q h. Since [v h ;q h ] satises equations in (1.5) like [u h ;p h ], we have, applying (2.20), [v h ;q h ] 6 C(h 1=2 [v h ;q h ] 9 + h [f;g] ): (2.21) Hence the inequality (2.18) follows from the following inequalities: u h = v h 6 Ch 1 v h and v h 9 6 Ch v ht 9 = Ch u ht 9. Similar for q h. From Theorem 2.1 we see that (1.5) is uniquely solvable on each triangle T h under the data [f;g] and [u h ;p h ]on9. In next section we will show that (1.5) is solvable in a global sense. 3. Global stability This section shows that (1.5) is solvable in a global sense and the approximate solution can be estimated by its values on the inowboundary, the data f and g (see Theorem 3.2). This section can be summarized into ve steps. Step 1: Lemma 3.1, an identity for the bilinear forms a 1 and c. Step 2: Lemma 3.2 is shown for a lower bound for the form B 1 given in (3.1) and Lemma 3.3 is for a lower bound for the form B 1 plus the viscous terms in each type triangle. Step 3: Lemma 3.4 and Lemma 3.5; local estimations for the solution [u h ;p h ]of(1.5), combining the results obtained in each type triangle. Step 4: Theorem 3.1, a stability result for the discrete solution on each layer. Step 5: Theorem 3.2, a global stability for the discrete solution [u h ;p h ]of (1.5). For simplicity, throughout this section, u and p will be used instead of u h and p h. Using the local stability result Theorem 2.1, we will establish a global stability for (1.5). To write in a compact form, we denote B 1 by a bilinear form on V n h M n h : B 1 ([u;p]; [v;]) := a 1 (u; v)+b (v;p)+c (p; )+d (u;); (3.1)

330 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 where a 1 (u; v)=( u; v). Note that a 1 (u; v) and c (p; ) can be analyzed in a usual way [3] but one has to be careful in computing b (v;p) and d (u;). Let Q be the centroid of the triangle and let 0 = (Q) be a unit constant vector. In order to control the interelement tangential derivatives of u and p we use the following test functions [8]: v 2u t 1 t 2 ( 0 n 1 )( 0 n 2 ) ; 2 p t 1 t 2 ( 0 n 1 )( 0 n 2 ) ; (3.2) where t i and n i denote the tangential and normal vectors to i (), respectively. Since 0 is constant, [v ; ] belongs to P n 2 () P n 2 () and is a valid test function for a triangle of either type. We next give some identity and inequality which are used in controlling the interelement tangential derivatives u t and p t (see [3,8]). Lemma 3.1. (a) Let be the unit vector obtained by rotating counterclockwise through =2. Then the function in (3.2) can be represented in the form = ( 2p + ) with 6 C(h 1 p + p ); (3.3) similar for v, and a 1 (u; v )+c (p; ) [ u t ; p t ] 2 n 9 ds +! [ u ; p ] 2 ds C [ u; p] 2 ; (3.4) 3 where and! are dened in (3.6) and (3.8) respectively, 3 := 9 3 denotes the inow (outow) side in type I (type II) triangles and C is a constant not depending on h. (b) If T h is a type II triangle and P n (), we have 6 C(h 1=2 9+ + P n 2 + h ): (3.5) Proof. Using the formula t i = t i + t i and t i = n i, the function can be written by [ ( t 1 = 2p 2 n + ) t1 p 1 n 1 2 ( ] t1 )( t 2 ) ( n 1 )( n 2 ) p 2 ( 0 n 1 )( 0 n 2 ) (p t 2 ( t 1 )+p t 2 ( t 1 )); where is dened by = ( n1 )( n 2 ) ( 0 n 1 )( 0 n 2 ) : (3.6) Since 1 is of order h, the function = ( 2p + ) where the function can be estimated as follows: under the assumptions on the triangulation, 6 C( p + p + p ) 6 C(h 1 p + p ):

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 331 Finally we show (3.4). Using [3, Lemma 4.2] and letting 12 =( 0 n 1 )( 0 n 2 ), we have a 1 (u; v )+c (p; ) = [ u t ; p t ] 2 9 n + ds +! [ u ; p ] 2 ds 9 3 1 12 ((z 1 ) t2 [ u t1 ; p t1 ] 2 +(z 2 ) t1 [ u t2 ; p t2 ] 2 ) dx( R); (3.7) where is given in (3.6) and! is dened by! = (t 1 n 3 )(t 2 n 3 ) 1 (3.8) ( 0 n 1 )( 0 n 2 ) n 3 and z 1 ;z 2 are dened by (z 1 ;z 2 )= 1 t 1 n 2 ( n2 ; n 1 ): Note that! has the same sign as n 3 since t 1 n 3 0 and t 2 n 3 0. So! 0( 0) for type I triangle (type II triangle). Since the quantity R in (3.7) can be estimated by C [ u; p] 2,(3.4) easily follows from (3.7). Finally (3.5) was given in [3, Lemma 3.3]. We next compute the term b (v ;p)+d (u; ) and combining it with (3.4), estimate the bilinear form B 1 below: Lemma 3.2. If (1.7) (1.8) hold, then the solution [u;p] of (1.5) satises B 1 ([u;p]; [v ; ]) 9 [ c 1 u t ; c 2 p t ] 2 n ds +! [ c 3 u ; c 4 p ] 2 ds C [ u; p] 2 ; (3.9) 3 where ;! are given in (3.6) and (3.8), respectively, and the constants c i (i =1;:::;4) are dened by c 1 = 2 ( 2 n ) ; c 3 = ( 2! n 2! c 2 = 16 n 123 2 sin2 0 ) ( n + ) ; c 4 =! 16 2 123! sin2 0 with 123 =( n 1 )( n 2 )( n 3 ) 2 and 0 the minimum angle in T h. ( n + ) ;! (3.10)

332 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 Proof. We rst write the gradient of p and the divergence of u in terms of the t i -directional derivatives as follows: ( ) ( ) px = 1 pt t 2 n 1 ( n2 ; n 1 1 ) and div u = 1 t 2 n (u 1 t 2 n1 u t 1 n 2 ): p y p t 2 Using the functions in (3.2) and letting 12 := 0 n 1 0 n 2, we have b (v ;p)+d (u; ) 2 = (u 12 t 2 n 1 t 1 t 2 (n1 p t 2 n 2 p t 1)+(u t 2 n 1 u t 1 n 2 )p t 1 t 2)dx = 2 12 t 2 n 1 (p t 2u t 2 n 1 t 1 n p t 1u t 1 n 2 t 2 n)ds: (3.11) 9 If is the unit vector obtained by rotating counterclockwise through =2, we have where t i = a i t 3 + b i (i =1; 2); a i = ti t 3 and b i = ti n 3 ; i=1; 2: n3 Using the above relations, for i =1; 2 we have p t iu t i n 3 i =((a 2 i p t 3 + a i b i p )u t 3 +(b 2 i p + a i b i p t 3)u ) n 3 i ; where the upper subscript of a k i written as follows: b (v ;p)+d (u; )= 2 12 where ij are given by means a 0 i =1; a 1 i = a i ; a 2 i =(a i ) 2, etc. Hence Eq. (3.11) can be ( 2 ij = a i 2b j 2 (t1 n 3 )n 1 a i 1b j 1 (t2 n 3 )n 2 : p t i u t i n 3 i ds i=1 9 i 1 ) [p t 2 n 1 t 3u t 3 20 +p u 02 +(p u t 3+p t 3 u ) 11 ]ds ; 9 3 (3.12) Since a i 6 1= t 3 =1= n 3 and b i 6 1= n 3,(i =1; 2), we have ij 6 2( n 3 ) 2 : (3.13)

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 333 Noting that 12 =( 0 n 1 )( 0 n 2 ) 0 and n 1 t 2 = sin 3 123 := ( n 1 )( n 2 )( n 3 ) 2 and 3 = 9 3, we have and using (3.12) (3.13), and letting b (v ;p)+d (u; ) 6 2 2 i=1 6 2 sin 0 + 3 p t i u t i ds + 4 9 i sin 3 ( 9 123 [ 1 u ; [ 1 u t ; 2 1 1 +2 1 2 p t] 2 ds 123 3 ) 2 1 1 +2 1 2 p ] 2 ds 123 ( u t 3 + u )( p t 3 + p )ds (3.14) for arbitrary positive numbers 1 and 2, where and! are dened in (3.6) and (3.8), respectively. Combining (3.4) and (3.14), we obtain B 1 ([u;p]; [v ; ]) [ c 1 u t ; c 2 p t ] 2 n 9 where t is the unit tangential vector to 9, and c 1 = c 3 = 4 n 1 ; c 2 = 2 n 123 sin 0 123 sin 0 4 123! sin 0 2 ; ds +! [ c 3 u ; c 4 p ] 2 ds C [ u; p] 2 ; 3 ( 1 ) ; + 1 1 2 ( 2 1 c 4 = + 1 123! sin 0 1 2 Finally, taking 1 =( 123 sin 0 =8 n ) and 2 =( 123! sin 0 =8), the required inequality (3.9) follows. It is seen that the constants c 2 and c 4 dened in (3.10) can be positive with a large condition on. So the condition (1.8) is needed in showing (3.9). We next compute the viscous terms and combine it with inequality (3.10) in each type of triangles. ) : Lemma 3.3. Let [u;p] be the solution of (1.5). Suppose (1.6) (1.8) hold. If 0 := 0=h is small enough with 0 := +, then for type I triangles, [ c 1 u t ; c 2 p t ] 2 ds + 0 u 2 n 9 6 C(! 1 h 1 [ c 3 u ; c 4 p ] 2 + [ u; p] 2 + [ h 1 P n 2 f; h 21 P n 2 g] 2 ) (3.15)

334 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 and for type II triangles, [ c 1 u t ; c 2 p t ] 2 ds +! 0 [ c 3 u ; c 4 p ] 2 9 n 3 + 0 u 2 9 6 C( 0h 1 u 2 + [ u; p] 2 + [ h 1 P n 2 f; h 2 P n 2 g] 2 ); (3.16) where! 0 ;! 1 are some positive numbers, and c 3 ;c 4 are given in (3.10) and C is a constant not depending on h. Proof. We rst compute the viscous term (u + div u; v ). Writing u = u + u div u = u + u, we write the viscous terms as follows: and u + div u = 0 (u + u + u (div + )+u (div + )): (3.17) Using (3.3) for v and an inverse inequality: u 6 Ch 1 u, we have ( u + div u; v ) C( 0 u 2 0h 1 u 2 0 u 2 ); (3.18) where C is a positive constant not depending on h. Again using (3.3) for v and, respectively, we have (f; v ) +(g; ) = (P n 2 f; v ) +(P n 2 g; ) 6 2 P n 2 f u + P n 2 f + Ch 1 P n 2 g p t1 6 1 u 2 + 2 h 1 u 2 + [ 2 u; p] 2 +C [ 1 1 + 1 2 h 1 P n 2 f; 1 h 2 P n 2 g] 2 (3.19) for 0; i 0. Since t 1 n 3 0 and t 2 n 3 0 from (3.8), we see that! has the same sign as n 3. That is,! 0( 0) for type I triangles (type II triangles), there exist constants! 0 ;! 1 such that {! 6!1 for a triangle of either type; (3.20)!! 0 0 for a type II triangle: For type I triangle we apply the inequality: v 9 6 Ch 1=2 v to the right-hand side of (3.9). So recalling that (u; v ) ( div u; v ) + B 1 ([u;p]; [v ; ])=(f; v ) +(g; ) ; (3.21) and using (3.18) (3.20), we obtain inequalities (3.15) (3.16). We next pick some similar test functions used in [3] to combine (3.15) and (3.16).

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 335 Lemma 3.4. Assume that (1.6) (1.8) hold. Then if 0 := 0=h is small enough, then the solution [u;p] of (1.5) satises the inequality 0 u 2 + h 1 [ 1 u ; 1 p ] 2 [ c 1 u t ; c 2 p t ] 2 + ds n 6 C(h 2r 1 [u;p] 2 9 + [ u; p] 2 + h 1 [f;g] 2 9 + [ h 1 P n 2 f; h 2 P n 2 g] 2 ); (3.22) where c 1 ;c 2 are given in (3.10) and 1 ; 1 are given in (3.28), and r 0 is to be chosen later and C is a generic constant not depending on h. Proof. We rst denote by P = P n &() for simplicity. Applying the test functions v = Pu 0 and = Pp 0 in (1.5), we have where [ Pu 0 ; Pp 0 ] 2 = A 1 + A 2 + A 3 ; (3.23) A 1 =(u + div u + f; Pu 0 ) +(g; Pp 0 ) ; A 2 = ( p; Pu 0 ) (div u; Pp 0 ) ; A 3 = (( 0 ) u; Pu 0 ) (( 0 ) p; Pp 0 ) : Using (3.17) and the inequality u 6 u 0 +ch u, the term A 1 can be estimated as follows: A 1 6 ( 0 u k + C( 0 h 1 u + 1 u )+ Pf ) u 0 + Pg p 0 6 C( 1 1 2 0 u 2 +( 1 + 0h) u 2 )+ [ C 0 + 1u 0 ; 2 p 0 ] 2 +C( 1 ; 2 ) [Pf; Pg] ; i 0: (3.24) Recall that the -directional derivatives u and p of the discrete solution [u;p]of(1.5) must not vanish on each triangle. Otherwise, (1.5) cannot be developed explicitly from triangle to triangle (see Theorem 2.1). In addition, if h is small, then u 0 and p 0 will not be zero. Hence, using an argument of norm equivalence, we can have, for any v Mh n with 0 v 0, v + h r v 9 6 C( 0 v + h r v 9 ); (3.25) where r will be chosen later and C is a constant not depending on h. Using the Schwartz s inequality, we obtain an upper bound for A 2 : A 2 6 C( p Pu 0 + u Pp 0 ) (using (3:26)) 6 C( [ 4 u 0 ; 1 3 p 0 ] 2 + [ 3 Pu 0 ; 1 4 Pp 0 ] 2 + h 2r [ 4 u; 1 3 p] 2 9 ); (3.26)

336 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 where r 0 is chosen later and i 0 is arbitrary. Since A 3 6 Ch [ u; p] 2, a combination of (3.23), (3.24) and (3.26) gives [ C 3 Pu 0 ; C= 4 Pp 0 ] 2 6 C( 1 1 2 0 u 2 + h 2r [ h + 4 + 1 2 1 1 u; h + 1 3 p] 2 9 + [ 2 1 Pf; 1 2 Pg] 2 ) + [ C(0 + 4 + h)+ 1 u 0 ; 2 + C( 1 3 + h)p 0 ] 2 : (3.27) For type I triangles, Pv 0 = v 0, while for type II triangles, v 0 6 C(h 1=2 v 0 9+ + Pv 0 ). Taking, in (3.27), we have 1 =( C( 0 + h)); 3 = 4 = 1 8C [ C( 0 + h)]; 2 =[ 16C 2 =( C( 0 + h)) Ch]; 1 = C( 0 + h) ; 1 = 1 4C 2 [ 16C2 =( C( 0 + h)) Ch]; (3.28) [ 1 Pu 0 ; 1 Pp 0 ] 2 6 C { A (type I triangle) A + h [u 0 ;p 0 ] 2 9 + (type II triangle); (3.29) where A =20 2 u 2 + h2r [u;p] 2 9 + [Pf; Pg] 2. Using v 6 v 0 + Ch v, and (3.15) and (3.29), we have, for type I triangles, (1 c 5 0) 0 u 2 [ c 1 u t ; c 2 p t ] 2 + ds 9 n 6 C( [ u; p] 2 + h 2r 1 [u;p] 2 9 + [ h 1 P n 2 f; h 2 P n 2 g] 2 ); (3.30) where c 5 = C! 1 max{c 3 ;c 4 }=min{ 1 ; 1 } and C is a constant not depending on h. Also we see that h 1 [ 1 u ; 1 p ] 2 can be bounded by the right-hand side of (3.30). So inequality (3.22) has been shown in the case of type I triangles. For type II triangle, since v 0 6 C(h 1=2 v 0 9 + P n 2 v 0 ), inequality (3.29) becomes 0h 1 u 2 6 C 0( u 0 2 9 + + h 1 P n 2 u 0 2 + u 2 ) 6 C 0( 0 0 u 2 + u 2 + h 2r 1 [u;p] 2 9 +h 1 [P n 2 f; P n 2 g] 2 + 0 [u 0 ;p 0 ] 2 9 + ):

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 337 Similar for 0 h 1 p 2. Finally, using v 0 9+ 6 v 9+ + Ch 1=2 v and inserting the above inequality into (3.16), we obtain (1 C(0) 2 ) 0 u 2 + [ 3 u 0 ; 3 p 0 ] 2 [ c 1 u t ; c 2 p t ] 2 9 + + ds 9 n 6 C( [ u; p] 2 + 0h 2r 1 [u;p] 2 9 + 0 [ h 1 P n 2 f; h 2 P n 2 g] 2 ); where 3 =! 0 c 3 C 0 and 3 =! 0 c 4 C 0 with c 3 and c 4 given in (3.10). So in the case of type II triangles (3.22) follows from (3.30) and the above inequality. It is seen that 3 and 3 can be positive if 0 is small enough and if = = is large. We next choose some test functions to bound u and p over the boundary of T h. Lemma 3.5. Suppose (1.6) (1.8) hold. Then the solution [u;p] of (1.5) satises [ u; p] 2 n ds 9 6 0 u 2 2 + [u ;p ] 2 [ ] h +C (h 2 [u t ;p t ] 2 9 + [u; p] 2 9 + h [f;g] 2 )+ [P n 2 f; P n 2 g] 2 ; (3.31) for 0; 0, where C is a generic constant not depending on h. Proof. We rst denote by P = P n 2 for simplicity. Considering the test functions v =2Pu and =2Pp, we have, from (1.5) [ u; p] 2 n ds =2(A 1 + A 2 ); (3.32) 9 where A 1 and A 2 are given by A 1 =(u + div u; u) + (u ; (I P)u) + (p ; (I P)p) +(f; Pu) +(g; Pp) ; A 2 = ( p; Pu) (div u; Pp) : Using (3.17) and an inverse inequality v 6 Ch 1 v, we have A 1 6 ( 0 u + 0 u + Ch 1 u ) u +h( u u k + p k p )+C [Pf; Pg] 2 + [u;p] 2 (using the Schwartz s inequality and (2:17) (2:18))

338 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 6 4( 2 0 u 2 +( 0) 2 u 2 + [u ;p ] 2 )+C [Pf; Pg] 2 +C 1 (h 3 [u t ;p t ] 2 9 + h [u;p] 2 9 + h 2 [f;g] 2 ); 0: Using (1.5) and (3.17), the second term A 2 is estimated as follows: A 2 6 C( 0 u + 0 u + 1h u )+ Pu + Pf ] u +( Pg + Pp ) p (using the Schwartz s inequality and (2:17) (2:18)) 6 4( 2 0 u 2 +( 0) 2 u 2 + C( 1) 2 h 3 [u t ;p t ] 2 9 + [Pu ; Pp ] 2 + [Pf; Pg] 2 )+C 1 (h 2 [f;g] 2 + h [u;p] 2 9 ); 0: Hence, we have that A 1 + A 2 6 0 u 2 2 + [u ;p ] 2 + C[ 1 (h 3 [u t ;p t ] 2 9 +h [u; p] 2 9 + h 2 [f;g] 2 )+ [Pf; Pg 2 ]; (3.33) where =4( + ) and =4((( 0 )2 +1)+(( 0 )2 + C). Thus, combining (3.32) and (3.33), the required inequality follows. The local results of Lemmas 3.4 and 3.5 lead to the global stability for u; p;u t and p t along interelement boundaries. To show this, we unify the two inequalities given in Lemmas 3.4 and 3.5, and use Theorem 2.1. First multiplying h to both sides of (3.22) and adding the resulted inequality to (3.31) and taking = 1 =2; = 1, we obtain 9 ds + [ 1 =2u ; 1 1 =2p ] 2 +(1 0) 0 h u 2 6 C(h 2r [u;p] 2 9 + h [ u; p] 2 + h 3 [u t ;p t ] 2 9 + h [u;p] 2 9 )+R (using (2:21) and taking r =1=2) 6 Ch ds + R ; 9 (3.34) where C = C(; ; c 1 ;c 2 ;c ), and 1 ; 1 are given in (3.28) and = h [ c 1 u t ; c 2 p t ] 2 n + [ u; p] 2 n; R = C(h [f;g] 2 + P n 2 f 2 + h 1 P n 2 g 2 ):

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 339 From (3.34) one can derive an inequality of the following type: [ ds + 1 9 + 2 u ; 1 ] 1 2 2 p 6 (1 + Ch) 9 ds + R ; (3.35) where C is a positive constant. We next dene fronts F j which describe the forward boundary of the solution after it has progressed through the rst j layers. More specically, they can be expressed as follows: F 0 = (); F j = F j 1 + (S j ) (S j ) ; j =1; 2;:::: Summing inequality (3.35) over all triangles S j, we obtain the following main result of this section. Theorem 3.1. Suppose (1.6) (1.8) hold. If [u;p] is the solution of (1.5), then [ ds + 1 F j 2 u ; 1 ] 1 2 2 p 6 C ds + R j ; (3.36) j F 0 where C is a generic constant and j = k6j S k with layers S j dened in (1.10). Proof. Summing (3.35) over all triangles S j, we have p j + a j 6 (1 + Ch)p j 1 + d j ; (3.37) where d j = R Sj and p j = ds; a j = F j [ 1 2 u ; 1 ] 1 2 2 p S j : The solution of inequality (3.37) is given by ( j ) ( p j 6 (1 + Ch) j d i a i + p (1 + Ch) i 0 6 (1 + Ch) j p 0 + i=1 ) j d j i=1 j a i : i=1 Since (1 + Ch) j 6 e Cjh follows. and there are only O(h 1 ) layers, the required inequality (3.36) easily Letting be any subtriangulation of such that ( ) (), summing (3.35) layer by layer, following the same procedures used in the proof of Theorem 3.1, it can be derived from (3.36) that +( ) ds + [ 1 2 u ; 1 ] 1 2 2 p

340 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 6 C ( ) ds + R 6 C(h [u t ;p t ] 2 ( ) + [u;p] 2 ( ) + h [f;g] 2 + [Pf; h 1 Pg] 2 ); (3.38) where P = P n 2, c 1 ;c 2 are given in (3.10), 1 and 1 are given in (3.28) and C is a constant not depending on h. Finally, combining Theorems 2.1 and 3.1, and setting u =u h and p=p h again and letting = h, we establish the following global stability. Theorem 3.2. Let be any subtriangulation of such that ( ) (). Suppose (1.6) (1.8) hold. Then the solution [u h ;p h ] satises [ [u h ;p h ] 2 + h [ u h; p h ] 2 + 1 2 u h; 1 ] 1 2 2 p h h ds + +( ) 6 C(h [u ht ;p ht ] 2 ( ) + [u h ;p h ] 2 ( ) + h [f;g] 2 + [Pf; h 1 Pg] 2 ); where P =P n 2, C =C(c 1 ;c 2 ;; ), c 1 and c 2 are given in (3.10), 1 and 1 are given in (3.28). Proof. Applying Theorem 2.1 over each layer S j, we have [u h ;p h ] 2 S j + h [ u h ; p h ] 2 S j 6 Ch(h [u ht ;p ht ] 2 F j 1 + [u h ;p h ] 2 F j 1 + [f;g] 2 S j ) 6 Ch h ds + Ch [f;g] 2 S j ; (3.39) F j 1 where C is the constant given in (3.34). Applying Theorem 3.1 to the right-hand side of (3.39), we obtain [u h ;p h ] 2 S j + h [ u h ; p h ] 2 S j 6 Ch h ds + R j ; (3.40) F 0 where R j is dened in (3.34). Summing over all layers S j and recalling that the number of S j is O(h 1 ), we have [u h ;p h ] 2 + h [ u h; p h ] 2 6 C h ds + R ; ( ) where C is a positive constant not depending on h. Hence the required inequality follows from the results of (3.38) and (3.34). 4. Error estimates In this section we are going to derive error estimates of the solution [u h ;p h ] for (1.5). In order to derive error estimates for (1.5), we dene an interpolation u I M n h = { C(): P n ()} as

J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 341 follows: in a triangle with vertices a i and sides 9 i, (u u I )(a i )=0; i=1; 2; 3; (u u I )v ds =0; v P n 2 (); i=1; 2; 3; 9 i (u u I )w ds =0; w P n 3 (): (4.1) Let [u;p] (H n+1 ( )) 2 H n ( ) where is given as in Theorem 3.2. We assume the following interpolation error estimates (cf. see [2, p. 124, 146] [8, (4.2)]): u u I m; 6 Ch n+1 m u n+1; ; u u I m;9 6 Ch n m+1=2 u n+1; ; (4.2) m =0; 1;:::;n. Furthermore the above interpolation u I has the additional properties P( 0 (u u I ))=0; P( (u u I ))=0; P(div(u u I ))=0; (4.3) where Pv denotes the L 2 projection over into P n 2 () and 0 = (Q) with Q the centroid of. We next dene the following quantities: r = (u u I ) div(u u I )+ (u u I )+ (p p I ); s = div(u u I )+ (p p I ); e u = u h u I ; e p = p h p I : Using the above quantities, for problem (1.5) becomes { a (e u ; v)+b (v;e p )=(r; v) ; v P n &() (); (4.4) b (e p ;)+d (e u ;)=(s; ) ; P n &() (): We are nowready to obtain the error estimate for the discrete solution of (1.5). We apply Theorem 3.2 with [u;p] replaced by [e u ;e p ], and [f;g] replaced by [r;s] and use the triangle inequality. Since u h = u I and p h = p I on ( ) and since v t = t v is the tangential derivative on ( ) we note that the following quantities: (u h u I ) t ( ); (p h p I ) t ( ); u h u I ( ); p h p I ( ) (4.5) are zero. The quantities r and s can be bounded by C( (u u I ) + (p p I ) ). Using (4.3) we have Ps = ( 0 ) (p p I ) and Pr = P((u u I )+ div(u u I )+ ( 0 ) (u u I )). So 6 Ch (p p I ) and Pr 6 C (u u I ) for a constant C. Consequently, applying the solution [e u ;e p ]of(4.4) to Theorem 3.2 and using the triangle inequality, we obtain [u u h ;p p h ] + [ 1 (u u h ) ; 0 (p p h ) ] + h 1=2 [ (u u h ); (p p h )] { + h 1=2 [ c 1 (u u h ) t ; } 1=2 c 2 (p p h ) t ] 2 n 1 ds +( )

342 J.R. weon, P. im / Journal of Computational and Applied Mathematics 156 (2003) 319 343 { + [ (u u h ); } 1=2 (p p h )] 2 n ds +( ) 6 C [ (u u I ); (p p I )] ; (4.6) where 0 := 1 1 =2, c 1 ;c 2 are given in (3.10) and 1 ; 1 are given in (3.28). Note that the error estimate (4.6) has been obtained independently regardless of the degree ( 2) of the polynomials. Finally, computing the right-hand side of (4.6) from (4.2), we obtain: Theorem 4.1. Let [u;p] be the solution of (1.1) and [u h ;p h ] the solution of (1.5), respectively. Let be given as in Theorem 3.2 and M = [ (u u I ); (p p I )]. Suppose (1.6) (1.8) hold. Then [u u h ;p p h ] 6 C M; [ (u u h ) ; 0 (p p h ) ] 6 C M; { [ (u u h ); 1=2 0 (p p h )] 2 n ds} 6 C M; +( ) [ (u u h ); (p p h )] 6 Ch 1=2 M; { [ (u u h ) t ; 1=2 0 (p p h ) t ] 2 n ds} 1 6 Ch 1=2 M; (4.7) +( ) where 0 := with a positive constant =C(; c ; 0 ), v t := t v is the tangential derivative along and n is the unit normal vector on and C is a constant not depending on h. Proof. The inequalities in (4.7) followfrom (4.6). Note that the constant is a positive constant depending on only ; c and 0 (see (3.10), (3.28), (4.6), etc.) and 0 can be positive under condition (1.8). 5. Conclusion In this paper we have applied to a compressible viscous Stokes system the (continuous) Galerkin nite element method which can be solved explicitly with respect to convection. When the viscous numbers are small, the considered system is regarded a degenerate hyperbolic one. However, with the method the nite element solution has not been completely determined in the regime near outow boundary because the boundary condition may cause a layer. In our forthcoming papers we will analyze this singular problem.

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