On convergence rate of the augmented Lagrangian algorithm for nonsymmetric saddle point problems

Transcription

1 Applied Numerical athematics ) wwwelseviercom/locate/apnum On convergence rate of the augmented Lagrangian algorithm for nonsymmetric saddle point problems G Awanou a,,jlai b a Institute for athematics and its Applications, University of innesota, USA b Department of athematics, University of Georgia, Athens, GA 30605, USA Available online 18 November 2004 Abstract We are interested in solving the system [ ][ ] [ ] A L T c F =, 1) L 0 λ G by a variant of the augmented Lagrangian algorithm This type of problem with nonsymmetric A typically arises in certain discretizations of the Navier Stokes equations Here A is a n, n) matrix, c, F R n, L is a m, n) matrix, and λ,g R m We assume that A is invertible on the kernel of L Convergence rates of the augmented Lagrangian algorithm are known in the symmetric case but the proofs in [R Glowinski, P LeTallec, Augmented Lagrangian and Operator Splitting ethods in Nonlinear echanics, SIA, 1989] used spectral arguments and cannot be extended to the nonsymmetric case The purpose of this paper is to give a rate of convergence of a variant of the algorithm in the nonsymmetric case We illustrate the performance of this algorithm with numerical simulations of the lid-driven cavity flow problem for the 2D Navier Stokes equations 2004 IACS Published by Elsevier BV All rights reserved Keywords: Augmented Lagrangian algorithm; Saddle point problems * Corresponding author Current address: IA, University of innesota, 400 Lind Hall, 207 Church Street SE, inneapolis, N , USA address: awanou@imaumnedu G Awanou) /$ IACS Published by Elsevier BV All rights reserved doi:101016/japnum

2 G Awanou, J Lai / Applied Numerical athematics ) Introduction We will use the same notations, ) and for the inner products and norms in R n and R m The particular inner product will be identified by the types of matrices appearing The augmented Lagrangian algorithm for symmetric problems can be derived by minimization arguments We refer to [8] for details It is described as follows: with r>0andρ l > 0 for all l as parameters, given λ 0) R m specified arbitrarily, with λ l) known, compute c l) then λ l+1) by { A + rl T L ) c l) + L T λ l) = F + rl T G, λ l+1) = λ l) + ρ l Lc l) G ) 2) In [2], we were interested in a variant of this algorithm for ρ l = ρ = 1 ε for all l, r = 1 where ε>0is ε fixed ore precisely, for this choice of the parameters, the algorithm reads { A + 1 ε LT L ) c l) + L T λ l) = F + 1 ε LT G, λ l+1) = λ l) + 1 ε Lc l) G ) 3) The variant we considered is the following algorithm { A + 1 ε LT L ) c l+1) + L T λ l) = F + 1 ε LT G, λ l+1) = λ l) + 1 ε Lc l+1) G ) 4), which can be easily shown to be equivalent to the following sequence of problems [ ][ ] [ ] A L T c l+1) F L ε λ l+1) = G ελ l), 5) for = I,whereI is the identity matrix of size m m In general is a suitably chosen matrix In [9], it was claimed that the later algorithm converges to the solution c of 1) and c c l+1) Cε c c l), for a constant C>0independent of ε We have not however been able to find a proof of this result in the literature The main objective of this article is to prove the convergence of the following algorithm { A + rl T 1 L ) c l+1) + L T λ l) = F + rl T 1 G, λ l+1) = λ l) + ρ 1 Lc l+1) G ) 6), ρ >0, which generalizes 5) and give a convergence rate similar to the one above A fine study of the convergence rate still appears to be difficult cf [8, Remark 212 p 64]) The algorithm 6) is the Uzawa algorithm applied to the augmented system { A + rl T 1 L ) c + L T λ = F + rl T 1 G, 7) Lc = G, and hence can have an improved convergence rate One other advantage of the augmented Lagrangian algorithm is to solve compared to 1), systems of smaller size However, following [7, p 15], A r = A + rl T 1 L has a condition number KA r ) asymptotically proportional to r, that is KA r ) r 1 L 2, when r, σ

3 124 G Awanou, J Lai / Applied Numerical athematics ) σ depending only on A and we again denote by the matrix norm associated with on R n This prevents the choice of extremely large values of r = ρ to get convergence in one step Our convergence result and numerical evidence also suggest that one does not have a good convergence rate for r small and ρ large We also see that a judicious choice of can balance the deterioration of the condition number It is reasonable to expect that an inexact Uzawa algorithm applied to 7) might perform as well However our numerical experiments for this inexact version with several choices of r and λ indicate no convergence This is an interesting question which is still under investigation Other references for the augmented Lagrangian algorithm are [5 8,3] The paper is organized as follows We first give a sufficient condition for solvability of 1) which leads to a Ladyzhenskaya Babuška Brezzi LBB) type condition We then prove the convergence rate Finally, we will give numerical experiments for the 2D Navier Stokes equations 2 Solvability In this section, we derive a sufficient condition for the solvability of 1) Let KerX) and ImX) denote the kernel and range of the operator XWefirstgiveafewlemmas: Lemma 1 R n = KerL) ImL T ) and R m = KerL T ) ImL) Proof It is enough to prove only one of the decompositions Since ImL) R m,wehaver m = ImL) ImL), where ImL) denotes the orthogonal of ImL) We need to show that ImL) = KerL T ) Let q ImL) Forw R n, Lw ImL),soq T Lw) = 0 Therefore w T L T q = 0 so that L T q is orthogonal to R n, that is L T q = 0, ie, q KerL T ) This argument also shows that KerL T ) ImL) and the result follows The following result can be found in [4], we give here a detailed proof for convenience Lemma 2 Suppose A is an invertible linear operator with positive definite symmetric part A s = 1 2 A + A T ) that satisfies Ax,y) αa s x,x) 1/2 A s y,y) 1/2, for all x,y R n and α 1, 8) then A 1 ) s is positive definite and satisfy A 1 ) s w,w) A s ) 1 w,w ) α 2 A 1) s w,w) for all w R n oreover A 1 x,y ) A s ) 1 x,x ) 1/2 As ) 1 y,y ) 1/2 Proof As ) 1 w,w ) = sup y R n w, y) 2 A s y,y),

4 G Awanou, J Lai / Applied Numerical athematics ) and w, y) 2 = w,a 1 Ay ) 2 ) = A 1 T ) 2 w,ay α 2 A s y,y) A ) s A 1 T ) w, A 1 T ) w, by 8) So w, y) 2 α 2 y ) 2 A s A A 1 T ) w, A 1 T ) w = y ) 2 A s A 1 T w,a T A 1) T ) ) w = y 2 As A 1 T ) w,w, where 2 A s = A s, ) and we used Aw,w) = A s w,w) for all w R n It follows that As ) 1 w,w ) α 2 A 1) s w,w) On the other hand using fractional power of symmetric positive definite matrices which can be defined via singular values decomposition) A 1 ) s w,w) = A 1 w,w ) = A 1/2 s A 1 w,a s ) 1/2 w ) A 1/2 s A 1 w As ) 1/2 w = A 1 w As w As ) 1 = AA 1 w,a 1 w ) 1/2 w As ) 1 = A 1 w,w ) 1/2 w As ) 1 It follows that ) A 1 s w,w) A s ) 1 w,w ) In addition A 1 x,y ) = A 1/2 s A 1 x,a s ) 1/2 y ) A 1 x,x ) 1/2 As ) 1 y,y ) 1/2 using the same arguments as above A s ) 1 x,x ) 1/2 As ) 1 y,y ) 1/2 using 9) 9) We can now prove the following theorem Theorem 3 Let A be a matrix which satisfies the condition 8) and has a symmetric part A s positive definite with respect to L in the sense that x T A s x 0 and x T A s x = 0 with Lx = 0 implies x = 0 In addition, assume that G ImL) Then 1) is solvable and moreover y, Lu) 2 sup u R n A r u, u) c 1 y 2 for all y ImL), 10) where c 1 > 0 is a positive constant which depends on r and A r = A + rl T 1 L Proof We have { Ac + L T λ = F, Lc = G, so rl T 1 Lc = rl T 1 G which gives A + rl T 1 L ) c + L T λ = F + rl T 1 G 11)

5 126 G Awanou, J Lai / Applied Numerical athematics ) We first show that A + rl T 1 L is invertible Since A is a square matrix, it is enough to show that A + rl T 1 L ) x = 0 x = 0 We have x T A + rl T 1 L ) x = x T A s + rl T 1 L ) x = x T A s x + rlx) T 1 Lx), so by the assumptions on A, x T A + rl T 1 L ) x = 0 x T A s x = 0 and Lx) T 1 Lx) = 0 It follows that x T A s x = 0andLx = 0 Since A s is assumed to be symmetric positive definite with respect to L,wegetx = 0 We can therefore write c = A + rl T 1 L ) 1 F + rl T 1 G ) A + rl T 1 L ) 1 L T λ Since Lc = G, we see that the solvability of 1) is equivalent to solving L A + rl T 1 L ) 1 L T λ = L A + rl T 1 L ) 1 F + rl T 1 G ) G, for λ By Lemma 1, we have λ = λ 0 + λ with λ 0 KerL T ) and λ ImL) Clearly it is enough to find λ We show that there is c 1 > 0 such that y T L A + rl T 1 L ) 1 L T y c 1 y 2, for all y ImL) This will imply that LA + rl T 1 L) 1 L T is invertible on ImL) and show that 1) is solvable Since A satisfies 8) and because A s x,x) A s + rl T 1 L ) x,x ) and rl T 1 Lx, x ) A s + rl T 1 L ) x,x ), we have A + rl T 1 L ) x,y ) α A s + rl T 1 L ) x,x ) 1/2 As + rl T 1 L ) y,y ) 1/2, for all x,y R n and α = 1 It follows from Lemma 2 that [ A + rl T 1 L ) 1] s w,w) 1 As + rl T 1 L ) 1 ) w,w, for all w R n α 2 Because A s is positive definite with respect to L, A s + rl T 1 L is symmetric positive definite and so LA s + rl T 1 L) 1 L T is symmetric positive definite on ImL) L T z = 0andz ImL) implies z = 0) so that we have y T L A s + rl T 1 L ) 1 L T y c 0 y 2, for all y ImL), 12) with c 0 > 0 depending on rthisgives y T L A + rl T 1 L ) 1 L T y = L T y ) T A + rl T 1 L ) 1 L T y ) = L T y ) T[ A + rl T 1 L ) 1]s L T y ) 1 L T y ) T As + rl T 1 L ) 1 L T y ) α 2 = 1 α 2 yt L A s + rl T 1 L ) 1 L T y c 0 α 2 y 2, Recall that A r = A + rl T 1 L) 1 and notice that for all y ImL)

6 G Awanou, J Lai / Applied Numerical athematics ) y T LA 1 r L T y = A 1 r L T y,l T y ) L T y,u) 2 = sup u R n A r u, u) = sup y, Lu) 2 u R n A r u, u) We have therefore proved that y, Lu) 2 sup u R n A r u, u) c 0 α 2 y 2, which is a LBB type condition for all y ImL), It would be desirable to have more information on the dependence on r of the constant c 1 in Theorem 3 Put E = A s + rl T 1 L, and recall from 12) that LE 1 L T is symmetric positive definite on ImL) We let D = LE 1 L T considered as a mapping from ImL) to ImL) and seek a lower bound of Ry) = yt Dy > 0, y ImL), y 0, y T y where Ry) is the Raleigh quotient We have So Ry) = yt LE 1 L T y y T y = yt LE 1 EE 1 )L T y y T y = yt LE 1 )EE 1 L T y) y T y Ry) = E 1 L T y 2 E, y T y where we have defined a norm E associated with the symmetric positive definite matrix E; 13) u 2 E = bu, u), with bu,v) = vt Eu We have E 1 L T y be 1 L T y,v) E = sup, v R n v E v 0 with So be 1 L T y,v) v E = vt EE 1 L T y v E E 1 L T y y T Lv E = sup v R n v E v 0 = vt L T y v E = yt Lv v E, v 0 R n We claim that L is an invertible mapping from ImL T ) to ImL), cf Lemma 5 below Since y ImL), there is v in ImL T ) such that y = Lv We write v 2 = v T v Then, 14)

7 128 G Awanou, J Lai / Applied Numerical athematics ) Ry) 1 Lv) T ) Lv) 2 = Lv 2 Lv 2 v E v 2 E Lv 2 = v T A s v + rv T L T 1 Lv = Lv 2 v T A s v + r Lv 2 1 For an operator X, µ X max denotes the greatest eigenvalue of X Finally, using Raleigh s principle, v T A s v µ As max v 2 On the other hand, since y = Lv 0, Lv v We therefore have L 1 Ry) We have 1/ L 1 2 µ As max + r L 2 > 0 c 1 = c 0 α 2 1 α 2 1/ L 1 2 µ As max + r L 2 3 Convergence In this section, we prove the convergence of the iterative algorithm 6) In the next section we give the convergence rate Theorem 4 Suppose that the linear system 1) has a unique solution c and that A s the symmetric part of A is positive definite with respect to L oreover, assume that is symmetric positive definite Then, the sequence c l) ) defined in 6) converges to the solution c of 1) for r ρ 2 Proof Clearly 6) is solvable since A r is invertible With λ l) given one computes successively c l+1) and λ l+1) The original problem 1), { Ac + L T λ = F, Lc = G, can be rewritten as { A + rl T 1 L ) c + L T λ = F + rl T 1 G, λ = λ + ρ 1 15) Lc G) Let u l+1) = c l+1) c and p l+1) = λ l+1) λ We have, using 15) and 6), A + rl T 1 L ) u l+1) + L T p l) = 0, 16) and p l+1) = p l) + ρ 1 Lu l+1) We deduce from 17) that p l+1) 2 = p l+1),p l+1)) = p l) + ρlu l+1),p l) + ρ 1 Lu l+1)), 17)

8 G Awanou, J Lai / Applied Numerical athematics ) which gives p l+1) 2 = p l) 2 + p l),ρ 1 Lu l+1)) + ρlu l+1),p l)) + ρ 2 Lu l+1), 1 Lu l+1)) = p l) 2 + 2ρ p l),lu l+1)) + ρ 2 Lu l+1), 1 Lu l+1)), since 1 is symmetric and hence p l) 2 p l+1) 2 = 2ρ p l),lu l+1)) ρ 2 Lu l+1), 1 Lu l+1)) 18) It follows from 16) that Au l+1),u l+1)) + r L T 1 Lu l+1),u l+1)) = p l),lu l+1)), and hence by substituting this into 18), we get p l) 2 p l+1) 2 = 2ρ Au l+1),u l+1)) + 2rρ ρ 2) Lu l+1), 1 Lu l+1)) = 2ρ Au l+1),u l+1)) + 2rρ ρ 2) Lu l+1) 2 1 Therefore for r ρ 2,sinceA s is nonnegative, p l) 2 p l+1) 2 0, ie, p l+1) p l), for all l, 19) and the sequence { p l) } is seen to be decreasing If for some l, p l) p l+1) = 0, then Au l+1),u l+1) ) = Lu l+1) = 0 and we have u l+1) = 0, since A s is symmetric positive definite with respect to L, and hence convergence Otherwise, the sequence being bounded below by 0 converges; hence p l) 2 p l+1) 2 converges to 0 which implies that A s u l),u l) ) and Lu l) 2 converge to 0 Since A s + rl T 1 L is positive definite, it follows that u l) converges to 0 and finally c l) converges to c 4 Convergence rate To show the convergence rate, we will need the following lemma Lemma 5 The mappings L :ImL T ) ImL) and L T : ImL) ImL T ) are bijections with bounded inverses Proof We show that L is one one on ImL T ) This is immediate since Lx = 0 with x ImL T ) implies x ImL T ) KerL) ={0} by Lemma 1 As a linear mapping between finite dimensional spaces, L has a bounded inverse on ImL) and there exists k 0 > 0 such that for any g ImL), there exists v g ImL T ) such that Lv g = g with v g 1 k 0 g 20) A similar proof applies to L T This completes the proof

9 130 G Awanou, J Lai / Applied Numerical athematics ) We would like to elaborate on this last inequality Let v R n and g = Lv ImL) so there is v g ImL T ) for which g = Lv g Thatis,Lv = Lv g It follows that v g = v + v 0 for some v 0 KerL) We therefore have v + v 0 1 k 0 Lv = 1 k 0 Lv + v 0 ), for any v + v 0 ImL T ) It follows that v 1 Lv = 1 q, Lv) sup k 0 k 0 q R n q, for all v ImLT ) 21) The same arguments applied to L T show that q 1 L T q = 1 v, L T q) sup, k 0 k 0 v R n v for all q ImL) 22) We have the following theorem: Theorem 6 Suppose that the linear system 1) has a unique solution c and that A s the symmetric part of A is positive definite with respect to L oreover, assume that is symmetric positive definite Then, c c l+1) C c c l), for a positive constant C which depends on r and ρ but independent of l oreover for r = ρ = 1 ε, c c l+1) Cε c c l), for a positive constant C independent of l and ε Proof We write u l+1) =û l+1) + u l+1) with û l+1) KerL) and u l+1) ImL T ) Using 21), we have k 0 u l+1) Lu l+1), where we used the inner product, ) instead of the canonical one Using 17), we have Lu l+1) = 1 ρ pl+1) 1 ρ pl), so k 0 u l+1) p l+1) ρ + ) p l), where we used the equivalence of norms on R m By 19) we have p l+1) p l) which gives u l+1) 2 p l) ρk 0 We next give a bound on û l+1) SinceA + rl T 1 L is invertible, A is invertible on KerL) Indeed if Ax = 0andLx = 0, then A + rl T 1 L)x = 0 which implies that x = 0 Therefore there is α 0 > 0 such that α 0 û l+1) sup v 0 KerL) v 0,Aû l+1) ) v 0 = sup v 0 KerL) v0 TAul+1) v0 TAul+1) v 0

10 G Awanou, J Lai / Applied Numerical athematics ) However, from 16), we have Au l+1) = L T p l) rl T 1 Lu l+1) which implies v0 T Aul+1) = v0 T LT p l) rv0 T LT 1 Lu l+1) = Lv 0 ) T p l) rlv 0 ) T 1 Lu l+1) = 0, for v 0 KerL) Thus, α 0 û l+1) sup v 0 KerL) v T 0 Aul+1) v 0 A u l+1) 2 A ρk 0 p l) We therefore have u l+1) u l+1) + û l+1) 2 ) A + 1 p l) ρk 0 α 0 We now give a bound on p l) in terms of u l) First we establish that p l) ImL) Solving for u l+1) in 16) and substituting in 17), we get p l+1) = I ρ 1 L A + rl T 1 L ) 1 ) L T p l) It follows that p l+1) p l) ) is in the range of L Since k+1 p l+1) = p j) p j 1) + p 0), j=1 we have p l) ImL) provided p 0) = λ λ 0 ImL) which is possible by a suitable choice of λ 0 One way to do this is to first notice that by 11), we may assume that λ ImL)Soif = I,where I is the identity matrix of R m, we may choose λ 0 in ImL) Otherwise, we can choose in such a way that maps ImL) into ImL) It follows from 22) that k 0 p l) 1 = k 0 p l) v T L T p l) sup v R n v Combining the equations in 16) and 17), we get A + rl T 1 L ρl T 1 L ) u l) + L T p l) = 0, thus v T L T p l) = v T A + rl T 1 L ρl T 1 L)u l) Therefore, p l) A r,ρ u l), k 0 where A r,ρ = A + r ρ)l T 1 L It follows that ) u l+1) 2 A + 1 A r,ρ u l) ρk 0 α 0 k 0 For r = ρ, A r,ρ = A and for r = ρ = 1 ε,weget c c l+1) Cε c c l), for a positive constant C independent of l and ε

11 132 G Awanou, J Lai / Applied Numerical athematics ) Numerical experiments In this section, we first present a spline discretization of the 2D Navier Stokes equations in velocitypressure formulation A simple iterative algorithm is used to linearize the nonlinear equations We have applied the algorithm described here to the solutions of the linear systems which arise and display the maximum number of the iterations that was necessary to fully solve the problem for various choices of the parameters r and ρ Let Ω = t t be a polygonal domain in R2 Given two integers d 0 and 0 r<d, we consider the spline space of degree d and smoothness r Sd r ) := { s C r Ω): s t P d, t }, where P d denotes the space of polynomials of degree less than or equal d It is possible to represent in a unique fashion such a spline by a vector of coefficients, the B-net of the spline We refer to [2] for the 3D case and [10] in the 2D case for additional details The weak form of the steady state Navier Stokes equations is: Find u H 1 Ω) 2 such that ν Ω u v + div u = 0 in Ω, u = g on Ω, where 3 j=1 Ω u u j v = x j Ω f v, v V 0, V 0 = { v H0 1 Ω)2, div v = 0 }, and Ω is the boundary of Ω We approximate elements of V 0 by vectors d in R 2N each component of u is approximated by a spline with coefficient vector in R N ), which represent smooth splines, H d = 0, are zero on the boundary, Rd = 0 and satisfy the divergence-free condition, Dd = 0 Let us describe these elements as vectors d in R 2N satisfying Ld = 0 for some matrix L Ifweletc encode the coefficients of the approximant of the velocity field, the discrete problem is: Find c in R 2N satisfying Lc = G with G encoding the boundary conditions and νc T Kd + Bc)c ) T d = d T F, for all d in R 2N with constraint Ld = 0 Here, K and are the stiffness and mass matrices, respectively; Bc)c) T d encodes the nonlinear term Using functional arguments, it can be shown that there exists a Lagrange multiplier λ such that: νkc + Bc)c + L T λ = F, Lc = G If we put A = νk +Bc), we showed in [1] that the previous problem is solvable and that A s is symmetric positive definite with respect to L for ν sufficiently large The system of nonlinear equations is linearized as follows: Let c 0), λ 0) ) be the solution of the linear problem ie, the associated Stokes equations) and for n = 0, 1,,define c n+1), λ n+1) ) as the solution of νkc n+1) + B c n)) c n+1) + L T λ n+1) = F, Lc n+1) = G

12 G Awanou, J Lai / Applied Numerical athematics ) Because we are not using a basis to represent the discrete solution, the stiffness matrix K is singular We have used the algorithm described in this paper to solve at each step the previous system of equations The termination criterion for these steps is to require the maximum norm of c n+1) c n) to be less than For the augmented Lagrangian iterations we use the same criterion over successive iterations We display in Tables 1 3 the maximum number of iterations over all steps) when this algorithm is applied Table 1 Number of iterations for r and ρ with r = ρ r ρ Iterations Table 2 Number of iterations for ρ = 2 r and various r r ρ = 2 r Iterations N/A N/A Table 3 Number of iterations for large r and various ρ r ρ Iterations Fig 1 2D cavity flow velocity profile

13 134 G Awanou, J Lai / Applied Numerical athematics ) to the lid driven cavity flow problem at Reynolds number 400 We used d = 7andr = 0 in which case A has size and L has size The actual mesh consists of 128 triangles obtained by uniformly refining 3 times a triangulation of the square into two triangles Each triangle was subdivided into 4 triangles by connecting the midpoint of the edges In Table 2, N/A stands for not available We did not get a fast convergence for these values The numerical results in Table 3 suggest that for this specific problem, the choice r = 10 6 and ρ = 10 5 is optimal We finally display the velocity profile cf Fig 1) when r = ρ = 10 3 was used 6 Conclusion In this paper, we have mainly given a convergence rate of a variant of the augmented Lagrangian algorithm We intend to undertake a study of the optimal choice of the parameters r and ρ when this algorithm is applied to the incompressible Navier Stokes equations filling a gap left by earlier researchers We also intend to investigate further inexact versions of this algorithm References [1] G Awanou, Energy ethods in 3D Spline Approximations of the Navier Stokes Equations, PhD Dissertation, University of Georgia, Athens, GA, 2003 [2] G Awanou, J Lai, Trivariate spline approximations of 3D Navier Stokes equations, ath Comp, in press [3] R Bramley, X Wang, D Pelletier, Orthogonalization based iterative methods for generalized Stokes problems, in: WG Habashi Ed), Solution Techniques for Large-Scale CFD Problems, Wiley, New York, 1995, pp [4] JH Bramble, Pasciak JE, Vassilev AT, Uzawa type-algorithms for nonsymmetric saddle point problems, ath Comp ) [5] SC Brenner, LR Scott, The athematical Theory of Finite Element ethods, Springer, New York, 1994 [6] F Brezzi, Fortin, ixed and Hybrid Finite Element ethods, Springer, New York, 1991 [7] Fortin, R Glowinski, Augmented Lagrangian ethods: Applications to the Numerical Solution of Boundary-Value Problems, Studies in athematics and Its Applications, vol 15, North-Holland, Amsterdam, 1983 [8] R Glowinski, P LeTallec, Augmented Lagrangian and Operator Splitting ethods in Nonlinear echanics, SIA, Philadelphia, PA, 1989 [9] Gunzburger, Finite Element ethods for Viscous Incompressible Flows, Academic Press, New York, 1989 [10] J Lai, P Wenston, Bivariate splines for fluid flows, Comput Fluids )