Stokes Equation. Chapter Mathematical Formulations

Transcription

1 Capter 5 Stokes Equation 5.1 Matematical Formulations We introduce some notations: Let v = (v 1,v 2 ) T or v = (v 1,v 2,v 3 ) T. Wen n = 2 we define curlv = v 2 x v 1 y, ) curlη =. ( η y, η x In fact, curlv is obtained by imbedding v into R 3, take 3D curl, ten take te tird component. For scalar function η, curlη te is same as imbedding η into R 3 as (0,0,η), take 3D curl, ten take te first two components. Wen n = 3 te curl of 3 -dim vector is defined as usual i j k curlv = v = x y z v 1 v 2 v 3 (5.1) curl(curlφ) = φ, n = 2 } curl(curl v) = v+grad(divv) curl(curl v) n = 3 n = 2 1

2 2 CHAPTER 5. STOKES EQUATION We define ( ) ( ) p/ x 1 τ 11 / x 1 + τ 12 / x 2 gradp =, divτ =, p/ x 2 τ 21 / x 1 + τ 22 / x 2 ( ) v 1 / x 1 v 1 / x 2 divv = v 1 / x 1 + v 2 / x 2, Gradv =. v 2 / x 1 v 2 / x 2 We also define ( ) ( ) curlv 1 u 1 curlv =, u =. curlv 2 u 2 Teorem Let Ω be simply connected. Ten u (L 2 (Ω)) n satisfies curlu = 0 iff tere exists p H 1 (Ω) s.t u = gradp. For any two square matrices A,B we write A : B = i,j a ij b ij. We also define special tensors ( ) ( ) δ =, χ =, and tr(τ) = τ : δ = τ 11 +τ Let Finally, Gradu = u = ( u1 u 1 x 1 x 2 u 2 u 2 x 1 x 2 ). (5.2) ǫ(v) = 1 2 [Gradv+(Gradv)t ], ǫ ij (u) = 1 2 ( u i x j + u j x i ) is called deviatoric or deformation tensor. Let Ω be a domain in R n (n = 2,3) wit its boundary Γ := Ω.

3 5.1. MATHEMATICAL FORMULATIONS 3 Te Navier-Stokes equations for a viscous fluid is are as follows: u i n t + u i u j 2ν x j=1 j j ǫ ij (u) x j + p x i = f i (1 i n) in Ω, (5.3) divu = 0 (incompressible), (5.4) u = g on Γ. (5.5) Here u is te velocity of te fluid, ν > 0 is te viscosity and p is te pressure; (Here we assume p and ν are normalized so we may assume ρ = 1) and te vector f represents body forces per unit mass. If we introduce te stress tensor σ ij := pδ ij +2νǫ ij (u) we ave a simpler form : u t +(u )u divσ = f in Ω, divu = 0 in Ω, u = g on Γ. (5.6) Here te first term is interpreted as (u )v = e i j u j v i x j = j u j v x j. Note tat if divu = 0, te following identity olds j ǫ ij (u) x j = 1 2 so tat te equation can be written as j ( ) 2 u i x u j = 1 j x i x j 2 u i, for eac i (5.7) u t +(u )u ν u+gradp = f in Ω, divu = 0 in Ω, u = g on Γ. (5.8) Remark If we define uv = u v = (u i v j ) i,j, ten wen divu = 0, we see tat te nonlinear term (u )u can be written as (u u). We only consider te steady-state case and assume tat u is so small tat we can ignore te non-linear convection term u j u i x j. Tus, we ave te Stokes equation:

4 4 CHAPTER 5. STOKES EQUATION ν u+gradp = f in Ω, divu = 0 in Ω, u = g on Γ. (5.9) A weak formulation Let V = {v H 1 0 (Ω)n, divv = 0} and L 2 0 (Ω) be te space of all L2 (Ω) functions q suc tat Ωqdx = 0. Multiply (5.9) by v H 1 0 (Ω)n and integrate by parts, we obtain (ν u, v) (p, divv) = (f,v). Define a(u,v) := ν n i,j=1 ( ui, v ) i = ν gradu : gradvdx (5.10) x j x j Ω b(v,q) := (q, divv). (5.11) Ten we ave te equivalent weak (abstract) form of (5.9) : Find u H 1 (Ω) n s.t. a(u,v)+b(v,p) = f,v for all v H 1 0 (Ω)n, b(u,q) = 0 for all q L 2 0 (Ω), u = g on Γ. (5.12) We can find a function u g H 1 (Ω) n suc tat divu g = 0 on Ω, u g = g on Γ so tat u can be decomposed as u = w+u g,w H 1 0 (Ω)n. Wit l,v := f,v a(u g,v) te problem (5.12) is equivalent to : Find a unique pair of functions (w,p)

5 5.1. MATHEMATICAL FORMULATIONS 5 H0 1(Ω)n L 2 0 (Ω) suc tat { a(w,v)+b(v,p) = l,v for all v H 1 0 (Ω)n b(w,q) = 0 for all q L 2 0 (Ω). (5.13) Tis problem actual as a unique solution by Corollary below. Furtermore, we ave te following : u 1 + p 0 C( f 1 + g 1/2,Γ ). (5.14) A General result Now let us put problem (5.13) into general framework of cap 4: We set X = H 1 0 (Ω)n, M = L 2 0 (Ω). Let X and M be two Hilbert spaces wit norms X and M and let X and M be teir dual spaces. As usual, we denote, be te duality pairing between X and X or M and M. Introduce bilinear forms a(, ) : X X R, b(, ) : X M R wit norms a = sup u,v a(u, v) u X v X, b = sup v X,µ M b(v, µ) v X µ M. We consider te following two variational problem called problem (Q): Given l X and χ M, find a pair (u,λ) X M suc tat a(u,v)+b(v,λ) = l,v for all v X (5.15) b(u,µ) = χ,µ for all µ M. (5.16) In order to study (Q), we associate two linear operators A L(X;X ) and B L(X;M ) defined by Au,v = a(u,v) for all u,v X (5.17) Bv,µ = b(v,µ) for all v X,µ M. (5.18)

6 6 CHAPTER 5. STOKES EQUATION Let B L(M;X ) be dual operators defined by B µ,v = µ,bv = b(v,µ),v X,µ M. (5.19) Wit tese, te problem can be written as: Find (u,λ) X M suc tat Au+B λ = l in X (5.20) Bu = χ in M. (5.21) We set V = Ker(B) and more generally define V(χ) = {v X;Bv = χ}. Note tat V = V(0). Now problem (Q) can be canged into equivalent form (P): Find u V(χ) suc tat a(u,v) = l,v, v V. (5.22) Define te polar set V 0 by V 0 = {g X ;< g,v >= 0, v V}. Lemma Te followings are equivalent: (i) Tere is a constant β > 0 suc tat inf sup b(v, µ) β > 0. (5.23) µ M v X, v X µ M (ii) Te operator B is an isomorpism from M onto V 0 and B µ X β µ M. (5.24) (ii) Te operator B is an isomorpism from V onto M and Bv M β v X. (5.25) Teorem Te problem (Q) as a unique solution if (i) πa is an isomorpism from V onto V (p. 59) and (ii) tere is a constant β > 0 suc

7 5.1. MATHEMATICAL FORMULATIONS 7 tat inf sup b(v, µ) β > 0. (5.26) µ M v X, v X µ M Corollary Assume tat a(, ) is coercive on V, i.e., tere exists a constant α > 0 suc tat a(v,v) α v 2 X, v V. (5.27) Ten problem (Q) as unique solution if and only if b(, ) satisfies inf-sup condition. Now let us put problem (5.9) into general framework of cap 4.: We set X = H 1 0(Ω) n, M = L 2 0(Ω). Te coice M = L 2 0 (Ω) is a matter of convenience and we can just as well take M = L 2 (Ω)/R. Finite dimensional problem p 123. Girault - Raviart, Finite element approximation of te Navier-Stokes equations. Now cange every space to finite dimensional one. Let X X, M M be finite dimensional subspace wit certain approximation properties. Asintecontinuouscase, weassociatetwolinearoperatorsa L(X;X ), B L(X;M ) and B L(M;X ) defined by A u,v = a(u,v ) for all v X,u X, (5.28) B v,µ = b(v,µ ) for all µ M,v X, (5.29) v,b µ = b(v,µ) for all v X,µ M. (5.30) We define te finite dimensional analogue of V: V (χ) = {v X ;b(v,µ ) =< χ,µ >,µ M }.

8 8 CHAPTER 5. STOKES EQUATION ŷ 2 O ˆK 1 0 Reference element ˆx 4 5 K 5 K K 4 K K 1 K K 0 K Local node number 0 1 Global nodes for CR FEM Figure 5.1: Nodes for CR nonconforming FEM We set V = V (0) = Ker(B ) X = {v X ;b(v,µ ) = 0,µ M }. Caution: V V and V (χ) V(χ), since M is a proper subspace of M. We now define te approximate problem. (Q ): For l given in X and χ M, find a pair (u,λ ) in X M suc tat a(u,v )+b(v,λ ) = l,v, v X (5.31) b(u,µ )+c(λ,µ ) = χ,µ, µ M. (5.32) Now problem (Q ) can be canged into te equivalent problem. (P ): Find u V (χ) suc tat a(u,v ) = l,v, v V. (5.33) Mapping to reference element If ˆφ is any scalar function defined over ˆK( ˆK), we associate a function on K(pull back) by φ = ˆφ F 1 K. (5.34) Hence φ = B T 1 K ˆφ F K, ˆφ = BK T φ F K. (5.35)

9 5.1. MATHEMATICAL FORMULATIONS Matrix representation A simplest coice (X,M ) is (P n 1,P 0). In tis case, te matrix A is just two copies of scalar Laplacian. For B, we test te following for v = (v 1,v 2 ) associated wit edges 2,3,7,10,12,13,14,15 only. It suffices to test v = (v 1,0) only. We test te equation for (0,φ 2 ), were φ 2 is te scalar basis function associated wit te node 2. Since φ 2 = 2y in K 1 3 2y in K 4 Or φ 2 (x,y) = ˆφ 2 F 1 K 1 on K 1 and φ 2 (x,y) = ˆφ 2 F 1 K 4 on K 4. Here We see ( )( ) ( ) ( )( ) ( ) 0 ˆx 0 ˆx 0 F K1 = +, F K4 = + 0 ŷ 0 ŷ Similarly, since < Bv,p > = b(v,p ) = (divv,p ) K1 K 4 2 = K 1 p 1 2 K 4 p 4 = (p 1 p 4 ) φ 12 = if we test te equation for v = (φ 12,0), 2x+2y in K 0 3 2x 2y in K 1 < Bv,p > = (divv,p ) K0 K 1 2 = K 0 p 0 2 K 1 p 1 = (p 0 p 1 ) Since φ 7 = 2x in K 1 3 2x in K 2 < Bv,p > = K 1 2 p 1 K 2 2 p 2 = (p 1 p 2 )

10 10 CHAPTER 5. STOKES EQUATION B t = (B 1,B 2 ) t is 16 8 matrix, see (5.53), but if we only see te rows of B t corresp to (φ,0) (x-component basis ftns) tey are like 8 8 matrix B1 t = exer. Write down entries of B corresponding to basis function of type (0,φ) Error estimate Teorem (A) Assume V (χ) is nonempty and tere exists a constant α > 0 suc tat (B) te discrete inf-sup condition old a(v,v ) α v 2 X, v V. (5.36) b(v,µ ) sup β µ M > 0, µ M. (5.37) v X v X Ten te problem (P ) as a unique solution u V (χ) and tere is a constant C > 0 suc tat Ten tere is a unique solution (u,λ ) of te problem (Q ) { } u u X + λ λ M C 2 inf u v X + inf λ µ M. (5.38) v X µ M Cecking te discrete inf-sup condition Lemma Te te discrete inf-sup condition (5.37) olds wit a constant β > 0 independent of if and only if tere exists an operator Π L(X;X ) satisfying b(v Π v,µ ) = 0, µ M,v X (5.39) and Π v X C v X, v X. (5.40)

11 5.1. MATHEMATICAL FORMULATIONS 11 Proof. Assume suc Π exists. Ten by Π v X C v X, we see b(v,µ ) sup v X v X b(π v,µ ) sup v X Π v X b(v,µ ) = sup v X Π v X 1 C sup v X b(v,µ ) v X β C µ M Error Estimate Hypotesis (1) Tere exists an operator r : H m+1 (Ω) n H0 1(Ω)n X suc tat v r v 1 C m v m+1, v H m+1 (Ω) n 1 m l. (5.41) (2) Tere exists an operator S : L 2 (Ω) M suc tat q S q 0 C m q m, q H m (Ω) n, 0 m l. (5.42) (3) (Uniform inf-sup condition) For eac q M tere exists v X suc tat (q, divv ) = q 2 0, (5.43) v 1 C q 0, (5.44) were C > 0 is independent of,q and v. Teorem Under te Hypotesis te solution of te problem(5.31) satisfies u u 1 + p p 0 C m { u m+1 + p m }. (5.45) Remark One can expect one iger order for L 2 error estimate by duality tecnique. u u 0 C{ u u 1 +inf p p 0 }. (5.46)

12 12 CHAPTER 5. STOKES EQUATION Stabilization. (Verfürtnote)Wemayaddtefollowingtermδ K 2 (div[ u+ p] divf) to te second equation for eac element K. 5.2 Solver Assembly of A and B One way to look at te eq. is u 1 +p x = f 1 (5.47) u 2 +p y = f 2 (5.48) (u 1 ) x +(u 2 ) y = 0. (5.49) Wit b 1 (u 1,p) = ((u 1 ) x,p), b 2 (u 2,p) = ((u 2 ) y,p), its weak form is a(u 1,v 1 )+b 1 (v 1,p) = f 1 (5.50) a(u 2,v 2 )+b 2 (v 2,p) = f 2 (5.51) b 1 (u 1,q)+b 2 (u 2,q) = 0 (5.52) In matrix, Let 2n v = dimx, n p = dimm. Ten for n v n v matrix A and n p n v matrix B A 0 B1 t U x F 1 0 A B2 t U y = F 2 (5.53) B 1 B 2 0 P 0 If we use Uzawa we do not need explicit B. Instead only need action of b(v,p ) for given v and p. Standard Uzawa Let p 0 given. Let small ǫ > 0 be fixed (ǫ = 1.5 in Verfüt note). Solve for m = 0,1,, until p m+1 p m is sufficiently small. a(u m+1,v )+b(v,p m ) = (f,v ) a(u g,v ), v X b(u m+1,q)+δc(p m,q) = 1 ǫ (pm+1 p m,q)+δχ (q), q M Normalize p m+1 eac step so tat it belongs to M = L 2 0 (Ω).

13 5.2. SOLVER 13 Au +B T p = l (5.54) Bu δcp = δχ. (5.55) Tus in matrix form we ave ( )( ) ( ) A B T u l = B δc p δχ (5.56) Standard Uzawa-Again (1) Given: an initial guess p 0 for te pressure, a tolerance Tol > 0 and a relaxation parameter ǫ > 0. (2) Apply a few Gauss-Seidel iterations (fix p m ) to te linear system Au m = l B T p m and denote te result by u m+1. Compute p m+1 = p m +ǫ(bum+1 δcp m +δχ ) (3) Stop if Au m+1 +B T p m+1 l + Bu m+1 δcp m+1 +δχ Tol Improved Uzawa-cg MG Solve te first equation for u (you may use multigrid) as u = A 1 (l B T p ) and insert into te second eq. Tis gives BA 1 (l B T p ) δcp = δχ Ap := [BA 1 B T +δc]p = BA 1 l +δχ := f. (5.57)

14 14 CHAPTER 5. STOKES EQUATION One can sow tat A is SPD and te condition number is O(1), ence we can Apply conjugate gradient metod to tis system Ap = f. Tis algoritm requires evaluation of A 1 f were one can use a fast algoritm suc as mutligrid metod. Te next task is ow to construct spaces X and M wic satisfy te ypoteses. Te CG-algoritm in general breaks down for non-symmetric or indefinite systems. However, tere are various variants of te CG-algoritm wic can be applied to tese problems. A naive approac consists in applying te CG-algoritm to te squared system L T k L kx k = L T k b k. Tis approac cannot be recommended since squaring te systems squares its condition number. A more efficient algoritm is te stabilized bi-conjugate gradient algoritm, sortly Bi-CG-stab. Te underlying idea rougly is to solve simultaneously te original problem L k x k = b k and its adjoint L T k y k = b T k. Stable pair for Stokes equation For Stokes equation, we need to coose pair of spaces so tat inf-sup condition olds. Assume T consists of triangles. Typically we use P 2 for te velocity and P 1 for te pressure. Anoter coice is P 1 -nonconforming for velocity and P 0 for pressure(called C-R(Crouzeix-Raviart-1973) element). (P 2,P 1 ) pair -Taylor Hood We can sow te inf-sup condition and we ave u u + p p 0 C 2 { u 2 + p 1 }. (5.58) (P1 n,p 0) pair- Crouzeix- Raviart Let P 0 be te space of all functions wic are piecewise constant on eac T. We ave u u + p p 0 C{ u 2 + p 1 }. (5.59) II.3. Petrov-Galerkin stabilization Te mini element revisited.

15 5.2. SOLVER 15 Table 5.1: Summary of 2D triangular elements velocity pressure Name Sketc LBB Order Remarks P 1 P 0 N 1 Rarely used P n 1 P 0 P + 1 P 1 (mini) P 1 P 1 (4 patc macro) P k P k 1 (Taylor-Hood) P k B k+1 P 1 k 1 Y 1 C-R, not for natural BC. Y 1 cubic bubble Y 1 iso P 2 -P 1 Y 2 P 2 P 1 engineer s favor Y 2 C-R P + 2 P 1 1 Table 5.2: Pressure given by circle in te interior means discontinuous Stabilization Motivated by mini element, we can solve te following eq. wit P 1 /P 1 pair. ν u+gradp = f in Ω, divu α p = αdivf in Ω, u = g on Γ. (5.60) Taking into account tat u T vanises elementwise (for mini), te discrete problem does not cange if we also add te term α divu to te left-and side of te second equation. Tis sows tat in total we may add te divergence of te momentum equation as a penalty. For te general form see, Verfurt note. (also my paper wit Kwon)

16 16 CHAPTER 5. STOKES EQUATION 5.3 Numerical metod for Navier Stokes equation Picard s iteration u m+1 + p m+1 = f (u m )u m in Ω, divu m+1 = 0 in Ω, u m+1 = g on Γ. (5.61) Newton s metod Consider u+(u )u+ p = f in Ω, divu = 0 in Ω, u = g on Γ. (5.62) Linearize (or correct wit) u m+1 = u m +δu to see u m+1 +(u m+1 )(u m +δu)+ p m+1 = f in Ω, u m+1 +(u m+1 )u m +(u m )δu+ p m+1 +(δu) 2 = f in Ω, u m+1 +(u m+1 )u m +(u m )(u m+1 u m )+ p m+1 = f in Ω. (5.63) Tus we define te Newton iteration as : Given initial guess u 0,p 0 solve te following for m = 0,1,,, say wit Uzawa for eac m until convergence. u m+1 + p m+1 +(u m+1 )u m +(u m )u m+1 = f +(u m )u m in Ω divu m+1 = 0 in Ω, u m+1 = g on Γ. (5.64) Te Newton iteration converges quadratically. However, te initial guess must be close to te sougt solution, oterwise te iteration may diverge. To avoid tis, one can use a damped Newton iteration Projection sceme for Navier-Stokes eq... Corin 68 See Jie Sen note IMS NUS.pdf. Consider te Full Navier Stokes problem: u t +(u )u divσ = f in Ω, divu = 0 in Ω, u = g on Γ. (5.65)

17 5.3. NUMERICAL METHOD FOR NAVIER STOKES EQUATION 17 Te original projection metod, proposed by Corin [12] and Temam [69], was motivated by te idea of operator splitting, its semi-discrete version. Remark Teabovescemeasonly1/2-orderforvelocity inl 2 (0,T;H 1 ) due to te nonpysical boundary condition pk+1 n Γ = 0. Te improved projection type sceme appears to be, te so called pressurecorrection sceme introduced in [26] (K. Goda. A multistep...cavity flows. J. C. P., 1979.). Its first-order version reads : Find ũ k+1 by solving ũ k+1 u k δt ν ũ k+1 +(u k )ũ k+1 + p k = f(t k+1 ) in Ω, ũ k+1 Γ = 0. (5.66) Ten find (p k+1,u k+1 ) from u k+1 ũ k+1 δt + (p k+1 p k ) = 0, u k+1 = 0, u k+1 n Γ = 0. (5.67) By taking tedivergence of tefirstequation in (5.67), we findtat te second step is equivalent to (p k+1 p k ) = 1 δt ũk+1 (p, k+1 p k ) n Γ = 0, u k+1 = ũ k+1 δt (p k+1 p k ). (5.68)... It is also sown tat te above sceme is first-order accurate for te velocity in L (0,T;L 2 ) L 2 (0,T;H 1 ) and pressurein L 2 (0,T;L 2 ). A popular second-order version reads 3ũ k+1 4u k +u k 1 2δt +(2u k u k 1 ) ũ k+1 ν ũ k+1 + p k = f(t k+1 ), ũ k+1 Ω = 0. (5.69) ten find (p k+1,u k+1 ) by 3u k+1 3ũ k+1 2δt + (p k+1 p k ) = 0, divu k+1 = 0 u k+1 n Ω = 0. (5.70) (To solve again take divergence as before) Te above sceme is second-order

18 18 CHAPTER 5. STOKES EQUATION for te velocity in te L 2 (0,T;L 2 (Ω))-norm. 5.4 Stream-function formulation Te discrete velocity fields computed by te metods of te previous sections in general are not exactly incompressible. It is only weakly incompressible. In tis section we will consider a formulation of te Stokes equations wic leads to conforming solenoidal discretizations. Tis advantage, of course, as to be paid for by oter drawbacks. Trougout tis section we assume tat Ω is a two dimensional, simply connected polygonal domain Te curl operators We need two curl-operators: curlφ = ( φ y, φ x ), curlv = ( v 1 y v 2 )(sgn is diffent from usual) x (Stokes-2D) Ω curlu ξdx = Ω u curlξdx u τξds (5.71) Ω Te following deep matematical result is fundamental: A vector-field v : Ω R 2 is solenoidal, i.e. divv = 0, if and only if tere is a unique stream-function φω R : suc tat v = curlφ in Ω and φ = 0 on Γ Stream-function formulation of te Stokes equations Let(u,p) betesolution oftestokes equationswitforcef andomogeneous boundary conditions and denote by ψ te stream function corresponding to u. Since we conclude tat in addition ψ n u t = 0 on Γ, = t curlψ = 0 on Γ.

19 5.4. STREAM-FUNCTION FORMULATION 19 Inserting tis representation of u in te momentum equation and applying te operator curl we obtain curl f = curl( u + p) (5.72) = (curlu) + ( p) (5.73) = (curl(curlψ)) = 2 ψ (5.74) Tis proves tat te stream function solves te biarmonic equation wit omo. BC. 2 ψ = curlf in Ω φ = 0 on Ω φ n = 0 on Ω How about two pase in tis form? Cange tis to MFVM Conversely, one can prove: If solves te above biarmonic equation, tere is a unique pressure p wit mean-value 0 suc tat u = curlψ and p solve te Stokes equations. In tis sense, te Stokes equations and te biarmonic equation are equivalent. Remark II.5.1. Given a solution ψ of te biarmonic equation and te corresponding velocity u = curl ψ te pressure is determined by te equation f + u = p. But tere is no constructive way to solve tis problem. Hence, te biarmonic equation is only capable to yield te velocity field of te Stokes equations.