A PROOF OF THE INF-SUP CONDITION FOR THE STOKES EQUATIONS ON LIPSCHITZ DOMAINS

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1 A PROOF OF THE INF-SUP CONDITION FOR THE STOKES EQUATIONS ON LIPSCHITZ DOMAINS JAMES H. BRAMBLE Abstact. The pupose of this pape is to pesent a athe simple poof of an inequality of Nečas [9] which is equivalent to the inf-sup condition. This inequality is fundamental in the study of the Stokes equations. The bounday of the domain is only assumed to be Lipschitz. 1. Intoduction One of the most impotant inequalities in the theoy of incompessible fluids is the so-called inf-sup condition, cf. [2] [3] [5] [10]. Let Ω be a bounded domain in R N with a Lipschitz bounday. Let L 2 (Ω) and H0 1 (Ω) be the standad Hilbet spaces of distibutions on Ω, with noms L 2 (Ω) and H 1 0 (Ω). The L 2 -inne poduct, as well as the paiing between H0(Ω) 1 and its dual, H 1 (Ω) is denoted by (, ). Note that the space D(Ω), of infinitely diffeentiable functions with compact suppot in Ω, is dense in both L 2 (Ω) and H0(Ω). 1 Let L 2 0 (Ω) = {u L2 (Ω) u dx = 0}. Denote by v vecto functions with Ω components v i, i = 1,..., N. In the following, C denotes a constant which depends only on Ω unless othewise stated. The impotant inf-sup condition is (1.1) inf u L 2 0 (Ω) (u, v) sup C > 0, v [H u 0 1(Ω)]N L 2 (Ω) v [H 1 (Ω)] N whee v = N v i is the standad divegence opeato and v 2 = [H0 1(Ω)]N N v i 2 H 1 (Ω). This is the same as (1.2) C u L 2 (Ω) sup v [H 1 0 (Ω)]N (u, v) v [H 1 (Ω)] N fo all u L 2 0(Ω). It is easy to see that both (1.1) and (1.2) ae equivalent to v x (1.3) C u L 2 (Ω) sup i ) fo all u L 2 0(Ω). The nom on H 1 (Ω) is given by (u, v [H0 1(Ω)] v H 1 (Ω) (u, v) (1.4) u H 1 (Ω) = sup. v [H0 1(Ω)] v H 1 (Ω) Date: Novembe 9, This wok was suppoted in pat unde the National Science Foundation Gant No. DMS

2 2 JAMES H. BRAMBLE Noting that, fo u L 2 (Ω), the distibutional deivative, u H 1 (Ω), (1.3) is the same as (1.5) C u L 2 (Ω) u H 1 (Ω), fo all u L 2 0 (Ω). The following inequality (1.6) ( N ) 1/2 C u L 2 (Ω) u 2 H 1 (Ω) + u 2 H x 1 (Ω) u X(Ω), fo all u L 2 (Ω), i was poved by Nečas [9]. It is not difficult to show that (1.6) is equivalent to (1.3) and theefoe equivalent to (1.2). We will show in Appendix I that existence and uniqueness in the Stokes poblem follows easily fom (1.2). Thus (1.6) is of fundamental impotance. To emphasize this point even moe, it is easy to see that Kon s second inequality follows fom (1.6), (cf. Duvault-Lions [3]) which is fundamental in the theoy of elasticity. Diect poofs of Kon s second inequality, in the case of nonsmooth domains ae given in, e.g., [4], [6] and [8]. The poofs ae not elementay. As is done in [3], we will show in Appendix II that Kon s second inequality is a coollay of (1.6). The pupose of this note is to povide a poof of (1.6). Moe pecisely we shall give a poof of the following. Theoem 1.1. Let Ω R N be a connected domain with a Lipschitz bounday. Then thee exists a constant C such that, fo all u L 2 (Ω), (1.7) C u L 2 (Ω) u X(Ω). Clealy if Ω = M Ω j and if (1.7) holds fo each Ω j, j = 1,...,M <, then it holds fo Ω. We will use the fact that any Lipschitz domain can be witten as the union of stongly sta shaped Lipschitz domains. Hence we may assume, without loss, that Ω is stongly sta shaped. A weake vesion of this statement is poved by Benadi [1]. The stonge vesion follows by a slight modification of he poof. A stongly sta shaped Lipschitz domain is a domain which can be epesented as follows. Ω = {x R N < g(x/)} and Ω = {x R N = g(x/)} whee = x and g W 1 with g W 1 Λ. Evidently we have assumed that Ω is sta shaped with espect to the oigin. By x/ we mean (x 1 /,...,x N /) and note that x i / is independent of away fom the oigin.

3 THE INF-SUP CONDITION 3 2. A tansfomation of B to Ω The plan is to map the unit ball B to Ω with B mapped to the bounday Ω and then to show that Theoem 1.1 holds if it holds fo Ω = B. We will use the vaiables y in B and set y = ρ. The bounday B = {y R N ρ = 1}. The simplest tansfomation would be defined by x i = y i g(y/ρ). This tansfomation is not smooth enough to do the job unless Ω is sufficiently smooth, so we follow the idea of Nečas and smooth g. We do this as follows. Let Φ D(R N ) with Φ(x) = 0 if x 1, Φ(x) 0 and Φ(x) dx = 1. Fo 0 < h < 1/4 and x > 1/2, define g(h, x) = 1 g(y/ρ)φ((x y)/h) dy. h N R N The following ae some impotant popeties of g(h, x). (1) lim h 0 g(h, y/ρ) = g(y/ρ) (2) g(h, x) C h (3) g(h, x) C (4) D α g(h, x) C/h, α = 2 (5) Fo x = 1, g(h, x) g(x) Ch, whee C depends only on Λ, the Lipschitz bound, and bounds fo g(x/). We next set h = ǫ(1 ρ). Fom the last inequality, 5, we obtain easily that, fo ǫ small enough g(ǫ(1 ρ), y/ρ) g(y/ρ) Cǫ C > 0. We ae now in a position to define the tansfomation T fom the unit ball to Ω as follows. Fo y B define T by (2.1) x i = y i g(ǫ(1 ρ), y/ρ), fo i = 1,..., N and ǫ small enough. We next check that T maps B to Ω. To this end notice that = ρg(ǫ(1 ρ), y/ρ) and set f(ρ) = ρg(ǫ(1 ρ), y/ y ) fo y B fixed. Now f(0) = 0 and f(1) = g(y/ y ). Now f (ρ) = g(ǫ(1 ρ), y/ y ) ǫ g(ǫ(1 ρ), y/ y ) > 0 h if ǫ is small enough. Thus T maps B to Ω. We next compute the Jacobian of T. Fo ease of notation, set G = g(ǫ(1 ρ), y/ρ). Fom (2.1) the Jacobian matix a ij is a ij = y j = δ ij G + y i G y j. Let v 1 be the vecto in R N whose components ae v 1 j = y j/ρ. Then a ij vj 1 = Gvi 1 + ρvi 1 G ρ = (G ǫρg h )v1 i,

4 4 JAMES H. BRAMBLE so that v 1 is an eigenvecto of a ij with eigenvalue (G ǫρ G ). To find the h est of the eigenvalues, let {v k } be an othonomal set of vectos. Then A kl = i,j a ijvi k vj l has the same eigenvalues as a ij. Now A kl = δ kl G + i v k i v 1 i ρ j v l j G. y j Hence A kl is an uppe diagonal matix with A kk = G if k > 1. Thus we have found that the Jacobian deteminant J = (G ǫρ G h )GN 1. Now fom Popety 4, D α g(1 ρ, y/ρ) C/(1 ρ), fo any α = 2. We have fo two constants c 0 and c 1 that c 0 g(y/ρ) c 1. Hence c 1 (1 ρ) (1 ρ)g(y/ρ) = g(x/) + ρ(g(ǫ(1 ρ), y/ρ) g(y/ρ)), whee we used the fact that y/ρ = x/. Thus, using Popety 5, Similaly c 1 (1 ρ) (1 ρ)g(y/ρ) g(x/) ρǫ C(1 ρ). c 0 (1 ρ) (1 ρ)g(y/ρ) g(x/) + ρǫ C(1 ρ). Hence, fo ǫ small enough we have It follows then that fo α = 2 and theefoe fo α = 1 c 0 (1 ρ) (g(x/) ) c 1 (1 ρ). D α g(1 ρ, y/ρ) C/(g(x/) ) D α J C/(g(x/) ), whee D α is any patial deivative of ode one with espect to x i o y i. We will need a few lemmas. Lemma 3.1. Let u L 2 (R N ). Then Poof. Using the Fouie tansfom, û 3. Some lemmas u L 2 (R N ) = u X(R N ). N u 2 L 2 (R N ) = û 2 L 2 (R N ) = (1+ ξ 2 ) 1/2 û 2 L 2 (R N ) + (1+ ξ 2 ) 1/2 ξ i û 2 L 2 (R N ) = u 2 X(R N ). Lemma 3.2. Let Ω R N be a bounded domain. Let γ D(Ω) be fixed. Then thee exists a constant C γ depending only on γ such that, fo all u L 2 (Ω), γu L 2 (Ω) C γ u X(Ω).

5 THE INF-SUP CONDITION 5 Poof. Using Lemma 3.1 γu L 2 (Ω) = γu L 2 (R N ) = γu X(R N ). The lemma follow fom the fact that multiplication by γ is a bounded opeato fom X(Ω) to X(R N ). We next give an elementay poof of a Hady-type inequality. Lemma 3.3. Let φ D(Ω). Then (3.1) (g(x/) ) dx 2 4 φ 2 H 1 (Ω). Ω Poof. Let = {x Ω x > δ > 0}. Then φ Ωδ 2 ( ) (g(x/) ) dx = dx 2 g(x/) Now ( ) g(x/) = (N 1) [ x i dx = (g(x/) ) dx Ωδ x =δ 1 g(x/) g(x/) ] dx g(x/) δ ds, ( ) whee the last integal is taken ove the suface of the N-sphee of adius δ. Hence, fo δ < min g(x/), ) Thus ( g(x/) dx 2 (g(x/) ) 2 dx 0. 1 g(x/) φφ dx. Using Schwaz s inequality and letting δ go to zeo we obtain Lemma 3.3. We now use this to pove the following. Lemma 3.4. Let f C 1 (Ω) C 0 ( Ω) and satisfy D α f C/(g(x/) ) α fo x Ω and α 1. Then thee is constant C depending only on C such that (3.2) fφ H 1 (Ω) C φ H 1 (Ω), fo all φ D(Ω). Poof. This follows immediately fom the Lemma Reduction to the unit ball Let B δ = {y B y > δ}. In view of Lemma 3.2 it suffices to pove (1.6) fo ṽ D(T(B δ )). Thus we want to pove the following poposition. Poposition 4.1. Let v D(B δ ) and ṽ(x) := v(t 1 (x)). Then dx. and v X(B) C ṽ X(Ω) ṽ L 2 (Ω) C v L 2 (B).

6 6 JAMES H. BRAMBLE Poof. Let η D(B \ B δ ) be such that η(y) = 1 fo y δ/2. Since v D(B δ ) we note that η v = 0. Hence Thus sup φ D(B) C, φ) = ((1 η) v, φ) =, (1 η)φ)., φ) φ H 1 (B) y sup i, ψ) φ H0 1(B ψ δ/2) H 1 (B) Next we make a change of vaiables. v, ψ) = ψ dy = y = sup i, (1 η)φ) (1 η)φ H 1 (B) φ D(B) (1 η)φ H 1 (B) φ H 1 (B) B = C sup φ D(B δ/2 ) Ω, ψ). ψ H 1 (B) ṽ ψj 1 dx, whee J = [g(ǫ(1 ρ), y/ρ) ǫρ g (ǫ(1 ρ), y/ρ)][g(ǫ(1 ρ), h y/ρ)]n 1. Set f ij = J 1. Then Since, ψ) ṽ H 1 (Ω) f ij ψ H 1 (Ω). D α f ij C/(g(x/) ) fo α = 1, it follows fom Lemma 3.4 that Hence f ij ψ H 1 (Ω) C ψ H 1 (Ω) C ψ H 1 (B). v H 1 (B) C Clealy it follows in the same way that ṽ H 1 (Ω). v H 1 (B) C ṽ H 1 (Ω). This poves the fist inequality of the poposition. Since J is bounded, the second inequality of the poposition follows. 5. The Lemma fo the unit ball We want to pove that fo the unit ball B in R N u L 2 (B) C( u H 1 (B) + u H 1 (B)). Let η D(B) be such that η(x) = 1 fo x 1/3 and η(x) = 0 fo x 1/2. Then ηu L 2 (B) = ηu L 2 (R N ) ηu X(R N ).

7 Now ηu X(R N ) THE INF-SUP CONDITION 7 (ηu, φ) sup + φ D(R N ) φ 1 sup φ D(R N ) ( (ηu), φ) φ 1 C u X(B). Hence we need to bound v = (1 η)u D(B 1/4 ) whee B 1/4 = {x 1/4 < x < 1}. In ode to pove that v L 2 (B) C u X(B), we define the following thee opeatos, each of which maps D(R N ) to HD 1 (B 1/4), whee HD 1 (B 1/4) = {v H 1 (B 1/4 ) v(x) = 0, x = 1}. Fo φ D(R N ), define ( ) ( ) (2 ) (3 2) Pφ = φ(x) + 3φ x 4φ x, ( ) ( ) (2 ) (3 2) Pφ = φ(x) 3φ x + 2φ x and ( ) ( ) (2 ) (3 2) ˆPφ = φ(x) + 3[/(2 )]φ x 4[/(3 2)]φ x. The following elations hold: Also (x i/) = 0, = x i By constuction we have N + ( ) x j x i = 0. ( ) x j x 2 j x i. P φ = ( Pφ) and [( ) ] ( ) P x j x i φ = x j x i ( x ˆPφ). j Define P t by (P t v, φ) := (v, Pφ) and note that fo v D(B 1/4 ) we have that P t v D(R N ) and P t v = v in B. In fact the suppot of P t v is contained in the annulus 1/4 < < 5/2. Now we have ( (P t v), φ) = (v, P( φ )) = ( v, P [ ( )]) xi φ ( [ ( )]) xj v, P x 2 j x i. Notice that if f is a function such that f(x) = h(x/) then fo any α > 0, f(αx) = f(x), i.e. f(x) is homogeneous of degee 0. Hence fo such a function P [fφ] = fpφ. Taking f(x) = x i we see that P [ x i φ] = x ipφ. Thus ( [ (xi ) ]) ( φ v, P = v, x ( )) i Pφ

8 8 JAMES H. BRAMBLE and ( [ ( xj xj v, P x i We now see that ( (P t v), φ) = ) ]) ( φ = v, x ( j xj x i ( xi ) v, ( Pφ ˆPφ) +, x ˆPφ). i ) ) ( ˆPφ). The mapping defined by y i = (2 ) x i, = x, fo x B 1/4 (and similaly the mapping y i = (3 2) x i, = x, fo x B 1/4 ) is smooth. We veify this at the end of this section by computing its Jacobian. Theefoe Pφ H 1 (B 1/4 ) + Pφ H 1 (B 1/4 ) + ˆPφ H 1 (B 1/4 ) C φ H 1 (R N ). Hence v L 2 (B) = P t v L 2 (B) P t v L 2 (R N ) P t v H 1 (R N ) + v H 1 (B 1/4 ) + Thus we finally conclude that v H 1 (B 1/4 ). (P t v) H 1 (R N ) v L 2 (B) C v X(B) C u X(B), which completes the poof povided we check the smoothness of the the abovementioned mappings. To this end let and Define by fo x B 0 and y B 1. Now and the Jacobian of M is B 0 = {x R N 0 < x < 1} B 1 = {x R N 1 < x < 2}. M : B 0 B 1 y i = b ij := = (2 ) x i, (2 ) δ ij 2 x ix j 3 J = det( ) = ((2 )/) N 1.

9 THE INF-SUP CONDITION 9 To see this, note that the vecto v 1, with components {vi 1 = x i /, i = 1,...,N}, is an eigenvecto with eigenvalue λ 1 = 1 of the matix with enties b ij ; i.e. b ij vj 1 (2 ) = vi 1 2/v1 i vj 1 v1 j = v1 i. Now let {v k, k = 1,...,N} be N othonomal vectos, with enties {vi k, i = 1,...,N} fo each k; i.e. vi k vi l = δ kl. Then fo k > 1 b ij vj k = (2 ) vi k 2/vi 1 Hence λ k = (2 ), fo k > 1 and theefoe vjv 1 j k = (2 ) vi k. J = det( ) = λ 1 λ N = ((2 )/) N Appendix I: Existence and uniqueness in the Stokes poblem We will show how (1.2) may be used to easily solve the following Stokes poblem: Fo f [H 1 (Ω)] N and g L 2 0 (Ω) find u [H1 0 (Ω)]N such that (6.1) D(u,v) + (p, v) = (f,v) fo all v [H 1 0(Ω)] N and (6.2) ( u, q) = (g, q) fo all q L 2 0 (Ω). Hee D(, ) is the Diichlet integal. Let T be the solution opeato fo the Diichlet poblem: T : H 1 (Ω) H0 1 (Ω) defined by D(Tf, φ) = (f, φ) fo all φ H 1 0 (Ω). The opeato T is extended to vectos component-wise. Then it is easy to check that, taking v = T q in the fist equation, a necessay condition is (6.3) ( T p, q) = (g Tf, q) fo all q L 2 0 (Ω). Noting that D 1/2 (v,v) is equivalent to the nom on [H 1 0 (Ω)]N it follows easily that ( T q, q) 1/2 = (T q, q) 1/2 = D 1/2 (T q, T q) = sup v [H 1 0 (Ω)]N (q, v) D 1/2 (v,v) C q L 2 (Ω) fo all q L 2 0(Ω). The bilinea fom on the left hand side of (6.3) is continuous on L 2 0 (Ω) L2 0 (Ω) and by the last inequality it is coecive. Since g Tf L 2 0(Ω) we may apply the Lax-Milgam Theoem and obtain the existence and uniqueness of p L 2 0(Ω)

10 10 JAMES H. BRAMBLE satisfying (6.3). With this p we can now solve (6.1) with u = Tf + T p and it follows that (6.2) is also satisfied. This solves the Stokes poblem (6.1), (6.2). 7. Appendix II: Kon s second inequality Let u [H 1 (Ω)] N with components u i and define the stain tenso ǫ ij = 1/2( u i + u j ). Theoem 7.1. Let Ω R N be a connected domain with a Lipschitz bounday. Then thee exists a constant C such that, C( u i 2 H 1 (Ω) )1/2 ( u i 2 L 2 (Ω) )1/2 + ( ǫ ij 2 L 2 (Ω) )1/2, fo all u [H 1 (Ω)] N. Poof. Since smooth functions ae dense in L 2 (Ω) and H 1 (Ω) it suffices to conside only smooth vectos u. The theoem follows by noting that, fo i, j fixed and any k, 2 u i = ǫ ik + ǫ ij ǫ jk x k x k and applying the Theoem 1.1 to u = u i. Refeences [1] C. Benadi, Méthode d éléments finis pou les équations de Navie-Stokes, Thése. Univ. Pais. (1979). [2] F. Bezzi and M. Fotin,Mixed and Hybid Finite Element Methods, Spinge, New Yok, [3] G. Duvaut and J.L. Lions, Inequalities in Mechanics and Physics, Seies of Compehensive Studies in Math., 219, Spinge, [4] G. Fichea, Sull esistenza e sul calcolo delle soluzioni dei poblemi al contono, elativi all equilibio di un copo elastico, Ann. Scuola Nomale Sup. Pisa s. III 4 (1950). [5] V. Giault and P.-A. Raviat, Finite Element Methods fo Navie-Stokes equations, Spinge Seies in Computational Mathematics, 5, Spinge, [6] J. Gobet, Une inégalité fondamentale de la théoie de l élasticité, Bull. Soc. Royale Science Liège, 31 année, No 3-4 (1962) [7] O.A. Ladyzhenskaya, The Mathematical Theoy of Viscous Incompessible Flows, Godon and Beach, London, [8] J. Nitsche, On Kon s second inequality, R.A.I.R.O. 15 (1981) [9] J. Nečas, Les Méthodes Diectes en Théoie des Équations Elliptiques, Masson, [10] R. Temam, Navie-Stokes Equations, Noth-Holland, New Yok, i, James H. Bamble, Depatment of Mathematics, Texas A&M Univesity, College Station, TX addess: bamble@math.tamu.edu

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