Uncoupled variational formulation of a vector Poisson problem Jiang Zhu 1, Luigi Quartapelle 2 and Abimael F. D. Loula 1 Abstract { This Note provides

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1 Uncoupled variational formulation of a vector Poisson problem Jiang Zhu 1, Luigi Quartapelle 2 and Abimael F. D. Loula 1 Abstract { his Note provides a rigorous analysis for the vector Poisson problem with the tangential component(s) of the unknown prescribed on the boundary together with the divergence of the unknown specied on it. his kind of boundary conditions implies a coupling between the Cartesian components of the unknown in two and three dimensions. A new uncoupled variational formulation of the problem is presented which is the basis of the inuence matrix uncoupling method proposed by Quartapelle and Muzzio [7] and which leads to an alternative nite element solution method. Formulation variationnelle decouplee d'un probleme de Poisson vectoriel Resume { Dans cette Note on analyse le probleme de Poisson vectoriel avec des conditions aux limites sur les composantes tangentielles de l'inconnue et sa divergence, la derniere condition impliquant un couplage entre les composantes cartesiennes de l'inconnue. Une formulation variationnelle decouplee du probleme est introduite qui est a l'origine de la methode de decouplage avec matrice d'inuence proposee par Quartapelle et Muzzio [7], et qui conduit a une methode de solution alternative par une approximation par elements nis. Version francaise abregee { Dans l'etude de certains problemes physiques il faut determiner un champ vectoriel regi par une'equation elliptique et sat- 1 Laboratorio Nacional de Computac~ao Cientca, CNPq, Rua Lauro Muller 455, Botafogo, Rio de Janeiro, Brazil. J.Z.: jiang@lncc.br, A.L.: aloc@lncc.br 2 Dipartimento di Fisica, Politecnico di Milano, Piazza Leonardo da Vinci, 32, Milano, Italy. 1

2 isfaisant les conditions aux limites suivantes: (P) (a)?u = f in (b) un = an on? (c) r u = b on? ou IR D, D = 2 ou 3,? est la frontiere de, f est un champ donne dans, an et b sont des fonctions resp. vectorielle et scalaire donnees sur?, et n est le vecteur unitaire exterieur perpendiculaire a?. Par exemple on rencontre ce systeme quand on employe le potentiel vectoriel pour representer le champ de vitesse d'un uide incompressible [2] ou la composante solenoidale d'un ecoulement potentiel transonique [3] ou encore le champ de vorticite dans les ecoulements incompressibles visqueux en trois dimensions [6]. Ce genre de problemes a ete examine par dierents chercheurs, voir par exemple les references dans [7]. La presence de l'operateur divergence dans la deuxieme condition aux limites implique un couplage entre les composantes cartesiennes de l'inconnue u, en sorte que le probleme de Poisson donne ne peut pas ^etre resolu comme un systeme de deux ou trois equations de Poisson scalaires independantes pour les composantes vectorielles de u. Par consequent, pour une solution numerique du probleme, il est interessant de developper des methodes qui permettent de reduire le probleme vectoriel couplee a la rsolution d'une ensemble de equations scalaires independants. Par example, Quartapelle et Muzzio ont montre que le probleme de Poisson vectoriel considere peut ^etre reduit, dans le cas discret, a un nombre ni de problemes de Dirichlet scalaires independantes plus un probleme lineaire pour une inconnue scalaire denie sur la frontiere [7]. La base de cette methode est la technique de decouplage appelee matrice d'inuence proposee par Glowinski et Pironneau in [5] et adoptee souvent dans la litterature (voire par example [1], [7], [], [9] et les references). Dans cette Note on donne la formulation variationnelle du probleme de Poisson considere et on demontre l'existence et l'unicite de sa solution dans les espaces convenables. Ensuite une formulation variationnelle decouplee du probleme est introduite dont la solution est exprimee en deux composantes que peuvent ^etre determinee en succesion: une composante vient de la solution d'un probleme de Dirichlet vectoriel classique tandis que l'autre est une 2

3 fonction vectorielle harmonique. L'equivalence de cette methode decouple avec celle propose dans [7] est demontree. Enn une nouvelle methode par elements nis pour la resolution approchee du probleme decouple est decrite. 1 Variational Formulation For the simplication, let us assume that is bounded and? is C 1;1 simply connected. In order to present a variational framework for system (P), we introduce some spaces: H 1 an() = n v 2 H 1 (); vn = an on? o ; (1) H 1 () = n v 2 H 1 (); vn = 0 on? o : (2) We can equip H 1 () with the norm j j 1 associated with the following inner product: (u ; v) 1 = (ru ; rv) 0 + (r u ; r v) 0 ; (3) where ( ; ) 0 denotes the standard inner product of [L 2 ()] D, D = 1; 2 or 3. he norm j j 1 is equivalent to the standard norm of H 1 () in H 1 () (see e.g. [4]). We denote by H?1 () the dual space of H 1 () with the following norm: kfk?1; = sup v 2 H 1 () hf ; vi 1; jvj 1 ; f 2 H?1 (); (4) v 6= 0 where h ; i 1; denotes the duality product between H?1 () and H 1 (). We also introduce a space: equiped with the norm: H?1 (r ; ) = n v 2 H?1 (); r v 2 H?1 () o (5) kvk H?1 (r ;) = kvk?1; + kr vk?1 : (6) 3

4 he following assumptions are needed: (H) (1) f 2 H?1 (r ; ); (2) an 2 H 1=2 (?); (3) b 2 H?1=2 (?): hen the (coupled) variational form of system (P) can be written as: Find u 2 H 1 an() such that (u; v) 1 = hf ; vi 1; + hb; v ni 1=2;? ; v 2 H 1 (); where h ; i 1=2;? denotes the duality product between H?1=2 (?) and H 1=2 (?). In problem (7), the tangential boundary conditions unj? = an are essential whereas the coupling condition r uj? = b is natural. heorem 1 Under the assumptions (H), system (P) is equivalent to variational problem (7). Proof: By standard arguments one can easily verify that any solution of system (P) is a solution of problem (7). Conversely, let u be a solution of problem (7). aking v 2 [D()] D, one can establish (Pa), by usual arguments. And from assumption (H1), we can see that r u 2 H?1=2 (?). We formally apply Green's formula to (Pa) premultiplied by v 2 H 1 (). Hence, (7) hr u? b ; v ni 1=2;? = 0; v 2 H 1 (): his implies that (Pc) holds. 2 Existence and Uniqueness Results heorem 2 Problem (7) has a unique solution. 4

5 Proof: Let u a 2 H 1 () be the solution of the following classical Dirichlet vector problem:?u a = 0 in u a n = an on? () u a n = 0 on?: hen, w = u? u a 2 H 1 () satises: (w; v) 1 = L(v); v 2 H 1 () (9) where L(v) = hf ; vi 1; + hb? r u a ; v ni 1=2;? : (10) It is easy to check that (v ; v) 1 = jvj 2 1 ; v 2 H 1 () (11) and jl(v)j kfk?1; jvj 1 + fkbk?1=2;? + kr u a k?1=2;? kv nk 1=2;? fkfk?1; + kbk?1=2;? + kr u a k?1=2;? g jvj 1 : (12) Since r u a 2 H() = fv 2 L 2 (); v = 0 in g, then r u a 2 H?1=2 (?). Hence, L(v) is a linear continuous functional over H 1 (). By the well-known Lax-Milgram heorem, we know that problem (9) has a unique solution w, then so does problem (7). 3 A New Uncoupled Formulation Both in two and in three dimensional cases, problem (7) is coupled through the boundary condition r uj? = b. he uncoupling technique proposed here is derived from the variational form (7) whereas that one obtained in [7] is based on (P). We can prove that they are equivalent. However, they lead to dierent nite element approximations (see Section 4). It is easy to know that 5

6 Lemma 1 We have the following decompositions: H 1 an() = H 1 an;n() H 1 () (13) and where H 1 () = H 1 0() H 1 () (14) H 1 an;n() = n v 2 H 1 (); vn = an and v n = 0 on? o ; (15) H 1 () = n v 2 H 1 (); v = 0 in and vn = 0 on? o : (16) By this Lemma, we can split u and v of the formulation (7) into two components, respectively: u = u an + u H ; v = v 0 + v H (17) where u an 2 H 1 an;n(), v 0 2 H 1 0 (), u H and v H 2 H 1 (). hen, problem (7) can be written in the following uncoupled form: Find (u an ; u H ) 2 H 1 an;n() H 1 () such that (a) (u an ; v 0 ) 1 = hf ; v 0 i 1; ; v 0 2 H 1 0 () (b) (u H ; v H ) 1 =?(u an ; v H ) 1 + hf ; v H i 1; (1) Actually, we have +hb ; v H ni 1=2;? ; v H 2 H 1 (): heorem 3 he uncoupled formulation (1) is equivalent to the one in [7]. Proof: We just need to check the harmonic component. Since (u H ; v H ) 1 = I? v H n r u H ; v H 2 H 1 (); 6

7 let us now introduce the auxiliary vector elds w 0, as in [7], which are arbitrary in and coincide with v H on the entire boundary (namely w 0 n = v n and w 0 n = 0 on?), we deduce (u H ; v H ) 1 = = = = = I Z Z Z Z? w 0 n r u H r (w 0 r u H ) [(r w 0 ) r u H + w 0 r(r u H )] [(r w 0 ) r u H + w 0 rru H ] [(r w 0 ) r u H + (rw 0 ) ru H ] = (u H ; w 0 ) 1 : his is exactly the expression for the surface linear operator introduced in [7]. he same argument applies to the right-hand side of the linear problem. Subproblem (1a) is a classical vector Dirichlet problem and is easily solved as D independent scalar Dirichlet problems. Let us now consider subproblem (1b). Similarly to [5], [1] and [9], we introduce an isomorphism between H 1=2 (?) and H 1 (),?! u H () dened by: 2 H 1=2 (?), u H () is the unique solution of the following classical vector Dirichlet problem: Find u H () 2 H 1 () such that u H () n = in H 1=2 (?) (19) hus, subproblem (1b) can be transformed into: Find 2 H 1=2 (?) such that (u H () ; u H ()) 1 =?(u an ; u H ()) 1 + hf ; u H ()i 1; (20) +hb ; i 1=2;? ; 2 H 1=2 (?) 7

8 Summarizing the above results, we have: heorem 4 he solution u of problem (7) can be split into: u = u an + u H () where u an is the solution of (1a) and is the unique solution of problem (20), u H () being the solution of problem (19). 4 Finite Element Approximation he main diculty to get a nite element scheme based on the uncoupled formulation in the last section is how to approximate problem (20). o overcome this diculty, an idea similar to [1] and [9] can be applied. hroughout this section, is assumed to be a polygonal (in two dimension) or polyhedronal (in three dimension) domain. Let h be a triangulation of such that n is a constant vector along s 2 h, where h is the set of segments contained in?, which are edges of an element of h. We introduce some nite element spaces: S h;k = n v h 2 C 0 ( ); v h j 2 P k ; 2 ho ; (21) S h;k 0 = S h;k \ H 1 0(); (22) S h;k an;n = n v h 2 [S h;k ] D ; v h nj s = I k fang and v h nj s = 0; s 2 ho ; (23) S h;k = n v h 2 [S h;k ] D ; v h nj s = 0; s 2 ho ; (24) where P k denotes the space of all polynomials dened in IR D, of degree less than or equal to k 0, I k denotes the standard P k {interpolation operator over s. According to the assumption of the triangulation h, it is easy to see that each vector function of spaces S h;k an;n scalar funtions of S h;k. and S h;k can be written into N

9 o approximate u H, we introduce a discrete vector harmonic function space, dened by H h;k = n v h 2 S h;k ; (v h ; w h ) 1 = 0; w h 2 [S h;k 0 ] Do : (25) hen, a nite element approximation to (1) can be proposed as follows: Find (u h ; an uh H) 2 S h;k an;n H h;k such that (a) (u h an ; vh 0 ) 1 = hf ; v h 0 i 1; ; v h 0 2 [S h;k 0 ] D (b) (u h H ; vh H) 1 =?(u h an ; vh H) 1 + hf ; v h Hi 1; (26) +hb ; v h H ni 1=2;? ; v h H 2 H h;k Since H h;k 6 H 1 (), (26) is a nonconforming approximation of (1). Similarly to [1] and [9], we can easily prove the following result: heorem 5 Problem (26) has a unique solution. is not explicitly known, o we shall dene an isomor- and H h;k, h?! u h H( h ) by: Since the space H h;k phism between? h = n v h nj? ; v h 2 H h;k h 2? h, u h H( h ) is the solution of u h H( h ) 2 H h;k satises u h H( h ) n = h on?: (27) Let a h ( h ; h ) = (u h H( h ) ; u h H( h )) 1 ; h ; h 2? h ; (2) b h ( h ) =?(u h an ; uh H( h )) 1 + hf ; u h H( h )i 1; +hb ; u h H( h ) ni 1=2;? ; h 2? h : (29) 9

10 hen (26) is equivalent to Find (u h an ; h ) 2 S h;k an;n? h such that (a) (u h an ; vh ) 1 = hf ; v h i 1; ; v h 2 [S h;k 0 ] D (b) a h ( h ; h ) = b h ( h ); h 2? h (30) An algorithm based on (30) would be: Solve problem (30a). Compute a discrete vector harmonic basis fu h H( i )g by (27) for a basis f i g of? h. Compute the bilinear form a h and the linear form b h from (2) and (29) respectively, by means of the functions fu h H( i )g. hen solve problem (30b). Remark 1 In the above algorithm, the determination and storage of the in- uence matrix associated with problem (30b) can be very expansive, especially for 3D problems. o avoid its explicit construction, one can resort to iterative solution methods, for example the conjugate gradient method ( see [5] and [7]). Acknowledgement. he authors would like to thank Professor Olivier Pironneau for his comments. he rst author's work was supported by CNPq of Brazil. References [1] Y. Achdou, R. Glowinski and O. Pironneau. uning the mesh of a mixed method for the stream function{vorticity formulation of the Navier{Stokes equations, Numer. Math. 63, 1992, p.145{163. [2] C. Bernardi. Methodes d'elements nis mixtes pour les equations de Navier{Stokes, hese de 3eme Cycle, Universite de Paris VI,

11 [3] F. El Dabaghi and O. Pironneau. Stream vectors in three dimensional aerodynamics, Numer. Math. 4, 196, p.561{59. [4] V. Girault and P.-A. Raviart. Finite Element Methods for Navier{ Stokes Equations, Springer-Verlag, Berlin, 196. [5] R. Glowinski and O. Pironneau. Numerical methods for the rst biharmonic equation and for the two-dimensional Stokes problem, SIAM Review 21, 1979, p.167{212. [6] L. Quartapelle. Numerical Solution of the Incompressible Navier{ Stokes Equations, Birkhauser, Basel, [7] L. Quartapelle and A. Muzzio. Decoupled solution of vector Poisson equations with boundary condition coupling, in Computional Fluid Dynamics, G. de Vahl Davis and C. Fletcher eds., Elsevier Science Publishers B. V., North-Holland, 19, p.609{619. [] V. Ruas and J. Zhu. Decoupled solution of the velocity{vorticity system for two-dimensional viscous incompressible ow, C. R. Acad. Sci. Paris, 31, I, 1994, p.293{29. [9] J. Zhu. On the velocity{vorticity formulation of the Stokes system and its nite element approximations, Doctoral hesis, Pontifcia Universidade Catolica do Rio de Janeiro, Brazil,