Approximation of the biharmonic problem using P1 finite elements

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1 Approximation of the biharmonic problem using P1 finite elements Robert Eymard, Raphaèle Herbin, Mohamed Rhoudaf To cite this version: Robert Eymard, Raphaèle Herbin, Mohamed Rhoudaf. Approximation of the biharmonic problem using P1 finite elements. Journal of Numerical Mathematics, De Gruyter, 2011, pp.volume 19, Issue 1, Pages <hal > HAL Id: hal Submitted on 24 May 2013 HAL is a multi-disciplinary open access archive for the deposit dissemination of scientific research documents, whether they are published or not. The documents may come from teaching research institutions in France or abroad, or from public or private research centers. L archive ouverte pluridisciplinaire HAL, est destinée au dépôt et à la diffusion de documents scientifiques de niveau recherche, publiés ou non, émanant des établissements d enseignement et de recherche français ou étrangers, des laboratoires publics ou privés.

2 Approximation of the biharmonic problem using P1 finite elements R. Eymard, R.Herbin M. Rhoudaf May 24, 2013 Abstract We study in this paper a P1 finite element approximation of the solution in H0 2 () of a biharmonic problem. Since the P1 finite element method only leads to an approximate solution in H0 1 (), a discrete Laplace operator is used in the numerical scheme. The convergence of the method is shown, for the general case of a solution with H0 2 () regularity, thanks to compactness results to the use of a particular interpolation of regular functions with compact supports. An error estimate is proved in the case where the solution is in C 4 (). The order of this error estimate is equal to 1 if the solution has a compact support, only 1/5 otherwise. Numerical results show that these orders are not sharp in particular situations. 1 Introduction This paper deals with the approximation of the following problem, called the biharmonic problem, which arises in various frameworks of fluid or solid mechanics: find u such that ( u)(x) = f(x) divg(x) + l(x), for x, u(x) = 0 u(x) n (x) = 0, for x. In this paper, Problem (1) is considered in the following weak sense: find u H0 2 () such that v H0 2 (), u(x) v(x)dx = (f(x)v(x) + g(x) v(x) + l(x) v(x))dx, where H0 2 () denotes the closure in H 2 () of the set Cc () of infinitely continuously differentiable functions with compact support in, where (1) (2) d N \ {0} denotes the space dimension, is an open polygonal bounded connected subset of R d, with Lipschitz-continuous boundary, (3) f L 2 (), l L 2 () g (L 2 ()) d. (4) Numerous discretization methods for Problem (2) have been proposed in the recent past. The most classical is probably the conforming finite element method; the finite element space must then be a finite dimensional subspace of the Sobolev space H 2 (). Hence elementary basis functions are sought such that the reconstructed global basis functions on belong to C 1 (). On Cartesian meshes, such basis functions are found by generalizing the one-dimensional P 3 Hermite finite element to the multi-dimensional framework. This task becomes much more difficult on more general meshes involves rather sophisticated finite elements such as the Argyris finite element on triangles in 2D, which unfortunately requires 21 degrees of freedom [8]. Hence non-conforming FEMs have also been widely studied: see e.g. [8, Section 49], 1

3 [9], references therein, [3, 4] for more recent works. Discontinuous Galerkin methods have also been recently developed analysed [12, 13, 14, 11]; error estimates have been derived for polynomials of degree greater or equal to two or three. Other methods which have been developed for fourth order problems include mixed methods [6, 12] (see also references therein), [15], compact finite difference methods [7, 2, 1]. All the above methods are high order methods, therefore, rather computationally expensive may not be so easy to implement. Recently, a cheaper low order method based on the discretization of the Laplace operator by a cell centred finite volume scheme was proposed [10]. The idea developed in the present paper is to use the discretization of the Laplace operator, which is naturally provided by the continuous piecewise linear finite element method (see Section 2). Such a method leads to a nonconforming method in H 2 (), since it only provides an approximate solution in H0 1 (). This discrete Laplace operator is then used in a discrete bilinear form, which is applied to the elements of the P1 finite element space which vanish at the boundary. Note that the condition u n = 0 at the boundary of the domain is satisfied by the limit of a sequence of approximate solutions thanks to the definition of the discrete Laplace operator (see Lemma 3.4 in Section 3). Then the convergence has to be proved, using a suitable interpolation of regular test functions. In the general case, the discrete Laplace operator applied to the stard interpolation of a regular function is not consistent with the continuous Laplace operator applied to this function, although it leads to a second order discrete operator (see Lemma 3.5). Therefore a non stard interpolation of regular functions in the P1 finite element space has to be derived (see Lemma 3.6). Error estimates are derived in the case where the continuous solution has some regularity (Section 4). An order 1 is shown in the case where the solution has a compact support in (Theorem 4.2), but only 1/5 for a general regular solution (Theorem 4.3). It is worth noticing that the order of these estimates is lower than that which is numerically observed in various situations (Section 5). A short conclusion of ongoing research is finally drawn in Section 6. 2 Definition of the scheme Let be an open polyhedral domain in R d, with d N. We consider a conforming simplicial mesh T of (in the stard sense provided for example in [8]). We denote by h T the maximum of the diameters h S of all S T. Let V be the finite set of the vertices ot the mesh; the set of the vertices of a simplex S T is denoted by V S, the subset of all S T such that z V S is denoted by T z. We denote by V ext the set of all z V such that z, we denote V int = V \ V ext the set of the interior vertices. Let (K z ) z V be a family of disjoint open connected subsets of such that: 1. for all z V, z K z K z S T z S, 2. z V K z =. We denote by θ T the minimum of (θ S ) S T, (θ z ) z V, where θ S is the ratio between the radius of the largest Euclidean ball contained in S diam(s), θ z =, z V. S T z S For any z V, we denote by V z the set of all y V such that there exists S T z with y V S (which means that V z = S T z V S ). In the following, we will also use the notation T for denoting the whole set of discrete geometric definitions. Remark 1 The results of this paper hold under the general requirements on K z given above. In the numerical examples given in this paper, we use the following definition of K z. For all S T all z V S, we denote by K S,z the subset of S of all points whose barycentric coordinate related to z is larger than that related to any z V S with z z (see figure 1). We then denote for all z V by K z the union of all K S,z, for all S T z. 2

4 K S,z K z z S Figure 1: Definition of K z For any z V, let ξ z H 1 () be the piecewise affine basis function of the P1 finite element, such that ξ z (z) = 1 ξ z (z ) = 0 for all z V \ {z}. We then denote by V T the vector space spanned by all functions ξ z, z V. We remark that the property v = z V v(z)ξ z holds for any v V T, we define the natural P1 interpolation I T ϕ V T of any continuous function ϕ by For any u V T I T ϕ(z) = ϕ(z), z V, ϕ C 0 (). (5) any S T, we denote by S u the constant value of u in S, we define T zy = ξ z (x) ξ y (x)dx = S S ξ z S ξ y, z, y V. (6) S T Using the property y V T zy = 0 since y V ξ y(x) = 1 for all x, let z : V T R, for all z V, be the linear form defined by z u = 1 u(y)t zy = 1 T zy (u(y) u(z)), u V T, z V. (7) y V y V z We then define T : V T L 2 () by T u(x) = z V z u 1 Kz (x), for a.e. x, u V T. (8) We define the discrete space V T,0 = {u V T, u = 0 on }. (9) Then the scheme for the approximation of Problem (2) consists in finding u V T,0 ; v V T,0, T u(x) T v(x)dx = (f(x)v(x) + g(x) v(x) + l(x) T v(x))dx. (10) An important relation for the mathematical analysis is u, v V T, u(x) v(x)dx = v(z) z u = z V P T v(x) T u(x)dx, (11) where we define the piecewise constant reconstruction of the elements of V T by P T v(x) = z V v(z) 1 Kz (x), for a.e. x, v V T. (12) 3

5 3 Convergence analysis Lemma 3.1 (Piecewise reconstruction) Let us assume Hypotheses (3). Let T be a conforming simplicial mesh of. We then have the following inequality: v P T v L2 () h T v L2 () d, v V T. (13) Proof. Let S T, z V S x S. We have v(x) v(z) = S v (x z), denoting by S v the constant gradient of v in S. This leads to v(x) v(z) S v h T. Therefore we get P T v(x)) (v(x) 2 dx = (v(x) v(z)) 2 dx h 2 T v(x) 2 dx, z V S T K z S which gives (13). Lemma 3.2 Let us assume Hypotheses (3). Let T be a conforming simplicial mesh of. Then the following inequalities hold: w L2 () d 2 diam() T w L 2 (), w V T,0, (14) Proof. Setting v = w in (11), we get w(x) 2 = w L 2 () 2 diam() 2 T w L 2 (), w V T,0. (15) Hence, by using the Cauchy-Schwarz inequality, we have P T w(x) T w(x)dx. w 2 L 2 () d P T w L 2 () T w L 2 (). (16) The Poincaré inequality [5], which holds since V T,0 H 1 0 (), reads By using (13) h T diam(), we have w L 2 () diam() w L 2 () d. P T w L 2 () P T w w L 2 () + w L 2 () (h T + diam()) w L 2 () d 2 diam() w L2 () d. Gathering the above results, we deduce (14) (15) from (16). Lemma 3.3 (Existence, uniqueness estimate on the solution of (10)) Let us assume Hypotheses (3) (4). Let T be a conforming simplicial mesh of. Then, for any u V T,0 satisfying (10), the following holds: u L2 () 4 diam() 4 f L2 () + 4 diam() 3 g L 2 () d + 2 diam()2 l L2 (), (17) u L2 () d 4 diam()3 f L 2 () + 4 diam() 2 g L2 () d + 2 diam() l L 2 (), (18) T u L 2 () 2 diam() 2 f L 2 () + 2 diam() g L2 () d + l L 2 (). (19) As a consequence, there exists one only one u V T,0 such that (10) holds. 4

6 Proof. Let u be given such that (10) holds. Let us take v = u in (10). We get ( T u(x)) 2 dx = (f(x)u(x) + g(x) u(x) + l(x) T u(x))dx, which leads to T u 2 L 2 () u L 2 () f L 2 () + u L2 () d g L 2 () d + l L 2 ()) T u L 2 (), Thanks to Lemma 3.2, the previous inequality provides T u L2 () 2 diam() 2 f L2 () + 2 diam() g L 2 () d + l L 2 (), which is (19). We then deduce (18) (17), using again Lemma 3.2. On the other h, we remark that (10) is equivalent to a square linear system. Setting f = 0, g = 0 l = 0, we get from (17) that u = 0, showing the invertibility of the matrix of the system. This implies the existence uniqueness of the discrete solution. Lemma 3.4 (Compactness of a sequence of approximate solutions) Let us assume Hypotheses (3). Let (T m ) m N be a sequence of conforming simplicial discretizations of such that h Tm tends to 0 as m. Assume that there exists θ > 0 with θ < θ Tm for all m N. Let (u m ) m N be a sequence of functions such that u m VT 0 m,0 for all m N. For simplicity, we shall denote the discrete operator Tm by m. Assume that the sequence ( m u m ) m N is bounded in L 2 () by C 0; then there exists a subsequence of (T m ) m N, again denoted (T m ) m N, u H0 2 (), such that the corresponding subsequence (u m ) m N satisfies: 1. u m u in L 2 (), 2. u m u in L 2 () d, 3. m u m u weakly in L 2 (), as m. Proof. Since the sequence ( m u m ) m N is bounded in L 2 (), we may extract a subsequence of (T m, u m ) m N, again denoted (T m, u m ) m N, such that ( m u m ) m N converges weakly in L 2 () to some w L 2 (). From Lemma 3.2, we get that u m L 2 () C, m N, where C R + only depends on C. Therefore, applying Rellich s theorem, we get the existence of some u H 1 0 () of a subsequence of (T m, u m ) m N, again denoted (T m, u m ) m N, such that u m u weakly in L 2 () d, u m u strongly in L 2 (), as m. Let us prove that u H 2 0 (). Let u (resp. u m ) denote the prolongement of u (resp. u m ) by 0 in R d \. Thanks to u H 1 0 () (resp. u m H 1 0 ()), we have u L 2 (R d ) d (resp. u m L 2 (R d ) d ), with the property u m u weakly in L 2 (R d ) d. (20) Let ϕ Cc (R d ); note that ϕ does not necessarily vanish at the boundary of. Let I m ϕ denote I Tm ϕ for short. Recall that I m ϕ tends to ϕ in H 1 () as m, thanks to the hypothesis that there exists θ > 0 with θ < θ Tm for all m N. We then define the approximation G m ϕ of ϕ by G m ϕ(x) = { Im ϕ(x) for a.e x, ϕ(x) for a.e x R d \. (21) 5

7 Let T m = u R d m (x) G m ϕ(x)dx. Using (20), the convergence of G m ϕ(x) to ϕ in L 2 (R d ) d, we get lim T m = u(x) ϕ(x)dx. m + R d On the other h, we have T m = u m (x) G m ϕ(x)dx = u m (x) I m ϕ(x)dx. R d Thanks to (11), we get T m = P Tm I m ϕ(x) m u(x)dx. Passing to the limit m in the above relation, since (13) shows that P Tm I m ϕ converges to ϕ in L 2 (), we get thanks to strong/weak convergence properties, u(x) ϕ(x)dx = ϕ(x)w(x) = ϕ(x)w(x), R d R d where we denote by w the prolongement of w by 0 in R d \. This proves that u H div (R d ) that u = w a.e. in R d, which means that u = 0 outside that u = w a.e. in. Since u H 1 (R d ) u L 2 (R d ), a classical result of regularity shows that u H 2 (R d ). Since u = 0 in R d \, we get that the trace of u on is equal to 0. Hence u H0 2 (). Let us now prove the strong convergence of u m to u. Using the weak convergence of this sequence, it suffices to prove the convergence of u m L 2 () to u d L 2 () d. To this aim, we write the relation obtained by setting u = v = u m in (11): u m (x) 2 = P Tm u m (x) m u m (x)dx, m N. Passing to the limit m in the above relation, we get, using strong/weak convergence properties in the right h side, lim u m (x) 2 = u(x) u(x)dx = u(x) 2, m hence concluding the proof. In order to conclude the convergence analysis, it is natural to examine the convergence of the discrete Laplace operator, when applied to I T ϕ V T,0, for any regular function ϕ C 2 () H 2 0 (). We show in the next lemma that this operator is indeed a second order discrete operator. Lemma 3.5 (Order of the discrete Laplace operator applied to stard interpolation) Let us assume Hypotheses (3). Let T be a conforming simplicial mesh of let θ > 0 such that θ θ T. Then there exists C 1 > 0, only depending on θ, such that S ξ z C 1 h S, z V S, S T, (22) z I T ϕ C 1 ϕ 2, ϕ C 2 (), z V int, (23) where z is defined by (7), I T ϕ is defined by (5) ϕ 2 = max i,j=1,d 2 ij ϕ L (). Proof. Inequality (22) results from the fact that the ratio, between the distance from any vertex of S to the opposite face h S, is larger than 2θ. We now consider z V int ϕ C 2 (). We can write, using (7), z I T ϕ = 1 (ϕ(y) ϕ(z))t zy. y V z 6

8 A Taylor expansion provides ϕ(x) ϕ(z) = G (x z) + D(x, z) x z 2, where D(x, z) d 2 ϕ 2 G = ϕ(z). Let us check that the discrete operator z vanishes on the affine function µ : x G (x z) (which is such that µ V T ). Indeed, we have, using (11) (12), z µ = T µ(x)1 Kz (x)dx = T µ(x)p T ξ z (x)dx = µ(x) ξ z (x)dx = G ξ z (x)dx = 0. We therefore get z I T ϕ = 1 T zy D(y, z) y z 2 = 1 y V z S T z y V S S S ξ z S ξ y D(y, z) y z 2. Using (22) the regularity condition θ S T z S y z h S, we conclude (23). Since the discrete Laplace operator is a second order discrete operator, the question of its strong convergence to the continuous Laplace operator arises. Indeed, the proof that T I T ϕ converges to ϕ for the weak topology of L 2 () results from the following property, the proof of which uses Lemma 3.1: ( T I T ϕ ϕ)p T vdx = ϕ(v P T v)dx ( I T ϕ ϕ) vdx Ch T v L2 () d, v V T,0. Nevertheless, whatever be the choice of (K z ) z V satisfying the hypotheses required above, it is not possible in the general case to obtain that T I T ϕ strongly converges to ϕ as h T 0, while θ θ T. Therefore it is not possible to conclude to the convergence of the scheme by letting v = I T ϕ in (10): another interpolation is necessary, which we introduce in the following Lemma 3.6. Lemma 3.6 (Interpolation of regular functions with compact support) Let us assume Hypotheses (3). Let T be a conforming simplicial discretization of, let θ > 0 be given such that θ < θ T. Let ϕ Cc 2 () let a = d(support(ϕ), ). Then there exists ĨT ϕ V T,0 C > 0 only depending on θ such that ĨT ϕ ϕ L2 () Ch T ϕ 2 a 2, (24) ĨT ϕ ϕ L2 () Ch d T ϕ 2 a 2, (25) T Ĩ T ϕ T ϕ L2 () Ch T ϕ 2 a 2, (26) where ϕ 2 = max i,j=1,d 2 ij ϕ L () T ϕ is the piecewise constant function defined by z ϕ = 1 ϕ(x)dx, z V, (27) K z T ϕ(x) = z V z ϕ 1 Kz (x), for a.e. x. (28) Proof. Let ρ C c (R d, R + ) be the function defined by ρ(x) = exp( 1/(1 x 2 )) B(0,1) exp( 1/(1, x B(0, 1), y 2 ))dy 7

9 ϕ 1 ψ a a Figure 2: Functions ϕ ψ ρ(x) = 0 for x / B(0, 1). Let ψ Cc (, [0, 1]) be the function defined by ( ) d ( ) 4 4 ψ(y) = ρ (y x) dx, y. (29) a a x,d(x, )> a 2 Then ψ(x) = 0 for all x such that d(x, ) < a 4 ψ(x) = 1 for all x such that d(x, ) > 3a 4 (see Figure 2). The idea of the construction of ĨT ϕ is to consider the approximation of ϕ in V T,0 obtained by the finite element method in the case where the right h side is given by ϕ; since ĨT ϕ must be equal to 0 on the boundary cells, we multiply this discrete solution by ψ. Then the proof mimics the identity (ψv) = v ψ + 2 ψ v + ψ v. We first suppose that T is such that h T < a 4. Let us define v ṽ V T,0 such that ṽ(x) v(x)dx = ϕ(x)p T v(x)dx, v V T,0, v(x) v(x)dx = ϕ(x)v(x)dx. We define w = ṽ v. By subtracting the second relation to the first one in (30), setting v = w, we get w(x) 2 dx = ϕ(x)(p T w(x) w(x))dx. Applying Lemma 3.1, we obtain w(x) 2 dx ϕ L2 () P T w(x) w(x) L2 () h T ϕ L2 () w L 2 () d. We then deduce ṽ v L2 () h d T ϕ L 2 (). A stard interpolation result gives the existence of C θ, only depending on θ, such that Thanks to we also get (30) ϕ I T ϕ L2 () C θ h T ϕ 2. (31) v(x) ϕ(x) 2 dx = ( v(x) ϕ(x)) ( I T ϕ(x) ϕ(x))dx, Hence, defining w = ṽ I T ϕ, we obtain v ϕ L 2 () d C θh T ϕ 2. w L2 () d (ṽ v) L 2 () + ( v ϕ) L 2 () + (ϕ I T ϕ) L 2 () (2C θ + 1)h T ϕ 2. (32) 8

10 The Poincaré inequality, which holds since w H 1 0 (),writes w L 2 () diam() w L2 () d diam()(2c θ + 1)h T ϕ 2. (33) Let us remark that, thanks to (30), ṽ satisfies z ṽ = z ϕ, z V int, (34) using the notation given by (27) (note that this equality is not a priori satisfied for z V ext ). We now observe that we have ψ(z) z ṽ = z ϕ, z V. (35) Indeed, if d(z, ) > 3a 4, we have ψ(z) = 1 z V int, which implies that (34) holds. Otherwise, if d(z, ) 3a 4 z V int, we have z ṽ = K z ϕ(x)dx = 0, if z V ext, we have ψ(z) = 0 K z ϕ(x)dx = 0. We now define ĨT ϕ V T,0 by ĨT ϕ(z) = ψ(z)ṽ(z), for all z V. From formula (7), we get z Ĩ T ϕ = T zy (ĨT ϕ(y) ĨT ϕ(z)) = T zy (ψ(y)ṽ(y) ψ(z)ṽ(z)). y V z y V z Thanks to the identity ab cd = c(b d) + d(a c) + (a c)(b d), we get, for z, y V, ψ(y)ṽ(y) ψ(z)ṽ(z) = ψ(z)(ṽ(y) ṽ(z)) + ṽ(z)(ψ(y) ψ(z)) + (ψ(y) ψ(z))(ṽ(y) ṽ(z)). This implies, from formula (7), that z Ĩ T ϕ = ψ(z) z ṽ + ṽ(z) z I T ψ + We remark that y V z T zy (ψ(y) ψ(z))(ṽ(y) ṽ(z)). (36) (ψ(y) ψ(z))ϕ(y) = (ψ(y) ψ(z))ϕ(z) = 0, z V, y V z. (37) Indeed, assuming ϕ(y) 0 or ϕ(z) 0, we have d(z, ) > a or d(y, ) > a. Since d(z, y) h T a/4, we get that d(z, ) > 3a 4 d(y, ) > 3a 4. This implies ψ(z) = 1 ψ(y) = 1. Therefore ψ(z) ψ(y) = 0. We then get from (37), for all z V y V z, (ψ(y) ψ(z))(ṽ(y) ṽ(z)) = (ψ(y) ψ(z))(ṽ(y) ϕ(y) ṽ(z) + ϕ(z)) ṽ(z) z I T ψ = ṽ(z) T zy (ψ(y) ψ(z)) = (ṽ(z) ϕ(z)) T zy (ψ(y) ψ(z)) = w(z) z I T ψ. y V z y V z Using (35) the two preceding relations in (36), we obtain ( ) z Ĩ T ϕ z ϕ = w(z) z I T ψ + y V z T zy (ψ(y) ψ(z))(w(y) w(z)). Taking the square of the previous relation applying the inequality (a + b) 2 2(a 2 + b 2 ) we get, ) 2 ( 2 z Ĩ T ϕ z ϕ 2 (w(z) Kz z I T ψ) T zy (ψ(y) ψ(z))(w(y) w(z)). y V z 9

11 Using the Cauchy-Schwarz inequality dividing by, we obtain ( ) 2 z Ĩ T ϕ z ϕ 2 w(z) 2 ( z I T ψ) T zy (ψ(y) ψ(z)) 2 T zy (w(y) w(z)) 2. y V z y V z We now use (22), which implies that ξ z (x) ξ y (x)dx S C2 1, y, z V S, S T. S Thanks to the definition (29) of ψ, we have the existence of a constant C 3 such that h 2 S ψ C 3 a. This leads, using ψ(y) ψ(z) C3 a h S, to the existence of C 4, only depending on θ, such that T zy (ψ(y) ψ(z)) 2 C 4 a 2 K z. y V z Applying (23) proved in Lemma 3.5, since ψ 2 C 5 /a 2, z I T ψ C 6 a 2, where C 6 only depends on θ (this inequality also holds for z V ext since in this case z I T ψ = 0). Hence we get ( ) 2 C 2 z Ĩ T ϕ z ϕ 2 6 a 4 w(z) 2 z V z V +2 C 4 a 2 T zy (w(y) w(z)) 2. z V y V z We now remark that, for S T such that y, z V S, we have T zy (w(y) w(z)) 2 Tzy ( S S w (y z)) 2 C1 2 S T,y,z V S where we denote by T S zy = S S ξ z S ξ y. Hence, we may write z V y V z T zy (w(y) w(z)) 2 2 d(d + 1)C2 1 2 S h 2 S T,y,z V S S S S w 2 S T S w 2 h 2 S, remarking that each edge of a simplex occurs two times in the above summation, that any simplex has d(d+1) 2 edges. Gathering the above results, we thus obtain z V ( z Ĩ T ϕ z ϕ ) 2 2 C 2 6 a 4 w 2 L 2 () + 2d(d + 1)C 4 C 2 1 a 2 w 2 L 2 () d. Using (33) (32) provides (26). We then remark that, for all v V T,0, we have ( z Ĩ T ϕ(x) z ϕ(x))p T v(x)dx = ĨT ϕ(x) v(x)dx + ṽ(x) v(x)dx, thanks to both (11) (30). Hence taking v = ĨT ϕ ṽ provides (25), as well as (24) using the Poincaré inequality. The proof of (26) in the case h T a 4 is obtained by defining ĨT ϕ = 0, using T ϕ L2 () ϕ 2 1/2, h T /a 1/4 1/a 1/diam(). Then (25) (24) follow in that case. 10

12 Theorem 3.7 (Convergence of the scheme) Let us assume Hypotheses (3) (4). Let u H0 2 () be the solution of Problem (2); let T be a conforming simplicial mesh of u T V T,0 be the solution of (10). Then, as h T tends to 0 with θ θ T, for a fixed value of θ > 0: 1. u T converges in L 2 () to u, 2. u T converges in L 2 () d to u, 3. T u T converges in L 2 () to u. Proof. Let (T m ) m N be a sequence of conforming simplicial meshes of such that h Tm tends to 0 as m θ < θ Tm for all m N. Let u m VT 0 m,0, for all m N, be the solution of (10). Thanks to Lemmas , we get the existence of a subsequence of (T m ) m N, again denoted (T m ) m N, of u H0 2 () such that the conclusion of Lemma 3.4 holds. Let ϕ Cc () be given. We take, in (10) with T = T m, v = ĨT m ϕ defined by Lemma 3.6. Passing to the limit as m in the resulting equation (thanks to the weak/strong convergence properties provided by Lemmas ) using the density of Cc () in H0 2 (), we get that u is the solution of Problem (2). By a classical uniqueness argument, we get that the whole sequence converges. Setting v = u m in (10), we get that Tm u m 2 L 2 () converges to (f(x)u(x) + g(x) u(x) + l(x) u(x)) dx = ( u(x))2 dx as m. Together with the weak convergence of Tm u m to u as m, this provides the convergence in L 2 () of Tm u m to u. 4 Error estimates Let us first prove a technical lemma used in the following error estimate results. Lemma 4.1 (An inequality for regular continuous solutions.) Under Hypotheses (3), let u C 4 () be given let f = ( u). Let T be a conforming simplicial mesh of let θ > 0 be such that θ < θ T. Let u T V T,0 be the solution of (10) in the case where f = ( u), g = 0 l = 0. Then there exists C > 0, only depending on θ, such that the following inequality holds: u T u T L 2 () Ch T u,4 + 2 T u T v L 2 (), v V T,0, (38) where, for all k N, u,k denotes the maximum value of the absolute value of u of its derivatives until order k T is defined by (28). Proof. Let v, w V T,0 be given. We have ( T v(x) T u T (x)) T w(x) dx = ( T v(x) T u(x)) T w(x) dx + ( T u(x) P T (I T u)(x)) T w(x) dx + (P T (I T u)(x) T u T (x)) T w(x) dx. Using (10), f = ( u) w H0 1 (), we may write T u T (x) T w(x) dx = ( u)(x) w(x) dx = We have, from (11), ( u)(x) w(x) dx. P T (I T u)(x) T w(x) dx = (I T u)(x) w(x) dx. 11

13 Hence we get ( T v(x) T u T (x)) T w(x) dx = ( T v(x) T u(x)) T w(x) dx + ( T u(x) P T (I T u)(x)) T w(x) dx + ( u I T u)(x) w(x) dx. Using (14) the Cauchy-Schwarz inequality, we obtain ( T v(x) T u T (x)) T w(x) dx ) ( T v T u L 2 () + T u P T (I T u) L 2 () + 2diam() ( u I T u) L 2 () d T w L 2 (). Taking w = v u T We now write in the above inequality, we get T v T u T L 2 () T v T u L 2 () + T u P T (I T u) L 2 () +2 diam() ( u I T u) L2 () d. u T u T L2 () u T u L2 () + T u T v L2 () + T v T u T L2 (). Thanks to the two above inequalities, we get u T u T L 2 () Thanks to the regularity of u, we have We may also write 2 T u T v L 2 () + u T u L 2 () + T u P T (I T u) L 2 () +2 diam() ( u I T u) L2 () d. u T u 2 L 2 () = z V K z T u P T (I T u) 2 L 2 () = z V ( ) 2 1 ( u(y) u(x))dx dy K z 4 h 2 T 4 u 2,3. ( ) 2 1 ( u(x) u(z))dx K z h 2 T 4 u 2,3. Applying stard results on the interpolation error of the regular function u in V T, we have the existence of C P 1, only depending on θ, such that Therefore the proof of (38) follows. We can then state the following result. ( u I T u) L2 () d C P 1h T u,4. Theorem 4.2 (Error estimate in the case where u C 4 c ()) Let us assume Hypotheses (3) (4), let u C 4 c () be given let f = ( u). Let T be a conforming simplicial mesh of let θ > 0 be such that θ < θ T. Let u T V T,0 be the solution of (10) in the case where f = ( u), g = 0 l = 0. Then there exists C > 0, only depending on, θ u such that (39) u T u L2 () Ch T, (40) u T u L2 () d Ch T, (41) T u T u L2 () Ch T. (42) 12

14 Proof. We apply Lemma 4.1 with v = ĨT u. Thanks to Lemma 3.6, we get the existence of C > 0, only depending on, θ u (by its derivatives, the distance of the support of u to the boundary of the domain) such that (42) holds. Then, writing we can apply Lemma 3.2. We get u T u L 2 () d u T ĨT u L 2 () d + ĨT u u L 2 () d, u T u L2 () d 2 diam() T u T T Ĩ T u L2 () + d ĨT u u L2 () d 2 diam()( T u T u L 2 () + u d T ĨT u L 2 () d) + ĨT u u L2 () d. Using (42) Lemma 3.6 provide (41). Then (40) results from the Poincaré inequality. Let us now state the result, without assuming that the solution has a compact support. Theorem 4.3 (Error estimate in the case where u C 4 () H0 2 ()) Let us assume Hypotheses (3) (4), let u C 4 () H0 2 () be given let f = ( u). Let T be a conforming simplicial mesh of let θ > 0 be such that θ < θ T. Let u T V T,0 be the solution of (10) in the case where f = ( u), g = 0 l = 0. Then there exists C > 0, only depending on, θ u such that 1 u T u L 2 () Ch 5 T, (43) u T u L ( ) Ch 1 d 5 T, (44) T u T u L 2 () Ch T 1 5. (45) Proof. For a given a > 0, we define the function ψ a by (29), the function u a by We remark that, for any i, j = 1,..., d, u a (x) = u(x)ψ a (x) x. 2 iju a (x) = ψ a (x) 2 iju(x) + i ψ a (x) j u(x) + j ψ a (x) i u(x) + u(x) 2 ijψ a (x). (46) Thanks to a Taylor expansion of u u from any point y such that x y = d(x, ), we get the existence of C u > 0, only depending on u such that, for all x, u(x) C u d(x, ) u(x) C u d(x, ) 2. Thanks to i ψ a (x) C/a 2 ij ψ a(x) C/a 2, we get from (46) the existence of C u, only depending on u, such that Hence we have the existence of C u, only depending on u, u a 2 C u, a [0, diam()]. (47) u a (x) u(x) C u, x such that d(x, ) a. Since u a (x) = u(x) for all x such that d(x, ) a, we get u a (x) u(x) 2 L 2 () Thanks to Hypotheses (3), there exists some C > 0 such that {x, d(x, ) a} (C u) 2. {x, d(x, ) a} C a, a [0, diam()]. 13

15 On the other h, since the distance between the support of u a Cc () is greater than a/4, Lemma 3.6 gives the existence of C 7 > 0, only depending on θ, such that We have T Ĩ T u a T u a L 2 () C 7 h T u a 2 a 2 C 7 h T C u a 2. T Ĩ T u a u L2 () T Ĩ T u a T u a L2 () + T u a T u L2 () + T u u L2 (). Thanks to the Cauchy-Schwarz inequality, we derive Since (39) provides T u a T u 2 L 2 () = ( ) 2 1 ( u a (x) u(x))dx z V K z ( u a (x) u(x)) 2 dx = u a u 2 L 2 (). z V K z T u u L2 () 4 1/2 h T u,3, we then get the existence of C 8 > 0 independent on the mesh, such that Choosing a 0 = h 2/5 T T Ĩ T u a u L 2 () C 7 h T C u a 2 + (C a) 1/2 (C u) + C 8 h T. leads to the existence of C 9 > 0, only depending on u, θ, such that T Ĩ T u a0 u L2 () C 9 h 1/5 T. Applying Lemma 4.1 with v = ĨT u a0, we conclude the proof of the theorem, following the proof of Theorem 4.2 for the derivation of (44) (43). 5 Numerical results Let us introduce the following error norms for the solution, its gradient its Laplacian: ( ) 1/2 E 0 = (u(z) u T (z)) 2 / u L2 (), z V ( ) 1/2 E 1 = S S u T u(x S ) 2 / u L 2 (), S T denoting by x S the center of gravity of S, ( ) 1/2 E 2 = ( z u T u(z)) 2 / u L 2 (). z V We are going to study these error norms for a one-dimensional a two-dimensional examples. 14

16 5.1 One-dimensional case We approximate the solution of the problem u(x) = (x(1 x))2, x [0, 1], 24 u (4) (x) = 1, x [0, 1], u(0) = u(1) = u (0) = u (1) = 0, using Scheme (10), with different 1D meshes with N interior points. In the following figure table, we show the numerical results obtained in the case where the mesh is uniform, i.e. the points z V are located at the abscissae i/n, for i = 0,..., N N E 0 order E 1 order E 2 order E E E E E E E E E E E E E E E E E E E E E E-6 2 An order 2 is numerically obtained for the solution, its gradient its discrete Laplacian, which is much more than the theoretical order proved in this case (1/5). In the following figure table, we show the numerical results obtained when the interior points z V are located at the abscissae (i + α i )/N, for i = 1,..., N 1, where α i is a rom value between N E 0 order E 1 order E 2 order E E E E E E E E E E E E E E E E E E E E E-6 2 Again, an order 2 is numerically obtained for the solution, its gradient its discrete Laplacian, which shows the robustness of the scheme in this less regular case. 5.2 Two-dimensional cases We consider Scheme (10) for the approximation of the 2D problem, where = (0, 1) 2 where the continuous solution is given by: u(x 1, x 2 ) = (1 cos(2πx 1 ))(1 cos(2πx 2 )), (x 1, x 2 ) [0, 1] 2, which satisfies (2) for the ad hoc data f = ( u), g = 0, l = 0, =]0, 1[ 2 ; hence we choose f as the function defined by: f(x 1, x 2 ) = ( u)(x 1, x 2 ) = (2π) 4( 4 cos(2πx 1 ) cos(2πx 2 ) (cos(2πx 1 ) + cos(2πx 2 )) ) 15

17 We first consider the case where the mesh is obtained by splitting N raf N raf squares in 2 triangles. The total number of triangles is then 2 N 2 raf, the size of the mesh is of order 1/N raf. In the figure below, we show one of the meshes used, in the table below, we present the results obtained using the scheme (10). N raf E 0 order E 1 order E 2 order u min u max E E E E E E E E E E E E E E In this 2D case, we again observe that the order of convergence is much better than 1/5. Turning to less regular meshes, we consider the case where the simplicial meshes are generated by the repetition of the same square pattern. In the figure below, one can see the repetition of N raf N raf times the initial pattern, with N raf = 5. The total number of triangles is then 14 Nraf 2, the size of the mesh is of order 1/N raf. The interest of such meshes is that no symmetry can increase the numerical order of convergence, whereas the regularity factor of the mesh remains constant. We then observe the results provided in the table below. N raf E 0 order E 1 order E 2 order u min u max E E E-2 > E E-2 < E-3 > E E-2 < E-4 > E E-2 < E-5 > E E-2 < We again observe that the numerical orders of convergence are much better than the theoretical ones for E 0 E 1, but only slightly better for E 2. 6 Conclusion We show in this paper that it is possible to approximate the solution in H 2 0 () of the biharmonic problem using a P1 finite element approximation, which results in a robust cheap scheme. Since the approximate solution only belongs to H 1 0 (), a discrete Laplace operator is used in the discrete variational formulation. This operator, applied to the natural interpolation of a regular function, is not consistent with the continuous Laplace operator, an adapted interpolation is provided. This allows to prove the convergence of the scheme in the general case, to derive error estimates. Numerical observations show that these error estimates are not sharp. Hence some further work should explore possible improvements of these error estimates. The problem of the approximation of more general fourth-order elliptic operators by P1 finite elements should also be examined. References [1] M. Ben-Artzi, I. Chorev, J.-P. Croisille, D. Fishelov. A compact difference scheme for the biharmonic equation in planar irregular domains. SIAM J. Numer. Anal., 47(4): , [2] M. Ben-Artzi, J.-P. Croisille, D. Fishelov. A fast direct solver for the biharmonic problem in a rectangular grid. SIAM J. Sci. Comput., 31(1): ,

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