Error estimates for nearly degenerate finite elements

Transcription

1 Error estimates for nearly degenerate finite elements Pierre Jamet In RAIRO: Analyse Numérique, Vol 10, No 3, March 1976, p Abstract We study a property which is satisfied by most commonly used finite elements and which can improve the interpolation error estimates in Sobolev spaces W m,p. This new estimate does not prohibit elements to be almost degenerate (nearly flat). 1 Introduction When using the finite element method to solve a partial differential equation on R n, the first step is the discretization the domain, commonly known as triangulation even when the elements used are not triangles. This discretization must meet certain to provide a satisfactory error estimate; conditions of regularity by Ciarlet and Raviart [5] and conditions of uniformity by Strang [10] are obtained by the standard error interpolation on a each element. Differrent authors have studied the interpolation error corresponding to various types of finite elements and deduced conditions of regularity for the triangulation (Zlamal [14], Bramble and Zlamal [4], Strang [10], Strang and Fix [11]); the most general results are given in Ciarlet and Raviart [5] [6]. The simplest case is that of straight elements, i.e. elements are the images by an affine transformation F of the element of reference. In this case Ciarlet and Raviart [5] established an estimate of the form: ( ) h k+1 u Πu m,p,k = O, (1.1) p where u is a function of sufficient regularity, Πu is the interpolant on element K, m,p,k is the norm on the Sobolev space W m,p (K), h is the diameter of K, ρ is the maximum diameter of a sphere contained in K and k is an integer determining the spaces of polynomials used for interpolation. From estimate (1.1), deduce that: u Πu m,p,k = O(h k+1 m ) (1.2) holds under the assumption that all elements K of the triangulation satisfy a condition of the form: ρ α > 0, (1.3) h 1

2 where α is an arbitrary constant. Condition (1.3) requires that elements K should not be too skinny; in the particular case of triangular elements, the angles of the triangle should not be too small. In this paper, we will show that for some finite elements used in practice, condition (1.3) can be weakened. We make an estimate of the form: u Πu m,p,k = O(h k+1 m /(cosθ) m ), (1.4) where θ is an angle between 0 and π which can approach π if all edges of K are almost 2 2 parallel to the same hyperplane. From (1.4), one deduces the condition: θ θ 0 < π 2 (1.5) This condition is weaker than (1.3) as it does not prohibit elements to be nearly flat. Thus, in the case of triangular elements, θ is equal to half of the largest angle of the triangle; for example, for an isosceles triangle that has an arbitrarily small angles and two angles near π, θ π and condition (1.5) is satisfied. An estimate identical to (1.4) was established by 2 4 Synge [12] in the case of linear interpolation on a triangle with k = 1, m = 1, and p =. For curved elements produced by the same reference element by means of a non-affine transformation F, it is also possible to weaken the regularity conditions given by Strange and Fix [11] and Ciarlet and Raviart [6]. Similarly, for some quadrilateral elements, we can allow quadrilaterals to degenerate into triangles as the length of one side goes to zero or an angle that tends to π. The study of these elements will be considered in a future paper. First, we define some notation (which is identical to that used by Ciarlet and Raviart [5]). Second, we prove results for a general linear operator Π possessing certain properties; the essential property is contained in the hypothesis H.2; although it is satisfied by most common finite elements, it seems to have been unnoticed. The main result is stated in Theorem 2.2; Theorem 2.3 is a variant. In the third section, we will apply these results to interpolation operators corresponding to various examples of finite elements. Notation : ξ = the Euclidean norm of vector ξ R n. (ξ 1, ξ 2 ) = the dot products of vectors ξ 1, ξ 2 R n. For a sufficiently regular function u defined on domain Ω R n : Du = the Fréchet derivative of order 1, D m u = the Fréchet derivative of order m, D m u (ξ 1, ξ 2,..., ξ m ) = the directional derivative of order m relative to the vectors ξ 1, ξ 2,..., ξ m, D m u(x) (ξ 1, ξ 2,..., ξ m ) = the value at point x Ω in the derivative above, D m u = max{ D m u(x) (ξ 1, ξ 2,..., ξ m ) ; x s = 1, 1 s m}. u p,ω = the norm of u in L p (Ω). u m,p,ω = D m u p,ω = ( D m u p) 1 p is the semi-norm in the Sobolev space W m,p (Ω), 1 p <. 2

3 ( m ) u m,p,ω = i=0 Di u p 1 p p,ω is a norm equivalent to the usual norm in W m,p (Ω). In the case p =, these last two definitions are modified in the usual fashion. L(X, Y) denotes the set of continuous, linear functions from normed space X into normed space Y. P k denotes the set of polynomials of n variables of degree k. 2 General Results Let ˆΩ be a bounded domain in R n and let Ω is any equivalent domain, i.e. there exists is an invertible affine transformation F such that: Ω = F(ˆΩ) = {x R n ; x = F(ˆx), ˆx ˆΩ} (2.1) Transformation F can be written in the form: F(ˆx) = Bˆx+b where B is an n n invertible matrixand b is an n-vector. Consider the following result (Lemma 2 in Ciarlet and Raviart [5]): B hˆρ (2.2) where h is the diameter of Ω, ˆρ is the diameter of the largest sphere contained in ˆΩ, and B is the norm on B induced by the vector Euclidean norm. Any function u defined on Ω corresponds uniquely to a function û defined on ˆΩ by the canonical relationship: û(ˆx) = u(x). Let u W l,p (Ω), integer l 0, and l p, the û W l,p (ˆΩ) and (equation (4.15) in Ciarlet and Raviart [5]): û l,p,ˆω B l det B 1 p u l,p,ω. (2.3) Estimates (2.2) and (2.3) will be used later. Let k, l, and m be three positive integers and consider an operator Π L(W l,p (Ω), W m,p (Ω)) with l p that satisfies the following two hypotheses: Πu = u, for all u P k. (H.1) There exists a set of vectors {ξ s } m (not necessarily distinct) such that D m Πu (ξ 1,..., ξ m ) = 0 (H.2) for all functions u such that D m u (ξ 1,..., ξ m ) = 0. Without loss of generality, we assume that the vectors ξ s are unit vectors. The operator Π corresponds to an equivalent operator ˆΠ L(W l,p (ˆΩ), W m,p (ˆΩ)) defined through the canonical relationship, ˆΠû(ˆx) = Πu(x). If Π satisfies properties H.1 and H.2, then ˆΠ satisfies the analogous properties with vectors ξ s replaced by ˆξ s = B 1 ξ s. 3

4 Next we show some preliminary lemmas. In what follows we will assume that the Ω satisfies the cone property (c.f. Courant and Hilbert [7], Agmon [2], Lions [8]), which is a very general property can always be verified in practical examples. All of the proofs will be given in the case of finite p; the p = case is similar but requires some modifications. Lemma 2.1. Let Ω be a bounded domain in R n satisfying the cone property and let ξ be an arbitrary vector. For any v W l,p (Ω), l 1, there exists a function u W l,p (Ω) such that Du (ξ) = v and u l,p,ω C v l,p,ω where C is a contains which only depends on Ω and l. Proof. We will show u r,p,ω C v l,p,ω for 0 r l. For simplicity, we consider only the particular case r = 1, n = 2, and p. The extension of the proof to the general case is immediate and merely complicates the notation. We choose rectangular coordinates x 1 and x 2 such that the x 1 axis is parallel to the vector ξ and Ω is contained in the square G = {(x 1, x 2 ); 0 x 1 h, 0 x 2 h} where h is the diameter of Ω. Following Lions [8] and the fact that the domain satisfies the cone property, we can extend the definition of v to all of R 2 by a continuous extension operator from W l,p (Ω) into W l,p (R 2 ). Denoting the extension with the same notation v, we have v 1,p,G v l,p,r 2 C 1 v l,p,ω where the constant C 1 only depends on Ω, l, and p. It follows that it is sufficient to prove Lemma 2.1 for the square G instead of Ω. Let v be an arbitrary function in C (Ḡ) and set: u(x 1, x 2 ) = x1 0 v(s, x 2 )ds. The function u satisfies Du (ξ) = u x 1 = v. So u x 1 = v p,g v l,p,g. (2.4) p,g Moreover: Applying Hölder s inequality: u x1 (x 1, x 2 ) = x 2 0 v x 2 (s, x 2 )ds. u h ( (x 1, x 2 ) x 2 v h (x 1, x 2 ) x 2 dx 1 h 1 q v (x 1, x 2 ) x 2 dx ) 1 p, for 1 p + 1 q = 1. Then deduce that h 0 u (x 1, x 2 ) x 2 p dx 2 h p q v x 2 p p,g, 4

5 and by integrating in x 1 : u x 2 h v p,g x 2 h v l,p,g. (2.5) p,g Using (2.4) and (2.5) conclude that: u 1,p,G 2max{1, h} v l,p,g. Through a density argument, this inequality extends to any function v W l,p (G) which completes the proof. Lemma 2.2. Let Ω be a bounded domain in R n satisfying the cone property. Let l and m be two positive integers and let {ξ s } m be a set of unit vectors which are not necessarily distinct. For all v W l,p (Ω), there exists a function u W l,p (Ω) such that: D m u (ξ 1,...,ξ m ) = v and u l,p,ω C v l,p,ω, (2.6) where C is a constant which depends only on Ω, m, and l. Proof. Proof is by induction using Lemma 2.1. By Lemma 2.1, there exists u 1 W l,p (Ω) such that Du 1 (ξ 1 ) = v and u 1 l,p,ω C v l,p,ω. Then, there exists u 2 W l,p (Ω) such that Du 2 (ξ 2 ) = u 1 and u 2 l,p,ω C u 1 l,p,ω. This function u 2 satisfies: Du 2 (ξ 1, ξ 2 ) = Du 1 (ξ 1 ) = v and u 2 l,p,ω C 2 v l,p,ω. Continuing inductively, we find function u = u m that satisfies the conditions in Lemma 2.2 wtih C = C m where C is the constant in Lemma 2.1. Lemma 2.3. Let Π L(W l,p (Ω), W m,p (Ω)) be an operator satisfying hypotheses H.1 and H.2. Then there exists a unique operator Q L(W l,p (Ω), L p (Ω)) such that D m Πu (ξ 1,...,ξ m ) = Q(D m u (ξ 1,...,ξ m )), (2.7) for all u W l+m,p (Ω). Moreover, this operator satisfies: Proof. Consider the following spaces: Qv = v, for all v P k m. (2.8) S 1 (l, p, Ω, ξ 1,...,ξ m ) = {u u W l,p (Ω), D m u (ξ 1,...,ξ m ) W l,p (Ω)} S 2 (l, p, Ω, ξ 1,...,ξ m ) = {u u W l,p (Ω), D m u (ξ 1,...,ξ m ) = 0} S(l, p, Ω, ξ 1,...,ξ m ) = S 1 (l, p, Ω, ξ 1,...,ξ m )/S 2 (l, p, Ω, ξ 1,..., ξ m ), with the W l,p (Ω) norm for the first two and the quotient norm for the third. 5

6 It follows from Lemma 2.2 that the relationship (2.6) defines a one-to-one continuous mapping from W l,p (Ω) into S(l, p, Ω, ξ 1,...,ξ m ). Let v be a function in W l,p (Ω) and let u be its image in S(l, p, Ω, ξ 1,...,ξ m ). As a result of hypothesis H.2, the operator Π maps u to a unique element of S(l, p, Ω, ξ 1,..., ξ m ) and therefore D m Πu (ξ 1,...,ξ m ) is determined uniquely in L p (Ω). This correspondence: v D m Πu (ξ 1,..., ξ m ) defines a linear, continuous operator from W l,p (Ω) into L p (Ω) which satisfies property (2.7). The uniqueness of this operator is immediate. Finally, (2.8) is a result of hypothesis H.1. Now we ll use Lemma 2.3 to demonstrate the following theorem which is similar to theorem 5 of Ciarlet and Raviart [5]. Theorem 2.1. Suppose k m and let Π L(W k+1 m,p (Ω), W m,p (Ω)) be an operator which satisfies hypotheses H.1 and H.2. Then, for any u W k+1,p (Ω), D m u (ξ 1,...,ξ m ) D m Πu (ξ 1,...,ξ m ) p,ω Ch k+1 m u k+1,p,ω (2.9) where C is a constant which depends only on Ω, Π, k, m, and p (i.e. C is the same for all equivalent domains Ω and equivalent operators Π). Proof. Applying Lemma 2.3 above and Lemma 7 of Ciarlet and Raviart [5] (extending the Bramble-Hilbert lemma [3]) in the reference ˆΩ yields: D m û (ˆξ 1,..., ˆξ m ) D m Πû (ˆξ 1,..., ˆξ m ) = (Î ˆQ)(D m û (ˆξ 1,..., ˆξ m )) (2.10) p,ˆω p,ˆω Ĉ Î ˆQ D m û (ˆξ 1,..., ˆξ m ), k+1 m,p,ˆω where Î denotes the canonical injection from W k+1 m,p (ˆΩ) into L p (Ω), Î ˆQ denotes the norm of the operator Î ˆQ in L(W k+1 m,p (ˆΩ), L p (ˆΩ)), and Ĉ is a constant. But, for any function w defined on Ω and any x Ω, we have: D m ŵ(ˆx) (ˆξ 1,..., ˆξ m ) = D m w(x) (ξ 1,...,ξ m ). Hence, taking w = u Πu: D m û (ˆξ 1,..., ˆξ m ) D m Πû (ˆξ 1,..., ˆξ m ) p,ˆω = det B 1 p D m u (ξ 1,...,ξ m ) D m Πu (ξ 1,...,ξ m ) p,ω (2.11) and also, by applying (2.3) to the function D m u (ξ 1,...,ξ m ): D m û (ˆξ 1,..., ˆξ m ) B k+1 m det B 1 p D m u (ξ 1,...,ξ m ) k+1 m,p,ω (2.12) k+1 m,p,ˆω 6

7 Hence, by combining (2.10), (2.11), and (2.12) and applying (2.2): D m u (ξ 1,...,ξ m ) D m Πu (ξ 1,...,ξ m ) p,ω Ĉ Î ˆQ B k+1 m D m u (ξ 1,...,ξ m ) k+1 m,p,ω Ĉ Î ˆQ ( ) k+1 m h u ˆρ k+1,p,ω (2.13) and this completes proof of the theorem. We will now consider a set E n = {e s } n of linearly independent unit vectors. Let ξ be any unit vector in R n. We can write: ξ = α s e s (2.14) Let θ s, 0 θ s π 2, be the angle between the vector ξ and the line containing e s and let: Lemma 2.4. For any unit vector ξ R n, we have: Equality occurs for vectors ξ when θ s = θ for all s. θ(e n ) = max ξ R n min e s E n {θ s }. (2.15) α s 1 cosθ, where θ = θ(e n). (2.16) Proof. Seek the maximum of φ(α 1,...,α n ) = n α s subject to the constraint: ( ) ξ 2 = α s e s, α r e r = Ψ(α 1,..., α n ) = 1. r=1 At the point where the maximum is realized, there exists a Lagrange multiplier (λ, µ) such that: φ λ gradφ + µ gradψ = 0, i.e. : λ φ α s + µ Ψ α s = 0, for s = 1, 2,..., n. However, α s = 1 and Ψ α s = 2(ξ, e s ) = 2 cosθs, where θ s is the angle between vectors ξ and e s (so 0 θs π; θs equals θ s or π θ s ). Thus, at a maximum of φ, we must have: (ξ, e s ) = cos θs = γ, 1 γ 1, γ independent of s. The maximum of φ is always positive since the sign of φ changes when we replace ξ with ξ; so taking the scalar product with (2.14) we obtain: 1 = α s (ξ, e s ) = γφ(α 1,...,α n ), 7

8 which implies γ must be positive. One can then ask for γ = cos θ with 0 < θ < π 2. We then get: θ s = θ for s = 1, 2,..., n and: α s = 1 cos θ. (2.17) The absolute maximum of the function φ is the largest angle θ such that there is a vector ξ for which θ s = θ for all s; this angle is given by: θ = max min θs. ξ s The lemma follows from the fact that for all vectors ξ, we can write ξ = n α se s, where e s = ±e s, αs = ±α s 0 and therefore n α s = n α s. Then the maximum of n α s is equal to the maximum of n α s for the set of vectors e s = ±e s, s = 1, 2,..., n. This maximum is given by (2.17) with angle θ replaced by the angle θ defined by (2.15) which is independent of direction of vectors e s or e s. Theorem 2.2. Let E n be a basis for R n and let Π L(W k+1 m,p (Ω), W m,p (Ω)), with k m be an operator which satisfies hypotheses H.1 and H.2 for all sets of vectors {ξ s } m (E n ) m. Then for all u W k+1,p (Ω), we have: u Πu m,p,ω C hk+1 m (cosθ) m u k+1,p,ω, (2.18) where C is a constant that depends only upon ˆΩ, ˆΠ, k, m, and p, and where θ = θ(e n ) is an angle which depends only on the set E n and is given by (2.15). Proof. Write E n = {e q,...,e n } and suppose the vectors e 1,..., e n have unit length. Let {ǫ r } m r=0 be m arbitrary unit vectors. We can write ǫ r = n α r,se s. Consider the interpolation error : w = u = Πu. We see that: D m w (ǫ 1,...,ǫ m ) = α 1,s1,..., α m,sm D m w (e s1,...,e sm ). Thus: D m w (ǫ 1,...,ǫ m ) p,ω s 1,...,s m=1 s 1,...,s m=1 α 1,s1 α m,sm D m w (e s1,...,e sm ) pω ( ) ( ) α 1,s α m,s max 1 s1...s Dm w (e s1,..., e sm ) pω m n C hk+1 m (cosθ) m u k+1,p,ω, using Lemma 2.4 and Theorem 2.1. Because the vectors ǫ r were arbitrary, we obtain (2.18). 8

9 We will now prove a variant of Theorem 2.2: we will weaken the assumption of continuity of the operator Π, but we must assume more regularity on the function u. We will use the following lemma which is a variant of Lemma 7 in Ciarlet-Raviart [5] and can be proved is very similar fashion. Lemma 2.5. Let Ω be an open bounded set in R n with a continuous boundary. Let l p and k, l, m be integers which are 0 with k+l k+1 m. Let Π L(W k+l,p (Ω), W m,p (Ω)) such that Πu = u for all u P k. Then there exists a constant C depending upon n, k, l, p and Ω such that for all u W k+l,p (Ω): u Πu m,p,ω C I Π l u k+s,p,ω, (2.19) where I is the identity function and I Π denotes the norm of the operator I Π in L(W k+l,p (Ω), W m,p (Ω)). Theorem 2.3. Let Π L(W k+l,p (Ω), W m,p (Ω)) with l 1 be an operator which satisfies the hypotheses for Theorem 2.2. Then for all u W k+l,p (Ω), we have: u Πu m,p,ω C hk+1 m (cosθ) m l 1 u k+1+r,p,ω. (2.20) Proof. Similar to Theorem 2.2. The operator Q in Lemma 2.3 is now continuous from W k+l,p (Ω) into L p (Ω). At equation (2.12) in the proof of Theorem 2.1, we apply Lemma 2.5 and everything else is unchanged. r=0 3 Applications We will successively consider several examples of finite elements that satisfy the conditions for applying Theorems 2.2 and 2.3, and then a counter-example. EXAMPLE 1 Let K be a non-degenerate n-simplex with vertices A 1, A 2,..., A n+1 and let λ 1, λ 2,...,λ n+1 denote the corresponding barycentric coordinates. Let k be a positive integer and let Σ be the set of points where each coordinate λ j is a multiple of 1. Now we consider the operator k Π which maps any continuous function u into the polynomial of degree less than k which equals u at all points in Σ. (For existence and uniqueness of this polynomial, see Nicolaides [9].) Let E N be a set of n(n+1)/2 unit vectors heading in the direction of the edges and with arbitrary orientation and let E n be a subset of E N which forms a basis for R n. We will show that the operator Π satisfies the hypothesis H.2 for any set of vectors {ξ s } m (E n) m. 9

10 Set E n = {e s } n and use that basis to define a coordinate system from any origin. For any function u, we can write: D m u (ξ 1,...,ξ m ) = α u where α = (α 1,...,α n ) is a multi-index with α = n j=1 α j = m, α j 0 and α u = α 1 ( y 1 ) αn u. α 1 αn ( y n ) Lemma 3.1. Let u be a function with α u = 0, then u can be written in the form: u = α j 1 (y j ) s f j,s (y j), (3.1) j=1 s=0 where y j = (y 1,...,y j 1, y j+1,...,y n ) and f j,s (y j) is an arbitrary function of y j. Proof. Straightforward, via successive integration. Now suppose that we have chosen the origin of the coordinate system at a vertex of n-simplex K and we chose the direction of vectors e s so that K is contained in the region: y j 0 for all j. Let h j denote the length of the side of K which is parallel to vector e j and let z j = y j h j. Substituting for new coordinates z j, the n=simplex K is contained in the hypercube: 0 z j 1, j = 1, 2,..., n. Consider the sequence of polynomials q s,k defined by: { q 0,k (t) = 1 q s,k (t) = s 1 i=0 ( ) t i k for s = 1, 2,..., k. (3.2) Since these polynomials are linearly independent, any single variable polynomial of degree less than k can be expressed as a linear combination of polynomials q s,k. We can write (3.1) in the form: u = j=1 α j 1 s=0 where q j, s are arbitrary functions of z j = y j h j. Lemma 3.2. Let u be a function of the form Then Πu has the form: q s,k (z j )g j,s (z j), (3.3) u = q s,k (z j )g j,s (z j ), 0 k. (3.4) Πu = q s,k (z j )gj,s (z j ), 0 k, (3.5) where g j,s(z j) si a polynomial of degree (k s) in variables z 1,..., z j 1, z j+1,...,z + n which interpolates the function g j,s at points of Σ located in the hyperplane z j = s k. 10

11 Proof. It is sufficient to check that functions (3.4) and (3.5) take the same value at all points of Σ. But in each of the hyperplans z j = i with 0 i s 1, we have u = Πu = 0. Now k consider a point P Σ which lies in the hyperplane z j = i ; the projection P of this point k onto the hyperplane z j = s parallel to vector e k j is also in Σ; thus, by the definition of gj,s, we have: gj,s(p) = gj,s(p ) = g j,s (P ) = g j,s (P), and it follows that Πu(P) = u(p). Using Lemma 3.1 and 3.2, we immediately deduce that: Corollary 1. Let α be an arbitrary multi-index 0 (sic.). If u is a function such that α u = 0, then the interpolant Πu satisfies: α Πu = 0. Now we can apply Theorem 2.2 and 2.3. Using the Sobolev embedding theorem, we obtain the following result. Theorem 3.1. Let Π be the interpolation operator considered above and let E n be the set of vectors in the direction of the N edges of the n-simplex K. Then: 1. If k + 1 m > n (or if k + 1 m 0 if p = ), estimate (2.18) is satisfied for any p function u W k+1,p (Ω), with θ = min En E N {θ(e n )}. 2. If l is a positive integer and k + l m > n, estimate (2.20) holds for any function p u W k+l,p (Ω). REMARK : Estimate (2.18) (or (2.20)) shows that the approximation is good provided that the angle θ is not too close to π, i.e. the directions of the edges are not too close to the same 2 hyperplane. On the other hand, there is not restriction on the flattening of n-simplex K, i.e. there is no restriction on h, where ρ is the diameter of the sphere inscribed in K. ρ In the particular case of n = 2, Theorem 3.1 gives θ = 1 { α 2 max {α, π α} max 2, π }, 3 where α is the largest angle in triangle in K. (This reasoning also shows that we can also take θ = α 2 ). When using the finite element method, we require that the angles α i of all triangles in the triangulation satisfy a condition of the form: α i γ < π (3.6) where γ is a fixed angle. Condition (3.6) is less restrictive than the condition of Zlamal [14] of the form: 0 < β α i, (3.7) for a fixed angle β. Indeed, (3.7) implies (3.6) with γ = π 2β (as long as β π 3 ). However, condition (3.6) does not exclude triangles K with an angle (a single angle) which 11

12 is arbitrarily small; the utilization of such triangles is obviously limited by the rounding errors of the computer used. EXAMPLE 2 Let {A s } n s=0 be a set of points in R n such that the vectors A 0 A s are linearly independent and let K be the parallelepiped: K = { P : A 0 P = s=0 } λ s A 0 A s, 0 λ s 1. Let k be a positive integer and let Σ be a set of points in K with coordinates λ s all equal a nonnegative integer multiple of 1. k Let Q k be the set of polynomials of degree less than nk in the variables λ s and with degree less than k in each variable. Now consider the operator Π which maps any continuous function u on K into a polynomial Πu Q k which interpolates u at the points in Σ. (The existence and uniqueness of this polynomial are well known.) Let E n be the set of vectors A 0 A s. It is easy to show using Lemma 3.1 that the operator Π satisfies hypothesis H.2 for any set of vectors {ξ s } m (E n) m and any positive integer m (the proof is similar to Example 1). In the following examples, K is a parallelogram constructed from two vectors A 0 A 1 and A 0 A 2. Our goal is to verify hypotheses H.1 and H.2 and these properties are invariant affine transformation, so we can assume that K is the unit square in the plane (x, y) (which is to confuse K with the reference element ˆK). The vertices are then the points : A 0 = (0, 0), A 1 = (1, 0), A 2 = (0, 1), A 3 = (1, 1). EXAMPLE 3 For any function u C 2 (K), let Πu be the polynomial in Q 3 such that for each vertex A s : Πu(A s ) = u(a s ), Πu(A x s) = u(a x s), Πu(A y s) = u(a y s), 2 Πu(A x y s) = 2 u(a x y s). (Hermite interpolation). It is simple to verify, using Lemma 3.1, that operator Π satisfies hypothesis H.2 for m 3 = k and ξ s { A 0 A 1, A 0 A 2 }. The following examples belong to the family of serendipity elements of Zienkiewicz [13] and based on the following property of polynomials in P 3. Let 0 < d < 1 and consider the points: A 4 = (0, d), A 5 = (d, 0), A 6 = (d, 1), A 7 = (d, 1), A 8 = (d, d). Lemma 3.3. For all polynomials u P 3, there exists a linear relationship between values of u at points in A s, 0 x 8. More precisely, we have: 7 u(a 8 ) = α s u(a s ), (3.9) s=0 12

13 with a 0 = (1 d) 2, α 1 = α 2 = d(1 d), α 3 = d 2, α 4 = α 5 = 1 d, α 6 = α 7 = d. Proof. Let u be any polynomial in P 3 and let a 1, a 2,...,a 10 be its coefficients. Let a denote the vector : a = (a 1, a 2,...,a 10 ). For each point A s, we have u(a s ) = L s (a), where L s (a) is a linear function of a. We will show that the support of functionals L s (α) are linearly dependent and calculate coefficients α such that: L 8 (a) 7 α s L s (a). (3.10) s=0 By identifying the coefficients of each component a in (3.10), we get a system of 10 equations for 7 unknowns α s. In fact, this system is solved easily and we find the values listed in Lemma 3.3. Corollary. Any polynomial u P 3 satisfies the relationship: 2 u x y (A 0) = u(a 0 )+u(a 1 )+u(a 2 ) u(a 3 ) u x (A 0) u y (A 0)+ u y (A 1)+ u x (A 2). (3.11) Proof. After division by d 2, the relationship (3.9) can be written as: u 0 + u 1 + u 2 u 3 u 4 u 0 d u 5 u 0 d u 6 u 1 d + u 7 u 2 d The relation (3.11) results by taking the limit as d goes to zero. u 8 u 4 u 5 + u 0 d 2 = 0. Lemma 3.6. Relations (3.9) and (3.11) are satisfied by polynomials x 3 y and xy 3. Proof. Immediate. EXAMPLE 4 Let K be the unit square with vertices A 0, A 1, A 2, and A 3, and let A 4, A 5,...,A 8, are the points defined as in Lemma 3.3 with d = 1. Π is the operator which interpolates any 2 function u C(K) at points A 0, A 1,...,A 7 with a polynomial in Q 2 which satisfies (3.9), i.e. a polynomial in Q 2 P 3. (See Zienkienvicz [13]). It is simple to verify that the operator Π satisfies hypothesis H.2 for m 2 = k and ξ s = { A 0 A 1, A 0 A 2 }. EXAMPLE 5 Let K be the unit square with vertices A 0, A 1, A 2, and A 3. Let 0 < d < 1 and consider the set of points Σ for which each coordinate is 0, d, 1 d, or 1. Let K denote the boundary 13

14 of K and let Σ 1 = Σ K, Σ 2 = Σ Σ 1. Operator Π maps any function u C(K) to its interpolant Πu which is a polynomial in Q 3 which equals u at points in Σ 1 and satisfies relationship (3.9) at points in Σ 2 (or similar relations deduced from a circular permutation of the vertices). It follow from the Lemma 3.5 and 3.6 that Πu is a polynomial of the form: p 3 (x, y) + a 11 x 3 y + a 12 xy 3, where p 3 is an arbitrary polynomial in P 3. Then we can verify that operator Π satisfies hypothesis H.2 for m 3 = k and ξ s = { A 0 A 1, A 0 A 2 }. Particular Cases : 1) d = 1 case. 3 This gives a classic element of Zienkiewicz [13]. 2) Limiting case at d = 0. This gives an element of Adini [1]. This gives a Hermite interpolant: the values of the function and its first derivative at the four vertices are used; this corresponds to the element in Example 3 in which by second derivatives 2 u have been eliminated through relation x y (3.11). COUNTER-EXAMPLE - The triangular serendipity element (see [13]). In the (x, y) plane, let triangle K have vertices A 1 = (0, 0), A 2 = (d, 0) and A 3 = (0, 1), where d is any positive number. Let A be the centroid of the triangle and let A ij = 1 3 (2A i + A j ), 1 i 3, 1 j 3. Consider operator Π which maps any continuous function u on K to a polynomial of degree 3 which equals u at points A ij and satisfies the relationship: Πu(A) = Πu(A i ) i=1 3 i, j = 1 i j Πu(A ij ) Consider the function u(x, y) = y 3. We have: u(a) = u(a 13 ) = u(a 23 ) = Πu(A 13 ) = Πu(A 2 3) = 1, and Πu(A) = Hypothesis H.2 does not hold. Moreover, as d 0, the derivative Πu tends to infinity x like 1 u but we have = 0! d x References [1] Adini A. and Clough R. W., Analysis of plate bending by the finite element method, N.S.F. report G. 7337, [2] Agmon S., Lecture on elliptic boundary value problems, Van Nostrand,

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