Error estimates for nearly degenerate finite elements
Pierre Jamet
In RAIRO: Analyse Num´erique, Vol 10, No 3, March 1976, p. 43–61
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 Wm,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 Rn, 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:
||u − Πu||m,p,K = O hk+1 p
, (1.1)
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 Wm,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(hk+1−m) (1.2)
holds under the assumption that all elements K of the triangulation satisfy a condition of
the form: ρ
h ≥ α > 0, (1.3)
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(hk+1−m/(cos θ)m), (1.4) where θ is an angle between 0 and π2 which can approach π2 if all edges of K are almost parallel to the same hyperplane. From (1.4), one deduces the condition:
θ ≤ θ0 < 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
2, θ ∼ π4 and condition (1.5) is satisfied. An estimate identical to (1.4) was established by 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 ∈ Rn. (ξ1, ξ2) = the dot products of vectors ξ1, ξ2 ∈ Rn.
For a sufficiently regular function u defined on domain Ω ⊂ Rn: Du = the Fr´echet derivative of order 1,
Dmu = the Fr´echet derivative of order m,
Dmu · (ξ1, ξ2, ..., ξm) = the directional derivative of order m relative to the vectors ξ1, ξ2, ..., ξm,
Dmu(x) · (ξ1, ξ2, ..., ξm) = the value at point x ∈ Ω in the derivative above,
||Dmu|| = max{|Dmu(x) · (ξ1, ξ2, ..., ξm)|; ||xs|| = 1, 1 ≤ s ≤ m}.
||u||p,Ω = the norm of u in Lp(Ω).
|u|m,p,Ω = ||Dmu||p,Ω = R ||Dmu||p1p
is the semi-norm in the Sobolev space Wm,p(Ω), 1 ≤ p < ∞.
||u||m,p,Ω = Pm
i=0||Diu||pp,Ω
1p
is a norm equivalent to the usual norm in Wm,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 .
Pk denotes the set of polynomials of n variables of degree ≤ k.
2 General Results
Let ˆΩ be a bounded domain in Rn and let Ω is any equivalent domain, i.e. there exists is an invertible affine transformation F such that:
Ω = F ( ˆΩ) = {x ∈ Rn; x = F (ˆx), ˆx ∈ ˆΩ} (2.1) Transformation F can be written in the form: F (ˆx) = B ˆx + 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|| ≤ h ˆ
ρ (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 ˆu defined on ˆΩ by the canonical relationship:
ˆ
u(ˆx) = u(x).
Let u ∈ Wl,p(Ω), integer l ≥ 0, and l ≤ p ≤ ∞, the ˆu ∈ Wl,p( ˆΩ) and (equation (4.15) in Ciarlet and Raviart [5]):
|ˆu|l,p, ˆΩ ≤ ||B||l|det B|−1p|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(Wl,p(Ω), Wm,p(Ω)) with l ≤ p ≤ ∞ that satisfies the following two hypotheses:
Πu = u, for all u ∈ Pk. (H.1)
There exists a set of vectors {ξs}ms=1 (not necessarily distinct) such that
DmΠu · (ξ1, ..., ξm) = 0 (H.2)
for all functions u such that Dmu · (ξ1, ..., ξm) = 0.
Without loss of generality, we assume that the vectors ξs are unit vectors.
The operator Π corresponds to an equivalent operator ˆΠ ∈ L(Wl,p( ˆΩ), Wm,p( ˆΩ)) defined through the canonical relationship, ˆΠˆu(ˆx) = Πu(x).
If Π satisfies properties H.1 and H.2, then ˆΠ satisfies the analogous properties with vectors ξs replaced by ˆξs = B−1ξs.
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 Rn satisfying the cone property and let ξ be an arbitrary vector. For any v ∈ Wl,p(Ω), l ≥ 1, there exists a function u ∈ Wl,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 x1 and x2 such that the x1 axis is parallel to the vector ξ and Ω is contained in the square G = {(x1, x2); 0 ≤ x1 ≤ h, 0 ≤ x2 ≤ 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 R2 by a continuous extension operator from Wl,p(Ω) into Wl,p(R2). Denoting the extension with the same notation v, we have
||v||1,p,G≤ ||v||l,p,R2 ≤ C1||v||l,p,Ω
where the constant C1 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∞( ¯G) and set:
u(x1, x2) = Z x1
0
v(s, x2)ds.
The function u satisfies Du · (ξ) = ∂x∂u1 = v. So
∂u
∂x1
p,G
= ||v||p,G ≤ ||v||l,p,G. (2.4) Moreover:
∂u
∂x2
(x1, x2) = Z x1
0
∂v
∂x2
(s, x2)ds.
Applying H¨older’s inequality:
∂u
∂x2
(x1, x2) ≤
Z h 0
∂v
∂x2
(x1, x2)
dx1 ≤ h1q
Z h 0
∂v
∂x2
(x1, x2)
dx1
1p ,
for 1p +1q = 1. Then deduce that Z h
0
∂u
∂x2
(x1, x2)
p
dx2 ≤ hpq
∂v
∂x2
p
p,G
,
and by integrating in x1:
∂u
∂x2
p,G
≤ h
∂v
∂x2
p,G
≤ h ||v||l,p,G. (2.5)
Using (2.4) and (2.5) conclude that:
|u|1,p,G≤√
2 max{1, h} ||v||l,p,G.
Through a density argument, this inequality extends to any function v ∈ Wl,p(G) which completes the proof.
Lemma 2.2. Let Ω be a bounded domain in Rn satisfying the cone property. Let l and m be two positive integers and let {ξs}ms=1 be a set of unit vectors which are not necessarily distinct. For all v ∈ Wl,p(Ω), there exists a function u ∈ Wl,p(Ω) such that:
Dmu · (ξ1, . . . , ξm) = 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 u1 ∈ Wl,p(Ω) such that Du1· (ξ1) = v and ||u1||l,p,Ω ≤ C ||v||l,p,Ω. Then, there exists u2 ∈ Wl,p(Ω) such that Du2· (ξ2) = u1 and ||u2||l,p,Ω ≤ C ||u1||l,p,Ω. This function u2 satisfies:
Du2· (ξ1, ξ2) = Du1· (ξ1) = v and ||u2||l,p,Ω≤ C2||v||l,p,Ω.
Continuing inductively, we find function u = um that satisfies the conditions in Lemma 2.2 wtih C′ = Cm where C is the constant in Lemma 2.1.
Lemma 2.3. Let Π ∈ L(Wl,p(Ω), Wm,p(Ω)) be an operator satisfying hypotheses H.1 and H.2. Then there exists a unique operator Q ∈ L(Wl,p(Ω), Lp(Ω)) such that
DmΠu · (ξ1, . . . , ξm) = Q(Dmu · (ξ1, . . . , ξm)), (2.7) for all u ∈ Wl+m,p(Ω). Moreover, this operator satisfies:
Qv = v, for all v ∈ Pk−m. (2.8)
Proof. Consider the following spaces:
S1(l, p, Ω, ξ1, . . . , ξm) = {u | u ∈ Wl,p(Ω), Dmu · (ξ1, . . . , ξm) ∈ Wl,p(Ω)}
S2(l, p, Ω, ξ1, . . . , ξm) = {u | u ∈ Wl,p(Ω), Dmu · (ξ1, . . . , ξm) = 0}
S(l, p, Ω, ξ1, . . . , ξm) = S1(l, p, Ω, ξ1, . . . , ξm)/S2(l, p, Ω, ξ1, . . . , ξm), with the Wl,p(Ω) norm for the first two and the quotient norm for the third.
It follows from Lemma 2.2 that the relationship (2.6) defines a one-to-one continuous mapping from Wl,p(Ω) into S(l, p, Ω, ξ1, . . . , ξm). Let v be a function in Wl,p(Ω) and let u be its image in S(l, p, Ω, ξ1, . . . , ξm). As a result of hypothesis H.2, the operator Π maps u to a unique element of S(l, p, Ω, ξ1, . . . , ξm) and therefore DmΠu · (ξ1, . . . , ξm) is determined uniquely in Lp(Ω). This correspondence: v → DmΠu · ·(ξ1, . . . , ξm) defines a linear, continu- ous operator from Wl,p(Ω) into Lp(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(Wk+1−m,p(Ω), Wm,p(Ω)) be an operator which satisfies hypotheses H.1 and H.2. Then, for any u ∈ Wk+1,p(Ω),
||Dmu · (ξ1, . . . , ξm) − DmΠu · (ξ1, . . . , ξm)||p,Ω ≤ Chk+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
mu · (ˆˆ ξ1, . . . , ˆξm) − DmΠˆu · (ˆξ1, . . . , ˆξm) p, ˆΩ =
(I − ˆˆ Q)(Dmu · (ˆˆ ξ1, . . . , ˆξm))
p, ˆΩ (2.10)
≤ ˆC
I − ˆˆ Q D
mu · (ˆˆ ξ1, . . . , ˆξm)
k+1−m,p, ˆΩ, where ˆI denotes the canonical injection from Wk+1−m,p( ˆΩ) into Lp(Ω),
I − ˆˆ Q
denotes the norm of the operator ˆI − ˆQ in L(Wk+1−m,p( ˆΩ), Lp( ˆΩ)), and ˆC is a constant. But, for any function w defined on Ω and any x ∈ Ω, we have:
Dmw(ˆˆ x) · (ˆξ1, . . . , ˆξm) = Dmw(x) · (ξ1, . . . , ξm).
Hence, taking w = u − Πu:
D
mu · (ˆˆ ξ1, . . . , ˆξm) − DmΠˆu · (ˆξ1, . . . , ˆξm) p, ˆΩ
= |det B|−p1 ||Dmu · (ξ1, . . . , ξm) − DmΠu · (ξ1, . . . , ξm)||p,Ω (2.11) and also, by applying (2.3) to the function Dmu · (ξ1, . . . , ξm):
D
mu · (ˆˆ ξ1, . . . , ˆξm)
k+1−m,p, ˆΩ ≤ ||B||k+1−m|det B|−1p|Dmu · (ξ1, . . . , ξm)|k+1−m,p,Ω (2.12)
Hence, by combining (2.10), (2.11), and (2.12) and applying (2.2):
||Dmu · (ξ1, . . . , ξm) − DmΠu · (ξ1, . . . , ξm)||p,Ω
≤ ˆC
I − ˆˆ Q ||B ||
k+1−m
|Dmu · (ξ1, . . . , ξm)|k+1−m,p,Ω
≤ ˆC
I − ˆˆ Q
h ˆ ρ
k+1−m
|u|k+1,p,Ω (2.13)
and this completes proof of the theorem.
We will now consider a set En = {es}ns=1 of linearly independent unit vectors. Let ξ be any unit vector in Rn. We can write:
ξ =
n
X
s=1
αses (2.14)
Let θs, 0 ≤ θs≤ π2, be the angle between the vector ξ and the line containing es and let:
θ(En) = max
ξ∈Rn min
es∈En{θs}. (2.15)
Lemma 2.4. For any unit vector ξ ∈ Rn, we have:
n
X
s=1
|αs| ≤ 1
cos θ, where θ = θ(En). (2.16)
Equality occurs for vectors ξ when θs = θ for all s.
Proof. Seek the maximum of φ(α1, . . . , αn) =Pn
s=1αs subject to the constraint:
||ξ||2 =
n
X
s=1
αses,
n
X
r=1
αrer
!
= Ψ(α1, . . . , αn) = 1.
At the point where the maximum is realized, there exists a Lagrange multiplier (λ, µ) such that:
λ gradφ + µ gradΨ = 0, i.e. : λ∂φ
∂αs
+ µ∂Ψ
∂αs
= 0, for s = 1, 2, . . . , n.
However, ∂α∂φ
s = 1 and ∂α∂Ψ
s = 2(ξ, es) = 2 cos θ∗s, where θ∗s is the angle between vectors ξ and es (so 0 ≤ θs∗ ≤ π; θ∗s equals θs or π − θs). Thus, at a maximum of φ, we must have:
(ξ, es) = cos θ∗s = 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 =
n
X
s=1
αs(ξ, es) = γφ(α1, . . . , αn),
which implies γ must be positive. One can then ask for γ = cos θ∗ with 0 < θ∗ < π2. We then get: θs∗ = θ∗ for s = 1, 2, . . . , n and:
n
X
s=1
αs = 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
ξ min
s θs∗.
The lemma follows from the fact that for all vectors ξ, we can write ξ = Pn
s=1αs∗e∗s, where e∗s = ±es, α∗s = ±αs ≥ 0 and therefore Pn
s=1αs∗ = Pn
s=1|αs|. Then the maximum of Pn
s=1|αs| is equal to the maximum ofPn
s=1α∗s for the set of vectors e∗s = ±es, 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 es or e∗s.
Theorem 2.2. Let En be a basis for Rn and let Π ∈ L(Wk+1−m,p(Ω), Wm,p(Ω)), with k ≥ m be an operator which satisfies hypotheses H.1 and H.2 for all sets of vectors {ξs}ms=1 ∈ (En)m. Then for all u ∈ Wk+1,p(Ω), we have:
|u − Πu|m,p,Ω ≤ C hk+1−m
(cos θ)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(En) is an angle which depends only on the set En and is given by (2.15).
Proof. Write En = {eq, . . . , en} and suppose the vectors e1, . . . , en have unit length. Let {ǫr}mr=0 be m arbitrary unit vectors. We can write ǫr =Pn
s=1αr,ses. Consider the interpola- tion error : w = u = Πu. We see that:
Dmw · (ǫ1, . . . , ǫm) =
n
X
s1,...,sm=1
α1,s1, . . . , αm,smDmw · (es1, . . . , esm).
Thus:
||Dmw · (ǫ1, . . . , ǫm)||p,Ω ≤
n
X
s1,...,sm=1
|α1,s1| · · · |αm,sm| ||Dmw · (es1, . . . , esm)||pΩ
≤
n
X
s=1
|α1,s|
!
· · ·
n
X
s=1
|αm,s|
!
1≤s1...smaxm≤n||Dmw · (es1, . . . , esm)||pΩ
≤ C hk+1−m
(cos θ)m |u|k+1,p,Ω,
using Lemma 2.4 and Theorem 2.1. Because the vectors ǫr were arbitrary, we obtain (2.18).
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 Rnwith a continuous boundary. Let l ≤ p ≤ ∞ and k, l, m be integers which are ≥ 0 with k +l ≥ k +1 ≥ m. Let Π ∈ L(Wk+l,p(Ω), Wm,p(Ω)) such that Πu = u for all u ∈ Pk. Then there exists a constant C depending upon n, k, l, p and Ω such that for all u ∈ Wk+l,p(Ω):
||u − Πu||m,p,Ω≤ C ||I − Π||
l
X
s=1
||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(Wk+l,p(Ω), Wm,p(Ω)).
Theorem 2.3. Let Π ∈ L(Wk+l,p(Ω), Wm,p(Ω)) with l ≥ 1 be an operator which satisfies the hypotheses for Theorem 2.2. Then for all u ∈ Wk+l,p(Ω), we have:
|u − Πu|m,p,Ω ≤ C hk+1−m (cos θ)m
l−1
X
r=0
|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 Wk+l,p(Ω) into Lp(Ω). At equation (2.12) in the proof of Theorem 2.1, we apply Lemma 2.5 and everything else is unchanged.
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 A1, A2, . . . , An+1and let λ1, λ2, . . . , λn+1
denote the corresponding barycentric coordinates. Let k be a positive integer and let Σ be the set of points where each coordinate λj is a multiple of 1k. Now we consider the operator 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 EN be a set of n(n + 1)/2 unit vectors heading in the direction of the edges and with arbitrary orientation and let En be a subset of EN which forms a basis for Rn. We will show that the operator Π satisfies the hypothesis H.2 for any set of vectors {ξs}ms=1 ∈ (En)m.
Set En = {es}ns=1 and use that basis to define a coordinate system from any origin.
For any function u, we can write: Dmu · (ξ1, . . . , ξm) = ∂αu where α = (α1, . . . , αn) is a multi-index with |α| =Pn
j=1αj = m, αj ≥ 0 and
∂αu = ∂α1
(∂y1)α1 · · · ∂αn (∂yn)αnu.
Lemma 3.1. Let u be a function with ∂αu = 0, then u can be written in the form:
u =
n
X
j=1 αj−1
X
s=0
(yj)sfj,s(y′j), (3.1)
where yj′ = (y1, . . . , yj−1, yj+1, . . . , yn) and fj,s(y′j) is an arbitrary function of yj′. 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 es so that K is contained in the region:
yj ≥ 0 for all j. Let hj denote the length of the side of K which is parallel to vector ej
and let zj = hyj
j. Substituting for new coordinates zj, the n=simplex K is contained in the hypercube: 0 ≤ zj ≤ 1, j = 1, 2, . . . , n. Consider the sequence of polynomials qs,k defined by:
(q0,k(t) = 1 qs,k(t) =Qs−1
i=0 t − ki
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 qs,k. We can write (3.1) in the form:
u =
n
X
j=1 αj−1
X
s=0
qs,k(zj)gj,s(z′j), (3.3)
where qj, s are arbitrary functions of zj′ = yh′j
j. Lemma 3.2. Let u be a function of the form
u = qs,k(zj)gj,s(zj′), 0 ≤ k. (3.4) Then Πu has the form:
Πu = qs,k(zj)gj,s∗ (zj′), 0 ≤ k, (3.5) where gj,s∗ (z′j) si a polynomial of degree ≤ (k − s) in variables z1, . . . , zj−1, zj+1, . . . , z + n which interpolates the function gj,s at points of Σ located in the hyperplane zj = ks.
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 zj = ki with 0 ≤ i ≤ s − 1, we have u = Πu = 0. Now consider a point P ∈ Σ which lies in the hyperplane zj = ki; the projection P′ of this point onto the hyperplane zj = sk parallel to vector ej is also in Σ; thus, by the definition of gj,s∗ , we have:
gj,s∗ (P ) = gj,s∗ (P′) = gj,s(P′) = gj,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
∂αu = 0, then the interpolant Πu satisfies: ∂αΠu = 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 En be the set of vectors in the direction of the N edges of the n-simplex K. Then:
1. If k + 1 − m > np (or if k + 1 − m ≥ 0 if p = ∞), estimate (2.18) is satisfied for any function u ∈ Wk+1,p(Ω), with θ = minEn⊂EN{θ(En)}.
2. If l is a positive integer and k + l − m > np, estimate (2.20) holds for any function u ∈ Wk+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 π2, i.e. the directions of the edges are not too close to the same hyperplane. On the other hand, there is not restriction on the flattening of n-simplex K, i.e.
there is no restriction on hρ, where ρ is the diameter of the sphere inscribed in K.
In the particular case of n = 2, Theorem 3.1 gives θ = 1
2max {α, π − α} ≤ maxnα 2,π
3 o,
where α is the largest angle in triangle in K. (This reasoning also shows that we can also take θ = α2). When using the finite element method, we require that the angles αi of all triangles in the triangulation satisfy a condition of the form:
αi ≤ 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 ≤ αi, (3.7)
for a fixed angle β. Indeed, (3.7) implies (3.6) with γ = π − 2β (as long as β ≤ π3).
However, condition (3.6) does not exclude triangles K with an angle (a single angle) which
is arbitrarily small; the utilization of such triangles is obviously limited by the rounding errors of the computer used.
EXAMPLE 2
Let {As}ns=0 be a set of points in Rnsuch that the vectors −−−→A0As are linearly independent and let K be the parallelepiped:
K = (
P : −−→A0P =
n
X
s=0
λs−−−→A0As, 0 ≤ λs≤ 1 )
.
Let k be a positive integer and let Σ be a set of points in K with coordinates λs all equal a nonnegative integer multiple of 1k.
Let Qk be the set of polynomials of degree less than nk in the variables λs 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 ∈ Qk which interpolates u at the points in Σ. (The existence and uniqueness of this polynomial are well known.) Let En be the set of vectors
−−−→A0As. It is easy to show using Lemma 3.1 that the operator Π satisfies hypothesis H.2 for any set of vectors {ξs}ms=1 ∈ (En)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 −−−→
A0A1 and
−−−→A0A2. 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 : A0 = (0, 0), A1 = (1, 0), A2 = (0, 1), A3 = (1, 1).
EXAMPLE 3
For any function u ∈ C2(K), let Πu be the polynomial in Q3 such that for each vertex As: Πu(As) = u(As), ∂x∂ Πu(As) = ∂x∂ u(As), ∂y∂Πu(As) = ∂y∂ u(As), ∂x∂y∂2 Πu(As) = ∂x∂y∂2 u(As).
(Hermite interpolation).
It is simple to verify, using Lemma 3.1, that operator Π satisfies hypothesis H.2 for m ≤ 3 = k and ξs ∈ {−−−→A0A1,−−−→A0A2}.
The following examples belong to the family of serendipity elements of Zienkiewicz [13]
and based on the following property of polynomials in P3. Let 0 < d < 1 and consider the points:
A4 = (0, d), A5 = (d, 0), A6 = (d, 1), A7 = (d, 1), A8 = (d, d).
Lemma 3.3. For all polynomials u ∈ P3, there exists a linear relationship between values of u at points in As, 0 ≤ x ≤ 8. More precisely, we have:
u(A8) =
7
X
s=0
αsu(As), (3.9)
with a0 = −(1 − d)2, α1 = α2 = −d(1 − d), α3 = −d2, α4 = α5 = 1 − d, α6 = α7 = d.
Proof. Let u be any polynomial in P3 and let a1, a2, . . . , a10 be its coefficients. Let a denote the vector : a = (a1, a2, . . . , a10). For each point As, we have
u(As) = Ls(a),
where Ls(a) is a linear function of a. We will show that the support of functionals Ls(α) are linearly dependent and calculate coefficients α such that:
L8(a) ≡
7
X
s=0
αsLs(a). (3.10)
By identifying the coefficients of each component a in (3.10), we get a system of 10 equations for 7 unknowns αs. In fact, this system is solved easily and we find the values listed in Lemma 3.3.
Corollary. Any polynomial u ∈ P3 satisfies the relationship:
∂2u
∂x∂y(A0) = −u(A0)+u(A1)+u(A2)−u(A3)−∂u
∂x(A0)−∂u
∂y(A0)+∂u
∂y(A1)+∂u
∂x(A2). (3.11) Proof. After division by d2, the relationship (3.9) can be written as:
−u0+ u1+ u2− u3− u4− u0
d − u5− u0 d
u6− u1
d + u7− u2
d −u8− u4− u5+ u0
d2 = 0.
The relation (3.11) results by taking the limit as d goes to zero.
Lemma 3.6. Relations (3.9) and (3.11) are satisfied by polynomials x3y and xy3. Proof. Immediate.
EXAMPLE 4
Let K be the unit square with vertices A0, A1, A2, and A3, and let A4, A5, . . . , A8, are the points defined as in Lemma 3.3 with d = 12. Π is the operator which interpolates any function u ∈ C(K) at points A0, A1, . . . , A7 with a polynomial in Q2 which satisfies (3.9), i.e. a polynomial in Q2∩ P3. (See Zienkienvicz [13]).
It is simple to verify that the operator Π satisfies hypothesis H.2 for m ≤ 2 = k and ξs= {−−−→A0A1,−−−→A0A2}.
EXAMPLE 5
Let K be the unit square with vertices A0, A1, A2, and A3. 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
of K and let Σ1 = Σ ∩ ∂K, Σ2 = Σ − Σ1. Operator Π maps any function u ∈ C(K) to its interpolant Πu which is a polynomial in Q3 which equals u at points in Σ1 and satisfies relationship (3.9) at points in Σ2 (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:
p3(x, y) + a11x3y + a12xy3, where p3 is an arbitrary polynomial in P3.
Then we can verify that operator Π satisfies hypothesis H.2 for m ≤ 3 = k and ξs = {−−−→
A0A1,−−−→
A0A2}.
Particular Cases : 1) d = 13 case.
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 ∂x∂y∂2u have been eliminated through relation (3.11).
COUNTER-EXAMPLE - The triangular serendipity element (see [13]).
In the (x, y) plane, let triangle K have vertices A1 = (0, 0), A2 = (d, 0) and A3 = (0, 1), where d is any positive number. Let A be the centroid of the triangle and let Aij = 13(2Ai+ Aj), 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 Aij and satisfies the relationship:
Πu(A) = −1 6
3
X
i=1
Πu(Ai) + 1 4
3
X i, j = 1
i 6= j
Πu(Aij)
Consider the function u(x, y) = y3. We have:
u(A) = u(A13) = u(A23) = Πu(A13) = Πu(A23) = 1
27, and Πu(A) = 0.
Hypothesis H.2 does not hold. Moreover, as d → 0, the derivative ∂x∂ Πu tends to infinity like 1d but we have ∂u∂x = 0!
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