Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
www.elsevier.com/locate/cma
Quadrilateral mesh revisited
Pingbing Ming *, Zhong-Ci Shi
Institute of Computational Mathematics, Chinese Academy of Sciences, A1 South 4 Street, Zhong Guan Cun,
P.O. Box 2719, Beijing 100080, China
Received 30 April 2002
Abstract
Several quadrilateral shape regular mesh conditions commonly used in the finite element method are proven to be
equivalent. The effect of the Bi-Section Condition on the degenerate mesh conditions is also checked.
Ó 2002 Elsevier Science B.V. All rights reserved.
Keywords: Quadrilateral mesh; 4-Node isoparametric element
1. Introduction
Quadrilateral mesh is widely used in the finite element method due to its simplicity and flexibility.
However, numerical accuracy cannot be achieved over an arbitrary mesh, certain mesh conditions have to
be imposed. There exist several mesh conditions in the literature, which can be classified into two groups.
One is the shape regular condition and the other is the degenerate condition. Roughly speaking, a shape
regular condition requires that the element cannot be too narrow on the one hand ((C–R)1 hereinafter) and
the interior angle of each vertex is neither too small nor too close to p on the other hand ((C–R)2 hereinafter). The first condition of this type belongs to Ciarlet–Raviart (C–R) [12,13] and another two are
attributed to Girault–Raviart (G–R) [6,16] and Arunakirinathar–Reddy (A–R) [5]. All these three conditions aim for the optimal interpolation error for the isoparametric element, whereas a similar condition of
Zhang (Z) [23] appeared in the study of the Wilson nonconforming element. It may be interesting to ask
whether such conditions are equivalent. One of the main results of this paper is a strict proof for the
equivalence of C–R, G–R, A–R and Z (see Theorem 3.3 below).
Meanwhile, there are several degenerate mesh conditions, which violate either (C–R)1 or (C–R)2 , and
sometimes even both. These conditions are scattered in the literature, most of them are mutatis mutandis.
We only consider three degenerate conditions, namely the Jamet condition (J ) [17,18], the Acosta–DuransÕ
regular decomposition property (RDP) [1] and the S€
uli condition (S) [22]. We will clarify their connections
to the shape regular mesh condition. In particular, we show by means of a counterexample that RDP is
*
Corresponding author. Tel.: +86-10-6262-6583; fax: +86-10-6254-2285.
E-mail addresses: [email protected] (P. Ming), [email protected] (Z.-C. Shi).
0045-7825/02/$ - see front matter Ó 2002 Elsevier Science B.V. All rights reserved.
PII: S 0 0 4 5 - 7 8 2 5 ( 0 2 ) 0 0 4 7 1 - 1
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P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
necessary for obtaining the optimal interpolation error in the H 1 -norm for the 4-node isoparametric element, thereby we solve the open problem proposed in [1].
On the other hand, the Bi-Section Condition and the Angle Condition are two mesh conditions which
quantify the deviation of an arbitrary quadrilateral from a parallelogram. S€
uli [22] proved that the Jamet
degenerate mesh condition plus the Bi-Section Condition actually imply the shape regular condition when
the mesh diameter tends to zero. We will give a new and shorter proof for S€
uliÕs result, and show that its
requirement of the Bi-Section Condition can be replaced by an even weaker condition. Moreover, we
proved that the Bi-Section Condition and the Angle Condition are equivalent if the mesh satisfies the Jamet
condition, which is shown to be also necessary for such equivalence by a counterexample.
The remaining part of this paper is organized as follows. In Section 2, we state all shape regular mesh
conditions mentioned above. Their equivalence is proven in Section 3. In Section 4, several degenerate mesh
conditions are reviewed and their connections to the shape regular mesh condition are elucidated with the
aid of the Bi-Section Condition and the Angle Condition. Some conclusions are drawn in the last section.
Throughout this paper, the generic constant C is independent of the element geometry unless otherwise
stated.
2. Shape regular mesh condition
Before introducing mesh conditions, we fix some notations. For any convex polygon X with Lipschitz
boundary, H k ðXÞ is defined as the standard Sobolev space [2] equipped with the norm k kk , and semi-norm
j jk , H0k ðXÞ is the corresponding homogeneous space. For any vectors x ¼ ðx1 ; x2 Þ and y ¼ ðy1 ; y2 Þ, x y is a
2 2 matrix with elements ðx yÞij :¼ xi yj . For any matrix A, kAk denotes its Euclidean norm.
2.1. Geometric facts of quadrilateral
Let Th be a partition of X by convex quadrilaterals with the mesh size h :¼ maxK2Th hK . We define hK
and hK the longest and shortest edge of K, respectively. qK is defined as the diameter of the largest circle
inscribed in K. As in Fig. 1 we denote the four vertices of K by Pi with the coordinates xi . Let their edges be
Pi Piþ1 with jPi Piþ1 j the corresponding lengths. The sub-triangle of K with vertices Pi1 , Pi and Piþ1 is denoted
by Ti , i.e., Ti :¼ hPi1 ; Pi ; Piþ1 i (i modulo 4). Similar to K, hi and qi are defined as the longest edge of Ti
and the diameter for the largest circle inscribed in Ti , respectively. Denote the interior angle of the vertex Pi
by hi , and lK :¼ max1 6 i 6 4 j cos hi j. Moreover, dK is denoted as the distance between midpoints of two
diagonals of K.
Fig. 1. (a) An arbitrary quadrilateral, (b) the dotted parallelogram is the equivalent parallelogram [5] of the quadrilateral.
P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
5673
We define by Pk the space of polynomials of degree no more than k, and by Qk the space of degree no
more than k in each variable.
b ¼ ½1; 12 be the reference square having the vertex Pbi with the coordinates x
^i ð1 6 i 6 4Þ, then
Let K
b Þ such that FK ð^
there exists a unique mapping FK ðn; gÞ 2 Q1 ð K
xi Þ ¼ xi , 1 6 i 6 4. The Jacobian of FK ðn; gÞ
is denoted by DFK ðn; gÞ which can be split as DFK ðn; gÞ ¼ DFK ð0; 0Þ þ k l, where k :¼ ðx1 x2 þ
T
T
x3 x4 Þ =4 and l :¼ ðg; nÞ . The determinant of DFK ðn; gÞ is JK ðn; gÞ ¼ det DFK ðn; gÞ ¼ J0 þ J1 n þ J2 g. It
can be shown that
max JK ðn; gÞ ¼ max jTi j=2;
ðn;gÞ2K^
16i64
min JK ðn; gÞ ¼ min jTi j=2
ðn;gÞ2K^
16i64
and
J0 ¼ JK ð0; 0Þ ¼ jKj=4;
JK ðn; gÞ > 0:
2.2. Shape regular mesh conditions
We mainly concern the following shape regular mesh conditions.
1. Ciarlet–Raviart (C–R) [12,13]. Th is regular [11, p. 247], if there exist two constants r0 P 1 and
0 < l < 1 such that
hK =hK 6 r0 ; 0 < lK 6 l < 1:
2. Girault–Raviart (G–R) [6,16].
Define qK :¼ 2 min1 6 i 6 4 qi . Th is regular if there exists r > 0 such that
ð2:1Þ
ð2:2Þ
qK 6 r:
max hK =
K2Th
3. Arunakirinathar–Reddy (A–R) [5].
Th is regular if there exist four constants C, C0 , C1 and c > 1 such that
kDFK ð0; 0Þk 6 ChK ;
1
kDF1
K ð0; 0Þk 6 ChK ;
C0 h2K 6 JK ð0; 0Þ 6 C1 h2K :
ðDF1
K ð0; 0Þk; lÞ P c:
4. Zhang (Z) [23].
Define q~K :¼ min1 6 i 6 4 qi . Th is regular if there exists r > 0 such that
ð2:3Þ
ð2:4Þ
ð2:5Þ
qK 6 r:
max hK =~
K2Th
A–R seems a bit strange, it is expressed in an analytic way instead of some geometric facts as C–R, G–R
and Z. We will show in the next lemma that it could be boiled down to a simpler condition.
Lemma 2.1. A–R is equivalent to the following mesh condition: there exists a constant C such that
JK ðn; gÞ P C h2K
b:
8ðn; gÞ 2 K
ð2:6Þ
To be more precise (A–R)1 , i.e. (2.3), is equivalent to
J0 P C0 h2K :
Proof. (A–R)2 , i.e. (2.4) implies that
ðDF1
K ð0; 0Þk; lÞ ¼ ðJK J0 Þ=J0 P c:
ð2:7Þ
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P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
In view of (A–R)1 , we obtain
JK P ð1 cÞJ0 P C0 ð1 cÞh2K ¼: C h2K ;
which is (2.6).
We remain to show that (2.6) implies A–R. An elementary manipulation shows that
kDFK ð0; 0Þk 6 hK =2 and
J0 6 h2K =8
and J0 ¼ JK ð0; 0Þ P C h2K yields
1
kDF1
K ð0; 0Þk 6 ðhK =2Þ=J0 6 1=ð2C ÞhK :
The above three inequalities comprise (A–R)1 . Further, it is seen that
ðDF1
K ð0; 0Þk; lÞ ¼ ðJK J0 Þ=J0 ¼ JK =J0 1 P 8C 1 ¼: c;
which proves the equivalence of A–R and (2.6).
The equivalence of (A–R)1 and (2.7) is trivial, we omit the details. Remark 2.2. (A–R)1 states that the equivalent parallelogram (see Fig. 1b) is C–R regular, and (A–R)2 states
that the difference between the original quadrilateral and the equivalent parallelogram is not too big.
Remark 2.3. It is seen that (2.6) is a part of the k-strongly regular condition introduced by Zlamal [25] for
investigating the superconvergence of isoparametric elements.
3. Equivalence of shape regular conditions
In this section, we will prove that all the aforementioned shape regular conditions are equivalent. Before
the presentation, we state two lemmas for later use.
Lemma 3.1. [24] For a family of triangular finite elements, if the ratio between the longest edge of the triangle
and the radius of the biggest circle inscribed into the triangle is uniformly bounded, then there exists a constant
h0 such that all interior angle hK of triangles satisfies
hK P h0 > 0:
ð3:1Þ
Lemma 3.2. For any triangle K, the diameter of the biggest circle inscribed into K is less than the shortest edge
of K.
Theorem 3.3. All the foregoing mentioned shape regular mesh conditions (C–R, G–R, A–R and Z) are
equivalent.
Proof. We only need to prove that all shape regular mesh conditions are equivalent to C–R. The equivalence between G–R and Z is obvious. So what we need to prove is the equivalence of C–R, G–R and A–R.
Firstly we prove that G–R implies C–R.
Given the regular condition G–R, in view of Lemma 3.2, we obtain
hK =hK 6 2hK =
qK 6 2r
which gives (C–R)1 with r0 ¼ 2r.
P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
5675
Notice that hi =qi 6 2hK =
qK 6 2r, invoking Lemma 3.1, we see that each interior angle of Ti is bounded
below by h0 , which in turn implies an upper bound for each angle with p 2h0 , so hi is bounded as
h0 6 hi 6 p 2h0 , thus we come to (C–R)2 with l ¼ maxðcos h0 ; j cos 2h0 jÞ. Therefore, C–R follows from
G–R.
We are in a position to prove that C–R implies G–R. For any triangle Ti , we have
qi ¼
2jTi j
h2 sin hi
ð1 l2 Þ1=2
P K
P
hi ;
jPi Piþ1 j þ jPiþ1 Pi1 j þ jPi1 Pi j
3hi
3r20
1=2
where jTi j is the area of the triangle Ti and sin hi P ð1 l2 Þ
immediately leads to
hi =qi 6 3r20 =ð1 l2 Þ1=2 :
since j cos hi j 6 l. The above inequality
ð3:2Þ
Let qK ¼ 2qi0 . Using (3.2), we get
qK 6 ðhK =hi0 Þðhi0 =2qi0 Þ 6 r0
hK =
3r20 =2
1=2
ð1 l2 Þ
¼: r
that is just the G–R condition.
We remain to prove that C–R and A–R are equivalent. In view of Lemma 2.1, we only need to show the
equivalence of C–R and (2.6). Firstly we show that C–R implies (2.6).
JK P min JK ðn; gÞ ¼ min jTi j=2 P ð1=4Þð1 l2 Þ
ðn;gÞ2K^
16i64
1=2 2
hK =r20 ;
ð3:3Þ
which is (2.6).
To deduce C–R from (2.6), without loss of generality, let T1 include the shortest edge hK , then (2.6)
implies
jT1 j P 2 min JK ðn; gÞ P 2C h2K :
ðn;gÞ2K^
ð3:4Þ
Noticing that jT1 j 6 hK hK =2, we obtain (C–R)1 with r0 ¼ ð4C Þ1 . Since the area of any triangle Ti can be
expressed as jTi j 6 1=2h2K sin hi , so repeating the above procedure and using jTi j 6 1=2h2K sin hi , we get
sin hi P 4C , which implies (C–R)2 with l ¼ ð1 16C2 Þ1=2 . We complete the proof. 4. Degenerate mesh condition
In what follows, we discuss some degenerate mesh conditions. As a preparation, we introduce the
ð1 þ aÞ-section condition.
Definition 4.1. ð1 þ aÞ-section condition ð0 6 a 6 1Þ
dK ¼ OðhK1þa Þ;
uniformly for all elements K as h ! 0.
If dK ¼ 0, K degenerates into a parallelogram. In case a > 0, we recover the condition A in [21]. In case a
equals 1, we obtain condition B, or the Bi-Section Condition [21].
The Angle Condition is another kind of degenerate mesh condition, which is introduced in [19] and used
to measure the deviation of a quadrilateral from a parallelogram. Define rK as
rK :¼ maxðjp a1 ; jp a2 jÞ:
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P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
Here a1 is the angle between the outward normal of two opposite sides of K and a2 is the angle between the
outward normal of other two sides. We call a mesh satisfying the Angle Condition if rK ¼ OðhK Þ, i.e., if
rK =hK is uniformly bounded for all elements. It is seen that 0 6 rK < p, and rK ¼ 0 iff K is a parallelogram.
Assuming that C–R holds, Chou et al. [9, Theorem 3.2] proved that the Angle Condition and the BiSection Condition are equivalent when h is sufficiently small. We will show later on that the assumption C–
R can be relaxed.
Remark 4.2. Observing that the h2 -parallelogram mesh condition in [15] is actually equivalent to the BiSection Condition.
Motivated by the Bi-Section Condition, we define a kind of mesh which will be shown to be quite useful
for the convergence analysis of finite elements.
Definition 4.3. We call Th an asymptotically regular parallelogram mesh if it satisfies C–R as well as the BiSection Condition.
Notice that any polygon can easily be meshed by asymptotically regular parallelograms with a mesh size
tending to zero. Indeed, if we begin with any mesh of convex quadrilaterals and refine it by dividing each
quadrilateral into four by connecting two midpoints of opposite edges, then the resulting mesh is an asymptotically regular parallelogram as that in Fig. 2 below.
There is also another kind of method for generating such an asymptotically regular parallelogram mesh.
For example, the C2 -grid in [3] which results from a mapping of uniform grids is an asymptotically regular
parallelogram mesh. This approach has already appeared in [25], see also [10, Remark 2.3].
As to the generation of a shape regular mesh satisfying the ð1 þ aÞ-section condition, we refer to [8] for
the work of WhitemanÕs school.
Degenerate mesh conditions
1. Jamet condition (J ) [17,18]. Th is regular if there exists a constant r > 0 such that
hK =qK 6 r:
2. Acosta–Duran RDP [1]. Th is regular with constant N 2 R and 0 < w < p, or shortly RDPðN ; wÞ, if we
can divide K into two triangles along one of its diagonals, which will always be called d1 , the other is d2 in
such a way that jd2 j=jd1 j 6 N and both triangles satisfy the maximum Angle Condition, i.e., each interior
angle of these two triangles is bounded from above by w.
3. S€
uli condition (S) [22]. Th is regular if it satisfies the J condition and the Bi-Section Condition simultaneously.
Fig. 2. (a) A partition of a square into four trapezoids, (b–d) meshes composed of the translated dilates of such partition.
P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
5677
Remark 4.4. The J condition is equivalent to (A–R)1 . To see this, invoking Lemma 2.1, we only need to
show that the J condition is equivalent to (2.7). Given the J condition, we obtain
J0 ¼ jKj=4 P ðp=16Þq2K P ðp=16Þh2K =r2 ;
ð4:1Þ
which gives (2.7). Using [9, Theorem 2.3], we deduce the J condition from (2.7).
Remark 4.5. S€
uliÕs condition firstly appeared in the convergence proof of a kind of cell vertex finite volume
method for hyperbolic problems and recently there is a resurgence of this condition in the mixed finite
volume method [10].
As addressed in [18], C–R implies the J condition with r ¼ ð1 l2 Þ1=2 =r0 . On the contrary, the J
condition allows for the degeneration of a quadrilateral K into a triangle since the ratio h=h can be arbitrarily large and the largest angle of K can be equal to p, i.e., the J condition may violate either (C–R)1 or
(C–R)2 . This is shown in Fig. 3 below.
pffiffiffi
Both elements in Fig. 3 satisfy thep
J ffifficondition
since hK =qK 6 2 þ 1. However, as to the left element, the
ffi
2 1=2
ratio hK =hK ¼ ða2 þ ða bÞ Þ =b 6 2a=b blows up as b tending to zero which obviously violates (C–R)1 .
As to the right element, the interior angle \ADC ¼ p x, which approaches p as x tending to zero, thereby
violates (C–R)2 . However, not the whole (C–R)2 is violated since the J condition excludes the interior angle
from becoming too small. This fact is hidden in [22, Lemma 1] which is stated as follows.
Lemma 4.6. If Th satisfies the S condition, then for sufficiently small h, Th is C–R shape regular.
Proceeding along the same line of the above lemma, we obtain
Corollary 4.7. If Th satisfies the J condition as well as the ð1 þ aÞ-section condition, then for sufficiently small
h Th is C–R shape regular.
Remark 4.8. Schmidt [20] replaced the Bi-Section Condition in Lemma 4.6 by the arbitrary smallness of the
ratio dK =hK .
Remark 4.9. Lemma 2.1 indicates that Zl
amalÕs 1-strongly regular condition is equivalent to C–R plus the
Bi-Section Condition, so it is also equivalent to the S condition. Notice that the 1-strongly regular condition
has already appeared in [14] but in a slightly different form.
Fig. 3. (a) Element satisfies the J condition but violates (C–R)1 , (b) element satisfies the J condition but violates (C–R)2 .
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P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
Notice that C–R does not imply the S condition. Indeed, considering the trapezoid mesh, it is seen that
the S condition is stronger than C–R. In fact, it is a strong shape regular mesh condition instead of a
degenerate one. Invoking [1, Remark 2.7], RDPðN ; wÞ is weaker than both the J condition and C–R,
therefore it is weaker than the S condition. Moreover, RDPðN ; wÞ together with the Bi-Section Condition
does not imply C–R. It is due to the fact that a rectangular element which satisfies both RDPð1; p=2Þ and
the Bi-Section Condition may still have its anisotropic aspect ratio arbitrarily large.
Remark 4.10. Braess [7, p. 99, Remark 2] proposed a new shape regular mesh condition which consists of
(C–R)1 , the J condition and the maximum interior Angle Condition. Obviously, it is equivalent to C–R.
However, due to its superfluous complexity this new condition is not advisable.
In what follows, we will give a new proof for S€
uliÕs result (cf. Lemma 4.6). Our proof also covers
Corollary 4.7 and Remark 4.8.
Theorem 4.11. If Th satisfies the J condition, and if there exists a constant h ð0 6 h < 1Þ such that
maxK2Th dK =hK 6 ðp=8Þh=r2 , then Th is C–R shape regular.
Proof. In view of Lemma 2.1, we only need to show that there exists a constant C such that
JK ðn; gÞ > C h2K :
Notice that
jJK J0 j ¼ jJ1 n þ J2 gj 6 dK hK =2 6 ðp=16Þhh2K =r2 ;
where we have used dK =hK 6 ðp=8Þh=r2 with 0 6 h < 1. The above inequality together with (4.1) yield
JK ðn; gÞ P J0 jJK J0 j P ð1 hÞðp=16Þh2K =r2 ¼: C h2K ;
which completes the proof.
The above proof is simpler than that in [22, Lemma 1] which heavily depends on some elementary
geometry arguments. Proceeding along the same line of the above proof, we obtain
Corollary 4.12. If Th satisfies the J condition, and if there exists a constant h ð0 6 h < 1Þ such that
maxK2Th rK 6 ðhp=4Þ=r2 , then Th is C–R shape regular.
Proof. Notice that jJ1 j and jJ2 j equal the area of the triangle hQ; M; N i and hP ; M; N i respectively, which can
further be expressed as
jJ1 j ¼ 12jQMjjQN j sin b1 ;
jJ2 j ¼ 12jPMjjPN j sin b2 :
Here b1 :¼ jp a1 j and b2 :¼ jp a2 j. Using the basic inequality sin x 6 x; 8x P 0, we bound J1 and J2 as
jJ1 j; jJ2 j 6 h2K rK =8:
The above inequality and (4.1) yield
JK P J0 h2K rK =4 P ð1 hÞðp=16Þh2K =r2 ;
provided that maxK2Th rK 6 ðhp=4Þ=r2 . The proof is completed.
As an application of the above results, we give a new proof for ChouÕs result, i.e., [9, Theorem 3.2], which
reads that the Bi-Section Condition and the Angle Condition are equivalent if the Th is C–R regular with h
P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
5679
tending to zero, this result was firstly stated in [4] without a proof. In what follows we state that for such
equivalence the C–R condition can be relaxed to the J condition which is also necessary.
Theorem 4.13. If Th satisfies the J condition, and if there exists a constant h ð0 6 h < 1Þ such that
maxK2Th dK =hK 6 ðph=8Þ=r2 and maxK2Th rK 6 ðph=4Þ=r2 , then dK =hK is of the same order with rK , thus the
Bi-Section Condition and the Angle Condition are equivalent. The J condition is even necessary for such
equivalence.
Proof. In the triangle hQ; M; N i, let b1 ¼ \MQN and b2 ¼ \MPN . It is seen that either \QMN or \QNM is
less than ðp b1 Þ=2. Without loss of generality, we assume that \QMN 6 ðp b1 Þ=2, therefore
dK ¼ jMN j ¼ jQN j sin b1 = sin \QMN
2 b1
h :
p 2 K
Here we have used the basic inequality sin x P 2x=p; 8x 2 ½0; p=2.
Noting that maxK2Th dK =hK 6 ðhp=8Þ=r2 , in view of Theorem 4.11, we obtain (2.7) with C :¼
ð1 hÞðp=16Þ=r2 . Invoking the last paragraph of Theorem 3.3, it follows that
P 2jQN j sin b1 =2 ¼ jADj sin b1 =2 P
hK P 4C hK :
Thus b1 6 ðp=4C ÞdK =hK . Similarly, b2 6 ðp=4C ÞdK =hK . The above two inequalities yield
0 6 rK < ð4r2 =ð1 hÞÞdK =hK :
ð4:2Þ
On the other hand, we extend NM and suppose that it intersects PQ with R. Notice that
jhP ; M; N ij þ jhQ; M; N ij h2K ðsin b1 þ sin b2 Þ=8 h2K rK =4
6
6
¼ rK =ð8C Þ;
jhP ; Q; N ij
jT1 j=4
C h2K =2
ð4:3Þ
where we have used
jT1 j P 2 min JK P 2C h2K ;
ðn;gÞ2K^
which is a direct consequence of Corollary 4.12.
The left side of (4.3) can be bounded from below as
jhP ; M; N ij þ jhQ; M; N ij
P dK =jNRj P 2dK =hK ;
jhP ; Q; N ij
ð4:4Þ
where we have used that jNRj 6 maxðjNP j; jNQjÞ 6 hK =2. A combination of (4.3) and (4.4) gives
rK P ðð1 hÞp=r2 ÞdK =hK :
ð4:5Þ
(4.2) and (4.5) lead to the bilateral estimate for rK as
ðð1 hÞp=r2 ÞdK =hK 6 rK 6 ð4r2 =ð1 hÞÞdK =hK ;
from which we have the equivalence of the Bi-Section Condition and the Angle Condition.
To show that the J condition is necessary, we consider the following two
counterexamples.
pffiffiffiffiffi
Consider the element K in the left of Fig. 4, the diameter of K is hK ¼ 10h=2. It is easily seen that K
satisfies the Bi-Section Condition with dK ¼ h2 =2. However, the Angle Condition is violated since rK ¼ p=4.
Notice that K does not satisfy the J condition since hK =qK ¼ 1 þ 1=h.
On the other hand, consider the element K in the right of Fig. 4, the diameter of K is hK ¼ 3h. A simple
manipulation shows that
h=ðð1 þ h2 Þð4 þ h2 ÞÞ
1=2
6 rK 6 ph=4;
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P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
Fig. 4. (a) A quadrilateral satisfies the Bi-Section Condition while violates the Angle Condition and the J condition, (b) a quadrilateral
satisfies the Angle Condition while violates the Bi-Section Condition and the J condition.
so the Angle Condition is satisfied. However, the Bi-Section Condition is violated since dK ¼ h=2. Notice
that
2 1=2
3=2ð1=4 þ ð1=hÞ Þ
2 1=2
6 hK =qK 6 3=2ð1 þ ð1=hÞ Þ
which indicates that the J condition is also violated.
;
Remark 4.14. Notice that the J condition in Theorem 4.13 cannot be further weakened to RDPðN ; wÞ.
Indeed, the element of Fig. 4a satisfies RDPð2; 3p=4Þ as well as the Bi-Section Condition, however, it does
not satisfy the Angle Condition.
As an application of the above mesh conditions, we consider the interpolation of the 4-node isoparametric element. Denoting Q the standard Lagrangian interpolate operator for the 4-node isoparametric
element, we look for a geometric condition under which the estimate
ku Quk0;K þ hK ju Quj1;K 6 Ch2K juj2;K
holds uniformly for K 2 Th .
In view of [1, Theorem 4.7], the optimal interpolation error estimate with respect to the L2 norm holds
with a constant independent of the geometry of K, so we only consider
ju Quj1;K 6 ChK juj2;K :
ð4:6Þ
Many authors derived the above estimate under various mesh conditions (see [1] for a review). With
RDPðN ; wÞ which seems the weakest mesh condition up to now, Acosta and Duran [1] obtained (4.6) (see
[1, Remark 2.4–Remark 2.7]). One may ask whether RDPðN ; wÞ is also necessary. Acosta and Duran put it
as an open problem in [1]. The following example shows that this condition is indeed necessary.
Counterexample
Consider the element K in Fig. 5. We assume that one diagonal BD is far longer than the other one AC,
i.e., 2h1 ¼ jBDj > jACj ¼ 2h2 . If we decompose K by the diagonal AC, then the triangles hA; B; Ci and
hA; C; Di indeed satisfy the maximal Angle Condition since all interior angles in these two triangles are less
than p=2. However, jBDj=jACj ¼ h1 =h2 which cannot be bounded by any constant as h2 =h1 tending to zero.
If we decompose K by the diagonal BD, a simple computation leads to sin \DAB ¼ 2h1 h2 =ðh21 þ h22 Þ, so
\DAB approaches p as h2 =h1 tending to zero, thus it also violates RDPðN ; wÞ. Consequently, this element
does not satisfy RDPðN ; wÞ for any N and w which are independent of the ratio h1 =h2 .
2
2
Let uðx; yÞ ¼ x2 , then juj2;K ¼ 8h1 h2 . A direct manipulation shows that koðu QuÞ=oyk0;K ¼ h51 =ð3h2 Þ, we
have
ju Quj21;K =juj22;K P koðu QuÞ=oyk20;K =juj22;K ¼ ð1=96Þðh1 =h2 Þ2 ð2h1 Þ2 :
Since the diameter of K is 2h1 , so (4.6) does not hold while h2 =h1 tending to zero.
ð4:7Þ
P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
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Fig. 5. A rhombus does not satisfy the RDPðN ; wÞ.
To sum up, we have the following interpolation result for the 4-node isoparametric element.
Theorem 4.15. For any u 2 H 2 ðKÞ, if the condition RDPðN ; wÞ holds, then there exists a constant C ¼
CðN ; wÞ such that
ju Quj1;K 6 ChK juj2;K :
ð4:8Þ
Moreover, RDPðN ; wÞ is also necessary for the validity of (4.8).
By CeaÕs Lemma, the error bound of the 4-node isoparametric element solution is governed by its interpolation error, i.e.,
ku uh k1 6 C inf ku vk1 :
v2Vh
ð4:9Þ
Here
Vh :¼ fv 2 H01 j vjK 2 Q1 ðKÞ; 8K 2 Th g:
However, the experience with triangle elements indicates that the interpolation error actually says nothing
about the approximation error of the finite element. Notice that the above counterexample shows that the
interpolation error diverges when the common 4-node isoparametric element is employed in the right-hand
side of (4.9), while K violates the RDPðN ; wÞ. However for this element K and uðx; yÞ ¼ x2 , if we define
another interpolant Pu as: PuðBÞ ¼ uðBÞ; PuðDÞ ¼ uðDÞ and PuðAÞ ¼ PuðCÞ ¼ ðuðBÞ þ uðDÞÞ=2, then a
simple manipulation shows that
pffiffiffi
ju Puj1;K =juj2;K ¼ 1=ð2 6Þh1 :
ð4:10Þ
Therefore, (4.7) and (4.10) substantiate the above observation for the 4-node isoparametric element.
5. Conclusions
In this paper, some commonly used shape regular mesh conditions are proven to be equivalent and their
connections to some degenerate mesh conditions are also clarified.
Added in the proof. After we completed this paper, we learned that a part of our results is proven by
Chou et al. [9]. i.e., the equivalence of C–R and G–R in Theorem 3.3.
Acknowledgements
This work was initiated when the first author was visiting the Faculty of Mathematics at the RuhrUniversity Bochum on a grant by the Alexander von Humboldt Foundation. He would like to thank AvH
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P. Ming, Z.-C. Shi / Comput. Methods Appl. Mech. Engrg. 191 (2002) 5671–5682
for the support. The authors would like to thank Professor F. Kikuchi for providing us with [17] and some
suggestive discussion.
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