some optimal error estimates of biharmonic problem using conforming finite element

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Applied Mathematics and Computation 194 (2007) 298–308

www.elsevier.com/locate/amc

Some optimal error estimates of biharmonic problemusing conforming finite element q

Yuan Li, Rong An *, Kaitai Li

Department of Mathematics, Xi’an Jiaotong University, Xi’an 710049, China

Abstract

In this paper, the W 2;p-error and Lp-error estimates of conforming finite element methods for biharmonic problem areestablished via weight function technique, where 2 6 p 6 +1. Finally, the numerical experiments are provided.� 2007 Elsevier Inc. All rights reserved.

Keywords: Optimal error estimates; Biharmonic problem; Finite element approximation

1. Introduction

About 30 yeas ago, many scholars began to investigate the max–norm error estimates of second orderPoisson equation

0096-3

doi:10

q Sup* Co

E-m

�Du ¼ f in X � R2

with Dirichlet boundary

u ¼ 0 on oX

when X was partitioned by regular mesh and irregular mesh, such as [1–11] and references therein. First,Nitsche in [1] gave

ku� uhk0;1 6 ch2j ln hj32juj2;1 ð1Þ

and

ku� uhk1;1 6 chj ln hkuj2;1 ð2Þ

by the method of weight function for linear finite element. But it is obvious that (1) and (2) are not of optimalorder. Subsequently, Nitsche in [2] improved the estimate (1) and showed

003/$ - see front matter � 2007 Elsevier Inc. All rights reserved.

.1016/j.amc.2007.04.040

ported by the Nature Science Foundation of China (Grant No. 50306019, No. 10571142, No. 10471110 and No. 10471109).rresponding author.ail address: [email protected] (R. An).

mailto:[email protected]

Y. Li et al. / Applied Mathematics and Computation 194 (2007) 298–308 299

ku� uhk0;1 6 ch2j ln hkuj2;1: ð3Þ

Fried in [3] gave an example based on radial symmetry which implied that the estimate (3) maybe was of opti-mal order. Hence, it is difficulty to remove the factor jln hj in (3). However, the logarithmic factor in (2) can beremoved. Rannacher and Scott in [4] improved the estimate (2) and obtained some optimal error estimates

ku� uhk1;p 6 chkuk2;p 2 6 p 6 þ1;ku� uhk0;q 6 ch2kuk2;q 2 6 q < þ1:

(ð4Þ

In particular, for high order finite element, the factor jln hj in (1) and (3) disappeared and optimal max–normerror estimates were established

ku� uhk0;1 6 ch2kuk2;1 ð5Þ

and

ku� uhk1;1 6 chkuk2;1: ð6Þ

See [1] for more details. Moreover, if the domain X was partitioned by irregular mesh, Scott in [5] showed that(5) was also true.

Similar results have been obtained for second order nonlinear problems. Frehse and Rannacher in [6]obtained

ku� uhk0;1 6 ch2j ln hkuj2;1
for linear finite element approximation and in [7]
ku� uhk0;1 6 chm

for high order m P 3 finite element approximation. If X � RN ;N P 3, they in [6] proved

ku� uhk0;1 6 ch2j ln hjN4þ1juj2;1:

We refer the reader to [8–12] for other similar results.However, to our best knowledge, it is vacant for max–norm error estimate of high order problem. There-

fore, in this paper, we deal with the biharmonic problem

D2u ¼ f ðxÞ in X � R2

with Dirichlet boundary

u ¼ ouon¼ 0 on oX:

We will extend the method in [4] to fourth order problem and show some optimal estimates

ku� uhk2;p 6 ch2kuk4;p; 2 6 p 6 þ1;ku� uhk0;q 6 ch4kuk4;q; 2 6 q < þ1

(ð7Þ

by conforming finite element approximation. If q = +1, we will show

ku� uhk0;1 6 ch3juj4;1; ð8Þ

which is not optimal. But the proof in regular triangular partition is much complicated than in quadrilateralpartition. In particular, if p ¼ q ¼ 2, it is well-known that the estimate (7) is true (cf. [13]).

This paper is organized as follows: in Section 2, we give some preliminary knowledge and define finite ele-ment subspace when X is partitioned by regular quadrilaterals; in Section 3, a stability theorem is established;in Section 4, we will prove the error estimates (7) and (8); in Section 5, the case when X is decomposed intoregular triangles is discussed; in Section 6, the numerical experiments are provided.

Throughout this paper, the symbol c always denotes a positive constant and is independent of h. Moreover,it maybe is different even in the same formulation.

300 Y. Li et al. / Applied Mathematics and Computation 194 (2007) 298–308

2. Preliminary

Let W m;pðXÞ be the Sobolev space with norm kukm;p. Let C10 ðXÞ be the space of infinite times continuouslydifferentiable functions with compact support in X. W m;p

0 ðXÞ denotes the closure of C10 ðXÞ in W m;pðXÞ. Inparticular, if p = 2, we write H mðXÞ ¼ W m;pðXÞ and H m

0 ðXÞ ¼ W m;p0 ðXÞ.

In this paper, we will study the standard biharmonic problem

D2u ¼ f ðxÞ in X;

u ¼ ouon ¼ 0 on oX;

(ð9Þ

where u : X! R is unknown, X � R2 is a bounded and convex polygonal domain with Lipschitz boundary oX,n denotes the unite outer normal vector to oX.

Introduce the linear space

V ¼ H 20ðXÞ ¼ v 2 H 2ðXÞ; v ¼ ov

on¼ 0 on X

� �
with the norm
kvkV ¼ aðv; vÞ12 8v 2 V ;

where

aðu; vÞ ¼Z

XDuDv dx:

Given f 2 L2(X), the weak variational formulation associated with (9) is

Find u 2 V such that

aðu; vÞ ¼R

X fvdx 8v 2 V :

�ð10Þ

By the regularity theory of biharmonic problem (cf. [14]), there admits a unique solution u 2 V \ H 4ðXÞ.Let Th be a family of regular quadrilateral partition of X into quadrilaterals of diameter not great than h.

We define finite dimensional space

W h ¼ fu 2 C1ðXÞ; 8s 2 T h; u 2 Q3ðsÞg;
where Q3(s) is the space of all polynomials in the reference space (x,y) of the form
Pcijxiyj, where the sum

ranges over all integers i and j such that 0 6 i, j 6 3.Let V h ¼ W h \ V be the conforming finite element subspace. Then finite element approximation solution

uh 2 Vh of u is defined by

aðuh; vhÞ ¼Z

Xfvh dx 8vh 2 V h: ð11Þ

Moreover, there exists a unique solution uh 2 Vh.Define a Rietz projection operator P : V! Vh by

aðPu; vhÞ ¼ aðu; vhÞ 8vh 2 V h: ð12Þ
It is well-known that
kPukV 6 kukV :

By the theory of polynomial approximation, there exists two interpolant operators Ih : V! Vh andJh : V! Vh such that (cf. [15])

kv� IhvkV 6 ch2½v�4;2 8v 2 V \ H 4ðXÞ ð13Þ

and

kv� J hvkk;p 6 ch4�kkvk4;p 8v 2 V \ W 4;pðXÞ; k ¼ 0; 1; 2; 2 6 p 6 þ1; ð14Þ


where

½v�4;2 ¼Z

X

o4vox4

1

�� 2 þ o4vox4

2

�� 2 dx

!1=2

:

It is obvious that

½vh�4;2;s � 0 8vh 2 V h; s 2 T h: ð15Þ

3. Stability theorem

The proof of error estimate (7) bases on the following stability theorem:

Theorem 1. Given f 2 L2(X). If u 2 V \ W 2;pðXÞ is a unique solution of (10), then we have

kPuk2;p 6 ckuk2;p; ð16Þ

where 2 6 p < +1, c > 0 is independent of h.

Use some notations and parameters in [4]. For any z 2 X such that z is contained in some sz 2 Th, there exists adz 2 C10 ðszÞ such that
Z
sz

dz dx ¼ 1; jDkdzj 6 ch�2�k; k ¼ 0; 1; 2; . . . ; ð17Þ

where dz is the Dirac function. By construction,

D/hðzÞ ¼ ðdz;D/hÞ 8/h 2 V h: ð18Þ
Define gz 2 V such that
aðgz;/Þ ¼ ðdz;D/Þ 8/ 2 V ; ð19Þ
then
D2gz ¼ Ddz

holds in the sense of V 0 ¼ H�2ðXÞ and following estimate holds:

kgzkV 6 kdzkL2 :

Introduce the weight function

rzðxÞ ¼ ðjx� zj2 þ j2h2Þ12; j P 1;

where the parameter j will be chosen sufficiently large and independent of h. From now on, we drop the sub-script z. The weight function r satisfies (cf. [4])

jDkrj 6 cr1�k6 cðjhÞ1�k

; k ¼ 0; 1; 2; . . . :

Moreover, for sufficiently large j,

maxs2T h

maxx2s

rðxÞ=minx2s

rðxÞ� �

6 c

holds uniformly for z 2 X. Then from (13), we have
ZX
rbjDðv� IhvÞj2 dx 6 ch4

Z 0rb j o

4vox4

1

j2 þ o4vox4

2

�� 2 !

dx ð20Þ

for v 2 V \ ðQ

s2T hH 4ðsÞÞ, where
Z 0
� � � dx ¼Xs2T h

Zs� � � dx:


Lemma 1. Let u 2 H10ðXÞ, then for each sufficiently large j, we have
Z
Xrb�2ju2jdx 6 c1

ZX

rbjDuj2 dx; ð21Þ

where c1 > 0 is some constant and b 2 R.

Proof. According to divergence formulation, one has

2

ZX

rb�2jujDijujxi dx ¼Z

Xrb�2Dijuj2xi dx ¼ �

ZXjuj2Dir

b�2xi dx� 2

ZXju2jrb�2 dx

¼ �ðb� 2ÞZ

Xrb�4jxj2juj2 dx� 2

ZXju2jrb�2 dx:

Thus
ZX
rb�2juj2 dx 6Z

Xrb�2jukDukxjdxþ c

ZX

rb�4jxj2juj2 dx:

For sufficiently large j, one has
ZX
rb�2juj2 dx 6 c1

ZX

rb�2jukDukxjdx 6 c1

ZX

rb�1jukDujdx 6 c1

ZX

rb�2juj2 dx� �1=2 Z

XrbjDuj2 dx

� �1=2

;

which gives (21). h

Lemma 2. For any given F 2 H 1ðXÞ, let v 2 V satisfy

D2v ¼ F in X;

v ¼ ovon ¼ 0 on oX;

(

then we have
ZX
r�4�ajD2vj2 dx 6 cðjhÞ�6

ZX

r4�ajDF j2 dx; ð22Þ

where c > 0 is independent of j and h and 0 < a < 1.

Proof. We argue as in [4]. Using Holder inequality and Sobolev inequality, we have
ZX
r�4�ajD2vj2 dx 6Z

Xr�ð4þaÞ=a dx

� �a

kD2vk22=1�a 6 cðjhÞa�4kDD2vk2

2=2�a:

Since

kDD2vk2

2�a2=2�a 6

ZX

r4�ajDD2vj2 dx� � 1

2�aZ

Xr�

4�a1�a dx

� �1�a2�a

:

Then

kDD2vk22=2�a 6 cðjhÞ�2�a

ZX

r4�ajDF j2 dx� �

;

which gives (22). h

Proof of Theorem 1. Let a 2 (0,1) and 2 6 p < +1. Setting / = Pu gives

DPuðzÞ ¼ aðg; PuÞ ¼ aðPg; PuÞ ¼ aðPg; uÞ ¼ ðDu; dÞ � aðu; g � PgÞ: ð23Þ


We estimate the second term of right hand side of (23). Using Holder inequality, one has

jðDu;Dðg � PgÞÞj 6Z

Xr�4�ajDujp dx

� �1pZ

Xr�4�a dx

� �p�22pZ

Xr4þajDðg � PgÞj2 dx

� �12

6 ch�ðp�2Þð2þaÞ

2p Mh

ZX

r�4�ajDujp dx� �1

p

;

where

Mh ¼Z


� �12

:

The first term of right hand side of (23) can be estimated by

ðDu; dÞ 6Z

sz

jDujp dx� �1

pZ

sz

jdjp

p�1

� �p�1p

6 ch�2p

Zsz

jDujp dx� �1

p

:

Hence

kDPukpp 6 ch�2

Z Zsz

jDujp dxdzþ cðh�2�aÞp�2

2 Mph

Z Zsz

r�4�ajDujp dxdz� �

6 ckDukpp þ h�

ð2þaÞp2 Mp

hkDukpp:

That is

kDPukp 6 c 1þ h�2�a

2 Mh

kDukp 2 6 p < þ1:

To obtain (16), it is enough that

maxz2X

ZX

r4þajDðg � PgÞj2 dx 6 ch2þa; ð24Þ

where c is independent of z and h. Denote w ¼ r4þaðg � PgÞ, then
ZX
r4þajDðg � PgÞj2 dx ¼Z

XDðg � PgÞDðw� IhwÞdx�

ZXðg � PgÞDðg � PgÞDr4þa dx

� 2

ZXrr4þa � rðg � PgÞDðg � PgÞdx:

Using Holder inequality, Young’s inequality, (15) and (20), one has
ZX
r4þajDðg � PgÞj2 dx 6Z

Xr�4�aDðw� IhwÞdxþ c

ZX

rajg � Pgj2 dx

6 ch4

Z 0r�4�a o

4wox4

1

�� 2 þ o4w

ox42

�� 2 !

dxþ cZ

Xrajg � Pgj2 dx:

Since
Z 0 o4wox4
1

�� 2 þ o4wox4

2

�� 2 dx 6Z 0jD2wj2 dx;

in view of Lemma 1, after a calculation, we obtain
ZX
r4þajDðg � PgÞj2 dx 6 ch4

Z 0r4þajD2gj2 dxþ c

ZX

rajg � Pgj2 dx

þ cj�4

ZX

r4þajDðg � PgÞj2 dxþZ

Xrajg � Pgj2 dx

� �:


Then for sufficiently large j, we have
ZX

Z 0r4þajD2gj2 dxþ c

ZX

rajg � Pgj2 dx

6 ch4

Zsz

r4þajDdj2 dxþ cZ

Xrajg � Pgj2 dx 6 ch2þa þ c

ZX

rajg � Pgj2 dx: ð25Þ

Consider the auxiliary problem

D2v ¼ raðg � PgÞ in X;

v ¼ ovon ¼ 0 on oX:

(ð26Þ

Then in view of Lemmas 1 and 2, we have
ZX
r�4�ajD2vj2 dx 6 cðjhÞ�6

ZX

r4�ajDðraðg � PgÞÞj2 dx 6 cðjhÞ�6

ZX

r2þajg � Pgj2 þ r4þajDðg � PgÞj2 dx

6 cðjhÞ�4

ZX

r4þajDðg � PgÞj2 þ rajg � Pgj2 dx:

Multiplying (26) by g � Pg and integrating over X, we obtain
ZX
rajg � Pgj2 dx ¼Z

XDðv� IhvÞDðg � PgÞdx 6

ZX

r4þajDðg � PgÞj2 dx� �1

2Z

Xr�4�ajDðv� IhvÞj2 dx

� �12

6 j�2

ZX

r4þajDðg � PgÞj2 dxþ cj2h4

ZX

r�4�ajD2vj2 dx

6 j�2

ZX

r4þajDðg � PgÞj2 dxþ cj�2

ZX

r4þajDðg � PgÞj2 þ rajg � Pgj2 dx� �

:

Then for sufficiently large j, we have
ZX
rajg � Pgj2 dx 6 cj�2

ZX

r4þajDðg � PgÞj2 dx:

Substituting above into (25), we obtain, for sufficiently large j,
ZX
r4þajDðg � PgÞj2 dx 6 ch2þa;

which gives (24). Hence we complete the proof of Theorem 1. h

4. Error estimates

By stability Theorem 1, we obtain some optimal error estimates.

Theorem 2. If u 2 V \ W 4;pðXÞ and uh 2 Vh satisfy (10) and (11), respectively, then the following error estimate

holds:

ku� uhk2;p 6 ch2kuk4;p;

ku� uhkp 6 ch4kuk4;p;

(
where c > 0 is independent of h and 2 6 p < +1.
Proof. Since the proof of first estimate is similar with Corollary 7.1.12 in [15] and another is similar with Cor-ollary in [4], so we omit it. h

Theorem 3. If u 2 V \ W 4;1ðXÞ and uh 2 Vh satisfy (10) and (11), respectively, then we have following stability

estimate:


kPuk2;1 6 ckuk2;1

and error estimate

ku� uhk2;1 6 ch2kuk4;1;

where c > 0 is independent of h.

Proof. Similar with the argument of Theorem 1, we estimate the right hand side of (23). In fact, if p = +1, wehave

jðDu;Dðg � PgÞÞj 6 kDuk1Z

XjDðg � PgÞjdx

6 kDuk1Z


� �12Z

Xr�4�a dx

� �12

6 ch�2þa

2 MhkDuk1

and

jðDu; dÞj 6 kDuk1Z

Xjdjdx 6 ckDuk1:

Hence, by (23),

kDPuk1 6 c 1þ h�2�a

2 Mh

kDuk1:

The rest follows the proofs of Theorems 1 and 2. h

Next, we prove the estimate (8).

Lemma 3. For all vh 2 Vh, we have

kvhk0;1 6 ch�1kvhk0;2:

Proof. By construction, we have

vhðzÞ ¼ ðdz; vhÞ;
where the Dirac function dz is defined in Section 3, then we have
kvhk0;1 6

Zsz

jdzkvhjdx 6Z

sz

jdzj2 dx� �1

2Z

Xjvhj2 dx

� �12

6 ch�1kvhk0;2: �

Theorem 4. If u 2 V \ W 4;1ðXÞ and uh 2 Vh satisfy (10) and (11), respectively, then we have

ku� uhk0;1 6 ch3kuk4;1;

where c > 0 is independent of h.

Proof. By Lemma 3, we have

ku� uhk0;1 6 ku� J huk0;1 þ kJ hu� uhk0;1 6 ch4kuk4;1 þ ch�1kJ hu� uhk0;2

6 ch4kuk4;1 þ ch�1ðku� J huk0;2 þ ku� uhk0;2Þ 6 ch4kuk4;1 þ ch3kuk4;2 6 ch3kuk4;1: �

5. Triangular partition

If the domain X is decomposed into regular triangles, Theorems 1–4 also hold. Define finite dimensionalspace


bW h ¼ fu 2 C1ðXÞ; 8s 2 T h; u 2 P 5ðsÞg;

where P5(s) is the space of all polynomials defined on s of degree less than or equal to 5. Let bV h ¼ bW h \ V bethe finite element subspace. In this case, the projection operator P is from V to bV h.

We assume that the following inverse inequality holds:
Z 0rbjD2vhj2 dx 6 ch�4
ZX

rbjDvhj2 dx 8vh 2 bV h: ð27Þ

Lemma 4 [16, Theorem A.8]. Let X be a bounded domain of R2 with a Lipschitz continuous boundary. For each

integer k P 0 and real p > 0, there exists a positive constant C such that

kvkk;p 6 Cðkvkp þ ½v�k;pÞ 8v 2 W k;pðXÞ:
By Lemma 4, one has
ku� IhukV 6 ch2ðkuk2 þ ½u�4;2Þ 8u 2 V \ H 4ðXÞ ð28Þ

or

ku� IhukV 6 ch2ðkukV þ ½u�4;2Þ 8u 2 V \ H 4ðXÞ; ð29Þ

where c is independent of h. Eqs. (28) (or (29)) plays the same role as (13). Moreover, the inequality (20)

becomes
ZX

ZX

rbjvj2 dxþ ch4

Z 0rb o4v

ox41

�� 2 þ o4vox4

2

�� 2 !

dx ð30Þ

or
ZX

ZX

rbjDvj2 dxþ ch4

Z 0rb j o

4vox4

1

j2 þ j o4v

ox42

j2� �

dx ð31Þ

for v 2 V \ ðQ

s2T hH 4ðsÞÞ.

Theorem 5. Given f 2 L2ðXÞ. If u 2 V \ W 2;pðXÞ and Pu 2 bV h satisfy (10) and (12), respectively, then we have

kPuk2;p 6 ckuk2;p;

where 2 6 p6 +1, c > 0 is independent of h.

Proof. From the argument of Theorems 1 and 3, it is enough to show (24). Moreover, according to (30), onehas
Z
Xr4þajDðg � PgÞj2 dx 6

ZX

r�4�aDðw� IhwÞdxþ cZ

Xrajg � Pgj2 dx

6 ch4

Z 0r�4�a o4w

ox41

�� 2 þ o4wox4

2

�� 2 !

dxþ ch4

ZX

r�4�ajwj2 dxþ cZ

Xrajg � Pgj2 dx:

Since

½wh�4;2;s 6¼ 0 8wh 2 V h;

then in terms of Lemma 1, one has
ZX

Z 0r4þajD2gj2 dxþ ch4

Z 0r4þajD2Pgj2 dxþ c

ZX

rajg � Pgj2 dx

þ ch4

Zr4þajg � Pgj2 dxþ cj�4

Zr4þajDðg � PgÞj2 dxþ

Zrajg � Pgj2 dx

� �:

X X X


Then for sufficiently large j, we obtain

TableW 2;p-er

H

8163264

TableLp-erro

H

8163264

ZX


Z 0r4þajD2gj2 dxþ ch4

Z 0r4þajD2Pgj2 dxþ c

ZX

rajg � Pgj2 dx

6 ch4

Zsz

r4þajDdj2 dxþ cZ

Xr4þajDPgj2 dxþ c

ZX

rajg � Pgj2 dx

6 ch2þa þ cZ

sz

r4þajdj2 dxþ cZ

Xrajg � Pgj2 dx 6 ch2þa þ c

ZX

rajg � Pgj2 dx; ð32Þ

which is (25) in the proof of Theorem 1. Let v 2 V satisfy (26). According to (31), we also have
ZX
rajg � Pgj2 dx 6Z


� �12Z

Xr�4�ajDðv� IhvÞj2 dx

� �12

6 j�2

ZX

r4þajDðg � PgÞj2 dxþ cj2h4

ZX

r�4�ajD2vj2 dxþ cj2h4

ZX

r�4�ajDvj2 dx

6 j�2

ZX


ZX


ZX

rajg � Pgj2 dx:

Hence, for sufficiently large j, we have
ZX
rajg � Pgj2 dx 6 cj�2

ZX

r4þajDðg � PgÞj2 dx;

which together with (32) gives (24). Therefore, we complete the proof of Theorem 5. Finally, by Theorem 5,we obtain the error estimates (7) and (8). h

6. Numerical experiments

The numerical experiments in this section are designed to verity the error estimate obtained in Section 4.Here, we use Bogner–Fox–Schmidt element (Q3 Element).

Let X ¼ fðx; yÞ; 0 6 x 6 1; 0 6 y 6 1g and choose a true solution

u ¼ sinðpxÞ2 sinðpyÞ2:
Denote M and N be the division numbers at x-axis and y-axis, respectively. Following table displays somerelative errors in different mesh partitions, from which we can conclude the convergent order.
Tables 1 and 2 display that the convergence rates of W 2;p and Lp, p = 2,4,8, errors are almost O(h2) andO(h4), which are consistent with Theorem 2.

1ror estimates under different norms when M = N = H

ku�uhkVkukV

Order ku�uhk2;4

kuk2;4

Orderku�uhk2;8

kuk2;8Order

7.900604 · 10�2 n 7.354391 · 10�2 n 6.911686 · 10�2 n1.987399 · 10�2 1.99 2.012550 · 10�2 1.87 2.148402 · 10�2 1.694.975050 · 10�3 2.00 5.044575 · 10�3 2.00 5.402413 · 10�3 1.991.244164 · 10�3 2.00 1.261903 · 10�3 2.00 1.352011 · 10�3 2.00

2r estimates under different norms when M = N = H

ku�uhk2

kuk2Order ku�uhk4

kuk4Order ku�uhk8

kuk8Order

7.074492 · 10�3 n 6.848307 · 10�3 n 7.128996 · 10�3 n4.568223 · 10�4 3.95 4.837706 · 10�4 3.82 5.557572 · 10�4 3.682.869772 · 10�5 3.99 3.037603 · 10�5 3.99 3.491115 · 10�5 3.991.795645 · 10�6 4.00 1.900341 · 10�6 4.00 2.183457 · 10�6 4.00


References

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some optimal error estimates of biharmonic problem using conforming finite element

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