The local ultraconvergence for high-degree Galerkin finite element methods

J. Math. Anal. Appl. 446 (2017) 62–86

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

The local ultraconvergence for high-degree Galerkin finite element methods Wen-ming He a,∗ , Junzhi Cui b , Jiangman Shen a a

Department of Mathematics, Wenzhou University, Wenzhou, Zhejiang, 320035, PR China Institute of Computational Mathematics and Scientific/Engineering Computing, Academy of Mathematics and Systems Science, Chinese Academy of Sciences, Beijing 100190, PR China b

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 18 July 2015 Available online 20 August 2016 Submitted by R.G. Durán Keywords: Elliptic problem Ultraconvergence Derivative recovery operator Displacement Derivative

In this paper, we investigate the local ultraconvergence of k-degree (k ≥ 3) finite element methods for the second order elliptic boundary value problem with constant coefficients over a family of uniform rectangular/triangular meshes Th on a bounded rectangular domain D. The k-degree finite element estimates are developed for the Green’s function and its derivatives. They are employed to explore the relationship among dist(x, ∂D), dist(x, M ) and the ultraconvergence of k-degree finite element methods at vertex x, where M is the set of corners of D. Numerical examples are conducted to demonstrate our theoretical results. © 2016 Elsevier Inc. All rights reserved.

1. Introduction, preliminaries and statement of main idea In this paper, we are interested in the ultraconvergence estimation of finite element method for the following elliptic problem with constant coefficients Lu ≡ −

∂ ∂u (aij ) + γu = f in ∂yi ∂yj

D,

u = 0 on ∂D.

(1.1)

Here D ⊂ 2 is a bounded rectangular domain, γ ≥ 0 and there exists m > 0 satisfying aij ξi ξj ≥ mξi2 . Note that throughout the paper, the Einstein convention is used: the summation will be taken over all repeated indices. We first introduce some related work. The local symmetric technique (see [19,18,16,20]), the two weak estimates (see [2,3,8,7,29]), the interpolation post-processing technique (see [12,14,11,13,10, 15]), the superconvergent patch recovery (SPR) technique (see [22,24,26,28,25,23,31,32]) and the derivative recovery technique presented by Q. Zhu, L. Meng and J. Wei (see [30,21]) are several important tools for study of finite element superconvergence, which have been research focuses for decades. We first introduce the local symmetric technique (see [19,18,16,20]). Suppose that x is a local symmetric point; that is, there * Corresponding author. E-mail addresses: [email protected] (W.-m. He), [email protected] (J. Cui). http://dx.doi.org/10.1016/j.jmaa.2016.08.017 0022-247X/© 2016 Elsevier Inc. All rights reserved.

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exists a neighborhood B(x, d) = {y : |x − y| ≤ d} ⊂ n of x such that the mesh is locally symmetric with respect to x. By the symmetric technique, A.H. Schatz et al. investigated the superconvergence of the Lagrange finite element method for the second order elliptic equations at arbitrary locally symmetric point r+1 x ∈ D. Suppose that u − uh satisfies a(u − uh , χ) = 0 ∀χ ∈ S0h (B2d (x)). Here u ∈ W∞ (B2d ) and a(., .) is  ∂ψ ∂φ defined by a(ψ, φ) = D (aij ∂yi ∂yj + γψφ)dy. Assume that the coefficients of a(., .) are smooth and there exists a constant m > 0 such that, m|ζ|2 ≤ aij ζi ζj ∀ζ ∈ N . Assume also that Assumptions A0 –A5 (see [19], pp. 519–520) hold for B2d (x) where d ≥ ch, for c > 0 sufficiently large. A.H. Schatz, I.H. Sloan, and L.B. Wahlbin (see [19], pp. 508–509) observed that, if r is odd, under the assumption that for some s ≥ 0 and 1 ≤ q ≤ ∞, u − uh Wq−s (D) ≤ chr+t , with t > 0, and that d = hδ with δ = t/(s + n/q + 1) ≤ 1, there exists a constant c independent of x and h such that |(u − uh )(x)| ≤ chr+δ | ln h|,

(1.2) 

if r is even, under the assumption that for some s ≥ 0 and 1 ≤ q ≤ ∞, u − uh Wq−s (D) ≤ chr−1+t , with 

t > 0, and that d = hδ with δ  = t /(s + n/q + 2) ≤ 1, there exists a constant c independent of x, β and h such that    ∂u(x) ∂ u  ˆh (x, β)   ≤ chr−1+δ | ln h|,  ∂xi − ∂xi  where β = (β1 , . . . , βN ) is any vector of unit length and

∂u ˆh (x,β) ∂yi

(1.3)

= lim 12 ( ∂u s→0

h

(x+sβ) ∂xi

+

∂uh (x−sβ) ). ∂xi

For more than thirty years, an impressive amount of work has been done concerning the postprocessing techniques for the ultraconvergence of the finite element method for the problem (1.1). In this paper, we mainly introduce three types of postprocessing techniques. The first is the interpolation post-processing which is presented by Q. Lin et al. (see [12,14,11,13,10,15]). We define the interpolation of projection Ihk u ∈ S h (see [12,14,11,13,10]) for a polygon, which is different from the Lagrange interpolation. Define 4  2k 2k I2h on a macro-element τ = Tih as follows: for all v ∈ Qk (Tih ) (i = 1, 2, 3, 4), I2h v ∈ Q2k (τ ) satisfies i=1

2k I2h v(x) = v(x) whenever x is a node of some Tih , where Qk is bi-k-degree polynomial functional space. Similarly, for the k-degree finite element defined on a family of isosceles right triangular triangulations, we 4  2k define the interpolation of projection I2h on a macro-element τ = Tih as follows: for all v ∈ Pk (Tih ) (i = i=1

2k 2k v ∈ P2k (τ ) satisfies I2h v(x) = v(x) whenever x is a node of some Tih . Assume that u is the 1, 2, 3, 4), I2h solution for the problem (1.1) and x ∈ Ω satisfies dist(x, ∂D) ≥ c where dist(x, ∂D) denotes the distance between x and ∂D. Let ∂l denote the directional derivative operator in the direction of l. Lin and Zhou (see [12]) observed that, if x is a local symmetric point of a family of uniform triangulations, 2k h |∇(u − I2h u )(x)| ≤ chk+1 | ln h| u W∞ k+2 (D) ,

|u(x) − uh (x)| ≤ ch2+k | ln h| u W∞ 2+k (D) ,

k k

is odd, is even,

(1.4) (1.5)

and if x is a local symmetric point of a family of uniform rectangular meshes, then 2k h |∇(u − I2h u )(x)| ≤ chk+2 | ln h| u W∞ k+3 (D) ,

k ≥ 3 is odd,

(1.6)

2k h u (x)| ≤ chk+3 | ln h| u W∞ |u(x) − I2h k+3 (D) ,

k ≥ 4 is even.

(1.7)

Furthermore, they observed that if (1.1) is the Poisson equation and x is a local symmetric point of a family of uniform rectangular meshes, then

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2k h |∇(u − I2h u )(x)| ≤ chk+3 | ln h| u W∞ k+4 (D) ,

|u(x) − uh (x)| ≤ chk+4 | ln h| u W∞ k+4 (D) ,

k ≥ 3 is odd, k ≥ 4 is even.

(1.8) (1.9)

The second is the superconvergence patch recovery (SPR) technique proposed by Zienkiewicz and Zhu (see [31,32]). Suppose that E is an element patch about an interelement vertex x = (y, z), consisting of four rectangular elements sharing x. The SPR operators are constructed by least square fitting from the sampling  point set G consisting of all k-order Gauss points in Ti ∈ E, G = {(y, z) ∈ Ti : lk (y) = 0, lk (z) = 0}. i=1,2,3,4

Zhang et al. (see [22,27,26,28,25,23]) have combined the SPR technique and the local symmetric technique to investigate the ultraconvergence for rectangular elements. Assume that x is a local symmetric point and k is even. Zhang et al. (see [26], p. 383) have proved that, for k-degree element of the Poisson problem, |Gh uh (x) − ∇u(x)| ≤ chk+2 | ln h| u W∞ k+3 (D) .

(1.10)

Here Gh is the superconvergence patch recovery operator. But they have not got any ultraconvergent result for odd-degree rectangular elements. Based on the interpolation postprocessing technique, Q. Zhu, L. Meng and J. Wei (see [30,21]) have presented a different derivative recovery technique to investigate the ultraconvergent result for rectangular elements. Assume that k ≥ 3 is odd. They have proved that, for the Poisson problem,     Ri uh (x) − ∂u(x)  ≤ chk+3 | ln h| u k+4 , (1.11) W∞ (D)  ∂xi  where Ri is the derivative recovery operator in the xi direction. We shall note that (1.4)–(1.11) all assume that dist(x, ∂D) ≥ c. The purpose of this work is to relax this assumption. We will trace precisely how the ultraconvergence in the interior of a rectangular domain depends on the distance from corners and boundary. Let x be a vertex and M = {Z1 , Z2 , Z3 , Z4 } be the set of corners of D. Assume that d0 = dist(x, M ) and R = dist(x, ∂D). We have, if x is a local symmetric point of a family of uniform triangulations, then   2+k h h2k 2k h 1+k | ln h|2 u W∞ |∇(u − I2h u )(x)| ≤ c + + h (1.12) 2+k (D) , if k ≥ 3 is odd, d20 Rk and  |u(x) − u (x)| ≤ c h

 h2k h3+k 2+k | ln h|2 u W∞ + k−1 + h 2+k (D) , d20 R

if k ≥ 4 is even,

and if x is a local symmetric point of a family of uniform rectangular partitions, then   hk+4 h2k+1 |u(x) − uh (x)| ≤ c hk+3 + 2 + k−1 | ln h|2 u W∞ k+3 (D) , if k ≥ 4 is even, d0 R

(1.13)

(1.14)

and   hk+3 h2k+1 2k h | ln h|2 u W∞ |∇(u − I2h u )(x)| ≤ c hk+2 + 2 + k+3 (D) , if k ≥ 3 is odd, d0 Rk

(1.15)

and if x is a local symmetric point of a family of uniform rectangular partitions for the Poisson problem, then   hk+5 h2k+2 |u(x) − uh (x)| ≤ c hk+4 + 2 + k−1 | ln h|2 u W∞ (1.16) k+4 (D) , if k ≥ 4 is even, d0 R

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and     h4+k h2+2k 2k h ∇(u − I2h | ln h|2 u W∞ u )(x) ≤ c h3+k + 2 + 4+k (D) , if k ≥ 3 is odd. d0 Rk

(1.17)

One observes that (1.12)–(1.17) are substantial development for (1.4)–(1.9). We state that (1.12)–(1.17) is challenging to obtain for the reason that we need to develop the k-degree finite element estimates for the Green’s function and its derivative that holds singularity at corners. The rest of this section is organized as follows. Some notations and preliminaries will be introduced in Subsection 1.1. In Subsection 1.2, we will describe our main idea of this work. 1.1. Notations and preliminaries We consider the usual variational form of (1.1). Define the bilinear form  AK (ψ, φ) =

aij

∂ψ(y) ∂φ(y) + γψ(y)φ(y)dy ∂yj ∂yi

∀ψ, φ ∈ H 1 (K),

(1.18)

K

and the linear functional  FK (φ) =

f φdy, K

where K ⊂ R2 . Assume that H01 (K) = {v ∈ H 1 (K) : v = 0 on the boundary of K}. Then the weak problem of (1.1) is to find u ∈ H01 (D) satisfying AD (u, φ) = FD (φ),

∀φ ∈ H01 (D).

1 For all z ∈ K, the Green’s function GK z ∈ Wp (K) ( 1 < p < 2) is defined by

AK (GK z , ψ) = ψ(z),

∀ψ ∈ W 1p (K) ∩ H01 (K), p−1

and GK z (y) = 0,

∀y ∈ ∂K.

(1.19)

K The derivative ∂zi GK z of the Green’s function Gz is introduced by the problem

AK (∂zi GK z , ψ) =

∂ψ(z) ∂zi

∀ψ ∈ W 2p (K) ∩ H01 (K), p−1

and ∂zi GK z (y) = 0,

∀y ∈ ∂K.

(1.20)

D K K In particular, we write Gz and ∂zi Gz for GD z and ∂zi Gz for simplicity. Assume that Th and Nh are uniform conforming partitions of K with grid size h and the set of all vertices of ThK , respectively. We use standard Lagrange nodal FEMs. Define FE spaces

Sh (K) = {v ∈ C(K) : v|e ∈ Pk (K) ∀e ∈ ThK } for simplicial elements and FE spaces Sh (K) = {v ∈ C(K) : v|e ∈ Qk (K) ∀e ∈ ThK } for rectangular elements. Here Pk is the space of polynomials of total degree up to k, and Qk is the space k of polynomials of degree up to k in each variable. Let Ih,K be the degree k Lagrange interpolation operator K over Th . Let S0h (K) = Sh (K) ∩ H01 (K).

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Define the FE projections Rh,K : H 1 (K) → S h (K) by AK (ψ − Rh,K ψ, φ) = 0,

∀φ ∈ S0h (K),

k and Rh,K ψ(y) = Ih,K ψ(y),

∀y ∈ ∂K.

In particular, we denote Th = ThD , Sh = Sh (D), S0h = S0h (D) and A = AD . And for all ψ ∈ H01 (D), we h denote ψ h = Rh,D ψ and Ih ψ = Ih,D ψ. The discrete Green’s function Rh,K GK z (y) ∈ S0 (K) satisfies the problem A(Rh,K GK z , w) = w(z),

∀w ∈ S0h (K)

(1.21)

h and the discrete derivative of the Green’s function Rh,K ∂zi GK z (y) ∈ S0 (K) is defined by

A(Rh,K ∂zi GK z , w) =

∂w(z) , ∂zi

∀w ∈ S0h (K).

(1.22)

In this paper, we will use gx , Ghx and gxh to denote ∂xi Gx , Rh,D Gx and Rh,D ∂xi Gx , respectively. In this article, we will use the letter c (with or without subscript) to denote a generic constant, which may not be the same in each occurrence, we also use the standard notations for the Sobolev spaces and norms. 1.2. The main idea Assume that u is the solution for the problem (1.1). In this subsection, we will use (1.4) as an example to show how we improve (1.4)–(1.9) briefly. In the process of proving (1.4), we mainly use the following result, which will be proved in Appendix. s Lemma 1.1. Assume that s is a positive integer, Ω ⊃⊃ D and u ∈ W∞ (D). Then there exists a function s v(y) ∈ W∞ (Ω) satisfying

(i) (ii) (iii)

v(y) = u(y) v(y) = 0

∀y ∈ D,

(1.23)

∀y ∈ ∂Ω,

(1.24)

s (Ω) ≤ c u W s (D) .

v W∞ ∞

(1.25)

∂ Let f1 (y) = − ∂y (aij ∂v(y) ∂yj ) + γv(y) ∀y ∈ Ω. One observes that v(y) satisfies the following problem i

−

∂ ∂v(y) (aij ) + γv(y) = f1 (y), ∂yi ∂yj

∀y ∈ Ω,

v(y) = 0,

∀y ∈ ∂Ω.

(1.26)

Assume that ThΩ is a family of uniform symmetric meshes with respect to vertex x such that, e ∈ ThΩ whenever e ∈ Th . We decompose (u − uh )(x) into (u − uh )(x) = (u − Rh,Ω v)(x) + (Rh,Ω v − uh )(x), where B(x, r) = {z : |x − z| ≤ r}.

(1.27)

k+2 |(u − Rh,Ω v)(x)| = |(v − Rh,Ω v)(x)| ≤ chk+2 | ln h| v W∞ | ln h| u W∞ k+2 k+2 (Ω) ≤ ch (D) .

(1.28)

By (1.5), we have

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We turn now to the estimation of (Rh,Ω v − uh )(x). Assume that R = 12 dist(x, ∂D) and GΩ x (y) is the Green’s function of the problem (1.26) at point x. Note that Ω (Rh,Ω v − v)(x) = AΩ (Ihk v − v, GΩ x − Rh,Ω Gx ) Ω k Ω Ω = AB(x,R) (Ihk v − v, GΩ x − Rh,Ω Gx ) + AΩB(x,R) (Ih v − v, Gx − Rh,Ω Gx ) Ω k Ω Ω = AB(x,R) (Ihk u − u, GΩ x − Rh,Ω Gx ) + AΩB(x,R) (Ih v − v, Gx − Rh,Ω Gx ),

(1.29)

and (u − uh )(x) = A(Ihk u − u, Gx − Ghx ) = AB(x,R) (Ihk u − u, Gx − Ghx ) + ADB(x,R) (Ihk u − u, Gx − Ghx ), where D and Ω are defined as in (1.1) and (1.26), respectively. Combining (1.29) and this estimate, we have that (Rh,Ω v − uh )(x) can be rewritten in the following variational form (Rh,Ω v − uh )(x) = (Rh,Ω v − v)(x) + (u − uh )(x) Ω h k Ω Ω = AB(x,R) (Ihk u − u, GΩ x − Rh,Ω Gx − Gx + Gx ) + AΩB(x,R) (Ih v − v, Gx − Rh,Ω Gx )

− ADB(x,R) (Ihk u − u, Gx − Ghx ).

(1.30)

By (1.28) and (1.30), we observe that the main difficulty of estimating (u − uh )(x) is in estimating GΩ x − h h Ω Ω 1 1 1 Rh,Ω GΩ

,

G − G

and

G − G + G − R G

. We will solve x W1 (DB(x,R)) x h,Ω x W1 (B(x,R)) x W1 (ΩB(x,R)) x x x these problems in Section 2. The rest of the paper is organized as follows. Assume that k ≥ 3. Section 2 will investigate the k-degree finite element estimates of the Green’s function and its derivative thoroughly. We will present our main results in Section 3. Some numerical examples will be given in Section 4. 2. The finite element estimates of Green’s function and the derivative of Green’s function We will discuss the k-degree finite element estimates for the derivative of Green’s function and Green’s function in this section. Assume that M = {Z1 , Z2 , Z3 , Z4 } is defined as in Section 1 and d0 denotes the distance between x ∈ D and M and R ≥ dist(x,∂D) . Set 2 B(M, d) =

4 

B(Zi , d).

(2.31)

i=1

For all z ∈ D, R > 0 and d > 0, define BzR,d = B(z, R) ∪ B(M, d).

(2.32)

Note that Gz (z) = +∞ for any z ∈ Ω. We shall define a special interpolation GIz (y) of Gz (y). Let νz (y) ∈ C ∞ (Ω) be the cutoff function such that νz (y) = 0 if y ∈ B(z, h/2), and νz (y) = 1 if y ∈ Ω \ B(z, h), and

∇νz L∞ (Ω) ≤ ch−1 for some constant c > 0. We define Gz (y) and GIz (y) by Gz (y) = νz (y)Gz (y),

k and GIz (y) = IN Gz (y).

(2.33)

h 1 (DB(x,R)) and G The rest of this section is organized as follows. We first estimate Ghx − Gx W∞ x − h h 1 (DB(x,R)) and g Gx W 1 (DBxR,d ) in Subsection 2.1. Next we estimate gx − gx W∞ − g

R,d x W 1 (DBx ) in x ∞ ∞ Subsection 2.2. Subsection 2.3 will give some corollaries.

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h 1 (DB(x,R)) and G − Gx

2.1. The estimation of Ghx − Gx W∞ R,d x W 1 (DBx ) ∞

We first discuss some lemmas. Assume that 2 ≤ p < +∞. Lemmas 2.1, 2.2 and 2.3 are used to estimate

gx Wp2 (DB(x,r)) and gx W∞ R,d . Lemma 2.4 is applied to give a local estimate for the k-degree k+1 (DBx ) Galerkin finite element method for the Green’s function. Lemma 2.1. Assume that r > 0 and 2 ≤ p < +∞. Then −1 1 (DB(x,r)) ≤ cr

Gx W∞ | ln r|,

(2.34)

Gx Wp2 (DB(x,r)) ≤ cpr p −2 ,

(2.35)

gx Wp2 (DB(x,r)) ≤ cpr p −3 .

(2.36)

and 2

and 2

Proof. Let Υ ⊃ D have a smooth boundary. Assume that Φ(y) ∈ C ∞ (D) is a cutoff function satisfying −l Φ(y) = 1 if y ∈ B(x, 2r ), and Φ(y) = 0 if y ∈ D  B(x, r), and |Φ|W∞ (0 ≤ l ≤ 2). Set l (D) ≤ cr Υ G1 (y) = Φ(y)Gx (y) and G2 (y) = Gx (y) − G1 (y). There holds (see [9]) −2 2 (ΥB(x,r/2)) ≤ cr

GΥ , x W ∞

and |GΥ x (y)| ≤ c| ln |x − y||.

(2.37)

Assume that the differential operator L is defined as in (1.1). Note that Υ (Gx − GΥ x )(y) = −Gx (y) = 0,

∀y ∈ ∂D,

and L(Gx − GΥ x )(y) = 0,

∀y ∈ D.

(2.38)

Combining (2.37) and (2.38), we have |Gx (y)| ≤ c| ln |x − y||.

(2.39)

Note that D is a bounded rectangular domain. Furthermore, by (2.39) and the theory of partial differential equation (see [5,6]), we get the desired result (2.34). We turn now to the proof of (2.35). Note that G2 (y) = r Gx (y) − GΥ x (y) for all y ∈ B(x, 2 ), we have LG2 (y) = L(Gx − GΥ x )(y) = 0,

r ∀y ∈ B(x, ). 2

(2.40)

Noticing the fact that G2 (y) = Gx (y) for all y ∈ D  B(x, r), we obtain LG2 (y) = LGx (y) = 0,

∀y ∈ D  B(x, r).

(2.41)

By (2.37), (2.40) and (2.41), we readily get, for any 2 ≤ p < +∞, 2

2

LG2 Lp (D) = LG2 Lp (B(x,r)B(x, r2 )) ≤ cr p LG2 L∞ (B(x,r)B(x, r2 )) = cr p LG2 L∞ (D) −2 2 (DB(x,r)) ≤ cr p ≤ cr p GΥ . x W ∞ 2

2

Note that D is a bounded rectangular domain. Furthermore, one observes that (see [5,6])

Gx Wp2 (DB(x,r)) ≤ G2 Wp2 (D) ≤ cp LG2 Lp (D) ≤ cpr p −2 . 2

Similarly, we get the desired result (2.36).

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Lemma 2.2. Assume that aij and γ are defined as in (1.1) and ψ(z) satisfies the following problem Lψ ≡ −

∂ ∂ψ(z) (aij ) + γψ(z) = 0 in ∂zi ∂zj

B(y, r).

(2.42)

Then, for any α = (α1 , α2 ) (|α| ≥ 1) and 2 ≤ p < +∞,  α   ∂ ψ(y)  2 −|α|− p  α 

ψ Lp (B(y,r)) .  ∂y 1 ∂y α2  ≤ cr 1

(2.43)

2

B(y,r  )

Proof. Assume that r ∈ [ r2 , r] and Gy (z) is the Green’s function of (2.42) in B(y, r ) at point y. One observes that  α   ∂ ψ(y)  B(y,r  )  

W∞

ψ L1 (∂B(y,r )) ≤ cr−|α|−1 ψ L1 (∂B(y,r )) . (2.44) |α|+1  ∂y α1 ∂y α2  ≤ c Gy (∂B(y,r  )) Integrating (2.44) from

r 2

to r gives,

 α   ∂ ψ(y)  2  r  α1 α2  ≤ cr−|α|−1 ψ L1 (B(y,r)B(y, r2 )) ≤ cr1−|α|− p ψ Lp (B(y,r)B(y, r2 )) . ∂y ∂y 1

2

This implies the desired result (2.43). Lemmas 2.1 and 2.2 can be combined to estimate gx Wpk+1 (DBxR,d ) and Gx Wpk+1 (DBxR,d ) . We have the following results. Lemma 2.3. There exists a constant c such that −3 1−k

gx W∞ d + R−k−2 ), R,d ≤ c(R k+1 (DBx )

(2.45)

−2 1−k d + R−k−1 ).

Gx W∞ R,d ≤ c(R k+1 (DBx )

(2.46)

2 Proof. We first consider (2.45). Assume that R1 = min{R, d} and β ∈ Z+ satisfies |β| = 2. Set ψ(z) = ∂ β gx (z) β β ∂z1 1 ∂z2 2

. Note that, for any y ∈ D  BxR,d , B(y, R21 ) ⊂ D  B(x, R2 ). Let p = | ln h|. Combining (2.36) and

2 (2.43), one has, for all α ∈ Z+ satisfying |α| = k − 1,

 α  2 2  ∂ ψ(y)  1−k− p  α  ≤ cR1−k− p ψ p R1 ≤ cR

gx W 2 (B(y, R1 )) α 1 1  ∂y 1 ∂y 2  L (B(y, 2 )) p 2 1

≤

2

2 1−k− p cR1

gx Wp2 (DB(x, R )) 2

−3 ≤ cR11−k gx W∞ ≤ cR11−k R−3 . 2 (DB(x, R )) ≤ cR 2

This implies 1−k −3

gx W∞ R ≤ c(R−3 d1−k + R−k−2 ). R,d ≤ cR k+1 1 (DBx )

Similarly, we get the desired result (2.46). Let θ(y) = (Gx − Ghx )(y). We have the following results. Lemma 2.4. Let r0 = min{r, d} and m ≥ 2 be an integer. Assume that k ≥ 3 and r0 ≥ c0 h where c0 is a constant large enough. Then there holds

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−2 1 (DB(x,mr)) ≤ ch| ln h|r

θ W∞ + cr−1 θ L∞ (DB(x,(m−1)r)) ,

(2.47)

θ W 1 (DBxmr,md ) ≤ chk (r−2 d1−k + r−k−1 ) + cr0−1 θ L∞ (DBx(m−1)r,(m−1)d ) .

(2.48)

and ∞

Proof. Let p = | ln h|. By (2.35) and Sobolev space theory, we have 1− p 1 (DB(x,(m−1)r)) ≤ ch

Gx − Ihk Gx W∞

Gx Wp2 (DB(x,(m−1)r)) ≤ ch| ln h|r−2 . 2

Then, by Theorem 1.2 (see [17]), we have k −1 1 (DB(x,mr)) ≤ c Gx − I Gx W 1 (DB(x,(m−1)r)) + cr

θ W∞

θ L∞ (DB(x,(m−1)r)) h ∞

≤ ch| ln h|r−2 + cr−1 θ L∞ (DB(x,(m−1)r)) . Similarly, by (2.46), we have

θ W 1 (DBxmr,md ) ≤ c Gx − Ihk Gx W 1 (DBx(m−1)r,(m−1)d ) + cr0−1 θ L∞ (Bx(m−1)r,(m−1)d ) ∞

∞

≤ ch (r k

−2 1−k

d

+ r−k−1 ) + cr0−1 θ L∞ (DBx(m−1)r,(m−1)d ) .

This ends the proof. To estimate θ L∞ (DB(x,mr)) , we need the following result. Lemma 2.5. Under the assumption that h is sufficiently small, there holds −1 1 (D) ≤ ch

Ghx W∞ | ln h|.

Proof. Using the inverse estimate, we have

Ghx L∞ (D) = max Ghx L∞ (e) ≤ c max Ghx L| ln h| (e) h−2/| ln h| e∈Th

e∈Th

1

≤ c Ghx L| ln h| (D) ≤ c| ln h| 2 Ghx H 1 (D) . Noticing the fact that

Ghx H 1 (D) =



Ghx 2H 1 (D) ≤ c A(Ghx , Ghx ) ≤ c Ghx L∞ (D) .

By the above two estimates, we have 1

Ghx L∞ (D) + | ln h| 2 Ghx H 1 (D) ≤ c| ln h|. This implies −1 1 (D) ≤ ch

Ghx W∞

Ghx L∞ (D) ≤ ch−1 | ln h|.

We complete the proof. In the following, we will use (2.49) to estimate θ L∞ (DB(x,mr)) .

(2.49)

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Lemma 2.6. Under the same assumptions of Lemma 2.4, there holds −2 1 (DB(x,(m−1)r)) + hr

θ L∞ (DB(x,mr)) ≤ c[h2 | ln h|r−2 + hr−1 θ W∞

θ W11 (D) ].

(2.50)

Proof. Assume that z ∈ D  B(x, mr). We have θ(z) = ADB(x,(m−1)r) (θ, Gz − GIz ) + AB(x,(m−1)r) (θ, Gz − GIz ) = K1 + K2 .

(2.51)

We first estimate K1 . From (2.33), (2.34) and the definition of Th it follows that

GIz − Gz W 1 (Bzh,h ) ≤ Gz W 1 (Bzh,h ) + GIz W 1 (Bzh,h ) ≤ c Gz W 1 (Bzh,h ) ≤ ch. 1

1

1

1

(2.52)

Note that (2.46) shows that, for any l ≥ 1 (l ∈ N ),

GIz − Gz W 1 (D∩(B 2l h,2l h B 2l−1 h,2l−1 h )) ≤ chk Gz W 1 1

z

l−1 h,2l−1 h 2l h,2l h Bz2 )) k+1 (D∩(Bz

z

≤ chk (2l−1 h)1−k ≤ ch2(l−1)(1−k) .

(2.53)

Furthermore, by (2.52) and (2.53), we have

GIz − Gz W11 (D) = GIz − Gz W 1 (Bzh,h ) + GIz − Gz W 1 (DBzh,h ) 1

≤ ch +

∞

1

ch2(l−1)(1−k) ≤ ch + ch ≤ ch.

(2.54)

l=1

Consequently, 1 (DB(x,(m−1)r)) ≤ ch θ W 1 (DB(x,(m−1)r)) . |K1 | ≤ c Gz − GIz W11 (D) θ W∞ ∞

(2.55)

To estimate K2 , we split it into K2 = AB(M,h)∩B(x,(m−1)r) (θ, Gz − GIz ) + AB(x,(m−1)r)B(M,h) (θ, Gz − GIz ) = K2,1 + K2,2 .

(2.56)

We need to estimate the two items of the right-hand side. By (2.35) and (2.49), we have 1 (B(M,h)B(z,r)) |K2,1 | ≤ c Gx W11 (B(M,h)) Gz − GIz W∞

I 1 (B(M,h)) Gz − G W 1 (B(M,h)B(z,r)) + c Ghx W∞ z 1

≤ ch2 | ln h|r−2 + ch−1 | ln h|h3 r−2 ≤ ch2 | ln h|r−2 .

(2.57)

Note that (2.46) shows −2 1 (D(B(z,r)∪B(M,h)) ≤ chr |K2,2 | ≤ c θ W11 (D) Gz − GIz W∞

θ W11 (D) .

(2.58)

By (2.56), (2.57) and (2.58), we have |K2 | ≤ ch2 | ln h|r−2 + chr−2 θ W11 (D) .

(2.59)

Then (2.50) follows by combining (2.51), (2.55) and (2.59). Based on Lemmas 2.4, 2.5 and 2.6, we are now in a position to give a proof of the main results of this section.

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Theorem 2.1. Assume that k ≥ 3. Then there holds

θ W11 (D) ≤ ch| ln h|,

(2.60)

and if R ≥ c0 h where c0 is a constant large enough, then −2 1 (DB(x,R)) ≤ ch| ln h|R

θ W∞ .

(2.61)

Proof. Assume that r ≥ c0 h where c0 is a constant large enough. Note that (2.49) implies

θ L∞ (DBxr,r ) ≤ Ghx L∞ (D) + Gx L∞ (DB(x,r)) ≤ c| ln h| + c| ln r| ≤ c| ln h|.

(2.62)

Combining Lemmas 2.4, 2.6 and (2.62), we have that there exists a constant γ0 , independent of h, such that

θ W 1 (DBx6r,6r ) ≤ chk r−k−1 + cr−1 θ L∞ (DBx5r,5r ) ∞

≤ chk r−k−1 + chr−1 θ W 1 (DBx4r,4r ) + γ0 hr−3 θ W11 (D) ∞

k −k−1

≤ ch r

k −k−1

+ ch r

+ chr−2 θ L∞ (DBx3r,3r ) + γ0 hr−3 θ W11 (D)

≤ chk r−k−1 + chk r−k−1 + ch2 r−2 θ W 1 (DBx2r,2r ) + γ0 hr−3 θ W11 (D) ∞

≤ chk r−k−1 + ch2 r−3 θ L∞ (DBxr,r ) + γ0 hr−3 θ W11 (D) ≤ chk r−k−1 + ch2 | ln h|r−3 + γ0 hr−3 θ W11 (D) .

(2.63)

Let r0 = c0 h. Note that (2.49) implies 1 (D) + ch| ln h|

θ W 1 (Bx6r0 ,6r0 ) ≤ Ghx W 1 (Bx6r0 ,6r0 ) + Gx W 1 (Bx6r0 ,6r0 ) ≤ ch2 Ghx W∞ 1

1

1

≤ ch| ln h| + ch| ln h| ≤ ch| ln h|.

(2.64)

By (2.63) and (2.64), we have

θ W11 (D) ≤ θ W 1 (DBx6r0 ,6r0 ) + θ W 1 (Bx6r0 ,6r0 ) ≤ ch| ln h| + γ0 hr0−1 θ W11 (D) 1

≤ ch| ln h| +

1

γ0 c−1 0 θ W11 (D) .

(2.65)

Note that c0 is a constant large enough. This implies the desired result (2.60). Similarly to (2.63), by (2.47), (2.49), (2.50) and (2.60), we have −2 1 (DB(x,4r)) ≤ ch| ln h|r

θ W∞ + chr−2 θ L∞ (DB(x,r)) ≤ ch| ln h|r−2 + chr−2 | ln h|

≤ ch| ln h|r−2 .

(2.66)

This implies the desired result (2.61). Remark 2.1. Under the assumption that the domain D is of class C 1,1 , which is not the case in Theorem 2.1, J. Freshe and R. Rannacher provided a proof of (2.60) (see [4]). To estimate θ W 1 (DBxR,d ) , we need the following result. ∞

Lemma 2.7. Assume that x ∈ D  B(M, d0 ). Under the same assumptions of Lemma 2.4, there holds

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θ L∞ (DBxmr,md ) −2 ≤ c[h4 | ln h|2 d−2 + r−2 ) + hk+1 | ln h|(r−k−1 + r−2 d01−k )] + ch θ W 1 (DBx(m−1)r,(m−1)d ) . 0 (d ∞

(2.67)

Proof. Assume that z ∈ D  Bxmr,md . We have θ(z) = ADBx(m−1)r,(m−1)d (θ, Gz − GIz ) + ABx(m−1)r,(m−1)d (θ, Gz − GIz ) = L1 + L2 .

(2.68)

We first estimate L1 . By (2.54), we have |L1 | ≤ c Gz − GIz W11 (D) θ W 1 (DBx(m−1)r,(m−1)d ) ∞

≤ ch θ W 1 (DBx(m−1)r,(m−1)d ) . ∞

(2.69)

To estimate L2 , we split it into L2 = AB(M,d0 /2)∩Bx(m−1)r,(m−1)d (θ, Gz − GIz ) + ABx(m−1)r,(m−1)d B(M,d0 /2) (θ, Gz − GIz ) = L2,1 + L2,2 .

(2.70)

We need to estimate the two items of the right-hand side. Let r0 = min{r, d}. Note that z ∈ D  Bxmr,md . (m−1)r,(m−1)d We have Bx ⊂ D  B(z, r0 ). Let p = | ln h|. Furthermore, by (2.35) and (2.46), we obtain

Gz − GIz W11 (DB(z,r0 )) ≤ Gz − GIz W11 ((DB(z,r0 ))∩B(M,h)) + Gz − GIz W11 (D(B(z,r0 )∪B(M,h))) 1 ≤ ch3 Gz Wp2 ((DB(z,r0 ))∩B(M,h)) + chk Gz Wk+1 (D(B(z,r0 )∪B(M,h)))

≤ ch3 | ln h|r0−2 + chk r0−2 h3−k | ln h| ≤ ch3 | ln h|r0−2 . This shows I 4 2 −2 −2 1 (B(M,d /2)) Gz − G W 1 (DB(z,r )) ≤ ch | ln h| d |L2,1 | ≤ c θ W∞ + r−2 ). z 0 (d 0 0 1

(2.71)

To estimate L2,2 , by (2.46), we have 1 (D(B(z,r)∪B(M,d /2))) |L2,2 | ≤ c θ W11 (D) Gz − GIz W∞ 0

≤ chk+1 | ln h|(r−k−1 + r−2 d01−k ).

(2.72)

Plugging the above two estimates into (2.70), we have −2 |L2 | ≤ ch4 | ln h|2 d−2 + r−2 ) + chk+1 | ln h|(r−k−1 + r−2 d01−k ). 0 (d

(2.73)

Then (2.67) follows by combining (2.68), (2.69) and (2.73). Similarly to (2.61), by (2.48) and (2.67), we have the following result. Theorem 2.2. Let min{d, d0 } ≥ c0 h where c0 is a constant large enough. Assume that x ∈ D  B(M, d0 ). Under the same assumptions of Theorem 2.1, −2

θ W 1 (DBxR,d ) ≤ c[h3 | ln h|2 d−2 + R−2 ) + hk | ln h|(R−k−1 + R−2 d01−k )]. 0 (d ∞

(2.74)

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h 1 (DB(x,R)) and g 2.2. The estimation of gxh − gx W∞ R,d x − gx W 1 (DBx ) ∞

Lemma 2.8. Assume that x ∈ e ⊂ D  B(y, R) satisfies (2.75) and (2.76): (i)

min |x − xl | ≥ ch,

(2.75)

dist(x, ∂e) ≥ ch,

(2.76)

1≤l≤s

(ii)

where x1 , . . . , xs are all vertices of mesh e. Then (gyh − gy )(x) can be rewritten in the following variational form (gyh − gy )(x) = A(gyh − gy , Gx − Ihk Gx ).

(2.77)

Proof. We decompose (−gy + gyh )(x) into (−gy + gyh )(x) = [−gy (x) + A(gy , Gx )] − [A(gy , Gx ) − A(gyh , Gx )] = [−gy (x) + A(gy , Gx )] + A(gyh − gy , Ihk Gx ) + A(gyh − gy , Gx − Ihk Gx ). This implies (gyh − gy )(x) − A(gyh − gy , Gx − Ihk Gx ) = −[gy (x) − A(gy , Gx )] + A(gyh − gy , Ihk Gx ).

(2.78)

Next we estimate gy (x) − A(gy , Gx ). Assume that ε > 0 and ψε ∈ Wq2 (D) ∩ H01 (D) (q > 2) satisfies ψε (z) = Gx (z) ∀z ∈ D  B(x, ε). Note that, for any x, x + Δx ∈ D, Gx (y) = Gy (x), and Gy (x + Δx) = Gx+Δx (y). One observes that gy (x) =

∂Gx (y) . ∂yi

Consequently, ∂Gx (y) − lim A(gy , ψε ) ε→0 ∂xi ∂Gx (y) ∂Gx (y) ∂Gx (y) ∂ψε (y) = − lim = − = 0. ε→0 ∂xi ∂xi ∂xi ∂xi

gy (x) − A(gy , Gx ) =

(2.79)

Assume that δ > 0. We turn now to the estimation of A(gyh − gy , Ihk Gx ). Suppose that με ∈ W02,q (D) (q > 2) 1 (D) = 0. We observe that satisfies με (z) = Ihk Gx (z) ∀z ∈ B(y, δh) and lim με − Ihk Gx W∞ ε→0

lim |A(gy , Ihk Gx − με )| = lim |ADB(x,δh) (gy , Ihk Gx − με )|

ε→0

ε→0

≤ c lim

ε→0

Ihk Gx

1 (D) gy W 1 (DB(x,δh)) = 0. − με W∞ 1

This gives ∂I k Gx (y) ∂με (y) = h . ε→0 ∂xi ∂xi

A(gy , Ihk Gx ) = lim A(gy , με ) + lim A(gy , Ihk Gx − με ) = lim ε→0

ε→0

By (2.78), (2.79), and (2.80), we get the desired result (2.77).

(2.80)

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Lemma 2.9. Let r0 = min{R, d}. Assume that k ≥ 3 and r0 ≥ c0 h where c0 is a constant large enough. Then there holds

gyh − gy L∞ (DB(y,R)) ≤ chR−2 | ln h|,

(2.81)

−2

gyh − gy L∞ (DByR,d ) ≤ c[h3 | ln h|2 d−2 + R−2 ) + hk | ln h|(R−k−1 + R−2 d01−k )]. 0 (d

(2.82)

and

Proof. Assume that x ∈ e ⊂ D  B(y, R) satisfies (2.75) and (2.76). By Lemma 2.6, we have A(gyh − gy , Gx − Ihk Gx ) = A(gyh − gy , Gx − Ghx ) = A(gy , Gx − Ghx ) =

∂(Gx − Ghx )(y) . ∂yi

(2.83)

Then we have (gyh − gy )(x) =

∂(Ghx − Gx )(y) . ∂yi

(2.84)

This, together with (2.61), gives −2 1 (DB(y,R)) ≤ ch| ln h|R |(gyh − gy )(x)| ≤ Ghx − Gx W∞ ,

(2.85)

and 1 (D(B(y,R)∪B(M,d))) |(gyh − gy )(x)| ≤ Ghx − Gx W∞

−2 ≤ c[h3 | ln h|2 d−2 + R−2 ) + hk | ln h|(R−k−1 + R−2 d01−k )]. 0 (d

(2.86)

Above we demand that x satisfies (2.75) and (2.76). Next we estimate gyh −gy L∞ (e) for all e ⊂ D B(y, R). Assume that S ⊂ e is a set of all points that satisfy (2.75) and (2.76). From above analysis one observes that S is an infinite set and, for any x ∈ S, 1 (DB(y,R−r)) ], |(gy − gyh )(x)| ≤ c[hr−2 | ln h| + h gyh − gy W∞

(2.87)

where c is independent of y. This indicates that, for any x ∈ S, |(Ihk gy − gyh )(x)| ≤ |(gy − gyh )(x)| + |(gy − Ihk gy )(x)| ≤ |(gy − gyh )(x)| + chr−2 | ln h| 1 (DB(y,R−r)) ]. ≤ c[hr−2 | ln h| + h gyh − gy W∞

Note that Ihk gy − gyh is a k-degree polynomial on e. One finds that there exists {x1 , . . . , xm } ⊂ S such that, for any x ∈ e, (Ihk gy

−

gyh )(x)

m

= (Ihk gy − gyh )(xl )Nl (x), l=1

where Nl (x) denotes the basis function of xl . The above two estimates give 1 (DB(y,R−r)) ).

Ihk gy − gyh L∞ (e) ≤ c(hr−2 | ln h| + h gyh − gy W∞

Consequently,

gyh − gy L∞ (e) ≤ Ihk gy − gyh L∞ (e) + Ihk gy − gy L∞ (e) ≤ chr−2 | ln h|. Therefore, we get the desired result (2.81). Similarly, by (2.74), we get the desired result (2.82).

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Similarly to Lemma 2.4, we have the following results. Lemma 2.10. Let r0 = min{r, d}. Assume that r0 ≥ c0 h where c0 is a constant large enough. Then there hold 2 −3 1 (DB 2r ) ≤ ch| ln h| r

gy − gyh W∞ + cr−1 gy − gyh L∞ (DByr ) , y

(2.88)

gy − gyh W 1 (DBy2r,2d ) ≤ chk (r−2 d−k + r−k−2 ) + cr0−1 gy − gyh L∞ (DByr,d ) .

(2.89)

and ∞

Let x = y. Using the same arguments in the proof of Theorem 2.1, by Lemmas 2.9 and 2.10, we have the following results. Theorem 2.3. Let r0 = min{R, d}. Assume that x ∈ D  B(M, d0 ) and r0 ≥ c0 h where c0 is a constant large enough. Then there hold −3 1 (DB(x,R)) ≤ ch| ln h|R

gxh − gx W∞ ,

(2.90)

−3

gxh − gx W 1 (DBxR,d ) ≤ ch3 | ln h|2 d−2 + R−3 ) + chk | ln h|(R−k−2 + R−3 d01−k ). 0 (d

(2.91)

and ∞

2.3. Some corollaries Assume that R =

dist(x,∂D) 2

and x ∈ D  B(M, d0 ). Based on Theorem 2.3, we have the following result.

Corollary 2.1. For a sufficiently small h, there holds k −k

gxh − gx W11 (DB(x,R)) ≤ c[ch2 d−2 ]| ln h|2 . 0 +h R

(2.92)

Proof. Let H = c0 h where c0 is a constant large enough. We decompose gxh − gx W11 (DB(x,R)) into

gxh − gx W11 (DB(x,R)) = gxh − gx W11 (B(M,H)B(x,R)) + gxh − gx W11 (D(B(M,H)∪B(x,R))) = J1 + J2 .

(2.93)

We will be concerned with the estimation of J1 and J2 . By (2.90), we have 2 3 2 −3 2 2 −2 J1 ≤ ch| ln h|2 d−3 0 × cH ≤ ch | ln h| d0 ≤ ch | ln h| d0 .

(2.94)

To estimate J2 , note that d0 = dist(x, M ) ≥ dist(x, ∂D) = 2R, from (2.91) it follows that −1 J2 ≤ ch3 | ln h|2 d−2 + chk | ln h|(R−k + R−1 d01−k + R−3 d03−k ) 0 H k −k ≤ ch2 | ln h|2 d−2 + R−1 d01−k + R−3 d03−k ). 0 + ch | ln h|(R

Note that R ≤ cd0 . Combining (2.93), (2.94), and (2.95), we derive k −k

gxh − gx W11 (DB(x,R)) ≤ c[h2 | ln h|2 d−2 + R−1 d01−k + R−3 d03−k )] 0 + h | ln h|(R k −k ≤ c[h2 d−2 ]| ln h|2 . 0 +h R

This ends the proof.

(2.95)

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Ω Assume that Ω is defined in Lemma 1.1. Let g x and Rh,Ω g x denote ∂xi GΩ x and Rh,Ω ∂xi Gx , respectively. We have the following results.

Corollary 2.2. Assume that B(x, R) ⊂ D. Under the same assumptions of Corollary 2.1, there hold k −k

g x − Rh,Ω g x W11 (ΩB(x,R)) ≤ c[h2 d−2 ]| ln h|2 , 0 +h R

(2.96)

k −k

g x − Rh,Ω g x − gx + gxh W11 (B(x,R)) ≤ c[h2 d−2 ]| ln h|2 . 0 +h R

(2.97)

Proof. Similarly to (2.92), one gets (2.96). Next we consider (2.97). One observes that g x − Rh,Ω g x − gx + gxh W11 (B(x,R)) can be split into

g x − Rh,Ω g x − gx + gxh W11 (B(x,R)) = g x − Rh,Ω g x − gx + gxh W11 (B(x,R)B(x,R/2)) + g x − Rh,Ω g x − gx + gxh W11 (B(x,R/2)) .

(2.98)

We first estimate g x − Rh,Ω g x − gx + gxh W11 (B(x,R)B(x,R/2)) . By using (2.92) and (2.96), we have

g x − Rh,Ω g x − gx + gxh W11 (B(x,R)B(x,R/2)) ≤ g x − Rh,Ω g x − gx + gxh W11 (DB(x,R/2)) k −k ≤ c[h2 d−2 ]| ln h|2 . 0 +h R

(2.99)

For the second item, let η(y) = (g x − gx )(y) and Rh,B(x,R) η(y) be the k-degree finite element approximation of η(y) on B(x, R). Then (g x − Rh,Ω g x − gx + gxh )(y) can be decomposed into (g x − Rh,Ω g x − gx + gxh )(y) = (η − Rh,B(x,R) η)(y) + (−Rh,Ω g x + gxh + Rh,B(x,R) η)(y).

(2.100)

To estimate η − Rh,B(x,R) η W11 (B(x,R/2)) , from −k−1

η W∞

η L∞ (B(x,R)) ≤ cd0−k−2 , k+1 (B(x, R )) ≤ cd0 2

we get 2 k −k−2

η − Rh,B(x,R) η W11 (B(x,R/2)) ≤ cR2 η − Rh,B(x,R) η W∞ ≤ chk d−k 1 (B(x, R )) ≤ cR h d0 0 . 2

(2.101)

We turn now to the estimation of − Rh,Ω g x + gxh + Rh,B(x,R) η W11 (B(x, R )) . Note that 2

η(y) = (g x − gx )(y). One observes that − Rh,Ω g x + gxh + Rh,B(x,R) η = −Rh,Ω g x + gxh + Rh,B(x,R) g x − Rh,B(x,R) gx = −(Rh,Ω g x − g x ) + (gxh − gx ) + (Rh,B(x,R) η − η). Consequently,

− Rh,Ω g x + gxh + Rh,B(x,R) η L∞ (∂B(x,R)) ≤ gxh − gx L∞ (∂B(x,R)) + − Rh,Ω g x + g x L∞ (∂B(x,R)) + η − Rh,B(x,R) η L∞ (∂B(x,R)) k −k ≤ c[h2 d−2 ]| ln h|2 . 0 +h R

(2.102)

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Note that −Rh,Ω g x + gxh + Rh,B(x,R) η satisfies, for any w ∈ S0h (B(x, R)), AB(x,R) (−Rh,Ω g x + gxh + Rh,B(x,R) η, w) = 0. Then by the arguments of Schatz and Wahlbin (see [17], Theorem 1.2), we have

− Rh,Ω g x + gxh + Rh,B(x,R) η W11 (B(x, R )) ≤ cR2 − Rh,Ω g x + gxh + Rh,B(x,R) η W∞ 1 (B(x, R )) 2

≤ cR − Rh,Ω g x +

gxh

+ Rh,B(x,R) η L∞ (∂B(x,R)) ≤

2

c[h2 d−2 0

k

+h R

−k

2

]| ln h| ,

(2.103)

where we have used Theorem 2.2. By using (2.100), (2.101), and (2.103), we get k −k

g x − Rh,Ω g x − gx + gxh W11 (B(x,R/2)) ≤ c[h2 d−2 ]| ln h|2 . 0 +h R

(2.104)

The combination of (2.99) and (2.104) gives the desired result (2.97). Similarly, we get the following results. Corollary 2.3. Under the same assumptions of Corollary 2.1, k 1−k

Gx − Rh Gx W11 (DB(x,R)) ≤ c[h3 d−2 ]| ln h|2 , 0 +h R

(2.105)

Ω 3 −2 k 1−k ]| ln h|2 ,

GΩ x − Rh,Ω Gx W11 (ΩB(x,R)) ≤ c[h d0 + h R

(2.106)

Ω h 3 −2 k 1−k ]| ln h|2 .

GΩ x − Rh,Ω Gx − Gx + Gx W11 (B(x,R)) ≤ c[h d0 + h R

(2.107)

Assume that k ≥ 3. We will apply Corollaries 2.1, 2.2 and 2.3 to investigate the local ultraconvergence for the k-degree Galerkin finite element method in Section 3. 3. The local ultraconvergence of the k-degree Galerkin finite element method This section is organized as follows. Subsection 3.1 will investigate the local ultraconvergence of the k-degree triangular finite elements for the problem (1.1). Subsection 3.2 will investigate the local ultraconvergence of the k-degree rectangular finite elements for the problem (1.1). Based on Subsections 3.1 and 3.2, we will give some corollaries in Subsection 3.3. In the following, we will use R and d0 to denote dist(x, ∂D) and dist(x, M ), respectively. 3.1. The local ultraconvergence of the k-degree triangular finite element method Assume that Th is a uniform family of triangulations and x is a vertex. Let uh (y) be the k-degree Galerkin FE solution over Th for the problem (1.1). We will study the ultraconvergence of the displacement and derivative for uh (y) at x. We first introduce the following lemma. Lemma 3.1. Assume that h is sufficiently small and Th on D is triangulations. If k ≥ 3 is odd, 2k k 2k −k |∇(I2h (Ih u − uh )(x)| ≤ c[h2+k d−2 + hk+1 ]| ln h|2 u W∞ 2+k 0 +h R (D) .

(3.108)

Proof. Assume that Ω is defined as in Subsection 1.2. By Lemma 1.1, one has that there exists v(y) ∈ k+2 W∞ (Ω) satisfying, v(y) = u(y) ∀y ∈ D, and v(y) = 0 ∀y ∈ ∂Ω, and v W∞ k+2 k+2 (Ω) ≤ c u W∞ (D) . We split 2k k h ∇I2h (Ih u − u )(x) into

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2k k 2k k ∇I2h (Ih u − uh )(x) = ∇I2h (Ih v − uh )(x) 2k k 2k = ∇I2h (Ih v − Rh,Ω v)(x) + ∇I2h (Rh,Ω v − uh )(x) = I1 + I2 .

(3.109)

We first estimate I1 . Set v(y) = v(2x − y),

and v˜(y) =

1 (v(y) − v(y)). 2

Note that, for any vertex y, Rh,Ω v(y) = Rh,Ω v(2x − y)

and Ihk v(y) = Ihk v(2x − y).

This implies 2k k 2k k ∇I2h (Ih v − Rh,Ω v)(x) = −∇I2h (Ih v − Rh,Ω v)(x).

Consequently, 2k k 2k k ∇I2h (Ih v − Rh,Ω v)(x) = ∇I2h (Ih v˜ − Rh,Ω v˜)(x).

(3.110)

Meanwhile, Q. Lin et al. (see [12], Theorem 3.3) showed that k+1 k+1 1 (τ ) ≤ ch

Ihk v˜ − Rh,Ω v˜ W∞ | ln h| v W∞ | ln h| u W∞ k+2 k+2 (Ω) ≤ ch (D) ,

where τ is defined as in Section 1. This, together with (3.110), implies,  2k k  2k k 1 (τ ) |I1 | = ∇I2h (Ih v − Rh,Ω v)(x) ≤ I2h (Ih v˜ − Rh,Ω v˜) W∞ k+1 1 (τ ) ≤ ch ≤ c Ihk v˜ − Rh,Ω v˜ W∞ | ln h| u W∞ k+2 (D) .

(3.111)

We turn now to the estimation of I2 . Assume that g x and Rh,Ω g x are defined as in Corollary 2.3. Similarly to (1.30), we split

∂(Rh,Ω v−uh )(x) ∂xi

into

∂(Rh,Ω v − uh )(x) = AB(x,R) (Ihk v − v, g x − Rh,Ω g x − gx + gxh ) + AΩB(x,R) (Ihk v − v, g x − Rh,Ω g x ) ∂xi − ADB(x,R) (Ihk u − u, gx − gxh ) = B1 + B2 + B3 .

(3.112)

Combining Corollaries 2.1, 2.2 and 2.3, we get 2k −k |B1 | ≤ c[h2+k d−2 )]| ln h|2 u W∞ k+2 0 +h R (D) ,

(3.113)

2k −k |B2 | ≤ c[h2+k d−2 )]| ln h|2 u W∞ k+2 0 +h R (D) ,

(3.114)

2k −k |B3 | ≤ c(h2+k d−2 )| ln h|2 u W∞ k+2 0 +h R (D) .

(3.115)

and

The above three estimates, together with (3.112), yield   2k −k |I2 | ≤ ∇(Rh,Ω v − uh )(x) ≤ c(h2+k d−2 )| ln h|2 u W∞ k+2 0 +h R (D) . Combining (3.109), (3.111), and (3.116), we derive the desired result (3.108).

(3.116)

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Theorem 3.1. Assume that k ≥ 3 is odd. Under the assumptions of Lemma 3.1, 2k h 2k −k |∇(u − I2h u )(x)| ≤ c(h2+k d−2 + hk+1 )| ln h|2 u W∞ k+2 0 +h R (D) .

(3.117)

2k h Proof. We decompose ∇(u − I2h u )(x) into 2k h 2k k 2k k ∇(u − I2h u )(x) = ∇(u − I2h Ih u)(x) + ∇(I2h (Ih u − uh ))(x).

(3.118)

2k From the definitions of I2h and Ihk u it follows that 2k k 2k (u − I2h Ih u)(x) = (u − I2h u)(x).

(3.119)

2k k 2k k+1 k+1 1 (τ ) ≤ ch |∇(u − I2h Ih u)(x)| ≤ c u − I2h u W∞

u W∞

u W∞ k+2 k+2 (τ ) ≤ ch (D) .

(3.120)

This implies

2k k 2k h We turn now to the estimation of ∇(I2h Ih u − I2h u )(x). By using (3.108), we have 2k k 2k −k |∇(I2h (Ih u − uh )(x)| ≤ c(h2+k d−2 )| ln h|2 u W∞ k+2 0 +h R (D) .

Combining the above estimate and (3.120), we get the desired result (3.117). Similarly, we get the following result. Theorem 3.2. Under the assumptions of Theorem 3.1, if k ≥ 4 is even, then 2k 1−k |u(x) − uh (x)| ≤ c[h3+k d−2 + hk+2 ]| ln h|2 u W∞ k+2 0 +h R (D) .

(3.121)

Remark 3.1. One observes that (3.117) and (3.121) are extensions of (1.4) and (1.5), respectively. 3.2. The local ultraconvergence of the k-degree rectangular finite element method for the problem (1.1) In this subsection, we use the local symmetric technique to investigate the local ultraconvergence of the k-degree rectangular finite element method for the problem (1.1). Theorem 3.3. Assume that h is sufficiently small. Then if k ≥ 4 is even, then 2k+1 1−k |u(x) − uh (x)| ≤ c(h4+k d−2 R + hk+3 )| ln h|2 u W∞ k+3 0 +h (D) ,

(3.122)

and if k ≥ 3 is odd, then 2k h 2k+1 −k |∇(u − I2h u )(x)| ≤ c(h3+k d−2 R + hk+2 )| ln h|2 u W∞ k+3 0 +h (D) .

(3.123)

Proof. Assume that Ω is defined as in Subsection 1.2. By Lemma 1.1, one has that there exists v(y) ∈ k+3 (Ω) satisfying, v(y) = u(y) ∀y ∈ D, and v(y) = 0 ∀y ∈ ∂Ω, and v W∞ W∞ k+3 k+3 (Ω) ≤ c u W∞ (D) . We decompose u(x) − uh (x) into u(x) − uh (x) = [u(x) − Ihk u(x)] + (Ihk u − uh )(x).

(3.124)

Note that x is a vertex. One finds that u(x) − Ihk u(x) = 0.

(3.125)

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To estimate (Ihk u − uh )(x), let u(y) be the k + 1-degree interpolation of u(y) over Th . Set u ˜(y) = u(y) − u(y).

(3.126)

Then Ihk u(x) − uh (x) can be decomposed into Ihk u(x) − uh (x) = [Ihk u(x) − uh (x)] + [Ihk u ˜(x) − u ˜h (x)] = J1 + J2 .

(3.127)

By the integral equality (see [12], Lemma 2.2), we have J1 = 0.

(3.128)

We turn now to the estimation of J2 . Assume that v(y), Ω and ThΩ are defined as in Subsection 1.2. Suppose that v(y) is the k + 1-degree interpolation of v(y) over ThΩ . Set v˜(y) = v(y) − v(y).

(3.129)

J2 = [Ihk v˜(x) − Rh,Ω v˜(x)] + [Rh,Ω v˜(x) − u ˜h (x)] = J2,1 + J2,2 .

(3.130)

We split J2 into

Similarly to (3.111) and (3.116), we have k+3 k+3 |J2,1 | ≤ chk+3 | ln h| ˜ v W∞ | ln h| v W∞ | ln h| u W∞ k+3 3+k 3+k (Ω) ≤ ch (Ω) ≤ ch (D) ,

(3.131)

2k+1 1−k R + hk+3 ]| ln h|2 u W∞ |J2,2 | ≤ c[h4+k d−2 k+3 0 +h (D) ,

(3.132)

respectively. By (3.130), (3.131), and (3.132), we obtain 2k+1 1−k |J2 | ≤ c(h4+k d−2 R + hk+3 )| ln h|2 u W∞ k+3 0 +h (D) .

(3.133)

Combining (3.127), (3.128), and (3.133), we get 2k+1 1−k |Ihk u(x) − uh (x)| ≤ c(h4+k d−2 R + hk+3 )| ln h|2 u W∞ k+3 0 +h (D) .

This, together with (3.125), gives the desired result (3.122). 2k h u )(x) into We turn now to the proof of (3.123). We split ∇(u − I2h 2k h 2k k 2k k ∇(u − I2h u )(x) = ∇(u − I2h Ih u)(x) + ∇I2h (Ih u − uh )(x).

(3.134)

2k k To estimate ∇(u − I2h Ih u)(x), one finds that 2k k |∇(u − I2h Ih u)(x)| ≤ chk+2 u W∞ k+3 (D) .

(3.135)

Similarly to Lemma 3.1, one finds that   2k k ∇I2h (Ih u − uh )(x) ≤ c(hk+3 d−2 + h2k+1 R−k )| ln h|2 u

0

k+3 W∞ (D) .

(3.136)

The combination of (3.134), (3.135), and (3.136) gives the desired result (3.123). Remark 3.2. One observes that (3.122) is an extension of (1.7), and (3.123) is an extension of (1.6) and (1.10).

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We now use the local symmetric technique to investigate the local ultraconvergence of the k-degree rectangular finite element method for the problem (1.1). We get the following results. Theorem 3.4. Assume that aij = 0 when i = j. Under the assumptions of Theorem 3.3, if k ≥ 4 is even, then 2k+2 1−k |u(x) − uh (x)| ≤ c(hk+5 d−2 R + hk+4 )| ln h|2 u W∞ k+4 0 +h (D) ,

(3.137)

and if k ≥ 3 is odd, then 2k h 2+2k −k |∇(u − I2h u )(x)| ≤ c[hk+4 d−2 R + hk+3 ]| ln h|2 u W∞ 4+k 0 +h (D) .

(3.138)

Proof. Similarly to (3.122), combining the integral equality (see [12], Lemma 2.1), Corollaries 2.1, 2.2 and 2.3, we get the desired result (3.137). Similarly to (3.123), we get the desired result (3.138). Remark 3.3. One observes that (3.137) is an extension of (1.9), and (3.138) is an extension of (1.8) and (1.11). 3.3. Some corollaries Assume that E is the set of all vertices over Th and the norm . ∞,h,R is defined by

ω ∞,h,R =

max

x∈E,dist(x,∂D)≥R

|ω(x)|.

(3.139)

Based on Theorems 3.1 and 3.2, we get the following corollary. 1

Corollary 3.1. Under the same assumptions of Theorem 3.2, if R ≥ h 2 , then

u − uh ∞,h,R ≤ chk+2 | ln h|2 u W∞ 2+k (D) ,

(3.140)

1

and under the same assumptions of Theorem 3.1, if R ≥ h 2 , then 2k h

∇(u − I2h u ) ∞,h,R ≤ chk+1 | ln h|2 u W∞ 2+k (D) .

(3.141)

1

Proof. Assume that x ∈ E and dist(x, ∂D) ≥ R. Note that d0 ≥ R ≥ h 2 , by using (3.121), we get |(u − uh )(x)| ≤ chk+2 | ln h|2 u W∞ 2+k (D) . This implies the desired result (3.140). Similarly, we get the desired result (3.141). Remark 3.4. One observes that (1.4) and (1.5) both assume that dist(x, ∂D) ≥ c, but Corollary 3.1 implies that the assumption is unnecessary. Similarly, based on Theorems 3.3 and 3.4, we have Corollaries 3.2 and 3.3. 1

Corollary 3.2. Under the same assumptions of Theorem 3.3, if k ≥ 4 is even and R ≥ h 2 , then

u − uh ∞,h,R ≤ chk+3 | ln h|2 u W∞ 3+k (D) ,

(3.142)

1

and if k ≥ 3 is odd and R ≥ h 2 , then 2k h

∇(u − I2h u ) ∞,h,R ≤ chk+2 | ln h|2 u W∞ 3+k (D) .

(3.143)

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Table 4.1 Error of the 3-rd element over the uniform triangular meshes. 1 25

h= 6

∇(u − I2h uh ) ∞,h,R 6 uh )∞,h,R ∇(u−I2h h4 | ln h|2

h=

1 36

h=

1 49

1 64

h=

1.224 × 10−4

3.25 × 10−5

1.120 × 10−5

4.59 × 10−6

4.61

4.25

4.26

4.45

Table 4.2 Error of the 4-th element over the uniform triangular meshes. h=

u − u ∞,h,R h

u − u ∞,h,R h h

−6

1 25

h=

4.36 × 10

−8

1.06 × 10

1

1 36

h=

4.91 × 10

−9

1.07 × 10

1

1 49

h=

1 64

7.73 × 10

−10

1.64 × 10−10

1.07 × 10

1

1.12 × 101

Remark 3.5. One observes that (1.6), (1.7) and (1.10) all assume that dist(x, ∂D) ≥ c, but Corollary 3.2 implies that the assumption is unnecessary. 1

Corollary 3.3. Under the same assumptions of Theorem 3.3, if k ≥ 4 is even and R ≥ h 2 , then

u − uh ∞,h,R ≤ chk+4 | ln h|2 u W∞ 4+k (D) ,

(3.144)

1

and if k ≥ 3 is odd and R ≥ h 2 , then 2k h

∇(u − I2h u ) ∞,h,R ≤ chk+3 | ln h|2 u W∞ 4+k (D) .

(3.145)

Remark 3.6. One observes that (1.8), (1.9) and (1.11) all assume that dist(x, ∂D) ≥ c, but Corollary 3.3 implies that the assumption is unnecessary. 4. Numerical examples We first use a numerical test to investigate Corollary 3.1. Consider the following second order elliptic equation with mixed second order derivative term −

∂2u ∂2u ∂2u − − 1.5 =f 2 2 ∂x1 ∂x2 ∂x1 ∂x2

in D,

u = 0 on ∂D,

(4.146)

with exact solution u(y) = sin(2πx1 )sin(2πx2 ). Assume that Th is a family of uniform isosceles triangulations with grid size h. (1a) We first consider the case k = 3. For h = 1/n3 and R = 1/n2 (n = 5, 6, 7, 8), we use the cubic element over Th to solve the problem (4.146). We obtain the numerical data shown in Table 4.1. By using Table 4.1, we have that, for the 3-rd element over the uniform triangular meshes, 6 h

∇(u − I2h u ) ∞,h,R ≤ ch4 | ln h|2 .

(4.147)

(4.147) shows that, for the cubic element, uh behaves better than our theoretical error estimate (3.141). (1b) Then we consider the case k = 4. For h = 1/n2 and R = 1/n (n = 5, 6, 7, 8), we use the 4-th element over the uniform triangular meshes to solve the problem (4.146). We obtain the numerical data shown in Table 4.2. By using Table 4.2, we have that, for the 4-th element over the uniform triangular meshes,

u − uh ∞,h,R ≤ ch6 .

(4.148)

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Table 4.3 Error of the bi-3 element over the uniform rectangular meshes. h= 6

∇(u − I2h uh ) ∞,h,R

∇(u −

6 I2h uh ) ∞,h,R h−6

1 25

h=

1 36

h=

1 49

h=

1 64

57.4 × 10−8

5.92 × 10−8

0.95 × 10−8

0.196 × 10−8

141.2

129.8

131.1

134.1

Table 4.4 Error of the bi-4 element over the uniform rectangular meshes. h=

u − uh ∞,h,R

u − u ∞,h,R h h

−8

1 25

h=

1 36

h=

1 49

h=

1 64

4.2318 × 10−10

2.1983 × 10−11

1.9164 × 10−12

1.0825 × 10−13

6.46 × 10

6.20 × 10

6.37 × 10

3.05 × 101

1

1

1

Table 4.5 Error of the bi-3 element over the uniform rectangular meshes. h=

1 25

h= −7

u − u ∞,h,R

8.41 × 10

u − uh ∞,h,R h−6

2.05 × 102

h

1 36

9.56 × 10

h= −8

1 49

1.48 × 10

2.08 × 102

h= −8

2.05 × 102

1 64

0.30 × 10−8 2.06 × 102

The equality (4.148) indicates that, for the 4-th element, uh (x) behaves better than our theoretical error estimate (3.140) by a factor | ln h|2 . Combining Tables 4.1 and 4.2, we have that Corollary 3.1 is reliable. We now use a numerical test to investigate Corollary 3.2. Consider the following Poisson equation −Δu = f

in

D,

u=0

on

∂D,

(4.149)

with exact solution u(y) = sin(πx1 )sin(πx2 ). (2a) We first consider the case k = 3. For h = 1/n2 and R = 1/n2 (n = 5, 6, 7, 8), we use the bi-3 element over the uniform rectangular meshes to solve the problem (4.149) and have the numerical data shown in Table 4.3. By Table 4.3, we have that, for the bi-3 element over the uniform rectangular meshes, 6 h

∇(u − I2h u ) ∞,h,R ≤ ch6 .

(4.150)

(2b) Then we consider the case k = 4. For h = 1/n2 and R = 1/n (n = 5, 6, 7, 8), we use the bi-4 element over the uniform rectangular meshes to solve the problem (4.149) and have the numerical data shown in Table 4.4. By using Table 4.4, we have that, for the bi-4 element over the uniform rectangular meshes,

u − uh ∞,h,R ≤ ch8 .

(4.151)

The equality (4.151) indicates that, for the bi-4 element, uh (x) behaves better than our theoretical error estimate (3.144) by a factor | ln h|2 . Combining (4.150) and (4.151), we have that Corollary 3.2 is reliable. We now use a numerical test to investigate Corollary 3.3. We consider the problem (4.146) and let the exact solution u(y) = sin(4πx1 )sin(4πx2 ). (3a) First we consider the case k = 3. For h = n12 and R = 1/n (n = 5, 6, 7, 8), we use the bi-3 element over the uniform rectangular meshes to solve the problem (4.146) and have the numerical data shown in Table 4.5. By using Table 4.5, we have that, for the bi-3 element over the uniform rectangular meshes,

u − uh ∞,h,R ≤ ch6 .

(4.152)

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85

Table 4.6 Error of the bi-4 element over the uniform rectangular meshes. h= 8

∇(u − I2h uh ) ∞,h,R

∇(u −

8 I2h uh ) ∞,h,R h−6 | ln h|−2

1 25

h=

1 36

h=

1 49

h=

1 64

1.16 × 10−4

2.21 × 10−5

0.49 × 10−5

0.77 × 10−6

2.88 × 10

3.19 × 10

3.34 × 10

3.24 × 103

3

3

3

The equality (4.152) indicates that, for the bi-3 element, uh (x) behaves better than our theoretical error estimate (3.142) by a factor | ln h|2 . Then we consider the case k = 4. For h = 1/n2 and R = 1/n (n = 5, 6, 7, 8), we use the bi-4 element over the uniform rectangular meshes to solve the problem (4.146) and have the numerical data shown in Table 4.6. Table 4.6 shows that, for the bi-4 element over the uniform rectangular meshes, 8 h |∇(u − I2h u )(x)| ≤ ch6 | ln h|2 .

(4.153)

Combining Tables 4.5 and 4.6, we have that Corollary 3.3 is reliable. Acknowledgments The authors would like to thank the anonymous referee for their careful reviews of the paper and valuable constructive comments and suggestions, which helped to improve the presentation and quality of this paper significantly. This work is supported in part by the National Natural Science Foundation of China under grants 11671304, 11301396, 11171257, and the Zhejiang Provincial Natural Science Foundation, China (No. LY15A010015). Appendix A. Proof of Lemma 1.1 Since D ⊂⊂ Ω, we get from the extension theorem (cf. [1]) that there exists w ∈ Ws∞ (Ω) satisfying w=u

and w Ws∞ (Ω) ≤ c u Ws∞ (D) .

in D

Assume that γ = min ρ(x, ∂Ω). Let ψ(x) be the cutoff function satisfying ψ(x) = 1 if x ∈ D, and ψ(x) = 0 x∈D

if ρ(x, ∂Ω) ≤ γ2 . Set

⎧ ⎪ ⎨ θ(x) =

⎪ ⎩

−1

e 1−|x|2

−1



|x|≤1

0,

e 1−|x|2 dx

,

|x − y| < 1, |x − y| ≥ 1.

 ∞ Let δ < γ4 and ψ1 (x) = Ω θ( x−y δ )ψ(y)dy. One observes that ψ1 ∈ C (Ω) satisfies ψ1 (x) = 1 if x ∈ D, and ψ1 (x) = 0 if x ∈ ∂Ω. Now v = ψ1 w is just the required function satisfying (1.23)–(1.25). References [1] R.A. Adams, Sobolev Spaces, Academic Press, New York, 1972. [2] C. Chen, W 1,∞ interior estimates for finite element methods on regular mesh, J. Comput. Math. 3 (1985) 1–7. [3] C. Chen, Y. Huang, High Accuracy Theory of Finite Element Methods, Hunan Science and Technology Press, P. R. China, 1995 (in Chinese). [4] J. Freshe, R. Rannacher, Eine L1 -Fehlerabschatzung diskreter Grundlosungen in der Methods der finiten Elemente, Tagungsband “Finite Elemente”, Bonner Math. Schriften 89 (1975) 92–114. [5] P. Grisvard, Behavior of the solutions of an elliptic boundary value problem in a polygonal or polyhedral domain, in: B. Hubbard (Ed.), Numerical Solution of Partial Differential Equations III, Academic Press, New York, 1976, pp. 207–274.

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