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http://dx.doi.org/10.4134/JKMS.j150547 pISSN: 0304-9914 / eISSN: 2234-3008

THE GRADIENT RECOVERY FOR FINITE VOLUME ELEMENT METHOD ON QUADRILATERAL MESHES

Yingwei Song and Tie Zhang

Abstract. We consider the finite volume element method for elliptic problems using isoparametric bilinear elements on quadrilateral meshes.

A gradient recovery method is presented by using the patch interpolation technique. Based on some superclose estimates, we prove that the recov- ered gradient R(∇u h ) possesses the superconvergence: k∇u−R(∇u h )k = O(h 2 )kuk 3 . Finally, some numerical examples are provided to illustrate our theoretical analysis.

1. Introduction

The derivative (gradient) recovery techniques are postprocess techniques that reconstruct the derivative from the discrete solution to achieve better derivative approximation, for example, to obtain the superconvergent result.

In the 1960s, the derivative recovery technique had been used to compute the derivative (stress) via the C 0 -finite elements [7]. In particular, the simple av- eraging or weighted averaging methods were employed by engineers for linear finite elements [16]. Subsequently, new derivative recovery techniques have been developed, for example, the L 2 -projection post-processing technique [6, 14], the well known Zienkiewicz-Zhu’s patch recovery technique (SPR) [22], the interpo- lation postprocess technique [10], the polynomial preserving recovery technique (PPR ) [13] and the derivative patch interpolation recovery technique [19], and so on. However, all these recovery techniques were presented for the finite element methods on triangle or rectangular meshes.

Finite volume element (FVE) method is a discrete technique for solving partial differential equations. In general, it represents the conservation of an interest quantity, such as mass, momentum, or energy in fluid mechanics, so that it can be expected to simulate corresponding physical phenomena more

Received September 7, 2015; Revised January 20, 2016.

2010 Mathematics Subject Classification. Primary 65N15, 65N30, 65M60.

Key words and phrases. finite volume method, elliptic problem, quadrilateral meshes, gradient recovery, superconvergence.

This work was supported by the National Natural Science Funds of China, No. 11371081;

and the State Key Laboratory of Synthetical Automation for Process Industries Fundamental Research Funds, No. 2013ZCX02.

c

2016 Korean Mathematical Society

1411

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effectively. Readers are referred to the monograph [8] for general presentation of the FVE method and to [1, 2, 3, 4, 9, 11, 12, 15, 17, 18, 20] and the references therein for details.

Compared with FVE methods on triangle and rectangular meshes, less works can be found for FVE methods on quadrilateral meshes. We know that finite element methods on quadrilateral meshes usually have better accuracy than that on triangle meshes, and quadrilateral meshes are more flexible than rect- angular meshes in handing complicated domain geometries. So quadrilateral meshes are also used frequently in practical applications. For isoparametric bi- linear FVE method solving elliptic problems on quadrilateral meshes, Schmidt, Li et al. [9, 15] first give the optimal H 1 -error estimate; Then, Lv and Li [11]

further obtain the optimal L 2 -error estimate if the quadrilateral mesh is h 2 - uniform; Recently, Lv and Li [12] also derive a superconvergence result in the average gradient form.

In this paper, we study the isoparametric bilinear FVE method to solve the following elliptic problem on quadrilateral meshes,

 −div(A∇u) + c u = f, in Ω, u = 0, on ∂Ω,

(1.1)

where Ω ⊂ R 2 is a polygonal domain with boundary ∂Ω, coefficient matrix A = (a ij ) 2×2 . Our main goal is to present a gradient recovery method for the FVE solution u h on quadrilateral meshes by using the patch interpolation technique [19]. This recovery method can provide a better approximation to the gradient of the exact solution. In fact, based on some superclose estimates, we prove that the recovered gradient Q(∇u h ) possesses the superconvergence:

(1.2) k∇u − Q(∇u h )k = O(h 2 )kuk 3 .

This paper is organized as follows. In Section 2, we introduce the FVE scheme and some related results. Section 3 is devoted to deriving some su- perclose estimates for the interpolation function. In Section 4, we present the gradient recovery method and give its superconvergence analysis. Finally, in Section 5, numerical experiments are provided to illustrate our theoretical analysis.

We shall use the standard notation for the Sobolev space W m,p (D) equipped

with the norm k · k m,p,D and the semi-norm | · | m,p,D . In order to simplify the

notations, we set W m,2 (D) = H m (D), k · k m,2,D = k · k m,D , and when D = Ω

we skip the index D. Furthermore, notations ( , ) and k · k denote the inner

product and norm in space L 2 (Ω), respectively. We use letter C to represent a

generic positive constant, independent of the mesh size h.

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2. Finite volume element method on quadrilateral meshes 2.1. Partition and isoparametric bilinear transformation

Let T h = S{K} be a convex quadrilateral mesh partition of domain Ω so that Ω = S

K∈T

h

{ K }, where h = max h K , h K is the diameter of element K.

We assume that partition T h is regular, that is, all the inner angles of any element in T h are uniformly bounded away from 0 and π, and there exists a positive constant γ > 0 such that

(2.1) h K /ρ K ≤ γ, ∀ K ∈ T h ,

where ρ K denotes the diameter of the biggest ball included in K.

The following strong regular and h 2 -uniform mesh conditions will be used in our analysis.

Definition 2.1. A regular quadrilateral partition T h is called strongly regular if for any element K = P 1 P 2 P 3 P 4 in T h , it holds (see Fig.1)

(2.2) | −−−→

P 1 P 2 + −−−→

P 3 P 4 | + | −−−→

P 1 P 4 + −−−→

P 3 P 2 | ≤ Ch 2 K .

Furthermore, a strongly regular quadrilateral partition T h is called h 2 -uniform, if for any two adjacent elements K and K 0 , it holds (see Fig. 1)

(2.3) | −−−→

P 1 P 2 + −−−→

P 1 P 6 | ≤ Ch 2 K .

K K

P6

P4

P3

P2 P5

P1

Figure 1. Two adjacent elements K and K 0 .

Let l = |P 1 P 4 | be the length of the common edge of K and K 0 (see Fig. 1).

Since

h K = h K

ρ K ρ K

l l

h K

0

h K

0

≤ γh K

0

, then h K in (2.3) may be replaced by h K

0

.

It is well known that a FVE method usually concerns two mesh partitions:

a basic partition T h and its dual partition T h . We here form the dual partition

T h in the following way. For each element K ∈ T h , we connect the center of

K to the midpoints of its edges by straight lines. Then, for each nodal point

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P in T h , there exists a polygonal region K P = G 1 G 2 G 3 G 4 surrounding P , K P is called the dual element at point P , and T h is the union of all such dual elements, see Fig. 2.

  

P

K

P*

M G4

G1

G2 G3

Figure 2. The dual element K P surrounding point P .

Let b K = [0, 1] × [0, 1] be the reference element in the ˆ x = (ξ, η) plane.

Then, for each convex quadrilateral K = P 1 P 2 P 3 P 4 , there exists an invertible isoparametric bilinear mapping F K : (ξ, η) ∈ b K → x = (x, y) ∈ K such that (see Fig. 3)

(2.4) x = P 1 + (P 2 − P 1 )ξ + (P 4 − P 1 )η + (P 1 − P 2 + P 3 − P 4 )ξη.

Figure 3. The isoparametric bilinear transformation.

Let P i = (x i , y i ). Transformation (2.4) also can be expressed as follows.

x = x 1 + a 1 ξ + a 2 η + a 3 ξη, (2.5)

y = y 1 + b 1 ξ + b 2 η + b 3 ξη, (2.6)

where

a 1 = x 2 − x 1 , a 2 = x 4 − x 1 , a 3 = x 1 − x 2 + x 3 − x 4 ,

b 1 = y 2 − y 1 , b 2 = y 4 − y 1 , b 3 = y 1 − y 2 + y 3 − y 4 .

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Denote the Jacobi matrix of mapping F K by J K and the determinant of J K by J K , then we have

J K =

∂x

∂ξ

∂x

∂η

∂y

∂ξ

∂y

∂η

!

=

 a 1 + a 3 η a 2 + a 3 ξ b 1 + b 3 η b 2 + b 3 ξ

 . Furthermore, by the differentiation law of inverse function we have (2.7) J K −1 =

∂ξ

∂x

∂ξ

∂y

∂η

∂x

∂η

∂y

!

= 1 J K

 b 2 + b 3 ξ −(a 2 + a 3

−(b 1 + b 3 η) a 1 + a 3 η

 . Under mesh conditions (2.2) and (2.3), by straightforward computation, one can derive the following results, see [9, 11].

Lemma 2.1. Assume that partition T h is strongly regular, then

|J K | 0,∞ = O(h 2 K ), |J K | 0,∞ = O(h K ), |J K −1 | 0,∞ = O(h −1 K ), (2.8)

|J K −1 | m,∞ ≤ Ch −1 K , m = 1, 2, 3, (2.9)

|J K −1 (ξ, η) − J K −10 , η 0 )| 0,∞ ≤ C, ∀ (ξ, η), (ξ 0 , η 0 ) ∈ b K.

(2.10)

Furthermore, if partition T h is h 2 -uniform, then for any two adjacent elements K and K 0 ,

(2.11) |J K −1 (ξ, η) − J K −1

0

0 , η 0 )| 0,∞ ≤ C, ∀ (ξ, η) ∈ b K, (ξ 0 , η 0 ) ∈ b K.

Let function ˆ u(ξ, η) = u(x(ξ, η), y(ξ, η)) = u ◦ F K (ξ, η), where u ◦ F K denotes the compound function of u(x, y) and the mapping F K (ξ, η). From Lemma 2.1 and noting that

∇ˆ b u = J K T ∇u, ∇u = J K −T ∇ˆ b u, we can derive the following estimates

|ˆ u| m,p, b K ≤ Ch m−

2 p

K kuk m,p,K , m = 0, 1, 2, 3, 1 ≤ p ≤ ∞, (2.12)

|u| m,p,K ≤ Ch −m+

2 p

K kˆ uk m,p, b K , m = 0, 1, 2, 3, 1 ≤ p ≤ ∞.

(2.13)

2.2. Finite volume element scheme

Consider problem (1.1). As usual, we assume that there exist positive con- stants C 1 and C 2 such that

(2.14) C 1 z T z ≤ z T A(x, y)z ≤ C 2 z T z, ∀ z ∈ R 2 , (x, y) ∈ Ω, We further assume that A ∈ [W 1,∞ (Ω)] 2×2 , c ∈ L ∞ (Ω) and c ≥ 0.

Associated with partition T h and T h , we introduce the trial function space U h and test function space V h , respectively,

U h = { u h ∈ C 0 (Ω) : u h | K = P

K b ◦ F K −1 , P

K b ∈ Q 11 ( b K), ∀ K ∈ T h , u h | ∂Ω = 0 }, V h = { v h ∈ L 2 (Ω) : v h | K

P

= constant, ∀ P ∈ N h , v h | K

P

= 0, ∀ P ∈ ∂Ω},

where Q 11 ( b K) is the set of all bilinear polynomials on b K and N h is the set of

all mesh points of T h .

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Let u be the solution of problem (1.1). Using the Green formula, we obtain (2.15) −

Z

∂K

P

n · (A∇u)vds + Z

K

P

cuv = Z

K

P

f v, K P ∈ T h , v ∈ V h , where n is the outward unit normal vector on the boundary concerned. Ac- cording to weak formula (2.15), we set the bilinear form a h (u, v) by

(2.16)

a h (u, v) = X

K

P

∈T

h

 − Z

∂K

P

n · (A∇u)vds + Z

K

P

cuv 

, u ∈ H 2 (Ω), v ∈ V h , and define the FVE approximation to problem (1.1): Find u h ∈ U h such that (2.17) a h (u h , v h ) = (f, v h ), ∀ v h ∈ V h .

Let Π h : U h → V h be the interpolation operator defined by Π h v h = X

P ∈N

h

v h (P )χ

P

, ∀ v h ∈ U h ,

where χ

P

is the characteristic function of the dual element K P . Obviously, Π h is a one to one mapping from U h onto V h . Then, we obtain the equivalent scheme of (2.17): Find u h ∈ U h such that

(2.18) a h (u h , Π h v h ) = (f, Π h v h ), ∀ v h ∈ U h ,

which is the FVE scheme actually used in our argument. From (2.15) and (2.18), we can derive the error equation

(2.19) a h (u − u h , Π h v h ) = 0, ∀ v h ∈ U h .

Let Π h u ∈ U h be the usual isoparametric bilinear interpolation of continuous function u. In our analysis, the following approximation property and trace inequality will be used frequently, see [5]. For 1 < p ≤ ∞, we have

ku − Π h uk m,p,K ≤ Ch 2−m K kuk 2,p,K , m = 0, 1, 2, (2.20)

kuk 0,p,∂K ≤ Ch

1 p

K kuk 0,p,K + h K k∇uk 0,p,K , u ∈ W 1,p (K).

(2.21)

Furthermore, the following two lemmas hold.

Lemma 2.2 ([8]). Let b Π h v ˆ h = d Π h v h . Then for v h ∈ U h , we have Z

K b

(ˆ v h − b Π h ˆ v h )d b K = 0, Z

b τ

(ˆ v h − b Π h v ˆ h )ds = 0, (2.22)

kv h − Π h v h k 0,q,K ≤ Ch K kv h k 1,q,K , 1 ≤ q ≤ ∞, (2.23)

where b τ ⊂ ∂ b K be any one edge of the reference element b K.

Lemma 2.3 ([9, 15]). Let partition T h be strongly regular, u and u h be the solutions of problems (1.1) and (2.18), respectively. Then, we have

a h (v h , Π h v h ) ≥ Ckv h k 2 1 , ∀ v h ∈ U h ,

ku − u h k 1 ≤ Chkuk 2 .

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3. Superclose estimate for the interpolation approximation The superclose estimate of interpolation function usually provides a useful analysis tool in the study of superconvergence of finite element method [10, 21].

In this section, we establish some superclose estimates for the finite volume element method.

Let w c be the piecewise constant approximation of function w on T h , (3.1) w c | K = 1

|K|

Z

K

w, K ∈ T h ; kw − w c k 0,p,K ≤ Ch K |w| 1,p,K , 1 ≤ p ≤ ∞.

Lemma 3.1. Let partition T h be strongly regular and a c be a constant vector, u ∈ W 3,p (Ω). Then, for v ∈ U h , we have

|(a c · ∇(u − Π h u) x , Π h v − v) K | + |(a c · ∇(u − Π h u) y , Π h v − v) K | (3.2)

≤ Ch 2 K kuk 3,p,K kv h k 1,q,K , 2 ≤ p, q ≤ ∞, 1/p + 1/q = 1.

Proof. Let w = u − Π h u and e h = Π h v − v. Using the isoparametric bilinear transformation, we have

(a c · ∇(u − Π h u) x , Π h v − v) K = Z

K b

a c · J K −T ∇( b w b ξ ξ x + w b η η x )ˆ e h J K d b K (3.3)

= Z

K b

a c · J K −T ( b ∇ w b ξ ξ x + b ∇ w b η η x )ˆ e h J K +

Z

K b

a c · J K −T ( w b ξ ∇ξ b x + w b η ∇η b x )ˆ e h J K

= E 1 + E 2 . From (2.7) we obtain

E 1 = Z

K b

a c · J K −T ( b ∇ w b ξ (b 2 + b 3 ξ) − b ∇ w b η (b 1 + b 3 η))ˆ e h

(3.4)

= Z

K b

a c · (J K −T − J K −T (0, 0))( b ∇ w b ξ (b 2 + b 3 ξ) − b ∇ w b η (b 1 + b 3 η))ˆ e h +

Z

K b

a c · J K −T (0, 0)( b ∇ w b ξ b 2 − b ∇ w b η b 1 )ˆ e h +

Z

K b

a c · J K −T (0, 0)( b ∇ w b ξ b 3 ξ − b ∇ w b η b 3 η)ˆ e h = F 1 + F 2 + F 3 . It follows from Lemma 2.1 and condition (2.2) that

|J K −T | ∞ ≤ Ch −1 K , |J K −T − J K −T (0, 0)| ≤ C, b 1 = b 2 = O(h K ), b 3 = O(h 2 K ), therefore, we have from (2.12), (2.20) and (2.23) that

F 1 + F 3 ≤ Ch K | w| b 2,p, b K kˆ e h k 0,q, b K ≤ Ch K h 2−

2 p

K kwk 2,p,K h

2 q

K ke h k 0,q,K

≤ Ch 2 K kuk 2,p,K kvk 1,q,K .

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For F 2 , noting that b ∇( b Π h u) ˆ ξ and b ∇( c Π h u) ˆ η are constants, then from (2.22), (3.1) and (2.12), we have

F 2 = Z

K b

a c · J K −T (0, 0)( b ∇ˆ u ξ b 2 − b ∇ˆ u η b 1 )ˆ e h

= Z

K b

a c · J K −T (0, 0)( b ∇ˆ u ξ b 2 − ( b ∇ˆ u ξ ) c b 2 − b ∇ˆ u η b 1 + ( b ∇ˆ u η ) c b 1 )ˆ e h

≤ C|J K −T | (|b 1 | + b 2 |)|ˆ u| 3,p, b K kˆ e h k 0,q, b K

≤ Ch 3−

2 p

K kuk 3,p,K h

2 q

K ke h k 0,q,K ≤ Ch 2 K kuk 3,p,K kvk 1,q,K . Combining estimates F 1 ∼ F 3 , we obtain from (3.4) that

E 1 ≤ Ch 2 K kuk 3,p,K kvk 1,q,K .

Next, we estimate E 2 . Since |J K −T | = O(h −1 K ), |J K | = O(h 2 K ) and (see (2.9))

| b ∇ξ x | ≤ |ξ xx x ξ + ξ xy y ξ | + |ξ xx x η + ξ xy y η | ≤ Ch K (|ξ xx | + |ξ xy |) ≤ C, then, we have from (2.12), (2.20) and (2.23) that

E 2 = Z

K b

a c · J K −T ( w b ξ ∇ξ b x + w b η ∇η b x )ˆ e h J K ≤ Ch K | w| b 1,p, b K kˆ e h k 0,q, b K

≤ Ch K h 1−

2 p

K ku − Π h uk 1,p,K h

2 q

K ke h k 0,q,K ≤ Ch 2 K kuk 2,p,K kvk 1,q,K . The proof is completed by substituting estimates E 1 and E 2 into (3.4). 

Let E h 0 be the union of all interior edges of elements in T h .

Lemma 3.2. Let partition T h be h 2 -uniform and matrix A M | τ be constant on each τ ∈ E h 0 , u ∈ W 3,p (Ω). Then, we have for 2 ≤ p, q ≤ ∞, 1/p + 1/q = 1,

X

K∈T

h

Z

∂K

n · A M ∇(u − Π h u)(Π h v − v)ds

≤ Ch 2 kuk 3,p kvk 1,q , v ∈ U h . Proof. Let w = u − Π h u and e h = Π h v − v. We need to estimate

X

K∈T

h

Z

∂K

n · A M ∇we h ds (3.5)

= X

K∈T

h

X

τ ∈∂K\∂Ω

Z

τ

n · A M ∇we h ds = X

K∈T

h

X

τ ∈∂K\∂Ω

F (τ )

= X

τ ∈E

h0

F (τ ∩ ∂K) + F (τ ∩ ∂K 0 ),

where K and K 0 be two adjacent elements with the common edge τ .

Let quadrilateral element K = P 1 P 2 P 3 P 4 , τ ∈ ∂K be an edge of K, for example, τ = P 1 P 4 , see Fig. 1. On edge τ (ˆ τ = { ξ = 0, 0 ≤ η ≤ 1}), we have from (2.5)-(2.6) that

ds = p

(dx) 2 + (dy) 2 | ξ=0 = q

x 2 η + y η 2 dη| ξ=0 = |P 1 P 4 |dη.

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Therefore, we can write (3.6) F (τ ) =

Z

τ

n · A M ∇we h ds = Z 1

0

|P 1 P 4 |ˆ n · A M J K −T (ˆ τ ) b ∇ wˆ b e h dη.

Let K 0 = P 6 P 1 P 4 P 5 be an adjacent element of K with the common edge τ = ∂K ∩ ∂K 0 = P 1 P 4 and τ 0 = τ ∩ ∂K 0 (see Fig. 1). Since the outward normal vector n 0 | τ

0

= −n| τ , then we have

F (τ ∩ ∂K) + F (τ ∩ ∂K 0 ) = Z 1

0

|P 1 P 4 |ˆ n · A M J K −T (ˆ τ ) b ∇ w(ˆ b τ )ˆ e h

− Z 1

0

|P 1 P 4 |ˆ n · A M J K −T

0

(ˆ τ 0 ) b ∇ w(ˆ b τ 0 )ˆ e h dη.

Set a h = |P 1 P 4 |ˆ n · A M = O(h K ). Noting that e h is continuous across edge τ = τ 0 (excepting at the midpoint of τ ), we obtain

F (τ ∩ ∂K) + F (τ ∩ ∂K 0 ) (3.7)

= Z 1

0

a h · (J K −T (ˆ τ ) − J K −1 (0, 0)) b ∇ w(ˆ b τ )ˆ e h

+ Z 1

0

a h · J K −T (0, 0)( b ∇ w(ˆ b τ ) − b ∇ w(ˆ b τ 0 ))ˆ e h

+ Z 1

0

a h · (J K −T (0, 0) − J K −T

0

(ˆ τ 0 )) b ∇ w(ˆ b τ 0 ))ˆ e h dη

= S 1 + S 2 + S 3 .

Using Lemma 2.1, trace inequality (2.21) and the finite element inverse inequal- ity, we obtain

S 1 + S 3 ≤ Ch K (k b ∇ wk b 0,p, b K + | b ∇ w| b 1,p, b K )kˆ e h k 0,q, b K + Ch K

0

(k b ∇ wk b 0,p, b K

0

+ | b ∇ w| b 1,p, b K

0

)kˆ e h k 0,q, ˆ K

0

≤ Ch K (h 1−

2 p

K kwk 1,p,K + h 2−

2 p

K kwk 2,p,K )h

2 q

K ke h k 0,q,K

+ Ch K

0

(h 1−

2 p

K

0

kwk 1,p,K

0

+ h 2−

2 p

K

0

kwk 2,p,K

0

)h

2 q

K

0

ke h k 0,q,K

0

≤ Ch 2 kuk 2,p,K∪K

0

kvk 1,q,K∪K

0

.

Next, we estimate S 2 . Let Π h | K = Π K , Π h | K

0

= Π K

0

. Since ∇u is continuous across edge τ = τ 0 , we have

(3.8) ∇ b w(ˆ b τ ) − b ∇ w(ˆ b τ 0 ) = b ∇ b Π

K b u(ˆ ˆ τ ) − b ∇ b Π

K b

0

u(ˆ ˆ τ 0 ).

Noting that the bilinear interpolation b Π

K b u can be written as ˆ Π b

K b u = u ˆ 1 + (u 2 − u 1 )ξ + (u 4 − u 1 )η + (u 1 − u 2 + u 3 − u 4 )ξη, (ξ, η) ∈ b K,

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where u i = u(P i ), and

(u 1 − u 2 + u 3 − u 4 ) = − Z 1

0

ˆ

u ξ (ξ, 0)dξ + Z 1

0

ˆ

u ξ (ξ, 1)dξ

= Z 1

0

Z 1 0

ˆ

u ξη dξdη = 1

| b K|

Z

K b

ˆ

u ξη d b K = (ˆ u ξη ) c ( b K), then, on edge ˆ τ = ˆ τ 0 = {ξ = 0, 0 ≤ η ≤ 1}, we have

∇ b b Π

K b u(ˆ ˆ τ ) − b ∇ b Π

K b

0

ˆ u(ˆ τ 0 )

= (u 2 − u 1 + (ˆ u ξη ) c ( b K)η, u 4 − u 1 ) T − (u 1 − u 6 + (ˆ u ξη ) c ( b K 0 )η, u 5 − u 6 ) T

= (u 2 − 2u 1 + u 6 , u 4 − u 1 − u 5 + u 6 ) T + ((ˆ u ξη ) c ( b K) − (ˆ u ξη ) c ( b K 0 ))ηε, where vector ε = (1, 0) T . Now, from (3.8) and (2.22) we obtain

S 2 = Z 1

0

a h · J K −T (0, 0)((ˆ u ξη ) c ( b K) − (ˆ u ξη ) c ( b K 0 ))ηεˆ e h

= Z 1

0

a h · J K −T (0, 0)((ˆ u ξη ) c ( b K) − ˆ u ξη + ˆ u ξη − (ˆ u ξη ) c ( b K 0 ))ηεˆ e h dη.

It follows from Lemma 2.1, trace inequality (2.21) and the finite element inverse inequality that

S 2 ≤ C(kˆ u ξη − (ˆ u ξη ) c k 0,p, b K + |ˆ u ξη | 1,p, b K )kˆ e h k 0,q, b K + C(kˆ u ξη − (ˆ u ξη ) c k 0,p, b K

0

+ |ˆ u ξη | 1,p, b K

0

)kˆ e h k 0,q, b K

0

≤ C|ˆ u| 3,p, b K kˆ e h k 0,q, b K + C|ˆ u| 3,p, b K

0

kˆ e h k 0,q, b K

0

≤ Ch 3−

2 p

K kuk 3,p,K h

2 q

K ke h k 0,q,K + Ch 3−

2 p

K

0

kuk 3,p,K

0

h

2 q

K

0

ke h k 0,q,K

0

≤ Ch 2 kuk 3,p,K∪K

0

kvk 1,q,K∪K

0

.

Substituting estimates S 1 ∼ S 3 into (3.7), it yields

(3.9) F (τ ∩ ∂K) + F (τ ∩ ∂K 0 ) ≤ Ch 2 kuk 3,p,K∪K

0

kvk 1,q,K∪K

0

.

The proof is completed by combining (3.9) with (3.5).  We known that the conventional bilinear form of finite element method for problem (1.1) reads as

(3.10) a(u, v) =

Z

A∇u · ∇v + cuv.

Under strongly regular mesh condition, the following interpolation weak esti- mate has been established [21] for 2 ≤ p ≤ ∞, 1/p + 1/q = 1 ,

(3.11) |a(u − Π h u, v)| ≤ Ch 2 kuk 3,p kvk 1,q , ∀ v ∈ U h .

Below we will prove that estimate (3.11) also holds for the bilinear form of

the FVE method a h (u, Π h v). In order to use the known result (3.11) in our

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argument, we need to give the difference between bilinear forms a(u, v) and a h (u, Π h v).

Let U h + H 2 (Ω) = {w : w = u h + u, u h ∈ U h , u ∈ H 2 (Ω)} be the algebraic sum space.

Lemma 3.3. It holds for w ∈ U h + H 2 (Ω) and v ∈ U h that a h (w, Π h v) − a(w, v) = X

K∈T

h

Z

∂K

n · A∇w(Π h v − v)ds (3.12)

+ X

K∈T

h

(−div(A∇w) + cw, Π h v − v) K .

Proof. Using the integration by parts, we obtain Z

K

A∇w · ∇v = − Z

K

div(A∇w)v + Z

∂K

n · (A∇w)vds, and (see Fig. 2)

X

K∈T

h

Z

K

div(A∇w)Π h v = X

K∈T

h

X

K

P

∈T

h

Z

K

P

∩K

div(A∇w)Π h v

= X

K∈T

h

Z

∂K

n · (A∇w)Π h vds + X

K

P

∈T

h

Z

∂K

P

n · (A∇w)Π h vds.

Then, from the definitions of a(w, v) and a h (w, Π h v) (see (2.16) and (3.10)),

the desired estimate is derived. 

Theorem 3.1. Let partition T h be h 2 -uniform, u ∈ W 3,p (Ω). Then, we have for 2 ≤ p, q ≤ ∞, 1/p + 1/q = 1 that

(3.13) |a h (u − Π h u, Π h v)| ≤ Ch 2 kuk 3,p kvk 1,q , ∀ v ∈ U h .

Proof. Denote by A M the value of matrix A at the centroid of edge τ ⊂ ∂K.

Then, it follows from Lemma 3.3 that

a h (u − Π h u, Π h v) − a(u − Π h u, v) (3.14)

= X

K∈T

h

Z

∂K

n · (A − A M )∇(u − Π h u)(Π h v − v)ds

+ X

K∈T

h

Z

∂K

n · (A M ∇(u − Π h u))(Π h v − v)ds

+ X

K∈T

h

(−div(A∇(u − Π h u)), Π h v − v) K

+ X

K∈T

h

(c(u − Π h u), Π h v − v) K

= R 1 + R 2 + R 3 + R 4 .

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Using trace inequality (2.21) and the approximation properties, we have R 1 ≤ C X

K∈T

h

h K |A| 1,∞ k∇(u − Π h u)k 0,p,∂K kv − Π h vk 0,q,∂K

≤ Ch 2 kuk 2,p kvk 1,q , R 4 ≤ C X

K∈T

h

ku − Π h uk 1,p,K kv − Π h vk 0,q,K ≤ Ch 2 kuk 2,p kvk 1,q .

Next, it follows from Lemma 3.2 that

R 2 ≤ Ch 2 kuk 3,p kvk 1,q . Now, we need to estimate R 3 . Set A = (a 1 , a 2 ). Since

(3.15) div(A∇w) = (div a 1 , div a 2 ) · ∇w + a 1 · ∇w x + a 2 · ∇w y , then we have

R 3 = X

K∈T

h

(−(div a 1 , div a 2 ) · ∇(u − Π h u), Π h v − v) K

− X

K∈T

h

((a 1 − a c 1 ) · ∇(u − Π h u) x + (a 2 − a c 2 ) · ∇(u − Π h u) y , Π h v − v) K

− X

K∈T

h

(a c 1 · ∇(u − Π h u) x + a c 2 · ∇(u − Π h u) y , Π h v − v) K .

Using Lemma 3.1 and the approximation properties, it yields R 3 ≤ Ch 2 kuk 3,p kvk 1,q .

Substituting estimates R 1 ∼ R 4 into (3.14), the proof is completed.  From Theorem 3.1, we immediately obtain the following superclose result.

(3.16) kΠ h u − u h k 1 ≤ Ch 2 kuk 3 .

In fact, from Lemma 2.3, error equation (2.19) and the interpolation weak estimate (3.13), we obtain

Cku h − Π h uk 2 1 ≤ a h (u h − Π h u, Π h (u h − Π h u))

= a h (u − Π h u, Π h (u h − Π h u)) ≤ Ch 2 kuk 3 ku h − Π h uk 1 . This gives estimate (3.16)

Remark 3.1. We note that Lv and Li in [12] derived estimate (3.13) for p =

q = 2 by a lengthy and complex argument. For this estimate, we here give an

alternative and more accessible argument, in particular, our result holds true for

2 ≤ p, q ≤ ∞. Such result will be useful in the study of W 1,p -superconvergence,

see [10, 21].

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4. Gradient recovery formula and superconvergence

In this section, we will construct the gradient recovery formula for the isoparametric bilinear element and make the superconvergence estimate for the recovered gradient.

Let P be an interior mesh point and elements K 1 , K 2 , K 3 , K 4 take point P as a vertex. Denote by S p = S 4

i=1 K i the patch recovery domain at point P , see Fig. 4. Let F K

i

: b K i → K i be the isoparametric bilinear mapping and S b p = S 4

i=1 K b i be the reference patch recovery domain, see Fig. 4.

Figure 4. The patch recovery domain S p around point P . We first give the gradient recovery formula on the reference domain b S p . Let b G i be the centroid of b K i (G i = F K

i

( b G i ) is the centroid of element K i ) and further let ϕ b i ∈ Q 11 ( b S P ) be the basis function corresponding to b G i such that ϕ b i ( b G j ) = δ ij . Then, functions { ϕ b 1 , ϕ b 2 , ϕ b 3 , ϕ b 4 } form the base of bilinear polynomial space Q 11 ( b S P ). Denote the piecewise smooth function space by

W h ( b S P ) = { w : b w| b

K b

i

= polynomial, b K i ∈ b S p }.

Now, let derivative operator b D = b D ξ or b D η . For w ∈ W b h ( b S P ), we define the derivative recovery operator b Q : b D w → b b Q( b D w) ∈ Q b 11 ( b S P ) such that

(4.1) Q( b b D w) = b

4

X

i=1

D b w( b b G i ) ϕ b i (ξ, η), (ξ, η) ∈ b S P , w ∈ W b h ( b S P ).

It is easy to see that b Q( b D w) is the bilinear interpolation of b b D w on b b S P with the interpolation nodes { b G 1 , b G 2 , b G 3 , b G 4 }. For ˆ w ∈ W h ( b S P ), function b D ˆ w may be discontinuous on b S P , while b Q( b D ˆ w) is a bilinear polynomial on b S P .

From (4.1), we also obtain the gradient recovery formula on b S p :

(4.2) Q( b b ∇ w) = ( b b Q( b D ξ w), b b Q( b D η w)) b T =

4

X

i=1

∇ b w( b b G i ) ϕ b i (ξ, η), (ξ, η) ∈ b S P .

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Lemma 4.1. Let b G 0 = (ξ 0 , η 0 ) be the centroid of b K. Then (4.3) ∇(ˆ b u − b Π h u)( b ˆ G 0 ) = 0, ∀ ˆ u ∈ P 2 ( b K),

where P 2 (S) represents the set of all quadratic polynomials on set S.

Proof. Let (ξ 0 , η 0 ) = ((ξ 1 + ξ 2 )/2, (η 1 + η 2 )/2) be the centroid of element b K = (ξ 1 , ξ 2 ) × (η 1 , η 2 ). Further let I ξ and I η be the linear interpolation operators with respect to variable ξ and η, respectively, such that I ξ u(ξ ˆ i , η) = ˆ u(ξ i , η) and I η u(ξ, η ˆ i ) = ˆ u(ξ, η i ), i = 1, 2. Then, the bilinear interpolation operator b Π h

can be represented as b Π h = I η I ξ . It follows from the Taylor expansion that for ˆ

u ∈ P 2 ( b K), ˆ

u − I ξ u = ˆ 1

2 (ξ − ξ 1 )(ξ − ξ 2 )ˆ u ξξ , ˆ u − I η u = ˆ 1

2 (η − η 1 )(η − η 2 )ˆ u ηη . Then, we have (note that ˆ u ξξ = constant)

ˆ

u − b Π h u = ˆ ˆ u − I η u + I ˆ η (ˆ u − I ξ u) ˆ

= 1

2 (η − η 1 )(η − η 2 )ˆ u ηη + I η

1

2 (ξ − ξ 1 )(ξ − ξ 2 )ˆ u ξξ



= 1

2 (η − η 1 )(η − η 2 )ˆ u ηη + 1

2 (ξ − ξ 1 )(ξ − ξ 2 )ˆ u ξξ . It yields

(ˆ u − b Π h u) ˆ ξ = (ξ − ξ 1 + ξ 2

2 )ˆ u ξξ , (ˆ u − b Π h ˆ u) η = (η − η 1 + η 2

2 )ˆ u ηη , ∀ ˆ u ∈ P 2 ( b K).

The proof is completed. 

From Lemma 4.1 we see that G i = F K

i

( b G i ) is the Gauss point, that is, for any u = ˆ u ◦ F K

i

, ˆ u ∈ P 2 ( b K i ), it holds

(4.4) ∇(u − Π h u)(G i ) = J K −T

i

∇(ˆ b u − b Π h u)( b ˆ G i ) = 0.

Lemma 4.2. The following properties hold true for operator b Q D ˆ b u = b Q( b D ˆ u), ∀ ˆ u ∈ P 2 ( b S P ),

(4.5)

D ˆ b u = b Q( b D b Π h u), ∀ ˆ ˆ u ∈ P 2 ( b S P ).

(4.6)

Proof. First, note that b D ˆ u ∈ Q 11 ( b S P ) if ˆ u ∈ P 2 ( b S P ). Then, equality (4.5) comes from (4.1) and the uniqueness of interpolation polynomial. Moreover, it follows from (4.3) that b D ˆ u( b G i ) = b D b Π h u( b ˆ G i ). Thus, using (4.1) we obtain Q( b b D ˆ u) = b Q( b D b Π h ˆ u). Together with (4.5), we complete the proof.  Lemma 4.3. Recovery operator b Q is bounded and

(4.7) k b Q( b D ˆ w)k 0, b S

P

≤ k b Qk k b D ˆ wk 0, b S

P

, ∀ ˆ w ∈ W h ( b S P ), where k b Qk represents the bound of operator b Q. Furthermore, we have (4.8) k b D ˆ u − b Q( b D b Π h u)k ˆ 0, b S

P

≤ C|ˆ u| 3, b S

P

, ˆ u ∈ H 3 ( b S P ).

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Proof. First, it is easy to see that

(4.9) k ϕ b i k 0, b K ≤ (meas( b K))

12

≤ 1, b K ∈ b S P , i = 1, . . . , 4.

Hence, it follows from (4.1) and the inverse inequality that k b Q( b D ˆ w)k 0, b S

P

=  X

K∈ b b S

P

k b Q( b D ˆ w)k 2

0, b K



12

≤  X

K∈ b b S

P

 X 4

i=1

| b D ˆ w( b G i )| k ϕ b i k 0, b K  2 

12

≤ C  X

K∈ b b S

P

k b D ˆ wk 2

0, b K



12

= Ck b D ˆ wk 0, b S

P

.

Estimate (4.7) is derived. Now, for given ˆ v ∈ L 2 ( b S P ), we introduce the linear functional

(4.10) F ( b u) = ( b D ˆ u − b Q( b D b Π h u), ˆ ˆ v)

S b

P

, ˆ u ∈ H 3 ( b S P ).

Using (4.7) and the boundness of interpolation operator b Π h , we obtain

|F (ˆ u)| ≤ k b D ˆ u − b Q( b D b Π h u)k ˆ 0, b S

P

kˆ vk 0, b S

P

≤ Ckˆ uk 3, b S

P

kˆ vk 0, b S

P

+ k b Qk  X

K∈ b b S

P

k b D b Π h uk ˆ 2 0, b K 

12

kˆ vk 0, b S

P

≤ Ckˆ uk 3, b S

P

kˆ vk 0, b S

P

.

Hence, F is a linear bounded functional on H 3 ( b S P ). Moreover, it follows from Lemma 4.2 that

F (ˆ u) = 0, ∀ ˆ u ∈ P 2 ( b S P ).

Thus, using the Bramble-Hilbert Lemma, we derive

|F (ˆ u)| ≤ C|ˆ u| 3, b S

P

kˆ vk 0, b S

P

, ∀ ˆ v ∈ L 2 ( b S P ), Together with (4.10), it yields

k b D ˆ u − b Q( b D b Π h u)k ˆ 0, b S

P

≤ C|ˆ u| 3, b S

P

.

This gives estimate (4.8). 

Now, for u h ∈ U h , we define its recovery gradient on the actual patch domain S P by the formula:

(4.11) Q(∇u h )(x, y) = J K −T

j

Q( b b ∇ˆ u h ), (x, y) ∈ K j , K j ∈ S P , where b Q( b ∇ˆ u h ) is given by (4.2). Since

Q(∇u h )(G j ) = J K −T

j

Q( b b ∇ˆ u h )( b G j ) = J K −T

j

∇ˆ b u h ( b G j ) = ∇u h (G j ),

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therefore, Q(∇u h ) is in fact an interpolation function of ∇u h on patch domain S P with the Gauss points {G i } as the interpolation nodes.

Theorem 4.1. Let quadrilateral partition T h be h 2 -uniform, u and u h be the solutions of problems (1.1) and (2.18), respectively, u ∈ H 3 (Ω). Then, the recovered gradient Q(∇u h ) satisfies the local estimate

(4.12) k∇u − Q(∇u h )k 0,S

P

≤ Ch 2 kuk 3,S

P

+ Ck∇(Π h u − u h )k 0,S

P

. Proof. From (4.11), Lemma 2.1 and Lemma 4.3, we obtain

k∇u − Q(∇u h )k 0,S

P

=  X 4

j=1

Z

K b

j

|J K −T

j

( b ∇ˆ u h − b Q( b ∇ˆ u h ))| 2 J K

j

d b K j



12

≤ C  X 4

j=1

Z

K b

j

| b ∇ˆ u − b Q( b ∇ˆ u h )| 2 d b K j



12

= Ck b ∇ˆ u − b Q( b ∇ˆ u h )k 0, b S

P

≤ k b ∇ˆ u − b Q( b ∇ b Π h u)k ˆ 0, b S

P

+ k b Q( b ∇ b Π h u − b ˆ ∇ˆ u h )k 0, b S

P

≤ C|ˆ u| 3, b S

P

+ Ck b ∇( b Π h ˆ u − ˆ u h )k 0, b S

P

≤ Ch 2 kuk 3,S

P

+ Ck∇(Π h u − u h )k 0,S

P

.

This gives estimate (4.12). 

The gradient recovery formula (4.11) is local, below we expand this formula to the whole domain Ω.

We will define the global recovery formula element-wise. For an interior node P j , let S P

j

be the patch recovery domain around point P j and Q j = Q the gradient recovery operator on S P

j

defined by (4.11). Note that for a given element K, there are four patch domains {S P

j

} covering K (or less than four if K is a boundary element). In order to balance the values of Q j (∇u h )| K∩S

Pj

for different S P

j

(or Q j ), we define the global recovery operator Q Ω by the average formula,

(4.13) Q (∇u h )| K = 1 N K

X

S

Pj

∩K6=Ø

Q j (∇u h )| S

Pj

∩K , K ∈ T h ,

where N K ≤ 4 is the total number of elements in set {S P

j

: S P

j

T K 6= Ø}.

Theorem 4.2. Under the conditions of Theorem 4.1, the following supercon- vergence estimate holds

k∇u − Q Ω (∇u h )k ≤ Ch 2 kuk 3 . (4.14)

Proof. It follows from (4.12) that k∇u − Q Ω (∇u h )k 2 = X

K∈T

h

k∇u − Q Ω (∇u h )k 2 0,K

(17)

≤ X

K∈T

h

k 1 N K

X

S

Pj

∩K6=Ø

(∇u − Q j (∇u h ))k 2 0,S

Pj

∩K

≤ X

K∈T

h

max

S

Pj

∩K6=Ø k∇u − Q j (∇u h )k 2 0,S

Pj

≤ X

K∈T

h

max

S

Pj

∩K6=Ø



Ch 4 kuk 2 3,S

Pj

+ Ck∇(Π h u − u h )k 2 0,S

Pj



≤ C h 4 kuk 2 3 + k∇(Π h u − u h )k 2 .

We complete the proof by using estimate (3.16). 

5. Numerical example

In this section, we will present some numerical results to illustrate our the- oretical analysis.

Let us consider problem (1.1) with the data:

A(x, y) =

 e 2x + y 3 + 1 e xy e xy e 2y + x 3 + 1



, c(x, y) = 1 + xy.

We take Ω = [0, 1] 2 and the exact solution u(x, y) = 2 sin(2πx) sin(3πy).

We present numerical results using sequence of meshes {T i }. This quadrilat- eral mesh sequence is generated in the following way. We first make an original quadrilateral mesh T 1 with mesh size h = h 1 . Then we connect the midpoints of each edge of elements in T i (i ≥ 1) to obtain the refined mesh T i+1 which has half mesh size of T i , see Fig. 5. It is easy to see that such bisection quadri- lateral meshes must be h 2 -uniform. Denote by e(T i ) the error on mesh T i in the corresponding norm, then the numerical convergence rate r is computed by the formula r = ln( e(T i )/e(T i+1 ) )/ ln 2. The numerical results are given in Table 5.1. We see that an O(h 2 ) order of convergence rate is achieved for the recovered gradient as the theoretical prediction.

Figure 5. Left: original mesh T 1 ; Right: refined mesh T 5 obtained by bisection partition.

Acknowledgements. The authors would like to thank the anonymous referees

for many helpful suggestions which improved the presentation of this paper.

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Table 5.1. Convergence rate and error estimator.

k∇u − Q Ω ∇u h k k∇u − ∇u h k mesh size error rate error rate h 1 = 0.373 0.3621 - 0.6424 -

h 1 /2 0.9166e-1 1.982 3.2276e-1 0.993 h 1 /4 0.2314e-1 1.986 1.6194e-1 0.995 h 1 /8 0.5821e-2 1.991 0.8108e-1 0.998 h 1 /16 1.4621e-3 1.993 4.0513e-2 1.001 h 1 /32 0.3670e-3 1.994 2.0257e-2 1.000

References

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Yingwei Song

Department of Mathematics and the State Key Laboratory of Synthetical Automation for Process Industries

Northeastern University Shenyang 110004, P. R. China

E-mail address: [email protected]

Tie Zhang

Department of Mathematics

Northeastern University

Shenyang 110004, P. R. China

E-mail address: [email protected]

수치

Figure 2. The dual element K P ∗ surrounding point P .
Figure 4. The patch recovery domain S p around point P . We first give the gradient recovery formula on the reference domain bS p
Figure 5. Left: original mesh T 1 ; Right: refined mesh T 5 obtained by bisection partition.
Table 5.1. Convergence rate and error estimator.

참조

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