THE POLYNOMIAL-PRESERVING RECOVERY FOR HIGHER ORDER FINITE ... · THE POLYNOMIAL-PRESERVING RECOVERY FOR HIGHER ORDER FINITE ELEMENT METHODS IN 2D AND 3D A. Naga and Z. Zhang 1 Department

DISCRETE AND CONTINUOUS Website: http://AIMsciences.orgDYNAMICAL SYSTEMS–SERIES BVolume 5, Number 30, August 2005 pp. 769–798

THE POLYNOMIAL-PRESERVING RECOVERY FOR HIGHERORDER FINITE ELEMENT METHODS IN 2D AND 3D

A. Naga and Z. Zhang 1

Department of MathematicsWayne State UniversityDetroit, MI 48202, USA

(Communicated by Jie Shen)

Abstract. The Polynomial-Preserving Recovery (PPR) technique is extendedto recover continuous gradients from C0 finite element solutions of an arbitraryorder in 2D and 3D problems. The stability of the PPR is theoretically in-vestigated in a general framework. In 2D, the stability is established undera simple geometric condition. The numerical experiments demonstrated thatthe PPR-recovered gradient enjoys superconvergence, and the Zienkiewicz-Zhuerror estimator based on the PPR-recovered gradient is asymptotically exact.

1. Introduction. Adaptive C0 Finite Element Methods (FEMs) are very efficienttools in approximating solutions of partial differential equations. A crucial partof this kind of adaptation is the error estimation. Since the pioneering work byBabuska and Rheinboldt [4], a posteriori error estimators have been the focus ofintensive research. For the history and the advances in this field, the reader isreferred to [1, 5, 8, 15, 23].

In 1987, Zienkiewicz and Zhu [28] introduced their simple error estimator, theZZ error estimator. This estimator relies on constructing a continuous gradientby a postprocessing operation. To complete their work, they [29] developed agradient recovery technique known as the Superconvergence Patch Recovery (SPR).According to [30], using the SPR-recovered gradient in the ZZ error estimatorproduced a robust estimator; namely the ZZ-SPR. Later, Babuska et al. establishedthe computer-based theory that enabled them to study and compare the knownerror estimators for a wide range of problems. One of their primary results [6, 7] isthat the ZZ-SPR is the most robust error estimator among the tested ones.

Basically, the SPR uses the gradient of the finite element solution and discreteleast-squares to recover the gradient at mesh nodes. Naturally, one asks: is itpossible to modify the SPR to use the finite element solution itself? If yes, is theZZ error estimator based upon the new recovery technique as robust as the ZZ-SPR?Fortunately, these questions have positive answers. In [26], the authors introducedthe Polynomial-Preserving Recovery (PPR) that uses the finite element solution.In [20], it was shown that the PPR-recovered gradient enjoys superconvergence.This result was established for C0 linear FEM in 2D meshes that meet some mild

1991 Mathematics Subject Classification. 65N30, 65N15, 65N12, 65D10,74S05, 41A10, 41A25.Key words and phrases. finite element method, SPR, PPR, discrete least-squares fitting, su-

perconvergence, a posteriori error estimator.1This research was partially supported by the National Science Foundation grants DMS-

0074301 and DMS-0311807.

769

770 A. NAGA AND Z. ZHANG

conditions. The ZZ-PPR (the ZZ error estimator based on the PPR) was validatedin [27] and it was shown to be as good as or better than the ZZ-SPR.

The goal of this paper is to extend the PPR to higher order C0 FEMs in 2Dand in 3D problems. The stability of the PPR recovery is studied in Section 3.The main result of that section shows that the PPR recovery is stable in regularmeshes satisfying a certain requirement. In Section 4, we show that this require-ment is guaranteed in 2D regular meshes satisfying a simple geometric condition.Although the stability of the PPR is studied in the context of regular meshes, weconjecture that the PPR is also stable in anisotropic meshes. This conjecture isverified through an example at the end of Section 4. Finally, Section 6 gives somenumerical experiments that illustrate the efficiency of the PPR and the ZZ-PPR.

1.1. Notations. Let s ∈ [0,∞], p ∈ [1,∞],m, k, l ∈ Z+ (the set of positive in-tegers), and A ⊂ Rd with d = 2 or 3. The spaces W s

p (A) and Hs(A) arethe classical Sobolev spaces equipped with the norms ‖ · ‖s,p,A and ‖ · ‖s,A, re-spectively, and the seminorms | · |s,p,A and | · |s,A, respectively. The measure ofA in Rd will be denoted by |A|. For vectors in Rm, ‖ · ‖ will denote the Eu-clidean length. If m = d, we may use | · | instead of ‖.‖. If p is a polyno-mial over A, then deg(p) will denote the total degree of p. The vector spacep : p is a polynomial over A and deg(p) ≤ m will be denoted by Pm(A) and nm

will denote the dimension of Pm(A) where nm =1d!

d∏

i=1

(m+i). Let P∗m(A) ⊂ Pm(A)

denote the subset p : p ∈ Pm(A), p is homogeneous, and deg(p) = m.The space of real matrices of order k×l is denoted by Rk×l. The identity of Rk×k

is denoted by Ik, while 1k(0k) denotes the k×1 column vector whose entries are allones (zeros). Let σ1(K) ≥ σ2(K) ≥ . . . ≥ σmax(k,l)(K) ≥ 0 denote the singularvalues of a matrix K ∈ Rk×l. Recall that σ2

m(K) = σm(KT K) = σm(KKT ) forall 1 ≤ m ≤ min(k, l) and σm(K) = 0 for all min(k, l) < m ≤ max(k, l). Also, ifK is symmetric positive semi-definite, then the set of its singular values is exactlythe same as the set of its eigenvalues. The null space of K and the range of K aredenoted by Null(K) and Range(K), respectively.

To avoid difficulties in naming the constants that will appear in the course ofthis work, we adopt the following notations. Let E1 and E2 be two mathematicalquantities and let C > 0 be some constant independent of any “essential” ingredi-ents of E1 and E2. Instead of E1 ≤ CE2, E1 ≥ CE2, and E2/C ≤ E1 ≤ CE2, wewill use the notations E1 . E2, E1 & E2, and E1

∼= E2, respectively.

1.2. Preliminaries. Consider the boundary value problem−∇(D∇u) + αu = f in Ων · (D∇u) = gN on ΓN

u = gD on ΓD

(1.1)

where Ω ⊂ Rd is a bounded domain with Lipschitz boundary ∂Ω = ΓN ∪ ΓD, theboundary segments ΓN and ΓD are disjoint, ν is the unit outward normal vectorto ∂Ω, and D is a d× d symmetric positive definite matrix. If ΓN = ∂Ω and α = 0,the compatibility condition

∫Ω

f +∫

∂Ωg = 0 must be satisfied and the condition∫

Ωu = 0 is used to ensure the uniqueness. For simplicity, Ω is assumed to be a

polygon when d = 2 or a polyhedron when d = 3.In the variational form of (1.1), we seek u ∈ V such that

B(u, v) = L(v) ∀ v ∈ V0 (1.2)

THE POLYNOMIAL-PRESERVING RECOVERY 771

where

V = v ∈ H1(Ω) : v|ΓD= gD, V0 = v ∈ H1(Ω) : v|ΓD

= 0,

B(u, v) =∫

Ω

[(D∇u)∇v + αuv], and L(v) =∫

Ω

fv +∫

ΓN

gNv.

If ΓD is empty, we take V = H1(Ω). Under suitable assumptions [13, pp. 35-53],the bilinear operator B is continuous and is V−elliptic, the linear operator L isbounded, and (1.2) has a unique solution u ∈ V .

Let Th be a conforming partition of Ω. The elements in Th are triangles whend = 2 and tetrahedrons when d = 3. To simplify notations, we assume the elementsin Th to be closed, i.e., T = T ∀T ∈ Th. For T ∈ Th, let hT denote the diameter ofT and set h = maxhT : T ∈ Th. Also, let %T denote the diameter of the largestball (circle when d=2 or sphere when d=3) inscribed in T . The regularity of T

is measured with the aspect ratiohT

%T. The mesh Th is said to be regular if there

exists a finite positive constant γ, independent of h, such thathT

%T≤ γ ∀T ∈ Th. (1.3)

For r ∈ Z+, define the finite element space

Sh = v ∈ C(Ω) : v|T ∈ Pr(T ) ∀T ∈ Th.Let Nh denote the set of the mesh nodes. A mesh node z ∈ Nh will be calledinternal (boundary) node if z ∈ Ω (z ∈ ∂Ω), and z will be called a mesh vertex if itis a vertex of an element T ∈ Th. The basis for Sh is the standard Lagrange basisφz : z ∈ Nh where φz(z) = δzz ∀ z, z ∈ Nh. Using this basis, a function v ∈ Sh

takes the form v =∑

z∈Nhv(z)φz. Let Ih : C0(Ω) → Sh denote the Lagrange

interpolation operator where

Ihw =∑

z∈Nh

w(z)φz ∀w ∈ C0(Ω).

The finite element approximation of u is the solution uh ∈ Sh of (1.2) when vvaries over Sh ∩ V , i.e.,

B(uh, v) = L(v)∀ v ∈ Sh ∩ V. (1.4)

2. The PPR and the ZZ-PPR. Let Gh : Sh →∏d

i=1 Sh denote the PPR op-erator. As in the SPR, the structure of Gh uh, the PPR-recovered gradient of uh,relies on the fact that every function in Sh is uniquely defined by its values at themesh nodes. If the values (Gh uh)(z) : z ∈ Nh are well-defined, then

Gh uh ,∑

z∈Nh

(Gh uh)(z)φz.

Before exploring the definitions of Gh uh at the mesh nodes, we need the followingnotations. If A is a union of mesh elements in Th and v ∈ Sh, let N(A) denote thenumber of mesh nodes in A, M(A) denote the number of mesh triangles in A, andvA denote the column vector whose entries are the values of v at the mesh nodes inA. If z is a mesh vertex and n ∈ Z+, let L(z, n) denote the union of mesh elementsin the first n layers around z, i.e.,

L(z, n) =⋃T : T ∈ Th, T ∩ L(z, n− 1) 6= ∅

where L(z, 0) , z.


2.1. Definition of the PPR. Let z ∈ Nh be a mesh vertex and let Kz denote apatch of mesh elements around z. Let pz ∈ Pr+1(Kz) be the polynomial that bestfits uh at the mesh nodes in Kz in discrete least-squares sense, i.e.,∑

z∈Nh∩Kz

|(uh − pz)(z)|2 = minp∈Pr+1(Kz)

∑

z∈Nh∩Kz

|(uh − p)(z)|2.

For easy referencing, pz will be called the least-squares polynomial approximation,LSPA, of uh at z. Then,

(Gh uh)(z) , ∇pz(z).To complete the definition of the PPR, we need to define Kz. Note that besides thedependence of its definition on the location of z, Kz must have at least nr+1 nodesdistributed around z in a way that leads to a unique pz.

For an internal mesh vertex z, we first define an initial patch Kz,0 such thatN(Kz,0) ≥ nr+1. This is achieved with the following definition:

Kz,0 ,

L(z, 1) if N(L(z, 1)) ≥ nr+1

L(z, 2) if d = 3, r = 1, and M(L(z, 1)) = 4L(z, 1) ∪ T ∈ Th : T ∩ L(z, 1) is a (d− 1)-simplex otherwise

.

The first part of this definition is illustrated in Fig. 1(b) while the third part isillustrated in Fig. 1(a). The formula in the third part means that we extend L(z, 1)by adding mesh elements having common (d−1)-simplices (edges if d = 2 and faceswhen d = 3) with L(z, 1). Fig. 2 depicts this idea. Unfortunately, if d = 3 andr = 1, extending L(z, 1) out does not work as the number of nodes after extensionis 9 < 10 = n2. The second part of the Kz,0 definition takes care of this situation.

Although N(Kz,0) ≥ nr+1, this does not ensure the uniqueness of pz. If Kz,0

does not lead to a unique pz and L(z, n) ⊆ Kz,0 ( L(z, n + 1), set Kz,0 toL(z, n+1) and recompute pz. The patch Kz is defined to be the smallest Kz,0 thatleads to a unique pz.

Next, we define Kz at a boundary mesh vertex z. Let n0 be the smallest positiveinteger such that L(z, n0) has at least one internal mesh vertex. Then,

Kz , L(z, n0) ∪ Kz : z ∈ L(z, n0) and z is an internal vertex.This definition ensures the uniqueness of pz as shown in Lemma 3.6. Examples forpatches corresponding to boundary nodes are shown in Fig. 1.

If r = 1, then all mesh nodes are vertices and Gh uh is completely defined. Ifr > 1, let z ∈ Nh be a non-vertex node. Then, z may lie on an edge between twovertices, inside a mesh element, or inside a face of a tetrahedron in 3D. Formally,z ∈ τ= τ \ ∂τ where τ ⊆ T for some T ∈ Th and τ is a d1-simplex whose verticesare in Nh and d1 ≤ d. Let z0, . . . , zd1 be the vertices of τ , then

(Gh uh)(z) , 1d1 + 1

d1∑

i=0

∇pzi(z)

where pzi is the LSPA of uh at zi. This completes the definition of Gh uh for r > 1.


(a) d=2, r=1

(b) d=2, r=2

Fig. 1. Examples for patches used in the PPR

Extend Out

(a) d=2, r=1

Extend Out

(b) d=3, r=1

Fig. 2. Examples for extending L(z, 1) out, where z is an internal mesh vertex


2.2. The computational aspects of the PPR. Consider a mesh vertex z whosepatch is Kz, let v ∈ Sh, and let pz be the LSPA of v at z. We now outline the detailsof computing pz. Let z1, z2, . . . , zN(Kz) denote the mesh nodes in Kz. Without lossof generality, let z = z1 and set hz = max|zi − z1| : 2 ≤ i ≤ N(Kz). To avoid thecomputational instability resulting from small hz, the computations will be carriedout on the patch ωz where

ωz , Fz(Kz) and Fz : x ∈ Kz → x =x− z1

hz. (2.5)

The patch ωz will be called the reference patch associated with z. For 1 ≤ i ≤N(Kz), let zi = F (zi) and set vi = v(zi). Let pz ∈ Pr+1(ωz) be such that pz =pz Fz. For x ∈ ωz, pz(x) can be written in the form

pz(x) = p(x)T cz (2.6)

where cz =[cz,1 cz,2 cz,3 · · · cz,nr+1

]T,p(x) =

[p1(x) p2(x) · · · pnr+1(x)

]T, and

the set pl : 1 ≤ l ≤ nr+1 is the monomials-basis of Pr+1(Rd). In other words, ify = (y1, . . . , yd) ∈ Rd and 1 ≤ l ≤ nr+1, then pl(y) = yαl =

∏dj=1 y

αl,j

j , where αl =(αl,1, . . . , αl,d) is a multi-index. For technical reasons (see the proof Lemma 3.6),an ordering of these monomials must satisfy the condition

|αl| ≤ |αm| and αl,1 ≤ αm,1 ∀ 1 ≤ l ≤ m ≤ nr+1.

From the definition of pz, one can show that cz is the solution of the linear system

ATz Azcz = AT

z vKz (2.7)

where

Az ,

p(z1)T

p(z2)T

...p(zN(Kz))T

and vKz ,

v1

v2

...vN(Kz)

. (2.8)

Setting Bz = ATz Az, we have

cz = B−1z AT

z vKz . (2.9)

Note that Bz is a real symmetric positive-definite matrix of order nr+1 × nr+1.The following lemma lists the basic facts about discrete least-squares. For the

proof of this lemma, see [22, pp. 162-164].

Lemma 2.1. Let z ∈ Th be a mesh vertex whose patch is Kz. Then, the followingare equivalent.

1. pz is unique.2. Rank(Az) = nr+1.3. The matrix Bz is invertible.4. There exists no q ∈ Pr+1(Kz) such that q(zi) = 0 for i = 1, 2, . . . , N(Kz).

The equivalence of the first three claims in Lemma 2.1 is somewhat obvious, atleast from the definitions of pz, Az, and Bz. The equivalence of the first and thefourth claims says that fitting, by discrete least-squares, the values of some functionat the nodes in Kz by a polynomial of degree r + 1 leads to a unique polynomialpz if and only if the nodes in Kz do not lie on a curve (in 2D), or surface (in 3D),corresponding to a polynomial q of degree r + 1. This equivalence is our main toolin analyzing the uniqueness of discrete least-squares best fitting by polynomials.


2.3. The ZZ-PPR error estimator. Following the standard notations, let ηh

denote the ZZ-PPR error estimator of eh , u− uh. As usual, ηh is obtained fromthe error indicators ηT,h : T ∈ Th where

ηT,h ,√∫

T

|Gh uh −∇uh|2 and ηh ,√ ∑

T∈Th

η2T,h.

Note that the ZZ-PPR, like the ZZ-SPR, is obtained from the ZZ error estimatorby using the PPR-recovered gradient. By definition, ηT,h is a good approximationof |eh|1,T if and only if Gh uh is a good approximation of of ∇u in T .

2.4. Properties of the PPR. The PPR has the following properties.1. By definition, Gh is linear.2. Gh satisfies the consistency condition, i.e.,

Gh(Ihp) = ∇p∀ p ∈ Pr+1(Ω). (2.10)

This fact is proved in [26]. Consequently, Gh is a polynomial-preserving op-erator and enjoys the approximation property

‖∇u−Gh(Ihu)‖L2(Ω) . hr+1|u|r+2,Ω ∀u ∈ Hr+2(Ω).

For more details about proving this property, see [13, pp. 121-133]. Basically,the PPR can be viewed as a generator of finite difference formulas for firstorder partial derivatives. The generated formulas recover the exact derivativesof polynomials in Pr+1(Ω).

3. The PPR-recovered gradient may enjoy superconvergence. Establishing thisproperty is not straightforward. Beside the previous properties, and accordingto the general framework in [1, Ch. 4], Gh uh superconverges to ∇u if thefollowing conditions are satisfied.(a) The recovery operator Gh is bounded in the following sense:

‖Gh v‖L2(T ) . |v|1,KT∀T ∈ Th and ∀ v ∈ Sh (2.11)

whereKT ,

⋃Kz : z is a vertex of T

is the patch corresponding to T . This condition is called the boundednesscondition.

(b) ∇uh enjoys superconvergence in the following sense:

‖∇(Ihu− uh)‖L2(Ω) ≤ C(u)hr+ρ (2.12)

for some ρ ∈ (0, 1], i.e.,∇uh is a “perturbation” of∇(Ihu). This conditionis called the superconvergence condition.

If the conditions in (2.11) and (2.12) are satisfied, it is straight forward toshow that Gh uh superconverges to ∇u as

∇u−Gh uh = ∇u−Gh(Ihu) + Gh(Ihu− uh).

4. As in any recovery technique, if Gh uh superconverges to ∇u, the ZZ-PPR isasymptotically exact.

Generally speaking, both the boundedness condition and the superconvergencecondition rely on the mesh geometry. The superconvergence phenomenon has beenin the focus of research and there are many results in this direction especially in2D. We will go briefly over some of these results in Section 6. The main concern inthis paper is establishing the boundedness condition.


3. The Boundedness of Gh.

Definition 3.1. Let Th be a partition of Ω. Then, Th is said to satisfy the finitelayer condition in Ω (on ∂Ω) if there exists a constant n ∈ Z+, independent of hand z, such that

Kz ⊆ L(z, n)∀ z ∈ Ω(∂Ω).

Also, Th is said to satisfy the minimum singular value condition in Ω (on ∂Ω) if itsatisfies the finite layer condition in Ω (on ∂Ω) and if

σnr+1(Bz) ≥ C ∀ z ∈ Ω(∂Ω)

for some constant C > 0 that depends only on r and γ. If Th satisfies the finitelayer (the minimum singular value) condition in Ω and on ∂Ω, then it is said tosatisfy the finite layer (the minimum singular value) condition in Ω.

Remark 3.2. In the rest of this chapter we will implicitly assume that Th satisfiesthe finite layer condition in Ω. In practical meshes, this condition is easily satisfied.Indeed, our numerical experiments showed that Kz ⊆ L(z, 2) for all z ∈ Ω and ford=2,3. Also, L(z, 1) corresponding to a boundary mesh vertex z always has at leastone internal mesh vertex. Hence, Kz ⊆ L(z, 3) for all z ∈ ∂Ω.

Remark 3.3. Obviously, the number of patterns of the PPR patches correspondingto internal vertices in a translation-invariant mesh is finite. For example, there isonly one pattern for all the PPR patches corresponding to internal vertices inthe uniform mesh with the regular pattern; see Fig. 15(a). Consequently, thetranslation-invariant meshes satisfy the minimum singular value condition in Ω.

Remark 3.4. The minimum singular value condition is directly linked to the meshgeometry. Hence, it is better to find a geometric condition that implies the minimumsingular value condition. This is possible in 2D regular meshes as we shall see inSection 4. Unfortunately, establishing that in 3D regular meshes is not clear yet.

The main product of this section is Theorem 3.10 which needs some preparation.The next lemma says that the PPR patches in regular meshes are quasi-uniform.

Lemma 3.5. Let Th be a regular partition of Ω and let z be a mesh vertex whosepatch is Kz. Then, there exists a constant C = C(γ) > 0 such that

hT ≥ Chz ∀T ⊂ Kz.

Moreover, the number of elements in a patch corresponding to a particular meshvertex and the number of patches that contain a particular mesh element are boundedby constants depending only on γ.

Proof. This is a direct corollary of Definition 3.1 and [1, Theorem 1.6].

The next lemma says that the minimum singular value condition is satisfied on∂Ω if it is satisfied in Ω. This result grants us a license to focus our efforts in Ω.

Lemma 3.6. Let Th be a regular partition of Ω that satisfies the minimum singularvalue condition in Ω. Then, there exists a constant C = C(r, γ) > 0 such that

σ1(Bz) ≤ C

where z is any mesh vertex. Moreover, Th satisfies the minimum singular valuecondition on ∂Ω.


Proof. Let z ∈ Nh be a mesh vertex whose patch is Kz and whose reference patchis ωz. From the definition of Bz and the fact that ωz is contained in the unitball, it is easy to verify that |Bz(i, j)| ≤ N(Kz) + 1 for 1 ≤ i, j ≤ nr+1. Hence,σ1(Bz) ≤ σ1(|Bz|) ≤ cN(Kz) for some constant c > 0 that depends only on r.Since Th is regular, Lemma 3.5 implies that N(Kz) ≤ N0 < ∞ for some constantpositive integer N0 = N0(γ), and the first claim is true.

To prove the second claim, let z ∈ ∂Ω be a mesh vertex. By construction, Kz hasat least one internal mesh vertex y such that Ky ⊆ Kz. Set y = (y1, . . . , yd) = Fz(y)and set ω = Fz(Ky). Without loss of generality, assume that y is on the positivex1-axis; otherwise rotate ωz. After reordering the nodes in Kz if necessary,

Az =[

Ay

Az

]

where the rows in Ay correspond to the mesh nodes in ωy. Hence,

Bz = ATz Az = AT

y Ay + ATz Az = By + Bz

where By = ATy Ay and Bz = AT

z Az. Since Bz, By, and Bz are positive semi-definite,

σnr+1(Bz) ≥ σnr+1(By). (3.13)

Since Fz(·) =hy

hzFy(·) + y,

ωy =hy

hzωy + y. (3.14)

This relation induces a linear transformation Q : Pr+1(ωy) → Pr+1(ωy). Using themonomial bases of Pr+1(ωy) and Pr+1(ωy), Q = Q2Q1 where Q1, Q2 ∈ Rnr+1×nr+1 .

The matrix Q1 is diag(

p

(hy

hz

))and it represents the scaling part of (3.14), while

Q2 is a lower triangular matrix whose diagonal entries are 1’s and it represents thetranslation part of (3.14). Obviously, the nonzero entries in Q2 are monomials in y1,i.e., σnr+1(Q2) is a continuous function of y1. Since det(Q2) = 1 for all y1 ∈ [0, 1],σnr+1(Q2) & 1. Using the definitions of Q, By, and Ay, one can verify that

ATy = Q2Q1A

Ty ⇒ By = Q2Q1By(Q2Q1)T .

Therefore,

bT Byb ≥ σnr+1(By)‖QT1 QT

2 b‖2 ≥ σnr+1(By)σ2nr+1

(Q1)‖QT2 b‖2

≥ σnr+1(By)σ2nr+1

(Q1)σ2nr+1

(Q2)‖b‖2

where b ∈ Rnr+1 . By virtue of (3.13), we get

σnr+1(Bz) ≥ σnr+1(By) ≥ σnr+1(By)σ2nr+1

(Q1)σ2nr+1

(Q2).

Since Th satisfies the minimum singular value condition in Ω, then σnr+1(Q1) =(hy

hz

)nr+1

, and hy & hz by Lemma 3.5, the second claim is true.

The following two lemmas are the key tools in establishing the boundedness ofGh. The following lemmas are the key tools in establishing the boundedness of Gh:


Lemma 3.7. Let Th be a regular partition of Ω that satisfies the minimum singularvalue condition in Ω. Let z be a mesh vertex whose patch is Kz and let v ∈ Sh. If∫Kz

v = 0, then there is a constant C > 0 that depends only on γ and r such that

|v|1,Kz∼= h(d/2−1)

z ‖vKz‖.

Proof. The proof uses three facts. Firstly, mesh regularity and Lemma 3.5 imply

%T∼= hz ∀T ⊂ Kz. (3.15)

Secondly, for any T ∈ Th, there exists an invertible affine mapping FT : T → Twhere T is a reference element. Set vT = (v|T ) FT and let JT = DFT whereDFT ∈ Rd×d is the Jacobian of FT . Since Th is regular, then

hT . σd(JT ) ≤ σ1(JT ) . hT

| det(JT )| ∼= hdT

(3.16)

for all T ∈ Th. Note that N(T ) = N(T ) = nr, vT = (vT )T φ, φ =[φ1 φ2 · · · φnr

]T

,

and φi is the standard Lagrange-basis function associated with the ith node in Tfor i = 1, 2, . . . , nr. Finally, the third fact says that

| det(JT )| 12 σd(J−1T )|vT |1,T . |v|1,T . | det(JT )| 12 σ1(J−1

T )|vT |1,T (3.17)

for any T ∈ Th and for any v ∈ Sh. For a comprehensive discussion about theinequalities in (3.16) and (3.17), the reader is referred to [13, pp. 121-131].

Let T1, T2, . . . , TM(Kz) denote the mesh elements in Kz. Using (3.15)-(3.17),

|v|1,Tk∼= h(d/2−1)

z |vTk|1,T for k = 1, 2, . . . , M(Kz). (3.18)

It is easy to verify that

|vTk|21,T

= (vTk)T QvTk

for k = 1, 2, . . . ,M(Kz)

where Q ∈ Rnr×nr and Q(i, j) =∫

T∇φi · ∇φj for i, j = 1, 2, . . . , nr. Using (3.18),

|v|21,Kz∼= h(d−2)

z

M(Kz)∑

k=1

vTTk

QvTk. (3.19)

For 1 ≤ k ≤ M(Kz), there exists a Boolean matrix Ek ∈ Rnr×N(Kz) such thatvTk

= EkvKz , where

Ek(i, j) =

1 if node i in Tk is node j in Kz

0 otherwise

for 1 ≤ i ≤ nr and 1 ≤ j ≤ N(Kz). Therefore, (3.19) takes the form

|v|21,Kz∼= h(d−2)

z

M(Kz)∑

k=1

vTKz

(ETk QEk)vKz = h(d−2)

z vTKz

QvKz (3.20)

where Q ,M(Kz)∑

k=1

ETk QEk is a symmetric positive semi-definite matrix of order

N(Kz)×N(Kz). Also, Q has only one zero eigenvalue corresponding to 1N(Kz) as

|v|1,Kz = 0 ⇔ v|Kz is identically constant.

Since the eigenvectors of Q form an orthonormal basis of RN(Kz), we can write

vKz = v(1)Kz

+ v(2)Kz


where v(1)Kz

, c1N(Kz) for some constant c and v(2)Kz

is a combination of the rest of

the eigenvectors of Q. Consequently, v|Kz= v

(1)Kz

+ v(2)Kz

where v(1)Kz

= c and the

nodal values of v(2)Kz

are the entries of v(2)Kz

. Note that |c| = ‖v(1)Kz‖/

√N(Kz), and

∣∣∣∣∫

Kz

v(1)Kz

∣∣∣∣ =∣∣∣∣∫

Kz

v(2)Kz

∣∣∣∣as

∫Kz

v = 0. From the definitions, it is straightforward to verify that∣∣∣∣∫

Kz

v(1)Kz

∣∣∣∣ =‖v(1)Kz‖|Kz|√

N(Kz)and

∣∣∣∣∫

Kz

v(2)Kz

∣∣∣∣ ≤ ‖v(2)Kz‖|Kz|.

Therefore,‖vKz

‖2 = ‖v(1)Kz‖2 + ‖v(2)

Kz‖2 ≤ (1 + N(Kz))‖v(2)

Kz‖2.

Thus,

σ(N(Kz)−1)(Q)1 + N(Kz)

‖vKz‖2 ≤ σ(N(Kz)−1)(Q)‖v(2)

Kz‖2

≤(v

(2)Kz

)T

Q(v

(2)Kz

)

= vTKz

QvKz ≤ σ1(Q)‖vKz‖2.

(3.21)

Note that σ(N(Kz)−1)(Q) and σ1(Q) depend only on r and N(Kz). Since N(Kz) isbounded by Lemma 3.5, combining (3.19)-(3.21) concludes the proof.

Remark 3.8. Note that |v|1,Kz and ‖vKz‖ in Lemma 3.7 define norms of v|Kz .Since Sh ∩C(Kz) has a finite dimension, the conclusion of that lemma is expected.The value of Lemma 3.7 stems from the fact that the equivalency constant doesnot depend on hz and depends only on r and the geometry of Kz.

Remark 3.9. There is another way to establish the equivalence in Lemma 3.7 forlinear elements [20]. The idea is quite simple. In linear elements, ∇v|T in anyelement T ∈ Th is constant. The equivalence in Lemma 3.7 is easily achieved if onecan show that the PPR-recovered gradient at a mesh vertex z is equivalent to alinear combination of ∇v|T : T ⊂ Kz and the coefficients in this combination arebounded uniformly by a constant independent of z, h, and v.

Lemma 3.10. Let Th be a partition of Ω and let Th satisfy the assumptions inLemma 3.6. Let z be a mesh vertex whose patch is Kz and consider v ∈ Sh. Ifpz ∈ Pr+1(Kz) is the LSPA of v at z, then

|pz|1,∞,Kz ≤ Ch−d/2z |v|1,Kz .

where C is a bounded constant that depends only on r and γ.

Proof. At first assume that∫Kz

v = 0. Let ωz be the reference patch correspondingto z and let pz ∈ Pr+1(ωz) be such that pz = pz Fz. Let x be a point in Kz andset x = Fz(x). From (2.6) and from the fact that x ∈ ωz,

|∇pz(x)| = h−1z |∇pz(x)| . h−1

z ‖cz‖Using (2.9),

‖cz‖ ≤ σ1(B−1z )σ1(AT

z )‖vz‖ = (σnr+1(Bz))−1√

σ1(Bz)‖vz‖Hence, and by Lemma 3.6,

‖cz‖ . ‖vz‖


and, consequently,|∇pz(y)| . h−1

z ‖vz‖.Therefore, and by virtue of Lemma 3.7,

|∇pz(y)| . h−d/2z |v|1,Kz

.

Thus, the lemma conclusion is true if∫Kz

v = 0. If∫Kz

v 6= 0, set v = v− 1|Kz|

∫Kz

v

and let pz be the LSPA of v at z. Since∫Kz

v = 0, |pz|1,∞,Kz. h

−d/2z |v|1,Kz

. Thisconcludes the proof as ∇p = ∇p and ∇v = ∇v.

The following theorem is the main result of this section.

Theorem 3.11. Let Th be a partition of Ω that satisfies the assumptions in Lemma3.6. Let T be a mesh element whose patch is KT and let v ∈ Sh. Then,

‖Gh v‖L2(T ) ≤ C|v|1,KT.

where C is a constant that depends only on r and γ.

Proof. Let zi : 1 ≤ i ≤ nr be the mesh nodes in T such that z1, . . . , zd+1 are thevertices of T . Since (Gh v)|T ∈ Pr(T ), then

‖Gh v‖L∞(T ) . max|(Gh v)(zi)| : 1 ≤ i ≤ nr.By Lemma 3.10 and the definitions of Gh v at mesh nodes,

‖Gh v‖L∞(T ) . maxh−d/2zi

|v|1,Kzi: 1 ≤ i ≤ d + 1.

Hence,

‖Gh v‖L2(T ) ≤ ‖Gh v‖L∞(T )

√|T |

. maxhd/2T h−d/2

zi|v|1,Kzi

: 1 ≤ i ≤ d + 1.Since T ⊂ Kzi for all 1 ≤ i ≤ d + 1, hT ≤ hzi . Therefore,

‖Gh v‖L2(T ) . max|v|1,Kzi: 1 ≤ i ≤ d + 1 .

√√√√d+1∑

i=1

|v|21,Kzi

and the proof is complete by virtue of Lemma 3.5.

4. Verifying the Minimum Singular Value Condition in 2D Meshes. Wehave seen that Gh is bounded in the sense of (2.11) given that Th satisfies theminimum singular value condition in Ω. Obviously, this condition can not be usedin mesh construction. It is better if it is replaced with a geometric condition. Thekey observation to achieve that is the equivalence between the first and the fourthitems in Lemma 2.1 which links the uniqueness of LSPAs and the distribution of thenodes used in constructing these LSPAs. The authors [20] used this equivalenceto establish the boundedness of Gh for the case r = 1. In fact, it was foundthat the PPR is bounded in meshes satisfying what is called the angle condition.Surprisingly, this condition leads to the same conclusion when r > 1.

Definition 4.1. Let Th be a partition of Ω and let z be a mesh vertex. The layerL(z, 1) is said to satisfy the angle condition if the sum of any two adjacent anglesin L(z, 1) is at most π. The partition Th is said to satisfy the angle condition in Ωif L(z, 1) satisfies the angle condition at every mesh vertex in Ω.


(a) (b)

Fig. 3. L(z, 1) violates the Angle Condition

According to this definition, if z is an internal mesh vertex and L(z, 1) satis-fies the angle condition, then M(L(z, 1)) ≥ 4. Moreover, if M(L(z, 1)) = 4, thetriangles in L(z, 1) form a quadrilateral whose diagonals intersect at z; see Fig. 3.

Remark 4.2. Nodes of the type shown in Fig. 3(a) rarely happen in practicalmeshes and, if they happen, they can be identified and removed. Also, a nodeof the type shown Fig. 3(b) may be fixed by moving it to the intersection of thequadrilateral diagonals.

Remark 4.3. If z is a mesh vertex, one can show that L(z, 1) satisfies the anglecondition if and only if every pair of adjacent triangles in L(z, 1) forms either atriangle or a convex quadrilateral. Indeed, a triangulation Th satisfies the anglecondition in Ω if and only if any pair of adjacent triangles in Ω forms a triangle ora convex quadrilateral. Indeed, robust mesh generators target this type of meshes.

Proposition 4.4. Let m ∈ Z+. The following are true.1. Let p ∈ P∗m(R2) and let z0 ∈ R2. If p(z0) = 0, then p is 0 at any point on the

line through z0 and the origin.2. Let p ∈ Pm(R2), and consider the distinct points zi ∈ R2 : i = 1, 2, . . . , nm.

Assume that these points are distributed on m + 1 non-overlapping parallellines in such a way that the kth line has exactly k points for k = 1, 2, . . . ,m+1.If p(zi) = 0 for i = 1, 2, . . . , nm, then p is identically zero.

Proof. For the first claim, note that any point on the line through z0 and the originis a multiple of z0. For the proof of the second claim, see [11, pp. 64].

Let z be an internal mesh node and let zi = (x1,i, x2,i) : i = 1, 2, . . . , N(L(z, 1))be the mesh nodes in L(z, 1) such that the first M(L(z, 1)) + 1 nodes are the meshvertices in L(z, 1) and z1 = z. For (x1, x2) ∈ L(z, 1), define the polynomials

`j(x1, x2) = (x1 − x1,1)(x2,j+1 − x2,1)− (x2 − x2,1)(x1,j+1 − x1,1) (4.22)

∀ 1 ≤ j ≤ M(L(z, 1)), and

pz,` =∏`j : 1 ≤ j ≤ M(L(z, 1)), `j is not a multiple of `k ∀ k < j.

Note that `j = 0 is the equation of the line passing through the nodes z1 and zj+1.Also, note that pz,` is the product of linear polynomials representing non-collineartriangles sides meeting at z1 = z and that 2 ≤ deg(pz,`) ≤ M(L(z, 1)). The proofof the following theorem is in [20].


Theorem 4.5. Let z be an internal mesh vertex and assume that L(z, 1) satisfiesthe angle condition and has no degenerate triangles. Let p ∈ P2(L(z, 1)) suchthat p(zi) = 0 for i = 1, 2, . . . ,M(L(z, 1)) + 1. Then, p is identically 0 whenM(L(z, 1)) > 4 and is a multiple of pz,` if M(L(z, 1)) = 4.

Before we continue, let us go over some properties of the matrix

D ,

rr+1 rr · · · r2 r1

(r − 1)r+1 (r − 1)r · · · (r − 1)2 (r − 1)1...

.... . .

......

2r+1 2r · · · 22 21

1 1 · · · 1 1

(4.23)

where r ≥ 2. Note that D is a Vandermonde matrix of order r × (r + 1). Thismatrix will play a key rule in Lemma 4.8.

Proposition 4.6. Let D1 and D2 be the sub-matrices of D obtained by omittingthe r + 1st and rth columns, respectively. Then, Rank(D1) = Rank(D2) = r.

Proof. We start with D1 and proceed by contradiction. If Rank(D1) 6= r, then

there exists a non-zero polynomial q ∈ Pr+1(R) of the form q(ξ) = ξ2r−1∑

i=0

aiξi such

that q(j) = 0 for 1 ≤ j ≤ r. This is impossible as q has only r − 1 non-zero roots.Next, consider D2. If Rank(D2) 6= r, then there exists a non-trivial polynomial

q ∈ Pr+1(R) of the form q(ξ) = ξ

r∑

i=0

aiξi such that q(j) = 0 for 1 ≤ j ≤ r and

a1 = 0. If ar = 0, one can proceed as in the previous case. So, assume that ar 6= 0.Without loss of generality, set ar = 1. Using the zeros of q, we get

q(ξ) = ξ

r∑

i=0

aiξi = ξ

r∏

j=1

(ξ − j).

Comparing the two expressions of q leads to a contradiction as

0 = a1 = (−1)r−1(r!)r∑

j=1

1j6= 0.

The following corollary is a direct result of Proposition 4.6.

Corollary 4.7. The reduced-row echelon form of D is

1 ∗ · · · ∗ ∗ ∗0 1 · · · ∗ ∗ ∗...

.... . .

......

...0 0 · · · 1 ∗ ∗0 0 · · · 0 1 β

.

for some nonzero constant β that depends only on r.

We are now ready for the next crucial lemma.

Lemma 4.8. Let z be an internal mesh vertex and assume that L(z, 1) satisfies theangle condition and has no degenerate triangles. Let p ∈ Pr+1(L(z, 1)) such thatp(zi) = 0 for 1 ≤ i ≤ N(L(z, 1)). If deg(p) = r + 1 and deg(pz,`) ≤ r + 1, thenthere exits a polynomial q such that p = qpz,`; otherwise p is identically 0.


Proof. By Theorem 4.5, our claim is true for r = 1. So, we need to considerr ≥ 2. Note that any triangle in L(z, 1) has nr nodes arranged as in the secondpart of Proposition 4.4. Hence, p is identically 0 if deg(p) < r + 1. So, let usassume that deg(p) = r + 1. Without loss of generality, assume that z = z1 =

(0, 0). Then, p =r+1∑

k=1

pk where pk ∈ P∗k(L(z, 1)). We claim that pk(zi) = 0 for

i = 2, 3, . . . , M(L(z, 1)) + 1 and for k = 1, 2, . . . , r + 1. To prove this claim, fix2 ≤ i ≤ M(L(z, 1)) + 1. Using the definition of the Lagrange element of order r,the edge z1zi has r +1 uniformly spaced nodes of the form jzi/r for j = 0, 1, . . . , r.Since p is zero at these nodes, we have

r+1∑

k=1

(j

r

)k

pk(zi) = 0 for j = 1, 2, . . . , r.

Setting pk = rkpk, we get

r+1∑

k=1

jkpk(zi) = 0 for j = 1, 2, . . . , r.

This system of equations is actually a linear system of the form

D

pr+1(zi)pr(zi)

...p2(zi)p1(zi)

= 0r (4.24)

where D is the Vandermonde matrix defined in (4.23). By virtue of Corollary 4.7,there exists a constant β = β(r) 6= 0 such that p2(zi) + βp1(zi) = 0. This is truefor all 1 ≤ i ≤ M(L(z, 1)) + 1. Hence, the quadratic polynomial q = p2 + βp1

passes through all the vertices in L(z, 1). By assumption, p2 is a homogeneousquadratic polynomial. Since L(z, 1) satisfies the angle condition, Theorem 4.5 im-plies that either q is identically 0 when M(L(z, 1)) > 4 or q is a homogeneousquadratic polynomial when M(L(z, 1)) = 4. Consequently, p1 is identically 0 forall M(L(z, 1)) ≥ 4. Hence, one can drop p1(zi) in the linear system in (4.24).This leads to a homogeneous linear system whose coefficients matrix is D1. ByProposition 4.6, D1 is non singular. Hence, pk(zi) = 0 for k = 2, 3, . . . , r + 1 andi = 2, 3, . . . , M(L(z, 1)) + 1. By the first part of Proposition 4.4, pz,l is a factor ofpk for k = 2, . . . , r + 1. Consequently, pk is identically zero for k < deg(pz,l) andpk = qkpz,l for some polynomial qk when k ≥ deg(pz,l).

Theorem 4.9. Let z be an internal mesh vertex, and assume that L(z, 1) satisfiesthe angle condition and has no degenerate triangles. Let p ∈ Pr+1(L(z, 1)) suchthat p(zi) = 0 for i = 1, 2, . . . , N(L(z, 1)). Then, p is identically 0 for r ≥ 2.

Proof. Using Lemma 4.8, we may assume that deg(p) = r+1 and deg(pz,`) ≤ r+1.Based on M(L(z, 1)), which is at least 4 by the angle condition, we have 3 cases:

Case 1: M(L(z, 1)) > 4.M(L(z, 1)) > 4. By Lemma 4.8, p = `1q for someq ∈ P∗r(L(z, 1)) and `1 is defined in (4.22). Since M(L(z, 1)) > 4, L(z, 1) musthave at least 2 vertices on one side of the line through z1 and z2 as depicted inFig. 4(a). Therefore, L(z, 1) has a triangle T that intersects with the line through


z1 and z2 only at z1. Hence, `1(x1, x2) 6= 0 for all (x1, x2) ∈ T \z1. Consequently,q(z) = 0 ∀ z ∈ T . By the part 2 of Proposition 4.4, q, and hence p, is identically 0.

Case 2: M(L(z, 1)) = 4 and r = 2. Since deg(pz,`) = 2, Lemma 4.8 implies thatp = qpz,` for some q ∈ P1(L(z, 1)). Note that q must be 0 at the four mid-sidesof the quadrilateral formed with the triangles in L(z, 1). Since these nodes arenon-collinear (see Fig. 4(b)), q must be identically 0.

Case 3: M(L(z, 1)) = 4 and r > 2. Using Fig. 4(c), one can show that L(z, 1)has nr+1 nodes distributed as in part 2 of Proposition 4.4, i.e., p must be 0.

(a) case 1 (b) case 2 (a) case 3

Fig. 4. Various cases in the proof of Theorem 4.9

Corollary 4.10. Let z be an internal mesh vertex, and assume that L(z, 1) satisfythe angle condition and has no degenerate triangles. Then, Kz = L(z, 1) for allr > 1, i.e., Bz is invertible.

Proof. This is obvious from Lemma 2.1 and Theorem 4.9.

Theorem 4.11. Let Th be a regular partition of Ω, and assume that the sum of anytwo adjacent angles in Th is at most π. Then, Th satisfies the minimum singularvalue condition in Ω for all r > 1.

Proof. Consider an internal mesh vertex z ∈ Nh. By assumption, L(z, 1) satisfiesthe angle condition and has no degenerate triangles. Hence, Kz = L(z, 1) byCorollary 4.10. The regularity of Th implies that any triangle angle is at leastθ0 ∈ (0, π) where θ0 is a constant that depends only on γ defined in (1.3). Since thesum of the angles in any triangle is π, we have 0 < θ0 ≤ θ ≤ π− 2θ0 where θ is anyangle in any triangle in Th. Let Θz ∈ R3M(Kz) denote the vector of all the trianglesangles in Kz. Note that Θz ∈ [θ0, π − 2θ0]3M(Kz). Consider the set Sz defined by

Sz = Θz ∈ [θ0, π − 2θ0]3M(Kz) : L(z, 1) satisfies the angle condition.We claim that Sz is compact. This is true if Sz is a closed subset of [θ0, π −2θ0]3M(Kz) as the latter is compact. For that, proceed by contradiction and assumethat Sz is not closed. Then, there exists a convergent sequence Θ(m)

z : m > 0such that Θ(0)

z = limm→∞

Θ(m)z corresponds to L(z, 1)(0) that does not satisfy the

angle condition. Let L(z, 1)(m) correspond to the angles in Θ(m)z for m ≥ 1. Since

L(z, 1)(0) does not satisfy the angle condition, it has two adjacent angles θ(0)1 and

θ(0)2 such that θ

(0)1 + θ

(0)2 > π. Hence, there exists m0 > 0 such that

θ(m0)1 + θ

(m0)2 > π.


This is a contradiction as L(z, 1)(m0) satisfies the angle condition by assumption.Let ωz be the reference patch corresponding to z. Since ωz has at least one

mesh vertex z on the unit circle, ωz is completely defined by z and the angles inωz. Without loss of generality, assume that z = (1, 0). Then, Bz is a continuousfunction of all the angles in ωz. By the compactness of Sz and the angle condition,σnr+1(Bz) achieves a minimum value σ0 > 0. Note that σ0 > 0 depends only onM(Kz) which is finite by the regularity of Th. Hence, σ0 is independent of z, andthe proof is complete.

Corollary 4.12. Under the conditions in Theorem 4.11, Gh is bounded. Also,uniform regular mesh-refinement preserves the minimum singular value condition.

Proof. The first claim is a direct consequence of theorems 3.11 and 4.11. Thesecond claim follows directly from the fact that uniform regular mesh-refinementpreserves the angle condition. The proof of this fact is elementary and is based onDefinition 4.1 and Fig. 5.

(a) (b)

θ

θ

θ

θθ

θ

θ

θ

Fig. 5. Regular refinement preserves the angle condition

Remark 4.13. It is straight forward to generalize the arguments in this section to3D meshes. However, one needs to establish a 3D version of Theorem 4.5.

Remark 4.14. It is possible to relax the angle condition, but we avoided that inorder to simplify our arguments. For example, the angle condition is not needed ifL(z, 1) has three vertices lying on one straight line. Also, it is possible to allow thesum of two adjacent angles in L(z, 1) to exceed π, at most, at two of its vertices.

5. Boundedness of the PPR in Anisotropic Meshes. In establishing theboundedness of Gh, it was always assumed that the mesh is regular. This assump-tion may not be necessary as we shall see in the following example. Indeed, webelieve that Gh is bounded in anisotropic meshes.

Example 1. Consider the patch of triangles T shown in Fig. 6(a). Note that Tis a union of translations of the parallelogram in Fig. 6(b). This parallelogram hasa horizontal side of length 1 and its other side is of length a and makes an angleθ ∈ (0, π/2) with the positive x1-axis. We set a1 = a cos θ and a2 = a sin θ. Let T bea triangle in T and let KT be its corresponding patch as in Fig. 6(a). The nodes inKT are labelled as z1, z2, . . . , z12. Let v be a continuous function whose restrictionto any triangle in T is linear, and let G v denote its PPR-recovered gradient. Inthe rest of this example, i is either 1 or 2.

Let Gxi v denote the xi-component of G v. Note that Gxi v at any node z is aweighted sum of the nodal values of v in Kz. Since T is translation invariant, the set


(a)

and

θθθθ

(b) Dimensions

1

61-a16a2

0

0

!

-1

6-1+a16a2

" #$% $% $% $% $% $$% $% $% $$%% $& '$

1

3-1-2a16a2

-1

62+a16a2" #$% $% $% $% $% $$% $% $% $$%% $& '$

-1

31+2a16a2

" #$% $% $% $% $% $$% $% $% $$%% $& '$

1

6-2-a16a2

(c) The PPR weights

(1

(2

Fig. 6. Geometry, patches, and the PPR weights for Example 1

of weights does not change from one node to another. The values of these weightsare shown in Fig. 6(c). Note that the weight at any node has two components; theith component is for Gxi v. We claim that

‖Gxi v‖L2(T ) ≤ Ci‖∂xiv‖L2(KT ) for i = 1, 2 (5.25)

where C1, C2 are constants independent of a and θ. Establishing this claim willimply that the mesh regularity is not necessary for the boundedness of Gh. Toprove this claim, we need the following proposition.


Proposition 5.1. If ∂xiv = 0 in KT , then Gxi

v = 0 in T .

Proof. Without loss of generality, and to simplify the notations, we consider thecase i = 2. Similarly, one can treat the case i = 1. If (∂x2v)|KT

= 0, then v isconstant along the fibers of KT that are parallel to x2-axis. By a fiber parallel tox2-axis, or simply x2-fiber, we mean a line segment ` ⊂ KT such that ` is parallelto x2-axis and ∂` ⊂ ∂KT . We have one of two cases.

Case 1: θ = π/2. In this case, KT has 4 of its x2-fibers composed of edges in KT .Denote these fibers by `1, `2, `3, and `4. In this case, v|KT

is a linear combinationof the functions v(1), v(2), v(3), and v(4) where v(m), 1 ≤ m ≤ 4, is a continuousfunction whose restriction to any triangle in KT is linear and

v(m)(z) =

1 if z ∈ `m

0 otherwise

for all z ∈ KT . Using the weights in Fig. 6(c), one can show that (Gx2 v(m))|T = 0.Therefore, (Gx2 v)|T = 0.

Case 2: θ ∈ (0, π/2). In this case, none of the x2-fibers of KT is a union of edgesin KT . Hence, v|KT

= c1 + c2x1 for some real constants c1 and c2. Since the PPRsatisfies the consistency condition, (Gx2 v)|T = ∂x2(c1 + c2x1) = 0.

Let vk = v(zk) for 1 ≤ k ≤ 12 and let v = [ v1 v2 · · · v12 ]T . After some lengthyalgebraic manipulations, one can show that

‖Gxi v‖2L2(T ) =(a2)3−2i

864vT Aiv and ‖∂xiv‖2L2(KT ) =

(a2)3−2i

2vT Biv.

The matrices A1 and B1 are constant while the entries of A2 and B2 are quadraticpolynomials in a1. Indeed,

A2 = a21A1 + a1A2 + PA1P

T

andB2 = a2

1B1 + a1B2 + PA1PT .

The constant matrices A1, B1, A2, and B2 are listed in [19, pp. 80-81], while thematrix P is a permutation matrix obtained from I12 by interchanging the rows 1with 12, 2 with 10, 3 with 7, 4 with 11, and 5 with 9. Note that KT is invariantunder the reflection on the line connecting z6 and z8. This explains why P ispart of the formulas of A2 and B2. Also, note that Ai is positive semi-definiteand ‖Gxi v‖L2(T ) = 0 if and only if v ∈ Null(Ai). Similarly, ‖∂xiv‖L2(KT ) = 0 ifand only if v ∈ Null(Bi). Let Ei : R12 → Null(Bi), let mi denote the dimensionof Null(Bi), and set vi = v − Eiv. Note that vi is a linear combination of theeigenvectors of Bi that correspond to the nonzero eigenvalues. Taking these vectorsas columns of a matrix Ei ∈ R12×(12−mi), we write vi = Eivi. Hence, and by virtueof Proposition 5.1,

‖Gxi v‖2L2(T ) =(a2)3−2i

864vT

i Aivi =(a2)3−2i

864vT

i Aivi

and

‖∂xiv‖2L2(KT ) =(a2)3−2i

2vT

i Bivi =(a2)3−2i

2vT

i Bivi

where Ai = ETi AiEi and Bi = ET

i BiEi. Hence,

‖Gxi v‖2L2(T )

‖∂xiv‖2L2(KT )

=1

432fi(vi)


where fi : R12−mi → R is defined by

fi(vi) =vT

i Aivi

vTi Bivi

.

By definition, it is clear that Bi is nonsingular. Moreover, the maximum value offi is the maximum eigenvalue, λmax, of the generalized eigenvalue problem

Aivi = λBivi. (5.26)

It is clear that λmax is a function of a1. Consequently, (5.25) is true if there existsa constant λ > 0 such that λmax ≤ λ and λ is independent of a1. According to iand θ, we have the following three cases.

Case 1: i = 1, θ ∈ [0, π/2]. Since A1 and B1 are constant, A1 and B1 are constanttoo, i.e., the eigenvalues of (5.26) are constant (indeed λmax = 38). Hence,

‖Gx1 v‖L2(T ) ≤ C‖∂x1v‖L2(KT )

where C is independent of a and θ.Case 2: i = 2, θ = π/2. In this case a1 = 0 and hence A2 and B2 are constant.

Proceeding as in Case 1, there is a constant C > 0, independent of a, such that

‖Gx2 v‖L2(T ) ≤ C‖∂x2v‖L2(KT ).

Case 3: i = 2, θ ∈ (0, π/2). Getting the eigenvectors of B2 is very difficult in thiscase and one has to pursue another direction. Since θ ∈ (0, π/2), the second case inthe proof of Proposition 5.1 leads to ‖∂x2v‖L2(KT ) = 0 if and only if v|KT

= c1+c2x1

for some constants c1, c2 ∈ R. Consequently, if∫

KT

v = 0 and∫

KT

∂x1v = 0, (5.27)

then‖∂x2v‖L2(KT ) = 0 ⇔ v|KT

= 0.

Hence, the null space of the equations in (5.27), after writing them in terms of v, isthe same as the space spanned by the eigenvectors of B2 corresponding to nonzeroeigenvalues. Hence, the members of a basis of this null space can serve as thecolumns of a matrix E that is similar to E2 (see [19, pp. 83] for one possible formof E). Without loss of generality, set E2 = E. After some algebraic manipulations,the characteristic equation of A2v2 = λB2v2 is

0 = λ7 [g1(λ) + g2(λ)g3(a1)]

whereg1(λ) = 6λ3 − 281λ2 + 1806λ− 3136,

g2(λ) = 2λ3 − 90λ2 + 486λ− 702,

and

g3(a1) =a21(1 + a1)2

a61 + 3a5

1 + 6a41 + 7a3

1 + 6a21 + 3a1 + 1

.

The nonzero roots of g1(λ) + g2(λ)g3(a1) are perturbations of the roots of g1(λ) as‖g3‖L∞(R1) is very small (indeed |g3(a1)| < 0.15∀ a1 > 0). The roots of g1(λ) are72 and 65±√2881

3 and the perturbations are within 0.1. Hence,

‖Gx2 v‖L2(T ) ≤ C‖∂x2v‖L2(KT )

where C is a finite constant that is independent of a and θ.


6. Numerical Experiments. In this section we will go over some numerical ex-periments which will show that the PPR-recovered gradient enjoys superconver-gence and that the ZZ-PPR accurately measures eh. The focus will be on quadraticelements in 2D, and linear and quadratic elements in 3D. Also, a comparison is heldbetween the PPR and the SPR. This comparison is limited to 2D examples.

The efficiency of many gradient recovery techniques deteriorates near ∂Ω. Thissuggests that we should separately study the performance of the PPR inside Ω andnear ∂Ω. To do that, Nh is partitioned into Nh,1 ∪Nh,2 where

Nh,1 = z ∈ Nh : disti(z, ∂Ω) ≥ Hi for 1 ≤ i ≤ dfor some positive constants H1, . . . , Hd, and disti(·) denotes the distance along xi-axis. Accordingly, Ω is partitioned into Ω1,h ∪ Ω2,h where

Ω1,h =⋃T ∈ Th : T has all of its vertices in Nh,1.

Let A ⊆ Ω be a union of a set of mesh elements in Th. Then, the ZZ-PPR errorindicator in A is defined by

ηh,A = ‖Gh uh −∇uh‖L2(A).

We use the effectivity index κh,A to measure the accuracy of ηh,A, where

κh,A =ηh,A

‖∇(u− uh)‖L2(A).

To trace the accuracy of the ZZ-PPR in each of the mesh elements in A, we willuse the mean, µh,A, and the standard deviation, σh,A, of the effectivity indices inthese elements. If the ZZ-PPR is asymptotically exact in each of the elements inA, then lim

h→0µh,A = 1 and lim

h→0σh,A = 0. Recall that

µh,A , 1M(A)

∑

T⊂Aκh,T and σ2

h,A , 1M(A)

∑

T⊂A(κh,T − µh,A)2.

In all of the following examples, the nodes in Nh are partitioned using H1 = H2 =H3 = 1

8 , except for the last example where we use H3 = 14 .

Example 2. In this example, we consider the model problem −4u + 100u = 0 in Ω = (0, 1)2

u = gD on ∂Ω

and u =cosh(10x1) + cosh(10x2)

2 cosh(10). This function has a peak at the corner (1, 1),

and this peak may reduce the accuracy of uh near (1, 1). The finite element solu-tion uh is computed using the quadratic element, and Th is obtained by Delaunaytriangulation. The initial mesh is shown in Fig. 7. In successive iterations, the newmesh is obtained from the previous one by uniform regular refinement. Regularrefinement of a single triangle is depicted in Fig. 8. Note that the initial mesh inFig. 7 satisfies the angle condition in Ω. By Corollary 4.12, Gh is bounded. Theresults in Fig. 9(a) shows that |Ihu − uh|1,Ω has superconvergence. This explainsthe superconvergence in Gh uh. It is clear that the PPR is better than the SPRand that the PPR performs well in both Ω1,h and Ω2,h. Finally, Fig. 9(b) showsthat the ZZ-PPR is almost exact everywhere.

Remark 6.1. The slopes, or convergence rates, in the figures corresponding to theprevious example, and the next ones, are computed using the last two points. Inuniform meshes, the slope is rounded to the nearest whole number.


Fig. 7. The initial Delaunay mesh used inExample 2

Regular refinement

(a)

Regular refinement

(b)

Fig. 8. Regular refinement

Remark 6.2. A triangular mesh Th satisfies the condition (ρ1, ρ2) for some ρ1, ρ2 >0 if Th admits a partition consisting of two sets that satisfy the following. In thefirst set, every pair of adjacent triangles form an O(h1+ρ1) parallelogram for someρ1 > 0 (An O(hδ) parallelogram, δ > 0, is a quadrilateral in which the distancebetween the mid-points of its diagonals is O(hδ).) In the second set, the triangles’total area is O(hρ2) for some ρ2 > 0. The ideal value for ρ1 or ρ2 is ∞. Notethat ρ1 = ρ2 = ∞ if any edge in any triangle in Th is parallel to one of threefixed directions. This ideal case was covered in [10] where the authors completelydetermined the expansion of u− uh at mesh vertices. The case ρ1 = 2 and ρ2 = ∞is another interesting case that was studied in [17, 16], while the general case wastreated in [25]. If a triangular mesh Th satisfies the condition (ρ1, ρ2) for someρ1, ρ2 > 0 and if Sh consists of piecewise linear polynomials, then |Ihu − uh|1,Ω isO(h1+ρ) where ρ = 1

2 min(1, 2ρ1, ρ2); see [25] for the details.

Example 3. Let us now consider a 3D example. The model problem is −4u = 3π2 sin(πx1) sin(πx2) sin(πx3) in Ω = (0, 1)3

u = 0 on ∂Ω .

The solution of this problem is u(x1, x2, x3) = sin(πx1) sin(πx2) sin(πx3). We willsolve this problem using linear and quadratic elements on Delaunay meshes, andthe initial mesh is shown in Fig. 10. In successive iterations, the new mesh isobtained from the previous one by uniform regular refinement. Regular refinementof a single tetrahedron is depicted in Fig. 8(b); see e.g. [18] for a comprehensivediscussion about this kind of refinement. The numerical results are shown in Fig. 11and Fig. 12. In both the linear element and the quadratic element, |Ihu − uh|1,Ω

has superconvergence, which may explain the the good performance for the PPRand the ZZ-PPR. Unfortunately, the superconvergence in |Ihu−uh|1,Ω has not beenjustified theoretically yet.


(a)

(b)

Fig. 9. Results for Example 2


(a) Cross section (b) Inside the mesh

Fig. 10. The initial Delaunay mesh used in Example 3

(a)

(b)

Fig. 11. Results for Example 3 - linear element case


Example 4. Our final example is about 3D problems in which u has a singularity.The model problem is −4u = 0 in Ω = (−1, 1)2 \ [0.5, 1)2 × (0, 0.5)

u = gD on ∂Ω .

Using a cylindrical coordinate system at ( 12 , 1

2 , 0), the boundary condition gD ischosen such that

u(r, θ, z) = zr13 sin

2θ − π

3, π/2 ≤ θ ≤ 2π.

Note that u has an edge singularity along the edge connecting the vertices ( 12 , 1

2 , 0)and ( 1

2 , 12 , 1

2 ). This singularity, beside affecting the smoothness of u, creates pollu-tion error if the mesh near this edge is not sufficiently refined. The initial mesh usedin computing uh is shown in Fig. 13. This mesh is finer near the edge singularity toreduce the effect of the pollution error. The numerical results are shown in Fig. 14where we can see that the PPR continues to perform good and ZZ-PPR is almostexact every where.

6.1. The Performance of the PPR in Uniform Meshes. All of our previousexamples focused on Delaunay partitions. Let us now study the performance of thePPR when we switch to uniform meshes.

In 2D, we consider solving the problem in Example 2 using uniform meshesconstructed by dividing Ω into m ×m equal squares and each of these squares isdivided into two triangles arranged as in Fig. 15(a). In successive iterations, we usem = 16, 32, 64.

For the 3D case, we consider solving the problem in Example 3 using uniformmeshes similar to the one depicted in Fig. 15. To construct one of such meshes,divide Ω into m ×m ×m equal cubes, and then divide each of these cubes into 6tetrahedrons using Kuhn’s partition illustrated in Fig. 16. For our purposes, we usem = 8, 16, 32 in successive iterations. Before we continue, note that the uniformmeshes used in the 2D and the 3D cases are translation invariant, and hence thePPR is bounded in both cases.

Various convergence rates of the PPR-recovered gradient are listed in Table 1.From this table, we may observe the following:

1. The rates of ‖∇u − Gh uh‖L2(Ω) indicate that the PPR enjoys superconver-gence. This is expected because Gh is bounded and |Ihu− uh|1,Ω has super-convergence in all of the four cases (see [21] for the linear 2D case, [2] for the2D quadratic case, [14] for the 3D linear case, and [12] for the 3D quadraticcase).

2. Gh uh superconverges to ∇u at all mesh nodes in the linear element cases. Aproof of this result in 2D is given in [26], and the argument in this proof canbe extended to cover the 3D linear case.

3. Gh uh ultra-converges to ∇u at all internal mesh nodes in the quadratic ele-ment cases. This result has no theoretical justification even in 2D.


(a)

(b)

Fig. 12. Results for Example 3 - quadratic element case


Fig. 13. The initial Delaunay mesh used in Example 4

(a)

(b)

Fig. 14. Results for Example 4


(a) Cross section (b) Inside the mesh

Fig. 15. A uniform partition of Ω = (0, 1)3

Partition the cube into two prisms

Partition each prism into 3 tetrahedrons

Fig. 16. Kuhn’s partition of the unit cube

Example Element

∞ Ω

∇ −

Ω∇ −

∞ Ω∇ − !

"#$ $ %& &

Ω∇ −

Linear 2 2 2 2 2

Quadratic 4 3 3 3

Linear 2 2 2 2 3

Quadratic 4 3 3 3

Table 1. Convergence rates for the PPR-recovered gradient in uniform meshes

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Received October 2004; revised February 2005.E-mail address: [email protected]

E-mail address: [email protected]

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