introduction to the conforming and nonconforming ﬁnite...

Introduction to the conforming and nonconformingfinite element methods

Dongwoo Sheen

Graduate School of InformaticsKyoto University

Kyoto 608–8501, Japan

Email: [email protected]://www.nasc.snu.ac.kr

July 14, 2015: 13:14

Finite element methods Section 0.0

c©2012–2015 Dr. Dongwoo Sheen

All rights reserved.

c©Dongwoo Sheen, Ph.D. (http://www.nasc.snu.ac.kr) Page 2 of 65

Contents

1 Introduction to conforming and nonconforming finite elements 51.1 The model problem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.1.1 Weak formulation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.1.2 Unique Solvability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.1.3 Elliptic regularity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.1.4 Equivalence of three formulations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

1.2 Finite Element Methods . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.1 Th: triangulation of Ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.2.2 Construction of FE space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.2.3 Compute FEM solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.2.4 Convergence study: Galerkin method . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.2.5 Finite elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3 Construction of basis functions and calculation of matrix components . . . . . . . . . . . . . 141.3.1 The 2–dimensional P1 triangular conforming element (the Courant element) and the Crouzeix–Raviart P1.3.2 Calculation of matrix components . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.3.3 Implementation of P1 conforming triangular element on a uniform grid . . . . . . . . . 16

1.4 Higher order finite elements in Rd . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4.1 Pm (conforming) simplicial element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4.2 Qm (conforming) d-linear element . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4.3 Unisolvency and optimality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191.4.4 Triangulations and reference elements . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

1.5 Assembly of matrix Akj =∑

K∈Thah(φj , φk) . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2 NC (Nonconforming) finite elements 252.1 P1 triangular NC element (Crouzeix-Raviart, 1973) and the rotated Q1-rectangular NC elements 262.2 The P1-NC quadrilateral and hexahedral elements . . . . . . . . . . . . . . . . . . . . . . . . 28

2.2.1 Error analysis for linear nonconforming Galerkin method . . . . . . . . . . . . . . . . 292.2.2 Implementation of P1–NC quadrilateral element . . . . . . . . . . . . . . . . . . . . . 30

2.3 Hermite–type finite elements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3 Preliminaries from functional analysis 393.1 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 403.2 Annihilators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.3 Big Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 423.4 Compact operators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

4 Sobolev Spaces 454.1 Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 454.2 Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47

4.2.1 Schwartz space S(Rd) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

3


4.2.2 Fourier transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 504.2.3 Prolongation (Extension) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 524.2.4 The definition of traces and trace theorems . . . . . . . . . . . . . . . . . . . . . . . . 534.2.5 Application of Trace Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56

4.3 The embedding theorems and fractional Sobolev spaces . . . . . . . . . . . . . . . . . . . . . 574.4 Application to elliptic problems and finite element subspace of H1(Ω) . . . . . . . . . . . . . 60

5 Appendix 615.1 Brief review on mathematical analysis and linear algebra . . . . . . . . . . . . . . . . . . . . . 61

c©Dongwoo Sheen, Ph.D. (http://www.nasc.snu.ac.kr) 4

Chapter 1

Introduction to conforming andnonconforming finite elements

1.1 The model problem

Throughout the lecture F denotes the real number field R or the complex number field C. Let Ω ⊂ Rdbe a bounded open region with smooth or polygonal boundary ∂Ω. Consider the following boundary valueproblem:given f ∈ H−1(Ω) find u ∈ H10 (Ω) such that

L u := −∇·A(x)∇u+ b(x)u = f, Ω, (1.1a)u = 0, ∂Ω. (1.1b)

• Uniform ellipticity assumption: A(·) is a uniformly positive–definite d × d matrix–valued functiondefined on Ω: i.e.,

〈A(x)ξ, ξ〉 ≥ c0|ξ|2 ∀ξ ∈ F d a.e. x ∈ Ω; (1.2a)b(x) ≥ 0 a.e. xΩ. (1.2b)

1.1.1 Weak formulation

Multiply (1.1a) by an arbitrary v ∈ H10 (Ω) and use the Divergence Theorem, keeping in mind (1.1b). Thenwe have the following weak problem: to find u ∈ H10 (Ω) such that

(A∇u,∇v) + (bu, v) = 〈f, v〉H−1,H10 ∀v ∈ H10 (Ω). (1.3)

Here, and in what follows, standard notations from functional analysis are used.

• The duality paring 〈·, ·〉V ′,V , where V and V ′ are a normed linear space V and its dual space V ′,respectively. The corresponding norms are denoted by ‖ · ‖V and ‖ · ‖V ′ . The duality is defined as〈f, v〉V ′,V = f(v) for all v ∈ V such that f : V 7→ F is a continuous linear functional on V. The dualnorm is given by

‖f‖V ′ = sup0 6=v∈V

〈f, v〉V ′,V‖v‖V

,

for all f ∈ V ′.• The space of square integrable functions L2(Ω) on Ω is denoted by

L2(Ω) = {v : Ω → F :∫

Ω

|v(x)|2 dx


with the L2(Ω)–inner product:

(u, v) =

∫

Ω

u(x)v(x) dx ∀u, v ∈ L2(Ω)

and the L2(Ω)–norm:

‖v‖ =√∫

Ω

|v(x)|2 dx ∀v ∈ L2(Ω).

• The Sobolev spaces, Hm(Ω) = {v ∈ L2(Ω) : ‖v‖m


Lemma 1.1 (The Lax-Milgram Lemma). Assume the following

1. V is a Hilbert space;

2. a(·, ·) is continuous on V , i.e. there exists C > 0 such that

|a(u, v)| ≤ C‖u‖V ‖v‖V ∀u, v ∈ V ;

3. a(·, ·) is V –coercive on V , i.e. there exists α > 0 such that

|a(u, u)| ≥ α‖u‖2V ∀u ∈ V ;

4. ℓ(·) is continuous on V , i.e. there exists M > 0 such that

|ℓ(v)| ≤M‖v‖V ∀v ∈ V.

Then, there exists a unique solution u ∈ V to (1.4) such that

‖u‖V ≤M

α‖ℓ‖V ′ .

Exercise 1.1. Prove the above Lax–Milgram Lemma.

1.1.3 Elliptic regularity

If the domain Ω and the coefficients A(x) and b(x) are smooth enough and f ∈ L2(Ω), then the abovesolution u ∈ H10 (Ω) to (1.4) belongs to H2(Ω) ∩H10 (Ω).

Exercise 1.2. Show that (c · ∇u, u) = − 12 ((∇·c)u, u) for all u ∈ H10 (Ω) and c ∈ L∞(div; Ω). Here,L∞(div; Ω) =

{f ∈ [L∞(Ω)]d | ∇·f ∈ [L∞(Ω)]

}.

Exercise 1.3. Consider the problem:

−∇·A(x)∇u+ c(x) · ∇u+ b(x)u = f, Ω, (1.5a)u = 0, ∂Ω, (1.5b)

where A and b satisfy the above assumptions (1.2) and c(x) fulfills the following assumption:

1

2∇·c(x) ≤ b(x) a.e. ∈ Ω.

Then, using the above Lax–Milgram Lemma, show that the above problem (1.5) has a unique solution inH10 (Ω).

1.1.4 Equivalence of three formulations

Introduce the quadratic functional E : V → R given by

E(v) =1

2a(v, v)− 〈f, v〉V ′,V =

1

2a(v, v)− ℓ(v), (1.6)

where E is called a complementary potential energy.Consider the minimization problem:

minv∈V

E(v), (1.7)

or equivalently, find u ∈ V such thatE(u) ≤ E(v) ∀v ∈ V.

The following lemma is useful.



Lemma 1.2. If w is continuous on Ω and

∫

Ω

w v dx = 0 ∀v ∈ H10 (Ω).

Then w(x) = 0 for all x ∈ Ω.

Proof. Exercise. Hint: use ǫ − δ definition of continuous functions and C∞0 (Ω) ⊂ H10 (Ω), where C∞0 (Ω)denotes the space of infinitely differentiable functions with compact support in Ω. The support supp(v) of vmeans the set {x ∈ Ω | v(x) 6= 0}. Since Ω is a subset of Rd, the compactness means supp(v) is bounded. �

Remark 1.1. Recall elliptic regularity for higher dimensional elliptic problems: Let u be the solution ofa(u, v) = (f, v), v ∈ V , where

a(v, w) =d∑

i,j=1

(aij

∂v

∂xi,∂w

∂xj

)+ (c v, w)

V = {v ∈ H1(Ω) : v = 0 on ∂Ω}If Ω is a convex polygonal domain and f ∈ L2(Ω), then the variational solution u ∈ H2(Ω)∩H10 (Ω), and

‖u‖2,Ω ≤ C ‖f‖0,Ω [Gri85] (1.8)

If ∂Ω is smooth, and f ∈ Hk(Ω), then the variational solution u ∈ Hk+2(Ω) ∩H10 (Ω), and

‖u‖k+2,Ω ≤ C(k,Ω) ‖f‖k,Ω for k = 0, 1, · · · (1.9)

Theorem 1.1. Let u, v ∈ H1(Ω) and A ∈ L∞(Ω). Moreover, assume that ∇·(A∇u) ∈ L2(Ω). Then thenormal trace ν · (A∇u) ∈ H−1/2(∂Ω), and the following integration by parts formula holds:

(A∇u,∇v) = −(∇·(A∇u), v) + 〈ν · (A∇u), v〉H−1/2(∂Ω),H1/2(∂Ω) ∀v ∈ H1(Ω). (1.10)

Theorem 1.2. The three problems (1.1), (1.4) and (1.7) are equivalent under a sufficient regularity conditionsuch as ∇·(A∇u) ∈ L2(Ω) and f ∈ L2(Ω).

Proof. Step 1. (1.1)⇒(1.4). This was already shown above.Step 2. (1.4)⇒(1.1). Assume that v ∈ V satisfies (1.4). Generally, this doesn’t hold. But, if ∇·A∇u ∈

L2(Ω), one can use the integration by parts Theorem 1.1. Thus,

∫

Ω

(L u− f) v dx = 0, ∀v ∈ V.

By the above Lemma 1.2,L u− f = 0 in V ′.

Step 3. (1.4)⇒(1.7). Suppose u is a solution of (1.4). Then, for an arbitrary v ∈ V , we have

E(v)− E(u) = 12a(v, v)− ℓ(v)−

[1

2a(u, u)− ℓ(u)

]

=1

2(a(u, u)− 2a(u, v) + a(v, v))

=1

2a(u− v, u− v) ≥ 0,

where we used a(u, u− v) = ℓ(u− v), since u solves (1.4).



Step 4. (1.7)⇒(1.4). Suppose u is a solution of (1.7). Then for any ǫ ∈ R,

E(u) ≤ E(u+ ǫv) ⇐⇒ g(ǫ).

Since g(ǫ) has a minimum at ǫ = 0, g′(0) = 0. Therefore,

g′(0) =d

dǫ

[1

2a(u+ ǫv, u+ ǫv)− 〈f, u+ εv〉V ′,V

]∣∣∣∣ǫ=0

= a(u, v)− 〈f, v〉V ′V = 0.

�

Theorem 1.3. A solution to (1.4) is unique if it exists, under the stated assumptions.

Proof. Suppose u1 and u2 ∈ V solve (1.4). Then

a(u1, v) = 〈f, v〉V ′,V , (1.11)a(u2, v) = 〈f, v〉V ′,V for all v ∈ V. (1.12)

By subtracting, a(u1 − u2, v) = 0 for all v ∈ V .Choose v = u1 − u2 ∈ V . Then

a(u1 − u2, u1 − u2) = 0, (1.13)

from which by the V –coercivity of a, ‖u1 − u2‖V ≤ Ca(u1 − u2, u1 − u2) = 0 Hence, u1(x) − u2(x) ⇐⇒constant for all x ∈ Ω. Since u1(0)− u2(0) = 0, u1(x)− u2(x) = 0, for all x ∈ Ω. �

1.2 Finite Element Methods

The finite element methods to solve (1.3) usually consist of the following procedure.

1. (Th)h: Triangulation of the domain Ω

2. (Vh)h: Construction of FE space: Vh → V as h→ 03. FE solution: Find uh ∈ Vh for each h4. Error Analysis: Convergence of uh → u as h→ 05. Numerical Simulation

1.2.1 Th: triangulation of Ω

Decompose Ω into a union of finite number of elementary geometry

y

Standard 2−dimensional FEM mesh−grid

(0,1)(1,1)

(0,0)

x

(1,0)

y

Standard 2−dimensional FEM mesh−grid

(0,1)(1,1)

(0,0)

x

(1,0)



Example of Two-Dimensional Mesh Example of Conforming Two−Dimensional Mesh

1.2.2 Construction of FE space

Based on Th construct finite dimensional subspace Vh

Piecewise linear basis function at (x , y ) i j

(x ,y )i j

(x ,y )i-1 j (x ,y ) i+1 j

(x ,y )i+1 j+1

(x ,y )i j+1

(x ,y )i j-1

(x ,y )i-1 j-1

1

2

3

4

5

6

Example of Conforming Two−Dimensional Mesh

• Let Vh ⊆ V = H10 (Ω) given by

Vh = {v ∈ C0(Ω) : v|K is piecewise linear for all K ∈ Th}

for triangular decomposition Th or

Vh = {v ∈ C0(Ω) : v|K is piecewise bilinear for all K ∈ Th}

for quadrilateral decomposition Th

• In general, the function spaces are based on piecewise polynomials such that v|K ∈ Pℓ(K) for someℓ ≥ 0.

• Let ϕj , j = 1 · · · , N, be the basis functions for Vh

Vh = Span{ϕj , j = 1, · · · , N}

= {v =N∑

j=1

cjϕj : cj ∈ F , j = 1, · · · , N}



• Let ψj , j = 1 · · · ,M, be the basis functions for Wh

Wh = {w =M∑

j=1

cjψj : cj ∈ F , j = 1, · · · ,M}

1.2.3 Compute FEM solution

1. Find uh ∈ Vh such that

a(uh, v) = ℓ(v) ∀v ∈Wh. (1.14)

2. Find uh =∑N

j=1 αjϕj such that

N∑

j=1

a(ϕj , v)αj = ℓ(v) ∀v ∈Wh.

3. Find uh =∑N

j=1 αjϕj such that

N∑

j=1

a(ϕj , ϕk)αj = ℓ(ϕk) ∀ϕk ∈Wh.

4. Denote b = (ℓ(ϕ1), · · · , ℓ(ϕM ))t α = (α1, · · · , αN )t; Akj = a(ϕj , ϕk), k = 1, · · · ,M, j = 1, · · · , N

5. Solve the M ×N linear system:

Aα = b. (1.15)

6. If M > N, α minimizes

‖Aα′ − b‖

• Vh: solution space, Wh: test function space

• Galerkin method (Ritz method) if Vh=Wh in (2)

• Petrov-Galerkin method: if Vh 6=Wh in (2)

• If either Vh or Wh is not a subspace of H10 (Ω), Nonconforming FEM.

Linear solvers

1. Direct method: Gaussian elimination (LU-decomposition) method, Cholesky decomposition method,QR-decomposition method

2. Iterative method: Gauss-Jacobi and Gauss-Seidel methods, SOR, Richardson iteration method, Gra-dient method, Conjugate Gradient Method, BiConjugate Gradient Method, Domain DecompositionMethod, MultiGrid solver,....



1.2.4 Convergence study: Galerkin method

Error equation

Subtracting (1.3) from (1.14), the error equation holds:

a(uh − u, ϕ) = 0 ∀ϕ ∈ Vh. (1.16)

Lemma 1.3 (Ceá’s lemma).

‖uh − u‖1 ≤C

αinf

v∈Vh‖v − u‖1. (1.17)

In particular, v = Phu, “the Vh-interpolation of u”, leads to

‖uh − u‖1 ≤C

α‖u− Phu‖1. (1.18)

Proof. From (1.16), coercivity, and boundedness; for any v ∈ Vh,

α‖uh − u‖21 ≤ a(uh − u,uh − u)= a(uh − v + (v − u),uh − u) = a(v − u,uh − u)≤ C‖v − u‖1‖u− uh‖1.

Therefore, by dividing by u − uh‖1, one gets (1.17). In particular, v = Phu, “the Vh-interpolation of u”,leads to (1.18) �

By the interpolation property ‖u− Phu‖1 ≤ C1h|u|2,

‖uh − u‖1 ≤ C1C

αh|u|2.

Hence, FE error : ellipticity * approximation properties in the FE space

Duality argument for L2(Ω) error estimate

The aim is to show

‖uh − u‖ ≤ Ch2|u|2,

under elliptic regularity conditions.

In order to obtain an L2-error estimate, the following inclusions and identifications are used:

H10 (Ω) →֒ L2(Ω) ⇐⇒ L2(Ω)′ →֒ H−1(Ω).

If φ ∈ L2(Ω), obviously φ ∈ H−1(Ω). Then, for all v ∈ H10 (Ω), one has

〈φ, v〉H−1(Ω),H10 (Ω) = (φ, v)L2(Ω).

We now state and prove the following lemma:



Lemma 1.4 (Aubin-Nitsche Lemma). Let H be Hilbert space with the norm ‖ · ‖H and V be a subspaceof H which is a Hilbert space with the norm ‖·‖V . Suppose V →֒ H is continuous. Then for any subspaceVh ⊂ V fulfilling

a(uh − u, v) = 0 v ∈ Vh, (1.19)

the following estimation is valid:

‖uh − u‖H ≤ C‖uh − u‖V supg∈H

infv∈Vh ‖v − ϕg‖V‖g‖H

,

where ϕg ∈ V is the unique solution of

a(w,ϕg) = 〈g, w〉V ′,V ∀w ∈ V, (1.20)

where the bilinear and liner forms satisfy the assumptions in the Lax-Milgram Lemma.

Proof. Let g ∈ H ⇐⇒ H ′ →֒ V ′. In (1.20), the choice of w = uh − u leads to for any v ∈ Vh, by (1.19)

| 〈g, uh − u〉V ′,V | = |a(uh − u, ϕg)| = |a(uh − u, ϕg − v)|≤ C ‖uh − u‖V ‖ϕg − v‖V .

Now, exploiting the Gelfand triplet: (g, uh − u)H = 〈g, uh − u〉V ′,V for g ∈ H, one has

‖uh − u‖H = supg∈H

(g, uh − u)H‖g‖H

≤ C ‖uh − u‖V supg∈H

infv∈Vh ‖ϕg − v‖V‖g‖H

.

This proves the lemma. �

Corollary 1.1. Let Th be shape-regular and u ∈ V ⇐⇒ H10 (Ω) be the solution of a(u, v) = 〈f, v〉V ′,V forevery v ∈ H10 (Ω). Then,

‖uh − u‖L2(Ω) ≤ Ch ‖uh − u‖H1(Ω) .

Moreover, assume that u ∈ H2(Ω) ∩H10 (Ω) satisfies a(u, v) = 〈f, v〉V ′,V for every v ∈ H2(Ω), then

‖uh − u‖0 ≤ Ch2 |u|2 .

Proof. Set H = L2(Ω) and V = H10 (Ω) in the Aubin-Nitsche Lemma, with ‖ · ‖H = ‖·‖0 and ‖·‖V = ‖·‖1respectively. Notice that H10 (Ω) →֒ L2(Ω) is continuous. Thus by the Aubin-Nitsche Lemma and ellipticregularity,

‖uh − u‖0 ≤ C ‖uh − u‖1 supg∈L2(Ω)

(infv∈Vh ‖ϕg − v‖1

‖g‖0

)

≤ Ch ‖uh − u‖1 ≤ Ch2|u|2.

�

1.2.5 Finite elements

Finite element:(K,PK ,ΣK)

1. K: “element”–geometric object, triangle, quadrilateral, simplex, hexahedron, ....

2. PK : “local finite element space”–vector space of polynomials defined on K



3. ΣK : “DOFs”–(Degrees of freedom) to determine an element p ∈ PK uniquely

• The Courant element or P1 triangular element and the bilinear element or the Q1 element.

��

��

��

V1

V2

V3

T

K = T

PK = Span{1, x1, x2}

ΣK = {φ(Vj), j = 1, 2, 3}e1

e2

e3

e2

V3

V1V2

V4

e3

e4

R e1

K = R

PK = Q1

ΣK = {φ(Vj), j = 1, · · · , 4}

• P2 triangular element and Q2-rectangular element

��

��

��

��

��

��

V1

V2

V3

T

K = T

PK = P2

ΣK = {φ(Vj), φ(mj), j = 1, 2, 3}

m3

e3

m1

e1

m2e2

e2m2(0, 1)

V3

V1V2

V4

e3

e4

m4(0,−1)

m3(−1, 0)m1(1, 0)

R e1

PK = Q2

K = RO

j = 1, · · · , 4;φ(O)}ΣK = {φ(Vj), φ(j),

Here, and in what follows, the following notations for polynomial spaces are used:

Pk = Span{xα11 · · ·xαdd | α1 + · · ·+ αd ≤ k, αj ≥ 0}, (1.21)Qk = Span{xα11 · · ·xαdd | 0 ≤ αj ≤ k}. (1.22)

1.3 Construction of basis functions and calculation of matrix com-ponents

1.3.1 The 2–dimensional P1 triangular conforming element (the Courant ele-ment) and the Crouzeix–Raviart P1 nonconforming element

Consider a triangle K with vertices Vj = (pj , qj), j = 1, 2, 3. Let Pj = (xj , yj), j = 1, 2, 3, be three pointson the triangle K. We consider the two cases:

1. Pj = Vj , j = 1, 2, 3 : the P1 triangular conforming element

2. Pj = Mj =Vj−1+Vj

2 , j = 1, 2, 3 : the Crouzeix–Raviart P1 nonconforming element

The basis function φ(j, x, y) associated with Pj is constructed as follows:

φ(j, x, y) =(x− xk)(yk − yl)− (xk − yl)(y − yk)(xj − xk)(yk − yl)− (xk − xl)(yj − yk)

, {j, k, l} = {1, 2, 3}. (1.23)



Then obviously, φ(j,Pk) = δjk, where δjk is the Kronecker delta. The gradient of φ(j, x, y) is given asfollows:

∇φ(j, x, y) = 1(xj − xk)(yk − yl)− (xk − yl)(yj − yk)

(yk − yl

−(xk − yl)

), {j, k, l} = {1, 2, 3}. (1.24)

The area of K is given by the cross product of vectors:

|K| = 12|(V1 −V3)× (V2 −V3)| =

1

2|V1 −V3| |V2 −V3|| sin θ| =

1

2

∣∣∣∣∣∣det

1 1 1p1 p2 p3q1 q2 q3

∣∣∣∣∣∣, (1.25)

where θ is the angle between the two vectors.

1.3.2 Calculation of matrix components

Now let us compute the matrix component Akj = a(φj , φk) in (1.15).

Akj = a(φj , φk) =

∫

Ω

(A∇φj) · ∇φk + b(x)φjφk dx

=∑

K∈Th

∫

K

(A(x)∇φj) · ∇φk + b(x)φjφk dx

=:∑

K∈Th

aK(φj , φk) =:∑

K∈Th

AKkj .

Exercise 1.4. (Due the start of class of July 3) Submit the program code with input data, if there is any,and analysis.

Consider the elliptic equation (1.1) with discontinuous coefficient. The domain is taken as Ω = (0, 1)2

with Ω− =(14 ,

34

)2and the coefficients A(x) = aχΩ−(x)I + χΩ\Ω−(x)I and b(x) = 1 for all x ∈ Ω. Here, χ

is the characteristic function and I denotes the identity matrix. Define

φ(x) =

{x− x2, x ∈ [0, 14 ] ∪ [ 34 , 1],316 +

1a

[116 − (x− 12 )2

], x ∈ [ 14 , 34 ].

Then set f(x, y) = −∇·(A(x, y)∇(φ(x)φ(y)) + φ(x)φ(y) Try with a = 1100 . Divide Ω into Nx ×Ny uniformrectangles and then divide each rectangle into two triangles with the diagonal from the top left to the bottomright. Try with Nx ×Ny = 50× 50, 100× 100, 200× 200. In each case, calculate the L2(Ω)–error

√∫

Ω

|uh(x, y)− u(x, y)|2 dxdy, h = 1/50, 1/100, 1/200.

The linear system can be solved by using the multi-grid method or the Conjugate Gradient Method.

Let u(x, y) = U(r, θ). Then we have

(∂U∂r

1r∂U∂θ

)=

(cos θ sin θ− sin θ cos θ

)(∂u∂x∂u∂y

)(1.26)

Hence,

(∂u∂x∂u∂y

)=

(cos θ − sin θsin θ cos θ

)(∂U∂r

1r∂U∂θ

). (1.27)



Thus,

∆u =

( ∂∂x∂∂y

)·(∂u

∂x∂u∂y

)=

[(cos θ − sin θsin θ cos θ

)(∂∂r1r

∂∂θ

)]·(cos θ − sin θsin θ cos θ

)(∂U∂r

1r∂U∂θ

)(1.28)

Or,

(∂u∂x∂u∂y

)=

(cos θ ∂U∂r − sin θr ∂U∂θsin θ ∂U∂r +

cos θr

∂U∂θ

)(1.29)

Consequently,

∆u = =

[cos θ

∂

∂r− sin θ

r

∂

∂θ

] [cos θ

∂U

∂r− sin θ

r

∂U

∂θ

]+

[sin θ

∂

∂r+

cos θ

r

∂

∂θ

] [sin θ

∂U

∂r+

cos θ

r

∂U

∂θ

](1.30)

=∂2U

∂r2+

1

r

∂U

∂r+

1

r2∂2U

∂θ2=

1

r

∂

∂r

[r∂U

∂r

]+

1

r2∂2U

∂θ2. (1.31)

1.3.3 Implementation of P1 conforming triangular element on a uniform grid

K5jk

K1jk

K2jk

K3jk

K4jk K6jkK6jk

(xj+1, yk)

(xj, yk)

(xj, yk+1)(xj−1, yk+1)

(xj−1, yk)

(xj, yk−1)

We begin with the following simplest example:

−∆u = f, Ω = (0, 1)2, (1.32a)u = 0, ∂Ω. (1.32b)

Let Th be the triangulation of Ω into the 2N × 2Nuniform triangles with vertices Vjk = (xj , yk) =h(j, k), j, k = 0, 1, · · · , N, with h = 1N .

In each triangle K1jk,K2jk, · · · ,K6jk which has Vjk as a vertex, we have

ϕjk(x, y) =

− 1h (x+ y − (j + k + 1)h) on K1jk,− 1h (y − (k + 1)h) on K2jk,1h (x− (j − 1)h) on K3jk,

1h (x+ y − (j + k − 1)h) on K4jk,

1h (y − (k − 1)h) on K5jk,− 1h (x− (j + 1)h) on K6jk.

Let uh(x, y) =∑N−1

j=1

∑N−1k=1 αjkϕjk(x, y) be an element of Vh.

Find αjk for j = 1, · · · , N − 1 and k = 1, · · · , N − 1 such thatN−1∑

j,k=1

αjk

∫

Ω

∇ϕjk · ∇ϕlmdxdy =∫

Ω

f ϕlmdxdy

for all l = 1, · · · , N − 1, m = 1, · · · , N − 1.Let

Alm,jk =

∫

Ω

∇ϕjk · ∇ϕlm dxdy and blm =∫

Ω

f ϕlm dxdy.

If we write α (and b) as the following order:

(α11, α21, · · · , xN−1,1, α12, α22, · · · , αN−1,2, · · · , α1,N−1, α2,N−1, · · · , αN−1,N−1)t,



then the matrix A satisfying Aα = b will be of the form:

A11,11 A11,21 · · · A11,N−1,1 · · · A11,1,N−1 · · · A11,N−1,N−1A21,11 A21,21 · · · A21,N−1,1 · · · A21,1,N−1 · · · A21,N−1,N−1

......

. . ....

. . ....

. . ....

AN−1,1,11 AN−1,1,21 · · · AN−1,1,N−1,1 · · · AN−1,1,1,N−1 · · · AN−1,1,N−1,N−1...

.... . .

.... . .

.... . .

...A1,N−1,11 A1,N−1,21 · · · A1,N−1,N−1,1 · · · A1,N−1,1,N−1 · · · A1,N−1,N−1,N−1

......

. . ....

. . ....

. . ....

AN−1,N−1,11 AN−1,N−1,21 · · · AN−1,N−1,N−1,1 · · · AN−1,N−1,1,N−1 · · · AN−1,N−1,N−1,N−1

.

In the above matrix, the nonzero components may be found only if |j−l|+|k−m| ≤ 1 or |j−l| = |k−m| = 1.To compute these nonzero terms, notice that ∇ϕjk in each region:

∇ϕjk =

1h (−1, 1) on K1jk,1h (0,−1) on K2jk,

1h (1, 0) on K

3jk,

1h (1,−1) on K4jk,

1h (0, 1) on K

5jk,

1h (−1, 0) on K6jk.

1. Then Ajk,jk =∑6

j=1

∫Kjjk

|∇ϕjk|2 dxdy = [( 1h )2 + ( 1h )2 + 2( 1h )2 + ( 1h )2 + ( 1h )2 + 2( 1h )2]× h2

2 = 4.

2. For (l,m) = (j − 1, k) [or (j + 1, k)], we can compute Alm,jk by calculating integrals only on K3jk andK4jk [or K

1jk and K

6jk]. Therefore, Alm,jk = −( 1h )2 · h

2

2 × 2 = −1.

3. Similarly, for (l,m) = (j, k − 1) [or (j, k + 1)], we can compute Alm,jk by calculating integrals only onK5jk and K

1jk [or K

2jk and K

4jk]. Therefore, Alm,jk = −( 1h )2 · h

2

2 × 2 = −1.

4. Finally, for (l,m) = (j− 1, k− 1) [or (j+1, k+1)], we can compute Alm,jk by calculating integral onlyon K3jk and K

5jk [or K

6jk and K

2jk]. In these cases, the gradients of ϕjk and ϕlm are perpendicular, so

Alm,jk = 0.

Combining the above calculations, we can see that

A =

B −I · · · O−I B . . . O...

. . .. . .

...O O · · · B

where B =

4 −1 · · · 0−1 4 . . . 0...

. . .. . .

...0 0 · · · 4

,

and I is the (N − 1)× (N − 1) identity matrix.Similarly, blm can be calculated by

∫

Ω

ϕlmf dxdy =∑

jk

∫

Kjk

ϕlm(x, y)f(x, y) dxdy. =

6∑

j=1

∫

Kjlm

ϕlm(x, y)f(x, y) dxdy.

The jk-th row of Aα = b is given by

4αjk − (αj,k+1 + αj,k−1 + αj−1,k + αj+1,k) =∫

Ω

ϕjkf dxdy.



For the constant force term, say, f(x, y) = 1 for all (x, y) ∈ Ω, one can calculate bjk =∫Ωf(x, y)ϕjk(x, y) dxdy

as the sum of volumes of six tetrahedrons:

bjk = 6 ·h2

6= h2.

Notice that this is identical to the linear system where finite difference method applied, by dividing bothsides by h2..

In particular, if f = 0, then

αjk =1

4(αj,k+1 + αj,k−1 + αj−1,k + αj+1,k),

which is called the Discrete Mean Value Property.

Generally, if u is harmonic (i.e. ∆u = 0), then u has the Mean Value Property:

u(x0) =1

2πr

∫

S(x0;r)

u(s)ds =1

πr2

∫

B(x0;r)

u(x)dx

where S(x0; r) and B(x0; r) denote circle and disk of radius r centered at x0.

1.4 Higher order finite elements in Rd

1.4.1 Pm (conforming) simplicial element

Let K be a non-degenerate simplex with vertices V1, · · · , Vd, Vd+1 with identification V0 = Vd+1. Denote byJ the set of indices α = (α1, · · · , αd+1) in Zd+1+ with

∑d+1j=1 αj = m Then consider the set of points

P =∑

α∈J

αjmVj .

• K = d− simplex

• PK = Pm(K) with dim(PK) =(d+mm

)

• ΣPK = {φ(V ) : V ∈ P} as the degrees of freedom

1.4.2 Qm (conforming) d-linear element

Similar as simplicial elements.

Let K be the d–cube (−1, 1)d. Then consider the set of points

P =

{2

m(α1, · · · , αd)− (1, · · · , 1) ∈ Rd | αj ∈ {0, 1, · · · ,m} ∀j

}

• K = d− cube = (−1, 1)d

• PK = Qm(K) with dim(PK) = (m+ 1)d

• ΣPK = {φ(V ) : V ∈ P} as the degrees of freedom.



1.4.3 Unisolvency and optimality

It is necessary to show that the DOFs determine the function in PK . Also in order to construct optimalfinite elements, one usually require that Pm(K) ⊂ PK . In this case the Bramble–Hilbert lemma shows theelement is optimal. We state the following results without proof at the moment.

Theorem 1.4. All the finite elements defined above have the “unisolvency” property. That is, if the DOFsof a given function f ∈ PK are zero, then the function f ⇐⇒ 0 in K.

Also we have the continuity property as follows:

Theorem 1.5. All the finite elements defined above have the “continuity” property. That is, if the DOFsrestricted on a face f of K of a given function f ∈ PK are zero, then the function f ⇐⇒ 0 on the face f .

1.4.4 Triangulations and reference elements

Let (Th)0 0 such that for every h,

hKρK

≤ κ for all K ∈ Th,

the family of triangulations is called “shape-regular”. If, in addition, there exists c > 0 such that

hK ≥ ch ∀K ∈ Th ∀h,

the family of triangulations is called “quasi-uniform”.We now introduce the following reference elements:

1. d-simplicial element: Let K̂ be the reference simplicial element in Rd with coordinates V̂j , j = 0, 1, · ∗dsuch that V̂0 = 0 and V̂j − V̂0 = êj be the jth unit vector for j = 1, · · · , d. Then, for each triangleor a simplex K ∈ Th, there is a surjective one–to–one affine map FK : K̂ → K such that FK(x̂) =BK x̂+ bK .

In R2 and R2 the elements are called triangular and simplicial elements, respectively.

2. d-rectangle-type element: Let K̂ be the d–cube (−1, 1)d. Then, for any d-rectangle K ∈ Th, there is asurjective one–to–one d-linear map FK : K̂ → K such that FK(x̂) = BK(x̂), where each componentof BK : K̂ → K is a d-linear map.In R2 and R2 the elements are called quadrilateral and hexahedral elements, respectively. Notice thatin two dimension the faces of a quadrilateral are straight lines segments, but in three dimension thefaces of a hexahedron may not be flat.

On the reference element K̂ the reference finite element is denoted by the triple:

(K̂, P̂K̂ , Σ̂K̂).

Remark 1.2. The DOFs(degrees of freedom), denoted by Σ̂K̂ or ΣK , are evaluations of function φ̂ ∈ P̂K̂or φ ∈ PK . Hence the set Σ̂K̂ or ΣK which denotes the DOFs is a subset of dual space of P̂K̂ or PK ,respectively. That is,

Σ̂K̂ ⊂ (P̂K̂)′ with #(Σ̂K̂) = dim(P̂K̂) or ΣK ⊂ (PK)′ with #(ΣK) = dim(PK)



If the DOFs consist of the evaluation of function at points or the integrations of function on a subset ofthe element only, the finite element is called “Lagrange–type finite element”. If they contain any evalua-tion of derivative of function or any integration of derivative, which is not included in the Lagrange–typeDOFs, the finite element is called “Hermite–type finite element”. All the finite elements introduced aboveare of Lagrange–type. There are a number of Hermite–type finite elements. We will introduce an importantHermite–type finite element, so–called the 21 DOFs “Argyris element” and the 6 DOFs “Morley noncon-forming element”

Using the reference finite element, one can define the finite element (K,PK ,ΣK) on each element K ∈ Thas follows:

• K = FK(K̂)

• PK = {p̂ ◦ F−1K | p̂ ∈ P̂K̂}

• ΣK = {f ∈ (PK)′ | Σ(φ) = Σ̂(FK), Σ̂ ∈ Σ̂K̂ ∀φ ∈ PK}

Notice that in ΣK there are at least three types of DOFs to distinguish as follows: with φ = φ̂ ◦F−1K , whichis equivalent to φ̂ = φ ◦ FK ,

1. Point–value type: Σ̂(φ̂) = φ̂(x̂j) ⇐⇒ Σ(φ) = (φ ◦ FK)(x̂j)

2. Integral type: Σ̂(φ̂) =∫êφ̂(x̂) dσ̂(x̂) ⇐⇒ Σ(φ) =

∫e(φ ◦ F−1K )(x)

∣∣∂x̂∂x

∣∣ dσ(x)

3. Derivative type: Σ̂(φ̂) = ∂Σ̂∂x̂k (x̂j) ⇐⇒ Σ(φ) =∑d

ℓ=1∂(φ◦FK)

∂xℓ(x̂j)

∂xℓ∂x̂k

Let {φ̂j | j = 1, · · · , J} be the standard basis functions on the reference element K̂. Indeed, they areconstructed as follows: denote

Σ̂K̂ = {Σ̂j ∈ (P̂K̂)′, j = 1, . . . , dim(P̂K̂)}.

Associated to each Σ̂j , find φ̂j ∈ P̂K̂ such that

Σ̂k(φ̂j) = δjk, δjkbeing the Kronecker delta function. (1.33)

Hence, the standard finite element basis functions are given by

φ̂j , j = 1, · · · , dim(P̂K̂). (1.34)The basis functions on each K ∈ Th is defined by the pull–back

{φK,j(x) = (φ̂j ◦ F−1K )(x) | j = 1, · · · , J}.Then the global finite element space Vh can be defined by

Vh = {vh ∈ C0(Ω) | vh|K =J∑

j=1

αK,jφK,j ∀K ∈ Th}. (1.35)

Notice that in order to meet the criterion that vh ∈ C0(Ω), one needs to haveαK,j = αK′,j′ ∀j′ such that xj(∈ K) = xj′(∈ K ′).

Definition 1.1. Consider a finite element (K,PK ,ΣK) and the global finite element space Vh. Let ℓ bethe maximum order of derivatives in the definition of the DOFs ΣK . For each K ∈ Th, define the localPK–interpolation operator IK : Cℓ(K) → PK such that

Σℓ(IKv − v) = 0 ∀Σℓ ∈ ΣK ∀v ∈ Cℓ(K). (1.36)Then the global Vh–interpolation operator Ih : Ω → Vh is defined piecewisely such that

Ihv|K(x) = IKv ∀K ∈ Th.



Ihv and IKv are called the global Vh– and local PK–interpolants of v, respectively.Let S ⊂ Rd be a bounded set with diam (S) = h. Bramble and Hilbert use the following normalized

norms on Wm,p(S) and Wm,∞(S), which are equivalent to the usual Sobolev norms:

|||u|||m,p,S =(

m∑

k=1

hkp−d|u|pk,p,S

) 1p

(1.37)

|||u|||m,∞,S =m∑

k=0

hk|u|k,∞,S . (1.38)

Theorem 1.6 (Bramble–Hilbert I.). Let Q = W k,p(S)/Pk−1 be a quotient space. Then the semi–normh−d/p|u|k,p,S is a norm on Q equivalent to the quotient norm

‖[u]‖Q = infv∈[u]

h−d/p‖v‖k,p,S , [u] denotes the equivalent class.

Furthermore, there exists C > 0 independent of h and u such that

hk−dp ‖u‖k,p,S ≤ ‖[u]‖Q ≤ Chk−

dp ‖u‖k,p,S ∀u ∈W k,p(S).

Theorem 1.7 (Bramble–Hilbert II.). Let S be as in the above theorem. Suppose that L : W k,p(S) → R bea bounded linear functional such that

1. |L(u)| ≤ C|||u|||k,p,S ∀u ∈W k,p(S) for C > 0 independent of h and u,

2. Pk−1(S) ⊂ Ker(L).

Then there exists C > 0 such that

|Lv| ≤ C1hk−d/p|u|k,p,S ∀u ∈W k,p(S) for C1 > 0 independent of h and u.

Theorem 1.8 (Bramble–Hilbert III.). Let S be as in the above theorem. Suppose that L :W k,p(S) → R bea bounded linear functional such that

1. |L(u)| ≤ C|||u|||j,∞,S ∀u ∈ Cj(S) for C > 0 independent of h and u,

2. Pk−1(S) ⊂ Ker(L).

Then there exists C1 > 0 such that, for all1p −

k−jd < 0, such that W

k,p(S) →֒ Cj(S) and

|Lv| ≤ C1hk−d/p|u|k,p,S ∀u ∈W k,p(S) for C1 > 0 independent of h and u.

Theorem 1.9 (Bramble–Hilbert IV.). Let S be as in the above theorem. Suppose u ∈ Cs, s = [s]+α ∈ (0, 1],where [s] denotes the Gauss integer which is the largest integer ≤ s. Suppose that L : C0(S) → R be a boundedlinear functional such that

1. |L(u)| ≤ C|u|0,∞,S ∀u ∈ C0(S) for C > 0 independent of h and u,

2. Pk−1(S) ⊂ Ker(L).

Then there exists C1 > 0 such that, for all1p −

k−jd < 0, such that W

k,p(S) →֒ Cj(S) and

|Lv| ≤ C1hs supx,y∈S

∑

|α|=[s]

∂αu(x)− ∂αu(y)||x− y|α 0 ≤ s < k ∀u ∈W

k,p(S) for C1 > 0 independent of h and u.



Applications of Bramble–Hilbert Theorems are as follows: In Theorem 1.7, suppose K ∈ Th be of sizediam (K) = h. Then, choose k = 2 and PK = P1(K). Consider the difference beween the following standardlocal P1–interpolant and a function :

Lu(x) = IKu(x)− u(x) =#(vertices of K)∑

j=1

u(xj)φK,j(x)− u(x)

Then, verify that|Lu(x)| = |IKu(x)− u(x)| ≤ C|||u|||2,p,K .

Thus from Theorem 1.7 it follows that

|Lu(x)| ≤ C1h2−dp |u|2,p,K .

Hence,

‖IKu− u‖0,K =(∫

K

|Lu(x)|p dx) 1

p

≤ C1h2−dp |u|2,p,K |K|

1p = C1h

2|u|2,p,K .

In the meanwhile, for the energy–norm estimate, set wj =∂u∂xj

choose k = 1 and PK = P0(K). Consider

the difference between the P0–interpolation (i.e., the mean–average operator on K) and a function:

Lwj(x) =1

|K|

∫

K

wj(x) dx− wj(x).

Then, verify that|Lwj(x)| ≤ C|||wj |||1,p,K .

Thus from Theorem 1.7 it follows that

|Lwj(x)| ≤ C1h1−dp |wj |1,p,K .

Hence,

|IKu− u|1,p,K =

d∑

j=1

∫

K

|Lwj(x)|p dx

1p

≤ C1h1−dp |u|2,p,K |K|

1p = C1h

1|u|2,p,K .

Definition 1.2. For m ≥ 1, the broken norms (or mesh-dependent norms) are defined as follows:

‖v‖m,h :=( ∑

T∈Th

‖v‖2m,T

) 12

∀v ∈ Vh. (1.39)

Summarizing the above, we have

Theorem 1.10. Let m = 2. Suppose that the finite element spaces PK contain P1, as all the finite elementsintroduced above. Then,

‖Ihu− u‖0,h ≤ Ch2|u|2,Ω,‖Ihu− u‖1,h ≤ Ch|u|2,Ω,

for all u ∈ H2(Ω). Invoking the Ceá lemma, we have the following error estimates between the finite elementsolution and the exact equation:

‖uh − u‖1,h ≤ C ‖Ihu− u‖1,h ≤ Ch|u|2,Ω.Theorem 1.11. Let m ≥ 2, {Th}h be a family of shape-regular triangulations of Ω, and Vh is a finiteelement space such that Pm−1(K̂) ⊂ P̂K̂ . Then, there exists a constant C > 0 such that

‖u− Ihu‖k,h ≤ Chm−k|u|m,Ω ∀u ∈ Hm(Ω) for 0 ≤ k ≤ min{1,m}.



1.5 Assembly of matrix Akj =∑

K∈Th ah(φj, φk)

For this topics see Long Chen’s implementation of iFEM.Long Chen’s lecture notes“http://www.math.uci.edu/ chenlong/226/Ch3FEMCode.pdf”


Chapter 2

NC (Nonconforming) finite elements

All the finite element spaces Vh introduced earlier are subspaces of H1(Ω) ∩ C0(Ω). This means that any

vh ∈ Vh are continuous across the interfaces Γℓm := ∂Kℓ ∩ ∂Km for all elements Kℓ and Km in Th. Denoteby [[·]]Γℓm the jump across the interface Γℓm defined by

[[u]]Γℓm = γΓℓm(u|Kℓ)− γΓℓm(u|Km), (2.1)

where u|Kℓ denotes the restriction of u ∈ L2(Ω) to Kℓ, and γΓℓmuKℓ the trace of uKℓ onto Γℓm. Recall thatKℓ’s are open sets.

Conforming finite elements require that

[[u]]Γℓm(x) = 0 ∀x ∈ Γℓm. (2.2)

Breaking this continuity restriction would substantially widen the realm of finite element spaces. However,there should be certain rules to relate functions defined on neighboring elements Kℓ and Km. These areessentially the degrees of freedom.

Let us classify DOFs (the degrees of freedom) in two categories:

1. Interface DOFs: the DOFs that transfer the information on the element to neighboring element

2. Interior DOFs: the DOFs that determine information only on the element

Observe that conforming finite elements have interior DOFs such that (2.2) holds.

Remark 2.1. DOFs are also classified into two kinds as follows:

1. Lagrange type of finite elements: DOFs consist of point values of a function (usual Pm−1 or Qm−1finite elements)

2. Hermitian type of finite elements: DOFs consist of point values of the derivatives of the function uptocertain order (Morley element, for instance)

Nonconforming finite elements may have two types of interface DOFs

1. Gauss points DOFs: ΣPK = {φ(gj) : gj are Gauss points on the faces of K}2. Orthogonality DOFs: ΣPK = {

∫fφq dσ : q ∈ Pm−1(f), f is a face of K}

which relax (2.2) as follows

1. [[u]]Γℓm(gj) = 0 ∀Gauss points on the faces Γℓm2.∫Γℓm

[[u]]Γℓmq dσ = 0 ∀q ∈ Pm−1(f), f is a face of K}respectively.

25


2.1 P1 triangular NC element (Crouzeix-Raviart, 1973) and therotated Q1-rectangular NC elements

1. P1 triangular NC element (Crouzeix-Raviart, 1973)

• K = triangle• PK = Span{1, x, y}• DOF: {φ(mj),mjmidpoints of edges, j = 1, 2, 3}

��

��

��

V1

V2

V3

T

K = T

e1

e2

e3

m2

m3m1

ΣK = {φ(mj), j = 1, 2, 3}

PK = Span{1, x, y}

Exercise 2.1. (Due to July 10) Let K be the triangle with vertices (0, 0), (1, 0), (0, 1), (0, 0). Find threebasis functions φj(x, y), j = 1, · · · , 3, for PK such that φj(mk) = δjk, where δjk is the Kronecker deltafunction.

2. The rotatedQ1-rectangular NC elements (Han (1984), Rannacher–Turek (1992), Chen (1992), Douglas–Santos–Sheen–Ye (1999)

e2

V3

V1V2

V4

m1

m2

m3

m4

K = RR

ΣIK = {∫ejφ dσ, j = 1, · · · , 4}

ΣMK = {φ(mj), j = 1, · · · , 4}e4

e1

e3

PK = Span{1, x, y, θℓ(x)− θℓ(y)}

(a) H. Han (1984):

• K = R = [−1, 1]2• PK = Span{1, x, y, x2 − 5/3x4, y2 − 5/3y4}• DOF: ΣMK = {φ(mj),mjmidpoints of edges, j = 1, 2, 3, 4} plus

∫Kφ

(b) Rannacher-Turek (1992, rotated Q1 element, also Z. Chen):

• K = [−1, 1]2• PK = Span{1, x, y, x2 − y2}



• DOF1 ΣPK = {φ(mj),mjmidpoints of edges, j = 1, 2, 3, 4};DOF2: ΣIK = {

∫ejφdσ, ej four edges, j = 1, 2, 3, 4};

(c) DSSY element(Douglas-Santos-Sheen-Ye, 1999): Cai-Douglas-Ye, 1999 stable Stokes element:

e2

V3

V1V2

V4

m1

m2

m3

m4

K = RR

ΣIK = {∫ejφ dσ, j = 1, · · · , 4}

ΣMK = {φ(mj), j = 1, · · · , 4}e4

e1

e3

PK = Span{1, x, y, θℓ(x)− θℓ(y)}

• K = [−1, 1]2• PK = Span{1, x, y, θℓ(x)− θℓ(y)}, ℓ = 0, 1, 2• DOF1 = DOF2; 1|ej |

∫ejφ dσ = φ(mj)

θℓ(t) =

t2, ℓ = 0;t2 − 5/3t4, ℓ = 1;t2 − 25/6t4 + 7/2t6, ℓ = 2.

Exercise 2.2. (Due to July 10) Let K = (−1, 1)2. Find four basis functions φj(x, y), j = 1, · · · , 4,for PK (DSSY element) such that φj(mk) = δjk, where δjk is the Kronecker delta function.

(d) For truly quadrilaterals, (Cai-Douglas-Santos-S.-Ye, CALCOLO, 2000):

• K = [−1, 1]2• PK = Span{1, x, y, θℓ(x)− θℓ(y), xy}, ℓ = 0, 1, 2• DOF: {φ(mj),mjmidpoints of edges, j = 1, 2, 3, 4} and

∫R̂φ(x, y)xy dxdy;

3. P2 triangular NC element (Fortin-Soulie, 1983; Lee-Sheen 2005) gj ’s denotes Gauss points on the face

��

��

��

��

��

��

��

��

��

��

V1

V2

V3

T

K = T

PK = P2

e3g1

g2

g3 g4

g5

g6

e2

e1

ΣK = {φ(gj), j = 1, · · · , 6;∫Kφ dx}

4. P2 rectangular NC element (or “rotated Q2-rectangular NC element) (Lee-Sheen 2005) gj ’s denotesGauss points on the face



V3

V1V2

V4

K = Re1

e2

g3g4

g5

g7 g8 ΣK = {φ(gj), j = 1, · · · , 8;∫Kφ dx}

PK = P2 ⊕ Span{x2y, xy2}g6 g1

g2

e3

e4

R

O

2.2 The P1-NC quadrilateral and hexahedral elements

(C. Park, Thesis 2002, Park-Sheen SIAM J. Numer. Anal. 2003)

e4V3 V4

V1

V2

m3

m2

m1

m4

e3

e2

e1

• K = Q: a quadrilateral

• PK = S (Q) = Span{1, x, y}

• ΣK = {φMj , j = 1, · · · , 4}

• Basis function

φj(m) =

{12 , if m =Mj ,Mj+1,0, if m =Mj+2,Mj+3

• S (Q) = Span{φ1, · · · , φ4}and dim(S (Q)) = 3

Lemma 2.1. If u ∈ S(Q), then u(M1) + u(M3) = u(M2) + u(M4). Conversely, if uj is a given value atMj , for j = 1, · · · , 4, satisfy u1 + u3 = u2 + u4, then there exists a unique function u ∈ S(Q) such thatu(Mj) = uj, j = 1, · · · , 4.

The P1-NC hexahedral element

��

��

��

��

��

��

��

cc

c

25

c

c6

v v

v

v

vv

v

v

1c 4

3

12

3 4

76

7 8

• K = hexahedron

• PK = S (Q) = Span{1, x, y, z}

• ΣK = {φ(Mj), j = 1, · · · , 6}

• S (Q) = Span{φ1, · · · , φ8}and dim(S (Q)) = 4

Lemma 2.2. If u ∈ S(Q), then u(M1) + u(M6) = u(M2) + u(M5) = u(M3) + u(M4). Conversely, if uj is agiven value at Mj, for 1 ≤ j ≤ 6, satisfying u1 + u6 = u2 + u5 = u3 + u4, then there exists a unique functionu ∈ S(Q) such that u(Mj) = uj, 1 ≤ j ≤ 6.

With these elements we can prove optimal convergence.

Theorem 2.1. The above element is unisolvent. Optimal error estimates holds for elliptic problems:



2.2.1 Error analysis for linear nonconforming Galerkin method

Define the global nonconforming finite element space N Ch0 on Th by

N Ch0 = {v ∈ L2(Ω) | v|K ∈ PK ∀K ∈ Th,〈[[v]]e, q〉e = 0, ∀q ∈ Pm−1(e), 〈ve, q〉e = 0, ∀e ∈ ∂}

where [[f ]]e means the jump of f across the interface e and 〈[[v]]e, q〉e =∫e[[v]]eq dσ.

Define the bilinear form ah : [N Ch0 +H

1(Ω)]2 → R by

ah(u, v) =∑

K∈Th

aK(u, v), where aK(u, v) denotes the integrations restricted to the domain K

Define the broken energy norm: for w ∈ H1(Ω) + N Ch0|w|1,h =

√ah(w,w)

Recall that the weak solution u ∈ H10 (Ω) satisfies

a(u, v) = 〈f, v〉 ∀v ∈ H10 (O). (2.3)

The nonconforming Galerkin method is to find uh ∈ N Ch0 such that

ah(uh, vh) = 〈f, vh〉 ∀vh ∈ N Ch0 . (2.4)

Lemma 2.3 (The second Strang Lemma). If u ∈ H10 (Ω) and uh ∈ NCh0 are the solutions of (2.3) and (2.4)respectively, then

ah(u− uh, u− uh) ≤ C(

infv∈NCh0

|u− v|1,h + supw∈NCh0

|ah(u,w)− (f, w)||w|1,h

)(2.5)

Remark 2.2. In the right-hand side of (2.5), the second term is called a consistency error term. Noticethat in the consistency error term,

ah(u,w)− (f, w) = ah(u,w)− ah(uh, w). (2.6)

If w ∈ C0(Ω), then the consistency error term become zero.Proof. First, by the triangle inequality

|u− uh|1,h ≤ |u− zh|1,h + |zh − uh|1,h ∀zh ∈ NCh0 .

Next, since ah(uh, uh − zh) = (f, uh − zh), one has

|uh − zh|21,h = ah(uh − zh, uh − zh)= ah(u− zh, uh − zh) + ah(uh − u, uh − zh)= ah(u− zh, uh − zh) + (f, uh − zh)− ah(u, uh − zh)≤ |u− zh|1,h|uh − zh|1,h + (f, uh − zh)− ah(u, uh − zh).

Then

|uh − zh|1,h ≤ |u− zh|1,h +(f, uh − zh)− ah(u, uh − zh)

|uh − zh|1,h,

which combined with the triangle inequality gives

|u− uh|1,h ≤ 2|u− zh|1,h +(f, uh − zh)− ah(u, uh − zh)

|uh − zh|1,h.



If we choose zh ∈ NCh such that

|u− zh|1,h = infv∈NCh0

|u− v|1,h,

then

|u− uh|1,h ≤ 2 infv∈NCh0

|u− v|1,h + supw∈NCh0

|(f, w)− ah(u,w)||w|1,h

.

�

For the nonconforming finite elements introduced above PK including Pm−1(K),m = 2, 3, by theBramble-Hilbert Theorem 1.7,

‖Ihu− u‖m,K̂ ≤ C|u|m,K̂ ∀u ∈ Hm(K̂),

from which the first part in the second Strang lemma is estimated as follows:

‖Ihu− u‖k,h ≤ Chm−k|u|m,Ω, 0 ≤ k ≤ m. (2.7)

2.2.2 Implementation of P1–NC quadrilateral element

Consider the special case: Ω = (0, 1) is decomposed into the uniform N × N uniform rectangles withcoordinates (xj , yk) = h(j, k), j, k = 0, 1, · · · , N, where h = 1N . The basis functions φjk(x, y) contain fourrectangles whose vertices contain (xj , yk). The basis function values are shown in the following, but in ashifted region (−h, h)2.

To find components of the mass matrix, using symmetry and translation invariance, one has the following:

• (l,m) = (j, k)

(φjk, φjk) =4

4h2

∫ h

0

∫ h

0

(−x− y + 3

2h

)2dx dy =

20

48h2,

• (l,m) = (j + 1, k)

(φjk, φj+1,k) =2

4h2

∫ h

0

∫ h

0

(−x− y + 32h2h

)(x− y + h

2) dx dy =

18

48h2,

and

• (l,m) = (j + 1, k + 1)

(φjk, φj+1,k+1) =1

4h2

∫ h

0

∫ h

0

(−x− y + 32h2h

)(x+ y − h

2) dx dy =

−1148

h2.



(h, h2)

(h, h)

(−h,−h)

φ(x, y)

(h2 , h)

(h, 0)(h2 , 0)

(0, h)

(0, 0)

(0,−h) (h,−h)

(−h, 0)

(−h, h)

x−y+32h2h

x+y+32h

2h−x+y+32h

2h

−x−y+32h2h

K3jk K4jk

K1jkK2jk

(h, h2)

(h2 , h)

(h2, 0)

K3jk K4jk

K1jkK2jk

18

−11 −11

−11−11

18

18

2018h2

48

(φjk, φℓm)

For the components of mass matrix, using symmetry and translation invariance,

• (l,m) = (j, k)

(∇φjk,∇φjk) =4

4h2

∫ h

0

∫ h

0

(−1−1

)·(−1−1

)dx dy = 2,

• (l,m) = (j + 1, k)

(∇φjk,∇φj+1,k) =2

4h2

∫ h

0

∫ h

0

(−1−1

)·(

1−1

)dx dy = 0,

and

• (l,m) = (j + 1, k + 1)

(∇φjk,∇φj+1,k+1) =1

4h2

∫ h

0

∫ h

0

(−1−1

)·(11

)dx dy = −1

2.



(h, h2)

(h, h)

(−h,−h)

(h2 , h)

(h, 0)(h2 , 0)

(0, h)

(0, 0)

(0,−h) (h,−h)

(−h, 0)

(−h, h)

K3jk K4jk

K1jkK2jk

∇φ(x, y)

12h

(−11

)

12h

(−1−1)

12h

(1−1)

12h

(11

)

(h, h2)

(h2 , 0)

K3jk K4jk

K1jkK2jk(0, h

2)

−1 0 −1

0

−1 0

40

(h2, h) −1

(∇φjk,∇φℓm)

12

2.3 Hermite–type finite elements

In this section we introduce Hermite–type finite elements for solving the fourth–order partial differentialequations (the biharmonic equation)

∆2u(x) = f(x), x ∈ Ω,u(x) = 0, x ∈ ∂Ω,∂u

∂ν= 0, x ∈ ∂Ω. (2.8)

Green’s second identity:

∫

Ω

∆u v dx−∫

Ω

u∆v dx =

∫

∂Ω

∂u

∂νv − u∂v

∂νdσ (2.9)

By Green’s second identity, we can have

(∆2u, v) = (f, v) −→∫

Ω

((∆2u)v dx−∫

Ω

∆u∆v dx =

∫

∂Ω

∂(∆u)

∂νv −∆u∂v

∂νdσ

∀v ∈ H20 (Ω) = C∞0 (Ω)H2(Ω)

(i.e. v|∂Ω = 0,∂v

∂ν

∣∣∣∣∂Ω

= 0)

The weak problem is given as finding u ∈ H20 (Ω) such that

(∆u,∆v) = (f, v) ∀v ∈ H20 (Ω) (2.10)

Finite element method is to find Vh ⊂ H20 (Ω) and then to find uh ∈ Vh such that

(∆uh,∆v) = (f, v) ∀v ∈ Vh (2.11)

In particular, we want to find a finite element space Vh ⊂ C1(Ω) ∩H20 (Ω).Argyris element (C1-quintic triangular element or 21-DOFs triangular element)



• T = triangle

• PT = P5(T )

• ΣT = {∂αp(ai), |α| ≤ 2, i = 1, 2, 3 : ∂νp(aij . 1 ≤ i ≤ j ≤ 3}={p(ai),

∂p

∂x1(ai),

∂p

∂x2(ai),

∂2p

∂x21(ai),

∂2p

∂x1∂x2(ai),

∂2

∂x

i = 1, 2, 3 ;∂p

∂ν(aij), 1 ≤ i ≤ j ≤ 3

}

• dim(P5(T )) = 21.

Argyris triangle

Figure 2.3.1: Argyris element

Let vh ∈ Vh, where Vh is constructed using Argyris element. We want to show that

vh|T1 − vh|T2 = 0 along T̄1 ∩ T̄2.

Let t denotes the variable along T̄1 ∩ T̄2 so that v(t) is polynomial of degree ≤ 5 fulfilling

v(t1) = v(t2) = v′(t1) = v

′(t2) = v′′(t1) = v

′′(t2) = 0 =⇒ v(t) has factors (t− t1)3 and (t− t2)3 =⇒ v = 0.

Of course, we have v′(t) = 0. Next, to show that the the normal directional derivative of v equals zero, let

w(t) =

(∂vh∂ν

∣∣∣∣T1

− ∂vh∂ν

∣∣∣∣T2

)(t),

where ν denotes the normal vector from T1 to T2. Note that w(t) is a polynomial of degree ≤ 4 andw(t1) = w(t2) = w(t12) = 0, where t12 represents the midpoint between a1 and a2.

w′(tj) =∂

∂t

(∂v

∂ν

∣∣∣∣T1

− ∂vh∂ν

∣∣∣∣T2

)(tj)

= ∇((∇vh|T1 · ν)− (∇vh|T2 · ν)) · τ = 0, j = 1, 2.

Thus w(t) = 0 along T̄1 ∩ T̄2. And hence vh ∈ C1(T 1 ∪ T 2). This proves that Vh ⊂ C1(Ω) ∩H20 (Ω).

Quadratic nonconforming elements on rectangles with Heejeong Lee

1. Triangular case



V2

V3 V1

m1

m3

m2

K = T

T

PK = P2

ΣK = {φ(Vj), ∂φ∂ν (mj), j = 1, 2, 3}



��

��

��

��

��

��

��

��

��

��

V1

V2

V3

T

K = T

PK = P2

e3g1

g2

g3 g4

g5

g6

e2

e1

ΣK = {φ(gj), j = 1, · · · , 6;∫Kφ dx}



R

Incomplete biquadratic element

a1a2

a3 a4

• Morley element• Fortin-Soulie element

2. Rectangular case

• Incomplete biquadratic element• Reduced incomplete biquadratic element (with Heejeong Lee)

Exercise 2.3 (Due July 24 (everything before the coding), and July 31 (coding)). Choose any nonconformingfinite element as you like. Then consider the same PDE as in the previous programming exercise.

1. Use N ×N rectangular meshes with N = 50, 100, 200.

2. Find an explicit formula for the standard local basis functions on a reference domain. Using thereference basis functions, find an explicit formula for the global basis functions φ′jk, where φjk is

associated with the point (xj , yk) = h(j, k), h =1N .

3. Give an explicit description how you would compute ah(φjk, φlm) and (f, φjk).

4. Formulate a linear system Ax = b, where x denotes the unknown coefficients α′jks. (Submit up to thisby July 24.)

5. Use any numerical linear algebra to solve Ax = b.



R

g3g4

g8 V4

V2

V3

V1

g1

g2

g7

g5

g6

Quadratic nonconforming rectangle



6. Compare the numerical solutions with the numerical solutions obtained by conforming finite elements.

7. Submit the codes and analysis and comments by July 31.


Chapter 3

Preliminaries from functional analysis

Let Ω be a nonempty open set in Rd, and set

• for 1 ≤ p


Definition 3.1. A function ‖·‖ : X 7→ R+ is called a norm if(i) ‖x+ y‖ ≤ ‖x‖+ ‖y‖ ∀x, y ∈ X(ii) ‖αx‖ = |α| ‖x‖ ∀x,∈ X ∀α ∈ R(iii) ‖x‖ > 0 ∀x, x 6= 0.

Definition 3.2. A function | · | : X 7→ R+ is called a semi-norm if (i) and (ii) hold in Definition 3.1.Theorem 3.1. [Rud91a, p. 30] A topological vector space X is normable if and only if the origin has aconvex bounded neighborhood.

For an open subset Ω of Rd,

Lp(Ω) := {f : Ω 7→ R |∫

Ω

|f(x)|pdx


Definition 3.6. [Yos95, p.120] A sequence (vn) in a normed linear space V is said to converge weakly tov∞ ∈ V if limn→∞ 〈v′, vn〉V ′,V exists and it is equal to 〈v′, v∞〉V ′,V for every v′ ∈ V ′.

Theorem 3.2. [Yos95, p.121] A sequence (vn) in a normed linear space V is said to converge weakly tov∞ ∈ V iff

1. supn≥1 ‖vn‖V


3.2 Annihilators

Definition 3.9. Suppose M ⊂ V is a subspace of a Banach space V , and N ⊂ V ′ is a subspace of V ′. Theannihilatos M⊥ and N⊥ are defined as follows:

M⊥ = {v′ ∈ V ′ : 〈v′, v〉V ′,V = 0 ∀v ∈M},⊥N = {v ∈ V : 〈v′, v〉V ′,V = 0 ∀v′ ∈ N}.

Theorem 3.8. Suppose M ⊂ V is a subspace of a Banach space V , and N ⊂ V ′ is a subspace of V ′. Then

1. ⊥(M⊥) =M‖·‖V

. (⊥N)⊥ = Nweak

∗V ′

.

2. In addidtion, if M is closed, then

M ′ = V ′/M⊥; (V/M)′ =M⊥.

3. For T ∈ L (V,W ),N (T ∗) = R(T )⊥; N (T ) = ⊥R(T ∗).

4. For T ∈ L (V,W ), N (T ∗) is weak∗-closed in W ′

5. For T ∈ L (V,W ), R(T ) is dense in W if and only if T ∗ is one-to-one.

6. For T ∈ L (V,W ), R(T ∗) is weak∗-dense in V ′ if and only if T is one-to-one.

3.3 Big Theorems

Theorem 3.9 (Bohnenblust-Sobczyk Extension Theorem). Let V be a linear space with a seminorm | · |V .Let M be a linear subspace of V and f is a linear functional defined on M with |f(v)| ≤ |v|V for all v ∈M.Then there exists a linearly extended fuctional f̃ defined on V such that |f̃(v)| ≤ |v|V for all v ∈ V.

Theorem 3.10 (Hahn-Banach Extension Theorem in Normed Linear Spaces). Let V be a normed linearspace with norm ‖ · ‖V . Let M be a linear subspace of V and f is a continuous linear functional defined onM with |f(v)| ≤ ‖v‖V for all v ∈M. Then there exists a continuous linearly extended fuctional f̃ defined onV such that ‖f̃‖V ′ = ‖f‖V ′ .

Theorem 3.11 (Banach Open Mapping Theorem in Banach Spaces). Let V and W be Banach spaces. LetT ∈ L (V,W ) with T (V ) =W . Then T maps every open set of V onto an open set of W.

Theorem 3.12 (Closed graph theorem). Let V,W be Banach spaces. Then if a map T : V → W iscontinuous, then the graph

G = {(v, Tv) : v ∈ V }

is closed in V ×W. Conversely, if a map T : V →W is linear and the graph

G = {(v, Tv) : v ∈ V }

is closed in V ×W, then the map T : V →W is continuous.



3.4 Compact operators

Definition 3.10. For Banach spaces V,W , a linear map T : V →W is said to be compact if T (B(0; 1)) iscompact in W, where B(0; 1) denotes the open unit ball in V.

Notice the following facts:

1. A compact operator T : V →W is evidently bounded.

2. T : V →W is compact if and only if every bounded sequence {vn} in V contains a subsequence {vnk}such that {Tvnk} converges to some w in W.

Definition 3.11. For a Banach space V , suppose that T ∈ L (V ). The spectrum σ(T ) is the set of all λ ∈ Csuch that T − λI is not invertible. Any λ ∈ σ(T ) should then satisfy at least one of the following statement:

1. R(T − λI) 6= V ;

2. N (T − λI) 6= {0}; in this case λ is called an eigenvalue of T with dim(N (T − λI)) ≥ 1, and anyv ∈ N (T − λI) is called an eigenvector of T associated with the eigenvalue λ.

The following theorems are standard. For instance see Rudin [Rud91b].

Theorem 3.13. Let V,W be Banach spaces.

1. If T ∈ L (V,W ) and R(T ) is a finite dimensional subspace of W , then T is compact.

2. If T ∈ L (V,W ) is compact and R(T ) is a closed subspace of W , then R(T ) is a finite dimensionalsubspace of W.

3. If T ∈ L (V ) is compact, then the null space N (T − λI)) is of finite dimension for all nonzero λ.

4. If T ∈ L (V ) is compact in an infinite dimensional Banach space V , then 0 ∈ σ(T ).

5. If S, T ∈ L (V ) and T is compact, the composition S ◦ T is compact.

6. If T ∈ L (V,W ), then T is compact if and only if T ∗ is compact. [Schauder’s Theorem]

7. If T ∈ L (V ) is compact, then T − λI has closed range for all nonzero λ.

8. Suppose that T ∈ L (V ) is compact, and r > 0. Let E be a set of eigenvalues λ of T such that |λ| > r.Then E is a finite set, and R(T − λI) 6= V for each λ ∈ E.

9. Suppose that T ∈ L (V ) is compact. Then for all nonzero λ,

dim(N (T − λI)) = dim(N (T ∗ − λI)) = dim(V/R(T − λI)) = dim(V ′/R(T ∗ − λI))


2. for each y ∈ V , the nonhomogeneous equation

x− Tx = y

is uniquely solvable. In this case, (I − T )−1 is also bounded.

Proof. See [GT83]. �

Theorem 3.15 (Fredholm Alternative in Hilbert spaces). Let T ∈ L (V ) be a compact operator in a (real)Hilbert space. Then there exists a countable set Λ ⊂ R with no limit point except possibly λ = 0, such that

1. for each nonzero λ /∈ Λ the equations

λx− Tx = y, λx− T ∗x = y (3.3)

hava a unique solution x ∈ V for each y ∈ V . Moreover (λI − T )−1 and (λI − T ∗)−1 are bounded.

2. for each λ ∈ Λ,λx− Tx = y is solvable if and only if y ⊥ N (λI − T ∗),

andλx− T ∗x = y is solvable if and only if y ⊥ N (λI − T );

moreover, N (λI − T ) and N (λI − T ∗) are of finite dimension.

Proof. See [GT83]. �

Definition 3.12. For more general linear operator T : V toV with D(T ) ⊂ V, set

Tλ = λI − T, λ ∈ C.

The resolvent set of T is defined by

ρ(T ) = {λ ∈ C | R(Tλ) is dense in V and (Tλ)−1 exists and is continuous}

For λ ∈ ρ(T ), the resolvent of T is defined by (Tλ)−1 and denoted by R(λ : T ). The spectrum of T is thecomplement of ρ(T ) in C, i.e.

σ(T ) = C \ ρ(T ).There are three kinds of spectra which are disjoint:

Point spectrum Pσ(T ) = {λ ∈ C | Tλ has no inverse}: = the set of eigenvalues of T ;

Continuous spectrum Cσ(T ) = {λ ∈ C | Tλ has discontinuous inverse T−1λ , D(T−1λ ) is dense in V };

Residual spectrum Rσ(T ) = {λ ∈ C | Tλ has an inverse T−1λ , D(T−1λ ) is not dense in V }.

Theorem 3.16. [Yos95, p.209] If T : V → V is a closed linear operator, for any λ ∈ ρ(T ) the resolventR(λ;T ) is an everywhere defined continuous linear operator, i.e. R(λ;T ) ∈ L (T ).


Chapter 4

Sobolev Spaces

Multi-index and partial derivatives Let α = (α1, · · · , αd), αi ≥ 0 integers, be a multi-index, and set |α| =α1 + · · ·+ αd.

Denote by ∂αϕ the partial derivative

Dαϕ =

(∂

∂x1

)α1· · ·(

∂

∂xd

)αdϕ =

∂|α|ϕ

∂xα11 · · · ∂xαdd.

4.1 Distributions

Definition of Distribution

Definition 4.1. (φj)j ⊂ C∞0 (Ω) is said to converge in the sense of the space D(Ω) to the function φ ∈ C∞0 (Ω)if

(i) there exists a compact subset K of Ω such that supp(φj) ⊂ K ∀j;(ii) for all multi-index α, ∂αφj → ∂αφ uniformly on K,

(i.e. ∀ ǫ > 0, ∃ an integer N(ǫ) such that n ≥ N(ǫ) −→ |∂αφm(x)− ∂αφ(x)| < ǫ ∀x ∈ K.)From now on, designate by D(Ω) the space C∞0 (Ω) equipped with the topology structure (complete locallyconvex topological vector space) given by (i) and (ii).

Definition 4.2. The dual space D ′(Ω) of D(Ω) is called the space of distributions.

The dual space D ′(Ω) is obviously the linear space (or vector space) of all continuous linear mappingT : D(Ω) → C. Thus if S, T ∈ D ′(Ω), c ∈ C, then

(T + S)(φ) = T (φ) + S(φ),

(cT )(φ) = c(T (φ)),

T (φ+ cψ) = T (φ) + cT (ψ)

for all φ, ψ ∈ D(Ω).T ∈ D ′(Ω) if and only if

limj→∞

T (φj) = T (φj) whenever D(Ω)− limj→∞

φj = φ in D(Ω).

T ∈ D ′(Ω) if and only if for any compact subset K ⊂ Ω there exist CK > 0 and NK > 0 such thatT (φ) ≤ CK sup

K|∂αφ| for all |α| ≤ NK , whenever φ ∈ D(Ω) with supp(φ) ⊂ K.

The topology structure on D ′(Ω) is given by the following convergence criterion: Tj → T in D ′(Ω) ⇐⇒Tj(φ) → T (φ) in C for all φ ∈ D(Ω). That is, the convergence of Tj → T in D ′(Ω) is weak* convergence.

45


Definition 4.3. A function u defined almost everywhere on Ω is said to be locally integrable on Ω if u ∈L1(K) for every measurable compact subset K of Ω. In this case, we denote by u ∈ L1loc(Ω).Example 4.1. f(x) = |x|, x ∈ (−∞,∞)f is not differentiable at 0, but f ′ is defined almost everywhere on (−∞,∞) and f ′ ∈ L1loc(−∞,∞).

Examples of distribution

Example 4.2. Regular distributions : For all u ∈ L1loc(Ω), there is an associated distribution Tu ∈ D ′(Ω).Indeed, let u ∈ L1loc(Ω), and define the map Tu : D(Ω) → C by

Tu(φ) =

∫

Ω

uφ dx, ∀φ ∈ D(Ω).

Then for c ∈ C, φ, ψ ∈ D(Ω),

Tu(φ+ cψ) =

∫

Ω

u(φ+ cψ)dx

=

∫

Ω

uφdx+ c

∫

Ω

uψdx

= Tu(φ) + cTu(ψ).

Therefore Tu : D(Ω) → C is a linear map. To show that Tu is continuous, suppose that φj → φ in D(Ω) asj → ∞. Then there exists a K ⋐ Ω such that supp(φj) ⊂ K, and supp(φ) ⊂ K. Thus,

|Tu(φj)− Tu(φ)| ≤ supx∈K

|φj(x)− φ(x)|∫

K

|u(x)|dx→ 0

as n→ ∞, since |φj − φ| → 0 uniformly. This shows that Tu is a distribution.Example 4.3. Dirac delta distribution. For a ∈ Ω, define a linear form δa : D(Ω) → R by

δa(φ) = φ(a) ∀φ ∈ D(Ω).Exercise 4.1. Check that δa is a distribution.

Example 4.4. Heaviside function Define H : R → R by

H(x) =

{1, x ≥ 00, x < 0.

Then H ∈ L1loc(R). Therefore, by Example 4.2 there is an element TH ∈ D ′(R). Indeed it is given by

TH(φ) =

∫

R

H(x)φ(x)dx =

∫ ∞

0

φ(x)dx.

Differentiation of distribution

Definition 4.4. Given a distribution T ∈ D ′(Ω), its weak or distributional derivative (or derivative in thedistribution sense) ∂T∂xj ∈ D

′(Ω) is defined by

∂T

∂xj(φ) = −T ( ∂φ

∂xj) ∀φ ∈ D(Ω).

In general, for T ∈ D ′(Ω), ∂αT ∈ D ′(Ω) is given by∂αT (φ) = (−1)|α|(∂αφ) ∀φ ∈ D(Ω). (4.1)

The partial differentiation ∂α : D ′(Ω) → D ′(Ω) defined by (4.1) is continuous in the following sense:if Tj → T in D ′(Ω), then ∂αTj → ∂αT in D ′(Ω).



Example 4.5. u(x) = 12 |x|, x ∈ (−∞,∞)(i) T ′u : D(R) 7→ R : let ϕ ∈ D(R) be arbitrary. Then set K = supp(ϕ), which is compact.

T ′u(ϕ) = −∫

K

1

2|x|ϕ′(x) dx = 1

2

[∫

(−∞,0)∩K

xϕ′(x) dx−∫

(0,∞)∩K

xϕ′(x) dx

]

=1

2

[−∫

(−∞,0)∩K

ϕ(x) dx+

∫

(0,∞)∩K

ϕ(x) dx

]

=

∫

K

(H(x)− 12)ϕ(x) dx.

Consequently, T ′u = TH− 12

in D ′(R)

(ii) T ′H : D(R) 7→ R : ∀ϕ ∈ D(R),

T ′H(ϕ) = (−1)TH(ϕ′) = −∫ ∞

0

ϕ′(x) dx

= ϕ(0) = δ0(ϕ)

where δ0 is Dirac delta distribution(function).

Example 4.6. If f : R → R has a continuous and bounded derivative in R \ {x1, · · · , xm}, with possiblejumps at xk, k = 1, · · · ,m. Let Jk = f(xk+)− f(xk−), k = 1, · · · ,m. Then

d

dxTf = T d̃f

dx

+m∑

k=1

Jkδxk ,

where d̃fdx is regarded as an L1loc function defined everywhere by the derivative of f except at xk’s. Indeed,

setting x0 = −∞, xm+1 = ∞, for all φ ∈ D(R),

d

dxTf (φ) = −

∫

R

f(x)dφ

dxdx =

m+1∑

k=1

∫ xkxk−1

df

dx(x)φ(x)dx+

m∑

k=1

Jkφ(xk)

=

∫

R

d̃f

dx(x)φ(x)dx+

m∑

k=1

Jkφ(xk).

Let v ∈ L2(Ω). Then since v ∈ L1loc(Ω), we identify it with its corresponding distribution Tv ∈ D ′(Ω)defined in Example 4.2.

From now on, all the derivatives are understood in the distributional sense.

4.2 Sobolev Spaces

Definition 4.5. For a nonnegative integer m and p ∈ [1,∞] the Sobolev norms are defined by

‖u‖m,p,Ω :=

∑

0≤|α|≤m

‖Dαu‖pp,Ω

1p

, 1 ≤ p


Theorem 4.1. In general,Hm,p(Ω) ⊂Wm,p(Ω).

The following salient theorem [MS64] removes pains to distinguish the above two definitions.

Theorem 4.2 (H=W, Meyers-Serrin (1964)). For p ∈ [1,∞),Hm,p(Ω) =Wm,p(Ω).

Exercise 4.2. Let Ω = (−1, 1) and u(x) = |x|. Then show that u ∈W 1,∞(Ω) but u 6∈ H1,∞(Ω).Due to Theorem 4.2, we will follow the definition of the original Sobolev spaces (due to Sobolev himself)

Wm,p(Ω). In particular if p = 2, we will use the notation Hm(Ω) for Wm,2(Ω) so that

Hm(Ω) =Wm,2(Ω).

Theorem 4.3. Wm,p(Ω) is a Banach space for p ∈ [1,∞] with the Sobolev norm ‖·‖m,p,Ω .Theorem 4.4. Wm,p(Ω) is a separable Banach space for p ∈ [1,∞) and a reflexive Banach space forp ∈ (1,∞)Corollary 4.1. Hm(Ω) is a separable Hilbert space for all nonnegative integer m with the inner product

(u, v)m,Ω =∑

0≤|α|≤m

∫

Ω

DαuDαv dx

Recall that

Definition 4.6. A topological vector space V is called separable if there exists a countable dense subset S ofV .

Thus Theorem 4.4 implies that there exists a countable subset (fn)n of Wm,p(Ω) such that for any

f ∈Wm,p(Ω), there exists (αn)n ∈ C such that∥∥∥∥∥f −

∞∑

n=1

αnn

∥∥∥∥∥m,p,Ω

= 0.

Denote by | · |m,p,Ω the Sobolev seminorm defined by

|f |m,p,Ω =

∑

|α|=m

‖∂αf‖0,p,Ω

1p

.

Notice that

|f |m,p,Ω = 0 if f(x) = polynomial of order m− 1 in Ω.

The spaces H1(Ω) and W 1,p(Ω)

Example 4.7. The Sobolev space of order (1, 2) on Ω is given by

H1(Ω) = {v ∈ L2(Ω) : ∂v∂xj

∈ L2(Ω), j = 1, · · · , d},

equipped with the inner product

(u, v)1,Ω =

∫

Ω

uvdx+d∑

k=1

∫

Ω

∂u

∂xk

∂v

∂xkdx, u, v ∈ H1(Ω).

and the norm||u||1,Ω = [(u, u)1,Ω]

12 .



Example 4.8. If m = 1,

‖f‖1,p,Ω =

‖f‖p0,p,Ω +

n∑

j=1

∥∥∥∥∂f

∂x

∥∥∥∥p

0,p,Ω

1p

=

∫

Ω

|f |p +n∑

j=1

∣∣∣∣∂f

∂x

∣∣∣∣p

dx

1p

,

where the derivatives are understood in the sense of distribution.

Example 4.9. H1(Ω) is a Hilbert space for the inner product (·, ·)1,Ω.

Proof. Suppose {vj} ⊂ H1(Ω) is a Cauchy sequence. Then {vj} and { ∂vj∂xk } for k = 1, · · · , d are Cauchysequences in L2(Ω). Since L2(Ω) is complete, there exist v ∈ L2(Ω) and v(k) ∈ L2(Ω), 1 ≤ k ≤ d, such thatvj → v in L2(Ω) and ∂vj∂xk → v

(k) in L2(Ω).

We have to prove that v(k) = ∂v∂xk in the sense of distribution on Ω. Since I : L2(Ω) → D ′(Ω) defined

by I (φ) = φ for all φ ∈ L2(Ω) is continuous,

I vj → I v in D ′(Ω)

I∂vj∂xk

→ I v(k) in D ′(Ω) k = 1, · · · , d.

Since ∂∂xk : D′(Ω) → D ′(Ω) is continuous,

I∂vj∂xk

=∂I vj∂xk

→ ∂I v∂xk

= I∂v

∂xkin D ′(Ω).

Since the limit is unique in D ′(Ω), we have I ∂v∂xk = I v(k) ∈ D ′(Ω); and hence ∂v∂xk = v

(k) ∈ L2(Ω) for eachk = 1, · · · , d.

Include a commuting diagram here.Therefore v ∈ H1(Ω), and

||vj − v||1,Ω =[∫

Ω

|vj − v|2dx+d∑

k=1

∫

Ω

| ∂vj∂xk

− ∂v∂xk

|2dx] 1

2

→ 0 asj → ∞,

which implies that vj → v in H1(Ω) as j → ∞. �

Example 4.10. H1(Ω) is separable (i.e. there exists a countable dense subset in H1(Ω).)

For a general domain Ω, D(Ω) is not dense in H1(Ω). The closed set Rd \Ω should satisfy (m, p′)-polarcondition, for example; i.e., a distribution T ∈ D(Ω) having support in Rd \ Ω is the zero distribution.However, we recall Definition 4.5, from which we have

Definition 4.7. H10 (Ω) = D(Ω)||·||1,Ω

Corollary 4.2. D(Rd) is dense in H1(Rd) i.e., H10 (Rd) = H1(Rd).

Proof. �

Corollary 4.3. For 1 ≤ p


4.2.1 Schwartz space S(Rd)

S(Rd) := {f ∈ C∞(Rd) | supx∈Rd

|xβ∂αf(x)|


Theorem 4.7 (Parseval formula).

‖f̂‖20 = (2π)d‖f‖20. (4.5)

Theorem 4.8 (Plancherel Theorem). The Fourier transform, defined originally on L1(Rd)∩L2(Rd) extendsuniquely to a map from L2(Rd) to L2(Rd) satisfying

(f̂ , ĝ) = (2π)d(f, g), ‖f̂‖20 = (2π)d‖f‖20, (4.6)

for all f, g ∈ L2(Rd).

We will extend the Fourier transformation to (temperate) distributions.The space S is endowed with the topology defined by the seminorm

|φ|S := supα,β,x

|xβ∂αφ(x)|

makes S a Fréchet space.A space V is called a Fréchet space if it is a locally convex space whose topology is induced by a complete

invariant metric dist(·, ·).

Definition 4.8. A temperate distribution S′ is the set of all continuous linear functional u on S.

We are now in a position to define the Fourier transformation for temperate distributions.

Definition 4.9. For u ∈ S′, the Fourier transform û is defined by

û(φ) = u(φ̂) ∀φ ∈ S. (4.7)

We then have the following theorems.

Theorem 4.9. The Fourier transformation F : S′ → S′ is an isomorphism (with the weak topology) withthe Fourier inversion of the form (4.4) for all u ∈ S′.

Example 4.11. [Rud91a, p.190]

1. All distributions with compact support are temperate distributions.

2. Lp(Rs) ⊂ S(Rd) for p ∈ [1,∞].

3. Pk(Rs) ⊂ S(Rd), k = 0, 1, · · · . The polynomials are temperate distributions.

4. A Borel measure µ on Rd such that∫

Rd

1

(1 + |x|2)k dµ(x)


1. d̂dxj u = iωj û.

2. x̂ju = −i ddωj û.

For any real number s, the Sobolev space can be defined through Fourier transformation:

Hs(Rd) := {f : Rd 7→ R |∫

Rd(1 + |ω|2)s|f̂(ω)|2 dω


Corollary 4.4 (Poincaré Inequality).

‖v‖1,2,Ω ≤ C

d∑

j=1

∥∥∥∥∂v

∂xj

∥∥∥∥2

0,2,Ω

12

∀v ∈ H10 (Ω), (4.8)

where C is independent of v, but depends only on Ω.

Remark 4.2. This corollary is a general version for Lemma 4.1 since

‖v‖21,Ω = ‖v‖20,Ω + ‖∇v‖

20,Ω ≤ C2 ‖∇v‖

20,Ω + ‖∇v‖

20,Ω = (C

2 + 1) ‖∇v‖20,Ω .

In H10 (Ω), by Corollary 4.5

‖·‖H1(Ω) ≃ | · |H1(Ω), i.e., ‖v‖H1(Ω) ≤ C1|v|H1(Ω) ≤ C2 ‖v‖H1(Ω) , ∀v ∈ H10 (Ω)

Corollary 4.5. If Ω is bounded, the seminorm

|v|1,Ω :=

d∑

j=1

‖ ∂v∂xj

‖20,Ω

1/2

is equivalent to the Sobolev norm || · ||1,Ω in H10 (Ω).(i.e. C‖v‖1 ≤ |v|1 ≤ C ′‖v‖1 for some C,C ′ ∈ R)

4.2.4 The definition of traces and trace theorems

For 1-dimensional case, let Ω = (a, b). Then if v ∈ H1(Ω) then v ∈ C0(Ω). In this case, it is not difficult todefine v|Γ i.e., the values v(a) and v(b) for all v ∈ H1(Ω).

However, for d-dimensional case, d ≥ 2, it is not trivial to define boundary values of v ∈ H1(Ω). Indeed,for R < 1, let Ω = {x ∈ R2 : 0 < |x| < R} and v(x) = | log |x||p. Then

∂v

∂xj= p

(log√x21 + x

22

)p−1xj

x21 + x22

and

||v||21,Ω =∫

|v|2 + |∇v|2dx

= 2π

∫ R

0

| log r|2prdr + 2p2π∫ R

0

| log r|2p−2 r2

r4r dr

< ∞ if p < 12

In order for v ∈ H1(Ω) it suffices to choose 0 < p < 12 ; however, the function v can not be represented by acontinuous function in Ω, since v is not bounded near the origin if p > 0.

Let Ω = Rd+. Then Γ = {x = (x′, 0);x′ ∈ Rd−1}.

Lemma 4.2. D(Rd+) is dense in H1(Rd+).

Proof. �

Definition 4.11. For an integer m ≥ 1, an open set Ω ⊂ Rd is said to be m-regular if its boundary Γ is amanifold of class Cm of dimension d− 1, Ω being locally at one side of Γ. Or equivalently, if there exists afinite open cover {Oj}Jj=1 of Ω and invertible ϕj = Oj → B(0; 1) such that ϕj and ϕ−1j are Cm maps and

ϕj(Oi ∩ Ω) = B(0; 1) ∩Rd+ = {y = (y′, yd) ∈ Rd : |y′| < 1, yd > 0}ϕj(Oi ∩ Γ) = {y = (y′, yd) ∈ Rd : |y′| < 1, yd = 0}.



Lemma 4.3. If Ω is 1-regular, then there exists a continuous linear extension operator E : H1(Ω) → H1(Rd)such that

E v(x) = v(x) ∀x ∈ Ω

Proof. (1st step) We begin by considering the case Ω = Rd+. If v ∈ D(Rd+), define E v on Rd by (reflection)

E v(x′, xd) =

{v(x′, xd), if xd ≥ 0v(x′,−xd), if xd ≤ 0.

Then E v is continuous, E v ∈ H1(Rd) and

∂

∂xjE v(x′, xd) =

∂v

∂xj(x′, xd), j = 1, · · · , d− 1,

∂v

∂xj(x′, xd), if xd > 0, j = d,

− ∂v∂xj

(x′,−xd), if xd < 0, j = d.

Then for all v ∈ D(Rd+), we have‖E v‖1,Rd =

√2 ‖v‖1,Rd

+.

Since D(Rd+) is dense inH1(Rd+), E can be extended by linearly and continuously to E : H

1(Rd+) → H1(Rd).(2nd step) Now assume that Ω is an open 1-regular domain in Rd. We know that there exists a partition

of unity {αj}Jj=0 subordinate to the cover {Oj}Jj=1 of the boundary Γ of Ω; i.e.,

αj ∈ D(Oj), 1 ≤ j ≤ J ;J∑

j=0

αj = 1 on Ω, 0 ≤ αj ≤ 1, and Γ ⊂ ∪Jj=1Oj .

Thus, for v ∈ H1(Ω) we can write

v =

J∑

j=0

αjv,

we can define E (αjv) for all j = 0, 1, · · · , J , and then define E v by

E v =J∑

j=0

E (αjv).

First, we haveE (α0v) = α̃0v : the extension of α0v by 0 to R

d \ Ω.Then for j = 1, · · · , J , consider

wj = (αjv) ◦ (ϕ−1j |B+), B+ = B(0; 1) ∩Rd+.

We have wj ∈ H1(B+) and wj = 0 in a neighborhood of {y ∈ ∂B+ : yd > 0}.One can extend wj by 0 in R

d+ \B+. Let w̃j denote this extension. Then w̃j ∈ H1(Rd+), which is again

extended to Rd by reflection (as in the first step)

˜̃wj ∈ H1(Rd) and supp(˜̃wj) ⊂ B(0; 1).

Then let˜̃w̃j ◦ ϕ ∈ H1(Rd): the extension of ˜̃wj ◦ ϕ by 0 in Rd \ Oj . We have

E (αjv) =˜̃w̃j ◦ ϕj , 1 ≤ j ≤ J.

Then v 7→∑Jj=0 E (αjv) is a 1-extension. �



Lemma 4.4. If Ω is 1-regular, D(Ω) is dense in H1(Ω).

Proof. Let v ∈ H1(Ω). Then by the Lemma 4.3, there exists a function E v ∈ H1(Rd) such that E v = v onΩ. By Theorem 4.10, H10 (R

d) = H1(Rd), there exists a sequence {wj}j ⊂ D(Rd) such that wj → E v inH1(Rd).

Let vj be the restriction of wj to Ω. Then

(vj)j ⊂ D(Ω) s.t. vj → v in H1(Ω).

Therefore D(Ω) is dense in H1(Ω). �

Define

‖v‖0,Γ =[∫

Γ

|v(x)|2dσ]1/2

. (4.9)

By using a partition of unity, {αk}Kk=0, we have

L2(Γ) = {v : Γ → R : ˜(αkv) ◦ ϕ−1k (·, 0) ∈ L2(Rd−1), 1 ≤ k ≤ K},

where ˜(αkv) ◦ ϕ−1k (·, 0) is the extension by 0 to Rd−1 \ {y′ ∈ Rd−1 : |y′| < 1}. Then, the map given by

v 7→[

K∑

k=0

∥∥∥∥ ˜(αkv) ◦ ϕ−1k

∥∥∥∥2

0,Rd−1

]1/2(4.10)

defines a norm which is equivalent to the norm given by ‖·‖0,Γ defined in (4.9).Consider the mapping γ0 : D(Ω) → L2(Γ) defined by

γ0v = v|G ∀v ∈ D(Ω).

Lemma 4.5. If Ω is 1-regular, then there exists C > 0 such that

|γ0v|0,Γ ≤ C||v||1,Ω ∀ v ∈ D(Ω).

where C is independent of v, but dependent only on Ω.

Proof. Let v ∈ D(Ω). By the partition of unity {αk}Kk=0, we put

wk = (αkv) ◦ ϕ−1k , 0 ≤ k ≤ K.

Since ||v(·, 0)||0,Rd−1 ≤ ||v||1,Rd+for all v ∈ D(Rd+),

‖w̃k(·, 0)‖0,Rd−1 ≤ ‖w̃k‖1,Rd+.

Since αk and ϕ−1k are smooth, we can find

‖w̃k‖1,Rd+≤ Ck ‖v‖1,Ω .

Thus||w̃k(·, 0)||0,Rd−1 ≤ Ck ‖v‖1,Ω .

By the equivalence of the two norms ‖·‖0,Γ given by (4.9) and (4.10), we are done. �

Theorem 4.12 (Trace Theorem). Suppose that Ω is 1-regular. Then D(Ω) is dense in H1(Ω) and γ0 : v 7→γ0v = v|Γ from D(Ω) to L2(Γ) can be extended linearly and continuously to a map, again denoted by γ0

γ0 : H1(Ω) → L2(Γ).

Remark 4.3. The trace theorem holds for domain Ω ⊂ Rd which is Lipschitz domain or piecewise 1-regular.



4.2.5 Application of Trace Theorem

Theorem 4.13. Suppose Ω is open bounded in Rd with its boundary Γ being piecewise C1. Then

H10 (Ω) = Ker(γ0) i.e.,

H10 (Ω) = {v ∈ H1(Ω) : γ0v = v|Γ = 0 on Γ}.

Proof. (1) H10 (Ω) ⊂ {v ∈ H1(Ω) : v|Γ = 0}. Indeed, let v ∈ H10 (Ω). Then there exists {ϕj} ⊂ D(Ω) suchthat ϕj → v in H1(Ω). Since γ0 is continuous,

L2(Γ)− limj→∞

γ0ϕj = γ0(H1(Ω)− lim

j→∞vj) = γ0v

Since γ0ϕj = 0 on Γ for all j, we have γ0v = 0.(2) To show H10 (Ω) ⊃ {v ∈ H1(Ω) : v|Γ = 0}. Using the local property and partition of unity, it suffices

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