introduction to numerical analysis for partial ﬀ equations · introduction to numerical analysis...

Introduction to Numerical Analysis forPartial Differential Equations

Norikazu SAITO (齊藤宣一)The University of Tokyo

norikazu[AT]ms.u-tokyo.ac.jp

June 25, 2015

Contents

Introduction. Modeling and analysis 1

Chapter I. Finite difference method for the heat equation 3

1 The heat equation 31.1 Initial-boundary value problems . . . . . . . . . . . . . . . . . . 31.2 Uniqueness and maximum principle . . . . . . . . . . . . . . . . 51.3 Construction of a solution and Fourier’s method . . . . . . . . 71.4 Duhamel’s principle . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Explicit finite difference scheme 112.1 Finite difference quotients . . . . . . . . . . . . . . . . . . . . . 112.2 Explicit scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Numerical experiments by Scilab . . . . . . . . . . . . . . . . . 15

3 Implicit finite difference schemes 253.1 Simple implicit scheme . . . . . . . . . . . . . . . . . . . . . . . 253.2 The implicit θ scheme . . . . . . . . . . . . . . . . . . . . . . . 283.3 Inhomogeneous problems . . . . . . . . . . . . . . . . . . . . . 303.4 Numerical experiments by Scilab . . . . . . . . . . . . . . . . . 31

4 Convergence and error estimates 374.1 ℓ∞ analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374.2 ℓ2 analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 414.3 Numerical examples . . . . . . . . . . . . . . . . . . . . . . . . 45

5 Nonlinear problems 495.1 Semilinear diffusion equation . . . . . . . . . . . . . . . . . . . 495.2 Explicit scheme . . . . . . . . . . . . . . . . . . . . . . . . . . . 505.3 Implicit schemes . . . . . . . . . . . . . . . . . . . . . . . . . . 555.4 An example: Gray-Scott model . . . . . . . . . . . . . . . . . . 57

6 Complement for FDM 616.1 Non-homogeneous Dirichlet boundary condition . . . . . . . . . 616.2 Neumann boundary condition . . . . . . . . . . . . . . . . . . . 626.3 ℓ∞ analysis revisited . . . . . . . . . . . . . . . . . . . . . . . . 68

Problems and further readings 76Problems for Chapter I . . . . . . . . . . . . . . . . . . . . . . . . . 76Further readings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

Chapter II. Finite element method for the Poisson equation 79

7 Variational approach for the Poisson equation 797.1 Dirichlet’s principle . . . . . . . . . . . . . . . . . . . . . . . . . 797.2 Galerkin’s approximation . . . . . . . . . . . . . . . . . . . . . 81

8 Finite element method (FEM) 83

9 Tools from Functional Analysis 899.1 Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 899.2 Lipschitz domain . . . . . . . . . . . . . . . . . . . . . . . . . . 919.3 Lemmas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

10 Weak solution and regularity 9710.1 Weak formulation . . . . . . . . . . . . . . . . . . . . . . . . . . 9710.2 Regularity of solutions . . . . . . . . . . . . . . . . . . . . . . . 9810.3 Galerkin’s approximation and Cea’s lemma . . . . . . . . . . . 99

11 Shape-regularity of triangulations 10111.1 Interpolation error estimates . . . . . . . . . . . . . . . . . . . 10111.2 Proof of Lemma 11.1 . . . . . . . . . . . . . . . . . . . . . . . . 104

12 Error analysis of FEM 109

13 Numerical experiments by FreeFem++ 113

Problems and further readings 12313.1 Problems for Chapter II . . . . . . . . . . . . . . . . . . . . . . 12313.2 Further readings . . . . . . . . . . . . . . . . . . . . . . . . . . 123

Chapter III. Abstract elliptic PDE and Galerkin method 127

14 Theory of Lax and Milgram 127

15 Galerkin approximation 131

16 Applications 13316.1 Convection-diffusion equation . . . . . . . . . . . . . . . . . . . 13316.2 Elliptic PDE of the second order . . . . . . . . . . . . . . . . . 136

Problems and further remark 139Problems for Chapter III . . . . . . . . . . . . . . . . . . . . . . . . 139Further remark . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139

References 141

Introduction. Modeling and analysis

The aim of this lecture is to give an introduction to the mathematical theoryof numerical methods for solving partial differential equations. To accomplishthis purpose, I shall concentrate my attention on the following two typicaltopics:

1. the finite difference method for the one dimensional heat equation; and

2. the finite element method for the two dimensional Poisson equation.

I believe that they contain the core idea of numerical methods for PDEs.

Anyway, why does one study Numerical Analysis?

approximation (discretization)

computer

real world(phenomena)

flow around buildingsnumerical weather forecastderivative (finance)etc.

mathematical model(differential equations)

continuous variables

data visualization

computational model

A partial answer (of mine) is as follows.

• Application of computer simulation of wide range phenomena is expand-ing to life sciences, clinical medicine, economics, and other areas beyondthe limited fields of science and technology.

• That expansion is bringing about wide and useful information for use inour life.

• The greater the degree to which computer simulations are used to addresscomplicated and large scale problems, the greater becomes the demandto find solutions to their related mathematical problems.

• Actually, the process of simulation is not completed inside of a computer.It encompasses processes of all kinds such as modeling of the targetedphenomenon, mathematical analysis of the model, approximation anddiscretization of differential equations, implementation of algorithms,program writing, visualization of computer output, validation of simula-tion by comparison with actual phenomena, and final evaluation of thesimulation reliability.

• All of these branches of processes are connected to the strong trunk ofmathematics.

• Summing up, numerical analysis demands the pursuit of mathematicaltruth and contributes to society through mathematics simultaneously —a quite rewarding activity.

Norikazu SaitoJune 25, 2015

I. Finite difference method for the heat equation

1 The heat equation

1.1 Initial-boundary value problems

The heat conduction phenomenon of a thin wire is described by the heatequation (heat conduction equation). The wire is assumed to be the unitlength. We suppose that a function u = u(x, t) denotes the temperature ofthe wire at a position x ∈ [0, 1] and a time t ≥ 0.The heat equation is expressed as

ut = kuxx + f(x, t),

where k is a positive constant, called the heat conduction coefficient, andf(x, t) denotes the supply/absorption of heat. Moreover, we write as

ut =∂u

∂t, uxx =

∂x2.

This equation should be considered together with one of the following bound-ary conditions:

• Dirichlet boundary condition:

u(0, t) = b0(t), u(1, t) = b1(t),

where b0(t) and b1(t) are given functions;

• Neumann (flux) boundary condition:

−kux(0, t) = b0(t), kux(1, t) = b1(t),

where b0(t) and b1(t) are given functions;

• Robin boundary condition:

−kux(0, t) = α(u(0, t)− β), kux(1, t) = α(u(1, t)− β),

where α and β are given positive constants.

Moreover, we solve the equation together with the initial condition

u(x, 0) = a(x) (0 ≤ x ≤ 1),

where a(x) denotes a given continuous function in 0 ≤ x ≤ 1.

Our first target problems of this chapter are the initial-boundary value prob-lems

ut = kuxx (0 < x < 1, t > 0)

u(0, t) = 0, u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1),

and ut = kuxx + f(x, t) (0 < x < 1, t > 0)

u(0, t) = 0, u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1).

Hereinafter, we make the following assumptions:

• k > 0 is a constant;

• a ∈ C[0, 1] with the compatibly condition a(0) = a(1) = 0;

• f ∈ C((0, 1)× (0,∞)).

Notation. For any subset Ω of Rn, we write as

C(Ω) = “the set of all continuous (real-valued) function defined in Ω”.

Moreover, we write as C[a, b] = C([a, b]) for abbreviation.

Remark 1.1Problem (1.1) is a special case of Problem (1.2). We, however, study theseproblems separately.

Remark 1.2We will treat only the homogeneous boundary condition

u(0, t) = 0, u(1, t) = 0.

If posing the inhomogeneous boundary condition

u(0, t) = b0(t), u(1, t) = b1(t)

instead of the homogeneous condition in (1.1) or (1.2), we can reduce theproblems to those of the form (1.2). In fact, set w(x, t) = b0(t)(1−x)+b1(t)xand consider a new unknown function v(x, t) = u(x, t) − w(x, t). Then, ifu(x, t) solves (1.2), the function v(x, t) satisfies vt−kvxx = f−(wt−kwxx) =f − wt and v(0, t) = v(1, t) = 0. Therefore, defining f(x, t) = f(x, t) −wt(x, t) and a(x) = a(x)−w(x, 0), we obtain the problem of the form (1.2)with f(x, t) and a(x). Thus, it suffices to consider only the homogeneousboundary condition.

Definition 1.3A function u = u(x, t) is a (classical) solution of (1.1) or (1.2) if and only if

1. u ∈ C([0, 1]× [0,∞));

2. ut, ux, uxx ∈ C((0, 1)× (0,∞));

3. it satisfies the heat equation ut = kuxx or ut = kuxx + f(x, t) for all0 < x < 1 and t > 0 together with the boundary condition u(0, t) =u(1, t) = for all t > 0;

4. limt→+0

∥u(·, t)− a∥∞ = 0.

Definition 1.4 (L∞ norm)For v ∈ C[0, 1], we let

∥v∥∞ = maxx∈[0,1]

|v(x)|,

which we call the L∞ norm or maximum norm of v.

Remark 1.5∥ · ∥ = ∥ · ∥∞ is a norm of X = C[0, 1]. That is,

(N1) ∥v∥ ≥ 0 (v ∈ X). Moreover, ∥v∥ = 0 ⇒ v ≡ 0;

(N2) ∥αv∥ = |α| · ∥v∥ (α ∈ R, v ∈ X);

(N3) ∥v + w∥ ≤ ∥v∥+ ∥w∥ (v, w ∈ X).

1.2 Uniqueness and maximum principle

For T > 0, we define

QT = (0, 1)× (0, T ], QT = [0, 1]× [0, T ],

ΓT = (x, t) | 0 ≤ x ≤ 1, t = 0 ∪ (x, t) | x = 0, 1, 0 ≤ t ≤ T= QT \QT (parabolic boundary).

Figure 1.1: QT and ΓT .

Theorem 1.6Let T > 0 be fixed. Suppose that a function u = u(x, t) is continuous in QT

and smooth in QT . Then, we have the following.

(i) ut − kuxx ≤ 0 in QT implies max(x,t)∈QT

u(x, t) = max(x,t)∈ΓT

u(x, t).

(ii) ut − kuxx ≥ 0 in QT implies min(x,t)∈QT

u(x, t) = min(x,t)∈ΓT

u(x, t).

Proof. (i) Assume that ut − kuxx ≤ 0 in QT . Let α = max(x,t)∈ΓT

u(x, t) and set

w = e−t(u−α). Clearly, we have w(x, t) ≤ 0 for (x, t) ∈ ΓT . Hence, it sufficesto prove

w(x, t) ≤ 0 for (x, t) ∈ QT , (∗)

since this implies that u(x, t) ≤ α for (x, t) ∈ QT , which is the desired inequal-ity.Inequality (∗) is proved in the following way. We suppose that the function

w achieves its positive maximum µ at (x0, t0) ∈ QT . Thus, we assume that

µ = w(x0, t0) = max(x,t)∈QT

w(x, t) > 0.

Then, we have wt(x0, t0) ≥ 0 and wxx(x0, t0) ≤ 0. Since w satisfies wt + w −kwxx ≤ 0 in QT , considering this inequality at x = x0 and t = t0, we obtain

0 < wt(x0, t0) + µ− kuxx(x0, t0) ≤ 0.

This is a contradiction; we have proved (∗).(ii) It is a consequence of (i), by considering −u instead of u itself.

Theorem 1.7Problem (1.1) (or (1.2)) has at most one solution.

Theorem 1.8The solution u of (1.1) satisfies the following:

(L∞ stability) ∥u(·, t)∥∞ ≤ ∥a∥∞;

(non-negativity) a(x) ≥ 0 ⇒ u(x, t) ≥ 0 for (x, t) ∈ [0, 1]× [0,∞).

Proof of Theorems 1.7 and 1.8. They are readily obtainable consequencesof Theorem 1.6.

Theorem 1.9

(positivity) a(x) ≥ 0, ≡ 0 ⇒ u(x, t) > 0 for (x, t) ∈ (0, 1)× (0,∞).

Proof. See, for example, the following textbook

[3] L. C. Evans: Partial Differential Equations, 2nd edition, Amer-ican Mathematical Society, 2010.

1.3 Construction of a solution and Fourier’s method

Our next task is to construct explicitly a solution of (1.1). To do so, we applyFourier’s method.

First, we assume that a solution u(x, t) of (1.1) is expressed as u(x, t) =φ(x)η(t), where φ(x) and η(t) are fictions only on x and t, respectively. Then,substituting them into the heat equation and the boundary condition, we have

(i) −φ′′(x) = λφ(x), φ(0) = φ(1) = 0;

(ii) −η′(t) = kλη(t).

The system (i) is nothing but the eigenvalue problem, and its solution is givenas

λn = (nπ)2 , φ(x) =√2 sin

√λnx =

√2 sin(nπx) (n = 1, 2, . . .).

Here, the coefficients of sin-functions have been defined as

(φn, φm) =

0φn(x)φm(x) dx = δnm.

On the other hand, a solution of (ii) is given as η(t) = e−kλnt. Therefore,functions

un(x, t) = e−kλntφn(x) (n = 1, 2, . . .)

solve the heat equation with the boundary condition.Then, we consider

u(x, t) =

∞∑n=1

αnun(x, t) =

∞∑n=1

αne−kλntφn(x)

and find coefficients αn∞n=1 to satisfy the initial condition. Setting t = 0 inthe expression above, we have (formally)

a(x) =

∞∑n=1

αnφn(x).

By using the orthonormal property of φn(x), we obtain

αm = (a, φm) =√2

0a(x) sin(mπx) dx.

Summing up, we get a solution of (1.1)

u(x, t) =

∞∑n=1

αne−kλntφn(x), αn = (a, φn). (1.3)

But, the derivation was somewhat formal and it should be justified. Actu-ally, we have the following theorem.

Theorem 1.10The function defined by the right-hand side of (1.3) is continuous in [0, 1]×[0,∞) and is of class C∞ in [0, 1]× (0,∞). Moreover, it is a solution of (1.1).

Proof. See Chapter 1 of

[6] 藤田宏，池部晃生，犬井鉄郎，高見穎郎：数理物理に現れる偏微分方程式 I [岩波講座基礎数学，解析学 II-iv]，岩波書店，1977年．

Remark 1.11The function defined by the right-hand side of (1.3) itself is well-defined fora ∈ L2(0, 1). Then, it is of class C∞ in [0, 1] × (0,∞), and it satisfies theheat equation with the boundary condition. Moreover, the initial conditionis satisfied in the following sense;

limt→+0

0|u(x, t)− a(x)|2 dx = 0.

1.4 Duhamel’s principle

The solution of (1.1) could be expressed as

u(x, t) =∞∑n=1

(a, φn)e−kλntφn(x),

=∞∑n=1

(∫ 1

0a(y)φn(y) dy

)e−kλntφn(x)

( ∞∑n=1

e−kλntφn(y)φn(x)

0G(x, y, t) a(y) dy,

G(x, y, t) =∞∑n=1

e−kλntφn(y)φn(x)

is the Green function associated with (1.1).We recall that the Green function G(x, y, t) satisfies the following:

1. G(x, y, t) is of class C∞ in [0, 1]× (0,∞);

2. G(x, y, t) satisfies the heat equation with the boundary condition as afunction of both (x, t) and (y, t);

3. G(x, y, t) = G(y, x, t), G(x, y, t) ≥ 0;

0G(x, y, t)dy ≤ 1.

For the proof of those facts, see Chapter 1 of [6] for example.

We now come to consider the heat equation with a source term:ut = kuxx + f(x, t) (0 < x < 1, t > 0)

u(0, t) = 0, u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1).

We are not able to apply Fourier’s method; So, we need another device. Wefirst decompose u into u(x, t) = v(x, t)+w(x, t), where v(x, t) and w(x, t) are,respectively, solutions of

vt = kvxx (0 < x < 1, t > 0)

v(0, t) = 0, v(1, t) = 0 (t > 0)

v(x, 0) = a(x) (0 ≤ x ≤ 1),wt = kwxx + f(x, t) (0 < x < 1, t > 0)

w(0, t) = 0, w(1, t) = 0 (t > 0)

u(x, 0) = 0 (0 ≤ x ≤ 1).

We know that v(x, t) is expressed as

v(x, t) =

0G(x, y, t)a(y) dy.

To derive the expression of w(x, t), we introduce a parameter s > 0 andsuppose that W (x, t; s) solves

Wt = kWxx (0 < x < 1, t > s)

W (0, t) = 0, W (1, t) = 0 (t > s)

W (x, s; s) = f(x, s) (0 ≤ x ≤ 1).

Actually, W (s, t; s) is expressed as

W (x, t; s) =

0G(x, y, t− s)f(y, s) dy (0 ≤ x ≤ 1, t ≥ s).

At this stage, we define

w(x, t) =

0W (x, t; s) ds

and this is actually the desired function. In fact, we have (formally)

∂t(x, t) = W (x, t; t) +

∂t(x, t; s) ds

= f(x, t) +

0k∂2W

∂x2(x, t; s) ds = f(x, t) + k

∂x2(x, t)

and w(x, t) satisfies the boundary and initial conditions.Summing up, the solution of (1.2) is given as

u(x, t) =

0G(x, y, t)u0(y) dy +

0W (x, t; s)ds

0G(x, y, t)u0(y) dy +

0G(x, y, t− s)f(y, s) dyds.

This formula is called Duhamel’s principle.

2 Explicit finite difference scheme

2.1 Finite difference quotients

Recall that the differentiation of a function u(x) at x = a is defined by

u′(a) =du

dx(a) = lim

u(a+ h)− u(a)

So, it is natural to expect that

u′(a) ≈ u(a+ h)− u(a)

h(|h| ≪ 1).

Actually, letting I be a bounded closed interval containing a and a ± h forh > 0, we have

u(a+ h) = u(a) + u′(a)h+1

2u′′(a+ θh)h2

with some θ ∈ (0, 1) and, hence,∣∣∣∣u′(a)− u(a+ h)− u(a)

∣∣∣∣ = ∣∣∣∣12u′′(a+ θh)h

∣∣∣∣ ≤ 1

2h∥u′′∥L∞(I). (2.1)

In the similar manner, we have

u(a− h) = u(a)− u′(a)h+1

2u′′(a− θh)h2

and ∣∣∣∣u′(a)− u(a)− u(a− h)

∣∣∣∣ = ∣∣∣∣12u′′(a− θh)h

∣∣∣∣ ≤ 1

2h∥u′′∥L∞(I). (2.2)

On the other hand, using

u(a+ h) = u(a) + u′(a)h+1

2u′′(a)h2 +

3!u(3)(a+ θ1h)h

u(a− h) = u(a)− u′(a)h+1

2u′′(a)h2 − 1

3!u(3)(a− θ2h)h

with some θ1, θ2 ∈ (0, 1), we can estimate as∣∣∣∣u′(a)− u(a+ h)− u(a− h)

∣∣∣∣≤∣∣∣∣ 1

2 · 3!u(3)(a+ θ1h)h

∣∣∣∣+ ∣∣∣∣ 1

2 · 3!u(3)(a− θ2h)h

∣∣∣∣ ≤ 1

6h2∥u(3)∥L∞(I).

Now, we propose a redefinition h = h/2 and obtain∣∣∣∣∣u′(a)− u(a+ h2 )− u(a− h

∣∣∣∣∣ ≤ 1

24h2∥u(3)∥L∞(I).

Name Definition Target Accuracy Assum.

forward Euleru(a+ h)− u(a)

hu′(a) O(h) u ∈ C2

backward Euleru(a)− u(a− h)

hu′(a) O(h) u ∈ C2

first order central differenceu(a+ h

2)− u(a− h

hu′(a) O(h2) u ∈ C3

second order central differenceu(a+ h)− 2u(a) + u(a− h)

h2u′′(a) O(h2) u ∈ C4

O(h) u ∈ C3

Table 2.1: finite difference quotients

Next, we consider the second derivatives. Thus, using

u(a+ h) = u(a) + u′(a)h+1

2u′′(a)h2 +

3!u(3)(a)h3 +

4!u(4)(a+ θ1h)h

u(a− h) = u(a)− u′(a)h+1

2u′′(a)h2 − 1

3!u(3)(a)h3 +

4!u(4)(a− θ2h)h

with some θ1, θ2 ∈ (0, 1), we can perform the estimation∣∣∣∣u′′(a)− u(a− h)− 2u(a) + u(a+ h)

∣∣∣∣≤∣∣∣∣ 14!u(4)(a+ θ1h)h

∣∣∣∣+ ∣∣∣∣ 14!u(4)(a− θ2h)h2

∣∣∣∣ ≤ 1

12h2∥u(4)∥L∞(I). (2.3)

We summarize those finite difference quotients in Table 2.1.

2.2 Explicit scheme

We return to the initial-boundary value problem for the heat equationut = kuxx (0 < x < 1, t > 0)

u(0, t) = u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1),

where k > 0 is a constant and a(x) is a continuous function satisfying a(0) =a(1) = 0.We introduce a set of grid points

Qτh = (xi, tn) | xi = ih, tn = nτ (0 ≤ i ≤ N + 1, n ≥ 0),

where h =1

N + 1with a positive integer N and τ > 0. We remark that x0 = 0

and xN+1 = 1. Moreover, let us denote by uni an approximate value of u(xi, tn)to be solved.

xx1 x2 xN xN+1

Figure 2.1: Qτh.

In view of the previous paragraph, ut and uxx are approximated by

ut(xi, tn) ≈un+1i − uni

τ(forward Euler) and

uxx(xi, tn) ≈uni−1 − 2uni + uni+1

h2(second order central difference).

The boundary and initial conditions are approximated by

un0 = unN+1 = 0 (n ≥ 1) and

u0i = a(xi) (0 ≤ i ≤ N + 1).

Thus, we arrive at a finite difference scheme to (1.1) as follows;un+1i − uni

uni−1 − 2uni + uni+1

h2(1 ≤ i ≤ N, n ≥ 0)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1).

We call this the explicit finite difference scheme (simply, explicit scheme) to(1.1).Set, throughout this chapter,

λ = kτ

It is useful to rewrite (2.4) as

u(n) = Kλu(n−1) (n ≥ 1), u(0) = a, (2.4)

u(n) =

un1...

∈ RN , a =

a(x1)...

a(xN )

∈ RN ,

1− 2λ λ

0. . .

λ 1− 2λ λ. . .

0 λ 1− 2λ

∈ RN×N . (2.5)

Definition 2.1(i) (The vector ∞ norm)

∥u∥∞(= ∥u∥ℓ∞) = max1≤i≤N

|ui| (u = (ui) ∈ RN ).

(ii) (The matrix ∞ norm)

∥G∥∞ = maxu∈RN

∥Gu∥∞∥u∥∞

(G ∈ RN×N ).

Lemma 2.2For G = (gij) ∈ RN×N , we have

(i) ∥Gu∥∞ ≤ ∥G∥∞∥u∥∞ (u ∈ RN );

(ii) ∥G∥∞ = max1≤i≤N

N∑j=1

|gij |.

Proof. (i) It is obvious from the definition.

(ii) Set M = max1≤i≤N

N∑j=1

|gij |. Let u ∈ RN be arbitrary. Then,

∥Gu∥∞ = max1≤i≤N

∣∣∣∣∣∣n∑

∣∣∣∣∣∣ ≤ max1≤i≤N

N∑j=1

|gij | · |uj |

≤ ∥u∥∞ max1≤i≤N

n∑j=1

|gij | = M∥u∥∞.

Hence, we have ∥G∥∞ ≤ M .

Next, we suppose M =

n∑j=1

|gkj |. We define v = (vj) by setting

1 (gkj ≥ 0)

−1 (gkj < 0).

Then, we have ∥v∥∞ = 1 and gkjvj ≥ 0 for any j. Therefore, we can estimateas

∥G∥∞ = maxu∈RN

∥Gu∥∞∥u∥∞

≥ ∥Gv∥∞∥v∥∞

= ∥Gv∥∞ = max1≤i≤N

∣∣∣∣∣∣N∑j=1

∣∣∣∣∣∣≥

∣∣∣∣∣∣N∑j=1

∣∣∣∣∣∣ =N∑j=1

|gkj | = max1≤i≤N

N∑j=1

|gij | = M.

Combining those inequalities, we obtain ∥G∥∞ = M .

Definition 2.3(i) For u = (ui) ∈ RN , u ≥ 0

def.⇐⇒ ui ≥ 0 (∀i).

(ii) For G = (gij) ∈ RN×N , G ≥ Odef.⇐⇒ gij ≥ 0 (∀i, j).

Theorem 2.4Let u(n) = (ui) ∈ RN be the solution of (2.4). Then, under the condition

(0 <)λ ≤ 1

we have for n ≥ 1

(ℓ∞ stability) ∥u(n)∥∞ ≤ ∥a∥∞;

(non-negativity) a ≥ 0 ⇒ u(n) ≥ 0.

Proof. We have Kλ ≥ O by 1− 2λ ≥ 0. Hence, a ≥ 0 implies u(n) = Knλa ≥

0. On the other hand, by Lemma 2.2,

∥Kλ∥∞ = |λ|+ |1− 2λ|+ |λ| = λ+ (1− 2λ) + λ = 1.

Therefore, ∥u(n)∥∞ ≤ ∥Kλ∥n∞∥a∥∞ ≤ ∥a∥∞.

2.3 Numerical experiments by Scilab

It is not difficult to implement the explicit scheme (2.4). Here, we presentseveral sample codes of Scilab. 1

A simple code for calculating the explicit scheme (2.4) is given below (Listing1, heat11.sci):1 It is explained in Wikipedia as follows: Scilab is an open source, cross-platform numericalcomputational package and a high-level, numerically oriented programming language. Itcan be used for signal processing, statistical analysis, image enhancement, fluid dynamicssimulations, numerical optimization, and modeling, simulation of explicit and implicitdynamical systems and (if the corresponding toolbox is installed) symbolic manipulations.MATLAB code, which is similar in syntax, can be converted to Scilab. Scilab is one ofseveral open source alternatives to MATLAB. (http://en.wikipedia.org/wiki/Scilab)

Listing 1: heat11.sci

// *** explicit scheme for heat equation without source ***// INPUTS:// N: division of space, lambda: parameter (<= 0.5),// Tmax: length of time inverval// coef: heat conduction coefficientfunction heat11(N, lambda, Tmax, coef)// [a, b]: space interval, u(a, t) = ua, u(b, t) = ub: boundary valuesa = 0.0; b = 1.0; ua = 0.0; ub = 0.0;// h: space mesh, x[ ]: vector for computation, xx[ ]: vector for

drawingh = (b - a)/(N + 1); x = [a + h: h: b - h]’; xx = [a; x; b];// tau: time increment, nmax: nmax*tau < Tmax <= (nmax+1)*tautau = lambda*h*h/coef; nmax = int(Tmax/tau) + 1;// step: parameter for drawingstep_num = 40; step = max(int(nmax/step_num), 1);// u: approximation of u(x, t), uu: vector for drawing// set initial valueu = func_a(x); uu = [ua; u; ub];scf(1); plot2d(xx, uu, style = 5);// def of matrices A and KA = 2*eye(N, N)-diag(ones(N-1 , 1), -1)-diag(ones(N-1 , 1), 1);K = eye(N, N) - lambda*A;// iterationfor n = 1:nmax

u = K*u;if modulo(n, step)==0

uu = [ua; u; ub]; plot2d(xx, uu, style = 2);end

end// label for scf(1)xlabel(’x’); ylabel(’u’);// pdf filexs2pdf(1,’heat11.pdf’);endfunction// *** local function ***// Initial valuesfunction [y] =func_a(x)

y=min(x, 1.0 - x);// y=x.*sin(3*%pi*x).*sin(3*%pi*x);// y=sin(%pi*x);

endfunction

The definitions of variables used above are given below:

variable definition variable definition

N N h h = 1/(N + 1)

Tmax T coef k

a left side of x-interval= 0 b right side of x-interval = 1

ua boundary value at x = a ub boundary value at x = b

tau τ nmax [T/τ ] + 1

lambda λ = kτ/h2

x x = (x1, . . . , xN ) xx x = (x0, x1, . . . , xN , xN+1)

u u(n) = (un1 , . . . , unN ) uu u = (un0 , u

n1 , . . . , u

nN , unN+1)

After running Scilab, we use heat11.sci in the following manner (see Fig.2.2):

In Scilab Window:

--> exec(’heat11.sci’);

--> heat_ex1(63, 0.5, 0.2, 1)

Figure 2.2: Results of computation. (left) a(x) = x sin2(3πx) (right) a(x) =sin(πx).

More sophisticated codes (Listing 2, heat12.sci and Listing 3, heat13.sci)are given below. We can get Fig. 2.3 and 2.4 by using heat12.sci andheat13.sci, respectively, however, skip the explanation.

// *** explicit scheme for heat equation without source ***// INPUTS:// N: division of space, lambda: parameter (<= 0.5),// Tmax: length of time inverval// coef: heat conduction coefficientfunction heat12(N, lambda, Tmax, coef)// [a, b]: space interval, u(a, t) = ua, u(b, t) = ub: boundary valuesa = 0.0; b = 1.0; ua = 0.0; ub = 0.0;// h: space mesh, x[ ]: vector for computation, xx[ ]: vector for

drawingh = (b - a)/(N + 1); x = [a + h: h: b - h]’; xx = [a; x; b];// tau: time increment, nmax: nmax*tau < Tmax <= (nmax+1)*tautau = lambda*h*h/coef; nmax = int(Tmax/tau) + 1;// step: parameter for drawingstep_num = 40; step = max(int(nmax/step_num), 1);// u: approximation of u(x, t), uu: vector for drawing// set initial valueu = func_a(x); uu = [ua; u; ub];// draw initial valuescf(1); plot2d(xx, uu, style = 5);// def of matrices A and KA = 2*eye(N, N)-diag(ones(N-1 , 1), -1)-diag(ones(N-1 , 1), 1);K = eye(N, N) - lambda*A;// set current timetnow = 0.0;// for 3D plottsp=tnow*ones(1,N); xsp=x.’; Z=u’;// iterationfor n = 1:nmax

u = K*u + tau*func_f(x, tnow);

tnow = n*tau;if modulo(n, step)==0

uu = [ua; u; ub]; plot2d(xx, uu, style = 2);tsp=[tsp;tnow*ones(1,N)]; xsp=[xsp;x’];Z=[Z;u’];

endend// label for scf(1)xlabel(’x’); ylabel(’u’);// 3D viewscf(2); mesh(xsp,tsp,Z);xset(’colormap’,coolcolormap(32));xlabel(’x’); ylabel(’time’);// pdf filexs2pdf(1,’heat12a.pdf’); xs2pdf(2,’heat12b.pdf’);endfunction// *** local functions ***// Initial valuesfunction [y] =func_a(x)

y=min(x, 1.0 - x);// y=x.*sin(3*%pi*x).*sin(3*%pi*x);// y=sin(%pi*x);// I = find(x<=0.3); J = find(x>0.3 & x<=0.6); K = find(x>0.6);

// y = zeros(size(x,1)); y(I) = 0.3; y(J) = -2*(x(J) - 0.6); y(K) = 0.6;

endfunction// sorce termsfunction [y] = func_f(x, t)

//y=0.0;y=exp(t+3)*(x.^2).*(1-x);

endfunction

Figure 2.3: A result of computation by heat12.sci

// *** explicit scheme for heat equation with source ***function heat13(N, lambda, Tmax, coef)a = 0.0; b = 1.0; ua = 0.0; ub = 0.0;h = (b - a)/(N + 1); x = [a + h: h: b - h]’; xx = [a; x; b];tau = lambda*h*h/coef; nmax = int(Tmax/tau) + 1;step_num = 30; step = max(int(nmax/step_num), 1); rate = 0.05;u = func_a(x); uu = [ua; u; ub]; tt = 0.0*ones(N+2,1);//scf() a new window for drawing,scf(10);set(’current_figure’,10); HF = get(’current_figure’); set(HF, ’

figure_size’,[800, 400]);utp = max(uu) + rate*(max(uu)-min(uu)); ubt = min(uu) - rate*(max(uu)-

min(uu));

subplot(1,2,1); plot2d(xx, uu, style = 5);subplot(1,2,2); param3d(xx, tt, uu, flag=[1,4], ebox=[min(xx),max(xx)

,-0.001,Tmax,ubt,utp]);// def of matrices A and KA = 2*eye(N, N)-diag(ones(N-1 , 1), -1)-diag(ones(N-1 , 1), 1);K = eye(N, N) - lambda*A;// iterationtpast = 0.0;for n = 1:nmax

u = K*u + tau*func_f(x, tpast); // tpast = t_n-1tnow = n*tau; // tnow = t_nif modulo(n, step)==0

uu = [ua; u; ub];tt = tnow*ones(N+2,1);utp = max(uu) + rate*(max(uu)-min(uu)); ubt = min(uu) - rate*(

max(uu)-min(uu));subplot(1,2,1); plot2d(xx, uu, style = 2);subplot(1,2,2);param3d(xx, tt, uu,-45,65,flag=[1,4], ebox=[min(xx),max(xx),min(

tt),max(tt),ubt,utp]);endtpast = tnow;

end// labelsubplot(1,2,1); xlabel(’x’); ylabel(’u’);subplot(1,2,2); xlabel(’x’); ylabel(’t’); zlabel(’u’);// pdf filexs2pdf(10,’heat13.pdf’);endfunction// *** local functions ***// Initial valuesfunction [y] =func_a(x)

//y=min(x, 1.0 - x);// y=x.*sin(3*%pi*x).*sin(3*%pi*x);y=sin(%pi*x);//I = find(x<=0.3); J = find(x>0.3 & x<=0.6); K = find(x>0.6);

// y = zeros(size(x,1)); y(I) = 0.3; y(J) = -2*(x(J) - 0.6); y(K) = 0.6; //

y=0.0;//y=exp(t+3)*(x.^2).*(1-x);

endfunction

Example 2.5We are going to examine the explicit scheme with the aid of Scilab. As-suming k = 1, we consider the initial-boundary value problem (1.1) and itsexplicit scheme (2.4) with the following four initial functions

a1(x) = x sin2(3πx), a2(x) =

0.3 (0 ≤ x ≤ 0.3)

−2(x− 0.6) (0.3 < x ≤ 0.6)

0.6 (0.6 < x ≤ 0.3),

a3(x) =

x (0 ≤ x ≤ 1/2)

1− x (1/2 < x ≤ 1),a4(x) = sin(πx).

• In Figures 2.5, 2.6 and 2.7, numerical solutions u(n) for a1(x), a2(x), and

Figure 2.4: A result of computation by heat13.sci

a3(x), respectively, are displayed.

• It should be noticed that a3(x) is discontinuous and a3(x) is not differ-entiable at x = 1/2. But, we observe that for tn > 0 both numericalsolutions actually approximate smooth functions.

• Figures 2.8 and 2.9 are numerical solutions for a3(x) with λ = 0.51 >1/2. In this case, since we cannot apply Theorem 2.4, the ℓ∞ stabilityand the non-negativity conservation are not guaranteed. In fact, a non-physical oscillation appears (Fig. 2.8) and then negative part of solutionsappears (Fig. 2.9). Moreover, we observe the following. A positive partof solutions at some tn > 0 becomes negative at the next time steptn+1. And, then, it becomes positive again at the next time step tn+2.This change of the sign occurs successively and, consequently, numericalsolution is violated.

• On the other hand, we take λ = 0.51 again and compute the explicitscheme for a4(x) (Fig. 2.10). Then, we do not observe the oscillation.However, if we take a larger λ, then the oscillation actually appears again(Fig. 2.11).

Figure 2.5: Initial value a1(x); λ = 0.49; N = 127; 0 ≤ t ≤ 0.1.

Figure 2.8: Initial value a3(x); λ = 0.51 ; N = 23; 0 ≤ t ≤ 0.1.

3 Implicit finite difference schemes

3.1 Simple implicit scheme

We continue to study finite difference schemes to approximate the initial-boundary value problem for the heat equation without a source term;

ut = kuxx (0 < x < 1, t > 0)

u(0, t) = u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1).

We recall that k > 0 is a constant and a(x) is a continuous function satisfyinga(0) = a(1) = 0.We now take

ut(xi, tn) ≈uni − un−1

τ(backward Euler)

as an approximation of ut(xi, tn). The resulting scheme reads asuni − un−1

h2(1 ≤ i ≤ N, n ≥ 1)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1).

It is equivalently written as

τu(n) − u(n−1) = − k

h2Au(n) (n ≥ 1), u(0) = a,

2 −1 0

0. . .. . .

−1 2 −1. . .

0 −1 2

, u(n) =

un1un2...

. (3.2)

Moreover, we set

Hλ = I + λA =

1 + 2λ −λ 0

0. . .. . .

−λ 1 + 2λ −λ. . .

0 −λ 1 + 2λ

with λ = kτ/h2. Then, (3.1) can be rewritten as

Hλu(n) = u(n−1) (n ≥ 1), u(0) = a. (3.1)

According to this expression, we successively obtain u(1),u(2), · · · by solvingthe linear system (3.1) with the initial vector u(0). We call (3.1) the implicit

finite difference scheme (or, simply, implicit scheme), because the solution isdetermined by solving the system of equations. (We note that the recursiveformula (2.4) is called the explicit scheme.) As described below, there are manyimplicit schemes. So, (3.1) is sometimes called the simple implicit scheme todistinguish it with other implicit schemes.The simple implicit scheme has fine mathematical properties. Thus, we have

the following result.

Theorem 3.1For any λ > 0 and n ≥ 1, there exists a solution u(n) = (uni ) ∈ RN of thesimple implicit scheme (3.1) with properties

(ℓ∞ stability) ∥u(n)∥∞ ≤ ∥a∥∞ for n ≥ 1; and

(positivity) a ≥ 0, = 0 ⇒ u(n) > 0 for n ≥ 1.

The key point is to examine the matrix Hλ very carefully.

Lemma 3.2The matrix Hλ defined by (3.3) is non-singular. Moreover, H−1

λ > O and

∥H−1λ ∥∞ ≤ 1.

The proof of this lemma depends on the following lemma.

Lemma 3.3If G ∈ RN×N satisfies ∥G∥ < 1 for a norm ∥ · ∥ of RN , then we have thefollowing:

(i) The matrix I −G is non-singular;

(ii) (I −G)−1 =

∞∑l=0

(iii) ∥(I −G)−1∥ ≤ 1

1− ∥G∥.

Proof. (i) We argue by contradiction. Thus, we assume that I−G is singularso that there exists u = 0 satisfying (I −G)u = 0 and, hence, Gu = u. Thisimplies ∥u∥ ≤ ∥G∥ · ∥u∥ and ∥G∥ ≥ 1, which is impossible. Hence, we haveverified that I −G is non-singular.(ii) We have

(I −G)(I +G+ · · ·+Gm) = I −Gm+1.

Thus,m∑l=0

Gl = (I −G)−1(I −Gm+1).

Therefore,∥∥∥∥∥(I −G)−1 −m∑l=0

∥∥∥∥∥ =∥∥(I −G)−1

[I − (I −Gm+1)

]∥∥≤ ∥(I −G)−1∥ · ∥G∥m+1 → 0 (m → ∞).

This gives (I −G)−1 =

∞∑l=1

(iii) Noting

(I −G)(I −G)−1 = I ⇔ (I −G)−1 = I +G(I −G)−1

we obtain

∥(I −G)−1∥ ≤ ∥I∥+ ∥G∥ · ∥(I −G)−1∥ ⇔ ∥(I −G)−1∥ ≤ 1

1− ∥G∥.

Proof of Lemma 3.2. We first observe

Hλ = (1 + 2λ)

1 −λ1+2λ. . .

. . .−λ

1+2λ 1 −λ1+2λ

. . .. . . −λ

1+2λ−λ

1+2λ 1

= (1 + 2λ) (I −G) with G =λ

1 + 2λ

. . .. . .

1 0 1. . .

. . . 11 0

Since ∥G∥∞ =λ

1 + 2λ· 2 < 1, we can apply Lemma 3.3 to obtain

H−1λ =

1 + 2λ(I −G)−1 =

1 + 2λ

∞∑l=0

∥H−1λ ∥∞ =

1 + 2λ∥(I −G)−1∥∞

1 + 2λ· 1

1− ∥G∥∞=

1 + 2λ· 1

1− 2λ1+2λ

It remains to show H−1λ > O. To do so, it suffices to verify (I − G)−1 > O.

Because of G ≥ O, we have (I − G)−1 ≥ O. We argue by contradiction toshow G > O. That is, we assume that there is a zero component of the jthcolumn v = (vi) of (I−G)−1. We note that we cannot have v = 0 since I−Gis non-singular. Thus, v ≥ 0, = 0. Without loss of generality, we suppose that

vk = 0 and vk+1 > 0. Then, noting that (I − G)v = ej = (δij) ≥ 0, the kthcomponent of v is

− λ

1 + 2λvk−1 + vk −

1 + 2λvk+1 ≥ 0 ⇔ −vk−1 ≥ vk+1.

Hence, we have vk−1 < 0 which implies a contradiction. Therefore, we haveverified (I −G)−1 > O.

Now, we can state the following proof.

Proof of Theorem 3.1. It is a readily obtainable consequence of Lemma3.2. Thus,

∥u(n+1)∥∞ = ∥H−1λ u(n)∥∞ ≤ ∥H−1

λ ∥∞∥u(n)∥∞ ≤ ∥u(n)∥∞,

andu(n) ≥ 0 ⇒ u(n+1) = H−1

λ u(n) > 0

3.2 The implicit θ scheme

We recall that the explicit and simple implicit schemes for (1.1) are given as

uni − un−1i

un−1i−1 − 2un−1

i + un−1i+1

uni − un−1i

At this stage, as an approximation of (1.1), we consider their average withweight θ ∈ [0, 1];

uni − un−1i

τ= (1− θ)k

un−1i−1 − 2un−1

i + un−1i+1

+θkuni−1 − 2uni + uni+1

h2(1 ≤ i ≤ N, n ≥ 1)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1).

This is called the implicit θ finite difference scheme or implicit θ scheme.Setting λ = kτ/h2 and θ′ = 1−θ, we can rewrite the first and second equalitiesas

τu(n) − u(n−1) = −θ′k

h2Au(n−1) − θk

h2Au(n).

Thus, (3.4) is equivalently expressed as

Hθλu(n) = Kθ′λu

(n−1) (n ≥ 1), u(0) = a. (3.4)

Hθλ = I + θλA =

1 + 2θλ −θλ 0

0. . .

−θλ 1 + 2θλ −θλ. . .

0 −θλ 1 + 2θλ

, (3.5)

Kθ′λ = I − θ′λA =

1− 2θ′λ θ′λ 0

0. . .

θ′λ 1− 2θ′λ θ′λ. . .

0 θ′λ 1− 2θ′λ

. (3.6)

We remark that (3.4) coincides with the explicit and simple implicit schemeswhen θ = 0 and θ = 1, respectively.The cases θ = 0 and θ = 1 have been described in Theorems 2.4 and 3.1.

We here deal with only the case 0 < θ < 1.

Theorem 3.4Assume that

0 < θ < 1, 1− 2(1− θ)λ ≥ 0. (3.7)

Then, for any n ≥ 1, there exists a unique solution u(n) = (uni ) ∈ RN of (3.4)with properties

(ℓ∞ stability) ∥u(n)∥∞ ≤ ∥a∥∞;

(non-negativity) a ≥ 0 ⇒ u(n) ≥ 0;

(positivity) λ = 12(1−θ) , a ≥ 0, = 0 ⇒ u(n) > 0.

Proof. Under the assumption (3.7), Hθλ is non-singular as is verified in theproof of Lemma 3.2. Moreover, we haveH−1

θλ > O and ∥H−1θλ ∥∞ ≤ 1 by Lemma

3.2. We also have Kθ′λ ≥ 0 and ∥Kθ′λ∥∞ = 1. Hence,

∥u(n+1)∥∞ = ∥H−1θλ Kθ′λu

(n)∥∞ ≤ ∥u(n)∥∞.

Next, we assume that a ≥ 0, = 0 and λ = 12(1−θ) . Then, we have Kθ′λa ≥ 0, =

0. Therefore, u(1) = H−1θλ Kθ′λa > 0 and, consequently, u(n) > 0 for n ≥ 1.

Remark 3.5The case θ = 1/2 is called the Crank-Nicolson scheme.

3.3 Inhomogeneous problems

We come to consider the initial-boundary value problem with a source term:ut = kuxx + f(x, t) (0 < x < 1, t > 0)

u(0, t) = u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1),

where f(x, t) is a given continuous function that represents a supply/absorp-tion of heat.The explicit scheme reads as

un+1i − uni

h2+ f(xi, tn) (0 < i < N, n ≥ 0)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1).

We set λ = kτ

f (n) =

f(x1, tn)f(x2, tn)

...f(xN , tn)

Since the first and second equalities could be expressed as

τu(n+1) − u(n) = − k

h2Au(n) + f (n),

the explicit scheme is equivalently written as

u(n) = Kλu(n−1) + τf (n−1) (n ≥ 1), u(0) = a,

where Kλ is defined as (2.5).On the other hand, the simple implicit scheme reads as

uni − un−1i

h2+ f(xi, tn) (0 < i < N, n ≥ 0)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1),

or, equivalently,

Hλu(n) = u(n−1) + τf (n) (n ≥ 1), u(0) = a,

where Hλ is defined as (3.3).Finally, the implicit θ scheme reads as

(n−1) + τ (1− θ)f (n−1) + θf (n)︸︷︷︸=f (n−1+θ)

, (n ≥ 1), u(0) = a,

(3.8)where Hθλ and Kθ′λ are those defined by (3.5) and (3.6), respectively.

0 ≤ θ ≤ 1, 2(1− θ)λ ≤ 1.

Then, for any n ≥ 1, there exists a unique solution u(n) = (uni ) ∈ RN of (3.8)that satisfies the following.

(ℓ∞ stability) ∥u(n)∥∞ ≤ ∥a∥∞ + τn∑

∥f (l−1+θ)∥∞;

(positivity)

θ = 0, λ = 1

2(1− θ), a ≥ 0, = 0, f (l) ≥ 0 (0 ≤ l ≤ n) ⇒ u(n) > 0;

(non-negativity)

θ = 0, a ≥ 0, f (l) ≥ 0 (0 ≤ l ≤ n− 1) ⇒ u(n) ≥ 0.

Proof. It is done in the similar way as that of Theorem 3.4.

3.4 Numerical experiments by Scilab

As is seen in the previous paragraphs, we often meet a system of linear equa-tions of the form

Hθλ︸︷︷︸=H

u(n) = Kθ′λu(n−1) + τf (n−1+θ)︸︷︷︸

where a tri-diagonal matrix H ∈ RN×N is defined as

1 + 2µ −µ 0

0. . .

−µ 1 + 2µ −µ. . .

0 −µ 1 + 2µ

, µ = (1− θ)λ.

Lemma 3.7If G = (gij) ∈ RN×N is a strictly diagonally dominant matrix, i.e.,

|gii| >N∑

j=1,j =i

|gij | (1 ≤ i ≤ N),

then G is non-singular and admits an LU factorization (without pivoting).

Proof. See, for example, Chapter 2 of [22].

SinceH above is a strictly diagonally dominant matrix, we can apply Lemma3.7 to obtain a unique LU factorization

H = LU with

L: the unit lower triangular matrix,U : the upper triangular matrix.

From this, we have

Hu(n) = g(n) ⇔

Lc(n) = g(n)

Uu(n) = c(n).

Thus, the implicit θ scheme is computed by the following steps:

• LU factorization. H is decomposed as H = LU ;

• For n = 1, 2, . . .,

Forward elimination Find c(n) by solving Lc(n) = g(n),

Backward substitution find u(n) by solving Uu(n) = c(n).

This procedure is done in Scilab as follows:

In Scilab Window:

// def of example

--> A = [3, -1, 0; 1, 4, 2; 1, 1, 3];

--> b = [1,2,3]’;

// LU factorization

--> [L, U] = lu(A)

3. - 1. 0.

0. 4.3333333 2.

0. 0. 2.3846154

1. 0. 0.

0.3333333 1. 0.

0.3333333 0.3076923 1.

// Solve Ly = b

--> y = L\b

1.6666667

2.1538462

// Solve Ux = y

--> x = U\y

0.3225806

- 0.0322581

0.9032258

// Check the residual

--> norm(b-A*x)

Moreover, the LU factorization for space matrices is available (the matrixH above is a sparse matrix!);

In Scilab Window>>>

// def of example

--> m = 10;

--> A = 2*eye(m,m)-diag(ones(m-1,1),-1)-diag(ones(m-1,1),1);

--> b = [1:1:10]’;

// redefinition of A as a sparse matrix

--> As = sparse(A);

// LU factorization for a space matrix

--> [Lh, rk] = lufact(As);

--> x=lusolve(Lh, b)

--> norm(b-A*x)

3.178D-14

The following Listing 4 (heat23.sci) is a Scilab code for computing theimplicit θ scheme (3.8). An example of computations is given in Fig. 3.1.

// *** implicit theta scheme for heat equation with source ***function heat23(N, lambda, theta, Tmax, coef)a = 0.0; b = 1.0; ua = 0.0; ub = 0.0;h = (b - a)/(N + 1); x = [a + h: h: b - h]’; xx = [a; x; b];tau = lambda*h*h/coef; nmax = int(Tmax/tau) + 1;step_num = 30; step = max(int(nmax/step_num), 1); rate = 0.05;u = func_a(x); uu = [ua; u; ub]; tt = 0.0*ones(N+2,1);//scf() a new window for drawing,scf(10);

set(’current_figure’,10); HF = get(’current_figure’); set(HF, ’figure_size’,[800, 400]);

utp = max(uu) + rate*(max(uu)-min(uu)); ubt = min(uu) - rate*(max(uu)-min(uu));

subplot(1,2,1); plot2d(xx, uu, style = 5);subplot(1,2,2); param3d(xx, tt, uu, flag=[1,4], ebox=[min(xx),max(xx)

,-0.001,Tmax,ubt,utp]);// def of matrices A and KA = 2*eye(N, N)-diag(ones(N-1 , 1), -1)-diag(ones(N-1 , 1), 1);K = eye(N, N) - (1-theta)*lambda*A; H = eye(N, N) + theta*lambda*A;// LU factorization of HHs = sparse(H); Ls = lufact(Hs);// iterationtpast = 0.0;for n = 1:nmax

tnow = n*tau;u = K*u + tau*((1.0-theta)*func_f(x,tpast) + theta*func_f(x,tnow))

;u = lusolve(Ls, u);if modulo(n, step)==0

uu = [ua; u; ub];tt = tnow*ones(N+2,1);utp = max(uu) + rate*(max(uu)-min(uu)); ubt = min(uu) - rate*(

max(uu)-min(uu));subplot(1,2,1); plot2d(xx, uu, style = 2);subplot(1,2,2);param3d(xx, tt, uu,-45,65,flag=[1,4], ebox=[min(xx),max(xx),min(

tt),max(tt),ubt,utp]);endtpast = tnow;

end// labelsubplot(1,2,1); xlabel(’x’); ylabel(’u’);subplot(1,2,2); xlabel(’x’); ylabel(’t’); zlabel(’u’);// eps filexs2pdf(10,’heat23.pdf’);endfunction// *** local functions ***// Initial valuesfunction [y] =func_a(x)

//y=min(x, 1.0 - x);y=x.*sin(3*%pi*x).*sin(3*%pi*x);//y=sin(%pi*x);//I = find(x<=0.3); J = find(x>0.3 & x<=0.6); K = find(x>0.6);

// y = zeros(size(x,1)); y(I) = 0.3; y(J) = -2*(x(J) - 0.6); y(K) = 0.6; //

//y=0.0;y=exp(t+3)*(x.^2).*(1-x);

endfunction

Figure 3.1: An example of computation of the implicit θ scheme (3.8) byheat23.sci, k = 1, a(x) = x sin2(3πx), f(x, t) = et+3x2(1 − x),N = 63, λ = 1.0, θ = 0.5.

4 Convergence and error estimates

4.1 ℓ∞ analysis

We consider the initial-boundary value problemut = kuxx + f(x, t) (0 < x < 1, t > 0)

u(0, t) = u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1)

and the implicit θ scheme

(n−1) + τf (n−1+θ) (n ≥ 1), u(0) = a, (3.8)

θ′ = 1− θ, f (n−1+θ) = (1− θ)f (n−1) + θf (n) = (fn−1+θi ).

In this section, we study the behavior of the error

e(n) = (eni ) ∈ RN , eni = u(xi, tn)− uni ,

where u(x, t) and u(n) = (uni ) are solutions of (1.2) and (3.8), respectively.We need some more notations. Let

Dτvni =

vni − vn−1i

τ, ∆hv

vni−1 − 2vni + vni+1

Then, the problem (3.8) is rewritten asDτu

ni = (1− θ)k∆hu

n−1i + θk∆hu

ni + fn−1+θ

i (0 < i < N, n ≥ 0)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1).

(3.8′)

Moreover, settingUni = u(xi, tn),

we have

Dτeni − (1− θ)k∆he

n−1i − θk∆he

= DτUni − (1− θ)k∆hU

n−1i − θk∆hU

ni − [Dτu

ni − (1− θ)k∆hu

n−1i − θk∆hu

= DτUni − (1− θ)k∆hU

n−1i − θk∆hU

ni − fn−1+θ

≡ rni .

Thus, e(n) = (eni ), which is called the error vector, is a solution ofDτe

ni = (1− θ)k∆he

n−1i + θk∆he

ni + rni (0 < i < N, n ≥ 0)

en0 = enN+1 = 0 (n > 0)

e0i = 0 (0 ≤ i ≤ N + 1).

At this stage, introducing r(n) = (rni ), which is called the residual vector, weobtain

Hθλe(n) = Kθ′λe

(n−1) + τr(n) (n ≥ 1), e(0) = 0. (4.1)

Therefore, in view of Theorem 3.6, we have the following lemma.

Lemma 4.1Assume that

0 ≤ θ ≤ 1, 2(1− θ)λ ≤ 1. (4.2)

Then, the error vector e(n) = (eni ) satisfies

∥e(n)∥∞ ≤ τ

n∑l=1

∥r(l)∥∞.

Let Q = [0, 1]× [0, T ] with a positive constant T . For a continuous functionv defined in Q, we write as

∥v(x, ·)∥L∞(0,T ) = max0≤t≤T

|v(x, t)| for 0 ≤ x ≤ 1,

∥v(·, t)∥L∞(0,1) = max0≤x≤1

|v(x, t)| for 0 ≤ t ≤ T ,

∥v∥L∞(Q) = max(x,t)∈Q

|v(x, t)|.

Then, ∥v∥L∞(Q) becomes a norm of C(Q). Moreover, for a sufficiently smoothfunction v = v(x, t) and a positive integer m, we write as

∂mt v =

∂tmv, ∂m

x v =∂m

∂xmv.

The following lemma plays a crucial role.

Lemma 4.2For arbitrary T > 0, set Q = [0, 1]× [0, T ]. Assume that

∂mx u ∈ C(Q) (0 ≤ m ≤ 4),

∂ltu ∈ C(Q)

(0 ≤ l ≤ 2) if θ = 1/2

(0 ≤ l ≤ 3) if θ = 1/2

is satisfied. Then, the residual vector r(n) = (rni ) admits an estimate

∥r(n)∥∞ ≤ αθ(T ) for any n satisfying 0 ≤ tn ≤ T,

αθ(T ) =

12h2∥∂4

xu∥L∞(Q) +τ

2∥∂2

t u∥L∞(Q) (θ = 1/2)

12h2∥∂4

xu∥L∞(Q) +5

12τ2∥∂3

t u∥L∞(Q) (θ = 1/2).

Theorem 4.3Let T > 0 be fixed. Assume that (4.2) and (4.3) are satisfied. Then, we havethe error estimate

max0≤tn≤T

∥e(n)∥∞ ≤

CT,θ(τ + h2) (θ = 1/2)

CT,1/2(τ2 + h2) (θ = 1/2),

CT,θ =

T ·max

12∥∂4

xu∥L∞(Q),1

2∥∂2

t u∥L∞(Q)

(θ = 1/2)

T ·max

12∥∂4

xu∥L∞(Q),5

12∥∂3

t u∥L∞(Q)

(θ = 1/2).

Proof of Theorem 4.3. It is a direct consequence of Lemmas 4.1 and 4.2.

Proof of Lemma 4.2. By considering the heat equation at (xi, tn) and (xi, tn−1),we have

f(xi, tn) = ut(xi, tn)− kuxx(xi, tn),

f(xi, tn−1) = ut(xi, tn−1)− kuxx(xi, tn−1).

Hence,

rni = Dτu(xi, tn)− (1− θ)ut(xi, tn−1)− θut(xi, tn)︸︷︷︸=R1

−(1− θ)k [∆hu(xi, tn−1)− uxx(xi, tn−1)]︸︷︷︸=R2

−θk [∆hu(xi, tn)− uxx(xi, tn)]︸︷︷︸=R3

First, we derive the estimations for space discretizations. The error estimate(2.3) gives

|∆hu(xi, tn−1)− uxx(xi, tn−1)| ≤1

12h2∥∂4

xu(·, tn−1)∥L∞(0,1),

|∆hu(xi, tn)− uxx(xi, tn)| ≤1

12h2∥∂4

xu(·, tn)∥L∞(0,1).

Therefore,

|R2|+ |R3| ≤k

12h2∥∂4

xu∥L∞(Q).

Next, we examine the time discretization. Suppose θ = 1/2. For the sakeof simplicity, setting v(t) = u(xi, t), we have

R1 = Dτv(tn)− (1− θ)v′(tn−1)− θv′(tn)

= (1− θ)[Dτv(tn)− v′(tn−1)] + θ[Dτv(tn)− v′(tn)].

Hence, by using (2.1) and (2.2), we obtain

|R1| ≤ (1− θ) · τ2∥∂2

t u(xi, ·)∥L∞(0,T ) + θ · τ2∥∂2

t u(xi, ·)∥L∞(0,T )

≤ (1− θ) · τ2∥∂2

t u∥L∞(Q) + θ · τ2∥∂2

t u∥L∞(Q)

2∥∂2

t u∥L∞(Q).

Combining those inequalities, we have |rni | ≤ |R1|+ |R2|+ |R3| ≤ αθ(T ).Now, we suppose θ = 1/2. By Taylor’s theorem,

v(tn) = v(tn−1) + v′(tn−1)τ +1

2v′′(tn−1)τ

3!v(3)(t)τ3,

v(tn−1) = v(tn)− v′(tn)τ +1

2v′′(tn)τ

2 − 1

3!v(3)(t)τ3

with tn−1 < t, t < tn. These imply

v(tn)− v(tn−1)

τ− v′(tn−1) =

2v′′(tn−1)τ +

3!v(3)(t)τ2,

v(tn)− v(tn−1)

τ− v′(tn) = −1

2v′′(tn)τ +

3!v(3)(t)τ2,

and, moreover,

R1 = Dτv(tn)−1

2v′(tn−1)−

2v′(tn)

2[Dτv(tn)− v′(tn−1)] +

2[Dτv(tn)− v′(tn)]

4[v′′(tn−1)− v′′(tn)] +

2 · 3![v(3)(t) + v(3)(t)]

= −τ

∫ tn

tn−1

v(3)(s) ds+τ2

2 · 3![v(3)(t) + v(3)(t)].

Therefore, we have

|R1| ≤ τ

∫ tn

tn−1

∥∂3t u∥L∞(Q) ds+

2 · 3!· 2∥∂3

t u∥L∞(Q)

4∥∂3

t u∥L∞(Q) +τ2

6∥∂3

t u∥L∞(Q)

12∥∂3

t u∥L∞(Q).

Thus, we deduce |rni | ≤ |R1|+ |R2|+ |R3| ≤ α1/2(T ).

Remark 4.4We extend u(n) = (uni ) to a function uh,τ (x, t) by a piecewise-constantinterpolation. Thus, we set

uh,τ (x, t) = uni (xi−1 ≤ x < xi, tn ≤ t < tn+1)

Let T > 0 be fixed. Assume that (4.2) and (4.3) are satisfied. Then, wehave the error estimate

∥u− uh,τ∥L∞(Q) ≤

CT,θ(τ + h2) (θ = 1/2)

CT,1/2(τ2 + h2) (θ = 1/2).

4.2 ℓ2 analysis

We continue to study the error e(n) of the implicit θ scheme for the heatequation.

Definition 4.5(i) (The vector 2 norm)

∥v∥2 (= ∥v∥ℓ2) =

(N∑i=1

|vi|2)1/2

(v = (vi) ∈ RN ).

(ii) (The matrix 2 norm)

∥G∥2 = maxv∈RN

∥Gv∥2∥v∥2

(G ∈ RN×N ).

Lemma 4.6Let λ1, . . . , λN be eigenvalues of a symmetric matrix G ∈ RN×N . Then, wehave ∥G∥2 = max

1≤i≤N|λi|．

Proof. Set ρ(G) = max1≤i≤N

|λi|, which is called the spectral radius. Since G is

a real symmetric matrix, it admits a diagonalization G = UΛUT, where

λ1 0. . .

, UTU = UUT = I.

By using a scalar product in RN ;

⟨x,y⟩ =N∑i=1

xiyi (x = (xi), y = (yi) ∈ RN ),

we have ∥x∥22 = ⟨x,x⟩ and ⟨Ax,x⟩ = ⟨x, ATx⟩ for any x ∈ RN and A ∈RN×N .Now we let 0 = v ∈ RN and set w = UTv. Then, we can calculate as

∥w∥22 = ⟨UTv, UTv⟩ = ⟨UUTv,v⟩ = ∥v∥22,∥Gv∥22 = ⟨Gv, Gv⟩ = ⟨UΛw, UΛw⟩ = ⟨UTUΛw,Λw⟩

= ⟨Λw,Λw⟩ =N∑i=1

|λivi|2 =N∑i=1

|λi|2|wi|2

≤ ρ(G)2N∑i=1

|wi|2 = ρ(G)2∥w∥22 = ρ(G)2∥v∥22.

Therefore,

∥G∥2 = maxv∈RN

∥Gv∥2∥v∥2

≤ maxv∈RN

ρ(G)∥v∥2∥v∥2

= ρ(G).

On the other hand, suppose that |λk| = ρ(G) and Gu = λku. Then,

∥G∥2 ≥∥Gu∥2∥u∥2

=|λk|∥u∥2∥u∥2

= ρ(G).

Combining those inequalities, we obtain ∥G∥2 = ρ(G).

We again consider the initial-boundary value problemut = kuxx (0 < x < 1, t > 0)

u(0, t) = u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1)

and its implicit θ scheme

(n−1) (n ≥ 1), u(0) = a. (3.4)

The crucial point of the ℓ2 analysis is to rewrite the finite difference schemesin terms of A as follows:

u(n+1) = ϕθ,λ(A)u(n), u(0) = a (3.4)

2 −1 0

0. . .. . .

−1 2 −1. . .

0 −1 2

ϕθ,λ(A) = (I + θλA)−1 (I − θ′λA), θ′ = 1− θ.

For example, we know:

• The explicit scheme: u(n+1) = (I − λA)u(n);

• The simple implicit scheme: u(n+1) = (I + λA)−1u(n);

• The Crank-Nicolson scheme : u(n+1) =

)−1(I − λ

)u(n).

Lemma 4.7The eigenpairs of the eigenvalue problem

Aϕ = µϕ, ϕ = (φi) = 0

are give as µ⟨m⟩ = 4 sin2(

2(N + 1)

φ⟨m⟩ = (φ⟨m⟩i ) =

(√2 sin(mπxi)

)(1 ≤ m ≤ N).

Proof. See Problem 4.

Lemma 4.8(i) For 1

2 ≤ θ ≤ 1, we have ∥ϕθ,λ(A)∥2 ≤ 1.

(ii) For 0 ≤ θ < 12 , we have ∥ϕθ,λ(A)∥2 ≤ 1 if 2λ(1− 2θ) ≤ 1.

Proof. We note that

1. ϕθ,λ(A) is a symmetric matrix;

2. ϕθ,λ(µ⟨m⟩)Nm=1 are all the eigenvalues of ϕθ,λ(A).

(See Problems 5 and 6.) Hence, Lemma 4.6 gives

∥ϕθ,λ(A)∥2 = max1≤m≤N

|ϕθ,λ(µ⟨m⟩)|,

where we have introduced a real-valued function ϕθ,λ(s) = (1 + θλs)−1(1− θ′λs)for s ≥ 0 associating with ϕθ,λ(A). The function ϕθ,λ(s) satisfies

ϕθ,λ(0) = 1;d

dsϕθ,λ(s) =

(1 + θ′λs)2< 0;

ϕθ,λ(s) = −1 ⇔ sλ(1− 2θ) = 2.

Therefore,

0 ≤ θ <1

2⇒ |ϕθ,λ(s)| ≤ 1 (∀s > 0 s.t. sλ(1− 2θ) ≤ 2)

2≤ θ ≤ 1 ⇒ |ϕθ,λ(s)| ≤ 1 (∀s > 0).

Thus, if 1/2 ≤ θ ≤ 1, we always have ∥ϕθ,λ(A)∥2 ≤ 1. On the other hand, if0 ≤ θ < 1/2, we have

(0 <)µ⟨m⟩ = 4 sin2(

2(N + 1)

)≤ 4 ≤ 2

λ(1− 2θ)(1 ≤ m ≤ N)

under the condition 2λ(1− 2θ) ≤ 1. This implies ∥ϕθ,λ(A)∥2 ≤ 1.

We then state stability and convergence results. We introduce

∥v∥h =√h∥v∥2 =

m∑i=1

Obviously, ∥v∥h ≤ ∥v∥∞.

2λ(1− 2θ) ≤ 1 (0 ≤ θ < 1/2)

no condition (1/2 ≤ θ ≤ 1).(4.8)

The, the solution u(n) of (3.4) satisfies

∥u(n)∥h ≤ ∥a∥h

for n ≥ 1.

Proof. It is a direct consequence of Lemma 4.8 and

∥u(n+1)∥h = ∥ϕθ,λ(A)u(n)∥h ≤ ∥ϕθ,λ(A)∥h∥u(n)∥h ≤ ∥ϕθ,λ(A)∥2∥u(n)∥h.

Now, we come to consider the initial-boundary value problem with inhomo-geneous source term:

ut = kuxx + f(x, t) (0 < x < 1, t > 0)

u(0, t) = u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1)

and its implicit θ scheme

(n−1) + τf (n−1+θ) (3.8)

or, equivalently,

u(n) = ϕθ,λ(A)u(n−1) + τ(I + θλA)−1f (n−1+θ), (3.8)

wheref (n−1+θ) = (1− θ)f (n−1) + θf (n).

Theorem 4.10Under the assumption (4.9), the solution u(n) of (3.8) satisfies

∥u(n)∥h ≤ ∥a∥h + τn∑

∥f (l−1+θ)∥h

for n ≥ 1.

Proof. We can prove ∥(I + θλA)−1∥2 ≤ 1 (θ, λ > 0) in the exactly same wayas the proof of Lemma 4.8. Hence, the result follows Lemma 4.8.

Theorem 4.11For arbitrary T > 0, we set Q = [0, 1] × [0, T ]. Assume that (4.9) and (4.3)are satisfied. Then, we have an error estimate

max0≤tn≤T

∥e(n)∥h ≤

CT,θ(τ + h2) (θ = 1/2)

CT,1/2(τ2 + h2) (θ = 1/2),

where CT,θ is a positive constant defined as (4.6).

Proof. In virtue of Theorem 4.10,

∥e(n)∥h ≤ τ

n∑l=1

∥r(l)∥h ≤ τ

n∑l=1

∥r(l)∥∞.

This, together with Lemma 4.2, implies (4.9).

4.3 Numerical examples

In this section, we offer some numerical examples in order to confirm thevalidity of Theorems 4.3 and 4.11. To this end, we use two Scilab functions(See Listing 5 and 6):

heat_error1.sci, error_plo1.sci.

Listing 5: heat error1.sci

// *** error observation 1: implicit theta scheme with source ***// Output: h: mesh size, err0: error in L^\infty, err1: error in L^2function [err0, err2, h] = heat_error1(N, lambda, theta, Tmax, coef)a = 0.0; b = 1.0; ua = 0.0; ub = 0.0;h = (b - a)/(N + 1); x = [a + h: h: b - h]’; xx = [a; x; b];tau = lambda*h*h/coef; nmax = int(Tmax/tau) + 1;u = func_a(x); uu = [ua; u; ub];A = 2*eye(N, N)-diag(ones(N-1 , 1), -1)-diag(ones(N-1 , 1), 1);K = eye(N, N) - (1-theta)*lambda*A; H = eye(N, N) + theta*lambda*A;// LU factorization of HHs = sparse(H); Ls = lufact(Hs);// iteration

tpast = 0.0; tnow = 0.0; err0 = -1.0; err2 = -1.0;for n = 1:nmax

tnow = n*tau;u = K*u + tau*((1.0-theta)*func_f(x,tpast) + theta*func_f(x,tnow))

;u = lusolve(Ls, u);errvect = u - func_sol(x, tnow);err0 = max(norm(errvect, %inf), err0);err2 = max(norm(errvect, 2)*sqrt(h), err2);tpast = tnow;

endendfunction// ********* local functions *********// Initial valuefunction [y] = func_a(x)

//y=sin(%pi*x); // case 1y=(x.^3).*(1-x); // case 2

endfunction// Heat source termfunction [y] = func_f(x, t)

//y=0.0; // case1y=exp(t)*(-x.^4+x.^3+12*x.^2-6*x); // case 2

endfunction// exact solutionfunction [y] = func_sol(x, t)

//y=exp(-%pi^2*t)*sin(%pi*x); // case 1y=exp(t)*(x.^3).*(1-x); // case 2

endfunction

Listing 6: error plot1.sci

// *** error observation: dependence on h (implicit theta scheme withsource) ***

function [errvect0, errvect2, hvect] = error_plot1(lambda, theta, Tmax, coef)

errvect0 = []; errvect2 = []; hvect=[]; N0 = 10; jmax = 5;for j = 1:jmax

N = N0*j;[err0, err2, h] = heat_error1(N, lambda, theta, Tmax, coef)errvect0 = [errvect0; err0]; errvect2 = [errvect2; err2]; hvect =

[hvect; h];end//plot errorsscf(20); set(’current_figure’,20); HF = get(’current_figure’); set(HF,

’figure_size’,[400, 800]);xset(’thickness’,2)plot2d(hvect, errvect0, style = 2, logflag="ll");plot2d(hvect, errvect2, style = 5, logflag="ll");xset(’thickness’,1)xgrid(); xtitle(’Mesh size h vs. E_infty and E_2 Errors’,’log (h)’,’

log (error)’)legend(’E_infty’,’E_2’,4);xs2pdf(20,’error1.pdf’)endfunction

Letting T = 1, we define as

E∞ = max0≤tn≤1

∥e(n)∥∞,

E2 = max0≤tn≤1

∥e(n)∥h = max0≤tn≤1

∥e(n)∥2√h.

Example 4.12Let f(x, t) = 0 and a(x) = sin(πx). Then, the solution of (1.2) is

u(x, t) = e−πt2 sin(πx).

In Fig. 4.1, we observe that E∞ ≈ Ch2 and E2 ≈ Ch2.

Example 4.13Let f(x, t) = et(−x4 + x3 + 12x − 6x) and a(x) = x3(1 − x). Then, thesolution of (1.2) is

u(x, t) = etx3(1− x).

In Fig. 4.2, we observe that E∞ ≈ Ch2 and E2 ≈ Ch2.

(a) θ = 0 (b) θ = 1/2 (c) θ = 1

Figure 4.1: log h vs. logE∞ and logE2 for u(x, t) = e−π2t sin(πx) (Example4.12) and λ = 1/2.

(a) θ = 0 (b) θ = 1/2 (c) θ = 1

Figure 4.2: log h vs. logE∞ and logE2 for u(x, t) = etx3(1 − x) (Example4.13) and λ = 1/2.

5 Nonlinear problems

5.1 Semilinear diffusion equation

Our next target is the initial-boundary value problems for a semilinear diffu-sion equation:

ut = kuxx + ε(1− u)u (0 < x < 1, t > 0)

u(0, t) = u(1, t) = 0 (t > 0)

u(x, t) = a(x) (0 ≤ x ≤ 1),

• k, ε are positive constants;

• a ∈ C[0, 1], a ≡ 0, 0 ≤ a ≤ 1, a(0) = a(1) = 0.

Theorem 5.1The problem (5.1) admits a unique (classical) solution u = u(x, t) in [0, 1]×[0,∞) and it satisfies

0 ≤ u(x, t) ≤ 1 (0 ≤ x ≤ 1, t ≥ 0).

Proof. See §6, Chapter 2 of

[9] 亀高惟倫：非線型偏微分方程式，産業図書，1977年.

We introduce the steady-state problem associated with (5.1): Find a functionw = w(x) (0 ≤ x ≤ 1) which depends only on x such that

0 = kw′′ + ε(1− w)w, w > 0 (0 < x < 1)

w(0) = w(1) = 0.(5.2)

The function w ≡ 0, which is called a trivial solution, clearly solves (5.2).

Theorem 5.2

(i) If ε > kπ2, then (5.2) admits a unique non-trivial solution w(x).

(ii) If ε ≤ kπ2, then (5.2) admits no non-trivial solution.

Proof. See §8, Chapter 1 of [9].

The asymptotic behavior of the solution u(x, t) of (5.1) is summarized asfollows.

Theorem 5.3Let u(x, t) be the solution of (5.1).

(i) If ε > kπ2, we have limt→∞

∥u(·, t)− w∥∞ = 0, where w(x) denotes the

non-trivial solution of (5.2).

(ii) If ε ≤ kπ2, we have limt→∞

∥u(·, t)∥∞ = 0.

Proof. See §8, Chapter 1 of [9].

We would like to verify Theorem 5.3 with the aid of numerical examples. Todo this, we introduce finite difference approximations in the following sections.

5.2 Explicit scheme

We introduce:

• h =1

N + 1with 0 < N ∈ Z;

• τ > 0;

• Qτh = (xi, tn)| xi = ih, tn = nτ (0 ≤ i ≤ N + 1, n ≥ 0);

• uni ≈ u(xi, tn).

The explicit scheme to (5.1) now reads asun+1i − uni

h2+ ε(1− uni )u

ni (1 ≤ i ≤ N, n ≥ 0)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1).

(5.3)Moreover, we let

λ = kτ

h2, r =

kπ2, ρ = τ

), q =

∈ RN .

Below we offer some numerical results in Fig. 5.1–5.6; Parameters are sum-marized in Tab. 5.1 .The cases r > 1 and r ≤ 1, respectively, correspond (i) and (ii) in Theorem

5.3. We observe from Fig. 5.1–5.3 that the solution decays if r ≤ 1 and that thesolution converges a non-trivial steady-state solution if r > 1. On the otherhand, Fig. 5.4–5.6 are results with the same parameters as Fig. 5.1–5.3 exceptfor λ (and thus ρ). We see that there are no differences between Fig. 5.1,5.2 and 5.4, 5.5, respectively. However, we observe an oscillation of numericalsolution in Fig. 5.6. As a matter of fact, the value of ρ plays an importantrole to obtain a stable numerical solution. We examine this issue next.

k ε λ r ρ

Fig. 5.1 10 10 0.4 0.0101 0.8002Fig. 5.2 1 10 0.4 1.0132 0.8015Fig. 5.3 0.1 10 0.4 101.32 0.8153Fig. 5.4 10 10 0.5 0.0101 1.0002Fig. 5.5 1 10 0.5 1.0132 1.0019Fig. 5.6 0.1 10 0.5 101.32 1.0192

Table 5.1: Parameters in Fig. 5.1–5.6. N is fixed as 50.

Figure 5.1: k = 10, ε = 10, N = 50, λ = 0.4, 0 ≤ t ≤ 0.2; r = 0.0101,ρ = 0.8002

Figure 5.2: k = 1, ε = 10, N = 50, λ = 0.4, 0 ≤ t ≤ 0.5; r = 1.0132,ρ = 0.8015

Figure 5.3: k = 0.1, ε = 10, N = 50, λ = 0.4, 0 ≤ t ≤ 1.6; r = 101.32,ρ = 0.8153

Figure 5.4: k = 10, ε = 10, N = 50, λ = 0.5, 0 ≤ t ≤ 0.2; r = 0.0101,ρ = 1.0002

Figure 5.5: k = 1, ε = 10, N = 50, λ = 0.5, 0 ≤ t ≤ 0.5; r = 1.0132,ρ = 1.0019

Figure 5.6: k = 0.1, ε = 10, N = 50, λ = 0.5, 0 ≤ t ≤ 1.6; r = 101.32,ρ = 1.0192

1− ετ − 2λ ≥ 0 ⇔ τ ≤(ε+

⇔ ρ ≤ 1. (5.4)

Then, we have0 ≤ u(n) ≤ q (n ≥ 0),

where u(n) = (uni ) denotes the solution of (5.3).

Proof. We argue by induction. First, by virtue of 0 ≤ a(x) ≤ 1, we have

0 ≤ u0i ≤ 1 (1 ≤ i ≤ N).

Suppose that 0 ≤ uni ≤ 1 for n ≥ 0. Since (5.3) gives

un+1i = λ(uni−1 + uni+1) + (1− 2λ)uni + ετuni (1− uni ),

we deduce un+1i ≥ 0. On the other hand, setting vni = 1 − uni , we have

0 ≤ vni ≤ 1 and

vn+1i = λ(vni−1 + vni+1) + (1− 2λ− ετuni )︸︷︷︸

≥1−2λ−ετ≥0

vni ≥ 0.

Therefore, we obtain un+1i ≤ 1.

The convergence is also guaranteed under the condition (5.4).

Theorem 5.5• Let T > 0 be fixed. Assume that (5.4) holds true.

• Suppose that the solution u(x, t) of (5.1) satisfies

u, ut, utt, ux, uxx, uxxx, uxxxx ∈ C(Q),

where Q = [0, 1]× [0, T ]. Define

2∥utt∥L∞(Q) +

12∥uxxxx∥L∞(Q).

• Let u(n) = (uni ) ∈ RN be the solution of (5.3).

• Define e(n) = (eni ) ∈ RN by setting eni = u(xi, tn)− uni .Then, we have

max0≤tn≤T

∥e(n)∥∞ ≤ eεT − 1

ε(τ + h2)ZT .

Proof. Setting Uni = u(xi, tn) and introducing the difference quotients by

Dτen+1i =

en+1i − eni

τ, ∆he

eni−1 − 2eni + eni+1

we have

Dτen+1i − k∆he

= DτUn+1i − k∆hU

ni − [Dτu

n+1i − k∆hu

ni − ε(1− uni )u

ni − [ut(xi, tn)− kuxx(xi, tn)]︸︷︷︸

+ ε(1− Uni )U

ni︸︷︷︸

That is, Dτe

n+1i = k∆he

ni + rni + gni (1 ≤≤ N, n ≥ 0)

en0 = enN+1 = 0 (n > 0)

e0i = 0 (0 ≤ i ≤ N).

e(n+1) = Kλe(n) + τr(n) + τg(n) (n ≥ 1), e(0) = 0,

where r(n) = (rni ), g(n) = (gni ) and Kλ is defined as (2.5).

Since, by (5.4), 1− 2λ > 0, we have

∥Kλ∥∞ ≤ 1, Kλ ≥ O.

In view of §22.1, we can estimate as

|rni | ≤∣∣DτU

n+1i − ut(xi, tn)

∣∣+ k |∆hUni − uxx(xi, tn)|

≤ τ1

2maxt∈[0,T ]

|utt(xi, t)|+ h2k

12maxx∈[0,1]

|uxxxx(x, tn)|

and, hence,∥r(n)∥∞ ≤ ZT (τ + h2).

Moreover, we can calculate as

gni = ε(1− Uni )U

= ε(1− Uni )U

ni − (1− Un

i )uni + (1− Un

i )uni − ε(1− uni )u

= ε(1− Uni )(U

ni − uni )− ε(Un

i − uni )uni

= ε(1− Uni )e

ni − εeni u

= ε(1− Uni − uni )e

This, together with −1 ≤ 1− Uni − uni ≤ 1, implies

∥g(n)∥∞ ≤ ε∥e(n)∥∞.

In conclusion, we have

∥e(n)∥∞≤ (1 + τε)∥e(n−1)∥∞ + τ(τ + h2)ZT

≤ (1 + τε)2∥e(n−2)∥∞ + [(1 + τε) + 1]τ(τ + h2)ZT

≤ · · ·≤ (1 + τε)n∥e(0)∥∞ + [(1 + τε)n−1 + · · ·+ (1 + τε) + 1]τ(τ + h2)ZT

≤ (1 + τε)n − 1

τε· τ(τ + h2)ZT

≤ enτε − 1

ε· (τ + h2)MT ≤ eεT − 1

ε(τ + h2)ZT .

5.3 Implicit schemes

The condition (5.4) may be strict in the practical computation. We can avoidthis by taking some implicit schemes. We still take the same grid points Qτ

introduced in the previous subsection. We consider semi-implicit schemes to(5.1);

uni − un−1i

h2+ ε(1− un−1

i )un−1i (1 ≤ i ≤ N, n ≥ 1)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1)

(5.5)and

uni − un−1i

h2+ ε(1− un−1

i )uni (1 ≤ i ≤ N, n ≥ 1)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1).

(5.6)They are, respectively, expressed as

(I + λA)u(n) = u(n−1) + ετD(u(n−1))u(n−1) (n ≥ 1), u(0) = a (5.5)

(I + λA)u(n) = u(n−1) + ετD(u(n−1))u(n) (n ≥ 1), u(0) = a, (5.6)

D(v) =

1− v1 0

1− v2. . .

0 1− vN

for v =

v1v2...vN

ε. (5.7)

Then, the solutions u(n) = (uni ) both (5.5) and (5.6) satisfy

0 ≤ u(n) ≤ q (n ≥ 0).

Proof. As before, we set Hλ = I + λA and recall that ∥H−1λ ∥∞ ≤ 1 and

H−1λ > O for any λ > 0.First, let u(n) = (uni ) be the solution of (5.5) and assume that 0 ≤ u(n−1) ≤

q. Then, u(n) ≥ 0 is obvious. Setting v(n) = q − u(n), we have

D(u(n−1))u(n−1) = D(v(n−1))v(n−1)

andHλq − q = (λ, 0, . . . , 0, λ)T ≥ 0.

Therefore,

Hλv(n) = Hλq −Hλu

= Hλq − u(n−1) − ετD(u(n−1))u(n−1)

= Hλq − q + v(n−1) − ετD(v(n−1))v(n−1)

≥[I − τεD(v(n−1))

]︸︷︷︸

v(n−1).

Since, in view of (5.7), the ith diagonal entry of W is estimated as

1− τε(1− vn−1i ) = 1− τε+ τεvn−1

i ≥ 1− τε > 0,

we have W ≥ O. Thus, we obtain v(n) ≥ 0 and, hence, u(n) ≤ q.Next, let u(n) = (uni ) be the solution of (5.6) and assume that 0 ≤ u(n−1) ≤

q. The scheme (5.6) is rewritten as[I + λA− τεD(un−1)

]︸︷︷︸=V

u(n) = u(n−1) (n ≥ 1), u(0) = a.

We observe the following:

• The ith diagonal entry of V is estimated as

1 + 2λ− τε(1− un−1i ) ≥ 1 + 2λ− τε > 1− τε ≥ 0.

On the other hand, all non-diagonal entries are non-positive.

• Writing V = (vi,j), we have, for 2 ≤ i ≤ N − 1,

N∑j=1

vi,j = 1 + 2λ− τε(1− un−1i )− 2λ = 1− τε > 0

and, for i = 1, N ,

N∑j=1

vi,j = 1 + 2λ− τε(1− un−1i )− λ > 0.

This indicates that ∥V ∥∞ < 1.

From these observations, we can get V −1 > O and ∥V −1∥∞ ≤ 1 in the samemanner as the proof of Lemma 3.2. Hence, we immediately deduce u(n) ≥ 0.Setting v(n) = q − u(n), we have

Hλv(n) = Hλq −Hλu

= Hλq − u(n−1) − ετD(u(n−1))u(n)

= Hλq − q + v(n−1) − ετD(v(n−1))v(n)

≥ v(n−1) − ετD(v(n−1))v(n).

Thus, v(n) satisfies[I + λA− τεD(vn−1)

]︸︷︷︸=V ′

v(n) ≥ v(n−1) ≥ 0.

Since V ′−1 > O as verified just above, we conclude v(n) ≥ 0 and, hence,u(n) ≤ q. This completes the proof.

5.4 An example: Gray-Scott model

As an example of nonlinear diffusion equations, we offer the Gray-Scott model:ut = kuuxx + u2v − (β + γ)u (0 < x < L, t > 0)

vt = kvvxx − u2v + β(1− v) (0 < x < L, t > 0)

ux(0, t) = ux(L, t) = vx(0, t) = vx(L, t) = 0 (t > 0)

u(x, 0) = u0(x), v(x, 0) = v0(x) (0 ≤ x ≤ L),

where L, ku, kv, β, γ are positive constants. This is a simple mathematicalmodel that describes a certain auto-catalytic reaction phenomenon. Here,the functions u = u(x, t) and v = v(x, t) are defined to be concentrations oftwo chemical substances. The diffusion coefficients are denoted by positiveconstants ku and kv. The rate of the supply of a chemical substance and theremoval of an intermediate product in a reaction are expressed by β and γ.See, for more concrete explanation,

[17] 三村昌泰 (編): パターン形成とダイナミクス (非線形・非平衡現象の数理 4)，東京大学出版会，2006年.

The explicit scheme to (5.8) is

un+1i − uni

τ= ku

uni+1 − 2uni + uni−1

h2+ (uni )

2vni − (β + γ)uni ,

vn+1i − vni

τ= kv

vni+1 − 2vni + vni−1

h2− (uni )

2vni + β(1− vni )

(1 ≤ i ≤ N, n ≥ 0),

un+10 = un+1

1 , un+1N = un+1

N+1, vn+10 = vn+1

1 , vn+1N = vn+1

N+1 (n ≥ 0),

u0i = u0(xi), v0i = u0(xi) (1 ≤ i ≤ N)(5.9)

• h = L/N (N ∈ N), τ > 0,

• Qτh =

(xi, tn)| xi =

(i− 1

)h, tn = nτ (0 ≤ i ≤ N + 1, n ≥ 0)

• uni ≈ u(xi, tn), vni ≈ v(xi, tn).

Some patterns created by this system are displayed in Fig. 5.7, where wesuppose 0 ≤ t ≤ 1000 and 0 ≤ x ≤ L = 0.5.

(a) (β, γ) = (0.1504, 0.1400)

(c) (β, γ) = (0.1504, 0.0308)

(e) (β, γ) = (0.0192, 0.0448)

(b) (β, γ) = (0.1504, 0.0392)

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9

(d) (β, γ) = (0.1504, 0.0056)

(f) (β, γ) = (0.0096, 0.0308)

Figure 5.7: Solutions uni of (5.9); ku = 10−5, kv = 2ku, N = 128, λ = 1/6,L = 0.5, T = 500.

6 Complement for FDM

6.1 Non-homogeneous Dirichlet boundary condition

So far, we have studied only the homogeneous Dirichlet boundary condition

u(0, t) = 0, u(1, t) = 0.

Non-homogeneous cases are treated similary. As an example, we considerut = kuxx + f(x, t) (0 < x < 1, t > 0)

u(0, t) = b0(t), u(1, t) = b1(t) (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1),

where b0(t) and b1(t) are given continuous function of t ≥ 0. We use the samenotation of Sections 2 and 3.First, we consider the explicit scheme. That is,

un+1i − uni

h2+ f(xi, tn) (1 ≤ i ≤ N, n ≥ 0).

For i = 1, we have

un+11 − un1

−2un1 + un2h2

+ f(x1, tn) + kb0(tn)

h2(n ≥ 0).

Consequently, we derive

u(n+1) = Kλu(n) + τf (n) + λb(n) (n ≥ 0),

b(n) =

b0(tn)

b1(tn)

∈ Rn.

In the similar manner, we derive the implicit θ scheme as

Hθλu(n) = K(1−θ)λu

(n) + τf (n−1+θ) + λb(n−1+θ) (n ≥ 0),

whereb(n−1+θ) = (1− θ)b(n−1) + θb(n).

Since the error vector e(n) satisfies (4.2), we are able to obtain exactly thesame error estimates in the previous section for non-homogeneous problems.

6.2 Neumann boundary condition

We move on to the initial-boundary value problem with the Neumann bound-ary condition:

ut = kuxx + f(x, t) (0 < x < 1, t > 0)

ux(0, t) = ux(1, t) = 0 (t > 0)

u(x, t) = a(x) (0 ≤ x ≤ 1)

We introduce:

• h =1

Nwith 0 < N ∈ Z;

• τ > 0;

• Qτh =

(xi, tn) | xi =

(i− 1

)h, tn = nτ (0 ≤ i ≤ N + 1, n ≥ 0)

• uni ≈ u(xi, tn).

The explicit scheme reads as

un+1i − uni

h2+ f(xi, tn) (1 ≤ i ≤ N, n ≥ 0).

However, un0 and unN+1 are not defined at this stage. We employ the Neumannboundary condition to treat those values. Thus,

ux(0, t) = 0 ⇒ u(x1, t)− u(x0, t)

h≈ 0,

ux(1, t) = 0 ⇒ u(xN+1, t)− u(xN , t)

h≈ 0.

From these, we have

un1 − un0h

= 0 ⇔ un0 = un1 ,

unN+1 − unNh

= 0 ⇔ unN+1 = unN .

The resulting scheme now reads asun+1i − uni

h2+ f(xi, tn) (1 ≤ i ≤ N, n ≥ 0)

un0 = un1 , unN+1 = unN (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1).

As before we set λ = kτ/h2. The first and second equalities are written as

τu(n+1) − u(n) = − k

h2Bu(n) + f (n),

1 −1 0

0. . .. . .

−1 2 −1. . .

0 −1 1

, u(n) =

un1...

, f (n) =

f(x1, tn)...

f(xN , tn)

Hence, the explicit scheme is expressed as

u(n) = (I − λB)︸︷︷︸=Lλ

u(n−1) + τf (n−1) (n ≥ 1), u(0) = a.

On the other hand, the simple implicit scheme isuni − un−1

h2+ f(xi, tn) (1 ≤ i ≤ N, n ≥ 1)

un0 = un1 , unN+1 = unN (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1)

and its matrix representation is

(I + λB)︸︷︷︸=Mλ

u(n) = u(n−1) + τf (n) (n ≥ 1), u(0) = a.

Moreover, the implicit θ scheme reads as

Mθλu(n) = Lθ′λu

(n−1) + τf (n−1+θ) (n ≥ 1), u(0) = a, (6.3)

where 0 ≤ θ ≤ 1, θ′ = 1− θ and

f (n−1+θ) = (1− θ)f (n−1) + θf (n) = (fn−1+θi ).

0 ≤ θ ≤ 1, 1− 2(1− θ)λ ≥ 0. (6.4)

Then, for any n ≥ 1, there exists a unique solution u(n) of (6.3) and itsatisfies

(ℓ∞ stability) ∥u(n)∥∞ ≤ ∥a∥∞ + τn∑

∥f (l−1+θ)∥∞.

Moreover, if f(x, t) ≥ 0 for x ∈ (0, 1) and t ≥ 0, u(n) satisfies the followingproperties:

(positivity) θ = 0, 1− 2(1− θ)λ > 0, a ≥ 0, = 0

⇒ u(n) > 0 for n ≥ 1;

(nonnegativity) θ = 0, a ≥ 0 ⇒ u(n) ≥ 0 for n ≥ 1.

Proof. We first note that we have Lθ′λ ≥ O under the assumption (6.4).(nonnegativity) When θ = 0，we have Mθλ = I and, hence, u(1) = Lθ′λa ≥0 for a ≥ 0.

(positivity) Assume that θ = 0. If 1 − 2(1 − θ)λ > 0, then the diagonalentries of Lθ′λ are positive. Hence, Lθ′λa ≥ 0, = 0 for a ≥ 0, = 0. To showM−1

θλ > O, we follow the method of the proof of Lemma 3.2. The matrix Mθλ

is represented as Ms = I + sB = D(I −G), where s = θλ and

01 + 2s

1 + 2s0 1 + s

0 µ1 0µ2 0 µ2

µ2 0 µ2

0 µ1 0

, µm =s

Since ∥G∥∞ = max

1 + s,

1 + 2s

1 + 2s< 1, we can apply Lemma 3.3

to obtain that I − G is non-singular. Hence, Ms = D(I − G) is also non-

singular and M−1s = (I −G)−1D−1 =

∞∑l=0

GlD−1 ≥ O. On the other hand, we

can verify that M−1s > O by the exactly same manner as the proof of Lemma

(ℓ∞ stability) Let n ≥ 1 be fixed. We set q = (1, . . . , 1)T ∈ RN ,

α = max

0, max

1≤i≤Nun−1i

, β = max

0, max

1≤i≤Nfn−1+θi

We will use the following facts:

• u(n−1) ≤ αq and f (n−1+θ) ≤ βq;

• Bq = 0;

• Ls′v ≥ Ls′v′ for v ≥ v′ because of Ls′ ≥ O.

We can calculate as

Ls′u(n−1) ≤ Ls′αq = α[I − (1− θ)λB]q

= α(I + θλB)q − αλBq = Ms(αq),

f (n−1+θ) ≤ βq = βq −Ms(βq) +Ms(βq)

= β[I − (I + λθB)]q +Ms(βq)

= −βλθBq +Ms(βq) = Ms(βq).

These inequalities, together with the equation (6.3), give

Ms(αq + τβq − u(n)) ≥ Ls′u(n−1) + τf (n−1+θ) −Msu

(n) = 0.

Therefore, by virtue of M−1s > O, we obtain

αq + τβq − u(n) ≥ 0.

In the similar manner, we deduce

Ms(u(n) − α′q − τβ′q) ≥ 0

and, hence,u(n) − α′q − τβ′q ≥ 0,

α′ = min

0, min

1≤i≤Nun−1i

, β′ = min

0, min

1≤i≤Nfn−1+θi

As a result of these results,

α′q + τβ′q ≤ u(n) ≤ αq + τβq.

Thus, we get∥u(n)∥∞ ≤ ∥u(n−1)∥∞ + τ∥f (n−1+θ)∥∞,

which implies the desired inequality.

Remark 6.2In the proof of Theorem 6.1, since

∥(I −G)−1∥∞ ≤ 1

1− ∥G∥∞=

1− 2s1+2s

= 1 + 2s,

we have

∥M−1s ∥∞ ≤ ∥(I −G)−1∥∞∥D−1∥∞ ≤ (1 + 2s) · 1

1 + s> 1.

Thus, we can not directly obtain the ℓ∞ stability, although ∥Lθ′λ∥∞ = 1.

Next, we proceed to a convergence analysis. We extend the solution u(x, t)of (6.2) to a function u(x, t) defined in [−1, 2]× [0,∞) by the reflection. Thus,we introduce

u(x, t) =

u(−x, t) (−1 ≤ x ≤ 0, t ≥ 0)

u(x, t) (0 ≤ x ≤ 1, t ≥ 0)

u(2− x, t) (1 ≤ x ≤ 2, t ≥ 0).

Moreover, the extension f(x, t) of f(x, t) is introduced similarly. Obviously,u(x, t) and f(x, t) are continuous. Furthermore, u(x, t) is a C1 function of x

because of the Neumann boundary condition ux(0, t) = ux(1, t) = 0. Combin-ing these facts and using the heat equation ut = kuxx+f(x, t), we can deducethat u(x, t) is a C2 function of x. This indicates that u(x, t) is a solution of theheat equation ut = kuxx + f(x, t) in (−1, 2)× (0,∞). In the similar manner,if we assume that

∂xm∈ C([0, 1]× [0, T ]) (0 ≤ m ≤ 4),

∂tl∈ C([0, 1]× [0, T ])

(0 ≤ l ≤ 2) if θ = 1/2

(0 ≤ l ≤ 3) if θ = 1/2,

we obtain∂mu

∂xm∈ C([−1, 2]× [0, T ]) (0 ≤ m ≤ 4),

∂tl∈ C([−1, 2]× [0, T ])

(0 ≤ l ≤ 2) if θ = 1/2

(0 ≤ l ≤ 3) if θ = 1/2,

where T > 0 denotes a positive constant and l,m positive integers.We set

M lm(T ) = max

t∈[0,T ]max

x∈[−1,2]

∣∣∣∣ ∂m

∂xm∂l

∂tlu(x, t)

∣∣∣∣ .We consider the error

eni = u(xi, tn)− uni (0 ≤ i ≤ N + 1, n ≥ 1)

and the error vectore(n) = (eni ) ∈ RN , (6.6)

where u(n) = (uni ) ∈ RN denotes the solution of (6.3) with un0 = un1 andunN+1 = unN . Note that eni = u(xi, tn)− uni for 1 ≤ i ≤ N so that e(n) actuallyimplies the error of the finite difference scheme.Then, since

u(x0, tn) = u(x1, tn), u(xN+1, tn) = u(xN , tn),

the error vector satisfies

Mθλe(n) = Lθ′λe

(n−1) + τr(n) (n ≥ 1), e(0) = 0,

where r(n) = (rni ) is defined as

rni = Dτ u(xi, tn)− (1− θ)k∆hu(xi, tn−1)− θk∆hu(xi, tn)

with the notation of §4.Hence, in the exactly same manner as the proof of Theorem 4.3, we can

prove the following result.

Theorem 6.3Let T > be fixed. Let u(x, t) and u(n) be solutions of (6.2) and (6.3), re-spectively. Assume that (6.4) and (4.3) are satisfied. Then, the error vector(6.6) admits an error estimate

max0≤tn≤T

∥e(n)∥∞ ≤

CT,θ(τ + h2) (θ = 1/2)

CT,1/2(τ2 + h2) (θ = 1/2),

CT,θ =

T ·max

4 (T ),1

0 (T )

(θ = 1/2)

T ·max

4 (T ),5

0 (T )

(θ = 1/2).

Remark 6.4 (conservation of the heat flux)The solution u(x, t) of (6.2) satisfies the conservation of the heat flux

J(t) = J(0) =

0ca(x) dx (t ≥ 0),

J(t) =

0cu(x, t) dx (c: the heat capacity)

denotes the total heat flux. Note that we are assuming

u ∈ C([0, 1]× [0,∞)),

ux ∈ C([0, 1]× (0,∞)),

ut, uxx ∈ C((0, 1)× (0,∞)).

Then, we can calculate as

dtJ(t) =

0cut(x, t) dx =

0cuxx(x, t) dx = [cux(x, t)]

x=1x=0 = 0.

Now, we introduce a discrete heat flux for the solution u(n) = (uni ) of (6.3):

N∑i=1

cuni h.

Then, we have the conservation of the discrete heat flux

N∑i=1

ca(xi)h.

6.3 ℓ∞ analysis revisited

Remark 6.5After having finished writing this paragraph, I noticed that the main result,Theorem 6.6 below, is exactly the same as Theorem 10.2 of Thomee [27,page 118]. Moreover, the proof is almost the same as that of Thomee’sone. However, my stability results, Theorem 6.9 and Lemma 6.10, are notdescribed in [27]; Those results may be out of scope of his interest. I shallgive an open question about the stability.

In this (rather long) paragraph, we revisit stability and convergence analysisof the FDM for a one-dimensional heat equation

ut = κuxx + f(x, t) (0 < x < d, t > 0)

u(0, t) = 0, u(d, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ d),

where κ, d are positive constants, f(x, t), a(x) are prescribed continuous func-tions with a(0) = a(d) = 0.In order to state the finite difference scheme, we introduce 0 < N ∈ Z and

τ > 0, and set h = d(1 + N)−1, xi = ih (0 ≤ i ≤ N + 1), and tn = nτ(n ≥ 0). We denote by uni the finite-difference approximation of u(xi, tn) tobe computed.Let 0 ≤ θ ≤ 1. Then, we consider the standard implicit θ scheme

Dτuni = κ∆hu

n−1+θi + fn−1+θ

i (1 ≤ i ≤ N, n ≥ 1)

un0 = unN+1 = 0 (n ≥ 1)

u0i = a(xi) (0 ≤ i ≤ N + 1),

un−1+θi = (1− θ)un−1

i + θuni ,

fn−1+θi = (1− θ)f(xi, tn−1) + θf(xi, tn),

Dτuni =

uni − un−1i

τ, ∆hu

We introduce A,H,K ∈ RN×N as

2 −1 0

0−1 2 −1. . .

0 −1 2

, H = (I + θλA), K = [I − (1− θ)λA],

where I ∈ RN×N denotes the identity matrix and λ = κτh−2. As is well-known, A and H are positive-definite real symmetric matrices so that theyare non-singular. Hence, we can set

G = H−1K.

Introducing

un1...

a(x1)...

a(xN )

, fn−1+θ =

fn−1+θ1...

fn−1+θN

we can rewrite (6.8) equivalently as

un = Gun−1 + τH−1fn−1+θ (n ≥ 1), u0 = a. (6.9)

As usual, we write as

∥v∥∞ = max1≤i≤N

|vi|, ∥v∥2 =

N∑j=1

|vj |2h

(v = (vi) ∈ RN ).

We use

(v,w)2 =N∑i=1

viwih (v = (vi),w = (wi) ∈ RN ).

Obviously, ∥v∥22 = (v,v)2. And, for any norm ∥ · ∥ in RN , we use the samesymbol to express the matrix norm corresponding to ∥ · ∥;

∥B∥ = maxv∈RN

∥Bv∥∥v∥

(B ∈ RN×N ).

The error vector en is defined as

en1...enN

u(x1, tn)− un1...

u(xN , tn)− unN

The main purpose of this paragraph is to prove the following result.

Theorem 6.6Let T > 0 and set Q = [0, d]× [0, T ].(i) Let 1/2 ≤ θ ≤ 1. Suppose that the solution u of (6.30) satisfies

u,∂l

∂tlu,

∂xmu ∈ C(Q) (1 ≤ l ≤ 2, 1 ≤ m ≤ 4). (6.10)

Then, there exists a positive constant C1 which depends only on κ and dsuch that

max0≤tn≤T

∥en∥∞ ≤ C1(τ + h2)√T(∥utt∥L∞(Q) + ∥uxxxx∥L∞(Q)

). (6.11)

(ii) Let θ = 1/2. Suppose that u satisfies

u,∂l

∂tlu,

∂xmu ∈ C(Q) (1 ≤ l ≤ 3, 1 ≤ m ≤ 4). (6.12)

Then, there exists a positive constant C2 which depends only on κ and dsuch that

max0≤tn≤T

∥en∥∞ ≤ C2(τ2 + h2)

√T(∥uttt∥L∞(Q) + ∥uxxxx∥L∞(Q)

). (6.13)

(iii) Let 0 ≤ θ < 1/2 . Take 0 < δ < 1 and assume

2λ(1− 2θ) ≤ 1− δ. (6.14)

Suppose that (6.10) is satisfied. Then, there exists a positive constant C3

which depends only on κ, d and δ such that

max0≤tn≤T

∥en∥∞ ≤ C3(τ + h2)√T(∥utt∥L∞(Q) + ∥uxxxx∥L∞(Q)

). (6.15)

Remark 6.7It is well-known (cf. Theorem 4.3) that, if θ = 1/2, we have

max0≤tn≤T

∥en∥∞ = O(τ2 + h2) (h, τ → 0)

provided withλ ≤ 1.

Theorem 6.6 claims that the condition λ ≤ 1 is not necessary to prove theconvergence (with the optimal-order) of the finite difference scheme.

The matrix

h2A ∈ RN×N

is positive-definite symmetric matrix so that its square root A1/2h is defined in

a natural way. We introduce

|||v||| = ∥A1/2h v∥2 (v ∈ RN ). (6.16)

Lemma 6.8(i) ∥v∥∞ ≤

√d |||v||| for v ∈ RN .

(ii) |||v||| ≤ 2h−1∥v∥2 for v ∈ RN .

Proof. (i) Let v = (v1, . . . , vN )T ∈ RN and set v0 = vN+1 = 0. Then,

∥A1/2h v∥22 = (Ahv,v)2 =

N∑i=1

−vi−1 + 2vi − vi−1

N+1∑i=1

(vi − vi−1

h. (6.17)

Now let 1 ≤ i ≤ N . We can write as vi =

i∑j=0

(vj − vj−1). Hence,

|vi| ≤N+1∑j=1

∣∣∣∣vj − vj−1

∣∣∣∣h1/2h1/2≤

N+1∑j=1

∣∣∣∣vj − vj−1

∣∣∣∣2 h1/2N+1∑

=√d ∥A1/2

h v∥2.

(ii) Again by using (6.17),

|||v|||2 = ∥A1/2h v∥22 =

N+1∑i=1

(vi − vi−1

N+1∑i=1

v2i + v2i−1

h2h ≤ 4

h2∥v∥22.

We introduce the truncation-error vectors rn = (rni ),Rn = (Rn

i ) ∈ RN as

rni = (1− θ)[Dτu(xi, tn)− ut(xi, tn−1)] + θ[Dτu(xi, t

n)− ut(xi, tn)],

Rnj = (1− θ)κ

[uxx(xi, tn−1)−∆hu(xi, t

n−1)]+ θκ [uxx(xi, tn)−∆hu(xi, t

We know

∥rn∥2 ≤√d ∥rn∥∞ ≤

C4τ∥utt∥L∞(Q)

C5τ2∥uttt∥L∞(Q) if θ = 1/2,

(6.18)

∥Rn∥2 ≤√d ∥Rn∥∞ ≤ C6κh

2∥uxxxx∥L∞(Q), (6.19)

where C4, C5, C6 are positive constants depending only on d.The error en solves

en − en−1

τ+ κAhe

n−1+θ = rn +Rn (n ≥ 1), e0 = 0,

where en−1+θ = (1− θ)en−1 + θen. From this, we have the identity(en − en−1

+ κ(Ahen−1+θ,v)2 = (rn +Rn,v)2 (6.20)

for any v ∈ RN . Moreover, the identity

en−θ+1 = τ

(θ − 1

)en − en−1

en + en−1

2(6.21)

is of use.

We can state the following proof.

Proof of Theorem 6.6. (i) and (ii). Let 1/2 ≤ θ ≤ 1. In view of (6.21), wecan calculate as(en − en−1

τ,Ahe

n−θ+1

(θ − 1

)(en − en−1

en − en−1

(en − en−1

τ, Ah

en + en−1

(θ − 1

)∥∥∥∥∥A1/2h (en − en−1)

∥∥∥∥∥2

(∥A1/2

h en∥22 − ∥A1/2h en−1∥22

)≥ 1

(|||en|||2 − |||en−1|||2

). (6.22)

Substituting v = Ahen−1+θ into (6.20), using (6.22) and Young’s inequality,

we have

(|||en|||2 − |||en−1|||2

)+ κ∥Ahe

n−1+θ∥22 ≤ (rn +Rn, Ahen−1+θ)2

2κ∥rn +Rn∥22 +

2∥Ahe

n−1+θ∥22.

Hence,

|||en|||2 ≤ |||en−1|||2 + τ

κ∥rn +Rn∥22

≤ τ

n∑k=1

∥rk +Rk∥22 ≤2τ

n∑k=1

(∥rk∥22 + ∥Rk∥22).

This, together with Lemma 6.8 (i), (6.18) and (6.19), gives the error estimate(6.11) and (6.13).

(iii) Let 0 ≤ θ < 1/2 and assume (6.14) with some 0 < δ < 1. We have(en − en−1

τ,Ahe

n− 12

∥∥∥A1/2h en

∥∥∥22− 1

∥∥∥A1/2h en−1

∥∥∥22.

Again, using (6.21),

(Ahen−1+θ, Ahe

n− 12 ) = ∥Ahe

n− 12 ∥22 + τ

(θ − 1

en − en−1

τ,Ahe

n− 12

)= ∥Ahe

n− 12 ∥22 +

(θ − 1

)(∥Ahe

n∥22 − ∥Ahen−1∥22

Substituting v = Ahen− 1

2 into (6.20) and using those identities, we deduce

(|||en|||2 − |||en−1|||2

)+ κ∥Ahe

n− 12 ∥22

(θ − 1

)(∥Ahe

n∥22 − ∥Ahen−1∥22

)≤ 1

2κ∥rn +Rn∥22 +

2∥Ahe

n− 12 ∥22.

Hence, setting

εn = |||en|||2 − κτ

2− θ

)∥Ahe

n∥22,

we obtain

εn ≤ εn−1 +τ

κ∥rn +Rn∥22 ≤

n∑k=1

(∥rk∥22 + ∥Rk∥22

In view of Lemma 6.8 (ii) and the condition (6.14), we estimate as

|||en|||2 − κτ

2− θ

)∥Ahe

n∥22 ≥ |||en|||2 − κτ

(1− 2θ

h2∥A1/2

h en∥22

≥ [1− 2λ (1− 2θ)] |||en|||2

≥ δ|||en|||2.

Therefore,

|||en|||2 ≤ 2δτ

n∑k=1

(∥rk∥22 + ∥Rk∥22

Summing up this, Lemma 6.8 (i), (6.18) and (6.19), we complete the proof ofTheorem 6.6 (iii).

Instead of the stability in the standard norms ∥ · ∥∞ and ∥ · ∥2, we have thefollowing result.

Theorem 6.9For the solution un of (6.8), we have the stability inequality

|||un||| ≤ |||a|||+ τ

n∑k=1

|||fk−1+θ|||

provided with2λ(1− 2θ) ≤ 1 (6.23)

if 0 ≤ θ < 1/2.

This is a direct consequence of the following lemma.

Lemma 6.10(i) |||G||| ≤ 1 provided with (6.23) if 0 ≤ θ < 1/2.(ii) |||H−1||| ≤ 1.

Proof. (i) It is well-known that (or it is a readily obtainable consequence of

the Spectral Mapping Theorem) ∥G∥2 ≤ 1. Hence, noting A1/2h G = GA

1/2h ,

we obtain |||G||| ≤ 1.

(ii) It follows ∥H−1∥2 ≤ 1 and A1/2h H−1 = H−1A

1/2h .

Remark 6.11According to the expression (6.9) and Lemma 6.10, we obtain

|||en||| ≤ τn∑

(|||rk|||+ |||Rk|||) (n ≥ 1). (6.24)

Therefore, we can deduce the error estimate if |||rn||| and |||Rn||| are esti-mated in terms of h and τ . Consequently, we will succeed in avoiding theconstant δ in the condition (6.14). As a matter of fact, the estimation of|||rn||| is not difficult. In view of (6.17), we can derive

|||rn||| ≤

C7τ∥uttx∥L∞(Q)

C8τ2∥utttx∥L∞(Q) (if θ = 1/2)

with positive constants C7 and C8 depending only on d. This is becausern naturally satisfies the “boundary condition” rn0 = rnN+1 = 0. (We meanthat, if we extend the original definition of rni for 1 ≤ i ≤ N to i = 0, N +1,we have rn0 = rnN+1 = 0.) However, we are able to derive only

|||Rn||| ≤ C9h3/2∥uxxxxx∥L∞(Q)

because of the lack of the “boundary condition” Rn0 = Rn

N+1 = 0. Fur-

thermore, we have observed that there are examples that |||Rn||| = O(h3/2)as h → 0 by numerical experiments. Therefore, it is impossible to deducethe optimal-order error estimates by means of the stability result (6.24).Instead, if we consider the periodic boundary condition, we obtain theoptimal-order error estimates by (6.24).

Remark 6.12For a C1 function w defined in [0, d] with w(0) = w(d) = 0, we have

|||w||| ≤ ∥wx∥L2(0,d),

where w = (w(xj)) ∈ RN . Therefore, under appropriate assumptions on aand f , we obtain, as a corollary of Theorem 6.9 and Lemma 6.8 (i),

∥un∥∞ ≤√d ∥ax∥L2(0,d) +

√d τ

n∑k=1

∥fx(·, tk)∥L2(0,d)

provided with (6.23) if 0 ≤ θ < 1/2.

In order to avoid unessential difficulties, we consider only the case θ = 1/2.Suppose now f ≡ 0. Then, (6.9) implies

un = Gna (n ≥ 1).

In view of Theorem 6.6, there exists τ0, h0 > 0 such that

∥Gna∥∞ ≤ 1 + ∥u∥L∞(Q) (0 ≤ tn ≤ T )

for τ ≤ τ0 and h ≤ h0. Therefore, by virtue of the uniform boundednessprinciple, there exists a potitive constant MT depending on T such that

∥Gn∥∞ ≤ MT (0 ≤ tn ≤ T ). (6.25)

(See also [20, §3.5].)In [14, §7], it is proved that

∥G∥∞ ≤ 1 ⇔ λ ≤ 3

2. (6.26)

λ ≤ 3

2⇒ ∥Gn∥∞ ≤ 1 (∀n ≥ 1). (6.27)

On the other hand, as is stated in Lemma 6.10, we always have

|||G||| ≤ 1 and |||Gn||| ≤ 1 (∀n). (6.28)

Therefore, it would be interesting to consider

∥Gn∥∞ or ∥G∥∞ for λ >3

Remark 6.13For the general θ, we have (cf. [14, §7])

∥G∥∞ ≤ 1 ⇔ λ ≤ 2− θ

4(1− θ). (6.29)

Problems and further readings

Problems for Chapter I

Problem 1Prove that eigenvalues λj and eigenvectors uj of a tri-diagonal matrix

b a0. . .

a b a. . .

∈ RN×N (6.30)

are given as

λj = b+ 2a cosjπ

N + 1and

N + 1, · · · , sin Njπ

)(1 ≤ j ≤ N).

Problem 2Consider the initial-boundary value problem for a heat equation

ut = uxx (0 < x < 1, t > 0)

u(0, t) = u(1, t) = 0 (t > 0)

u(x, 0) = a(x) (0 ≤ x ≤ 1),

(6.31)

where a(x) is a sufficiently smooth function satisfying a(0) = a(1) = 0. (Con-sequently, the solution u(x, t) becomes a sufficiently smooth function of x andt.) Define a set of grid points (xi, tn) = (ih, nτ)| 0 ≤ i ≤ N +1, n ≥ 0 withh = 1/(N + 1) and τ > 0. Put Un

i = u(xi, tn). Then, determine the value ofλ = τ/h2 which gives an estimate of the form∣∣∣∣Un+1

i − Uni

Uni−1 − 2Un

i + Uni+1

∣∣∣∣ = O(τ2 + h4)

as h, τ → 0.

Problem 3Consider the explicit scheme for the initial-boundary value problem (6.31).Suppose the initial values are given as u0i = (−1)i sin(iπh). Prove that asolution of the finite difference scheme is expressed as

uni = (1− 2λ− 2λ cos(πh))nu0i .

Moreover, prove that ∥u(n)∥∞ → ∞ (n → ∞) for λ > 1/2.

Problem 4Prove that (4.7).

Problem 5In the proof of Lemma 4.8, prove that ϕθ,λ(A) is a symmetric matrix.

Problem 6In the proof of Lemma 4.8, prove that ϕθ,λ(µ

⟨m⟩)Nm=1 are all the eigenvaluesof ϕθ,λ(A).

Problem 7Recall that

µm = (mπ)2, φm(x) =√2 sin(mπx) (m = 1, 2, . . .).

are the eigenpairs of the eigenvalue problem

−φ′′(x) = µφ(x) (0 < x < 1), φ(0) = φ(1) = 0, φ ≡ 0.

We introduce a finite difference approximation:−φi−1 − 2φi + φi+1

h2= µφi (1 ≤ i ≤ N)

φ0 = φN+1 = 0, φ = (φi) = 0.(6.32)

Prove that µ⟨m⟩ =4

h2sin2

2(N + 1)

φ⟨m⟩ = (φ⟨m⟩i ) =

(√2 sin(mπxi)

)(1 ≤ m ≤ N).

are the eigenpairs of (6.32). Moreover, prove that

µ⟨m⟩h → µm (h → 0), φm(xi) = φ

⟨m⟩i (0 ≤ i ≤ N + 1).

Problem 8 (machine assignment)Compute numerically (by the finite difference method) the Gray-Scott model

ut = kuuxx + u2v − (β + γ)u (0 < x < 1, 0 < t < T )

vt = kvvxx − u2v + β(1− v) (0 < x < 1, 0 < t < T )

ux(0, t) = ux(1, t) = 0 (0 < t < T )

vx(0, t) = vx(1, t) = 0 (0 < t < T )

u(x, 0) = u0(x), v(x, 0) = v0(x) (0 ≤ x ≤ 1)

for several (β, γ)’s. Report non-trivial shapes of solutions. (recommendation:ku = 10−5，kv = 2ku，T = 500，0.009 ≤ β ≤ 0.151, and 0.03 ≤ γ ≤ 0.141.)

Further readings

In this chapter, I explained only FDM for the one space-dimensional heatequation. But, FDM can be applied to higher space-dimensional partial dif-ferential equations of parabolic, elliptic and hyperbolic types. For those topics,the following standard textbooks are useful:

[16] K. W. Morton and D. F. Mayers: Numerical Solution of Par-tial Differential Equations (2nd ed.), Cambridge University Press,2005.

[25] G. D. Smith: Numerical Solution of Partial Differential Equa-tions, Oxford University Press, 1965. (藤川洋一郎訳：コンピュータによる偏微分方程式の解法，新訂版，サイエンス社，1996年)

[15] S. Larsson and V. Thomee: Partial Differential Equations withNumerical Methods, Springer, 2009.

[26] 田端正久：偏微分方程式の数値解析，岩波書店, 2010.

Therein, important topics including von Neumann condition and Fourier anal-ysis are also described. Moreover, the implementation of FDM is describedin

[24]齊藤宣一：線形・非線形拡散方程式の差分解法と解の可視化，講義ノート (http://www.infsup.jp/saito/modules/mydownloads/)，2011年．

An interesting example of nonlinear problems and its analysis are given in

[18]三村昌泰：微分方程式と差分方程式—数値解は信用できるか？—，「数値解析と非線形現象 (山口昌哉編)」の第 3章，日本評論社，1996年 (オリジナルは 1981年)．

The following old articles are also worth reading:

[20] R. D. Richtmyer and K. W. Morton: Difference Methods forInitial-Value Problems, Interscience Publishers, 1967.

[4] 藤田宏：初期値問題の差分法による近似解法 (B: 微分方程式の近似解法)，自然科学者のための数学概論 [応用編]，寺沢寛一 (編)，岩波書店，1960年．

[5] 藤田宏：境界値問題の差分法による近似解法 (B: 微分方程式の近似解法)，自然科学者のための数学概論 [応用編]，寺沢寛一 (編)，岩波書店，1960年．

II. Finite element method for the Poisson equation

7 Variational approach for the Poisson equation

7.1 Dirichlet’s principle

As a typical model of elliptic partial differential equations, we consider theboundary value problem for the Poisson equation in a two-dimensional boundeddomain Ω ⊂ R2 with the boundary ∂Ω,

−∆u = −(

∂x21+

∂x22

)u = f in Ω, u = 0 on ∂Ω, (7.1)

where u = u(x) = u(x1, x2) denotes the unknown function to be solved andf = f(x) is a given continuous function in Ω.As will be stated in Theorem 7.2, Problem (7.1) and the following mini-

mization problem (variational problem) are closely related:

Find u ∈ U s.t. J(u) = minv∈U

J(v), (7.2)

J(v) =1

∫Ω|∇v|2 dx−

∫Ωfv dx

[(∂v

dx−∫Ωfv dx,

U = v ∈ C1(Ω) | v|∂Ω = “the boundary value of v on Γ” = 0.

Moreover, the following problem is called the Euler-Lagrange equation for(7.2):

Find u ∈ U s.t.

∫Ω∇u · ∇ϕ dx =

∫Ωfϕ dx (∀ϕ ∈ U). (7.3)

Theorem 7.1A function u is a solution of (7.2), if and only if it is a solution of (7.3).

Proof. Let u be a solution of (7.2). Let ϕ ∈ U be arbitrary. Consider a realvalued function j(t) = J(u+ tϕ) for t ∈ R. Then,

j(t) = J(u+ tϕ) =1

∫Ω∇(u+ tϕ) · ∇(u+ tϕ) dx−

∫Ωf(u+ tϕ) dx

∫Ω|∇u|2 dx−

∫Ωfu dx

∫Ω|∇ϕ|2 dx+

∫Ω(∇u · ∇ϕ+∇ϕ · ∇u) dx− t

∫Ωfuϕ dx.

The function j(t) achieves its minimum at t = 0. Hence, j′(t)|t=0 = 0. Thisimplies (7.3).

Conversely, let u be a solution (7.3). Let v ∈ U be arbitrary, and set ϕ =v − u ∈ U . Then,

J(v)− J(u) =1

∫Ω∇(u+ ϕ) · ∇(u+ ϕ) dx−

∫Ωf(u+ ϕ) dx− J(u)

∫Ω|∇ϕ|2 dx+

∫Ω(∇u · ∇ϕ+∇ϕ · ∇u) dx−

∫Ωfϕ dx

∫Ω|∇ϕ|2 dx+

[∫Ω(∇u · ∇ϕ) dx−

∫Ωfϕ dx

]≥ 1

∫Ω|∇ϕ|2 dx ≥ 0.

Theorem 7.2 (Dirichlet’s principle)Suppose that Ω is a bounded Lipschitz domain (See §9 for the definition).

(i) If u ∈ C2(Ω) is a solution of (7.1), then it solves (7.2).

(ii) If u ∈ C2(Ω) is a solution of (7.2)，then it solves (7.1).

Proof. (i) Let u ∈ C2(Ω) be a solution of (7.1). Multiplying by ϕ ∈ U theboth sides of −∆u = f and integrating them over Ω, we have by the integrationby parts∫

Ωfϕ dx =

∫Ω(−∆u)ϕ dx

∫Ω∇u · ∇ϕ dx−

∫Γ[(∇u) · n]ϕ dS =

∫Ω∇u · ∇ϕ dx,

where n = n(s) (s ∈ Γ) denotes the unit outward vector to Γ and dS the lineelement of Γ. Since ϕ is arbitrary, u solves (7.3). Hence, in view of Theorem7.1, it also solves (7.2).(ii) We deduce ∫

Ω(f +∆u)ϕ dx = 0 (∀ϕ ∈ U)

in the similar manner as (i). We argue by contradiction to show that w =f + ∆u ≡ 0. Assume that there exists z ∈ Ω such that w(z) > 0. Then, bycontinuity of w, there exists δ satisfying

w(x) > 0 (x ∈ B(z; δ) = x ∈ R2 | |z − x| < δ ⊂ Ω).

At this stage, we define ϕ ∈ U as

ϕ ≥ 0 (x ∈ Ω), ϕ(x) > 0 (x ∈ B(z; δ/2)), supp ϕ ⊂ B(z; δ).

Then, we have∫Ωwϕ dx =

∫B(z;δ)

wϕ dx ≥∫B(z;δ/2)

wϕ dx > 0,

which is a contradiction. Hence, we have w ≤ 0. By considering −w, wededuce w ≥ 0. Therefore, w ≡ 0, which completes the proof.

Remark 7.3At this stage, we do not know whether a solution exists or not.

We say that v ∈ C(Ω) is a piecewise C1 function if and only if there existsa decomposition Ω = Ω1 ∪ · · · ∪ ΩN with Ωi ∩ Ωj = ∅ (i = j) and that v isof class C1 in each Ωi (i = 1, . . . , N). Then, we can take

V = v ∈ C(Ω) | v is a piecewise C1 function, v|Γ = 0

instead of U in Theorems 7.1 and 7.2. Note that U ⊂ V .

7.2 Galerkin’s approximation

We move to finite dimensional approximations to (7.2) and (7.3). To thispurpose, we introduce

vN ∈ V

∣∣∣∣∣ vN (x) =N∑i=1

ciϕi(x), ciNi=1 ⊂ R

⊂ V,

where ϕ1, ϕ2, . . . , ϕN ∈ V with 0 < N ∈ Z.A finite dimensional approximation to (7.3) reads as

Find uN ∈ VN s.t. J(uN ) = minvN∈VN

J(vN ), (7.4)

which is called the Ritz approximation.On the other hand, a finite dimensional approximation to (7.3) reads as

Find uN ∈ VN s.t.

∫Ω∇uN · ∇vN dx =

∫ΩfvN dx (∀vN ∈ VN ) (7.5)

which is called the Galerkin approximation.Both (7.4) and (7.5) can be represented in the vector-matrix form as follows:

Au = f , (7.6)

A = (aij) ∈ RN×N , aij =

∫Ω∇ϕj · ∇ϕi dx;

f = (fi) ∈ RN , fi =

∫Ωfϕi dx;

u = (ui) ∈ RN with uh =N∑i=1

uiϕi.

Remark 7.4(7.4) ⇔ (7.5).

8 Finite element method (FEM)

We give a concrete example VN by the finite element method (FEM). Weassume that Ω is a polygonal domain for the sake of simplicity. Below wecollect some notions of FEM.

• T (= triangulation of Ω) is defined as follows.

1. T is a set of closed triangles, and Ω =∪T∈T

2. any two triangles of T meet only in entire common faces or sidesor in vertices.

Those triangles are prohibited.

• The size (mesh, granularity) parameter of T is defined as

h = maxT∈T

wherehT = diam (T ) (≡ the diameter of T ).

In what follows, we write T = Th.

• T ∈ Th is called a element (要素)，A vertex of T is called a node (節点).

• Let T ∈ Th be arbitrary. Suppose that Pi = (xi, yi) (i = 1, 2, 3) arevertices of T and |T | the area of T . We set

λi(x, y) =PPjPk

∣∣∣∣∣∣1 1 1x xj xky yj yk

∣∣∣∣∣∣=

2|T |(xjyk − xkyj) + (yj − yk)x− (xj − xk)y

for P = (x, y) ∈ T . Hereinafter, we write as (i, j, k) = (1, 2, 3), (2, 3, 1),and (3, 1, 2). Obviously, λ1, λ2, λ3 are affine functions defined in T andwe have

λi(Pj) = δij . (8.1)

See Fig. 8.5. We call λi3i=1 the barycentric coordinate of T and λi

that corrsponding to Pi.

We have

λ1(x, y) + λ3(x, y) + λ3(x, y) = 1 (x, y) ∈ T. (8.2)

Figure 8.1: Triangulation of Ω = (0, 1)× (0, 1) by a uniform division. Numberof nodes: 81 (left), 289 (center) and 1089 (right). Number ofelements: 128 (left), 512 (center), 2048 (right).

Figure 8.2: Triangulations of Ω = (0, 1) × (0, 1) by a non-uniform division(freefem++ http://www.freefem.org). Number of nodes: 95(left), 333 (center) and 1267 (right). Number of elements: 156(left), 600 (center), 2404 (right).

Figure 8.3: Triangulations of a polygonal domain.

Figure 8.4: Triangulations of a domain with the piecewise smooth boundary.

Figure 8.5: The barycentric coordinate λi(x, y) of T corresponding to Pi.

• N = N +NB is the total number of nodes, where

N is the number of nodes located on the interior of Ω and

NB is the number of nodes located on the boundary Γ.

• PiNi=1 is the set of all nodes, where

PiNi=1 is the set of all nodes located on Ω and

Pi+NINBi=1 is the set of all nodes located on Γ.

• Let Λi = T ∈ Th | Pi ∈ T.

• For 1 ≤ i ≤ N , we define ϕi = ϕh,i(x, y) ∈ C(Ω) by setting

ϕi(x, y) =

λT,i(x, y) (x, y) ∈ T, T ∈ Λi,

0 otherwise,

where λT,i denotes the barycentric coordinate of T corresponding to Pi.See Fig 8.6. From the definition,

suppϕi =∪

T∈Λi

T. (8.3)

In view of (8.1) and (8.2), we have

ϕi(Pj) =

1 (i = j)

0 (i = j),

N∑i=1

ϕi = 1 (8.4)

• The space

Figure 8.6: (left) Λi. (right) ϕi.

Xh = spanϕiNi=1 (8.5)

is called the P1 element on Th. Each vh ∈ Xh is characterized by thefollowing two conditions:

– vh is a continuous function in Ω;

– For each T ∈ Th, there exist α, β, γ ∈ R such that vh|T = α+ βx+γy.

Below we also use

Vh = spanϕiNi=1 (8.6)

which is also called the P1 element (with the zero Dirichlet boundarycondition) on Th. Each vh ∈ Vh is characterized by the following threeconditions:

– vh is a continuous function in Ω;

– For each T ∈ Th, there exist α, β, γ ∈ R such that vh|T = α+ βx+γy;

– vh|∂Ω = 0.

Then, ϕiNi=1 is called the standard basis of Xh and ϕiNi=1 that of Vh.

At this stage, we can state the finite element approximation to (7.3), whichreads as

Find uh ∈ Vh s.t.

∫Ω∇uh · ∇vh dx =

∫Ωfvh dx (∀vh ∈ Vh). (8.7)

Figure 8.7: (left) An example of vh ∈ Xh. (right) An example of vh ∈ Vh.

Au = f ,

A = (aij) ∈ RN×N , aij =

∫Ω∇ϕj · ∇ϕi dx;

f = (fi) ∈ RN , fi =

∫Ωfϕi dx;

u = (ui) ∈ RN , ui = uh(Pi).

Remark 8.1Let T ∈ Th with their vertices Pi = (xi, yi) (i = 1, 2, 3). We have

∇λi =1

(yj − yk

−(xj − xk)

)(constant vector).

Consequently,

∇λi ·−−−→PjPk = 0, ∇λi ·

−−→PjPi = ∇λi ·

−−−→PkPi = 1.

Moreover, we deduce

|∇λi| =1

2|T |PjPk =

κi · PjPk

PjPk =1

where κi denotes the perpendicular length from the vertex Pi to the segmentPjPk; see Fig. ??.

Remark 8.2It is a hard task to construct a triangulation of a given polygonal domain.Some useful soft-wares are available; freefem++ [8].

9 Tools from Functional Analysis

In order to establish a mathematical justification of FEM, we need to introducesome concepts from Functional Analysis. In particular, Sobolev spaces H1(Ω)and H1

0 (Ω) play important roles. We collect some notions below.

9.1 Sobolev spaces

In what follows, Ω is assumed to be a domain (open and connected subset) inR2. A generic point is denoted by x = (x1, x2).

• We shall treat only real-valued functions. Sets of continuous functionsC(R2) = C0(R2) and C(Ω) = C0(Ω) are well-known. For a nonnegativeinteger k, the set of Ck functions defined in R2 and Ω are denoted,respectively, by Ck(R2) and Ck(Ω). The spaces

Ck0 (R2) = v ∈ Ck(R2) | v has a compact support,Ck0 (Ω) = v|Ω | v ∈ Ck

0 (R2), the support of v is contained in Ω

play important roles. Therein, the support of v is defined as

supp v = x ∈ Ω | v(x) = 0.

We introduce, for k ≥ 0,

Ck(Ω) = v|Ω | v ∈ Ck0 (R2).

Moreover, as usual, set

C∞0 (R2) =

∩k≥0

Ck0 (R2), C∞(Ω) =

∩k≥0

Ck(Ω).

• For a continuous function v in Ω, we write

∥v∥∞ = ∥v∥L∞(Ω) = maxx∈Ω

|v(x)|,

which is called the L∞ norm or maximum norm of v.

• The L2 space defined in Ω, which is denoted by L2(Ω), is a Hilbert spaceequipped with the scalar product

(u, v) = (u, v)Ω = (u, v)L2(Ω) =

∫Ωu(x)v(x) dx.

The induced norm is

∥u∥ = ∥u∥Ω = ∥u∥L2(Ω) = (u, u)1/2 =

√∫Ω|u(x)|2 dx.

• We recall partial derivatives in the L2 sense. Let v ∈ L2(Ω). If there isa function g ∈ L2(Ω) such that∫

Ωv∂φ

∂x1dx = −

∫Ωgφ dx (∀φ ∈ C∞

0 (Ω)),

then g is called a generalized partial derivative (in L2(Ω)) of v withrespect to x1 and is denoted by

g =∂v

∂x1, ∂1v, · · · .

We define successively

∂x2= ∂2v,

∂x21= ∂2

1v,∂2v

∂x1x2= ∂1∂2v, · · · .

Those generalized derivatives are unique. If v ∈ L2(Ω) ∩ C1(Ω), then ageneralized derivative implies the usual one. In what follows, we consideronly generalized derivatives.

• Now we can introduce

H1(Ω) =u ∈ L2(Ω) | ∃∂1u, ∂2u ∈ L2(Ω)

together with

(u, v)H1(Ω) = (u, v) + (∂1u, ∂1v) + (∂2u, ∂2v)︸︷︷︸=(∇u,∇v)

∥u∥H1(Ω) =√

(u, u)H1(Ω).

• We also introduce

H2(Ω) =u ∈ L2(Ω) | ∂1u, ∂2u, ∂2

1u, ∂1∂2u, ∂22u ∈ L2(Ω)

together with

(u, v)H2(Ω) = (u, v) +∑i=1,2

(∂iu, ∂iv) +∑

i,j=1,2

(∂i∂ju, ∂i∂jv),

∥u∥H2(Ω) =√

(u, u)H2(Ω).

• The space Hm(Ω), m = 1, 2, is a Hilbert space equipped with the scalarproduct (·, ·)Hm(Ω) and the norm ∥ · ∥Hm(Ω). The space Hm(Ω) is calledthe Sobolev space.

• In the space H1(Ω)

∥∇u∥ =√

(∇u,∇u) =

[∫Ω

(|∂1u|2 + |∂2u|2

defines a semi-norm, where

(∇u,∇v) = (∂1u, ∂1v) + (∂2u, ∂2v).

• We use Schwarz’s inequalities:

|(u, v)| ≤ ∥u∥ · ∥v∥ (u, v ∈ L2(Ω)),

|(∇u,∇v)| ≤ ∥∇u∥ · ∥∇v∥ (u, v ∈ H1(Ω)),

|(u, v)H1(Ω)| ≤ ∥u∥H1(Ω)∥v∥H1(Ω) (u, v ∈ H1(Ω)).

• Let Hm0 (Ω) be the closure of C∞

0 (Ω) in the norm ∥ · ∥Hm(Ω). That is,

Hm0 (Ω) = v ∈ Hm(Ω) | ∃φn ⊂ C∞

0 (Ω) s.t. ∥v − φn∥Hm(Ω) → 0.

9.2 Lipschitz domain

At this stage, we state the definition of a Lipschitz domain Ω.First, we say that the boundary Γ = ∂Ω is a Lipschitz continuous if and

only if, for every x = (x1, x2) ∈ Γ, there exists a neighborhood U of x in R2

and new orthogonal coordinates y = (y1, y2) such that

(i) U = y ∈ R2y | −ai < yi < ai (i = 1, 2) with some a1, a2 > 0.

(ii) There exists a Lipschitz continuous function ϕ(y1) defined in U1 =−a1 < y1 < a1 that satisfies

|ϕ(y1)| ≤ a2/2 (y1 ∈ U1),

Ω ∩ U = (y1, y2) | y2 < ϕ(y1), y1 ∈ U1,Γ ∩ U = (y1, y2) | y2 = ϕ(y1), y1 ∈ U1.

Then, we say that Ω is a Lipschitz domain, if and only if its boundary Γ = ∂Ωis Lipschitz continuous.Similarly, we define a Ck domain for k ≥ 0. A C0 domain is often called a

domain with the continuous boundary.

9.3 Lemmas

Lemmas 9.1–9.3 below are valid for a bounded C0 domain Ω.

Figure 9.1: Lipschitz domain

Lemma 9.1 (Density)C∞(Ω) is dense in Hm(Ω). That is, for any u ∈ Hm(Ω), there exists φn ⊂C∞(Ω) such that ∥u− φn∥Hm(Ω) → 0 (n → ∞).

Lemma 9.2 (Poincare’s inequality)There exists a domain constant CP such that ∥v∥ ≤ CP ∥∇v∥ for any v ∈H1

0 (Ω).

Here and hereafter, by the domain constant, we mean a positive constantdepending only on Ω.

Lemma 9.3 (Rellich’s theorem)A bounded sequence vnn≥1 in H1(Ω) admits a sub-sequence vk ⊂ vnthat converges to some v ∈ H1(Ω) in L2(Ω); ∥vk − v∥ → 0 as k → ∞.

The following Lemmas 9.4–9.6 below hold true for a bounded Lipschitz do-main Ω.

Lemma 9.4 (Sobolev’s inequality)For every u ∈ H2(Ω), there exists a continuous function u such that u = ua.e. in Ω and ∥u∥L∞(Ω) ≤ C∥u∥H2(Ω) with a domain constant C. Below, wewill identify u with u.

Lemma 9.5 (Trace theorem)The mapping v 7→ v|Γ of C∞(Ω) → C∞(Γ) is extended by continuity to acontinuous linear mapping γ : H1(Ω) → L2(Γ), which is called the trace

operator. Thus, we have

γu = u|∂Ω (v ∈ C(Ω)),

∥γv∥L2(Γ) ≤ CT ∥v∥H1(Ω) (v ∈ H1(Ω))

with the domain constant CT .

Lemma 9.6The space H1

0 (Ω) is characterized by

H10 (Ω) = u ∈ H1(Ω) | γu = 0,

where γ : H1(Ω) → L2(Γ) denotes the trace operator described in Lemma9.5.

We skip the proofs of those lemmas, which could be found in

[19] J. Necas: Direct Methods in the Theory of Elliptic Equations,Springer, 2011.

For the reader’s convenience, the correspondence is given in Tab. 9.1.

this note Necas’ book assumption on Ω

Lemma 9.1 Theorem 3.1 (Chapter 2) bounded C0

Lemma 9.4 Theorem 3.8 (Chapter 2) bounded LipschitzLemma 9.5 Theorem 1.2 (Chapter 1) bounded LipschitzLemma 9.6 Theorem 4.10 (Chapter 2) bounded Lipschitz

Table 9.1: The correspondence between Lemmas in this note and Necas’ book.

Lemma 9.7 (Poincare and Wirtinger’s inequality)There exists a domain constant CW such that

∥v∥ ≤ CW ∥∇v∥

for all v ∈ H1(Ω) satisfying

∫Ωv dx = 0.

Proof. We argue by contradiction. Set H =

v ∈ H1(Ω) |

∫Ωv dx = 0

• Assume that, for every n ≥ 1, there is vn ∈ H such that ∥vn∥ > n∥∇vn∥.

• Define wn = vn/∥vn∥. Then, we have wn ∈ H, ∥wn∥ = 1 and ∥∇wn∥ <1/n.

• The sequence wk is bounded in H1(Ω). By virtue of Rellich’s theorem(Lemma 9.3), there exists a sub-sequence wk ⊂ wn which converges(strongly) in L2(Ω) to some element w ∈ H1(Ω).

• Let φ ∈ C∞0 (Ω) be arbitrary. Noting that∣∣∣∣∫Ω

∂xiφ dx

∣∣∣∣ ≤ ∥∇wk∥ · ∥φ∥ ≤ 1

k∥φ∥ → 0 (k → ∞),

we observe∫Ωw∂φ

∂xidx = lim

k→∞

∫Ωwk

∂xidx = − lim

k→∞

∂xiφ dx = 0

for i = 1, 2.

• Consequently, we have w ∈ H1(Ω) and ∇w = 0 a.e. Ω．Hence, w is aconstant function.

• On the other hand, we have

∫Ωw dx = 0 and ∥w∥ = 1.

• This contradiction establishes the result.

Lemma 9.8Suppose that a bounded Lipschitz domain Ω is decomposed to two (disjoint)Lipschitz subdomains Ω1 and Ω2 by a simple smooth curve S; Ω = Ω1∪S∪Ω2

and Ω1 ∩ Ω2 = ∅. Then, v ∈ C(Ω), v1 = v|Ω1 ∈ C1(Ω1) and v2 = v|Ω2 ∈C1(Ω2) implies v ∈ H1(Ω).

Proof. We define g ∈ L2(Ω) by setting

∂v1/∂xj ∈ C(Ω1) in Ω1

∂v2/∂xj ∈ C(Ω2) in Ω2,(j = 1, 2).

Let nk = (n1,k, n2,k) be the unit normal vector to S outgoing from Ωk. Ob-viously, we have n1 = −n2. For an arbitrary φ ∈ C∞

0 (Ω), we have by theintegration by parts∫

Ωgjφ dx =

∫Ω1

∂v1∂xj

φ dx+

∫Ω2

∂v2∂xj

= −∫Ω1

v1∂φ

∂xjdx−

∫Ω2

v2∂φ

∂xjdx+

∫Sv(nj,1 + nj,2)φ dS

= −∫Ωv∂φ

∂xjdx.

Hence, we obtain v ∈ H1(Ω) and ∇v = (g1, g2).At this stage, we recall an important application of the projection theorem;

For the proof, see a text book of Functional Analysis.

Lemma 9.9 (Riesz’s representation theorem)Let X be a Hilbert space equipped with the scalar product (·, ·)X and thenorm ∥ · ∥X . Further, let F be a bounded linear functional on X, that is,F : X → R is assumed to be a linear mapping satisfying

∥F∥X′ = supv∈X

|F (v)|∥v∥X

< ∞.

Then, there exists a unique a ∈ X such that F (v) = (a, v)X for all v ∈ X.

10 Weak solution and regularity

10.1 Weak formulation

We return to consider the Dirichlet BVP for the Poisson equation:

−∆u = f in Ω, u = 0 on ∂Ω, (10.1)

where Ω ⊂ R2 is a bounded Lipschitz domain with the boundary Γ = ∂Ω, andf ∈ L2(Ω) is a given function.The basic function space of our consideration is

V = H10 (Ω)

which is a Hilbert space equipped with the standard scalar product and normof H1(Ω). As scalar product and norm in V , however, we take

(u, v)V = (∇u,∇v) =

∫Ω∇u · ∇v dx, ∥u∥V =

√(u, u)V .

In view of the Poincare inequality (Lemma 9.2), ∥ · ∥V is an equivalent normof ∥ · ∥H1(Ω) in V. That is, we have

∥v∥V ≤ ∥v∥H1(Ω) ≤ [C2P + 1]1/2∥v∥V (v ∈ V )

with the Poincare constant CP appearing in Lemma 9.2. Consequently, thespace V forms a Hiblert space equipped with (·, ·)V and ∥ · ∥V .We derive a reformulation of (10.1) in the function space V . To this end,

supposing that u is smooth, multiplying both sides of (10.1) by φ ∈ C∞0 (Ω),

and integrating them over Ω, we have by the integration by parts∫Ω∇u · ∇φ dx =

∫Ωfφ dx.

By density, this implies

∫Ω∇u · ∇v dx︸︷︷︸=(u,v)V

∫Ωfv dx︸︷︷︸

=(f,v)

(∀v ∈ V ).

Notice that the left hand-side is meaningful for u ∈ H1(Ω).Now we can state the reformulation form of (10.1);

Find u ∈ V s.t. (u, v)V = (f, v) (∀v ∈ V ). (10.2)

We call (10.2) a weak form of (10.1). The solution u of (10.2) called a weaksolution (generalized solution) of (10.1). The solution u ∈ C2(Ω) ∩ C(Ω) of(10.1) is called a classical solution of (10.1).

Theorem 10.1Suppose that Ω is a bounded Lipschitz domain and f ∈ L2(Ω). Then, wehave the following.

(i) There exists a unique u satisfying (10.2).

(ii) ∥u∥V ≤ C∥f∥ with a domain constant C.

(iii) The function u ∈ V is characterized by

J(u) = minv∈V

J(v), J(v) =1

2∥u∥2V − (f, v).

Proof. (i) We apply Riesz’s representation theorem (Lemma 9.9). In doingso, F ∈ V ′ is defined by setting F (v) = (f, v). We have F ∈ V ′, since |F (v)| ≤∥f∥ · ∥v∥ ≤ CP ∥f∥ · ∥v∥V in view of the Schwarz and Poincare inequalities.Hence, there exists a unique u ∈ V satisfying (u, v)V = F (v) = (f, v) (∀v ∈ V ).(ii) Choosing v = u, we have by Schwarz and Poincare inequalities,

∥u∥2V = (f, u) ≤ ∥f∥ · ∥u∥ ≤ ∥f∥ · CP ∥∇u∥ = CP ∥f∥ · ∥u∥V .

(iii) See the proof of Theorem 7.1.

10.2 Regularity of solutions

Theorem 10.2 (Elliptic regularity)(i) Let k ≥ 0 be an integer. Assume that Γ = ∂Ω is the boundary of class

Ck+2 and f ∈ Hk(Ω). Then, the solution u ∈ V of (10.2) satisfies

u ∈ Hk+2(Ω), ∥u∥Hk+2(Ω) ≤ Ck∥f∥Hk(Ω)

with a domain constant Ck. In particular, if Ω and f are smooth enough,then u ∈ C2(Ω) ∩ C(Ω).

(ii) Assume that Ω is a convex polygon and f ∈ L2(Ω). Then, the solutionu ∈ V of (10.2) satisfies

u ∈ H2(Ω), ∥u∥H2(Ω) ≤ C∥f∥

with a domain constant C. Moreover, if the polygon Ω has a non-convexcorner, there exist a lot of f ∈ L2(Ω) such that the solution u of (10.2)cannot belong to H2(Ω).

Proof. The proof of (i) is described in the standard monographs of PDEs.The proof of (ii) is given, for example, in the following references:

• M. Dauge: Elliptic Boundary Value Problems on Corner Domains.Smoothness and Asymptotics of Solutions, Lecture Notes in Mathe. 1341,Springer, 1988.

• P. Grisvard: Elliptic Problems in Nonsmooth Domains, Pitman, 1985.

• P. Grisvard: Behavior of the solutions of an elliptic boundary valueproblem in a polygonal or polyhedral domain, Numerical solution of par-tial differential equations, III (Proc. Third Sympos. (SYNSPADE),Univ. Maryland, College Park, Md., 1975), pp. 207–274, AcademicPress, New York, 1976.

• A. Kufner and A. M. Sandig: Some applications of weighted Sobolevspaces, Teubner Texts in Mathematics 100, Teubner Verlagsgesellschaft,Leipzig, 1987.

10.3 Galerkin’s approximation and Cea’s lemma

Let Vh, h > 0 being a parameter, be a finite dimensional subspace of V withdimVh = N . The Galerkin approximation of (10.2) reads as

Find uh ∈ Vh s.t. (uh, vh)V = (f, vh) (∀vh ∈ Vh). (10.3)

Theorem 10.3Suppose that Ω is a bounded Lipschitz domain and f ∈ L2(Ω). Then, wehave the following.

(i) There exists a unique uh satisfying (10.3).

(ii) ∥uh∥V ≤ C∥f∥ with the same domain constant C appearing Theorem10.1.

(iii) The function uh ∈ Vh is characterized by

J(uh) = minvh∈Vh

J(vh), J(vh) =1

2∥vh∥2V − (f, vh).

Proof. It is the exactly same as that of Theorem 10.1.

Theorem 10.4 (Galerkin’s orthogonality)Let u ∈ V and uh ∈ Vh be solutions of (10.2) and (10.3), respectively. Then,we have

(u− uh, vh)V = 0 (∀vh ∈ Vh).

Proof. Let vh ∈ Vh be arbitrary. Subtracting (10.3) from (10.2) with v = vh,we obtain the desired equality.

Theorem 10.5 (Cea’s lemma)Let u ∈ V and uh ∈ Vh be solutions of (10.1) and (10.3), respectively. Then,we have

∥u− uh∥V = minvh∈Vh

∥u− vh∥V .

Proof. By Galerkin’s orthogonality,

∥u− uh∥2V = (u− uh, u− uh)V

= (u− uh, u)V − (u− uh, uh)V

= (u− uh, u)V − (u− uh, vh)V

= (u− uh, u− vh)V ≤ ∥u− uh∥V ∥u− vh∥V ;∥u− uh∥V ≤ ∥u− vh∥V

for any vh ∈ Vh. Hence,

∥u− uh∥V ≤ infvh∈Vh

∥u− vh∥V .

Thus, we obtain the desired relation.

11 Shape-regularity of triangulations

11.1 Interpolation error estimates

Throughout this section, we assume that Ω is a polygonal domain with theboundary Γ = ∂Ω. We recall the following.

• Th = Thh↓0 is a family of triangulations of Ω, PiNi=1 is the set ofall vertices of Th,

• Xh and Vh are sets of continuous piecewise affine functions defined as

Xh = vh ∈ C(Ω) | vh is an affine function on every T ∈ Th,Vh = vh ∈ Xh | vh|Γ = 0,

and ϕiNi=1 is the standard basis of Xh,

• According to Lemma 9.8, Xh and Vh are subspaces of H1(Ω) and H10 (Ω),

respectively.

The aim of this section is to show that every function v ∈ H1(Ω) can beapproximated by using functions of Xh.

We use the H2 semi-norm defined as

|u|2,ω = |u|H2(ω) =

(∫ω

[|∂1u|2 + 2|∂1∂2u|2 + |∂2

2u|2]dx

where ω ⊂ R2. Then, the H2 norm is given as

∥u∥2,ω = ∥u∥H2(ω) =(∥u∥2ω + ∥∇u∥2ω + |u|22,ω

where ∥u∥ω = ∥u∥L2(ω).

Lemma 11.1 (local interpolation error)Let T be a closed triangle. Then, for every u ∈ H2(T ), we have

∥Πu− u∥T ≤ C1h2T |u|2,T , (11.1)

∥∇(Πu− u)∥T ≤ C2h2TρT

|u|2,T . (11.2)

Therein,

• Πu is the affine function defined on T such that the values at verticesof T coincide with those of u. Thus, (Πu)(Pi) = u(Pi) for i = 1, 2, 3,where P ′

is are vertices of T . (Recall u is a continuous function in viewof Lemma 9.4.)

• C1 and C2 are absolute positive constants which are independent of Tand v.

• hT is the diameter of the circumscribed circle of T , ρT is the diameterof the inscribed circle of T .

Example 11.2We consider the triangle T whose vertices are (0, 0), (L, 0) and (L/2, Lα)with L > 0 and α > 0. Obviously, u(x1, x2) = u(x, y) = x2 is in H2(T )and (Πu)(x, y) = Lx− 1

4L2−αy. Hence, we can calculate as

(u−Πu)x = 2x− L, (u−Πu)y =1

4L2−α, uxx = 2, uxy = uyy = 0.

Therefore, we have

∥∇(Πu− u)∥2T|u|22,T

32· 1

L2α−4.

Hence, if α > 2, Inequality (11.2) is meaningless as L → 0.

Definition 11.3 (shape regularity of triangulations)A family of triangulations Th = Thh>0 is of shape-regular

def.⇐⇒ ∃ν1 > 0 s.t.hTρT

≤ ν1 (∀T ∈ Th ∈ Th). (11.3)

Lemma 11.4 (Zlamal’s minimum angle condition)The condition (11.3) is equivalent to

∃θ1 > 0 s.t. θT ≥ θ1 (∀T ∈ Th ∈ Th), (11.4)

where θT is the minimum angle of T .

Proof. EXERCISE (→ Problem 9).

We introduce the Lagrange interpolation operator Πh : C(Ω) → Xh definedas

(Πhu)(x) =N∑i=1

u(Pi)ϕi(x) (u ∈ C(Ω)). (11.5)

By Lemma 9.4, Πhu ∈ Xh can be defined for every u ∈ H2(Ω).

Theorem 11.5 (global interpolation error)If Th is of shape-regular, then we have

∥Πhu− u∥ ≤ C1h2|u|2,Ω (u ∈ H2(Ω)), (11.6)

∥∇(Πhu− u)∥ ≤ C2ν1h|u|2,Ω (u ∈ H2(Ω)), (11.7)

where C1 are C2 are constants appearing in Lemma 11.1.

Proof. It is a direct consequence of the shape-regularity of Thhand Lemma11.1. In fact, for u ∈ H2(Ω),

∥∇(Πhu− u)∥2 =∑T∈Th

∥∇(Πhu− u)∥2T

≤∑T∈Th

(h2TρT

|u|2H2(T )

≤∑T∈Th

C22 (hT ν1)

2 |u|2H2(T )

≤ (C2hν1)2∑T∈Th

|u|2H2(T ) = (C2hν1)2|u|22,Ω;

Thus we obtain (11.7).

Theorem 11.6 (approximation property of Xh and Vh)If Th is of shape-regular,

limh↓0

infvh∈Xh

∥u− vh∥ = 0 (u ∈ L2(Ω)), (11.8)

limh↓0

infvh∈Xh

∥∇(u− vh)∥ = 0 (u ∈ H1(Ω)), (11.9)

limh↓0

infvh∈Vh

∥∇(u− vh)∥ = 0 (u ∈ H10 (Ω)). (11.10)

Proof. Let u ∈ H1(Ω). Then, by virtue of (11.7), for any v ∈ C∞(Ω),

infvh∈Xh

∥∇(u− vh)∥ ≤ infvh∈Xh

[∥∇(u− v)∥+ ∥∇(v − vh)∥]

≤ ∥∇(u− v)∥+ ∥∇(v − πhv)∥≤ ∥∇(u− v)∥+ C3h|v|2,Ω (C3 = C2ν1) .

At this stage, let ε > 0 be arbitrary. By the density, there is v ∈ C∞(Ω)satisfying ∥∇(u− v)∥ ≤ ε/2. Hence, if putting δ = ε/(2C3|v|2,Ω), we have

0 < h ≤ δ ⇒ infvh∈Xh

∥∇(u− vh)∥ ≤ ε

2= ε.

This implies (11.9). The proofs of (11.8) and (11.10) are similar.

11.2 Proof of Lemma 11.1

We proceed to the proof of Lemma 11.1. The following notation is employed.

• Let T be a closed triangle:

– |T | is the area of T ;

– Pi are vertices of T (i = 1, 2, 3);

– pi =−−→OPi (i = 1, 2, 3).

• T is the reference elementdef.⇐⇒ T is the triangle with vertices P1 = (0, 0), P2 = (1, 0), and P3 =(0, 1).

• Set B = [b1, b2], b1 = p2 − p1, b2 = p3 − p1, and consider

x = Φ(ξ) ≡ Bξ + p1 (ξ ∈ T )

which is a mapping from T onto T ; T = Φ(T ).

Figure 11.1: The reference element T and Φ(ξ) = Bξ + p1.

Remark 11.7We will make use of Poincare and Wirtinger’s inequality (Lemma 9.7):

∥v∥T ≤ C1∥∇v∥T

(v ∈ H1(T ),

∫Tv dξ = 0

)and Sobolev’s inequality (Lemma 9.4):

∥v∥L∞(T ) ≤ C2∥v∥2,T (v ∈ H2(T )).

We note that C1 and C2 are absolute positive constants.

Lemma 11.8There exists an absolute positive constant C3 such that

infq∈P1

∥v + q∥2,T ≤ C3|v|2,T

for any v ∈ H2(T ).

Proof. We write, for example, L2 = L2(T ) and ∥ · ∥ = ∥ · ∥T for short.Let v ∈ H2. Then, we can take p ∈ P1 such that∫

T(v + p) dξ =

∂ξ1(v + p) dξ =

∂ξ2(v + p) dξ = 0.

We apply Poincare and Wirtinger to v + p, and obtain

∥v + p∥ ≤ C1∥∇(v + p)∥.

Again, we apply Poincare and Wirtinger to ∂(v+ p)/∂ξi (i = 1, 2) and deduce∥∥∥∥ ∂

∂ξi(v + p)

∥∥∥∥ ≤ C1

∥∥∥∥∇ ∂

∂ξi(v + p)

∥∥∥∥ ≤ C1|v + p|2,T = C1|v|2,T .

Combining these inequalities, we have

∥v + p∥2 ≤ C21

(∥∥∥∥ ∂

∂ξ1(v + p)

∥∥∥∥2 + ∥∥∥∥ ∂

∂ξ2(v + p)

∥∥∥∥2)

≤ 2C21 |v|22,T .

Hence,

∥v + p∥22,T

= ∥v + p∥2 + ∥∇(v + p)∥21,T

+ |v|22,T

≤ (3C21 + 1)︸︷︷︸C2

|v|22,T

This completes the proof.

Lemma 11.9Let S and T are closed triangles. Let

Φ : S ∋ ξ 7→ x = Bξ + b ∈ T,

where B ∈ R2×2 is a non-singular matrix and b ∈ R2 (thus, Φ is an affinemapping from S to T ). Then, we have the following.

(i) ∥B∥ ≡ sup|ξ|=1

|Bξ| ≤ hTρS，∥B−1∥ ≤ hS

(iii) For v ∈ H1(T ), we have w ≡ v Φ ∈ H1(S) and

∥∇w∥S ≤ ∥B∥ · | detB|−1/2∥∇v∥T .

(iii) For v ∈ H2(T ), we have w ≡ v Φ ∈ H2(S) and

|w|2,S ≤ C0∥B∥2 · | detB|−1/2|v|2,Twith some absolute positive constant C0.

Proof. (i) |ξ| = ρS implies |Bξ| ≤ hT . Hence, ∥B∥ = ρS−1 sup|Bξ| | |ξ| =

ρS ≤ hT /ρS .(ii) Let JΦ and JΦ−1 be determinants of Jacobi matrices of Φ and Φ−1. Then,|JΦ| = |detB|(= |T |/|S| = 0) and |JΦ−1 | = | detB−1| = | detB|−1(= |S|/|T |)．By the density, it suffices to consider the case v ∈ C1(T ). Then, w ∈ C1(S)．Since Φ is affine,

∇ξw = BT∇xv

Hence, by ∥B∥ = ∥BT∥,

∥∇w∥2S =

∫S|∇ξw|2 dξ

∫T|BT∇xv|2 JΦ−1dx

≤∫T∥BT∥2 · |∇xv|2 | detB−1| dx

= ∥B∥2| detB|−1∥∇v∥2T .

(iii) By the density, it suffices to consider the case v ∈ C2(T ). Then, w ∈ C2(S)and ∣∣∣∣∂2w

∂ξ21

∣∣∣∣ =∣∣∣∣∣∣

2∑i,j=1

Bi1Bj1∂2v

∂xi∂xj

∣∣∣∣∣∣ ≤ ∥B∥22∑

∣∣∣∣ ∂2v

∂xi∂xj

∣∣∣∣ .Since we obtain similar estimations for other second derivatives, we have 2∑

|α|=2

∣∣Dαξ w∣∣2 ≤ 16∥B∥4

∑|α|=2

|Dαxv|

Hence,

|w|2H2(S) ≤ 16∥B∥4∫T

∑|α|=2

|Dαxv|

2 JΦ−1dx

= 16∥B∥4| detB|−1|v|2H2(T ).

Thus, we have showed the desired inequality with C0 = 4. 3

Lemma 11.10There exists an absolute positive constant C such that

∥v −Πv∥1,T ≤ C|v|2,T

for all v ∈ H2(T ).

Proof. We write, for example, H1 = H1(T ) and (·, ·) = (·, ·)T for short.

2We use a rough estimate: maxi,j |Bij | ≤ ∥B∥, where B = (Bij). In fact, setting η = (1, 0),we have |B11| ≤

√|B11|2 + |B21|2 = |Bη|/|η| ≤ ∥B∥.

3However, we can take C0 = 1 by another method of analysis.

• Let w ∈ H1 be arbitrary. Consider a functional F : H2 → R defined by

F (v) = (v −Πv, w) + (∇(v −Πv),∇w) (v ∈ H2).

(Recall that Πv is well-defined; see Lemma 9.4.)

• By Sobolev’s inequality, we have for v ∈ H2

∥Πv∥L∞ ≤ ∥v∥L∞ ≤ C1∥v∥2,T

∥∇(Πv)∥L∞

|v(P1)− v(P2)|,

|v(P2)− v(P3)|√2

, |v(P3)− v(P1)|

≤ 2∥v∥L∞ ≤ 2C1∥v∥2,T .

• Therefore,

∥v −Πv∥21,T

≤ ∥v −Πv∥2 + ∥∇(v −Πv)∥2

≤ 2(∥v∥2 + |T |2∥Πv∥2L∞) + 2(∥∇v∥2 + |T |2∥∇Πv∥2L∞)

≤ 2∥v∥21,T

2(∥Πv∥2L∞ + ∥∇Πv∥2L∞)

≤ 2∥v∥21,T

2C22∥v∥22,T

≤ max

∥v∥2

2,T≡ C2

4∥v∥22,T .

Thus,|F (v)| ≤ ∥v −Πv∥1,T ∥w∥1,T ≤ C4∥v∥2,T ∥w∥1,T .

• At this stage, let q ∈ P1 be arbitrary. Then, since F (v + q) = F (v), wehave

|F (v)| = |F (v + q)| ≤ C4∥v + q∥2,T ∥w∥1,T .

This gives|F (v)| ≤ C4 inf

q∈P1

∥v + q∥2,T ∥w∥1,T .

We apply Lemma 11.8 to obtain

|F (v)| ≤ C3C4|v|2,T ∥w∥1,T .

• Finally, choosing w = v −Πv, we arrive at

∥v −Πv∥21,T

≤ C3C4|v|2,T ∥v −Πv∥1,T ;

which implies the desired inequality.

As a result of lemmas described above, we can state the following proof.

Proof of Lemma 11.1. Let T be any triangle, and let Φ(ξ) = Bξ + b bethe affine mapping which maps the reference triangle T onto T . Set h = hTand ρ = ρT . Fix v ∈ H2(T ), and define v = v Φ. Suppose that Πv is the

affine function defined on T whose values at vertices coincide with those of v.(The meaning of Πv is the same as described in Lemma.) First, by Lemma11.9 (i) and (iii),

|v|22,T

≤ C20∥B∥4| detB|−1|v|22,T ≤ C2

ρ4h4T |detB|−1|v|22,T .

This, together with Lemma 11.10, gives

∥v −Πv∥2T ≤ | detB| · ∥v − Πv∥2T≤ | detB| · C2|v|2

≤ | detB| · (C20 C

2/ρ4) · h4T | detB|−1|v|22,T .

Thus, we deduce (11.1). Similarly,

∥∇x(v −Πv)∥2T ≤ ∥B−1∥2| detB| · ∥∇ξ(v − Πv)∥2L2(T )

≤ h2

ρ2T| detB| · C2|v|2

≤ h2

ρ2T| detB| · (C2

0 C2/ρ4) · h4T | detB|−1|v|22,T

(C20 h

)h4Tρ2T

|v|22,T .

This implies (11.2) and hence completes the proof.

12 Error analysis of FEM

We are now ready to study the convergence of FEM. We recall that the weakform of the Poisson equation is described as

Find u ∈ V s.t.

∫Ω∇u · ∇v dx︸︷︷︸=(u,v)V

∫Ωfv dx︸︷︷︸

=(f,v)

(∀v ∈ V ) (12.1)

and the finite element approximation is given as

Find uh ∈ Vh s.t. (uh, vh)V = (f, v) (∀vh ∈ Vh). (12.2)

We recall the following:

• Ω ⊂ R2 is a polygonal domain;

• f ∈ L2(Ω) is a given function;

• V = H10 (Ω) is a Hilbert space equipped with

∥v∥V = ∥∇v∥, (u, v)V =

∫Ω∇u · ∇v dx;

• ∥u∥ = ∥u∥L2(Ω)，(u, v) =

∫Ωuv dx;

• |u|22,Ω = ∥∂21u∥2 + 2∥∂1∂2u∥2 + ∥∂2

2u∥2;

• Thh>0 is a family of triangulations of Ω;

• Vh ⊂ V is the set of continuous piecewise linear functions defined in Th;

• The Lagrange interpolation Πh : C(Ω) → Xh is defined as

(Πhv)(x) =

N∑i=1

v(Pi)ϕi(x) (v ∈ C(Ω)).

(Recall that Πhv is well-defined for any v ∈ H2(Ω); See Lemma 9.4.)

Remark 12.1Problem (12.1) is the weak form of the Dirichlet BVP for the Poisson equa-tion

−∆u = f in Ω, u = 0 on ∂Ω.

There exists a unique u satisfying (12.1), and we have ∥u∥V = ∥∇u∥V ≤C∥f∥ (cf. Theorem 10.1). If Ω is a convex polygon, we further obtainu ∈ H2(Ω) with ∥u∥H2(Ω) ≤ C∥f∥ (cf. Theorem 10.2).

Theorem 12.2 (Convergence)Suppose that Thh>0 is of shape-regular. Let u ∈ V and uh ∈ Vh besolutions of (12.1) and (12.2), respectively. Then, we have

limh↓0

∥∇u−∇uh∥ = 0.

Thus, uh converges to u as h ↓ 0 in H1(Ω).

Proof. Recall Theorem 10.5 (Cea’s lemma):

∥u− uh∥V = minvh∈Vh

∥u− vh∥V . (12.3)

This, together with Theorem 11.6, implies the conclusion.

Theorem 12.3 (H1 error estimate)Suppose that Thh>0 is of shape-regular. Let u ∈ V and uh ∈ Vh besolutions of (12.1) and (12.2), respectively. Further, assume u ∈ H2(Ω).Then, we have

∥∇u−∇uh∥ ≤ Ch|u|2,Ωwith C = C2ν1 (C2 > 0 is the constant appearing in Lemma 11.1).

Proof. The equality (12.3) implies

∥u− uh∥V ≤ ∥u− vh∥V (∀vh ∈ Vh).

We choose vh = Πhu ∈ Vh (→ Problem 11) and apply Theorem 11.5 (globalinterpolation error) to obtain

∥u− uh∥V ≤ ∥u−Πhu∥V ≤ C2ν1h|u|2,Ω.

Theorem 12.4 (L2 error estimate)Assume that Thh>0 is of shape-regular. Let u ∈ V and uh ∈ Vh besolutions of (12.1) and (12.2), respectively. Further, assume that Ω is aconvex polygon. Then, we have

∥u− uh∥ ≤ C ′h2|u|2,Ω

with C ′ = ν21C22CR > 0, where CR denotes a domain constant appearing

Theorem 10.2 (ii).

Remark 12.5There is a big difference between Theorems 12.3 and 12.4. In Theorem 12.3,we assume that the weak solution u of the Poisson equation −∆u = f withu|∂Ω = 0 is in H2(Ω) for some f ∈ L2(Ω). On the other hand, if Ω is aconvex polygon, the weak solution w of the Poisson equation −∆w = g withw|∂Ω = 0 belongs to H2(Ω) for all g ∈ L2(Ω).

Proof of Theorem 12.4. Define eh = u − uh and consider the variationalproblem

find w ∈ V s.t. (v, w)V = (eh, v) (∀v ∈ V ),

which we call the adjoint problem of (12.1). In view of Theorems 10.1 and10.2, there exists a unique solution w ∈ V satisfying ∥w∥H2(Ω) ≤ CR∥g∥ witha domain constant CR. Moreover, we know ∥eh∥V ≤ ν1C2h|u|2,Ω by Theorem12.3. Choosing v = eh, we deduce

∥eh∥2 = (eh, w)V

= (eh, w −Πhw)V (by Galerkin orthogonality, (eh,Πhw)V = 0.)

≤ ∥eh∥V ∥w −Πhw∥V≤ ∥eh∥V · C2ν1h|w|2,Ω≤ C2ν1h|u|2,Ω · C2ν1h · CR∥eh∥.

Hence, we have ∥u− uh∥ ≤ C22CRν

2|u|2,Ω.

Remark 12.6The method of analysis in the proof of Theorem 12.4 is said to be dualityargument or Aubin-Nitsche’s trick.

13 Numerical experiments by FreeFem++

So far we have studied a mathematical theory of FEM. Unfortunately, theimplementation of FEM is not an easy task. Actually, Professor O. Pironneauwrote in his famous book “Finite Element Methods for Fluids (Wiley, 1989)”that

Numerical analysis is somewhat dry if it is taught without test-ing the methods. Unfortunately experience shows that a simplefinite element solution of a Laplace equation with the P 1 conform-ing element requires at least 20 hours of programming time; so itis difficult to reach the more interesting applications discussed inthis book in the time allotted to a Master course. [page 197]

In order to avoid these difficulties, we are able to utilize a free softwareFreefem++

http://www.freefem.org/ff++/index.htm

or Freefem++-cs

http://www.ann.jussieu.fr/~lehyaric/ffcs/index.htm

Figure 13.1: Freefem++-cs

Example 13.1In the unit square Ω = (0, 1)× (0, 1), we consider the Poisson equation

−∆u = x2 + 2y in Ω, u = 0 on ∂Ω.

stop run

Figure 13.2: Useful buttons

List 7 is a freefrem++ code to solve this problem, and Fig. 13.3 is output.

Listing 7: Example 13.1

// parameters and functionsfunc g = 0; // boundary conditionfunc f = x*x + 2.0*y; // righthand functionint n = 20; // division number

// domainmesh Th=square(n, n);// finite element spacefespace Vh(Th, P1);Vh u,v;

// Poisson equationsolve poisson(u, v) =int2d(Th)(dx(u)*dx(v) + dy(u)*dy(v))- int2d(Th)(f*v) + on(1, 2, 3, 4, u = g);

// plot dataplot(u, wait = true, ps = "prog1.eps");

Example 13.2Consider the same problem as Example 13.1 and apply gnuplot to displaythe shape of the solution. List 8 is a freefrem++ code. After running this,we get prog2.data. Then, in gnuplot, type as follows:

gnuplot> set pm3d

gnuplot> set palette rgbformulae 33,13,10

gnuplot> set ticslevel 0

gnuplot> splot "prog2.data" with lines pal

Results are Figs 13.4 and 13.5. To save those results as eps file, type asfollows:

Figure 13.3: Example 13.1

gnuplot> set term pdf

gnuplot> set output "prog2.pdf"

gnuplot> replot

// parameters and functionsfunc g = 0; // boundary conditionfunc f = x*x + 2.0*y; // righthand functionint n = 10; // division number

// domainmesh Th=square(n, n);// finite element spacefespace Vh(Th, P1);Vh u,v;

// Poisson equationsolve poisson(u, v) =int2d(Th)(dx(u)*dx(v) + dy(u)*dy(v))- int2d(Th)(f*v) + on(1, 2, 3, 4, u = g);

// plot Thplot(Th, ps = "prog2.eps");

// gnuplot data fileofstream ff("prog2.data");for(int i = 0; i < Th.nt ; i++)

for(int j = 0; j < 3; j++)ff << Th[i][j].x << " "<< Th[i][j].y << " " << u[][Vh(i,j)] <<

ff << Th[i][0].x << " " << Th[i][0].y << " " << u[][Vh(i,0)] <<endl << endl << endl;

"prog2_10.data"

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

Figure 13.4: Example 13.2 (n = 10)

"prog2_30.data"

0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.1

Example 13.3In a “complex-shaped” domain as in Figs 13.6 and 13.7, we solve the Pois-son equation

−∆u = x2y in Ω, u = 0 on ∂Ω.

List 9 is a freefrem++ code to solve this problem, and Figs 13.6 and 13.7are outputs.

// parametersfunc g = 0; // boundary value

func f = x*x*y; // righthand functionint n = 30; // division number

// domainborder G1(t = 0, 3) x = t; y = 0;border G2(t = 0, pi/2) x = 3*cos(t); y = 3*sin(t);border G3(t = 0, 3) x = 0; y = 3 - t;border G4(t = 0, 2*pi) x = 1.9 - 0.8*cos(t); y = 0.9 + 0.8*sin(t);border G5(t = 0, 2*pi) x = 0.7 - 0.5*cos(t); y = 2.3 + 0.5*sin(t);

// triangulationmesh Th = buildmesh(G1(n)+G2(2*n)+G3(n)+G4(2*n)+G5(n));

// finite element spacefespace Vh(Th,P1);Vh u,v;

// Poisson equationsolve poisson(u,v)= int2d(Th)(dx(u)*dx(v) + dy(u)*dy(v))- int2d(Th)(f*v) + on(G1,G2,G3,G4,G5,u = g);

// Plot Thplot(Th, ps = "prog3m.eps");

endl;ff << Th[i][0].x << " " << Th[i][0].y << " " << u[][Vh(i,0)] <<

endl << endl << endl;

"prog3_10.data"

Example 13.4In an oval-shaped domain Ω = x2/(1.5)2 + y2 < 1, we solve the Poissonequation

−∆u = 1 in Ω, u = x2 + y2 on ∂Ω.

"prog3_30.data"

List 10 is a freefrem++ code to solve this problem, and Figs 13.8 and 13.9are outputs.

// parametersfunc g = x*x + y*y; // boundary valuefunc f = 1.0; // righthand functionint n = 60; // division number

// domainborder G1(t = 0, 2*pi) x = 1.5*cos(t); y = sin(t);

// triangulationmesh Th = buildmesh(G1(n));

// finite element spacefespace Vh(Th,P1);Vh u,v;

// Poisson equationsolve poisson(u,v)= int2d(Th)(dx(u)*dx(v) + dy(u)*dy(v))- int2d(Th)(f*v) + on(G1, u = g);

// Plot Thplot(Th, ps = "prog4m.eps");

endl;ff << Th[i][0].x << " " << Th[i][0].y << " " << u[][Vh(i,0)] <<

endl << endl << endl;

-1.5-1

-0.5 0

1.5-1-0.8

-0.6-0.4

-0.2 0

0.2 0.4

0.6 0.8

"prog4_30.data"

-1.5-1

-0.5 0

"prog4_60.data"

Example 13.5We examine the rate of convergence of FEM. To this end, we take

−∆u = 2π2 sin(πx) sin(πy) in Ω, u = 0 on ∂Ω

in the unit square Ω = (0, 1)× (0, 1). The exact solution is given as

u(x, y) = sin(πx) sin(πy).

We calculate

eh = ∥∇u−∇uh∥, Eh = ∥u− uh∥,

ρh =log e2h − log ehlog(2h)− log(h)

, Rh =logE2h − logEh

log(2h)− log(h).

We take uniform meshes illustrated in Fig. 13.10. Results are reported inFig. 13.11 and Tab. 13.1. We observe from those results that errors areexpressed as

eh ≈ C1h, Eh ≈ C2h2 (C1, C2 some constants),

and they are consistent with theoretical results (Theorems 12.3 and 12.4).List 11 is a freefrem++ code used for this calculation.

Figure 13.10: Example 13.5

// error1.edpfunc exact = sin(pi*x)*sin(pi*y); // exact solutionfunc f = 2.0*pi*pi*exact; // rifht-hand side functionfunc g = exact; // Dirichlet boundary conditionreal hsize, hold; // mesh sizereal errh1, errh1old, errl2, errl2old, rateh1, ratel2;int n, nn;

// fine triangulationnn = 256;mesh Th2 = square(nn, nn);fespace Vh2(Th2, P1);Vh2 uproj, w, uex;uex = exact;

// output fileofstream f1("error1.dat");

// n=4,8,16,32,64n = 2;errh1old = errl2old = 1.0;hold = 1.0;for (int i = 1; i < 6; i++)

n = 2*n;mesh Th = square(n, n);plot(Th, ps="error1.eps");fespace Vh(Th, P1);Vh u, v, hh = hTriangle;hsize = hh[].max;solve Poisson(u,v) =int2d(Th)( dx(u)*dx(v) + dy(u)*dy(v)) - int2d(

Th) ( f*v ) + on(1,2,3,4,u = g);// computation of error using the fine triangulationuproj = u; // projection of u into the fine triangulationw = uproj - uex; // error functionerrh1 = sqrt( int2d(Th2)(dx(w)*dx(w) + dy(w)*dy(w)) ); // H1-errorerrl2 = sqrt( int2d(Th2)(w^2) ); // L2-error// computation of ratesrateh1 = (log(errh1) - log(errh1old))/(log(hsize) - log(hold));ratel2 = (log(errl2) - log(errl2old))/(log(hsize) - log(hold));errh1old = errh1;errl2old = errl2;hold = hsize;// output resultsf1 << hsize << " "<< errh1 << " " << rateh1 << " "<< errl2 << " " << ratel2<< " " << endl;

0.0001

0.01 0.1 1

mesh size h

H1 errorL2 error

Figure 13.11: Behavior of errors (Example 13.5)

h eh ρh Eh Rh

0.353553 0.838452 — 0.079064 —0.176777 0.431591 0.96 0.021119 1.900.088388 0.217113 0.99 0.005364 1.980.044194 0.108122 1.01 0.001337 2.000.022097 0.052783 1.03 0.000325 2.04

Table 13.1: eh, ρh, eh，Rh

Problems and further readings

13.1 Problems for Chapter II

Problem 9Prove Lemma 11.4.

Problem 10Under the same assumptions of Problem ??, prove the following equalities.

3∑i=1

(x− ai)λi = 0,

3∑i=1

(y − bi)λi = 0,

3∑i=1

(x− ai)∂λi

∂x= −1,

3∑i=1

(y − bi)∂λi

∂y= −1,

3∑i=1

(x− ai)∂λi

∂y= 0,

3∑i=1

(y − bi)∂λi

∂x= 0.

Problem 11Prove that Πhu ∈ Vh for u ∈ H2(Ω)∩H1

0 (Ω), where Πh : C(Ω) → Xh denotesthe Lagrange interpolation operator defined as (11.5) and Vh is defined as(8.6).

Problem 12Under the same assumptions of Problem ??, prove the following equalities.∫ ∫

Tλiλj dxdy =

S/6 (i = j),

S/12 (i = j),∫ ∫T∇λi∇λj dxdy

4S[(aj − ak)

2 + (bj − bk)2] (i = j),

4S[(aj − ak)(ak − ai) + (bj − bk)(bk − bi)] (i = j).

Problem 13 (machine assignment)Make FEM meshes for domains illustrated in Fig. 13.12. Moreover, in thoseΩ, solve numerically the following PDEs and represent shapes of solutions bygnuplot.

−∆u = 1 in Ω, u = 0 on Γ,

−∂21u− 10−2∂2

2u = 1 in Ω, u = 0 on Γ.

13.2 Further readings

There are many excellent textbooks devoted to the mathematical theory ofFEM. For example, I recommend students

Figure 13.12:

[12] 菊地文雄: 有限要素法の数理 (数学的基礎と誤差解析)，培風館，1994年．

[15] S. Larsson and V. Thomee: Partial Differential Equations withNumerical Methods, Springer, 2009.

For researchers, the following books might be valuable:

[2] S. C. Brenner and L. R. Scott: The Mathematical Theory ofFinite Element Methods (3rd ed.), Springer, 2007.

[13] P. Knabner and L. Angermann: Numerical Methods for Ellip-tic and Parabolic Partial Differential Equations, Springer, 2003.

[21] P. A. Raviart and J. M. Thomas: Introduction a l’AnalyseNumerique des Equations aux Derivees Partielles, Masson, Paris,1983.

My favorite one is [21]; But, unfortunately，it is written in French.Another important topic of FEM is the discrete maximum principle. For

example, we refer to [13] and

[23] 齊藤宣一：発展方程式の数値解析，第 33回発展方程式若手セミナー報告集 (http://www.infsup.jp/saito/materials/00Fnsaito.pdf)，1–24，2011.

FEM for parabolic PDEs is explained, for example, in [13], [15], [21], [26]and

[7] H. Fujita, N. Saito and T. Suzuki: Operator Theory and Nu-merical Methods, Elsevier, 2001.

[28] V. Thomee: Galerkin Finite Element Methods for ParabolicProblems (2nd ed.), Springer, 2006.

In particular, [7] is based on the analytical semigroup theory.The implementation of FEM is not an easy task. But, I recommend the

readers to try it by following

[11] 菊地文雄: 有限要素法概説 (理工学における基礎と応用)，サイエンス社，1980年．

The following article on Variational Methods by Professor Kato is worthreading:

[10] 加藤敏夫：変分法，自然科学者のための数学概論 [応用編]，寺沢寛一 (編)，岩波書店，1960年．

III. Abstract elliptic PDE and Galerkin method

14 Theory of Lax and Milgram

In Chapter II, we studied only the Poisson equation. However, the theorycould be extended to more general PDEs of elliptic type. We introduce thefollowing notions.

• Let V be a (real) Hilbert space with the scalar product (·, ·)V and thenorm ∥ · ∥V .

• The space V ′ denotes the dual space of V (= the set of all boundedlinear functionals over V ). Thus,

F ∈ V ′ def.⇐⇒ F : V → R,F (v + w) = F (v) + F (w) (v, w ∈ V )

F (αv) = αF (v) (v ∈ V, α ∈ R)

∥F∥V ′ ≡ supv∈V

∥v∥V= sup

|F (v)|∥v∥V

< ∞.

Hereinafter, we write as ⟨F, v⟩ = F (v) (v ∈ V ); ⟨·, ·⟩ = ⟨·, ·⟩V ′,V denotesthe duality product between V ′ and V .

• a : V × V → R is a bilinear form on V × Vdef.⇐⇒

a(αu+ βv,w) = α · a(u,w) + β · a(v, w),

a(u, αv + βw) = α · a(u, v) + β · a(u, v) (u, v, w ∈ V, α, β ∈ R)

• A bilinear form a : V × V → R is bounded (or continuous)def.⇐⇒

∥a∥ ≡ supu,v∈V

a(u, v)

∥u∥V ∥v∥V< ∞.

Theorem 14.1 (Lax-Milgram)Suppose that a bounded bilinear form a : V × V → R satisfies the followingcondition:

[Coercivity] ∃α > 0 s.t. a(v, v) ≥ α∥v∥2V (v ∈ V ). (14.1)

Then, for any F ∈ V ′, there exists a unique u ∈ V satisfying

a(u, v) = ⟨F, v⟩ (∀v ∈ V ). (14.2)

Remark 14.2The function u appearing Theorem 14.1 satisfies a priori estimate

∥u∥V ≤ 1

α∥F∥V ′ .

In fact, choosing v = u, we have α∥u∥2V ≤ a(u, u) = ⟨F, u⟩ ≤ ∥F∥V ′∥u∥V .Thus, ∥u∥V ≤ (1/α)∥F∥V ′ .

We use the Riesz mapping σ = σV from V ′ to V that is a bijective operatorfrom V ′ to V defined as 4

(σF, v)V = ⟨F, v⟩ (∀v ∈ V )

for F ∈ V ′. It satisfies

∥σ∥V ′,V = supF∈V ′

∥σF∥V∥f∥V ′

∥σ−1∥V,V ′ = supv∈V

∥σ−1v∥V ′

∥v∥V= 1.

Note that V ′ forms a Hiblert space equipped with the norm ∥ · ∥V ′ . Its scalarproduct is defined by

(F,G)V ′ = (σF, σG)V (F,G ∈ V ′).

We shall present two different proofs.

Proof of Theorem 14.1, I. For short, set ∥ · ∥ = ∥ · ∥V , (·, ·) = (·, ·)V , andσ = σV . In view of Riesz’s representation theorem (Lemma 9.9), there existsa linear operator A : V → V such that 5

(Au, v) = a(u, v) (u, v ∈ V ).

In particular,

(Au, u) ≥ α∥u∥2, ∥Au∥ ≤ ∥a∥ · ∥u∥ (u ∈ V ).

4For an arbitrary F ∈ V ′, there exists a unique w ∈ V satisfying ⟨F, v⟩ = (w, v)V (v ∈ V ).This correspondence is denoted by σ : F 7→ w. Obviously, σ is a linear operator fromV ′ to V . Combining ∥F∥V ′ = supv∈V |⟨F, v⟩|/∥v∥V ≤ ∥w∥V = ∥σF∥V and ∥F∥V ′ ≥|⟨F,w⟩|/∥w∥V = ∥w∥V = ∥σF∥V , we have ∥F∥V ′ = ∥σF∥V . Hence, we obtain ∥σ∥V ′,V =1 (isometry). On the other hand, since, for any w ∈ V , we have (w, ·)V ∈ V ′, σ issubjective. Moreover, for F ∈ V with σF = 0, we have F = 0 by the isometry. Thus,the operator σ is bijective.

5Fix u ∈ V and consider φu(v) = a(u, v) (v ∈ V ). φu(v) is a linear functional on V . Since|φu(v)| ≤ (∥a∥·∥u∥)·∥v∥, φu(v) is bounded. By virtue of Riesz (Lemma 9.9), there exists aunique w ∈ V satisfying (w, v)V = φu(v) = a(u, v) (∀v ∈ V ). The correspondence u 7→ wis denoted by w = Au. Then, the operator A is linear on V . Choosing v = w = Au, wehave ∥Au∥2 = a(u,Au) ≤ ∥a∥ · ∥Au∥ · ∥u∥. Hence, ∥A∥ = supu∈V ∥Au∥/∥u∥ ≤ ∥a∥. Thisimplies that A : V → V is a bounded linear operator.

Since F is expressed as ⟨F, v⟩ = (σF, v)V (∀v ∈ V ), the equation (14.2) isequivalently written as

(Au, v) = (σF, v) (∀v ∈ V ) ⇔ Au− σF = 0 in V. (14.3)

We will show that this equation admits a unique solution by the contractionmapping principle 6. To this end, we introduce E : V → V and B : V → V by

Eu = u− ρ(Au− σF ) = Bu+ ρσF (Bu = u− ρAu)

with a constant ρ. Then,

(Bu,Bu) = ∥u∥2 − 2ρ(Au, u) + ρ2∥Au∥2

≤ ∥u∥2 − 2ρα∥u∥2 + ρ2∥a∥2∥u∥2

= (1− 2ρα+ ρ2∥a∥2)︸︷︷︸k

∥u∥2 (u ∈ V ).

Now we take ρ such that 0 < ρ < 2α/∥a∥2; Then, 0 < k < 1 and ∥Bu∥ ≤√k ∥u∥ (u ∈ V ). Therefore,

∥Eu− Ev∥ = ∥B(u− v)∥ ≤√k ∥u− v∥ (u, v ∈ V ).

This implies that E is contraction on V . So, we can apply the contractionmapping principle to obtain a unique fixed point u ∈ V satisfying

u = Eu ⇔ u = u− ρ(Au− σF ).

The function u is a unique solution of the operator equation Au − σF = 0.

Proof of Theorem 14.1, II. We prove that the linear operator A definedabove is bijective on V . First, since

(Av, v) ≥ α∥v∥2V (v ∈ V ),

the oprerator A is injective. Next, R(A), the range of A, is a closed set in V .Indeed, let wn ⊂ R(A) and wn → w in V with some w ∈ V . We can takevn ⊂ V such that Avn = wn. Then, since

∥vn − vm∥ ≤ ∥Avv −Avm∥ = ∥wn − wm∥,

we see that vn is a Cauchy sequence in V . Hence, there is v ∈ V suchthat vn → v in V and we have Av = w. This gives that w ∈ R(A) and,consequently, that R(A) is closed. Finaly, R(A) is dense in V . To verify this,suppose that v ∈ V satisfies (Au, v) = 0 for all u ∈ V . Taking u = v, weobtain 0 = (Av, v) ≥ α∥v∥2; hence v = 0. This imples that R(A) is dense.As a consequence, we have verified that A is bijevtive on V . Therefore, forσF ∈ V , there exists a unique u ∈ V such that Au = σF .

6Suppose that T is a contraction operator on a Hilbert space H. That is, T satisfies∥Tu− Tv∥H ≤ λ∥u− v∥H (u, v ∈ H) with 0 < λ < 1. Then, there exists a unique u ∈ Hsuch that u = Tu. Such u ∈ H is called the fixed point.

Theorem 14.3 (variational principle)In addition to assumptions of Theorem 14.1, we assume that a is symmetric,that is,

a(u, v) = a(v, u) (u, v ∈ V ). (14.4)

Then, u ∈ V is a solution of (14.2) if and only if u is a solution of

J(u) = minv∈V

J(v), J(v) =1

2a(v, v)− ⟨F, v⟩. (14.5)

Proof. It is the exactly same as that of Theorem 7.1.

15 Galerkin approximation

We follow the notation of the previous section. Further, we introduce a finitedimensional subspace Vh of V , h being the discretization parameter such thath ↓ 0.For a given F ∈ V ′, we consider an abstract elliptic problem:

Find u ∈ V s.t. a(u, v) = ⟨F, v⟩ (∀v ∈ V ) (15.1)

and its Galerkin approximation:

Find uh ∈ Vh s.t. a(uh, vh) = ⟨F, vh⟩ (∀vh ∈ Vh). (15.2)

Let ϕiNi=1 be a basis of Vh, N being the dimension of Vh. We write uh as

N∑i=1

Uiϕi.

Then, we have the matrix representation of (15.2)

Au = f ,

where we have set

A = (a(ϕj , ϕi)) ∈ RN×N , u = (Ui) ∈ RN , f = (⟨F, ϕi⟩) ∈ RN .

As a consequence of Lax-Milgram’s theorem, Problems (15.1) and (15.2)admit the unique solutions. Hence, the matrix A is non-singular.

Theorem 15.1 (Galerkin orthogonality)Let u ∈ V and uh ∈ Vh be solutions of (15.1) and (15.2), respectively. Then,we have

a(u− uh, vh) = 0 (∀vh ∈ Vh).

Theorem 15.2 (Cea’s lemma)Let u ∈ V and uh ∈ VN be solutions of (15.1) and (15.2), respectively. Then,we have

∥u− uh∥V ≤ ∥a∥α

infvh∈Vh

∥u− vh∥V . (15.3)

Proof. Let vh ∈ Vh. Then a(u−uh, vh) = 0 by Galerkin orthogonality. Hence,

α∥u− uh∥2V ≤ a(u− uh, u− uh)

= a(u− uh, u− vh)

≤ ∥a∥ · ∥u− uh∥V ∥u− vh∥V .

Therefore,

∥u− uh∥V ≤ ∥a∥α

infvh∈Vh

∥u− vh∥V .

Remark 15.3Let Ph be the orthogonal projection operator from V to Vh. Then, foru ∈ V , we have

∥u− Phu∥V = minvh∈Vh

∥u− vh∥V .

(This is nothing but the projection theorem.) Hence, we can replace inf bymin in Theorem 15.2.

Remark 15.4If we assume, in addition assumptions of Theorem 15.2, that a is symmetric,then we have

∥u− uh∥V ≤√

∥a∥α

minvh∈Vh

∥u− vh∥V .

In fact, we can now define a scalar product of V by

((u, v)) = a(u, v) (u, v ∈ V ). (15.4)

Then, u ∈ V satisfies

((u− uh, wh)) = 0 (∀wh ∈ Vh).

This means that the operator defined as u 7→ uh is the projection operator ofV → Vh with respect to the scalar product (15.4). Hence, by the projectiontheorem, we obtain

|||u− uh||| = minvh∈Vh

|||u− vh|||

with |||u||| =√

((u, u)). Finally, noting α∥u∥2V ≤ |||u|||2 ≤ ∥a∥ · ∥u∥2V , wededuce the desired inequality.

Remark 15.5Let S be a closed subspace of V . If, for u ∈ S, there exists uS ∈ S satisfying∥u− uS∥V = min

v∈S∥u− v∥V , then uS is called the best approximation of u in

S. If the bilinear form a is symmetric, the solution uh of (15.2) is actuallythe best approximation of the solution u of (15.1) in Vh with ((·, ·)).

16 Applications

16.1 Convection-diffusion equation

Assume that Ω ⊂ R2 is a bounded Lipschitz domain and that its boundaryΓ = ∂Ω consists of two parts Γ1,Γ2 ⊂ Γ such that Γ = Γ1 ∪ Γ2. Since Ω isa Lipschitz domain, the unit outer normal vector to Γ, which is denoted byn = n(s) = (n1(s), n2(s)), is well-defined for almost every s ∈ Γ.

Differential problem. Suppose that we are given

ν > 0, f, c : Ω → R, b : Ω → R2, g1 : Γ1 → R, g2 : Γ2 → R.

The function u = u(x) is defined to be a concentration/density of a certainsubstance. We consider the equations:

j = −ν∇u+ bu in Ω (flux of u),∇ · j + cu = f in Ω (conservation law),u = g1 on Γ1 (Dirichlet B.C.),−j · n = g2 on Γ2 (Non-flux B.C.).

Thus, we consider

− ν∆u+∇ · (bu) + cu = f in Ω, (16.1a)

u = g1 on Γ1, (16.1b)

ν∂u

∂n− (b · n)u = g2 on Γ2, (16.1c)

where ∂u/∂n = ∇u · n. Multiplying the both sides of (16.1a) of the form∇ · j + cu = f by v ∈ C∞(Ω) with v|Γ1 = 0 and integrating them over Ω, wehave by the integration by parts∫

Ω∇ · jv dx+

∫Ωcuv dx =

∫Ωfv dx

⇔∫Γ(j · n)v dS −

∫Ωj · ∇v dx+

∫Ωcuv dx =

∫Ωfv dx

⇔ ν

∫Ω∇u · ∇v dx−

∫Ω(bu) · ∇v dx+

∫Ωcuv dx︸︷︷︸

=a(u,v)

∫Ωfv dx+

∫Γ2

g2v dS.

Hence, the solution u of (16.1) must satisfy

a(u, v) =

∫Ωfv dx+

∫Γ2

g2v dS (∀v ∈ C∞(Ω), v|Γ1 = 0).

Next, we take g1 : Ω → R satisfying g1|Γ1 = g1 and put u = u − g1. Then,u|Γ = 0 and

a(u, v) =

∫Ωfv dx+

∫Γ2

g2v dS + a(g1, v)︸︷︷︸=F (v)=⟨F,v⟩

Function spaces and forms.

V = v ∈ H1(Ω) | v|Γ1 = 0, ∥ · ∥V = ∥ · ∥1,2 = ∥ · ∥H1(Ω),

a(u, v) = ν

∫Ω∇u · ∇v dx−

∫Ω(bu) · ∇v dx+

∫Ωcuv dx,

⟨F, v⟩ = ⟨F, v⟩V ′,V =

∫Ωfv dx+

∫Γ2

g2v dS + a(g1, v),

∥v∥ = ∥v∥L2(Ω), ∥v∥Γk= ∥v∥L2(Γk) (k = 1, 2),

∥b∥∞ = supx∈Ω

√b1(x)2 + b2(x)2, ∥v∥∞ = sup

x∈Ω|v(x)|.

Weak formulation.

Find u = u+ g1 ∈ H1(Ω) s.t. u ∈ V and a(u, v) = ⟨F, v⟩ (∀v ∈ V ). (16.2)

Assumptions.

(A1) b ∈ C1(Ω)2，c ∈ C(Ω)，f ∈ L2(Ω);

2∇ · b+ c ≥ 0 (x ∈ Ω)，b · n ≤ 0 (x ∈ Γ2);

(A3) g1 ∈ H1(Ω)，g1 = g1|Γ1 ∈ L2(Γ1)，g2 ∈ L2(Γ2).

Continuity of a. For u, v ∈ V ,

|a(u, v)| ≤ ν

∫Ω|∇u| · |∇v| dx+

∫Ω∥b∥∞|u| · |∇v| dx+

∫Ω∥c∥∞ · |u| · |v| dx

≤ ν∥∇u∥ · ∥∇v∥+ ∥b∥∞∥∇u∥ · ∥v∥+ ∥c∥∞∥u∥ · ∥v∥≤ maxν, ∥b∥∞, ∥c∥∞︸︷︷︸

∥u∥V ∥v∥V .

Continuity of F . According to Lemma 9.5 (trace theorem), for v ∈ H1(Ω),we have η = v|Γ2 ∈ L2(Γ2) and ∥η∥L2(Γ2) ≤ C2∥v∥1,2 with a domain constantC2. Hence, we have

|⟨F, v⟩| ≤∫Ω|fv| dx+

∫Γ2

|g2v| dS + |a(g1, v)|

≤ ∥f∥∥v∥+ ∥g2∥Γ2∥v∥Γ2 + C1∥g1∥V ∥v∥V≤ (∥f∥+ C2∥g2∥Γ2 + C1∥g1∥V )︸︷︷︸

∥v∥V

Coercivity of a. We first note that

−∫Ωb(∇u)u dx = −

2∑i=1

bi∂u

∂xiu dx = −

2∑i=1

bi ·1

∂xi(u2) dx

= −1

∫∂Ω

2∑i=1

2ni dS +1

2∑i=1

∂bi∂xi

· u2 dx

= −1

∫Γ2

(b · n)u2 dS +1

2∑i=1

∂bi∂xi

· u2 dx

∫Ω(∇ · b)u2 dx.

Therefore, for u ∈ V ,

a(u, u) = ν∥∇u∥2 +∫Ωb(∇u)u dx+

∫Ωcu2 dx

≥ ν∥∇u∥2 − 1

∫Ω(∇ · b)u2 dx+

∫Ωcu2 dx

≥ ν∥∇u∥2 +∫Ω

2∇ · b

)u2 dx

≥ ν∥∇u∥2

2∥∇u∥2 + ν

2∥∇u∥2

≥ ν

2∥∇u∥2 + ν

∥u∥2

≥ C24∥u∥2V .

Therein, we have used the following lemma.

Lemma 16.1(Poincare’s inequality, II) If the Lebesgue measure of Γ1 is positive, thenthere exists a domain constant CP such that

∥v∥ ≤ CP ∥∇v∥ (v ∈ H1(Ω), v|Γ1 = 0).

Proof. It is the exactly same as that of Lemma 9.7.

Well-posedness.

Theorem 16.2Under assumptions (A1), (A2) and (A3), there exists a unique solution u of(16.2). Moreover, it satisfies

∥u∥1,2 ≤ C5(∥f∥+ ∥g1∥1,2 + ∥g∥2,Γ2).

Proof. We can apply Lax-Milgram’s theorem to obtain the unique existenceof a solution. Substituting v = u into (16.2), we have

C4∥u∥2V ≤ a(u, u) = ⟨F, u⟩ ≤ (∥f∥+ C2∥g2∥2,Γ2 + C1∥g1∥V )∥u∥V .

Hence,

∥u∥V ≤ 1

C4(∥f∥+ C2∥g2∥2,Γ2 + C1∥g1∥V ).

This, together with ∥u∥V ≥ ∥u∥V −∥g1∥V , implies the desired inequality.

16.2 Elliptic PDE of the second order

Assume that Ω ⊂ RN is a bounded smooth domain and that its boundaryΓ = ∂Ω consists of two parts Γ1,Γ2 ⊂ Γ such that Γ = Γ1 ∪ Γ2. Since Ωis a smooth domain, the unit outer normal vector to Γ, which is denoted byn = n(s) = (n1(s), n2(s)), is well-defined for almost every s ∈ Γ.

Elliptic PDE of the second order

Lu = f in Ω, u = g1 on Γ1,∂u

∂nL= g2 on Γ2, (16.3)

Lv = −N∑

∂xjaij(x)

∂xi+

N∑i=1

bi(x)∂v

∂xi+ c(x)v,

∂nL=

N∑i,j=1

aij(x)∂v

∂xinj .

SettingA = (aij(x)), b = (bi(x)),

the equation Lu = f is expressed as

−∇ ·A∇u+ b · ∇u+ cu = f.

For v ∈ C∞(Ω) with v|Γ1 = 0,∫Ω(Lu)v dx = −

N∑i,j=1

∂xjaij

)v dx+

N∑i=1

∫Ωbi

∂xiv dx+

∫Ωcuv dx

(−∫Γaij

∂xinjv dx+

∫Ωaij

∂xjdx

N∑i=1

∫Ωbi

∂xiv dx+

∫Ωcuv dx

N∑i,j=1

aij∂u

∂xj+

N∑i=1

bi∂u

∂xiv + cuv

︸︷︷︸=a(u,v)

−∫Γ2

∂nLv dx.

Consequently, the solution u of (16.3) satisfies

a(u, v) =

∫Ωfv dx+

∫Γ2

g2v dx

for any v ∈ C∞(Ω) with v|Γ1 = 0. Suppose g1 : Ω → R is such that g1|Γ1 = g1.Put u = u− g1. Then, we have u|Γ = 0 and

a(u, v) =

∫Ωfv dx+

∫Γ2

g2v dx+ a(g1, v)︸︷︷︸=F (v)=⟨F,v⟩

Function spaces and forms.

V = v ∈ H1(Ω) | v|Γ1 = 0, ∥ · ∥V = ∥ · ∥1,2 = ∥ · ∥H1(Ω),

a(u, v) =

N∑i,j=1

aij∂u

∂xj+

N∑i=1

bi∂u

∂xiv + cuv

⟨F, v⟩ = ⟨F, v⟩V ′,V =

∫Ωfv dx+

∫Γ2

g2v dS + a(g1, v),

∥v∥ = ∥v∥L2(Ω), ∥v∥Γk= ∥v∥L2(Γk) (k = 1, 2).

Weak formulation.

Find u = u− g1 ∈ H1(Ω) s.t. u ∈ V, a(u, v) = ⟨F, v⟩ (∀v ∈ V ). (16.4)

Assumptions.

(B1) aij = aji, bi, c ∈ C(Ω), f ∈ L2(Ω);

α = max1≤i,j≤N

supx∈Ω

|aij(x)|, β = max1≤i≤N

supx∈Ω

|bi(x)|,

γ = supx∈Ω

|c(x)|, γ′ = infx∈Ω

(B2) ∃λ0 > 0 s.t.∑

1≤i,j≤N

aij(x)ξiξj ≥ λ0|ξ|2 (∀x ∈ Ω, ∀ξ ∈ RN );

(B3) γ′ ≥ β2

(B4) g1 ∈ H1(Ω), g1 = g1|Γ1 ∈ L2(Γ1), g2 ∈ L2(Γ2).

Continuity of a.

|a(u.v)| ≤ α∥∇u∥ · ∥∇v∥+ β∥∇u∥ · ∥v∥+ γ∥u∥ · ∥v∥≤ maxα, β, γ︸︷︷︸

∥u∥V ∥v∥V .

Continuity of F . For v ∈ V ,

|⟨F, v⟩| ≤∫Ω|fv| dx+

∫Γ2

|g2v| dS + |a(g1, v)|

≤ ∥f∥∥v∥+ ∥g2∥Γ2∥v∥Γ2 + C1∥g1∥V ∥v∥V≤ (∥f∥+ C2∥g2∥Γ2 + C1∥g1∥V )︸︷︷︸

∥v∥V .

Coercivity of a. For u ∈ V ,

a(u, u) ≥ λ0∥∇u∥2 − β∥∇u∥ · ∥u∥+ γ′∥u∥2

2∥∇u∥2 +

(√λ0

2∥∇u∥ − β√

2λ0∥u∥

(γ′ − β2

)∥u∥2

≥ λ0

2∥∇u∥2 ≥ C2

4∥u∥2V .

Well-posedness

Theorem 16.3Under the assumptions (B1)–(B4), there exists a unique solution u of (16.4).Moreover, it satisfies

∥u∥1,2 ≤ C5(∥f∥+ ∥g1∥1,2 + ∥g∥2,Γ2).

Proof. It is exactly the same as that of the previous theorem.

Problems and further remark

Problems for Chapter III

Problem 14Let Ω be a bounded domain in R2 with the smooth boundary Γ. Let f ∈ L2(Ω)and g ∈ L2(Γ). Give a variational formulation of the problem

−∆u = f in Ω, with∂u

∂n+ u = g on Γ,

where ∆ denotes the Laplacian and ∂/∂n differentiation along the outer unitnormal vector n to Γ. Prove existence and uniqueness of a weak solution. Givean interpretation of the boundary condition in connection with some problemin mechanics or physics.

Problem 15Let Ω be a bounded domain in R2 with the smooth boundary Γ. Suppose thata bounded smooth domain Ω1 is strictly contained in Ω. That is, we assumethat Ω1 ⊂ Ω. Define Ω2 = Ω\Ω1 and S = ∂Ω1 ∩ ∂Ω2. Moreover, n denotesthe unit normal vector to S outgoing from Ω1. Then, we seek the functions

u1 : Ω1 → R, u2 : Ω2 → R

such that

−∆u1 = f in Ω1, −ε∆u2 = f in Ω2, u2 = 0 on Γ

together with the continuity condition:

u1 = u2,∂u1∂n

= ε∂u2∂n

where f ∈ L2(Ω) and 0 < ε < 1 is a constant. This problem is called theinterface problem. Introducing the function

u1 in Ω1

u2 in Ω2,

give a variational formulation in H10 (Ω) of this interface problem and discuss

the unique existence of a weak solution.

Further remark

Without the coercivity condition (14.1), we can prove the unique existence ofthe variational problem (14.2). Actually, the Lax-Milgram theorem (Theorem14.1) is generalized as follows:

Theorem (The generalized Lax-Milgram theorem). Suppose thatU and V are Hilbert spaces equipped with the norms ∥ · ∥U and ∥ · ∥V ,respectively. Assume that we are given a bounded bilinear form b : U ×V →R. Then, the following (i) and (ii) are equivalent.

(i) The following two conditions are satisfied:

(H1) ∃β > 0 s.t. infu∈U

supv∈V

b(u, v)

∥u∥U∥v∥V≥ β,

(H2) supu∈U

|b(u, v)| > 0 (∀v ∈ V, v = 0).

(ii) For any F ∈ V ′, there exists a unique u ∈ U satisfying

b(u, v) = ⟨F, v⟩V ′,V (∀v ∈ V ),

where ⟨·, ·⟩V ′,V denotes the duality pairing between V ′ and V .

Supposing V = U in this theorem, we deduce that the variational problem(14.2) admits a unique solution, if and only if the following two conditions aresatisfied:

(A1) ∃β > 0 s.t. infu∈V

supv∈V

a(u, v)

∥u∥V ∥v∥V≥ β,

(A2) supu∈V

|a(u, v)| > 0 (∀v ∈ V, v = 0).

Clearly, the coercivity condition (14.1) implies (A1) and (A2); However, theconditions (A1) and (A2) does not always lead to (14.1).

The proof of the generalized Lax-Milgram theorem is explained in

[1] I. Babuska and A. K. Aziz: Survey lectures on the mathematicalfoundations of the finite element method, In; A.K. Aziz (ed.), TheMathematical Foundations of the Finite Element Method with Ap-plications to Partial Differential Equations, pp. 1–359, AcademicPress, New York, 1972.

References

[1] I. Babuska and A. K. Aziz: Survey lectures on the mathematical founda-tions of the finite element method, In; A.K. Aziz (ed.), The MathematicalFoundations of the Finite Element Method with Applications to PartialDifferential Equations, pp. 1–359, Academic Press, New York, 1972.

[2] S. C. Brenner and L. R. Scott: The Mathematical Theory of Finite Ele-ment Methods (3rd ed.), Springer, 2007.

[3] L. C. Evans: Partial Differential Equations (2nd ed.), American Mathe-matical Society, 2010.

[4] 藤田宏：初期値問題の差分法による近似解法 (B: 微分方程式の近似解法)，自然科学者のための数学概論 [応用編]，寺沢寛一 (編)，岩波書店，1960年．

[5] 藤田宏：境界値問題の差分法による近似解法 (B: 微分方程式の近似解法)，自然科学者のための数学概論 [応用編]，寺沢寛一 (編)，岩波書店，1960年．

[6] 藤田宏，池部晃生，犬井鉄郎，高見穎郎：数理物理に現れる偏微分方程式I [岩波講座基礎数学，解析学 II-iv]，岩波書店，1977年．

[7] H. Fujita, N. Saito and T. Suzuki: Operator Theory and Numerical Meth-ods, Elsevier, 2001.

[8] F. Hecht, O. Pironneau, A. Le Hyaric and K. Ohtsuka: freefem++, http://www.freefem.org/ff++/index.htm, Laboratoire Jacques-Louis Lions(LJLL), University of Paris VI. (See also freefem++-cs http://www.ann.jussieu.fr/~lehyaric/ffcs/index.htm)

[9] 亀高惟倫：非線型偏微分方程式，産業図書，1977年.

[10] 加藤敏夫：変分法，自然科学者のための数学概論 [応用編]，寺沢寛一 (編)，岩波書店，1960年．

[11] 菊地文雄: 有限要素法概説 (理工学における基礎と応用)，サイエンス社，1980年．

[12] 菊地文雄: 有限要素法の数理 (数学的基礎と誤差解析)，培風館，1994年．

[13] P. Knabner and L. Angermann: Numerical Methods for Elliptic andParabolic Partial Differential Equations, Springer, 2003.

[14] J. F. B. M. Kraaijevanger: Maximum norm contractivity of discretizationschemes for the heat equation, Appl. Numer. Math. 9 (1992) 475–492.

[15] S. Larsson and V. Thomee: Partial Differential Equations with NumericalMethods, Springer, 2009.

[16] K. W. Morton and D. F. Mayers: Numerical Solution of Partial Differen-tial Equations (2nd ed.), Cambridge University Press, 2005.

[17] 三村昌泰 (編): パターン形成とダイナミクス (非線形・非平衡現象の数理4)，東京大学出版会，2006年.

[18] 三村昌泰：微分方程式と差分方程式—数値解は信用できるか？—，「数値解析と非線形現象 (山口昌哉編)」の第 3章，日本評論社，1996年 (オリジナルは 1981年)．

[19] J. Necas: Direct Methods in the Theory of Elliptic Equations, Springer,2011. ( The original French version was published in 1967.)

[20] R. D. Richtmyer and K. W. Morton: Difference Methods for Initial-ValueProblems, Interscience Publishers, 1967.

[21] P. A. Raviart and J. M. Thomas: Introduction a l’Analyse Numeriquedes Equations aux Derivees Partielles, Masson, Paris, 1983.

[22] 齊藤宣一：数値解析入門，東京大学出版会，2012年.

[23] 齊藤宣一：発展方程式の数値解析，第 33回発展方程式若手セミナー報告集 (http://www.infsup.jp/saito/materials/00Fnsaito.pdf)，1–24，2011.

[24] 齊藤宣一：線形・非線形拡散方程式の差分解法と解の可視化，講義ノート (http://www.infsup.jp/saito/materials/fdm_heat11b.pdf/)，2011年．

[25] G. D. Smith: Numerical Solution of Partial Differential Equations, OxfordUniversity Press, 1965. (藤川洋一郎訳：コンピュータによる偏微分方程式の解法，新訂版，サイエンス社，1996年)

[27] V. Thomee: Finite Difference Methods for Linear Parabolic Equations,In Handbook of Numerical Analysis, Vol. I, 5–196, North-Holland, 1990.

[28] V. Thomee: Galerkin Finite Element Methods for Parabolic Problems(2nd ed.), Springer, 2006.

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