Multidimensional Integral Equations and Inequalities

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ATLANTIS S TUDIES IN M ATHEMATICS FOR E NGINEERING AND S CIENCE VOLUME 9 S ERIES E DITOR : C.K. C HUI

Atlantis Studies in Mathematics for Engineering and Science Series Editor: C.K. Chui Stanford University, USA (ISSN: 1875-7642)

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A MSTERDAM – PARIS c ATLANTIS PRESS

Multidimensional Integral Equations and Inequalities B.G. Pachpatte 57, Shri Niketen Coloney, Aurangabad, India

A MSTERDAM – PARIS

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Dedicated to the memory of my parents.

Preface

Integral equations that involve functions of two or more independent variables occur frequently in the study of many problems in partial differential equations which arise from various dynamic models. The study of various multidimensional integral equations and inequalities attracted the attention of many researchers and there exists a very vast literature. The aim of this monograph is to provide the readers, representative overview of the important recent developments, focusing on some selected multidimensional integral equations and inequalities. It is assessable to any one having a reasonable background in real analysis, partial differential equations and acquaintance with their related areas. The material included in the monograph is recent and hard to find in other books. It is self-contained and all results are presented in an easy-to-read, informal style and it could also serve as a textbook for an advanced graduate course. It will be an invaluable reading for pure and applied mathematicians, physists, engineers, computer scientists and will also be most valuable as a source of reference in the field. It is impossible to thank all the individuals who have influenced me directly or indirectly during the writing of this book, without their constant encouragement it would still have been remained no more than an idea. In particular, I wish to express my deep and sincere gratitude to Professor Charles Chui, Editor AMES who offered invaluable suggestions for the improvement of the presentation. Also, I am grateful to Professor Jan van Mill and Arjen Sevenster for their support and interest in the present work. It is a pleasure to acknowledge the fine collaboration and assistance provided by the editorial and production staff of Atlantis Press. Last, but not least, I would like to thank to my family members for their understanding, patience and long-lasting inspiration. B.G. Pachpatte

vii

Contents Preface

vii

Introduction

1

1. Integral equations in two variables 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 2.

9

Introduction . . . . . . . . . . . . . . . . . . . Basic integral inequalities . . . . . . . . . . . . Volterra-type integral equation . . . . . . . . . Volterra-Fredholm-type integral equation . . . Integrodifferential equations of hyperbolic-type Fredholm-type integrodifferential equation . . . Miscellanea . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . .

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59 59 65 73 78 83 89 95

Integral inequalities and equations in two and three variables 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8

Introduction . . . . . . . . . . . . . . . . . . . . . . . Integral inequalities in two variables . . . . . . . . . . Integral inequalities in three variables . . . . . . . . . Integral equation in two variables . . . . . . . . . . . . Integral equation in three variables . . . . . . . . . . . Hyperbolic-type Fredholm integrodifferential equation Miscellanea . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . .

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59

3. Mixed integral equations and inequalities 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 4.

97

Introduction . . . . . . . . . . . . . . . . . . . . . Volterra-Fredholm-type integral inequalities I . . . Volterra-Fredholm-type integral inequalities II . . . Integral equation of Volterra-Fredholm-type . . . . Volterra-Fredholm-type integral equations . . . . . General Volterra-Fredholm-type integral equations Miscellanea . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . .

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Parabolic-type integrodifferential equations 4.1 4.2 4.3 4.4

Introduction . . . . . . . . . . . . . . . . . . Basic integral inequalities . . . . . . . . . . . Integrodifferential equation of Barbashin-type General integral equation of Barbashin-type . ix

97 97 108 116 122 128 135 141 143

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143 143 147 155

x

Multidimensional Integral Equations and Inequalities

4.5 4.6 4.7 4.8 5.

Integrodifferential equation of the type arising in reactor dynamics Initial-boundary value problem for integrodifferential equations . Miscellanea . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Multivariable sum-difference inequalities and equations 5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

Introduction . . . . . . . . . . . . . . . . . . . . Sum-difference inequalities in two variables . . . Sum-difference inequalities in three variables . . Multivariable sum-difference inequalities . . . . Sum-difference equations in two variables . . . . Volterra-Fredholm-type sum-difference equations Miscellanea . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . .

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161 169 179 189 191

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191 191 198 204 211 222 228 235

Bibliography

237

Subject Index

243

Introduction

In studying mathematical models of various dynamic equations, it is often desirable not only to prove the existence of a solution satisfying the given initial or boundary conditions but also to ensure that the solution in question possesses certain qualitative properties. It is well known that the beginning of the qualitative theory of differential equations is directly connected with the classical works of H. Poincaré, A.M. Lyapunov and G.D. Birkhoff on problems of ordinary and classical mechanics. The theory of partial differential equations and many physical, chemical and biological phenomena give rise to multidimensional integral and integrodifferential equations and their study provide results of enormous importance, revealing deep and fundamental connections. The classical book: Partial integral operators and integro-differential equations by J.M. Appell, A.S. Kalitvin and P.P. Zabrejko [5] contains an overview of many contributions to such equations, including comprehensive list of references. First, it will be helpful to summarize briefly certain important multivariable integral and integrodifferential equations arising while studying some specific problems, which greatly stimulated the present work. However, the sample results we are going to discuss are certainly far from being exhaustive. In [28, p. 20] C. Corduneanu pointed out that, by means of the substitution u = v exp (−

x 0

b0 (y,t) dy), the following hyperbolic equation uxt + a0 (x,t)ux + b0 (x,t)ut = c0 (x,t, u),

(1)

considered on the semi-strip 0 x , 0 t < ∞, with the given characteristic data u(x, 0) = u1 (x), u(0,t) = u0 (t),

(2)

vxt + a(x,t)vx = c(x,t, v),

(3)

takes the form

1

2


where a(x,t) and c(x,t, v) are like a0 (x,t) and c0 (x,t, u) and the data on the characteristics preserve their form

v(x, 0) = u1 (x) exp

x

0

b0 (y, 0) dy = v1 (x),

v(0,t) = u0 (t).

(4)

Furthermore, by taking z(x,t) = vxt (x,t) it is easy to observe that the equation (3) with characteristic data (4) takes the form

z(x,t) + a(x,t)

∂ v1 (x) + ∂x

t 0

z(x, τ ) d τ

x t z(y, τ ) d τ dy . = c x,t, u0 (t) + v1 (x) − u1 (0) + 0

(5)

0

In [22], J.R. Cannon and Y. Lin described and analyzed a typical boundary value problem with pseudo-parabolic equation: uxxt = Au + F

in QT ,

(6)

u(x, 0) = φ (x), 0 x 1,

(7)

u(0,t) = f (t), 0 t T,

(8)

ux (0,t) = g(t), 0 t T,

(9)

Au = a1 ut + a2 uxt + a3 uxx + a4 ux + a5 u,

(10)

where

QT = (0, 1) × (0, T ], T > 0 and f , g, φ , F and ai (i = 1, . . . , 5) are given functions. In [22], the above problem is studied by reducing it to the following equivalent integrodifferential equation u(x,t) = G(x,t) +

x η t 0

0

0

[Au(ξ , τ ) + F(ξ , τ )]d τ d ξ d η ,

(11)

in QT , where G(x,t) = φ (x) − φ (0) − φ (0)x + g(t)x + f (t).

(12)

For more details, see [22] and the references therein. In [4], G. Andrews studied the partial differential equation of the form utt (x,t) − auxxt (x,t) = F(x,t, u(x,t)),

(13)

for x ∈ [0, L], t ∈ [0, T ]; L > 0, T > 0, with the initial conditions u(x, 0) = φ (x),

ut (x, 0) = ψ (x),

x ∈ [0, L],

(14)

Introduction

3

and boundary conditions u(0,t) = u(L, T ) = 0,

t ∈ [0, T ].

(15)

Under appropriate conditions, in [4] it is shown that the problem (13)–(15) can be reduced to an integral equation of the form t s L

u(x,t) = f (x,t) + 0

0

0

G(x, y, s − τ )F(y, τ , u(y, τ )) dy d τ ds,

(16)

where G(x, y,t) is the Green’s function for the heat equation wt (x,t) = auxx (x,t) with zero Dirichlet boundary data, a is a positive constant and L > 0, T > 0 are finite but can be arbitrarily large constants. For more details, see [4]. D.L. Lovelady [64] studied the hyperbolic-type Fredholm integrodifferential equation ∂2 ∂ ∂ u(s,t, z) = f (s,t, z) + H z, u(s,t, z), u(s,t, z), u(s,t, z) ∂ s∂ t ∂s ∂t 1 ∂ ∂ (17) K z, r, u(s,t, r), u(s,t, r), u(s,t, r) dr, + ∂s ∂t 0 with the given data u(s, 0, z) = σ (s, z),

u(0,t, z) = τ (t, z),

(18)

for s, t ∈ [0, ∞), z ∈ [0, 1]. It is easy to observe that the problem (17)–(18) contains as a special case the integral equation of the form s t 1

u(s,t, z) = h(s,t, z) +

L(z, v, w, r, u(v, w, r)) dr dw dv. 0

0

(19)

0

In [31], O. Diekmann analyzed a model of spatio-temporal development of an epidemic. The model considered leads to the following nonlinear integral equation of the form t

u(t, x) = f (t, x) + 0

Ω

S0 (ξ )A(τ , x, ξ )g(u(t − τ )) d ξ d τ ,

(20)

for (t, x) ∈ [0, ∞) × Ω, where Ω is a bounded domain in Rn . Detailed descriptions and analysis of the above model and of related ones may be found in Diekmann [31,32] and Thieme [128] which contain additional relevant references. The integral equation (20) appears to be Volterra-type in t, and of Fredholm-type with respect to x and hence it can be viewed as a mixed Volterra-Fredholm-type integral equation. In [8], E.A. Barbashin first initiated the study of the integrodifferential equations of the form

∂ u(t, x) = c(t, x)u(t, x) + ∂t

b

k(t, x, y)u(t, y) dy + f (t, x), a

(21)

4


which arise in mathematical modeling of many applied problems (see [5]). The equation (21) has been studied by many authors and is now known in the literature as integrodifferential equation of Barbashin-type or simply Barbashin equation (see [5, p. 1]). For detailed account on the study of such equations, see [5] and the references cited therein. Another significant source of parabolic-type integrodifferential equations is provided by the study of C.V. Pao [121,122] related to an integrodifferential system arising in reactor dynamics of the form p (t) = p(t)

Ω

β (x)u(t, x) dx (t > 0, x ∈ Ω),

∂u − Lu = f (t, x, p(t) − p∗ ), ∂t

(22) (23)

p∗ 0 is a constant, with the given boundary and initial conditions

α1 (x)

∂u + α2 (x)u = 0 (t > 0, x ∈ ∂ Ω), ∂τ

(24)

u(0, x) = u0 (x) (x ∈ Ω),

(25)

p(0) = p0 ,

(26)

where n

Lu =

∑

i, j=1

n

ai j (x)uxi x j + ∑ ai (x)uxi ,

(27)

i=1

on the bounded domain Ω in Rn . For detailed account on the study of such equations, see [37,39,57,69,125,134]. Solving the equation (22) by using (26) and then substituting it into (23), we get

t ∂u − Lu = f t, x, p0 exp β (x)u(s, x) dx ds − p∗ , ∂t 0 Ω

(28)

for t > 0, x ∈ Ω. We note that the study of equations like (28) with (24), (25) is interesting in itself. Integral and integrodifferential equations of the type (5), (11), (16), (19), (20), (21), (28) are remarkable in terms of simplicity, the large number of results to which they lead, and the variety of applications which can be related to them. When dealing with the above noted equations, the basic considerations give rise to the questions to be answered are: (i) under what conditions the equations under considerations have solutions? (ii) how can we find the solutions or closely approximate them? (iii) what are their nature?

Introduction

5

The study of such questions related to the above noted equations is a challenging task and requires special attention for handling such problems. Although, there is an enormous literature on all these equations, some of them are still in an elementary stage of development. In practice, it is often difficult to obtain explicitly the solutions to nonlinear equations and thus need a new insight to handle the qualitative properties of their solutions. In general, existence theorems for equations of the above noted forms are proved by the use of one of the three fundamental methods: the method of successive approximations, the method based on the theory of nonexpansive and monotone mappings and the theory exploiting the compactness of the operator often by the use of the well known fixed point theorems. The method of inequalities which provides explicit estimates on unknown functions has been a significant source in the study of many qualitative properties of solutions of various differential, integral and finite difference equations. It enable us to obtain valuable information about solutions without the need to know in advance the solutions explicitly. In many cases while studying the behavior of solutions, the method which works very effectively to establish existence does not yield other properties of the solutions in ready fashion and one often needs some new ideas and methods in the analysis. It is easy to observe that the explicit estimates available on various inequalities in [82,85,87,134] and the references cited therein are not directly applicable to study the qualitative behavior of solutions of the equations of the above noted forms and their discrete versions. Moreover, in [22, p. 378] J.R. Cannon and Y. Lin pointed out that in the study of certain basic qualitative properties of solutions, the equation (11) can be dealt with in a more satisfactory manner than dealing directly with the equations (6)–(9). During the last decade or so the above noted facts inspired the author a new line of thought, which resulted in a series of recent papers [88–118] dealing with the qualitative theory related to the equations of the above noted types. The literature related to the above types of cited integral equations and inequalities is now very extensive, it is scattered in various journals encompassing different subject areas. There is thus an urgent need of a book that brings readers to the forefront of current research in this prosperous field. This monograph is an attempt to organize in a systematic way the recent progress related to the equations and inequalities of the above noted types, in the hope that it will further broaden developments and the scope of applications. The field is vast and has not stabilized as yet so that it is extremely difficult to produce a work that traces all the relevant contributions. We mostly focus on certain recent advances not covered in earlier monographs, which reflect our taste and as well as those we consider potentially applicable in a wide

6


range of problems. The exposition consists of five chapters and references. The first chapter presents a large number of basic results related to certain integral and integrodifferential equations in two variables. The tools employed in the analysis are based on the applications of Banach fixed point theorem; with Bielecki-type norm and integral inequalities with explicit estimates. In the second chapter, we consider some integral inequalities with explicit estimates involving functions of two and three independent variables. In this chapter, the reader will also find the study of some important qualitative properties of solutions of certain integral and integrodifferential equations in two and three independent variables. Chapter 3 is devoted to present some fundamental mixed Volterra-Fredholm-type integral inequalities which can be used as tools in certain applications. It also contains the results on existence, uniqueness and other properties related to certain mixed Volterra-Fredholm-type integral equations. Chapter 4 is concerned with certain parabolic type integrodifferential equations which arise in mathematical modeling of many applied problems. The basic problems of existence, uniqueness and other qualitative properties of solutions are dealt with by using different techniques. Chapter 5 is dedicated to the theory of multivariable sum-difference inequalities and equations in the hope that it will provide a clue to effective methods for dealing with the discrete dynamics theory for its treatment. Each chapter contains a section on miscellanea, indicating adequate sources, intended to stimulate the reader’s interest. Throughout, we let R and N denote the set of real and natural numbers respectively and Ia = [0, a] (a > 0), R+ = [0, ∞), R1 = [1, ∞), N0 = {0, 1, 2, . . .}, Nα ,β = {α , α + 1, . . . , α + n = β } (α ∈ N0 , n ∈ N) are the given subsets of R, and Rn the real n-dimensional Euclidean space with appropriate norm denoted by | · |. The derivatives of a function u(t), t ∈ R are denoted by u(i) (t) for i = 1, . . . , n. The partial derivatives of a function z(x, y) for x, y ∈ R with respect to x, y and xy are denoted by ∂ ∂ z(x, y) (or zx (x, y)), D2 z(x, y) or z(x, y) (or zy (x, y)), D1 z(x, y) or ∂x ∂y and

∂2 z(x, y) (or (or zxy (x, y)) ∂ x∂ y respectively. For any function w(n), n ∈ N0 , we define the operator Δ by D1 D2 z(x, y) = D2 D1 z(x, y) or

Δw(n) = w(n + 1) − w(n) and Δi w(n) = Δ(Δi−1 w(n)) for i 2. For any function z(m, n), m, n ∈ N0 , we define the operators Δ1 z(m, n) = z(m + 1, n) − z(m, n),

Δ2 z(m, n) = z(m, n + 1) − z(m, n)

Introduction

7

and Δ2 Δ1 z(m, n) = Δ2 (Δ1 z(m, n)). The class of continuous functions and the class of discrete functions from the set S1 to the set S2 are denoted by C(S1 , S2 ) and D(S1 , S2 ) respectively. We use the usual conventions that the empty sums and products are taken to be 0 and 1 respectively. Furthermore, we shall assume that all the integrals, sums and products involved exist on the respective domains of their definitions and are finite, and hence converge. We note that the results we establish for scalar equations can be extended without any difficulty to the case of vector valued functions. The notation, definitions and symbols used in the text are standard or otherwise explained.

Chapter 1

Integral equations in two variables

1.1

Introduction

Integral equations in two and more variables have been treated by many investigators and various methods have been proposed for the study of different aspects of their solutions. This intensively investigated area is in a process of continuous development, reflected in the great number of books and papers dedicated to it, see [5–10,13–18,20,23–27,33,34,36,38– 45,49,50,52–55,59–66,68,74–140] and the references cited therein. Nevertheless, there are still many aspects related to certain such equations, which we believe need to be enlightened. In response to the growing use of such equations in many applications, in this chapter we study some important qualitative properties of solutions of certain integral and integrodifferential equations in two variables. The fundamental tools employed in the analysis are based on applications of the Banach fixed point theorem and certain recent integral inequalities with explicit estimates on the unknown functions.

1.2

Basic integral inequalities

In this section we present some basic integral inequalities with explicit estimates needed in the sequel. In our considerations here and subsequent chapters we shall use the notation E = R+ × R+ , E0 = Ia × Ib , E1 = (x, y, s) : 0 s x < ∞, y ∈ R+ and E2 = (x, y, s,t) ∈ E 2 : 0 s x < ∞, 0 t y < ∞ . We start with the following inequality established in [111]. Theorem 1.2.1.

Let u ∈ C(E, R+ ); q, D1 q ∈ C(E1 , R+ ); r, D1 r, D2 r, D2 D1 r ∈ C(E2 , R+ )

and c 0 is a constant. If u(x, y) c +

x 0

q(x, y, ξ )u(ξ , y) d ξ +

x y 0

9

0

r(x, y, σ , τ )u(σ , τ ) d τ d σ ,

(1.2.1)

10


for x, y ∈ R+ , then u(x, y) c P(x, y) exp

x 0

y

R(s,t) dt ds ,

(1.2.2)

0

for x, y ∈ R+ , where P(x, y) = exp(Q(x, y)), in which

x

q(η , y, η ) +

Q(x, y) = 0

and

x

R(x, y) = r(x, y, x, y)P(x, y) +

0

D1 q(η , y, ξ ) d ξ d η ,

D1 r(x, y, σ , y)P(σ , y) d σ +

0 x y

+ 0

Proof.

η

0

(1.2.3)

y 0

(1.2.4)

D2 r(x, y, x, τ )P(x, τ ) d τ

D2 D1 r(x, y, σ , τ )P(σ , τ ) d τ d σ .

(1.2.5)

Define a function z(x, y) by

x y

z(x, y) = c + 0

0

r(x, y, σ , τ )u(σ , τ ) d τ d σ ,

then (1.2.1) can be restated as u(x, y) z(x, y) +

x 0

q(x, y, ξ )u(ξ , y) d ξ .

(1.2.6)

(1.2.7)

From the hypotheses, it is easy to observe that z(x, y) is nonnegative and nondecreasing for x, y ∈ R+ . Treating (1.2.7) as a one-dimensional integral inequality for any fixed y ∈ R+ and a suitable application of the inequality given in [87, Theorem 1.2.1, Remark 1.2.1, p. 11] yields u(x, y) P(x, y)z(x, y).

(1.2.8)

From (1.2.6) and (1.2.8), we have z(x, y) c +

x y 0

0

r(x, y, σ , τ )P(σ , τ )z(σ , τ ) d τ d σ .

(1.2.9)

Now a suitable application of the inequality given in [87, Theorem 2.2.1, Remark 2.2.1, p. 66] to (1.2.9) yields z(x, y) c exp

x 0

y

R(s,t) dt ds .

(1.2.10)

0

Using (1.2.10) in (1.2.8), we get the required inequality in (1.2.2). Next, we shall state the following versions of the inequalities in Theorems 2.5.7 and 2.5.1 given in [87] for completeness.


11

Theorem 1.2.2. Let u, a, b, c, f , g ∈ C(E, R+ ) and u(x, y) a(x, y) + b(x, y) + c(x, y)

x y

f (s,t)u(s,t) dt ds 0∞ 0 ∞

g(s,t)u(s,t) dt ds, 0

(1.2.11)

0

for x, y ∈ R+ . If p=

∞ ∞ 0

g(s,t) d(s,t) dt ds < 1,

(1.2.12)

0

then u(x, y) B(x, y) + M D(x, y),

(1.2.13)

for x, y ∈ R+ , where x y

f (s,t)a(s,t) dt ds,

(1.2.14)

D(x, y) = c(x, y) + b(x, y)A(x, y) f (s,t)c(s,t) dt ds, 0 0 x y f (s,t)b(s,t) dt ds , A(x, y) = exp

(1.2.15)

B(x, y) = a(x, y) + b(x, y)A(x, y) 0

0

x y

0

(1.2.16)

0

and M=

1 1− p

∞ ∞

g(s,t)B(s,t) dt ds. 0

(1.2.17)

0

As a consequence of Theorem 1.2.2, we have the following inequality. Theorem 1.2.3. Let u, p, q, r ∈ C(E0 , R+ ). Suppose that u(x, y) p(x, y) + q(x, y)

a b

r(s,t)u(s,t) dt ds, 0

(1.2.18)

0

for (x, y) ∈ E0 . If a b

d=

r(s,t)q(s,t) dt ds < 1, 0

(1.2.19)

0

then

u(x, y) p(x, y) + q(x, y) for (x, y) ∈ E0 .

1 1−d

a b 0

0

r(s,t)p(s,t) dt ds ,

(1.2.20)

12


Theorem 1.2.4. Let u, p, q, r ∈ C(E0 , R+ ) and x y s t p(s,t) u(s,t) + q(σ , τ )u(σ , τ ) d τ d σ u(x, y) c + 0

0

0

a b

+ 0

0

0

r(σ , τ )u(σ , τ ) d τ d σ dt ds,

for (x, y) ∈ E0 , where c 0 is a constant. If σ τ a b d= r(σ , τ ) exp [p(s,t) + q(s,t)]dt ds d τ d σ < 1, 0

0

0

(1.2.22)

0

then u(x, y)

(1.2.21)

c exp 1−d

x 0

y

[p(s,t) + q(s,t)]dt ds ,

(1.2.23)

0

for (x, y) ∈ E0 . Another useful inequality proved in [92] is embodied in the following theorem. Theorem 1.2.5. Let u, a ∈ C(E, R+ ), b, D1 b, D2 b, D2 D1 b, e ∈ C(E2 , R+ ) and c 0 is a constant. If u(x, y) c +

x y

a(s,t)u(s,t) + b(x, y, s,t)u(s,t)

0

0

s t

+

e(s,t, m, n)u(m, n) dn dm dt ds, 0

for x, y ∈ R+ , then u(x, y) c exp for x, y ∈ R+ , where x y

+ 0

Proof.

x 0

y

[a(s,t) + K(s,t)] dt ds ,

0

0

(1.2.25)

0

x

K(x, y) = b(x, y, x, y) +

(1.2.24)

0

y

D1 b(x, y, m, y) dm +

D2 b(x, y, x, n) dn

0

x y

D2 D1 b(x, y, m, n) dn dm +

e(x, y, m, n) dn dm. 0

(1.2.26)

0

Define a function z(x, y) by the right hand side of (1.2.21). Then z(x, 0) = z(0, y) =

c, u(x, y) z(x, y), z(x, y) is nondecreasing in x and y and (see [87, p. 65]) D2 D1 z(x, y) = a(x, y)u(x, y) + b(x, y, x, y)u(x, y) x

+ 0

y

D1 b(x, y, m, y)u(m, y) dm +

0

x y

+ 0

0

D2 b(x, y, x, n)u(x, n) dn

x y

D2 D1 b(x, y, m, n)u(m, n) dn dm +

e(x, y, m, n)u(m, n) dn dm 0

0


13

a(x, y)z(x, y) + b(x, y, x, y)z(x, y) x

+

y

D1 b(x, y, m, y)z(m, y) dm +

0

0

x y

+ 0

0

x y

D2 D1 b(x, y, m, n)z(m, n) dn dm +

a(x, y) + b(x, y, x, y) +

+ 0

e(x, y, m, n)z(m, n) dn dm 0

0

x

y

D1 b(x, y, m, y) dm +

0

x y 0

D2 b(x, y, x, n)z(x, n) dn

0

D2 b(x, y, x, n) dn

x y

D2 D1 b(x, y, m, n) dn dm +

e(x, y, m, n) dn dm z(x, y) 0

0

= [a(x, y) + K(x, y)]z(x, y),

(1.2.27)

where K(x, y) is given by (1.2.26). Now by following the proof of Theorem 4.2.1 given in [82], from (1.2.27), we get z(x, y) c exp

x 0

y

[a(s,t) + K(s,t)] dt ds .

(1.2.28)

0

Using (1.2.28) in u(x, y) z(x, y), we get the required inequality in (1.2.25). 1.3

Volterra-type integral equation

Consider the integral equation of the form x

u(x, y) = f (x, y) + 0

g(x, y, ξ , u(ξ , y)) d ξ +

x y 0

0

h(x, y, σ , τ , u(σ , τ )) d τ d σ ,

(1.3.1)

for x, y ∈ R+ , where f ∈ C(E, Rn ), g ∈ C(E1 × Rn , Rn ), h ∈ C(E2 × Rn , Rn ) are the given functions and u is the unknown function to be fond. The origin of equation (1.3.1) can be traced back to the important observation in [28, p. 20] and the driven equation in (5). In this section, we present some basic qualitative properties of solutions of equation (1.3.1) under some suitable conditions on the functions f , g, h (see [115]). Let S be the space of functions z ∈ C(E, Rn ) which fulfil the condition |z(x, y)| = O(exp(λ (x + y))),

(1.3.2)

where λ > 0 is a constant. In the space S we define the norm |z|S = sup [|z(x, y)| exp (−λ (x + y))] . (x,y)∈E

(1.3.3)

14


It is easy to see that S with norm defined in (1.3.3) is a Banach space. We note that the condition (1.3.2) implies that there exists a constant N 0 such that |z(x, y)| N exp(λ (x + y)). Using this fact in (1.3.3), we observe that |z|S N.

(1.3.4)

To carry out the proof of the existence and uniqueness of solutions of equation (1.3.1) and some other equations considered in the subsequent chapters, we shall make use of the above class S of functions without further mention. We start with the following theorem which ensures the existence of a unique solution to equation (1.3.1). Theorem 1.3.1. Suppose that (i) the functions g, h in equation (1.3.1) satisfy the conditions |g(x, y, ξ , u) − g(x, y, ξ , u)| a(x, y, ξ )|u − u|, |h(x, y, σ , τ , u) − h(x, y, σ , τ , u)| b(x, y, σ , τ )|u − u|,

(1.3.5) (1.3.6)

where a ∈ C(E1 , R+ ), b ∈ C(E2 , R+ ), (ii) for λ as in (1.3.2), (a1 ) there exists a nonnegative constant α such that α < 1 and x 0

a(x, y, ξ ) exp(λ (ξ + y)) d ξ +

x y 0

0

b(x, y, σ , τ ) exp(λ (σ + τ )) d τ d σ

α exp(λ (x + y)), (a2 ) there exists a nonnegative constant β such that x x y f (x, y) + β exp(λ (x + y)), g(x, y, ξ , 0) d ξ + h(x, y, σ , τ , 0) d τ d σ 0 0 0

(1.3.7)

(1.3.8)

where f , g, h are the functions in equation (1.3.1). Under the assumptions (i) and (ii) the equation (1.3.1) has a unique solution u(x, y) on E in S. Proof. Let u ∈ S and define the operator T by x

(Tu)(x, y) = f (x, y) + 0

g(x, y, ξ , u(ξ , y)) d ξ +

x y 0

0

h(x, y, σ , τ , u(σ , τ )) d τ d σ . (1.3.9)

Now we shall show that T maps S into itself. Evidently, Tu is continuous on E and Tu ∈ Rn . We verify that (1.3.2) is fulfilled. From (1.3.9) and using the hypotheses and (1.3.4), we have

x x y g(x, y, ξ , 0) d ξ + h(x, y, σ , τ , 0) d τ d σ |(Tu)(x, y)| f (x, y) + 0

0

0


x

+

15

|g(x, y, ξ , u(ξ , y)) − g(x, y, ξ , 0)|d ξ +

0

β exp(λ (x + y)) +

x y

|h(x, y, σ , τ , u(σ , τ )) − h(x, y, σ , τ , 0)|d τ d σ

0 0

x

x y

a(x, y, ξ )|u(ξ , y)|d ξ + b(x, y, σ , τ )|u(σ , τ )|d τ d σ 0 0 0 x β exp(λ (x + y)) + |u|S a(x, y, ξ ) exp(λ (ξ + y)) d ξ x y

+ 0

0

0

b(x, y, σ , τ ) exp(λ (σ + τ ))d τ d σ

[β + N α ] exp(λ (x + y)).

(1.3.10)

From (1.3.10) it follows that Tu ∈ S. This proves that T maps S into itself. Now, we verify that the operator T is a contraction map. Let u, v ∈ S. From (1.3.9) and using the hypotheses, we have |(Tu)(x, y) − (T v)(x, y)| x y

+ 0

x

0

x 0

|g(x, y, ξ , u(ξ , y)) − g(x, y, ξ , v(ξ , y))|d ξ

|h(x, y, σ , τ , u(σ , τ )) − h(x, y, σ , τ , v(σ , τ ))|d τ d σ x y

a(x, y, ξ )|u(ξ , y) − v(ξ , y)|d ξ + b(x, y, σ , τ )|u(σ , τ ) − v(σ , τ )|d τ d σ 0 0 0 x x y |u − v|S a(x, y, ξ ) exp(λ (ξ + y)) d ξ + b(x, y, σ , τ ) exp(λ (σ + τ ))d τ d σ

0

0

0

α |u − v|S exp(λ (x + y)).

(1.3.11)

From (1.3.11), we obtain |Tu − T v|S α |u − v|S . Since α < 1, it follows from Banach fixed point theorem (see [28, p. 37] and [151, p. 372]) that T has a unique fixed point in S. The fixed point of T is however a solution of equation (1.3.1). The proof is complete. Remark 1.3.1.

We note that, Theorem 1.3.1 given above provides a simple way to estab-

lish the existence and uniqueness for solutions of equation (1.3.1) in the space of continuous functions. The norm defined by (1.3.3) is a variant of Bielecki’s norm [12,29], first used for the study of solutions of ordinary differential equations and has the role of improving the length of the interval on which the existence is assured. For a number results on the existence and uniqueness of solutions of special and general versions of equation (1.3.1) by using various techniques, see [5,134] and the references therein. A slight variant of Theorem 1.3.1 is given in the following theorem.

16


Theorem 1.3.2.

Suppose that the functions g, h in equation (1.3.1) satisfy the conditions

(1.3.5), (1.3.6) and assume that x x y a(x, y, ξ ) d ξ + b(x, y, σ , τ ) d τ d σ α < 1. sup x,y∈R+

0

0

(1.3.12)

0

Then the equation (1.3.1) has a unique solution u ∈ C(E, Rn ). Proof.

Consider the space C(E, Rn ) endowed with a norm · defined by u = sup |u(x, y)|,

for u ∈ C(E, Rn ). It is well known any u,

v ∈ C(E, Rn ),

x,y∈R+ that C(E, Rn )

(1.3.13)

is a Banach space with norm (1.3.13). For

one can easily verify from the hypotheses that, the operator T defined

by (1.3.9) for any u ∈ C(E, Rn ) satisfies Tu − T v α u − v.

(1.3.14)

This shows that T is a contraction. Therefore, (1.3.1) has a unique solution u ∈ C(E, Rn ). The following theorem deals with the estimate on the solution of equation (1.3.1). Theorem 1.3.3.

Suppose that the functions f , g, h in equation (1.3.1) satisfy the condi-

tions |g(x, y, ξ , u) − g(x, y, ξ , u)| q(x, y, ξ )|u − u|, |h(x, y, σ , τ , u) − g(x, y, σ , τ , u)| r(x, y, σ , τ )|u − u|, where q, D1 q ∈ C(E1 , R+ ) and r, D1 r, D2 r, D2 D1 r ∈ C(E2 , R+ ) and x x y g(x, y, ξ , 0) d ξ + h(x, y, σ , τ , 0) d τ d σ < ∞. c = sup f (x, y) + 0 0 0

(1.3.15) (1.3.16)

(1.3.17)

x,y∈R+

If u(x, y) is any solution of equation (1.3.1) on E, then x y R(s,t) dt ds , |u(x, y)| cP(x, y) exp 0

(1.3.18)

0

for (x, y) ∈ E, where P(x, y) and R(x, y) are given by (1.2.3) and (1.2.5). Proof.

x

+

Using the fact that u(x, y) is a solution of equation (1.3.1) and hypotheses, we have x x y g(x, y, ξ , 0) d ξ + h(x, y, σ , τ , 0) d τ d σ |u(x, y)| f (x, y) + 0 0 0

|g(x, y, ξ , u(ξ , y)) − g(x, y, ξ , 0)|d ξ +

0

c+

x 0

x y

|h(x, y, σ , τ , u(σ , τ )) − h(x, y, σ , τ , 0)|d τ d σ

0 0

q(x, y, ξ )|u(ξ , y)|d ξ +

x y 0

0

r(x, y, σ , τ )|u(σ , τ )|d τ d σ .

Now, an application of Theorem 1.2.1 to (1.3.19) yields (1.3.18).

(1.3.19)


17

Remark 1.3.2. In Theorem 1.3.3, if we assume that (i) the Q(x, y) given by (1.2.4) is such that Q(x, y) < ∞ and function ∞

∞

0

0

(ii)

R(s,t) dt ds < ∞,

then the solution u(x, y) of equation (1.3.1) is bounded on E. A variant of Theorem 1.3.3 is embodied in the following theorem. Suppose that the functions g, h in equation (1.3.1) satisfy the conditions

Theorem 1.3.4.

(1.3.15), (1.3.16) and assume that x 0

|g(x, y, ξ , f (ξ , y))|d ξ +

x y 0

0

|h(x, y, σ , τ , f (σ , τ ))|d τ d σ d,

(1.3.20)

for x, y ∈ R+ , where f is the function involved in (1.3.1) and d 0 is a constant. If u(x, y) is any solution of equation (1.3.1) on E, then |u(x, y) − f (x, y)| d P(x, y) exp

x

y

R(s,t) dt ds , 0

(1.3.21)

0

for (x, y) ∈ E, where P(x, y) and R(x, y) are given by (1.2.3) and (1.2.5). Proof.

Let w(x, y) = |u(x, y) − f (x, y)| for (x, y) ∈ E. Using the fact that u(x, y) is a solu-

tion of equation (1.3.1) and the hypotheses, we have w(x, y)

x 0

|g(x, y, ξ , f (ξ , y))|d ξ + x

+ 0

x y

+ 0

d+

x 0

0

x y 0

0

|h(x, y, σ , τ , f (σ , τ ))|d τ d σ

|g(x, y, ξ , u(ξ , y)) − g (x, y, ξ , f (ξ , y))| d ξ

|h(x, y, σ , τ , u(σ , τ )) − h(x, y, σ , τ , f (σ , τ ))|d τ d σ

q(x, y, ξ )w(ξ , y) d ξ +

x y 0

0

r(x, y, σ , τ )w(σ , τ ) d τ d σ .

(1.3.22)

Now, an application of Theorem 1.2.1 to (1.3.22) yields (1.3.21). We call the function u ∈ C(E, Rn ) an ε -approximate solution to equation (1.3.1), if there exists a constant ε 0 such that

x x y u(x, y) − f (x, y) + ε, g(x, y, ξ , u( ξ , y)) d ξ + h(x, y, σ , τ , u( σ , τ )) d τ d σ 0 0 0 for x, y ∈ R+ . The next theorem deals with the estimate on the difference between two approximate solutions of equation (1.3.1).

18


Let u1 (x, y) and u2 (x, y) be respectively, ε1 - and ε2 -approximate solutions

Theorem 1.3.5.

of equation (1.3.1) on E. Suppose that the functions g, h in equation (1.3.1) satisfy the conditions (1.3.15), (1.3.16). Then |u1 (x, y) − u2 (x, y)| (ε1 + ε2 ) P(x, y) exp

x 0

y

R(s,t) dt ds ,

(1.3.23)

0

for (x, y) ∈ E, where P(x, y) and R(x, y) are given by (1.2.3) and (1.2.5). Since u1 (x, y) and u2 (x, y) are respectively, ε1 - and ε2 -approximate solutions to

Proof.

equation (1.3.1), we have x ui (x, y) − f (x, y) + g(x, y, ξ , ui (ξ , y)) d ξ 0

h(x, y, σ , τ , ui (σ , τ )) d τ d σ εi , 0 0 for i = 1, 2. From (1.3.24) and using the elementary inequalities x y

+

|v − z| |v| + |z|,

|v| − |z| |v − z|,

(1.3.24)

(1.3.25)

we observe that

ε1 + ε2

x x y g(x, y, ξ , u1 (ξ , y)) d ξ + h(x, y, σ , τ , u1 (σ , τ )) d τ d σ u1 (x, y) − f (x, y) + 0 0 0

x x y + u2 (x, y) − f (x, y) + g(x, y, ξ , u2 (ξ , y)) d ξ + h(x, y, σ , τ , u2 (σ , τ )) d τ d σ 0 0 0 [u1 (x, y) − u2 (x, y)]

x

−

f (x, y) + 0

− f (x, y) +

x 0

g(x, y, ξ , u1 (ξ , y)) d ξ +

g(x, y, ξ , u2 (ξ , y)) d ξ +

x

y

0

0

x y 0

0

h(x, y, σ , τ , u1 (σ , τ )) d τ d σ

h(x, y, σ , τ , u2 (σ , τ )) d τ d σ

x |u1 (x, y) − u2 (x, y)| − {g(x, y, ξ , u1 (ξ , y)) − g(x, y, ξ , u2 (ξ , y))} d ξ 0 x y {h(x, y, σ , τ , u1 (σ , τ )) − h(x, y, σ , τ , u2 (σ , τ ))} d τ d σ . (1.3.26) + 0 0

Let w(x, y) = |u1 (x, y) − u2 (x, y)|, (x, y) ∈ E. From (1.3.26) and using conditions (1.3.15), (1.3.16), we have w(x, y) (ε1 + ε2 ) + x y

+ 0 0 x

(ε1 + ε2 ) +

0

x 0

|g(x, y, ξ , u1 (ξ , y)) − g(x, y, ξ , u2 (ξ , y))| d ξ

|h(x, y, σ , τ , u1 (σ , τ )) − h(x, y, σ , τ , u2 (σ , τ ))| d τ d σ

q(x, y, ξ )w(ξ , y) d ξ +

x y 0

0

r(x, y, σ , τ )w(σ , τ ) d τ d σ .

Now an application of Theorem 1.2.1 to (1.3.27) yields (1.3.23).

(1.3.27)


19

In case, if u1 (x, y) is a solution of equation (1.3.1), then we have ε1 = 0

Remark 1.3.3.

and from (1.3.23) we see that u1 (x, y) → u2 (x, y) as ε2 → 0. Moreover, from (1.3.23), the uniqueness of solutions of equation (1.3.1) follows if εi = 0 (i = 1, 2). We next consider the following variants of equation (1.3.1) x

u(x, y) = f (x, y) + 0

g(x, y, ξ , u(ξ , y), μ ) d ξ

x y

+ 0

0

h(x, y, σ , τ , u(σ , τ ), μ ) d τ d σ ,

(1.3.28)

and x

u(x, y) = f (x, y) +

g(x, y, ξ , u(ξ , y), μ0 ) d ξ

0 x y

+ 0

0

h(x, y, σ , τ , u(σ , τ ), μ0 ) d τ d σ ,

(1.3.29)

for (x, y) ∈ E, where f ∈ C(E, Rn ), g ∈ C(E1 × Rn × R, Rn ), h ∈ C(E2 × Rn × R, Rn ), and

μ , μ0 are parameters. The following theorem shows the dependency of solutions of equations (1.3.28) and (1.3.29) on parameters. Theorem 1.3.6.

Suppose that the functions g, h in equations (1.3.28), (1.3.29) satisfy the

conditions |g(x, y, ξ , u, μ ) − g(x, y, ξ , u, μ )| q(x, y, ξ )|u − u|,

(1.3.30)

|g(x, y, ξ , u, μ ) − g(x, y, ξ , u, μ0 )| p1 (x, y, ξ )|μ − μ0 |,

(1.3.31)

|h(x, y, σ , τ , u, μ ) − h(x, y, σ , τ , u, μ )| r(x, y, σ , τ )|u − u|,

(1.3.32)

|h(x, y, σ , τ , u, μ ) − h(x, y, σ , τ , u, μ0 )| p2 (x, y, σ , τ )|μ − μ0 |,

(1.3.33)

where p1 , q, D1 q ∈ C(E1 , R+ ), p2 , r, D1 r, D2 r, D2 D1 r ∈ C(E2 , R+ ) and x

x y 0

0

p1 (x, y, ξ ) d ξ M1 ,

(1.3.34)

p2 (x, y, σ , τ ) d τ d σ M2 ,

(1.3.35)

0

in which M1 , M2 are nonnegative constants. Let u1 (x, y) and u2 (x, y) be the solutions of equations (1.3.28) and (1.3.29) respectively. Then |u1 (x, y) − u2 (x, y)| (M1 + M2 )|μ − μ0 |P(x, y) exp

x

y

R(s,t) dt ds , 0

0

for (x, y) ∈ E, where P(x, y) and R(x, y) are given by (1.2.3) and (1.2.5).

(1.3.36)

20

Proof.


Let w(x, y) = |u1 (x, y) − u2 (x, y)|, (x, y) ∈ E. Using the facts that u1 (x, y) and

u2 (x, y) are the solutions of equations (1.3.28) and (1.3.29) and hypotheses, we have w(x, y) x

+ 0

x y

+ 0

0

x y

+ 0

0

x y

+ 0

0

x 0

|g(x, y, ξ , u1 (ξ , y), μ ) − g(x, y, ξ , u2 (ξ , y), μ )|d ξ

|g(x, y, ξ , u2 (ξ , y), μ ) − g(x, y, ξ , u2 (ξ , y), μ0 )| d ξ

|h(x, y, σ , τ , u1 (σ , τ ), μ ) − h(x, y, σ , τ , u2 (σ , τ ), μ )| d τ d σ

|h(x, y, σ , τ , u2 (σ , τ ), μ ) − h(x, y, σ , τ , u2 (σ , τ ), μ0 )| d τ d σ x 0

q(x, y, ξ )w(ξ , y) d ξ +

r(x, y, σ , τ )w(σ , τ ) d τ d σ +

(M1 + M2 )|μ − μ0 | x

+ 0

q(x, y, ξ )w(ξ , y) d ξ +

x y 0

0

x 0

p1 (x, y, ξ )|μ − μ0 |d ξ

x y 0

0

p2 (x, y, σ , τ )|μ − μ0 |d τ d σ

r(x, y, σ , τ )w(σ , τ ) d τ d σ .

(1.3.37)

Now an application of Theorem 1.2.1 to (1.3.37) yields (1.3.36), which shows the dependency of solutions of equations (1.3.28) and (1.3.29) on parameters. Remark 1.3.4

We note that the method employed in this section can be extended to study

the integrodifferential equation of the form D2 D1 u(x, y) = f (x, y, u(x, y), Gu(x, y), Hu(x, y)),

(1.3.38)

u(x, 0) = σ (x), u(0, y) = τ (y),

(1.3.39)

with the given data

for x, y ∈ R+ , where

x

Gu(x, y) = 0

g(x, y, ξ , u(ξ , y)) d ξ ,

(1.3.40)

x y

Hu(x, y) =

h(x, y, m, n, u(m, n)) dn dm, 0

(1.3.41)

0

under some suitable conditions on the functions f , g, h, σ , τ and by making use of a suitable variant of the inequality given in [87, Theorem 2.5.1, p. 96].


1.4

21

Volterra-Fredholm-type integral equation

We shall deal in this section with the Volterra-Fredholm-type integral equation in the form x y

F(x, y, s,t, u(s,t))dt ds +

u(x, y) = h(x, y) + 0

0

for x, y ∈ R+ , where h ∈ C(E, Rn ), F ∈ C(E2

∞ ∞

G(x, y, s,t, u(s,t))dt ds, (1.4.1)

0 0 ×Rn , Rn ), G ∈ C(E 2 ×Rn , Rn ).

The equations

of the form (1.4.1) are of particular interest, since the special versions of the same arise in a variety of applications, see [5,134]. It is our aim here to give some important qualitative properties of solutions of equation (1.4.1) under various assumptions on the functions involved therein (see [89]). Our first result concerning the existence and uniqueness of solutions of equation (1.4.1) is given in the following theorem. Theorem 1.4.1. Assume that (i) the functions F, G in equation (1.4.1) satisfy the conditions |F(x, y, s,t, u) − F(x, y, s,t, u)| k(x, y, s,t)|u − u|,

(1.4.2)

|G(x, y, s,t, u) − G(x, y, s,t, u)| r(x, y, s,t)|u − u|,

(1.4.3)

where k ∈ C(E2 , R+ ), r ∈ C(E 2 , R+ ), (ii) for λ as in (1.3.2), (b1 ) there exist nonnegative constants α1 , α2 such that α1 + α2 < 1 and x y

0∞ 0∞ 0

0

k(x, y, s,t) exp(λ (s + t)) dt ds α1 exp(λ (x + y)),

(1.4.4)

r(x, y, s,t) exp(λ (s + t)) dt ds α2 exp(λ (x + y)),

(1.4.5)

(b2 ) there exists a nonnegative constant β such that |h(x, y)| +

x y 0

0

|F(x, y, s,t, 0)| dt ds +

∞ ∞ 0

0

|G(x, y, s,t, 0)| dt ds

β exp(λ (x + y)),

(1.4.6)

where h, F, G are the functions in equation (1.4.1). Then the equation (1.4.1) has a unique solution u(x, y) on E in S. Proof. Let u ∈ S and define the operator T by x y

(Tu)(x, y) = h(x, y) + +

F(x, y, s,t, u(s,t)) dt ds 0 ∞ 0 ∞

G(x, y, s,t, u(s,t)) dt ds. 0

(1.4.7)

0

for (x, y) ∈ E. The proof that T maps S into itself and is a contraction map can be completed by closely looking at the proof of Theorem 1.3.1 with suitable modifications. Here, we omit the details.

22


Remark 1.4.1.

We note that Theorem 1.4.1 yields existence and uniqueness of solutions

of equation (1.4.1) in S. One can also formulate existence and uniqueness result similar to that of Theorem 1.3.2 for the solution u ∈ C(E, Rn ) of equation (1.4.1). The following theorem which asserts only the uniqueness of solution of equation (1.4.1) on E in Rn can be obtained by using the inequality in Theorem 1.2.2. Theorem 1.4.2. Suppose that the functions F, G in equation (1.4.1) satisfy the conditions |F(x, y, s,t, u) − F(x, y, s,t, u)| b(x, y) f (s,t)|u − u|,

(1.4.8)

|G(x, y, s,t, u) − G(x, y, s,t, u)| c(x, y)g(s,t)|u − u|,

(1.4.9)

where b, f , c, g ∈ C(E, R+ ). Let p, D(x, y) be as in (1.2.12), (1.2.15) respectively. Then the equation (1.4.1) has at most one solution on E in Rn . Proof.

Let u1 (x, y) and u2 (x, y) be two solutions of equation (1.4.1). Then, we have

u1 (x, y) − u2 (x, y) = +

x y

{F(x, y, s,t, u1 (s,t)) − F(x, y, s,t, u2 (s,t))}dt ds

0∞ 0 ∞ 0

0

{G(x, y, s,t, u1 (s,t)) − G(x, y, s,t, u2 (s,t))}dt ds.

(1.4.10)

From (1.4.10) and using hypotheses, we have |u1 (x, y) − u2 (x, y)| b(x, y) + c(x, y)

x y

0∞ 0∞ 0

0

f (s,t)|u1 (s,t) − u2 (s,t)|dt ds g(s,t)|u1 (s,t) − u2 (s,t)|dt ds.

(1.4.11)

Here, it is easy to observe that B(x, y) and M defined in (1.2.14) and (1.2.17) reduces to B(x, y) = 0 and M = 0. Now an application of Theorem 1.2.2 (with a(x, y) = 0) to (1.4.11) yields |u1 (x, y) − u2 (x, y)| 0, and hence u1 (x, y) = u2 (x, y). Thus there is at most one solution to equation (1.4.1) on E in Rn . The following theorem concerning the estimate on the solution of equation (1.4.1) holds. Theorem 1.4.3.

Assume that the functions F, G in equation (1.4.1) satisfy the conditions |F(x, y, s,t, u)| b(x, y) f (s,t)|u|,

(1.4.12)

|G(x, y, s,t, u)| c(x, y)g(s,t)|u|,

(1.4.13)

where b, f , c, g ∈ C(E, R+ ). Let p, D(x, y) be as in (1.2.12), (1.2.15) respectively and ∞ ∞ 1 g(s,t)B1 (s,t) dt ds, (1.4.14) M1 = 1− p 0 0 where B1 (x, y) is defined by the right hand side of (1.2.14) by replacing a(x, y) by |h(x, y)|. If u(x, y) is any solution of equation (1.4.1), then |u(x, y)| B1 (x, y) + M1 D(x, y), for (x, y) ∈ E.

(1.4.15)


Proof.

23

Using the fact that u(x, y) is a solution of equation (1.4.1) and the hypotheses, we

have |u(x, y)| |h(x, y)| +

x y

|h(x, y)| + b(x, y)

0

0

|F(x, y, s,t, u(s,t))| dt ds +

x y

∞ ∞ 0

f (s,t)|u(s,t)| dt ds + c(x, y) 0

0

0

|G(x, y, s,t, u(s,t))| dt ds

∞ ∞

g(s,t)|u(s,t)| dt ds. 0

0

(1.4.16) Now an application of Theorem 1.2.2 to (1.4.16) yields (1.4.15). In the next theorem we give an estimate on the solution of equation (1.4.1) assuming that the functions F, G satisfy the Lipschitz type conditions (1.4.8), (1.4.9). Theorem 1.4.4. Suppose that the functions F, G in equation (1.4.1) satisfy the conditions (1.4.8), (1.4.9) respectively. Let p, D(x, y) be as in (1.2.12), (1.2.15) respectively and x y

h0 (x, y) =

0

0

|F(x, y, s,t, h(s,t))| dt ds + M2 =

1 1− p

∞ ∞ 0

0

∞ ∞ 0

0

|G(x, y, s,t, h(s,t))| dt ds,

g(s,t)B2 (s,t) dt ds,

(1.4.17)

(1.4.18)

where B2 (x, y) is defined by the right hand side of (1.2.14) by replacing a(x, y) by h0 (x, y). If u(x, y) is any solution of equation (1.4.1), then |u(x, y) − h(x, y)| B2 (x, y) + M2 D(x, y),

(1.4.19)

for (x, y) ∈ E. The proof can be completed by following the proof of Theorem 1.3.4. Here, we omit the details. The following theorem deals with the estimate on the difference between the solutions of equation (1.4.1) and the system of Volterra integral equations x y

v(x, y) = h(x, y) +

F(x, y, s,t, v(s,t)) dt ds, 0

(1.4.20)

0

for (x, y) ∈ E, where the functions h, F are as given in equation (1.4.1). Theorem 1.4.5. Suppose that the functions F, G in equation (1.4.1) satisfy the conditions (1.4.8), (1.4.9) respectively and G(x, y, s,t, 0) = 0. Let v(x, y) be a solution of equation (1.4.20) such that |v(x, y)| Q, where Q 0 is a constant. Let p, D(x, y) be as in (1.2.12), (1.2.15) respectively and a(x, y) = Q c(x, y)

∞ ∞

g(s,t) dt ds, 0

0

(1.4.21)

24


M3 =

∞ ∞

1 1− p

0

0


(1.4.22)

where B3 (x, y) is defined by the right hand side of (1.2.14) by replacing a(x, y) by a(x, y). If u(x, y) is a solution of equation (1.4.1), then |u(x, y) − v(x, y)| B3 (x, y) + M3 D(x, y),

(1.4.23)

for (x, y) ∈ E. Proof.

Using the facts that u(x, y) and v(x, y) are the solutions of equations (1.4.1) and

(1.4.20) on E, we observe that u(x, y) − v(x, y) = +

x y

{F(x, y, s,t, u(s,t)) − F(x, y, s,t, v(s,t))} dt ds

0∞ 0 ∞ 0

0

G(x, y, s,t, u(s,t)) − G(x, y, s,t, v(s,t))

+ G(x, y, s,t, v(s,t)) − G(x, y, s,t, 0) dt ds.

(1.4.24)

From (1.4.24) and using the hypotheses, we have |u(x, y) − v(x, y)| a(x, y) + b(x, y) + c(x, y)

x y

f (s,t)|u(s,t) − v(s,t)| dt ds

0∞ 0 ∞ 0

0

g(s,t)|u(s,t) − v(s,t)| dt ds.

(1.4.25)

Now, an application of Theorem 1.2.2 to (1.4.25) yields (1.4.23). The following theorems provide conditions for continuous dependence of solutions of equation (1.4.1) on functions involved on the right hand side and continuous dependence of solutions of equations of the form (1.4.1) on parameters. Consider the equation (1.4.1) and the system of Volterra-Fredholm-type integral equations x y

w(x, y) = h(x, y) + +

F(x, y, s,t, w(s,t)) dt ds 0 ∞ 0 ∞

G(x, y, s,t, w(s,t)) dt ds, 0

(1.4.26)

0

for (x, y) ∈ E, where h ∈ C(E, Rn ), F ∈ C(E2 × Rn , Rn ), G ∈ C(E 2 × Rn , Rn ). Theorem 1.4.6. Suppose that the functions F, G in equation (1.4.1) satisfy the conditions (1.4.8), (1.4.9) respectively. Let w(x, y) be a given solution of equation (1.4.26) on E and assume that |h(x, y) − h(x, y)| + +

∞ ∞ 0

0

x y 0

0

|F(x, y, s,t, w(s,t)) − F(x, y, s,t, w(s,t))| dt ds

|G(x, y, s,t, w(s,t)) − G(x, y, s,t, w(s,t))| dt ds ε ,

(1.4.27)


25

where h, F, G and h, F, G are functions in equations (1.4.1) and (1.4.26) and ε > 0 is an arbitrary small constant. Let p, D(x, y) be as in (1.2.12), (1.2.15) respectively and 1 1− p

M4 =

∞ ∞ 0

0


(1.4.28)

where B4 (x, y) is defined by the right hand side of (1.2.14) by replacing a(x, y) by ε . Then the solution u(x, y) of equation (1.4.1) on E depends continuously on the functions involved on the right hand side of equation (1.4.1). Proof.

Let u(x, y) be a solution of (1.4.1) on E and w(x, y) be a given solution of equation

(1.4.26). Then u(x, y) − w(x, y) = h(x, y) − h(x, y)

+

x y 0

F(x, y, s,t, u(s,t)) − F(x, y, s,t, w(s,t))

0

+F(x, y, s,t, w(s,t)) − F(x, y, s,t, w(s,t)) dt ds +

∞ ∞ 0

0

{G(x, y, s,t, u(s,t)) − G(x, y, s,t, w(s,t))

+G(x, y, s,t, w(s,t)) − G(x, y, s,t, w(s,t)) dt ds.

(1.4.29)

From (1.4.29) and using the hypotheses, we have |u(x, y) − w(x, y)| |h(x, y) − h(x, y)|

x y

|F(x, y, s,t, u(s,t)) − F(x, y, s,t, w(s,t))| dt ds

+ 0

0

x y

+ + +

|F(x, y, s,t, w(s,t)) − F(x, y, s,t, w(s,t))| dt ds

0∞ 0 ∞ 0∞ 0∞ 0

0

|G(x, y, s,t, u(s,t)) − G(x, y, s,t, w(s,t))| dt ds |G(x, y, s,t, w(s,t)) − G(x, y, s,t, w(s,t))| dt ds

ε + b(x, y) +c(x, y)

x y

∞ 0 ∞ 0 0

0

f (s,t)|u(s,t) − w(s,t)|dt ds

g(s,t)|u(s,t) − w(s,t)|dt ds.

(1.4.30)

Now, an application of Theorem 1.2.2 to (1.4.30) yields |u(x, y) − w(x, y)| B4 (x, y) + M4 D(x, y),

(1.4.31)

for (x, y) ∈ E. From (1.4.31), it follows that the solutions of equation (1.4.1) depends continuously on the functions involved on the right hand side of equation (1.4.1).

26

Remark 1.4.2.


From (1.4.31), it is easy to observe that if B4 (x, y) and D(x, y) are bounded

for (x, y) ∈ E and ε → 0, then |u(x, y) − w(x, y)| → 0 on E. We next consider the following systems of Volterra-Fredholm-type integral equations x y

z(x, y) = h(x, y) + +

0

and

0

x y

z(x, y) = h(x, y) + +

F(x, y, s,t, z(s,t), μ ) dt ds

0 ∞ 0 ∞

0

(1.4.32)

F(x, y, s,t, z(s,t), μ0 ) dt ds

0 ∞ 0 ∞ 0

G(x, y, s,t, z(s,t), μ ) dt ds,

G(x, y, s,t, z(s,t), μ0 ) dt ds,

(1.4.33)

for (x, y) ∈ E, where h ∈ C(E, Rn ), F ∈ C(E2 × Rn × R, Rn ), G ∈ C(E 2 × Rn × R, Rn ) and

μ , μ0 are parameters. Theorem 1.4.7.

Suppose that the functions F, G in equations (1.4.32), (1.4.33) satisfy

the conditions |F(x, y, s,t, z, μ ) − F(x, y, s,t, z, μ )| b(x, y) f (s,t)|z − z|,

(1.4.34)

|F(x, y, s,t, z, μ ) − F(x, y, s,t, z, μ0 )| r1 (x, y, s,t)|μ − μ0 |,

(1.4.35)

|G(x, y, s,t, z, μ ) − G(x, y, s,t, z, μ )| c(x, y)g(s,t)|z − z|,

(1.4.36)

|G(x, y, s,t, z, μ ) − G(x, y, s,t, z, μ0 )| r2 (x, y, s,t)|μ − μ0 |,

(1.4.37)

where b, f , c, g ∈ C(E, R+ ), and r1 , r2 ∈

C(E 2 , R+ ).

(1.2.15) respectively and x y r1 (x, y, s,t) dt ds + a0 (x, y) = |μ − μ0 | 0

0

0

Let p, D(x, y) be as in (1.2.12),

∞ ∞ 0

r2 (x, y, s,t) dt ds ,

(1.4.38)

∞ ∞ 1 g(s,t)B5 (s,t) dt ds, (1.4.39) 1− p 0 0 where B5 (x, y) is defined by the right hand side of (1.2.14) by replacing a(x, y) by a0 (x, y).

M5 =

Let z1 (x, y) and z2 (x, y) be the solutions of equations (1.4.32) and (1.4.33) respectively. Then |z1 (x, y) − z2 (x, y)| B5 (x, y) + M5 D(x, y),

(1.4.40)

for (x, y) ∈ E, which shows the dependency of solutions of equations (1.4.32) and (1.4.33) on parameters. The proof can be completed by following the proof of Theorem 1.3.6 and using Theorem 1.2.2. Here, we leave the details to the reader.


Remark 1.4.3.

27

We note that, one can use our approach here, to study the equations of the

form (1.4.1) involving more than two independent variables. The details of the formulation of such results are very close to those of given above with suitable modifications. We omit the details. 1.5

Integrodifferential equations of hyperbolic-type

In this section, first we shall deal with the following Volterra-Fredholm-type integrodifferential equation D2 D1 u(x, y) = f (x, y, u(x, y), Gu(x, y), Hu(x, y)),

(1.5.1)

u(x, 0) = α (x),

(1.5.2)

with the given data

for x ∈ Ia , y ∈ Ib , where

u(0, y) = β (y),

u(0, 0) = 0,

x y

Gu(x, y) = 0

0

0

0

g(x, y, s,t, u(s,t)) dt ds,

(1.5.3)

h(x, y, s,t, u(s,t)) dt ds,

(1.5.4)

a b

Hu(x, y) =

f ∈ C(E0 × R3 , R), g, h ∈ C(E02 × R, R), α ∈ C(Ia , R), β ∈ C(Ib , R). The problem of existence and uniqueness of solutions to (1.5.1)–(1.5.2) can be dealt with the method employed in earlier sections. Here, we shall discuss the error evaluation of two approximate solutions of (1.5.1)–(1.5.2) and convergence properties of solutions of approximate problems (see [117]). Let u ∈ C(E0 , R); D2 D1 u(x, y) exists and satisfies the inequality |D2 D1 u(x, y) − f (x, y, u(x, y), Gu(x, y), Hu(x, y))| ε , for a given constant ε 0, where it is supposed that (1.5.2) holds. Then we call u(x, y) an

ε -approximate solution of equation (1.5.1) with the given data (1.5.2). The following theorem deals with the estimate on the difference between the two approximate solutions of equation (1.5.1). Theorem 1.5.1.


tions | f (x, y, u, v, w) − f (x, y, u, v, w)| k(x, y) [|u − u| + |v − v| + |w − w|] ,

(1.5.5)

28


|g(x, y, σ , τ , u) − g(x, y, σ , τ , u)| e(x, y)q(σ , τ )|u − u|,

(1.5.6)

|h(x, y, σ , τ , u) − h(x, y, σ , τ , u)| m(x, y)r(σ , τ )|u − u|,

(1.5.7)

where k, e, q, m, r ∈ C(E0 , R+ ). Let p(x, y) = max{k(x, y), e(x, y), m(x, y)} for (x, y) ∈ E0 and d be as in (1.2.22). Let ui (x, y) (i = 1, 2) be respectively εi -approximate solutions of equation (1.5.1) on E0 with ui (x, 0) = αi (x),

ui (0, y) = βi (y),

ui (0, 0) = 0,

(1.5.8)

where αi ∈ C(Ia , R), βi ∈ C(Ib , R) such that |α1 (x) − α2 (x) + β1 (y) − β2 (y)| δ , where δ 0 is a constant. Then |u1 (x, y) − u2 (x, y)|

(ε ab + δ ) exp 1−d

x 0

y


(1.5.9)

(1.5.10)

0

for (x, y) ∈ E0 , where ε = ε1 + ε2 . Proof.

Since ui (x, y) (i = 1, 2) for (x, y) ∈ E0 are respectively, εi -approximate solutions

of equation (1.5.1) with (1.5.8), we have |D2 D1 ui (x, y) − f (x, y, ui (x, y), Gui (x, y), Hui (x, y))| εi .

(1.5.11)

Integrating both sides of (1.5.11) on E0 and using (1.5.8), we have

εi xy

x y

|D2 D1 ui (s,t) − f (s,t, ui (s,t), Gui (s,t), Hui (s,t))|dt ds 0 x 0 y {D2 D1 ui (s,t) − f (s,t, ui (s,t), Gui (s,t), Hui (s,t))}dt ds 0 0 x y = ui (x, y) − αi (x) − βi (y) − f (s,t, ui (s,t), Gui (s,t), Hui (s,t)) . 0

(1.5.12)

0

From (1.5.12) and using the elementary inequalities in (1.3.25), we observe that x y f (s,t, u1 (s,t), Gu1 (s,t), Hu1 (s,t)) dt ds (ε1 + ε2 )xy u1 (x, y) − [α1 (x) + β1 (y)] − 0

x + u2 (x, y) − [α2 (x) + β2 (y)] − 0

y 0

0

f (s,t, u2 (s,t), Gu2 (s,t), Hu2 (s,t)) dt ds

x y u1 (x, y) − [α1 (x) + β1 (y)] − f (s,t, u1 (s,t), Gu1 (s,t), Hu1 (s,t)) dt ds 0 0

x − u2 (x, y) − [α2 (x) + β2 (y)] − 0

y 0

f (s,t, u2 (s,t), Gu2 (s,t), Hu2 (s,t)) dt ds


29

|u1 (x, y) − u2 (x, y)| − |[α1 (x) + β1 (y)] − [α2 (x) + β2 (y)]| x y − f (s,t, u1 (s,t), Gu1 (s,t), Hu1 (s,t)) dt ds 0

−

0

x y 0

0

f (s,t, u2 (s,t), Gu2 (s,t), Hu2 (s,t)) dt ds .

(1.5.13)

Let z(x, y) = |u1 (x, y) − u2 (x, y)| for (x, y) ∈ E0 . From (1.5.13) and using the hypotheses, we observe that z(x, y) ε xy + |α1 (x) − α2 (x) + β1 (y) − β2 (y)| x y

+ 0

0

| f (s,t, u1 (s,t), Gu1 (s,t), Hu1 (s,t))

− f (s,t, u2 (s,t), Gu2 (s,t), Hu2 (s,t))|dt ds x y s t ε xy + δ + k(s,t) z(s,t) + e(s,t) q(σ , τ )z(σ , τ ) d τ d σ 0

0

0

a b

+m(s,t) 0

[ε ab + δ ] +

x y 0

0

r(σ , τ )z(σ , τ ) d τ d σ dt ds

s t p(s,t) z(s,t) + q(σ , τ )z(σ , τ ) d τ d σ 0

a b

+ 0

0

0

0

0

r(σ , τ )z(σ , τ ) d τ d σ dt ds.

(1.5.14)

Now an application of Theorem 1.2.4 to (1.5.14) yields (1.5.10). Remark 1.5.1.

In case if u1 (x, y) is a solution of problem (1.5.1)–(1.5.2), then we have

ε1 = 0 and from (1.5.10), we see that u2 (x, y) → u1 (x, y) as ε2 → 0 and δ → 0. Furthermore, if we put ε1 = ε2 = 0, α1 (x) = α2 (x), β1 (y) = β2 (y) in (1.5.10), then the uniqueness of solutions of problem (1.5.1)–(1.5.2) is established. Now we consider the problem (1.5.1)–(1.5.2) together with the following problem D2 D1 v(x, y) = f (x, y, v(x, y), Gv(x, y), Hv(x, y)),

(1.5.15)

v(x, 0) = α (x),

(1.5.16)

v(0, y) = β (y),

v(0, 0) = 0,

where G, H are as defined in (1.5.3), (1.5.4) and f ∈ C(E0 × R3 , R), α ∈ C(Ia , R), β ∈ C(Ib , R). In the next theorem we provide conditions concerning the closeness of solutions of problems (1.5.1)–(1.5.2) and (1.5.15)–(1.5.16).

30


Theorem 1.5.2.


tions (1.5.5)–(1.5.7) and there exist constants ε 0, δ 0 such that | f (x, y, u, v, w) − f (x, y, u, v, w)| ε ,

(1.5.17)

|α (x) − α (x) + β (y) − β (y)| δ ,

(1.5.18)

where f , α , β and f , α , β are as in problems (1.5.1)–(1.5.2) and (1.5.15)–(1.5.16). Let p(x, y) and d be as in Theorem 1.5.1. Let u(x, y) and v(x, y) be respectively the solutions of problems (1.5.1)–(1.5.2) and (1.5.15)–(1.5.16) on E0 . Then x y (ε ab + δ ) exp |u(x, y) − v(x, y)| [p(s,t) + q(s,t)]dt ds , 1−d 0 0

(1.5.19)

for (x, y) ∈ E0 . Proof.

Let w(x, y) = |u(x, y) − v(x, y)| for (x, y) ∈ E0 . Using the facts that u(x, y) and

v(x, y) are the solutions of problems (1.5.1)–(1.5.2) and (1.5.15)–(1.5.16) and hypotheses, we observe that x y

+ 0

0

0

0

x y

+

w(x, y) |α (x) − α (x) + β (y) − β (y)| | f (x, y, u(s,t), Gu(s,t), Hu(s,t)) − f (x, y, v(s,t), Gv(s,t), Hv(s,t))|dt ds | f (x, y, v(s,t), Gv(s,t), Hv(s,t)) − f (x, y, v(s,t), Gv(s,t), Hv(s,t))|dt ds s t k(s,t) w(s,t) + e(s,t) q(σ , τ )w(σ , τ ) d τ d σ 0 0 0 0 a b x y +m(s,t) r(σ , τ )w(σ , τ ) d τ d σ + ε dt ds

δ+

x y

0

ε ab + δ +

0

0

x y

0

s t

q(σ , τ )w(σ , τ ) d τ d σ r(σ , τ )w(σ , τ ) d τ d σ dt ds.

p(s,t) w(s,t) + 0

0

a b

+ 0

0

0

0

(1.5.20)

Now an application of Theorem 1.2.4 to (1.5.20) yields (1.5.19). Remark 1.5.2.

The result given in Theorem 1.5.2 relates the solutions of problems

(1.5.1)–(1.5.2) and (1.5.15)–(1.5.16) in the sense that if f is close to f , α is close to α ,

β is close to β , then the solutions to problems (1.5.1)–(1.5.2) and (1.5.15)–(1.5.16) are also close together. Next we consider the problem (1.5.1)–(1.5.2) together with the sequence of problems D2 D1 u(x, y) = f k (x, y, u(x, y), Gu(x, y), Hu(x, y)),

(1.5.21)


31

u(x, 0) = αk (x), u(0, y) = βk (y), u(0, 0) = 0,

(1.5.22)

where G, H are as defined in (1.5.3), (1.5.4) and fk ∈ C(E0 × R3 , R), αk ∈ C(Ia , R), βk ∈ C(Ib , R) for k = 1, 2, . . .. The following result is an immediate consequence of Theorem 1.5.2. Theorem 1.5.3.


tions (1.5.5)–(1.5.7) and there exist constants εk 0, δk 0 (k = 1, 2, . . .) such that | f (x, y, u, v, w) − fk (x, y, u, v, w)| εk ,

(1.5.23)

|α (x) − αk (x) + β (y) − βk (y)| δk ,

(1.5.24)

with εk → 0 and δk → 0 as k → ∞, where f , α , β and f k , αk , βk are as in problems (1.5.1)–(1.5.2) and (1.5.21)–(1.5.22). Let p(x, y) and d be as in Theorem 1.5.1. If uk (x, y) (k = 1, 2, . . .) and u(x, y) are respectively the solutions of problems (1.5.21)–(1.5.22) and (1.5.1)–(1.5.2) on E0 , then uk (x, y) → u(x, y) as k → ∞. Proof.

For k = 1, 2, . . ., the conditions of Theorem 1.5.2 hold. As an application of The-

orem 1.5.2 yields |uk (x, y) − u(x, y)|

(εk ab + δk ) exp 1−d

x 0

y


(1.5.25)

0

for (x, y) ∈ E0 . The required result follows from (1.5.25). Remark 1.5.3.

We note that the result obtained in Theorem 1.5.3 provide sufficient con-

ditions that ensures solutions to problems (1.5.21)–(1.5.22) will converge to solutions to problem (1.5.1)–(1.5.2). Next, we shall study some fundamental properties of solutions related to the following neutral type hyperbolic integrodifferential equation (see [27]) D2 D1 u(x, y) = f (x, y, u(x, y), D2 D1 u(x, y), Mu(x, y)),

(1.5.26)

with the given data u(x, 0) = σ (x), for x, y ∈ R+ , where

u(0, y) = τ (y),

u(0, 0) = 0,

(1.5.27)

x y

Mu(x, y) = 0 n

0 n

g(x, y, m, n, u(m, n), D2 D1 u(m, n))dn dm,

(1.5.28)

f ∈ C(E × Rn × Rn × R , R ), g ∈ C(E2 × Rn × Rn , Rn ), σ , τ ∈ C(R+ , Rn ). Obviously, M0 =

xy 0 0

g(x, y, m, n, 0, 0) dn dm. By a solution of equation (1.5.26) with the given data

32


(1.5.27) (IBVP (1.5.26)–(1.5.27) for short), we mean a function u ∈ C(E, Rn ) which satisfy the equations (1.5.26) and (1.5.27). For z, D2 D1 z ∈ C(E, Rn ), we denote by |z(x, y)|0 = |z(x, y)| + |D2 D1 z(x, y)|. Let S0 be the space of functions z, D2 D1 z ∈ C(E, Rn ) which fulfil the condition |z(x, y)|0 = O(exp(λ (x + y))),

(1.5.29)

for (x, y) ∈ E, where λ > 0 is a constant. As in Section 1.3, in the space S0 we define the norm |z|S0 = sup [|z(x, y)|0 exp(−λ (x + y))] .

(1.5.30)

(x,y)∈E

It is easy to see that S0 is a Banach space and |z|S0 N,

(1.5.31)

where N 0 is a constant. Now we formulate the following theorem concerning the existence of a unique solution to IBVP (1.5.26)–(1.5.27). Theorem 1.5.4. Assume that (i) the functions f , g in equation (1.5.26) satisfy the conditions | f (x, y, u, v, w) − f (x, y, u, v, w)| k(x, y) [|u − u| + |v − v|] + |w − w|,

(1.5.32)

|g(x, y, m, n, u, v) − g(x, y, m, n, u, v)| h(x, y, m, n) [|u − u| + |v − v|] ,

(1.5.33)

where k ∈ C(E, R+ ), h ∈ C(E2 , R+ ), (ii) for λ as in (1.5.29) (c1 ) there exists a nonnegative constant α such that α < 1 and x y

L(x, y) + 0

0

for (x, y) ∈ E, where L(x, y) = k(x, y) exp(λ (x + y)) +

L(s,t) dt ds α exp(λ (x + y)), x y 0

0

(1.5.34)

h(x, y, m, n) exp(λ (m + n)) dn dm, (1.5.35)

(c2 ) there exists a nonnegative constant β such that |σ (x)| + |τ (y)| + | f (x, y, 0, 0, M0)| +

x y 0

0

| f (s,t, 0, 0, M0)|dt ds β exp(λ (x + y)).

Then the IBVP (1.5.26)–(1.5.27) has a unique solution u(x, y) on E in S0 .

(1.5.36)


Proof.

33

Let u(x, y) ∈ S0 and define the operator T by

(Tu)(x, y) = σ (x) + τ (y) +

x y 0

f (s,t, u(s,t), D2 D1 u(s,t), Mu(s,t)) dt ds.

0

(1.5.37)

From (1.5.37), we observe that D2 D1 (Tu)(x, y) = f (x, y, u(x, y), D2 D1 u(x, y), Mu(x, y)).

(1.5.38)

Evidently Tu is continuous on E and Tu ∈ Rn . From (1.5.37), (1.5.38), using the hypotheses and (1.5.31), we have |(Tu)(x, y)|0 |σ (x)| + |τ (y)| x y

+ 0

0

| f (s,t, u(s,t), D2 D1 u(s,t), Mu(s,t)) − f (s,t, 0, 0, M0)|dt ds x y

+ 0

0

| f (s,t, 0, 0, M0)|dt ds

+| f (x, y, u(x, y), D2 D1 u(x, y), Mu(x, y)) − f (x, y, 0, 0, M0)| + | f (x, y, 0, 0, M0)|

β exp(λ (x + y)) +

x

0

y

0

s

k(s,t)|u(s,t)|0 +

t

0

0

h(s,t, m, n)|u(m, n)|0 dn dm dt ds

x y

+k(x, y)|u(x, y)|0 +

0

0

h(x, y, m, n)|u(m, n)|0 dn dm

β exp(λ (x + y)) + |u|S0 k(x, y) exp(λ (x + y)) + x y

+ 0

0

k(s,t) exp(λ (s + t)) +

s t 0

0

x y 0

0

h(x, y, m, n) exp(λ (m + n))dn dm

h(s,t, m, n) exp(λ (m + n))dn dm dt ds

x y L(s,t) dt ds β exp(λ (x + y)) + N L(x, y) + 0

0

[β + N α ] exp(λ (x + y)).

(1.5.39)

From (1.5.39), it follows that Tu ∈ S0 . This proves that the operator T maps S0 into itself. Let u(x, y), v(x, y) ∈ S0 . From (1.5.37), (1.5.38) and using the hypotheses, we have |(Tu)(x, y) − (T v)(x, y)|0

x y 0

0

| f (s,t, u(s,t), D2 D1 u(s,t), Mu(s,t)) − f (s,t, v(s,t), D2 D1 v(s,t), Mv(s,t))|dt ds

+| f (x, y, u(x, y), D2 D1 u(x, y), Mu(x, y)) − f (x, y, v(x, y), D2 D1 v(x, y), Mv(x, y))|

34


x y 0

k(s,t)|u(s,t) − v(s,t)|0 +

0

+k(x, y)|u(x, y) − v(x, y)|0 +

s t 0

0

x y 0

0

|u − v|S0 k(x, y) exp(λ (x + y)) +

x y

+ 0

0

k(s,t) exp(λ (s + t)) +

h(s,t, m, n)|u(m, n) − v(m, n)|0 dn dm dt ds

h(x, y, m, n)|u(m, n) − v(m, n)|0 dn dm

x y

s t 0

0

0

0

h(x, y, m, n) exp(λ (m + n)) dn dm

h(s,t, m, n) exp(λ (m + n))dn dm dt ds

x y L(s,t) dt ds α |u − v|S0 exp(λ (x + y)). = |u − v|S0 L(x, y) + 0

(1.5.40)

0

From (1.5.40), we have |Tu − T v|S0 α |u − v|S0 . Since α < 1, it follows from Banach fixed point theorem (see [9, p. 37], [16, p. 372]) that T has a fixed point in S0 . The fixed point of T is however a solution of IBVP (1.5.26)– (1.5.27). The proof is complete. The following theorem contains the estimate on the solution of IBVP (1.5.26)–(1.5.27). Theorem 1.5.5. Assume that | f (x, y, u, v, w)| γ [|u| + |v|] + |w|, |g(x, y, m, n, u, v)| q(x, y, m, n) [|u| + |v|] , |σ (x)| + |τ (y)| δ ,

(1.5.41) (1.5.42) (1.5.43)

where γ , δ are nonnegative constants such that γ < 1 and q, D1 q, D2 q, D2 D1 q ∈ C(E2 , R+ ). If u(x, y) is any solution of IBVP (1.5.26)–(1.5.27) on E, then x y δ γ |u(x, y)|0 exp + K(s,t) dt ds , 1−γ 1−γ 0 0

(1.5.44)

for (x, y) ∈ E, where K(x, y) is defined by the right hand side of (1.2.26), replacing b and e by

1 1−γ q.


35

Using the fact that u(x, y) is a solution of IBVP (1.5.26)–(1.5.27) and the hypothe-

Proof.

ses, we have x y

+ 0

0

|u(x, y)|0 |σ (x)| + |τ (y)| | f (s,t, u(s,t), D2 D1 u(s,t), Mu(s,t))|dt ds +| f (x, y, u(x, y), D2 D1 u(x, y), Mu(x, y))| x y

s t δ+ γ |u(s,t)|0 + q(s,t, m, n)|u(m, n)|0 dn dm dt ds 0

0

0

+γ |u(x, y)|0 +

x y 0

From (1.5.45), we observe that 1 δ + |u(x, y)|0 1−γ 1−γ

0

0

q(x, y, m, n)|u(m, n)|0 dn dm.

x y 0

0

γ u(s,t)|0 + q(x, y, s,t)|u(s,t)|0

s t

+ 0

0

(1.5.45)

q(s,t, m, n)|u(m, n)|0 dn dm ds dt.

(1.5.46)

Now, a suitable application of Theorem 1.2.5 to (1.5.46) yields (1.5.44). Remark 1.5.4. We note that, if the estimate obtained in (1.5.44) is bounded, then the solution u(x, y) of IBVP (1.5.26)–(1.5.27) is bounded on E. The next result deals with the continuous dependence of solutions of equation (1.5.26) on given initial boundary values. Theorem 1.5.6. Assume that the functions f , g in equation (1.5.26) satisfy the conditions | f (x, y, u, v, w) − f (x, y, u, v, w)| d [|u − u| + |v − v|] + |w − w|,

(1.5.47)

|g(x, y, m, n, u, v) − g(x, y, m, n, u, v)| p(x, y, m, n) [|u − u| + |v − v|] ,

(1.5.48)

where d is a nonnegative constant such that d < 1 and p, D1 p, D2 p, D2 D1 p ∈ C(E2 , R+ ). Let u1 (x, y) and u2 (x, y) be the solutions of equation (1.5.26) with the given initial boundary conditions u1 (x, 0) = σ1 (x),

u1 (0, y) = τ1 (y),

u1 (0, 0) = 0,

(1.5.49)

u2 (x, 0) = σ2 (x),

u2 (0, y) = τ2 (y),

u2 (0, 0) = 0,

(1.5.50)

respectively, where σ1 , σ2, τ1 , τ2 ∈ C(R+

, Rn )

and

|σ1 (x) + τ1 (y) − σ2 (x) − τ2 (y)| μ , where μ 0 is a constant. Then

(1.5.51)

x y μ d exp + K0 (s,t) dt ds , (1.5.52) 1−d 1−d 0 0 for (x, y) ∈ E, where K0 (x, y) is defined by the right hand side of (1.2.26), replacing b and |u1 (x, y) − u2 (x, y)|0

e by

1 1−d

p.

36


Proof.

From the hypotheses, we have |u1 (x, y) − u2 (x, y)|0 |σ1 (x) + τ1 (y) − σ2 (x) − τ2 (y)| x y

+ 0

0

| f (s,t, u1 (s,t), D2 D1 u1 (s,t), Mu1 (s,t))

− f (s,t, u2 (s,t), D2 D1 u2 (s,t), Mu2 (s,t))|dt ds +| f (x, y, u1 (x, y), D2 D1 u1 (x, y), Mu1 (x, y)) − f (x, y, u2 (x, y), D2 D1 u2 (x, y), Mu2 (x, y))| μ+ s t

+ 0

0

x y 0

0

d|u1 (s,t) − u2 (s,t)|0

p(s,t, m, n)|u1 (m, n) − u2 (m, n)|0 dn dm ds dt +d|u1 (x, y) − u2 (x, y)|0

x y

+ 0

0

p(x, y, m, n)|u1 (m, n) − u2 (m, n)|0 dn dm.

(1.5.53)

From (1.5.53), we observe that |u1 (x, y) − u2 (x, y)|0

μ 1 + 1−d 1−d

x y 0

s t

+ 0

0

0

d|u1 (s,t) − u2 (s,t)|0 + p(x, y, s,t)|u1 (s,t) − u2 (s,t)|0

p(x, y, m, n)|u1 (m, n) − u2 (m, n)|0 dn dm dt ds.

(1.5.54)

Now, a suitable application of Theorem 1.2.5 to (1.5.54) yields the bound (1.5.52), which shows the dependency of solutions of equation (1.5.26) on given initial boundary conditions. Remark 1.5.5.

We note that the inequalities in Theorems 1.2.4 and 1.2.5 can be used

to study some other properties related to the solutions of equations (1.5.1)–(1.5.2) and (1.5.26)–(1.5.27), similar to those obtained in earlier sections. Here, we do not discuss the details.


1.6

37

Fredholm-type integrodifferential equation

The main objective of the present section is to give some fundamental qualitative properties of solutions of the Fredholm-type integral equation a b

u(x, y) = f (x, y) + 0

0

g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t))dt ds,

(1.6.1)

where f , g are given functions and u is the unknown function to be fond (see [96]). Throughout, we assume that f ∈ C(E0 , R), g ∈ C(E02 × R3 , R), and Di f ∈ C(E0 , R), Di g ∈ C(E02 × R3 , R) for i = 1, 2. By a solution of equation (1.6.1) we mean a function u ∈ C(E0 , R) which is continuously differentiable with respect to x and y for (x, y) ∈ E0 and satisfies the equation (1.6.1). For z, D1 z, D2 z ∈ C(E0 , R) we denote by |z(x, y)|1 = |z(x, y)| + |D1 z(x, y)| + |D2 z(x, y)|. Let S1 be the space of functions z, D1 z, D2 z ∈ C(E0 , R) which fulfil the condition |z(x, y)|1 = O(exp(λ (x + y))),

(1.6.2)

for (x, y) ∈ E0 , where λ is a positive constant. In the space S1 we define the norm |z|S1 = max [|z(x, y)|1 exp(−λ (x + y))]. (x,y)∈E0

(1.6.3)

It is easy to see that S1 is a Banach space and |z|S1 N,

(1.6.4)

where N 0 is a constant. In the following theorem we give conditions under which a solution of equation (1.6.1) exists on E0 in S1 . Theorem 1.6.1. Suppose that (i) the function g in equation (1.6.1) and its derivatives Di g for i = 1, 2 satisfy the conditions |g(x, y, s,t, u, v, w) − g(x, y, s,t, u, v, w)| r(x, y, s,t)[|u − u| + |v − v| + |w − w|],

(1.6.5)

and |Di g(x, y, s,t, u, v, w) − Di g(x, y, s,t, u, v, w)| ri (x, y, s,t)[|u − u| + |v − v| + |w − w|], (ii) for λ as in (1.6.2),

(1.6.6)

38


(e1 ) there exists a nonnegative constant α such that α < 1 and a b 0

0

[r(x, y, s,t) + r1 (x, y, s,t) + r2 (x, y, s,t)] exp(λ (s + t)) dt ds α exp(λ (s + t)),

(1.6.7)

for (x, y) ∈ E0 , (e2 ) there exists a nonnegative constant β such that | f (x, y)|1 +

a b 0

0

|g(x, y, s,t, 0, 0, 0)|1 dt ds β exp(λ (x + y)),

(1.6.8)

for (x, y) ∈ E0 . Then the equation (1.6.1) has a unique solution u(x, y) on E0 in S1 . Let u(x, y) ∈ S1 and define the operator T by

Proof.

a b

(Tu)(x, y) = f (x, y) + 0

0

g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t)) dt ds.

(1.6.9)

Differentiating both sides of (1.6.9) partially with respect to x and y, we have a b

Di (Tu)(x, y) = Di f (x, y) +

0

Di g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t)) dt ds, (1.6.10)

0

for i = 1, 2. Evidently Tu, Di (Tu) for i = 1, 2 are continuous on E0 and Tu, Di (Tu) ∈ R. From (1.6.9), (1.6.10), (1.6.4) and using the hypotheses, we have |(Tu)(x, y)|1 | f (x, y)|1 a b

+ 0

0

|g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t)) − g(x, y, s,t, 0, 0, 0)|dt ds a b

+ 0

a b

+ 0

0

0

|D1 g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t)) − D1 g(x, y, s,t, 0, 0, 0)|dt ds a b

+ 0

a b

+ 0

0

|g(x, y, s,t, 0, 0, 0)|dt ds

0

|D1 g(x, y, s,t, 0, 0, 0)|dt ds

|D2 g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t)) − D2 g(x, y, s,t, 0, 0, 0)|dt ds a b

+ 0

| f (x, y)|1 +

a b 0

0

0

|D2 g(x, y, s,t, 0, 0, 0)|dt ds

|g(x, y, s,t, 0, 0, 0)|1 dt ds +

a b 0

0

r(x, y, s,t)|u(s,t)|1 dt ds


39

a b

+ 0

a b

r1 (x, y, s,t)|u(s,t)|1 dt ds +

0

0

0

r2 (x, y, s,t)|u(s,t)|1 dt ds

β exp(λ (x + y)) a b

+|u|S1

0

0

[r(x, y, s,t) + r1 (x, y, s,t) + r2 (x, y, s,t)] exp(λ (s + t)) dt ds [β + N α ] exp(λ (x + y)).

(1.6.11)

From (1.6.11), it follows that Tu ∈ S1 . This proves that the operator T maps S1 into itself. Let u(x, y), v(x, y) ∈ S1 . From (1.6.9), (1.6.10) and using the hypotheses, we have |(Tu)(x, y) − (T v)(x, y)|1

a b 0

0

|g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t))

−g(x, y, s,t, v(s,t), D1 v(s,t), D2 v(s,t))|ds dt a b

+ 0

0

|D1 g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t))

−D1 g(x, y, s,t, v(s,t), D1 v(s,t), D2 v(s,t))|ds dt a b

+ 0

0


−D2 g(x, y, s,t, v(s,t), D1 v(s,t), D2 v(s,t))|ds dt a b

+ 0

0

a b 0

0

r(x, y, s,t)|u(s,t) − v(s,t)|1 dt ds

r1 (x, y, s,t)|u(s,t) − v(s,t)|1 dt ds +

|u − v|S1

a b 0

0

a b 0

0

r2 (x, y, s,t)|u(s,t) − v(s,t)|1 dt ds

[r(x, y, s,t) + r1 (x, y, s,t) + r2 (x, y, s,t)] exp(λ (s + t)) dt ds |u − v|S1 α exp(λ (x + y)).

(1.6.12)

From (1.6.12), we obtain |Tu − T v|S1 α |u − v|S1 . Since α < 1, it follows from Banach fixed point theorem (see [51, p. 372]) that T has a unique fixed point in S1 . The fixed point of T is however a solution of equation (1.6.1). Using the inequality in Theorem 1.2.3, we present the following result which deals with the uniqueness of solutions of equation (1.6.1) on E0 in R.

40


Theorem 1.6.2. Suppose that the function g in equation (1.6.1) and its derivatives Di g for i = 1, 2 satisfy the conditions (1.6.5) and (1.6.6) with r(x, y, s,t) = c(x, y)h(s,t), ri (x, y, s,t) = c(x, y)hi (s,t) for i = 1, 2, where c, h, hi ∈ C(E0 , R+ ) and a b

d1 =

0

[h(s,t) + h1 (s,t) + h2 (s,t)]c(s,t) dt ds < 1.

0

(1.6.13)

Then the equation (1.6.1) has at most one solution on E0 in R. Let u(x, y) and v(x, y) be two solutions of equation (1.6.1). Then from the hy-

Proof.

potheses, we have |u(x, y) − v(x, y)|1

a b 0

0


−g(x, y, s,t, v(s,t), D1 v(s,t), D2 v(s,t))|dt ds

a b

+ 0

0


−D1 g(x, y, s,t, v(s,t), D1 v(s,t), D2 v(s,t))|dt ds a b

+ 0

0


−D2 g(x, y, s,t, v(s,t), D1 v(s,t), D2 v(s,t))|dt ds a b

+ 0

0

a b 0

0

c(x, y)h(s,t)|u(s,t) − v(s,t)|1 dt ds

c(x, y)h1 (s,t)|u(s,t) − v(s,t)|1 dt ds + a b

= c(x, y) 0

0

a b 0

0

c(x, y)h2 (s,t)|u(s,t) − v(s,t)|1 dt ds

[h(s,t) + h1 (s,t) + h2 (s,t)]|u(s,t) − v(s,t)|1 dt ds.

(1.6.14)

Now, an application of Theorem 1.2.3 (when p(x, y) = 0) to (1.6.14) yields |u(x, y) − v(x, y)|1 0, and hence u(x, y) = v(x, y), which proves the uniqueness of solutions of equation (1.6.1) on E0 in R. The following theorem concerning the estimate on the solution of equation (1.6.1) holds. Theorem 1.6.3. Suppose that the function g in equation (1.6.1) and its derivatives Di g for i = 1, 2 satisfy the conditions |g(x, y, s,t, u, v, w)| k(x, y)e(s,t)[|u| + |v| + |w|],

(1.6.15)

|Di g(x, y, s,t, u, v, w)| k(x, y)ei (s,t)[|u| + |v| + |w|],

(1.6.16)

where k, e, ei ∈ C(E0 , R) and

a b

d2 =

0

0

[e(s,t) + e1 (s,t) + e2 (s,t)]k(s,t) dt ds < 1.

(1.6.17)

Then for every solution u ∈ C(E0 , R) of equation (1.6.1), the estimate

1 × 1 − d2 holds for (x, y) ∈ E0 .

a b 0

0

|u(x, y)|1 | f (x, y)|1 + k(x, y)

[e(s,t) + e1 (s,t) + e2 (s,t)]| f (s,t)|1 dt ds ,

(1.6.18)


Proof.

41

Let u ∈ C(E0 , R) be a solution of equation (1.6.1). Then from the hypotheses, we

have |u(x, y)|1 | f (x, y)| +

a b 0

a b

+|D1 f (x, y)| +

0

0

a b

+|D2 f (x, y)| +

0

0

0

|g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t))|dt ds

|D1 g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t))|dt ds

|D2 g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t))|dt ds

| f (x, y)|1 +

a b 0

0

k(x, y)e(s,t)|u(s,t)|1 dt ds

a b

+ 0

0

a b

k(x, y)e1 (s,t)|u(s,t)|1 dt ds +

= | f (x, y)|1 + k(x, y)

0

0

k(x, y)e2 (s,t)|u(s,t)|1 dt ds

a b 0

0

[e(s,t) + e1 (s,t) + e2 (s,t)]|u(s,t)|1 dt ds.

(1.6.19)

Now, an application of Theorem 1.2.3 to (1.6.19) gives the desired estimate in (1.6.18). We note that the estimate obtained in (1.6.18) yields not only the bound

Remark 1.6.1.

on the solution u(x, y) of equation (1.6.1), but also the bound on their derivatives Di u(x, y) for i = 1, 2. Next, we shall obtain the estimate on the solution of equation (1.6.1) assuming that the function g and its derivatives Di g for i = 1, 2 satisfy Lipschitz type conditions. Theorem 1.6.4.

Suppose that the function g in equation (1.6.1) and its derivatives Di g

for i = 1, 2 satisfy the conditions in Theorem 1.6.2 and the condition (1.6.13) holds. If u ∈ C(E0 , R) is any solution of equation (1.6.1) on E0 , then |u(x, y) − f (x, y)|1 Q(x, y) + c(x, y)

×

1 1 − d1

[h(s,t) + h1 (s,t) + h2 (s,t)]Q(s,t) dt ds ,

(1.6.20)

|g(x, y, σ , τ , f (σ , τ ), D1 f (σ , τ ), D2 f (σ , τ ))|1 d τ d σ ,

(1.6.21)

a b 0

0

for (x, y) ∈ E0 , where a b

Q(x, y) = 0

for (x, y) ∈ E0 .

0

42


Proof.

Since u(x, y) is a solution of equation (1.6.1), by using the hypotheses, we have |u(x, y) − f (x, y)|1

a b 0

0


−g(x, y, s,t, f (s,t), D1 f (s,t), D2 f (s,t))|dt ds a b

+ 0

0

|g(x, y, s,t, f (s,t), D1 f (s,t), D2 f (s,t))|dt ds

a b

+ 0

0


−D1 g(x, y, s,t, f (s,t), D1 f (s,t), D2 f (s,t))|dt ds a b

+ 0

0

|D1 g(x, y, s,t, f (s,t), D1 f (s,t), D2 f (s,t))|dt ds

a b

+ 0

0


−D2 g(x, y, s,t, f (s,t), D1 f (s,t), D2 f (s,t))|dt ds a b

+ 0

0

|D2 g(x, y, s,t, f (s,t), D1 f (s,t), D2 f (s,t))|dt ds

Q(x, y) +

a b 0

a b

+ 0

0

a b

+ 0

0

a b

= Q(x, y) + c(x, y) 0

0

0

c(x, y)h(s,t)|u(s,t) − f (s,t)|1 dt ds

c(x, y)h1 (s,t)|u(s,t) − f (s,t)|1 dt ds c(x, y)h2 (s,t)|u(s,t) − f (s,t)|1 dt ds

[h(s,t) + h1 (s,t) + h2 (s,t)]|u(s,t) − f (s,t)|1 dt ds.

(1.6.22)

Now an application of Theorem 1.2.3 to (1.6.22) yields (1.6.20). We next consider the equation (1.6.1) and the following Fredholm-type integral equation a b

z(x, y) = F(x, y) + 0

0

G(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t)) dt ds,

(1.6.23)

for (x, y) ∈ E0 , where F ∈ C(E0 , R), G ∈ C(E02 ×R3 , R) and Di F ∈ C(E0 , R), Di G ∈ C(E02 × R3 , R) for i = 1, 2. The following theorem holds.


43

Theorem 1.6.5. Suppose that the function g in equation (1.6.1) and its derivatives Di g for i = 1, 2 satisfy the conditions as in Theorem 1.6.2 and the condition (1.6.13) holds. Then for every given solution z ∈ C(E0 , R) of equation (1.6.23) and any solution u ∈ C(E0 , R) of equation (1.6.1), the estimate |u(x, y) − z(x, y)|1 [| f (x, y) − F(x, y)|1 + M(x, y)] + c(x, y)

1 × 1 − d1

a b 0

0

[h(s,t) + h1 (s,t) + h2 (s,t)][| f (s,t) − F(s,t)|1 + M(s,t)]dt ds , (1.6.24)

holds for (x, y) ∈ E0 , where

a b

M(x, y) = 0

0

|g(x, y, σ , τ , z(σ , τ ), D1 z(σ , τ ), D2 z(σ , τ )).

−G(x, y, σ , τ , z(σ , τ ), D1 z(σ , τ ), D2 z(σ , τ ))|1 d τ d σ ,

(1.6.25)

for (x, y) ∈ E0 . From the hypotheses, we have

Proof.

|u(x, y) − z(x, y)|1 | f (x, y) − F(x, y)|1 a b

+ 0

0

a b

+ 0

0

|g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t))−g(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t))|dt ds, |g(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t)) − G(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t))|dt ds a b

+ 0

0

a b

+ 0

0

a b

+ 0

0

a b

+ 0

0

|D1 g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t)) − D1 g(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t))|dt ds |D1 g(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t)) − D1 G(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t))|dt ds |D2 g(x, y, s,t, u(s,t), D1 u(s,t), D2 u(s,t)) − D2 g(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t))|dt ds |D2 g(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t)) − D2 G(x, y, s,t, z(s,t), D1 z(s,t), D2 z(s,t))|dt ds

[| f (x, y) − F(x, y)|1 + M(x, y)] + c(x, y) ×

a b 0

0

[h(s,t) + h1 (s,t) + h2 (s,t)]|u(x, y) − z(x, y)|1 dt ds.

Now, an application of Theorem 1.2.3 to (1.6.26) yields (1.6.24).

(1.6.26)

44


Remark 1.6.2. We note that, one can formulate results on the continuous dependence of solutions of equation (1.6.1) and its variants by closely looking at the corresponding results given in Theorems 1.4.6 and 1.4.7. Furthermore, the idea used in this section can be very easily extended to study the version of equation (1.6.1) involving functions of more than two variables. Moreover, the results established in Theorems 1.6.1-1.6.5 can be extended for equations of the form (1.6.1) when the function g is of the form ∂ n u(s,t) ∂ m u(s,t) , , g x, y, s,t, u(s,t), ∂ sn ∂ sm or ∂ n u(s,t) ∂ m u(s,t) ∂ n+m u(s,t) g x, y, s,t, u(s,t), , , , ∂ sn ∂ sm ∂ sn ∂ t m under some suitable conditions. Naturally, these considerations will make the analysis more complicated and we leave it to the reader to fill in where needed. The generality of the equation (1.6.1) allow us to obtain results similar to the ones given above, concerning the following equation a b

u(x, y) = f (x, y) + 0

0

K(x, y, s,t)h(s,t, u(s,t), D1 u(s,t), D2 u(s,t)) dt ds,

(1.6.27)

where f ∈ C(E0 , R), K ∈ C(E02 , R), h ∈ C(E0 × R3 , R) and assume that Di f ∈ C(E0 , R), Di K ∈ C(E02 , R), for i = 1, 2. Below we present a result on the existence of a unique solution of equation (1.6.27) by using the Banach fixed point theorem coupled with maximum norm. One can formulate other results given above in Theorems 1.6.2–1.6.5 for the equation (1.6.27). Theorem 1.6.6. Suppose that (i) f , Di f ∈ C(E0 , R) for i = 1, 2, (ii) h ∈ C(E0 × R3 , R), h(s,t, 0, 0, 0) = 0 and there is a constant L > 0 such that |h(s,t, u, v, w) − h(s,t, u, v, w)| L [|u − u| + |v − v| + |w − w|] ,

(1.6.28)

(iii) K, Di K ∈ C(E02 , R) for i = 1, 2 and a b

L 0

0

|K(x, y, s,t)|1 dt ds α < 1.

Then the equation (1.6.27) has a unique solution u ∈ C(E0 , R).

(1.6.29)


Proof.

45

Let B be a Banach space of bounded functions u ∈ C(E0 , R), which are con-

tinuously differentiable with respect to x and y on E0 with maximum norm · , where u = max(x,y)∈E0 |u(x, y)|1 . Let u ∈ B and define the operator F by a b

(Fu)(x, y) = f (x, y) + 0

0

K(x, y, s,t)h(s,t, u(s,t), D1 u(s,t), D2 u(s,t)) dt ds. (1.6.30)

Differentiating both sides of (1.6.30) partially with respect to x and y, we have Di (Fu))(x, y) = Di f (x, y) a b

+ 0

0

Di K(x, y, s,t)h(s,t, u(s,t), D1 u(s,t), D2 u(s,t)) dt ds,

(1.6.31)

for i = 1, 2. From the hypotheses, it follows that Fu, Di (Fu) (i = 1, 2) are continuous on E0 and |(Fu))(x, y)|1 | f (x, y)|1 +

a b 0

0

|K(x, y, s,t)|1

×|h(s,t, u(s,t), D1 u(s,t), D2 u(s,t)) − h(s,t, 0, 0, 0)|dt ds | f (x, y)|1 + L

a b 0

0

| f (x, y)|1 + Lu

|K(x, y, s,t)|1 |u(s,t)|1 dt ds

a b 0

0

|K(x, y, s,t)|1 dt ds < ∞.

(1.6.32)

Here, we have used the fact that | f (x, y)|1 is bounded, since f , Di f ∈ C(E0 , R) and the condition (1.6.29). This proves that the operator F maps B into itself. Let u, v ∈ B. From (1.6.30), (1.6.31) and the hypotheses, we have |(Fu)(x, y) − (Fv)(x, y)|1

a b 0

0

|K(x, y, s,t)|1 |h(s,t, u(s,t), D1 u(s,t), D2 u(s,t))

−h(s,t, v(s,t), D1 v(s,t), D2 v(s,t))|dt ds L

a b 0

0

|K(x, y, s,t)|1 |u(s,t) − v(s,t)|1 dt ds

Lu − v

a b 0

0

|K(x, y, s,t)|1 dt ds

α u − v.

(1.6.33)

From (1.6.33), we have Fu − Fv α u − v. Since α < 1, it follows that F is a contraction mapping, which proves that the equation (1.6.27) has a unique solution u in B on E0 .

46


Remark 1.6.3.

We note that the equation (1.6.27) includes as a special case, the study of

the following important integral equation

a b

u(x, y) = f (x, y) +

K(x, y, s,t)h(s,t, u(s,t)) dt ds, 0

(1.6.34)

0

which may be considered as a two independent variable generalization of the well-known Hammerstein type integral equation studied by many authors in the literature.

1.7

Miscellanea

1.7.1

Haçia [42]

(e1 ) Let f and k be continuous functions in E and E2 respectively. If k is nonnegative, and the continuous function u defined on E satisfies the inequality u(x, y) f (x, y) +

for (x, y) ∈ E, then u(x, y) f (x, y) + for (x, y) ∈ E, where

x

y

k(x, y, s,t)u(s,t) ds dt, 0

0

x y

r(x, y, s,t) f (s,t) ds dt, 0

0 ∞

r(x, y, s,t) =

∑ kn (x, y, s,t),

n=1

is the resolvent kernel defined by formulas k (x, y, s,t) = k(x, y, s,t),

1x y

kn (x, y, s,t) =

0

0

k(x, y, ξ , η )kn−1 (ξ , η , s,t) d ξ d η ,

for n = 2, 3, . . .. Moreover, if f is nondecreasing with respect to every variable, then u(x, y) f (x, y) 1 +

x y

for (x, y) ∈ E.

r(x, y, s,t) ds dt , 0

0

(e2 ) Let a, b and f be nonnegative continuous functions in E. If the continuous function u defined on E satisfies the inequality u(x, y) f (x, y) + a(x, y) for (x, y) ∈ E, then u(x, y) f (x, y) + a(x, y) for (x, y) ∈ E.

x y

b(s,t)u(s,t) ds dt, 0

0

x

x y

y

b(s,t) exp 0

0

s

t

a(ξ , η )b(ξ , η ) d ξ d η f (s,t)ds dt,

Moreover, if f is nondecreasing with respect to every variable, then u(x, y) f (x, y) 1 + a(x, y) for (x, y) ∈ E.

x y

x y

b(s,t) exp 0

0

s

t

a(ξ , η )b(ξ , η ) d ξ d η ds dt ,


1.7.2

47

Haçia [42]

Consider the system of integral equations n

ui (x, y) = wi (x, y) + ∑

x y

j=1 0

0

ki j (x, y, s,t)ui (s,t) ds dt,

(1.7.1)

for i = 1, 2, . . . , n, where functions wi and ki j are continuous in E and E2 respectively. (e3 ) If n

∑ max

|ki j (x, y, s,t)| : 1 j n a(s,t),

i=1

where a ∈ C(E, R+ ), then the estimate n

n

i=1

i=1

∑ |ui (x, y)| sup ∑ |wi (s,t)| :

0 s x, 0 t y exp

x

holds for (x, y) ∈ E. Moreover, if a ∈ L(R2+ ) and

0

n

∑ |wi (s,t)| :

sup

y

a(s,t) ds dt ,

0

0 s x < ∞, 0 t y < ∞

< ∞,

i=1

then the solution {ui (x, y)}, i = 1, 2, . . . , n, of (1.7.1) is bounded in E. (e4 ) If n

∑ max

|ki j (x, y, s,t)| : 1 j n = b(x, y)a(s,t),

i=1

where a, b ∈ C(E, R+ ) (b = 0), then the estimate

n n |wi (s,t)| ∑ |ui (x, y)| b(x, y) sup ∑ b(s,t) : 0 s x < ∞, 0 t y < ∞ i=1 i=1

× exp

x 0

y

a(s,t)b(s,t) ds dt ,

0

holds for (x, y) ∈ E. Moreover, if ab ∈ L(R2+ ) and

n |wi (s,t)| : 0 s x < ∞, 0 t y < ∞ < ∞, sup ∑ i=1 b(s,t) then the solution {ui (x, y)} (i = 1, 2, . . . , n), of (1.7.1) is bounded in E.

48

1.7.3


Pachpatte [76]

Consider the integral equation z(x, y) = F(x, y, (T1 z)(x, y), (T2 z)(x, y), (T3 z)(x, y), z(x, y)), for x ∈ Iα , y ∈ Iβ , where (T1 z)(x, y) =

x y 0

x

(T2 z)(x, y) =

(1.7.2)

0

0

f1 (x, y, s,t, z(s,t)) ds dt,

(1.7.3)

f2 (x, y, s, z(s, y)) ds,

(1.7.4)

f3 (x, y,t, z(x,t)) dt,

(1.7.5)

y

(T3 z)(x, y) =

0

f1 , f2 , f3 , F are given functions and z is unknown function. Let (B, · ) be a Banach space, Δ = Iα × Iβ , Δ1 = {(x, y, s,t) : 0 s x α , 0 t y β }, Δ2 = {(x, y, s) : 0 s x α , 0 y β }, Δ3 = {(x, y,t) : 0 x α , 0 t y β }, fi ∈ C(Δi × B, B) (i = 1, 2, 3), F ∈ C(Δ × B4 , B). (H1 ) Suppose that (i) there exist functions wi ∈ C(Δi × R+ , R+ ) (i = 1, 2, 3), W ∈ C(Δ × R4+ , R+ ) such that f1 (x, y, s,t, z) − f1 (x, y, s,t, z) w1 (x, y, s,t, z − z), f2 (x, y, s, z) − f2 (x, y, s, z) w2 (x, y, s, z − z), f3 (x, y,t, z) − f 3 (x, y,t, z) w3 (x, y,t, z − z), and F(x, y, u1 , u2 , u3 , u4 ) − F(x, y, u1 , u2 , u3 , u4 ) W (x, y, u1 − u1 , u2 − u2 , u3 − u3 , u4 − u4 ), (ii) if u ∈ C(Δ, R+ ) and v(x, y) = W (x, y, (M1 u)(x, y), (M2 u)(x, y), (M3 u)(x, y), u(x, y)), for (x, y) ∈ Δ, where for i = 1, 2, 3, Mi u are defined by the right hand sides in (1.7.3)–(1.7.5) by replacing fi by wi and z by u, then v ∈ C(Δ, R+ ), (iii) If u, v ∈ C(Δ, R+ ) and u v, then W (x, y, M1 u, M2 u, M3 u, u) W (x, y, M1 v, M2 v, M3 v, v), (iv) if un ∈ C(Δ, R+ ), un+1 un , n = 0, 1, 2, . . ., un → u, then W (x, y, M1 un , M2 un , M3 un , un ) → W (x, y, M1 u, M2 u, M3 u, u).


49

(H2 ) Suppose that (i) there exists a solution u ∈ C(Δ, R+ ) of the inequality W (x, y, M1 u, M2 u, M3 u, u) + h(x, y) u, where h(x, y) = sup F(ξ , η , (T1 0)(ξ , η ), (T2 0)(ξ , η ), (T3 0)(ξ , η ), 0), 0ξ x 0η y

(ii) in the class of functions satisfying the condition 0 u(x, y) u(x, y), the function u(x, y) ≡ 0 is the only solution of the inequality u(x, y) W (x, y, (M1 u)(x, y), (M2 u)(x, y), (M3 u)(x, y), u(x, y)). Define the sequence {zn } by z0 (x, y) ≡ 0, zn+1 (x, y) = F(x, y, (T1 zn )(x, y), (T2 zn )(x, y), (T3 zn )(x, y), zn (x, y)),

(1.7.6)

for n = 0, 1, 2, . . .. Also define the sequence {un } by u0 (x, y) = u(x, y), un+1 (x, y) = W (x, y, (M1 un )(x, y), (M2 un )(x, y), (M3 un )(x, y), un (x, y)),

(1.7.7)

for n = 0, 1, 2, . . ., where u(x, y) is as in hypotheses (H2 ). (e5 ) If hypothesis (H2 ) and the conditions (ii), (iii), (iv) of hypothesis (H1 ) are satisfied, then 0 un+1 (x, y) un (x, y) u(x, y), for n = 0, 1, 2, . . . and lim un (x, y) = 0,

n→∞

where the convergence is uniform in each bounded set. (e6 ) If hypotheses (H1 ) and (H2 ) are satisfied, then there exists a continuous solution z of equation (1.7.2) on Δ. The sequence {zn } defined by (1.7.6) converges uniformly on Δ to z, and the following estimates z(x, y) − zn (x, y) un (x, y),

(1.7.8)

for (x, y) ∈ Δ, n = 0, 1, 2, . . ., and z(x, y) u(x, y),

(1.7.9)

for (x, y) ∈ Δ hold true. The solution z of equation (1.7.2) is unique in the class of functions satisfying the condition (1.7.9). (e7 ) If hypothesis (H1 ) is satisfied and the function r(x, y) ≡ 0, (x, y) ∈ Δ is the only nonnegative continuous solution of the inequality r(x, y) W (x, y, (M1 r)(x, y), (M2 r)(x, y), (M3 r)(x, y), r(x, y)), for (x, y) ∈ Δ, then the equation (1.7.2) has at most one solution on Δ.

50

1.7.4


Pachpatte [78]

Consider the nonlinear functional integral equation z(x, y) = F(x, y, (L1 z)(x, y), (L2 z)(x, y), (L3 z)(x, y), z(g(x, y), h(x, y)), μ ),

(1.7.10)

for x ∈ Iα , y ∈ Iβ , where a(x,y) b(x,y)

(L1 z)(x, y) =

0

c(x,y)

(L2 z)(x, y) =

0

0

f1 (x, y, s,t, z(s,t)) dt ds,

f 2 (x, y, s, z(s, p(x, y))) ds,

d(x,y)

(L3 z)(x, y) =

0

f 3 (x, y,t, z(q(x, y),t)) dt,

f1 , f2 , f3 , F, a, b, c, d, p, q, g, h are given functions, μ is a real parameter and z is the unknown function. Let (B, · ) be a Banach space and Δ, Δi (i = 1, 2, 3) be as defined below the equation (1.7.2). The functions a, b, c, d, p, q, g, h are defined and continuous on Δ and a, c, q, g ∈ Iα ; b, d, p, h ∈ Iβ , fi ∈ C(Δi × B, B) (i = 1, 2, 3), F ∈ C(Δ × B4 × R, B). (H3 ) Suppose that (i) there exist nonnegative constants Ki , Mi (i = 1, 2, 3) and M such that f1 (x, y, s,t, z) − f1 (x, y, s,t, z) M1 z − z, f2 (x, y, s, z) − f2 (x, y, s, z) M2 z − z, f3 (x, y,t, z) − f 3 (x, y,t, z) M3 z − z, and F(x, y, z1 , z2 , z3 , z4 , μ ) − F(x, y, z1 , z2 , z3 , z4 , μ ) K1 z1 − z1 + K2 z2 − z2 + K3 z3 − z3 + Mz4 − z4 , (ii) for every fixed μ ∈ R, there exist constants λ > 0, Q 0 such that F(x, y, (L1 0)(x, y), (L2 0)(x, y), (L3 0)(x, y), 0, μ ) Q exp(λ (x + y)), (iii) there exist constants N1 , N2 such that 0 N1 + N2 < 1 and K1 M1 K2 M2 exp(λ [a(x, y) + b(x, y)]) + exp(λ [c(x, y) + p(x, y)]) λ2 λ +

K3 M3 exp(λ [q(x, y) + d(x, y)]) N1 exp(λ (x + y)), λ M exp (λ [g(x, y) + h(x, y)]) N2 exp(λ (x + y)).


51

(H4 ) suppose that there exists a constant K 0 and a function P ∈ C(Δ, R+ ) such that F(x, y, z1 , z2 , z3 , z4 , μ1 ) − F(x, y, z1 , z2 , z3 , z4 , μ2 ) P(x, y)| μ1 − μ2 |, and max [P(x, y) exp(−λ (x + y))] K,

(x,y)∈Δ

where λ > 0 is a constant. (e8 ) Let S be the space defined as in Section 1.3. If hypothesis (H3 ) is satisfied, then for every μ ∈ R, there exists exactly one solution z ∈ S of equation (1.7.10) on Δ. (e9 ) Let S be the space defined as in Section 1.3. If hypotheses (H3 ) and (H4 ) are satisfied, then the solution z(x, y, μ ) of equation (1.7.10) belonging to S is continuous with respect to the variables (x, y, μ ) on Δ × R. 1.7.5

Pachpatte [97]

Consider the integrodifferential equation of the form D2 D1 u(x, y) = f (x, y, u(x, y), (Ku)(x, y)),

(1.7.11)

with the given initial boundary conditions u(x, 0) = σ (x), for x, y ∈ R+ , where

u(0, y) = τ (y),

u(0, 0) = 0,

(1.7.12)

x y

(Ku)(x, y) =

k(x, y, m, n, u(m, n)) dn dm, 0

k ∈ C(E2 × Rn , Rn ), f

0 ∈ C(E × Rn × Rn , Rn ),

σ , τ ∈ C(R+ , Rn ).

(e10 ) Suppose that the functions f , k in (1.7.11) satisfy the conditions | f (x, y, u, v) − f (x, y, u, v)| p(x, y) [|u − u| + |v − v|] ,

(1.7.13)

|k(x, y, m, n, u) − k(x, y, m, n, u)| r(x, y, m, n)|u − u|,

(1.7.14)

where p ∈ C(E, R+ ) and r, D1 r, D2 r, D2 D1 r ∈ C(E2 , R+ ). Let x y f (s,t, 0, (K0)(s,t)) dt ds < ∞, c = sup σ (x) + τ (y) + 0 0 (x,y)∈E

where σ , τ are as in (1.7.12). If u(x, y) is any solution of problem (1.7.11)–(1.7.12) on E, then

s t x y p(s,t) exp [p(m, n) + A(m, n)]dn dm dt ds , |u(x, y)| c 1 + 0

0

0

0

52


for (x, y) ∈ E, where

x

A(x, y) = r(x, y, x, y) + 0

y

+ 0

D2 r(x, y, x, η ) d η +

x y 0

0

D1 r(x, y, ξ , y) d ξ

D2 D1 r(x, y, ξ , η ) d η d ξ .

(1.7.15)

(e11 ) Suppose that the functions f , k in (1.7.11) satisfy the conditions (1.7.13), (1.7.14) respectively. Let ui (x, y) (i = 1, 2) be respectively εi -approximate solutions of (1.7.11) on E, i.e., |D2 D1 ui (x, y) − f (x, y, ui (x, y), (Kui )(x, y))| εi , on E with ui (x, 0) = αi (x),

ui (0, y) = βi (y),

ui (0, 0) = 0,

where αi , βi ∈ C(R+ , Rn ) and assume that |α1 (x) − α2 (x) + β1 (y) − β1 (y)| δ , where δ 0 is a constant. Then

x y |u1 (x, y) − u2 (x, y)| e(x, y) 1 + p(s,t) 0

× exp

s 0

t

0

[p(m, n) + A(m, n)]dn dm dt ds ,

0

for (x, y) ∈ E, where e(x, y) = (ε1 + ε2 )xy + δ , and A(x, y) is given by (1.7.15). 1.7.6

Brzychczy and Janus [20]

Consider the nonlinear integrodifferential equation of the form zxy (x, y) = f (x, y, z(x, y), zx (x, y), zy (x, y), (Mz)(x, y)),

(1.7.16)

with the given initial boundary conditions z(x, 0) = σ (x),

z(0, y) = τ (y),

for x ∈ Ia , y ∈ Ib , where

y

(Mz)(x, y) = 0

σ (0) = τ (0) = c0 ,

m(y − s)g(z(x, s)) ds,

(1.7.17)

(1.7.18)


53

f ∈ C(E0 × R4 , R), g ∈ C(R, R), m ∈ L2 (Ib , R), σ ∈ C1 (Ia , R), τ ∈ C1 (Ib , R). Denote by C1,∗ (E0 , R) the space of functions u ∈ C(E0 , R) such that ux , uy , uxy exist and are continuous on E0 with the norm

u := max max |u(x, y)|, max |ux (x, y)|, max |uy (x, y)| . E0

E0

E0

The notation (u, ux , uy ) (v, vx , vy ) (respectively (u, ux , uy ) < (v, vx , vy )) means that u(x, y) v(x, y), ux (x, y) vx (x, y), uy (x, y) vy (x, y) (respectively u(x, y) < v(x, y), ux (x, y) < vx (x, y), uy (x, y) < vy (x, y) for each (x, y) ∈ E0 ) for each (x, y) ∈ E0 . A function u ∈ C1,∗ (E0 , R) is called a lower function of problem (1.7.16)–(1.7.17) on E0 if uxy (x, y) f (x, y, u(x, y), ux (x, y), uy (x, y), (Mu)(x, y)), ux (x, 0) σ (x), uy (0, y) τ (y), u(0, 0) c0 , and an upper function of problem (1.7.16)–(1.7.17) on E0 if the reversed inequalities hold. (e12 ) Assume that (i) u, v ∈ C1,∗ (E0 , R) and uxy (x, y) < f (x, y, u(x, y), ux (x, y), uy (x, y), (Mu)(x, y)), vxy (x, y) > f (x, y, v(x, y), vx (x, y), vy (x, y), (Mv)(x, y)), for (x, y) ∈ E0 , ux (x, 0) < vx (x, 0), x ∈ Ia , uy (0, y) < vy (0, y), y ∈ Ib , u(0, 0) < v(0, 0); (ii) f (x, y, p1 , p2 , p3 , p4 ) is nondecreasing with respect to p1 , p2 , p3 , p4 for all (x, y) ∈ E0 ; (iii) g is nondecreasing; (iv) m(ξ ) 0 for ξ ∈ Ib . Then (u, ux , uy ) < (v, vx , vy ) on E0 . (e13 ) Assume that (i) u, v ∈ C1,∗ (E0 , R) and uxy (x, y) f (x, y, u(x, y), ux (x, y), uy (x, y), (Mu)(x, y)), vxy (x, y) f (x, y, v(x, y), vx (x, y), vy (x, y), (Mv)(x, y)),

54


for (x, y) ∈ E0 , ux (x, 0) vx (x, 0), x ∈ Ia , uy (0, y) vy (0, y),

y ∈ Ib ,

u(0, 0) v(0, 0); (ii) f (x, y, p1 , p2 , p3 , p4 ) is nondecreasing with respect to p1 , p2 , p3 , p4 for all (x, y) ∈ E0 ; (iii) f (x, y, p1 , p2 , p3 , p4 ) − f (x, y, p1 , p2 , p3 , p4 ) L ∑4i=1 (pi − pi ), whenever pi , pi ∈ R, pi pi (i = 1, 2, 3, 4) for (x, y) ∈ E0 , where L > 0 is a constant; (iv) 0 g(ξ ) − g(ξ ) K(ξ − ξ ), whenever ξ ξ for constant K > 0; (v) m(ξ ) 0 for ξ ∈ Ib . Then (u, ux , uy ) (v, vx , vy ) on E0 . (e14 ) let u, v be respectively, the lower and upper functions for problem (1.7.16)–(1.7.17) and f , g, m as in (e13 ). Then for any solution z of problem (1.7.16)–(1.7.17), the inequalities (u, ux , uy ) (z, zx , zy ) (v, vx , vy ) hold on E0 . Brzychczy and Janus [20]

1.7.7

Define the operator G : C1,∗ (E0 , R) → C1,∗ (E0 , R) as, for u ∈ C1,∗ (E0 , R), let v = Gu be the unique solution of the following linear problem (called (LP)) vxy (x, y) = F[u](x, y),

(x, y) ∈ E0 ,

v(x, 0) = σ (x),

x ∈ Ia ,

v(0, y) = τ (y),

y ∈ Ib ,

σ (0) = τ (0) = c0 , where F[u](x, y) = f (x, y, u(x, y), ux (x, y), uy (x, y), (Mu)(x, y)), the operator M is defined by (1.7.18). It is easy to see that the solution of problem (LP) is given by the formula

x y

v(x, y) = 0

0

F[u](ξ ,t) d ξ dt + σ (x) + τ (y) − c0 .

(e15 ) Assume that (i) there exist u0 , v0 , the lower and upper functions for problem (1.7.16)–(1.7.17) on E0 respectively;


55

(ii) f (x, y, p1 , p2 , p3 , p4 ) is nondecreasing with respect to p1 , p2 , p3 , p4 for all (x, y) ∈ E0 ; (iii) f (x, y, p1 , p2 , p3 , p4 ) satisfies the Lipschitz condition 4

| f (x, y, p1 , p2 , p3 , p4 ) − f (x, y, p1 , p2 , p3 , p4 )| L ∑ |pi − pi |, i=1

for (x, y) ∈ E0 , pi , pi ∈ R, where L > 0 is a constant; (iv) 0 g(ξ ) − g(ξ ) K(ξ − ξ ), whenever ξ ξ for a constant K > 0; (v) m(ξ ) 0 for ξ ∈ Ib . Then, the sequences {un } and {vn } defined inductively by u1 = Gu0 ,

un = Gun−1 ,

v1 = Gv0 ,

vn = Gvn−1 ,

for n = 1, 2, . . ., satisfy the following conditions: 1o ) un−1 (x, y) un (x, y), vn (x, y) vn−1 (x, y), for (x, y) ∈ E0 ; 2o ) the functions un and vn for n = 1, 2, . . . are the lower and upper functions for problem (1.7.16)–(1.7.17) on E0 , respectively; 3o ) the following inequalities hold 0 vn (x, y) − un (x, y) N0

Cn (x + y)n , n!

0 vnx (x, y) − unx (x, y) N0

Cn (x + y)n , n!

0 vny (x, y) − uny (x, y) N0

Cn (x + y)n , n!

for n = 1, 2, . . . and (x, y) ∈ E0 , where N0 := v0 − u0 and C is a some constant. Moreover, a function z defined by z(x, y) := lim un (x, y) = lim vn (x, y). n→∞

n→∞

is the unique solution of problem (1.7.16)–(1.7.17) on E0 . 1.7.8

Pachpatte [84]

Consider the Darboux problem for the hyperbolic partial integrodifferential equation of the form uxy (x, y) = f (x, y, u(x, y), ux (x, y), uy (x, y), (Ωu)(x, y), μ ),

(1.7.19)

with the given initial boundary conditions u(x, 0) = σ (x),

u(0, y) = τ (y),

u(0, 0) = 0,

(1.7.20)

56


for x, y ∈ R+ , where

x y

(Ωu)(x, y) = 0

0 n

g(x, y, ξ , η , u(ξ , η ), uξ (ξ , η ), uη (ξ , η )) d ξ d η ,

f ∈ C(E × Rn × Rn × Rn × R × R, Rn ), g ∈ C(E2 × Rn × Rn × Rn , Rn ), σ , τ ∈ C(R+ , Rn ) and μ is a parameter. Let S1 be the space defined as in Section 1.6. (e16 ) Assume that (i) the functions f , g in (1.7.19) satisfy the conditions | f (x, y, u1 , u2 , u3 , u4 , μ ) − f (x, y, u1 , u2 , u3 , u4 , μ )| L |u1 − u1 | + |u2 − u2 | + |u3 − u3 | + |u4 − u4 | , |g(x, y, s,t, u1 , u2 , u3 ) − g(x, y, s,t, u1 , u2 , u3 )| M |u1 − u1 | + |u2 − u2 | + |u3 − u3 | , where L 0, M 0 are constants, (ii) for every μ ∈ R there exist constants Q j 0 ( j = 1, 2, 3) such that |σ (x)| + |τ (y)| + |σx (x)| + |τy (y)| +

x y 0

y

| f (x,t, 0, 0, 0, (Ω0)(x,t), μ )|dt Q2 exp(λ (x + y)),

0

x 0

| f (s,t, 0, 0, 0, (Ω0)(s,t), μ )|ds dt Q1 exp(λ (x + y)),

0

| f (s, y, 0, 0, 0, (Ω0)(s, y), μ )|ds Q3 exp(λ (x + y)),

where λ > 0 is a constant and 0 is the zero element in Rn . If α = M 1 2 + λ < 1, then there exists a unique solution u ∈ S1 of the problem L 1 + λ2 λ2 (1.7.19)–(1.7.20) on E. (e17 ) Assume that (i) the conditions in (e16 ) hold, (ii) there exist constants N j 0 ( j = 1, 2, 3) and the function P ∈ C(E, R+ ) such that | f (x, y, u1 , u2 , u3 , u4 , μ1 ) − f (x, y, u1 , u2 , u3 , u4 , μ2 )| P(x, y)|μ1 − μ2 |, for μ1 , μ2 ∈ R and

x y 0

0

P(s,t) ds dt N1 exp(λ (x + y)),

y 0

x 0

P(x,t) dt N2 exp(λ (x + y)), P(s, y) ds N3 exp(λ (x + y)).

Then the solution u(x, y, μ ) of problem (1.7.19)–(1.7.20) belonging to S1 is continuous with respect to the variables (x, y, μ ) in E × R.


1.8

57

Notes

The literature concerning the Volterra type equations and inequalities is particularly rich, see [2,5,7,9,12,22,24,42,45] and the references given therein. Section 1.2 deals with some basic integral inequalities with explicit estimates, needed in our subsequent discussion, which are recently appeared in the literature. All the results are due to Pachpatte and are taken from [24,27,40]. The results given in sections 1.3 and 1.4 deals with some important qualitative properties of solutions of certain integral equations in two variables and adapted from [40,26]. Sections 1.5 and 1.6 contains some basic qualitative properties concerning certain partial integrodifferential and integral equations and are taken from [18,27,28]. Section 1.7 is written mostly to provide results on certain aspects related to some topics, which we did not cover in our exposition. Moreover, many generalizations, extensions and variants of the results are possible and progress is to be expected in the future.

Chapter 2

Integral inequalities and equations in two and three variables

2.1

Introduction

The inequalities with explicit estimates are among the most powerful and widely used analytic tools in the study of various dynamical systems. The extensive surveys of such inequalities may be found in monographs [82,85,87,134]. It is easy to check that the inequalities available in the literature are not directly applicable to study the qualitative behavior of solutions of many dynamical systems arising in natural phenomena. For instance, see the equations of the forms (11), (16), (19) to name a few. Motivated by the needs of diverse applications and the desire to widen the scope of such inequalities, recently in [108,,99,104,102,95,112] the present author investigated new explicit estimates on some integral inequalities, which are equally important to achieve the diversity of desired goals. In this chapter, we offer some such fundamental inequalities established in [41,30,37,32,35,36], which can be used as tools for handling equations like (11), (16), (19) and their variants. Some important qualitative aspects of the general forms of equations (11), (19) are also studied in a simple and unified way. In what follows, we shall use the notations given earlier without further mention.

2.2

Integral inequalities in two variables

In this section we present some basic integral inequalities with explicit estimates which can be used in the study of qualitative properties of solutions of equations of the forms (11), (16) and (19). The inequalities established in [108] are embodied in the following theorems.

Theorem 2.2.1.

Let u, r, n ∈ C(E, R+ ) and c 0 is a constant. 59

60


(a1 ) If u(x,t) c +

x s t 0

for (x,t) ∈ E, then u(x,t) c exp

0

r(σ , τ )u(σ , τ ) d τ d σ ds,

0

x s 0

0

(2.2.1)

r(σ , τ ) d τ d σ ds ,

t 0

(2.2.2)

for (x,t) ∈ E (a2 ) Let n(x,t) be nondecreasing in each variable x, t ∈ R+ . If u(x,t) n(x,t) +

x s t 0

for (x,t) ∈ E, then u(x,t) n(x,t) exp

0

0


x s 0

0

t 0

(2.2.3)

r(σ , τ ) d τ d σ ds ,

(2.2.4)

for (x,t) ∈ E. Theorem 2.2.2. Let u, p, q, r ∈ C(E, R+ ). (a3 ) Let L ∈ C(E × R+ , R+ ) be such that 0 L(x,t, u) − L(x,t, v) M(x,t, v)(u − v),

(2.2.5)

for u v 0, where M ∈ C(E × R+ , R+ ). If u(x,t) p(x,t) + q(x,t) for (x,t) ∈ E, then u(x,t) p(x,t) + q(x,t) × exp

x s 0

0

t 0

x s t 0

0

0

x s 0

0

L(σ , τ , u(σ , τ )) d τ d σ ds,

(2.2.6)

t

L(σ , τ , p(σ , τ )) d τ d σ ds

0

M(σ , τ , p(σ , τ ))q(σ , τ ) d τ d σ ds ,

(2.2.7)

for (x,t) ∈ E. (a4 ) If u(x,t) p(x,t) + q(x,t) for (x,t) ∈ E, then u(x,t) p(x,t) + q(x,t) × exp for (x,t) ∈ E.

0

0

0

0

x s 0

x s 0

x s t

t 0

0

t 0


(2.2.8)

r(σ , τ )p(σ , τ ) d τ d σ ds

r(σ , τ )q(σ , τ ) d τ d σ ds ,

(2.2.9)


61

Theorem 2.2.3. Let u, r ∈ C(E, R+ ) and c 0 is a constant. (a5 ) If u2 (x,t) c +

x s t 0

0

0


(2.2.10)

for (x,t) ∈ E, then u(x,t)

√

c+

1 2

x s t 0

0

0

r(σ , τ ) d τ d σ ds,

(2.2.11)

for (x,t) ∈ E. (a6 ) If u ∈ C (E, R1 ), c 1 and u(x,t) c +

x s t 0

0

0

r(σ , τ )u(σ , τ ) log u(σ , τ ) d τ d σ ds,

(2.2.12)

for (x,t) ∈ E, then u(x,t) cA(x,t) , for (x,t) ∈ E, where

x s

t

A(x,t) = exp 0

0

0

(2.2.13)

r(σ , τ ) d τ d σ ds .

(2.2.14)

In the following theorem we present the inequality proved in [104]. Theorem 2.2.4.

Let IL = [0, L], IT = [0, T ] (L > 0, T > 0) are the given subsets of R and

D = IL × IT . (a7 ) Let u, p, q, r ∈ C(D, R+ ). If u(x,t) p(x,t) + q(x,t) for (x,t) ∈ D, then

t s L 0

0

0

t s

L

r(y, τ )u(y, τ ) dy d τ ds,

r(y, τ )p(y, τ ) dy d τ ds u(x,t) p(x,t) + q(x,t) 0 0 0 t s L r(y, τ )q(y, τ ) dy d τ ds , × exp 0

0

(2.2.15)

(2.2.16)

0

for (x,t) ∈ D. Proofs of Theorems 2.2.1–2.2.4.

To prove (a1 ) and (a5 ) it is sufficient to assume that

c > 0, since the standard limiting argument can be used to treat the remaining case, see [82, p. 108].

62


(a1 ) Let c > 0 and define a function z(x,t) by the right hand side of (2.2.1), then z(x, 0) = z(0,t) = c, u(x,t) z(x,t),

x s

zt (x,t) =

0

0

0

0

x t

zx (x,t) =

t

zxx (x,t) =

0

r(σ , τ )u(σ , τ ) d σ d τ , r(σ , τ )u(σ , τ ) d τ d σ ,

r(x, τ )u(x, τ ) d τ ,

and zxxt (x,t) = r(x,t)u(x,t) r(x,t)z(x,t).

(2.2.17)

From (2.2.17) and using the facts that zxx (x,t) 0, zt (x,t) 0, z(x,t) > 0, we observe that (see [22, p. 346]) zxxt (x,t) zxx (x,t)zt (x,t) r(x,t) + , z(x,t) z2 (x,t) i.e.,

∂ ∂t

zxx (x,t) z(x,t)

r(x,t).

(2.2.18)

By keeping x fixed in (2.2.18), we set t = τ and then integrating with respect to τ from 0 to t and using the fact that zxx (x, 0) = 0, we have zxx (x,t) z(x,t)

t 0

r(x, τ ) d τ .

(2.2.19)

Again as above, from (2.2.19) and the facts that zx (x,t) 0, z(x,t) > 0, we observe that t ∂ zx (x,t) r(x, τ ) d τ . (2.2.20) ∂ x z(x,t) 0 By taking t fixed in (2.2.20), set x = σ and then integrating with respect to σ from 0 to x and using the fact that zx (0,t) = 0, we have zx (x,t) z(x,t) From (2.2.21), we get z(x,t) c exp

x t 0

0

r(σ , τ ) d τ d σ .

x s 0

0

t

0

r(σ , τ ) d τ d σ ds .

(2.2.21)

(2.2.22)

Using (2.2.22) in u(x,t) z(x,t), we get (2.2.2). (a2 ) First we assume that n(x,t) > 0 for (x,t) ∈ E. From (2.2.3), it is easy to observe that u(x,t) 1+ n(x,t)

x s t 0

0

0

r(σ , τ )

u(σ , τ ) d τ d σ ds. n(σ , τ )

(2.2.23)


63

Now an application of the inequality in part (a1 ) to (2.2.23) yields (2.2.4). The proof of the case when n(x,t) = 0 can be completed as in [82, p. 326]. (a3 ) Define a function z(x,t) by

x s t

z(x,t) = 0

0

0

L(σ , τ , u(σ , τ ))d τ d σ ds,

(2.2.24)

then z(0,t) = z(x, 0) = 0 and (2.2.6) can be restated as u(x,t) p(x,t) + q(x,t)z(x,t).

(2.2.25)

From (2.2.24), (2.2.25) and (2.2.5), we observe that z(x,t)

x s t 0

0

0

{L (σ , τ , p(σ , τ ) + q(σ , τ )z(σ , τ ))

−L(σ , τ , p(σ , τ )) + L(σ , τ , p(σ , τ ))} d τ d σ ds

x s t 0

x s t

+ 0

0

0

0

0

L(σ , τ , p(σ , τ )) d τ d σ ds

M(σ , τ , p(σ , τ ))q(σ , τ )z(σ , τ ) d τ d σ ds.

(2.2.26)

Clearly, the first term on the right hand side in (2.2.26) is nonnegative and nondecreasing in x, t ∈ R+ . Now a suitable application of the inequality in part (a2 ) to (2.2.26) yields x s t L(σ , τ , p(σ , τ )) d τ d σ ds z(x,t) 0

× exp

0

x s 0

0

t 0

0

M(σ , τ , p(σ , τ ))q(σ , τ ) d τ d σ ds ,

(2.2.27)

for (x,t) ∈ E. Using (2.2.27) in (2.2.25), we get the required inequality in (2.2.7). (a4 ) By taking L(σ , τ , u(σ , τ )) = r(σ , τ )u(σ , τ ) in part (a3 ) for r, u ∈ C(E, R+ ), we get the required inequality. (a5 ) Let c > 0 and define a function z(x,t) by the right hand side of (2.2.10), then z(x, 0) = z(0,t) = c, u(x,t) z(x,t) and zxxt (x,t) = r(x,t)u(x,t) r(x,t) z(x,t). (2.2.28) Now by following the same arguments as in the proof of part (a1 ) below (2.2.17) upto (2.2.21) with suitable modifications (see [82, p. 528]), from (2.2.28), we get z (x,t) x z(x,t)

x t 0

0

r(σ , τ ) d τ d σ .

(2.2.29)

64


From (2.2.29), we obtain √ 1 x s t z(x,t) c + r(σ , τ ) d τ d σ ds. 2 0 0 0 Using (2.2.30) in u(x,t) z(x,t), we get (2.2.11).

(2.2.30)

(a6 ) Define a function z(x,t) by the right hand side of (2.2.12), then z(x, 0) = z(0,t) = c, u(x,t) z(x,t) and zxxt (x,t) = r(x,t)u(x,t) log u(x,t) r(x,t)z(x,t) log z(x,t).

(2.2.31)

The remaining proof can be completed by following the proof of part (a1 ) and closely looking at the proof of Theorem 5.10.1 given in [82, p. 544]. (a7 ) Introducing the notation e(τ ) =

L 0

r(y, τ )u(y, τ ) dy,

in (2.2.15), we get u(x,t) p(x,t) + q(x,t) for (x,t) ∈ D. Define

t s

z(t) = 0

for t ∈ IT , then it is easy to see that

0

t s 0

0

(2.2.32)

e(τ ) d τ ds,

e(τ ) d τ ds,

z(0) = 0, z (0) = 0

(2.2.33)

(2.2.34)

and

u(x,t) p(x,t) + q(x,t)z(t).

(2.2.35)

From (2.2.34), (2.2.32), (2.2.35), we observe that z (t) = e(t) =

L 0

r(y,t)u(y,t)dy

L

L

r(y,t)[p(y,t) + q(y,t)z(t)]dy 0

L

r(y,t)p(y,t) dy + z(t)

= 0

r(y,t)q(y,t) dy.

(2.2.36)

0

From (2.2.36) and the fact that z(t) is nondecreasing for t ∈ IT , it is easy to see that z(t)

t s L 0

0

t

+ 0

r(y, τ )p(y, τ ) dy d τ ds

s L z(s) r(y, τ )q(y, τ ) dy d τ ds. 0

0

(2.2.37)

0

Clearly, the first term on the right hand side in (2.2.37) is nonnegative and nondecreasing in t ∈ IT . Now a suitable application of the inequality in Theorem 1.3.1 in [82] to (2.2.37) yields z(t)

t s 0

× exp

0

L 0

r(y, τ )p(y, τ ) dy d τ ds

t s 0

0

0

L

r(y, τ )q(y, τ ) dy d τ ds ,

for t ∈ IT . Using (2.2.38) in (2.2.35), we get the required inequality in (2.2.16).

(2.2.38)


2.3

65

Integral inequalities in three variables

Our main goal in this section is to present some integral inequalities with explicit estimates in three variables established in [95,111,102], which can be used in certain new applications for which the earlier inequalities do not apply directly. In what follows I = [a, b] (a < b) denotes a given subset of R and G = R2+ × I. First we shall give the following theorems which contains the inequalities investigated in [95]. Theorem 2.3.1. Let u, p, q, f ∈ C(G, R+ ) and k 0 is a constant. (b1 ) If u(x, y, z) k +

x y b

f (s,t, r)u(s,t, r) dr dt ds, 0

for (x, y, z) ∈ G, then u(x, y, z) k exp

0

(2.3.1)

a

x y 0

0

b

f (s,t, r) dr dt ds ,

(2.3.2)

f (s,t, r)u(s,t, r) dr dt ds,

(2.3.3)

a

for (x, y, z) ∈ G. (b2 ) If x y b

u(x, y, z) p(x, y, z) + q(x, y, z)

0

for (x, y, z) ∈ G, then

x y 0

0

a

x y

u(x, y, z) p(x, y, z) + q(x, y, z)

× exp

0

b

f (s,t, r)p(s,t, r) dr dt ds 0

0

a

f (s,t, r)q(s,t, r) dr dt ds ,

b

(2.3.4)

a

for (x, y, z) ∈ G. Theorem 2.3.2. Let u, f ∈ C(G, R+ ) and k 0, c 1 are constants. (b3 ) If u2 (x, y, z) k +

x y b

f (s,t, r)u(s,t, r) dr dt ds, 0

0

(2.3.5)

a

for (x, y, z) ∈ G, then u(x, y, z) for (x, y, z) ∈ G.

√

k+

1 2

x y b

f (s,t, r) dr dt ds, 0

0

a

(2.3.6)

66


(b4 ) If u(x, y, z) 1 and u(x, y, z) c +

x y b

f (s,t, r)u(s,t, r) log u(s,t, r) dr dt ds, 0

0

(2.3.7)

a

for (x, y, z) ∈ G, then u(x, y, z) cexp(

xyb

f (s,t,r) dr dt ds)

0 0 a

,

(2.3.8)

for (x, y, z) ∈ G. In the following theorems we present the inequalities established in [102]. Theorem 2.3.3. Let u, p, q, f ∈ C(G, R+ ). (b5 ) If u(x, y, z) p(x, y, z) + q(x, y, z)

x ∞ b 0

for (x, y, z) ∈ G, then

× exp

0

0

(2.3.9)

∞ b y

f (s,t, r)p(s,t, r) dr dt ds

a


∞ b y


a

x

u(x, y, z) p(x, y, z) + q(x, y, z) x

y

(2.3.10)

a

for (x, y, z) ∈ G. (b6 ) If u(x, y, z) p(x, y, z) + q(x, y, z)

∞ ∞ b x

for (x, y, z) ∈ G, then u(x, y, z) p(x, y, z) + q(x, y, z)

× exp

x

y

∞ ∞ b y


(2.3.11)

a

∞ ∞ b

x

y


a


(2.3.12)

a

for (x, y, z) ∈ G. Theorem 2.3.4. Let u, p, q, c, f , g ∈ C(G, R+ ). (b7 ) suppose that u(x, y, z) p(x, y, z) + q(x, y, z) + c(x, y, z)

x y b

f (s,t, r)u(s,t, r) dr dt ds 0

0

0

0

a

∞ ∞ b

g(s,t, r)u(s,t, r) dr dt ds, a

(2.3.13)


for (x, y, z) ∈ G. If

α1 =

∞ ∞ b 0

0

67

g(s,t, r)B1 (s,t, r) dr dt ds < 1,

a

(2.3.14)

then u(x, y, z) A1 (x, y, z) + D1 B1 (x, y, z),

(2.3.15)

for (x, y, z) ∈ G, where

x y b A1 (x, y, z) = p(x, y, z) + q(x, y, z) f (s,t, r)p(s,t, r) dr dt ds 0 0 a x y b f (s,t, r)q(s,t, r) dr dt ds (2.3.16) × exp 0 0 a x y b f (s,t, r)c(s,t, r) dr dt ds B1 (x, y, z) = c(x, y, z) + q(x, y, z) 0 0 a x y b f (s,t, r)q(s,t, r) dr dt ds (2.3.17) × exp 0

and D1 =

1 1 − α1

0

a

∞ ∞ b 0

0

(b8 ) Suppose that u(x, y, z) p(x, y, z) + q(x, y, z) + c(x, y, z) for (x, y, z) ∈ G. If

α2 =

∞ ∞ b 0

0

a

a

g(s,t, r)A1 (s,t, r) dr dt ds.

∞ ∞ b x

y

0

0

∞

(2.3.18)

f (s,t, r)u(s,t, r) dr dt ds

a ∞ b

g(s,t, r)u(s,t, r) dr dt ds,

(2.3.19)

a


(2.3.20)

then u(x, y, z) A2 (x, y, z) + D2 B2 (x, y, z),

(2.3.21)


∞ ∞ b A2 (x, y, z) = p(x, y, z) + q(x, y, z) f (s,t, r)p(s,t, r) dr dt ds x y a ∞ ∞ b f (s,t, r)q(s,t, r) dr dt ds , (2.3.22) × exp x y a ∞ ∞ b f (s,t, r)c(s,t, r) dr dt ds B2 (x, y, z) = c(x, y, z) + q(x, y, z) x y a ∞ ∞ b f (s,t, r)q(s,t, r) dr dt ds , (2.3.23) × exp x

and D2 =

1 1 − α2

y

a

∞ ∞ b 0

0

a


The next theorem deals with a slight variant of the inequality proved in [111].

(2.3.24)

68


Theorem 2.3.5. Let u, p ∈ C(G, R+ ), q ∈ C(G × I, R+ ) and c 0 is a constant. If x

u(x, y, z) c +

y

b

p(s,t, z)u(s,t, z) + 0

q(s,t, z, r)u(s,t, r) dr dt ds,

0

(2.3.25)

a

for (x, y, z) ∈ G, then u(x, y, z) c H(x, y, z) exp

x y

b

q(s,t, z, r)H(s,t, r) dr dt ds , 0

0


(2.3.26)

a

x

y

H(x, y, z) = exp 0

0

p(σ , τ , z) d τ d σ .

(2.3.27)

We give the details of the proofs of (b2 ), (b3 ), (b5 ), (b8 )

Proofs of Theorems 2.3.1–2.3.5.

and the inequality in Theorem 2.3.5 only. The proofs of other inequalities can be completed by following the proofs of these inequalities and closely looking at the proofs of the similar results given in [82, 87]. (b2 ) Introducing the notation

b

E(s,t) =

f (s,t, r)u(s,t, r) dr,

(2.3.28)

a

the inequality (2.3.3) can be restated as u(x, y, z) p(x, y, z) + q(x, y, z) Define

x y

E(s,t) dt ds. 0

(2.3.29)

0

x y

w(x, y) =

E(s,t) dt ds, 0

(2.3.30)

0

then w(x, 0) = w(0, y) = 0 and u(x, y, z) p(x, y, z) + q(x, y, z)w(x, y).

(2.3.31)

From (2.3.30), (2.3.28), (2.3.31), we observe that b

wxy (x, y) = E(x, y) =

f (x, y, r)u(x, y, r) dr a

b

f (x, y, r)[p(x, y, z) + q(x, y, z)w(x, y)]dr a

b

= w(x, y)

b

f (x, y, r)q(x, y, r) dr + a

f (x, y, r)p(x, y, r) dr.

(2.3.32)

a

Now, by following the similar arguments as in the proof of Theorem 4.3.1 given in [82, p. 328] with suitable modifications, from (2.3.32), we obtain x y b w(x, y) f (s,t, r)p(s,t, r) dr dt ds 0

0

a


× exp

x y 0

0

b

69

f (s,t, r)q(s,t, r) dr dt ds .

(2.3.33)

a

Now, using (2.3.33) in (2.3.31), we get the required inequality in (2.3.4). (b3 ) Introducing the notation (2.3.28) in (2.3.5), we get u2 (x, y, z) k + Let k > 0 and define

x y

E(s,t) dt ds. 0

(2.3.34)

0

x y

m(x, y) = k +

E(s,t) dt ds, 0

(2.3.35)

0

then m(x, 0) = m(0, y) = k and u2 (x, y, z) m(x, y).

(2.3.36)

From (2.3.35), (2.3.28) and (2.3.36), we observe that b f (x, y, r)u(x, y, r) dr m(x, y) mxy (x, y) = E(x, y) = a

b

f (x, y, r) dr.

(2.3.37)

a

The inequality (2.3.37) implies (see [82, Theorem 5.8.1, p. 527]) √ 1 x y b m(x, y) k + f (s,t, r) dr dt ds. 2 0 0 a

(2.3.38)

Using (2.3.38) in (2.3.36), we get the required inequality in (2.3.6). If k 0 we carry out the above procedure with k + ε instead of k, where ε > 0 is an arbitrary small constant, and subsequently pass to the limit as ε → 0 to obtain (2.3.6). (b5 ) Using the notation (2.3.28) in (2.3.9), we get u(x, y, z) p(x, y, z) + q(x, y, z) Define w(x, y) =

x ∞

E(s,t) dt ds. 0

(2.3.39)

y

x ∞

E(s,t) dt ds, 0

(2.3.40)

y

then w(0, y) = 0 and u(x, y, z) p(x, y, z) + q(x, y, z)w(x, y). From (2.3.40), (2.3.28) and (2.3.41), we observe that ∞ ∞ b wx (x, y) = E(x,t) dt = f (x,t, r)u(x,t, r) dr dt y

∞ b y

a

y

a

f (x,t, r) [p(x,t, r) + q(x,t, r)w(x,t)] dr dt

(2.3.41)

70


=

∞ b y

f (x,t, r)p(x,t, r) dr dt +

a

∞

b

w(x,t) y

f (x,t, r)q(x,t, r) dr dt. (2.3.42)

a

By taking x = s in (2.3.42) and integrating both sides with respect to s from 0 to x, we get x ∞ b f (s,t, r)q(s,t, r) dr w(s,t) dt ds, (2.3.43) w(x, y) e1 (x, y) + 0

y

where e1 (x, y) =

a

x ∞ b 0

y

f (s,t, r)p(s,t, r) dr dt ds.

(2.3.44)

a

Clearly e1 (x, y) is nonnegative, continuous, nondecreasing in x and nonincreasing in y for x, y ∈ R+ . Now a suitable application of the inequality in Theorem 1.2.4 given in [87, p. 110] (see also [82, p. 440]) to (2.3.43) yields x ∞ b f (s,t, r)q(s,t, r) dr dt ds . w(x, y) e1 (x, y) exp 0

y

(2.3.45)

a

Using (2.3.45) in (2.3.41), we get (2.3.10). (b8 ) Let E(s,t) be as in (2.3.28) and

λ=

∞ ∞ b

g(s,t, r)u(s,t, r) dr dt ds. 0

0

(2.3.46)

a

Then (2.3.19) can be restated as u(x, y, z) p(x, y, z) + q(x, y, z) Let v(x, y) =

∞ ∞ x

y

E(s,t) dt ds + c(x, y, z)λ .

(2.3.47)

E(s,t) dt ds,

(2.3.48)

∞ ∞ x

y

then v(∞, y) = 0 and from (2.3.47), we have u(x, y, z) p(x, y, z) + q(x, y, z)v(x, y) + c(x, y, z)λ . From (2.3.48), (2.3.28) and (2.3.49), we have vx (x, y) = − −

∞ b y

a

∞ y

E(x,t) dt = −

∞ b y

y

f (x,t, r)u(x,t, r)dr dt

a

f (x,t, r)[p(x,t, r) + q(x,t, r)v(x,t) + c(x,t, r)λ ]dr dt =−

−

∞ b

(2.3.49)

a

∞ b y

f (x,t, r)p(x,t, r) dr dt

a

f (x,t, r)[q(x,t, r)v(x,t) + c(x,t, r)λ ]dr dt.

(2.3.50)


71

By taking x = s in (2.3.50) and integrating both sides with respect to s from x to ∞ for x ∈ R+ , we have v(x, y) e2 (x, y) + where e2 (x, y) =

∞ ∞ b x

y

∞ ∞ b x

y

a

f (s,t, r)q(s,t, r) dr v(s,t) dt ds,

(2.3.51)

a

f (s,t, r)[p(s,t, r) + c(s,t, r)λ ] dr dt ds.

(2.3.52)

Clearly, e2 (x, y) is nonnegative, continuous, nonincreasing in each variable x, y ∈ R+ . Now, a suitable application of the inequality in Theorem 1.2.3 given in [87, p. 110] (see also [82, p. 440]) to (2.3.51) yields v(x, y) e2 (x, y) exp

f (s,t, r)q(s,t, r) dr dt ds .

∞ ∞ b x

y

Using (2.3.53), (2.3.52) in (2.3.49), we get ∞ ∞ u(x, y, z) p(x, y, z) + q(x, y, z) x

× exp

y

y

b

a

f (s,t, r){p(s,t, r) + c(s,t, r)λ } dr dt ds

f (s,t, r)q(s,t, r) dr dt ds + c(x, y, z)λ

∞ ∞ b x

(2.3.53)

a

a

= A2 (x, y, z) + λ B2 (x, y, z).

(2.3.54)

From (2.3.46) and (2.3.54), we have

λ=

∞ ∞ b

g(s,t, r)u(s,t, r) dr dt ds 0

∞ ∞ b 0

0

a

0

a

g(s,t, r)[A2 (s,t, r) + λ B2 (s,t, r)] dr dt ds,

which implies

λ D2 .

(2.3.55)

Using (2.3.55) in (2.3.54), we get (2.3.21). To prove the inequality in Theorem 2.3.5, let x y b

m(x, y, z) = c +

q(s,t, z, r)u(s,t, r) dr dt ds, 0

0

then (2.3.25) can be restated as u(x, y, z) m(x, y, z) +

(2.3.56)

a

x y

p(s,t, z)u(s,t, z) dt ds. 0

0

(2.3.57)

72


It is easy to observe that m(x, y, z) is nonnegative for (x, y, z) ∈ G and nondecreasing in each variable x, y ∈ R+ and for z ∈ I. Treating (2.3.57) as two-dimensional integral inequality in x, y ∈ R+ for every z ∈ I and a suitable application of the inequality given in [82, Theorem 4.2.2, p. 325] to (2.3.57) yields u(x, y, z) m(x, y, z)H(x, y, z).

(2.3.58)

From (2.3.56) and (2.3.58), we observe that m(x, y, z) c +

x y b 0

0

q(s,t, z, r)H(s,t, r)m(s,t, r) dr dt ds.

(2.3.59)

q(s,t, z, r)H(s,t, r)m(s,t, r) dr,

(2.3.60)

a

Setting b

e(s,t) = a

for every z ∈ I, the inequality (2.3.59) can be restated as m(x, y, z) c +

x y

e(s,t) dt ds. 0

(2.3.61)

0

Let x y

n(x, y) = c +

e(s,t) dt ds, 0

(2.3.62)

0

then n(x, 0) = n(0, y) = c and m(x, y, z) n(x, y)

(2.3.63)

for x, y ∈ R+ and for every z ∈ I. From (2.3.62), (2.3.60) and (2.3.63), we observe that b

nxy (x, y) = e(x, y) = n(x, y)

q(x, y, z, r)H(x, y, r)m(x, y, r) dr a

b

q(x, y, z, r)H(x, y, r) dr.

(2.3.64)

a

The inequality (2.3.64) implies (see [82, p. 325]) n(x, y) c exp

x y 0

0

b

q(s,t, z, r)H(s,t, r) dr dt ds ,

(2.3.65)

a

for x, y ∈ R+ and for every z ∈ I. The required inequality in (5.3.26) follows from (2.3.65), (2.3.63) and (2.3.58).


2.4

73

Integral equation in two variables

Consider the integral equation of the form t s L

u(x,t) = f (x,t) + 0

0

0

K(x,t, s, y, τ , u(y, τ )) dy d τ ds,

(2.4.1)

for (x,t) ∈ D = J × I; J = [0, L], I = [0, T ], where L > 0, T > 0 are finite but can be arbitrarily large constants and f ∈ C(D, R), K ∈ C(D × I × D × R, R). Such equations arises, in the study of partial differential equations of the forms (13)–(15), see [4]. This section is devoted to address some basic results related to the solution of equation (2.4.1) given in [104]. Let U be the space of those functions φ ∈ C(D, R) which fulfil the condition |φ (x,t)| = O(exp(λ (x + t))),

(2.4.2)

where λ > 0 is a constant. In the space U, we define the norm |φ |U = max [|φ (x,t)| exp(−λ (x + t))]. (x,t)∈D

(2.4.3)

It is easy to see that U is a Banach space and |φ |U N,

(2.4.4)

where N 0 is a constant. First, we formulate the following theorem which provide conditions for the existence of a unique solution to equation (2.4.1). Theorem 2.4.1. Suppose that (i) the function K in equation (2.4.1) satisfies the condition |K(x,t, s, y, τ , u) − K(x,t, s, y, τ , v)| h(x,t, s, y, τ )|u − v|,

(2.4.5)

where h ∈ C (D × I × D, R+ ), (ii) for λ as in (2.4.2), ( j1 ) there exists a nonnegative constant α < 1 such that t s L 0

0

0

h(x,t, s, y, τ ) exp(λ (y + τ )) dy d τ ds α exp(λ (x + t)),

(2.4.6)

( j2 ) there exists a nonnegative constant β such that | f (x,t)| +

t s L 0

0

0

|K(x,t, s, y, τ , 0)|dy d τ ds β exp(λ (x + t)),

where f , K are as defined in equation (2.4.1). Then the equation (2.4.1) has a unique solution u(x,t) in u on D.

(2.4.7)

74


Proof.

Let u ∈ U and define the operator F by t s L

(Fu)(x,t) = f (x,t) + 0

0

0


(2.4.8)

for (x,t) ∈ D. The proof that F maps U into itself and is a contraction map can be completed by closely looking at the proof of Theorem 1.3.1 given in Chapter 1 with suitable modifications. We leave the details to the reader. The next theorem deals with the uniqueness of solutions of equation (2.4.1) on D. Theorem 2.4.2. Suppose that the function K in equation (2.4.1) satisfies the condition |K(x,t, s, y, τ , u) − K(x,t, s, y, τ , v)| q(x,t)g(y, τ )|u − v|,

(2.4.9)

where q, g ∈ C(D, R+ ). Then the equation (2.4.1) has at most one solution on D. Proof.

Let u(x,t) and v(x,t) be two solutions of equation (2.4.1) on D. Using these facts

and hypotheses, we have |u(x,t) − v(x,t)|

t s L 0

0

q(x,t)

0

|K(x,t, s, y, τ , u(y, τ )) − K(x,t, s, y, τ , v(y, τ ))|dy d τ ds

t s L 0

0

0

g(y, τ )|u(y, τ ) − v(y, τ )|dy d τ ds.

(2.4.10)

Now a suitable application of Theorem 2.2.4 part (a7 ) (with p(x,t) = 0) to (2.4.10) yields |u(x,t) − v(x,t)| 0, which implies u(x,t) = v(x,t). Thus there is at most one solution to equation (2.4.1) on D. The following theorems provide estimates on the solution of equation (2.4.1). Theorem 2.4.3. Suppose that the function K in equation (2.4.1) satisfies the condition |K(x,t, s, y, τ , u)| q(x,t)g(y, τ )|u|,

(2.4.11)

where q, g ∈ C(D, R+ ). If u(x,t) is any solution of equation (2.4.1) on D, then t s L g(y, τ )| f (y, τ )|dy d τ ds |u(x,t)| | f (x,t)| + q(x,t) × exp

0

t s 0

0

L 0

0

0

g(y, τ )q(y, τ ) dy d τ ds ,

(2.4.12)

for (x,t) ∈ D. Proof.

Using the fact that u(x,t) is a solution of equation (2.4.1) and hypotheses, we have |u(x,t)| | f (x,t)| +

t s L 0

0

| f (x,t)| + q(x,t)

0

|K(x,t, s, y, τ , u(y, τ ))|dy d τ ds

t s L 0

0

0

g(y, τ )|u(y, τ )|dy d τ ds.

Now an application of Theorem 2.2.4 part (a7 ) to (2.4.13) yields (2.4.12).

(2.4.13)


Theorem 2.4.4.

75

Suppose that the function K in equation (2.4.1) satisfies the condition

(2.4.9). If u(x,t) is any solution of equation (2.4.1) on D, then t s L g(y, τ )Q(y, τ ) dy d τ ds |u(x,t) − f (x,t)| Q(x,t) + q(x,t) × exp for (x,t) ∈ D, where

0

t s 0

0

0

t s L 0

0

0


(2.4.14)

|K (x,t, s, y, τ , f (y, τ ))| dy d τ ds,

Q(x,t) = 0

L

0

(2.4.15)

for (x,t) ∈ D. Proof.

From the fact that u(x,t) is a solution of equation (2.4.1) and the condition (2.4.9),

we have |u(x,t) − f (x,t)|

t s L 0

0

0

0

0

|K(x,t, s, y, τ , u(y, τ )) − K (x,t, s, y, τ , f (y, τ ))| dy d τ ds

t s L

+ Q(x,t) + q(x,t)

0

|K (x,t, s, y, τ , f (y, τ ))| dy d τ ds

t s L 0

0

0

g(y, τ ) |u(y, τ ) − f (y, τ )| dy d τ ds.

(2.4.16)

An application of Theorem 2.2.4 part (a7 ) to (2.4.16) yields (2.4.14). We call the function u ∈ C(D, R) an ε -approximate solution of equation (2.4.1), if there exists a constant ε 0 such that t s L u(x,t) − f (x,t) − K(x,t, s, y, τ , u(y, τ )) dy d τ ds ε , 0

0

0

for (x,t) ∈ D. The relation between an ε -approximate solution and a solution of equation (2.4.1) is shown in the following theorem. Theorem 2.4.5. Suppose that (i) the function K in equation (2.4.1) satisfies the condition (2.4.9), (ii) the functions uε (x,t), u(x,t) ∈ C(D, R) are respectively, an ε -approximate solution and any solution of equation (2.4.1). Then

t s L g(y, τ ) dy d τ ds |uε (x,t) − u(x,t)| ε 1 + q(x,t) × exp

for (x,t) ∈ D.

t s 0

0

0

0

L

0

0

g(y, τ )q(y, τ ) dy d τ ds

,

(2.4.17)

76

Proof.


Let z(x,t) = |uε (x,t) − u(x,t)|, (x,t) ∈ D. From the hypotheses, we observe that t s L z(x,t) = uε (x,t) − f (x,t) − K(x,t, s, y, τ , uε (y, τ )) dy d τ ds 0

t s L

+ 0

0

0

0

0

{K(x,t, s, y, τ , uε (y, τ )) − K(x,t, s, y, τ , u(y, τ ))}dy d τ ds

t s L uε (x,t) − f (x,t) − K(x,t, s, y, τ , uε (y, τ )) dy d τ ds 0

t s L

+ 0

0

0

0

0

|K(x,t, s, y, τ , uε (y, τ )) − K(x,t, s, y, τ , u(y, τ ))|dy d τ ds

ε + q(x,t)

t s L 0

0

0

g(y, τ )z(y, τ ) dy d τ ds.

(2.4.18)

Now an application of Theorem 2.2.4 part (a7 ) to (2.4.18) yields (2.4.17). In order to establish the dependency of solutions on parameters, we consider the equations t s L

u(x,t) = fi (x,t) +

0

0

0


(2.4.19)

for (x,t) ∈ D, i = 1, 2; where fi ∈ C(D, R), K ∈ C(D × I × D × R, R). In the following theorem we formulate conditions for continuous dependence of solutions of equations (2.4.19) on the functions involved therein. Theorem 2.4.6. Suppose that (i) the function K in (2.4.19) satisfies the condition (2.4.9), (ii) for i = 1, 2 the functions ui ∈ C(D, R) are respectively the εi -approximate solutions of (2.4.19) and let f (x,t) = | f1 (x,t) − f 2 (x,t)| + ε1 + ε2 . Then |u1 (x,t) − u2 (x,t)| f (x,t) + q(x,t)

× exp for (x,t) ∈ D.

t s 0

0

L 0

t s 0

0

L 0

g(y, τ ) f (y, τ ) dy d τ ds


(2.4.20)


Proof.

77

Let z(x,t) = |u1 (x,t) − u2 (x,t)|, (x,t) ∈ D. Following the proof of Theorem 1.3.5

in Chapter 1 and using the hypotheses, we obtain t s L

+ 0

0

0

z(x,t) | f1 (x,t) − f2 (x,t)| + ε1 + ε2 |K(x,t, s, y, τ , u1 (y, τ )) − K(x,t, s, y, τ , u2 (y, τ ))|dy d τ ds

f (x,t) + q(x,t)

t s L 0

0

0

g(y, τ )z(y, τ ) dy d τ ds.

(2.4.21)

Applying Theorem 2.2.4 part (a7 ) to (2.4.21) yields (2.4.20), which shows that the solutions of (2.4.19) depends continuously on functions on the right hand side of (2.4.19). We next consider the equation (2.4.1) and the integral equation t s L

w(x,t) = f (x,t) + 0

0

0

K(x,t, s, y, τ , w(y, τ )) dy d τ ds,

(2.4.22)

for (x,t) ∈ D, where f ∈ C(D, R), K ∈ C(D × I × D × R, R). The following theorem holds. Theorem 2.4.7.


(2.4.9). Then for every given solution w ∈ C(D, R) of equation (2.4.22) and any solution u ∈ C(D, R) of equation (2.4.1), the estimation t s L g(y, τ )h(y, τ ) dy d τ ds |u(x,t) − w(x,t)| h(x,t) + q(x,t) 0

× exp

t s 0

0

L 0

0

0

g(y, τ )q(y, τ ) dy d τ ds

(2.4.23)

holds for (x,t) ∈ D, where h(x,t) = | f (x,t) − f (x,t)| +

t η L 0

0

0

|K(x,t, η , z, σ , w(z, σ )) − K(x,t, η , z, σ , w(z, σ ))|dz d σ d η ,

(2.4.24)

for (x,t) ∈ D. Proof. Using the facts that u(x,t) and w(x,t) are respectively the solutions of equations (2.4.1) and (2.4.22) and hypotheses, we have |u(x,t) − w(x,t)| | f (x,t) − f (x,t)| t s L

+ 0

0

0

|K(x,t, s, y, τ , u(y, τ )) − K(x,t, s, y, τ , w(y, τ ))|dy d τ ds | f (x,t) − f (x,t)|

78


t s L

+ 0

0

0

t s L

+ 0

0

0

|K(x,t, s, y, τ , u(y, τ )) − K(x,t, s, y, τ , w(y, τ ))|dy d τ ds |K(x,t, s, y, τ , w(y, τ )) − K(x,t, s, y, τ , w(y, τ ))|dy d τ ds

h(x,t) + q(x,t)

t s L 0

0

0

g(y, τ )|u(y, τ ) − w(y, τ )|dy d τ ds.

(2.4.25)

Now applying Theorem 2.2.4 part (a7 ) to (2.4.25) yields (2.4.23). We note that, one can use the approach here to establish the results similar to those given above to study the equation of the form

x s t

u(x,t) = h(x,t) + 0

0

0

F(x,t, σ , τ , u(σ , τ )) d τ d σ ds,

(2.4.26)

under some suitable conditions on the functions involved in (2.4.26) and using Theorem 2.2.2 part (a4 ). Here, we do not discuss the details. 2.5

Integral equation in three variables

Consider the integral equation of the form u(x, y, z) = e(x, y, z) + (Lu)(x, y, z) + (Mu)(x, y, z),

(2.5.1)

for x, y ∈ R+ , z ∈ I = [a, b] (a < b), where x y b

(Lu)(x, y, z) =

F(x, y, z, s,t, r, u(s,t, r)) dr dt ds, 0

0

(2.5.2)

a

∞ ∞ b

(Mu)(x, y, z) =

H(x, y, z, s,t, r, u(s,t, r)) dr dt ds, 0

0

(2.5.3)

a

e, F, H are the given functions and u is the unknown function. The roots of the special version of equation (2.5.1) can be found in the work of Lovelady [64] in the field of partial differential equations, see also [5,18,132]. The main aim of the present section is to offer some basic properties of solutions of equation (2.5.1), recently studied in [112]. Let G be as defined in Section 2.3 and E3 = {(x, y, z, s,t, r) ∈ G2 : 0 s x < ∞, 0 t y < ∞, z, r ∈ I}. Throughout, we assume that e ∈ C(G, R), F ∈ C(E3 × R, R), H ∈ C(G2 × R, R). Let Ω be the space of functions φ ∈ C(G, R) which fulfil the condition |φ (x, y, z)| = O(exp(λ (x + y + |z|))),

(2.5.4)

for (x, y, z) ∈ G, where λ > 0 is a constant. In the space Ω we define the norm |φ |Ω = sup [|φ (x, y, z)| exp(−λ (x + y + |z|))]. (x,y,z)∈G

(2.5.5)


79

It is easy to see that Ω with the norm defined in (2.5.5) is a Banach space and |φ |Ω N,

(2.5.6)

where N 0 is a constant. First, we formulate the following theorem concerning the existence of a unique solution of equation (2.5.1). Theorem 2.5.1. Suppose that (i) the functions F, H in equation (2.5.1) satisfy the conditions |F(x, y, z, s,t, r, u) − F(x, y, z, s,t, r, u)| p(x, y, z, s,t, r)|u − u|,

(2.5.7)

|H(x, y, z, s,t, r, u) − H(x, y, z, s,t, r, u)| q(x, y, z, s,t, r)|u − u|,

(2.5.8)

where p ∈ C(E3 , R+ ), q ∈ C(G2 , R+ ), (ii) for λ as in (2.5.4), (j1 ) there exists a nonnegative constant α such that α < 1 and x y b 0

+

0

a

p(x, y, z, s,t, r) exp(λ (s + t + |r|)) dr dt ds

∞ ∞ b 0

0

a

q(x, y, z, s,t, r) exp(λ (s + t + |r|)) dr dt ds α exp(λ (x + y + |z|)),

(2.5.9)

(j2 ) there exists a nonnegative constant β such that |e(x, y, z) + (L0)(x, y, z) + (M0)(x, y, z)| β exp(λ (x + y + |z|)),

(2.5.10)

where e, L, M are as in equation (2.5.1). Then the equation (2.5.1) has a unique solution u(x, y, z) on G in Ω. The proof is analogous to the proof of Theorem 1.3.1. Here, we omit the details. Next, we shall give the following theorem concerning the uniqueness of solutions of equation (2.5.1). Theorem 2.5.2.

Suppose that the functions F, H in equation (2.5.1) satisfy the conditions

|F(x, y, z, s,t, r, u) − F(x, y, z, s,t, r, u)| q(x, y, z) f (s,t, r)|u − u|,

(2.5.11)

|H(x, y, z, s,t, r, u) − H(x, y, z, s,t, r, u)| c(x, y, z)g(s,t, r)|u − u|,

(2.5.12)

where q, f , c, g ∈ C(G, R+ ). Let α1 , D1 , A1 , B1 be as in Theorem 2.3.4 part (b7 ). Then the equation (2.5.1) has at most one solution on G.

80

Proof.


Let u1 (x, y, z) and u2 (x, y, z) be two solutions of equation (2.5.1). Then by using

the hypotheses, we have

+

|u1 (x, y, z) − u2 (x, y, z)|

x y b 0

0

a

0

a

|F(x, y, z, s,t, r, u1 (s,t, r)) − F(x, y, z, s,t, r, u2 (s,t, r))| dr dt ds

∞ ∞ b 0

|H(x, y, z, s,t, r, u1 (s,t, r)) − H(x, y, z, s,t, r, u2 (s,t, r))| dr dt ds

q(x, y, z) +c(x, y, z)

x y b 0

0

∞ ∞ b 0

0

a

a

f (s,t, r)|u1 (s,t, r) − u2 (s,t, r)| dr dt ds

g(s,t, r)|u1 (s,t, r) − u2 (s,t, r)| dr dt ds.

(2.5.13)

Here, it is easy to see that A1 (x, y, z) and D1 defined in (2.3.16) and (2.3.18) reduces to A1 (x, y, z) = 0 and D1 = 0. Now a suitable application of Theorem 2.3.4 part (b7 ) to (2.5.13) yields |u1 (x, y, z) − u2 (x, y, z)| 0, and hence u1 (x, y, z) = u2 (x, y, z). Thus, there is at most one solution to equation (2.5.1) on G. The following theorems deal with the estimates on the solution of equation (2.5.1). Theorem 2.5.3.

Suppose that the functions F, H in equation (2.5.1) satisfy the conditions |F(x, y, z, s,t, r, u)| q(x, y, z) f (s,t, r)|u|,

(2.5.14)

|H(x, y, z, s,t, r, u)| c(x, y, z)g(s,t, r)|u|,

(2.5.15)

where q, f , c, g ∈ C(G, R+ ). Let α1 , B1 be as in Theorem 2.3.4 part (b7 ) and ∞ ∞ b 1 D2 = g(s,t, r)A2 (s,t, r) dr dt ds, (2.5.16) 1 − α1 0 0 a where A2 (x, y, z) is defined by the right hand side of (2.3.16) by replacing p(x, y, z) by |e(x, y, z)|. If u(x, y, z) is any solution of equation (2.5.1) on G, then |u(x, y, z)| A2 (x, y, z) + D2 B1 (x, y, z),

(2.5.17)

for (x, y, z) ∈ G. Proof.

Using the fact that u(x, y, z) is a solution of equation (2.5.1) and hypotheses, we

have |u(x, y, z)| |e(x, y, z)| + +

∞ ∞ b 0

0

a

x y b 0

a

|F(x, y, z, s,t, r, u(s,t, r))| dr dt ds

|H(x, y, z, s,t, r, u(s,t, r))| dr dt ds

|e(x, y, z)| + q(x, y, z) +c(x, y, z)

0

x y b

f (s,t, r)|u(s,t, r)| dr dt ds 0

∞ ∞ b

0

a

g(s,t, r)|u(s,t, r)| dr dt ds. 0

0

a

Now, an application of Theorem 2.3.4 part (b7 ) to (2.5.18) yields (2.5.17).

(2.5.18)


81

Suppose that the functions F, H in equation (2.5.1) satisfy the conditions

Theorem 2.5.4.

(2.5.11), (2.5.12). Let α1 , B1 be as in Theorem 2.3.4 part (b7 ) and x y b

e0 (x, y, z) = +

0

0

∞ ∞ b 0

0

a

a

|F(x, y, z, s,t, r, e(s,t, r))| dr dt ds

|H(x, y, z, s,t, r, e(s,t, r))| dr dt ds,

(2.5.19)

∞ ∞ b 1 g(s,t, r)A3 (s,t, r) dr dt ds, (2.5.20) 1 − α1 0 0 a where A3 (x, y, z) is defined by the right hand side of (2.3.16) by replacing p(x, y, z) by

D3 =

e0 (x, y, z). If u(x, y, z) is any solution of equation (2.5.1) on G, then |u(x, y, z) − e(x, y, z)| A3 (x, y, z) + D3 B1 (x, y, z),

(2.5.21)


Using the fact that u(x, y, z) is a solution of equation (2.5.1) and the hypotheses,

we have |u(x, y, z) − e(x, y, z)|

x y b 0

0

a

|F(x, y, z, s,t, r, u(s,t, r)) − F(x, y, z, s,t, r, e(s,t, r))| dr dt ds x y b

+ 0

+

∞ ∞ b 0

0

a

0

a

|F(x, y, z, s,t, r, e(s,t, r))| dr dt ds

|H(x, y, z, s,t, r, u(s,t, r)) − H(x, y, z, s,t, r, e(s,t, r))| dr dt ds ∞ ∞ b

+ 0

0

a

e0 (x, y, z) + q(x, y, z) +c(x, y, z)

|H(x, y, z, s,t, r, e(s,t, r))| dr dt ds

x y b 0

0

∞ ∞ b 0

0

a

a

f (s,t, r)|u(s,t, r) − e(s,t, r)| dr dt ds

g(s,t, r)|u(s,t, r) − e(s,t, r)| dr dt ds.

(2.5.22)

Applying Theorem 2.3.4 part (b7 ) to (2.5.22), we get (2.5.21). Now, we present the following theorem which deals with the estimate on the difference between the solutions of equation (2.5.1) and the equation of the form x y b

v(x, y, z) = e(x, y, z) +

F(x, y, z, s,t, r, v(s,t, r)) dr dt ds, 0

0

a

for (x, y, z) ∈ G, where the functions e, F are as in equation (2.5.1).

(2.5.23)

82


Suppose that the functions F, H in equations (2.5.1), (2.5.23) satisfy

Theorem 2.5.5.

the conditions (2.5.11), (2.5.12) and H(x, y, z, s,t, r, 0) = 0. Let v(x, y, z) be a solution of equation (2.5.23) on G such that |v(x, y, z)| Q, where Q 0 is a constant. Let α1 , B1 be as in Theorem 2.3.4 part (b7 ) and p(x, y, z) = Qc(x, y, z)

∞ ∞ b

g(s,t, r) dr dt ds, 0

0

(2.5.24)

a

∞ ∞ b 1 g(s,t, r)A4 (s,t, r) dr dt ds, (2.5.25) 1 − α1 0 0 a where A4 (x, y, z) is defined by the right hand side of (2.3.16) by replacing p(x, y, z) by

D4 =

p(x, y, z). If u(x, y, z) is a solution of equation (2.5.1) on G, then |u(x, y, z) − v(x, y, z)| A4 (x, y, z) + D4 B1 (x, y, z),

(2.5.26)


Using the facts that u(x, y, z) and v(x, y, z) are the solutions of equations (2.5.1) and

(2.5.23) and hypotheses, we observe that |u(x, y, z) − v(x, y, z)| +

x y b 0

0

a

∞ ∞ b 0

+

0

a

|F(x, y, z, s,t, r, u(s,t, r)) − F(x, y, z, s,t, r, v(s,t, r))| dr dt ds |H(x, y, z, s,t, r, u(s,t, r)) − H(x, y, z, s,t, r, v(s,t, r))| dr dt ds

∞ ∞ b 0

0

a

|H(x, y, z, s,t, r, v(s,t, r)) − H(x, y, z, s,t, r, 0)| dr dt ds

q(x, y, z) +c(x, y, z)

x y b 0

0

∞ ∞ b 0

+c(x, y, z)

0

a

p(x, y, z) + q(x, y, z)

0

0

a

x y b 0

0

∞ ∞ b 0

g(s,t, r)|u(s,t, r) − v(s,t, r)| dr dt ds

∞ ∞ b 0

+c(x, y, z)

f (s,t, r)|u(s,t, r) − v(s,t, r)| dr dt ds

a

a

a

g(s,t, r) |v(s,t, r)| dr dt ds f (s,t, r)|u(s,t, r) − v(s,t, r)| dr dt ds

g(s,t, r)|u(s,t, r) − v(s,t, r)| dr dt ds.

Applying Theorem 2.3.4 part (b7 ) to (2.5.27), we get (2.5.26).

(2.5.27)


83

In concluding we note that the idea used in this section can be used to formulate the results similar to those given above for the equations of the form (2.5.1) when the operator (Lu)(x, y, z) defined in (2.5.2) is replaced by ∞ ∞ b

F(x, y, z, s,t, r, u(s,t, r)) dr dt ds, x

or

y

a

x ∞ b

F(x, y, z, s,t, r, u(s,t, r)) dr dt ds. 0

y

a

Furthermore, one can formulate results on the continuous dependence of solutions and ε approximate solutions of equations of the form (2.5.1) by making use of suitable inequalities given in Section 2.3 or their variants. We leave the details of such results to the reader to fill in where needed. 2.6

Hyperbolic-type Fredholm integrodifferential equation

In this section we shall be concerned with the hyperbolic type Fredholm integrodifferential equation D2 D2 u(x, y, z) = F(x, y, z, u(x, y, z), D1 u(x, y, z), D2 u(x, y, z), (Hu)(x, y, z)),

(2.6.1)

with the given data u(x, 0, z) = σ (x, z),

u(0, y, z) = τ (y, z),

(2.6.2)

for x, y ∈ R+ , z ∈ I = [a, b] ⊂ R (a < b), where b

(Hu)(x, y, z) = a

K(x, y, z, r, u(x, y, r), D1 u(x, y, r), D2 u(x, y, r)) dr,

F, K are the given functions and u is the unknown function. Obviously, (H0)(x, y, z) =

b a

K(x, y, z, r, 0, 0, 0) dr. The origin of problem (2.6.1)–(2.6.2) can be traced back in

the work of Lovelady [64], who studied the existence and uniqueness of solutions of special form of equation (2.6.1) with the given data in (2.6.2). Inspired by the results in [64], recently in [111] the present author has studied some basic aspects of solutions of problem (2.6.1)–(2.6.2). Our main goal here is to present the results given in [111], which deals with some important qualitative properties of solutions of problem (2.6.1)–(2.6.2) and its special version. Throughout, we assume that F ∈ C(G × R4 , R), K ∈ C(G × I × R3 , R), σ , σx , τ , τy ∈ C(R+ × I, R), in which G is defined as in Section 1.3. By a solution of problem (2.6.1)–(2.6.2) we mean a function u ∈ C(G, R) which satisfy the equations (2.6.1), (2.6.2). For u, D1 u, D2 u ∈ C(G, R), we denote by

84


|u(x, y, z)|0 = |u(x, y, z)| + |D1 u(x, y, z)| + |D2 u(x, y, z)|. Let V be the space of functions u, D1 u, D2 u ∈ C(G, R) which fulfil the condition |u(x, y, z)|0 = O exp(λ (x + y + |z|)) ,

(2.6.3)

for (x, y, z) ∈ G, where λ > 0 is a constant. In the space V we define the norm |u|V = sup [|u(x, y, z)|0 exp(−λ (x + y + |z|))].

(2.6.4)

(x,y,z)∈G

It is easy to see that V with norm defined in (2.6.4) is a Banach space and |u|V N,

(2.6.5)

where N 0 is a constant. The following theorem ensures the existence of a unique solution to problem (2.6.1)– (2.6.2). Theorem 2.6.1. Suppose that (i) the functions F, K in (2.6.1) satisfy the conditions |F(x, y, z, u1 , u2 , u3 , u4 ) − F(x, y, z, u1 , u2 , u3 , u4 )| L(x, y, z) [|u1 − u1 | + |u2 − u2 | + |u3 − u3 | + |u4 − u4 |] ,

(2.6.6)

|K(x, y, z, r, u1 , u2 , u3 ) − K(x, y, z, r, u1 , u2 , u3 )| M(x, y, z, r)[|u1 − u1 | + |u2 − u2 | + |u3 − u3 |],

(2.6.7)

where L ∈ C(G, R+ ), M ∈ C(G × I, R+ ), (ii) for λ as in (2.6.3), ( j1 ) there exist nonnegative constants αi (i = 1, 2, 3) such that x y 0

0

y 0

x 0

where

P(s,t, z) dt ds α1 exp(λ (x + y + |z|)),

(2.6.8)

P(x,t, z) dt α2 exp(λ (x + y + |z|)),

(2.6.9)

P(s, y, z) ds α3 exp(λ (x + y + |z|)),

(2.6.10)

P(x, y, z) = L(x, y, z) exp(λ (x + y + |z|)) +

b a

M(x, y, z, r) exp(λ (x + y + |r|))dr ,


85

(j2 ) there exist nonnegative constants βi (i = 1, 2, 3) such that |σ (x, z) + τ (y, z) − σ (0, z)| +

x y 0

0

|F(s,t, z, 0, 0, 0, (H0)(s,t, z))|dt ds

β1 exp(λ (x + y + |z|)),

|σx (x, z)| + |τy (y, z)| +

y 0x 0

(2.6.11)

|F(x,t, z, 0, 0, 0, (H0)(x,t, z))|dt β2 exp(λ (x + y + |z|)),

(2.6.12)

|F(s, y, z, 0, 0, 0, (H0)(s, y, z))|ds β3 exp(λ (x + y + |z|)),

(2.6.13)

where σ , τ are as in (2.6.2). If α = α1 + α2 + α3 < 1, then the problem (2.6.1)–(2.6.2) has a unique solution u(x, y, z) on G in V . Proof. Let u ∈ V and define the operator T by (Tu)(x, y, z) = σ (x, z) + τ (y, z) − σ (0, z) x y

+ 0

0

F(s,t, z, u(s,t, z), D1 u(s,t, z), D2 u(s,t, z), (Hu)(s,t, z)) dt ds.

(2.6.14)

First, we shall show that Tu maps V into itself. Evidently, Tu is continuous on G and Tu ∈ R. We verify that (2.6.3) is fulfilled. From (2.6.14), hypotheses and (2.6.5), we have |(Tu)(x, y, z)| |σ (x, z) + τ (y, z) − σ (0, z)| x y

+ 0

0

|F(s,t, z, u(s,t, z), D1 u(s,t, z), D2 u(s,t, z), (Hu)(s,t, z)) −F(s,t, z, 0, 0, 0, (H0)(s,t, z))|dt ds x y

+ 0

x y

+ 0

0

0

|F(s,t, z, 0, 0, 0, (H0)(s,t, z))|dt ds

β1 exp(λ (x + y + |z|)) b L(s,t, z) |u(s,t, z)|0 + M(s,t, z, r)|u(s,t, r)|0 dr dt ds a

β1 exp(λ (x + y + |z|)) + |u|V

x y

P(s,t, z) dt ds 0

0

[β1 + N α1 ] exp(λ (x + y + |z|)).

(2.6.15)

Differentiating both sides of (2.6.14) with respect to x, using hypotheses and (2.6.5), we have |D1 (Tu)(x, y, z)| |σx (x, z)| +

y 0

|F(x,t, z, u(x,t, z), D1 u(x,t, z), D2 u(x,t, z), (Hu)(x,t, z))

86


−F(x,t, z, 0, 0, 0, (H0)(x,t, z))|dt +

y 0

|F(x,t, z, 0, 0, 0, (H0)(x,t, z))|dt

β2 exp(λ (x + y + |z|)) + |u|V

y

P(x,t, z) dt 0

[β2 + N α2 ] exp(λ (x + y + |z|)).

(2.6.16)

|D2 (Tu)(x, y, z)| [β3 + N α3 ] exp(λ (x + y + |z|)).

(2.6.17)

Similarly, we obtain

From (2.6.15)–(2.6.17), we observe that |Tu|V [β1 + β1 + β1 + N α ]. This shows that T maps V into itself. Next, we verify that the operator T is a contraction map. Let u, u ∈ V . From (2.6.14) and using the hypotheses, we have |(Tu)(x, y, z) − (T u)(x, y, z)|

x y 0

0

|F(s,t, z, u(s,t, z), D1 u(s,t, z), D2 u(s,t, z), (Hu)(s,t, z))

−F(s,t, z, u(s,t, z), D1 u(s,t, z), D2 u(s,t, z), (Hu)(s,t, z))|dt ds |u − u|V

x y

P(s,t, z) dt ds 0

0

|u − u|V α1 exp(λ (x + y + |z|)).

(2.6.18)

Similarly, differentiating both sides of (2.6.14), with respect to x and with respect to y and using hypotheses, we obtain D1 (Tu)(x, y, z) − D1 (T u)(x, y, z) |u − u|V α2 exp(λ (x + y + |z|))

(2.6.19)

and |D2 (Tu)(x, y, z) − D2 (T u)(x, y, z)| |u − u|V α3 exp(λ (x + y + |z|)).

(2.6.20)

From (2.6.18)–(2.6.20), we obtain |Tu − T u|V α |u − u|V .

(2.6.21)

Since α < 1, from (2.6.21), it follows from Banach fixed point theorem (see [51, p. 372]) that T has a unique fixed point in V . The fixed point of T is however a solution of problem (2.6.1)–(2.6.2). The proof is complete.


Remark 2.6.1.

87

We note that in [64], the existence and uniqueness of solutions to problem

(2.6.1)–(2.6.2) have been analyzed as an application of the fixed point theorem established therein to study perturbed differential equations. The result in Theorem 2.6.1 ensure the existence of a unique solution to more general problem (2.6.1)–(2.6.2) under different conditions from those used in [64]. Below, we study some fundamental qualitative properties of solutions of a hyperbolic type Fredholm integrodifferential equation of the form (see [111]) D2 D1 u(x, y, z) = f (x, y, z, u(x, y, z), (hu)(x, y, z)),

(2.6.22)

with the given data (2.6.2) for x, y ∈ R+ , z ∈ I, where b

(hu)(x, y, z) =

k(x, y, z, r, u(x, y, r)) dr, a

f ∈ C(G × R2 , R), k ∈ C(G × I × R, R). First, we shall give the following theorem which deals with the uniqueness of solutions of problem (2.6.22)–(2.6.2) on G in R. Theorem 2.6.2. Assume that the functions f , k in (2.6.22) satisfy the conditions | f (x, y, z, u, v) − f (x, y, z, u, v)| p(x, y, z)|u − u| + |v − v|,

(2.6.23)

|k(x, y, z, r, u) − k(x, y, z, r, u)| q(x, y, z, r)|u − u|,

(2.6.24)

where p ∈ C(G, R+ ), q ∈ C(G × I, R+ ) and ∞ ∞ 0

∞ ∞ b 0

0

a

0

p(s,t, z) dt ds < ∞,

q(s,t, z, r)H(s,t, r) dr dt ds < ∞,

(2.6.25)

for z ∈ I, where H(x, y, z) is given by (2.3.27). Then the problem (2.6.22)–(2.6.2) has at most one solution on G in R. Proof. Let u1 (x, y, z) and u2 (x, y, z) be two solutions of problem (2.6.22)–(2.6.2) on G. Using these facts and the hypotheses, we have x y f (s,t, z, u1 (s,t, z), (hu1 )(s,t, z)) |u1 (x, y, z) − u2 (x, y, z)| 0

0

− f (s,t, z, u2 (s,t, z), (hu2 )(s,t, z))dt ds

x y 0

0

p(s,t, z)|u1 (s,t, z) − u2 (s,t, z)| + |(hu1 )(s,t, z) − (hu2 )(s,t, z| dt ds

88


x y 0

0

p(s,t, z)|u1 (s,t, z) − u2 (s,t, z)| +

b a

q(s,t, z, r)|u1 (s,t, r) − u2 (s,t, r)|dr dt ds. (2.6.26)

Now an application of Theorem 2.3.5 (when c = 0) to (2.6.26) yields |u1 (x, y, z) − u2 (x, y, z)| 0, which implies u1 (x, y, z) = u2 (x, y, z). Thus there is at most one solution to the problem (2.6.22)–(2.6.2) on G in R. Next, we shall give the following theorem which deals with the bound on the solution of problem (2.6.22)–(2.6.2). Theorem 2.6.3.

Suppose That the functions f , k, σ , τ in (2.6.22)–(2.6.2) satisfy the con-

ditions | f (x, y, z, u, v)| p(x, y, z)|u| + |v|,

(2.6.27)

|k(x, y, z, r, u)| q(x, y, z, r)|u|

(2.6.28)

|σ (x, z) + τ (y, z) − σ (0, z)| c,

(2.6.29)

where p ∈ C(G, R+ ), q ∈ C(G × I, R+ ), c 0 is a constant. If u(x, y, z) is any solution of problem (2.6.22)–(2.6.2) on G, then x y b q(s,t, z, r)H(s,t, r) dr dt ds , |u(x, y, z)| cH(x, y, z) exp 0

0

(2.6.30)

a

for (x, y, z) ∈ G, where H(x, y, z) is given by (2.3.27). Proof.

Using the fact that u(x, y, z) is a solution of problem (2.6.22)–(2.6.2) and hypothe-

ses, we have |u(x, y, z)| |σ (x, z) + τ (y, z) − σ (0, z)| + c+

x y

b

p(s,t, z)|u(s,t, z)| + 0

0

x y 0

0

| f (s,t, z, u(s,t, z), (hu)(s,t, z))|dt ds

q(s,t, z, r)|u(s,t, r)|dr dt ds.

(2.6.31)

a

Applying Theorem 2.3.5 to (2.6.31) yields (2.6.30). The following theorem deals with the dependency of solutions of equation (2.6.22) on given data. Theorem 2.6.4. Suppose that the functions f , k in (2.6.22) satisfy the conditions (2.6.23), (2.6.24). Let u(x, y, z) and v(x, y, z) be the solutions of equation (2.6.22) with the given data (2.6.2) and v(x, 0, z) = σ (x, z),

v(0, y, z) = τ (x, z),

(2.6.32)


89

respectively, where σ , τ , σ x , τ y ∈ C(R+ × I, R) and σ (x, z) + τ (y, z) − σ (0, z) − {σ (x, z) + τ (y, z) − σ (0, z)} d, where d 0 is a constant. Then |u(x, y, z) − v(x, y, z)| dH(x, y, z) exp

x y 0

0

b

(2.6.33)

q(s,t, z, r)H(s,t, r)dr dt ds , (2.6.34)

a

for (x, y, z) ∈ G, where H(x, y, z) is given by (2.3.27). Proof.

Using the facts that u(x, y, z) and v(x, y, z) are the solutions of problems (2.6.22)–

(2.6.2) and (2.6.22)–(2.6.32) and the hypotheses, we have |u(x, y, z) − v(x, y, z)| |σ (x, z) + τ (y, z) − σ (0, z) − {σ (x, z) + τ (y, z) − σ (0, z)}| x y

+ 0

0

| f (s,t, z, u(s,t, z), (hu)(s,t, z)) − f (s,t, z, v(s,t, z), (hv)(s,t, z))|dt ds d+ b

+ a

x y 0

0

p(s,t, z)|u(s,t, z) − v(s,t, z)|

q(s,t, z, r)|u(s,t, r) − v(s,t, r)|dr dt ds.

(2.6.35)

Now an application of Theorem 2.3.5 to (2.6.35) yields the estimate (2.6.34), which shows the dependency of solutions of equation (2.6.22) on given data. Remark 2.6.2. In general one may expect that the analysis used to study the properties of solutions of problem (2.6.22)–(2.6.2) in Theorems 2.6.2–2.6.4 will also be equally useful in the study of general problem (2.6.1)–(2.6.2). In fact, it involves the task of designing a new inequality, similar to the one given in Theorem 2.3.5, which will allow applications in the discussion of problem (2.6.1)–(2.6.2). Indeed, it is not an easy matter and any attempt to extend the analysis to general problem (2.6.1)–(2.6.2) is of great interest and importance and its detailed treatment is desired.

2.7 2.7.1

Miscellanea Haçia and Kaczmarek [41]

(h1 ) Let u, w, a, b ∈ C(E, R+ ) and b(x, y) > 0. If u(x, y) satisfies u(x, y) w(x, y) + b(x, y) then u(x, y) b(x, y)h(x, y) exp

x y

a(s,t)u(s,t) ds dt, 0

0

x 0

y 0

a(s,t)b(s,t) ds dt ,

90


where

h(x, y) = sup

w(s,t) : 0 s x, 0 t y . b(s,t)

(h2 ) Let u, w ∈ C(E, R+ ), K ∈ C(E2 , R+ ) and p(x, y) = sup K(x, y, s,t) > 0. 0sx 0ty

If u(x, y) satisfies u(x, y) w(x, y) + then

u(x, y) p(x, y) sup

2.7.2

x y

K(x, y, s,t)u(s,t) ds dt, 0

0

x y w(s,t) : 0 s x, 0 t y exp p(s,t) ds dt . p(s,t) 0 0

Haçia [40]

(h3 ) Let f be a continuous function in D = {(x,t) : a x b, t 0} and K be nonnegative and continuous in Ω = {(x,t, y, s) : a x, y b, 0 s t < ∞}. If the continuous function u satisfies the inequality u(x,t) f (x,t) + for (x,t) ∈ D, then u(x,t) f (x,t) +

t b

K(x,t, y, s)u(y, s) dy ds, 0

a

t b

r(x,t, y, s) f (y, s) dy ds, 0

a

where

∞

r(x,t, y, s) =

∑ Kn (x,t, y, s),

n=0

is the resolvent kernel defined by formulas K0 (x,t, y, s) = K(x,t, y, s), t b

Kn (x,t, y, s) =

s

a

K(x,t, p, q)Kn−1 (p, q, y, s) d p dq,

for n = 1, 2, . . .. (h4 ) Let u, f , A, B be continuous in D and A · B is nonnegative. If u satisfies the inequality u(x,t) f (x,t) + A(x,t) for (x,t) ∈ D, then u(x,t) f (x,t) + A(x,t) for (x,t) ∈ D.

t b

B(y, s)u(y, s) dy ds, 0

a

t

t b

b

B(y, s) exp 0

a

s

a

A(z, τ )B(z, τ ) dz d τ f (y, s) dy ds,


2.7.3

91

Haçia [40]

Consider the following nonlinear integral equation of the Volterra-Fredholm-type t b

u(x,t) = f (x,t) +

K(x,t, y, s, u(y, s)) dy ds, 0

(2.7.1)

a

with assumptions: (γ1 ) f and K are continuous in D and Ω × R respectively, (γ2 ) |K(x,t, y, s, u)| B(y, s)|u|, (γ3 ) |K(x,t, y, s, u) − K(x,t, y, s, u)| B(y, s)|u − u|, where B is continuous and integrable in D. (h5 ) If assumptions (γ1 ) and (γ2 ) are satisfied, then a solution u(x,t) of equation (2.7.1) is bounded in D and |u(x,t)| ψ (t) exp

t 0

b

B(y, s) dy ds ,

a

where ψ (t) = sup{| f (x,t)| : a x b, 0 s t}. (h6 ) If assumptions (γ1 ) and (γ3 ) are satisfied, then the equation (2.7.1) has at most one solution, which is stable. 2.7.4

Pachpatte [108]

Consider the integral equation of the form

x s t

u(x,t) = h(x,t) + 0

0

0

F(x,t, σ , τ , u(σ , τ )) d τ d σ ds,

(2.7.2)

where h ∈ C(R2+ , R), F ∈ C(R4+ × R, R) and u is the unknown function. (h7 ) Suppose that the functions F, h in equation (2.7.2) satisfy the conditions |F(x,t, σ , τ , u) − F(x,t, σ , τ , v)| q(x,t)r(σ , τ )|u − v|, x s t h(x,t) + p(x,t), F(x,t, σ , τ , 0) d τ d σ ds 0

where p, q, r ∈

C(R2+ , R+ ).

0

0

If u(x,t) is any solution of equation (2.7.2) for (x,t) ∈ R2+ ,

then |u(x,t)| p(x,t) + q(x,t) × exp for (x,t) ∈ R2+ .

(2.7.3)

x s

x s 0

0

0 t

0

0

0

t

r(σ , τ )p(σ , τ ) d τ d σ ds

r(σ , τ )q(σ , τ ) d τ d σ ds ,

92


(h8 ) Let ui (x,t) (i = 1, 2) be respectively εi -approximate solutions of equation (2.7.2) for (x,t) ∈ R2+ i.e.,

x s t ui (x,t) − h(x,t) + εi , F(x,t, σ , τ , u ( σ , τ )) d τ d σ ds i 0 0 0 for (x,t) ∈ R2+ . Suppose that the function F in equation (2.7.2) satisfies the condition (2.7.3). Then

|u1 (x,t) − u2 (x,t)| (ε1 + ε2 ) 1 + q(x,t) × exp

x s 0

0

t 0

x s 0

0

t 0

r(σ , τ ) d τ d σ ds

r(σ , τ )q(σ , τ ) d τ d σ ds

,

for (x,t) ∈ R2+ . 2.7.5

Pachpatte [95]

Consider the integral equation of the form x y b

u(x, y, z) = h(x, y, z) +


0

(2.7.4)

a

for (x, y, z) ∈ G, where h ∈ C(G, R), F ∈ C(G2 × R, R) and G is as defined in section 2.3. (h9 ) Suppose that the function F in equation (2.7.4) satisfies the condition |F(x, y, z, s,t, r, u) − F(x, y, z, s,t, r, v)| q(x, y, z) f (s,t, r)|u − v|,

(2.7.5)

where q, f ∈ C(G, R+ ). If u(x, y, z) is any solution of equation (2.7.4) on G, then x y b |u(x, y, z) − h(x, y, z)| d(x, y, z) + q(x, y, z) f (s,t, r) d(s,t, r) dr dt ds 0

× exp for (x, y, z) ∈ G, where

x y 0

0

b

x y b 0

a


a

d(x, y, z) = 0

0

a

|F(x, y, z, s,t, r, h(s,t, r))| dr dt ds.

(h10 ) Suppose that the function F in equation (2.7.4) satisfies the condition (2.7.5). Then for every given solution v ∈ C(G, R) of equation x y b

v(x, y, z) = g(x, y, z) +

L(x, y, z, s,t, r, v(s,t, r)) dr dt ds, 0

0

a

where g ∈ C(G, R), L ∈ C(G2 × R, R) and any solution u ∈ C(G, R) of equation (2.7.4), the estimation |u(x, y, z) − v(x, y, z)| [h(x, y, z) + d(x, y, z)]


x y

93

b

+q(x, y, z)

f (s,t, r)[h(s,t, r) + d(s,t, r)] dr dt ds 0

× exp

0

a

x y 0

0

b


a

holds for (x, y, z) ∈ G, in which h(x, y, z) = |h(x, y, z) − g(x, y, z)|, x y b

d(x, y, z) = 0

0

a

|F(x, y, z, s,t, r, v(s,t, r)) − L(x, y, z, s,t, r, v(s,t, r))| dr dt ds,

for (x, y, z) ∈ G. 2.7.6

Pachpatte [112]

(h11 ) Let u, p, q ∈ C(G, R+ ) and L ∈ C (G × R+ , R+ ) be such that 0 L(x, y, z, u) − L(x, y, z, v) M(x, y, z, v)(u − v),

(2.7.6)

for u v 0, where M ∈ C(G × R+ , R+ ) and G is as defined in section 2.3. If u(x, y, z) p(x, y, z) + q(x, y, z) for (x, y, z) ∈ G, then u(x, y, z) p(x, y, z) + q(x, y, z) × exp

x y 0

0

x y b

L(s,t, r, u(s,t, r)) dr dt ds, 0

0

a

x y

b

L(s,t, r, p(s,t, r)) dr dt ds 0

0

a

M(s,t, r, p(s,t, r))q(s,t, r) dr dt ds ,

b

a

for (x, y, z) ∈ G. (h12 ) Let u, f , g ∈ C(G, R+ ) and L is as defined in (h11 ), which verifies the condition (2.7.6) and k 0 is a real constant. If u (x, y, z) k + 2 2

x y b

2

[ f (s,t, r)u(s,t, r)L(s,t, r, u(s,t, r)) + g(s,t, r)u(s,t, r)]dr dt ds, 0 0 a

for (x, y, z) ∈ G, then u(x, y, z) n(x, y) + × exp

x y

f (s,t, r)L(s,t, r, n(s,t)) dr dt ds 0

x y 0

0

b

b

0

a

f (s,t, r)M(s,t, r, n(s,t)) dr dt ds ,

a

for (x, y, z) ∈ G, where M ∈ C(G × R+ , R+ ) and x y b

n(x, y) = k + 0

for x, y ∈ R+ .

0

a

g(σ , τ , η ) d η d τ d σ ,

94

2.7.7


Pachpatte [102]

(h13 ) Let u, p, q, c, f , g ∈ C(G, R+ ) and suppose that u(x, y, z) p(x, y, z) + q(x, y, z)

+c(x, y, z)

x ∞ b 0

y

f (s,t, r)u(s,t, r) dr dt ds

a

∞ ∞ b

g(s,t, r)u(s,t, r) dr dt ds, 0

0

a

for (x, y, z) ∈ G, where G is as defined in section 2.3. If

α3 =

∞ ∞ b 0

0


a

then u(x, y, z) A3 (x, y, z) + D3 B3 (x, y, z), for (x, y, z) ∈ G, where

x

A3 (x, y, z) = p(x, y, z) + q(x, y, z) × exp

x 0

∞ b y

0

∞ b y

a

x x 0

and D3 =

1 1 − α3

a


B3 (x, y, z) = c(x, y, z) + q(x, y, z) × exp


∞ b y

0

∞ b y

f (s,t, r)c(s,t, r) dr dt ds

a


a

∞ ∞ b 0

0

a


(h14 ) Let u, p, c, g ∈ C(G, R+ ) and suppose that u(x, y, z) p(x, y, z) + c(x, y, z) for (x, y, z) ∈ G. If

α0 =

∞ ∞ b 0

0

then

for (x, y, z) ∈ G.

g(s,t, r)u(s,t, r) dr dt ds, 0

0

a

g(s,t, r)c(s,t, r) dr dt ds < 1,

a

u(x, y, z) p(x, y, z) + c(x, y, z)

∞ ∞ b

1 1 − α0

∞ ∞ b 0

0

a

g(s,t, r)p(s,t, r) dr dt ds ,


2.7.8

95

Pachpatte [102]

Consider the following integral equation u(x, y, z) = h(x, y, z) +

x ∞ b


y

(2.7.7)

a

with assumptions (δ1 ) h ∈ C(G, R), F ∈ C(G2 × R, R), (δ2 ) |F(x, y, z, s,t, r, u)| q(x, y, z) f (s,t, r)|u|, (δ3 ) |F(x, y, z, s,t, r, u) − F(x, y, z, s,t, r, v)| q(x, y, z) f (s,t, r)|u − v|, where q, f ∈ C(G, R+ ) and G is as defined in section 2.3. (h15 ) Suppose that the assumptions (δ1 ) and (δ2 ) are satisfied. If u(x, y, z) is any solution of equation (2.7.7) on G, then |u(x, y, z)| |h(x, y, z)| + q(x, y, z) × exp

x 0

∞ b y

x 0

y

∞ b

f (s,t, r)|h(s,t, r)| dr dt ds

a


a

for (x, y, z) ∈ G. (h16 ) If assumptions (δ1 ) and (δ3 ) are satisfied, then the equation (2.7.7) has at most one solution on G.

2.8

Notes

Various approaches are developed by different researchers for studying multidimensional integral equations.

Sections 2.2 and 2.3 deals with some basic integral inequalities

with explicit estimates in two and three variables, recently established by Pachpatte [108,98,104,95,111,102] which will be equally important in handling the dynamic equations of various forms, when the earlier inequalities in the literature do not apply directly. The material in sections 2.4 and 2.5 contains some fundamental qualitative properties of solutions of various types of integral equations in two and three variables and is adapted from Pachpatte [104,112]. The results included in section 2.6 are recently obtained by Pachpatte in [111], which are motivated by the work of Lovelady in [64]. A detailed account including a comprehensive list of references related to such equations can be found in books by Walter [134] and Appell, Kalitvin and Zabrejko [5]. Section 2.7 is devoted to miscellanea related to the results given in earlier sections, which we hope will stimulate the reader’s interest.

Chapter 3

Mixed integral equations and inequalities

3.1

Introduction

In the study of many basic models in parabolic differential equations which describe diffusion or heat transfer phenomena and epidemiology, the integral equation of the form t

u(t, x) = f (t, x) +

k(t, x, s, y)g(u(s, y)) dy ds, 0

(3.1.1)

B

where B is a bounded domain in Rn , t ∈ R+ ; f , k, g are given functions and u is the unknown function, occur in a natural way, see [5,8,28,31,32,37-39,69,134 ]. The integral equation (3.1.1) appears to be Volterra-type in t, and of Fredholm-type with respect to x and hence it can be viewed as a mixed Volterra-Fredholm-type integral equation. In the general case solving integral equation (3.1.1) is highly nontrivial problem and handling the study of its qualitative properties need a fresh outlook. The method of integral inequalities with explicit estimates serve as an important tool which provides valuable information of various dynamic equations, without the need to know in advance the solutions explicitly. Recently in [99,103,105,107,116,119] explicit estimates on a number of new integral inequalities are considered and used in various applications. In the present chapter, we offer some fundamental mixed integral inequalities with explicit estimates established in the above noted papers and also focus our attention on some basic qualitative aspects of solutions of equations of the form (3.1.1). A particular feature of our approach here is that it is elementary and provide some basic results for future advanced studies in the field.

3.2

Volterra-Fredholm-type integral inequalities I

In this section we present some basic integral inequalities with explicit estimates investigated in [99,103,105,116], which can be used as tools for handling the equations 97

98


like (3.1.1). In what follows, we denote by B a bounded domain in Rn defined by n

B = ∏ [ai , bi ] (ai < bi ), i=1

x = (x1 , . . . , xn ) (xi ∈ R) is a variable point in B and dx = dx1 · · · dxn . For any continuous function z : B → R, we denote by bn an

···

b1 a1

B z(x) dx

the n-fold integral

z(x1 , . . . , xn ) dx1 · · · dxn .

Let D0 = R+ × B, Ω = {(t, x, s, y) : 0 s t < ∞; x, y ∈ B} and denote by D1 h(t, x, s, y) the partial derivative of a function h(t, x, s, y) defined on Ω with respect to the first variable. The first theorem deals with the inequalities given in [99]. Theorem 3.2.1. Let u, p, q, f ∈ C (D0 , R+ ). (c1 ) Let L ∈ C (D0 × R+ , R+ ) be such that 0 L(t, x, u) − L (t, x, v) M (t, x, v) (u − v),

(3.2.1)

for u v 0, where M ∈ C (D0 × R+ , R+ ). If u(t, x) p(t, x) + q(t, x)

t

L(s, y, u(s, y)) dy ds, 0

(3.2.2)

B

for (t, x) ∈ D0 , then t

u(t, x) p(t, x) + q(t, x) L(s, y, p(s, y)) 0 B t M(τ , z, p(τ , z))q(τ , z) dz d τ dy ds, × exp s

(3.2.3)

B

for (t, x) ∈ D0 . (c2 ) If u(t, x) p(t, x) + q(t, x)

t

f (s, y)u(s, y) dy ds, 0

(3.2.4)

B


u(t, x) p(t, x) + q(t, x) f (s, y)p(s, y) 0 B t f (τ , z)q(τ , z) dz d τ dy ds, × exp s

B

for (t, x) ∈ D0 . The inequalities established in [105] are given in the following theorems.

(3.2.5)


99

Theorem 3.2.2. Let u, p, q, f , g ∈ C(D0 , R+ ). (c3 ) If u(t, x) p(t, x) + q(t, x)

t

[ f (s, y)u(s, y) + g(s, y)] dy ds,

(3.2.6)

B

0


u(t, x) p(t, x) + q(t, x) [ f (s, y)p(s, y) + g(s, y)] 0 B t f (τ , z)q(τ , z) dz d τ dy ds, × exp s

(3.2.7)

B

for (t, x) ∈ D0 . (c4 ) Let c 0 and 0 < α < 1 be real constants. If u(t, x) c +

t 0

[ f (s, y)u(s, y) + g(s, y)uα (s, y)]dy ds,

(3.2.8)

B

for (t, x) ∈ D0 , then u(t, x) exp

t 0

B

f (τ , z)dz d τ

c1−α + (1 − α )

t

g(s, y) 0

1 s 1−α f (τ , z)dz d τ dy ds , × exp −(1 − α ) 0

B

(3.2.9)

B

for (t, x) ∈ D0 . Remark 3.2.1.

If we take g = 0 in (3.2.6), then the bound obtained in (3.2.7) reduces to

(3.2.5). In this case, we observe that the obtained result is a new variant of the inequality in Corollary 4.3.1 given in [82, p. 329]. We note that the bound on the unknown function u(t, x) involved in (3.2.8) when α = 1, 1 < α < ∞ can be obtained by closely looking at the proof of the inequality given in [82, Theorem 2.7.4, p. 153]. By taking (i) g = 0 and (ii) f = 0 in (3.2.8), it is easy to see that the bound obtained in (3.2.9) reduces respectively to u(t, x) c exp

t 0

B

f (τ , z) dz d τ ,

(3.2.10)

and 1 t 1−α u(t, x) c1−α + (1 − α ) g(s, y) dy ds , 0

for (t, x) ∈ D0 .

B

(3.2.11)

100


Theorem 3.2.3. Let u, f , g ∈ C(D0 , R+ ) and k 0, c 1, β > 1 be real constants. (c5 ) If uβ (t, x) kβ + β for (t, x) ∈ D0 , then u(t, x) exp

t

f (s, y)u(s, y) + g(s, y)uβ (s, y) dy ds,

(3.2.12)

B

0

t 0

B

g(τ , z) dz d τ

kβ −1 + (β − 1)

t

f (s, y) 0

B

1 s β −1 g(τ , z) dz d τ dy ds , × exp −(β − 1) 0

(3.2.13)

B

for (t, x) ∈ D0 . (c6 ) If u ∈ C(D0 , R1 ) and u(t, x) c +

t

f (s, y)u(s, y) log u(s, y) dy ds, 0

(3.2.14)

B

for (t, x) ∈ D0 , then u(t, x) cexp(

t

0 B

f (s,y) dy ds)

,

(3.2.15)

for (t, x) ∈ D0 . Remark 3.2.2. to

If we take g = 0 in (3.2.12), then the bound obtained in (3.2.13) reduces 1 t β −1 f (s, y) , u(t, x) kβ −1 + (β − 1) 0

(3.2.16)

B

for (t, x) ∈ D0 . By taking g = 0 and β = 2 in part (c5 ), we get a new variant of the inequality given in Theorem 5.8.1 in [82, p. 527]. The inequalities in the following theorem are established in [99]. Theorem 3.2.4. Let u, p, q, r, f , g ∈ C(D0 , R+ ). (c7 ) Suppose that u(t, x) p(t, x) + q(t, x) + r(t, x)

t

f (s, y)u(s, y) dy ds 0∞ B

g(s, y)u(s, y) dy ds, 0

for (t, x) ∈ D0 . If d=

∞ 0

B

(3.2.17)

B

g(s, y)K2 (s, y) dy ds < 1,

(3.2.18)

then u(t, x) K1 (t, x) + DK2 (t, x),

(3.2.19)


101

for (t, x) ∈ D0 , where

t

K1 (t, x) = p(t, x) + q(t, x) × exp

B

t B

s

f (s, y)p(s, y) 0

f (τ , z)q(τ , z) dz d τ

dy ds,

(3.2.20)

t

K2 (t, x) = r(t, x) + q(t, x) × exp

B

and D=

1 1−d


∞ B

0

u(t, x) p(t, x) + r(t, x)

d0 =

∞ 0

then

u(t, x) p(t, x) + r(t, x)

dy ds,

g(s, y)K1 (s, y) dy ds.

(c8 ) Suppose that

for (t, x) ∈ D0 . If

B

t s

f (s, y)r(s, y) 0

(3.2.21)

(3.2.22)

∞

g(s, y)u(s, y) dy ds,

(3.2.23)

B

0

g(s, y)r(s, y) dy ds < 1,

(3.2.24)

B

1 1 − d0

g(s, y)p(s, y) dy ds ,

∞ 0

(3.2.25)

B

for (t, x) ∈ D0 . Remark 3.2.3.

By taking g = 0 in Theorem 3.2.4 part (c7 ), we get the inequality given in

Theorem 3.2.1 part (c2 ). The next two theorems deal with the integral inequalities established in [116]. Theorem 3.2.5. Let u ∈ C(D0 , R+ ), k, D1 k ∈ C(Ω, R+ ) and c 0 is a real constant. (c9 ) If u(t, x) c +

t 0

for (t, x) ∈ D0 , then u(t, x) c exp for (t, x) ∈ D0 , where

k(t, x, s, y)u(s, y) dy ds,

(3.2.26)

A(σ , x) d σ ,

(3.2.27)

B

t 0

t

A(t, x) =

k(t, x,t, y) dy + B

0

B

D1 k(t, x, s, y) dy ds,

(3.2.28)

102


for (t, x) ∈ D0 . (c10 ) Let g ∈ C(R+ , R+ ) be a nondecreasing function, g(u) > 0 on (0, ∞). If u(t, x) c +

t

k(t, x, s, y)g(u(s, y)) dy ds,

(3.2.29)

B

0

for (t, x) ∈ D0 , then for 0 t t1 ; t, t1 ∈ R+ , x ∈ B, t u(t, x) W −1 W (c) + A(σ , x) d σ ,

(3.2.30)

0

where W (r) =

r ds r0

g(s)

,

r > 0,

(3.2.31)

r0 > 0 is arbitrary and W −1 is the inverse of W and A(t, x) is given by (3.2.28) and t1 ∈ R+ is chosen so that

t

W (c) + 0

A(σ , x) d σ ∈ Dom W −1 ,

for all t ∈ R+ lying in the interval 0 t t1 and x ∈ B. Theorem 3.2.6. Let u ∈ C(D0 , R+ ); k, D1 k ∈ C(Ω, R+ ) and c 0 is a real constant. (c11 ) If

t

u2 (t, x) c +

k(t, x, s, y)u(s, y) dy ds, 0

for (t, x) ∈ D0 , then u(t, x)

√

(3.2.32)

B

1 2

t

A(σ , x) d σ ,

(3.2.33)

k(t, x, s, y)u(s, y)g(u(s, y)) dy ds,

(3.2.34)

c+

0

for (t, x) ∈ D0 , where A(t, x) is given by (3.2.28). (c12 ) Let g(u) be as in part (c10 ). If u2 (t, x) c +

t 0

B

for (t, x) ∈ D0 , then for 0 t t2 ; t, t2 ∈ R+ , x ∈ B, √ 1 t A(σ , x) d σ , u(t, x) W −1 W c + 2 0

(3.2.35)

where W, W −1 , A(t, x) are as in part (c10 ) and t2 ∈ R+ is chosen so that W

√ 1 t c + A(σ , x) d σ ∈ Dom W −1 , 2 0

for all t ∈ R+ lying in the interval 0 t t2 and x ∈ B. The integral inequality established in [103] and its variant are given in the following theorem.


103

Theorem 3.2.7. Let u, p, q, f , g ∈ C(D0 , R+ ). (c13 ) If u(t, x) p(t, x)

t

+q(t, x)

s

f (s, y) u(s, y) + q(s, y) 0

0

B

for (t, x) ∈ D0 , then u(t, x) p(t, x) + q(t, x) × exp

(3.2.36)

t

p(s, y) [ f (s, y) + g(s, y)] 0

B

t s

B

g(τ , z)u(τ , z) dz d τ dy ds,

B

q(τ , z)[ f (τ , z) + g(τ , z)]dz d τ

dy ds,

(3.2.37)

for (t, x) ∈ D0 . (c14 ) If

+q(t, x) 0 0

u(t, x) p(t, x)

t s B

τ σ

f (τ , y) u(τ , y) + q(τ , y)

for (t, x) ∈ D0 , then u(t, x) p(t, x) + q(t, x) × exp

0

0

0

B

0

B

g(ξ , z)u(ξ , z) dz d ξ dy d τ ,

(3.2.38)

t s

t s 0

0

B

p(τ , y)[ f (τ , y) + g(τ , y)]dy d τ

q(τ , y)[ f (τ , y) + g(τ , y)]dy d τ ,

(3.2.39)

for (t, x) ∈ D0 . Proofs of Theorems 3.2.1–3.2.7.

The proofs resemble one another, we give the details for

(c1 ), (c3 ), (c5 ), (c7 ), (c9 ), (c11 ), (c13 ) only; the proofs of other inequalities can be completed by following the proofs of the above inequalities, see also [82,87]. (c1 ) Setting

e(s) =

L(s, y, u(s, y)) dy,

(3.2.40)

B

the inequality (3.2.2) can be restated as u(t, x) p(t, x) + q(t, x) Define

t

e(s) ds.

(3.2.41)

0

t

m(t) =

e(s) ds, 0

(3.2.42)

104


then m(0) = 0 and u(t, x) p(t, x) + q(t, x)m(t).

(3.2.43)

From (3.2.42), (3.2.40), (3.2.43) and (3.2.1), we observe that m (t) = e(t) =

B

=

B

L(t, y, u(t, y)) dy

L(t, y, p(t, y) + q(t, y)m(t)) dy B

{L(t, y, p(t, y) + q(t, y)m(t)) − L(t, y, p(t, y)) + L(t, y, p(t, y))}dy

M(t, y, p(t, y))q(t, y)m(t) dy + B

L(t, y, p(t, y)) dy B

= m(t)

M(t, y, p(t, y))q(t, y) dy +

L(t, y, p(t, y)) dy.

B

The inequality (3.2.44) implies (see [82, Theorem 1.3.2]) t t L(s, y, p(s, y)) exp M(τ , z, p(τ , z))q(τ , z) dz d τ dy ds, m(t) 0

(3.2.44)

B

s

B

(3.2.45)

B

for (t, x) ∈ D0 . Using (3.2.45) in (3.2.43), we get the required inequality in (3.2.3). (c3 ) Setting

e1 (s) =

[ f (s, y)u(s, y) + g(s, y)]dy, B

and following the proof of part (c1 ) with suitable changes we get the required inequality in (3.2.7). (c5 ) Introducing the notation e2 (s) =

f (s, y)u(s, y) + g(s, y)uβ (s, y) dy,

(3.2.46)

B

in (3.2.12), we get uβ (t, x) kβ + β Let k > 0 and define z(t) = kβ + β

t 0

e2 (s) ds.

t 0

e2 (s),

(3.2.47)

then z(0) = kβ and uβ (t, x) z(t),

(3.2.48)

for (t, x) ∈ D0 . From (3.2.47), (3.2.46), (3.2.48), we observe that z (t) = β e2 (t) f (t, y)u(t, y) + g(t, y)uβ (t, y) dy =β B 1 f (t, y)(z(t)) β + g(t, y)z(t) dy β B 1 f (t, y) dy . = β z(t) g(t, y) dy + (z(t)) β B

B

(3.2.49)


105

The inequality (3.2.49) implies (see [82, Theorem 3.5.5]) t t g(τ , z) dz d τ kβ −1 + (β − 1) f (s, y) z(t) exp β 0

0

B

× exp −(β − 1)

s 0

B

g(τ , z) dz d τ

B

β β −1 dy ds .

(3.2.50)

Using (3.2.50) in (3.2.48), we get the required inequality in (3.2.13). If k 0, we carry out the above procedure with k + ε instead of k, where ε > 0 is an arbitrary small constant, and then pass to the limit as ε → 0 to obtain (3.2.13). (c7 ) Let

t

w(t) =

λ=

f (s, y)u(s, y) dy ds,

(3.2.51)

g(s, y)u(s, y) dy ds,

(3.2.52)

0 ∞ B 0

B

then (3.2.17) can be restated as u(t, x) p(t, x) + q(t, x)w(t) + r(t, x)λ . Introducing the notation

(3.2.53)

e3 (s) =

f (s, y)u(s, y) dy,

(3.2.54)

B

in (3.2.51), we get

t

w(t) = 0

e3 (s) ds.

(3.2.55)

From (3.2.55), (3.2.54) and (3.2.53), we have w (t) = e3 (t) =

= w(t)

B

f (t, y)u(t, y) dy

f (t, y)q(t, y) dy + B

B

B

f (t, y)[p(t, y) + q(t, y)w(t) + r(t, y)λ ]dy

f (t, y) [p(t, y) + r(t, y)λ ] dy.

(3.2.56)

The inequality (3.2.56) implies (see [82, Theorem 1.3.2]) t t w(t) f (s, y)[p(s, y) + r(s, y)λ ] exp f (τ , z)q(τ , z) dz d τ dy ds 0 B s B t t = f (s, y)p(s, y) exp f (τ , z)q(τ , z) dz d τ dy ds 0 B s B t t +λ (3.2.57) f (s, y)r(s, y) exp f (τ , z)q(τ , z) dz d τ dy ds. 0

s

B

B

From (3.2.53) and (3.2.57), we get u(t, x) p(t, x) + q(t, x)

106


× +λ

t 0

s

B

f (s, y)r(s, y) exp B

B

s

B


t

t 0

t f (s, y)p(s, y) exp


dy ds

dy ds + r(t, x)λ

= K1 (t, x) + λ K2 (t, x).

(3.2.58)

From (3.2.52) and (3.2.58), we observe that

λ

∞

B

0

g(s, y)[K1 (s, y) + λ K2 (s, y)]dy ds,

which implies

λ D.

(3.2.59)

Using (3.2.59) in (3.2.58), we get (3.2.19). (c9 ) Setting

E(t, s) =

k(t, x, s, y)u(s, y) dy,

(3.2.60)

B

for every x ∈ B, the inequality (3.2.26) can be restated as t

u(t, x) c + Define

E(t, s) ds.

(3.2.61)

0

t

z(t) = c +

E(t, s) ds,

(3.2.62)

0

then z(0) = c and u(t, x) z(t).

(3.2.63)

From (3.2.62), (3.2.60), (3.2.63) and the fact that z(t) is nondecreasing in t ∈ R+ , we observe that z (t) = E (t,t) +

0

D1 E (t, s) ds

D1

0

B

t

k(t, x,t, y)u(t, y) dy +

=

t

t

k(t, x,t, y)z(t) dy + B

0

B

k(t, x, s, y)u(s, y) dy ds B

D1 k(t, x, s, y)z(s) dy ds

A(t, x)z(t). The inequality (3.2.64) implies z(t) c exp

(3.2.64)

t 0

A(σ , x) d σ .

(3.2.65)


107

Using (3.2.65) in (3.2.63), we get the required inequality in (3.2.27). (c11 ) Let E(t, s) be given by (3.2.60). Then (3.2.32) can be restated as u2 (t, x) c +

t

E (t, s) ds.

(3.2.66)

0

Let c > 0 and define by z(t) the right hand side of (3.2.66), then z(0) = c and u(t, x) z(t). Following the proof of part (c9 ), we get z (t) A(t, x) z(t). (3.2.67) The inequality (3.2.67) implies

z(t)

√

c+

1 2

t 0

A(σ , x) d σ .

(3.2.68)

The required inequality in (3.2.33) follows by using (3.2.68) in u(t, x) z(t). The proof of the case when c 0 can be completed as mentioned in the proof of part (c5 ). (c13 ) Introducing the notation s g(τ , z)u(τ , z) dz d τ dy, E0 (s) = f (s, y) u(s, y) + q(s, y) 0

B


(3.2.69)

B

t 0

E0 (s) ds.

(3.2.70)

t

m(t) = 0

E0 (s) ds,

(3.2.71)

for t ∈ R+ , then m(0) = 0 and from (3.2.70), we get u(t, x) p(t, x) + q(t, x)m(t),

(3.2.72)

for (t, x) ∈ D0 . From (3.2.71), (3.2.69) and (3.2.72), we observe that t g(τ , z)u(τ , z) dz d τ dy m (t) = E0 (t) = f (t, y) u(t, y) + q(t, y) 0

B

f (t, y) p(t, y) + q(t, y)m(t) B

t

+q(t, y) 0

B

B

g(τ , z)[p(τ , z) + q(τ , z)m(τ )]dz d τ dy,

(3.2.73)

for t ∈ R+ Introducing the notation r(τ ) =

B

g(τ , z) [p(τ , z) + q(τ , z)m(τ )] dz,

(3.2.74)

108


the inequality (3.2.73) can be written as

t m (t) f (t, y) p(t, y) + q(t, y) m(t) + r(τ ) d τ dy, B

(3.2.75)

0

for t ∈ R+ . Define

t

v(t) = m(t) + 0

r(τ ) d τ ,

(3.2.76)

then m(t) v(t), v(0) = m(0) = 0 and m (t)

f (t, y) [p(t, y) + q(t, y)v(t)]dy.

(3.2.77)

B

From (3.2.76), (3.2.77), (3.2.74) and the fact that m(t) v(t), t ∈ R+ , we observe that v (t) = m (t) + r(t)

f (t, y)[p(t, y) + q(t, y)v(t)]dy + B

v(t)

g(t, z)[p(t, z) + q(t, z)m(t)]dz B

q(t, y)[ f (t, y) + g(t, y)]dy + B

p(t, y)[ f (t, y) + g(t, y)]dy.

(3.2.78)

B

The inequality (3.2.78) implies

× exp

v(t) t s

B

t

p(s, y)[ f (s, y) + g(s, y)] q(τ , z)[ f (τ , z) + g(τ , z)]dz d τ dy ds. 0

B

(3.2.79)

Using the fact that m(t) v(t), t ∈ R+ in (3.2.79) and then using the bound on m(t) in (3.2.72), we get the required inequality in (3.2.37).

3.3

Volterra-Fredholm-type integral inequalities II

The main objective of this section is to present some more mixed Volterra-Fredholm-type integral inequalities established in [107,119 ] which can be used as tools in certain new situations. In what follows we shall use the notation as given in section 3.2. Furthermore, let x = (x1 , . . . , xn ) be any point in Rn+ , Di =

∂ ∂ xi ,

Dn · · · D1 =

∂ ∂ xn

· · · ∂∂x for 1 i n, 1

m n B0,x = {x ∈ Rn+ : 0 < x < ∞}, Ha,b = ∏m i=1 [ai , bi ] ⊂ R (ai < bi ) and H = R+ × Ha,b . For

the functions u(s) and v(t) defined on B0,x and Ha,b we denote by the n-fold and m-fold integrals x1 0

···

xn 0

u(s1 , . . . , sn ) dsn · · · ds1 ,

b1 a1

···

bm am

B0,x u(s) ds, Ha,b v(t) dt

v(t1 , . . . ,tm ) dtm · · · dt1

respectively. The mixed Volterra-Fredholm-type integral inequalities established in [107] are given in the following theorems.


109

Theorem 3.3.1. Let u, p, q, f ∈ C(D0 , R+ ) and c 0 be a constant. (d1 ) Let L ∈ C(D0 × R+ , R+ ) satisfies the condition (3.2.1) in Theorem 3.2.1 part (c1 ). If u(t, x) p(t, x) + q(t, x) for (t, x) ∈ D0 , then u(t, x) p(t, x) + q(t, x) × exp

t s 0

0

B

t s 0

0

B

t s 0

0

B

L(τ , y, u(τ , y)) dy d τ ds,

(3.3.1)

L(τ , y, p(τ , y)) dy d τ ds

M(τ , y, p(τ , y))q(τ , y) dy d τ ds ,

(3.3.2)

for (t, x) ∈ D0 . (d2 ) Let g ∈ C(R+ , R+ ) be a nondecreasing function, g(u) > 0 on (0, ∞). If u(t, x) c +

t s 0

0

B

f (τ , y)g(u(τ , y)) dy d τ ds,

(3.3.3)

for (t, x) ∈ D0 , then for 0 t t1 ; t, t1 ∈ R+ , x ∈ B, t s u(t, x) W −1 W (c) + f (τ , y) dy d τ ds ,

(3.3.4)

where W, W −1 are as in Theorem 3.2.5 part (c10 ) and t1 ∈ R+ is chosen so that t s W (c) + f (τ , y) dy d τ ds ∈ Dom W −1 ,

(3.3.5)

0

0

B

0

B

0

for all t ∈ R+ lying in 0 t t1 and x ∈ B. Theorem 3.3.2. Let u, f ∈ C(D0 , R+ ) and c 0 be a constant. (d3 ) If u2 (t, x) c +

t s 0

for (t, x) ∈ D0 , then u(t, x)

√

c+

B

0

1 2

f (τ , y)u(τ , y) dy d τ ds,

t s 0

0

B

f (τ , y) dy d τ ds,

(3.3.6)

(3.3.7)

for (t, x) ∈ D0 . (d4 ) Let g(u) be as in part (d2 ). If u2 (t, x) c +

t s 0

0

B

f (τ , y)u(τ , y)g(u(τ , y)) dy d τ ds,

for (t, x) ∈ D0 , then for 0 t t2 ; t, t2 ∈ R+ , x ∈ B, √ 1 t s u(t, x) W −1 W c + f (τ , y) dy d τ ds , 2 0 0 B where W, W −1 are as in part (d2 ) and t2 ∈ R+ is chosen so that √ 1 t s W c + f (τ , y) dy d τ ds ∈ Dom W −1 , 2 0 0 B for all t ∈ R+ lying in 0 t t2 and x ∈ B.

(3.3.8)

(3.3.9)

110


Theorem 3.3.3. Let u ∈ C(D0 , R1 ), f ∈ C(D0 , R+ ) and c 1 be a constant. (d5 ) If u(t, x) c +

t s 0

f (τ , y)u(τ , y) log u(τ , y) dy d τ ds,

B

0

(3.3.10)

for (t, x) ∈ D0 , then u(t, x) cexp(

t s

0 0 B

f (τ ,y) dy d τ ds)

,

(3.3.11)

for (t, x) ∈ D0 . (d6 ) Let g(u) be as in part (d2 ). If u(t, x) c +

t s 0

B

0

f (τ , y)u(τ , y)g(log u(τ , y)) dy d τ ds,

for (t, x) ∈ D0 , then for 0 t t3 ; t, t3 ∈ R+ , x ∈ B, t s u(t, x) exp W −1 W (log c) + f (τ , y) dy d τ ds , 0

0

(3.3.12)

(3.3.13)

B

where W, W −1 are as in part (d2 ) and t3 ∈ R+ be chosen so that t s W (log c) + f (τ , y) dy d τ ds ∈ Dom W −1 , 0

0

B

for all t ∈ R+ lying in the interval 0 t t3 and x ∈ B. The following theorems contain the inequalities given in [119]. Theorem 3.3.4. Let u, f ∈ C(H, R+ ) and c 0 is a constant. (d7 ) If u(x, y) c +

f (s,t)u(s,t) dt ds,

(3.3.14)

f (s,t) dt ds ,

(3.3.15)

B0,x Ha,b

for (x, y) ∈ H, then u(x, y) c exp

B0,x Ha,b

for (x, y) ∈ H. (d8 ) If c 1, u 1 and u(x, y) c +

f (s,t)u(s,t) log u(s,t) dt ds,

(3.3.16)

B0,x Ha,b

for (x, y) ∈ H, then u(x, y) c for (x, y) ∈ H.

exp B

0,x Ha,b

f (s,t) dt ds

,

(3.3.17)


111

Theorem 3.3.5. (d9 ) Let u, f ∈ C(H, R+ ); c, p, q be positive constants and suppose that u p (x, y) c +

f (s,t)uq (s,t) dt ds,

(3.3.18)

B0,x Ha,b

for (x, y) ∈ H. If 0 < q < p, then 1 p−q p−q p−q p + f (s,t) dt ds , u(x, y) (c) p B0,x Ha,b

(3.3.19)

for (x, y) ∈ H. (d10 ) Let u, p, q, f ∈ C(H, R+ ) and L ∈ C(H × R+ , R+ ) be such that 0 L(x, y, u) − L(x, y, v) M(x, y, v)(u − v), for u v 0, where M ∈ C(H × R+ , R+ ). If u(x, y) p(x, y) + q(x, y) for (x, y) ∈ H, then u(x, y) p(x, y) + q(x, y) × exp

L(s,t, u(s,t)) dt ds,

(3.3.21)

B0,x Ha,b

L (s,t, p(s,t)) dt ds

B0,x Ha,b

M(s,t, p(s,t))q(s,t) dt ds ,

(3.3.20)

(3.3.22)

B0,x Ha,b

for (x, y) ∈ H. Proofs of Theorems 3.3.1–3.3.5.

To prove (d2 )–(d4 ), (d7 ), it is sufficient to assume that

c > 0, since the standard limiting argument can be used to treat the remaining case, see [82, p. 108]. (d1 ) Setting r(τ ) =

B

L(τ , y, u(τ , y)) dy,


t s

z(t) = 0

0

t s 0

0

r(τ ) d τ ds.

r(τ ) d τ ds,

(3.3.23)

(3.3.24)

(3.3.25)

then, it is easy to see that z(0) = 0, z (0) = 0 and u(t, x) p(t, x) + q(t, x)z(t). From (3.3.25), (3.3.23), (3.3.26) and (3.2.1), we observe that z (t) = r(t) =

L(t, y, u(t, y)) dy B

(3.3.26)

112


B

L(t, y, p(t, y) + q(t, y)z(t)) − L(t, y, p(t, y)) dy + L(t, y, p(t, y)) dy B

z(t)

M(t, y, p(t, y))q(t, y) dy + B

L(t, y, p(t, y)) dy.

(3.3.27)

B

From (3.3.27) and using the fact that z(t) is nondecreasing in t ∈ R+ , it is easy to see that z(t)

0

0

z(s) 0

0

B


M(τ , y, p(τ , y))q(τ , y) dy d τ ds.

s

t

+

t s

B

(3.3.28)

Clearly, the first term on the right hand side of (3.3.28) is continuous, nonnegative and nondecreasing in t ∈ R+ . Now a suitable application of the inequality in [82, Theorem 1.3.1] to (3.3.28) yields z(t) × exp

0

0

B

0

t s 0

t s

B


M(τ , y, p(τ , y))q(τ , y) dy d τ ds .

(3.3.29)

Using (3.3.29) in (3.3.26), we get the required inequality in (3.3.2). (d2 ) Setting r1 (τ ) =

B

f (τ , y)g(u(τ , y)) dy,

the inequality (3.3.3) can be restated as u(t, x) c +

t s 0

0

(3.3.30)

r1 (τ )d τ ds.

(3.3.31)

Let c > 0 and define by z(t) the right hand side of (3.3.31). Following the proof of part (d1 ) given above, we get z (t) = r1 (t) =

B

f (t, y)g(u(t, y)) dy g(z(t))

f (t, y) dy.

(3.3.32)

B

From (3.3.32) and using the fact that z(t) is nondecreasing in t ∈ R+ , it is easy to see that

s t (3.3.33) f (τ , y) dy d τ ds. z(t) c + g (z(s)) 0

0

B

Now a suitable application of the inequality in [82, Theorem 2.3.1] to (3.3.33) yields t s z(t) W −1 W (c) + f (τ , y) dy d τ ds . (3.3.34) 0

0

B

Using (3.3.34) in u(t, x) z(t) we get the required inequality in (3.3.4). The proof of the case when c 0 can be completed as mentioned in the proof of part (c5 ). The subinterval 0 t t1 is obvious.


113

(d3 ) setting r2 (τ ) =

B

f (τ , y)u(τ , y) dy,

the inequality (3.3.6) can be restated as u2 (t, x) c +

t s 0

0

(3.3.35)

r2 (τ )d τ ds.

(3.3.36)

Let c > 0 and define by z(t) the right hand side of (3.3.36), then z(0) = c, z (0) = 0 and u(t, x) z(t). Following the proof of part (d1 ), we get (3.3.37) z (t) = r2 (t) = f (t, y)u(t, y) dy z(t) f (t, y) dy. B

B

By taking t = τ in (3.3.37) and integrating it over τ from 0 to t and using the fact that z(t) is nondecreasing in t ∈ R+ , we get z (t)

t z(t) f (τ , y) dy d τ . 0

(3.3.38)

B

The inequality (3.3.38) implies (see [82, p. 233]) √ 1 t s z(t) c + f (τ , y) dy d τ ds. 2 0 0 B

(3.3.39)

The required inequality in (3.3.7) follows by using (3.3.39) in u(t, x) z(t). (d4 ) Setting r3 (τ ) =

B

f (τ , y)u(τ , y)g(u(τ , y)) dy,

the inequality (3.3.8) can be restated as u2 (t, x) c +

t s 0

0

r3 (τ ) d τ ds.

(3.3.40)

(3.3.41)

Let c > 0 and define by z(t) the right hand side of (3.3.41). Following the proof of part (d1 ), we get z (t) = r3 (t)

z(t) g

z(t)

f (t, y) dy.

(3.3.42)

B

From (3.3.42) and using the fact that z(t) is nondecreasing in t ∈ R+ , it is easy to observe that

t z (t) f (τ , y) dy d τ . g z(t) 0 B z(t)

From (3.3.43), we get

s √ 1 t z(t) c + g z(s) f (τ , y) dy d τ ds. 2 0 0 B

(3.3.43)

(3.3.44)

Now a suitable application of the inequality in [82, Theorem 2.3.1] to (3.3.44) yields √ 1 t s z(t) W −1 W c + f (τ , y) dy d τ ds . (3.3.45) 2 0 0 B

114


Using (3.3.45) in u(t, x)

z(t), we get (3.3.9).

(d5 ) Setting r4 (τ ) =

B

f (τ , y)u(τ , y) log u(τ , y) dy,

the inequality (3.3.10) can be restated as u(t, x) c +

t s 0

0

(3.3.46)

r4 (τ ) d τ ds.

(3.3.47)

Let c > 0 and define by z(t) the right hand side of (3.3.47). Following the proof of (d1 ) and using the fact that u(t, x) z(t), we have z (t) = r4 (t) =

B

f (t, y)u(t, y) log u(t, y) dy z(t) log z(t)

f (t, y) dy.

(3.3.48)

B

From (3.3.48) and using the fact that z(t) is nondecreasing in t ∈ R+ , it is easy to observe that z (t) z(t) log z(t)

t 0

which implies log z(t) log c +

B

f (τ , y) dy d τ ,

s

t

log z(s) 0

0

(3.3.49)

B

f (τ , y) dy d τ ds.

(3.3.50)

Now by following the proof of Theorem 3.8.2 given in [82, p. 269] we get z(t) cexp(

t s

0 0 B

f (τ ,y) dy d τ ds)

.

(3.3.51)

Using (3.3.51) in u(t, x) z(t), we get the required inequality in (3.3.11). (d6 ) The proof can be completed by following the proof of (d5 ) and closely looking at the proof of Theorem 3.9.1 given in [82, p. 270]. Here, we omit the details. Next, we will give the proofs of (d7 ) and (d9 ); the proofs of (d8 ) and (d10 ) can be completed by following the proofs of (d7 ) and (d9 ) and closely looking at the proofs of (d5 ) and (d1 ). (d7 ) Introducing the notation

R1 (s) =

f (s,t)u(s,t) dt,

(3.3.52)

Ha,b

in (3.3.14), we get u(x, y) c + for (x, y) ∈ H. Let c > 0 and define

B0,x

R1 (s) ds,

(3.3.53)

z(x) = c + B0,x

R1 (s) ds,

(3.3.54)


115

for x ∈ Rn+ , then z(0) = c and u(x, y) z(x).

(3.3.55)

From (3.3.54), (3.3.52), (3.3.55), we observe that (see [82, Theorem 4.9.1]) Dn · · · D1 z(x) = R1 (x) =

Ha,b

f (x,t)u(x,t) dt z(x)

From (3.3.56), it is easy to observe that z(x) c +

f (x,t) dt.

(3.3.56)

Ha,b

f (s,t) dt ds.

z(s) B0,x

(3.3.57)

Ha,b

Now a suitable application of Theorem 4.9.1 part (i) in [82, p. 397] to (3.3.57) yields f (s,t) dt ds . (3.3.58) z(x) c exp B0,x Ha,b

Using (3.3.58) in (3.3.55) yields (3.3.15). (d9 ) Introducing the notation

f (s,t)uq (s,t) dt,

R2 (s) =

(3.3.59)

Ha,b

the inequality (3.3.18) can be restated as u p (x, y) c + Define

B0,x

R2 (s) ds.

(3.3.60)

z(x) = c + B0,x

R2 (s) ds,

(3.3.61)

for x ∈ Rn+ , then z(0) = c and u p (x, y) z(x),

(3.3.62)

for (x, y) ∈ H. From (3.3.61), (3.3.59) and (3.3.62), we observe that Dn · · · D1 z(x) = R2 (x) =

q

f (x,t)uq (x,t) dt (z(x)) p

Ha,b

i.e., Dn · · · D1 z(x)

f (x,t) dt. Ha,b

f (x,t) dt. (3.3.63) Ha,b (z(x)) Now by following the similar arguments as in the proofs of Theorems 4.9.1 and 5.9.1 given q p

in [82] with suitable modifications, from (3.3.63), we obtain p−q p−q p−q f (s,t) dt ds, (z(x)) p − (c) p p B0,x Ha,b for x ∈ Rn+ , which implies

p p−q p−q p−q z(x) (c) p + f (s,t) dt ds . p B0,x Ha,b

The assertion (3.3.19) follows by using (3.3.64) in (3.3.62).

(3.3.64)

116

3.4


Integral equation of Volterra-Fredholm-type

The mixed Volterra-Fredholm integral equations of the form t

u(t, x) = f (t, x) +

F(t, x, s, y, u(s, y)) dy ds, 0

(3.4.1)

B

often arise from mathematical modeling of many physical and biological phenomena, see [5,16–18,28,31,32] and the related references given therein. In (3.4.1) f , F are given functions, B is as defined in section 3.2 and u is the unknown function. We assume that f ∈ C(D0 , Rn ), F ∈ C(Ω × Rn , Rn ), where D0 , Ω are as defined in section 3.2. Let Z be the space of functions φ ∈ C(D0 , Rn ) which fulfil the condition |φ (t, x)| = O(exp(λ (t + |x|))),

(3.4.2)

where λ > 0 is a constant. In the space Z we define the norm |φ |Z = sup [|φ (t, x)| exp(−λ (t + |x|))].

(3.4.3)

(t,x)∈D0

It is easy to see that Z with norm defined in (3.4.3) is a Banach space and |φ |Z M,

(3.4.4)

where M 0 is a constant. The main objective of this section is to present some qualitative aspects of solutions of equation (3.4.1) developed in [79,99,105,116]. The following result guarantees the existence and uniqueness of solutions of equation (3.4.1). Theorem 3.4.1. Suppose that (i) the function F in equation (3.4.1) satisfies the condition |F(t, x, s, y, u) − F(t, x, s, y, v)| h(t, x, s, y)|u − v|,

(3.4.5)

where h ∈ C(D20 , R+ ), (ii) for λ as in (3.4.2) (a1 ) there exists a nonnegative constant α < 1 such that t 0

B

h(t, x, s, y) exp(λ (s + |y|)) dy ds α exp(λ (t + |x|)),

(3.4.6)

(a2 ) there exists a nonnegative constant β such that | f (t, x)| +

t 0

B

|F(t, x, s, y, 0)|dy ds β exp(λ (t + |x|)),

where f , F are as defined in equation (3.4.1). Then the equation (3.4.1) has a unique solution u(t, x) in Z on D0 .

(3.4.7)


Proof.

117

Let u ∈ Z and define the operator T by t

(Tu)(t, x) = f (t, x) +

F(t, x, s, y, u(s, y)) dy ds,

(3.4.8)

B

0

for (t, x) ∈ D0 . From (3.4.8) and using the hypotheses, we have |(Tu)(t, x)| | f (t, x)| +

t 0

B

t

+ 0

B

β exp(λ (t + |x|)) + |u|Z

|F(t, x, s, y, u(s, y)) − F(t, x, s, y, 0)|dy ds

|F(t, x, s, y, 0)|dy ds t 0

B

h(t, x, s, y) exp(λ (s + |y|)) dy ds

[M α + β ] exp(λ (t + |x|)).

(3.4.9)

From (3.4.9), it follows that Tu ∈ Z. Let u, v ∈ Z. From (3.4.8) and using the hypotheses, we have |(Tu)(t, x) − (T v)(t, x)| |u − v|Z

t

B

0

t 0

B

|F(t, x, s, y, u(s, y)) − F(t, x, s, y, v(s, y))|dy ds

h(t, x, s, y) exp(λ (s + |y|)) dy ds

α |u − v|Z exp(λ (t + |x|)).

(3.4.10)

From (3.4.10), it follows that |Tu − T v|Z α |u − v|Z . Since α < 1, it follows from Banach fixed point theorem (see [28] and [51]) that T has a unique fixed point in Z. The fixed point of T is a solution of equation (3.4.1). Remark 3.4.1.

We note that one can formulate existence and uniqueness result similar

to that of Theorem 1.3.2 for the solution u ∈ C (D0 , Rn ) of equation (3.4.1). Furthermore, Theorem 3.4.1 can also be extended to more general mixed Volterra-Fredholm integral equation of the form

t

u(t, x) = f (t, x) +

G(t, x, s, u(s, x)) ds 0

t

+

H(t, x, y, u(t, y)) dy + B

F(t, x, s, y, u(s, y)) dy ds, 0

(3.4.11)

B

under some suitable conditions on the functions involved in (3.4.11). We leave the precise formulations of such results to the reader. The next result concerning the estimate on the solution of equation (3.4.1) holds.

118


Theorem 3.4.2. Suppose that the function F in equation (3.4.1) satisfies the condition |F(t, x, s, y, u) − F(t, x, s, y, v)| k(t, x, s, y)|u − v|,

(3.4.12)

where k, D1 k ∈ C(Ω, R+ ). Let

t c = sup f (t, x) + F(t, x, s, y, 0) dy ds < ∞, 0 B

(3.4.13)

(t,x)∈D0

where f , F are the functions in equation (3.4.1). If u(t, x) is any solution of equation (3.4.1) on D0 , then |u(t, x)| c exp

t

A(σ , x) d σ ,

0

(3.4.14)

for (t, x) ∈ D0 , where A(t, x) is given by (3.2.28). Proof.

Using the fact that u(t, x) is a solution of equation (3.4.1) on D0 and hypotheses,

we have

t |u(t, x)| f (t, x) + F(t, x, s, y, 0) dy ds 0

t

+ 0

B

B

|F(t, x, s, y, u(s, y)) − F(t, x, s, y, 0)|dy ds

c+

t

k(t, x, s, y)|u(s, y)|dy ds. 0

(3.4.15)

B

Now an application of the inequality in Theorem 3.2.5 part (c9 ) to (3.4.15) yields (3.4.14). A variant of Theorem 3.4.2 is given in the following theorem. Theorem 3.4.3.

Suppose that the function F in equation (3.4.1) satisfies the condition

(3.4.12). Let t

d = sup (t,x)∈D0 0

B

|F(t, x, s, y, f (s, y))|dy ds < ∞,

(3.4.16)

where f , F are the functions in equation (3.4.1). If u(t, x) is any solution of equation (3.4.1) on D0 , then |u(t, x) − f (t, x)| d exp

for (t, x) ∈ D0 , where A(t, x) is given by (3.2.28).

t 0

A(σ , x) d σ ,

(3.4.17)


Proof.

119

Let z(t, x) = |u(t, x) − f (t, x)| for (t, x) ∈ D0 . Using the fact that u(t, x) is a solution

of equation (3.4.1) and hypotheses, we have z(t, x)

t 0

B

|F(t, x, s, y, u(s, y)) − F(t, x, s, y, f (s, y))|dy ds t

+ 0

d+

B

|F(t, x, s, y, f (s, y))|dy ds

t

k(t, x, s, y)z(s, y) dy ds. 0

(3.4.18)

B

Now an application of the inequality in Theorem 3.2.5 part (c9 ) to (3.4.18) yields (3.4.17). We call the function u ∈ C(D0 , Rn ) an ε -approximate solution of equation (3.4.1) if there exists a constant ε 0 such that

t u(t, x) − f (t, x) + ε, F(t, x, s, y, u(s, y)) dy ds 0 B for (t, x) ∈ D0 . The next theorem deals with the estimate on the difference between the two approximate solutions of equation (3.4.1). Theorem 3.4.4.

Let u1 (t, x) and u2 (t, x) be respectively ε1 - and ε2 -approximate solutions

of equation (3.4.1) on D0 . Suppose that the function F in equation (3.4.1) satisfies the condition (3.4.12). Then |u1 (t, x) − u2 (t, x)| (ε1 + ε2 ) exp

t 0

A(σ , x) d σ ,

(3.4.19)

for (t, x) ∈ D0 , where A(t, x) is given by (3.2.28). Proof.

Since u1 (t, x) and u2 (t, x) are respectively, ε1 - and ε2 -approximate solutions of

equation (3.4.1) on D0 , we have

t ui (t, x) − f (t, x) + εi , F(t, x, s, y, u (s, y)) dy ds i 0 B

(3.4.20)

for i = 1, 2. From (3.4.20) and using the elementary inequalities in (1.3.25), we observe that

t ε2 + ε2 u1 (t, x) − f (t, x) + F(t, x, s, y, u1 (s, y)) dy ds 0

B

t F(t, x, s, y, u2 (s, y)) dy ds + u2 (t, x) − f (t, x) + 0

B

t f (t, x) + F(t, x, s, y, u1 (s, y)) dy ds [u1 (t, x) − u2 (t, x)] − 0

B

120


−

F(t, x, s, y, u2 (s, y)) dy ds B

t

f (t, x) + 0

t |u1 (t, x) − u2 (t, x)| − {F(t, x, s, y, u1 (s, y)) − F(t, x, s, y, u2 (s, y))}dy ds. (3.4.21) 0

B

Let w(t, x) = |u1 (t, x) − u2 (t, x)| for (t, x) ∈ D0 . From (3.4.21) and using (3.4.12), we have w(t, x) (ε2 + ε2 ) +

t 0

B

|F(t, x, s, y, u1 (s, y)) − F(t, x, s, y, u2 (s, y))|dy ds

(ε2 + ε2 ) +

t

k(t, x, s, y)w(s, y) dy ds.

(3.4.22)

B

0

Now an application of the inequality in Theorem 3.2.5 part (c9 ) to (3.4.22) yields (3.4.19). Remark 3.4.2.

In case u1 (t, x) is a solution of equation (3.4.1), then we have ε1 = 0 and

from (3.4.19), we see that u2 (t, x) → u1 (t, x) as ε2 → 0. Moreover, from (3.4.19) it follows that if ε1 = ε2 = 0, then the uniqueness of solutions of equation (3.4.1) is established. Consider the equation (3.4.1) and the following Volterra-Fredholm integral equation t

v(t, x) = f (t, x) +

F(t, x, s, y, v(s, y)) dy ds, 0

(3.4.23)

B

where f ∈ C(D0 , Rn ), F ∈ C(Ω × Rn , Rn ). The following result that relates the solutions of equations (3.4.1) and (3.4.23) holds. Theorem 3.4.5.


(3.4.12) and there exist constants δi 0 (i = 1, 2) such that | f (t, x) − f (t, x)| δ1 , t 0

B

|F(t, x, s, y, p) − F(t, x, s, y, p)|dy ds δ2 ,

(3.4.24)

(3.4.25)

where f , F and f , F are as given in equations (3.4.1) and (3.4.23). Let u(t, x) and v(t, x) be respectively, solutions of (3.4.1) and (3.4.23) on D0 , then t |u(t, x) − v(t, x)| (δ1 + δ2 ) exp A(σ , x) d σ , 0

for (t, x) ∈ D0 , where A(t, x) is given by (3.2.28).

(3.4.26)


Proof.

121

Let r(t, x) = |u(t, x) − v(t, x)|, (t, x) ∈ D0 . Using the facts that u(t, x), v(t, x) are

respectively the solutions of equations (3.4.1), (3.4.23) on D0 and hypotheses, we have r(t, x) | f (t, x) − f (t, x)| t

+ 0

B

t

+ 0

B

|F(t, x, s, y, u(s, y)) − F(t, x, s, y, v(s, y))|dy ds |F(t, x, s, y, v(s, y)) − F(t, x, s, y, v(s, y))|dy ds

(δ1 + δ2 ) +

t

k(t, x, s, y)r(s, y) dy ds. 0

(3.4.27)

B

Now an application of the inequality in Theorem 3.2.5 part (c9 ) to (3.4.27) yields (3.4.26). We now consider the system of nonlinear Volterra-Fredholm integral equations of the form t

[k(t, x, s, y)u(s, y) + F(t, x, s, y, u(s, y))]dy ds,

u(t, x) = f (t, x) +

(3.4.28)

B

0

as a perturbation of the linear system of Volterra-Fredholm integral equations t

v(t, x) = f (t, x) +

k(t, x, s, y)v(s, y) dy ds, 0

(3.4.29)

B

where f , F are as given in equation (3.4.1) and k ∈ C(Ω, R). The following theorem deals with the estimate on the difference between the solutions of equations (3.4.28) and (3.4.29). Theorem 3.4.6. Suppose that the functions F and k in equation (3.4.28) satisfy respectively the conditions |F(t, x, s, y, u1 ) − F(t, x, s, y, u2 )| q(t, x)h(s, y)|u1 − u2 |,

(3.4.30)

F(t, x, s, y, 0) = 0 and |k(t, x, s, y)| q(t, x)g(s, y),

(3.4.31)

where q, h, g ∈ C(D0 , R+ ). Let v(t, x) be a solution of equation (3.4.29) on D0 such that |v(t, x)| M, where M 0 is a constant and

t

p(t, x) = Mq(t, x)

h(s, y) dy ds. 0

(3.4.32)

B

If u(t, x) is any solution of equation (3.4.28) on D0 , then |u(t, x) − v(t, x)| p(t, x) + q(t, x) × exp for (t, x) ∈ D0 .

t s

B

t

[h(s, y) + g(s, y)]p(s, y) 0

B

[h(τ , z) + g(τ , z)]q(τ , z) dz d τ

dy ds,

(3.4.33)

122


Proof.

From the hypotheses, we observe that |u(t, x) − v(t, x)| t

+ + 0

0

B

B

|k(t, x, s, y)| |u(s, y) − v(s, y)|dy ds

|F(t, x, s, y, v(s, y)) − F(t, x, s, y, 0)|dy ds

q(t, x)

+q(t, x)

B

0

|F(t, x, s, y, u(s, y)) − F(t, x, s, y, v(s, y))|dy ds

0 B t

t

t

t 0

B

g(s, y)|u(s, y) − v(s, y)|dy ds

h(s, y)|u(s, y) − v(s, y)|dy ds + q(t, x)

p(t, x) + q(t, x)

t 0

B

t

h(s, y)|v(s, y)|dy ds 0

B

[h(s, y) + g(s, y)]|u(s, y) − v(s, y)|dy ds.

(3.4.34)

Now an application of the inequality in Theorem 3.2.1 part (c2 ) to (3.4.34) yields (3.4.33). Remark 3.4.3.

We note that the inequality in Theorem 3.2.5 part (c9 ) can be used to

formulate results on the uniqueness and continuous dependence of solutions of equation (3.4.1) by following the corresponding results given in section 1.4.

3.5

Volterra-Fredholm-type integral equations

Inspired by the equations like (16) and (19) (see [4,64]), we consider the VolterraFredholm-type integral equations ofthe forms t

s

u(t, x) = h(t, x) + 0

and

0

B

F(t, x, τ , y, u(τ , y)) dy d τ ds,

(3.5.1)

K(x, y, s,t, u(s,t)) dt ds,

(3.5.2)

u(x, y) = f (x, y) + B0,x Ha,b

where h, F and f , K are given functions and u is the unknown function. In this section we will make use of the notations and definitions given in sections 3.2 and 3.4 and the concept of the ε -approximate solution extended in a natural way to equations (3.5.1) and (3.5.2). We assume that h ∈ C(D0 , Rn ), F ∈ C(Ω × Rn , Rn ), f ∈ C(H, Rn ), K ∈ C(H 2 × Rn , Rn ). Owing to the importance of equations (3.5.1), (3.5.2), we believe that the qualitative theory of such equations needs to be developed in various directions. The problems of existence of solutions of equations (3.5.1) and (3.5.2) can be dealt with the method employed in section 3.4. In this section, we present conditions under which we can offer simple, unified and concise proofs of some of the important qualitative properties of solutions of equations (3.5.1) and (3.5.2). In our discussion, we use the following special forms of the inequalities given in (d1 ) and (d10 ).


123

Lemma 3.5.1. Let u, p, q, g ∈ C (D0 , R+ ) . If u(t, x) p(t, x) + q(t, x) for (t, x) ∈ D0 , then u(t, x) p(t, x) + q(t, x)

× exp

0

0

B

0

0

B

0

g(τ , y)u(τ , y) dy d τ ds,

t s

t s 0

t s

B

g(τ , y)p(τ , y) dy d τ ds

g(τ , y)q(τ , y) dy d τ ds ,

for (t, x) ∈ D0 . Lemma 3.5.2. Let u, p, q, g ∈ C(H, R+ ). If u(x, y) p(x, y) + q(x, y) for (x, y) ∈ H, then u(x, y) p(x, y) + q(x, y)

× exp

g(s,t)u(s,t) dt ds, B0,x Ha,b

g(s,t)p(s,t) dt ds

B0,x Ha,b

g(s,t)q(s,t) dt ds ,

B0,x Ha,b

for (x, y) ∈ H. We start with the following theorem which deals with the estimate on the difference between the two approximate solutions of equation (3.5.1). Theorem 3.5.1.

Let ui (t, x) (i = 1, 2) be respectively εi -approximate solutions of equation

(3.5.1) on D0 . Suppose that the function F in equation (3.5.1) satisfies the condition |F(t, x, τ , y, u) − F(t, x, τ , y, v)| q(t, x)g(τ , y)|u − v|,

(3.5.3)

where q, g ∈ C(D0 , R+ ). Then

t s g(τ , y) dy d τ ds |u1 (t, x) − u2 (t, x)| (ε1 + ε2 ) 1 + q(t, x) 0

× exp for (t, x) ∈ D0 .

0

B

t s 0

0

B

g(τ , y)q(τ , y) dy d τ ds

,

(3.5.4)

124

Proof.


Since ui (t, x)(i = 1, 2) are respectively εi -approximate solutions of equation

(3.5.1), we have

t s εi . ui (t, x) − h(t, x) + F(t, x, τ , y, u ( τ , y)) dy d τ ds i 0 0 B From the above inequality and using the elementary inequalities in (1.3.25) and following the proof of Theorem 3.4.4, we get |u1 (t, x) − u2 (t, x)| (ε1 + ε2 ) t s

+ 0

0

B

|F(t, x, τ , y, u1 (τ , y)) − F(t, x, τ , y, u2 (τ , y))|dy d τ ds.

(3.5.5)

Let w(t, x) = |u1 (t, x) − u2 (t, x)|, (t, x) ∈ D0 . From (3.5.5) and using (3.5.3), we have w(t, x) (ε1 + ε2 ) + q(t, x)

t s 0

0

B

g(τ , y)w(τ , y) dy d τ ds.

(3.5.6)

Now an application of Lemma 3.5.1 to (3.5.6) yields (3.5.4). Remark 3.5.1.

When u1 (t, x) is a solution of equation (3.5.1), we have ε1 = 0 and from

(3.5.4), we see that u2 (t, x) → u1 (t, x) as ε2 → 0. Furthermore, if we put ε1 = ε2 = 0 in (3.5.4), then the uniqueness of solutions of equation (3.5.1) is established. Consider the equation (3.5.1) together with the equation t s

v(t, x) = h(t, x) + 0 B n n ∈ C(Ω × R , R ). 0

where h ∈ C(D0 , Rn ), F

F(t, x, τ , y, v(τ , y)) dy d τ ds,

(3.5.7)

In the next theorem we provide conditions concerning the closeness of solutions of equations (3.5.1) and (3.5.7). Theorem 3.5.2.

Suppose that the function F in equation (3.5.1) satisfies (3.5.3) and there

exist constants δi 0 (i = 1, 2) such that |h(t, x) − h(t, x)| δ1 , t s 0

0

B

(3.5.8)

|F(t, x, τ , y, z) − F(t, x, τ , y, z)|dy d τ ds δ2 ,

(3.5.9)

where h, F and h, F are given as in (3.5.1) and (3.5.7). Let u(t, x) and v(t, x) for (t, x) ∈ D0 , be solutions of equations (3.5.1) and (3.5.7) respectively. Then t s g(τ , y) dy d τ ds |u(t, x) − v(t, x)| (δ1 + δ2 ) 1 + q(t, x) 0

× exp for (t, x) ∈ D0 .

0

0

0

B

t s B


,

(3.5.10)


Proof.

125

Let r(t, x) = |u(t, x) − v(t, x)|, (t, x) ∈ D0 . Using the facts that u(t, x), v(t, x) are

solutions of equations (3.5.1), (3.5.7) and the hypotheses, we have r(t, x) |h(t, x) − h(t, x)|

t s

+ 0

0

B

0

0

B

t s

+

|F(t, x, τ , y, u(τ , y)) − F(t, x, τ , y, v(τ , y))|dy d τ ds |F(t, x, τ , y, v(τ , y)) − F(t, x, τ , y, v(τ , y))|dy d τ ds t s

(δ1 + δ2 ) + q(t, x)

0

0

B

g(τ , y)r(τ , y) dy d τ ds.

(3.5.11)

Now, an application of Lemma 3.5.1 to (3.5.11) yields (3.5.10). Remark 3.5.2.

The result given in Theorem 3.5.2 relates the solutions of equations

(3.5.1) and (3.5.7) in the sense that if F is close to F and h is close to h, then the solutions of equations (3.5.1) and (3.5.7) are also close to each other. A slight variation of Theorem 3.5.2 is embodied in the following theorem. Theorem 3.5.3. Suppose that the functions F and F in equations (3.5.1) and (3.5.7) satisfies the condition |F(t, x, τ , y, u) − F(t, x, τ , y, v)| q(t, x)g(τ , y)|u − v|,

(3.5.12)

where q, g ∈ C(D0 , R+ ) and (3.5.8) holds. Let u(t, x) and v(t, x) be solutions of equations (3.5.1) and (3.5.7), respectively, on D0 . Then t s g(τ , y) dy d τ ds |u(t, x) − v(t, x)| δ1 1 + q(t, x) × exp

0

t s 0

0

B

0

B


,

(3.5.13)

for (t, x) ∈ D0 . The proof of the above theorem is similar to that of Theorem 3.5.2, with suitable modifications, and hence we omit the details. We next consider the equations

t s

F(t, x, τ , y, u(τ , y), μ ) dy d τ ds,

(3.5.14)

F(t, x, τ , y, u(τ , y), μ0 ) dy d τ ds,

(3.5.15)

u(t, x) = h(t, x) + 0

0

t s

u(t, x) = h(t, x) + for (t, x) ∈ D0 , where h ∈ C(D0

B

0 0 B , Rn ), F ∈ C(Ω × Rn × R, Rn )

and μ , μ0 are parameters.

The following theorem shows the dependency of solutions of equations (3.5.14) and (3.5.15) on parameters.

126


Theorem 3.5.4. Suppose that the function F in (3.5.14), (3.5.15) satisfy the conditions |F(t, x, τ , y, u, μ ) − F(t, x, τ , y, u, μ )| q0 (t, x)g0 (τ , y)|u − u|, |F(t, x, τ , y, u, μ ) − F(t, x, τ , y, u, μ0 )| N|μ − μ0 |,

(3.5.16) (3.5.17)

where q0 , g0 ∈ C(D0 , R+ ) and N 0 is a constant. Let u1 (t, x) and u2 (t, x) be the solutions of equations (3.5.14) and (3.5.15) respectively. Then t s g0 (τ , y) dy d τ ds |u1 (t, x) − u2 (t, x)| N|μ − μ0 | 1 + q0 (t, x) 0

× exp

t s 0

0

0

B

B

g0 (τ , y)q0 (τ , y) dy d τ ds

,

(3.5.18)

for (t, x) ∈ D0 . Proof.

Let z(t, x) = |u1 (t, x)−u2 (t, x)|, (t, x) ∈ D0 . Using the facts that u1 (t, x) and u2 (t, x)

are respectively the solutions of equations (3.5.14) and (3.5.15) and the hypotheses, we have z(t, x)

t s 0

0

t s

+ 0

0

B

B

|F(t, x, τ , y, u1 (τ , y), μ ) − F(t, x, τ , y, u2 (τ , y), μ )|dy d τ ds

|F(t, x, τ , y, u2 (τ , y), μ ) − F(t, x, τ , y, u2 (τ , y), μ0 )|dy d τ ds

N|μ − μ0 | + q0 (t, x)

t s 0

0

B

g0 (τ , y)z(τ , y) dy d τ ds.

(3.5.19)

Now an application of Lemma 3.5.1 to (3.5.19) yields (3.5.18), which shows the dependency of solutions of equations (3.5.14) and (3.5.15) on parameters. Below, we apply the inequality in Lemma 3.5.2 to obtain the uniqueness and explicit estimates on the solutions of equation (3.5.2). One can formulate results similar to those given in Theorems 3.5.1–3.5.4 for the equation (3.5.2) by using Lemma 3.5.2. The following theorem deals with the uniqueness of solutions of equation (3.5.2). Theorem 3.5.5. Suppose that the function K in equation (3.5.2) satisfies the condition |K(x, y, s,t, u) − K(x, y, s,t, v)| q(x, y)g(s,t)|u − v|, where q, g ∈ C(H, R+ ).Then the equation (3.5.2) has at most one solution on H.

(3.5.20)


Proof.

127

Let u(x, y) and v(x, y) be two solutions of equation (3.5.2) on H. Using these facts

and the hypotheses (3.5.20), we have |u(x, y) − v(x, y)|

B0,x Ha,b

q(x, y)

|K(x, y, s,t, u(s,t)) − K(x, y, s,t, v(s,t))|dt ds

B0,x Ha,b

g(s,t)|u(s,t) − v(s,t)|dt ds.

(3.5.21)

Now a suitable application of Lemma 3.5.2 (when p(x, y) = 0) to (3.5.21) yields |u(x, y) − v(x, y)| 0, which implies u(x, y) = v(x, y). Thus there is at most one solution to equation (3.5.2) on H. The next theorem deals with the estimate on the solution of equation (3.5.2). Theorem 3.5.6.

Suppose that the function K in equation (3.5.2) satisfies the condition |K(x, y, s,t, u)| q(x, y)g(s,t)|u|,

(3.5.22)

where q, g ∈ C(H, R+ ). If u(x, y) is any solution of equation (3.5.2) on H, then g(s,t)| f (s,t)|dt ds |u(x, y)| | f (x, y)| + q(x, y) × exp

B0,x Ha,b


(3.5.23)

B0,x Ha,b

for (x, y) ∈ H. Proof.

Using the fact that u(x, y) is a solution of equation (3.5.2) and hypotheses, we have |u(x, y)| | f (x, y)| +

B0,x Ha,b

| f (x, y)| + q(x, y)

|K(x, y, s,t, u(s,t))|dt ds

g(s,t)|u(s,t)|dt ds.

(3.5.24)

B0,x Ha,b

Now an application of Lemma 3.5.2 to (3.5.24) yields (3.5.23). A slight variation of Theorem 3.5.6 is given in the following theorem. Theorem 3.5.7.


(3.5.20). If u(x, y) is any solution of equation (3.5.2) on H, then g(s,t)Q(s,t) dt ds |u(x, y) − f (x, y)| Q(x, y) + q(x, y) × exp for (x, y) ∈ H, where

Q(x, y) = B0,x Ha,b

for (x, y) ∈ H.

B0,x Ha,b


(3.5.25)

B0,x Ha,b

|K(x, y, σ , τ , f (σ , τ ))|d τ d σ ,

(3.5.26)

128


Proof.

Using the fact that u(x, y) is a solution of equation (3.5.2) and hypotheses, we

observe that

|u(x, y) − f (x, y)|

+ B0,x Ha,b

B0,x Ha,b

|K(x, y, s,t, f (s,t))|dt ds

|K(x, y, s,t, u(s,t)) − K(x, y, s,t, f (s,t))|dt ds

Q(x, y) + q(x, y)

B0,x Ha,b

g(s,t)|u(s,t) − f (s,t)|dt ds,

(3.5.27)

for (x, y) ∈ H. Now an application of Lemma 3.5.2 to (3.5.27) gives the required estimate in (3.5.25). Remark 3.5.3.

We note that the equation (3.5.1) contains as a special case, the study of

integral equation (16). Moreover, the generality of the equation (3.5.2) allow us to include in the special cases (i) n = 1, m = 1, (ii) n = 2, m = 1, (iii) n = 2, m = 2; respectively the study of integral equations of the forms u(x1 , y1 ) = f (x1 , y1 ) +

x 1 b1 0

a1

K(x1 , s1 , y1 , u(s1 , y1 )) dy1 ds1 ,

(3.5.28)

u(x1 , x2 , y1 ) = f (x1 , x2 , y1 ) +

x 1 x 2 b1 0

0

a1

K(x1 , x2 , s1 , s2 , y1 , u(s1 , s2 , y1 )) dy1 ds2 ds1 ,

(3.5.29)

u(x1 , x2 , y1 , y2 ) = f (x1 , x2 , y1 , y2 ) +

x1 x2 b1 b2 0

3.6

0

a1

a2

K(x1 , x2 , s1 , s2 , y1 , y2 , u(s1 , s2 , y1 , y2 ))dy2 dy1 ds2 ds1 .

(3.5.30)

General Volterra-Fredholm-type integral equations

Consider the general nonlinear Volterra-Fredholm-type integral equations of the forms: t

u(t, x) = h(t, x) +

F(t, x, s, y, u(s, y), (Tu)(s, y)) dy ds,

(3.6.1)

B

0

and

t

u(t, x) = e(t, x) +

G(t, x, s, y, u(s, y)) dy ds 0

B

∞

+

L(t, x, s, y, u(s, y)) dy ds, 0

where

(3.6.2)

B

t

(Tu)(t, x) = 0

B

K(t, x, τ , z, u(τ , z)) dz d τ ,

(3.6.3)


129

h, F, K and e, G, L are given functions and u is the unknown function. In this section we shall use the notations and definitions given in sections 3.2 and 3.4 without further mention. We assume that h, e ∈ C(D0 , Rn ), K ∈ C(Ω × Rn , Rn ), F ∈ C(Ω × Rn × Rn , Rn ), G ∈ C(Ω × Rn , Rn ), L ∈ C(D20 × Rn , Rn ). The present section is devoted to study some fundamental qualitative properties of solutions of equations (3.6.1) and (3.6.2), which we hope will serve as a source for further investigations. First, we formulate the following theorem concerning the existence of a unique solution of equation (3.6.1). Theorem 3.6.1. Suppose that (i) the functions F and K in equation (3.6.1) satisfy the conditions |F(t, x, s, y, u, v) − F(t, x, s, y, u, v)| r(t, x, s, y) [|u − u| + |v − v|] ,

(3.6.4)

and |K(t, x, τ , z, u) − K(t, x, τ , z, u)| m(t, x, τ , z)|u − u|,

(3.6.5)

where r, m ∈ C(Ω, R+ ), (ii) for λ as in (3.4.2) (b1 ) there exists a nonnegative constant α < 1 such that t t r(t, x, s, y) exp(λ (s + |y|)) + m(s, y, τ , z) exp(λ (τ + |z|)) dz d τ dy ds 0

0

B

B

α exp(λ (t + |x|)),

(3.6.6)

(b2 ) there exists a nonnegative constant β such that |h(t, x)| +

t 0

B

|F(t, x, s, y, 0, (T 0)(s, y))|dy ds β exp(λ (t + |x|)),

(3.6.7)

where h, F are as defined in equation (3.6.1). Under the assumptions (i) and (ii) the equation (3.6.1) has a unique solution u(t, x) on D0 in Z. The proof follows by the similar arguments as in the proof of Theorem 3.4.1. We omit the details. The following theorem concerning the estimate on the solution of equation (3.6.1) holds.

130


Theorem 3.6.2. Suppose that the functions F, K in equation (3.6.1) satisfy the conditions |F(t, x, s, y, u, v) − F(t, x, s, y, u, v)| q(t, x) f (s, y) [|u − u| + |v − v|] ,

(3.6.8)

|K(t, x, τ , z, u) − K(t, x, τ , z, u)| q(t, x)g(τ , z)|u − u|,

(3.6.9)

where q, f , g ∈ C(D0 , R+ ). Let t c = sup h(t, x) + F(t, x, s, y, 0, (T 0)(s, y)) dy ds < ∞,

(3.6.10)

0

(t,x)∈D0

B

where h, F are the functions in equation (3.6.1). If u(t, x) is any solution of equation (3.6.1) on D0 , then

|u(t, x)| c 1 + q(t, x)

× exp

[ f (s, y) + g(s, y)] 0

B

t s

t

B

q(τ , z)[ f (τ , z) + g(τ , z)]dz d τ

dy ds ,

(3.6.11)

for (t, x) ∈ D0 . Proof.

Using the fact that u(t, x) is a solution of equation (3.6.1) and hypotheses, we have t F(t, x, s, y, 0, (T 0)(s, y)) dy ds |u(t, x)| h(t, x) + B

0

t

+ 0

B

|F(t, x, s, y, u(s, y), (Tu)(s, y)) − F(t, x, s, y, 0, (T 0)(s, y))| dy ds

c + q(t, x)

t

f (s, y) 0

B

s × |u(s, y)| + q(s, y) g(τ , z)|u(τ , z)|dz d τ dy ds. 0

(3.6.12)

B

Now applying Theorem 3.2.7 part (c13 ) to (3.6.12) yields (3.6.11). In the next theorem we will employ Theorem 3.2.7 part (c13 ) to obtain the uniqueness of solutions of equation (3.6.1). Theorem 3.6.3. Suppose that the functions F, K in equation (3.6.1) satisfy the conditions (3.6.8), (3.6.9) respectively. Then the equation (3.6.1) has at most one solution on D0 .


131

Let u1 (t, x) and u2 (t, x) be two solutions of equation (3.6.1) on D0 and w(t, x) =

Proof.

|u1 (t, x) − u2 (t, x)|, (t, x) ∈ D0 . From the hypotheses, we have w(t, x)

t

B

0

q(t, x)

|F(t, x, s, y, u1 (s, y), (Tu1 )(s, y)) − F(t, x, s, y, u2 (s, y), (Tu2 )(s, y))|dy ds

t 0

B

s f (s, y) w(s, y) + q(s, y) g(τ , z)w(τ , z) dz d τ dy ds. 0

(3.6.13)

B

Now applying Theorem 3.2.7 part (c13 ) (with p(t, x) = 0) to (3.6.13) yields |u1 (t, x) − u2 (t, x)| 0, which implies u1 (t, x) = u2 (t, x). Thus there is at most one solution to equation (3.6.1) on D0 . Now, we present a result on the continuous dependence of solution of equation (3.6.1) on the functions involved therein. Consider the equation (3.6.1) and the corresponding equation v(t, x) = h(t · x) +

t

F(t, x, s, y, v(s, y), (T v)(s, y)) dy ds, 0

for (t, x) ∈ D0 , where

(3.6.14)

B

t

(T v)(t, x) = 0

B

K(t, x, τ , z, v(τ , z)) dz d τ ,

(3.6.15)

h ∈ C(D0 , Rn ), K ∈ C(Ω × Rn , Rn ), F ∈ C(Ω × Rn × Rn , Rn ). Theorem 3.6.4. Suppose that the functions F, K in equation (3.6.1) satisfy the conditions (3.6.8), (3.6.9) respectively. Furthermore, suppose that v(t, x) is a given solution of equation (3.6.14) on D0 and |h(t, x) − h(t, x)| +

t 0

B

|F(t, x, s, y, v(s, y), (T v)(s, y))

−F(t, x, s, y, v(s, y), (T v)(s, y))|dy ds ε ,

(3.6.16)

where h, F, Tu and h, F, T v are as in equations (3.6.1) and (3.6.14) respectively and ε > 0 is an arbitrary small constant. Then the solution u(t, x) of equation (3.6.1) on D0 depends continuously on the functions involved in equation (3.6.1). Proof.

Let w(t, x) = |u(t, x) − v(t, x)|, for (t, x) ∈ D0 . Using the hypotheses, we have w(t, x) |h(t, x) − h(t, x)| t

+ 0

B

t

+ 0

B

|F(t, x, s, y, u(s, y), (Tu)(s, y)) − F(t, x, s, y, v(s, y), (T v)(s, y))|dy ds |F(t, x, s, y, v(s, y), (T v)(s, y)) − F(t, x, s, y, v(s, y), (T v)(s, y))|dy ds

132


ε + q(t, x)

t 0

B

s f (s, y) w(s, y) + q(s, y) g(τ , z)w(τ , z) dz d τ dy ds. 0

(3.6.17)

B

Now using Theorem 3.2.7 part (c13 ) to (3.6.17) yields t [ f (s, y) + g(s, y)] |u(t, x) − v(t, x)| ε 1 + q(t, x) 0

× exp

t s

B

B

q(τ , z)[ f (τ , z) + g(τ , z)]dz d τ

dy ds ,

(3.6.18)

for (t, x) ∈ D0 . From (3.6.18) it follows that the solution u(t, x) of equation (3.6.1) depends continuously on the functions involved therein. We now turn our attention to some basic qualitative properties of solutions of equation (3.6.2). Here, we note that one can obtain the existence and uniqueness result for the solution of equation (3.6.2) by using the idea employed in Theorem 3.4.1. Now we formulate the following theorem which estimates the difference between the two approximate solutions of equation (3.6.2). Theorem 3.6.5. Suppose that (i) The functions G, L in equation (3.6.2) satisfy the conditions |G(t, x, s, y, u) − G(t, x, s, y, v)| q(t, x) f (s, y)|u − v|,

(3.6.19)

|L(t, x, s, y, u) − L(t, x, s, y, v)| r(t, x)g(s, y)|u − v|,

(3.6.20)

where q, f , r, g ∈ C(D0 , R+ ), (ii) the functions ui (t, x) ∈ C(D0 , Rn ) (i = 1, 2) be respectively εi -approximate solutions of equation (3.6.2) on D0 , (iii) let d, K2 (t, x) be as in (3.2.18), (3.2.21) respectively and D1 =

1 1−d

∞ 0

B

g(s, y)A1 (s, y) dy ds,

(3.6.21)

where A1 (t, x) is defined by the right hand side of (3.2.20) by replacing p(t, x) by ε1 + ε2 . Then |u1 (t, x) − u2 (t, x)| A1 (t, x) + D1 K2 (t, x), for (t, x) ∈ D0 .

(3.6.22)


Proof.

133

Since ui (t, x) (i = 1, 2) are respectively εi -approximate solutions of equation

(3.6.2), we have

t ui (t, x) − e(t, x) + G(t, x, s, y, ui (s, y)) dy ds B

0

+

∞ 0

L(t, x, s, y, ui (s, y)) dy ds εi . B

(3.6.23)

From (3.6.23) and following the proof of Theorem 3.4.4, we get |u1 (t, x) − u2 (t, x)| ε1 + ε2 t

+ 0

+

B

∞ 0

B

|g(t, x, s, y, u1 (s, y)) − g(t, x, s, y, u2 (s, y))|dy ds |L(t, x, s, y, u1 (s, y)) − L(t, x, s, y, u2 (s, y))|dy ds.

(3.6.24)

Let w(t, x) = |u1 (t, x) − u2 (t, x)|, (t, x) ∈ D0 . From (3.6.24) and using (3.6.19), (3.6.20), we have w(t, x) (ε1 + ε2 ) + q(t, x)

+r(t, x)

t

f (s, y)w(s, y) dy ds 0

B

∞

g(s, y)w(s, y) dy ds. 0

(3.6.25)

B

Now an application of Theorem 3.2.4 part (c7 ) to (3.6.25) yields (3.6.22). The next result is a consequence of Theorem 3.6.5. Theorem 3.6.6. Suppose that (i) the functions G, L in equation (3.6.2) satisfy the conditions (3.6.19), (3.6.20) respectively, (ii) the functions uε (t, x) and u(t, x) be respectively an ε -approximate solution and any solution of equation (3.6.2) on D0 , (iii) let d, K2 (t, x) be as in (3.2.18), (3.2.21) respectively and D2 =

1 1−d

∞ 0

B


(3.6.26)

where A2 (t, x) is defined by the right hand side of (3.2.20) by replacing p(t, x) by ε . Then |uε (t, x) − u(t, x)| A2 (t, x) + D2 K2 (t, x), for (t, x) ∈ D0 .

(3.6.27)

134


Proof.

Let w(t, x) = |uε (t, x) − u(t, x)|, (t, x) ∈ D0 . From the hypotheses, we observe that t w(t, x) = uε (t, x) − e(t, x) − G(t, x, s, y, uε (s, y)) dy ds 0

− t

+ B

0

∞

ε + q(t, x)

0

B

L(t, x, s, y, uε (s, y)) dy ds

|G(t, x, s, y, uε (s, y)) − G(t, x, s, y, u(s, y))|dy ds

+ 0

∞

B

B

|L(t, x, s, y, uε (s, y)) − L(t, x, s, y, u(s, y))|dy ds

t

f (s, y)w(s, y) dy ds + r(t, x) B

0

∞

g(s, y)w(s, y) dy ds. 0

(3.6.28)

B

Now an application of Theorem 3.2.4 part (c7 ) to (3.6.28) yields (3.6.27). In the following theorem we present conditions for continuous dependence on parameters of solutions of equations t

u(t, x) = ei (t, x) +

G(t, x, s, y, u(s, y)) dy ds + 0

∞

L(t, x, s, y, u(s, y)) dy ds, (3.6.29) 0

B

B

for (t, x) ∈ D0 , i = 1, 2; where ei ∈ C(D0 , Rn ) and G, L are as in equation (3.6.2). Theorem 3.6.7. Suppose that (i) the functions G, L in (3.6.29) satisfy the conditions (3,6,19), (3.6.20) respectively, (ii) the functions ui (t, x)(i = 1, 2) are respectively the εi -approximate solutions of equations in (3.6.29) corresponding to i = 1, 2, (iii) let d, K2 (t, x) be as in (3.2.18), (3.2.21) respectively and D3 =

1 1−d

∞ 0

B


(3.6.30)

where A3 (t, x) is defined by the right hand side of (3.2.20) by replacing p(t, x) by ε1 + ε2 + |e1 (t, x) − e2 (t, x)|. Then |u1 (t, x) − u2 (t, x)| A3 (t, x) + D3 K2 (t, x),

(3.6.31)

for (t, x) ∈ D0 . The proof follows by the similar arguments as in the proof of Theorem 3.6.6 with suitable modifications. Here, we omit the details.


3.7 3.7.1

135

Miscellanea Bacot¸iu [7]

Let (X, X ) be a Banach space. Consider the nonlinear integral equation of VolterraFredholm-type

t

u(t, x) = g(t, x, u(t, x)) + t b

+ 0

a

H(t, x, s, u(s, x)) ds 0

K(t, x, s, y, u(s, y), u(φ1 (s, y), φ2 (s, y))) dy ds,

(3.7.1)

for all (t, x) ∈ [0, T ] × [a, b] := D; u ∈ C(D, X ), where b > a > 0 and T > 0. Assume that 2

(i) g ∈ C(D × X , X), H ∈ C(D × [0, T ] × X , X), K ∈ C(D × X 2 , X), φ1 ∈ C(D, [0, T ]) and

φ2 ∈ C(D, [a, b]); (ii) there exists Lg > 0 such that g(t, x, u) − g(t, x, v)X Lg u − vX , for all (t, x) ∈ D and u, v ∈ X; (iii) there exists LH > 0 such that H(t, x, s, u) − H(t, x, s, v)X LH u − vX , for all (t, x, s) ∈ D × [0, T ] and u, v ∈ X; (iv) there exists LK > 0 such that K(t, x, s, y, u, u) − K(t, x, s, y, v, v)X LK [u − vX + u − vX ] , 2

for all (t, x, s, y) ∈ D and u, v, u, v ∈ X; (v) Lg < 1; (vi) there exists τ > 0 such that 1 b−a α = Lg + LH + LK + max τ τ

t 0

b a

exp(τ [φ1 (s, y) − t])dy ds : t ∈ [0, T ] LK < 1.

Then (3.7.1) has a unique solution u ∈ C(D, X). 3.7.2

Kauthen [50]

Consider the linear Volterra-Fredholm integral equation t

u(t, x) = f (t, x) + 0

Ω

K(t, s, x, ξ )u(s, ξ ) d ξ ds,

where Ω is a closed subset of Rn (with n = 1, 2, 3) with assumptions (i) f ∈ C(I × Ω), where I = [0, T ], (ii) K ∈ C(D × Ω2 ), where D = {(t, s) : 0 s t T } and Ω2 = Ω × Ω.

(3.7.2)

136


Then the equation (3.7.2) has a unique solution u ∈ C(I × Ω) and this solution is given by t

u(t, x) = f (t, x) + 0

Ω

R(t, s, x, ξ ) f (s, ξ ) d ξ ds,

for (t, x) ∈ I × Ω, where the resolvent kernel R ∈ C(D × Ω2 ) associated with the kernel K(t, s, x, ξ ) is the limit of the Neumann series for K and solves the resolvent equation R(t, s, x, ξ ) = K(t, s, x, ξ ) +

t

Ω

0

K(t, v, x, z)R(v, s, z, ξ )dz dv,

on D × Ω2 . 3.7.3

Diekmann [31]

Consider the nonlinear integral equation t

u(t, x) = 0

Ω

g(u(t − τ , ξ ))S0 (ξ )A(τ , x, ξ ) d ξ d τ + f (t, x),

(3.7.3)

where Ω is a closed subset of Rn . Let BC(Ω) denote the Banach space of the bounded continuous functions on Ω equipped with supremum norm and CT = C([0, T ], BC(Ω)) be the Banach space of continuous functions on [0, T ] with values in BC(Ω), equipped with the norm zCT = sup z[t]BC(Ω) . 0tT

When looking at u as an element of CT , write u[t](x) instead of u(t, x). With this convention (3.7.3) can be written as u[t] + Qu[t] + f [t], where Q is defined by

t

Qu[t](x) = 0

Ω

(3.7.4)

g(u[t − τ ](ξ ))S0 (ξ )A(τ , x, ξ ) d ξ d τ .

Assume that (i) S0 ∈ L∞ (Ω); S0 is nonnegative, (ii) g : R → R is continuous; g(0) = 0, (iii) A(·, ·, ·) is defined and nonnegative on [0, ∞) × Ω × Ω; for every x ∈ Ω and every T > 0, A(·, x, ·) ∈ L1 ([0, T ] × Ω), (iv) let

η (t, x) =

t 0

Ω

A(τ , x, ξ )d ξ d τ

and T > 0 be arbitrary, then the family of functions on [0, T ], {η (·, x) : x ∈ Ω} is uniformly bounded and equicontinuous.


137

(v) For every T > 0 and every ε > 0 there exists δ = δ (ε , T ) > 0 such that if x1 , x2 ∈ Ω and |x1 − x2 | < δ , then

T Ω

0

|A(τ , x1 , ξ ) − A(τ , x2 , ξ )|d ξ d τ < ε ,

(vi) f : [0, ∞) → BC(Ω) is continuous. Then 1. For every T > 0 and every u ∈ CT , Qu ∈ CT . 2. Suppose g is locally Lipschitz continuous, then there exists a T > 0 such that (3.7.3) has a unique solution u in CT and the mapping f → u is continuous from CT into CT . 3. Suppose g is uniformly Lipschitz continuous, then (3.7.3) has a unique continuous solution u : [0, ∞) → BC(Ω). 3.7.4

Diekmann [31]

Suppose that the functions g, f in (3.7.3) satisfy 1. g(y) > 0 for y > 0 and f [t] 0 for all t 0, then u[t] 0 on the domain of definition of u. 2. In addition, g is monotone nondecreasing and f [t + h] f [t] for all h 0, then u[t +h] u[t] for all h 0 and t 0 such that t + h is in the domain of definition of u. 3. In addition to the assumptions 1, 2, suppose that g is bounded and uniformly Lipschitz continuous on [0, ∞), that the subset { f [t] : t 0} of BC(Ω) is uniformly bounded and equicontinuous and that A satisfies (vii) for each x ∈ Ω

t 0

A(τ , x, ·) d τ →

∞ 0

A(τ , x, ·) d τ ,

in L1 (Ω) as t → ∞, and for some C > 0, ∞

sup

x∈Ω Ω 0

A(τ , x, ξ ) d τ d ξ < C.

(viii) For each ε > 0 there exists δ = δ (ε ) > 0 such that if x1 , x2 ∈ Ω and |x1 − x2 | < δ , then

∞ Ω 0

|A(τ , x1 , ξ ) − A(τ , x2 , ξ )|d τ d ξ < ε .

Then the solution u of (3.7.3) is defined on [0, ∞) and there exists u[∞] ∈ BC(Ω) such that, as t → ∞, u[t] → u[∞] in BC(Ω) if Ω is compact, and uniformly on compact subsets of Ω if Ω is not compact. Moreover, u[∞] satisfies the limit equation

u[∞](x) =

Ω

g(u[∞](ξ ))S0 (ξ )

∞ 0

A(τ , x, ξ ) d τ d ξ + f [∞](x).

138

3.7.5


Pachpatte [116]

Under the notations as in section 3.2, let u ∈ C(D0 , R1 ), k, D1 k ∈ C(Ω, R+ ) and c 1 is a constant. (f1 ) If u(t, x) c +

t

k(t, x, s, y)u(s, y) log u(s, y) dy ds, B

0

for (t, x) ∈ D0 , then

u(t, x) cexp(

t

0 A(σ ,x) d σ

),

for (t, x) ∈ D0 , where A(t, x) is given by (3.2.28). (f2 ) Let g(u) be as in Theorem 3.2.5 part (c10 ). If u(t, x) c +

t

k(t, x, s, y)u(s, y)g(log u(s, y)) dy ds, 0

B

for (t, x) ∈ D0 , then for 0 t t1 ; t, t1 ∈ R+ , x ∈ B, u(t, x) W where W,

W −1 ,

−1

t

W (log c) + 0

A(σ , x) d σ ,

A(t, x) are as in Theorem 3.2.5 and t1 ∈ R+ is chosen so that t W (log c) + A(σ , x) d σ ∈ Dom W −1 , 0

for all t ∈ R+ lying in the interval 0 t t1 , x ∈ B. 3.7.6

Pachpatte [119]

Under the notations as in section 3.3, let u, p, q, f ∈ C(H, R+ ) and r > 1 be a real constant. (f3 ) Let L ∈ C(H × R+ , R+ ) satisfies the condition (3.3.20) in Theorem 3.3.5. If ur (x, y) p(x, y) + q(x, y)

for (x, y) ∈ H, then

L(s,t, u(s,t)) dt ds, B0,x Ha,b

r − 1 p(s,t) + dt ds L s,t, r r B0,x Ha,b 1 r q(s,t) r − 1 p(s,t) + dt ds M s,t, , × exp r r r B0,x Ha,b for (x, y) ∈ H, where M is the function given in Theorem 3.3.5. u(x, y) p(x, y) + q(x, y)

(f4 ) If ur (x, y) p(x, y) + q(x, y) for (x, y) ∈ H, then u(x, y) p(x, y) + q(x, y)

for (x, y) ∈ H.

f (s,t)u(s,t) dt ds, B0,x Ha,b

r − 1 p(s,t) + dt ds r r B0,x Ha,b 1 r q(s,t) dt ds f (s,t) , × exp r B0,x Ha,b

f (s,t)


3.7.7

139

Pachpatte [105]

Under the notations as in section 3.2, consider the Volterra-Fredholm-type integral equations

t

v(t, x) = h1 (t, x) +

0

L(t, x, s, y, v(s, y)) dy ds,

(3.7.5)

M(t, x, s, y, w(s, y)) dy ds,

(3.7.6)

B

t

w(t, x) = h2 (t, x) +

0

B

with assumptions (i) h1 , h2 ∈ C(D0 , R), L, M ∈ C(Ω × R, R), (ii) the function L in equation (3.7.5) satisfies the condition |L(t, x, s, y, v) − L(t, x, s, y, w)| q(t, x) f (s, y)|v − w|, where q, f ∈ C(D0 , R+ ). Then for every given solution w ∈ C(D0 , R) of equation (3.7.6) and v ∈ C(D0 , R) a solution of equation (3.7.5), the estimate |v(t, x) − w(t, x)| [h(t, x) + r(t, x)]

t

t

+q(t, x)

f (s, y)[h(s, y) + r(s, y)] exp 0

s

B

B


dy ds,

holds for (t, x) ∈ D0 , in which h(t, x) = |h1 (t, x) − h2 (t, x)|, t

r(t, x) = 0

B

|L(t, x, s, y, w(s, y)) − M(t, x, s, y, w(s, y))|dy ds,

for (t, x) ∈ D0 . 3.7.8

Pachpatte [79]

Consider the Volterra-Fredholm integral equation of the form t

u(t, x) = 0

t

+ 0

∂Ω

Ω

G(t, x; s, y)F(s, y, u(s, y)) dy ds

G(t, x; s, y)ψ (s, y) dy ds +

Ω

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

(3.7.7)

arising in the study of nonlinear parabolic system

∂u − Lu = F(t, x, u), ∂t

t ∈ (0, T ],

x ∈ Ω.

(3.7.8)

140


with the given boundary and initial conditions

∂u + b(x)u = c0 (x), ∂ν

t ∈ (0, T ],

u(0, x) = u0 (x),

x ∈ ∂ Ω,

(3.7.9)

x ∈ Ω,

(3.7.10)

where Ω is a bounded domain in Rn and ∂ Ω is a smooth boundary of Ω and L is the uniformly elliptic operator of the form n

L=

∑

ai j (x)

i, j=1

n ∂2 ∂ + ∑ ai (x) . ∂ xi ∂ x j i=1 ∂ xi

For detailed derivation of (3.7.7) and the assumptions on the functions involed in (3.7.8)– (3.7.10), see [37]. Let D = (0, T ] × Ω and D is the closure of D and in (3.7.7), G is the fundamental solution of

∂w ∂t

− Lw = 0. The second integral in (3.7.7) is a single layer

potential with the density ψ (t, x) which can be determined from the following Volterratype integral equation

ψ (t, x) = where

t 0

∂Ω

R(t, x; s, y)ψ (s, y) dy ds + H(t, x, u(t, x)),

∂ R(t, x; s, y) = 2 G(t, x; s, y) + b(x)G(t, x; s, y) , ∂ν

H(t, x, u(t, x)) =

t

Ω

R(t, x; 0, y)u0 (y) dy +

In fact ψ is given by (see [37, p. 145])

ψ (t, x) = H(t, x, u(t, x)) + 2

0

Ω

R(t, x; s, y)F(s, y, u(s, y)) dy ds − c0 (x).

t 0

∂Ω

R(t, x; s, y)H(s, y, u(s, y)) dy ds.

Consider the equation (3.7.7), under the following assumptions: (H1 ) the function F satisfies the condition |F(t, x, u) − F(t, x, v)| g(t, x)|u − v|, where g ∈ C(D, R+ ), (H2 ) there exist nonnegative constants Qi (i = 1, 2, 3) and a constant μ > 0 such that t 0

t 0 ∂Ω

Ω

|G(t, x; s, y)|g(s, y) exp( μ (s + |y|)) dy ds Q1 exp( μ (t + |x|)),

|G(t, x; s, y)| exp(μ (s + |y|)) + 2

s

|R(s, y; τ , z)| exp(μ (τ + |z|))dz d τ dyds

0 ∂Ω

Q2 exp(μ (t + |x|)),


t 0

Ω

141

|R(t, x; s, y)|g(s, y) exp(μ (s + |y|)) dy ds Q3 exp(μ (t + |x|)),

(H3 ) there exist nonnegative constants N1 , N2 such that

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

Ω

|R(t, x; 0, y)||u0 (y)|dy +

Ω

t

|G(t, x; s, y)||F(s, y, 0)|dyds N1 exp( μ (t + |x|)),

0 Ω

t

|R(t, x; s, y)||F(s, y, 0)|dyds + |c0 (x)| N2 exp( μ (t + |x|)).

0 Ω

(f5 ) Suppose that the assumptions (H1 )–(H3 ) hold and assume that 0 Q1 + Q2 Q3 < 1. Then equation (3.7.7) has a unique solution u(t, x) in Z on D, where Z is the space of functions as defined in section 3.4. 3.8

Notes

Mixed Volterra-Fredholm-type integral equations occur in a natural way while studying parabolic equations which describe diffusion or heat transfer phenomena and other areas of science and technology. Most of the basic problems for various types of such equations are still in a very early stage of development. The material included in sections 3.2 and 3.3 is adapted from the recent work of Pachpatte [99,103,105,107,116,119], which in fact is motivated by the desire to widen the scope of the integral inequalities with explicit estimates. Section 3.4 contains some basic results on the theory of nonlinear mixed Volterra-Fredholm-type integral equation and are adapted from Pachpatte [79,116]. The results in section 3.5 deals with the qualitative properties of solutions of certain general Volterra-Fredholm-type integral equations, which are inspired by the integral equations arising while studying certain partial differential and integrodifferential equations and they are adapted from Pachpatte [107,119]. Section 3.6 is devoted to the study of some basic qualitative properties of solutions of general nonlinear mixed Volterra-Fredholm-type integral equations arising from the study of certain physical models and are taken from [103,99]. Section 3.7 is dedicated to the presentation of a few results related to certain selected topics, which may stimulate the interest of the readers in pursuing further developments.

Chapter 4

Parabolic-type integrodifferential equations

4.1

Introduction

The study of many physical processes arising in science and engineering leads to the models of parabolic integrodifferential equations with different initial boundary conditions. These equations occur in several applications, such as in heat flow in materials with memory, control and optimization theories, reaction diffusion processes, epidemic models and various other fields of science and technology.

Considerable re-

search on the special kinds of parabolic integrodifferential equations has been carried out by using new mathematical ideas, approaches, and theories, see [1,5,8,9,31–33,59– 62,74,80,98,110,121,122,124,126,127,132,140,141] and the references given therein. The main goal of this chapter is to discuss the wellposedness related to certain nonlinear parabolic integrodifferential equations studied in [109,110,75,77,140]. There are other topics related to such equations, which we did not include here due to space limitation e.g., see the references quoted in [1,5,25,26,39,45,60,61,80,121,122,124,126,137,141]. A detailed survey, including comprehensive list of references of the early developments to the topic can be found in the monograph [5].

4.2

Basic integral inequalities

This section is concerned with some fundamental integral inequalities with explicit estimates which play a vital role in certain applications. In what follows, we use some of the notations and definitions in section 3.2 without further mention. Moreover, we denote by I = [a, b] ⊂ R (a < b), G = R+ × I and Ω0 = {(t, x, s) : 0 s t < ∞, x ∈ B}. In the following theorems we present the useful variants of the integral inequality given in [98] and also the integral inequalities given in [109]. 143

144


Theorem 4.2.1. Let u, f ∈ C(G, R+ ), h ∈ C(G × I, R+ ) and c 0 is a constant. (p1 ) If

t

u(t, x) c +

b

f (s, x)u(s, x) + 0

h(s, x, y)u(s, y) dy ds,

(4.2.1)

a

for (t, x) ∈ G, then u(t, x) cF(t, x) exp

t

for (t, x) ∈ G, where

0

h(s, x, y)F(s, y) dy ds ,

(4.2.2)

f (σ , x) d σ .

(4.2.3)

b a

t

F(t, x) = exp 0

(p2 ) Let g ∈ C(R+ , R+ ) be nondecreasing submultiplactive function and g(u) > 0 on (0, ∞). If

t

h(s, x, y)g(u(s, y)) dy ds,

(4.2.4)

for (t, x) ∈ G, then for 0 t t1 ; t, t1 ∈ R+ , x ∈ I, t b −1 W (c) + h(s, x, y)g(F(s, y)) dy ds , u(t, x) F(t, x)W

(4.2.5)

u(t, x) c +

b

f (s, x)u(s, x) + 0

a

0

a

where F(t, x) is given by (4.2.3) and W, W −1 are as in Theorem 3.2.5 part (c10 ) and t1 ∈ R+ is chosen so that

t b

W (c) + 0

a

h(s, x, y)g(F(s, y)) dy ds ∈ Dom W −1 ,

for all t ∈ R+ lying in the interval 0 t t1 and x ∈ I. Theorem 4.2.2.

Let u ∈ C(D0 , R+ ); r, D1 r ∈ C(Ω0 , R+ ), k, D1 k ∈ C(Ω, R+ ) and c 0 is

a constant. (p3 ) If u(t, x) c +

t

t

r(t, x, s)u(s, x) ds + 0

for (t, x) ∈ D0 , then u(t, x) cP(t, x) exp

0

B

t

0

k(s, x, s, y)u(s, y) dy ds,

(4.2.6)

A(σ , x) d σ ,

(4.2.7)

for (t, x) ∈ D0 , where P(t, x) = exp(Q(t, x)), in which

t

Q(t, x) = 0

r(η , x, η ) +

η 0

D1 r(η , x, ξ ) d ξ d η ,

(4.2.8)

(4.2.9)


and

145

t

k(t, x,t, y)P(t, y) dy +

A(t, x) =

0

B

B

D1 k(t, x, s, y)P(s, y) dy ds,

(4.2.10)

for (t, x) ∈ D0 . (p4 ) Let g(u) be as in Theorem 4.2.1 part (p2 ). If u(t, x) c +

t

t

r(t, x, s)u(s, x) ds +

k(t, x, s, y)g(u(s, y)) dy ds,

0

0

(4.2.11)

B

for (t, x) ∈ D0 , then for 0 t t2 ; t, t2 ∈ R+ , x ∈ B, t u(t, x) P(t, x)W −1 W (c) + A(σ , x) d σ ,

(4.2.12)

0

where P(t, x) is given by (4.2.8); W, W −1 are as in part (p2 ) and t2 ∈ R+ is chosen so that t W (c) + A(σ , x) d σ ∈ Dom W −1 , 0

for all t ∈ R+ lying in the interval 0 t t2 and x ∈ B, where A(t, x) is given by the right hand side of (4.2.10) by replacing k(t, x, s, y)P(s, y) by k(t, x, s, y)g(P(s, y)). We shall give the details of the proofs of (p2 ), (p3 )

Proofs of Theorems 4.2.1 and 4.2.2.

only, the proofs of (p1 ), (p4 ) can be completed by following the proofs of these inequalities and the results given in Chapter 3, sections 3.2 and 3.3. (p2 ) Define a function m(t, x) by

t b

m(t, x) = c +

h(s, x, y)g(u(s, y)) dy ds, 0

(4.2.13)

a

then (4.2.4) can be restated as u(t, x) m(t, x) +

t

f (s, x)u(s, x) ds.

(4.2.14)

0

It is easy to observe that m(t, x) is nonnegative for (t, x) ∈ G and nondecreasing in t ∈ R+ for every x ∈ I. Treating (4.2.14) as one-dimensional integral inequality in t ∈ R+ for every x ∈ I and a suitable application of the inequality given in [82, Theorem 1.3.1, p. 12] yields u(t, x) m(t, x)F(t, x).

(4.2.15)

From (4.2.13) and (4.2.15), we observe that m(t, x) c + c+ Setting

t b

h(s, x, y)g (F(s, y)m(s, y)) dy ds 0

a

0

a

t b

h(s, x, y)g(F(s, y))g (m(s, y)) dy ds.

(4.2.16)

b

e(s) =

h(s, x, y)g(F(s, y))g (m(s, y)) dy, a

(4.2.17)

146


for every x ∈ I, the inequality (4.2.16) can be restated as t

m(t, x) c + Let

e(s) ds.

(4.2.18)

0

t

z(t) = c +

e(s) ds,

(4.2.19)

0

then z(0) = c and m(t, x) z(t),

(4.2.20)

for t ∈ R+ and for every x ∈ I. From (4.2.19), (4.2.17) and (4.2.20), we observe that z (t) = e(t) b

=

h(t, x, y) g(F(t, y)) g(m(t, y)) dy a

g(z(t))

b

h(t, x, y) g(F(t, y)) dy.

(4.2.21)

a

Now by following the proof of Theorem 2.3.1 given in [82, p. 107], we get t b z(t) W −1 W (c) + h(s, x, y)g(F(s, y)) dy ds . 0

(4.2.22)

a

The desired inequality in (4.2.5) follows from (4.2.22), (4.2.20) and (4.2.15). (p3 ) Define a function n(t, x) by

t

n(t, x) = c +

k(t, x, s, y)u(s, y) dy ds, 0

(4.2.23)

B

then (4.2.6) can be restated as u(t, x) n(t, x) +

t

r(t, x, s)u(s, x) ds.

(4.2.24)

0

From the hypotheses, it is easy to observe that n(t, x) is nonnegative for (t, x) ∈ D0 and nondecreasing in t ∈ R+ for every x ∈ B. Treating (4.2.24) as one-dimensional integral inequality for every x ∈ B and a suitable application of the inequality given in [87, Theorem 1.2.1, Remark 1.2.1, p. 11] yields u(t, x) n(t, x)P(t, x).

(4.2.25)

From (4.2.23) and (4.2.25), we obtain n(t, x) c +

t

k(t, x, s, y)P(s, y)n(s, y) dy ds. 0

(4.2.26)

B

Now a suitable application of Theorem 3.2.5 part (c9 ) to (4.2.26) yields t n(t, x) c exp A(σ , x) d σ . 0

Using (4.2.27) in (4.2.25), we get the required inequality in (4.2.7).

(4.2.27)


4.3

147

Integrodifferential equation of Barbashin-type

E.A. Barabashin [8] first initiated the study of the integrodifferential equation of the form

b ∂ u(t, x) = c(t, x)u(t, x) + k(t, x, y)u(t, y) dy + f (t, x), (B) ∂t a which arise in mathematical modeling of many applied problems (see [5, section 19]). This

equation attracted the attention of many researchers and is now known in the literature as integrodifferential equation of Barbashin-type or simply Barbashin equation, see [5, p. 1]. For a detailed account on the study of such equations by using various techniques, see [5] and the references cited therein. In this section, we consider the nonlinear integrodifferential equation of Barbashin-type (see [110])

∂ u(t, x) = f (t, x, u(t, x)) + ∂t which satisfies the initial condition

b

g(t, x, y, u(t, y)) dy + h(t, x),

(4.3.1)

a

u(0, x) = u0 (x),

(4.3.2)

for (t, x) ∈ Δ, where f ∈ C(Δ × R, R), g ∈ C(Δ × I × R, R), h ∈ C (Δ, R), u0 ∈ C(I, R) are given functions and u is the unknown function to be found, in which I = [a, b] ⊂ R (a < b) and Δ = R+ ×I. Here, we focus our attention to study some fundamental qualitative properties of solutions of problem (4.3.1)–(4.3.2). Let E be the space of functions z ∈ C(Δ, R) which fulfil the condition |z(t, x)| = O(exp(λ (t + |x|))),

(4.3.3)

where λ > 0 is a constant. In the space E we define the norm |z|E = sup |z(t, x)|(exp(−λ (t + |x|))).

(4.3.4)

(t,x)∈Δ

It is easy to see that E is a Banach space and |z|E N, where N 0 is a constant. The existence and uniqueness of solutions of problem (4.3.1)–(4.3.2) is given in the following theorem. Theorem 4.3.1. Suppose that (i) the functions f , g in equation (4.3.1) satisfy the conditions | f (t, x, u) − f (t, x, u)| c(t, x)|u − u|,

(4.3.5)

|g(t, x, y, u) − g(t, x, y, u)| k(t, x, y)|u − u|,

(4.3.6)

where c ∈ C(Δ, R+ ), k ∈ C(Δ × I, R+ ),

148


(ii) for λ as in (4.3.3) (l1 ) there exists a nonnegative constant α such that α < 1 and t b c(s, x) exp(λ (s + |x|)) + k(s, x, y) exp(λ (s + |y|)) dy ds 0

a

α exp(λ (t + |x|)),

(4.3.7)

(l2 ) there exists a nonnegative constant β such that t b | f (s, x, 0)| + |φ (t, x)| + |g(s, x, y, 0)|dy ds β exp(λ (t + |x|)), 0

(4.3.8)

a

for (t, x) ∈ Δ, where

φ (t, x) = u0 (x) +

t

h(s, x) ds,

(4.3.9)

0

in which h, u0 are as in (4.3.1), (4.3.2). Under the assumptions (i) and (ii), the problem (4.3.1)–(4.3.2) has a unique solution u(t, x) on Δ in E. Proof.

Let u ∈ E and define the operator T by t b f (s, x, u(s, x)) + g(s, x, y, u(s, y)) dy ds, (Tu)(t, x) = φ (t, x) + 0

a

for (t, x) ∈ Δ, where φ (t, x) is given by (4.3.9). The proof that T maps E into itself and is a contraction map can be completed by following the proof of Theorem 1.3.1 in Chapter 1 with suitable modifications. We leave the details to the reader. The following theorem deals with the estimate on the solution of problem (4.3.1)–(4.3.2). Theorem 4.3.2. Suppose that the functions f , g in equation (4.3.1) satisfy the conditions (4.3.5), (4.3.6) and d = sup

|φ (t, x)| +

(t,x)∈Δ

t 0

b

| f (s, x, 0)| +

a

|g(s, x, y, 0)|dy ds < ∞,

(4.3.10)

where φ (t, x) is given by (4.3.9). If u(t, x) is any solution of problem (4.3.1)–(4.3.2) on Δ, then |u(t, x)| dC(t, x) exp

t

for (t, x) ∈ Δ, where

0

k(s, x, y)C(s, y) dy ds ,

(4.3.11)

c(s, x) ds .

(4.3.12)

b

a

C(t, x) = exp 0

t


149

Using the fact that u(t, x) is a solution of problem (4.3.1)–(4.3.2) and hypotheses,

Proof.

we observe that |u(t, x)| |φ (t, x)| + t

+ 0

t 0

| f (s, x, 0)| +

| f (s, x, u(s, x)) − f (s, x, 0)| +

b a

b a

|g (s, x, y, 0)| dy ds

|g(s, x, y, u(s, y)) − g (s, x, y, 0)| dy ds

t b c(s, x)|u(s, x)| + d+ k(s, x, y)|u(s, y)|dy ds. 0

(4.3.13)

a

Now an application of the inequality in Theorem 4.2.1 part (p1 ) to (4.3.13) yields (4.3.11). We note that the estimate obtained in (4.3.11) provides the bound on the

Remark 4.3.1.

solution u(t, x) of problem (4.3.1)–(4.3.2) on Δ. In Theorem 4.3.2 in addition, if we assume that

∞ 0

c(s, x) ds < ∞,

∞ b 0

a

k(s, x, y)C(s, y) dy ds < ∞,

for every x ∈ I, then the solution u(t, x) of problem (4.3.1)–(4.3.2) is bounded on Δ. A slight variant of Theorem 4.3.2 is embodied in the following theorem. Theorem 4.3.3. Suppose that the functions f , g, h, u0 in (4.3.1)–(4.3.2) satisfy the conditions (4.3.5), (4.3.6) and t

d = sup (t,x)∈Δ

0

| f (s, x, φ (s, x))| +

b a

|g(s, x, y, φ (s, y))|dy ds < ∞,

(4.3.14)

where φ (t, x) is defined by (4.3.9). If u(t, x) is any solution of problem (4.3.1)–(4.3.2) on Δ, then |u(t, x) − φ (t, x)| dC(t, x) exp

t 0

b


(4.3.15)

a

for (t, x) ∈ Δ, where C(t, x) is given by (4.3.12). Proof.

Let e(t, x) = |u(t, x) − φ (t, x)|, (t, x) ∈ Δ. Using the fact that u(t, x) is a solution of

problem (4.3.1)–(4.3.2) and hypotheses, we observe that e(t, x) b

+ a

t

0

| f (s, x, u(s, x)) − f (s, x, φ (s, x)) + f (s, x, φ (s, x))|

|g(s, x, y, u(s, y)) − g(s, x, y, φ (s, y)) + g(s, x, y, φ (s, y))|dy ds

150


t 0

| f (s, x, φ (s, x))| +

|g (s, x, y, φ (s, y))| dy ds

b a

t | f (s, x, u(s, x)) − f (s, x, φ (s, x))| + 0

b

+ a

d+

|g(s, x, y, u(s, y)) − g(s, x, y, φ (s, y))|dy

t

k(s, x, y)u(s, y) dy ds.

b

c(s, x)e(s, x) + 0

(4.3.16)

a

Now an application of the inequality in Theorem 4.2.1 part (p1 ) to (4.3.16) yields (4.3.15). Let u(t, x) ∈ C (Δ, R) ; ∂∂t u(t, x) exists on Δ and satisfies the inequality b ∂ u(t, x) − f (t, x, u(t, x)) − g(t, x, y, u(t, y)) dy − h(t, x) ε , ∂t a for a given constant ε 0, where it is supposed that (4.3.2) holds. Then we call u(t, x) the

ε -approximate solution with respect to the equation (4.3.1). The following theorem estimates the difference between the two approximate solutions of problem (4.3.1)–(4.3.2). Theorem 4.3.4.

Suppose that the functions f , g in (4.3.1) satisfy the conditions (4.3.5),

(4.3.6). Let ui (t, x) (i = 1, 2), (t, x) ∈ Δ be respectively εi -approximate solutions of (4.3.1) with ui (0, x) = ui (x), where ui ∈ C (I, R) and

φi (t, x) = ui (x) +

(4.3.17)

t

h(s, x) ds.

(4.3.18)

0

Suppose that |φ1 (t, x) − φ2 (t, x)| δ ,

(4.3.19)

M = sup [(ε1 + ε2 )t + δ ] < ∞.

(4.3.20)

where δ 0 is a constant and t∈R+

Then |u1 (t, x) − u2 (t, x)| MC(t, x) exp

t 0

for (t, x) ∈ Δ, where C(t, x) is given by (4.3.12).

a

b


(4.3.21)


Proof.

151

Since ui (t, x) (i = 1, 2), (t, x) ∈ Δ are respectively εi -approximate solutions of

(4.3.1) with (4.3.17), we have b ∂ ui (t, x) − f (t, x, ui (t, x)) − εi . g(t, x, y, u (t, y)) dy − h(t, x) i ∂t a

(4.3.22)

By taking t = s in (4.3.22) and integrating both sides with respect to s from 0 to t for t ∈ R+ , we get

εi t

t

b

0

a

∂ ui (s, x) − f (s, x, ui (s, x)) − ∂s

g(s, x, y, ui (s, y)) dy − h(t, x) ds

t

b ∂ ui (s, x) − f (s, x, ui (s, x)) − g (s, x, y, ui (s, y)) dy − h(t, x) ds ∂ s 0 a

t b = ui (t, x) − φi (t, x) − f (s, x, ui (s, x)) + |g(s, x, y, ui (s, y))|dy ds . 0

(4.3.23)

a

From (4.3.23) and using the elementary inequalities in (1.3.25), we observe that (ε1 + ε2 )t t b f (s, x, u1 (s, x)) + u1 (t, x) − φ1 (t, x) − g (s, x, y, u1 (s, y)) dy ds 0

a

t b + u2 (t, x) − φ2 (t, x) − f (s, x, u2 (s, x)) + g (s, x, y, u2 (s, y)) dy ds 0

a

t b u1 (t, x) − φ1 (t, x) − f (s, x, u1 (s, x)) + g (s, x, y, u1 (s, y)) dy ds 0

− u2 (t, x) − φ2 (t, x) −

a

t 0

b

f (s, x, u2 (s, x)) +

a

g (s, x, y, u2 (s, y)) dy ds

|u1 (t, x) − u2 (t, x)| − |φ1 (t, x) − φ2 (t, x)| t b f (s, x, u1 (s, x)) + − g (s, x, y, u1 (s, y)) dy ds 0 a −

t 0

b

f (s, x, u2 (s, x)) +

a

g (s, x, y, u2 (s, y)) dy ds .

(4.3.24)

Let u(t, x) = |u1 (t, x) − u2 (t, x)|, (t, x) ∈ Δ. From (4.3.24) and using the hypotheses, we observe that u(t, x) (ε1 + ε2 )t + δ + M+

t

c(s, x)u(s, x) + 0

t

k(s, x, y)u(s, y) dy ds a

b

c(s, x)u(s, x) + 0

b

k(s, x, y)u(s, y) dy ds.

(4.3.25)

a

Now an application of the inequality in Theorem 4.2.1 part (p1 ) to (4.3.25) yields (4.3.21).

152


Remark 4.3.2.

In case u1 (t, x) is a solution of (4.3.1) with u1 (0, x) = u1 (x), then we have

ε1 = 0 and from (4.3.21), we see that u2 (t, x) → u1 (t, x) as ε2 → 0 and δ → 0. Furthermore, if we put (i) ε1 = ε2 = 0, u1 (x) = u2 (x) in (4.3.21), then the uniqueness of solutions of (4.3.1) is established and (ii) ε1 = ε2 = 0 in (4.3.21), then we get the bound which shows the dependency of solutions of (4.3.1) on given initial values. Consider (4.3.1)–(4.3.2) together with the following integrodifferential equation

∂ v(t, x) = f (t, x, v(t, x)) + ∂t with the initial condition

b

g(t, x, y, v(t, y)) dy + h(t, x),

(4.3.26)

a

v(0, x) = v0 (x),

(4.3.27)

for (t, x) ∈ Δ, where f ∈ C(Δ × R, R), g ∈ C(Δ × I × R, R), h ∈ C(Δ, R), v0 ∈ C(I, R). In the following theorem we provide conditions concerning the closeness of solutions of problems (4.3.1)–(4.3.2) and (4.3.26)–(4.3.27). Theorem 4.3.5.


(4.3.6) and there exist constants ε i 0, δ i 0 (i = 1, 2) such that | f (t, x, u) − f (t, x, u)| ε 1 ,

(4.3.28)

|g(t, x, y, u) − g(t, x, y, u)| ε 2 ,

(4.3.29)

|h(t, x) − h(t, x)| δ 1 ,

(4.3.30)

|u0 (x) − v0 (x)| δ 2 ,

(4.3.31)

where f , g, h, u0 and f , g, h, v0 are the functions in (4.3.1)–(4.3.2) and (4.3.26)–(4.3.27) and

M = sup

δ 1 + ε 1 + ε 2 (b − a) t + δ 2 < ∞.

(4.3.32)

t∈R+

Let u(t, x) and v(t, x) be respectively the solutions of (4.3.1)–(4.3.2) and (4.3.26)–(4.3.27) for (t, x) ∈ Δ. Then |u(t, x) − v(t, x)| MC(t, x) exp

t 0

b a

for (t, x) ∈ Δ, where C(t, x) is given by (4.3.12).


(4.3.33)


Proof.

153

Let z(t, x) = |u(t, x) − v(t, x)|, (t, x) ∈ Δ. Using the facts that u(t, x), v(t, x) are

respectively the solutions of (4.3.1)–(4.3.2), (4.3.26)–(4.3.27) and hypotheses, we observe that

t t z(t, x) u0 (x) + h(s, x) ds − v0 (x) − h(s, x) ds 0

0

t | f (s, x, u(s, x)) − f (s, x, v(s, x))| + | f (s, x, v(s, x)) − f (s, x, v(s, x))| + 0

b

+ a

{|g(s, x, y, u(s, y)) − g(s, x, y, v(s, y))| + |g(s, x, y, v(s, y)) − g(s, x, y, v(s, y))|} dy ds

|u0 (x) − v0 (x)| + t

+ 0

c(s, x)z(s, x) + ε 1 +

t 0

b a

|h(s, x) − h(s, x)|ds

{k(s, x, y)z(s, y) + ε 2 } dy ds

b t c(s, x)z(s, x) + k(s, x, y)z(s, y) dy ds δ 2 + δ 1 t + ε 1 t + ε 2 (b − a)t + 0

M+

a

t

b

c(s, x)z(s, x) + 0

k(s, x, y)z(s, y) dy ds.

(4.3.34)

a

Now an application of the inequality in Theorem 4.2.1 part (p1 ) to (4.3.34) yields (4.3.33). Remark 4.3.3.

We note that the result given in Theorem 4.3.5 relates the solutions of

(4.3.1)–(4.3.2) and (4.3.26)–(4.3.27) in the sense that if f , g, h, u0 are respectively close to f , g, h, v0 ; then the solutions of (4.3.1)–(4.3.2) and (4.3.26)–(4.3.27) are also close together. Next, we consider the following integrodifferential equations of Barbashin-type

∂ u(t, x) = f (t, x, u(t, x), μ ) + ∂t ∂ u(t, x) = f (t, x, u(t, x), μ0 ) + ∂t

b a

b a

g (t, x, y, u(t, y), μ ) dy + h(t, x),

(4.3.35)

g (t, x, y, u(t, y), μ0 ) dy + h(t, x),

(4.3.36)

with the given initial condition (4.3.2), where f ∈ C(Δ×R×R, R), g ∈ C(Δ×I ×R×R, R), h ∈ C(Δ, R) and μ , μ0 are parameters. The following theorem deals with the dependency of solutions of (4.3.35)–(4.3.2) and (4.3.36)–(4.3.2) on parameters.

154


Suppose that the functions f , g in (4.3.35), (4.3.36) satisfy the conditions

Theorem 4.3.6.

| f (t, x, u, μ ) − f (t, x, u, μ )| c(t, x)|u − u|,

(4.3.37)

| f (t, x, u, μ ) − f (t, x, u, μ0 )| n1 (t, x)|μ − μ0 |,

(4.3.38)

|g(t, x, y, u, μ ) − g(t, x, y, u, μ )| k(t, x, y)|u − u|,

(4.3.39)

|g(t, x, y, u, μ ) − g(t, x, y, u, μ0 )| n2 (t, x, y)|μ − μ0 |,

(4.3.40)

where c, n1 ∈ C(Δ, R+ ), k, n2 ∈ C(Δ × I, R+ ) and t b N = sup n1 (s, x) + n2 (s, x, y) dy ds < ∞. (t,x)∈Δ 0

a

Let u1 (t, x) and u2 (t, x) be respectively, the solutions of (4.3.35)–(4.3.2) and (4.3.36)– (4.3.2) on Δ. Then |u1 (t, x) − u2 (t, x)| |μ − μ0 |NC(t, x) exp

t 0

b


(4.3.41)

a

for (t, x) ∈ Δ, where C(t, x) is given by the right hand side of (4.3.12) by replacing c(t, x) by c(t, x). Proof.

Let w(t, x) = |u1 (t, x)− u2 (t, x)|, (t, x) ∈ Δ. Using the facts that u1 (t, x) and u2 (t, x)

are respectively the solutions of (4.3.35)–(4.3.2) and (4.3.36)–(4.3.2) and hypotheses, we observe that w(t, x)

t 0

[| f (s, x, u1 (s, x), μ ) − f (s, x, u2 (s, x), μ )|

+| f (s, x, u2 (s, x), μ ) − f (s, x, u2 (s, x), μ0 )| b

+ a

{|g(s, x, y, u1 (s, y), μ ) − g(s, x, y, u2 (s, y), μ )|

+|g(s, x, y, u2 (s, y), μ ) − g(s, x, y, u2 (s, y), μ0 )|} dy] ds +

t 0

[c(s, x)|u1 (s, x) − u2 (s, x)| + n1 (s, x)|μ − μ0 |

b a

k(s, x, y)|u1 (s, y) − u2 (s, y)| + n2 (s, x, y)|μ − μ0 | dy ds

|μ − μ0 |N +

t

b

c(s, x)w(s, x) + 0

k(s, x, y)w(s, y) dy ds.

(4.3.42)

a

Now an application of the inequality in Theorem 4.2.1 part (p1 ) to (4.3.42) yields (4.3.41), which shows the dependency of solutions of (4.3.35)–(4.3.2) and (4.3.36)–(4.3.2) on parameters.


Remark 4.3.4.

155

We note that by using the inequality in Theorem 4.2.2 part (p3 ), one can

obtain results similar to those given above to the following Barbashin-type integrodifferential equation

∂ u(t, x) = f (t, x, u(t, x)) + ∂t

k(t, x, y, u(t, y)) dy + h(t, x),

(4.3.43)

B

with the given initial condition, under some suitable conditions on the functions involved in (4.3.43) and the given initial condition, where B is a compact subset of Rn . As far as equations of the forms (4.3.1)–(4.3.2) and (4.3.43) is concerned, we believe that the numerous models (see [5]) whose detailed treatment is desired. 4.4

General integral equation of Barbashin-type

Integrodifferential equations of Barbashin-type (B) and partial integral equations are connected to each other in several ways. For example, suppose we are intersted in findining a solution u of equation (B) satisfying the initial condition (4.3.2). Putting

∂ ∂ t u(t, x) := w(t, x)

we arrive at the equation t

w(t, x) = g(t, x) + 0

c(t, x)w(τ , x) d τ +

t b 0

where

a

k(t, x, y)w(τ , y) dy d τ ,

(4.4.1)

k(t, x, y)u0 (y) dy.

(4.4.2)

b

g(t, x) = f (t, x) + c(t, x)u0 (x) +

a

Throughout this section, we shall use the notations and definitions given in section 3.2 without further mention. The monograph [5] contains the study of many variants and generalizations of equations (B) and (4.4.1) by using different techniques. In this section we consider the following general partial integral equation of the form (see [109]) t

u(t, x) = h(t, x) +

t

f (t, x, s, u(s, x)) ds + 0

g(t, x, s, y, u(s, y)) dy ds, 0

(4.4.3)

B

for (t, x) ∈ D0 , where h ∈ C(D0 , R), f ∈ C(Ω0 × R, R), g ∈ C(Ω × R, R) are given functions and u is the unknown function to be found, in which Ω0 = {(t, x, s) : 0 s t < ∞, x ∈ B}. The problems of existence and uniqueness of solutions of equation (4.4.3) can be dealt with the method employed in Chapter 3, section 3.4. Here, we present some basic qualitative properties of solutions of (4.4.3) which provide simple and elegant results and thus have a wider scope of applicability. First we shall give the following theorem concerning the estimate on the solution of equation (4.4.3).

156


Theorem 4.4.1. Suppose that the functions f , g, h in (4.4.3) satisfy the conditions

and

| f (t, x, s, u) − f (t, x, s, u)| r(t, x, s)|u − u|,

(4.4.4)

|g(t, x, s, y, u) − g(t, x, s, y, u)| k(t, x, s, y)|u − u|,

(4.4.5)

|h(t, x)| +

d = sup (t,x)∈D0

t 0

t

| f (t, x, s, 0)|ds +

0

B

|g(t, x, s, y, 0)|dy ds < ∞,

(4.4.6)

where r ∈ C(Ω0 , R+ ), k ∈ C(Ω, R+ ); D1 r, D1 k exist and D1 r ∈ C(Ω0 , R+ ), D1 k ∈ C(Ω, R+ ). If u(t, x) is any solution of equation (4.4.3) on D0 , then t A(σ , x) d σ , |u(t, x)| dP(t, x) exp

(4.4.7)

0

for (t, x) ∈ D0 , where P and A are given by (4.2.8) and (4.2.10). Proof.

Using the fact that u(t, x) is a solution of (4.4.3) and hypotheses, we observe that |u(t, x)| |h(t, x)| + t

+ 0

B

t 0

0

| f (t, x, s, u(s, x)) − f (t, x, s, 0) + f (t, x, s, 0)|ds

|g(t, x, s, y, u(s, y)) − g(t, x, s, y, 0) + g(t, x, s, y, 0)|dy ds

|h(t, x)| + +

t

t 0

| f (t, x, s, 0)|ds +

| f (t, x, s, u(s, x)) − f (t, x, s, 0)|ds + d+

t 0

t

B

t 0

B

|g(t, x, s, y, 0)|dy ds

|g(t, x, s, y, u(s, y)) − g(t, x, s, y, 0)|dy ds

t

r(t, x, s)|u(s, x)|ds + 0

k(t, x, s, y)|u(s, y)|dy ds. 0

(4.4.8)

B

Now an application of Theorem 4.2.2 part (p3 ) to (4.4.8) yields (4.4.7). The following theorem deals with a slight variant of Theorem 4.4.1. Theorem 4.4.2.

Suppose that the functions f , g and h in (4.4.3) satisfy the conditions

(4.4.4), (4.4.5) and t t d = sup | f (t, x, s, h(s, x))|ds + |g(t, x, s, y, h(s, y))|dy ds < ∞. (t,x)∈D0

0

0

(4.4.9)

B

If u(t, x) is any solution of equation (4.4.3) on D0 , then t A(σ , x) d σ , |u(t, x) − h(t, x)| dP(t, x) exp 0

for (t, x) ∈ D0 , where P and A are given by (4.2.8) and (4.2.10).

(4.4.10)


Proof.

157

Let z(t, x) = |u(t, x) − h(t, x)|, (t, x) ∈ D0 . Using the fact that u(t, x) is a solution

of (4.4.3) and hypotheses, we observe that z(t, x) t

+ 0

B

t 0

| f (t, x, s, u(s, x)) − f (t, x, s, h(s, x)) + f (t, x, s, h(s, x))|ds

|g(t, x, s, y, u(s, y)) − g(t, x, s, y, h(s, y)) + g(t, x, s, y, h(s, y))|dy ds

t 0

| f (t, x, s, h(s, x))|ds + t

+ 0

t

+ 0

d+

B

t 0

|g(t, x, s, y, h(s, y))|dy ds

B

| f (t, x, s, u(s, x)) − f (t, x, s, h(s, x))|ds

|g(t, x, s, y, u(s, y)) − g(t, x, s, y, h(s, y))|dy ds

t

t

r(t, x, s)z(s, x) ds + 0


(4.4.11)

B

Now an application of Theorem 4.2.2 part (p3 ) to (4.4.11) yields (4.4.10). In the next theorem we formulate conditions on the functions involved in (4.4.3) which shows that the solutions of (4.4.3) tends to zero as t → ∞. Theorem 4.4.3. Suppose that (i) the functions f , g in (4.4.3) satisfy the conditions (4.4.4), (4.4.5) and f (t, x, s, 0) = 0, g(t, x, s, y, 0) = 0, (ii) the function h in (4.4.3) satisfies the condition |h(t, x)| M exp(−α t),

(4.4.12)

for (t, x) ∈ D0 , where α > 0, M 0 are constants, (iii) the functions r, D1 r, k, D1 k be as in Theorem 4.4.1 and sup Q(t, x) < ∞, (t,x)∈D0

∞ 0

A(σ , x) d σ < ∞,

(4.4.13)

where Q and A are given respectively by the right hand sides of (4.2.9) and (4.2.10) by replacing r and k by r exp(α (t − s)) and k exp(α (t − s)). If u(t, x) is any solution of (4.4.3) on D0 , then it tends exponentially toward zero as t → ∞.

158


Proof.

From the hypotheses, we observe that |u(t, x)| |h(t, x)| + t

+ 0

M exp(−α t) +

B

t 0

| f (t, x, s, u(s, x)) − f (t, x, s, 0)|ds

|g(t, x, s, y, u(s, y)) − g(t, x, s, y, 0)|dy ds

t

t

r(t, x, s)|u(s, x)|ds + 0

k(t, x, s, y)|u(s, y)|dy ds.

(4.4.14)

B

0

From (4.4.14), we observe that |u(t, x)| exp(α t) M + t

+ 0

B

t 0

r(t, x, s) exp(α (t − s))|u(s, x)| exp(α s) ds

k(t, x, s, y) exp(α (t − s))|u(s, y)| exp(α s) dy ds.

Now an application of Theorem 4.2.2 part (p3 ) to (4.4.15) yields t A(σ , x) d σ , |u(t, x)| exp(α t) MP(t, x) exp

(4.4.15)

(4.4.16)

0

where P is given by the right hand side of (4.2.8) by replacing Q by Q. Multiplying both sides of (4.4.16) by exp(−α t) and in view of (4.4.13), the solution u(t, x) tends to zero as t → ∞. The following theorem deals with the estimate on the difference between the two approximate solutions of equation (4.4.3). Theorem 4.4.4.


(4.4.5) respectively. Let ui (t, x) (i = 1, 2) be respectively εi -approximate solutions of equation (4.4.3) on D0 , i.e., t t ui (t, x) − h(t, x) − f (t, x, s, ui (s, x)) ds − g(t, x, s, y, ui (s, y)) dy ds εi , (4.4.17) 0

0

B

on D0 for constants εi 0. Then |u1 (t, x) − u2 (t, x)| (ε1 + ε2 )P(t, x) exp

t 0

A(σ , x) d σ ,

(4.4.18)

for (t, x) ∈ D0 , where P and A are given by (4.2.8) and (4.2.10). The proof follows by the arguments as in the proof Theorem 3.4.4 and making use of the inequality in Theorem 4.2.2 part (p3 ). We leave the details to the reader.


Remark 4.4.1.

159

In case u1 (t, x) is a solution of equation (4.4.3), then we have ε1 = 0 and

from (4.4.18) we see that u2 (t, x) → u1 (t, x) on D0 as ε2 → 0. Moreover, from (4.4.18) it follows that if ε1 = ε2 = 0, then the uniqueness of solutions of (4.4.3) is established. Consider the equation (4.4.3) together with the following integral equation t

v(t, x) = h(t, x) +

t

f (t, x, s, v(s, x)) ds + 0

g(t, x, s, y, v(s, y)) dy ds, 0

(4.4.19)

B

for (t, x) ∈ D0 , where h ∈ C(D0 , R), f ∈ C(Ω0 ×R, R), g ∈ C(Ω×R, R), are given functions and v is the unknown function. In the following theorem we provide conditions concerning the closeness of solutions of equations (4.4.3) and (4.4.19). Theorem 4.4.5.


(4.4.5) respectively and there exist constants ε i 0, (i = 1, 2), δ 0 such that | f (t, x, s, u) − f (t, x, s, u)| ε 1 ,

(4.4.20)

|g(t, x, s, y, u) − g(t, x, s, y, u)| ε 2 ,

(4.4.21)

|h(t, x) − h(t, x)| δ ,

(4.4.22)

where f , g, h and f , g, h are functions involved in (4.4.3) and (4.4.19) and

n

M = sup δ + t ε 1 + ε 2 ∏(bi − ai ) t∈R+

< ∞.

(4.4.23)

i=1

Let u(t, x) and v(t, x) be respectively the solutions of (4.4.3) and (4.4.19) on D0 . Then t (4.4.24) A(σ , x) d σ , |u(t, x) − v(t, x)| MP(t, x) exp 0

for (t, x) ∈ D0 , where P and A are given by (4.2.8) and (4.2.10). Proof.

Let z(t, x) = |u(t, x) − v(t, x)|, (t, x) ∈ D0 . Using the hypotheses, we observe that z(t, x) |h(t, x) − h(t, x)| t

+ 0

| f (t, x, s, u(s, x)) − f (t, x, s, v(s, x)).

+ f (t, x, s, v(s, x)) − f (t, x, s, v(s, x))|ds t

+ 0

B

g(t, x, s, y, u(s, y)) − g(t, x, s, y, v(s, y))

+g(t, x, s, y, v(s, y)) − g(t, x, s, y, v(s, y))|dy ds

160


δ+

t

t

+ 0

t

+ B

0

t

+ 0

δ+

B

0

| f (t, x, s, u(s, x)) − f (t, x, s, v(s, x))|ds

| f (t, x, s, v(s, x)) − f (t, x, s, v(s, x))|ds

|g(t, x, s, y, u(s, y)) − g(t, x, s, y, v(s, y))| dy ds

|g(t, x, s, y, v(s, y)) − g(t, x, s, y, v(s, y))| dy ds

t

t

r(t, x, s)z(s, x) ds + 0

0

M+

ε 1 ds +

t

t

k(t, x, s, y)z(s, y) dy ds + 0

t

0

B

B

ε 2 dy ds

t

r(t, x, s)z(s, x) ds + 0


(4.4.25)

B

Now an application of Theorem 4.2.2 part (p3 ) to (4.4.25) yields (4.4.24). Remark 4.4.2.

The result given in Theorem 4.4.5 relates the solutions of (4.4.3) and

(4.4.19) in the sense that if f , g, h are respectively close to f , g, h; then the solutions of (4.4.3) and (4.4.19) are also close together. In the following theorem we formulate conditions for continuous dependence of solutions of equations t

u(t, x) = hi (t, x) +

t

f (t, x, s, u(s, x)) ds + 0

g(t, x, s, y, u(s, y)) dy ds, 0

(4.4.26)

B

for (i = 1, 2), (t, x) ∈ D0 , where hi ∈ C(D0 , R) and the functions f , g are as defined in (4.4.3). Theorem 4.4.6. Suppose that the functions f , g in (4.4.26) satisfy the conditions (4.4.4), (4.4.5) and there exists a constant δ 0 such that |h1 (t, x) − h2 (t, x)| δ ,

(4.4.27)

for (t, x) ∈ D0 . Let ui (t, x) (i = 1, 2) be respectively εi – approximate solutions of (4.4.26). Then |u1 (t, x) − u2 (t, x)| (δ + ε1 + ε2 )P(t, x) exp

t 0

A(σ , x) d σ ,

for (t, x) ∈ D0 , where P and A are given by (4.2.8) and (4.2.10).

(4.4.28)


Proof.

161

Let z(t, x) = |u1 (t, x) − u2 (t, x)|, (t, x) ∈ D0 . Following the proof of Theorem 3.4.4

and using the hypotheses, we obtain z(t, x) |h1 (t, x) − h2 (t, x)| + ε1 + ε2 + t

+ 0

δ + ε1 + ε2 +

B

t 0

| f (t, x, s, u1 (s, x)) − f (t, x, s, u2 (s, x))|ds

|g(t, x, s, y, u1 (s, y)) − g(t, x, s, y, u2 (s, y))|dy ds

t

t

r(t, x, s)z(s, x) ds + 0


(4.4.29)

B

Now applying Theorem 4.2.2 part (p3 ) to (4.4.29) yields (4.4.28), which shows that the solutions of (4.4.26) depends continuously on the functions on the right hand side of (4.4.26). Remark 4.4.3.

As noted in [5, p. 12 and p. 476], another significant source of parabolic

type integrodifferential equations is provided by the study of equation of the form t

P(t, x, s, u(s, x)) ds +

u(t, x) = f (t, x) + 0

Q(t, x, y, u(s, y)) dy B

t

+

F(t, x, s, y, u(s, y)) dy ds, 0

(4.4.30)

B

and variants thereof under some suitable conditions. It is hoped and expected that the analysis presented above in Theorems 4.4.1–4.4.6 will also be equally useful in the study of equations like (4.4.30). Indeed, it involves the task of desiging a new inequality similar to the one given in Theorem 4.2.2 part (p3 ), which will allow applications in the discussion of equation (4.4.30). In fact, it is not an easy matter but a challenging problem for future investigations. 4.5

Integrodifferential equation of the type arising in reactor dynamics

Parabolic integrodifferential equations of various types occurring in reactor dynamics have been extensively investigated in the literature by using different techniques. In this section we offer some basic results from [75, 77] concerning the existence, uniqueness and asymptotic behavior of solutions of the following general nonlinear diffusion system

∂u − Lu = f (x,t, u) + (Hu)(x,t) (t > 0, x ∈ Ω), ∂t with the given boundary conditions B[u] = d(x)

∂u + u = 0 (t > 0, x ∈ ∂ Ω), ∂τ

u(x, 0) = u0 (x),

x ∈ Ω,

(4.5.1)

(4.5.2) (4.5.3)

162


where Ω is a bounded domain in Rn , ∂ Ω the sufficiently smooth boundary of Ω, τ is the outward normal unit vector on the boundary ∂ Ω and n

L=

∑

ai j (x)

i, j=1

n ∂2 ∂ + ∑ ai (x) − a0 (x), ∂ xi ∂ x j i=1 ∂ xi

(4.5.4)

is the uniformly elliptic operator. Let D = Ω × (0, T ], D = Ω × [0, T ], S = ∂ Ω × (0, T ], where T > 0 is a finite but can be arbitrarily large and Ω is the closure of Ω. Denote by C2 (D) the class of functions which are continuous in D, continuously differentiable in t and twice continuously differentiable in x and for (x,t) ∈ D and

∂u ∂τ

exists on ∂ Ω. Throughout assume that the coefficients of L

and the first order derivatives of ai j are Hölder continuous in Ω (of exponent α , 0 < α < 1), the matrix ai j is symmetric positive definite in Ω, a0 (x) 0 in Ω, f is Hölder continuous in (x,t, u) in every bounded subset of Ω × R+ × Rn , the operator Hu is uniformly Hölder continuous with respect to u ∈ Rn and for (x,t) ∈ Ω × R+ , the boundary ∂ Ω is of class C2+α , d ∈ H 1+α (∂ Ω) and u0 ∈ H 2+α Ω and satisfies boundary condition (4.5.2) at t = 0, where C2+α , H 2+α Ω , H 1+α (∂ Ω) are the function spaces defined in Chapter 1 of [57] (see also [37,69,122]). The above smoothness assumptions are required only for ensuring the existence of a solution for the corresponding linear problem. A function u ∈ C2 (D) which satisfies (4.5.1)–(4.5.3) is called a solution of problem (4.5.1)–(4.5.3). System (4.5.1)– (4.5.3) occur in nuclear reactor dynamics with the operator Hu may in general be of the form

t (Hu)(x,t) = u(x,t)h x,t, k(x,t, s)u(x, s) ds , 0

or

t (Hu)(x,t) = h x,t, k(x,t, s)u(x, s) ds , 0

or variants thereof (see [59,60,75 ]), where all the functions are Hölder continuous and defined on the respective domains of their definitions. Below we employ the notion of upper and lower solutions and monotone method to establish the existence of maximal and minimal solutions of (4.5.1)–(4.5.3) which also yields the upper and lower estimates on the exact solution of the problem (4.5.1)–(4.5.3). In what follows, we assume that all inequalities between vectors are componentwise. The function w ∈ C2 (D) is called an upper solution of (4.5.1)–(4.5.3) if

∂w − Lw f (x,t, w) + (Hw)(x,t), ∂t

(x,t) ∈ D,

(4.5.5)


163

B [w] 0,

(x,t) ∈ S,

w(x, 0) u0 (x), Similarly, the function v

∈ C2 (D)

x ∈ Ω.

(4.5.6) (4.5.7)

is called a lower solution of (4.5.1)–(4.5.3) if it satisfies

all the reversed inequalities in (4.4.5)–(4.4.7). The functions u, u ∈ C2 (D) are called maximal and minimal solutions of (4.5.1)–(4.5.3), respectively, if every other solution u ∈ C2 (D) of (4.5.1)–(4.5.3) satisfies the relation u(x,t) u(x,t) u(x,t) for (x,t) ∈ D. We list for convenience the following assumptions: (H1 ) the functions v, w ∈ C2 (D) with v w on D are lower and upper solutions of (4.5.1)– (4.5.3), (H2 ) for each i, fi (x,t, u1 )− fi (x,t, u2 ) −M(u1i −u2i ) whenever v(x,t) u2 u1 w(x,t) for (x,t) ∈ D, where M 0 is a constant, (H3 ) for each i, the operator Hi u is monotone nondecreasing in u, whenever v(x,t) u w(x,t) for (x,t) ∈ D. For any η ∈ C2 (D) such that v η w on D, consider the linear system ∂ ui − Lui = fi (x,t, η ) + (Hi η )(x,t) − M(ui − ηi ), (x,t) ∈ D, ∂t B[ui ] = 0,

(x,t) ∈ S,

ui (x, 0) = u0i (x),

x ∈ Ω.

(4.5.8) (4.5.9) (4.5.10)

For a known η the assumptions on f and H ensures (see [57,69,122]) the existence of a unique solution of (4.5.8)–(4.5.10). For each η ∈ C2 (D) such that v η w on D, define the mapping A by Aη = u,

(4.5.11)

where u is the unique solution of (4.5.8)–(4.5.10). This mapping will be used to define the sequences that converge to the minimal and maximal solutions of (4.5.1)–(4.5.3). To achieve this, we first prove the following lemma. Lemma 4.5.1. Assume that the hypotheses (H1 )–(H3 ) hold. Then (i) the unique solution u of (4.5.8)–(4.5.10) satisfies v(x,t) u(x,t) w(x,t), (x,t) ∈ D; (ii) v Av, w Aw, for (x,t) ∈ D; (iii) A is a monotone operator on the set of functions U(v, w) = u ∈ C2 (D) : v(x,t) u w(x,t), (x,t) ∈ D .

164


Proof.

(i) We shall first show that v(x,t) u(x,t) for (x,t) ∈ D. Setting pi = vi − ui , we

get

∂ pi Lvi + fi (x,t, v) + (Hi v)(x,t) − Lui − fi (x,t, η ) − (Hi η )(x,t) + M(ui − ηi ) ∂t Lpi + M(ηi − vi ) + (Hi η )(x,t) − (Hi η )(x,t) + M(ui − ηi ) = Lpi − M pi . This implies

∂ pi − Lpi + M pi 0, ∂t and also we observe that B[pi ] 0, pi (x, 0) 0. Hence by the maximum principle (see [125]) pi 0 for i = 1, 2, . . . , n. This proves that v(x,t) u(x,t) for (x,t) ∈ D. Similarly, we can show that u(x,t) w(x,t) for (x,t) ∈ D. This proves (i). (ii) Let Av = u, where u is the unique solution of (4.5.8)–(4.5.10) with η = v. Now, we shall prove that v u for (x,t) ∈ D. Setting pi = vi − ui , we get

∂ pi Lvi + fi (x,t, v) + (Hi v)(x,t) − Lui − f i (x,t, v) − (Hi v)(x,t) + M(ui − vi ) ∂t = Lpi − M pi . This implies

∂ pi − Lpi + M pi 0, ∂t and also we observe that B[pi ] 0, pi (x, 0) 0. Hence by the maximum principle pi 0, i = 1, 2, . . . , n. This implies v(x,t) u(x,t) for (x,t) ∈ D. This proves v Av. In a similar way we can prove w Aw. (iii) Let η1 , η2 ∈ C2 (D) be such that η1 , η2 ∈ U(v, w) and η1 η2 . Let Aη1 = u1 and Aη2 = u2 , where u1 and u2 are the unique solutions of (4.5.8)–(4.5.10) corresponding to η1 and η2 , respectively. Then setting pi = u1i − u2i , we get

∂ pi = Lu1i + f i (x,t, η1 ) + (Hi η1 )(x,t) − M(u1i − η1i ) ∂t −Lu2i − fi (x,t, η2 ) − (Hi η2 )(x,t) + M(u2i − η2i ) Lpi − { fi (x,t, η2 ) − fi (x,t, η1 )} + (Hi η2 )(x,t) − (Hi η2 )(x,t) −M(u1i − η1i ) + M(u2i − η2i ) Lpi + M(η2i − η1i ) − M(u1i − η1i ) + M(u2i − η2i )


165

= Lpi − M pi . This implies

∂ pi − Lpi + M pi 0, ∂t and also we have B[pi ] = 0, pi (x, 0) = 0. Hence, by the maximum principle pi 0, i = 1, 2, . . . , n, which implies u1 (x,t) u2 (x,t) for (x,t) ∈ D. It therefore follows that the mapping A is monotone on U(v, w). The proof is complete. In view of Lemma 4.5.1, we can define the sequences vn = Avn−1 , wn = Awn−1 , with v0 = v and w0 = w. It is easy to observe that the sequences {vn } and {wn } are monotone nondecreasing and nonincreasing respectively, and v vn wn w on D. Furthermore, using the standard arguments, it follows that these sequences converge uniformly and monotonically to solutions u and u of (4.5.1)–(4.5.3). Let u be any solution of (4.5.1)–(4.5.3) such that u ∈ U(v, w). Then, by the induction argument, it is easily seen that u wn and u vn for every n = 0, 1, 2, . . .. Hence, we have u u u. This shows that u is a maximal solution and u is a minimal solution of (4.5.1)–(4.5.3) on D. Thus we have proved the following theorem. Theorem 4.5.1.

Assume that the hypotheses (H1 )–(H3 ) hold. Then the sequence {wn }

converges uniformly from above to a maximal solution u of (4.5.1)–(4.5.3), while the sequence {vn } converges uniformly from below to a minimal solution u of (4.5.1)–(4.5.3). Furthermore, if u is any solution of (4.5.1)–(4.5.3) such that u ∈ U(v, w), then v v1 · · · vn · · · u u u · · · wn · · · w1 w, on D. We note that the maximal and minimal solutions established in Theorem 4.5.1 are not necessarily the same. However, if f and Hu satisfy the conditions (H4 ) f (x,t, ξ1 ) − f (x,t, ξ2 ) c1 (ξ1 − ξ2 ), (H ξ1 )(x,t) − (H ξ1 )(x,t) c2 (ξ1 − ξ2 ), whenever v ξ2 ξ1 w on D, where c1 and c2 are nonnegative constants, then we have the uniqueness result. Theorem 4.5.2.

Assume that the hypotheses (H1 )–(H4 ) hold. Then the maximal and

minimal solutions of (4.5.1)–(4.5.3) coincide. Furthermore, if (H2 ) and (H4 ) hold for v = 0 and every finite w, then (4.5.1)–(4.5.3) has a unique nonnegative solution.

166


Proof.

Let λ be a constant satisfying λ > c + c1 + c2 and let z = e−λ t (u − u). Then

∂z − Lz + (λ − c1 − c2 )z ∂t = e−λ t [ f (x,t, u) − f (x,t, u) + (Hu)(x,t) − (Hu)(x,t)] − (c1 + c2 )z e−λ t [c1 (u − u) + c2 (u − u)] − (c1 + c2 )z = 0, and B[z] = 0, z(x, 0) = 0. Hence by the maximum principle z = 0, that is u = u. Let u be any nonnegative solution of (4.5.1)–(4.5.3). Then u is also an upper solution and thus by Theorem 4.5.1 with w = u we have u u. By replacing u by u in the definition of z the same argument leads immediately to u = u. Hence, problem (4.5.1)–(4.5.3) cannot have more than one nonnegative solution. The proof is complete. Consider the eigenvalue problem Lφ + λ φ = 0,

B [φ ] = 0,

x ∈ Ω;

x ∈ ∂ Ω,

(4.5.12)

where L and B are defined in (4.5.4) and (4.5.2). The next result depends on the construction of the upper solution which depends upon the use of the eigenfunction φ corresponding to the least eigenvalue λ0 of (4.5.12). It is well known that λ0 is real and positive and φ is (or can be chosen) positive in Ω (see [125]). When d(x) > 0 the maximum principle implies that φ (x) > 0 on Ω. We normalize φ so that max{φ (x) : x ∈ Ω} = 1 and write

φm ≡ min{φ (x) : x ∈ Ω}. Notice that φm > 0 when d > 0 on ∂ Ω. Theorem 4.5.3.

Assume that the hypotheses of Theorem 4.5.1 hold. Let u(x,t) be

nonnegative solution of (4.5.1)–(4.5.3) and d(x) > 0. Let β > 0, b < λ0 , α =

δ=

(λ0 4β

−b)2

(λ0 −b) 2 ,

be such that for t > 0, x ∈ Ω, f (x,t, ξ ) bξ , (H ξ )(x,t) β δ ξ 2

t 0

ξ 0,

e−α s ds,

(4.5.13)

ξ 0.

(4.5.14)

Then for any constant k > 0, 0 u(x,t) ke−α t φ (x) (t 0, x ∈ Ω), whenever 0 u0 (x) kφ (x).

(4.5.15)


Proof.

167

To prove (4.5.15) it suffices to show that w(x,t) = δ e−α t φ (x) is an upper solution.

Clearly, the inequalities (4.5.6), (4.5.7) are satisfied. We observe that

∂w − Lw = δ e−α t [−αφ (x) − Lφ (x)] , ∂t and in view of the assumptions (4.5.13) and (4.5.14) we need only to show that −αφ (x) − Lφ (x) bφ (x) + β δ φ 2 (x) Since φ 1, −Lφ = λ0 φ , the inequality holds if βδ 1 − e−α t , (λ0 − b − α ) α An optimal choice of α , δ is given by α =

(λ0 −b) 2 ,

δ =

t

e−α s ds.

0

t > 0. (λ0 −b)2 4β .

Hence w is an upper

solution of (4.5.1)–(4.5.3). Conclusion (4.5.15) is now a consequence of Theorem 4.5.1. We note that the result in Theorem 4.5.3 gives the rate of the exponential decay of the solutions of (4.5.1)–(4.5.3) and u(x,t) → 0 as t → ∞. In [124,77] the authors have studied the integrodifferential equation of the form

∂u − Lu = Φ(x,t, u, Fu), ∂t B[u] = a

∂u + bu = h(x,t), ∂τ

u (x.0) = u0 (x),

(x,t) ∈ D,

(4.5.16)

(x,t) ∈ S,

(4.5.17)

x ∈ Ω,

(4.5.18)

where n

L=

∑

ai j (x,t)

i, j=1

n ∂2 ∂ + ∑ ai (x,t) , ∂ xi ∂ x j i=1 ∂ xi

is the uniformly elliptic operator whose coefficients are as defined above with suitable modifications (see also [37,57,122,125]), a, b are nonnegative constants, h ∈ H 1+α (s), Φ is Hölder continuous in (x,t, ω , ξ ) in every bounded subset of Ω × R+ × Rn × Rn and the operator F may in particular be of the form t

Fu(x,t) = 0

or

k1 (x,t, s, u(x, s)) ds,

Fu(x,t) =

Ω

k2 (x,t, y, u(y,t)) dy,

or variants thereof, in which k1 , k2 are defined on the respective domains of their definitions. In [77] the problem of existence of maximal and minimal solution is investigated by using

168


the concept of upper and lower solutions and monotone method. Below we explore in brief the results given in [77]. The function σ + ∈ C2 (D) is called an upper solution of problem (4.5.16)–(4.5.18) if

∂σ+ − Lσ + Φ(x,t, σ + , F σ + ), ∂t B[σ + ] h(x,t),

(x,t) ∈ D,

(4.5.19)

(x,t) ∈ S,

(4.5.20)

x ∈ Ω.

(4.5.21)

σ + (x, 0) u0 (x),

Similarly, the function σ − ∈ C2 (D) is called a lower solution of (4.5.16)–(4.5.18) if the inequalities in (4.5.19)–(4.5.21) are reversed. We list the following assumptions for convenience: (G1 ) the functions σ − , σ + ∈ C2 (D) with σ − (x,t) σ + (x,t) on D are lower and upper solutions of (4.5.16)–(4.5.18), (G2 ) for each i, Φi (x,t, u, Fu)− Φi (x,t, u, Fu) −N(ui −ui ), for every (x,t) ∈ D, whenever

σ − (x,t) u u σ + (x,t), where N 0 is a constant, (G3 ) the operator Fu is monotone increasing on U (σ − , σ + ) and for each i, Φi (x,t, u, v) is monotone nondecreasing in v. For any η ∈ C2 (D) such that σ − η σ + on D, consider the following linear system

∂ ui − Lui = Φi (x,t, η , F η ) − N(ui − ηi ), ∂t B [ui ] = h(x,t),

(x,t) ∈ D,

(4.5.22)

(x,t) ∈ S,

(4.5.23)

x ∈ Ω.

(4.5.24)

ui (x, 0) = u0i (x),

For a known function η , the assumptions on Φ and F ensure (see [37,57,122]) the existence of a unique solution of (4.5.22)–(4.5.24). For each η ∈ C2 (D) such that σ − η σ + on D, define the mapping P by Pη = u,

(4.5.25)

where u is the unique solution of (4.5.22)–(4.5.24). Using the techniques of proof of Lemma 4.5.1 and Theorem 4.5.1, one can easily prove the following results.


169

Lemma 4.5.2. Assume that the hypotheses (G1 )–(G3 ) hold. Then (i) the unique solution u of (4.5.22)–(4.5.24) satisfies σ − (x,t) u(x,t) σ + (x,t), (x,t) ∈ D; (ii) σ − Pσ − , σ + Pσ + on D; (iii) P is monotone operator on the set of functions U(σ − , σ + ). From Lemma 4.5.2, we can define the sequences − σn− = Pσn−1 ,

+ σn+ = Pσn−1 ,

with σ0− = σ − and σ0+ = σ + . By following the similar observations below Lemma 4.5.1 on the sequences {σn− }, {σn+ }, we have the following theorem on the existence of maximal and minimal solutions of (4.5.16)–(4.5.18). Theorem 4.5.4.

Assume that the hypotheses (G1 )–(G3 ) hold. Then the sequence {σn+ },

converges uniformly from above to the maximal solution u+ of (4.5.16)–(4.5.18), while the sequence {σn− } converges uniformly from below to a minimal solution u− of (4.5.16)– (4.5.18). Furthermore, if u is any solution of (4.5.16)–(4.5.18) such that u ∈ U(σ − , σ + ), then

σ − σ1− · · · σn− · · · u− u u+ · · · σn+ · · · σ1+ σ + , on D. 4.6

Initial-boundary value problem for integrodifferential equations

In this section, we deal with solvability in the classical sense of a nonlinear integrodifferential initial-boundary value problem:

t

ut = a(x,t, u, ux )uxx + b(x,t, u, ux ) +

0

c(x, τ , u, ux ) d τ ,

(4.6.1)

in QT , u(0,t) = f1 (t),

0 t T,

(4.6.2)

u(1,t) = f2 (t),

0 t T,

(4.6.3)

u(x, 0) = u0 (x), 0 x 1,

(4.6.4)

where QT = [0, 1] × [0, T ] with T > 0 arbitrary. One of the characteristics of this kind of problem is that the maximum principle is no longer valid in general. In dealing with this problem in [140], integral estimates in conjunction with Schauder estimate theory to derive an a priori bound for the solution of (4.6.1)–(4.6.4) in the norm of the Banach space

170

Multidimensional Integral Equations and Inequalities α

C2+α ,1+ 2 (QT ) is used. The notations of the norms in Banach spaces C(QT ), C2,1 (QT ), etc. are those of Ladyzenskaya et.al. [57]. The method of continuity, which is similar to that applicable for a regular parabolic boundary value problem is then applied in [140] to establish a global solvability of (4.6.1)–(4.6.4) in the classical sense. Below, the main goal is to present the results obtained in [140] concerning the solvability of (4.6.1)–(4.6.4). The following hypotheses are assumed throughout the discussion. (H1 ) the functions a(x,t, u, p), b(x,t, u, p) and c(x,t, u, p) are differentiable with respect to all of their arguments. Furthermore, (i) a(x,t, u, p) A1 > 0, (ii) |b(x,t, u, p)| A2 [1 + |u| + |p|], (iii) |c(x,t, u, p)| A3 [1 + |u| + |p|], for (x,t, u, p) ∈ QT × R2 , where A1 , A2 and A3 are three absolute constants. (H2 ) the functions f1 (t), f2 (t) ∈ C2 [0, T ], u0 (x) ∈ C2+α [0, 1] and the consistency conditions f1 (0) = u0 (0),

f 2 (0) = u0 (1),

f 1 (0) = a(0, 0, u0 (0), u0 (0))u0 (0) + b(0, 0, u0 (0), u0 (0)), and f 2 (0) = a(1, 0, u0 (1), u0 (1))u0 (1) + b(1, 0, u0 (1), u0 (1)), are satisfied. We need the following well known inequalities to establish the results. 1. Young’s inequality: If a 0, b 0, then, for any η > 0, s s b ar ab η + η − r , r s where r > 1, s > 1 and 1r + 1s = 1. 2. Interpolation inequalities: If u(x) ∈ H 1 (0, 1), then 2

1

uL∞ (0,1) CuH3 1 (0,1) uL31 (0,1) . α

We first establish a global a priori bound for the solution u(x,t) in C2+α ,1+ 2 (QT ) based on an integral calculation, imbedding inequalities and Schauder estimates under the hypotheses (H1 )–(H2 ). Let T > 0 be arbitrary and assume that u(x,t) is an arbitrary solution of the problem (4.6.1)– (4.6.4). We first deduce the following result.


171

Lemma 4.6.1. Under the assumptions (H1 )–(H2 ), u(x,t) satisfies the following inequality: QT

u2xx dx dt + sup

1

0tT 0

u2x (x,t) dx C1 ,

(4.6.5)

where C1 depends only on the Ai (i = 1, 2, 3), the known data and the upper bound of T . Proof.

In what follows, various constants which appear during the process of the proof

will be denoted by C; their dependency is the same as final constants unless stated otherwise. Let v(x,t) = (1 − x) f 1 (t) + x f 2 (t) and w(x,t) = u(x,t) − v(x,t), (x,t) ∈ QT . Then w(x,t) is a solution of the following problem: wt = awxx + b − vt +

t 0

c (x, τ , w + v, wx + vx ) d τ ,

(4.6.6)

0 t T,

(4.6.7)

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

w(x, 0) = u0 (x) − [(1 − x) f 1 (0) + x f2 (0)] = w0 (x) (say), 0 x 1.

(4.6.8)

Multiplying equation (4.6.6) by wxx and integrating it over QT , we obtain, employing the Cauchy–Schwarz inequality with small parameter ε > 0 and the assumption (H1 ) that

A1

ε

QT

QT

t

+ 0

−

wt wxx dx dt =

QT

QT

t 0

1 2

1

1 + w2 + w2x

QT

0

2 dx dt

T 1 t 0

0

0

dx dt.

wx (x, T )2 dx −

w2 dx dt C

(1 + |w| + |wx |) d τ 2T

wt wxx dx dt

A3 (1 + |w| + |wx |) d τ

and that

QT

2

QT

w2xx dx dt +C(ε )

Observe that

w2xx dx dt −

1 2

1 0

(4.6.9)

w0 (x)2 dx,

(4.6.10)

QT

w2x dx dt,

T 1

t

2t 0

0

0

(4.6.11)

(1 + w2 + w2x ) d τ dx dt

[1 + w2 + w2x ]d τ dx dt

172


≡ 2T

T 1 0

0

2T 2

(T − τ )[1 + w2 + w2x ]dx dt

T 1 0

0

[1 + w2 + w2x ]dx dt.

Combining (4.6.10)–(4.6.12) by choosing ε = A1 2

QT

w2xx dx dt +

1 0

1 4A1 ,

(4.6.12)

we have from (4.6.9) that

wx (x, T )2 dx (1 + T 2 )C

QT

w2x dx dt + (1 + T 2 )C.

Since T is arbitrary, Gronwall’s inequality (see [82]) implies that 1 0

Therefore,

wx (x,t)2 dx C(T ).

QT

and

wx (x,t)2 dx dt C,

QT

w2xx dx dt + sup

1

0tT 0

(4.6.13)

w2x (x,t) dx C.

(4.6.14)

This concludes the estimate (4.6.5) since u(x,t) = w(x,t) + v(x,t) on QT . Corollary 4.6.1.

There exists a positive constant C2 such that u(x,t)C(QT ) C2 ,

(4.6.15)

where C2 depends on the same quantities as C1 . Proof.

This can be obtained directly from the estimate (4.6.5).

Next we establish the following lemma which deals with the estimate on the norm of ux . Lemma 4.6.2. There exists a constant C3 such that ux C(QT ) C3 ,

(4.6.16)

where the dependency of C3 is the same as C1 . Proof. Let p > 2 be an arbitrary even integer. Since 1 T T 1 d p wx dx dt = pwxp−1 wxt dx dt 0 dt 0 0 0 =−

T 1 0

0

p(p − 1)wxp−2 wxx wt dx dt +

T 0

x=1 pwxp−1 wt dt x=0


=−

t p(p − 1)wxp−2 wxx awxx + b − vt + c d τ dx dt,

T 1 0

it follows that

173

0

(4.6.17)

0

1

T 1

wxp (x, T ) dx + A1 p(p − 1)wxp−2 w2xx dx dt 0 0 0 1 T t p(p − 1)wxp−2 wxx b − vt + cd τ dx dt w0 (x) p dx + 0

1

0

+C(ε ) Choosing ε =

A2 2

0

w0 (x) p dx + ε

0

0

(4.6.18)

0

and using (H1 ), we find 1 A1 T 1 wxp (x, T ) dx + p(p − 1)wxp−2 w2xx dx dt 2 0 0 0 1 T 1 p p−2 1 + w2 + w2x w0 (x) dx +C p(p − 1)wx 0

0

0

Let

T 1

I= 0

0

0

2 (1 + |w| + |wx |) d τ dx dt.

t

+

Then

0

0

p(p − 1)wxp−2 w2xx dx dt

2 t p(p − 1)wxp−2 b − vt + cd τ dx dt.

T 1 0

T 1

wxp−2

t 0

(4.6.19)

2 (1 + |w| + |wx |) d τ

dx dt.

t 2 2 I 2T T +C2 + wx d τ dx dt 0 0 0 t T 1 T 1 CT (1 + T ) wxp−2 dx dt + 2T wxp−2 w2x d τ dx dt T 1

0

wxp−2

0

0

0

0

≡ CT (1 + T )I1 + 2T I2 . p p p−2 , s = 2 and η = 1, we have t I2 = wxp−2 w2x d τ dx dt 0 0 0 t p T 1 2 p−2 p 2 2 wx + dx dt wx d τ p p 0 0 0 t T 1 T 1 p−2 t 2 wxp dx dt + wxp d τ dx dt

Using Young’s inequality with r = T 1

0

0

T 1 0

0

T 1 0

0

0

wxp dx dt + T

p−2 2

wxp dx dt + T

p−2 2

p 1+T 2

0

T

1 0

0

T 1 t 0

0

0

0

T 1

wxp dx dt.

0

0

wxp d τ dx dt

(T − τ )wxp dx d τ (4.6.20)

174


For the moment, we restrict T by 0 < T T0 = 1 (say). Under this condition, it follows from (4.6.19)–(4.6.20) and T ∈ [0, T0 ] arbitrary that 1

sup 0tT 0

1 0

wxp (x,t) dx +

A1 2

QT

T 1

w0 (x) p dx +C

0

p(p − 1)wxp−2 wxx dx dt

p(p − 1)wxp−2 1 + w2x dx dt,

0

(4.6.21)

where C depends only on C2 and known data. Assume that w(x,t)L∞ (QT ) max 1, T1 w0 (x)0 (here 0 < T T0 = 1 is a fixed number). Otherwise, we already have the estimate (4.6.16) on the interval [0, T0 ]. Then 1

sup 0tT 0

wxp dx +

A1 2

T 1 0

p(p − 1)wxp−2 w2xx dx dt C

0

C

T 0

T 1 0

0

p(p − 1)wxp dx dt

p(p − 1)wx (·,t)Lp∞ (0,1) dt.

If the interpolation inequality is employed, we have p 13 p p 23 2 2 C wx2 1 wx 1 wx L∞ (0,1)

H (0,1)

L (0,1)

(4.6.22)

,

i.e., p 43 wx Lp∞ (0,1) C wx2 1

H (0,1)

p 2 Cη wx2 1

p 23 2 wx 1

H (0,1)

L (0,1)

+Cη −2 wx p p

,

L 2 (0,1)

where the last inequality is obtained from Young’s inequality for r = 32 and s = 3. Note that 1 p 1 p 2 2 p −1 2 wx2 wxx dx + = wxp dx. wx 1 2 H (0,1) 0 0 As a consequence, it follows that 1

sup 0tT 0

wxp dx +

p2 Cp(p − 1) η 4

A1 2

T 1

T 1 0

0

0

1

sup 0tT 0

wxp dx + p(p − 1)

A1 1 2CT0 , p2C ,

T 1 0

0

p(p − 1)wxp−2 w2xx dx dt

wxp−2 w2xx dx dt + η

+Cp(p − 1)η −2 If now η is chosen as η = min

0

T 0

wx

T 1

p p

0

0

wxp dx dt

dt.

L 2 (0,1)

then

wxp−2 w2xx dx dt Cp(p − 1)η −2 T sup wx p p 0tT

L 2 (0,1)


175

Cp4 sup wx 0tT

p

,

p

L 2 (0,1)

where C is constant which depends only on known data. To complete the proof we want to consider large value of p. Let

1 p1 k wxpk dx . p = pk = 2k and αk = sup 0

0tT

If we take the pk -th root of both sides of above inequality, we obtain 1 αk Cp4k pk αk−1 . Now +∞

1

∏ C pk

=C

1 ∑+∞ k=1 p

k

+∞

1 k

= C∑k=1 2 C,

k=1

and +∞

4 p

∏ pk k

+∞ 4k

= 2∑k=1 2k C,

k=1

since +∞

+∞

4k

=4∑

k , k 2 k=1 k=1 1 is convergent. Thus it follows that, for dk = Cp4k pk ,

αk dk αk−1

∏ d

∑ 2k

k

α1 C α1 .

=1

As lim αk = wx L∞ (QT ) ,

k→+∞

and α1 C1 by Lemma 4.6.1, it follows that Q

wx 0 T Cα1 C.

(4.6.23)

Note that for the interval [T0 , 2T0 ], we can repeat the above procedure and obtain previously the same inequality (4.6.23). After finitely many steps, one has the estimate (4.6.16). Lemma 4.6.3. There exist constants C4 and α (0 < α < 1), which depend on the same quantities as Ci (i = 1, 2, 3), such that u

C

α 1+α , 1+ 2

(QT )

C4 ,

(4.6.24)

and hence ux

C

α, α 2

(QT )

C4 .

(4.6.25)

176

Proof.


Let

t μ = max |a(x,t, u, ux )| + |b(x,t, u, ux )| + |c(x, τ , u, ux )|d τ . 0

(x,t)∈QT |u|C2 |ux |C3

Lemma 4.6.2 implies that μ is uniformly bounded and that the bound depends only on the known data. The desired result then follows from Theorem 5.1 in Ladyzenskaya et.al. [57, p. 561] as a regular parabolic equation case. Lemma 4.6.4. There exists a constant C5 such that u

α

C2+α ,1+ 2 (QT )

C5 ,

(4.6.26)

where C5 depends only on the same quantities as Ci , i = 1, 2, 3, 4. Proof.

By Lemma 4.6.3, we know that a(x,t, u(x,t), ux (x,t)) and b(x,t, u(x,t), ux (x,t)) α 2

are uniformly Hölder continuous in QT with exponents α and

with respect to x and t,

respectively. Considering equation (4.6.1) as a linear equation t

ut = auxx + b +

0

cd τ ,

with initial-boundary conditions (4.6.2)–(4.6.4), we employ the Schauder estimate to obtain t . (4.6.27) C 1+ u 2+α ,1+ α2 cd τ α (QT )

C

0

C

α, 2

(QT )

α

Note that, for any function g(x,t) ∈ Cα , 2 (QT ), we have the property t 1− α2 g(x, τ ) d τ α g . g(x, 0) + T + T α , C[0,1] α, α 0 C 2 (QT ) C

2

As a consequence t c(x, τ , u, ux ) d τ 0

(4.6.28)

(QT )

α

Cα , 2 (QT )

α C 1 + T + T 1− 2 c(x,t, u, ux ) α C 1 + T + T 1− 2 u

C

C

α 1+α , 1+ 2

α, α 2

(QT )

(QT )

,

is uniformly bounded by Lemma 4.6.3, and the bound depends only on the known data. Hence the estimate (4.6.26) follows from (4.6.27) and the above inequality. From Lemmas 4.6.1–4.6.4, we establish the following existence theorem. Theorem 4.6.1. α

Under hypotheses (H1 )–(H2 ), there exists a solution u(x,t) ∈

C2+α ,1+ 2 (QT ) to the problem (4.6.1)–(4.6.4).


Proof.

177

Define the operator Lλ by Lλ u = ut − [auxx + b + λ

t 0

c d τ ].

Let

∑(λ ) = {λ ∈ [0, 1] :

the problem (4.6.1)λ –(4.6.4) is solvable},

where (4.6.1)λ is the equation Lλ u = 0. By the standard continuation method (∑(λ ) is not empty, ∑(λ ) is open and also closed), it follows that ∑(λ ) ≡ [0, 1]. The following theorem deals with the regularity of the solution for the problem (4.6.1)– (4.6.4). Theorem 4.6.2.

Assume that a(x,t, u, p), b(x,t, u, p) and c(x,t, u, p) are infinitely differ-

entiable in all of their arguments and that the boundary values f1 (t) and f 2 (t) belong to C∞ (0, T ]. Then the solution u(x,t) is infinitely differentiable with respect to x and t on the region QT ∩ {(x,t) : t > 0}. Proof.

α

Since u ∈ C2+α ,1+ 2 (QT ), we can differentiate equation (4.6.1) with respect to t

and then v = ut satisfies vt = avxx + [a p uxx + b p ]vx + [au uxx + bu]v + [at uxx + bt + c(x,t, u, ux )] in QT ,

(4.6.29)

v(0,t) = f1 (t),

0 t T,

(4.6.30)

f2 (t),

0 t T.

(4.6.31)

v(1,t) =

Since the coefficients of equation (4.6.29) are Hölder continuous with respect to x and t, α

the Schauder estimate for a parabolic equation implies that the solution v ∈ C2+α ,1+ 2 (QT ). α

Hence u ∈ C4+α ,2+ 2 (QT ). We can repeat this procedure and obtain v ∈ C+∞,+∞ (QT ∩ {t : t > 0}). It follows that u(x,t) ∈ C+∞,+∞ (QT ∩ {t : t > 0}). Next, we give the following theorem on continuous dependence of the solution of (4.6.1)– (4.6.4) upon the known data. Theorem 4.6.3. Assume that ( f1 (t), f2 (t), u0 (x)) and ( f1∗ (t), f2∗ (t), u∗0 (x)) are two known sets of initial-boundary values which satisfy (H2 ). Let u(x,t) and u∗ (x,t) be two solutions of the problem (4.6.1)–(4.6.4) corresponding, respectively, to the above data. Then u(x,t) − u∗ (x,t)

α

C2+α ,1+ 2 (QT )

C f 1 (t) − f1∗ (t)

+ f2 (t) − f2∗ (t) 1+ α2 C [0,T ] ∗ + u0 (x) − u0 (x)C2+α [0,1] , α

C1+ 2 [0,T ]

where C depends only on known data.

(4.6.32)

178

Proof.


Let w(x,t) = u(x,t) − u∗ (x,t), (x,t) ∈ QT . Then w(x,t) satisfies wt = awxx + b∗ (x,t)wx + c∗ (x,t)w + d ∗ (x,t) in QT ,

(4.6.33)

w(0,t) = f1 (t) − f1∗ (t),

0 t T,

(4.6.34)

f2∗ (t),

0 t T,

(4.6.35)

w(x, 0) = u0 (x) − u∗0 (x), 0 x 1,

(4.6.36)

w(1,t) = f2 (t) −

where b∗ (x,t) =

1 0

1

+ 0

a p (x,t, u∗ , zux + (1 − z)u∗x ) dz u∗xx ,

c∗ (x,t) =

1 0

1

+ 0

b p (x,tu∗ , zux + (1 − z)u∗x ) dz

bu (x,t, zu + (1 − z)u∗ , ux ) dz

au (x,t, zu + (1 − z)u∗ , ux ) dz u∗xx ,

d ∗ (x,t) =

t 0

1

d1 (x,t) =

0

c p (x,t, u∗ , zux + (1 − z)u∗x ) dz,

1

d2 (x,t) =

0

[d1 (x, τ )wx + d2 (x, τ )w] d τ ,

cz (x,t, zu + (1 − z)u∗ , ux ) dz.

The estimate (4.6.24) implies that all the Hölder moduli of the coefficients in (4.6.33) are dominated by known data. From the Schauder estimate for the linear parabolic equation (4.6.33), we have u

C

2

2+α ,1+ α 2

(QT )

C ∑ fi (t) − fi∗ (t)

C

i=1

+u0 (x) − u∗0 (x)C2+α [0,1] + d ∗ (x,t)

C

1+ α 2

α, α 2

[0,T ]

(QT )

.

The inequalities (4.6.26) and (4.6.28) yield α . d ∗ (x,t) α , α2 C w(x, 0)C[0,1] + T + T 1− 2 u 2+α ,1+ α2 C (QT ) C (QT ) α Therefore, when T is small enough so that T + T 1− 2 C 12 we have the desired result. By taking a finite number of steps, we establish (4.6.32) for arbitrary T .


The solution of the problem (4.6.1)–(4.6.4) is unique.

Corollary 4.6.2. 4.7

179

Miscellanea

4.7.1

Corduneanu [28]

Consider the integrodifferential equation of the form 1 ∂ f (t, x) = ∂t 2

x 0

φ (x − y, y) f (t, x − y) f (t, y) dy − f (t, x)

∞ 0

φ (x, y) f (t, y) dy,

(4.7.1)

which is encountered in the mathematical description of coagulation processes, with the initial condition f (0, x) = f 0 (x),

x 0.

(4.7.2)

For the interpretation of equation (4.7.1) and detailed meanings of the functions therein, see [28, p. 274]. Assume that (H1 ) the function φ (x, y) is continuous and symmetric in the quadrant x 0, y 0, and verifies 0 φ (x, y) A, x 0, y 0, where A is a positive number, (H2 ) f0 (x) is continuous, bounded and integrable on x 0, (H3 ) f0 (x) 0 for x 0. Under the hypotheses (H1 )–(H3 ), there exists a solution f (t, x) of equation (4.7.1), satisfying (4.7.2), which is defined for x 0, t 0, is continuous, bounded, nonnegative, analytic in t for each fixed x 0 and integrable in x for each t 0. Moreover, the solution is unique. 4.7.2

Pachpatte [98]

Consider the parabolic-type Fredholm integral equation b ∂ u(x,t) = F x,t, u(x,t), K(x,t, y, u(y,t)) dy , ∂t a

(4.7.3)

with the given initial condition u(x, 0) = φ (x),

(4.7.4)

for (x,t) ∈ Δ, where φ ∈ C(I, R), K ∈ C(Δ × I × R, R), F ∈ C(Δ × R2 , R) in which I = [a, b] (a < b), Δ = I × R+ . Assume that (G1 ) the functions F, K in equation (4.7.3) satisfy the conditions |F(x,t, u, v) − F(x,t, u, v)| [p(x,t)|u − u| + |v − v|], |K(x,t, y, u) − K(x,t, y, u)| r(x,t, y)|u − u|,

180


where p ∈ C(Δ, R+ ), r ∈ C(Δ × I, R+ ); (G2 ) for λ as in section 4.3: (i) there exists a nonnegative constant α such that α < 1 and t b p(x, s) exp(λ (|x| + s)) + r(x, s, y) exp(λ (|y| + s)) dy ds α exp(λ (|x| + t)), 0

a

(ii) there exists a nonnegative constant β such that t b K(x, s, y, 0) dy ds β exp(λ (|x| + t)), |φ (x)| + F x, s, 0, 0

a

for (x,t) ∈ Δ. Under assumptions (G1 )–(G2 ), the problem (4.7.3)–(4.7.4) has a unique solution on Δ in E, where E is the space of functions as defined in section 4.3. 4.7.3

Pachpatte [74]

Under the notations as in section 4.5, consider the following nonlinear integrodifferential system of the form

t ∂u − Lu = F x,t, u, K(x,t, s, u(x, s)) dx ds ∂t 0 Ω

(t > 0, x ∈ Ω),

(4.7.5)

with the given boundary and initial conditions B[u] = α1 (x)

∂u + α2 (x)u = 0 (t > 0, x ∈ ∂ Ω), ∂τ

u(x, 0) = u0 (x),

x ∈ Ω,

(4.7.6) (4.7.7)

where L denote the uniformly elliptic operator defined by n

L=

∑

i, j=1

ai j (x)

n ∂2 ∂ + ∑ ai (x) , ∂ xi ∂ x j i=1 ∂ xi

(4.7.8)

on the bounded domain Ω in Rn , τ is the outward normal unit vector on the boundary

∂ Ω; αi (x) 0 (i = 1, 2) with α1 (x) + α2 (x) = 0 on ∂ Ω; the coefficients of L and the first partial derivatives of ai j are Hölder continuous (of exponent α ∈ (0, 1)) in Ω; the matrix (ai j ) is symmetric positive definite in Ω; K and F are Hölder continuous in every bounded subset of Ω × R2+ × R and Ω × R+ × R2 respectively; the boundary ∂ Ω is of class of C2+α , αi ∈ H 1+α (∂ Ω), and u0 ∈ H 1+α Ω satisfies (4.7.6) at t = 0. Assume that (H1 ) there exist pair of functions v, w ∈ C2 (D) with v w on D such that t ∂w − Lw F x,t, w, K(x,t, s, w(x, s)) dx ds , (x,t) ∈ D, ∂t 0 Ω


181

B[w] 0,

(x,t) ∈ S,

w(x, 0) u0 (x), and

x ∈ Ω;

t ∂v − Lv F x,t, v, K(x,t, s, v(x, s)) dx ds , ∂t 0 Ω B[v] 0,

(x,t) ∈ D,

(x,t) ∈ S,

v(x, 0) u0 (x),

x ∈ Ω;

(H2 ) the function K(x,t, s, u) is monotone nondecreasing in u for fixed x, t, s and the function F(x,t, u, ξ ) is monotone nondecreasing in ξ for fixed x, t, u; (H3 ) for a given constant M 0, F(x,t, u, ξ ) − F(x,t, u, ξ ) −M(u − u), whenever v(x,t) u u w(x,t) for (x,t) ∈ D. For any η ∈ C2 (D) such that v η w on D, consider the following linear system t ∂u − Lu = F x,t, η , K(x,t, s, η ) dx ds − M(u − η ), (x,t) ∈ D, (4.7.9) ∂t 0 Ω B[u] = 0,

(x,t) ∈ S,

u(x, 0) = u0 (x),

x ∈ Ω.

(4.7.10) (4.7.11)

For a known η the above assumptions ensures the existence of a unique solution of (4.7.9)– (4.7.11) (see [37,57 ]). For each η ∈ C2 (D) such that v η w on D, define the mapping A by Aη = u, where u is the unique solution of (4.7.9)–(4.7.11). (g1 ) Assume that the hypotheses (H1 )–(H3 ) hold. Then (i) the unique solution u of (4.7.9)–(4.7.11) satisfies v(x,t) u(x,t) w(x,t), for (x,t) ∈ D; (ii) v Av, w Aw for (x,t) ∈ D; (iii) A is monotone operator on the set of functions U(v, w) = {u ∈ C2 (D) : v(x,t) u w(x,t), (x,t) ∈ D}. In view of (g1 ), we can define the sequences vn = Avn−1 ,

wn = Awn−1 ,

182


with v0 = v and w0 = w. (g2 ) Assume that the hypotheses (H1 )–(H3 ) hold. Then the sequence {wn } converges uniformly from above to a maximal solution u of (4.7.5)–(4.7.7), while the sequence {vn } converges uniformly from below to a minimal solution u of (4.7.5)–(4.7.7). Furthermore, if u is any solution of (4.7.5)–(4.7.7) such that u ∈ U (v, w), then v v1 · · · vn · · · u u u · · · wn · · · w1 w, on D 4.7.4

Pachpatte [80]

Under the notations as in section 4.5, consider the following nonlinear coupled parabolic integrodifferential equations of the form

∂u − Lu = F(x,t, u, v, M[u, v](x,t)), ∂t ∂v = H(x,t, u, v, N[u, v](x,t)), ∂t for (x,t) ∈ D, with the given boundary and initial conditions B[u] =

∂u + b(x)u = c0 (x,t), ∂τ

u(x, 0) = u0 (x), where

(4.7.12) (4.7.13)

(x,t) ∈ S,

(4.7.14)

x ∈ Ω,

(4.7.15)

v(x, 0) = v0 (x),

t

M[u, v](x,t) = 0

Ω

0

Ω

t

N[u, v](x,t) =

f (x,t, y, s, u(y, s), v(y, s)) dy ds, h(x,t, y, s, u(y, s), v(y, s)) dy ds,

and L is the uniformly elliptic operator as defined in (4.7.8); f , h and F, H are real valued Hölder continuous in (x,t, y, s, u, v) and (x,t, u, v, r) in every bounded subset of Ω × R+ × Ω × R+ × R2 and Ω × R+ × R3 respectively; b(x) 0 on ∂ Ω, b ∈ H 1+α (∂ Ω), c0 ∈ H 1+α (s), u0 , v0 ∈ H 2+α Ω and u0 satisfies the boundary condition u(x, 0) = u0 (x) at t = 0. Assume that (H4 ) there exist pair of functions σ = (u, v), σ = (u, v); u, u ∈ C2 (D) and v, v ∈ C(D) with

σ σ on D such that ∂u − Lu F(x,t, u, v, M[u, v](x,t)), ∂t ∂v H(x,t, u, v, N[u, v](x,t)), ∂t

(x,t) ∈ D, (x,t) ∈ D,


183

B[u] c0 (x,t), u(x, 0) u0 (x),

(x,t) ∈ S,

v(x, 0) v0 (x),

x ∈ Ω,

and

∂u − Lu F(x,t, u, v, M[u, v](x,t)), (x,t) ∈ D, ∂t ∂v H(x,t, u, v, N[u, v](x,t)), (x,t) ∈ D, ∂t B[u] c0 (x,t), u(x, 0) u0 (x),

(x,t) ∈ S,

v(x, 0) v0 (x),

x ∈ Ω;

(H5 ) the functions f (x,t, y, s, u, v), h(x,t, y, s, u, v) both are monotone nondecreasing in u and v; (H6 ) the function F(x,t, u, v, r) is monotone nondecreasing in v and r, and the function H(x,t, u, v, r) is monotone nondecreasing in u and r; (H7 ) the functions F and H satisfy F(x,t, u1 , v, r) − F(x,t, u2 , v, r) −Q(u1 − u2 ), whenever u(x,t) u2 u1 u(x,t) for (x,t) ∈ D, and H(x,t, u, v1 , r) − H(x,t, u, v2 , r) −Q(v1 − v2 ), whenever v(x,t) v2 v1 v(x,t) for (x,t) ∈ D, where Q 0 is a constant. For a given function z = (ξ , η ); ξ ∈ C2 (D), η ∈ C(D) such that σ z σ on (x,t) ∈ D, where σ = (u, v), σ = (u, v), consider the following linear system ∂u − Lu = F(x,t, ξ , η , M[ξ , η ](x,t)) − Q(u − ξ ), ∂t ∂v = H(x,t, ξ , η , N[ξ , η ](x,t)) − Q(v − η ), ∂t for (x,t) ∈ D, with the given boundary and initial conditions B[u] = c0 (x,t), u(x, 0) = u0 (x),

(x,t) ∈ S,

v(x, 0) = v0 (x),

(4.7.16) (4.7.17)

(4.7.18) x ∈ Ω.

(4.7.19)

For a known function z = (ξ , η ) the above assumptions ensures (see [37, 57]) the existence of a unique solution of (4.7.16)–(4.7.19). For each z = (ξ , η ), ξ ∈ C2 (D), η ∈ C(D) such that σ z σ on D, define the mapping A by Az = σ , where σ = (u, v) is the solution of (4.7.16)–(4.7.19). (g3 ) Assume that the hypotheses (H4 )–(H7 ) hold. Then

184


(i) the unique solution σ = (u, v) of (4.7.16)–(4.7.19) satisfies σ (x,t) σ (x,t) σ (x,t), (x,t) ∈ D; (ii) σ Aσ ,

σ Aσ on D;

(iii) A is monotone operator on the set of functions Δ = σ = (u, v) : σ σ σ ; u ∈ C2 (D), v ∈ C(D) , on D. In view of (g3 ), we can define the sequences

σ n = Aσ n−1 ,

σ n = Aσ n−1 ,

where σ n = (un , vn ), σ n = (un , vn ) with σ 0 = σ and σ 0 = σ . (g4 ) Assume that the hypotheses (H4 )–(H7 ) hold. Then the sequence {(un , vn )} converges uniformly from above to a maximal solution (α , β ) of (4.7.12)–(4.7.15), while the sequence {(un , vn )} converges uniformly from below to a minimal solution (α , β ) of (4.7.12)–(4.7.15). Furthermore, if σ = (u, v) is any solution of (4.7.12)–(4.7.15) such that

σ ∈ Δ, then u u1 · · · un · · · α u α · · · un · · · u1 u, v v1 · · · vn · · · β v β · · · vn · · · v1 v, on D. 4.7.5

Cannon and Lin [23]

Under the standard notations for Hölder classes defined in Chapter 1 of [57], consider the linear integrodifferential equation of parabolic type t

ut (x,t) = A(x,t)u(x,t) +

0

B(x,t, τ )u(x, τ ) d τ + f (x,t),

(4.7.20)

in QT = Ω × (0, T ], where T > 0 and Ω is a bounded open subset of Rn with regular boundary ∂ Ω and QT is the closure of QT , n

Au =

∑

i, j=1

n

ai j (x,t)uxi x j (x,t) + ∑ ai (x,t)uxi (x,t) + a(x,t)u(x,t),

(4.7.21)

i=1

B(x,t, τ )u(x, τ ) =

n

∑

bi j (x,t, τ )uxi x j (x, τ )

i, j=1 n

+ ∑ bi (x,t, τ )uxi (x, τ ) + b(x,t, τ )u(x, τ ). i=1

(4.7.22)


185

The equation (4.7.20) is considered together with the following three types of boundary conditions: (L) u(x, 0) = u0 (x), x ∈ Ω, u(x,t) = φ (x,t),

(x,t) ∈ ST = ∂ Ω × [0, T ];

(M) u(x, 0) = u0 (x), x ∈ Ω, n

∑ ci (x,t)uxi (x,t) cos(xi , ν ) + c(x,t)u(x,t) = Φ(x,t),(x,t) ∈ ST ;

i=1

(N) u(x, 0) = u0 (x), x ∈ Ω, n

∑ ci (x,t)uxi (x,t) cos(xi , ν ) + c(x,t)u(x,t)

i=1

t

+ 0

n

∑ di (x,t, τ )uxi (x, τ ) + d(x,t, τ )u(x, τ )

d τ = Φ(x,t),

(x,t) ∈ ST ;

i=1

where

n ∑ ci (x,t) cos(xi , ν ) δ , i=1

for some positive constant δ and ν (x) = (ν1 (x), . . . , νn (x)) is the outer normal direction to

∂ Ω. The equation (4.7.20) will be treated as a parabolic equation with integral term as a perturbation and employing the basic classical theory of parabolic partial differential equation ut (x,t) = A(x,t)u(x,t) + f (x,t).

(4.7.23)

Assume that

α (H1 ) u0 ∈ H 2+α Ω ; ai j , ai , a, f ∈ H α , 2 (QT ) and

α0 |ξ |2 ai j ξi ξ j α1 |ξ |2 , for (x,t) ∈ QT , α0 , α1 are positive constants and ξ ∈ Rn ; α (H2 ) Φ ∈ H 2+α ,1+ 2 ST ; (H3 ) bi j (x,t, ·), bi (x,t, ·), b(x,t, ·) ∈ C([0, T ]) uniformly for (x,t) ∈ QT and bi j (·, ·, τ ), α

bi (·, ·, τ ), b(·, ·, τ ) ∈ H α , 2 (QT ) uniformly for τ ∈ [0, T ]; (1+α ) (H4 ) ci , c, Φ ∈ H 1+α , 2 ST ; (H5 ) di (x,t, ·), d(x,t, ·) ∈ C([0, T ]) uniformly on ST and di (·, ·, τ ), d(·, ·, τ ) ∈ H 1+α ,

(1+α ) 2

(ST ) uniformly for τ ∈ [0, T ].

Problem (4.7.20)–(L) or (4.7.23)–(L) is said to satisfy condition (C1 ) if (C1 ) u0 (x) = Φ(x, 0), Φt (x, 0) = Au0 + f (x, 0),

186


and problem (4.7.20)–(M) or (4.7.20)–(N) or (4.7.23)–(M) is said to satisfy condition (C2 ) if (C2 ) ∑ni=1 ci (x, 0)u0xi (x) cos(xi , ν ) + c(x, 0)u0 (x) = Φ(x, 0). (g5 ) For 0 < α < 1, assume that ∂ Ω ∈ H 2+α . Under assumptions (H1 )–(H3 ) and (C1 ), there α

exists a unique smooth solution u in H 2+α ,1+ 2 (QT ) for problem (4.7.20)–(L). (g6 ) For 0 < α < 1, assume that ∂ Ω ∈ H 2+α . Under assumptions (H1 ), (H3 )–(H4 ) and (C2 ), α

there exists a unique smooth solution u in H 2+α ,1+ 2 (QT ) for problem (4.7.20)–(M). (g7 ) For 0 < α < 1, assume that ∂ Ω ∈ H 2+α . Under assumptions (H1 ), (H3 )–(H5 ) and (C2 ), α

there exists a unique smooth solution u in H 2+α ,1+ 2 (QT ) for problem (4.7.20)–(N). 4.7.6

Roux and Thomée [126]

Consider the semilinear parabolic integrodifferential equation

∂ u(x,t) + Au(x,t) = ∂t

t

f (t, s, x, u(x, s)) ds,

(4.7.24)

0

for x ∈ Ω, 0 t T , where A is a self-adjoint elliptic second order partial differential operator with smooth, time-independent coefficients of the form d ∂ ∂u ai j + a0 u, a0 0, Au = ∑ ∂xj i, j=1 ∂ xi in a domain Ω ⊂ Rd with smooth boundary ∂ Ω, and f is a given smooth function of its arguments that is bounded together with a sufficient number of its derivatives. The equation (4.7.24) is considered together with the Dirichlet type boundary condition u = 0 on ∂ Ω for 0 t T,

(4.7.25)

u(x, 0) = v(x) in Ω.

(4.7.26)

and with the initial condition

For the numerical solution of this problem by the Galerkin finite element method we shall assume that we are given a family of subspaces Sh of H01 (Ω) with the approximation property that, for some r 2, inf {v − χ + hv − χ 1 } C hs vs ,

χ ∈Sh

for v ∈ H s (Ω) ∩ H01 , 1 s r. Here · denotes the norm in L2 (Ω) and · s that in H s (Ω) (see [57]). With (·, ·) the standard inner product in L2 (Ω) and A(·, ·) the positive


187

definite bilinear form on H01 (Ω) × H01 (Ω) defined by the operator A, we may then pose the semidiscrete problem of finding uh : [0, T ] → Sh such that, with f (t, s, u) = f (t, s, x, u(x, s)), t ∂ uh , χ + A(uh , χ ) = ( f (t, s, uh ), χ ) ds, (4.7.27) ∂t 0 for all χ ∈ Sh , 0 t T , uh (0) = vh ,

(4.7.28)

where vh is a suitable approximation of v in Sh . (g8 ) Let uh and u be the solutions of (4.7.27)–(4.7.28) and (4.7.24)–(4.7.26), respectively. Then, if u is appropriately smooth, the estimate uh (t) − u(t) Cvh − v +Chr vr +

t 0

ut r ds,

holds for 0 t T . 4.7.7

Yin [141]

Let T > 0 and QT = Ω × (0, T ], where Ω is an open bounded region in Rn with a smooth boundary S = ∂ Ω and ST = S × [0, T ]. Consider the following integrodifferential equation

∂ ai (x,t, u, ux ) − a(x,t, u, ux ) ∂ xi t ∂ = bi (x,t, τ , u, ux ) + b(x,t, τ , u, ux ) d τ , (x,t) ∈ QT , 0 ∂ xi ut −

(4.7.29)

with the given boundary and initial conditions u(x,t) = 0,

(x,t) ∈ ST ,

(4.7.30)

x ∈ Ω.

(4.7.31)

u(x, 0) = u0 (x),

A function u(x,t) in V2 (QT ) = L∞ (0, T ; L2 (Ω)) ∩ L2 (0, T ; H01 (Ω)) is said to be a weak solution of (4.7.29)–(4.7.31), if u(x,t) satisfies T t t −uφt + ai + bi d τ φxi − a + bd τ φ dx dt 0

Ω

0

0

u0 (x)φ (x, 0) dx − u(x, T )φ (x, T ) dx, Ω Ω H 1 (0, T ; H01 (Ω)). =

for any φ (x,t) ∈ Assume that

(H1 ) the functions ai , a and bi , b are differentiable with respect to all of their arguments in QT × R × Rn and QT × R+ × R × Rn , respectively;

188


(H2 ) the following ellipticity assumption and the growth conditions hold: n

∑ [ai (x,t, u, p1 ) − ai (x,t, u, p2 )](p1 − p2 ) a0 (p1 − p2 )2 ,

i=1

n

∑ [|ai (x,t, u, p)| + |bi (x,t, τ , u, p)|] + |a| + |b| A0 [1 + |u| + |p|] ;

i=1

n

∑ |bip (x,t, τ , u, p)| B0 ,

i=1

where a0 , A0 and B0 are positive constants; (H3 ) u0 (x) ∈ H01 (Ω). Under the hypotheses (H1 )–(H3 ), the problem (4.7.29)–(4.7.31) admits a unique weak solution. 4.7.8

Yin [141]

Let T > 0 and QT = Ω × (0, T ], where Ω is an open bounded region on Rn with a smooth boundary S = ∂ Ω and ST = S × [0, T ]. Consider the following initial-boundary value problem: ut − Δu − f (u) =

u(x,t) = 0,

t 0

[b(t, τ )Δu]d τ ,

(x,t) ∈ ST ,

(x,t) ∈ QT ,

u(x, 0) = u0 (x),

x ∈ Ω,

(4.7.32)

(4.7.33)

with the assumptions (G1 ) the function b(t, τ ) is differentiable on R2 ; (G2 ) f (u) is differentiable and there exist positive constants A0 and M such that f (u) A0 for |u| M; (G3 ) u0 (x) ∈ C2+α Ω (Ω is the closure of Ω) and the following compatibility conditions hold u0 (x) = 0,

Δu0 (x) + f (u0 ) = 0 on S.

Under the hypotheses (G1 )–(G3 ), the problem (4.7.32)–(4.7.33) admits a unique classical solution on [0, T ] for any T > 0.


4.8

189

Notes

Many phenomena in physical, chemical, biological sciences and engineering can be modeled via parabolic integrodifferential equations, for which elegant theories and powerful techniques have been developed, see [1,5,27,31–33,59–63,121,122,124,126,131– 134,140,141] and the references cited therein. The results in section 4.2 provides some basic integral inequalities with explicit estimates which can be used as tools in handling the study of qualitative properties of solutions of certain parabolic partial integrodifferential and integral equations and are adapted from Pachpatte [98,109]. Sections 4.3 and 4.4 are concerned with the study of some fundamental qualitative properties of solutions of an integrodifferential equation of Barbashin-type and the general integral equation of Barbashintype, and are taken from Pachpatte [110,109]. Here, the treatment of results is essentially different from those used in [5] to study such equations. Section 4.5 deals with the existence, uniqueness and asymptotic behavior of solutions of an integrodifferential equation of the type arising in reactor dynamics. The notion of upper and lower solutions based on the monotone iterative method is used to study the problem and the results are taken from Pachpatte [75,77]. The section 4.6 is devoted to the solvability in the classical sense of a class of nonlinear one-dimensional integrodifferential equation of parabolic type and the results are due to Yin [140]. Section 4.7 contains results related to certain aspects of some selected equations, which we hope will motivate the reader’s interest in further study of related topics.

Chapter 5

Multivariable sum-difference inequalities and equations

5.1

Introduction

Multivariable difference equations occur in numerous settings and forms in various applications. By computing the value of a unknown function recursively for a set of equidistant points from a given set of values, leads to the equation which may be viewed as sumdifference equation on the fixed region of summation, see [2,3,14,16,46,47,51,67,68,85]. The problems of existence of solutions for these equations can be dealt with the use of the well known fixed point theorems, see [51,54]. Besides the existence problems, there are many basic questions which are significant with respect to the theory itself or to applications to it. In practice it is often difficult to obtain explicitly the solutions, and thus need a new insight to handle the qualitative properties of their solutions. The method of finite difference inequalities with explicit estimates provides the most powerful and widely used analytic tool in the study of various discrete dynamic equations. It enable us to obtain valuable information about solutions without the need to know in advance the solution explicitly. In this chapter, we focus our attention to present some fundamental sum-difference inequalities with explicit estimates recently established in [108,104,98,95,102,111,116,114,106], which can be used as tools for handling the qualitative properties of solutions of various types of multivariable sum-difference equations . Furthermore, some important basic qualitative aspects related to the solutions of certain sum-difference equations investigated in [100,103,116,101,99,107,109] are also given.

5.2

Sum-difference inequalities in two variables

In this section we offer some basic sum-difference inequalities in two variables which can be used as tools in certain applications when the earlier inequalities in the literature do not 191

192


apply directly. The inequalities given in the following theorems are adapted from [108,104,98]. Theorem 5.2.1. Let u, f , e ∈ D N20 , R+ and c 0 is a real constant. (q1 ) If n−1 s−1 m−1

u(n, m) c + ∑

∑∑

s=0 σ =0 τ =0

for (n, m) ∈ N20 , then n−1

u(n, m) c ∏ 1 + s=0

f (σ , τ )u(σ , τ ),

s−1 m−1

∑∑

σ =0 τ =0

(5.2.1)

f (σ , τ ) ,

(5.2.2)

for (n, m) ∈ N20 . (q2 ) Let e(n, m) be nondecreasing in each variable n, m ∈ N0 . If n−1 s−1 m−1

u(n, m) e(n, m) + ∑

∑∑

s=0 σ =0 τ =0

for (n, m) ∈ N20 , then n−1

u(n, m) e(n, m) ∏ 1 + s=0

f (σ , τ )u(σ , τ ),

s−1 m−1

∑∑

σ =0 τ =0

(5.2.3)

f (σ , τ ) ,

(5.2.4)

for (n, m) ∈ N20 . Theorem 5.2.2. Let u, p, q, f ∈ D(N20 , R+ ). (q3 ) Let L ∈ D(N20 × R+ , R+ ) be such that 0 L (n, m, u) − L(n, m, v) M(n, m, v)(u − v),

(5.2.5)

for u v 0, where M ∈ D(N20 × R+ , R+ ). If n−1 s−1 m−1

u(n, m) p(n, m) + q(n, m) ∑

∑ ∑ L(σ , τ , u(σ , τ )),

(5.2.6)

s=0 σ =0 τ =0

for (n, m) ∈ N20 , then

u(n, m) p(n, m) + q(n, m) n−1

×∏ 1+ s=0

for (n, m) ∈ N20 .

s−1 m−1

n−1 s−1 m−1

∑ ∑ ∑ L(σ , τ , p(σ , τ ))

s=0 σ =0 τ =0

∑ ∑ M(σ , τ , p(σ , τ ))q(σ , τ )

σ =0 τ =0

,

(5.2.7)


193

(q4 ) If n−1 s−1 m−1

u(n, m) p(n, m) + q(n, m) ∑

∑∑

s=0 σ =0 τ =0

for (n, m) ∈ N20 , then

×∏ 1+ s=0

n−1 s−1 m−1

∑∑∑

u(n, m) p(n, m) + q(n, m)

n−1

f (σ , τ )u(σ , τ ),

s=0 σ =0 τ =0

(5.2.8)

f (σ , τ )p(σ , τ )

s−1 m−1

∑∑

f (σ , τ )q(σ , τ ) ,

σ =0 τ =0

(5.2.9)

for (n, m) ∈ N20 . Theorem 5.2.3. Let u, f ∈ D(E, R+ ) and c 0 is a real constant, where E = N0 × Nα ,β . (q5 ) If n−1 β

u(n, m) c + ∑

∑ f (s,t)u(s,t),

(5.2.10)

s=0 t=α

for (n, m) ∈ E, then n−1

β

u(n, m) c ∏ 1 + ∑ f (s,t) ,

(5.2.11)

t=α

s=0

for (n, m) ∈ E. (q6 ) Let g ∈ C(R+ , R+ ) be nondecreasing function, g(u) > 0 on (0, ∞). If n−1 β

u(n, m) c + ∑

∑ f (s,t)g(u(s,t)),

(5.2.12)

s=0 t=α

for (n, m) ∈ E, then for 0 n n1 ; n, n1 ∈ N0 , m ∈ Nα ,β , u(n, m) W

n−1 β

W (c) + ∑

−1

∑ f (s,t)

,

where W (r) =

r dz r0

g(z)

,

r > 0,

(5.2.14)

r0 > 0 is arbitrary and W −1 is the inverse of W and n1 ∈ N0 is chosen so that n−1 β

W (c) + ∑

(5.2.13)

s=0 t=α

∑ f (s,t) ∈ Dom W −1

s=0 t=α

for all n ∈ N0 lying in 0 n n1 and m ∈ Nα ,β .

,

194


Theorem 5.2.4. Let u, f , c be as in Theorem 5.2.3. (q7 ) If n−1 β

u2 (n, m) c + ∑

∑ f (s,t)u(s,t),

(5.2.15)

s=0 t=α

for (n, m) ∈ E, then u(n, m)

√

c+

for (n, m) ∈ E

1 n−1 β ∑ ∑ f (s,t), 2 s=0 t=α

(5.2.16)

(q8 ) Let g be as in part (q6 ). If n−1 β

u2 (n, m) c + ∑

∑ f (s,t)u(s,t)g(u(s,t)),

(5.2.17)

s=0 t=α

for (n, m) ∈ E, then for 0 n n2 ; n, n2 ∈ N0 , m ∈ Nα ,β , √ 1 n−1 β −1 W ( c) + ∑ ∑ f (s,t) , u(n, m) W 2 s=0 t=α

(5.2.18)

where W, W −1 are as in part (q6 ) and n2 ∈ N0 is chosen so that √ 1 n−1 β W c + ∑ ∑ f (s,t) ∈ Dom W −1 , 2 s=0 t=α for all n ∈ N0 lying in 0 n n2 and m ∈ Nα ,β . Theorem 5.2.5. (q9 ) Let u, p, q, f ∈ D(E, R+ ). If n−1 s−1 β

u(n, m) p(n, m) + q(n, m) ∑

∑∑

s=0 σ =0 τ =α

for (n, m) ∈ E, then

f (σ , τ )u(σ , τ ),

u(n, m) p(n, m) + q(n, m) n−1

× ∏ 1+ s=0

for (n, m) ∈ E.

s−1 β

∑∑

σ =0 τ =α

n−1 s−1 β

∑∑∑

s=0 σ =0 τ =α

f (σ , τ )p(σ , τ )

f (σ , τ )q(σ , τ ) ,

(5.2.20)

(q10 ) Let u, f , c be as in Theorem 5.2.3 and h ∈ D(E × Nα ,β , R+ ). If n−1

u(n, m) c + ∑ f (s, m)u(s, m) + s=0

for (n, m) ∈ E, then n−1

u(n, m) cF(n, m) ∏ 1 + s=0

n−1

∑ h(s, m, y)u(s, y)

,

(5.2.21)

y=α

β

∑ h(s, m, y)F(s, m)

y=α

∏ [1 + f (σ , m)].

σ =0

β

for (n, m) ∈ E, where F(n, m) =

(5.2.19)

,

(5.2.22)


Proofs of Theorems 5.2.1-5.2.5.

195

We give the proofs of (q1 ), (q3 ), (q6 ), (q7 ), (q10 ) only;

the proofs of other inequalities can be completed by following the proofs of the above noted inequalities and closely looking at the similar results given in [85,87]. To prove (q1 ), (q5 )– (q7 ), (q10 ), it is sufficient to assume that c > 0, since the standard limiting arguments can be used to treat the remaining case, see [85, p. 300]. (q1 ) Let c > 0 and define a function z(n, m) by the right hand side of (5.2.1). Then z(n, 0) = z(0, m) = c, u(n, m) z(n, m), Δ2 z(n, m) =

Δ1 z(n, m) =

n−1 s−1

∑∑

s=0 σ =0

f (σ , m)u(σ , m),

n−1 m−1

∑∑

σ =0 τ =0

Δ21 z(n, m) =

m−1

∑

τ =0

f (σ , τ )u(σ , τ ),

f (n, τ )u(n, τ ),

and Δ2 Δ21 z(n, m) = f (n, m)u(n, m) f (n, m)z(n, m).

(5.2.23)

From (5.2.23) and in view of the facts that z(n, m) z(n, m + 1) and Δ21 z(n, m) 0 we observe that (see [85, p. 324]) Δ21 z(n, m + 1) Δ21 z(n, m) − f (n, m). z(n, m + 1) z(n, m)

(5.2.24)

Keeping n fixed in (5.2.24), set m = τ and sum over τ from 0 to m − 1 and use the fact that Δ21 z(n, 0) = 0, to obtain the estimate Δ21 z(n, m) m−1 ∑ f (n, τ ). z(n, m) τ =0

(5.2.25)

From (5.2.25) and in view of the facts that z(n, m) z(n + 1, m) and Δ1 z(n, m) 0, we observe that Δ1 z(n + 1, m) Δ1 z(n, m) m−1 − ∑ f (n, τ ). z(n + 1, m) z(n, m) τ =0

(5.2.26)

Keeping m fixed in (5.2.26), set n = σ and sum over σ from 0 to n − 1 and use the fact that Δ1 z(0, m) = 0, to obtain the estimate Δ1 z(n, m) n−1 m−1 ∑ ∑ f (σ , τ ). z(n, m) σ =0 τ =0

(5.2.27)

196


From (5.2.27), we observe that

z(n + 1, m) 1 +

n−1 m−1

∑∑

σ =0 τ =0

f (σ , τ ) z(n, m).

(5.2.28)

Now keeping m fixed in (5.2.28), set n = s and substitute s = 0, 1, 2, . . . , n − 1 successively to obtain the estimate n−1

z(n, m) c ∏ 1 +

s−1 m−1

∑∑

f (σ , τ ) .

(5.2.29)

∑ ∑ ∑ L(σ , τ , u(σ , τ )),

(5.2.30)

s=0

σ =0 τ =0

Using (5.2.29) in u(n, m) z(n, m), we get (5.2.2). (q3 ) Define a function z(n, m) by n−1 s−1 m−1

z(n, m) =

s=0 σ =0 τ =0

then z(0, m) = z(n, 0) = 0 and (5.2.6) can be restated as u(n, m) p(n, m) + q(n, m)z(n, m).

(5.2.31)

From (5.2.30), (5.2.31) and (5.2.5), we observe that z(n, m)

n−1 s−1 m−1

∑∑∑

s=0 σ =0 τ =0

L(σ , τ , p(σ , τ ) + q(σ , τ )z(σ , τ ))

−L(σ , τ , p(σ , τ )) + L(σ , τ , p(σ , τ ))

n−1 s−1 m−1

∑ ∑ ∑ L(σ , τ , p(σ , τ ))

s=0 σ =0 τ =0

n−1 s−1 m−1

+∑

∑ ∑ M(σ , τ , p(σ , τ ))q(σ , τ )z(σ , τ ).

(5.2.32)

s=0 σ =0 τ =0

Clearly the first term on the right hand side in (5.2.32) is nonnegative and nondecreasing in n, m ∈ N0 . Now a suitable application of the inequality in part (q2 ) to (5.2.32) yields z(n, m) n−1

×∏ 1+ s=0

n−1 s−1 m−1

∑ ∑ ∑ L(σ , τ , p(σ , τ ))

s=0 σ =0 τ =0

s−1 m−1

∑ ∑ M(σ , τ , p(σ , τ ))q(σ , τ )

.

(5.2.33)

σ =0 τ =0

Using (5.2.33) in (5.2.31), we get the required inequality in (5.2.7). (q6 ) Setting β

e(s) =

∑ f (s,t)g(u(s,t)),

t=α

(5.2.34)


197

the inequality (5.2.12) can be restated as n−1

u(n, m) c + ∑ e(s).

(5.2.35)

s=0

Let c > 0 and define n−1

z(n) = c + ∑ e(s),

(5.2.36)

u(n, m) z(n).

(5.2.37)

s=0

then z(0) = c and

From (5.2.36), (5.2.34), (5.2.37) and using the fact that z(n) is nondecreasing in n ∈ N0 , we observe that Δz(n) = e(n) =

β

β

t=α

t=α

∑ f (n,t)g(u(n,t)) g(z(n)) ∑ f (n,t).

Now, by following the proof of Theorem 2.3.1 given in [85], we get n−1 β

z(n) W −1 W (c) + ∑

∑ f (s,t)

,

(5.2.38)

s=0 t=α

for 0 n n1 . Using (5.2.38) in (5.2.37), we get (5.2.13). (q7 ) Setting β

E(s) =

∑ f (s,t)u(s,t),

(5.2.39)

t=α

the inequality (5.2.15) can be restated as n−1

u2 (n, m) c + ∑ E(s).

(5.2.40)

s=0

Let c > 0 and define by z(n) the right hand side of (5.2.40), then z(0) = c, u(n, m)

z(n)

and we observe that Δz(n) = E(n) =

β

∑ f (n,t)u(n,t)

t=α

β

z(n) ∑ f (n,t). t=α

Now by following the proof of Theorem 3.3.1 given in [85], we obtain

z(n)

√

for n ∈ N0 . Using (5.2.41) in u(n, m)

c+

1 n−1 β ∑ ∑ f (s,t), 2 s=0 t=α

(5.2.41)

z(n), we get the required inequality in (5.2.16).

(q10 ) The proof can be completed by following the arguments as in the proof of Theorem 4.2.1 in Chapter 4 and closely looking at the proof of (q6 ) given above with suitable modifications, see also [85,87].

198

5.3


Sum-difference inequalities in three variables

In this section, we shall deal with some fundamental sum-difference inequalities in three variables which will also be equally important in the situations for which other available inequalities fail to apply directly. In the following theorems we present the inequalities established by Pachpatte in [95,102,111]. Theorem 5.3.1.

Let u, p, q, f ∈ D(H, R+ ) and c 0 is a real constant, where H = N02 ×

Nα ,β . (r1 ) If u(m, n, k) c +

m−1 n−1 β

∑∑∑

f (s,t, r)u(s,t, r),

(5.3.1)

s=0 t=0 r=α

for (m, n, k) ∈ H, then n−1

n−1 β

u(m, n, k) c ∏ 1 + ∑

∑

f (s,t, r) ,

(5.3.2)

t=0 r=α

s=0

for (m, n, k) ∈ H. (r2 ) If u(m, n, k) p(m, n, k) + q(m, n, k) for (m, n, k) ∈ H, then

m−1 n−1 β

∑∑∑


u(m, n, k) p(m, n, k) + q(m, n, k) m−1

s=0

m−1 n−1 β

∑∑∑

∑

f (s,t, r)p(s,t, r)

s=0 t=0 r=α

n−1 β

× ∏ 1+ ∑

(5.3.3)

s=0 t=0 r=α

f (s,t, r)q(s,t, r) ,

(5.3.4)

t=0 r=α

for (m, n, k) ∈ H. Theorem 5.3.2. Let u, f ∈ D(H, R+ ) and c 0, d 1 are real constants. (r3 ) If u2 (m, n, k) c +

m−1 n−1 β

∑∑∑


(5.3.5)

s=0 t=0 r=α

for (m, n, k) ∈ H, then u(m, n, k)

√

c+

1 m−1 n−1 β ∑ ∑ ∑ f (s,t, r), 2 s=0 t=0 r=α

(5.3.6)


199

for (m, n, k) ∈ H. (r4 ) If u(m, n, k) 1 and u(m, n, k) d +

m−1 n−1 β

∑∑∑

f (s,t, r)u(s,t, r) log u(s,t, r),

(5.3.7)

s=0 t=0 r=α

for (m, n, k) ∈ H, then u(m, n, k) d

m−1 n−1 β 1+∑t=0 ∏s=0 ∑r=α f (s,t,r)

,

(5.3.8)

for (m, n, k) ∈ H. Theorem 5.3.3. Let u, p, q, f ∈ D(H, R+ ). (r5 ) If u(m, n, k) p(m, n, k) + q(m, n, k)

m−1

∞

β

∑ ∑ ∑


(5.3.9)

s=0 t=n+1 r=α

for (m, n, k) ∈ H, then

m−1

s=0

∞

β

∞

∑ ∑ ∑

u(m, n, k) p(m, n, k) + q(m, n, k)

× ∏ 1+

m−1

f (s,t, r)p(s,t, r)

s=0 t=n+1 r=α

β

∑ ∑

f (s,t, r)q(s,t, r) ,

(5.3.10)

t=n+1 r=α

for (m, n, k) ∈ H. (r6 ) If u(m, n, k) p(m, n, k) + q(m, n, k)

∞

∞

β

∑ ∑ ∑


(5.3.11)

s=m+1 t=n+1 r=α

for (m, n, k) ∈ H, then

×

∞

∏

s=m+1

for (m, n, k) ∈ H.

1+

∞

∞

∞

β

∑ ∑ ∑

u(m, n, k) p(m, n, k) + q(m, n, k)

f (s,t, r)p(s,t, r)

s=m+1 t=n+1 r=α

β

∑ ∑

t=n+1 r=α

f (s,t, r)q(s,t, r) ,

(5.3.12)

200


Theorem 5.3.4. Let u, p, q, c, f , g ∈ D(H, R+ ). (r7 ) Suppose that u(m, n, k) p(m, n, k) + q(m, n, k) ∞

∞

+c(m, n, k) ∑ ∑

m−1

∞

β

∑ ∑ ∑

f (s,t, r)u(s,t, r)

s=0 t=n+1 r=α

β

∑ g(s,t, r)u(s,t, r),

(5.3.13)

∑ ∑ ∑ g(s,t, r)B1 (s,t, r) < 1,

(5.3.14)

s=0 t=0 r=α

for (m, n, k) ∈ H. If

β1 =

∞

∞

β

s=0 t=0 r=α

then u(m, n, k) A1 (m, n, k) + D1 B1 (m, n, k), for (m, n, k) ∈ H, where

m−1

× ∏ 1+ s=0

∑ ∑

f (s,t, r)q(s,t, r) , m−1

s=0

∞

f (s,t, r)c(s,t, r)

s=0 t=n+1 r=α

β

∞

β

∑ ∑ ∑

B1 (m, n, k) = c(m, n, k) + q(m, n, k) m−1

(5.3.16)

t=n+1 r=α

× ∏ 1+

f (s,t, r)p(s,t, r)

s=0 t=n+1 r=α

β

∞

β

∑ ∑ ∑

A1 (m, n, k) = p(m, n, k) + q(m, n, k) m−1

∞

(5.3.15)

∑ ∑

f (s,t, r)q(s,t, r) ,

(5.3.17)

t=n+1 r=α

and D1 =

∞ ∞ β 1 ∑ ∑ ∑ g(s,t, r)A1 (s,t, r). 1 − β1 s=0 t=0 r=α

(5.3.18)

(r8 ) Suppose that u(m, n, k) p(m, n, k) + q(m, n, k) ∞

∞

+c(m, n, k) ∑ ∑

∞

∞

β

∑ ∑ ∑

f (s,t, r)u(s,t, r)

s=m+1 t=n+1 r=α

β

∑ g(s,t, r)u(s,t, r),

(5.3.19)

∑ ∑ ∑ g(s,t, r)B2 (s,t, r) < 1,

(5.3.20)

s=0 t=0 r=α

for (m, n, k) ∈ H. If

β2 =

∞

∞

β

s=0 t=0 r=α


201

then u(m, n, k) A2 (m, n, k) + D2 B2 (m, n, k), for (m, n, k) ∈ H, where

×

∏

∑ ∑

f (s,t, r)q(s,t, r) ,

∏

∞

(5.3.22)

β

f (s,t, r)c(s,t, r)

s=m+1 t=n+1 r=α

β

∞

∑ ∑

1+

∞

∑ ∑ ∑

B2 (m, n, k) = c(m, n, k) + q(m, n, k)

×

f (s,t, r)p(s,t, r)

s=m+1 t=n+1 r=α

t=n+1 r=α

s=m+1

∞

β

β

∞

1+

∞

∑ ∑ ∑

A2 (m, n, k) = p(m, n, k) + q(m, n, k) ∞

∞

(5.3.21)

f (s,t, r)q(s,t, r) ,

(5.3.23)

∞ ∞ β 1 ∑ ∑ ∑ g(s,t, r)A2 (s,t, r). 1 − β2 s=0 t=0 r=α

(5.3.24)

t=n+1 r=α

s=m+1

and D2 = Theorem 5.3.5.

Let u, p ∈ D(H, R+ ), q ∈ D(H × Nα ,β , R+ ), and c 0 is a real constant.

If u(m, n, z) c +

m−1 n−1

∑∑

β

p(s,t, z)u(s,t, z) +

for (m, n, z) ∈ H, then m−1

,

(5.3.25)

∑ q(s,t, z, r)L(s,t, r)

,

(5.3.26)

t=0 r=α

s=0

for (m, n, z) ∈ H, where m−1

L(m, n, z) =

n−1 β

u(m, n, z) cL(m, n, z) ∏ 1 + ∑

∏

σ =0

Proofs of Theorems 5.3.1–5.3.5.

∑ q(s,t, z, r)u(s,t, r)

r=α

s=0 t=0

n−1

1 + ∑ p(σ , τ , z) . τ =0

We give the details of the proofs for (r1 ), (r4 ) and (r5 )

only; the proofs of other inequalities can be completed by following the proofs of these inequalities and the ideas employed in the proofs of the results in Chapter 2, section 2.3 with suitable modifications, see also [85,87]. (r1 ) Introducing the notation β

e(s,t) =

∑

r=α


(5.3.27)

202


in (5.3.1), we get m−1 n−1

∑ ∑ e(s,t),

u(m, n, k) c +

(5.3.28)

s=0 t=0

for (m, n, k) ∈ H. Let c > 0 and define m−1 n−1

v(m, n) = c +

∑ ∑ e(s,t),

(5.3.29)

s=0 t=0

then v(m, 0) = v(0, n) = c and u(m, n, k) v(m, n).

(5.3.30)

From (5.3.29), (5.3.27) and (5.3.30), we observe that (see [85, p. 299]) Δ2 Δ1 v(m, n) = e(m, n) β

∑

=

f (m, n, r)u(m, n, r)

r=α

v(m, n)

β

∑

f (m, n, r).

(5.3.31)

r=α

The rest of the proof can be completed by following similar arguments as in the proof of Theorem 4.2.1 given in [85] with suitable modifications. (r4 ) Introducing the notation β

e(s,t) =

∑

f (s,t, r)u(s,t, r) log u(s,t, r),

(5.3.32)

r=α

in (5.3.7), we get u(m, n, k) d +

m−1 n−1

∑ ∑ e(s,t).

(5.3.33)

s=0 t=0

Define m−1 n−1

w(m, n) = d +

∑ ∑ e(s,t),

(5.3.34)

s=0 t=0

then w(m, 0) = w(0, n) = d and u(m, n, k) w(m, n). From (5.3.34), (5.3.32) and (5.3.35), we observe that (see [85, p. 437]) Δ2 Δ1 w(m, n) = e(m, n) β

=

∑

r=α

f (m, n, r)u(m, n, r) log u(m, n, r)

(5.3.35)


w(m, n)

203

β

∑

f (m, n, r) log w(m, n).

(5.3.36)

r=α

The remaining proof can be completed by following the proof of Theorem 5.5.1 given in [85]. (r5 ) Introducing the notation

β

∑

e0 (s,t) =


(5.3.37)

r=α

the inequality (5.3.9) can be restated as

u(m, n, k) p(m, n, k) + q(m, n, k)

∞

m−1

∑ ∑

e0 (s,t).

(5.3.38)

s=0 t=n+1

Define

ψ (m, n) =

∞

m−1

∑ ∑

e0 (s,t),

(5.3.39)

s=0 t=n+1

then ψ (0, n) = 0 and

u(m, n, k) p(m, n, k) + q(m, n, k)ψ (m, n).

(5.3.40)

From (5.3.39), (5.3.37) and (5.3.40), we observe that ∞

Δ1 ψ (m, n) = ∑ e0 (m,t) t=n+1

β

∞

∑

=

∞

∑

t=n+1 ∞

=

∑

β

∑

r=α

f (m,t, r)[p(m,t, r) + q(m,t, r)ψ (m,t)]

β

∑ ∑

f (m,t, r)u(m,t, r)

r=α

t=n+1

∞

f (m,t, r)p(m,t, r) +

t=n+1 r=α

∑

t=n+1

β

∑

r=α

f (m,t, r)q(m,t, r)ψ (m,t) .

(5.3.41)

By taking m = s in (5.3.41) and then taking sum over s from 0 to m − 1, m ∈ N0 , we get

ψ (m, n)

m−1

∞

β

∑ ∑ ∑

s=0 t=n+1 r=α m−1

= E(m, n) +

∑ ∑

s=0 t=n+1 β

∞

∑ ∑

∑

ψ (s,t)

m−1

E(m, n) =

∞

∑

r=α

f (s,t, r)q(s,t, r)ψ (s,t)

f (s,t, r)q(s,t, r) ,

(5.3.42)

r=α

s=0 t=n+1

where

β

∞

m−1

f (s,t, r)p(s,t, r) +

β

∑ ∑ ∑

f (s,t, r)p(s,t, r).

(5.3.43)

s=0 t=n+1 r=α

Clearly, E(m, n) is nonnegative, nondecreasing in m and nonincreasing in n for m, n ∈ N0 . Now a suitable application of the inequality given in Theorem 5.4.1 part (a1 ) in [87, p. 266] to (5.3.42) yields m−1

ψ (m, n) E(m, n) ∏ 1 + s=0

∞

β

∑ ∑

t=n+1 r=α

Using (5.3.44), (5.3.43) in (5.3.40), we get (5.3.10).

f (s,t, r)q(s,t, r) .

(5.3.44)

204

5.4


Multivariable sum-difference inequalities

Let Ni [αi , βi ] = {αi , αi + 1, . . . , βi } (αi < βi ), αi , βi ∈ N0 for i = 1, . . . , m and G = m ∏m i=1 Ni [αi , βi ] ⊂ R . Let E = N0 × G, Ω = {(n, x, s · y) : 0 s n < ∞, x, y ∈ G} for

s, n ∈ N0 and for the function r defined on Ω, we define Δ1 r(n, x, s, y) = r(n + 1, x, s, y) − r(n, x, s, y). For any function w defined on G we denote the m-fold sum over G with respect β

β

to the variable y = (y1 , . . . , ym ) ∈ G by ∑G w(y) = ∑y11=α1 · · · ∑ymm=αm w(y1 , . . . , ym ). Clearly, ∑G w(y) = ∑G w(x) for x, y ∈ G. The main objective of this section is to provide some basic finite difference inequalities which can be used as tools in the development of the theory of certain new classes of sum-difference equations. The following theorems deals with the inequalities recently established by Pachpatte [116,114,106]. Theorem 5.4.1. Let u ∈ D(E, R+ ), h, Δ1 h ∈ D (Ω, R+ ) and c 0 is a real constant. (s1 ) If n−1

u(n, x) c + ∑ ∑ h(n, x, s, y)u(s, y),

(5.4.1)

n−1 u(n, x) c ∏ 1 + A(σ , x) ,

(5.4.2)

s=0 G

for (n, x) ∈ E, then

σ =0

for (n, x) ∈ E, where

n−1

A(n, x) = ∑ h(n + 1, x, n, y) + ∑ ∑ Δ1 h(n, x, s, y), G

for (n, x) ∈ E.

(5.4.3)

s=0 G

(s2 ) Let g ∈ C(R+ , R+ ) be nondecreasing function, g(u) > 0 on (0, ∞). If n−1

u(n, x) c + ∑ ∑ h(n, x, s, y)g(u(s, y)),

(5.4.4)

s=0 G

for (n, x) ∈ E, then for 0 n n1 ; n, n1 ∈ N0 , x ∈ G, u(n, x) W

−1

where

W (c) +

∑ A(σ , x)

,

(5.4.5)

σ =0

r ds

, r > 0, (5.4.6) g(s) is the inverse of W , A(n, x) is given by (5.4.3) and n1 ∈ N0 be W (r) =

r0 > 0 is arbitrary and W −1

n−1

r0

chosen so that n−1

W (c) +

∑ A(σ , x) ∈ Dom

σ =0

for all n ∈ N0 lying in 0 n n1 and x ∈ G.

−1 , W


205

Theorem 5.4.2. Let u, h, Δ1 h and c be as in Theorem 5.4.1. (s3 ) If n−1

u2 (n, x) c + ∑ ∑ h(n, x, s, y)u(s, y),

(5.4.7)

s=0 G

for (n, x) ∈ E, then u(n, x)

√

c+

1 n−1 A(σ , x), 2 σ∑ =0

(5.4.8)

for (n, x) ∈ E, where A(n, x) is given by (5.4.3). (s4 ) Let g be as in Theorem 5.4.1 part (s2 ). If n−1

u2 (n, x) c + ∑ ∑ h(n, x, s, y)u(s, y)g(u(s, y)),

(5.4.9)

s=0 G

for (n, x) ∈ E, then for 0 n n2 ; n, n2 ∈ N0 , x ∈ G, √ 1 n−1 −1 W c + ∑ A(σ , x) , u(n, x) W 2 σ =0

(5.4.10)

where W, W −1 , A(n, x) are as in part (s2 ) and n2 ∈ N0 be chosen so that √ 1 n−1 W c + ∑ A(σ , x) ∈ Dom W −1 , 2 σ =0


Theorem 5.4.3. Let u ∈ D(E, R1 ), h, Δ1 h ∈ D(Ω, R+ ) and c 1 be a real constant. (s5 ) If n−1

u(n, x) c + ∑ ∑ h(n, x, s, y)u(s, y) log u(s, y),

(5.4.11)

s=0 G

for (n, x) ∈ E, then n−1

u(n, x) c∏σ =0 [1+A(σ ,x)] ,

(5.4.12)

for (n, x) ∈ E, where A(n, x) is given by (5.4.3). (s6 ) Let g be as in Theorem 5.4.1 part (s2 ). If n−1

u(n, x) c + ∑ ∑ h(n, x, s, y)u(s, y)g(log u(s, y)),

(5.4.13)

s=0 G

for (n, x) ∈ E, then for 0 n n3 ; n, n3 ∈ N0 , x ∈ G, u(n, x) exp W

−1

W (log c) +

n−1

∑ A(σ , x)

,

σ =0

where W, W −1 , A(n, x) are as in part (s2 ) and n3 ∈ N0 be chosen so that n−1 W (log c) + ∑ A(σ , x) ∈ Dom W −1 , σ =0


(5.4.14)

206


Theorem 5.4.4. (s7 ) Let u, p, q, f , g ∈ D(E, R+ ) and n−1

s−1

s=0 G

τ =0 G

u(n, x) p(n, x) + q(n, x) ∑ ∑ f (s, y) u(s, y) + q(s, y) ∑ ∑ g(τ , z)u(τ , z) ,

(5.4.15)

for (n, x) ∈ E. Then n−1

u(n, x) p(n, x) + q(n, x) ∑ ∑ p(s, y)[ f (s, y) + g(s, y)] s=0 G

×

n−1

∏

σ =s+1

1 + ∑ q(σ , z)[ f (σ , z) + g(σ , z)] ,

(5.4.16)

G

for (n, x) ∈ E. (s8 ) Let u, p, q, f , g ∈ D(E, R+ ) where E = N02 × G and suppose that n−1 m−1

u(n, m, x) p(n, m, x) + q(n, m, x) ∑

∑ ∑ f (s,t, y)

s=0 t=0 G

× u(s,t, y) + q(s,t, y)

s−1 t−1

∑ ∑ ∑ g(σ , τ , z)u(σ , τ , z)

,

(5.4.17)

σ =0 τ =0 G

for (n, m, x) ∈ E. Then u(n, m, x) p(n, m, x) + q(n, m, x) ×

n−1 m−1

∑ ∑ ∑ p(s,t, y)[ f (s,t, y) + g(s,t, y)]

s=0 t=0 G n−1

× ∏ 1+ s=0

m−1

∑ ∑ q(s,t, y)[ f (s,t, y) + g(s,t, y)]

,

(5.4.18)

t=0 G

for (n, m, x) ∈ E. Theorem 5.4.5. Let u, a, b, c, p, q ∈ D(E, R+ ). (s9 ) Suppose that ∞

n−1

u(n, x) a(n, x) + b(n, x) ∑ ∑ p(s, y)u(s, y) + c(n, x) ∑ ∑ q(s, y)u(s, y), s=0 G

(5.4.19)

s=0 G

for (n, x) ∈ E. If ∞

∑ ∑ q(s, y)B(s, y) < 1,

(5.4.20)

u(n, x) A(n, x) + DB(n, x),

(5.4.21)

d=

s=0 G

then


207

for (n, x) ∈ E, where n−1

A(n, x) = a(n, x) + b(n, x) ∑ ∑ p(s, y)a(s, y) s=0 G

n−1

∏

s=0 G

n−1

1 + ∑ p(σ , y)b(σ , y) ,

σ =s+1

n−1

B(n, x) = c(n, x) + b(n, x) ∑ ∑ p(s, y)c(s, y)

(5.4.22)

G

∏

1 + ∑ p(σ , y)b(σ , y) ,

σ =s+1

(5.4.23)

G

for (n, x) ∈ E and D=

1 ∞ ∑ ∑ q(s, y)A(s, y). 1 − d s=0 G

(5.4.24)

(s10 ) If n−1

u(n, x) a(n, x) + b(n, x) ∑ ∑ p(s, y)u(s, y),

(5.4.25)

s=0 G


u(n, x) a(n, x) + b(n, x) ∑ ∑ p(s, y)a(s, y) s=0 G

n−1

∏

σ =s+1

1 + ∑ p(σ , y)b(σ , y) ,

(5.4.26)

G

for (n, x) ∈ E. Proofs of Theorems 5.4.1–5.4.5.

Below, we give the proofs of the inequalities in (s1 ),

(s2 ), (s7 ), (s9 ) only; the proofs of other inequalities can be completed by following the proofs of these inequalities and closely looking at the proofs of the inequalities in Chapter 3, see also [85,87]. To prove (s1 )–(s4 ), it is sufficient to assume that c > 0, since the standard limiting argument can be used to treat the remaining case, see [85, p. 300]. (s1 ) Setting e(n, s) = ∑ h(n, x, s, y)u(s, y),

(5.4.27)

G

for every x ∈ G, the inequality (5.4.1) can be restated as n−1

u(n, x) c + ∑ e(n, s).

(5.4.28)

s=0

Let c > 0 and define n−1

z(n) = c + ∑ e(n, s),

(5.4.29)

u(n, x) z(n).

(5.4.30)

s=0

then z(0) = c and

208


From (5.4.29), (5.4.27), (5.4.30) and the fact that z(n) is nondecreasing in n ∈ N0 , we observe that n−1

Δz(n) = e(n + 1, n) + ∑ Δ1 e(n, s) s=0

n−1

= ∑ h(n + 1, x, n, y)u(n, y) + ∑ Δ1

∑ h(n, x, s, y)u(s, y)

s=0

G

G

n−1

∑ h(n + 1, x, n, y)z(n) + ∑ ∑ Δ1 h(n, x, s, y)z(s) G

s=0 G

n−1

∑ h(n + 1, x, n, y) + ∑ ∑ Δ1 h(n, x, s, y) z(n) G

s=0 G

= A(n, x)z(n).

(5.4.31)

Now a suitable application of Theorem 1.2.1 given in [85, p. 11] to (5.4.31) yields n−1 z(n) c ∏ 1 + A(σ , x) .

(5.4.32)

σ =0

Using (5.4.32) in (5.4.30), we get the required inequality in (5.4.2). (s2 ) Setting e(n, s) = ∑ h(n, x, s, y)g(u(s, y)),

(5.4.33)

G

for every x ∈ G, the inequality (5.4.4) can be restated as n−1

u(n, x) c + ∑ e(n, s).

(5.4.34)

s=0

Let c > 0 and define by z(n) the right hand side of (5.4.34). Following the proof of (s1 ) given above, we get Δz(n) A(n, x)g(z(n)).

(5.4.35)

Now by following the proof of Theorem 2.3.1 given in [85, p. 104] from (5.4.35), we get z(n) W −1 W (c) +

n−1

∑ A(σ , x)

,

(5.4.36)

σ =0

on 0 n n1 . Using (5.4.36) in u(n, x) z(n), we get the desired inequality in (5.4.5). (s7 ) Introducing the notation

s−1

e(s) = ∑ f (s, y) u(s, y) + q(s, y) ∑ ∑ g(τ , z)u(τ , z) , G

τ =0 G

(5.4.37)


209

in (5.4.15), we get n−1

u(n, x) p(n, x) + q(n, x) ∑ e(s),

(5.4.38)

s=0

for (n, x) ∈ E. Define n−1

r(n) =

∑ e(s),

(5.4.39)

s=0

for n ∈ N0 , then r(0) = 0 and from (5.4.38), we get u(n, x) p(n, x) + q(n, x)r(n)

(5.4.40)

for (n, x) ∈ E. From (5.4.39), (5.4.37) and (5.4.40), we observe that Δr(n) = e(n)

n−1

= ∑ f (n, y) u(n, y) + q(n, y) ∑ ∑ g(τ , z)u(τ , z) τ =0 G

G

∑ f (n, y) p(n, y) + q(n, y)r(n) G

n−1

+q(n, y) ∑ ∑ g(τ , z)[p(τ , z) + q(τ , z)r(τ )] τ =0 G

n−1

= ∑ f (n, y) p(n, y) + q(n, y) + r(n) + ∑ ∑ g(τ , z)[p(τ , z) + q(τ , z)r(τ )]

. (5.4.41)

τ =0 G

G

Define n−1

v(n) = r(n) + ∑ ∑ g(τ , z)[p(τ , z) + q(τ , z)r(τ )],

(5.4.42)

τ =0 G

for n ∈ N0 , then v(0) = r(0) = 0, r(n) v(n) and from (5.4.41), we get Δr(n) ∑ f (n, y)[p(n, y) + q(n, y)v(n)].

(5.4.43)

G

From (5.4.42), (5.4.43) and the fact that r(n) v(n), n ∈ N0 , we observe that Δv(n) = Δr(n) + ∑ g(n, z)[p(n, z) + q(n, z)r(n)] G

v(n) ∑ q(n, y)[ f (n, y) + g(n, y)] + ∑ p(n, y)[ f (n, y) + g(n, y)]. G

G

(5.4.44)

210


Now a suitable application of Theorem 1.2.1 given in [85, p. 11] with v(0) = 0 to (5.4.44) yields v(n)

n−1

∑ ∑ p(s, y)[ f (s, y) + g(s, y)]

s=0 G

×

n−1

1 + ∑ q(σ , z)[ f (σ , z) + g(σ , z)] .

∏

σ =s+1

(5.4.45)

G

Using the fact that r(n) v(n) and (5.4.40), (5.4.45) we get the required inequality in (5.4.16). (s9 ) Let n−1

z(n) =

λ=

∑ ∑ p(s, y)u(s, y),

(5.4.46)

∑ ∑ q(s, y)u(s, y).

(5.4.47)

s=0 G ∞ s=0 G

Then (5.4.19) can be restated as u(n, x) a(n, x) + b(n, x)z(n) + c(n, x)λ .

(5.4.48)

e0 (s) = ∑ p(s, y)u(s, y),

(5.4.49)

Introducing the notation G

in (5.4.46), we get n−1

z(n) =

∑ e0 (s).

(5.4.50)

s=0

From (5.4.50), (5.4.49) and (5.4.48), we observe that Δz(n) = e0 (n) = ∑ p(n, y)u(n, y) G

∑ p(n, y)[a(n, y) + b(n, y)z(n) + c(n, y)λ ] G

= z(n) ∑ p(n, y)b(n, y) + ∑ p(n, y)[a(n, y) + c(n, y)λ ]. G

(5.4.51)

G

Now, applying the inequality in Theorem 1.2.1 given in [85, p. 11] with z(0) = 0 to (5.4.51) yields z(n)

n−1

n−1

∑ ∑ p(s, y)[a(s, y) + c(s, y)λ ] ∏

s=0 G

σ =s+1

1 + ∑ p(σ , y)b(σ , y) G


n−1

=

n−1

∑ ∑ p(s, y)a(s, y) ∏

σ =s+1

s=0 G

+λ

n−1

∑ ∑ p(s, y)c(s, y)

s=0 G

n−1

211

1 + ∑ p(σ , y)b(σ , y) G

1 + ∑ p(σ , y)b(σ , y) .

∏

σ =s+1

From (5.4.48) and (5.4.52), we have

n−1

∑ ∑ p(s, y)a(s, y)

u(n, x) a(n, x) + b(n, x)

s=0 G

+λ

n−1

n−1

∏

σ =s+1

1 + ∑ p(σ , y)b(σ , y) G

1 + ∑ p(σ , y)b(σ , y)

σ =s+1

s=0 G

n−1

∑ ∑ p(s, y)c(s, y) ∏

(5.4.52)

G

+ c(n, x)λ

G

= A(n, x) + λ B(n, x).

(5.4.53)

From (5.4.47) and (5.4.53), we observe that

λ

∞

∑ ∑ q(s, y)[A(s, y) + λ B(s, y)],

s=0 G

which implies

λ D.

(5.4.54)

Using (5.4.54) in (5.4.53), we get (5.4.21). 5.5

Sum-difference equations in two variables

In this section, first we shall deal with the following initial value problem (see [100]) Δ2 Δ1 u(m, n) = f (m, n, u(m, n), Gu(m, n)),

(5.5.1)

with u(m, 0) = α (m),

u(0, n) = β (n),

α (0) = β (0),

(5.5.2)

for m, n ∈ N0 , where m−1 n−1

Gu(m, n) :=

∑ ∑ g(m, n, σ , τ , u(σ , τ )),

(5.5.3)

σ =0 τ =0

f , g are given functions and u is the unknown function. The equations of the form (5.5.1) arise naturally in the approximation of solutions of partial integrodifferential equations by finite difference methods and also appear in their own right. We assume that f ∈ D(N20 × R2 , R), g ∈ D(N40 × R, R), α , β ∈ D(N0 , R). Here, it is to be noted that the

212


problem (5.5.1)–(5.5.2) under some suitable conditions admits a unique solution. Below, we offer the conditions for the error evaluation of approximate solutions of equation (5.5.1) and convergence properties of solutions of approximate problems and also dependency of solutions of equations of the form (5.5.1) on parameters, by employing a certain finite difference inequality with explicit estimate given in [87]. Let u ∈ D(N20 , R) and Δ2 Δ1 u(m, n) for m, n ∈ N0 exist and satisfy the inequality |Δ2 Δ1 u(m, n) − f (m, n, u(m, n), Gu(m, n))| ε , for a given constant ε 0, where it is assumed that (5.5.2) holds. Then we call u(m, n) an

ε -approximate solution of (5.5.1). The following finite difference inequality given in [87] (see also [85, Theorem 5.3.2]) is crucial in the study of problem (5.5.1)–(5.5.2). Lemma 5.5.1.

Let u, a, p ∈ D(N20 , R+ ); q, Δ1 q, Δ2 q, Δ2 Δ1 q ∈ D(N40 , R+ ). If a(m, n) is

nondecreasing in each variable m, n ∈ N0 and u(m, n) a(m, n) +

m−1 n−1

∑ ∑ p(s,t)

s−1 t−1

u(s,t) +

u(m, n) a(m, n) 1 +

∑ ∑ q(s,t · σ , τ )u(σ , τ )

,

(5.5.4)

σ =0 τ =0

s=0 t=0

for m, n ∈ N0 , then

m−1 n−1

s−1

s=0 t=0

ξ =0

∑ ∑ p(s,t) ∏

1+

t−1

∑ [p(ξ , η ) + T q(ξ , η )]

,

(5.5.5)

η =0

for m, n ∈ N0 , where m−1

T q(m, n) := q(m + 1, n + 1, m, n) +

∑ Δ1 q(m, n + 1, σ , n)

σ =0 n−1

m−1 n−1

τ =0

σ =0 τ =0

+ ∑ Δ2 q(m + 1, n, m, τ ) +

∑ ∑ Δ2Δ1 q(m, n, σ , τ ).

(5.5.6)

The following theorem estimates the difference between the two approximate solutions of (5.5.1). Theorem 5.5.1. Suppose that f , g in (5.5.1) satisfy the conditions | f (m, n, u, v) − f (m, n, u, v)| p(m, n) [|u − u| + |v − v|] ,

(5.5.7)

|g(m, n, σ , τ , u) − g(m, n, σ , τ , u)| q(m, n, σ , τ )|u − u|,

(5.5.8)


213

where p ∈ D(N20 , R+ ), q ∈ D(N40 , R+ ) with Δ1 q, Δ2 q, Δ2 Δ1 q ∈ D(N40 , R+ ). For i = 1, 2, let ui (m, n) (m, n ∈ N0 ) be respectively εi -approximate solutions of (5.5.1) with ui (m, 0) = αi (m),

ui (0, n) = βi (n),

αi (0) = βi (0),

(5.5.9)

for m, n ∈ N0 , where αi , βi ∈ D(N0 , R) satisfy |α1 (m) − α2 (m) + β1 (n) − β2 (n)| δ ,

(5.5.10)

in which δ 0 is a constant. Then |u1 (m, n) − u2 (m, n)| ((ε1 + ε2 )mn + δ )

× 1+

m−1 n−1

s−1

∑ ∑ p(s,t) ∏

t−1

1+

ξ =0

s=0 t=0

∑ [p(ξ , η ) + T q(ξ , η )]

,

(5.5.11)

η =0

for m, n ∈ N0 , where T q(m, n) is given by (5.5.6). Proof.

Since ui (m, n) (i = 1, 2) for m, n ∈ N0 , are respectively εi -approximate solutions

of equation (5.5.1) with (5.5.9), we have |Δ2 Δ1 ui (m, n) − f (m, n, ui (m, n), Gui (m, n))| εi . Now keeping m fixed in the above inequality, setting n = t and taking sum on both sides over t from 0 to n − 1, then keeping n fixed in the resulting inequality and setting m = s and taking sum over s from 0 to m − 1 and using (5.5.9), we observe that

εi mn

m−1 n−1

∑ ∑ |Δ2 Δ1 ui (s,t) − f (s,t, ui (s,t), Gui (s,t))|

s=0 t=0

m−1 n−1 ∑ ∑ {Δ2 Δ1 ui (s,t) − f (s,t, ui (s,t), Gui (s,t))} s=0 t=0

m−1 n−1 = ui (m, n) − [αi (m) + βi (n)] − ∑ ∑ f (s,t, ui (s,t), Gui (s,t)) . s=0 t=0

(5.5.12)

From this inequality and using the elementary inequalities in (1.3.25), we observe that

m−1 n−1 (ε1 + ε2 )mn u1 (m, n) − [α1 (m) + β1 (n)] − ∑ ∑ f (s,t, u1 (s,t), Gu1 (s,t)) s=0 t=0

m−1 n−1 + u2 (m, n) − [α2 (m) + β2 (n)] − ∑ ∑ f (s,t, u2 (s,t), Gu2 (s,t)) s=0 t=0

m−1 n−1 u1 (m, n) − [α1 (m) + β1 (n)] − ∑ ∑ f (s,t, u1 (s,t), Gu1 (s,t)) s=0 t=0

214


− u2 (m, n) − [α2 (m) + β2 (n)] −

m−1 n−1

∑∑

s=0 t=0

f (s,t, u2 (s,t), Gu2 (s,t))

|u1 (m, n) − u2 (m, n)| − |α1 (m) + β1 (n) − {α2 (m) + β2 (n)}| m−1 n−1 m−1 n−1 − ∑ ∑ f (s,t, u1 (s,t), Gu1 (s,t)) − ∑ ∑ f (s,t, u2 (s,t), Gu2 (s,t)) . s=0 t=0 s=0 t=0

(5.5.13)

Let u(m, n) = |u1 (m, n) − u2 (m, n)| for m, n ∈ N0 . From (5.5.13) and using the hypotheses, we observe that u(m, n) (ε1 + ε2 )mn + δ +

m−1 n−1

∑ ∑ | f (s,t, u1 (s,t), Gu1 (s,t)) − f (s,t, u2 (s,t), Gu2 (s,t))|

s=0 t=0

(ε1 + ε2 )mn + δ +

m−1 n−1

∑ ∑ p(s,t)

s=0 t=0

u(s,t) +

s−1 t−1

∑ ∑ q(s,t, σ , τ )u(σ , τ )

.

(5.5.14)

σ =0 τ =0


In case u1 (m, n) is a solution of (5.5.1)–(5.5.2), then we have ε1 = 0 and

from (5.5.11), we see that u2 (m, n) → u1 (m, n) as ε2 → 0 and δ → 0. Furthermore, if we put (i) ε1 = ε2 = 0, α1 (m) = α2 (m), β1 (n) = β2 (n) in (5.5.11), then the uniqueness of solutions of (5.5.1)–(5.5.2) is established, and (ii) ε1 = ε2 = 0 in (5.5.11), then we get the bound which shows the dependency of solutions of (5.5.1) on given initial values. Consider the problem (5.5.1)–(5.5.2) together with Δ2 Δ1 v(m, n) = f (m, n, v(m, n), Gv(m, n)),

(5.5.15)

v(m, 0) = α (m), v(0, n) = β (n), α (0) = β (0),

(5.5.16)

for m, n ∈ N0 , where G is given by (5.5.3) and f ∈ D(N20 × R2 , R), α , β ∈ D(N0 , R). The next theorem concerns the closeness of solutions of (5.5.1)–(5.5.2) and of (5.5.15)– (5.5.16).


Theorem 5.5.2.

215

Suppose that f , g in (5.5.1) satisfy (5.5.7), (5.5.8) and there exist con-

stants ε 0, δ 0 such that | f (m, n, u, w) − f (m, n, u, w)| ε ,

(5.5.17)

|α (m) − α (m) + β (n) − β (n)| δ ,

(5.5.18)

where f , α , β and f , α , β are as in (5.5.1)–(5.5.2) and (5.5.15)–(5.5.16). Let u(m, n) and v(m, n) be respectively the solutions of (5.5.1)–(5.5.2) and of (5.5.15)–(5.5.16) for m, n ∈ N0 . Then (ε mn + δ ) 1 +

m−1 n−1

|u(m, n) − v(m, n)|

∑ ∑ p(s,t) ∏

1+

ξ =0

s=0 t=0

t−1

s−1

∑ [p(ξ , η ) + T q(ξ , η )]

,

(5.5.19)

η =0

for m, n ∈ N0 , where T q(m, n) is given by (5.5.6). Proof.

Let e(m, n) = |u(m, n) − v(m, n)| for m, n ∈ N0 . Using the facts that u(m, n),

v(m, n) are the solutions of (5.5.1)–(5.5.2), (5.5.15)–(5.5.16) and the hypotheses, we observe that e(m, n) |α (m) − α (m) + β (n) − β (n)| m−1 n−1

+

∑ ∑ | f (s,t, u(s,t), Gu(s,t)) − f (s,t, v(s,t), Gv(s,t))|

s=0 t=0

m−1 n−1

+

∑ ∑ | f (s,t, v(s,t), Gv(s,t)) − f (s,t, v(s,t), Gv(s,t))|

s=0 t=0

(ε mn + δ ) +

m−1 n−1

∑ ∑ p(s,t)

s=0 t=0

e(s,t) +

s−1 t−1

∑ ∑ q(s,t, σ , τ )e(σ , τ )

.

(5.5.20)

σ =0 τ =0


The result given in Theorem 5.5.2 relates the solutions of (5.5.1)–(5.5.2)

and of (5.5.15)–(5.5.16) in the sence that if f is close to f , α is close to α , β is close to β , then the solutions of (5.5.1)–(5.5.2) and of (5.5.15)–(5.5.16) are also close to each other. Now we consider (5.5.1)–(5.5.2) and sequence of initial value problems Δ2 Δ1 w(m, n) = f k (m, n, w(m, n), Gw(m, n)), w(m, 0) = αk (m),

w(0, n) = βk (n),

αk (0) = βk (0),

for m, n ∈ N0 , k = 1, 2, . . ., where G is given by (5.5.3) and f k ∈

D(N20 × R2 , R),

D(N0 , R). As an immediate consequence of Theorem 5.5.2 we have the following corollary.

(5.5.21) (5.5.22)

αk , β k ∈

216


Corollary 5.5.1.

Suppose that f , g in (5.5.1) satisfy (5.5.7), (5.5.8) and there exist con-

stants εk 0, δk 0 (k = 1, 2, . . .) such that | f (m, n, u, v) − f k (m, n, u, v)| εk ,

(5.5.23)

|α (m) − αk (m) + β (n) − βk (n)| δk ,

(5.5.24)

with εk → 0 and δk → 0 as k → ∞, where f , α , β and fk , αk , βk are respectively as in (5.5.1)–(5.5.2) and in (5.5.21)–(5.5.22). If wk (m, n) (k = 1, 2, . . .) and u(m, n) are respectively the solutions of (5.5.21)–(5.5.22) and (5.5.1)–(5.5.2) for m, n ∈ N0 , then wk (m, n) → u(m, n) as k → ∞. Proof.

For k = 1, 2, . . ., the conditions of Theorem 5.5.2 hold. An application of Theo-

rem 5.5.2 yields (εk mn + δk ) 1 +

m−1 n−1

|wk (m, n) − u(m, n)|

∑ ∑ p(s,t) ∏

s=0 t=0

t−1

s−1

1+

ξ =0

∑ [p(ξ , η ) + T q(ξ , η )]

,

(5.5.25)

η =0

for m, n ∈ N0 , where T q(m, n) is given by (5.5.6) and k = 1, 2, . . .. The required result follows from (5.5.25). Remark 5.5.3.

We note that the result obtained in Corollary 5.5.1 provides sufficient

conditions that ensures, solutions of (5.5.21)–(5.5.22) will converge to the solutions to (5.5.1)–(5.5.2). We now consider the sum-difference equations Δ2 Δ1 u(m, n) = f (m, n, u(m, n), Gu(m, n), μ ),

(5.5.26)

Δ2 Δ1 u(m, n) = f (m, n, u(m, n), Gu(m, n), μ0 ),

(5.5.27)

with the initial conditions (5.5.2), where Gu(m, n) is given by (5.5.3), f ∈ D(N20 × R2 × R, R) and μ , μ0 are parameters. The next theorem shows the dependency of solutions of (5.5.26)–(5.5.2) and (5.5.27)– (5.5.2) on the parameters μ , μ0 . Theorem 5.5.3. Suppose that g and f in (5.5.26), (5.5.27) satisfy respectively (5.5.8) and | f (m, n, u, v, μ ) − f (m, n, u, v, μ )| p(m, n) [|u − u| + |v − v|] ,

(5.5.28)

| f (m, n, u, v, μ ) − f (m, n, u, v, μ0 )| r(m, n)|μ − μ0 |,

(5.5.29)


217

where p, r ∈ D(N20 , R+ ). Let u1 (m, n) and u2 (m, n) be the solutions of (5.5.26)–(5.5.2) and of (5.5.27)–(5.5.2) respectively. Then a(m, n) 1 +

|u1 (m, n) − u2 (m, n)|

m−1 n−1

s−1

∑ ∑ p(s,t) ∏

t−1

1+

∑ [p(ξ , η ) + T q(ξ , η )]

,

(5.5.30)

η =0

ξ =0

s=0 t=0

for m, n ∈ N0 , where a(m, n) = |μ − μ0 |

m−1 n−1

∑ ∑ r(s,t),

(5.5.31)

s=0 t=0

for m, n ∈ N0 . Proof.

Let e(m, n) = |u1 (m, n) − u2 (m, n)| for m, n ∈ N0 . Using the facts that u1 (m, n)

and u2 (m, n) are respectively the solutions of (5.5.26)–(5.5.2) and of (5.5.27)–(5.5.2) and the hypotheses, we observe that e(m, n)

m−1 n−1

∑ ∑ | f (s,t, u1 (s,t), Gu1 (s,t), μ ) − f (s,t, u2(s,t), Gu2(s,t), μ )|

s=0 t=0 m−1 n−1

+

∑ ∑ | f (s,t, u2(s,t), Gu2 (s,t), μ ) − f (s,t, u2(s,t), Gu2 (s,t), μ0 )|

s=0 t=0

a(m, n) +

m−1 n−1

∑ ∑ p(s,t)

s−1 t−1

e(s,t) +

∑ ∑ q(s,t, σ , τ )e(σ , τ )

.

(5.5.32)

σ =0 τ =0

s=0 t=0

Now an application of Lemma 5.5.1 to (5.5.32) yields (5.5.30), which shows the dependency of solutions of (5.5.26)–(5.5.2) and (5.5.27)–(5.5.2) on the parameters μ , μ0 . Remark 5.5.4.

We note that the results given above can be extended very easily to study

the sum-difference equation Δ2 Δ1 u(m, n) + Δ2 (b(m, n)u(m, n)) = f (m, n, u(m, n), Gu(m, n), Hu(m, n)),

(5.5.33)

with the given initial conditions in (5.5.2), where G is defined by (5.5.3) and H is given by ∞

Hu(m, n) :=

∞

∑ ∑ h(m, n, σ , τ , u(σ , τ )),

(5.5.34)

σ =0 τ =0

under some suitable conditions on b, f , g, h involved in (5.5.33), (5.5.34), (5.5.2) by making use of the finite difference inequality given in [87, Theorem 5.2.3]. Next, we shall study some fundamental qualitative properties of solutions of the following Fredholm type sum-difference equation (see [113]) a

b

u(m, n) = f (m, n) + ∑ ∑ g(m, n, s,t, u(s,t), Δ1 u(s,t), Δ2 u(s,t)), s=0 t=0

(5.5.35)

218


where f , g are given functions and u is the unknown function. Let H = N0,a × N0,b and for any function w : H → R, we denote by |w(m, n)|0 = |w(m, n)| + |Δ1 w(m, n)| + |Δ2 w(m, n)| and assume that w(m, n) = 0 for (m, n) ∈ / H. It is assumed that f , Δi f ∈ D(H, R) and g, Δi g ∈ D(H 2 × R3 , R) for i = 1, 2. The origin of equation (5.5.35) can be traced back to the study of its one variable integral analogue in [11] (see also [90,91]). By a solution of equation (5.5.35) we mean a function u : H → R for which Δi u(m, n) (i = 1, 2) exist and satisfies the equation (5.5.35). It is easy to observe that the solution u(m, n) of equation (5.5.1) satisfy for i = 1, 2 the following sum-difference equations a

b

Δi u(m, n) = Δi f (m, n) + ∑ ∑ Δi g(m, n, s,t, u(s,t), Δ1 u(s,t), Δ2 u(s,t)),

(5.5.36)

s=0 t=0

for (m, n) ∈ H. The problem of existence of solutions for equations of the forms (5.5.35) can be dealt with the method employed in in Chapter 1, section 1.6, see also [51,54]. Here we present some basic qualitative aspects of solutions of equation (5.5.35) under some suitable conditions on the functions involved therein. We recall the following special version of the finite difference inequality given in [87, Theorem 5.5.1, p. 286] that will be needed to establish the results. Lemma 5.5.2. Let z, p, q, r ∈ D(H, R+ ) and a

b

z(m, n) p(m, n) + q(m, n) ∑ ∑ r(s,t)z(s,t),

(5.5.37)

s=0 t=0

for (m, n) ∈ H. If a

d=

b

∑ ∑ r(s,t)q(s,t) < 1,

(5.5.38)

s=0 t=0

then

z(m, n) p(m, n) + q(m, n)

1 a b ∑ ∑ r(s,t)p(s,t) , 1 − d s=0 t=0

(5.5.39)

for (m, n) ∈ H. The results concerning estimates, uniqueness, and dependency of solutions of (5.5.35) are given in the following theorems. Theorem 5.5.4.

Suppose that the function g in (5.5.35) and Δi g for i = 1, 2 in (5.5.36)

satisfy the conditions |g(m, n, s,t, u, v, w) − g(m, n, s,t, u, v, w)| c(m, n)h(s,t) [|u − u| + |v − v| + |w − w|] ,

(5.5.40)


219

and |Δi g(m, n, s,t, u, v, w) − Δi g(m, n, s,t, u, v, w)| c(m, n)hi (s,t) [|u − u| + |v − v| + |w − w|] ,

(5.5.41)

where c, h, hi ∈ D(H, R+ ) and a

d1 =

b

∑ ∑ [h(s,t) + h1(s,t) + h2 (s,t)] c(s,t) < 1.

(5.5.42)

s=0 t=0

If u(m, n) is any solution of equation (5.5.35) on H, then ×

|u(m, n) − f (m, n)|0 Q(m, n) + c(m, n)

a b 1 ∑ ∑ [h(s,t) + h1 (s,t) + h2(s,t)] Q(s,t) , 1 − d1 s=0 t=0

(5.5.43)

for (m, n) ∈ H, where a

Q(m, n) =

b

∑ ∑ |g(m, n, σ , τ , f (σ , τ ), Δ1 f (σ , τ ), Δ2 f (σ , τ ))|0 ,

(5.5.44)

s=0 t=0

for (m, n) ∈ H. Theorem 5.5.5.


satisfy the conditions (5.5.40) and (5.5.41) and the condition (5.5.42) holds. Then equation (5.5.35) has at most one solution on H. We next consider the equation (5.5.35) together with the following Fredholm type sumdifference equation a

b

v(m, n) = F(m, n) + ∑ ∑ G(m, n, s,t, v(s,t), Δ1 v(s,t), Δ2 v(s,t)),

(5.5.45)

s=0 t=0

for (m, n) ∈ H, where F, Δi F ∈ D(H, R); G, Δi G ∈ D(H 2 × R3 , R) for i = 1, 2. Theorem 5.5.6.


satisfy the conditions (5.5.40) and (5.5.41) and the condition (5.5.42) holds. Then for every given solution v ∈ D(H, R) of (5.5.45) and any solution u ∈ D(H, R) of (5.5.35), the estimate |u(m, n) − v(m, n)|0 [| f (m, n) − F(m, n)|0 + M(m, n)] a b 1 +c(m, n) ∑ ∑ [h(s,t) + h1 (s,t) + h2 (s,t)] 1 − d1 s=0 t=0

× [| f (s,t) − F(s,t)|0 + M(s,t)] ,

(5.5.46)

220


holds for (m, n) ∈ H, where b

a

M(m, n) =

∑ ∑ |g(m, n, σ , τ , v(σ , τ ), Δ1v(σ , τ ), Δ2 v(σ , τ ))

σ =0 τ =0

− G(m, n, σ , τ , v(σ , τ ), Δ1 v(σ , τ ), Δ2 v(σ , τ ))|0 ,

(5.5.47)

for (m, n) ∈ H. We next consider the following Fredholm type sum-difference equations a

b

z(m, n) = f (m, n) + ∑ ∑ g(m, n, s,t, z(s,t), Δ1 z(s,t), Δ2 z(s,t), μ ),

(5.5.48)

s=0 t=0 a

b

z(m, n) = f (m, n) + ∑ ∑ g(m, n, s,t, z(s,t), Δ1 z(s,t), Δ2 z(s,t), μ0 ),

(5.5.49)

s=0 t=0

for (m, n) ∈ H, where f , Δi f ∈ D(H, R); g, Δi g ∈ D(H 2 × R3 × R, R) for i = 1, 2 and μ , μ0 are parameters. Theorem 5.5.7. Suppose that the function g in (5.5.48), (5.5.49) and Δi g for i = 1, 2 satisfy the conditions |g(m, n, s,t, u, v, w, μ ) − g(m, n, s,t, u, v, w, μ )| c(m, n)h(s,t) [|u − u| + |v − v| + |w − w|] , |g(m, n, s,t, u, v, w, μ ) − g(m, n, s,t, u, v, w, μ0 )| γ (m, n, s,t)|μ − μ0 |,

(5.5.50) (5.5.51)

and |Δi g(m, n, s,t, u, v, w, μ ) − Δi g(m, n, s,t, u, v, w, μ )| c(m, n)hi (s,t) [|u − u| + |v − v| + |w − w|] , |Δi g(m, n, s,t, u, v, w, μ ) − Δi g(m, n, s,t, u, v, w, μ0 )| γi (m, n, s,t)|μ − μ0 |,

(5.5.52) (5.5.53)

where c, h, hi ∈ D(H, R+ ); γ , γi ∈ D(H , R+ ). Let 2

P(m, n) = |μ − μ0 |

a

b

∑ ∑ [γ (m, n, σ , τ ) + γ1(m, n, σ , τ ) + γ2 (m, n, σ , τ )] ,

(5.5.54)

h(s,t) + h1 (s,t) + h2 (s,t) c(s,t) < 1.

(5.5.55)

σ =0 τ =0

and suppose that a

d2 =

b

∑∑

s=0 t=0

Let z1 (m, n) and z2 (m, n) be the solutions of (5.5.48) and (5.5.49) respectively on H. Then × for (m, n) ∈ H.

|z1 (m, n) − z2 (m, n)|0 P(m, n) + c(m, n)

a b 1 ∑ ∑ [h(s,t) + h1(s,t) + h2 (s,t)]P(s,t) , 1 − d2 s=0 t=0

(5.5.56)


221

Proofs of Theorems 5.5.4–5.5.7. Here we present the proof of Theorem 5.5.7 only; the proofs of Theorems 5.5.4–5.5.6 can be completed by following the proof of Theorem 5.5.7 and closely looking at the proofs of the results given in Chapter 1, section 1.6. Let w(m, n) = z1 (m, n) − z2 (m, n). Using the facts that z1 (m, n) and z2 (m, n) are the solutions of (5.5.48) and (5.5.49) on H and the hypotheses, we have |w(m, n)|0

a

b

∑ ∑ |g(m, n, s,t, z1 (s,t), Δ1z1 (s,t), Δ2 z1 (s,t), μ )

s=0 t=0

−g(m, n, s,t, z2 (s,t), Δ1 z2 (s,t), Δ2 z2 (s,t), μ )| a

b

+ ∑ ∑ |g(m, n, s,t, z2 (s,t), Δ1 z2 (s,t), Δ2 z2 (s,t), μ ) s=0 t=0

−g(m, n, s,t, z2 (s,t), Δ1 z2 (s,t), Δ2 z2 (s,t), μ0 )| a

b

+ ∑ ∑ |Δ1 g(m, n, s,t, z1 (s,t), Δ1 z1 (s,t), Δ2 z1 (s,t), μ ) s=0 t=0

−Δ1 g(m, n, s,t, z2 (s,t), Δ1 z2 (s,t), Δ2 z2 (s,t), μ )| a

b


−Δ1 g(m, n, s,t, z2 (s,t), Δ1 z2 (s,t), Δ2 z2 (s,t), μ0 )| a

b


−Δ2 g(m, n, s,t, z2 (s,t), Δ1 z2 (s,t), Δ2 z2 (s,t), μ )| b

a


−Δ2 g(m, n, s,t, z2 (s,t), Δ1 z2 (s,t), Δ2 z2 (s,t), μ0 )|

a

a

b

b

∑ ∑ c(m, n)h(s,t)|w(s,t)|0 + ∑ ∑ γ (m, n, s,t)|μ − μ0 |

s=0 t=0 a

s=0 t=0 a

b

b

+ ∑ ∑ c(m, n)h1 (s,t)|w(s,t)|0 + ∑ ∑ γ1 (m, n, s,t)|μ − μ0 | s=0 t=0 a

s=0 t=0 a

b

b

+ ∑ ∑ c(m, n)h2 (s,t)|w(s,t)|0 + ∑ ∑ γ2 (m, n, s,t)|μ − μ0 | s=0 t=0

s=0 t=0

= P(m, n) + c(m, n) ∑ ∑ h(s,t) + h1 (s,t) + h2 (s,t) |w(s,t)|0 . a

b

(5.5.57)

s=0 t=0

Now an application of Lemma 5.5.2 to (5.5.57) yields (5.5.56), which shows the dependency of solutions of equations (5.5.48) and (5.5.49) on parameters.

222

5.6


Volterra-Fredholm-type sum-difference equations

The numerical methods are often used very effectively in studying the behavior of solutions of equations of the forms (3.3.1) which often leads to the study of Volterra-Fredholm-type sum-difference equation of the form n−1

u(n, x) = h(n, x) + ∑ ∑ F(n, x, s, y, u(s, y)),

(5.6.1)

s=0 G

where h, F are known functions and u is the unknown function. In this section, we adopt the notations as given earlier in section 5.4 without further mention and assume that h ∈ D(E, R), F ∈ D E 2 × R, R . Here we note that by modifying the idea employed in Theorem 3.4.1, Chapter 3, one can formulate existence result for the solution of equation (5.6.1), see also [51,54]. The main goal here is to study some fundamental qualitative properties of solutions of equation (5.6.1) under some suitable conditions on the functions involved therein. We call the function u ∈ D(E, R) an ε -approximate solution to equation (5.6.1), if there exists a constant ε 0 such that

n−1 u(n, x) − h(n, x) + ∑ ∑ F(n, x, s, y, u(s, y)) ε , s=0 G for (n, x) ∈ E. First we shall give the following theorem which deals with the relation between an ε approximate solution and a solution of equation (5.6.1). Theorem 5.6.1. Suppose that (i) the function F in (5.6.1) satisfies the condition |F(n, x, s, y, u) − F(n, x, s, y, v)| b(n, x)p(s, y)|u − v|,

(5.6.2)

where b, p ∈ D(E, R+ ); (ii) the functions uε (n, x), u(n, x) ∈ D(E, R) are respectively, an ε -approximate solution and any solution of equation (5.6.1). Then

n−1

|uε (n, x) − u(n, x)| ε 1 + b(n, x) ∑ ∑ p(s, y) s=0 G

for (n, x) ∈ E.

n−1

∏

σ =s+1

1 + ∑ p(σ , y)b(σ , y) G

, (5.6.3)


Proof.

223

Let e(n, x) = |uε (n, x) − u(n, x)| for (n, x) ∈ E. From the hypotheses, we observe

that

n−1 e(n, x) = uε (n, x) − h(n, x) − ∑ ∑ F(n, x, s, y, uε (s, y)) s=0 G + ∑ ∑ {F(n, x, s, y, uε (s, y)) − F(n, x, s, y, u(s, y))} s=0 G n−1

n−1 uε (n, x) − h(n, x) + ∑ ∑ F(n, x, s, y, uε (s, y)) s=0 G n−1

+ ∑ ∑ |F(n, x, s, y, uε (s, y)) − F(n, x, s, y, u(s, y))| s=0 G

n−1

ε + b(n, x) ∑ ∑ p(s, y)e(s, y).

(5.6.4)

s=0 G

Now an application of the inequality in Theorem 5.4.5 part (s10 ) to (5.6.4) yields (5.6.3). The next theorem deals with the estimate on the difference between the two approximate solutions of equation (5.6.1). Theorem 5.6.2. Suppose that (i) the function F in (5.6.1) satisfies the condition (5.6.2); (ii) the functions ui (n, x) ∈ D(E, R) for i = 1, 2 are the εi -approximate solutions of (5.6.1). Then

n−1

|u1 (n, x) − u2 (n, x)| (ε1 + ε2 ) 1 + b(n, x) ∑ ∑ p(s, y) s=0 G

×

n−1

∏

σ =s+1

1 + ∑ p(σ , y)b(σ , y)

,

(5.6.5)

G

for (n, x) ∈ E. The proof follows by the similar arguments as in the proof of Theorem 3.4.4, Chapter 3 and using the inequality in Theorem 5.4.5 part (s10 ). We omit the details. Remark 5.6.1.

If we take ε1 = ε2 = 0 in Theorem 5.6.2, then the uniqueness of solutions

of equation (5.6.1) follows. The following theorem deals with the estimate on the solution of equation (5.6.1) by assuming that the function F satisfies the Lipschitz type condition (5.6.2).

224


Theorem 5.6.3.


(5.6.2). If u(n, x) is any solution of equation (5.6.1) on E, then n−1

|u(n, x) − h(n, x)| a(n, x) + b(n, x) ∑ ∑ p(s, y)a(s, y) × for (n, x) ∈ E, where

s=0 G

n−1

∏

σ =s+1

1 + ∑ p(σ , y)b(σ , y)

(5.6.6)

G

n−1

a(n, x) = for (n, x) ∈ E. Proof.

∑ ∑ |F(n, x, s, y, h(s, y))| ,

s=0 G

Using the fact that u(n, x) is any solution of equation (5.6.1) and the condition

(5.6.2), we have |u(n, x) − h(n, x)|

n−1

∑ ∑ |F(n, x, s, y, u(s, y)) − F(n, x, s, y, h(s, y))|

s=0 G n−1

+ ∑ ∑ |F(n, x, s, y, h(s, y))| s=0 G

n−1

a(n, x) + b(n, x) ∑ ∑ p(s, y)|u(s, y) − h(s, y)|,

(5.6.7)

s=0 G

for (n, x) ∈ E. Now an application of the inequality in Theorem 5.4.5 part (s10 ) to (5.6.7) yields (5.6.6). Consider the equation (5.6.1) and the following Volterra-Fredholm-type sum-difference equation n−1

v(n, x) = h(n, x) + ∑ ∑ F(n, x, s, y, v(s, y)), for (n, x) ∈ E where h ∈ D(E, R), F ∈

(5.6.8)

s=0 G D(E 2 × R, R).

The following theorem deals with the continuous dependence of solution of equation (5.6.1) on the functions involved therein. Theorem 5.6.4.

Suppose that the function F in (5.6.1) satisfies the condition (5.6.2).

Furthermore, suppose that n−1

|h(n, x) − h(n, x)| + ∑ ∑ |F(n, x, s, y, v(s, y)) − F(n, x, s, y, v(s, y))| ε ,

(5.6.9)

s=0 G

where h, F and h, F are as in equations (5.6.1) and (5.6.8) respectively, v(n, x) ∈ D(E, R) is a given solution of equation (5.6.8) and ε > 0 is an arbitrary small constant. Then the solution u(n, x) ∈ D(E, R) of equation (5.6.1) depends continuously on the functions involved in equation (5.6.1).


Proof.

225

Let z(n, x) = |u(n, x)−v(n, x)| for (n, x) ∈ E. Using the facts that u(n, x) and v(n, x)

are the solutions of equations (5.6.1) and (5.6.8) respectively and the hypotheses, we have z(n, x) |h(n, x) − h(n, x)| n−1

+ ∑ ∑ |F(n, x, s, y, u(s, y)) − F(n, x, s, y, v(s, y))| s=0 G

n−1

+ ∑ ∑ |F(n, x, s, y, v(s, y)) − F(n, x, s, y, v(s, y))| s=0 G

n−1

ε + b(n, x) ∑ ∑ p(s, y)z(s, y).

(5.6.10)

s=0 G

Now an application of the inequality in Theorem 5.4.5 part (s10 ) to (5.6.10) yields n−1

|u(n, x) − v(n, x)| ε 1 + b(n, x) ∑ ∑ p(s, y) s=0 G

×

n−1

∏

σ =s+1

1 + ∑ p(σ , y)b(σ , y) ,

(5.6.11)

G

for (n, x) ∈ E. From (5.6.11), it follows that the solution of equation (5.6.1) depends continuously on the functions involved therein. We next consider the following Volterra-Fredholm-type sum-difference equations n−1

z(n, x) = g(n, x) + ∑ ∑ F(n, x, s, y, u(s, y), μ ),

(5.6.12)

z(n, x) = g(n, x) + ∑ ∑ F(n, x, s, y, u(s, y), μ0 ),

(5.6.13)

s=0 G n−1 s=0 G

for (n, x) ∈ E, where g ∈ D(E, R), F ∈ D(E 2 × R2 , R) and μ , μ0 are parameters. The following theorem shows the dependency of solutions of equations (5.6.12), (5.6.13) on parameters. Theorem 5.6.5.

Suppose that the function F in equations (5.6.12), (5.6.13) satisfy the

conditions |F(n, x, s, y, u, μ ) − F(n, x, s, y, v, μ )| b(n, x)p(s, y)|u − v|, |F(n, x, s, y, u, μ ) − F(n, x, s, y, u, μ0 )| c(s, y)|μ − μ0 |,

(5.6.14) (5.6.15)

where b, p, c ∈ D(E, R+ ). Let z1 (n, x) and z2 (n, x) be the solutions of equations (5.6.12) and (5.6.13) respectively on E. Assume that n−1

∑ ∑ c(s, y) M,

s=0 G

(5.6.16)

226


where M 0 is a constant. Then

n−1

|z1 (n, x) − z2 (n, x)| M|μ − μ0 | 1 + b(n, x) ∑ ∑ p(s, y) s=0 G

×

n−1

∏

σ =s+1

1 + ∑ p(σ , y)b(σ , y)

,

(5.6.17)

G

for (n, x) ∈ E. Proof.

Let z(n, x) = |z1 (n, x) − z2 (n, x)| for (n, x) ∈ E. Using the facts that z1 (n, x) and

z2 (n, x) are the solutions of equations (5.6.12) and (5.6.13) and hypotheses, we have z(n, x)

n−1

∑ ∑ |F(n, x, s, y, z1(s, y), μ ) − F(n, x, s, y, z2 (s, y), μ )|

s=0 G n−1

+ ∑ ∑ |F(n, x, s, y, z2 (s, y), μ ) − F(n, x, s, y, z2 (s, y), μ0 )| s=0 G

n−1

n−1

b(n, m) ∑ ∑ p(s, y)|z1 (s, y) − z2 (s, y)| + ∑ ∑ c(s, y)|μ − μ0 | s=0 G

s=0 G

n−1

M|μ − μ0 | + b(n, m) ∑ ∑ p(s, y)z(s, y).

(5.6.18)

s=0 G

Now an application of the inequality in Theorem 5.4.5 part (s10 ) to (5.6.18) yields (5.6.17), which shows the dependency of solutions of equations (5.6.12) and (5.6.13) on parameters. Next we study some basic qualitative ascepts of solutions of general Volterra-Fredholmtype sum-difference equation n−1

∞

w(n, x) = f (n, x) + ∑ ∑ F(n, x, s, y, w(s, y)) + ∑ ∑ H(n, x, s, y, w(s, y)), s=0 G

(5.6.19)

s=0 G

for (n, x) ∈ E, where f , F, H are known functions and u is the unknown function. We assume that f ∈ D(E, R) and F, H ∈ D(E 2 × R, R). The following theorems hold. Theorem 5.6.6.

Suppose that the functions F, H in equation (5.6.19) satisfy the condi-

tions |F(n, x, s, y, u) − F(n, x, s, y, v)| b(n, x)p(s, y)|u − v|,

(5.6.20)

|H(n, x, s, y, u) − H(n, x, s, y, v)| c(n, x)q(s, y)|u − v|,

(5.6.21)

where b, p, c, q ∈ D(E, R+ ). Let d, D, A(n, x), B(n, x) be defined as in Theorem 5.4.5 part (s9 ). Then the equation (5.6.19) has at most one solution on E.


Theorem 5.6.7.

227

Suppose that the functions F, H in equation (5.6.19) satisfy the condi-

tions (5.6.20), (5.6.21). Let d, B(n, x) be defined as in Theorem 5.4.5 part (s9 ) and ∞

n−1

r(n, x) =

∑ ∑ |F(n, x, s, y, f (s, y))| + ∑ ∑ |H(n, x, s, y, f (s, y))|,

s=0 G

(5.6.22)

s=0 G

D2 =

1 ∞ ∑ ∑ q(s, y)A2(s, y), 1 − d s=0 G

(5.6.23)

where A2 (n, x) is defined by the right hand side of (5.4.22) by replacing a(n, x) by r(n, x). If w(n, x) is any solution of equation (5.6.19), then |w(n, x) − f (n, x)| A2 (n, x) + D2 B(n, x),

(5.6.24)

for (n, x) ∈ E. Theorem 5.6.8.

Suppose that the functions F, H in equations (5.6.1), (5.6.19) satisfy the

conditions (5.6.20), (5.6.21) and H(n, x, s, y, 0) = 0. Let u ∈ D(E, R) be a solution of equation (5.6.1) such that |u(n, x)| Q for (n, x) ∈ E, where Q 0 is a constant. Let d, B(n, x) be defined as in Theorem 5.4.5 part (s9 ) and ∞

a(n, x) = | f (n, x) − h(n, x)| + Qc(n, x) ∑ ∑ q(s, y),

(5.6.25)

s=0 G

D3 =

1 ∞ ∑ ∑ q(s, y)A3(s, y), 1 − d s=0 G

(5.6.26)

where A3 (n, x) is defined by the right hand side of (5.4.22) by replacing a(n, x) by a(n, x). If w ∈ D(E, R) is any solution of equation (5.6.19), then |w(n, x) − u(n, x)| A3 (n, x) + D3 B(n, x),

(5.6.27)

for (n, x) ∈ E. Proofs of Theorems 5.6.6–5.6.8.

Below, we give the proof of Theorem 5.6.8 only; the

proofs of Theorems 5.6.6 and 5.6.7 can be completed by following the ideas used to prove the Theorems 1.4.2 and 1.4.4 in Chapter 1. Using the facts that w(n, x) and u(n, x) are the solutions of equations (5.6.19) and (5.6.1) and the hypotheses, we observe that |w(n, x) − u(n, x)| | f (n, x) − h(n, x)| n−1

+ ∑ ∑ |F(n, x, s, y, w(s, y)) − F(n, x, s, y, u(s, y))| s=0 G ∞

+ ∑ ∑ |H(n, x, s, y, w(s, y)) − H(n, x, s, y, u(s, y))| s=0 G

228

Multidimensional Integral Equations and Inequalities ∞

+ ∑ ∑ |H(n, x, s, y, u(s, y)) − H(n, x, s, y, 0)| s=0 G

n−1

| f (n, x) − h(n, x)| + b(n, x) ∑ ∑ p(s, y)|w(s, y) − u(s, y)| s=0 G

∞

∞

+c(n, x) ∑ ∑ q(s, y)|w(s, y) − u(s, y)| + c(n, x) ∑ ∑ q(s, y)|u(s, y)| s=0 G

s=0 G

n−1

a(n, x) + b(n, x) ∑ ∑ p(s, y)|w(s, y) − u(s, y)| s=0 G

∞

+c(n, x) ∑ ∑ q(s, y)|w(s, y) − u(s, y)|.

(5.6.28)

s=0 G

Now an application of the inequality in Theorem 5.4.5 part (s9 ) to (5.6.28) yields (5.6.27). Remark 5.6.2.

We note that, one can use the inequality in Theorem 5.4.4 part (s7 ) to for-

mulate results similar to those given in Theorems 5.6.1–5.6.5 for the solutions of VolterraFredholm-type sum-difference equation of the form n−1

u(n, x) = h(n, x) + ∑ ∑ F(n, x, s, y, u(s, y), Tu(s, y)),

(5.6.29)

s=0 G

for (n, x) ∈ E, where n−1

Tu(n, x) :=

∑ ∑ L(n, x, τ , z, u(τ , z)),

τ =0 G

under some suitable conditions on the functions involved in (5.6.29). 5.7

Miscellanea

5.7.1

Pachpatte [108]

Let u, r ∈ D(N20 , R+ ) and c 0 is a real constant. (p1 ) If n−1 s−1 m−1

u2 (n, m) c + ∑

∑ ∑ r(σ , τ )u(σ , τ ),

s=0 σ =0 τ =0

for (n, m) ∈ N20 , then u(n, m) for (n, m) ∈ N20 .

√ 1 n−1 s−1 m−1 c + ∑ ∑ ∑ r(σ , τ ), 2 s=0 σ =0 τ =0

(5.6.30)


229

(p2 ) Suppose that u 1 and c 1. If n−1 s−1 m−1

u(n, m) c + ∑

∑ ∑ r(σ , τ )u(σ , τ ) log u(σ , τ ),

s=0 σ =0 τ =0

for (n, m) ∈ N20 , then u(n, m) cB(n,m) , for (n, m) ∈ N20 , where n−1

B(n, m) = ∏ 1 + s=0

5.7.2

s−1 m−1

∑ ∑ r(σ , τ )

.

σ =0 τ =0

Pachpatte [112]

Let u, p, q, f , g ∈ D(H, R+ ), where H = N20 × Nα ,β ; k 0 is a real constant and L ∈ D(H × R+ , R+ ) be such that 0 L(x, y, z, u) − L(x, y, z, v) M(x, y, z, v)(u − v), for u v 0, where M ∈ D(H × R+ , R+ ). (p3 ) If x−1 y−1 β

u(x, y, z) p(x, y, z) + q(x, y, z) ∑

∑ ∑ L(s,t, r, u(s,t, r)),

s=0 t=0 r=α

for (x, y, z) ∈ H, then

u(x, y, z) p(x, y, z) + q(x, y, z) x−1

∑ ∑ ∑ L(s,t, r, p(s,t, r))

s=0 t=0 r=α

y−1 β

×∏ 1+ ∑

x−1 y−1 β

∑ M(s,t, r, p(s,t, r))q(s,t, r)

,

t=0 r=α

s=0

for (x, y, z) ∈ H. (p4 ) If x−1 y−1 β

u2 (x, y, z) k2 + 2 ∑ for (x, y, z) ∈ H, then

∑ ∑ [g(s,t, r)u(s,t, r) + f (s,t, r)u(s,t, r)L(s,t, r, u(s,t, r))],

s=0 t=0 r=α

u(x, y, z) n0 (x, y) + x−1

∑∑∑

s=0 t=0 r=α

f (s,t, r)L(s,t, r, n0 (s,t))

y−1 β

×∏ 1+ ∑ s=0

x−1 y−1 β

∑

t=0 r=α

f (s,t, r)M(s,t, r, n0 (s,t)) ,

for (x, y, z) ∈ H, where x−1 y−1 β

n0 (x, y) = k + for x, y ∈ N0 .

∑ ∑ ∑ g(σ , τ , ω ),

σ =0 τ =0 ω =α

230

5.7.3


Pachpatte [107]

Under the notations as in section 5.4, let u, p, q, f ∈ D(E, R+ ) and c 0 is a real constant. (p5 ) Let L ∈ D(E × R+ , R+ ) be such that 0 L(n, x, u) − L(n, x, v) M(n, x, v)(u − v), for u v 0, where M ∈ D(E × R+ , R+ ). If n−1 s−1

u(n, x) p(n, x) + q(n, x) ∑

∑ ∑ L(τ , y, u(τ , y)),

s=0 τ =0 G

for (n, x) ∈ E, then

u(n, x) p(n, x) + q(n, x) n−1

n−1 s−1

∑

∑ ∑ L(τ , y, p(τ , y))

s=0 τ =0 G

s−1

× ∏ 1 + ∑ ∑ M(τ , y, p(τ , y))q(τ , y) , τ =0 G

s=0

for (n, x) ∈ E. (p6 ) Let g be as in Theorem 5.4.1 part (s2 ). If n−1 s−1

u(n, x) c + ∑

∑ ∑ f (τ , y)g(u(τ , y)),

s=0 τ =0 G

for (n, x) ∈ E, then for 0 n n1 ; n, n1 ∈ N0 , x ∈ G u(n, x) W

−1

n−1 s−1

W (c) + ∑

∑ ∑ f (τ , y)

,

s=0 τ =0 G

where W, W −1 are as in Theorem 5.4.1 part (s2 ) and n1 ∈ N0 be chosen so that n−1 s−1

W (c) + ∑

∑ ∑ f (τ , y) ∈ Dom

−1 W ,

s=0 τ =0 G

for all n ∈ N0 lying in 0 n n1 and x ∈ G. 5.7.4

Pachpatte [101]

Under the notations as in section 5.4, let u, p, q, f , g ∈ D(E, R+ ) and k 0 is a real constant. (p7 ) Let L ∈ D(E × R+ , R+ ) be such that 0 L(n, x, u) − L(n, x, v) M(n, x, v)(u − v), for u v 0, where M ∈ D(E × R+ , R+ ). If n−1

u(n, x) p(n, x) + q(n, x) ∑ ∑ L(s, y, u(s, y)), s=0 G


231


u(n, x) p(n, x) + q(n, x) ∑ ∑ L(s, y, p(s, y)) s=0 G

n−1

∏

σ =s+1

1 + ∑ M(σ , y, p(σ , y))q(σ , y) , G

for (n, x) ∈ E. (p8 ) If n−1 u2 (n, x) k2 + 2 ∑ ∑ f (s, y)u2 (s, y) + g(s, y)u(s, y) , s=0 G

for (n, x) ∈ E, then n−1

n−1

u(n, x) k ∏ 1 + ∑ f (s, y) + ∑ ∑ g(s, y) s=0

G

s=0 G

n−1

∏

σ =s+1

1 + ∑ f (σ , y) , G

for (n, x) ∈ E. 5.7.5

Pachpatte [88]

Consider the initial value problem Δ2 Δ1 u(m, n) = f (m, n, u(m, n), Δ1 u(m, n)),

(5.7.1)

u(m, 0) = σ (m), u(0, n) = τ (n), u(0, 0) = 0,

(5.7.2)

with

for m, n ∈ N0 , where f , σ , τ are given functions and u is the unknown function and f ∈ D N20 × R2 , R , σ , τ ∈ D(N0 , R). (p9 ) Assume that | f (m, n, u, v) − f (m, n, u, v)| p(m, n) [|u − u| + |v − v|] , where p ∈ D(N20 , R+ ). Then the problem (5.7.1)–(5.7.2) has at most one solution on N20 . (p10 ) Assume that | f (m, n, u, v)| r(m, n) [|u| + |v|] , |σ (m)| + |τ (n)| + |Δσ (m)| k, where r ∈ D(N20 , R+ ) and k 0 is a constant. If u(m, n) is any solution of problem (5.7.1)– (5.7.2), then m−1

n−1

|u(m, n)| + |Δ1 u(m, n)| kq(m, n) ∏ 1 + ∑ r(s,t)q(s,t) , s=0

t=0

for m, n ∈ N0 , where m−1

q(m, n) =

∏ [1 + r(m,t1 )],

t1 =0

for m, n ∈ N0 .

232

5.7.6


Pachpatte [118]

Consider the initial value problem Δ2 Δ1 u(m, n) = F(m, n, u(m, n), Δ2 Δ1 u(m, n)),

(5.7.3)

with u(m, 0) = α (m),

u(0, n) = β (n),

u(0, 0) = 0,

(5.7.4)

for m, n ∈ N0 , where f , α , β are given functions, u is the unknown function and F ∈ D(N20 × R2 , R), α , β ∈ D(N0 , R). (p11 ) Let ui (m, n) ∈ D(N20 , R) (i = 1, 2) be respectively εi -approximate solutions of equation (5.7.3) i.e. |Δ2 Δ1 ui (m, n) − F(m, n, ui (m, n), Δ2 Δ1 ui (m, n))| εi for given constants εi 0, with ui (m, 0) = αi (m),

ui (0, n) = βi (n),

ui (0, 0) = 0,

where αi , βi ∈ D(N0 , R) and such that |α1 (m) − α2 (m) + β1 (n) − β2 (n)| δ , where δ 0 is a constant. Suppose that the function F in (5.7.3) satisfies the condition |F(m, n, u, v) − F(m, n, u, v)| p(m, n) [|u − u| + |v − v|] ,

(5.7.5)

where p ∈ D(N20 , R+ ) satisfies p(m, n) < 1 for m, n ∈ N0 . Then |u1 (m, n) − u2 (m, n)| + |Δ2 Δ1 u1 (m, n) − Δ2 Δ1 u2 (m, n)|

L(m, n) + E(m, n)

m−1 n−1

n−1

∑ ∑ p(s,t)L(s,t) ∏

s=0

s=0 t=0

n−1

1 + ∑ p(s,t)E(s,t) , t=0

for m, n ∈ N0 , where L(m, n) =

(ε1 + ε2 )(mn + 1) + δ , 1 − p(m, n)

E(m, n) =

1 . 1 − p(m, n)

(5.7.6)

Consider the initial value problem (5.7.3)–(5.7.4) together with the following initial value problem Δ2 Δ1 v(m, n) = G(m, n, v(m, n), Δ2 Δ1 v(m, n)),

(5.7.7)

with v(m, 0) = α (m),

v(0, n) = β (n),

v(0, 0) = 0,

(5.7.8)


233

for m, n ∈ N0 , where G ∈ D(N20 × R2 , R), α , β ∈ D(N0 , R). (p12 ) Suppose that the function F in (5.7.3) satisfies the condition (5.7.5) and there exist constants ε 0, δ 0 such that |F(m, n, u, v) − G(m, n, u, v)| ε , |α (m) − α (m) + β (n) − β (n)| δ , where F, α , β and G, α , β are as in (5.7.3)–(5.7.4) and (5.7.7)–(5.7.8). Let u(m, n) and v(m, n) be respectively the solutions of (5.7.3)–(5.7.4) and (5.7.7)–(5.7.8) for m, n ∈ N0 . Then |u(m, n) − v(m, n)| + |Δ2 Δ1 u(m, n) − Δ2 Δ1 v(m, n)| L(m, n) + E(m, n)

m−1 n−1

n−1

s=0 t=0

s=0

∑ ∑ p(s,t)L(s,t) ∏

n−1

1 + ∑ p(s,t)E(s,t) , t=0

for m, n ∈ N0 , where L(m, n) =

ε (mn + 1) + δ , 1 − p(m, n)

and E(m, n) is as in (5.7.6). 5.7.7

Pachpatte [104]

Consider the sum-difference equation m−1 s−1 β

v(m, x) = f (m, x) +

∑ ∑ ∑ K(m, x, s, τ , y, v(τ , y)),

(5.7.9)

s=0 τ =0 y=α

for (m, x) ∈ H = Na,b × Nα ,β , where f , K are given functions, v is the unknown function and f ∈ D(H, R), K ∈ D(H × Na,b × H × R, R). (p13 ) Assume that the function K in (5.7.9) satisfies the condition |K(m, n, s, τ , y, v) − K(m, n, s, τ , y, w)| q(m, x)g(τ , y)|v − w|, where q, g ∈ D(H, R+ ). Then the equation (5.7.9) has at most one solution on H. (p14 ) Assume that the function K in (5.7.9) satisfies the condition |K(m, n, s, τ , y, v)| q(m, x)g(τ , y)|v|, where q, g ∈ D(H, R+ ). If v(m, x) is any solution of equation (5.7.9) on H, then |v(m, n)| | f (m, x)|+q(n, x) for (m, x) ∈ H.

m−1 s−1 β

m−1

s=0 τ =0 y=α

s=0

∑ ∑ ∑ g(τ , y)| f (τ , y)| ∏

s−1 β

1+ ∑

∑ g(τ , y)q(τ , y)

τ =0 y=α

,

234


5.7.8

Pachpatte [101]

Under the notations as in section 5.4, consider the sum-difference equation of the form n−1

u(n, x) = h(n, x) + ∑ ∑ H(n, x, s, y, u(s, y)),

(5.7.10)

s=0 G

for (n, x) ∈ E, where h, H are given functions, u is the unknown function and h ∈ D(E, R), H ∈ D(E 2 × R, R). (p15 ) Suppose that the function H in equation (5.7.10) satisfies the condition |H(n, x, s, y, u)| b(n, x)L(s, y, |u|), where b ∈ D(E, R+ ) and L is as in part (p7 ). Then for every solution u ∈ D(E, R) of equation (5.7.10), the estimate n−1

|u(n, x)| |h(n, x)|+b(n, x) ∑ ∑ L(s, y, |h(s, y)|) s=0 G

n−1

∏

σ =s+1

1 + ∑ M(σ , y, |h(σ , y)|)b(σ , y) , G

holds for (n, x) ∈ E, where M is as in (p7 ). Consider the equation (5.7.10) together with the following sum-difference equation n−1

w(n, x) = h(n, x) + ∑ ∑ K(n, x, s, y, w(s, y)),

(5.7.11)

s=0 G

for (n, x) ∈ E, where h ∈ D(E, R), K ∈ D(E 2 × R, R). (p16 ) Suppose that the function H in (5.7.10) satisfies the condition |H(n, x, s, y, u) − H(n, x, s, y, w)| b(n, x)L(s, y, |u − w|), where b ∈ D(E, R+ ) and L is as in part (p7 ). Then for every given solution w ∈ D(E, R) of equation (5.7.11) and u ∈ D(E, R) any solution of equation (5.7.10), the estimate |u(n, x) − w(n, x)| [h0 (n, x) + r(n, x)] n−1

+b(n, x) ∑ ∑ L (s, y, [h0 (s, y) + r(s, y)]) s=0 G

n−1

∏

σ =s+1

[1 + M(σ , y, [h0 (σ , y) + r(σ , y)])] ,

holds for (n, x) ∈ E, where h0 (n, x) = |h(n, x) − h(n, x)|, n−1

r(n, x) =

∑ ∑ |H(n, x, s, y, w(s, y)) − K[h0 (s, y) + r(s, y)]|,

s=0 G

and M is as in part (p7 ).


5.8

235

Notes

Although some books and papers contains some basic results on partial finite difference equations (see [3,17,46,47,51,67,85,87]), it seems, much of the qualitative and quantitative theory of these equations remain to be developed in various directions. The results in sections 3.2–3.4 deals with a large number of fundamental finite difference inequalities with explicit estimates involving functions of two, three and many variables recently investigated, as a response to the needs of diverse applications and are adapted from Pachpatte [98,104,108,95,102,111,116,114,106]. Section 5.5 contains results related to error evaluation of approximate solutions of a certain sum-difference equation in two variables and also some basic results on Fredholm-type sum-difference equation in two variables recently obtained in [100,113]. Section 5.6 is devoted to present the basic theory of mixed sum-difference equations typically arise while studying some initial boundary value problems for partial differential equations of parabolic type by using discretization methods and are taken from Pachpatte [116,101,99,107,109]. Section 5.7 addresses some selected results which, we hope, will provide a clue to effective methods for future important developments.

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Subject Index

A analytic tools, 59 approximate solutions, 17, 18, 27, 28, 52, 75, 76, 83, 92, 119, 122–124, 132–134, 150, 151, 158, 160, 212, 213, 222, 223, 232, 235 approximation of solutions, 211

D Darboux problem, 55 dependency of solutions, 19, 20, 26, 36, 76, 88, 89, 125, 126, 152–154, 212, 214, 216–218, 221, 225, 226 difference equation, 5, 191, 217–220, 222, 224–226, 228, 233–235 differential equations, 1, 15, 87, 97 difference between two approximate solutions, 17, 27, 119, 123, 132, 150, 158, 212, 223 diffusion, 97, 141, 143, 161 Dirichlet boundary data, 3 type boundary condition, 186 dynamic equations, 1, 95, 97, 191 dynamical systems, 59

B Banach fixed point theorem, 6, 9, 15, 34, 39, 44, 86, 117 space of bounded functions, 45 spaces, 14, 16, 32, 37, 48, 50, 73, 79, 84, 116, 135, 136, 147, 169, 170 Barbashin-type, 4 equation, 4, 147, 155, 189 integrodifferential equation, 147, 153, 155, 189 basic integral inequalities, 9, 57, 59, 95, 97, 143, 189 behavior of solutions, 5, 59, 161, 189, 222 Bielecki-type norm, 6, 15 boundary and initial conditions, 4, 35, 36, 51, 52, 55, 140, 143, 176, 180, 182, 183, 187

E elementary inequalities, 18, 28, 119, 124, 151, 213 elliptic second order partial differential operator, 186 empty sums and products, 7 epidemic models, 143 epidemiology, 97 error evaluation, 27, 212, 235 estimate on the solution, 16, 22, 23, 34, 40, 41, 74, 80, 117, 127, 129, 148, 155, 223 Euclidean space, 6 existence of a solution, 1, 122, 162, 191, 218 and uniqueness of solutions, 14, 15, 21, 22, 27, 83, 87, 116, 147, 155 of a unique solution, 14, 32, 44, 73, 79, 84, 87, 129, 163, 168, 181, 183

C characteristic data, 1, 2 classical mechanics, 1 closeness of solutions, 29, 124, 152, 159, 214 compactness of the operator, 5 continuous dependence of solutions, 24, 35, 44, 76, 83, 122, 131, 160, 177, 224 continuously differentiable, 37, 45, 162 contraction mapping, 15, 16, 21, 45, 74, 86, 148 convergence properties, 27, 212 coupled parabolic integrodifferential equations, 182 243

244


explicit estimates, 5, 6, 9, 57, 59, 65, 95, 97, 126, 141, 143, 189, 191, 212, 235

lower solution, 162, 163, 168, 189

F finite difference equations, 5, 235 difference methods, 211 Fredholm type, 3, 97 integral equation, 37, 42 integrodifferential equation, 3, 37, 83, 87 type integral equation, 179 type sum-difference equations, 217, 219, 220 G Galerkin finite element method, 186 Green’s function, 3 H Hölder continuous, 162, 167, 176, 177, 180, 182 classes, 184 Hammerstein type integral equation, 46 hyperbolic equation, 1 integrodifferential equation, 31 partial integrodifferential equation, 55 type, 3 I inequalities, 5, 18, 28, 54, 119, 124, 213 with explicit estimates, 6, 9, 59, 97, 141, 191, 235 initial value problem, 211, 215, 231, 232 boundary conditions, 1, 4, 35, 36, 51, 52, 55, 140, 143, 176, 187 boundary value problem, 169, 188, 235 integral equation, 97, 128, 135, 136, 141, 189 equation of Barbashin-type, 189 inequalities, 97, 101, 102, 141 integrodifferential equations, 1–4, 6, 9, 20, 31, 51, 52, 83, 87, 141, 143, 147, 152, 153, 155, 161, 167, 169, 179, 182, 184, 186, 187, 189, 211 equation of Barbashin-type, 4, 147, 153, 155, 189 system, 4, 180 interpolation inequalities, 174 L Lipschitz condition, 55 type conditions, 23, 41, 223

M mathematical models, 1 maximal solution, 165, 169, 182, 184 and minimal solution, 162, 163, 165, 167, 169 method of continuity, 170 of integral inequalities, 97 of successive approximations, 5 minimal solution, 162, 163, 165, 167, 169, 182, 184 mixed Volterra-Fredholm integral equations Volterra-Fredholm type integral inequalities, 108 Volterra-Fredholm-type integral equations, 3, 6, 97, 116, 117, 141 Volterra-Fredholm-type integral inequalities, 6 monotone method, 162, 168 multidimensional, 1, 95 multivariable integral and integrodifferential equations, 1 sum-difference equations, 191 sum-difference inequalities, 6, 204 sum-difference inequalities and equations, 6 N Neumann series, 136 neutral type hyperbolic integrodifferential equation, 31 nonexpansive and monotone mappings, 5 numerical methods, 222 O outward normal unit vector, 162, 180 P parabolic differential equations, 97 boundary value problem, 170 integrodifferential equations, 143, 161, 182, 184, 186, 189 type Fredholm integral equation, 179 type integrodifferential equations, 4, 6 partial differential equations, 141, 185, 235 derivative, 6, 98, 180 integral equation, 155 integral operators, 1 integrodifferential equations, 141, 189, 211 physical and biological phenomena, 1, 116, 189

Subject Index

pseudo-parabolic equation, 2 Q qualitative theory, 1, 5, 122, 235 behavior, 5, 59 properties, 1, 5, 6, 9, 13, 21, 37, 57, 59, 83, 87, 95, 97, 122, 129, 132, 141, 147, 155, 189, 191, 217, 222 R reaction diffusion processes, 143 reactor dynamics, 4, 161, 162, 189 S Schauder estimate, 169, 170, 176–178 science and engineering, 143, 189 self-adjoint, 186 sum-difference equations, 6, 191, 204, 216–220, 233–235 inequalities, 6, 191, 204 inequalities in three variables, 198 inequalities in two variables, 191 T time-independent coefficients, 186

245

two variables, 6, 9, 27, 44, 46, 57, 191, 235 and three independent variables, 6, 95 U uniqueness of solutions, 14, 15, 19, 21, 22, 27, 29, 39, 40, 74, 79, 83, 87, 116, 120, 124, 126, 130, 147, 152, 155, 159, 214, 223 upper solution, 162, 166–168 and lower solutions, 162, 163, 168, 189 V Volterra type, 3, 57, 97 -Fredholm-type integral equation, 3, 6, 21, 24, 26, 91, 97, 116, 117, 120–122, 128, 135, 139–141 -Fredholm-type integral inequalities, 6, 108 -Fredholm-type integrodifferential equations, 27 -Fredholm-type sum-difference equations, 222, 224–226, 228 type integral equation, 23 Y Young’s inequality, 170, 173, 174

Multidimensional Integral Equations and Inequalities

Multidimensional Integral Equations and Inequalities

Multidimensional weakly singular integral equations

Multidimensional Singular Integrals and Integral Equations

Multidimensional singular integrals and integral equations

Inequalities for differential and integral equations

Differential and integral inequalities: - Ordinary differential equations

Integral inequalities and applications

Integral Equations and Applications

Differential and integral equations

Integral equations and applications

Differential and Integral Equations

Differential and Integral Equations

Integral Equations

Integral and Finite Difference Inequalities and Applications

Integral and finite difference inequalities and applications

Volterra integral and functional equations

Volterra Integral and Functional Equations

Integral Equations and their Applications

Volterra integral and differential equations

Integral equations and their applications

Volterra Integral and Differential Equations

Evolutionary Integral Equations and Applications

Evolutionary Integral Equations and Applications

Introduction to multidimensional integrable equations

Volterra Stieltjes-integral equations

Boundary Integral Equations

Handbook of Integral Equations

Handbook of integral equations

Random integral equations

Handbook of Integral Equations

Multidimensional Integral Equations and Inequalities