Hyers-Ulam Stability of Ordinary Differential Equations 0367636670, 9780367636678

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Hyers-Ulam Stability of Ordinary Differential Equations

Hyers-Ulam Stability of Ordinary Differential Equations

Arun Kumar Tripathy

First edition published 2021 by CRC Press 6000 Broken Sound Parkway NW, Suite 300, Boca Raton, FL 33487-2742 and by CRC Press 2 Park Square, Milton Park, Abingdon, Oxon, OX14 4RN © 2021 Arun Kumar Tripathy CRC Press is an imprint of Taylor & Francis Group, LLC The right of Arun Kumar Tripathy to be identified as author of this work has been asserted by him in accordance with sections 77 and 78 of the Copyright, Designs and Patents Act 1988. Reasonable efforts have been made to publish reliable data and information, but the author and publisher cannot assume responsibility for the validity of all materials or the consequences of their use. The authors and publishers have attempted to trace the copyright holders of all material reproduced in this publication and apologize to copyright holders if permission to publish in this form has not been obtained. If any copyright material has not been acknowledged please write and let us know so we may rectify in any future reprint. Except as permitted under U.S. Copyright Law, no part of this book may be reprinted, reproduced, transmitted, or utilized in any form by any electronic, mechanical, or other means, now known or hereafter invented, including photocopying, microfilming, and recording, or in any information storage or retrieval system, without written permission from the publishers. For permission to photocopy or use material electronically from this work, access www.copyright. com or contact the Copyright Clearance Center, Inc. (CCC), 222 Rosewood Drive, Danvers, MA 01923, 978-750-8400. For works that are not available on CCC please contact [email protected] Trademark notice: Product or corporate names may be trademarks or registered trademarks and are used only for identification and explanation without intent to infringe. Library of Congress Cataloging‑in‑Publication Data ISBN: [978-0-367-63667-8] (hbk) ISBN: [978-0-367-63668-5] (pbk) ISBN: [978-1-003-12017-9] (ebk) Typeset in CMR10 font by KnowledgeWorks Global Ltd.

To my family for their endless love, constant support and encouragement.

Contents

Preface

ix

Author Biography

xi

1 Introduction and Preliminaries 1.1 Stability of Functional Equations on Banach Space . . . . . 1.1.1 Stability of f (x + y) = f (x) + f (y) . . . . . . . . . . . 1.1.2 Stability of f (x + y) = f (x) + f (y) . . . . . . . . . . . 1.1.3 Stability of f (x + y) = g(x) + h(y) . . . . . . . . . . . 1.1.4 Stability of f (xy) + f (x + y) = f (xy + x) + f (y) . . . 1.1.5 Stability of f (x + iy) + f (x − iy) = 2f (x) − 2f (y) . . 1.2 Stability of Some Functional Equations on Abelian Groups . 1.2.1 Stability of f (x+y +z)+f (x−y)+f (y −z)+f (x−z) = 3f (x) + 3f (y) + 3f (z) . . . . . . . . . . . . . . . . . . 1.2.2 Stability of f (rx + sy) = ((r + s)/2)f (x + y) + ((r − s)/2)f (x − y) . . . . . . . . . . . . . . . . . . . . 1.3 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1 3 4 6 10 20 30 41

2 Stability of First Order Linear Differential Equations 2.1 Stability of f 0 (x) = f (x) . . . . . . . . . . . . . . . . . 2.2 Stability of y 0 (t) = λy(t) . . . . . . . . . . . . . . . . . 2.3 Stability of ϕ(t)y 0 (t) = y(t) . . . . . . . . . . . . . . . . 2.4 Stability of p(x)y 0 − q(x)y − r(x) = 0 . . . . . . . . . . 2.5 Stability of y 0 = λy on Banach Spaces . . . . . . . . . . 2.6 Stability of y 0 = F (x, y) . . . . . . . . . . . . . . . . . 2.7 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . .

42 53 61

. . . . . . .

. . . . . . .

. . . . . . .

63 63 73 86 94 97 103 111

3 Stability of Second Order Linear Differential Equations 3.1 Hyers-Ulam Stability of y 00 + αy 0 + βy = 0 . . . . . . . . 3.2 Hyers-Ulam Stability of y 00 + β(x)y = 0 . . . . . . . . . . 3.3 Hyers-Ulam Stability of y 00 + β(x)y = f (x) . . . . . . . . 3.4 Hyers-Ulam Stability of y 00 + p(x)y 0 + q(x)y + r(x) = 0 . 3.5 Hyers-Ulam Stability of y 00 + p(x)y 0 + q(x)y = f (x) . . . 3.6 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . .

. . . . . .

113 113 120 122 124 128 135

vii

viii

Contents

4 Hyers-Ulam Stability of Exact Linear Differential Equations 137 4.1 Hyers-Ulam Stability of p0 (x)y 00 + p1 (x)y 0 + p2 (x)y + f (x) = 0 137 4.2 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 147 5 Hyers-Ulam Stability of Euler’s Differential Equations 5.1 Hyers-Ulam Stability of ty 0 (t) + αy(t) + βtr x0 = 0 . . . . . . 5.2 Hyers-Ulam Stability of t2 y 00 (t) + αty 0 (t) + βy(t) = 0 . . . . 000 00 0 5.3 Hyers-Ulam Stability of t3 y + αt2 y + βty + γy = 0 . . . . 000 00 0 5.4 Hyers-Ulam Stability of t4 y (iv) + αt3 y + βt2 y + γty + δy = 0 5.5 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

149 149 153 156 163 178

6 Generalized Hyers-Ulam Stability of Differential Equations in Complex Banach Space 179 6.1 Hyers-Ulam Stability of First Order Differential Equations . 179 6.2 Hyers-Ulam Stability of Second Order Differential Equations 185 6.3 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 7 Hyers-Ulam Stability of Difference Equations 193 7.1 Hyers-Ulam Stability of Second Order Difference Equations-I 193 7.2 Hyers-Ulam Stability of Second Order Difference Equations-II 197 7.3 Hyers-Ulam Stability First Order Difference Operators . . . 200 7.4 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 Bibliography

209

Index

215

Preface

The study of stability problems for various functional equations originated from a talk given by S. M. Ulam in 1940. In that talk Ulam discussed a number of important unsolved problems. Among such problems, a problem concerning the stability of functional equations is one of them. In 1941, D. Hyers gave an answer to the problem on Banach space. Furthermore, the result of Hyers has been generalized by Th. M. Rassias. After that many authors have extended the Ulam’s stability problems to other functional equations and generalized Hyer’s result in various directions. Thereafter, Ulam’s stability problem for functional equations was replaced by stability of differential equations. S. M. Jung has investigated the Hyers-Ulam stability of linear differential equations of different classes. Keeping in view of the definition of Hyers-Ulam stability for differential equations, we can view the corresponding definition of HyersUlam stability for difference equations as well. Because difference equations can be generated from differential equations by using the Euler’s method, the study of Hyers-Ulam stability for differential and difference equations is interesting. The purpose of this book is to reflect the new developments in the HyersUlam Stability Theory for ordinary differential equations and difference equations including some contributions of the author as well as some collections from the published articles. In the proposed topic of interest, first chapter deals with an insight of Hyers-Ulam stability of some functional equations and its origin and subsequent development to ordinary differential and difference equations. In the remaining chapters a discussion has been made comprising of Hyers-Ulam stability of first order, second order, exact and Euler’s differential equations. Also, the Hyers-Ulam stability of some complex differential equations is presented. In the last chapter the Hyers-Ulam stability of some difference equations are discussed. A. K. Tripathy

ix

Author Biography

Dr. Arun Kumar Tripathy, Reader, Department of Mathematics, Sambalpur University, Sambalpur, Odisha, India, is a known name in the literature of Oscillation Theory since past two decades. His contribution basically deals with the linear and nonlinear neutral equations in difference equations, differential equations as well as in time scales of first, second, fourth and higher order equations. It is important that this theory is a part of so-called Dynamical Systems based on Qualitative Behaviour of Solutions Differential and Difference equations. To his credit, he has published 70 research papers in peer-reviewed journals of international repute. Apart from that, he has several international collaborators Prof. S. Pinelas (Portugal), Prof. E. Schmeidal (Poland), Prof. T. G. Baskar (USA), etc. He has been invited to give talks at several international conferences. He is a potential reviewer of many international journals. After completing successfully the two years Post-Doctoral Fellowship offered by National Board for Higher Mathematics, Department of Atomic Energy, Mumbai, Govt of India, Dr Tripathy started his teaching career in 2003, and till now it is a primary job. Aside from this, research is his special interest, and this interest has been continued since 19 years.

xi

Chapter 1 Introduction and Preliminaries

The subject functional equations generalizes a modern branch of mathematics (see, e.g., [55]). During 1747–1750, J. d’Alembert published three papers and these three papers were the first on functional equations. Many mathematicians like N. H. Abel, J. Bolyai, A. L. Cauchy, J. d’Alembert, L. Euler, M. Frechat, C. F. Gauss, J. L. W. V. Jensen, A. N. Kolmogorov, N. I. Lobacevskii, J. V. Pexider and S. D. Poisson have studied functional equations because of their apparent simplicity and harmonic nature. In the theory of functional equations we have a question: When is it true that a function which approximately satisfies a functional equation  must be close to an exact solution ? If the problem accepts a solution, we say that the equation  is stable. In 1940, S. M. Ulam [63] has studied the problem for the stability of functional equations in Banach space. Then Ulam questioned that: Let G1 be a group and G2 a metric group with a metric d(., .). Given ε > 0, does there exist a δ > 0 such that if a mapping f : G1 → G2 satisfies d(f (xy), f (x)f (y)) ≤ δ for all x, y ∈ G1 , then there is a homomorphism g : G1 → G2 with d(f (x), g(x)) ≤ ε for all x ∈ G1 ? It implies that the homomorphism is stable, i.e. if a mapping is almost a homomorphism, then there exists a true homomorphism near it. A certain formula or equation is applicable for solution of a problem. A small change in the formula or equation gives rise to a small change in the corresponding result. In this case we call this formula or equation is stable. The quadratic equation f (x + y) = f (x) + f (y) is not always true for all x, y ∈ R but it may be true approximately, that is, f (x + y) − f (x) − f (y) ≈ 0, for all x, y ∈ R. In particular, if we choose f (x) = sinx, then the above equation doesn’t hold. This can be stated mathematically as |f (x + y) − f (x) − f (y)| ≤ ε for small positive ε and for all x, y ∈ R. This is called additive Cauchy functional equation. If there is a small change on Cauchy additive functional equation, we take only small effects on its solution. This is called the Stability theory. 1

2

Hyers-Ulam Stability of Ordinary Differential Equations

Definition 1 Let E1 be a group and E2 a quasi-normed space. If the functions F, G : E1 → E2 fulfil the inequality d[F (x + y + z) + F (x − y) + F (y − z) + F (x − z), G(x) + G(y) + G(z)] ≤ h(x, y, z) and h is constant, then it is called Hyers-Ulam Stability. Rassias introduced the inequality of the type kf (x + y) − f (x) − f (y)k ≤ θ(kxkp + kykp ), θ ≥ 0 is called Hyers-Ulam-Rassias stability (generalized Hyers-Ulam stability). If the inequality kf (x + y) − f (x) − f (y)k ≤ ε, is replaced by kf (x + y) − f (x) − f (y)k ≤ ε(kxkp + kykp ), ε > 0, it is called generalized Hyers-Ulam-Rassias stability. Definition 2 An additive mapping or additive function that preserves the addition operation if f (x + y) = f (x) + f (y). Definition 3 For a given mapping f : X → Y is called additive if it satisfies f (rx + sy) =

r−s r+s f (x + y) + f (x − y) 2 2

r, s ∈ R and r 6= s. Definition 4 Let A be a Banach algebra. An algebraic left A-module X is said to be a normed (Banach) left A-module if X is a normed (Banach) space and the outer multiplication is jointly continuous, that is, if there is a nonnegative number M such that kaxk ≤ M kakkxk, a ∈ A, x ∈ X. If A has an identity e, then X is called unital if ex = x for all x ∈ X. For example, every closed left ideal I of a normed algebra A can be regarded as a Banach left A-module with the product of A giving the module multiplication. Definition 5 A C ∗ algebra is a complex algebra A of continuous linear operator on a complex Hilbert space with following two properties : 1. A is a topologically closed set in the norm topology of operators. 2.A is closed under the operation of taking adjoints of operators. Definition 6 Given two groups (G, ∗) and (H, .), a group homomorphism from (G, ∗) to (H, .) is a function h : G → H such that for all u and v in G it holds that h(u ∗ v) = h(u).h(v), where the group operation on the left hand side of the equations is that of G and on the right hand side that of H.

Introduction and Preliminaries

3

Definition 7 An additive mapping f : X → Y is called A-algebra if f (ax) = af (x) for all x ∈ X and all a ∈ A. Similarly if a ∈ C, then it is called C-linear. Theorem 1 Let (X, d) be a complete generalized metric space and let J : X → X be a strictly contractive mapping with Lipschitz constant L < 1. Then for each element x ∈ X, either d(J n x, J n+1 x) = ∞, for all nonnegative integers n or there exists a positive integer n0 such that (1) d(J n x, J n+1 x) < ∞, for all n ≥ n0 ; (2) the sequence {J n x} converges to a fixed point y ∗ of J; (3) y ∗ is the unique fixed point of J in the set Y = {y ∈ X|d(J n0 x, y) < ∞}; (4) d(y, y ∗ ) ≤

1 1−L d(y, Jy)

for all y ∈ Y .

Theorem 2 Let E be a (real or complex) linear space, E and F are Banach spaces, and ( 2, i = 0 qi = 1/2, i = 1. Suppose that the mapping f : E → F satisfies the condition f (0) = 0 and an inequality

of the
form

(Jϕ ) 2f x+y − f (x) − f (y) F ≤ ϕ(x, y), 2 for all x, y ∈ E, where ϕ : E × E → [0, ∞) is a given function. If there exists L = L(i) < 1 such that the mapping x → ψ(x) = ϕ(x, 0) has the property (Hi ) ψ(x) ≤ L.qi .ψ(x/qi ) f or all x ∈ E and the mapping ϕ has property (Hi∗ ) limn→∞ ϕ(2qin x, 2qin y) = 0 f or all x, y ∈ E then there exists a unique additive mapping j : E → F such that (Esti ) kf (x) − j(x)kF ≤

L1−i ψ(x) 1−L

for all x ∈ E.

1.1

Stability of Functional Equations on Banach Space

Ulam proposed the problem: When does a linear transformation near an “approximately linear” transformation exist?

4

Hyers-Ulam Stability of Ordinary Differential Equations

This problem was solved partially by D. H. Hyers on Banach spaces. Let E and E 0 be Banach spaces. Then a transformation f (x) from E into E 0 is called approximately linear if there exist K ≥ 0 and p(0 ≤ p < 1) such that kf (x + y) − f (x) − f (y)k ≤ K(kxkp + kykp ) f or any x, y ∈ E. If there exist two transformations f (x) and ϕ(x) from E into E 0 and K ≥ 0 and p (0 ≤ p < 1) such that kf (x) − ϕ(x)k ≤ Kkxkp , then these are called near.

1.1.1

Stability of f (x + y) = f (x) + f (y)

Theorem 3 If f (x) is an approximately linear transformation from E into E 0 then there is a linear transformation ϕ(x) near f(x) and such ϕ(x) is unique. Proof 1 Let us assume that K0 ≥ 0 and p(0 ≤ p < 1) such that kf (x + y) − f (x) − f (y)k ≤ K0 (kxkp + kykp ).

(1.1)

Putting x = y in (1.1), we get kf (2x) − 2f (x)k ≤ 2K0 (kxkp ), that is, kf (2x)/2 − f (x)k ≤ K0 kxkp .

(1.2)

We claim that kf (2n x)/2n − f (x)k ≤ K0 kxkp

n−1 X

2i(p−1) , n ∈ N.

(1.3)

i=0

Clearly n = 1 holds by (1.2). Assume that (1.3) holds for n. We shall show that (1.3) holds for n = n + 1. Replacing x by 2x in (1.3), we obtain kf (2n+1 x)/2n+1 − f (2x)k ≤ K0 k2xkp

n−1 X

2i(p−1)

i=0

= K0 kxkp 2p

n−1 X

2i(p−1) .

i=0

Therefore, kf (2n+1 x)/2n+1 − f (2x)/2k ≤ K0 kxkp 2p−1

n−1 X

2i(p−1)

i=0

= K0 kxkp

n X i=1

2i(p−1) .

Introduction and Preliminaries

5

Now, kf (2n+1 x)/2n+1 − f (x)k ≤ kf (2n+1 x)/2n+1 − f (2x)/2k + kf (2x)/2 − f (x)k n X ≤ K0 kxkp [1 + 2i(p−1) ] i=1

= K0 kxkp

n X

2i(p−1)

i=0

implies that (1.3) is true for all n. Indeed, kf (2n x)/2n − f (x)k ≤ K0 kxkp

n−1 X

2i(p−1)

i=0

≤ K0 kxkp

∞ X

2i(p−1)

i=0

≤ Kkxkp 2/2 − 2p = Kkxkp , where K = 2K0 /2 − 2p . Next, we claim that {f (2n x)/2n } is a Cauchy sequence. Therefore, 1 kf (2m x)/2m−n − f (2n x)k 2n 1 = n kf (2m−n 2n x)/2m−n − f (2n x)k 2 K ≤ n k2n xkp 2 = 2n(p−1) Kkxkp → 0(n → ∞).

kf (2m x)/2m − f (2n x)/2n k =

So, {f (2n x)/2n } is a Cauchy sequence. Since E 0 is complete, then {f (2n x)/2n } converges. Let ϕ(x) ≡ lim f (2n x)/2n . n→∞

We assert that ϕ(x) is linear. Since f (x) is approximate linear, then kf (2n (x + y)) − f (2n x) − f (2n y)k ≤ K0 (k2n xkp + k2n ykp ) ≤ 2np K0 (kxkp + kykp ), that is, kf (2n (x + y))/2n − f (2n x)/2n − f (2n y)/2n k ≤ 2n(p−1) K0 (kxkp + kykp ). Taking limit as n → ∞, we get ϕ(x + y) = ϕ(x) + ϕ(y).

6

Hyers-Ulam Stability of Ordinary Differential Equations

So, ϕ is linear. Taking limn→∞ in (1.3), we find kϕ(x) − f (x)k ≤ Kkxkp which shows that ϕ(x) is near f (x). Finally, we have to show that ϕ(x) is unique. If not, then there exists ψ(x) near f (x) such that for K 0 (≥ 0), 0 ≤ p0 < 1, we have 0

kψ(x) − f (x)k ≤ K 0 kxkp . By the triangle inequality kϕ(x) − ψ(x)k

≤ kϕ(x) − f (x)k + kψ(x) − f (x)k 0

≤ Kkxkp + K 0 kxkp . Since ϕ and ψ are linear, then kϕ(x) − ψ(x)k

= kϕ(nx) − ψ(nx)k/n 0

= (Kknxkp + K 0 knxkp )/n 0

= Kkxkp /n1−p + K 0 kxkp /n1−p

0

→ 0 as n → ∞. This completes the proof of the theorem. In the above Theorem 3, we proved the Ulam’s problem. Now we prove a generalization of the stability of approximately additive mappings in the spirit of Hyers-Ulam and Rassias. Denote (G, +) an abelian group, (X, k.k) a Banach space and ϕ : G × G → [0, ∞) be a mapping such that ϕ(x, e y) =

∞ X

2−k ϕ(2k x, 2k y) < ∞ f or all x, y ∈ G.

(1.4)

k=0

1.1.2

Stability of f (x + y) = f (x) + f (y)

Theorem 4 Let f : G → X be such that kf (x + y) − f (x) − f (y)k ≤ ϕ(x, y)

f or all x, y ∈ G.

(1.5)

Then there exists a unique mapping T : G → X such that T (x + y) = T (x) + T (y), and kf (x) − T (x)k ≤

1 ϕ(x, e x) 2

f or all x, y ∈ G.

(1.6)

f or all x ∈ G.

(1.7)

Introduction and Preliminaries

7

Proof 2 Putting x = y in (1.5), we get kf (2x) − 2f (x)k ≤ ϕ(x, x). Thus, k2−1 f (2x) − f (x)k ≤

1 ϕ(x, x) 2

f or all x ∈ G.

(1.8)

Replacing x by 2x in (1.8), we get k2−1 f (22 x) − f (2x)k ≤

1 ϕ(2x, 2x) 2

f or all x ∈ G.

(1.9)

Hence, k2−2 f (22 x) − f (x)k ≤ k2−2 f (22 x) − 2−1 f (2x)k +k2−1 f (2x) − f (x)k ≤ 2−1 k2−1 f (22 x) − f (2x)k + k2−1 f (2x) − f (x)k 1 1 ≤ 2−1 ϕ(2x, 2x) + ϕ(x, x). 2 2

(1.10)

Replacing x by 2x in (1.10), we find k2−2 f (23 x) − f (2x)k ≤

1 1 [ϕ(2x, 2x) + ϕ(22 x, 22 x)], 2 2

and therefore k2−3 f (23 x) −f (x)k ≤ k2−3 f (23 x) − 2−1 f (2x)k +k2−1 f (2x) − f (x)k 1 1 1 ≤ 2−1 [ϕ(2x, 2x) + ϕ(22 x, 22 x)] + ϕ(x, x). 2 2 2

(1.11)

Proceeding inductively, we obtain k2−n f (2n x) − f (x)k ≤

n−1 1 X −k 2 ϕ(2k x, 2k x) 2

f or all x ∈ G.

(1.12)

k=0

The process will be complete if, we show that (1.12) holds for n = n + 1. Clearly, k2−(n+1) f (2n+1 x) − f (x)k ≤ k2−(n+1) f (2n+1 x) − 2−1 f (2x)k +k2−1 f (2x) − f (x)k ≤ 2−1 k2−n f (2n+1 x) − f (2x)k + k2−1 f (2x) − f (x)k ≤ 2−1

n−1 1 X −k 1 2 ϕ(2k+1 x, 2k+1 x) + ϕ(x, x) 2 2 k=0

=

n 1X

2

k=0

2−k ϕ(2k x, 2k x).

8

Hyers-Ulam Stability of Ordinary Differential Equations

We claim that {2−n f (2n x)} is a Cauchy sequence. For n > m, we have k2−n f (2n x) − 2−m f (2m x)k = 2−m k2−(n−m) f (2n−m 2m x) − f (2m x)k ≤ 2−m

n−m−1 1 X −k 2 ϕ(2k+m x, 2k+m x) 2 k=0

≤

=

1 2

n−m−1 X

2−(k+m) ϕ(2k+m x, 2k+m x)

k=0 n−1 X

1 2−p ϕ(2p x, 2p x), 2 p=m

where p = k + m. When k = 0, p = m and k = n − m − 1, p = n − 1. Taking limit as n → ∞ in the above result and keeping m as fixed, we obtain lim k2−n f (2n x) − 2−m f (2m x)k = 0.

n→∞

Since X is a Banach space, then it follows that the sequence {2−n f (2n x)} converges in X. Let f (2n x) T (x) = lim . n→∞ 2n We claim that T satisfies (1.6). From (1.5), we have kf (2n x + 2n y) − f (2n x) − f (2n y)k ≤ ϕ(2n x, 2n y) f or all x, y ∈ G and k2−n f (2n x + 2n y) − 2−n f (2n x) − 2−n f (2n y)k ≤ 2−n ϕ(2n x, 2n y). Therefore, lim k2−n f (2n x + 2n y) − 2−n f (2n x) − 2−n f (2n y)k ≤ lim 2−n ϕ(2n x, 2n y)

n→∞

n→∞

(1.13) implies that kT (x + y) − T (x) − T (y)k = 0 due to (1.4). Taking limit as n → ∞ in (1.12), we obtain kT (x) − f (x)k ≤

1 ϕ(x, e x) 2

f or all x ∈ G.

Next, we show that T is unique. If not, then there exists F : G → X with F (x + y) = F (x) + F (y)

Introduction and Preliminaries

9

and satisfies (1.7). Then, kT (x) − F (x)k = k2−n T (2n x) − 2−n F (2n x)k ≤ k2−n T (2n x) − 2−n f (2n x)k + k2−n f (2n x) − 2−n F (2n x)k 1 1 ≤ 2−n ϕ(2 e n x, 2n x) + 2−n ϕ(2 e n x, 2n x) 2 2 = 2−n ϕ(2 e n x, 2n x) ∞ X = 2−n 2−k ϕ(2k+n x, 2k+n x) k=0

=

∞ X

2−p ϕ(2p x, 2p x).

p=n

Hence, kT (x) − F (x)k ≤

∞ X

2−p ϕ(2p x, 2p x) f or all x ∈ G.

(1.14)

p=n

Taking limit as n → ∞ in (1.14), we obtain T (x) = F (x) f or all x ∈ G. Example 1 Let G be a normed linear space and define H : R+ × R+ → R+ and ϕ0 : R+ → R+ such that ϕ0 (λ) > 0 f or all λ > 0. ϕ0 (2) < 2, ϕ0 (2λ) ≤ ϕ0 (2)ϕ0 (λ) f or all λ > 0. H(λt, λs) ≤ ϕ0 (λ)H(t, s) f or all t, s ∈ R+ , λ > 0. In Theorem 4, let ϕ(x, y) = H(kxk, kyk). Then, ϕ(2k x, 2k y) = H(2k kxk, 2k kkyk) ≤ ϕ0 (2k )H(kxk, kyk) ≤ [ϕ0 (2)ϕ0 (2)ϕ0 (2) · · · k − times]H(kxk, kyk) ≤ (ϕ0 (2))k H(kxk, kyk). But, ϕ0 (2) < 2. So, ϕ(x, e y) = ≤ =

∞ X k=0 ∞ X k=0 ∞ X

2−k ϕ(2k x, 2k y) 2−k (ϕ0 (2))k H(kxk, kyk) (ϕ0 (2)/2)k H(kxk, kyk)

k=0

=

1 H(kxk, kyk). 1 − (ϕ0 (2)/2)

10

Hyers-Ulam Stability of Ordinary Differential Equations

Therefore, kf (x) − T (x)k ≤ 12 ϕ(x, e x) for all x ∈ G implies that 1 H(kxk, kxk) 2 − ϕ0 (2) 1 ≤ ϕ0 (kxk)H(1, 1). 2 − ϕ0 (2)

kf (x) − T (x)k ≤

Remark 1 The above result generalizes that if f (tx) is continuous in t for each fixed x and T (x) = lim 2−n f (2n x), n→∞

then T is a linear mapping.

1.1.3

Stability of f (x + y) = g(x) + h(y)

Let (X, +), (Y, +) be abelian groups and f, g, h: X → Y be mappings. If f, g and h satisfy the functional equation f (x + y) − g(x) − h(y) = 0 f or all x, y ∈ X, then it is called as Pexider equation. Let (G, +) be an abelian group and (X, k.k) be a Banach space.We define ϕ : G × G → [0, ∞) be such that ε(x) =

∞ X ϕ(2j−1 x, 0) + ϕ(0, 2j−1 x) + ϕ(2j−1 x, 2j−1 x) j=1

and

2j

< ∞,

(1.15)

φ(2n x, 2n y) → 0 as n → ∞ f or all x, y ∈ G. 2n

Theorem 5 Let f, g, h: G → X be mappings satisfying the inequality kf (x + y) − g(x) − g(y)k ≤ ϕ(x, y)

(1.16)

for all x, y ∈ G. Then there exists a unique additive mapping T : G → X such that kf (x) − T (x)k ≤ kg(0)k + kh(0)k + ε(x), (1.17) kg(x) − T (x)k ≤ kg(0)k + 2kh(0)k + ϕ(x, 0) + ε(x),

(1.18)

kh(x) − T (x)k ≤ 2kg(0)k + kh(0)k + ϕ(0, x) + ε(x)

(1.19)

Proof 3 Putting x = y in (1.16), we find kf (2x) − g(x) − h(x)k ≤ ϕ(x, x)

(1.20)

for all x ∈ G. For y = 0 in (1.16), we get kf (x) − g(x) − h(0)k ≤ ϕ(x, 0) f or all x ∈ G.

(1.21)

Introduction and Preliminaries

11

From (1.21), it follows that kg(x) − f (x)k − kh(0)k ≤ ϕ(x, 0) f or all x ∈ G.

(1.22)

Let x = 0 in (1.16). Then, kf (y) − g(0) − h(y)k ≤ ϕ(0, y) f or all y ∈ G.

(1.23)

Consequently, (1.23) becomes kh(x) − f (x)k ≤ kg(0)k + ϕ(0, x) f or all x ∈ G.

(1.24)

Assume that u(x) := kh((0)k + kg(0)k + ϕ(0, x) + ϕ(x, 0) + ϕ(x, x), x ∈ G. Since, kf (2x) − 2f (x)k ≤ kf (2x) − g(x) − h(x)k + kg(x) − f (x)k +kh(x) − f (x)k, then using (1.20),(1.22) and (1.24), we obtain kf (2x) − 2f (x)k ≤ kh(0)k + kg(0)k + ϕ(0, x) + ϕ(x, 0) + ϕ(x, x) = u(x) f or all x ∈ G.

(1.25)

Replacing x by 2x in (1.25), it happens that kf (22 x) − 2f (2x)k ≤ u(2x) f or all x ∈ G

(1.26)

and hence kf (22 x) − 22 f (x)k ≤ kf (22 x) − 2f (2x)k + 2kf (2x) − 2f (x)k ≤ u(2x) + 2u(x) f or all x ∈ G.

(1.27)

We claim that kf (2n x) − 2n f (x)k ≤

n X

2j−1 u(2n−j x) f or all x ∈ G.

(1.28)

j=1

By the method of induction, (1.28) is true for n = 1 due to (1.25). Substituting 2x for x in (1.28), we obtain kf (2n+1 x) − 2n f (2x)k ≤

n X j=1

2j−1 u(2n+1−j x) f or all x ∈ G.

(1.29)

12

Hyers-Ulam Stability of Ordinary Differential Equations

Hence, using (1.29) kf (2n+1 x) − 2n+1 f (x)k ≤ kf (2n+1 x) − 2n f (2x)k + 2n kf (2x) − 2f (x)k n X ≤ 2j−1 u(2n+1−j x) + 2n u(x) j=1

=

n+1 X

2j−1 u(2n+1−j x)

(1.30)

j=1

for all x ∈ G. So, our claim is true. It follows from (1.28) that k2−n f (2n x) − f (x)k ≤

n X

2j−1−n u(2n−j x) f or all x ∈ G.

(1.31)

j=1

We show that {2−n f (2n x)} is a Cauchy sequence in X. For m < n, k2−n f (2n x) −2−m f (2m x)k = k2−n f (2n x) − 2−m f (2m x) + 2−(m+1) f (2m+1 x) −2−(m+1) f (2m+1 x) + · · · + 2−(n−1) f (2n−1 x) − 2−(n−1) f (2n−1 x)k ≤ k2−m f (2m x) − 2−(m+1) f (2m+1 x)k + · · · + 2−(n−1) f (2n−1 x) − 2−n f (2n x)k ≤

n−1 X

k2−j f (2j x) − 2−(j+1) f (2j+1 x)k

j=m

≤

n−1 X

2−(j+1) kf (2j+1 x) − 2f (2j x)k

j=m

≤

n−1 X

2−(j+1) u(2j x)

j=m

≤

=

=

+

n−1 X

u(2j x) 2j+1 j=m n−1 X

kg(0)k + kh(0)k + ϕ(2j x, 0) + ϕ(0, 2j x) + ϕ(2j x, 2j x) 2j+1 j=m n−1 X

kg(0)k + kh(0)k 2j+1 j=m n−1 X

ϕ(2j x, 0) + ϕ(0, 2j x) + ϕ(2j x, 2j x) 2j+1 j=m

= kg(0)k + kh(0)k(1/2m+1 + 1/2m+2 + · · · + 1/2n ) +

n−1 X

ϕ(2j x, 0) + ϕ(0, 2j x) + ϕ(2j x, 2j x) 2j+1 j=m

Introduction and Preliminaries = +

13

kg(0)k + kh(0)k (1 + 1/2 + 1/22 + · · · + 1/2n−m−1 ) 2m+1 n−1 X ϕ(2j x, 0) + ϕ(0, 2j x) + ϕ(2j x, 2j x) 2j+1

j=m

Since 2m+1 > 2m , then kg(0)k + kh(0)k 2m ∞ j X ϕ(2 x, 0) + ϕ(0, 2j x) + ϕ(2j x, 2j x) + 2j+1 j=1

k2−n f (2n x) − 2−m f (2m x)k ≤

(1.32)

for all x ∈ G. Taking limit as n → ∞, and keeping m as fixed, we find lim k2−n f (2n x) − 2−m f (2m x)k = 0

n→∞

for all x ∈ G. Since X is a Banach space, then it follows that the sequence {2−n f (2n x)} converges. Define T : G → X by T (x) = lim 2−n f (2n x). n→∞

We claim that T satisfies (1.17). From (1.16), we have k

f (2n x + 2n y) g(2n x) h(2n y) ϕ(2n x, 2n y) − − k ≤ 2n 2n 2n 2n

(1.33)

for all x, y ∈ G. From(1.22) we get, k

g(2n x) f (2n x) kh(0)k + ϕ(2n x, 0) − k ≤ 2n 2n 2n

(1.34)

for all x ∈ G. Since ∞ X ϕ(2n x, 0) ϕ(2j x, 0) + ϕ(0, 2j x) + ϕ(2j x, 2j x) ≤2 n 2 2j+1 j=n

(1.35)

→ 0 as n → ∞, then we obtain from (1.34) that f (2n x) g(2n x) = lim n→∞ n→∞ 2n 2n lim

(1.36)

for all x ∈ G. Therefore, from (1.24) k

h(2n x) f (2n x) kg(0)k + ϕ(0, 2n x) − k ≤ 2n 2n 2n

(1.37)

14

Hyers-Ulam Stability of Ordinary Differential Equations

for all x ∈ G and hence f (2n x) h(2n x) = lim n n→∞ n→∞ 2 2n lim

(1.38)

for all x ∈ G. Due to (1.36), (1.38), and G is commutative, we conclude that

 

f (2n x + 2n y) g(2n x) h(2n y)

lim 0 = − −

n→∞

2n 2n 2n f (2n x) f (2n y) = kT (x + y) − lim − lim k n n→∞ n→∞ 2 2n = kT (x + y) − T (x) − T (y)k f or all x, y ∈ G.

(1.39)

Taking limit as n → ∞ in (1.31), we have kT (x) − f (x)k ≤ lim

n→∞

n X

= lim

n→∞

n X

2j−1−n u(2n−j x)

j=1

2j−1−n [kg(0)k + kh(0)k + ϕ(0, 2n−j x)

j=1

+ ϕ(2n−j x, 0) + ϕ(2n−j x, 2n−j x)] n X 2j−1−n (kg(0)k + kh(0)k) = lim n→∞

j=1 n X ϕ(2j−1 x, 0) + ϕ(0, 2j−1 x) + ϕ(2j−1 x, 2j−1 x)

+ lim

n→∞

j=1

2j

1 )(kg(0)k + kh(0)k) 2n ϕ(2j−1 x, 0) + ϕ(0, 2j−1 x) + ϕ(2j−1 x, 2j−1 x) } 2j

= lim {(1 − +

n→∞ n X j=1

= kg(0)k + kh(0)k + ε(x) f or all x ∈ G.

(1.40)

Finally, we show that T is unique. If not, there exists U : G → X another such mapping with U (x + y) = U (x) + U (y) and (1.17) is satisfied. Then, kT (x) − U (x)k = k2−n T (2n x) − 2−n U (2n x)k ≤ k2−n T (2n x) − 2−n f (2n x)k + k2−n f (2n x) − 2−n U (2n x)k ≤ 2−n (kg(0)k + kh(0)k + ε(2n x)) + 2−n (kg(0)k + kh(0)k + ε(2n x))

Introduction and Preliminaries

15

kg(0)k + kh(0)k 2n−1 ∞ X ϕ(2n+j−1 x, 0) + ϕ(0, 2n+j−1 x) + ϕ(2n+j−1 x, 2n+j−1 x) +2 2n+j j=1 =

=

kg(0)k + kh(0)k 2n−1

+2

∞ X ϕ(2k−1 x, 0) + ϕ(0, 2k−1 x) + ϕ(2k−1 x, 2k−1 x) 2k

(1.41)

k=n+1

for all x ∈ G. Taking the limit in (1.41) as n → ∞, we have T (x) = U (x) due to (1.35) for all x ∈ G. This completes the proof of the theorem. Corollary 1 Let f, g, h : G → X be such that g(0) = 0, h(0) = 0, kf (x + y) − g(x) − h(y)k ≤ ϕ(x, y) for all x, y ∈ G. Then there exists a unique mapping T : G → X such that T (x + y) = T (x) + T (y), kf (x) − T (x)k ≤ ε(x), kg(x) − T (x)k ≤ ϕ(x, 0) + ε(x) and kh(x) − T (x)k ≤ ϕ(0, x) + ε(x) f or all x, y ∈ G. Proof 4 Given that kf (x + y) − g(x) − h(y)k ≤ ϕ(x, y). Putting x = 0 and y = 0, we obtain kf (y) − h(y)k ≤ ϕ(0, y), and kf (x) − g(x)k ≤ ϕ(x, 0) respectively. Clearly, kf (2n y)/2n − h(2n y)/2n k ≤ ϕ(0, 2n y)/2n . Taking limit as n → ∞ on both sides, we find lim kf (2n y)/2n − h(2n y)/2n k ≤ lim ϕ(0, 2n y)/2n = 0,

n→∞

n→∞

16

Hyers-Ulam Stability of Ordinary Differential Equations

that is, h(2n y) f (2n y) = lim n→∞ n→∞ 2n 2n n due to (1.35). By Theorem 5, the sequence {f (2 x)/2n } is a Cauchy sequence. Let f (2n x) lim = T (x). n→∞ 2n Using (1.36) and (1.39), it follows that lim

T (x + y) = T (x) + T (y) f or all x, y ∈ G. Hence T is additive. As kf (x) − T (x)k ≤ kg(0)k + kh(0)k + ε(x), then g(0) = h(0) = 0 implies that kf (x) − T (x)k ≤ ε(x), ∀ x ∈ G. Similarly, kg(x) − T (x)k ≤ ϕ(x, 0) + ε(x), ∀ x ∈ G and kh(x) − T (x)k ≤ ϕ(0, x) + ε(x), ∀ x ∈ G. T is unique due to (1.41). This complete the proof of the Corollary. Corollary 2 Let δ > 0, and let f, g, h : G → X be such that g(0) = 0, h(0) = 0, kf (x + y) − g(x) − h(y)k ≤ δ for all x, y ∈ G. Then there exists a unique mapping T : G → X such that T (x + y) = T (x) + T (y), kf (x) − T (x)k ≤ 3δ, kg(x) − T (x)k ≤ 4δ, and kh(x) − T (x)k ≤ 4δ f or all x, y ∈ G. Proof 5 Since such that

P∞

j=1

2−j ϕ(2j x, 2j y) is convergent, then we can find δ > 0 ∞ X

2−j ϕ(2j x, 2j y) ≤ δ.

j=1

Given that kf (x + y) − g(x) − h(y)k ≤ δ.

Introduction and Preliminaries

17

Putting x = 0 and y = 0, we obtain kf (y) − h(y)k ≤ δ and kf (x) − g(x)k ≤ δ respectively. Clearly, kf (2n y)/2n − h(2n y)/2n k ≤ δ/2n . Taking limit as n → ∞ on both sides, we find lim kf (2n y)/2n − h(2n y)/2n k ≤ lim δ/2n = 0,

n→∞

n→∞

that is, lim f (2n y)/2n = lim h(2n y)/2n

n→∞

n→∞

due to (1.35). By Theorem 5, the sequence {f (2n x)/2n } is a Cauchy sequence. Let f (2n x) = T (x). lim n→∞ 2n Using (1.36) and (1.39), it follows that T (x + y) = T (x) + T (y) f or all x, y ∈ G. Hence T is additive. Indeed, kf (x) − T (x)k ≤ ε(x) ∞ ∞ ∞ X X X −j j j −j j = 2 ϕ(2 x, 2 y) + 2 ϕ(2 x, 0) + 2−j ϕ(0, 2j y) j=1

j=1

j=1

= 3δ. Similarly, kg(x) − T (x)k ≤ ϕ(x, 0) + ε(x) f or all x ∈ G and kh(x) − T (x)k ≤ ϕ(0, x) + ε(x), ∀ x ∈ G. The proof of T is unique is similar as proved in Theorem 5. Corollary 3 Let V be a real or complex vector space and X a Banach space. And let F, G, H : V → X be mappings satisfying the inequality k2F (x + y/2) − G(x) − H(y)k ≤ ϕ(x, y)

(1.42)

for all x, y ∈ V. Then there exists a unique additive mapping T : V → X such that k2F (x/2) − T (x)k ≤ kG(0)k + kH(0)k + ε(x), kG(x) − T (x)k ≤ kG(0)k + 2kH(0)k + ϕ(x, 0) + ε(x)

18

Hyers-Ulam Stability of Ordinary Differential Equations

and kH(x) − T (x)k ≤ 2kG(0)k + kH(0)kϕ(0, x) + ε(x) f or all x, y ∈ V. Proof 6 Putting y = 0 and x = 0 in(1.42), we obtain k2F (x/2) − G(x) − H(0)k ≤ ϕ(x, 0), k2F (y/2) − G(0) − H(y)k ≤ ϕ(0, y). Let 2F (x/2) = f (x). Then 2F (x + y/2) = f (x + y) and hence kf (x) − G(x) − H(0)k ≤ ϕ(x, 0). Consequently, kf (2n x)/2n − G(2n x)/2n k ≤ [kH(0)k + ϕ(2n x, 0)]/2n . Taking limit as n → ∞ to the above relation, it follows that lim f (2n x)/2n = lim G(2n x)/2n

n→∞

n→∞

and similarly, lim f (2n y)/2n = lim H(2n y)/2n f or all x, y ∈ V

n→∞

n→∞

due to (1.35). Using the fact that limn→∞ f (2n x/2n ) = T (x), it happens that 0 = lim kf (2n x + 2n y)/2n − G(2n x)/2n − H(2n y)/2n k n→∞

= T (x + y) − T (x) − T (y). Following (1.40), it concludes that k2F (x/2) − T (x)k ≤ kG(0)k + kH(0)k + ε(x). Similarly, kG(x) − T (x)k ≤ kG(0)k + 2kH(0)k + ϕ(x, 0) + ε(x) and kH(x) − T (x)k ≤ 2kG(0)k + kH(0)k + ϕ(0, x) + ε(x) because of (1.18) and (1.19), respectively. Uniqueness of T is similar as proved in Theorem 5. This complete the proof of the Corollary. Corollary 4 Let V be a real or complex vector space and X a Banach space. Let a 6= 0, b 6= 0 be two real or complex numbers. And let F, G, H : V → X be mappings satisfying the inequality kF (ax + by) − aF (x) − bF (y)k ≤ ϕ(x, y)

Introduction and Preliminaries

19

for all x, y ∈ V. Then there exists a unique additive mapping T : V → X such that kF (x) − T (x)k ≤ (|a| + |b|)kF (0)k + ε(x), kaF (x/a) − T (x)k ≤ (|a| + 2|b|)kF (0)k + ϕ(x, 0) + ε(x),

and kbF (x/b) − T (x)k ≤ (2|a| + |b|)kF (0)k + ϕ(0, x) + ε(x) for all x, y ∈ V . Proof 7 Given that kF (ax + by) − aF (x) − bF (y)k ≤ ϕ(x, y) for all x, y ∈ V. Let ax = m, by = n and aF (m/a) = g(m/a), bF (n/b) = h(n/b). Then the last inequality becomes kF (m + n) − g(m/a) − h(n/b)k ≤ ϕ(m/a, n/b). When m = 0, we obtain kF (n) − g(0) − h(n/b)k ≤ ϕ(0, n/b) and when n = 0, we obtain kF (m) − g(m/a) − h(0)k ≤ ϕ(m/a, 0). Proceeding as in Theorem 5, it follows that kF (m/a) − T (m/a)k ≤ kg(0)k + kh(0)k + ε(x/a). Hence, kF (x) − T (x)k ≤ |a|kF (0)k + |b|kF (0)k + ε(x). Moreover, kaF (x/a) − T (x)k ≤ (|a| + 2|b|)kF (0)k + ϕ(x, 0) + ε(x), and kbF (x/b) − T (x)k ≤ (2|a| + |b|)kF (0)k + ϕ(0, x) + ε(x) f or all x, y ∈ V. Corollary 5 Consider E1 , E2 to be two Banach spaces, and let f, g, h : E1 → E2 be mappings. Assume that there exists θ ≥ 0 and p ∈ [0, 1) such that kf (x + y) − g(x) − h(y)k ≤ θ(kxkp + kykp )

20

Hyers-Ulam Stability of Ordinary Differential Equations

for all x, y ∈ E1 .Then there exists a unique linear mapping T : E1 → E2 such that 4θ kxkp , 2 − 2p 6 − 2p kg(x) − T (x)k ≤ kg(0)k + 2kh(0)k + kxkp , 2 − 2p 6 − 2p kh(x) − T (x)k ≤ 2kg(0)k + kh(0)k + kxkp 2 − 2p

kf (x) − T (x)k ≤ kg(0)k + kh(0)k +

for all x ∈ E1 . Corollary 6 Let 0 ≤ p + q < 1, where p and q are the non-negative real numbers, and let f, g, h : G → X be such that g(0) = 0, h(0) = 0, kf (x + y) − g(x) − h(y)k ≤ θkxkp kykq for all x, y ∈ G.Then there exists a unique mapping T : G → X such that T (x + y) = T (x) + T (y), θ kf (x) − T (x)k ≤ kxkp+q , 2 − 2p+q θ kg(x) − T (x)k ≤ kxkp+q , 2 − 2p+q

kh(x) − T (x)k ≤

θ kxkp+q 2 − 2p+q

for all x, y ∈ G.

1.1.4

Stability of f (xy) + f (x + y) = f (xy + x) + f (y)

In this section, we denote by F a ring with the unit element 1 and by E a Banach space. Let ϕ : F × F → [0, ∞) be such that ∞ X

2−n ϕ(2n−1 x, 2n−1 y + z) < ∞, f or all x, y, z ∈ F.

(1.43)

n=1

Hyers-Ulam-Rassias stability of the Davison functional equation f (xy) + f (x + y) = f (xy + x) + f (y) for a class of functions from a ring to a Banach space and also, we investigate the Davison equation of Pexider type. Theorem 6 If a function f : F → E satisfies the inequality kf (xy) + f (x + y) − f (xy + x) − f (y)k ≤ ϕ(x, y)

(1.44)

Introduction and Preliminaries

21

for all x, y ∈ F , then there exists a unique additive function A : F → E such that ∞ X M (2n x) kf (6x) − A(x) − f (0)k ≤ , ∀ x ∈ F, (1.45) 2n n=0 where 1 [ϕ(4x, −4x) + ϕ(4x, −4x + 1) + ϕ(8x, −2x) + ϕ(3x, 0) 2 + ϕ(3x, 1) + ϕ(6x, 0) + ϕ(7x, −x) + ϕ(7x, −4x + 1) + ϕ(14x, −2x)].

M (x) =

Proof 8 If we replace y by y + 1 in (1.44), we get kf (xy + x) + f (x + y + 1) − f (xy + 2x) − f (y + 1)k ≤ ϕ(x, y + 1)

(1.46)

for any x, y ∈ F . Hence, kf (xy) + f (x + y) + f (x + y + 1) − f (y) − f (xy + 2x) − f (y + 1)k ≤ kf (xy) + f (x + y) − f (xy + x) − f (y)k +kf (xy + x) + f (x + y + 1) − f (xy + 2x) − f (y + 1)k ≤ ϕ(x, y) + ϕ(x, y + 1), f or all x, y ∈ F due to (1.44) and (1.46). Replacing y by 4y in the last inequality, we obtain kf (4xy) + f (x + 4y) + f (x + 4y + 1) − f (4y) − f (4xy + 2x) − f (4y + 1)k ≤ ϕ(x, 4y) + ϕ(x, 4y + 1), f or all x, y ∈ F. Therefore, kf (x + 4y) + f (x + 4y + 1) − f (2x + 2y) − f (4y) − f (4y + 1) + f (2y)k ≤ kf (4xy) + f (x + 4y) + f (x + 4y + 1) − f (4y) − f (4xy + 2x) −f (4y + 1)k + kf (4xy) + f (2x + 2y) − f (4xy + 2x) − f (2y)k ≤ ϕ(x, 4y) + ϕ(x, 4y + 1) + ϕ(2x, 2y) f or all x, y ∈ F because of (1.44). If we replace x by x − y in the last inequality, we find kf (x + 3y) + f (x + 3y + 1) − f (2x) − f (4y) − f (4y + 1) + f (2y)k ≤ ϕ(x − y, 4y) + ϕ(x − y, 4y + 1) + ϕ(2x − 2y, 2y) for all x, y ∈ F . Consequently, kf (3x + 3y) + f (3x + 3y + 1) − f (6x) − f (4y) −f (4y + 1) + f (2y)k ≤ ϕ(3x − y, 4y) + ϕ(3x − y, 4y + 1) + ϕ(6x − 2y, 2y)

(1.47)

22

Hyers-Ulam Stability of Ordinary Differential Equations

for any x, y ∈ F and substituting 3x for x.If we replace y by −x in (1.47), it yields that kf (0) + f (1) − f (6x) − f (−4x) − f (−4x + 1) + f (−2x)k ≤ ϕ(4x, −4x) + ϕ(4x, −4x + 1) + ϕ(8x, −2x)

(1.48)

for every x ∈ F . If we replace y by 0 in (1.47), we have kf (3x) + f (3x + 1) − f (6x) − f (1)k ≤ ϕ(3x, 0) + ϕ(3x, 1) + ϕ(6x, 0) (1.49) for every x ∈ F . If we replace x, y by 2x, −x respectively in (1.47), we have kf (3x) + f (3x + 1) − f (12x) − f (−4x) − f (−4x + 1) + f (−2x)k ≤ ϕ(7x, −4x) + ϕ(7x, −4x + 1) + ϕ(14x, −2x)

(1.50)

for every x ∈ F . From (1.48),(1.49) and (1.50) it follows that f (12x) − f (0) − (f (6x) − f (0))k 2 1 = k2f (6x) − f (12x) − f (0)k 2 1 ≤ kf (0) + f (1) − f (6x) − f (−4x) − f (−4x + 1) + f (−2x)k 2 + kf (3x) + f (3x + 1) − f (6x) − f (1)k k

 + kf (3x) + f (3x + 1) − f (12x) − f (−4x) − f (−4x + 1) + f (−2x)k  1 ≤ ϕ(4x, −4x) + ϕ(4x, −4x + 1) + ϕ(8x, −2x) + ϕ(3x, 0) 2  + ϕ(3x, 1) + ϕ(6x, 0) + ϕ(7x, −x) + ϕ(7x, −4x + 1) + ϕ(14x, −2x) = M (x) f or every x ∈ F

(1.51)

implies that k

f (2n .6x) − f (0) f (2n−1 .6x) − f (0) M (2n−1 x) − k ≤ 2n 2n−1 2n−1

for all n ∈ N and x ∈ F . Replacing x by 2x in (1.51), we find kf (2.6x) − f (0) −

f (22 .6x) − f (0) k ≤ M (2x). 2

Hence, k[f (2.6x) − f (0)]/2 −

f (22 .6x) − f (0) k ≤ M (2x)/2 22

Introduction and Preliminaries

23

implies that kf (6x) − f (0) −

f (22 .6x) − f (0) f (2.6x) − f (0) k − kf (6x) − f (0) − k 22 2 M (2x) . ≤ 2

Consequently, kf (6x) − f (0) −

M (2x) f (22 .6x) − f (0) k≤ + M (x) 22 2 2 X M (2i−1 x) . = 2i−1 i=1

Further, kf (6x) − f (0) −

f (23 .6x) − f (0) f (22 .6x) − f (0) k ≤ kf (6x) − f (0) − k 3 2 22 f (22 .6x) − f (0) f (23 .6x) − f (0) +k − k 22 23 2 3 X M (2i−1 x) M (22 x) X M (2i−1 x) = + = 2i−1 22 2i−1 i=1 i=1

implies that n

kf (6x) − f (0) −

X M (2i−1 x) f (2n .6x) − f (0) k ≤ 2n 2i−1 i=1

(1.52)

inductively, for all n ∈ N and x ∈ F . Now, k

f (2n .6x) − f (0) f (2m+n .6x) − f (0) − k 2n 2m+n n f (2 .6x) − f (0) f (2n+1 .6x) − f (0) ≤k − k 2n 2n+1 f (2n+1 .6x) − f (0) f (2n+2 .6x) − f (0) +k − k 2n+1 2n+2 f (2m+n−1 .6x) − f (0) f (2m+n .6x) − f (0) +··· + k − k 2m+n−1 2m+n M (2n x) M (2m+n x) = + ··· + n 2 2m+n m+n X M (2i−1 x) 0 ∞ X

2−n ϕ(2n−1 x, 2n−1 y + z) ≤ ε.

n=1

Hence, kf (xy) + f (x + y) − f (xy + x) − f (y)k ≤ ε. From Theorem 6, we know that kf (6x) − A(x) − f (0)k ≤

=

∞ X n=0

∞ X M (2n x) 2n n=0

 ϕ(2n .4x, −2n .4x) + ϕ(2n .4x, −2n .4x + 1) + ϕ(2n .8x, −2n .2x) n+1 1

2

+ ϕ(3.2n x, 0) + ϕ(2n .3x, 1) + ϕ(2n .6x, 0) + ϕ(2n .7x, −2n .x) + ϕ(2n .7x, −2n .4x + 1) + ϕ(2n .14x, −2n .2x) = 9ε f or all x ∈ F. This completes the proof of the Corollary.



26

Hyers-Ulam Stability of Ordinary Differential Equations

Corollary 8 Let F be a normed algebra and let 0 ≤ p < 1. If a function f : F → E satisfies the inequality kf (xy) + f (x + y) − f (xy + x) − f (y)k ≤ θ(kxkp + kykp ) for all x, y ∈ F , then there exists a unique additive function A : F → E such that kf (6x) − A(x) − f (0)k ≤

1 p 2−2p θ[(2.2 p p

+ 2.3p + 5.4p + 6p

+2.7 + 8 + 14p + 1)kxkp ] + 3.k1kp for all x ∈ F . Now we prove the stability of Davison functional equation of Pexider type. Let ϕ : F × F → [0, ∞) be a function and let ψ : F × F → [0, ∞) be a function defined by ψ(x, y) = ϕ(x, y) + ϕ(xy + x, 0) + ϕ(0, y) + ϕ(1, xy) + ϕ(xy + 1, 0) + ϕ(0, xy) for all x, y ∈ F . Theorem 7 If the function f, g, h, k : F → E satisfy the inequality kf (xy) + g(x + y) − h(xy + x) − k(y)k ≤ ϕ(x, y)

(1.58)

for all x, y ∈ F and ψ is a function such that ∞ X

2−n ψ(2n−1 x, 2n−1 y + z) < ∞

n=1

for all x, y, z ∈ F , then there exists a unique additive function A : F → E such that ∞ X M (2n x) kf (6x) − A(x) − f (0)k ≤ + ϕ(1, 6x) (1.59) 2n n=0 + ϕ(6x + 1, 0) + ϕ(0, 6x) + ϕ(0, 0) kg(6x) − A(x) − g(0)k ≤

∞ X M (2n x) 2n n=0

(1.60)

kh(6x) − A(x) − h(0)k ≤

∞ X M (2n x) + ϕ(0, 0) + ϕ(6x, 0) 2n n=0

(1.61)

kk(6x) − A(x) − k(0)k ≤

∞ X M (2n x) + ϕ(0, 0) + ϕ(0, 6x) 2n n=0

(1.62)

Introduction and Preliminaries

27

for all x ∈ F , where 1 [4ϕ(0, 0) + ϕ(0, 1) + ϕ(0, −x) + ϕ(0, −2x) 2 + ϕ(0, −2x) + ϕ(0, 3x) + ϕ(0, −4x) + 2ϕ(0, −4x + 1)

M (x) =

+ ϕ(0, −7x2 ) + 2ϕ(0, −16x2 ) + ϕ(0, −16x2 + 4x) + ϕ(0, −28x2 ) + ϕ(0, −28x2 + 7x) + 4ϕ(1, 0) + ϕ(1, 3x) + ϕ(1, −7x2 ) + 2ϕ(1, −16x2 ) + ϕ(1, −16x2 + 4x) + ϕ(1, −28x2 ) + ϕ(1, −28x2 + 7x) + 2ϕ(3x, 0) + ϕ(3x, 1) + ϕ(3x + 1, 0) + ϕ(4x, −4x) + ϕ(4x, −4x + 1) + 3ϕ(6x, 0) + ϕ(7x, −x) + ϕ(7x, −4x + 1) + ϕ(8x, −2x) + ϕ(14x, −2x) + ϕ(−7x2 + 1, 0) + ϕ(−7x2 + 7x, 0) + 2ϕ(−16x2 + 4x + 1, 0) + ϕ(−16x2 + 4x, 0) + ϕ(−16x2 + 4x + 1, 0) + 2ϕ(−16x2 + 8x, 0) + ϕ(−28x2 + 1, 0) + ϕ(−28x2 + 7x + 1, 0) + 2ϕ(−28x2 + 14x, 0)]. Proof 10 Assume that the functions f, g, h, k : F → E satisfy (1.58). Putting y = 0, x = 0 and x = 1 separately in (1.58), we find kf (0) + g(x) − h(x) − k(0)k ≤ ϕ(x, 0), kf (0) + g(y) − h(0) − k(y)k ≤ ϕ(0, y)

(1.63)

and kf (y) + g(y + 1) − h(y + 1) − k(y)k ≤ ϕ(1, y). Now, kf (y) − g(y) + h(0) + k(0) − 2f (0)k ≤ kf (y) + g(y + 1) − h(y + 1) − k(y)k + k − g(y + 1) + h(y + 1) + k(0) − f (0)k + k − g(y) + k(y) + h(0) − f (0)k ≤ ϕ(1, y) + ϕ(y + 1, 0) + ϕ(0, y)

(1.64)

due to (1.63). Hence, kg(xy) + g(x + y) − g(xy + x) − g(y)k ≤ kf (xy) + g(x + y) − h(xy + x) − k(y)k + k − f (0) − g(xy + x) + h(xy + x) + k(0)k + k − f (0) − g(y) + h(0) + k(y)k + k − f (xy) + g(xy) − h(0) − k(0) + 2f (0)k ≤ ϕ(x, y) + ϕ(xy + x, 0) + ϕ(0, y) + ϕ(1, xy) + ϕ(xy + 1, 0) + ϕ(0, xy)

(1.65)

28

Hyers-Ulam Stability of Ordinary Differential Equations

due to (1.58),(1.63) and (1.64). Furthermore, if we replace y by y+1 in (1.65), we get kg(xy + x) + g(x + y + 1) − g(xy + 2x) − g(y + 1)k ≤ ϕ(x, y + 1) + ϕ(xy + 2x, 0) + ϕ(0, y + 1) + ϕ(1, xy + x) + ϕ(xy + x + 1, 0) + ϕ(0, xy + x) kg(xy) + g(x + y) + g(x + y + 1) − g(y) − g(xy + 2x) − g(y + 1)k ≤ kg(xy) + g(x + y) − g(xy + x) − g(y)k + kg(xy + x) + g(x + y + 1) − g(xy + 2x) − g(y + 1)k ≤ ϕ(x, y) + ϕ(xy + x, 0) + ϕ(0, y) + ϕ(1, xy) + ϕ(xy + 1, 0) + ϕ(0, xy) + ϕ(x, y + 1) + ϕ(xy + 2x, 0) + ϕ(0, y + 1) + ϕ(1, xy + x) + ϕ(xy + x + 1, 0) + ϕ(0, xy + x) (1.66) for all x, y ∈ F . Lastly, if we replace y by 4y in (1.65), we get kg(4xy) + g(x + 4y) + g(x + 4y + 1) − g(4y) − g(4xy + 2x) − g(4y + 1)k ≤ ϕ(x, 4y) + ϕ(4xy + x, 0) + ϕ(0, 4y) + ϕ(1, 4xy) + ϕ(4xy + 1, 0) + ϕ(0, 4xy) + ϕ(x, 4y + 1) + ϕ(4xy + 2x, 0) + ϕ(0, 4y + 1) + ϕ(1, 4xy + x) + ϕ(4xy + x + 1, 0) + ϕ(0, 4xy + x) for all x, y ∈ F . Using (1.65) and the last relation we find, kg(x + 4y) + g(x + 4y + 1) − g(2x + 2y) − g(4y) − g(4y + 1) + g(2y)k ≤ kg(4xy) + g(x + 4y) + g(x + 4y + 1) − g(4y) − g(4xy + 2x) − g(4y + 1)k + kg(4xy) + g(2x + 2y) − g(4xy + 2x) − g(2y)k ≤ ϕ(x, 4y) + ϕ(4xy + x, 0) + ϕ(0, 4y) + ϕ(1, 4xy) + ϕ(4xy + 1, 0) + ϕ(0, 4xy) + ϕ(x, 4y + 1) + ϕ(4xy + 2x, 0) + ϕ(0, 4y + 1) + ϕ(1, 4xy + x) + ϕ(4xy + x + 1, 0) + ϕ(0, 4xy + x) + ϕ(2x, 2y) + ϕ(4xy + 2x, 0) + ϕ(0, 2y) + ϕ(1, 4xy) + ϕ(4xy + 1, 0) + ϕ(0, 4xy). Proceeding as in the proof of Theorem 6, we find g(12x) − g(0) − (g(6x) − g(0))k 2 1 = k2g(6x) − g(12x) − g(0)k 2 1 ≤ [kg(0) + g(1) − g(6x) − g(−4x) − g(−4x + 1) + g(−2x)k 2 + kg(3x) + g(3x + 1) − g(6x) − g(1)k k

+ kg(3x) + g(3x + 1) − g(12x) − g(−4x) − g(−4x + 1) + g(−2x)k] 1 ≤ [4ϕ(0, 0) + ϕ(0, 1) + ϕ(0, −x) + ϕ(0, −2x) 2

Introduction and Preliminaries

29

+ ϕ(0, −2x) + ϕ(0, 3x) + ϕ(0, −4x) + 2ϕ(0, −4x + 1) + ϕ(0, −7x2 ) + 2ϕ(0, −16x2 ) + ϕ(0, −16x2 + 4x) + ϕ(0, −28x2 ) + ϕ(0, −28x2 + 7x) + 4ϕ(1, 0) + ϕ(1, 3x) + ϕ(1, −7x2 ) + 2ϕ(1, −16x2 ) + ϕ(1, −16x2 + 4x) + ϕ(1, −28x2 ) + ϕ(1, −28x2 + 7x) + 2ϕ(3x, 0) + ϕ(3x, 1) + ϕ(3x + 1, 0) + ϕ(4x, −4x) + ϕ(4x, −4x + 1) + 3ϕ(6x, 0) + ϕ(7x, −x) + ϕ(7x, −4x + 1) + ϕ(8x, −2x) + ϕ(14x, −2x) + ϕ(−7x2 + 1, 0) + ϕ(−7x2 + 7x, 0) + 2ϕ((−16x2 + 1, 0) + ϕ(−16x2 + 4x, 0) + ϕ(−16x2 + 4x + 1, 0) + 2ϕ(−16x2 + 8x, 0) + ϕ(−28x2 + 1, 0) + ϕ(−28x2 + 7x + 1, 0) + 2ϕ(−28x2 + 14x, 0) = M (x).

(1.67)

Using 2n−1 x for x in (1.67) and then dividing by 2n−1 , the resulting inequality becomes k

g(2n .6x) − g(0) g(2n−1 .6x) − g(0) M (2n−1 x) − k ≤ , 2n 2n−1 2n−1 n

kg(6x) − g(0) − and thus { g(2 by

n

.6x)−g(0) } 2n

X M (2i−1 x) g(2n .6x) − g(0) k ≤ 2n 2i−1 i=1

(1.68)

is a Cauchy sequence for x ∈ F . If A : F → E defined g(2n .6x) − g(0) , n→∞ 2n

A(x) = lim then

kg(6x) − A(x) − g(0)k ≤

∞ X M (2n x) . 2n n=0

From (1.60) and (1.63), we have kh(6x) − h(0) − A(x)k ≤ kh(6x) − g(6x) − f (0) + k(0)k + kg(6x) − A(x) − g(0)k + kf (0) + g(0) − h(0) − k(0)k ∞ X M (2n x) ≤ + ϕ(0, 0) + ϕ(6x, 0) 2n n=0 and kk(6x) − A(x) − k(0)k ≤ kk(6x) − g(x) − f (0) + h(0)k + kg(6x) − A(x) − g(0)k + kg(0) + f (0) − h(0) − k(0)k ∞ X M (2n x) ≤ + ϕ(0, 0) + ϕ(0, 6x) 2n n=0 for all x ∈ F . This completes the proof of the theorem.

30

Hyers-Ulam Stability of Ordinary Differential Equations

Corollary 9 If a function f : F → E satisfies the inequality kf (xy) + g(x + y) − h(xy + x) − k(y)k ≤ ε for all x, y ∈ F , then there exists a unique additive function A : F → E such that kf (6x) − A(x) − f (0)k ≤ 57ε for all x ∈ F .

1.1.5

Stability of f (x + iy) + f (x − iy) = 2f (x) − 2f (y)

Let X, Y be complex vector spaces. It is shown that if a mapping f : X → Y satisfies f (x + iy) + f (x − iy) = 2f (x) − 2f (y) (1.69) or f (x + iy) − f (ix + y) = 2f (x) − 2f (y)

(1.70)

for all x, y ∈ X, then the mapping f : X → Y satisfies f (x + y) + f (x − y) = 2f (x) + 2f (y) f or all x, y ∈ X. Again, we prove the generalized Hyers-Ulam stability of the functional equations (1.69) and (1.70) in complex Banach spaces. In this section, we solve the functional equations (1.69) and (1.70) and prove the generalized Hyers-Ulam stability by using the fixed point method. Here, we assume that X and Y are complex vector spaces. Proposition 1 If a mapping f : X → Y satisfies f (x + iy) + f (x − iy) = 2f (x) − 2f (y) f or all x, y ∈ X,

(1.71)

then the mapping f : X → Y is quadratic, that is, f (x + y) + f (x − y) = 2f (x) + 2f (y) holds for all x ∈ X.If a mapping f : X → Y is quadratic and f (ix) = −f (x) holds for all x ∈ X, then the mapping f : X → Y satisfies (1.71). Proof 11 Suppose that f : X → Y satisfies (1.71). Putting x = y in (1.69), we find f ((1 + i)x) + f ((1 − i)x) = 0 f or all x ∈ X. Replacing x by (1 + i)x in the above result, we find f ((1 + i)(1 + i)) + f ((1 − i)(1 + i)x) = 0 implies that f (2ix) + f (2x) = 0 f or all x ∈ X.

Introduction and Preliminaries

31

Replacing x by x/2 in the above result, we find f (ix) + f (x) = 0 f or all x ∈ X. Then f (ix) = −f (x). Therefore, f (x + iy) + f (x − iy) = 2f (x) − 2f (y) = 2f (x) + 2f (iy) f or all x, y ∈ X.

(1.72)

Putting z = iy in (1.72), we get f (x + z) + f (x − z) = 2f (x) + 2f (z) f or all x, z ∈ X. Conversely assume that a quadratic mapping f : X → Y satisfies f (ix) = −f (x) for all x ∈ X. Then f (x + iy) + f (x − iy) = 2f (x) + 2f (iy) = 2f (x) − 2f (y) for all x, y ∈ X. So the mapping f : X → Y satisfies (1.72). Proposition 2 If a mapping f : X → Y satisfies f (0) = 0 and f (x + iy) − f (ix + y) = 2f (x) − 2f (y)

(1.73)

for all x, y ∈ X, then the mapping f : X → Y is quadratic. If a mapping f : X → Y is quadratic and f (ix) = −f (x) holds for all x ∈ X, then the mapping f : X → Y satisfies f (x + iy) − f (ix + y) = 2f (x) − 2f (y). Proof 12 Assuming that f : X → Y satisfies the equation (1.73). Putting y = 0 in (1.73), we find f (x) − f (ix) = 2f (x) f or all x ∈ X and hence f (ix) = −f (x) f or all x ∈ X. Therefore, f (x + iy) + f (x − iy) = f (x + iy) − f (ix + y) = 2f (x) − 2f (y) = 2f (x) − 2f (iy) f or all x, y ∈ X.

(1.74)

Putting z = iy in (1.74), we get f (x + z) + f (x − z) = 2f (x) + 2f (z) f or all x, z ∈ X. Suppose that a quadratic mapping f : X → Y satisfies f (ix) = −f (x) for all x ∈ X. Then f (x + iy) − f (ix + y) = f (x + iy) + f (x − iy) = 2f (x) + 2f (iy) = 2f (x) − 2f (y) for all x, y ∈ X. Hence, the mapping f : X → Y satisfies f (x+iy)−f (ix+y) = 2f (x) − 2f (y).

32

Hyers-Ulam Stability of Ordinary Differential Equations

In the following we assume that X is a normed vector space and Y is a Banach space. For a given mapping f : X → Y , we define Cf (x, y) = f (x + iy) + f (x − iy) − 2f (x) + 2f (y), ∀ x, y ∈ X. Theorem 8 Let p < 2, and θ be positive numbers, and f : X → Y be a mapping satisfying f (ix) = −f (x) and kCf (x, y)k ≤ θ(kxkp + kykp )

(1.75)

for all x, y ∈ X. Then there exists a unique quadratic mapping Q : X → Y such that 2θ kxkp f or all x, y ∈ X. (1.76) kf (x) − Q(x)k ≤ 4 − 2p Proof 13 Since f (ix) = −f (x) for all x ∈ X, then f (0) = 0. So, f (−x) = f (i2 x) = −f (ix) = f (x) for all x ∈ X. Putting y = −ix in (1.75), we find kf (x + x) + f (x − x) − 2f (x) − 2f (−ix)k ≤ θ(kxkp + k − ixkp ), that is, kf (2x) − 4f (x)k ≤ 2θkxkp

(1.77)

for all x ∈ X. Hence, 1 θ kf (x) − f (2x)k ≤ kxkp 4 2 for all x ∈ X. So, k

1 1 1 1 1 f (2n x) − m f (2m x)k = k n f (2n x) − m f (2m x) + m+1 f (2m+1 x) n 4 4 4 4 4 1 1 1 m+1 n−1 − m+1 f (2 x) + · · · − n−1 f (2 x) + n−1 f (2n−1 x)k 4 4 4 1 1 m m+1 x)k + · · · ≤ k m f (2 x) − m+1 f (2 4 4 1 1 +k n−1 f (2n−1 x) − n f (2n x)k 4 4 n−1 n−1 X 1 X 1 1 1 = k j f (2j x) − j+1 f (2j+1 x)k = kf (2j x) − f (2j+1 x)k j 4 4 4 4 j=m j=m n−1 X 2pj θ 1 θ j p k2 xk = . kxkp j 2j+1 4 2 2 j=m j=m  pm  p(n−1) 2 θ 2 θ = + · · · + 2(n−1)+1 kxkp 22m+1 2   pm 2 θ = 2m+1 1 + 2p /22 + · · · + 2p(n−m−1) /22(n−m−1) kxkp 2 2pm 4θkxkp 1 − (2p /22 )n−m = 22m+1 4 − 2p

≤

n−1 X

(1.78)

Introduction and Preliminaries

33

for all nonnegative integers m and n with n > m. Taking limit as n → ∞ and m as fixed in (1.78), the right side converges due to p < 2 and hence { 41n f (2n x)} is a Cauchy sequence for all x ∈ X. Let there be a mapping Q : X → Y such that 1 f (2n x), ∀ x ∈ X. n→∞ 4n

Q(x) = lim From (1.75), it follows that

kCf (2n x, 2n y)k

≤ θ(k2n xk + k2n ykp ) = θ2np (kxkp + kykp )

and

2pn θ 1 n n Cf (2 x, 2 y)k ≤ (kxkp + kykp ), 4n 4n 1 kCQ(x, y)k = lim n kCf (2n x, 2n y)k, ∀ x, y ∈ X n→∞ 4 k

implies that 2pn θ (kxkp + kykp ) n→∞ 4n = 0, ∀ x, y ∈ X.

kCQ(x, y)k ≤ lim

So, CQ(x, y) = 0. By Proposition 1, the mapping Q : X → Y is quadratic. Putting m = 0 and taking limit as n → ∞ in (1.78), we get lim kf (x) −

n→∞

n−1 X 2pj θ 1 n f (2 x)k ≤ lim kxkp , n 2j+1 n→∞ 4 2 j=0

that is, ∞ X 2pj θ kf (x) − Q(x)k ≤ kxkp 2j+1 2 j=0

θ (1 + 2p /22 + · · · )kxkp 2 4θkxkp 2θ = = kxkp . 2(4 − 2p ) 4 − 2p

=

Next, to show that Q is unique. If not, there exists T : X → Y such that T (x + iy) + T (x − iy) = 2T (x) − 2T (y) and kf (x) − T (x)k ≤

2θ kxkp , ∀ x, y ∈ X. 4 − 2p

34

Hyers-Ulam Stability of Ordinary Differential Equations

Since Q is quadratic, then by definition (Proposition 1) 4Q(x) = Q(2x) and hence by the induction method 4n Q(x) = Q(2n x), for all x ∈ X. Hence, 1 kQ(2n x) − T (2n x)k 4n 1 ≤ n (kQ(2n x) − f (2n x)k + kT (2n x) − f (2n x)k 4 1 2θ2pn 4θ 2pn ≤ n 2. kxkp = kxkp p 4 4−2 4 − 2p 4n

kQ(x) − T (x)k =

implies that limn→∞ kQ(x) − T (x)k = 0 f or all x ∈ X. This completes the proof of the theorem. Theorem 9 Let p > 2, and θ be positive numbers, and f : X → Y be a mapping satisfying (1.75) and f (ix) = −f (x) for all x, y ∈ X. Then there exists a unique quadratic mapping Q : X → Y such that kf (x) − Q(x)k ≤

2θ kxkp , ∀ x, y ∈ X. −4

2p

(1.79)

Proof 14 Given that kCf (x, y)k ≤ θ(kxkp + kykp ), that is, kf (x + iy) + f (x − iy) − 2f (x) + 2f (y)k ≤ θ(kxkp + kykp ). Putting y = −ix, in the last inequality, we find kf (2x) − 4f (x)k ≤ 2θkxkp .

(1.80)

Replacing x by x/2 in (1.80), we get kf (x) − 4f (x/2)k ≤

2θ kxkp , f or x ∈ G. 2p

Hence, k4n f (x/2n ) − 4m f (x/2m )k = k4n f (x/2n ) − 4m f (x/2m ) + 4m+1 f (x/2m+1 ) − 4m+1 f (x/2m+1 ) + · · · + 4n−1 f (x/2n−1 ) − 4n−1 f (x/2n−1 )k ≤ k4m f (x/2m ) − 4m+1 f (x/2m+1 )k + · · · + k4n−1 f (x/2n−1 ) − 4n f (x/2n )k ≤

n−1 X j=m

k4j f (x/2j ) − 4j+1 f (x/2j+1 )k

Introduction and Preliminaries =

n−1 X

35

4j kf (x/2j ) − 4f (x/2j+1 )k

j=m

=

n−1 X

n−1 X 2.4j θ 4j 2θ p kxk kxkp = p 2pj pj+p 2 2 j=m j=m

= 2θkxkp (4m /2p(m+1) + · · · + 4n−1 /2pn ) 1 − (1/2p−2 )n−m = 2θ2p kxkp 4m /2p(m+1) 2p − 4

(1.81)

for all non-negative integers n and m with n > m. Taking limit n → ∞ and keeping m as fixed in (1.81), it follows that {4n f (x/2n )} is a Cauchy sequence for all x ∈ X. Let there be a mapping Q : X → Y such that Q(x) = lim 4n f (x/2n ) f or all x ∈ X. n→∞

From (1.75), it follows that 4n θ (kxkp + kykp ) = 0, n→∞ 2pn

kCQ(x, y)k = lim 4n kCf (x/2n , y/2n )k ≤ lim n→∞

for all x, y ∈ X. So CQ(x, y) = 0. Hence, Q : X → Y is quadratic. Taking m = 0 and n → ∞ in (1.81), we obtain lim kf (x) − 4n f (x/2n )k ≤ limn→∞

n→∞

=

2.4j θ p j=0 2pj+p kxk

Pn−1

2θ 1 p 2p 1−4/2p kxk

=

2θ 2p p 2p 2p −4 kxk ,

that is, 2θ kxkp . −4 The uniqueness of Q is similar to Theorem 8. This completes the proof of the theorem. kf (x) − Q(x)k ≤

2p

Theorem 10 Let p < 1 and θ be a positive real number, and let f : X → Y be a mapping satisfying f (ix) = −f (x) and kCf (x, y)k ≤ θ.kxkp .kykp

(1.82)

for all x, y ∈ X. Then there exists a unique quadratic mapping Q : X → Y such that θ kf (x) − Q(x)k ≤ kxk2p f or all x ∈ X. (1.83) 4 − 4p Proof 15 Putting y = −ix in (1.82), we find that kf (2x) − 4f (2x)k ≤ θkxkp .k − ixkp = θkxk2p ,

(1.84)

36

Hyers-Ulam Stability of Ordinary Differential Equations

for all x ∈ X. Hence, kf (x) − 1/4f (2x)k ≤ (θ/4)kxk2p , for all x ∈ X. Now, k1/4n f (2n x) − 1/4m f (2m x)k = k1/4n f (2n x) − 1/4m f (2m x) + 1/4m+1 f (2m+1 x) + · · · − 1/4n−1 f (2n−1 x) + 1/4n−1 f (2n−1 x)k 1 1 ≤ k m f (2m x) − m+1 f (2m+1 x)k + · · · 4 4 1 1 + k n−1 f (2n−1 x) − n f (2n x)k 4 4 n−1 X k1/4k f (2k x) − 1/4k+1 f (2k+1 x)k = k=m

=

n−1 X k=m

≤

n−1 X k=m

1 kf (2k x) − 1/4f (2k+1 x)k 4k n−1 X 4pk θ 1 θ k 2p k2 xk = kxk2p 4k 4 4k+1

(1.85)

k=m

for all non-negative integers n and m with n > m. Taking limit n → ∞ and m as fixed in (1.85), the right side converges and hence {4n f (x/2n )} is a Cauchy sequence for all x ∈ X. Let there be a mapping Q : X → Y such that Q(x) = limn→∞ f (2n x), for all x ∈ X. From (1.82), kCQ(x, y)k

1 kCf (2n x, 2n y)k 4n 4pn θ kxkp kykp = 0 ≤ lim n→∞ 4n = lim

n→∞

for all x, y ∈ X. So CQ(x, y) = 0. By Proposition 1, the mapping Q : X → Y is quadratic. Again taking m = 0 and limit n → ∞ in (1.85), we find (1.83). Uniqueness of Q is similar to that of Theorem 8. Theorem 11 Let p > 1 and θ be positive real numbers, and let F : X → Y be a mapping satisfying kCf (x, y)k ≤ θ.kxkp .kykp and f (ix) = −f (x) for all x ∈ X. Then there exists a unique quadratic mapping Q : X → Y such that kf (x) − Q(x)k ≤ for all x ∈ X.

θ kxk2p 4p − 4

(1.86)

Introduction and Preliminaries

37

Proof 16 From (1.84), it follows that kf (x) − 4f (x/2)k ≤

θ kxk2p 4p

for all x ∈ X. Therefore, k4n f (x/2n ) − 4m f (x/2m )k ≤

m−1 X j=1

4j θ kxk2p 4pj+p

(1.87)

for all non-negative n and m with n > m and all x ∈ X. As n → ∞ in (1.87), it follows that {4n f (x/2n )} is a Cauchy sequence for all x ∈ X. If Q : X → Y is defined by lim 4n f (x/2n ) = Q(x), ∀ x ∈ X, n→∞

then 4n θ .kxkp .kykp = 0 n→∞ n→∞ 4pn for all x ∈ X. Thus CQ(x, y) = 0. By Proposition 1, the mapping Q : X → Y is quadratic. Also, putting m = 0 and taking limit n → ∞ in (1.87),we find (1.86). Uniqueness of Q is similar to that of Theorem 8. This completes the proof of the theorem. kCQ(x, y)k = lim 4n kCf (x/2n , y/2n )k ≤ lim

Let X be a normed vector space and Y be a Banach space. There is a mapping f : X → Y such that Df (x, y) = f (x + iy) − f (ix + y) − 2f (x) + 2f (y) for all x, y ∈ X. In the following we discuss the generalized Hyers-Ulam stability of the quadratic functional equation Df (x, y) = 0 by applying the fixed point method. Theorem 12 Let f : X → Y be a mapping with f (ix) = −f (x) for all x ∈ X for which there exists a function ϕ : X × X → [0, ∞) such that ∞ X

4−j ϕ(2j x, 2j y) < ∞

(1.88)

j=0

and kDf (x, y)k ≤ ϕ(x, y)

(1.89)

for all x, y ∈ X. If there exists an L < 1 such that ϕ(x, −ix) ≤ 4Lϕ(x/2, −ix/2) for all x ∈ X, then there exists a unique quadratic mapping Q : X → Y satisfying f (x + iy) − f (ix + y) = 2f (x) − 2f (y) and kf (x) − Q(x)k ≤ for all x ∈ X.

1 ϕ(x, −ix) 4 − 4L

(1.90)

38

Hyers-Ulam Stability of Ordinary Differential Equations

Proof 17 Because of f (ix) = −f (x) for all x ∈ X, f (0) = 0. f (−x) = f (i2 x) = −f (ix) = f (x) for all x ∈ X. Therefore, Df (x, y) = f (x + iy) − f (ix + y) − 2f (x) + 2f (y) = f (x + iy) + f (i(ix + y)) − 2f (x) + 2f (y) = f (x + iy) + f (−(x − iy)) − 2f (x) + 2f (y) = f (x + iy) + f (x − iy) − 2f (x) + 2f (y) = Cf (x, y), ∀ x, y ∈ X. Let us define the set S := {g : X → Y } and develop the generalized metric on S as d(g, h) = inf {K ∈ R+ : kg(x) − h(x)k ≤ Kϕ(x, −ix), ∀x ∈ X}. Hence d is finite and Y is a Banach space. Let {gn } be a Cauchy sequence. Then kgn − gk → 0 as n → ∞. Now d(gn , g) < ∞. So, g ∈ S. Therefore, (S, d) is complete. Define the linear mapping J : S → S such that Jg(x) =

1 g(2x), ∀ x ∈ S. 4

By Theorem 2 and for g, h ∈ S we have d(Jg, Jh) ≤ Ld(g, h) f or all g, h ∈ S, that is, J is a strictly contractive self-mapping of S with the Lipschitz constant L, and it follows that d(f, Jf ) ≤ L for all g, h ∈ S. Putting y = −ix in (1.89), we find kf (x + x) − f (0) − 2f (x) + 2f (−ix)k ≤ ϕ(x, −ix), that is, kf (2x) − 4f (x)k ≤ ϕ(x, −ix), for all x ∈ X. Therefore, kf (x) − 1/4f (2x)k ≤ 1/4ϕ(x, −ix) for all x ∈ X. Thus d(f, Jf ) ≤ 1/4. By Theorem 1, there exists a mapping Q : X → Y such that (1) Q is a fixed point of J, that is, Q(2x) = 4Q(x), f or all x ∈ X. The mapping Q is a unique fixed point of J in the set M = {g ∈ S : d(f, g) < ∞}.

(1.91)

Introduction and Preliminaries

39

It follows that Q is a unique mapping satisfying (1.91) such that there exists K ∈ (0, ∞) satisfying kf (x) − Q(x)k ≤ Kϕ(x, −ix) for all x ∈ X. (2) d(J n f, Q) → 0 as n → ∞. Therefore f (2n x) = Q(x) f or all x ∈ X. n→∞ 4n lim

(3) d(f, Q) ≤

1 1−L d(f, Jf )

(1.92)

which implies that d(f, Q) ≤

1 4 − 4L

which is (1.90). Now, kDQ(x, y)k = lim (1/4n )kDf (2n x, 2n y) ≤ lim (1/4n )ϕ(2n x, 2n y) = 0 n→∞

n→∞

for all x, y ∈ X due to (1.88), (1.89) and (1.92). Hence DQ(x, y) = 0 for all x, y ∈ X. Clearly, Q(ix) = −Q(x) is equivalent to f (ix) = −f (x) for all x ∈ X. By Proposition 2, the mapping Q : X → Y is quadratic. This completes the proof of the theorem. Corollary 10 Let p < 2 and θ be positive real numbers, and let f : X → Y be a mapping satisfying f (ix) = −f (x) and kDf (x, y)k ≤ θ(kxkp + kykp )

(1.93)

for all x, y ∈ X. Then there exists a unique quadratic mapping Q : X → Y satisfying f (x + iy) − f (ix + y) = 2f (x) − 2f (y) and kf (x) − Q(x)k ≤

2θ kxkp 4 − 2p

for all x ∈ X. Proof 18 Let ϕ(x, y) = θ(kxkp + kykp ) for all x, y ∈ X. Replacing y by −ix, we obtain ϕ(x, −ix) = θ(kxkp + k − ixkp ) = θ(kxkp + kxkp ) = 2θkxkp . Therefore, as in Theorem 12 kf (x) − (1/4)f (2x)k ≤ (1/2)θkxkp .

40

Hyers-Ulam Stability of Ordinary Differential Equations

Since f (x + iy) − f (ix + y) − 2f (x) + 2f (y) = θ(kxkp + kykp ), then replacing 2x by x and y = 0, we find f (2x) − f (i2x) − 2f (2x) = θk2xkp implies that k4f (x) − f (2x)k ≤ 2p θkxkp , where we have used the fact that f is quadratic due to Proposition 2. Therefore, kf (x) − (1/4)f (2x)k ≤ 2p−2 θkxkp . For L = 2p−2 and by Theorem 12, it follows that kf (x) − Q(x)k ≤

2θ kxkp 4 − 2p

for all x ∈ X. Q is quadratic as proved in Theorem 12. This completes the proof of the corollary. Theorem 13 Let f : X → Y be a mapping with f (ix) = −f (x) for all x ∈ X for which there exists a function ϕ : X × X → [0, ∞) satisfying kDf (x, y)k ≤ ϕ(x, y) such that ∞ X j=0

4j ϕ(

x y , ) r and x ∈ E1 , kAn (x) − Ar (x)k = k −2n kF (k n x) − k 2(n−r) F (k r x)k ≤ k −2r kF (0)k + k −2n kF (k n−r k r x) + (k 2(n−r) − 1)F (0) − k 2(n−r) F (k r x)k = k −2r kF (0)k + k −2

k−1 X n−r−1 X i=1

= k −2r kF (0)k + k −2

bi H((k − i)k j+r x, k j+r x, 0)k −2(j+r)

j=0

k−1 X n−1 X

bi H((k − i)k j x, k j x, 0)k −2j

i=1 j=r

→ 0 as r → ∞. Since E2 is complete, then let lim An (x) = A(x)

n→∞

for all x ∈ E1 . Similarly, we can show that k −2n G(k n x) is a Cauchy sequence. Now, we prove that, 3A(x) = lim k −2n G(k n x) (1.118) n→∞

for all x ∈ E1 . Indeed, k3A(x) − k −2n G(k n x)k ≤ k3A(x) − 3k −2n F (k n x)k + k3k −2n F (k n x) + k −2n F (0) − k −2n G(k n x) + 2k −2n G(0)k + kk −2n G(k n x) + 2k −2n G(0) − k −2n G(k n x)k + k −2n kF (0)k ≤ 3kA(x) − An (x)k + k −2n h(k n x, 0, 0) + 2k −2n kG(0)k + k −2n kF (0)k due to (1.98) and kxk := d(x.o), x ∈ X implies that lim k3A(x) − k −2n G(k n x)k = 0

n→∞

for all x ∈ E1 . Next, we show that the function A is quadratic. Now, kAn (x + y + z) + An (x − y) + An (y − z) + An (x − z) − k −2n G(k n x) + k −2n G(k n y) + k −2n G(k n z)k ≤ k −2n h(k n x, k n y, k n z),

Introduction and Preliminaries

49

and so taking limit as n → ∞, we get A(x + y + z) + A(x − y) + A(y − z) + A(x − z) = 3A(x) + 3A(y) + 3A(z) for all x, y, z ∈ E1 because of (1.118). From (1.108), it follows that d[k −2n F (k n x) + (1 − k −2n )F (0), F (x)] ≤ k −2

k−1 X n−1 X

bi H((k − i)k j x, k j x, 0)k −2j ,

i=1 j=0

that is, d[An (x) + (1 − k −2n )F (0), F (x)] ≤ k −2

k−1 X n−1 X

bi H((k − i)k j x, k j x, 0)k −2j .

i=1 j=0

Upon taking limit as n → ∞ in the above inequality, we find that kA(x) + F (0) − F (x)k ≤ k −2

k−1 ∞ XX

bi H((k − i)k j x, k j x, 0)k −2j

i=1 j=0

which is (3.20). On the other hand, (1.109) gives d[k −2n G(k n x) + (1 − k −2n )G(0), G(x)] ≤ k −2

k−1 X n−1 X

bi K((k − i)k j x, k j x, 0)k −2j ,

i=1 j=0

and taking limit as n → ∞, we find k3A(x) + G(0) − G(x)k ≤ k −2

k−1 ∞ XX

bi K((k − i)k j x, k j x, 0)k −2j

i=1 j=0

which is (1.116). For uniqueness, assume that there exist two quadratic functions Cm : E1 → E2 , m = 1, 2 such that kCm (x) + F (0) − F (x)k ≤ k −2 am

k−1 ∞ XX

bi H((k − i)k j x, k j x, 0)k −2j ,

i=1 j=0

where x ∈ E1 , n ∈ N and am ≥ 0. So, k −2n kC1 (k n x) − C2 (k n x)k ≤ k −2n (kC1 (k n x) + F (0) − F (k n x)k + kF (k n x) − C2 (k n x) + F (0)k)

50

Hyers-Ulam Stability of Ordinary Differential Equations ≤ (a1 + a2 )k −2

k−1 ∞ XX

bi H((k − i)k j+n x, k j+n x, 0)k −2(j+n)

i=1 j=0

= (a1 + a2 )k −2

k−1 ∞ XX

bi H((k − i)k j x, k j x, 0)k −2j

i=1 j=n

→ 0 as n → ∞. Hence, C1 (x) = C2 (x) for all x ∈ E1 . To complete the proof of the theorem, let L be any continuous liner functional defined on the space E2 . Let ϕ : R → R be given by ϕ(t) := L[A(tx)], f or all x ∈ E1 , t ∈ R, where x is fixed. Then ϕ is quadratic function and moreover, as the pointwise limit of the sequence ϕn (t) = k −2n L[F (k n tx)], t ∈ R is also measurable and hence has the form ϕ(t) = t2 ϕ(1) for t ∈ R. Therefore, for all t ∈ R and all x ∈ E1 L[A(tx)] = ϕ(t) = t2 ϕ(1) = L[t2 A(x)] which proves (1.117). This completes the proof of the theorem. Lemma 3 Let E1 be a group divisible by k and E2 a quasi-normed space. If F, G : E1 → E2 satisfy the inequality (1.98), then d[F (x) + (k 2n − 1)F (0), k 2n F (k −n x)] ≤ k −2

k−1 n XX

bi H((k − i)k −j x, k −j x, 0)k 2j

(1.119)

i=1 j=1

and d[G(x) + (k 2n − 1)G(0), k 2n G(k −n x)] ≤ k −2

k−1 n XX

bi K((k − i)k −j x, k −j x, 0)k 2j

(1.120)

i=1 j=1

for all x ∈ E1 and n, k ∈ N, where k ≥ 2, bi = 2bi−1 + bi−2 (b1 = 1, b2 = 2). Proof 23 Putting x = k −n t into (1.108), we get d[F (k n .k −n t) + (k 2n − 1)F (0), k 2n F (k −n t)] ≤ k 2(n−1)

k−1 X n−1 X i=1 j=0

bi H((k − i)k j−n t, k j−n t, 0)k −2j .

Introduction and Preliminaries

51

Let n − j = m. When j = 0, m = n and when j = n − 1, m = 1. Hence, d[F (t) + (k 2n − 1)F (0), k 2n F (k −n t)] ≤ k 2(n−1)

k−1 X

n X

bi H((k − i)k −m t, k −m t, 0)k 2(m−n)

i=1 m=1

= k 2(n−1) .k −2n

k−1 X

n X

bi H((k − i)k −m t, k −m t, 0)k 2m

i=1 m=1

= k −2

k−1 X

n X

bi H((k − i)k −m t, k −m t, 0)k 2m

i=1 m=1

which verifies (1.119). Similarly, (1.120). Theorem 15 Let E1 be an abelian group divisible by k, k ≥ 2 and E2 a complete quasi-normed space. Suppose that the series ∞ X

h((k − i)k −j x, k −j x, 0)k 2j and

j=1

∞ X

h(k −j x, 0, 0)k 2j

j=1

are convergent for all x ∈ E1 and the condition lim inf h(k −n x, k −n y, k −n z) = 0 n→∞

for all x, y, z ∈ E1 , then there exists exactly one quadratic function A : E1 → E2 such that d[B(x), F (x)] ≤ k −2

∞ k−1 XX

bi H((k − i)k −j x, k −j x, 0)k 2j

(1.121)

bi K((k − i)k −j x, k −j x, 0)k 2j

(1.122)

i=1 j=1

and d[B(x), F (x)] ≤ k −2

k−1 ∞ XX i=1 j=1

for all x ∈ E1 . P∞ −j Proof 24 Given that x, 0, 0)k 2j is convergent. Then it follows j=1 h(k P∞ P∞ 2j that j=1 h(0, 0, 0)k is convergent, that is, j=0 h(0, 0, 0)k 2j . Since lim inf h(k −n x, k −n y, k −n z) = 0, n→∞

then h(0, 0, 0) = 0. So H(0, 0, 0) = 0 and 4F (0) = 3G(0) due to (1.98). Thus, 9d[F (0), F (0)] = 0 implies that F (0) = 0 and F (0) = G(0) = 0. Let

52

Hyers-Ulam Stability of Ordinary Differential Equations

Bn (x) = k 2n F (k −n x) for all x ∈ E1 and n ∈ N. We claim that Bn (x) is a Cauchy sequence for all x ∈ E1 . kBn (x), Bm (x)k = kk 2n F (k −n x) − k 2m F (k −m x)k = k 2n kF (k −n x) − k 2(m−n) F (k −m x)k ≤ k 2m kF (0)k + k 2n kF (k m−n .k −m x) + (k 2(m−n) )F (0) − k 2(m−n) F (k −m x)k ≤ k 2m kF (0)k + k −2

k−1 X m−n−1 X i=1

=k

2m

+ k −2

bi H((k − i)k −(j+m) x, k −(j+m) x, 0)k 2(j+m)

j=0

kF (0)k k−1 X X n−1

bi H((k − i)k −j x, k −j x, 0)k 2j

i=1 j=r

→ 0 as n → ∞. Since E2 is complete, then we let B(x) = lim Bn (x) n→∞

(1.123)

for all x ∈ E1 . Now, k3B(x) − k 2n G(k −n x)k ≤ k3B(x) − 3.k 2n F (k −n x)k + k3.k 2n F (k −n x) + k 2n F (0) − k 2n G(k −n x) + 2k 2n G(0)k + kk 2n G(k −n x) + 2k 2n G(0) − k 2n G(k −n x)k + k 2n kF (0)k ≤ 3kB(x) − Bn (x)k + k 2n h(k −n x, 0, 0) + 2k 2n kG(0)k + k 2n kF (0)k implies that limn→∞ k3B(x) − k 2n G(k −n x)k = 0 due to (1.123) and lim inf n→∞ h(k −n x, k −n y, k −n z)k 2n = 0. Therefore, 3B(x) = lim k 2n G(k −n x) n→∞

(1.124)

for all x ∈ E1 . Since d[Bn (x + y + z) + Bn (x − y) + Bn (y − z) + Bn (x − z), k 2n G(k −n x)+k 2n G(k −n y)+k 2n G(k −n z)] ≤ k 2n h(k −n x, k −n y, k −n z), then using (1.123) and (1.124), it follows that B(x + y + z) + B(x − y) + B(y − z) + B(x − z) = 3B(x) + 3B(y) + 3B(z) for all x, y, z ∈ E1 by taking limits of both sides as n → ∞. Thus, B is a quadratic function. Using F (0) = 0 and Bn (x) = k 2n F (k −n ) in (1.119), we get (1.121) as n → ∞. Similarly, (1.122). Uniqueness is similar to that of Theorem14. This completes the proof of the theorem.

Introduction and Preliminaries

1.2.2

53

Stability of f (rx+sy) = ((r+s)/2)f (x+y)+((r−s)/2)f (x−y)

Throughout this section, we assume that A C ∗ -algebra is a unital with norm k.k, unit 1. Also we assume that X and Y are normed left A-module and Banach left A-module with norms k.kX and k.kY , respectively. Let U (A) be the set of unitary elements in A, and let r, s ∈ R and r 6= s. For a given mapping f : X → Y, u ∈ U (A) and a given µ ∈ C, we define Du f, Dµ f : X 2 → Y by Du f (x, y) := f (rux + suy) −

r−s r+s uf (x + y) − uf (x − y) 2 2

(1.125)

Dµ f (x, y) := f (rµx + sµy) −

r+s r−s µf (x + y) − µf (x − y), 2 2

(1.126)

for all x, y ∈ X. Proposition 3 Let L : X → Y be a mapping with L(0) = 0 such that Du L(x, y) = 0 for all x, y ∈ X, for all u ∈ U (A). Then L is A-linear. Proof 25 Du L(x, y) = L(rux + suy) −

r+s r−s uL(x + y) − uL(x − y). 2 2

Given that Du L(x, y) = 0. Hence, for additive mapping L r−s r+s uL(x + y) + uL(x − y) 2 2 ru su ru su =( + )L(x + y) + ( − )L(x − y) 2 2 2 2 ru su = (L(x + y) + L(x − y)) + (L(x + y) − L(x − y)) 2 2 ru = (L(x) + L(y) + L(x) − L(y)) 2 su + (L(x) + L(y) − L(x) + L(y)), 2

L(rux + suy) =

implies that L(rux + suy) = ruL(x) + suL(y). Thus, L is A-linear. Corollary 11 Let L : X → Y be such that L(0) = 0 satisfy D1 L(x, y) = 0 for all x, y ∈ X. Then L is additive. Proof 26 Since D1 L(rx + sy) = L(rx + sy) − then L(rx + sy) = So, L is additive by (3).

r+s 2 L(x

+ y) −

r+s r−s L(x + y) + L(x − y). 2 2

r−s 2 L(x

− y),

54

Hyers-Ulam Stability of Ordinary Differential Equations

Corollary 12 A mapping L : X → Y such that L(0) = 0 satisfies Dµ L(x, y) = 0 for all x, y ∈ X and all µ ∈ T := µ ∈ C : |µ| = 1, if and only if L is C-linear. Proof 27 Given that Dµ L(x, y) = L(rµx + sµy) −

r−s r+s µL(x + y) − µL(x − y). 2 2

Since Dµ L(x, y) = 0, then r+s r−s µL(x + y) + µL(x − y) 2 2 rµ = (L(x + y) + L(x − y)) 2 sµ + (L(x + y) − L(x − y)) 2 = rµL(x) + sµL(y).

L(rµx + sµy) =

So L is C-linear. Theorem 16 Let f ; X → Y be a mapping satisfying f (0) = 0 for which there exist a function ϕ : X 2 → [0, ∞) such that 1 ϕ(2k x, 2k y) = 0, k→∞ 2k lim

(1.127)

and ϕ(x) ˜ =

  k+1  ∞ X 1 2 rx −2k+1 sx ϕ , 2k r2 − s2 r2 − s2 k=0  k   k  2 x 2k x 2 x −2k x +ϕ , , +ϕ < ∞, r+s r+s r−s r−s kD1 f (x, y)kY ≤ ϕ(x, y)

(1.128) (1.129)

for all x, y ∈ X. Then, there exists a unique additive mapping L : X → Y such that 1 ˜ (1.130) kf (x) − L(x)kY ≤ ϕ(x) 2 for all x ∈ X. Proof 28 Given that kD1 f (x, y)kY ≤ ϕ(x, y).So,

   

D1 f (x, y) − D1 f x + y , x + y − D1 f x − y , y − x

2 2 2 2 Y

   

x + y y−x x + y x − y

, , ≤ kD1 f (x, y)kY + D f + D f

1

1

2 2 2 2 Y Y     x−y y−x x+y x+y , +ϕ , ≤ ϕ(x, y) + ϕ 2 2 2 2

Introduction and Preliminaries

55

for all x, y ∈ X. Taking u = 1 in (1.125), it follows that

 

f (rx + sy) − r + s f (x + y) − r − s f (x − y) − f r(x + y) + s(x + y)

2 2 2 2     r+s x+y x+y r−s x+y x+y + f + + f − 2 2 2 2 2 2   x−y y−x − f r( ) + s( ) 2 2     r+s x−y y−x r−s x−y y−x + f + + f −( )

2 2 2 2 2 2 Y



r + s r − s

≤

f (rx + sy) − 2 f (x + y) − 2 f (x − y) Y

   

r(x + y) x + y s(x + y) r + s x+y

+ − f + + f 2 2 2 2 2  

r−s x+y x+y − f −

2 2 2 Y

      

x − y y − x r+s x−y y−x

+ f r +s − f + 2 2 2 2 2   

x−y y−x r−s

− f −

2 2 2 Y

   

x + y x−y y−x x+y



≤ kD1 f (x, y)kY + D1 f , ,

+ D1 f

2 2 2 2 Y Y     x−y y−x x+y x+y , +ϕ , , ≤ ϕ(x, y) + ϕ 2 2 2 2 that is,

 

f (rx + sy) − r + s f (x + y) − r − s f (x − y) − f r + s (x + y)

2 2 2   r+s r−s r−s + f (x + y) + f (0) − f (x − y) 2 2 2

r+s r−s + f (0) + f (x − y)

2 2  Y   x+y x+y x−y y−x , +ϕ , . ≤ ϕ(x, y) + ϕ 2 2 2 2 Consequently,

   

f (rx + sy) − f r + s (x + y) − f r − s (x − y)

2 2    Y  x−y y−x x+y x+y , +ϕ , . (1.131) ≤ ϕ(x, y) + ϕ 2 2 2 2

56

Hyers-Ulam Stability of Ordinary Differential Equations

Replacing x by

1 r+s x

+

1 r−s y,

y by

1 r+s x

−

1 r−s y

in (1.131), we obtain

    

1 1 1 1

f r x + y + s x − y

r+s r−s r+s r−s    r+s 1 1 1 1 −f x+ y+ x− y 2 r+s r−s r+s r−s   

r−s 1 1 1 1

−f x+ y− x+ y

2 r+s r−s r+s r−s Y   1 1 1 1 ≤ϕ x+ y, x− y r+s r−s r+s r−s      1 x 1 x y x y y x y +ϕ , + − − + + − 2 r+s r−s r+s r−s 2 r+s r−s r+s r−s      1 x y x y 1 x y x y +ϕ + − + , − − − , 2 r+s r−s r+s r−s 2 r+s r−s r+s r−s

that is, kf (x + y) − f (x) − f (y)kY   y x y x + , − ≤ϕ r+s r−s r+s r−s     x x y y +ϕ , +ϕ ,− r+s r+s r−s r−s

(1.132)

for all x, y ∈ X. Hence, for y = x in (1.132), we get     2rx −2sx x x kf (2x) − 2f (x)kY ≤ ϕ , , +ϕ r2 − s2 r2 − s2 r+s r+s   x −x +ϕ , (1.133) r−s r−s for all x ∈ X. Denote       2rx −2sx x x x −x Ψ(x) = ϕ , + ϕ , + ϕ , , r2 − s2 r2 − s2 r+s r+s r−s r−s for all x ∈ X. Then, it follows from (1.128) that  ∞  X 1 Ψ(2k x) = ϕ(x) ˜ 0, I be an open interval and for any function f satisfying the differential inequality 0 an (t)y (n) (t) + an−1 (t)y (n−1) (t) + · · · + a1 (t)y (t) + a0 (t)y(t) + h(t) ≤ , then there exists a solution f0 of the above differential equation such that |f (t) − f0 (t)| ≤ K() and lim K() = 0, for t ∈ I. If the preceding statement →0

is also true when we replace  and K() by φ(t) and ψ(t) respectively, where φ, ψ: I → [0, ∞) are functions not depending on f and f0 explicitly, then we say that the corresponding differential equation has the generalized Hyers-Ulam stability. S. M. Jung has investigated the Hyers-Ulam stability of linear differential equations of different classes. Keeping in view of the above definition of HyersUlam stability for differential equations, we can view the corresponding definition of Hyers-Ulam stability for difference equations as follows: Definition 9 The difference equation ak (n)y(n + k) + ak−1 (n)y(n + k − 1) + · · · + a1 (n)y(n + 1) +a0 (n)y(n) + h(n) = 0 has the Hyers-Ulam stability, if for given  > 0, I be an open interval and for any real valued function f satisfying the inequality | ak (n)y(n + k) + ak−1 (n)y(n + k − 1) + · · · + a1 (n)y(n + 1) +a0 (n)y(n) + h(n) | ≤ , then there exists a solution f0 of the above difference equation such that | f (n) − f0 (n) | ≤ K() and lim K() = 0, for n ∈ I ⊂ N (0) = {0, 1, 2, 3, · · · }. →0

Because difference equations can be generated from differential equations by using the Euler’s method, the above definitions exist simultaneously. Hence, the study of Hyers-Ulam stability for differential and difference equations is interesting.

Introduction and Preliminaries

1.3

61

Notes

In order to have an insight on Hyers-Ulam stability, Section 1.2 is devoted for some functional equations on Banach space while Section 1.3 deals with some more functional equations on abelian groups. For details, we refer [1], [2], [4], [7], [8], [12] and [14]. In addition, suggested references [17], [54] and [55] are provided for exclusive study of Hyers-Ulam stability of some other kind of functional equations.

Chapter 2 Stability of First Order Linear Differential Equations

The real exponential function f (x) = ex is the only nontrivial solution of 0 the differential equation f = f and the objective is to go through the the Hyers-U lam stability of this equation, i.e. to solve for a given ε > 0 the inequality 0

|f (x) − f (x)| ≤ ε and also to study the related inequality: f (x + y) f (y) − f (x) − ≤ ε, 2 y−x

(2.1)

(2.2)

for all x 6= y. By using (2.1) and (2.2) several other inequalities can be solved. Here I = any real interval and R+ for the set of all nonnegative real numbers. A function f will be termed if f satisfies the inequality   f (x) + f (y) x+y ≥ f 2 2 and f will be said to be k − Lipschitz, whenever |f (x) − f (y)| ≤ k|x − y|, ∀ x, y in the (convex) domain of f. A monotonic Jensen concave function f : I → R has to be necessarily concave in the usual sense, that is, to satisfy the inequality f (λx + (1 − λ)y) ≥ λf (x) + (1 − λ)f (y), for all x, y ∈ I and all λ ∈ [0, 1].

2.1

Stability of f 0 (x) = f (x)

Lemma 4 Let g : I → R be a differentiable function. Then: 63

64

Hyers-Ulam Stability of Ordinary Differential Equations

(i) the inequality g(x) ≤ g 0 (x) holds for all x in I if and only if g can be represented in the form g(x) = i(x)ex , x ∈ I,

(2.3)

where i : I → R is an arbitrary nondecreasing differentiable function; (ii) the inequality g 0 (x) ≤ g(x) holds for all x in I if and only if g admits the representation g(x) = d(x)ex , x ∈ I, (2.4) where d : I → R is an arbitrary nonincreasing differentiable function. Proof 32 Let g(x) ≤ g 0 (x), x ∈ I. Then the function i : I → R is defined by i(x) = g(x)e−x , x ∈ I, is differentiable and can be written i0 (x) = g 0 (x)e−x − g(x)e−x = (g 0 (x) − g(x))e−x ≥ 0, ∀x ∈ I. Hence i is nondecreasing. Conversely, suppose that g(x) = i(x)ex , x ∈ I. Now, g 0 (x) − g(x) = i0 (x)ex ≥ 0. Hence, g(x) ≤ g 0 (x). So, part(i) is proved. For part (ii), let g 0 (x) ≤ g(x), x ∈ I. Then the function d : I → R is defined by d(x) = g(x)e−x , x ∈ I, is differentiable and can be written d0 (x) = g 0 (x)e−x − g(x)e−x = (g 0 (x) − g(x))e−x ≤ 0, ∀x ∈ I. Hence d is nonincreasing. Conversely, if g(x) = d(x)ex , x ∈ I, then g 0 (x) − g(x) = d0 (x)ex ≤ 0. Hence, g 0 (x) ≤ g(x). This completes the proof of the lemma.

Stability of First Order Linear Differential Equations

65

Lemma 5 For every real number x 6= y the exponential function satisfies the inequalities (x+y) ey − ex ex + ey e 2 ≤ ≤ . (2.5) y−x 2 Proof 33 For t ≥ 0 1+

t2 t3 t t2 t3 t + 2 + 3 ··· ≤ 1 + + + + ··· 2 2 2! 2 3! 2! 3! 4!

implies that

    t 2 t 3 1 t2 t3 t4 t 2 2 + + ··· ≤ t + + + + ··· . 1+ 2 2! 3! t 2! 3! 4! Therefore,

    t 3 t 2 t 1 t3 t4 t2 2 2 1+ + ··· ≤ + 1 + t + + + + ··· − 1 2 2! 3! t 2! 3! 4! and



t 2

e ≤

et − 1 t

 .

(2.6)

Again 1+

t t2 t3 t t2 t3 + + + ··· ≤ 1 + + + + ··· 2! 3! 4! 2 2(2!) 2(3!)

implies that     t2 t3 t4 1 t2 t3 t4 1 t + + + + ··· ≤ 2 + t + + + + ··· . t 2! 3! 4! 2 2! 3! 4! Hence,     1 t2 t3 t4 1 t2 t3 t4 1 + t + + + + ··· − 1 ≤ 1 + t + + + + ··· + 1 , t 2! 3! 4! 2 2! 3! 4! that is, 

et − 1 t



 ≤

et + 1 2

 .

(2.7)

From (2.6) and (2.7) we obtain t

e2 ≤

et − 1 et + 1 ≤ . t 2

Put y − x = t in (2.8). Then e

y−x 2

≤

ey−x − 1 ey−x + 1 ≤ . y−x 2

(2.8)

66

Hyers-Ulam Stability of Ordinary Differential Equations

Multiplying ex to the last expression, it follows that e

x+y 2

≤

ey − ex ey + ex ≤ . y−x 2

This completes the proof of the lemma. Lemma 6 A function f : I → R+ satisfies the inequality   x+y f (y) − f (x) f ≤ 2 y−x

(2.9)

for all x 6= y, in I, if and only if f can be represented in the form f (x) = i(x)ex , where i : I → R+ is an arbitrary nondecreasing function. Proof 34 Suppose that f satisfies (2.9), x in I, h > 0, and x + h ∈ I. To prove that f can be represented in the form f (x) = i(x)ex , where i : I → R+ is an arbitrary nondecreasing function. Now,   f (y) − f (x) x+y ≥f y−x 2 implies that x+y 2



x+y 2



 f (y) − f (x) ≥ (y − x)f or

 f (y) − f (x) ≤ (y − x)f

, ∀y ≥ x

(2.10)

, ∀y ≤ x.

(2.11)

Hence from (2.10), f (y) − f (x) ≥ 0, ∀y ≥ x, that is, f (y) ≥ f (x), ∀y ≥ x implies that f is a nondecreasing function and   x+y f (y) ≥ f (x) + (y − x)f . 2 Put x + h = y in last inequality. Then  f (x + h) ≥ f (x) + ((x + h) − x)f

x + (x + h) 2

 ,

that is,  f (x + h) ≥ f (x) + hf

h x+ 2

 .

Therefore,  f (x + h) ≥ f (x) + hf

h x+ 2

 ≥ f (x) + hf (x) = (1 + h)f (x).

We claim that f (x + ih) ≥ (1 + h)i f (x),

Stability of First Order Linear Differential Equations

67

for x + ih ∈ I and i ∈ N. Our claim is true for i = 1. Let’s assume that it is true for i = k, that is, f (x + kh) ≥ (1 + h)k f (x). For i = k + 1, f (x + (k + 1)h) = f ((x + h) + kh) ≥ (1 + h)k f (x + h) ≥ (1 + h)k (1 + h)f (x), implies that f (x + (k + 1)h) ≥ (1 + h)k+1 f (x). Hence, our claim is true. So, it is true for ∀i ∈ N. Thus, for an arbitrarily fixed n ∈ N, for every x ≤ y in I, we have   n  y−x (y − x) f (x). ≥ 1+ f (y) = f x + n n n For n → ∞, f (y) ≥ ey−x f (x) implies that f (y)e−y ≥ e−x f (x), ∀y ≥ x. So, the function i : I → R+ is defined by i(x) = f (x)e−x is nondecreasing. Again from (2.11), f (y) − f (x) ≤ 0, ∀y ≤ x, that is, f (y) ≤ f (x), ∀y ≤ x, implies that f is a nondecreasing function and   x+y f (y) ≤ f (x) + (y − x)f . 2 Put x − h = y in last inequality. Then  f (x − h) ≤ f (x) + (x − h − x)f

x+x−h 2

 ,

that is,  f (x − h) ≤ f (x) − hf

x−

h 2

 ≤ f (x) − hf (x − h)

implies that f (x) ≥ (1 + h)f (x − h), that is, f (x − h) ≤ f (x − ih) ≤

f (x) 1+h .

We claim that

f (x) (1 + h)i

for x − ih ∈ I and i ∈ N. Our claim is true for i = 1. Let’s assume that it is f (x) true for i = k, that is, f (x − kh) ≤ (1+h) k . For i = k + 1, f (x − (k + 1)h) = f ((x − h) − kh) ≤

f (x − h) f (x) ≤ k (1 + h) (1 + h)(1 + h)k

implies that f (x − (k + 1)h) ≤

f (x) . (1 + h)k+1

Hence, our claim holds. So, it is true ∀i ∈ N. Thus, for an arbitrarily fixed n ∈ N and for every y ≤ x in I, we have     (y − x) (x − y) f (x) f (y) = f x + n =f x−n ≤
n . n n 1 + x−y n

68

Hyers-Ulam Stability of Ordinary Differential Equations

(x) For n → ∞, f (y) ≤ efx−y , that is, ex−y f (y) ≤ f (x) implies that e−y f (y) ≤ −x e f (x), ∀y ≤ x. So, the function i : I → R+ , is defined by i(x) = f (x)e−x , is nondecreasing. Conversly, let f (x) = i(x)ex , x ∈ I with i : I → R+ is nondecreasing. To prove that   f (y) − f (x) x+y ≤ . f 2 y−x
For x ≤ y we have i(x) ≤ i x+y ≤ i(y) implies that 2     x+y x+y y−x y−x i(y)e −i e ≥ 0 ≥ i(x) − i . 2 2

Hence, y−x

i(y)e

 − i(x) ≥ i

x+y 2



(ey−x − 1).

Multiplying the above inequality by ex /(y − x), we get   x + y ey − ex i(y)ey − i(x)ex ≥i . y−x 2 y−x Upon using Lemma 5, we obtain     x + y ey − ex x+y i(y)ey − i(x)ex ≥i ≥i e(x+y)/2 y−x 2 y−x 2 which in turn

  x+y f (y) − f (x) ≥f . y−x 2
Again for y ≤ x, we have i(y) ≤ i x+y ≤ i(x) and we can deduce that 2     x+y x+y x−y i(y) − i ≤ 0 ≤ i(x)e −i ex−y . 2 2

Hence,  i

x+y 2



(ex−y − 1) ≤ i(x)ex−y − i(y).

Multiplying the last inequality by ey /(x − y) we have   x + y ex − ey i(x)ex − i(y)ey i ≤ , 2 x−y x−y that is,  i

x+y 2



ey − ex i(y)ey − i(x)ex ≤ y−x y−x

Stability of First Order Linear Differential Equations

69

implies that  i

x+y 2



ey − ex f (y) − f (x) ≤ . y−x y−x

Using Lemma 5, it follows that     x+y x+y x + y ey − ex f (y) − f (x) 2 i ≤i e ≤ , 2 2 y−x y−x that is,  f

x+y 2

 ≤

f (y) − f (x) . y−x

This completes the proof of the lemma. Lemma 7 If a function f : I → R+ is nondecreasing and satisfies the inequality   x+y f (y) − f (x) ≤f (2.12) y−x 2 for all x < y in I, then f can be represented in the form f (x) = d(x)ex , x ∈ I, with a nonincreasing function d : I → R+ . Proof 35 From (2.12), we get  f (y) − f (x) ≤ (y − x)f

x+y 2



2x + h 2



.

Put x + h = y in last inequality. Then  f (x + h) ≤ f (x) + hf

.

Therefore, for h ∈ (0, 1)  0 ≤ f (x + h) − f (x) ≤ hf

h x+ 2

 ≤ hf (x + h)

implies that (1 − h)f (x + h) ≤ f (x) and hence f (x + h) ≤ f (x + ih) ≤

f (x) 1−h .

We claim that

f (x) (1 − h)i

for x + ih ∈ I and i ∈ N. Our claim is true for i = 1. Assume that it is true f (x) for i = k, that is, f (x + kh) ≤ (1−h) k . For i = k + 1, f (x + (k + 1)h) = f ((x + h) + kh) ≤

f (x + h) f (x) ≤ (1 − h)k (1 − h)(1 − h)k

70

Hyers-Ulam Stability of Ordinary Differential Equations

implies that f (x + (k + 1)h) ≤

f (x) . (1 − h)k+1

So, our claim holds and it is true ∀i ∈ N. If x ≤ y in I and n is sufficiently large, then (y−x) ∈ (0, 1). Hence we have n   f (x) (y − x) ≤ f (y) = f (x + (y − x)) = f x + n
n . n 1 − y−x n When n → ∞, f (y) ≤ f (x)ey−x , that is, f (y)e−y ≤ f (x)e−x , ∀x < y. If, d(x) = f (x)e−x , x ∈ I, then it is a decreasing function. Hence the lemma is proved. Lemma 8 A nondecreasing Jensen concave function f : I → R+ satisfies the inequality (2.12) if and only if there exists a nonincreasing function d : I → R+ such that I 3 x 7→ d(x)ex is concave and f (x) = d(x)ex , x ∈ I. Proof 36 Necessary part is same as Lemma 7. To prove the sufficient part, assume that f (x) = d(x)ex for x ∈ I is Jensen concave and nondecreasing with d nonincreasing. If x < y in I then d(x) = f (x)e−x ≥ f (y)e−y = d(y), that is, f (x) ≥ f (y)ex−y . Applying Lemma 5, we get f (y) − f (x) f (y) − f (y)ex−y f (y) ey − ex f (y) ex + ey ≤ = y . ≤ y . y−x y−x e y−x e 2   f (y)ex−y + f (y) f (x) + f (y) x+y = ≤ ≤f 2 2 2 which is the inequality (2.12). Theorem 18 Given an ε > 0 let f : I −→ R be a differentiable function. Then |f 0 (x) − f (x)| ≤ ε, x ∈ I holds for all x in I if and only if f can be represented in the form f (x) = ε + ex l(e−x ), x ∈ I, where l is an arbitrary differentiable function defined on the interval J = {e−x |x ∈ I} nonincreasing and 2ε − Lipschitz.

(2.13)

Stability of First Order Linear Differential Equations

71

Proof 37 By (2.1) f (x) − ε ≤ f 0 (x) ≤ ε + f (x), ∀x ∈ I. Let g(x) = f (x) − ε. Then g(x) ≤ g 0 (x) and by Lemma 4(i), we can write f (x) − ε = i(x)ex , x ∈ I, that is, f (x) = i(x)ex + ε, x ∈ I,

(2.14)

where i is differentiable and nondecreasing. On the other hand, if h(x) = 0 f (x) + ε with h (x) ≤ h(x), then we can write f (x) + ε = d(x)ex , x ∈ I, that is, f (x) = d(x)ex − ε,

(2.15)

where d is differentiable and nonincreasing. From (2.14) and (2.15), it follows that i(x)ex + ε = d(x)ex − ε, x ∈ I. (2.16) Consequently, i0 (x)ex + i(x)ex = d0 (x)ex + ex d(x) = d0 (x)ex + i(x)ex + 2ε, x ∈ I. Using the fact that d is a nonincreasing function and so d0 ≤ 0 and d0 (x) =

i0 (x)ex − 2ε = i0 (x) − 2εe−x ≤ 0, x ∈ I ex

implies 0 ≤ i0 (x) ≤ 2εe−x , x ∈ I. Define J = {e−x |x ∈ I} and l : J → R be such that l(z) = i(−Inz), z ∈ J.

(2.17)

then l is differentiable and l0 (z) = −i0 (−lnz)/z ≤ 0, z ∈ J. Hence l is nonincreasing and also, l is 2ε-Lipschitz. Indeed ∀z1 , z2 in J, z1 6= z2 , by the mean value theorem there exists z3 in (min(z1 , z2 ), max(z1 , z2 )) such that −i0 (− ln z ) 3 0 |l(z1 ) − l(z2 )| = |l (z3 )||z1 − z2 | = |z1 − z2 | z3 = i0 (− ln z3 )e− ln z3 |z1 − z2 | ≤ 2ε|z1 − z2 |. Thus by (2.17) we have the required expression: f (x) = ε + i(x)ex = ε + l(e−x )ex .

72

Hyers-Ulam Stability of Ordinary Differential Equations

Theorem 19 Given an ε > 0, let f : I → R+ be a function such that f (x) ≥ ε for all x in I. Then f satisfies the inequality   x+y f (y) − f (x) f ≤ +ε (2.18) 2 y−x for all x < y in I, if and only if f can be represented in the form f (x) = ε + i(x)ex , x ∈ I, where i : I → R+ is a nondecreasing function. Proof 38 Applying the Lemma 6 to the function f − ε, we obtain a function f : I → R+ satisfies   (f (y) − ε) − (f (x) − ε) x+y −ε≤ , f 2 y−x for all x 6= y in I, if and only if f −ε can be represented in the form f (x)−ε = i(x)ex , where i : I → R+ is an arbitrary nondecreasing function. Hence   f (y) − f (x) x+y −ε≤ , f 2 y−x that is,  f

x+y 2

 ≤

f (y) − f (x) +ε y−x

if and only if f (x) = i(x)ex +ε, x ∈ I which completes the proof of the theorem. Theorem 20 Given an ε > 0, a nondecreasing Jensen concave function f : I → R satisfying f (x) ≥ −ε for all x ∈ I, is a solution of the inequality   f (y) − f (x) x+y −ε≤f , (2.19) y−x 2 if and only if f (x) = d(x)ex − ε, where d : I → R+ is nonincreasing and I 3 x 7→ d(x)ex is Jensen concave. Proof 39 Applying the Lemma 8 to the function f + ε, we obtain a function f : I → R+ satisfying   (f (y) + ε) − (f (x) + ε) x+y ≤f +ε y−x 2 for all x < y in I if and only if there exists a nonincreasing function d : I → R+ such that I 3 x 7→ d(x)ex is concave and f (x) + ε = d(x)ex , x ∈ I. Hence,   f (y) − f (x) x+y ≤f + ε, y−x 2 that is, f (y) − f (x) −ε≤f y−x



x+y 2



if and only if f (x) = d(x)ex − ε, where d : I → R+ is nonincreasing and I 3 x 7→ d(x)ex is Jensen concave. Hence the theorem is proved.

Stability of First Order Linear Differential Equations

2.2

73

Stability of y 0 (t) = λy(t)

In this section, the Hyers-Ulam-Rassias stability of the following linear differential equation: y 0 (t) = λy(t) (2.20) is provided which generalizes a theorem of Alsina and Ger[1]. Moreover, it is proved that if λ is a non-zero real number and I = (a, b) is an arbitrary open interval, and also if a continuously differentiable function y : I → R satisfies the inequality n X |y 0 (t) − λy(t)| ≤ ak tk , ∀t ∈ I, k=0

then there exist real numbers c and αk , k ∈ {0, 1, ..., n}, such that ( P Pn n | k=0 αk tk − lims→b− k=0 αk sk eλ(t−s) | (f orλ > 0), λt |y(t) − ce | ≤ Pn Pn | k=0 αk tk − lims→a+ k=0 αk sk eλ(t−s) | (f orλ < 0), for any t ∈ I. Let I = (a, b) be an open real interval, where −∞ ≤ a ≤ b ≤ ∞. Suppose, real numbers a0 , a1 , · · · , an are given which satisfy the property: n X

ak tk ≥ 0 (t ∈ I),

(2.21)

k=0

where n is a fixed non-negative integer. Set αk =

n X i=k

i!ai (k = 0, 1, · · · , n) k!λi+1−k

(2.22)

and λ is true for some non-zero real number. Indeed, λα0 − α1 =

n X i!ai i=0

λα1 − 2α2 =

λi

−

n X i!ai

λi

i=1

= a0 ,

n n X i!ai X i!ai − = a1 λi−1 i=2 λi−1 i=1

and hence λαn−1 − nαn =

n X

n

X i!ai i!ai − = an−1 i−n+1 i+1−n (n − 1)!λ (n − 1)!λ i=n−1 i=n

74

Hyers-Ulam Stability of Ordinary Differential Equations

implies that n X k=0

λαk tk −

n X

kαk tk−1 = (λα0 − α1 ) + (λα1 − 2α2 )t + · · ·

k=1

+ (λαn−1 − nαn )tn−1 + λαn tn =

n X

ak tk , (t ∈ R).

k=0

(2.23) Lemma 9 Let I be an arbitrary non-degenerate open interval. Assume that z : I → R is a continuously differentiable function and λ is a real number. (a) The inequality λz(t) ≤ z 0 (t) holds true for all t ∈ I if and only if there exists an increasing continuously differentiable function i : I → R such that z(t) = i(t)eλt for all t ∈ I; (b) The inequality λz(t) ≥ z 0 (t) holds true for any t ∈ I if and only if there exists a decreasing continuously differentiable function d : I → R such that z(t) = d(t)eλt for all t ∈ I. Proof 40 The proof of the lemma follows from the proof of Lemma 4. Hence the details are omitted. Lemma 10 For any non-zero real number λ and for any real sequence {ai }ni=0 , the equality m X m X k=0 i=k

m

k

X X k!tk−i i!ai k t = a , (t ∈ R) k k!λi+1−k (k − i)!λi+1 i=0 k=0

holds for any non-negative integer m. Proof 41 Our claim is true for m = 0. Let’s assume that, it is true for m = n ≥ 0, that is, n X n X k=0 i=k

n

k

X X i!ai k!tk−i k t = a . k k!λi+1−k (k − i)!λi+1 i=0 k=0

Stability of First Order Linear Differential Equations

75

For m = n + 1, n+1 X n+1 X k=0 i=k

n+1 n n+1 X X X i!ai tk i!ai tk i!ai tn+1 = + i+1−k i+1−k k!λ k!λ (n + 1)!λi+1−(n+1) i=n+1 k=0 i=k n n+1 X X

(n + 1)!an+1 tn+1 i!ai tk + k!λi+1−k (n + 1)!λ(n+2)−(n+1) k=0 i=k ! n n X X (n + 1)!an+1 tn+1 (n + 1)!an+1 tk i!ai tk + = + i+1−k n+2−k k!λ k!λ (n + 1)!λ =

k=0

=

i=k

n X n X k=0 i=k

=

n X n X k=0 i=k

i!ai tk + k!λi+1−k

n X (n + 1)!an+1 tn+1 (n + 1)!an+1 tk + n+2−k k!λ (n + 1)!λ

!

k=0

n+1

X (n + 1)!an+1 tk i!ai tk + , i+1−k k!λ k!λn+2−k k=0

that is, n+1 X n+1 X k=0 i=k

n

k

n+1

X X X (n + 1)!tk k!tk−i i!ai tk = a + a . k n+1 k!λi+1−k (k − i)!λi+1 k!λn+2−k i=0 k=0

(2.24)

k=0

Putting k = n + 1 − i in the second sum of the equality (2.24), we obtain n+1 X n+1 X k=0 i=k

n

k

X X i!ai tk k!tk−i = a + an+1 k k!λi+1−k (k − i)!λi+1 i=0 k=0

=

=

=

n X

ak

k X

k=0

i=0

n X

k X

ak

k=0

i=0

n+1 X

k X

k=0

ak

i=0

n+1−i=n+1 X n+1−i=0

(n + 1)!tn+1−i (n + 1 − i)!λi+1

i=0 X

(n + 1)!tn+1−i k!t + a n+1 (k − i)!λi+1 (n + 1 − i)!λi+1 i=n+1 k−i

n+1

X (n + 1)!tn+1−i k!tk−i + a n+1 (k − i)!λi+1 (n + 1 − i)!λi+1 i=0 k!tk−i . (k − i)!λi+1

So, our claim holds for any non-negative integer m. Hence the lemma is proved. Theorem 21 Let I be an arbitrary open interval and let λ be a non-zero real number. A continuously differentiable function y : I → R satisfies the following inequality: n X |y 0 (t) − λy(t)| ≤ a k tk (2.25) k=0

for all t ∈ I if and only if there exists an increasing continuously differentiable

76

Hyers-Ulam Stability of Ordinary Differential Equations

function i : I → R such that y(t) = i(t)eλt +

n X

αk tk

(2.26)

ak tk e−λt

(2.27)

k=0

and 0 ≤ i0 (t) ≤ 2

n X k=0

for any t ∈ I. Proof 42 Let’s assume that a continuously differentiable function y : I → R, satisfies (2.25). So, −

n X

ak tk ≤ y 0 (t) − λy(t) ≤

k=0

n X

ak tk ,

k=0

that is, λy(t) −

n X

k

0

ak t ≤ y (t) ≤ λy(t) +

k=0

n X

ak tk , (∀t ∈ I).

(2.28)

k=0

(2.28) implies following two inequalities 0

y (t) − λy(t) −

n X

ak tk ≤ 0

(2.29)

ak tk ≥ 0

(2.30)

k=0

and y 0 (t) − λy(t) +

n X k=0

for each t ∈ I. Let’s define a function z1 : I → R such that z1 (t) = y(t) +

n X

αk tk .

k=0

Indeed, z1 (t) is continuously differentiable and z10 (t) = y 0 (t) +

n X

kαk tk−1 .

k=1

Therefore, z10 (t) − λz1 (t) = y 0 (t) + 0

= y (t) + = y 0 (t) +

n X k=1 n X k=1 n X k=1

kαk tk−1 − λz1 (t) kαk t

k−1

− λ y(t) +

kαk tk−1 − λy(t) − λ

n X k=0 n X

! αk t

k

αk tk .

k=0

(2.31)

Stability of First Order Linear Differential Equations

77

Applying (2.23) to (2.31), we get z10 (t) − λz1 (t) = (y 0 (t) − λy(t)) −

n X

a k tk

k=0

and hence (2.29) becomes z10 (t) − λz1 (t) ≤ 0, t ∈ I. By Lemma 9(b), there exists a decreasing continuously differentiable function d : I → R such that n X

z1 (t) = y(t) +

αk tk = d(t)eλt ,

k=0

that is, y(t) = d(t)eλt −

n X

αk tk , ∀t ∈ I.

(2.32)

k=0

Consider, z2 : I → R which is continuously differentiable function defined by z2 (t) = y(t) −

n X

αk tk .

k=0

Therefore, n X

z20 (t) = y 0 (t) −

kαk tk−1

k=1

implies that z20 (t)

0

− λz2 (t) = y (t) −

n X

kαk t

k−1

− λ y(t) −

k=1 0

= (y (t) − λy(t)) +

n X

k

λαk t −

k=0

n X

! αk t

k=0 n X

k

kαk t

, (2.33)

! k−1

.

k=1

Applying (2.23), (2.33) becomes z20 (t) − λz2 (t) = (y 0 (t) − λy(t)) +

n X

ak tk ,

k=0

that is, z20 (t) − λz2 (t) ≥ 0, t ∈ I. By Lemma 9(a), there exists an increasing continuously differentiable function i : I → R such that z2 (t) = y(t) −

n X k=0

αk tk = i(t)eλt

78

Hyers-Ulam Stability of Ordinary Differential Equations

implies that λt

y(t) = i(t)e

+

n X

αk tk

(2.34)

k=0

which is (2.27). Comparing (2.32) and (2.34), we obtain n X

y(t) = d(t)eλt −

αk tk = i(t)eλt +

k=0

n X

αk tk .

(2.35)

k=0

Differentiating (2.35) with respect to t, we have n X

d0 (t)eλt + λd(t)eλt −

kαk tk−1 = i0 (t)eλt + λi(t)eλt +

k=1

n X

kαk tk−1 ,

k=1

that is, 0

d (t)e

λt

+ λ y(t) +

= i0 (t)eλt + λ y(t) −

n X k=0 n X

! αk t

k

− !

αk tk

+

k=0

n X k=1 n X

kαk tk−1 kαk tk−1

k=1

due to (2.32) and (2.34). Consequently, d0 (t)eλt +

n X

λαk tk −

n X

λαk tk +

n X k=1

k=0

k=1

k=0

n X

kαk tk−1 = i0 (t)eλt −

implies that d0 (t)eλt = i0 (t)eλt − 2

n X

λαk tk + 2

n X

kαk tk−1 ,

k=1

k=0

that is, 0

n X

0

d (t) = i (t) + 2

kαk t

k−1

k=1

−

n X

! k

λαk t

e−λt .

k=0

0

Since d(t) is decreasing, then d (t) ≤ 0. Hence, n X

0

i (t) − 2

k

λαk t −

k=0

n X

! kαk t

k−1

e−λt ≤ 0.

k=1

By (2.23), 0

i (t) − 2

n X k=0

! k

ak t

e−λt ≤ 0.

kαk tk−1 ,

Stability of First Order Linear Differential Equations

79

Because i(t) is an increasing function, so !

n X

0

0 ≤ i (t) ≤ 2

k

ak t

e−λt

k=0

which is the inequality (2.27). Conversly, suppose that a continuously differentiable function y : I → R is such that n X −λt y(t) = i(t)e + αk tk , k=0

where i : I → R is a continuously differentiable function satisfying (2.27). Now, n X y 0 (t) = i0 (t)eλt + λi(t)eλt + kαk tk−1 , ∀t ∈ I k=1

implies that y 0 (t) − λy(t) = i0 (t)eλt + λi(t)eλt +

n X

kαk tk−1 − λ i(t)eλt +

= i (t)e

λt

−

n X

k

λαk t −

! α k tk

,

k=0

k=1 0

n X

!

n X

k−1

kαk t

.

k=1

k=0

Upon using (2.23), it follows that 0

0

y (t) − λy(t) = i (t)e

λt

−

n X

ak tk .

k=0

By (2.27), 0 ≤ i0 (t)eλt ≤ 2

n X

a k tk .

k=0

Ultimately, −

n X k=0

ak tk ≤ i0 (t)eλt −

n X

ak tk ≤

k=0

n X

ak tk

k=0

and therefore, −

n X k=0

ak tk ≤ y 0 (t) − λy(t) ≤

n X

ak tk ∀t ∈ I

k=0

due to (2.36). This completes the proof of the theorem.

(2.36)

80

Hyers-Ulam Stability of Ordinary Differential Equations

Theorem 22 Let λ be a non-zero real constant. Assume that I = (a, b) is an arbitrary open interval with −∞ ≤ a < b ≤ ∞. If a continuously differentiable function y : I → R satisfies the inequality (2.25) for all t ∈ I, then there exists a real number c such that ( P Pn n k λ(t−s) | k=0 αk tk − lims→b− | (f orλ > 0), λt k=0 αk s e |y(t) − ce | ≤ Pn P n k k λ(t−s) | k=0 αk t − lims→a+ | (f orλ < 0), k=0 αk s e (2.37) for all t ∈ I. Proof 43 Consider the case when λ > 0. Define ( ) n X k −λs αk s e , c = lim i(s) + s→b−

k=0

where i : I → R is an increasing continuously differentiable function given in Theorem 21. Integrating (2.27) from t to b we obtain 0≤

b (i(τ ))t

≤2

n X

k X k!τ k−i e−λτ − (k − i)!λi+1 i=0

ak

k=0

!b ,

(2.38)

t

where we have used the formula Z

τ k e−λτ dτ = −

k X i=0

k!τ k−i e−λτ (k − i)!λi+1

for k = 0, 1, 2, · · · . Now, (2.38) implies that 0 ≤ i(b) − i(t) ≤ −2

n X k=0

ak

k X k!τ k−i e−λτ (k − i)!λi+1 i=0

!b , t

that is, 0 ≤ lim i(s) − i(t) ≤ −2 s→b−

≤ −2 lim

s→b−

n X

ak

k=0 n X k=0

k k X X k!bk−i e−λb k!tk−i e−λt − (k − i)!λi+1 i=0 (k − i)!λi+1 i=0

!

k n k X X X k!sk−i e−λs k!tk−i e−λt ak + 2 a . k (k − i)!λi+1 (k − i)!λi+1 i=0 i=0 k=0

Applying Lemma 10, to the above inequality, we find 0 ≤ lim i(s)−i(t) ≤ −2 lim s→b−

s→b−

n X n X k=0 i=k

i!ai sk k!λi+1−k

! e

−λs

+2

n X n X k=0 i=k

i!ai tk k!λi+1−k

! e−λt ,

Stability of First Order Linear Differential Equations

81

that is, by using (2.22) n X

0 ≤ lim i(s) − i(t) ≤ −2 lim s→b−

s→b−

n X

≤ c − lim

s→b−

n X

αk sk e−λs + 2

k=0

αk tk e−λt

k=0

αk sk e−λs − i(t) ≤ −2 lim

s→b−

k=0

n X

n X

αk sk e−λs + 2

k=0

αk tk e−λt .

k=0

(2.39) Multiplying eλt to (2.39), we get n X

0 ≤ ceλt − lim

s→b−

αk sk e−λs eλt − i(t)eλt ≤ −2 lim

s→b−

k=0

that is, by adding lims→b− equality, we get lim

s→b−

n X

αk sk eλ(t−s) −

Pn

k=0

n X

αk tk ≤ ceλt − ≤ − lim

n X

αk tk e0 ,

k=0

Pn

n X

k=0

αk tk to the above in-

αk tk − i(t)eλt

k=0 n X

s→b−

αk sk e−λs eλt + 2

k=0

αk sk eλ(t−s) −

k=0

k=0

n X

αk sk eλ(t−s) +

k=0

n X

α k tk .

k=0

Therefore, −

n X

αk tk − lim

s→b−

k=0

n X

! αk sk eλ(t−s)

≤ ceλt −

k=0

n X

! αk tk + i(t)eλt

k=0

≤

n X

αk tk − lim

s→b−

k=0

n X

αk sk eλ(t−s) .

k=0

(2.40) By (2.26) and (2.40) we obtain −

n X

k

αk t − lim

k=0

s→b−

n X

! k λ(t−s)

αk s e

≤ ceλt − y(t)

k=0

≤

n X

αk tk − lim

s→b−

k=0

n X

αk sk eλ(t−s) .

k=0

Consequently, λt

|ce

− y(t)| ≤

n X k=0

k

αk t − lim

s→b−

n X

αk sk eλ(t−s) ,

k=0

that is, n n X X |y(t) − ceλt | ≤ αk tk − lim αk sk eλ(t−s) . k=0

s→b−

k=0

82

Hyers-Ulam Stability of Ordinary Differential Equations

Let λ be a negative real constant and define ( ) n X k −λs , c = lim i(s) + αk s e s→a+

k=0

where i is the increasing continuously differentiable function defined by i : I → R. Now, integrating the inequality in (2.27) from a to t, we find !t n k X X k!τ k−i e−λτ t 0 ≤ (i(τ ))a ≤ 2 ak − (2.41) (k − i)!λi+1 i=0 k=0

a

implies that 0 ≤ i(t) − i(a) ≤ −2

n X

k X k!τ k−i e−λτ (k − i)!λi+1 i=0

ak

k=0

!t , a

that is, 0 ≤ i(t) − lim i(s) ≤ −2 s→a+

n X

k k X X k!tk−i e−λt k!ak−i e−λa − (k − i)!λi+1 (k − i)!λi+1 i=0 i=0

ak

k=0

≤ −2

n X

ak

!

k n k X X X k!sk−i e−λs k!tk−i e−λt a + 2 lim . k s→a+ (k − i)!λi+1 (k − i)!λi+1 i=0 i=0 k=0

k=0

Applying Lemma 10 to the above inequality, we find n X n X

0 ≤ i(t)− lim i(s) ≤ −2 s→a+

k=0 i=k

i!ai tk k!λi+1−k

! −λt

e

n X n X

+2 lim

s→a+

k=0 i=k

i!ai sk k!λi+1−k

! e−λs ,

that is, by using (2.22) 0 ≤ i(t) − lim i(s) ≤ −2 s→a+

n X

αk tk e−λt + 2 lim

s→a+

k=0

≤ i(t) − c + lim

s→a+

n X

αk sk e−λs ≤ −2

k=0

n X

n X

αk sk e−λs

k=0

αk tk e−λt + 2 lim

n X

s→a+

k=0

αk sk e−λs .

k=0

(2.42) Multiplying eλt to (2.42), we get 0 ≤ i(t)eλt − ceλt + lim

n X

s→a+

that is, by adding equality, we get n X k=0

αk tk − lim

k=0

Pn

s→a+

αk sk e−λs eλt ≤ −2

n X

αk tk e0 + 2 lim

αk t − lims→a+

Pn

k=0

k λ(t−s)

αk s e

αk sk eλ(t−s) ≤ i(t)eλt − ceλt +

k=0

n X

s→a+

k=0

k

k=0

n X

n X

αk sk e−λs eλt ,

k=0

to the above in-

αk tk

k=0

≤−

n X k=0

αk tk + lim

s→a+

n X k=0

αk sk eλ(t−s) .

Stability of First Order Linear Differential Equations

83

Therefore, −

lim

s→a+

n X

k λ(t−s)

αk s e

−

k=0

n X

! αk t

k

≤

i(t)e

λt

+

k=0

n X

! αk t

k

− ceλt

k=0

≤ lim

s→a+

n X

αk sk eλ(t−s) −

k=0

n X

αk tk .

k=0

(2.43) By (2.26) and (2.43) we obtain −

lim

s→a+

n X

k λ(t−s)

αk s e

−

k=0

n X

! αk t

k

≤ y(t) − ceλt

k=0

≤ lim

s→a+

n X

αk sk eλ(t−s) −

n X

αk tk .

k=0

k=0

Consequently, |y(t) − ceλt | ≤ lim

s→a+

n X

αk sk eλ(t−s) −

n X

αk tk ,

k=0

k=0

that is, X n X n αk sk eλ(t−s) . αk tk − lim |y(t) − ceλt | ≤ k=0

s→a+

k=0

Hence, the theorem is proved. Corollary 15 Let λ be a non-zero real constant and define ( (a, ∞) (f or λ > 0), I= (−∞, b) (f or λ < 0), where a ∈ R∪{−∞} and b ∈ R∪{∞} are fixed. If a continuously differentiable function y : I → R satisfies the inequality (2.25) for all t ∈ I, then there exists a unique real number c such that n X λt k |y(t) − ce | ≤ αk t (2.44) k=0

for all t ∈ I. Proof 44 According to Theorem 22, there exists a constant c ∈ R such that the inequality (2.44) holds for all t ∈ I. Hence, we only prove the uniqueness

84

Hyers-Ulam Stability of Ordinary Differential Equations

of c. Let c1 is another real number with which the inequality (2.44) is satisfied. Now, |c − c1 | = |ce−λt eλt − c1 e−λt eλt + e−λt y(t) − e−λt y(t)| = |e−λt y(t) − c1 e−λt eλt + ce−λt eλt − e−λt y(t)| ≤ e−λt |y(t) − c1 e−λt | + e−λt |ceλt − y(t)| = e−λt |y(t) − c1 e−λt | + e−λt |y(t) − ceλt | n X k −λt αk t ≤ 2e k=0

→ 0, when t → ∞ for λ > 0 or t → −∞ for λ < 0, which completes the proof. Theorem 23 Let a be a real constant and let I = (a, b) be an arbitrary nondegenerate open interval. Assume that a function y : I ∪ {a} → R is continuously differentiable on I and continuous at a on the right. If y satisfies the inequality (2.25) for any t ∈ I, then ! n n X X k y(a) − αk a eλ(t−a) + αk tk ≤ y(t) k=0

k=0

≤

y(a) +

n X

! k

αk a

e

λ(t−a)

k=0

−

n X

αk tk

k=0

for all t ∈ I. Proof 45 Let c = lims→a+ i(s), where i : I → R is from Theorem 21. Integrating (2.27) from a to t, we obtain 0≤

t (i(τ ))a

≤2

n X

−

ak

k X i=0

k=0

k!τ k−i e−λτ (k − i)!λi+1

!t , a

that is, 0 ≤ i(t) − i(a) ≤ −2

n X k=0

ak

k n k X X X k!tk−i e−λt k!ak−i e−λa + 2 a . k i+1 (k − i)!λ (k − i)!λi+1 i=0 i=0 k=0

Consequently, 0 ≤ i(t) − c ≤ −2

n X n X i!ai tk e−λt k=0 i=k

k!λi+1−k

+2

n X n X i!ai ak e−λa k=0 i=k

k!λi+1−k

due to Lemma 10. Hence, 0 ≤ i(t) − c ≤ −2

n X k=0

αk tk e−λt + 2

n X k=0

αk ak e−λa

Stability of First Order Linear Differential Equations

85

due to (2.22). Thus, 0 ≤ i(t)eλt − ceλt ≤ −2

n X

αk tk + 2

k=0

n X

αk ak eλ(t−a)

k=0

implies that n X

αk tk ≤ i(t)eλt +

k=0

n X

αk tk − ceλt ≤ −

k=0

n X

αk tk + 2

k=0

n X

αk ak eλ(t−a) .

k=0

Upon using (2.26), the above inequality yields n X

αk tk ≤ y(t) − ceλt ≤ −

k=0

n X

αk tk + 2

k=0

n X

αk ak eλ(t−a) ,

k=0

that is, ceλt +

n X

αk tk ≤ y(t) ≤ ceλt + 2

k=0

n X

αk ak eλ(t−a) −

k=0

n X

αk tk

k=0

implies that λt

ce

+

n X

k

αk t ≤ y(t) ≤

λa

ce

+2

k=0

n X

! k

αk a

eλ(t−a) −

k=0

n X

αk tk .

(2.45)

k=0

Taking t → a+ in (2.26), we obtain y(a) = i(a)e

λa

+

n X

k

λa

αk a = ce

+

n X

αk ak .

(2.46)

k=0

k=0

As t → a+, (2.45) becomes ceλa +

n X

αk ak ≤ y(a) ≤ ceλa + 2

k=0

n X

αk ak −

k=0

n X

αk ak ,

k=0

that is, λa

y(a) = ce

+

n X

αk ak

k=0

which is same as (2.46). Hence, the theorem is proved. Example 2 Let y : I → R be a continuously differentiable function that satisfy (2.25). Choose a0 = ε and ak = 0, for k ∈ N. Then |y 0 (t) − λy(t)| ≤ ε, ∀t ∈ I. By (2.22), α0 = ελ−1 and αk = 0, for k ∈ N. By Corollary 15, there exists a unique real number c such that |y(t) − ceλt | ≤ ε|λ|−1 , ∀t ∈ I, which is the Hyers-Ulam stability of the differential equation (2.20).

86

Hyers-Ulam Stability of Ordinary Differential Equations

Example 3 Assume that a0 and a1 satisfy a0 + a1 t ≥ 0, ∀t ∈ I. By (2.22), α0 = a0 λ−1 + a1 λ−2 and α1 = a1 λ−1 and ak = 0, k = 2, 3, 4, .... Let a continuously differentiable function y : I → R be satisfy (2.25), that is, |y 0 (t) − λy(t)| ≤ a0 + a1 t ∀t ∈ I. Then by Corollary 15, there exists a unique real number c such that |y(t) − ceλt | ≤ |a0 λ−1 + a1 λ−2 + a1 λ−1 t|∀t ∈ I, which is the Hyers-Ulam stability of the differential equation (2.20).

2.3

Stability of ϕ(t)y 0 (t) = y(t)

This section deals with the Hyers-Ulam-Rassias stability of the following linear differential equation: ϕ(t)y 0 (t) = y(t), t ∈ I,

(2.47)

where I = (a, b) is an arbitrary open interval and we assume that a and b satisfy −∞ ≤ a < Rb < +∞. Also, we assume that ϕ : I → R is a given t function for which a dτ /ϕ(τ ) exists for any t ∈ I. In brief, we prove that if either ϕ(t) ≥ 0 satisfies ∀t ∈ I or ϕ(t) < 0 holds ∀t ∈ I, and also if a differentiable function y : I → R satisfies the inequality |ϕ(t)y 0 (t) − y(t)| ≤ ε for all t ∈ I, then there exists a real number c such that (Z ) t dτ ≤ ε, ∀t ∈ I. y(t) − c exp a ϕ(τ ) Remark 2 If ϕ(t) = λ1 , λ ∈ R \ {0}, t ∈ I, then (2.47) becomes (2.20). Otherwise, (2.47) is more general than (2.20). Hence, the results followed by (2.47) are more general results than (2.20). Lemma 11 Suppose a differentiable function z : I → R is given. (a) The inequality z(t) ≤ ϕ(t)z 0 (t) is true for all t ∈ I if and only if there exists a differentiable function α : I → R such that α0 (t)ϕ(t) ≥ 0 and (Z ) t dτ z(t) = α(t) exp , a ϕ(τ ) for all t ∈ I.

Stability of First Order Linear Differential Equations

87

(b) The inequality z(t) ≥ ϕ(t)z 0 (t) holds true for any t ∈ I if and only if there exists a differentiable function β : I → R such that β 0 (t)ϕ(t) ≤ 0 and (Z ) t dτ z(t) = β(t) exp a ϕ(τ ) for all t ∈ I. Proof 46 (a) Suppose that z(t) ≤ ϕ(t)z 0 (t) is true for all t ∈ I. To prove α0 (t)ϕ(t) ≥ 0, ∀t ∈ I. Define a function α : I → R such that ( Z ) t dτ α(t) = exp − z(t). a ϕ(τ ) Then, 1 α0 (t) = exp ϕ(t)

( Z −

) dτ (ϕ(t)z 0 (t) − z(t)). ϕ(τ )

t

a

So, α0 (t)ϕ(t) ≥ 0 which is true for all t ∈ I. Conversely, let there exist a differentiable function α : I → R with α0 (t)ϕ(t) ≥ 0, for each t ∈ I. To prove that z(t) ≤ ϕ(t)z 0 (t) is true for all t ∈ I. If (Z ) t dτ z(t) = α(t) exp , a ϕ(τ ) then z 0 (t) = α0 (t) exp

(Z

t

a

dτ ϕ(τ )

) +

z(t) , ϕ(t)

that is, 0

(Z

0

ϕ(t)z (t) = α (t)ϕ(t) exp a

t

dτ ϕ(τ )

) + z(t) ≥ z(t)

for all t ∈ I. (b) Suppose that z(t) ≥ ϕ(t)z 0 (t) is true for all t ∈ I. To prove β 0 (t)ϕ(t) ≤ 0 ∀t ∈ I. Define a function β : I → R such that ( Z ) t dτ β(t) = exp − z(t). a ϕ(τ ) Then 1 β 0 (t) = exp ϕ(t)

(

Z − a

t

) dτ (ϕ(t)z 0 (t) − z(t)), ϕ(τ )

So, β 0 (t)ϕ(t) ≤ 0 which is true for all t ∈ I.

88

Hyers-Ulam Stability of Ordinary Differential Equations

Conversely, assume that there exists a differentiable function β : I → R with β 0 (t)ϕ(t) ≤ 0, for each t ∈ I. To prove z(t) ≥ ϕ(t)z 0 (t) is true for all t ∈ I. If (Z ) t dτ z(t) = β(t) exp , a ϕ(τ ) then, 0

0

(Z

t

dτ ϕ(τ )

z (t) = β (t) exp a

) +

z(t) , ϕ(t)

that is, ϕ(t)z 0 (t) = β 0 (t)ϕ(t) exp

(Z

t

a

dτ ϕ(τ )

) + z(t) ≤ z(t)

for all t ∈ I. This completes the proof of the lemma. Theorem 24 Given an ε > 0, a differentiable function y : I → R is a solution of the following inequality: |ϕ(t)y 0 (t) − y(t)| ≤ ε

(2.48)

for all t ∈ I if and only if there exists a differentiable function α : I → R such that ) (Z t dτ (2.49) y(t) = ε + α(t) exp a ϕ(τ ) and

( 0

0 ≤ α (t)ϕ(t) ≤ 2ε exp

Z − a

t

dτ ϕ(τ )

) (2.50)

for any t ∈ I. Proof 47 Suppose that a differentiable function y : I → R is a solution of (2.48), that is, y satisfies the inequality y(t) − ε ≤ ϕ(t)y 0 (t) ≤ y(t) + ε

(2.51)

for all t ∈ I. To prove that there exists a differentiable function α : I → R such that the function α(t) satisfies (2.49) and (2.50). Define z1 (t) = y(t) − ε. Hence from (2.51), z1 (t) ≤ ϕ(t)z10 (t), ∀t ∈ I. By Lemma 11(a), there exists a differentiable function α : I → R such that ) (Z t dτ , z1 (t) = α(t) exp a ϕ(τ ) that is, (Z y(t) − ε = α(t) exp a

t

) dτ . ϕ(τ )

Stability of First Order Linear Differential Equations

89

So, (2.49) holds for all t ∈ I, where α satisfies α0 (t)ϕ(t) ≥ 0 for all t ∈ I due to Lemma 11(a). If we choose, z2 (t) = y(t) + ε, then (2.51) implies that z2 (t) ≥ ϕ(t)z20 (t), ∀t ∈ I. By Lemma 11(b), there exists a differentiable function β : I → R such that ) (Z t dτ , z2 (t) = β(t) exp a ϕ(τ ) that is, (Z

t

y(t) + ε = β(t) exp a

dτ ϕ(τ )

) (2.52)

and β 0 (t)ϕ(t) ≤ 0, ∀t ∈ I due to Lemma 11(b). Using (2.49) and (2.52), we obtain ) (Z ) (Z t t α(t) dτ dτ 0 0 + exp (2.53) y (t) = α (t) exp ϕ(t) a ϕ(τ ) a ϕ(τ ) and 0

(Z

0

t

y (t) = β (t) exp a

= β 0 (t) exp

(Z

t

a

dτ ϕ(τ )

)

dτ ϕ(τ )

)

(Z

β(t) + ϕ(t) +

t

exp a

dτ ϕ(τ )

)!

1 (y(t) + ε) ϕ(t)

respectively. Again, we use (2.49) in the last relation to get ) (Z )! (Z t t 1 dτ dτ 0 0 + 2ε + α(t) exp . (2.54) y (t) = β (t) exp ϕ(t) a ϕ(τ ) a ϕ(τ ) Comparing (2.53) and (2.54), it follows that (Z ) (Z ) t t dτ dτ 2ε 0 0 α (t) exp = β (t) exp + , ϕ(τ ) ϕ(τ ) ϕ(t) a a that is, ( 0

0

0 ≥ β (t)ϕ(t) = α (t)ϕ(t) − 2ε exp

Z − a

t

dτ ϕ(τ )

)

implies that ( 0 ≤ α0 (t)ϕ(t) ≤ 2ε exp

Z − a

t

) dτ . ϕ(τ )

Conversely, assume that there exists a differentiable function α : I → R such that the function α(t) satisfies (2.49) and (2.50). To prove that a

90

Hyers-Ulam Stability of Ordinary Differential Equations

differentiable function y : I → R is a solution of (2.48). Clearly, (2.53) holds, that is, (Z ) (Z ) t t dτ dτ 0 0 y (t)ϕ(t) = α (t)ϕ(t) exp + α(t) exp . a ϕ(τ ) a ϕ(τ ) Consequently, y 0 (t)ϕ(t) − y(t) = α0 (t)ϕ(t) exp

(Z a

t

dτ ϕ(τ )

) −ε

due to (2.49), that is, ( Z (y (t)ϕ(t) − y(t) + ε) exp − 0

a

t

dτ ϕ(τ )

) = α0 (t)ϕ(t) ( Z ≤ 2ε exp − a

t

dτ ϕ(τ )

)

due to (2.50). Therefore, 0 ≤ (y 0 (t)ϕ(t) − y(t) + ε) ≤ 2ε and |ϕ(t)y 0 (t) − y(t)| ≤ ε. Hence, the theorem is proved. Theorem 25 If either ϕ(t) > 0 holds for all t ∈ I or ϕ(t) < 0 holds for all t ∈ I, and if a differentiable function y : I → R satisfies inequality (2.48) for all t ∈ I, then there exists a real number c such that (Z ) t dτ (2.55) y(t) − c exp ≤ε a ϕ(τ ) for any t ∈ I. Proof 48 Suppose that ϕ(t) > 0, for all t ∈ I and consider a differentiable function y : I → R which satisfies (2.48) ∀ t ∈ I. Define c = limt→b− α(t), where α : I → R. Using the fact that ϕ0 (t) > 0, (2.50) becomes ( Z ) t dτ 2ε 0 exp − . 0 ≥ −α (t) ≥ − ϕ(t) a ϕ(τ ) Integrating the above inequality from t to b, we obtain ( Z )  Z b s −1 dτ b 0 ≥ − (α(s))t ≥ 2ε exp − ds, ϕ(s) t a ϕ(τ ) that is, ( 0 ≥ α(t) − α(b) ≥ 2ε exp

Z − a

s

dτ ϕ(τ )

)!b . t

Stability of First Order Linear Differential Equations

91

Thus, ( 0 ≥ α(t) − c ≥ 2ε exp

Z

b

− a

dτ ϕ(τ )

)

( − 2ε exp

t

Z

dτ ϕ(τ )

− a

)

implies that (Z

t

0 ≥ (α(t) − c) exp a

dτ ϕ(τ )

)

(Z

t

≥ 2ε exp b

dτ ϕ(τ )

) − 2ε,

that is, (Z ε ≥ (α(t) − c) exp a

t

dτ ϕ(τ )

)

( + ε ≥ 2ε exp

Z

b

− t

dτ ϕ(τ )

) − ε ≥ −ε.

Consequently, the above inequality becomes ) (Z ) (Z t t dτ dτ − c exp ε ≥ ε + α(t) exp a ϕ(τ ) a ϕ(τ ) ( Z ) b dτ ≥ 2ε exp − − ε ≥ −ε t ϕ(τ ) which is because of (2.49) and equivalent to (Z t

ε ≥ y(t) − c exp a

dτ ϕ(τ )

) ≥ −ε.

Next, we assume that ϕ(t) < 0 and consider c = limt→a+ α(t). Dividing the inequalities (2.50) by ϕ(t), we get ( Z ) t 2ε dτ 0 ≥ α0 (t) ≥ exp − . ϕ(t) a ϕ(τ ) Integrating the above inequality from a to t, we obtain ( Z ) Z t s 1 dτ t 0 ≥ (α0 (t))a ≥ 2ε exp − ds, a ϕ(s) a ϕ(τ ) that is, ( 0 ≥ α(t) − α(a) ≥ −2ε exp

Z − a

s

dτ ϕ(τ )

)!t . a

The rest of the proof follows from the above case. This completes the proof of the theorem.

92

Hyers-Ulam Stability of Ordinary Differential Equations

Corollary 16 Assume that a is a real number and that ϕ(t) > 0 holds for all t ∈ I. Let a function y : I ∪ {a} → R be differentiable on I and continuous at a on the right. If y satisfies (2.48) for all t ∈ I and for some ε > O, then (Z ) (Z ) t t dτ dτ (y(a) − ε) exp + ε ≤ y(t) ≤ (y(a) + ε) exp − ε (2.56) a ϕ(τ ) a ϕ(τ ) for any t ∈ I. Proof 49 If y satisfies (2.48) for any t ∈ I, then Theorem 24 implies that there exists a differentiable function α : I → R satisfying (2.49) and (2.50) for all t ∈ I. Define c = limt→a+ α(t), where α : I → R is given in Theorem 24. Since ϕ(t) > 0 for t ∈ I, (2.50) becomes ( Z ) t dτ 2ε 0 exp − . (2.57) 0 ≤ α (t) ≤ ϕ(t) a ϕ(τ ) Integrating (2.57) from a to t, we obtain ) ( Z  Z t s −1 dτ t 0 ≤ (α(s))a ≤ (−2ε) ds, exp − ϕ(s) a a ϕ(τ ) that is, ( 0 ≤ α(t) − c ≤ (−2ε) exp

Z − a

t

dτ ϕ(τ )

) + 2ε.

Hence, (Z

t

0 ≤ (α(t) − c) exp a

dτ ϕ(τ )

)

(Z

t

≤ 2ε exp a

dτ ϕ(τ )

) − 2ε

implies that (Z 0 ≤ y(t) − ε − c exp a

t

dτ ϕ(τ )

)

(Z ≤ 2ε exp a

t

dτ ϕ(τ )

) − 2ε

due to (2.49), that is, (Z ) (Z ) (Z ) t t t dτ dτ dτ + ε ≤ y(t) ≤ 2ε exp + c exp − ε. c exp a ϕ(τ ) a ϕ(τ ) a ϕ(τ ) If t → a+ in (2.49), then c = y(a) − ε. Therefore, (Z ) (Z ) t t dτ dτ (y(a) − ε) exp + ε ≤ y(t) ≤ (2ε + c) exp −ε a ϕ(τ ) a ϕ(τ ) (Z ) t dτ ≤ (y(a) + ε) exp − ε. a ϕ(τ ) Hence, the corollary is proved.

Stability of First Order Linear Differential Equations

93

Corollary 17 Assume that a is a real number and that ϕ(t) < 0 is true for any t ∈ I. Moreover, assume that y : I ∪ {a} → R is a function which is differentiable on I and continuous at a on the right. If y satisfies (2.48) for all t ∈ I and for some ε > 0, then (Z ) (Z ) t t dτ dτ (y(a) + ε) exp − ε ≤ y(t) ≤ (y(a) − ε) exp + ε (2.58) a ϕ(τ ) a ϕ(τ ) for each t ∈ I. Proof 50 If y satisfies (2.48) for any t ∈ I, then Theorem 24 implies that there exists a differentiable function α : I → R, satisfying (2.49) and (2.50) for all t ∈ I. Define c = limt→a+ α(t), where α : I → R is given in Theorem 24. Since ϕ(t) < 0 for t ∈ I, then (2.50) becomes ( Z ) t 2ε dτ 0 0 ≥ α (t) ≥ exp − . (2.59) ϕ(t) a ϕ(τ ) Integrating (2.59) from a to t, we obtain 0≥

(α(s))ta

Z t ≥ (−2ε) a

−1 ϕ(s)

(



) dτ ds, ϕ(τ )

s

Z −

exp

a

that is, ( 0 ≥ α(t) − c ≥ 2ε − 2ε exp

t

Z − a

) dτ . ϕ(τ )

Hence, (Z

t

0 ≥ (α(t) − c) exp a

dτ ϕ(τ )

)

(Z

t

dτ ϕ(τ )

≥ 2ε exp a

) − 2ε

implies that (Z 0 ≥ y(t) − ε − c exp a

t

dτ ϕ(τ )

)

(Z

t

≥ 2ε exp a

dτ ϕ(τ )

) − 2ε

due to (2.49), that is, (Z ) (Z ) t t dτ dτ c exp + ε ≥ y(t) ≥ (2ε + c) exp −ε a ϕ(τ ) a ϕ(τ ) implies that (Z (2ε + c) exp a

t

dτ ϕ(τ )

)

(Z − ε ≤ y(t) ≤ c exp a

t

dτ ϕ(τ )

) + ε.

94

Hyers-Ulam Stability of Ordinary Differential Equations

If t → a+ in (2.49), then c = y(a) − ε. Therefore, (Z ) (Z ) t t dτ dτ (y(a) + ε) exp − ε ≤ y(t) ≤ (y(a) − ε) exp + ε. a ϕ(τ ) a ϕ(τ ) Hence, the corollary is proved.

2.4

Stability of p(x)y 0 − q(x)y − r(x) = 0

Definition 10 We say that p(x)y 0 − q(x)y − r(x) = 0 has the Hyers-Ulam stability if there exists a constant K ≥ 0 with the following property: for every  > 0, y ∈ C 1 (I), if |p(x)y 0 − q(x)y − r(x)| ≤ , then there exists some z ∈ C 1 (I) satisfying p(x)z 0 − q(x)z − r(x) = 0 such that |y(x) − z(x)| ≤ K, where I = (a, b), −∞ ≤ a ≤ b ≤ ∞. We call such K a Hyers-Ulam stability constant for the given equation. Theorem 26 Let p(x), q(x) and r(x) be continuous real functions defined on the interval I = (a, b) such that p(x) 6= 0 and |q(x)| ≥ δ for all x ∈ I and some δ > 0 independent of x. Then p(x)y 0 − q(x)y − r(x) = 0 has the Hyers-Ulam stability. Proof 51 Let  > 0 and y : I → R be a continuously differentiable function such that |p(x)y 0 − q(x)y − r(x)| ≤  (2.60) holds for all x ∈ I. We show that there exists a constant K > 0 independent of , y and x such that |y(x) − z(x)| ≤ K for all x ∈ I, and for some z ∈ C 1 (I) satisfying p(x)z 0 − q(x)z − r(x) = 0, x ∈ I. Without loss of generality, we assume that q(x) ≥ 1 for all x ∈ I. Let p(x) > 0 for all x ∈ I. In view of (2.60), we have − ≤ p(x)y 0 − q(x)y − r(x) ≤  and hence −

n 1 exp − p(x)

Z a

x

n Z x q(s) o q(s) o 1 ds ≤ exp − ds [p(x)y 0 − q(x)y − r(x)] p(s) p(x) p(s) a n Z x q(s) o 1 ≤ exp − ds , p(x) a p(s)

Stability of First Order Linear Differential Equations

95

that is, −

n q(x) exp − p(x)

Z

x

a

n Z x q(s) o q(s) o 1 ds ≤ exp − ds [p(x)y 0 − q(x)y − r(x)] p(s) p(x) p(s) a n Z x q(s) o q(x) ≤ exp − ds , p(x) a p(s)

that is, −

n q(x) exp − p(x)

Z

x

a

n Z x q(s) o q(s) o 1 ds ≤ exp − ds [p(x)y 0 − q(x)y] p(s) p(x) a p(s) n Z x q(s) o r(x) exp − ds − p(x) a p(s) Z x n q(x) q(s) o ≤ exp − ds . (2.61) p(x) a p(s)

If we choose b1 ∈ [a, b] such that y(b1 ) < ∞, then for any x ∈ (a, b1 ] and integrating (2.61) from x to b1 , we get ! n Z b1 q(s) o n Z x q(s) o ds − exp − ds −  exp − p(s) a a p(s) n Z b1 q(s) o n Z x q(s) o ≤ y(b1 )exp − ds − y(x)exp − ds p(s) a a p(s) Z b1 n Z s q(t) o r(s) − exp − dt ds p(s) x a p(t) ! n Z x q(s) o n Z b1 q(s) o ≤  exp − ds − exp − ds , p(s) a p(s) a that is, n Z −  exp −

x

q(s) o ds a p(s) n Z b1 q(s) o n Z x q(s) o ≤ (y(b1 ) − )exp − ds − y(x)exp − ds p(s) a a p(s) Z b1 n Z s q(t) o r(s) − exp − dt ds p(s) x a p(t) ! n Z x q(s) o n Z b1 q(s) o ≤  exp − ds − 2exp − ds . p(s) a p(s) a

96

Hyers-Ulam Stability of Ordinary Differential Equations

Therefore, n Z −  exp −

x

q(s) o ds a p(s) n Z b1 q(s) o n Z x q(s) o ≤ (y(b1 ) − )exp − ds − y(x)exp − ds p(s) a a p(s) Z b1 n Z s q(t) o n Z x q(s) o r(s) exp − dt ds ≤  exp − ds − p(s) a p(t) a p(s) x implies that " # n Z b1 q(s) o q(s) o −  ≤ exp ds (y(b1 ) − )exp − ds p(s) a p(s) a " # n Z s q(t) o n Z x q(s) o Z b1 r(s) ds exp − dt ds − y(x) ≤ . − exp p(s) x a p(t) a p(s) nZ

x

(2.62)

Similarly, for any x ∈ [b1 , b) and integrating (2.61) from b1 to x, we obtain " # n Z b1 q(s) o n Z x q(s) o ds (y(b1 ) − )exp − ds  ≤ y(x) − exp p(s) a a p(s) " # n Z x q(s) o Z b1 r(s) n Z s q(t) o + exp ds exp − dt ds p(s) a p(s) x a p(t) !   n Z x q(s) o n Z b q(s) o ≤  2exp ds − 1 ≤  2exp ds − 1 = (2A − 1), b1 p(s) a p(s) (2.63) where A = exp

nR

nZ

b a

o q(s) ds . If we select p(s)

x

q(s) o ds × a p(s) n Z × (y(b1 ) − )exp −

z1 (x) = exp "

a

b1

# Z b1 n Z s q(t) o q(s) o r(s) ds − exp − dt ds , p(s) p(s) x a p(t)

then combining (2.62) and (2.63), it follows that |y(x) − z1 (x)| ≤ (2A − 1), x ∈ I. We note that p(x)z10 − q(x)z1 − r(x) = 0 x ∈ I. For the case p(x) < 0, we can apply the above similar argument and obtain that |y(x) − z2 (x)| ≤ (2B − 1), x ∈ I,

Stability of First Order Linear Differential Equations 97 n R o b q(s) where B = exp − a p(s) ds and n Z x q(s) o z2 (x) = exp ds × a p(s) " # n Z b1 q(s) o Z b1 r(s) n Z s q(t) o × (y(b1 ) − )exp − ds − exp − dt ds . p(s) p(s) a x a p(t) Indeed, p(x)z20 − q(x)z2 − r(x) = 0 x ∈ I.

2.5

Stability of y 0 = λy on Banach Spaces

In this section, we consider the Hyers-Ulam-Rassias stability of the Banach space valued differential equation y 0 = λy, where λ is a complex constant. Let (X, ||.||) be a non-zero complex Banach space and I = (a, b) be an open interval in extended real line system. A function f (t) ∈ X is said to be strongly differentiable, if for all t ∈ I there exists a function f 0 (t) ∈ X, ∀t ∈ I such that f (t + s) − f (t) lim − f 0 (t) = 0. s→0 s Consider the following two equivalent statements: (i) f 0 (t) = λf (t), ∀t ∈ I. (ii) There is an x ∈ X such that f (t) = eλt x, ∀t ∈ I. Theorem 27 Suppose λ is a complex number, ε : I → [0, ∞) is a continuous function and f : I → X is a strongly differentiable function such that ||f 0 (t) − λf (t)|| ≤ (t)

(2.64)

for all t ∈ I. (a) If (t)e−Re(λ)t is integrable on (a, ta ] for some ta ∈ I, then there is a unique xa ∈ X such that Z t ||f (t) − eλt xa || ≤ eRe(λ)t (σ)e−Re(λ)σ dσ a

for all t ∈ I. (b) If (t)e−Re(λ)t is integrable on [tb , b) for some tb ∈ I, then there is a unique xb ∈ X such that Z b λt Re(λ)t ||f (t) − e xb || ≤ e (σ)e−Re(λ)σ dσ t

for all t ∈ I.

98

Hyers-Ulam Stability of Ordinary Differential Equations

Proof 52 Let X ∗ be the dual space of X and fϕ : I → C, defined by fϕ (t) = ϕ(f (t)), ∀t ∈ I, ϕ ∈ X ∗ and ||x|| = sup{|ψ(x)| : ψ ∈ X ∗ , ||ψ|| = 1, ∀x ∈ X}. (fϕ )0 (t) = ϕf 0 (t) and hence ϕ is continuous. Now, |(fϕ )0 (t) − λfϕ (t)| = |ϕf 0 (t) − ϕ(λf (t))| ≤ ||ϕ|| ||((f )0 (t)) − (λf (t))|| ≤ ||ϕ||(t) implies that Z t Z t 0 −λσ −λσ −λσ 0 {(f ) (σ)e − λf (σ)e }dσ {e = f (σ)} dσ ϕ ϕ ϕ s s Z t = {(fϕ )0 (σ) − λfϕ (σ)}e−λσ dσ s Z t −Re(λ)σ (σ)e dσ , ∀s, t ∈ I. ≤ ||ϕ||

(2.65)

s

Hence, |ϕ(e

−λt

f (t)−e

−λs

f (s))| = |e

−λt

fϕ (t)−e

−λs

Z t −Re(λ)σ (σ)e dσ fϕ (s))| ≤ ||ϕ|| s

(2.66) for all s, t ∈ I. Therefore, ||e−λt f (t) − e−λs f (s)|| = sup{|ψ(e−λt f (t) − e−λs f (s))| : ψ ∈ X ∗ , ||ψ|| = 1} Z t   ∗ −Re(λ)σ ≤ sup ||ψ|| (σ)e dσ : ψ ∈ X , ||ψ|| = 1 s Z t (2.67) (σ)e−Re(λ)σ dσ , ∀s, t ∈ I. = s

(a) By (2.67) it follows that {e−λs f (s)}s∈I is a Cauchy net and so f (s)} converges to an element xa ∈ X when s → a +. Now,

−λs

{e

||f (t) − eλt xa || = ||f (t) − eλ(t−s) f (s) + eλ(t−s) f (s) − eλt xa || ≤ ||f (t) − eλ(t−s) f (s)|| + ||eλ(t−s) f (s) − eλt xa || = ||eλt (e−λt f (t) − e−λs f (s))|| + ||eλt (e−λs f (s) − xa )|| = |eλt | ||(e−λt f (t) − e−λs f (s))|| + |eλt | ||(e−λs f (s) − xa )|| Z t Re(λ)t −Re(λ)σ ≤e (σ)e dσ + eRe(λ)t ||e−λs f (s) − xa ||, ∀s, t ∈ I s

(2.68) due to (2.67). As s → a+, e−λs f (s) → xa . Hence, Z t ||f (t) − eλt xa || ≤ eRe(λ)t (σ)e−Re(λ)σ dσ, ∀t ∈ I. a

(2.69)

Stability of First Order Linear Differential Equations

99

For any x ∈ X, let’s suppose that λt

||f (t) − e x|| ≤ e

Re(λ)t

Z

t

(σ)e−Re(λ)σ dσ, ∀t ∈ I.

a

Then ||xa − x|| = ||xa − e−λt f (t) + e−λt f (t) − x|| ≤ ||xa − e−λt f (t)|| + ||e−λt f (t) − x|| = ||e−λt (eλt xa − f (t))|| + ||e−λt (f (t) − eλt x)|| = |e−λt | ||eλt xa − f (t)|| + |e−λt | ||f (t) − eλt x|| Z t ≤ e−Re(λ)t eRe(λ)t (σ)e−Re(λ)σ dσ + e−Re(λ)t eRe(λ)t a Z t × (σ)e−Re(λ)σ dσ a Z t =2 (σ)e−Re(λ)σ dσ a

→ 0 when t → a +. Consequently, xa is unique. (b) Now suppose that (t)e−Re(λ)t is integrable on [tb , b) for some tb ∈ I. By (a), (t)e−Re(λ)t is integrable on [t0 , b), ∀t0 ∈ I. By (2.67), e−λs f (s) converges to xb such that xb ∈ X when s → b − . By (2.65), we obtain Z s Z s −Re(λ)(σ) −λσ 0 (2.70) (σ)e dσ , ∀s, t ∈ I. {e fϕ (σ)} dσ ≤ ||ϕ|| t

t

Since

Z

s

{e−λσ fϕ (σ)}0 dσ = ϕ(e−λs f (s) − e−λt f (t))

(2.71)

t

for all s, t ∈ I, then from (2.70) and (2.71) we find Z s −λs −λt −Re(λ)σ kϕ(e f (s) − e f (t))k ≤ ||ϕ|| (σ)e dσ

(2.72)

t

for all s, t ∈ I. As a result, Z ||e−λt f (t) − e−λs f (s)|| ≤

t

s

(σ)e−Re(λ)σ dσ

for all s, t ∈ I. Taking s → b− and proceeding as in (2.68), it follows that Z s ||f (t)−eλt xb || ≤ eRe(λ)t (σ)e−Re(λ)σ dσ +eRe(λ)t ||e−λs f (s)−xb ||, ∀s, t ∈ I. t

100

Hyers-Ulam Stability of Ordinary Differential Equations

Since e−λs f (s) → xb as s → b+, then Z λt Re(λ)t ||f (t) − e xb || ≤ e

b

(σ)e

−Re(λ)σ

t

dσ , ∀t ∈ I.

Using same type of reasoning as above, we can prove that xb is unique. Hence, the theorem is proved. Corollary 18 Let f : I → X be a strongly differentiable function that satisfies the inequality ||f 0 (t) − λf (t)|| ≤ ε (2.73) for all t ∈ I and for some ε > 0. If Re(λ) 6= 0 and m = 0, there exists a unique x0 ∈ X such that sup ||f (t) − eλt x0 || < ∞. t∈I

Moreover, for the above x0 ∈ X, the following estimate ε ||f (t) − eλt x0 || ≤ |Re(λ)|

(2.74)

holds for all t ∈ I, where m = inf{e−Re(λ)t : t ∈ I} = limσ→a+ e−Re(λ)σ . Proof 53 For Re(λ) < 0, εe−Re(λ)t is integrable on (a, ta ] for any ta ∈ I. By Theorem 27(a), there is an xa ∈ X such that Z t ε ε ||f (t) − eλt xa || ≤ εeRe(λ)t −m= e−Re(λ)σ dσ = |Re(λ)| |Re(λ)| a for all t ∈ I. Again, for Re(λ) > 0, εe−Re(λ)t is integrable on [tb , b) for any tb ∈ I. By Theorem 27(b), there exists an xb ∈ X such that Z b ε ε λt Re(λ)t ||f (t) − e xb || ≤ εe e−Re(λ)σ dσ = m − = Re(λ) |Re(λ)| t for all t ∈ I. Define ( x0 =

xa , f or Re(λ) < 0, xb , f or Re(λ) > 0.

Clearly, x0 satisfies (2.74) for all t ∈ I. Let x1 ∈ X such that ||f (t)−eλt x1 || ≤ L, ∀t ∈ I for some 0 < L < ∞. By (2.74), ||x0 − x1 || = ||x0 − e−λt f (t) + e−λt f (t) − x1 || ≤ ||x0 − e−λt f (t)|| + ||e−λt f (t) − x1 || = ||e−λt || ||eλt x0 − f (t)|| + ||e−λt || ||f (t) − eλt x1 ||   ε ≤ e−Re(λ)t +L |Re(λ)| for all t ∈ I. As m = inf{e−Re(λ)t : t ∈ I} = 0, x0 is unique.

Stability of First Order Linear Differential Equations

101

Corollary 19 Assume that a strongly differentiable function f : I → X satisfies (2.73) for all t ∈ I and for some ε > 0. If Re(λ) 6= 0 and m > 0, then there are infinitely many x0 ∈ X for which the inequality  m ε 1− (2.75) ||f (t) − eλt x0 || ≤ |Re(λ)| M holds for all t ∈ I. More explicitly, if S is the set of all x0 ∈ X satisfying (2.75), then the cardinal number of S is at least c, where c denotes that of the continuum and M = sup{e−Re(λ)t : t ∈ I}. Proof 54 As in Corollary 18 and Theorem 27, if we define ( lims→a+ e−λs f (s), f or Re(λ) < 0, x0 = lims→b− e−λs f (s), f or Re(λ) > 0, then (2.75) is true for all t ∈ I. Since ( lims→a+ e−λs s, f or Re(λ) < 0, m= lims→b− e−λs s, f or Re(λ) > 0, then we define J = {e−Re(λ)t : m < e−Re(λ)s ≤ e−Re(λ)t < M implies e−λs f (s) = x0 }. We may note that either J = φ or J 6= φ with supJ < M. Define ( supJ, f or J 6= φ, α= m, f or J = φ. By the definitions of J and α, e−λs e−Re(λ)s → α, and there
f (s) → x0 when −Re(λ)s m < α + δ0 implies fore there is 0 < δ0 < α 1 − M such that α < e that |e−Re(λ)s − α| ≤ |δ0 |. Hence, ||x0 − e

where α =

mε |Re(λ)| .

−λs

 m  f (s)|| ≤ α 1 − M  mε m ≤ 1− , |Re(λ)| M

(2.76)

By the definition of J, there exists an s0 ∈ I such that α < e−Re(λ)s0 < α + δ0

(2.77)

and x0 6= e−λs0 f (s0 ). Consider r0 = ||x0 − e−λs0 f (s0 )|| > 0. The function

102

Hyers-Ulam Stability of Ordinary Differential Equations

s 7→ ||x0 − e−λs f (s)|| is continuous so, by the intermediate value theorem, there exists to each r ∈ (0, r0 ) an sr ∈ I such that α < e−Re(λ)sr < e−Re(λ)s0

(2.78)

||x0 − e−λsr f (sr )|| = r.

(2.79)

and Let xr = e−λsr f (sr ), ∀r ∈ (0, r0 ). If r1 , r2 ∈ (0, r0 ) and r1 6= r2 , then xr1 6= xr2 by (2.79). We obtain from (2.76), (2.77) and (2.78) that mε  m ||x0 − e−λsr f (sr )|| ≤ 1− , ∀r ∈ (0, r0 ). (2.80) |Re(λ)| M Choose t ∈ I arbitrarily. Now, we have two possibilities: either e−Re(λ)t ≤ α or e−Re(λ)t > α. In the former case and by the definition of J, we obtain e−λt f (t) = x0 . Therefore, (2.80) becomes ||f (t) − eλt xr || = eRe(λ)t ||x0 − e−λsr f (sr )||  m ε ≤ 1− , ∀ r ∈ (0, r0 ). |Re(λ)| M
m In the latter case as 0 < δ0 < α 1 − M , (2.77) and (2.78) give rise to  m α < e−Re(λ)sr < α 2 − , ∀r ∈ (0, r0 ). (2.81) M Ultimately, (2.81) becomes  α e−Re(λ)sr α m , < < 2 − M e−Re(λ)t e−Re(λ)t e−Re(λ)t that is, e−Re(λ)sr m m < −Re(λ)t < 2 − M M e due to m ≤ α < e−Re(λ)t < M. Therefore, −Re(λ)s r e ≤ 1 − m , ∀ r ∈ (0, r0 ). (2.82) − 1 e−Re(λ)t M By (2.67), ||e−λt f (t) − e−λsr || ≤

ε |e−Re(λ)t − e−Re(λ)sr |, ∀ r ∈ (0, r0 ). |Re(λ)|

(2.83)

From (2.82) and (2.83), it follows that ε |e−Re(λ)t − e−Re(λ)sr | |Re(λ)|  ε m ≤ 1− , ∀ r ∈ (0, r0 ). |Re(λ)| M

||f (t) − eλt xr || ≤ eRe(λ)t

Hence, all elements of {xr : r ∈ (0, r0 )} satisfy (2.75) for each t ∈ I. Therefore, by (2.79), the cardinal number of the set {xr : r ∈ (0, r0 )} is that of (0, r0 ), and hence c. Thus, the corollary is proved.

Stability of First Order Linear Differential Equations

103

Corollary 20 Assume that a strongly differentiable function f : I → X satisfies the inequality (2.73), for all t ∈ I and for some ε > 0. If Re(λ) = 0 and the diameter δ(I) of I is finite, then there exist unique xa ∈ X and xb ∈ X such that ||f (t) − eλt xa || ≤ ε(t − a) and ||f (t) − eλt xb || ≤ ε(b − t) for all t ∈ I respectively. Proof 55 As δ(I) is finite, ε is integrable on (a, b). Here Re(λ) = 0. Hence, by (a) and (b) of Theorem 27, there exist unique xa ∈ X and xb ∈ X such that ||f (t) − eλt xa || ≤ ε(t − a) and ||f (t) − eλt xb || ≤ ε(b − t) for all t ∈ I respectively. Hence, the corollary is proved.

2.6

Stability of y 0 = F (x, y)

In this section, for a bounded and continuous function F (x, y), we will prove the Hyers-Ulam-Rassias stability as well as the Hyers-Ulam stability of the differential equations of the form y 0 (x) = F (x, y(x)).

(2.84)

Theorem 28 Let (X, d) be a generalized complete metric space. Assume that Λ : X → X is a strictly contractive operator with the Lipschitz constant L < 1. If there exists a nonnegative integer k such that d(Λk+1 x, Λk x) < ∞ for some x ∈ X, then the followings are true: (a) The sequence Λn x converges to a fixed point x∗ of Λ; (b) x∗ is the unique fixed point of Λ in X ∗ = {y ∈ X | d(Λk x, y) < ∞}; (c) If y ∈ X ∗ , then d(y, x∗ ) ≤

1 d(Λ, y). 1−L

Proof 56 Let x0 ∈ X, and consider that: sequence of successive approximations with initial element x0 : x0 , Λx0 , Λ2 x0 , · · · , Λl x0 , · · · . Then the following alternative holds such that either (A) for each integer l = 0, 1, 2, · · · , we have d(Λl x0 , Λl+1 x0 ) = ∞,

104

Hyers-Ulam Stability of Ordinary Differential Equations

or (B) the sequence of successive approximations x0 , Λx0 , Λ2 x0 , · · · , Λl x0 , · · · , is d−convergent to a fixed point of Λ. (a) Assume that sequence of numbers d(x0 , Λx0 ), d(Λx0 , Λ2 x0 ), · · · , d(Λl x0 , Λl+1 x0 ), · · · , the sequence of distances between consecutive neighbors of the sequence of successive approximations with initial element x0 . There are two mutually exclusive possibilities: either (i) for each integer l = 0, 1, 2, · · · , we have either d(Λl x0 , Λl+1 x0 ) = ∞, or (ii) for some integer l = 0, 1, 2, · · · , one has d(Λl x0 , Λl+1 x0 ) < ∞. In case(ii), let N = N(x0 ) denote a particular one of all the integers l = 0, 1, 2, · · · such that d(Λl x0 , Λl+1 x0 ) < ∞. Since d(ΛN x0 , ΛN+1 x0 ) < ∞ and Λ is a strictly contractive operator with the Lipschitz constant, so by the given theorem we obtain that d(Λn+l x0 , ΛN+l+1 x0 ) = d(ΛΛn+l−1 x0 , ΛΛN+l x0 ) Theorem 29 For given real numbers a and b with a < b, let I = [a, b] be a closed interval and choose a c ∈ I. Let K and L be positive constants with 0 < KL < 1. Assume that F : I × R → R is a continuous function which satisfies a Lipschitz condition |F (x, y) − F (x, z)| ≤ L|y − z|

(2.85)

for any x ∈ I and y, z ∈ R. If a continuously differentiable function y : I → R satisfies |y 0 (x) − F (x, y(x))| ≤ ϕ(x) (2.86) for all x ∈ I, where ϕ : I → (0, ∞) is a continuous function with Z x ϕ(τ )dτ ≤ Kϕ(x)

(2.87)

c

for each x ∈ I, then there exists a unique continuous function y0 : I → R such that Z x y0 (x) = y(c) + F (τ, y0 (τ ))dτ (2.88) c

(consequently, y0 is a solution to (2.84)) and |y(x) − y0 (x)| ≤ for all x ∈ I.

K ϕ(x) 1 − KL

(2.89)

Stability of First Order Linear Differential Equations

105

Proof 57 Define a set X of all continuous functions f : I → R by X = {f : I → R : f is continuous}

(2.90)

and also introduce a generalized metric on X as given below: d(f, g) = inf{C ∈ [0, ∞] : |f (x) − g(x)| ≤ Cϕ(x), ∀x ∈ I}.

(2.91)

Assume that d(f, g) > d(f, h) + d(h, g) is true for some f, g, h ∈ X. Hence, by (2.91), there exists an x0 ∈ I such that |f (x0 ) − g(x0 )| > {d(f, h) + d(h, g)}ϕ(x0 ) = d(f, h)ϕ(x0 ) + d(h, g)ϕ(x0 ) ≥ |f (x0 ) − h(x0 )| + |h(x0 ) − g(x0 )|, which is a contradiction to the triangle law. Let’s assume that (X, d) be complete. Suppose {hn } be a Cauchy sequence in (X, d). So, for any ε > 0, there exists an integer N > 0 such that d(hm , hn ) ≤ ε for each m, n ≥ Nε . By (2.91) we obtain that |hm (x) − hn (x)| ≤ εϕ(x)

(2.92)

for all ε > 0 and there exists Nε ∈ N for all m, n ≥ Nε ∀x ∈ I. By fixing x, (2.92) implies that {hn (x)} is a Cauchy sequence in R. Since R is complete, then hn (x) converges for all x ∈ I. Hence, we can define a function h : I → R by h(x) = lim hn (x). n→∞

For m → ∞, (2.92) implies that ∀ ε > 0 ∃ Nε ∈ N ∀n ≥ Nε ∀x ∈ I : |h(x) − hn (x)| ≤ εϕ(x).

(2.93)

But ϕ is bounded on I, so {hn } converges uniformly to h. Therefore, h is continuous and h ∈ X. Again (2.91) and (2.93) imply that ∀ ε > 0 ∃ Nε ∈ N ∀n ≥ Nε ∀x ∈ I : d(h, hn ) ≤ ε. So, the Cauchy sequence {hn } converges to h in (X, d). Hence, (X, d) is complete. Consider, an operator Λ : X → X defined by Z x (Λf )(x) = y(c) + F (τ, f (τ ))dτ, x ∈ I, ∀f ∈ X. (2.94) c

For each f, g ∈ X, suppose Cf g ∈ [0, ∞] be an arbitrary constant such that d(f, g) ≤ Cf g . Hence, (2.91) implies that |f (x) − g(x)| ≤ Cf g ϕ(x) ∀x ∈ I.

(2.95)

106

Hyers-Ulam Stability of Ordinary Differential Equations

By (2.85), (2.87), (2.91), (2.94) and (2.95) we obtain that Z x {F (τ, f (τ )) − F (τ, g(τ ))}dτ |(Λf )(x) − (Λg)(x)| = Zc x |F (τ, f (τ )) − F (τ, g(τ ))|dτ ≤ cZ x |f (τ ) − g(τ )|dτ ≤ L c Z x ϕ(τ )dτ ≤ LCf g c

≤ KLCf g ϕ(x), ∀x ∈ I implies that d(Λf, Λg) ≤ KLCf g . Ultimately, d(Λf, Λg) ≤ KLd(f, g), ∀f, g ∈ X, where 0 < KL < 1. For an arbitrary g0 ∈ X, there exists a constant 0 < C < ∞ such that it can be concluded from (2.90) and (2.94) that Z x (2.96) |(Λg0 )(x) − g0 (x)| = y(c) + F (τ, g0 (τ ))dτ − g0 (x) , ∀x ∈ I. c

Rx But by (2.95), |(Λg0 )(x) − g0 (x)| ≤ Cϕ(x). Hence, y(c) + c F (τ, g0 (τ ))dτ − g0 (x) ≤ Cϕ(x) < ∞ ∀x ∈ I by (2.96), that is, d(Λg0 , g0 ) < ∞. Hence, by Theorem 28 (a), there exists a continuous function y0 : I → R such that Λn g0 → y0 in (X, d) and Λy0 = y0 , and hence, y0 satisfies equation (2.88) for all x ∈ I. Let g ∈ X be arbitrary. As g and g0 are bounded on I and minx∈I ϕ(x) > 0, so there exists a constant 0 < Cg < ∞ such that |g0 (x) − g(x)| ≤ Cg ϕ(x), ∀x ∈ I. Hence, d(g0 , g) < ∞, for each g ∈ X and {g ∈ X|d(g0 , g) < ∞} = X. By Theorem 28(b), we conclude that y0 is the unique continuous function by using the property of (2.88). (2.88) implies that −ϕ(x) ≤ y 0 (x) − F (x, y(x)) ≤ ϕ(x), ∀x ∈ I. Integrating (2.97) from c to x, we get Z Z x y(x) − y(c) − F (τ, y(τ ))dτ ≤ c

c

From (2.87) and (2.94), we obtain Z |y(x) − (Λy)(x)| ≤

c

x

x

ϕ(τ ))dτ , ∀x ∈ I.

ϕ(τ )dτ ≤ Kϕ(x), ∀x ∈ I

(2.97)

Stability of First Order Linear Differential Equations

107

implies that d(y, Λy) ≤ K.

(2.98)

Ultimately, from Theorem 28(c) and (2.98) we obtain d(y, y0 ) ≤

d(Λy, y) K ≤ , 1 − KL 1 − KL

which is (2.89) for all x ∈ I. Theorem 30 For given real numbers a and b, let I denote either (−∞, b] or R or [a, ∞). Set either c = a for I = [a, ∞) or c = b for I = (−∞, b] or c is a fixed real number if I = R. Let K and L be positive constants with 0 < KL < 1. Assume that F : I ×R → R is a continuous function which satisfies a Lipschitz condition (2.85) for all x ∈ I and all y, z ∈ R. If a continuously differentiable function y : I → R satisfies the differential inequality (2.86) for all x ∈ I, where ϕ : I → (0, ∞) is a continuous function satisfying the condition (2.87) for any x ∈ I, then there exists a unique continuous function y0 : I → R which satisfies (2.88) and (2.89) for all x ∈ I. Proof 58 Define In = [c − n, c + n], ∀n ∈ N. By Theorem 29, there exists a unique continuous function yn : In → R such that Z x yn (x) = y(c) + F (τ, yn (τ ))dτ (2.99) c

and

K ϕ(x), ∀x ∈ In . 1 − KL By the uniqueness of yn if x ∈ In , then |y(x) − yn (x)| ≤

(2.100)

yn (x) = yn+1 (x) = yn+2 (x) = ... .

(2.101)

For each x ∈ R, define n(x) ∈ N by n(x) = min{n ∈ N : x ∈ In }. Define a function y0 : R → R as y0 (x) = yn(x) (x).

(2.102)

We claim that y0 is continuous. Let x1 ∈ R be arbitrary and choose the integer n1 = n(x1 ). Hence, x1 belongs to the interior of In1 +1 and there exists an ε > 0 such that y0 (x) = yn1 +1 (x) for each x with x1 − ε < x < x1 + ε. But yn1 +1 is continuous at x1 , so is y0 . Therefore, y0 is continuous at x1 for all x1 ∈ R. Let x → R be arbitrary and choose the integer n(x). Then, for x ∈ In(x) , it follows from (2.99) and (2.102) that Z x Z x y0 (x) = yn(x) (x) = y(c) + F (τ, yn(x) (τ ))dτ = y(c) + F (τ, y0 (τ ))dτ. c

c

108

Hyers-Ulam Stability of Ordinary Differential Equations

By (2.101) and (2.102) we obtain that yn(x) (τ ) = yn(τ ) (τ ) = y0 (τ ). By (2.100) and (2.102), we get |y(x) − y0 (x)| = |y(x) − yn(x) (x)| ≤

K ϕ(x), 1 − KL

where x ∈ In(x) for any x ∈ R. For uniqueness of y0 assume that z0 : R → R is another continuous function which satisfies (2.88) and (2.89), by taking z0 in place of y0 , for any x ∈ R. Let x be an arbitrary real number. As the restrictions y0 I (= yn(x) ) n(x) and z0 I both satisfy (2.88) as well as (2.89) for all x ∈ In(x) , the uniquen(x) ness of yn(x) = y0 I implies that n(x)

y0 (x) = y0 I

n(x)

(x) = z0 I

n(x)

(x) = z0 (x).

Hence, the theorem is proved. Theorem 31 Given c ∈ R and r > 0, let I denote a closed ball of radius r and centered at c, that is, I = {x ∈ R : c − r ≤ x ≤ c + r} and let F : I × R → R be a continuous function which satisfies a Lipschitz condition (2.85) for all x ∈ I and y, z ∈ R, where L is a constant with 0 < Lr < 1. If a continuously differentiable function y : I → R satisfies the differential inequality |y 0 (x) − F (x, y(x))| ≤ ε

(2.103)

for all x ∈ I and for some ε ≥ 0, then there exists a unique continuous function y0 : I → R satisfying equation (2.88); y0 is a solution to (2.84) and |y(x) − y0 (x)| ≤

r ε 1 − Lr

(2.104)

for any x ∈ I. Proof 59 Define a set X of all continuous functions f : I → R by X = {f : I → R : f is continuous} and also introduce a generalized metric on X as given below: d(f, g) = inf{C ∈ [0, ∞] : |f (x) − g(x)| ≤ C, ∀x ∈ I}. By Theorem 29, it can be proved that (X, d) is a generalized complete metric space. Consider, an operator Λ : X → X defined by Z x (Λf )(x) = y(c) + F (τ, f (τ ))dτ, x ∈ I, ∀f ∈ X. (2.105) c

Stability of First Order Linear Differential Equations

109

Assume that Λ is strictly contractive on X. For each f, g ∈ X, suppose Cf g ∈ [0, ∞] be an arbitrary constant such that d(f, g) ≤ Cf g . Let’s assume that |f (x) − g(x)| ≤ Cf g ∀x ∈ I. (2.106) By (2.85), (2.105) and (2.106) it follows that Z x {F (τ, f (τ )) − F (τ, g(τ ))}dτ |(Λf )(x) − (Λg)(x)| = Zc x |F (τ, f (τ )) − F (τ, g(τ ))|dτ ≤ cZ x |f (τ ) − g(τ )|dτ ≤ L c

≤ LrCf g , ∀x ∈ I implies that d(Λf, Λg) ≤ LrCf g . Therefore, d(Λf, Λg) ≤ Lrd(f, g), ∀f, g ∈ X, where 0 < Lr < 1. By Theorem 29, for an arbitrary g0 ∈ X, we can prove that d(Λg0 , g0 ) < ∞. Hence, by Theorem 28(a), there exists a continuous function y0 : I → R such that Λn g0 → y0 in (X, d) as n → ∞ and Λy0 = y0 , and hence, y0 satisfies equation (2.88), for all x ∈ I. Let g ∈ X be arbitrary. As g and g0 are bounded on a compact interval I, so there exists a constant C > 0 such that |g0 (x) − g(x)| ≤ C, ∀x ∈ I. Hence, d(g0 , g) < ∞ for each g ∈ X and {g ∈ X|d(g0 , g) < ∞} = X. By Theorem 28(b), we conclude that y0 is the unique continuous function by using the property of (2.88). (2.103) implies that −ε ≤ y 0 (x) − F (x, y(x)) ≤ ε, ∀x ∈ I.

(2.107)

Integrating (2.107) from c to x, we get |(Λy)(x) − y(x)| ≤ εr, ∀x ∈ I which implies that d(Λy, y) ≤ εr. Ultimately, from Theorem 28(c) we obtain d(y, y0 ) ≤

d(Λy, y) rε ≤ , 1 − Lr 1 − Lr

which is (2.104), for all x ∈ I. Example 4 Choose positive constants K and L such that KL < 1. Let I = [0, 2K − ε] be a closed interval for a positive number ε < 2K. For a given

110

Hyers-Ulam Stability of Ordinary Differential Equations

polynomial p(x), assume that a continuously differentiable function y : I → R satisfies |y 0 (x) − Ly(x) − p(x)| ≤ x + ε, ∀x ∈ I. Set F (x, y) = Ly + p(x) and ϕ(x) = x + ε. Hence, above inequality has the identical form of (2.86). So, we obtain Z x 1 ϕ(τ )dτ = x2 + εx, ∀x ∈ I. (2.108) 2 0 Rx By (2.87), 0 ϕ(τ )dτ ≤ Kϕ(x). Hence, (2.108) can be written as Z

0

x

1 ϕ(τ )dτ = x2 + εx ≤ Kϕ(x), ∀x ∈ I. 2

By Theorem 29, there exists a unique continuous function y0 : I → R such that Z x y0 (x) = y(0) + {Ly0 (τ ) + p(τ )}dτ 0

and |y(x) − y0 (x)| ≤

K (x + ε) 1 − KL

for all x ∈ I. Example 5 Let a be a constant greater than 1 and choose a constant L with 0 < L < lna. Given an interval I = [0, ∞) and a polynomial p(x), suppose y : I → R is a continuously differentiable function satisfying |y 0 (x) − Ly(x) − p(x)| ≤ ax , ∀x ∈ I. Setting ϕ(x) = ax , we obtain Z

0

x

1 ϕ(τ )dτ ≤ ϕ(x) lna

for all x ∈ I. By Theorem 30, there exists a unique continuous function y0 : I → R such that Z x y0 (x) = y(0) + {Ly0 (τ ) + p(τ )}dτ 0

and |y(x) − y0 (x)| ≤ for all x ∈ I.

1 ax lna − L

Stability of First Order Linear Differential Equations

111

Example 6 Assume that r and L are positive constants such that 0 < Lr < 1 and define a closed interval I = {x ∈ R : c − r ≤ x ≤ c + r} for some real number c. Assume that a continuously differentiable function y : I → R satisfies |y 0 (x) − Ly(x) − p(x)| ≤ ε for each x ∈ I and for some ε ≥ 0, where p(x) is a polynomial. Then by Theorem 31, there exists a unique continuous function y0 (x) : I → R such that Z x {Ly0 (τ ) + p(τ )}dτ

y0 (x) = y(c) + c

and |y(x) − y0 (x)| ≤

r ε 1 − Lr

for all x ∈ I.

2.7

Notes

To the best of our knowledge, S. M. Jung, T. Miura and M. Obloza should be credited with their contribution works [1], [5], [17], [18], [19], [20], [24], [37], [38], [42] and [65] on the Hyers-Ulam stability of first order differential equations of both linear and nonlinear type. Section 2.6 is especially meant for fixed point approach of Hyers-Ulam stability.

Chapter 3 Stability of Second Order Linear Differential Equations

In this chapter, we present the Hyers-Ulam stability of the most basic linear second order differential equations of the form: y 00 + αy 0 + βy = 0,

(3.1)

y 00 + αy 0 + βy = f (x),

(3.2)

00

00

y + α(x)y = 0,

(3.3)

y 00 + β(x)y = f (x),

(3.4)

0

y + p(x)y + q(x)y + r(x) = 0

(3.5)

y 00 + p(x)y 0 + q(x)y = f (x),

(3.6)

and 2

where α(x), β(x) ∈ R, p, q, r ∈ C(R, R), y ∈ C [a, b] and f ∈ C[a, b], −∞ < a < b < +∞. We can not ignore such equations as long as their association with the physical problems are concerned.

3.1

Hyers-Ulam Stability of y 00 + αy 0 + βy = 0

Theorem 32 If the characteristic equation λ2 + αλ + β = 0 has two different positive roots, then (3.1) has the Hyers-Ulam stability. Proof 60 Let  > 0 and y ∈ C 2 [a, b] be such that | y 00 + αy 0 + βy | ≤ . We show that there exists a constant K independent of  and y such that |y(x) − u(x)| < K, for some u ∈ C 2 [a, b] satisfying u00 + αu0 + βu = 0. Let λ1 and λ2 be the roots of characteristic equation λ2 + αλ + β = 0. Define

113

114

Hyers-Ulam Stability of Ordinary Differential Equations

g(x) = y 0 (x) − λ1 y(x). Then g 0 (x) = y 00 (x) − λ1 y 0 (x). Now, | g 0 (x) − λ2 g(x) | = | y 00 (x) − λ1 y 0 (x) − λ2 (y 0 (x) − λ1 y(x)) | = | y 00 (x) − λ1 y 0 (x) − λ2 y 0 (x) + λ1 λ2 y(x) | = | y 00 (x) + (−λ1 − λ2 )y 0 (x) + λ1 λ2 y(x) | = | y 00 + αy 0 (x) + βy(x) | ≤  implies that − ≤ g 0 (x) − λ2 g(x) ≤ . Therefore, for any x ∈ [a, b] −e−λ2 (x−a) ≤ g 0 (x)e−λ2 (x−a) − λ2 g(x)e−λ2 (x−a) ≤ e−λ2 (x−a) , that is, i d h g(x)e−λ2 (x−a) ≤ e−λ2 (x−a) . dx Integrating (3.7) from x to b, we get −e−λ2 (x−a) ≤

Z

b

−

e−λ2 (y−a) dy ≤

Z

x

b

x

(3.7)

Z b i d h g(y)e−λ2 (y−a) dy ≤ e−λ2 (y−a) dy, dy x

that is, 

e−λ2 (y−a) − −λ2

b

h

≤ g(y)e x

−λ2 (y−a)

ib x

 λ2 (y−a) b e . ≤  −λ2 x

(3.8)

Assume that λ2 > 1. If 0 < λ2 ≤ 1, then there exists M > 0 such that M λ2 > 1 and the procedure can similarly be dealt with. Using λ2 > 1, (3.8) can be written as [e−λ2 (y−a) ]bx ≤ [g(y)e−λ2 (y−a) ]bx ≤ −[e−λ2 (y−a) ]bx . Hence, e−λ2 (b−a) − e−λ2 (x−a) ≤ g(b)e−λ2 (b−a) − g(x)e−λ2 (x−a) ≤ −e−λ2 (b−a) + e−λ2 (x−a) , that is, −e−λ2 (x−a) ≤ (g(b) − )e−λ2 (b−a) − g(x)e−λ2 (x−a) ≤ e−λ2 (x−a) − 2e−λ2 (b−a) implies that −e−λ2 (x−a) ≤ (g(b) − )e−λ2 (b−a) − g(x)e−λ2 (x−a) ≤ e−λ2 (x−a) .

Stability of Second Order Linear Differential Equations

115

Consequently, − ≤ (g(b) − )e−λ2 (b−x) − g(x) ≤ . If z(x) = (g(b) − )eλ2 (x−b) , then z 0 (x) = λ2 (g(b) − )eλ2 (x−b) and z 0 (x) − λ2 z(x) = λ2 (g(b) − )eλ2 (x−b) − λ2 (g(b) − )eλ2 (x−b) = 0.

(3.9)

So, z(x) satisfying z 0 (x) − λ2 z(x) = 0 and | g(x) − z(x) | ≤ . Since g(x) = y 0 (x) − λ1 y(x), then − ≤ y 0 (x) − λ1 y(x) − z(x) ≤ . Therefore, for any x ∈ [a, b] −e−λ1 (x−a) ≤ y 0 (x)e−λ1 (x−a) − λ1 y(x)e−λ1 (x−a) − z(x)e−λ1 (x−a) ≤ e−λ1 (x−a) . Consequently, −e−λ1 (x−a) ≤

d [y(x)e−λ1 (x−a) ] − z(x)e−λ1 (x−a) ≤ e−λ1 (x−a) . dx

(3.10)

Integrating (3.10) from x to b, we get Z

b −λ1 (s−a)

−

e x

Z du ≤

b

d [y(s)e−λ1 (s−a) ]ds − ds

x −λ1 (s−a)

≤ e

Z

b

z(s)e−λ1 (s−a) ds

x

du,

that is,  b h ib Z b eλ1 (s−a) − ≤ y(s)e−λ1 (s−a) − z(s)e−λ1 (s−a) du −λ1 x x x  −λ1 (s−a) b e . ≤  −λ1 x

(3.11)

Using λ1 > 1, (3.11) can be written as h

−λ1 (s−a)

e

ib x

h ib Z −λ1 (s−a) ≤ y(s)e − x

b

h ib z(s)e−λ1 (s−a) ds ≤ −e−λ1 (s−a) ,

x

that is, h i e−λ1 (b−a) − e−λ1 (x−a) ≤ y(b)e−λ1 (b−a) − y(x)e−λ1 (x−a) Z b − z(s)e−λ1 (s−a) ds x

≤ −e−λ1 (b−a) + e−λ1 (x−a) .

x

116

Hyers-Ulam Stability of Ordinary Differential Equations

Therefore, −e

−λ1 (x−a)

≤ (y(b) − )e

−λ1 (b−a)

− y(x)e

−λ1 (x−a)

Z

b

−

z(s)e−λ1 (s−a) ds

x

≤ e−λ1 (x−a) implies that − ≤ (y(b) − )e

λ1 (x−b)

− y(x) − e

λ1 (x−a)

Z

b

z(s)e−λ1 (s−a) ds ≤ .

x

Rb If we choose, u(x) = (y(b) − )eλ1 (x−b) − eλ1 (x−a) x z(s)e−λ1 (s−a) ds, then Rb u0 (x) = λ1 (y(b) − )eλ1 (x−b) − λ1 eλ1 (x−a) x z(s)e−λ1 (s−a) ds + z(x) and u0 (x) − λ1 u(x) − z(x) = λ1 (y(b) − )eλ1 (x−b) − λ1 eλ1 (x−a)

Z

b

z(s)e−λ1 (s−a) ds

x

+ z(x) − λ1 (y(b) − )eλ1 (x−b) + λ1 eλ1 (x−a)

Z

b

z(s)e−λ1 (s−a) − z(x) = 0.

x

Hence, z(x) = u0 (x) − λ1 u(x) and z 0 (x) = u00 (x) − λ1 u0 (x), | y(x) − u(x) | ≤ . Using this fact in (3.9), we get u00 (x) − λ1 u0 (x) − λ2 {u0 (x) − λ1 u(x)} = 0, that is, u00 (x) + (−λ1 − λ2 )u0 (x) + λ1 λ2 u(x) = 0. As a result, u00 (x) + αu0 (x) + βu(x) = 0. This completes the proof of the theorem . Theorem 33 Assume that the characteristic equation λ2 + αλ + β = 0 has two different positive roots. Then for every  > 0, f ∈ C[a, b], y ∈ C 2 [a, b], if | y 00 + αy 0 + βy − f (x) | ≤ , there exists some u ∈ C 2 [a, b] and K > 0 satisfying u00 + αu0 + βu = f (x), such that | y(x) − u(x) | < K. Proof 61 Let  > 0, f ∈ C[a, b] and y ∈ C 2 [a, b] be such that | y 00 + αy 0 + βy − f (x) | ≤ .

Stability of Second Order Linear Differential Equations

117

Let λ1 and λ2 be the two roots of the characteristic equation λ2 + αλ + β = 0. We may assume that λ1 > 1 and λ2 > 1. Define g(x) = y 0 (x) − λ1 y(x). Then g 0 (x) = y 00 (x) − λ1 y 0 (x). Now, | g 0 (x) − λ2 g(x) − f (x) | = | y 00 (x) − λ1 y 0 (x) − λ2 (y 0 (x) − λ1 y(x)) − f (x) | = | y 00 (x) − λ1 y 0 (x) − λ2 y 0 (x) + λ1 λ2 y(x) − f (x) | = | y 00 (x) + (−λ1 − λ2 )y 0 (x) + λ1 λ2 y(x) − f (x) | = | y 00 (x) + αy 0 (x) + βy(x) − f (x) | ≤  implies that | g 0 (x) − λ2 g(x) − f (x) | ≤ . Therefore, for any x ∈ [a, b] −e−λ2 (x−a) ≤ g 0 (x)e−λ2 (x−a) −λ2 g(x)e−λ2 (x−a) −f (x)e−λ2 (x−a) ≤ e−λ2 (x−a) , that is, −e−λ2 (x−a) ≤

i d h g(x)e−λ2 (x−a) − f (x)e−λ2 (x−a) ≤ e−λ2 (x−a) . dx

(3.12)

Integrating (3.12) from x to b, we get Z

b

−

e−λ2 (s−a) ds ≤

Z

x

b

x b

Z

i d h g(s)e−λ2 (s−a) ds − ds

Z

b

f (s)e−λ2 (s−a) ds

x

e−λ2 (s−a) ds,

≤ x

that is, b h  ib Z b e−λ2 (s−a) e−λ2 (s−a) −λ2 (s−a) ≤ g(s)e ds − f (s) − −λ2 −λ2 x x x  λ2 (s−a) b e ≤  . −λ2 x

(3.13)

Using λ2 > 1, (3.13) becomes [e−λ2 (s−a) ]bx

≤

[g(s)e−λ2 (s−a) ]bx

Z

b

f (s)e−λ2 (s−a) ds

− x

≤ −[e−λ2 (s−a) ]bx . Hence, e

−λ2 (b−a)

− e

−λ2 (x−a)

≤ g(b)e

−λ2 (b−a)

− g(x)e

−λ2 (x−a)

Z

b

− x

≤ −e−λ2 (b−a) + e−λ2 (x−a) ,

f (s)e−λ2 (s−a) ds

118

Hyers-Ulam Stability of Ordinary Differential Equations

that is, −e

−λ2 (x−a)

≤ (g(b) − )e

−λ2 (b−a)

− g(x)e

−λ2 (x−a)

Z

b

−

f (s)e−λ2 (s−a) ds

x

≤ e−λ2 (x−a) . Consequently, − ≤ (g(b) − )e−λ2 (b−x) − g(x) − eλ2 (x−a)

Z

b

f (s)e−λ2 (s−a) ds ≤ .

x

R (x−a) b

If we choose z(x) = (g(b) − )eλ2 (x−b) − eλ2 f (s)e−λ2 (s−a) ds, then x R 0 λ2 (x−b) λ2 (x−a) b −λ2 (s−a) z (x) = λ2 (g(b) − )e − λ2 e f (s)e ds + f (x) and x z 0 (x) − λ2 z(x) − f (x) = λ2 (g(b) − )eλ2 (x−b) − λ2 eλ2 (x−a) Z

b

×

f (s)e−λ2 (s−a) ds + f (x) − λ2 (g(b) − )eλ2 (x−b) + λ2 eλ2 (x−a)

x

Z

b

×

f (s)e−λ2 (s−a) ds − f (x) = 0.

(3.14)

x

So, z(x) satisfing z 0 (x) − λ2 z(x) − f (x) = 0, and | g(x) − z(x) | ≤ . Since g(x) = y 0 (x) − λ1 y(x), then − ≤ y 0 (x) − λ1 y(x) − z(x) ≤ . Therefore, for any x ∈ [a, b] −e−λ1 (x−a) ≤ y 0 (x)e−λ1 (x−a) − λ1 y(x)e−λ1 (x−a) − z(x)e−λ1 (x−a) ≤ e−λ1 (x−a) . Consequently, −e−λ1 (x−a) ≤

d [y(x)e−λ1 (x−a) ] − z(x)e−λ1 (x−a) ≤ e−λ1 (x−a) . dx

(3.15)

Integrating (3.15) from x to b, we get Z b Z b Z b d −λ1 (s−a) −λ1 (s−a) − e ds ≤ [y(s)e ]ds − z(s)e−λ1 (s−a) ds x x ds x ≤ e−λ1 (s−a) ds, that is,  b h ib Z b eλ1 (s−a) − ≤ y(s)e−λ1 (s−a) − z(s)e−λ1 (s−a) ds −λ1 x x x  −λ1 (s−a) b e ≤  . −λ1 x

(3.16)

Stability of Second Order Linear Differential Equations

119

Using λ1 > 1, (3.16) can be written as h

e−λ1 (s−a)

ib

ib Z h ≤ y(s)e−λ1 (s−a) − x

x

b

ib h z(s)e−λ1 (s−a) ds ≤ −e−λ1 (s−a) , x

x

that is, i h e−λ1 (b−a) − e−λ1 (x−a) ≤ y(b)e−λ1 (b−a) − y(x)e−λ1 (x−a) Z b − z(s)e−λ1 (s−a) ds ≤ −e−λ1 (b−a) + e−λ1 (x−a) . x

Therefore, −e−λ1 (x−a) ≤ (y(b) − )e−λ1 (b−a) − y(x)e−λ1 (x−a) −

Z

b

z(s)e−λ1 (s−a) ds

x

≤ e−λ1 (x−a) implies that − ≤ (y(b) − )e

λ1 (x−b)

− y(x) − e

λ1 (x−a)

Z

b

z(s)e−λ1 (s−a) ds ≤ .

x

Rb If we choose u(x) = (y(b) − )eλ1 (x−b) − eλ1 (x−a) x z(s)e−λ1 (s−a) ds, then Rb u0 (x) = λ1 (y(b) − )eλ1 (x−b) − λ1 eλ1 (x−a) x z(s)e−λ1 (s−a) ds + z(x), and 0

u (x) − λ1 u(x) − z(x) = λ1 (y(b) − )e

λ1 (x−b)

− λ1 e

λ1 (x−a)

Z

b

z(s)e−λ1 (s−a) ds

x

+z(x) − λ1 (y(b) − )eλ1 (x−b) + λ1 eλ1 (x−a)

Z

b

z(s)e−λ1 (s−a) − z(x) = 0.

x

Hence, z(x) = u0 (x) − λ1 u(x) and z 0 (x) = u00 (x) − λ1 u0 (x), | y(x) − u(x) | ≤ . Using this fact in (3.14), we get u00 (x) − λ1 u0 (x) − λ2 (u0 (x) − λ1 u(x)) − f (x) = 0, that is, u00 (x) + (−λ1 − λ2 )u0 (x) + λ1 λ2 u(x) − f (x) = 0. As a result, u00 (x) + αu0 (x) + βu(x) = f (x). Thus the proof is complete.

120

3.2

Hyers-Ulam Stability of Ordinary Differential Equations

Hyers-Ulam Stability of y 00 + β(x)y = 0

In this section, Hyers-Ulam stability of (3.3) is established by using the boundary conditions y(a) = 0 = y(b), −∞ < a < b < ∞ and initial conditions y(a) = 0 = y 0 (a), −∞ < a < ∞. 8 Theorem 34 If max | β(x) |< (b−a) 2 . Then (3.3) has the Hyers-Ulam stability with the boundary conditions y(a) = 0 = y(b).

Proof 62 For every  > 0, y ∈ C 2 [a, b], let | y 00 + β(x)y | ≤  with boundary conditions y(a) = 0 = y(b). Let M = max{| y(x) |: x ∈ [a, b]}. Since y(a) = 0 = y(b), then there exists x0 ∈ (a, b), such that |y(x0 )| = M . By Taylor’s theorem, y 00 (ξ) y(a) = y(x0 ) + y 0 (x0 )(x0 − a) + (x0 − a)2 , 2 y 00 (η) y(b) = y(x0 ) + y 0 (x0 )(x0 − b) + (x0 − b)2 . 2 By Rolle’s theorem y 0 (x0 ) = 0 and therefore, |y 00 (ξ)| =

2M , (x0 − a)2

|y 00 (η)| =

2M . (x0 − b)2

If x0 ∈ (a, a+b 2 ], then a < x0 ≤

a+b . 2

Therefore,  and thus

2 a+b − a ≥ (x0 − a)2 2

2M 2M 8M ≥ (b−a)2 = . 2 (x0 − a) (b − a)2 4

If x0 ∈

[ a+b 2 , b),then

a+b ≤ x0 < b. 2

Therefore,  and so is

a+b −b 2

2

≥ (x0 − b)2

2M 2M 8M ≥ (b−a)2 = . (x0 − b)2 (b − a)2 4

Stability of Second Order Linear Differential Equations

121

For ξ, η ∈ (a, b), |y 00 (ξ)| ≥

8M , (b − a)2

|y 00 (η)| ≥

8M . (b − a)2

Therefore, for all ξ, η ∈ (a, b) max |y 00 (x)| ≥

8 8M = max |y(x)|, (b − a)2 (b − a)2

that is, max |y(x)| ≤

(b − a)2 max |y 00 (x)|. 8

Thus, max |y(x)| ≤

(b − a)2 [max |y 00 (x)| − max |β(x)| max |y(x)| 8 + max |β(x)| max |y(x)|].

Since max |y 00 (x) − β(x)y(x)| ≥ max |y 00 (x)| − max |β(x)| max |y(x)|, then for max |y 00 (x) − β(x)y| ≤ max |y 00 (x) + β(x)y|, we find (b − a)2 [max |y 00 (x) + β(x)y| + max |β(x)| max | y(x)|] 8 (b − a)2 (b − a)2 ≤ ε+ [max |β(x)| max | y(x)|]. 8 8

max |y(x)| ≤

2

2

|β(x)| (b−a) Let η = (b−a) max , K = (8(1−η)) . Then max |y(x)| ≤ K, that is, |y(x)| ≤ 8 max |y(x)| ≤ K implies that |y(x)| ≤ K and hence by Definition 8, |y(x) − 0| ≤ K. Therefore, z0 (x) = 0 is a solution of y 00 −β(x)y = 0 with z0 (a) = 0 = z0 (b) and satisfying the relation |y−z0 | ≤ K. Hence (3.3) has the Hyers-Ulam stability. 2 Theorem 35 If max |β(x)| < (b−a) 2 . Then (3.3) has the Hyers-Ulam stability with initial conditions y(a) = 0 = y 0 (a).

Proof 63 For every  > 0 and y ∈ C 2 [a, b], let’s assume that |y 00 + β(x)y| ≤  with initial conditions y(a) = 0 = y 0 (a). By Taylor’s theorem, y(x) = y(a) + y 0 (a)(x − a) +

y 00 (ξ) (x − a)2 . 2

So, |y(x)| = |

y 00 (ξ) (b − a)2 (x − a)2 | ≤ max |y 00 (x) | 2 2

122

Hyers-Ulam Stability of Ordinary Differential Equations

implies that max |y(x)| ≤ max |y 00 (x)

(b − a)2 |. 2

Therefore, max |y(x)| ≤

(b − a)2 [max |y 00 (x)| − max |β(x)| max |y(x)| 2 + max |β(x)| max |y(x)|].

Since, max |y 00 (x) − β(x)y(x)| ≥ max |y 00 (x)| − max |β(x)| max |y(x)|, then using max |y 00 (x) − β(x)y| ≤ max |y 00 (x) + β(x)y|, we obtain (b − a)2 [max |y 00 (x) + β(x)y| + max |β(x)| max |y(x)|] 2 (b − a)2 (b − a)2 ≤ + [max |β(x)| max |y(x)|]. 2 2

max |y(x)| ≤

2

2

|β(x)| (b−a) , K = (2(1−η)) . Then max |y(x)| ≤ K, that is, |y(x)| ≤ Let η = (b−a) max 2 max |y(x)| ≤ K, implies that |y(x)| ≤ K and hence by definition, |y(x)−0| ≤ K. Therefore, z0 (x) = 0 is a solution of y 00 −β(x)y = 0 with z0 (a) = 0 = z00 (a) and satisfying the relation |y − z0 | ≤ K. Hence (3.3) has the Hyers-Ulam stability.

3.3

Hyers-Ulam Stability of y 00 + β(x)y = f (x)

In this section, Hyers-Ulam stability of (3.4) is established by using the initial conditions y(a) = 0 = y 0 (a). 2 Theorem 36 Suppose | β(x) | < M , where M = (b−a) 2 , ϕ : [a, b] → [0, ∞) in an increasing function. The equation (3.4) has the Generalized Hyers-Ulam stability if for θϕ ∈ C(R+ , R+ ) and for each approximate solution y ∈ C 2 [a, b] of (3.4) satisfying | y 00 − β(x)y − f (x) | ≤ ϕ(x), (3.17)

there exists a solution z0 ∈ C 2 [a, b] of (3.4) with condition y(a) = 0 = y 0 (a) such that | y(x) − z0 (x) | ≤ θϕ (x) Proof 64 Given that | β(x) | < M , where M = (3.17), we have

2 (b−a)2 ,

from the inequality

ϕ(x) ≤ y 00 − β(x)y − f (x) ≤ ϕ(x) ≤ ϕ(x).

(3.18)

Stability of Second Order Linear Differential Equations Integrating (3.18) from a to x, Z Z x Z x β(t)y(t)dt − ϕ(t)dt ≤ y 0 (x) −

Z f (t)dt ≤

x

ϕ(t)dt.

(3.19)

a

a

a

a

x

123

Again integrating (3.19) from a to r, r

Z

x

Z

Z

r

x

Z

a

a

a

r

Z

x

Z

r

Z

Z

x

f (t)dtdr ≤

β(t)y(t)dtdr −

ϕ(t)dtdr ≤ y(x) −

a

a

ϕ(t)dtdr. a

a

a

By using Replacement Lemma, we obtain Z x Z x (x − t)ϕ(t)dt ≤ y(x) − (x − t)β(t)y(t)dt a a Z x Z x − (x − t)f (t)dt ≤ (x − t)ϕ(t)dt. a

a

Hence, Z x Z x Z x y(x) − (x − t)β(t)y(t)dt − (x − t)f (t)dt ≤ (x − t)ϕ(t)dt, a

a

a

that is, Z x Z x y(x) − (x − t){β(t)y(t)dt + f (t)dt} ≤ (x − t)ϕ(t)dt. a

a

Rx

Let us consider z0 (x) = a (x − t){β(t)z0 (t)dt + f (t)dt}, then Z x |y(x) − z0 (x)| = y(x) − (x − t){β(t)y(t)dt + f (t)dt} a Z x Z x + (x − t){β(t)y(t)dt + f (t)dt} − (x − t){β(t)z0 (t)dt + f (t)dt} a a Z x ≤ y(x) − (x − t){β(t)y(t)dt + f (t)dt} a Z x + (x − t){β(t)y(t)dt + f (t)dt} a Z x − (x − t){β(t)z0 (t)dt + f (t)dt} , a

that is, x

Z |y(x) − z0 (x)| ≤ a

Z x (x − t)ϕ(t)dt + (x − t)β(t){y(t) − z0 (t)}dt a

implies that Z |y(x) − z0 (x)| ≤

x

Z (x − t)ϕ(t)dt+ | β(t) |

a

x

(x − t) | y(t) − z0 (t) | dt a

124

Hyers-Ulam Stability of Ordinary Differential Equations

that is, x

Z |y(x) − z0 (x)| ≤

Z (x − t)ϕ(t)dt + M

x

(x − t) | y(t) − z0 (t) | dt.

a

a

Applying Gronwall inequality, we have Z x Rx |y(x) − z0 (x)| ≤ (x − t)ϕ(t)dteM a (x−t)dt a

that is, Z

x

(x − t)ϕ(t)dte

|y(x) − z0 (x)| ≤

M (x−a)2 2

.

a

Therefore, Z |y(x) − z0 (x)| ≤ c

x

(x − t)ϕ(t)dt, a

x−a

2

where c = e( b−a ) . Remark 3 This is to be noted that as x → b, the above system considered is the Hyers-Ulam stable.

3.4

Hyers-Ulam Stability of y 00 + p(x)y 0 + q(x)y + r(x) = 0

In this section, Hyers-Ulam stability of (3.5) is presented by means of Riccati differential equation. Theorem 37 Let p(x), q(x) and r(x) be real continuous functions on I = (a, b) such that p(x) 6= 0, r(x) 6= 0, q(x) 6= 0. If a twice continuously differentiable function y : I → R satisfies the differential inequality | y 00 + p(x)y 0 + q(x)y + r(x) | ≤ ε for all t ∈ I and for ε > 0 and the Riccati equation u0 (x) + p(x)u(x) − u2 (x) = q(x) has a particular solution c(x), then there exists a solution v : I → R of y 00 + p(x)y 0 + q(x)y + r(x) = 0 such that |y(x) − v(x)| ≤ Kε, where K > 0 is a constant, p(x) − c(x) ≥ 1 for all x ∈ I. Proof 65 Let ε > 0 and y : I → R be a continuously differentiable function satisfy the differential inequality | y 00 + p(x)y 0 + q(x)y + r(x) | ≤ ε.

Stability of Second Order Linear Differential Equations

125

We show that there exists a constant K independent of ε and y such that |y(x)−v(x)| < Kε for some v ∈ C 2 (I) satisfying v 00 +p(x)v 0 +q(x)v+r(x) = 0. Let c(x) be a particular solution of the Riccati equation u0 (x) + p(x)u(x) − u2 (x) = q(x). Then c0 (x) + p(x)c(x) − c2 (x) = q(x). Define g(x) = y 0 (x) + c(x)y(x), then g 0 (x) = y 00 (x) + c(x)y 0 (x) + c0 (x)y(x) and let d(x) = p(x) − c(x). Therefore, q(x) = c0 (x) + d(x)c(x). Now, | g 0 (x) + d(x)g(x) + r(x) | = | y 00 (x) + c(x)y 0 (x) + c0 (x)y(x) + d(x){y 0 (x) + c(x)y(x)} + r(x) | = | y 00 (x) + {c(x) + d(x)}y 0 (x) + {c0 (x) + d(x)c(x)}y(x) + r(x) | = | y 00 + p(x)y 0 + q(x)y + r(x) | ≤ . Using the same technique as in Section 2.4, we have " # Z b R Rb Rs − ax d(s)ds d(x)ds d(t)dt w(x) = e (g(b) − )e a + r(s)e a ds x

and w0 (x) = −d(x)e−

Rx

Rb

(g(b) − )e a d(x)ds − r(x) Z b Rx Rs − d(x)e− a d(s)ds r(s)e a d(t)dt ds. a

d(s)ds

x

Hence, w0 (x) + d(x)w(x) + r(x) = −d(x)e−

Rx

Rb

(g(b) − )e a d(x)ds − r(x) Z b Rx Rs − d(x)e− a d(s)ds r(s)e a d(t)dt ds a

d(s)ds

x −

Rx

Rb

(g(b) − )e a d(x)ds Z b Rx Rs + d(x)e− a d(s)ds r(s)e a d(t)dt ds + r(x)

+ d(x)e

a

d(s)ds

x

= 0.

(3.20)

0

So, w(x) satisfying w (x) + d(x)w(x) + r(x) = 0, and | g(x) − w(x) |≤ . Since g(x) = y 0 (x) + c(x)y(x), then − ≤ y 0 (x) + c(x)y(x) − w(x) ≤ . Using the same technique as above, we get " v(x) = e−

Rx a

c(s)ds

Rb

(y(b) − )e

a

c(x)ds

Z

b

−

# Rs

w(s)e x

a

c(t)dt

ds

126

Hyers-Ulam Stability of Ordinary Differential Equations

and v 0 (x) = −c(x)e−

Rx

Rb

(g(b) − )e a c(x)ds + w(x) Z b Rs Rx w(s)e a c(t)dt ds. + c(x)e− a c(s)ds a

c(s)ds

x

So, v 0 (x) + c(x)v(x) − w(x) = 0. Hence w(x) = v 0 (x) + c(x)v(x) and w0 (x) = v 00 (x) + c0 (x)v(x) + v 0 (x)c(x) + d(x)v 0 (x), |y(x) − v(x)| ≤ . Using this fact in (3.20), we get v 00 (x) + c0 (x)v(x) + v 0 (x)c(x) + d(x)v 0 (x) + d(x)c(x)v(x) + r(x) = 0 implies that v 00 (x) + (c(x) + d(x))v 0 (x) + (c0 (x) + d(x)c(x))v(x) + r(x) = 0, that is, v 00 (x) + p(x)v 0 (x) + q(x)v(x) + r(x) = 0. This completes the proof of the theorem. Theorem 38 Let p(x), q(x) and r(x) be continuous real functions defined on the interval I = (a, b) such that p(x) 6= 0 and y0 (x) is a non-zero bounded particular solution p(x)y 00 + q(x)y 0 + r(x)y = 0. If y : I → R is a twice continuously differentiable function,which satisfies the differential inequality |p(x)y 00 + q(x)y 0 + r(x)y| ≤  for all t ∈ I and for some  > 0, then there exists a solution v : I → R such that |y(x) − v(x)| ≤ K, where K > 0 is a constant and v satisfies p(x)v 00 + q(x)v 0 + r(x)v = 0. Proof 66 Let  > 0 and y : I → R be a twice continuously differentiable function, satisfy the differential inequality |p(x)y 00 + q(x)y 0 + r(x)y| ≤ . Since y0 (x) is a particular solution of p(x)y 00 + q(x)y 0 + r(x)y R x = 0, then p(x)y000 (x)+q(x)y00 (x)+r(x)y0 (x) = 0. Consider y(x) = y0 (x) a z(s)ds. Then | p(x)y 00 (x) + q(x)y 0 (x) + r(x)y(x) | Z x = |p(x)y000 (x) z(x)ds + 2p(x)y00 (x)z(x) + p(x)y0 (x)z 0 (x) a Z x Z x 0 + q(x)y0 (x) z(s)ds + q(x)y0 (x)z(x) + r(x)y0 (x) z(s)ds| a

a

Stability of Second Order Linear Differential Equations

127

= | p(x)y0 (x)z 0 (x) + 2p(x)y00 (x)z(x) + q(x)y0 (x)z(x) Z x + z(x)ds {p(x)y000 (x) + q(x)y00 (x) + r(x)y0 (x)} | a

= | p(x)y0 (x)z 0 (x) + [2p(x)y00 (x) + q(x)y0 (x)]z(x) | ≤ . Using the same technique as in Section 2.4, we have  R 2p(s)y00 (s)+q(s) R b 2p(s)y00 (s)+q(s)  − x ds ds y0 (s) y0 (s) z1 (x) = e a (z(b) − ε)e a So, z1 (x) satisfying p(x)y1 (x)z10 (x) + [2p(x)y10 (x) + q(x)y1 (x)]z1 (x) = 0 and | z(x) − z1 (x) | ≤ ε implies that −ε ≤ z(x) − z1 (x) ≤ ε. Using the same technique as above, we get  Z b  y(b) −ε − z2 (x) = z1 (s)ds. y1 (b) x So, z2 (x) satisfies z20 (x) − z1 (x) = 0 and | z(x) − z2 (x) | ≤ ε. Consequently, | z(x) − z2 (x)y0 (x) | ≤ ε. If we define v(x) = z2 (x)y0 (x), then the above inequality is | y(x) − v(x) | ≤ Kε, where K > 0 is a constant and v satisfies p(x)v 00 + q(x)v 0 + r(x)v = 0. Corollary 21 Let p(x), q(x) and r(x) be continuous real functions defined on the interval I = (a, b) such that p(x) 6= 0 and r2 + p(x)r + q(x) = 0. If y : I → R is a twice continuously differentiable function, which satisfies the differential inequality | p(x)y 00 + q(x)y 0 + r(x)y | ≤ ε for all t ∈ I and for some ε > 0, then there exists a solution v : I → R such that | y(x) − v(x) | ≤ Kε, where K > 0 is a constant and v satisfies p(x)v 00 + q(x)v 0 + r(x)v = 0. Corollary 22 Let p(x), q(x), r(x), and s(x) be continuous real functions defined on the interval I = (a, b) such that p(x) 6= 0 and y0 (x) is a nonzero bounded particular solution of p(x)y 000 + q(x)y 00 + r(x)y 0 + s(x)y = 0. If

128

Hyers-Ulam Stability of Ordinary Differential Equations

y : I → R is a twice continuously differentiable function, which satisfies the differential inequality | p(x)y 000 + q(x)y 00 + r(x)y 0 + s(x)y |≤ ε for all t ∈ I and for some ε > 0, then there exists a solution v : I → R such that | y(x) − v(x) |≤ Kε, where K > 0 is a constant and v satisfies p(x)v 000 +q(x)v 00 +r(x)v 0 +s(x)v = 0.

3.5

Hyers-Ulam Stability of y 00 + p(x)y 0 + q(x)y = f (x)

In this section, the generalized Hyers-Ulam stability of differential equation (3.6) is presented. Theorem 39 Let X be a complex Banach space. Assume that p, q : I → C and f : I → X are continuous functions and y1 : I → X is a non-zero twice continuously differentiable function which satisfies the differential equation y100 + p(x)y10 + q(x)y1 = 0.

(3.21)

If a twice continuously differentiable function y : I → X satisfies ky 00 + p(x)y 0 + q(x)y − f (x)k ≤ ϕ(x)

(3.22)

for all x ∈ I and ϕ : I → (0, ∞) is a continuous function, then there exists a unique x0 ∈ X such that x Rs

Z ky(x) − y1 (x).



a

e "a

0 (u) 2y1 +p(u) y1 (u)

Z

. x0 + a

Z ≤ ky1 (x)k.

x Rs

e



a



s

f (t) e y1 (t)

0 (u) 2y1 +p(u) y1 (u)

Rt a

du



0 (u) 2y1 +p(u) y1 (u)

 du

# dt ds + kk

 du

 aZ R  0 b t 2y1 (u) +p(u) du ϕ(t) y1 (u) a · dt ds, e s ky1 (t)k

where k =

y(a) y1 (a)

∈ C and e


0 such that | z(x) − z(t) |<  for x, t ≥ x0 , that is, z(t)t∈I is a Cauchy net and hence

140

Hyers-Ulam Stability of Ordinary Differential Equations

there exists a z ∈ R such that z(t) converges to z as t → b.   Z x R R v p(u)du y(x) − e− ax p(u)du z − a e h(v)dv a Rx − a p(u)du [z(x) − z] = e Rx

Rx

|z(x) − z(t)| + e− a p(u)du |z(t) − z| Z x R Rx Rx v e a p(u)du ϕ1 (v)dv + e− a p(u)du |z(t) − z| ≤ e− a p(u)du Zt x R Rx v e a p(u)du ϕ1 (v)dv , ≤ e− a p(u)du

≤ e−

a

p(u)du

b

where z(t) → z as t → b. If y0 (x) = e−

Rx

p(u)du

a

 Z z−

x R v

e

a

p(u)du

 h(v)dv ,

a

then the last inequality becomes |y(x) − y0 (x)| Z b Rx Rv − a p(u)du p(u)du ≤ e ϕ1 (v)e a dv x   R Z Z Rx b v v − a p(u)du ≤ e | p0 (v) |−1 ϕ(t)dt e a p(u)du dv x a for all x ∈ I. Corollary 23 Let p0 , p1 , p2 , µ : I → R be continuous functions with p0 (x) 6= 0 and µ(x) 6= 0 for all x ∈ I,and let ϕ : I → [0, ∞) be a function.Assume that y : I → R is a twice continuously differentiable function satisfying the differential inequality | µ(x)[p0 (x)y 00 + p1 (x)y 0 + p2 (x)y + f (x)] |≤ ϕ(x)

(4.7)

for all x ∈ I and {µ(x)p0 (x)}00 − {p1 (x)µ(x)}0 + p2 (x)µ(x) = 0 is true. Then there exists a solution y0 : I → R of µ(x)[p0 (x)y 00 +p1 (x)y 0 +p2 (x)y+f (x)] = 0 such that |y(x) − y0 (x)| ≤e

−

Rx a

p(u)du

Z

b



−1

Z

|µ(v)p0 (v)| x

v

 R v ϕ(t)dt e a p(u)du dv

a

for x ∈ I, where p(x) = {µ(x)p0 (x)}−1 [µ(x)p1 (x) − {µ(x)p0 (x)}0 ].

Hyers-Ulam Stability of Exact Linear Differential Equations

141

Proof 69 Indeed, the equation µ(x)[p0 (x)y 00 − p1 (x)y 0 + p2 (x)y + f (x)] = 0 is said to be exact if {µ(x)p0 (x)}00 − {p1 (x)µ(x)}0 + p2 (x)µ(x) = 0.

(4.8)

It follows from (4.7) and (4.8) that ϕ(x) ≥ | µ(x)p0 (x)y 00 + µ(x)p1 (x)y 0 + µ(x)p2 (x)y + µ(x)f (x) | = | {µ(x)p0 (x)}y 00 + {µ(x)p1 (x)}y 0 + {µ(x)p2 (x)}y + µ(x)f (x) {µ(x)p0 (x)}0 y 0 − {µ(x)p0 (x)}0 y 0 + {µ(x)p0 (x)}00 y − {µ(x)p0 (x)}00 y + {µ(x)p1 (x)}0 y − {µ(x)p1 (x)}0 y | = | [{µ(x)p0 (x)}y 00 + {µ(x)p0 (x)}0 y 0 − {µ(x)p0 (x)}00 y − {µ(x)p0 (x)}0 y + [{µ(x)p1 (x)}0 y + {µ(x)p1 (x)}y 0 ] + [{µ(x)p0 (x)}00 y − {µ(x)p1 (x)}0 y + {µ(x)p2 (x)}y] + µ(x)f (x) | = | [{µ(x)p0 (x)}y 0 − {µ(x)p0 (x)}0 y]0 + [{µ(x)p1 (x)}y]0 + [{µ(x)p0 (x)}00 − {µ(x)p1 (x)}0 + {µ(x)p2 (x)}]y + µ(x)f (x) | = | [{µ(x)p0 (x)}y 0 − {µ(x)p0 (x)}0 y]0 + [{µ(x)p1 (x)}y]0 + µ(x)f (x) |, that is, −ϕ(x) ≤ [{µ(x)p0 (x)}y 0 − {µ(x)p0 (x)}0 y]0 + [{µ(x)p1 (x)}y]0 + µ(x)f (x) ≤ ϕ(x).

(4.9)

Integrating (4.9) from a to x for each x ∈ I, we get Z x Z x Z x − ϕ(x)dx ≤ [{µ(x)p0 (x)}y 0 − {µ(x)p0 (x)}0 y]0 dx + {µ(x)p1 (x)}y]0 dx a a a Z x Z x + µ(x)f (x)dx ≤ ϕ(x)dx, a

a

that is, Z x − ϕ(t)dt ≤ [{µ(x)p0 (x)}y 0 − {µ(x)p0 (x)}0 y]xa + [{µ(x)p1 (x)}y]xa a Z x Z x + µ(t)f (t)dt ≤ ϕ(t)dt a

a

implies that Z x − ϕ(t)dt ≤ {µ(x)p0 (x)}y 0 − {µ(x)p0 (x)}0 y − {µ(a)p0 (a)}y 0 + {µ(a)p0 (a)}0 y a Z x Z x + {µ(x)p1 (x)}y − {µ(a)p1 (a)}y + µ(t)f (t)dt ≤ ϕ(t)dt. a

a

142

Hyers-Ulam Stability of Ordinary Differential Equations

Therefore, Z x − ϕ(t)dt ≤ {µ(x)p0 (x)}y 0 − {µ(x)p0 (x)}0 y + {µ(x)p1 (x)y} + {µ(a)p0 (a)}0 y a Z x Z x − {µ(a)p0 (a)}y 0 − {µ(a)p1 (a)}y} + µ(t)f (t)dt ≤ ϕ(t)dt. a

a

Hence, Z {µ(x)p0 (x)}y 0 − {µ(x)p0 (x)}0 y + {µ(x)p1 (x)}y + k +

x

µ(t)f (t)dt a Z x ≤ ϕ(t)dt, a

that is,  {µ(x)p0 (x)} y 0 + {µ(x)p0 (x)}−1 [µ(x)p1 (x) − {µ(x)p0 (x)}0 ]y    Z x Z x −1 ≤ ϕ(t)dt + {µ(x)p0 (x)} k+ µ(t)f (t)dt a

a

and hence 0 y + {µ(x)p0 (x)}−1 [µ(x)p1 (x) − {µ(x)p0 (x)}0 ]y + {µ(x)p0 (x)}−1   Z x Z x −1 k+ µ(t)f (t)dt ≤ |µ(x)p0 (x)| ϕ(t)dt. a

(4.10)

a

0 If we choose, Rp(x) = {µ(x)p0 (x)}−1 [µ(x)p1 (x) − {µ(x)p R x 0 (x)} ], ϕ1
(x) = −1 x −1 | µ(x)p0 (x) | ϕ(t)dt and h(x) = {µ(x)p0 (x)} k + a µ(t)f (t)dt , a where k = −[{µ(a)p0 (a)}y 0 (a) − {µ(a)p0 (a)}0 y(a) + {µ(a)p1 (a)}y(a)], then (4.10) becomes | y 0 + p(x)y + h(x) | ≤ ϕ1 (x).

By using Theorem 40, there exists a unique z ∈ R such that   Z x R Rx v y0 (x) = e− a p(u)du z − e a p(u)du h(v)dv . a

Thus, | y(x) − y0 (x) | Z Rx ≤ e− a p(u)du

b

x

 | µ(v)p0 (v) |

−1

Z

v

 R v ϕ(t)dt e a p(u)du dv.

a

Corollary 24 Consider the equation y 00 +cy 0 +by +f (x) = 0. Let c2 −4b ≥ 0, √ c± c2 −4b m= , f : I → R be a continuous function and let ϕ : I → [0, ∞) 2

Hyers-Ulam Stability of Exact Linear Differential Equations

143

be a function. Assume that y : I → R is a twice continuously differentiable function satisfying the differential inequality | y 00 + cy 0 + by + f (x) | ≤ ϕ(x)

(4.11)

for all x ∈ I. Then there exists a solution y0 : I → R of y 00 + cy 0 + by + f (x) = 0 such that  Z b Z v | y(x) − y0 (x) | ≤ e(m−c)(x−a) e(−mv) e(mt) ϕ(t) e(c−m)(x−a) dv x

a

for all x ∈ I. Proof 70 For the second order linear differential equation y 00 + cy 0 + by + f (x) = 0,

(4.12)

if we choose µ(x) is an integrating factor of (4.12), then it follows that µ00 (x) − cµ0 (x) + bµ(x) = 0. Consider µ(x) = emx is an integrating factor of (4.12) when c2 − 4b ≥ 0 with √ 2 m = c± c2 −4b , then (4.11) becomes emx | y 00 + cy 0 + by + f (x) |≤ emx ϕ(x) for all x ∈ I. Using Corollary 23 with ϕ1 (x) = emx ϕ(x),    Z x Z h(x) = {µ(x)p0 (x)}−1 k + µ(t)f (t)dt = (emx )−1 k + a

x

 emt f (t)dt

a

and p(x) = {µ(x)p0 (x)}−1 [µ(x)p1 (x) − {µ(x)p0 (x)}0 ] = (emx )−1 [emx c − (emx . 1)0 ] 1 = mx (emx c − memx ) e = c − m, there exists a unique z ∈ R such that   Z x R Rx v y0 (x) = e− a p(u)du z − e a p(u)du h(v)dv a     Z x R Z v R v − ax (c−m)du =e z− e a (c−m)du {emv }−1 k + emx f (x)dx dv a a     Z x Z v (m−c)(x−a) (c−m)(v−a) −mv =e z− e .e k+ emx f (x)dx dv a a     Z x Z v (m−c)(x−a) v(c−2m)−a(c−m) k+ emx f (x)dx dv , =e z− e a

a

144

Hyers-Ulam Stability of Ordinary Differential Equations

where k = −[ema y 0 (a) − mema y(a) + cema y(a)] for all x ∈ I and hence  R Z v Z b Rx v ϕ(t)dt e a p(u)du dv | µ(v)p0 (v) |−1 | y(x) − y0 (x) |≤ e− a p(u)du a

x

≤e ≤e

−

Rx a

(c−m)du

(m−c)(x−a)

Z

b

 |e

Z

mv −1

Z

v

 R v ϕ(t)dt e a (c−m)du dv

|

a

x b

e

mv

Z

v mt

e

x

 ϕ(t) e(c−m)(v−a) dv.

a √

2

Corollary 25 Consider (4.12). Let c2 − 4b < 0, m = c± c2 −4b = α ± iβ, f : I → R be a continuous function and let ϕ : I → [0, ∞) be a function. Assume that y : I → R is a twice continuously differentiable function satisfying the differential inequality (4.11) for all x ∈ I. Then there exists a solution y0 : I → R of (4.12) such that | y(x) − y0 (x) | Z Rx ≤ e− a p(u)du

b



| µ(v) |−1

x

for all x ∈ I, where µ(x) = e

Z

v

 R v eαt ϕ(t)dt e a p(u)du dv

a

αx

cos βx and p(x) = [c − α + β tan βx]. √ c ± c2 − 4b 2 Proof 71 Consider (4.12) when c −4b < 0 with m = = α ± iβ, 2 αx µ(x) = e cos βx is an integrating factor. Then the inequality (4.11) becomes eαx | cos βx{y 00 + cy 0 + by + f (x)} | ≤ eαx | y 00 + cy 0 + by + f (x) | ≤ eαx ϕ(x) for all x ∈ I. By Corollary 24 with   Z x h(x) = {µ(x)p0 (x)}−1 k + µ(t)f (t)dt a   Z x αx −1 = (e cos βx) k+ eαt cos βtf (t)dt , a

and p(x) = {µ(x).1}−1 [µ(x)c − {µ(x).1}0 ] = (eαx cos βx)−1 [eαx cos βx.c − (eαx cos βx)0 ] 1 = αx [eαx cos βx.c − α cos βxeαx + βeαx sin βx] e cos βx = c − α + β tan βx, there exists a unique z ∈ R such that   Z x R Rx v y0 (x) = e− a p(u)du z − e a p(u)du h(v)dv a   Z x R Z R v − ax p(u)du =e z− e a p(u)du (eαx cos βx)−1 k + a

a

v

eαx cos βxf (x)dx

 ,

Hyers-Ulam Stability of Exact Linear Differential Equations

145

where k = [eαa cos βay 0 (a) − (eαa cos βa)0 y(a) + ceαa cos βay(a)] for all x ∈ I and hence  R Z v Z b R v −1 − ax p(u)du ϕ(t)dt e a p(u)du dv | µ(v) | | y(x) − y0 (x) | ≤ e a

x

≤ e−

Rx a

p(u)du

Z

b



| µ(v) |−1

Z

v

 R v eαt ϕ(t)dt e a p(u)du dv.

a

x

Example 9 Consider the x2 y 00 + αxy 0 + βy + f (x) = 0, where α and β are real constants. It is exact when α−β = 2. By using Theorem 40, it has the Hyers-Ulam stability. Let us consider µ(x) be an integrating factor of the Euler differential equation. It is said to be exact if {x2 µ(x)}00 − {αxµ(x)}0 + βµ(x) = 0, that is, 2µ(x) + 2xµ0 (x) + x2 µ00 (x) + 2xµ0 (x) − αxµ0 (x) − αµ(x) + βµ(x) = 0 implies that x2 µ00 (x) + (4 − α)xµ0 (x) + (2 − α + β)µ(x) = 0.

(4.13)

Let µ(x) = xm , µ0 (x) = mxm−1 and µ00 (x) = m(m − 1)xm−2 . Then (4.13) becomes x2 (m2 − m)xm−2 + (4 − α)xmxm−1 + (2 − α + β)xm = 0, that is, m2 + (3 − α)m + (2 − α + β) = 0. From (4.14), we obtain that p

(3 − α)2 − 4(2 − α + β) 2 p −(3 − α) ± (1 − α)2 − 4β = . 2

m=

−(3 − α) ±

When (1 − α)2 − 4β ≥ 0, by Corollary 24 with p(x) = {µ(x)p0 (x)}−1 [µ(x)p1 (x) − {µ(x)p0 (x)}0 ] = (emx .x2 )−1 [emx .αx − (emx .x2 )0 ] 1 = mx 2 [emx .αx − 2xemx − memx .x2 ] e .x α 2 = − −m x x

(4.14)

146

Hyers-Ulam Stability of Ordinary Differential Equations

and   Z x µ(t)f (t)dt h(x) = {µ(x)p0 (x)} k+ a   Z x mx 2 −1 mt = (e .x ) k+ e f (t)dt , −1

a

there exists a unique z ∈ R such that   Z x R Rx v e a p(u)du h(v)dv y0 (x) = e− a p(u)du z − a   Z x R Z R v − ax p(u)du =e z− e a p(u)du (emv .v 2 )−1 k + a

v

emx f (x)dx



a

for all x ∈ I and hence | y(x) − y0 (x) |≤ e

−

≤ e−

Rx a

Rx a

p(u)du

p(u)du

Z

Z

b

 Z | µ(v)p0 (v) |−1

x b

| µ(v)p0 (v) |−1

x

Z

v

a v

 R v ϕ(t)dt e a p(u)du dv

 R v emt ϕ(t) e a p(u)du dv.

a

When (1 − α)2 − 4β < 0, by Corollary 25, p(x) = {µ(x)p0 (x)}−1 [µ(x)p1 (x) − {µ(x)p0 (x)}0 ] = (eαx cos βx.x2 )−1 [eαx cos βx.αx − ((eαx cos βx.x2 )0 ] 1 = αx [eαx cos βx.αx − eαx cos βx.2x e cos βx.x2 − x2 (αeαx cos βx − βeαx sin βx)] 1 = αx [eαx cos βx.αx − 2xeαx cos βx e cos βx.x2 − αx2 eαx cos βx + βx2 eαx sin βx] α 2 = − − α + β tan βx, x x and   Z x h(x) = {µ(x)p0 (x)}−1 k + µ(t)f (t)dt  aZ x  αx 2 −1 αt = (e cos βx.x ) k+ e cos βtf (t)dt , a

there exists a unique z ∈ R such that   Z x R Rx v y0 (x) = e− a p(u)du z − e a p(u)du h(v)dv a  Z x R R v − ax p(u)du =e z− e a p(u)du (emv cos βv.v 2 )−1 a   Z v k+ emx cos βxf (x)dx a

Hyers-Ulam Stability of Exact Linear Differential Equations

147

for all x ∈ I and hence | y(x) − y0 (x) |≤ e

−

Rx a

p(u)du

b

Z

≤ e−

a

p(u)du

Z

 R v ϕ(t)dt e a p(u)du dv

a

x Rx

v

 Z | µ(v)p0 (v) |−1

b

x



| µ(v)p0 (v) |−1

Z

v

 R v eαt ϕ(t)dt e a p(u)du dv.

a



4.2

Notes

The results in this chapter are adopted from the work [9].

Chapter 5 Hyers-Ulam Stability of Euler’s Differential Equations

This chapter deals with the Hyers-Ulam stability of Euler’s differential equations of the following forms: ty 0 (t) + αy(t) + βtr x0 = 0, t2 y 00 (t) + αty 0 (t) + βy(t) = 0, t3 y

000

00

0

+ αt2 y + βty + γy = 0

t4 y (iv) + αt3 y

000

00

0

+ βt2 y + γty + δy = 0,

where t ∈ I = (a, b), −∞ < a < b < +∞ and α, β, γ, δ ∈ C.

5.1

Hyers-Ulam Stability of ty 0 (t) + αy(t) + βtr x0 = 0

Consider ty 0 (t) + αy(t) + βtr x0 = 0.

(5.1)

Let I = (a, b) be an open interval with either 0 < a < b ≤ ∞ or ∞ < a < b < 0. If x0 = 1 in the equation (5.1), then the function  c β r   − t (for α + r 6= 0), α α+r y(t) = t   c − β ln | t | (for α + r = 0), tα where c is an arbitrary real number, is the general solution of (5.1). Theorem 41 Let X be a complex Banach space and let I = (a, b) be an open interval as above. Assume that a function ϕ : I → [0, ∞) is given, α, β, r are complex constants,and that x0 is a fixed element of X. Furthermore, suppose a continuously differentiable function f : I → Xsatisfies the differential inequality |ty 0 (t) + αy(t) + βtr x0 | ≤ ϕ(t) (5.2) 149

150

Hyers-Ulam Stability of Ordinary Differential Equations

for all t ∈ I. If both tα+r−1 and tα−1 ϕ(t) are integrable on (a, c) for any c with a < c ≤ b, then there exists a unique solution f0 : I → X of the differential equation (5.1) such that Z b α−1 −α (5.3) v ϕ(v)dv kf (t) − f0 (t)k ≤| t | t for any t ∈ I. Proof 72 (a) For case α + r 6= 0. Let X ∗ be the dual space of X, i.e., the set of all continuous linear functionals λ : X → C. For λ ∈ X ∗ , fλ : I → C defined by fλ (t) = λf (t) and (fλ )0 (t) = λ{f 0 (t)}. Therefore |t(fλ )0 (t) + αfλ (t) + λ(βtr x0 )| = |λ(tf 0 (t) + αf (t) + βtr x0 )| ≤ kλkktf 0 (t) + αf (t) + βtr x0 k ≤ kλkϕ(t), for all t ∈ I. Consider  α t β z(t) = (tα+r − aα+r )x0 f (t) + a (α + r)aα for all s, t ∈ I. Now,  α  t β α+r α+r |λ{z(t) − z(s)}| = λ f (t) + (t −a )x0 a (α + r)aα    s α β α+r α+r (s − a )x −λ f (s) + 0 a (α + r)aα  α  s α t β α+r α+r = fλ (t) − fλ (s) + (t −s )λx0 a a (α + r)aα Z t  Z t  i d h v α β α+r−1 = fλ (v) dv + v x0 dv λ α a a s dv s Z t   α−1 α v v = fλ0 (v) + αfλ (v) α dv a a s  Z t  β α+r−1 + λ v x0 dv α a s Z t   n o α v α 0 r−1 = fλ (v) + fλ (v) + λ(βv x0 ) dv a v s Z t α−1 v 0 r = {v(fλ ) (v) + αfλ (v) + λ(βv x0 )}dv α a s Z t α−1 v ≤ kλk ϕ(v)dv . (5.4) α a s

Hyers-Ulam Stability of Euler’s Differential Equations

151

Since λ ∈ X ∗ is selected arbitrarily, then the inequality (5.4) becomes Z t α−1 v , ϕ(v)dv kz(t) − z(s)k ≤ α a s for all s, t ∈ I. Since tα−1 ϕ(t) is integrable on (a, c) for any c with a < c ≤ b, then we can find t0 ∈ I such that kz(t)−z(s)k <  for s, t ≥ t0 , that is, z(s)s∈I is a Cauchy net and hence there exists an x ∈ X such that z(s) converges to x as s → b, since X is complete. Indeed,

 α

β α+r a+r

f (t) − a x +

(t − a )x 0

t (α + r)tr = kaα t−α {z(t) − x}k ≤ |aα t−α |kz(t) − z(s)k + |aα t−α |kz(s) − xk Z t α−1 v + |aα t−α |kz(s) − xk ≤ |aα t−α | ϕ(v)dv α a s Z t α−1 v = |aα t−α | ϕ(v)dv α a b as s → b, z(s) → x and thus

 α

β α+r a+r

f (t) − a x +

(t − a )x 0

t (α + r)tr Z t v α−1 ϕ(v)dv . ≤ |t−α |

(5.5)

b

If, f0 (t) =

 a α t

x−

β (tα+r − aa+r )x0 , (α + r)tr

(5.6)

then (5.5) becomes, kf (t) − f0 (t)k ≤ |t

−α

Z t α−1 | v ϕ(v)dv . b

Now, it remains to show the uniqueness of f0 . Assume that x1 ∈ X and there exists another solution  a α β f1 (t) = x1 − (tα+r − aα+r )x0 , t ∈ I. t (α + r)tr So,

 

 α

a α

β

= f (t) − a x + (x − x) (tα+r − aα+r )x0 1

t

t (α + r)tr

 a α

β α+r α+r − f (t) + x1 − (t −a )x0

t (α + r)tr

152

Hyers-Ulam Stability of Ordinary Differential Equations

 a α

β α+r α+r

≤ f (t) − x+ (t −a )x0

r t (α + r)t

   

a α β α+r a+r

+ − f (t) − x1 + (t − a )x0

t (α + r)tr Z t Z t v α−1 ϕ(v)dv v α−1 ϕ(v)dv + |t−α | ≤ |t−α | b b Z t v α−1 ϕ(v)dv ≤ 2|t−α | b

implies that Z 2 b α−1 kx1 − xk ≤ α v ϕ(v)dv → 0, |a | t as t → b. Hence x1 = x. (b) For the case α + r = 0. Consider  α t β z(t) = f (t) + α (ln | t | − ln | a |)x0 a a for all t ∈ I. Now,   α t β |λ{z(t) − z(s)}| = λ f (t) + α (ln |t| − ln |a|)x0 a a   s α β f (s) − α (ln |s| − ln |a|)x0 − a a  α  s α t β = fλ (t) − fλ (s) + α (ln |t| − ln |s|)λx0 a a a Z t   Z t i h  α d β 1 v = fλ (v) dv + λ α x0 dv dv a a v s

s

Z t    v α 0 v α−1 = fλ (v) + αfλ (v) α dv a a s   Z t β 1 x0 dv + λ α a v s Z t     v α β v α 0 vfλ (v) + αfλ (v) + λ α x0 dv = a a a s Z t   v α 0 −α {vfλ (v) + αfλ (v) + λ(βv x0 )}dv = a s Z t   v α 0 r = {vfλ (v) + αfλ (v) + λ(βv x0 )}dv a s Z t   v α ϕ(v)dv . (5.7) ≤ kλk a s

Hyers-Ulam Stability of Euler’s Differential Equations

153

Since λ ∈ X ∗ is selected arbitrarily, then the inequality (5.7) becomes Z t α−1 v , kz(t) − z(s)k ≤ ϕ(v)dv α a s for s, t ∈ I. Proceeding as in case (a), we choose  a α β f0 (t) = x − α (ln | t | − ln | a |)x0 . t t Therefore, kf (t) − f0 (t)k ≤ |t

−α

Z t α−1 v ϕ(v)dv | b

for any t ∈ I. Uniqueness is same as in case(a). This completes the proof of theorem. Theorem 42 Let X be a Complex Banach space and let I = (a, b) be an open interval such that 0 < a < b ≤ ∞ or −∞ < a < b < 0. Assume that a function φ : I → [0, ∞) is given and h : I → X is a continuous function. Furthermore, suppose a continuously differentiable function f : I → X satisfies

0

t ∈ I.

ty (t) + αy(t) + h(t) ≤ φ(t), If both tα−1 φ(t) and tα−1 h(t) are integrable on (a, c), for any c with a < c ≤ b, then there exists a unique solution f0 : I → X of the differential equation 0

ty (t) + αy(t) + h(t) = 0

(5.8)

such that (5.3) holds for any t ∈ I, where α is a complex constant. Proof 73 The proof of the theorem follows from the proof of Theorem 41, if we set  α Z t α−1 t u z(t) = f (t) + h(u)du, a aα a where the unique solution is given by Z t  a α 1 f0 (t) = − α uα−1 h(u)du. t t a The details are omitted.

5.2

Hyers-Ulam Stability of t2 y 00 (t) + αty 0 (t) + βy(t) = 0

Let I = (a, b) be an open interval with 0 < a < b ≤ ∞ or −∞ < a < b < 0. Assume that α, β are real constants satisfying either β < 0 or β > 0, α < 1

154

Hyers-Ulam Stability of Ordinary Differential Equations

and (1 − α)2 − 4β > 0. Consider, p α − 1 − (1 − α)2 − 4β c= , 2

d=

α−1+

p

(1 − α)2 − 4β . 2

Then the function

c2 c1 + d, c t t is the general solution of the Euler differential equation y(t) =

t2 y 00 (t) + αty 0 (t) + βy(t) = 0,

(5.9)

(5.10)

where c1 and c2 are arbitrary real numbers. Theorem 43 If a twice continuously differentiable function f : I → R satisfies the differential inequality | t2 f 00 (t) + αtf 0 (t) + βf (t) | ≤ 

(5.11)

for all t ∈ I and for some  > 0, then there exists a solution f0 : I → R of the Euler equation (5.10) such that  c  b |f (t) − f0 (t)| ≤ − 1 |β| t for any t ∈ I. In particular,if I = (a, ∞) with a > 0, then |f (t) − f0 (t)| ≤

 |β|

for all t > a. Proof 74 Define g(t) = tf 0 (t) + cf (t) and g 0 (t) = tf 00 (t) + f 0 (t) + cf 0 (t) for any t ∈ I, then |tg 0 (t) + dg(t)| = |t{tf 00 (t) + f 0 (t) + cf 0 (t)} + d{tf 0 (t) + cf (t)}| = |t2 f 00 (t) + tf 0 (t) + ctf 0 (t) + tdf 0 (t) + cdf (t)| = |t2 f 00 (t) + t(1 + c + d)f 0 (t) + cdf (t)| = |t2 f 00 (t) + αtf 0 (t) + βf (t)| ≤ ,

(5.12)

for all t ∈ I, if h(t) = −dg(t) and h0 (t) = −dg 0 (t), then inequality (5.12) becomes t 0 h (t) − h(t) ≤ . d By using Theorem 2.3.3, there exists a real number c0 such that R t dT h(t) − c0 ed a T ≤ 

Hyers-Ulam Stability of Euler’s Differential Equations

155

implies that t −dg(t) − c0 ed[ln T ]a ≤ , that is, t d −d{tf 0 (t) + cf (t)} − c0 eln( a ) ≤ . Hence, d 0 tf (t) + cf (t) + c0 a t−d ≤  |d| d for every t ∈ I, where c < 0 and c < d. Consider  c t c0 f (t) + (tc−d − ac−d )ad z(t) = a d(c − d)ac for all t ∈ I. Proceeding as in Theorem 41, Z t c−1 v  kz(t) − z(s)k ≤ dv ac d s for all s, t ∈ I. By Theorem 41, there exists a solution f0 : I → R of the differential equation, ty 0 (t) + cy(t) +

c0 ad −d t =0 d

 kf (t) − f0 (t)k ≤ |β

 c b − 1 | t

such that

for any t ∈ I. Indeed due to (5.6), there exists a real number c3 such that f0 has the following form  c  a c0 c0 ad − . f0 (t) = c3 + d(c − d) tc d(c − d) td Hence, f0 (t) is a solution of the Euler equation (5.10). Example 10 Consider the equation t2 f 00 (t) − 6tf 0 (t) + 10f (t) = 0. The function f : I → R satisfies the differential inequality |t2 f 00 (t) − 6tf 0 (t) + 10f (t)| ≤ .

(5.13)

156

Hyers-Ulam Stability of Ordinary Differential Equations

Assume that f (t) = tr , f 0 (t) = rtr−1 and f 00 (t) = r(r − 1)tr−2 . Then (5.13) becomes 0 = t2 r(r − 1)tr−2 − t6rtr−1 + 10tr = r(r − 1)tr − 6rtr + 10tr = (r2 − r − 6r + 10)tr = (r2 − 7r + 10)t2 implies that r2 − 7r + 10 = 0, that is, (r − 2)(r − 5) = 0. Let’s define g(t) = tf 0 (t) − 2f (t) and g 0 (t) = tf 00 (t) + f 0 (t) − 2f 0 (t), then |tg 0 (t) − 5g(t)| = |t{tf 00 (t) + f 0 (t) − 2f 0 (t)} − 5{tf 0 (t) − 2f (t)}| = |t2 f 00 (t) + tf 0 (t) − 2tf 0 (t) − 5tf 0 (t) + 10f (t)| = |t2 f 00 (t) + t(1 − 2 − 5)f 0 (t) + (−2. − 5)f (t)| = |t2 − 6f 0 (t) + 10f (t)| ≤ . Using Theorem 41, there exists a solution f0 : I → R of the differential equation (5.13), which is of the form   c a c0 ad c0 − f0 (t) = c3 + d(c − d) tc d(c − d) td  c0  −2 5 c0 −5 5 = c3 − a t + a t . 35 35 Hence f0 (t) is the solution of the Euler equation (5.13).

5.3

000

00

0

Hyers-Ulam Stability of t3 y + αt2 y + βty + γy = 0

The objective of this section is to investigate the generalized Hyers-Ulam stability of the following Euler’s differential equations of the form: 00

0

t2 y (t) + αty (t) + βy(t) + γtr x0 = 0 and

000

00

0

t3 y (t) + αt2 y (t) + βty (t) + γy(t) = 0,

(5.14)

(5.15)

where α, β, γ and r are complex constants. Intuitively, we shall prove that if a twice continuously differentiable function f : I → X satisfies the differential inequality

00

0

2

(5.16)

t y (t) + αty (t) + βy(t) + γtr x0 ≤ φ(t)

Hyers-Ulam Stability of Euler’s Differential Equations

157

for all t ∈ I, where x0 is fixed element of the Complex Banach space X and I = (a, b) with 0 < a < b ≤ ∞ or −∞ < a < b < 0, then there exists a twice continuously differentiable solution f0 (t) of (5.14) such that Z l b −l−1 (5.17) u ψ(u)du kf (t) − f0 (t)k ≤ t t for any t ∈ I, where Z b −m−1 u φ(u)du . ψ(t) = |t | t m

We also apply this result to the investigation of the Hyers-Ulam stability of (5.15). Let I = (a, b) with 0 < a < b ≤ ∞ or −∞ < a < b < 0. For fixed x0 6= 0, the general solution of (5.14) in the class of real valued functions defined on I is given by  r 0t  c1 tl + c2 tm − γx g(r) 6= 0, l 6= m,  g(r) ,       0  c1 tl + c2 tm − γx0 t0 r ln|t| , g(r) = 0 6= g (r), l 6= m, g (r) y(t) = r 0t  (c1 + c2 ln |t|) tl − γx g(r) 6= 0, l = m,  g(r) ,        r 2 0  , g(r) = 0 = g (r), l = m, (c1 + c2 ln |t|) tl − γx0 t (ln|t|) 2 where g(r) = r2 + (α − 1)r + β, and c1 , c2 are arbitrary constants. 0

Remark 6 Indeed, g (r) = 2r + α − 1. If g(r) = 0, then either r − l = 0 or r −m = 0. Therefore, r −l = 0 implies that 2r +α−1 = 2r −l −m = r −m 6= 0 and r − m = 0 implies that 2r + α − 1 = r − l 6= 0. Hence, the second solution could be any one of the following: c1 tl + c2 tm −

γx0 tr ln |t| ; r − l = 0, r − m 6= 0, l 6= m, r−m

c1 tl + c2 tm −

γx0 tr ln |t| ; r − l 6= 0, r − m = 0, l 6= m. r−l

Theorem 44 Let X be a Complex Banach space and let I = (a, b) be an open interval with either 0 < a < b ≤ ∞ or −∞ < a < b < 0. Assume that a function φ : I → [0, ∞) is given, that α, β, γ, r are complex constants and that x0 is a fixed element of X . Furthermore, suppose a twice continuously differentiable function f : I → X satisfies the differential inequality (5.16). If tr−l−1 , tr−m−1 , tm−l−1 , t−m−1 φ(t) and t−l−1 ψ(t) are integrable on (a, c), for any c with a < c ≤ b, then there exists a unique solution f0 : I → X (which is twice continuously differentiable) of (5.14) such that (5.17) holds for any t ∈ I, where l and m are the roots of g(r) = 0.

158

Hyers-Ulam Stability of Ordinary Differential Equations

Proof 75 We prove the theorem for five possible cases, viz., (i) (r − l)(r − m) 6= 0, l 6= m; (ii) r − l = 0 6= r − m, l 6= m; (iii) r − l 6= 0 = r − m, l 6= m; (iv) r − l 6= 0, l = m; (v) r − l = 0, l = m. Case (i) Suppose that X is a Complex Banach space and f : I → X is a twice continuously differentiable function satisfying the differential inequality 0 (5.16). Define h : I → X such that h(t) = tf (t) − lf (t). Then it is easy to verify that

0

t h (t) − m h(t) + γ tr x0

00 0

= t2 f (t) + (1 − l − m)t f (t) + lm f (t) + γ tr x0

00 0

= t2 f (t) + αt f (t) + β f (t) + γ tr x0 ≤ φ(t). Hence by Theorem 41, it follows that there exists a unique solution h0 : I → X 0 of the differential equation ty − my + γ tr x0 = 0 such that Z b m −m−1 kh(t) − h0 (t)k ≤ |t | v φ(v)dv (5.18) t provided tr−m−1 and t−m−1 φ(t) are integrable on (a, c), for any c with a < c ≤ b. Indeed, h0 (t) =

 a −m t

x−


γtm tr−m − ar−m x0 , r−m

where x is a limit point in X. If we denote Z b m ψ(t) = t v −m−1 φ(v)dv , t then (5.18) becomes

0



tf (t) − lf (t) − h0 (t) ≤ ψ(t).

(5.19)

Clearly, ψ : I → [0, ∞). By Theorem 42, there exists a unique solution f1 : 0 I → X of the differential equation ty (t) − ly(t) − h0 (t) = 0 such that Z l b −l−1 kf (t) − f1 (t)k ≤ t u ψ(u)du (5.20) t provided t−l−1 ψ(t) and t−l−1 h0 (t) are integrable on (a, c), for any c with a
0. Consider the Euler’s equation 00

0

t4 y (iv) (t) + 2t2 y (t) − 4ty (t) + 4y(t) = 0

(5.39)

If we compare (5.39) with (5.27), then α = 0, β = 2, γ = −4 and δ = 4, and l = 1, m = 1, n = 2, p = 2 are the characteristic roots of (5.39). Let f : I → X be satisfy the differential inequality

00 0

4 (iv)

t f (t) + 2f 2 y (t) − 4tf (t) + 4f (t) ≤  for any  > 0 and for any t ∈ I. Then, by Theorem 47, there exists a unique solution f0 : I → X such that Z 2 b −3 kf (t) − f0 (t)k ≤ t u θ(u)du , t where I = (a, b) and   Z t 2 2 b −3  − 1 . φ(t) = t u du = t 2 b When b → ∞, φ(t) = 2 . Also Z   b  t  −2  du = − 1 = ψ(t) = |t| u t 2 2 b 2 and

Z b   θ(t) = |t| u−2 du = t 2 2

as b → ∞. Hence, Z   t 2 2 b −3   kf (t) − f0 (t)k ≤ t u du = − 1 . t 2 4 b When b → ∞, kf (t) − f0 (t)k ≤

 4

in which     t 2x0 − x − x2 x0 − x2 f0 (t) = + ln t a a a      x0 x ¯ + x + x2 − 2x0 t t2 + + ln a2 a2 a 

and x, x ¯, x0 , x2 are the unique elements of X.

178

5.5

Hyers-Ulam Stability of Ordinary Differential Equations

Notes

The results of Sections 5.2 and 5.3 for Hyers-Ulam stability of first and second order Euler’s differential equations are extracted from the work of S. M. Jung [18] except Theorem 42. It is noticed that Theorem 42 plays an important role to go for the Hyers-Ulam stability of third and fourth order Euler’s differential equations and these results are depicted from the works of A. K. Tripathy and A. Satapathy [58] and [59]. It would be interesting to extend the works of Tripathy and Satapathy to fifth order and sixth order Euler’s differential equations of the form: t5 y (v) + αt4 y (iv) + βt3 y

000

00

+ γt2 y + ηty 0 + σy = 0

t6 y (vi) + αt5 y (v) + βt4 y iv + γt3 y

000

+ ηt2 y 00 + σty 0 + τ y = 0.

Chapter 6 Generalized Hyers-Ulam Stability of Differential Equations in Complex Banach Space

This chapter deals with the Hyers-Ulam stability of linear first and second order complex differential equations f 0 (z) + p(z)f (z) + q(z) = 0,

(6.1)

y 00 (t) + αy 0 (t) + βy(t) = 0

(6.2)

and y 00 (t) + αy 0 (t) + βy(t) = σ(x) 2

(6.3)

1

in Banach space, where y ∈ C (I), α, β ∈ C, f ∈ C (I), σ ∈ C(I), I = (a, b), −∞ < a < b < +∞.

6.1

Hyers-Ulam Stability of First Order Differential Equations

In this section, we present the Hyers-Ulam stability of the Banach spacevalued differential equations of the form 6.1. Let X be a complex Banach space and let Ω be an open set of C. A mapping f : Ω −→ X is said to be holomorphic if and only if lim

w−→z

f (w) − f (z) w−z

exists in the norm-topology of X for each z ∈ Ω. We know that f : Ω −→ X is holomorphic if and only if φ of : Ω −→ C is holomorphic (as a complexvalued function) for each φ ∈ X ∗ , the dual space of X. We denote H(Ω, X), the set of all holomorphic mappings f : Ω −→ X. In short, H(Ω) = H(Ω, C). Let f ∈ H(Ω, X). Let’s consider the case where 0 ∈ Ω. For z ∈ Ω, we write Rz R1 f (ζ)dζ for 0 zf (zt)dt, the integral of f over the path γ defined by 0 γ(t) = zt, ∀ ∈ [0, 1]. 179

180

Hyers-Ulam Stability of Ordinary Differential Equations

For each p ∈ H(Ω), define p˜ as z

Z

p(ζ)dζ, ∀z ∈ Ω.

p˜(z) = exp 0

If Ω is convex, then p˜ ∈ H(Ω) and p˜0 (z) = p(z)˜ p(z), ∀z ∈ Ω. Lemma 12 Let Ω be a convex open set of C with 0 ∈ Ω. Let X be a complex Banach space and p ∈ H(Ω). For f, q, u ∈ H(Ω, X), each of the following conditions are equivalent. (a) f 0 (z) + p(z)f (z) + q(z) = u(z), for all z ∈ Ω. Rz 1 (b) f (z) = p(z) {f (0) + 0 p˜(ζ)(u(ζ) − q(ζ))dζ}, for all z ∈ Ω. ˜ Proof 80 Consider the case: (a) implies (b). Let f 0 (z) + p(z)f (z) + q(z) − u(z) = 0 for every z ∈ Ω. Since, p˜0 (z) = p(z)˜ p(z), z ∈ Ω, then, {˜ p(z)f (z)}0 = p˜0 (z)f (z) + p˜(z)f 0 (z) = p(z)˜ p(z)f (z) + p˜(z)f 0 (z) = p˜(z){p(z)f (z) + f 0 (z)} = p˜(z)(u(z) − q(z)), ∀z ∈ Ω. Integrating the above relation from 0 to z, we obtain Z z Z z {f (ζ)˜ p(ζ)}0 dz = p˜(ζ)(u(ζ) − q(ζ))dζ, 0

0

that is, Z f (z)˜ p(z) − f (0)˜ p(0) =

z

p˜(ζ)(u(ζ) − q(ζ))dζ, ∀z ∈ Ω. 0

By definition, p˜(0) = 1 and hence   Z z 1 f (0) + p˜(ζ)(u(ζ) − q(ζ))dζ , ∀z ∈ Ω. f (z) = p˜(z) 0 Next, we consider the case for (b) implies (a). Let   Z z 1 f (z) = f (0) + p˜(ζ)(u(ζ) − q(ζ))dζ , ∀z ∈ Ω. p˜(z) 0

Generalized Hyers-Ulam Stability

181

Then f 0 (z) =

 0 Z z 1 p˜(z) f (0) + p ˜ (ζ)(u(ζ) − q(ζ))dζ p˜2 (z) 0    Z z − f (0) + p˜(ζ)(u(ζ) − q(ζ))dζ p˜0 (z) , ∀z ∈ Ω. 0

Hence, f 0 (z) =

1 p˜2 (z)

{˜ p2 (z)(u(z) − q(z)) − p˜(z)f (z)˜ p0 (z)},

that is, f 0 (z) + p(z)f (z) + q(z) = u(z), ∀z ∈ Ω. Hence, the lemma is proved. Theorem 48 Let Ω be a convex open set of C with 0 ∈ Ω. Let X be a complex Banach space and q ∈ H(Ω, X). Suppose that p ∈ H(Ω) satisfies that Z z 1 | |˜ p(ζ)|dζ| < ∞. Cp = sup ˜(z) 0 z∈Ω p For each ε ≥ 0 and f ∈ H(Ω, X) satisfying ||f 0 (z) + p(z)f (z) + q(z)|| ≤ ε, ∀z ∈ Ω

(6.4)

there exists g ∈ H(Ω, X) such that g 0 (z) + p(z)g(z) + q(z) = 0, ∀z ∈ Ω and that ||f (z) − g(z)|| ≤ Cp ε, ∀z ∈ Ω. Proof 81 Suppose that ε ≥ 0 and f ∈ H(Ω, X), such that ||f 0 (z) + p(z)f (z) + q(z)|| ≤ ε, ∀z ∈ Ω. If u(z) = f 0 (z) + p(z)f (z) + q(z), ∀z ∈ Ω, then by Lemma 12,   Z z 1 f (0) + p˜(ζ)(u(ζ) − q(ζ))dζ , ∀z ∈ Ω. f (z) = p˜(z) 0 For all z ∈ Ω, define g(z) =

  Z z 1 f (0) − {˜ p(ζ)q(ζ))dζ . p˜(z) 0

Then g ∈ H(Ω, X) satisfies g 0 (z) + p(z)g(z) + q(z) = 0, ∀z ∈ Ω

(6.5)

182

Hyers-Ulam Stability of Ordinary Differential Equations

due to Lemma 12. Now, Z 1 z p˜(ζ)u(ζ)dζ ||f (z) − g(z)|| ≤ |˜ p(z)| Z 0 ε z , ∀z ∈ Ω, ≤ |˜ p (ζ)|dζ |˜ p(z)| 0 because ||u(z)|| ≤ ε, ∀z ∈ Ω. Hence, Z 1 z = Cp ε, ∀z ∈ Ω. |˜ p (ζ)|dζ p(z)| 0 z∈Ω |˜

||f (z) − g(z)|| ≤ ε sup Thus, the proof is complete.

Theorem 49 Let Ω be a convex open set of C with 0 ∈ Ω. Let X be a complex Banach space and q ∈ H(Ω, X) and p ∈ H(Ω). Suppose that there exists λ ∈ ∂Ω, the boundary of Ω such that Z 1 z Dp (λ) = sup |˜ p(ζ)|dζ < ∞, |˜ p (z)| z∈Ω λ where Z

z

Z |˜ p(ζ)|dζ = lim

a→0

λ

1

|˜ p(λ + t(z − λ))|(z − λ)dt. a

For each ε ≥ 0 and f ∈ H(Ω, X) satisfying (6.4), there exists g ∈ H(Ω, X) such that g 0 (z) + p(z)g(z) + q(z) = 0, (∀z ∈ Ω) and that kf (z) − g(z)k ≤ Dp (λ)ε, ∀z ∈ Ω. Proof 82 Let there exist λ ∈ ∂Ω such that Z 1 z |˜ p(ζ)|dζ < ∞. Dp (λ) = sup p(z)| λ z∈Ω |˜ For ε ≥ 0 and f ∈ H(Ω, X), let ||f 0 (z) + p(z)f (z) + q(z)|| ≤ ε, (∀z ∈ Ω). For all z ∈ Ω, let’s set v(z) = f 0 (z) + p(z)f (z) + q(z). Then kv(z)k ≤ ε, ∀z ∈ Ω. By Lemma 12,   Z z 1 f (z) = f (0) + {˜ p(ζ)(v(ζ) − q(ζ))dζ , ∀z ∈ Ω. p˜(z) 0

(6.6)

Generalized Hyers-Ulam Stability

183

¯ and the closure of Ω. Hence for 0 < t ≤ 1, But Ω is convex, so is Ω λ + t(z − λ) ∈ Ω, ∀z ∈ Ω. By the hypothesis, p˜v ∈ H(Ω, X) and by using Cauchy theorem Z z Z λ Z z p˜(ζ)v(ζ)dζ = p˜(ζ)v(ζ)dζ + p˜(ζ)v(ζ)dζ, ∀z ∈ Ω. 0

0

(6.7)

λ

Setting   Z z Z λ 1 p˜(ζ)q(ζ)dζ , ∀z ∈ Ω, g(z) = p˜(ζ)v(ζ)dζ − f (0) + p˜(z) 0 0

(6.8)

it follows that g 0 (z) + p(z)g(z) + q(z) = 0, (∀z ∈ Ω). By (6.6),(6.7) and (6.8), we obtain Z 1 z kf (z) − g(z)k ≤ p ˜ (ζ)v(ζ)dζ |˜ p(z)| λ Z z 1 = Dp (λ)ε, ∀z ∈ Ω. ≤ ε sup p ˜ (ζ)dζ p(z)| λ z∈Ω |˜ This completes the proof of the theorem. Example 13 Let Ω be a bounded convex open set of C, say |z| ≤ M for all z ∈ Ω. If 0 ∈ Ω and p ∈ H(Ω) is bounded, then Z 1 z Cp = sup |˜ p(ζ)|dζ < ∞. |˜ p(z)| 0 z∈Ω If p ∈ H(Ω) is bounded such that |p(z)| ≤ K for all z ∈ Ω, then Z 1 Z z |p(ζt)||z|dt ≤ KM, ∀z ∈ Ω. p(ζ)dζ ≤ 0

0

Hence,  Z |˜ p(ζ)| = exp Re

z

 p(ζ)dζ

≥ e−KM , ∀z ∈ Ω.

(6.9)

0

As |˜ p(ζ)| ≤ eKM , ∀z ∈ Ω, then we obtain Z z |˜ p(z)|dζ ≤ M eKM , ∀z ∈ Ω. 0

It follows from (6.9) and (6.10) that Z 1 z ≤ M e2KM < ∞. Cp = sup |˜ p (ζ)|dζ |˜ p(z)| 0 z∈Ω

(6.10)

184

Hyers-Ulam Stability of Ordinary Differential Equations

Example 14 Let Ω = {z ∈ C : |z| < 1}. Let p ∈ H(Ω) be defined by, 1 p(z) = z+1 , ∀z ∈ Ω. Then log(z + 1), ∀z ∈ Ω is well-defined such that Z z p˜(z) = exp p(ζ)dζ = elog(z+1) = z + 1, ∀z ∈ Ω. (6.11) 0

Hence, Z 1 |˜ p(−1 + t(z + 1))|(z + 1)dt |˜ p(ζ)|dζ = lim a→0 a −1 Z 1 |t(z + 1)|(z + 1)dt = lim

Z

z

a→0

(6.12)

a

= |z + 1|2

Z

1

tdt = 0

|z + 1|2 , ∀z ∈ Ω. 2

By (6.11) and (6.12), we conclude that Z |z + 1| 1 z Dp (−1) = supz∈ω |˜ p(ζ)|dζ = supz∈Ω = 1. |˜ p| −1 2 We note that Dp (λ) = supz∈ω Therefore,

Z 1 z = ∞, ∀λ ∈ ∂Ω\{−1}. |˜ p (ζ)|dζ |˜ p| λ

Z 1 z Cp = supz∈ω |˜ p(ζ)|dζ = ∞. |˜ p| 0

(6.13)

(6.14)

Consider that λ ∈ ∂Ω\{−1}. For all n ∈ N, we can associate corresponding zn ∈ Ω such that |λ + 1| ≥ |λ − zn | and |zn + 1|
0 and u, v ∈ D(N(0), X) with kTp u − vk ≤ , there exists u0 ∈ D(N(0), X) such that Tp u0 = v and ku − u0 k∞ ≤ L0 (x). If we put w = Tp u − v for u, v ∈ D(N(0), X), then kwk∞ ≤ 1. Hence for w + v = Tp u and for any arbitrary x1 ∈ X, we have # " n−1 n−1 X X 1 v(s)P (s + 1) w(s)P (s + 1) + x1 +  u(n) = P (n) s=0 s=0 for any n ∈ N(0). Let " # n−1 X 1 u0 (n) = x1 − x + v(s)P (s + 1) , n ∈ N(0). P (n) s=0 Then u0 ∈ D(N(0), X) and Tp u0 = v. Consequently, ku − u0 k∞ = sup ku(n) − u0 (n)k n≥0

" # n−1

 X

x+ w(s)P (s + 1) ≤ L0 (x) = sup

P (n) n≥0 s=0

due to the fact that kwk∞ ≤ 1. Hence, L0 (x) is a HUS constant for Tp . Thus, LTp ≤ L0 (x). Since x ∈ X is arbitrary, then it follows that LTp ≤ inf x∈X L0 (x). Theorem 57 Let Tp : D(N(0), X) → D(N(0), X) be the linear operator defined by (7.12). If βp < ∞, then Tp has the Hyers-Ulam stability with HUS constant LTp . Proof 94 Suppose that βp < ∞. Then by Remark 7.4.1, Tp has the HyersUlam stability with HUS constant βp . Because the uniqueness doesn’t hold in case when βp < ∞ due to Remark 7.4.2 and LTp < ∞ exists due to Remark 7.4.3, then it is sufficient to show that LTp = βp . By definition, LTp ≤ βp . Hence, we need to show that LTp ≥ βp only. Define a linear operator S : D(N(0), X) → D(N(0), X) by n−1 1 X w(s)P (s + 1), ∀ n ∈ N(0), w ∈ D(N(0), X). (Sw)(n) = P (n) s=0

Hyers-Ulam Stability of Difference Equations

207

Then for all w ∈ D(N(0), X), kSwk∞



1 n−1

X

= sup k(Sw)(n)k = sup w(s)P (s + 1)

P (n) n≥0 n≥0 s=0

kwk∞ n≥0 |P (n)|

≤ sup

n−1 X

|P (s + 1)| = βp kwk∞ < ∞.

s=0

Hence, S is a bounded linear operator with kSk ≤ βp . Moreover, if x0 is a (n)| unit element of X and u0 = |P P (n) x0 for n ∈ N(0), then u0 ∈ D(N(0), X) and ku0 k∞ = 1. Consequently, kSu0 k∞ = βp and hence kSk ≥ βp . Therefore, kSk = βp . Since |P (0)| = 1, then we can find n1 > 0 such that |P (n)| ≥ 21 for n ≥ n1 + 1. Thus, βp ≥

n−1 n−1 X 1 (n − n1 ) 1 X |P (s + 1)| ≥ |P (s + 1)| ≥ , |P (n)| s=0 |P (n)| s=n 2|P (n)| 1

1 that is, |P (n)| ≤ 2βp < ∞. Therefore, if we choose x ∈ X arbitrary, then it x follows that |P (n)| ∈ D(N(0), X). We notice that the set ({w ∈ D(N(0), X) : kwk∞ ≤ 1}) is a symmetric set of D(N(0), X). Applying Lemma 7.4.2, we obtain

# " n−1

1 X

w(s)P (s + 1) sup sup x+

P (n) w∈D(N(0),X) n≥0 s=0 kwk∞ ≤1

=



x

sup + (Sw)(n)

w∈D(N(0),X) n≥0 P (n) sup

kwk∞ ≤1

=

x

sup k(Sw)k∞ = kSk.

+ (Sw) ≥ ∞ w∈D(N(0),X) P w∈D(N(0),X) sup

kwk∞ ≤1

kwk∞ ≤1

which holds for all x ∈ X. Ultimately,

" # n−1

1 X

inf sup sup x+ w(s)P (s + 1) ≥ kSk = βp . x∈X w∈D(N(0),X) n≥0 P (n)

s=0 kwk∞ ≤1

This completes the proof of the theorem. Example 16 Consider (Tp u)(n) = 4u(n) − (1 + (−1)n )u(n)

208

Hyers-Ulam Stability of Ordinary Differential Equations Q −1 n−1 i such that 1 + p(n) = 2 + (−1)n and P (n) = . Indeed, i=0 (2 + (−1) ) ∞ 1 X P (m + 1) P (n) m=n   1 1 1 = 1+ + + ··· (2 + (−1)n ) (2 + (−1)n ) (2 + (−1)n )(2 − (−1)n )   1 1 + ··· = 3 ≤2 1+ + 1.3 12 .32

implies that αp ≤ 3. Hence by Theorem7.4.3, Tp has the Hyers-Ulam stability with HUS constant αp ≤ 3. 

7.4

Notes

In Sections 7.2 and 7.3, the contribution work [60] of A. K. Tripathy have been kept for Hyers-Ulam stability of second order linear difference equations. Section 7.4 provides the exclusive work of Tripathy et al. [61] for first order difference operators on Banach space.

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Index

A-linear mapping 58 additive mapping 2 Banach algebra 2 Borel set 48 Cauchy net 98 Cauchy sequence 5, 8, 12, 16, 17 characteristic equation 113, 196 complete generalized metric space 3 complex algebra 2 complex vector spaces 30 continuously differentiable 84 Davison equation 19 difference operator 200 differential inequality 60 Euler’s differential equations 149 Gronwall inequality 124

Hyers-Ulam-Rassias stability 2 integrating factor 143,144 Jensen concave function 63 k − Lipschitz 63 left A-module 2 Lipschitz constant 38 mean value theorem 71 normed algebra 26 quadratic mapping 31 quasi-normed space 44 Riccati equation 125 strictly contractive operator 103 successive approximations 104 symmetric set 201

215