Integral Inequalities on Time Scales 9783110705553, 9783110705508

Table of contents :
Preface
Contents
1. Linear integral inequalities on time scales
2. Linear integro-dynamic inequalities
3. Nonlinear integral inequalities
4. Diamond-α integral inequalities
5. Fractional integral inequalities
6. Two-dimensional linear integral inequalities
7. Snow’s inequalities
8. Two-dimensional linear integro-dynamic inequalities
9. Two-dimensional nonlinear integral inequalities
10. Delay integral inequalities
11. Applications
Bibliography
Index

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Svetlin G. Georgiev Integral Inequalities on Time Scales

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Svetlin G. Georgiev

Integral Inequalities on Time Scales |

Mathematics Subject Classification 2010 39A10, 34A60, 34B37, 34C10, 34K20, 34K30 Author Prof. Dr. Svetlin G. Georgiev Kliment Ohridski University of Sofia Department of Differential Equations Faculty of Mathematics and Informatics 1126 Sofia Bulgarien [email protected]

ISBN 978-3-11-070550-8 e-ISBN (PDF) 978-3-11-070555-3 e-ISBN (EPUB) 978-3-11-070566-9 Library of Congress Control Number: 2020939558 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2020 Walter de Gruyter GmbH, Berlin/Boston Cover image: Vipul Umretiya / DigitalVision Vectors / gettyimages.de Typesetting: VTeX UAB, Lithuania Printing and binding: CPI books GmbH, Leck www.degruyter.com

Preface Time scale theory was first initiated by Stefan Hilger in 1988 in his PhD thesis to unify both approaches of dynamic modeling: difference and differential equations. Similar ideas have been used before and go back in the introduction of the Riemann–Stieltjes integral which unifies sums and integrals. Many results to differential equations carry over easily to corresponding results for difference equations, while other results seem to be totally different in nature. Because of these reasons, the theory of dynamic equations is an active area of research. The time scale calculus can be applied to any fields in which dynamic processes are described by discrete or continuous time models. So, the calculus of time scales has various applications involving noncontinuous domains such as certain bug populations, phytoremediation of metals, wound healing, maximization problems in economics, and traffic problems. This book is devoted on recent developments of some linear and nonlinear integral inequalities on time scales. The book is intended for the use in the field of dynamic calculus on time scales, dynamic and integral equations on time scales. It is also suitable for graduate courses in the above fields. The book is designed for those who have mathematical background on time scales calculus. This book contains eight chapters. The basic Gronwall’s and Bellman’s inequalities are considered in Chapter 1, as well as some linear Volterra-type inequalities, namely the inequalities of Gamidov and Rodrigues. We also investigate simultaneous and Pachapatte’s inequalities. Chapter 2 introduces linear integro-dynamic inequalities of Pachpatte-type and linear integro-dynamic inequalities with several iterated integrals. Chapter 3 is concerned with Dragomir- and Pachpatte-type nonlinear integral inequalities. In Chapter 4, Diamond-α integral inequalities are investigated, such as Steffensen, Jensen, Radon, and Schlömilch inequalities. Chapter 5 is devoted to fractional integral inequalities on time scales. We consider the Poincaré-, Opial-, Sobolev-, and Ostrowskitype inequalities. Chapter 6 is devoted to two-dimensional linear integral inequalities. We investigate Wendroff- and Pachpatte-type two-dimensional linear integral inequalities. Chapter 7 deals with Snow-type two-dimensional linear integral inequalities. In Chapter 8, some two-dimensional Pachpatte-type linear integro-dynamic inequalities are investigated, as well as some of their modifications. Chapter 9 is devoted to Wendroff- and Pachpatte-type two-dimensional nonlinear integral inequalities. Chapter 10 introduces delay integral inequalities with one and two independent variables. Chapter 11 deals with some applications of certain linear integral inequalities and some linear integro-dynamic inequalities. We investigate the existence and uniqueness of the solutions of first order dynamic equations, as well as continuous dependence on initial conditions of the solutions of first order dynamic equations. We give applications for some second order integro-dynamic equations and Volterra integral equations. Some bounds on the solutions of delay dynamic equations are deduced. https://doi.org/10.1515/9783110705553-201

VI | Preface This book is addressed to a wide audience of specialists such as mathematicians, physicists, engineers, and biologists. It can be used as a textbook at the graduate level and as a reference book for several disciplines. The author welcomes any suggestions for the improvement of the text. Paris August 2019

Svetlin Georgiev

Contents Preface | V 1 1.1 1.2 1.3 1.4 1.5

Linear integral inequalities on time scales | 1 The inequalities of Gronwall and Bellman | 1 Volterra-type integral inequalities | 17 The inequalities of Gamidov and Rodrigues | 23 Simultaneous inequalities | 32 Pachpatte’s inequalities | 37

2 2.1 2.2 2.3

Linear integro-dynamic inequalities | 59 Pachpatte’s inequalities | 59 Modifications of Pachpatte’s inequalities | 69 Inequalities with several iterated integrals | 73

3 3.1 3.2

Nonlinear integral inequalities | 93 Dragomir’s inequalities | 93 Pachpatte’s inequalities | 95

4 4.1 4.2 4.3 4.4 4.5

Diamond-α integral inequalities | 113 Diamond-α calculus on time scales | 113 Steffensen’s inequalities | 122 The Jensen inequality | 130 The Radon inequality | 132 The Schlömilch Inequality | 134

5 5.1 5.2 5.3 5.4 5.5

Fractional integral inequalities | 135 Integral inequalities for the Chebyshev functional | 135 A Poincaré-type inequality | 138 A Sobolev-type inequality | 142 An Opial-type inequality | 143 Ostrowski-type inequalities | 144

6 6.1 6.2

Two-dimensional linear integral inequalities | 147 Wendroff’s inequality | 147 Pachpatte’s inequalities | 150

7 7.1 7.2

Snow’s inequalities | 189 Existence of solutions of some partial dynamic equations | 189 Snow’s inequalities | 194

VIII | Contents 8 8.1 8.2

Two-dimensional linear integro-dynamic inequalities | 219 Pachpatte’s inequalities | 219 Modifications of Pachpatte’s inequalities | 235

9 9.1 9.2

Two-dimensional nonlinear integral inequalities | 249 Wendroff’s inequality | 249 Pachpatte’s inequality | 251

10 10.1 10.2

Delay integral inequalities | 255 Linear Gronwall–Bellman-type delay integral inequalities | 255 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay | 256 Gronwall–Bellman-type nonlinear delay integral inequalities with several delays | 273 Delay integral inequalities in two independent variables | 277 Delay Volterra–Fredholm-type integral inequalities | 282

10.3 10.4 10.5 11 11.1 11.2 11.3 11.4 11.5

Applications | 289 Existence and uniqueness of the solution of first order dynamic equations | 289 Continuous dependence on initial conditions of the solutions of first order dynamic equations | 292 Second order integro-dynamic equations | 293 Volterra integral equations | 298 Applications to nonlinear delay dynamic equations | 299

Bibliography | 305 Index | 307

1 Linear integral inequalities on time scales In this chapter are investigated Gronwall’s and Bellman’s inequalities, some linear Volterra type inequalities, the inequalities of Gamidov and Rodrigues. They are investigated simultaneous inequalities and Pachapatte’s inequalities. The presentation in this chapter follows some results in [1, 2, 5–7, 16, 24–26, 29]. Suppose that 𝕋 is a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let also, a, b ∈ 𝕋, a < b, and J = [a, b] be a time scale interval. The set of all rd-continuous functions on 𝕋 will be denoted with 𝒞rd . The set of all rd-continuous and regressive functions on 𝕋 will be denoted by ℛ, i. e., ℛ is the set of all rd-continuous functions f : 𝕋 → ℝ such that 1 + μ(t)f (t) ≠ 0,

t ∈ 𝕋.

If A ⊂ 𝕋, by ℛ(A) we will denote the set of all functions f : A → ℝ which are rd-continuous on A and 1 + μ(t)f (t) ≠ 0

for any t ∈ A.

We define the sets ℛ = {f ∈ ℛ : 1 + μ(t)f (t) > 0 for all t ∈ 𝕋}, +

ℛ (A) = {f ∈ ℛ(A) : 1 + μ(t)f (t) > 0 for all t ∈ A}. +

1.1 The inequalities of Gronwall and Bellman Theorem 1.1.1 (Gronwall’s Inequality). Let x ∈ 𝒞rd (J) be a nonnegative function and c, p be nonnegative constants. Then the inequality t

x(t) ≤ ∫(px(s) + c)Δs, a

t ∈ J,

implies the inequality t

x(t) ≤ c ∫ e⊖p (σ(s), t)Δs, a

t ∈ J.

Proof. Let t

g(t) = ∫(px(s) + c)Δs, a

https://doi.org/10.1515/9783110705553-001

t ∈ J.

2 | 1 Linear integral inequalities on time scales Then g(a) = 0,

x(t) ≤ g(t),

t ∈ J,

and g Δ (t) = px(t) + c

≤ pg(t) + c,

t ∈ J,

g Δ (t) − pg(t) ≤ c,

t ∈ J.

or

We have Δ

(g(⋅)e⊖p (⋅, a)) (t) = g Δ (t)e⊖p (σ(t), a)

+ (⊖p)(t)g(t)e⊖p (t, a)

= g Δ (t)e⊖p (σ(t), a) −

pg(t) e (t, a) 1 + pμ(t) ⊖p

= g Δ (t)e⊖p (σ(t), a) −

pg(t) (1 + pμ(t))e⊖p (σ(t), a) 1 + pμ(t)

= g Δ (t)e⊖p (σ(t), a)

− pg(t)e⊖p (σ(t), a)

= (g Δ (t) − pg(t))e⊖p (σ(t), a) ≤ ce⊖p (σ(t), a),

t ∈ J.

Hence, t

g(t)e⊖p (t, a) ≤ c ∫ e⊖p (σ(s), a)Δs, a

t ∈ J,

and x(t) ≤ g(t) t

≤ c∫ a

e⊖p (σ(s), a) e⊖p (t, a)

Δs

t

= c ∫ e⊖p (σ(s), t)Δs, a

This completes the proof.

t ∈ J.

1.1 The inequalities of Gronwall and Bellman

| 3

Example 1.1.2. Let 𝕋 = lℤ, l > 0, x, c and p be as in Theorem 1.1.1. Then σ(t) = t + l,

t ∈ 𝕋,

and t

t−l

∫(px(s) + c)Δs = l ∑ (px(n) + c),

t ∈ J, t > a,

n=a

a

p 1 + μ(t)p p , t ∈ 𝕋, =− 1 + lp

(⊖p)(t) = −

e⊖p (σ(s), t) = e⊖p (s + l, t) t−l

= ∏ (1 + l(⊖p))

−1

n=s+l t−l

lp ) 1 + lp

−1

= ∏ (1 − n=s+l t−l

= ∏ (1 + lp) n=s+l

= (1 + lp)

t−s−l l

,

t, s ∈ J, t − l ≥ s + l.

Hence, by Theorem 1.1.1, it follows that the inequality t−l

x(t) ≤ l ∑ (px(n) + c),

t ∈ J, t > a,

n=a

implies the inequality t−l

x(t) ≤ cl ∑ (1 + lp)

t−s−l l

s=a

t ∈ J, t ≥ a + l.

,

Example 1.1.3. Let 𝕋 = qℕ0 , q > 0. Let also, x, p and c be as in Theorem 1.1.1. Suppose that a = qr1 ,

t = qr2 ,

s = qr3 ,

We have σ(t) = qt,

t ∈ 𝕋,

and t

qr2

∫(px(s) + c)Δs = ∫ (px(s) + c)Δs a

qr1

r2 > r3 ≥ r1 .

4 | 1 Linear integral inequalities on time scales qr2 −1

= ∑ (q − 1)s(px(s) + c), s=qr1

p 1 + μ(t)p p =− , 1 + (q − 1)tp

(⊖p)(t) = −

e⊖p (σ(s), t) = e⊖p (qs, t)

= e⊖p (qr3 +1 , qr2 ) qr2 −1

= ∏ (1 − (q − 1)z( z=qr3 +1

p )) 1 + (q − 1)sp

qr2 −1

= ∏ (1 + (q − 1)zp),

−1

r2 ≥ r3 + 2.

z=qr3 +1

Then, by Theorem 1.1.1, the inequality qr2 −1

x(t) ≤ ∑ (q − 1)s(px(s) + c), s=qr1

r2 ≥ r3 + 1,

implies the inequality qr2 −1

qr2 −1

s=qr1

z=qs

x(t) ≤ c ∑ (q − 1)s ∏ (1 + (q − 1)zp),

r2 ≥ r1 + 2.

Theorem 1.1.4 (Bellman’s Inequality). Let x, f ∈ 𝒞rd (J) be nonnegative functions and c be a nonnegative constant. Then the inequality t

x(t) ≤ c + ∫ f (s)x(s)Δs, a

t ∈ J,

implies the inequality x(t) ≤ cef (t, a),

t ∈ J.

Proof. Let t

y(t) = c + ∫ f (s)x(s)Δs, a

Then y(a) = c,

x(t) ≤ y(t),

t ∈ J,

t ∈ J.

1.1 The inequalities of Gronwall and Bellman

| 5

and Δ

t

Δ

y (t) = (c + ∫ f (s)x(s)Δs) a

= f (t)x(t)

≤ f (t)y(t),

(1.1) t ∈ J.

Hence, Δ

(ye⊖f (⋅, a)) (t) = yΔ (t)e⊖f (σ(t), a)

+ y(t)(⊖f )(t)e⊖f (t, a)

= yΔ (t)e⊖f (σ(t), a) −

y(t)f (t) e (t, a) 1 + μ(t)f (t) ⊖f

= yΔ (t)e⊖f (σ(t), a)

− y(t)f (t)e⊖f (σ(t), a)

= (yΔ (t) − y(t)f (t))e⊖f (σ(t), a). Since f ∈ ℛ+ (J), then 1 + μ(t)f (t) > 0,

t ∈ J,

and μ(t)f (t) 1 + μ(t)f (t) 1 = 1 + μ(t)f (t) > 0, t ∈ J,

1 + μ(t)(⊖f )(t) = 1 −

i. e., ⊖f ∈ ℛ+ (J). Therefore e⊖f (t, a) > 0,

e⊖f (σ(t), a) > 0,

t ∈ J.

Hence, applying inequality (1.1) to inequality (1.2), we get Δ

(ye⊖f (⋅, a)) (t) ≤ 0,

t ∈ J,

whereupon, integrating from a to t, t ∈ J, we get y(t)e⊖f (t, a) − y(a)e⊖f (a, a) = y(t)e⊖f (t, a) − c ≤ 0,

t ∈ J.

(1.2)

6 | 1 Linear integral inequalities on time scales From here, y(t)e⊖f (t, a) ≤ c,

t ∈ J,

or c e⊖f (t, a)

y(t) ≤

= cef (t, a),

t ∈ J.

This completes the proof. Example 1.1.5. Let 𝕋 = lℤ, l > 0, x, f and c be as in Theorem 1.1.4. We have σ(t) = t + l,

t ∈ 𝕋,

and t

t−l

∫ f (s)x(s)Δs = l ∑ f (s)x(s), s=a

a

t−l

ef (t, a) = ∏(1 + lf (s)),

t ∈ J, t > a.

s=a

Then, by Theorem 1.1.4, it follows that the inequality t−l

x(t) ≤ c + l ∑ f (s)x(s), s=a

t ∈ J, t > a,

implies the inequality t−l

x(t) ≤ c ∏(1 + lf (s)), s=a

t ∈ J, t > a.

Example 1.1.6. Let 𝕋 = qℕ0 , q > 1, x, f and c be as in Theorem 1.1.4. Let also a = qr1 ,

t = qr2 ,

r1 , r2 ∈ ℕ0 ,

We have σ(t) = qt,

t ∈ 𝕋,

and t

qr2 −1

∫ f (s)x(s)Δs = ∑ μ(s)f (s)x(s) a

s=qr1

r2 > r1 .

1.1 The inequalities of Gronwall and Bellman

| 7

qr2 −1

= (q − 1) ∑ sf (s)x(s), s=qr1

qr2 −1

ef (t, a) = ∏ (1 + (q − 1)sf (s)). s=qr1

Then, by Theorem 1.1.4, it follows that the inequality qr2 −1

r2

x(q ) ≤ c ∑ sf (s)x(s),

r2 > r1 ,

s=qr1

implies the inequality r2

qr2 −1

x(q ) ≤ c ∏ (1 + (q − 1)sf (s)),

r2 > r1 .

s=qr1

Theorem 1.1.7. Let x, f ∈ 𝒞rd (J) be nonnegative functions. Let also g ∈ 𝒞rd (J) be a positive and nondecreasing function. Then the inequality t

x(t) ≤ g(t) + ∫ x(s)f (s)Δs, a

t ∈ J,

implies the inequality x(t) ≤ g(t)ef (t, a),

t ∈ J.

Proof. Let y(t) =

x(t) , g(t)

t ∈ J.

Then, using (1.3), we get t

x(t) x(s)f (s) ≤1+∫ Δs g(t) g(t) a

t

x(s)f (s) Δs, g(s)

t ∈ J,

y(t) ≤ 1 + ∫ f (s)y(s)Δs,

t ∈ J.

≤1+∫ a

whereupon t a

(1.3)

8 | 1 Linear integral inequalities on time scales Hence, by Theorem 1.1.4, we obtain y(t) ≤ ef (t, a),

t ∈ J.

x(t) ≤ ef (t, a), g(t)

t ∈ J,

Therefore

or x(t) ≤ g(t)ef (t, a),

t ∈ J.

This completes the proof. Example 1.1.8. Let 𝕋 = lℤ, l > 0, x, f and g be as in Theorem 1.1.4. Then σ(t) = t + l,

t ∈ 𝕋,

and t

t−l

∫ f (s)x(s)Δs = l ∑ f (s)x(s), s=a

a

t−l

ef (t, a) = ∏(1 + lf (s)),

t ∈ J, t > a.

s=a

Then, by Theorem 1.1.4, it follows that the inequality t−l

x(t) ≤ g(t) + l ∑ f (s)x(s), s=a

t ∈ J, t > a,

implies the inequality t−l

x(t) ≤ g(t) ∏(1 + lf (s)), s=a

t ∈ J, t > a.

Example 1.1.9. Let 𝕋 = qℕ0 , q > 1, x, f and g be as in Theorem 1.1.4. Let also a = qr1 ,

t = qr2 ,

r1 , r2 ∈ ℕ0 ,

We have σ(t) = qt,

t ∈ 𝕋,

and t

qr2 −1

∫ f (s)x(s)Δs = ∑ μ(s)f (s)x(s) a

s=qr1

r2 > r1 .

1.1 The inequalities of Gronwall and Bellman

| 9

qr2 −1

= (q − 1) ∑ sf (s)x(s), s=qr1

qr2 −1

ef (t, a) = ∏ (1 + (q − 1)sf (s)). s=qr1

Then, by Theorem 1.1.4, it follows that the inequality qr2 −1

x(qr2 ) ≤ g(qr2 ) + ∑ sf (s)x(s),

r2 > r1 ,

s=qr1

implies the inequality qr2 −1

x(qr2 ) ≤ g(qr2 ) ∏ (1 + (q − 1)sf (s)),

r2 > r1 .

s=qr1

Theorem 1.1.10. Let x, f , g, h ∈ 𝒞rd (J) be nonnegative functions. Then the inequality t

x(t) ≤ f (t) + g(t) ∫ h(s)x(s)Δs,

t ∈ J,

a

(1.4)

implies the inequality t

x(t) ≤ f (t) + g(t) ∫ h(s)f (s)e⊖(hg) (σ(s), t)Δs, a

t ∈ J.

Proof. Let t

y(t) = ∫ h(s)x(s)Δs, a

t ∈ J.

Then, using (1.4), we get x(t) ≤ f (t) + g(t)y(t),

t ∈ J.

We have yΔ (t) = h(t)x(t)

≤ h(t)(f (t) + g(t)y(t)),

t ∈ J,

or yΔ (t) − y(t)g(t)h(t) ≤ h(t)f (t),

t ∈ J,

(1.5)

10 | 1 Linear integral inequalities on time scales and y(a) = 0, Δ

(y(⋅)e⊖(hg) (⋅, a)) (t) = yΔ (t)e⊖(hg) (σ(t), a)

+ y(t)(⊖(hg))(t)e⊖(hg) (t, a)

= yΔ (t)e⊖(hg) (σ(t), a) −

y(t)h(t)g(t) e (t, a) 1 + μ(t)h(t)g(t) ⊖(hg)

= yΔ (t)e⊖(hg) (σ(t), a)

− y(t)h(t)g(t)e⊖(hg) (σ(t), a)

= (yΔ (t) − y(t)h(t)g(t))e⊖(hg) (σ(t), a)

≤ h(t)f (t)e⊖(hg) (σ(t), a),

t ∈ J.

We integrate the last inequality from a to t and get t

y(t)e⊖(hg) (t, a) ≤ ∫ h(s)f (s)e⊖(hg) (σ(s), a)Δs, a

t ∈ J.

Since hg ∈ ℛ+ (J), we have e⊖(hg) (t, a) > 0,

t ∈ J.

Therefore t

y(t) ≤

1 ∫ h(s)f (s)e⊖(hg) (σ(s), a)Δs e⊖(hg) (t, a) a

t

= ∫ h(s)f (s)e⊖(hg) (a, t)e⊖(hg) (σ(s), a)Δs a

t

= ∫ h(s)f (s)e⊖(hg) (σ(s), t)Δs, a

t ∈ J.

Hence, using (1.5), we obtain x(t) ≤ f (t) + g(t)y(t) t

≤ f (t) + g(t) ∫ h(s)f (s)e⊖(hg) (σ(s), t)Δs, a

This completes the proof.

t ∈ J.

1.1 The inequalities of Gronwall and Bellman | 11

Example 1.1.11. Let 𝕋 = lℤ, l > 0, x, f , g and h be as in Theorem 1.1.10. Then σ(t) = t + l,

t ∈ 𝕋,

and t

t−l

∫ h(s)x(s)Δs = l ∑ h(s)x(s),

t ∈ J, t > a,

s=a

a

⊖(hg)(t) = −

h(t)g(t) , 1 + lh(t)g(t)

t ∈ J,

e⊖(hg) (σ(s), t) = e⊖(hg) (s + l, t) t−l

= ∏ (1 + l(− z=s+l

h(z)g(z) )) 1 + lh(z)g(z)

t−l

= ∏ (1 + lh(z)g(z)), z=s+l

t

t−l

t−l

s=a

z=s+l

−1

t ≥ s + 2l,

∫ h(s)f (s)e⊖(hg) (σ(s), t)Δs = l ∑ h(s)f (s) ∏ (1 + lh(z)g(z)), a

t ≥ s + 2l.

Then, by Theorem 1.1.10, it follows that the inequality t−l

x(t) ≤ f (t) + lg(t) ∑ h(s)x(s), s=a

t > a,

implies the inequality t−l

t−l

s=a

z=s+l

x(t) ≤ f (t) + lg(t) ∑ h(s)f (s) ∏ (1 + lh(z)g(z)),

t ∈ J, t ≥ a + 2l.

Theorem 1.1.12. Let x, g, h ∈ 𝒞rd (J) be nonnegative functions, f ∈ 𝒞rd (J) be a positive nondecreasing function. Then the inequality t

x(t) ≤ f (t) + g(t) ∫ h(s)x(s)Δs, a

t ∈ J,

(1.6)

implies the inequality t

x(t) ≤ f (t)(1 + g(t) ∫ h(s)e⊖(hg) (σ(s), t)Δs), a

Proof. Let y(t) =

x(t) , f (t)

t ∈ J.

t ∈ J.

12 | 1 Linear integral inequalities on time scales Then, by inequality (1.6), we get t

x(t) g(t) ≤1+ ∫ h(s)x(s)Δs f (t) f (t) a t

≤ 1 + g(t) ∫ h(s) a

x(s) Δs f (s)

t

= 1 + g(t) ∫ y(s)h(s)Δs, a

t ∈ J,

or t

y(t) ≤ 1 + g(t) ∫ y(s)h(s)Δs, a

t ∈ J.

Hence, using Theorem 1.1.10, we get t

y(t) ≤ 1 + g(t) ∫ h(s)e⊖(hg) (σ(s), t)Δs, a

t ∈ J.

Therefore t

x(t) ≤ 1 + g(t) ∫ h(s)e⊖(hg) (σ(s), t)Δs, f (t) a

t ∈ J.

This completes the proof. Example 1.1.13. Let 𝕋 = lℤ, l > 0. Then σ(t) = t + l,

μ(t) = l,

t ∈ 𝕋.

Suppose that x, g, h ∈ 𝒞rd (J) are nonnegative functions and f ∈ 𝒞rd (J) is a positive nondecreasing function. Then h(t)g(t) 1 + μ(t)h(t)g(t) h(t)g(t) , =− 1 + lh(t)g(t)

⊖(hg)(t) = −

t−l

e⊖(hg) (σ(s), t) = ∏ (1 − z=s+l t−l

lh(z)g(z) ) 1 + lh(z)g(z)

= ∏ (1 + lh(z)g(z)). z=s+l

−1

1.1 The inequalities of Gronwall and Bellman | 13

Then the inequality t−l

x(t) ≤ f (t) + g(t) ∑ lh(s)x(s), s=a

t ∈ J,

implies the inequality t−l

x(t) ≤ f (t)(1 + g(t) ∑ lh(s)e⊖(hg) (σ(s), t)) s=a

t−l

t−l

s=a

z=s+l

= f (t)(1 + lg(t) ∑ h(s) ∏ (1 + lh(z)g(z))), t ∈ J, t ≥ a + 2l. Example 1.1.14. Let a = 1, 𝕋 = qℕ0 , q > 1. Then σ(t) = qt,

μ(t) = (q − 1)t,

t ∈ 𝕋.

Suppose that x, g, h ∈ 𝒞rd (J) are nonnegative functions and f ∈ 𝒞rd (J) is a positive nondecreasing function. Then the inequality t

t

t

qt−1

x(q ) ≤ f (q ) + g(q ) ∑ (q − 1)sh(s)x(s), s=qr

t ∈ ℕ0 , t ≥ r + 1, qt ∈ J,

implies the inequality qt−1

x(qt ) ≤ f (qt )(1 + g(qt ) ∑ (q − 1)sh(s) s=qr

q

t−1

× ∏ (1 + (q − 1)zh(z)g(z))) z=qr+1

t ∈ ℕ, t ≥ r + 2, qt ∈ J. Theorem 1.1.15. Let f , g, h, p ∈ 𝒞rd (J) be nonnegative functions. Then the inequality t

x(t) ≤ f (t) + g(t) ∫(h(s)x(s) + p(s))Δs, a

t ∈ J,

(1.7)

implies the inequality t

x(t) ≤ f (t) + g(t) ∫(h(s)f (s) + p(s))e⊖(hg) (σ(s), t)Δs, a

t ∈ J.

14 | 1 Linear integral inequalities on time scales Proof. Let t

y(t) = ∫(h(s)x(s) + p(s))Δs, a

t ∈ J.

Then y(a) = 0 and x(t) ≤ f (t) + g(t)y(t),

t ∈ J,

(1.8)

and yΔ (t) = h(t)x(t) + p(t)

≤ h(t)(f (t) + g(t)y(t)) + p(t)

= h(t)f (t) + h(t)g(t)y(t) + p(t),

t ∈ J.

yΔ (t) − h(t)g(t)y(t) ≤ h(t)f (t) + p(t),

t ∈ J.

Therefore

Hence, Δ

(y(⋅)e⊖(hg) (⋅, a)) (t) = yΔ (t)e⊖(hg) (σ(t), a)

+ (⊖(hg))(t)y(t)e⊖(hg) (t, a)

= yΔ (t)e⊖(hg) (σ(t), a)

− h(t)g(t)y(t)e⊖(hg) (σ(t), a)

= (yΔ (t) − h(t)g(t)y(t))e⊖(hg) (σ(t), a) ≤ (h(t)f (t) + p(t))e⊖(hg) (σ(t), a),

t ∈ J.

Integrating the last inequality from a to t, we get t

y(t)e⊖(hg) (t, a) ≤ ∫(h(s)f (s) + p(s))e⊖(hg) (σ(s), a)Δs, a

t ∈ J.

From here, t

y(t) ≤ e⊖(hg) (a, t) ∫(h(s)f (s) + p(s))e⊖(hg) (σ(s), a)Δs t

a

= ∫(h(s)f (s) + p(s))e⊖(hg) (σ(s), t)Δs, a

Now, using (1.8), we obtain x(t) ≤ f (t) + g(t)y(t)

t ∈ J.

1.1 The inequalities of Gronwall and Bellman | 15 t

≤ f (t) + g(t) ∫(h(s)f (s) + p(s))e⊖(hg) (σ(s), t)Δs, a

t ∈ J.

This completes the proof. Theorem 1.1.16. Let x, f , g, h ∈ 𝒞rd (J) be nonnegative functions. Then the inequality t

x(t) ≥ f (z) + g(t) ∫ h(s)x(s)Δs, z

a ≤ z ≤ t ≤ b,

implies the inequality x(t) ≥ f (z)e⊖p (z, t),

a ≤ z ≤ t ≤ b,

where p(z) = g(t)h(z),

(⊖p)(z) = −

g(t)h(z) , 1 + μ(z)g(t)h(z)

a ≤ z ≤ t ≤ b.

Proof. Let t

y(z) = x(t) − g(t) ∫ h(s)x(s)Δs, z

a ≤ z ≤ t ≤ b.

Then y(t) = x(t),

t ∈ J,

y(t) ≤ x(t),

a ≤ z ≤ t ≤ b,

y(t) ≥ f (z),

and yΔ (z) = g(t)h(z)x(z) ≥ g(t)h(z)y(z) = p(z)y(z),

a ≤ z ≤ t ≤ b.

Then yΔ (z) − p(z)y(z) ≥ 0,

a ≤ z ≤ t ≤ b.

We have Δ

(y(⋅)e⊖p (⋅, a)) (z) = yΔ (z)e⊖p (σ(z), a)

+ y(z)(⊖p)(z)e⊖p (z, a)

16 | 1 Linear integral inequalities on time scales = yΔ (z)e⊖p (σ(z), a)

− y(z)p(z)e⊖p (σ(z), a)

= (yΔ (z) − p(z)y(z))e⊖p (σ(z), a)

≥ 0,

a ≤ z ≤ t ≤ b.

Hence, y(t)e⊖p (t, a) ≥ y(z)e⊖p (z, a)

≥ f (z)e⊖p (z, a),

or x(t) ≥ f (z)

a ≤ z ≤ t ≤ b,

e⊖p (z, a) e⊖p (t, a)

= f (z)e⊖p (z, t),

a ≤ z ≤ t ≤ b.

This completes the proof. Example 1.1.17. Let 𝕋 = lℤ, l > 0, a, b, z, t ∈ 𝕋, a ≤ z ≤ t ≤ b. Suppose that x, f , g, h and p are as in Theorem 1.1.16. Then t

t−l

∫ h(s)x(s)Δs = ∑ lh(n)x(n), n=z

z

(⊖p)(z) = −

g(t)h(z) , 1 + lg(t)h(z)

t−l

e⊖p (z, t) = ∏ n=z t−1

1−

1

g(t)h(n) l 1+lg(t)h(n)

= ∏(1 + lg(t)h(n)). n=z

Then, by Theorem 1.1.16, it follows that the inequality t−l

x(t) ≥ f (z) + lg(t) ∑ h(n)x(n), n=z

a ≤ z < t ≤ b,

implies the inequality t−l

x(t) ≥ f (z) ∏(1 + lg(t)h(z)), n=z

a ≤ z < t ≤ b.

Example 1.1.18. Let 𝕋 = qℕ0 , q > 0, a, b, z, t ∈ 𝕋, a ≤ z ≤ t ≤ b, z = qr1 , t = qr2 , r1 < r2 , r1 , r2 ∈ ℕ. Let also, x, f , g, h, and p be as in Theorem 1.1.16. Then t

qr2

∫ h(s)x(s)Δs = ∫ h(s)x(s)Δs z

qr1

1.2 Volterra-type integral inequalities | 17 qr2 −1

= ∑ (q − 1)nh(n)x(n), n=qr1

(⊖p)(qr1 ) = −

g(qr2 )h(qr1 ) , 1 − (q − 1)ng(qr2 )h(qr1 )

qr2 −1

e⊖p (z, t) = ∏ (1 + (q − 1)n(− n=qr1

g(qr2 )h(n) )) 1 + (q − 1)ng(qr2 )h(n)

−1

qr2 −1

= ∏ (1 + (q − 1)ng(qr2 )h(n)). n=qr1

Then, by Theorem 1.1.16, it follows that the inequality qr2 −1

x(qr2 ) ≥ f (qr1 ) + g(qr2 ) ∑ (q − 1)nh(n)x(n), n=qr1

implies the inequality qr2 −1

x(qr2 ) ≥ f (qr1 ) ∏ (1 + (q − 1)ng(qr2 )h(n)). n=qr1

1.2 Volterra-type integral inequalities Theorem 1.2.1. Let x, f ∈ 𝒞rd (J), k ∈ 𝒞rd (J × J) be nonnegative functions on J × J, and t

x(t) ≤ f (t) + ∫ k(t, s)x(s)Δs, a

t ∈ J.

(1.9)

Let also k1 (t, s) = k(t, s), t

kl (t, s) = ∫ k(t, s1 )kl−1 (s1 , s)Δs1 , s

l ∈ ℕ, l ≥ 2, a ≤ s ≤ t ≤ b.

Suppose that ∞

H(t, s) = ∑ kn (t, s) n=1

is an uniformly convergent series on a ≤ s ≤ t ≤ b. Then t

x(t) ≤ f (t) + ∫ H(t, s)f (s)Δs, a

t ∈ J.

(1.10)

18 | 1 Linear integral inequalities on time scales Proof. We have t

x(t) ≤ f (t) + ∫ k(t, s)x(s)Δs a

t

s

≤ f (t) + ∫ k(t, s)(f (s) + ∫ k(s, s1 )x(s1 )Δs1 )Δs a

a

t

= f (t) + ∫ k(t, s)f (s)Δs t s

a

+ ∫ ∫ k(t, s)k(s, s1 )x(s1 )Δs1 Δs a a

t

= f (t) + ∫ k(t, s)f (s)Δs t

a t

+ ∫(∫ k(t, s)k(s, s1 )Δs)x(s1 )Δs1 a

s1

t

t

= f (t) + ∫ k1 (t, s)f (s)Δs + ∫ k2 (t, s)x(s)Δs, a

a

t ∈ J.

Hence, for t ∈ J, we get t

x(t) ≤ f (t) + ∫ k(t, s)x(s)Δs a

t

s

s

≤ f (t) + ∫ k(t, s)(f (s) + ∫ k1 (s, s1 )f (s1 )Δs1 + ∫ k2 (s, s1 )x(s1 )Δs1 )Δs a

t

a

= f (t) + ∫ k(t, s)f (s)Δs t s

a

+ ∫ ∫ k(t, s)k1 (s, s1 )f (s1 )Δs1 Δs a a

t s

+ ∫ ∫ k(t, s)k2 (s, s1 )x(s1 )Δs1 Δs a a

t

= f (t) + ∫ k1 (t, s)f (s)Δs a

a

1.2 Volterra-type integral inequalities | 19 t

t

+ ∫(∫ k(t, s)k1 (s, s1 )Δs)f (s1 )Δs1 a

s1

t

t

+ ∫(∫ k(t, s)k2 (s, s1 )Δs)x(s1 )Δs1 a

s1

t

t

= f (t) + ∫ k1 (t, s)f (s)Δs + ∫ k2 (t, s)f (s)Δs a

t

a

+ ∫ k3 (t, s)x(s)Δs a

t

= f (t) + ∫(k1 (t, s) + k2 (t, s))f (s)Δs a

t

+ ∫ k3 (t, s)x(s)Δs. a

Assume that n

t

t

x(t) ≤ f (t) + ∑ ∫ kl (t, s)f (s)Δs + ∫ kn+1 (t, s)x(s)Δs, l=1 a

a

t ∈ J,

for some n ∈ ℕ. We will prove that n+1 t

t

l=1 a

a

x(t) ≤ f (t) + ∑ ∫ kl (t, s)f (s)Δs + ∫ kn+2 (t, s)x(s)Δs,

t ∈ J.

Indeed, for t ∈ J, we have t

x(t) ≤ f (t) + ∫ k(t, s)x(s)Δs a

t

n

s

s

≤ f (t) + ∫ k(t, s)(f (s) + ∑ ∫ kl (s, s1 )f (s1 )Δs1 + ∫ kn+1 (s, s1 )x(s1 )Δs1 )Δs a

t

l=1 a

= f (t) + ∫ k(t, s)f (s)Δs a

n

t s

+ ∑ ∫ ∫ k(t, s)kl (s, s1 )f (s1 )Δs1 Δs l=1 a a t s

+ ∫ ∫ k(t, s)kn+1 (s, s1 )x(s1 )Δs1 Δs a a

a

(1.11)

20 | 1 Linear integral inequalities on time scales t

= f (t) + ∫ k(t, s)f (s)Δs a

n

t

t

+ ∑ ∫(∫ k(t, s)kl (s, s1 )Δs)f (s1 )Δs1 l=1 a t

s1

t

+ ∫(∫ k(t, s)kn+1 (s, s1 )Δs)x(s1 )Δs1 a

s1

t

t

n

= f (t) + ∫ k1 (t, s)f (s)Δs + ∑ ∫ kl+1 (t, s)f (s)Δs t

l=1 a

a

+ ∫ kn+2 (t, s)x(s)Δs a

t

= f (t) + ∫ k1 (t, s)f (s)Δs a

n+1 t

+ ∑ ∫ kl (t, s)f (s)Δs l=2 a t

+ ∫ kn+2 (t, s)x(s)Δs a

n+1 t

t

l=1 a

a

= f (t) + ∑ ∫ kl (t, s)f (s)Δs + ∫ kn+2 (t, s)x(s)Δs. Therefore (1.11) holds for any n ∈ ℕ. Letting n → ∞ in (1.11) and using that H(t, s) = ∑∞ n=1 kn (t, s) is a uniformly convergent series on a ≤ s ≤ t ≤ b, we get inequality (1.10). This completes the proof. Theorem 1.2.2. Let x ∈ 𝒞rd (J) be a nonnegative function, k ∈ 𝒞rd (J × J) be a nonnegative function and k(t, s) be nondecreasing in t for each s ∈ J. If t

x(t) ≤ c + ∫ k(t, s)x(s)Δs, a

t ∈ J,

(1.12)

for a nonnegative constant c, then x(t) ≤ cek(t,⋅) (t, a),

t ∈ J.

Proof. Fix a T ∈ J. Then for a ≤ t ≤ T, using (1.12), we get t

x(t) ≤ c + ∫ k(T, s)x(s)Δs. a

(1.13)

1.2 Volterra-type integral inequalities | 21

We apply Theorem 1.1.4 for t ∈ [a, T] and get x(t) ≤ cek(T,⋅) (t, a),

t ∈ [a, T].

In particular, x(T) ≤ cek(T,⋅) (T, a). Since T ∈ J was arbitrarily chosen, we get inequality (1.13). This completes the proof. Theorem 1.2.3. Let x ∈ 𝒞rd (J) be a nonnegative function, k ∈ 𝒞rd (J × J) be a nonnegative function and k(t, s) be nondecreasing in t for each s ∈ J. Let also, g ∈ 𝒞rd (J) be a positive and nondecreasing function. If t

x(t) ≤ g(t) + ∫ k(t, s)x(s)Δs,

t ∈ J,

a

(1.14)

then x(t) ≤ g(t)ek(t,⋅) (t, a),

t ∈ J.

(1.15)

Proof. Let T ∈ J be arbitrarily chosen and fixed. Then, by (1.14), we obtain t

x(t) ≤ g(t) + ∫ k(T, s)x(s)Δs, a

a ≤ t ≤ T.

We apply Theorem 1.1.7 for a ≤ t ≤ T and get x(t) ≤ g(t)ek(T,⋅) (t, a). In particular, when t = T, we obtain x(T) ≤ g(T)ek(T,⋅) (T, a). Because T ∈ J was arbitrarily chosen, we obtain inequality (1.15). This completes the proof. Theorem 1.2.4. Let x, f , g, h, p ∈ 𝒞rd (J) be nonnegative functions, k ∈ 𝒞rd (J × J) be a nonnegative function such that ktΔ (t, s) exists for any t, s ∈ J, and ktΔ (t, s) ≥ 0 for any t, s ∈ J. Let also t

a(t) = k(σ(t), t)h(t)g(t) + ∫ ktΔ (t, s)h(s)g(s)Δs, a

22 | 1 Linear integral inequalities on time scales b(t) = k(σ(t), t)(h(t)f (t) + p(t)) t

+ ∫ ktΔ (t, s)(h(s)f (s) + p(s))Δs, t

a

B(t) = ∫ b(s)Δs, a

t ∈ J.

Suppose that a ∈ ℛ+ (J) and t

x(t) ≤ f (t) + g(t) ∫ k(t, s)(h(s)x(s) + p(s))Δs, a

t ∈ J.

(1.16)

Then t

x(t) ≤ B(t) + ∫ a(s)B(s)e⊖a (σ(s), t)Δs, a

t ∈ J.

(1.17)

Proof. Let t

z(t) = ∫ k(t, s)(h(s)x(s) + p(s))Δs, a

t ∈ J.

Then z(a) = 0 and z Δ (t) = k(σ(t), t)(h(t)x(t) + p(t)) t

+ ∫ ktΔ (t, s)(h(s)x(s) + p(s))Δs ≥ 0,

a

t ∈ J.

Therefore z is nonnegative and nondecreasing on J. By (1.16), we obtain x(t) ≤ f (t) + g(t)z(t),

t ∈ J.

From (1.18), we get z Δ (t) ≤ k(σ(t), t)(h(t)f (t) + h(t)g(t)z(t) + p(t)) t

+ ∫ ktΔ (t, s)(h(s)f (s) + h(s)g(s)z(s) + p(s))Δs a

(1.18)

1.3 The inequalities of Gamidov and Rodrigues |

23

≤ k(σ(t), t)(h(t)f (t) + p(t)) + k(σ(t), t)h(t)g(t)z(t) t

+ ∫ ktΔ (t, s)(h(s)f (s) + p(s))Δs a

t

+ (∫ ktΔ (t, s)h(s)g(s)Δs)z(t) a

t

= (k(σ(t), t)h(t)g(t) + ∫ ktΔ (t, s)h(s)g(s)Δs)z(t) a

+ k(σ(t), t)(h(t)f (t) + p(t)) t

+ ∫ ktΔ (t, s)(h(s)f (s) + p(s))Δs a

= a(t)z(t) + b(t),

t ∈ J.

Integrating the last inequality, we get t

t

z(t) ≤ ∫ b(s)Δs + ∫ a(s)z(s)Δs a

a

t

= B(t) + ∫ a(s)z(s)Δs, a

t ∈ J.

Hence, using Theorem 1.1.10, we obtain t

x(t) ≤ B(t) + ∫ a(s)B(s)e⊖a (σ(s), t)Δs, a

t ∈ J,

i. e., we get (1.17). This completes the proof.

1.3 The inequalities of Gamidov and Rodrigues Theorem 1.3.1 (Gamidov’s Inequality). Let x, f , gi , hi ∈ 𝒞rd (J), i ∈ {1, . . . , n}, be nonnegative functions, and g(t) = sup gi (t), i∈{1,...,n}

n

h(t) = ∑ hi (t), i=1

t ∈ J.

Then the inequality n

t

i=1

a

x(t) ≤ f (t) + ∑ gi (t) ∫ hi (s)x(s)Δs,

t ∈ J,

(1.19)

24 | 1 Linear integral inequalities on time scales implies the inequality t

x(t) ≤ f (t) + g(t) ∫ h(s)f (s)e⊖(hg) (σ(s), t)Δs, a

t ∈ J.

(1.20)

Proof. By inequality (1.19), we get t n

x(t) ≤ f (t) + g(t) ∫ ∑ hi (s)x(s)Δs a i=1 t

= f (t) + g(t) ∫ h(s)x(s)Δs, a

t ∈ J.

Hence, from Theorem 1.1.10, we get inequality (1.20). This completes the proof. Theorem 1.3.2 (Gamidov’s Inequality). Let n ∈ ℕ, n ≥ 2, x, f , g1 , g2 , hi ∈ 𝒞rd (J), i ∈ {1, . . . , n}, be nonnegative functions, a = t1 ≤ t2 ≤ ⋅ ⋅ ⋅ ≤ tn = b, ci , i ∈ {1, . . . , n}, be nonnegative constants, ti

mi = ci ∫ hi (s)x(s)Δs, t1

i ∈ {2, . . . , n},

n

F(t) = f (t) + g2 (t) ∑ mi , i=2 t

p1 (t) = f (t) + g1 (t) ∫ h1 (s)f (s)e⊖(h1 g1 ) (σ(s), t)Δs, a

t

p2 (t) = g2 (t) + g1 (t) ∫ h1 (s)g2 (s)e⊖(h2 g1 ) (σ(s), t)Δs. a

Suppose that M

−1

n

tj

= 1 − ∑ cj ∫ hj (s)p2 (s)Δs > 0. j=2

t1

Then the inequality t

x(t) ≤ f (t) + g1 (t) ∫ h1 (s)x(s)Δs, t1

t ∈ J,

(1.21)

1.3 The inequalities of Gamidov and Rodrigues |

25

implies the inequality ti

n

x(t) ≤ p1 (t) + Mp2 (t) ∑ ci ∫ hi (s)p1 (s)Δs, i=2

t1

t ∈ J.

(1.22)

Proof. Using inequality (1.21), we obtain n

t

i=2

t1

x(t) ≤ (f (t) + g2 (t) ∑ mi ) + g1 (t) ∫ h1 (s)x(s)Δs t

= F(t) + g1 (t) ∫ h1 (s)x(s)Δs, t1

t ∈ J.

Hence, by Theorem 1.1.10, we obtain t

x(t) ≤ F(t) + g1 (t) ∫ h1 (s)F(s)e⊖(h1 g1 ) (σ(s), t)Δs t1

n

= f (t) + g2 (t) ∑ mi i=2

t

n

+ g1 (t) ∫ h1 (s)(f (s) + g2 (s) ∑ mi )e⊖(h1 g1 ) (σ(s), t)Δs i=2

t1

(1.23)

t

= f (t) + g1 (t) ∫ h1 (s)f (s)e⊖(h1 g1 ) (σ(s), t)Δs t1

t

n

+ ∑ mi (g2 (t) + g1 (t) ∫ h1 (s)g2 (s)e⊖(h1 g1 ) (σ(s), t)Δs) i=2

t1

n

= p1 (t) + p2 (t) ∑ mi , i=2

t ∈ J.

Note that n

n

i=2

i=2

ti

∑ mi = ∑ ci ∫ hi (s)x(s)Δs n

t1

ti

n

≤ ∑ ci ∫ hi (s)(p1 (s) + p2 (s) ∑ mi )Δs i=2 n

t1

ti

= ∑ ci ∫ hi (s)p1 (s)Δs i=2

t1

i=2

26 | 1 Linear integral inequalities on time scales

n

n

i=2

j=2

tj

+ ∑ mi (∑ cj ∫ hj (s)p2 (s)Δs), t1

or tj

n

n

i=2

j=2

n

ti

∑ mi (1 − ∑ cj ∫ hj (s)p2 (s)Δs) ≤ ∑ ci ∫ hi (s)p1 (s)Δs. i=2

t1

t1

Therefore n

n

i=2

i=2

ti

∑ mi ≤ M ∑ ci ∫ hi (s)p1 (s)Δs. t1

Hence from (1.23), we get inequality (1.22). This completes the proof. We denote ℝ+ = [0, ∞). Below we suppose that ℝ+ ∩ 𝕋 ≠ 0 and sup 𝕋 = ∞. Theorem 1.3.3. Let x, f ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions such that 1 − μ(t)f (t) > 0,

t ∈ ℝ+ ∩ 𝕋,

and 󵄨󵄨 ∞ 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 ∫ log(1 − μ(t)f (t))Δt 󵄨󵄨󵄨 < ∞. 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨0 󵄨 Let also c be a nonnegative constant. Then the inequality ∞

x(t) ≤ c + ∫ f (s)x(s)Δs, t

t ∈ ℝ+ ∩ 𝕋,

implies the inequality x(t) ≤ cf (t)eg (∞, t),

t ∈ ℝ+ ∩ 𝕋,

f (t) , 1 − μ(t)f (t)

t ∈ ℝ+ ∩ 𝕋.

where g(t) = Proof. Let ∞

y(t) = c + ∫ f (s)x(s)Δs, t

t ∈ ℝ+ ∩ 𝕋.

(1.24)

1.3 The inequalities of Gamidov and Rodrigues |

27

Then lim y(t) = c

t→∞

and, using (1.24), we have x(t) ≤ y(t),

t ∈ ℝ+ ∩ 𝕋.

(1.25)

Also, we have yΔ (t) = −f (t)x(t) ≥ −f (t)y(t), or yΔ (t) + f (t)y(t) ≥ 0,

t ∈ ℝ+ ∩ 𝕋.

Using the last inequality, we get Δ

(y(t)eg (t, ∞)) = yΔ (t)eg (σ(t), ∞)

+ y(t)g(t)eg (t, ∞)

= yΔ (t)eg (σ(t), ∞) +

y(t)g(t) (1 + μ(t)g(t))eg (t, ∞) 1 + μ(t)g(t)

= yΔ (t)eg (σ(t), ∞) +

y(t)g(t) e (σ(t), ∞) 1 + μ(t)g(t) g

= (yΔ (t) + = (yΔ (t) +

g(t) y(t))eg (σ(t), ∞) 1 + μ(t)g(t) 1

f (t) 1−μ(t)f (t) μ(t)f (t) + 1−μ(t)f (t)

y(t))eg (σ(t), ∞)

= (yΔ (t) + f (t)y(t))eg (σ(t), ∞). Observe that σ(t)

1 μ(s)

log(1+μ(s)g(s))Δs

σ(t)

1 μ(s)

log(1+ 1−μ(s)f (s) )Δs

σ(t)

1 μ(s)

1 log( 1−μ(s)f )Δs (s)

eg (σ(t), ∞) = e

∫∞

=e

∫∞

=e

∫∞

=e

− ∫∞

=e

σ(t)

1 μ(s)

∞ 1 ∫σ(t) μ(s)

≥ 0,

μ(s)f (s)

log(1−μ(s)f (s))Δs

log(1−μ(s)f (s))Δs

t ∈ ℝ+ ∩ 𝕋.

(1.26)

28 | 1 Linear integral inequalities on time scales Hence, by (1.26), we obtain that Δ

(y(t)eg (t, ∞)) ≥ 0,

t ∈ ℝ+ ∩ 𝕋.

We integrate the last inequality from t to ∞ and get lim (y(t)eg (t, ∞)) − y(t)eg (t, ∞) ≥ 0,

t→∞

or y(t)eg (t, ∞) ≤ c, or y(t) ≤ ceg (∞, t),

t ∈ ℝ+ ∩ 𝕋.

From here and from (1.25), we obtain x(t) ≤ ceg (∞, t),

t ∈ ℝ+ ∩ 𝕋.

This completes the proof. Theorem 1.3.4. Let x, g ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions. Let also f ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive decreasing function and 1 − μ(t)g(t) > 0, t ∈ ℝ+ ∩ 𝕋, 󵄨󵄨 󵄨󵄨󵄨 ∞ 󵄨󵄨 󵄨󵄨 󵄨󵄨 ∫ log(1 − μ(s)g(s))Δs󵄨󵄨󵄨 < ∞. 󵄨󵄨 󵄨󵄨 󵄨 󵄨t Then the inequality ∞

x(t) ≤ f (t) + ∫ g(s)x(s)Δs, t

t ∈ ℝ+ ∩ 𝕋,

implies the inequality x(t) ≤ f (t)g(t)eh (∞, t),

t ∈ ℝ+ ∩ 𝕋,

where h(t) =

g(t) , 1 − μ(t)g(t)

t ∈ ℝ+ ∩ 𝕋.

Proof. Using that f is decreasing and nonnegative on ℝ+ ∩ 𝕋, we have ∞

x(t) g(s) ≤1+∫ x(s)Δs f (t) f (t) t

1.3 The inequalities of Gamidov and Rodrigues |

∞

≤1+∫ t

g(s) x(s)Δs, f (s)

29

t ∈ ℝ+ ∩ 𝕋.

Hence, by Theorem 1.3.3, we obtain x(t) ≤ g(t)eh (∞, t), f (t) or x(t) ≤ f (t)g(t)eh (∞, t),

t ∈ ℝ+ ∩ 𝕋.

This completes the proof. Theorem 1.3.5. Let x, f , g ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, γ ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive decreasing function, c be a nonnegative constant, 1 − μ(t)g(t) > 0,

1 − β − μ(t)g(t) > 0, 󵄨󵄨 󵄨󵄨 ∞ 󵄨 󵄨󵄨 󵄨󵄨 ∫ log(1 − β − μ(s)g(s))Δs󵄨󵄨󵄨 < ∞, 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 t ∞ 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 ∫ log(1 − μ(t)g(t))Δt 󵄨󵄨󵄨 < ∞, 󵄨󵄨 󵄨󵄨 󵄨0 󵄨

t ∈ ℝ+ ∩ 𝕋,

and γx be a bounded function on ℝ+ ∩ 𝕋. Suppose also that there exists a sufficiently large a ∈ ℝ+ ∩ 𝕋 such that ∞

∞

a

a

β = ∫ f (s)Δs + ∫ g(s)Δs < 1. Then the inequality t

∞

a

t

1 x(t) ≤ c + ∫ f (s)x(s)Δs + ∫ γ(s)g(s)x(s)Δs, γ(t)

t ∈ ℝ+ ∩ 𝕋,

implies the inequality x(t) ≤ c

g(t) e (∞, t), (1 − β)2 a

h(t) =

g(t) , 1 − β − μ(t)g(t)

t ∈ ℝ+ ∩ 𝕋, t ≥ a,

where t ∈ ℝ+ ∩ 𝕋, t ≥ a,

30 | 1 Linear integral inequalities on time scales and x(t) ≤ cg(t)eh1 (∞, t),

t ∈ ℝ+ ∩ 𝕋, t ≤ a,

g(t) , 1 − μ(t)g(t)

t ∈ ℝ+ ∩ 𝕋, t ≤ a.

where h1 (t) =

Proof. 1. Let t ≥ a, t ∈ ℝ+ ∩ 𝕋. We set y(t) = max x(s). a≤s≤t

Then y is an increasing function on [a, ∞) ∩ 𝕋, y ∈ 𝒞rd ([a, ∞) ∩ 𝕋) and x(t) ≤ y(t),

t ∈ ℝ+ ∩ 𝕋, t ≥ a.

Also, γy is a bounded function on (ℝ+ ∩ 𝕋) ∩ [a, ∞). For a given t ≥ a, there exists t1 ∈ [a, t] such that y(t) = x(t1 ). Hence, y(t) = x(t1 )

t1

∞

a

t1

t1

∞

a

t1

t1

t

1 ≤ c + ∫ f (s)x(s)Δs + ∫ γ(s)g(s)x(s)Δs γ(t1 ) 1 ≤ c + ∫ f (s)y(s)Δs + ∫ γ(s)g(s)y(s)Δs γ(t1 ) 1 = c + ∫ f (s)y(s)Δs + ∫ γ(s)g(s)y(s)Δs γ(t1 ) a

t1

∞

+

1 ∫ γ(s)g(s)y(s)Δs γ(t1 ) t1

t

t

≤ c + ∫ f (s)y(s)Δs + a

∞

+

1 ∫ γ(t1 )g(s)y(s)Δs γ(t1 ) t1

1 ∫ γ(s)g(s)y(s)Δs γ(t1 ) t

1.3 The inequalities of Gamidov and Rodrigues | t1

t

≤ c + ∫ f (s)y(s)Δs + ∫ g(s)y(s)Δs a

t1

∞

+

1 ∫ γ(s)g(s)y(s)Δs γ(t) t1

t

t

≤ c + ∫ f (s)y(s)Δs + y(t) ∫ g(s)Δs a

t1

∞

+

1 ∫ γ(s)g(s)y(s)Δs γ(t) t1

t

∞

≤ c + ∫ f (s)y(s)Δs + y(t) ∫ g(s)Δs a

a

∞

+

1 ∫ γ(s)g(s)y(s)Δs γ(t) t

t1

∞

≤ c + y(t) ∫ f (s)Δs + y(t) ∫ g(s)Δs a ∞

+

a

1 ∫ γ(s)g(s)y(s)Δs γ(t) t

∞

∞

≤ c + y(t)( ∫ f (s)Δs + ∫ g(s)Δs) + a

= c + βy(t) +

a

∞

1 ∫ γ(s)g(s)y(s)Δs, γ(t) t

∞

1 ∫ γ(s)g(s)y(s)Δs γ(t) t

t ∈ ℝ+ ∩ 𝕋, t ≥ a.

Hence, (1 − β)y(t) ≤ c +

∞

1 ∫ γ(s)g(s)y(s)Δs γ(t) t

and ∞

c 1 y(t)γ(t) ≤ γ(t) + ∫ γ(s)g(s)y(s)Δs, 1−β 1−β t

t ∈ ℝ+ ∩ 𝕋, t ≥ a.

Hence, by Theorem 1.3.4, it follows that y(t)γ(t) ≤ cγ(t)

g(t) e (∞, t), (1 − β)2 h

31

32 | 1 Linear integral inequalities on time scales or y(t) ≤ c

g(t) e (∞, t), (1 − β)2 h

t ∈ ℝ+ ∩ 𝕋, t ≥ a.

x(t) ≤ c

g(t) e (∞, t), (1 − β)2 h

t ∈ ℝ+ ∩ 𝕋, t ≥ a.

Therefore

2.

Let t ∈ ℝ+ ∩ 𝕋, t ≤ a. Then t

∫ f (s)x(s)Δs ≤ 0 a

and x(t) ≤ c +

∞

1 ∫ γ(s)g(s)x(s)Δs γ(t) t

∞

≤ c + ∫ g(s)x(s)Δs,

t ∈ ℝ+ ∩ 𝕋, t ≤ a.

t

Hence, by Theorem 1.3.3, it follows that x(t) ≤ cg(t)eh1 (∞, t),

t ∈ ℝ+ ∩ 𝕋, t ≤ a.

This completes the proof.

1.4 Simultaneous inequalities In this section we suppose that 0 ∈ 𝕋. Theorem 1.4.1. Let x, y, h1 , h2 , h3 , h4 ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, λ, k1 , and k2 be nonnegative constants, and t

t

x(t) ≤ k1 + ∫ h1 (s)x(s)Δs + ∫ eλ (s, 0)h2 (s)y(s)Δs, 0

t

0

(1.27)

t

y(t) ≤ k2 + ∫ e⊖λ (s, 0)h3 (s)x(s)Δs + ∫ h4 (s)y(s)Δs, 0

0

Then x(t) ≤ (k1 M + k2 )eλ⊕h (t, 0), y(t) ≤ (k1 M + k2 )eh (t, 0),

t ∈ ℝ+ ∩ 𝕋,

t ∈ ℝ+ ∩ 𝕋.

1.4 Simultaneous inequalities | 33

where h(t) = max{h1 (t) + h3 (t), M = max e⊖λ (t, 0), t≥0

h2 (t) + h4 (t)},

t ∈ ℝ+ ∩ 𝕋.

Proof. Since λ ≥ 0, we have eλ (s, 0) ≤ eλ (t, 0),

t ≥ s, s, t ∈ ℝ+ ∩ 𝕋,

and e⊖λ (t, 0) ≤ e⊖λ (s, 0),

t ≥ s, s, t ∈ ℝ+ ∩ 𝕋.

Hence, t

t

x(t) ≤ k1 + ∫ h1 (s)x(s)Δs + ∫ eλ (t, 0)h2 (s)y(s)Δs 0

0

t

t

= k1 + ∫ h1 (s)x(s)Δs + eλ (t, 0) ∫ h2 (s)y(s)Δs, 0

0

whereupon t

e⊖λ (t, 0)x(t) ≤ k1 e⊖λ (t, 0) + e⊖λ (t, 0) ∫ h1 (s)x(s)Δs 0

t

+ ∫ h2 (s)y(s)Δs 0

t

≤ k1 e⊖λ (t, 0) + ∫ e⊖λ (s, 0)h1 (s)x(s)Δs 0

t

(1.28)

+ ∫ h2 (s)y(s)Δs 0

t

≤ Mk1 + ∫ e⊖λ (s, 0)h1 (s)x(s)Δs t

0

+ ∫ h2 (s)y(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

Let F(t) = e⊖λ (t, 0)x(t) + y(t),

t ∈ ℝ+ ∩ 𝕋.

34 | 1 Linear integral inequalities on time scales Hence, after adding (1.28) and the second inequality of (1.27), we obtain t

F(t) ≤ (k1 M + k2 ) + ∫ e⊖λ (s, 0)(h1 (s) + h3 (s))x(s)Δs t

0

+ ∫(h2 (s) + h4 (s))y(s)Δs 0

t

≤ (k1 M + k2 ) + ∫ e⊖λ (s, 0)x(s)h(s)Δs t

0

+ ∫ h(s)y(s)Δs 0

t

= (k1 M + k2 ) + ∫ F(s)h(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

From the last inequality and Theorem 1.1.4, we get F(t) ≤ (k1 M + k2 )eh (t, 0),

t ∈ ℝ+ ∩ 𝕋.

Therefore e⊖λ (t, 0)x(t) ≤ (k1 M + k2 )eh (t, 0), or x(t) ≤ (k1 M + k2 )eλ (t, 0)eh (t, 0) = (k1 M + k2 )eλ⊕h (t, 0),

t ∈ ℝ+ ∩ 𝕋,

and y(t) ≤ (k1 M + k2 )eh (t, 0),

t ∈ ℝ+ ∩ 𝕋.

This completes the proof. Theorem 1.4.2. Let x, y, a, b, h1 , h2 , h3 , h4 ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, λ be a nonnegative constant, and t

t

x(t) ≤ a(t) + p(t) ∫ h1 (s)x(s)Δs + p(t) ∫ eλ (s, 0)h2 (s)y(s)Δs, 0

t

0

y(t) ≤ b(t) + p(t) ∫ e⊖λ (s, 0)h3 (s)x(s)Δs + p(t) ∫ h4 (s)y(s)Δs, 0

(1.29)

t 0

t ∈ ℝ+ ∩ 𝕋.

1.4 Simultaneous inequalities | 35

Then t

x(t) ≤ (a(t) + b(t)eλ (t, 0)) + p(t)eλ (t, 0) ∫ h(s)(a(s)e⊖λ (s, 0) + b(s))e⊖(hp) (σ(s), t)Δs, 0

and t

y(t) ≤ (a(t)e⊖λ (t, 0) + b(t)) + p(t) ∫ h(s)(a(s)e⊖λ (s, 0) + b(s))e⊖(hp) (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋, where h(t) = max{h1 (t) + h3 (t),

h2 (t) + h4 (t)},

t ∈ ℝ+ ∩ 𝕋.

Proof. Since λ ≥ 0, we have eλ (s, 0) ≤ eλ (t, 0),

t ≥ s, s, t ∈ ℝ+ ∩ 𝕋,

and e⊖λ (t, 0) ≤ e⊖λ (s, 0),

t ≥ s, s, t ∈ ℝ+ ∩ 𝕋.

Hence, t

t

x(t) ≤ a(t) + p(t) ∫ h1 (s)x(s)Δs + p(t) ∫ eλ (t, 0)h2 (s)y(s)Δs 0

0

t

t

= a(t) + p(t) ∫ h1 (s)x(s)Δs + eλ (t, 0)p(t) ∫ h2 (s)y(s)Δs, 0

0

whereupon t

e⊖λ (t, 0)x(t) ≤ a(t)e⊖λ (t, 0) + e⊖λ (t, 0)p(t) ∫ h1 (s)x(s)Δs 0

t

+ p(t) ∫ h2 (s)y(s)Δs 0

(1.30)

t

≤ a(t)e⊖λ (t, 0) + p(t) ∫ e⊖λ (s, 0)h1 (s)x(s)Δs t

0

+ p(t) ∫ h2 (s)y(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

36 | 1 Linear integral inequalities on time scales Let F(t) = e⊖λ (t, 0)x(t) + y(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, after adding (1.30) and the second inequality of (1.29), we obtain t

F(t) ≤ (a(t)e⊖λ (t, 0) + b(t)) + p(t) ∫ e⊖λ (s, 0)(h1 (s) + h3 (s))x(s)Δs 0

t

+ p(t) ∫(h2 (s) + h4 (s))y(s)Δs 0

t

≤ (a(t)e⊖λ (t, 0) + b(t)) + p(t) ∫ e⊖λ (s, 0)x(s)h(s)Δs 0

t

+ p(t) ∫ h(s)y(s)Δs 0

t

= (a(t)e⊖λ (t, 0) + b(t)) + p(t) ∫ F(s)h(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

From the last inequality and Theorem 1.1.10, we get t

F(t) ≤ (a(t)e⊖λ (t, 0)+b(t))+p(t) ∫ h(s)(a(s)e⊖λ (s, 0)+b(s))e⊖(hp) (σ(s), t)Δs, 0

t ∈ ℝ+ ∩𝕋.

Therefore t

e⊖λ (t, 0)x(t) ≤ (a(t)e⊖λ (t, 0) + b(t)) + p(t) ∫ h(s)(a(s)e⊖λ (s, 0) + b(s))e⊖(hp) (σ(s), t)Δs, 0

or t

x(t) ≤ (a(t) + b(t)eλ (t, 0)) + p(t)eλ (t, 0) ∫ h(s)(a(s)e⊖λ (s, 0) + b(s))e⊖(hp) (σ(s), t)Δs, 0

and t

y(t) ≤ (a(t)e⊖λ (t, 0) + b(t)) + p(t) ∫ h(s)(a(s)e⊖λ (s, 0) + b(s))e⊖(hp) (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋. This completes the proof.

1.5 Pachpatte’s inequalities | 37

1.5 Pachpatte’s inequalities Theorem 1.5.1 (Pachpatte’s Inequality). Let x, f , g ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions and t

t

s

x(t) ≤ c + ∫ f (s)x(s)Δs + ∫ f (s)(∫ g(y)x(y)Δy)Δs, 0

0

0

t ∈ ℝ+ ∩ 𝕋, where c is a nonnegative constant. Then x(t) ≤ cef +g (t, 0) and t

x(t) ≤ c(1 + ∫ f (s)ef +g (s, 0)Δs), 0

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t

z(t) = c + ∫ f (s)x(s)Δs 0

t

s

+ ∫ f (s)(∫ g(y)x(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

Then z(0) = c and x(t) ≤ z(t),

t

Δ

z (t) = f (t)x(t) + f (t) ∫ g(y)x(y)Δy t

0

(1.31)

= f (t)(x(t) + ∫ g(y)x(y)Δy) 0

t

≤ f (t)(z(t) + ∫ g(y)z(y)Δy), 0

t ∈ ℝ+ ∩ 𝕋.

Let t

m(t) = z(t) + ∫ g(y)z(y)Δy, 0

t ∈ ℝ+ ∩ 𝕋.

38 | 1 Linear integral inequalities on time scales Then z(t) ≤ m(t),

z Δ (t) ≤ f (t)m(t),

m(0) = z(0) Δ

= c,

m (t) = z Δ (t) + g(t)z(t)

≤ f (t)m(t) + g(t)m(t) = (f (t) + g(t))m(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, t

m(t) ≤ m(0) + ∫(f (y) + g(y))m(y)Δy 0

t

= c + ∫(f (y) + g(y))m(y)Δy, 0

t ∈ ℝ+ ∩ 𝕋.

From here and Theorem 1.1.4, we conclude that m(t) ≤ cef +g (t, 0), x(t) ≤ z(t)

≤ m(t)

≤ cef +g (t, 0),

t ∈ ℝ+ ∩ 𝕋.

Hence, by (1.31), we get z Δ (t) ≤ cf (t)ef +g (t, 0),

t ∈ ℝ+ ∩ 𝕋.

Then t

z(t) − z(0) ≤ c ∫ f (s)ef +g (s, 0)Δs, 0

t

z(t) ≤ c(1 + ∫ f (s)ef +g (s, 0)Δs), 0

t ∈ ℝ+ ∩ 𝕋.

Therefore x(t) ≤ z(t)

t

≤ c(1 + ∫ f (s)ef +g (s, 0)Δs), 0

This completes the proof.

t ∈ ℝ+ ∩ 𝕋.

1.5 Pachpatte’s inequalities | 39

Theorem 1.5.2 (Pachpatte’s Inequality). Let x, f , p, g ∈ 𝒞rd (ℝ+ ∩𝕋) be nonnegative functions, c be a nonnegative constant, and t

t

s

x(t) ≤ c + ∫(f (s)x(s) + p(s))Δs + ∫ f (s)(∫ g(y)x(y)Δy)Δs, 0

0

0

t ∈ ℝ+ ∩ 𝕋. Then t

x(t) ≤ c + ∫(p(s) + q(s)f (s))Δs 0

t

s

+ ∫ f (s) ∫(f (y) + g(y))q(y)e⊖(f +g) (σ(y), s)ΔyΔs, 0

0

t ∈ ℝ+ ∩ 𝕋, where t

q(t) = c + ∫ p(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Let t

t

s

z(t) = c + ∫(f (s)x(s) + p(s))Δs + ∫ f (s)(∫ g(y)x(y)Δy)Δs, 0

0

0

t ∈ ℝ+ ∩ 𝕋. Then z(0) = c,

x(t) ≤ z(t),

t

z Δ (t) = f (t)x(t) + p(t) + f (t) ∫ g(y)x(y)Δy 0

t

≤ z(t)f (t) + p(t) + f (t) ∫ g(y)z(y)Δy t

0

= p(t) + f (t)(z(t) + ∫ g(y)z(y)Δy), 0

t ∈ ℝ+ ∩ 𝕋.

Let t

v(t) = z(t) + ∫ g(y)z(y)Δy, 0

t ∈ ℝ+ ∩ 𝕋.

40 | 1 Linear integral inequalities on time scales Then z(t) ≤ v(t),

z Δ (t) ≤ p(t) + f (t)v(t), v(0) = z(0) = c,

Δ

v (t) = z Δ (t) + g(t)z(t)

≤ p(t) + f (t)v(t) + g(t)v(t) ≤ p(t) + (f (t) + g(t))v(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, t

t

v(t) ≤ c + ∫ p(s)Δs + ∫(f (s) + g(s))v(s)Δs 0

0

t

= q(t) + ∫(f (s) + g(s))v(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

From here and Theorem 1.1.10, we get t

v(t) ≤ q(t) + ∫(f (s) + g(s))q(s)e⊖(f +g) (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋. Therefore z Δ (t) ≤ p(t) + q(t)f (t) t

+ f (t) ∫(f (s) + g(s))q(s)e⊖(f +g) (σ(s), t)Δs 0

and t

z(t) ≤ c + ∫(p(s) + q(s)f (s))Δs 0

t

s

+ ∫ f (s) ∫(f (y) + g(y))q(y)e⊖(f +g) (σ(y), s)ΔyΔs, 0

t ∈ ℝ+ ∩ 𝕋. Consequently, x(t) ≤ z(t)

0

1.5 Pachpatte’s inequalities | 41 t

≤ c + ∫(p(s) + q(s)f (s))Δs 0

t

s

+ ∫ f (s) ∫(f (y) + g(y))q(y)e⊖(f +g) (σ(y), s)ΔyΔs, 0

0

t ∈ ℝ+ ∩ 𝕋. This completes the proof. Corollary 1.5.3. Let x, f , g, p ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, c be a nonnegative constant, and t

t

s

x(t) ≤ c + ∫ f (s)x(s)Δs + ∫ f (s)(∫(g(y)x(y) + p(y))Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

0

Then t

x(t) ≤ c + ∫(q(s) + l(s)f (s))Δs 0

t

s

+ ∫ f (s) ∫(f (y) + g(y))l(y)e⊖(f +g) (σ(y), s)ΔyΔs, 0

0

t ∈ ℝ+ ∩ 𝕋,

where t

q(t) = f (t) ∫ p(y)Δy, 0 t

l(t) = c + ∫ q(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

Proof. We have t

t

s

x(t) ≤ c + ∫ f (s)x(s)Δs + ∫ f (s) ∫ p(y)ΔyΔs 0

t

s

0

0

+ ∫ f (s)(∫ g(y)x(y)Δy)Δs 0

t

0

= c + ∫(f (s)x(s) + q(s))Δs 0

t

s

+ ∫ f (s)(∫ g(y)x(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

Hence, by Theorem 1.5.2, we get the desired result. This completes the proof.

42 | 1 Linear integral inequalities on time scales Corollary 1.5.4. Let x, f , g, h ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, c be a nonnegative constant, and t

x(t) ≤ c + ∫ f (s)x(s)Δs 0

t

s

+ ∫ g(s)(x(s) + ∫ h(y)x(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

Then t

x(t) ≤ c + c ∫(f (s) + g(s))Δs 0

t

s

+ c ∫(f (s) + g(s)) ∫(f (y) + g(y) + h(y))e⊖(f +g+h) (σ(y), s)ΔyΔs, 0

0

t ∈ ℝ+ ∩ 𝕋. Proof. We have t

x(t) ≤ c + ∫(f (s) + g(s))x(s)Δs 0

t

s

+ ∫ g(s)(∫ h(y)x(y)Δy)Δs 0

0

t

≤ c + ∫(f (s) + g(s))x(s)Δs 0

t

s

+ ∫(f (s) + g(s))(∫ h(y)x(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

Hence, by Theorem 1.5.2, we get the desired result. This completes the proof. Theorem 1.5.5 (Pachpatte’s Inequality). Let x, f , g, h, p ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, and t

x(t) ≤ h(t) + p(t) ∫ f (s)x(s)Δs t

0

s

+ p(t) ∫ f (s)p(s)(∫ g(y)x(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

1.5 Pachpatte’s inequalities | 43

Then x(t) ≤ h(t) + p(t)l(t) t

+ p(t) ∫ q(s)h(s)e⊖q (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋,

where t

l(t) = ∫(f (s) + g(s))h(s)Δs, 0

q(t) = p(t)(f (t) + g(t)),

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t

z(t) = ∫ f (s)x(s)Δs 0

s

t

+ ∫ f (s)p(s)(∫ g(y)x(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

Then z(0) = 0,

x(t) ≤ h(t) + p(t)z(t),

and t

z Δ (t) = f (t)x(t) + f (t)p(t) ∫ g(s)x(s)Δs 0

≤ f (t)h(t) + f (t)p(t)z(t) t

+ f (t)p(t) ∫ g(s)(h(s) + p(s)z(s))Δs 0

= f (t)(h(t) + p(t)(z(t) t

+ ∫ g(s)(h(s) + p(s)z(s))Δs)), 0

t ∈ ℝ+ ∩ 𝕋.

We set t

m(t) = z(t) + ∫ g(s)(h(s) + p(s)z(s))Δs, 0

t ∈ ℝ+ ∩ 𝕋.

44 | 1 Linear integral inequalities on time scales Then z(t) ≤ m(t),

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

and z Δ (t) ≤ f (t)(h(t) + p(t)m(t)),

t ∈ ℝ+ ∩ 𝕋.

Also, mΔ (t) = z Δ (t) + g(t)h(t) + g(t)p(t)z(t) ≤ f (t)(h(t) + p(t)m(t))

+ g(t)h(t) + g(t)p(t)m(t)

= (f (t) + g(t))h(t)

+ p(t)(f (t) + g(t))m(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, t

m(t) ≤ ∫(f (s) + g(s))h(s)Δs 0

t

+ ∫ p(s)(f (s) + g(s))m(s)Δs 0

t

= l(t) + ∫ q(s)m(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

Hence, by Theorem 1.1.10, we get t

m(t) ≤ l(t) + ∫ q(s)l(s)e⊖q (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

Therefore x(t) ≤ h(t) + p(t)z(t)

≤ h(t) + p(t)m(t) ≤ h(t) + p(t)l(t) t

+ p(t) ∫ q(s)l(s)e⊖q (σ(s), t)Δs, 0

This completes the proof.

t ∈ ℝ+ ∩ 𝕋.

1.5 Pachpatte’s inequalities | 45

Theorem 1.5.6 (Pachpatte’s Inequality). Let x, f , g, h ∈ 𝒞rd (ℝ+ ∩𝕋) be nonnegative functions, c be a nonnegative constant, and t

t

s

x(t) ≤ c + ∫ f (s)x(s)Δs + ∫ f (s)(∫ g(y)x(y)Δy)Δs 0

t

0 y

s

0

+ ∫ f (s)(∫ g(y)(∫ h(τ)x(τ)Δτ)Δy)Δs, 0

0

0

t ∈ ℝ+ ∩ 𝕋. Then t

x(t) ≤ c(1 + ∫(f (s) + g(s))ef +g+h (s, 0)Δs), 0

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t

t

s

z(t) = c + ∫ f (s)x(s)Δs + ∫ f (s)(∫ g(y)x(y)Δy)Δs 0

0 y

s

t

0

+ ∫ f (s)(∫ g(y)(∫ h(τ)x(τ)Δτ)Δy)Δs, 0

0

0

t ∈ ℝ+ ∩ 𝕋. Then x(t) ≤ z(t),

z(0) = c,

t

Δ

z (t) = f (t)x(t) + f (t) ∫ g(y)x(y)Δy 0

t

y

+ f (t) ∫ g(y)(∫ h(s)x(s)Δs)Δy 0

t

0

= f (t)(x(t) + ∫ g(y)x(y)Δy t

0 y

+ ∫ g(y)(∫ h(s)x(s)Δs)Δy) 0

0

t

≤ f (t)(z(t) + ∫ g(y)z(y)Δy t

0 y

+ ∫ g(y)(∫ h(s)z(s)Δs)Δy), 0

0

t ∈ ℝ+ ∩ 𝕋.

46 | 1 Linear integral inequalities on time scales Let

t

v(t) = z(t) + ∫ g(y)z(y)Δy t

0

y

+ ∫ g(y)(∫ h(s)z(s)Δs)Δy, 0

0

t ∈ ℝ+ ∩ 𝕋.

Then v(0) = z(0) Δ

= c,

z (t) ≤ f (t)v(t), z(t) ≤ v(t),

vΔ (t) = z Δ (t) + g(t)z(t) t

+ g(t) ∫ h(s)z(s)Δs 0

≤ f (t)v(t) + g(t)v(t) t

+ g(t) ∫ h(s)v(s)Δs 0

t

= (f (t) + g(t))v(t) + g(t) ∫ h(s)v(s)Δs 0

t

≤ (f (t) + g(t))v(t) + (f (t) + g(t)) ∫ h(s)v(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

Hence, t

v(t) ≤ c + ∫(f (s) + g(s))v(s)Δs 0

t

s

+ ∫(f (s) + g(s))(∫ h(y)v(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

Hence, by Theorem 1.5.1, we get t

v(t) ≤ c(1 + ∫(f (s) + g(s))ef +g+h (s, 0)Δs), 0

Therefore x(t) ≤ z(t)

t ∈ ℝ+ ∩ 𝕋.

1.5 Pachpatte’s inequalities | 47

≤ v(t)

t

≤ c(1 + ∫(f (s) + g(s))ef +g+h (s, 0)Δs),

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 1.5.7. Let x, k, p, f , g, h ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, c be a nonnegative constant, f (t) ≤ g(t), t ∈ ℝ+ ∩ 𝕋, and s

t

t

x(t) ≤ k(t) + p(t)(∫ f (s)x(s)Δs + ∫ f (s)p(s)(∫ g(τ)x(τ)Δτ)Δs 0

t

0

0

s

τ

+ ∫ f (s)p(s)(∫ g(τ)p(τ)(∫ h(y)x(y)Δy)Δτ)Δs), 0

0

0

t ∈ ℝ+ ∩ 𝕋. Then x(t) ≤ k(t) + p(t)l(t) + 2p(t)r(t) t

+ 2p(t) ∫ c(s)l(s)e⊖c (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋,

where t

q(t) = f (t)k(t) + f (t)p(t) ∫ g(s)k(s)Δs 0

t

τ

+ f (t)p(t) ∫ g(τ)p(τ)(∫ h(y)k(y)Δy)Δτ, t

0

0

r(t) = ∫(p(s)g(s) + h(s)p(s))l(s)Δs, 0

c(t) = 2(p(t)g(t) + h(t)p(t)), t

l(t) = ∫ q(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Let t

t

s

z(t) = ∫ f (s)x(s)Δs + ∫ f (s)p(s)(∫ g(τ)x(τ)Δτ)Δs 0

t

0 s

0

τ

+ ∫ f (s)p(s)(∫ g(τ)p(τ)(∫ h(y)x(y)Δy)Δτ)Δs, 0

0

0

48 | 1 Linear integral inequalities on time scales t ∈ ℝ+ ∩ 𝕋. Then

z(0) = 0,

x(t) ≤ k(t) + p(t)z(t),

t

z Δ (t) = f (t)x(t) + f (t)p(t) ∫ g(s)x(s)Δs 0

t

τ

+ f (t)p(t) ∫ g(τ)p(τ)(∫ h(y)x(y)Δy)Δτ 0

≤ f (t)k(t) + f (t)p(t)z(t)

0

t

+ f (t)p(t) ∫ g(s)k(s)Δs 0

t

+ f (t)p(t) ∫ p(s)g(s)z(s)Δs 0

t

τ

+ f (t)p(t) ∫ g(τ)p(τ)(∫ h(y)k(y)Δy)Δτ 0

0

t

τ

+ f (t)p(t) ∫ g(τ)p(τ)(∫ h(y)p(y)z(y)Δy)Δτ 0

= q(t) + f (t)p(t)z(t)

0

t

+ f (t)p(t) ∫ p(s)g(s)z(s)Δs 0

t

τ

+ f (t)p(t) ∫ g(τ)p(τ)(∫ h(y)p(y)z(y)Δy)Δτ 0

0

t

= q(t) + f (t)p(t)(z(t) + ∫ p(s)g(s)z(s)Δs t

τ

0

+ ∫ g(τ)p(τ)(∫ h(y)p(y)z(y)Δy)Δτ), 0

t ∈ ℝ+ ∩ 𝕋. Let

0

t

v(t) = z(t) + ∫ p(s)g(s)z(s)Δs t

0

τ

+ ∫ g(τ)p(τ)(∫ h(y)p(y)z(y)Δy)Δτ, 0

0

t ∈ ℝ+ ∩ 𝕋.

1.5 Pachpatte’s inequalities | 49

Then v(0) = z(0) Δ

= 0,

z (t) ≤ q(t) + f (t)p(t)v(t), z(t) ≤ v(t),

vΔ (t) = z Δ (t) + p(t)g(t)z(t) t

+ p(t)g(t) ∫ h(y)p(y)z(y)Δy 0

≤ q(t) + f (t)p(t)v(t) + p(t)g(t)v(t) t

+ p(t)g(t) ∫ h(y)p(y)v(y)Δy 0

t

= q(t) + g(t)p(t)(2v(t) + ∫ h(y)p(y)v(y)Δy), 0

t ∈ ℝ+ ∩ 𝕋.

Hence, t

t

v(t) ≤ ∫ q(s)Δs + 2 ∫ p(s)g(s)v(s)Δs 0

0

t

s

+ ∫ g(s)p(s) ∫ h(y)p(y)v(y)ΔyΔs 0

t

0

≤ l(t) + 2 ∫ p(s)g(s)v(s)Δs t

0

s

+ 4 ∫ g(s)p(s) ∫ h(y)p(y)v(y)ΔyΔs, 0

0

t ∈ ℝ+ ∩ 𝕋.

From here and Theorem 1.5.5, we obtain t

v(t) ≤ l(t) + 2r(t) + 2 ∫ c(s)l(s)e⊖c (σ(s), t)Δs, 0

Therefore x(t) ≤ k(t) + p(t)z(t)

≤ k(t) + p(t)v(t)

t ∈ ℝ+ ∩ 𝕋.

50 | 1 Linear integral inequalities on time scales ≤ k(t) + p(t)l(t) + 2p(t)r(t) t

+ 2p(t) ∫ c(s)l(s)e⊖c (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 1.5.8. Let x, f , g ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, h ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive and nondecreasing function, and s

t

t

x(t) ≤ h(t) + ∫ f (s)x(s)Δs + ∫ f (s)(∫ g(y)x(y)Δy)Δs, 0

0

0

t ∈ ℝ+ ∩ 𝕋. Then t

x(t) ≤ h(t)(1 + ∫ f (s)ef +g (s, 0)Δs), 0

t ∈ ℝ+ ∩ 𝕋.

Proof. Since h is positive and nondecreasing on ℝ+ ∩ 𝕋, we get t

t

s

0

0

x(t) 1 1 ≤1+ ∫ f (s)x(s)Δs + ∫ f (s)(∫ g(y)x(y)Δy)Δs h(t) h(t) h(t) 0

t

≤ 1 + ∫ f (s) 0

t

s

0

0

x(s) x(y) Δs + ∫ f (s)(∫ g(y) Δy)Δs, h(s) h(y)

t ∈ ℝ+ ∩ 𝕋. Hence, by Theorem 1.5.1, we get t

x(t) ≤ 1 + ∫ f (s)ef +g (s, 0)Δs, h(t) 0

t ∈ ℝ+ ∩ 𝕋,

or t

x(t) ≤ h(t)(1 + ∫ f (s)ef +g (s, 0)Δs), 0

t ∈ ℝ+ ∩ 𝕋.

This completes the proof. 2 Theorem 1.5.9. Let x, h, p ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, f ∈ 𝒞rd (ℝ+ ∩ 𝕋) and 1 g ∈ 𝒞rd (ℝ+ ∩ 𝕋) be positive functions, c be a positive constant, and t

s

τ

x(t) ≤ c + ∫ f (s)(h(s) + ∫ g(τ)(∫ p(y)x(y)Δy)Δτ)Δs, 0

0

0

1.5 Pachpatte’s inequalities | 51

t ∈ ℝ+ ∩ 𝕋. Then t

t

x(t) ≤ (c + ∫ f (s)h(s)Δs)(1 + ∫ q(s)e⊖q (σ(s), t)Δs), 0

0

t ∈ ℝ+ ∩ 𝕋, where y

t

q(t) = f (t) ∫ g(y) ∫ p(τ)ΔτΔy, 0

0

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t

l(t) = c + ∫ f (s)h(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

Then t

s

τ

x(t) ≤ l(t) + ∫ f (s) ∫ g(τ)(∫ p(y)x(y)Δy)ΔτΔs, 0

0

t ∈ ℝ+ ∩ 𝕋.

0

Note that l is a positive and nondecreasing function on ℝ+ ∩ 𝕋. Then τ

s

t

x(t) 1 ≤1+ ∫ f (s) ∫ g(τ)(∫ p(y)x(y)Δy)ΔτΔs l(t) l(t) 0

t

s

0

τ

0

≤ 1 + ∫ f (s) ∫ g(τ)(∫ p(y) 0

0

0

t

s

τ

x(y) Δy)ΔτΔs, l(y)

t ∈ ℝ+ ∩ 𝕋.

x(y) Δy)ΔτΔs, l(y)

t ∈ ℝ+ ∩ 𝕋.

Let z(t) = 1 + ∫ f (s) ∫ g(τ)(∫ p(y) 0

0

0

Then x(t) ≤ z(t), l(t) z(0) = 1, Δ

t

τ

z (t) = f (t) ∫ g(τ)(∫ p(y) 0

0

x(y) Δy)Δτ, l(y)

52 | 1 Linear integral inequalities on time scales t ∈ ℝ+ ∩ 𝕋. Hence, t

τ

z Δ (t) x(y) = ∫ g(τ)(∫ p(y) Δy)Δτ, f (t) l(y) 0

(

0

t

Δ

x(y) z Δ (t) ) = g(t) ∫ p(y) Δy, f (t) l(y) t

Δ

Δ

0

1 z (t) x(y) ( ) = ∫ p(y) Δy, g(t) f (t) l(y) (

0

Δ Δ

Δ

x(t) 1 z (t) ( ) ) = p(t) g(t) f (t) l(t) ≤ p(t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Therefore Δ

1 z (t) Δ Δ ( g(t) ( f (t) ) )

≤ p(t),

z(t)

t ∈ ℝ+ ∩ 𝕋.

Note that z Δ (t) ≥ 0,

Δ

1 z Δ (t) ( ) ≥ 0. g(t) f (t)

z(t) ≥ 0,

Therefore Δ

1 z (t) Δ Δ ( f (t) ) ) ( g(t)

z(t)

≤ p(t) +

1 z Δ (t) Δ Δ ( ) z (t) g(t) f (t)

z(t)z(σ(t))

,

t ∈ ℝ+ ∩ 𝕋. Since z is nondecreasing on ℝ+ ∩ 𝕋, we have Δ

1 z (t) Δ Δ ( f (t) ) ) z(t) ( g(t)

(z(t))2

Δ

≥

1 z (t) Δ Δ ( f (t) ) ) z(t) ( g(t)

z(t)z(σ(t))

,

t ∈ ℝ+ ∩ 𝕋.

Therefore Δ

1 z (t) Δ Δ ( f (t) ) ) z(t) ( g(t)

z(t)z(σ(t))

≤ p(t) +

1 z Δ (t) Δ Δ ( ) z (t) g(t) f (t)

z(t)z(σ(t))

or (

1 z Δ (t) Δ ( ) g(t) f (t)

z(t)

Δ

) ≤ p(t),

t ∈ ℝ+ ∩ 𝕋.

,

1.5 Pachpatte’s inequalities | 53

From here, 1 z Δ (t) Δ ( ) g(t) f (t)

t

≤ ∫ p(s)Δs,

z(t)

0

or Δ

t

(t) Δ ( zf (t) )

≤ g(t) ∫ p(s)Δs,

z(t)

0

or Δ

(t) Δ ( zf (t) ) z(t)

(z(t))2

t

≤ g(t) ∫ p(s)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

Since z is nondecreasing, we obtain Δ

Δ

(t) Δ ( zf (t) ) z(t)

z(t)z(σ(t))

≤

(t) Δ ( zf (t) ) z(t)

(z(t))2 t

≤ g(t) ∫ p(s)Δs + 0

z Δ (t) Δ z (t) f (t)

z(t)z(σ(t))

or (

z Δ (t) f (t)

z(t)

t

Δ

) ≤ g(t) ∫ p(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

From here, z Δ (t) f (t)

z(t)

t

s

≤ ∫ g(s) ∫ p(y)ΔyΔs, 0

0

or t

s

0

0

z Δ (t) ≤ z(t) ∫ g(s) ∫ p(y)ΔyΔs, f (t) or Δ

t

s

z (t) ≤ z(t)f (t) ∫ g(s) ∫ p(y)ΔyΔs, 0

0

,

54 | 1 Linear integral inequalities on time scales or t

y

s

z(t) ≤ 1 + ∫(f (s) ∫ g(y) ∫ p(τ)ΔτΔy)z(s)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

0

Hence, by Theorem 1.1.10, we get t

z(t) ≤ 1 + ∫ q(s)e⊖q (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

Consequently, t

x(t) ≤ l(t)(1 + ∫ q(s)e⊖q (σ(s), t)Δs) t

0

t

= (c + ∫ f (s)h(s)Δs)(1 + ∫ q(s)e⊖q (σ(s), t)Δs), 0

0

t ∈ ℝ+ ∩ 𝕋. This completes the proof. 1 Theorem 1.5.10. Let x, h, p ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, f ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive function, c be a positive constant, and t

s

x(t) ≤ c + ∫ f (s)(h(s) + ∫ p(y)x(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

Then t

x(t) ≤ g(t) + ∫ l(s)g(s)e⊖l (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋,

where t

g(t) = c + ∫ f (s)h(s)Δs, 0

t

l(t) = f (t) ∫ p(y)Δy, 0

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t

s

z(t) = c + ∫ f (s)(h(s) + ∫ p(y)x(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

1.5 Pachpatte’s inequalities | 55

Then x(t) ≤ z(t),

z(0) = c,

t

Δ

z (t) = f (t)h(t) + f (t) ∫ p(y)x(y)Δy,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, z Δ (t) ≥ 0, t

z Δ (t) − f (t)h(t) = ∫ p(y)x(y)Δy f (t) 0

≥ 0,

t ∈ ℝ+ ∩ 𝕋,

and Δ

(

z Δ (t) − f (t)h(t) ) = p(t)x(t) f (t) ≤ p(t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Therefore (z

Δ

(t)−f (t)h(t) Δ ) f (t)

z(t)

≤ p(t) ≤ p(t) +

z Δ (t)−f (t)h(t) Δ z (t) f (t)

z(t)z(σ(t))

t ∈ ℝ+ ∩ 𝕋.

,

Consequently, (z

Δ

(t)−f (t)h(t) Δ ) z(t) f (t)

z(t)z(σ(t))

≤

(z

Δ

(t)−f (t)h(t) Δ ) z(t) f (t) (z(t))2

≤ p(t) +

z Δ (t)−f (t)h(t) Δ z (t) f (t)

z(t)z(σ(t))

Hence, (

z Δ (t)−f (t)h(t) f (t)

z(t)

Δ

) ≤ p(t)

and t

z Δ (t) − f (t)h(t) ≤ ∫ p(s)Δs, f (t)z(t) 0

,

t ∈ ℝ+ ∩ 𝕋.

56 | 1 Linear integral inequalities on time scales or t

z Δ (t) − f (t)h(t) ≤ f (t) ∫ p(s)Δs, z(t) 0

or t

Δ

z (t) ≤ f (t)h(t) + z(t)f (t) ∫ p(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Therefore t

t

s

z(t) ≤ c + ∫ f (s)h(s)Δs + ∫ f (s)(∫ p(y)Δy)z(s)Δs 0

0

t

0

= g(t) + ∫ l(s)z(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, by Theorem 1.1.10, we get t

z(t) ≤ g(t) + ∫ l(s)g(s)e⊖l (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

whereupon t

x(t) ≤ g(t) + ∫ l(s)g(s)e⊖l (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 1.5.11. Let x, v, p, f , g ∈ 𝒞rd (J) be nonnegative functions, p be decreasing on J, 1 − μ(s)h(s) > 0 for a ≤ s ≤ t, and t

t

t

x(t) ≥ v(s) − p(t)(∫ f (y)v(y)Δy + ∫ f (y)(∫ g(τ)v(τ)Δτ)Δy), s

s

y

a ≤ s ≤ t ≤ b. Then t

x(t) ≥ v(s)(1 + p(t) ∫ f (y)e s

−1

h 1−μh

(t, y)Δy) ,

a ≤ s ≤ t ≤ b,

where h(s) = p(s)f (s) + g(s),

a ≤ s ≤ t ≤ b.

1.5 Pachpatte’s inequalities | 57

Proof. Let t ∈ J be fixed and define, for a ≤ s ≤ t ≤ b, t

t

t

z(s) = x(t) + p(t)(∫ f (y)v(y)Δy + ∫ f (y)(∫ g(τ)v(τ)Δτ)Δy), s

y

s

a ≤ s ≤ t ≤ b. Then z(t) = x(t),

v(s) ≤ z(s),

t

z Δ (s) = −p(t)f (s)(v(s) + ∫ g(y)v(y)Δy) s

t

≥ −p(t)f (s)(z(s) + ∫ g(y)z(y)Δy), s

a ≤ s ≤ t ≤ b. Let

t

r(s) = z(s) + ∫ g(y)z(y)Δy, s

a ≤ s ≤ t ≤ b.

Then z(s) ≤ r(s),

r Δ (s) = z Δ (s) − g(s)z(s)

≥ −p(s)f (s)r(s) − g(s)z(s)

≥ −p(s)f (s)r(s) − g(s)r(s), or r Δ (s) + (p(s)f (s) + g(s))r(s) ≥ 0, or r Δ (s) + h(s)r(s) ≥ 0,

a ≤ s ≤ t ≤ b.

Note that (r(s)e

h 1−μh

Δ

(s, a)) = r Δ (s)e + r(s)

h 1−μh

(σ(s), a)

h(s) e h (s, a) 1 − μ(s)h(s) 1−μh

μ(s)h(s) )e h (s, a) 1 − μ(s)h(s) 1−μh h(s) + r(s) e h (s, a) 1 − μ(s)h(s) 1−μh

= r Δ (s)(1 +

58 | 1 Linear integral inequalities on time scales 1 (r Δ (s) + h(s)r(s))e h (s, a) 1−μh 1 − μ(s)h(s) ≥ 0, a ≤ s ≤ t ≤ b. =

Therefore r(t)e or

h 1−μh

(t, a) ≥ r(s)e

r(t) ≥ r(s)e or r(s) ≤ e

h 1−μh

h 1−μh

h 1−μh

(s, a),

(s, t),

(t, s)r(t)

≤ x(t)e

h 1−μh

(t, s),

a ≤ s ≤ t ≤ b.

Therefore z Δ (s) ≥ −p(t)f (s)x(t)e

h 1−μh

(t, s)

and t

z(t) − z(s) ≥ −p(t)x(t) ∫ f (y)e s

or

h 1−μh

(t, y)Δy,

t

z(t) ≥ z(s) − p(t)x(t) ∫ f (y)e s

h 1−μh

(t, y)Δy

t

≥ v(s) − p(t)x(t) ∫ f (y)e s

or

h 1−μh

(t, y)Δy,

t

x(t) ≥ v(s) − p(t)x(t) ∫ f (y)e s

or

h 1−μh

(t, y)Δy,

t

(1 + p(t) ∫ f (y)e or

s

(t, y)Δy)x(t) ≥ v(s),

h 1−μh

t

x(t) ≥ v(s)(1 + p(t) ∫ f (y)e s

a ≤ s ≤ t ≤ b. This completes the proof.

−1

h 1−μh

(t, y)Δy) ,

2 Linear integro-dynamic inequalities This chapter introduces linear integro-dynamic inequalities of Pachpatte type and linear integro-dynamic inequalities with several iterated integrals. The material in this chapter is based on some results in [24] and [25]. Let 𝕋 be a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Suppose that 0 ∈ 𝕋.

2.1 Pachpatte’s inequalities We will start with the following useful lemma. Lemma 2.1.1. Let x, xΔ , a, b ∈ 𝒞rd (ℝ+ ∩ 𝕋), b ∈ ℛ+ and xΔ (t) ≤ a(t) + b(t)x(t),

t ∈ ℝ+ ∩ 𝕋.

Then t

x(t) ≤ x(0)eb (t, 0) + ∫ a(s)e⊖b (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Since b ∈ ℛ+ (ℝ+ ∩ 𝕋), we have that eb (t, 0) > 0

and

e⊖b (t, 0) > 0,

t ∈ ℝ+ ∩ 𝕋.

We have Δ

(x(t)e⊖b (t, 0)) = xΔ (t)e⊖b (σ(t), 0)

+ x(t)(⊖b)(t)e⊖b (t, 0)

= xΔ (t)e⊖b (σ(t), 0) x(t)b(t) e (t, 0) − 1 + μ(t)b(t) ⊖b = xΔ (t)e⊖b (σ(t), 0)

− x(t)b(t)e⊖b (σ(t), 0)

= (xΔ (t) − x(t)b(t))e⊖b (σ(t), 0)

≤ a(t)e⊖b (σ(t), 0),

t ∈ ℝ+ ∩ 𝕋.

Hence, t

x(t)e⊖b (t, 0) ≤ x(0) + ∫ a(s)e⊖b (σ(s), 0)Δs, 0 https://doi.org/10.1515/9783110705553-002

t ∈ ℝ+ ∩ 𝕋,

60 | 2 Linear integro-dynamic inequalities or t

x(t) ≤ x(0)eb (t, 0) + ∫ a(s)e⊖b (σ(s), 0)e⊖b (0, t)Δs 0

t

= x(0)eb (t, 0) + ∫ a(s)e⊖b (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 2.1.2 (Pachpatte’s Inequality). Let x, x Δ , a, b, c ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions. If b(t) ≥ 1, t ∈ ℝ+ ∩ 𝕋, and t

xΔ (t) ≤ a(t) + b(t) ∫ c(s)(x(s) + x Δ (s))Δs, a

t ∈ ℝ+ ∩ 𝕋,

then t

xΔ (t) ≤ a(t) + b(t) ∫ c(s)(A(s) + b(s)B(s))Δs,

t ∈ ℝ+ ∩ 𝕋,

0

where t

A(t) = x(0) + ∫ a(s)Δs + a(t), 0

f (t) = b(t)(c(t) + 1), t

B(t) = ∫ c(s)A(s)e⊖f (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Let t

m(t) = ∫ c(s)(x(s) + x Δ (s))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then x Δ (t) ≤ a(t) + b(t)m(t),

t ∈ ℝ+ ∩ 𝕋,

(2.1)

and mΔ (t) = c(t)(x(t) + xΔ (t))

≤ c(t)x(t) + c(t)(a(t) + b(t)m(t))

= a(t)c(t) + c(t)x(t) + b(t)c(t)m(t),

(2.2) t ∈ ℝ+ ∩ 𝕋.

2.1 Pachpatte’s inequalities | 61

By (2.1), we get t

x(t) ≤ x(0) + ∫(a(s) + b(s))m(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, from (2.2), we obtain t

Δ

m (t) ≤ c(t)x(0) + c(t) ∫ a(s)Δs 0

t

+ c(t) ∫ b(s)m(s)Δs 0

(2.3)

+ c(t)a(t) + b(t)c(t)m(t)

t

= c(t)A(t) + c(t)(b(t)m(t) + ∫ b(s)m(s)Δs) t

0

≤ c(t)(A(t) + b(t)(m(t) + ∫ b(s)m(s)Δs)), 0

t ∈ ℝ+ ∩ 𝕋. Let t

r(t) = m(t) + ∫ b(s)m(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then r(0) = m(0) = 0,

m(t) ≤ r(t), Δ

t ∈ ℝ+ ∩ 𝕋,

m (t) ≤ c(t)(A(t) + b(t)r(t)),

t ∈ ℝ+ ∩ 𝕋,

and r Δ (t) = mΔ (t) + b(t)m(t)

≤ c(t)(A(t) + b(t)r(t)) + b(t)r(t) = c(t)A(t) + b(t)(c(t) + 1)r(t) = c(t)A(t) + f (t)r(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, by Lemma 2.1.1, we get t

r(t) ≤ r(0)ef (t, 0) + ∫ c(s)A(s)e⊖f (σ(s), t)Δs 0

(2.4)

62 | 2 Linear integro-dynamic inequalities t

= ∫ c(s)A(s)e⊖f (σ(s), t)Δs 0

= B(t),

t ∈ ℝ+ ∩ 𝕋.

Thus, from (2.4), we obtain mΔ (t) ≤ c(t)(A(t) + b(t)B(t)),

t ∈ ℝ+ ∩ 𝕋,

whereupon t

m(t) ≤ ∫ c(s)(A(s) + b(s)B(s))Δs,

t ∈ ℝ+ ∩ 𝕋,

0

and, using (2.1), we get xΔ (t) ≤ a(t) + b(t)m(t) t

≤ a(t) + b(t) ∫ c(s)(A(s) + b(s)B(s))Δs, 0

t ∈ ℝ+ ∩ 𝕋. This completes the proof. Theorem 2.1.3. Let x, xΔ , a, b, c ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, and t

Δ

x (t) ≤ a(t) + b(t)(x(t) + ∫ c(s)(x(s) + x Δ (s))Δs),

t ∈ ℝ+ ∩ 𝕋.

0

Then xΔ (t) ≤ a(t) + b(t)x(0)ef (t, 0) t

+ b(t) ∫ a(s)(1 + c(s))e⊖f (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋, where f (t) = c(t) + b(t)(1 + c(t)),

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t

z(t) = x(t) + ∫ c(s)(x(s) + xΔ (s))Δs, 0

Then x(0) = z(0),

t ∈ ℝ+ ∩ 𝕋.

2.1 Pachpatte’s inequalities | 63

x(t) ≤ z(t),

t ∈ ℝ+ ∩ 𝕋,

Δ

x (t) ≤ a(t) + b(t)z(t),

t ∈ ℝ+ ∩ 𝕋,

(2.5)

and hence, z Δ (t) = xΔ (t) + c(t)(x(t) + xΔ (t)) = c(t)x(t) + (1 + c(t))xΔ (t)

≤ c(t)z(t) + (1 + c(t))(a(t) + b(t)z(t))

= a(t)(1 + c(t)) + (c(t) + b(t)(1 + c(t)))z(t) = a(t)(1 + c(t)) + f (t)z(t),

t ∈ ℝ+ ∩ 𝕋.

From Lemma 2.1.1, we now get t

z(t) ≤ z(0)ef (t, 0) + ∫ a(s)(1 + c(s))e⊖f (σ(s), t)Δs 0

t

= x(0)ef (t, 0) + ∫ a(s)(1 + c(s))e⊖f (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋. From the last inequality and (2.5), we obtain xΔ (t) ≤ a(t) + b(t)x(0)ef (t, 0) t

+ b(t) ∫ a(s)(1 + c(s))e⊖f (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋. This completes the proof. Theorem 2.1.4 (Pachpatte’s Inequality). Let x, xΔ , a, b ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions and t

Δ

x (t) ≤ x(0) + ∫ a(s)(x(s) + xΔ (s))Δs t

0

s

+ ∫(∫ b(y)xΔ (y)Δy)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

0

Then t

xΔ (t) ≤ x(0)(1 + 2 ∫ a(s)ec (s, 0)Δs),

t ∈ ℝ+ ∩ 𝕋,

0

where c(t) = 1 + a(t) + b(t),

t ∈ ℝ+ ∩ 𝕋.

64 | 2 Linear integro-dynamic inequalities Proof. Let t

m(t) = x(0) + ∫ a(s)(x(s) + x Δ (s))Δs t

0

s

+ ∫ a(s)(∫ b(y)xΔ (y)Δy)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

0

Then m(0) = x(0) and xΔ (t) ≤ m(t),

t ∈ ℝ+ ∩ 𝕋.

(2.6)

We have t

mΔ (t) = a(t)(x(t) + xΔ (t)) + a(t) ∫ b(y)xΔ (y)Δy t

0

= a(t)(x(t) + xΔ (t) + ∫ b(y)x Δ (y)Δy)

(2.7)

0 t

≤ a(t)(x(t) + m(t) + ∫ b(y)m(y)Δy),

t ∈ ℝ+ ∩ 𝕋.

0

By (2.6), we get t

x(t) ≤ x(0) + ∫ m(y)Δy 0

t

= m(0) + ∫ m(y)Δy,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, by (2.7), we obtain Δ

t

t

m (t) ≤ a(t)(m(0) + ∫ m(y)Δy + m(t) + ∫ b(y)m(y)Δy) 0

t

0

= a(t)(m(0) + m(t) + ∫(1 + b(y))m(y)Δy),

t ∈ ℝ+ ∩ 𝕋.

0

Let t

z(t) = m(0) + m(t) + ∫(1 + b(y))m(y)Δy, 0

t ∈ ℝ+ ∩ 𝕋.

2.1 Pachpatte’s inequalities | 65

Then z(0) = 2m(0) and m(t) ≤ z(t),

mΔ (t) ≤ a(t)z(t),

t ∈ ℝ+ ∩ 𝕋.

By the definition of function z, we obtain z Δ (t) = mΔ (t) + (1 + b(t))m(t)

≤ a(t)z(t) + (1 + b(t))z(t) = (1 + a(t) + b(t))z(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, using Lemma 2.1.1, we get z(t) ≤ z(0)ec (t, 0)

= 2x(0)ec (t, 0),

Δ

m (t) ≤ a(t)z(t)

≤ 2x(0)a(t)ec (t, 0),

t ∈ ℝ+ ∩ 𝕋.

From here, t

m(t) ≤ m(0) + 2x(0) ∫ a(s)ec (s, 0)Δs 0

t

= x(0) + 2x(0) ∫ a(s)ec (s, 0)Δs t

0

= x(0)(1 + 2 ∫ a(s)ec (s, 0)Δs), 0

Δ

x (t) ≤ m(t)

t

≤ x(0)(1 + 2 ∫ a(s)ec (s, 0)Δs),

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 2.1.5 (Pachpatte’s Inequality). Let x, xΔ , a, b ∈ 𝒞rd (ℝ+ ∩ 𝕋) and t

xΔ (t) ≤ x(0) + ∫ a(s)(x(s) + xΔ (s))Δs t

0

s

+ ∫ a(s)(∫ b(y)(x(y) + xΔ (y))Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

66 | 2 Linear integro-dynamic inequalities Then

t

xΔ (t) ≤ x(0)(1 + 2 ∫ a(s)ec (s, 0)Δs 0

t

s

+ ∫ a(s)(∫(1 + 2b(y))e⊖c (σ(y), s)Δy)Δs),

t ∈ ℝ+ ∩ 𝕋,

0

0

where c(t) = 2 + 2a(t) + b(t),

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t

m(t) = x(0) + ∫ a(s)(x(s) + xΔ (s))Δs t

0

s

+ ∫ a(s)(∫ b(y)(x(y) + xΔ (y))Δy)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

0

Then m(0) = x(0),

xΔ (t) ≤ m(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, t

x(t) ≤ x(0) + ∫ m(s)Δs 0

t

= m(0) + ∫ m(s)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

and mΔ (t) = a(t)(x(t) + xΔ (t)) t

+ a(t) ∫ b(y)(x(y) + x Δ (y))Δy 0

t

= a(t)(x(t) + xΔ (t) + ∫ b(y)(x(y) + x Δ (y))Δy) t

0

≤ a(t)(m(0) + ∫ m(s)Δs + m(t) 0

2.1 Pachpatte’s inequalities | 67 t

s

+ ∫ b(s)(x(0) + ∫ m(y)Δy)Δs 0

0

t

+ ∫ b(y)m(y)Δy) 0

t

t

= a(t)(m(0) + m(0) ∫ b(s)Δs + ∫(1 + b(y))m(y)Δy t

0

s

0

+ m(t) + ∫ b(s) ∫ m(y)ΔyΔs), 0

t ∈ ℝ+ ∩ 𝕋.

0

Let t

t

z(t) = m(0) + m(0) ∫ b(s)Δs + ∫(1 + b(y))m(y)Δy 0

t

0

s

+ m(t) + ∫ b(s) ∫ m(y)ΔyΔs, 0

t ∈ ℝ+ ∩ 𝕋.

0

Then m(t) ≤ z(t),

mΔ (t) ≤ a(t)z(t), z(0) = 2m(0),

t ∈ ℝ+ ∩ 𝕋.

Hence, t

m(t) ≤ m(0) + ∫ a(s)z(s)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

and z Δ (t) = mΔ (t) + m(0)b(t)

t

+ (1 + b(t))m(t) + b(t) ∫ m(y)Δy ≤ a(t)z(t) + m(0)b(t)

0

t

+ (1 + b(t)) ∫ a(s)z(s)Δs + (1 + b(t))m(0) t

0

+ b(t) ∫ z(y)Δy 0

68 | 2 Linear integro-dynamic inequalities t

≤ m(0)(1 + 2b(t)) + a(t)z(t) + (1 + b(t)) ∫(1 + a(s))z(s)Δs 0

t

≤ m(0)(1 + 2b(t)) + (1 + a(t) + b(t))(z(t) + ∫(1 + a(s))z(s)Δs), 0

t ∈ ℝ+ ∩ 𝕋. Let t

v(t) = z(t) + ∫(1 + a(s))z(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then v(0) = z(0) Δ

= 2m(0),

z (t) ≤ m(0)(1 + 2b(t)) + (1 + a(t) + b(t))v(t), z(t) ≤ v(t),

vΔ (t) = z Δ (t) + (1 + a(t))z(t)

≤ m(0)(1 + 2b(t)) + (1 + a(t) + b(t))v(t) + (1 + a(t))v(t)

≤ m(0)(1 + 2b(t)) + (2 + 2a(t) + b(t))v(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, employing Lemma 2.1.1, we get t

v(t) ≤ 2m(0)ec (t, 0) + m(0) ∫(1 + 2b(s))e⊖c (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Therefore t

z(t) ≤ x(0)(2ec (t, 0) + ∫(1 + 2b(s))e⊖c (σ(s), t)Δ), 0 Δ

t

m (t) ≤ x(0)a(t)(2ec (t, 0) + ∫(1 + 2b(s))e⊖c (σ(s), t)Δs),

t ∈ ℝ+ ∩ 𝕋.

0

Consequently, t

m(t) ≤ x(0) + 2x(0) ∫ a(s)ec (s, 0)Δs t

0

s

+ x(0) ∫ a(s)(∫(1 + 2b(y))e⊖c (σ(y), t)Δy)Δs, 0

0

2.2 Modifications of Pachpatte’s inequalities | 69

t ∈ ℝ+ ∩ 𝕋, and t

Δ

x (t) ≤ x(0)(1 + 2 ∫ a(s)ec (s, 0)Δs 0

t

s

+ ∫ a(s)(∫(1 + 2b(y))e⊖c (σ(y), t)Δy)Δs), 0

0

t ∈ ℝ+ ∩ 𝕋. This completes the proof.

2.2 Modifications of Pachpatte’s inequalities We will start with the following useful lemma. Lemma 2.2.1. Let f ∈ 𝒞rd (ℝ+ ∩ 𝕋). Then t s

t

∫ ∫ f (y)ΔyΔs = ∫(t − σ(y))f (y)Δy, 0 0

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Let t s

t

g(t) = ∫ ∫ f (y)ΔyΔs − ∫(t − σ(y))f (y)Δy, 0 0

t ∈ ℝ+ ∩ 𝕋.

0

Then g(0) = 0 and t

t

g Δ (t) = ∫ f (y)Δy − ∫ f (y)Δy − (σ(t) − σ(t))f (t) 0

= 0,

0

t ∈ ℝ+ ∩ 𝕋.

Therefore g(t) = 0,

t ∈ ℝ+ ∩ 𝕋.

This completes the proof. 2

Theorem 2.2.2. Let x, xΔ , xΔ , f , g, h ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions and Δ2

t

x (t) ≤ f (t) + h(t) ∫ g(s)(x(s) + xΔ (s))Δs, 0

t ∈ ℝ+ ∩ 𝕋.

70 | 2 Linear integro-dynamic inequalities Then t

Δ2

x (t) ≤ f (t) + h(t) ∫ p(s)e⊖q (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

where t

Δ

p(t) = g(t)(x(0) + (t + 1)x (0) + (t + 1) ∫ f (s)Δs), 0

t

q(t) = (t + 1)g(t) ∫ h(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Let t

z(t) = ∫ g(s)(x(s) + xΔ (s))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then 2

x Δ (t) ≤ f (t) + h(t)z(t),

t ∈ ℝ+ ∩ 𝕋,

and z(0) = 0. Hence, using Lemma 2.2.1, we get t

x Δ (t) ≤ xΔ (0) + ∫(f (s) + h(s)z(s))Δs, 0 Δ

t s

x(t) ≤ x(0) + tx (0) + ∫ ∫(f (y) + h(y)z(y))ΔyΔs 0 0 t

= x(0) + tx Δ (0) + ∫(t − σ(s))(f (s) + h(s)z(s))Δs 0 Δ

t

≤ x(0) + tx (0) + ∫(t − s)(f (s) + h(s)z(s))Δs, 0

Next, z(t) is nondecreasing on ℝ+ ∩ 𝕋 and z Δ (t) = g(t)(x(t) + xΔ (t))

t ∈ ℝ+ ∩ 𝕋.

2.2 Modifications of Pachpatte’s inequalities | 71 t

Δ

≤ g(t)(x(0) + tx (0) + ∫(t − s)(f (s) + h(s)z(s))Δs 0

t

+ xΔ (0) + ∫(f (s) + h(s)z(s))Δs) 0

t

Δ

t

≤ g(t)(x(0) + tx (0) + t ∫ f (s)Δs + t ∫ h(s)z(s)Δs t

t

0

0

0

0

+ xΔ (0) + ∫ f (s)Δs + ∫ h(s)z(s)Δs) Δ

t

t

≤ g(t)(x(0) + tx (0) + t ∫ f (s)Δs + (t ∫ h(s)Δs)z(t) 0

t

0

t

+ xΔ (0) + ∫ f (s)Δs + (∫ h(s)Δs)z(t)) 0

0

= p(t) + q(t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, using Lemma 2.1.1, we obtain

t

z(t) ≤ z(0)eq (t, 0) + ∫ p(s)e⊖q (σ(s), t)Δs 0

t

= ∫ p(s)e⊖q (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

and

2

xΔ (t) ≤ f (t) + h(t)z(t) t

≤ f (t) + h(t) ∫ p(s)e⊖q (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. 2

Theorem 2.2.3. Let x, xΔ , xΔ , f , g, h ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions and Δ2

t

Δ

x (t) ≤ f (t) + h(t)(x (t) + ∫ g(s)(x(s) + x Δ (s))Δs),

t ∈ ℝ+ ∩ 𝕋.

0

Then

2

x Δ (t) ≤ f (t) + xΔ (0)h(t)eq (t, 0) t

+ h(t) ∫ p(s)e⊖q (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋,

72 | 2 Linear integro-dynamic inequalities where

t

p(t) = f (t) + g(t)x(0) + (1 + t)g(t)x Δ (0) + (1 + t)g(t) ∫ f (s)Δs, 0

t

q(t) = h(t) + (1 + t)g(t) ∫ h(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Let t

z(t) = xΔ (t) + ∫ g(s)(x(s) + x Δ (s))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then

2

x Δ (t) ≤ f (t) + h(t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Applying Lemma 2.2.1, we get t

x Δ (t) ≤ xΔ (0) + ∫(f (s) + h(s)z(s))Δs, 0

t s

Δ

x(t) ≤ x(0) + tx (0) + ∫ ∫(f (y) + h(y)z(y))ΔyΔs 0 0 t

= x(0) + tx Δ (0) + ∫(t − σ(s))(f (s) + h(s)z(s))Δs 0

t

Δ

≤ x(0) + tx (0) + ∫(t − s)(f (s) + h(s)z(s))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Therefore, since z(t) is a nondecreasing function on ℝ+ ∩ 𝕋, we obtain 2

z Δ (t) = x Δ (t) + g(t)(x(t) + xΔ (t)) ≤ f (t) + h(t)z(t)

t

Δ

+ g(t)x(0) + tg(t)x (0) + tg(t) ∫ f (s)Δs 0

t

+ tg(t)(∫ h(s)Δs)z(t) + g(t)xΔ (0) t

0

t

+ g(t) ∫ f (s)Δs + g(t)(∫ h(s)Δs)z(t) 0

= p(t) + q(t)z(t),

0

t ∈ ℝ+ ∩ 𝕋.

2.3 Inequalities with several iterated integrals | 73

Hence, by Lemma 2.1.1, we obtain t

z(t) ≤ z(0)eq (t, 0) + ∫ p(s)e⊖q (σ(s), t)Δs 0

t

= xΔ (0)eq (t, 0) + ∫ p(s)e⊖q (σ(s), t)Δs, 0

Δ2

Δ

x (t) ≤ f (t) + x (0)h(t)eq (t, 0) t

+ h(t) ∫ p(s)e⊖q (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof.

2.3 Inequalities with several iterated integrals Theorem 2.3.1. Let x, h ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, r ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive function, x0 be a positive constant, and t

t1

x(t) ≤ x0 + ∫ r(t1 ) ∫ h(t2 )x(t2 )Δt2 Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0

Then x(t) ≤ x0 eg (t, 0),

t ∈ ℝ+ ∩ 𝕋,

where t

g(t) = r(t) ∫ h(t1 )Δt1 ,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Let t

t1

v(t) = x0 + ∫ r(t1 ) ∫ h(t2 )x(t2 )Δt2 Δt1 , 0

0

Then v(0) = x0 ,

x(t) ≤ v(t),

t ∈ ℝ+ ∩ 𝕋,

t ∈ ℝ+ ∩ 𝕋.

74 | 2 Linear integro-dynamic inequalities and t

Δ

v (t) = r(t) ∫ h(t2 )x(t2 )Δt2 ,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, vΔ (0) = 0, t

vΔ (t) = ∫ h(t2 )x(t2 )Δt2 r(t) 0

t

≤ ∫ h(t2 )v(t2 )Δt2 ,

t ∈ ℝ+ ∩ 𝕋.

0

Note that v is a nondecreasing function on ℝ+ ∩ 𝕋. Therefore Δ

(

vΔ (t) ) ≤ h(t)v(t), r(t)

or Δ

(t) Δ ( vr(t) ) v(t)

v(t)v(σ(t))

≤ h(t) ≤ h(t) +

(vΔ (t))2 r(t)

v(t)v(σ(t))

,

t ∈ ℝ+ ∩ 𝕋.

From here, Δ

(t) Δ ( vr(t) ) v(t) −

(vΔ (t))2 r(t)

v(t)v(σ(t))

or

Δ

(

vΔ (t) ) ≤ h(t), r(t)v(t)

≤ h(t),

t ∈ ℝ+ ∩ 𝕋.

Consequently, t

vΔ (t) ≤ ∫ h(t1 )Δt1 , r(t)v(t) 0

or t

Δ

v (t) ≤ v(t)r(t) ∫ h(t1 )Δt1 0

= v(t)g(t),

t ∈ ℝ+ ∩ 𝕋.

2.3 Inequalities with several iterated integrals | 75

Hence, by Lemma 2.1.1, we obtain v(t) ≤ v(0)eg (t, 0) = x0 eg (t, 0)

and x(t) ≤ v(t)

≤ x0 eg (t, 0),

t ∈ ℝ+ ∩ 𝕋.

This completes the proof. Theorem 2.3.2. Let x, h ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, g ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive function, x0 be a positive constant, and t

x(t) ≤ x0 + ∫ 0

t1

1 ∫ h(t2 )x(t2 )Δt2 Δt1 , g(t1 )

t ∈ ℝ+ ∩ 𝕋.

0

Then x(t) ≤ x0 ep (t, 0),

t ∈ ℝ+ ∩ 𝕋,

where t

p(t) =

∫0 h(s)Δs g(t)

,

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t

z(t) = x0 + ∫ 0

t1

1 ∫ h(t2 )x(t2 )Δt2 Δt1 , g(t1 )

t ∈ ℝ+ ∩ 𝕋.

0

Then z(0) = x0 and x(t) ≤ z(t), z Δ (t) =

t

1 ∫ h(t2 )x(t2 )Δt2 , g(t) t

0

g(t)z Δ (t) = ∫ h(t2 )x(t2 )Δt2 , 0

t ∈ ℝ+ ∩ 𝕋.

76 | 2 Linear integro-dynamic inequalities Hence, Δ

(g(t)z Δ (t)) = h(t)x(t)

≤ h(t)z(t),

or (g(t)z Δ (t))Δ ≤ h(t), z(t)

t ∈ ℝ+ ∩ 𝕋.

Note that z(t) is a positive nondecreasing function on ℝ+ ∩ 𝕋. Therefore (g(t)z Δ (t))Δ z(t) (g(t)z Δ (t))Δ ≤ z(t)z(σ(t)) z(t) ≤ h(t) +

g(t)(z Δ (t))2 , z(t)z(σ(t))

t ∈ ℝ+ ∩ 𝕋.

Consequently, Δ

(

g(t)z Δ (t) ) ≤ h(t), z(t)

t ∈ ℝ+ ∩ 𝕋.

From here, t

g(t)z Δ (t) ≤ ∫ h(s)Δs, z(t) 0

or Δ

t

z (t) ≤

∫0 h(s)Δs

g(t) = p(t)z(t),

z(t) t ∈ ℝ+ ∩ 𝕋.

Hence, employing Lemma 2.1.1, we obtain z(t) ≤ z(0)ep (t, 0) = x0 ep (t, 0),

t ∈ ℝ+ ∩ 𝕋.

Consequently, x(t) ≤ z(t)

≤ x0 ep (t, 0),

This completes the proof.

t ∈ ℝ+ ∩ 𝕋.

2.3 Inequalities with several iterated integrals | 77

Theorem 2.3.3. Let x, h ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, g ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive function, x0 be a nonnegative constant, and t

t1 τ1

0

0 0

1 x(t) ≤ x0 + ∫ ∫ ∫ h(t2 )x(t2 )Δt2 Δt1 , g(t1 )

t ∈ ℝ+ ∩ 𝕋.

Then t

t1

s

0

0

0

t s x(t) ≤ x0 (1 + ∫ 1 ∫ h(s)(1 + ∫ h(y)e⊖f (σ(y), s)Δy)Δs1 Δt1 + 1), g(t1 ) g(s) t ∈ ℝ+ ∩ 𝕋, where f (t) =

t (h(t) + 1), g(t)

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t

z(t) = x0 + ∫ 0

t1 τ1

1 ∫ ∫ h(t2 )x(t2 )Δt2 Δt1 , g(t1 )

t ∈ ℝ+ ∩ 𝕋.

0 0

Then x(t) ≤ z(t),

z(0) = x0 , and t t1

1 z (t) = ∫ ∫ h(t2 )x(t2 )Δt2 Δt1 g(t) Δ

0 0

t t1

1 ≤ ∫ ∫ h(t2 )z(t2 )Δt2 Δt1 g(t) 0 0 t

1 = ∫(t − σ(t1 ))h(t1 )z(t1 )Δt1 g(t) 0

t

≤

t ∫ h(t1 )z(t1 )Δt1 g(t) 0

t

t ≤ ∫ h(t1 )(z Δ (t1 ) + z(t1 ))Δt1 , g(t) 0

t ∈ ℝ+ ∩ 𝕋.

78 | 2 Linear integro-dynamic inequalities Hence, by Theorem 2.1.2, we get t

s

t s z (t) ≤ z(0) ∫ h(y)e⊖f (σ(y), s)Δy)Δs ∫ h(s)(z(0) + g(t) g(s) Δ

0

= z(0)

t

t s ∫ h(s)(1 + ∫ h(y)e⊖f (σ(y), s)Δy)Δs g(t) g(s) 0

t

= x0

0

s

s

0

t s ∫ h(s)(1 + ∫ h(y)e⊖f (σ(y), s)Δy)Δs, g(t) g(s) 0

t ∈ ℝ+ ∩ 𝕋.

0

Then t

t1

s

0

0

0

t

t1

s

0

0

0

t s z(t) ≤ x0 (1 + ∫ 1 ∫ h(s)(1 + ∫ h(y)e⊖f (σ(y), s)Δy)ΔsΔt1 ) g(t1 ) g(s) and x(t) ≤ z(t)

t s ≤ x0 (1 + ∫ 1 ∫ h(s)(1 + ∫ h(y)e⊖f (σ(y), s)Δy)ΔsΔt1 ), g(t1 ) g(s) t ∈ ℝ+ ∩ 𝕋. This completes the proof. Theorem 2.3.4. Let x, p, q ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, r ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive function, x0 be a nonnegative constant, and t1

t

t3

t2

x(t) ≤ x0 + ∫ r(t1 ) ∫ p(t2 )(x(t2 ) + ∫ r(t3 ) ∫ q(t4 )x(t4 )Δt4 Δt3 )Δt2 Δt1 , 0

0

0

0

t ∈ ℝ+ ∩ 𝕋. Then t

t1

t2

x(t) ≤ x0 (1 + ∫ r(t1 ) ∫ p(t2 )(1 + ∫ g(t3 )Δt3 )Δt2 Δt1 ), 0

0

t ∈ ℝ+ ∩ 𝕋,

0

where f (t) = r(t)(p(t) + q(t) + 1), t

s

g(t) = 1 + r(t) ∫(p(s) + q(s))(1 + r(s) ∫(p(τ) + q(τ))e⊖f (σ(τ), s)Δτ)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

0

2.3 Inequalities with several iterated integrals | 79

Proof. Let t1

t

t3

t2

z(t) = x0 + ∫ r(t1 ) ∫ p(t2 )(x(t2 ) + ∫ r(t3 ) ∫ q(t4 )x(t4 )Δt4 Δt3 )Δt2 Δt1 , 0

0

0

t ∈ ℝ+ ∩ 𝕋.

0

Then z(0) = x0 ,

x(t) ≤ z(t),

t ∈ ℝ+ ∩ 𝕋,

and t1

t

t2

z Δ (t) = r(t) ∫ p(t1 )(x(t1 ) + ∫ r(t2 ) ∫ q(t3 )x(t3 )Δt3 Δt2 )Δt1 , 0

0

(2.8)

0

and t1

t

t2

z Δ (t) = ∫ p(t1 )(x(t1 ) + ∫ r(t2 ) ∫ q(t3 )x(t3 )Δt3 Δt2 )Δt1 , r(t) 0

0

Δ

(

t

t1

0

0

0

z Δ (t) ) = p(t)(x(t) + ∫ r(t1 ) ∫ q(t2 )x(t2 )Δt2 Δt1 ) r(t) t1

t

≤ p(t)(z(t) + ∫ r(t1 ) ∫ q(t2 )z(t2 )Δt2 Δt1 ), 0

t ∈ ℝ+ ∩ 𝕋.

0

Let t

t1

v(t) = z(t) + ∫ r(t1 ) ∫ q(t2 )z(t2 )Δt2 Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0

Hence, v(0) = z(0) = x0 ,

z(t) ≤ v(t),

t ∈ ℝ+ ∩ 𝕋,

and Δ

Δ

t

v (t) = z (t) + r(t) ∫ q(t1 )z(t1 )Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

(2.9)

80 | 2 Linear integro-dynamic inequalities By (2.8), we get t1

t

Δ

t2

z (t) ≤ r(t) ∫ p(t1 )(z(t1 ) + ∫ r(t2 ) ∫ q(t3 )z(t3 )Δt3 Δt2 )Δt1 0

0

t

= r(t) ∫ p(t1 )v(t1 )Δt1 ,

0

t ∈ ℝ+ ∩ 𝕋.

0

Hence, from (2.9), we obtain t

t

0

0

vΔ (t) ≤ r(t) ∫ p(t1 )v(t1 )Δt1 + r(t) ∫ q(t1 )v(t1 )Δt1 t

= r(t) ∫(p(t1 ) + q(t1 ))v(t1 )Δt1 0

t

≤ r(t) ∫(p(t1 ) + q(t1 ))(v(t1 ) + vΔ (t1 ))Δt1 ,

t ∈ ℝ+ ∩ 𝕋.

0

Hence Theorem 2.1.2 now yields s

t

vΔ (t) ≤ v(0) + r(t) ∫(p(s) + q(s))(v(0) + r(s)v(0) ∫(p(τ) + q(τ))e⊖f (σ(τ), s)Δτ) 0

t

s

0

= v(0)(1 + r(t) ∫(p(s) + q(s))(1 + r(s) ∫(p(τ) + q(τ))e⊖f (σ(τ), s)Δτ)Δs) t

0

s

0

= x0 (1 + r(t) ∫(p(s) + q(s))(1 + r(s) ∫(p(τ) + q(τ))e⊖f (σ(τ), s)Δτ)Δs) = x0 g(t),

0

t ∈ ℝ+ ∩ 𝕋.

Hence, t

v(t) ≤ v(0) + x0 ∫ g(s)Δs t

0

= x0 (1 + ∫ g(s)Δs), 0 Δ

Δ

t

z (t) ( ) ≤ x0 p(t)(1 + ∫ g(s)Δs), r(t) 0

0

2.3 Inequalities with several iterated integrals | 81 t1

t

z Δ (t) ≤ x0 ∫ p(t1 )(1 + ∫ g(t2 )Δt2 )Δt1 , r(t) 0

0

t

Δ

t1

z (t) ≤ x0 r(t) ∫ p(t1 )(1 + ∫ g(t2 )Δt2 )Δt1 , 0

t1

t

0

t2

z(t) ≤ x0 + x0 ∫ r(t1 ) ∫ p(t2 )(1 + ∫ g(t3 )Δt3 )Δt2 Δt1 0

0

0

t1

t

t2

= x0 (1 + ∫ r(t1 ) ∫ p(t2 )(1 + ∫ g(t3 )Δt3 )Δt2 Δt1 ), x(t) ≤ z(t)

0

0

0

t

t1

t2

≤ x0 (1 + ∫ r(t1 ) ∫ p(t2 )(1 + ∫ g(t3 )Δt3 )Δt2 Δt1 ) t ∈ ℝ+ ∩ 𝕋. 0

0

0

This completes the proof. Theorem 2.3.5. Let x, q ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, p, r ∈ 𝒞rd (ℝ+ ∩ 𝕋) be positive functions, x0 be a nonnegative constant, and t1 t2

t

x(t) ≤ x0 + ∫ r(t1 ) ∫ ∫ p(t3 )(x(t3 ) + t3

0

0 0

t4 t5

+ ∫ r(t4 ) ∫ ∫ q(t6 )x(t6 )Δt6 Δt5 Δt4 )Δt3 Δt2 Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0 0

Then t

t1

t2

x(t) ≤ x0 (1 + ∫ r(t1 ) ∫(t1 − t2 )p(t2 )(1 + ∫ g(t3 )Δt3 )Δt2 Δt1 ), 0

0

0

t ∈ ℝ+ ∩ 𝕋, where f (t) = tr(t)(p(t) + q(t) + 1), t

t1

g(t) = tr(t) ∫(p(t1 ) + q(t1 ))(1 + t1 r(t1 ) ∫(p(t2 ) + q(t2 ))e⊖f (σ(t2 ), t1 )Δt2 )Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0

82 | 2 Linear integro-dynamic inequalities Proof. Let t1 t2

t

z(t) = x0 + ∫ r(t1 ) ∫ ∫ p(t3 )(x(t3 ) + 0 0

0

t3

t4 t5

+ ∫ r(t4 ) ∫ ∫ q(t6 )x(t6 )Δt6 Δt5 Δt4 )Δt3 Δt2 Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0 0

Then z(0) = x0 and t t1

Δ

t3 t4

t2

z (t) = r(t) ∫ ∫ p(t2 )(x(t2 ) + ∫ r(t3 ) ∫ ∫ q(t5 )x(t5 )Δt5 Δt4 Δt3 )Δt2 Δt1 , 0 0

0

(2.10)

0 0

t ∈ ℝ+ ∩ 𝕋. Also, t t1

t2

t3 t4

0

0 0

z Δ (t) = ∫ ∫ p(t2 )(x(t2 ) + ∫ r(t3 ) ∫ ∫ q(t5 )x(t5 )Δt5 Δt4 Δt3 )Δt2 Δt1 , r(t) 0 0

Δ2

Δ

(

z (t) ) r(t)

t1 t2

= p(t)(x(t) + ∫ r(t1 ) ∫ ∫ q(t3 )x(t3 )Δt3 Δt2 Δt1 )Δt, 0 t

2

Δ

1 z Δ (t) ( ) p(t) r(t)

t

t1 t2

0 0

= x(t) + ∫ r(t1 ) ∫ ∫ q(t3 )x(t3 )Δt3 Δt2 Δt1 0 t

0 0

t1 t2

≤ z(t) + ∫ r(t1 ) ∫ ∫ q(t3 )z(t3 )Δt3 Δt2 Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0 0

Let t

t1 t2

v(t) = z(t) + ∫ r(t1 ) ∫ ∫ q(t3 )z(t3 )Δt3 Δt2 Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0 0

Then Δ

Δ

t t1

v (t) = z (t) + r(t) ∫ ∫ q(t2 )z(t2 )Δt2 Δt1 ,

t ∈ ℝ+ ∩ 𝕋.

0 0

By (2.10), we get Δ

t t1

t2

t3 t4

z (t) ≤ r(t) ∫ ∫ p(t2 )(z(t2 ) + ∫ r(t3 ) ∫ ∫ q(t5 )z(t5 )Δt5 Δt4 Δt3 )Δt2 Δt1 0 0

0

0 0

(2.11)

2.3 Inequalities with several iterated integrals | 83 t t1

= r(t) ∫ ∫ p(t2 )v(t2 )Δt2 Δt1 ,

t ∈ ℝ+ ∩ 𝕋.

0 0

Hence, from (2.11), we obtain t t1

vΔ (t) ≤ r(t) ∫ ∫ p(t2 )v(t2 )Δt2 Δt1 0 0

t t1

+ r(t) ∫ ∫ q(t2 )v(t2 )Δt2 Δt1 0 0

t t1

= r(t) ∫ ∫(p(t2 ) + q(t2 ))v(t2 )Δt2 Δt1 0 0 t

= r(t) ∫(t − σ(t1 ))(p(t1 ) + q(t1 ))v(t1 )Δt1 0

t

≤ tr(t) ∫(p(t1 ) + q(t1 ))v(t1 )Δt1 0

t

≤ tr(t) ∫(p(t1 ) + q(t1 ))(v(t1 ) + vΔ (t1 ))Δt1 ,

t ∈ ℝ+ ∩ 𝕋.

0

Theorem 2.1.2 now implies t1

t

Δ

v (t) ≤ tr(t) ∫(p(t1 ) + q(t1 ))(v(0) + t1 r(t1 )v(0) ∫(p(t2 ) + q(t2 ))e⊖f (σ(t2 ), t1 )Δt2 )Δt1 0

t1

t

0

= x0 tr(t) ∫(p(t1 ) + q(t1 ))(1 + t1 r(t1 ) ∫(p(t2 ) + q(t2 ))e⊖f (σ(t2 ), t1 )Δt2 )Δt1 = x0 g(t),

0

0

t ∈ ℝ+ ∩ 𝕋.

Therefore t

v(t) ≤ x0 + x0 ∫ g(t1 )Δt1 0

t

= x0 (1 + ∫ g(t1 )Δt1 ), 1 z Δ (t) ( ) p(t) r(t)

Δ2

0

≤ v(t)

84 | 2 Linear integro-dynamic inequalities t

≤ x0 (1 + ∫ g(t1 )Δt1 ), 0 Δ2

(

z Δ (t) ) r(t)

t

≤ x0 p(t)(1 + ∫ g(t1 )Δt1 ), 0

t t1

Δ

t2

z (t) ≤ x0 ∫ ∫ p(t2 )(1 + ∫ g(t3 )Δt3 )Δt2 Δt1 r(t) 0 0

0

t1

t

= x0 ∫(t − σ(t1 ))p(t1 )(1 + ∫ g(t2 )Δt2 )Δt1 , 0

0

t1

t

Δ

z (t) ≤ x0 r(t) ∫(t − t1 )p(t1 )(1 + ∫ g(t2 )Δt2 )Δt1 , 0

0

t1

t

t2

z(t) ≤ x0 + x0 ∫ r(t1 ) ∫(t1 − t2 )p(t2 )(1 + ∫ g(t3 )Δt3 )Δt2 Δt1 , 0

0

0

t ∈ ℝ+ ∩ 𝕋. Then x(t) ≤ z(t)

t1

t

t2

≤ x0 (1 + ∫ r(t1 ) ∫(t1 − t2 )p(t2 )(1 + ∫ g(t3 )Δt3 )Δt2 Δt1 ), 0

0

0

t ∈ ℝ+ ∩ 𝕋. This completes the proof. Theorem 2.3.6. Let x, f , g, h ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, r ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive function such that r(t) ≥ 1, t ∈ ℝ+ ∩ 𝕋, and t

t1

x(t) ≤ f (t) + g(t) ∫ r(t1 ) ∫ h(t2 )x(t2 )Δt2 Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0

Then t

t1

x(t) ≤ f (t) + g(t) ∫(h(t1 ) + r(t1 ) ∫ h4 (t2 )Δt2 )Δt1 , 0

where t

h1 (t) = r(t) ∫ h(t1 )f (t1 )Δt1 , 0

h2 (t) = h(t)g(t),

0

t ∈ ℝ+ ∩ 𝕋,

2.3 Inequalities with several iterated integrals | 85

h3 (t) = r(t)(h2 (t) + 1), t

t

s

h4 (t) = h2 (t)(∫ h1 (s)Δs + h1 (t) + r(t) ∫ h2 (s)(∫ h1 (y)Δy 0

0

+ h1 (s))e⊖h3 (σ(s), t)Δs),

0

t ∈ ℝ+ ∩ 𝕋.

Proof. Let t1

t

z(t) = ∫ r(t1 ) ∫ h(t2 )x(t2 )Δt2 Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0

Then z(0) = 0 and x(t) ≤ f (t) + g(t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Next we can estimate t

z Δ (t) = r(t) ∫ h(t1 )x(t1 )Δt1 0

t

≤ r(t) ∫ h(t1 )(f (t1 ) + g(t1 )z(t1 ))Δt1 0

t

t

= r(t) ∫ h(t1 )f (t1 )Δt1 + r(t) ∫ h(t1 )g(t1 )z(t1 )Δt1 0

0

t

≤ h1 (t) + r(t) ∫ h2 (t1 )(z(t1 ) + z Δ (t1 ))Δt1 ,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, using Theorem 2.1.2, we obtain t1

t

Δ

t1

z (t) ≤ h1 (t) + r(t) ∫ h2 (t1 )(z(0) + ∫ h1 (s)Δs + h1 (t1 ) + r(t1 ) ∫ h2 (s)(z(0) s

0

0

+ ∫ h1 (y)Δy + h1 (s))e⊖h3 (σ(s), t1 )Δs)Δt1 0

t

t1

= h1 (t) + r(t) ∫ h2 (t1 )(∫ h1 (s)Δs + h1 (t1 ) 0

0

0

86 | 2 Linear integro-dynamic inequalities t1

s

+ r(t1 ) ∫ h2 (s)(∫ h1 (y)Δy + h1 (s))e⊖h3 (σ(s), t1 )Δs)Δt1 0

0

t

= h1 (t) + r(t) ∫ h4 (t1 )Δt1 , 0

t1

t

z(t) ≤ ∫(h1 (t1 ) + r(t1 ) ∫ h4 (t2 )Δt2 )Δt1 , 0

0

t1

t

x(t) ≤ f (t) + g(t) ∫(h(t1 ) + r(t1 ) ∫ h4 (t2 )Δt2 )Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 2.3.7. Let f , q, p, x ∈ 𝒞rd (ℝ+ ∩ 𝕋) be nonnegative functions, g ∈ 𝒞rd (ℝ+ ∩ 𝕋) be a positive function such that tg(t) ≥ 1, t ∈ ℝ+ ∩ 𝕋, and t1 τ1

t

x(t) ≤ f (t) + q(t) ∫ g(t1 ) ∫ ∫ p(τ2 )x(τ2 )Δτ2 Δτ1 Δt1 , 0

t ∈ ℝ+ ∩ 𝕋.

0 0

Then t

x(t) ≤ f (t) + g(t) ∫ h4 (s)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

where t t1

h1 (t) = g(t) ∫ ∫ p(t2 )f (t2 )Δt2 Δt1 , 0 0

h2 (t) = p(t)g(t),

h3 (t) = tg(t)(h2 (t) + 1),

t1

t

h4 (t) = h1 (t) + tg(t) ∫ h2 (t1 )(∫ h1 (s)Δs + h1 (t1 ) t1

0

0 s

+ t1 g(t1 ) ∫ h2 (s)(∫ h1 (y)Δy + h1 (s))e⊖h3 (σ(s), t)Δs)Δt1 , 0

0

Proof. Let t

t1 τ1

z(t) = ∫ g(t1 ) ∫ ∫ p(τ2 )x(τ2 )Δτ2 Δτ1 Δt1 , 0

0 0

t ∈ ℝ+ ∩ 𝕋.

t ∈ ℝ+ ∩ 𝕋.

2.3 Inequalities with several iterated integrals | 87

Then z(0) = 0 and x(t) ≤ f (t) + g(t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Next we can write t t1

z Δ (t) = g(t) ∫ ∫ p(t2 )x(t2 )Δt2 Δt1 0 0

t t1

≤ g(t) ∫ ∫ p(t2 )f (t2 )Δt2 Δt1 0 0

t t1

+ g(t) ∫ ∫ p(t2 )g(t2 )z(t2 )Δt2 Δt1 0 0

t

= h1 (t) + g(t) ∫(t − σ(t1 ))p(t1 )g(t1 )z(t1 )Δt1 0

t

≤ h1 (t) + tg(t) ∫ p(t1 )g(t1 )z(t1 )Δt1 0

t

= h1 (t) + tg(t) ∫ h2 (t1 )z(t1 )Δt1 0

t

≤ h1 (t) + tg(t) ∫ h2 (t1 )(z(t1 ) + z Δ (t1 ))Δt1 ,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, by Theorem 2.1.2, we get t

Δ

t1

z (t) ≤ h1 (t) + tg(t) ∫ h2 (t1 )(∫ h1 (s)Δs + h1 (t1 ) t1

0

0

s

+ t1 g(t1 ) ∫(∫ h1 (y)Δy + h1 (s))e⊖h3 (σ(s), t)Δs)Δt1 = h4 (t),

0

0

t ∈ ℝ+ ∩ 𝕋.

Therefore t

z(t) ≤ ∫ h4 (s)Δs, 0

88 | 2 Linear integro-dynamic inequalities x(t) ≤ f (t) + g(t)z(t) t

≤ f (t) + g(t) ∫ h4 (s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 2.3.8. Let n ≥ 2, x, fk ∈ 𝒞rd (ℝ+ ∩ 𝕋), k ∈ {1, . . . , n}, be nonnegative functions, pl ∈ 𝒞rd (ℝ+ ∩ 𝕋), l ∈ {1, . . . , n}, be positive functions, u0 be a positive constant, and n

t

k=1

0

sk−2

s1

x(t) ≤ u0 + ∑ (∫ p1 (s1 ) ∫ p2 (s2 ) ⋅ ⋅ ⋅ ∫ pk−1 (sk−1 ) sk−1

0

0

× ∫ pk (sk )fk (sk )x(sk )Δsk Δsk−1 ⋅ ⋅ ⋅ Δs2 Δs1 ), 0

t ∈ ℝ+ ∩ 𝕋. Then x(t) ≤ u0 eq1 (t, 0),

t ∈ ℝ+ ∩ 𝕋,

where n

t

k=2

0

sk−2

s2

sk−1

q(t) = ∑ (∫ p2 (s2 ) ∫ ⋅ ⋅ ⋅ ∫ pk−1 (sk−1 ) ∫ pk (sk )fk (sk )Δsk Δsk−1 ⋅ ⋅ ⋅ Δs3 Δs2 ), 0

0

0

q1 (t) = f1 (t)p1 (t) + p1 (t)q(t),

t ∈ ℝ+ ∩ 𝕋.

Proof. Let n

t

k=1

0

sk−2

s1

z(t) = u0 + ∑ (∫ p1 (s1 ) ∫ p2 (s2 ) ⋅ ⋅ ⋅ ∫ pk−1 (sk−1 ) sk−1

0

0

× ∫ pk (sk )fk (sk )x(sk )Δsk Δsk−1 ⋅ ⋅ ⋅ Δs2 Δs1 ), 0

t ∈ ℝ+ ∩ 𝕋. Then z(0) = u0 > 0, and x(t) ≤ z(t),

z Δ (t) = p1 (t)f1 (t)x(t) n

t

k=2

0

sk−2

+ p1 (t) ∑ (∫ p2 (s2 ) ⋅ ⋅ ⋅ ∫ pk−1 (sk−1 ) 0

2.3 Inequalities with several iterated integrals | 89 sk−1

× ∫ pk (sk )fk (sk )x(sk )Δsk Δsk−1 ⋅ ⋅ ⋅ Δs2 ), 0

t ∈ ℝ+ ∩ 𝕋. Let n

t

k=2

0

sk−2

sk−1

z1 (t) = ∑ (∫ p2 (s2 ) ⋅ ⋅ ⋅ ∫ pk−1 (sk−1 ) ∫ pk (sk )fk (sk )x(sk )Δsk Δsk−1 ⋅ ⋅ ⋅ Δs2 ), 0

0

t ∈ ℝ+ ∩ 𝕋. Hence, z Δ (t) = p1 (t)f1 (t)x(t) + p1 (t)z1 (t),

t ∈ ℝ+ ∩ 𝕋,

or z Δ (t) − f (t)x(t) = z1 (t), p1 (t) 1

t ∈ ℝ+ ∩ 𝕋.

z1Δ (t) − f (t)x(t) = z2 (t), p2 (t) 2

t ∈ ℝ+ ∩ 𝕋,

Similarly,

where n

t

k=3

0

sk−1

z2 (t) = ∑ (∫ p3 (s3 ) ⋅ ⋅ ⋅ ∫ pk (sk )fk (sk )x(sk )Δsk Δsk−1 ⋅ ⋅ ⋅ Δs3 ), 0

And so on, Δ zn−2 (t) − f (t)x(t) = zn−1 (t), pn−1 (t) n−1

t ∈ ℝ+ ∩ 𝕋,

where t

zn−1 (t) = ∫ pn (sn )fn (sn )x(sn )Δsn ,

t ∈ ℝ+ ∩ 𝕋.

0

Therefore Δ zn−1 (t) = pn (t)fn (t)x(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, Δ zn−1 (t) pn (t)fn (t)x(t) = z(t) z(t) ≤ pn (t)fn (t), t ∈ ℝ+ ∩ 𝕋.

t ∈ ℝ+ ∩ 𝕋.

90 | 2 Linear integro-dynamic inequalities Now, using that t

t

∫ 0

Δ

Δ zn−1 (s) z (s)z Δ (s) z (s) Δs = ∫(( n−1 ) + n−1 )Δs z(s) z(s) z(s)z(σ(s)) 0

t

Δ

≥ ∫( 0

=

zn−1 (s) ) Δs z(s)

zn−1 (t) , z(t)

t ∈ ℝ+ ∩ 𝕋,

we get t

z Δ (s) zn−1 (t) ≤ ∫ n−1 Δs z(t) z(s) 0

t

≤ ∫ pn (s)fn (s)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

and t

z Δ (s ) zn−2 (t) ≤ ∫ n−2 n−1 Δsn−1 z(t) z(sn−1 ) 0

t

=∫ 0

zn−1 (sn−1 )pn−1 (sn−1 ) + fn−1 (sn−1 )pn−1 (sn−1 )x(sn−1 ) Δsn−1 z(sn−1 )

t

≤ ∫ fn−1 (sn−1 )pn−1 (sn−1 )Δsn−1 0

t

+ ∫ pn−1 (sn−1 ) t

0

zn−1 (sn−1 ) Δsn−1 z(sn−1 )

≤ ∫ fn−1 (sn−1 )pn−1 (sn−1 )Δsn−1 0

t

sn−1

+ ∫ pn−1 (sn−1 ) ∫ pn (sn )fn (sn )Δsn Δsn−1 , 0

t ∈ ℝ+ ∩ 𝕋.

0

Similarly, t

s2

sk−2

0

0

0

n z Δ (t) − f1 (t)x(t) ≤ z(t) ∑ (∫ p2 (s2 ) ∫ p3 (s3 ) ⋅ ⋅ ⋅ ∫ pk−1 (sk−1 ) p1 (t) k=2

2.3 Inequalities with several iterated integrals | 91 sk−1

× ∫ pk (sk )fk (sk )Δsk Δsk−1 ⋅ ⋅ ⋅ Δs3 Δs2 ) 0

= z(t)q(t),

t ∈ ℝ+ ∩ 𝕋,

i. e., z Δ (t) x(t) ≤ f1 (t)p1 (t) + p1 (t)q(t) z(t) z(t) ≤ f1 (t)p1 (t) + p1 (t)q(t) = q1 (t),

t ∈ ℝ+ ∩ 𝕋,

or z Δ (t) ≤ q1 (t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, due to Lemma 2.1.1, we get z(t) ≤ z(0)eq1 (t, 0) = u0 eq1 (t, 0),

t ∈ ℝ+ ∩ 𝕋.

Consequently, x(t) ≤ z(t)

≤ u0 eq1 (t, 0),

This completes the proof.

t ∈ ℝ+ ∩ 𝕋.

3 Nonlinear integral inequalities This chapter is concerned with Dragomir type and Pachpatte type nonlinear integral inequalities. The results contained in this chapter can be found in [3, 9, 11, 17] and [18]. Let 𝕋 be a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively.

3.1 Dragomir’s inequalities Suppose that a, b ∈ 𝕋, a < b, J = [a, b] is a time scale interval. Theorem 3.1.1 (Dragomir’s Inequality). Let x, f , g ∈ 𝒞 (J) be nonnegative functions, L ∈ 𝒞 (J × ℝ+ ) be a nonnegative function such that 0 ≤ L(t, x) − L(t, y) ≤ M(t, y)(x − y) for any t ∈ J and for any x ≥ y ≥ 0, where M ∈ 𝒞 (J × ℝ) is a nonnegative function. If t

x(t) ≤ f (t) + g(t) ∫ L(s, x(s))Δs, a

t ∈ J,

then t

x(t) ≤ f (t) + g(t) ∫ h1 (s)e⊖h2 (σ(s), t)Δs, a

t ∈ J,

where h1 (t) = L(t, f (t)),

h2 (t) = g(t)M(t, f (t)),

t ∈ J.

Proof. Let t

z(t) = ∫ L(s, x(s))Δs,

t ∈ J.

x(t)≤ f (t) + g(t)z(t),

t ∈ J,

a

Then

and z(0) = 0. https://doi.org/10.1515/9783110705553-003

(3.1)

94 | 3 Nonlinear integral inequalities We have z Δ (t) = L(t, x(t))

≤ L(t, f (t) + g(t)z(t))

= L(t, f (t) + g(t)z(t)) − L(t, f (t)) + L(t, f (t))

≤ M(t, f (t))(f (t) + g(t)z(t) − f (t)) + L(t, f (t))

= L(t, f (t)) + g(t)M(t, f (t))z(t) ≤ h1 (t) + h2 (t)z(t),

t ∈ J.

Hence, by Lemma 2.1.1, we get t

z(t) ≤ ∫ h1 (s)e⊖h2 (σ(s), t)Δs, a

t ∈ J.

From here and (3.1), we obtain t

x(t) ≤ f (t) + g(t) ∫ h1 (s)e⊖h2 (σ(s), t)Δs,

t ∈ J.

a

This completes the proof. Theorem 3.1.2 (Dragomir’s Inequality). Let x, f , g ∈ 𝒞 (J) be nonnegative functions, L ∈ 𝒞 (J × ℝ+ ) be a nonnegative function, ψ ∈ 𝒞 (ℝ+ ) be a nonnegative strictly increasing

function with ψ(0) = 0 and

0 ≤ L(t, x) − L(t, y) ≤ M(t, y)ψ−1 (x − y) for any t ∈ J and for any x ≥ y ≥ 0, where M ∈ 𝒞 (J × ℝ+ ) is a nonnegative function and ψ−1 is the inverse of ψ. If t

x(t) ≤ f (t) + ψ(g(t) ∫ L(s, x(s))Δs), a

t ∈ J,

then t

x(t) ≤ f (t) + ψ(g(t) ∫ h1 (s)e⊖h2 (σ(s), t)Δs), a

where h1 and h2 are defined as in Theorem 3.1.1. Proof. Let t

z(t) = ∫ L(s, x(s))Δs, a

t ∈ J.

t ∈ J,

3.2 Pachpatte’s inequalities | 95

Then x(t) ≤ f (t) + ψ(g(t)z(t)),

t ∈ J,

(3.2)

and z(0) = 0. We have z Δ (t) = L(t, x(t))

≤ L(t, f (t) + ψ(g(t)z(t)))

= L(t, f (t) + ψ(g(t)z(t))) − L(t, f (t)) + L(t, f (t))

≤ M(t, f (t))ψ−1 (f (t) + ψ(g(t))z(t) − f (t)) + L(t, f (t)) = M(t, f (t))ψ−1 (ψ(g(t)z(t))) + h1 (t) = h1 (t) + M(t, f (t))g(t)z(t) = h1 (t) + h2 (t)z(t),

t ∈ J.

Hence, by Lemma 2.1.1, we get t

z(t) ≤ ∫ h1 (s)e⊖h2 (σ(s), t)Δs, a

t ∈ J.

From here and from (3.2), we obtain t

x(t) ≤ f (t) + ψ(g(t) ∫ h1 (s)e⊖h2 (σ(s), t)Δs), a

t ∈ J.

This completes the proof.

3.2 Pachpatte’s inequalities In this section we suppose that 0 ∈ 𝕋. Theorem 3.2.1 (Pachpatte’s Inequality). Let x, f , g ∈ 𝒞 (ℝ+ ∩ 𝕋) be nonnegative functions, F ∈ 𝒞 (ℝ+ ) be nonnegative, strictly increasing, convex and submultiplicative function on (0, ∞), and lim F(u) = ∞,

u→∞

let F −1 denote the inverse function of F, and α, β ∈ 𝒞 (ℝ+ ∩ 𝕋) be positive functions such that α(t) + β(t) = 1,

t ∈ ℝ+ ∩ 𝕋.

96 | 3 Nonlinear integral inequalities Let also L ∈ 𝒞 ((ℝ+ ∩ 𝕋) × ℝ+ ) be a nonnegative function such that 0 ≤ L(t, x) − L(t, y) ≤ M(t, y)(x − y),

t ∈ ℝ+ ∩ 𝕋,

for any x ≥ y ≥ 0, where M ∈ 𝒞 ((ℝ+ ∩ 𝕋) × ℝ) is a nonnegative function. If t

x(t) ≤ f (t) + g(t)F (∫ L(s, F(x(s)))Δs),

t ∈ ℝ+ ∩ 𝕋,

−1

(3.3)

0

then t

x(t) ≤ f (t) + g(t)F −1 (∫ L(s, h5 (s))Δs),

t ∈ ℝ+ ∩ 𝕋,

0

where h1 (t) = α(t)F(f (t)(α(t))−1 ),

h2 (t) = β(t)F(g(t)(β(t))−1 ),

h3 (t) = L(t, h1 (t)),

h4 (t) = h2 (t)M(t, h1 (t)), t

h5 (t) = h1 (t) + h2 (t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. We rewrite inequality (3.3) in the following way: t

x(t) ≤ α(t)(f (t)(α(t))−1 ) + β(t)(g(t)(β(t))−1 )F −1 (∫ L(s, F(x(s)))Δs), 0

t ∈ ℝ+ ∩ 𝕋. Hence, since F is strictly increasing, convex and submultiplicative on (0, ∞), we get t

F(x(t)) ≤ F(α(t)(f (t)(α(t)) ) + β(t)(g(t)(β(t)) )F (∫ L(s, F(x(s)))Δs)) −1

−1

−1

0 t

≤ α(t)F(f (t)(α(t))−1 ) + β(t)F(y(t)(β(t))−1 )F −1 (∫ L(s, F(x(s)))Δs) 0

≤ α(t)F(f (t)(α(t))−1 )

t

+ β(t)F(g(t)(β(t)) )F(F (∫ L(s, F(x(s)))Δs)) −1

−1

0

3.2 Pachpatte’s inequalities | 97 t

= h1 (t) + h2 (t) ∫ L(s, F(x(s)))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Now we apply Theorem 3.1.1 for F(x(t)), t ∈ ℝ+ ∩ 𝕋, and get t

F(x(t)) ≤ h1 (t) + h2 (t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs 0

= h5 (t),

t ∈ ℝ+ ∩ 𝕋.

Hence, t

x(t) ≤ f (t) + g(t)F (∫ L(s, h5 (s))Δs), −1

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 3.2.2 (Pachpatte’s Inequality). Let x, f , g, L and M be as in Theorem 3.2.1. Let also H ∈ 𝒞 (ℝ+ ) be a positive, strictly increasing, subadditive and submultiplicative function on (0, ∞) with limy→∞ H(y) = ∞. If t

x(t) ≤ f (t) + g(t)H (∫ L(s, H(x(s)))Δs), −1

t ∈ ℝ+ ∩ 𝕋,

(3.4)

0

then t

x(t) ≤ H (h1 (t) + h2 (t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs), −1

t ∈ ℝ+ ∩ 𝕋,

0

where h1 (t) = H(f (t)),

h2 (t) = H(g(t)),

h3 (t) = L(t, h1 (t)),

h4 (t) = h2 (t)M(t, h1 (t)),

t ∈ ℝ+ ∩ 𝕋.

Proof. Since H is monotonic, subadditive and submultiplicative on (0, ∞), by inequality (3.4), we get t

H(x(t)) ≤ H(f (t) + g(t)H (∫ L(s, H(x(s)))Δs)) −1

0

98 | 3 Nonlinear integral inequalities t

≤ H(f (t)) + H(g(t)H (∫ L(s, H(x(s)))Δs)) −1

0

t

≤ H(f (t)) + H(g(t))H(H −1 (∫ L(s, H(x(s)))Δs)) t

0

= H(f (t)) + H(g(t)) ∫ L(s, H(x(s)))Δs t

0

= h1 (t) + h2 (t) ∫ L(s, H(x(s)))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Now we apply Theorem 3.1.1 for H(x(t)), t ∈ ℝ+ ∩ 𝕋, and get t

H(x(t)) ≤ h1 (t) + h2 (t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Since H −1 is increasing, we get t

x(t) ≤ H (h1 (t) + h2 (t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs), −1

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 3.2.3 (Pachpatte’s Inequality). Let x, f , g, L, and M be as in Theorem 3.2.1. Let also k ∈ 𝒞 ((ℝ+ ∩ 𝕋) × (ℝ+ ∩ 𝕋)) and suppose there exists ktΔ ∈ 𝒞 ((ℝ+ ∩ 𝕋) × (ℝ+ ∩ 𝕋)), such that k(t, s) and ktΔ (t, s) are nonnegative functions for 0 ≤ s ≤ t < ∞. If t

x(t) ≤ f (t) + g(t) ∫ k(t, s)(x(s) + L(s, x(s)))Δs,

t ∈ ℝ+ ∩ 𝕋,

0

then t

x(t) ≤ f (t) + g(t) ∫ A(s)e⊖B (σ(s), t)Δs, 0

where A(t) = k(σ(t), t)(f (t) + L(t, f (t))) t

+ ∫ ktΔ (t, s)(f (s) + L(s, f (s)))Δs, 0

t ∈ ℝ+ ∩ 𝕋,

3.2 Pachpatte’s inequalities | 99

B(t) = k(σ(t), t)(M(t, f (t)) + 1)g(t) t

+ ∫ ktΔ (t, s)g(s)(M(s, f (s)) + 1)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Let t

z(t) = ∫ k(t, s)(x(s) + L(s, x(s)))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then z(t) is a nondecreasing function on ℝ+ ∩ 𝕋, x(t) ≤ f (t) + g(t)z(t),

t ∈ ℝ+ ∩ 𝕋,

and z(0) = 0. We have z Δ (t) = k(σ(t), t)(x(t) + L(t, x(t))) t

+ ∫ ktΔ (t, s)(x(s) + L(s, x(s)))Δs 0

≤ k(σ(t), t)(f (t) + g(t)z(t) + L(t, f (t) + g(t)z(t)) − L(t, f (t)) + L(t, f (t))) t

+ ∫ ktΔ (t, s)(f (s) + g(s)z(s) + L(s, f (s) + g(s)z(s)) 0

− L(s, f (s)z(s)) + L(s, f (s)z(s)))Δs

≤ k(σ(t), t)(f (t) + g(t)z(t) + M(t, f (t))g(t)z(t) + L(t, f (t))) t

+ ∫ ktΔ (t, s)(f (s) + g(s)z(s) + M(s, f (s))g(s)z(s) + L(s, f (s)))Δs 0

= k(σ(t), t)(f (t) + L(t, f (t))) t

+ ∫ ktΔ (t, s)(f (s) + L(s, f (s)))Δs 0

+ k(σ(t), t)(M(t, f (t)) + 1)g(t)z(t) t

+ ∫ ktΔ (t, s)g(s)(M(s, f (s)) + 1)z(s)Δs 0

100 | 3 Nonlinear integral inequalities ≤ A(t) + k(σ(t), t)(M(t, f (t)) + 1)g(t)z(t) t

+ (∫ ktΔ (t, s)g(s)(M(s, f (s)) + 1)Δs)z(t) 0

= A(t) + B(t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, by Lemma 2.1.1, we get t

z(t) ≤ ∫ A(s)e⊖B (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

and t

x(t) ≤ f (t) + g(t) ∫ A(s)e⊖B (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 3.2.4. Let p ∈ (1, ∞), f ∈ 𝒞 (ℝ+ ∩ 𝕋) be a nonnegative function, x ∈ 𝒞 (ℝ+ ∩ 𝕋) be a positive function such that x(t) ≥ 1,

t

p

(x(t)) ≤ c1 + c2 ∫ f (y)x(y)Δy,

t ∈ ℝ+ ∩ 𝕋,

0

for some positive constants c1 and c2 . Then 1

x(t) ≤ (c1 ec2 f (t, 0)) p ,

t ∈ ℝ+ ∩ 𝕋.

Proof. Since x(t) ≥ 1, t ∈ ℝ+ ∩ 𝕋, we get p

x(t) ≤ (x(t)) ,

t ∈ ℝ+ ∩ 𝕋.

Let t

z(t) = c1 + c2 ∫ f (y)x(y)Δy,

t ∈ ℝ+ ∩ 𝕋.

0

Then z(0) = c1 , p

(x(t)) ≤ z(t),

t ∈ ℝ+ ∩ 𝕋,

3.2 Pachpatte’s inequalities | 101

and z Δ (t) = c2 f (t)x(t)

≤ c2 f (t)(x(t))

p

≤ c2 f (t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, by Lemma 2.1.1, we get z(t) ≤ z(0)ec2 f (t, 0) = c1 ec2 f (t, 0),

t ∈ ℝ+ ∩ 𝕋.

Therefore p

(x(t)) ≤ c1 ec2 f (t, 0),

t ∈ ℝ+ ∩ 𝕋,

whereupon 1

x(t) ≤ (c1 ec2 f (t, 0)) p ,

t ∈ ℝ+ ∩ 𝕋.

This completes the proof. Theorem 3.2.5. Let p ∈ (1, ∞), f ∈ 𝒞 (ℝ+ ∩ 𝕋) be a positive function, g, h1 , h2 ∈ 𝒞 (ℝ+ ∩ 𝕋) be nonnegative functions, x ∈ 𝒞 (ℝ+ ∩ 𝕋) be a positive function such that x(t) ≥ 1,

t

p

p

(x(t)) ≤ f (t) + g(t) ∫(h1 (s)(x(s)) + h2 (s)x(s))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then 1 p

t

x(t) ≤ (f (t) + g(t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs) ,

t ∈ ℝ+ ∩ 𝕋,

0

where h3 (t) = f (t)(h1 (t) + h2 (t)),

h4 (t) = g(t)(h1 (t) + h2 (t)),

t ∈ ℝ+ ∩ 𝕋.

Proof. Since x(t) ≥ 1, t ∈ ℝ+ ∩ 𝕋, we get p

x(t) ≤ (x(t)) ,

t ∈ ℝ+ ∩ 𝕋.

Let t

p

z(t) = ∫(h1 (s)(x(s)) + h2 (s)x(s))Δs, 0

t ∈ ℝ+ ∩ 𝕋.

102 | 3 Nonlinear integral inequalities Then p

(x(t)) ≤ f (t) + g(t)z(t),

t ∈ ℝ+ ∩ 𝕋,

and z(0) = 0. We have p

z Δ (t) = h1 (t)(x(t)) + h2 (t)x(t) ≤ (h1 (t) + h2 (t))(x(t))

p

≤ (h1 (t) + h2 (t))(f (t) + g(t)z(t))

= f (t)(h1 (t) + h2 (t)) + g(t)(h1 (t) + h2 (t))z(t)

= h3 (t) + h4 (t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, from Lemma 2.1.1, we get t

z(t) ≤ ∫ h3 (s)e⊖h4 (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then t

p

(x(t)) ≤ f (t) + g(t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

whereupon 1 p

t

x(t) ≤ (f (t) + g(t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs) ,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 3.2.6. Let p ∈ (1, ∞), x, f ∈ 𝒞 (ℝ+ ∩ 𝕋) be nonnegative functions, c1 and c2 be nonnegative constants. If x(t) ≤ 1, 1

t

(x(t)) p ≤ c1 + c2 ∫ f (s)x(s)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

then p

x(t) ≤ (c1 ec2 f (t, 0)) ,

t ∈ ℝ+ ∩ 𝕋.

3.2 Pachpatte’s inequalities | 103

Proof. Since 0 ≤ x(t) ≤ 1,

t ∈ ℝ+ ∩ 𝕋,

we have 1

x(t) ≤ (x(t)) p ,

t ∈ ℝ+ ∩ 𝕋.

Let t

z(t) = c1 + c2 ∫ f (s)x(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then z(0) = c1 , 1

(x(t)) p ≤ z(t),

z Δ (t) = c2 f (t)x(t)

1

≤ c2 f (t)(x(t)) p

≤ c2 f (t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, using Lemma 2.1.1, we get z(t) ≤ c1 ec2 f (t, 0),

t ∈ ℝ+ ∩ 𝕋.

Therefore 1

(x(t)) p ≤ c1 ec2 f (t, 0),

t ∈ ℝ+ ∩ 𝕋,

whereupon p

x(t) ≤ (c1 ec2 f (t, 0)) ,

t ∈ ℝ+ ∩ 𝕋.

This completes the proof. Theorem 3.2.7. Let p ∈ (1, ∞), x, f , g, h1 , h2 ∈ 𝒞 (ℝ+ ∩ 𝕋) be nonnegative functions. If x(t) ≤ 1, 1

t

1

(x(t)) p ≤ f (t) + g(t) ∫(h1 (s)(x(s)) p + h2 (s)x(s))Δs,

t ∈ ℝ+ ∩ 𝕋,

0

then t

p

x(t) ≤ (f (t) + g(t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs) , 0

t ∈ ℝ+ ∩ 𝕋,

104 | 3 Nonlinear integral inequalities where h3 (t) = f (t)(h1 (t) + h2 (t)),

h4 (t) = g(t)(h1 (t) + h2 (t)),

t ∈ ℝ+ ∩ 𝕋.

Proof. Since 0 ≤ x(t) ≤ 1,

t ∈ ℝ+ ∩ 𝕋,

we have 1

x(t) ≤ (x(t)) p ,

t ∈ ℝ+ ∩ 𝕋.

Let t

1

z(t) = ∫(h1 (s)(x(s)) p + h2 (s)x(s))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then z(0) = 0 and 1

(x(t)) p ≤ f (t) + g(t)z(t), 1

z Δ (t) = h1 (t)(x(t)) p + h2 (t)x(t) 1

≤ (h1 (t) + h2 (t))(x(t)) p

≤ (h1 (t) + h2 (t))(f (t) + g(t)z(t))

= (h1 (t) + h2 (t))f (t) + (h1 (t) + h2 (t))g(t)z(t)

= h3 (t) + h4 (t)z(t),

t ∈ ℝ+ ∩ 𝕋.

Hence, by Lemma 2.1.1, we conclude that t

z(t) ≤ ∫ h3 (s)e⊖h4 (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

and 1

(x(t)) p ≤ f (t) + g(t)z(t) t

≤ f (t) + g(t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs, 0

t ∈ ℝ+ ∩ 𝕋,

3.2 Pachpatte’s inequalities | 105

whereupon p

t

x(t) ≤ (f (t) + g(t) ∫ h3 (s)e⊖h4 (σ(s), t)Δs) ,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 3.2.8. Let p ∈ (1, ∞), c1 and c2 be positive constants, x ∈ 𝒞 (ℝ+ ∩ 𝕋) be a positive function, f , g ∈ 𝒞 (ℝ+ ∩ 𝕋) be nonnegative functions, L ∈ 𝒞 ((ℝ+ ∩ 𝕋) × ℝ+ ) be a function that satisfies 0 ≤ L(t, x) − L(t, y) ≤ k(t, y)(x − y) for t ∈ ℝ+ ∩ 𝕋, x ≥ y ≥ 0, where k ∈ 𝒞 ((ℝ+ ∩ 𝕋) × ℝ+ ) is a nonnegative function. If x(t) ≥ 1,

t

p

(x(t)) ≤ c1 + c2 ∫(f (s)L(s, x(s)) + g(s)x(s))Δs,

t ∈ ℝ+ ∩ 𝕋,

0

then 1 p

t

t ∈ ℝ+ ∩ 𝕋,

x(t) ≤ (c1 eh1 (t, 0) + ∫ h2 (s)e⊖h1 (σ(s), t)Δs) , 0

where h1 (t) = c2 (f (t)k(t, g(t)) + g(t)),

h2 (t) = c2 f (t)L(t, g(t)),

t ∈ ℝ+ ∩ 𝕋.

Proof. Since x(t) ≥ 1,

t ∈ ℝ+ ∩ 𝕋,

then p

x(t) ≤ (x(t)) ,

t ∈ ℝ+ ∩ 𝕋.

Let t

z(t) = c1 + c2 ∫(f (s)L(s, x(s)) + g(s)x(s))Δs, 0

Then z(0) = c1

t ∈ ℝ+ ∩ 𝕋.

106 | 3 Nonlinear integral inequalities and p

(x(t)) ≤ z(t),

t ∈ ℝ+ ∩ 𝕋.

We have z Δ (t) = c2 (f (t)L(t, x(t)) + g(t)x(t)) p

p

≤ c2 (f (t)L(t, (x(t)) ) + g(t)(x(t)) ) ≤ c2 (f (t)L(t, z(t)) + g(t)z(t))

≤ c2 (f (t)L(t, z(t) + g(t)) + g(t)z(t))

= c2 (f (t)L(t, z(t) + g(t)) − f (t)L(t, g(t)) + f (t)L(t, g(t)) + g(t)z(t))

≤ c2 (f (t)k(t, g(t))z(t) + f (t)L(t, g(t)) + g(t)z(t)) = c2 ((f (t)k(t, g(t)) + g(t))z(t) + f (t)L(t, g(t))) = h1 (t)z(t) + h2 (t),

t ∈ ℝ+ ∩ 𝕋.

Hence, by Lemma 2.1.1, we get t

z(t) ≤ c1 eh1 (t, 0) + ∫ h2 (s)e⊖h1 (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Therefore t

p

(x(t)) ≤ c1 eh1 (t, 0) + ∫ h2 (s)e⊖h1 (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋,

0

whereupon 1 p

t

x(t) ≤ (c1 eh1 (t, 0) + ∫ h2 (s)e⊖h1 (σ(s), t)Δs) ,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 3.2.9. Let x, y, gi , hi ∈ 𝒞 (ℝ+ ∩ 𝕋), i ∈ {1, 2, 3, 4}, be nonnegative functions, c1 and c2 be nonnegative constants, and t

2

2

(x(t)) ≤ c1 + ∫(g1 (s)(x(s)) + h1 (s)x(s))Δs t

0 2

+ ∫(g2 (s)(y(s)) + h2 (s)y(s))Δs, 0

3.2 Pachpatte’s inequalities | 107 t

2

2

(y(t)) ≤ c2 + ∫(g3 (s)(x(s)) + h3 (s)x(s))Δs t

0 2

+ ∫(g4 (s)(y(s)) + h4 (s)y(s))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Then 1 2

t

x(t) + y(t) ≤ (c3 + c3 ∫ p(s)e⊖p (σ(s), t)Δs) ,

t ∈ ℝ+ ∩ 𝕋,

0

if x(t) + y(t) ≥ 1, t ∈ ℝ+ ∩ 𝕋, and 1 2

t

x(t) + y(t) ≤ (c3 + ∫ p(s)Δs) ,

t ∈ ℝ+ ∩ 𝕋,

0

if x(t) + y(t) ≤ 1, t ∈ ℝ+ ∩ 𝕋. Here G(t) = max{g1 (t) + g3 (t), g2 (t) + g4 (t)},

H(t) = max{h1 (t) + h3 (t), h2 (t) + h4 (t)}, p(t) = 2(G(t) + H(t)),

t ∈ ℝ+ ∩ 𝕋,

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

t ∈ ℝ+ ∩ 𝕋.

c3 = 2(c1 + c2 ).

Proof. Let

Then 2

t

2

2

(x(t)) + (y(t)) ≤ c1 + c2 + ∫((g1 (s) + g3 (s))(x(s)) + (h1 (s) + h3 (s))x(s))Δs t

0 2

+ ∫((g2 (s) + g4 (s))(y(s)) + (h2 (s) + h4 (s))y(s))Δs 0

t

2

≤ c1 + c2 + ∫(G(s)(x(s)) + H(s)x(s))Δs t

0 2

+ ∫(G(s)(y(s)) + H(s)y(s))Δs 0

t

2

2

= c1 + c2 + ∫ G(s)((x(s)) + (y(s)) )Δs 0

108 | 3 Nonlinear integral inequalities t

+ ∫ H(s)(x(s) + y(s))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, using that 1 a2 + b2 ≥ (a + b)2 , 2 a2 + b2 ≤ (a + b)2 , a ≥ 0,

b ≥ 0,

we get 1 2 2 2 (x(t) + y(t)) ≤ (x(t)) + (y(t)) 2 t

2

≤ c1 + c2 + ∫ G(s)(x(s) + y(s)) Δs 0

t

+ ∫ H(s)(x(s) + y(s))Δs,

t ∈ ℝ+ ∩ 𝕋,

0

or t

t

0

0

1 2 2 (z(t)) ≤ c1 + c2 + ∫ G(s)(z(s)) Δs + ∫ H(s)z(s)Δs, 2

t ∈ ℝ+ ∩ 𝕋,

or t

2

t

2

(z(t)) ≤ 2(c1 + c2 ) + 2 ∫ G(s)(z(s)) Δs + 2 ∫ H(s)z(s)Δs t

0

0 2

= c3 + ∫(2G(s)(z(s)) + 2H(s)z(s))Δs, 0

1.

Let x(t) + y(t) ≥ 1,

t ∈ ℝ+ ∩ 𝕋.

Then z(t) ≥ 1,

t ∈ ℝ+ ∩ 𝕋,

and 2

z(t) ≤ (z(t)) ,

t ∈ ℝ+ ∩ 𝕋,

t ∈ ℝ+ ∩ 𝕋.

3.2 Pachpatte’s inequalities | 109

and t

2

2

(z(t)) ≤ c3 + ∫(2G(s) + 2H(s))(z(s)) Δs 0

t

2

= c3 + ∫ p(s)(z(s)) Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, from Theorem 1.1.10, we obtain t

2

(z(t)) ≤ c3 + c3 ∫ p(s)e⊖p (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Therefore 1 2

t

x(t) + y(t) ≤ (c3 + c3 ∫ p(s)e⊖p (σ(s), t)Δs) ,

t ∈ ℝ+ ∩ 𝕋.

0

2.

Let x(t) + y(t) ≤ 1,

t ∈ ℝ+ ∩ 𝕋.

Then z(t) ≤ 1,

t ∈ ℝ+ ∩ 𝕋.

Therefore t

2

(z(t)) ≤ c3 + ∫(2G(s) + 2H(s))z(s)Δs 0

t

= c3 + ∫ p(s)z(s)Δs 0

t

≤ c3 + ∫ p(s)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

From here, t

1 2

z(t) ≤ (c3 + ∫ p(s)Δs) , 0

or

t ∈ ℝ+ ∩ 𝕋,

110 | 3 Nonlinear integral inequalities t

1 2

x(t) + y(t) ≤ (c3 + ∫ p(s)Δs) ,

t ∈ ℝ+ ∩ 𝕋.

0

This completes the proof. Theorem 3.2.10. Let p ∈ (1, ∞), g, h1 , h2 ∈ 𝒞 (ℝ+ ∩ 𝕋) be nonnegative functions, x, f ∈ 𝒞 (ℝ+ ∩ 𝕋) be positive functions, and x(t) ≥ 1,

t

p

s

(x(t)) ≤ f (t) + g(t) ∫ h1 (s)(∫ h2 (y)x(y)Δy)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

0

Then t

1 p

s

x(t) ≤ (f (t) + g(t) ∫(h3 (s) + h1 (s) ∫ h4 (y)(A(y) + h1 (y)B(y))Δy)Δs) , 0

0

where t

h3 (t) = h1 (t) ∫ h2 (s)f (s)Δs, 0

h4 (t) = h2 (t)g(t), t

A(t) = ∫ h3 (s)Δs + h3 (t), 0

h5 (t) = h1 (t)(h4 (t) + 1), t

B(t) = ∫ h4 (s)A(s)e⊖h5 (σ(s), t)Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Proof. Since x(t) ≥ 1,

t ∈ ℝ+ ∩ 𝕋,

we have that p

x(t) ≤ (x(t)) ,

t ∈ ℝ+ ∩ 𝕋.

Let t

s

z(t) = ∫ h1 (s)(∫ h2 (y)x(y)Δy)Δs, 0

0

t ∈ ℝ+ ∩ 𝕋.

t ∈ ℝ+ ∩ 𝕋,

3.2 Pachpatte’s inequalities | 111

Then z(0) = 0, p

(x(t)) ≤ f (t) + g(t)z(t), and

t ∈ ℝ+ ∩ 𝕋,

t

z Δ (t) = h1 (t) ∫ h2 (s)x(s)Δs 0

t

p

≤ h1 (t) ∫ h2 (s)(x(s)) Δs 0

t

≤ h1 (t) ∫ h2 (s)(f (s) + g(s)z(s))Δs 0

t

t

= h1 (t) ∫ h2 (s)f (s)Δs + h1 (t) ∫ h2 (s)g(s)z(s)Δs 0

0

t

= h3 (t) + h1 (t) ∫ h4 (s)z(s)Δs 0

t

≤ h3 (t) + h1 (t) ∫ h4 (s)(z(s) + z Δ (s))Δs,

t ∈ ℝ+ ∩ 𝕋.

0

Hence, by Theorem 2.1.2, we get t

Δ

z (t) ≤ h3 (t) + h1 (t) ∫ h4 (s)(A(s) + h1 (s)B(s))Δs, 0

t

s

z(t) ≤ ∫(h3 (s) + h1 (s) ∫ h4 (y)(A(y) + h1 (y)B(y))Δy)Δs, 0

t ∈ ℝ+ ∩ 𝕋.

0

Therefore t

p

s

(x(t)) ≤ f (t) + g(t) ∫(h3 (s) + h1 (s) ∫ h4 (y)(A(y) + h1 (y)B(y))Δy)Δs, 0

t ∈ ℝ+ ∩ 𝕋,

0

whereupon t

s

1 p

x(t) ≤ (f (t) + g(t) ∫(h3 (s) + h1 (s) ∫ h4 (y)(A(y) + h1 (y)B(y))Δy)Δs) , 0

This completes the proof.

0

t ∈ ℝ+ ∩ 𝕋.

4 Diamond-α integral inequalities In this chapter are investigated Diamond-α integral inequalities such as Steffensen, Jensen, Radon and Schlömilch inequalities. This material in this chapter cna be found in [4, 12, 23, 24] and [25]. Suppose that 𝕋 is a time scale with forward jump operator, backward jump operator, delta differentiation operator, and nabla differentiation operator σ, ρ, Δ, and ∇, respectively.

4.1 Diamond-α calculus on time scales Definition 4.1.1. Let α ∈ [0, 1] and f : 𝕋 → ℝ be Δ and ∇ differentiable at t ∈ 𝕋. Define the diamond-α dynamic derivative f ⬦α of f at t as follows: f ⬦α (t) = αf Δ (t) + (1 − α)f ∇ (t). Thus, f is diamond-α differentiable at t ∈ 𝕋 if and only if f is Δ and ∇ differentiable at t. When α = 1, we have f ⬦α (t) = f Δ (t) and for α = 0, we get f ⬦α (t) = f ∇ (t). Example 4.1.2. Let 𝕋 = 2ℕ0 , f : 𝕋 → ℝ be defined as follows: f (t) = t 3 − 2t 2 + t + 1,

t ∈ 𝕋.

Then σ(t) = 2t,

2

f Δ (t) = (σ(t)) + tσ(t) + t 2 − 2(σ(t) + t) + 1 = 4t 2 + 2t 2 + t 2 − 6t + 1 = 7t 2 − 6t + 1,

t ∈ 𝕋,

and 1 ρ(t) = t, 2

2

f ∇ (t) = (ρ(t)) 1 = t2 + 4 https://doi.org/10.1515/9783110705553-004

+ tρ(t) + t 2 − 2(ρ(t) + t) + 1 1 2 2 t + t − 3t + 1 2

114 | 4 Diamond-α integral inequalities

=

7 2 t − 3t + 1, 4

t ∈ 𝕋.

Hence, f

⬦1

f

⬦1

2

1 1 (t) = f Δ (t) + f ∇ (t) 2 2 1 7 1 = (7t 2 − 6t + 1) + ( t 2 − 3t + 1) 2 2 4 1 7 2 3 1 7 2 = t − 3t + + t − t + 2 2 8 2 2 35 7 = t 2 − t + 1, 8 2

and 3

2 1 (t) = f Δ (t) + f ∇ (t) 3 3 1 2 7 = (7t 2 − 6t + 1) + ( t 2 − 3t + 1) 3 3 4 7 2 1 7 2 2 = t − 2t + + t − 2t + 3 3 6 3 7 2 = t − 4t + 1, t ∈ 𝕋. 2

Below we will list some of the properties of the diamond-α derivative. Let f , g : 𝕋 → ℝ be diamond-α differentiable at t ∈ 𝕋. Theorem 4.1.3. f + g is diamond-α differentiable at t and (f + g)⬦α (t) = f ⬦α (t) + g ⬦α (t). Proof. We have (f + g)⬦α (t) = α(f + g)Δ (t) + (1 − α)(f + g)∇ (t)

= α(f Δ (t) + g Δ (t)) + (1 − α)(f ∇ (t) + g ∇ (t)) = αf Δ (t) + (1 − α)f ∇ (t)

+ αg Δ (t) + (1 − α)g ∇ (t)

= f ⬦α (t) + g ⬦α (t). This completes the proof.

Theorem 4.1.4. For any c ∈ ℂ, we have that cf is diamond-α differentiable at t and (cf )⬦α (t) = cf ⬦α (t). Proof. We have (cf )⬦α (t) = α(cf )Δ (t) + (1 − α)(cf )∇ (t)

4.1 Diamond-α calculus on time scales | 115

= α(cf Δ (t)) + (1 − α)(cf ∇ (t))

= c(αf Δ (t) + (1 − α)f ∇ (t)) = cf ⬦α (t).

This completes the proof. Theorem 4.1.5. fg is diamond-α differentiable at t and (fg)⬦α (t) = f ⬦α (t)g(t) + αf σ (t)g Δ (t) + (1 − α)f ρ (t)g ∇ (t)

= f (t)g ⬦α (t) + αf Δ (t)g σ (t) + (1 − α)f ∇ (t)g ρ (t).

Proof. We have (fg)⬦α (t) = α(fg)Δ (t) + (1 − α)(fg)∇ (t)

= α(f Δ (t)g(t) + f σ (t)g Δ (t)) + (1 − α)(f ∇ (t)g(t) + f ρ (t)g ∇ (t))

= (αf Δ (t) + (1 − α)f ∇ (t))g(t) + αf σ (t)g Δ (t) + (1 − α)f ρ (t)g ∇ (t) = f ⬦α (t)g(t) + αf σ (t)g Δ (t) + (1 − α)f ρ (t)g ∇ (t)

= α(f Δ (t)g σ (t) + f (t)g Δ (t)) + (1 − α)(f ∇ (t)g ρ (t) + f (t)g ∇ (t))

= (αg Δ (t) + (1 − α)g ∇ (t))f (t) + αf Δ (t)g σ (t) + (1 − α)f ∇ (t)g ρ (t)

= f (t)g ⬦α + αf Δ (t)g σ (t) + (1 − α)f ∇ (t)g ρ (t). This completes the proof.

Definition 4.1.6. Let α ∈ [0, 1], a, b ∈ 𝕋, a < b, t ∈ [a, b], and suppose f is Δ and ∇ integrable over [a, b]. Define diamond-α integral from a to t of f as follows: t

t

t

∫ f (s)⬦α s = α ∫ f (s)Δs + (1 − α) ∫ f (s)∇s. a

a

a

Example 4.1.7. Let 𝕋 = 3ℕ0 ∪ { 31 }, f (t) = t 2 + t,

t ∈ 𝕋.

We will compute 9

∫ f (s)⬦ 1 s. 1

3

Let h1 (t) = t 3 ,

h2 (t) = t 2 ,

t ∈ 𝕋.

116 | 4 Diamond-α integral inequalities We have σ(t) = 3t,

2

hΔ1 (t) = (σ(t)) + tσ(t) + t 2 = 9t 2 + 3t 2 + t 2 = 13t 2 ,

hΔ2 (t) = σ(t) + t = 3t + t = 4t, t ∈ 𝕋, 1 ρ(t) = t, 3 2

h∇1 (t) = (ρ(t)) + tρ(t) + t 2 1 1 = t2 + t2 + t2 9 3 13 = t2, 9 ∇ h2 (t) = ρ(t) + t 1 = t+t 3 4 = t, t ∈ 𝕋, t ≥ 1. 3 Hence, 9

∫ f (s)⬦ 1 s = 1

3

9

9

1

1

2 1 ∫ f (s)Δs + ∫ f (s)∇s 3 3 9

9

2 1 = ∫(s2 + s)Δs + ∫(s2 + s)∇s 3 3 1

1

9

=

1 1 1 ∫( hΔ1 (s) + hΔ2 (s))Δs 3 13 4 1

9

9 3 2 + ∫( h∇1 (s) + h∇2 (s))∇s 3 13 4 1

1 1 = (h (9) − h1 (1)) + (h2 (9) − h2 (1)) 39 1 12 18 1 + (h1 (9) − h1 (1)) + (h2 (9) − h2 (1)) 39 2 19 7 = (729 − 1) + (81 − 1) 39 12

4.1 Diamond-α calculus on time scales | 117

1064 140 + 3 3 1204 . = 3

=

Remark 4.1.8. Note that t

Δ

t

⬦α

(∫ f (s)⬦α s)

t

∇

= α(∫ f (s)⬦α s) + (1 − α)(∫ f (s)⬦α s)

a

a

t

a

t

Δ

= α(α ∫ f (s)Δs + (1 − α) ∫ f (s)∇s) a

a

t

t

∇

+ (1 − α)(α ∫ f (s)Δs + (1 − α) ∫ f (s)∇s) a

2

a

= α f (t) + α(1 − α)f (σ(t)) + α(1 − α)f (ρ(t)) + (1 − α)2 f (t)

= (2α2 − 2α + 1)f (t) + α(1 − α)(f (σ(t)) + f (ρ(t))),

t ∈ [a, b].

Thus, in the general case we do not have t

⬦α

(∫ f (s)⬦α s)

= f (t).

a

Below we suppose that f , g : 𝕋 → ℝ are diamond-α integrable over [a, b]. Theorem 4.1.9. For any c ∈ ℂ, the function cf is diamond-α integrable over [a, b] and b

b

∫(cf )(s)⬦α s = c ∫ f (s)⬦α s. a

a

Proof. We have b

b

b

∫(cf )(s)⬦α s = α ∫(cf )(s)Δs + (1 − α) ∫(cf )(s)∇s a

a

a

b

b

= c(α ∫ f (s)Δs + (1 − α) ∫ f (s)∇s) a

b

= c ∫ f (s)⬦α s. a

This completes the proof.

a

118 | 4 Diamond-α integral inequalities Theorem 4.1.10. f + g is diamond-α integrable over [a, b] and b

b

b

∫(f + g)(s)⬦α s = ∫ f (s)⬦α s + ∫ g(s)⬦α s. a

a

a

Proof. We have b

b

b

∫(f + g)(s)⬦α s = α ∫(f + g)(s)Δs + (1 − α) ∫(f + g)(s)∇s a

a

a

b

b

= α ∫ f (s)Δs + (1 − α) ∫ f (s)∇s a

a

b

b

+ α ∫ g(s)Δs + (1 − α) ∫ g(s)∇s a

a

b

b

= ∫ f (s)⬦α s + ∫ g(s)⬦α s. a

a

This completes the proof. Theorem 4.1.11. We have b

t

b

∫ f (s)⬦α s = ∫ f (s)⬦α s + ∫ f (s)⬦α s a

a

t

for any t ∈ [a, b]. Proof. We have b

b

b

∫ f (s)⬦α s = α ∫ f (s)Δs + (1 − α) ∫ f (s)∇s a

a

a

t

b

= α(∫ f (s)Δs + ∫ f (s)Δs) a

t

t

b

+ (1 − α)(∫ f (s)∇s + ∫ f (s)∇s) a

t

t

b

= ∫ f (s)⬦α s + ∫ f (s)⬦α s, a

This completes the proof.

t

t ∈ [a, b].

4.1 Diamond-α calculus on time scales | 119

Theorem 4.1.12. Let a, b ∈ 𝕋κκ with a < b and suppose f , g, h : [a, b] → ℝ are ⬦α -integrable functions. Suppose that l, γ ∈ [a, b] are such that γ

b

b

∫ h(t)⬦α t = ∫ g(t)⬦α t = ∫ h(t)⬦α t. a

a

l

Then γ

b

∫ f (t)g(t)⬦α t = ∫(f (t)h(t) − (f (t) − f (γ))(h(t) − g(t)))⬦α t a

a

b

+ ∫(f (t) − f (γ))g(t)⬦α t, γ

b

l

∫ f (t)g(t)⬦α t = ∫(f (t) − f (l))g(t)⬦α t a

a

b

+ ∫(f (t)h(t) − (f (t) − f (l))(h(t) − g(t)))⬦α t. l

Proof. We have γ

b

∫(f (t)h(t) − (f (t) − f (γ))(h(t) − g(t)))⬦α t − ∫ f (t)g(t)⬦α t a

a

γ

= ∫(f (t)h(t) − f (t)g(t) − (f (t) − f (γ))(h(t) − g(t)))⬦α t a

γ

b

+ ∫ f (t)g(t)⬦α t − ∫ f (t)g(t)⬦α t γ

a

a

b

= ∫ f (γ)(h(t) − g(t))⬦α t − ∫ f (t)g(t)⬦α t γ

a

γ

γ

b

= f (γ)(∫ h(t)⬦α t − ∫ g(t)⬦α t) − ∫ f (t)g(t)⬦α t a

a

b

γ

γ

b

= f (γ)(∫ g(t)⬦α t − ∫ g(t)⬦α t) − ∫ f (t)g(t)⬦α t a

b

a

b

= f (γ) ∫ g(t)⬦α t − ∫ f (t)g(t)⬦α t γ

γ

γ

120 | 4 Diamond-α integral inequalities b

= ∫(f (γ) − f (t))g(t)⬦α t. γ

Next we get b

b

∫(f (t)h(t) − (f (t) − f (l))(h(t) − g(t)))⬦α t − ∫ f (t)g(t)⬦α t a

l b

= ∫(f (t)h(t) − f (t)g(t) − (f (t) − f (l))(h(t) − g(t)))⬦α t l b

b

+ ∫ f (t)g(t)⬦α t − ∫ f (t)g(t)⬦α t a

l

l

b

= f (l) ∫(h(t) − g(t))⬦α t − ∫ f (t)g(t)⬦α t a

l

l

b

b

= f (l) ∫ h(t)⬦α t − f (l) ∫ g(t)⬦α t − ∫ f (t)g(t)⬦α t l

l

a

b

b

l

= f (l) ∫ g(t)⬦α t − f (l) ∫ g(t)⬦α t − ∫ f (t)g(t)⬦α t a

a

l b

l

b

= f (l)(∫ g(t)⬦α t − ∫ g(t)⬦α t) − ∫ f (t)g(t)⬦α t a

a

l

l

l

= f (l) ∫ g(t)⬦α t − ∫ f (t)g(t)⬦α t a

a

l

= ∫(f (l) − f (t))g(t)⬦α t. a

This completes the proof. Theorem 4.1.13. Assume (A1) k is a positive ⬦α -integrable function on [a, b], (A2) ϕ, ψ, h : [a, b] → ℝ are ⬦α -integrable functions on [a, b], (A3) [c, d] ⊆ [a, b] with d

b

∫ h(t)k(t)⬦α t = ∫ ψ(t)k(t)⬦α t, c

(A4) z ∈ [a, b].

a

4.1 Diamond-α calculus on time scales | 121

Then d

b

c

∫ ϕ(t)h(t)⬦α t − ∫ ϕ(t)ψ(t)⬦α t = ∫( c

a

a

ϕ(z) ϕ(t) − )ψ(t)k(t)⬦α t k(z) k(t)

d

+ ∫( c

b

+ ∫( d

ϕ(t) ϕ(z) − )k(t)(h(t) − ψ(t))⬦α t k(t) k(z) ϕ(z) ϕ(t) − )ψ(t)k(t)⬦α t. k(z) k(t)

Proof. We have d

b

∫ ϕ(t)h(t)⬦α t − ∫ ϕ(t)ψ(t)⬦α t c

a

d

b

c

a

ϕ(t) ϕ(t) ⬦α t − ∫ k(t)ψ(t)⬦α t = ∫ k(t)h(t) k(t) k(t) d

d

= ∫ k(t)(h(t) − ψ(t)) c

b

−∫ a

ϕ(t) ϕ(t) ⬦ t + ∫ k(t)ψ(t) ⬦ t k(t) α k(t) α c

ϕ(t) k(t)ψ(t)⬦α t k(t)

d

= ∫ k(t)(h(t) − ψ(t)) c

ϕ(t) ⬦ t k(t) α

c

b

d

a

ϕ(t) ϕ(t) ⬦ t+∫ k(t)ψ(t)⬦α t) − (∫ k(t)ψ(t) k(t) α k(t) d

= ∫ k(t)(h(t) − ψ(t)) c

b

c

− (∫ k(t)ψ(t) a

c

= ∫( a

ϕ(t) ϕ(t) ⬦α t + ∫ k(t)ψ(t)⬦α t) k(t) k(t) d

ϕ(z) ϕ(t) − )k(t)ψ(t)⬦α t k(z) k(t)

d

+ ∫( c

ϕ(t) ⬦ t k(t) α

ϕ(t) ϕ(z) − )k(t)(h(t) − ψ(t))⬦α t k(t) k(z)

122 | 4 Diamond-α integral inequalities b

+ ∫( d

ϕ(z) ϕ(t) − )k(t)ψ(t)⬦α t k(z) k(t) d

c

c

a

ϕ(z) (∫ k(t)(h(t) − ψ(t))⬦α t − ∫ k(t)ψ(T)⬦α t + k(z) b

− ∫ k(t)ψ(t)⬦α t) c

d

= ∫( a

ϕ(z) ϕ(t) − )k(t)ψ(t)⬦α t k(z) k(t)

d

+ ∫( c

b

+ ∫( d

+

ϕ(t) ϕ(z) − )k(t)(h(t) − ψ(t))⬦α t k(t) k(z) ϕ(z) ϕ(t) − )k(t)ψ(t)⬦α t k(z) k(t) d

b

c

a

ϕ(z) (∫ k(t)h(t)⬦α t − ∫ k(t)ψ(t)⬦α t) k(z)

c

= ∫( a

ϕ(z) ϕ(t) − )k(t)ψ(t)⬦α t k(z) k(t)

d

+ ∫( c

b

+ ∫( d

ϕ(t) ϕ(z) − )k(t)(h(t) − ψ(t))⬦α t k(t) k(z) ϕ(z) ϕ(t) − )k(t)ψ(t)⬦α t. k(z) k(t)

This completes the proof.

4.2 Steffensen’s inequalities Theorem 4.2.1. Let a, b ∈ 𝕋κκ with a < b and f , g, h : [a, b] → ℝ be ⬦α -integrable with f of one sign and decreasing, and 0 ≤ g(t) ≤ h(t),

t ∈ [a, b].

Assume that l, γ ∈ [a, b] and b

b

γ

∫ h(t)⬦α t ≤ ∫ g(t)⬦α t ≤ ∫ h(t)⬦α t l

a

a

if

f ≥ 0,

4.2 Steffensen’s inequalities | 123 γ

b

b

∫ h(t)⬦α t ≤ ∫ g(t)⬦α t ≤ ∫ h(t)⬦α t a

a

if

f ≤ 0.

l

Then b

b

b

∫ f (t)h(t)⬦α t ≤ ∫ f (t)g(t)⬦α t ≤ ∫ f (t)h(t)⬦α t. a

l

a

Proof. Assume that f ≥ 0 on [a, b]. Then b

b

l

b

∫ f (t)g(t)⬦α t − ∫ f (t)h(t)⬦α t = ∫ f (t)g(t)⬦α t + ∫ f (t)g(t)⬦α t a

l

a

l b

− ∫ f (t)h(t)⬦α t l l

b

= ∫ f (t)g(t)⬦α t − ∫ f (t)(h(t) − g(t))⬦α t a

l

l

b

≥ ∫ f (t)g(t)⬦α t − f (l) ∫(h(t) − g(t))⬦α t a

l

l

b

= ∫ f (t)g(t)⬦α t − f (l) ∫ h(t)⬦α t a

l b

+ f (l) ∫ g(t)⬦α t l l

b

≥ ∫ f (t)g(t)⬦α t − f (l) ∫ g(t)⬦α t a

a

b

+ f (l) ∫ g(t)⬦α t l l

= ∫ f (t)g(t)⬦α t a

b

b

− f (l)(∫ g(t)⬦α t − ∫ g(t)⬦α t) a

l

l l

= ∫ f (t)g(t)⬦α t − f (l) ∫ g(t)⬦α t a

a

124 | 4 Diamond-α integral inequalities l

= ∫(f (t) − f (l))g(t)⬦α t a

≥ 0, i. e.,

b

b

∫ f (t)g(t)⬦α t ≥ ∫ f (t)h(t)⬦α t. a

l

Next we have γ

b

γ

γ

∫ f (t)h(t)⬦α t − ∫ f (t)g(t)⬦α t = ∫ f (t)h(t)⬦α t − ∫ f (t)g(t)⬦α t a

a

a

a

b

− ∫ f (t)g(t)⬦α t γ

γ

b

= ∫ f (t)(h(t) − g(t))⬦α t − ∫ f (t)g(t)⬦α t γ

a

γ

b

≥ f (γ) ∫(h(t) − g(t))⬦α t − ∫ f (t)g(t)⬦α t a

γ

γ

γ

= f (γ) ∫ h(t)⬦α t − f (γ) ∫ g(t)⬦α t a

a

b

− ∫ f (t)g(t)⬦α t γ

γ

γ

≥ f (γ) ∫ h(t)⬦α t − f (γ) ∫ g(t)⬦α t a

a

b

− f (γ) ∫ g(t)⬦α t γ

γ

= f (γ) ∫ h(t)⬦α t a

γ

b

− f (γ)(∫ g(t)⬦α t + ∫ g(t)⬦α t) a

γ

γ

b

= f (γ) ∫ h(t)⬦α t − f (γ) ∫ g(t)⬦α t a

a

4.2 Steffensen’s inequalities | 125 b

b

≥ f (γ) ∫ g(t)⬦α t − f (γ) ∫ g(t)⬦α t = 0,

a

a

i. e., γ

b

∫ f (t)h(t)⬦α t ≥ ∫ f (t)g(t)⬦α t. a

a

The case when f ≤ 0 we leave to the reader as an exercise. This completes the proof. Theorem 4.2.2. Let a, b ∈ 𝕋κκ with a < b and f , g, h : [a, b] → ℝ be ⬦α -integrable functions, f of one sign and decreasing, and 0 ≤ g(t) ≤ h(t),

t ∈ [a, b].

Assume that l, γ ∈ [a, b] are such that γ

b

b

∫ h(t)⬦α t = ∫ g(t)⬦α t = ∫ h(t)⬦α t. a

a

l

Then b

b

∫ f (t)h(t)⬦α t ≤ ∫(f (t)h(t) − (f (t) − f (l))(h(t) − g(t)))⬦α t l

l

b

≤ ∫ f (t)g(t)⬦α t a γ

≤ ∫(f (t)h(t) − (f (t) − f (γ))(h(t) − g(t)))⬦α t a γ

≤ ∫ f (t)h(t)⬦α t. a

Proof. From the assumptions on the functions f , g, h, we get l

∫(f (t) − f (l))g(t)⬦α t ≥ 0, a

b

∫(f (l) − f (t))(h(t) − g(t))⬦α t ≥ 0. l

126 | 4 Diamond-α integral inequalities By the second integral identity of Theorem 4.1.12, we have b

l

∫ f (t)g(t)⬦α t = ∫(f (t) − f (l))g(t)⬦α t a

a

b

+ ∫(f (t)h(t) − (f (t) − f (l))(h(t) − g(t)))⬦α t, l

whereupon b

b

∫ f (t)h(t)⬦α t ≤ ∫(f (t)h(t) − (f (t) − f (l))(h(t) − g(t)))⬦α t l

l

b

≤ ∫ f (t)g(t)⬦α t. a

By the first integral identity of Theorem 4.1.12, we have γ

b

∫ f (t)g(t)⬦α t = ∫(f (t)h(t) − (f (t) − f (γ))(h(t) − g(t)))⬦α t a

a

b

+ ∫(f (t) − f (γ))g(t)⬦α t γ

γ

≤ ∫(f (t)h(t) − (f (t) − f (γ))(h(t) − g(t)))⬦α t a γ

≤ ∫ f (t)h(t)⬦α t. a

This completes the proof. Theorem 4.2.3. Let a, b ∈ 𝕋κκ with a < b and f , g, h, ϕ : [a, b] → ℝ be ⬦α -integrable functions, f of one sign and decreasing and 0 ≤ ϕ(t) ≤ g(t) ≤ h(t) − ϕ(t),

t ∈ [a, b].

Assume l, γ ∈ [a, b] are such that γ

b

b

∫ h(t)⬦α t = ∫ g(t)⬦α t = ∫ h(t)⬦α t. a

a

l

4.2 Steffensen’s inequalities | 127

Then b

b

󵄨 󵄨 ∫ f (t)h(t)⬦α t + ∫󵄨󵄨󵄨f (t) − f (l)󵄨󵄨󵄨ϕ(t)⬦α t a

l

b

≤ ∫ f (t)g(t)⬦α t a

γ

b

󵄨 󵄨 ≤ ∫ f (t)h(t)⬦α t − ∫󵄨󵄨󵄨f (t) − f (γ)󵄨󵄨󵄨ϕ(t)⬦α t. a

a

Proof. From the assumptions on the functions f , g, h, and ϕ, we get γ

b

∫(f (t) − f (γ))(h(t) − g(t))⬦α t + ∫(f (γ) − f (t))g(t)⬦α t γ

a

γ

b

≥ ∫(f (t) − f (γ))ϕ(t)⬦α t + ∫(f (γ) − f (t))ϕ(t)⬦α t γ

a

b

󵄨 󵄨 = ∫󵄨󵄨󵄨f (t) − f (γ)󵄨󵄨󵄨ϕ(t)⬦α t a

and l

b

∫(f (t) − f (l))g(t)⬦α t + ∫(f (l) − f (t))(h(t) − g(t))⬦α t a

l b

l

≥ ∫(f (t) − f (l))ϕ(t)⬦α t + ∫(f (l) − f (t))ϕ(t)⬦α t a

l

b

󵄨 󵄨 = ∫󵄨󵄨󵄨f (t) − f (l)󵄨󵄨󵄨ϕ(t)⬦α t. a

Hence, due to the identities of Theorem 4.1.12, we find b

γ

∫ f (t)g(t)⬦α t = ∫(f (t)h(t) − (f (t) − f (γ))(h(t) − g(t)))⬦α t a

a

b

+ ∫(f (t) − f (γ))g(t)⬦α t γ

γ

γ

= ∫ f (t)h(t)⬦α t − ∫(f (t) − f (γ))(h(t) − g(t))⬦α t a

a

128 | 4 Diamond-α integral inequalities b

− ∫(f (γ) − f (t))g(t) ⋄α t γ

γ

b

󵄨 󵄨 ≤ ∫ f (t)h(t)⬦α t − ∫󵄨󵄨󵄨f (t) − f (γ)󵄨󵄨󵄨ϕ(t)⬦α t a

a

and b

l

∫ f (t)g(t)⬦α t = ∫(f (t) − f (l))g(t)⬦α t a

a

b

+ ∫(f (t)h(t) − (f (t) − f (l))(h(t) − g(t)))⬦α t l b

b

= ∫ f (t)h(t)⬦α t − ∫(f (t) − f (l))(h(t) − g(t))⬦α t l

l l

+ ∫(f (t) − f (l))g(t) ⋄α t a

b

b

= ∫ f (t)h(t) ⋄α t + ∫(f (l) − f (t))(h(t) − g(t)) ⋄α t l

l l

+ ∫(f (t) − f (l))g(t) ⋄α t a

b

b

󵄨 󵄨 ≥ ∫ f (t)h(t)⬦α t + ∫󵄨󵄨󵄨f (t) − f (l)󵄨󵄨󵄨ϕ(t)⬦α t. l

a

This completes the proof. Theorem 4.2.4. Suppose (A1)–(A3) and ϕ (A5) k is nonincreasing on [a, b], (A6) 0 ≤ ψ(t) ≤ h(t), t ∈ [a, b]. Then b

d

c

∫ ϕ(t)ψ(t)⬦α t ≤ ∫ ϕ(t)ψ(t)⬦α t + ∫( c

a

Proof. Since

ϕ k

a

ϕ(t) ϕ(d) − )ψ(t)k(t)⬦α t. k(t) k(d)

is nonincreasing, k is positive, and 0 ≤ ψ ≤ h on [a, b], we have d

∫( c

ϕ(t) ϕ(d) − )k(t)(h(t) − ψ(t))⬦α t ≥ 0, k(t) k(d)

4.2 Steffensen’s inequalities | 129 b

∫( d

ϕ(d) ϕ(t) − )ψ(t)k(t)⬦α t ≥ 0. k(d) k(t)

Now, we apply Theorem 4.1.13 for z = d and get d

b

c

∫ ϕ(t)h(t)⬦α t − ∫ ϕ(t)ψ(t)⬦α t − ∫( c

a

a

ϕ(d) ϕ(t) − )ψ(t)k(t)⬦α t k(d) k(t)

d

b

c

d

ϕ(d) ϕ(t) ϕ(t) ϕ(d) − )k(t)(h(t) − ψ(t))⬦α t + ∫( − )ψ(t)k(t)⬦α t = ∫( k(t) k(d) k(d) k(t) ≥ 0. This completes the proof. Theorem 4.2.5. Suppose (A1)–(A3) and (A5). Then d

b

b

c

d

a

ϕ(c) ϕ(t) − )ψ(t)k(t)⬦α t ≤ ∫ ϕ(t)ψ(t)⬦α t. ∫ ϕ(t)ψ(t)⬦α t − ∫( k(c) k(t) Proof. Since

ϕ k

is nonincreasing and 0 ≤ ψ ≤ h on [a, b], we have d

∫( c

ϕ(c) ϕ(t) − )k(t)(h(t) − ψ(t))⬦α t ≥ 0, k(c) k(t) c

∫( a

ϕ(t) ϕ(c) − )ψ(t)k(t)⬦α t ≥ 0. k(t) k(c)

Hence, by Theorem 4.1.13 for z = c, we find d

b

∫ ϕ(t)h(t)⬦α t − ∫( c

d

ϕ(c) ϕ(t) − )ψ(t)k(t)⬦α t k(c) k(t)

b

= ∫ ϕ(t)ψ(t)⬦α t a

d

+ ∫(

ϕ(t) ϕ(c) − )k(t)(h(t) − ψ(t))⬦α t k(t) k(c)

+ ∫(

ϕ(c) ϕ(t) − )ψ(t)k(t)⬦α t k(c) k(t)

c c

a

b

≤ ∫ ϕ(t)ψ(t)⬦α t. a

This completes the proof.

130 | 4 Diamond-α integral inequalities

4.3 The Jensen inequality Theorem 4.3.1 (Jensen’s Inequality). Let g ∈ 𝒞 ([a, b]) and let Φ : ℝ → ℝ be convex. Then b

b

a

a

1 1 Φ( ∫ g(t)⬦α t) ≤ ∫ Φ(g(t))⬦α t. b−a b−a Proof. Since Φ is convex, we have b

b

a

a

b

1 α 1−α Φ( ∫ g(t)⬦α t) = Φ( ∫ g(t)Δt + ∫ g(t)∇t) b−a b−a b−a ≤ αΦ(

a

b

b

a

a

1 1 ∫ g(t)Δt) + (1 − α)Φ( ∫ g(t)∇t). b−a b−a

(4.1)

Let b

x0 =

∫a g(t)Δt

b−a x = g(t).

,

Since Φ is convex, there exists a β ∈ ℝ such that Φ(x) − Φ(x0 ) ≥ β(x − x0 ),

(4.2)

or b

b

a

a

1 1 Φ(g(t)) − Φ( ∫ g(t)Δt) ≥ β(g(t) − ∫ g(t)Δt). b−a b−a We integrate the last inequality with respect to t from a to b and find b

b

∫ Φ(g(t))Δt − (b − a)Φ( a

b

b

a

a

1 ∫ g(t)Δt) ≥ β(∫ g(t)Δt − ∫ g(t)Δt) b−a a

= 0,

i. e., b

b

a

a

b

b

a

a

1 1 Φ( ∫ g(t)Δt) ≤ ∫ Φ(g(t))Δt. b−a b−a Similarly, 1 1 Φ( ∫ g(t)∇t) ≤ ∫ Φ(g(t))∇t. b−a b−a

(4.3)

4.3 The Jensen inequality | 131

By the last inequality and (4.1), (4.3), we find Φ(

b

b

b

a

a

a

1 α 1−α ∫ g(t)⬦α t) ≤ ∫ Φ(g(t))Δt + ∫ Φ(g(t))∇t b−a b−a b−a b

=

1 ∫ Φ(g(t))⬦α t. b−a a

This completes the proof. Theorem 4.3.2 (The Generalized Jensen Inequality). Let g, h ∈ 𝒞 ([a, b]) with b

󵄨 󵄨 ∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨⬦α t > 0, a

and let Φ : ℝ → ℝ be convex. Then b

Φ(

∫a |h(t)|g(t)⬦α t b

∫a |h(t)|⬦α t

b

)≤

∫a |h(t)|Φ(g(t))⬦α t b

∫a |h(t)|⬦α t

.

Proof. Take b

x0 =

∫a |h(t)|g(t)⬦α t b

∫a |h(t)|⬦α t

,

x = g(t) in (4.2) and find b

b

b

∫ |h(t)|g(t)⬦α t 󵄨 󵄨 󵄨 󵄨 ) ∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨Φ(g(t))⬦α t − (∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨⬦α t)Φ( a b ∫a |h(t)|⬦α t a a b

b

󵄨 󵄨 󵄨 󵄨 = ∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨Φ(g(t))⬦α t − (∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨⬦α t)Φ(x0 ) a

a

b

󵄨 󵄨 = ∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨(Φ(g(t)) − Φ(x0 ))⬦α t a

b

󵄨 󵄨 ≥ β ∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨(g(t) − x0 )⬦α t a

b

b

󵄨 󵄨 󵄨 󵄨 = β(∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨g(t)⬦α t − ∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨x0 ⬦α t) a

a

132 | 4 Diamond-α integral inequalities b

b

󵄨 󵄨 󵄨 󵄨 = β(∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨g(t)⬦α t − ∫󵄨󵄨󵄨h(t)󵄨󵄨󵄨g(t)⬦α t) a

a

= 0. This completes the proof.

4.4 The Radon inequality Theorem 4.4.1. Let w, f , g ∈ 𝒞 ([a, b]) be positive and ⬦α -integrable functions and c1 , c2 , c3 , c4 be positive constants. If β ≥ 1, γ, ζ , η, λ ∈ [0, ∞), and λ

b

λ 󵄨 󵄨 c3 (∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨g(t)⬦α t) ≥ c4 sup (g(t)) , t∈[a,b]

a

then b

b

b

(c1 (∫a |w(t)|⬦α t)η (∫a |w(t)|f (t)⬦α t)β+ζ + c2 (∫a |w(t)|f (t)⬦α t)β+η )γ+1 b

b

(c3 (∫a |w(t)|⬦α t)λ − c4 )(∫a |w(t)|g(t)⬦α t)γ(λ+1) ×

≤

1

b (∫a |w(t)|⬦α t)(γ+1)(β+η−1)−γλ b b ζ η γ+1 β(γ+1) 󵄨󵄨 󵄨󵄨 (c1 (∫a |w(t)|f (t)⬦α t) + c2 |f (t)| ) |f (t)| ⬦α t. ∫󵄨󵄨w(t)󵄨󵄨 b (c3 (∫a |w(t)|g(t)⬦α t)λ − c4 |g(t)|λ )γ (g(t))γ a

Proof. Set b

󵄨 󵄨 Λ = ∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨⬦α t, a

b

󵄨 󵄨 Y = ∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨f (t)⬦α t, a

b

󵄨 󵄨 Ω = ∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨g(t)⬦α t, a

󵄨 󵄨η 󵄨 󵄨β J(t) = (c1 Y ζ + c2 󵄨󵄨󵄨f (t)󵄨󵄨󵄨 )󵄨󵄨󵄨f (t)󵄨󵄨󵄨 , λ

k(t) = (c3 Ωλ − c4 (g(t)) )g(t).

Then the right-hand side of inequality (4.4) can be rewritten in the form b

γ+1

󵄨 󵄨 (J(t)) ⬦ t ∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨 (k(t))γ α a

(4.4)

4.4 The Radon inequality | 133 b

γ+1

J(t) 󵄨 󵄨 = ∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨k(t)( ) k(t) a

b

b

󵄨 󵄨 = (∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨k(t)⬦α t) ∫ a

a

⬦α t γ+1

J(t) ) b k(t) ∫a |w(t)|k(t)⬦α t |w(t)|k(t)

(

⬦α t.

Let Φ(x) = xγ+1 ,

x ∈ (0, ∞).

Then Φ is convex and by the Jensen inequality, we get b

γ+1

J(t) (∫ b )⬦α t) ( k(t) a ∫a |w(t)|k(t)⬦α t |w(t)|k(t)

b

≤∫ a

γ+1

J(t) ) b ∫a |w(t)|k(t)⬦α t k(t) |w(t)|k(t)

(

⬦α t,

whereupon γ+1

b

( ∫a |w(t)|J(t)⬦α t)

b

γ+1

󵄨 󵄨 (J(t)) ⬦ t. ≤ ∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨 (k(t))γ α

b

(∫a |w(t)|k(t)⬦α t)γ

a

Now we put the values of J and k in the right-hand side of the last inequality and find b

ζ η β γ+1 󵄨 󵄨 ((c Y + c |f (t)| )|f (t)| ) ⬦α t ∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨 1 λ 2 ((c3 Ω − c4 |g(t)|λ )|g(t)|)γ a

b

≥

(∫a |w(t)|(c1 Y ζ + c2 |f (t)|η )|f (t)|β ⬦α t)γ+1 b

(∫a |w(t)|(c3 Ωλ − c4 |g(t)|λ )|g(t)|⬦α t)γ b

=

≥ =

b

(c1 Y ζ ∫a |w(t)||f (t)|β ⬦α t + c2 ∫a |w(t)||f (t)|β+η ⬦α t)γ+1 b

(c3 Ωλ+1 − c4 ∫a |w(t)||g(t)|λ+1 ⬦α t)γ

(c1 Y β+ζ ( Λ1 )β−1 + c2 Y β+η ( Λ1 )β+η−1 )γ+1 (c3 Ωλ+1 − c4 Ωλ+1 ( Λ1 )λ )γ

(c1 Y β+ζ Λη + c2 Y β+η )γ+1 . (c3 Λλ − c4 )γ Ωγ(λ+1) Λ(γ+1)(β+η−1)−γλ

This completes the proof.

134 | 4 Diamond-α integral inequalities

4.5 The Schlömilch Inequality Theorem 4.5.1 (The Schlömilch Inequality). Let w, f ∈ 𝒞 ([a, b]) and b

󵄨 󵄨 ∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨⬦α t = 1. a

If η2 ≥ η1 > 0, then b

󵄨 󵄨󵄨 󵄨η (∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨󵄨󵄨󵄨f (t)󵄨󵄨󵄨 1 ⬦α t)

1 η1

a

1 η2

b

󵄨 󵄨󵄨 󵄨η ≤ (∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨󵄨󵄨󵄨f (t)󵄨󵄨󵄨 2 ⬦α t) .

(4.5)

a

Proof. Let β = 1, γ > 0, c1 = 0,

ζ = η = λ = 0, η γ + 1 = 2 , g(t) = 1, η1 c2 = c3 = 1, c4 = 0.

Then the Radon inequality takes the form b

󵄨 󵄨󵄨 󵄨 (∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨󵄨󵄨󵄨f (t)󵄨󵄨󵄨⬦α t) a

η2 η1

b

η

󵄨 󵄨󵄨 󵄨 2 ≤ ∫󵄨󵄨󵄨w(t)󵄨󵄨󵄨󵄨󵄨󵄨f (t)󵄨󵄨󵄨 η1 ⬦α t.

Replacing |f (t)| by |f (t)|η1 and taking power

a

1 , we get (4.5). This completes the proof. η2

5 Fractional integral inequalities This chapter is devoted on fractional integral inequalities on time scales. They are considered the Poincaré type inequality, opial type inequalities, Sobolev type inequality and Ostrowski type inequalities. Some of the results in this chapter can be found in [4, 27] and [28]. Suppose that 𝕋 is an unbounded time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let also 0 ∈ 𝕋. For α ≥ 0, by hα : 𝕋×𝕋 → ℝ we will denote the generalized polynomials on time scales defined as follows: h0 (t, s) = 1, t

hα (t, s) = ∫ hα−1 (t, σ(τ))Δτ, s

t, s ∈ 𝕋.

For α ≥ 1 and f ∈ 𝒞rd (𝕋), by Dα0 we denote the delta-Riemann–Liouville operator defined by t

Dα0 f (t) = ∫ hα−1 (t, σ(τ))f (τ)Δτ, 0

D00 f (t) = f (t),

t ∈ 𝕋.

5.1 Integral inequalities for the Chebyshev functional Definition 5.1.1. Let t ∈ 𝕋, t > 0, and f , g be two integrable functions defined on [0, t]. If for any x, y ∈ [0, t], (f (x) − f (y))(g(x) − g(y)) ≥ 0, then the functions f and g are called synchronous functions on [0, t]. Theorem 5.1.2. Let f and g be two synchronous functions on [0, ∞). Then for any t ≥ 0, α ≥ 1, we have Dα0 (fg)(t) ≥ (hα (t, 0)) Dα0 f (t)Dα0 g(t). −1

Proof. Take t > 0 and α ≥ 1 arbitrarily. Since f and g are two synchronous functions on [0, ∞), for any ϕ, τ ≥ 0, we have (f (τ) − f (ϕ))(g(τ) − g(ϕ)) ≥ 0, or f (τ)g(τ) − f (τ)g(ϕ) − f (ϕ)g(τ) + f (ϕ)g(ϕ) ≥ 0, https://doi.org/10.1515/9783110705553-005

136 | 5 Fractional integral inequalities whereupon f (τ)g(τ) + f (ϕ)g(ϕ) ≥ f (τ)g(ϕ) + f (ϕ)g(τ). Hence, hα−1 (t, σ(τ))f (τ)g(τ) + hα−1 (t, σ(τ))f (ϕ)g(ϕ)

≥ hα−1 (t, σ(τ))f (τ)g(ϕ) + hα−1 (t, σ(τ))f (ϕ)g(τ)

for any τ, ϕ ≥ 0. We integrate the last inequality with respect to τ from 0 to t and, using that τ and ϕ are independent, find t

t

∫ hα−1 (t, σ(τ))f (τ)g(τ)Δτ + (∫ hα−1 (t, σ(τ))Δτ)f (ϕ)g(ϕ) 0

0

t

t

≥ (∫ hα−1 (t, σ(τ))f (τ)Δτ)g(ϕ) + (∫ hα−1 (t, σ(τ))g(τ)Δτ)f (ϕ), 0

0

or Dα0 (fg)(t) + f (ϕ)g(ϕ)hα (t, 0)

≥ g(ϕ)Dα0 f (t) + f (ϕ)Dα0 g(t)

for any ϕ ≥ 0. Now we multiply both sides of the last inequality with hα−1 (t, σ(ϕ)) and get Dα0 (fg)(t)hα−1 (t, σ(ϕ)) + f (ϕ)g(ϕ)hα−1 (t, σ(ϕ))hα (t, 0)

≥ hα−1 (t, σ(ϕ))g(ϕ)Dα0 f (t) + hα−1 (t, σ(ϕ))f (ϕ)Dα0 g(t),

which we integrate with respect to ϕ from 0 to t and find t α D0 (fg)(t) ∫ hα−1 (t, σ(ϕ))Δϕ t

0

t

+ (∫ f (ϕ)g(ϕ)hα−1 (t, σ(ϕ))Δϕ)hα (t, 0) 0

t

≥ (∫ hα−1 (t, σ(ϕ))g(ϕ)Δϕ)Dα0 f (t) + (∫ hα−1 (t, σ(ϕ))f (ϕ)Δϕ)Dα0 g(t) 0

0

and Dα0 (fg)(t)hα (t, 0) + hα (t, 0)Dα0 (fg)(t) ≥ Dα0 f (t)Dα0 g(t) + Dα0 f (t)Dα0 g(t), or hα (t, 0)Dα0 (fg)(t) ≥ Dα0 f (t)Dα0 g(t). This completes the proof.

5.1 Integral inequalities for the Chebyshev functional | 137

Theorem 5.1.3. Let f and g be two synchronous functions on [0, ∞). Then for any t > 0, α, β ≥ 1, we have β

β

β

hα (t, 0)D0 (fg)(t) + hβ (t, 0)Dα0 (fg)(t, 0) ≥ Dα0 f (t)D0 g(t) + D0 f (t)Dα0 g(t). Proof. Take t > 0 and α, β ≥ 1 arbitrarily. As in the proof of Theorem 5.1.2, for any ϕ, τ ≥ 0, we have f (τ)g(τ) + f (ϕ)g(ϕ) ≥ f (τ)g(ϕ) + f (ϕ)g(τ) and Dα0 (fg)(t) + f (ϕ)g(ϕ)hα (t, 0)

≥ g(ϕ)Dα0 f (t) + f (ϕ)Dα0 g(t).

Now we multiply both sides of the last inequality with hβ−1 (t, σ(ϕ)) and get Dα0 (fg)(t)hβ−1 (t, σ(ϕ)) + f (ϕ)g(ϕ)hβ−1 (t, σ(ϕ))hα (t, 0)

≥ hβ−1 (t, σ(ϕ))g(ϕ)Dα0 f (t) + hβ−1 (t, σ(ϕ))f (ϕ)Dα0 g(t),

which we integrate with respect to ϕ from 0 to t and find t

t

0

0

Dα0 (fg)(t) ∫ hβ−1 (t, σ(ϕ))Δϕ + (∫ f (ϕ)g(ϕ)hβ−1 (t, σ(ϕ))Δϕ)hα (t, 0) t

≥

(∫ hβ−1 (t, σ(ϕ))g(ϕ)Δϕ)Dα0 f (t) 0

t

+ (∫ hβ−1 (t, σ(ϕ))f (ϕ)Δϕ)Dα0 g(t) 0

and β

β

β

Dα0 (fg)(t)hβ (t, 0) + hα (t, 0)D0 (fg)(t) ≥ Dα0 f (t)D0 g(t) + D0 f (t)Dα0 g(t). This completes the proof. Theorem 5.1.4. Let n ∈ ℕ be arbitrarily chosen and fj , j ∈ {1, . . . , n}, be positive increasing functions. Then for any t > 0, α ≥ 1, we have n

Dα0 (∏ fj )(t) ≥ (hα (t, 0)) j=1

1−n

n

∏ Dα0 fj (t). j=1

Proof. We will use the principle of mathematical induction. 1. For n = 1 the assertion is evident. 2. Assume that the assertion is true for some n ∈ ℕ.

(5.1)

138 | 5 Fractional integral inequalities 3.

We will prove n+1

Dα0 (∏ fj )(t) ≥ (hα (t, 0))

−n

j=1

n+1

∏ Dα0 fj (t). j=1

We apply Theorem 5.1.2 for the functions f (t) = fn+1 (t), n

g(t) = (∏ fj )(t) j=1

and, using (5.1), get n+1

Dα0 (∏ fj )(t) = Dα0 (fg)(t) j=1

≥ (hα (t, 0)) Dα0 fn+1 (t)Dα0 g(t) −1

1−n

≥ (hα (t, 0)) Dα0 fn+1 (t)(hα (t, 0)) −1

= (hα (t, 0))

−n

n

∏ Dα0 fj (t) j=1

n+1

∏ Dα0 fj (t). j=1

Hence, by the principal of mathematical induction, we conclude that inequality (5.1) is true for any n ∈ ℕ. This completes the proof.

5.2 A Poincaré-type inequality Let a ∈ 𝕋, a > 0. We will start with the following useful auxiliary results. Lemma 5.2.1. Let α, β > 1, f ∈ 𝒞rd ([0, a]). Then α+β

β

Dα0 D0 f (t) = D0 f (t) t

+ ∫ f (u)μ(u)hα−1 (t, σ(u))hβ−1 (u, σ(u))Δu, 0

Proof. Let t ∈ [0, a]. Using Fubini’s Theorem, we get β

t

τ

Dα0 D0 f (t) = ∫ hα−1 (t, σ(τ)) ∫ hβ−1 (τ, σ(u))f (u)ΔuΔτ 0

t

t

0

= ∫ f (u)(∫ hα−1 (t, σ(τ))hβ−1 (τ, σ(u))Δτ)Δu 0

u

t ∈ [0, a].

5.2 A Poincaré-type inequality | 139 σ(u)

t

= ∫ f (u)( ∫ hα−1 (t, σ(τ))hβ−1 (τ, σ(u))Δτ u

0

t

+ ∫ hα−1 (t, σ(τ))hβ−1 (τ, σ(u))Δτ)Δu σ(u) t

= ∫ f (u)(hα+β−1 (t, σ(u)) + μ(u)hα−1 (t, σ(u))hβ−1 (u, σ(u)))Δu 0

t

= ∫ hα+β−1 (t, σ(u))f (u)Δu 0

t

+ ∫ μ(u)hα−1 (t, σ(u))hβ−1 (u, σ(u))Δu 0

t

α+β

= D0 f (t) + ∫ μ(u)hα−1 (t, σ(u))hβ−1 (u, σ(u))Δu. 0

This completes the proof. Definition 5.2.2. Let α, β > 1, f ∈ 𝒞rd ([0, a]). The integral t

E(f , α, β, t) = ∫ f (u)μ(u)hα−1 (t, σ(u))hβ−1 (u, σ(u))Δu,

t ∈ [0, a],

0

is called the forward graininess deviation functional of f . By Lemma 5.2.1, we have β

α+β

Dα0 D0 f (t) = D0 f (t) + E(f , α, β, t),

t ∈ [0, a].

Definition 5.2.3. Let α > 2 and m − 1 < α < m, m ∈ ℕ, ν = m − α. For a function m f ∈ 𝒞rd ([0, a]), define m

ν+1 Δ Δα−1 0 f (t) = D0 f (t) t

m

= ∫ hν (t, σ(u))f Δ (u)Δu,

t ∈ [0, a].

0 m Lemma 5.2.4. Let α > 2, m − 1 < α < m, m ∈ ℕ, ν = m − α, f ∈ 𝒞rd ([0, a]). Then t

m

t

m

∫ hm−1 (t, σ(τ))f Δ (τ)Δτ = − ∫ f Δ (u)μ(u)hα−2 (t, σ(u))hν (u, σ(u))Δu 0

0

t

+ ∫ hα−2 (t, σ(τ))Δm−1 0 f (τ)Δτ, 0

t ∈ [0, a].

140 | 5 Fractional integral inequalities Proof. We have t

m−1 m−1 Dα−1 0 Δ0 f (t) = ∫ hα−2 (t, σ(τ))Δ0 f (τ)Δτ 0

m

ν+1 Δ = Dα−1 0 D0 f (t) m

Δ = Dα+ν 0 f (t) t

m

+ ∫ f Δ (u)μ(u)hα−1 (t, σ(u))hν−1 (u, σ(u))Δu 0

m

Δ = Dm 0 f (t) t

m

+ ∫ f Δ (u)μ(u)hα−1 (t, σ(u))hν−1 (u, σ(u))Δu t

0 m

= ∫ hm−1 (t, σ(u))f Δ (u)Δu 0

t

m

+ ∫ f Δ (u)μ(u)hα−1 (t, σ(u))hβ−1 (u, σ(u))Δu, 0

t ∈ [0, a]. This completes the proof. Lemma 5.2.5 (Fractional Taylor Formula). Let α > 2, m − 1 < α < m, m ∈ ℕ, ν = m − α, m f ∈ 𝒞rd ([0, a]). Then m−1

k

f (t) = ∑ hk (t, 0)f Δ (0) k=0

t

m

− ∫ f Δ (u)μ(u)hα−2 (t, σ(u))hν (u, σ(u))Δu 0

t

+ ∫ hα−2 (t, σ(τ))Δα−1 0 f (τ)Δτ,

t ∈ [0, a].

0

Proof. By the Taylor formula on time scales and Lemma 5.2.4, we have m−1

k

t

m

f (t) = ∑ hk (t, 0)f Δ (0) + ∫ hm−1 (t, σ(τ))f Δ (τ)Δτ k=0

0

m−1

k

= ∑ hk (t, 0)f Δ (0) k=0

t

m

− ∫ f Δ (u)μ(u)hα−2 (t, σ(u))hν (u, σ(u))Δu 0

5.2 A Poincaré-type inequality | 141 t

+ ∫ hα−2 (t, σ(τ))Δα−1 0 f (τ)Δτ,

t ∈ [0, a].

0

This completes the proof. m Definition 5.2.6. For α > 2, m − 1 < α < m, m ∈ ℕ, ν = m − α, f ∈ 𝒞rd ([0, a]), define m

B(t) = f (t) + E(f Δ , α − 1, ν + 1, t),

t ∈ [0, a]. k

m Lemma 5.2.7. Let α > 2, m − 1 < α < m, m ∈ ℕ, ν = m − α, f ∈ 𝒞rd ([0, a]), f Δ (0) = 0, k ∈ {0, . . . , m − 1}. Then t

B(t) = ∫ hα−2 (t, σ(τ))Δα−1 0 f (τ)Δτ,

t ∈ [0, a].

0

Proof. By the fractional Taylor formula, we get m−1

k

f (t) = ∑ hk (t, 0)f Δ (0) k=0

t

m

− ∫ f Δ (u)μ(u)hα−2 (t, σ(u))hν (u, σ(u))Δu 0

t

+ ∫ hα−2 (t, σ(τ))Δα−1 0 f (τ)Δτ 0

t

m

= − ∫ f Δ (u)μ(u)hα−2 (t, σ(u))hν (u, σ(u))Δu 0

t

+ ∫ hα−2 (t, σ(τ))Δα−1 0 f (τ)Δτ,

t ∈ [0, a].

0

Hence, m

B(t) = f (t) + E(f Δ , α − 1, ν + 1, t) t

m

= f (t) + ∫ μ(u)hα−2 (t, σ(u))hν (u, σ(u))f Δ (u)Δu t

0

= ∫ hα−2 (t, σ(τ))Δα−1 0 f (τ)Δτ, 0

This completes the proof.

t ∈ [0, a].

142 | 5 Fractional integral inequalities Theorem 5.2.8 (Poincaré-type Inequality). Let α > 2, m − 1 < α < m, m ∈ ℕ, k hα−2 (s, σ(τ)), hν (s, σ(τ)) ∈ 𝒞 ([0, a] × [0, a]), and f Δ (0) = 0, k ∈ {0, . . . , m − 1}. Let also p, q > 1, p1 + q1 = 1. Then a

a

t

0

0

0

q p

a

󵄨 󵄨󵄨q 󵄨p 󵄨 󵄨q 󵄨 ∫󵄨󵄨󵄨B(t)󵄨󵄨󵄨 Δt ≤ (∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) Δt)(∫󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨 Δt). 0

Proof. By Lemma 5.2.7, we have t

B(t) = ∫ hα−2 (t, σ(τ))Δα−1 0 f (τ)Δτ,

t ∈ [0, a].

0

Hence, by Hölder’s inequality, we obtain 󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 α−1 󵄨󵄨B(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ hα−2 (t, σ(τ))Δ0 f (τ)Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 󵄨 0 t

󵄨 󵄨󵄨 󵄨󵄨 ≤ ∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨󵄨󵄨󵄨Δα−1 0 f (τ)󵄨󵄨Δτ 0

t

󵄨p

1 p

t

1 q

󵄨q

󵄨 󵄨 󵄨 ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) (∫󵄨󵄨󵄨Δα−1 0 f (τ)󵄨󵄨 Δτ) ,

t ∈ [0, a].

0

0

Hence, q p

t

t

󵄨󵄨 󵄨q 󵄨 α−1 󵄨 󵄨q 󵄨p 󵄨󵄨B(t)󵄨󵄨󵄨 ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) (∫󵄨󵄨󵄨Δ0 f (τ)󵄨󵄨󵄨 Δτ) 0

q p

t

0

a

󵄨 󵄨󵄨q 󵄨 󵄨p ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) (∫󵄨󵄨󵄨Δα−1 0 f (τ)󵄨󵄨 Δτ),

t ∈ [0, a],

0

0

and a

a

t

0

0

0

q p

a

󵄨 󵄨󵄨q 󵄨 󵄨q 󵄨 󵄨p ∫󵄨󵄨󵄨B(t)󵄨󵄨󵄨 Δt ≤ (∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) Δt)(∫󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨 Δt). 0

This completes the proof.

5.3 A Sobolev-type inequality Theorem 5.3.1 (Sobolev-type Inequality). Let α > 2, m−1 < α < m, m ∈ ℕ, hα−2 (s, σ(τ)), k

hν (s, σ(τ)) ∈ 𝒞 ([0, a] × [0, a]), and f Δ (0) = 0, k ∈ {0, . . . , m − 1}. Let also p, q > 1, r ≥ 1,

5.4 An Opial-type inequality | 143

1 p

+

1 q

= 1. Then a

1 r

󵄨r

a

t

0

0

r p

󵄨p

1 r

a

󵄨q

1 q

󵄨 󵄨 󵄨 󵄨 (∫󵄨󵄨󵄨B(t)󵄨󵄨󵄨 Δt) ≤ (∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) Δt) (∫󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨 Δt) . 0

0

Proof. As in the proof of Theorem 5.2.8, we have 1 p

t

1 q

t

󵄨 α−1 󵄨q 󵄨󵄨 󵄨 󵄨 󵄨p 󵄨󵄨B(t)󵄨󵄨󵄨 ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) (∫󵄨󵄨󵄨Δ0 f (τ)󵄨󵄨󵄨 Δτ) ,

t ∈ [0, a].

0

0

Hence, r p

t

r q

a

󵄨 α−1 󵄨q 󵄨󵄨 󵄨r 󵄨 󵄨p 󵄨󵄨B(t)󵄨󵄨󵄨 ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) (∫󵄨󵄨󵄨Δ0 f (τ)󵄨󵄨󵄨 Δτ) ,

t ∈ [0, a],

0

0

and 1 r

a

a

t

0

0

r p

1 r

1 q

a

󵄨 󵄨p 󵄨 󵄨󵄨q 󵄨 󵄨r (∫󵄨󵄨󵄨B(t)󵄨󵄨󵄨 Δt) ≤ (∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) Δt) (∫󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨 Δt) . 0

0

This completes the proof.

5.4 An Opial-type inequality Theorem 5.4.1. Let α > 2, m − 1 < α < m, m ∈ ℕ, hα−2 (s, σ(τ)), hν (s, σ(τ)) ∈ 𝒞 ([0, a] × k

[0, a]), and f Δ (0) = 0, k ∈ {0, . . . , m − 1}. Let also p, q > 1, is an increasing function on [0, a]. Then a

a

t

0

0

0

1 p

+

1 q

= 1, and suppose |Δα0 f |

1 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨p 󵄨󵄨 q ∫󵄨󵄨󵄨B(t)󵄨󵄨󵄨󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨Δt ≤ a (∫(∫󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨 Δτ)Δt)

a

×

2q (∫(Δα−1 0 f (t)) Δt)

1 p

1 q

.

0

Proof. By the proof of Theorem 5.2.8, we get t

1 p

t

󵄨󵄨 󵄨 󵄨 󵄨p 󵄨 α−1 󵄨q 󵄨󵄨B(t)󵄨󵄨󵄨 ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) (∫󵄨󵄨󵄨Δ0 f (τ)󵄨󵄨󵄨 Δτ) 0

t

1 p

0

󵄨 󵄨󵄨 q1 󵄨 󵄨p ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) 󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨t , 0

1 q

t ∈ [0, a].

144 | 5 Fractional integral inequalities Hence, t

1 p

󵄨p

2 1 󵄨 󵄨 󵄨󵄨 α−1 󵄨󵄨 α−1 󵄨󵄨B(t)󵄨󵄨󵄨󵄨󵄨󵄨Δ0 f (t)󵄨󵄨󵄨 ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) (Δ0 f (t)) t q ,

t ∈ [0, a].

0

Then a

a

t

0

0

0

󵄨󵄨p 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 ∫󵄨󵄨󵄨B(t)󵄨󵄨󵄨󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨Δt ≤ ∫((∫󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨 Δτ) 2

1 p

1

q × (Δα−1 0 f (t)) t )Δt

t

a

󵄨 󵄨p ≤ (∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ)Δt) 0

0

a

×

2q (∫(Δα−1 0 f (t)) tΔt)

1 p

1 q

0

a

t

0

0

󵄨 󵄨p ≤ a (∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ)Δt) 1 q

a

2q

1 p

1 q

× (∫(Δα−1 0 f (t)) Δt) . 0

This completes the proof.

5.5 Ostrowski-type inequalities Theorem 5.5.1. Let α > 2, m − 1 < α < m, m ∈ ℕ, hα−2 (s, σ(τ)), hν (s, σ(τ)) ∈ 𝒞 ([0, a] × k

[0, a]), and f Δ (0) = 0, k ∈ {1, . . . , m − 1}. Then

a t 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 1 󵄨 󵄨󵄨 ∫ B(t)Δt − f (0)󵄨󵄨󵄨 ≤ 1 (∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨Δτ)Δt) sup 󵄨󵄨󵄨Δα−1 f (t)󵄨󵄨󵄨. 󵄨󵄨 a 󵄨󵄨 a 󵄨 󵄨 󵄨 0 󵄨 󵄨󵄨 󵄨󵄨 t∈[0,a] 0 0 0

Proof. By the fractional Taylor formula, we get t

m

f (t) = f (0) − ∫ f Δ (u)μ(u)hα−2 (t, σ(u))hν (u, σ(u))Δu t

0

+ ∫ hα−2 (t, σ(τ))Δα−1 0 f (τ)Δτ, 0

t ∈ [0, a].

5.5 Ostrowski-type inequalities | 145

Hence, by Lemma 5.2.7, we get t

B(t) − f (0) = ∫ hα−2 (t, σ(τ))Δα−1 0 f (τ)Δτ,

t ∈ [0, a],

0

and 󵄨󵄨 󵄨󵄨 t 󵄨󵄨 󵄨 󵄨󵄨󵄨 󵄨󵄨 α−1 󵄨 󵄨󵄨B(t) − f (0)󵄨󵄨 = 󵄨󵄨∫ hα−2 (t, σ(τ))Δ0 f (τ)Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨0 t

󵄨 󵄨󵄨 󵄨󵄨 ≤ ∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨󵄨󵄨󵄨Δα−1 0 f (τ)󵄨󵄨Δτ 0

t

󵄨 󵄨 󵄨 󵄨󵄨 ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨Δτ) sup 󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨 t∈[0,a]

0

Thus, 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 1 󵄨 󵄨 󵄨 󵄨󵄨 ∫ B(t)Δt − f (0)󵄨󵄨󵄨 = 󵄨󵄨󵄨 1 ∫(B(t) − f (0))Δt 󵄨󵄨󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 0 0 a

≤

1 󵄨󵄨 󵄨 ∫󵄨B(t) − f (0)󵄨󵄨󵄨Δt a 󵄨 0

a

t

0

0

1 󵄨 󵄨󵄨 󵄨 󵄨 ≤ ( ∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨Δτ)Δt) sup 󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨. a t∈[0,a] Theorem 5.5.2. Suppose all conditions of Theorem 5.5.1 hold. Let also p, q > 1, p1 + q1 = 1. Then 1

a t 󵄨󵄨 󵄨󵄨 a p 󵄨 󵄨󵄨 1 󵄨󵄨 ∫ B(t)Δt − f (0)󵄨󵄨󵄨 ≤ 1 (∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨p Δτ) Δt) 󵄨󵄨 a 󵄨 󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 0 0 0 a

1 q

󵄨 󵄨󵄨q × (∫󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨 Δt) . 0

Proof. By the proof of Theorem 5.5.1, it follows that t

󵄨󵄨 󵄨 󵄨 󵄨󵄨 α−1 󵄨 󵄨󵄨B(t) − f (0)󵄨󵄨󵄨 ≤ ∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨󵄨󵄨󵄨Δ0 f (τ)󵄨󵄨󵄨Δτ 0

t

󵄨 󵄨p ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) 0

1 p

146 | 5 Fractional integral inequalities t

󵄨󵄨q 󵄨 × (∫󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨 Δt)

1 q

0

t

󵄨p 󵄨 ≤ (∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) 0

1 p

1 q

a

󵄨 󵄨󵄨q × (∫󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨 Δt) , 0

t ∈ [0, a]. Thus 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 1 󵄨 󵄨 󵄨 󵄨󵄨 ∫ B(t)Δt − f (0)󵄨󵄨󵄨 = 󵄨󵄨󵄨 1 ∫(B(t) − f (0))Δt 󵄨󵄨󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 a 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 0 0 a

1 󵄨 󵄨 ≤ ∫󵄨󵄨󵄨B(t) − f (0)󵄨󵄨󵄨Δt a 0

a

t

0

0

1 p

1 󵄨 󵄨p ≤ (∫(∫󵄨󵄨󵄨hα−2 (t, σ(τ))󵄨󵄨󵄨 Δτ) Δt) a a

1 q

󵄨 󵄨󵄨q × (∫󵄨󵄨󵄨Δα−1 0 f (t)󵄨󵄨 Δt) . 0

This completes the proof.

6 Two-dimensional linear integral inequalities This chapter is devoted on two dimensional linear integral inequalities. They are investigated Wendroff type and Pachpatte type two dimensional linear integral inequalities. Some of the results in this chapter can be found in [6, 24] and [25]. Let 𝕋1 and 𝕋2 be time scales with forward jump operators and delta differentiation operators σ1 , σ2 and Δ1 , Δ2 , respectively. Suppose that 0 ∈ 𝕋1 and 0 ∈ 𝕋2 .

6.1 Wendroff’s inequality Theorem 6.1.1 (Wendroff’s Inequality). Let u, c ∈ 𝒞 ((ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 )), a ∈ 𝒞 1 (ℝ+ ∩ 𝕋1 ) and b ∈ 𝒞 1 (ℝ+ ∩ 𝕋2 ) be nonnegative functions such that a(t1 ) + b(0) ≠ 0,

t1 ∈ ℝ+ ∩ 𝕋1 ,

Δ1

a (t1 ) ≥ 0,

t1 ∈ ℝ+ ∩ 𝕋1 ,

Δ2

b (t2 ) ≥ 0,

t2 ∈ ℝ+ ∩ 𝕋2 .

If t1 t2

u(t1 , t2 ) ≤ a(t1 ) + b(t2 ) + ∫ ∫ c(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ (a(0) + b(t2 ))ef (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where t2

f (t1 , t2 ) =

aΔ1 (t1 ) + ∫ c(t1 , s2 )Δ2 s2 , a(t1 ) + b(0)

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0

Proof. Let t1 t2

z(t1 , t2 ) = a(t1 ) + b(t2 ) + ∫ ∫ c(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). We have that u(t1 , t2 ) ≤ z(t1 , t2 ), z(t1 , t2 ) ≥ 0,

z(t1 , 0) = a(t1 ) + b(0),

https://doi.org/10.1515/9783110705553-006

148 | 6 Two-dimensional linear integral inequalities

Δ zt11 (t1 , t2 )

t2

Δ1

= a (t1 ) + ∫ c(t1 , s2 )u(t1 , s2 )Δ2 s2 ,

Δ zt11 (t1 , 0)

= a (t1 ),

Δ zt22 (t1 , t2 )

Δ2

0

Δ1

z(0, t2 ) = a(0) + b(t2 ),

Δ zt22 (0, t2 )

t1

= b (t2 ) + ∫ c(s1 , t2 )u(s1 , t2 )Δ1 s1 , 0

Δ2

= b (t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Next, ΔΔ

zt11t2 2 (t1 , t2 ) = c(t1 , t2 )u(t1 , t2 )

≤ c(t1 , t2 )z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, ΔΔ

zt11t2 2 (t1 , t2 ) z(t1 , t2 )

≤ c(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

or ΔΔ

zt11t2 2 (t1 , t2 )z(t1 , t2 ) (z(t1 , t2 ))2

≤ c(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Since aΔ1 (t1 ) ≥ 0

and

bΔ2 (t2 ) ≥ 0

for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

we conclude that Δ

zt11 (t1 , t2 ) ≥ 0

Δ

and zt22 (t1 , t2 ) ≥ 0

for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Therefore z(t1 , σ2 (t2 )) ≥ z(t1 , t2 )

for

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Also, ΔΔ

zt11t2 2 (t1 , t2 ) ≥ 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

From here and (6.1), we get ΔΔ

ΔΔ

zt11t2 2 (t1 , t2 )z(t1 , t2 )

z(t1 , t2 )z(t1 , σ2 (t2 ))

≤

zt11t2 2 (t1 , t2 )z(t1 , t2 ) (z(t1 , t2 ))2

(6.1)

6.1 Wendroff’s inequality | 149

≤ c(t1 , t2 ) ≤ c(t1 , t2 ) +

Δ

Δ

zt11 (t1 , t2 )zt22 (t1 , t2 )

z(t1 , t2 )z(t1 , σ2 (t2 ))

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, Δ

(

zt11 (t1 , t2 ) z(t1 , t2 )

Δ2

)

t2

≤ c(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Therefore Δ

Δ

zt11 (t1 , t2 )

−

z(t1 , t2 )

zt11 (t1 , 0) z(t1 , 0)

t2

≤ ∫ c(t1 , s2 )Δ2 s2 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

0

or t2

Δ

zt11 (t1 , t2 ) z(t1 , t2 )

−

aΔ1 (t1 ) ≤ ∫ c(t1 , s2 )Δ2 s2 , a(t1 ) + b(0)

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

0

or t2

Δ

zt11 (t1 , t2 )

aΔ1 (t1 ) ≤ + ∫ c(t1 , s2 )Δ2 s2 z(t1 , t2 ) a(t1 ) + b(0) = f (t1 , t2 ),

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

or Δ

zt11 (t1 , t2 ) ≤ f (t1 , t2 )z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Applying Lemma 2.1.1, we obtain z(t1 , t2 ) ≤ z(0, t2 )ef (t1 , 0)

= (a(0) + b(t2 ))ef (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Consequently, u(t1 , t2 ) ≤ z(t1 , t2 )

≤ (a(0) + b(t2 ))ef (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

This completes the proof. Theorem 6.1.2. Let u, a, b ∈ 𝒞 ((ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 )) be nonnegative functions, a(t1 , t2 ) be nondecreasing in each variable t1 , t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). If t1 t2

u(t1 , t2 ) ≤ a(t1 , t2 ) + ∫ ∫ b(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

150 | 6 Two-dimensional linear integral inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ a(t1 , t2 )ec (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where t2

c(t1 , t2 ) = ∫ b(t1 , s2 )Δ2 s2 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0

Proof. We have t1 t2

u(t1 , t2 ) 1 ≤1+ ∫ ∫ b(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 a(t1 , t2 ) a(t1 , t2 ) 0 0

t1 t2

= 1 + ∫ ∫ b(s1 , s2 ) 0 0

t1 t2

≤ 1 + ∫ ∫ b(s1 , s2 ) 0 0

u(s1 , s2 ) ΔsΔs a(t1 , t2 ) 2 2 1 1 u(s1 , s2 ) ΔsΔs, a(s1 , s2 ) 2 2 1 1

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, by Theorem 6.1.1, we get u(t1 , t2 ) ≤ ec (t1 , 0), a(t1 , t2 )

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

whereupon u(t1 , t2 ) ≤ a(t1 , t2 )ec (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

This completes the proof.

6.2 Pachpatte’s inequalities Theorem 6.2.1 (Pachpatte’s Inequality). Let u, p, q ∈ 𝒞 ((ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 )) be nonnegative functions, k(t1 , t2 , s1 , s2 ) and its partial derivatives Δ

Δ

kt11 (t1 , t2 , s1 , s2 ),

kt11 (σ1 (t1 ), t2 , s1 , s2 ),

kt22 (t1 , t2 , s1 , s2 ),

kt22 (t1 , σ2 (t2 ), s1 , s2 ),

Δ

ΔΔ

kt11t2 2 (t1 , t2 , s1 , s2 ),

Δ

(t1 , t2 ), (s1 , s2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

be nonnegative continuous functions. If t1 t2

u(t1 , t2 ) ≤ p(t1 , t2 ) + q(t1 , t2 ) ∫ ∫ k(t1 , t2 , s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

6.2 Pachpatte’s inequalities | 151

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ p(t1 , t2 ) + q(t1 , t2 )A(t1 , t2 )ec (t2 , 0), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where a(t1 , t2 ) = k(σ1 (t1 ), σ2 (t2 ), t1 , t2 )p(t1 , t2 ) t2

Δ

+ ∫ kt22 (σ1 (t1 ), t2 , t1 , s2 )p(t1 , s2 )Δ2 s2 0

t1

Δ

+ ∫ kt11 (t1 , σ2 (t2 ), s1 , t2 )p(s1 , t2 )Δ1 s1 0

t1 t2

ΔΔ

+ ∫ ∫ kt11t2 2 (t1 , t2 , s1 , s2 )p(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

b(t1 , t2 ) = k(σ1 (t1 ), σ2 (t2 ), t1 , t2 )q(t1 , t2 ) t2

Δ

+ ∫ kt22 (σ1 (t1 ), t2 , t1 , s2 )q(t1 , s2 )Δ2 s2 0

t1

Δ

+ ∫ kt11 (t1 , σ2 (t2 ), s1 , t2 )q(s1 , t2 )Δ1 s1 0

t1 t2

ΔΔ

+ ∫ ∫ kt11t2 2 (t1 , t2 , s1 , s2 )q(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

A(t1 , t2 ) = ∫ ∫ a(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0 t2

c(t1 , t2 ) = ∫ b(t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

z(t1 , t2 ) = ∫ ∫ k(t1 , t2 , s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0 0

Then u(t1 , t2 ) ≤ p(t1 , t2 ) + q(t1 , t2 )z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

152 | 6 Two-dimensional linear integral inequalities and z(t1 , t2 ) is a nondecreasing function in each variable t1 , t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) ×

(ℝ+ ∩ 𝕋2 ). Then

z(0, 0) = 0, Δ zt11 (t1 , t2 )

t2

= ∫ k(σ1 (t1 ), t2 , t1 , s2 )u(t1 , s2 )Δ2 s2 0

t1 t2

Δ

+ ∫ ∫ kt11 (t1 , t2 , s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 , ΔΔ zt11t2 2 (t1 , t2 )

0 0

= k(σ1 (t1 ), σ2 (t2 ), t1 , t2 )u(t1 , t2 ) t2

Δ

+ ∫ kt22 (σ1 (t1 ), t2 , t1 , s2 )u(t1 , s2 )Δ2 s2 0

t1

Δ

+ ∫ kt11 (t1 , σ2 (t2 ), s1 , t2 )u(s1 , t2 )Δ1 s1 0

t1 t2

ΔΔ

+ ∫ ∫ kt11t2 2 (t1 , t2 , s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

≤ k(σ1 (t1 ), σ2 (t2 ), t1 , t2 )p(t1 , t2 ) + k(σ1 (t1 ), σ2 (t2 ), t1 , t2 )q(t1 , t2 )z(t1 , t2 ) t2

Δ

+ ∫ kt22 (σ1 (t1 ), t2 , t1 , s2 )p(t1 , s2 )Δ2 s2 0

t2

Δ

+ ∫ kt22 (σ1 (t1 ), t2 , t1 , s2 )q(t1 , s2 )z(t1 , s2 )Δ2 s2 0

t1

Δ

+ ∫ kt11 (t1 , σ2 (t2 ), s1 , t2 )p(s1 , t2 )Δ1 s1 0

t1

Δ

+ ∫ kt11 (t1 , σ2 (t2 ), s1 , t2 )q(s1 , t2 )z(s1 , t2 )Δ1 s1 0

t1 t2

ΔΔ

+ ∫ ∫ kt11t2 2 (t1 , t2 , s1 , s2 )p(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

ΔΔ

+ ∫ ∫ kt11t2 2 (t1 , t2 , s1 , s2 )q(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 0 0

6.2 Pachpatte’s inequalities | 153

≤ a(t1 , t2 )

+ k(σ1 (t1 ), σ2 (t2 ), t1 , t2 )q(t1 , t2 )z(t1 , t2 ) t2

Δ

+ (∫ kt22 (σ1 (t1 ), t2 , t1 , s2 )q(t1 , s2 )Δ2 s2 )z(t1 , t2 ) 0

t1

Δ

+ (∫ kt11 (t1 , σ2 (t2 ), s1 , t2 )q(s1 , t2 )Δ1 s1 )z(t1 , t2 ) 0

t1 t2

ΔΔ

+ (∫ ∫ kt11t2 2 (t1 , t2 , s1 , s2 )q(s1 , s2 )Δ2 s2 Δ1 s1 )z(t1 , t2 ) 0 0

= a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, using that Δ

zt11 (t1 , 0) = 0, z(0, t2 ) = 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

we obtain Δ zt11 (t1 , t2 )

−

Δ zt11 (t1 , 0)

Δ zt11 (t1 , t2 )

t2

t2

≤ ∫ a(t1 , s2 )Δ2 s2 + ∫ b(t1 , s2 )z(t1 , s2 )Δ2 s2 , 0

0

t2

t2

≤ ∫ a(t1 , s2 )Δ2 s2 + ∫ b(t1 , s2 )z(t1 , s2 )Δ2 s2 , 0

0

t1 t2

z(t1 , t2 ) − z(0, t2 ) ≤ ∫ ∫ a(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ ∫ ∫ b(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

z(t1 , t2 ) ≤ A(t1 , t2 ) + ∫ ∫ b(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Now we apply Theorem 6.1.2 and get z(t1 , t2 ) ≤ A(t1 , t2 )ec (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

whereupon u(t1 , t2 ) ≤ p(t1 , t2 ) + q(t1 , t2 )z(t1 , t2 )

≤ p(t1 , t2 ) + q(t1 , t2 )A(t1 , t2 )ec (t2 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof.

154 | 6 Two-dimensional linear integral inequalities Corollary 6.2.2. Let u, p, q, k ∈ 𝒞 ((ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 )) be nonnegative functions. If t1 t2

u(t1 , t2 ) ≤ p(t1 , t2 ) + q(t1 , t2 ) ∫ ∫ k(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ p(t1 , t2 ) + q(t1 , t2 )A(t1 , t2 )ec (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where a(t1 , t2 ) = k(t1 , t2 )p(t1 , t2 ),

b(t1 , t2 ) = k(t1 , t2 )q(t1 , t2 ), t1 t2

A(t1 , t2 ) = ∫ ∫ a(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0 t2

c(t1 , t2 ) = ∫ b(t1 , s2 )Δ2 s2 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0

Theorem 6.2.3. Let c1 and c2 be nonnegative constants, u, v, hi ∈ 𝒞 ((ℝ+ ∩𝕋1 )×(ℝ+ ∩𝕋2 )), i ∈ {1, 2, 3, 4}, be nonnegative functions. If t1 t2

u(t1 , t2 ) ≤ c1 + ∫ ∫ h1 (s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ ∫ ∫ h2 (s1 , s2 )v(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

v(t1 , t2 ) ≤ c2 + ∫ ∫ h3 (s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ ∫ ∫ h4 (s1 , s2 )v(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) + v(t1 , t2 ) ≤ c3 + A(t1 , t2 )ec (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where c3 = c1 + c2 ,

H(t1 , t2 ) = max{h1 (t1 , t2 ) + h3 (t1 , t2 ), h2 (t1 , t2 ) + h4 (t1 , t2 )},

6.2 Pachpatte’s inequalities | 155 t1 t2

A(t1 , t2 ) = c3 ∫ ∫ H(s1 , s2 )Δ2 s2 Δ1 s1 , t2

0 0

c(t1 , t2 ) = ∫ H(t1 , s2 )Δ2 s2 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0

Proof. Let f (t1 , t2 ) = u(t1 , t2 ) + v(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Then f (t1 , t2 ) = u(t1 , t2 ) + v(t1 , t2 ) t1 t2

≤ c1 + ∫ ∫ h1 (s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ ∫ ∫ h2 (s1 , s2 )v(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ c2 + ∫ ∫ h3 (s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 t1 t2

0 0

+ ∫ ∫ h4 (s1 , s2 )v(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

= c3 + ∫ ∫(h1 (s1 , s2 ) + h3 (s1 , s2 ))u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ ∫ ∫(h2 (s1 , s2 ) + h4 (s1 , s2 ))v(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

≤ c3 + ∫ ∫ H(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ ∫ ∫ H(s1 , s2 )v(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

= c3 + ∫ ∫ H(s1 , s2 )(u(s1 , s2 ) + v(s1 , s2 ))Δ2 s2 Δ1 s1 0 0

t1 t2

= c3 + ∫ ∫ H(s1 , s2 )f (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

156 | 6 Two-dimensional linear integral inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, using Corollary 6.2.2, we obtain u(t1 , t2 ) + v(t1 , t2 ) = f (t1 , t2 )

≤ c3 + A(t1 , t2 )ec (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

This completes the proof. ΔΔ

Theorem 6.2.4 (Pachpatte’s Inequality). Let u ∈ 𝒞 2 ((ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 )), ut11,t22 (t1 , t2 ) and c(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and u(0, t2 ) = u(t1 , 0) = 0

for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Let also a ∈ 𝒞 1 (ℝ+ ∩ 𝕋1 ), b ∈ 𝒞 1 (ℝ+ ∩ 𝕋2 ) be positive functions having derivatives such that aΔ1 (t1 ) ≥ 0,

t1 ∈ ℝ+ ∩ 𝕋1 ,

bΔ2 (t2 ) ≥ 0,

t2 ∈ ℝ+ ∩ 𝕋2 .

If ΔΔ ut11t2 2 (t1 , t2 )

t1 t2

ΔΔ

≤ a(t1 ) + b(t2 ) + ∫ ∫ c(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then t1 t2

u(t1 , t2 ) ≤ ∫ ∫ h(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

0 0

where t2

aΔ1 (t1 ) p(t1 , t2 ) = + ∫(1 + c(t1 , s2 ))Δ2 s2 , a(t1 ) + b(0) 0

q(t1 , t2 ) = (a(0) + b(t2 ))ep (t1 , 0)c(t1 , t2 ), t1 t2

h(t1 , t2 ) = a(t1 ) + b(t2 ) + ∫ ∫ q(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

ΔΔ

z(t1 , t2 ) = a(t1 ) + b(t2 ) + ∫ ∫ c(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

6.2 Pachpatte’s inequalities | 157

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

z(t1 , 0) = a(t1 ) + b(0),

t1 ∈ ℝ+ ∩ 𝕋1 ,

z(0, t2 ) = a(0) + b(t2 ),

t2 ∈ ℝ+ ∩ 𝕋2 ,

t2

Δ

(6.2)

ΔΔ

zt11 (t1 , t2 ) = aΔ1 (t1 ) + ∫ c(t1 , s2 )(u(t1 , s2 ) + ut11t2 2 (t1 , s2 ))Δ2 s2 , 0

and ΔΔ

ΔΔ

zt11t2 2 (t1 , t2 ) = c(t1 , t2 )(u(t1 , t2 ) + ut11t2 2 (t1 , t2 )),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Since u(0, t2 ) = u(t1 , 0) = 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

we have Δ

Δ

ut22 (0, t2 ) = ut11 (t1 , 0) = 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, by (6.2), we obtain Δ ut11 (t1 , t2 )

−

Δ ut11 (t1 , 0)

t2

≤ ∫ z(t1 , s2 )Δ2 s2 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

0

or Δ ut11 (t1 , t2 )

t2

≤ ∫ z(t1 , s2 )Δ2 s2 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

0

whereupon t1 t2

u(t1 , t2 ) − u(0, t2 ) ≤ ∫ ∫ z(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

0 0

or t1 t2

u(t1 , t2 ) ≤ ∫ ∫ z(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0 0

Using the last inequality and (6.3), we get ΔΔ zt11t2 2 (t1 , t2 )

t1 t2

≤ c(t1 , t2 )(z(t1 , t2 ) + ∫ ∫ z(s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(6.3)

158 | 6 Two-dimensional linear integral inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Let t1 t2

v(t1 , t2 ) = z(t1 , t2 ) + ∫ ∫ z(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0 0

Then v(t1 , 0) = z(t1 , 0),

v(0, t2 ) = z(0, t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

and z(t1 , t2 ) Δ1 Δ2 zt1 t2 (t1 , t2 )

≤ v(t1 , t2 ),

≤ c(t1 , t2 )v(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Next, Δ vt11 (t1 , t2 ) ΔΔ vt11t2 2 (t1 , t2 )

t2

=

Δ zt11 (t1 , t2 )

=

ΔΔ zt11t2 2 (t1 , t2 )

+ ∫ z(t1 , s2 )Δ2 s2 , 0

+ z(t1 , t2 )

≤ c(t1 , t2 )v(t1 , t2 ) + v(t1 , t2 ) = (1 + c(t1 , t2 ))v(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

From here, ΔΔ

vt11t2 2 (t1 , t2 ) v(t1 , t2 )

≤ 1 + c(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

or ΔΔ

vt11t2 2 (t1 , t2 )v(t1 , t2 ) (v(t1 , t2 ))2

≤ 1 + c(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Since Δ zt22 (t1 , t2 )

Δ2

t1

ΔΔ

= b (t2 ) + ∫ c(s1 , t2 )(u(s1 , t2 ) + ut11t2 2 (s1 , t2 ))Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and because bΔ2 (t2 ) ≥ 0,

c(t1 , t2 ) ≥ 0,

t2 ∈ ℝ+ ∩ 𝕋2 ,

u(t1 , t2 ) ≥ 0,

ΔΔ

ut11t2 2 (t1 , t2 ) ≥ 0,

(6.4)

6.2 Pachpatte’s inequalities | 159

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), we conclude that z(t1 , t2 ) is a nondecreasing function with respect to t2 . Therefore v(t1 , t2 ) is a nondecreasing function with respect to t2 . From here, v(t1 , t2 ) ≤ v(t1 , σ2 (t2 )),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, using (6.4), we obtain ΔΔ

vt11t2 2 (t1 , t2 )v(t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

≤ 1 + c(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Observe that Δ

vt11 (t1 , t2 ) ≥ 0, Δ vt22 (t1 , t2 )

=

Δ zt22 (t1 , t2 )

≥ 0,

t1

+ ∫ z(s1 , t2 )Δ1 s1 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

From here and (6.5), we obtain ΔΔ

vt11t2 2 (t1 , t2 )v(t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

≤ 1 + c(t1 , t2 ) ≤ 1 + c(t1 , t2 ) +

Δ

Δ

vt11 (t1 , t2 )vt22 (t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or ΔΔ

vt11t2 2 (t1 , t2 )v(t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

Δ

−

Δ

vt11 (t1 , t2 )vt22 (t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

≤ 1 + c(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or Δ

(

vt11 (t1 , t2 ) v(t1 , t2 )

Δ2

)

t2

≤ 1 + c(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

From here, Δ

vt11 (t1 , t2 ) v(t1 , t2 )

Δ

−

vt11 (t1 , 0) v(t1 , 0)

t2

≤ ∫(1 + c(t1 , s2 ))Δ2 s2 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

0

or Δ

vt11 (t1 , t2 ) v(t1 , t2 )

Δ

−

zt11 (t1 , 0) z(t1 , 0)

t2

≤ ∫(1 + c(t1 , s2 ))Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

(6.5)

160 | 6 Two-dimensional linear integral inequalities or t2

Δ

vt11 (t1 , t2 )

aΔ1 (t1 ) − ≤ ∫(1 + c(t1 , s2 ))Δ2 s2 , v(t1 , t2 ) a(t1 ) + b(0)

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

0

or t2

Δ

vt11 (t1 , t2 ) v(t1 , t2 )

≤

aΔ1 (t1 ) + ∫(1 + c(t1 , s2 ))Δ2 s2 a(t1 ) + b(0) 0

= p(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

or Δ

vt11 (t1 , t2 ) ≤ p(t1 , t2 )v(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

From the last inequality and Lemma 2.1.1, we obtain v(t1 , t2 ) ≤ v(0, t2 )ep (t1 , 0) = z(0, t2 )ep (t1 , 0)

= (a(0) + b(t2 ))ep (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Therefore ΔΔ

zt11t2 2 (t1 , t2 ) ≤ c(t1 , t2 )v(t1 , t2 )

≤ (a(0) + b(t2 ))ep (t1 , 0)c(t1 , t2 ) = q(t1 , t2 ),

Δ

Δ

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

t2

zt11 (t1 , t2 ) − zt11 (t1 , 0) ≤ ∫ q(t1 , s2 )Δ2 s2 , 0

Δ

t2

zt11 (t1 , t2 ) ≤ aΔ1 (t1 ) + ∫ q(t1 , s2 )Δ2 s2 , 0

t1 t2

z(t1 , t2 ) − z(0, t2 ) ≤ a(t1 ) − a(0) + ∫ ∫ q(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

z(t1 , t2 ) ≤ a(0) + b(t2 ) + a(t1 ) − a(0) + ∫ ∫ q(s1 , s2 )Δ2 s2 Δ1 s1 t1 t2

0 0

= a(t1 ) + b(t2 ) + ∫ ∫ q(s1 , s2 )Δ2 s2 Δ1 s1 = h(t1 , t2 ),

0 0

6.2 Pachpatte’s inequalities | 161

ΔΔ

ut11t2 2 (t1 , t2 ) ≤ h(t1 , t2 ), Δ ut11 (t1 , t2 )

−

Δ ut11 (t1 , 0)

Δ ut11 (t1 , t2 )

t2

≤ ∫ h(t1 , s2 )Δ2 s2 , 0

t2

≤ ∫ h(t1 , s2 )Δ2 s2 , 0

t1 t2

u(t1 , t2 ) − u(0, t2 ) ≤ ∫ ∫ h(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

u(t1 , t2 ) ≤ ∫ ∫ h(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0 0

This completes the proof. ΔΔ

Theorem 6.2.5 (Pachpatte’s Inequality). Let u ∈ 𝒞 2 ((ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 )), ut11,t22 (t1 , t2 ) and c(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and u(0, t2 ) = u(t1 , 0) = 0 for

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Let also a ∈ 𝒞 1 (ℝ+ ∩ 𝕋1 ), b ∈ 𝒞 1 (ℝ+ ∩ 𝕋2 ) be positive functions having derivatives such that aΔ1 (t1 ) ≥ 0,

bΔ2 (t2 ) ≥ 0,

t1 ∈ ℝ+ ∩ 𝕋1 ,

t2 ∈ ℝ+ ∩ 𝕋2 ,

and assume that M is a nonnegative constant. If ΔΔ

ut11t2 2 (t1 , t2 ) ≤ a(t1 ) + b(t2 )

t1 t2

ΔΔ

+ M(u(t1 , t2 ) + ∫ ∫ c(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ q(t1 , t2 )eh1 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where q(t1 , t2 ) = a(t1 ) + b(t2 ),

p(t1 , t2 ) = c(t1 , t2 )(1 + M) + M, t2

h1 (t1 , t2 ) = ∫ p(t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

162 | 6 Two-dimensional linear integral inequalities Proof. Let z(t1 , t2 ) = a(t1 ) + b(t2 )

t1 t2

ΔΔ

+ M(u(t1 , t2 ) + ∫ ∫ c(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then Mu(t1 , t2 ) ≤ z(t1 , t2 ),

z(0, t2 ) = a(0) + b(t2 ) + Mu(0, t2 ) = a(0) + b(t2 ),

z(t1 , 0) = a(t1 ) + b(0) + Mu(t1 , 0) ΔΔ ut11t2 2 (t1 , t2 )

= a(t1 ) + b(0), ≤ z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Next, Δ

zt11 (t1 , t2 ) = aΔ1 (t1 )

t2

Δ

ΔΔ

+ M(ut11 (t1 , t2 ) + ∫ c(t1 , s2 )(u(t1 , s2 ) + ut11t2 2 (t1 , s2 ))Δ2 s2 ), ΔΔ

0

ΔΔ

ΔΔ

zt11t2 2 (t1 , t2 ) = M(ut11t2 2 (t1 , t2 ) + c(t1 , t2 )(u(t1 , t2 ) + ut11t2 2 (t1 , t2 ))) ΔΔ

= M(1 + c(t1 , t2 ))ut11t2 2 (t1 , t2 ) + Mc(t1 , t2 )u(t1 , t2 )

≤ M(1 + c(t1 , t2 ))z(t1 , t2 ) + c(t1 , t2 )z(t1 , t2 ) = (c(t1 , t2 )(1 + M) + M)z(t1 , t2 )

= p(t1 , t2 )z(t1 , t2 ), Δ zt11 (t1 , t2 )

−

t2

Δ zt11 (t1 , 0)

= ∫ p(t1 , s2 )z(t1 , s2 )Δ2 s2 ,

Δ zt11 (t1 , t2 )

= aΔ1 (t1 ) + Mut11 (t1 , 0)

0

Δ

t2

+ ∫ p(t1 , s2 )z(t1 , s2 )Δ2 s2 , 0 Δ1

t2

= a (t1 ) + ∫ p(t1 , s2 )z(t1 , s2 )Δ2 s2 , 0

t1 t2

z(t1 , t2 ) − z(0, t2 ) = a(t1 ) − a(0) + ∫ ∫ p(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

6.2 Pachpatte’s inequalities | 163

z(t1 , t2 ) = a(0) + b(t2 ) + a(t1 ) − a(0) t1 t2

+ ∫ ∫ p(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

= a(t1 ) + b(t2 ) + ∫ ∫ p(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 t1 t2

0 0

= q(t1 , t2 ) + ∫ ∫ p(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, by Theorem 6.1.2, we get z(t1 , t2 ) ≤ q(t1 , t2 )eh1 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ q(t1 , t2 )eh1 (t1 , 0), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 6.2.6 (Pachpatte’s Inequality). Let u(t1 , t2 ), f (t1 , t2 ), and g(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), a(t1 ) and b(t2 ) be positive continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) such that aΔ1 (t1 ) ≥ 0,

bΔ2 (t2 ) ≥ 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

If u(t1 , t2 ) ≤ a(t1 ) + b(t2 ) t1 t2

s1 s2

+ ∫ ∫ f (s1 , s2 )(u(s1 , s2 ) + ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then t1 t2

u(t1 , t2 ) ≤ a(t1 ) + b(t2 ) + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where h1 (t1 , t2 ) = f (t1 , t2 ) + g(t1 , t2 ), t2

h2 (t1 , t2 ) =

aΔ1 (t1 ) + ∫ h1 (t1 , s2 )Δ2 s2 , a(t1 ) + b(0) 0

h3 (t1 , t2 ) = f (t1 , t2 )(a(0) + b(t2 ))eh2 (t1 , 0), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

164 | 6 Two-dimensional linear integral inequalities Proof. Let z(t1 , t2 ) = a(t1 ) + b(t2 )

s1 s2

t1 t2

+ ∫ ∫ f (s1 , s2 )(u(s1 , s2 ) + ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then u(t1 , t2 ) ≤ z(t1 , t2 ),

z(0, t2 ) = a(0) + b(t2 ),

z(t1 , 0) Δ1 zt1 (t1 , t2 )

= a(t1 ) + b(0), = aΔ1 (t1 ) t2

t1 s2

+ ∫ f (t1 , s2 )(u(t1 , s2 ) + ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0

t1 t2

ΔΔ

0 0

zt11t2 2 (t1 , t2 ) = f (t1 , t2 )(u(t1 , t2 ) + ∫ ∫ g(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 ) 0 0

t1 t2

≤ f (t1 , t2 )(z(t1 , t2 ) + ∫ ∫ g(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Let t1 t2

v(t1 , t2 ) = z(t1 , t2 ) + ∫ ∫ g(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then ΔΔ

zt11t2 2 (t1 , t2 ) ≤ f (t1 , t2 )v(t1 , t2 ), z(t1 , t2 ) ≤ v(t1 , t2 ), v(t1 , 0) = z(t1 , 0)

= a(t1 ) + b(0),

v(0, t2 ) = z(0, t2 )

= a(0) + b(t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Next, Δ vt11 (t1 , t2 )

=

Δ zt11 (t1 , t2 )

t2

+ ∫ g(t1 , s2 )z(t1 , s2 )Δ2 s2 , 0

(6.6)

6.2 Pachpatte’s inequalities | 165

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and ΔΔ

ΔΔ

vt11t2 2 (t1 , t2 ) = zt11t2 2 (t1 , t2 ) + g(t1 , t2 )z(t1 , t2 )

≤ f (t1 , t2 )v(t1 , t2 ) + g(t1 , t2 )v(t1 , t2 )

= (f (t1 , t2 ) + g(t1 , t2 ))v(t1 , t2 ) = h1 (t1 , t2 )v(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩𝕋1 )×(ℝ+ ∩𝕋2 ). Since a(t1 ) and b(t2 ) are positive continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩𝕋1 )×(ℝ+ ∩𝕋2 ), we get that z(t1 , t2 ) is a positive function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and then v(t1 , t2 ) is a positive function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, ΔΔ

vt11t2 2 (t1 , t2 )v(t1 , t2 ) (v(t1 , t2 ))2

ΔΔ

=

vt11t2 2 (t1 , t2 ) v(t1 , t2 )

≤ h1 (t1 , t2 ),

(6.7) (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Also since Δ vt22 (t1 , t2 )

=

Δ zt22 (t1 , t2 )

t1

+ ∫ g(s1 , t2 )z(s1 , t2 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and Δ

zt22 (t1 , t2 ) = bΔ2 (t2 ) t1

s1 t2

+ ∫ f (s1 , t2 )(u(s1 , t2 ) + ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 ≥ 0,

0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

we get Δ

vt22 (t1 , t2 ) ≥ 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

By (6.6), we obtain that Δ

vt11 (t1 , t2 ) ≥ 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Consequently, v(t1 , t2 ) is a nondecreasing function with respect to t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then v(t1 , t2 ) ≤ v(t1 , σ2 (t2 )),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, by (6.7), we obtain ΔΔ

ΔΔ

vt11t2 2 (t1 , t2 )v(t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

≤

vt11t2 2 (t1 , t2 )v(t1 , t2 ) (v(t1 , t2 ))2

166 | 6 Two-dimensional linear integral inequalities ≤ h1 (t1 , t2 )

≤ h1 (t1 , t2 ) Δ

+

Δ

vt11 (t1 , t2 )vt22 (t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or ΔΔ

Δ

Δ

vt11t2 2 (t1 , t2 )v(t1 , t2 ) − vt11 (t1 , t2 )vt22 (t1 , t2 ) v(t1 , t2 )v(t1 , σ2 (t2 ))

≤ h1 (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or Δ

(

vt11 (t1 , t2 ) v(t1 , t2 )

Δ2

)

t2

≤ h1 (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From here, Δ

vt11 (t1 , t2 ) v(t1 , t2 )

Δ

−

vt11 (t1 , 0) v(t1 , 0)

t2

≤ ∫ h1 (t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or Δ

vt11 (t1 , t2 ) v(t1 , t2 )

Δ

−

zt11 (t1 , 0) z(t1 , 0)

t2

≤ ∫ h1 (t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or t2

Δ

vt11 (t1 , t2 )

aΔ1 (t1 ) − ≤ ∫ h1 (t1 , s2 )Δ2 s2 , v(t1 , t2 ) a(t1 ) + b(0) 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or t2

Δ

vt11 (t1 , t2 )

aΔ1 (t1 ) ≤ + ∫ h1 (t1 , s2 )Δ2 s2 v(t1 , t2 ) a(t1 ) + b(0) = h2 (t1 , t2 ),

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or Δ

vt11 (t1 , t2 ) ≤ h2 (t1 , t2 )v(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, by Lemma 2.1.1, we obtain v(t1 , t2 ) ≤ v(0, t2 )eh2 (t1 , 0)

6.2 Pachpatte’s inequalities | 167

= z(0, t2 )eh2 (t1 , 0) ΔΔ zt11t2 2 (t1 , t2 )

Δ zt11 (t1 , t2 )

−

Δ zt11 (t1 , 0)

Δ zt11 (t1 , t2 )

= (a(0) + b(t2 ))eh2 (t1 , 0),

≤ f (t1 , t2 )(a(0) + b(t2 ))eh2 (t1 , 0) = h3 (t1 , t2 ), t2

≤ ∫ h3 (t1 , s2 )Δ2 s2 , 0

t2

Δ1

≤ a (t1 ) + ∫ h3 (t1 , s2 )Δ2 s2 , 0

t1 t2

z(t1 , t2 ) − z(0, t2 ) ≤ a(t1 ) − a(0) + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

z(t1 , t2 ) ≤ a(0) + b(t2 ) + a(t1 ) − a(0) t1 t2

+ ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

= a(t1 ) + b(t2 ) + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 , u(t1 , t2 ) ≤ z(t1 , t2 )

0 0 t1 t2

≤ a(t1 ) + b(t2 ) + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 6.2.7. Let u(t1 , t2 ), f (t1 , t2 ), and g(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) and let c(t1 , t2 ) be a positive continuous function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) and nondecreasing in each variable t1 , t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). If t1 t2

u(t1 , t2 ) ≤ c(t1 , t2 ) + ∫ ∫ f (s1 , s2 )(u(s1 , s2 ) s1 s2

0 0

+ ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then t1 t2

u(t1 , t2 ) ≤ c(t1 , t2 )(1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

168 | 6 Two-dimensional linear integral inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where h1 (t1 , t2 ) = f (t1 , t2 ) + g(t1 , t2 ), t2

h2 (t1 , t2 ) = ∫ h1 (t1 , s2 )Δ2 s2 , 0

h3 (t1 , t2 ) = f (t1 , t2 )eh2 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Proof. Since c(t1 , t2 ) is positive for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) and nondecreasing in each variable t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), we get t1 t2

u(t1 , t2 ) u(s , s ) ≤ 1 + ∫ ∫ f (s1 , s2 )( 1 2 c(t1 , t2 ) c(t1 , t2 ) 0 0 s1 s2

+ ∫ ∫ g(τ1 , τ2 ) 0 0

u(τ1 , τ2 ) Δ τ Δ τ )Δ s Δ s c(t1 , t2 ) 2 2 1 1 2 2 1 1

t1 t2

≤ 1 + ∫ ∫ f (s1 , s2 )( 0 0 s1 s2

+ ∫ ∫ g(τ1 , τ2 ) 0 0

u(s1 , s2 ) c(s1 , s2 )

u(τ1 , τ2 ) Δ τ Δ τ )Δ s Δ s , c(τ1 , τ2 ) 2 2 1 1 2 2 1 1

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Let z(t1 , t2 ) =

u(t1 , t2 ) , c(t1 , t2 )

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Then t1 t2

z(t1 , t2 ) ≤ 1 + ∫ ∫ f (s1 , s2 )(z(s1 , s2 ) 0 0 s1 s2

+ ∫ ∫ g(τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, by Theorem 6.2.6, we obtain t1 t2

z(t1 , t2 ) ≤ 1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), whereupon t1 t2

u(t1 , t2 ) ≤ 1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 , c(t1 , t2 ) 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

6.2 Pachpatte’s inequalities | 169

or t1 t2

u(t1 , t2 ) ≤ c(t1 , t2 )(1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 6.2.8. Let u(t1 , t2 ), f (t1 , t2 ), g(t1 , t2 ), and p(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and u0 be a positive constant. If t1 t2

u(t1 , t2 ) ≤ u0 + ∫ ∫(f (s1 , s2 )u(s1 , s2 ) + p(s1 , s2 ))Δ2 s2 Δ1 s1 0 0

t1 t2

s1 s2

+ ∫ ∫ f (s1 , s2 )(∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then t1 t2

t1 t2

u(t1 , t2 ) ≤ (u0 + ∫ ∫ p(s1 , s2 )Δ2 s2 Δ1 s1 )(1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where h1 (t1 , t2 ), h2 (t1 , t2 ) and h3 (t1 , t2 ) are defined as in Theorem 6.2.7. Proof. Let t1 t2

c(t1 , t2 ) = u0 + ∫ ∫ p(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0 0

Then c(t1 , t2 ) is a positive continuous function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and c(t1 , t2 ) is a nondecreasing function in each variable t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then t1 t2

u(t1 , t2 ) ≤ c(t1 , t2 ) + ∫ ∫ f (s1 , s2 )(u(s1 , s2 ) s1 s2

0 0

+ ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, by Theorem 6.2.7, we get t1 t2

u(t1 , t2 ) ≤ c(t1 , t2 )(1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof.

170 | 6 Two-dimensional linear integral inequalities Theorem 6.2.9. Let u(t1 , t2 ), f (t1 , t2 ), g(t1 , t2 ), and p(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and u0 be a positive constant. If t1 t2

u(t1 , t2 ) ≤ u0 + ∫ ∫ f (s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

s1 s2

+ ∫ ∫ f (s1 , s2 )(∫ ∫(g(τ1 , τ2 )u(τ1 , τ2 ) + p(τ1 , τ2 ))Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then t1 t2

s1 s2

u(t1 , t2 ) ≤ (u0 + ∫ ∫ f (s1 , s2 ) ∫ ∫ p(σ1 , σ2 )Δ2 σ2 Δ1 σ1 Δ2 s2 Δ1 s1 ) 0 0

0 0

t1 t2

× (1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where h1 (t1 , t2 ), h2 (t1 , t2 ), and h3 (t1 , t2 ) are defined as in Theorem 6.2.7. Proof. Let t1 t2

s1 s2

c(t1 , t2 ) = u0 + ∫ ∫ f (s1 , s2 ) ∫ ∫ p(τ1 , τ2 )Δ2 τ2 Δ1 τ1 Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then c(t1 , t2 ) is a positive continuous function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and c(t1 , t2 ) is a nondecreasing function in each variable t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then t1 t2

u(t1 , t2 ) ≤ c(t1 , t2 ) + ∫ ∫ f (s1 , s2 )(u(s1 , s2 ) s1 s2

0 0

+ ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, by Theorem 6.2.7, we get t1 t2

u(t1 , t2 ) ≤ c(t1 , t2 )(1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof.

6.2 Pachpatte’s inequalities | 171

Theorem 6.2.10. Let u(t1 , t2 ), f (t1 , t2 ), g(t1 , t2 ), and h(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), u0 be a positive constant. If t1 t2

u(t1 , t2 ) ≤ u0 + ∫ ∫ h(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ ∫ ∫ f (s1 , s2 )(u(s1 , s2 ) 0 0 s1 s2

+ ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then t1 t2

u(t1 , t2 ) ≤ u0 (1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where f1 (t1 , t2 ) = 2(f (t1 , t2 ) + g(t1 , t2 )),

h1 (t1 , t2 ) = f1 (t1 , t2 ) + g(t1 , t2 ), t2

h2 (t1 , t2 ) = ∫ h1 (t1 , s2 )Δ2 s2 , 0

h3 (t1 , t2 ) = f1 (t1 , t2 )eh2 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Proof. We have t1 t2

u(t1 , t2 ) ≤ u0 + ∫ ∫(h(s1 , s2 ) + f (s1 , s2 ))u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ ∫ ∫(f (s1 , s2 ) + h(s1 , s2 ))(u(s1 , s2 ) 0 0 s1 s2

+ ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 0 0

t1 t2

≤ u0 + ∫ ∫ 2(h(s1 , s2 ) + f (s1 , s2 ))(u(s1 , s2 ) 0 0 s1 s2

+ ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 0 0

172 | 6 Two-dimensional linear integral inequalities t1 t2

= u0 + ∫ ∫ f1 (s1 , s2 )(u(s1 , s2 ) 0 0 s1 s2

+ ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, by Theorem 6.2.7, we get t1 t2

u(t1 , t2 ) ≤ u0 (1 + ∫ ∫ h3 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 6.2.11. Let u(t1 , t2 ), h(t1 , t2 ), p(t1 , t2 ), f (t1 , t2 ), and g(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and t1 t2

u(t1 , t2 ) ≤ h(t1 , t2 ) + p(t1 , t2 ) ∫ ∫ f (s1 , s2 )(u(s1 , s2 ) + p(s1 , s2 ) 0 0

s1 s2

× ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then u(t1 , t2 ) ≤ h(t1 , t2 ) + p(t1 , t2 )(1 + h(t1 , t2 )) t1 t2

× (1 + ∫ ∫ h4 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where t1 t2

h1 (t1 , t2 ) = ∫ ∫(f (s1 , s2 )h(s1 , s2 ) 0 0

s1 s2

+ p(s1 , s2 ) ∫ ∫ g(τ1 , τ2 )h(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

p1 (t1 , t2 ) = g(t1 , t2 )p(t1 , t2 ),

f1 (t1 , t2 ) = (1 + f (t1 , t2 ))p(t1 , t2 ),

h2 (t1 , t2 ) = f1 (t1 , t2 ) + p1 (t1 , t2 ), t2

h3 (t1 , t2 ) = ∫ h2 (t1 , s2 )Δ2 s2 , 0

6.2 Pachpatte’s inequalities | 173

h4 (t1 , t2 ) = f1 (t1 , t2 )eh3 (t1 , 0), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

z(t1 , t2 ) = ∫ ∫ f (s1 , s2 )(u(s1 , s2 ) + p(s1 , s2 ) 0 0 s1 s2

× ∫ ∫ g(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then u(t1 , t2 ) ≤ h(t1 , t2 ) + p(t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, t1 t2

z(t1 , t2 ) ≤ ∫ ∫ f (s1 , s2 )(h(s1 , s2 ) + p(s1 , s2 )z(s1 , s2 ) 0 0

s1 s2

+ p(s1 , s2 ) ∫ ∫ g(τ1 , τ2 )(h(τ1 , τ2 ) 0 0

+ p(τ1 , τ2 )z(τ1 , τ2 ))Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 t1 t2

= ∫ ∫(f (s1 , s2 )h(s1 , s2 ) + p(s1 , s2 ) 0 0 s1 s2

× ∫ ∫ g(τ1 , τ2 )h(τ1 , τ2 )Δ2 τ2 Δ1 τ1 ) 0 0

t1 t2

+ ∫ ∫(f (s1 , s2 )p(s1 , s2 )z(s1 , s2 ) + p(s1 , s2 ) 0 0 s1 s2

× ∫ ∫ g(τ1 , τ2 )p(τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 0 0

t1 t2

= h1 (t1 , t2 ) + ∫ ∫(f (s1 , s2 ) + 1)p(s1 , s2 )(z(s1 , s2 ) s1 s2

0 0

+ ∫ ∫ p1 (τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 0 0

(6.8)

174 | 6 Two-dimensional linear integral inequalities t1 t2

= 1 + h1 (t1 , t2 ) + ∫ ∫ f1 (s1 , s2 )(z(s1 , s2 ) 0 0

s1 s2

+ ∫ ∫ p1 (τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩𝕋1 )×(ℝ+ ∩𝕋2 ). Since the function 1+h1 (t1 , t2 ) is a nondecreasing function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), we can apply Theorem 6.2.7 and obtain t1 t2

z(t1 , t2 ) ≤ (1 + h1 (t1 , t2 ))(1 + ∫ ∫ h4 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From here and (6.8), we get u(t1 , t2 ) ≤ h(t1 , t2 ) + p(t1 , t2 )(1 + h(t1 , t2 )) t1 t2

× (1 + ∫ ∫ h4 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 6.2.12 (Pachpatte’s Inequality). Let u(t1 , t2 ), h(t1 , t2 ), and p(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), f (t1 , t2 ) be a positive continuous function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and suppose u0 is a nonnegative constant. If u(t1 , t2 ) ≤ u0

t1 t2

s1 s2

+ ∫ ∫ f (s1 , s2 )(h(s1 , s2 ) + ∫ ∫ p(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ q(t1 , t2 )eq2 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where t1 t2

q(t1 , t2 ) = u0 + ∫ ∫ f (s1 , s2 )h(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

q1 (t1 , t2 ) = f (t1 , t2 ) ∫ ∫ p(τ1 , τ2 )Δ2 τ2 Δ1 τ1 , t2

0 0

q2 (t1 , t2 ) = ∫ q1 (t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

6.2 Pachpatte’s inequalities | 175

Proof. We can rewrite the given inequality in the form t1 t2

u(t1 , t2 ) ≤ u0 + ∫ ∫ f (s1 , s2 )h(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

s1 s2

+ ∫ ∫ f (s1 , s2 ) ∫ ∫ p(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 Δ2 s2 Δ1 s1 0 0

0 0

t1 t2

s1 s2

= q(t1 , t2 )

+ ∫ ∫ f (s1 , s2 ) ∫ ∫ p(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Note that q(t1 , t2 ) is a positive nondecreasing function in each variable t1 , t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then t1 t2

s1 s2

0 0

0 0

u(t1 , t2 ) u(τ , τ ) ≤ 1 + ∫ ∫ f (s1 , s2 ) ∫ ∫ p(τ1 , τ2 ) 1 2 Δ2 τ2 Δ1 τ1 Δ2 s2 Δ1 s1 q(t1 , t2 ) q(t1 , t2 ) t1 t2

s1 s2

≤ 1 + ∫ ∫ f (s1 , s2 ) ∫ ∫ p(τ1 , τ2 ) 0 0

0 0

u(τ1 , τ2 ) ΔτΔτΔsΔs, q(τ1 , τ2 ) 2 2 1 1 2 2 1 1

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Let z(t1 , t2 ) =

u(t1 , t2 ) , q(t1 , t2 )

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Then t1 t2

s1 s2

z(t1 , t2 ) ≤ 1 + ∫ ∫ f (s1 , s2 ) ∫ ∫ p(τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). We set t1 t2

s1 s2

v(t1 , t2 ) = 1 + ∫ ∫ f (s1 , s2 ) ∫ ∫ p(τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, z(t1 , t2 ) ≤ v(t1 , t2 ), v(t1 , 0) = v(0, t2 ) = v(0, 0) = 1,

176 | 6 Two-dimensional linear integral inequalities t1 s2

t2

Δ vt11 (t1 , t2 )

= ∫ f (t1 , s2 ) ∫ ∫ p(τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 Δ2 s2 ,

Δ vt11 (t1 , 0)

= 0,

Δ vt22 (t1 , t2 )

= ∫ f (s1 , t2 ) ∫ ∫ p(τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 Δ1 s1 ,

0 0

0

t1

s1 t2

0

Δ

0 0

vt22 (0, t2 ) = 0, ΔΔ vt11t2 2 (t1 , t2 ) ΔΔ vt11t2 2 (t1 , t2 )

f (t1 , t2 )

(

ΔΔ vt11t2 2 (t1 , t2 )

f (t1 , t2 )

t2

t1

Δ1 Δ2

)

t1 t2

Δ Δ

v(t1 , t2 )

Δ Δ

= ∫ ∫ p(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

v 1 2 (t1 ,t2 ) Δ Δ ( t1ft2(t ,t ) )t 1t 2 1 2 1 2

(

0 0

t1 t2

) = ∫ p(t1 , s2 )z(t1 , s2 )Δ2 s2 ,

ΔΔ

(

= f (t1 , t2 ) ∫ ∫ p(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 ,

Δ1

f (t1 , t2 )

vt11t2 2 (t1 , t2 )

t1 t2

vt 1t 2 (t1 ,t2 ) Δ1 Δ2 1 2 )t t v(t1 , t2 ) f (t1 ,t2 ) 1 2

v(t1 , t2 )v(t1 , σ2 (t2 ))

0

= p(t1 , t2 )z(t1 , t2 ) ≤ p(t1 , t2 )v(t1 , t2 ), ≤ p(t1 , t2 ), Δ Δ

≤

(

vt 1t 2 (t1 ,t2 ) Δ1 Δ2 1 2 )t t v(t1 , t2 ) f (t1 ,t2 ) 1 2 (v(t1 , t2 ))2

≤ p(t1 , t2 )

≤ p(t1 , t2 ) Δ Δ

+ Δ Δ

(

(

vt 1t 2 (t1 ,t2 ) Δ1 1 2 )t f (t1 ,t2 ) 1

v(t1 , t2 )

Δ2

)

Δ Δ

(

t2

vt 1t 2 (t1 ,t2 ) Δ1 1 2 )t f (t1 ,t2 ) 1

v(t1 , t2 )

v(t1 , t2 )v(σ1 (t1 ), t2 )

vt 1t 2 (t1 ,t2 ) Δ1 Δ 1 2 )t vt22 (t1 , t2 ) f (t1 ,t2 ) 1

v(t1 , t2 )v(t1 , σ2 (t2 ))

≤ p(t1 , t2 ), t2

≤ ∫ p(t1 , s2 )Δ2 s2 , 0

Δ Δ

v 1 2 (t1 ,t2 ) Δ ( t1ft2(t ,t ) )t 1 v(t1 , t2 ) 1 1 2

(

≤

(

Δ Δ

vt 1t 2 (t1 ,t2 ) Δ1 1 2 )t v(t1 , t2 ) f (t1 ,t2 ) 1 (v(t1 , t2 ))2

,

6.2 Pachpatte’s inequalities | 177 t2

≤ ∫ p(t1 , s2 )Δ2 s2 0

t2

≤ ∫ p(t1 , s2 )Δ2 s2 0

Δ Δ

+ ΔΔ

(

vt11t2 2 (t1 , t2 )

f (t1 , t2 )v(t1 , t2 )

t2

Δ1

vt 1t 2 (t1 ,t2 ) Δ 1 2 vt11 (t1 , t2 ) f (t1 ,t2 )

v(t1 , t2 )v(σ1 (t1 ), t2 )

,

) ≤ ∫ p(t1 , s2 )Δ2 s2 , t1

0

t1 t2

ΔΔ vt11t2 2 (t1 , t2 )

f (t1 , t2 )v(t1 , t2 ) ΔΔ vt11t2 2 (t1 , t2 )

v(t1 , t2 )

≤ ∫ ∫ p(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

≤ f (t1 , t2 ) ∫ ∫ p(s1 , s2 )Δ2 s2 Δ1 s1 , ΔΔ

ΔΔ

vt11t2 2 (t1 , t2 )v(t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

t1 t2

≤

0 0

vt11t2 2 (t1 , t2 )v(t1 , t2 ) (v(t1 , t2 ))2 t1 t2

≤ f (t1 , t2 ) ∫ ∫ p(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

≤ f (t1 , t2 ) ∫ ∫ p(s1 , s2 )Δ2 s2 Δ1 s1

+ Δ

(

vt11 (t1 , t2 ) v(t1 , t2 )

Δ2

)

t2

Δ

vt11 (t1 , t2 ) v(t1 , t2 ) Δ

0 0 Δ1 Δ vt1 (t1 , t2 )vt22 (t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

,

≤ q1 (t1 , t2 ), t2

≤ ∫ q1 (t1 , s2 )Δ2 s2 0

= q2 (t1 , t2 ),

vt11 (t1 , t2 ) ≤ q2 (t1 , t2 )v(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From the last inequality and Lemma 2.1.1, we conclude that

v(t1 , t2 ) ≤ eq2 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

178 | 6 Two-dimensional linear integral inequalities Therefore z(t1 , t2 ) ≤ eq2 (t1 , 0),

u(t1 , t2 ) ≤ q(t1 , t2 )eq2 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

This completes the proof. Theorem 6.2.13. Let u(t1 , t2 ), h(t1 , t2 ), and p(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), f (t1 , t2 ) and g(t1 , t2 ) be positive continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and suppose u0 is a positive constant. If t1 t2

s1 s2

u(t1 , t2 ) ≤ u0 + ∫ ∫ f (s1 , s2 )(h(s1 , s2 ) + ∫ ∫ g(r1 , r2 ) 0 0 r1 r2

0 0

× (∫ ∫ p(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 ) 0 0

Δ2 r2 Δ1 r1 )Δ2 s2 Δ1 s1 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ q(t1 , t2 )eh3 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where t1 t2

q(t1 , t2 ) = u0 + ∫ ∫ f (s1 , s2 )h(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

h1 (t1 , t2 ) = g(t1 , t2 ) ∫ ∫ p(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

h2 (t1 , t2 ) = f (t1 , t2 ) ∫ ∫ h1 (s1 , s2 )Δ2 s2 Δ1 s1 , t2

0 0

h3 (t1 , t2 ) = ∫ h2 (t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. We can rewrite the given inequality in the following form: t1 t2

u(t1 , t2 ) ≤ u0 + ∫ ∫ f (s1 , s2 )h(s1 , s2 )Δ2 s2 Δ1 s1 0 0

6.2 Pachpatte’s inequalities | 179 t1 t2

s1 s2

r1 r2

+ ∫ ∫ f (s1 , s2 ) ∫ ∫ g(r1 , r2 )(∫ ∫ p(τ1 , τ2 )u(τ1 , τ2 ) 0 0

0 0

0 0

Δ2 τ2 Δ1 τ1 )Δ2 r2 Δ1 r1 Δ2 s2 Δ1 s1 = q(t1 , t2 ) t1 t2

s1 s2

r1 r2

+ ∫ ∫ f (s1 , s2 ) ∫ ∫ g(r1 , r2 )(∫ ∫ p(τ1 , τ2 )u(τ1 , τ2 ) 0 0

0 0

0 0

Δ2 τ2 Δ1 τ1 )Δ2 r2 Δ1 r1 Δ2 s2 Δ1 s1 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Note that q(t1 , t2 ) is a positive nondecreasing function in each variable t1 , t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then u(t1 , t2 ) ≤1 q(t1 , t2 )

t1 t2

s1 s2

r1 r2

+ ∫ ∫ f (s1 , s2 ) ∫ ∫ g(r1 , r2 )(∫ ∫ p(τ1 , τ2 ) 0 0

0 0

0 0

u(τ1 , τ2 ) q(t1 , t2 )

Δ2 τ2 Δ1 τ1 )Δ2 r2 Δ1 r1 Δ2 s2 Δ1 s1 ≤1

t1 t2

s1 s2

r1 r2

+ ∫ ∫ f (s1 , s2 ) ∫ ∫ g(r1 , r2 )(∫ ∫ p(τ1 , τ2 ) 0 0

0 0

0 0

u(τ1 , τ2 ) q(τ1 , τ2 )

Δ2 τ2 Δ1 τ1 )Δ2 r2 Δ1 r1 Δ2 s2 Δ1 s1 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Let z(t1 , t2 ) =

u(t1 , t2 ) , q(t1 , t2 )

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, z(t1 , t2 ) ≤ 1

t1 t2

s1 s2

r1 r2

+ ∫ ∫ f (s1 , s2 ) ∫ ∫ g(r1 , r2 )(∫ ∫ p(τ1 , τ2 )z(τ1 , τ2 ) 0 0

0 0

Δ2 τ2 Δ1 τ1 )Δ2 r2 Δ1 r1 Δ2 s2 Δ1 s1 ,

0 0

180 | 6 Two-dimensional linear integral inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Now we set v(t1 , t2 ) = 1

r1 r2

s1 s2

t1 t2

+ ∫ ∫ f (s1 , s2 ) ∫ ∫ g(r1 , r2 )(∫ ∫ p(τ1 , τ2 )z(τ1 , τ2 ) 0 0

0 0

0 0

Δ2 τ2 Δ1 τ1 )Δ2 σ2 Δ1 r1 Δ2 s2 Δ1 s1 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then z(t1 , t2 ) ≤ v(t1 , t2 ), v(t1 , 0) = v(0, t2 ) = v(0, 0)

= 1, Δ

t2

t1 s2

r1 r2

vt11 (t1 , t2 ) = ∫ f (t1 , s2 ) ∫ ∫ g(r1 , r2 )(∫ ∫ p(τ1 , τ2 )z(τ1 , τ2 ) 0

0 0

0 0

Δ2 τ2 Δ1 τ1 )Δ2 r2 Δ1 r1 Δ2 s2 , Δ

vt11 (t1 , 0) = 0, Δ vt22 (t1 , t2 )

t1

s1 t2

r1 r2

= ∫ f (s1 , t2 ) ∫ ∫ g(r1 , r2 )(∫ ∫ p(τ1 , τ2 )z(τ1 , τ2 ) 0

0 0

0 0

Δ2 τ2 Δ1 τ1 )Δ2 r2 Δ1 r1 Δ1 s1 , Δ

vt22 (0, t2 ) = 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Note that Δ

vt11 (t1 , t2 ) ≥ 0,

Δ

vt22 (t1 , t2 ) ≥ 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Therefore v(t1 , t2 ) is a positive nondecreasing function in each variable t1 , t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Consequently, v(t1 , t2 ) ≤ v(t1 , σ2 (t2 )), v(t1 , t2 ) ≤ v(σ1 (t1 ), t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Next we get ΔΔ vt11t2 2 (t1 , t2 )

t1 t2

s1 s2

= f (t1 , t2 ) ∫ ∫ g(s1 , s2 ) ∫ ∫ p(r1 , r2 )z(r1 , r2 ) 0 0

Δ2 r2 Δ1 r1 Δ2 s2 Δ1 s1 ,

0 0

(6.9)

6.2 Pachpatte’s inequalities | 181 ΔΔ

vt11t2 2 (t1 , t2 ) f (t1 , t2 )

= ∫ ∫ g(s1 , s2 ) ∫ ∫ p(r1 , r2 )z(r1 , r2 ) Δ2 r2 Δ1 r1 Δ2 s2 Δ1 s1 ,

ΔΔ

f (t1 , 0)

ΔΔ

= =

ΔΔ

Δ1

vt11t2 2 (t1 , t2 ) f (t1 , t2 )

vt11t2 2 (t1 , t2 ) f (t1 , t2 )

Δ1 Δ2

)

t1 t2

0

(

(

t1 t2

= g(t1 , t2 ) ∫ ∫ p(r1 , r2 )z(r1 , r2 )Δ2 r2 Δ1 r1 , 0 0

f (t1 ,t2 )

v(t1 , t2 )

ΔΔ )t 1t 2 1 2

g(t1 , t2 )

t1 t2

= ∫ ∫ p(r1 , r2 )z(r1 , r2 )Δ2 r2 Δ1 r1 , t2

Δ1

) = ∫ p(t1 , r2 )z(t1 , r2 )Δ2 r2 ,

g(t1 , t2 )

Δ Δ vt 1t 2 (t1 ,t2 ) 1 2

)

g(t1 ,t2 )

0 0

0 0

Δ Δ

v 1 2 (t1 ,t2 ) Δ Δ ( t1ft2(t ,t ) )t 1t 2 1 2 1 2

Δ Δ v 1 2 (t1 ,t2 ) Δ1 Δ2 t1 t2 f (t1 ,t2 ) t1 t2

t1 s2

Δ2 r2 Δ1 r1 Δ2 s2 ,

g(t1 , t2 )

(

f (0, 0)

t2

Δ Δ

(

f (0, t2 ) Δ1 Δ2 vt1 t2 (0, 0)

= 0,

t1

v 1 2 (t1 ,t2 ) Δ Δ ( t1ft2(t ,t ) )t 1t 2 1 2 1 2

(

vt11t2 2 (0, t2 )

) = ∫ g(t1 , s2 ) ∫ ∫ p(r1 , r2 )z(r1 , r2 )

ΔΔ

(

0 0

0 0

vt11t2 2 (t1 , 0)

(

s1 s2

t1 t2

t1

Δ1 Δ2

)

t1 t2

0

= p(t1 , t2 )z(t1 , t2 ) ≤ p(t1 , t2 )v(t1 , t2 ),

Δ1 Δ2

)

t1 t2

≤ p(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, using (6.9), we get

(

(

Δ Δ v 1 2 (t1 ,t2 ) Δ1 Δ2 t1 t2 f (t1 ,t2 ) t1 t2

)

g(t1 ,t2 )

Δ1 Δ2

)

t1 t2

v(t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

(

(

Δ Δ v 1 2 (t1 ,t2 ) Δ1 Δ2 t1 t2 f (t1 ,t2 ) t1 t2

)

g(t1 ,t2 )

≤

v(t1 , t2 ) ≤ p(t1 , t2 )

Δ1 Δ2

)

t1 t2

≤ p(t1 , t2 ) ( +

(

Δ Δ v 1 2 (t1 ,t2 ) Δ1 Δ2 t1 t2 f (t1 ,t2 ) t1 t2

)

g(t1 ,t2 )

Δ1

Δ

) vt22 (t1 , t2 ) t1

v(t1 , t2 )v(t1 , σ2 (t2 ))

,

182 | 6 Two-dimensional linear integral inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), whereupon (

(

Δ Δ v 1 2 (t1 ,t2 ) Δ1 Δ2 t1 t2 f (t1 ,t2 ) t1 t2

)

g(t1 ,t2 )

(

(

(

v(t1 , t2 ) Δ Δ v 1 2 (t1 ,t2 ) Δ1 Δ2 t1 t2 f (t1 ,t2 ) t1 t2

)

g(t1 ,t2 )

Δ1 t1

)

Δ1

≤ p(t1 , t2 ),

t2

t2

)

v(t1 , t2 )

Δ2

)

t1

≤ ∫ p(t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From here, using (6.9), we obtain (

(

Δ Δ v 1 2 (t1 ,t2 ) Δ1 Δ2 t1 t2 f (t1 ,t2 ) t1 t2

)

g(t1 ,t2 )

Δ1

) v(t1 , t2 ) t1

(

(

Δ Δ v 1 2 (t1 ,t2 ) Δ1 Δ2 t1 t2 f (t1 ,t2 ) t1 t2

)

g(t1 ,t2 )

≤

v(t1 , t2 )v(r1 (t1 ), t2 )

v(t1 , t2 )

t2

Δ1

)

t1

≤ ∫ p(t1 , s2 )Δ2 s2 0

t2

≤ ∫ p(t1 , s2 )Δ2 s2 0

+

(

(

Δ Δ v 1 2 (t1 ,t2 ) Δ1 Δ2 t1 t2 f (t1 ,t2 ) t1 t2

)

g(t1 ,t2 )

Δ

)vt11 (t1 , t2 )

v(t1 , t2 )v(σ1 (t1 ), t2 )

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore Δ Δ

(

(

vt 1t 2 (t1 ,t2 ) Δ1 Δ2 1 2 )t t f (t1 ,t2 ) 1 2

g(t1 , t2 )v(t1 , t2 ) (

Δ Δ vt 1t 2 (t1 ,t2 ) 1 2

f (t1 ,t2 )

t2

Δ1

) ≤ ∫ p(t1 , s2 )Δ2 s2 , t1

ΔΔ )t 1t 2 1 2

g(t1 , t2 )v(t1 , t2 )

0 t1 t2

≤ ∫ ∫ p(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

Δ Δ

v 1 2 (t1 ,t2 ) Δ Δ ( t1ft2(t ,t ) )t 1t 2 1 2 1 2

v(t1 , t2 )

t1 t2

≤ g(t1 , t2 ) ∫ ∫ p(s1 , s2 )Δ2 s2 Δ1 s1 0 0

= h1 (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Using (6.9), we obtain Δ Δ

(

vt 1t 2 (t1 ,t2 ) Δ1 Δ2 1 2 )t t v(t1 , t2 ) f (t1 ,t2 ) 1 2

v(t1 , t2 )v(t1 , σ2 (t2 ))

Δ Δ

≤

(

vt 1t 2 (t1 ,t2 ) Δ1 Δ2 1 2 )t t f (t1 ,t2 ) 1 2

v(t1 , t2 )

,

6.2 Pachpatte’s inequalities | 183

≤ h1 (t1 , t2 )

≤ h1 (t1 , t2 ) Δ Δ

+

(

vt 1t 2 (t1 ,t2 ) Δ1 Δ 1 2 )t vt22 (t1 , t2 ) f (t1 ,t2 ) 1

v(t1 , t2 )v(t1 , σ2 (t2 ))

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, Δ Δ

(

(

vt 1t 2 (t1 ,t2 ) Δ1 1 2 )t f (t1 ,t2 ) 1

v(t1 , t2 )

Δ2 t2

Δ Δ

(

≤ h1 (t1 , t2 ),

)

vt 1t 2 (t1 ,t2 ) Δ1 1 2 )t f (t1 ,t2 ) 1

t2

≤ ∫ h1 (t1 , s2 )Δ2 s2 ,

v(t1 , t2 )

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Now we use (6.9) and find Δ Δ

(

vt 1t 2 (t1 ,t2 ) Δ1 1 2 )t v(t1 , t2 ) f (t1 ,t2 ) 1

v(t1 , t2 )v(σ1 (t1 ), t2 )

Δ Δ

(

≤

vt 1t 2 (t1 ,t2 ) Δ1 1 2 )t f (t1 ,t2 ) 1

t2

v(t1 , t2 )

≤ ∫ h1 (t1 , s2 )Δ2 s2 0

t2

≤ ∫ h1 (t1 , s2 )Δ2 s2 0

Δ Δ

+

(

vt 1t 2 (t1 ,t2 ) 1 2

f (t1 ,t2 )

Δ

)vt11 (t1 , t2 )

v(t1 , t2 )v(σ1 (t1 ), t2 )

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then ΔΔ

(

vt11t2 2 (t1 , t2 )

v(t1 , t2 )f (t1 , t2 )

t2

Δ1

) ≤ ∫ h1 (t1 , s2 )Δ2 s2 , t1

ΔΔ vt11t2 2 (t1 , t2 )

f (t1 , t2 )v(t1 , t2 ) ΔΔ vt11t2 2 (t1 , t2 )

v(t1 , t2 )

0

t1 t2

≤ ∫ ∫ h1 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

≤ f (t1 , t2 ) ∫ ∫ h1 (s1 , s2 )Δ2 s2 Δ1 s1 0 0

= h2 (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Using (6.9), we get ΔΔ

ΔΔ

vt11t2 2 (t1 , t2 )v(t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

≤

vt11t2 2 (t1 , t2 ) v(t1 , t2 )

184 | 6 Two-dimensional linear integral inequalities ≤ h2 (t1 , t2 ) ≤ h2 (t1 , t2 ) Δ

+

Δ

vt11 (t1 , t2 )vt22 (t1 , t2 )

v(t1 , t2 )v(t1 , σ2 (t2 ))

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore Δ

(

vt11 (t1 , t2 ) v(t1 , t2 )

Δ2

)

t2

Δ

vt11 (t1 , t2 ) v(t1 , t2 )

≤ h2 (t1 , t2 ), t2

≤ ∫ h2 (t1 , s2 )Δ2 s2 0

= h3 (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

or Δ

vt11 (t1 , t2 ) ≤ h3 (t1 , t2 )v(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, by Lemma 2.1.1, we obtain v(t1 , t2 ) ≤ eh3 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

and u(t1 , t2 ) = z(t1 , t2 ) q(t1 , t2 ) ≤ v(t1 , t2 )

≤ eh3 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

or u(t1 , t2 ) ≤ q(t1 , t2 )eh3 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

This completes the proof. Theorem 6.2.14. Let u(t1 , t2 ) and h(t1 , t2 ) be positive continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), b(t2 ), p(t1 ) and q(t2 ) be positive twice continuously-differentiable functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), aΔ1 (t1 ), pΔ1 (t1 ) and bΔ2 (t2 ), qΔ2 (t2 ) be nonnegative functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and aΔ1 Δ1 (t1 ) and pΔ1 Δ1 (t1 ) be positive functions for t1 ∈ ℝ+ ∩ 𝕋1 . If u(t1 , t2 ) ≤ a(t1 ) + b(t2 ) + t2 p(t1 ) + t1 q(t2 ) t1 s1 t2 s2

+ ∫ ∫ ∫ ∫ h(r1 , r2 )u(r1 , r2 )Δ2 r2 Δ2 s2 Δ1 r1 Δ1 s1 , 0 0 0 0

6.2 Pachpatte’s inequalities | 185

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ (a(0) + b(t2 ) + t2 p(0))eh3 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where t2

h1 (t1 , t2 ) =

pΔ1 Δ1 (t1 ) + ∫ h(t1 , s2 )Δ2 s2 , a(t1 ) + b(0) + t1 q(0) 0

t2

h2 (t1 , t2 ) = ∫ h1 (t1 , s2 )Δ2 s2 + 0

h3 (t1 , t2 ) =

aΔ1 Δ1 (t1 ) , a(t1 ) + b(0) + t1 q(0)

aΔ1 (0) + t2 pΔ1 (0) + q(t2 ) a(0) + b(t2 ) + t1 q(0) t1

+ ∫ h2 (s1 , t2 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let z(t1 , t2 ) = a(t1 ) + b(t2 ) + t2 p(t1 ) + t1 q(t2 ) t1 s1 t2 s2

+ ∫ ∫ ∫ ∫ h(r1 , r2 )u(r1 , r2 )Δ2 r2 Δ2 s2 Δ1 r1 Δ1 s1 , 0 0 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then u(t1 , t2 ) ≤ z(t1 , t2 ),

u(t1 , 0) = a(t1 ) + b(0) + t1 q(0),

u(0, t2 ) Δ1 zt1 (t1 , t2 )

= a(0) + b(t2 ) + t2 p(0),

= aΔ1 (t1 ) + t2 pΔ1 (t1 ) + q(t2 ) t1 t2 s2

+ ∫ ∫ ∫ h(r1 , r2 )u(r1 , r2 )Δ2 r2 Δ2 s2 Δ1 r1 , Δ

0 0 0

zt11 (t1 , 0) = aΔ1 (t1 ) + q(0), Δ

zt11 (0, t2 ) = aΔ1 (0) + t2 pΔ1 (0) + q(t2 ),

ΔΔ

zt11t1 1 (t1 , t2 ) = aΔ1 Δ1 (t1 ) + t2 pΔ1 Δ1 (t1 ) t2 s2

+ ∫ ∫ h(t1 , r2 )u(t1 , r2 )Δ2 r2 Δ2 s2 , 0 0

186 | 6 Two-dimensional linear integral inequalities ΔΔ

zt11t1 1 (t1 , 0) = aΔ1 Δ1 (t1 ), ΔΔ

zt11t1 1 (0, t2 ) = aΔ1 Δ1 (0) + t2 pΔ1 Δ1 (0) t2 s2

+ ∫ ∫ h(0, r2 )u(0, r2 )Δ2 r2 Δ2 s2 , 0 0 ΔΔΔ zt11t1 t12 2 (t1 , t2 ) ΔΔΔ zt11t1 t12 2 (t1 , 0) ΔΔΔΔ zt11t1 t12 t22 2 (t1 , t2 ) ΔΔΔΔ zt11t1 t12 t22 2 (t1 , t2 )

z(t1 , t2 )

=p

Δ1 Δ1

=p

Δ1 Δ1

t2

(t1 ) + ∫ h(t1 , r2 )u(t1 , r2 )Δ2 r2 , 0

(t1 ),

= h(t1 , t2 )u(t1 , t2 ) ≤ h(t1 , t2 )z(t1 , t2 ), ≤ h(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Since z(t1 , t2 ) is a positive nondecreasing function in each variable t1 , t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), we get ΔΔΔΔ

zt11t1 t12 t22 2 (t1 , t2 )z(t1 , t2 ) z(t1 , t2 )z(t1 , σ2 (t2 ))

ΔΔΔΔ

≤

zt11t1 t12 t22 2 (t1 , t2 )

z(t1 , t2 ) ≤ h(t1 , t2 ) ≤ h(t1 , t2 )

ΔΔΔ

+

Δ

zt11t1 t12 2 (t1 , t2 )zt22 (t1 , t2 ) z(t1 , t2 )z(t1 , σ2 (t2 ))

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, ΔΔΔ

(

zt11t1 t12 2 (t1 , t2 ) z(t1 , t2 )

Δ2

)

t2

≤ h(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then ΔΔΔ

zt11t1 t12 2 (t1 , t2 ) z(t1 , t2 )

ΔΔΔ

−

zt11t1 t12 2 (t1 , 0) z(t1 , 0)

t2

≤ ∫ h(t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or ΔΔΔ

zt11t1 t12 2 (t1 , t2 ) z(t1 , t2 )

t2

pΔ1 Δ1 (t1 ) − ≤ ∫ h(t1 , s2 )Δ2 s2 , a(t1 ) + b(0) + t1 q(0) 0

6.2 Pachpatte’s inequalities | 187

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or t2

ΔΔΔ

zt11t1 t12 2 (t1 , t2 ) z(t1 , t2 )

pΔ1 Δ1 (t1 ) + ∫ h(t1 , s2 )Δ2 s2 a(t1 ) + b(0) + t1 q(0)

≤

0

= h1 (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Now, since z(t1 , t2 ) is a positive nondecreasing function in t2 , t2 ∈ ℝ+ ∩ 𝕋2 , we get ΔΔΔ

ΔΔΔ

zt11t1 t12 2 (t1 , t2 )z(t1 , t2 ) z(t1 , t2 )z(t1 , σ2 (t2 ))

≤

zt11t1 t12 2 (t1 , t2 )

z(t1 , t2 ) ≤ h1 (t1 , t2 ) ≤ h1 (t1 , t2 ) ΔΔ

Δ

zt11t1 1 (t1 , t2 )zt22 (t1 , t2 )

+

z(t1 , t2 )z(t1 , σ2 (t2 ))

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), whereupon ΔΔ

(

zt11t1 1 (t1 , t2 ) z(t1 , t2 )

Δ2

)

t2

≤ h1 (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or ΔΔ

zt11t1 1 (t1 , t2 ) z(t1 , t2 )

ΔΔ

−

zt11t1 1 (t1 , 0) z(t1 , 0)

t2

≤ ∫ h1 (t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or ΔΔ

zt11t1 1 (t1 , t2 ) z(t1 , t2 )

t2

aΔ1 Δ1 (t1 ) − ≤ ∫ h1 (t1 , s2 )Δ2 s2 , a(t1 ) + b(0) + t1 q(0) 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or ΔΔ

zt11t1 1 (t1 , t2 ) z(t1 , t2 )

t2

≤ ∫ h1 (t1 , s2 )Δ2 s2 + 0

aΔ1 Δ1 (t1 ) a(t1 ) + b(0) + t1 q(0)

= h2 (t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Now we use that z(t1 , t2 ) is a positive nondecreasing function in t1 , t1 ∈ ℝ+ ∩ 𝕋1 , and find ΔΔ

zt11t1 1 (t1 , t2 )z(t1 , t2 )

z(t1 , t2 )z(σ1 (t1 ), t2 )

ΔΔ

≤

zt11t1 1 (t1 , t2 ) z(t1 , t2 )

188 | 6 Two-dimensional linear integral inequalities ≤ h2 (t1 , t2 ) ≤ h2 (t1 , t2 )

Δ

+

(zt11 (t1 , t2 ))2

z(t1 , t2 )z(σ1 (t1 ), t2 )

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), whereupon Δ

(

zt11 (t1 , t2 ) z(t1 , t2 )

Δ1

)

t1

≤ h2 (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or Δ

zt11 (t1 , t2 ) z(t1 , t2 )

Δ

−

zt11 (0, t2 ) z(0, t2 )

t1

≤ ∫ h2 (s1 , t2 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or t1

Δ

zt11 (t1 , t2 )

aΔ1 (0) + t2 pΔ1 (0) + q(t2 ) − ≤ ∫ h2 (s1 , t2 )Δ1 s1 , z(t1 , t2 ) a(0) + b(t2 ) + t1 q(0) 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) t1

Δ

zt11 (t1 , t2 )

aΔ1 (0) + t2 pΔ1 (0) + q(t2 ) ≤ + ∫ h2 (s1 , t2 )Δ1 s1 z(t1 , t2 ) a(0) + b(t2 ) + t1 q(0) = h3 (t1 , t2 ),

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or Δ

zt11 (t1 , t2 ) ≤ h3 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). By the last inequality and Lemma 2.1.1, we obtain z(t1 , t2 ) ≤ z(0, t2 )eh3 (t1 , 0)

= (a(0) + b(t2 ) + t2 p(0))eh3 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Consequently, u(t1 , t2 ) ≤ z(t1 , t2 )

= (a(0) + b(t2 ) + t2 p(0))eh3 (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof.

7 Snow’s inequalities This chapter deals with Snow type two dimensional linear integral inequalities. The material in this chapter is based on some results in [9, 24] and [25]. Let 𝕋1 and 𝕋2 be time scales with forward jump operators and delta differentiation operators σ1 , σ2 and Δ1 , Δ2 , respectively.

7.1 Existence of solutions of some partial dynamic equations Let D and D̃ be open bounded domains in 𝕋1 × 𝕋2 such that D ⊂ D̃ and {(σ1 (t1 ), σ2 (t2 )) : (t1 , t2 ) ∈ D} ⊂ D.̃ By 𝜕D we will denote the boundary of D and let D = D ∪ 𝜕D. Let also (t10 , t20 ) ∈ D, 2 (t10 , t20 ) ∉ 𝜕D. By 𝒞1,2 (D) we will denote the space of all functions u ∈ 𝒞 (D) such that Δ

Δ

ΔΔ

ut11 (t1 , t2 ), ut22 (t1 , t2 ), and ut11t2 2 (t1 , t2 ) exist and are continuous for (t1 , t2 ) ∈ D. Consider the problem ΔΔ

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 ,

vt11t2 2 − b(t1 , t2 )v = 0,

v(t1 , t20 )

Δ vt22 (t10 , t2 )

= 1,

= 0,

(t1 , t2 ) ∈ D, t1 ≥

t10 , t2

≥

(7.1) t20 .

(7.2)

2 Theorem 7.1.1. Let b be a continuous function on D. Then v ∈ 𝒞1,2 (D) is a solution of the problem (7.1), (7.2) if and only if it satisfies the integral equation t1 t2

v(t1 , t2 ) = 1 + ∫ ∫ b(s1 , s2 )v(s1 , s2 )Δ2 s2 Δ1 s1 ,

(7.3)

t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Proof. 2 1. Let v ∈ 𝒞1,2 (D) be a solution of the problem (7.1), (7.2). We integrate (7.1) from t10 to Δ

t1 and using that vt22 (t10 , t2 ) = 0, t2 ≥ t20 , (t10 , t2 ) ∈ D, get Δ vt22 (t1 , t2 )

−

Δ vt22 (t10 , t2 )

t1

= ∫ b(s1 , t2 )v(s1 , t2 )Δ1 s1 , t10

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , or Δ vt22 (t1 , t2 )

t1

= ∫ b(s1 , t2 )v(s1 , t2 )Δ1 s1 , t10

https://doi.org/10.1515/9783110705553-007

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 .

190 | 7 Snow’s inequalities Now we integrate the last equation from t20 to t2 and using that v(t1 , t20 ) = 1, t1 ≥ t10 , (t1 , t20 ) ∈ D, get v(t1 , t2 ) −

v(t1 , t20 )

t1 t2

= ∫ ∫ b(s1 , s2 )v(s1 , s2 )Δ2 s2 Δ1 s1 , t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , or t1 t2

v(t1 , t2 ) = 1 + ∫ ∫ b(s1 , s2 )v(s1 , s2 )Δ2 s2 Δ1 s1 , t10 t20

2.

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , i. e., v satisfies equation (7.3). 2 Let v ∈ 𝒞1,2 (D) satisfy equation (7.3). Then v(t1 , t20 ) = 1,

(t1 , t20 ) ∈ D, t1 ≥ t10 .

We differentiate (7.3) with respect to t2 and find Δ vt22 (t1 , t2 )

t1

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 ,

= ∫ b(s1 , t2 )v(s1 , t2 )Δ1 s1 ,

(7.4)

t10

whereupon Δ

vt22 (t10 , t2 ) = 0,

(t10 , t2 ) ∈ D, t2 ≥ t20 .

Now we differentiate (7.4) with respect to t1 and get ΔΔ

vt11t2 2 (t1 , t2 ) = b(t1 , t2 )v(t1 , t2 ),

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 ,

i. e., v satisfies equation (7.1). Therefore v is a solution of the problem (7.1), (7.2). This completes the proof. Lemma 7.1.2. Let b ∈ 𝒞 (D) and u be a nonnegative continuous function on D. If t1 t2

u(t1 , t2 ) ≤ ∫ ∫ b(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 ,

t10 t20

then u(t1 , t2 ) = 0,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 .

7.1 Existence of solutions of some partial dynamic equations | 191

Proof. Let t1 t2

z(t1 , t2 ) = ∫ ∫ b(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 .

t10 t20

Then u(t1 , t2 ) ≤ z(t1 , t2 ),

z(t10 , t2 ) = 0, Δ zt11 (t1 , t2 )

t2

= ∫ b(t1 , s2 )u(t1 , s2 )Δ2 s2 t20 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 .

≤ ∫ b(t1 , s2 )z(t1 , s2 )Δ2 s2 , t20

From here, using Gronwall’s inequality, we get z(t1 , t2 ) = 0,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 .

u(t1 , t2 ) = 0,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 .

Hence,

This completes the proof. Theorem 7.1.3. Let b ∈ 𝒞 (D). Then the problem (7.1), (7.2) has a unique solution v ∈ 2 𝒞1,2 (D). Proof. Define the sequence {vl }l∈ℕ0 as follows: v0 (t1 , t2 ) = 1, l

t1 t2

v (t1 , t2 ) = 1 + ∫ ∫ b(s1 , s2 )vl−1 (s1 , s2 )Δ2 s2 Δ1 s1 , t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , l ∈ ℕ. Let 󵄨 󵄨 M = sup 󵄨󵄨󵄨b(t1 , t2 )󵄨󵄨󵄨. (t1 ,t2 )∈D

Then t t

󵄨󵄨 1 2 󵄨󵄨 󵄨󵄨 󵄨󵄨 1 󵄨 󵄨󵄨 0 󵄨󵄨v (t1 , t2 ) − v (t1 , t2 )󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ ∫ b(s1 , s2 )Δ2 s2 Δ1 s1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t 0 t 0 󵄨 1

2

192 | 7 Snow’s inequalities t1 t2

󵄨 󵄨 ≤ ∫ ∫󵄨󵄨󵄨b(s1 , s2 )󵄨󵄨󵄨Δ2 s2 Δ1 s1 t10 t20

≤ Mh1 (t1 , t10 )h1 (t2 , t20 ), (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Assume that 󵄨 󵄨󵄨 l l 0 0 l−1 󵄨󵄨v (t1 , t2 ) − v (t1 , t2 )󵄨󵄨󵄨 ≤ M hl (t1 , t1 )hl (t2 , t2 ),

(7.5)

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , for some l ∈ ℕ. We will prove that 󵄨󵄨 l+1 󵄨 l l+1 0 0 󵄨󵄨v (t1 , t2 ) − v (t1 , t2 )󵄨󵄨󵄨 ≤ M hl+1 (t1 , t1 )hl+1 (t2 , t2 ), (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . In fact, t1 t2 󵄨󵄨 󵄨󵄨 l+1 󵄨󵄨 󵄨󵄨󵄨 l l 󵄨󵄨v (t1 , t2 ) − v (t1 , t2 )󵄨󵄨 = 󵄨󵄨1 + ∫ ∫ b(s1 , s2 )v (s1 , s2 )Δ2 s2 Δ1 s1 󵄨󵄨 󵄨 t0 t0 1

2

t1 t2

󵄨󵄨 󵄨󵄨 − 1 − ∫ ∫ b(s1 , s2 )vl−1 (s1 , s2 )Δ2 s2 Δ1 s1 󵄨󵄨󵄨 󵄨󵄨 󵄨 t0 t0 1

2

󵄨󵄨 t1 t2 󵄨󵄨 󵄨󵄨 󵄨󵄨 = 󵄨󵄨󵄨∫ ∫ b(s1 , s2 )(vl (s1 , s2 ) − vl−1 (s1 , s2 ))Δ2 s2 Δ1 s1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t 0 t 0 󵄨 1

2

t1

󵄨 󵄨󵄨 󵄨 ≤ ∫ ∫󵄨󵄨󵄨b(s1 , s2 )󵄨󵄨󵄨󵄨󵄨󵄨vl (s1 , s2 ) − vl−1 (s1 , s2 )󵄨󵄨󵄨Δ2 s2 Δ1 s1 t10 t20

≤M

l+1

t1 t2

∫ ∫ hl (s1 , t10 )hl (s2 , t20 )Δ2 s2 Δ1 s1 t10 t20

= M l+1 hl+1 (t1 , t10 )hl+1 (t2 , t20 ), (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Therefore (7.5) holds for any l ∈ ℕ and (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Since D is a bounded domain, there is a constant P > 0 such that (σ1 (t1 ) − t10 )(σ2 (t2 ) − t20 ) ≤ P,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 .

We have 󵄨󵄨 ∞ 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨 lim 󵄨󵄨󵄨vl (t1 , t2 ) − 1󵄨󵄨󵄨 = 󵄨󵄨󵄨 ∑ (vl+1 (t1 , t2 ) − vl (t1 , t2 ))󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 l→∞ 󵄨l=0 󵄨 ∞

󵄨 󵄨 ≤ ∑ 󵄨󵄨󵄨vl+1 (t1 , t2 ) − vl (t1 , t2 )󵄨󵄨󵄨 l=0

7.1 Existence of solutions of some partial dynamic equations | 193 ∞

≤ ∑ M l+1 hl+1 (t1 , t10 )hl+1 (t2 , t20 ) l=0 ∞

≤ ∑ M l+1

(σ1 (t1 ) − t10 )l+1 (σ2 (t2 ) − t20 )l+1 ((l + 1)!)2

≤ ∑ M l+1

P l+1 ((l + 1)!)2

≤ ∑ M l+1

P l+1 (l + 1)!

l=0 ∞

l=0 ∞

l=0

≤ eMP < ∞, (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Therefore vl (t1 , t2 ) converges uniformly to a solution v(t1 , t2 ), (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , of the problem (7.1), (7.2). Suppose that v1 (t1 , t2 ) and v2 (t1 , t2 ) are two solutions of the problem (7.1), (7.2), (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then t1 t2 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨v1 (t1 , t2 ) − v2 (t1 , t2 )󵄨󵄨󵄨 = 󵄨󵄨󵄨1 + ∫ ∫ b(s1 , s2 )v1 (s1 , s2 )Δ2 s2 Δ1 s1 󵄨󵄨󵄨 t0 t0 1

2

t1 t2

󵄨󵄨 󵄨󵄨 − 1 − ∫ ∫ b(s1 , s2 )v2 (s1 , s2 )Δ2 s2 Δ1 s1 󵄨󵄨󵄨 󵄨󵄨 󵄨 t0 t0 1

2

󵄨󵄨 t1 t2 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 = 󵄨󵄨∫ ∫ b(s1 , s2 )(v1 (s1 , s2 ) − v2 (s1 , s2 ))Δ2 s2 Δ1 s1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t 0 t 0 󵄨 1

2

t1 t2

󵄨 󵄨󵄨 󵄨 ≤ ∫ ∫󵄨󵄨󵄨b(s1 , s2 )󵄨󵄨󵄨󵄨󵄨󵄨v1 (s1 , s2 ) − v2 (s1 , s2 )󵄨󵄨󵄨Δ2 s2 Δ1 s1 , t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Hence, by Lemma 7.1.2, we get that v1 (t1 , t2 ) − v2 (t1 , t2 ) = 0,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 .

This completes the proof. As in above one can prove the following theorem. Theorem 7.1.4. Let b ∈ 𝒞 (D). Then the problem ΔΔ

vt11t2 2 − b(t1 , t2 )vσ1 σ2 = 0,

v

σ1

(t1 , t20 )

= 1,

v(t10 , t2 )

= 1,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , (t1 , t2 ) ∈ D, t1 ≥

t10 , t2

≥

t20 ,

(7.6) (7.7)

194 | 7 Snow’s inequalities 2 has a unique solution v ∈ 𝒞1,2 (D) which can be represented in the form t1 t2

v(t1 , t2 ) = ∫ ∫ b(s1 , s2 )Δ2 s2 Δ1 s1 + 1. t10 t20

7.2 Snow’s inequalities Theorem 7.2.1 (Snow’s Inequality). Let u, a, and b be nonnegative continuous functions on D. For any (t10 , t20 ) by v(t1 , t2 ) we denote the nonnegative solution of the problem (7.1), (7.2) in D. If, for (t10 , t20 ) ∈ D, u(t10 , t20 )

≤

a(t10 , t20 )

t1 t2

+ ∫ ∫ b(s1 , s2 )uσ1 σ2 (s1 , s2 )Δ2 s2 Δ1 s1 , t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , then u(t10 , t20 )

≤

a(t10 , t20 )

t1 t2

+ ∫ ∫ v(s1 , s2 )aσ1 σ2 (s1 , s2 )b(s1 , s2 )Δ2 s2 Δ1 s1 , t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Proof. Let t10 t20

z(t10 , t20 ) = ∫ ∫ b(s1 , s2 )uσ1 σ2 (s1 , s2 )Δ2 s2 Δ1 s1 , t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then u(t10 , t20 ) ≤ a(t10 , t20 ) + z(t10 , t20 ), z(t1 , t20 ) = 0,

Δ

t20

zt 01 (t10 , t20 ) = ∫ b(t10 , s2 )uσ1 σ2 (t10 , s2 )Δ2 s2 , 1

Δ zt 01 (t10 , t2 ) 1

t2

= 0,

ΔΔ

zt 01t 02 (t10 , t20 ) = b(t10 , t20 )uσ1 σ2 (t10 , t20 ) 1 2

≤ aσ1 σ2 (t10 , t20 )b(t10 , t20 ) + b(t10 , t20 )z σ1 σ2 (t10 , t20 ),

whereupon ΔΔ

zt 01t 02 (t10 , t20 ) − b(t10 , t20 )z σ1 σ2 (t10 , t20 ) ≤ aσ1 σ2 (t10 , t20 )b(t10 , t20 ). 1 2

7.2 Snow’s inequalities | 195

We fix (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let R = {(s1 , s2 ) ∈ D : t1 ≥ s1 ≥ t10 , t2 ≥ s2 ≥ t20 } 2 and 𝜕R be its boundary. For (s1 , s2 ) ∈ R and x ∈ 𝒞1,2 (D), we define the operators

L(x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − b(s1 , s2 )x σ1 σ2 (s1 , s2 ),

M(x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − b(s1 , s2 )x(s1 , s2 ). 2 Then, for x, y ∈ 𝒞1,2 (D) and (s1 , s2 ) ∈ D, we have

y(s1 , s2 )L(x)(s1 , s2 ) − xσ1 σ2 (s1 , s2 )M(y)(s1 , s2 )

= y(s1 , s2 )(xsΔ11sΔ22 (s1 , s2 ) − b(s1 , s2 )x σ1 σ2 (s1 , s2 ))

− xσ1 σ2 (s1 , s2 )(ysΔ11sΔ22 (s1 , s2 ) − b(s1 , s2 )y(s1 , s2 ))

= y(s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − xσ1 σ2 (s1 , s2 )ysΔ11sΔ22 (s1 , s2 ), and Δ

Δ

2

1

(yxsΔ11 )s 2 (s1 , s2 ) − (x σ2 ysΔ22 )s 1 (s1 , s2 ) =

=

y(s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) + ysΔ22 (s1 , s2 )xsΔ11 σ2 (s1 , s2 ) − xσ1 σ2 (s1 , s2 )ysΔ11sΔ22 (s1 , s2 ) − xsΔ11 σ2 (s1 , s2 )ysΔ22 (s1 , s2 ) y(s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − xσ1 σ2 (s1 , s2 )ysΔ11sΔ22 (s1 , s2 ),

i. e., y(s1 , s2 )L(x)(s1 , s2 ) − xσ1 σ2 (s1 , s2 )M(y)(s1 , s2 ) Δ

Δ

2

1

= (yxsΔ11 )s 2 (s1 , s2 ) − (x σ2 ysΔ22 )s 1 (s1 , s2 ).

Applying Green’s formula, we get ∫∫ v(s1 , s2 )L(z)(s1 , s2 )Δ2 s2 Δ1 s1 R

= ∫∫(v(s1 , s2 )L(z)(s1 , s2 ) − z σ1 σ2 (s1 , s2 )M(v)(s1 , s2 ))Δ2 s2 Δ1 s1 R Δ

= ∫∫((vzsΔ11 )s 2 (s1 , s2 ) 2

R

Δ

− (z σ2 vsΔ22 )s 1 (s1 , s2 ))Δ2 s2 Δ1 s1 1

σ2

= − ∫ z (s1 , s2 )vsΔ22 (s1 , s2 )Δ2 s2 + v(s1 , s2 )zsΔ11 (s1 , s2 )Δ1 s1 𝜕R

196 | 7 Snow’s inequalities t2

= − ∫ z σ2 (t1 , s2 )vsΔ22 (t1 , s2 )Δ2 s2 t20

t2

+ ∫ z σ2 (t10 , s2 )vsΔ22 (t10 , s2 )Δ2 s2 t20

t1

− ∫ v(s1 , t20 )zsΔ11 (s1 , t20 )Δ1 s1 t10

t1

+ ∫ v(s1 , t2 )zsΔ11 (s1 , t2 )Δ1 s1 t10

t1

= − ∫ v(s1 , t20 )zsΔ11 (s1 , t20 )Δ1 s1 t10

t1

= − ∫ zsΔ11 (s1 , t20 )Δ1 s1 t10

= −z(t1 , t20 ) + z(t10 , t20 ) = z(t10 , t20 ), i. e., z(t10 , t20 ) = ∫∫ v(s1 , s2 )L(z)(s1 , s2 )Δ2 s2 Δ1 s1 R

t10 t20

≤ ∫ ∫ v(s1 , s2 )aσ1 σ2 (s1 , s2 )b(s1 , s2 )Δ2 s2 Δ1 s1 . t1 t2

Therefore u(t10 , t20 ) ≤ a(t10 , t20 ) + z(t10 , t20 ) ≤ a(t10 , t20 ) t10 t20

+ ∫ ∫ v(s1 , s2 )aσ1 σ2 (s1 , s2 )b(s1 , s2 )Δ2 s2 Δ1 s1 . t1 t2

This completes the proof. Theorem 7.2.2 (Snow’s Inequality). Let u, a, b, c, and f be nonnegative continuous functions on D. For (t10 , t20 ) ∈ D, by v we denote a nonnegative solution of the problem ΔΔ

vt11t2 2 − (b(t1 , t2 ) + f (t1 , t2 ))vσ1 σ2 = 0,

7.2 Snow’s inequalities | 197

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , Δ

v(t1 , t20 ) = 1,

vt22 (t10 , t2 ) = 0,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . If u(t10 , t20 )

≤

a(t10 , t20 )

t1 t2

+ ∫ ∫ b(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 t10 t20

t1 t2

t1 t2

+ ∫ ∫ b(s1 , s2 )(c(s1 , s2 ) + ∫ ∫ f (ξ1 , ξ2 )u(ξ1 , ξ2 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 , s1 s2

t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , then u(t10 , t20 ) ≤ a(t10 , t20 ) t10 t20

+ ∫ ∫ v(s1 , s2 )(b(s1 , s2 )(a(s1 , s2 ) + c(s1 , s2 )) t1 t2

+ f (s1 , s2 )a(s1 , s2 ))Δ2 s2 Δ1 s1 , (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Proof. Let t1 t2

ϕ(t10 , t20 ) = ∫ ∫ b(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 t10 t20 t1 t2

t1 t2

+ ∫ ∫ b(s1 , s2 )(c(s1 , s2 ) + ∫ ∫ f (ξ1 , ξ2 )u(ξ1 , ξ2 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 , s1 s2

t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then u(t10 , t20 ) ≤ a(t10 , t20 ) + ϕ(t10 , t20 ), Δ

t20

ϕt 01 (t10 , t20 ) = ∫ b(t10 , s2 )u(t10 , s2 )Δ2 s2 1

t2

t20

+

∫ b(t10 , s2 )(c(t10 , s2 )

t2 ΔΔ ϕt 01t 02 (t10 , t20 ) 1 2

= b(t10 , t20 )u(t10 , t20 )

t1 t2

+ ∫ ∫ f (ξ1 , ξ2 )u(ξ1 , ξ2 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 , t10 s2

198 | 7 Snow’s inequalities

+ b(t10 , t20 )(c(t10 , t20 ) t1 t2

+ ∫ ∫ f (s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 ) t10 t20

≤ b(t10 , t20 )(a(t10 , t20 ) + ϕ(t10 , t20 )) +

b(t10 , t20 )(c(t10 , t20 )

t1 t2

+ ∫ ∫ f (s1 , s2 )(a(s1 , s2 ) t10 t20

+ ϕ(s1 , s2 ))Δ2 s2 Δ1 s1 ) = b(t10 , t20 )(a(t10 , t20 ) + c(t10 , t20 ) + ϕ(t10 , t20 ) t1 t2

+ ∫ ∫ f (s1 , s2 )(a(s1 , s2 ) + ϕ(s1 , s2 ))Δ2 s2 Δ1 s1 ), ϕ(t10 , t2 ) ϕ(t1 , t20 )

t10 t20

= 0, = 0,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let z(t10 , t20 )

=

ϕ(t10 , t20 )

t1 t2

+ ∫ ∫ f (s1 , s2 )(a(s1 , s2 ) + ϕ(s1 , s2 ))Δ2 s2 Δ1 s1 , t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then z(t10 , t2 ) = ϕ(t10 , t2 ) z(t1 , t20 ) Δ zt 01 (t10 , t20 ) 1

= 0,

= ϕ(t1 , t20 ) = 0,

Δ

= ϕt 01 (t10 , t20 ) 1

t20

+ ∫ f (t10 , s2 )(a(t10 , s2 ) + ϕ(t10 , s2 ))Δ2 s2 , ΔΔ zt 01t 02 (t10 , t20 ) 1 2

t2

=

ΔΔ ϕt 01t 02 (t10 , t20 ) 1 2

+ f (t10 , t20 )(a(t10 , t20 ) + ϕ(t10 , t20 )),

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Note that ϕ(t10 , t20 ) is a nonnegative nonincreasing function for (t10 , t20 ) ∈ D. Hence, z(t10 , t20 ) is a nonnegative nonincreasing function for (t10 , t20 ) ∈ D,

7.2 Snow’s inequalities | 199

and ΔΔ

ϕt 01t 02 (t10 , t20 ) ≤ b(t10 , t20 )(a(t10 , t20 ) + c(t10 , t20 ) + z(t10 , t20 )), 1 2

ϕ(t10 , t20 ) ≤ z(t10 , t20 ),

(t10 , t20 ) ∈ D. Then ΔΔ

zt 01t 02 (t10 , t20 ) ≤ b(t10 , t20 )(a(t10 , t20 ) + c(t10 , t20 ) + z(t10 , t20 )) 1 2

+ f (t10 , t20 )(a(t10 , t20 ) + z(t10 , t20 ))

= (b(t10 , t20 ) + f (t10 , t20 ))z(t10 , t20 )

+ b(t10 , t20 )(a(t10 , t20 ) + c(t10 , t20 )) + f (t10 , t20 )a(t10 , t20 ),

(t10 , t20 ) ∈ D. We set g(t10 , t20 ) = b(t10 , t20 ) + f (t10 , t20 ),

h(t10 , t20 ) = b(t10 , t20 )(a(t10 , t20 ) + c(t10 , t20 )) + f (t10 , t20 )a(t10 , t20 ),

(t10 , t20 ) ∈ D. Therefore ΔΔ

zt 01t 02 (t10 , t20 ) ≤ g(t10 , t20 )z(t10 , t20 ) + h(t10 , t20 ), 1 2

or ΔΔ

zt 01t 02 (t10 , t20 ) − g(t10 , t20 )z(t10 , t20 ) ≤ h(t10 , t20 ), 1 2

(t10 , t20 ) ∈ D. We fix (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 and let R be as in the proof of Theorem 7.2.1. 2 For (s1 , s2 ) ∈ R and x ∈ 𝒞1,2 (D), we define the operators L(x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − g(s1 , s2 )x(s1 , s2 ),

M(x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − g(s1 , s2 )x σ1 σ2 (s1 , s2 ). 2 Then, for x, y ∈ 𝒞1,2 (D) and (s1 , s2 ) ∈ D, we have

yσ1 σ2 (s1 , s2 )L(x)(s1 , s2 ) − x(s1 , s2 )M(y)(s1 , s2 )

= yσ1 σ2 (s1 , s2 )(xsΔ11sΔ22 (s1 , s2 ) − g(s1 , s2 )x(s1 , s2 ))

− x(s1 , s2 )(ysΔ11sΔ22 (s1 , s2 ) − g(s1 , s2 )yσ1 σ2 (s1 , s2 ))

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ),

200 | 7 Snow’s inequalities and Δ

Δ

2

1

(yσ1 xsΔ11 )s 2 (s1 , s2 ) − (xysΔ22 )s 1 (s1 , s2 ) =y

σ1 σ2

−

(s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) + ysΔ22 σ1 (s1 , s2 )xsΔ11 (s1 , s2 ) x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ) − xsΔ11 (s1 , s2 )ysΔ22 σ1 (s1 , s2 )

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ), i. e., yσ1 σ2 (s1 , s2 )L(x)(s1 , s2 ) − x(s1 , s2 )M(y)(s1 , s2 ) Δ

Δ

2

1

= (yσ1 xsΔ11 )s 2 (s1 , s2 ) − (xysΔ22 )s 1 (s1 , s2 ).

Applying Green’s formula, we get ∫∫ vσ1 σ2 (s1 , s2 )L(z)(s1 , s2 )Δ2 s2 Δ1 s1 R

= ∫∫(vσ1 σ2 (s1 , s2 )L(z)(s1 , s2 ) − z(s1 , s2 )M(v)(s1 , s2 ))Δ2 s2 Δ1 s1 R Δ

= ∫∫((vσ1 zsΔ11 )s 2 (s1 , s2 ) 2

R

Δ

− (zvsΔ22 )s 1 (s1 , s2 ))Δ2 s2 (s1 , s2 )Δ1 s1 1

= − ∫ z(s1 , s2 )vsΔ22 (s1 , s2 )Δ2 s2 + vσ1 (s1 , s2 )zsΔ11 (s1 , s2 )Δ1 s1 𝜕R t2

= − ∫ z(t1 , s2 )vsΔ22 (t1 , s2 )Δ2 s2 t20

t2

+ ∫ z(t10 , s2 )vsΔ22 (t10 , s2 )Δ2 s2 t20

t1

− ∫ vσ1 (s1 , t20 )zsΔ11 (s1 , t20 )Δ1 s1 t10

t1

+ ∫ vσ1 (s1 , t2 )zsΔ11 (s1 , t2 )Δ1 s1 t10

t1

= − ∫ vσ1 (s1 , t20 )zsΔ11 (s1 , t20 )Δ1 s1 t10

7.2 Snow’s inequalities | 201 t1

= − ∫ zsΔ11 (s1 , t20 )Δ1 s1 t10

= −z(t1 , t20 ) + z(t10 , t20 ) = z(t10 , t20 ),

i. e., z(t10 , t20 ) = ∫∫ v(s1 , s2 )L(z)(s1 , s2 )Δ2 s2 Δ1 s1 R

t10 t20

≤ ∫ ∫ v(s1 , s2 )h(s1 , s2 )Δ2 s2 Δ1 s1 t1 t2 t10 t20

≤ ∫ ∫ v(s1 , s2 )(b(s1 , s2 )(a(s1 , s2 ) + c(s1 , s2 )) t1 t2

ϕ(t10 , t20 )

+ f (s1 , s2 )a(s1 , s2 ))Δ2 s2 Δ1 s1 ,

≤ z(t10 , t20 ) t10 t20

≤ ∫ ∫ v(s1 , s2 )(b(s1 , s2 )(a(s1 , s2 ) + c(s1 , s2 )) t1 t2

u(t10 , t20 )

+ f (s1 , s2 )a(s1 , s2 ))Δ2 s2 Δ1 s1 ,

≤ a(t10 , t20 ) + ϕ(t10 , t20 ) ≤ a(t10 , t20 ) t10 t20

+ ∫ ∫ v(s1 , s2 )(b(s1 , s2 )(a(s1 , s2 ) + c(s1 , s2 )) t1 t2

+ f (s1 , s2 )a(s1 , s2 ))Δ2 s2 Δ1 s1 , (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . This completes the proof. Theorem 7.2.3 (Snow’s Inequality). Let a, b, c, p, q, and u be nonnegative continuous functions on D. Let, for (t10 , t20 ) ∈ D, v and w be the nonnegative solutions of the problems ΔΔ

vt11t2 2 − (p(t1 , t2 ) + b(t1 , t2 )(c(t1 , t2 ) + q(t1 , t2 )))v = 0, (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , vσ1 (t1 , t20 ) = 1,

Δ

vt22 (t10 , t2 ) = 0,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , ΔΔ

wt11t2 2 − b(t1 , t2 )c(t1 , t2 )w = 0,

202 | 7 Snow’s inequalities (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , wσ1 (t1 , t20 ) = 1,

Δ

wt22 (t10 , t2 ) = 0,

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , respectively. If u(t10 , t20 )

≤

a(t10 , t20 )

t10 t20

+

b(t10 , t20 )(∫ ∫ c(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 t1 t2

t10 t20

s1 s2

+ ∫ ∫ p(s1 , s2 )(∫ ∫ q(ξ1 , ξ2 )u(ξ1 , ξ2 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 ), t1 t2

t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , then u(t10 , t20 )

≤

a(t10 , t20 )

t10 t20

+

b(t10 , t20 ) ∫ ∫ p(s1 , s2 )w(s1 , s2 ) t1 t2

s1 s2

× (∫ ∫ v(ξ1 , ξ2 )a(ξ1 , ξ2 )(c(ξ1 , ξ2 ) + q(ξ1 , ξ2 ))Δ2 ξ2 Δ1 ξ1 + a(s1 , s2 )c(s1 , s2 ))Δ2 s2 Δ1 s1 , t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Proof. Let t10 t20

ϕ(t10 , t20 ) = ∫ ∫ c(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 t1 t2

t10 t20

s1 s2

+ ∫ ∫ p(s1 , s2 )(∫ ∫ q(ξ1 , ξ2 )u(ξ1 , ξ2 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 , t1 t2

t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then u(t10 , t20 ) ≤ a(t10 , t20 ) + b(t10 , t20 )ϕ(t10 , t20 ), ϕ(t1 , t20 ) = 0,

ϕ(t10 , t2 ) = 0, Δ ϕt 01 (t10 , t20 ) 1

t20

= ∫ c(t10 , s2 )u(t10 , s2 )Δ2 s2 t2

7.2 Snow’s inequalities | 203 t20

+

0

t1 s2 ∫ p(t10 , s2 )(∫ ∫ q(ξ1 , ξ2 )

t2

t1 t2

× u(ξ1 , ξ2 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 , ΔΔ

ϕt 01t 02 (t10 , t20 ) = c(t10 , t20 )u(t10 , t20 ) 1 2

t10 t20

+

p(t10 , t20 ) ∫ ∫ q(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 t1 t2

≤

c(t10 , t20 )(a(t10 , t20 ) t10 t20

+ b(t10 , t20 )ϕ(t10 , t20 ))

+ p(t10 , t20 ) ∫ ∫ q(s1 , s2 )(a(s1 , s2 ) t1 t2

+ b(s1 , s2 )ϕ(s1 , s2 ))Δ2 s2 Δ1 s1

≤ c(t10 , t20 )(a(t10 , t20 ) + b(t10 , t20 )ϕ(t10 , t20 )) +

p(t10 , t20 )ϕ(t10 , t20 )

t10 t20

+ ∫ ∫ q(s1 , s2 ) t1 t2

× (a(s1 , s2 ) + b(s1 , s2 )ϕ(s1 , s2 ))Δ2 s2 Δ1 s1 , (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let z(t10 , t20 ) = ϕ(t10 , t20 ) t10 t20

+ ∫ ∫ q(s1 , s2 )(a(s1 , s2 ) + b(s1 , s2 )ϕ(s1 , s2 ))Δ2 s2 Δ1 s1 , t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then ϕ(t10 , t20 ) ≤ z(t10 , t20 ),

ΔΔ

ϕt 01t 02 (t10 , t20 ) ≤ c(t10 , t20 )(a(t10 , t20 ) + b(t10 , t20 )ϕ(t10 , t20 )) 1 2

+ p(t10 , t20 )z(t10 , t20 )

≤ c(t10 , t20 )(a(t10 , t20 ) + b(t10 , t20 )z(t10 , t20 )) + p(t10 , t20 )z(t10 , t20 )

= a(t10 , t20 )c(t10 , t20 )

+ z(t10 , t20 )(p(t10 , t20 ) + b(t10 , t20 )c(t10 , t20 )),

z(t1 , t20 ) = ϕ(t1 , t20 ) = 0,

204 | 7 Snow’s inequalities z(t10 , t2 ) = ϕ(t10 , t2 ) Δ zt 01 (t10 , t20 ) 1

= 0,

Δ

= ϕt 01 (t10 , t20 ) 1

t20

+ ∫ q(t10 , s2 )(a(t10 , s2 ) + b(t10 , s2 )ϕ(t10 , s2 ))Δ2 s2 , ΔΔ zt 01t 02 (t10 , t20 ) 1 2

t2

=

ΔΔ ϕt 01t 02 (t10 , t20 ) 1 2

+ q(t10 , t20 )(a(t10 , t20 ) + b(t10 , t20 )ϕ(t10 , t20 ))

≤ a(t10 , t20 )c(t10 , t20 )

+ z(t10 , t20 )(p(t10 , t20 ) + b(t10 , t20 )c(t10 , t20 ))

+ q(t10 , t20 )(a(t10 , t20 ) + b(t10 , t20 )z(t10 , t20 ))

= a(t10 , t20 )(c(t10 , t20 ) + q(t10 , t20 ))

+ z(t10 , t20 )(p(t10 , t20 ) + b(t10 , t20 ) × (c(t10 , t20 ) + q(t10 , t20 ))).

Let g1 (t10 , t20 ) = a(t10 , t20 )(c(t10 , t20 ) + q(t10 , t20 )),

h1 (t10 , t20 ) = p(t10 , t20 ) + b(t10 , t20 )

× (c(t10 , t20 ) + q(t10 , t20 )).

Then ΔΔ

zt 01t 02 (t10 , t20 ) ≤ g1 (t10 , t20 ) + h1 (t10 , t20 )z(t10 , t20 ), 1 2

or ΔΔ

zt 01t 02 (t10 , t20 ) − h1 (t10 , t20 )z(t10 , t20 ) ≤ g1 (t10 , t20 ). 1 2

We fix (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let R and 𝜕R be as in the proof of Theorem 7.2.1. For 2 (s1 , s2 ) ∈ R and x ∈ 𝒞1,2 (D), we define the operators L(x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − h1 (s1 , s2 )x(s1 , s2 ),

M(x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − h1 (s1 , s2 )x σ1 σ2 (s1 , s2 ). 2 Then, for x, y ∈ 𝒞1,2 (D) and (s1 , s2 ) ∈ D, we have

yσ1 σ2 (s1 , s2 )L(x)(s1 , s2 ) − x(s1 , s2 )M(y)(s1 , s2 )

= yσ1 σ2 (s1 , s2 )(xsΔ11sΔ22 (s1 , s2 ) − h1 (s1 , s2 )x(s1 , s2 ))

7.2 Snow’s inequalities | 205

− x(s1 , s2 )(ysΔ11sΔ22 (s1 , s2 ) − h1 (s1 , s2 )yσ1 σ2 (s1 , s2 ))

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ), and Δ

Δ

2

1

(yσ1 (s1 , s2 )xsΔ11 (s1 , s2 ))s 2 (s1 , s2 ) − (x(s1 , s2 )ysΔ22 (s1 , s2 ))s 1 (s1 , s2 ) = yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) + ysΔ22 σ1 (s1 , s2 )xsΔ11 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ) − xsΔ11 (s1 , s2 )ysΔ22 σ1 (s1 , s2 )

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ), i. e., yσ1 σ2 (s1 , s2 )L(x)(s1 , s2 ) − x(s1 , s2 )M(y)(s1 , s2 ) Δ

Δ

2

1

= (yσ1 xsΔ11 )s 2 (s1 , s2 ) − (xysΔ22 )s 1 (s1 , s2 ). Applying Green’s formula, we get ∫∫ vσ1 σ2 (s1 , s2 )L(z)(s1 , s2 )Δ2 s2 Δ1 s1 R

= ∫∫(vσ1 σ2 (s1 , s2 )L(z)(s1 , s2 ) − z(s1 , s2 )M(v)(s1 , s2 ))Δ2 s2 Δ1 s1 R Δ

= ∫∫((vσ1 zsΔ11 )s 2 (s1 , s2 ) 2

R

Δ

− (zvsΔ22 )s 1 (s1 , s2 ))Δ2 s2 Δ1 s1 1

= − ∫ z(s1 , s2 )vsΔ22 (s1 , s2 )Δ2 s2 + vσ1 (s1 , s2 )zsΔ11 (s1 , s2 )Δ1 s1 𝜕R t2

= − ∫ z(t1 , s2 )vsΔ22 (t1 , s2 )Δ2 s2 t20

t2

+ ∫ z(t10 , s2 )vsΔ22 (t10 , s2 )Δ2 s2 t20

t1

− ∫ vσ1 (s1 , t20 )zsΔ11 (s1 , t20 )Δ1 s1 t10

t1

+ ∫ vσ1 (s1 , t2 )zsΔ11 (s1 , t2 )Δ1 s1 t10

(7.8)

206 | 7 Snow’s inequalities t1

= − ∫ vσ1 (s1 , t20 )zsΔ11 (s1 , t20 )Δ1 s1 t10

t1

= − ∫ zsΔ11 (s1 , t20 )Δ1 s1 t10

= −z(t1 , t20 ) + z(t10 , t20 ) = z(t10 , t20 ), i. e., z(t10 , t20 ) = ∫∫ v(s1 , s2 )L(z)(s1 , s2 )Δ2 s2 Δ1 s1 R

t10 t20

≤ ∫ ∫ v(s1 , s2 )g1 (s1 , s2 )Δ2 s2 Δ1 s1 , t1 t2

and from here ΔΔ

ϕt 01t 02 (t10 , t20 ) ≤ c(t10 , t20 )(a(t10 , t20 ) + b(t10 , t20 )ϕ(t10 , t20 )) 1 2

+ p(t10 , t20 )z(t10 , t20 )

≤ b(t10 , t20 )c(t10 , t20 )ϕ(t10 , t20 ) + a(t10 , t20 )c(t10 , t20 ) t10 t20

+ p(t10 , t20 ) ∫ ∫ v(s1 , s2 )g1 (s1 , s2 )Δ2 s2 Δ1 s1 . t1 t2

Let g2 (t10 , t20 ) = b(t10 , t20 )c(t10 , t20 ), t10 t20

h2 (t10 , t20 ) = p(t10 , t20 ) ∫ ∫ v(s1 , s2 )g1 (s1 , s2 )Δ2 s2 Δ1 s1 t1 t2

+

a(t10 , t20 )c(t10 , t20 ).

Then ΔΔ

ϕt 01t 02 (t10 , t20 ) ≤ g2 (t10 , t20 )z(t10 , t20 ) + h2 (t10 , t20 ), 1 2

or ΔΔ

ϕt 01t 02 (t10 , t20 ) − g2 (t10 , t20 )z(t10 , t20 ) ≤ h2 (t10 , t20 ). 1 2

7.2 Snow’s inequalities | 207

We fix (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let R and 𝜕R be as in the proof of Theorem 7.2.1. For 2 (s1 , s2 ) ∈ R and x ∈ 𝒞1,2 (D), we define the operators L1 (x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − g2 (s1 , s2 )x(s1 , s2 ),

M1 (x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − g2 (s1 , s2 )x σ1 σ2 (s1 , s2 ). 2 Then, for x, y ∈ 𝒞1,2 (D) and (s1 , s2 ) ∈ D, we have

yσ1 σ2 (s1 , s2 )L1 (x)(s1 , s2 ) − x(s1 , s2 )M1 (y)(s1 , s2 )

= yσ1 σ2 (s1 , s2 )(xsΔ11sΔ22 (s1 , s2 ) − g2 (s1 , s2 )x(s1 , s2 ))

− x(s1 , s2 )(ysΔ11sΔ22 (s1 , s2 ) − g2 (s1 , s2 )yσ1 σ2 (s1 , s2 ))

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ), and using (7.8), we get yσ1 σ2 (s1 , s2 )L1 (x)(s1 , s2 ) − x(s1 , s2 )M1 (y)(s1 , s2 ) Δ

Δ

2

1

= (yσ1 xsΔ11 )s 2 (s1 , s2 ) − (xysΔ22 )s 1 (s1 , s2 ).

Applying Green’s formula, we get ∫∫ wσ1 σ2 (s1 , s2 )L1 (ϕ)(s1 , s2 )Δ2 s2 Δ1 s1 R

= ∫∫(wσ1 σ2 (s1 , s2 )L1 (ϕ)(s1 , s2 ) − ϕ(s1 , s2 )M1 (w)(s1 , s2 ))Δ2 s2 Δ1 s1 R Δ

= ∫∫((wσ1 ϕΔs11 )s 2 (s1 , s2 ) 2

R

Δ

− (ϕwsΔ22 )s 1 (s1 , s2 ))Δ2 s2 Δ1 s1 1

= − ∫ ϕ(s1 , s2 )wsΔ22 (s1 , s2 )Δ2 s2 + wσ1 (s1 , s2 )ϕΔs11 (s1 , s2 )Δ1 s1 𝜕R t2

= − ∫ ϕ(t1 , s2 )wsΔ22 (t1 , s2 )Δ2 s2 t20

t2

+ ∫ ϕ(t10 , s2 )wsΔ22 (t10 , s2 )Δ2 s2 t20

t1

− ∫ wσ1 (s1 , t20 )ϕΔs11 (s1 , t20 )Δ1 s1 t10

208 | 7 Snow’s inequalities t1

+ ∫ wσ1 (s1 , t2 )ϕΔs11 (s1 , t2 )Δ1 s1 t10

t1

= − ∫ wσ1 (s1 , t20 )ϕΔs11 (s1 , t20 )Δ1 s1 t10

t1

= − ∫ ϕΔs11 (s1 , t20 )Δ1 s1 t10

= −ϕ(t1 , t20 ) + ϕ(t10 , t20 )

= ϕ(t10 , t20 ), i. e.,

ϕ(t10 , t20 ) = ∫∫ w(s1 , s2 )L1 (ϕ)(s1 , s2 )Δ2 s2 Δ1 s1 R

t10 t20

≤ ∫ ∫ w(s1 , s2 )h2 (s1 , s2 )Δ2 s2 Δ1 s1 , t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then u(t10 , t20 ) ≤ a(t10 , t20 ) + b(t10 , t20 )ϕ(t10 , t20 ) ≤ a(t10 , t20 ) +

t10 t20

b(t10 , t20 ) ∫ ∫ p(s1 , s2 )w(s1 , s2 ) s1 s2

t1 t2

× (∫ ∫ v(ξ1 , ξ2 )a(ξ1 , ξ2 )(c(ξ1 , ξ2 ) t1 t2

+ q(ξ1 , ξ2 ))Δ2 ξ2 Δ1 ξ1 + a(s1 , s2 )c(s1 , s2 ))Δ2 s2 Δ1 s1 , (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . This completes the proof. Theorem 7.2.4 (Snow’s Inequality). Let a, b, c, h, p, q, and u be nonnegative continuous functions on D. Let also, for (t10 , t20 ) ∈ D, v1 , v2 , and v3 be the nonnegative solutions of the problems ΔΔ

v1t1 t 2 − g1 (t1 , t2 )v1 = 0, 1 2

(t1 , t2 ) ∈ D, t1 ≥

t10 , t2

≥

t20 , v1 (t1 , t20 ) = 1,

Δ

v1t2 (t10 , t2 ) = 0, 2

7.2 Snow’s inequalities | 209

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , ΔΔ

v2t1 t 2 − g2 (t1 , t2 )v2 = 0, 1 2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , v2 (t1 , t20 ) = 1,

Δ

v2t2 (t10 , t2 ) = 0, 2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , ΔΔ

v3t1 t 2 − b(t1 , t2 )v3 = 0, 1 2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , v3 (t1 , t20 ) = 1, (t1 , t2 ) ∈ D, t1 ≥

t10 , t2

≥

u(t10 , t20 )

t20 ,

≤

Δ

v3t2 (t10 , t2 ) = 0, 2

respectively. If

a(t10 , t20 )

t10 t20

t10 t20

+ ∫ ∫ b(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 t1 t2 s1 s2

+ ∫ ∫ c(s1 , s2 )(∫ ∫ h(ξ1 , ξ2 )u(ξ1 , ξ2 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 t1 t2

t1 t2

t10 t20

s1 s2

+ ∫ ∫ c(s1 , s2 )(∫ ∫ p(ξ1 , ξ2 ) t1 t2

t1 t2

ξ1 ξ2

× (∫ ∫ q(η1 , η2 )u(η1 , η2 )Δ2 η2 Δ1 η1 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 , t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , then u(t10 , t20 )

≤

a(t10 , t20 )

t10 t20

+ ∫ ∫ v3 (s1 , s2 )h3 (s1 , s2 )Δ2 s2 Δ1 s1 , t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 , where h1 (t10 , t20 ) = a(t10 , t20 )b(t10 , t20 ) + h(t10 , t20 )a(t10 , t20 ) + q(t10 , t20 )a(t10 , t20 ),

g1 (t10 , t20 ) = b(t10 , t20 ) + c(t10 , t20 ) + h(t10 , t20 ) + p(t10 , t20 ) + q(t10 , t20 ),

210 | 7 Snow’s inequalities h2 (t10 , t20 ) = a(t10 , t20 )(b(t10 , t20 ) + h(t10 , t20 )) + p(t10 , t20 ) t10 t20

× ∫ ∫ v1 (s1 , s2 )h1 (s1 , s2 )Δ2 s2 Δ1 s1 , g2 (t10 , t20 ) h3 (t10 , t20 )

t1 t2

= b(t10 , t20 ) + c(t10 , t20 ) + h(t10 , t20 ), = a(t10 , t20 )b(t10 , t20 ) + c(t10 , t20 ) t10 t20

× ∫ ∫ v2 (s1 , s2 )h2 (s1 , s2 )Δ2 s2 Δ1 s1 , t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Proof. Let ϕ(t10 , t20 )

t10 t20

= ∫ ∫ b(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 t1 t2

t10 t20

s1 s2

+ ∫ ∫ c(s1 , s2 )(∫ ∫ h(ξ1 , ξ2 )u(ξ1 , ξ2 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 t1 t2

t1 t2

t10 t20

s1 s2

+ ∫ ∫ c(s1 , s2 )(∫ ∫ p(ξ1 , ξ2 ) t1 t2

t1 t2

ξ1 ξ2

× (∫ ∫ q(η1 , η2 )u(η1 , η2 )Δ2 η2 Δ1 η1 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 , t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then u(t10 , t20 ) ≤ a(t10 , t20 ) + ϕ(t10 , t20 ), ϕ(t1 , t20 ) = 0,

ϕ(t10 , t2 ) = 0,

ΔΔ

ϕt 01t 02 (t10 , t20 ) = b(t10 , t20 )u(t10 , t20 ) 1 2

t10 t20

+

c(t10 , t20 )(∫ ∫ h(s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 ) t1 t2

7.2 Snow’s inequalities | 211 0

0

t1 t2 s1 s2 c(t10 , t20 )∫ ∫ p(s1 , s2 )(∫ ∫ q(ξ1 , ξ2 )

+

t1 t2

t1 t2

× u(ξ1 , ξ2 )Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 ≤ b(t10 , t20 )(a(t10 , t20 ) + ϕ(t10 , t20 )) + c(t10 , t20 )(ϕ(t10 , t20 ) t10 t20

+ ∫ ∫ h(s1 , s2 )(a(s1 , s2 ) + ϕ(s1 , s2 ))Δ2 s2 Δ1 s1 t1 t2 t10 t20

s1 s2

+ ∫ ∫ p(s1 , s2 )(∫ ∫ q(ξ1 , ξ2 ) t1 t2

t1 t2

× (a(ξ1 , ξ2 ) + ϕ(ξ1 , ξ2 ))Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 ), (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let ψ(t10 , t20 ) = ϕ(t10 , t20 ) t10 t20

+ ∫ ∫ h(s1 , s2 )(a(s1 , s2 ) + ϕ(s1 , s2 ))Δ2 s2 Δ1 s1 t1 t2 t10 t20

s1 s2

+ ∫ ∫ p(s1 , s2 )(∫ ∫ q(ξ1 , ξ2 ) t1 t2

t1 t2

× (a(ξ1 , ξ2 ) + ϕ(ξ1 , ξ2 ))Δ2 ξ2 Δ1 ξ1 )Δ2 s2 Δ1 s1 , (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then ψ(t1 , t20 ) = ϕ(t1 , t20 ) ψ(t10 , t2 ) ϕ(t10 , t20 ) ΔΔ ϕt 01t 02 (t10 , t20 ) 1 2

= 0,

= ϕ(t10 , t2 ) = 0,

≤ ψ(t10 , t20 ),

≤ b(t10 , t20 )(a(t10 , t20 ) + ψ(t10 , t20 )) + c(t10 , t20 )ψ(t10 , t20 )

212 | 7 Snow’s inequalities = a(t10 , t20 )b(t10 , t20 )

+ (b(t10 , t20 ) + c(t10 , t20 ))ψ(t10 , t20 ),

ΔΔ

ΔΔ

ψt 01t 02 (t10 , t20 ) = ϕt 01t 02 (t10 , t20 ) 1 2

1 2

+ h(t10 , t20 )(a(t10 , t20 ) + ϕ(t10 , t20 )) t10 t20

+

p(t10 , t20 ) ∫ ∫ q(s1 , s2 ) t1 t2

× (a(s1 , s2 ) + ϕ(s1 , s2 ))Δ2 s2 Δ1 s1

≤ a(t10 , t20 )b(t10 , t20 )

+ (b(t10 , t20 ) + c(t10 , t20 ))ψ(t10 , t20 )

+ h(t10 , t20 )(a(t10 , t20 ) + ψ(t10 , t20 )) + p(t10 , t20 )(ψ(t10 , t20 ) t10 t20

+ ∫ ∫ q(s1 , s2 )(a(s1 , s2 ) + ψ(s1 , s2 ))Δ2 s2 Δ1 s1 ), t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let z(t10 , t20 ) = ψ(t10 , t20 ) t10 t20

+ ∫ ∫ q(s1 , s2 )(a(s1 , s2 ) + ψ(s1 , s2 ))Δ2 s2 Δ1 s1 , t1 t2

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Then ψ(t10 , t20 ) ≤ z(t10 , t20 ), z(t1 , t20 ) = ψ(t1 , t20 )

z(t10 , t2 ) ΔΔ ψt 01t 02 (t10 , t20 ) 1 2

= 0,

= ψ(t10 , t2 ) = 0,

≤ a(t10 , t20 )b(t10 , t20 ) + (b(t10 , t20 ) + c(t10 , t20 ))z(t10 , t20 )

+ h(t10 , t20 )(a(t10 , t20 ) + z(t10 , t20 )) ΔΔ

+ p(t10 , t20 )z(t10 , t20 ), ΔΔ

zt 01t 02 (t10 , t20 ) = ψt 01t 02 (t10 , t20 ) 1 2

1 2

+ q(t10 , t20 )(a(t10 , t20 ) + ψ(t10 , t20 ))

7.2 Snow’s inequalities | 213

≤ a(t10 , t20 )b(t10 , t20 ) + h(t10 , t20 )a(t10 , t20 ) + q(t10 , t20 )a(t10 , t20 )

+ (b(t10 , t20 ) + c(t10 , t20 ) + h(t10 , t20 ) + p(t10 , t20 ) + q(t10 , t20 ))z(t10 , t20 )

= g1 (t10 , t20 )z(t10 , t20 ) + h1 (t10 , t20 ), or ΔΔ

zt 01t 02 (t10 , t20 ) − g1 (t10 , t20 )z(t10 , t20 ) ≤ h1 (t10 , t20 ). 1 2

We fix (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let R and 𝜕R be as in the proof of Theorem 7.2.1. For 2 (s1 , s2 ) ∈ R and x ∈ 𝒞1,2 (D), we define the operators L(x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − g1 (s1 , s2 )x(s1 , s2 ),

M(x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − g1 (s1 , s2 )x σ1 σ2 (s1 , s2 ). 2 Then, for x, y ∈ 𝒞1,2 (D) and (s1 , s2 ) ∈ D, we have

yσ1 σ2 (s1 , s2 )L(x)(s1 , s2 ) − x(s1 , s2 )M(y)(s1 , s2 )

= yσ1 σ2 (s1 , s2 )(xsΔ11sΔ22 (s1 , s2 ) − g1 (s1 , s2 )x(s1 , s2 ))

− x(s1 , s2 )(ysΔ11sΔ22 (s1 , s2 ) − g1 (s1 , s2 )yσ1 σ2 (s1 , s2 ))

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ), and Δ

Δ

2

1

(yσ1 xsΔ11 )s 2 (s1 , s2 ) − (xysΔ22 )s 1 (s1 , s2 )

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) + ysΔ22 σ1 (s1 , s2 )xsΔ11 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ) − xsΔ11 (s1 , s2 )ysΔ22 σ1 (s1 , s2 )

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ), i. e., yσ1 σ2 (s1 , s2 )L(x)(s1 , s2 ) − x(s1 , s2 )M(y)(s1 , s2 ) Δ

Δ

2

1

= (yσ1 xsΔ11 )s 2 (s1 , s2 ) − (xysΔ22 )s 1 (s1 , s2 ).

Applying Green’s formula, we get σσ

∫∫ v1 1 2 (s1 , s2 )L(z)(s1 , s2 )Δ2 s2 Δ1 s1 R

σσ

= ∫∫(v1 1 2 (s1 , s2 )L(z)(s1 , s2 ) − z(s1 , s2 )M(v1 )(s1 , s2 ))Δ2 s2 Δ1 s1 R

(7.9)

214 | 7 Snow’s inequalities Δ

σ

= ∫∫((v1 1 zsΔ11 )s 2 (s1 , s2 ) 2

R

Δ

Δ

− (zv1s2 )s 1 (s1 , s2 ))Δ2 s2 Δ1 s1 2

1

σ

= − ∫ z(s1 , s2 )vsΔ22 (s1 , s2 )Δ2 s2 + v1 1 (s1 , s2 )zsΔ11 (s1 , s2 )Δ1 s1 𝜕R t2

Δ

= − ∫ z(t1 , s2 )v1s2 (t1 , s2 )Δ2 s2 2

t20 t2

Δ

+ ∫ z(t10 , s2 )v1s2 (t10 , s2 )Δ2 s2 2

t20 t1

σ

− ∫ v1 1 (s1 , t20 )zsΔ11 (s1 , t20 )Δ1 s1 t10

t1

σ

+ ∫ v1 1 (s1 , t2 )zsΔ11 (s1 , t2 )Δ1 s1 t10

t1

σ

= − ∫ v1 1 (s1 , t20 )zsΔ11 (s1 , t20 )Δ1 s1 t10

t1

= − ∫ zsΔ11 (s1 , t20 )Δ1 s1 t10

= −z(t1 , t20 ) + z(t10 , t20 ) = z(t10 , t20 ), i. e., z(t10 , t20 ) = ∫∫ v1 (s1 , s2 )L(z)(s1 , s2 )Δ2 s2 Δ1 s1 R

t10 t20

≤ ∫ ∫ v1 (s1 , s2 )h1 (s1 , s2 )Δ2 s2 Δ1 s1 . t1 t2

Hence, ΔΔ

ψt 01t 02 (t10 , t20 ) ≤ a(t10 , t20 )(b(t10 , t20 ) + h(t10 , t20 )) 1 2

+ p(t10 , t20 )z(t10 , t20 )

+ (b(t10 , t20 ) + c(t10 , t20 ) + h(t10 , t20 ))ψ(t10 , t20 )

7.2 Snow’s inequalities | 215

≤ a(t10 , t20 )(b(t10 , t20 ) + h(t10 , t20 )) + p(t10 , t20 ) t10 t20

× ∫ ∫ v1 (s1 , s2 )h1 (s1 , s2 )Δ2 s2 Δ1 s1 t1 t2

+ (b(t10 , t20 ) + c(t10 , t20 ) + h(t10 , t20 ))ψ(t10 , t20 )

= g2 (t10 , t20 )ψ(t10 , t20 ) + h2 (t10 , t20 ), or ΔΔ

ψt 01t 02 (t10 , t20 ) − g2 (t10 , t20 )ψ(t10 , t20 ) ≤ h2 (t10 , t20 ). 1 2

We fix (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let R and 𝜕R be as in the proof of Theorem 7.2.1. For 2 (s1 , s2 ) ∈ R and x ∈ 𝒞1,2 (D), we define the operators L1 (x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − g2 (s1 , s2 )x(s1 , s2 ),

M1 (x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − g2 (s1 , s2 )x σ1 σ2 (s1 , s2 ). 2 Then, for x, y ∈ 𝒞1,2 (D) and (s1 , s2 ) ∈ D, we have

yσ1 σ2 (s1 , s2 )L1 (x)(s1 , s2 ) − x(s1 , s2 )M1 (y)(s1 , s2 )

= yσ1 σ2 (s1 , s2 )(xsΔ11sΔ22 (s1 , s2 ) − g2 (s1 , s2 )x(s1 , s2 ))

− x(s1 , s2 )(ysΔ11sΔ22 (s1 , s2 ) − g2 (s1 , s2 )yσ1 σ2 (s1 , s2 ))

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ), and, using (7.9), get yσ1 σ2 (s1 , s2 )L1 (x)(s1 , s2 ) − x(s1 , s2 )M1 (y)(s1 , s2 ) Δ

Δ

2

1

= (yσ1 xsΔ11 )s 2 (s1 , s2 ) − (xysΔ22 )s 1 (s1 , s2 ).

Applying Green’s formula, we get σσ

∫∫ v2 1 2 (s1 , s2 )L1 (ψ)(s1 , s2 )Δ2 s2 Δ1 s1 R

σσ

= ∫∫(v2 1 2 (s1 , s2 )L1 (ψ)(s1 , s2 ) − ψ(s1 , s2 )M1 (v2 )(s1 , s2 ))Δ2 s2 Δ1 s1 R

Δ

σ

= ∫∫((v2 1 ψΔs11 )s 2 (s1 , s2 ) R

2

Δ

Δ

− (ψv2s2 )s 1 (s1 , s2 ))Δ2 s2 Δ1 s1 2

1

216 | 7 Snow’s inequalities Δ

σ

= − ∫ ψ(s1 , s2 )v2s2 (s1 , s2 )Δ2 s2 + v2 1 (s1 , s2 )ψΔs11 (s1 , s2 )Δ1 s1 2

𝜕R t2

Δ

= − ∫ ψ(t1 , s2 )v2s2 (t1 , s2 )Δ2 s2 2

t20 t2

Δ

+ ∫ ψ(t10 , s2 )v2s2 (t10 , s2 )Δ2 s2 2

t20 t1

σ

− ∫ v2 1 (s1 , t20 )ψΔs11 (s1 , t20 )Δ1 s1 t10

t1

σ

+ ∫ v2 1 (s1 , t2 )ψΔs11 (s1 , t2 )Δ1 s1 t10

t1

σ

= − ∫ v2 1 (s1 , t20 )ψΔs11 (s1 , t20 )Δ1 s1 t10

t1

= − ∫ ψΔs11 (s1 , t20 )Δ1 s1 t10

= −ψ(t1 , t20 ) + ψ(t10 , t20 ) = ψ(t10 , t20 ), i. e., ψ(t10 , t20 ) = ∫∫ v2 (s1 , s2 )L1 (ψ)(s1 , s2 )Δ2 s2 Δ1 s1 R

t1 t2

≤ ∫ ∫ v2 (s1 , s2 )h2 (s1 , s2 )Δ2 s2 Δ1 s1 . t10 t20

From here,

ΔΔ

ϕt 01t 02 (t10 , t20 ) ≤ b(t10 , t20 )(a(t10 , t20 ) + ϕ(t10 , t20 )) 1 2

+ c(t10 , t20 )ψ(t10 , t20 )

≤ a(t10 , t20 )b(t10 , t20 ) + c(t10 , t20 ) t1 t2

× ∫ ∫ v2 (s1 , s2 )h2 (s1 , s2 )Δ2 s2 Δ1 s1 t10 t20

+ b(t10 , t20 )ϕ(t10 , t20 )

7.2 Snow’s inequalities | 217

= b(t10 , t20 )ϕ(t10 , t20 ) + h3 (t10 , t20 ), or ΔΔ

ϕt 01t 02 (t10 , t20 ) − b(t10 , t20 )ϕ(t10 , t20 ) ≤ h3 (t10 , t20 ). 1 2

We fix (t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Let R and 𝜕R be as in the proof of Theorem 7.2.1. For 2 (s1 , s2 ) ∈ R and x ∈ 𝒞1,2 (D), we define the operators L2 (x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − b(s1 , s2 )x(s1 , s2 ),

M2 (x)(s1 , s2 ) = xsΔ11sΔ22 (s1 , s2 ) − b(s1 , s2 )x σ1 σ2 (s1 , s2 ). 2 Then, for x, y ∈ 𝒞1,2 (D) and (s1 , s2 ) ∈ D, we have

yσ1 σ2 (s1 , s2 )L2 (x)(s1 , s2 ) − x(s1 , s2 )M2 (y)(s1 , s2 )

= yσ1 σ2 (s1 , s2 )(xsΔ11sΔ22 (s1 , s2 ) − b(s1 , s2 )x(s1 , s2 ))

− x(s1 , s2 )(ysΔ11sΔ22 (s1 , s2 ) − b(s1 , s2 )yσ1 σ2 (s1 , s2 ))

= yσ1 σ2 (s1 , s2 )xsΔ11sΔ22 (s1 , s2 ) − x(s1 , s2 )ysΔ11sΔ22 (s1 , s2 ), and, using (7.9), get yσ1 σ2 (s1 , s2 )L2 (x)(s1 , s2 ) − x(s1 , s2 )M2 (y)(s1 , s2 ) Δ

Δ

2

1

= (yσ1 xsΔ11 )s 2 (s1 , s2 ) − (xysΔ22 )s 1 (s1 , s2 ).

Applying Green’s formula, we get σσ

∫∫ v3 1 2 (s1 , s2 )L2 (ϕ)(s1 , s2 )Δ2 s2 Δ1 s1 R

σσ

= ∫∫(v3 1 2 (s1 , s2 )L2 (ϕ)(s1 , s2 ) − ϕ(s1 , s2 )M2 (v3 )(s1 , s2 ))Δ2 s2 Δ1 s1 R

Δ

σ

= ∫∫((v3 1 ϕΔs11 )s 2 (s1 , s2 ) 2

R

Δ

Δ

− (ϕv3s2 )s 1 (s1 , s2 ))Δ2 s2 Δ1 s1 2

1

Δ

σ2

= − ∫ ϕ (s1 , s2 )v3s2 (s1 , s2 )Δ2 s2 + v3 (s1 , s2 )ϕΔs11 (s1 , s2 )Δ1 s1 2

𝜕R t2

Δ

= − ∫ ϕ(t1 , s2 )v3s2 (t1 , s2 )Δ2 s2 t20 t2

2

Δ

+ ∫ ϕ(t10 , s2 )v3s2 (t10 , s2 )Δ2 s2 t20

2

218 | 7 Snow’s inequalities t1

σ

− ∫ v3 1 (s1 , t20 )ϕΔs11 (s1 , t20 )Δ1 s1 t10

t1

σ

+ ∫ v3 1 (s1 , t2 )ϕΔs11 (s1 , t2 )Δ1 s1 t10

t1

σ

= − ∫ v3 1 (s1 , t20 )ϕΔs11 (s1 , t20 )Δ1 s1 t10

t1

= − ∫ ϕΔs11 (s1 , t20 )Δ1 s1 t10

= −ϕ(t1 , t20 ) + ϕ(t10 , t20 )

= ϕ(t10 , t20 ), i. e.,

ϕ(t10 , t20 ) = ∫∫ v3 (s1 , s2 )L2 (ϕ)(s1 , s2 )Δ2 s2 Δ1 s1 R

t1 t2

≤ ∫ ∫ v3 (s1 , s2 )h3 (s1 , s2 )Δ2 s2 Δ1 s1 , t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . Therefore u(t10 , t20 ) ≤ a(t10 , t20 ) + ϕ(t10 , t20 ) ≤ a(t10 , t20 ) t1 t2

+ ∫ ∫ v3 (s1 , s2 )h3 (s1 , s2 )Δ2 s2 Δ1 s1 , t10 t20

(t1 , t2 ) ∈ D, t1 ≥ t10 , t2 ≥ t20 . This completes the proof.

8 Two-dimensional linear integro-dynamic inequalities In this chapter are investigated some two dimensional Pachpatte type linear integrodynamic inequalities and some of their modifications. The material in this chapter is based on some results in [24] and [25]. Let 𝕋1 and 𝕋2 be time scales with forward jump operators and delta differentiation operators σ1 , σ2 and Δ1 , Δ2 , respectively. Suppose that 0 ∈ 𝕋1 , 0 ∈ 𝕋2 .

8.1 Pachpatte’s inequalities Theorem 8.1.1. Let u(t1 , t2 ) be a nonnegative twice continuously-differentiable function ΔΔ for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), ut11t2 2 (t1 , t2 ) be a nonnegative function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), a(t1 , t2 ) be a positive nondecreasing function in each variable t1 , t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and b(t1 , t2 ) be a nonnnegative continuous function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). If ΔΔ ut11t2 2 (t1 , t2 )

t1 t2

ΔΔ

≤ a(t1 , t2 ) + ∫ ∫ b(s1 , s2 )ut11t2 2 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ a(t1 , t2 )ef (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where t2

f (t1 , t2 ) = ∫ b(t1 , s2 )Δ2 s2 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0

Proof. Let ΔΔ

z(t1 , t2 ) = ut11t2 2 (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Then we can rewrite the given inequality in the following way: t1 t2

z(t1 , t2 ) ≤ a(t1 , t2 ) + ∫ ∫ b(s1 , s2 )z(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0 0

Hence, t1 t2

z(t1 , t2 ) z(s , s ) ≤ 1 + ∫ ∫ b(s1 , s2 ) 1 2 Δ2 s2 Δ1 s1 a(t1 , t2 ) u(t1 , t2 ) 0 0

https://doi.org/10.1515/9783110705553-008

220 | 8 Two-dimensional linear integro-dynamic inequalities t1 t2

≤ 1 + ∫ ∫ b(s1 , s2 ) 0 0

z(s1 , s2 ) ΔsΔs, a(s1 , s2 ) 2 2 1 1

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Let v(t1 , t2 ) =

z(t1 , t2 ) , a(t1 , t2 )

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Then t1 t2

v(t1 , t2 ) ≤ 1 + ∫ ∫ b(s1 , s2 )v(s1 , s2 )Δ2 s2 Δ1 s1 ,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

0 0

Now we apply Wendroff’s inequality, Theorem 6.1.1, and get v(t1 , t2 ) ≤ ef (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

whereupon z(t1 , t2 ) ≤ a(t1 , t2 )ef (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

This completes the proof. Lemma 8.1.2. Let u(t1 , t2 ) be a nonnegative twice continuously-differentiable function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) such that u(t1 , t2 ) is a nondecreasing function with Δ respect to t2 , ut22 (t1 , t2 ) be a nonnegative function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), Δ

ut11 (t1 , 0) = a(t1 ), u(0, t2 ) = c(t2 ),

u(t1 , 0) = b(t1 ),

t2 ∈ ℝ+ ∩ 𝕋2 ,

t ∈ ℝ+ ∩ 𝕋1 ,

where a(t1 ) is a nonnegative continuous function for t1 ∈ ℝ+ ∩ 𝕋1 , b(t1 ) is a positive continuous function for t1 ∈ ℝ+ ∩ 𝕋1 , c(t2 ) is a nonnegative continuous function for t2 ∈ ℝ+ ∩ 𝕋2 , and b(0) = c(0). Let also f (t1 , t2 ) be a nonnegative continuous function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). If ΔΔ

ut11t2 2 (t1 , t2 ) ≤ f (t1 , t2 )u(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

then u(t1 , t2 ) ≤ c(t2 )eg (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ),

where t2

a(t1 ) + ∫ f (t1 , s2 )Δ2 s2 , g(t1 , t2 ) = b(t1 ) 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

8.1 Pachpatte’s inequalities | 221

Proof. Since u(t1 , t2 ) is a nondecreasing function in t2 , t2 ∈ ℝ+ ∩ 𝕋2 , we have u(t1 , t2 ) ≤ u(t1 , σ2 (t2 )),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, ΔΔ

ΔΔ

ut11t2 2 (t1 , t2 )u(t1 , t2 )

≤

u(t1 , t2 )u(t1 , σ2 (t2 ))

ut11t2 2 (t1 , t2 )

u(t1 , t2 ) ≤ f (t1 , t2 ) ≤ f (t1 , t2 ) +

u(t1 , t2 )u(t1 , σ2 (t2 ))

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore Δ

(

ut11 (t1 , t2 ) u(t1 , t2 )

Δ

ut11 (t1 , t2 ) u(t1 , t2 )

Δ2

)

t2

Δ

−

ut11 (t1 , 0) u(t1 , 0)

Δ ut11 (t1 , t2 )

u(t1 , t2 )

Δ

ut22 (t1 , t2 )

,

≤ f (t1 , t2 ), t2

≤ ∫ f (t1 , s2 )Δ2 s2 , 0

≤

t2

a(t1 ) + ∫ f (t1 , s2 )Δ2 s2 b(t1 ) 0

= g(t1 , t2 ),

Δ

ut11 (t1 , t2 ) ≤ g(t1 , t2 )u(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From the last inequality and Lemma 2.1.1, we obtain u(t1 , t2 ) ≤ u(0, t2 )eg (t1 , 0) = c(t2 )eg (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

This completes the proof. Theorem 8.1.3 (Pachpatte’s Inequality). Let u(t1 , t2 ) be a nonnegative twice contiΔΔ nuously-differentiable function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), ut11t2 2 (t1 , t2 ), a(t1 , t2 ), b(t1 , t2 ), and c(t1 , t2 ) be nonnegative continuous functions, u(t1 , 0) = f (t1 ), f (0) = g(0),

t1 ∈ ℝ+ ∩ 𝕋1 ,

u(0, t2 ) = g(t2 ),

t2 ∈ ℝ+ ∩ 𝕋2 ,

where f (t1 ) is a nonnegative continuously-differentiable function for t1 ∈ ℝ+ ∩ 𝕋1 , g(t2 ) is a nonnegative continuously-differentiable function for t2 ∈ ℝ+ ∩ 𝕋2 . If ΔΔ

ut11t2 2 (t1 , t2 ) ≤ a(t1 , t2 )

t1 t2

ΔΔ

+ b(t1 , t2 ) ∫ ∫ c(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

222 | 8 Two-dimensional linear integro-dynamic inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then ΔΔ ut11t2 2 (t1 , t2 )

t1

≤ a(t1 , t2 ) + b(t1 , t2 ) ∫ h3 (s1 , s2 )e⊖h4 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where t1 t2

h1 (t1 , t2 ) = f (t1 ) + g(t2 ) − g(0) + ∫ ∫ a(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

t1 t2

h2 (t1 , t2 ) = ∫ ∫ b(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0 t2

h3 (t1 , t2 ) = ∫ c(t1 , s2 )(h1 (t1 , s2 ) + a(t1 , s2 ))Δ2 s2 , 0

t2

h4 (t1 , t2 ) = ∫ c(t1 , s2 )(h2 (t1 , s2 ) + b(t1 , s2 ))Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

ΔΔ

z(t1 , t2 ) = ∫ ∫ c(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, Δ ut11 (t1 , t2 )

−

Δ ut11 (t1 , 0)

t2

≤ ∫(a(t1 , s2 ) + b(t1 , s2 )z(t1 , s2 ))Δ2 s2 , 0

t1 t2

u(t1 , t2 ) − u(0, t2 ) − u(t1 , 0) + u(0, 0) ≤ ∫ ∫(a(s1 , s2 ) + b(s1 , s2 )z(s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

u(t1 , t2 ) ≤ f (t1 ) + g(t2 ) − g(0) t1 t2

+ ∫ ∫(a(s1 , s2 ) + b(s1 , s2 )z(s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

8.1 Pachpatte’s inequalities | 223

(t1 , t2 ) ∈ (ℝ+ ∩𝕋1 )×(ℝ+ ∩𝕋2 ). Note that z(t1 , t2 ) is a nondecreasing function with respect to t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore t1 t2

u(t1 , t2 ) ≤ f (t1 ) + g(t2 ) − g(0) + ∫ ∫ a(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

+ (∫ ∫ b(s1 , s2 )Δ2 s2 Δ1 s1 )z(t1 , t2 ) 0 0

= h1 (t1 , t2 ) + h2 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). We have Δ zt11 (t1 , t2 )

t2

ΔΔ

= ∫ c(t1 , s2 )(u(t1 , s2 ) + ut11t2 2 (t1 , s2 ))Δ2 s2 0

t2

≤ ∫ c(t1 , s2 )(h1 (t1 , s2 ) + h2 (t1 , s2 )z(t1 , s2 ))Δ2 s2 0

t2

+ ∫ c(t1 , s2 )(a(t1 , s2 ) + b(t1 , s2 )z(t1 , s2 ))Δ2 s2 t2

0

= ∫ c(t1 , s2 )(h1 (t1 , s2 ) + a(t1 , s2 ))Δ2 s2 0

t2

+ ∫ c(t1 , s2 )(h2 (t1 , s2 ) + b(t1 , s2 ))z(t1 , s2 )Δ2 s2 0

≤ h3 (t1 , t2 ) t2

+ (∫ c(t1 , s2 )(h2 (t1 , s2 ) + b(t1 , s2 ))Δ2 s2 )z(t1 , t2 ) 0

= h3 (t1 , t2 ) + h4 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). By Lemma 2.1.1, we obtain z(t1 , t2 ) ≤ z(0, t2 )eh4 (t1 , 0) t1

+ ∫ h3 (s1 , t2 )e⊖h4 (σ1 (s1 ), t1 )Δ1 s1 t1

0

= ∫ h3 (s1 , t2 )e⊖h4 (σ1 (s1 ), t1 )Δ1 s1 , 0

224 | 8 Two-dimensional linear integro-dynamic inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore ΔΔ ut11t2 2 (t1 , t2 )

t1

≤ a(t1 , t2 ) + b(t1 , t2 ) ∫ h3 (s1 , t2 )e⊖h4 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 8.1.4 (Pachpatte’s Inequality). Let u(t1 , t2 ) be a positive twice continuouslydifferentiable function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), Δ

ut11 (t1 , t2 ) ≥ 0, u(t1 , t2 ) Δ1 ut1 (t1 , 0)

≥ u0 ,

Δ

ut22 (t1 , t2 ) ≥ 0,

= q(t1 ),

u(t1 , 0) = p(t1 ),

u(0, t2 ) = r(t2 ),

p(0) = r(0),

where u0 is a positive constant, p(t1 ) and q(t1 ) are positive continuous functions for t1 ∈ ℝ+ ∩ 𝕋1 , r(t2 ) is a positive continuous function for t2 ∈ ℝ+ ∩ 𝕋2 . Let also a(t1 , t2 ), b(t1 , t2 ), and c(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). If ΔΔ

0 ≤ ut11t2 2 (t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )(u(t1 , t2 ) t1 t2

ΔΔ

+ ∫ ∫ c(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )eg (t1 , 0), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where f (t1 , t2 ) =

a(t1 , t2 )(1 + c(t1 , t2 )) + b(t1 , t2 ) + c(t1 , t2 ) + b(t1 , t2 )c(t1 , t2 ), u0 t2

q(t1 ) g(t1 , t2 ) = + ∫ f (t1 , s2 )Δ2 s2 , p(t1 ) 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

ΔΔ

z(t1 , t2 ) = u(t1 , t2 ) + ∫ ∫ c(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

8.1 Pachpatte’s inequalities | 225

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then u(t1 , t2 ) Δ1 Δ2 ut1 t2 (t1 , t2 )

≤ z(t1 , t2 ),

≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, Δ zt11 (t1 , t2 ) ΔΔ zt11t2 2 (t1 , t2 )

t2

=

Δ ut11 (t1 , t2 )

=

ΔΔ ut11t2 2 (t1 , t2 )

ΔΔ

+ ∫ c(t1 , s2 )(u(t1 , s2 ) + ut11t2 2 (t1 , s2 ))Δ2 s2 , 0

ΔΔ

+ c(t1 , t2 )(u(t1 , t2 ) + ut11t2 2 (t1 , t2 ))

≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 ) + c(t1 , t2 )z(t1 , t2 ) + a(t1 , t2 )c(t1 , t2 ) + b(t1 , t2 )c(t1 , t2 )z(t1 , t2 )

= a(t1 , t2 )(1 + c(t1 , t2 ))

+ (b(t1 , t2 ) + c(t1 , t2 ) + b(t1 , t2 )c(t1 , t2 ))z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then ΔΔ

zt11t2 2 (t1 , t2 ) z(t1 , t2 )

≤

a(t1 , t2 )(1 + c(t1 , t2 )) z(t1 , t2 )

+ b(t1 , t2 ) + c(t1 , t2 ) + b(t1 , t2 )c(t1 , t2 ) ≤

a(t1 , t2 )(1 + c(t1 , t2 )) u0

(8.1)

+ b(t1 , t2 ) + c(t1 , t2 ) + b(t1 , t2 )c(t1 , t2 )

= f (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩𝕋1 )×(ℝ+ ∩𝕋2 ). By the definition of the function z(t1 , t2 ), using that u(t1 , t2 ) is a nondecreasing function with respect to t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), it follows that z(t1 , t2 ) is a nondecreasing function with respect to t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore z(t1 , t2 ) ≤ z(t1 , σ2 (t2 )),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

From here and (8.1), we get ΔΔ

zt11t2 2 (t1 , t2 )z(t1 , t2 )

z(t1 , t2 )z(t1 , σ2 (t2 ))

ΔΔ

≤

zt11t2 2 (t1 , t2 ) z(t1 , t2 )

≤ f (t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Note that Δ zt22 (t1 , t2 )

=

Δ ut22 (t1 , t2 )

≥ 0,

t1

ΔΔ

+ ∫ c(s1 , t2 )(u(s1 , t2 ) + ut11t2 2 (s1 , t2 ))Δ1 s1 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

(8.2)

226 | 8 Two-dimensional linear integro-dynamic inequalities Also, Δ

zt11 (t1 , t2 ) ≥ 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

Hence, by (8.2), we obtain ΔΔ

zt11t2 2 (t1 , t2 )z(t1 , t2 )

≤ f (t1 , t2 )

z(t1 , t2 )z(t1 , σ2 (t2 ))

≤ f (t1 , t2 ) Δ

Δ

zt11 (t1 , t2 )zt22 (t1 , t2 )

+

z(t1 , t2 )z(t1 , σ2 (t2 ))

,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Consequently, Δ

(

zt11 (t1 , t2 ) z(t1 , t2 )

Δ

zt11 (t1 , t2 ) z(t1 , t2 )

Δ zt11 (t1 , t2 )

z(t1 , t2 )

Δ2

)

t2

Δ

−

−

zt11 (t1 , 0) z(t1 , 0)

Δ ut11 (t1 , 0)

u(t1 , 0)

≤ f (t1 , t2 ), t2

≤ ∫ f (t1 , s2 )Δ2 s2 , 0

t2

≤ ∫ f (t1 , s2 )Δ2 s2 , 0

t2

Δ zt11 (t1 , t2 )

q(t1 ) ≤ + ∫ f (t1 , s2 )Δ2 s2 z(t1 , t2 ) p(t1 ) 0

= g(t1 , t2 ),

Δ

zt11 (t1 , t2 ) ≤ g(t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From the last inequality and Lemma 2.1.1, we obtain z(t1 , t2 ) ≤ z(0, t2 )eg (t1 , 0)

= u(0, t2 )eg (t1 , 0) = r(t2 )eg (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore u(t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 )

≤ a(t1 , t2 ) + b(t1 , t2 )r(t2 )eg (t1 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 8.1.5 (Pachpatte’s Inequality). Let u(t1 , t2 ) be a nonnegative twice contiΔΔ nuously-differentiable function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), ut11t2 2 (t1 , t2 ) be a

8.1 Pachpatte’s inequalities | 227

nonnegative function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), a(t1 , t2 ), b(t1 , t2 ) and c(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), u0 be a positive constant. If t1 t2

ΔΔ

ΔΔ

ut11t2 2 (t1 , t2 ) ≤ u0 + ∫ ∫ a(s1 , s2 )ut11t2 2 (s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

s1 s2

ΔΔ

+ ∫ ∫ b(s1 , s2 )(∫ ∫ c(τ1 , τ2 )ut11t2 2 (τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ u0 eg (t1 , 0), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where t1 t2

f (t1 , t2 ) = a(t1 , t2 ) + b(t1 , t2 ) ∫ ∫ c(τ1 , τ2 )Δ2 τ2 Δ1 τ1 , 0 0

t2

g(t1 , t2 ) = ∫ f (t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

ΔΔ

z(t1 , t2 ) = u0 + ∫ ∫ a(s1 , s2 )ut11t2 2 (s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

s1 s2

ΔΔ

+ ∫ ∫ b(s1 , s2 )(∫ ∫ c(τ1 , τ2 )ut11t2 2 (τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Note that z(t1 , t2 ) is a positive continuous nondecreasing function in the variables t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). We have Δ zt11 (t1 , t2 )

t2

ΔΔ

= ∫ a(t1 , s2 )ut11t2 2 (t1 , s2 )Δ2 s2 0

t2

t1 s2

ΔΔ

+ ∫ b(t1 , s2 )(∫ ∫ c(τ1 , τ2 )ut11t2 2 (τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0

0 0

228 | 8 Two-dimensional linear integro-dynamic inequalities ΔΔ

ΔΔ

zt11t2 2 (t1 , t2 ) = a(t1 , t2 )ut11t2 2 (t1 , t2 ) t1 t2

ΔΔ

+ b(t1 , t2 ) ∫ ∫ c(τ1 , τ2 )ut11t2 2 (τ1 , τ2 )Δ2 τ2 Δ1 τ1 0 0

≤ a(t1 , t2 )z(t1 , t2 ) t1 t2

+ b(t1 , t2 ) ∫ ∫ c(τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 0 0

≤ a(t1 , t2 )z(t1 , t2 )

t1 t2

+ b(t1 , t2 )(∫ ∫ c(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )z(t1 , t2 ) 0 0

t1 t2

= (a(t1 , t2 ) + b(t1 , t2 )(∫ ∫ c(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )) 0 0

× z(t1 , t2 )

= f (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), i. e., ΔΔ

zt11t2 2 (t1 , t2 ) ≤ f (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Note that Δ

t1

ΔΔ

zt22 (t1 , t2 ) = ∫ a(s1 , t2 )ut11t2 2 (s1 , t2 )Δ1 s1 0

s1 t2

t1

ΔΔ

+ ∫ b(s1 , t2 )(∫ ∫ c(τ1 , τ2 )ut11t2 2 (τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ1 s1 , Δ zt11 (t1 , 0)

0

0 0

= 0,

z(t1 , 0) = u0 ,

z(0, t2 ) = u0 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, using (8.3) and Lemma 8.1.2, we get z(t1 , t2 ) ≤ u0 eg (t1 , 0), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Consequently, ΔΔ

ut11t2 2 (t1 , t2 ) ≤ u0 eg (t1 , 0), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof.

(8.3)

8.1 Pachpatte’s inequalities | 229

Theorem 8.1.6. Let u(t1 , t2 ) be nonnegative twice continuously-differentiable function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), u(0, t2 ) = p(t2 ),

u(t1 , 0) = q(t1 ),

p(0) = q(0),

where p(t2 ) is a nonnegative continuous function for t2 ∈ ℝ+ ∩ 𝕋2 , q(t1 ) is a nonnegative continuous function for t1 ∈ ℝ+ ∩ 𝕋1 . Let a(t1 , t2 ), b(t1 , t2 ) and c(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), u0 a positive constant. If ΔΔ ut11t2 2 (t1 , t2 )

t1 t2

ΔΔ

≤ u0 + ∫ ∫ a(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 0 0

t1 t2

s1 s2

ΔΔ

+ ∫ ∫ b(s1 , s2 )(∫ ∫ c(τ1 , τ2 )ut11t2 2 (τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then t1

u(t1 , t2 ) ≤ u0 eg (t1 , 0) + ∫ f (s1 , t2 )e⊖g (τ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where t2

f (t1 , t2 ) = ∫ a(t1 , s2 )(p(s2 ) + q(t1 ) − q(0))Δ2 s2 , 0

t2

g(t1 , t2 ) = ∫ a(t1 , s2 )(t1 s2 + 1)Δ2 s2 0

t2

t1 s2

+ ∫ b(t1 , s2 )(∫ ∫ c(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 , 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

ΔΔ

z(t1 , t2 ) = u0 + ∫ ∫ a(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 0 0

t1 t2

s1 s2

ΔΔ

+ ∫ ∫ b(s1 , s2 )(∫ ∫ c(τ1 , τ2 )ut11t2 2 (τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

230 | 8 Two-dimensional linear integro-dynamic inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Note that z(t1 , t2 ) is a positive nondecreasing function with respect to the variables t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). We have ΔΔ

ut11t2 2 (t1 , t2 ) ≤ z(t1 , t2 ), Δ ut11 (t1 , t2 )

≤

Δ ut11 (t1 , 0)

t2

+ ∫ z(t1 , s2 )Δ2 s2 , 0

u(t1 , t2 ) ≤ u(0, t2 ) + u(t1 , 0) − u(0, 0) t1 t2

+ ∫ ∫ z(s1 , s2 )Δ2 s2 Δ1 s1 0 0

≤ (p(t2 ) + q(t1 ) − q(0)) + t1 t2 z(t1 , t2 ), Δ zt11 (t1 , t2 )

t2

ΔΔ

= ∫ a(t1 , s2 )(u(t1 , s2 ) + ut11t2 2 (t1 , s2 ))Δ2 s2 0

t2

t1 s2

ΔΔ

+ ∫ b(t1 , s2 )(∫ ∫ c(τ1 , τ2 )ut11t2 2 (τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 t2

0

0 0

≤ ∫ a(t1 , s2 )(p(s2 ) + q(t1 ) − q(0) + t1 s2 z(t1 , s2 ) + z(t1 , s2 ))Δ2 s2 0

t2

t1 s2

+ ∫ b(t1 , s2 )(∫ ∫ c(τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 t2

0

0 0

≤ ∫ a(t1 , s2 )(p(s2 ) + q(t1 ) − q(0))Δ2 s2 0

t2

(∫ a(t1 , s2 )(t1 s2 + 1)Δ2 s2 )z(t1 , t2 ) 0

t2

t1 s2

+ (∫ b(t1 , s2 )(∫ ∫ c(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 ) 0

× z(t1 , t2 )

0 0

= f (t1 , t2 ) + g(t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, by Lemma 2.1.1, we get t1

z(t1 , t2 ) ≤ z(0, t2 )eg (t1 , 0) + ∫ f (s1 , t2 )e⊖g (τ1 (s1 ), t1 )Δ1 s1 0

8.1 Pachpatte’s inequalities | 231 t1

= u0 eg (t1 , 0) + ∫ f (s1 , t2 )e⊖g (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Consequently, ΔΔ

ut11t2 2 (t1 , t2 ) ≤ z(t1 , t2 )

t1

≤ u0 eg (t1 , 0) + ∫ f (s1 , t2 )e⊖g (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 8.1.7 (Pachpatte’s Inequality). Let u(t1 , t2 ) be a nonnegative twice contiΔΔ nuously-differentiable function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), ut11t2 2 (t1 , t2 ) be a nonnegative function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), Δ

ut11 (t1 , 0) = f (t1 ),

u(0, t2 ) = g(t2 ),

f (t1 ) be a nonnegative continuous function for t1 ∈ ℝ+ ∩𝕋1 , g(t2 ) be a nonnegative continuous function for t2 ∈ ℝ+ ∩ 𝕋2 , a(t1 , t2 ) and b(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), u0 be a nonnegative constant. If ΔΔ ut11t2 2 (t1 , t2 )

t1 t2

ΔΔ

≤ u0 + ∫ ∫ a(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s1 0 0

t1 t2

s1 s2

ΔΔ

+ ∫ ∫ a(s1 , s2 )(∫ ∫ b(τ1 , τ2 )(u(τ1 , τ2 ) + ut11t2 2 (τ1 , τ2 ))Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then ΔΔ ut11t2 2 (t1 , t2 )

t1

≤ u0 eh2 (t1 , 0) + ∫ h1 (s1 , t2 )e⊖h2 (σ1 (s1 ), t2 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and t1

u(t1 , t2 ) ≤ g(t2 ) + ∫ f (s1 )Δ1 s1 t1 t2

0

s1

+ ∫ ∫(u0 eh2 (s1 , 0) + ∫ h1 (y1 , s2 )e⊖h2 (σ1 (y1 ), s2 )Δ1 y1 )Δ2 s2 Δ1 s1 , 0 0

0

232 | 8 Two-dimensional linear integro-dynamic inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where t2

t1

h1 (t1 , t2 ) = ∫ a(t1 , s2 )(g(s2 ) + ∫ f (s1 )Δ1 s1 )Δ2 s2 0

t2

t1 s2

0

τ1

+ ∫ a(t1 , s2 )(∫ ∫ b(τ1 , τ2 )(g(τ2 ) + ∫ f (y1 )Δ1 y1 )Δ2 τ2 Δ1 τ1 )Δ2 s2 , 0

0 0

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and t2

h2 (t1 , t2 ) = ∫ a(t1 , s2 )(t1 s2 + 1)Δ2 s2 0

t2

t1 s2

+ ∫ a(t1 , s2 )(∫ ∫(τ1 τ2 + 1)b(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 , 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

ΔΔ

z(t1 , t2 ) = u0 + ∫ ∫ a(s1 , s2 )(u(s1 , s2 ) + ut11t2 2 (s1 , s2 ))Δ2 s2 Δ1 s2 0 0

t1 t2

s1 s2

ΔΔ

+ ∫ ∫ a(s1 , s2 )(∫ ∫ b(τ1 , τ2 )(u(τ1 , τ2 ) + ut11t2 2 (τ1 , τ2 ))Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 , 0 0

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ z(t1 , t2 ), z(t1 , 0) = u0 ,

z(0, t2 ) = u0 , and Δ ut11 (t1 , t2 )

−

Δ ut11 (t1 , 0)

Δ ut11 (t1 , t2 )

t2

≤ ∫ z(t1 , s2 )Δ2 s2 , 0

t2

≤ f (t1 ) + ∫ z(t1 , s2 )Δ2 s2 , t1

0

t1 t2

u(t1 , t2 ) − u(0, t2 ) ≤ ∫ f (s1 )Δ1 s1 + ∫ ∫ z(s1 , s2 )Δ2 s2 Δ1 s1 , 0

0 0

8.1 Pachpatte’s inequalities | 233 t1

u(t1 , t2 ) ≤ g(t2 ) + ∫ f (s1 )Δ1 s1 t1 t2

0

+ ∫ ∫ z(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩𝕋1 )×(ℝ+ ∩𝕋2 ). Note that z(t1 , t2 ) is a nonnegative nondecreasing function in each variable t1 , t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Next, Δ zt11 (t1 , t2 )

t2

ΔΔ

= ∫ a(t1 , s2 )(u(t1 , s2 ) + ut11t2 2 (t1 , s2 ))Δ2 s2 0

t2

t1 s2

ΔΔ

+ ∫ a(t1 , s2 )(∫ ∫ b(τ1 , τ2 )(u(τ1 , τ2 ) + ut11t2 2 (τ1 , τ2 ))Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 t2

0

0 0

≤ ∫ a(t1 , s2 )u(t1 , s2 )Δ2 s2 0

t2

+ ∫ a(t1 , s2 )z(t1 , s2 )Δ2 s2 0

t2

t1 s2

+ ∫ a(t1 , s2 )(∫ ∫ b(τ1 , τ2 )u(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 0

0 0

t2

t1 s2

+ ∫ a(t1 , s2 )(∫ ∫ b(τ1 , τ2 )z(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 Δ1 s1 t2

0

0 0

t1

t1 s2

≤ ∫ a(t1 , s2 )(g(s2 ) + ∫ f (s1 )Δ1 s1 + ∫ ∫ z(s1 , τ2 )Δ2 τ2 Δ1 s1 )Δ2 s2 0

0

t2

0 0

+ (∫ a(t1 , s2 )Δ2 s2 )z(t1 , t2 ) t2

0

t1 s2

τ1

+ ∫ a(t1 , s2 )(∫ ∫ b(τ1 , τ2 )(g(τ2 ) + ∫ f (y1 )Δ1 y1 0 τ1 τ2

0 0

0

+ ∫ ∫ z(y1 , y2 )Δ2 y2 Δ1 y1 )Δ2 τ2 Δ1 τ1 )Δ2 s2 0 0

t2

t1 s2

+ (∫ a(t1 , s2 )(∫ ∫ b(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 )z(t1 , t2 ) 0

0 0

234 | 8 Two-dimensional linear integro-dynamic inequalities t2

t1

≤ ∫ a(t1 , s2 )(g(s2 ) + ∫ f (s1 )Δ1 s1 )Δ2 s2 0

0

t2

+ (∫ a(t1 , s2 )t1 s2 Δ2 s2 )z(t1 , t2 ) 0

t2

+ (∫ a(t1 , s2 )Δ2 s2 )z(t1 , t2 ) t2

0

t1 s2

τ1

+ ∫ a(t1 , s2 )(∫ ∫ b(τ1 , τ2 )(g(τ2 ) + ∫ f (y1 )Δ1 y1 )Δ2 τ2 Δ1 τ1 )Δ2 s2 0

0 0

t2

0

t1 s2

+ (∫ a(t1 , s2 )(∫ ∫ τ1 τ2 b(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 )z(t1 , t2 ) 0

0 0

t2

t1 s2

+ (∫ a(t1 , s2 )(∫ ∫ b(τ1 , τ2 )Δ2 τ2 Δ1 τ1 )Δ2 s2 )z(t1 , t2 ) 0

0 0

= h1 (t1 , t2 ) + h2 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), i. e., Δ

zt11 (t1 , t2 ) ≤ h1 (t1 , t2 ) + h2 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From the last inequality and Lemma 2.1.1, we get t1

z(t1 , t2 ) ≤ z(0, t2 )eh2 (t1 , 0) + ∫ h1 (s1 , t2 )e⊖h2 (σ1 (s1 ), t1 )Δ1 s1 t1

0

= u0 eh2 (t1 , 0) + ∫ h1 (s1 , t2 )e⊖h2 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore ΔΔ

ut11t2 2 (t1 , t2 ) ≤ z(t1 , t2 )

t1

≤ u0 eh2 (t1 , 0) + ∫ h1 (s1 , t2 )e⊖h2 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and Δ ut11 (t1 , t2 )

−

Δ ut11 (t1 , 0)

t2

t1

≤ ∫(u0 eh2 (t1 , 0) + ∫ h1 (s1 , s2 )e⊖h2 (σ1 (s1 ), s1 )Δ1 s1 )Δ2 s2 , 0

0

8.2 Modifications of Pachpatte’s inequalities |

235

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or Δ

ut11 (t1 , t2 ) ≤ f (t1 ) t2

t1

+ ∫(u0 eh2 (t1 , 0) + ∫ h1 (s1 , s2 )e⊖h2 (σ1 (s1 ), s1 )Δ1 s1 )Δ2 s2 , 0

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and t1

u(t1 , t2 ) − u(0, t2 ) ≤ ∫ f (s1 )Δ1 s1 0

t1 t2

s1

+ ∫ ∫(u0 eh2 (s1 , 0) + ∫ h1 (y1 , s2 )e⊖h2 (σ1 (y1 ), s1 )Δ1 y1 )Δ2 s2 Δ1 s1 , 0 0

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or t1

u(t1 , t2 ) ≤ g(t2 ) + ∫ f (s1 )Δ1 s1 t1 t2

0

s1

+ ∫ ∫(u0 eh2 (s1 , 0) + ∫ h1 (y1 , s2 )e⊖h2 (σ1 (y1 ), s1 )Δ1 y1 )Δ2 s2 Δ1 s1 , 0 0

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof.

8.2 Modifications of Pachpatte’s inequalities Theorem 8.2.1. Let u(t1 , t2 ) be a nonnegative twice continuously-differentiable function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), Δ

ut11 (t1 , t2 ) ≥ 0,

Δ

ut22 (t1 , t2 ) ≥ 0,

ΔΔ

ut11t2 2 (t1 , t2 ) ≥ 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), Δ

ut11 (t1 , 0) = f (t1 ),

u(0, t2 ) = g(t2 ),

where f (t1 ) is a nonnegative continuous function for t1 ∈ ℝ+ ∩ 𝕋1 and g(t2 ) is a nonnegative continuous function for t2 ∈ ℝ+ ∩ 𝕋2 . Let a(t1 , t2 ) and b(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). If ΔΔ ut11t2 2 (t1 , t2 )

t1 t2

Δ

≤ a(t1 , t2 ) + ∫ ∫ b(s1 , s2 )ut11 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

236 | 8 Two-dimensional linear integro-dynamic inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ h3 (t1 , t2 ), Δ ut11 (t1 , t2 )

t2

≤ f (t1 ) + ∫ h3 (t1 , s2 )Δ2 s2 , 0

t1

t2

u(t1 , t2 ) ≤ g(t2 ) + ∫(f (s1 ) + ∫ h3 (s1 , s2 )Δ2 s2 )Δ1 s1 , 0

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where t2

f1 (t1 , t2 ) = f (t1 ) + ∫ a(t1 , s2 )Δ2 s2 , 0

t2

h1 (t1 , t2 ) = ∫ f1 (t1 , s2 )b(t1 , s2 )Δ2 s2 , 0

t2

h2 (t1 , t2 ) = ∫ b(t1 , s2 )s2 Δ2 s2 , 0

h3 (t1 , t2 ) = a(t1 , t2 ) t1

+ ∫ h1 (s1 , t2 )e⊖h2 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

Δ

z(t1 , t2 ) = ∫ ∫ b(s1 , s2 )ut11 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Note that z(t1 , t2 ) is a nonnegative continuous nondecreasing function in each variable t1 and t2 , (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ z(t1 , t2 ) + a(t1 , t2 ), Δ ut11 (t1 , t2 )

−

Δ ut11 (t1 , 0)

Δ ut11 (t1 , t2 )

t2

t2

≤ ∫ z(t1 , s2 )Δ2 s2 + ∫ a(t1 , s2 )Δ2 s2 , 0

t2

0

t2

≤ f (t1 ) + ∫ z(t1 , s2 )Δ2 s2 + ∫ a(t1 , s2 )Δ2 s2 0

≤ f1 (t1 , t2 ) + t2 z(t1 , t2 ),

0

8.2 Modifications of Pachpatte’s inequalities |

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Next, Δ zt11 (t1 , t2 )

t2

Δ

= ∫ b(t1 , s2 )ut11 (t1 , s2 )Δ2 s2 0

t2

≤ ∫ b(t1 , s2 )(f1 (t1 , s2 ) + s2 z(t1 , s2 ))Δ2 s2 0

t2

= ∫ b(t1 , s2 )f1 (t1 , s2 )Δ2 s2 0

t2

+ ∫ b(t1 , s2 )s2 z(t1 , s2 )Δ2 s2 t2

0

≤ ∫ f1 (t1 , s2 )b(t1 , s2 )Δ2 s2 0

t2

+ (∫ b(t1 , s2 )s2 Δ2 s2 )z(t1 , t2 ) 0

= h1 (t1 , t2 ) + h2 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), i. e., Δ

zt11 (t1 , t2 ) ≤ h1 (t1 , t2 ) + h2 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From the last inequality and Lemma 2.1.1, we get t1

z(t1 , t2 ) ≤ ∫ h1 (s1 , t2 )e⊖h2 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore ΔΔ

ut11t2 2 (t1 , t2 ) ≤ a(t1 , t2 ) + z(t1 , t2 ) t1

≤ a(t1 , t2 ) + ∫ h1 (s1 , t2 )e⊖h2 (σ1 (s1 ), t1 )Δ1 s1 = h3 (t1 , t2 ), Δ ut11 (t1 , t2 )

−

Δ ut11 (t1 , 0)

Δ ut11 (t1 , t2 )

0

t2

≤ ∫ h3 (t1 , s2 )Δ2 s2 , 0

t2

≤ f (t1 ) + ∫ h3 (t1 , s2 )Δ2 s2 , 0

237

238 | 8 Two-dimensional linear integro-dynamic inequalities t1

t2

u(t1 , t2 ) − u(0, t2 ) ≤ ∫(f (s1 ) + ∫ h3 (s1 , s2 )Δ2 s2 )Δ1 s1 , 0

u(t1 , t2 ) ≤ g(t2 )

0

t1

t2

+ ∫(f (s1 ) + ∫ h3 (s1 , s2 )Δ2 s2 )Δ1 s1 , 0

0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 8.2.2. Let u(t1 , t2 ) be a nonnegative twice continuously-differentiable function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) such that Δ

Δ

ut11 (t1 , t2 ) ≥ 0,

ut22 (t1 , t2 ) ≥ 0,

ΔΔ

ut11t2 2 (t1 , t2 ) ≥ 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and Δ

ut11 (t1 , 0) = f1 (t1 ),

u(t1 , 0) = f3 (t1 ), f3 (0) = f4 (0),

Δ

ut22 (0, t2 ) = f2 (t2 ), u(0, t2 ) = f4 (t2 ),

where f1 (t1 ) and f3 (t1 ) are nonnegative continuous functions for t1 ∈ ℝ+ ∩ 𝕋1 , f2 (t2 ) and f4 (t2 ) are nonnegative continuous functions for t2 ∈ ℝ+ ∩ 𝕋2 . Let also b1 (t1 , t2 ), b2 (t1 , t2 ), and b3 (t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). If t1 t2

ΔΔ

ut11t2 2 (t1 , t2 ) ≤ (∫ ∫ b1 (s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

Δ

+ ∫ ∫ b2 (s1 , s2 )ut11 (s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

Δ

+ ∫ ∫ b3 (s1 , s2 )ut22 (s1 , s2 )Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then ΔΔ

t1

ut11t2 2 (t1 , t2 ) ≤ ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , Δ

0

ut11 (t1 , t2 ) ≤ f1 (t1 )

t1

+ t2 ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , 0

8.2 Modifications of Pachpatte’s inequalities |

Δ

ut22 (t1 , t2 ) ≤ f2 (t2 )

t1

+ t1 ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , 0

t1

u(t1 , t2 ) ≤ f4 (t2 ) + ∫ f1 (s1 )Δ1 s1 t1

0

+ t1 t2 ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where t2

t1

F(t1 , t2 ) = max{∫(f4 (s2 ) + ∫ f1 (s1 )Δ1 s1 )b1 (t1 , s2 )Δ2 s2 , 0

t2

0

s2

∫ b1 (t1 , s2 )(f3 (t1 ) + ∫ f2 (y2 )Δ2 y2 )Δ2 s2 }, 0

0

t2

G(t1 , t2 ) = t1 ∫ b1 (t1 , s2 )s2 Δ2 s2 , 0

t2

h1 (t1 , t2 ) = f1 (t1 ) ∫ b2 (t1 , s2 )Δ2 s2 , 0

t2

h2 (t1 , t2 ) = ∫ b(t1 , s2 )s2 Δ2 s2 , 0

t2

h3 (t1 , t2 ) = ∫ b3 (t1 , s2 )f2 (s2 )Δ2 s2 , 0

t2

h4 (t1 , t2 ) = t1 ∫ b3 (t1 , s2 )Δ2 s2 , 0

h5 (t1 , t2 ) = F(t1 , t2 ) + h1 (t1 , t2 ) + h3 (t1 , t2 ),

h6 (t1 , t2 ) = G(t1 , t2 ) + h2 (t1 , t2 ) + h4 (t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

z(t1 , t2 ) = ∫ ∫ b1 (s1 , s2 )u(s1 , s2 )Δ2 s2 Δ1 s1 0 0

239

240 | 8 Two-dimensional linear integro-dynamic inequalities t1 t2

Δ

+ ∫ ∫ b2 (s1 , s2 )ut11 (s1 , s2 )Δ2 s2 Δ1 s1 0 0

t1 t2

Δ

+ ∫ ∫ b3 (s1 , s2 )ut22 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Note that z(t1 , t2 ) is a nonnegative continuous nondecreasing function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and z(0, t2 ) = 0,

z(t1 , 0) Δ1 Δ2 ut1 t2 (t1 , t2 )

= 0,

≤ z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). We have Δ ut11 (t1 , t2 )

−

Δ ut11 (t1 , 0)

t2

≤ ∫ z(t1 , s2 )Δ2 s2 , 0

t2

Δ

ut11 (t1 , t2 ) ≤ f1 (t1 ) + ∫ z(t1 , s2 )Δ2 s2 0

≤ f1 (t1 ) + t2 z(t1 , t2 ), Δ ut22 (t1 , t2 )

−

Δ ut22 (0, t2 )

t1

≤ ∫ z(s1 , t2 )Δ1 s1 , 0

t1

Δ

ut22 (t1 , t2 ) ≤ f2 (t2 ) + ∫ z(s1 , t2 )Δ1 s1 0

≤ f2 (t2 ) + t1 z(t1 , t2 ), t1

t1

u(t1 , t2 ) − u(0, t2 ) ≤ ∫ f1 (s1 )Δ1 s1 + t2 ∫ z(s1 , t2 )Δ1 s1 , 0

t1

0

u(t1 , t2 ) ≤ f4 (t2 ) + ∫ f1 (s1 )Δ1 s1 t1

0

+ t2 ∫ z(s1 , t2 )Δ1 s1 0

t1

≤ f4 (t2 ) + ∫ f1 (s1 )Δ1 s1 0

+ t1 t2 z(t1 , t2 ),

8.2 Modifications of Pachpatte’s inequalities | 241 t2

t2

u(t1 , t2 ) − u(t1 , 0) ≤ ∫ f2 (s2 )Δ2 s2 + t1 ∫ z(t1 , s2 )Δ2 s2 , 0

0

t2

u(t1 , t2 ) ≤ f3 (t1 ) + ∫ f2 (s2 )Δ2 s2 0

t2

+ t1 ∫ z(t1 , s2 )Δ2 s2 0

t2

≤ f3 (t1 ) + ∫ f2 (s2 )Δ2 s2 0

+ t1 t2 z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). 1. Let t1

u(t1 , t2 ) ≤ f4 (t2 ) + ∫ f1 (s1 )Δ1 s1 + t1 t2 z(t1 , t2 ), 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then t2

t2

t1

∫ b1 (t1 , s2 )u(t1 , s2 )Δ2 s2 ≤ ∫ b1 (t1 , s2 )(f4 (s2 ) + ∫ f1 (s1 )Δ1 s1 0

0

0

+ t1 s2 z(t1 , s2 ))Δ2 s2 t2

t1

= ∫(f4 (s2 ) + ∫ f1 (s1 )Δ1 s1 )b1 (t1 , s2 )Δ2 s2 0

0

t2

+ t1 ∫ b1 (t1 , s2 )s2 z(t1 , s2 )Δ2 s2 0

≤ F(t1 , t2 )

t2

+ (t1 ∫ b1 (t1 , s2 )s2 Δ2 s2 )z(t1 , t2 ) 0

= F(t1 , t2 ) + G(t1 , t2 )z(t1 , t2 ), 2.

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Let t2

u(t1 , t2 ) ≤ f3 (t1 ) + ∫ f2 (s2 )Δ2 s2 + t1 t2 z(t1 , t2 ), 0

242 | 8 Two-dimensional linear integro-dynamic inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then t2

t2

s2

∫ b1 (t1 , s2 )u(t1 , s2 )Δ2 s2 ≤ ∫ b1 (t1 , s2 )(f3 (t1 ) + ∫ f2 (y2 )Δ2 y2 0

0

0

+ t1 s2 z(t1 , s2 ))Δ2 s2 t2

s2

= ∫ b1 (t1 , s2 )(f3 (t1 ) + ∫ f2 (y2 )Δ2 y2 )Δ2 s2 0

0

t2

+ t1 ∫ b1 (t1 , s2 )s2 z(t1 , s2 )Δ2 s2 0

≤ F(t1 , t2 ) + G(t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Next, t2

Δ

t2

∫ b2 (t1 , s2 )ut11 (t1 , s2 )Δ2 s2 ≤ ∫ b2 (t1 , s2 )(f1 (t1 ) + s2 z(t1 , s2 ))Δ2 s2 0

0

t2

= f1 (t1 ) ∫ b2 (t1 , s2 )Δ2 s2 t2

0

+ ∫ b2 (t1 , s2 )s2 z(t1 , s2 )Δ2 s2 0

≤ h1 (t1 , t2 ) t2

+ (∫ b2 (t1 , s2 )s2 Δ2 s2 )z(t1 , t2 ) 0

t2

Δ ∫ b3 (t1 , s2 )ut22 (t1 , s2 )Δ2 s2 0

= h1 (t1 , t2 ) + h2 (t1 , t2 )z(t1 , t2 ), t2

≤ ∫ b3 (t1 , s2 )(f2 (s2 ) + t1 z(t1 , s2 ))Δ2 s2 0

t2

= ∫ b3 (t1 , s2 )f2 (s2 )Δ2 s2 0

t2

+ t1 ∫ b3 (t1 , s2 )z(t1 , s2 )Δ2 s2 0

8.2 Modifications of Pachpatte’s inequalities | 243

≤ h3 (t1 , t2 ) t2

+ (t1 ∫ b3 (t1 , s2 )Δ2 s2 )z(t1 , t2 ) 0

= h3 (t1 , t2 ) + h4 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then Δ zt11 (t1 , t2 )

t2

= ∫ b1 (t1 , s2 )u(t1 , s2 )Δ2 s2 0

t2

Δ

+ ∫ b2 (t1 , s2 )ut11 (t1 , s2 )Δ2 s2 0

t2

Δ

+ ∫ b3 (t1 , s2 )ut22 (t1 , s2 )Δ2 s2 0

≤ F(t1 , t2 ) + G(t1 , t2 )z(t1 , t2 )

+ h1 (t1 , t2 ) + h2 (t1 , t2 )z(t1 , t2 )

+ h3 (t1 , t2 ) + h4 (t1 , t2 )z(t1 , t2 )

= h5 (t1 , t2 ) + h6 (t1 , t2 )z(t1 , t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From the last inequality and Lemma 2.1.1, we obtain t1

z(t1 , t2 ) ≤ ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, Δ

ut11 (t1 , t2 ) ≤ f1 (t1 ) + t2 z(t1 , t2 ) ≤ f1 (t1 )

t1

+ t2 ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , Δ

0

ut22 (t1 , t2 ) ≤ f2 (t2 ) + t1 z(t1 , t2 ) ≤ f2 (t2 )

t1

+ t1 ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , 0

t1

u(t1 , t2 ) ≤ f4 (t2 ) + ∫ f1 (s1 )Δ1 s1 0

+ t1 t2 z(t1 , t2 )

244 | 8 Two-dimensional linear integro-dynamic inequalities t1

≤ f4 (t2 ) + ∫ f1 (s1 )Δ1 s1 t1

0

+ t1 t2 ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof. Theorem 8.2.3. Let a(t1 , t2 ), b(t1 , t2 ) and c(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), u(t1 , t2 ) be a nonnegative twice continuouslydifferentiable function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) such that Δ

Δ

ut11 (t1 , t2 ) ≥ 0,

ut22 (t1 , t2 ) ≥ 0,

ΔΔ

ΔΔ

ut11t2 2 (t1 , t2 ) ≥ 0,

ut22t2 2 (t1 , t2 ) ≥ 0,

Δ

ut11 (t1 , 0) = f (t1 ),

Δ

ut22 (0, t2 ) = g(t2 ),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where f (t1 ) is a nonnegative continuous function for t1 ∈ ℝ+ ∩ 𝕋1 , g(t2 ) is a nonnegative continuous function for t2 ∈ ℝ+ ∩ 𝕋2 . If ΔΔ

Δ

ut11t2 2 (t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )(ut22 (t1 , t2 ) t1 t2

Δ

Δ

+ ∫ ∫ c(s1 , s2 )(ut11 (s1 , s2 ) + ut22 (s1 , s2 ))Δ2 s2 Δ1 s1 ), 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then ΔΔ

ut11t2 2 (t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )g(t2 )eh6 (t1 , 0) t1

+ b(t1 , t2 ) ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where t2

h1 (t1 , t2 ) = f (t1 ) + ∫ a(t1 , s2 )Δ2 s2 , t2

0

h2 (t1 , t2 ) = ∫ b(t1 , s2 )Δ2 s2 , 0

t1

h3 (t1 , t2 ) = g(t2 ) + ∫ a(s1 , t2 )Δ1 s1 , 0

8.2 Modifications of Pachpatte’s inequalities |

245

t1

h4 (t1 , t2 ) = ∫ b(s1 , t2 )Δ1 s1 , 0

t2

h5 (t1 , t2 ) = a(t1 , t2 ) + ∫ c(t1 , s2 )h1 (t1 , s2 )Δ2 s2 0

t2

+ ∫ c(t1 , s2 )h3 (t1 , s2 )Δ2 s2 , 0

t2

h6 (t1 , t2 ) = b(t1 , t2 ) + ∫ c(t1 , s2 )h2 (t1 , s2 )Δ2 s2 0

t2

+ ∫ c(t1 , s2 )h4 (t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let Δ

z(t1 , t2 ) = ut22 (t1 , t2 ) t1 t2

Δ

Δ

+ ∫ ∫ c(s1 , s2 )(ut11 (s1 , s2 ) + ut22 (s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Since ΔΔ

ut11t2 2 (t1 , t2 ) ≥ 0,

ΔΔ

ut22t2 2 (t1 , t2 ) ≥ 0,

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), we have that z(t1 , t2 ) is a nonnegative nondecreasing continuous function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). We have ΔΔ

ut11t2 2 (t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, Δ

t2

Δ

ut11 (t1 , t2 ) − ut11 (t1 , 0) ≤ ∫(a(t1 , s2 ) + b(t1 , s2 )z(t1 , s2 ))Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or Δ ut11 (t1 , t2 )

t2

≤ f (t1 ) + ∫(a(t1 , s2 ) + b(t1 , s2 )z(t1 , s2 ))Δ2 s2 0

246 | 8 Two-dimensional linear integro-dynamic inequalities t2

= f (t1 ) + ∫ a(t1 , s2 )Δ2 s2 0

t2

+ ∫ b(t1 , s2 )z(t1 , s2 )Δ2 s2 0

t2

≤ f (t1 ) + ∫ a(t1 , s2 )Δ2 s2 0

t2

+ (∫ b(t1 , s2 )Δ2 s2 )z(t1 , t2 ) 0

= h1 (t1 , t2 ) + h2 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), and Δ ut22 (t1 , t2 )

−

Δ ut22 (0, t2 )

t1

≤ ∫(a(s1 , t2 ) + b(s1 , t2 )z(s1 , t2 ))Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), or Δ ut22 (t1 , t2 )

t1

≤ g(t2 ) + ∫(a(s1 , t2 ) + b(s1 , t2 )z(s1 , t2 ))Δ1 s1 0

t1

= g(t2 ) + ∫ a(s1 , t2 )Δ1 s1 0

t1

+ ∫ b(s1 , t2 )z(s1 , t2 )Δ1 s1 0

t1

≤ g(t2 ) + ∫ a(s1 , t2 )Δ1 s1 t1

0

+ (∫ b(s1 , t2 )Δ1 s1 )z(t1 , t2 ) 0

= h3 (t1 , t2 ) + h4 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Now, using the definition of function z(t1 , t2 ), we get Δ zt11 (t1 , t2 )

=

ΔΔ ut11t2 2 (t1 , t2 )

t2

Δ

Δ

+ ∫ c(t1 , s2 )(ut11 (t1 , s2 ) + ut22 (t1 , s2 ))Δ2 s2 0

8.2 Modifications of Pachpatte’s inequalities | 247

=

ΔΔ ut11t2 2 (t1 , t2 ) t2

t2

Δ

+ ∫ c(t1 , s2 )ut11 (t1 , s2 )Δ2 s2 0

Δ

+ ∫ c(t1 , s2 )ut22 (t1 , s2 )Δ2 s2 0

≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 ) t2

+ ∫ c(t1 , s2 )(h1 (t1 , s2 ) + h2 (t1 , s2 )z(t1 , s2 ))Δ2 s2 0

t2

+ ∫ c(t1 , s2 )(h3 (t1 , s2 ) + h4 (t1 , s2 )z(t1 , s2 ))Δ2 s2 0

= a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 ) t2

+ ∫ c(t1 , s2 )h1 (t1 , s2 )Δ2 s2 0

t2

+ ∫ c(t1 , s2 )h2 (t1 , s2 )z(t1 , s2 )Δ2 s2 0

t2

+ ∫ c(t1 , s2 )h3 (t1 , s2 )Δ2 s2 0

t2

+ ∫ c(t1 , s2 )h4 (t1 , s2 )z(t1 , s2 )Δ2 s2 0

≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 ) t2

+ ∫ c(t1 , s2 )h1 (t1 , s2 )Δ2 s2 0

t2

+ (∫ c(t1 , s2 )h2 (t1 , s2 )Δ2 s2 )z(t1 , t2 ) t2

0

+ ∫ c(t1 , s2 )h3 (t1 , s2 )Δ2 s2 0

t2

+ (∫ c(t1 , s2 )h4 (t1 , s2 )Δ2 s2 )z(t1 , t2 ) 0

= h5 (t1 , t2 ) + h6 (t1 , t2 )z(t1 , t2 ),

248 | 8 Two-dimensional linear integro-dynamic inequalities (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), i. e., Δ

zt11 (t1 , t2 ) ≤ h5 (t1 , t2 ) + h6 (t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). From the last inequality and Lemma 2.1.1, we find z(t1 , t2 ) ≤ z(0, t2 )eh6 (t1 , 0) t1

+ ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 0

= g(t2 )eh6 (t1 , 0) t1

+ ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Consequently, ΔΔ

ut11t2 2 (t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 )

≤ a(t1 , t2 ) + b(t1 , t2 )g(t2 )eh6 (t1 , 0) t1

+ b(t1 , t2 ) ∫ h5 (s1 , t2 )e⊖h6 (σ1 (s1 ), t1 )Δ1 s1 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof.

9 Two-dimensional nonlinear integral inequalities This chapter is devoted on Wendroff type and Pachpatte type two dimensional nonlinear integral inequalities. Some of the results in this chapter can be found in [8, 24] and [25]. Let 𝕋1 and 𝕋2 be time scales with forward jump operators and delta differentiation operators σ1 , σ2 and Δ1 , Δ2 , respectively. Suppose that 0 ∈ 𝕋1 and 0 ∈ 𝕋2 .

9.1 Wendroff’s inequality Theorem 9.1.1 (Wendroff’s Inequality). Let u(t1 , t2 ), a(t1 , t2 ), and b(t1 , t2 ) be nonnegative continuous functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Let also L : (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) × ℝ+ → ℝ+ be a continuous function such that 0 ≤ L(t1 , t2 , x) − L(t1 , t2 , y) ≤ M(t1 , t2 , y)(x − y) for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) and x ≥ y ≥ 0, where M : (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) × ℝ+ → ℝ+ is a continuous function. If u(t1 , t2 ) ≤ a(t1 , t2 )

t1 t2

+ b(t1 , t2 ) ∫ ∫ L(s1 , s2 , u(s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )h1 (t1 , t2 )

+ b(t1 , t2 )A(t1 , t2 )eB (t2 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where t1 t2

h1 (t1 , t2 ) = ∫ ∫ L(s1 , s2 , a(s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

h2 (t1 , t2 ) = b(t1 , t2 )M(t1 , t2 , a(t1 , t2 ))h1 (t1 , t2 ),

h3 (t1 , t2 ) = b(t1 , t2 )M(t1 , t2 , a(t1 , t2 )), https://doi.org/10.1515/9783110705553-009

250 | 9 Two-dimensional nonlinear integral inequalities t1 t2

A(t1 , t2 ) = ∫ ∫ h2 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0 t2

B(t1 , t2 ) = ∫ h3 (t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Proof. Let t1 t2

z(t1 , t2 ) = ∫ ∫ L(s1 , s2 , u(s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Then z(t1 , t2 ) is a nonnegative continuous function for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) and u(t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 ), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, using the conditions for the function L, we get t1 t2

z(t1 , t2 ) ≤ ∫ ∫ L(s1 , s2 , a(s1 , s2 ) + b(s1 , s2 )z(s1 , s2 ))Δ2 s2 Δ1 s1 0 0

t1 t2

= ∫ ∫ L(s1 , s2 , a(s1 , s2 ))Δ2 s2 Δ1 s1 0 0

t1 t2

+ ∫ ∫(L(s1 , s2 , a(s1 , s2 ) + b(s1 , s2 )z(s1 , s2 )) 0 0

− L(s1 , s2 , a(s1 , s2 )))Δ2 s2 Δ1 s1

≤ h1 (t1 , t2 ) t1 t2

+ ∫ ∫ b(s1 , s2 )M(s1 , s2 , a(s1 , s2 ))z(s1 , s2 )Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, by Corollary 6.2.2, we get z(t1 , t2 ) ≤ h1 (t1 , t2 ) + A(t1 , t2 )eB (t2 , 0), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Therefore u(t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 )z(t1 , t2 )

≤ a(t1 , t2 ) + b(t1 , t2 )h1 (t1 , t2 )

+ b(t1 , t2 )A(t1 , t2 )eB (t2 , 0),

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof.

9.2 Pachpatte’s inequality | 251

9.2 Pachpatte’s inequality Theorem 9.2.1 (Pachpatte’s Inequality). Let functions u(t1 , t2 ), a(t1 , t2 ). and L(t1 , t2 ) satisfy all conditions of Theorem 9.1.1. Let also F(u) be a continuous, strictly increasing, convex and submultiplicative function for u > 0, lim F(u) = ∞,

u→∞

suppose F −1 denotes the inverse function of F, also let α(t1 , t2 ) and β(t1 , t2 ) be continuous and positive functions for (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ) and α(t1 , t2 ) + β(t1 , t2 ) = 1, (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). If u(t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 ) t1 t2

× F (∫ ∫ L(s1 , s2 , F(u(s1 , s2 )))Δ2 s2 Δ1 s1 ), −1

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), then u(t1 , t2 ) ≤ a(t1 , t2 ) + b(t1 , t2 ) t1 t2

× F (∫ ∫ L(s1 , s2 , h1 (s1 , s2 ))Δ2 s2 Δ1 s1 −1

0 0

+ A(t1 , t2 )eB (t2 , 0)), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ), where h1 (t1 , t2 ) = α(t1 , t2 )F(a(t1 , t2 )(α(t1 , t2 )) ), −1

h2 (t1 , t2 ) = β(t1 , t2 )F(b(t1 , t2 )(β(t1 , t2 )) ), −1

t1 t2

h3 (t1 , t2 ) = ∫ ∫ L(s1 , s2 , h1 (s1 , s2 ))Δ2 s2 Δ1 s1 , 0 0

h4 (t1 , t2 ) = h2 (t1 , t2 )M(t1 , t2 , h1 (t1 , t2 )),

h5 (t1 , t2 ) = h2 (t1 , t2 )M(t1 , t2 , h1 (t1 , t2 )), t1 t2

A(t1 , t2 ) = ∫ ∫ h4 (s1 , s2 )Δ2 s2 Δ1 s1 , 0 0 t2

B(t1 , t2 ) = ∫ h5 (t1 , s2 )Δ2 s2 , 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ).

(9.1)

252 | 9 Two-dimensional nonlinear integral inequalities Proof. We rewrite inequality (9.1) in the following form: u(t1 , t2 ) ≤ α(t1 , t2 )a(t1 , t2 )(α(t1 , t2 ))

−1

+ β(t1 , t2 )b(t1 , t2 )(β(t1 , t2 ))

−1

t1 t2

× F (∫ ∫ L(s1 , s2 , F(u(s1 , s2 )))Δ2 s2 Δ1 s1 ), −1

0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Now, since F is strictly increasing, convex and submultiplicative, we get F(u(t1 , t2 )) ≤ F(α(t1 , t2 )a(t1 , t2 )(α(t1 , t2 ))

−1

+ β(t1 , t2 )b(t1 , t2 )(β(t1 , t2 ))

−1

t1 t2

× F −1 (∫ ∫ L(s1 , s2 , F(u(s1 , s2 )))Δ2 s2 Δ1 s1 )) 0 0

≤ α(t1 , t2 )F(a(t1 , t2 )(α(t1 , t2 )) ) −1

+ β(t1 , t2 )F(b(t1 , t2 )(β(t1 , t2 ))

−1

t1 t2

× F (∫ ∫ L(s1 , s2 , F(u(s1 , s2 )))Δ2 s2 Δ1 s1 )) −1

0 0

≤ α(t1 , t2 )F(a(t1 , t2 )(α(t1 , t2 )) ) −1

+ β(t1 , t2 )F(b(t1 , t2 )(β(t1 , t2 )) ) −1

t1 t2

× F(F (∫ ∫ L(s1 , s2 , F(u(s1 , s2 )))Δ2 s2 Δ1 s1 )) −1

= h1 (t1 , t2 )

0 0 t1 t2

+ h2 (t1 , t2 ) ∫ ∫ L(s1 , s2 , F(u(s1 , s2 )))Δ2 s2 Δ1 s1 , 0 0

(t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Now we apply Theorem 9.1.1, and get F(u(t1 , t2 )) ≤ h1 (t1 , t2 ) + h2 (t1 , t2 )h3 (t1 , t2 ) + h2 (t1 , t2 )A(t1 , t2 )eB (t2 , 0)

= α(t1 , t2 )F(a(t1 , t2 )(α(t1 , t2 )) ) −1

+ β(t1 , t2 )F(b(t1 , t2 )(β(t1 , t2 )) ) −1

9.2 Pachpatte’s inequality | 253 t1 t2

× (∫ ∫ L(s1 , s2 , h1 (s1 , s2 ))Δ2 s2 Δ1 s1 0 0

+ A(t1 , t2 )eB (t2 , 0)), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). Hence, u(t1 , t2 ) ≤ F −1 (h1 (t1 , t2 ) + h2 (t1 , t2 )h3 (t1 , t2 ) + h2 (t1 , t2 )A(t1 , t2 )eB (t2 , 0))

= F −1 (α(t1 , t2 )F(a(t1 , t2 )(α(t1 , t2 )) ) −1

+ β(t1 , t2 )F(b(t1 , t2 )(β(t1 , t2 )) ) −1

t1 t2

× (∫ ∫ L(s1 , s2 , h1 (s1 , s2 ))Δ2 s2 Δ1 s1 0 0

+ A(t1 , t2 )eB (t2 , 0))) ≤ α(t1 , t2 )F −1 (F(a(t1 , t2 )(α(t1 , t2 )) )) −1

+ β(t1 , t2 )F −1 (F(b(t1 , t2 )(β(t1 , t2 )) ) −1

t1 t2

× (∫ ∫ L(s1 , s2 , h1 (s1 , s2 ))Δ2 s2 Δ1 s1 0 0

+ A(t1 , t2 )eB (t2 , 0))) ≤ α(t1 , t2 )a(t1 , t2 )(α(t1 , t2 ))

−1

+ β(t1 , t2 )F −1 (F(b(t1 , t2 )(β(t1 , t2 )) )) −1

t1 t2

× F −1 (∫ ∫ L(s1 , s2 , h1 (s1 , s2 ))Δ2 s2 Δ1 s1 0 0

+ A(t1 , t2 )eB (t2 , 0)) = a(t1 , t2 )

+ β(t1 , t2 )b(t1 , t2 )(β(t1 , t2 ))

−1

t1 t2

× F (∫ ∫ L(s1 , s2 , h1 (s1 , s2 ))Δ2 s2 Δ1 s1 −1

0 0

254 | 9 Two-dimensional nonlinear integral inequalities

+ A(t1 , t2 )eB (t2 , 0)) = a(t1 , t2 ) + b(t1 , t2 ) t1 t2

× F (∫ ∫ L(s1 , s2 , h1 (s1 , s2 ))Δ2 s2 Δ1 s1 −1

0 0

+ A(t1 , t2 )eB (t2 , 0)), (t1 , t2 ) ∈ (ℝ+ ∩ 𝕋1 ) × (ℝ+ ∩ 𝕋2 ). This completes the proof.

10 Delay integral inequalities This chapter introduces delay integral inequalities with one and two independent variables. The material in this chapter is based on some results in [10, 14, 15, 19–22] and [30]. Let 𝕋 be an unbounded time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let also t0 ∈ 𝕋.

10.1 Linear Gronwall–Bellman-type delay integral inequalities In this section we will investigate the inequality t

x(t) ≤ a(t) + c(t) ∫(f (s)x(τ(s)) + g(s)x(s))Δs, t0

t ∈ [t0 , ∞),

(10.1)

with the initial conditions x(t) = ϕ(t),

t ∈ [α, t0 ],

ϕ(τ(t)) ≤ a(t),

(10.2)

t ∈ [t0 , ∞), τ(t) ≤ t0 ,

where (A1) x, a, c, f , g ∈ 𝒞rd ([t0 , ∞)) are nonnegative functions, a and c are nondecreasing functions on [t0 , ∞). (A2) τ : [t0 , ∞) → 𝕋, τ(t) ≤ t, t ∈ [t0 , ∞), −∞ < α = inf{τ(t) : t ∈ [t0 , ∞)}, ϕ ∈ 𝒞rd ([α, t0 ]) is a nonnegative function. Theorem 10.1.1. Suppose (A1) and (A2) hold. Then t

x(t) ≤ a(t) + c(t) ∫ e⊖(f +g)c (σ(s), t)a(s)(f (s) + g(s))Δs, t0

t ∈ [t0 , ∞).

Proof. Let t ∗ ∈ [t0 , ∞). Define t

z(t) = a(t ∗ ) + c(t) ∫(f (s)x(τ(s)) + g(s)x(s))Δs, t0

t ∈ [t0 , t ∗ ].

Then z is a nondecreasing function on [t0 , t ∗ ] and t

x(t) ≤ a(t ∗ ) + c(t) ∫(f (s)x(τ(s)) + g(s)x(s))Δs t0

https://doi.org/10.1515/9783110705553-010

(10.3)

256 | 10 Delay integral inequalities = z(t),

t ∈ [t0 , t ∗ ].

Next, x(τ(t)) ≤ z(τ(t)) ≤ z(t),

t ∈ [t0 , t ∗ ], τ(t) ≥ t0 ,

and x(τ(t)) = ϕ(τ(t)) ≤ a(t)

≤ a(t ∗ ) ≤ z(t),

t ∈ [t0 , t ∗ ], τ(t) ≤ t0 .

Hence, by (10.3), we get t

z(t) ≤ a(t ∗ ) + c(t) ∫(f (s)z(s) + g(s)z(s))Δs t0

t

= a(t ∗ ) + c(t) ∫(f (s) + g(s))z(s)Δs,

t ∈ [t0 , t ∗ ].

t0

Because t ∗ ∈ [t0 , ∞) was arbitrarily chosen, by the last inequality, we obtain t

z(t) ≤ a(t) + c(t) ∫(f (s) + g(s))z(s)Δs, t0

t ∈ [t0 , ∞).

Then, applying Theorem 1.1.10, we arrive at x(t) ≤ z(t)

t

≤ a(t) + c(t) ∫(f (s) + g(s))a(s)e⊖(f +g)c (σ(s), t)Δs, t0

t ∈ [t0 , ∞).

This completes the proof.

10.2 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay Consider the integral inequality p

t

(x(t)) ≤ a(t) + c(t) ∫(f (s)x(τ(s)) + g(s))Δs, t0

t ∈ [t0 , ∞),

(10.4)

10.2 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay | 257

with the initial conditions x(t) = ϕ(t),

1 p

t ∈ [α, t0 ],

ϕ(τ(t)) ≤ (a(t)) ,

t ∈ [t0 , ∞), τ(t) ≤ t0 ,

where (A3) x, a, c, f , g ∈ 𝒞rd ([t0 , ∞)) are nonnegative functions, (A4) p ≥ 1 is a constant, τ : [t0 , ∞) → 𝕋, τ(t) ≤ t, t ∈ [t0 , ∞), −∞ < α = inf{τ(t) : t ∈ [t0 , ∞)}, ϕ ∈ 𝒞rd ([t0 , ∞)) is a nonnegative function. We will start our investigations with the following useful lemma. Lemma 10.2.1. Let p ≥ q ≥ 0, p ≠ 0, and a ≥ 0. Then, for any k > 0, we have q

ap ≤

q q−p p − q pq k p a+ k . p p

Proof. 1. Let a = 0. Then 0≤ 2.

p − q pq k . p

Let a > 0. Set f (k) =

q q−p p − q pq k p a+ k , p p

k > 0.

Then q(q − p) q−2p p − q q−p k p a+q 2 k p 2 p p q(p − q) q−2p = k p (k − a), k > 0. p2

f 󸀠 (k) =

Hence, we conclude that f (k) ≥ f (a),

k > 0,

and since f (a) =

p − q pq q q−p (a p )a + a p p

(10.5)

258 | 10 Delay integral inequalities

=

q pq p − q pq a + a p p q

= ap , we get the desired result. This completes the proof. Theorem 10.2.2. Suppose (A3), (A4) hold and a, c are nondecreasing on [t0 , ∞). Then inequality (10.4) with the initial data (10.5) implies t

x(t) ≤ (a(t) + c(t)(h(t) + ∫ eB (t, σ(s))h(s)B(s)Δs))

1 p

t0

1

≤ (a(t) + c(t)h(t)eB (t, t0 )) p ,

t ∈ [t0 , ∞),

for any k > 0, where t

h(t) = ∫(f (s)( t0

B(t) =

c(t)f (t) pk

p−1 p

,

a(s) p − 1 p1 ) + g(s))Δs, k + p−1 p pk p t ∈ [t0 , ∞).

Proof. Take t ∗ arbitrarily. Define 1 p

t

z(t) = (a(t ) + c(t) ∫(f (s)x(τ(s)) + g(s))Δs) , ∗

t0

t ∈ [t0 , t ∗ ].

We have that z is a nonnegative and nondecreasing function on [t0 , t ∗ ], and x(t) ≤ z(t),

t ∈ [t0 , t ∗ ].

If t ∈ [t0 , t ∗ ] and τ(t) ≤ t0 , then x(τ(t)) = ϕ(τ(t)) 1

≤ (a(t)) p

1

≤ (a(t ∗ )) p ≤ z(t).

If t ∈ [t0 , t ∗ ] and τ(t) ≥ t0 , then x(τ(t)) ≤ z(t).

10.2 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay | 259

Therefore t

p

(z(t)) = a(t ∗ ) + c(t) ∫(f (s)x(τ(s)) + g(s))Δs t0

t

≤ a(t ∗ ) + c(t) ∫(f (s)z(s) + g(s))Δs,

t ∈ [t0 , t ∗ ].

t0

In particular, t∗

p

(z(t ∗ )) ≤ a(t ∗ ) + c(t ∗ ) ∫(f (s)z(s) + g(s))Δs. t0

Therefore t

p

(z(t)) ≤ a(t) + c(t) ∫(f (s)z(s) + g(s))Δs, t0

t ∈ [t0 , ∞),

(10.6)

and x(t) ≤ z(t),

t ∈ [t0 , ∞).

(10.7)

Let t

u(t) = ∫(f (s)z(s) + g(s))Δs, t0

t ∈ [t0 , ∞).

Then, for any k > 0, applying Lemma 10.2.1, we get 1

z(t) ≤ (a(t) + c(t)u(t)) p p − 1 p1 a(t) c(t)u(t) ≤ k + + , p−1 p−1 p pk p pk p

t ∈ [t0 , ∞).

Thus, t

u(t) ≤ ∫(f (s)( t0

t

= ∫(f (s)( t0

+

1 pk

p−1 p

a(s) c(s)u(s) p − 1 p1 k + + ) + g(s))Δs p−1 p−1 p pk p pk p p − 1 p1 a(s) k + ) + g(s))Δs p−1 p pk p t

∫ f (s)c(s)u(s)Δs

t0

260 | 10 Delay integral inequalities t

= h(t) + ∫ B(s)u(s)Δs, t0

t ∈ [t0 , ∞).

Now, applying Theorem 1.1.10, we obtain t

u(t) ≤ h(t) + ∫ B(s)h(s)eB (σ(s), t)Δs, t0

t ∈ [t0 , ∞),

and x(t) ≤ z(t)

1

≤ (a(t) + c(t)u(t)) p

1 p

t

≤ (a(t) + c(t)h(t) + c(t) ∫ B(s)h(s)eB (σ(s), t)Δs) , t0

t ∈ [t0 , ∞). Note that h is a nonnegative and nondecreasing function on [t0 , ∞). Then t

x(t) ≤ (a(t) + c(t)h(t)(1 + ∫ B(s)eB (σ(s), t)Δs))

1 p

t0

t

Δ

= (a(t) + c(t)h(t)(1 − ∫ eBs (t, s)Δs))

1 p

t0

1

= (a(t) + c(t)h(t)(1 − eB (t, t) + eB (t, t0 ))) p 1

= (a(t) + c(t)h(t)eB (t, t0 )) p ,

t ∈ [t0 , ∞).

This completes the proof. Next, consider the inequality t

η(x(t)) ≤ a(t) + ∫(f (t, s)ψ(x(τ(s)))w(x(τ(s))) + g(t, s)ψ(x(τ(s))))Δs t0

(10.8)

t s

+ ∫ ∫ h(ξ )ψ(x(τ(ξ )))Δξ Δs t0 t0

with the initial conditions η(x(t) = ϕ(t),

ϕ(τ(t)) ≤ a(t), where

t ∈ [α, t0 ],

t ∈ [t0 , ∞), τ(t) ≤ t0 ,

(10.9)

10.2 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay | 261

(A6) x, a ∈ 𝒞rd ([t0 , ∞)) are nonnegative and nondecreasing functions, f (t, s), ftΔ (t, s) ∈ 𝒞rd ([t0 , ∞) × [t0 , ∞)) are nonnegative functions, w ∈ 𝒞 ([t0 , ∞)) is a nonnegative and nondecreasing function, p > 0 is a constant, τ : [t0 , ∞) → 𝕋, τ(t) ≤ t, t ∈ [t0 , ∞), −∞ < α = inf{τ(t) : t ∈ [t0 , ∞)}, ϕ ∈ 𝒞rd ([α, t0 ]) is a nonnegative function. Theorem 10.2.3. Suppose (A6) holds. Let also g(t, s), gtΔ (t, s) ∈ 𝒞rd ([t0 , ∞)) be nonnegative functions, h ∈ 𝒞rd ([t0 , ∞)) be a nonnegative function, ψ, η ∈ 𝒞 ([t0 , ∞)) be nonnegative functions, and suppose ψ is nondecreasing, while η is strictly increasing. If x satisfies inequality (10.8) with the initial data (10.9), then t

s

̃−1 (H ̃ ̃ −1 (H( ̃ G(a(t)) x(t) ≤ η−1 (G + ∫(g(t, s) + ∫ h(ξ )Δξ )Δs) t0

t0

t

+ ∫ f (t, s)Δs))), t0

t ∈ [t0 , ∞),

̃ and H ̃ are increasing, bijective functions, and where G Δ

̃ (G(v(t))) = Δ

̃ (H(z(t))) =

vΔ (t) , ψ(η−1 (v(t)))

z Δ (t) , ̃−1 (z(t)))) w(η−1 (G

t ∈ [t0 , ∞).

Proof. Let t ∗ ∈ [t0 , ∞) be arbitrarily chosen. Define t

z(t) = a(t ) + ∫(f (t, s)ψ(x(τ(s)))w(x(τ(s))) + g(t, s)ψ(x(τ(s))))Δs ∗

t0

t s

+ ∫ ∫ h(ξ )ψ(x(τ(ξ )))Δξ Δs, t0 t0

t ∈ [t0 , ∞).

Then z is a nondecreasing and nonnegative function on [t0 , t ∗ ]. Next, t

η(x(t)) ≤ a(t) + ∫(f (t, s)ψ(x(τ(s)))w(x(τ(s))) + g(t, s)ψ(x(τ(s))))Δs t0

t s

+ ∫ ∫ h(ξ )ψ(x(τ(ξ )))Δξ Δs t0 t0

262 | 10 Delay integral inequalities t

≤ a(t ) + ∫(f (t, s)ψ(x(τ(s)))w(x(τ(s))) + g(t, s)ψ(x(τ(s))))Δs ∗

t0

t s

+ ∫ ∫ h(ξ )ψ(x(τ(ξ )))Δξ Δs t0 t0

= z(t),

t ∈ [t0 , t ∗ ],

whereupon x(t) ≤ η−1 (z(t)),

t ∈ [t0 , t ∗ ].

If t ∈ [t0 , t ∗ ] and τ(t) ≤ t0 , then η(x(τ(t))) = ϕ(τ(t)) ≤ a(t)

≤ a(t ∗ )

≤ z(t). If t ∈ [t0 , t ∗ ] and τ(t) ≥ t0 , then

η(x(τ(t))) ≤ z(τ(t)) ≤ z(t).

Therefore η(x(τ(t))) ≤ z(t),

t ∈ [t0 , t ∗ ],

and x(τ(t)) ≤ η−1 (z(t)),

t ∈ [t0 , t ∗ ].

Next, Δ

t

z (t) = ∫(ftΔ (t, s)ψ(x(τ(s)))w(x(τ(s))) + gtΔ (t, s)ψ(x(τ(s))))Δs t0

+ f (σ(t), t)ψ(x(τ(t)))w(x(τ(t))) t

+ g(σ(t), t)ψ(x(τ(t))) + ∫ h(ξ )ψ(x(τ(ξ )))Δξ t0

t

≤ (∫(ftΔ (t, s)w(η−1 (z(s))) + gtΔ (t, s))Δs t0

10.2 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay | 263 t

+ f (σ(t), t)w(η (z(t))) + g(σ(t), t) + ∫ h(ξ )Δξ ) −1

t0

× ψ(η (z(t))), −1

t ∈ [t0 , t ], ∗

and t

z Δ (t) ≤ ∫(ftΔ (t, s)w(η−1 (z(s))) + gtΔ (t, s))Δs ψ(η−1 (z(t))) t0

t

+ f (σ(t), t)w(η (z(t))) + g(σ(t), t) + ∫ h(ξ )Δξ −1

t0

s

t

Δ

= (∫(f (t, s)w(η (z(s))) + g(t, s) + ∫ h(ξ )Δξ )Δs) , −1

t0

t0

t ∈ [t0 , t ∗ ]. Now we integrate the last inequality from t0 to t and get t −1 ̃ ̃ G(z(t)) − G(z(t 0 )) ≤ ∫(f (t, s)w(η (z(s))) + g(t, s) t0

s

+ ∫ h(ξ )Δξ )Δs, t0

t ∈ [t0 , t ∗ ].

Hence, using that z(t0 ) = a(t ∗ ), we find t ∗ ̃ ̃ G(z(t)) ≤ G(a(t )) + ∫(f (t, s)w(η−1 (z(s))) + g(t, s) s

t0

+ ∫ h(ξ )Δξ )Δs, t0

t ∈ [t0 , t ∗ ],

and t ∗ ̃−1 (G(a(t ̃ z(t) ≤ G )) + ∫(f (t, s)w(η−1 (z(s))) + g(t, s) s

t0

+ ∫ h(ξ )Δξ )Δs), t0

Let

t ∈ [t0 , t ∗ ].

264 | 10 Delay integral inequalities t∗ ∗ ̃ v(t) = G(a(t )) + ∫(f (t, s)w(η−1 (z(s))) + g(t, s) s

t0

+ ∫ h(ξ )Δξ )Δs t0

t

+ ∫ f (t, s)w(η−1 (z(s)))Δs,

t ∈ [t0 , t ∗ ].

t0

Then ̃−1 (v(t)), z(t) ≤ G

t ∈ [t0 , t ∗ ],

and Δ

t

v (t) = ∫ ftΔ (t, s)w(η−1 (z(s)))Δs + f (σ(t), t)w(η−1 (z(t))) t0

t

̃−1 (v(t)))), ≤ (∫ ftΔ (t, s)Δs + f (σ(t), t))w(η−1 (G t0

t ∈ [t0 , t ∗ ], and t

vΔ (t) ≤ ∫ ftΔ (t, s)Δs + f (σ(t), t) −1 −1 ̃ w(η (G (v(t)))) t0

Δ

t

= (∫ f (t, s)Δs) , t0

t ∈ [t0 , t ∗ ].

Integrating the last inequality from t0 to t, t ∈ [t0 , t ∗ ], we get t

̃ ̃ H(v(t)) − H(v(t 0 )) ≤ ∫ f (t, s)Δs, t0

t ∈ [t0 , t ∗ ],

and t

̃ −1 (H(v(t ̃ v(t) ≤ H 0 )) + ∫ f (t, s)Δs) t0

t∗

̃ ̃ G(a(t = H (H( )) + ∫(f (t, s)w(η−1 (z(s))) + g(t, s) ̃ −1

∗

t0

10.2 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay | 265 s

+ ∫ h(ξ )Δξ )Δs) t0

t

+ ∫ f (t, s)Δs),

t ∈ [t0 , t ∗ ].

t0

Hence, x(t) ≤ η−1 (z(t)) ̃−1 (v(t))) ≤ η−1 (G

t∗

∗ ̃−1 (H ̃ ̃ −1 (H( ̃ G(a(t ≤ η−1 (G )) + ∫(f (t, s)w(η−1 (z(s))) + g(t, s) t0

s

t

+ ∫ h(ξ )Δξ )Δs) + ∫ f (t, s)Δs))), t0

t0

t ∈ [t0 , t ]. In particular, for t = t , we find ∗

∗

t∗ ∗ ̃−1 (H ̃ ̃ −1 (H( ̃ G(a(t x(t ) ≤ η (G )) + ∫(f (t, s)w(η−1 (z(s))) + g(t, s) ∗

−1

t0

s

t

∗

+ ∫ h(ξ )Δξ )Δs) + ∫ f (t ∗ , s)Δs))). t0

t0

Because t ∈ [t0 , ∞) was arbitrarily chosen, we get the desired result. This completes the proof. ∗

Next, consider the delay integral inequality p

t

t

p

q

r

(x(t)) ≤ a(t) + ∫ b(s)(x(s)) Δs + c(t) ∫(f (s)(x(τ(s))) + g(s)(x(s)) )Δs, t0

t0

(10.10)

with the initial data x(t) = ϕ(t),

1 p

t ∈ [t0 , ∞),

t ∈ [α, t0 ],

ϕ(τ(t)) ≤ (a(t)) ,

t ∈ [t0 , ∞),

τ(t) ≤ t0 ,

(10.11)

where (A7) p, q, r are constants, p ≠ 0, p ≥ q ≥ 0, p ≥ r ≥ 0, τ : [t0 , ∞) → 𝕋, τ(t) ≤ t, t ∈ [t0 , ∞), −∞ < α = inf{τ(t) : t ∈ [t0 , ∞)}, ϕ ∈ 𝒞rd ([α, t0 ]) is a nonnegative function,

266 | 10 Delay integral inequalities (A8) x, a, b, c, f , g ∈ 𝒞rd ([t0 , ∞)) are nonnegative functions, a and c are nondecreasing functions on [t0 , ∞). Theorem 10.2.4. If x satisfies (10.10) with the initial data (10.11), then t

x(t) ≤ (eb (t, t0 )(a(t) + c(t)(F(t) + ∫ eG (t, σ(s))F(s)G(s)Δs)))

1 p

t0

1

≤ (eb (t, t0 )(a(t) + c(t)F(t)eG (t, t0 ))) p ,

t ∈ [t0 , ∞),

for any k > 0, where t

q

F(t) = ∫(

f (s)(eb (s, t0 )) p (k(p − q) + qa(s)) pk

t0

+

p−q p

r

g(s)(eb (s, t0 )) p (k(p − r) + ra(s))

G(t) = c(t)(

pk

p−q p q

qf (t)(eb (t, t0 )) p pk

p−q p

)Δs, r

+

rg(t)(eb (t, t0 )) p pk

p−q p

),

t ∈ [t0 , ∞). Proof. Define the function t

t

p

q

r

1 p

z(t) = (a(t) + ∫ b(s)(x(s)) Δs + c(t) ∫(f (s)(x(τ(s))) + g(s)(x(s)) )Δs) , t0

t0

t ∈ [t0 , ∞). We have that z is a nonnegative and nondecreasing function on [t0 , ∞) and x(t) ≤ z(t),

t ∈ [t0 , ∞).

Also, if t ∈ [t0 , ∞), τ(t) > t0 , then x(τ(t)) ≤ z(τ(t)) ≤ z(t).

If t ∈ [t0 , ∞) and τ(t) ≤ t0 , then x(τ(t)) = ϕ(τ(t)) 1

≤ (a(t)) p

≤ z(t).

10.2 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay | 267

Therefore x(τ(t)) ≤ z(t),

t ∈ [t0 , ∞).

From here, t

p

t

p

q

r

(z(t)) = a(t) + ∫ b(s)(x(s)) Δs + c(t) ∫(f (s)(x(τ(s))) + g(s)(x(s)) )Δs t0

t0

t

t

p

q

r

≤ a(t) + ∫ b(s)(z(s)) Δs + c(t) ∫(f (s)(z(s)) + g(s)(z(s)) )Δs, t0

t0

t ∈ [t0 , ∞). Define t

q

r

u(t) = ∫(f (s)(z(s)) + g(s)(z(s)) )Δs, t0

w(t) = a(t) + c(t)u(t),

t ∈ [t0 , ∞).

Then t

p

p

(z(t)) ≤ w(t) + ∫ b(s(z(s)) Δs, t0

t ∈ [t0 , ∞).

Hence, by Theorem 1.1.10, we arrive at t

p

(z(t)) ≤ w(t) + ∫ eb (t, σ(s))w(s)b(s)Δs, t0

t ∈ [t0 , ∞).

Note that w is a nondecreasing function on [t0 , ∞). Then t

p

(z(t)) ≤ w(t) + w(t) ∫ eb (t, σ(s))b(s)Δs t0

t

Δ

= w(t)(1 − ∫ ebs (t, s)Δs) t0

= w(t)(1 − eb (t, t) + eb (t, t0 ))

= w(t)eb (t, t0 ),

t ∈ [t0 , ∞).

Therefore p

(z(t)) ≤ (a(t) + c(t)u(t))eb (t, t0 ),

t ∈ [t0 , ∞).

(10.12)

268 | 10 Delay integral inequalities Taking k > 0 arbitrarily and applying Lemma 10.2.1, we get q

q

q

(z(t)) = (eb (t, t0 )) p (a(t) + c(t)u(t)) p

q p − q pq q q−p ≤ (eb (t, t0 )) p ( k p (a(t) + c(t)u(t)) + k ) p p q qa(t) + qc(t)u(t) (p − q)k = (eb (t, t0 )) p ( + ) p−q p−q pk p pk p q γ(p − q) + qa(t) qc(t)u(t) = (eb (t, t0 )) p ( + ), p−q p−q pk p pk p

t ∈ [t0 , ∞), and r

r

r

r

k(p − r) + ra(t)

(z(t)) ≤ (eb (t, t0 )) p (a(t) + c(t)u(t)) p ≤ (eb (t, t0 )) p (

pk

rc(t)u(t)

+

p−q p

pk

p−q p

),

t ∈ [t0 , ∞). By the last two inequalities and the definition of function u, we find t

t

q

r

u(t) = ∫ f (s)(z(s)) Δs + ∫ g(s)(z(s)) Δs t0

t0

t

q

≤ ∫ f (s)(eb (s, t0 )) p (

k(p − q) + qa(s) pk

t0

t

r

+ ∫ g(s)(eb (s, t0 )) p ( q

= ∫(f (s)(eb (s, t0 )) p

pk

p−q p

k(p − q) + qa(s) pk

t0

t

q

+ ∫(f (s)(eb (s, t0 )) p t0

+

k(p − r) + ra(s)

t0

t

p−q p

qc(s) pk

p−q p

p−q p

qc(s)u(s) pk +

p−q p

)Δs

rc(s)u(s) pk

p−q p

)Δs r

+ g(s)(eb (s, t0 )) p r

+ g(s)(eb (s, t0 )) p

rc(s) pk

p−q p

r(p − r) + ra(s) pk

)u(s)Δs

t

= F(t) + ∫ G(s)u(s)Δs,

t ∈ [t0 , ∞).

t0

Applying Theorem 1.1.10, we find t

u(t) ≤ F(t) + ∫ eG (t, σ(s))F(s)G(s)Δs, t0

t ∈ [t0 , ∞),

p−q p

)Δs

10.2 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay | 269

and then w(t) = a(t) + c(t)u(t)

t

≤ a(t) + c(t)(F(t) + ∫ eG (t, σ(s))F(s)G(s)Δs), t0

t ∈ [t0 , ∞),

and, using (10.12), we obtain p

p

(x(t)) ≤ (z(t))

≤ w(t)eb (t, t0 )

t

≤ eb (t, t0 )(a(t) + c(t)(F(t) + ∫ eG (t, σ(s))F(s)G(s)Δs)) t0

t

≤ eb (t, t0 )(a(t) + c(t)F(t)(1 + ∫ eG (t, σ(s))G(s)Δs)) t0

t

Δ

= eb (t, t0 )(a(t) + c(t)F(t)(1 − ∫ eGs (t, s)Δs)) t0

= eb (t, t0 )(a(t) + c(t)F(t)(1 − eG (t, t) + eG (t, t0 )))

= eb (t, t0 )(a(t) + c(t)F(t)eG (t, t0 )),

t ∈ [t0 , ∞).

This completes the proof. Now, we consider a special case of the inequality (10.10) t

p

t

p

p−1

(x(t)) ≤ c + ∫ b(s)(x(s)) Δs + ∫ f (s)(x(τ(s)))

Δs,

t0

t0

t ∈ [t0 , ∞),

(10.13)

with initial data x(t) = ϕ(t), 1 p

ϕ(τ(t)) ≤ c ,

t ∈ [α, t0 ], t ∈ [t0 , ∞),

τ(t) ≤ t0 ,

(10.14)

where (A9) c is a positive constant, p > 0, τ, α and ϕ are as in (A7). (A10) x, b, f ∈ 𝒞rd ([t0 , ∞)) be nonnegative functions. Theorem 10.2.5. Suppose (A9), (A10) hold. If x satisfies (10.13) with initial data (10.14), then t

1

x(t) ≤ c p e b (t, t0 ) + p

1 ∫ e b (t, σ(s))f (s)Δs, p p t0

t ∈ [t0 , ∞).

270 | 10 Delay integral inequalities Proof. Define the function t

p

t

p

p−1

(z(t)) = c + ∫ b(s)(x(s)) Δs + ∫ f (s)(x(τ(s)))

Δs,

t0

t0

t ∈ [t0 , ∞).

As in the proof of Theorem 10.2.4, we have x(t) ≤ z(t),

x(τ(t)) ≤ z(t),

t ∈ [t0 , ∞).

By Pötzsche’s chain rule, for any t ∈ [t0 , ∞), there exists a θ ∈ [t, σ(t)] such that p−1 Δ

p(z(θ))

p

p−1

z (t) = b(t)(x(t)) + f (t)(x(τ(t))) p

p−1

≤ b(t)(z(t)) + f (t)(z(t))

t ∈ [t0 , ∞).

,

Since z is a nondecreasing function and θ ∈ [t, σ(t)] for any t ∈ [t0 , ∞), we have p(z(t))

p−1 Δ

p

p−1

z (t) ≤ b(t)(z(t)) + f (t)(z(t))

t ∈ [t0 , ∞),

,

whereupon z Δ (t) ≤

b(t) f (t) z(t) + , p p

t ∈ [t0 , ∞).

Hence, using the Gronwall inequality, we get x(t) ≤ z(t)

t

1 p

≤ c e b (t, t0 ) + ∫ e b (t, σ(s)) p

t0

p

f (s) Δs, p

t ∈ [t0 , ∞).

This completes the proof. Next, consider the delay integral inequality of the form p

t

q

(x(t)) ≤ a(t) + c(t) ∫(f (s)(x(s)) + L(s, x(τ(s))))Δs, t0

t ∈ [t0 , ∞),

(10.15)

subject to the initial condition (10.11), where q, τ, α satisfy (A7), x, a, c, f satisfy (A8) and (A11) L ∈ 𝒞 ([t0 , ∞) × ℝ) is a nonnegative function such that 0 ≤ L(t, x) − L(t, y) ≤ K(t, y)(x − y), t ∈ [t0 , ∞), x ≥ y ≥ 0, K ∈ 𝒞 ([t0 , ∞) × ℝ).

10.2 Nonlinear Gronwall–Bellman-type delay integral inequalities with one delay | 271

Theorem 10.2.6. If x is a solution of inequality (10.15) with the initial data (10.11), then t

x(t) ≤ (a(t) + c(t)(H(t) + ∫ eJ (t, σ(s))H(s)J(s)Δs)) t0

≤ a(t) + c(t)H(t)eJ (t, t0 ),

1 p

t ∈ [t0 , ∞),

where t

H(t) = ∫(

f (s)(k(p − q) + qa(s)) pk

t0

J(t) =

qc(t)f (t) pk

p−q p

p−q p

+ K(t,

+ L(s,

p − 1 a(s) + ))Δs, p p

p − 1 a(t) c(t) + ) , p p p

t ∈ [t0 , ∞).

Proof. Define the function t

q

z(t) = ∫(f (s)(x(s)) + L(s, x(τ(s))))Δs, t0

t ∈ [t0 , ∞).

We have that z is a nonnegative and nondecreasing function on [t0 , ∞) and then, using Lemma 10.2.1 for k = 1, obtain 1

x(t) ≤ (a(t) + c(t)z(t)) p p − 1 a(t) c(t)z(t) ≤ + + , p p p

t ∈ [t0 , ∞).

If t ∈ [t0 , ∞), τ(t) > t0 , then p − 1 a(τ(t)) c(τ(t))z(τ(t)) + + p p p p − 1 a(t) c(t)z(t) + + . ≤ p p p

x(τ(t)) ≤

If t ∈ [t0 , ∞) and τ(t) ≤ t0 , then x(τ(t)) = ϕ(τ(t)) 1

≤ (a(t)) p

≤ x(t) p − 1 a(t) c(t)z(t) ≤ + + . p p p Thus, x(τ(t)) ≤

p − 1 a(t) c(t)z(t) + + , p p p

t ∈ [t0 , ∞).

272 | 10 Delay integral inequalities Therefore, applying Lemma 10.2.1 for k > 0, we have t

q

z(t) = ∫(f (s)(x(s)) + L(s, x(τ(s))))Δs t0

t

q

≤ ∫ f (s)(a(s) + c(s)z(s)) p Δs t0

t

+ ∫ L(s, t0

t

≤ ∫ f (s)(

p − 1 a(s) c(s)z(s) + + )Δs p p p

k(p − q) + qa(s) pk

t0

t

+ ∫(L(s, t0

+ L(s, t

≤ ∫(

p−q p

+

pk

t0

p−q p

)Δs

p − 1 a(s) + ))Δs p p

t0

+ ∫(

pk

p − 1 a(s) c(s)z(s) p − 1 a(s) + + ) − L(s, + ) p p p p p

f (s)(k(p − q) + qa(s)) t

qc(s)z(s)

p−q p

qc(s)f (s) pk

p−q p

+ K(s,

+ L(s,

p − 1 a(s) + ))Δs p p

p − 1 a(s) c(s) + ) )z(s)Δs p p p

t

= H(t) + ∫ J(s)z(s)Δs,

t ∈ [t0 , ∞).

t0

Hence, by Theorem 1.1.10, we get t

z(t) ≤ H(t) + ∫ eJ (t, σ(s))H(s)J(s)Δs t0

t

≤ H(t)(1 + ∫ eJ (t, σ(s))J(s)Δs) t0

t

Δ

= H(t)(1 − ∫ eJ s (t, s)Δs) t0

= H(t)(1 − eJ (t, t) + eJ (t, t0 )) = H(t)eJ (t, t0 ),

t ∈ [t0 , ∞),

10.3 Gronwall–Bellman-type nonlinear delay integral inequalities with several delays | 273

and 1

x(t) ≤ (a(t) + c(t)z(t)) p t

≤ (a(t) + c(t)(H(t) + ∫ eJ (t, σ(s))H(s)J(s)Δs))

1 p

t0

1

t ∈ [t0 , ∞).

≤ (a(t) + c(t)H(t)eJ (t, t0 )) p , This completes the proof.

10.3 Gronwall–Bellman-type nonlinear delay integral inequalities with several delays Consider the inequality p

t

q

q

(x(t)) ≤ c + ∫(f (s)(x(τ1 (s))) + g(s)(x(τ2 (s))) w(x(τ2 (s))))Δs, t0

t ∈ [t0 , ∞), (10.16)

with the initial conditions x(t) = ϕ(t),

1 p

t ∈ [α, t0 ],

ϕ(τi (t)) ≤ (a(t)) ,

t ∈ [t0 , ∞),

τi (t) ≤ t0 ,

i ∈ {1, 2},

(10.17)

where (A12) a, x, f , g ∈ 𝒞rd ([t0 , ∞)) are nonnegative and nondecreasing functions, w ∈ 𝒞 ([0, ∞)) is a nonnegative function, (A13) τi ∈ 𝒞 ([t0 , ∞)), τi (t) ≤ t, t ∈ [t0 , ∞), i ∈ {1, 2}, −∞ < α = inf{ min τi (t) : t ∈ [t0 , ∞)}, i∈{1,2}

ϕ ∈ 𝒞rd ([α, t0 ]), p, q, c are positive constants, p > q. Theorem 10.3.1. Suppose (A12), (A13) hold. If x satisfies (10.16) with the initial condition (10.17), then t

t

1 p

x(t) ≤ (G (H (H(G(c) + ∫ f (s)Δs) + ∫ g(s)Δs))) , −1

−1

t0

t0

where G and H are increasing, bijective functions and v

G(v) = ∫ 1

1

q

rp

dr,

v > 0,

t ∈ [t0 , ∞),

274 | 10 Delay integral inequalities z

H(z) = ∫ 1

1

1

w((G−1 (r)) p )

dr,

z > 0,

H(∞) = ∞.

Proof. Let t

q

q

z(t) = c + ∫(f (s)(x(τ1 (s))) + g(s)(x(τ2 (s))) w(x(τ2 (s))))Δs, t0

t ∈ [t0 , ∞).

Then 1

x(t) ≤ (z(t)) p ,

t ∈ [t0 , ∞),

(10.18)

and, as in the proof of Theorem 10.2.6, we have 1

x(τi (t)) ≤ (z(t)) p ,

t ∈ [t0 , ∞),

i ∈ {1, 2}.

Next, q

q

z Δ (t) = f (t)(x(τ1 (t))) + g(t)(x(τ2 (t))) w(x(τ2 (t))) q

q

1

≤ f (t)(z(t)) p + g(t)(z(t)) p w((z(t)) p ), On the other hand, if t ∈ [t0 , ∞) and σ(t) > t, then Δ

(G(z(t))) = =

G(z(σ(t))) − G(z(t)) σ(t) − t 1 σ(t) − t

z(σ(t))

∫ z(t)

1

q

rp

dr

z(σ(t)) − z(t) 1 ≤ q σ(t) − t (z(t)) p =

z Δ (t)

.

q

(z(t)) p

If t ∈ [t0 , ∞), σ(t) = t, then Δ

(G(z(t))) = lim s→t

= lim s→t

= lim s→t

=

G(z(t)) − G(z(s)) t−s z(t)

1 1 ∫ q dr t−s rp z(s)

z(t) − z(s) 1 q t − s (z(ξ )) p

z Δ (t)

q

(z(t)) p

.

t ∈ [t0 , ∞).

10.3 Gronwall–Bellman-type nonlinear delay integral inequalities with several delays | 275

Thus, z Δ (t)

Δ

(G(z(t))) ≤

q

(z(t)) p

1

t ∈ [t0 , ∞).

≤ f (t) + g(t)w((z(t)) p ), Hence, t

1

G(z(t)) − G(z(t0 )) ≤ ∫(f (s) + g(s)w((z(s)) p ))Δs,

t ∈ [t0 , ∞),

t0

or t

1

G(z(t)) ≤ G(c) + ∫(f (s) + g(s)w((z(s)) p ))Δs, t0

t ∈ [t0 , ∞).

Therefore t

1

t ∈ [t0 , ∞).

z(t) ≤ G (G(c) + ∫(f (s) + g(s)w((z(s)) p ))Δs), −1

t0

Let t ∗ ∈ [t0 , ∞) be arbitrarily chosen. Set t∗

t

1

v(t) = G(c) + ∫ f (s)Δs + ∫ g(s)w((z(s)) p )Δs, t0

t0

t ∈ [t0 , t ∗ ].

Then t

q

z(t) ≤ G−1 (G(c) + ∫(f (s) + g(s)w((z(s)) p ))Δs) t0

t∗

t

1

≤ G (G(c) + ∫ f (s)Δs + ∫ g(s)w((z(s)) p )Δs) −1

t0

= G (v(t)), −1

t0

t ∈ [t0 , t ]. ∗

Next, 1

vΔ (t) = g(t)w((z(t)) p )

1

≤ g(t)w((G−1 (v(t))) p ),

t ∈ [t0 , t ∗ ].

(10.19)

276 | 10 Delay integral inequalities Thus, vΔ (t)

1

w((G−1 (v(t))) p )

≤ g(t),

t ∈ [t0 , t ∗ ].

If t ∈ [t0 , t ∗ ], σ(t) > t, then Δ

(H(v(t))) = =

H(v(σ(t))) − H(v(t)) σ(t) − t 1 σ(t) − t

v(σ(t))

∫

1

1

w((G−1 (v(r))) p )

v(t)

dr

v(σ(t)) − v(t) 1 = σ(t) − t w((G−1 (v(t))) p1 ) =

vΔ (t)

1

w((G−1 (v(t))) p )

.

If t ∈ [t0 , t ∗ ], σ(t) = t, then Δ

(H(v(t))) = lim s→t

H(v(t)) − H(v(s)) t−s v(t)

1 1 = lim dr ∫ 1 s→t t − s −1 p) w((G (v(r))) v(s) = lim s→t

=

1 v(t) − v(s) t − s w((G−1 (v(ξ ))) p1 ) vΔ (t)

1

w((G−1 (v(t))) p )

.

Therefore vΔ (t)

Δ

(H(v(t))) ≤

1

w((G−1 (v(t))) p ) ≤ g(t), t ∈ [t0 , t ∗ ],

and t

H(v(t)) − H(v(t0 )) ≤ ∫ g(s)Δs, t0

t ∈ [t0 , t ∗ ],

or t∗

t

H(v(t)) ≤ H(G(c) + ∫ f (s)Δs) + ∫ g(s)Δs, t0

t0

t ∈ [t0 , t ∗ ],

10.4 Delay integral inequalities in two independent variables | 277

or t∗

t

v(t) ≤ H (H(G(c) + ∫ f (s)Δs) + ∫ g(s)Δs), −1

t0

t0

t ∈ [t0 , t ∗ ].

By the last inequality and (10.18), (10.19), we get 1

x(t) ≤ (z(t)) p

1

≤ (G−1 (v(t))) p t∗

1 p

t

≤ (G (H (H(G(c) + ∫ f (s)Δs) + ∫ g(s)Δs))) , −1

−1

t0

t0

t ∈ [t0 , t ]. In particular, ∗

t∗

1 p

t∗

x(t ) ≤ (G (H (H(G(c) + ∫ f (s)Δs) + ∫ g(s)Δs))) . ∗

−1

−1

t0

t0

Because t ∗ ∈ [t0 , ∞) was arbitrarily chosen, we conclude that t

1 p

t

x(t) ≤ (G−1 (H −1 (H(G(c) + ∫ f (s)Δs) + ∫ g(s)Δs))) , t0

t0

t ∈ [t0 , ∞).

This completes the proof.

10.4 Delay integral inequalities in two independent variables ̃ = [y0 , ∞). Consider the integral inequality Let x0 , y0 ∈ 𝕋, 𝕋̂ = [x0 , ∞), 𝕋 n

x y

q

η(u(x, y)) ≤ a(x, y) + b(x, y) ∑ ∫ ∫(f (s, t)(u(τ1j (s), τ2j (t))) w(u(τ1j (s), τ2j (t))) j=1 x y 0 0

q

(10.20)

+ gj (s, t)(u(τ1j (s), τ2j (t))) s t

q

+ ∫ ∫ hj (ξ , ζ )(u(τ1j (ξ ), τ2j (ζ ))) Δζ Δξ )ΔtΔs, x0 y0

̃ subject to the initial condition (x, y) ∈ 𝕋̂ × 𝕋, η(u(x, y)) = ϕ(x, y),

ϕ(τ1j (x), τ2j (y)) ≤ a(x, y)

(x, y) ∈ [α, x0 ] × [β, y0 ],

̃ j ∈ {1, . . . , n}, where if τ1j (x) ≤ x0 or τ2j (y) ≤ y0 , (x, y) ∈ 𝕋̂ × 𝕋,

(10.21)

278 | 10 Delay integral inequalities ̃ are nonnegative functions, j ∈ {1, . . . , n}, a and b (A14) u, a, b, fj , gj , hj ∈ 𝒞rd (𝕋̂ × 𝕋) are nondecreasing, η ∈ 𝒞 ([0, ∞)) is a nonnegative strictly increasing function, q > 0, (A15) ̂ −∞ < α = inf{ min τ1j (x) : x ∈ 𝕋}, j∈{1,...,n}

̃ −∞ < β = inf{ min τ2j (y) : y ∈ 𝕋}, j∈{1,...,n}

τ1j ∈ 𝒞rd ([α, x0 ]), τ2j ∈ 𝒞rd ([β, y0 ]), j ∈ {1, . . . , n}, τ1j (x) ≤ x, x ∈ 𝕋,̂ τ2j (y) ≤ y, ̃ j ∈ {1, . . . , n}. y ∈ 𝕋, Theorem 10.4.1. Suppose (A14) and (A15) hold. If u satisfies the integral inequality (10.20) with the initial condition (10.21), then u(x, y) ≤ η−1 (G−1 (H −1 (H(G(a(x, y)) x y

n

s t

+ b(x, y) ∑ ∫ ∫(gj (s, t) + ∫ ∫ hj (ξ , ζ )Δζ Δξ )ΔtΔs) j=1 x y 0 0

x0 y0

x y

n

+ b(x, y) ∑ ∫ ∫ fj (s, t)ΔtΔs))),

(10.22)

̃ (x, y) ∈ 𝕋̂ × 𝕋,

j=1 x y 0 0

where G and H are increasing, bijective functions and Δ

(G(v(x, y)))x = Δ

(H(z(x, y)))x =

vxΔ (x, y) , (η−1 (v(x, y)))q

zxΔ (x, y) , −1 w(η (G1 (z(x, y))))

̃ (x, y) ∈ 𝕋̂ × 𝕋.

̃ be arbitrarily chosen. Define Proof. Let X ∈ 𝕋̂ and Y ∈ 𝕋 n

x y

q

v(x, y) = a(X, Y) + b(X, Y) ∑ ∫ ∫(fj (s, t)(u(τ1j (s), τ2j (t))) w(u(τ1j (s), τ2j (t))) j=1 x y 0 0

q

+ gj (s, t)(u(τ1j (s), τ2j (t))) s t

q

+ ∫ ∫ hj (ξ , ζ )(u(τ1j (ξ ), τ2j (ζ ))) Δζ Δξ )ΔtΔs, x0 y0

(x, y) ∈ [x0 , X] × [y0 , Y]. Then n

x y

q

u(x, y) ≤ η (a(x, y) + b(x, y) ∑ ∫ ∫(fj (s, t)(u(τ1j (s), τ2j (t))) w(u(τ1j (s), τ2j (t))) −1

j=1 x0 y

0

10.4 Delay integral inequalities in two independent variables | 279

q

+ gj (s, t)(u(τ1j (s), τ2j (t))) s t

q

+ ∫ ∫ hj (ξ , ζ )(u(τ1j (ξ ), τ2j (ζ ))) Δζ Δξ )ΔtΔs) x0 y0

n

x y

q

≤ η−1 (a(X, Y) + b(X, Y) ∑ ∫ ∫(fj (s, t)(u(τ1j (s), τ2j (t))) w(u(τ1j (s), τ2j (t))) j=1 x y 0 0 q

+ gj (s, t)(u(τ1j (s), τ2j (t))) s t

q

+ ∫ ∫ hj (ξ , ζ )(u(τ1j (ξ ), τ2j (ζ ))) Δζ Δξ )ΔtΔs) x0 y0

= η (v(x, y)), −1

(x, y) ∈ [x0 , X] × [y0 , Y].

As in the proof of Theorem 10.2.6, we have that u(τ1j (x), τ2j (y)) ≤ η−1 (v(x, y)),

x ∈ [x0 , X], y ∈ [y0 , Y].

Next, vxΔ (x, y)

y

n

q

= b(X, Y) ∑ ∫(fj (x, t)(u(τ1j (x), τ2j (t))) w(u(τ1j (x), τ2j (t))) j=1 y

0

q

+ gj (x, t)(u(τ1j (x), τ2j (t))) x t

q

+ ∫ ∫ hj (ξ , ζ )(u(τ1j (ξ ), τ2j (ζ ))) Δζ Δξ )Δt x0 y0

y

n

≤ (b(X, Y) ∑ ∫(fj (x, t)w(η−1 (v(x, t))) + gj (x, t) j=1 y

0

x t

q

+ ∫ ∫ hj (ξ , ζ )Δζ Δξ )Δt)(η−1 (v(x, y))) , x0 y0

(x, y) ∈ [x0 , X] × [y0 , Y], which implies that y

n vxΔ (x, y) ≤ b(X, Y) ∑ ∫(fj (x, t)w(η−1 (v(x, t))) + gj (x, t) −1 q (η (v(x, y))) j=1 y0

x t

+ ∫ ∫ hj (ξ , ζ )Δζ Δξ )Δt, x0 y0

(x, y) ∈ [x0 , X] × [y0 , Y]. As in the proof of Theorem 10.3.1, we get Δ

(G(v(x, y)))x =

vxΔ (x, y) , (η−1 (v(x, y)))q

(x, y) ∈ [x0 , X] × [y0 , Y].

280 | 10 Delay integral inequalities Thus, y

n

Δ

(G(v(x, y)))x ≤ (b(X, Y) ∑ ∫(fj (x, t)w(η−1 (v(x, t))) + gj (x, t) j=1 y

0

x t

+ ∫ ∫ hj (ξ , ζ )Δζ Δξ )Δt), x0 y0

(x, y) ∈ [x0 , X] × [y0 , Y],

whereupon n

x y

G(v(x, y)) − G(v(x0 , y)) ≤ b(X, Y) ∑ ∫ ∫(fj (s, t)w(η−1 (v(s, t))) + gj (s, t) j=1 x y 0 0

s t

+ ∫ ∫ hj (ξ , ζ )Δζ Δξ )ΔtΔs, x0 y0

(x, y) ∈ [x0 , X] × [y0 , Y],

or n

x y

G(v(x, y)) ≤ G(a(X, Y)) + b(X, Y) ∑ ∫ ∫(fj (s, t)w(η−1 (v(s, t))) + gj (s, t) j=1 x y 0 0

s t

+ ∫ ∫ hj (ξ , ζ )Δζ Δξ )ΔtΔs,

(x, y) ∈ [x0 , X] × [y0 , Y],

x0 y0

or n

x y

v(x, y) ≤ G (G(a(X, Y)) + (b(X, Y) ∑ ∫ ∫(fj (s, t)w(η−1 (v(s, t))) + gj (s, t) −1

j=1 x y 0 0

s t

+ ∫ ∫ hj (ξ , ζ )Δζ Δξ )ΔtΔs)), x0 y0

(x, y) ∈ [x0 , X] × [y0 , Y].

Let z(x, y) = G(a(X, Y)) n

x y

+ b(X, Y) ∑ ∫ ∫ fj (s, t)w(η−1 (v(s, t)))ΔtΔs j=1 x0 y

0

n X Y

+ b(X, Y) ∑ ∫ ∫(gj (s, t) j=1 x y 0 0

n

s t

+ ∑ ∫ ∫ hj (ξ , ζ )Δζ Δξ )ΔtΔs, j=1 x0 y

0

(x, y) ∈ [x0 , X] × [y0 , Y].

10.4 Delay integral inequalities in two independent variables | 281

We have v(x, y) ≤ G−1 (z(x, y)),

(x, y) ∈ [x0 , X] × [y0 , Y],

and furthermore, y

n

z Δ (x, y) ≤ b(X, Y) ∑ ∫ fj (x, t)w(η−1 (v(x, t)))Δt j=1 y

0

y

n

≤ (b(X, Y) ∑ ∫ fj (x, t)Δt)w(η−1 (G−1 (z(x, y)))), j=1 y

0

(x, y) ∈ [x0 , X] × [y0 , Y], which is followed by y

n z Δ (x, y) ≤ b(X, Y) ∑ ∫ fj (x, t)Δt, w(η−1 (G−1 (z(x, y)))) j=1 y0

(x, y) ∈ [x0 , X] × [y0 , Y]. As in the proof of Theorem 10.3.1, we have Δ

(H(z(x, y)))x =

zxΔ (x, y) w(η−1 (G−1 (z(x, y)))) y

n

≤ b(X, Y) ∑ ∫ fj (x, t)Δt, j=1 y

0

(x, y) ∈ [x0 , X] × [y0 , Y]. Hence, n

x y

H(z(x, y)) − H(z(x0 , y)) ≤ b(X, Y) ∑ ∫ ∫ fj (x, t)ΔtΔs, j=1 x y 0 0

(x, y) ∈ [x0 , X] × [y0 , Y], whereupon n

x y

z(x, y) ≤ H (H(z(x0 , y)) + b(X, Y) ∑ ∫ ∫ fj (x, t)ΔtΔs), −1

j=1 x y 0 0

(x, y) ∈ [x0 , X] × [y0 , Y]. Then u(x, y) ≤ η−1 (v(x, y))

≤ η−1 (G−1 (z(x, y))) ≤ η−1 (G−1 (H −1 (H(z(x0 , y)) n

x y

+ b(X, Y) ∑ ∫ ∫ fj (x, t)ΔtΔs))), j=1 x0 y

0

282 | 10 Delay integral inequalities (x, y) ∈ [x0 , X] × [y0 , Y]. In particular, u(X, Y) ≤ η−1 (G−1 (H −1 (H(z(x0 , Y)) n X Y

+ b(X, Y) ∑ ∫ ∫ fj (x, t)ΔtΔs))). j=1 x y 0 0

̂ 𝕋 ̃ was arbitrarily chosen, from the last inequality, we get inequalBecause (X, Y) ∈ 𝕋× ity (10.22). This completes the proof.

10.5 Delay Volterra–Fredholm-type integral inequalities Let T ∈ [t0 , ∞), I = [t0 , T]. In this section we will investigate the inequality α(t)

l

s

q1

r1

θ1

(u(t)) ≤ a(t) + b(t) ∫ f1 (s)((u(s)) + f2 (s) ∫ g1 (τ)(u(τ)) Δτ) Δs α(t0 ) β(T)

α(t0 ) q2

s

r2

θ2

+ ∫ f3 (s)((u(s)) + f4 (s) ∫ q2 (τ)(u(τ)) Δτ) Δs

(10.23)

β(t0 )

β(t0 ) γ(T)

q3

s

r3

θ3

+ ∫ f5 (s)((u(s)) + f6 (s) ∫ q3 (τ)(u(τ)) Δτ) Δs, γ(t0 )

t ∈ I,

γ(t0 )

where (A16) α : I → 𝕋 is continuous and strictly increasing such that α(t) ≤ t, t ∈ I, αΔ ∈ 𝒞rd (I), (A17) β, γ ∈ 𝒞 (I), β(t) ≤ t, γ(t) ≤ t, t ∈ I, (A18) u, a, b ∈ 𝒞rd (I) are nonnegative functions, a is nondecreasing and bΔ ≥ 0 on I, (A19) fi , gi ∈ 𝒞rd (I), i ∈ {1, . . . , 6}, (A20) 0 ≤ qi ≤ l, 0 ≤ r0 ≤ l, l ≠ 0, 0 ≤ θi ≤ 1, i ∈ {1, 2, 3}. Define β(T)

s

β(t0 )

β(t0 )

̃ = ∫ (θ2 k θ2 −1 f3 (s)( l − q2 k q2 + f4 (s) ∫ g2 (τ)( l − r2 k r2 )Δτ) K 5 4 6 l l θ

+ (1 − θ2 )k42 f3 (s))Δs

10.5 Delay Volterra–Fredholm-type integral inequalities | 283 γ(T)

+ ∫ γ(t0 )

l θ −1 (θ3 k7 3 f5 (s)(

s

l − r3 r3 − q3 q3 k8 + f6 (s) ∫ g3 (τ)( k7 )Δτ) l l β(t0 )

θ

+ (1 − θ3 )k7 3 f5 (s))Δs, α(t)

̃ = a(t) + b(t) ∫ (θ1 k θ1 −1 f1 (s)( l − q1 k q1 V(t) 1 2 l s

α(t0 )

+ f2 (s) ∫ g1 (τ)( α(t0 )

l − r1 r1 θ k )Δτ) + (1 − θ1 )k1 1 f1 (s))Δs, l 1

α(σ(t))

s

α(t0 )

α(t0 )

θ −1 A(t) = bΔ (t) ∫ (θ1 k1 1 ̃f1 (s)(1 + f2 (s) ∫ g̃1 (τ)Δτ))Δs,

A(t) , B(t) = 1 − μ(t)A(t) C(t) = ̃f (t) = 1 ̃f (t) = 3 ̃f (t) = 5 g̃1 (t) = g̃2 (t) = g̃3 (t) =

b(t)(θ1 k1θ−1 ̃f1 (t)(1 q1 q1 −l k f (t), l 2 1 q2 q2 −l k f (t), l 3 3 q3 q3 −l k f (t), l 8 5 r1 l−q1 r1 −l k k3 g1 (t), l 2 r2 l−q2 r2 −l k k6 g2 (t), l 5 r3 l−q3 r3 −l k k9 g3 (t), l 8

t

+ f2 (t) ∫ g̃1 (τ)Δτ))αΔ (t), α(t0 )

t ∈ I.

Suppose that (A21) there exist positive constants ki , i ∈ {1, . . . , 9}, such that β(T)

θ −1

s

λ = ∫ (θ2 k42 f3 (s)(eB⊕C (s, t0 ) + f4 (s) ∫ g2 (τ)eB⊕C (τ, t0 )Δτ))Δs β(t0 )

γ(T)

+ ∫ < 1,

γ(t0 )

β(t0 )

θ −1 (θ3 k7 3 f5 (s)(eB⊕C (s, t0 )

(A22) μ(t)A(t) < 1, t ∈ I.

s

+ f6 (s) ∫ g3 (τ)eB⊕C (τ, t0 )Δτ))Δs β(t0 )

284 | 10 Delay integral inequalities Theorem 10.5.1. Assume (A16)–(A22) hold. If u satisfies inequality (10.23), then 1

l ̃ + V(T) ̃ K u(t) ≤ ( eB⊕C (t, t0 )) , 1−λ

t ∈ I.

Proof. Let α(t)

s

q

r

θ1

z(t) = a(t) + b(t) ∫ f1 (s)((u(s)) 1 + f2 (s) ∫ g1 (τ)(u(τ)) 1 Δτ) Δs α(t0 ) β(T)

α(t0 ) s

q2

r2

θ2

+ ∫ f3 (s)((u(s)) + f4 (s) ∫ q2 (τ)(u(τ)) Δτ) Δs β(t0 )

β(t0 )

γ(T)

s

q3

r3

θ3

+ ∫ f5 (s)((u(s)) + f6 (s) ∫ q3 (τ)(u(τ)) Δτ) Δs,

t ∈ I.

γ(t0 )

γ(t0 )

Then z is a nondecreasing function on I and l

(u(t)) ≤ z(t),

t ∈ I.

Furthermore, using Lemma 10.2.1, we get α(t)

θ −1

z(t) ≤ a(t) + b(t) ∫ (θ1 k1 1 f1 (s)( α(t0 )

+

q1 q1 −l l k (u(s)) l 2

s

l − q1 q1 r r −l l − r1 r1 l k2 + f2 (s) ∫ g1 (τ)( 1 k31 (u(τ)) + k )Δτ) l l l 3 α(t0 )

θ

+ (1 − θ1 )k1 1 f1 (s))Δs β(T)

θ −1

+ ∫ (θ2 k42 f3 (s)( β(t0 )

l − q2 q2 q2 q2 −l l k3 (u(s)) + k5 l l

s

l − r2 r2 r r −l l k6 )Δτ) + f4 (s) ∫ q2 (τ)( 2 k62 (u(τ)) + l l β(t0 )

θ

+ (1 − θ2 )k42 f3 (s))Δs γ(T)

θ −1

+ ∫ (θ3 k7 3 f5 (s)( γ(t0 )

q3 q3 −l l − q3 q3 l k (u(s)) + k8 l 8 l

10.5 Delay Volterra–Fredholm-type integral inequalities | 285 s

+ f6 (s) ∫ g3 (τ)( γ(t0 )

r3 r3 −l l − r3 r3 l k (u(τ)) + k9 )Δτ) l 9 l

θ

+ (1 − θ3 )k7 3 f5 (s))Δs ̃ + V(t) ̃ =K

α(t)

s

θ −1 + b(t) ∫ (θ1 k1 1 ̃f1 (s)(z(s) + f2 (s) ∫ g̃1 (τ)z(τ)Δτ))Δs α(t0 )

α(t0 )

β(T)

s

β(t0 )

β(t0 )

γ(T)

s

θ −1 + ∫ (θ2 k42 ̃f3 (s)(z(s) + f4 (s) ∫ g̃2 (τ)z(τ)Δτ))Δs

θ −1 + ∫ (θ3 k7 3 ̃f5 (s)(z(s) + f6 (s) ∫ g̃3 (τ)z(τ)Δτ))Δs,

t ∈ I.

γ(t0 )

γ(t0 )

̃ is nondecreasing, from the last inequality, we get Since V ̃ + V(T) ̃ z(t) ≤ K

α(t)

s

θ −1 + b(t) ∫ (θ1 k1 1 ̃f1 (s)(z(s) + f2 (s) ∫ g̃1 (τ)z(τ)Δτ))Δs α(t0 )

α(t0 )

β(T)

s

β(t0 )

β(t0 )

γ(T)

s

θ −1 + ∫ (θ2 k42 ̃f3 (s)(z(s) + f4 (s) ∫ g̃2 (τ)z(τ)Δτ))Δs

θ −1 + ∫ (θ3 k7 3 ̃f5 (s)(z(s) + f6 (s) ∫ g̃3 (τ)z(τ)Δτ))Δs, γ(t0 )

t ∈ I.

γ(t0 )

Let ̃ + V(T) ̃ M=K β(T)

+ ∫ β(t0 )

θ −1 (θ2 k42 ̃f3 (s)(z(s)

s

+ f4 (s) ∫ g̃2 (τ)z(τ)Δτ))Δs β(t0 )

γ(T)

s

γ(t0 )

γ(t0 )

θ −1 + ∫ (θ3 k7 3 ̃f5 (s)(z(s) + f6 (s) ∫ g̃3 (τ)z(τ)Δτ))Δs.

Therefore α(t)

s

θ −1 z(t) ≤ M + b(t) ∫ (θ1 k1 1 ̃f1 (s)(z(s) + f2 (s) ∫ g̃1 (τ)z(τ)Δτ))Δs, α(t0 )

α(t0 )

t ∈ I.

286 | 10 Delay integral inequalities Let α(t)

s

θ −1 w(t) = M + b(t) ∫ (θ1 k1 1 ̃f1 (s)(z(s) + f2 (s) ∫ g̃1 (τ)z(τ)Δτ))Δs α(t0 )

t ∈ I.

α(t0 )

Then w is nondecreasing on I and Δ

Δ

α(σ(t))

w (t) = b (t) ∫ α(t0 )

θ −1 (θ1 k1 1 ̃f1 (s)(z(s)

s

+ f2 (s) ∫ g̃1 (τ)z(τ)Δτ))Δs α(t0 ) t

θ −1 + b(t)(θ1 k1 1 ̃f1 (t)(z(t) + f2 (t) ∫ g̃1 (τ)z(τ)Δτ))αΔ (t) α(t0 )

α(σ(t))

s

θ −1 ≤ bΔ (t) ∫ (θ1 k1 1 ̃f1 (s)(w(s) + f2 (s) ∫ g̃1 (τ)w(τ)Δτ))Δs α(t0 )

α(t0 )

t

θ −1 + b(t)(θ1 k1 1 ̃f1 (t)(w(t) + f2 (t) ∫ g̃1 (τ)w(τ)Δτ))αΔ (t) α(t0 )

α(σ(t))

s

θ −1 ≤ w(σ(t))bΔ (t) ∫ (θ1 k1 1 ̃f1 (s)(1 + f2 (s) ∫ g̃1 (τ)Δτ))Δs α(t0 )

α(t0 )

t

θ −1 + w(t)b(t)(θ1 k1 1 ̃f1 (t)(1 + f2 (t) ∫ g̃1 (τ)Δτ))αΔ (t) α(t0 )

= A(t)w(σ(t)) + w(t)C(t) B(t) = w(σ(t)) + w(t)C(t) 1 + μ(t)B(t) B(t) (w(t) + μ(t)wΔ (t)) + w(t)C(t), = 1 + μ(t)B(t)

t ∈ I.

Hence,

or

wΔ (t) B(t) ≤( + C(t))w(t), 1 + μ(t)B(t) 1 + μ(t)B(t)

t ∈ I,

wΔ (t) ≤ (B(t) + C(t) + μ(t)B(t)C(t))w(t) = (B ⊕ C)(t)w(t),

t ∈ I.

Now we apply the Gronwall inequality and obtain w(t) ≤ w(t0 )eB⊕C (t, t0 ) = MeB⊕C (t, t0 ),

t ∈ I,

10.5 Delay Volterra–Fredholm-type integral inequalities | 287

and z(t) ≤ MeB⊕C (t, t0 ),

t ∈ I.

Now, by the definition of M, we get β(T)

s

β(t0 )

β(t0 )

θ −1 ̃ + V(T) ̃ M≤K + M( ∫ (θ2 k42 f3 (s)(eB⊕C (s, t0 ) + f4 (s) ∫ g2 (τ)eB⊕C (τ, t0 )Δτ)Δs γ(T)

s

θ −1

+ ∫ (θ3 k7 3 f5 (s)(eB⊕C (s, t0 ) + f6 (s) ∫ g3 (τ)eB⊕C (τ, t0 )Δτ)Δs))) γ(t0 )

β(t0 )

̃ + V(T) ̃ =K + Mλ, whereupon M≤

̃ + V(T) ̃ K . 1−λ

Consequently, l

(u(t)) ≤ z(t) ̃ + V(T) ̃ K ≤ eB⊕C (t, t0 ), 1−λ This completes the proof.

t ∈ I.

11 Applications This chapter deals with some applications of some linear integral inequalities and some linear integro-dynamic inequalities. They are investigated the existence and uniqueness of the solutions of first order dynamic equations, continuous dependence on initial conditions of the solutions of first order dynamic equations. They are given applications for some second order integro-dynamic equations and Volterra integral equations. THey are deducted some bounds of the solutions of delay dynamic equations. Some of the results in this chapter can be found in [6, 7, 11, 12, 20, 21] and [23]. Let 𝕋 be a time scale with forward jump operator and delta differentiation operator σ and Δ, respectively. Let also t0 , a ∈ 𝕋, a > t0 , J = [t0 , a].

11.1 Existence and uniqueness of the solution of first order dynamic equations Consider the following IVP: yΔ (t) = f (t, y(t)),

y(t0 ) = y0 ,

t ∈ J,

(11.1) (11.2)

where y0 is a given constant. Theorem 11.1.1. Let f : J × ℝ → ℝ, f ∈ 𝒞 (J × ℝ), 󵄨󵄨 󵄨 󵄨󵄨f (t, y(t))󵄨󵄨󵄨 ≤ M for any t ∈ J, y ∈ 𝒞 (J), 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨f (t, y1 (t)) − f (t, y2 (t))󵄨󵄨󵄨 ≤ L󵄨󵄨󵄨y1 (t) − y2 (t)󵄨󵄨󵄨, t ∈ J, y1 , y2 ∈ 𝒞 (J), where L and M are some positive constants. Then the problem (11.1), (11.2) has a unique solution y ∈ 𝒞 (J). Proof. Let y(t), t ∈ J, be a solution of the IVP (11.1), (11.2). We integrate equation (11.1) from t0 to t and, using (11.2), get t

y(t) = y0 + ∫ f (τ, y(τ))Δτ, t0

t ∈ J.

(11.3)

Let now y ∈ 𝒞 (J) be a solution of the integral equation (11.3). We put t = t0 in (11.3) and get y(t0 ) = y0 . Differentiating equation (11.3) with respect to t, we obtain that y(t), t ∈ J, satisfies equation (11.1). Therefore y(t), t ∈ J, is a solution of the problem (11.1), (11.2). Hence, https://doi.org/10.1515/9783110705553-011

290 | 11 Applications the IVP (11.1), (11.2) is equivalent to the integral equation (11.3). Define the sequence {yl (t)}l∈ℕ0 as follows: y0 (t) = y0 ,

t

yl (t) = y0 + ∫ f (τ, yl−1 (τ))Δτ, t0

l ∈ ℕ, t ∈ J.

We have t 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨y1 (t) − y0 (t)󵄨󵄨󵄨 = 󵄨󵄨󵄨y0 + ∫ f (τ, y0 (τ))Δτ − y0 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 󵄨 t0

󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨󵄨 = 󵄨󵄨󵄨∫ f (τ, y0 (τ))Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t0 󵄨 t

󵄨 󵄨 ≤ ∫󵄨󵄨󵄨f (τ, y0 (τ))󵄨󵄨󵄨Δτ t0

≤ M(t − t0 )

= Mh1 (t, t0 ),

t ∈ J.

Assume that 󵄨󵄨 󵄨 l−1 󵄨󵄨yl (t) − yl−1 (t)󵄨󵄨󵄨 ≤ ML hl (t, t0 ),

t ∈ J,

for some l ∈ ℕ. We will prove that 󵄨󵄨 󵄨 l 󵄨󵄨yl+1 (t) − yl (t)󵄨󵄨󵄨 ≤ ML hl+1 (t, t0 ),

t ∈ J.

In fact, we have t 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨yl+1 (t) − yl (t)󵄨󵄨 = 󵄨󵄨y0 + ∫ f (τ, yl (τ))Δτ 󵄨󵄨 󵄨 t0 t 󵄨󵄨 󵄨󵄨 − y0 − ∫ f (τ, yl−1 (τ))Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨 t0

󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨󵄨 = 󵄨󵄨󵄨∫(f (τ, yl (τ)) − f (τ, yl−1 (τ)))Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t0 󵄨 t

󵄨 󵄨 ≤ ∫󵄨󵄨󵄨f (τ, yl (τ)) − f (τ, yl−1 (τ))󵄨󵄨󵄨Δτ t0

(11.4)

11.1 Existence and uniqueness of the solution of first order dynamic equations | 291 t

󵄨 󵄨 ≤ L ∫󵄨󵄨󵄨yl (τ) − yl−1 (τ)󵄨󵄨󵄨Δτ t0

l

t

≤ ML ∫ hl (τ, t0 )Δτ l

t0

= ML hl+1 (t, t0 ),

t ∈ J.

Consequently, (11.4) holds for any l ∈ ℕ. Note that 󵄨󵄨 ∞ 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 lim 󵄨󵄨yl (t) − y0 (t)󵄨󵄨 = 󵄨󵄨∑(yr (t) − yr−1 (t))󵄨󵄨󵄨 󵄨󵄨r=1 󵄨󵄨 l→∞ 󵄨 󵄨 ∞

󵄨 󵄨 ≤ ∑󵄨󵄨󵄨yr (t) − yr−1 (t)󵄨󵄨󵄨 r=1

∞

≤ M ∑ Lr−1 hr (t, t0 ) r=1

M ≤ eL (t, t0 ), L

t ∈ J.

Therefore yl (t) converges uniformly on J to a solution y(t), t ∈ J, of the IVP (11.1), (11.2). Suppose that the problem (11.1), (11.2) has two solutions y(t) and z(t), t ∈ J. For z(t), t ∈ J, we have the following representation: t

z(t) = y0 + ∫ f (τ, z(τ))Δτ, t0

t ∈ J.

Then t t 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨y(t) − z(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨y0 + ∫ f (τ, y(τ))Δτ − y0 − ∫ f (τ, z(τ))Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 t0 t0

󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨󵄨 = 󵄨󵄨󵄨∫(f (τ, y(τ)) − f (τ, z(τ)))Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t0 󵄨 t

󵄨 󵄨 ≤ ∫󵄨󵄨󵄨f (τ, y(τ)) − f (τ, z(τ))󵄨󵄨󵄨Δτ t0

t

󵄨 󵄨 ≤ L ∫󵄨󵄨󵄨y(τ) − z(τ)󵄨󵄨󵄨Δτ, t0

t ∈ J.

From the last inequality and Theorem 1.1.10, we get 󵄨󵄨 󵄨 󵄨󵄨y(t) − z(t)󵄨󵄨󵄨 = 0,

i. e., y = z on J. This completes the proof.

t ∈ J,

292 | 11 Applications

11.2 Continuous dependence on initial conditions of the solutions of first order dynamic equations Let D ⊂ J × ℝ, t1 ∈ J, t1 ≥ t0 , (t0 , y0 ), (t1 , z1 ) ∈ D. Theorem 11.2.1. Let f ∈ 𝒞 (D), |f (t, y)| ≤ M, (t, y) ∈ D, 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨f (t, y1 (t)) − f (t, y2 (t))󵄨󵄨󵄨 ≤ L󵄨󵄨󵄨y1 (t) − y2 (t)󵄨󵄨󵄨,

t ∈ J,

y1 , y2 ∈ 𝒞 (J).

Let also g ∈ 𝒞 (D) and 󵄨󵄨 󵄨 󵄨󵄨g(t, z)󵄨󵄨󵄨 ≤ M1 ,

(t, z) ∈ D,

where M, M1 , and L are some positive constants. If y and z are solutions of the IVP (11.1), (11.2), and z Δ (t) = f (t, z(t)) + g(t, z(t)),

z(t1 ) = z1 ,

respectively, then 󵄨󵄨 󵄨 󵄨󵄨y(t) − z(t)󵄨󵄨󵄨 ≤ |y0 − z1 | + M(t1 − t0 ) + M1 |t − t1 | t

+ L ∫(|y0 − z1 | + M(t1 − t0 ) + M1 |s − t1 |)e⊖L (σ(s), t)Δs, t0

Proof. We have t

y(t) = y0 + ∫ f (τ, y(τ))Δτ, t0 t

t

z(t) = z1 + ∫ f (τ, z(τ))Δτ + ∫ g(τ, z(τ))Δτ t1

t1

t

t1

= z1 + ∫ f (τ, z(τ))Δτ − ∫ f (τ, z(τ))Δτ t0

t0

t

+ ∫ g(τ, z(τ))Δτ, t1

t t 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨y(t) − z(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨y0 + ∫ f (τ, y(τ))Δτ − z1 − ∫ f (τ, z(τ))Δτ 󵄨󵄨 󵄨 t0 t0 t1 t 󵄨󵄨 󵄨󵄨 + ∫ f (τ, z(τ))Δτ − ∫ g(τ, z(τ))Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨 t0 t1

t ∈ J.

11.3 Second order integro-dynamic equations |

293

t1 t 󵄨󵄨 󵄨󵄨 󵄨 = 󵄨󵄨y0 − z1 + ∫ f (τ, z(τ))Δτ − ∫ g(τ, z(τ))Δτ 󵄨󵄨 󵄨 t0 t1 t

󵄨󵄨 󵄨󵄨 + ∫(f (τ, y(τ)) − f (τ, z(τ)))Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨 t0 󵄨󵄨 t1 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨󵄨 t 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 ≤ |y0 − z1 | + 󵄨󵄨∫ f (τ, z(τ))Δτ󵄨󵄨󵄨 + 󵄨󵄨󵄨∫ g(τ, z(τ))Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨t1 󵄨t0 󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨󵄨 + 󵄨󵄨󵄨∫(f (τ, y(τ)) − f (τ, z(τ)))Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t0 󵄨 t1

󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 ≤ |y0 − z1 | + ∫󵄨󵄨f (τ, z(τ))󵄨󵄨Δτ + 󵄨󵄨󵄨∫󵄨󵄨󵄨g(τ, z(τ))󵄨󵄨󵄨Δτ󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t1 󵄨 t0 t

󵄨 󵄨 + ∫󵄨󵄨󵄨f (τ, y(τ)) − f (τ, z(τ))󵄨󵄨󵄨Δτ t0

≤ |y0 − z1 | + M(t1 − t0 ) + M1 |t − t1 | t

󵄨 󵄨 + L ∫󵄨󵄨󵄨y(τ) − z(τ)󵄨󵄨󵄨Δτ, t0

t ∈ J.

Hence, by Theorem 1.1.10, we get 󵄨󵄨 󵄨 󵄨󵄨y(t) − z(t)󵄨󵄨󵄨 ≤ |y0 − z1 | + M(t1 − t0 ) + M1 |t − t1 | t

+ L ∫(|y0 − z1 | + M(t1 − t0 ) + M1 |s − t1 |)e⊖L (σ(s), t)Δs, t0

t ∈ J.

This completes the proof.

11.3 Second order integro-dynamic equations Let f , g ∈ 𝒞 (J), k ∈ 𝒞 (J × J × ℝ), h ∈ 𝒞 (J × ℝ × ℝ). Suppose that f (t) ≠ 0,

t ∈ J,

f σ (a) ≠ 0.

(11.5)

Consider the following second order integro-dynamic equation: Δ

t

(f (t)xΔ ) + g(t)x = h(t, x, ∫ k(t, s, x(s))Δs), t0

t ∈ J,

294 | 11 Applications which we can rewrite in the form Δ2

Δ

t

Δσ

f (t)x (t) + f (t)x (t) + g(t)x(t) = h(t, x(t), ∫ k(t, s, x(s))Δs),

t ∈ J.

(11.6)

0

Let x1 (t) and x2 (t), t ∈ J, be two linearly independent solutions of the equation 2

f σ (t)xΔ (t) + f Δ (t)xΔ (t) + g(t)x(t) = 0, Then

t ∈ J.

(11.7)

2

f σ (t)x1Δ (t) + f Δ (t)x1Δ (t) + g(t)x1 (t) = 0, 2

f σ (t)x2Δ (t) + f Δ (t)x2Δ (t) + g(t)x2 (t) = 0,

t ∈ J.

(11.8)

We will search for a solution x(t) of equation (11.6) in the form x(t) = c1 (t)x1 (t) + c2 (t)x2 (t),

t ∈ J,

where c1 , c2 ∈ 𝒞 1 (J) will be determined below. We have xΔ (t) = c1Δ (t)x1σ (t) + c1 (t)x1Δ (t)

+ c2Δ (t)x2σ (t) + c2 (t)x2Δ (t),

We want Then and

c1Δ (t)x1σ (t) + c2Δ (t)x2σ (t) = 0,

t ∈ J. t ∈ J.

xΔ (t) = c1 (t)x1Δ (t) + c2 (t)x2Δ (t), 2

(11.9)

t ∈ J,

2

xΔ (t) = c1Δ (t)x1Δσ (t) + c1 (t)x1Δ (t) 2

+ c2Δ (t)x2Δσ (t) + c2 (t)x2Δ (t),

t ∈ J.

Hence, using (11.6) and (11.8), we obtain 2

f σ (t)xΔ (t) + f Δ (t)xΔ (t) + g(t)x(t)

2

= f σ (t)c1Δ (t)x1Δσ (t) + f σ (t)c1 (t)x1Δ (t) 2

+ f σ (t)c2Δ (t)x2Δσ (t) + f σ (t)c2 (t)x2Δ (t)

+ f Δ (t)c1 (t)x1Δ (t) + f Δ (t)c2 (t)x2Δ (t) + g(t)c1 (t)x1 (t) + g(t)c2 (t)x2 (t)

= f σ (t)(c1Δ (t)x1Δσ (t) + c2Δ (t)x2Δσ (t)) t

= h(t, x(t), ∫ k(t, s, x(s))Δs), t0

whereupon

t ∈ J,

11.3 Second order integro-dynamic equations |

c1Δ (t)x1Δσ (t)

+

c2Δ (t)x2Δσ (t)

295

t

1 = σ h(t, x(t), ∫ k(t, s, x(s))Δs), f (t) t0

t ∈ J.

From the last equation and (11.9), we get the following system: c1Δ (t)x1σ (t) + c2Δ (t)x2σ (t) = 0, { { { { t 1 { Δ Δσ Δ Δσ { h(t, x(t), k(t, s, x(s))Δs), c (t)x (t) + c (t)x (t) = ∫ { 1 1 2 2 { f σ (t) t0 {

t ∈ J.

Hence, t

x2σ (t) { Δ Δσ σ σ Δσ { { c (t)(x (t)x (t) − x (t)x (t)) = h(t, x(t), ∫ k(t, s, x(s))Δs), { 1 1 2 1 2 { f σ (t) { { { t0 { t { { { x1σ (t) { Δ Δσ σ σ Δσ { c (t)(−x (t)x (t) + x (t)x (t)) = h(t, x(t), k(t, s, x(s))Δs), ∫ { 2 1 2 1 2 { f σ (t) t0 {

t ∈ J,

or t

x2σ (t) { Δ { { c (t) = h(t, x(t), ∫ k(t, s, x(s))Δs), { 1 { { f σ (t)(x1Δσ (t)x2σ (t) − x1σ (t)x2Δσ (t)) { { t0 { t { { { x1σ (t) { Δ { { {c2 (t) = f σ (t)(−x Δσ (t)xσ (t) + xσ (t)xΔσ (t)) h(t, x(t), ∫ k(t, s, x(s))Δs), 1 1 2 2 t0 {

t ∈ J.

We take t

{ x2σ (y) { { { c1 (t) = ∫ σ { { f (y)(x1Δσ (y)x2σ (y) − x1σ (y)x2Δσ (y)) { { t0 { { { { y { { { { { { × h(y, x(y), k(y, s, x(s))Δs)Δy, ∫ { { { { { t0 { t { { { x1σ (y) { { {c2 (t) = ∫ σ { Δσ { f (y)(−x1 (y)x2σ (y) + x1σ (y)x2Δσ (y)) { { t0 { { { { y { { { { { { × h(y, x(y), k(y, s, x(s))Δs)Δy, t ∈ J. ∫ { { t0 { Suppose that f σ (t)(x1Δσ (t)x2σ (t) − x1σ (t)x2Δσ (t)) = 1,

t ∈ J.

(11.10)

296 | 11 Applications Then y

t

{ { { { c1 (t) = ∫ x2σ (y)h(y, x(y), ∫ k(y, s, x(s))Δs)Δy, { { { { t0 t0 y t { { { { σ { { c (t) = − ∫ x1 (y)h(y, x(y), ∫ k(y, s, x(s))Δs)Δy, { { 2 t0 t0 {

t ∈ J,

and x(t) = c1 x1 (t) + c2 x2 (t) + c1 (t)x1 (t) + c2 (t)x2 (t) = c1 x1 (t) + c2 x2 (t)

y σ ∫ x1 (t)x2 (y)h(y, x(y), ∫ k(y, s, x(s))Δs)Δy t

+

t0

t0

y σ ∫ x2 (t)x1 (y)h(y, x(y), ∫ k(y, s, x(s))Δs)Δy t

−

t0

= c1 x1 (t) + c2 x2 (t) t

+

∫(x1 (t)x2σ (y)

t0

−

t0

y σ x2 (t)x1 (y))h(y, x(y), ∫ k(y, s, x(s))Δs)Δy, t0

t ∈ J,

is a solution of equation (11.6). Below, using Pachpatte’s inequalities, we will deduct some of the properties of the solution of equation (11.6). Theorem 11.3.1. Let f , g ∈ 𝒞 (𝕋), k ∈ 𝒞 (𝕋 × 𝕋 × ℝ), h ∈ 𝒞 (𝕋 × ℝ × ℝ), f (t) ≠ 0, t ∈ 𝕋, 󵄨󵄨 󵄨󵄨 󵄨󵄨k(t, s, x)󵄨󵄨 ≤ p(s)|x|, (t, s, x) ∈ 𝕋 × 𝕋 × ℝ, 󵄨󵄨 󵄨 󵄨󵄨h(t, x, z)󵄨󵄨󵄨 ≤ q(t)(|x| + |z|), (t, x, z) ∈ 𝕋 × ℝ × ℝ, where p, q ∈ 𝒞 (𝕋) are nonnegative functions such that a

∫ p(s)Δs < ∞,

t0

a

∫ q(s)Δs < ∞.

t0

If x1 (t) and x2 (t) are bounded solutions of equation (11.7) so that (11.10) holds, then the corresponding solution x(t) of equation (11.6) is bounded on J. Proof. Let c1 , c2 ∈ ℝ be arbitrarily chosen. Let also c > 0 and M > 0 be chosen so that 󵄨󵄨 󵄨 󵄨󵄨c1 x1 (t) + c2 x2 (t)󵄨󵄨󵄨 ≤ c

for any t ∈ J,

11.3 Second order integro-dynamic equations |

and 󵄨 󵄨󵄨 σ σ 󵄨󵄨x1 (t)x2 (y) − x2 (t)x1 (y)󵄨󵄨󵄨 ≤ M

for any t, y ∈ J.

Then 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨c1 x1 (t) + c2 x2 (t) 󵄨󵄨󵄨 t

+

∫(x1 (t)x2σ (y)

t0

−

y 󵄨󵄨 󵄨󵄨 σ x2 (t)x1 (y))h(y, x(y), ∫ k(y, s, x(s))Δs)Δy󵄨󵄨󵄨 󵄨

󵄨󵄨

t0

󵄨 󵄨 ≤ 󵄨󵄨󵄨c1 x1 (t) + c2 x2 (t)󵄨󵄨󵄨

y 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨󵄨󵄨 󵄨󵄨 σ σ + ∫󵄨󵄨x1 (t)x2 (y) − x2 (t)x1 (y)󵄨󵄨󵄨󵄨h(y, x(y), ∫ k(y, s, x(s))Δs)󵄨󵄨󵄨Δy 󵄨󵄨 󵄨󵄨 󵄨 󵄨 t0 t0 t

󵄨󵄨 y 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨 ≤ c + M ∫ q(y)(󵄨󵄨x(y)󵄨󵄨 + 󵄨󵄨∫ k(y, s, x(s))Δs󵄨󵄨󵄨)Δy 󵄨󵄨 󵄨󵄨 󵄨t0 󵄨 t0 t

t

󵄨 󵄨 = c + M ∫ q(y)󵄨󵄨󵄨x(y)󵄨󵄨󵄨Δy t0

t

y

󵄨 󵄨 + M ∫ q(y) ∫󵄨󵄨󵄨k(y, s, x(s))󵄨󵄨󵄨ΔsΔy t0

t0

t

󵄨 󵄨 ≤ c + M ∫ q(y)󵄨󵄨󵄨x(y)󵄨󵄨󵄨Δy t0

t

y

󵄨 󵄨 + M ∫ q(y) ∫ p(s)󵄨󵄨󵄨x(s)󵄨󵄨󵄨ΔsΔy, t0

t0

t ∈ J.

Hence, by Theorem 1.5.1, we get t

󵄨󵄨 󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ c(1 + M ∫ q(y)eMq+p (y, 0)Δy), t0

t ∈ J.

From here we conclude that x(t) is bounded on J. This completes the proof.

297

298 | 11 Applications

11.4 Volterra integral equations Consider the following Volterra integral equation: t

x(t) = f (t) + ∫ F(t, s, x(s))Δs, t0

t ∈ J,

(11.11)

where f ∈ 𝒞 (J), F ∈ 𝒞 (J × J × ℝ) are given functions, x is unknown. Theorem 11.4.1. Let f ∈ 𝒞 (J), F ∈ 𝒞 (J × J × ℝ), 󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨F(t, s, x(s))󵄨󵄨󵄨 ≤ g(t)(h(s) + 󵄨󵄨󵄨x(s)󵄨󵄨󵄨),

t, s ∈ J,

where g, h ∈ 𝒞 (J) are nonnegative functions which are bounded on J. If x(t) is a solution of equation (11.11), then it is bounded on J. Proof. We have t 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨f (t) + ∫ F(t, s, x(s))Δs󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 t0

󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨 ≤ 󵄨󵄨󵄨f (t)󵄨󵄨󵄨 + 󵄨󵄨󵄨∫ F(t, s, x(s))Δs󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t0 󵄨 t

󵄨 󵄨 󵄨 󵄨 ≤ 󵄨󵄨󵄨f (t)󵄨󵄨󵄨 + ∫󵄨󵄨󵄨F(t, s, x(s))󵄨󵄨󵄨Δs t0

t

󵄨 󵄨 󵄨 󵄨 ≤ 󵄨󵄨󵄨f (t)󵄨󵄨󵄨 + g(t) ∫(h(s) + 󵄨󵄨󵄨x(s)󵄨󵄨󵄨)Δs t0

t

t

󵄨 󵄨 󵄨 󵄨 = 󵄨󵄨󵄨f (t)󵄨󵄨󵄨 + g(t) ∫ h(s)Δs + g(t) ∫󵄨󵄨󵄨x(s)󵄨󵄨󵄨Δs, t0

t0

Let t

󵄨 󵄨 p(t) = 󵄨󵄨󵄨f (t)󵄨󵄨󵄨 + g(t) ∫ h(s)Δs, t0

t ∈ J.

Then t

󵄨󵄨 󵄨 󵄨 󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ p(t) + g(t) ∫󵄨󵄨󵄨x(s)󵄨󵄨󵄨Δs, t0

Hence, by Theorem 1.1.10, we obtain

t ∈ J.

t ∈ J.

11.5 Applications to nonlinear delay dynamic equations | 299 t

󵄨 󵄨󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ p(t) + g(t) ∫ p(s)e⊖g (σ(s), t)Δs, t0

t ∈ J.

Therefore x(t) is bounded on J. This completes the proof.

11.5 Applications to nonlinear delay dynamic equations Suppose that the time scale 𝕋 is unbounded above. Consider the following IVP: p Δ

((x(t)) ) = M(t, x(τ(t))),

t ∈ [t0 , ∞),

(11.12)

with the initial data x(t) = ϕ(t), 1 p

t ∈ [α, t0 ],

(11.13)

t ∈ [t0 , ∞), τ(t) ≤ t0 ,

ϕ(τ(t)) = c ,

where (A1) M : [t0 , ∞) × ℝ → ℝ is a continuous function, (A2) p ≥ 1, c = (x(t0 ))p are constants, τ : [t0 , ∞) → 𝕋, τ(t) ≤ t, t ∈ [t0 , ∞), −∞ < α = inf{τ(t) : t ∈ [t0 , ∞)}, ϕ ∈ 𝒞rd ([t0 , ∞)). Theorem 11.5.1. Suppose (A1), (A2) and 󵄨󵄨 󵄨 󵄨󵄨M(t, x)󵄨󵄨󵄨 ≤ f (t)|x| + g(t),

t ∈ [t0 , ∞), x ∈ ℝ,

where f , g ∈ 𝒞rd ([t0 , ∞)) are nonnegative functions. If x is a solution of the IVP (11.12), (11.13), then t

󵄨󵄨 󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ (|c| + h(t) ∫ eB (t, σ(s))h(s)B(s)Δs)

1 p

t0

1 p

≤ (|c| + h(t)eB (t, t0 )) ,

t ∈ [t0 , ∞),

for any k > 0, where t

h(t) = ∫(f (s)( t0

B(t) =

f (t)

pk

p−1 p

,

p − 1 p1 |c| k + ) + g(s))Δs, p−1 p pk p t ∈ [t0 , ∞).

(11.14)

300 | 11 Applications Proof. The solution x of the IVP (11.12), (11.13) satisfies the following integral equation: t

󵄨p 󵄨󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 = c + ∫ M(s, x(τ(s)))Δs, t0

whereupon

t ∈ [t0 , ∞),

t

󵄨 󵄨󵄨 󵄨 󵄨p 󵄨󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ |c| + ∫(󵄨󵄨󵄨f (s)󵄨󵄨󵄨󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨 + g(s))Δs,

t ∈ [t0 , ∞).

t0

Hence, by Theorem 10.2.2, we get inequality (11.14). This completes the proof. Consider the following delay dynamic equation: t

xΔ (t) = F(t, x(τ(t)), ∫ M(ξ , x(τ(ξ )))Δξ ), t0

with initial data

x(t0 ) = c,

x(t) = ϕ(t),

ϕ(τ(t)) ≤ |c|,

t ∈ [t0 , ∞),

(11.15)

(11.16)

t ∈ [α, t0 ],

t ∈ [t0 , ∞), τ(t) ≤ t0 ,

where (A3) c is a constant, c ≠ 0, ϕ ∈ 𝒞rd ([α, t0 ]) is a nonnegative function, τ : [t0 , ∞) → 𝕋, τ ∈ 𝒞 ([t0 , ∞)), τ(t) ≤ t, t ∈ [t0 , ∞), −∞ < α = inf{τ(t) : t ∈ [t0 , ∞)}, (A4) F ∈ 𝒞 ([t0 , ∞) × ℝ × ℝ), M ∈ 𝒞 ([t0 , ∞) × ℝ). Theorem 11.5.2. Assume (A3), (A4) and 󵄨󵄨 󵄨 󵄨󵄨F(t, u, v)󵄨󵄨󵄨 ≤ f (t)ψ(|u|)w(|u|) + g(t)ψ(|u|) + |v|, 󵄨󵄨 󵄨 󵄨󵄨M(t, u)󵄨󵄨󵄨 ≤ h(t)ψ(|u|), t ∈ [t0 , ∞), u, v ∈ ℝ, where f , g, h ∈ 𝒞rd ([t0 , ∞)) are nonnegative functions with f not equivalent to zero, w, ψ ∈ 𝒞 ([t0 , ∞)) are nonnegative and nondecreasing functions. Then the solution x of the IVP (11.15), (11.16) satisfies the following estimate: t

s

󵄨󵄨 󵄨 ̃−1 ̃ −1 ̃ ̃ (H (H(G(|c| + ∫(g(s) + ∫ h(ξ )Δξ )Δs) 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ G t0

t0

t

+ ∫ f (s)Δs))), t0

t ∈ [t0 , ∞),

̃ and H ̃ are increasing and bijective functions and where G

11.5 Applications to nonlinear delay dynamic equations | 301

Δ

̃ (G(v(t))) = Δ

̃ (H(z(t))) =

vΔ (t) , ψ(v(t))

z Δ (t) , ̃−1 (z(t))) w(G

t ∈ [t0 , ∞).

Proof. We have that the solution of the IVP (11.15), (11.16) satisfies the following integral equation: t

s

x(t) = c + ∫ F(s, x(τ(s)), ∫ M(ξ , x(τ(ξ )))Δξ )Δs, t0

t0

t ∈ [t0 , ∞).

Then t󵄨 s 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ |c| + ∫󵄨󵄨󵄨F(s, x(τ(s)), ∫ M(ξ , x(τ(ξ )))Δξ )󵄨󵄨󵄨Δs 󵄨󵄨 󵄨󵄨󵄨 t0 󵄨 t0 t

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ≤ |c| + ∫(f (s)ψ(󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨)w(󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨) + g(s)ψ(󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨) s

t0

󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 + 󵄨󵄨󵄨∫ M(ξ , x(τ(ξ )))Δξ 󵄨󵄨󵄨)Δs 󵄨󵄨 󵄨󵄨 󵄨t0 󵄨 t

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ≤ |c| + ∫(f (s)ψ(󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨)w(󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨) + g(s)ψ(󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨))Δs t0

t s

󵄨 󵄨 + ∫ ∫󵄨󵄨󵄨M(ξ , x(τ(ξ )))󵄨󵄨󵄨Δξ Δs t0 t0

t

󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 ≤ |c| + ∫(f (s)ψ(󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨)w(󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨) + g(s)ψ(󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨))Δs t0

t s

󵄨 󵄨 + ∫ ∫ h(ξ )ψ(󵄨󵄨󵄨x(τ(ξ ))󵄨󵄨󵄨)Δξ Δs, t0 t0

t ∈ [t0 , ∞).

Hence, by Theorem 10.2.3, it follows that t

s

󵄨󵄨 󵄨 ̃−1 ̃ −1 ̃ ̃ (H (H(G(|c| + ∫(g(s) + ∫ h(ξ )Δξ )Δs) 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ G t0

t

+ ∫ f (s)Δs))), t0

This completes the proof.

t ∈ [t0 , ∞).

t0

302 | 11 Applications Now, consider the delay dynamic equation p Δ

((x(t)) ) = M(t, x(t), x(τ(t))),

t ∈ [t0 , ∞),

(11.17)

subject to the initial condition x(t) = ϕ(t), 1 p

ϕ(τ(t)) = c ,

t ∈ [α, t0 ], t ∈ [t0 , ∞),

(11.18)

τ(t) ≤ t0 ,

where (A5) p > 0, c = (x(t0 ))p , τ ∈ 𝒞rd ([t0 , ∞)), τ(t) ≤ t, t ∈ [t0 , ∞), −∞ < α = inf{τ(t) : t ∈ [t0 , ∞)}. (A6) M ∈ 𝒞 ([t0 , ∞) × ℝ2 ) and 󵄨󵄨 󵄨 q r 󵄨󵄨M(t, u, v)󵄨󵄨󵄨 ≤ f (t)|v| + g(t)|u| , t ∈ [t0 , ∞), u, v ∈ ℝ, q and r are constants such that p ≥ q ≥ 0 and p ≥ r ≥ 0. Theorem 11.5.3. Assume (A5), (A6). Then the solution x of the IVP (11.17), (11.18) satisfies the following estimate: t

󵄨󵄨 󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ (|c| + F(t) + ∫ eG (t, σ(s))F(s)G(s)Δs)

1 p

(11.19)

t0

1

≤ (|c| + F(t)eG (t, t0 )) p ,

t ∈ [t0 , ∞),

for any k > 0, where t

F(t) = ∫(

f (s)(k(p − q) + q|c|) pk

t0

G(t) =

qf (t)

pk

p−q p

+

p−q p

rg(t) pk

p−q p

,

+

g(s)(k(p − r) + r|c|) pk

p−q p

)Δs,

t ∈ [t0 , ∞).

Proof. We have that the solution x of the IVP (11.17), (11.18) satisfies the integral equation p

t

(x(t)) = c + ∫ M(s, x(s), x(τ(s)))Δs, t0

Hence, t 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨p 󵄨󵄨 󵄨󵄨x(t)󵄨󵄨󵄨 = 󵄨󵄨󵄨c + ∫ M(s, x(s), x(τ(s)))Δs󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 t0 󵄨

t ∈ [t0 , ∞).

11.5 Applications to nonlinear delay dynamic equations | 303

󵄨󵄨 󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨 ≤ |c| + 󵄨󵄨∫ M(s, x(s), x(τ(s)))Δs󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨t0 t

󵄨 󵄨 ≤ |c| + ∫󵄨󵄨󵄨M(s, x(s), x(τ(s)))󵄨󵄨󵄨Δs t0

t

󵄨r 󵄨 󵄨q 󵄨 ≤ |c| + ∫(f (s)󵄨󵄨󵄨x(τ(s))󵄨󵄨󵄨 + g(s)󵄨󵄨󵄨x(s)󵄨󵄨󵄨 )Δs, t0

t ∈ [t0 , ∞).

By the last inequality and Theorem 10.2.4, we get the estimate (11.19). This completes the proof. Consider the IVP t

p Δ

((x(t)) ) = F(t, x(τ1 (t)), ∫ M(ξ , x(τ2 (ξ )))Δξ ),

t ∈ [t0 , ∞),

t0

(11.20)

subject to the initial condition x(t) = ϕ(t),

1 󵄨󵄨 󵄨 󵄨󵄨ϕ(τi (t))󵄨󵄨󵄨 ≤ |c| p ,

t ∈ [α, t0 ], t ∈ [t0 , ∞), τi (t) ≤ t0 , i ∈ {1, 2},

(11.21)

where (A6) x ∈ 𝒞rd ([t0 , ∞)), c = (x(t0 ))p , p ≥ 1, τi ∈ 𝒞 ([t0 , ∞)), τi (t) ≤ t, t ∈ [t0 , ∞), i ∈ {1, 2}, −∞ < α = inf{ min τi (t) : t ∈ [t0 , ∞)}, i∈{1,2}

ϕ ∈ 𝒞rd ([t0 , ∞)), (A7) F ∈ 𝒞 ([t0 , ∞) × ℝ2 ), M ∈ 𝒞 ([t0 , ∞) × ℝ), 󵄨󵄨 󵄨 󵄨󵄨F(t, u, v)󵄨󵄨󵄨 ≤ f (t)|u| + |v|, 󵄨󵄨 󵄨 󵄨󵄨M(t, u)󵄨󵄨󵄨 ≤ h(t)|u|, t ∈ [t0 , ∞), u, v ∈ ℝ, f , h ∈ 𝒞rd ([t0 , ∞)) are nonnegative functions. Theorem 11.5.4. If x is a solution to the IVP (11.20), (11.21), then t

1 p

s

󵄨󵄨 󵄨 −1 󵄨󵄨x(t)󵄨󵄨󵄨 ≤ (G (G(|c|) + ∫(f (s) + ∫ h(ξ )Δξ )Δs)) , t0

t0

where G is an increasing, bijective function and v

G(v) = ∫ 1

1

q

rp

dr,

v > 0.

t ∈ [t0 , ∞),

(11.22)

304 | 11 Applications Proof. Every solution of the IVP (11.20), (11.21), satisfies the integral equation p

t

s

(x(t)) = c + ∫ F(s, x(τ1 (s)), ∫ M(ξ , x(τ2 (ξ )))Δξ )Δs, t0

t0

t ∈ [t0 , ∞).

Then t s 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨p 󵄨󵄨󵄨 󵄨󵄨x(t)󵄨󵄨 = 󵄨󵄨c + ∫ F(s, x(τ1 (s)), ∫ M(ξ , x(τ2 (ξ )))Δξ )Δs󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 t0 󵄨 t0 s 󵄨󵄨 t 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 ≤ |c| + 󵄨󵄨∫ F(s, x(τ1 (s)), ∫ M(ξ , x(τ2 (ξ )))Δξ )Δs󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨t0 󵄨 t0 t󵄨

s 󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨 󵄨 ≤ |c| + ∫󵄨󵄨F(s, x(τ1 (s)), ∫ M(ξ , x(τ2 (ξ )))Δξ )󵄨󵄨󵄨Δs 󵄨󵄨 󵄨󵄨 󵄨 t0 󵄨 t0 t

󵄨󵄨 s 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 ≤ |c| + ∫(f (s)󵄨󵄨x(τ1 (s))󵄨󵄨 + 󵄨󵄨∫ M(ξ , x(τ2 (ξ )))Δξ 󵄨󵄨󵄨)Δs 󵄨󵄨 󵄨󵄨 󵄨t0 󵄨 t0 t

s

󵄨 󵄨 󵄨 󵄨 ≤ |c| + ∫(f (s)󵄨󵄨󵄨x(τ1 (s))󵄨󵄨󵄨 + ∫󵄨󵄨󵄨M(ξ , x(τ2 (ξ )))󵄨󵄨󵄨Δξ )Δs t0

t0

t

s

󵄨 󵄨 󵄨 󵄨 ≤ |c| + ∫(f (s)󵄨󵄨󵄨x(τ1 (s))󵄨󵄨󵄨 + ∫ h(ξ )󵄨󵄨󵄨x(τ2 (ξ ))󵄨󵄨󵄨Δξ )Δs, t0

t0

t ∈ [t0 , ∞). Hence, by Theorem 10.3.1, we get the estimate (11.22). This completes the proof.

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Index 2 𝒞1,2 (D) 189 𝒞rd 1

diamond-α derivative 113 diamond-α integral 115 Dragomir’s inequality 93, 94 forward graininess deviation functional 139 fractional Taylor formula 140 Gamidov’s inequality 24, 25 generalized Jensen inequality 131 Gronwall’s inequality 1 Jensen inequality 130

Pachpatte’s inequality 60, 63, 66, 96–98, 156, 161, 163, 174, 222, 224, 232, 251 ℛ 1 ℛ+ 1 ℛ+ (A) 1 Schlömilch inequality 134 Snow’s inequality 194, 197 Sobolev type inequality 143 synchronous functions 135 Wendroff’s inequality 147, 250