On Some Non-Linear Integral Inequalities for Two-Variable Functions with Delay Involving Integration over Infinite Intervals

On Some Non-Linear Integral Inequalities for Two-Variable Functions with Delay Involving Integration over Infinite Intervals

Dalila BITAT University of Constantine Department of Mathematics Route d’Ain el Bay, 25017 Constantine

ALGERIA bitat.dalila@umc.edu.dz

Hassane KHELLAF University of Constantine Department of Mathematics Route d’Ain el Bay, 25017 Constantine

ALGERIA

khellafhassane@umc.edu.dz

Abstract: In this paper, we present some non-linear integral inequalities with a term of delay for functions of two independent variables that can be used in the theory of differential and integral equations with time delay. Also, we generalize these inequalities by integration over infinite intervals. An application is given as an illustration.

Keywords: Non-linear integral inequalities with delay; Two independent variables; Non-decreasing functions;

Differential and integral equations with delay; Integration over infinite intervals

1 Introduction

The integral inequalities with a term of delay are used in the study of some partial differential equations with time delay. Many researches have established their basic properties, namely generalizations to the uni-dimensional case, applications to some partial differential equations, and the existence and unique- ness of solutions [6, 9, 4].

Recently, a number of research papers have been written on retarded integral inequalities. For instance, Mi [5] investigated some generalized Gronwall-Bellman type impulsive integral inequal- ities containing integrations over infinite intervals.

Likewise, Xu and Ma [8] established some non-linear retarded Volterra-Fredholm type integral inequalities in two independent variables. Also, Boudeliou and Khellaf [1] obtained some useful generalizations of two-variable integral inequalities with delay.

In 2013, Qingling and Zhonghua [7] gave the fol- lowing inequality:

Lemma 1 ([7]) Let a ∈ C(R⁺₀, R⁺) and α ∈ C¹(R⁺₀, R⁺₀) be non-decreasing functions in each variable, with α(x) ≤ x on R⁺₀, f, g, h ∈ C(R⁺₀ R⁺₀). If u ∈ C(R⁺₀, R₀⁺) satisfies

u(x) ≤ a(x) + Zx

0

h(s)u(s)ds

+

α(x)Z

0

f (s)^³ Zs

0

g(τ )u(τ )dτ^´ds,

for x ∈ R⁺₀, then

u(x) ≤ a(x) exp

"Z_x

0

h(s)ds +

α(x)Z

0

f (s)^³ Zs

0

g(τ )dτ^´ds

# ,

for all x ∈ R⁺₀.

In 2017, Ghrissi and Hammami [2] obtained some generalized retarded integral inequalities of the Gronwall type for one-variable functions; their results were based on the results proved in [7]. For example:

(i) If u ∈ C(R₀⁺, R⁺₀) satisfies

u^p(x) ≤ a(x) + Zx

0

h(s)u^p(s)ds

+

α(x)Z

0

f (s) ÃZs

0

g(τ )u(τ )dτ

! ds,

for x ∈ R⁺₀, then

u(x) ≤

· a(x) +

α(x)Z

0

m₂f (s) ÃZ_s

0

g(τ )dτ

! ds

#¹

p

exp1 p

"Z_x

0

h(s)ds +

α(x)Z

0

m₁f (s) ÃZ_s

0

g(τ )dτ

! ds

# ,

for all x ∈ R⁺₀.

(ii) If u ∈ C(R⁺₀, R⁺₀) satisfies

u^p(x) ≤ 1 + Zx

0

h(s)u^p(s)ds

+

α(x)Z

0

f (s) ÃZ_s

0

g(τ )u^q(τ )dτ

! ds,

for x ∈ R⁺₀, then

u(x) ≤ exp1 p

"Z_x

0

h(s)ds +

α(x)Z

0

f (s) ÃZ_s

0

g(τ )dτ

! ds

# ,

for all x ∈ R₀⁺.

(iii) If u ∈ C(R⁺₀, R⁺₀) satisfies

u(x) ≤ a(x) + b(x) Zx

0

h(s)u(s)ds

+

α(x)Z

0

f (s) ÃZ_s

0

g(τ )u(τ )dτ

! ds,

for x ∈ R⁺₀, then

u(x) ≤ a(x) exp

"

b(x) Zx

0

h(s)ds

+

α(x)Z

0

m₁f (s) ÃZ_s

0

g(τ )dτ

! ds

# ,

for all x ∈ R₀⁺.

The objective of this paper is to establish some non-linear retarded integral inequalities in two inde- pendent variables, and also to generalize those in- equalities by integration over infinite intervals. These new inequalities present generalizations of the results proved by Ghrissi and Hammami [2]. An application is given to illustrate the efficiency of the obtained re- sults in the study of the boundedness of solutions of certain hyperbolic partial differential equations.

2 Main Results

We denote R⁺₀ = [0, ∞] and R⁺ = (0, ∞). The first-order partial derivatives of the function u(x, y) for x, y ∈ R with respect to x and y are denoted by

∂

∂xu(x, y) and _∂y^∂ u(x, y) respectively.

2.1 Some Retarded Non-linear Integral In- equalities in Two Independent Variables In this section, we present some generalizations of the results proved in [2] and [7].

Theorem 2 Let a ∈ C(R⁺₀ × R⁺₀, R⁺) and α, β ∈ C¹(R⁺₀, R₀⁺) be non-decreasing functions in each variable, with α(x) ≤ x on R⁺₀ and β(y) ≤ y on R⁺₀, f, g, h ∈ C(R⁺₀ × R⁺₀, R⁺₀). If u ∈ C(R⁺₀ ×R⁺₀, R⁺₀) satisfies

u(x, y) ≤ a(x, y) + Zx

0

Zy

0

h(s, t)u(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)u(τ, σ)dτ dσ^´dsdt,

(1) for (x, y) ∈ R⁺₀ × R⁺₀, then

u(x, y) ≤ a(x, y) exp

"Z_x

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# , (2) for all (x, y) ∈ R⁺₀ × R⁺₀.

Proof: Since a(x, y) is positive and non-decreasing, from (1) we have

u(x, y)

a(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)u(s, t) a(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t)

³Z^s

0

Zt

0

g(τ, σ)u(τ, σ)

a(τ, σ)dτ dσ^´dsdt.

Define a function z(x, y) on R⁺₀ × R⁺₀ by

z(x, y) = 1 + Zx

0

Zy

0

h(s, t)u(s, t) a(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t)

³Z^s

0

Zt

0

g(τ, σ)u(τ, σ)

a(τ, σ)dτ dσ^´dsdt,

then z(x, y) is positive and non-decreasing, z(0, y) = z(x, 0) = 1, ^u(x,y)_a(x,y) ≤ z(x, y), and

∂

∂x

∂

∂y z(x, y) = h(x, y)u(x, y)

a(x, y)+ f (α(x), β(y))

α⁰(x)β⁰(y)

α(x)Z

0 β(y)Z

0

g(s, t)u(s, t) a(s, t)dsdt.

i.e.

∂

∂x

∂

∂y z(x, y) ≤ h(x, y)z(x, y) + f (x, y)α⁰(x)β⁰(y) z(x, y)

α(x)Z

0 β(y)Z

0

g(s, t)dsdt,

hence

∂x∂ ∂

∂y z(x, y)

z(x, y) ≤ h(x, y) + f (x, y)α⁰(x)β⁰(y)

α(x)Z

0 β(y)Z

0

g(s, t)dsdt.

Therefore

∂

∂y

" _∂

∂xz(x, y) z(x, y)

#

≤ h(x, y) + f (x, y)α⁰(x)β⁰(y)

α(x)Z

0 β(y)Z

0

g(s, t)dsdt.

(3) By keeping y fixed, setting x = s, and integrating from 0 to x in (3), and again by keeping x fixed, set- ting y = t, and integrating from 0 to y in the resulting inequality, we obtain

z(x, y) ≤ exp

"Z_x

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# .

Since ^u(x,y)_a(x,y) ≤ z(x, y), we have

u(x, y) ≤ a(x, y) exp

"Z_x

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# .

So the relation (2) is true.

Remark 3 If we replace the functions in Theorem 2 with one-variable functions, we obtain Proposition 2 in [2].

Now we give two lemmas that would be used in the proofs of the coming theorems.

Lemma 4 ([2]) Assume that a ≥ 0, p ≥ 1. Then, for any k > 0, we have

a^1p ≤ 1

pk^1−pp a +p − 1

p k^p1. (4)

Or equivalently a

1

p ≤ m₁ a + m₂, where m₁ =

p1k^1−pp and m₂ = ^p−1_p k^1p.

Lemma 5 ([3]) Assume that a ≥ 0, p ≥ q ≥ 0, and p 6= 0. Then, for any k > 0, we have

a^qp ≤ q

pk^q−pp a +p − q

p k^qp. (5)

Or equivalently a

q

p ≤ m₃ a + m₄, where m₃ =

q

pk^q−pp and m₄ = ^p−q_p k^qp.

Theorem 6 Let a ∈ C(R₀⁺× R⁺₀, R⁺) and α, β ∈ C¹(R₀⁺, R⁺₀) be non-decreasing functions in each variable, with α(x) ≤ x on R⁺₀ and β(y) ≤ y on R₀⁺, f, g, h ∈ C(R⁺₀ × R⁺₀, R⁺₀), p ≥ 1. If u ∈ C(R⁺₀ × R⁺₀, R⁺₀) satisfies

u^p(x, y) ≤ a(x, y) + Zx

0

Zy

0

h(s, t)u^p(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)(

Zs

0

Zt

0

g(τ, σ)u(τ, σ)dτ dσ)dsdt, (6) for (x, y) ∈ R⁺₀ × R₀⁺, then

u(x, y) ≤ A(x, y) exp1 p

"Z_x

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

m₁f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

# , (7) for all (x, y) ∈ R⁺₀ × R⁺₀, where

A(x, y) =

"

a(x, y) +

α(x)Z

0 β(y)Z

0

m₂f (s, t) ÃZs

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

#¹

p

. (8)

Proof: Let u^p(x, y) = z(x, y), then z(x, y) satisfies z(x, y) ≤ a(x, y) +

Zx

0

Zy

0

h(s, t)z(s, t)dsdt+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)z^1p(τ, σ)dτ dσ^´dsdt.

Using (4) we get z(x, y) ≤ a(x, y) +

Zx

0

Zy

0

h(s, t)z(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)

[m₁z(τ, σ) + m₂]dτ dσ^´dsdt.

i.e.

z(x, y) ≤ a(x, y) + Zx

0

Zy

0

h(s, t)z(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

m₂f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

+

α(x)Z

0 β(y)Z

0

m₁f (s, t)^³ Zs

0

Zt

0

g(τ, σ)z(τ, σ)dτ dσ^´dsdt.

Using Theorem 2 we get

z(x, y) ≤

"

a(x, y) +

α(x)Z

0 β(y)Z

0

m₂f (s, t)

³Z^s

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# exp

"Z_x

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

m₁f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

# .

Since u(x, y) = z^1p(x, y), we have u(x, y) ≤ A(x, y) exp1

p

"Zx 0

Zy

0

h(s, t)dsdt+

α(x)Z

0 β(y)Z

0

m₁f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

# ,

where A(x, y) is defined in (8).

Corollary 7 Let a ∈ C(R⁺₀ × R⁺₀, R⁺) and α, β ∈ C¹(R⁺₀, R₀⁺) be non-decreasing functions in each variable, with α(x) ≤ x on R⁺₀ and β(y) ≤ y on R⁺₀, f, g, h ∈ C(R⁺₀ × R⁺₀, R⁺₀), p ≥ 1. If u ∈ C(R⁺₀ × R⁺₀, R⁺₀) satisfies

u^p(x, y) ≤ a(x, y) + Zx

0

Zy

0

h(s, t)u(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)u^p(τ, σ)dτ dσ^´dsdt, (9) for (x, y) ∈ R⁺₀ × R⁺₀, then

u(x, y) ≤ B(x, y) exp1 p

"Z_x

0

Zy

0

m₁h(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# , (10) for all (x, y) ∈ R⁺₀ × R⁺₀, where

B(x, y) =

"

a(x, y) + Zx

0

Zy

0

m₂h(s, t)dsdt

#¹

p

. (11)

Proof: Let u^p(x, y) = z(x, y), then z(x, y) satisfies z(x, y) ≤ a(x, y) +

Zx

0

Zy

0

h(s, t)z^1p(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)z(τ, σ)dτ dσ^´dsdt.

Using (4) we get

z(x, y) ≤ a(x, y) + Zx

0

Zy

0

h(s, t)[m₁z(s, t) + m₂]dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)z(τ, σ)dτ dσ^´dsdt.

i.e.

z(x, y) ≤ a(x, y) + Zx

0

Zy

0

m₂h(s, t)dsdt

+ Zx

0

Zy

0

m₁h(s, t)z(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)z(τ, σ)dτ dσ^´dsdt.

Using Theorem 2 we get

z(x, y) ≤

"

a(x, y) + Zx

0

Zy

0

m₂h(s, t)dsdt

# exp

"Z_x

0

Zy

0

m₁h(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t)

³Z^s

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# .

Since u(x, y) = z

1

p(x, y), we have

u(x, y) ≤ B(x, y) exp1 p

"

m₁ Zx

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# ,

where B(x, y) is defined in (11).

Remark 8 (i) If we replace the functions in Theorem 6 and 7 with one-variable functions, we obtain Propo- sitions 9 and 11 in [2].

(ii) If we take p = 1, then Theorem 6 and Corollary 7 reduce to Theorem 2.

Theorem 9 Let α, β ∈ C¹(R⁺₀, R⁺₀) be non- decreasing functions in each variable, with α(x) ≤ x on R⁺₀ and β(y) ≤ y on R⁺₀, f, g, h ∈ C(R⁺₀ × R⁺₀, R⁺₀), p ≥ q ≥ 1. If u ∈ C(R⁺₀ × R₀⁺, R⁺₀) satisfies

u^p(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)u^p(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)u^q(τ, σ)dτ dσ

! dsdt,

(12) for (x, y) ∈ R⁺₀ × R⁺₀, then

u(x, y) ≤ C(x, y) exp1 p

"Z_x

0

Zy

0

h(s, t)dsdt +

α(x)Z

0 β(y)Z

0

m₃f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

# ,

(13) for all (x, y) ∈ R⁺₀ × R⁺₀, where

C(x, y) =

"

1 +

α(x)Z

0 β(y)Z

0

m₄f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

#¹

p

. (14) Proof: Define a function z(x, y) by the right side of (12), then z(x, y) > 0, u^p(x, y) ≤ z(x, y), u(x, y) ≤ z^1p(x, y), u^q(x, y) ≤ z^qp(x, y), and

z(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)z(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)z^qp(τ, σ)dτ dσ

! dsdt.

Using (5) we get

z(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)z(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t) ÃZs

0

Zt

0

g(τ, σ)[m₃z(τ, σ) + m₄]dτ dσ

! dsdt.

i.e.

z(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)z(s, t)dsdt +

α(x)Z

0 β(y)Z

0

m₄f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt +

α(x)Z

0 β(y)Z

0

m₃f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)z(τ, σ)dτ dσ

! dsdt.

Using Theorem 2 we get

z(x, y) ≤

"

1 +

α(x)Z

0 β(y)Z

0

m₄f (s, t)^³ Zs

0

Zt

0

g(τ, σ)

dτ dσ^´dsdt

# exp

"Z_x

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

m₃f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# .

Finally,

u(x, y) ≤ C(x, y) exp1 p

"Zx 0

Zy

0

h(s, t)dsdt +

α(x)Z

0 β(y)Z

0

m₃f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# ,

where C(x, y) is defined in (14).

Remark 10 If we take p = q = 1 and a(x, y) = 1, then Theorem 9 reduces to Theorem 2.

Corollary 11 Let a ∈ C(R⁺₀ × R⁺₀, R⁺) and α, β ∈ C¹(R⁺₀, R⁺₀) be non-decreasing functions in each variable, with α(x) ≤ x on R⁺₀ and β(y) ≤ y on R⁺₀, f, g, h ∈ C(R⁺₀ × R⁺₀, R⁺₀), p ≥ q ≥ 1. If u ∈ C(R⁺₀ × R⁺₀, R⁺₀) satisfies

u^p(x, y) ≤ a(x, y) + Zx

0

Zy

0

h(s, t)u^p(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)

u^q(τ, σ)dτ dσ

! dsdt,

(15) for (x, y) ∈ R⁺₀ × R⁺₀, then

u(x, y) ≤ a^1p(x, y)C(x, y) exp1 p

"

Zx

0

Zy

0

h(s, t)dsdtr +

α(x)Z

0 β(y)Z

0

m₃f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

# ,

(16) for all (x, y) ∈ R⁺₀ × R⁺₀, where C(x, y) is defined in (14).

Proof: Since a(x, y) is positive and non-decreasing, from (15) we have

u^p(x, y)

a(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)u^p(s, t) a(s, t) dsdt +

α(x)Z

0 β(y)Z

0

f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)u^q(τ, σ) a(τ, σ)dτ dσ

! dsdt.

i.e.

u^p(x, y)

a(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)u^p(s, t) a(s, t) dsdt +

α(x)Z

0 β(y)Z

0

f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)u^q(τ, σ) a^qp(τ, σ)dτ dσ

! dsdt.

Or equivalently à u(x, y)

a^1p(x, y)

!_p

≤ 1 + Zx

0

Zy

0

h(s, t)

à u(s, t) a^1p(s, t)

!_p dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t) ÃZ_s

0

Zt

0

g(τ, σ) Ã u(τ, σ)

a^1p(τ, σ)

!_q dτ dσ

! dsdt.

Using Theorem 9 we get u(x, y)

a^1p(x, y) ≤ C(x, y) exp1 p

"Z_x

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

m₃f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

# .

Finally,

u(x, y) ≤ a^1p(x, y)C(x, y) exp1 p

"Z_x

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

m₃f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

# ,

where C(x, y) is defined in (14).

Remark 12 If we take p = q = 1, then Corollary 11 reduces to Theorem 2.

Corollary 13 Let α, β ∈ C¹(R⁺₀, R⁺₀) be non- decreasing functions in each variable, with α(x) ≤ x on R⁺₀ and β(y) ≤ y on R⁺₀, f, g, h ∈ C(R⁺₀ × R⁺₀, R⁺₀), p ≥ q ≥ 1. If u ∈ C(R⁺₀ × R⁺₀, R⁺₀) satisfies

u^p(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)u^q(s, t)dsdt +

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)u(τ, σ)dτ dσ^´dsdt, (17) for (x, y) ∈ R⁺₀ × R⁺₀, then

u(x, y) ≤ K(x, y) exp1 p

"Z_x

0

Zy

0

m₃h(s, t)dsdt

α(x)Z

0 β(y)Z

0

m₁f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# ,

for all (x, y) ∈ R⁺₀ × R⁺₀, where

K(x, y) =

"

1 + Zx

0

Zy

0

m₄h(s, t)dsdt +

α(x)Z

0 β(y)Z

0

m₂f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

#¹

p

.

(18) Proof: Define a function z(x, y) by the right side of (17), then z(x, y) > 0, u^p(x, y) ≤ z(x, y), u(x, y) ≤ z^1p(x, y), u^q(x, y) ≤ z^qp(x, y), and

z(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)z^qp(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)z^1p(τ, σ)dτ dσ^´dsdt.

Using (4) and (5), we get

z(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)[m₃z(s, t) + m₄]dsdt +

α(x)Z

0 β(y)Z

0

f (s, t)(

Zs

0

Zt

0

g(τ, σ)[m₁z(τ, σ)+m₂]dτ dσ)dsdt.

i.e.

z(x, y) ≤ 1 + Zx

0

Zy

0

m₄h(s, t)dsdt +

α(x)Z

0 β(y)Z

0

m₂

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt +

Zx

0

Zy

0

m₃h(s, t)z(s, t)dsdt +

α(x)Z

0 β(y)Z

0

m₁

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)z(τ, σ)dτ dσ^´dsdt.

Using Theorem 2, we get

z(x, y) ≤

"

1+

Zx

0

Zy

0

m₄h(s, t)dsdt+

α(x)Z

0 β(y)Z

0

m₂f (s, t)

³Z^s

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# exp

"Z_x

0

Zy

0

m₃h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

m₁f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# .

Since u(x, y) ≤ z

1

p(x, y), we have u(x, y) ≤ K(x, y) exp1

p

"Z_x

0

Zy

0

m₃h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

m₁f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# ,

where K(x, y) is defined in (18).

Corollary 14 Let α, β ∈ C¹(R⁺₀, R⁺₀) be non- decreasing functions in each variable, with α(x) ≤ x on R⁺₀ and β(y) ≤ y on R⁺₀, f, g, h ∈ C(R⁺₀ × R₀⁺, R⁺₀), p ≥ q ≥ 1. If u ∈ C(R₀⁺× R⁺₀, R⁺₀) satisfies

u(x, y) ≤ 1 + Zx

0

Zy

0

h(s, t)u^p(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t) ÃZs

0

Zt

0

g(τ, σ)u^q(τ, σ)dτ dσ

! dsdt,

(19) for (x, y) ∈ R⁺₀ × R⁺₀, then

u(x, y) ≤

"

1 + (1 − p) ÃZ_x

0

Zy

0

h(s, t)dsdt +

α(x)Z

0 β(y)Z

0

f (s, t) ÃZ_s

0

Zt

0

g(τ, σ)dτ dσ

! dsdt

!# ¹

1−p

, (20) for all (x, y) ∈ R⁺₀ × R⁺₀.

Proof: Define a function z(x, y) by the right side of (19), then

u(x, y) ≤ z(x, y), z(0, y) = z(x, 0) = 1,

∂

∂xz(x, 0) = ∂

∂yz(0, y) = 0,

(21) and

∂

∂x

∂

∂y z(x, y) ≤ h(x, y)z^p(x, y) + α⁰(x)β⁰(y)

f (x, y) Ã α(x)Z

0 β(y)Z

0

g(s, t)z^q(s, t)dsdt

! .

i.e.

∂

∂x

∂

∂y z(x, y) ≤ z^p(x, y)

"

h(x, y) + α⁰(x)β⁰(y)

f (x, y) Ã α(x)Z

0 β(y)Z

0

g(s, t)dsdt

!#

,

hence

∂x∂ ∂

∂y z(x, y)z(x, y) z^p+1(x, y) −

∂x∂ z(x, y)_∂y^∂ z(x, y)

z^p+1(x, y) ≤ h(x, y)

+α⁰(x)β⁰(y)f (x, y) Ãα(x)Z

0 β(y)Z

0

g(s, t)dsdt

! .

Therefore

∂

∂y

" _∂

∂xz(x, y) z^p(x, y)

#

≤ h(x, y) + α⁰(x)β⁰(y)f (x, y) Ã α(x)Z

0 β(y)Z

0

g(s, t)dsdt

! .

By keeping y fixed, setting x = s, and integrating from 0 to x in the above inequality and using (21), and again by keeping x fixed, setting y = t, and in- tegrating from 0 to y in the resulting inequality and using (21), we obtain the inequality (20).

Remark 15 (i) If we replace the functions in Theorem 9 and Corollaries 11, 13, and 14 with one-variable functions, we obtain Propositions 13, 15, 17, and 19 in [2] respectively.

(ii) If we take p = q = 1 and a(x, y) = 1, then Corol- laries 13 and 14 reduce to Theorem 2.

Theorem 16 Let a ∈ C(R⁺₀ × R⁺₀, R⁺) and α, β ∈ C¹(R⁺₀, R₀⁺) be non-decreasing functions in each variable, with α(x) ≤ x on R⁺₀ and β(y) ≤ y on R⁺₀, f, g, h ∈ C(R⁺₀ × R⁺₀, R⁺₀), b ∈ C(R⁺₀ × R⁺₀, R⁺₀).

If u ∈ C(R⁺₀ × R₀⁺, R⁺₀) satisfies

u(x, y) ≤ a(x, y) + b(x, y) Zx

0

Zy

0

h(s, t)u(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)u(τ, σ)dτ dσ^´dsdt, (22) for (x, y) ∈ R⁺₀ × R⁺₀, then

u(x, y) ≤ a(x, y) exp

"

b(x, y) Zx

0

Zy

0

h(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)dτ dσ^´dsdt

# , (23) for all (x, y) ∈ R⁺₀ × R⁺₀.

Proof: Since a(x, y) is positive and non-decreasing, from (22) we have

u(x, y)

a(x, y) ≤ 1 + b(x, y) Zx

0

Zy

0

h(s, t)u(s, t) a(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)u(τ, σ)

a(τ, σ)dτ dσ^´dsdt.

Fix any (X, Y ) ∈ R⁺₀ × R⁺₀. Then, for 0 ≤ x ≤ X, 0 ≤ y ≤ Y, we have

u(x, y)

a(x, y) ≤ 1 + b(X, Y ) Zx

0

Zy

0

h(s, t)u(s, t) a(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)u(τ, σ)

a(τ, σ)dτ dσ^´dsdt.

Define a function z(x, y) by

z(x, y) = 1 + b(X, Y ) Zx

0

Zy

0

h(s, t)u(s, t) a(s, t)dsdt

+

α(x)Z

0 β(y)Z

0

f (s, t)^³ Zs

0

Zt

0

g(τ, σ)u(τ, σ)

a(τ, σ)dτ dσ^´dsdt,