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volume 7, issue 2, article 48, 2006.

Received 24 September, 2005;

accepted 09 November, 2005.

Communicated by:D. Hinton

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Journal of Inequalities in Pure and Applied Mathematics

EXPLICIT BOUNDS ON SOME NONLINEAR INTEGRAL INEQUALITIES

YOUNG-HO KIM

Department of Applied Mathematics Changwon National University Changwon 641-773, Korea.

EMail:yhkim@sarim.changwon.ac.kr

c

2000Victoria University ISSN (electronic): 1443-5756 288-05

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Explicit Bounds On some Nonlinear Integral Inequalities

Young-Ho Kim

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Abstract

In this paper, some new nonlinear integral inequalities involving functions of one and two independent variables which provide explicit bounds on unknown func- tions are established. We also present some of its applications to the qualitative study of retarded differential equations.

2000 Mathematics Subject Classification:26D10, 26D15, 34K10, 35B35.

Key words: Integral inequalities, Retarded integral inequality, Retarded differential equations, Partial delay differential equations.

This research is financially supported by Changwon National University in 2005.

Contents

1 Introduction. . . 3 2 The Integral Inequalities. . . 5 3 Applications. . . 17

References

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1. Introduction

Nonlinear differential equations whose solutions cannot be found explicitly arise in essentially every branch of modern science, engineering and mathemat- ics. One of the most useful methods available for studying a nonlinear system of ordinary differential equations is to compare it with a single first-order equa- tion derived naturally from some bounds on the system. However, the bounds provided by the comparison method are sometimes difficult or impossible to cal- culate explicitly. In fact, in many applications explicit bounds are more useful while studying the behavior of solutions of such systems. Another basic tool, which is typical among investigations on this subject, is the use of nonlinear integral inequalities which provide explicit bounds on the unknown functions.

Over the last scores of years several new nonlinear integral inequalities have been developed in order to study the behavior of solutions of such systems.

One of the most useful inequalities in the development of the theory of dif- ferential equations is given in the following theorem. If u, f are nonnegative continuous functions onR+ = [0,∞), u0 ≥0is a constant and

u2(t)≤u20+ 2 Z t

0

f(s)u(s)ds fort∈R+,then

u(t)≤u0 Z t

0

f(s)ds, t∈R+.

It appears that this inequality was first considered by Ou-Iang [5], while investi- gating the boundedness of certain solutions of certain second-order differential equations.

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In the past few years this inequality given in [5] has been used considerably in the study of qualitative properties of the solutions of certain abstract differ- ential, integral and partial differential equations.

Recently, Pachpatte in [9] obtained a useful upper bound on the following inequality:

(1.1) up(t)≤c+p

n

X

i=1

Z αi(t) αi(t0)

[ai(s)up(s) +bi(s)u(s)]ds,

and its variants, under some suitable conditions on the functions involved in (1.1), including the constantp > 1. In fact, the results given in [9] are gener- alized versions of the inequalities established by Lipovan in [4], Qu-Iang in [5]

and Pachpatte in [6].

The main purpose of this paper is to obtain explicit bounds on the following retarded integral inequality

(1.2) u(t)≤c+

n

X

i=1

"

Z t t0

ai(s)up(s) + Z αi(t)

αi(t0)

bi(s)up(s)ds

# ,

and its variants, under some suitable conditions on the functions involved in (1.2), including the constantp ≥ 0, p 6= 1, orp > 1. We also prove the two independent variable generalization of the result and present some applications of those to the global existence of solutions of differential equations with time delay.

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2. The Integral Inequalities

We shall introduce some notation,Rdenotes the set of real numbers andR+ = [0,∞), I = [t0, T) are the given subsets ofR. The first order derivative of a function z(t) with respect to t will be denoted by z0(t). LetC(M, N) denote the class of continuous functions from the setM to the setN.In the following theorems we prove some nonlinear integral inequalities.

Theorem 2.1. Let u, ai, bi ∈ C(I,R+), αi ∈ C1(I, I) be nondecreasing with αi(t)≤tonI,fori= 1,2, . . . , n.Letp≥0, p6= 1,andc≥0be constants. If

(2.1) u(t)≤c+

n

X

i=1

"

Z t t0

ai(s)up(s)ds+ Z αi(t)

αi(t0)

bi(s)up(s)ds

#

fort∈I,then

(2.2) u(t)≤

"

cq+q

n

X

i=1

Z t t0

ai(s)ds+ Z αi(t)

αi(t0)

bi(s)ds

!#1q

fort ∈[t0, T1),whereq = 1−pandT1 is chosen so that the expression inside [. . .]is positive on the subinterval[t0, T1).

Proof. From the hypotheses we observe thatα0(t)≥0fort ∈I.Letc≥0and define a function z(t)by the right-hand side of (2.1). Then, z(t0) = c, z(t)is

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nondecreasing fort∈I, u(t)≤z(t),and z0(t) =

n

X

i=1

[ai(t)up(t) +bii(t))upi(t))α0i(t)]

n

X

i=1

[ai(t)zp(t) +bii(t))zpi(t))α0i(t)]

n

X

i=1

[ai(t) +bii(t))α0i(t)][z(t)]p.

By making the constant q = 1−pand using the functionz1(t) = zq(t)/qwe get

(2.3) z10(t)≤

n

X

i=1

[ai(t) +bii(t))α0i(t)].

By takingt =sin (2.3) and integrating it with respect tosfromt0 tot, t ∈I, we obtain

(2.4)

Z t t0

z10(s)ds≤

n

X

i=1

Z t t0

ai(s)ds+ Z t

t0

bii(s))α0i(s)ds

.

Integrating, making the change of the function on the left side in (2.4) and rewriting yields

zq(t) q ≤ cq

q +

n

X

i=1

"

Z t t0

ai(s)ds+ Z αi(t)

αi(t0)

bi(s)ds

#

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fort∈I.It follows that (2.5) z(t)≤

"

cq+q

n

X

i=1

Z t t0

ai(s)ds+ Z αi(t)

αi(t0)

bi(s)ds

!#1q

fort ∈[t0, T1),whereT1 is chosen so that the expression inside[. . .]is positive on the subinterval [t0, T1).Using (2.5) in u(t) ≤ z(t)we get the inequality in (2.2).

Corollary 2.2. Let u, bi ∈ C(I,R+), αi ∈ C1(I, I) be nondecreasing with αi(t)≤tonI fori= 1, . . . , n,and letp≥0, p6= 1,andc≥0be constants. If

u(t)≤c+

n

X

i=1

Z αi(t) αi(t0)

bi(s)up(s)ds

fort∈I,then

u(t)≤

"

cq+q

n

X

i=1

Z αi(t) αi(t0)

bi(s)ds

#1q

fort ∈[t0, T1),whereq = 1−pandT1 is chosen so that the expression inside [. . .]is positive on the subinterval[t0, T1).

The following theorem deals with the constant1 < p < ∞versions of the inequalities established in Theorem2.1.

Theorem 2.3. Letu, a, bi, ci ∈C(I,R+), αi ∈C1(I, I)be nondecreasing with αi(t) ≤ t onI,fori = 1,2, . . . , n.Also, leta(t)be nondecreasing int, t ∈ I

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andp >1be constants. If (2.6) u(t)≤a(t) +

n

X

i=1

"

Z t t0

bi(s)up(s)ds+ Z αi(t)

αi(t0)

ci(s)up(s)ds

#

fort∈I,then (2.7) u(t)≤a(t)

"

1−(p−1)ap−1(t)

×

n

X

i=1

Z t t0

bi(s)ds+ Z αi(t)

αi(t0)

ci(s)ds

!#1−p1

fort∈[t0, T),where T = sup

(

t∈I : (p−1)ap−1(t)

n

X

i=1

Z t t0

bi(s)ds+ Z αi(t)

αi(t0)

ci(s)ds

!

<1 )

. Proof. From the hypotheses we observe that α0(t) ≥ 0 for t ∈ I. Define a functionv(t)by

v(t) =

n

X

i=1

"

Z t t0

bi(s)up(s)ds+ Z αi(t)

αi(t0)

ci(s)up(s)ds

# . Then,v(t0) = 0, v(t)is nondecreasing fort∈I, u(t)≤a(t) +v(t),and

v0(t)≤

n

X

i=1

[bi(t) +cii(t))α0i(t)][a(t) +z(t)]p

≤R(t)[a(t) +v(t)],

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where

R(t) =

n

X

i=1

[bi(t) +cii(t))α0i(t)][a(t) +z(t)]p−1.

Now by the comparison result for linear differential inequalities(see [1, Lemma 1.1., p. 2]), this implies that

v(t)≤ Z t

t0

R(s)a(s) exp Z t

s

R(τ)dτ

ds

≤a(t) Z t

t0

R(s) exp Z t

s

R(τ)dτ

ds (2.8)

fors≥t0.By integrating on the right hand side in (2.8) we get (2.9) v(t) +a(t)≤a(t) exp

Z t t0

R(τ)dτ

. From (2.9) we successively obtain

[v(t) +a(t)]p−1 ≤ap−1(t) exp Z t

t0

(p−1)R(τ)dτ

,

R(t)≤ap−1(t) exp Z t

t0

(p−1)R(τ)dτ n

X

i=1

[bi(t) +cii(t))α0i(t)]

and

A(t) = (p−1)R(t)

≤(p−1)ap−1(t) exp Z t

t0

A(τ)dτ n

X

i=1

[bi(t) +cii(t))α0i(t)].

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Consequently, we get A(t) exp

− Z t

t0

A(τ)dτ

≤(p−1)ap−1(t)

n

X

i=1

[bi(t) +cii(t))α0i(t)], or

(2.10) d dt

−exp

− Z t

t0

A(τ)dτ

≤(p−1)ap−1(t)

n

X

i=1

[bi(t) +cii(t))α0i(t)], By takingt=sin (2.10) and integrating it with respect tosfromt0 tot, t∈I, we obtain

(2.11) 1−exp

− Z t

t0

A(τ)dτ

≤(p−1)ap−1(t)

n

X

i=1

Z t t0

bi(s)ds+ Z t

t0

cii(s))α0i(s)ds

. Making the change of the variables on the right side in (2.11) and rewriting yields

exp Z t

t0

R(τ)dτ

"

1−(p−1)ap−1(t)

n

X

i=1

Z t t0

bi(s)ds+ Z αi(t)

αi(t0)

ci(s)ds

!#1−p1

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fort ∈ [t0, T),whereT is chosen so that the expression inside[. . .]is positive in the subinterval[t0, T).This, together with (2.9) andu(t)≤a(t) +v(t),gives the inequality in (2.7).

In the following theorem we establish two independent-variable versions of Theorem2.1which can be used in the qualitative analysis of hyperbolic partial differential equations with retarded arguments. Let 4 = I1×I2,whereI1 = [x0, X), I2 = [y0, Y) are the given subsets of the real numbers, R. The first order partial derivatives of a functionz(x, y)defined forx, y ∈ Rwith respect toxandyare denoted by∂z(x, y)/∂xand∂z(x, y)/∂yrespectively.

Theorem 2.4. Let u, ai, bi ∈ C(4,R+), αi ∈ C1(I1, I1), βi ∈ C1(I2, I2)be nondecreasing withαi(x) ≤ xon I1, βi(y) ≤ y onI2,fori = 1,2, . . . , n.Let p≥0, p6= 1,andc≥0be constants. If

(2.12) u(x, y)≤c+

n

X

i=1

Z x x0

Z y y0

ai(s, t)up(s, t)dt ds

+

Z αi(x) αi(x0)

Z βi(y) βi(y0)

bi(s, t)up(s, t)dt ds

!

for(x, y)∈I1×I2,then (2.13) u(x, y)

"

cq+q

n

X

i=1

Z x x0

Z y y0

ai(s, t)dt ds +

Z αi(x) αi(x0)

Z βi(y) βi(y0)

bi(s, t)dt ds

!#1q

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for(x, y)∈[x0, X)×[y0, Y),whereq = 1−pandX, Y are chosen so that the expression inside[. . .]is positive on the subintervals[x0, X)and[y0, Y).

Proof. The details of the proof of Theorem 2.4 follow by an argument similar to that in the proof of Theorem2.1with suitable changes. From the hypotheses we observe that α0(x) ≥ 0 forx ∈ I1 and β0(y) ≥ 0 for y ∈ I2. Let c ≥ 0 and define a functionz(x, y)by the right-hand side of (2.12). Then,z(x0, y) = z(x, y0) =c, z(x, y)is nondecreasing for(x, y)∈ 4, u(x, y)≤z(x, y),and

∂xz(x, y)≤

n

X

i=1

"

Z y y0

ai(x, t)dt+ Z βi(y)

βi(y0)

bii(x), t)α0i(x)dt

#

[z(x, y)]p.

By making the constantq = 1−pand using the functionv(x, y) = zq(x, y)/q we get

(2.14) ∂

∂xv(x, y)≤

n

X

i=1

"

Z y y0

ai(x, t)dt+ Z βi(y)

βi(y0)

bii(x), t)α0i(x)dt

# .

By taking x = s in (2.14) and integrating it with respect to s from x0 to x, x∈I1,we obtain

(2.15) Z x

x0

∂sv(s, y)ds

n

X

i=1

"

Z x x0

Z y y0

ai(s, t)dt ds+

Z αi(x) αi(x0)

Z βi(y) βi(y0)

bi(s, t)dt ds

# .

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Integrating with respect tosfromx0 tox,making the change of the function on the left side in (2.15) and rewriting yields

zq(x, y) q ≤ cq

q +

n

X

i=1

"

Z x x0

Z y y0

ai(s, t)dt ds+

Z αi(x) αi(x0)

Z βi(y) βi(y0)

bi(s, t)dt ds

#

for(x, y)∈ 4.This implies (2.16) z(x, y)

"

cq+q

n

X

i=1

Z x x0

Z y y0

ai(s, t)dt ds +

Z αi(x) αi(x0)

Z βi(y) βi(y0)

bi(s, t)dt ds

!#1q

for (x, y) ∈ [x0, X)×[y0, Y), where X, Y are chosen so that the expression inside [. . .]is positive on the subintervals [x0, X)and[y0, Y). Using (2.16) in u(x, y)≤z(x, y)we get the inequality in (2.13).

The following theorem deals with the constant1 < p < ∞versions of the inequalities established in Theorem2.4.

Theorem 2.5. Let u, ai, bi ∈ C(4,R+), αi ∈ C1(I1, I1), βi ∈ C1(I2, I2)be nondecreasing withαi(x) ≤ xon I1, βi(y) ≤ y onI2,fori = 1,2, . . . , n.Let a(x, y)be nondecreasing in(x, y)∈ 4and1< p <∞be constant. If

(2.17) u(x, y)≤a(x, y) +

n

X

i=1

Z x x0

Z y y0

bi(s, t)up(s, t)dt ds

+

Z αi(x) αi(x0)

Z βi(y) βi(y0)

ci(s, t)up(s, t)dt ds

#

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for(x, y)∈I1×I2,then

(2.18) u(x, y)≤a(x, y)

"

1−(p−1)ap−1(x, y)

n

X

i=1

Ψi(x, y)

#1−p1

for(x, y)∈ 41,where Ψi(x, y) =

Z x x0

Z y y0

bi(s, t)dt ds+

Z αi(x) αi(x0)

Z βi(y) βi(y0)

ci(s, t)dt ds and

41 = sup (

(x, y)∈ 4: (p−1)ap−1(x, y)

n

X

i=1

ξi(x, y)<1 )

.

Proof. The details of the proof of Theorem 2.5follows by an argument similar to that in the proof of Theorem2.3with suitable changes. From the hypotheses we observe that α0(x) ≥ 0 for x ∈ I1 and β0(y) ≥ 0 for y ∈ I2. Define a functionv(x, y)by

v(x, y) =

n

X

i=1

Z x x0

Z y y0

bi(s, t)up(s, t)dt ds+ Z αi(x)

αi(x0)

Z βi(y) βi(y0)

ci(s, t)up(s, t)dt ds

# .

Then, v(x0, y) = v(x, y0) = 0, v(x, y) is nondecreasing for (x, y) ∈ 4, u(x, y)≤a(x, y) +v(x, y),and

∂xv(x, y)≤R(x, y)[a(x, y) +v(x, y)],

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where R(x, y) =

n

X

i=1

"

Z y y0

bi(x, t)dt+ Z βi(y)

βi(y0)

ci(α(x), t)α0(x)dt

#

[a(x, y)+v(x, y)]p−1.

Now by keeping yfixed and using the comparison result for linear differential inequalities (see [1, Lemma 1.1., p. 2]), this implies that

(2.19) v(x, y)≤ Z x

x0

R(s, y)ai(s, y) exp Z x

s

R(τ, y)dτ

ds fors≥x0.By integrating on the right hand side in (2.19) we get (2.20) v(x, y) +a(x, y)≤a(x, y) exp

Z x x0

R(τ, y)dτ

. From (2.20) we successively obtain

[v(x, y) +a(x, y)]p−1 ≤ap−1(x, y) exp Z x

x0

(p−1)R(τ, y)dτ

,

A(x, y) = (p−1)R(x, y)

≤(p−1)ap−1(x, y) exp Z x

x0

R(τ, y)dτ n

X

i=1

ψi(x, y),

where

ψi(x, y) = Z y

y0

bi(x, t)dt+ Z βi(y)

βi(y0)

ci(α(x), t)α0(x)dt.

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Consequently, we have (2.21) ∂

∂x

−exp

− Z x

x0

A(τ, y)dτ

≤(p−1)ap−1(x, y)

n

X

i=1

ψi(x, y).

By taking x = s in (2.21) and integrating it with respect to s from x0 to x, x∈I1,we obtain

(2.22) 1−exp

− Z x

x0

A(τ, y)dτ

≤(p−1)ap−1(x, y)

n

X

i=1

Ψi(x, y),

where

Ψi(x, y) = Z x

x0

Z y y0

bi(s, t)dt ds+

Z αi(x) αi(x0)

Z βi(y) βi(y0)

ci(s, t)dt ds.

Making the change of the function on the inequality (2.22) and rewriting yields

exp Z x

x0

R(τ, y)dτ

"

1−(p−1)ap−1(x, y)

n

X

i=1

Ψi(x, y)

#1−p1

for(x, y) ∈ 41,where41 is chosen so that the expression inside[. . .]is pos- itive in the subinterval41.This, together with (2.20) andu(x, y) ≤ a(x, y) + v(x, y),gives the inequality in (2.18).

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3. Applications

In this section we will show that our results are useful in proving the global ex- istence of solutions to certain differential equations with time delay. First con- sider the functional differential equation involving several retarded arguments with the initial condition

(3.1)

(x0(t) = F(t, x(t), x(t−h1(t)), . . . , x(t−hn(t)), t∈I, x(t0) =x0,

where x0 is constant, F ∈ C(I × Rn+1,R) and for i = 1, . . . , n, let hi ∈ C1(I,R+)be nonincreasing and such thatt−hi(t)≥0, x−hi(t)∈C1(I, I), h0i(t) < 1,and h(t0) = 0. The following theorem deals with a bound on the solution of the problem (3.1).

Theorem 3.1. Assume that F : I ×Rn+1 → R is a continuous function for which there exist continuous nonnegative functions ai(t), bi(t)for t ∈ I such that

(3.2) |F(t, v, u1, . . . , un)| ≤

n

X

i=1

[ai(t)|v|p+bi(t)|ui|p],

wherep≥0, p6= 1is constant, and let

(3.3) Mi = max

t∈I

1

1−h0i(x), i= 1, . . . , n.

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Ifx(t)is any solution of the problem (3.1), then

|x(t)| ≤

"

|x0|q+q

n

X

i=1

Z t t0

ai(σ)dσ+

Z t−hi(t) t0

bi(σ)dσ

!#1q

fort∈I,wherebi(σ) = Mibi(σ+hi(s)), σ, s∈I.

Proof. The solutionx(t)of the problem (3.1) can be written as (3.4) x(t) =x0+

Z t t0

F(s, x(s), x(s−h1(s)), . . . , x(s−hn(s))ds.

From (3.2), (3.3), (3.4) and making the change of variables we have

|x(t)| ≤ |x0|+

n

X

i=1

Z t t0

ai(s)|x(s)|pds+ Z t

t0

bi(s)|x(s−hi(s))|pds

≤ |x0|+

n

X

i=1

Z t t0

ai(s)|x(s)|pds+

Z t−hi(t) t0

bi(σ)|x(σ)|p

! (3.5)

fort∈I,wherebi(σ) =Mibi(σ+hi(s)), σ, s∈I.Now a suitable application of the inequality in Theorem2.1to (3.5) yields the result.

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Remark 1.

(i) For1< p <∞,a suitable application of the inequality in Theorem2.3to (3.5) yields the following result

|x(t)| ≤

"

|x0| −(p−1)|x0|p

n

X

i=1

Z t t0

ai(s)ds+

Z t−hi(t) t0

bi(σ)dσ

!#1−p1 .

This shows in particular that the solutionx(t)is bounded on[t0, T),where T is chosen so that the expression inside[. . .]is positive in the subinterval [t0, T).

(ii) Consider the functional differential equation

(3.6)

( x0(t) =F(t, x(t−h1(t)), . . . , x(t−hn(t)), t∈I, x(t0) = x0.

Assume thatF : I ×Rn+1 → Ris a continuous function for which there exist continuous nonnegative functionsbi(t)fort∈I such that

(3.7) |F(t, u1, . . . , un)| ≤

n

X

i=1

bi(t)|ui|,

wherep ≥ 0, p 6= 1 is constant. LetMi be a function defined by (3.3). If x(t)is any solution of (3.6), the solutionx(t)can be written as

(3.8) x(t) = x0+ Z t

t0

F(s, x(s−h1(s)), . . . , x(s−hn(s))ds.

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Explicit Bounds On some Nonlinear Integral Inequalities

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From (3.7), (3.8) and making the change of variables we have

(3.9) |x(t)| ≤ |x0|+p

n

X

i=1

Z t−hi(t) t0

bi(σ)|x(σ)|p

for t ∈ I, where bi(σ) = Mibi(σ +hi(s)), σ, s ∈ I. Now a suitable application of the inequality in Corollary2.2to (3.8) yields

|x(t)| ≤

"

|x0|q+q

n

X

i=1

Z t−hi(t) t0

bi(σ)dσ

#1q

fort∈I,whereq= 1−p, bi(σ) = Mibi(σ+hi(s)), σ, s∈I.

In the following we present an application of the inequality given in Section2 to study the boundedness of the solutions of the initial boundary value problem for hyperbolic partial delay differential equations of the form

(3.10)





∂y

∂xz(x, y)

=F(x, y, z(x, y), z(x−h1(x), y−g1(y)), . . . , z(x−hn(x), y−gn(y)),

z(x, y0) = e1(x), z(x0, y) = e2(y), e1(x0) =e2(y0) = 0, where p > 0, p 6= 1 is constant,F ∈ C(4 ×Rn+1,R), e1 ∈ C1(I1,R), e2 ∈ C1(I2,R),andhi ∈ C1(I1,R+), gi ∈ C1(I2,R+) are nonincreasing and such thatx−hi(x)≥0, x−hi(x)∈C1(I1, I1), y−gi(y)≥0, y−hi(y)∈C1(I2, I2), h0i(t)<1, g0i(t)<1,andhi(x0) =gi(y0) = 0fori= 1, . . . , n.

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Explicit Bounds On some Nonlinear Integral Inequalities

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Theorem 3.2. Assume that F : 4 ×Rn+1 → Ris a continuous function for which there exists continuous nonnegative functions ai(x, y), bi(x, y) for i = 1, . . . , n;x∈I1, y ∈I2such that

(3.11)

(|F(x, y, v, u1, . . . , un)| ≤Pn

i=1[ai(x, y)|v|p+bi(x, y)|ui|p],

|e1(x) +e2(y)| ≤c forc≥0, p≥0, p6= 1,and let (3.12) Mi = max

x∈I1

1

1−h0i(x), Ni = max

y∈I2

1

1−gi0(y), i= 1, . . . , n.

Ifz(x, y)is any solution of the problem (3.10), then

|z(x, y)| ≤

"

cq+q

n

X

i=1

Z x x0

Z y y0

ai(σ, τ)dτ dσ+

Z φ(x) x0

Z ψ(y) y0

bi(σ, τ)]dτ dσ

!#1q

for (x, y) ∈ 41,where q = 1−p, φ(x) = x−hi(x), ψ(y) = y−gi(y) and b(σ, τ) =MiNibi(σ+hi(s), τ +gi(t)), σ, s∈I1, τ, t∈I2.

Proof. It is easy to see that the solution z(x, y)of the problem (3.10) satisfies the equivalent integral equation

(3.13) z(x, y) =e1(x) +e2(y) +

Z x x0

Z y y0

F(s, t, z(s, t), z(s−h1(s), t−g1(t)),

. . . , z(s−hn(s), t−gn(t))dt ds.

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Explicit Bounds On some Nonlinear Integral Inequalities

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From (3.11), (3.13) and making the change of variables, we have (3.14) |z(x, y)|

≤c+ Z x

x0

Z y y0

n

X

i=1

ai|z(s, t)|p+bi|z(s−hi(s), t−gi(t))|p ds dt.

Now a suitable application of the inequality given in Theorem 2.4 to (3.14) yields the desired result.

Remark 2. For1< p <∞,a suitable application of the inequality in Theorem 2.5to (3.14) yields the following result

|z(x, y)| ≤

"

c−cp(p−1)

n

X

i=1

Ψi(x, y)

#1−p1 ,

where

Ψi(x, y) = Z x

x0

Z y y0

ai(σ, τ)dτ dσ+ Z φ(x)

x0

Z ψ(y) y0

bi(σ, τ)dτ dσ.

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Explicit Bounds On some Nonlinear Integral Inequalities

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References

[1] D. BAINOV AND P. SIMEONOV, Integral Inequalities and Applications, Kluwer Academic Publishers, Dordrecht, 1992.

[2] S.S. DRAGOMIR, The Gronwall Type Lemmas and Applica- tions,Monografii Mathematica, Univ. Timisoara, No. 29, 1987.

[3] S.S. DRAGOMIR, On some nonlinear generalizations of Gronwall’s in- equality, Coll. Sci. Fac. Sci. Kraqujevac, Sec. Math. Tech. Sci., 13 (1992), 23–28.

[4] O. LIPOVAN, A retarded integral inequality and its applications, J. Math.

Anal. Appl., 285 (2003), 336–443.

[5] L. QU-IANG, The boundedness of solutions of linear differential equations y00+A(t)y = 0, Shuxue Jinzhan, 3 (1957), 409–415.

[6] B.G. PACHPATTE, On some new inequalities related to certain inequalities in the theory of differential equations, J. Math. Anal. Appl., 189 (1995), 128–144.

[7] B.G. PACHPATTE, Inequalities for Differential and Integral Equations, Academic Press, New York, 1998.

[8] B.G. PACHPATTE, Explicit bounds on certain integral inequalities, J. Math.

Anal. Appl., 267 (2002), 48–61.

[9] B.G. PACHPATTE, On some new nonlinear retarded integral inequalities, J. Inequal. Pure and Appl. Math., 5(3) (2004), Art. 80. [ONLINE:http:

//jipam.vu.edu.au/article.php?sid=436].

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