ON SOME NEW NONLINEAR RETARDED INTEGRAL INEQUALITIES WITH ITERATED INTEGRALS AND THEIR APPLICATIONS

J. Korean Math. Soc. 45 (2008), No. 2, pp. 331–353

ON SOME NEW NONLINEAR RETARDED INTEGRALINEQUALITIES WITH ITERATED INTEGRALS AND THEIR

APPLICATIONS

Qing-Hua Ma and Josip Pecaric

Reprinted from theJournal of the Korean Mathematical Society

Vol. 45, No. 2, March 2008

c©2008 The Korean Mathematical Society

J. Korean Math. Soc. 45 (2008), No. 2, pp. 331–353

ON SOME NEW NONLINEAR RETARDED INTEGRALINEQUALITIES WITH ITERATED INTEGRALS AND THEIR

APPLICATIONS

Qing-Hua Ma and Josip Pecaric

Abstract. Some new nonlinear retarded integral inequalities of Gronwa-ll-like type are established, which mainly generalized some results givenby Cho, Dragomir and Kim (J. Korean Math. Soc. 43 (2006), No. 3,pp. 563–578) and can be used in the analysis of various problems in thetheory of certain classes of differential equations and integral equations.Applications examples are also indicated.

1. Introduction

Integral inequalities are indispensable for us in the quantitative study of vari-ous differential equations and integral equations. Besides the famous Gronwall-Bellman inequality and its first nonlinear generalization by Bihari (see [1], [2],[13], [20], [21]), there are several other very useful Gronwall-like inequality.Haraux [12, Corollary 16, p.139] derived a Gronwall-like inequality and used itto prove the existence of solutions of wave equations with logarithmic nonlin-earities. On the other hand, Engler [11] utilized the following slight variant ofinequality due to Haraux [12] in the study of global regular solutions for thedynamic antiplane shear problem in nonlinear viscoelasticity.

Lemma 1.1. Let c > 1 and a ∈ C(I = [0, T ], R+), R+ = [0,∞), and assumethat the function w → [1,∞) satisfies

w(t) ≤ c

(1 +

∫ t

0

a(s)w(s) log w(s)ds

), t ∈ I.

Then

w(t) ≤ c exp(∫ t

0

a(s)ds

), t ∈ I.

Received July 16, 2006.2000 Mathematics Subject Classification. 26D15, 35A05.Key words and phrases. Gronwall-like integral inequalities, retarded, iterated integrals,

priori bound.The first author’s research was supported by the Research Group Grants Council of the

Guangdong University of Foreign Studies(Project No. GW2006-TB-002) of China.

c©2008 The Korean Mathematical Society

331

332 QING-HUA MA AND JOSIP PECARIC

Recently, Cho, Dragomir and Kim [6] proved the following Haraux-Engler-Bykov type inequality and other interesting related results:

Lemma 1.2. Let u(t), b(t), k(t, s) and h(t, s, τ) be nonnegative continuous forα ≤ σ ≤ τ ≤ s ≤ t ≤ β with u(t) ≥ 1. Suppose that

(1.1)u(t) ≤ a +

∫ t

α

b(s)u(s) log u(s)ds +∫ t

α

(∫ s

α

k(s, τ)u(τ) log u(τ)dτ

)ds

+∫ t

α

(∫ s

α

(∫ τ

α

h(s, τ, σ)u(σ) log u(σ)dσ

)dτ

)ds

for t ∈ I = [α, β], where a ≥ 1 is a constant. Then, for any t ∈ I,

u(t) ≤ aexp(A(t)),

where

A(t) =∫ t

α

b(s)ds +∫ t

α

(∫ s

α

k(s, τ)dτ

)ds

+∫ t

α

(∫ s

α

(∫ τ

α

h(s, τ, σ)dσ

)dτ

)ds.

The aim of the present paper is to establish some new nonlinear retardedinequalities, which generalize the inequalities discussed in the above lemmasand other results appeared in [6] and [14, 18]. Application examples are alsoincluded.

2. Some new nonlinear retarded inequalities

In what follows, R denotes the set of all real numbers, R+ = [0, +∞), R1 =[1, +∞), I = [α, β] are the give subsets of R. Ci(M, S) denotes the class ofall i-times continuously differentiable functions defined on set M with range inthe set S(i = 1, 2, . . .) and C0(M, S) = C(M, S).

Lemma 2.1. Let u(t), b(t), k(t, s) and h(t, s, τ) be nonnegative continuous func-tions for α ≤ σ ≤ τ ≤ s ≤ t ≤ β. Let φ(t) ∈ C1(I, I) with φ(t) ≤ t and g(u)be a nonnegative and nondecreasing continuous functions for u ∈ R+ withg(u) > 0 for u > 0. Suppose that

(2.1)

u(t) ≤ a +∫ φ(t)

φ(α)

b(s)g(u(s))ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)g(u(τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)g(u(σ))dσ

)dτ

)ds

ON SOME NEW NONLINEAR RETARDED INTEGRAL INEQUALITIES 333

for t ∈ I, where a ≥ 0 is a constant. Then

(2.2)

u(t) ≤ G−1

[G(a) +

∫ φ(t)

φ(α)

b(s)ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

]

for t ∈ [α, T1], where

(2.3) G(r) =∫ r

r0

ds

g(s), r ≥ r0 > 0

G−1 denotes the inverse function of G, and T1 ∈ I is chosen so that the quantityin the square brackets of (2.2) is in the range of G.

Proof. Let a > 0 and define a function v1(t) by the right-hand side of (2.1).Then v1(α) = a, v1(t) is positive and nondecreasing, u(t) ≤ v1(t) for t ∈ I and

v′1(t) = b(φ(t))g(u(φ(t)))φ′(t) +

(∫ φ(t)

φ(α)

k(φ(t), τ)g(u(τ))dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)g(u(σ))dσ

)dτ

)φ′(t)

≤ b(φ(t))g(v1(φ(t)))φ′(t) +

(∫ φ(t)

φ(α)

k(φ(t), τ)g(v1(τ))dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)g(v1(σ))dσ

)dτ

)φ′(t)

≤ g(v1(t))

[b(φ(t))φ′(t) +

(∫ φ(t)

φ(α)

k(φ(t), τ)dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)dσ

)dτ

)φ′(t)

],

i.e.,

v′1(t)g(v1(t))

≤ b(φ(t))φ′(t) +

(∫ φ(t)

φ(α)

k(φ(t), τ)dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)dσ

)dτ

)φ′(t).


Setting t = s and then integrating from α to t, t ∈ I, changing the variable andusing the definition of function G we get

G(v1(t)) ≤ G(a) +∫ φ(t)

φ(α)

b(s)ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

or

v1(t) ≤ G−1

[G(a) +

∫ φ(t)

φ(α)

b(s)ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

]

for t ∈ [α, T1]. Now by u(t) ≤ v1(t) and the last inequality, we get the desiredinequality (2.2) for a > 0. By continuity, (2.2) also holds for any a ≥ 0. ¤

Theorem 2.2. Let a, u, b, k, h, g and φ be as in Lemma 2.1. Let ϕ ∈ C1(R+,R+) with ϕ′(u) > 0 and ϕ′(u) is increasing for u > 0. Suppose that

(2.4)

ϕ(u(t)) ≤ a +∫ φ(t)

φ(α)

b(s)ϕ′(u(s))g(u(s))ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)ϕ′(u(τ))g(u(τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)ϕ′(u(σ))g(u(σ))dσ

)dτ

)ds

for t ∈ I. Then

(2.5)

u(t) ≤ G−1

[G(ϕ−1(a)) +

∫ φ(t)

φ(α)

b(s)ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

]

for t ∈ [α, T2], where T2 ∈ I is chosen so that the quantity in the square bracketsof (2.5) is in the range of G, G and G−1 are as in Lemma 2.1.

Proof. Let a > 0 and define a function v2(t) by the right-hand side of (2.4).Then v2(α) = a, v2(t) is positive and nondecreasing, u(t) ≤ ϕ−1(v2(t)) for t ∈ I


and

v′2(t) = b(φ(t))ϕ′(u(φ(t)))g(u(φ(t)))φ′(t)

+

(∫ φ(t)

φ(α)

k(φ(t), τ)ϕ′(u(τ))g(u(τ))dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)ϕ′(u(σ))g(u(σ))dσ

)dτ

)φ′(t)

≤ b(φ(t))ϕ′[ϕ−1(v2(φ(t))]g[ϕ−1(v2(φ(t)))]φ′(t)

+

(∫ φ(t)

φ(α)

k(φ(t), τ)ϕ′[ϕ−1(v2(τ))]g[ϕ−1(v2(τ))]dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)ϕ′[ϕ−1(v2(σ))]g[ϕ−1(v2(σ))]dσ

)dτ

)φ′(t)

≤ ϕ′[ϕ−1(v2(t))][b(φ(t))g[ϕ−1(v2(φ(t)))]φ′(t)

+

(∫ φ(t)

φ(α)

k(φ(t), τ)g[ϕ−1(v2(τ))]dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)g[ϕ−1(v2(σ))]dσ

)dτ

)φ′(t)

],

i.e.,

(2.6)

v′2(t)ϕ′[ϕ−1(v2(t))]

≤ b(φ(t))g[ϕ−1(v2(φ(t)))]φ′(t)

+

(∫ φ(t)

φ(α)

k(φ(t), τ)g[ϕ−1(v2(τ))]dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)g[ϕ−1(v2(σ))]dσ

)dτ

)φ′(t).

Observe that for any continuously differentiable and invertible function f(ξ),by a change of variable η = f−1(ξ), we have

(2.7)∫

dξ

f ′[f−1(ξ)]=

∫f ′(η)f ′(η)

dη = η + c = f−1(ξ) + c.

Setting t = s in (2.6), integrating with respect to s from α to t and using (2.7)to the left-hand side and changing variable in the right-hand side we obtain

ϕ−1(v2(t)) ≤ ϕ−1(a) +∫ φ(t)

φ(α)

b(s)g[ϕ−1(v2(s))]ds


+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)g[ϕ−1(v2(τ))]dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)g[ϕ−1(v2(σ))]dσ

)dτ

)ds.

Now, using Lemma 2.1 to the last inequality we have

ϕ−1(v2(t)) ≤ G−1

[G(ϕ−1(a)) +

∫ φ(t)

φ(α)

b(s)ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

]

for t ∈ [α, T2]. By u(t) ≤ ϕ−1(v2(t)), we get the desired inequality (2.5). Thecase a ≥ 0 follows as mentioned in the proof of Lemma 2.1. ¤

Letting ϕ(u) = up(p ≥ 1 is a constant) in Theorem 2.2, we obtain the follow-ing Ou-Iang type retarded integral inequality with iterated integrals. AboutOu-Iang type inequalities and their generalizations and applications, one cansee [21].

Corollary 2.3. Let a, u, b, k, h, g and φ be as in Lemma 2.1, p ≥ 1 be a con-stant. Suppose that

(2.8)

up(t) ≤ a +∫ φ(t)

φ(α)

b(s)up−1(s)g(u(s))ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)up−1(τ)g(u(τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)up−1(σ)g(u(σ))dσ

)dτ

)ds

for t ∈ I. Then

(2.9)

u(t) ≤ G−1

[G(a1/p) +

∫ φ(t)

φ(α)

b(s)p

ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)p

dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)p

dσ

)dτ

)ds

]



Corollary 2.4. Let b, k, h, g and φ be as in Lemma 2.1, a > 1 and p > 0 beconstants and u ∈ C(I, R1). Suppose that

(2.10)

up(t) ≤ a +∫ φ(t)

φ(α)

b(s)up(s)g(log u(s))ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)up(τ)g(log u(τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)up(σ)g(log u(σ))dσ

)dτ

)ds

for t ∈ I. Then(2.11)

u(t) ≤ exp

(G−1

[G(

1p

log a) +∫ φ(t)

φ(α)

b(s)p

ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)p

dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)p

dσ

)dτ

)ds

])


Proof. Taking v(t) = log u(t), then inequality (2.10) reduces to

exp(pv(t)) ≤ a +∫ φ(t)

φ(α)

b(s) exp(pv(s))g(v(s))ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ) exp(pv(τ))g(v(τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ) exp(pv(σ))g(v(σ))dσ

)dτ

)ds

which is a special case of inequality (2.4) when ϕ = exp(pv). By Theorem 2.2,we get the desired inequality (2.11) directly. ¤

Remark 2.1. (i) When p = 1, g(u) = u and φ(t) = t, from Corollary 2.4 we canderive Theorem 2.1 [6].

(ii) If constant a is placed by a nondecreasing continuous function a(t) onI, all conclusions in Lemma 2.1 to Corollary 2.4 are still valid.

Theorem 2.5. Let u, b, k, h and φ be as in Corollary 2.4, g be as in Lemma2.1 with the subadditivity and multiplicity and π ∈ C(I, R1). Suppose that

up(t) ≤ π(t)

{a +

∫ φ(t)

φ(α)

b(s)uq(s)g(log u(s))ds


(2.12)

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)uq(τ)g(log u(τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)uq(σ)g(log u(σ))dσ

)dτ

)ds

}

for t ∈ I, where p ≥ q > 0 and a ≥ 1 are a constant. Then, if p = q

(2.13)

u(t) ≤ π1/p(t) exp

{1pG−1

[G(B1(t)) +

∫ φ(t)

φ(α)

b(s)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

]}

for t ∈ [α, T5], where

(2.14)

B1(t) = log a +∫ φ(t)

φ(α)

b(s)g(log π1/p(s))ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)g(log π1/p(τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)g(log π1/p(σ))dσ

)dτ

)ds

for t ∈ [α, T5]; if p > q,

(2.15)

u(t) ≤ π1/p(t) exp

{1

p− qG−1

[G(B1(t)) +

∫ φ(t)

φ(α)

b(s)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

]}

for t ∈ [α, T6], where T5, T6 ∈ I are chosen so that the quantity in the squarebrackets of (2.13) and (2.15) are in the range of G respectively, G and G−1 areas in Lemma 2.1.


Proof. Define a function v3(t) by

v3(t) = a +∫ φ(t)

φ(α)

b(s)g(log u(s))ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)g(log u(τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)g(log u(σ))dσ

)dτ

)ds.

Then v3(t) is nondecreasing for t ∈ I, v3(α) = a and

(2.16) u(t) ≤ π1/p(t)v1/p3 (t)

for t ∈ I. Differentiating v3(t), we get

v′3(t) = b(φ(t))uq(φ(t))g(log u(φ(t)))φ′(t)

+

(∫ φ(t)

φ(α)

k(φ(t), τ)uq(τ)g(log u(τ))dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)uq(σ)g(log u(σ))dσ

)dτ

)φ′(t)

≤ b(φ(t))π1/p(φ(t))vq/p3 (φ(t))g(log[π1/p(φ(t))vq/p

3 (φ(t))])φ′(t)

+

(∫ φ(t)

φ(α)

k(φ(t), τ)π1/p(τ)vq/p3 (τ)g(log[π1/p(τ)vq/p

3 (τ)])dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)π1/p(σ)vq/p3 (σ)g(log[π1/p(σ)vq/p

3 (σ)])dσ

)dτ

)φ′(t)

≤ vq/p3 (t)

[b(φ(t))π1/p(φ(t))g(log[π1/p(φ(t))vq/p

3 (φ(t))])φ′(t)

+

(∫ φ(t)

φ(α)

k(φ(t), τ)π1/p(τ)g(log[π1/p(τ)vq/p3 (τ)])dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)π1/p(σ)g(log[π1/p(σ)vq/p3 (σ)])dσ

)dτ

)φ′(t)

].

It follows that(2.17)

v′3(t)

vq/p3 (t)

≤ b(φ(t))π1/p(φ(t))g(log[π1/p(φ(t))vq/p3 (φ(t))])φ′(t)

+

(∫ φ(t)

φ(α)

k(φ(t), τ)π1/p(τ)g(log[π1/p(τ)vq/p3 (τ)])dτ

)φ′(t)

+

(∫ φ(t)

φ(α)

(∫ τ

φ(α)

h(φ(t), τ, σ)π(σ)g(log[π1/p(σ)vq/p3 (σ)])dσ

)dτ

)φ′(t).


If p = q, setting t = s in (2.17) and then integrating from α to t, t ∈ I, changingthe variables and using the subadditivity of function g we get

log v3(t) ≤ B1(t) +∫ φ(t)

φ(α)

b(s)g(log v3(s))ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)g(log v3(τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)g(log v3(σ))dσ

)dτ

)ds,

where B1(t) is defined by (2.14) and obviously B1(t) is nondecreasing contin-uous on I. By Lemma 2.1 and Remark 2.1 we get(2.18)

v3(t) ≤ exp

{G−1

[G(B1(t)) +

∫ φ(t)

φ(α)

b(s)ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

]}

for t ∈ [α, T5].If p > q, setting t = s in (2.17) and then integrating from α to t, t ∈ I,

changing the variables and using the multiplicity of function g we get

vp−q

p

3 (t) ≤ B1(t) +∫ φ(t)

φ(α)

b(s)g(q

p− q)g(log v

p−qp

3 (s))ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)g(q

p− q)g(log v

p−qp

3 (τ))dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)g(q

p− q)g(log v

p−qp

3 (σ))dσ

)dτ

)ds,

where B1(t) is defined by (2.14). By Lemma 2.1 and Remark 2.1 we get

vp−q

p

3 (t) ≤ exp

{G−1

[G(B1(t)) +

∫ φ(t)

φ(α)

b(s)ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds

+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

]}

or

v3(t) ≤ exp

{p

p− qG−1

[G(B1(t)) +

∫ φ(t)

φ(α)

b(s)ds +∫ φ(t)

φ(α)

(∫ s

φ(α)

k(s, τ)dτ

)ds


+∫ φ(t)

φ(α)

(∫ s

φ(α)

(∫ τ

φ(α)

h(s, τ, σ)dσ

)dτ

)ds

]}

for t ∈ [α, T6]. Now, by using (2.18) and the last inequality in (2.16), weget the desired inequalities (2.13) and (2.15), respectively. This completes theproof. ¤

3. Some more generalized inequalities

In this section, we give some more generalizations of the results obtained insection 2. Let α < β and set Ji = {(t1, t2, . . . , ti) ∈ Ri : α ≤ ti ≤ · · · ≤ t1 ≤ β}for i = 1, . . . , n.

Theorem 3.1. Let a, u(t), φ(t) and g be as in Lemma 2.1, ϕ be as in Theorem2.2. Suppose that(3.1)

ϕ(u(t))

≤ a +∫ φ(t)

φ(α)

k1(t, t1)ϕ′(u(t1))g(u(t1))dt1 + · · ·

+∫ φ(t)

φ(α)

(∫ t1

φ(α)

· · ·(∫ tn−1

φ(α)

kn(t, t1, . . . , tn)ϕ′(u(tn))g(u(tn))dtn

)· · ·

)dt1

for t ∈ I, where ki(t, t1, . . . , ti) are nonnegative, continuous functions in Ji+1

and the partial derivative ∂ki

∂t (t, t1, . . . , ti) exist and are nonnegative, continuousin Ji+1 for i = 1, 2, . . . , n. Then

(3.2) u(t) ≤ G−1

{G(ϕ−1(a)) +

∫ φ(t)

φ(α)

R[1](τ, τ)dτ +∫ t

α

Q[1](τ)dτ

}

for t ∈ [α, T7], where T7 is chosen so that the quality in the braces of (3.2) inthe range of G,(3.3)

R[x](t, t) = k1(t, t)x(t) +∫ φ(t)

φ(α)

k2(t, t, t2)x(t2)dt2

+n∑

i=3

∫ φ(t)

φ(α)

(∫ t2

φ(α)

· · ·(∫ ti−1

φ(α)

ki(t, t, t2, . . . , ti)x(ti)dti

)· · ·

)dt2,

Q[x](t) =∫ φ(t)

φ(α)

∂k1

∂t(t, t1)x(t)dt1

+n∑

i=2

∫ φ(t)

φ(α)

(∫ t2

φ(α)

· · ·(∫ ti−1

φ(α)

∂ki

∂t(t, t1, . . . , ti)x(t)dti

)· · ·

)dt1

for any x(t) ∈ C(I, I) and t, t ∈ I,G and G−1 are defined as in Lemma 2.1.


Proof. Define a function v(t) by the right side of (3.1). Then v(t) is nonde-creasing continuous, u(t) ≤ ϕ−1(v(t)) for t ∈ I and v(α) = a. Taking derivativeto v(t), we have

v′(t) = R[ϕ′(u)g(u)](t, φ(t))φ′(t) + Q[ϕ′(u)g(u)](t)

≤{

R[g(ϕ−1(v)](t, φ(t))φ′(t) + Q[g(ϕ−1(v)](t)}

ϕ′(ϕ−1(v(t))),

orv′(t)

ϕ′(ϕ−1(v(t)))≤ R[g(ϕ−1(v)](t, φ(t))φ′(t) + Q[g(ϕ−1(v)](t).

As the procedures in the proof of Theorem 2.2, setting t = s in the last in-equality and then integrating from α to t, t ∈ I, changing the variables to theright-hand side first integral we get

(3.4) ϕ−1(v(t)) ≤ ϕ−1(a)+∫ φ(t)

φ(α)

R[g(ϕ−1(v)](τ, τ)dτ +∫ t

α

Q[g(ϕ−1(v)](τ)dτ.

Now, we denote the right-hand side of (3.4) by v(t). Then v(α) = ϕ−1(a),ϕ−1(v(t)) ≤ v(t), the function v(t) is nondecreasing in t ∈ [α, β] and

v′(t) ≤{

R[1](t, φ(t))φ′(t) + Q[1](t)}

g(v(t)).

Integrating the last inequality from α to t, t ∈ I and changing the variables tothe first integral we get

G(v(t)) ≤ G(v(α)) +∫ φ(t)

φ(α)

R[1](τ, τ)dτ +∫ t

α

Q[1](τ)dτ

or

(3.5) v(t) ≤ G−1

{G(ϕ−1(a)) +

∫ φ(t)

φ(α)

R[1](τ, τ)dτ +∫ t

α

Q[1](τ)dτ

}

for t ∈ [α, T ], where T is chosen so that the quality in the braces of (3.5) inthe range of G. Now the desired inequality (3.2) follows by the inequalitiesu(t) ≤ ϕ−1(v(t)) ≤ v(t) and (3.5). This completes the proof. ¤

Let ϕ(u) = up in Theorem 3.1, we get the following Ou-Iang type retardedintegral inequality with iterated integrals immediately.

Corollary 3.2. Let a, u, g, φ and ki(i = 1, 2, . . . , n) be as in Theorem 3.1,p ≥ 1 is a constant. Suppose that(3.6)

up(t)

≤ a +∫ φ(t)

φ(α)

k1(t, t1)up−1(t1)g(u(t1))dt1 + · · ·

+∫ φ(t)

φ(α)

(∫ t1

φ(α)

· · ·(∫ tn−1

φ(α)

kn(t, t1, . . . , tn)up−1(tn)g(u(tn))dtn

)· · ·

)dt1


for t ∈ I. Then

(3.7) u(t) ≤ G−1

{G(a1/p) +

1p

∫ φ(t)

φ(α)

R[1](τ, τ)dτ +1p

∫ t

α

Q[1](τ)dτ

}

for t ∈ [α, T8], where T8 is chosen so that the quality in the braces of (3.7) inthe range of G,R, Q, G and G−1 are defined as in Theorem 3.1.

As similar argument as in Corollary 2.4, we can get the following interestingresult from Theorem 3.1:

Corollary 3.3. Let g, φ and ki(i = 1, 2, . . . , n) be as in Theorem 3.1, a >1, p ≥ 0 be constants and u ∈ C(I,R1). Suppose that(3.8)

up(t)

≤ a +∫ φ(t)

φ(α)

k1(t, t1)up(t1)g(log u(t1))dt1 + · · ·

+∫ φ(t)

φ(α)

(∫ t1

φ(α)

· · ·(∫ tn−1

φ(α)

kn(t, t1, . . . , tn)up(tn)g(log u(tn))dtn

)· · ·

)dt1

for t ∈ I. Then(3.9)

u(t) ≤ exp

(G−1

{G(a1/p) +

1p

∫ φ(t)

φ(α)

R[1](τ, τ)dτ +1p

∫ t

α

Q[1](τ)dτ

})

for t ∈ [α, T9], where T9 is chosen so that the quality in the braces of (3.9) inthe range of G, R,Q, G and G−1 are defined as in Theorem 3.1.

Remark 3.1. Let p = 1, φ(t) ≡ t and g(u) = u in Corollary 3.3, we can get asimilar result as Theorem 3.1 in [6].

Theorem 3.4. Let u and φ be as in Lemma 2.1, a ∈ Cn−1(R+, R+), p1, . . . , pn,f1, . . . , fn be nonnegative, continuous functions in I. Let g(u) be a continuouslydifferentiable function defined on R+, g(u) > 0 on (0,∞), g′(u) ≥ 0 for u ∈ R+

and ϕ ∈ C2(R+, R+) with ϕ′(u) > 0 for u > 0. Suppose that(3.10)

ϕ(u(t))

≤ a(t) +∫ φ(t)

φ(α)

p1(s)f1(s)ϕ′(u(s))g(u(s))ds

+∫ φ(t)

φ(α)

p1(t1)∫ t1

φ(α)

p2(s)f2(s)ϕ′(u(s))g(u(s))dsdt1 + · · ·

+∫ φ(t)

φ(α)

p1(t1)

(∫ t1

φ(α)

p2(t2) · · ·(∫ tn−1

φ(α)

pn(s)fn(s)ϕ′(u(s))g(u(s))ds · · ·)

dt2

)dt1


for t ∈ I. Then

(3.11) u(t) ≤ G−1{G[ϕ−1(a(t))] + F (t)

}

for t ∈ [α, T10], where(3.12)

F (t) =∫ φ(t)

φ(α)

p1(s)f1(s)ds +∫ φ(t)

φ(α)

p1(t1)∫ t1

φ(α)

p2(s)f2(s)dsdt1 + · · ·

+∫ φ(t)

φ(α)

p1(t1)

(∫ t1

φ(α)

p2(t2) · · ·(∫ tn−1

φ(α)

pn(s)fn(s)ds · · ·)

dt2

)dt1,

T10 is chosen so that the quality in the braces of (3.11) in the range of G, Gand G−1 are same as in Lemma 2.1.

Proof. Define a function v(t) by the right side of (3.10). Then v(t) is non-decreasing continuous, u(t) ≤ ϕ−1(v(t)) for t ∈ I and v(α) = a(t). Thendifferentiating v(t) and rewriting we have

(3.13)v′(t)− a′(t)φ′(t)p1(φ(t))

− f1(φ(t))ϕ′(u)g(u) = v1(t),

wherev1(t)

=

∫ φ(t)

φ(α)

p2(s)f2(s)ϕ′(u(s))g(u(s))ds + · · ·

+

∫ φ(t)

φ(α)

p2(t2)

(∫ t2

φ(α)

p3(t3) · · ·(∫ tn−1

φ(α)


dt3

)dt2.

From the definition of v1(t) we get

(3.14)v′1(t)

φ′(t)p2(φ(t))− f2(φ(t))ϕ′(u)g(u) = v2(t),

wherev2(t)

=

∫ φ(t)

φ(α)

p3(s)f3(s)ϕ′(u(s))g(u(s))ds + · · ·

+

∫ φ(t)

φ(α)

p3(t3)

(∫ t3

φ(α)

p4(t4) · · ·(∫ tn−1

φ(α)


dt4

)dt3.

Continuing in this way, we obtain

(3.15)v′n−2(t)

φ′(t)pn−1(φ(t))− fn−1(φ(t))ϕ′(u)g(u) = vn−1(t),

where

vn−1(t) =∫ φ(t)

φ(α)

pn(s)fn(s)ϕ′(u(s))g(u(s))ds,


from the definition of vn−1(t) and using u(t) ≤ ϕ−1(v(t)) we can get

v′n−1

ϕ′(ϕ−1(v(t)))≤ φ′(t)pn(φ(t))fn(φ(t))g(ϕ−1(v(φ(t)))).

Integrating the last inequality and changing the variable to the right-hand sideintegral we have

(3.16)∫ t

α

v′n−1

ϕ′(ϕ−1(v(s)))ds ≤

∫ φ(t)

φ(α)

pn(s)fn(s)g(ϕ−1(v(s)))ds.

By integrating by parts to the left-hand side of (3.16) we obtain

(3.17)

∫ t

α

v′n−1(s)ϕ′(ϕ−1(v(s)))

ds

=vn−1

ϕ′(ϕ−1(v(t)))

+∫ t

α

vn−1(s)ϕ′2(ϕ−1(v(s)))

× ϕ′′(ϕ−1(v(s)))ϕ′((v(s)))

× v′(s)ds

≥ vn−1

ϕ′(ϕ−1(v(t))).

From (3.16) and (3.17) we have

(3.18)vn−1

ϕ′(ϕ−1(v(t)))≤

∫ φ(t)

φ(α)

pn(s)fn(s)g(ϕ−1(v(s)))ds.

Now by (3.15) and (3.18) we observe that

∫ t

α

v′n−2(s)ϕ′(ϕ−1(v(s)))

ds

≤ v′n−1(t)ϕ′(ϕ−1(v(t)))

pn−1(φ(t))φ′(t)

+ pn−1(φ(t))fn−1(φ(t))g(ϕ−1(v(φ(t))))φ′(t).

Using the same procedures from (3.16) to (3.17) to the last inequality we have

vn−2

ϕ′(ϕ−1(v(t)))≤

∫ φ(t)

φ(α)

pn−1(s)fn−1(s)g(ϕ−1(v(s)))ds

+∫ φ(t)

φ(α)

pn−1(t1)∫ t1

φ(α)

pn(s)fn(s)g(ϕ−1(v(s)))dsdt1.


Proceeding in this way we can get(3.19)

v1

ϕ′(ϕ−1(v(t)))

≤∫ φ(t)

φ(α)

p2(s)f2(s)g(ϕ−1(v(s)))ds + · · ·

+∫ φ(t)

φ(α)

p2(t2)

(∫ t2

φ(α)

p3(t3) · · ·(∫ tn−1

φ(α)

pn(s)fn(s)g(ϕ−1(v(s)))ds · · ·

)dt3

)dt2.

From (3.19), (3.13) and using u(t) ≤ ϕ−1(v(t)), we observe that

(3.20)

v′

ϕ′(ϕ−1(v(t)))− a′

ϕ′(ϕ−1(a(t)))

≤ v′

ϕ′(ϕ−1(v(t)))− a′

ϕ′(ϕ−1(v(t)))

= p1(φ(t))f1(φ(t))ϕ′(u)g(u)φ′(t) +v1

ϕ′(ϕ−1(v(t)))p1(φ(t))φ′(t).

Integrating (3.20) from α to t, and then using (2.7) to the left-hand side inte-grals and using (3.19) to the right-hand side integrals and changing variableswe can get(3.21)

ϕ−1(v(t))− ϕ−1(a(t))

≤∫ φ(t)

φ(α)

p1(s)f1(s)g(ϕ−1(v(s)))ds

+∫ φ(t)

φ(α)

p1(t1)∫ t1

φ(α)

p2(s)f2(s)g(ϕ−1(v(s)))dsdt1 + · · ·

+∫ φ(t)

φ(α)

p1(t1)

(∫ t1

φ(α)

p2(t2) · · ·(∫ tn−1

φ(α)

pn(s)fn(s)g(ϕ−1(v(s)))ds · · ·

)dt2

)dt1.

Denoting z(t) = ϕ−1(v(t)) and l(t) = ϕ−1(a(t)), from (3.21) we have(3.22)

z(t) ≤ l(t) +∫ φ(t)

φ(α)

p1(s)f1(s)g(z(s))ds

+∫ φ(t)

φ(α)

p1(t1)∫ t1

φ(α)

p2(s)f2(s)g(z(s))dsdt1 + · · ·

+∫ φ(t)

φ(α)

p1(t1)

(∫ t1

φ(α)

p2(t2) · · ·(∫ tn−1

φ(α)

pn(s)fn(s)g(z(s))ds · · ·

)dt2

)dt1.


Now, let V (t) be the right-hand side of (3.22), then z(t) ≤ V (t), using samesteps as used from (3.13) to (3.20) we can get(3.23)

V1

g(V (t))≤

∫ φ(t)

φ(α)

p2(s)f2(s)ds + · · ·

+∫ φ(t)

φ(α)

p2(t2)

(∫ t2

φ(α)

p3(t3) · · ·(∫ tn−1

φ(α)


dt3

)dt2.

and

(3.24)

V ′

g(V (t))− l′

g(l(t))

≤ V ′

g(V (t))− l′

g(V (t))

= p1(φ(t))f1(φ(t))φ′(t) +V1

g(V (t))p1(φ(t))φ′(t),

where

(3.25) V1(t) =V ′(t)− l′(t)φ′(t)p1(φ(t))

− f1(φ(t))g(z).

Integrating (3.24) from α to t, using the definition of G and z(t) ≤ V (t) weobtain

G(V (t))−G(l(t))

≤∫ φ(t)

φ(α)


φ(α)

p1(t1)∫ t1

φ(α)

p2(s)f2(s)dsdt1 + · · ·

+∫ φ(t)

φ(α)

p1(t1)

(∫ t1

φ(α)

p2(t2) · · ·(∫ tn−1

φ(α)


dt2

)dt1

or(3.26)

V (t)

≤ G−1

{G(l(t)) +

∫ φ(t)

φ(α)


φ(α)

p1(t1)∫ t1

φ(α)

p2(s)f2(s)dsdt1 + · · ·

+∫ φ(t)

φ(α)

p1(t1)

(∫ t1

φ(α)

p2(t2) · · ·(∫ tn−1

φ(α)


dt2

)dt1

}

for t ∈ [α, T10], T10 is chosen so that the quality in the braces of (3.26) in therange of G. Now from u(t) ≤ ϕ−1(v(t)) = z(t) ≤ V (t) and (3.26) we can getthe desired inequality (3.11). This completed the proof. ¤

Remark 3.2. (i) When ϕ(u) = u, φ(t) = t in Theorem 3.4, we can get Theorem1 which was given by Medved [18]; (ii) When ϕ(u) = u, a slightly different


version of Theorem 3.4 was also given in [18]; (iii) When f1 = f2 = · · · =fn−2 = 0, φ(t) = t, form Theorem 3.4 we can get Theorem 2.1 given in [14].

By a similar argument as in the proof of Corollary 2.4, we have an interestingresult as follows:

Corollary 3.5. Let φ, pi, and fi(i = 1, 2, . . . , n) be as in Theorem 3.4, u ∈C(R+, R1), a ∈ C1(R+, R1) and p > 0 be a constant. Suppose that(3.27)

up(t)

≤ a(t) +∫ φ(t)

φ(α)

p1(s)f1(s)up(s)g(log u(s))ds

+∫ φ(t)

φ(α)

p1(t1)∫ t1

φ(α)

p2(s)f2(s)up(s)g(log u(s))dsdt1 + · · ·

+∫ φ(t)

φ(α)

p1(t1)

(∫ t1

φ(α)

p2(t2) · · ·(∫ tn−1

φ(α)

pn(s)fn(s)up(s)g(log u(s))ds · · ·

)dt2

)dt1

for t ∈ I, then

(3.28) u(t) ≤ exp(

G−1

{G[

1p

log a(t))] + Fp(t)})

,

for t ∈ [α, T10], where T10 is chosen so that the quality in the braces of (3.28)in the range of G, G and G−1 are same as in Lemma 2.1, Fp(t) = 1

pF (t), F (t)is defined as in (3.12).

Let p = 1, g(u) = u in Corollary 3.5 we have

Corollary 3.6. Let u, a, φ, pi, and fi(i = 1, 2, . . . , n) be as in Corollary 3.5.Suppose that(3.29)

u(t)

≤ a(t) +∫ φ(t)

φ(α)

p1(s)f1(s)u(s) log u(s)ds

+∫ φ(t)

φ(α)

p1(t1)∫ t1

φ(α)

p2(s)f2(s)u(s)) log u(s)dsdt1 + · · ·

+∫ φ(t)

φ(α)

p1(t1)

(∫ t1

φ(α)

p2(t2) · · ·(∫ tn−1

φ(α)

pn(s)fn(s)u(s) log u(s)ds · · ·

)dt2

)dt1

for t ∈ I. Then

(3.30) u(t) ≤ (a(t))exp(F (t))

for t ∈ I, where F (t) is defined in (3.12).


Remark 3.3. When a(t) = a (a is a constant) fi = 1(i = 1, 2, . . . , n) andφ(t) = t in Corollary 3.6, [6] established a explicit bound of the solutions of(3.29), which is not only different to the bound given in (3.30) but also is validonly on a subset of I.

4. Applications

(A) Consider the following retarded integro-differential equation

(4.1) (h(t)x′)′ = F

(t, x(t− τ(t)),

∫ t

α

G(σ, x(σ − τ(σ)))dσ

)

for t ∈ I, where h is positive continuous on I, τ ∈ C1(I, I) is nonincreasing witht− τ(t) ≥ 0, τ(α) = 0, t− τ(t) ∈ C1(I, I), τ ′(t) < 1, F ∈ C(I × R × R, R), G ∈C(I ×R,R).

Theorem 4.1. Assume that the conditions

(4.2) |F (t, u, v)| ≤ b1(t)|u|g(log |u|) + b2(t)|v|,

(4.3) |G(s, v)| ≤ b3(s)|v|g(log |v|),

are satisfied, where bi(t) ∈ C(I, R+)(i = 1, 2, 3) and function g is defined inLemma 2.1. If x(t) is any solution of (4.1), then(4.4)

|x(t)| ≤ exp

(G−1

[G(log a) +

∫ t−τ(t)

α

MH(t, η + τ(s))b1(η + τ(s))dη

+∫ t−τ(t)

α

∫ η

α

M2H(t, η + τ(s))b2(η + τ(s))b3(ζ + τ(ξ))dζdη

])− 1

for t, s, ξ ∈ [α, T ∗1 ], where M = maxt∈I

11− τ ′(t)

, T ∗1 ∈ I is chosen so that the

quantity in the square brackets of (4.4) is in the range of G, G and G−1 are asin Lemma 2.1.

Proof. Let x(t) be any solution of (4.1) which is passing through the initialpoint (x(α), x′(α)). Integrating the both side of (4.1) two times and using thewell-known Dirichlet formula we obtain

(4.5)x(t) = x(α) + h(α)x′(α)H(t, α)

+∫ t

α

H(t, s)F[s, x(s− τ(s)),

∫ s

α

G(σ, x(σ − τ(σ)))dσ

]ds,


Since T is arbitrary, from (4.7) with T replaced by t we have

|x(t)| ≤ exp

(G−1

[G(log a) +

∫ t−τ(t)

α

MH(t, η + τ(s))b1(η + τ(s))dη

+∫ t−τ(t)

α

∫ η

α

M2H(t, η + τ(s))b2(η + τ(s))b3(ζ + τ(ξ))dζdη

])− 1,

which is the desired inequality (4.4). ¤

(B) Consider the retarded differential equation

(4.8) Dnx(t) = F (t, x(t− τ(t))), t ∈ I,

(4.9) Dix(α) = Ci−1, i = 1, 2, . . . , n,

where

D0x(t) = x(t), Dix(t) =1

pi(t)d

dt(Di−1x(t)), 1 ≤ i ≤ n,

with pn(t) = 1, pi(t) > 0(i = 1, 2, . . . , n − 1) are some continuous functionsdefined on I; Ci−1, 1 ≤ i ≤ n are constants; function τ is defined as in Theorem4.1. For convenience, for t ∈ I and some continuous functions qj(t) > 0, j =1, 2, . . . , n− 1 and q ≥ 0 defined on I, we define(4.10)

N [t, q1, q2, . . . , qn−1] =∫ t

α

q1(t1)∫ t1

α

q2(t2) · · ·∫ tn−1

α

q(tn)dtndtn−1 · · · dt1,

where t0 = t.

Theorem 4.2. Assume that the condition

(4.11) |F (t, u)| ≤ f(t)|u|g(log |u|)is satisfied, where f(t) ∈ C(I, R+), F ∈ C(I × R, R) and function g is definedin Lemma 2.1. If x(t) is any solution of (4.1), then

(4.12) |x(t)| ≤ exp(G−1{G[log(1 + |b(t)|)] + N(t)}

)− 1,

for t, t1, . . . , tn−1 ∈ [α, T ∗2 ], where

N(t) = Mn−1

∫ t−τ(t)

α

q1(s1 + τ(t1))∫ s1

α

q2(s2 + τ(t2)) · · ·

×∫ sn−2

α

qn−1(sn−1 + τ(tn−1))∫ sn−1

α

f(sn + τ(tn))dsndsn−1 · · · ds1,

T ∗2 is chosen so that the quality in the braces of (4.12) in the range of G, Gand G−1 are same as in Lemma 2.1.


Proof. Let x(t) be any solution of (4.8) with initial value (4.9). Then it is easyto observe that x(t) satisfies

(4.13) x(t) = b(t) + N [t, p1, p2, . . . , pn−1, F (t, x(t− τ(t)))], t ∈ I,

where b(t) = C0 +n−1∑

i=1

CiN [t, p1, . . . , pi]. Using the condition (4.11) to (4.13)

and changing the variables to N [t, p1, p2, . . . , pn−1, F (t, x(t− τ(t)))] we can get

|x(t)|+ 1

≤ 1 + |b(t)|+ N [t, p1, p2, . . . , |F (t, x(t− τ(t)))|]≤ 1 + |b(t)|+ N [t, p1, p2, . . . , f(t)(|x(t− τ(t)|+ 1)g(log |x(t− τ(t))|+ 1))]

≤ 1 + |b(t)|+ N [t− τ(t), p1, p2, . . . , pn−1, f · (|x|+ 1)g(log(|x|+ 1))],

whereN [t− τ(t), p1, p2, . . . , pn−1, f · (|x|+ 1)g(log(|x|+ 1))]

= Mn−1

∫ t−τ(t)

α

q1(s1 + τ(t1))∫ s1

α

q2(s2 + τ(t2)) · · ·∫ sn−2

α

qn−1(sn−1 + τ(tn−1))

×∫ sn−1

α

f(sn + τ(tn))(|x(sn)|+ 1)g(log(|x(sn)|+ 1))dsndsn−1 · · · ds1.

An application of Corollary 3.5 to the last inequality yields the desired inequal-ity (4.12). ¤

Obviously, under our assumptions to equations (4.1) and (4.8)-(4.9), a suffi-cient condition for boundedness of all solutions defined on suitable intervals isthat the integrals in (4.4) and (4.12) are boundedness.

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Qing-Hua MaDepartment of Applied MathematicsFaculty of Information Science and TechnologyGuangdong University of Foreign StudiesGuangzhou 510420, P. R. ChinaE-mail address: [email protected]

Josip PecaricFaculty of Textile TechnologyUniversity of ZagrebPierottijeva 6, 10000 Zagreb, CroatiaE-mail address: [email protected]

ON SOME NEW NONLINEAR RETARDED INTEGRAL INEQUALITIES WITH ITERATED INTEGRALS AND THEIR APPLICATIONS

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