Sen 1 1 Department of Mathematical Sciences, Florida Institute of Technology, 150 West University Boulevard, Melbourne, FL 32901, USA 2 Department of Applied Mathematics, College of Natu
Trang 1Volume 2008, Article ID 908784, 15 pages
doi:10.1155/2008/908784
Research Article
New Retarded Integral Inequalities
with Applications
Ravi P Agarwal, 1 Young-Ho Kim, 2 and S K Sen 1
1 Department of Mathematical Sciences, Florida Institute of Technology, 150 West University Boulevard, Melbourne, FL 32901, USA
2 Department of Applied Mathematics, College of Natural Sciences Changwon National University Changwon, Kyeongnam 641-773, South Korea
Correspondence should be addressed to Young-Ho Kim, yhkim@changwon.ac.kr
Received 29 January 2008; Accepted 24 April 2008
Recommended by Yeol Je Cho
Some new nonlinear integral inequalities of Gronwall type for retarded functions are established, which extend the results Lipovan 2003 and Pachpatte 2004 These inequalities can be used as basic tools in the study of certain classes of functional differential equations as well as integral equations A existence and a uniqueness on the solution of the functional differential equation involving several retarded arguments with the initial condition are also indicated.
Copyright q 2008 Ravi P Agarwal et al This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1 Introduction
t
a
a
hence furnishes a handy tool in the study of solutions of differential equations Because of its
Trang 2fundamental importance, several generalizations and analogous results of the inequality1.2 have been established over years Such generalizations are, in general, referred to as Gronwall
of differential equations, integral equations, and inequalities of various types Many authors
It is useful in the study of boundedness of certain second-order differential equations
Theorem A Ou-Yang 15 If u and f are nonnegative continuous functions on 0, ∞ such that
u2t ≤ u2
0 2
t
0
for all t ∈ 0, ∞, where u0≥ 0 is a constant, then
u t ≤ u0
t
0
The Ou-Yang inequality prompted researchers to devote considerable time for its
Theorem B Lipovan 14 Let u, f, and g be continuous nonnegative functions on Rand c be a nonnegative constant Also, let w ∈ CR, R be a nondecreasing function with wu > 0 on 0, ∞
and α ∈ C1R, R be nondecreasing with αt ≤ t on R If
u2t ≤ c2 2
α t
0
for all t ∈ R, then for 0≤ t ≤ t1,
u t ≤ Ω−1
Ω
c
α t
0
α t
0
Ωr
r
1
ds
w s , r > 0,
1.6
Ω−1 is the inverse function of Ω, and t1 ∈ R is chosen so that Ωc α t
0 g sds α t
DomΩ−1 for all t ∈ Rlying in the interval 0 ≤ t ≤ t1.
equations and functional differential equations
Theorem C Pachpatte 20 Let u, a i , b i ∈ CI, R and let α i ∈ C1I, I be nondecreasing with
α i t ≤ t on I for i 1, , n Let p > 1 and c ≥ 0 be constants and w ∈ CR, R be nondecreasing
with w u > 0 on 0, ∞ If for t ∈ I,
u p t ≤ c p n
i 1
α i t
Trang 3then for t0 ≤ t ≤ t1,
i 1
α i t
α i t0 a i σdσ
1/p−1
where
r
r0
ds
w
s 1/p−1 , r ≥ r0> 0,
A t c p−1/p p − 1 n
i 1
α i t
α i t0 b i σdσ,
1.9
r0 > 0 is arbitrary, G−1is the inverse function of G and t1∈ I is so chosen that
i 1
α i t
α i t0 a i σdσ ∈ DomG−1
The present paper establishes some nonlinear retarded inequalities which extend the foregoing theorems In addition, it illustrates the use/application of these inequalities
2 Main results
Theorem 2.1 Let u, f i , g i ∈ CI, R, i 1, , n, and let α i ∈ C1I, I be nondecreasing with
α i t ≤ t, i 1, , n Suppose that c ≥ 0 and q > 0 are constants, ϕ ∈ C1R, R is an increasing
function with ϕ ∞ ∞ on I, and ψu is a nondecreasing continuous function for u ∈ R with
ψ u > 0 for u > 0 If
ϕ
u t≤ c n
i 1
α i t
for t ∈ I, then
u t ≤ ϕ−1
G−1 Ψ−1
Ψk
t0
i 1
α i t
α i t0 f i sds
2.2
for t ∈ t0, t1, where
r
r0
ds
ϕ−1s q , r ≥ r0> 0,
Ψr
r
r0
ds
ψ ϕ−1G−1s , r ≥ r0 > 0,
k
t0
i 1
α i t
α i t0 g i sds,
2.3
Trang 4G−1andΨ−1denote the inverse functions of G and Ψ, respectively, for t ∈ I t1∈ I is so chosen
that
Ψk
t0
i 1
α i t
α i t0 f i sds ∈ DomΨ−1
Proof Assume that c > 0 Define a function z t by the right-hand side of 2.1 Clearly, zt is
zt n
i 1
u
α i tq
f i
α i tψ
u
α i t g i
α i tαi t
≤ϕ−1
z tq n
i 1
f i
α i tψ
ϕ−1
z
α i t g i
α i tαi t.
2.5
ϕ−1ztq ≥ϕ−1zt0q ϕ−1cq
> 0. 2.6 That is
zt
ϕ−1
z tq ≤ n
i 1
f i
α i tψ
ϕ−1
z
α i t g i
left-hand side, and changing variable in the right-hand side, we obtain
G
i 1
α i t
α i t0
f i sψϕ−1
G
z t≤ pt n
i 1
α i t
α i t0 f i sψϕ−1
where
i 1
α i t
G
z t≤ pt1
i 1
α i t
α i t0 f i sψϕ−1
kt n
i 1
f i
α i tψ
ϕ−1
z
α i tαi t ≤ ψϕ−1
G−1
k t n
i 1
f i
α i tαi t 2.12
Trang 5Using the monotonicity of ψ, ϕ−1, G−1, and k, we deduce
kt
ψ
ϕ−1
G−1
n
i 1
f i
left-hand side, and changing variable in the right-hand side, we obtain
Ψk t≤ Ψk
t0
i 1
α i t
Ψp t1
i 1
α i t
α i t0 f i sds
2.15
produces the required inequality
retarded integral inequality for nonlinear functions
Corollary 2.2 Let u, f i , g i ∈ CI, R, i 1, , n, and let α i ∈ C1I, I be nondecreasing with
α i t ≤ t, i 1, , n Suppose that c ≥ 0 and p > q > 0 are constants, and ψu is a nondecreasing
continuous function for u ∈ Rwith ψ u > 0 for u > 0 If
u p t ≤ c n
i 1
α i t
for t ∈ I, then
0
Ψ0
k1 t0
p
n
i 1
α i t
α i t0 f i sds
1/p−q
2.17
for t ∈ t0, t, where
Ψ0r
r
r0
ds
ψ
s 1/p−q , r ≥ r0> 0, k1
t0
c p−q/p p − q
p
n
i 1
α i t
α i t0 g i sds,
2.18
Ψ−1
0 denotes the inverse function ofΨ0for t ∈ I t ∈ I is so chosen that
Ψ0
k1
t0
p
n
i 1
α i t
α i t0 f i sds ∈ DomΨ−1
0
Trang 6
Proof The proof follows by an argument similar to that in the proof of Theorem 2.1 with suitable modification We omit the details here
Remark 2.3 When q 1, fromCorollary 2.2, one derives Theorem C When p 2, q 1, from
Corollary 2.2, one derives Theorem B
Theorem 2.1can easily be applied to generate other useful nonlinear integral inequalities
Theorem 2.4 Let u ∈ CI, R1, f i , g i ∈ CI, R, i 1, , n, and let α i ∈ C1I, I be nondecreasing
with α i t ≤ t, i 1, , n Suppose that c ≥ 1 is a constant, ϕ ∈ C1R, R is an increasing function
with ϕ ∞ ∞ and ψ j u, j 1, 2 are nondecreasing continuous functions for u ∈ Rwith ψ j u > 0
for u > 0 If
ϕ
u t≤ c n
i 1
α i t
α i t0 u q sf i sψ1
u s g i sψ2
for t ∈ I, then
i as the case ψ1u ≥ ψ2logu,
u t ≤ ϕ−1
G−1 Ψ−1 1
Ψ1
G c n
i 1
α i t
α i t0
f i s g i sds
2.21
for t ∈ t0, t1,
ii as the case ψ1u < ψ2logu,
u t ≤ ϕ−1
G−1 Ψ−1 2
Ψ2
G c n
i 1
α i t
α i t0
f i s g i sds
2.22
for t ∈ t0, t2, where
Ψj r
r
r0
ds
ψ j
ϕ−1
G−1, Ψ−1
G t is as defined in Theorem 2.1 for t ∈ I, and t j ∈ I, j 1, 2 are so chosen that
Ψj
G c n
i 1
α i t
α i t0
f i s g i sds∈ DomΨ−1
j
Proof Let c > 0 Define a function z t by the right-hand side of 2.20 Clearly, zt is
zt n
i 1
u
α i tq f i
α i tψ1
u
α i t g i
α i tψ2
u
α i tαi t
≤ϕ−1
z tq n
i 1
f i
α i tψ1
ϕ−1
z
α i t g i
α i tψ2
ϕ−1
z
α i tαi t.
2.25
Trang 7Using the monotonicity of ϕ−1and z, we deduce
ϕ−1
z tq≥ϕ−1
z
t0q ϕ−1cq
> 0. 2.26 That is
zt
ϕ−1zt q ≤ n
i 1
f i α i tψ1ϕ−1zα i t g i α i tψ2logϕ−1zα i tαi t 2.27
left-hand side, and changing variable in the right-hand side, we obtain
G
i 1
α i t
α i t0
f i sψ1
ϕ−1
z s g i sψ2
G
i 1
α i t
α i t0
f i s g i sψ1
ϕ−1
kt n
i 1
f i
α i t g i
α i tψ1
ϕ−1
z
α i tαi t
≤ ψ1
ϕ−1
G−1
k t n
i 1
f i
α i t g i
α i tαi t.
2.30
kt
ψ1
ϕ−1
G−1
n
i 1
f i
left-hand side, and changing variable in the right-hand side, we obtain
Ψ1
k t≤ Ψ1
k
t0
i 1
α i t
α i t0
1
Ψ1
G c n
i 1
α i t
α i t0
f i s g i sds
2.33
inequality in2.21
Trang 8When ψ1u < ψ1logu, from the inequality 2.28, we find
G
i 1
α i t
α i t0
f i s g i sψ1
ϕ−1
conclude that
2
Ψ2
G c n
i 1
α i t
α i t0
f i s g i sds
2.35
inequality in2.22
retarded integral inequality for nonlinear functions
Corollary 2.5 Let u ∈ CI, R1, f i , g i ∈ CI, R, i 1, , n, and let α i ∈ C1I, I be nondecreasing
with α i t ≤ t, i 1, , n Suppose that c ≥ 0 and p > q > 0 are constants, and ψ j u, j 1, 2 are
nondecreasing continuous functions for u ∈ Rwith ψ j u > 0 for u > 0 If
u p t ≤ c n
i 1
α i t
α i t0 u q sf i sψ1
u s g i sψ2
for t ∈ I, then
i as the case ψ1u ≥ ψ2logu,
u t ≤ G−11
G1
c p−q/p
p
n
i 1
α i t
α i t0
f i s g i sds
1/p−q
2.37
for t ∈ t0, t1,
ii as the case ψ1u < ψ2logu,
u t ≤ G−12
G2
c p−q/p
p
n
i 1
α i t
α i t0
f i s g i sds
1/p−q
2.38
for t ∈ t0, t2, where G−1
j , j 1, 2, denote the inverse functions of G j , j 1, 2,
G j r
r
r0
ds
ψ j
s 1/p−q , r ≥ r0> 0, 2.39
for t ∈ I, and t j ∈ I, j 1, 2, are chosen so that
G j
c p−q/p
p
n
i 1
α i t
α i t0
f i s g i sds ∈ DomG−1
Trang 9Proof The proof follows by an argument similar to that in the proof of Theorem 2.4 with suitable modification We omit the details here
Theorem 2.1can easily be applied to generate another useful nonlinear integral
Theorem 2.6 Let u, f i , g i ∈ CI, R, i 1, , n, and let α i ∈ C1I, I be nondecreasing with
α i t ≤ t, i 1, , n Suppose that c ≥ 0 and q > 0 are constants, ϕ ∈ C1R, R is an increasing
function with ϕ ∞ ∞ on I, and L, M ∈ CR2
, R satisfy
for t, v, w ∈ Rwith v ≥ w ≥ 0 If
ϕ
u t≤ c n
i 1
α i t
α i t0 u q sf i sLs, u s g i susds 2.42
for t ∈ I, then
u t ≤ ϕ−1
G−1 Ω−1
Ωk2t0
i 1
α i t
α i t0
f i sMs g i sds
2.43
for t ∈ t0, t1, where
Ωr
r
r0
ds
ϕ−1
G−1s , r ≥ r0> 0,
k2
t0
i 1
α i t
α i t0 f i sLsds,
2.44
G−1 and Ω−1 denote the inverse function of G and Ω, respectively, the function G is as defined in Theorem 2.1 for t ∈ I and t1∈ I is so chosen that
Ωk2
t0
i 1
α i t
α i t0 f i sds ∈ DomΩ−1
Proof Let c > 0 Define a function z t by the right-hand side of 2.42 Clearly, zt is
zt n
i 1
u
α i tq
f i
α i tL
α i t, uα i t g i
α i tu
α i tαi t
≤ϕ−1
z tq n
i 1
f i
α i tL
α i t, ϕ−1
z
α i t g i
α i tϕ−1
z
α i tαi t.
2.46
zt
ϕ−1
z tq ≤ n
i 1
f i
α i tL
α i t, ϕ−1
z
α i t g i
α i tϕ−1
z
α i tαi t 2.47
Trang 10Setting t s in the inequality 2.47, integrating it from t0 to t, using the function G in the
left-hand side, and changing variable in the right-hand side, we obtain
G
i 1
α i t
α i t0
f i sLs, ϕ−1
G
i 1
α i t1
α i t0 f i sL s ds n
i 1
α i t
α i t0
f i sMs g i sϕ−1
k2t n
i 1
f i α i tMα i t g i α i tϕ−1zsα
i t
≤ ϕ−1G−1k2t n
i 1
f i α i tMα i t g i α i tα
i t.
2.50
k2t
ψ
ϕ−1
G−1
k2t ≤
n
i 1
f i
α i tM
α i t g i
left-hand side, and changing variable in the right-hand side, we obtain
Ωk2 t≤ Ωk2
t0
i 1
α i t
α i t0
Ωk2
t0
i 1
α i t
α i t0
f i sMs g i sds
2.53
retarded integral inequality for nonlinear functions
Corollary 2.7 Let u, f i , g i , and α i be as defined in Theorem 2.6 Suppose that c ≥ 0 and p > q > 0 are
constants, and L, M ∈ CR2
, R satisfy
Trang 11for t, v, w ∈ Rwith v ≥ w ≥ 0 If
u p t ≤ c n
i 1
α i t
for t ∈ I, then
1
k3
t0
p
n
i 1
α i t
α i t0
f i sMs g i sds
1/p−q
2.56
for t ∈ t0, t1, where
r
r0
ds
s 1/p−q , r ≥ r0> 0, k3
t0
c p−q/p p − q
p
n
i 1
α i t
α i t0 f i sLsds,
2.57
Ω−1
1 denotes the inverse function ofΩ1for t ∈ I and t1∈ I is so chosen that
k3
t0
p
n
i 1
α i t
α i t0
f i sMs g i sds∈ DomΩ−1
1
Proof The proof follows by an argument similar to that in the proof of Theorem 2.6 with suitable modification We omit the details here
3 Applications
We will show that our results are useful in proving the global existence of solutions to certain differential equations with time delay Consider the functional differential equation involving several retarded arguments with the initial condition:
ϕ
x txt Ft, x
t − h1t, , x
t − h n t, t ∈ I,
x
t0
t − h i t ≥ 0, t − h i t ∈ C1I, I, h
Theorem 3.1 Assume that F : I × R n →R is a continuous function for which there exist continuous
nonnegative functions f i t, g i t, i 1, , n for t ∈ I such that
F
t, u1, , u n ≤ n
i 1
u iq
Trang 12where q > 0 is a constant and ψ is as in Theorem 2.1 Let
t ∈I
1
1− h
If x t is any solution of the problem 3.1, then
x t ≤ ϕ−1
G−1 Ψ−1
Ψk
t0
i 1
t −h i t
t0−h i t0 f i σdσ
3.4
for t ∈ I, where G, Ψ are as in Theorem 2.1 and
k
t0
Gϕ
x0 n
i 1
t −h i t
f i σ Q i f i σ h i s, g i σ Q i g i σ h i s for s, σ ∈ I.
Proof It is easy to see that the solution x t of the problem 3.1 satisfies the equivalent integral equation:
ϕ
x t ϕx
t0
t
t0
F
s, x
s − h1s, , x
ϕ
x t ≤ ϕxt0 t
t0
F
s, x
s − h1s, , x
s − h n sds
≤ϕ
x0 t
t0
n
i 1
x
s − h i sq
f i tψx
s − h i s g i tds
3.7
rewriting, we have
ϕx t ≤ ϕx0 n
i 1
t −h i t
t0−h i t0
x σq
Trang 13Remark 3.2 Consider the functional differential equation involving several retarded arguments with the initial condition:
px p−1txt Ft, x
t − h1t, , x
t − h n t, t ∈ I,
x
t0
such that t − h i t ≥ 0, t − h i t ∈ C1I, I, h
i t < 1.
nonnegative functions f i t, g i t, i 1, , n for t ∈ I such that
F
t, u1, , u n ≤ n
i 1
u iq
x p t x p
t0
t
t0
F
s, x
s − h1s, , x
x tp ≤x0p
t
t0
F
s, x
s − h1s, , x
s − h n sds
≤x0p
t
t0
n
i 1
x
s − h i sq
f i tψx
s − h i s g i tds
3.12
x tp ≤x0p n
i 1
t −h i t
t0−h i t0
x σq
x t ≤ Ψ−1
0
Ψ0
k1
t0
p
n
i 1
t −h i t
t0−h i t0 f i σdσ
1/p−q
3.14
k1 t0 x p −q
0 p − q
p
n
i 1
t −h i t
f i σ Q i f i σ h i s, and g i σ Q i g i σ h i s for s, σ ∈ I.
... present paper establishes some nonlinear retarded inequalities which extend the foregoing theorems In addition, it illustrates the use/application of these inequalities2 Main results... of solutions to certain differential equations with time delay Consider the functional differential equation involving several retarded arguments with the initial condition:
ϕ... i sds
2.15
produces the required inequality
retarded integral inequality for nonlinear functions
Corollary 2.2 Let u, f i