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RESEARCH Open Access
Existence results of Brezis-Browder type for
systems of Fredholm integral equations
Ravi P Agarwal
1,2*
, Donal O’Regan
3
and Patricia JY Wong
4
* Correspondence:

1
Department of Mathematics, Texas
A&M University - Kingsville,
Kingsville, TX 78363, USA
Full list of author information is
available at the end of the article
Abstract
In this article, we consider the following systems of Fredholm integral equations:
u
i
(t )=h
i
(t )+
T

0
g
i
(t , s)f
i

(s, u
1
(s), u
2
(s), , u
n
(s))ds, t ∈ [0, T], 1 ≤ i ≤ n,
u
i
(t )=h
i
(t )+
∞

0
g
i
(t , s)f
i
(s, u
1
(s), u
2
(s), , u
n
(s))ds, t ∈ [0, ∞), 1 ≤ i ≤ n
.
Using an argument originating from Brezis and Browder [Bull. Am. Math. Soc. 81, 73-
78 (1975)] and a fixed point theorem, we establish the existence of solutions of the
first system in (C[0, T])

n
, whereas for the second system, the existence criteria are
developed separately in (C
l
[0,∞))
n
as well as in (BC[0,∞))
n
. For both systems, we
further seek the existence of constant-sign solutions, which include positive solution s
(the usual consideration) as a special case. Several examples are also included to
illustrate the results obtained.
2010 Mathematics Subject Classification: 45B05; 45G15; 45M20.
Keywords: system of Fredholm integral equations, Brezis-Browder arguments, con-
stant-sign solutions
1 Introduction
In this article, we shall consider the system of Fredholm integral equations:
u
i
(t )=h
i
(t )+
T

0
g
i
(t , s)f
i
(s, u

1
(s), u
2
(s), , u
n
(s))ds, t ∈ [0, T], 1 ≤ i ≤
n
(1:1)
where 0 <T<∞, and also the following system on the half-line
u
i
(t )=h
i
(t )+
∞

0
g
i
(t , s)f
i
(s, u
1
(s), u
2
(s), , u
n
(s))ds, t ∈ [0, ∞), 1 ≤ i ≤ n
.
(1:2)

Throughout, let u =(u
1
, u
2
, , u
n
). We are interested in establishing the existence of
solutions u of the system (1.1) in (C[0, T])
n
=C[0, T]×C[0, T]×ℙ ×C[0, T](n
times), whereas for the system (1.2), we shall seek a solution in (C
l
[0, ∞))
n
as well as in
(BC[0, ∞))
n
. Here, BC[0, ∞) denotes the space of functions that are bounded and con-
tinuous on [0, ∞) and C
l
[0, ∞)={x Î BC[0, ∞) : lim
t®∞
x(t) exists}.
We shall also tackle the existence of constant-sign solutions of (1.1) and (1.2). A
solution u of (1.1) (or (1.2)) is said to be of constant sign if for each 1 ≤ i ≤ n, we have
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>© 2011 Agarwal et al; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons
Attribution License ( which permits unrestricted use, distribution, and reproduction in
any medium, provided the or iginal work is properly cited .
θ

i
u
i
(t) ≥ 0 for all t Î [0, T] (or t Î [0,∞)), where θ
i
Î {-1, 1} is fixed. Note that when θ
i
=1forall1≤ i ≤ n, a constan t-sign solution reduces to a positive solution, which is
the usual consideration in the literature.
In the literature, there is a vast amount of research on th e existence of positive solu-
tions of the nonlinear Fredholm integral equations:
y(t)=h(t)+
T

0
g(t, s)f (y(s ))ds, t ∈ [0, T
]
(1:3)
and
y(t)=h(t)+
∞

0
g(t, s)f (y(s ))ds, t ∈ [0, ∞)
.
(1:4)
Particular cases of (1.3) are also considered in [1-3]. The reader is referred to the
monographs [[4,5], and the references cited therein] for the related literature. Recently,
a generalization of (1.3) and (1.4) to systems similar to (1.1) and (1.2) have been made,
and the existence of single and multiple constant-sign solutions has been established

for these systems in [6-10].
The technique used in these articles has relied heavily on various fixed point results
such as Krasnosel’skii’s fixed point theorem in a cone, Leray-Schauder alternative, Leg-
gett-Williams’ fixed point theorem, five-functional fixed point theorem, Schauder fixed
point theo rem, and Schauder-Tychonoff fixed point theorem. In the current study, we
will make use of an argument that originates from Brezis and Browder [11]; therefore,
the technique is different from those of [6-10] and the results subsequently obtained
are also different. The present article also extends, improves, and complements the stu-
dies of [5,12-23]. Indeed, we have generalized the problems to (i) systems;(ii)more
general form of nonlinearities f
i
,1≤ i ≤ n,; and (iii) existence of constant-sign solutions.
The outline of the article is as follows. In Section 2, we shall state the necessary fixed
point theorem and compactnes s criterion, which are used later. In Section 3, we tackle
the existence of solution s of system (1.1) in (C[0, T])
n
, while Sections 4 and 5 deal
with the existence of solutions of system (1.2) in (C
l
[0, ∞))
n
and (BC[0, ∞))
n
, respec-
tively. In Section 6, we seek the existence of constant-sign solutions of (1.1) and (1.2)
in (C[0, T])
n
,(C
l
[0, ∞))

n
and (BC[0, ∞))
n
. Finally, several examples are presented in
Section 7 to illustrate the results obtained.
2 Preliminaries
In this se ction, we shall state the theorems that ar e used later to develop the existence
criteria–Theorem 2.1 [24] is Schauder’s nonlinear alternative for continuo us and com-
pact maps, whereas Theorem 2.2 is the criterion of compactness on C
l
[0, ∞)[[16],p.
62].
Theorem 2.1 [24]Let B be a Banach space with E ⊆ B closed and convex. Assume U
is a relatively open subset of E with 0 Î Uand
S : U
→
E
is a continuous and compact
map. Then either
(a) S has a fixed point in
U
, or
(b) there exist u Î ∂U and l Î (0, 1) such that u = lSu.
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 2 of 35
Theorem 2.2 [[16], p. 62] Let P ⊂ C
l
[0, ∞). Then P is compact in C
l
[0, ∞) if the fol-

lowing hold:
(a) P is bounded in C
l
[0, ∞).
(b) Any y Î P is equicontinuous on any compact interval of [0, ∞).
(c) P is equiconvergent, i.e., g iven ε >0, there exi sts T(ε) >0 such that |y(t)-y(∞ )| <ε
for any t ≥ T(ε) and y Î P.
3 Existence results for (1.1) in (C[0, T])
n
Let the Banach space B =(C[0, T])
n
be equipped with the norm:
||u|| =max
1≤i≤n
sup
t∈
[
0,T
]
|u
i
(t ) | =max
1≤i≤n
|u
i
|
0
where we let |u
i
|

0
= sup
tÎ[0,T]
|u
i
(t)|, 1 ≤ i ≤ n . Throughout, for u Î B and t Î [0, T],
we shall denote
|
|u(t)|| =max
1
≤
i
≤
n
|u
i
(t ) |
.
Moreover, for each 1 ≤ i ≤ n,let1≤ p
i
≤ ∞ be an integer and q
i
be such that
1
p
i
+
1
q
i

=1
. For
x ∈ L
p
i
[
0, T
]
, we shall define

x

p
i
=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩

T

0

|x(s)|
p
i
ds

1
p
i
,1≤ p
i
<
∞
ess sup
s∈[0,T]
|x(s)|, p
i
= ∞.
Our first existence result uses Theorem 2.1.
Theorem 3.1 For each 1 ≤ i ≤ n, assume (C1)- (C4) hold where
(C1) h
i
Î C[0, T], denote H
i
≡ sup
tÎ [0, T]
|h
i
(t)|,
(C2) f
i

: [0, T]×ℝ
n
® ℝ is a
L
q
i
-Carathéodory function:
(i) the map u a f
i
(t, u) is continuous for almost all t Î [0, T],;
(ii) the map t a f
i
(t, u) is measurable for all u Î ℝ
n
;
(iii) for any r >0,there exists
μ
r
,
i
∈ L
q
i
[
0, T
]
such that |u| ≤ rimplies|f
i
(t, u)| ≤ μ
r,i

(t) for almost all t Î [0, T];
(C3)
g
t
i
(s)=g
i
(t , s) ∈ L
p
i
[0, T
]
for each t Î [0, T];
(C4) the map
t → g
t
i
is continuous from [0, T] to
L
p
i
[
0, T
]
.
In addition, suppose there is a const ant M > 0, independent of l,with||u|| ≠ M for
any solution u Î (C[0, T])
n
to
u

i
(t )=λ
⎛
⎝
h
i
(t )+
T

0
g
i
(t , s)f
i
(s, u(s))ds
⎞
⎠
, t ∈ [0, T], 1 ≤ i ≤ n (3.1)
λ
for each l Î (0, 1). Then, (1.1) has at least one solution in (C[0, T])
n
.
Proof Let the operator S be defined by
Su
(
t
)
=
(
S

1
u
(
t
)
, S
2
u
(
t
)
, , S
n
u
(
t
))
, t ∈ [0, T
]
(3:2)
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 3 of 35
where
S
i
u(t )=h
i
(t )+
T


0
g
i
(t , s)f
i
(s, u(s))ds , t ∈ [0, T], 1 ≤ i ≤ n
.
(3:3)
Clearly, the system (1.1) is equivalent to u = Su, and (3.1)
l
is the same as u = lSu.
Note that S maps (C[0, T])
n
into (C[0, T])
n
,i.e.,S
i
:(C[0, T])
n
® C[0, T], 1 ≤ i ≤ n.
To see this, note that for any u Î (C[0, T])
n
, there exits r > 0 su ch that ||u|| <r. Since
f
i
is a
L
q
i
-Carathéodory function, there exists

(t
1
) − h
i
(t
2
)| +

T
∫
0
|g
t
1
i
(s) − g
t
2
i
(s)|
p
i
ds

1
p
i


μ

r,i


q
i
→
0
(3:4)
as t
1
® t
2
, where we have used (C1) and (C3). This shows that S :(C[0, T])
n
® (C[0,
T])
n
.
Next,weshallprovethatS :(C[0, T])
n
® ( C[0, T])
n
is continuous. Let
u
m
=(u
m
1
, u
m

2
, , u
m
n
) →
u
in (C[0, T])
n
, i.e.,
u
m
i
→ u
i
in C[0, T], 1 ≤ i ≤ n.Weneedto
show that Su
m
® Su in (C[0, T])
n
, or equivalently S
i
u
m
® S
i
u in C[0, T], 1 ≤ i ≤ n.
There exists r > 0 such that ||u
m
||, ||u|| <r.Sincef
i

is a
L
q
i
-Carathéodory function,
there exists
μ
r
,
i
∈ L
q
i
[
0, T
]
such that |f
i
(s, u
m
)|, |f
i
(s, u)| ≤ μ
r,i
(s) for almost all s Î [0,
T]. Using a similar argument as in (3.4), we get for any t
1
, t
2
Î [0, T] and 1 ≤ i ≤ n:

|
S
i
u
m
(
t
1
)
− S
i
u
m
(
t
2
)
|→0and|S
i
u
(
t
1
)
− S
i
u
(
t
2

)
|→
0
(3:5)
as t
1
® t
2
. Furthermore, S
i
u
m
(t) ® S
i
u(t) pointwise on [0, T], since, by the Lebesgue-
dominated convergence theorem,
|
S
i
u
m
(t ) − S
i
u(t ) |≤ sup
t∈
[
0,T
]
||g
t

i
||
p
i

T
∫
0
|f
i
(s, u
m
(s)) − f
i
(s, u(s))|
q
i
ds

1
q
i
→
0
(3:6)
as m ® ∞. Combining (3.5) and (3.6) and using the fact that [0, T] is compact, gives
for all t Î [0, T],
|S
i
u

m
(
t
)
− S
i
u
(
t
)
|≤|S
i
u
m
(
t
)
− S
i
u
m
(
t
1
)
| + |S
i
u
m
(

t
1
)
− S
i
u
(
t
1
)
| + |S
i
u
(
t
1
)
− S
i
u
(
t
)
|→
0
(3:7)
as m ® ∞. Hence, we have proved that S :(C[0, T])
n
® (C[0, T])
n

is continuous.
Finally, we shall show that S :(C[0, T])
n
® (C[0, T])
n
is completely continuous. Let
Ω be a bounded set in (C[0, T])
n
with ||u|| ≤ r for all u Î Ω. We need to show that
S
i
Ω is relatively compact for 1 ≤ i ≤ n.Clearly,S
i
Ω is uniforml y bounded, sin ce there
exists
μ
r
,
i
∈ L
q
i
[
0, T
]
such that |f
i
(s, u)| ≤ μ
r,i
(s)forallu Î Ω and a. e. s Î [0, T], and

hence
|
S
i
u|
0
≤ H
i
+sup
t∈
[
0,T
]


g
t
i


p
i
·


μ
r,i


q

i
≡ K
i
, u ∈ 
.
(3:8)
Further, using a similar argument as in (3.4), we see that S
i
Ω is equicontinuous. It
follows from the Arzéla-Ascoli theorem [[5], Theorem 1 .2.4] that S
i
Ω is relatively
compact.
We now apply Theorem 2.1 with U ={u Î (C[0, T])
n
:||u|| <M}andB = E =(C[0,
T])
n
to obtain the conclusion of the theorem. □
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 4 of 35
Our subsequent results will apply Theorem 3.1. To do so, we shall show that any
solution u of (3.1)
l
is bounded above. This is achieved by bounding the integral of |f
i
(t,
u(t))| (or
|f
i

(
t, u
(
t
))
|
ρ
i
) on two complementary subsets of [0, T], namely {t Î [0, T]:||u
(t)|| ≤ r}and{t Î [0, T]:||u(t)|| >r}, where r
i
and r are some constants–this techni-
que originates from the study of Brezis and Bro wder [11]. In the next four theorems
(Theorems 3.2-3.5), we shall apply Theorem 3.1 to the case p
i
= ∞ and q
i
=1,1≤ i ≤
n.
Theorem 3.2. Let the following conditions be satisfied for each 1 ≤ i ≤ n : (C1)-(C4)
with p
i
= ∞ and q
i
= 1, (C5) and (C6) where
(C5) there exist B
i
>0such that for any u Î (C[0, T])
n
,

T
∫
0

f
i
(t , u(t ))
T
∫
0
g
i
(t , s)f
i
(s, u(s))ds

dt ≤ B
i
,
(C6) there exist r >0and a
i
>0with ra
i
>H
i
such that for any u Î (C[0, T])
n
,
u
i

(
t
)
f
i
(
t, u
(
t
))
≥ rα
i
|f
i
(
t, u
(
t
))
| for ||u
(
t
)
|| > randa.e. t ∈ [0, T]
.
Then, (1.1) has at least one solution in (C[0, T])
n
.
Proof We shall employ Theorem 3.1, and so let u =(u
1

, u
2
, l , u
n
) Î (C[0, T])
n
be
any solution of (3.1)
l
where l Î (0, 1).
Define
I = {t ∈ [0, T]:||u
(
t
)
|| ≤ r} and J = {t ∈ [0, T]:||u
(
t
)
|| > r}
.
(3:9)
Clearly, [0, T]=I ∪ J, and hence

T
0
=

I
+


J
.
Let 1 ≤ i ≤ n.Ift Î I, then by (C2), there exists μ
r,i
Î L
1
[0, T] such that |f
i
(t, u(t))| ≤
μ
r,i
(t). Thus, we get

I
|f
i
(t , u(t ))|dt ≤

I
μ
r,i
(t )dt ≤
T

0
μ
r,i
(t )dt = ||μ
r,i

||
1
.
(3:10)
On the oth er hand, if t Î J, then it is clear from (C6) that u
i
(t)f
i
(t, u(t)) ≥ 0fora.e. t
Î [0, T]. It follows that

J
u
i
(t ) f
i
(t , u(t ))dt ≥ rα
i

J
|f
i
(t , u(t ))|dt
.
(3:11)
We now multiply (3.1)
l
by f
i
(t, u(t)), then integrate from 0 to T to get

T

0
u
i
(t)f
i
(t, u(t))dt = λ
T

0
h
i
(t)f
i
(t, u(t))dt + λ
T

0
⎡
⎣
f
i
(t, u(t))
T

0
g
i
(t, s)f

i
(s, u(s))ds
⎤
⎦
dt
.
(3:12)
Using (C5) in (3.12) yields
T

0
u
i
(t ) f
i
(t , u(t ))dt ≤ H
i
T

0
|f
i
(t , u(t ))|dt + B
i
.
(3:13)
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 5 of 35
Splitting the integrals in (3.13) and applying (3.11), we get


)

J
|f
i
(t , u(t ))|dt ≤ H
i

I
|f
i
(t , u(t ))|dt +

I
|u
i
(t ) f
i
(t , u(t ))|dt + B
i
≤
(
H
i
+ r
)
||μ
r,i
||
1

+ B
i
where we have used (3.10) in the last inequality. It follows that

J
|f
i
(t , u(t ))|dt ≤
(H
i
+ r)||μ
r,i
||
1
+ B
i
rα
i
− H
i
≡ k
i
.
(3:14)
Finally, it is clear from (3.1)
l
that for t Î [0, T] and 1 ≤ i ≤ n,


u

i
(t )


≤ H
i
+
T

0
|g
i
(t , s)f
i
(s, u(s))|ds
= H
i
+

∫
I
+
∫
J

|g
i
(t , s)f
i
(s, u(s))|ds

≤ H
i
+

sup
t∈[0,T]
||g
t
i
||
∞

(||μ
r,i
||
1
+ k
i
) ≡ l
i
(3:15)
where we have applied (3.10) and (3.14) in the last inequality. Thus, |u
i
|
0
≤ l
i
for 1 ≤
i ≤ n and ||u|| ≤ max
1≤i≤n

l
i
≡ L. It follows from Theorem 3.1 (with M = L +1)that
(1.1) has a solution u* Î (C[0, T])
n
. □
Theorem 3.3 Let the following conditions be s atisfied for each 1 ≤ i ≤ n : (C1)-(C4)
with p
i
= ∞ and q
i
=1,(C7) and (C8) where
(C7) there exist constants a
i
≥ 0 and b
i
such that for any u Î (C[0, T])
n
,
T

0
⎡
⎣
f
i
(t , u(t ))
T

0

g
i
(t , s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ a
i
T

0
|f
i
(t , u(t ))|dt + b
i
,
(C8) there exist r >0 and a
i
>0 with ra
i
>H
i
+a
i
such that for any u Î (C[0, T])
n
,
u
i

(
t
)
f
i
(
t, u
(
t
))
≥ rα
i
|f
i
(
t, u
(
t
))
| for ||u
(
t
)
|| > randa.e. t ∈ [0, T]
.
Then, (1.1) has at least one solution in (C[0, T])
n
.
Proof The proof follows that of The orem 3.2 un til (3.12). Let 1 ≤ i ≤ n.Weuse(C7)
in (3.12) to get

T

0
u
i
(t)f
i
(t, u(t))dt ≤
T

0
|h
i
(t)f
i
(t, u(t))|dt + λ
T

0
⎡
⎣
f
i
(t, u(t))
T

0
g
i
(t, s)f

i
(s, u(s))ds
⎤
⎦
d
t
≤ (H
i
+ a
i
)
T

0
|f
i
(t, u(t))|dt + |b
i
|.
(3:16)
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 6 of 35
Splitting the integrals in (3.16) and applying (3.11) gives
(rα
i
− H
i
− a
i
)


J
|f
i
(t, u(t))|dt ≤ (H
i
+ a
i
)

I
|f
i
(t, u(t))|dt +

I
|u
i
(t)f
i
(t, u(t))|dt + |b
i
|
≤
(
H
i
+ a
i
+ r

)
||μ
r,i
||
1
+ |b
i
|
where we have also used (3.10) in the last inequality. It follows that

J
|f
i
(t, u(t))|dt ≤
(H
i
+ a
i
+ r)||μ
r,i
||
1
+ |b
i
|
rα
i
− H
i
− a

i
≡ k
i
.
(3:17)
The rest of the proof follows that of Theorem 3.2. □
Theorem 3.4 Let the following conditions be s atisfied for each 1 ≤ i ≤ n : (C1)-(C4)
with p
i
= ∞ and q
i
= 1, (C9) and (C10) where
(C9) there exist constants a
i
≥ 0, 0 < τ
i
≤ 1 and b
i
such that for any u Î (C[0, T])
n
,
T

0
⎡
⎣
f
i
(t, u(t))
T


0
g
i
(t, s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ a
i
⎡
⎣
T

0
|f
i
(t, u(t))|dt
⎤
⎦
τ
i
+ b
i
,
(C10) there exist r >0and b
i
>0such that for any u Î ( C[0, T])
n

,
u
i
(
t
)
f
i
(
t, u
(
t
))
≥ β
i
||u
(
t
)
|| · |f
i
(
t, u
(
t
))
| for ||u
(
t
)

|| > randa.e. t ∈ [0, T]
.
Then, (1.1) has at least one solution in (C[0, T])
n
.
Proof Let u =(u
1
, u
2
, , u
n
) Î (C[0, T])
n
be any solution of (3.1)
l
where l Î (0, 1).
Define
r
0
=max

r,max
1≤i≤n
H
i
+ a
i
2
τ
i

+1
β
i

,
I
0
= {t ∈ [0, T]:||u
(
t
)
|| ≤ r
0
} and J
0
= {t ∈ [0, T]:||u
(
t
)
|| > r
0
}
.
(3:18)
Clearly, [0, T]=I
0
∪ J
0
and hence
T


0
=

I
0
+

J
0
.
Let 1 ≤ i ≤ n.Ift Î I
0
, then by (C2) there exists
μ
r
0
,i
∈ L
1
[0, T
]
such that
|f
i
(t , u(t))|≤μ
r
0
,i
(t

)
and

I
0
|f
i
(t, u(t))|dt ≤

I
0
μ
r
0
,i
(t)dt ≤
T

0
μ
r
0
,i
(t)dt = ||μ
r
0
,i
||
1
.

(3:19)
Further, if t Î J
0
, then by (C10) we have

J
0
u
i
(t)f
i
(t, u(t))dt ≥ β
i

J
0
||u(t)|| · |f
i
(t, u(t))|dt ≥ β
i
r
0

J
0
|f
i
(t, u(t))|dt
.
(3:20)

⎤
⎦
τ
i
⎫
⎬
⎭
+ |b
i
|
(3:21)
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 7 of 35
where in the last inequality, we have made use of the inequality:
(
x + y
)
α
≤ 2
α
(
x
α
+ y
α
)
, x, y ≥ 0, α ≥ 0
.
Now, noting (3.19) we find that


I
0
|u
i
(t)f
i
(t, u(t))|dt +

I
0
|h
i
(t)f
i
(t, u(t))|dt + a
i
2
τ
i
⎡
⎣

I
0
|f
i
(t, u(t))|dt
⎤
⎦
τ

i
+ |b
i
|
≤ (r
0
+ H
i
)||μ
r
0
,i
||
1
+ a
i
2
τ
i
(||μ
r
0
,i
||
1
)
τ
i
+ |b
i

i
(t, u(t))|dt
⎤
⎦
τ
i
+ k

i
.
Since τ
i
≤ 1, there exists a constant
k


i
such that
(β
i
r
0
− H
i
− a
i
2
τ
i
)


J
0
|f
i
(t, u(t))|dt ≤ k


i
which leads to

J
0
|f
i
(t, u(t))|dt ≤
k

i
β
i
r
0
− H
i
− a
i
2
τ
i

≡ k
i
.
(3:23)
Finally, it is clear from (3.1)
l
that for t Î [0, T] and 1 ≤ i ≤ n,
|u
i
(t)|≤H
i
+
T

0
|g
i
(t, s)f
i
(s, u(s))|ds
= H
i
+

∫
I
0
+
∫
J

0

|g
i
(t, s)f
i
(s, u(s))|ds
≤ H
i
+

sup
t∈[0,T]
||g
t
i
||
∞

(||μ
r
0
,i
||
1
+ k
i
) ≡ l
i
(3:24)

where we hav e applied (3.19) and (3.23) in the last inequality. The conclusion now
follows from Theorem 3.1. □
Theorem 3.5 Let the following conditions be satisfied for each 1 ≤ i ≤ n : (C1), (C2)-
(C4) with p
i
= ∞ and q
i
=1,(C10), (C11) and (C12) where
(C11) there exist r >0,h
i
>0,g
i
>0and
φ
i
∈ L
γ
i
+1
γ
i
[
0, T
]
such that for any u Î (C[0,
T])
n
,
|
|u

(
t
)
|| ≥ η
i
|f
i
(
t, u
(
t
)
|
γ
i
+ φ
i
(
t
)
for ||u
(
t
)
|| > randa.e. t ∈ [0, T]
,
(C12) there exist a
i
≥ 0, 0 <τ
i

<g
i
+1,b
i
, and
ψ
i
∈ L
γ
i
+1
γ
i
[
0, T
]
with ψ
i
≥ 0 almost every-
where on [0, T], such that for any u Î (C[0, T])
n
,
T

0
⎡
⎣
f
i
(t , u(t ))

T

0
g
i
(t , s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ a
i
⎡
⎣
T

0
ψ
i
(t ) |f
i
(t , u(t ))|dt
⎤
⎦
τ
i
+ b
i
.
Agarwal et al. Advances in Difference Equations 2011, 2011:43

/>Page 8 of 35
Also, j
i
Î C[0, T],
h
i
∈ L
γ
i
+1
γ
i
[0, T
]
, ψ
i
Î C[0, T] and
T

0
|g
i
(t , s)|
γ
i
+1
γ
i
ds ∈ C[0, T
]

.
Then, (1.1) has at least one solution in (C[0, T])
n
.
Proof Let u =(u
1
, u
2
, , u
n
) Î (C[0, T])
n
be any solution of (3.1)
l
where l Î (0, 1).
Define the sets I and J as in (3.9). Let 1 ≤ i ≤ n. Applying (C10) and (C11), we get

J
u
i
(t)f
i
(t, u(t))dt ≥ β
i

J
||u(t)|| · |f
i
(t, u(t))|dt
≥ β

|
≤ (r + H
i
)||μ
r,i
||
1
+ a
i
2
τ
i
⎡
⎣

I
ψ
i
(t)μ
r,i
(t)d t
⎤
⎦
τ
i
+ |b
i
|≡k
i
.

γ
i
+1
γ
i
dt
⎤
⎦
γ
i
+1
γ
i
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1

+
⎡
⎣
T

0
|h
i
(t)|
γ
i
+1
γ
i
dt
⎤
⎦
γ
i
γ
i
+1
⎡
⎣

J
|f
i
(t, u(t))|
γ

i
+1
dt
⎤
⎦
1
γ
i
+1
+a
i
2
τ
i
⎡
⎣
T

0
|ψ
i
(t)|
γ
i
+1
γ
i
dt
⎤
⎦

τ
i
γ
i
γ
i
+1
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
τ
i
γ
i
+1
+ k
i
.
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 9 of 35

Since
1
γ
i
+1
<
1
and
τ
i
γ
i
+1
<
1
, there exists a constant k
i
such that

J
|f
i
(t , u(t ))|
γ
i
+1
dt ≤ k
i
.
(3:28)

Finally, it is clear from (3.1)
l
that for t Î [0, T] and 1 ≤ i ≤ n,
|u
i
(t)|≤H
i
+
⎛
⎝

I
+
∫
J
⎞
⎠


g
i
(t, s)f
i
(s, u(s))|ds
≤ H
i
+

sup
t∈[0,T]

||g
t
i
||
∞

||μ
r,i
||
1
+
⎧
⎪
⎪
⎨
⎪
⎪
⎩
sup
t∈[0,T]
⎡
⎣
T

0
|g
i
(t, s)|
γ
i

+1
γ
i
ds
⎤
⎦
γ
i
γ
i
+1
⎫
⎪
⎪
⎬
⎪
⎪
⎭
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤

⎦
1
γ
i
+1
≤ l
i
(3:29)
where we have used (3.28) and (C12) in the last inequality, and l
i
is some constant.
The conclusion is now immediate by Theorem 3.1. □
In the next six results (Theorem 3.6-3.11), we shall apply Theorem 3.1 for general p
i
and q
i
.
Theorem 3.6 Let the following conditions be satisfied for each 1 ≤ i ≤ n : (C1)-(C4),
(C5), (C10) and (C13) where
(C13) there exist r >0,h
i
>0,g
i
>0and
φ
i
∈ L
p
i
[

0, T
]
such that for any u Î (C[0, T])
n
,
|
|u
(
t
)
|| ≥ η
i
|f
i
(
t, u
(
t
)
|
γ
i
+ φ
i
(
t
)
for ||u
(
t

)
|| > randa.e. t ∈ [0, T]
.
Then, (1.1) has at least one solution in (C[0, T])
n
.
Proof Let u =(u
1
, u
2
, , u
n
) Î (C[0, T])
n
be any solution of (3.1)
l
where l Î (0, 1).
Define the sets I and J as in (3.9). Let 1 ≤ i ≤ n.Ift Î I,thenby(C2),thereexists
μ
r
,
i
∈ L
q
i
[
0, T
]
such that |f
i

(t, u(t))| ≤ μ
r,i
(t). Consequently, we have

I
|f
i
(t , u(t ))|dt ≤

I
μ
r,i
(t )dt ≤
T

0
μ
r,i
(t )dt ≤ T
1
p
i
||μ
r,i
||
q
i
.
(3:30)
On the other hand, using (C10) and (C13), we derive at (3.25).

Next, applying (C5) in (3.12) leads to (3.13). Splitting the integrals in (3.13) and using
(3.25), we find that
β
i
η
i

J
|f
i
(t, u(t))|
γ
i
+1
dt
≤ β
i

J


φ
i
(t)f
i
(t, u(t)) |dt + H
i

J
|f

(t, u(t))|dt + B
i
+(r + H
i
)T
1
p
i
||μ
r,i
||
q
i
= β
i

J


φ
i
(t)f
i
(t, u(t)) |dt + H
i

J
|f
i
(t, u(t))|dt + B


i
(3:31)
where (3.30) has been used in the last inequality and
B

i
≡ B
i
+(r + H
i
)T
1
p
i
||μ
r,i
||
q
i
.
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 10 of 35
Now, an application of Hölder’s inequality gives

J
|φ
i
(t ) f
i

(t , u(t ))|dt ≤
⎡
⎣
T

0
|φ
i
(t ) |
γ
i
+1
γ
i
dt
⎤
⎦
γ
i
γ
i
+1
·
⎡
⎣

J
|f
i
(t , u(t ))|

γ
i
+1
dt
⎤
⎦
1
γ
i
+1
.
(3:32)
Another application of Hölder’s inequality yields
T

0
|φ
i
(t ) |
γ
i
+1
γ
i
dt ≤ T
γ
i
p
i
−γ

i
−1
p
i
γ
i
⎡
⎣
T

0
|φ
i
(t ) |
p
i
dt
⎤
⎦
γ
i
+
1
γ
i
p
i
.
(3:33)
Substituting (3.33) into (3.32) then leads to


J
|φ
i
(t ) f
i
(t , u(t ))|dt ≤ T
γ
i
p
i
−γ
i
−1
p
i
(γ
i
+1)
||φ
i
||
p
i
⎡
⎣

J
|f
i

(t , u(t ))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1
.
(3:34)
Further, using Hölder’s inequality again, we get

J
|f
i
(t , u(t ))|dt ≤ T
γ
i
γ
i
+1
⎡
⎣

J
|f
i

(t , u(t ))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1
.
(3:35)
Substituting (3.34) and (3.35) into (3.31), we obtain
β
i
η
i

J
|f
i
(t , u(t ))|
γ
i
+1
dt ≤ A
i
⎡
⎣


J
|f
i
(t , u(t ))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1
+ B

i
(3:36)
where
A
i
≡ T
γ
i
p
i
−γ
i
−1

p
i
(γ
i
+1)
β
i
||φ
i
||
p
i
+ H
i
T
γ
i
γ
i
+1
.Since
1
γ
i
+1
<
1
, from (3.36), there exists a
constant k
i

such that

J
|f
i
(t , u(t ))|
γ
i
+1
dt ≤ k
i
.
(3:37)
Finally, it is clear from (3.1)
l
that for t Î [0, T] and 1 ≤ i ≤ n,
|
u
i
(t)|≤H
i
+
⎛
⎝

I
+

J
⎞

⎠


g
i
(t, s)f
i
(s ,u(s))


ds
≤ H
i
+

sup
t∈[0,T]
||g
t
i
||
p
i

||μ
r,i
||
q
i
+ T

γ
i
p
i
−γ
i
−1
p
i
(γ
i
+1)

sup
t∈[0,T]
||g
t
i
||
p
i



J
|f
i
(s, u(s))|
γ
i

+1
ds

1
γ
i
+1
≤ l
i
(
aconstant
)
,
(3:38)
where in the second last inequality a similar argument as in (3.34) is used, and in the
last inequality we have used (3.37). An application of Theorem 3.1 completes the
proof. □
Theorem 3.7 Let the following conditions be satisfied for each 1 ≤ i ≤ n : (C1)-(C4),
(C7), (C10) and (C13). Then, (1.1) has at least one solution in (C[0, T])
n
.
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 11 of 35
Proof Let u =(u
1
, u
2
, , u
n
) Î (C[0, T])

n
be any solution of (3.1)
l
where l Î (0, 1).
Define the sets I and J as in (3.9). Let 1 ≤ i ≤ n. As in the proof of Theorems 3.3 and
3.6, respectively, (C7) leads to (3.16), whereas (C10) and (C13) yield (3.25).
Splitting the integrals in (3.16) and applying (3.25), we find that
β
i
η
i

J
|f
i
(t, u(t))|
γ
i
+1
dt
≤ β
i

J
|φ
i
(t)f
i
(t, u(t))|dt +(H
i

+ a
i
)

J
|f
i
(t, u(t))|dt + |b
i
| +

I
(|u
i
(t)| + H
i
+ a
i
)|f
i
(t, u(t))|d
t
≤ β
i

J
|φ
i
(t)f
i

(t, u(t))|dt +(H
i
+ a
i
)

J
|f
i
(t, u(t))|dt + |b
i
| +(r + H
i
+ a
i
)T
1
p
i
||μ
r,i
||
q
i
= β
i
J

J
|φ

||μ
r,i
||
q
i
. Substituting (3.34) and (3.35) into (3.39)
then leads to
β
i
η
i

J
|f
i
(t , u(t ))|
γ
i
+1
dt ≤ A

i
⎡
⎣

J
|f
i
(t , u(t ))|
γ

i
+1
dt
⎤
⎦
1
γ
i
+1
+ B


i
(3:40)
where
A

i
≡ T
γ
i
p
i
−γ
i
−1
p
i
(γ
i

+1)
β
i
||φ
i
||
p
i
+(H
i
+ a
i
)T
γ
i
γ
i
+1
.Since
1
γ
i
+1
<
1
, from (3.40), we can
obtain (3.37) where k
i
is some constant. The rest of the proof proceeds as that of The-
orem 3.6. □

Theorem 3.8 Let the following conditions be satisfied for each 1 ≤ i ≤ n : (C1)-(C4),
(C10), (C13), and (C14) where
(C14) there exist constants a
i
≥ 0, 0 <τ
i
<g
i
+1and b
i
such that for any u Î (C[0, T])
n
,
T

0
⎡
⎣
f
i
(t, u(t))
T

0
g
i
(t, s)f
i
(s, u(s))ds
⎤

⎦
dt ≤ a
i
⎡
⎣
T

0
|f
i
(t, u(t))|dt
⎤
⎦
τ
i
+ b
i
.
Then, (1.1) has at least one solution in (C[0, T])
n
.
Proof Let u =(u
1
, u
2
, , u
n
) Î (C[0, T])
n
be any solution of (3.1)

l
where l Î (0, 1).
Define the sets I and J as in (3.9). Let 1 ≤ i ≤ n.FromtheproofofTheorem3.6,we
see that (C10) and (C13) lead to (3.25).
Using (3.25) and (C14) in (3.12), we obtain
β
i
η
i

J
|f
i
(t, u(t))|
γ
i
+1
dt
≤

I
|u
i
(t)f
i
(t, u(t))|dt + β
i

J
|φ

i
|.
(3:41)
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 12 of 35
Note that

I
|u
i
(t)f
i
(t, u(t))|dt +

I
|h
i
(t)f
i
(t, u(t))|dt + a
i
2
τ
i
⎡
⎣

I
|f
i

i
+ |b
i
|
≤ (r + H
i
)T
1
p
i
||μ
r,i
||
q
i
+ a
i
2
τ
i
T
τ
i
p
i
(||μ
r,i
||
q
i

)
τ
i
+ |b
i
|≡k

i
(3:42)
where we have used (3.30) in the last i nequality. Substituting (3.42) into (3.41) and
using (3.34) and (3.35) then provides
β
i
η
i

J
|f
i
(t, u(t))|
γ
i
+1
dt
≤ β
i

J
|φ
i

T
γ
i
p
i
−γ
i
−1
p
i
(γ
i
+1)
||φ
i
||
p
i
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤

⎦
1
γ
i
+1
+ H
i
T
γ
i
γ
i
+1
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1

+a
i
2
τ
i
T
τ
i
γ
i
γ
i
+1
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
τ
i
γ
i

+1
+ k

i
.
(3:43)
Since
1
γ
i
+1
<
1
and
τ
i
γ
i
+1
<
1
, there exists a constant k
i
such that (3.37) holds. The rest
of the proof is similar to that of Theorem 3.6. □
Theorem 3.9 Let the following conditions be satisfied for each 1 ≤ i ≤ n : (C1)-(C4),
(C10), (C13), and (C15) where
(C15) there exist constants d
i
≥ 0, 0 < τ

i
< g
i
+1and e
i
such that for any u Î (C[0, T])
n
,
T

0
⎡
⎣
f
i
(t, u(t))
T

0
g
i
(t, s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ d
i
⎡
⎣

T

0
|f
i
(t, u(t))|
q
i
dt
⎤
⎦
τ
i
q
i
+ e
i
.
Then, (1.1) has at least one solution in (C[0, T])
n
.
Proof Let u =(u
1
, u
2
, , u
n
) Î (C[0, T])
n
be any solution of (3.1)

q
i
+ |e
i
|
≤ β
i

J
|φ
i
(t)f
i
(t, u(t))|dt +

I
|u
i
(t)f
i
(t, u(t))|dt + H
i
T

0
|f
i
(t, u(t))|dt
+d
i

2
τ
i
q
i
⎧
⎪
⎨
⎪
⎩
⎡
⎣

I
|f
i
(t, u(t))|
q
i
dt
⎤
⎦
τ
i
q
i
+
⎡
⎣


+ |e
i
|
≤ (r + H
i
)

I
μ
r,i
(t )dt + d
i
2
τ
i
q
i
⎡
⎣

I
(μ
r,i
(t ))
q
i
dt
⎤
⎦
τ

i
q
i
+ |e
i
|
≤ (r + H
i
)
T

0
μ
r,i
(t )dt + d
i
2
τ
i
q
i
⎡
⎣
T

0
(μ
r,i
(t ))
q

i
dt
⎤
⎦
τ
i
q
i
+ |e
i
|≡
ˆ
k
i
.
(3:45)
Moreover, an application of Hölder’s inequality gives

J
|f
i
(t , u(t ))|
q
i
dt ≤ T
γ
i
+1−q
i
γ

i
+1
⎡
⎣

J
|f
i
(t , u(t ))|
γ
i
+1
dt
⎤
⎦
q
i
γ
i
+1
.
(3:46)
Substituting (3.45) into (3.44) and using (3.34), (3.35) and (3.46) then leads to
β
i
η
i

J
|f

J
|f
i
(t, u(t))|
q
i
dt
⎤
⎦
τ
i
q
i
+
ˆ
k
i
≤ β
i
T
γ
i
p
i
−γ
i
−1
p
i
(γ

i
+1)
||φ
i
||
p
i
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1
+ H
i
T
γ
i
γ

i
+1
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1
+d
i
2
τ
i
q
i
T
τ
i
(γ

i
+1−q
i
)
q
i
(γ
i
+1)
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
τ
i
γ
i
+1
+
ˆ
k

i
.
(3:47)
Noting
1
γ
i
+1
<
1
and
τ
i
γ
i
+1
<
1
, there exists a constant k
i
such that (3.37) holds. The
rest of the proof follows that of Theorem 3.6. □
Theorem 3.10 Let the following conditions be satisfied for eac h 1 ≤ i ≤ n : (C1)-(C4),
(C10), (C13) and (C16) where
(C16) there exist constants c
i
≥ 0, d
i
≥ 0, 0 < τ
i

<g
i
+1and e
i
with
β
i
η
i
> 2c
i
(
2T
)
γ
i
+1−q
i
q
i
such that for any u Î (C[0, T])
n
,
T

0
⎡
⎣
f
i

(t , u(t ))
T

0
g
i
(t , s)f
i
(s, u(s))ds
⎤
⎦
dt
≤ c
i
⎡
⎣
T

0
|f
i
(t , u(t ))|
q
i
dt
⎤
⎦
γ
i
+1

q
i
+ d
i
⎡
⎣
T

0
|f
i
(t , u(t ))|
q
i
dt
⎤
⎦
τ
i
q
i
+ e
i
.
Then, (1.1) has at least one solution in (C[0, T])
n
.
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 14 of 35
Proof Let u =(u

1
, u
2
, , u
n
) Î (C[0, T])
n
be any solution of (3.1)
l
where l Î (0, 1).
Define the sets I and J as in (3.9). Let 1 ≤ i ≤ n.Asbefore,weseethat(C10)and
(C13) lead to (3.25).
Using (3.25) and (C16) in (3.12) gives
β
i
η
i

J
|f
i
(t , u(t ))|
γ
i
+1
dt
≤ β
i

J

|φ
i
(t ) f
i
(t , u(t ))|dt +

I
|u
i
(t ) f
i
(t , u(t ))|dt +
T

0
|h
i
(t ) f
i
(t , u(t ))|d
t
+c
i
⎡
⎣
T

0
|f
i

(t , u(t ))|
q
i
dt
⎤
⎦
γ
i
+1
q
i
+ d
i
⎡
⎣
T

0
|f
i
(t , u(t ))|
q
i
dt
⎤
⎦
τ
i
q
i

+ |e
i
|
≤ β
i

J
|φ
i
(t ) f
i
(t , u(t ))|dt +

I
|u
i
(t ) f
i
(t , u(t ))|dt + H
i
T

0
|f
i
(t , u(t ))|dt
+c
i
2
γ

i
+1
q
i
⎧
⎪
⎪
⎨
⎪
⎪
⎩
⎡
⎣

I
|f
i
(t , u(t ))|
q
i
dt
⎤
⎦
γ
i
+1
q
i
+
⎡

⎣

J
|f
i
(t , u(t ))|
q
i
dt
⎤
⎦
γ
i
+1
q
i
⎫
⎪
⎪
⎬
⎪
⎪
⎭
+d
i
2
τ
i
q
i

⎧
⎪
⎨
⎪
⎩
⎡
⎣

I
|f
i
(t , u(t ))|
q
i
dt
⎤
⎦
τ
i
q
i
+
⎡
⎣

J
|f
i
(t , u(t ))|
q

(t , u(t ))|dt + c
i
2
γ
i
+1
q
i
⎡
⎣

I
|f
i
(t , u(t ))|
q
i
dt
⎤
⎦
γ
i
+1
q
i
+d
i
2
τ
i

q
i
⎡
⎣

I
|f
i
(t , u(t ))|
q
i
dt
⎤
⎦
τ
i
q
i
+ |e
i
|
≤ (r + H
i
)

I
μ
r,i
(t )dt + c
i

2
γ
i
+1
q
i
⎡
⎣

I
(μ
r,i
(t ))
q
i
dt
⎤
⎦
γ
i
+1
q
i
+d
i
2
τ
i
q
i

⎡
⎣

I
(μ
r,i
(t ))
q
i
dt
⎤
⎦
τ
i
q
i
+ |e
i
|
≤ (r + H
i
)
T

0
μ
r,i
(t )dt + c
i
2

γ
i
+1
q
i
⎡
⎣
T

0
(μ
r,i
(t ))
q
i
dt
⎤
⎦
γ
i
+1
q
i
+d
i
2
τ
i
q
i

⎡
⎣
T

0
(μ
r,i
(t ))
q
i
dt
⎤
⎦
τ
i
q
i
+ |e
i
|≡k

i
.
(3:49)
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 15 of 35
Substituting (3.49) into (3.48) and then using (3.34), (3.35) and (3.46) leads to
β
i
η

i

J
|f
i
(t, u(t))|
γ
i
+1
dt
≤ β
i
T
γ
i
p
i
−γ
i
−1
p
i
(γ
i
+1)
||φ
i
||
p
i

⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1
+ H
i
T
γ
i
γ
i
+1
⎡
⎣

J
|f

i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1
+c
i
2
γ
i
+1
q
i
T
γ
i
+1−q
i
q
i

J
|f

i
(t, u(t))|
γ
i
+1
dt
+d
i
2
τ
i
q
i
T
τ
i
(γ
i
+1−q
i
)
q
i
(γ
i
+1)
⎡
⎣

J

|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
τ
i
γ
i
+1
+ k

i
.
(3:50)
Noting
1
γ
i
+1
<
1
,
τ
i
γ

i
+1
<
1
as well as
β
i
η
i
> 2c
i
(
2T
)
γ
i
+1−q
i
q
i
, from (3.50) there exists a con-
stant k
i
such that (3.37) holds. The rest of the proof proceeds as that of Theorem 3.6. □
Theorem 3.11 Let the following conditions be satisfied for eac h 1 ≤ i ≤ n : (C1)-(C4),
(C10), (C13) and (C17) where
(C17) there exist a
i
≥ 0, 0 <τ
i

< g
i
+1,b
i
, and
ψ
i
∈ L
p
i
[
0, T
]
with ψ
i
≥ 0 almost every-
where on [0, T], such that for any u Î (C[0, T])
n
,
T

0
⎡
⎣
f
i
(t, u(t))
T

0

g
i
(t, s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ a
i
⎡
⎣
T

0
ψ
i
(t)


f
i
(t, u(t))


dt
⎤
⎦
τ
i
+ b

i
.
Then, (1.1) has at least one solution in (C[0, T])
n
.
Proof Let u =(u
1
, u
2
, , u
n
) Î (C[0, T])
n
be any solution of (3.1)
l
where l Î (0, 1).
Define the sets I and J as in (3.9). Let 1 ≤ i ≤ n. Once again, conditions (C10) and
(C13) give rise to (3.25).
Similar to the proof of Theorem 3.5, we apply (3.25) and (C17) in (3.12) to get (3.26).
Next, using (3.30) and Hölder’s inequality, we find that

I
|u
i
(t)f
i
(t, u(t))|dt +

I
|h

||μ
r,i
||
q
i
+ a
i
2
τ
i
⎡
⎣

I
ψ
i
(t)μ
r,i
(t)d t
⎤
⎦
τ
i
+ |b
i
|
≤ (r + H
i
)T
1

p
i
||μ
r,i
||
q
i
+ a
i
2
τ
i
(||ψ
i
||
p
i
||μ
r,i
||
q
i
)
τ
i
+ |b
i
|≡k

i

i
(t, u(t))|dt + a
i
2
τ
i
⎡
⎣

J
ψ
i
(t)|f
i
(t, u(t))|dt
⎤
⎦
τ
i
+ k

i
≤ β
i
T
γ
i
p
i
−γ

i
−1
p
i
(γ
i
+1)
||φ
i
||
p
i
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1
+ H

i
T
γ
i
γ
i
+1
⎡
⎣

J
|f
i
(t, u(t))|
γ
i
+1
dt
⎤
⎦
1
γ
i
+1
+a
i
2
τ
i
T

τ
i
(γ
i
p
i
−γ
i
−1)
p
i
(γ
i
+1)
(||ψ
i
||
p
i
)
τ
i
⎡
⎣

J
|f
i
(t, u(t))|
γ

i
+1
dt
⎤
⎦
τ
i
γ
i
+1
+ k

i
.
(3:52)
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 16 of 35
Since
1
γ
i
+1
<
1
and
τ
i
γ
i
+1

<
1
, from (3.52), there exists a constant k
i
such that (3.37)
holds. The rest of the proof proceeds as that of Theorem 3.6. □
Rem ark 3.1 In Theorem 3.5, the conditions (C10) and (C11) can be replaced by the
following, which is evident from the proof.
(C10)’ There exist r>0 and b
i
>0 such that for any u Î (C[0, T])
n
,
u
i
(
t
)
f
i
(
t, u
(
t
))
≥ β
i
|u
i
|

0
·|f
i
(
t, u
(
t
))
| for ||u
(
t
)
|| > r and a.e. t ∈ [0, T]
,
where we denote
|
u
i
|
0
=sup
t∈
[
0,T
]
|u
i
(t )
|
.

(C11)’ There exist r>0, h
i
>0, g
i
>0and
φ
i
∈ L
γ
i
+1
γ
i
[
0, T
]
such that for any u Î (C[0,
T])
n
,
|
u
i
|
0
≥ η
i
|f
i
(

t, u
(
t
)
|
γ
i
+ φ
i
(
t
)
for ||u
(
t
)
|| > r and a.e. t ∈ [0, T]
.
Remark 3.2 In Theorems 3.6-3.11, the conditions (C10) and (C13) can be replaced by
(C10)’ and (C13)’ below, and the proof will be similar.
(C13)’ There exist r>0, h
i
>0, g
i
>0, and
φ
i
∈ L
p
i

[
0, T
]
such that for any u Î (C[0, T])
n
,
|u
i
|
0
≥ η
i
|f
i
(
t, u
(
t
))
|
γ
i
+ φ
i
(
t
)
for ||u
(
t

)
|| > r and a.e . t ∈ [0, T]
.
4 Existence results for (1.2) in (C
l
[0, ∞))
n
Let the Banach space B =(C
l
[0, ∞))
n
be equipped with the norm:
|
|u|| =max
1≤i≤n
sup
t∈[0,∞
)
|u
i
(t ) | =max
1≤i≤n
|u
i
|
0
wherewelet|u
i
|
0

=sup
tÎ[0,∞)
|u
i
(t)|, 1 ≤ i ≤ n. Throughout, for u Î B and t Î [0,
∞), we shall denote that
|
|u(t)|| =max
1
≤
i
≤
n
|u
i
(t ) |
.
Moreover, for each 1 ≤ i ≤ n,let1≤ p
i
≤∞be an integer and q
i
be such that
1
p
i
+
1
q
i
=1

. For
x ∈ L
p
i
[0, ∞
)
, we shall define that
|
|x||
p
i
=
⎧
⎪
⎪
⎨
⎪
⎪
⎩

∞

0
|x(s)|
p
i
ds

1
p

i
,1≤ p
i
< ∞
ess sup
s∈[0,∞
)
|x(s)|, p
i
= ∞.
We shall apply Theorem 2.1 to obtain the first existence result for (1.2) in (C
l
[0, ∞))
n
.
Theorem 4.1 For each 1 ≤ i ≤ n, assume (D1)-(D5) hold where
(D1) h
i
Î C
l
[0, ∞ ), denote H
i
≡ sup
tÎ[0,∞)
|h
i
(t)|,
(D2) f
i
: [0, ∞)×ℝ

n
® ℝ is a L
1
-Carathéodory function, i.e.,
(i) the map u a f
i
(t, u) is continuous for almost all t Î [0, ∞),
(ii) the map t a f
i
(t, u) is measurable for all u Î ℝ
n
,
(iii) for any r >0, there exists μ
r,i
Î L
1
[0, ∞) such that |u| ≤ rimplies|f
i
(t, u)| ≤ μ
r,i
(t) for almost all t Î [0, ∞).
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 17 of 35
(D3)
g
t
i
(s)=g
i
(t , s) ∈ L

∞
[0, ∞
)
for each t Î [0, ∞),
(D4) the map
t → g
t
i
is continuous from [0, ∞) to L
∞
[0, ∞),
(D5) there exists
˜
g
i
∈ L
∞
[0, ∞
)
such that
g
t
i
→
˜
g
i
in L
∞
[0, ∞) as t ® ∞, i.e.,

lim
t→∞
||g
t
i
−
˜
g
i
||
∞
= lim
t→∞
ess sup
s∈[0,∞
)
|g
i
(t , s) −
˜
g
i
(s)| =0
.
In addition, suppose there is a constant M>0, independent of l,with||u|| ≠ M for
any solution u Î (C
l
[0, ∞))
n
to

u
i
(t )=λ
⎛
⎝
h
i
(t )+
∞

0
g
i
(t , s)f
i
(s, u(s))ds
⎞
⎠
, t ∈ [0, ∞), 1 ≤ i ≤ n (4.1)
λ
for each l Î (0, 1). Then, (1.2) has at least one solution in (C
l
[0, ∞ ))
n
.
Proof To begin, let the operator S be defined by
Su
(
t
)

=
(
S
1
u
(
t
)
, S
2
u
(
t
)
, , S
n
u
(
t
))
, t ∈ [0, ∞
)
(4:2)
where
S
i
u(t )=h
i
(t )+
∞


0
g
i
(t , s)f
i
(s, u(s))ds, t ∈ [0, ∞), 1 ≤ i ≤ n
.
(4:3)
Clearly, the system (1.2) is equivalent to u = Su, and (4.1)
l
is the same as u = lSu.
First, we shall show that S :(C
l
[0, ∞))
n
® (C
l
[0, ∞))
n
, or equiv alently S
i
:(C
l
[0, ∞))
n
® C
l
[0, ∞), 1 ≤ i ≤ n.Letu Î (C
l

[0, ∞))
n
.Then,thereexistsr>0 such that ||u|| ≤ r,
and from (D2) there exis ts μ
r,i
Î L
1
[0, ∞) such that |f
i
(s, u)| ≤ μ
r,i
(s) for almost all s Î
[0, ∞). Let t
1
, t
2
Î [0, ∞). Together with (D1) and (D4), we find that
|S
i
u(t
1
) − S
i
u(t
2
)|≤|h
i
(t
1
) − h

i
(t
2
)| +
∞

0
|g
t
1
i
(s) − g
t
2
i
(s)|μ
r,i
(s)d
s
≤|h
i
(t
1
) − h
i
(t
2
)| + ||g
t
1

i
− g
t
2
i
||
∞
||μ
r,i
||
1
→ 0
(4:4)
as t
1
® t
2
. Hence, S
i
u Î C[0, ∞).
To see that S
i
u is bounded, we have for t Î [0, ∞),
|S
i
u(t ) |≤H
i
+
∞


0
|g
i
(t , s)|μ
r,i
(s)ds ≤ H
i
+ ||g
t
i
||
∞
||μ
r,i
||
1
.
(4:5)
By (D5), there exists T
1
>0 such that for t>T
1
,
|
|g
t
i
||
∞
≤||

˜
g
i
||
∞
+1.
On the other hand, for t Î [0, T
1
], we have
|
|g
t
i
||
∞
≤ sup
t∈
[
0,T
1
]
||g
t
i
||
∞
.
Hence,
sup
t∈[0,∞)

||g
t
i
||
∞
≤ max

sup
t∈[0,T
1
]
||g
t
i
||
∞
, ||
˜
g
i
||
∞
+1

≡ K
i
.
(4:6)
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 18 of 35

It follows from (4.5) that for t Î [0, ∞),
|
S
i
u
(
t
)
|≤H
i
+ K
i
||μ
r,i
||
1
≡ M
i
.
(4:7)
Hence, S
i
u is bounded.
It remains to check the existence of the limit lim
t®∞
S
i
u(t). We claim that
lim
t→∞

S
i
u(t )=h
i
(∞)+
∞

0
˜
g
i
(s)f
i
(s, u(s))d
s
(4:8)
where h
i
(∞) ≡ lim
t®∞
h
i
(t). In fact, it follows from (D5) that
∞

0


[g
t

i
(s) −
˜
g
i
(s)]f
i
(s, u(s))


ds ≤||g
t
i
−
˜
g
i
||
∞
||μ
r,i
||
1
→
0
as t ® ∞. This implies
lim
t→∞
∞


0
g
t
i
(s)f
i
(s, u(s))ds =
∞

0
˜
g
i
(s)f
i
(s, u(s))d
s
and so (4.8) is proved. We have hence shown that S :(C
l
[0, ∞))
n
® (C
l
[0, ∞))
n
.
Next,weshallprovethatS :(C
l
[0, ∞))
n

® (C
l
[0, ∞))
n
is continuous. Let {u
m
}bea
sequence in (C
l
[0, ∞))
n
and
u
m
=(u
m
1
, u
m
2
, , u
m
n
) →
u
.In(C
l
[0, ∞))
n
, i.e.,

u
m
i
→ u
i
,in
C
l
[0, ∞), 1 ≤ i ≤ n. We need to show that Su
m
® Su in (C
l
[0, ∞))
n
,orequivalently
S
i
u
m
® S
i
u in C
l
[0, ∞), 1 ≤ i ≤ n.Thereexistsr>0suchthat||u
m
||, ||u|| <r,Noting
(D2), there exists μ
r,i
Î L
1

m
(∞) − S
i
u(∞)|≤
∞

0
|
˜
g
i
(s)[f
i
(s, u
m
(s)) − f
i
(s, u(s))]|ds
.
(4:9)
Since
|
˜
g
i
(
s
)
[f
i

(
s, u
m
(
s
))
− f
i
(
s, u
(
s
))
]|→0asm →∞for almost every s ∈ [0, ∞
)
and
|
˜
g
i
(
s
)
[f
i
(
s, u
m
(
s

))
− f
i
(
s, u
(
s
))
]|≤2μ
r,i
(
s
)
|
˜
g
i
(
s
)
|∈L
1
[0, ∞
),
by the Lebesgue-dominated convergence theorem, it is clear from (4.9) that
|
S
i
u
m

(
∞
)
− S
i
u
(
∞
)
|→0asm →∞
.
(4:10)
Further, using (4.8) again we find that
|S
i
u(t ) − S
i
u(∞)|≤|h
i
(t ) − h
i
(∞)| +
∞

0


g
t
i

(s) −
˜
g
i
(s)


μ
r,i
(s)d
s
≤|h
i
(t ) − h
i
(∞)| + ||g
t
i
−
˜
g
i
||
∞
||μ
r,i
||
1
→ 0
(4:11)

Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 19 of 35
as t ® ∞. Similarly, we also have that
|S
i
u
m
(
t
)
− S
i
u
m
(
∞
)
|→0ast →∞
.
(4:12)
Combining (4.10)-(4.12), we have
|
S
i
u
m
(
t
)
− S

i
u
(
t
)
|→0ast →∞and m →
∞
or equivalently, there exist
ˆ
T >
0
such that
|
S
i
u
m
(
t
)
− S
i
u
(
t
)
|→0asm →∞,forallt >
ˆ
T
.

(4:13)
It remains to check the convergence in
[
0,
ˆ
T
]
. As in (4.4), we find for any
|S
i
u
m
(
t
1
)
− S
i
u
m
(
t
2
)
|→0and|S
i
u
(
t
1

)
− S
i
u
(
t
2
)
|→
0
,
|
S
i
u
m
(
t
1
)
− S
i
u
m
(
t
2
)
|→0and|S
i

u
(
t
1
)
− S
i
u
(
t
2
)
|→
0
(4:14)
as t
1
® t
2
. Furthermore, S
i
u
m
(t) ® S
i
u(t) pointwise on
[
0,
ˆ
T

]
, since, by the Lebesgue-
dominated convergence theorem,
|
S
i
u
m
(t ) − S
i
u(t ) |≤ sup
t∈[0,
ˆ
T]
||g
t
i
||
∞
∞

0
|f
i
(s, u
m
(s)) − f
i
(s, u(s))|ds →
0

(4:15)
as m ® ∞. Combining (4.14) and (4.15) and the fact that
[
0,
ˆ
T
]
is compact yields
|
S
i
u
m
(
t
)
− S
i
u
(
t
)
|→0asm →∞,forallt ∈ [0,
ˆ
T
]
(4:16)
Coupling (4.13) and (4.16), we see that S
i
u

m
® S
i
u in C
l
[0, ∞).
Finally, we shall show that S :(C
l
[0, ∞))
n
® (C
l
[0, ∞))
n
is completely continuous. Let
Ω beaboundedsetin(C
l
[0, ∞))
n
with ||u|| ≤ r for all u Î Ω We need to show that
S
i
Ω is relatively compact for 1 ≤ i ≤ n.First,weseethatS
i
Ω is bounded; in fact, this
follows from an earlier argument in (4.7). Next, using a similar argument as in (4.4),
we see that S
i
Ω is equicontinuous. Moreover, S
i

Ω is equiconvergent follows as in
(4.11). By Theorem 2.2, we conclude that S
i
Ω is relatively compact. Hence, S :(C
l
[0,
∞))
n
® (C
l
[0, ∞))
n
is completely continuous.
We now apply Theorem 2.1 with U ={u Î (C
l
[0, ∞))
n
:||u|| <M} and B = E =(C
l
[0,
∞))
n
to obtain the conclusion of the theorem. □
Remark 4.1 In Theorem 4.1, the conditions (D2)-(D5) can be stated in terms of gen-
eral p
i
and q
i
as follows, and the proof will be similar:
(D2)’ f

i
: [0, ∞)×ℝ
n
® ℝ is a
L
q
i
-Carathéodory function, i.e.,
(i) the map u a f
i
(t, u) is continuous for almost all t Î [0, ∞),
(ii) the map t a f
i
(t, u) is measurable for all u Î ℝ
n
,
(iii) for any r>0, there exists
μ
r,i
∈ L
q
i
[0, ∞
)
such that |u| ≤ r implies |f
i
(t, u)| ≤ μ
r,i
(t) for almost all t Î [0, ∞),
(D3)’

g
t
i
(s)=g
i
(t , s) ∈ L
p
i
[0, ∞
)
, for each t Î [0, ∞),
(D4)’ the map
t → g
t
i
is continuous from [0, ∞)to
L
p
i
[0, ∞
)
,
(D5)’ there exists
˜
g
i
∈ L
p
i
[0, ∞

)
such that
g
t
i
→
˜
g
i
,in
L
p
i
[0, ∞
)
as t ® ∞, i.e.,
lim
t→∞
||g
t
i
−
˜
g
i
||
p
i
= lim
t→∞

⎛
⎝
∞

0
|g
i
(t , s) −
˜
g
i
(s)|
p
i
ds
⎞
⎠
1
p
i
=0
.
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 20 of 35
Our subsequent Theorems 4.2-4.5 use an argument originating from Brezis and
Browder [11]. These results are parallel to Theorems 3.2-3.5 for system (1.1).
Theorem 4.2 Let the following conditions be satisfied for each 1 ≤ i ≤ n : (D1)-(D5),
(C5)
∞
, and (C6)

∞
where
(C5)
∞
there exist B
i
>0 such that for any u Î (C
l
[0, ∞ ))
n
,
∞

0
⎡
⎣
f
i
(t , u(t ))
∞

0
g
i
(t , s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ B

i
,
(C6)
∞
there exist r >0 and a
i
>0 with ra
i
>H
i
such that for any u Î (C
l
[0, ∞ ))
n
,
u
i
(
t
)
f
i
(
t, u
(
t
))
≥ rα
i
|f

i
(
t, u
(
t
))
| for ||u
(
t
)
|| > randa.e. t ∈ [0, ∞
).
Then, (1.2) has at least one solution in (C
l
[0, ∞))
n
.
Proof We shall employ Theorem 4.1, so let u =(u
1
, u
2
, , u
n
) Î (C
l
[0, ∞))
n
be any
solution of (4.1)
l

where l Î (0, 1). The rest of the proof is similar to that of Theorem
3.2 with the obvious modification that [0, T] be replaced by [0, ∞). Also, noting (4.6)
we see that the analog of (3.15) holds. □
In view of the proof of Theorem 4.2, we see that the proof of subsequent Theorems
4.3-4.5 will also be similar to that of Theorems 3.3-3.5 with the appropriate modifica-
tion. As such, we shall present the results and omit the proof.
Theorem 4.3 Let the following conditions be satisfied for each 1 ≤ i ≤ n : (D1)-(D5),
(C7)
∞
and (C8)
∞
where
(C7)
∞
there exist constants a
i
≥ 0 and b
i
such that for any u Î (C
l
[0, ∞ ))
n
,
∞

0
⎡
⎣
f
i

(t , u(t ))
∞

0
g
i
(t , s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ a
i
∞

0
|f
i
(t , u(t ))|dt + b
i
,
(C8)
∞
there exist r >0 and a
i
>0 with ra
i
>H
i
+ a

i
such that for any u Î (C
l
[0, ∞ ))
n
,
u
i
(
t
)
f
i
(
t, u
(
t
))
≥ rα
i
|f
i
(
t, u
(
t
))
| for ||u
(
t

)
|| > randa.e. t ∈ [0, ∞
).
Then, (1.2) has at least one solution in (C
l
[0, ∞))
n
.
Theorem 4.4 Let the following conditions be satisfied for each 1 ≤ i ≤ n : (D1)-(D5),
(C9)
∞
and (C10)
∞
where
(C9)
∞
there exist constants a
i
≥ 0, 0 < τ
i
≤ 1 and b
i
such that for any u Î (C
l
[0, ∞))
n
,
∞

0

⎡
⎣
f
i
(t , u(t ))
∞

0
g
i
(t , s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ a
i
⎡
⎣
∞

0
|f
i
(t , u(t ))|dt
⎤
⎦
τ
i
+ b

i
,
(C10)
∞
there exist r >0 and b
i
>0 such that for any u Î (C
l
[0, ∞))
n
,
u
i
(
t
)
f
i
(
t, u
(
t
))
≥ β
i
||u
(
t
)
|| · |f

i
(
t, u
(
t
))
| for ||u
(
t
)
|| > randa.e. t ∈ [0, ∞
).
Then, (1.2) has at least one solution in (C
l
[0, ∞))
n
.
Theorem 4.5 Let the following conditions be satisfied for each 1 ≤ i ≤ n : (D1)-(D5),
(C10)
∞
, (C11)
∞
and (C12)
∞
where
(C11)
∞
there exist r >0, h
i
>0, g

i
>0 and
φ
i
∈ L
γ
i
+1
γ
i
[0, ∞
)
such that for any u Î (C
l
[0,
∞))
n
,
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 21 of 35
|
|u
(
t
)
|| ≥ η
i
|f
i
(

t, u
(
t
)
|
γ
i
+ φ
i
(
t
)
for ||u
(
t
)
|| > randa.e. t ∈ [0, ∞
),
(C12)
∞
there exist a
i
≥ 0, 0 <τ
i
< g
i
+1,b
i
, and
ψ

i
∈ L
γ
i
+1
γ
i
[0, ∞
)
with ψ
i
≥ 0 almost
everywhere on [0, ∞), such that for any u Î (C
l
[0, ∞))
n
,
∞

0
⎡
⎣
f
i
(t , u(t ))
∞

0
g
i

(t , s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ a
i
⎡
⎣
∞

0
ψ
i
(t ) |f
i
(t , u(t ))|dt
⎤
⎦
τ
i
+ b
i
.
Also, j
i
Î BC[0, ∞),
h
i
∈ L

γ
i
+1
γ
i
[0, ∞
)
, ψ
i
Î BC[0, ∞) and
∞

0
|g
i
(t , s)|
γ
i
+1
γ
i
ds ∈ BC[0, ∞
)
.
Then, (1.2) has at least one solution in (C
l
[0, ∞))
n
.
We also have a remark similar to Remark 3.1.

Remark 4.2 In Theorem 4.5 the conditions (C10)
∞
and (C11)
∞
can be replaced by the
following; this is evident from the proof.
(C10)

∞
There exist r>0 and b
i
>0 such that for any u Î (C
l
[0, ∞))
n
,
u
i
(
t
)
f
i
(
t, u
(
t
))
≥ β
i

|u
i
|
0
·|f
i
(
t, u
(
t
))
| for ||u
(
t
)
|| > r and a.e . t ∈ [0, ∞
),
where we denote
|
u
i
|
0
=sup
t∈[0,∞
)
|u
i
(t )
|

.
(C11)

∞
There exist r >0,h
i
>0,g
i
>0and
φ
i
∈ L
γ
i
+1
γ
i
[0, ∞
)
such that f or any u Î
(C
l
[0, ∞))
n
,
|
u
i
|
0

≥ η
i
|f
i
(
t, u
(
t
))
|
γ
i
+ φ
i
(
t
)
for ||u
(
t
)
|| > r and a.e. t ∈ [0, ∞
).
5 Existence results for (1.2) in (BC[0, ∞))
n
Let the Banach space B =(BC[0, ∞))
n
be equipped with the norm:
||u|| =max
1≤i≤n

sup
t∈[0,∞
)
|u
i
(t ) | =max
1≤i≤n
|u
i
|
0
where we let |u
i
|
0
= sup
tÎ[0,∞)
|u
i
(t)|, 1 <i<n. Throughout, for u Î B and t Î [0, ∞)
we shall denote
|
|u(t)|| =max
1
≤
i
≤
n
|u
i

(t ) |
.
Moreover, for each 1 ≤ i ≤ n,let1≤ p
i
≤∞be an integer and q
i
be such that
1
p
i
+
1
q
i
=1
. For
x ∈ L
p
i
[0, ∞
)
, we shall define
||
x
||
p
i
as in Section 4.
Our first result is a variation of an existence principle of Lee and O’Regan [25].
Theorem 5.1 For each 1 ≤ i ≤ n, assume (D2)’-(D4)’ and (D6) hold where

(D6) h
i
Î BC[0, ∞), denote H
i
≡ sup
tÎ[0, ∞)
|h
i
(t)|.
For each k = 1, 2, , suppose there exists
u
k
=(u
k
1
, u
k
2
, , u
k
n
) ∈ (C[0, k])
n
that satis-
fies
u
k
i
(t )=h
i

(t )+
k

0
g
i
(t , s)f
i
(s, u
k
1
(s), u
k
2
(s), , u
k
n
(s))ds, t ∈ [0, k], 1 ≤ i ≤ n
.
(5:1)
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 22 of 35
Further, for 1 ≤ i ≤ n and k = 1, 2, , there is a bounded set B ⊆ ℝ such that
u
k
i
(t ) ∈
B
for each t Î [0, k]. Then, (1.2) has a solution u* Î (BC[0, ∞))
n

such that for
1 ≤ i ≤ n,
u
∗
i
(t ) ∈
¯
B
for all t Î [0, ∞).
Proof First we shall show that

for each 1 ≤ i ≤ n and  =1,2, , the sequence {u
k
i
}
k≥

is uniformly bounded and equicontinuous on [0, ].
(5:2)
The uniform boundedness of
{u
k
i
}
k≥

follows immediately from the hypotheses; there-
fore,weonlyneedtoprovethat
{u
k

i
}
k≥
is equi continuous. Let 1 ≤ i ≤ n.Since
u
k
i
(t ) ∈
B
for each t Î [0, k], there exists
μ
B
∈ L
q
i
[0, ∞
)
such that |f
i
(s,u
k
(s))| ≤ μ
B
( s)
for almost every s Î [0, k].Fix t, t’ Î [0, l]. Then, from (5.1) we find that



u
k

i
(t ) − u
k
i
(t

)



≤|h
i
(t ) − h
i
(t

)| +
k

0



g
t
i
(s) − g
t

i

(s)



·|f
i
(s, u
k
(s))|ds
= |h
i
(t ) − h
i
(t

)| +
∞

0
1
[0,k]



g
t
i
(s) − g
t


i
(s)



·|f
i
(s, u
k
(s))|d
s
≤|h
i
(t ) − h
i
(t

)| + ||g
t
i
− g
t

i
||
p
i
·||μ
B
||

q
i
→ 0
as t ® t’. Therefore,
{u
k
i
}
k≥

is equicontinuous on [0, l].
Let 1 ≤ i ≤ n. Now, (5.2) and the Arzéla-Ascoli theorem yield a subsequence N
1
of N
= {1, 2, } and a function
z
1
i
∈ C[0, 1
]
such that
u
k
i
→ z
1
i
uniformly on [0,1] as k ® ∞
in N
1

. Let
N
∗
2
= N
1
\
{
1
}
. Then, (5.2) and the Arzéla-Ascoli theorem yield a subsequence
N
2
of
N
∗
2
and a function
z
2
i
∈ C[0, 2
]
such that
u
k
i
→ z
2
i

uniformly on [0,2] as k ® ∞ in
N
2
. Note that
z
2
i
= z
1
i
on [0,1] since N
2
⊆ N
1
. Continuing this process, we obtain subse-
quences of integers N
1
, N
2
, with
N
1
⊇ N
2
⊇···⊇N

⊇··· ,whereN

⊆
{

,  +1,
},
(5:3)
and functions
z

i
∈ C[0, 
]
such that
u
k
i
→ z

i
uniformly on [0, ]ask →∞in N

,
and z
+1
i
= z

i
on [0, ],  =1,2,
(5:4)
Let 1 ≤ i ≤ n. Define a function
u
∗

i
:[0,∞] →
R
by
u
∗
i
(t )=z

i
(t ), t ∈ [0, ]
.
(5:5)
Clearly,
u
∗
i
∈ C[0, ∞
)
and
u
∗
i
(t ) ∈
¯
B
for each t Î [0, l]. It remains to prove that
u
∗
=(u

∗
1
, u
∗
2
, , u
∗
n
)
solves (1.2). Fix t Î [0, ∞). Then, choose and fix l such that t Î
[0, l]. Take k ≥ l. Now, from (5.1) we have
u
k
i
(t )=h
i
(t )+
k

0
g
i
(t , s)f
i
(s, u
k
1
(s), u
k
2

(s), , u
k
n
(s))ds, t ∈ [0, 
]
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 23 of 35
or equivalently
u
k
i
(t ) − h
i
(t ) −


0
g
i
(t , s)f
i
(s, u
k
1
(s), u
k
2
(s), , u
k
n

(s))ds
=
k

l
g
i
(t , s)f
i
(s, u
k
1
(s), u
k
2
(s), , u
k
n
(s))ds, t ∈ [0, ]
.
(5:6)
Since f
i
is a
L
q
i
-Carathéodory function and
u
k

i
(t ) ∈
B
for each t Î [0, k], there exists
μ
B
∈ L
q
i
[0, ∞
)
such that
|g
i
(t , s)f
i
(s, u
k
1
(s), u
k
2
(s), , u
k
n
(s))|≤|g
t
i
(s)|μ
B

(s), a.e. s ∈ [0, k
]
and
|g
t
i
|μ
B
∈ L
1
[0, ∞
)
. Let k ® ∞ (k Î N
ℓ
) in (5.6). Since
u
k
i
→ z

i
uniformly on [0, ℓ],
an application of Lebesgue-dominated convergence theorem gives






z


i
(t) − h
i
(t) −


0
g
i
(t, s)f
i
(s, z

1
(s), z

2
(s), , z

n
(s))ds






≤
∞


l
|g
t
i
(s)|μ
B
(s)ds, t ∈ [0, 
]
or equivalently (noting (5.5))






u
∗
i
(t) − h
i
(t) −


0
g
i
(t, s)f
i
(s, u

∗
1
(s), u
∗
2
(s), , u
∗
n
(s))ds






≤
∞

l
|g
t
i
(s)|μ
B
(s)ds, t ∈ [0, ]
.
(5:7)
Finally, letting ℓ ® ∞ in (5.7) and use the fact
|
g

t
i
|μ
B
∈ L
1
[0, ∞
)
to get
u
∗
i
(t ) − h
i
(t ) −
∞

0
g
i
(t , s)f
i
(s, u
∗
1
(s), u
∗
2
(s), , u
∗

n
(s))ds =0, t ∈ [0, ∞)
.
Hence,
u
∗
=(u
∗
1
, u
∗
2
, , u
∗
n
)
is a solution of (1.2). □
It is noted that one of the conditions in Theo rem 5.1, namely, (5.1) has a solution in
( C[0, k])
n
, which has already been discussed in Section 3. As such, our subsequent
Theorems 5.2-5.5 will make use of Theorem 5.1 and the technique used in S ection 3.
These results are parallel to Theorems 3.2-3.5 and 4.2-4.5.
Theorem 5.2 Let (D2)-(D4) and (D6) be satisfied for each 1 ≤ i ≤ n. Moreover, sup-
pose the following conditions hold for each 1 ≤ i ≤ n and each w Î {1, 2, }:
(C5)
w
there exist B
i
>0such that for any u Î ( C[0, w])

n
,
w

0

f
i
(t , u(t ))
w

0
g
i
(t , s)f
i
(s, u(s))ds

dt ≤ B
i
,
(C6)
w
there exist r >0 and a
i
>0 with ra
i
>H
i
(H

i
as in (D6)) such that for any u Î
(C[0, w])
n
,
u
i
(
t
)
f
i
(
t, u
(
t
))
≥ rα
i
|f
i
(
t, u
(
t
))
| for ||u
(
t
)

|| randa.e. t ∈ [0, w]
.
Then, (1.2) has at least one solution in (BC[0, ∞))
n
.
Proof We shall apply Theorem 5.1. To do so, for w = 1, 2, , we shall show that the
system
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 24 of 35
u
i
(t )=h
i
(t )+
w

0
g
i
(t , s)f
i
(s, u(s))ds , t ∈ [0, w], 1 ≤ i ≤
n
(5:8)
has a solution in (C[0, w])
n
. Obviously, (5.8) is just (1.1) with T = w. Let w Î {1, 2, }
be fixed.
Let u =(u
1

, u
2
, , u
n
) Î (C[0,w])
n
be any solution of (3.1)
l
(with T = w)wherel Î
(0, 1). We shall model after the proof of Theorem 3.2 with T = w and H
i
given in
(D6). As in (3.9), define
I = {t ∈ [0, w]:||u
(
t
)
|| ≤ r} and J = {t ∈ [0, w]:||u
(
t
)
|| > r}
.
Let 1 ≤ i ≤ n.Ift Î I, then by (D2) there exists μ
r,i
Î L
1
[0, ∞ ) such that

I

|f
i
(t , u(t ))|dt ≤

I
μ
r,i
(t )dt ≤
∞

0
μ
r,i
(t )dt = ||μ
r,i
||
1
[which is the analog of (3.10)]. Proceeding as in the proof of Theorem 3.2, we then
obtain the analog of (3.14) as

J
|f
i
(t , u(t ))|dt ≤
(H
i
+ r)||μ
r,i
||
1

+ B
i
rα
i
− H
i
≡ k
i
(independent of w)
.
Further, the analog of (3.15) appears as
|
u
i
(t)|≤ sup
t∈[0,w]
|h
i
(t)| +

sup
t∈[0,w]
ess sup
s∈[0,w]
|g
i
(t, s)|

(||μ
r,i

||
1
+ k
i
)
≤ H
i
+

sup
t∈[0,∞)
ess sup
s∈[0,∞)
|g
i
(t, s)|

(||μ
r,i
||
1
+ k
i
) ≡ l
i
(independent of w), t ∈ [0, w]
.
(5:9)
Hence, ||u|| ≤ max
1≤i≤n

l
i
= L and we conclude from Theorem 3.1 that (5.8) has a
solution u*in(C[0, w])
n
. Using similar arguments as in getting (5.9), we find
|u
∗
i
(t ) |≤l
i
for each t Î [0, w]. All the conditions of Theorem 5.1 are now satisfied, it
follows that (1.2) has at least one solution in (BC[0, ∞))
n
. □
The proof of subsequent Theorems 5.3-5.5 will model after the proof of Theorem
5.2, and will employ similar arguments as in the proof of Theorems 3.3-3.5. As such,
we shall present the results and omit the proof.
Theorem 5.3 Let (D2)-(D4) and (D6) be satisfied for each 1 ≤ i ≤ n. Moreover, sup-
pose the following conditions hold for each 1 ≤ i ≤ n and each w Î {1, 2, } :
(C7)
w
there exist constants a
i
≥ 0 and b
i
such that for any u Î (C[0, w])
n
,
w


0
⎡
⎣
f
i
(t , u(t))
w

0
g
i
(t , s)f
i
(s, u(s))ds
⎤
⎦
dt ≤ a
i
w

0


f
i
(t , u(t ))


dt + b

i
,
(C8)
w
there exist r >0 and a
i
>0with ra
i
>H
i
+ a
i
(H
i
as in (D6)) such that for any
u Î (C[0, w])
n
,
u
i
(
t
)
f
i
(
t, u
(
t
))

≥ rα
i
|f
i
(
t, u
(
t
))
| for ||u
(
t
)
|| > randa.e. t ∈ [0, w]
.
Then, (1.2) has at least one solution in (BC[0, ∞))
n
.
Theorem 5.4 Let (D2)-(D4) and (D6) be satisfied for each 1 ≤ i ≤ n. Moreover, sup-
pose the following conditions hold for each 1 ≤ i ≤ n and each w Î {1, 2, } :
Agarwal et al. Advances in Difference Equations 2011, 2011:43
/>Page 25 of 35

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