# Existence results of Brezis-Browder type for systems of Fredholm integral equations

## Abstract

In this article, we consider the following systems of Fredholm integral equations:

$u i ( t ) = h i ( t ) + ∫ 0 T g i ( t , s ) f i ( s , u 1 ( s ) , u 2 ( s ) , … , u n ( s ) ) d s , t ∈ [ 0 , T ] , 1 ≤ i ≤ n , u i ( t ) = h i ( t ) + ∫ 0 ∞ g i ( t , s ) f i ( s , u 1 ( s ) , u 2 ( s ) , … , u n ( s ) ) d s , t ∈ [ 0 , ∞ ) , 1 ≤ i ≤ n .$

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,∞))nas well as in (BC[0,∞))n. For both systems, we further seek the existence of constant-sign solutions, which include positive solutions (the usual consideration) as a special case. Several examples are also included to illustrate the results obtained.

2010 Mathematics Subject Classification: 45B05; 45G15; 45M20.

## 1 Introduction

In this article, we shall consider the system of Fredholm integral equations:

$u i ( t ) = h i ( t ) + ∫ 0 T g i ( t , s ) f i ( s , u 1 ( s ) , u 2 ( s ) , … , u n ( s ) ) d s , t ∈ [ 0 , T ] , 1 ≤ i ≤ n$
(1.1)

where 0 < T <∞, and also the following system on the half-line

$u i ( t ) = h i ( t ) + ∫ 0 ∞ g i ( t , s ) f i ( s , u 1 ( s ) , u 2 ( s ) , … , u n ( s ) ) d s , t ∈ [ 0 , ∞ ) , 1 ≤ i ≤ n .$
(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] × × C[0, T] (n times), whereas for the system (1.2), we shall seek a solution in (C l [0, ∞))nas well as in (BC[0, ∞))n. Here, BC[0, ∞) denotes the space of functions that are bounded and continuous on [0, ∞) and C l [0, ∞) = {x 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 ≤ in, we have θ i u i (t) ≥ 0 for all t [0, T] (or t [0,∞)), where θ i {-1, 1} is fixed. Note that when θ i = 1 for all 1 ≤ in, a constant-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 the existence of positive solutions of the nonlinear Fredholm integral equations:

$y ( t ) = h ( t ) + ∫ 0 T g ( t , s ) f ( y ( s ) ) d s , t ∈ [ 0 , T ]$
(1.3)

and

$y ( t ) = h ( t ) + ∫ 0 ∞ g ( t , s ) f ( y ( s ) ) d s , t ∈ [ 0 , ∞ ) .$
(1.4)

Particular cases of (1.3) are also considered in . 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 .

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, Leggett-Williams' fixed point theorem, five-functional fixed point theorem, Schauder fixed point theorem, and Schauder-Tychonoff fixed point theorem. In the current study, we will make use of an argument that originates from Brezis and Browder ; therefore, the technique is different from those of  and the results subsequently obtained are also different. The present article also extends, improves, and complements the studies of [5, 1223]. Indeed, we have generalized the problems to (i) systems; (ii) more general form of nonlinearities f i , 1 ≤ in,; 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 compactness criterion, which are used later. In Section 3, we tackle the existence of solutions 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 , respectively. 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 section, we shall state the theorems that are used later to develop the existence criteria--Theorem 2.1  is Schauder's nonlinear alternative for continuous and compact maps, whereas Theorem 2.2 is the criterion of compactness on C l [0, ∞) [, p. 62].

Theorem 2.1Let B be a Banach space with E B closed and convex. Assume U is a relatively open subset of E with 0 U and $S: U ¯ →E$ is a continuous and compact map. Then either

1. (a)

S has a fixed point in $U ¯$, or

2. (b)

there exist u U and λ (0, 1) such that u = λSu.

Theorem 2.2 [, p. 62] Let P C l [0, ∞). Then P is compact in C l [0, ∞) if the following hold:

1. (a)

P is bounded in C l [0, ∞).

2. (b)

Any y P is equicontinuous on any compact interval of [0, ∞).

3. (c)

P is equiconvergent, i.e., given ε > 0, there exists T(ε) > 0 such that |y(t) - y(∞)| < ε for any tT(ε) and y 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 t[0,T]|u i (t)|, 1 ≤ in. Throughout, for u B and t [0, T], we shall denote

$∥ u ( t ) ∥ = max 1 ≤ i ≤ n ∣ u i ( t ) ∣ .$

Moreover, for each 1 ≤ in, let 1 ≤ 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

$x p i = ∫ 0 T ∣ x ( s ) ∣ p i d s 1 p i , 1 ≤ p i < ∞ ess sup s ∈ [ 0 , T ] ∣ x ( s ) ∣ , p i = ∞ .$

Our first existence result uses Theorem 2.1.

Theorem 3.1 For each 1 ≤ in, assume (C1)- (C4) hold where

(C1) h i C[0, T], denote H i ≡ sup t [0, T]|h i (t)|,

(C2) f i : [0, T] × n is a $L q i$-Carathéodory function:

1. (i)

the map u α f i (t, u) is continuous for almost all t [0, T],;

2. (ii)

the map t α f i (t, u) is measurable for all u n ;

3. (iii)

for any r > 0, there exists $μ r , i ∈ L q i [ 0 , T ]$ such that |u| ≤ r implies |f i (t, u)| ≤ μ r,i (t) for almost all t [0, T];

(C3) $g i t ( s ) = g i ( t , s ) ∈ L p i [ 0 , T ]$ for each t [0, T];

(C4) the map $t↦ g i t$ is continuous from [0, T] to $L p i [ 0 , T ]$.

In addition, suppose there is a constant M > 0, independent of λ, with ||u|| ≠ M for any solution u (C[0, T])nto

$u i ( t ) = λ h i ( t ) + ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s , t ∈ [ 0 , T ] , 1 ≤ i ≤ n ( 3 . 1 ) λ$

for each λ (0, 1). Then, (1.1) has at least one solution in (C[0, T]) n .

Proof Let the operator S be defined by

$S u ( t ) = ( S 1 u ( t ) , S 2 u ( t ) , … , S n u ( t ) ) , t ∈ [ 0 , T ]$
(3.2)

where

$S i u ( t ) = h i ( t ) + ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s , t ∈ [ 0 , T ] , 1 ≤ i ≤ n .$
(3.3)

Clearly, the system (1.1) is equivalent to u = Su, and (3.1) λ is the same as u = λSu.

Note that S maps (C[0, T])ninto (C[0, T]) n , i.e., S i : (C[0, T]) nC[0, T], 1 ≤ in. To see this, note that for any u (C[0, T]) n , there exits r > 0 such that ||u|| < r. Since f i is a $L q i$-Carathéodory function, there exists $μ r , i ∈ L q i [ 0 , T ]$ such that |f i (s, u)| ≤ μ r,i (s) for almost all s [0, T]. Hence, for any t 1, t 2 [0, T], we find for 1 ≤ in,

$∣ S i u ( t 1 ) - S i u ( t 2 ) ∣ ≤ ∣ h i ( t 1 ) - h i ( t 2 ) ∣ + ∫ 0 T ∣ g i t 1 ( s ) - g i t 2 ( s ) ∣ p i d s 1 p i μ r , i q i → 0$
(3.4)

as t 1t 2, where we have used (C1) and (C3). This shows that S : (C[0, T]) n → (C[0, T]) n .

Next, we shall prove that S : (C[0, T])n→ (C[0, T])nis continuous. Let $u m = ( u 1 m , u 2 m , … , u n m ) →u$ in (C[0, T]) n , i.e., $u i m → u i$ in C[0, T], 1 ≤ in. We need to show that Su mSu in (C[0, T]) n , or equivalently S i u mS i u in C[0, T], 1 ≤ in. There exists r > 0 such that ||u m ||, ||u|| < r. Since f 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 [0, T]. Using a similar argument as in (3.4), we get for any t 1, t 2 [0, T] and 1 ≤ in:

$∣ S i u m ( t 1 ) - S i u m ( t 2 ) ∣ → 0 and ∣ S i u ( t 1 ) - S i u ( t 2 ) ∣ → 0$
(3.5)

as t 1t 2. Furthermore, S i u m (t) → S i u(t) pointwise on [0, T], since, by the Lebesgue-dominated convergence theorem,

$∣ S i u m ( t ) - S i u ( t ) ∣ ≤ sup t ∈ [ 0 , T ] ∥ g i t ∥ p i ∫ 0 T ∣ f i ( s , u m ( s ) ) - f i ( s , u ( s ) ) ∣ q i d s 1 q i → 0$
(3.6)

as m → ∞. Combining (3.5) and (3.6) and using the fact that [0, T] is compact, gives for all t [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])nis continuous.

Finally, we shall show that S : (C[0, T])n→ (C[0, T])nis completely continuous. Let Ω be a bounded set in (C[0, T])nwith ||u|| ≤ r for all u Ω. We need to show that S i Ω is relatively compact for 1 ≤ in. Clearly, S i Ω is uniformly bounded, since there exists $μ r , i ∈ L q i [ 0 , T ]$ such that |f i (s, u)| ≤ μ r,i (s) for all u Ω and a.e. s [0, T], and hence

$∣ S i u ∣ 0 ≤ H i + sup t ∈ [ 0 , T ] g i t p i ⋅ μ r , i q i ≡ K i , u ∈ Ω .$
(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 [, Theorem 1.2.4] that S i Ω is relatively compact.

We now apply Theorem 2.1 with U = {u (C[0, T])n: ||u|| < M} and B = E = (C[0, T])nto obtain the conclusion of the theorem. □

Our subsequent results will apply Theorem 3.1. To do so, we shall show that any solution u of (3.1) λ 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 [0, T] : ||u(t)|| ≤ r} and {t [0, T] : ||u(t)|| > r}, where ρ i and r are some constants--this technique originates from the study of Brezis and Browder . 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 ≤ in.

Theorem 3.2. Let the following conditions be satisfied for each 1 ≤ in : (C1)-(C4) with p i =and q i = 1, (C5) and (C6) where

(C5) there exist B i > 0 such that for any u (C[0, T]) n ,

$∫ 0 T f i ( t , u ( t ) ) ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s d t ≤ B i ,$

(C6) there exist r > 0 and α i > 0 with i > H i such that for any u (C[0, T])n,

$u i ( t ) f i ( t , u ( t ) ) ≥ r α i ∣ f i ( t , u ( t ) ) ∣ f o r ∥ u ( t ) ∥ > r and a . e . t ∈ [ 0 , T ] .$

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 ) (C[0, T]) n be any solution of (3.1) λ where λ (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 $∫ 0 T = ∫ I + ∫ J$.

Let 1 ≤ in. If t I, then by (C2), there exists μ r,i L 1[0, T] such that |f i (t, u(t))| ≤ μ r,i (t). Thus, we get

$∫ I ∣ f i ( t , u ( t ) ) ∣ d t ≤ ∫ I μ r , i ( t ) d t ≤ ∫ 0 T μ r , i ( t ) d t = ∥ μ r , i ∥ 1 .$
(3.10)

On the other hand, if t J, then it is clear from (C6) that u i (t)f i (t, u(t)) ≥ 0 for a.e. t [0, T]. It follows that

$∫ J u i ( t ) f i ( t , u ( t ) ) d t ≥ r α i ∫ J ∣ f i ( t , u ( t ) ) ∣ d t .$
(3.11)

We now multiply (3.1) λ by f i (t, u(t)), then integrate from 0 to T to get

$∫ 0 T u i ( t ) f i ( t , u ( t ) ) d t = λ ∫ 0 T h i ( t ) f i ( t , u ( t ) ) d t + λ ∫ 0 T f i ( t , u ( t ) ) ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s d t .$
(3.12)

Using (C5) in (3.12) yields

$∫ 0 T u i ( t ) f i ( t , u ( t ) ) d t ≤ H i ∫ 0 T ∣ f i ( t , u ( t ) ) ∣ d t + B i .$
(3.13)

Splitting the integrals in (3.13) and applying (3.11), we get

$∫ I u i ( t ) f i ( t , u ( t ) ) d t + r α i ∫ J ∣ f i ( t , u ( t ) ) ∣ d t ≤ H i ∫ I ∣ f i ( t , u ( t ) ) ∣ d t + H i ∫ J ∣ f i ( t , u ( t ) ) ∣ d t + B i$

or

$( r α i - H i ) ∫ J ∣ f i ( t , u ( t ) ) ∣ d t ≤ H i ∫ I ∣ f i ( t , u ( t ) ) ∣ d t + ∫ I ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + B i ≤ ( H i + r ) ∥ μ r , i ∥ 1 + B i$

where we have used (3.10) in the last inequality. It follows that

$∫ J ∣ f i ( t , u ( t ) ) ∣ d t ≤ ( H i + r ) ∥ μ r , i ∥ 1 + B i r α i - H i ≡ k i .$
(3.14)

Finally, it is clear from (3.1) λ that for t [0, T] and 1 ≤ in,

$u i ( t ) ≤ H i + ∫ 0 T ∣ g i ( t , s ) f i ( s , u ( s ) ) ∣ d s = H i + ∫ I + ∫ J ∣ g i ( t , s ) f i ( s , u ( s ) ) ∣ d s ≤ H i + sup t ∈ [ 0 , T ] ∥ g i t ∥ ∞ ( ∥ μ r , i ∥ 1 + k i ) ≡ l i$
(3.15)

where we have applied (3.10) and (3.14) in the last inequality. Thus, |u i |0l i for 1 ≤ in and ||u|| ≤ max1≤in l i L. It follows from Theorem 3.1 (with M = L + 1) that (1.1) has a solution u* (C[0, T])n. □

Theorem 3.3 Let the following conditions be satisfied for each 1 ≤ in : (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 (C[0, T])n,

$∫ 0 T f i ( t , u ( t ) ) ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s d t ≤ a i ∫ 0 T ∣ f i ( t , u ( t ) ) ∣ d t + b i ,$

(C8) there exist r > 0 and α i > 0 with rα i > H i + a i such that for any u (C[0, T])n,

$u i ( t ) f i ( t , u ( t ) ) ≥ r α i ∣ f i ( t , u ( t ) ) ∣ f o r ∥ u ( t ) ∥ > r and a . e . t ∈ [ 0 , T ] .$

Then, (1.1) has at least one solution in (C[0, T]) n .

Proof The proof follows that of Theorem 3.2 until (3.12). Let 1 ≤ in. We use (C7) in (3.12) to get

$∫ 0 T u i ( t ) f i ( t , u ( t ) ) d t ≤ ∫ 0 T ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + λ ∫ 0 T f i ( t , u ( t ) ) ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s d t ≤ ( H i + a i ) ∫ 0 T ∣ f i ( t , u ( t ) ) ∣ d t + ∣ b i ∣ .$
(3.16)

Splitting the integrals in (3.16) and applying (3.11) gives

$( r α i - H i - a i ) ∫ J ∣ f i ( t , u ( t ) ) ∣ d t ≤ ( H i + a i ) ∫ I ∣ f i ( t , u ( t ) ) ∣ d t + ∫ I ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + ∣ b i ∣ ≤ ( H i + a i + r ) ∥ μ r , i ∥ 1 + ∣ b i ∣$

where we have also used (3.10) in the last inequality. It follows that

$∫ J ∣ f i ( t , u ( t ) ) ∣ d t ≤ ( H i + a i + r ) ∥ μ r , i ∥ 1 + ∣ b i ∣ r α i - H i - a i ≡ k i .$
(3.17)

The rest of the proof follows that of Theorem 3.2. □

Theorem 3.4 Let the following conditions be satisfied for each 1 ≤ in : (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 (C[0, T])n,

$∫ 0 T f i ( t , u ( t ) ) ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s d t ≤ a i ∫ 0 T ∣ f i ( t , u ( t ) ) ∣ d t τ i + b i ,$

(C10) there exist r > 0 and β i > 0 such that for any u (C[0, T])n,

$u i ( t ) f i ( t , u ( t ) ) ≥ β i ∥ u ( t ) ∥ ⋅ ∣ f i ( t , u ( t ) ) ∣ f o r ∥ u ( t ) ∥ > r and a . e . t ∈ [ 0 , T ] .$

Then, (1.1) has at least one solution in (C[0, T]) n .

Proof Let u = (u 1, u 2,..., u n ) (C[0, T]) n be any solution of (3.1) λ where λ (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 $∫ 0 T = ∫ I 0 + ∫ J 0$.

Let 1 ≤ in. If t 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

$∫ I 0 ∣ f i ( t , u ( t ) ) ∣ d t ≤ ∫ I 0 μ r 0 , i ( t ) d t ≤ ∫ 0 T μ r 0 , i ( t ) d t = ∥ μ r 0 , i ∥ 1 .$
(3.19)

Further, if t J 0, then by (C10) we have

$∫ J 0 u i ( t ) f i ( t , u ( t ) ) d t ≥ β i ∫ J 0 ∥ u ( t ) ∥ ⋅ ∣ f i ( t , u ( t ) ) ∣ d t ≥ β i r 0 ∫ J 0 ∣ f i ( t , u ( t ) ) ∣ d t .$
(3.20)

Now, using (3.20) and (C9) in (3.12) gives

$β i r 0 ∫ J 0 ∣ f i ( t , u ( t ) ) ∣ d t ≤ ∫ I 0 ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ 0 T ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i ∫ 0 T ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣ ≤ ∫ I 0 u i ( t ) f i ( t , u ( t ) ) d t + ∫ 0 T ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ I 0 ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∫ J 0 ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣$
(3.21)

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

$∫ I 0 ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ I 0 ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ I 0 ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣ ≤ ( r 0 + H i ) ∥ μ r 0 , i ∥ 1 + a i 2 τ i ( ∥ μ r 0 , i ∥ 1 ) τ i + ∣ b i ∣ ≡ k ′ i$
(3.22)

Substituting (3.22) in (3.21) then yields

$β i r 0 ∫ J 0 ∣ f i ( t , u ( t ) ) ∣ d t ≤ ∫ J 0 ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ J 0 ∣ f i ( t , u ( t ) ) ∣ d t τ i + k ′ i ≤ H i ∫ J 0 ∣ f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ J 0 ∣ f i ( t , u ( t ) ) ∣ d t τ i + k ′ i .$

Since τ i ≤ 1, there exists a constant $k i ″$ such that

$( β i r 0 - H i - a i 2 τ i ) ∫ J 0 ∣ f i ( t , u ( t ) ) ∣ d t ≤ k i ″$

$∫ J 0 ∣ f i ( t , u ( t ) ) ∣ d t ≤ k ″ i β i r 0 - H i - a i 2 τ i ≡ k i .$
(3.23)

Finally, it is clear from (3.1) λ that for t [0, T] and 1 ≤ in,

$∣ u i ( t ) ∣ ≤ H i + ∫ 0 T ∣ g i ( t , s ) f i ( s , u ( s ) ) ∣ d s = H i + ∫ I 0 + ∫ J 0 ∣ g i ( t , s ) f i ( s , u ( s ) ) ∣ d s ≤ H i + sup t ∈ [ 0 , T ] ∥ g i t ∥ ∞ ( ∥ μ r 0 , i ∥ 1 + k i ) ≡ l i$
(3.24)

where we have 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 ≤ in : (C1), (C2)-(C4) with p i = ∞ and q i = 1, (C10), (C11) and (C12) where

(C11) there exist r > 0, η i > 0, γ i > 0 and $ϕ i ∈ L γ i + 1 γ i [ 0 , T ]$ such that for any u (C[0, T])n,

$∥ u ( t ) ∥ ≥ η i ∣ f i ( t , u ( t ) ∣ γ i + ϕ i ( t ) f o r ∥ u ( t ) ∥ > r and a . e . t ∈ [ 0 , T ] ,$

(C12) there exist a i ≥ 0, 0 < τ i < γ i + 1, b i , and $ψ i ∈ L γ i + 1 γ i [ 0 , T ]$ with ψ i ≥ 0 almost everywhere on [0, T], such that for any u (C[0, T])n,

$∫ 0 T f i ( t , u ( t ) ) ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s d t ≤ a i ∫ 0 T ψ i ( t ) ∣ f i ( t , u ( t ) ) ∣ d t τ i + b i .$

Also, ϕ i C[0, T], $h i ∈ L γ i + 1 γ i [ 0 , T ]$, ψ i C[0, T] and $∫ 0 T ∣ g i ( t , s ) ∣ γ i + 1 γ i d s∈C [ 0 , T ]$.

Then, (1.1) has at least one solution in (C[0, T]) n .

Proof Let u = (u 1, u 2,..., u n ) (C[0, T]) n be any solution of (3.1) λ where λ (0, 1). Define the sets I and J as in (3.9). Let 1 ≤ in. Applying (C10) and (C11), we get

$∫ J u i ( t ) f i ( t , u ( t ) ) d t ≥ β i ∫ J ∥ u ( t ) ∥ ⋅ ∣ f i ( t , u ( t ) ) ∣ d t ≥ β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t + β i ∫ J ϕ i ( t ) ∣ f i ( t , u ( t ) ) ∣ d t .$
(3.25)

Using (3.25) and (C12) in (3.12), we obtain

$β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ ∫ I ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ 0 T ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i ∫ 0 T ψ i ( t ) ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣ ≤ ∫ I ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ 0 T ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ I ψ i ( t ) ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∫ J ψ i ( t ) ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣ .$
(3.26)

Now, in view of (3.10) and (C12), we have

$∫ I ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ I ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ I ψ i ( t ) ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣ ≤ ( r + H i ) ∥ μ r , i ∥ 1 + a i 2 τ i ∫ I ψ i ( t ) μ r , i ( t ) d t τ i + ∣ b i ∣ ≡ k ¯ i .$
(3.27)

Substituting (3.27) into (3.26) and using Hölder's inequality, we find

$β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ J ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ J ψ i ( t ) ∣ f i ( t , u ( t ) ) ∣ d t τ i + k ¯ i ≤ β i ∫ 0 T ∣ ϕ i ( t ) ∣ γ i + 1 γ i d t γ i + 1 γ i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 + ∫ 0 T ∣ h i ( t ) ∣ γ i + 1 γ i d t γ i γ i + 1 ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 + a i 2 τ i ∫ 0 T ∣ ψ i ( t ) ∣ γ i + 1 γ i d t τ i γ i γ i + 1 ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t τ i γ i + 1 + k ¯ i .$

Since $1 γ i + 1 <1$ and $τ i γ i + 1 <1$, there exists a constant k i such that

$∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ k i .$
(3.28)

Finally, it is clear from (3.1) λ that for t [0, T] and 1 ≤ in,

$∣ u i ( t ) ∣ ≤ H i + ∫ I + ∫ J g i ( t , s ) f i ( s , u ( s ) ) ∣ d s ≤ H i + sup t ∈ [ 0 , T ] ∥ g i t ∥ ∞ ∥ μ r , i ∥ 1 + sup t ∈ [ 0 , T ] ∫ 0 T ∣ g i ( t , s ) ∣ γ i + 1 γ i d s γ i γ i + 1 ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 ≤ l i$
(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 ≤ in : (C1)-(C4), (C5), (C10) and (C13) where

(C13) there exist r > 0, η i > 0, γ i > 0 and $ϕ i ∈ L p i [ 0 , T ]$ such that for any u (C[0, T])n,

$∥ u ( t ) ∥ ≥ η i ∣ f i ( t , u ( t ) ∣ γ i + ϕ i ( t ) f o r ∥ u ( t ) ∥ > r and a . e . t ∈ [ 0 , T ] .$

Then, (1.1) has at least one solution in (C[0, T]) n .

Proof Let u = (u 1, u 2,..., u n ) (C[0, T]) n be any solution of (3.1) λ where λ (0, 1). Define the sets I and J as in (3.9). Let 1 ≤ in. If t I, then by (C2), there exists $μ r , i ∈ L q i [ 0 , T ]$ such that |f i (t, u(t))| ≤ μ r,i (t). Consequently, we have

$∫ I ∣ f i ( t , u ( t ) ) ∣ d t ≤ ∫ I μ r , i ( t ) d t ≤ ∫ 0 T μ r , i ( t ) d t ≤ T 1 p i ∥ μ r , i ∥ q i .$
(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

$β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ β i ∫ J ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + H i ∫ J ∣ f i ( t , u ( t ) ) ∣ d t + B i + ∫ I ( ∣ u i ( t ) ∣ + H i ) ∣ f i ( t , u ( t ) ) ∣ d t ≤ β i ∫ J ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + H i ∫ J ∣ f i ( t , u ( t ) ) ∣ d t + B i + ( r + H i ) T 1 p i ∥ μ r , i ∥ q i = β i ∫ J ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + H i ∫ J ∣ f i ( t , u ( t ) ) ∣ d t + B ′ i$
(3.31)

where (3.30) has been used in the last inequality and $B i ′ ≡ B i + ( r + H i ) T 1 p i ∥ μ r , i ∥ q i$.

Now, an application of Hölder's inequality gives

$∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t ≤ ∫ 0 T ∣ ϕ i ( t ) ∣ γ i + 1 γ i d t γ i γ i + 1 ⋅ ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 .$
(3.32)

Another application of Hölder's inequality yields

$∫ 0 T ∣ ϕ i ( t ) ∣ γ i + 1 γ i d t ≤ T γ i p i - γ i - 1 p i γ i ∫ 0 T ∣ ϕ i ( t ) ∣ p i d t γ i + 1 γ i p i .$
(3.33)

Substituting (3.33) into (3.32) then leads to

$∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t ≤ T γ i p i - γ i - 1 p i ( γ i + 1 ) ∥ ϕ i ∥ p i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 .$
(3.34)

Further, using Hölder's inequality again, we get

$∫ J ∣ f i ( t , u ( t ) ) ∣ d t ≤ T γ i γ i + 1 ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 .$
(3.35)

Substituting (3.34) and (3.35) into (3.31), we obtain

$β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ A i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 + B i ′$
(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

$∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ k i .$
(3.37)

Finally, it is clear from (3.1) λ that for t [0, T] and 1 ≤ in,

$∣ u i ( t ) ∣ ≤ H i + ( ∫ I + ∫ J ) | g i ( t , s ) f i ( s , u ( s ) ) | d s ≤ H i + ( sup t ∈ [ 0 , T ] ∥ g i t ∥ p i ) ∥ μ r , i ∥ q i + T γ i p i − γ i − 1 p i ( γ i + 1 ) ( sup t ∈ [ 0 , T ] ∥ g i t ∥ p i ) [ ∫ J ∣ f i ( s , u ( s ) ) ∣ γ i + 1 d s ] 1 γ i + 1 ≤ l i ( a constant ) ,$
(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 ≤ in : (C1)-(C4), (C7), (C10) and (C13). Then, (1.1) has at least one solution in (C[0, T]) n .

Proof Let u = (u 1, u 2,..., u n ) (C[0, T]) n be any solution of (3.1) λ where λ (0, 1). Define the sets I and J as in (3.9). Let 1 ≤ in. 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

$β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ( H i + a i ) ∫ J ∣ f i ( t , u ( t ) ) ∣ d t + ∣ b i ∣ + ∫ I ( ∣ u i ( t ) ∣ + H i + a i ) ∣ f i ( t , u ( t ) ) ∣ d t ≤ β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ( H i + a i ) ∫ J ∣ f i ( t , u ( t ) ) ∣ d t + ∣ b i ∣ + ( r + H i + a i ) T 1 p i ∥ μ r , i ∥ q i = β i ∫ J J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ( H i + a i ) ∫ J ∣ f i ( t , u ( t ) ) ∣ d t + B ″ i$
(3.39)

where $B i ″ ≡∣ b i ∣+ ( r + H i + a i ) T 1 p i ∥ μ r , i ∥ q i$. Substituting (3.34) and (3.35) into (3.39) then leads to

$β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ A i ′ ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 + B i ″$
(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 Theorem 3.6. □

Theorem 3.8 Let the following conditions be satisfied for each 1 ≤ in : (C1)-(C4), (C10), (C13), and (C14) where

(C14) there exist constants a i ≥ 0, 0 < τ i < γ i + 1 and b i such that for any u (C[0, T])n,

$∫ 0 T f i ( t , u ( t ) ) ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s d t ≤ a i ∫ 0 T ∣ f i ( t , u ( t ) ) ∣ d t τ i + b i .$

Then, (1.1) has at least one solution in (C[0, T])n.

Proof Let u = (u 1, u 2,..., u n ) (C[0, T]) n be any solution of (3.1) λ where λ (0, 1). Define the sets I and J as in (3.9). Let 1 ≤ in. From the proof of Theorem 3.6, we see that (C10) and (C13) lead to (3.25).

Using (3.25) and (C14) in (3.12), we obtain

$β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ ∫ I ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ 0 T ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i ∫ 0 T ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣ ≤ ∫ I ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ 0 T ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ I ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∫ J ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣ .$
(3.41)

Note that

$∫ I ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ I ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ I ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣ ≤ ( r + H i ) ∫ I ∣ f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ I ∣ f i ( t , u ( t ) ) ∣ d t τ i + ∣ b i ∣ ≤ ( r + H i ) T 1 p i ∥ μ r , i ∥ q i + a i 2 τ i T τ i p i ( ∥ μ r , i ∥ q i ) τ i + ∣ b i ∣ ≡ k ′ i$
(3.42)

where we have used (3.30) in the last inequality. Substituting (3.42) into (3.41) and using (3.34) and (3.35) then provides

$β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ J ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + a i 2 τ i ∫ J ∣ f i ( t , u ( t ) ) ∣ d t τ i + k ′ i ≤ β i T γ i p i - γ i - 1 p i ( γ i + 1 ) ∥ ϕ i ∥ p i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 + H i T γ i γ i + 1 ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t 1 γ i + 1 + a i 2 τ i T τ i γ i γ i + 1 ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t τ i γ i + 1 + k ′ i .$
(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 ≤ in : (C1)-(C4), (C10), (C13), and (C15) where

(C15) there exist constants d i ≥ 0, 0 < τ i < γ i + 1 and e i such that for any u (C[0, T])n,

$∫ 0 T f i ( t , u ( t ) ) ∫ 0 T g i ( t , s ) f i ( s , u ( s ) ) d s d t ≤ d i ∫ 0 T ∣ f i ( t , u ( t ) ) ∣ q i d t τ i q i + e i .$

Then, (1.1) has at least one solution in (C[0, T]) n .

Proof Let u = (u 1, u 2,..., u n ) (C[0, T]) n be any solution of (3.1) λ where λ (0, 1). Define the sets I and J as in (3.9). Let 1 ≤ in. As before, we see that (C10) and (C13) lead to (3.25).

Using (3.25) and (C15) in (3.12), we obtain

$β i η i ∫ J ∣ f i ( t , u ( t ) ) ∣ γ i + 1 d t ≤ β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ I ∣ u i ( t ) f i ( t , u ( t ) ) ∣ d t + ∫ 0 T ∣ h i ( t ) f i ( t , u ( t ) ) ∣ d t + d i ∫ 0 T ∣ f i ( t , u ( t ) ) ∣ q i d t τ i q i + ∣ e i ∣ ≤ β i ∫ J ∣ ϕ i ( t ) f i ( t , u ( t )$