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Solutions and nonnegative solutions for a weighted variable exponent impulsive integro-differential system with multi-point and integral mixed boundary value problems

Boundary Value Problems20132013:161

DOI: 10.1186/1687-2770-2013-161

Received: 19 March 2013

Accepted: 18 June 2013

Published: 5 July 2013

Abstract

This paper investigates the existence of solutions for a weighted p ( t ) -Laplacian impulsive integro-differential system with multi-point and integral mixed boundary value problems via Leray-Schauder’s degree; sufficient conditions for the existence of solutions are given. Moreover, we get the existence of nonnegative solutions.

MSC:34B37.

Keywords

weighted p ( t ) -Laplacian impulsive integro-differential system Leray-Schauder’s degree

1 Introduction

In this paper, we consider the existence of solutions and nonnegative solutions for the following weighted p ( t ) -Laplacian integro-differential system:
p ( t ) u + f ( t , u , ( w ( t ) ) 1 p ( t ) 1 u , S ( u ) , T ( u ) ) = 0 , t ( 0 , 1 ) , t t i ,
(1)
where u : [ 0 , 1 ] R N , f ( , , , , ) : [ 0 , 1 ] × R N × R N × R N × R N R N , t i ( 0 , 1 ) , i = 1 , , k , with the following impulsive boundary value conditions:
lim t t i + u ( t ) lim t t i u ( t ) = A i ( lim t t i u ( t ) , lim t t i ( w ( t ) ) 1 p ( t ) 1 u ( t ) ) , i = 1 , , k ,
(2)
lim t t i + w ( t ) | u | p ( t ) 2 u ( t ) lim t t i w ( t ) | u | p ( t ) 2 u ( t ) = B i ( lim t t i u ( t ) , lim t t i ( w ( t ) ) 1 p ( t ) 1 u ( t ) ) , i = 1 , , k ,
(3)
u ( 0 ) = 0 1 g ( t ) u ( t ) d t , u ( 1 ) = = 1 m 2 α u ( ξ ) 0 1 h ( t ) u ( t ) d t ,
(4)

where p C ( [ 0 , 1 ] , R ) and p ( t ) > 1 , p ( t ) u : = ( w ( t ) | u | p ( t ) 2 u ) is called the weighted p ( t ) -Laplacian; 0 < t 1 < t 2 < < t k < 1 , 0 < ξ 1 < < ξ m 2 < 1 ; α 0 ( = 1 , , m 2 ); g L 1 [ 0 , 1 ] is nonnegative, 0 1 g ( t ) d t = σ [ 0 , 1 ] ; h L 1 [ 0 , 1 ] , 0 1 h ( t ) d t = δ ; A i , B i C ( R N × R N , R N ) ; T and S are linear operators defined by ( S u ) ( t ) = 0 1 h ( t , s ) u ( s ) d s , ( T u ) ( t ) = 0 t k ( t , s ) u ( s ) d s , t [ 0 , 1 ] , where k , h C ( [ 0 , 1 ] × [ 0 , 1 ] , R ) .

If σ < 1 and = 1 m 2 α δ 1 , we say the problem is nonresonant, but if σ = 1 or = 1 m 2 α δ = 1 , we say the problem is resonant.

Throughout the paper, o ( 1 ) means functions which are uniformly convergent to 0 (as n + ); for any v R N , v j will denote the j th component of v; the inner product in R N will be denoted by , , | | will denote the absolute value and the Euclidean norm on R N . Denote J = [ 0 , 1 ] , J = ( 0 , 1 ) { t 1 , , t k } , J 0 = [ t 0 , t 1 ] , J i = ( t i , t i + 1 ] , i = 1 , , k , where t 0 = 0 , t k + 1 = 1 . Denote by J i o the interior of J i , i = 0 , 1 , , k . Let
P C ( J , R N ) = { x : J R N | x C ( J i , R N ) , i = 0 , 1 , , k and  lim t t i + x ( t )  exists for  i = 1 , , k } ,
w P C ( J , R ) satisfy 0 < w ( t ) , t ( 0 , 1 ) { t 1 , , t k } , and ( w ( t ) ) 1 p ( t ) 1 L 1 ( 0 , 1 ) ,
P C 1 ( J , R N ) = { x P C ( J , R N ) | x C ( J i o , R N ) , lim t t i + ( w ( t ) ) 1 p ( t ) 1 x ( t ) and  lim t t i + 1 ( w ( t ) ) 1 p ( t ) 1 x ( t )  exists for  i = 0 , 1 , , k } .

For any x = ( x 1 , , x N ) P C ( J , R N ) , denote | x i | 0 = sup { | x i ( t ) | t J } .

Obviously, P C ( J , R N ) is a Banach space with the norm x 0 = ( i = 1 N | x i | 0 2 ) 1 2 , and P C 1 ( J , R N ) is a Banach space with the norm x 1 = x 0 + ( w ( t ) ) 1 p ( t ) 1 x 0 . Denote L 1 = L 1 ( J , R N ) with the norm
x L 1 = ( i = 1 N | x i | L 1 2 ) 1 2 , x L 1 ,  where  | x i | L 1 = 0 1 | x i ( t ) | d t .
In the following, P C ( J , R N ) and P C 1 ( J , R N ) will be simply denoted by PC and P C 1 , respectively. We denote
u ( t i + ) = lim t t i + u ( t ) , u ( t i ) = lim t t i u ( t ) , w ( 0 ) | u | p ( 0 ) 2 u ( 0 ) = lim t 0 + w ( t ) | u | p ( t ) 2 u ( t ) , w ( 1 ) | u | p ( 1 ) 2 u ( 1 ) = lim t 1 w ( t ) | u | p ( t ) 2 u ( t ) , A i = A i ( lim t t i u ( t ) , lim t t i ( w ( t ) ) 1 p ( t ) 1 u ( t ) ) , i = 1 , , k , B i = B i ( lim t t i u ( t ) , lim t t i ( w ( t ) ) 1 p ( t ) 1 u ( t ) ) , i = 1 , , k .
The study of differential equations and variational problems with nonstandard p ( t ) -growth conditions has attracted more and more interest in recent years (see [14]). The applied background of these kinds of problems includes nonlinear elasticity theory [4], electro-rheological fluids [1, 3], and image processing [2]. Many results have been obtained on these kinds of problems; see, for example, [515]. Recently, the applications of variable exponent analysis in image restoration have attracted more and more attention [1619]. If p ( t ) p (a constant), (1)-(4) becomes the well-known p-Laplacian problem. If p ( t ) is a general function, one can see easily p ( t ) c u c p ( t ) 1 ( p ( t ) u ) in general, but p c u = c p 1 ( p u ) , so p ( t ) represents a non-homogeneity and possesses more nonlinearity, thus p ( t ) is more complicated than p . For example:
  1. (a)
    If Ω R N is a bounded domain, the Rayleigh quotient
    λ p ( x ) = inf u W 0 1 , p ( x ) ( Ω ) { 0 } Ω 1 p ( x ) | u | p ( x ) d x Ω 1 p ( x ) | u | p ( x ) d x
     
is zero in general, and only under some special conditions λ p ( x ) > 0 (see [9]), when Ω R ( N = 1 ) is an interval, the results show that λ p ( x ) > 0 if and only if p ( x ) is monotone. But the property of λ p > 0 is very important in the study of p-Laplacian problems, for example, in [20], the authors use this property to deal with the existence of solutions.
  1. (b)

    If w ( t ) 1 and p ( t ) p (a constant) and p u > 0 , then u is concave, this property is used extensively in the study of one-dimensional p-Laplacian problems (see [21]), but it is invalid for p ( t ) . It is another difference between p and p ( t ) .

     

In recent years, many results have been devoted to the existence of solutions for the Laplacian impulsive differential equation boundary value problems; see, for example, [2229]. There are some methods to deal with these problems, for example, sub-super-solution method, fixed point theorem, monotone iterative method, coincidence degree. Because of the nonlinear property of p , results on the existence of solutions for p-Laplacian impulsive differential equation boundary value problems are rare (see [3033]). In [34], using the coincidence degree method, the present author investigates the existence of solutions for p ( r ) -Laplacian impulsive differential equation with multi-point boundary value conditions, when the problem is nonresonant. Integral boundary conditions for evolution problems have various applications in chemical engineering, thermo-elasticity, underground water flow and population dynamics. There are many papers on the differential equations with integral boundary value problems; see, for example, [3538].

In this paper, when p ( t ) is a general function, we investigate the existence of solutions and nonnegative solutions for the weighted p ( t ) -Laplacian impulsive integro-differential system with integral and multi-point boundary value conditions. Results on these kinds of problems are rare. Our results contain both of the cases of resonance and nonresonance. Our method is based upon Leray-Schauder’s degree. The homotopy transformation used in [34] is unsuitable for this paper. Moreover, this paper will consider the existence of (1) with (2), (4) and the following impulsive condition:
lim t t i + ( w ( t ) ) 1 p ( t ) 1 u ( t ) lim t t i ( w ( t ) ) 1 p ( t ) 1 u ( t ) = D i ( lim t t i u ( t ) , lim t t i ( w ( t ) ) 1 p ( t ) 1 u ( t ) ) , i = 1 , , k ,
(5)

where D i C ( R N × R N , R N ) , the impulsive condition (5) is called a linear impulsive condition (LI for short), and (3) is called a nonlinear impulsive condition (NLI for short). In general, p-Laplacian impulsive problems have two kinds of impulsive conditions, including LI and NLI; but Laplacian impulsive problems only have LI in general. It is another difference between p-Laplacian impulsive problems and Laplacian impulsive problems. Moreover, since the Rayleigh quotient λ p ( x ) = 0 in general and the p ( t ) -Laplacian is non-homogeneity, when we deal with the existence of solutions of variable exponent impulsive problems like (1)-(4), we usually need the nonlinear term that satisfies the sub- ( p 1 ) growth condition, but for the p-Laplacian impulsive problems, the nonlinear term only needs to satisfy the sub- ( p 1 ) growth condition.

Let N 1 , the function f : J × R N × R N × R N × R N R N is assumed to be Caratheodory, by which we mean:
  1. (i)

    For almost every t J , the function f ( t , , , , ) is continuous;

     
  2. (ii)

    For each ( x , y , s , z ) R N × R N × R N × R N , the function f ( , x , y , s , z ) is measurable on J;

     
  3. (iii)
    For each R > 0 , there is a α R L 1 ( J , R ) such that, for almost every t J and every ( x , y , s , z ) R N × R N × R N × R N with | x | R , | y | R , | s | R , | z | R , one has
    | f ( t , x , y , s , z ) | α R ( t ) .
     

We say a function u : J R N is a solution of (1) if u P C 1 with w ( t ) | u | p ( t ) 2 u absolutely continuous on J i o , i = 0 , 1 , , k , which satisfies (1) a.e. on J.

In this paper, we always use C i to denote positive constants, if it cannot lead to confusion. Denote
z = inf t J z ( t ) , z + = sup t J z ( t ) for any  z P C ( J , R ) .
We say f satisfies the sub- ( p 1 ) growth condition if f satisfies
lim | u | + | v | + | s | + | z | + f ( t , u , v , s , z ) ( | u | + | v | + | s | + | z | ) q ( t ) 1 = 0 for  t J  uniformly,

where q ( t ) P C ( J , R ) and 1 < q q + < p .

We will discuss the existence of solutions for system (1)-(4) or (1) with (2), (4) and (5) in the following three cases:

Case (i): σ < 1 , = 1 m 2 α δ = 1 ;

Case (ii): σ = 1 , = 1 m 2 α δ 1 ;

Case (iii): σ < 1 , = 1 m 2 α δ < 1 .

This paper is organized as five sections. In Section 2, we present some preliminaries and give the operator equation which has the same solutions of (1)-(4) in the three cases, respectively. In Section 3, we give the existence of solutions for system (1)-(4) or (1) with (2), (4) and (5) when σ < 1 , = 1 m 2 α δ = 1 . In Section 4, we give the existence of solutions for system (1)-(4) or (1) with (2), (4) and (5) when σ = 1 , = 1 m 2 α δ 1 . Finally, in Section 5, we give the existence of solutions and nonnegative solutions for system (1)-(4) or (1) with (2), (4) and (5) when σ < 1 , = 1 m 2 α δ < 1 .

2 Preliminary

For any ( t , x ) J × R N , denote φ ( t , x ) = | x | p ( t ) 2 x . Obviously, φ has the following properties.

Lemma 2.1 (see [34])

φ is a continuous function and satisfies:
  1. (i)
    For any t [ 0 , 1 ] , φ ( t , ) is strictly monotone, i.e.,
    φ ( t , x 1 ) φ ( t , x 2 ) , x 1 x 2 > 0 for any x 1 , x 2 R N , x 1 x 2 .
     
  2. (ii)
    There exists a function α : [ 0 , + ) [ 0 , + ) , α ( s ) + as s + such that
    φ ( t , x ) , x α ( | x | ) | x | for all x R N .
     
It is well known that φ ( t , ) is a homeomorphism from R N to R N for any fixed t J . Denote
φ 1 ( t , x ) = | x | 2 p ( t ) p ( t ) 1 x for  x R N { 0 } , φ 1 ( t , 0 ) = 0 , t J .

It is clear that φ 1 ( t , ) is continuous and sends bounded sets to bounded sets.

In this section, we will do some preparation and give the operator equation which has the same solutions of (1)-(4) in three cases, respectively. At first, let us now consider the following simple impulsive problem with boundary value condition (4):
( w ( t ) φ ( t , u ( t ) ) ) = f ( t ) , t ( 0 , 1 ) , t t i , lim t t i + u ( t ) lim t t i u ( t ) = a i , i = 1 , , k , lim t t i + w ( t ) | u | p ( t ) 2 u ( t ) lim t t i w ( t ) | u | p ( t ) 2 u ( t ) = b i , i = 1 , , k , }
(6)

where a i , b i R N ; f L 1 .

Denote a = ( a 1 , , a k ) , b = ( b 1 , , b k ) . Obviously, a , b R k N .

We will discuss it in three cases, respectively.

2.1 Case (i)

Suppose that σ < 1 and = 1 m 2 α δ = 1 . If u is a solution of (6) with (4), we have
w ( t ) φ ( t , u ( t ) ) = w ( 0 ) φ ( 0 , u ( 0 ) ) + t i < t b i + 0 t f ( s ) d s , t J .
(7)
Denote ρ 1 = w ( 0 ) φ ( 0 , u ( 0 ) ) . It is easy to see that ρ 1 is dependent on a, b and f ( ) . Define the operator F : L 1 P C as
F ( f ) ( t ) = 0 t f ( s ) d s , t J , f L 1 .
By solving for u in (7) and integrating, we find
u ( t ) = u ( 0 ) + t i < t a i + F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( f ) ( t ) ) ] } ( t ) , t J ,
which together with boundary value condition (4) implies
u ( 0 ) = 1 ( 1 σ ) 0 1 g ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( f ) ( t ) ) ] } ( t ) + t i < t a i ) d t ,
and
= 1 m 2 α { t i < ξ a i + 0 ξ φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( f ) ( t ) ) ] d t } i = 1 k a i 0 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( f ) ( t ) ) ] d t 0 1 h ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( f ) ( t ) ) ] } ( t ) + t i < t a i ) d t = 0 .
Denote W = R 2 k N × L 1 with the norm
ω = i = 1 k | a i | + i = 1 k | b i | + ϑ L 1 , ω = ( a , b , ϑ ) W ,

then W is a Banach space.

For any ω W , we denote
Λ ω ( ρ 1 ) = = 1 m 2 α { t i < ξ a i + 0 ξ φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( ϑ ) ( t ) ) ] d t } i = 1 k a i 0 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( ϑ ) ( t ) ) ] d t 0 1 h ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) + t i < t a i ) d t .
Denote ξ m 1 = 1 . Then
Λ ω ( ρ 1 ) = = 1 m 2 α { ξ t i a i + ξ 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( ϑ ) ( t ) ) ] d t } + 0 1 h ( t ) ( t 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( ϑ ) ( t ) ) ] d t + t i t a i ) d t = = 1 m 2 ( α ξ ξ + 1 h ( t ) d t ) ξ 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t b i + F ( ϑ ) ( t ) ) ] d t = 1 m 2 ξ ξ + 1 h ( t ) ξ t φ 1 [ s , ( w ( s ) ) 1 ( ρ 1 + s i < s b i + F ( ϑ ) ( s ) ) ] d s d t + 0 ξ 1 h ( t ) t 1 φ 1 [ s , ( w ( s ) ) 1 ( ρ 1 + s i < s b i + F ( ϑ ) ( s ) ) ] d s d t = 1 m 2 α ξ t i a i + 0 1 h ( t ) t i t a i d t .
Throughout the paper, we denote
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equs_HTML.gif
Lemma 2.2 Suppose that h ( t ) 0 on [ ξ 1 , 1 ] , α ξ ξ + 1 h ( t ) d t ( = 1 , , m 2 ) and h ( t ) 0 on [ 0 , ξ 1 ] . Then the function Λ ω ( ) has the following properties:
  1. (i)
    For any fixed ω W , the equation
    Λ ω ( ρ 1 ) = 0
    (8)
     
has a unique solution ρ 1 ˜ ( ω ) R N .
  1. (ii)
    The function ρ 1 ˜ : W R N , defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, for any ω = ( a , b , ϑ ) W , we have
    | ρ 1 ˜ ( ω ) | 3 N [ ( 2 N ) p + ( δ E + 1 E i = 1 k | a i | ) p # 1 + i = 1 k | b i | + ϑ L 1 ] ,
     
where the notation M p # 1 means
M p # 1 = { M p + 1 , M > 1 , M p 1 , M 1 .
Proof (i) From Lemma 2.1, it is immediate that
Λ ω ( x 1 ) Λ ω ( x 2 ) , x 1 x 2 < 0 for  x 1 x 2 , x 1 , x 2 R N ,

and hence, if (8) has a solution, then it is unique.

Set R 0 = 3 N [ ( 2 N ) p + ( δ E + 1 E i = 1 k | a i | ) p # 1 + i = 1 k | b i | + ϑ L 1 ] .

Suppose that | ρ 1 | > R 0 , it is easy to see that there exists some j 0 { 1 , , N } such that the absolute value of the j 0 th component https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq130_HTML.gif of ρ 1 satisfies
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equw_HTML.gif
Thus the j 0 th component of ρ 1 + t i < t b i + F ( ϑ ) ( t ) keeps sign on J, namely, for any t J , we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equx_HTML.gif
Obviously, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equy_HTML.gif
then it is easy to see that the j 0 th component of Λ ω ( ρ 1 ) keeps the same sign of https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq130_HTML.gif . Thus,
Λ ω ( ρ 1 ) 0 .
Let us consider the equation
λ Λ ω ( ρ 1 ) + ( 1 λ ) ρ 1 = 0 , λ [ 0 , 1 ] .
(9)
According to the preceding discussion, all the solutions of (9) belong to b ( R 0 + 1 ) = { x R N | x | < R 0 + 1 } . Therefore
d B [ Λ ω ( ρ 1 ) , b ( R 0 + 1 ) , 0 ] = d B [ I , b ( R 0 + 1 ) , 0 ] 0 ,

it means the existence of solutions of Λ ω ( ρ 1 ) = 0 .

In this way, we define a function ρ 1 ˜ ( ω ) : W R N , which satisfies Λ ω ( ρ 1 ˜ ( ω ) ) = 0 .
  1. (ii)
    By the proof of (i), we also obtain ρ 1 ˜ sends bounded sets to bounded sets, and
    | ρ 1 ˜ ( ω ) | 3 N [ ( 2 N ) p + ( δ E + 1 E i = 1 k | a i | ) p # 1 + i = 1 k | b i | + ϑ L 1 ] .
     

It only remains to prove the continuity of ρ 1 ˜ . Let { ω n } be a convergent sequence in W and ω n ω , as n + . Since { ρ 1 ˜ ( ω n ) } is a bounded sequence, it contains a convergent subsequence { ρ 1 ˜ ( ω n j ) } . Suppose that ρ 1 ˜ ( ω n j ) ρ 0 as j + . Since Λ ω n j ( ρ 1 ˜ ( ω n j ) ) = 0 , letting j + , we have Λ ω ( ρ 0 ) = 0 , which together with (i) implies ρ 0 = ρ 1 ˜ ( ω ) , it means ρ 1 ˜ is continuous. This completes the proof. □

Now we denote by N f ( u ) : [ 0 , 1 ] × P C 1 L 1 the Nemytskii operator associated to f defined by
N f ( u ) ( t ) = f ( t , u ( t ) , ( w ( t ) ) 1 p ( t ) 1 u ( t ) , S ( u ) , T ( u ) ) on  J .
(10)
We define ρ 1 : P C 1 R N as
ρ 1 ( u ) = ρ 1 ˜ ( A , B , N f ) ( u ) ,
(11)

where A = ( A 1 , , A k ) , B = ( B 1 , , B k ) .

It is clear that ρ 1 ( ) is continuous and sends bounded sets of P C 1 to bounded sets of R N , and hence it is compact continuous.

If u is a solution of (6) with (4), we have
u ( t ) = u ( 0 ) + t i < t a i + F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 ˜ ( ω ) + t i < t b i + F ( f ) ( t ) ) ] } ( t ) , t [ 0 , 1 ] .
For fixed a , b R k N , we denote K ( a , b ) : L 1 P C 1 as
K ( a , b ) ( ϑ ) ( t ) = F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 ˜ ( a , b , ϑ ) + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) , t J .
Define K 1 : P C 1 P C 1 as
K 1 ( u ) ( t ) = F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 ( u ) + t i < t B i + F ( N f ( u ) ) ( t ) ) ] } ( t ) , t J .
Lemma 2.3 (i) The operator K ( a , b ) is continuous and sends equi-integrable sets in L 1 to relatively compact sets in P C 1 .
  1. (ii)

    The operator K 1 is continuous and sends bounded sets in P C 1 to relatively compact sets in P C 1 .

     
Proof (i) It is easy to check that K ( a , b ) ( ϑ ) ( ) P C 1 , ϑ L 1 , a , b R k N . Since https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq161_HTML.gif and
K ( a , b ) ( ϑ ) ( t ) = φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 ˜ ( a , b , ϑ ) + t i < t b i + F ( ϑ ) ) ] , t [ 0 , 1 ] ,

it is easy to check that K ( a , b ) ( ) is a continuous operator from L 1 to P C 1 .

Let now U be an equi-integrable set in L 1 , then there exists α L 1 such that
| u ( t ) | α ( t ) a.e. in  J  for any  u L 1 .

We want to show that K ( a , b ) ( U ) ¯ P C 1 is a compact set.

Let { u n } be a sequence in K ( a , b ) ( U ) , then there exists a sequence { ϑ n } U such that u n = K ( a , b ) ( ϑ n ) . For any t 1 , t 2 J , we have
| F ( ϑ n ) ( t 1 ) F ( ϑ n ) ( t 2 ) | = | 0 t 1 ϑ n ( t ) d t 0 t 2 ϑ n ( t ) d t | = | t 1 t 2 ϑ n ( t ) d t | | t 1 t 2 α ( t ) d t | .

Hence the sequence { F ( ϑ n ) } is uniformly bounded and equi-continuous. By the Ascoli-Arzela theorem, there exists a subsequence of { F ( ϑ n ) } (which we rename the same) which is convergent in PC. According to the bounded continuity of the operator ρ 1 ˜ , we can choose a subsequence of { ρ 1 ˜ ( a , b , ϑ n ) + F ( ϑ n ) } (which we still denote { ρ 1 ˜ ( a , b , ϑ n ) + F ( ϑ n ) } ) which is convergent in PC, then https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq174_HTML.gif is convergent in PC.

Since
K ( a , b ) ( ϑ n ) ( t ) = F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 ˜ ( a , b , ϑ n ) + t i < t b i + F ( ϑ n ) ) ] } ( t ) , t [ 0 , 1 ] ,
it follows from the continuity of φ 1 and the integrability of https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq176_HTML.gif in L 1 that K ( a , b ) ( ϑ n ) is convergent in PC. Thus { u n } is convergent in P C 1 .
  1. (ii)

    It is easy to see from (i) and Lemma 2.2.

     

This completes the proof. □

Let us define P 1 : P C 1 P C 1 as
P 1 ( u ) = 0 1 g ( t ) [ K 1 ( u ) ( t ) + t i < t A i ] d t 1 σ .

It is easy to see that P 1 is compact continuous.

Lemma 2.4 Suppose that σ < 1 , = 1 m 2 α δ = 1 ; h ( t ) 0 on [ ξ 1 , 1 ] , α ξ ξ + 1 h ( t ) d t ( = 1 , , m 2 ) and h ( t ) 0 on [ 0 , ξ 1 ] . Then u is a solution of (1)-(4) if and only if u is a solution of the following abstract operator equation:
u = P 1 ( u ) + t i < t A i + K 1 ( u ) .
(12)
Proof Suppose that u is a solution of (1)-(4). By integrating (1) from 0 to t, we find that
w ( t ) φ ( t , u ( t ) ) = ρ 1 ( u ) + t i < t B i + F ( N f ( u ) ) ( t ) , t ( 0 , 1 ) , t t 1 , , t k .
(13)
It follows from (13) and (4) that
u ( t ) = u ( 0 ) + t i < t A i u ( t ) = + F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 ( u ) + t i < t B i + F ( N f ( u ) ) ) ] } ( t ) , t [ 0 , 1 ] , u ( 0 ) = 1 ( 1 σ ) u ( 0 ) = × 0 1 g ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 ( u ) + t i < t B i + F ( N f ( u ) ) ) ] } ( t ) + t i < t A i ) d t u ( 0 ) = 0 1 g ( t ) [ K 1 ( u ) ( t ) + t i < t A i ] d t 1 σ = P 1 ( u ) .
(14)
Combining the definition of ρ 1 , we can see
u = P 1 ( u ) + t i < t A i + K 1 ( u ) .
Conversely, if u is a solution of (12), then (2) is satisfied. It is easy to check that
u ( 0 ) = P 1 ( u ) = 0 1 g ( t ) [ K 1 ( u ) ( t ) + t i < t A i ] d t 1 σ , u ( 0 ) = σ u ( 0 ) + 0 1 g ( t ) [ K 1 ( u ) ( t ) + t i < t A i ] d t = 0 1 g ( t ) u ( t ) d t ,
(15)
and
u ( 1 ) = P 1 ( u ) + i = 1 k A i + K 1 ( u ) ( 1 ) .
By the condition of the mapping ρ 1 , we have
= 1 m 2 α { t i < ξ A i + 0 ξ φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t B i + F ( N f ( u ) ) ( t ) ) ] d t } i = 1 k A i 0 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t B i + F ( N f ( u ) ) ( t ) ) ] d t 0 1 h ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 + t i < t B i + F ( N f ( u ) ) ( t ) ) ] } ( t ) + t i < t A i ) d t = 0 .
Thus
u ( 1 ) = = 1 m 2 α u ( ξ ) 0 1 h ( t ) u ( t ) d t .
(16)

It follows from (15) and (16) that (4) is satisfied.

From (12), we have
w ( t ) φ ( t , u ( t ) ) = ρ 1 ( u ) + t i < t B i + F ( N f ( u ) ) ( t ) , t ( 0 , 1 ) , t t i , ( w ( t ) φ ( t , u ) ) = N f ( u ) ( t ) , t ( 0 , 1 ) , t t i .
(17)

It follows from (17) that (3) is satisfied.

Hence u is a solution of (1)-(4). This completes the proof. □

2.2 Case (ii)

Suppose that σ = 1 and = 1 m 2 α δ 1 . If u is a solution of (6) with (4), we have
w ( t ) φ ( t , u ( t ) ) = w ( 0 ) φ ( 0 , u ( 0 ) ) + t i < t b i + 0 t f ( s ) d s , t J .
Denote ρ 2 = w ( 0 ) φ ( 0 , u ( 0 ) ) . It is easy to see that ρ 2 is dependent on a, b and f ( ) . Boundary value condition (4) implies that
0 1 g ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 2 + t i < t b i + F ( f ) ( t ) ) ] } ( t ) + t i < t a i ) d t = 0 , u ( 0 ) = = 1 m 2 α { t i < ξ a i + 0 ξ φ 1 [ t , ( w ( t ) ) 1 ( ρ 2 + t i < t b i + F ( f ) ( t ) ) ] d t } 1 i = 1 m 2 α + δ u ( 0 ) = i = 1 k a i + 0 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 2 + t i < t b i + F ( f ) ( t ) ) ] d t 1 = 1 m 2 α + δ u ( 0 ) = 0 1 h ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 2 + t i < t b i + F ( f ) ( t ) ) ] } ( t ) + t i < t a i ) d t 1 = 1 m 2 α + δ .
For any ω W , we denote
Γ ω ( ρ 2 ) = 0 1 g ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 2 + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) + t i < t a i ) d t .

Throughout the paper, we denote https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq185_HTML.gif .

Lemma 2.5 The function Γ ω ( ) has the following properties:
  1. (i)

    For any fixed ω W , the equation Γ ω ( ρ 2 ) = 0 has a unique solution ρ 2 ˜ ( ω ) R N .

     
  2. (ii)
    The function ρ 2 ˜ : W R N , defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, for any ω = ( a , b , ϑ ) W , we have
    | ρ 2 ˜ ( ω ) | 3 N [ ( 2 N ) p + ( E 1 + 1 E 1 i = 1 k | a i | ) p # 1 + i = 1 k | b i | + ϑ L 1 ] ,
     
where the notation M p # 1 means
M p # 1 = { M p + 1 , M > 1 , M p 1 , M 1 .

Proof Similar to the proof of Lemma 2.2, we omit it here. □

We define ρ 2 : P C 1 R N as ρ 2 ( u ) = ρ 2 ˜ ( A , B , N f ) ( u ) , where A = ( A 1 , , A k ) , B = ( B 1 , , B k ) .

It is clear that ρ 2 ( ) is continuous and sends bounded sets of P C 1 to bounded sets of R N , and hence it is compact continuous.

For fixed a , b R k N , we denote K ( a , b ) : L 1 P C 1 as
K ( a , b ) ( ϑ ) ( t ) = F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 2 ˜ ( a , b , ϑ ) + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) , t J .
Define K 2 : P C 1 P C 1 as
K 2 ( u ) ( t ) = F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 2 ( u ) + t i < t B i + F ( N f ( u ) ) ( t ) ) ] } ( t ) , t J .

Similar to the proof of Lemma 2.3, we have the following.

Lemma 2.6 (i) The operator K ( a , b ) is continuous and sends equi-integrable sets in L 1 to relatively compact sets in P C 1 .
  1. (ii)

    The operator K 2 is continuous and sends bounded sets in P C 1 to relatively compact sets in P C 1 .

     
Let us define P 2 : P C 1 P C 1 as
P 2 ( u ) = = 1 m 2 α [ t i < ξ A i + K 2 ( u ) ( ξ ) ] i = 1 k A i 1 = 1 m 2 α + δ K 2 ( u ) ( 1 ) + 0 1 h ( t ) [ K 2 ( u ) ( t ) + t i < t A i ] d t 1 = 1 m 2 α + δ .

It is easy to see that P 2 is compact continuous.

Lemma 2.7 Suppose that σ = 1 , = 1 m 2 α δ 1 , then u is a solution of (1)-(4) if and only if u is a solution of the following abstract operator equation:
u = P 2 ( u ) + t i < t A i + K 2 ( u ) .

Proof Similar to the proof of Lemma 2.4, we omit it here. □

2.3 Case (iii)

Suppose that σ < 1 and = 1 m 2 α δ < 1 . If u is a solution of (6) with (4), we have
w ( t ) φ ( t , u ( t ) ) = w ( 0 ) φ ( 0 , u ( 0 ) ) + t i < t b i + 0 t f ( s ) d s , t J .

Denote ρ 3 = w ( 0 ) φ ( 0 , u ( 0 ) ) . It is easy to see that ρ 3 is dependent on a, b and f ( ) .

From u ( 0 ) = 0 1 g ( t ) u ( t ) d t , we have
u ( 0 ) = 1 ( 1 σ ) × 0 1 g ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( f ) ( t ) ) ] } ( t ) + t i < t a i ) d t .
(18)
From u ( 1 ) = = 1 m 2 α u ( ξ ) 0 1 h ( t ) u ( t ) d t , we obtain
u ( 0 ) = = 1 m 2 α { t i < ξ a i + 0 ξ φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( f ) ( t ) ) ] d t } 1 = 1 m 2 α + δ i = 1 k a i + 0 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( f ) ( t ) ) ] d t 1 = 1 m 2 α + δ 0 1 h ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( f ) ( t ) ) ] } ( t ) + t i < t a i ) d t 1 = 1 m 2 α + δ .
(19)
For fixed ω W , we denote
ϒ ω ( ρ 3 ) = 1 ( 1 σ ) 0 1 g ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) + t i < t a i ) d t = 1 m 2 α { t i < ξ a i + 0 ξ φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] d t } 1 = 1 m 2 α + δ + i = 1 k a i + 0 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] d t 1 = 1 m 2 α + δ + 0 1 h ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) + t i < t a i ) d t 1 = 1 m 2 α + δ , ρ 3 R N .

From (18) and (19), we have ϒ ω ( ρ 3 ) = 0 .

Obviously, ϒ ω ( ρ 3 ) can be rewritten as
ϒ ω ( ρ 3 ) = 1 ( 1 σ ) 0 1 g ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) + t i < t a i ) d t + = 1 m 2 α { ξ t i a i + ξ 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] d t } 1 = 1 m 2 α + δ + ( 1 = 1 m 2 α ) 0 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] d t 1 = 1 m 2 α + δ + i = 1 k a i ( 1 = 1 m 2 α ) 1 = 1 m 2 α + δ + 0 1 h ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) + t i < t a i ) d t 1 = 1 m 2 α + δ .
Denote ξ m 1 = 1 . Moreover, we also have
ϒ ω ( ρ 3 ) = 1 ( 1 σ ) 0 1 g ( t ) ( F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) + t i < t a i ) d t + = 1 m 2 α ξ t i a i 1 = 1 m 2 α + δ + = 1 m 2 ( α ξ ξ + 1 h ( t ) d t ) ξ 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] d t 1 = 1 m 2 α + δ + = 1 m 2 ξ ξ + 1 h ( t ) ξ t φ 1 [ s , ( w ( s ) ) 1 ( ρ 3 + s i < s b i + F ( ϑ ) ( s ) ) ] d s d t 1 = 1 m 2 α + δ 0 ξ 1 h ( t ) t 1 φ 1 [ s , ( w ( s ) ) 1 ( ρ 3 + s i < s b i + F ( ϑ ) ( s ) ) ] d s d t + 0 1 h ( t ) t i t a i d t 1 = 1 m 2 α + δ + 0 1 φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 + t i < t b i + F ( ϑ ) ( t ) ) ] d t + i = 1 k a i .

Lemma 2.8 Suppose that α , g, h satisfy one of the following:

(10) = 1 m 2 α 1 , g ( t ) ( 1 = 1 m 2 α + δ ) + h ( t ) ( 1 σ ) 0 ;

(20) h ( t ) 0 on [ ξ 1 , 1 ] , α ξ ξ + 1 h ( t ) d t ( = 1 , , m 2 ) and h ( t ) 0 on [ 0 , ξ 1 ] .

Then the function ϒ ω ( ) has the following properties:
  1. (i)

    For any fixed ω W , the equation ϒ ω ( ρ 3 ) = 0 has a unique solution ρ 3 ˜ ( ω ) R N .

     
  2. (ii)
    The function ρ 3 ˜ : W R N , defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, for any ω = ( a , b , ϑ ) W , we have
    | ρ 3 ˜ ( ω ) | 3 N { ( 2 N ) p + [ ( E 1 + 1 ( 1 σ ) E 1 + ( δ + 1 ) E + 1 ( 1 = 1 m 2 α + δ ) E ) i = 1 k | a i | ] p # 1 + i = 1 k | b i | + ϑ L 1 } ,
     
where the notation M p # 1 means
M p # 1 = { M p + 1 , M > 1 , M p 1 , M 1 .

Proof Similar to the proof of Lemma 2.2, we omit it here. □

We define ρ 3 : P C 1 R N as ρ 3 ( u ) = ρ 3 ˜ ( A , B , N f ) ( u ) , where A = ( A 1 , , A k ) , B = ( B 1 , , B k ) .

It is clear that ρ 3 ( ) is continuous and sends bounded sets of P C 1 to bounded sets of R N , and hence it is compact continuous.

For fixed a , b R k N , we denote K ( a , b ) : L 1 P C 1 as
K ( a , b ) ( ϑ ) ( t ) = F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 ˜ ( a , b , ϑ ) + t i < t b i + F ( ϑ ) ( t ) ) ] } ( t ) , t J .
Define K 3 : P C 1 P C 1 as
K 3 ( u ) ( t ) = F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 3 ( u ) + t i < t B i + F ( N f ( u ) ) ( t ) ) ] } ( t ) , t J .

Similar to the proof of Lemma 2.3, we have

Lemma 2.9 (i) The operator K ( a , b ) is continuous and sends equi-integrable sets in L 1 to relatively compact sets in P C 1 .
  1. (ii)

    The operator K 3 is continuous and sends bounded sets in P C 1 to relatively compact sets in P C 1 .

     
Let us define P 3 : P C 1 P C 1 as
P 3 ( u ) = 0 1 g ( t ) [ K 3 ( u ) ( t ) + t i < t A i ] d t 1 σ .

It is easy to see that P 3 is compact continuous.

Lemma 2.10 Suppose that σ < 1 , = 1 m 2 α δ < 1 and α , g, h satisfy one of the following:

(10) = 1 m 2 α 1 , g ( t ) ( 1 = 1 m 2 α + δ ) + h ( t ) ( 1 σ ) 0 ;

(20) h ( t ) 0 on [ ξ 1 , 1 ] , α ξ ξ + 1 h ( t ) d t ( = 1 , , m 2 ) and h ( t ) 0 on [ 0 , ξ 1 ] .

Then u is a solution of (1)-(4) if and only if u is a solution of the following abstract operator equation:
u = P 3 ( u ) + t i < t A i + K 3 ( u ) .

Proof Similar to the proof of Lemma 2.4, we omit it here. □

3 Existence of solutions in Case (i)

In this section, we apply Leray-Schauder’s degree to deal with the existence of solutions for system (1)-(4) or (1) with (2), (4) and (5) when σ < 1 , = 1 m 2 α δ = 1 .

When f satisfies the sub- ( p 1 ) growth condition, we have the following theorem.

Theorem 3.1 Suppose that σ < 1 , = 1 m 2 α δ = 1 ; h ( t ) 0 on [ ξ 1 , 1 ] , α ξ ξ + 1 h ( t ) d t ( = 1 , , m 2 ) and h ( t ) 0 on [ 0 , ξ 1 ] ; f satisfies the sub- ( p 1 ) growth condition; and operators A and B satisfy the following conditions:
i = 1 k | A i ( u , v ) | C 1 ( 1 + | u | + | v | ) q + 1 p + 1 , i = 1 k | B i ( u , v ) | C 2 ( 1 + | u | + | v | ) q + 1 , ( u , v ) R N × R N ,
(20)

then problem (1)-(4) has at least a solution.

Proof First we consider the following problem:
( S 1 ) { p ( t ) u = λ N f ( u ) ( t ) , t ( 0 , 1 ) , t t i , lim t t i + u ( t ) lim t t i u ( t ) = λ A i ( lim t t i u ( t ) , lim t t i ( w ( t ) ) 1 p ( t ) 1 u ( t ) ) , i = 1 , , k , lim t t i + w ( t ) | u | p ( t ) 2 u ( t ) lim t t i w ( t ) | u | p ( t ) 2 u ( t ) = λ B i ( lim t t i u ( t ) , lim t t i ( w ( t ) ) 1 p ( t ) 1 u ( t ) ) , i = 1 , , k , u ( 0 ) = 0 1 g ( t ) u ( t ) d t , u ( 1 ) = = 1 m 2 α u ( ξ ) 0 1 h ( t ) u ( t ) d t .
Denote
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbh_HTML.gif

where N f ( u ) is defined in (10).

Obviously, ( S 1 ) has the same solution as the following operator equation when λ = 1 :
u = Ψ f ( u , λ ) .
(21)

It is easy to see that the operator https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq228_HTML.gif is compact continuous for any λ [ 0 , 1 ] . It follows from Lemma 2.2 and Lemma 2.3 that Ψ f ( , λ ) is compact continuous from P C 1 to P C 1 for any λ [ 0 , 1 ] .

We claim that all the solutions of (21) are uniformly bounded for λ [ 0 , 1 ] . In fact, if it is false, we can find a sequence of solutions { ( u n , λ n ) } for (21) such that u n 1 + as n + and u n 1 > 1 for any n = 1 , 2 ,  .

From Lemma 2.2, we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbi_HTML.gif
Thus
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ22_HTML.gif
(22)
From ( S 1 ), we have
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbj_HTML.gif
It follows from (11) and Lemma 2.2 that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbk_HTML.gif
Denote α = q + 1 p 1 . If the above inequality holds then
( w ( t ) ) 1 p ( t ) 1 u n ( t ) 0 C 8 u n 1 α , n = 1 , 2 , .
(23)
It follows from (14), (20) and (22) that
| u n ( 0 ) | C 9 u n 1 α , where  α = q + 1 p 1 .
For any j = 1 , , N , we have
| u n j ( t ) | = | u n j ( 0 ) + t i < t A i + 0 t ( u n j ) ( s ) d s | | u n j ( 0 ) | + | t i < t A i | + | 0 t ( w ( s ) ) 1 p ( s ) 1 sup t ( 0 , 1 ) | ( w ( t ) ) 1 p ( t ) 1 ( u n j ) ( t ) | d s | u n 1 α [ C 10 + C 8 E ] + | t i < t A i | C 11 u n 1 α , t J , n = 1 , 2 , ,
which implies that
| u n j | 0 C 12 u n 1 α , j = 1 , , N ; n = 1 , 2 , .
Thus
u n 0 N C 12 u n 1 α , n = 1 , 2 , .
(24)

It follows from (23) and (24) that { u n 1 } is uniformly bounded.

Thus, we can choose a large enough R 0 > 0 such that all the solutions of (21) belong to B ( R 0 ) = { u P C 1 u 1 < R 0 } . Therefore the Leray-Schauder degree d L S [ I Ψ f ( , λ ) , B ( R 0 ) , 0 ] is well defined for λ [ 0 , 1 ] , and
d L S [ I Ψ f ( , 1 ) , B ( R 0 ) , 0 ] = d L S [ I Ψ f ( , 0 ) , B ( R 0 ) , 0 ] .
It is easy to see that u is a solution of u = Ψ f ( u , 0 ) if and only if u is a solution of the following usual differential equation:
( S 2 ) { p ( t ) u = 0 , t ( 0 , 1 ) , u ( 0 ) = 0 1 g ( t ) u ( t ) d t , u ( 1 ) = = 1 m 2 α u ( ξ ) 0 1 h ( t ) u ( t ) d t .
Obviously, system ( S 2 ) possesses a unique solution u 0 . Since u 0 B ( R 0 ) , we have
d L S [ I Ψ f ( , 1 ) , B ( R 0 ) , 0 ] = d L S [ I Ψ f ( , 0 ) , B ( R 0 ) , 0 ] 0 ,

which implies that (1)-(4) has at least one solution. This completes the proof. □

Theorem 3.2 Suppose that σ < 1 , = 1 m 2 α δ = 1 ; h ( t ) 0 on [ ξ 1 , 1 ] , α ξ ξ + 1 h ( t ) d t ( = 1 , , m 2 ) and h ( t ) 0 on [ 0 , ξ 1 ] ; f satisfies the sub- ( p 1 ) growth condition; and operators A and D = ( D 1 , , D k ) satisfy the following conditions:
i = 1 k | A i ( u , v ) | C 1 ( 1 + | u | + | v | ) q + 1 p + 1 , i = 1 k | D i ( u , v ) | C 2 ( 1 + | u | + | v | ) α i + , ( u , v ) R N × R N ,

where α i q + 1 p ( t i ) 1 , and p ( t i ) 1 q + α i , i = 1 , , k .

Then problem (1) with (2), (4) and (5) has at least a solution.

Proof Obviously, B i ( u , v ) = φ ( t i , v + D i ( u , v ) ) φ ( t i , v ) .

From Theorem 3.1, it suffices to show that
i = 1 k | B i ( u , v ) | C 2 ( 1 + | u | + | v | ) q + 1 , ( u , v ) R N × R N .
(25)
  1. (a)
    Suppose that | v | M | D i ( u , v ) | , where M is a large enough positive constant. From the definition of D, we have
    | B i ( u , v ) | C 1 | D i ( u , v ) | p ( t i ) 1 C 2 ( 1 + | u | + | v | ) α i ( p ( t i ) 1 ) .
     
Since α i < q + 1 p ( t i ) 1 , we have α i ( p ( t i ) 1 ) q + 1 . Thus (25) is valid.
  1. (b)
    Suppose that | v | > M | D i ( u , v ) | , we can see that
    | B i ( u , v ) | C 3 | v | p ( t i ) 1 | D i ( u , v ) | | v | = C 4 | v | p ( t i ) 2 | D i ( u , v ) | .
     

There are two cases: Case (i): p ( t i ) 1 1 ; Case (ii): p ( t i ) 1 < 1 .

Case (i): Since p ( t i ) 1 q + α i , we have p ( t i ) 2 + α i q + 1 , and
| B i ( u , v ) | C 5 | v | p ( t i ) 2 | D i ( u , v ) | C 6 ( 1 + | u | + | v | ) p ( t i ) 2 + α i C 6 ( 1 + | u | + | v | ) q + 1 .

Thus (25) is valid.

Case (ii): Since α i < q + 1 p ( t i ) 1 , we have α i ( p ( t i ) 1 ) q + 1 , and
| B i ( u , v ) | C 7 | v | p ( t i ) 2 | D i ( u , v ) | C 8 | D i ( u , v ) | p ( t i ) 1 C 9 ( 1 + | u | + | v | ) α i ( p ( t i ) 1 ) .

Thus (25) is valid.

Thus problem (1) with (2), (4) and (5) has at least a solution. This completes the proof. □

Let us consider
( w ( t ) | u | p ( t ) 2 u ) + ϕ ( t , u , ( w ( t ) ) 1 p ( t ) 1 u , S ( u ) , T ( u ) , ε ) = 0 , t ( 0 , 1 ) , t t i ,
(26)
where ε is a parameter, and
ϕ ( t , u , ( w ( t ) ) 1 p ( t ) 1 u , S ( u ) , T ( u ) , ε ) = f ( t , u , ( w ( t ) ) 1 p ( t ) 1 u , S ( u ) , T ( u ) ) + ε h ( t , u , ( w ( t ) ) 1 p ( t ) 1 u , S ( u ) , T ( u ) ) ,

where h , f : J × R N × R N × R N × R N R N are Caratheodory. We have the following theorem.

Theorem 3.3 Suppose that σ < 1 , = 1 m 2 α δ = 1 ; h ( t ) 0 on [ ξ 1 , 1 ] , α ξ ξ + 1 h ( t ) d t ( = 1 , , m 2 ) and h ( t ) 0 on [ 0 , ξ 1 ] ; f satisfies the sub- ( p 1 ) growth condition; and we assume that
i = 1 k | A i ( u , v ) | C 1 ( 1 + | u | + | v | ) q + 1 p + 1 , i = 1 k | B i ( u , v ) | C 2 ( 1 + | u | + | v | ) q + 1 , ( u , v ) R N × R N ,

then problem (26) with (2)-(4) has at least one solution when parameter ε is small enough.

Proof Denote
ϕ λ ( t , u , ( w ( t ) ) 1 p ( t ) 1 u , S ( u ) , T ( u ) , ε ) = f ( t , u , ( w ( t ) ) 1 p ( t ) 1 u , S ( u ) , T ( u ) ) + λ ε h ( t , u , ( w ( t ) ) 1 p ( t ) 1 u , S ( u ) , T ( u ) ) .
We consider the existence of solutions of the following equation with (2)-(4)
( w ( t ) | u | p ( t ) 2 u ) + ϕ λ ( t , u , ( w ( t ) ) 1 p ( t ) 1 u , S ( u ) , T ( u ) , ε ) = 0 , t ( 0 , 1 ) , t t i .
(27)
Denote
ρ 1 , λ # ( u , ε ) = ρ 1 ˜ ( A , B , N ϕ λ ) ( u ) , K 1 , λ # ( u , ε ) = F { φ 1 [ t , ( w ( t ) ) 1 ( ρ 1 , λ # ( u , ε ) + t i < t B i + F ( N ϕ λ ( u ) ) ( t ) ) ] } , P 1 , λ # ( u , ε ) = 0 1 g ( t ) [ K 1 , λ # ( u , ε ) ( t ) + t i < t A i ] d t ( 1 σ ) , Φ ε ( u , λ ) = P 1 , λ # ( u , ε ) + t i < t A i + K 1 , λ # ( u , ε ) ,

where N ϕ λ ( u ) is defined in (10).

We know that (27) with (2)-(4) has the same solution of u = Φ ε ( u , λ ) .

Obviously, ϕ 0 = f . So Φ ε ( u , 0 ) = Ψ f ( u , 1 ) . As in the proof of Theorem 3.1, we know that all the solutions of u = Φ ε ( u , 0 ) are uniformly bounded, then there exists a large enough R 0 > 0 such that all the solutions of u = Φ ε ( u , 0 ) belong to B ( R 0 ) = { u P C 1 u 1 < R 0 } . Since Φ ε ( , 0 ) is compact continuous from P C 1 to P C 1 , we have
inf u B ( R 0 ) u Φ ε ( u , 0 ) 1 > 0 .
(28)
Since f and h are Caratheodory, we have
F ( N ϕ λ ( u ) ) F ( N ϕ 0 ( u ) ) 0 0 for  ( u , λ ) B ( R 0 ) ¯ × [ 0 , 1 ]  uniformly, as  ε 0 , | ρ 1 , λ # ( u , ε ) ρ 1 , 0 # ( u , ε ) | 0 for  ( u , λ ) B ( R 0 ) ¯ × [ 0 , 1 ]  uniformly, as  ε 0 , K 1 , λ # ( u , ε ) K 1 , 0 # ( u , ε ) 1 0 for  ( u , λ ) B ( R 0 ) ¯ × [ 0 , 1 ]  uniformly, as  ε 0 , | P 1 , λ # ( u , ε ) P 1 , 0 # ( u , ε ) | 0 for  ( u , λ ) B ( R 0 ) ¯ × [ 0 , 1 ]  uniformly, as  ε 0 .
Thus
Φ ε ( u , λ ) Φ 0 ( u , λ ) 1 0 for  ( u , λ ) B ( R 0 ) ¯ × [ 0 , 1 ]  uniformly, as  ε 0 .
Obviously, Φ 0 ( u , λ ) = Φ ε ( u , 0 ) = Φ 0 ( u , 0 ) . We obtain
Φ ε ( u , λ ) Φ ε ( u , 0 ) 1 0 for  ( u , λ ) B ( R 0 ) ¯ × [ 0 , 1 ]  uniformly, as  ε 0 .
Thus, when ε is small enough, from (28), we can conclude that
inf ( u , λ ) B ( R 0 ) × [ 0 , 1 ] u Φ ε ( u , λ ) 1 inf u B ( R 0 ) u Φ ε ( u , 0 ) 1 sup ( u , λ ) B ( R 0 ) ¯ × [ 0 , 1 ] Φ ε ( u , 0 ) Φ ε ( u , λ ) 1 > 0 .
Thus u = Φ ε ( u , λ ) has no solution on B ( R 0 ) for any λ [ 0 , 1 ] , when ε is small enough. It means that the Leray-Schauder degree d L S [ I Φ ε ( , λ ) , B ( R 0 ) , 0 ] is well defined for any λ [ 0 , 1 ] , and
d L S [ I Φ ε ( u , λ ) , B ( R 0 ) , 0 ] = d L S [ I Φ ε ( u , 0 ) , B ( R 0 ) , 0 ] .

Since Φ ε ( u , 0 ) = Ψ f ( u , 1 ) , from the proof of Theorem 3.1, we can see that the right-hand side is nonzero. Thus (26) with (2)-(4) has at least one solution when ε is small enough. This completes the proof. □

Theorem 3.4 Suppose that σ < 1 , = 1 m 2 α δ = 1 ; h ( t ) 0 on [ ξ 1 , 1 ] , α ξ ξ + 1 h ( t ) d t ( = 1 , , m 2 ) and h ( t ) 0 on [ 0 , ξ 1 ] ; f satisfies the sub- ( p 1 ) growth condition; and we assume that
i = 1 k | A i ( u , v ) | C 1 ( 1 + | u | + | v | ) q + 1 p + 1 , i = 1 k | D i ( u , v ) | C 2 ( 1 + | u | + | v | ) α i + , ( u , v ) R N × R N ,

where α i q + 1 p ( t i ) 1 , and p ( t i ) 1 q + α i , i = 1 , , k , then problem (26) with (2), (4) and (5) has at least one solution when parameter ε is small enough.

Proof Similar to the proof of Theorem 3.2 and Theorem 3.3, we omit it here. □

4 Existence of solutions in Case (ii)

In this section, we apply Leray-Schauder’s degree to deal with the existence of solutions for system (1)-(4) or (1) with (2), (4) and (5) when σ = 1 , = 1 m 2 α δ 1 .

When f satisfies the sub- ( p 1 ) growth condition, we have the following.

Theorem 4.1 Suppose that σ = 1 , = 1 m 2 α δ 1 ; f satisfies the sub- ( p 1 ) growth condition; and operators A and B satisfy the following conditions:
i = 1 k | A i ( u , v ) | C 1 ( 1 + | u | + | v | ) q + 1 p + 1 , i = 1 k | B i ( u , v ) | C 2 ( 1 + | u | + | v | ) q + 1 , ( u , v ) R N × R N ,

then problem (1)-(4) has at least a solution.

Proof Similar to the proof of Theorem 3.1, we omit it here. □

Theorem 4.2 Suppose that σ = 1 , = 1 m 2 α δ 1 ; f satisfies the sub- ( p 1 ) growth condition; and operators A and D = ( D 1 , , D k ) satisfy the following conditions:
i = 1 k | A i ( u , v ) | C 1 ( 1 + | u | + | v | ) q + 1 p + 1 , i = 1 k | D i ( u , v ) | C 2 ( 1 + | u | + | v | ) α i + , ( u , v ) R N × R N ,
where
α i q + 1 p ( t i ) 1 and p ( t i ) 1 q + α i , i = 1 , , k ,

then problem (1) with (2), (4) and (5) has at least a solution.

Proof Similar to the proof of Theorem 3.2, we omit it here. □

Theorem 4.3 Suppose that σ = 1 , = 1 m 2 α δ 1 ; f satisfies the sub- ( p 1 ) growth condition; and we assume that
i = 1 k | A i ( u , v ) | C 1 ( 1 + | u | + | v | ) q + 1 p + 1 , i = 1 k | B i ( u , v ) | C 2 ( 1 + | u | + | v | ) q + 1 , ( u , v ) R N × R N ,

then problem (26) with (2)-(4) has at least one solution when parameter ε is small enough.

Proof Similar to the proof of Theorem 3.3, we omit it here. □

Theorem 4.4 Suppose that σ = 1 , = 1 m 2 α δ 1 ; f satisfies the sub- ( p 1 ) growth condition; and we assume that
i = 1 k | A i ( u , v ) | C 1 ( 1 + | u | + | v | ) q + 1 p + 1 , i = 1 k | D i ( u , v ) | C 2 ( 1 + | u | + | v | ) α i + , ( u , v ) R N × R N ,

where α i q + 1 p ( t i ) 1 , and p ( t i ) 1 q + α i , i = 1 , , k , then problem (26) with (2), (4) and (5) has at least one solution when parameter ε is small enough.

Proof Similar to the proof of Theorem 3.2 and Theorem 3.3, we omit it here. □

5 Existence of solutions in Case (iii)

In this section, we apply Leray-Schauder’s degree to deal with the existence of solutions and nonnegative solutions for system (1)-(4) or (1) with (2), (4) and (5) when σ < 1 , = 1 m 2 α δ < 1 .

When f satisfies the sub- ( p 1 ) growth condition, we have the following theorem.

Theorem 5.1 Suppose that σ < 1 , = 1 m 2 α δ < 1 and α , g, h satisfy one of the following:

(10) = 1 m 2 α 1 , g ( t ) ( 1 = 1 m 2 α + δ ) + h ( t ) ( 1 σ ) 0 ;

(20) h ( t ) 0 on [ ξ 1 , 1 ] , α ξ ξ + 1 h ( t ) d t ( = 1 , , m 2