Open Access

Existence of solutions and nonnegative solutions for a class of p ( t ) -Laplacian differential systems with multipoint and integral boundary value conditions

Boundary Value Problems20132013:106

DOI: 10.1186/1687-2770-2013-106

Received: 12 November 2011

Accepted: 12 April 2013

Published: 26 April 2013

Abstract

This paper explores the existence of solutions for a class of p ( t ) -Laplacian differential systems with multipoint and integral boundary value conditions via Leray-Schauder’s degree. Moreover, the existence of nonnegative solutions is discussed.

MSC:34B10.

Keywords

p ( t ) -Laplacian Leray-Schauder degree fixed point

1 Introduction

In this paper, we consider the existence of solutions for the following system:
( P ) { p 1 ( t ) u = δ 1 f 1 ( t , u , u , v , v ) , t ( 0 , 1 ) , p 2 ( t ) v = δ 2 f 2 ( t , u , u , v , v ) , t ( 0 , 1 ) , u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 , lim t 1 | u | p 1 ( t ) 2 u ( t ) = 0 1 k ( t ) | u | p 1 ( t ) 2 u ( t ) d t + e 2 , v ( 0 ) k 1 v ( 0 ) = 0 1 e ( t ) v ( t ) d t , v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) ,

where p l C ( [ 0 , 1 ] , R ) , p l ( t ) > 1 ( l = 1 , 2 ); p ( t ) γ : = ( | γ | p ( t ) 2 γ ) is called p ( t ) -Laplacian; 0 < ξ 1 < < ξ m 2 < 1 , 0 < η 1 < < η m 2 < 1 ; α i 0 , β i 0 ( i = 1 , , m 2 ) and 0 < i = 1 m 2 α i < 1 , 0 < i = 1 m 2 β i < 1 ; k ( t ) , e ( t ) L 1 ( 0 , 1 ) , they are both nonnegative, σ 1 = 0 1 k ( t ) d t ( 0 , 1 ) , σ 2 = 0 1 e ( t ) d t ( 0 , 1 ) ; e 1 , e 2 R N ; k 1 and k 2 are nonnegative constants; δ 1 and δ 2 are positive parameters.

The study of differential equations and variational problems with variable exponent growth conditions has attracted more and more attention in recent years. Many results have been obtained on these problems, for example, [116]. We refer to [3, 12, 16] for the applied background of these problems. If p ( t ) p (a constant), p ( t ) becomes the well-known p-Laplacian. If p ( t ) is a general function, p ( t ) represents a non-homogeneity and possesses more nonlinearity, thus p ( t ) is more complicated than p (see [7]).

In recent years, because of the wide mathematical and physical background (see [1719]), the existence of positive solutions for the p-Laplacian equation group has received extensive attention. Especially, when p = 2 , the existence of positive solutions for the equation group boundary value problems has been obtained (see [2025]). On the integral boundary value problems, we refer to [2630]. But as for the p ( t ) -Laplacian equation group, there are few papers dealing with the existence of solutions, especially the existence of solutions for the systems with multipoint and integral boundary value problems. Therefore, when p ( t ) is a general function, this paper mainly investigates the existence of solutions for a class of p ( t ) -Laplacian differential systems with multipoint and integral boundary value conditions. Moreover, we discuss the existence of nonnegative solutions.

Let N 1 and J = [ 0 , 1 ] , the function f l = ( f l 1 , , f l N ) : J × R N × R N × R N × R N R N , ( l = 1 , 2 ) is assumed to be Carathéodory, by which we mean:
  1. (i)

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

     
  2. (ii)

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

     
  3. (iii)
    For each R > 0 , there are β R , ρ R L 1 ( J , R ) such that, for almost every t J and every ( x , y , z , w ) R N × R N × R N × R N with | x | R , | y | R , | z | R , | w | R , one has
    | f 1 ( t , x , y , z , w ) | β R ( t ) , | f 2 ( t , x , y , z , w ) | ρ R ( t ) .
     
Throughout the paper, we denote
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The inner product in R N will be denoted by , , | | will denote the absolute value and the Euclidean norm on R N . For N 1 , we set C = C ( J , R N ) , C 1 = { γ C γ C } ; W = { ( u , v ) u , v C 1 } . For any γ ( t ) = ( γ 1 ( t ) , , γ N ( t ) ) C , we denote | γ i | 0 = max t [ 0 , 1 ] | γ i ( t ) | , γ 0 = ( i = 1 N | γ i | 0 2 ) 1 2 and γ 1 = γ 0 + γ 0 . For any ( u , v ) W , we denote ( u , v ) = u 1 + v 1 . Spaces C, C 1 and W will be equipped with the norm 0 , 1 and , respectively. Then ( C , 0 ) , ( C 1 , 1 ) and ( W , ) are Banach spaces. Denote L 1 = L 1 ( J , R N ) with the norm γ L 1 = [ i = 1 N ( 0 1 | γ i | d t ) 2 ] 1 2 .

We say a function ( u , v ) : J R N is a solution of (P) if ( u , v ) W satisfies the differential equation in (P) a.e. on J and the boundary value conditions.

In this paper, we always use C i to denote positive constants if this does not lead to confusion. Denote
b = inf t J b ( t ) , b + = sup t J b ( t ) for any  b C ( J , R ) .
We say f l ( l = 1 , 2 ) satisfies a sub- ( p l 1 ) growth condition if f l satisfies
lim | x | + | y | + | z | + | w | + f l ( t , x , y , z , w ) ( | x | + | y | + | z | + | w | ) q l ( t ) 1 = 0 for  t J  uniformly ,

where q l ( t ) C ( J , R ) , and 1 < q l q l + < p l . We say f l satisfies a general growth condition if f l does not satisfy a sub- ( p l 1 ) growth condition.

We will discuss the existence of solutions for (P) in the following two cases:
  1. (i)

    f l satisfies a sub- ( p l 1 ) growth condition for l = 1 , 2 ;

     
  2. (ii)

    f l satisfies a general growth condition for l = 1 , 2 .

     

This paper is organized as follows. In Section 2, we do some preparation. In Section 3, we discuss the existence of solutions of (P). Finally, in Section 4, we discuss the existence of nonnegative solutions for (P).

2 Preliminary

For any ( t , x ) J × R N , denote φ p l ( t , x ) = | x | p l ( t ) 2 x ( l = 1 , 2 ). Obviously, φ p l has the following properties.

Lemma 2.1 (see [5])

φ p l is a continuous function and satisfies the following:
  1. (i)
    For any t [ 0 , 1 ] , φ p l ( t , ) is strictly monotone, that is,
    φ p l ( t , x 1 ) φ p l ( 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 β l : [ 0 , + ) [ 0 , + ) , β l ( s ) + as s + , such that
    φ p l ( t , x ) , x β l ( | x | ) | x | for all x R N .
     
It is well known that φ p l ( t , ) is a homeomorphism from R N to R N for any fixed t [ 0 , 1 ] . For any t J , denote by φ p l 1 ( t , ) the inverse operator of φ p l ( t , ) , then
φ p l 1 ( t , x ) = | x | 2 p l ( t ) p l ( t ) 1 x , for  x R N { 0 } , φ p l 1 ( t , 0 ) = 0 .

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

Let us now consider the following problem:
( φ p 1 ( t , u ( t ) ) ) = g 1 ( t ) , t ( 0 , 1 ) ,
(1)
with the boundary value condition
u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 , lim t 1 | u | p 1 ( t ) 2 u ( t ) = 0 1 k ( t ) | u | p 1 ( t ) 2 u ( t ) d t + e 2 ,
(2)
where g 1 L 1 . If u is a solution of (1) with (2), by integrating (1) from 0 to t, we find that
φ p 1 ( t , u ( t ) ) = φ p 1 ( 0 , u ( 0 ) ) 0 t g 1 ( s ) d s .
(3)
Denote a 1 = φ p 1 ( 0 , u ( 0 ) ) . It is easy to see that a 1 is dependent on g 1 ( ) . Define operator F : L 1 C as
F ( g 1 ) ( t ) = 0 t g 1 ( s ) d s , t J , g 1 L 1 .
From (3), we have
u ( t ) = φ p 1 1 [ t , a 1 F ( g 1 ) ] .
(4)
By integrating (4) from 0 to t, we find that
u ( t ) = u ( 0 ) + F { φ p 1 1 [ t , a 1 F ( g 1 ) ] } ( t ) , t J .
From (2), we have
a 1 = 0 1 g 1 ( t ) d t 0 1 k ( t ) 0 t g 1 ( s ) d s d t + e 2 1 σ 1 ,
and
u ( 0 ) = i = 1 m 2 α i 0 ξ i φ p 1 1 [ t , a 1 F ( g 1 ) ( t ) ] d t 0 1 φ p 1 1 [ t , a 1 F ( g 1 ) ( t ) ] d t + e 1 1 i = 1 m 2 α i .
For fixed h 1 L 1 , we define a 1 : L 1 R N as
a 1 ( h 1 ) = 0 1 h 1 ( t ) d t 0 1 k ( t ) 0 t h 1 ( s ) d s d t + e 2 1 σ 1 .
(5)

It is easy to obtain the following lemma.

Lemma 2.2 a 1 : L 1 R N is continuous and sends bounded sets of L 1 to bounded sets of R N . Moreover,
| a 1 ( h 1 ) | 2 N 1 σ 1 ( h 1 L 1 + | e 2 | ) .
(6)

It is clear that a 1 ( ) is a compact continuous mapping.

Let us now consider another problem
( φ p 2 ( t , v ( t ) ) ) = g 2 ( t ) , t ( 0 , 1 ) ,
(7)
with the boundary value condition
v ( 0 ) k 1 v ( 0 ) = 0 1 e ( t ) v ( t ) d t , v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) ,
(8)
where g 2 L 1 . Similar to the discussion of the solutions of (1) with (2), we have
v ( t ) = φ p 2 1 [ t , a 2 F ( g 2 ) ] ,
and
v ( t ) = v ( 0 ) + F { φ p 2 1 [ t , a 2 F ( g 2 ) ] } ( t ) , t J ,

where a 2 : = φ p 2 ( 0 , v ( 0 ) ) , F ( g 2 ) ( t ) = 0 t g 2 ( s ) d s for any t J .

From v ( 0 ) k 1 v ( 0 ) = 0 1 e ( t ) v ( t ) d t , we have
v ( 0 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 F ( g 2 ) ( s ) ] d s d t 1 σ 2 .
(9)
From v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) , we have
v ( 0 ) = i = 1 m 2 β i 0 η i φ p 2 1 [ t , a 2 F ( g 2 ) ( t ) ] d t 0 1 φ p 2 1 [ t , a 2 F ( g 2 ) ( t ) ] d t 1 i = 1 m 2 β i k 2 φ p 2 1 [ 1 , a 2 F ( g 2 ) ( 1 ) ] 1 i = 1 m 2 β i .
(10)
From (9) and (10), we have
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For fixed h 2 C , we denote
Λ h 2 ( a 2 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 h 2 ( s ) ] d s d t 1 σ 2 i = 1 m 2 β i 0 η i φ p 2 1 [ t , a 2 h 2 ( t ) ] d t 0 1 φ p 2 1 [ t , a 2 h 2 ( t ) ] d t 1 i = 1 m 2 β i + k 2 φ p 2 1 [ 1 , a 2 h 2 ( 1 ) ] 1 i = 1 m 2 β i .
Lemma 2.3 The function Λ h 2 ( ) has the following properties:
  1. (i)
    For any fixed h 2 C , the equation
    Λ h 2 ( a 2 ) = 0
    (11)
     
has a unique solution a 2 ˜ ( h 2 ) R N .
  1. (ii)
    The function a 2 ˜ : C R N , defined in (i), is continuous and sends bounded sets to bounded sets. Moreover,
    | a 2 ˜ ( h 2 ) | 3 N h 2 0 .
     
Proof (i) It is easy to see that
Λ h 2 ( a 2 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 h 2 ( s ) ] d s d t 1 σ 2 + i = 1 m 2 β i η i 1 φ p 2 1 [ t , a 2 h 2 ( t ) ] d t + k 2 φ p 2 1 [ 1 , a 2 h 2 ( 1 ) ] 1 i = 1 m 2 β i + 0 1 φ p 2 1 [ t , a 2 h 2 ( t ) ] d t .
From Lemma 2.1, it is immediate that
Λ h 2 ( x ) Λ h 2 ( y ) , x y > 0 for  x , y R N  with  x y ,

and hence, if (11) has a solution, then it is unique.

Let t 0 = 3 N h 2 0 . Suppose | a 2 | > t 0 . Since h 2 C , it is easy to see that there exists an i { 1 , , N } such that the i th component a 2 i of a 2 satisfies
| a 2 i | | a 2 | N > 3 h 2 0 .
Thus ( a 2 i h 2 i ( t ) ) keeps sign on J and
| a 2 i h 2 i ( t ) | | a 2 i | h 2 0 2 | a 2 | 3 N > 2 h 2 0 , t J .
Obviously, | a 2 h 2 ( t ) | 4 | a 2 | 3 2 N | a 2 i h 2 i ( t ) | , then
| a 2 h 2 ( t ) | 2 p 2 ( t ) p 2 ( t ) 1 | a 2 i h 2 i ( t ) | > 1 2 N | a 2 i h 2 i ( t ) | 1 p 2 ( t ) 1 > 1 2 N [ 2 h 2 0 ] 1 p 2 ( ζ ) 1 , ζ J , t J .
Thus the i th component Λ h 2 i ( a 2 ) of Λ h 2 ( a 2 ) is nonzero and keeps sign, and then we have
Λ h 2 ( a 2 ) 0 .
Let us consider the equation
λ Λ h 2 ( a 2 ) + ( 1 λ ) a 2 = 0 , λ [ 0 , 1 ] .
(12)
It is easy to see that all the solutions of (12) belong to b ( t 0 + 1 ) = { x R N | x | < t 0 + 1 } . So, we have
d B [ Λ h 2 ( a 2 ) , b ( t 0 + 1 ) , 0 ] = d B [ I , b ( t 0 + 1 ) , 0 ] 0 ,

which implies the existence of solutions of Λ h 2 ( a 2 ) = 0 .

In this way, we define a function a 2 ˜ ( h 2 ) : C [ 0 , 1 ] R N , which satisfies
Λ h 2 ( a 2 ˜ ( h 2 ) ) = 0 .
  1. (ii)
    By the proof of (i), we also obtain that a 2 ˜ sends bounded sets to bounded sets, and
    | a 2 ˜ ( h 2 ) | 3 N h 2 0 .
     

It only remains to prove the continuity of a 2 ˜ . Let { v n } be a convergent sequence in C and v n v as n + . Since { a 2 ˜ ( v n ) } is a bounded sequence, then it contains a convergent subsequence { a 2 ˜ ( v n j ) } . Let a 2 ˜ ( v n j ) a 0 as j + . Since Λ v n j ( a 2 ˜ ( v n j ) ) = 0 , letting j + , we have Λ v ( a 0 ) = 0 . From (i), we get a 0 = a 2 ˜ ( v ) , it means that a 2 ˜ is continuous. The proof is completed. □

Now, we define the operator a 2 : L 1 R N as
a 2 ( v ) = a 2 ˜ ( F ( v ) ) .
(13)

It is clear that a 2 ( ) is continuous and sends bounded sets of L 1 into bounded sets of R N , and hence it is a compact continuous mapping.

If u is a solution of (1) with (2), we have
u ( t ) = u ( 0 ) + F { φ p 1 1 [ t , a 1 F ( g 1 ) ] } ( t ) , t J ,
and
u ( 0 ) = i = 1 m 2 α i 0 ξ i φ p 1 1 [ t , a 1 F ( g 1 ) ( t ) ] d t 0 1 φ p 1 1 [ t , a 1 F ( g 1 ) ( t ) ] d t + e 1 1 i = 1 m 2 α i .
If u is a solution of (7) with (8), we have
v ( t ) = v ( 0 ) + F { φ p 2 1 [ t , a 2 F ( g 2 ) ] } ( t ) , t J ,
and
v ( 0 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 F ( g 2 ) ( s ) ] d s d t 1 σ 2 .
We denote
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Lemma 2.4 The operators K l ( l = 1 , 2 ) are continuous and send equi-integrable sets in L 1 to relatively compact sets in C 1 .

Proof We only prove that the operator K 1 is continuous and sends equi-integrable sets in L 1 to relatively compact sets in C 1 , the rest is similar.

It is easy to check that K 1 ( h 1 ) ( t ) C 1 for all h 1 L 1 . Since
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equaf_HTML.gif

it is easy to check that K 1 is a continuous operator from L 1 to C 1 .

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

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

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

Hence the sequence { F ( h n ) } is uniformly bounded and equicontinuous. By the Ascoli-Arzela theorem, there exists a subsequence of { F ( h n ) } (which we still denote by { F ( h n ) } ) convergent in C. According to the bounded continuous of the operator a 1 , we can choose a subsequence of { a 1 ( h n ) F ( h n ) } (which we still denote by { a 1 ( h n ) F ( h n ) } ) which is convergent in C, then φ p 1 ( t , K 1 ( h n ) ( t ) ) = a 1 ( h n ) F ( h n ) is convergent in C.

From the definition of K 1 ( h n ) ( t ) and the continuity of φ p 1 1 , we can see that K 1 ( h n ) is convergent in C. Thus, { u n } is convergent in C 1 . This completes the proof. □

Let us define P 1 , P 2 : C 1 C 1 as
P 1 ( h 1 ) = i = 1 m 2 α i K 1 ( h 1 ) ( ξ i ) K 1 ( h 1 ) ( 1 ) + e 1 1 i = 1 m 2 α i , P 2 ( h 2 ) = k 1 φ p 2 1 ( 0 , a 2 ( h 2 ) ) + 0 1 e ( t ) K 2 ( h 2 ) ( t ) d t 1 σ 2 .

It is easy to see that P 1 and P 2 are both compact continuous.

We denote N f l ( u , v ) : [ 0 , 1 ] × C 1 L 1 ( l = 1 , 2 ) the Nemytskii operator associated to f l defined by
N f l ( u , v ) ( t ) = f l ( t , u ( t ) , u ( t ) , v ( t ) , v ( t ) ) a.e. on  J .
Lemma 2.5 ( u , v ) is a solution of (P) if and only if ( u , v ) is a solution of the following abstract equation:
( S ) { u = P 1 ( δ 1 N f 1 ( u , v ) ) + K 1 ( δ 1 N f 1 ( u , v ) ) , v = P 2 ( δ 2 N f 2 ( u , v ) ) + K 2 ( δ 2 N f 2 ( u , v ) ) .

Proof If ( u , v ) is a solution to (P), according to the proof before Lemma 2.5, it is easy to obtain that ( u , v ) is a solution to (S).

Conversely, if ( u , v ) is a solution to (S), then
u ( 1 ) = P 1 ( δ 1 N f 1 ( u , v ) ) + K 1 ( δ 1 N f 1 ( u , v ) ) ( 1 ) = i = 1 m 2 α i K 1 ( δ 1 N f 1 ( u , v ) ) ( ξ i ) K 1 ( δ 1 N f 1 ( u , v ) ) ( 1 ) + e 1 1 i = 1 m 2 α i + K 1 ( δ 1 N f 1 ( u , v ) ) ( 1 ) = i = 1 m 2 α i K 1 ( δ 1 N f 1 ( u , v ) ) ( ξ i ) i = 1 m 2 α i K 1 ( δ 1 N f 1 ( u , v ) ) ( 1 ) + e 1 1 i = 1 m 2 α i = i = 1 m 2 α i [ u ( ξ i ) u ( 0 ) ] i = 1 m 2 α i [ u ( 1 ) u ( 0 ) ] + e 1 1 i = 1 m 2 α i = i = 1 m 2 α i u ( ξ i ) i = 1 m 2 α i u ( 1 ) + e 1 1 i = 1 m 2 α i ,
which implies
u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 .
It follows from (S) that
φ p 1 ( t , u ( t ) ) = a 1 ( δ 1 N f 1 ( u , v ) ) F ( δ 1 N f 1 ( u , v ) ) ( t ) ,
then
φ p 1 ( 1 , u ( 1 ) ) = a 1 ( δ 1 N f 1 ( u , v ) ) F ( δ 1 N f 1 ( u , v ) ) ( 1 ) .
By the condition of the mapping a 1 , we have
φ p 1 ( 1 , u ( 1 ) ) = 0 1 δ 1 N f 1 ( u , v ) ( t ) d t 0 1 k ( t ) 0 t δ 1 N f 1 ( u , v ) ( s ) d s d t + e 2 1 σ 1 0 1 δ 1 N f 1 ( u , v ) ( t ) d t = σ 1 0 1 δ 1 N f 1 ( u , v ) ( t ) d t 0 1 k ( t ) 0 t δ 1 N f 1 ( u , v ) ( s ) d s d t + e 2 1 σ 1 = σ 1 [ a 1 φ p 1 ( 1 , u ( 1 ) ) ] 0 1 k ( t ) [ a 1 φ p 1 ( t , u ( t ) ) ] d t + e 2 1 σ 1 = σ 1 φ p 1 ( 1 , u ( 1 ) ) + 0 1 k ( t ) φ p 1 ( t , u ( t ) ) d t + e 2 1 σ 1 ,
and then
φ p 1 ( 1 , u ( 1 ) ) = 0 1 k ( t ) φ p 1 ( t , u ( t ) ) d t + e 2 .
From (S), we have
v ( t ) = φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ] ,
and
v ( 0 ) = P 2 ( δ 2 N f 2 ( u , v ) ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) K 2 ( δ 2 N f 2 ( u , v ) ) ( t ) d t 1 σ 2 = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) v ( t ) d t σ 2 v ( 0 ) 1 σ 2 ,
then
v ( 0 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) v ( t ) d t .
Thus
v ( 0 ) k 1 v ( 0 ) = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) v ( t ) d t k 1 φ p 2 1 ( 0 , a 2 ) = 0 1 e ( t ) v ( t ) d t .
From (S), we have
v ( 1 ) = P 2 ( δ 2 N f 2 ( u , v ) ) + K 2 ( δ 2 N f 2 ( u , v ) ) ( 1 ) .
By the condition of the mapping a 2 , we have
v ( 1 ) = P 2 ( δ 2 N f 2 ( u , v ) ) + K 2 ( δ 2 N f 2 ( u , v ) ) ( 1 ) = i = 1 m 2 β i η i 1 φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ( t ) ] d t + k 2 φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] 1 i = 1 m 2 β i = i = 1 m 2 β i [ v ( 1 ) v ( η i ) ] + k 2 φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] 1 i = 1 m 2 β i ,
which implies that
v ( 1 ) = i = 1 m 2 β i v ( η i ) k 2 φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] .
Since v ( 1 ) = φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] , then we have
v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) .
Moreover, from (S), it is easy to obtain
( φ p 1 ( t , u ( t ) ) ) = δ 1 N f 1 ( u , v )
and
( φ p 2 ( t , v ( t ) ) ) = δ 2 N f 2 ( u , v ) .

Hence ( u , v ) is a solution of (P).

This completes the proof. □

3 Existence of solutions

In this section, we apply Leray-Schauder’s degree to deal with the existence of solutions for (P), when f l satisfies a sub- ( p l 1 ) growth condition or a general growth condition ( l = 1 , 2 ).

We denote (S) as
( u , v ) = A ( u , v ) = ( Ψ f 1 ( u , v ) , Φ f 2 ( u , v ) ) ,
where
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Theorem 3.1 If f l satisfies a sub- ( p l 1 ) growth condition, then the problem (P) has at least one solution for any fixed parameter δ l ( l = 1 , 2 ).

Proof Denote
A λ ( u , v ) = ( Ψ λ f 1 ( u , v ) , Φ λ f 2 ( u , v ) ) ,
where
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According to Lemma 2.5, we know that (P) has the same solution of
( u , v ) = A λ ( u , v )
(14)

when λ = 1 .

It is easy to see that the operators P 1 and P 2 are compact continuous. According to Lemma 2.2, Lemma 2.3 and Lemma 2.4, we can see that Ψ λ f 1 ( u , v ) and Φ λ f 2 ( u , v ) are compact continuous from C 1 × [ 0 , 1 ] to C 1 , thus A λ ( u , v ) is compact continuous from W × [ 0 , 1 ] to W.

We claim that all the solutions of (14) are uniformly bounded for λ [ 0 , 1 ] . In fact, if it is false, we can find a sequence of solutions { ( ( u n , v n ) , λ n ) } for (14) such that ( u n , v n ) + as n + .

From Lemma 2.2, we have
| a 1 ( λ n δ 1 N f 1 ( u n , v n ) ) | C 1 ( N f 1 ( u n , v n ) L 1 + | e 2 | ) C 2 ( 1 + ( u n , v n ) ) q 1 + 1 ,
which together with the sub- ( p 1 1 ) growth condition of f 1 implies that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equ15_HTML.gif
(15)
From (14), we have
| u n ( t ) | p 1 ( t ) 2 u n ( t ) = a 1 ( λ n δ 1 N f 1 ( u n , v n ) ) F ( λ n δ 1 N f 1 ( u n , v n ) ) , t J ,
then
| u n ( t ) | p 1 ( t ) 1 | a 1 ( λ n δ 1 N f 1 ( u n , v n ) ) | + | F ( λ n δ 1 N f 1 ( u n , v n ) ) | C 4 ( 1 + ( u n , v n ) ) q 1 + 1 .
Denote α 1 = q 1 + 1 p 1 1 . From the above inequality we have
u n ( t ) 0 C 5 ( 1 + ( u n , v n ) ) α 1 .
(16)
It follows from (14) and (15) that
| u n ( 0 ) | C 6 ( 1 + ( u n , v n ) ) α 1 , where  α 1 = q 1 + 1 p 1 1 .
For any j = 1 , , N , we have
| u n j ( t ) | = | u n j ( 0 ) + 0 t ( u n j ) ( r ) d r | | u n j ( 0 ) | + | 0 t ( u n j ) ( r ) d r | [ C 7 + C 5 ] ( 1 + ( u n , v n ) ) α 1 C 8 ( 1 + ( u n , v n ) ) α 1 ,
which implies that
| u n j | 0 C 9 ( 1 + ( u n , v n ) ) α 1 , j = 1 , , N ; n = 1 , 2 , .
Thus
u n 0 C 10 ( 1 + ( u n , v n ) ) α 1 , n = 1 , 2 , .
(17)

It follows from (16) and (17) that u n 1 C 11 ( 1 + ( u n , v n ) ) α 1 .

Similarly, we have v n 1 C 12 ( 1 + ( u n , v n ) ) α 2 , where α 2 = q 2 + 1 p 2 1 .

Thus, { ( u n , v n ) } is bounded.

Thus, we can choose a large enough R 0 > 0 such that all the solutions of (14) belong to B ( R 0 ) = { ( u , v ) W ( u n , v n ) < R 0 } . Therefore, the Leray-Schauder degree d L S [ I A λ ( u , v ) , B ( R 0 ) , 0 ] is well defined for each λ [ 0 , 1 ] , and
d L S [ I A 1 ( u , v ) , B ( R 0 ) , 0 ] = d L S [ I A 0 ( u , v ) , B ( R 0 ) , 0 ] .
Denote
u 0 = i = 1 m 2 α i 0 ξ i φ p 1 1 [ t , a 1 ( 0 ) ] d t 0 1 φ p 1 1 [ t , a 1 ( 0 ) ] d t + e 1 1 i = 1 m 2 α i + 0 r φ p 1 1 [ t , a 1 ( 0 ) ] d t , v 0 = k 1 φ p 2 1 [ 0 , a 2 ( 0 ) ] + 0 1 e ( t ) 0 t φ p 2 1 [ r , a 2 ( 0 ) ] d r d t 1 σ 2 + 0 r φ p 2 1 [ t , a 2 ( 0 ) ] d t , }
(18)

where a 1 ( 0 ) and a 2 ( 0 ) are defined in (5) and (13), then ( u 0 , v 0 ) is the unique solution of ( u , v ) = A 0 ( u , v ) .

It is easy to see that ( u , v ) is a solution of ( u , v ) = A 0 ( u , v ) if and only if ( u , v ) is a solution of the following system:
p 1 ( t ) u = 0 , t ( 0 , 1 ) , p 2 ( t ) v = 0 , t ( 0 , 1 ) , u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 , lim t 1 | u | p 1 ( t ) 2 u ( t ) = 0 1 k ( t ) | u | p 1 ( t ) 2 u ( t ) d t + e 2 , v ( 0 ) k 1 v ( 0 ) = 0 1 e ( t ) v ( t ) d t , v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) . }
(19)
Obviously, (19) possesses a unique solution ( u 0 , v 0 ) . Note that ( u 0 , v 0 ) B ( R 0 ) , we have
d L S [ I A 1 ( u , v ) , B ( R 0 ) , 0 ] = d L S [ I A 0 ( u , v ) , B ( R 0 ) , 0 ] 0 .

Therefore (P) has at least one solution. This completes the proof. □

In the following, we investigate the existence of solutions for (P) when f l satisfies a general growth condition.

Denote
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equbn_HTML.gif

Assume the following.

(A1) Let a positive constant ε be such that ( u 0 , v 0 ) Ω ε , | P 1 ( 0 ) | < θ , | P 2 ( 0 ) | < θ and | a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 , | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 , where ( u 0 , v 0 ) is defined in (18), a 1 ( ) and a 2 ( ) are defined in (5) and (13), respectively.

It is easy to see that Ω ε is an open bounded domain in W. We have the following theorem.

Theorem 3.2 Assume that (A1) is satisfied. If positive parameters δ 1 and δ 2 are small enough, then the problem (P) has at least one solution on Ω ε ¯ .

Proof Similarly, we denote A λ ( u , v ) = ( Ψ λ f 1 ( u , v ) , Φ λ f 2 ( u , v ) ) . By Lemma 2.5, ( u , v ) is a solution of
{ p 1 ( t ) u = λ δ 1 f 1 ( t , u , u , v , v ) , t ( 0 , 1 ) , p 2 ( t ) v = λ δ 2 f 2 ( t , u , u , v , v ) , t ( 0 , 1 ) ,
with (2) and (8) if and only if ( u , v ) is a solution of the following abstract equation:
( u , v ) = A λ ( u , v ) .
(20)

From the proof of Theorem 3.1, we can see that A λ ( u , v ) is compact continuous from W × [ 0 , 1 ] to W. According to Leray-Schauder’s degree theory, we only need to prove that

(1) ( u , v ) = A λ ( u , v ) has no solution on Ω ε for any λ [ 0 , 1 ] ,

(2) d L S [ I A 0 ( u , v ) , Ω ε , 0 ] 0 ,

then we can conclude that the system (P) has a solution on Ω ε ¯ .

(1) If there exists a λ [ 0 , 1 ] and ( u , v ) Ω ε is a solution of (20), then ( u , v ) and λ satisfy
u ( t ) = φ p 1 1 [ t , a 1 F ( λ δ 1 N f 1 ( u , v ) ) ]
and
v ( t ) = φ p 2 1 [ t , a 2 F ( λ δ 2 N f 2 ( u , v ) ) ] .
Since ( u , v ) Ω ε , there exists an i such that | u i | 0 + | ( u i ) | 0 = ε or | v i | 0 + | ( v i ) | 0 = ε .
  1. (i)

    If | u i | 0 + | ( u i ) | 0 = ε .

     

( i ) Suppose that | u i | 0 > 2 θ , then | ( u i ) | 0 < ε 2 θ = θ . On the other hand, for any t , t J , we have

| u i ( t ) u i ( t ) | = | t t ( u i ) ( r ) d r | 0 1 | ( u i ) ( r ) | d r < θ .
This implies that | u i ( t ) | > θ for each t J .
Note that ( u , v ) Ω ε ¯ , then | f 1 ( t , u , u , v , v ) | β N ε ( t ) , holding | F ( N f 1 ) | 0 1 β N ε ( t ) d t . Since P 1 ( ) is continuous, when 0 < δ 1 is small enough, from (A1), we have
| u ( 0 ) | = | P 1 ( λ δ 1 N f 1 ( u , v ) ) | < θ .

It is a contradiction to | u i ( t ) | > θ for each t J .

( ii ) Suppose that | u i | 0 2 θ , then θ | ( u i ) | 0 ε . This implies that | ( u i ) ( t 2 ) | θ for some t 2 J , and we can find
θ | ( u i ) ( t 2 ) | | ( u ) ( t 2 ) | = | φ p 1 1 [ t 2 , a 1 F ( λ δ 1 N f 1 ( u , v ) ) ( t 2 ) ] | .
(21)
Since ( u , v ) Ω ε ¯ and f 1 is Carathéodory, it is easy to see that
| f 1 ( t , u , u , v , v ) | β N ε ( t ) ,
thus
| δ 1 F ( N f 1 ) | δ 1 0 1 β N ε ( t ) d t .
From Lemma 2.2, a 1 ( ) is continuous, then we have
| a 1 ( λ δ 1 N f 1 ) | | a 1 ( 0 ) | as  δ 1 0 .
When 0 < δ 1 is small enough, from (A1) and (21), we can conclude that
θ | φ p 1 1 [ t , a 1 F ( λ δ 1 N f 1 ( u , v ) ) ( t ) ] | < θ 3 .
It is a contradiction. Thus | u i | 0 + | ( u i ) | 0 ε .
  1. (ii)

    If | v i | 0 + | ( v i ) | 0 = ε . Similar to the proof of (i), we get a contradiction. Thus | v i | 0 + | ( v i ) | 0 ε .

     

Summarizing this argument, for each λ [ 0 , 1 ) , ( u , v ) = A λ ( u , v ) has no solution on Ω ε when positive parameters δ 1 and δ 2 are small enough.

(2) Since ( u 0 , v 0 ) (where ( u 0 , v 0 ) is defined in (18)) is the unique solution of ( u , v ) = A 0 ( u , v ) , and (A1) holds ( u 0 , v 0 ) Ω ε , we can see that the Leray-Schauder degree
d L S [ I A 0 ( u , v ) , Ω ε , 0 ] 0 .

This completes the proof. □

As applications of Theorem 3.2, we have the following.

Corollary 3.3 Assume that f l ( t , u , u , v , v ) = μ l ( t ) | u | m l ( t ) 2 u ( t ) + γ l ( t ) | u | n l ( t ) 2 u ( t ) + μ ˜ l ( t ) | v | m ˜ l ( t ) 2 v ( t ) + γ ˜ l ( t ) | v | n ˜ l ( t ) 2 v ( t ) , where l = 1 , 2 ; m l , n l , m ˜ l , n ˜ l , μ l , γ l , μ ˜ l , γ ˜ l C ( J , R ) satisfy max t J p l ( t ) < m l , n l , m ˜ l , n ˜ l , t J . If | e 1 | and | e 2 | are small enough, then the problem (P) possesses at least one solution.

Proof It is easy to have
| f l ( t , u , u , v , v ) | | μ l ( t ) | | u | m l ( t ) 1 + | γ l ( t ) | | u | n l ( t ) 1 + | μ ˜ l ( t ) | | v | m ˜ l ( t ) 1 + | γ ˜ l ( t ) | | v | n ˜ l ( t ) 1 .
From μ l , γ l , μ ˜ l , γ ˜ l C ( J , R ) and the definition of Ω ε , we have
| f l ( t , u , u , v , v ) | C 13 ε m l ( t ) 1 + C 14 ε n l ( t ) 1 + C 15 ε m ˜ l ( t ) 1 + C 16 ε n ˜ l ( t ) 1 .
Since max t J p l ( t ) < m l , n l , m ˜ l , n ˜ l , then there exists a small enough ε such that
| f l ( t , u , u , v , v ) | 1 σ 1 4 N ( θ 3 ) p l ( t ) 1 .
From Lemma 2.2 and the small enough | e 2 | , we have
| a 1 ( δ 1 f 1 ) | 2 N 1 σ 1 ( δ 1 f 1 L 1 + | e 2 | ) < ( θ 3 ) p 1 ( t ) 1 ,

then | a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 is valid.

Similarly, we have | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 .

Obviously, it follows from | a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 , | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 and the small enough | e 1 | that ( u 0 , v 0 ) Ω ε , | P 1 ( 0 ) | < θ , and | P 2 ( 0 ) | < θ .

Thus, the conditions of (A1) are satisfied, then the problem (P) possesses at least one solution. □

Corollary 3.4 Assume that f l ( t , u , u , v , v ) = μ l ( t ) | u | m l ( t ) 2 u ( t ) + γ l ( t ) | u | n l ( t ) 2 u ( t ) + μ ˜ l ( t ) | v | m ˜ l ( t ) 2 v ( t ) + γ ˜ l ( t ) | v | n ˜ l ( t ) 2 v ( t ) , where l = 1 , 2 ; m l , n l , m ˜ l , n ˜ l , μ l , γ l , μ ˜ l , γ ˜ l C ( J , R ) satisfy min t J p l ( t ) m l , n l , m ˜ l , n ˜ l max t J p l ( t ) . If | e 1 | , | e 2 | and δ l are small enough, then the problem (P) possesses at least one solution.

Proof From Lemma 2.2, we have
| a 1 ( δ 1 f 1 ) | 2 N 1 σ 1 ( δ 1 f 1 L 1 + | e 2 | ) .
Since a 1 ( δ 1 f 1 ) is dependent on the small enough δ 1 and | e 2 | , then it follows from the continuity of a 1 that | a 1 ( 0 ) | is small enough, which implies that
| a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 .

Similarly, we have | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 .

From | a 1 ( 0 ) | < min t J ( θ 3 ) p 1 ( t ) 1 , | a 2 ( 0 ) | < min t J ( θ 2 ) p 2 ( t ) 1 and the small enough | e 1 | and | e 2 | , it is easy to have that ( u 0 , v 0 ) Ω ε , | P 1 ( 0 ) | < θ , and | P 2 ( 0 ) | < θ .

Thus, the conditions of (A1) are satisfied, then the problem (P) possesses at least one solution. □

We denote
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equce_HTML.gif

Assume the following.

(A2) Let positive constants ε 1 and ε 2 be such that ( u 0 , v 0 ) Ω ε 1 , ε 2 , | P 1 ( 0 ) | < θ 1 , | P 2 ( 0 ) | < θ 2 and | a 1 ( 0 ) | < min t J ( θ 1 3 ) p 1 ( t ) 1 , | a 2 ( 0 ) | < min t J ( θ 2 2 ) p 2 ( t ) 1 , where ( u 0 , v 0 ) is defined in (18), a 1 ( ) and a 2 ( ) are defined in (5) and (13), respectively.

It is easy to see that Ω ε 1 , ε 2 is an open bounded domain in W. We have the following.

Corollary 3.5 Assume that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcf_HTML.gif

where ϵ, ϵ ˜ are positive constants; m , n , m ˜ , n ˜ , ϱ , ϱ ˜ , μ , γ , μ ˜ , γ ˜ , ϰ , ϰ ˜ C ( J , R ) satisfy 1 < m , n , m ˜ , n ˜ < min t J p 1 ( t ) , and max t J p 2 ( t ) < ϱ , ϱ ˜ , t J . Then the problem (P) possesses at least one solution.

Proof Similar to the proof of Theorem 3.2, we only need to prove that (A2) is satisfied, then we can conclude that the problem (P) possesses at least one solution.

From μ , γ , μ ˜ , γ ˜ C ( J , R ) and the definition of Ω ε 1 , ε 2 , it is easy to have that
| f 1 ( t , u , u , v , v ) | C 17 ε 1 m ( t ) 1 + C 18 ε 1 n ( t ) 1 + C 19 ε 2 m ˜ ( t ) 1 + C 20 ε 2 n ˜ ( t ) 1 ,
where we suppose ε 2 < 1 < ε 1 . Since 1 < m , n , m ˜ , n ˜ < min t J p 1 ( t ) , then there exists a big enough ε 1 such that
| f 1 ( t , u , u , v , v ) | 1 σ 1 3 N ( θ 1 3 ) p 1 ( t ) 1 .
From Lemma 2.2, we have
| a 1 ( δ 1 f 1 ) | 2 N 1 σ 1 ( δ 1 f 1 L 1 + | e 2 | ) < ( θ 1 3 ) p 1 ( t ) 1 ,

then | a 1 ( 0 ) | < min t J ( θ 1 3 ) p 1 ( t ) 1 is valid.

From ϰ , ϰ ˜ C ( J , R ) and the definition of Ω ε 1 , ε 2 , we have
| f 2 ( t , u , u , v , v ) | C 21 ε 1 ϵ ε 2 ϱ ( t ) 1 + C 22 ε 1 ϵ ˜ ε 2 ϱ ˜ ( t ) 1 .
Since max t J p 2 ( t ) < ϱ , ϱ ˜ , then there exists a ε 2 such that ε 2 < ( C 23 C 21 ε 1 ϵ + C 22 ε 1 ϵ ˜ ) 1 ϱ p 2 + (where ϱ = min { ϱ , ϱ ˜ } ), which implies that
| f 2 ( t , u , u , v , v ) | 1 4 N ( θ 2 2 ) p 2 ( t ) 1 .
From Lemma 2.3, we have
| a 2 ( δ 2 f 2 ) | 3 N δ 2 f 2 0 < ( θ 2 2 ) p 2 ( t ) 1 ,

then | a 2 ( 0 ) | < min t J ( θ 2 2 ) p 2 ( t ) 1 is valid.

Obviously, it follows from | a 1 ( 0 ) | < min t J ( θ 1 3 ) p 1 ( t ) 1 and | a 2 ( 0 ) | < min t J ( θ 2 2 ) p 2 ( t ) 1 that ( u 0 , v 0 ) Ω ε 1 , ε 2 , | P 1 ( 0 ) | < θ 1 , and | P 2 ( 0 ) | < θ 2 .

Thus, the conditions of (A2) are satisfied, then the problem (P) possesses at least one solution. □

Corollary 3.6 Assume that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equcm_HTML.gif

where ϵ, ϵ ˜ are positive constants; ϱ , ϱ ˜ , m , n , m ˜ , n ˜ , ϰ , ϰ ˜ , μ , γ , μ ˜ , γ ˜ C ( J , R ) satisfy max t J p 1 ( t ) < ϱ , ϱ ˜ , and 1 < m , n , m ˜ , n ˜ < min t J p 2 ( t ) , t J . If | e 1 | and | e 2 | are small enough, then the problem (P) possesses at least one solution.

Proof Similar to the proof of Corollary 3.5, we conclude that (A2) is satisfied. Then the problem (P) possesses at least one solution. □

4 Existence of nonnegative solutions

In the following, we deal with the existence of nonnegative solutions of (P). For any x = ( x 1 , , x N ) R N , the notation x 0 ( x > 0 ) means x j 0 ( x j > 0 ) for any j = 1 , , N . For any x , y R N , the notation x y means x y 0 , the notation x > y means x y > 0 .

Theorem 4.1 We assume that

(10) δ 1 f 1 ( t , x , y , z , w ) 0 , ( t , x , y , z , w ) J × R N × R N × R N × R N ;

(20) δ 2 f 2 ( t , x , y , z , w ) 0 , ( t , x , y , z , w ) J × R N × R N × R N × R N ;

(30) e 1 0 ;

(40) e 2 0 .

Then every solution of (P) is nonnegative.

Proof (i) We shall show that u ( t ) is nonnegative.

If ( u , v ) is a solution of (P), from Lemma 2.5, we have
φ p 1 ( t , u ( t ) ) = a 1 ( δ 1 N f 1 ( u , v ) ) 0 t δ 1 f 1 ( s , u , u , v , v ) d s , t J ,
which together with (5), (10) and (40) implies that
https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-106/MediaObjects/13661_2011_Article_367_Equco_HTML.gif

Thus u ( t ) 0 for any t J . Holding u ( t ) is decreasing, namely u ( t 1 ) u ( t 2 ) for any t 1 , t 2 J with t 1 < t 2 .

According to the boundary value condition (2) and condition (30), we have
u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 i = 1 m 2 α i u ( 1 ) + e 1 ,
then
u ( 1 ) e 1 1 i = 1 m 2 α i 0 .
Thus u ( t ) is nonnegative.
  1. (ii)

    We shall show that v ( t ) is nonnegative.

     
If ( u , v ) is a solution of (P), From Lemma 2.5, we have
v ( t ) = v ( 0 ) + F { φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ] } ( t ) .
We claim that a 2 ( δ 2 N f 2 ( u , v ) ) 0 . If it is false, then there exists some j { 1 , , N } such that a 2 j ( δ 2 N f 2 ( u , v ) ) < 0 , which together with condition (20) implies that
[ a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( t ) ] j < 0 , t J .
(22)
Similar to the proof of Lemma 2.3, the boundary value condition (8) implies
0 = k 1 φ p 2 1 ( 0 , a 2 ) + 0 1 e ( t ) 0 t φ p 2 1 [ s , a 2 F ( δ 2 N f 2 ( u , v ) ) ( s ) ] d s d t 1 σ 2 + i = 1 m 2 β i η i 1 φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ( t ) ] d t + k 2 φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] 1 i = 1 m 2 β i + 0 1 φ p 2 1 [ t , a 2 F ( δ 2 N f 2 ( u , v ) ) ( t ) ] d t .
(23)

From (22) and a 2 j ( δ 2 N f 2 ( u , v ) ) < 0 , we get a contradiction to (23).

Thus a 2 ( δ 2 N f 2 ( u , v ) ) 0 .

We claim that
a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( 1 ) 0 .
(24)
If it is false, then there exists some j { 1 , , N } such that
[ a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] j > 0 ,
which together with condition (20) implies
[ a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( t ) ] j > 0 , t J .
(25)

From (25) and a 2 ( δ 2 N f 2 ( u , v ) ) 0 , we get a contradiction to (23). Thus (24) is valid.

Denote
Γ ( t ) = a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( t ) , t J .
Obviously, Γ ( 0 ) = a 2 ( δ 2 N f 2 ( u , v ) ) 0 , Γ ( 1 ) 0 , and Γ ( t ) is decreasing, i.e., Γ ( t ) Γ ( t ) for any t , t J with t t . For any j = 1 , , N , there exist ζ j J such that
Γ j ( t ) 0 , t [ 0 , ζ j ] and Γ j ( t ) 0 , t [ ζ j , T ) .
We can conclude that v j ( t ) is increasing on [ 0 , ζ j ] , and v j ( t ) is decreasing on [ ζ j , T ] . Thus
min { v j ( 0 ) , v j ( 1 ) } = inf t I v j ( t ) , j = 1 , , N .
For any fixed j { 1 , , N } , if
v j ( 0 ) = inf t I v j ( t ) ,
which together with (8) implies that
v j ( 0 ) = 0 1 e ( t ) v j ( t ) d t + k 1 ( v ) j ( 0 ) 0 1 e ( t ) v j ( 0 ) d t + k 1 ( v ) j ( 0 ) .
(26)
From a 2 ( δ 2 N f 2 ( u , v ) ) 0 , we have
( v ) j ( 0 ) = ( φ p 2 1 [ 0 , a 2 ] ) j 0 .
(27)
It follows from (26) and (27) that
v j ( 0 ) k 1 ( v ) j ( 0 ) 1 σ 2 0 .
If
v j ( 1 ) = inf t I v j ( t ) ,
(28)
from (8) and (28), we have
v j ( 1 ) = i = 1 m 2 β i v j ( η i ) k 2 ( v ) j ( 1 ) i = 1 m 2 β i v j ( 1 ) k 2 ( v ) j ( 1 ) .
(29)
Since a 2 ( δ 2 N f 2 ( u , v ) ) F ( δ 2 N f 2 ( u , v ) ) ( 1 ) 0 , we have
( v ) j ( 1 ) = ( φ p 2 1 [ 1 , a 2 F ( δ 2 N f 2 ( u , v ) ) ( 1 ) ] ) j 0 .
(30)
Combining (29) and (30), we have
v j ( 1 ) k 2 ( v ) j ( 1 ) 1 i = 1 m 2 β i 0 .

Thus v ( t ) is nonnegative.

Combining (i) and (ii), we find that every solution of (P) is nonnegative. □

Corollary 4.2 We assume that

(10) δ 1 f 1 ( t , x , y , z , w ) 0 , ( t , x , y , z , w ) J × R N × R N × R N × R N with x , z 0 ;

(20) δ 2 f 2 ( t , x , y , z , w ) 0 , ( t , x , y , z , w ) J × R N × R N × R N × R N with x , z 0 ;

(30) e 1 0 ;

(40) e 2 0 .

Then we have
  1. (a)

    Under the conditions of Theorem 3.1, (P) has at least one nonnegative solution ( u , v ) ;

     
  2. (b)

    Under the conditions of Theorem 3.2, (P) has at least one nonnegative solution ( u , v ) .

     
Proof (a) Define
L 1 ( u ) = ( L ( u 1 ) , , L ( u N ) ) , L 2 ( v ) = ( L ( v 1 ) , , L ( v N ) ) ,
where
L ( t ) = { t , t 0 , 0 , t < 0 .
Denote
f l ˜ ( t , u , u , v , v ) = f l ( t , L 1 ( u ) , u , L 2 ( v ) , v ) , ( t , u , u , v , v ) J × R N × R N × R N × R N ,

where l = 1 , 2 , then f l ˜ ( t , u , u , v , v ) satisfies the Carathéodory condition, f 1 ˜ ( t , u , u , v , v ) 0 and f 2 ˜ ( t , u , u , v , v ) 0 .

We assume the following.

(A2) lim | u | + | v | + ( f l ˜ ( t , u , u , v , v ) / ( | u | + | v | ) q l ( t ) 1 ) = 0 for t J uniformly, where q l ( t ) C ( I , R ) and 1 < q l q l + < p l .

Obviously, f l ˜ ( t , , , , ) satisfies a sub- ( p l 1 ) growth condition.

Let us consider the existence of solutions of the following system:
p 1 ( t ) u = δ 1 f 1 ˜ ( t , u , u , v , v ) , t ( 0 , 1 ) , p 2 ( t ) v = δ 2 f 2 ˜ ( t , u , u , v , v ) , t ( 0 , 1 ) , u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 , lim t 1 | u | p 1 ( t ) 2 u ( t ) = 0 1 k ( t ) | u | p 1 ( t ) 2 u ( t ) d t + e 2 , v ( 0 ) k 1 v ( 0 ) = 0 1 e ( t ) v ( t ) d t , v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) . }
(31)
According to Theorem 3.1, (31) has at least a solution ( u , v ) . From Theorem 4.1, we can see that ( u , v ) is nonnegative. Thus, ( u , v ) is a nonnegative solution of (P).
  1. (b)

    It is similar to the proof of (a).

     

This completes the proof. □

5 Examples

Example 5.1 Consider the following problem:
( S 1 ) { p 1 ( t ) u = f 1 ( t , u , u , v , v ) = e 2 t ( | u | q ( t ) 2 u + | u | q ( t ) 2 u ) + | v | q ( t ) 2 v + | v | q ( t ) 2 v + ( t + 1 ) 2 , t ( 0 , 1 ) , p 2 ( t ) v = f 2 ( t , u , u , v , v ) = | u | q ( t ) 2 u + | u | q ( t ) 2 u + t 2 ( | v | q ( t ) 2 v + | v | q ( t ) 2 v ) + ( t + 2 ) 2 , t ( 0 , 1 ) , u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + e 1 , lim t 1 | u | p 1 ( t ) 2 u ( t ) = 0 1 1 1 + t | u | p 1 ( t ) 2 u ( t ) d t + e 2 , v ( 0 ) k 1 v ( 0 ) = 0 1 e t v ( t ) d t , v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) ,

where p 1 ( t ) = 7 + 3 t cos 3 t , p 2 ( t ) = 7 + 3 t sin 3 t , q ( t ) = 3 + 2 t cos t .

Obviously, f 1 and f 2 are Caratheodory, q ( t ) 4 < 5 min { p 1 ( t ) , p 2 ( t ) } , i = 1 m 2 α i < 1 , i = 1 m 2 β i < 1 , then the conditions of Theorem 3.1 are satisfied, then ( S 1 ) has a solution.

Example 5.2 Consider the following problem
( S 2 ) { p 1 ( t ) u = f 1 ( t , u , u , v , v ) = e 2 t ( | u | q ( t ) 2 u + | u | q ( t ) 1 ) | v | q ( t ) 1 | v | q ( t ) 1 ( t + 1 ) 2 , t ( 0 , 1 ) , p 2 ( t ) v = f 2 ( t , u , u , v , v ) = | u | q ( t ) 2 u + | u | q ( t ) 1 + t 2 ( | v | q ( t ) 2 v + | v | q ( t ) 1 ) + ( t + 2 ) 2 , t ( 0 , 1 ) , u ( 1 ) = i = 1 m 2 α i u ( ξ i ) + 1 , lim t 1 | u | p 1 ( t ) 2 u ( t ) = 0 1 1 1 + t | u | p 1 ( t ) 2 u ( t ) d t 2 , v ( 0 ) k 1 v ( 0 ) = 0 1 e t v ( t ) d t , v ( 1 ) + k 2 v ( 1 ) = i = 1 m 2 β i v ( η i ) ,

where N = 1 , p 1 ( t ) = 7 + 3 t cos 3 t , p 2 ( t ) = 7 + 3 t sin 3 t , q ( t ) = 4 + e 2 t sin 2 t .

Obviously, f 1 and f 2 are Caratheodory, q ( t ) 5 < 6 min { p 1 ( t ) , p 2 ( t ) } , i = 1 m 2 α i < 1 , i = 1 m 2 β i < 1 , the conditions of Corollary 4.2 are satisfied, then ( S 2 ) has a nonnegative solution.

Declarations

Acknowledgements

Partly supported by the National Science Foundation of China (10701066 & 10971087).

Authors’ Affiliations

(1)
Department of Mathematics and Information Science, Zhengzhou University of Light Industry
(2)
Department of Mathematics, Henan Institute of Science and Technology

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© Zhi et al.; licensee Springer. 2013

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