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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

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.

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)dt,u(1)= = 1 m 2 α u( ξ ) 0 1 h(t)u(t)dt,
(4)

where pC([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,,m2); g L 1 [0,1] is nonnegative, 0 1 g(t)dt=σ[0,1]; h L 1 [0,1], 0 1 h(t)dt=δ; A i , B i C( R N × R N , R N ); T and S are linear operators defined by (Su)(t)= 0 1 h (t,s)u(s)ds, (Tu)(t)= 0 t k (t,s)u(s)ds, t[0,1], where k , h C([0,1]×[0,1],R).

If σ<1 and = 1 m 2 α δ1, 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

PC ( 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 } ,

wPC(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 )PC(J, R N ), denote | x i | 0 =sup{| x i (t)|t J }.

Obviously, PC(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 ) | dt.

In the following, PC(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 ) cu c p ( t ) 1 ( p ( t ) u) in general, but p cu= 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 ΩR (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-(p1) growth condition.

Let N1, 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 tJ, 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 tJ 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 uP 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 zPC(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 =0for tJ uniformly,

where q(t)PC(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 δ1;

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 δ1. 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 tJ. Denote

φ 1 (t,x)=|x | 2 p ( t ) p ( t ) 1 xfor x R N {0}, φ 1 (t,0)=0,tJ.

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)ds,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 PC as

F(f)(t)= 0 t f(s)ds,tJ,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),tJ,

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 ) dt,

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 ωW, 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

Lemma 2.2 Suppose that h(t)0 on [ ξ 1 ,1], α ξ ξ + 1 h(t)dt (=1,,m2) and h(t)0 on [0, ξ 1 ]. Then the function Λ ω () has the following properties:

  1. (i)

    For any fixed ωW, 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)|3N [ ( 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 <0for  x 1 x 2 , x 1 , x 2 R N ,

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

Set R 0 =3N[ ( 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 of ρ 1 satisfies

Thus the j 0 th component of ρ 1 + t i < t b i +F(ϑ)(t) keeps sign on J, namely, for any tJ, we have

Obviously, we have

then it is easy to see that the j 0 th component of Λ ω ( ρ 1 ) keeps the same sign of . Thus,

Λ ω ( ρ 1 )0.

Let us consider the equation

λ Λ ω ( ρ 1 )+(1λ) 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)|3N [ ( 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),tJ.

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),tJ.

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 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)dt 0 t 2 ϑ n (t)dt|=| t 1 t 2 ϑ n (t)dt|| t 1 t 2 α(t)dt|.

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 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 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)dt (=1,,m2) 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)dt.
(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 δ1. 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)ds,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 ωW, 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 ) dt.

Throughout the paper, we denote .

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

  1. (i)

    For any fixed ωW, 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)|3N [ ( 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),tJ.

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),tJ.

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 δ1, 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)ds,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)dt, 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)dt, 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 ωW, 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)(1σ)0;

(20) h(t)0 on [ ξ 1 ,1], α ξ ξ + 1 h(t)dt (=1,,m2) and h(t)0 on [0, ξ 1 ].

Then the function ϒ ω () has the following properties:

  1. (i)

    For any fixed ωW, 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),tJ.

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),tJ.

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)(1σ)0;

(20) h(t)0 on [ ξ 1 ,1], α ξ ξ + 1 h(t)dt (=1,,m2) 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)dt (=1,,m2) 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

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 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

Thus

(22)

From ( S 1 ), we have

It follows from (11) and Lemma 2.2 that

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 )={uP 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)dt (=1,,m2) 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 )11; 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)dt (=1,,m2) 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 )={uP 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