Open Access

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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq1_HTML.gif-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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq1_HTML.gif-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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq1_HTML.gif-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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ1_HTML.gif
(1)
where u : [ 0 , 1 ] R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq2_HTML.gif, f ( , , , , ) : [ 0 , 1 ] × R N × R N × R N × R N R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq3_HTML.gif, t i ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq4_HTML.gif, i = 1 , , k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq5_HTML.gif, 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ2_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ3_HTML.gif
(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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ4_HTML.gif
(4)

where p C ( [ 0 , 1 ] , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq6_HTML.gif and p ( t ) > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq7_HTML.gif, p ( t ) u : = ( w ( t ) | u | p ( t ) 2 u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq8_HTML.gif is called the weighted p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq1_HTML.gif-Laplacian; 0 < t 1 < t 2 < < t k < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq9_HTML.gif, 0 < ξ 1 < < ξ m 2 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq10_HTML.gif; α 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq11_HTML.gif ( = 1 , , m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq12_HTML.gif); g L 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq13_HTML.gif is nonnegative, 0 1 g ( t ) d t = σ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq14_HTML.gif; h L 1 [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq15_HTML.gif, 0 1 h ( t ) d t = δ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq16_HTML.gif; A i , B i C ( R N × R N , R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq17_HTML.gif; T and S are linear operators defined by ( S u ) ( t ) = 0 1 h ( t , s ) u ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq18_HTML.gif, ( T u ) ( t ) = 0 t k ( t , s ) u ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq19_HTML.gif, t [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq20_HTML.gif, where k , h C ( [ 0 , 1 ] × [ 0 , 1 ] , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq21_HTML.gif.

If σ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif and = 1 m 2 α δ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq23_HTML.gif, we say the problem is nonresonant, but if σ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq24_HTML.gif or = 1 m 2 α δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq25_HTML.gif, we say the problem is resonant.

Throughout the paper, o ( 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq26_HTML.gif means functions which are uniformly convergent to 0 (as n + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq27_HTML.gif); for any v R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq28_HTML.gif, v j https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq29_HTML.gif will denote the j th component of v; the inner product in R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq30_HTML.gif will be denoted by , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq31_HTML.gif, | | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq32_HTML.gif will denote the absolute value and the Euclidean norm on R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq30_HTML.gif. Denote J = [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq33_HTML.gif, J = ( 0 , 1 ) { t 1 , , t k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq34_HTML.gif, J 0 = [ t 0 , t 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq35_HTML.gif, J i = ( t i , t i + 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq36_HTML.gif, i = 1 , , k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq5_HTML.gif, where t 0 = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq37_HTML.gif, t k + 1 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq38_HTML.gif. Denote by J i o https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq39_HTML.gif the interior of J i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq40_HTML.gif, i = 0 , 1 , , k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq41_HTML.gif. 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 } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equa_HTML.gif
w P C ( J , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq42_HTML.gif satisfy 0 < w ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq43_HTML.gif, t ( 0 , 1 ) { t 1 , , t k } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq44_HTML.gif, and ( w ( t ) ) 1 p ( t ) 1 L 1 ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq45_HTML.gif,
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 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equb_HTML.gif

For any x = ( x 1 , , x N ) P C ( J , R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq46_HTML.gif, denote | x i | 0 = sup { | x i ( t ) | t J } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq47_HTML.gif.

Obviously, P C ( J , R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq48_HTML.gif is a Banach space with the norm x 0 = ( i = 1 N | x i | 0 2 ) 1 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq49_HTML.gif, and P C 1 ( J , R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq50_HTML.gif is a Banach space with the norm x 1 = x 0 + ( w ( t ) ) 1 p ( t ) 1 x 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq51_HTML.gif. Denote L 1 = L 1 ( J , R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq52_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equc_HTML.gif
In the following, P C ( J , R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq48_HTML.gif and P C 1 ( J , R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq53_HTML.gif will be simply denoted by PC and P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equd_HTML.gif
The study of differential equations and variational problems with nonstandard p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq1_HTML.gif-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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq55_HTML.gif (a constant), (1)-(4) becomes the well-known p-Laplacian problem. If p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq1_HTML.gif is a general function, one can see easily p ( t ) c u c p ( t ) 1 ( p ( t ) u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq56_HTML.gif in general, but p c u = c p 1 ( p u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq57_HTML.gif, so p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq58_HTML.gif represents a non-homogeneity and possesses more nonlinearity, thus p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq58_HTML.gif is more complicated than p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq59_HTML.gif. For example:
  1. (a)
    If Ω R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq60_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Eque_HTML.gif
     
is zero in general, and only under some special conditions λ p ( x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq61_HTML.gif (see [9]), when Ω R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq62_HTML.gif ( N = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq63_HTML.gif) is an interval, the results show that λ p ( x ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq64_HTML.gif if and only if p ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq65_HTML.gif is monotone. But the property of λ p > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq66_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq67_HTML.gif and p ( t ) p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq55_HTML.gif (a constant) and p u > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq68_HTML.gif, 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq58_HTML.gif. It is another difference between p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq59_HTML.gif and p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq58_HTML.gif.

     

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq59_HTML.gif, 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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq69_HTML.gif-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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq1_HTML.gif is a general function, we investigate the existence of solutions and nonnegative solutions for the weighted p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq1_HTML.gif-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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ5_HTML.gif
(5)

where D i C ( R N × R N , R N ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq70_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq71_HTML.gif in general and the p ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq1_HTML.gif-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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq72_HTML.gif growth condition, but for the p-Laplacian impulsive problems, the nonlinear term only needs to satisfy the sub- ( p 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq73_HTML.gif growth condition.

Let N 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq74_HTML.gif, the function f : J × R N × R N × R N × R N R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq75_HTML.gif is assumed to be Caratheodory, by which we mean:
  1. (i)

    For almost every t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq76_HTML.gif, the function f ( t , , , , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq77_HTML.gif is continuous;

     
  2. (ii)

    For each ( x , y , s , z ) R N × R N × R N × R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq78_HTML.gif, the function f ( , x , y , s , z ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq79_HTML.gif is measurable on J;

     
  3. (iii)
    For each R > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq80_HTML.gif, there is a α R L 1 ( J , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq81_HTML.gif such that, for almost every t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq76_HTML.gif and every ( x , y , s , z ) R N × R N × R N × R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq78_HTML.gif with | x | R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq82_HTML.gif, | y | R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq83_HTML.gif, | s | R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq84_HTML.gif, | z | R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq85_HTML.gif, one has
    | f ( t , x , y , s , z ) | α R ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equf_HTML.gif
     

We say a function u : J R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq86_HTML.gif is a solution of (1) if u P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq87_HTML.gif with w ( t ) | u | p ( t ) 2 u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq88_HTML.gif absolutely continuous on J i o https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq39_HTML.gif, i = 0 , 1 , , k https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq41_HTML.gif, which satisfies (1) a.e. on J.

In this paper, we always use C i https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq89_HTML.gif 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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equg_HTML.gif
We say f satisfies the sub- ( p 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq72_HTML.gif 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, https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equh_HTML.gif

where q ( t ) P C ( J , R ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq90_HTML.gif and 1 < q q + < p https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq91_HTML.gif.

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif, = 1 m 2 α δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq92_HTML.gif;

Case (ii): σ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq24_HTML.gif, = 1 m 2 α δ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq93_HTML.gif;

Case (iii): σ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif, = 1 m 2 α δ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq94_HTML.gif.

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif, = 1 m 2 α δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq25_HTML.gif. In Section 4, we give the existence of solutions for system (1)-(4) or (1) with (2), (4) and (5) when σ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq24_HTML.gif, = 1 m 2 α δ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq93_HTML.gif. 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif, = 1 m 2 α δ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq95_HTML.gif.

2 Preliminary

For any ( t , x ) J × R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq96_HTML.gif, denote φ ( t , x ) = | x | p ( t ) 2 x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq97_HTML.gif. 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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq20_HTML.gif, φ ( t , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq98_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equi_HTML.gif
     
  2. (ii)
    There exists a function α : [ 0 , + ) [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq99_HTML.gif, α ( s ) + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq100_HTML.gif as s + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq101_HTML.gif such that
    φ ( t , x ) , x α ( | x | ) | x | for all x R N . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equj_HTML.gif
     
It is well known that φ ( t , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq98_HTML.gif is a homeomorphism from R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq30_HTML.gif to R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq30_HTML.gif for any fixed t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq76_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equk_HTML.gif

It is clear that φ 1 ( t , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq102_HTML.gif 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 , } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ6_HTML.gif
(6)

where a i , b i R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq103_HTML.gif; f L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq104_HTML.gif.

Denote a = ( a 1 , , a k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq105_HTML.gif, b = ( b 1 , , b k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq106_HTML.gif. Obviously, a , b R k N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq107_HTML.gif.

We will discuss it in three cases, respectively.

2.1 Case (i)

Suppose that σ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif and = 1 m 2 α δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq92_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ7_HTML.gif
(7)
Denote ρ 1 = w ( 0 ) φ ( 0 , u ( 0 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq108_HTML.gif. It is easy to see that ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq109_HTML.gif is dependent on a, b and f ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq110_HTML.gif. Define the operator F : L 1 P C https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq111_HTML.gif as
F ( f ) ( t ) = 0 t f ( s ) d s , t J , f L 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equl_HTML.gif
By solving for u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq112_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equm_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equn_HTML.gif
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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equo_HTML.gif
Denote W = R 2 k N × L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq113_HTML.gif with the norm
ω = i = 1 k | a i | + i = 1 k | b i | + ϑ L 1 , ω = ( a , b , ϑ ) W , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equp_HTML.gif

then W is a Banach space.

For any ω W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq114_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equq_HTML.gif
Denote ξ m 1 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq115_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equr_HTML.gif
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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq116_HTML.gif on [ ξ 1 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq117_HTML.gif, α ξ ξ + 1 h ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq118_HTML.gif ( = 1 , , m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq12_HTML.gif) and h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq119_HTML.gif on [ 0 , ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq120_HTML.gif. Then the function Λ ω ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq121_HTML.gif has the following properties:
  1. (i)
    For any fixed ω W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq114_HTML.gif, the equation
    Λ ω ( ρ 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ8_HTML.gif
    (8)
     
has a unique solution ρ 1 ˜ ( ω ) R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq122_HTML.gif.
  1. (ii)
    The function ρ 1 ˜ : W R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq123_HTML.gif, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, for any ω = ( a , b , ϑ ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq124_HTML.gif, 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 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equt_HTML.gif
     
where the notation M p # 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq125_HTML.gif means
M p # 1 = { M p + 1 , M > 1 , M p 1 , M 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equu_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equv_HTML.gif

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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq126_HTML.gif.

Suppose that | ρ 1 | > R 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq127_HTML.gif, it is easy to see that there exists some j 0 { 1 , , N } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq128_HTML.gif such that the absolute value of the j 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq129_HTML.gifth 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq109_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq129_HTML.gifth component of ρ 1 + t i < t b i + F ( ϑ ) ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq131_HTML.gif keeps sign on J, namely, for any t J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq76_HTML.gif, 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq129_HTML.gifth component of Λ ω ( ρ 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq132_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equz_HTML.gif
Let us consider the equation
λ Λ ω ( ρ 1 ) + ( 1 λ ) ρ 1 = 0 , λ [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ9_HTML.gif
(9)
According to the preceding discussion, all the solutions of (9) belong to b ( R 0 + 1 ) = { x R N | x | < R 0 + 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq133_HTML.gif. Therefore
d B [ Λ ω ( ρ 1 ) , b ( R 0 + 1 ) , 0 ] = d B [ I , b ( R 0 + 1 ) , 0 ] 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equaa_HTML.gif

it means the existence of solutions of Λ ω ( ρ 1 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq134_HTML.gif.

In this way, we define a function ρ 1 ˜ ( ω ) : W R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq135_HTML.gif, which satisfies Λ ω ( ρ 1 ˜ ( ω ) ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq136_HTML.gif.
  1. (ii)
    By the proof of (i), we also obtain ρ 1 ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq137_HTML.gif 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equab_HTML.gif
     

It only remains to prove the continuity of ρ 1 ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq137_HTML.gif. Let { ω n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq138_HTML.gif be a convergent sequence in W and ω n ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq139_HTML.gif, as n + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq27_HTML.gif. Since { ρ 1 ˜ ( ω n ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq140_HTML.gif is a bounded sequence, it contains a convergent subsequence { ρ 1 ˜ ( ω n j ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq141_HTML.gif. Suppose that ρ 1 ˜ ( ω n j ) ρ 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq142_HTML.gif as j + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq143_HTML.gif. Since Λ ω n j ( ρ 1 ˜ ( ω n j ) ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq144_HTML.gif, letting j + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq143_HTML.gif, we have Λ ω ( ρ 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq145_HTML.gif, which together with (i) implies ρ 0 = ρ 1 ˜ ( ω ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq146_HTML.gif, it means ρ 1 ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq137_HTML.gif is continuous. This completes the proof. □

Now we denote by N f ( u ) : [ 0 , 1 ] × P C 1 L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq147_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ10_HTML.gif
(10)
We define ρ 1 : P C 1 R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq148_HTML.gif as
ρ 1 ( u ) = ρ 1 ˜ ( A , B , N f ) ( u ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ11_HTML.gif
(11)

where A = ( A 1 , , A k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq149_HTML.gif, B = ( B 1 , , B k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq150_HTML.gif.

It is clear that ρ 1 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq151_HTML.gif is continuous and sends bounded sets of P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif to bounded sets of R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq30_HTML.gif, 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 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equac_HTML.gif
For fixed a , b R k N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq152_HTML.gif, we denote K ( a , b ) : L 1 P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq153_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equad_HTML.gif
Define K 1 : P C 1 P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq154_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equae_HTML.gif
Lemma 2.3 (i) The operator K ( a , b ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq155_HTML.gif is continuous and sends equi-integrable sets in L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq156_HTML.gif to relatively compact sets in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif.
  1. (ii)

    The operator K 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq157_HTML.gif is continuous and sends bounded sets in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif to relatively compact sets in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif.

     
Proof (i) It is easy to check that K ( a , b ) ( ϑ ) ( ) P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq158_HTML.gif, ϑ L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq159_HTML.gif, a , b R k N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq160_HTML.gif. 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 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equaf_HTML.gif

it is easy to check that K ( a , b ) ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq162_HTML.gif is a continuous operator from L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq156_HTML.gif to P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif.

Let now U be an equi-integrable set in L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq156_HTML.gif, then there exists α L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq163_HTML.gif such that
| u ( t ) | α ( t ) a.e. in  J  for any  u L 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equag_HTML.gif

We want to show that K ( a , b ) ( U ) ¯ P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq164_HTML.gif is a compact set.

Let { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq165_HTML.gif be a sequence in K ( a , b ) ( U ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq166_HTML.gif, then there exists a sequence { ϑ n } U https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq167_HTML.gif such that u n = K ( a , b ) ( ϑ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq168_HTML.gif. For any t 1 , t 2 J https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq169_HTML.gif, 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 | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equah_HTML.gif

Hence the sequence { F ( ϑ n ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq170_HTML.gif is uniformly bounded and equi-continuous. By the Ascoli-Arzela theorem, there exists a subsequence of { F ( ϑ n ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq170_HTML.gif (which we rename the same) which is convergent in PC. According to the bounded continuity of the operator ρ 1 ˜ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq171_HTML.gif, we can choose a subsequence of { ρ 1 ˜ ( a , b , ϑ n ) + F ( ϑ n ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq172_HTML.gif (which we still denote { ρ 1 ˜ ( a , b , ϑ n ) + F ( ϑ n ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq173_HTML.gif) 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 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equai_HTML.gif
it follows from the continuity of φ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq175_HTML.gif 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq156_HTML.gif that K ( a , b ) ( ϑ n ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq177_HTML.gif is convergent in PC. Thus { u n } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq165_HTML.gif is convergent in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif.
  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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq178_HTML.gif as
P 1 ( u ) = 0 1 g ( t ) [ K 1 ( u ) ( t ) + t i < t A i ] d t 1 σ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equaj_HTML.gif

It is easy to see that P 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq179_HTML.gif is compact continuous.

Lemma 2.4 Suppose that σ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif, = 1 m 2 α δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq25_HTML.gif; h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq116_HTML.gif on [ ξ 1 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq117_HTML.gif, α ξ ξ + 1 h ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq180_HTML.gif ( = 1 , , m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq181_HTML.gif) and h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq119_HTML.gif on [ 0 , ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq120_HTML.gif. 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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ12_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ13_HTML.gif
(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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ14_HTML.gif
(14)
Combining the definition of ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq109_HTML.gif, we can see
u = P 1 ( u ) + t i < t A i + K 1 ( u ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equak_HTML.gif
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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ15_HTML.gif
(15)
and
u ( 1 ) = P 1 ( u ) + i = 1 k A i + K 1 ( u ) ( 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equal_HTML.gif
By the condition of the mapping ρ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq109_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equam_HTML.gif
Thus
u ( 1 ) = = 1 m 2 α u ( ξ ) 0 1 h ( t ) u ( t ) d t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ16_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ17_HTML.gif
(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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq24_HTML.gif and = 1 m 2 α δ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq182_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equan_HTML.gif
Denote ρ 2 = w ( 0 ) φ ( 0 , u ( 0 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq183_HTML.gif. It is easy to see that ρ 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq184_HTML.gif is dependent on a, b and f ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq110_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equao_HTML.gif
For any ω W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq114_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equap_HTML.gif

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 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq186_HTML.gif has the following properties:
  1. (i)

    For any fixed ω W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq114_HTML.gif, the equation Γ ω ( ρ 2 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq187_HTML.gif has a unique solution ρ 2 ˜ ( ω ) R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq188_HTML.gif.

     
  2. (ii)
    The function ρ 2 ˜ : W R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq189_HTML.gif, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, for any ω = ( a , b , ϑ ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq124_HTML.gif, 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 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equaq_HTML.gif
     
where the notation M p # 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq125_HTML.gif means
M p # 1 = { M p + 1 , M > 1 , M p 1 , M 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equar_HTML.gif

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

We define ρ 2 : P C 1 R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq190_HTML.gif as ρ 2 ( u ) = ρ 2 ˜ ( A , B , N f ) ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq191_HTML.gif, where A = ( A 1 , , A k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq149_HTML.gif, B = ( B 1 , , B k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq150_HTML.gif.

It is clear that ρ 2 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq192_HTML.gif is continuous and sends bounded sets of P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif to bounded sets of R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq30_HTML.gif, and hence it is compact continuous.

For fixed a , b R k N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq152_HTML.gif, we denote K ( a , b ) : L 1 P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq193_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equas_HTML.gif
Define K 2 : P C 1 P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq194_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equat_HTML.gif

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

Lemma 2.6 (i) The operator K ( a , b ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq195_HTML.gif is continuous and sends equi-integrable sets in L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq156_HTML.gif to relatively compact sets in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif.
  1. (ii)

    The operator K 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq196_HTML.gif is continuous and sends bounded sets in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif to relatively compact sets in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif.

     
Let us define P 2 : P C 1 P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq197_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equau_HTML.gif

It is easy to see that P 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq198_HTML.gif is compact continuous.

Lemma 2.7 Suppose that σ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq24_HTML.gif, = 1 m 2 α δ 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq93_HTML.gif, 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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equav_HTML.gif

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

2.3 Case (iii)

Suppose that σ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif and = 1 m 2 α δ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq94_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equaw_HTML.gif

Denote ρ 3 = w ( 0 ) φ ( 0 , u ( 0 ) ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq199_HTML.gif. It is easy to see that ρ 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq200_HTML.gif is dependent on a, b and f ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq110_HTML.gif.

From u ( 0 ) = 0 1 g ( t ) u ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq201_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ18_HTML.gif
(18)
From u ( 1 ) = = 1 m 2 α u ( ξ ) 0 1 h ( t ) u ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq202_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ19_HTML.gif
(19)
For fixed ω W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq114_HTML.gif, 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equax_HTML.gif

From (18) and (19), we have ϒ ω ( ρ 3 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq203_HTML.gif.

Obviously, ϒ ω ( ρ 3 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq204_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equay_HTML.gif
Denote ξ m 1 = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq115_HTML.gif. 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equaz_HTML.gif

Lemma 2.8 Suppose that α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq205_HTML.gif, g, h satisfy one of the following:

(10) = 1 m 2 α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq206_HTML.gif, g ( t ) ( 1 = 1 m 2 α + δ ) + h ( t ) ( 1 σ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq207_HTML.gif;

(20) h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq116_HTML.gif on [ ξ 1 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq117_HTML.gif, α ξ ξ + 1 h ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq208_HTML.gif ( = 1 , , m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq12_HTML.gif) and h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq119_HTML.gif on [ 0 , ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq120_HTML.gif.

Then the function ϒ ω ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq209_HTML.gif has the following properties:
  1. (i)

    For any fixed ω W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq114_HTML.gif, the equation ϒ ω ( ρ 3 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq210_HTML.gif has a unique solution ρ 3 ˜ ( ω ) R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq211_HTML.gif.

     
  2. (ii)
    The function ρ 3 ˜ : W R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq212_HTML.gif, defined in (i), is continuous and sends bounded sets to bounded sets. Moreover, for any ω = ( a , b , ϑ ) W https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq124_HTML.gif, 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 } , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equba_HTML.gif
     
where the notation M p # 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq125_HTML.gif means
M p # 1 = { M p + 1 , M > 1 , M p 1 , M 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbb_HTML.gif

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

We define ρ 3 : P C 1 R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq213_HTML.gif as ρ 3 ( u ) = ρ 3 ˜ ( A , B , N f ) ( u ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq214_HTML.gif, where A = ( A 1 , , A k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq149_HTML.gif, B = ( B 1 , , B k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq150_HTML.gif.

It is clear that ρ 3 ( ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq215_HTML.gif is continuous and sends bounded sets of P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif to bounded sets of R N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq30_HTML.gif, and hence it is compact continuous.

For fixed a , b R k N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq152_HTML.gif, we denote K ( a , b ) : L 1 P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq216_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbc_HTML.gif
Define K 3 : P C 1 P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq217_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbd_HTML.gif

Similar to the proof of Lemma 2.3, we have

Lemma 2.9 (i) The operator K ( a , b ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq218_HTML.gif is continuous and sends equi-integrable sets in L 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq156_HTML.gif to relatively compact sets in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif.
  1. (ii)

    The operator K 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq219_HTML.gif is continuous and sends bounded sets in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif to relatively compact sets in P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif.

     
Let us define P 3 : P C 1 P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq220_HTML.gif as
P 3 ( u ) = 0 1 g ( t ) [ K 3 ( u ) ( t ) + t i < t A i ] d t 1 σ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Eqube_HTML.gif

It is easy to see that P 3 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq221_HTML.gif is compact continuous.

Lemma 2.10 Suppose that σ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif, = 1 m 2 α δ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq222_HTML.gif and α https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq205_HTML.gif, g, h satisfy one of the following:

(10) = 1 m 2 α 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq223_HTML.gif, g ( t ) ( 1 = 1 m 2 α + δ ) + h ( t ) ( 1 σ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq207_HTML.gif;

(20) h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq116_HTML.gif on [ ξ 1 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq117_HTML.gif, α ξ ξ + 1 h ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq208_HTML.gif ( = 1 , , m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq12_HTML.gif) and h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq119_HTML.gif on [ 0 , ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq120_HTML.gif.

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 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbf_HTML.gif

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif, = 1 m 2 α δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq25_HTML.gif.

When f satisfies the sub- ( p 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq224_HTML.gif growth condition, we have the following theorem.

Theorem 3.1 Suppose that σ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif, = 1 m 2 α δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq25_HTML.gif; h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq116_HTML.gif on [ ξ 1 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq117_HTML.gif, α ξ ξ + 1 h ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq180_HTML.gif ( = 1 , , m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq181_HTML.gif) and h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq119_HTML.gif on [ 0 , ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq120_HTML.gif; f satisfies the sub- ( p 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq72_HTML.gif 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ20_HTML.gif
(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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbg_HTML.gif
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 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq225_HTML.gif is defined in (10).

Obviously, ( S 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq226_HTML.gif) has the same solution as the following operator equation when λ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq227_HTML.gif:
u = Ψ f ( u , λ ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ21_HTML.gif
(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 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq229_HTML.gif. It follows from Lemma 2.2 and Lemma 2.3 that Ψ f ( , λ ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq230_HTML.gif is compact continuous from P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif to P C 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq54_HTML.gif for any λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq229_HTML.gif.

We claim that all the solutions of (21) are uniformly bounded for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq229_HTML.gif. In fact, if it is false, we can find a sequence of solutions { ( u n , λ n ) } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq231_HTML.gif for (21) such that u n 1 + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq232_HTML.gif as n + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq233_HTML.gif and u n 1 > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq234_HTML.gif for any n = 1 , 2 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq235_HTML.gif .

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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq226_HTML.gif), 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 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq236_HTML.gif. 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 , . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ23_HTML.gif
(23)
It follows from (14), (20) and (22) that
| u n ( 0 ) | C 9 u n 1 α , where  α = q + 1 p 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbl_HTML.gif
For any j = 1 , , N https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq237_HTML.gif, 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 , , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbm_HTML.gif
which implies that
| u n j | 0 C 12 u n 1 α , j = 1 , , N ; n = 1 , 2 , . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbn_HTML.gif
Thus
u n 0 N C 12 u n 1 α , n = 1 , 2 , . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equ24_HTML.gif
(24)

It follows from (23) and (24) that { u n 1 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq238_HTML.gif is uniformly bounded.

Thus, we can choose a large enough R 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq239_HTML.gif such that all the solutions of (21) belong to B ( R 0 ) = { u P C 1 u 1 < R 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq240_HTML.gif. Therefore the Leray-Schauder degree d L S [ I Ψ f ( , λ ) , B ( R 0 ) , 0 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq241_HTML.gif is well defined for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq229_HTML.gif, and
d L S [ I Ψ f ( , 1 ) , B ( R 0 ) , 0 ] = d L S [ I Ψ f ( , 0 ) , B ( R 0 ) , 0 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbo_HTML.gif
It is easy to see that u is a solution of u = Ψ f ( u , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq242_HTML.gif 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 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbp_HTML.gif
Obviously, system ( S 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq243_HTML.gif) possesses a unique solution u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq244_HTML.gif. Since u 0 B ( R 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq245_HTML.gif, 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 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_Equbq_HTML.gif

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

Theorem 3.2 Suppose that σ < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq22_HTML.gif, = 1 m 2 α δ = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq25_HTML.gif; h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq116_HTML.gif on [ ξ 1 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq117_HTML.gif, α ξ ξ + 1 h ( t ) d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq180_HTML.gif ( = 1 , , m 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq181_HTML.gif) and h ( t ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq119_HTML.gif on [ 0 , ξ 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq120_HTML.gif; f satisfies the sub- ( p 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq224_HTML.gif growth condition; and operators A and D = ( D 1 , , D k ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-161/MediaObjects/13661_2013_Article_463_IEq246_HTML.gif 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