Open Access

Solvability for p-Laplacian boundary value problem at resonance on the half-line

Boundary Value Problems20132013:207

DOI: 10.1186/1687-2770-2013-207

Received: 5 April 2013

Accepted: 23 August 2013

Published: 11 September 2013

Abstract

The existence of solutions for p-Laplacian boundary value problem at resonance on the half-line is investigated. Our analysis relies on constructing the suitable Banach space, defining appropriate operators and using the extension of Mawhin’s continuation theorem. An example is given to illustrate our main result.

MSC:70K30, 34B10, 34B15.

Keywords

p-Laplacian resonance half-line multi-point boundary value problem continuation theorem

1 Introduction

A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a non-trivial solution. Resonance problems can be expressed as an abstract equation L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq1_HTML.gif, where L is a noninvertible operator. When L is linear, Mawhin’s continuation theorem [1] is an effective tool in finding solutions for these problems, see [210] and references cited therein. But it does not work when L is nonlinear, for instance, p-Laplacian operator. In order to solve this problem, Ge and Ren [11] proved a continuation theorem for the abstract equation L x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq1_HTML.gif when L is a noninvertible nonlinear operator and used it to study the existence of solutions for the boundary value problems with a p-Laplacian:
{ ( φ p ( u ) ) + f ( t , u ) = 0 , 0 < t < 1 , u ( 0 ) = 0 = G ( u ( η ) , u ( 1 ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equa_HTML.gif

where φ p ( s ) = | s | p 2 s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq2_HTML.gif, p > 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq3_HTML.gif, 0 < η < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq4_HTML.gif. φ p ( s ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq5_HTML.gif is nonlinear when p 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq6_HTML.gif.

As far as the boundary value problems on unbounded domain are concerned, there are many excellent results, see [1215] and references cited therein.

To the best of our knowledge, there are few papers that study the p-Laplacian boundary value problem at resonance on the half-line. In this paper, we investigate the existence of solutions for the boundary value problem
{ ( φ p ( u ) ) + f ( t , u , u ) = 0 , 0 < t < + , u ( 0 ) = 0 , φ p ( u ( + ) ) = i = 1 n α i φ p ( u ( ξ i ) ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equ1_HTML.gif
(1.1)

where α i > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq7_HTML.gif, i = 1 , 2 , , n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq8_HTML.gif, i = 1 n α i = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq9_HTML.gif.

In order to obtain our main results, we always suppose that the following conditions hold.

(H1) 0 < ξ 1 < ξ 2 < < ξ n < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq10_HTML.gif, α i > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq7_HTML.gif, i = 1 n α i = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq9_HTML.gif.

(H2) f : [ 0 , + ) × R 2 R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq11_HTML.gif is continuous, f ( t , 0 , 0 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq12_HTML.gif, t ( 0 , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq13_HTML.gif and for any r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq14_HTML.gif, there exists a nonnegative function h r ( t ) L 1 [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq15_HTML.gif such that
| f ( t , x , y ) | h r ( t ) , a.e.  t [ 0 , + ) , x , y R , | x | 1 + t r , | y | r . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equb_HTML.gif

2 Preliminaries

For convenience, we introduce some notations and a theorem. For more details, see [11].

Definition 2.1 [11]

Let X and Y be two Banach spaces with the norms X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq16_HTML.gif, Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq17_HTML.gif, respectively. A continuous operator M : X dom M Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq18_HTML.gif is said to be quasi-linear if
  1. (i)

    Im M : = M ( X dom M ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq19_HTML.gif is a closed subset of Y,

     
  2. (ii)

    Ker M : = { x X dom M : M x = 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq20_HTML.gif is linearly homeomorphic to R n https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq21_HTML.gif, n < https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq22_HTML.gif, where domM denote the domain of the operator M.

     

Let X 1 = Ker M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq23_HTML.gif and X 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq24_HTML.gif be the complement space of X 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq25_HTML.gif in X, then X = X 1 X 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq26_HTML.gif. On the other hand, suppose that Y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq27_HTML.gif is a subspace of Y, and that Y 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq28_HTML.gif is the complement of Y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq27_HTML.gif in Y, i.e., Y = Y 1 Y 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq29_HTML.gif. Let P : X X 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq30_HTML.gif and Q : Y Y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq31_HTML.gif be two projectors and Ω X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq32_HTML.gif an open and bounded set with the origin θ Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq33_HTML.gif.

Definition 2.2 [11]

Suppose that N λ : Ω ¯ Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq34_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq35_HTML.gif is a continuous operator. Denote N 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq36_HTML.gif by N. Let Σ λ = { x Ω ¯ : M x = N λ x } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq37_HTML.gif. N λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq38_HTML.gif is said to be M-compact in Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq39_HTML.gif if there exist a vector subspace Y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq27_HTML.gif of Y satisfying dim Y 1 = dim X 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq40_HTML.gif and an operator R : Ω ¯ × [ 0 , 1 ] X 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq41_HTML.gif being continuous and compact such that for λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq35_HTML.gif,
  1. (a)

    ( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq42_HTML.gif,

     
  2. (b)

    Q N λ x = θ , λ ( 0 , 1 ) Q N x = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq43_HTML.gif,

     
  3. (c)

    R ( , 0 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq44_HTML.gif is the zero operator and R ( , λ ) | Σ λ = ( I P ) | Σ λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq45_HTML.gif,

     
  4. (d)

    M [ P + R ( , λ ) ] = ( I Q ) N λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq46_HTML.gif.

     

Theorem 2.1 [11]

Let X and Y be two Banach spaces with the norms X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq16_HTML.gif, Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq17_HTML.gif, respectively, and Ω X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq32_HTML.gif an open and bounded nonempty set. Suppose that
M : X dom M Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equc_HTML.gif

is a quasi-linear operator and N λ : Ω ¯ Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq34_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq35_HTML.gif M-compact. In addition, if the following conditions hold:

(C1) M x N λ x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq47_HTML.gif, x Ω dom M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq48_HTML.gif, λ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq49_HTML.gif,

(C2) deg { J Q N , Ω Ker M , 0 } 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq50_HTML.gif,

then the abstract equation M x = N x https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq51_HTML.gif has at least one solution in dom M Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq52_HTML.gif, where N = N 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq53_HTML.gif, J : Im Q Ker M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq54_HTML.gif is a homeomorphism with J ( θ ) = θ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq55_HTML.gif.

3 Main result

Let X = { u | u C 1 [ 0 , + ) , u ( 0 ) = 0 , sup t [ 0 , + ) | u ( t ) | 1 + t < + , lim t + u ( t )  exists } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq56_HTML.gif with norm u = max { u 1 + t , u } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq57_HTML.gif, where u = sup t [ 0 , + ) | u ( t ) | https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq58_HTML.gif. Y = L 1 [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq59_HTML.gif with norm y 1 = 0 + | y ( t ) | d t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq60_HTML.gif. Then ( X , ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq61_HTML.gif and ( Y , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq62_HTML.gif are Banach spaces.

Define operators M : X dom M Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq63_HTML.gif and N λ : X Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq64_HTML.gif as follows:
M u = ( φ p ( u ) ) , N λ u = λ f ( t , u , u ) , λ [ 0 , 1 ] , t [ 0 , + ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equd_HTML.gif
where
dom M = { u X | φ p ( u ) A C [ 0 , + ) , ( φ p ( u ) ) L 1 [ 0 , + ) , φ p ( u ( + ) ) = i = 1 n α i φ p ( u ( ξ i ) ) } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Eque_HTML.gif

Then the boundary value problem (1.1) is equivalent to M u = N u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq65_HTML.gif.

Obviously,
Ker M = { a t | a R } , Im M = { y | y Y , i = 1 n α i ξ i + y ( s ) d s = 0 } . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equf_HTML.gif

It is clear that KerM is linearly homeomorphic to , and Im M Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq66_HTML.gif is closed. So, M is a quasi-linear operator.

Define P : X X 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq30_HTML.gif, Q : Y Y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq31_HTML.gif as
( P u ) ( t ) = u ( + ) t , ( Q y ) ( t ) = i = 1 n α i ξ i + y ( s ) d s i = 1 n α i e ξ i e t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equg_HTML.gif

where X 1 = Ker M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq23_HTML.gif, Y 1 = Im Q = { b e t | b R } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq67_HTML.gif. We can easily obtain that P : X X 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq30_HTML.gif, Q : Y Y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq31_HTML.gif are projectors. Set X = X 1 X 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq26_HTML.gif, Y = Y 1 Y 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq29_HTML.gif.

Define an operator R : X × [ 0 , 1 ] X 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq68_HTML.gif:
R ( u , λ ) ( t ) = 0 t φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ u ( + ) t , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equh_HTML.gif

where 1 p + 1 q = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq69_HTML.gif, φ q = φ p 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq70_HTML.gif. By (H1) and (H2), we get that R : X × [ 0 , 1 ] X 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq68_HTML.gif is continuous.

Lemma 3.1 [15]

V X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq71_HTML.gif is compact if { u ( t ) 1 + t | u V } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq72_HTML.gif and { u ( t ) | u V } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq73_HTML.gif are both equicontinuous on any compact intervals of [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq74_HTML.gif and equiconvergent at infinity.

Lemma 3.2 R : X × [ 0 , 1 ] X 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq68_HTML.gif is compact.

Proof Let Ω X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq32_HTML.gif be nonempty and bounded. There exists a constant r > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq14_HTML.gif such that u r https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq75_HTML.gif, u Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq76_HTML.gif. It follows from (H2) that there exists a nonnegative function h r ( t ) L 1 [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq15_HTML.gif such that
| f ( t , u ( t ) , u ( t ) ) | h r ( t ) , a.e.  t [ 0 , + ) , u Ω ¯ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equi_HTML.gif
For any T > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq77_HTML.gif, t 1 , t 2 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq78_HTML.gif, u Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq79_HTML.gif, λ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq35_HTML.gif, we have
| R ( u , λ ) ( t 1 ) 1 + t 1 R ( u , λ ) ( t 2 ) 1 + t 2 | | 1 1 + t 1 0 t 1 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ 1 1 + t 2 0 t 2 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ | + | t 1 1 + t 1 t 2 1 + t 2 | | u ( + ) | | 1 1 + t 1 t 2 t 1 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ | + | 1 1 + t 1 1 1 + t 2 | × | 0 t 2 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ | + | t 1 1 + t 1 t 2 1 + t 2 | r φ q [ h r 1 ( 1 + 1 i = 1 n α i e ξ i ) + φ p ( r ) ] [ | t 1 t 2 | + T | 1 1 + t 1 1 1 + t 2 | ] + | t 1 1 + t 1 t 2 1 + t 2 | r . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equj_HTML.gif
Since { t , 1 1 + t , t 1 + t } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq80_HTML.gif are equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq81_HTML.gif, we get that { R ( u , λ ) ( t ) 1 + t , u Ω ¯ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq82_HTML.gif are equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq81_HTML.gif.
| R ( u , λ ) ( t 1 ) R ( u , λ ) ( t 2 ) | = | φ q [ t 1 + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] φ q [ t 2 + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] | . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equk_HTML.gif
Let
g ( t , u ) = t + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equl_HTML.gif
Then
| g ( t , u ) | h r 1 ( 1 + 1 i = 1 n α i e ξ i ) + φ p ( r ) : = k , t [ 0 , T ] , u Ω ¯ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equ2_HTML.gif
(3.1)
For t 1 , t 2 [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq78_HTML.gif, t 1 < t 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq83_HTML.gif, u Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq79_HTML.gif, we have
| g ( t 1 , u ) g ( t 2 , u ) | = | t 1 t 2 λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s | t 1 t 2 h r ( s ) + h r 1 i = 1 n α i e ξ i e s d s . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equm_HTML.gif

It follows from the absolute continuity of integral that { g ( t , u ) , u Ω ¯ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq84_HTML.gif are equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq81_HTML.gif. Since φ q ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq85_HTML.gif is uniformly continuous on [ k , k ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq86_HTML.gif, by (3.1), we can obtain that { R ( u , λ ) ( t ) , u Ω ¯ } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq87_HTML.gif are equicontinuous on [ 0 , T ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq81_HTML.gif.

For u Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq79_HTML.gif, since
| τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s | τ + h r ( s ) + h r 1 i = 1 n α i e ξ i e s d s , lim τ + τ + h r ( s ) + h r 1 i = 1 n α i e ξ i e s d s = 0 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equn_HTML.gif
and φ q ( x ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq85_HTML.gif is uniformly continuous on [ r r p 1 , r + r p 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq88_HTML.gif, for any ε > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq89_HTML.gif, there exists a constant T 1 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq90_HTML.gif such that if τ T 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq91_HTML.gif, then
| φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) | < ε 4 , u Ω ¯ . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equ3_HTML.gif
(3.2)
Since
| 0 T 1 φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] d τ u ( + ) T 1 | { φ q [ h r 1 ( 1 + 1 i = 1 n α i e ξ i ) + φ p ( r ) ] + r } T 1 , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equ4_HTML.gif
(3.3)
there exists a constant T > T 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq92_HTML.gif such that if t > T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq93_HTML.gif, then
1 1 + t { φ q [ h r 1 ( 1 + 1 i = 1 n α i e ξ i ) + φ p ( r ) ] + r } T 1 < ε 4 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equ5_HTML.gif
(3.4)
For t 2 > t 1 > T https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq94_HTML.gif, by (3.2), (3.3) and (3.4), we have
| R ( u , λ ) ( t 1 ) 1 + t 1 R ( u , λ ) ( t 2 ) 1 + t 2 | = | 1 1 + t 1 0 t 1 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ 1 1 + t 2 0 t 2 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | | 1 1 + t 1 0 T 1 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | + | 1 1 + t 1 T 1 t 1 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | + | 1 1 + t 2 0 T 1 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | + | 1 1 + t 2 T 1 t 2 { φ q [ τ + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) } d τ | < ε , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equo_HTML.gif
and
| R ( u , λ ) ( t 1 ) R ( u , λ ) ( t 2 ) | | φ q [ t 1 + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) | + | φ q [ t 2 + λ ( f ( s , u ( s ) , u ( s ) ) i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e s ) d s + φ p ( u ( + ) ) ] u ( + ) | < ε . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equp_HTML.gif

By Lemma 3.1, we get that { R ( u , λ ) | u Ω ¯ , λ [ 0 , 1 ] } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq95_HTML.gif is compact. The proof is completed. □

In the spaces X and Y, the origin θ = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq96_HTML.gif. In the following sections, we denote the origin by 0.

Lemma 3.3 Let Ω X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq32_HTML.gif be nonempty, open and bounded. Then N λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq38_HTML.gif is M-compact in Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq39_HTML.gif.

Proof By (H2), we know that N λ : Ω ¯ Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq34_HTML.gif is continuous. Obviously, dim X 1 = dim Y 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq97_HTML.gif. For u Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq76_HTML.gif, since Q ( I Q ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq98_HTML.gif is a zero operator, we get ( I Q ) N λ ( u ) Im M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq99_HTML.gif. For y Im M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq100_HTML.gif, y = Q y + ( I Q ) y = ( I Q ) y ( I Q ) Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq101_HTML.gif. So, we have ( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq42_HTML.gif. It is clear that
Q N λ u = 0 , λ ( 0 , 1 ) Q N u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equq_HTML.gif
and R ( u , 0 ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq102_HTML.gif, u X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq103_HTML.gif. u Σ λ = { u Ω ¯ : M u = N λ u } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq104_HTML.gif means that N λ u Im M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq105_HTML.gif and ( φ p ( u ) ) + λ f ( t , u , u ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq106_HTML.gif, thus,
R ( u , λ ) ( t ) = 0 t φ q [ τ + ( φ p ( u ) ) d s + φ p ( u ( + ) ) ] d τ u ( + ) t = u ( t ) u ( + ) t = ( I P ) u ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equr_HTML.gif
For u X https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq107_HTML.gif, we have
M [ P + R ( u , λ ) ] ( t ) = λ f ( t , u ( t ) , u ( t ) ) + i = 1 n α i ξ i + λ f ( r , u ( r ) , u ( r ) ) d r i = 1 n α i e ξ i e t = ( I Q ) N λ u ( t ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equs_HTML.gif

These, together with Lemma 3.2, mean that N λ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq38_HTML.gif is M-compact in Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq39_HTML.gif. The proof is completed. □

In order to obtain our main results, we need the following additional conditions.

(H3) There exist nonnegative functions a ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq108_HTML.gif, b ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq109_HTML.gif, c ( t ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq110_HTML.gif with ( 1 + t ) p 1 a ( t ) , b ( t ) , c ( t ) Y https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq111_HTML.gif and ( 1 + t ) p 1 a ( t ) 1 + b ( t ) 1 < 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq112_HTML.gif such that
| f ( t , x , y ) | a ( t ) | φ p ( x ) | + b ( t ) | φ p ( y ) | + c ( t ) , a.e.  t [ 0 , + ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equt_HTML.gif
(H4) There exists a constant d 0 > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq113_HTML.gif such that if | d | > d 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq114_HTML.gif, then one of the following inequalities holds:
d f ( t , x , d ) < 0 , ( t , x ) [ 0 , + ) × R ; d f ( t , x , d ) > 0 , ( t , x ) [ 0 , + ) × R . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equu_HTML.gif
Lemma 3.4 Assume that (H3) and (H4) hold. The set
Ω 1 = { u | u dom M , M u = N λ u , λ [ 0 , 1 ] } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equv_HTML.gif

is bounded in X.

Proof If u Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq115_HTML.gif, then Q N λ u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq116_HTML.gif, i.e., i = 1 n α i ξ i + f ( r , u ( r ) , u ( r ) ) d r = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq117_HTML.gif. By (H4), there exists t 0 [ 0 , + ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq118_HTML.gif such that | u ( t 0 ) | d 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq119_HTML.gif. It follows from M u = N λ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq120_HTML.gif that
φ p ( u ( t ) ) = t 0 t λ f ( s , u ( s ) , u ( s ) ) d s + φ p ( u ( t 0 ) ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equw_HTML.gif
Considering (H3), we have
| φ p ( u ( t ) ) | 0 + [ a ( t ) | φ p ( u ( t ) ) | + b ( t ) | φ p ( u ( t ) ) | + c ( t ) ] d t + φ p ( d 0 ) a ( t ) ( 1 + t ) p 1 1 φ p ( u 1 + t ) + b 1 φ p ( u ) + c 1 + φ p ( d 0 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equ6_HTML.gif
(3.5)
Since u ( t ) = 0 t u ( s ) d s https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq121_HTML.gif, we get
| u ( t ) 1 + t | t 1 + t u u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equx_HTML.gif
Thus,
u 1 + t u . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equ7_HTML.gif
(3.6)
By (3.5), (3.6) and (H3), we get
φ p ( u ) c 1 + φ p ( d 0 ) 1 a ( t ) ( 1 + t ) p 1 1 b 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equy_HTML.gif
So,
u φ q ( c 1 + φ p ( d 0 ) 1 a ( t ) ( 1 + t ) p 1 1 b 1 ) . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equz_HTML.gif

This, together with (3.6), means that Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq122_HTML.gif is bounded. The proof is completed. □

Lemma 3.5 Assume that (H4) holds. The set
Ω 2 = { u | u Ker M , Q N u = 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equaa_HTML.gif

is bounded in X.

Proof u Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq123_HTML.gif means that u = a t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq124_HTML.gif, a R https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq125_HTML.gif and Q N u = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq126_HTML.gif, i.e.,
i = 1 n α i ξ i + f ( s , a s , a ) d s = 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equab_HTML.gif

By (H4), we get that | a | d 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq127_HTML.gif. So, Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq128_HTML.gif is bounded. The proof is completed. □

Theorem 3.1 Suppose that (H1)-(H4) hold. Then problem (1.1) has at least one solution.

Proof Let Ω = { u X | u < d 0 } https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq129_HTML.gif, where d 0 = max { d 0 , sup u Ω 1 u , sup u Ω 2 u } + 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq130_HTML.gif. It follows from the definition of Ω 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq122_HTML.gif and Ω 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq128_HTML.gif that M u N λ u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq131_HTML.gif, λ ( 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq49_HTML.gif, u Ω https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq132_HTML.gif and Q N u 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq133_HTML.gif, u Ω Ker M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq134_HTML.gif.

Define a homeomorphism J : Im Q Ker M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq135_HTML.gif as J ( k e t ) = k t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq136_HTML.gif. If d f ( t , x , d ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq137_HTML.gif for | d | > d 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq114_HTML.gif, take the homotopy
H ( u , μ ) = μ u + ( 1 μ ) J Q N u , u Ω ¯ Ker M , μ [ 0 , 1 ] . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equac_HTML.gif
For u Ω ¯ Ker M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq138_HTML.gif, we have u = k t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq139_HTML.gif. Then
H ( u , μ ) = μ k t ( 1 μ ) i = 1 n α i ξ i + f ( s , k s , k ) d s i = 1 n α i e ξ i t . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equad_HTML.gif
Obviously, H ( u , 1 ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq140_HTML.gif, u Ω Ker M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq134_HTML.gif. For μ [ 0 , 1 ) https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq141_HTML.gif, u = k t Ω Ker M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq142_HTML.gif, if H ( u , μ ) = 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq143_HTML.gif, we have
i = 1 n α i ξ i + k f ( s , k s , k ) d s i = 1 n α i e ξ i = μ 1 μ k 2 0 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equae_HTML.gif
A contradiction with d f ( t , x , d ) < 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq137_HTML.gif, | d | > d 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq114_HTML.gif. If d f ( t , x , d ) > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq144_HTML.gif, | d | > d 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq114_HTML.gif, take
H ( u , μ ) = μ u ( 1 μ ) J Q N u , u Ω ¯ Ker M , μ [ 0 , 1 ] , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equaf_HTML.gif

and the contradiction follows analogously. So, we obtain H ( u , μ ) 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq145_HTML.gif, μ [ 0 , 1 ] https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq146_HTML.gif, u Ω Ker M https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq134_HTML.gif.

By the homotopy of degree, we get that
deg ( J Q N , Ω Ker M , 0 ) = deg ( H ( , 0 ) , Ω Ker M , 0 ) = deg ( H ( , 1 ) , Ω Ker M , 0 ) = deg ( I , Ω Ker M , 0 ) = 1 . https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equag_HTML.gif

By Theorem 2.1, we can get that M u = N u https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq65_HTML.gif has at least one solution in Ω ¯ https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq39_HTML.gif. The proof is completed. □

4 Example

Let us consider the following boundary value problem at resonance
{ ( | u | 1 2 u ) + e 4 t 1 + t sin | u | + e 4 t | u | 1 2 u + 1 4 e 4 t = 0 , 0 < t < + , u ( 0 ) = 0 , | u ( + ) | 1 2 u ( + ) = i = 1 n α i | u ( ξ i ) | 1 2 u ( ξ i ) , https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_Equ8_HTML.gif
(4.1)

where 0 < ξ 1 < ξ 2 < < ξ n < + https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq10_HTML.gif, α i > 0 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq7_HTML.gif, i = 1 n α i = 1 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq9_HTML.gif.

Corresponding to problem (1.1), we have p = 3 2 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq147_HTML.gif, f ( t , x , y ) = e 4 t 1 + t sin | x | + e 4 t | y | 1 2 y + 1 4 e 4 t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq148_HTML.gif.

Take a ( t ) = e 4 t 1 + t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq149_HTML.gif, b ( t ) = e 4 t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq150_HTML.gif, c ( t ) = 1 4 e 4 t https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq151_HTML.gif, d 0 = 4 https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-207/MediaObjects/13661_2013_Article_605_IEq152_HTML.gif. By simple calculation, we can get that conditions (H1)-(H4) hold. By Theorem 3.1, we obtain that problem (4.1) has at least one solution.

Declarations

Acknowledgements

This work is supported by the National Science Foundation of China (11171088) and the Natural Science Foundation of Hebei Province (A2013208108). The author is grateful to anonymous referees for their constructive comments and suggestions, which led to the improvement of the original manuscript.

Authors’ Affiliations

(1)
College of Sciences, Hebei University of Science and Technology

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