Open Access

Solvability of boundary value problem with p-Laplacian at resonance

Boundary Value Problems20142014:36

DOI: 10.1186/1687-2770-2014-36

Received: 29 October 2013

Accepted: 24 January 2014

Published: 7 February 2014

Abstract

By generalizing the extension of the continuous theorem of Ge and Ren and constructing suitable Banach spaces and operators, we investigate the existence of solutions for p-Laplacian boundary value problems at resonance. An example is given to illustrate our results.

MSC:34B15.

Keywords

continuous theorem resonance p-Laplacian boundary value problem

1 Introduction

In this paper, we will study the boundary value problem
{ ( φ p ( u ) ) ( t ) = f ( t , u ( t ) , u ( t ) , u ( t ) ) , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = 0 1 k ( t ) u ( t ) d t ,
(1.1)
and
{ ( φ p ( u ) ) ( t ) = f ( t , u ( t ) , u ( t ) , u ( t ) ) , u ( 0 ) = 0 , u ( 0 ) = 0 1 g ( t ) u ( t ) d t , u ( 1 ) = 0 1 h ( t ) u ( t ) d t ,
(1.2)

where φ p ( s ) = | s | p 2 s , p > 1 , 0 1 k ( t ) d t = 1 , 0 1 g ( t ) d t = 1 , 0 1 h ( t ) d t = 1 .

A boundary value problem is said to be a resonance one if the corresponding homogeneous boundary value problem has a non-trivial solution. Mawhin’s continuous theorem [1] is an effective tool to solve this kind of problems when the differential operator is linear, see [210] and references cited therein. But it does not work for nonlinear cases such as boundary value problems with a p-Laplacian, which attracted the attention of mathematicians in recent years [1115]. Ge and Ren extended Mawhin’s continuous theorem [15] and many authors used their results to solve boundary value problems with a p-Laplacian, see [16, 17]. In this new theorem, two projectors P and Q must be constructed. But it is difficult to give the projector Q in many boundary value problems with a p-Laplacian. In this paper, we generalize the extension of the continuous theorem and show that the p-Laplacian problem is solvable when Q is not a projector. And we will use this new theorem to discuss problems (1.1) and (1.2), respectively.

In this paper, we will always suppose that

(H1) k ( t ) , g ( t ) , h ( t ) L 1 [ 0 , 1 ] are nonnegative and k 1 = g 1 = h 1 = 1 , where k 1 : = 0 1 | k ( t ) | d t .

(H2) f ( t , u , v , w ) is continuous in [ 0 , 1 ] × R 3 .

2 Preliminaries

Definition 2.1 [15]

Let X and Y be two Banach spaces with norms X , Y , respectively. A continuous operator M : X dom M Y is said to be quasi-linear if
  1. (i)

    Im M : = M ( X dom M ) is a closed subset of Y,

     
  2. (ii)

    Ker M : = { x X dom M : M x = 0 } is linearly homeomorphic to R n , n < ,

     

where domM denote the domain of the operator M.

Let X 1 = Ker M and X 2 be the complement space of X 1 in X, then X = X 1 X 2 . Let P : X X 1 be a projector and Ω X an open and bounded set with the origin θ Ω .

Definition 2.2 Suppose N λ : Ω ¯ Y , λ [ 0 , 1 ] is a continuous and bounded operator. Denote N 1 by N. Let Σ λ = { x Ω ¯ dom M : M x = N λ x } . N λ is said to be M-quasi-compact in Ω ¯ if there exists a vector subspace Y 1 of Y satisfying dim Y 1 = dim X 1 and two operators Q, R with Q : Y Y 1 , Q Y = Y 1 , being continuous, bounded, and satisfying Q ( I Q ) = 0 , R : Ω ¯ × [ 0 , 1 ] X 2 dom M continuous and compact such that for λ [ 0 , 1 ] ,
  1. (a)

    ( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Y ,

     
  2. (b)

    Q N λ x = θ , λ ( 0 , 1 ) Q N x = θ ,

     
  3. (c)

    R ( , 0 ) is the zero operator and R ( , λ ) | Σ λ = ( I P ) | Σ λ ,

     
  4. (d)

    M [ P + R ( , λ ) ] = ( I Q ) N λ .

     
Theorem 2.1 Let X and Y be two Banach spaces with the norms X , Y , respectively, and let Ω X be an open and bounded nonempty set. Suppose
M : X dom M Y

is a quasi-linear operator and that N λ : Ω ¯ Y , λ [ 0 , 1 ] is M-quasi-compact. In addition, if the following conditions hold:

(C1) M x N λ x , x Ω dom M , λ ( 0 , 1 ) ,

(C2) deg { J Q N , Ω Ker M , 0 } 0 ,

then the abstract equation M x = N x has at least one solution in dom M Ω ¯ , where N = N 1 , J : Im Q Ker M is a homeomorphism with J ( θ ) = θ .

Proof The proof is similar to the one of Lemma 2.1 and Theorem 2.1 in [15]. □

We can easily get the following inequalities.

Lemma 2.1 For any u , v 0 , we have
  1. (1)

    φ p ( u + v ) φ p ( u ) + φ p ( v ) , 1 < p 2 .

     
  2. (2)

    φ p ( u + v ) 2 p 2 ( φ p ( u ) + φ p ( v ) ) , p 2 .

     

In the following, we will always suppose that q satisfies 1 / p + 1 / q = 1 .

3 The existence of a solution for problem (1.1)

Let X = C 2 [ 0 , 1 ] with norm u = max { u , u , u } , Y = C [ 0 , 1 ] × C [ 0 , 1 ] with norm ( y 1 , y 2 ) = max { y 1 , y 2 } , where y = max t [ 0 , 1 ] | y ( t ) | . We know that ( X , ) and ( Y , ) are Banach spaces.

Define operators M : X dom M Y , N λ : X Y as follows:
M u = [ ( φ p ( u ) ) ( t ) T ( φ p ( u ) ) ( t ) ] , N λ u = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) 0 ] ,
where T y = c , y C [ 0 , 1 ] , c satisfying
0 1 k ( t ) t 1 φ q ( 0 s y ( r ) c d r ) d s d t = 0 , dom M = { u X φ p ( u ) C 1 [ 0 , 1 ] , u ( 0 ) = u ( 0 ) = 0 } .
(3.1)

Lemma 3.1 For y C [ 0 , 1 ] , there is only one constant c R such that T y = c with | c | y and that T : C [ 0 , 1 ] R is continuous.

Proof For y C [ 0 , 1 ] , let
F ( c ) = 0 1 k ( t ) t 1 φ q ( 0 s ( y ( r ) c ) d r ) d s d t .

Obviously, F ( c ) is continuous and strictly decreasing in . Take a = min t [ 0 , 1 ] y ( t ) , b = max t [ 0 , 1 ] y ( t ) . It is easy to see that F ( a ) 0 , F ( b ) 0 . Thus, there exists a unique constant c [ a , b ] such that F ( c ) = 0 , i.e. there is only one constant c R such that T y = c with | c | y .

For y 1 , y 2 C [ 0 , 1 ] , assume T y 1 = c 1 , T y 2 = c 2 . By k ( t ) 0 , 0 1 k ( t ) d t = 1 and φ q being strictly increasing, we obtain, if c 2 c 1 > max t [ 0 , 1 ] ( y 2 ( t ) y 1 ( t ) ) , then
0 = 0 1 k ( t ) t 1 φ q ( 0 s ( y 2 ( r ) c 2 ) d r ) d s d t = 0 1 k ( t ) t 1 φ q ( 0 s [ ( y 1 ( r ) c 1 ) + ( y 2 ( r ) y 1 ( r ) ( c 2 c 1 ) ) d r ] ) d s d t < 0 1 k ( t ) t 1 φ q ( 0 s ( y 1 ( r ) c 1 ) d r ) d s d t = 0 .
This is a contradiction. On the other hand, if c 2 c 1 < min t [ 0 , 1 ] ( y 2 ( t ) y 1 ( t ) ) , then
0 = 0 1 k ( t ) t 1 φ q ( 0 s ( y 2 ( r ) c 2 ) d r ) d s d t = 0 1 k ( t ) t 1 φ q ( 0 s [ ( y 1 ( r ) c 1 ) + ( y 2 ( r ) y 1 ( r ) ( c 2 c 1 ) ) d r ] ) d s d t > 0 1 k ( t ) t 1 φ q ( 0 s ( y 1 ( r ) c 1 ) d r ) d s d t = 0 .

This is a contradiction, too. So, we have min t [ 0 , 1 ] ( y 2 ( t ) y 1 ( t ) ) c 2 c 1 max t [ 0 , 1 ] ( y 2 ( t ) y 1 ( t ) ) , i.e. | c 2 c 1 | y 2 y 1 . So, T : C [ 0 , 1 ] R is continuous. The proof is completed. □

It is clear that u dom M is a solution if and only if it satisfies M u = N u , where N = N 1 . For convenience, let ( a , b ) L : = [ a b ] .

Lemma 3.2 M is a quasi-linear operator.

Proof It is easy to see that Ker M = { b t b R } : = X 1 .

For u X dom M , if M u = ( y , c ) L , then c satisfies (3.1). On the other hand, if y C [ 0 , 1 ] , T y = c , take
u ( t ) = 0 t ( t s ) φ q ( 0 s y ( r ) d r ) d s .
By a simple calculation, we get u X dom M and M u = ( y , c ) L . Thus
Im M = { ( y , c ) L y C [ 0 , 1 ] , c  satisfies (3.1) } .

By the continuity of T, we find that Im M Y is closed. So, M is quasi-linear. The proof is completed. □

Lemma 3.3 T ( c ) = c , T ( y + c ) = T ( y ) + c , T ( c y ) = c T ( y ) , c R , y C [ 0 , 1 ] .

Proof The proof is simple. Therefore, we omit it. □

Take a projector P : X X 1 and an operator Q : Y Y 1 as follows:
( P u ) ( t ) = u ( 0 ) t , Q ( y , y 1 ) L = ( 0 , T y 1 T y ) L ,

where Y 1 = { ( 0 , c ) L c R } . Obviously, Q Y = Y 1 , and dim Y 1 = dim X 1 .

By the continuity and boundedness of T, we can easily see that Q is continuous and bounded in Y. It follows from Lemma 3.3 that Q ( I Q ) ( y , y 1 ) L = ( 0 , 0 ) L , y , y 1 C [ 0 , 1 ] .

Define an operator R : X × [ 0 , 1 ] X 2 as
R ( u , λ ) ( t ) = 0 t ( t s ) φ q ( 0 s λ f ( r , u ( r ) , u ( r ) , u ( r ) ) d r ) d s ,

where Ker M X 2 = X . By (H2) and the Arzela-Asscoli theorem, we can easily see that R : Ω ¯ × [ 0 , 1 ] X 2 dom M is continuous and compact, where Ω X is an open bounded set.

Lemma 3.4 Assume that Ω X is an open bounded set. Then N λ is M-quasi-compact in  Ω ¯ .

Proof It is clear that Im P = Ker M , Q N λ x = θ , λ ( 0 , 1 ) Q N x = θ and R ( , 0 ) = 0 . For u Ω ¯ ,
( I Q ) N λ u = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) 0 ] [ 0 T [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] ] = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] ] Im M .

Since Im M Ker Q and y = Q y + ( I Q ) y , we obtain Im M ( I Q ) Y . Thus, ( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Y .

For u Σ λ = { u Ω ¯ dom M : M u = N λ u } , we get
R ( u , λ ) = 0 t ( t s ) φ q ( 0 s λ f ( r , u ( r ) , u ( r ) , u ( r ) ) d r ) d s = 0 t ( t s ) φ q ( 0 s ( φ p ( u ) ) ) d s = u ( t ) u ( 0 ) t = ( I P ) u ,
i.e. Definition 2.2(c) holds. For u Ω ¯ , we have
M [ P u + R ( u , λ ) ] = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] ] = ( I Q ) N λ u .

So, Definition 2.2(d) holds. Therefore, N λ is M-quasi-compact in Ω ¯ . The proof is completed. □

Theorem 3.1 Assume that the following conditions hold.

(H3) There exists a nonnegative constant K such that one of (1) and (2) holds:
  1. (1)

    B f ( t , A , B , C ) > 0 , t [ 0 , 1 ] , | B | > K , A , C R ,

     
  2. (2)

    B f ( t , A , B , C ) < 0 , t [ 0 , 1 ] , | B | > K , A , C R .

     
(H4) There exist nonnegative functions a ( t ) , b ( t ) , c ( t ) , e ( t ) L 1 [ 0 , 1 ] such that
| f ( t , x , y , z ) | a ( t ) φ p ( | x | ) + b ( t ) φ p ( | y | ) + c ( t ) φ p ( | z | ) + e ( t ) , t [ 0 , 1 ] , x , y , z R ,

where φ q ( a 1 + b 1 + c 1 ) < 2 2 q , if 1 < p 2 ; φ q ( 2 p 2 a 1 + 2 p 2 b 1 + c 1 ) < 1 , if p 2 .

Then boundary value problem (1.1) has at least one solution.

In order to prove Theorem 3.1, we show two lemmas.

Lemma 3.5 Suppose (H3) and (H4) hold. Then the set
Ω 1 = { u dom M M u = N λ u , λ ( 0 , 1 ) }

is bounded in X.

Proof For u Ω 1 , we have Q N λ u = 0 , i.e. T f ( t , u ( t ) , u ( t ) , u ( t ) ) = 0 . By (H3), there exists a constant t 0 [ 0 , 1 ] such that | u ( t 0 ) | K . Since u ( t ) = 0 t u ( s ) d s , u ( t ) = u ( t 0 ) + t 0 t u ( s ) d s , we have
| u ( t ) | u , | u ( t ) | K + u , t [ 0 , 1 ] .
(3.2)
It follows from M u = N λ u , (H4), and (3.2) that
| u ( t ) | = | φ q ( 0 t λ f ( s , u ( s ) , u ( s ) , u ( s ) ) d s ) | φ q ( 0 1 a ( t ) φ p ( | u | ) + b ( t ) φ p ( | u | ) + c ( t ) φ p ( | u | ) + e ( t ) d t ) φ q [ ( a 1 + b 1 ) φ p ( K + u ) + c 1 φ p ( u ) + e 1 ] .
If 1 < p 2 , by Lemma 2.1, we get
| u ( t ) | φ q ( B 1 + A 1 φ p ( u ) ) 2 q 2 [ φ q ( B 1 ) + φ q ( A 1 ) u ] ,
thus
u 2 q 2 φ q ( B 1 ) 1 2 q 2 φ q ( A 1 ) ,

where B 1 = ( a 1 + b 1 ) φ p ( K ) + e 1 , A 1 = a 1 + b 1 + c 1 .

If p > 2 , by Lemma 2.1, we get
| u ( t ) | φ q ( B 2 + A 2 φ p ( u ) ) [ φ q ( B 2 ) + φ q ( A 2 ) u ] ,
thus
u φ q ( B 2 ) 1 φ q ( A 2 ) ,

where B 2 = 2 p 2 ( a 1 + b 1 ) φ p ( K ) + e 1 , A 2 = 2 p 2 ( a 1 + b 1 ) + c 1 .

These, together with (3.2), mean that Ω 1 is bounded in X. □

Lemma 3.6 Assume (H3) holds. Then
Ω 2 = { u Ker M Q N u = 0 }

is bounded in X, where N = N 1 .

Proof For u Ω 2 , we have u = b t and T f ( t , b t , b , 0 ) = 0 . By (H3), we get | b | K . So, Ω 2 is bounded. The proof is completed. □

Proof of Theorem 3.1 Let Ω = { u X u < r } , where r is large enough such that K < r < + and Ω Ω 1 ¯ .

By Lemmas 3.5 and 3.6, we know M u N λ u , u dom M Ω and Q N u 0 , u Ker M Ω .

Let H ( u , δ ) = ρ δ u + ( 1 δ ) J Q N u , δ [ 0 , 1 ] , u Ker M Ω ¯ , where J : Im Q Ker M is a homeomorphism with J ( 0 , b ) L = b t , ρ = { 1 , if (H 3 )(1) holds , 1 , if (H 3 )(2) holds .

Define a function Sgn ( x ) = { 1 , if  x > 0 , 1 , if  x < 0 .

For u Ker M Ω , we have u = b t 0 . Thus
H ( u , δ ) = ρ δ b t + ( 1 δ ) ( T f ( t , b t , b , 0 ) ) t .
If δ = 1 , H ( u , 1 ) = ρ b t 0 . If δ = 0 , by Q N u 0 , we get H ( u , 0 ) = J Q N ( b t ) 0 . For 0 < δ < 1 , we now prove that H ( u , δ ) 0 . Otherwise, if H ( u , δ ) = 0 , then
T f ( t , b t , b , 0 ) = ρ δ 1 δ b .
(3.3)

Since u = r > K , we have | b | > K . Thus, T [ b f ( t , b t , b , 0 ) ] = b T f ( t , b t , b , 0 ) = ρ δ 1 δ b 2 . So, we have Sgn ( b f ( t , b t , b , 0 ) ) = Sgn { T [ b f ( t , b t , b , 0 ) ] } = Sgn ( ρ δ 1 δ b 2 ) = Sgn ( ρ ) . A contradiction with the definition of ρ. So, H ( u , δ ) 0 , u Ker M Ω , δ [ 0 , 1 ] .

By the homotopy of degree, we get
deg ( J Q N , Ω Ker M , 0 ) = deg ( H ( , 0 ) , Ω Ker M , 0 ) = deg ( H ( , 1 ) , Ω Ker M , 0 ) = deg ( ρ I , Ω Ker M , 0 ) 0 .

By Theorem 2.1, we can see that M u = N u has at least one solution in Ω ¯ . The proof is completed. □

Example Let us consider the following boundary value problem at resonance:
{ ( φ p ( u ) ) ( t ) = 1 8 t sin x 3 + 1 16 y 3 + t 3 sin z 3 + cos t , u ( 0 ) = u ( 0 ) = 0 , u ( 1 ) = 2 0 1 t u ( t ) d t ,
(3.4)

where p = 4 .

Corresponding to problem (1.1), we have q = 4 3 , a ( t ) = 1 8 t , b ( t ) = 1 16 , c ( t ) = t 3 , e ( t ) = cos t , k ( t ) = 2 t .

Take K = 4 . By a simple calculation, we find that the conditions (H1)-(H4) hold. By Theorem 3.1, we obtain the result that problem (3.4) has at least one solution.

4 The existence of a solution for problem (1.2)

Let X = C 2 [ 0 , 1 ] with norm u = max { u , u , u } , Y = C [ 0 , 1 ] × C [ 0 , 1 ] × C [ 0 , 1 ] with norm ( y 1 , y 2 , y 3 ) = max { y 1 , y 2 , y 3 } , where y = max t [ 0 , 1 ] | y ( t ) | . We know that ( X , ) and ( Y , ) are Banach spaces.

Define operators M : X dom M Y , N λ : X Y as follows:
M u = [ ( φ p ( u ) ) ( t ) T 1 ( φ p ( u ) ) ( t ) T 2 ( φ p ( u ) ) ( t ) ] , N λ u = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) 0 0 ] ,
where T 1 y = c 1 , T 2 y = c 2 , y C [ 0 , 1 ] , c 1 , c 2 satisfy
0 1 g ( t ) 0 t φ q ( 0 s y ( r ) c 1 d r ) d s d t = 0 , 0 1 h ( t ) t 1 φ q ( 0 s y ( r ) c 2 d r ) d s d t = 0 , dom M = { u X φ p ( u ) C 1 [ 0 , 1 ] , u ( 0 ) = 0 } .
(4.1)

Lemma 4.1 For y C [ 0 , 1 ] , there is only one constant c i R such that T i y = c i with | c i | y . And T i : C [ 0 , 1 ] R are continuous, i = 1 , 2 .

The proof is similar to Lemma 3.1.

It is clear that u dom M is a solution if and only if it satisfies M u = N u , where N = N 1 . For convenience, let ( a , b , c ) T : = [ a b c ] .

Lemma 4.2 M is a quasi-linear operator.

Proof It is easy to get Ker M = { a + b t a , b R } : = X 1 .

For u X dom M , if M u = ( y , c 1 , c 2 ) T , then c 1 , c 2 satisfy (4.1). On the other hand, if y C [ 0 , 1 ] , T 1 y = c 1 , T 2 y = c 2 , take
u ( t ) = 0 t ( t s ) φ q ( 0 s y ( r ) d r ) d s .
By simple calculation, we get u X dom M and M u = ( y , c 1 , c 2 ) T . Thus
Im M = { ( y , c 1 , c 2 ) T y C [ 0 , 1 ] , c 1 , c 2  satisfy (4.1) } .

By the continuity of T i , i = 1 , 2 , we see that Im M Y is closed. So, M is quasi-linear. The proof is completed. □

Take a projector P : X X 1 and an operator Q : Y Y 1 as follows:
( P u ) ( t ) = u ( 0 ) + u ( 0 ) t , Q ( y , y 1 , y 2 ) T = ( 0 , T 1 y 1 T 1 y , T 2 y 2 T 2 y ) T ,

where Y 1 = { ( 0 , c 1 , c 2 ) T c i R , i = 1 , 2 } . Obviously, Q Y = Y 1 , and dim Y 1 = dim X 1 .

By the continuity and boundedness of T i , i = 1 , 2 , we can easily see that Q is continuous and bounded in Y. It follows from Lemma 3.3 that Q ( I Q ) ( y , y 1 , y 2 ) T = ( 0 , 0 , 0 ) T , y , y 1 , y 2 C [ 0 , 1 ] .

Define an operator R : X × [ 0 , 1 ] X 2 as
R ( u , λ ) ( t ) = 0 t ( t s ) φ q ( 0 s λ f ( r , u ( r ) , u ( r ) , u ( r ) ) d r ) d s ,

where Ker M X 2 = X . By (H2) and the Arzela-Asscoli theorem, we can easily see that R : Ω ¯ × [ 0 , 1 ] X 2 dom M is continuous and compact, where Ω X is an open bounded set.

Lemma 4.3 Assume that Ω X is an open bounded set. Then N λ is M-quasi-compact in  Ω ¯ .

Proof It is clear that Im P = Ker M , Q N λ x = θ , λ ( 0 , 1 ) Q N x = θ and R ( , 0 ) = 0 . For u Ω ¯ ,
( I Q ) N λ u = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) 0 0 ] [ 0 T 1 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 2 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 1 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 2 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] Im M .

Since Im M Ker Q and y = Q y + ( I Q ) y , we obtain Im M ( I Q ) Y . Thus, ( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Y .

For u Σ λ = { u Ω ¯ dom M : M u = N λ u } , we get
R ( u , λ ) = 0 t ( t s ) φ q ( 0 s λ f ( r , u ( r ) , u ( r ) , u ( r ) ) d r ) d s = 0 t ( t s ) φ q ( 0 s ( φ p ( u ) ) ) d s = u ( t ) u ( 0 ) u ( 0 ) t = ( I P ) u ,
i.e. Definition 2.2(c) holds. For u Ω ¯ , we have
M [ P u + R ( u , λ ) ] = [ λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 1 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) T 2 λ f ( t , u ( t ) , u ( t ) , u ( t ) ) ] = ( I Q ) N λ u .

Thus, Definition 2.2(d) holds. Therefore, N λ is M-quasi-compact in Ω ¯ . The proof is completed. □

Theorem 4.1 Assume that the following conditions hold:

(H5) There exists a nonnegative constant L such that if | u ( t ) | > L , t [ 0 , 1 ] then either
T 1 f ( t , u ( t ) , u ( t ) , u ( t ) ) 0
or
T 2 f ( t , u ( t ) , u ( t ) , u ( t ) ) 0 .
(H6) There exist nonnegative constants K 1 , K 2 such that one of (1) and (2) holds:
  1. (1)
    B f ( t , A , B , C ) > 0 , t [ 0 , 1 ] , | B | > K 1 , A , C R ,
     
and
A f ( t , A , B , C ) > 0 , t [ 0 , 1 ] , | B | K 1 , | A | > K 2 , C R .
  1. (2)
    B f ( t , A , B , C ) < 0 , t [ 0 , 1 ] , | B | > K 1 , A , C R ,
     
and
A f ( t , A , B , C ) < 0 , t [ 0 , 1 ] , | A | > K 2 , | B | K 1 , C R .
(H7) There exist nonnegative functions a ( t ) , b ( t ) , c ( t ) , e ( t ) L 1 [ 0 , 1 ] such that
| f ( t , x , y , z ) | a ( t ) φ p ( | x | ) + b ( t ) φ p ( | y | ) + c ( t ) φ p ( | z | ) + e ( t ) , t [ 0 , 1 ] , x , y , z R ,

where φ q ( a 1 + b 1 + c 1 ) < 2 2 q , if 1 < p 2 ; φ q ( 2 p 2 a 1 + 2 p 2 b 1 + c 1 ) < 1 , if p 2 .

Then boundary value problem (1.2) has at least one solution.

In order to prove Theorem 4.1, we show two lemmas.

Lemma 4.4 Suppose (H5)-(H7) hold. Then the set
Ω 1 = { u dom M M u = N λ u , λ ( 0 , 1 ) }

is bounded in X.

Proof For u Ω 1 , we have Q N λ u = 0 , i.e. T i f ( t , u ( t ) , u ( t ) , u ( t ) ) = 0 , i = 1 , 2 . By (H5) and (H6), there exist constants t 0 , t 1 [ 0 , 1 ] such that | u ( t 0 ) | L , | u ( t 1 ) | K 1 . Since u ( t ) = u ( t 0 ) + t 0 t u ( s ) d s , u ( t ) = u ( t 1 ) + t 1 t u ( s ) d s , then
| u ( t ) | L + u , | u ( t ) | K 1 + u , t [ 0 , 1 ] .
(4.2)
It follows from M u = N λ u , (H7), and (4.2) that
| u ( t ) | = | φ q ( 0 t λ f ( s , u ( s ) , u ( s ) u ( s ) ) d s ) | φ q ( 0 1 a ( t ) φ p ( | u | ) + b ( t ) φ p ( | u | ) + c ( t ) φ p ( | u | ) + e ( t ) d t ) φ q ( a 1 φ p ( K 1 + L + u ) + b 1 φ p ( K 1 + u ) + c 1 φ p ( u ) + e 1 ) .
If 1 < p 2 , by Lemma 2.1, we get
| u ( t ) | φ q ( B 1 + A 1 φ p ( u ) ) 2 q 2 [ φ q ( B 1 ) + φ q ( A 1 ) u ] ,
thus
u 2 q 2 φ q ( B 1 ) 1 2 q 2 φ q ( A 1 ) ,

where B 1 = a 1 φ p ( K 1 + L ) + b 1 φ p ( K 1 ) + e 1 , A 1 = a 1 + b 1 + c 1 .

If p > 2 , by Lemma 2.1, we get
| u ( t ) | φ q ( B 2 + A 2 φ p ( u ) ) [ φ q ( B 2 ) + φ q ( A 2 ) u ] ,
thus
u φ q ( B 2 ) 1 φ q ( A 2 ) ,

where B 2 = 2 p 2 a 1 φ p ( K 1 + L ) + 2 p 2 b 1 φ p ( K 1 ) + e 1 , A 2 = 2 p 2 a 1 + 2 p 2 b 1 + c 1 .

These, together with (4.2), mean that Ω 1 is bounded in X. □

Lemma 4.5 Assume (H6) holds. Then
Ω 2 = { u Ker M Q N u = 0 }

is bounded in X, where N = N 1 .

Proof For u Ω 2 , we have u = a + b t and Q ( N u ) = 0 . By (H6), we see that there exists a constant t 0 [ 0 , 1 ] such that | u ( t 0 ) | = | a + b t 0 | K 2 , | u ( t ) | = | b | K 1 . So, Ω 2 is bounded. The proof is completed. □

Proof of Theorem 4.1 Let Ω = { u X u < r } , where r is large enough such that K 1 + K 2 < r < + and Ω Ω 1 ¯ Ω 2 ¯ .

By Lemmas 4.4 and 4.5, we know M u N λ u , u dom M Ω and Q N u 0 , u Ker M Ω .

Let H ( u , δ ) = ρ δ u + ( 1 δ ) J Q N u , δ [ 0 , 1 ] , u Ker M Ω ¯ , where J : Im Q Ker M is a homeomorphism with J ( 0 , a , b ) T = a + b t , ρ = { 1 , if (H 6 )(1) holds , 1 , if (H 6 )(2) holds .

Take the function Sgn ( x ) is the same as the one in Proof of Theorem 3.1.

For u Ker M Ω , we have u = a + b t 0 . Thus
H ( u , δ ) = ρ δ ( a + b t ) + ( 1 δ ) ( T 1 f ( t , a + b t , b , 0 ) T 2 f ( t , a + b t , b , 0 ) t ) .
If δ = 1 , H ( u , 1 ) = ρ ( a + b t ) 0 . If δ = 0 , by Q N u 0 , we get H ( u , 0 ) = J Q N ( a + b t ) 0 . For 0 < δ < 1 , we now prove that H ( u , δ ) 0 . Otherwise, if H ( u , δ ) = 0 , then
T 1 f ( t , a + b t , b , 0 ) = ρ δ 1 δ a , T 2 f ( t , a + b t , b , 0 ) = ρ δ 1 δ b .
(4.3)

Since u = max { a + b t , | b | } = r > K 1 + K 2 , we have either | b | > K 1 or a + b t > K 1 + K 2 . If | b | > K 1 , then T 2 b f ( t , a + b t , b , 0 ) = b T 2 f ( t , a + b t , b , 0 ) = ρ δ 1 δ b 2 . So, we have Sgn ( b f ( t , a + b t , b , 0 ) ) = Sgn ( T 2 b f ( t , a + b t , b , 0 ) ) = Sgn ( ρ δ 1 δ b 2 ) = Sgn ( ρ ) . This is a contradiction with the definition of ρ. If | b | K 1 , then a + b t > K 1 + K 2 . Thus min t [ 0 , 1 ] | a + b t | > K 2 and Sgn ( a ) = Sgn ( a + b t ) . By T 1 a f ( t , a + b t , b , 0 ) = a T 1 f ( t , a + b t , b , 0 ) = ρ δ 1 δ a 2 , we get Sgn ( T 1 ( a + b t ) f ( t , a + b t , b , 0 ) ) = Sgn ( T 1 a f ( t , a + b t , b , 0 ) ) = Sgn ( ρ δ 1 δ a 2 ) = Sgn ( ρ ) . This is a contradiction with the definition of ρ, too. So, H ( u , δ ) 0 , u Ker M Ω , δ [ 0 , 1 ] .

By the homotopy of degree, we get
deg ( J Q N , Ω Ker M , 0 ) = deg ( H ( , 0 ) , Ω Ker M , 0 ) = deg ( H ( , 1 ) , Ω Ker M , 0 ) = deg ( ρ I , Ω Ker M , 0 ) 0 .

By Theorem 2.1, we find that (1.2) has at least one solution in Ω ¯ . The proof is completed. □

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 improvement of the original manuscript.

Authors’ Affiliations

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

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© Jiang; licensee Springer. 2014

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