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Solvability for second-order nonlocal boundary value problems with dim ( ker M ) = 2

Boundary Value Problems20142014:255

https://doi.org/10.1186/s13661-014-0255-7

Received: 1 August 2014

Accepted: 28 November 2014

Published: 20 December 2014

Abstract

The existence of at least one solution to the second-order nonlocal boundary value problems on the real line is investigated by using an extension of Mawhin’s continuation theorem.

MSC: 34B10, 34B40, 34B15.

Keywords

resonancenonlocal boundary conditionsolvabilityinfinite interval

1 Introduction

Boundary value problems on an infinite interval arise quite naturally in the study of radially symmetric solutions of nonlinear elliptic equations and in various applications such as an unsteady flow of gas through a semi-infinite porous medium, theory of drain flows and plasma physics. For an extensive collection of results to boundary value problems on unbounded domains, we refer the reader to a monograph by Agarwal and O’Regan [1]. The study of nonlocal elliptic boundary value problems was investigated by Bicadze and Samarskiĭ [2], and later continued by Il’in and Moiseev [3] and Gupta [4]. Since then, the existence of solutions for nonlocal boundary value problems has received a great deal of attention in the literature. For more recent results, we refer the reader to [5]–[22] and the references therein.

In this paper, we consider the following second-order nonlinear differential equation with integral boundary conditions:
{ ( c φ p ( u ) ) ( t ) = f ( t , u ( t ) , u ( t ) ) , a.e.  t ( , ) , lim t ( c φ p ( u ) ) ( t ) = g ( s ) ( c φ p ( u ) ) ( s ) d s , lim t ( c φ p ( u ) ) ( t ) = h ( s ) ( c φ p ( u ) ) ( s ) d s ,
(1)

where φ p ( s ) : = | s | p 2 s , p > 1 , f : ( , ) 3 ( , ) is a Carathéodory function, i.e., f = f ( t , u , v ) is Lebesgue measurable in t for all ( u , v ) ( , ) 2 and continuous in ( u , v ) for almost all t ( , ) . Throughout this paper, we assume that the following assumptions hold:

(H1) g , h L 1 ( , ) satisfy g ( s ) d s = h ( s ) d s = 1 ;

(H2) c : ( , ) ( 0 , ) is a continuous function which satisfy φ p 1 ( 1 c ) L loc 1 ( , ) L 1 ( , ) ;

(H3) let w ( t ) : = 0 t φ p 1 ( 1 c ( s ) ) d s , and there exist nonnegative measurable functions α, β and γ such that ( 1 + | w | ) p 1 α , β / c , γ L 1 ( , ) and
| f ( t , u , v ) | α ( t ) | u | p 1 + β ( t ) | v | p 1 + γ ( t ) , a.e.  t ( , ) ;
(H4) there exists a function k ( t ) such that ( 1 + | w ( ) | ) e k ( ) L 1 ( , ) and
: = a 11 a 22 a 12 a 21 0 ,

where a 11 : = Q 2 ( w ( ) e k ( ) ) , a 12 : = Q 1 ( w ( ) e k ( ) ) , a 21 : = Q 2 ( e k ( ) ) , a 22 : = Q 1 ( e k ( ) ) , and Q 1 , Q 2 : L 1 ( , ) ( , ) will be defined in Section 3.

A boundary value problem is called 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 , where L is a noninvertible operator. When L is linear, Mawhin’s continuation theorem [23] is an efficient tool in finding solutions for these problems. However, it is not suitable for the case L is nonlinear. Recently, Ge and Ren [24] extended Mawhin’s continuation theorem from the case of linear L to the case of quasi-linear L. The purpose of this paper is to establish the sufficient conditions for the existence of solutions to the problem (1) on the real line at resonance with dim ( ker L ) = 2 by using an extension of Mawhin’s continuation theorem [24].

2 Preliminaries

In this section, we recall some definitions and theorems. Let X and Y be two Banach spaces with the norms X and Y , respectively.

Definition 2.1

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 ( , ) n for some n < .

     

Definition 2.2

Let M : X dom M Y be a quasi-linear operator. Let X 1 = Ker M and Ω X be an open and bounded set with the origin θ X Ω . Then N λ : Ω ¯ Y , λ [ 0 , 1 ] is said to be M-compact in Ω ¯ if N λ : Ω ¯ Y , λ [ 0 , 1 ] is a continuous operator, and there exist a vector subspace Y 1 of Y satisfying dim Y 1 = dim X 1 and an operator R : Ω ¯ × [ 0 , 1 ] X 2 being continuous and compact such that, for λ [ 0 , 1 ] ,
  1. (i)

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

     
  2. (ii)

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

     
  3. (iii)

    R ( , 0 ) is the zero operator and R ( , λ ) | Σ λ = ( I P ) | Σ λ , where Σ λ = { x Ω ¯ : M x = N λ x } ;

     
  4. (iv)

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

     

Here, X 2 is a complement space of X 1 in X, θ Y is the origin of Y and P : X X 1 , Q : Y Y 1 are projections.

Now, we give an extension of Mawhin’s continuation theorem [24].

Theorem 2.3

Let Ω X be an open and bounded set with θ X Ω . Suppose that M : X dom M Y is a quasi-linear operator and N λ : Ω ¯ Y , λ [ 0 , 1 ] is M-compact. In addition, if the following conditions hold:

(A1) M x N λ x for every ( u , λ ) ( dom M Ω ) × ( 0 , 1 ) ;

(A2) deg { J Q N 1 , Ω Ker M , 0 } 0 , where J : Im Q Ker M is a homeomorphism with J ( θ X ) = θ Y ,

then the abstract equation M x = N 1 x has at least one solution in Ω ¯ .

Finally, we give a theorem which is useful to show the compactness of operators defined on an infinite interval.

Theorem 2.4

[1]

Let Z be the space of all bounded continuous functions on ( , ) and S Z . Then S is relatively compact in Z if the following conditions hold:
  1. (i)

    S is bounded in Z;

     
  2. (ii)

    S is equicontinuous on any compact interval of ( , ) ;

     
  3. (iii)

    S is equiconvergent at ±∞, that is, given ϵ > 0 , there exists a constant T = T ( ϵ ) > 0 such that | ϕ ( t ) ϕ ( ) | < ϵ (respectively, | ϕ ( t ) ϕ ( ) | < ϵ ) for all t > T (respectively, t < T ) and all ϕ S .

     

3 Main result

Let X be the set of the functions u C 1 ( , ) such that
u 1 + | w | , φ p 1 ( c ) u L ( , ) ,
where w is the function in the assumption (H3). Then X is a Banach space equipped with a norm u X = u 1 + u 2 , where
u 1 = sup t ( , ) | u ( t ) | 1 + | w ( t ) | and u 2 = sup t ( , ) | ( φ p 1 ( c ) u ) ( t ) | .
Let Y denote the Banach space L 1 ( , ) equipped with a usual norm
h Y = | h ( s ) | d s .

Remark 3.1

  1. (1)
    It is well known that, for any u , v ( , ) and q > 0 ,
    | u + v | q max { 1 , 2 q 1 } ( | u | q + | v | q ) .
     
Thus, φ p 1 ( u + v ) α p ( φ p 1 ( u ) + φ p 1 ( v ) ) for all u , v 0 , where α p : = max { 1 , 2 2 p p 1 } .
  1. (2)

    Since φ p 1 ( 1 c ) L loc 1 ( , ) Y , then w is a continuous function which satisfies lim t w ( t ) = and lim t w ( t ) = .

     
  2. (3)

    For any continuous functions w ( t ) , we can choose a function k ( t ) which satisfies ( 1 + | w ( ) | ) e k ( ) Y . For example, put k ( t ) = 0 t ( 1 + | w ( s ) | ) d s , then ( 1 + | w ( ) | ) e k ( ) Y .

     
Define M : X dom M Y by M u = ( c φ p ( u ) ) , where
dom M = { u : ( c φ p ( u ) ) Y , lim t ( c φ p ( u ) ) ( t ) = g ( s ) ( c φ p ( u ) ) ( s ) d s , and lim t ( c φ p ( u ) ) ( t ) = h ( s ) ( c φ p ( u ) ) ( s ) d s } .
Then M : X dom M Y is continuous. Let Ω be an open bounded subset of X such that dom M Ω ¯ . For λ [ 0 , 1 ] , define N λ : Ω Y by N λ x = λ f ( , x , x ) . By (H3) and the Lebesgue dominated convergence theorem, N λ is continuous. Denote N 1 by N. Then problem (1) is equivalent to M x = N x , x dom M . Define Q 1 , Q 2 : Y ( , ) by
Q 1 ( y ) : = g ( s ) s y ( τ ) d τ d s , Q 2 ( y ) : = h ( s ) s y ( τ ) d τ d s .

Then Q 1 , Q 2 : Y ( , ) are continuous.

Lemma 3.2

Assume that (H1) and (H2) hold. Then the operator M : X dom M Y is quasi-linear. Moreover, Ker M = { a + b w : a , b ( , ) } and Im M = { y Y : Q 1 ( y ) = Q 2 ( y ) = 0 } .

Proof

Clearly, Ker M = { a + b w : a , b ( , ) } , and it is linearly homeomorphic to ( , ) 2 . Next, we show that
Im M = { y Y : Q 1 ( y ) = Q 2 ( y ) = 0 } .
Let y Im M . Then there exists x X dom M such that
( c φ p ( x ) ) ( t ) = y ( t ) , t ( , ) .
For t ( , ) ,
( c φ p ( x ) ) ( t ) = ( c φ p ( x ) ) ( ) + t y ( s ) d s
and
g ( s ) ( c φ p ( x ) ) ( s ) d s = ( c φ p ( x ) ) ( ) + g ( s ) s y ( τ ) d τ d s .

Thus Q 1 ( y ) = 0 . In a similar manner, Q 2 ( y ) = 0 .

On the other hand, let y Y satisfying Q 1 ( y ) = Q 2 ( y ) = 0 . Take
x ( t ) = 0 t φ p 1 ( 1 c ( s ) ) φ p 1 ( 0 s y ( τ ) d τ ) d s .

Then x X dom M , and ( c φ p ( x ) ) = y Im M . Thus, Im M = { y Y : Q 1 ( y ) = Q 2 ( y ) = 0 } . Since Q 1 , Q 2 : Y ( , ) are continuous, ImM is closed in Y. Consequently, M is a quasi-linear operator.

Let T 1 , T 2 : Y Y be linear operators which are defined as follows:
T 1 y = 1 ( a 11 Q 1 ( y ) + a 12 Q 2 ( y ) ) e k ( )
and
T 2 y = 1 ( a 21 Q 1 ( y ) + a 22 Q 2 ( y ) ) e k ( ) ,
where a i j ( i , j = 1 , 2 ) are the constants in the assumption in (H4). Then, by direct calculations,
T 1 ( T 1 y ) = T 1 y , T 1 ( ( T 2 y ) w ) = 0 , T 2 ( T 1 y ) = 0 and T 2 ( ( T 2 y ) w ) = T 2 y .
Define the bounded linear operators Q : Y Y 1 and P : X X 1 by
Q ( y ) = T 1 y + ( T 2 y ) w , P ( x ) = x ( 0 ) + ( φ p 1 ( c ) x ) ( 0 ) w ,

where X 1 : = Ker M and Y 1 : = Im Q = { ( a + b w ( ) ) e k ( ) : a , b ( , ) } . Then Q : Y Y 1 , P : X X 1 are projections, and dim Y 1 = dim X 1 = 2 . By (H4), 0 , and it follows from Lemma 3.2 that Im M = Ker Q . □

Lemma 3.3

Assume that (H1)-(H4) hold. Assume that Ω is an open bounded subset of X such that dom M Ω ¯ . Then N λ : Ω ¯ Y , λ [ 0 , 1 ] is M-compact on Ω ¯ .

Proof

Let X 2 : = Ker P . Then X 2 is a complement space of X 1 in X, i.e., X = X 1 X 2 . Define R : Ω ¯ × [ 0 , 1 ] X 2 , for t ( , ) , by
R ( x , λ ) ( t ) = 0 t φ p 1 ( 1 c ( s ) ) [ φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 s ( I Q ) N x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) ] d s .
Since Ω is bounded, there exists a constant r > 0 such that x X r for any x Ω ¯ . For x Ω ¯ , and for almost all t ( , ) , by (H3)
| ( N x ) ( t ) | = | f ( t , x ( t ) , x ( t ) ) | ( 1 + | w ( t ) | ) p 1 α ( t ) ( | x ( t ) | 1 + w ( t ) ) p 1 + β c ( t ) | ( φ p 1 ( c ) x ) ( t ) | p 1 + γ ( t ) ( ( 1 + | w ( t ) | ) p 1 α ( t ) + β c ( t ) ) x X p 1 + γ ( t ) ,
(2)
which implies that
N x Y r p 1 ( 1 + | w | ) p 1 α + β c Y + γ Y = : l r .
Since | Q 1 ( N x ) | N x Y and | Q 2 ( N x ) | N x Y , for x Ω ¯ ,
| Q ( N x ) ( t ) | | T 1 ( N x ) ( t ) | + | ( T 2 ( N x ) ( t ) ) w ( t ) | 1 | | ( | a 11 | | Q 1 ( N x ) | + | a 12 | | Q 2 ( N x ) | + ( | a 21 | | Q 1 ( N x ) | + | a 22 | | Q 2 ( N x ) | ) | w ( t ) | ) e k ( t ) N x Y ( M 1 + M 2 | w ( t ) | ) e k ( t ) l r ( M 1 + M 2 | w ( t ) | ) e k ( t ) .
(3)
Here,
M 1 : = 1 | | [ | a 11 | + | a 12 | ] and M 2 : = 1 | | [ | a 21 | + | a 22 | ] .
Thus
Q ( N x ) Y D N x Y ,
(4)
where
D : = ( M 1 + M 2 | w ( τ ) | ) e k ( τ ) d τ .
(5)
First, we prove that R : Ω ¯ × [ 0 , 1 ] X 2 is compact by using Theorem 2.4. Let Z = C ( , ) L ( , ) with the usual sup norm. For x Ω ¯ ,
sup t ( , ) | R ( x , λ ) ( t ) | 1 + | w ( t ) | ( φ p 1 ( | ( c φ p ( x ) ) ( 0 ) | + | N x ( τ ) | + | Q ( N x ) ( τ ) | d τ ) + | ( φ p 1 ( c ) x ) ( 0 ) | ) ( α p + 1 ) r + α p ( ( 1 + D ) N x Y ) 1 p 1 ( α p + 1 ) r + α p ( ( 1 + D ) l r ) 1 p 1
and
sup t ( , ) | ( φ p 1 ( c ) R ( x , λ ) ) ( t ) | = sup t ( , ) | φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 t ( I Q ) N x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) | ( α p + 1 ) r + α p ( ( 1 + D ) N x Y ) 1 p 1 ( α p + 1 ) r + α p ( ( 1 + D ) l r ) 1 p 1 .

Here α p is the constant in Remark 3.1(1). Thus { R ( x , λ ) 1 + | w ( x ) | : x Ω ¯ } and { φ p 1 ( c ) R ( x , λ ) : x Ω ¯ } are bounded in Z.

Let T > 0 and let ϵ > 0 be given. First, for any t 1 , t 2 [ 0 , T ] with t 1 < t 2 , we have
| R ( x , λ ) ( t 1 ) 1 + w ( t 1 ) R ( x , λ ) ( t 2 ) 1 + w ( t 2 ) | w ( t 2 ) w ( t 1 ) ( 1 + w ( t 1 ) ) ( 1 + w ( t 2 ) ) | w ( t 1 ) | ( ( α p + 1 ) r + α p ( ( 1 + D ) N x Y ) 1 p 1 ) + 1 1 + w ( t 2 ) ( w ( t 2 ) w ( t 1 ) ) ( ( α p + 1 ) r + α p ( ( 1 + D ) N x Y ) 1 p 1 ) 2 ( w ( t 2 ) w ( t 1 ) ) ( ( α p + 1 ) r + α p ( ( 1 + D ) N x Y ) 1 p 1 )
and
| ( φ p 1 ( c ) R ( x , λ ) ) ( t 1 ) ( φ p 1 ( c ) R ( x , λ ) ) ( t 2 ) | = | φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 t 1 ( I Q ) N x ( τ ) d τ ) φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 t 2 ( I Q ) N x ( τ ) d τ ) | .
By (2) and (3), there exists z Y such that
| ( I Q ) N x | z for all  x Ω ¯ ,
(6)
and since φ p 1 and w are uniformly continuous on a compact interval in ( , ) , there exists δ 1 > 0 such that if | t 1 t 2 | < δ 1 with t 1 , t 2 [ 0 , T ] , then
| R ( x , λ ) ( t 1 ) 1 + | w ( t 1 ) | R ( x , λ ) ( t 2 ) 1 + | w ( t 2 ) | | < ϵ 2
and
| ( φ p 1 ( c ) R ( x , λ ) ) ( t 1 ) ( φ p 1 ( c ) R ( x , λ ) ) ( t 2 ) | < ϵ 2 .
In a similar manner, there exists δ 2 > 0 such that if | t 1 t 2 | < δ 2 with t 1 , t 2 [ T , 0 ] , then
| R ( x , λ ) ( t 1 ) 1 + | w ( t 1 ) | R ( x , λ ) ( t 2 ) 1 + | w ( t 2 ) | | < ϵ 2
and
| ( φ p 1 ( c ) R ( x , λ ) ) ( t 1 ) ( φ p 1 ( c ) R ( x , λ ) ) ( t 2 ) | < ϵ 2 .
Letting δ = min { δ 1 , δ 2 } > 0 , if | t 1 t 2 | < δ with t 1 , t 2 [ T , T ] , then
| R ( x , λ ) ( t 1 ) 1 + | w ( t 1 ) | R ( x , λ ) ( t 2 ) 1 + | w ( t 2 ) | | < ϵ
and
| ( φ p 1 ( c ) R ( x , λ ) ) ( t 1 ) ( φ p 1 ( c ) R ( x , λ ) ) ( t 2 ) | < ϵ .

Consequently, { R ( x , λ ) 1 + | w ( x ) | : x Ω ¯ } and { φ p 1 ( c ) R ( x , λ ) : x Ω ¯ } are equicontinuous on any compact intervals in ( , ) .

For x Ω ¯ , by L’Hôspital’s rule,
lim t R ( x , λ ) ( t ) 1 + | w ( t ) | = lim t 1 1 + | w ( t ) | 0 t φ p 1 ( 1 c ( s ) ) [ φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 s ( I Q ) N x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) ] d s = lim t φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 t ( I Q ) N x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) = φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 ( I Q ) N x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 )
and
lim t ( φ p 1 ( c ) R ( x , λ ) ) ( t ) = φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 ( I Q ) N x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) .
In a similar manner,
lim t R ( x , λ ) ( t ) 1 + | w ( t ) | = φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 ( I Q ) N x ( τ ) d τ ) + ( φ p 1 ( c ) x ) ( 0 )
and
lim t ( φ p 1 ( c ) R ( x , λ ) ) ( t ) = φ p 1 ( ( c φ p ( x ) ) ( 0 ) λ 0 ( I Q ) N x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) .

By (6), we conclude that { R ( x , λ ) 1 + | w ( x ) | : x Ω } and { φ p 1 ( c ) R ( x , λ ) : x Ω } are equiconvergent at ±∞. Thus, R : Ω ¯ × [ 0 , 1 ] X 2 is compact in view of Theorem 2.4.

Next, we prove that R : Ω ¯ × [ 0 , 1 ] X 2 is continuous. Let { ( x n , λ n ) } be a sequence in Ω ¯ × [ 0 , 1 ] such that x n x in X and λ n λ in ( , ) as n . Then { x n } is bounded in X and x n ( t ) x ( t ) pointwise as n . Since R is compact, there exists a subsequence { ( x n k , λ n k ) } of { ( x n , λ n ) } such that R ( x n k , λ n k ) ( t ) L in X as n k . By the Lebesgue dominated convergence theorem, R ( x n , λ n ) ( t ) R ( x , λ ) ( t ) as n . Thus, L R ( x , λ ) . By a standard argument, R : Ω ¯ × [ 0 , 1 ] X 2 is continuous.

Finally, we show that (i)-(iv) hold in Definition 2.2. Since Q ( I Q ) N λ ( Ω ¯ ) = 0 , ( I Q ) N λ ( Ω ¯ ) Ker Q = Im M . For y Im M , Q y = 0 , and y = ( I Q ) y ( I Q ) Y . Consequently, ( I Q ) N λ ( Ω ¯ ) Im M ( I Q ) Y . Since N λ x = λ N x for any x Ω ¯ ,
Q N λ x = 0 , λ ( 0 , 1 ) Q N x = 0 ,
and R ( , 0 ) = θ X . For x Σ λ = { x Ω ¯ : M x = N λ x } , N λ x = ( c φ p ( x ) ) Im M = Ker Q . Then, for x X and t ( , ) ,
R ( x , λ ) ( t ) = 0 t φ p 1 ( 1 c ( s ) ) [ φ p 1 ( ( c φ p ( x ) ) ( 0 ) + 0 s ( I Q ) N λ x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) ] d s = 0 t φ p 1 ( 1 c ( s ) ) [ φ p 1 ( ( c φ p ( x ) ) ( 0 ) + 0 s N λ x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) ] d s = 0 t φ p 1 ( 1 c ( s ) ) [ ( φ p 1 ( c ) x ) ( s ) ( φ p 1 ( c ) x ) ( 0 ) ] d s = x ( t ) ( x ( 0 ) + ( φ p 1 ( c ) x ) ( 0 ) w ( t ) ) = ( I P ) x ( t ) .
On the other hand, for x X and t ( , ) ,
M [ P x + R ( x , λ ) ] ( t ) = M [ x ( 0 ) + ( φ p 1 ( c ) x ) ( 0 ) w ( t ) + 0 t φ p 1 ( 1 c ( s ) ) [ φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 s ( I Q ) N x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) ] d s ] = ( I Q ) N λ x ( t ) .

Thus, N λ is M-compact on Ω ¯ . □

Now, we give the main result in this paper.

Theorem 3.4

Assume that (H1)-(H4) hold. Assume also that the following hold:

(H5)there exist positive constants A and B such that if | x ( t ) | > A for every t [ B , B ] or | ( c φ p ( x ) ) ( t ) | > A for every t ( , ) , then Q ( N x ) 0 , i.e., either Q 1 ( N x ) 0 or Q 2 ( N x ) 0 ;

(H6)there exists a positive constant C such that if | a | > C or | b | > C , then either
  1. (1)

    a Q 1 ( N ( a + b w ( ) ) ) + b Q 2 ( N ( a + b w ( ) ) ) < 0 or

     
  2. (2)

    a Q 1 ( N ( a + b w ( ) ) ) + b Q 2 ( N ( a + b w ( ) ) ) > 0 .

     
Then problem (1) has at least one solution in X provided that
( 1 + | w | ) p 1 α Y + β c Y < ( 1 2 α p 3 + ( 4 + | w ( B ) | + 2 ( 1 + D ) 1 p 1 ) α p 2 ) p 1 .
(7)

Here, D is the constant defined in (5).

Proof

We divide the proof into three steps.

Step 1. Let
Ω 1 = { x dom M : M x = N λ x ,  for some  λ ( 0 , 1 ) } .

We will prove that Ω 1 is bounded. For x Ω 1 , M x = N λ x Im M = Ker Q .

Thus
Q 1 ( N x ) = Q 2 ( N x ) = 0 .
By (H5), there exist t 0 [ B , B ] and t 1 ( , ) such that
| x ( t 0 ) | A and | ( c φ p ( x ) ) ( t 1 ) | A ,
which imply that
| ( c φ p ( x ) ) ( 0 ) | = | ( c φ p ( x ) ) ( t 1 ) ( c φ p ( x ) ) ( t 1 ) + ( c φ p ( x ) ) ( 0 ) | | ( c φ p ( x ) ) ( t 1 ) | + | 0 t 1 ( c φ p ( x ) ) ( s ) d s | A + M x Y A + N x Y ,
and | ( φ p 1 ( c ) x ) ( 0 ) | ( A + N x Y ) 1 p 1 α p A 1 p 1 + α p N x Y 1 p 1 . Then we have
| x ( 0 ) | = | x ( t 0 ) 0 t 0 φ p 1 ( 1 c ( s ) ) φ p 1 [ ( c φ p ( x ) ) ( t 1 ) + t 1 s ( c φ p ( x ) ) ( τ ) d τ ] d s | | x ( t 0 ) | + | w ( B ) | [ φ p 1 ( | ( c φ p ( x ) ) ( t 1 ) | + 0 | ( c φ p ( x ) ) ( τ ) | d τ ) ] A + | w ( B ) | ( α p A 1 p 1 + α p N x Y 1 p 1 ) .
Thus,
P x X = P x 1 + P x 2 | x ( 0 ) | + 2 | ( φ p 1 ( c ) x ) ( 0 ) | A + α p ( 2 + | w ( B ) | ) A 1 p 1 + α p ( 2 + | w ( B ) | ) N x Y 1 p 1 .
On the other hand, by (4),
| R ( x , λ ) ( t ) | 1 + | w ( t ) | = 1 1 + | w ( t ) | | 0 t φ p 1 ( 1 c ( s ) ) [ φ p 1 ( ( c φ p ( x ) ) ( 0 ) + λ 0 s ( I Q ) N x ( τ ) d τ ) ( φ p 1 ( c ) x ) ( 0 ) ] d s | ( α p + 1 ) φ p 1 ( | ( c φ p ( x ) ) ( 0 ) | ) + α p φ p 1 ( N x Y + Q N x Y ) ( α p + 1 ) ( A + N x Y ) 1 p 1 + α p ( 1 + D ) 1 p 1 N x Y 1 p 1 = α p ( α p + 1 ) A 1 p 1 + α p ( α p + 1 + ( 1 + D ) 1 p 1 ) N x Y 1 p 1
and
| ( φ p 1 ( c ) ( R ( x , λ ) ) ) ( t ) | α p ( α p + 1 ) A 1 p 1 + α p ( α p + 1 + ( 1 + D ) 1 p 1 ) N x Y 1 p 1 .
Thus,
R ( x , λ ) X 2 α p ( α p + 1 ) A 1 p 1 + 2 α p ( α p + 1 + ( 1 + D ) 1 p 1 ) N x Y 1 p 1 .
It follows that
x X = P x + ( I P ) x X P x X + ( I P ) x X = P x X + R ( x , λ ) X A + α p ( 2 + | w ( B ) | ) A 1 p 1 + α p ( 2 + | w ( B ) | ) N x Y 1 p 1 + 2 α p ( α p + 1 ) A 1 p 1 + 2 α p ( α p + 1 + ( 1 + D ) 1 p 1 ) N x Y 1 p 1 A + α p ( 2 α p + 4 + | w ( B ) | ) A 1 p 1 + α p 2 ( 2 α p + 4 + | w ( B ) | + 2 ( 1 + D ) 1 p 1 ) γ Y 1 p 1 + α p 2 ( 2 α p + 4 + | w ( B ) | + 2 ( 1 + D ) 1 p 1 ) ( ( 1 + | w | ) p 1 α Y + β c Y ) 1 p 1 x X .

By (7), Ω 1 is bounded.

Step 2. Define a homeomorphism J : Im Q Ker M by
J ( ( a + b w ( ) ) e k ( ) ) = a 22 a a 12 b + ( a 21 a + a 11 b ) w ( ) .
Assume (H6)(1) holds, i.e., there exists a positive constant C such that if | a | > C or | b | > C , then a Q 1 ( N ( a + b w ( ) ) ) + b Q 2 ( N ( a + b w ( ) ) ) < 0 . Let
Ω 2 = { x ker M : λ x + ( 1 λ ) J Q N x = 0 ,  for some  λ [ 0 , 1 ] } .

Let x Ω 2 . Then x = a + b w for some a , b ( , ) . If λ = 0 , J Q N ( a + b w ( ) ) = 0 . Since J is homeomorphism, Q N ( a + b w ( ) ) = 0 . By (H6), we obtain | a | C and | b | C . If λ = 1 , then a = b = 0 .

For λ ( 0 , 1 ) , by λ x = ( 1 λ ) J Q N x , we obtain
λ a = ( 1 λ ) Q 1 ( N ( a + b w ( ) ) ) , λ b = ( 1 λ ) Q 2 ( N ( a + b w ( ) ) ) .
If | a | > C or | b | > C , then, by (H6)(1), we obtain
λ ( a 2 + b 2 ) = ( 1 λ ) ( a Q 1 N ( a + b w ( ) ) + b Q 2 N ( a + b w ( ) ) ) < 0 ,

which is a contradiction. Thus, Ω 2 is bounded.

In the case that (H6)(2) holds, we take
Ω 2 = { x ker M : λ x + ( 1 λ ) J Q N x = 0  for some  λ [ 0 , 1 ] } ,

and it follows that Ω 2 is bounded in a similar manner.

Step 3. Take an open bounded set Ω Ω ¯ 1 Ω ¯ 2 { 0 } in X. By Step 1,

(A1) M u N λ u for every ( u , λ ) ( dom M Ω ) × ( 0 , 1 ) .

Now we will show that

(A2) deg ( J Q N , Ω ker M , 0 ) 0 .

Let H ( x , λ ) = ± λ x + ( 1 λ ) J Q N x . By Step 2, we know that H ( x , λ ) 0 , for every ( x , λ ) ( ker M Ω ) × [ 0 , 1 ] . Thus, by the homotopy property of the degree, we obtain
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 0 .

By Theorem 2.3, M x = N x has at least one solution in dom M Ω , and consequently problem (1) has at least one solution in X. □

4 Example

Consider the following second-order nonlinear differential equation:
{ ( | u | u ) ( t ) = f ( t , u ( t ) , u ( t ) ) , a.e.  t ( , ) , lim t ( | u | u ) ( t ) = g ( s ) ( | u | u ) ( s ) d s , lim t ( | u | u ) ( t ) = h ( s ) ( | u | u ) ( s ) d s ,
(8)
where
g ( t ) = { 2 t , t [ 1 , 0 ] , 0 , otherwise , h ( t ) = { 2 t , t [ 0 , 1 ] , 0 , otherwise .
Define f : ( , ) 3 ( , ) by f ( t , u , v ) = α ( t ) u + β ( t ) v + γ ¯ ( t ) , where
α ( t ) = { 10 3 e t , t [ 1 , 0 ] , 0 , otherwise , β ( t ) = { 10 4 e t , t 1 , 0 , otherwise , γ ¯ ( t ) = { e t + 15 e 1 6 , t [ 1 , 0 ] , 0 , otherwise .
Then
| f ( t , u , v ) | α ( t ) | u | + β ( t ) | v | + | γ ¯ ( t ) | α ( t ) ( | u | 2 + 1 ) + β ( t ) ( | v | 2 + 1 ) + | γ ¯ ( t ) | = α ( t ) | u | 2 + β ( t ) | v | 2 + γ ( t ) ,

where γ ( t ) = α ( t ) + β ( t ) + | γ ¯ ( t ) | . Since c ( t ) = 1 for t ( , ) and p = 3 , w ( t ) = t , and thus (H1), (H2), and (H3) hold.

For y Y ,
Q 1 ( y ) = 1 y ( τ ) d τ + 1 0 y ( τ ) τ 2 d τ
and
Q 2 ( y ) = 1 y ( τ ) d τ + 0 1 y ( τ ) τ 2 d τ .
Take k ( t ) = | t | , then
a 11 = 14 e 1 + 6 , a 12 = 14 e 1 + 6 , a 21 = 4 e 1 2 , a 22 = 4 e 1 + 2 ,
and
= a 11 a 22 a 12 a 21 = ( 14 e 1 + 6 ) ( 8 e 1 + 4 ) > 0 .

Thus, (H4) holds.

Take B = 1 and A = 1 . If | x ( t ) | > 1 , for t [ 1 , 1 ] , then | Q 1 ( N x ) | = | 10 3 1 0 t 2 e | t | x ( t ) d t | > | 10 3 1 0 t 2 e | t | d t | = 10 3 ( 5 e 1 + 2 ) > 0 . If | ( | x | x ) ( t ) | > 1 for t ( , ) , then | x ( t ) | > 1 for t ( , ) , and
| Q 2 ( N x ) | = | 10 4 1 e | t | x ( t ) d t | > | 10 4 1 e | t | d t | = 10 4 e 1 > 0 .

Thus, (H5) holds.

For any C > 0 , if | a | > C or | b | > C , then
a Q 1 ( N ( a + b t ) ) + b Q 2 ( N ( a + b t ) ) = 10 3 ( 5 e 1 + 2 ) a 2 + 10 3 ( 16 e 1 6 ) a b + 10 4 e 1 b 2 = 10 3 ( 5 e 1 + 2 ) ( a + 8 e 1 3 5 e 1 + 2 b ) 2 + 10 4 ( e 1 10 ( 8 e 1 3 ) 2 5 e 1 + 2 ) b 2 > 0 .

Thus, (H6)(2) is satisfied.

Since p = 3 , α p = 1 , B = 1 , and D = 1 2 e 1 + 1 + 1 7 e 1 + 3 , we have
( 1 + | w | ) 2 α Y + β c Y < 4 × 10 4 < ( 1 2 + ( 5 + 2 ( 1 + 1 2 e 1 + 1 + 1 7 e 1 + 3 ) 1 2 ) ) 2

and (7) holds. Consequently, there exists at least one solution to problem (8) in view of Theorem 3.4.

Declarations

Acknowledgements

This research was supported by Basic Science Research Program through the National Research Foundation of Korea (NRF) funded by the Ministry of Education, Science and Technology (2012R1A1A1011225).

Authors’ Affiliations

(1)
Department of Mathematics, Pusan National University, Busan, South Korea
(2)
Department of Mathematics Education, Pusan National University, Busan, South Korea

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© Jeong et al.; licensee Springer. 2014

This article is published under license to BioMed Central Ltd.Open Access This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/4.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly credited.

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