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Structure of positive solution sets of differential boundary value problems

Xu Xian and Zhu Xunxia*

Author Affiliations

Department of Mathematics, Jiangsu Normal University, Xuzhou, Jiangsu, 221116, P.R. China

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Boundary Value Problems 2013, 2013:100  doi:10.1186/1687-2770-2013-100

The electronic version of this article is the complete one and can be found online at: http://www.boundaryvalueproblems.com/content/2013/1/100


Received:26 September 2012
Accepted:8 March 2013
Published:22 April 2013

© 2013 Xian and Xunxia; licensee Springer

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

In this paper, we first obtain some results on the structure of positive solution sets of differential boundary value problems. Then by using the results, we obtain an existence result for differential boundary value problems. The method used to show the main result is the global bifurcation theory.

Keywords:
structure of positive solution sets; differential boundary value problems; bifurcation theory

1 Introduction

This paper considers the differential boundary value problem

{ ( p ( t ) ϕ ( u ) ) + λ p ( t ) f ( t , u ) = 0 , a < t < b , u ( a ) = u ( b ) = 0 , (1.1)

where f is ϕ-superlinear at ∞ and f ( t , 0 ) maybe negative and p is a positive continuous function, λ > 0 is a parameter.

Equations of form (1.1) occur in the study for the p-Laplacian equation, non-Newtonial fluid theory and the turbulent flow of a gas in a porous medium. The case where

α ( | u | 2 ) | u | = | u | p 2 | u | , p > 1 ,

i.e., perturbations of the p-Laplacian, has received much attention in the recent literature. Also, problem (1.1) with f ( t , 0 ) > 0 has been studied by several authors in recent years (see [1] and the references therein). Here, we are interested in the case when f ( t , 0 ) may be negative (the so-called semipositone case) (see [2] and its references for a review). As pointed out by Lions in [3], semi-positone problems are mathematically very challenging. During the last ten years, finding positive solutions to semi-positone problems has been actively pursued and significant progress on semi-positone problems has taken place; see [4-8] and the references therein. For instance, Hai et al.[9] considered the existence positive solution of (1.1). Under some super-linear conditions on the non-linear term f, they proved that there exists λ > 0 such that (1.1) has one positive solution for 0 < λ < λ . The main method in [9] used to show the main result are the fixed-point theorems.

The main purpose of this paper is going to study the structure of the positive set of (1.1). Rabinowitz [10] gave the first important results on the structure of the solution sets of non-linear equations and obtained by the degree theoretic method. Amamn [11] studied the structure of the positive solution set of non-linear equations; the reader is referred to [12,13] for other results concerning the structure of solution sets of non-linear equations. In our paper, we will study the existence results for an unbounded connected component of a positive solution set for the differential boundary value problem of (1.1). This paper generalizes some results from the literature [9]. The paper is arranged as follows. In Section 2, we will give some preliminary lemmas. The main results will be given in Section 3.

2 Some lemmas

For convenience, we make the following assumptions:

(A1) ϕ is an odd, increasing homeomorphism on R with ϕ 1 concave on R + .

(A2) For each c > 0 , there exists A c > 0 such that ϕ 1 ( c u ) A c ϕ 1 ( u ) , u R + and lim c A c = + (note that (A2) implies the existence of B c > 0 such that ϕ 1 ( c u ) B c ϕ 1 ( u ) , u R + and lim c B c = 0 ).

(A3) p : [ a , b ] ( 0 , + ) is continuous.

(A4) f : [ a , b ] × R 1 R 1 is continuous and

lim u + f ( t , u ) ϕ ( u ) = +

uniformly for t [ a , b ] .

Let E = C [ 0 , 1 ] , the usual real Banach space of continuous functions with the maximum norm . Let e ( t ) = ( t a ) ( b t ) ( b a ) 2 for t [ a , b ] and P = { x E : u ( t ) 0 , t [ a , b ] } . Then P is a cone of E. Define

f ( t , u ) = { f ( t , u ) , u 0 , f ( t , 0 ) , u < 0 .

Then f ( t , u ) m for u R 1 (here m > 0 is a constant). For u E , λ R + , define

A ( λ , u ) = a t ϕ 1 ( C p ( s ) λ p ( s ) a s p ( r ) f ( r , u ) d r ) d s ,

here C is a constant such that

a b ϕ 1 ( C p ( s ) λ p ( s ) a s p ( r ) f ( r , u ) d r ) d s = 0 .

We know that C exists and is unique for every u P (see [14]). Then u = A ( λ , u ) if and only if u is a solution of

{ ( p ( t ) ϕ ( u ) ) + λ p ( t ) f ( t , u ) = 0 , a < t < b , u ( a ) = u ( b ) = 0 .

From [9], we have the following Lemmas 2.1 and 2.2.

Lemma 2.1Letϕsatisfy (A1) and (A2). Then for each d > 0 , there exist constants K d and L d such that

K d ϕ ( x ) ϕ ( d x ) L d ϕ ( x ) , x 0 .

Further, L d 0 as d 0 and K d as d .

Lemma 2.2Letϕbe as in Lemma 2.1. Then there exist 0 < α 1 and β 1 such that ϕ 1 ( x y ) α ϕ 1 ( x ) β ϕ 1 ( y ) for x 0 , y 0 .

Lemma 2.3Let p 0 , p 1 > 0 such that p 0 p ( t ) p 1 for t [ a , b ] . Let λ > 0 andωbe a solution of

{ ( p ( t ) ϕ ( ω ) ) + λ p ( t ) h ( t ) = 0 , t [ a , b ] , ω ( a ) = ω ( b ) = 0 , (2.1)

here h ( t ) is continuous function with h ( t ) m (here m > 0 is a constant) for t [ a , b ] , if ω β ( b a ) α ϕ 1 ( λ m δ ) , then

ω ( t ) ( α ω β ϕ 1 ( λ m δ ) ( b a ) ) e ( t ) for   t [ a , b ] ,

where δ = p 1 ( b a ) p 0 .

Proof By integrating, it follows that (3.1) has the unique solution given by

ω ( t ) = a t ϕ 1 { 1 p ( s ) ( C λ a s p ( r ) h ( r ) d r ) } d s ,

where C is such that ω ( b ) = 0 . Let ω = | ω ( t 0 ) | for some t 0 ( a , b ) . Then

ω ( t ) = a t ϕ 1 { λ p ( s ) ( s t 0 p ( r ) h ¯ ( r ) d r λ m p ( s ) s t 0 p ( r ) d r ) } d s ,

where h ¯ ( t ) = h ( t ) + m 0 . By Lemma 2.2, we get

ω ( t ) α a t ϕ 1 [ λ p ( s ) s t 0 p ( r ) h ¯ ( r ) d r ] d s β a t ϕ 1 ( λ m p ( s ) s t 0 p ( r ) d r ) d s .

Now

a t ϕ 1 ( λ m p ( s ) s t 0 p ( r ) d r ) d s a t ϕ 1 ( λ m p 0 s t 0 p 1 d r ) d s a t ϕ 1 ( λ m p 1 p 0 ( b a ) ) d s = ϕ 1 ( λ m δ ) ( t a ) .

And so

ω ( t ) α ω ¯ ( t ) β ϕ 1 ( λ m δ ) ( t a ) β ϕ 1 ( λ m δ ) ( t a ) for  t [ a , t 0 ] , (2.2)

here

ω ¯ ( t ) = a t ϕ 1 [ λ p ( s ) s t 0 p ( r ) h ¯ ( r ) d r ] d s for  t [ a , t 0 ] .

Note that ω ¯ satisfies

{ ( p ( t ) ϕ ( ω ¯ ) ) + λ p ( t ) h ¯ ( t ) = 0 , t [ a , t 0 ] , ω ¯ ( a ) = 0 , ω ¯ ( t 0 ) | ω ( t 0 ) | .

In fact, ω ¯ ( t ) ω ( t ) for t [ a , t 0 ] . We next prove that ω ¯ ( t ) ν ( t ) for t [ a , t 0 ] , here ν satisfies

{ ( p ( t ) ϕ ( ν ) ) = 0 , t [ a , t 0 ] , ν ( a ) = 0 , ν ( t 0 ) = ω .

Suppose it is not true, then ω ¯ ν has a negative absolute minimum at τ ( a , t 0 ) . Since

ω ¯ ( a ) ν ( a ) = 0 , ω ¯ ( t 0 ) ν ( t 0 ) 0 ,

there exist τ 0 , τ 1 [ a , t 0 ] such that

ω ¯ ( τ 0 ) ν ( τ 0 ) = ω ¯ ( τ 1 ) ν ( τ 1 ) = 0 ,

and

ω ¯ ( t ) ν ( t ) < 0 , t ( τ 0 , τ 1 ) .

Then

( p ( t ) ϕ ( ω ( t ) ) ) ( p ( t ) ϕ ( ν ( t ) ) ) = λ p ( t ) h ¯ ( t ) 0 for  t ( τ 0 , τ 1 ) .

Let u ( t ) = ω ¯ ( t ) ν ( t ) , t ( τ 0 , τ 1 ) , then

τ 0 τ 1 ( ( p ( t ) ϕ ( ω ¯ ( t ) ) ) ( p ( t ) ϕ ( ν ( t ) ) ) ) u ( t ) d t 0 .

On the other hand, using the inequality

( ϕ ( b ) ϕ ( a ) ) ( b a ) 0 , a , b R 1 ,

and the fact that there exists τ [ τ 0 , τ 1 ] such that ω ¯ ( τ ) ν ( τ ) , we have

τ 0 τ 1 ( ( p ( t ) ϕ ( ω ¯ ( t ) ) ) ( p ( t ) ϕ ( ν ( t ) ) ) ) u ( t ) d t = τ 0 τ 1 ( ( p ( t ) ϕ ( ω ¯ ( t ) ) ) ( p ( t ) ϕ ( ν ( t ) ) ) ) ( ω ¯ ν ) d t < 0 ,

which is a contradiction. So, ω ¯ ( t ) ν ( t ) for t [ a , t 0 ] . Obviously, ν ( t ) = ω t 0 a ( t a ) , t [ a , t 0 ] , since ω ¯ θ for each t [ a , t 0 ] . From (2.2), we have

ω ( t ) ( α ω β ϕ 1 ( λ m δ ) ( b a ) ) t a t 0 a , t [ a , t 0 ] .

Similarly,

ω ( t ) ( α ω β ϕ 1 ( λ m δ ) ( b a ) ) b t b t 0 , t [ t 0 , b ] .

If ω β ( b a ) α ϕ 1 ( λ m δ ) , then

ω ( t ) ( α ω β ϕ 1 ( λ m δ ) ( b a ) ) e ( t ) for  t [ a , b ] .

The proof is complete. □

Let Q λ = { u E u ( α u β ϕ 1 ( λ m δ ) ( b a ) ) e ( t ) , t [ a , b ] } for each λ R + , where α , β > 0 and δ > 0 are defined as that in Lemma 2.2 and Lemma 2.3, respectively. Then Q λ is also a cone of E. From Lemma 2.3, we know that A : P Q λ is completely continuous.

Let

L ( P ) = { ( λ , u ) | λ R + , u P { θ }  is a solution of  (1.1) } ¯ .

From [[15], Lemma 29.1], we have Lemma 2.4.

Lemma 2.4LetXbe a compact metric space. Assume thatAandBare two disjoint closed subsets ofX. Then either there exist a connected component ofXmeeting bothAandBor X = Ω A Ω B where Ω A , Ω B are disjoint compact subsets ofXcontainingAandB, respectively.

LetUbe an open and bounded subset of the metric space [ a , b ] × E . We set U ( λ ) = { x E : ( λ , x ) U } , whose boundary is denoted by U ( λ ) . Consider a map h ( λ , x ) = x k ( λ , x ) , such that k ( λ , ) is compact and θ h ( U ) . Such a maphwill also be called an admissible homotopy onU. Ifhis an admissible homotopy, for every λ [ a , b ] and every x U ( λ ) , one has that h λ ( x ) : = h ( λ , x ) θ and it makes sense to evaluate deg ( h λ , U ( λ ) , θ ) .

Lemma 2.5Ifhis an admissible homotopy on U [ a , b ] × E , the deg ( h λ , U ( λ ) , θ ) is constant for all λ [ a , b ] .

Lemma 2.6Let h E { θ } such that h α h e ( t ) . Then for arbitrary λ ( 0 , 1 ] , there exists R λ > 0 such that for each λ [ λ , 1 ] , R R λ and μ 0 ,

u A ( λ , u ) + μ h , u B ( θ , R ) ,

where B ( θ , R ) = { u : u < R } .

Proof From (A4), for m 1 > 0 , such that

α ( b a ) 16 ϕ 1 ( m 1 λ 0 p 0 ( b a ) 16 p 1 ) α ( b a ) 8 > 1 ,

there exists m 0 > 0 such that f ( t , u ) m 1 ϕ ( u ) for u m 0 . Let

R λ = max { 1 α [ m 0 min t [ b + 3 a 4 , 3 b + a 4 ] e ( t ) + β ϕ 1 ( λ 0 m δ ( b a ) ) ] , β ( b a ) α ϕ 1 ( λ 0 m δ ) } + 1 .

Assume by contradiction that

u 0 = A ( λ 0 , u 0 ) + μ 0 h (2.3)

for some R R λ , u 0 B ( θ , R ) , λ 0 [ λ , 1 ] and μ 0 0 . Let

y 0 ( t ) = A ( λ 0 , u 0 ) = a t ϕ 1 ( C p ( s ) λ 0 p ( s ) a s p ( r ) f ( r , u 0 ) d r ) d s .

From Lemma 2.3, we know that y 0 ( t ) Q λ 0 . Namely,

y 0 ( t ) ( α y 0 β ϕ 1 ( λ 0 m δ ) ( b a ) ) e ( t ) . (2.4)

From (2.3) and (2.4), we have

u 0 = y 0 ( t ) + μ 0 h ( α y 0 β ϕ 1 ( λ 0 m δ ( b a ) ) ) e ( t ) + μ 0 α h e ( t ) ( α y 0 + μ 0 h β ϕ 1 ( λ 0 m δ ( b a ) ) e ( t ) = ( α u 0 β ϕ 1 ( λ 0 m δ ) ( b a ) ) e ( t ) ,

so u 0 Q λ 0 . For λ 0 [ λ , 1 ] , we have

u 0 ( t ) ( α u 0 β ϕ 1 ( λ 0 m δ ( b a ) ) ) e ( t ) m 0 ,

for t ( 3 a + b 4 , a + b 2 ) . Therefore, let u 0 = u ( t u 0 ) , t u 0 ( 3 a + b 4 , a + b 2 ) , assume that t u 0 5 a + 3 b 8 , then

u 0 = u ( t u 0 ) = 3 a + b 4 t u 0 ϕ 1 ( λ 0 p ( s ) s t u 0 p ( τ ) f ( t , u 0 ) d τ ) d s + μ 0 h α 3 a + b 4 t u 0 ϕ 1 ( λ 0 p 0 p 1 s t u 0 p ( τ ) ( f ( t 0 , u 0 ) + m ) d τ ) d s β 3 a + b 4 t u 0 ϕ 1 ( λ 0 p 1 m p 0 ( t u 0 s ) ) d s ,

where

3 a + b 4 t u 0 ϕ 1 ( λ 0 p 0 p 1 s t u 0 ( f ( τ , u 0 ) + m ) d τ ) d s 3 a + b 4 11 a + 5 b 16 ϕ 1 ( λ 0 p 0 p 1 s t u 0 ( f ( τ , u 0 ) + m ) d τ ) d s 3 a + b 4 11 a + 5 b 16 ϕ 1 ( λ 0 p 0 p 1 11 a + 5 b 16 5 a + 3 b 8 m 1 ϕ ( u 0 ) d τ ) d s = 3 a + b 4 11 a + 5 b 16 ϕ 1 ( m 1 λ 0 p 0 ( b a ) 4 p 1 ) u 0 d s = b a 16 ϕ 1 ( m 1 λ 0 p 0 ( b a ) 16 p 1 ) u 0 ,

and

3 a + b 4 t u 0 ϕ 1 ( λ 0 p 1 m p 0 ( t u 0 s ) ) d s 3 a + b 4 t u 0 ϕ 1 ( λ 0 p 1 m p 0 ( b a ) ) d s = ϕ 1 ( λ 0 p 1 m p 0 ( b a ) ) ( t u 0 3 a + b 4 ) ( b a ) 8 ϕ 1 ( λ 0 m δ ) .

Thus,

u 0 α ( b a ) 16 ϕ 1 ( m 1 λ 0 p 0 ( b a ) 16 p 1 ) u 0 β ( b a ) 8 ϕ 1 ( λ 0 m δ ) α ( b a ) 16 ϕ 1 ( m 1 λ 0 p 0 ( b a ) 16 p 1 ) u 0 α ( b a ) 8 u 0 = ( α ( b a ) 16 ϕ 1 ( m 1 λ 0 p 0 ( b a ) 16 p 1 ) α ( b a ) 8 ) u 0 ,

so α ( b a ) 16 ϕ 1 ( m 1 λ 0 p 0 ( b a ) 16 p 1 ) α ( b a ) 8 1 , which is a contradiction. Then (2.3) holds. The proof is complete. □

3 Main results

For convenience, let us introduce the following symbols. For any r,

M ( r ) = { ( λ ˜ , u ) [ 0 , 1 ] × E : u = r } , M [ r , + ) = { ( λ ˜ , u ) [ 0 , 1 ] × E : r u < + } , M [ 0 , r ) = { ( λ ˜ , u ) [ 0 , 1 ] × E : 0 u < r } .

Now we give our main results of this paper.

Theorem 3.1Suppose (A1) to (A4) hold. Then L ( P ) ( [ 0 , 1 ] × P ) possesses an unbounded connected component D such that Proj λ D ( 0 , λ ] for some λ > 0 and

lim λ 0 + , ( λ , u ) D u = + ,

where Proj λ D denotes the projection of D onto theλ-axis.

Proof We divide our proof into four steps.

Step 1. Let

T ( λ , u ) = { A ( λ , u ) , ( [ 0 , 1 ] × E ) M [ 2 , + ) , θ , ( [ 0 , 1 ] × B ( θ , 1 ) ¯ ) ( { 0 } × E ) , (3.1)

where B ( θ , 1 ) ¯ = { u E : u 1 } . Obviously, T ( , ) is a completely continuous operator.

Note that T ( λ , u ) = A ( λ , u ) for all λ [ 0 , 1 ] and u E with u 2 . From Lemma 2.5, there exists R 1 > 2 large enough such that Fix T ( 1 , ) B ( θ , R 1 ) and

deg ( I T ( 1 , ) , B ( θ , R 1 ) , θ ) = 0 . (3.2)

Obviously,

deg ( I T ( 1 , ) , B ( θ , 1 ) , θ ) = 1 . (3.3)

Therefore,

deg ( I T ( 1 , ) , B ( θ , R 1 ) B ( θ , 1 ) ¯ , θ ) = deg ( I T ( 1 , ) , B ( θ , R 1 ) , θ ) deg ( I T ( 1 , ) , B ( θ , 1 ) , θ ) = 0 1 = 1 . (3.4)

So, Fix T ( 1 , ) and Fix T ( 1 , ) U : = B ( θ , R 1 ) B ( θ , 1 ) ¯ .

Step 2. Let

S = { ( λ , u ) R + × E : u = T ( λ , u ) } .

From Lemma 2.6, we have

L ( P ) [ 0 , 1 ] × { u E : 0 < u 1 } = .

This implies that T has no bifurcation point on [ 0 , 1 ] . From step 1,we have L ( P ) ( { 1 } × E ) , then for each ( 1 , u ) L ( P ) ( { 1 } × E ) , denote by D u the connected component of the metric space L ( P ) ( { 1 } × E ) emitting from ( 1 , u ) . Now we will show that, there must exist ( 1 , u 0 ) L ( P ) ( { 1 } × E ) such that D u 0 is unbounded. Assume on the contrary that D u is bounded for each ( 1 , u ) L ( P ) ( { 1 } × E ) . Take a bounded open neighborhood U u 1 in L ( P ) ( { 1 } × E ) for each D u such that

C l ( { 1 } × E ) U u 1 ( [ 0 , 1 ] × { θ } ) = (3.5)

and U u 1 ( { 0 } × E ) = , where C l ( { 1 } × E ) U u 1 denotes the closure of U u 1 in the metric space [ 0 , 1 ] × E . Let [ 0 , 1 ] × E U u 1 denote the boundary of U u 1 in the metric space [ 0 , 1 ] × E . Obviously, [ 0 , 1 ] × E U u 1 L ( P ) is a compact subset. Assume that [ 0 , 1 ] × E U u 1 L ( P ) . From the maximal connectedness of D u , there is no connected subset of [ 0 , 1 ] × E U u 1 L ( P ) meeting both [ 0 , 1 ] × E U u 1 L ( P ) and D u . From Lemma 2.4, there exist compact disjoint subsets Ω u ( 1 ) and Ω u ( 2 ) of C l [ 0 , 1 ] × E Ω u ( 1 ) L ( P ) such that D u Ω u ( 1 ) and [ 0 , 1 ] × E U u ( 1 ) L ( P ) Ω u ( 2 ) , and C l [ 0 , 1 ] × E U u ( 1 ) L ( P ) = Ω u ( 1 ) Ω u ( 2 ) . Let d = d ( Ω u ( 1 ) , Ω u ( 2 ) ) > 0 and U u ( 2 ) be the d 3 -neighborhood of Ω u ( 1 ) in the metric space [ 0 , 1 ] × E . Set

U u = { U u ( 1 ) U u ( 2 ) , when  [ 0 , 1 ] × E U u ( 1 ) L ( P ) , U u ( 1 ) , when  [ 0 , 1 ] × E U u ( 1 ) L ( P ) = . (3.6)

Then we have D u U u and

[ 0 , 1 ] × E U u L ( P ) = . (3.7)

Obviously, the collection of the subsets

{ U u { 1 } × E : ( 1 , u ) L ( P ) ( { 1 } × E ) }

is an open cover of L ( P ) ( { 1 } × E ) . Since L ( P ) ( { 1 } × E ) is compact, then there exist finite points, namely

( 1 , u 1 ) , ( 1 , u 2 ) , , ( 1 , u n ) ( { 1 } × E ) L ( P ) ,

such that

( { 1 } × E ) L ( P ) i = 1 n ( U u i ( { 1 } × E ) ) .

Let U = i = 1 n U u i . Then U is a bounded open subset of [ 0 , 1 ] × E . From (3.7), we have

[ 0 , 1 ] × E U L ( P ) i = i n ( ( [ 0 , 1 ] × E U u i ) L ( P ) ) = . (3.8)

Thus,

[ 0 , 1 ] × E U L ( P ) = .

From (3.5) and (3.8), we have

[ 0 , 1 ] × E U S = .

Then from Lemma 2.5, we have

deg ( I T ( λ , ) , U ( 1 ) , θ ) = deg ( I T ( λ , ) , U ( 0 ) , θ ) . (3.9)

Since U ( 0 ) = , then

deg ( I T ( λ , ) , U ( 0 ) , θ ) = 0 . (3.10)

Since Fix T B ( θ , R ) , then from (3.4) we have

deg ( I T ( λ , ) , U ( 1 ) , θ ) = 1 , (3.11)

which contradicts to (3.10) and (3.11). Therefore, there must exist ( 1 , u 0 ) ( { 1 } × E ) L ( P ) such that D u 0 is bounded.

Step 3. Obviously, the projection of D u is a interval, denote it by [ 0 , λ ] , then λ 1 . Then D u is a bounded connected component of ( [ 0 , λ ] × E ) . Take r 0 = β ( b a ) α ϕ ( λ m δ ) + 2 , let

Y 1 = ( { 1 } × E ) M [ r 0 , + ) , Y 2 = ( [ 0 , 1 ] × E ) M ( r 0 ) , Y = ( [ 0 , 1 ] × E ) M [ r 0 , + ) .

Obviously, D u ( Y 1 Y 2 ) . For each p D u ( Y 1 Y 2 ) , denote by E ( p ) the connected component of the metric space D u Y , which passes the point p. Now we shall prove that there must exist a p 0 D u ( Y 1 Y 2 ) such that E ( p 0 ) is an unbounded connected component of the metric space D u Y . On the contrary, assume that E ( p ) is bounded for each p D u ( Y 1 Y 2 ) . Then, for each p D u ( Y 1 Y 2 ) , in the same way as in the construction of U in (3.6) we can show that there exists a neighborhood U ( p ) of E ( p ) in Y such that

Y U ( p ) D u = . (3.12)

Obviously, the set of { U ( p ) ( Y 1 Y 2 ) | p D u ( Y 1 Y 2 ) } is an open cover of the set D u ( Y 1 Y 2 ) and D u ( Y 1 Y 2 ) is a compact set. Thus, there exist finite subsets of { U ( p ) ( Y 1 Y 2 ) | p D u ( Y 1 Y 2 ) } , say

U ( p 1 ) ( Y 1 Y 2 ) , U ( p 2 ) ( Y 1 Y 2 ) , , U ( p n ) ( Y 1 Y 2 ) ,

which is also an open cover of D u ( Y 1 Y 2 ) , that is,

i = 1 n ( U ( p i ) ( Y 1 Y 2 ) ) D u ( Y 1 Y 2 ) . (3.13)

Let U 0 = i = 1 n U ( p i ) , then U 0 is a bounded open subset of Y . Since

Y U 0 i = 1 n Y U ( p i ) ,

then by (3.12) we have

Y U 0 D u = . (3.14)

Let

W 1 = ( ( [ 0 , 1 ] × E ) M [ 0 , r 0 ) ) U 0 ,

and W 2 = ( [ 0 , 1 ] × E ) C l [ 0 , 1 ] × E W 1 . It is easy to see that

[ 0 , 1 ] × E W 1 Y U 0 [ ( M ( r 0 ) ( [ 0 , 1 ] × E ) ) ( U 0 ( Y 1 Y 2 ) ) ] .

From (3.12) and (3.13), we see that [ 0 , 1 ] × E W 1 D u = . Obviously, W 1 D u and W 1 W 2 = . Note the unboundedness of D u , then W 2 D u . Now we have D u = ( W 1 D u ) ( W 2 D u ) , which is a contradiction of the connectedness of D u . Therefore, there must exist p 0 D u ( Y 1 Y 2 ) such that E ( p 0 ) is an unbounded connected component of D u Y .

Step 4. Since u = T ( λ , u ) Q λ for each ( λ , u ) E ( p 0 ) , we have

u ( α u β ϕ 1 ( λ m δ ) ) e ( t ) > θ .

So, u = A ( λ , u ) for each ( λ , u ) E ( p 0 ) . This implies E ( p 0 ) L ( P ) is an unbounded subset. Let D be the connected component of L ( P ) containing E ( p 0 ) . Obviously, there exists λ > 0 such that Proj D ( 0 , λ ] . As D is unbounded, we easily see that

lim λ 0 + , ( λ , u ) D u = + .

The proof is complete. □

Corollary 3.1Let (A1) to (A4) hold. Then there exists λ > 0 such that problem (1.1) has a positive solution u λ for 0 < λ < λ with u λ as λ 0 .

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

The authors declare that the study was realized in collaboration with the same responsibility. All authors read, checked and approved the final manuscript.

Acknowledgements

This paper is supported by Innovation Project of Jiangsu Province postgraduate training project (CXLX12_0979).

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