Existence and Multiplicity of Positive Solutions for a Singular Third-Order Three-Point Boundary Value Problem with a Parameter

In this paper, we investigate the existence of positive solutions for a singular third-order three-point boundary value problem with a parameter. By using fixed point index theory, some existence, multiplicity and nonexistence results for positive solutions are derived in terms of different values of λ.

• Three-Point Boundary Value Problem
• Fixed Point Index
• Positive Solution
• Existence
• Multiplicity

1. Introduction

In this paper, we are concerned with the existence, multiplicity and nonexistence of positive solutions for the following third-order boundary value problem (BVP for short):

u ‴ ( t ) = λ q ( t ) f ( t , u ( t ) ) ,     t ∈ ( 0 , 1 ) , (1)

u ( 0 ) = α u ( η ) ,     u ′ ( η ) = 0 ,     u ″ ( 1 ) = 0 , (2)

where α ∈ ( 0,1 ) , η ∈ [ 1 2 ,1 ) are constants, λ is a positive parameter, q : ( 0,1 ) → [ 0, ∞ ) , f : [ 0,1 ] × [ 0, ∞ ) → [ 0, ∞ ) are continuous and q ( t ) may be singular at t = 0 and 1.

Third-order differential equations arise in a variety of different areas of applied mathematics and physics, e.g., in the deflection of curved beam having a constant or varying cross section, a three-layer beam, electromagnetic waves or gravity driven flows and so on [1]. In recent years, third-order boundary value problems have been studied by many methods [2] - [10], such as upper and lower solutions method, monotone iterative method and the different fixed point theory, etc.

In [11], Sun proved the existence of triple positive solutions to the following BVP by using a fixed-point theorem due to Avery and Peterson:

u ‴ ( t ) = a ( t ) f ( t , u ( t ) , u ′ ( t ) , u ″ ( t ) ) ,     t ∈ ( 0 , 1 ) ,

u ( 0 ) = δ u ( η ) ,     u ′ ( η ) = 0 ,     u ″ ( 1 ) = 0 ,

where δ ∈ ( 0,1 ) , η ∈ [ 1 2 ,1 ) are constants. a : ( 0,1 ) → [ 0, ∞ ) and f : [ 0,1 ] × [ 0, ∞ ) × R × R → [ 0, ∞ ) are continuous.

By applying the Krasnoselskii’s fixed point theorem, Sun [12] established the existence of infinitely many solutions to the following BVP, which is the special case for α = 0 in BVP (1) and (2):

u ‴ ( t ) = λ a ( t ) F ( t , u ( t ) ) ,     t ∈ ( 0,1 ) ,

u ( 0 ) = u ′ ( η ) = u ″ ( 1 ) = 0 ,

with λ > 0 , η ∈ [ 1 2 ,1 ) , where a ( t ) is nonnegative continuous function defined on ( 0,1 ) and F : [ 0,1 ] × [ 0, ∞ ) → [ 0, ∞ ) is continuous, a ( t ) may be singular at t = 0 and/or t = 1 .

Motivated by the above works, here we study the third order BVP (1) and (2). Under certain suitable conditions, we establish the results of existence, multiplicity and nonexistence of positive solutions for BVP (1) and (2) via the fixed point index theory.

2. Preliminaries

In this section, we present some notation and Lemmas that will be used in subsequent sections.

Lemma 2.1. [11] Let α ≠ 1 , h ∈ C [ 0,1 ] , then the BVP

u ‴ ( t ) = h ( t ) ,       t ∈ ( 0 , 1 ) , (3)

u ( 0 ) = α u ( η ) ,     u ′ ( η ) = 0 ,     u ″ ( 1 ) = 0 , (4)

has a unique solution

u ( t ) = ∫ 0 1     G ( t , s ) h ( s ) d s ,

where

G ( t , s ) = { s 2 2 ( 1 − α ) ,     s ≤ t ,     s ≤ η , − 1 2 t 2 + t s + α s 2 2 ( 1 − α ) ,     t ≤ s ≤ η , 1 2 s 2 − t s + η t + α η 2 2 ( 1 − α ) ,     η ≤ s ≤ t , − 1 2 t 2 + η t + α η 2 2 ( 1 − α ) ,     η ≤ s ,     t ≤ s . (5)

Lemma 2.2. [11] Suppose 0 < α < 1 , 1 2 ≤ η < 1 , g ( s ) = 1 2 ( 1 − α ) min { s 2 , η 2 } , then

α g ( s ) ≤ G ( t , s ) ≤ g ( s ) ,     t , s ∈ [ 0 , 1 ] .

Let E = C [ 0,1 ] be equipped with norm ‖ u ‖ = max t ∈ [ 0,1 ] | u ( t ) | , then ( E , ‖   ⋅   ‖ ) is a real Banach space.

From Lemma 2.2, we know that if 0 < α < 1 , 1 2 ≤ η < 1 , then for h ∈ C + [ 0,1 ] = { x ∈ C [ 0,1 ] : x ( t ) ≥ 0,   t ∈ [ 0,1 ] } , the unique solution u ( t ) of BVP (2.1) and (2.2) is nonnegative and satisfies

min t ∈ [ 0,1 ] u ( t ) ≥ α ‖ u ‖ .

Define the cone P by

P = { u ∈ C [ 0,1 ] : min t ∈ [ 0,1 ] u ( t ) ≥ α ‖ u ‖ } ,

then P is a non-empty closed and convex subset of E.

For u , v ∈ E , we write u ≤ v if u ( t ) ≤ v ( t ) for any t ∈ [ 0,1 ] . For any r > 0 , let K r = { u ∈ E : ‖ u ‖ < r } and ∂ K r = { u ∈ E : ‖ u ‖ = r } .

Define the operator T : P → E by

( T u ) ( t ) = ∫ 0 1     G ( t , s ) q ( s ) f ( s , u ( s ) ) d s . (6)

In view of the Lemma 2.1, it is easy to see that u is a positive solution BVP (1) and (2) if and only if u is a fixed point of the operator λ T .

In the following, we assume that:

(H₁) q ( t ) ≥ 0 , q ( t ) ≡ 0 and ∫ 0 1     q ( s ) d s < ∞ .

(H₂) f ∈ C ( [ 0,1 ] × [ 0, ∞ ) , [ 0, ∞ ) ) , f ( t , u ) is non-decreasing in u and f ( t , u ) > 0 for any t ∈ [ 0 , 1 ] , u > 0 .

Lemma 2.3. Assume (H₁)-(H₂) hold, then the operator T : P → P is completely continuous.

Proof. For u ∈ P , according to the definition of T and Lemma 2.2, it is easy to prove that T ( P ) ⊂ P . By the Ascoli-Arzela theorem, it is easy to show T : P → P is completely continuous.

The proofs of our main theorems are based on the fixed index theory. The following three well-known Lemmas in [13] [14].

Lemma 2.4. Let E be a Banach space and P ⊂ E be a cone in E. Assume that Ω is a bounded open subset of E. Suppose that T : P ∩ Ω ¯ → P is a completely continuous operator. If there exists x 0 ∈ P \ { θ } such that x − T x ≠ μ x 0 , for all x ∈ P ∩ ∂ Ω and μ ≥ 0 , then the fixed point index i ( T , P ∩ Ω , P ) = 0 .

Lemma 2.5. Let E be a Banach space and P ⊂ E be a cone in E. Assume that Ω is a bounded open subset of E. Suppose that T : P ∩ Ω ¯ → P is a completely continuous operator. If inf x ∈ P ∩ ∂ Ω ‖ T x ‖ > 0 and μ T x ≠ x for x ∈ P ∩ ∂ Ω and μ ≥ 1 , then the fixed point index i ( T , P ∩ Ω , P ) = 0 .

Lemma 2.6. Let E be a Banach space and P ⊂ E be a cone in E. Assume that Ω is a bounded open subset of E with θ ∈ Ω . Suppose that T : P ∩ Ω ¯ → P is a completely continuous operator. If T x ≠ μ x for all x ∈ P ∩ ∂ Ω and μ ≥ 1 , then the fixed point index i ( T , P ∩ Ω , P ) = 1 .

Now for convenience we use the following notations. Let

f 0 = l i m x → 0 + max t ∈ [ 0,1 ] f ( t , x ) x ,   f ∞ = l i m x → + ∞ max t ∈ [ 0,1 ] f ( t , x ) x ,

f 0 = l i m x → 0 + min t ∈ [ 0,1 ] f ( t , x ) x ,   f ∞ = l i m x → + ∞ min t ∈ [ 0,1 ] f ( t , x ) x ,

A = ∫ 0 1     g ( s ) q ( s ) d s ,

Φ = { ( λ , u ) : λ > 0 , u ∈ P isapositivesolutionofBVP   ( 1.1 )   and   ( 1.2 ) } ;

Λ = { λ > 0 : thereexists   u ∈ P suchthat   ( λ , u ) ∈ Φ } ;

λ * = sup Λ ,       λ * = inf Λ .

3. The Main Results and Proofs

Lemma 3.1. Suppose (H₁) holds and f 0 = ∞ , then Φ ≠ ∅ .

Proof. Let R > 0 be fixed, then we can choose λ 0 > 0 small enough such that λ 0 sup u ∈ P ∩ K ¯ R ‖ T u ‖ < R . It is easy to see that

λ 0 T u ≠ μ u ,       ∀ u ∈ P ∩ ∂ K R ,   μ ≥ 1.

By Lemma 2.6, it follows that

i ( λ 0 T , P ∩ K R , P ) = 1. (7)

From f 0 = ∞ , it follows that there exists r ∈ ( 0, R ) such that

f ( t , x ) ≥ 1 λ 0 α 2 A x ,       ∀ x ∈ [ 0, r ] ,     t ∈ [ 0,1 ] . (8)

We may suppose that λ 0 T has no fixed point on P ∩ ∂ K r . Otherwise, the proof is finished. Let e ( t ) ≡ 1 for t ∈ [ 0,1 ] , Then e ∈ ∂ K 1 . We claim that

u ≠ λ 0 T u + μ e ,       ∀ u ∈ P ∩ ∂ K r ,     μ ≥ 0. (9)

In fact, if not, there exist u 1 ∈ P ∩ ∂ K r and μ 1 ≥ 0 such that u 1 = λ 0 T u 1 + μ 1 e , then μ 1 > 0 . For u 1 ∈ P ∩ ∂ K r and μ 1 > 0 , by Lemma 2.2 and (8), we have

‖ u 1 ‖ ≥ u 1 ( t ) = ( λ 0 T u 1 ) ( t ) + μ 1 e ( t ) = λ 0 ∫ 0 1     G ( t , s ) q ( s ) f ( s , u 1 ( s ) ) d s + μ 1 ≥ α λ 0 ∫ 0 1     g ( s ) q ( s ) f ( s , u 1 ( s ) ) d s + μ 1 ≥ α λ 0 1 α 2 λ 0 A ∫ 0 1     g ( s ) q ( s ) u 1 ( s ) d s + μ 1 ≥ 1 A ‖ u 1 ‖ ∫ 0 1     g ( s ) q ( s ) d s + μ 1 = ‖ u 1 ‖ + μ 1 = r + μ 1 ,

we get r ≥ r + μ 1 , which is a contradiction. Hence by Lemma 2.4, it follows that

i ( λ 0 T , P ∩ K r , P ) = 0. (10)

By virtue of the additivity of the fixed point index, by (7) and (10), we have

i ( λ 0 T , P ∩ ( K R \ K ¯ r ) , P ) = i ( λ 0 T , P ∩ K R , P ) − i ( λ 0 T , P ∩ K r , P ) = 1 ,

which implies that the nonlinear operator λ 0 T has one fixed point u 0 ∈ P ∩ ( K R \ K ¯ r ) . Therefore, ( λ 0 , u 0 ) ∈ Φ . The proof is complete.

Lemma 3.2. Suppose (H₁) and (H₂) hold, f ∞ = 0 , then Φ ≠ ∅ .

Proof. Let r > 0 be fixed. From (H₂) and the definition of cone P, it follows that there exists C > 0 such that f ( t , u ( t ) ) ≥ C for all t ∈ [ 0,1 ] and u ∈ P ∩ ∂ K r . Then for sufficiently large λ with λ > r A C α and u ∈ P ∩ ∂ K r , we have

( λ T u ) ( t ) = λ ∫ 0 1     G ( t , s ) q ( s ) f ( s , u ( s ) ) d s ≥ λ α C ∫ 0 1     g ( s ) q ( s ) d s > r ,     t ∈ [ 0 , 1 ] .

This implies that inf u ∈ P ∩ ∂ K r ‖ λ T u ‖ > 0 and μ λ T u ≠ u for u ∈ P ∩ ∂ K r , μ ≥ 1 . By Lemma 2.5, it follows that

i ( λ T , P ∩ K r , P ) = 0. (11)

From f ∞ = 0 , there exists R > r such that

f ( t , u ) ≤ 1 2 A λ u ,         ∀ u ∈ [ α R , ∞ ) ,     t ∈ [ 0,1 ] .

Then for u ∈ P ∩ ∂ K R , by the definition of cone P, we get min t ∈ [ 0 , 1 ] u ( t ) ≥ α ‖ u ‖ = α R , and so

( λ T u ) ( t ) = λ ∫ 0 1     G ( t , s ) q ( s ) f ( s , u ( s ) ) d s ≤ λ 1 2 λ A ∫ 0 1     g ( s ) q ( s ) u ( s ) d s < R ,       t ∈ [ 0 , 1 ] .

We obtain λ T u ≠ μ u for u ∈ P ∩ ∂ K R , μ ≥ 1 . It follows from Lemma 2.6 that

i ( λ T , P ∩ K R , P ) = 1. (12)

According to the additivity of the fixed point index, by (11) and (12), we have

i ( λ T , P ∩ ( K R \ K ¯ r ) , P ) = i ( λ T , P ∩ K R , P ) − i ( λ T , P ∩ K r , P ) = 1,

which implies that the nonlinear operator λ T has at least one fixed point u ∈ P ∩ ( K R \ K ¯ r ) . Therefore, ( λ , u ) ∈ Φ . The proof is complete.

Lemma 3.3. Suppose (H₁) and (H₂) hold, f 0 = f ∞ = 0 , then 0 < λ * < ∞ .

Proof. By Lemma 3.1, it is easy to see that λ * > 0 . It follows from (H₂) and f 0 = f ∞ = ∞ that there exists C > 0 such that f ( t , u ) ≥ C u for all u ≥ 0 and t ∈ [ 0,1 ] . Let ( λ ,   u ) ∈ Φ , by the definition of cone P and Lemma 2.1, we obtain that

u ( t ) = ( λ T u ) ( t ) = λ ∫ 0 1     G ( t , s ) q ( s ) f ( s , u ( s ) ) d s ≥ λ α C ∫ 0 1     g ( s ) q ( s ) u ( s ) d s ≥ λ α 2 C ‖ u ‖ ∫ 0 1     g ( s ) q ( s ) d s = λ α 2 A C ‖ u ‖ ,

so ‖ u ‖ ≥ λ α 2 A C ‖ u ‖ , thus λ ≤ ( α 2 A C ) − 1 . This completes the proof of Lemma 3.3.

Lemma 3.4. Suppose (H₁) and (H₂) hold, hold and f 0 = f ∞ = 0 , then 0 < λ * < ∞ .

Proof. By Lemma 3.2, it is easy to see that λ * < ∞ . It follows from (H₂) and f 0 = f ∞ = 0 that there exists C 1 > 0 such that f ( t , u ) ≤ C 1 u for all u ≥ 0 and t ∈ [ 0,1 ] . Let ( λ ,   u ) ∈ Φ , from the definition of cone P and Lemma 2.2, we have

u ( t ) = ( λ T u ) ( t ) = λ ∫ 0 1     G ( t , s ) q ( s ) f ( s , u ( s ) ) d s ≤ λ C 1 ∫ 0 1     g ( s ) q ( s ) u ( s ) d s ≤ λ C 1 ‖ u ‖ ∫ 0 1     g ( s ) q ( s ) d s = λ A C 1 ‖ u ‖ ,     t ∈ [ 0 , 1 ] ,

so ‖ u ‖ ≤ λ A C 1 ‖ u ‖ , thus λ ≥ 1 A C 1 . This completes the proof of Lemma 3.4.

Lemma 3.5. Suppose (H₁) and (H₂) hold, f 0 = f ∞ = ∞ , then ( 0, λ * ) ⊂ Λ . Moreover, for any λ ∈ ( 0, λ * ) , BVP (1) and (2) has at least two positive solutions.

Proof. For any fixed λ ∈ ( 0, λ * ) , we prove that λ ∈ Λ . By the definition of λ * , there exists λ 2 ∈ Λ , such that λ < λ 2 ≤ λ * and ( λ 2 , u 2 ) ∈ Φ . Let R < min t ∈ [ 0 , 1 ] u 2 ( t ) be fixed. From the proof of Lemma 3.1, we see that there exist λ 1 < λ ,   r < R and u 1 ( t ) ∈ P ∩ ( K R \ K ¯ r ) such that ( λ 1 , u 1 ) ∈ Φ . It is easy to see that 0 < u 1 ( t ) < u 2 ( t ) for all t ∈ [ 0,1 ] . Then we have

u ‴ 1 ( t ) = λ 1 q ( t ) f ( t , u 1 ( t ) ) ,     t ∈ ( 0,1 ) ,

and

u ‴ 2 ( t ) = λ 2 q ( t ) f ( t , u 2 ( t ) ) ,     t ∈ ( 0 , 1 ) .

Consider now the modified BVP:

u ‴ ( t ) = λ q ( t ) f 1 ( t , u ( t ) ) ,     t ∈ ( 0 , 1 ) , (13)

u ( 0 ) = α u ( η ) ,     u ′ ( η ) = 0 ,     u ″ ( 1 ) = 0 , (14)

where

f 1 ( t , u ( t ) ) = { f ( t , u 1 ( t ) ) ,     u ( t ) ≤ u 1 ( t ) , f ( t , u ( t ) ) ,     u 1 ( t ) < u ( t ) < u 2 ( t ) , f ( t , u 2 ( t ) ) ,     u ( t ) ≥ u 2 ( t ) .

Clearly, the function λ f 1 is bounded for t ∈ [ 0,1 ] , u ∈ P and is continuous in u . Define the operator T 1 : E → E by

( T 1 u ) ( t ) = ∫ 0 1     G ( t , s ) q ( s ) f 1 ( s , u ( s ) ) d s ,     u ∈ E ,     t ∈ [ 0 , 1 ] .

Then T 1 : P → P is completely continuous and all the fixed points of operator λ T 1 are the solutions for BVP (13) and (14). It is easy to see that there exists r 0 > ‖ u 2 ‖ such that ‖ λ T 1 u ‖ < r 0 for any u ∈ P . From Lemma 2.6, we have

i ( λ T 1 , P ∩ K r 0 , P ) = 1. (15)

Let

U = { u ∈ P : u 1 ( t ) < u ( t ) < u 2 ( t ) , ∀ t ∈ [ 0 , 1 ] } .

We claim that if u ∈ P is a fixed point of operator λ T 1 , then u ∈ U . In fact, if u = λ T 1 u , then

u ( t ) = ( λ T 1 u ) ( t ) = λ ∫ 0 1     G ( t , s ) q ( s ) f 1 ( s , u ( s ) ) d s < λ 2 ∫ 0 1     G ( t , s ) q ( s ) f ( s , u 2 ( s ) ) d s = ( λ 2 T u 2 ) ( t ) = u 2 ( t ) ,

and

u ( t ) = ( λ T 1 u ) ( t ) = λ ∫ 0 1     G ( t , s ) q ( s ) f 1 ( s , u ( s ) ) d s > λ 1 ∫ 0 1     G ( t , s ) q ( s ) f ( s , u 1 ( s ) ) d s = ( λ 1 T u 1 ) ( t ) = u 1 ( t ) .

From the excision property of the fixed point index and (15), we obtain that

i ( λ T 1 , U , P ) = i ( λ T 1 , P ∩ K r 0 , P ) = 1.

From the definition of T 1 , we know that T 1 = T on U ¯ , then

i ( λ T , U , P ) = 1. (16)

Hence, the nonlinear operator λ T has at least fixed point v 1 ∈ U . Then v 1 is one positive solution of BVP (1) and (2). This gives λ ∈ Λ , ( λ ,   v 1 ) ∈ Φ and ( 0, λ ) ⊂ Λ .

We now find the second positive solution of BVP (1) and (2). By f ∞ = ∞ and the continuity of f ( t , u ) with respect to u , there exists C > 0 such that

f ( t , u ) ≥ 2 u λ α 2 A − C α A ,       ∀ u ≥ 0,   t ∈ [ 0,1 ] . (17)

For e ( t ) ≡ 1 , let

Ω = { u ∈ P : thereexists   τ ≥ 0 suchthat   u = λ T u + τ e } .

We claim that Ω is bounded in E. In fact, for any u ∈ Ω , it follows from Lemma 2.2 and (17) that

u ( t ) = ( λ T u ) ( t ) + τ e ( t ) = ( λ T u ) ( t ) + τ ≥ λ ∫ 0 1     G ( t , s ) q ( s ) f ( s , u ( s ) ) d s ≥ λ α ∫ 0 1     g ( s ) q ( s ) [ 2 u ( t ) λ α 2 A − C α A ] d s

≥ λ α ∫ 0 1     g ( s ) q ( s ) [ 2 α ‖ u ‖ λ α 2 A − C α A ] d s = 2 ‖ u ‖ − λ C .

This implies ‖ u ‖ ≤ λ C . Thus Ω is bounded in E. Therefore there exists R 1 > ‖ u 2 ‖ such that

u ≠ λ T u + τ e ,     ∀ u ∈ P ∩ ∂ K R 1 ,   τ ≥ 0.

By Lemma 2.4, we get that

i ( λ T , P ∩ K R 1 , P ) = 0. (18)

Using a similar argument as in deriving (10), we have that

i ( λ T , P ∩ K r 1 , P ) = 0 , (19)

where 0 < r 1 < min t ∈ [ 0 , 1 ] u 1 ( t ) . According to the additivity of the fixed point index, by (16), (18) and (19), we have

i ( λ T , P ∩ ( K R 1 \ ( U ¯ ∪ K ¯ r 1 ) ) , P ) = i ( λ T , P ∩ K R 1 , P ) − i ( λ T , U , P ) − i ( λ T , P ∩ K r 1 , P ) = − 1 ,

which implies that the nonlinear operator λ T has at least one fixed point v 2 ∈ P ∩ ( K R 1 \ ( U ¯ ∪ K ¯ r 1 ) ) . Thus, BVP (1)-(2) has another positive solution. The proof is complete.

Lemma 3.6. Suppose (H₁) and (H₂) hold, f 0 = f ∞ = 0 , then ( λ * , + ∞ ) ⊂ Λ . Moreover, for any λ ∈ ( λ * , + ∞ ) , BVP (1)-(2) has at least two positive solutions.

Proof. For any fixed λ ∈ ( λ * , + ∞ ) , we prove that λ ∈ Λ . By the definition of λ * , there exists λ 1 ∈ Λ , such that λ * ≤ λ 1 < λ and ( λ 1 , u 1 ) ∈ Φ . Let r > 1 α ‖ u 1 ‖ be fixed. From the proof of Lemma 3.2, we see that there exist λ 2 > λ , R > r and u 2 ( t ) ∈ P ∩ ( K R \ K ¯ r ) such that ( λ 2 , u 2 ) ∈ Φ . By the definition of cone P, it is easy to see that 0 < u 1 ( t ) < u 2 ( t ) for all t ∈ [ 0,1 ] . Define

V = { u ∈ P : u 1 ( t ) < u ( t ) < u 2 ( t ) , ∀ t ∈ [ 0 , 1 ] } .

Using the method similar to get (16), we yield

i ( λ T , V , P ) = 1, (20)

Hence, the nonlinear operator λ T has at least fixed point v 1 ∈ V . Then v 1 is one positive solution of BVP (1) and (2). This gives λ ∈ Λ , ( λ , v 1 ) ∈ Φ and ( λ * , + ∞ ) ⊂ Λ .

We now find the second positive solution of BVP (1) and (2). From f 0 = 0 , there exists 0 < r 0 < min t ∈ [ 0 , 1 ] u 1 ( t ) such that

f ( t , u ) ≤ u 2 λ A ,     ∀ u ∈ [ 0, r 0 ] ,   t ∈ [ 0,1 ] .

Then for u ∈ P ∩ ∂ K r 0 , we have

( λ T u ) ( t ) = λ ∫ 0 1     G ( t , s ) q ( s ) f ( s , u ( s ) ) d s ≤ λ ∫ 0 1     g ( s ) q ( s ) ‖ u ‖ 2 λ A d s = ‖ u ‖ 2 < r 0 ,     t ∈ [ 0 , 1 ] .

This implies λ T u ≠ μ u for u ∈ P ∩ ∂ K r 0 , μ ≥ 1 . It follows from Lemma 2.6 that

i ( λ T , P ∩ K r 0 , P ) = 1. (21)

Using a similar argument as in deriving (12), we have that

i ( λ T , P ∩ K R 0 , P ) = 1. (22)

where R 0 > ‖ u 2 ‖ . According to the additivity of the fixed point index, by (20), (21) and (22), we have

i ( λ T , P ∩ ( K R 0 \ ( V ¯ ∪ K ¯ r 0 ) ) , P ) = i ( λ T , P ∩ K R 0 , P ) − i ( λ T , V , P ) − i ( λ T , P ∩ K r 0 , P ) = − 1 ,

which implies that the nonlinear operator λ T has at least one fixed point v 2 ∈ P ∩ ( K R 0 \ ( V ¯ ∪ K ¯ r 0 ) ) . Thus, BVP (1)-(2) has another positive solution. The proof is complete.

Lemma 3.7. Suppose (H₁) and (H₂) hold, f 0 = f ∞ = ∞ , then Λ = ( 0, λ * ] .

Proof. In view of Lemma 3.5, it suffices to prove that λ * ∈ Λ . By the definition of λ * , we can choose { λ n } ⊂ Λ with λ n ≥ λ * 2 ( n = 1 , 2 , ⋯ ) such that

λ n → λ * as n → ∞ . By the definition of Λ , there exists { u n } ⊂ P \ { θ } such that ( λ n , u n ) ∈ Φ . We now show that { u n } is bounded. Suppose the contrary, then there exists a subsequence of { u n } (still denoted by { u n } ) such that ‖ u n ‖ → ∞ as n → ∞ . It follows from { u n } ⊂ P \ { θ } that u n ≥ α ‖ u n ‖ for all t ∈ [ 0,1 ] . Choose sufficiently large τ such that

λ * α 2 A τ 2 > 1.

By f ∞ = ∞ , there exists R > 0 such that f ( t , u ) ≥ τ u for all u > α R and t ∈ [ 0,1 ] . Since ‖ u n ‖ → ∞ as n → ∞ , there exists sufficiently large n 0 such that ‖ u n 0 ‖ ≥ R . Thus, we have

‖ u n 0 ‖ ≥ u n 0 ( t ) = ( λ n 0 T u n 0 ) ( t ) = λ n 0 ∫ 0 1     G ( t , s ) q ( s ) f ( s , u n 0 ( s ) ) d s ≥ λ * 2 α τ ∫ 0 1     g ( s ) q ( s ) u n 0 ( s ) d s ≥ λ * 2 α 2 τ ‖ u n 0 ‖ ∫ 0 1     g ( s ) q ( s ) d s = λ * 2 α 2 τ A ‖ u n 0 ‖ .

This gives

λ * α 2 A τ 2 ≤ 1, (23)

which contradicts the choice of τ . Hence, { u n } is bounded. It follows from the completely continuous of T that { T u n } is equicontinuous, i.e., for each ε > 0 , there is a δ > 0 such that

| u n ( t 1 ) − u n ( t 2 ) | = λ n | ( T u n ) ( t 1 ) − ( T u n ) ( t 2 ) | < λ n ε ≤ λ * ε ,

where n = 1 , 2 , ⋯ , t 1 , t 2 ∈ [ 0 , 1 ] and | t 1 − t 2 | < δ . Then { u n } is equicontinuous. According to the Ascoli-Arzela theorem, { u n } is relatively compact. Hence, there exists a subsequence of { u n } (still denoted by { u n } ) and u * ∈ P such that u n → u * as n → ∞ . By u n = λ n T u n , letting n → ∞ , we obtain that u * = λ * T u * . If u * = θ , using a similar argument as in deriving (23), by f 0 = ∞ , we also get a contradiction. Then u * ∈ P \ { θ } , and so λ * ∈ Λ . This completes the proof.

Lemma 3.8. Suppose (H₁) and (H₂) hold, f 0 = f ∞ = 0 , then Λ = [ λ * , + ∞ ) .

Proof. In view of Lemma 3.6, it suffices to prove that λ * ∈ Λ . By the definition of λ * , we can choose { λ n } ⊂ Λ with λ n ≤ 2 λ * ( n = 1 , 2 , ⋯ ) such that λ n → λ * as n → ∞ . By the definition of Λ , there exists { u n } ⊂ P \ { θ } such that ( λ n , u n ) ∈ Φ . We now show that { u n } is bounded. Suppose the contrary, then there exists a subsequence of { u n } (still denoted by { u n } ) such that ‖ u n ‖ → ∞ as n → ∞ . It follows from { u n } ⊂ P \ { θ } that u n ≥ α ‖ u n ‖ for all t ∈ [ 0,1 ] . Choose τ small enough such that

2 λ * A τ < 1.

By f ∞ = 0 , there exists R > 0 such that f ( t , u ) ≤ τ u for all u > α R and t ∈ [ 0,1 ] . Since ‖ u n ‖ → ∞ as n → ∞ , there exists sufficiently large n 0 such that ‖ u n 0 ‖ ≥ R . Thus, we have

u n 0 ( t ) = ( λ n 0 T u n 0 ) ( t ) = λ n 0 ∫ 0 1     G ( t , s ) q ( s ) f ( s , u n 0 ( s ) ) d s ≤ 2 λ * τ ∫ 0 1     g ( s ) q ( s ) u n 0 ( s ) d s ≤ 2 λ * τ ‖ u n 0 ‖ ∫ 0 1     g ( s ) q ( s ) d s = 2 λ * τ A ‖ u n 0 ‖ .

This gives

2 λ * τ A ≥ 1, (24)

which contradicts the choice of τ . Hence, { u n } is bounded. It follows from the completely continuous of T that { T u n } is equicontinuous, i.e., for each ε > 0 , there is a δ > 0 such that

| u n ( t 1 ) − u n ( t 2 ) | = λ n | ( T u n ) ( t 1 ) − ( T u n ) ( t 2 ) | < λ n ε ≤ 2 λ * ε ,

where n = 1 , 2 , ⋯ , t 1 , t 2 ∈ [ 0 , 1 ] and | t 1 − t 2 | < δ . Then { u n } is equicontinuous. According to the Ascoli-Arzela theorem, { u n } is relatively compact. Hence, there exists a subsequence of { u n } (still denoted by { u n } ) and u * ∈ P such that u n → u * as n → ∞ . By u n = λ n T u n , letting n → ∞ , we obtain that u * = λ * T u * . If u * = θ , using a similar argument as in deriving (24), by f 0 = 0 , we also get a contradiction. Then u * ∈ P \ { θ } , and so λ * ∈ Λ . This completes the proof.

From Lemmas 3.1, 3.3, 3.5 and 3.7, we get the main result as follows.

Theorem 3.1. Let (H₁), (H₂) be fulfilled and suppose that f 0 = f ∞ = ∞ , then there exists λ * > 0 such that BVP (1)-(2) has at least two positive solutions for λ ∈ ( 0, λ * ) , at least one positive solution for λ = λ * and no positive solution for λ > λ * .

By Lemmas 3.2, 3.4, 3.6 and 3.8, we obtain the main result as follows.

Theorem 3.2. Let (H₁), (H₂) be fulfilled and suppose that f 0 = f ∞ = 0 , then there exists λ * > 0 such that BVP (1)-(2) has at least two positive solutions for λ > λ * , at least one positive solution for λ = λ * and no positive solution for λ ∈ ( 0, λ * ) .

Founding

Supported by the Foundation for Basic Disciplines of Army Engineering University of PLA (KYSZJQZL2013).

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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