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ISSN Online: 2327-4379 ISSN Print: 2327-4352

DOI: 10.4236/jamp.2018.612217 Dec. 26, 2018 2600 Journal of Applied Mathematics and Physics

The Existence and Uniqueness of Positive

Solutions for a Singular Nonlinear Three-Point

Boundary Value Problems

Yao Dong, Baoqiang Yan

*

School of Mathematical Sciences, Shandong Normal University, Jinan, China

Abstract

Using the method of lower and upper solutions, we study the following sin-gular nonlinear three-point boundary value problems:

( ) ( ) ( ) ( ), (0,1),

(0) 0, (1) ( ),

q p

x t K t x t x t t

x x ax

λ

η

 ′′− + = ∈

= =

 , where K C∈ [0,1], 0< <a 1,

0< <η 1 and λ is a positive parameter and present the existence,

unique-ness, and the dependency on parameters of the positive solutions under vari-ous assumptions. Our result improves those in the previvari-ous literatures.

Keywords

Three-Point Boundary Value Problem, Positive Solution, Lower and Upper Solutions, Eigenvalue and Eigenfunction

1. Introduction and Main Results

In this paper, we consider the three-point boundary value problem

( ) ( ) ( ) ( ), (0,1),

(0) 0, (1) ( ),

q p

x t K t x t x t t

x x ax

λ

η

 ′′− + = ∈

= =

 (1.1)

where K C∈ [0,1], 0< <a 1, 0< <η 1, and λ is a positive parameter.

The m-point boundary value problem for linear second-order ordinary diffe-rential equations was initiated by Ilin and Moiseev [1] [2]. Since then, there are many results on the existence of general nonlinear multi-point boundary value problems, see [3] [4] [5] [6] and their references. For examples, in [6], Rynne studied the $m$-point boundary value problem

How to cite this paper: Dong, Y. and Yan, B.Q. (2018) The Existence and Uniqueness of Positive Solutions for a Singular Nonli-near Three-Point Boundary Value Prob-lems. Journal of Applied Mathematics and Physics, 6, 2600-2620.

https://doi.org/10.4236/jamp.2018.612217

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DOI: 10.4236/jamp.2018.612217 2601 Journal of Applied Mathematics and Physics 2

1

( ), (0,1), ,

(0) 0, (1) m i ( ),i

i

u f u on u R X

u u − α ηu

= ′′

− = ∈ ×

 

= =



where m≥3, η ∈i (0,1), α >i 0 with

2

1 1

m i i

α

− =

<

and presented the existence

of the sign changing solutions by Rabinowitz bifurcation theorem. Especially, Rynne ([7]) discussed the three-point boundary value problem

( ) , (0,1),

(0) 0, (1) ( ),

u f u h on u u α ηu

′′

− = +

= =

and showed the solvability and non-solvability results from either the half-eigenvalue or the Fucik spectrum approach. As we known, the method of upper and lower solutions is very important for the study of the boundary val-ue problems, see [8]-[18]. Therefore, establishing the method of upper and lower solutions for three-point boundary value problems is necessary and im-portant.

In [19], when f is nondecreasing on x, Du and Zhao got the methods of upper and lower solutions of

( ) ( , ( )), (0,1),

(0) ( ), (1) 0,

x t f t x t t x axη x

′′

− = ∈

= =

and used iterative techniques to study the existence of positive solutions. And in [3] when f is decreasing on u, Du and Zhao considered the existence and uni-queness of positive solutions of the problem

2

1

( ) ( , ( )), (0,1),

(0) m i ( ),i (1) 0

i

u t f t u t t

u − α ηu u

= ′′

− = ∈

 

= =



by constructing lower and upper solutions. Wei ([15]) constructed the method of upper and lower solutions for three-point boundary value problems and gave the sufficient and necessary conditions for the existence of positive solutions of the problem

( ) ( , ( )), (0,1),

(0) ( ), (1) 0.

x t f t x t t x axη x

′′

− = ∈

= =

On the other hand, singular boundary problems arise in the contexts of chemical heterogeneous catalysts, non-Newtonian fluids and also the theory of heat conduction in electrically conducting materials, see [20]-[25] for a detailed discussion. An interesting result comes from [25], in which, using method of upper and lower solutions, Shi and Yao discussed the following problem

( ) , ,

( ) 0, ,

| 0,

q p

u K x u u x u x x

u

λ

∂Ω

−∆ + = ∈ Ω

> ∀ ∈ Ω

=

where K C 2,β( ) , p q, (0,1) and λ is a positive parameter. Under

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DOI: 10.4236/jamp.2018.612217 2602 Journal of Applied Mathematics and Physics

uniqueness of classical solutions.

Motivated by above works, under various appropriate assumptions on p, q

and K t( ), we will obtain the existence and uniqueness of positive solution of problem (1.1) for λ in different circumstances. In our proof, the upper and

lower solutions theorem (see [16]) plays an important role in the paper. Define

[0,1]

*

* [0,1]

max ( ), min ( ).

t t

K K t K K t

∈ ∈

= =

The main results of this paper are stated in the following theorems.

Theorem 1.1. When K*>0,

1) If 0< p q, <1,thereexists λ >0 suchthattheproblem (1.1) hasatleast oneC[0,1]positivesolution x tλ( ) for λ λ> .

2) For λ λ> , (1.1) has amaximal solution x tλ( ) and x tλ( ) is increasing

withrespectto λ.

Theorem1.2. When K*<0,

1) If 0< p<1,0<q, (1.1) has at least one C[0,1] positive solution for all

0

λ> .

2) If 0< p q, <1, (1.1) hasanunique C1[0,1] positivesolution x t( ) λ forall

0

λ> .

3) x tλ( ) in (2) isincreasingwithrespectto λ.

Theorem1.3. When *

* 0

K < <K ,

1) If 0< p q, <1, there exists a λ >* 0 such that the problem (1.1) has at

leastoneC[0,1] posit-ivesolution x tλ( ) for λ λ> *.

2) For λ λ> *, x tλ( ) in (1) isincreasingwithrespectto λ.

Remark 1.1: Note K t( ) 0> in Theorem (1.1). This is different from the

conditions in [3] [15] [19] because K t( ) 0< in these references.

Remark 1.2: The unique result in Theorem 1.2 is different from that in [3] because we remove the monotonicity of nonlinearity f in x.

Remark 1.3: Note K t( ) is sigh-changing in Theorem 1.3. This is different from the conditions in [3] [15] [19] because K t( ) 0< in these references and is

different from conditions in [1] [2] [4] [5] [6] [7] [26] because f is continuous at 0

x= in these references.

This paper is organised as follows. Some preliminary lemmas are stated and proved in Section 2. And Section 3 is devoted to prove the results.

2. Preliminaries

In this section, we first consider the following problem ( ) ( , ( ), ( )), (0,1),

(0) 0, ( ) (1),

x t f t x t x t t x xη ax

′′ ′

− = ∈

= =

 (2.1)

where η ∈(0,1), 0< <a 1 and f ∈[0,1]× × .

Let C1[0,1] { :[0,1]= x | ( )x t is differential continuous on [0,1]} with

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DOI: 10.4236/jamp.2018.612217 2603 Journal of Applied Mathematics and Physics

|| || max{| | ,| | }x = xx′∞ ,

where | | max | ( ) |[0,1] t

x′ =∞ x t . Obviously, C1[0,1] is a Banach space. Now we give the definitions of lower and upper solutions for problem (2.1).

Definition 2.1. A function α( )t is called a lower solution to the problem (2.1), if α( )t C[0,1]C2(0,1) and satisfies

( ) ( , ( ), ( )), (0,1),

(0) 0, (1) ( ).

t f t t t t a

α α α

α α α η

′′ ′

− ≤ ∈

 (2.2)

Upper solution is defined by reversing the above inequality signs in problem (2.2).

If there exists a lower solution α( )t and an upper solution β( )t to problem (2.1) such that α( )t ≤β( )t , then ( ( ), ( ))α t β t is called a couple of upper and lower solutions of problem (2.1).

Set Dβ {( , ) (0,1)t x , ( )t x ( ),t t (0,1)}.

α = ∈ ×+ α ≤ ≤β ∈

We list a lemma for the eigenvalues and eigenfunctions for the following li-near problem

( ) ( ), (0,1),

(0) 0, (1) ( ).

x t x t t x x ax

λ

η

′′

− = ∈

= =

 (2.3)

Lemma 2.1. (see [6]) The spectrum σ( )L of problem (2.3) consists of a

strictly increasing sequence ofeigenvalues λ >k 0, k=1, 2,, with

eigenfuc-tions sin( 12 )

k kt

φ = λ . Inaddition,

1) lim k

k→+∞λ = +∞;

2) φk( )t has exact k−1 simple zeros in (0,1), k=2,3, and φ1 is

strictlypositiveon (0,1).

Lemma 2.2. Supposethat h L 1(0,1). Then, foreach λ>0, theproblem

( ) ( ), (0,1),

(0) 0, ( ) (1)

x t x h t t x x x

λ

η α

′′

− + = ∈

= =

 (2.4)

hasanuniquesolutioninC[0,1].

Proof. Assume that v t1( ) and v t2( ) satisfies that

( ) ( ), (0,1),

(0) 0, (0) 1

x t x h t t x x

λ

′′

− + = ∈

= =

 and

( ) ( ), (0,1),

(1) 0, (1) 1

x t x h t t x x

λ

′′

− + = ∈

= = −

 respectively. Define

2 1

1 1

( ) ( ), 0 1,

1 ( , )

( ) ( ), 0 1,

v t v s s t G t s

v t v s t s

ω

≤ ≤ ≤ 

=  ≤ ≤ ≤

 and

1 1 1

0 0

1 1

( )

( ) ( , ) ( ) ( , ) ( ) , [0,1].

(1) ( )

e t

x t G t s h s ds G s h s ds s

e α ηe α η

= + ∈

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DOI: 10.4236/jamp.2018.612217 2604 Journal of Applied Mathematics and Physics

( ) ( )

( ) ( )

1 0

2 1 1 2

1

2 1 1 2

1 1

0

1 1

2 1 1 2

1

1 1

( ) ( ) ( ) ( ) ( ) ( ) ]''

''( ) ( , ) ( )

(1) ( )

( ) ( ,

1

''( ) ( ) [

( )

1 [ '( ) ( ) '( ) ( )] ( )

1 [ ( ) (

) (1) ( ) ] ) 0 t t t

x t x t

x t

v t v t v t v t h t t

v t v s h s ds v t v s

v t v s h s ds v t v

h s d s h s ds

e t G s h s ds

e e

e t G

e s s h e λ α ω λ ω λ λ ω η α η

λ α η

α η − + = − = − − − + − + − + − −

1 0

2 1 1 2

1 1 0 1 0 1 1 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ] ( ) ( , ) ( ) ( ) 1 ( ) [ ( ) ( ),

(1) ( )

(0,1)

t t

s ds

v t v s h s ds v t v s h s ds

x t

h t

x t

h t

e t G s h s ds

e e

t

λ

λ ω

λ α η λ

α η + + − = − − = ∈ +

and

1 1 1

0 0

1 1

1 1 1

0 0 1 1 1 1 0 1 1

1 1 1

0 0

1 1

(1)

(1) ( ) (1, ) ( ) ( , ) ( )

(1) ( )

( )

[ ( , ) ( ) ( , ) ( ) ]

(1) ( )

(1) ( , ) ( )

(1) ( )

( )

[ ( , ) ( ) ( , ) ( ) ]

(1) ( )

e

x x G s h s ds G s h s ds

e e

e

G s h s ds G s h s ds

e e

e G s h s ds

e e

e

G s h s ds G s h s ds

e e

α η α η

α η η

α η α η

α η

α η

α η

η

α η α η

α η − = + − − + − = − − + −

0. =

Hence, x t( ) is a C[0,1] solution to problem(2.4). Since λ>0, Lemma 2.1 guarantees that problem (2.4) has an unique C[0,1] solution. The proof is com-plete. 

Theorem 2.1. Let

α

and βC([0,1])C1(0,1) be lower andupper

solu-tions of (2.1) such that α β≤ . Let ψ∈L1[0,1] and φ : + +0

  be a conti-nuousfunctionthatsatisfies

0

1 .

( )s ds

φ ∞

= +∞

(2.5)

Suppose f D: β

α× →  isan L1-Carathéodory-functionsuchthat

| ( , , ) |f t x v ( ) (| |), ( , )t v t x Dβ, v .

α ψ φ

≤ ∀ ∈ ∈ (2.6)

Thentheproblem (2.1) hasatleastonesolution x C 1[0,1] suchthatforall [0,1]

t∈ ,

( )t x t( ) ( ).t

α ≤ ≤β

Proof. The proof proceeds in five steps.

Step 1. We consider a new modified problem. From (2.5), there is an R>0 be large enough so that

1 0

1 || || .

( ) R

ds

s ψ

φ >

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DOI: 10.4236/jamp.2018.612217 2605 Journal of Applied Mathematics and Physics

And (2.6) guarantees that there is an N t( ) with N L 1[0,1] such that

| ( , , ) |f t x v N t( ), ( , )t x Dβ, | |v R.

α

≤ ∀ ∈ ≤ (2.8)

Define then

( ), ( ),

( , ) , ( ) ( ),

( ), ( )

t if x t t x x if t x t

t if x t

α α

χ α β

β β

< 

= ≤ ≤

>

(2.9)

and

( , , ) max{min{ ( , ( , ), ), ( )}, ( )}.

g t x v = f t χ t x v N tN t (2.10) Choose a λ >0 and consider the new boundary value problem

( ) ( , ( ), ( )) ( , ( )), (0,1),

(0) 0, (1) ( ),

x t x g t x t x t t x t t x x ax

λ λχ

η

′′ ′

− + = + ∈

= =

 (2.11)

where 0< <a 1, 0< <η 1.

Step 2. We discuss the existence of a C1[0,1] solution of (2.11).

Now Lemma 2.2 guarantees that for each h L 1[0,1], the linear problem

( ) , (0,1),

(0) 0, (1) ( )

x t x h t x x ax

λ

η

′′

− + = ∈

= =

 has an unique C[0,1] solution

1 1 1

0 0

1 1

( )

( ) ( , ) ( ) ( , ) ( ) , [0,1].

(1) ( )

e t

v t G t s h s ds a G s h s ds s

e ae η η

= + ∈

For x C 1[0,1], define

( )( )Fx t =g t x t x t( , ( ), ( ))′ +λχ( , ( )),t x t t∈[0,1] and

1 1 1

0 1 1 0

( )

( )( ) ( , )( )( ) ( , )( )( ) , [0,1].

(1) ( )

η

η

= + ∈

e t

Tx t G t s Fx s ds a G s Fx s ds s

e ae

From (2.9) and (2.10), we have

[0,1] [0,1]

| ( , ( ), ( )) ( , ( )) | ( ) max{sup | ( ) |, sup | ( ) |},

t t

g t x t x t

λχ

t x t N t

λ

α

t

β

t

∈ ∈

′ + ≤ + which

implies that the functions belonging to {( )( ) :Tx t x C 1[0,1]} and 1

{( ) ( ) :Tx t x C′ ∈ [0,1]} are bounded and equicontinuous. The Arzela-Ascoli Theorem guarantees that TC1[0,1] is relatively compact. The proof of the

con-tinuity of T is standard. Using the Schauder’s fixed point theorem, we assert that

T has at least one fixed point x C 1[0,1].

Step 3. The solution x of (2.11) is such that α( )tx t( )≤β( )t .

We prove that x t( )≤β( )t for t∈[0,1] only. In fact, suppose that there ex-ist a t0∈[0,1) such that x t( )0 >β( )t0 . Since x(0) 0= ≤β(0), t0>0. Let

( ) ( ) ( )

w t =x t −β t , t∈[0,1]. Then w(0) 0≤ and w t( ) 00 > .

Let t∗=sup{∣t w s( ) 0,> s t t∈[ , ]}0 , t∗=inf{∣t w s( ) 0,> s t t∈[ , ]}.0

It is obvious that w t( ) 0> for all * *

( , )

tt t , w t( ) 0* = and w t( ) 0* ≥ . If *

( ) 0

w t = , then there exists a *

*

( , )

t′∈ t t such that *

*

[ , ]

( ) max ( )

t t t

w t w t

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DOI: 10.4236/jamp.2018.612217 2606 Journal of Applied Mathematics and Physics *

( ) 0

w t > , obviously t*=1 and w(1)=x(1)β(1 0)> . Since

1 1

( ) ( ) ( ) ( (1) (1)) (1) (1)

w x x w w

a a

η = η −β η = −β = > , there exists *

*

( , )

t′∈ t t

such that *

*

[ , ]

( ) max ( )

t t t

w t w t

′ = also. Hence, w t′ ′ =( ) 0 (i.e., β ′ ′( )t =x t′ ′( )) and ( ) 0

w t′′

− ≥ . On the other hand, since

( ) ( ) ( )

( , ( ), ( )) ( , ( ), ( )) ( , ( )) ( )

( , ( ), ( )) max{min{ ( , ( ), ( )), ( )}, ( )}

( ) ( )

( , ( ), ( )) ( , ( ), ( )) ( )

w t t x t

f t t t g t x t x t t x t x t f t t t f t t t N t N t

t x t

f t t t f t t t t

β

β β λχ λ

β β β β

λβ λ

β β β β λβ

′′ ′′ ′′

− = −

′ ′ ′ ′ ′ ′ ′ ′ ′

≤ − + + −

′ ′ ′ ′ ′ ′ ′

= − + −

′ ′

+ −

′ ′ ′ ′ ′ ′ ′ ′ ′

= − + + ( )

( ( ) ( )) 0.

x t t x t

λ λ β

′ −

′ ′

= − <

This is a contradiction.

A similar argument holds to prove x t( )≤β( )t for all t∈[0,1]. Hence, from (2.10), one know that x satisfies that

( ) ( , ( ), ( )) max{min{ ( , ( ), ( )), ( )}, (0,1),

(0) 0, (1) ( ).

x t g t x t x t f t x t x t N t t

x x ax

η

′′ ′ ′

− = = ∈

= =

 (2.12)

Step 4. The solution x of (2.11) is such that | |x′ ≤∞ R.

On the contrary, suppose that there is a t′∈(0,1) such that | ( ) |x t′ ′ >R. Without loss of generality, we assume that x t′ ′ >( ) R. Since x(0) 0= and

(1) ( )

x =axη with 0< <a 1, there is a t0∈(0,1) such that x t′( ) 00 = . Without

loss of generality, we assume that x t′( ) 0> for all ( , )t t0 . Observe that, for all

( , )t x Dβ α

∈ , v,

max{min{ ( , , ), ( )},f t x v N tN t( )}≤ψ φ( ) (| |).t v

Then, from (2.12), one has

0

0

0 0

0 0

( )

0 ( )

1

1 | 1 | | 1 ( ) |

( ) ( ) ( ( ))

( ) ( , ( ), ( ))

| | | |

( ( )) ( ( ))

( ) ( ( )) ( ) || || .

( ( ))

R x t t

x t t

t t

t t

t t

t t

ds ds dx t

s s x t

x t dt g t x t x t dt x t x t t x t dt t dt

x t

φ φ φ

φ φ

ψ φ ψ ψ

φ

′ ′

′ ′

′ ′

′ ′

= =

′′ ′

= =

′ ′

= = =

This contradicts to (2.7).

Hence | ( , ( ), ( )) |f t x t x t′ ≤N t( ), which together with u∈[ , ]α β guarantees that

( , ( ), ( )) ( , ( ), ( )), (0,1).

g t x t x t′ = f t x t x t′ ∀ ∈t

Step 5. We claim that x t( ) satisfies (2.1).

Since | |x′ ≤∞ R and α( )tx t( )≤β( )t , by (2.8), (2.10) and (2.12), we have

( ) max{min{ ( , ( ), ( )), ( )} ( , ( ), ( )), (0,1),

(0) 0, (1) ( ),

x t f t x t x t N t f t x t x t t

x x ax

η

′′ ′ ′

− = = ∈

= =

that is, x t( ) is a C1[0,1] solution of (2.1). The proof is complete.

Now we consider the following problem

( ) ( , ( )), (0,1),

(0) 0, ( ) (1),

x t f t x t t x xη ax

′′

− = ∈

= =

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DOI: 10.4236/jamp.2018.612217 2607 Journal of Applied Mathematics and Physics

where η ∈(0,1), 0< <a 1 and f ∈[0,1]× × .

Now we give the definitions of lower and upper solutions for problem (2.13). Definition 2.2. (see [16]) A function α( )t is called a lower solution to the problem (2.13), if α( )t C[0,1]C2(0,1) and satisfies

( ) ( , ( )), (0,1),

(0) 0, (1) ( ).

t f t t t a

α α

α α α η

′′

− ≤ ∈

 (2.14)

Upper solution is defined by reversing the above inequality signs in problem (2.14).

By Theorem 2.1, we have following result.

Corollary 2.1. Suppose that there exists a lower solution α( )t and an upper solution β( )t of problem (2.1) such that α( )t ≤β( )t , t∈[0,1] and there ex-ists F L 1[0,1] such that | ( , ) |f t x F t( ) for all ( , )t x Dβ

α

. Then the problem

(2.13) has at least one C[0,1] solution x t( ) satisfies α( )tx t( )≤β( )t ,

[0,1]

t∈ .

Remark 2.1: This result can be found in [15]. So our theorem improves the works in the previous literature.

Lemma 2.3. Suppose that f : (0,1) [0,× +∞ →) is a continuous functions such that s f t s−1 ( , ) is strictly decreasing for s>0 at each t(0,1). Let

2

, [0,1] (0,1)

w v C∈ ∩C satisfies:

1) w′′+ f t w( , ) 0≤ ≤ +v′′ f t v( , ), t∈(0,1);

2) w v, >0, t∈(0,1) and w(0)≥v(0), w(1)≥aw( )η , v(1)≤av( )η ;

3) v′′∈L1[0,1].

Then w t( )≥v t( ), t∈[0,1].

Proof. By v′′∈L1(0,1), we know that v′ +(0 ) and v′ −(1 ) exist and then 1[0,1]

v C∈ .

Suppose conversely v t( )≤/w t( ) on [0,1]. We may assume without loss of ge-nerality that there exists t0∈(0,1) such that v t( )0 −w t( ) max( ( )0 = 0≤ ≤t 1 v tw t( )) 0> .

Let

* inf{ | 01 1 0, ( ) ( ), ( , )},1 0

t = t ≤ <t t v t >w t tt t

*

2 0 2 0 2

sup{ | 1, ( ) ( ), ( , )}.

t = t t ≤ <t v t >w t tt t

It’s obvious that *

*

0≤ < ≤t t 1 and v t( )* =w t( )* , v t′ + ≥( )* D w t+ ( )*+ , where

D+ denote Dini derivatives.

For t*1, there are three cases.

1) t*<1. Then v t( )* =w t( )* , v t( )* w t( )* , v t( )>w t( ) for all * *

( , )

tt t .

2) t*=1 and v t( )* =w t( )* , v t(* ) D w t(* )

′ − ≤ − , v t( )>w t( ) for all

* *

( , )

tt t , where D− denotes Dini derivatives.

3) t*=1 and v t( )* >w t( )* , v t( )>w t( ) for all * *

( , ]

tt t . Since

(1) (1) ( ( ) ( )) ( ) ( )

vwa vη −wη <vη −wη , then there is t′∈[ ,1]η such that

( ) ( ) 0, ( ( ) ( )) 0.

v t w t > v t w t<

Combining above (1), (2) and (3), there is a t′ >t* such that

* * * *

( ) ( ), ( ) ( ), ( ) ( ), ( ) ( ),

v t w t v t D w t+ v t w t v t D w t

′ ′ ′ ′ ′ ′

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and

*

( ) ( ), ( , ).

v t >w t ∀ ∈t t t

Let y t( )=v t w t w t v t′( ) ( )− ′( ) ( ), t∈( , )t t*. Then we have

*

lim inf ( ) 0 lim sup ( ).

t t→ + y t ≥ ≥t t→ −′ y t (2.15)

On the other hand,

( ) ( ) ( ) ( ) ( )

( ) ( , ( ) ( ) ( , ( ))

( , ( )) ( , ( ))

( ) ( )( )

( ) ( )

0

y t w t v t w t v t

w t f t v t v t f t w t

f t w t f t v t

w t v t

w t v t

′ = ′′ − ′′

= − +

= −

for t∈( , )t t* ′ and y t′( ) 0≡/ on ( , )α β . This implies y t( )′ > y t( )* . This

con-tradicts (2.15), so v t( )≤w t( ). The proof is complete.

By analogous methods in [19], we establish the following maximal theorem, which can be used in the proof of the uniqueness of positive solutions.

Lemma 2.4. (maximal theorem) Suppose that 0< <η 1, and

2

{ [0,1] (0,1), (1) ( ) 0, (0) 0}

F = x C∈ ∩C xaxη ≥ x ≥ , if x t( )∈F such that

''( ) 0 x t

− ≥ for t∈(0,1), then x t( ) 0≥ for t∈[0,1].

3. Proofs of Main Theorems

In this section, we’ll always assume that f t x( , )=λxpK t x( ) −q.

(A) The proof of Theorem 1.1. Proof.

1) We consider the problem

( ) ( ) ( ) ( ), (0,1),

(0) 0, (1) ( ),

q p

x t K t x t x t t

x x ax

λ

η

 ′′− + = ∈

= =

 (3.1.1)

where 0<q p, <1, K C∈ [0,1], K*>0, 0< <a 1, 0< <η 1 and λ is a

positive parameter.

In [19], when f t x( , ) is increasing in x, the problem

( ) ( , ), (0,1),

(0) ( ), (1) 0

x t f t x t x axη x

′′

− = ∈

= =

has an unique C1[0,1] positive solution. From that, suppose that *( )

x t is an unique C1[0,1] positive solution of the problem

( ) ( ), (0,1),

(0) 0, (1) ( ),

p

x t x t t

x x ax

η

 ′′− = ∈

= =

 (3.1.2)

where 0< <a 1, 0< <η 1. Set

1 1

*

( )t px t( )

β

=

λ

. Then

1

1 1

* *

1

1 1

* * *

1 1

*

( ) ( ) ( ) ( ) ( ) ( )

( ) ( )

( ),

q

q p p q

q q

p p

p p

t K t t x t K t x t

x t K x t

x t

β β λ λ

λ λ

λ

− − − −

− −

− −

− ′′

− + = +

> +

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1 *

( ) ( ).

p t px tp

λβ

=

λ

Thus β′′( )t +K t( )βq( )t >λβp( ).t Combining it with (3.1.2) we obtain

( ) ( ) ( ) ( ), (0,1),

(0) 0, (1) ( ).

q p

t K t t t t

a

β

β

λβ

β

β

β η

− ′′ + > ∈

= =

Consequently, β( )t is a upper solution of (3.1.1). Set

2 1 1

( )t M q

α

=

ϕ

+ , where M is a positive constant and 1

ϕ is the first eigen-function. Then 1 1 2 1 2 2 1 1 2 1 2 1 2 1

1 1 2

2 1 1 1

2 1

1 1 2

2 1 1 2 * 1 1 2 2 1 1 1

2(1 ) | ' |

2 ( )

''( ) ( ) ( ) ( ) ''( )

1

(1 )

2(1 ) | ' |

2 ( )

1

(1 )

2(1 ) | ' |

2 . (1 ) q q q q q q q q q q q q q q q q q q q q q M

M K t

t K t t t t

q

q M

q M

M K t

q M q q M M K M q ϕ

α α ϕ ϕ

ϕ ϕ

ϕ

λ ϕ ϕ

ϕ ϕ λ ϕ ϕ ϕ − + + + + + + − + + + − − + = − + − + + − = + − + + −

< + −

+

By Lemma 2.1 we have

ϕ

1( ) sin(t =

λ

1t),

ϕ

1( )t =

λ

1cos(

λ

1t). Thus there

exists δ >0 0 and b∈(0,1) such that

1' 1 1 0

| ( ) | |

ϕ

t =

λ

cos(

λ

t) |>

δ

, t∈[0, ),b

1 1 0

| ( ) | | sin(

ϕ

t =

λ

t) |>

δ

, t b∈[ ,1].

a) On [0, )b , choosing

1 2 * 1 1 2 0 (1 ) [ ] 2(1 ) q q K M M q

δ

+ + ≥ =

− , then we have

2 * 1 1 1 2 1 1 1 q q q q M K q M λ ϕ ϕ + + ≤ + .

b) On [ ,1]b , choosing

1 2 * 1 2 2 0 (1 ) [ ] 2(1 ) q q K M M q

δ

+ + ≥ =

− , then we have

2 * 1 1 1 2 1 1 . 1 q q q q M K q M λ ϕ ϕ + + ≤ +

Fixing M=max{ ,M M1 2}, then

2 1 1 1 3 ( ) ( ) ( ) 1

q M q

t K t t

q

λ

α

′′

α

ϕ

+

− + ≤ + and 2 1 1 ( ) . q

p t Mp q

λα

=

λ

ϕ

+

Set

2 2 1

1 0 31 | |1

p q q M q

λ

ϕ

− − + ∞ =

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2 2

1 1

1

1 1 0

3 , .

1

p p

q q

M M

q

λ

ϕ

+ <

λ

ϕ

+ ∀ >

λ λ

+

Hence, α′′( )t +K t( )αq( )t <λαp( )t ,

0 λ λ

∀ > .

It follows from Lemma (2.1) that

2 1 1

(0) M q(0) 0

α

=

ϕ

+ =

and

2 2 2 2 2

1 1 1 1 1

1 1 1 1

(1) M q(1) M a[ ( )] q Ma q q( ) aM q( ) a ( ).

α

=

ϕ

+ =

ϕ η

+ = +

ϕ

+

η

<

ϕ

+

η

=

α η

Set

1 1 1 1

2 1

*

( | | | | )

q p q M

x

ϕ

λ

ϕ

− − +

∞ ∞

= . Then 12 11

1 *

( )t M q( )t px t( ) ( )t

α

=

ϕ

+

λ

=

β

for

all λ λ> 2. Thus we choose λ =max{ , }λ λ0 2 and λ λ> , then ( ( ), ( ))α t β t is

a couple of upper and lower solutions of (3.1.1).

We choose F t( )=λβp+K*βq, then | ( , ) |f t x F t( ) for all ( , )t x Dβ α

∈ . It’s easy to see that F t( )L1[0,1]. From Corollary 2.1, the problem (3.1.1) has at

least one C[0,1] positive solution x t( ) satisfying α( )tx t( )≤β( )t for λ λ> . 2) (Existence of the maximal solution) We observe the problem

( ) ( ), (0,1),

(0) 0, (1) ( ).

p

x t x t t

x x ax

λ

η

 ′′− = ∈

= =

 (3.1.3)

From [19], we note the unique solution of (3.1.3) is w tλ( ) for any λ>0. In (1) we obtained the solution x tλ( ) of (3.1.1) then we have

( ) ( ) 0 ( ) (

' ' )

' p ' p

w tλ +λw tλ = <x tλ +λx tλ

and x f t x1 ( , ) xp 1( )t λ λ

= is decreasing in x. Noting that x t( ) L1[0,1]

λ ∈ by (1).

From Lemma 2.3, we have x tλ( )≤w tλ( ). Let

0

1

[ ,1)

j i j

Ω =

+ , j=1, 2, and w tj( ) be the solution of

1 1

1

0

( ) ( ) ( ) ( ), ,

1

( ) ( ), [0, ),

(1) ( )

q p

j j j

j

x t K t w t w t t

x t w t t

i j

x ax

λ

η −

− −

 ′′− + = ∈ Ω

=

+

=

(3.1.4)

for j=1, 2,, with w t0( )=w tλ( ) defined in (3.1.3). Let x tλ( ) be a solution

of (3.1.1).

In (3.1.4), letting j=1 we have

1 1

1

0

1 1

( ) ( ) ( ) ( ), ,

1

( ) ( ), [0, ),

(1) ( ).

q p

w t K t w t w t t

w t w t t

i j

w aw

λ λ

λ

λ

η −

− ′′ + = ∈ Ω

=

+

=

(3.1.5)

Combining (3.1.5) with (3.1.3) we have w t w t1''( )− λ''( ) 0≥ for t∈ Ω1. By

maximum principle, we have w t1( )≤w t0( )=w tλ( ). Similarly, we can obtain that

1( ) ( ) ( )

j j

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Furthermore, we observe problem (3.1.1)

( ) ( ) ( ) ( ), (0,1),

(0) 0, (1) ( ).

q p

x t K t x t x t t

x x ax

λ

η

 ′′− + = ∈

= =

Combining it with (3.1.5) we have

1''( ) ''( ) ( )( q( ) q( )) ( ( )p p( )) 0,

w t x tλ K t w tλ− x tλ− λ w tλ x tλ

− + + − = − ≥

thus x t w tλ''( )− 1''( )≥0 for t∈ Ω1. It’s easy to verify that x tλ( )≤w t1( ) for [0,1]

t∈ by maximum principle. By similar method we can obtain

1

( ) j ( ) j( ) ( )

x tλ ≤w+ tw tw tλ for t∈[0,1].

Furthermore, we have { ( )}w tj j N∈ is bounded from below by x tλ( ).

Because w tj( ) is a solution to (3.1.3),

1 1

1 * 1

1 * 1

1 *

( ) ( ) ( ) ( )

( ) ( )

[ ( ) ] ( )

[ ( )

''

] ( ).

p q

j j j

p q

j j

p q q

j j

p q q

j j

w t w t K t w t

w t K w t

w t K w t

w t K w t

λ λ λ λ

− −

− −

+ −

− −

+ −

− = −

≤ −

≤ −

≤ −

Suppose that t0∈(0,1), w tj( ) max ( )0 = 0≤ ≤t 1w tj , then w tj'( )0 =0 and w tj( ) is

increasing on ( , )t t0 . By integration of −w tj''( ) from t to t0, we have

0 0

1 *

''( ) [ ( ) ] ( )

t t p q q

j j j

t w s ds t λw s K w s ds

+ −

− ≤ −

.

So w t w tj'( ) ( )qj

λ

wjp q−+1( )t0 −K*. Similarly, by integration of −w tj''( ) from

0

t to t, we can obtain |w t w tj'( ) ( ) |j

λ

wjp q−+1( )t0 −K*. For giving t t1 2, ∈[0,1],

we have

2 2 2

1 1 1

'

1 0 *

'( ) ( ) | ( ) ( ) | [ ( ) ] .

t q t q t p q

j j j j j

t w s w s ds t w s w s ds t

λ

w t K ds

+ −

≤ ≤ −

We can find K large such that |

λ

wp qj−+1 ( )t0 −K*|<K. Then

2

1 '( ) ( ) | 2 1|

t q

j j

t w s w s ds K t t≤ −

, 1 1

2 1 2 1

| q ( ) q ( ) | | | .

j j

w + t w + t K t t

(3.1.4)

We define an operator I w( )=wq+1, then

1 1( ) q 1

I w=w + . It follows from

(3.1.4) that { ( ( ))}I w tj j N∈ is a uniformly bounded and equicontinuous

func-tions in [0,1]. Obviously, I−1 is uniformly continuous in a bounded and closed

domain Ω, i.e., for all ε >0, there exists a δ >0 such that when w1,

2

w ∈ Ω , |w w12|<δ , we have 1 1

1 2

|I w− ( )I w− ( ) |<ε . Since 0

0<w tj( )<w t( ), there exists a M >0 such that w tj( ) (0, ]∈ M . From (3.1.4),

for the above δ >0, there exists δ′ >0 such that when |t t1− <2| δ ′, we have

1 1

2 1

| q ( ) q ( ) |

j j

w + t w+ t <

δ

.

Therefore, for all ε >0, there exists δ′ >0 such that when |t t1− <2| δ ′, we

have

1 1 1 1

2 1 2 1

| ( ) ( ) | | ( q ( )) ( q ( )) |

j j j j

w t w t = I w− + t I w− + t <

ε

.

Thus { ( )}w tj j N∈ is equicontinuous. Using Arzela-Ascoli theorem, there

ex-ists a subsequence {w tjk( )}jk∈{ }i such that jklim→+∞w tjk( )=x tλ( ). Without loss of

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DOI: 10.4236/jamp.2018.612217 2612 Journal of Applied Mathematics and Physics

lim j( ) ( ), [0,1].

j→+∞w t =x tλ t∈ (3.1.5)

In the following, we shall show that x tλ( ) is a C[0,1] positive solution of (3.1.1).

Fixing (0,1)( 1)

2

tt≠ , then w tj( ) can be stated

1 1 1

2

1 1 1

( ) ( ) ( )( ) ( )[ ( ) ( ) ( )] .

2 ' 2 2

t q p

j j j j j

w t w w t s t K s ws λw s ds

− −

= + − +

− − (3.1.6)

Fixing j N∈ , by Lagrange mean value theorem, there exists ( ,1)1 2

n

t ∈ such

that xλ(1)−wj( )12wj(1)−wj( )21 =w tj′( )(1n21)<w0(1).

So there exists M1>0 such that |w tj′( ) | 2n < M1 . Since { ( )}w tj j N∈ is

bounded in [0,1], we may assume that m w t< j( )<M2, t∈[ , ]12 tn ,

''

1 1 1 1

2 2

1 1 * 1

2

*

| ( ) | | [ ( ) ( ) ( )] |

| [ ( ) ( )] |

.

n n

n

t t p q

j j j

t p q

j j

p q

w s ds w s K s w s ds

w s K w s ds

M K m

λ

λ

λ

− −

− −

− = −

≤ −

≤ −

Thus

* 2

1 1

| '( ) | | '( ) | | '( ) '( ) |

2 2 p q

j j n j j n

w w t w w t λM K m

i.e.,

*

1 2

1

| ( ) | 2 .

2

' p q

j

w M +λM K m

Thus both { ( )}1 2 '

j j N

w ∈ and { ( )}wj 12 j N∈ are bounded. Then they all have a

convergence subsequence. Without loss of generality, we note the subsequences are 1

{ ( )} 2

j j N

w ∈ and{wj'( )}12 j N∈ . And fixing j N∈ , we assume limj→∞wj' 1( )2 =r0.

In equation (3.1.6), letting j→ ∞ we have

1 0

2

1 1

( ) ( ) ( ) ( )[ ( ) ( ) ( )]

2 2

t q p

x tλ =xλ +r t− +

s t K s x− λ− s −λx s dsλ

for t∈(0,1), i.e., x t''( ) K t x t( ) q( ) x tp( )

λ λ− λ λ

− + = . Therefore x tλ( ) is a C[0,1]

positive solution of (3.1.1). Therefore x tλ( ) is the maximal solution of (3.1.1). Next we shall verify the dependence on λ of maximal solution x tλ( ). Let H = {µ>0: (3.1.1) has a C[0,1] positive solution with λ µ= }.

Obviously, by (1), H ≠ ∅. Let λ ∈1 H. and x tλ( ) be the corresponding maximal solution of (3.1.1) for λ λ= 1 . Then for any λ2 >λ1>λ ,

1''( ) 1 1( ) 0, (0,1)

p

xλ tx tλ ≥ t∈ . By Lemma (2.3), x tλ1( )≤w tλ2( ) in [0,1]. Just

re-placing x tλ( ) by x tλ1( ) in above proof. We can easily find that

1 1 1 1

2 2 2

''

1 2

''

2

( ) ( ) ( ) , (0,1),

( ) ( ) ( ) ( ).

q p p

q p

x t K t x t x x t w t K t w t w t

λ λ λ λ

λ λ λ

λ λ

λ

− −

− + = ≤ ∈

 

− + ≥

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Combining it with boundary conditions, we can obtain that ( ( ),x t w tλ1 λ2( )) is a couple of lower and upper solutions of (3.1.1) for λ λ= 2 >λ1. One can be

prove that there is a solution x tλ2( ) of (3.1.1) with λ λ= 2 such that

1( ) 2( ) 2( ).

x tλ ≤x tλ ≤w tλ

Therefore λ ∈2 H. Moreover, by (ii), for any λ2 >λ1≥λ, x tλ2( )≥x tλ1( ). This completes the proof of Theorem 1.1. 

(B) The proof of Theorem 1.2. Proof. 1) We consider the problem

''( ) ( ) ( ) ( ), (0,1),

(0) 0, (1) ( ),

q p

x t K t x t x t t

x x ax

λ

η

− + = ∈

= =

 (3.2.1)

where q>0,0< p<1, K t( )∈C[0,1], K*<0, 0< <a 1, 0< <η 1 and λ

is a positive parameter.

Now we consider an approximate problem of (3.2.1) as follows

''( ) ( ) ( ) ( ), (0,1),

1 1

(0) , (1) ( ) ,

q p

x t K t x t x t t

x x ax

n n

λ

η −

− + = ∈

 

= = +

 (3.2.2)

where 0< <a 1, 0< <η 1, n≥1.

Let

ε

very small. We’ll verify that n( )t 1( )t 1 n

α =εϕ + is a lower solution of

(3.2.2). Indeed, when n is big enough, we can obtain that εϕ1( )t +1n is close to

0. Since 1 (0, 2) π

λ ∈ (see [6]), we can deduce

1 1 1 1

1 1 1

1

1 1 1

( ) ( ) ( ) ( )

1 1

( ) ( )( ( ) ) ( (

'

) )

1

( ) ( ( ) )

1

( )[ ( ( ) ) ]

0,

' q p

n n n

q p

p

p

t K t t t

t K t t t

n n

t t

n

t t

n

α α λα

λ εϕ εϕ λ εϕ

λ εϕ λ εϕ

εϕ λ λ εϕ

− + −

= + + − +

< − +

< − +

<

1

1

(0) (0) 0

n n

α − =εϕ =

and αn(1) [− aα ηn( )+n1]=ε ϕ ηa 1( )+ −1n aεϕ η1( )− − <an n1 0, which imply that

( ) n t

α is a lower solutions of (3.2.2).

In the following, we’ll construct an upper solution of (3.2.2). Let

2

( )t Mt (M aM t M)

β = − + + + ,

where M is big enough for

1

1 1

{(2 ) , }

(1 )

p M

n a

λ

− >

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*

''( ) ( ) ( ) 2 ( )[ ( ) ]

2 ,

q q

q

t K t t M K t Mt M aM t M M K M

M

β β− −

− + = + − + + +

> +

>

2

2

( ) [ ( ) ]

(1 )

[ ]

4 (2 ) ,

p p

p

p

t Mt M aM t M M a M

M

λβ λ

λ

λ

= − + + +

+

< +

<

( )t K t( ) q( )t p( ),t

β′′ β− λβ

− + ≥

2

1 1

(1) ( ( ) ) ( 1) [ ( ) ]

1

( 1) 2

1

0

a a M a M M aM M

n n

a M aM n M aM

n

β − β η + = + − − η + + η+ −

> + − −

= − −

>

and (0) 1 M 1 0

n n

β − = − > . It’s easy to see that β( )t is na upper solution of

(3.2.2).

Choosing F tn( )=λβpKnq, then | ( , ) |f t xF tn( ), for all ( , )t x Dn β α

∈ . It’s

easy to verify that ( ) 1[0,1]

n

F tL . Because that

ε

is small and n is big enough,

( ) ( )

n t t

α ≤β . From Corollary 2.1, ( ( ), ( ))αn t β t is a couple of upper and lower solutions of (3.2.2). And for all n N, (3.2.2) has at least one C[0,1] positive solution x tn( ) such that αn( )tx tn( )≤β( )t .

In the following, we shall obtain a result as follows, there exists a subsequence { ( )}x tnk and x t( ) such that lim k( ) ( )

k n

n→∞x t =x t .

Since β( )t C[0,1]C2(0,1), β( )t is bounded. Therefore { ( )}

n n N

x t ∈ is a

uniformly bounded sequence of functions in [0,1]. Because x tn( ) is a C[0,1] positive solution of (3.2.2), x tn( ) satisfies

*

*

( ) ( ) ( ) ( )

( ) ( )

[ ( ) ] (

''

).

p q

n n n

p q

n n

p q q

n n

x t x t K t x t

x t K x t

x t K x t

λ λ λ

+ −

− = −

≤ −

≤ −

Suppose that t0∈(0,1),

0 1

0

( ) max ( )

t

n n

x t x t

≤ ≤

= , then x tn'( )0 =0 and x tn( ) is increasing on ( , )t t0 . By integration of −x tn''( ) from t to t0, we have

0 0

*

''( ) [ ( ) ] ( ) .

t t p q q

n n n

t x s ds t λx s K x s ds

+ −

− ≤ −

So n'( ) q1( )[ np q( )0 *]

n

x t x t K

x t λ

+

≤ − . We can find a K>0 such that

) ( '( q )

n n

x t x tK. And by integration of −x tn( ) from t0 to t, we have

0 0

*

''( ) [ ( ) ] ( ) .

t t p q q

n n n

t x s ds t λx s K x s ds

+ −

− ≤ −

So n'( ) q1()[ np q( )0 *]

n

x t x t K

x t λ

+

− ≤ − . For above K, we have | '( ) ( )q |

n n

x t x t K

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DOI: 10.4236/jamp.2018.612217 2615 Journal of Applied Mathematics and Physics i.e., | ' ) ( )( q |

n n

x t x tK.

For giving t t1 2, ∈[0,1], we have

2 2 2

1 '( ) ( ) 1| ( ) ( ) |' 1 .

t q t q t

n n n n

t x s x s dst x s x s dst Kds

Then 2

1 '( ) ( ) | 2 1|

t q

n n

t x s x s ds K t t≤ −

. The above inequality can be rewritten as

2

1

( ) 1 1

2 1 2 1 2 1

( )

| n ( ) ( ) | | |, | ( ) ( ) | | | .

n

x t q q q

n n n n

x t x s dx s K t t x t x t K t t

+ +

≤ − − ≤ −

(3.2.3)

We now define an operator I x( )=xq+1, then

1 1( ) q1

I x=x + . It follows from

(3.2.3) that { ( ( ))}I x tn n N∈ is a uniformly bounded and equicontinuous

func-tions in [0,1]. Obviously, I−1 is uniformly continuous in a bounded and closed

domain Ω , i.e., for all ε >0 , there exists a δ >0 such that

1 1

1 2

|I x−( )I x− ( ) |<ε for

1 2

|x x− |<δ , x x1, 2∈ Ω . Since 0<x tn( )<β( )t , there exists a M>0 such that x tn( ) (0, ]∈ M . From (3.2.3), for the above

0

δ > , there exists δ′ >0 such that 1 1

2 1

| q ( ) q ( ) |

n n

x + t x+ t <δ for 1 2

|t t− <| δ ′.

Therefore, for all ε >0, there exists δ′ >0 such that

1 1 1 1

2 1 2 1

| ( ) ( ) | | ( q ( )) ( q ( )) |

n n n n

x t x t = I x− + t I x− + t <ε

for |t t1− <2| δ ′. Consequently, { ( )}x tn n N∈ is equicontinuous. Using

Arze-la-Ascoli theorem, there exists a subsequence { ( )}x tnk such that nklim→+∞x tnk( )=x t( ).

Without loss of generality, we assume that

lim ( )n ( ), [0,1].

n→+∞x t =x t t∈ (3.2.4)

In the following, we shall show that x t( ) is a C[0,1] positive solution of (3.2.1). Fixing (0,1)( 1)

2

tt≠ , x tn( ) can be stated

1 2

1 1 1

( ) ( ) '( )( ) ( )[ ( ) ( ) ( )] .

2 2 2

t q p

n n n n n

x t =x +x t +

s t K s x sλx s ds (3.2.5)

Fixing n N, by Lagrange mean value theorem, there exists ( ,1)1 2

n

t ∈ such

that (1) ( )1 (1) ( )1 '( )(1 1) (1)

2 2 2

n xn xn xn x tn n

α − ≤ − = − ≤β .

So there exists M1>0 such that |x tn'(n) |≤2M1. Since { ( )}x tn n N∈ is

bounded in [0,1], we may assume that m x tn( )≤M2, t∈[ , ]12 tn .

''

1 1

2 2

| tn ( ) | | [tn p( ) ( ) q( )] | .

n n n

x s ds λx s K s x s ds

− = −

We can obtain

2 * 2

1

| ( ) '( ) |

2

' p q

n n n

x t x λM K M

− + ≤ − and |xn'( ) | 212M1+λM2pK M* 2−q.

Therefore both { ( )}1 2

n n N

x ∈ and { ( )}xn' 12 n N∈ are bounded. They all have a

convergence subsequence. Without loss of generality, we note the subsequences are { ( )}1

2

n n N

(17)

DOI: 10.4236/jamp.2018.612217 2616 Journal of Applied Mathematics and Physics

From (3.2.5), letting n→ ∞, we obtain

1 0

2

1 1

( ) ( ) ( ) ( )[ ( ) ( ) ( )] .

2 2

t q p

x t =x +r t +

s t K s x sλx s ds

By derivation twice of x t( ), we have

( ) ( ) q( ) p( ).

x t′′ K t x t− λx t

− + =

Combining it with (3.2.4), we can obtain that x t( ) is a C[0,1] positive solu-tion of (3.2.1).

2) We study the uniqueness of C1[0,1] positive solution of problem (3.2.1).

Let F t( )=λβpK*(εϕ1)−q. Obviously, when 0< <q 1, F t( ) is integrable

over (0,1). Since | ( ) |x t′′ ≤F t( ), x t( ) is absolutely integrable over (0,1). Then both x′ +(0 ) and x′ −(1 ) exist, i.e., x t( )C1[0,1].

Suppose conversely that x t1( ), x t2( ) are two C1[0,1] positive solutions of

the problem (3.2.1), x t1( )≡/x t2( ) on [0,1]. We may assume without loss of

ge-nerality that there exists t*(0,1) such that

* *

2( ) 1( ) max( ( )0t1 2 1( )) 0

x t x t x t x t

≤ ≤

− = − > . Let

* *

1 1 2 1 1

inf{ | 0t t t x t, ( ) x t t( ), ( , )},t t

α= ≤ < > ∈

* *

2 2 2 1 2

sup{ |t t t 1, ( )x t x t t( ), ( , )}.t t

β = ≤ < > ∈

It’s obvious that 0≤α β< ≤1 and

1 2 1 2 1 2

1 2 1 2

( ) ( ), ( ) ( ), ( ) ( ),

( ) ( ), ( ) ( ),

' '

' ' ( , ).

x x x x x x x x x t x t t

α α α α β β

β β α β

= ≤ ≤

+ ≥ + < ∈

Let y t( )=x t x t1( ) ( )2' −x t x t t2( ) ( ),1' ∈( , )α β . Then we have

lim inf ( ) 0 lim sup ( ).

t→ +α y t ≥ ≥t→ +β y t (3.2.6)

On the other hand,

'' '' 1 2 2 1

1 2 2 2 1 1

1 2 1 2 1 2 1 2

1 1 1 1

1 2 2 1 1 2 1 2

'( )

( ) ( )

( ) ( )

0

q p p q

q p p q

q q p p

y t x x x x

x Kx x x x Kx

Kx x x x x x Kx x

Kx x x x x x x x

λ λ

λ λ

λ

− −

− −

− − − − − −

= −

= − + −

= − + −

= − + −

for t∈( , )α β and y t′( ) 0≡/ on ( , )α β . This implies y( )β− > y( )α+ ,

contra-dicts (3.2.6), so x t1( )≡x t2( ). Thus the C1[0,1] positive solution of (3.2.6) is

unique.

3) We assume that 0<λ1<λ2 and x tλ1( ), x tλ2( ) are the corresponding unique C1[0,1] positive solutions to (3.2.1). Obviously,

1

1

( ) [

'' 0, ]1

xλ tL . In

(3.2.1), f t x( , )=λx tp( )K t x t( ) −q( ) is continuous. Since p q, ∈(0,1), K*<0, it’s easy to see that

1 ( , ) p1( ) ( ) q1( )

x f t x=λxt K t x− − t is decreasing for x>0 at each t[0,1].

2''( ) ( ) 2( ) 2 2( ) 0 1''( ) ( ) 1( ) 2 1()

q p q p

xλ tK t x tλ− +λ x tλ = <xλ tK t x tλ− +λ x tλ

for t∈(0,1),

2(0) 1(0)

References

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