Research

# Existence results of positive solutions for boundary value problems of fractional differential equations

Guoqing Chai

Author Affiliations

College of Mathematics and Statistics, Hubei Normal University, Hubei, 435002, P.R. China

Boundary Value Problems 2013, 2013:109  doi:10.1186/1687-2770-2013-109

 Received: 15 October 2012 Accepted: 15 April 2013 Published: 29 April 2013

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 are concerned with the following fractional equation:

D 0 + α C u ( t ) = f ( t , u ( t ) , u ( t ) ) , t ( 0 , 1 )

with the boundary value conditions

u ( 1 ) = u ( 1 ) = 0 , δ u ( 0 ) = u ( 1 ) , γ u ( 0 ) = u ( 1 ) ,

where D 0 + α C is the standard Caputo derivative with 3 < α 4 and δ, γ are constants with δ > 1 , γ > 1 . By applying a new fixed point theorem on cone and Krasnoselskii’s fixed point theorem, some existence results of positive solution are obtained.

MSC: 34A08, 34B15, 34B18.

##### Keywords:
fractional differential equations; existence results; fixed point theorem; positive solution

### 1 Introduction

In this paper, we are concerned with the existence of positive solutions for the fractional equation

D 0 + α C u ( t ) = f ( t , u ( t ) , u ( t ) ) , t ( 0 , 1 ) (1.1)

with the boundary value conditions

u ( 1 ) = u ( 1 ) = 0 , δ u ( 0 ) = u ( 1 ) , γ u ( 0 ) = u ( 1 ) , (1.2)

where D 0 + α C is the standard Caputo derivative with 3 < α 4 and δ, γ are constants with δ > 1 , γ > 1 .

Differential equations of fractional order have recently proved to be valuable tools in the modeling of many phenomena in various fields of science and engineering. Indeed, we can find numerous applications in viscoelasticity, electrochemistry, control, porous media, electromagnetism, etc. (see [1-5]). There has been a significant development in the study of fractional differential equations and inclusions in recent years, see the monographs of Podlubny [5], Kilbas et al.[6], Lakshmikantham et al.[7], Samko et al.[8], Diethelm [9], and the survey by Agarwal et al.[10]. For some recent contributions on fractional differential equations, see [9-25] and the references therein.

On the other hand, it is well known that the fourth-order boundary value problem describes the deformations of an elastic beam in equilibrium state. Owing to its importance in physics, the existence of solutions to this problem has been studied by many authors; see, for example, [26-30] and references therein. Recently, there have been a few papers dealing with the existence of solutions for fractional equations of order α ( 3 , 4 ] .

In [14], Xu et al. discussed the problem

D 0 + α u ( t ) = f ( t , u ( t ) ) , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = u ( 1 ) = u ( 1 ) = 0 ,

where 3 < α 4 and f C ( [ 0 , 1 ] × ( 0 , ) ) is nonnegative, D 0 + α is the Riemann-Liouville fractional derivative of order α. The existence results of positive solutions are obtained by applying the Leray-Schauder nonlinear alternative theorem.

In [15], Liang and Zhang studied the following nonlinear fractional boundary value problem:

D 0 + α u ( t ) = f ( t , u ( t ) ) , t ( 0 , 1 ) , u ( 0 ) = u ( 0 ) = u ( 0 ) = u ( 1 ) = 0 ,

where 3 < α 4 , f ( t , u ) C ( [ 0 , 1 ] × [ 0 , ) , [ 0 , ) ) is nondecreasing relative to u, D 0 + α is the Riemann-Liouville fractional derivative of order α. By means of the lower and upper solution method and fixed point theorems, some results on the existence of positive solutions were obtained.

In [16], Agarwal and Ahmad studied the solvability of the following anti-periodic boundary value problem for a nonlinear fractional differential equation:

{ D 0 + α u ( t ) = f ( t , u ( t ) ) , t ( 0 , T ) , u ( 0 ) + u ( T ) = 0 , u ( 0 ) + u ( T ) = 0 , u ( 0 ) + u ( T ) = 0 , u ( 0 ) + u ( T ) = 0 ,

where 3 < α 4 . The existence results were obtained by the nonlinear alternative theorem.

Inspired by above work, the author will be concerned with the boundary value problem (BVP for short in the sequel) (1.1)-(1.2). To the best of our knowledge, no contribution exists concerning the existence of solutions for BVP (1.1)-(1.2). In the present paper, by applying a new fixed point theorem on cone and Krasnoselskii’s fixed point theorem, some existence results of positive solution for BVP (1.1)-(1.2) are obtained. It is worth to point out that the results in this paper are also new even for α = 4 relative to the corresponding literature with regard to the fourth-order boundary value problem. In addition, the conditions imposed in this paper are easily verified.

The organization of this paper is as follows. In Section 2, we present some necessary definitions and preliminary results that will be used to prove our main results. In Section 3, we put forward and prove our main results. Finally, we give two examples to demonstrate our main results.

### 2 Preliminaries

In this section, we introduce some preliminary facts which are useful throughout this paper.

Let ℕ be the set of positive integers, ℝ be the set of real numbers, R + = [ 0 , + ) , and R = ( , 0 ] . Let I = [ 0 , 1 ] . Denote by C 1 ( I , R ) the Banach space endowed with the norm u 1 = u 0 + u 0 , where u 0 = max t I | u ( t ) | for u C ( I , R ) .

Definition 2.1[6]

The Riemann-Liouville fractional integral of order α > 0 of a function y : ( a , b ] R is given by

I a + α y ( t ) = 1 Γ ( α ) 0 t ( t s ) α 1 y ( s ) d s , t ( a , b ] .

Definition 2.2[6]

The Riemann-Liouville fractional derivative of order α > 0 of a function y : ( a , b ] R is given by

D a + α y ( t ) = 1 Γ ( n α ) ( d d t ) n 0 t y ( s ) ( t s ) α n + 1 d s , t ( a , b ] ,

where n = [ α ] + 1 , [ α ] denotes the integer part of α.

Definition 2.3[6]

The Caputo fractional derivative of order α > 0 of a function y on ( a , b ] is defined via the above Riemann-Liouville derivatives by

( C D a + α y ) ( x ) = ( D a + α [ y ( t ) k = 0 n 1 y ( k ) ( a ) k ! ( t a ) k ] ) ( x ) , x ( a , b ] .

Lemma 2.1[6]

Let α > 0 and y C [ a , b ] . Then

( C D a + α I a + α y ) ( x ) = y ( x )

holds on [ a , b ] .

Lemma 2.2[17]

Let n N with n 2 , α ( n 1 , n ] . If y C n 1 [ a , b ] and D a + α C y C ( a , b ) , then

I a + α C D a + α y ( t ) = y ( t ) k = 0 n 1 y ( k ) ( a ) k ! ( t a ) k

holds on ( a , b ) .

For convenience, we first list some hypotheses which will be used throughout this paper.

( H 1 ) f C ( I × R + × R , R + ) .

( H 2 ) 3 < α 4 , γ > 1 , δ > 1 .

For h C [ 0 , 1 ] , consider the following BVP:

D 0 + α C u ( t ) = h ( t ) , t ( 0 , 1 ) , (2.1)

u ( 1 ) = u ( 1 ) = 0 , δ u ( 0 ) = u ( 1 ) , γ u ( 0 ) = u ( 1 ) . (2.2)

We have the following lemma, which is important in this paper.

Lemma 2.3Let ( H 2 ) hold. Then u C 3 [ 0 , 1 ] is a solution of BVP (2.1)-(2.2) iff u C [ 0 , 1 ] has the expression as follows:

u ( t ) = 0 1 G ( t , s ) h ( s ) d s , t I , (2.3)

where

G ( t , s ) = { G 1 ( t , s ) , 0 s t < 1 , G 2 ( t , s ) , 0 t s < 1 , (2.4)

and

G 1 ( t , s ) = [ ( t s ) α 1 ( 1 s ) α 1 ] Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) G 1 ( t , s ) = + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 , (2.5)

G 2 ( t , s ) = ( 1 s ) α 1 Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) G 2 ( t , s ) = + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 . (2.6)

Proof Let u C 3 [ 0 , 1 ] be a solution of (2.1)-(2.2). Then by Lemma 2.2, we have

u ( t ) = c 0 + c 1 t + c 2 t 2 + c 3 t 3 + I 0 + α h ( t ) , t I , (2.7)

and so

u ( t ) = c 1 + 2 c 2 t + 3 c 3 t 2 + I 0 + α 1 h ( t ) , t I , u ( t ) = 2 c 2 + 6 c 3 t + I 0 + α 2 h ( t ) , t I , u ( t ) = 6 c 3 + I 0 + α 3 h ( t ) , t I .

Thus, by the boundary value condition (2.2), we can obtain

c 0 + c 1 + c 2 + c 3 + I 0 + α h ( 1 ) = 0 , (2.8)

c 1 + 2 c 2 + 3 c 3 + I 0 + α 1 h ( 1 ) = 0 , (2.9)

2 ( 1 δ ) c 2 + 6 c 3 + I 0 + α 2 h ( 1 ) = 0 , (2.10)

6 ( 1 γ ) c 3 + I 0 + α 3 h ( 1 ) = 0 . (2.11)

From (2.11), we have

c 3 = 1 6 ( γ 1 ) I 0 + α 3 h ( 1 ) . (2.12)

Substituting (2.12) into (2.10), we get

c 2 = 1 2 ( γ 1 ) ( δ 1 ) I 0 + α 3 h ( 1 ) + 1 2 ( δ 1 ) I 0 + α 2 h ( 1 ) . (2.13)

So, by (2.13), (2.12), and (2.9), we have

c 1 = 1 + δ 2 ( γ 1 ) ( δ 1 ) I 0 + α 3 h ( 1 ) 1 δ 1 I 0 + α 2 h ( 1 ) I 0 + α 1 h ( 1 ) . (2.14)

Thus, from (2.8), we have

c 0 = 1 + 2 δ 6 ( γ 1 ) ( δ 1 ) I 0 + α 3 h ( 1 ) + 1 2 ( δ 1 ) I 0 + α 2 h ( 1 ) + I 0 + α 1 h ( 1 ) I 0 + α h ( 1 ) . (2.15)

Hence, from (2.7) together with (2.12)-(2.15), it follows that

u ( t ) = c 0 + c 1 t + c 2 t 2 + c 3 t 3 + I 0 + α h ( t ) = I 0 + α h ( 1 ) + ( 1 t ) I 0 + α 1 h ( 1 ) + 1 2 ( δ 1 ) ( 1 t ) 2 I 0 + α 2 h ( 1 ) + [ ( 2 δ + 1 ) 3 ( 1 + δ ) t + 3 t 2 + ( δ 1 ) t 3 ] 6 ( γ 1 ) ( δ 1 ) I 0 + α 3 h ( 1 ) + I 0 + α h ( t ) . (2.16)

Noticing that

( 2 δ + 1 ) 3 ( 1 + δ ) t + 3 t 2 + ( δ 1 ) t 3 = ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] ,

by Definition 2.1, we have

u ( t ) = 0 t [ ( t s ) α 1 ( 1 s ) α 1 Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 ] h ( s ) d s + t 1 [ ( 1 s ) α 1 Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 ] h ( s ) d s = 0 1 G ( t , s ) h ( s ) d s , t I .

Conversely, if u has the expression (2.3), then from the fact that h C [ 0 , 1 ] , we can easily verify that

u ( t ) = 0 t [ ( t s ) α 2 ( 1 s ) α 2 Γ ( α 1 ) + ( t 1 ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( t 1 ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] h ( s ) d s u ( t ) = + t 1 [ ( 1 s ) α 2 Γ ( α 1 ) + ( t 1 ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( t 1 ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] h ( s ) d s , (2.17)

u ( t ) = 0 t [ ( t s ) α 3 Γ ( α 2 ) + ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + 1 + ( δ 1 ) t ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] h ( s ) d s u ( t ) = + t 1 [ ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + 1 + ( δ 1 ) t ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] h ( s ) d s , (2.18)

u ( t ) = 0 t [ ( t s ) α 4 Γ ( α 3 ) + 1 ( γ 1 ) Γ ( α 3 ) ] h ( s ) d s + t 1 1 ( γ 1 ) Γ ( α 3 ) h ( s ) d s (2.19)

hold for t I , and u satisfies the boundary condition (2.2).

Again, from (2.16) and Lemma 2.1, we have that D 0 + α C x ( t ) = h ( t ) , t ( 0 , 1 ) . In addition, noting that h C [ 0 , 1 ] , it is easy to see that x C 3 [ 0 , 1 ] from (2.19). □

For the forthcoming analysis, we need to introduce some new notations.

Let η 1 = min { 2 α 5 3 ( α 1 ) , 4 ( 1 + 2 δ ) 27 δ } , and η 2 = min { 2 ( α 2 ) 3 ( α + δ 3 ) , 1 + δ 3 δ } . Denote

w 1 ( s ) = ( 1 s ) α 2 Γ ( α 1 ) + ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) , s [ 0 , 1 ) , w 2 ( s ) = ( α + δ 3 ) ( 1 s ) α 3 ( α 2 ) ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) , s [ 0 , 1 ) .

It is easy to verify that η 1 , η 2 ( 0 , 1 ) , and w 1 , w 2 C ( [ 0 , 1 ) , ( 0 , + ) ) noting that 3 < α 4 , γ > 1 , δ > 1 .

We also need the following lemma, which will play an important role in obtaining our main results in Section 3.

Lemma 2.4Under the assumption ( H 2 ), Green’s function G ( t , s ) has the following properties:

(1) Gis continuous on [ 0 , 1 ) × [ 0 , 1 ) ;

(2) G ( t , s ) > 0 , t , s [ 0 , 1 ) ; G t ( t , s ) < 0 , t , s [ 0 , 1 ) , s t ;

(3) G ( t , s ) w 1 ( s ) , t , s [ 0 , 1 ) ; G ( t , s ) η 1 w 1 ( s ) , t [ 1 6 , 1 3 ] , s [ 0 , 1 ) ;

0 < G t ( t , s ) w 2 ( s ) , t , s [ 0 , 1 ) , s t ; G t ( t , s ) η 2 w 2 ( s ) , t [ 1 6 , 1 3 ] , s [ 0 , 1 ) , s t .

Proof (1) Observing the expression of Green’s function given by (2.4)-(2.6), the conclusion (1) of Lemma 2.4 is obvious.

(2) We first show that G t ( t , s ) < 0 , t , s [ 0 , 1 ) , s t .

In fact, if 0 s < t < 1 , then by (2.5) we have

G 1 t = [ ( t s ) α 2 ( 1 s ) α 2 ] Γ ( α 1 ) ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 . (2.20)

Owing to the fact that 3 < α 4 and 0 s < t < 1 , we have that ( t s ) α 2 < ( 1 s ) α 2 . Thus, we immediately obtain that G 1 t < 0 for 0 s < t < 1 from (2.20) together with the condition δ > 1 , γ > 1 .

Similarly, we can deduce that

G 2 t = ( 1 s ) α 2 Γ ( α 1 ) ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 (2.21)

for 0 t < s < 1 , and so G 2 t < 0 for 0 t < s < 1 .

To summarize, G t < 0 for all s , t [ 0 , 1 ) with s t .

Now, since G ( 1 , s ) = G 1 ( 1 , s ) = 0 for 0 s < 1 , and G t < 0 for t , s [ 0 , 1 ) with t s , it follows that G ( t , s ) > 0 for all t , s [ 0 , 1 ) .

(3) The proof is divided into four steps.

Step 1. We show that G ( t , s ) w 1 ( s ) for t , s [ 0 , 1 ) .

(i) If 0 s t < 1 , then by (2.5) and the assumption that 3 < α 4 , γ > 1 , and δ > 1 , we have

G 1 ( t , s ) ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 < ( 1 s ) α 2 Γ ( α 1 ) + ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = Δ w 1 ( s ) . (2.22)

(ii) If 0 t s < 1 , then by an argument similar to (2.22), we have

G 2 ( t , s ) w 1 ( s ) .

Summing up the above analysis (i)-(ii), we obtain

G ( t , s ) w 1 ( s ) for  t , s [ 0 , 1 ) .

Step 2. We show that G ( t , s ) η 1 w 1 ( s ) for t [ 1 6 , 1 3 ] and s [ 0 , 1 ) .

In fact, if 0 s t and t [ 1 6 , 1 3 ] , then by (2.5) combined with the assumption that 3 < α 4 , γ > 1 , δ > 1 , we have

G 1 ( t , s ) ( 1 s ) α 1 Γ ( α ) + ( 1 t ) ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) 2 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) 2 [ 1 + 2 δ + ( δ 1 ) t ] 6 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 ( 1 s ) α 2 Γ ( α ) + 2 ( 1 s ) α 2 3 Γ ( α 1 ) + 2 ( 1 s ) α 3 9 ( δ 1 ) Γ ( α 2 ) + 2 ( 1 + 2 δ ) ( 1 s ) α 4 27 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = ( 2 α 5 ) 3 ( α 1 ) ( 1 s ) α 2 Γ ( α 1 ) + 4 9 ( 1 s ) α 3 2 ( δ 1 ) Γ ( α 2 ) + 4 ( 1 + 2 δ ) 27 δ δ ( 1 s ) α 4 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) η 1 w 1 ( s ) , (2.23)

where η 1 = min { 2 α 5 3 ( α 1 ) , 4 9 , 4 ( 1 + 2 δ ) 27 δ } = min { 2 α 5 3 ( α 1 ) , 4 ( 1 + 2 δ ) 27 δ } , because 2 α 5 3 ( α 1 ) < 4 9 .

If t [ 1 6 , 1 3 ] with t s < 1 , then by an argument similar to (2.23), we have

G 2 ( t , s ) η 1 w 1 ( s ) . (2.24)

So, by (2.23)-(2.24), we have

G ( t , s ) η 1 w 1 ( s ) for s [ 0 , 1 ) , and t [ 1 6 , 1 3 ] .

Step 3. Now, we show that

0 < G ( t , s ) t w 2 ( s ) for  s , t [ 0 , 1 ) , s t .

(i) If 0 s < t < 1 , then by (2.20) and keeping in mind that 3 < α 4 , γ > 1 , δ > 1 , it follows that

G 1 t = ( t s ) α 2 Γ ( α 1 ) + ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 ( 1 s ) α 2 Γ ( α 1 ) + ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 3 Γ ( α 1 ) + ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = ( α + δ 3 ) ( 1 s ) α 3 ( α 2 ) ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = Δ w 2 ( s ) . (2.25)

(ii) If 0 t < s < 1 , then by an argument similar to (2.25), from (2.21), we have

G 2 ( t , s ) t = ( 1 s ) α 2 Γ ( α 1 ) + ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 w 2 ( s ) .

Summing up the above analysis (i)-(ii), and noting Step 2 of the proof as before, it follows that

0 < G ( t , s ) t w 2 ( s ) for  s , t [ 0 , 1 ) , s t .

Step 4. It remains to show that

G ( t , s ) t η 2 w 2 ( s ) for  t [ 1 6 , 1 3 ] , s [ 0 , 1 )  with  s t .

(i) If 0 s < t < 1 , then by (2.20) and the fact that ( 1 s ) α 2 > ( t s ) α 2 , we know that the relations

G 1 t > ( 1 t ) ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( 1 t ) [ 1 + δ + ( δ 1 ) t ] 2 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ( 1 s ) α 4 2 3 ( 1 s ) α 3 ( δ 1 ) Γ ( α 2 ) + ( 1 + δ ) ( 1 s ) α 4 3 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) = 2 ( α 2 ) 3 ( α + δ 3 ) ( α + δ 3 ) ( 1 s ) α 3 ( α 2 ) ( δ 1 ) Γ ( α 2 ) + 1 + δ 3 δ δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) η 2 w 2 ( s )

hold for t [ 1 6 , 1 3 ] , s [ 0 , 1 ) with s < t , where η 2 = min { 2 ( α 2 ) 3 ( α + δ 3 ) , 1 + δ 3 δ } , and

w 2 ( s ) = ( α + δ 3 ) ( 1 s ) α 3 ( α 2 ) ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) , s [ 0 , 1 ) .

Similarly, we can obtain that

G 2 t η 2 w 2 ( s ) for  t [ 1 6 , 1 3 ]  with  t < s < 1 .

The proof is complete. □

Now, we introduce a cone as follows:

P = { u C 1 [ I , R ] : u ( t ) 0 , u ( t ) 0 , t I ; u ( 1 ) = 0 ; u ( t ) η 1 u 0 , u ( t ) η 2 u 0 , t [ 1 6 , 1 3 ] } .

It is easy to check that the above set P is a cone in the space C 1 [ I , R ] , which will be used in the sequel.

We define an operator T on P as follows:

T u = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s , u P . (2.26)

Obviously, under the assumption ( H 1 )-( H 2 ), the operator T is well defined. Moreover,

( T u ) ( t ) = 0 1 G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s , t I , u P , (2.27)

where G t ( t , s ) = t G 1 ( t , s ) , 0 s < t < 1 , G t ( t , s ) = t G 2 ( t , s ) , 0 t < s < 1 , and t G 1 ( t , s ) , t G 2 ( t , s ) are given by (2.20)-(2.21), respectively.

A function u C 3 [ 0 , 1 ] is a positive solution of BVP (1.1)-(1.2) if u ( t ) 0 , t I , u 0 , and u satisfies BVP (1.1)-(1.2).

By Lemma 2.3, it is easy to know that a function u C 3 [ 0 , 1 ] is a positive solution of BVP (1.1)-(1.2) iff u P is a nonzero fixed point of T. So, we can focus on seeking the existence of a nonzero fixed point of T in P.

Finally, for the remainder of this section, we give the following two theorems, which are fundamental in the proof of our main results.

Let X be a Banach space, and let P X be a cone. Suppose that the functions α, β satisfy the following condition:

(D) α , β : X R + are continuous convex functionals satisfying α ( λ u ) = | λ | α ( u ) , β ( λ u ) = | λ | ( u ) for u X , λ R ; u k max { α ( u ) , β ( u ) } for u X , and α ( u 1 ) α ( u 2 ) for u 1 , u 2 P with u 1 u 2 , where k > 0 is a constant.

Lemma 2.5[31]

Assume that r 1 , r 2 , Lare constants with r 2 > r 1 > 0 , L > 0 , and

Ω i = { u X : α ( u ) < r i , β ( u ) < L } , i = 1 , 2   are two bounded open sets in   X .

Set D i = { u X : α ( u ) = r i } . Suppose that T : P P is a completely continuous operator satisfying

( C 1 ) α ( T u ) < r 1 , u D 1 P ; α ( T u ) > r 2 , u D 2 P ;

( C 2 ) β ( T u ) < L , u P ;

( C 3 ) there is a p ( Ω 2 P ) { 0 } such that α ( p ) 0 and α ( u + λ p ) α u for all u P and λ 0 .

ThenThas at least one fixed point in ( Ω 2 Ω ¯ 1 ) P .

Lemma 2.6[32]

Assume that Ω 1 , Ω 2 are two open subsets ofXwith 0 Ω 1 Ω ¯ 1 Ω 2 , and let T : P P be a completely continuous operator such that either

(i) T u u , u P Ω 1 ; T u u , u P Ω 2 , or

(ii) T u u , u P Ω 1 ; T u u , u P Ω 2 .

ThenThas a fixed point in P ( Ω ¯ 1 Ω 1 ) .

### 3 Main results

We first prove the following lemma to obtain our main results.

Lemma 3.1Suppose that ( H 1 )-( H 2 ) hold. Then the operatorTdefined by (2.26) mapsPintoP, andTis completely continuous.

Proof It is well known that the norms u 1 = u 0 + u 0 and u 2 = max { u 0 , u 0 } are equivalent on C 1 [ 0 , 1 ] . So, we can consider that the Banach space C 1 [ 0 , 1 ] is equipped with the norm u 2 = max { u 0 , u 0 } in the following proof.

For any u P , in view of the conclusion (1)-(2) of Lemma 2.4 and the hypotheses ( H 1 )-( H 2 ), it is easy to see that T u C 1 ( I , R ) , ( T u ) ( t ) 0 , t I , and ( T u ) ( t ) 0 , t I observing (2.26)-(2.27). Moreover, the conclusion (3) of Lemma 2.4 implies that

( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s 0 1 w 1 ( s ) f ( s , u ( s ) , u ( s ) ) d s , t I , (3.1)

and

( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s η 1 0 1 w 1 ( s ) f ( s , u ( s ) , u ( s ) ) d s (3.2)

for t [ 1 6 , 1 3 ] .

From (3.1)-(3.2), it follows that ( T u ) ( t ) η 1 ( T u ) ( τ ) , t [ 1 6 , 1 3 ] , τ I . Thus,

( T u ) ( t ) η 1 T u 0 , t [ 1 6 , 1 3 ] .

Similarly, we can obtain

( T u ) ( t ) η 2 ( T u ) , t [ 1 6 , 1 3 ] .

In addition, ( T u ) ( 1 ) = 0 . Thus, T : P P .

Now, we show that the operator T is compact on P.

In fact, let U be an arbitrary bounded set in P. Then there exists a positive number L such that u 2 L for all u U , and so M > 0 such that 0 f ( t , u ( s ) , u ( s ) ) M , t I for all u U .

In terms of Lemma 2.4, it follows from (2.26)-(2.27) that

0 ( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s M 0 1 w 1 ( s ) d s , t I , (3.3)

0 ( T u ) ( t ) = 0 1 ( G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s M 0 1 w 2 ( s ) d s , t I . (3.4)

Because the functions w 1 and w 2 are integrable on I, the formulae (3.3)-(3.4) yield that T u 0 M 0 , ( T u ) 0 M 0 , where M 0 = M max { 0 1 w 1 ( s ) d s , 0 1 w 2 ( s ) d s } . So, T u 1 M 0 . That is, TU is uniformly bounded.

On the other hand, for any t 1 , t 2 I with t 1 < t 2 , by setting h = f ( t , u , u ) , the formula (2.26) implies that

| ( T u ) ( t 2 ) ( T u ) ( t 1 ) | = | 0 t 1 ( G 1 ( t 2 , s ) G 1 ( t 1 , s ) ) h ( s ) d s + t 1 t 2 ( G 1 ( t 2 , s ) G 2 ( t 1 , s ) ) h ( s ) d s + t 2 1 ( G 2 ( t 2 , s ) G 2 ( t 1 , s ) ) h ( s ) d s | M 0 t 1 | G 1 ( t 2 , s ) G 1 ( t 1 , s ) | d s + 2 M t 1 t 2 w 1 ( s ) d s + M t 2 1 | G 2 ( t 2 , s ) G 2 ( t 1 , s ) | d s . (3.5)

According to (2.4)-(2.5) and by applying the mean value theorem, we have

| G 1 ( t 2 , s ) G 1 ( t 1 , s ) | [ 2 Γ ( α 1 ) + 1 ( δ 1 ) Γ ( α 2 ) + δ ( 1 s ) α 4 ( γ 1 ) ( δ 1 ) Γ ( α 3 ) ] | t 2 t 1 | ,

and so

0 t 1 | G 1 ( t 2 , s ) G 1 ( t 1 , s ) | d s N 1 | t 2 t 1 | , (3.6)

where N 1 = 2 Γ ( α 1 ) + 1 ( δ 1 ) Γ ( α 2 ) + δ ( γ 1 ) ( δ 1 ) Γ ( α 2 ) .

Similarly, there is another constant N 2 such that

t 2 1 | G 2 ( t 2 , s ) G 2 ( t 1 , s ) | d s N 2 | t 1 t 2 | . (3.7)

Again, because the function w 1 is integrable on I, the absolute continuity of integral of w 1 on [ t 1 , t 2 ] ensures that there exists a constant N 3 such that

t 1 t 2 w 1 ( s ) d s < N 3 ( t 2 t 1 ) . (3.8)

So, (3.5) together with (3.6)-(3.8) implies that there exists a constant N such that the inequality

| ( T u ) ( t 2 ) ( T u ) ( t 1 ) | N ( t 2 t 1 )

holds for any u U and t 1 , t 2 I with t 1 < t 2 . That is, the set TU is equicontinuous.

Similarly, we can deduce that the set { ( T u ) | u U } is also equicontinuous in terms of (2.27).

So, as a consequence of the Arzelà-Ascoli theorem, we have that TU is a compact set.

Now, we come to prove the operator T is continuous on P.

Let { u n } be an arbitrary sequence in P with u n u 0 P . Then there exists an L > 0 such that

u n ( t ) [ 0 , L ] , u n ( t ) [ L , 0 ] , t I , n = 0 , 1 , 2 , .

According to the uniform continuity of f on I × [ 0 , L ] × [ L , 0 ] , for an arbitrary number ε > 0 , there is a number N 1 such that

| f ( t , u n ( t ) , u n ( t ) ) f ( t , u 0 ( t ) , u 0 ( t ) ) | < ε (3.9)

for all t I , whenever n N .

Thus, in view of Lemma 2.4, from (2.26)-(2.27) and (3.9), it follows that

| ( T u n ) ( t ) ( T u 0 ) ( t ) | ε 0 1 w 1 ( s ) d s , t I ,

and

| ( T u n ) ( t ) ( T u ) ( t ) | ε 0 1 w 2 ( s ) d s , t I ,

whenever n N . That is, T is continuous on P. □

We are now in a position to state and prove the first theorem in the article. Let constants k 1 , k 2 satisfy k 1 ( 0 , ( 0 1 w 2 ( s ) d s ) 1 ) , k 2 > ( η 2 1 6 1 3 w 2 ( s ) d s ) 1 .

Theorem 3.1Suppose that ( H 1 )-( H 2 ) hold. In addition, there are two constants r 1 , r 2 with 1 6 η 1 η 2 r 2 > r 1 > 0 such thatfsatisfies the following condition:

( H 3 ) f ( t , x , y ) k 1 r 1 , for ( t , x , y ) I × [ 0 , r 1 ] × [ r 1 , 0 ] ;

f ( t , x , y ) k 2 r 2 , for ( t , x , y ) I × [ 1 6 η 1 η 2 r 2 , r 2 ] × [ r 2 , η 2 r 2 ] .

Then BVP (1.1)-(1.2) has at least one positive solutionusatisfying r 1 < max t I | u ( t ) | < r 2 and max t I | u ( t ) | < r 2 .

Proof We already know that T : P P is completely continuous by Lemma 3.1.

Let α ( u ) = max t I | u ( t ) | , β ( u ) = max t I | u ( t ) | for u P . It is easy to verify that the functions α, β satisfy the condition (D).

Choose a constant L large enough so that L > max { M 0 1 w 1 ( s ) d s , r 2 } , where M = max ( t , x , y ) I × J × Δ f ( t , x , y ) , and J = [ 0 , r 2 ] , Δ = [ r 2 , 0 ] . Set D i = { u C 1 ( I , R ) : α ( u ) = r i } , Ω i = { u C 1 ( I , R ) : α ( u ) < r i , β ( u ) < L } , i = 1 , 2 . Define the function f ¯ on I × R + × R as f ¯ ( t , x , y ) = f ( t , ϕ ( x ) , φ ( y ) ) , ( t , x , y ) I × R + × R , where ϕ ( x ) = min { x , r 2 } , φ ( y ) = max { y , r 2 } .

Consider the following ancillary BVP:

{ D 0 + α C x ( t ) = f ¯ ( t , x ( t ) , x ( t ) ) , t ( 0.1 ) , x ( 1 ) = x ( 1 ) = 0 , δ x ( 0 ) = x ( 1 ) , γ x ( 0 ) = x ( 1 ) . (3.10)

Obviously, the function f ¯ is continuous on I × R + × R according to the continuity of f. Thus, by an argument similar to that in Lemma 3.1, the operator T ¯ given by T ¯ u = 0 1 G ( t , s ) f ¯ ( s , u ( s ) , u ( s ) ) d s is also completely continuous on P and maps P into P.

We will prove that T has at least one nonzero fixed point in P by applying Lemma 2.5. The approach is divided into four steps.

Step 1. We first show that

u 0 u 0 , u ( t ) 1 6 η 1 η 2 u 0 , t [ 1 6 , 1 3 ] for any  u P . (3.11)

In fact, for any u P , owing to the condition u ( 1 ) = 0 , we have that u ( t ) = t 1 u ( s ) d s , t I , and so u 0 u 0 .

On the other hand, applying the mean value theorem, for any u P , we have that u 0 | u ( 1 3 ) u ( 1 6 ) | = 1 6 | u ( ξ ) | for some ξ ( 1 6 , 1 3 ) . Therefore, we have that u 0 1 6 η 2 u 0 from the fact that u ( ξ ) η 2 u 0 , ξ [ 1 6 , 1 3 ] because u P . So,

u ( t ) η 1 u 0 1 6 η 1 η 2 u 0 , t [ 1 6 , 1 3 ] ,

keeping in mind that u ( t ) η 1 u 0 , t [ 1 6 , 1 3 ] for u P .

Step 2. Now, we come to verify that the conditions corresponding to ( C 1 ) in Lemma 2.5 hold.

For any u P with α ( u ) = r 1 , we have that u ( t ) 0 , u ( t ) 0 , t I , and u 0 = r 1 . Thus, in view of (3.11), we have that 0 u ( t ) r 1 , r 1 u ( t ) 0 , t I . So, according to ( H 3 ), we have

f ¯ ( t , u ( t ) , u ( t ) ) = f ( t , u ( t ) , u ( t ) ) k 1 r 1 , t I .

Thus, from (2.27) and Lemma 2.4, it follows that

( T ¯ u ) ( t ) = 0 1 G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s k 1 r 1 0 1 ( G t ( t , s ) ) d s k 1 r 1 0 1 w 2 ( s ) d s , t I .

Thus, max t I | ( T ¯ u ) ( t ) | k 1 r 1 0 1 w 2 ( s ) d s < r 1 , noting that the assumption k 1 < ( 0 1 w 2 ( s ) d s ) 1 . That is, α ( T u ) < r 1 .

For any u P with α ( u ) = r 2 , then u ( t ) 0 , u ( t ) 0 , t I , and max t I | u ( t ) | = r 2 . Thus, from (3.11), we obtain

1 6 η 1 η 2 r 2 u ( t ) r 2 , r 2 u ( t ) η 2 r 2 , t [ 1 6 , 1 3 ] ,

and so

f ¯ ( t , u ( t ) , u ( t ) ) = f ( t , u ( t ) , u ( t ) ) k 2 r 2 , t [ 1 6 , 1 3 ] ,

from the condition ( H 3 ). Therefore, in view of Lemma 2.4, we have

( T ¯ u ) ( t ) = 0 1 G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s 1 6 1 3 G t ( t , s ) f ( s , u ( s ) , u ( s ) ) d s k 2 r 2 1 6 1 3 ( G t ( t , s ) ) d s k 2 r 2 η 2 1 6 1 3 w 2 ( s ) d s , t I .

Thus, max t I | T u ( t ) | k 2 η 2 r 2 1 6 1 3 w 2 ( s ) d s > r 2 , noting that k 2 > ( η 2 1 6 1 3 w 2 ( s ) d s ) 1 . That is, α ( T u ) > r 2 .

Step 3. We verify that the conditions corresponding to ( C 2 ) in Lemma 2.5 hold.

For any u P , owing to the fact that 0 ϕ ( u ( t ) ) r 2 , r 2 φ ( u ( t ) ) 0 , from the meaning of M, we have immediately that

f ¯ ( t , u ( t ) , u ( t ) ) = f ( t , ϕ ( u ( t ) ) , φ ( u ( t ) ) ) M , t I .

Thus,

( T ¯ u ) ( t ) = 0 1 G ( t , s ) f ¯ ( s , u ( s ) , u ( s ) ) d s M 0 1 G ( t , s ) d s M 0 1 w 1 ( s ) d s , t I .

Hence, ( T ¯ u ) 0 M 0 1 w 1 ( s ) d s , and so ( T ¯ u ) 0 < L from the choice of L. Thus, β ( T ¯ u ) < L .

Step 4. Finally, take p = σ 0 1 G ( t , s ) d s with 0 < σ < r 2 ( 0 1 w 2 ( s ) d s ) 1 . Then, by an argument similar to that in Lemma 3.1, we can know that p P . Moreover, 0 p ( t ) = σ 0 1 ( G t ( t , s ) ) d s σ 0 1 w 2 ( s ) d s < r 2 , t I , and 0 p ( t ) < r 2 , t I from (3.11). Thus, p ( Ω 2 P ) { 0 } , α ( p ) 0 . Again,

α ( u + λ p ) = max t I | u + λ p ( t ) | = max ( u ( t ) + λ ( p ( t ) ) λ max ( u ( t ) ) = λ α ( u )

for any u P , λ 0 .

So, the conditions corresponding to ( C 3 ) in Lemma 2.5 hold.

Summing up the above steps 1-4 and applying Lemma 2.5, we obtain that BVP (3.10) has at least one positive solution u ( Ω 2 Ω ¯ 1 ) P . That is, r 1 < u 0 < r 2 . u 0 < L , and so u 0 < r 2 from the fact that r 2 < L and u 0 u 0 by (3.13). Thus, 0 u ( t ) < r 2 , t I , 0 u ( t ) < r 2 , t I , and u 0 . Hence, f ¯ ( t , u ( t ) , u ( t ) ) = f ( t , u ( t ) , u ( t ) ) , t I , and so u is a positive solution of BVP (1.1)-(1.2). The proof is complete. □

Now, we state another theorem in this paper. Let us begin with introducing some notations.

Let B = R + × R . Denote f = lim inf x + | y | min t I f ( t , x , y ) x + | y | with ( x , y ) B , and f 0 = lim sup x + | y | 0 max t I f ( t , x , y ) x + | y | with ( x , y ) B . Put λ 1 = ( η 0 μ 1 ) 1 , λ 2 = μ 2 1 , where μ 1 = 1 6 1 3 [ G ( 0 , s ) G t ( 0 , s ) ] d s , μ 2 = 0 1 ( w 1 ( s ) + w 2 ( s ) ) d s , η 0 = min { η 1 , η 2 } , and η i ( i = 1 , 2 ) are given in Lemma 2.4.

Theorem 3.2Assume that ( H 1 )-( H 2 ) hold. If f > λ 1 , f 0 < λ 2 , then BVP (1.1)-(1.2) has at least one positive solution.

Proof As described in the proof of Theorem 3.1, T : P P is completely continuous. Again, from f > λ 1 , it follows that there exists an R 1 > 0 such that

f ( t , x , y ) > λ 1 ( x + | y | ) , t I (3.12)

holds when x + | y | R 1 with x 0 , y 0 .

Take R η 0 1 R 1 . Set Ω R = { u C 1 [ 0 , 1 ] : u 1 < R } . Now, we show that the following relation holds:

T u 1 u 1 , u Ω R P . (3.13)

In fact, for any u Ω R P , we have that u P with u 1 = R . Then

u ( t ) 0 , u ( t ) 0 , t I ; u ( t ) η 1 u 0 , u ( t ) η 2 u 0 , t [ 1 6 , 1 3 ] .

Thus,

u ( t ) + | u ( t ) | η 1 u 0 + η 2 u 0 η 0 ( u 0 + u 0 ) = η 0 R R 1 , for  t [ 1 6 , 1 3 ] . (3.14)

Thus, from (3.12), (3.14), it follows that

f ( t , u ( t ) , u ( t ) ) > λ 1 η 0 R , t [ 1 6 , 1 3 ] . (3.15)

Hence, (2.26) together with (3.15) implies that

( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s 1 6 1 3 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s λ 1 η 0 R 1 6 1 3 G ( t , s ) d s , t I .

So,

T u 0 ( T u ) ( 0 ) λ 1 η 0 R 1 6 1 3 G ( 0 , s ) d s . (3.16)

Similarly, we can obtain

( T u ) 0 ( T u ) ( 0 ) λ 1 η 0 R 1 6 1 3 [ G t ( 0 , s ) ] d s . (3.17)

Therefore, noting that λ 1 = ( η 0 μ 1 ) 1 , from (3.16)-(3.17), we have

T u 1 = T u 0 + ( T u ) 0 λ 1 η 0 R 1 6 1 3 [ G ( 0 , s ) G t ( 0 , s ) ] d s = λ 1 η 0 μ 1 R = R = u 1 .

So, the relation (3.13) holds.

Now, from f 0 < λ 2 , it follows that there exists an r 2 > 0 such that

f ( t , x , y ) < λ 2 ( x + | y | ) , t I , (3.18)

whenever x + | y | r 2 with x 0 , y 0 .

Take 0 < r < min { r 2 , R } . Set Ω r = { u C 1 [ 0 , 1 ] : u 1 < r } . We prove the following relation holds:

T u 1 u 1 , u Ω r P . (3.19)

In fact, for any u Ω r P , we have u P and u 1 = r . Thus, u ( t ) 0 , u ( t ) 0 , t I , and u ( t ) + | u ( t ) | r , t I . So, by (3.18), it follows that

f ( t , u ( t ) , u ( t ) ) < λ 2 ( u ( t ) + | u ( t ) | ) λ 2 r .

Therefore, from (2.26) and in view of Lemma 2.4, we have

( T u ) ( t ) = 0 1 G ( t , s ) f ( s , u ( s ) , u ( s ) ) d s λ 2 r 0 1 G ( t , s ) d s λ 2 r 0 1 w 1 ( s ) d s , t I . (3.20)

So,

T u 0 λ 2 r 0 1 w 1 ( s ) d s .