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Existence results for m-point boundary value problems of nonlinear fractional differential equations with p-Laplacian operator

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Abstract

In this paper, we discuss the existence and multiplicity of positive solutions to m-point boundary value problems of nonlinear fractional differential equations with p-Laplacian operator

{ D 0 + β ( φ p ( D 0 + α u ( t ) ) ) + φ p ( λ ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + γ u ( 1 ) = i = 1 m 2 ξ i D 0 + γ u ( η i ) , D 0 + α u ( 0 ) = 0 ,

where D 0 + α , D 0 + β and D 0 + γ are the standard Riemann-Liouville fractional derivatives with 1<α2, 0<β,γ1, 0αβ1, λ(0,+), 0< ξ i , η i <1, i=1,2,,m2, i = 1 m 2 ξ i η i α β 1 <1, 0αγ1, fC([0,1]×[0,+),[0,+)), and φ p (s)= | s | p 2 s, p>1, φ p 1 = φ q , 1 p + 1 q =1. Our results are based on the monotone iterative technique and the theory of the fixed point index in a cone. Furthermore, two examples are also given to illustrate the results.

1 Introduction

Fractional differential equations arise in various areas of science and engineering. Due to their applications, fractional differential equations have gained considerable attention (see, e.g., [126] and the references therein).

Recently, there have been some papers dealing with the existence of solutions for nonlinear fractional differential equations with p-Laplacian operator. In [1], Wang et al. investigated the following boundary value problem for fractional differential equations with p-Laplacian operator:

{ D 0 + β ( φ p ( D 0 + α u ( t ) ) ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) = a u ( ξ ) , D 0 + α u ( 0 ) = 0 , D 0 + α u ( 1 ) = b D 0 + α u ( η ) ,

where D 0 + α , D 0 + β are the standard Riemann-Liouville fractional derivatives, 1<α,β2, 0a,b1, 0<ξ,η<1, f(t,u)C[(0,1)×(0,+),[0,+)].

In [2], Chai studied the existence of positive solutions of the following fractional differential equations with p-Laplacian operator:

{ D 0 + β ( φ p ( D 0 + α u ( t ) ) ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , u ( 1 ) + σ D 0 + γ u ( 1 ) = 0 , D 0 + α u ( 0 ) = 0 ,

where D 0 + α , D 0 + β and D 0 + γ are the standard Riemann-Liouville fractional derivatives with 1<α2, 0<β1, 0<γ1, 0αγ1, the constant σ is a positive number, f(t,u)C(I× R + , R + ).

In [3], Chen and Liu studied the following fractional differential equations with p-Laplacian operator:

{ D 0 + β ( φ p ( D 0 + α x ( t ) ) ) = f ( t , x ( t ) ) , t [ 0 , 1 ] , x ( 0 ) = x ( 1 ) , D 0 + α x ( 0 ) = D 0 + α x ( 1 ) ,

where 0<α,β1, 1<α+β2, D 0 + α , D 0 + β are Caputo fractional derivatives, and f:[0,1]×RR is continuous.

In [4], Lu et al. studied the following fractional differential equations with p-Laplacian operator:

{ D 0 + β ( φ p ( D 0 + α u ( t ) ) ) = f ( t , u ( t ) ) , t [ 0 , 1 ] , u ( 0 ) = u ( 0 ) = u ( 1 ) = 0 , D 0 + α u ( 0 ) = D 0 + α u ( 1 ) = 0 ,

where 2<α3, 1<β2, D 0 + α , D 0 + β are the standard Riemann-Liouville fractional derivatives, and f(t,u)C([0,1]×[0,+),[0,+)).

On the other hand, in [5], Bai studied an eigenvalue interval of the following fractional boundary problem:

{ D 0 + α c u ( t ) + λ h ( t ) f ( u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ( 1 ) = u ( 0 ) = 0 ,

where 2<α3, D 0 + α c is the standard Caputo fractional derivative, λ>0.

In [6], Zhang et al. studied the following singular eigenvalue problem for a higher order fractional differential equation:

{ D α x ( t ) = λ f ( x ( t ) , D μ 1 x ( t ) , D μ 2 x ( t ) , , D μ n 1 x ( t ) ) , 0 < t < 1 , x ( 0 ) = 0 , D μ i x ( 0 ) = 0 , D μ x ( 1 ) = j = 1 p 2 a j D μ x ( ξ j ) , 1 i n 1 ,

where n3, n1<αn, nl1<α μ l <n1 for l=1,2,,n2, and μ μ n 1 >0, α μ n 1 2, αμ>1. D 0 + α is the standard Riemann-Liouville fractional derivative.

Moreover, in recent years, we have done some work on fractional differential equations [79]. In [7], we considered the following m-point boundary value problem for fractional differential equations:

{ D 0 + α u ( t ) + f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = i = 1 m 2 ξ i D 0 + β u ( η i ) ,

where D 0 + α is the standard Riemann-Liouville fractional derivative, n=[α]+1, f:[0,1]×[0,)[0,) is continuous, 1<α2, 0β1, 0αβ1, 0< ξ i , η i <1, i=1,2,,m2, and i = 1 m 2 ξ i η i α β 1 <1.

Combining our work, in this paper, we discuss the existence of positive solutions for the following fractional differential equations with p-Laplacian operator:

{ D 0 + β ( φ p ( D 0 + α u ( t ) ) ) + φ p ( λ ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + γ u ( 1 ) = i = 1 m 2 ξ i D 0 + γ u ( η i ) , D 0 + α u ( 0 ) = 0 ,
(1.1)

where D 0 + α , D 0 + β and D 0 + γ are the standard Riemann-Liouville fractional derivatives with 1<α2, 0<β,γ1, 0αβ1, λ(0,+), 0< ξ i , η i <1, i=1,2,,m2, i = 1 m 2 ξ i η i α β 1 <1, 0αγ1, fC([0,1]×[0,+),[0,+)), and φ p (s)= | s | p 2 s, p>1, φ p 1 = φ q , 1 p + 1 q =1.

Our work presented in this paper has the following features. Firstly, to the best of the author’s knowledge, there are few results on the existence of solutions for nonlinear fractional p-Laplacian differential equations with m-point boundary value problems. Secondly, we transform (1.1) into an equivalent integral equation and discuss the eigenvalue interval for the existence of multiplicity of positive solutions. The paper is organized as follows. In Section 2, we present some background materials and preliminaries. Section 3 deals with some existence results. In Section 4, two examples are given to illustrate the results.

2 Background materials and preliminaries

Definition 2.1 ([10, 11])

The fractional integral of order α with the lower limit t 0 for a function f is defined as

I α f(t)= 1 Γ ( α ) t 0 t ( t s ) α 1 f(s)ds,t> t 0 ,α>0,

where Γ is the gamma function.

Definition 2.2 ([10, 11])

The Riemann-Liouville derivative of order α with the lower limit t 0 for a function f is

D t 0 α f(t)= 1 Γ ( n α ) ( d d t ) n t 0 t ( t s ) n α 1 f(s)ds,t> t 0 ,α>0,n=[α]+1.

Lemma 2.1 ([12])

Assume that uC(0,1) L 1 (0,1) with a fractional derivative of order α>0 that belongs to C(0,1) L 1 (0,1). Then

I 0 + α D 0 + α u(t)=u(t)+ C 1 t α 1 + C 2 t α 2 ++ C N t α N for some  C i R,i=1,2,,N,

where N is the smallest integer greater than or equal to α.

Lemma 2.2 ([7])

Let yC[0,1]. Then the fractional differential equation

{ D 0 + α u ( t ) + y ( t ) = 0 , 0 < t < 1 , 1 < α 2 , u ( 0 ) = 0 , D 0 + β u ( 1 ) = i = 1 m 2 ξ i D 0 + β u ( η i )

has a unique solution which is given by

u(t)= 0 1 G(t,s)y(s)ds,

where

G(t,s)= G 1 (t,s)+ G 2 (t,s),

in which

G 1 ( t , s ) = { t α 1 ( 1 s ) α β 1 ( t s ) α 1 Γ ( α ) , 0 s t 1 , t α 1 ( 1 s ) α β 1 Γ ( α ) , 0 t s 1 , G 2 ( t , s ) = { 1 A Γ ( α ) [ 0 s η i ( ξ i η i α β 1 t α 1 ( 1 s ) α β 1 ξ i t α 1 ( η i s ) α β 1 ) ] , t [ 0 , 1 ] , 1 A Γ ( α ) ( η i s 1 ξ i η i α β 1 t α 1 ( 1 s ) α β 1 ) , t [ 0 , 1 ] ,

where

A=1 i = 1 m 2 ξ i η i α β 1 .

Lemma 2.3 ([7])

If i = 1 m 2 ξ i η i α β 1 <1, then the function G(t,s) in Lemma  2.2 satisfies the following conditions:

  1. (i)

    G(t,s)>0, for s,t(0,1),

  2. (ii)

    G(t,s) G (s,s), for s,t[0,1],

where

G (s,s)= 1 Γ ( α ) ( 1 s ) α β 1 + 1 A Γ ( α ) i = 1 m 2 ξ i η i α β 1 ( 1 s ) α β 1 .

Lemma 2.4 G(t,s) in [7]has the following property:

G(t,s) t α 1 G(1,s).

Proof For 0st1, we conclude that

t α 1 ( 1 s ) α β 1 ( t s ) α 1 = t α 1 [ ( 1 s ) α β 1 ( 1 s t ) α 1 ] t α 1 [ ( 1 s ) α β 1 ( 1 s ) α 1 ] .

Thus

G 1 (t,s) t α 1 G 1 (1,s).

It is obvious that

G 2 (t,s) t α 1 G 2 (1,s).

Therefore

G(t,s) t α 1 G(1,s).

 □

Lemma 2.5 Let fC([0,1]×[0,+),[0,+)), then BVP (1.1) has a unique solution

u(t)=λ 0 1 G(t,s) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) ds.

Proof Let w= D 0 + α u, v= φ p (w). From (1.1), we have

{ D 0 + β v ( t ) + φ p ( λ ) f ( t , u ( t ) ) = 0 , 0 < t < 1 , v ( 0 ) = 0 .

By Lemma 2.1, we have

v(t)= c 1 t β 1 I 0 + β ( φ p ( λ ) f ( t , u ( t ) ) ) ,0<t<1.

It follows from v(0)=0 that

v(t)= I 0 + β ( φ p ( λ ) f ( t , u ( t ) ) ) ,0<t<1.

Thus, from (1.1) we know that

{ D 0 + α u ( t ) = φ p 1 ( I 0 + β ( φ p ( λ ) f ( t , u ( t ) ) ) ) , 0 < t < 1 , u ( 0 ) = 0 , D 0 + γ u ( 1 ) = i = 1 m 2 ξ i D 0 + γ u ( η i ) .

By Lemma 2.2, (1.1) has a unique solution

u(t)=λ 0 1 G(t,s) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) ds.

It follows from fC([0,1]×[0,+),[0,+)) that

0 1 G(t,s) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) ds= 0 1 G(t,s) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) ds.

Thus

u(t)=λ 0 1 G(t,s) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) ds.

 □

Lemma 2.6 ([27])

Let E be a real Banach space, PE be a cone, Ω r ={uP:ur}. Let the operator T:P Ω r P be completely continuous and satisfy Txx, x Ω r . Then

( i ) If T x x , x Ω r , then i ( T , Ω r , P ) = 1 , ( ii ) If T x x , x Ω r , then i ( T , Ω r , P ) = 0 .

3 Main results

We consider the Banach space E=C([0,1],R) endowed with the norm defined by u= sup 0 t 1 |u(t)|. Let P={uE|u(t)0}, then P is a cone in E. Define an operator T:PP as

(Tu)(t)=λ 0 1 G(t,s) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) ds.
(3.1)

Then T has a solution if and only if the operator T has a fixed point.

Lemma 3.1 If fC([0,1]×[0,+),[0,+)), then the operator T:PP is completely continuous.

Proof From the continuity and non-negativeness of G(t,s) and f(t,u(t)), we know that T:PP is continuous.

Let ΩP be bounded. Then, for all t[0,1] and uΩ, there exists a positive constant M such that |f(t,u(t))|M. Thus,

| ( T u ) ( t ) | = | λ 0 1 G ( t , s ) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) d s | λ 0 1 G ( s , s ) ( 0 s ( s τ ) β 1 d τ ) q 1 d s M q 1 ( Γ ( β ) ) q 1 = λ M q 1 ( Γ ( β + 1 ) ) q 1 0 1 G ( s , s ) s ( q 1 ) β d s λ M q 1 ( Γ ( β + 1 ) ) q 1 0 1 G ( s , s ) d s = λ M q 1 L ( Γ ( β + 1 ) ) q 1 ,

where

L= 0 1 G (s,s)ds.

This means that T(Ω) is uniformly bounded.

On the other hand, from the continuity of G(t,s) on [0,1]×[0,1], we see that it is uniformly continuous on [0,1]×[0,1]. Thus, for fixed s[0,1] and for any ε>0, there exists a constant δ>0 such that t 1 , t 2 [0,1] and | t 1 t 2 |<δ,

|G( t 1 ,s)G( t 2 ,s)|< ( Γ ( β + 1 ) ) q 1 λ M q 1 ε.

Hence, for all uΩ,

| ( T u ) ( t 2 ) ( T u ) ( t 1 ) | λ 0 1 | G ( t 2 , s ) G ( t 1 , s ) | φ p 1 ( I 0 + β f ( s , u ( s ) ) ) d s λ 0 1 | G ( t 2 , s ) G ( t 1 , s ) | ( 0 s ( s τ ) β 1 d τ ) q 1 d s M q 1 ( Γ ( β ) ) q 1 = λ M q 1 ( Γ ( β + 1 ) ) q 1 0 1 | G ( t 2 , s ) G ( t 1 , s ) | s ( q 1 ) β d s λ M q 1 ( Γ ( β + 1 ) ) q 1 0 1 | G ( t 2 , s ) G ( t 1 , s ) | d s = ε ,

which implies that T(Ω) is equicontinuous. By the Arzela-Ascoli theorem, we obtain that T:PP is completely continuous. The proof is complete. □

Theorem 3.2 If fC([0,1]×[0,+),[0,+)), f(t,u) is nondecreasing in u and λ(0,+), then BVP (1.1) has a minimal positive solution v ¯ in B r and a maximal positive solution w ¯ in B r . Moreover, v m (t) v ¯ (t), w m (t) w ¯ (t) as m uniformly on [0,1], where

v m (t)=λ 0 1 G(t,s) φ p 1 ( I 0 + β f ( s , v m 1 ( s ) ) ) ds
(3.2)

and

w m (t)=λ 0 1 G(t,s) φ p 1 ( I 0 + β f ( s , w m 1 ( s ) ) ) ds.
(3.3)

Proof Let

B r = { u P : u r } ,

where

r λ M 1 q 1 ( Γ ( β + 1 ) ) q 1 0 1 G (s,s)ds.

Step 1: Problem (1.1) has at least one solution.

For u B r , there exists a positive constant M 1 such that |f(t,u(t))| M 1 ,

| ( T u ) ( t ) | = | λ 0 1 G ( t , s ) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) d s | λ ( Γ ( β ) ) q 1 0 1 G ( s , s ) ( 0 s ( s τ ) β 1 f ( τ , u ( τ ) ) d τ ) q 1 d s λ M 1 q 1 ( Γ ( β ) ) q 1 0 1 G ( s , s ) ( 0 s ( s τ ) β 1 d τ ) q 1 d s = λ M 1 q 1 ( Γ ( β + 1 ) ) q 1 0 1 G ( s , s ) s β ( q 1 ) d s λ M 1 q 1 ( Γ ( β + 1 ) ) q 1 0 1 G ( s , s ) d s .

Thus

T: B r B r .

By Lemma 3.1, we can see that T: B r B r is completely continuous. Hence, by means of the Schauder fixed point theorem, the operator T has at least one fixed point, and BVP (1.1) has at least one solution in B r .

Step 2: BVP (1.1) has a positive solution in B r , which is a minimal positive solution.

From (3.1) and (3.2), one can see that

v m (t)=(T v m 1 )(t),t[0,1],m=1,2,3,.
(3.4)

This, together with f(t,u) being nondecreasing in u, yields that

0= v 0 (t) v 1 (t) v m (t),t[0,1].

Since T is compact, we obtain that { v m } is a sequentially compact set. Consequently, there exists v ¯ B r such that v m v ¯ (m).

Let u(t) be any positive solution of BVP (1.1) in B r . It is obvious that 0= v 0 (t)u(t)=(Tu)(t).

Thus,

v m (t)u(t)(m=0,1,2,3,).
(3.5)

Taking limits as m in (3.5), we get v ¯ (t)u(t) for t[0,1].

Step 3: BVP (1.1) has a positive solution in B r , which is a maximal positive solution.

Let w 0 (t)=r, t[0,1] and w 1 (t)=T w 0 (t). From T: B r B r , we have w 1 B r . Thus

0 w 1 (t)r= w 0 (t).

This, together with f(t,u) being nondecreasing in u, yields that

w m (t) w 1 (t) w 0 (t),t[0,1].

Using a proof similar to that of Step 2, we can show that

w m (t) w ¯ (t)(m)

and

w ¯ (t)= 0 1 G(t,s)f ( s , w ¯ ( s ) ) ds.

Let u(t) be any positive solution of BVP (1.1) in B r .

Obviously,

u(t) w 0 (t).

Thus

u(t) w m (t).
(3.6)

Taking limits as m in (3.6), we obtain u(t) w ¯ (t) for t[0,1].

The proof is complete. □

Define

f 0 = lim u 0 + sup t [ 0 , 1 ] f ( t , u ) φ p ( l 1 u ) , f 0 = lim u 0 + inf t [ 0 , 1 ] f ( t , u ) φ p ( l 2 u ) , f = lim u + sup t [ 0 , 1 ] f ( t , u ) φ p ( l 3 u ) , f = lim u + inf t [ 0 , 1 ] f ( t , u ) φ p ( l 4 u ) .

Let

B= 0 1 G (s,s) s β ( q 1 ) dsand B 1 = 0 1 G(1,s) s β ( q 1 ) ds.

Theorem 3.3 Assume that fC([0,1]×[0,+),[0,+)), and the following conditions hold:

(H1) f 0 = f =+.

(H2) There exists a constant ρ 1 >0 such that f(t,u) φ p ( l 5 u) for t[0,1], u[0, ρ 1 ].

Then BVP (1.1) has at least two positive solutions u 1 and u 2 such that

0< u 1 < ρ 1 < u 2

for

λ ( ( Γ ( β + 1 ) ) q 1 l 2 B 1 , ( Γ ( β + 1 ) ) q 1 l 5 B ) ( ( Γ ( β + 1 ) ) q 1 l 4 B 1 , ( Γ ( β + 1 ) ) q 1 l 5 B ) ,
(3.7)

where

l 2 B 1 > l 5 Band l 4 B 1 > l 5 B.

Proof Since

f 0 = lim u 0 + inf t [ 0 , 1 ] f ( t , u ) φ p ( l 2 u ) =+,

there is ρ 0 (0, ρ 1 ) such that

f(t,u) φ p ( l 2 u ) for t[0,1],u[0, ρ 0 ].

Let

Ω ρ 0 = { u P : u ρ 0 } .

Then, for any u Ω ρ 0 , it follows from Lemma 2.4 that

( T u ) ( t ) = λ 0 1 G ( t , s ) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) d s λ 0 1 t α 1 G ( 1 , s ) φ p 1 ( I 0 + β ( φ p ( l 2 u ) ) ) d s = λ l 2 0 1 t α 1 G ( 1 , s ) ( 1 Γ ( β ) 0 s ( s τ ) β 1 d τ ) q 1 d s u = λ l 2 ( Γ ( β + 1 ) ) q 1 0 1 t α 1 G ( 1 , s ) s β ( q 1 ) d s u .

Thus

Tu λ l 2 B 1 ( Γ ( β + 1 ) ) q 1 u.

This, together with (3.7), yields that

Tuu,u Ω ρ 0 .

By Lemma 2.6, we have

i(T, Ω ρ 0 ,P)=0.
(3.8)

In view of

f = lim u + inf t [ 0 , 1 ] f ( t , u ) φ p ( l 4 u ) =+,

there is ρ 0 , ρ 0 > ρ 1 , such that

f(t,u) φ p ( l 4 u ) for t[0,1],u[ ρ 0 ,+).

Let

Ω ρ 0 = { u P : u ρ 0 } .

Then, for any u Ω ρ 0 , it follows from Lemma 2.4 that

( T u ) ( t ) = λ 0 1 G ( t , s ) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) d s λ 0 1 t α 1 G ( 1 , s ) φ p 1 ( I 0 + β ( φ p ( l 4 u ) ) ) d s = λ l 4 0 1 t α 1 G ( 1 , s ) ( 1 Γ ( β ) 0 s ( s τ ) β 1 d τ ) q 1 d s u = λ l 4 ( Γ ( β + 1 ) ) q 1 0 1 t α 1 G ( 1 , s ) s β ( q 1 ) d s u .

Thus

Tu λ l 4 B 1 ( Γ ( β + 1 ) ) q 1 u.

This, together with (3.7), yields that

Tuu,u Ω ρ 0 .

By Lemma 2.6, we have

i(T, Ω ρ 0 ,P)=0.
(3.9)

Finally, let Ω ρ 1 ={uP:u ρ 1 }. For any u Ω ρ 1 , it follows from Lemma 2.3 and (H2) that

( T u ) ( t ) = λ 0 1 G ( t , s ) φ p 1 ( I 0 + β f ( s , u ( s ) ) ) d s λ 0 1 G ( s , s ) φ p 1 ( I 0 + β ( φ p ( l 5 u ) ) ) d s = λ l 5 0 1 G ( s , s ) ( 1 Γ ( β ) 0 s ( s τ ) β 1 d τ ) q 1 d s u = λ l 5 ( Γ ( β + 1 ) ) q 1 0 1 G ( s , s ) s β ( q 1 ) d s u .

Thus

Tu λ l 5 B ( Γ ( β + 1 ) ) q 1 u.

This, together with (3.7), yields that

Tuu,u Ω ρ 1 .

Using Lemma 2.6, we get

i(T, Ω ρ 1 ,P)=1.
(3.10)

From (3.8)-(3.10) and ρ 0 < ρ 1 < ρ 0 , we have

i(T, Ω ρ 0 Ω ¯ ρ 1 ,P)=1,i(T, Ω ρ 1 Ω ¯ ρ 0 ,P)=1.

Therefore, T has a fixed point u 1 Ω ρ 1 Ω ¯ ρ 0 and a fixed point u 2 Ω ρ 0 Ω ¯ ρ 1 . Clearly, u 1 , u 2 are both positive solutions of BVP (1.1) and 0< u 1 < ρ 1 < u 2 . The proof of Theorem 3.3 is completed. □

In a similar way, we can obtain the following result.

Corollary 3.4 Assume that fC([0,1]×[0,+),[0,+)), and the following conditions hold:

(H1) f 0 = f =0.

(H2) There exists a constant ρ 2 >0 such that f(t,u) φ p ( l 6 u) for t[0,1], u[0, ρ 2 ].

Then BVP (1.1) has at least two positive solutions u 1 and u 2 such that

0< u 1 < ρ 2 < u 2

for

λ ( ( Γ ( β + 1 ) ) q 1 l 6 B 1 , ( Γ ( β + 1 ) ) q 1 l 3 B ) ( ( Γ ( β + 1 ) ) q 1 l 6 B 1 , ( Γ ( β + 1 ) ) q 1 l 1 B ) ,

where

l 6 B 1 > l 3 Band l 6 B 1 > l 1 B.

4 Examples

Example 4.1 Consider the following boundary value problem:

{ D 0 + 1 2 ( φ 2 ( D 0 + 3 2 u ( t ) ) ) + φ 2 ( λ ) ( ( t + 1 ) π | u ( t ) | 1 + | u ( t ) | ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + 1 2 u ( 1 ) = i = 1 2 ξ i D 0 + 1 2 u ( η i ) , D 0 + 3 2 u ( 0 ) = 0 ,
(4.1)

where

α = 3 2 , β = 1 2 , γ = 1 2 , m = 4 , p = q = 2 , ξ 1 = η 1 = 1 4 , ξ 2 = η 2 = 1 2 , λ ( 0 , + ) , f ( t , u ) = ( t + 1 ) π | u ( t ) | 1 + | u ( t ) | .

Thus

fC ( [ 0 , 1 ] × [ 0 , + ) , [ 0 , + ) ) and|f(t,u)|= | ( t + 1 ) π | u ( t ) | 1 + | u ( t ) | | 2π.

By computation, we deduce that

i = 1 2 ξ i η i α β 1 = ξ i + ξ 2 = 3 4 , A = 1 i = 1 2 ξ i η i α β 1 = 1 4

and

αγ10.

On the other hand,

0 1 G ( s , s ) d s = 1 Γ ( α ) 0 1 ( 1 s ) α β 1 d s + 1 A Γ ( α ) i = 1 2 ξ i η i α β 1 0 1 ( 1 s ) α β 1 d s = ( 1 Γ ( α ) + 1 A Γ ( α ) i = 1 2 ξ i η i α β 1 ) 0 1 ( 1 s ) α β 1 d s = 1 Γ ( α ) + 1 A Γ ( α ) i = 1 2 ξ i = 1 π 2 + 1 1 4 π 2 3 4 = 2 π + 6 π = 8 π .

Take

r λ M 1 q 1 ( Γ ( β + 1 ) ) q 1 0 1 G ( s , s ) d s = λ 2 π π 2 8 π = 32 λ .

Hence, by Theorem 3.2, BVP (4.1) has a minimal positive solution v ¯ in B r and a maximal positive solution w ¯ in B r .

Example 4.2 Consider the following boundary value problem:

{ D 0 + 1 2 ( φ 3 2 ( D 0 + 3 2 u ( t ) ) ) + φ 3 2 ( λ ) ( 1 + t ) ( 1 2 | u ( t ) | 1 3 + 1 2 u 1 3 + u 2 ) = 0 , 0 < t < 1 , u ( 0 ) = 0 , D 0 + 1 2 u ( 1 ) = i = 1 2 ξ i D 0 + 1 2 u ( η i ) , D 0 + 3 2 u ( 0 ) = 0 ,
(4.2)

where

α = 3 2 , β = 1 2 , γ = 1 2 , p = 3 2 , q = 3 , m = 4 , ξ 1 = η 1 = 1 4 , ξ 2 = η 2 = 1 2 , α γ 1 = 0 , α β 1 = 0

and

f(t,u)=(1+t) ( 1 2 | u ( t ) | 1 3 + 1 2 u 1 3 + u 2 ) .

It follows from Example 4.1 that

i = 1 2 ξ i η i α β 1 = ξ i + ξ 2 = 3 4 ,A=1 i = 1 2 ξ i η i α β 1 = 1 4 .

By computation, we deduce that

B = 0 1 G ( s , s ) s β ( q 1 ) d s = 1 Γ ( α ) 0 1 ( 1 s ) α β 1 s β ( q 1 ) d s + 1 A Γ ( α ) i = 1 2 ξ i η i α β 1 0 1 ( 1 s ) α β 1 s β ( q 1 ) d s = 1 Γ ( α ) 0 1 s 1 2 × ( 3 1 ) d s + 1 A Γ ( α ) i = 1 2 ξ i 0 1 s 1 2 × ( 3 1 ) d s = 1 Γ ( α ) 0 1 s d s + 1 A Γ ( α ) i = 1 2 ξ i 0 1 s d s = 1 2 ( 1 Γ ( α ) + 1 A Γ ( α ) i = 1 2 ξ i ) = 1 2 ( 2 π + 8 π ( 1 2 + 1 4 ) ) = 1 2 × 8 π = 4 π

and

B 1 = 0 1 G ( 1 , s ) s β ( q 1 ) d s = 0 1 G 1 ( 1 , s ) s β ( q 1 ) d s + 0 1 G 2 ( 1 , s ) s β ( q 1 ) d s = 1 Γ ( 3 2 ) 0 1 [ 1 ( 1 s ) 1 2 ] s β ( q 1 ) d s + 1 A Γ ( 3 2 ) 0 1 4 [ ξ 1 η 1 0 ( 1 s ) 0 ξ 1 ( η 1 s ) 0 ] s β ( q 1 ) d s + 1 A Γ ( 3 2 ) 1 4 1 ξ 1 η 1 0 ( 1 s ) 0 s β ( q 1 ) d s + 1 A Γ ( 3 2 ) 0 1 2 [ ξ 2 η 2 0 ( 1 s ) 0 ξ 2 ( η 2 s ) 0 ] s β ( q 1 ) d s + 1 A Γ ( 3 2 ) 1 2 1 ξ 2 η 2 0 ( 1 s ) 0 s β ( q 1 ) d s = 1 Γ ( 3 2 ) 0 1 [ s s ( 1 s ) 1 2 ] d s + 1 A Γ ( 3 2 ) 1 4 1 ξ 1 s d s + 1 A Γ ( 3 2 ) 1 2 1 ξ 2 s d s = 2 π 0 1 [ s s ( 1 s ) 1 2 ] d s + 2 π 1 4 1 s d s + 4 π 1 2 1 s d s = 2 π ( 1 2 4 15 ) + 2 π 1 2 s 2 | 1 4 1 + 4 π 1 2 s 2 | 1 2 1 = 697 240 π .

Taking

ρ 1 =8, l 5 =2,178,

we have

f(t,u)(1+1)(2+64)=132= φ p ( l 5 u ) = φ 3 2 (2,178×8)for t[0,1],u[0, ρ 1 ].

Thus, condition (H2) is satisfied. It is obvious that condition (H1) holds.

On the other hand, let l 2 =4,000, l 4 =3,600, we have l 2 B 1 > l 5 B, l 4 B 1 > l 5 B and

λ ( ( Γ ( β + 1 ) ) q 1 l 2 B 1 , ( Γ ( β + 1 ) ) q 1 l 5 B ) ( ( Γ ( β + 1 ) ) q 1 l 4 B 1 , ( Γ ( β + 1 ) ) q 1 l 5 B ) = ( ( Γ ( β + 1 ) ) q 1 l 4 B 1 , ( Γ ( β + 1 ) ) q 1 l 5 B ) = ( ( π 2 ) 2 3 , 600 × 697 240 π , ( π 2 ) 2 2 , 178 × 4 π ) = ( π 3 2 41 , 820 , π 3 2 34 , 848 ) .

Hence, by Theorem 3.3, BVP (4.2) has at least two solutions u 1 and u 2 such that 0< u 1 <8< u 2 for

λ ( π 3 2 41 , 820 , π 3 2 34 , 848 ) .

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Acknowledgements

This research is supported by Henan Province College Youth Backbone Teacher Funds (2011GGJS-213) and the National Natural Science Foundation of China (11271336).

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Correspondence to Zhi-Wei Lv.

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Keywords

  • fractional differential equation
  • m-point boundary value problems
  • p-Laplacian operator