1 Introduction

Recently, some authors have paid much attention to the existence of positive solutions for functional dynamic equations on time scales [18], especially for the p-Laplacian functional dynamic equations on time scales [1, 38]. For convenience, throughout this paper we denote by Φ p (s) the p-Laplacian operator, i.e., Φ p (s)=|s | p 2 s, p>1, ( Φ p ) 1 = Φ q , 1 p + 1 q =1.

In [2], Kaufmann and Raffoul considered a nonlinear functional dynamic equation on time scales and obtained sufficient conditions for the existence of positive solutions. In [5], by using a double fixed point theorem due to Avery et al. [9], Song and Gao considered the existence of at least twin positive solutions to the following p-Laplacian functional dynamic equations on time scales:

{ [ Φ p ( u ( t ) ) ] + a ( t ) f ( u ( t ) , u ( μ ( t ) ) ) = 0 , t ( 0 , T ) T , u ( t ) = φ ( t ) , t [ r , 0 ] T , u ( 0 ) = u ( T ) = 0 , u ( T ) + B 0 ( u ( η ) ) = 0 ,
(1.1)

where η ( 0 , ρ ( T ) ) T , −r, 0, TT.

In [8], Wang and Guan considered the existence of positive solutions to problem (1.1) by applying the well-known Leggett-Williams fixed point theorem [10].

Motivated by [2, 5] and [8], we shall show that problem (1.1) has at least three positive solutions by means of the fixed point theorem due to Avery and Peterson [11].

In this article, we always assume that:

(C1) f: [ 0 , + ) 2 (0,+) is continuous;

(C2) a:T(0,+) is left dense continuous (i.e., a C ld (T,(0,+))) and does not vanish identically on any closed subinterval of [ 0 , T ] T , where C ld (T,(0,+)) denotes the set of all left dense continuous functions from T to (0,+);

(C3) φ: [ r , 0 ] T [0,+) is continuous and r>0;

(C4) μ: [ 0 , T ] T [ r , T ] T is continuous, μ(t)t for all t;

(C5) B 0 (v) is a continuous function defined on ℝ and satisfies that there exist B1 and A1 such that

Bx B 0 (x)Axfor all R.

Remark 1.1 Although the Banach space in this paper is the same as that of [8], i.e., E= C ld ( [ 0 , T ] T ,R) with u=max{ max t [ 0 , T ] T |u(t)|, max t [ 0 , T ] T k | u (t)|}, the hypotheses utilized in the existence theorem in this paper differ from those of [8] where the effect of a(t) was imposed.

Throughout this work, we assume the knowledge of time scales and time-scale notation, first introduced by Hilger [12]. For more on time scales, please see the texts by Bohner and Peterson [13, 14].

In the remainder of this section, we state the following theorem which is crucial to our proof.

Let γ and θ be nonnegative continuous convex functions on P, α be a nonnegative continuous concave function on P, and ψ be a nonnegative continuous function on P. Then, for positive real numbers a, b, c and d, we define the following convex sets:

P ( γ , d ) = { x P : γ ( x ) < d } , P ( γ , α , b , d ) = { x P : b α ( x ) , γ ( x ) d } , P ( γ , θ , α , b , c , d ) = { x P : b α ( x ) , θ ( x ) c , γ ( x ) d } ,

and a closed set

R(γ,ψ,a,d)= { x P : a ψ ( x ) , γ ( x ) d } .

To prove our main results, we need the following fixed point theorem due to Avery and Peterson in [11].

Theorem 1.1 Let P be a cone in a real Banach space E. Let γ and θ be nonnegative continuous convex functionals on P, α be a nonnegative continuous concave functional on P and ψ be a nonnegative continuous functional on P satisfying ψ(λx)λψ(x) for 0λ1, such that for some positive numbers h and d,

α(x)ψ(x)andxhγ(x)

for all x P ( γ , d ) ¯ . Suppose that

F: P ( γ , d ) ¯ P ( γ , d ) ¯

is completely continuous and there exist positive numbers a, b and c with a<b such that:

  1. (i)

    {xP(γ,θ,α,b,c,d):α(x)>b}ϕ and α(Fx)>b for xP(γ,θ,α,b,c,d);

  2. (ii)

    α(Fx)>b for xP(γ,α,b,d) with θ(Fx)>c;

  3. (iii)

    0R(γ,ψ,a,d) and ψ(Fx)<a for xR(γ,ψ,a,d) with ψ(x)=a.

Then F has at least three fixed points x 1 , x 2 , x 3 P ( γ , d ) ¯ such that

γ ( x i ) d  for  i = 1 , 2 , 3 , b < α ( x 1 ) , a < ψ ( x 2 )  with  α ( x 2 ) < b  and  ψ ( x 3 ) < a .

2 Main result

In this section we consider the existence of three positive solutions for problem (1.1).

We say that u is concave on [ 0 , T ] T if u (t)0 for t [ 0 , T ] T k T k .

We note that u(t) is a solution of problem (1.1) if and only if

u(t)={ 0 T ( T s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s B 0 ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) + 0 t ( t s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s , t [ 0 , T ] T , φ ( t ) , t [ r , 0 ] T .

Let E= C ld ( [ 0 , T ] T ,R) with u=max{ max t [ 0 , T ] T |u(t)|, max t [ 0 , T ] T k | u (t)|}, P={uE:u is nonnegative, decreasing and concave on  [ 0 , T ] T }. So E is a Banach space with the norm u and P is a cone in E. For each uE, extend u(t) to [ r , T ] T with u(t)=φ(t) for t [ r , 0 ] T .

Define F:PE by

( F u ) ( t ) = 0 T ( T s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s B 0 ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) + 0 t ( t s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s , t [ 0 , T ] T .

It is well known that this operator F is completely continuous.

We seek a fixed point u 1 of F in the cone P. Define

u(t)={ φ ( t ) , t [ r , 0 ] T , u 1 ( t ) , t [ 0 , T ] T .

Then u(t) denotes a positive solution of problem (1.1).

Lemma 2.1 If uP, then

  1. (i)

    FuP, i.e., F:PP.

  2. (ii)

    u(t) T t T max t [ 0 , T ] T |u(t)|, t [ 0 , T ] T .

  3. (iii)

    u(t) is decreasing on t [ 0 , T ] T .

Proof This is easy, so we omit it here. □

Let lT be fixed such that 0<l<η<T, and set

Y 1 = { t [ 0 , T ] T : μ ( t ) 0 } ; Y 2 = { t [ 0 , T ] T : μ ( t ) > 0 } ; Y 3 = Y 1 [ 0 , η ] T .

Throughout this paper, we assume Y 3 ϕ and Y 3 a(r)r>0.

Define the nonnegative continuous concave functional α, the nonnegative continuous convex functionals θ, γ, and the nonnegative continuous functional ψ on the cone P, respectively, as

γ ( u ) = u , θ ( u ) = max t [ 0 , l ] T k | u ( t ) | , α ( u ) = min t [ l , η ] T u ( t ) , ψ ( u ) = min t [ 0 , η ] T u ( t ) .

In addition, by Lemma 2.1, we have α(u)=ψ(u)=u(η) for each uP.

For convenience, we define

ρ = T ( B + 2 T ) Φ q ( 0 T a ( r ) r ) , δ = A Y 3 Φ q ( 0 s a ( r ) s ) s , λ = T ( B + T + η ) Φ q ( 0 T a ( r ) r ) .

We now state growth conditions on f so that BVP (1.1) has at least three positive solutions.

Theorem 2.1 Let 0< T η a<b<d, ρb<δd, and suppose that f satisfies the following conditions:

(H1) f(u,φ(s)) Φ p ( d ρ ) if 0ud uniformly in s [ r , 0 ] T ; f( u 1 , u 2 ) Φ p ( d ρ ) if 0 u i d, i=1,2,

(H2) f(u,φ(s))> Φ p ( b δ ) if bud uniformly in s [ r , 0 ] T ,

(H3) f(u,φ(s))< Φ p ( a λ ) if 0u T η a uniformly in s [ r , 0 ] T ; f( u 1 , u 2 )< Φ p ( a λ ) if 0 u i T η a, i=1,2.

Then BVP (1.1) has at least three positive solutions of the form

u(t)={ u i ( t ) , t [ 0 , T ] T , i = 1 , 2 , 3 , φ ( t ) , t [ r , 0 ] T ,

where γ( u i )d for i=1,2,3, b<α( u 1 ), a<ψ( u 2 ) with α( u 2 )<b and ψ( u 3 )<a.

Proof We first assert that F: P ( γ , d ) ¯ P ( γ , d ) ¯ .

Let u P ( γ , d ) ¯ , then γ(u)=ud, consequently, 0u(t)d for t [ 0 , T ] T .

From (H1), we have

| F u ( t ) | = 0 T ( T s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s B 0 ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) + 0 t ( t s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s < 0 T T Φ q ( 0 T a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s + B 0 T Φ q ( 0 T a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s + 0 T T Φ q ( 0 T a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s = T ( B + 2 T ) Φ q ( Y 1 a ( r ) f ( u ( r ) , φ ( μ ( r ) ) ) r + Y 2 a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) T ( B + 2 T ) Φ q ( 0 T a ( r ) r ) d ρ = d , | ( F u ) ( t ) | = | 0 t Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s | 0 T Φ q ( 0 T a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s = T Φ q ( Y 1 a ( r ) f ( u ( r ) , φ ( μ ( r ) ) ) r + Y 2 a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) T Φ q ( 0 T a ( r ) r ) d ρ = d B + 2 T d .

Therefore F(u) P ( γ , d ) ¯ , i.e., F: P ( γ , d ) ¯ P ( γ , d ) ¯ .

Secondly, we assert that {uP(γ,θ,α,b,c,d):α(u)>b}ϕ and α(Fu)>b for uP(γ,θ,α,b,c,d).

Let u(t)=kb with k= ρ δ >1, then u(t)=kb>b and θ(u)=0<b. Furthermore, by ρb<δd we have γ(u)d. Let c=kb, then {uP(γ,θ,α,b,c,d):α(u)>b}ϕ.

Moreover, uP(γ,θ,α,b,kb,d), we have bu(t)d, t [ 0 , η ] T .

From (H2), we see that

α ( F u ) = ( F u ) ( η ) = 0 T ( T s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s B 0 ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) + 0 η ( t s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s B 0 ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) A ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) A ( Y 3 Φ q ( 0 s a ( r ) f ( u ( r ) , φ ( μ ( r ) ) ) r ) s ) > A Y 3 Φ q ( 0 s a ( r ) s ) s b δ = b ,

as required.

Thirdly, we assert that α(Fu)>b for uP(γ,α,b,d) with θ(Fu)>c.

uP(γ,α,b,d) with θ(Fu)>kb, from Lemma 2.1 we have

θ ( F u ) = | ( F u ) ( l ) | = 0 l Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s > k b .

So,

α ( F u ) = ( F u ) ( η ) = 0 T ( T s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s B 0 ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) + 0 η ( t s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s B 0 ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) A ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) A ( 0 l Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) > A k b = A ρ δ b ( A + T ) b > b .

This implies that α(Fu)>b for uP(γ,α,b,d) with θ(Fu)>c.

Finally, we assert that 0R(γ,ψ,a,d) and ψ(Fu)<a for uR(γ,ψ,a,d) with ψ(u)=a.

As ψ(0)=0<a, we have 0R(γ,ψ,a,d). uR(γ,ψ,a,d) with ψ(u)= min t [ 0 , η ] T u(t)=u(η)=a, by Lemma 2.1 we have 0u(t) T T η a for t [ 0 , T ] T .

From (H3), we have

ψ ( F u ) = ( F u ) ( η ) = 0 T ( T s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s B 0 ( 0 η Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s ) + 0 η ( t s ) Φ q ( 0 s a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s < 0 T T Φ q ( 0 T a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s + B 0 T Φ q ( 0 T a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s + 0 T η Φ q ( 0 T a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) s = T ( B + T + η ) Φ q ( Y 1 a ( r ) f ( u ( r ) , φ ( μ ( r ) ) ) r + Y 2 a ( r ) f ( u ( r ) , u ( μ ( r ) ) ) r ) T ( B + T + η ) Φ q ( 0 T a ( r ) r ) a λ = a ,

which shows that condition (iii) of Theorem 1.1 is fulfilled.

Thus, all the conditions of Theorem 1.1 are satisfied. Hence, F has at least three fixed points u 1 , u 2 , u 3 satisfying

γ( u i )d for i=1,2,3,b<α( u 1 ),a<ψ( u 2 ) with α( u 2 )<b and ψ( u 3 )<a.

Let

u(t)={ u i ( t ) , t [ 0 , T ] T , i = 1 , 2 , 3 , φ ( t ) , t [ r , 0 ] T ,

which are three positive solutions of BVP (1.1) □