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Positive solutions of m-point integral boundary value problems for second-order p-Laplacian dynamic equations on time scales

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Abstract

In this article, we use the Krasnosel’skii fixed point theorem, the Avery-Henderson fixed point theorem and the Leggett-Williams fixed point theorem to obtain some results for the existence of at least one, two or three positive solutions of m-point integral boundary value problems for nonlinear second-order p-Laplacian dynamic equations on time scales. Two examples are presented to illustrate the applications of the results.

MSC:34B15, 34N05.

1 Introduction

Analysis on measure chains was initiated by Stefan Hilger [1] as a bridge between continuous and discrete calculus. Dynamic equations on time scales have been a component of applied analysis on measure chains to describe the processes that feature both continuous and discrete elements [26]. This subject not only gives a unified approach to the study of differential and difference equations, but also gives an extended approach to the study of dynamic equations with nonuniform step size or a combination of real and discrete domains. Further, the study of time scale equations has led to several important applications, e.g., in the study of economics, insect population models, heat transfer, stock market and epidemic models (see [710]), etc. Integral boundary value problems occur in the study of nonlocal phenomena in many different areas of applied mathematics, physics and engineering, e.g., in heat conduction, chemical engineering, underground water flow, thermo-elasticity, plasma physics, etc. (see [1115] and the references therein).

Throughout this paper, we denote the one-dimensional p-Laplacian operator by φ p (u), i.e., φ p (u)= | u | p 2 u for p>1 with φ p 1 = φ q , where 1/p+1/q=1. For convenience, we make the blanket assumption that 0, T are points in a time scale ; for an interval ( 0 , T ) T , we always mean (0,T)T. Other types of an interval are defined similarly.

In 2007, Sun and Li [16] discussed the existence of at least one, two or three positive solutions of the following boundary value problem:

( φ p ( u ( t ) ) ) +h(t)f ( u σ ( t ) ) =0,t [ a , b ] T ,
(1.1)
u(a) B 0 ( u ( a ) ) =0, u ( σ ( b ) ) =0.
(1.2)

They used the Krasnosel’skii fixed point theorem, the Avery-Henderson fixed point theorem and the Leggett-Williams fixed point theorem to prove the existence of multiple positive solutions to problem (1.1)-(1.2).

In 2009, Zhang and Qiao [17] studied the existence criteria for the m-point boundary value problem:

( φ p ( u ( t ) ) ) +a(t)f ( t , u ( t ) ) =0,t [ 0 , 1 ] T ,
(1.3)
u(0)=0,u(1)= i = 1 m 2 a i u( ξ i ).
(1.4)

They obtained some results for the existence of multiple positive solutions of problem (1.3)-(1.4) by using the Krasnosel’skii fixed point theorem, the Avery-Henderson fixed point theorem and the Leggett-Williams fixed point theorem.

In 2011, Li and Zhang [18] considered the existence of at least three positive solutions for the boundary value problem with integral boundary conditions:

( φ p ( x ( t ) ) ) +λf ( t , x ( t ) , x ( t ) ) =0,t ( 0 , T ) T ,
(1.5)
x (0)=0,αx(T)βx(0)= 0 T g(s)x(s)s.
(1.6)

They established some sufficient conditions for the existence of positive solutions to problem (1.5)-(1.6) by using the Legget-Williams fixed point theorem. For some recent results on the existence of positive solutions for p-Laplacian dynamic equations on time scales, see [1927]. However, to the best of the authors’ knowledge, existence results for positive solutions of m-point integral boundary value problems for nonlinear p-Laplacian dynamic equations on time scales have not been studied.

In this article, we are concerned with the existence of multiple positive solutions to the m-point integral boundary value problem for a second-order p-Laplacian dynamic equation on time scale :

( φ p ( u ( t ) ) ) +a(t)f ( t , u ( t ) ) =0,t [ 0 , 1 ] T ,
(1.7)
u Δ (0)=0,u(1)= i = 1 m 1 α i ξ i 1 ξ i u(s)Δs,
(1.8)

where is a time scale, 0= ξ 0 < ξ 1 < ξ 2 << ξ m 2 < ξ m 1 =1 and

(H1) 0< i = 1 m 1 α i ( ξ i ξ i 1 )<1 such that α i 0 for i{1,2,,m3}{m1}, α m 2 >0;

(H2) f C r d ( [ 0 , 1 ] T ×[0,),[0,));

(H3) a C r d ( [ 0 , 1 ] T ,[0,)) and there exists t 0 ( ξ m 2 , 1 ) T such that a( t 0 )>0.

The rest of the paper is organized as follows. In Section 2, we state and prove some lemmas which are used later. In Section 3, we use the Krasnosel’skii [28] fixed point theorem to obtain the existence of at least one positive solution of problem (1.7)-(1.8). In Section 4, by using the Avery-Henderson [29] fixed point theorem, we establish sufficient conditions for the existence of at least two positive solutions of problem (1.7)-(1.8). In Section 5, the existence of at least three positive solutions of problem (1.7)-(1.8) are proved by using the Leggett-Williams [30] fixed point theorem. Two illustrative examples are given in Section 6.

For convenience, we list the following well-known definitions which can be found in [4] and the references therein.

Definition 1.1 A time scale is an arbitrary nonempty closed subset of the real set with topology and ordering inherited from .

The forward and backward jump operators σ,ρ:TT and the graininess μ:T R + are defined, respectively, by

σ(t):=inf{sT|s>t},ρ(t):=sup{sT|s<t},μ(t):=σ(t)t,

for all tT. If σ(t)>t, t is said to be right scattered, and if ρ(t)<t, t is said to be left scattered; if σ(t)=t, t is said to be right dense, and if ρ(t)=t, t is said to be left dense. If has a left-scattered maximum M, define T k =T{M}; otherwise set T k =T.

Definition 1.2 A function f:TR is rd-continuous (rd-continuous is short for right-dense continuous) provided it is continuous at each right-dense point in and has a left-sided limit at each left-dense point in . The set of rd-continuous functions f:TR will be denoted by C r d (T)= C r d (T,R).

Definition 1.3 For f:TR and t T k , the delta derivative of f at the point t is defined to be the number f (t) (provided it exists), with the property that for each ϵ>0, there is a neighborhood U of t such that

| f ( σ ( t ) ) f ( s ) f ( t ) ( σ ( t ) s ) | ϵ | σ ( t ) s |

for all sU.

Definition 1.4 For a function f:TR, the delta derivative is defined at the point t by

f (t)= f ( σ ( t ) ) f ( t ) σ ( t ) t ,

if f is continuous at t and t is right-scattered. If t is not right-scattered, then the derivative is defined by

f (t)= lim s t f ( σ ( t ) ) f ( s ) σ ( t ) s = lim s t f ( t ) f ( s ) t s

provided this limit exists.

Definition 1.5 If F (t)=f(t), then we define the delta integral by

a t f(s)Δs=F(t)F(a).

2 Preliminaries

In this section, we first prove and recall some lemmas which are used in what follows.

Lemma 2.1 Let i = 1 m 1 α i ( ξ i ξ i 1 )1. Then, for y C r d ( [ 0 , 1 ] T ,R), the problem

( φ p ( u ( t ) ) ) +y(t)=0,t [ 0 , 1 ] T ,
(2.1)
u Δ (0)=0,u(1)= i = 1 m 1 α i ξ i 1 ξ i u(s)Δs,
(2.2)

has a unique solution

u ( t ) = 0 t φ q ( 0 τ y ( s ) Δ s ) Δ τ 1 1 i = 1 m 1 α i ( ξ i ξ i 1 ) i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ y ( s ) Δ s ) Δ τ Δ η + 1 1 i = 1 m 1 α i ( ξ i ξ i 1 ) 0 1 φ q ( 0 τ y ( s ) Δ s ) Δ τ .
(2.3)

Proof Integrating (2.1) from 0 to t and using the first condition of (2.2), one gets

u (t)= φ q ( 0 t y ( s ) s ) .
(2.4)

Integrating (2.4) from 0 to t, we obtain

u(t)=u(0) 0 t φ q ( 0 τ y ( s ) s ) τ.
(2.5)

In particular, for t=1, we have

u(1)=u(0) 0 1 φ q ( 0 τ y ( s ) s ) τ.

Using the second condition of (2.2), we get that

u ( 0 ) 0 1 φ q ( 0 τ y ( s ) s ) τ = u ( 0 ) i = 1 m 1 α i ( ξ i ξ i 1 ) i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ y ( s ) s ) τ η .

Hence,

u ( 0 ) = 1 1 i = 1 m 1 α i ( ξ i ξ i 1 ) [ 0 1 φ q ( 0 τ y ( s ) s ) τ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ y ( s ) s ) τ η ] .

Substituting the value of u(0) in (2.5), we obtain the solution (2.3). □

Lemma 2.2 Let i = 1 m 1 α i ( ξ i ξ i 1 )1. If y C r d ( [ 0 , 1 ] T ,[0,)), then the unique solution u of problem (2.1)-(2.2) satisfies

u Δ (t)0, u Δ Δ (t)0,t [ 0 , 1 ] T .

Proof From (2.4), we have u Δ (t)0 for t [ 0 , 1 ] T . In fact, φ q (x) is a monotone increasing continuously differentiable function and

( 0 t y ( s ) Δ s ) Δ =y(t)0.

Then, by the chain rule [4], we get u Δ Δ (t)0 for t [ 0 , 1 ] T . □

Lemma 2.3 Let 0< i = 1 m 1 α i ( ξ i ξ i 1 )<1. If y C r d ( [ 0 , 1 ] T ,[0,)), then the unique solution u of problem (2.1)-(2.2) satisfies

u(t)0,t [ 0 , 1 ] T .

Proof From Lemma 2.2, u Δ (t)0 for t [ 0 , 1 ] T , we know that u is nonincreasing on [ 0 , 1 ] T . Consequently, for each t 1 , t 2 T and t 1 t 2 , it holds that u( t 1 )u( t 2 ).

Therefore,

u(0)u( ξ 1 )u( ξ i 1 )u( ξ i )u( ξ m 2 )u(1).
(2.6)

If u(1)<0, then the second condition of (2.2) together with (2.6) implies that

u ( 1 ) = i = 1 m 1 α i ξ i 1 ξ i u ( s ) Δ s i = 1 m 1 α i u ( ξ i ) ( ξ i ξ i 1 ) u ( 1 ) i = 1 m 1 α i ( ξ i ξ i 1 ) .

This contradicts the fact that 0< i = 1 m 1 α i ( ξ i ξ i 1 )<1.

If u(0)<0, it follows that u(1)<0 since u is nonincreasing. Hence, we get a contradiction. Indeed, if u(0)<0 and u(1)<0, we again obtain a contradiction. □

Lemma 2.4 Let i = 1 m 1 α i ( ξ i ξ i 1 )>1. If y C r d ( [ 0 , 1 ] T ,[0,)), then problem (2.1)-(2.2) has no positive solutions.

Proof Suppose that problem (2.1)-(2.2) has a positive solution u satisfying u(t)0 for t [ 0 , 1 ] T . Then u( ξ i )0 for all i=1,,m1. By the second condition of (2.2) and (2.6), we have

u ( 1 ) = i = 1 m 1 α i ξ i 1 ξ i u ( s ) Δ s i = 1 m 1 α i u ( ξ i ) ( ξ i ξ i 1 ) u ( 1 ) i = 1 m 1 α i ( ξ i ξ i 1 ) > u ( 1 ) ,

getting a contradiction. □

Let E denote the Banach space C r d [ 0 , 1 ] T with the norm u= sup t [ 0 , 1 ] T |u(t)|. Define the cone PE, by

P = { u E | u ( t ) 0 , u Δ ( t ) 0 , u Δ Δ ( t ) 0  for  t [ 0 , 1 ] T , and  u Δ ( 0 ) = 0 , u ( 1 ) = i = 1 m 1 α i ξ i 1 ξ i u ( s ) Δ s } .
(2.7)

Define the operator A:PE by

A u ( t ) = 0 t φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η + 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ ,
(2.8)

where a positive constant Λ= i = 1 m 1 α i ( ξ i ξ i 1 )<1. In view of Lemma 2.1, the solutions of problem (1.7)-(1.8) are given by the operator equation, u(t)=Au(t).

From (2.8), we claim that for each uP, AuP and satisfies (1.8). In fact, for t [ 0 , 1 ] T , we get

A u ( t ) A u ( 1 ) = 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η + 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ = Λ 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η 0 .

This implies that Au(t)0 for t [ 0 , 1 ] T . As in Lemma 2.2, we can prove that ( A u ) Δ (t)0, ( A u ) Δ Δ (t)0 for t [ 0 , 1 ] T . In addition, we find that ( A u ) Δ (0)=0 and (Au)(1)= i = 1 m 1 α i ξ i 1 ξ i Au(s)Δs. So, A:PP. It is also easy to check that A:PP is completely continuous.

Lemma 2.5 Let (H1) hold. If uP, then

min t [ 0 , 1 ] T u(t)γu,
(2.9)

where

γ= α m 2 ( ξ m 2 ξ m 3 ) ( 1 ξ m 2 ) 1 α m 2 ξ m 2 ( ξ m 2 ξ m 3 ) ,
(2.10)

which γ>0.

Proof Since u Δ (t)0 for t [ 0 , 1 ] T , we have u=u(0), min t [ 0 , 1 ] T u(t)=u(1).

Thus,

u(1)= i = 1 m 1 α i ξ i 1 ξ i u(s)Δs i = 1 m 1 α i u( ξ i )( ξ i ξ i 1 ) α m 2 u( ξ m 2 )( ξ m 2 ξ m 3 ).
(2.11)

From u Δ Δ (t)0 for t [ 0 , 1 ] T and (2.11), we get

u ( 0 ) u ( 1 ) + u ( 1 ) u ( ξ m 2 ) 1 ξ m 2 ( 0 1 ) u ( 1 ) [ 1 1 1 ξ m 2 + 1 α m 2 ( ξ m 2 ξ m 3 ) ( 1 ξ m 2 ) ] = u ( 1 ) [ 1 α m 2 ξ m 2 ( ξ m 2 ξ m 3 ) α m 2 ( ξ m 2 ξ m 3 ) ( 1 ξ m 2 ) ] .

This implies that

min t [ 0 , 1 ] T u(t) α m 2 ( ξ m 2 ξ m 3 ) ( 1 ξ m 2 ) 1 α m 2 ξ m 2 ( ξ m 2 ξ m 3 ) u.

Note that (H1) yields

0<1 i = 1 m 1 α i ( ξ i ξ i 1 )<1 α m 2 ( ξ m 2 ξ m 3 )<1 α m 2 ξ m 2 ( ξ m 2 ξ m 3 ).

Thus we have γ>0. The proof of Lemma 2.5 is complete. □

In the following, for the sake of convenience, we set constants

L= 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ ,
(2.12)
M= 1 Λ γ ξ m 2 ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ ,
(2.13)
N= 1 Λ γ Λ ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ .
(2.14)

3 Existence of at least one positive solution

Now we are in a position to establish the main result. Our first result is based on the Krasnosel’skii fixed point theorem.

Theorem 3.1 (see [28])

Let E be a Banach space, and let PE be a cone. Assume that Ω 1 and Ω 2 are bounded open subsets of E with 0 Ω 1 , Ω ¯ 1 Ω 2 , and let A:P( Ω ¯ 2 Ω 1 )P be a completely continuous operator such that either

  1. (i)

    Auu for uP Ω 1 , Auu for uP Ω 2 ; or

  2. (ii)

    Auu for uP Ω 1 , Auu for uP Ω 2 hold.

Then A has a fixed point in P( Ω ¯ 2 Ω 1 ).

Theorem 3.2 Assume that (H1)-(H3) hold. In addition, suppose that there exist numbers 0<r<R< such that

(A1) f(t,u) φ p (L) φ p (r) for t [ 0 , 1 ] T and 0ur;

(A2) f(t,u) φ p (Mr) φ p (R) for t [ ξ m 2 , 1 ] T and Ru<,

where constants L, M are defined by (2.12) and (2.13), respectively.

Then problem (1.7)-(1.8) has at least one positive solution.

Proof Firstly, we define a cone P and a completely continuous operator A:PP as in (2.7) and (2.8), respectively.

Let Ω 1 ={u C r d ( [ 0 , 1 ] T ):u<r}. For any uP Ω 1 with u=r, from condition (A1), we obtain

A u ( t ) = 0 t φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η + 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ φ q ( φ p ( L ) φ p ( r ) ) 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = r L 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = r = u .

This implies that Auu for uP Ω 1 .

Set Ω 2 ={u C r d ( [ 0 , 1 ] T ):u<R}. Since uP Ω 2 , it follows that min t [ 0 , 1 ] T u(t)γu=γR. Hence from condition (A2), for any uP Ω 2 , we have

A u A u ( ξ m 2 ) = 0 ξ m 2 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η + 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ = 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 0 ξ m 2 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 Λ + 1 1 Λ [ Λ 0 ξ m 2 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η ] 1 1 Λ ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ ξ m 2 1 ξ m 2 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η 1 1 Λ ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ ξ m 2 1 ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η = ξ m 2 1 Λ ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ ξ m 2 1 Λ ξ m 2 1 φ q ( ξ m 2 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ φ q ( φ p ( M γ ) φ p ( R ) ) ξ m 2 ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ 1 Λ = M γ R 1 Λ ξ m 2 ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ = u .

Therefore, Auu for uP Ω 2 .

Thus, from Theorem 3.1, it follows that A has a fixed point u in P( Ω ¯ 2 Ω 1 ) such that ruR. Therefore, problem (1.7)-(1.8) has at least one positive solution. □

4 Existence of at least two positive solutions

In this section, we obtain the existence of at least two positive solutions of problem (1.7)-(1.8) by using the Avery-Henderson fixed point theorem which is as follows.

Theorem 4.1 (see [29])

Let P be a cone in a real Banach space E. Set

P(Φ, ρ 3 )= { u P | Φ ( u ) < ρ 3 } .

Let ν and Φ be increasing nonnegative continuous functionals on P, and let θ be a nonnegative continuous functional on P with θ(0)=0 such that, for some ρ 3 >0 and N>0,

Φ(u)θ(u)ν(u)anduNΦ(u)

for all u P ( Φ , ρ 3 ) ¯ . Suppose there exist a completely continuous operator A: P ( Φ , ρ 3 ) ¯ P and 0< ρ 1 < ρ 2 < ρ 3 such that

θ(λu)=λθ(u) for 0λ1 and uP(θ, ρ 2 ),

and

  1. (i)

    Φ(Au)> ρ 3 for all uP(Φ, ρ 3 );

  2. (ii)

    θ(Au)< ρ 2 for all uP(θ, ρ 2 );

  3. (iii)

    P(ν, ρ 1 ) and ν(Au)> ρ 1 for all uP(ν, ρ 1 ).

Then A has at least two fixed points u 1 and u 2 belonging to P ( Φ , ρ 3 ) ¯ satisfying

ρ 1 <ν( u 1 ) with θ( u 1 )< ρ 2 ,and ρ 2 <θ( u 2 ) with Φ( u 2 )< ρ 3 .

Define a constant l ( 0 , 1 ) T such that 0< ξ m 2 <l<1. Let Φ, θ and ν be increasing, nonnegative and continuous functionals on P, defined by

Φ(u)=u( ξ m 2 ),θ(u)=u( ξ m 2 ),ν(u)=u(l).

Obviously, Φ(u)=θ(u)ν(u) for each uP. Moreover, Lemma 2.5 implies Φ(u)=u( ξ m 2 )γu for each uP. It is easy to see that θ(0)=0 and θ(λu)=λθ(u) for all 0λ1 and uP(θ, ρ 2 ).

We can now prove the following theorem.

Theorem 4.2 Assume that (H1)-(H3) hold, and suppose that there exist positive numbers ρ 1 < ρ 2 < ρ 3 such that the function f satisfies the following conditions:

(B1) f(t,u)> φ p (Nγ) φ p ( ρ 1 ) for t [ ξ m 2 , l ] T and u[γ ρ 1 , ρ 1 ];

(B2) f(t,u)< φ p (L) φ p ( ρ 2 ) for t [ ξ m 2 , 1 ] T and u[0, ρ 2 ];

(B3) f(t,u)> φ p (Mγ) φ p ( ρ 3 ) for t [ ξ m 2 , l ] T and u[ ρ 3 ,(1/γ) ρ 3 ],

where constants L, M, N are defined by (2.12), (2.13) and (2.14), respectively.

Then problem (1.7)-(1.8) has at least two positive solutions u 1 and u 2 such that ρ 1 < u 1 (l) with u 1 ( ξ m 2 )< ρ 2 and ρ 2 < u 2 ( ξ m 2 ) with u 2 ( ξ m 2 )< ρ 3 .

Proof We now wish to prove that all of the conditions of Theorem 4.1 are satisfied. For this purpose, we define the cone P as (2.7) and a completely continuous operator A:PP by (2.8).

To check condition (i) of Theorem 4.1, we choose uP(Φ, ρ 3 ), then Φ(u)= ρ 3 . This implies that ρ 3 u(1/γ)Φ(u)=(1/γ) ρ 3 . For t [ ξ m 2 , 1 ] T , we have ρ 3 u(t)(1/γ) ρ 3 . From condition (B3), we get that f(t,u)> φ p (Mγ) φ p ( ρ 3 ) for t [ ξ m 2 , l ] T . Since AuP, we obtain

Φ ( A u ) = ( A u ) ( ξ m 2 ) = 0 ξ m 2 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η + 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ = 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 0 ξ m 2 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 Λ + 1 1 Λ [ Λ 0 ξ m 2 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η ] 1 1 Λ ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ ξ m 2 1 ξ m 2 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η 1 1 Λ ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ ξ m 2 1 ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η = ξ m 2 1 Λ ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ ξ m 2 1 Λ ξ m 2 1 φ q ( ξ m 2 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ > φ q ( φ p ( M γ ) φ p ( ρ 3 ) ) ξ m 2 ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ 1 Λ = M γ ρ 3 1 Λ ξ m 2 ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ = ρ 3 .

Hence, condition (i) of Theorem 4.1 holds.

We now prove that condition (ii) in Theorem 4.1 holds. In fact, for uP(θ, ρ 2 ), we have θ(u)= ρ 2 . This implies that 0u(t)u(1/γ) ρ 2 for t [ ξ m 2 , 1 ] T . From condition (B2), we have

θ ( A u ) = ( A u ) ( ξ m 2 ) 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ < φ q ( φ p ( L ) φ p ( ρ 2 ) ) 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = L ρ 2 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = ρ 2 = u .

This shows that condition (ii) of Theorem 4.1 is satisfied.

Now, we assert that condition (iii) of Theorem 4.1 also holds. If u(t)= ρ 1 /2 for t [ 0 , 1 ] T , then ν(u)= ρ 1 /2. Thus P(ν, ρ 1 ). Let uP(ν, ρ 1 ), then ν(u)=u(l)= ρ 1 . So that γ ρ 1 u(t)u ρ 1 . From condition (B1), for any AuP, we have

ν ( A u ) = ( A u ) ( l ) ( A u ) ( 1 ) = 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η + 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ = 1 1 Λ [ Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η ] = 1 1 Λ [ i = 1 m 1 α i ξ i 1 ξ i 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η ] = 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i η 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η Λ 1 Λ ξ m 2 1 φ q ( ξ m 2 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ > φ q ( φ p ( N γ ) φ p ( ρ 1 ) ) Λ ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ 1 Λ = N γ ρ 1 Λ 1 Λ ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ = ρ 1 .

Therefore, condition (iii) of Theorem 4.1 is satisfied.

Thus, by Theorem 4.1, problem (1.7)-(1.8) has at least two positive solutions u 1 and u 2 such that ρ 1 < u 1 (l) with u 1 ( ξ m 2 )< ρ 2 and ρ 2 < u 2 ( ξ m 2 ) with u 2 ( ξ m 2 )< ρ 3 . □

5 Existence of at least three positive solutions

In this section, we use the Leggett-Williams fixed point theorem to prove the existence of at least three positive solutions to problem (1.7)-(1.8). The Leggett-Williams fixed point theorem is as follows.

Theorem 5.1 (see [30])

Let P be a cone in the real Banach space E. Set

P r = { x P | x < r } ,P(Ψ,a,b)= { x P | a Ψ ( x ) , x b } .

Let A: P ¯ r P ¯ r be a completely continuous operator and let Ψ be a nonnegative continuous concave functional on P with Ψ(u)u for all u P ¯ r . Suppose that there exists 0< ρ 1 < ρ 2 <(1/γ) ρ 2 < ρ 3 such that the following conditions hold:

  1. (i)

    {uP(Ψ, ρ 2 ,(1/γ) ρ 2 )|Ψ(u)> ρ 2 } and Ψ(Au)> ρ 2 for all uP(Ψ, ρ 2 ,(1/γ) ρ 2 );

  2. (ii)

    Au< ρ 1 for u ρ 1 ;

  3. (iii)

    Ψ(Au)> ρ 2 for uP(Ψ, ρ 2 , ρ 3 ) with Au>(1/γ) ρ 2 .

Then A has at least three fixed points u 1 , u 2 and u 3 in P ¯ r satisfying u 1 < ρ 1 , Ψ( u 2 )> ρ 2 , ρ 1 < u 3 with Ψ( u 3 )< ρ 2 .

We now prove the following result.

Theorem 5.2 Assume that (H1)-(H3) hold. Suppose that there exist constants 0< ρ 1 < ρ 2 <(1/γ) ρ 2 ρ 3 such that

(C1) f(t,u) φ p (L) φ p ( ρ 3 ) for t [ ξ m 2 , 1 ] T and u[0, ρ 3 ];

(C2) f(t,u)> φ p (Mγ) φ p ( ρ 2 ) for t [ ξ m 2 , 1 ] T and u[ ρ 2 ,(1/γ) ρ 2 ];

(C3) f(t,u)< φ p (L) φ p ( ρ 1 ) for t [ ξ m 2 , 1 ] T and u[0, ρ 1 ],

where constants L, M are defined by (2.12) and (2.13), respectively.

Then problem (1.7)-(1.8) has at least three positive solutions u 1 , u 2 and u 3 such that u 1 < ρ 1 , u 2 ( ξ m 2 )> ρ 2 , u 3 > ρ 1 with u 3 ( ξ m 2 )< ρ 2 .

Proof We will show that all the conditions of Leggett-Williams Theorem 5.1 hold with respect to the operator A defined in (2.8).

At first, we define a nonnegative continuous concave functional Ψ:P[0,) by Ψ(u)=u( ξ m 2 ), where the cone P is defined by (2.7). In fact, for uP, we get Ψ(u)u. If u P ¯ ρ 3 , then u ρ 3 . From condition (C1), we obtain

A u ( t ) = 0 t φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η + 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ φ q ( φ p ( L ) φ p ( ρ 3 ) ) 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = L ρ 3 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = ρ 3 .

This implies that Au ρ 3 . Therefore, we have A: P ¯ ρ 3 P ¯ ρ 3 . Since ( ρ 2 /γ)P(Ψ, ρ 2 ,( ρ 2 /γ)) and Ψ(( ρ 2 /γ))=( ρ 2 /γ)> ρ 2 , then {uP(Ψ, ρ 2 ,( ρ 2 /γ))|Ψ(u)> ρ 2 }. For uP(Ψ, ρ 2 ,( ρ 2 /γ)), we get ρ 2 u( ξ m 2 )u( ρ 2 /γ). By using condition (C2), we obtain

Ψ ( A u ) = ( A u ) ( ξ m 2 ) = 0 ξ m 2 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η + 1 1 Λ 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ = 0 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 0 ξ m 2 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 Λ + 1 1 Λ [ Λ 0 ξ m 2 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ i = 1 m 1 α i ξ i 1 ξ i 0 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η ] 1 1 Λ ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ ξ m 2 1 ξ m 2 η φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η 1 1 Λ ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ 1 1 Λ ξ m 2 1 ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ Δ η = ξ m 2 1 Λ ξ m 2 1 φ q ( 0 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ ξ m 2 1 Λ ξ m 2 1 φ q ( ξ m 2 τ a ( s ) f ( s , u ( s ) ) Δ s ) Δ τ > φ q ( φ p ( M γ ) φ p ( ρ 2 ) ) ξ m 2 ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ 1 Λ = M γ ρ 2 1 Λ ξ m 2 ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ = ρ 2 .

Hence, condition (i) of Theorem 5.1 is satisfied.

Indeed, if u ρ 1 , then condition (C3) implies that

( A u ) ( t ) < φ q ( φ p ( L ) φ p ( ρ 1 ) ) 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = L ρ 1 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = ρ 1 .

Thus Au< ρ 1 . Therefore, condition (ii) of Theorem 5.1 holds.

We finally show that condition (iii) of Theorem 5.1 also holds. Assume that uP(Ψ, ρ 2 , ρ 3 ), with Au>(1/γ) ρ 2 . Then we obtain

Ψ ( A u ) = ( A u ) ( ξ m 2 ) ( A u ) ( 1 ) γ A u > ρ 2 .

So, condition (iii) of Theorem 5.1 is satisfied. Therefore, an application of Theorem 5.1 implies that problem (1.7)-(1.8) has at least three positive solutions u 1 , u 2 and u 3 such that u 1 < ρ 1 , u 2 ( ξ m 2 )> ρ 2 and u 3 > ρ 1 with u 3 ( ξ m 2 )< ρ 2 . □

6 Numerical examples

In this section, we present some examples to illustrate our results.

Example 6.1 Consider the following six-point integral boundary value problem with p=3 and T=R:

( φ p ( u ( t ) ) ) +f ( t , u ( t ) ) =0,t [ 0 , 1 ] T ,
(6.1)
u Δ (0)=0,u(1)= 1 4 0 1 / 5 u(s)Δs+ 1 5 2 / 5 3 / 5 u(s)Δs+2 3 / 5 4 / 5 u(s)Δs,
(6.2)

where

f(t,u)={ 1 100 t + u 3 , t [ 0 , 1 ] , u [ 0 , 1 5 ] , 1 100 t + u 3 + 100 ( u 1 5 ) 1 / 4 , t [ 0 , 1 ] , u [ 1 5 , 3 5 ] , 1 100 t + u 3 + 100 ( u 1 5 ) 1 / 4 + 10 ( u 3 5 ) , t [ 0 , 1 ] , u [ 3 5 , ) .

Set α 1 =1/4, α 3 =1/5, α 4 =2, α 2 = α 5 =0, ξ 0 =0, ξ 1 =1/5, ξ 2 =2/5, ξ 3 =3/5, ξ 4 =4/5, ξ 5 =1 and a(t)=1. We can show that

Λ= i = 1 5 α i ( ξ i ξ i 1 )= 49 100 <1.

Through a simple calculation we can get

γ = α m 2 ( ξ m 2 ξ m 3 ) ( 1 ξ m 2 ) 1 α m 2 ξ m 2 ( ξ m 2 ξ m 3 ) = 2 17 , M = 1 Λ γ ξ m 2 ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ = 2601 64 5 , L = 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = 153 200 .

Choose r=1/5 and R=3/5, then f(t,u) satisfies

f(t,u) 1 100 + ( 1 5 ) 3 < ( 153 200 × 1 5 ) 2 = φ 3 (Lr),t[0,1],u [ 0 , 1 5 ] ,

and

f ( t , u ) 1 100 ( 4 5 ) + ( 3 5 ) 3 + 100 ( 3 5 1 5 ) 1 / 4 > ( 2601 5 64 × 2 17 × 3 5 ) 2 = φ 3 ( M γ R ) , t [ 4 5 , 1 ] , u [ 3 5 , ) .

By Theorem 3.2, we have that boundary value problem (6.1)-(6.2) has at least one positive solution.

Example 6.2 Consider the following six-point integral boundary value problem with p=2 and T={0}{1/ 2 n :nN}( 1 2 ,1] ( stands for the natural number set).

( φ p ( u ( t ) ) ) +f ( t , u ( t ) ) =0,t [ 0 , 1 ] T ,
(6.3)
u Δ (0)=0,u(1)= 1 4 0 1 / 16 u(s)Δs+ 1 6 1 / 8 1 / 4 u(s)Δs+3 1 / 4 1 / 2 u(s)Δs,
(6.4)

where

f(t,u)={ 1 50 t + 1 100 u , t [ 1 2 , 1 ] , u [ 0 , 1 ] , 1 50 t + 1 100 u + 7 ( u 1 ) 1 / 6 , t [ 1 2 , 1 ] , u [ 1 , 2 ] , 1 50 t + 1 100 u + 7 ( u 1 ) 1 / 6 + 1 20 ( u 2 ) 1 / 2 , t [ 1 2 , 1 ] , u [ 2 , 4096 585 ] , 1 50 t + 1 100 u + 7 ( u 1 ) 1 / 6 + 1 20 ( u 2 ) 1 / 2 + 1 40 ( u 4096 585 ) , t [ 1 2 , 1 ] , u [ 4096 585 , 30 ] .

Set α 1 =1/4, α 3 =1/6, α 4 =3, α 2 = α 5 =0, ξ 0 =0, ξ 1 =1/16, ξ 2 =1/8, ξ 3 =1/4, ξ 4 =1/2, ξ 5 =1 and a(t)=1. We can show that

Λ= i = 1 5 α i ( ξ i ξ i 1 )= 151 192 <1.

Through a simple calculation we can get

γ = α m 2 ( ξ m 2 ξ m 3 ) ( 1 ξ m 2 ) 1 α m 2 ξ m 2 ( ξ m 2 ξ m 3 ) = 3 5 , M = 1 Λ γ ξ m 2 ξ m 2 1 φ q ( ξ m 2 τ a ( s ) Δ s ) Δ τ = 205 36 , L = 1 Λ 0 1 φ q ( 0 τ a ( s ) Δ s ) Δ τ = 41 88 .

Choose ρ 1 =1, ρ 2 =2 and ρ 3 =30, then f(t,u) satisfies

f(t,u) 1 50 + 1 100 < 41 88 ×1= φ 2 (L ρ 1 ),t [ 1 2 , 1 ] ,u[0,1],

and

f ( t , u ) 1 50 ( 1 2 ) + 1 100 ( 2 ) + 7 ( 2 1 ) 1 / 6 > 205 36 × 3 5 × 2 = φ 2 ( M γ ρ 2 ) , t [ 1 2 , 1 ] , u [ 2 , 4096 585 ] ,

and

f ( t , u ) 1 50 + 1 100 ( 30 ) + 7 ( 30 1 ) 1 / 6 + 1 20 ( 30 2 ) 1 / 2 + 1 40 ( 30 4096 585 ) < 41 88 × 30 = φ 2 ( L ρ 3 ) , t [ 1 2 , 1 ] , u [ 0 , 30 ] .

By Theorem 5.2, we get that problem (6.3)-(6.4) has at least three positive solutions u 1 , u 2 and u 3 such that u 1 <1, u 2 ( 1 2 )>2 and u 3 >1 with u 3 ( 1 2 )<2.

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Acknowledgements

We would like to thank the reviewers for their valuable comments and suggestions on the manuscript. This research is supported by King Mongkut’s University of Technology North Bangkok, Thailand.

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Correspondence to Jessada Tariboon.

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Keywords

  • positive solution
  • p-Laplacian
  • time scales
  • fixed point theorem
  • integral boundary condition