1 Introduction

In this paper, we are concerned with the existence of solutions of the following nonlocal p-Laplacian dynamic equation on a time scale T:

( ϕ p ( u ( t ) ) ) = λ a ( t ) f ( u ( t ) ) ( 0 T f ( u ( s ) ) s ) k ,t ( 0 , T ) T ,
(1.1)

with integral initial value

u ( 0 ) = 0 T g ( s ) u ( s ) s , u ( 0 ) = A ,
(1.2)

where ϕ p () is the p-Laplace operator defined by ϕ p (s)= | s | p 2 s, p>1, ϕ p 1 = ϕ q with q the Hölder conjugate of p, i.e., 1 p + 1 q =1, λ>0, k>0, f: [ 0 , T ] T R + is continuous ( R + denotes positive real numbers), a: [ 0 , T ] T R + is left dense continuous, g(s) L 1 ( [ 0 , T ] T ) and A is a real constant.

This model arises in ohmic heating phenomena, which occur in shear bands of metals which are deformed at high strain rates [1, 2], in the theory of gravitational equilibrium of polytropic stars [3], in the investigation of the fully turbulent behavior of real flows, using invariant measures for the Euler equation [4], in modeling aggregation of cells via interaction with a chemical substance (chemotaxis) [5]. For the one-dimensional case, problems with the nonlocal initial condition appear in the investigation of diffusion phenomena for a small amount of gas in a transparent tube [6, 7]; nonlocal initial value problems in higher dimension are important from the point of view of their practical applications to modeling and investigating of pollution processes in rivers and seas, which are caused by sew-age [8].

The study of dynamic equations on time scales has led to some important applications [911], and an amount of literature has been devoted to the study the existence of solutions of second-order nonlinear boundary value problems (e.g., see [1218]).

Motivated by the above works, in this paper, we study the existence of solutions to Problem (1.1), (1.2). Compared with the works mentioned above, this article has the following new features: firstly, the main technique used in this paper is the topological degree method; secondly, Problem (1.1), (1.2) involves the integral initial condition.

The paper is organized as follows. We introduce some necessary definitions and lemmas in the rest of this section. In Section 2, we provide some necessary preliminaries, and in Section 3, the main results are stated and proved.

Definition 1.1 For t<supT and r>infT, define the forward jump operator σ and the backward jump operator ρ, respectively,

σ(t)=inf{τTτ>t}T,ρ(r)=sup{τTτ<r}T

for all t,rT. If σ(t)>t, t is said to be right scattered, and if ρ(r)<r, r is said to be left scattered. If σ(t)=t, t is said to be right dense, and if ρ(r)=r, r is said to be left dense. If T has a right scattered minimum m, define ; otherwise, set . If T has a left scattered maximum M, define ; otherwise, set .

Definition 1.2 For x:TR and , we define the delta derivative of x(t), x (t), to be the number (when it exists) with the property that for any ε>0, there is a neighborhood U of t such that

| [ x ( σ ( t ) ) x ( s ) ] x ( t ) [ σ ( t ) s ] | <ε | σ ( t ) s |

for all sU. For x:TR and , we define the nabla derivative of x(t), x (t), to be the number (when it exists) with the property that for any ε>0, there is a neighborhood V of t such that

| [ x ( ρ ( t ) ) x ( s ) ] x ( t ) [ ρ ( t ) s ] | <ε | ρ ( t ) s |

for all sV.

Definition 1.3 If F (t)=f(t), then we define the delta integral by

a t f(s)s=F(t)F(a).

If Φ (t)=f(t), then we define the nabla integral by

a t f(s)s=Φ(t)Φ(a).

Throughout this paper, we assume that T is a nonempty closed subset of ℝ with 0 T k , T T k .

Lemma 1.1 (Alternative theorem)

Suppose that X is a Banach space and A is a completely continuous operator from X to X. Then for any λ0, only one of the following statements holds:

  1. (i)

    For any yX, there exists a unique xX, such that

    (AλI)x=y;
  2. (ii)

    There exists an xX, x0, such that

    (AλI)x=0.

2 Preliminaries

Let E= C l d ( [ 0 , T ] T ,R) be a Banach space equipped with the maximum norm u=max lim [ 0 , T ] T |u(t)|.

Consider the following problem:

(2.1)
(2.2)

where yC( [ 0 , T ] T ), 0 T g(s)s1.

Integrating Eq. (2.1) from 0 to t, one obtains

ϕ p ( x ( t ) ) ϕ p ( x ( 0 ) ) = 0 t y(s)s.

Using the initial condition (2.2), we have

x (t)= ϕ p 1 ( ϕ p ( A ) 0 t y ( s ) s ) .

Integrating the above equality from 0 to t again, we obtain

x(t) 0 T g(s)x(s)s= 0 t ϕ p 1 ( ϕ p ( A ) 0 τ y ( s ) s ) τ.
(2.3)

Let F(t):= 0 t ϕ p 1 ( ϕ p (A) 0 τ y(s)s)τ.

Define an operator K: C l d ( [ 0 , T ] T ) C l d ( [ 0 , T ] T ) by

(Kx)= 0 T g(s)x(s)s,

then (2.3) can be rewritten as

(IK)x(t)=F(t).
(2.4)

Thus, x(t) is a solution to (2.1), (2.2) if and only if it is a solution to (2.4).

Lemma 2.1 IK is a Fredholm operator.

Proof To prove that IK is a Fredholm operator, we need only to show that K is completely continuous.

It is easy to see from the definition of K that K is a bounded linear operator from C l d ( [ 0 , T ] T ) to C l d ( [ 0 , T ] T ). Obviously, dimR(K)=1. So, K is a completely continuous operator. This completes the proof. □

Lemma 2.2 Problem (2.1), (2.2) admits a unique solution.

Proof Since Problem (2.1), (2.2) is equivalent to Problem (2.4), we need only to show that Problem (2.4) has a unique solution.

Using Lemma 2.1 and the alternative theorem, it is sufficient to prove that

(IK)x(t)=0
(2.5)

has a trivial solution x0 only.

On the contrary, suppose (2.5) has a nontrivial solution μ, then μ is a constant, and we have

Iμ=Kμ=μ.

The definition of K and the above equality yield

[ 1 0 T g ( s ) s ] μ=0,

which is a contradiction to the assumptions 0 T g(s)s1 and μ0.

Thus, we complete the proof. □

3 Main results

Throughout this section, we assume that the following conditions hold.

(H1) 0 T |g(s)|s=M<1;

(H2) f: [ 0 , T ] T R + is continuous;

(H3) a: [ 0 , T ] T R + is left dense continuous and max t [ 0 , T ] T a(t) M 1 ;

(H4) f(y) [ c 1 ϕ p ( | y | ) + c 2 ] 1 1 k , c 1 , c 2 >0 and c 1 < ϕ p ( 1 M 2 q 1 T ) λ M 1 T 1 k , when k<1;

(H5) f(y) [ c 3 ϕ p ( | y | ) ] 1 1 k , c 3 >0 and c 3 < ϕ p ( 1 M 2 q 1 T ) λ M 1 T 1 k , when k>1.

From Lemma 2.2 we know that u(t) is a solution to Problem (1.1), (1.2) if and only if it is a solution to the following integral equation:

(IK)u(t)= 0 t ϕ p 1 ( ϕ p ( A ) 0 τ λ a ( s ) f ( u ( s ) ) ( 0 T f ( u ( s ) ) s ) k s ) τ.
(3.1)

Define an operator F: C l d ( [ 0 , T ] T ) C l d ( [ 0 , T ] T ) by

(Fu)(t)= 0 t ϕ p 1 ( ϕ p ( A ) 0 τ λ a ( s ) f ( u ( s ) ) ( 0 T f ( u ( s ) ) s ) k s ) τ,

then (3.1) can be rewritten as

(IK)u(t)=(Fu)(t).

In order to prove the existence of solutions to (3.1), we need the following lemmas.

Lemma 3.1 F is completely continuous.

Proof Let R 1 be an arbitrary positive real number and denote B 1 ={u C l d ( [ 0 , T ] T );u R 1 }. Then we have for any u B 1 ,

| ( F u ) ( t ) | 0 t | ϕ p 1 ( ϕ p ( A ) 0 τ λ a ( s ) f ( u ( s ) ) ( 0 T f ( u ( s ) ) s ) k s ) | τ 0 T ϕ p 1 ( | ϕ p ( A ) | + | 0 T λ a ( s ) sup u B 1 f ( T inf u B 1 f ) k s | ) τ ϕ p 1 ( | ϕ p ( A ) | + M 1 T λ sup u B 1 f ( T inf u B 1 f ) k ) T .

This shows that F( B 1 ) is uniformly bounded.

Moreover, for any t [ 0 , T ] T , we have

| ( F u ) ( t ) | = | ϕ p 1 ( ϕ p ( A ) 0 t λ a ( s ) f ( u ( s ) ) ( 0 T f ( u ( s ) ) s ) k s ) | ϕ p 1 ( | ϕ p ( A ) | + M 1 T λ sup u B 1 f ( T inf u B 1 f ) k ) .

Thus, it is easy to prove that F( B 1 ) is equicontinuous. This together with the Ascoli-Arzelà theorem guarantees that F( B 1 ) is relatively compact in C l d ( [ 0 , T ] T ).

Therefore, F is completely continuous. The proof of Lemma 3.1 is completed. □

Theorem 3.1 Assume that conditions (H1)-(H5) hold. Then Problem (1.1), (1.2) has at least one solution.

Proof Lemma 2.1 and Lemma 3.1 imply that the operator K+F is completely continuous. It suffices for us to prove that the equation

( I ( K + F ) ) u=0
(3.2)

has at least one solution.

Define H:[0,1]× C l d ( [ 0 , T ] T ) C l d ( [ 0 , T ] T ) as

H(σ,u)=(K+σF)u,

and it is clear that H is completely continuous.

Set h σ (u)=uH(σ,u), then we have

To apply the Leray-Schauder degree to h σ , we need only to show that there exists a ball B R (θ) in C l d ( [ 0 , T ] T ), whose radius R will be fixed later, such that θ h σ ( B R (θ)).

If k<1, choosing R> 2 q 1 ϕ p 1 ( | ϕ p ( A ) | + λ M 1 T 1 k c 2 ) T 1 M 2 q 1 ϕ p 1 ( λ M 1 T 1 k c 1 ) T , then for any fixed u B R (θ), there exists a t 0 [ 0 , T ] T such that |u( t 0 )|=R. By direct calculation, we have

(3.3)

From (H4), we have

| ( h σ u ) ( t 0 ) | ( 1 M ) R 0 T ϕ p 1 ( | ϕ p ( A ) | + λ M 1 ( 0 T f ( u ( s ) ) s ) 1 k ) τ ( 1 M ) R 0 T ϕ p 1 [ | ϕ p ( A ) | + λ M 1 T 1 k ( c 1 ϕ p ( u ) + c 2 ) ] τ > 0 .
(3.4)

If k>1, choosing R> 2 q 1 | A | T 1 M 2 q 1 ϕ p 1 ( λ M 1 T 1 k c 3 ) T , then for any fixed u B R (θ), there exists a t 0 [ 0 , T ] T such that |u( t 0 )|=R. From (H5), we have

| ( h σ u ) ( t 0 ) | ( 1 M ) R 0 T ϕ p 1 ( | ϕ p ( A ) | + λ M 1 ( 0 T f ( u ( s ) ) s ) k 1 ) τ ( 1 M ) R 0 T ϕ p 1 [ | ϕ p ( A ) | + λ M 1 T 1 k c 3 ϕ p ( u ) ] τ > 0 .
(3.5)

If k=1, choosing R> ϕ p 1 ( | ϕ p ( A ) | + λ M 1 ) T 1 M , then for any fixed u B R (θ), there exists a t 0 [ 0 , T ] T such that |u( t 0 )|=R. By direct calculation, we have

| ( h σ u ) ( t 0 ) | ( 1 M ) R 0 T ϕ p 1 ( | ϕ p ( A ) | + λ M 1 ) τ > 0 .
(3.6)

This implies h σ uθ and hence we obtain θ h σ ( B R (θ)).

Since deg( h 1 , B R (θ),θ)=deg( h 0 , B R (θ),θ)=±10, we know that (3.2) admits a solution u B R (θ), which implies that (1.1), (1.2) also admits a solution in B R (θ). □