1 Introduction

Let T be a time scale, i.e., T is a nonempty closed subset of R. Let 0, T be points in T, an interval (0, T) T denoting time scales interval, that is, (0, T) T : = (0, T) ⋂ T. Other types of intervals are defined similarly.

The theory of impulsive differential equations is emerging as an important area of investigation, since it is a lot richer than the corresponding theory of differential equations without impulse effects. Moreover, such equations may exhibit several real world phenomena in physics, biology, engineering, etc. (see [13]). At the same time, the boundary value problems for impulsive differential equations and impulsive difference equations have received much attention [418]. On the other hand, recently, the theory of dynamic equations on time scales has become a new important branch (see, for example, [1921]). Naturally, some authors have focused their attention on the boundary value problems of impulsive dynamic equations on time scales [2236]. However, to the best of our knowledge, few papers concerning PBVPs of impulsive dynamic equations on time scales with semi-position condition.

In this article, we are concerned with the existence of positive solutions for the following PBVPs of impulsive dynamic equations on time scales with semi-position condition

x Δ ( t ) + f ( t , x ( σ ( t ) ) ) = 0 , t J : = [ 0 , T ] T , t t k , k = 1 , 2 , , m , x ( t k + ) - x ( t k - ) = I k ( x ( t k - ) ) , k = 1 , 2 , , m , x ( 0 ) = x ( σ ( T ) ) ,
(1.1)

where T is an arbitrary time scale, T > 0 is fixed, 0, TT, fC (J × [0, ∞), (-∞, ∞)), I k C([0, ∞), [0, ∞)), t k ∈ (0, T) T , 0 < t1 < ⋯ < t m < T, and for each k = 1, 2,..., m, x ( t k + ) = lim h 0 + x ( t k + h ) and x ( t k - ) = lim h 0 - x ( t k + h ) represent the right and left limits of x(t) at t = t k . We always assume the following hypothesis holds (semi-position condition):

  1. (H)

    There exists a positive number M such that

    M x - f ( t , x ) 0 for x [ 0 , ) , t [ 0 , T ] T .

By using a fixed point theorem for operators on cone [37], some existence criteria of positive solution to the problem (1.1) are established. We note that for the case T = R and I k (x) ≡ 0, k = 1, 2,..., m, the problem (1.1) reduces to the problem studied by [38] and for the case I k (x) ≡ 0, k = 1, 2,..., m, the problem (1.1) reduces to the problem (in the one-dimension case) studied by [39].

In the remainder of this section, we state the following fixed point theorem [37].

Theorem 1.1. Let X be a Banach space and KX be a cone in X. Assume Ω1, Ω2 are bounded open subsets of X with 0 Ω 1 Ω ̄ 1 Ω 2 and Φ: K ( Ω ̄ 2 \ Ω 1 ) K is a completely continuous operator. If

  1. (i)

    There exists u 0K\{0} such that u - Φu ≠ λu 0, uK ⋂ ∂ Ω2, λ ≥ 0; Φuτu, uK ⋂ ∂Ω1, τ ≥ 1, or

  2. (ii)

    There exists u 0K\{0} such that u - Φu ≠ λu 0, uK ⋂ ∂Ω1, λ ≥ 0; Φuτu, uK ⋂ ∂Ω2, τ ≥ 1.

Then Φ has at least one fixed point in K ( Ω ̄ 2 \ Ω 1 ) .

2 Preliminaries

Throughout the rest of this article, we always assume that the points of impulse t k are right-dense for each k = 1, 2,...,m.

We define

P C = { x [ 0 , σ ( T ) ] T R : x k C ( J k , R ) , k = 0 , 1 , 2 , , m  and there exist x ( t k + ) and x ( t k ) with x ( t k ) = x ( t k ) , k = 1 , 2 , , m } ,

where x k is the restriction of x to Jk = (t k , tk+1] T ⊂ (0, σ(T)] T , k = 1, 2,..., m and J0 = [0, t1] T , tm +1= σ(T).

Let

X = { x : x P C , x ( 0 ) = x ( σ ( T ) ) }

with the norm x = sup t [ 0 , σ ( T ) ] T x ( t ) , then X is a Banach space.

Lemma 2.1. Suppose M > 0 and h: [0, T] T R is rd-continuous, then x is a solution of

x ( t ) = 0 σ ( T ) G ( t , s ) h ( s ) Δ s + k = 1 m G ( t , t k ) I k ( x ( t k ) ) , t [ 0 , σ ( T ) ] T ,

where G ( t , s ) = e M ( s , t ) e M ( σ ( T ) , 0 ) e M ( σ ( T ) , 0 ) - 1 , 0 s t σ ( T ) , e M ( s , t ) e M ( σ ( T ) , 0 ) - 1 , 0 t < s σ ( T ) ,

if and only if x is a solution of the boundary value problem

x Δ ( t ) + M x ( σ ( t ) ) = h ( t ) , t J : = [ 0 , T ] T , t t k , k = 1 , 2 , , m , x ( t k + ) - x ( t k - ) = I k ( x ( t k - ) ) , k = 1 , 2 , , m , x ( 0 ) = x ( σ ( T ) ) .

Proof. Since the proof similar to that of [34, Lemma 3.1], we omit it here.

Lemma 2.2. Let G(t, s) be defined as in Lemma 2.1, then

1 e M ( σ ( T ) , 0 ) - 1 G ( t , s ) e M ( σ ( T ) , 0 ) e M ( σ ( T ) , 0 ) - 1 for all t , s [ 0 , σ ( T ) ] T .

Proof. It is obviously, so we omit it here.

Remark 2.1. Let G(t, s) be defined as in Lemma 2.1, then 0 σ ( T ) G ( t , s ) Δs= 1 M .

For uX, we consider the following problem:

{ x Δ ( t ) + M x ( σ ( t ) ) = M u ( σ ( t ) ) f ( t , u ( σ ( t ) ) , t [ 0 , T ] T , t t k , k = 1 , 2 , , m , x ( t k + ) x ( t k ) = I k ( x ( t k ) ) , k = 1 , 2 , , m , x ( 0 ) = x ( σ ( T ) ) .
(2.1)

It follows from Lemma 2.1 that the problem (2.1) has a unique solution:

x ( t ) = 0 σ ( T ) G ( t , s ) h u ( s ) Δ s + k = 1 m G ( t , t k ) I k ( x ( t k ) ) , t [ 0 , σ ( T ) ] T ,

where h u (s) = Mu(σ(s)) - f(s, u(σ(s))), s ∈ [0, T] T .

We define an operator Φ: XX by

Φ ( u ) ( t ) = 0 σ ( T ) G ( t , s ) h u ( s ) Δ s + k = 1 m G ( t , t k ) I k ( u ( t k ) ) , t [ 0 , σ ( T ) ] T .

It is obvious that fixed points of Φ are solutions of the problem (1.1).

Lemma 2.3. Φ: XX is completely continuous.

Proof. The proof is divided into three steps.

Step 1: To show that Φ: XX is continuous.

Let { u n } n = 1 be a sequence such that u n u (n → ∞) in X. Since f(t, u) and I k (u) are continuous in x, we have

h u n ( t ) - h u ( t ) = M ( u n - u ) - ( f ( t , u n ) - f ( t , u ) ) 0 ( n ) , I k ( u n ( t k ) ) - I k ( u ( t k ) ) 0 ( n ) .

So

Φ ( u n ) ( t ) - Φ ( u ) ( t ) = 0 σ ( T ) G ( t , s ) [ h u n ( s ) - h u ( s ) ] Δ s + k = 1 m G ( t , t k ) [ I k ( u n ( t k ) ) - I k ( u ( t k ) ) ] e M ( σ ( T ) , 0 ) e M ( σ ( T ) , 0 ) - 1 0 σ ( T ) h u n ( s ) - h u ( s ) Δ s + k = 1 m I k ( u n ( t k ) ) - I k ( u ( t k ) ) 0 ( n ) ,

which leads to ||Φu n - Φu|| → 0 (n → ∞). That is, Φ: XX is continuous.

Step 2: To show that Φ maps bounded sets into bounded sets in X.

Let BX be a bounded set, that is, ∃ r > 0 such that ∀ uB we have ||u|| ≤ r. Then, for any uB, in virtue of the continuities of f(t, u) and I k (u), there exist c > 0, c k > 0 such that

f ( t , u ) c , I k ( u ) c k , k = 1 , 2 , , m .

We get

Φ ( u ) ( t ) = 0 σ ( T ) G ( t , s ) h u ( s ) Δ s + k = 1 m G ( t , t k ) I k ( u ( t k ) ) 0 σ ( T ) G ( t , s ) h u ( s ) Δ s + k = 1 m G ( t , t k ) I k ( u ( t k ) ) e M ( σ ( T ) , 0 ) e M ( σ ( T ) , 0 ) - 1 σ ( T ) ( M r + c ) + k = 1 m c k .

Then we can conclude that Φu is bounded uniformly, and so Φ(B) is a bounded set.

Step 3: To show that Φ maps bounded sets into equicontinuous sets of X.

Let t1, t2 ∈ (t k , tk+1] T ⋂ [0, σ(T)] T , uB, then

Φ ( u ) ( t 1 ) - Φ ( u ) ( t 2 ) 0 σ ( T ) G ( t 1 , s ) - G ( t 2 , s ) h u ( s ) Δ s + k = 1 m G ( t 1 , t k ) - G ( t 2 , t k ) I k ( u ( t k ) ) .

The right-hand side tends to uniformly zero as |t1 - t2| → 0.

Consequently, Steps 1-3 together with the Arzela-Ascoli Theorem shows that Φ: XX is completely continuous.

Let

K = { u X : u ( t ) δ u , t [ 0 , σ ( T ) ] T } ,

where δ= 1 e M ( σ ( T ) , 0 ) ( 0 , 1 ) . It is not difficult to verify that K is a cone in X.

From condition (H) and Lemma 2.2, it is easy to obtain following result:

Lemma 2.4. Φ maps K into K.

3 Main results

For convenience, we denote

f 0 = lim u 0 + sup max t [ 0 , T ] T f ( t , u ) u , f = lim u sup max t [ 0 , T ] T f ( t , u ) u , f 0 = lim u 0 + inf min t [ 0 , T ] T f ( t , u ) u , f = lim u inf min t [ 0 , T ] T f ( t , u ) u .

and

I 0 = lim u 0 + I k ( u ) u , I = lim u I k ( u ) u .

Now we state our main results.

Theorem 3.1. Suppose that

(H1) f0 > 0, f < 0, I0 = 0 for any k; or

(H2) f > 0, f0 < 0, I = 0 for any k.

Then the problem (1.1) has at least one positive solutions.

Proof. Firstly, we assume (H1) holds. Then there exist ε > 0 and β > α > 0 such that

f ( t , u ) ε u , t [ 0 , T ] T , u ( 0 , α ] ,
(3.1)
I k ( u ) [ e m ( σ ( T ) , 0 ) - 1 ] ε 2 M m e M ( σ ( T ) , 0 ) u , u ( 0 , α ] , for any k ,
(3.2)

and

f ( t , u ) - ε u , t [ 0 , T ] T , u [ β , ) .
(3.3)

Let Ω1 = {uX: ||u|| < r1}, where r1 = α. Then uK ⋂ ∂Ω1, 0 < δα = δ ||u|| ≤ u(t) ≤ α, in view of (3.1) and (3.2) we have

Φ ( u ) ( t ) = 0 σ ( T ) G ( t , s ) h u ( s ) Δ s + k = 1 m G ( t , t k ) I k ( u ( t k ) ) 0 σ ( T ) G ( t , s ) ( M - ε ) u ( σ ( s ) ) Δ s + k = 1 m G ( t , t k ) [ e M ( σ ( T ) , 0 ) - 1 ] ε 2 M m e M ( σ ( T ) , 0 ) u ( t k ) ( M - ε ) M u + e M ( σ ( T ) , 0 ) e M ( σ ( T ) , 0 ) - 1 k = 1 m [ e M ( σ ( T ) , 0 ) - 1 ] ε 2 M m e M ( σ ( T ) , 0 ) u = M - ε 2 M u < u , t [ 0 , σ ( T ) ] T ,

which yields ||Φ(u)|| < ||u||.

Therefore

Φ u τ u , u K Ω 1 , τ 1 .
(3.4)

On the other hand, let Ω2 = {uX: ||u|| < r2}, where r 2 = β δ .

Choose u0 = 1, then u0K\{0}. We assert that

u - Φ u λ u 0 , u K Ω 2 , λ 0 .
(3.5)

Suppose on the contrary that there exist ūK Ω 2 and λ ̄ 0 such that

ū - Φ ū = λ ̄ u 0 .

Let ς= min t [ 0 , σ ( T ) ] T ū ( t ) , then ςδ ū =δ r 2 =β, we have from (3.3) that

ū ( t ) = Φ ( ū ) ( t ) + λ ̄ = 0 σ ( T ) G ( t , s ) h ū ( s ) Δ s + k = 1 m G ( t , t k ) I k ( ū ( t k ) ) + λ ̄ 0 σ ( T ) G ( t , s ) h ū ( s ) Δ s + λ ̄ ( M + ε ) M ς + λ ̄ , t [ 0 , σ ( T ) ] T .

Therefore,

ς = min t [ 0 , σ ( T ) ] T ū ( t ) ( M + ε ) M ς + λ ̄ > ς ,

which is a contradiction.

It follows from (3.4), (3.5) and Theorem 1.1 that Φ has a fixed point u * K ( Ω ̄ 2 \ Ω 1 ) , and u* is a desired positive solution of the problem (1.1).

Next, suppose that (H2) holds. Then we can choose ε' > 0 and β' > α' > 0 such that

f ( t , u ) ε u , t [ 0 , T ] T , u [ β , ) ,
(3.6)
I k ( u ) [ e M ( σ ( T ) , 0 ) - 1 ] ε 2 M m e M ( σ ( T ) , 0 ) u , u [ β , ) for any k ,
(3.7)

and

f ( t , u ) - ε u , t [ 0 , T ] T , u ( 0 , α ] .
(3.8)

Let Ω3 = {uX: ||u|| < r3}, where r3 = α'. Then for any uK ⋂ ∂Ω3, 0 < δ ||u|| ≤ u(t) ≤ ||u|| = α'.

It is similar to the proof of (3.5), we have

u - Φ u λ u 0 , u K Ω 3 , λ 0 .
(3.9)

Let Ω4 = {uX: ||u|| < r4}, where r 4 = β δ . Then for any uK ⋂ ∂Ω4, u(t) ≥ δ ||u|| = δr4 = β', by (3.6) and (3.7), it is easy to obtain

Φ u τ u , u K Ω 4 , τ 1 .
(3.10)

It follows from (3.9), (3.10) and Theorem 1.1 that Φ has a fixed point u * K ( Ω ̄ 4 \ Ω 3 ) , and u* is a desired positive solution of the problem (1.1).

Theorem 3.2. Suppose that

(H3) f0 < 0, f < 0;

(H4) there exists ρ > 0 such that

min { f ( t , u ) - u | t [ 0 , T ] T , δ ρ u ρ } > 0 ;
(3.11)
I k ( u ) [ e M ( σ ( T ) , 0 ) - 1 ] M m e M ( σ ( T ) , 0 ) u , δ ρ u ρ , for any k .
(3.12)

Then the problem (1.1) has at least two positive solutions.

Proof. By (H3), from the proof of Theorem 3.1, we should know that there exist β" > ρ > α" > 0 such that

u - Φ u λ u 0 , u K Ω 5 , λ 0 ,
(3.13)
u - Φ u λ u 0 , u K Ω 6 , λ 0 ,
(3.14)

where Ω5 = {uX: ||u|| < r5}, Ω6 = {uX: ||u|| < r6}, r 5 = α , r 6 = β δ .

By (3.11) of (H4), we can choose ε > 0 such that

f ( t , u ) ( 1 + ε ) u , t [ 0 , T ] T , δ ρ u ρ .
(3.15)

Let Ω7 = {uX: ||u|| < ρ}, for any uK ⋂ ∂Ω7, δρ = δ ||u|| ≤ u(t) ≤ ||u|| = ρ, from (3.12) and (3.15), it is similar to the proof of (3.4), we have

Φ u τ u , u K Ω 7 , τ 1 .
(3.16)

By Theorem 1.1, we conclude that Φ has two fixed points u * * K ( Ω ̄ 6 \ Ω 7 ) and u * * * K ( Ω ̄ 7 \ Ω 5 ) , and u** and u*** are two positive solution of the problem (1.1).

Similar to Theorem 3.2, we have:

Theorem 3.3. Suppose that

(H4) f0 > 0, f > 0, I0 = 0, I = 0;

(H5) there exists ρ > 0 such that

max { f ( t , u ) | t [ 0 , T ] T , δ ρ u ρ } < 0 .

Then the problem (1.1) has at least two positive solutions.

4 Examples

Example 4.1. Let T = [0, 1] ∪ [2, 3]. We consider the following problem on T

x Δ ( t ) + f ( t , x ( σ ( t ) ) ) = 0 , t [ 0 , 3 ] T , t 1 2 , x 1 2 + - x 1 2 - = I x 1 2 , x ( 0 ) = x ( 3 ) ,
(4.1)

where T = 3, f(t, x) = x - (t + 1)x2, and I(x) = x2

Let M = 1, then, it is easy to see that

M x - f ( t , x ) = ( t + 1 ) x 2 0 for x [ 0 , ) , t [ 0 , 3 ] T ,

and

f 0 1 , f = - , and I 0 = 0 .

Therefore, by Theorem 3.1, it follows that the problem (4.1) has at least one positive solution.

Example 4.2. Let T = [0, 1] ∪ [2, 3]. We consider the following problem on T

x Δ ( t ) + f ( t , x ( σ ( t ) ) ) = 0 , t [ 0 , 3 ] T , t 1 2 , x 1 2 + - x 1 2 - = I x 1 2 , x ( 0 ) = x ( 3 ) ,
(4.2)

where T=3,f ( t , x ) =4 e 1 - 4 e 2 x- ( t + 1 ) x 2 e - x , and I(x) = x2e-x.

Choose M = 1, ρ = 4e2, then δ= 1 2 e 2 , it is easy to see that

M x - f ( t , x ) = x ( 1 - 4 e 1 - 4 e 2 ) + ( t + 1 ) x 2 e - x 0 for x [ 0 , ) , t [ 0 , 3 ] T , f 0 4 e 1 - 4 e 2 > 0 , f 4 e 1 - 4 e 2 > 0 , I 0 = 0 , I = 0 ,

and

max ( f ( t , u ) | t [ 0 , T ] T , δ ρ u ρ } = max { f ( t , u ) | t [ 0 , 3 ] T , 2 u 4 e 2 } = 16 e 3 4 e 2 ( 1 e ) < 0.

Therefore, together with Theorem 3.3, it follows that the problem (4.2) has at least two positive solutions.