Open Access

Periodic boundary value problems for nonlinear first-order impulsive dynamic equations on time scales

Advances in Difference Equations20122012:12

DOI: 10.1186/1687-1847-2012-12

Received: 23 August 2011

Accepted: 15 February 2012

Published: 15 February 2012

Abstract

By using the classical fixed point theorem for operators on cone, in this article, some results of one and two positive solutions to a class of nonlinear first-order periodic boundary value problems of impulsive dynamic equations on time scales are obtained. Two examples are given to illustrate the main results in this article.

Mathematics Subject Classification: 39A10; 34B15.

Keywords

time scale periodic boundary value problem positive solution fixed point impulsive dynamic equation

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, T T, f C (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 K X 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 0 K\{0} such that u - Φu ≠ λu 0, u K ∂ Ω2, λ ≥ 0; Φuτu, u K ∂Ω1, τ ≥ 1, or

     
  2. (ii)

    There exists u 0 K\{0} such that u - Φu ≠ λu 0, u K ∂Ω1, λ ≥ 0; Φuτu, u K ∂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 ) Δ s = 1 M .

For u X, 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 B X be a bounded set, that is, r > 0 such that u B we have ||u|| ≤ r. Then, for any u B, 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 , u B, 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 = {u X: ||u|| < r1}, where r1 = α. Then u K ∂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 = {u X: ||u|| < r2}, where r 2 = β δ .

Choose u0 = 1, then u0 K\{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 ū K Ω 2 and λ ̄ 0 such that
ū - Φ ū = λ ̄ u 0 .
Let ς = min t [ 0 , σ ( T ) ] T ū ( t ) , then ς δ 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 = {u X: ||u|| < r3}, where r3 = α'. Then for any u K ∂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 = {u X: ||u|| < r4}, where r 4 = β δ . Then for any u K ∂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 = {u X: ||u|| < r5}, Ω6 = {u X: ||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 = {u X: ||u|| < ρ}, for any u K ∂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.

Declarations

Acknowledgements

The author thankful to the anonymous referee for his/her helpful suggestions for the improvement of this article. This work is supported by the Excellent Young Teacher Training Program of Lanzhou University of Technology (Q200907)

Authors’ Affiliations

(1)
Department of Applied Mathematics, Lanzhou University of Technology

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Copyright

© Wang; licensee Springer. 2012

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