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 equation1 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 [1-3]). At the same time, the boundary value problems for impulsive differential equations and impulsive difference equations have received much attention [4-18]. On the other hand, recently, the theory of dynamic equations on time scales has become a new important branch (see, for example, [19-21]). Naturally, some authors have focused their attention on the boundary value problems of impulsive dynamic equations on time scales [22-36]. 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
where T is an arbitrary time scale, T > 0 is fixed, 0, T ∈ T, f ∈ C (J × [0, ∞), (-∞, ∞)), Ik ∈ C([0, ∞), [0, ∞)), tk ∈ (0, T)T, 0 < t1 < ⋯ < tm < T, and for each k = 1, 2,..., m,
and
represent the right and left limits of x(t) at t = tk. We always assume the following hypothesis holds (semi-position condition):
(H) There exists a positive number M such that
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 Ik(x) ≡ 0, k = 1, 2,..., m, the problem (1.1) reduces to the problem studied by [38] and for the case Ik(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
and Φ:
is a completely continuous operator. If
(i) There exists u0 ∈ K\{0} such that u - Φu ≠ λu0, u ∈ K ⋂ ∂ Ω2, λ≥ 0; Φu ≠ τu, u ∈ K ⋂ ∂Ω1, τ ≥ 1, or
(ii) There exists u0 ∈ K\{0} such that u - Φu ≠ λu0, u ∈ K ⋂ ∂Ω1, λ≥ 0; Φu ≠ τu, u ∈ K ⋂ ∂Ω2, τ ≥ 1.
2 Preliminaries
Throughout the rest of this article, we always assume that the points of impulse tk are right-dense for each k = 1, 2,...,m.
We define
where xk is the restriction of x to Jk = (tk, tk+1]T ⊂ (0, σ(T)]T, k = 1, 2,..., m and J0 = [0, t1]T, tm +1 = σ(T).
Let
with the norm
, 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
if and only if x is a solution of the boundary value problem
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
Proof. It is obviously, so we omit it here.
Remark 2.1. Let G(t, s) be defined as in Lemma 2.1, then
.
For u ∈ X, we consider the following problem:
It follows from Lemma 2.1 that the problem (2.1) has a unique solution:
where hu(s) = Mu(σ(s)) - f(s, u(σ(s))), s ∈ [0, T]T.
We define an operator Φ: X → X by
It is obvious that fixed points of Φ are solutions of the problem (1.1).
Lemma 2.3. Φ: X → X is completely continuous.
Proof. The proof is divided into three steps.
Step 1: To show that Φ: X → X is continuous.
Let
be a sequence such that un → u (n → ∞) in X. Since f(t, u) and Ik(u) are continuous in x, we have
So
which leads to ||Φun - Φu|| → 0 (n → ∞). That is, Φ: X → X 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 Ik(u), there exist c > 0, ck > 0 such that
We get
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 ∈ (tk, tk+1]T ⋂ [0, σ(T)]T, u ∈ B, then
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: X → X is completely continuous.
Let
where
. 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
and
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
and
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
which yields ||Φ(u)|| < ||u||.
Therefore
On the other hand, let Ω2 = {u ∈ X: ||u|| < r2}, where
.
Choose u0 = 1, then u0 ∈ K\{0}. We assert that
Suppose on the contrary that there exist
and
such that
Let
, then
, we have from (3.3) that
Therefore,
which is a contradiction.
It follows from (3.4), (3.5) and Theorem 1.1 that Φ has a fixed point
, 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
and
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
Let Ω4 = {u ∈ X: ||u|| < r4}, where
. Then for any u ∈ K ⋂ ∂Ω4, u(t) ≥ δ ||u|| = δr4 = β', by (3.6) and (3.7), it is easy to obtain
It follows from (3.9), (3.10) and Theorem 1.1 that Φ has a fixed point
, 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
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
where Ω5 = {u ∈ X: ||u|| < r5}, Ω6 = {u ∈ X: ||u|| < r6}, 
By (3.11) of (H4), we can choose ε > 0 such that
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
By Theorem 1.1, we conclude that Φ has two fixed points
and
, 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
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
where T = 3, f(t, x) = x - (t + 1)x2, and I(x) = x2
Let M = 1, then, it is easy to see that
and
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
Choose M = 1, ρ = 4e2, then
, it is easy to see that
and
Therefore, together with Theorem 3.3, it follows that the problem (4.2) has at least two positive solutions.
Competing interests
The author declares that they have no competing interests.
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)
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