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 firstorder 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 [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 semiposition 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 semiposition condition
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 < t_{1 }< ⋯ < t_{m }< T, and for each k = 1, 2,..., m,
(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 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 onedimension 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 Ω_{1}, Ω_{2 }are bounded open subsets of X with
(i) There exists u_{0 }∈ K\{0} such that u  Φu ≠ λu_{0}, u ∈ K ⋂ ∂ Ω_{2}, λ≥ 0; Φu ≠ τu, u ∈ K ⋂ ∂Ω_{1}, τ ≥ 1, or
(ii) There exists u_{0 }∈ K\{0} such that u  Φu ≠ λu_{0}, u ∈ K ⋂ ∂Ω_{1}, λ≥ 0; Φu ≠ τu, u ∈ K ⋂ ∂Ω_{2}, τ ≥ 1.
Then Φ has at least one fixed point in
2 Preliminaries
Throughout the rest of this article, we always assume that the points of impulse t_{k }are rightdense for each k = 1, 2,...,m.
We define
where x_{k }is the restriction of x to J_{k }= (t_{k}, t_{k+1}]_{T }⊂ (0, σ(T)]_{T}, k = 1, 2,..., m and J_{0 }= [0, t_{1}]_{T}, t_{m +1 }= σ(T).
Let
with the norm
Lemma 2.1. Suppose M > 0 and h: [0, T]_{T }→ R is rdcontinuous, then x is a solution of
where
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 h_{u}(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
So
which leads to Φu_{n } 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 I_{k}(u), there exist c > 0, c_{k }> 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 t_{1}, t_{2 }∈ (t_{k}, t_{k+1}]_{T }⋂ [0, σ(T)]_{T}, u ∈ B, then
The righthand side tends to uniformly zero as t_{1 } t_{2} → 0.
Consequently, Steps 13 together with the ArzelaAscoli Theorem shows that Φ: X → X is completely continuous.
Let
where
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
(H_{1}) f_{0 }> 0, f^{∞ }< 0, I_{0 }= 0 for any k; or
(H_{2}) f_{∞ }> 0, f^{0 }< 0, I_{∞ }= 0 for any k.
Then the problem (1.1) has at least one positive solutions.
Proof. Firstly, we assume (H_{1}) holds. Then there exist ε > 0 and β > α > 0 such that
and
Let Ω_{1 }= {u ∈ X: u < r_{1}}, where r_{1 }= α. Then u ∈ K ⋂ ∂Ω_{1}, 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 Ω_{2 }= {u ∈ X: u < r_{2}}, where
Choose u_{0 }= 1, then u_{0 }∈ K\{0}. We assert that
Suppose on the contrary that there exist
Let
Therefore,
which is a contradiction.
It follows from (3.4), (3.5) and Theorem 1.1 that Φ has a fixed point
Next, suppose that (H_{2}) holds. Then we can choose ε' > 0 and β' > α' > 0 such that
and
Let Ω_{3 }= {u ∈ X: u < r_{3}}, where r_{3 }= α'. Then for any u ∈ K ⋂ ∂Ω_{3}, 0 < δ u ≤ u(t) ≤ u = α'.
It is similar to the proof of (3.5), we have
Let Ω_{4 }= {u ∈ X: u < r_{4}}, where
It follows from (3.9), (3.10) and Theorem 1.1 that Φ has a fixed point
Theorem 3.2. Suppose that
(H_{3}) f^{0 }< 0, f^{∞ }< 0;
(H_{4}) there exists ρ > 0 such that
Then the problem (1.1) has at least two positive solutions.
Proof. By (H_{3}), from the proof of Theorem 3.1, we should know that there exist β" > ρ > α" > 0 such that
where Ω_{5 }= {u ∈ X: u < r_{5}}, Ω_{6 }= {u ∈ X: u < r_{6}},
By (3.11) of (H_{4}), we can choose ε > 0 such that
Let Ω_{7 }= {u ∈ X: u < ρ}, for any u ∈ K ⋂ ∂Ω_{7}, 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
Similar to Theorem 3.2, we have:
Theorem 3.3. Suppose that
(H_{4}) f_{0 }> 0, f_{∞ }> 0, I_{0 }= 0, I_{∞ }= 0;
(H_{5}) 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)x^{2}, and I(x) = x^{2}
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
where
Choose M = 1, ρ = 4e^{2}, then
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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