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Multiple positive solutions for first-order impulsive integral boundary value problems on time scales
Boundary Value Problems volume 2011, Article number: 12 (2011)
Abstract
In this paper, we first present a class of first-order nonlinear impulsive integral boundary value problems on time scales. Then, using the well-known Guo-Krasnoselskii fixed point theorem and Legget-Williams fixed point theorem, some criteria for the existence of at least one, two, and three positive solutions are established for the problem under consideration, respectively. Finally, examples are presented to illustrate the main results.
MSC: 34B10; 34B37; 34N05.
1 Introduction
In fact, continuous and discrete systems are very important in implementing and applications. It is well known that the theory of time scales has received a lot of attention, which was introduced by Stefan Hilger in order to unify continuous and discrete analyses. Therefore, it is meaningful to study dynamic systems on time scales, which can unify differential and difference systems.
In recent years, a great deal of work has been done in the study of the existence of solutions for boundary value problems on time scales. For the background and results, we refer the reader to some recent contributions [1–5] and references therein. At the same time, boundary value problems for impulsive differential equations and impulsive difference equations have received much attention [6–12], since such equations may exhibit several real-world phenomena in physics, biology, engineering, etc. see [13–15] and the references therein.
In paper [16], Sun studied the first-order boundary value problem on time scales
where 0 < β < 1. By means of the twin fixed point theorem due to Avery and Henderson, some existence criteria for at least two positive solutions were established.
Tian and Ge [17] studied the first-order three-point boundary value problem on time scales
Using several fixed point theorems, the existence of at least one positive solution and multiple positive solutions is obtained.
However, except BVP of differential and difference equations, that is, for particular time scales ( or ), there are few papers dealing with multi-point boundary value problems more than three-point for first-order systems on time scales. In addition, problems with integral boundary conditions arise naturally in thermal conduction problems [18], semiconductor problems [19], hydrodynamic problems [20]. In continuous case, since integral boundary value problems include two-point, three-point,..., n-point boundary value problems, such boundary value problems for continuous systems have received more and more attention and many results have worked out during the past ten years, see Refs. [21–27] for more details. To the best of authors' knowledge, up to the present, there is no paper concerning the boundary value problem with integral boundary conditions on time scales. This paper is to fill the gap in the literature.
In this paper, we are concerned with the following first-order nonlinear impulsive integral boundary value problem on time scales:
where is a time scale which is a nonempty closed subset of ℝ with the topology and ordering inherited from ℝ, 0, and T are points in , an interval which has finite right-scattered points, , and p is regressive, ℝ+), I i (1 ≤ i ≤ m) ∈ C([0, +∞), [0, +∞)), g is a nonnegative integrable function on and , ep(0,σ(T)) is the exponential function on time scale , which will be introduced in the next section, , 0 < t1 < · · · < t m < T, and for each and represent the right and left limits of x(t) at .
Remark 1.1. Let denote the set of right-scattered points in interval , 0 ≤ θ1< · · · < θ q ≤ T, σ(θ0) = 0, θq+1= T. By some basic concepts and time scale calculus formulae in the book by Bohner and Peterson[28], we have
The main purpose of this paper is to establish some sufficient conditions for the existence of at least one, two, or three positive solutions for BVP (1.3) using Guo-Krasnoselskii and Legget-Williams fixed point theorem, respectively.
For convenience, we introduce the following notation:
where i = 1, 2,..., m.
This paper is organized as follows. In Section 2, some basic definitions and lemmas on time scales are introduced without proofs. In Section 3, some useful lemmas are established. In particular, Green's function for BVP (1.3) is established. We prove the main results in Sections 4-6.
2 Preliminaries
In this section, we shall first recall some basic definitions, lemmas that are used in what follows. For the details of the calculus on time scales, we refer to books by Bohner and Peterson [28, 29].
Definition 2.1. [28]A time scaleis an arbitrary nonempty closed subset of the real set ℝ with the topology and ordering inherited from ℝ. The forward and backward jump operatorsand the graininessare defined, respectively, by
In this definition, we put (i.e., σ(t) = t ifhas a maximum t) and (i.e., ρ(t) = t ifhas a minimum t). The pointis called left-dense, left-scattered, right-dense, or right-scattered if ρ(t) = t, ρ(t) < t, σ(t) = t, or σ(t) > t, respectively. Points that are right-dense and left-dense at the same time are called dense. Ifhas a left-scattered maximum m1, defined; otherwise, set. Ifhas a right-scattered minimum m2, defined, otherwise, set.
Definition 2.2. [28]A functionis rd continuous provided it is continuous at each right-dense point inand has a left-sided limit at each left-dense point in. The set of rd-continuous functionswill be denoted by.
Definition 2.3. [28]Ifis a function and, then the delta derivative of f at the point t is defined to be the number fΔ(t) (provided it exists) with the property that for each ε > 0 there is a neighborhood U of t such that
Definition 2.4. [28]For a function (the range ℝ of f may be actually replaced by Banach space), the (delta) derivative is defined by
if f is continuous at t and t is right-scattered. If t is not right-scattered, then the derivative is defined by
provided this limit exists.
Definition 2.5. [28]If FΔ(t) = f(t), then we define the delta integral by
Definition 2.6. [28]A functionis said to be regressive provided 1 + μ(t)p(t) ≠ 0 for all, where μ(t) = σ(t) - t is the graininess function. The set of all regressive rd-continuous functionsis denoted by, while the setis given byfor all. Let. The exponential function is defined by
where ξh(z)is the so-called cylinder transformation.
Lemma 2.1. [28]Let p, . Then
-
(1)
e 0(t, s) ≡ 1 and e p (t, t) ≡ 1;
-
(2)
e p (σ(t), s) = (1 + μ(t)p(t))e p (t, s);
-
(3)
, where ;
-
(4)
e p (t, s)e p (s, r) = e p (t, r),
-
(5)
.
Lemma 2.2. [28]Assume thatare delta differentiable at. Then
Lemma 2.3. [28]Let, , and assume thatis continuous at (t, t), wherewith t > a. Also, assume that fΔ(t, ·) is rd-continuous on [a, σ(t)]. Suppose that for each ε > 0 there exists a neighborhood U of t, independent of τ ∈ [a, σ(t)], such that
where f Δ denotes the derivative of f with respect to the first variable. Then
-
(1)
;
-
(2)
.
3 Foundational lemmas
In this section, we first introduce some background definitions, fixed point theorems in Banach space, then present basic lemmas that are very crucial in the proof of the main results.
We define is a piecewise continuous map with first-class discontinuous points in and at each discontinuous point it is continuous on the left} with the norm , then PC is a Banach Space.
Definition 3.1. A function x is said to be a positive solution of problem (1.3) if x ∈ PC satisfying problem (1.3) and x(t) > 0 for all.
Definition 3.2. Let X be a real Banach space, the nonempty set K ⊂ X is called a cone of X, if it satisfies the following conditions.
-
(1)
x ∈ K and λ ≥ 0 implies λx ∈ K;
-
(2)
x ∈ K and -x ∈ K implies x = 0.
Every cone K ⊂ X induces an ordering in X, which is given by x ≤ y if and only if y - x ∈ K.
Definition 3.3. An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.
Lemma 3.1. (Guo-Krasnoselskii[30]) Let X be a Banach space and K ⊂ X be a cone in X. Assume that Ω1, Ω2are bounded open subsets of X withandis a completely continuous operator such that, either
-
(1)
||Φx|| ≤ ||x||, x ∈ K ∩ ∂Ω1, and ||Φx|| ≥ ||x||, x ∈ K ∩ ∂Ω2; or
-
(2)
||Φx|| ≥ ||x||, x ∈ K ∩ ∂Ω1, and ||Φx|| ≤ ||x||, x ∈ K ∩ ∂Ω2.
Then Φ has at least one fixed point in.
Lemma 3.2. Suppose, ν i ∈ ℝ, then x is a solution of
where
if and only if x is a solution of the boundary value problem
Proof. Assume that x(t) is a solution of (3.2). By the first equation in (3.2), we have
If t ∈ [0, t1], integrating (3.3) from 0 to t, we get
while t → t1, we have
then
Now, let t ∈ (t1, t2], integrating (3.3) from t1 to t, we obtain
For t ∈ (t k , tk+1], repeating the above process, we can get
that is
It follows from that
where . Then
This means that if x is a solution of (3.2) then x satisfies (3.1).
On the other hand, if x satisfies (3.1), we have
Then
where
Notice that
Similarly,
Hence, we get from (3.5) that
that is
Finally, we can obtain from (3.1) that
and
So the proof of this lemma is completed.
Lemma 3.3. Let G(t, s) be defined the same as that in Lemma 3.2, then the following properties hold.
-
(1)
G(t, s) > 0 for all ;
-
(2)
A ≤ G(t, s) ≤ B for all , where
Proof. Since , then it is clear that (1) holds. Now we will show that (2) holds.
Hence, the left-hand side of (2) holds. And it is easy to show that the right-hand side of (2) also holds. The proof is complete. ■
Define an operator Φ : PC → PC by
By Lemma 3.2, the fixed points of Φ are solutions of problem (1.3).
Lemma 3.4. The operator Φ : PC → PC is completely continuous.
Proof. The first step we will show that Φ : PC → PC is continuous. Let be a sequence such that in PC. Then
Since f(t, x) and I i (x)(1 ≤ i ≤ m) are continuous in x, we have |(Φx n )(t) - (Φx)(t)| → 0, which leads to ||Φx n - Φx|| PC → 0, as n → ∞. That is, Φ : PC → PC is continuous.
Next, we will show that Φ : PC → PC is a compact operator by two steps.
Let U ⊂ PC be a bounded set.
Firstly, we will show that {Φx : x ∈ U}is bounded. For any x ∈ U, we have
In virtue of the continuity of f(t, x) and I i (x)(1 ≤ i ≤ m), we can conclude that {Φx : x ∈ U} is bounded from above inequality.
Secondly, we will show that {Φx : x ∈ U} is the set of equicontinuous functions. For any x, y ∈ U, then
In virtue of the continuity of f(t, x) and I i (x)(1 ≤ i ≤ m), the right-hand side tends to zero uniformly as |x - y| → 0. Consequently, {Φx : x ∈ U} is the set of equicontinuous functions.
By Arzela-Ascoli theorem on time scales [31], {Φx : x ∈ U} is a relatively compact set. So Φ maps a bounded set into a relatively compact set, and Φ is a compact operator.
From above three steps, it is easy to see that Φ : PC → PC is completely continuous. The proof is complete. ■
Let , where . It is not difficult to verify that K is a cone in PC.
Lemma 3.5. Φ maps K into K.
Proof. Obviously, Φ(K) ⊂ PC. ∀x ∈ K, we have
which implies
Therefore,
Hence, Φ(K) ⊂ K. The proof is complete. ■
4 Existence of at least one positive solution
In this section, we will state and prove our main result about the existence of at least one positive solution of problem (1.3).
Theorem 4.1. Assume that one of the following conditions is satisfied:
(H1) max f0 = 0, min f∞ = ∞, and Ii 0= 0, i = 1, 2,..., m; or
(H2) max f∞ = 0, min f0 = ∞, and Ii∞= 0, i = 1, 2,..., m.
Then, problem (1.3) has at least one positive solution.
Proof. Firstly, we assume that (H1) holds. In this case, since max f0 = 0 and Ii 0= 0, i = 1, 2,..., m, for ε ≤ (Bσ(T) + Bm)-1, there exists a positive constant r1 such that
In view of min f∞ = ∞, we have that for M ≥ (Aσ(T)δ)-1, there exists a constant such that
Let Ω i = {x ∈ PC : ||x|| < r i }, i = 1, 2.
On the one hand, if x ∈ K ∩ ∂Ω1, we have
which yields
On the other hand, if x ∈ K ∩ ∂Ω2, we have
which implies
Therefore, by (4.1), (4.2), and Lemma 3.1, it follows that Φ has a fixed point in .
Next, we assume that (H2) holds. In this case, since max f∞ = 0 and Ii∞= 0, i = 1, 2,..., m, for ε' ≤ (Bσ(T) + Bm)-1, there exists a positive constant r3 such that
In view of min f∞ = ∞, we have that for M' ≥ (Aσ(T)δ)-1, there exists a positive constant r4< δr3 such that
Let Ω i = {x ∈ PC : ||x|| < r i }, i = 3, 4.
On the one hand, if x ∈ K ∩ ∂Ω3, we have
which yields
On the other hand, if x ∈ K ∩ ∂Ω4, we have
which implies
Hence, from (4.3) and (4.4) and Lemma 3.1, we conclude that Φ has a fixed point in , that is, problem (1.3) has at least one positive solution. The proof is complete. ■
5 Existence of at least two positive solutions
In this section, we will state and prove our main results about the existence of at least two positive solutions to problem (1.3).
Theorem 5.1. Assume that the following conditions hold.
(H3) min f0 = +∞, min f∞ = +∞.
(H4) There exists a positive constant R such thatfor all 0 < x ≤ R.
(H5) , x ∈ (0, ∞), i = 1, 2,..., m.
Then, problem (1.3) has at least two positive solutions.
Proof. Let Ω R = {x ∈ PC : ||x|| < R}. From (H4) and (H5), for x ∈ K ∩ ∂Ω R , we get
So
Since min f0 = +∞, for M ≥ (Aσ(T)δ)-1, there exists a positive constant R1< δ R such that
Let . For any , we have
Hence,
Similarly, since min f∞ = +∞, for M' ≥ (Aσ(T)δ)-1, there exists a positive constant such that
Let . For any , we have
Hence,
Equations 5.1 and 5.2 imply that Φ has at least one fixed point in , which is a positive solution of problem (1.3). Besides, (5.1) and (5.3) imply that Φ has at least one fixed point in , which is a positive solution of problem (1.3). Therefore, problem (1.3) has at least two positive solutions x1 and x2 satisfying 0 < R1 ≤ ||x1|| < R < ||x2|| ≤ R2. The proof is complete. ■
Theorem 5.2. Assume that the following conditions hold.
(H6) max f0 = 0, max f∞ = 0, Ii 0= 0, Ii∞= 0, i = 1, 2,..., m.
(H7) There exists a positive constant r such thatfor all 0 < x ≤ r.
Then problem (1.3) has at least two positive solutions.
Proof. Let Ω r = {x ∈ PC : ||x|| < r}. From (H7), for x ∈ K ∩ ∂Ω r , we get
So
Since max f0 = 0 and Ii 0= 0, i = 1, 2,..., m, for ε ≤ (Bσ(T) + Bm)-1, there exists a positive constant r1< δ r such that
Let . For any , we have
Hence,
Similarly, since max f∞ = 0 and Ii∞= 0, i = 1, 2,..., m, for ε' ≤ (Bσ(T) + Bm)-1, there exists a positive constant such that
Let . For any , we have
Hence,
Equations 5.4 and 5.5 imply that Φ has at least one fixed point in , which is a positive solution of problem (1.3). Besides, (5.4) and (5.6) imply that Φ has at least one fixed point in , which is a positive solution of problem (1.3). Therefore, problem (1.3) has at least two positive solutions x1 and x2 satisfying 0 < r1 ≤ ||x1|| < r < ||x2|| ≤ r2. The proof is complete. ■
Similar to Theorems 5.1 and 5.2, one can easily obtain the following corollary:
Corollary 5.1. Assume that (H7) and the following conditions hold.
(H8) max f0 = 0, max f∞ = 0, Ii 0= 0, i = 1, 2,..., m.
(H9) There exists a positive constant d such thatfor all x ≥ d, i = 1, 2,..., m.
Then, problem (1.3) has at least two positive solutions.
6 Existence of at least three positive solutions
In this section, we will state and prove our multiplicity result of positive solutions to problem (1.3) via Legget-Williams fixed point theorem. For readers' convenience, we first illustrate Legget-Williams fixed point theorem.
Let be a real Banach space with cone K. A map α : K → [0, +∞) is said to be a continuous concave functional on K if α is continuous and
for all x, y ∈ K and t ∈ [0, 1]. Let a, b be two numbers such that 0 < a < b and α be a nonnegative continuous concave functional on K. We define the following convex sets:
Lemma 6.1. (Legget-Williams fixed point theorem[32]). Letbe completely continuous and α be a nonnegative continuous concave functional on K such that α(x) ≤ ||x|| for all x ∈ K c . Suppose that there exist 0 < d < a < b ≤ c such that
-
(1)
{x ∈ K(α, a, b) : α(x) > a} ≠ ∅, and α(Φ(x)) > a for all x ∈ K(α, a, b);
-
(2)
||Φx|| < d for all ||x|| ≤ d;
-
(3)
α(Φ(x)) > a for all x ∈ K(α, a, c) with ||Φ(x)|| > b.
Then, Φ has at least three fixed points x1, x2, x3insatisfying ||x1|| < d, a < α(x2), ||x3|| > d, and α(x3) < a.
Theorem 6.1. Assume that there exist numbers d, a, and c withsuch that
Then, problem (1.3) has at least three positive solutions.
Proof. For x ∈ K, we define
It is easy to verify that α is a nonnegative continuous concave functional on K with α(x) < ||x|| for all x ∈ K.
We first claim that if there exists a positive constant r such that , , i = 1, 2,..., m, for x ∈ (0, r], then .
Indeed, if ,
Thus, ||Φx|| < r, that is Φx ∈ K r . Hence, we have shown that (6.1) or (6.2) hold, then Φ maps into K d or into K c , respectively. So condition (2) of Lemma 6.1 holds.
Let . Next, we will show that {x ∈ K(α, a, b) : α(x) > a} ≠ ∅, and α(Φ(x)) > a for x ∈ K(α, a, b). In fact, , then the constant function .
Since (6.3) holds, for x ∈ K(α, a, b), we obtain