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EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS FOR THE SYSTEMS OF HIGHER ORDER BOUNDARY VALUE PROBLEMS ON TIME SCALES
EXISTENCE OF MULTIPLE POSITIVE SOLUTIONS FOR THE SYSTEMS OF HIGHER ORDER BOUNDARY VALUE PROBLEMS ON TIME SCALES
Journal of Applied Mathematics & Informatics. 2015. Jan, 33(1_2): 1-12
Copyright © 2015, Korean Society of Computational and Applied Mathematics
  • Received : September 10, 2014
  • Accepted : November 07, 2014
  • Published : January 30, 2015
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A. KAMESWARA Rao

Abstract
This paper is concerned with boundary value problems for systems of n -th order dynamic equations on time scales. Under the suitable conditions, the existence and multiplicity of positive solutions are established by using abstract fixed-point theorems. AMS Mathematics Subject Classification : 34B15, 39B10, 34B18.
Keywords
1. Introduction
Let
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be a time scale with
PPT Slide
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Given an interval J of ℝ, we will use the interval notation
PPT Slide
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In this paper we are concerned with the existence and multiplicity of positive solutions for dynamic equation on time scales
PPT Slide
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satisfying the boundary conditions
PPT Slide
Lager Image
where
PPT Slide
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ƒ ( t , 0) ≡ 0, g ( t , 0) ≡ 0.
Recently, existence and multiplicity of solutions for boundary value problems of dynamic equations have been of great interest in mathematics and its applications to engineering sciences. To our knowledge, most existing results on this topic are concerned with the single equation and simple boundary conditions.
It should be pointed out that Eloe and Henderson [6] discussed the boundary value problem as follows
PPT Slide
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By using a Krasnosel’skii fixed point theorem, the existence of solutions are obtained in the case when, either ƒ is superlinear, or ƒ is sublinear. Yang and Sun [15] considered the boundary value problem of the system of differential equations
PPT Slide
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By appealing to the degree theory, the existence of solutions are established. Note, that, there is only one differential equation in [4] and BVP in [15] contains the system of second order differential equations.
The arguments for establishing the existence of solutions of the BVP (1)-(2) involve properties of Green’s function that play a key role in defining some cones. A fixed point theorem due to Krasnosel’skii [10] is applied to yield the existence of positive solutions of the BVP (1)-(2). Another fixed point theorem about multiplicity is applied to obtain the multiplicity of positive solutions of BVP (1)-(2).
The rest of this paper is organized as follows. In Section 2, we shall provide some properties of certain Green’s functions and preliminaries which are needed later. For the sake of convenience, we also state Krasnosel’skii fixed point theorem in a cone. In Section 3, we establish the existence and multiplicity of positive solutions of the BVP (1)-(2). In Section 4, some examples are given to illustrate our main results.
2. Preliminaries
In this section, we will give some lemmas which are useful in proving our main results.
To obtain solutions of the BVP (1)-(2), we let G ( t , s ) be the Green’s function for the boundary value problem
PPT Slide
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PPT Slide
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Using the Cauchy function concept G ( t , s ) is given by
PPT Slide
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Lemma 1. For
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we have
PPT Slide
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Proof . For t s , we have
PPT Slide
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Similarly, for σ ( s ) ≤ t and σi ( s ) ≤ t , i = 1, · · · , n − 1, we have
PPT Slide
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Thus, we have
PPT Slide
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Lemma 2. Let
PPT Slide
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we have
PPT Slide
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Proof . The Green’s function for the BVP (3)-(4) is given in (5) shows that
PPT Slide
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For t s , t I and σ i−1 ( a ) ≤ σ n−2 ( a ), i = 1, 2, · · · , n − 1, we have
PPT Slide
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For σ ( s ) ≤ t , t I and σ i−1 ( a ) ≤ σ n−2 ( a ), i = 1, 2, · · · , n − 1, we have
PPT Slide
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Therefore
PPT Slide
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We note that a pair ( u ( t ), v ( t )) is a solution of the BVP (1)-(2) if and only if
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Assume throughout that
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is such that
PPT Slide
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both exist and satisfy
PPT Slide
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Next, let
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be defined by
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Finally, we define
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and let
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For our construction, let
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with supremum norm
PPT Slide
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Then ( E , ∥·∥) is a Banach space. Define
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It is obivious that P is a positive cone in E . Define an integral operator T : P E by
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Lemma 3. If the operator T is defined as (10), then T : P → P is completely continuous .
Proof . From the continuity of ƒ and g , and (8) that, for u P , Tu ( t ) ≥ 0 on
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Also, for u P , we have from (6) that
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so that
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Next, if u P , we have from (7), (9), and (10) that
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Therefore T : P P . Since G ( t , s ), ƒ ( t , u ) and g ( t , u ) are continuous, it is easily known that T : P P is completely continuous. The proof is complete. □
From above arguments, we know that the existence of positive solutions of (1)-(2) can be transferred to the existence of positive fixed points of the operator T .
Lemma 4 ( [3 , 4 , 10] ). Let ( E , ∥ · ∥) be a Banach space, and let P E be a cone in E. Assume that Ω 1 and Ω 2 are open bounded subsets of E such that 0 ∈ Ω 1 .
PPT Slide
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If
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is a completely continuous operator such that either
  • (i) ∥Tu∥ ≤ ∥u∥,u∈P∩∂Ω1, and ∥Tu∥ ≥ ∥u∥,u∈P∩∂Ω2, or
  • (ii) ∥Tu∥ ≥ ∥u∥,u∈P∩∂Ω1, and ∥Tu∥ ≤ ∥u∥,u∈P∩∂Ω2,
then T has a fixed point in
PPT Slide
Lager Image
Lemma 5 ( [3 , 4 , 10] ). Let ( E , ∥ · ∥) be a Banach space, and let P E be a cone in E. Assume that Ω 1 , Ω 2 and Ω 3 are open bounded subsets of E such that
PPT Slide
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If
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is a completely continuous operator such that:
PPT Slide
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then T has at least two fixed points
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and furthermore
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3. Main Results
First we give the following assumptions:
(A1)
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(A2)
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(A3)
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(A4)
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(A5) ƒ ( t , u ), g ( t , u ) are increasing functions with respect to u and, there is a number N > 0, such that
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Theorem 1. If ( A 1) and ( A 2) are satisfied, then (1)-(2) has at least one positive solution
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satisfying u ( t ) > 0, v ( t ) > 0.
Proof . From ( A 1) there is a number H 1 ∈ (0; 1) such that for each
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one has
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where η > 0 satisfies
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For every u P and ∥ u ∥ = H 1 /2, note that
PPT Slide
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thus
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So, ∥ Tu ∥ ≤ ∥ u ∥. If we set
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then
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On the other hand, from ( A 2) there exist four positive numbers μ , μ ′, C 1 and C 2 such that
PPT Slide
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where μ and μ ′ satisfy
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For u P , we have
PPT Slide
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where
PPT Slide
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Therefore
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from which it follows that ∥ Tu ∥ ≥ Tu ( τ ) ≥ ∥ u ∥ as ∥ u ∥ → ∞.
Let Ω 2 = { u E : ∥ u ∥ < H 2 }. Then for u P and ∥ u ∥ = H 2 > 0 sufficient by large, we have
PPT Slide
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Thus, from (11), (12) and Lemma 4, we know that the operator T has a fixed point in
PPT Slide
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The proof is complete. □
Theorem 2. If ( A 3) and ( A 4) are satisfied, then (1)-(2) has at least one positive solution
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satisfying u ( t ) > 0, v ( t ) > 0.
Proof . From ( A 3) there is a number
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such that for each
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one has
PPT Slide
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where λ > and λ ′ satisfy
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From g ( t , 0) ≡ 0 and the continuity of g ( t , u ), we know that there exists a number
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small enough such that
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For every u P and ∥ u ∥ = H 3 , note that
PPT Slide
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thus
PPT Slide
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So, ∥ Tu ∥ ≥ ∥ u ∥. If we set
PPT Slide
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then
PPT Slide
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On the other hand, we know from ( A 4) that there exist three positive numbers η ′, C 4 , and C 5 such that for every
PPT Slide
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PPT Slide
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where
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Thus we have
PPT Slide
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where
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from which it follows that Tu ( t ) ≤ ∥ u ∥ as ∥ u ∥ → ∞. Let Ω 4 = { u E :∥ u ∥< H 4 }. For each u P and ∥ u ∥ = H 4 > 0 large enough, we have
PPT Slide
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From (13), (14) and Lemma 4, we know that the operator T has a fixed point in
PPT Slide
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The proof is complete. □
Theorem 3. If ( A 2), ( A 3) and ( A 5) are satisfied, then (1)-(2) has at least two distinct positive solutions
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satisfying ui ( t ) > 0, vi ( t ) > 0 ( i = 1, 2).
Proof . Note that
PPT Slide
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Then from ( A 5), for every
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we have
PPT Slide
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Thus
PPT Slide
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And from ( A 2) and ( A 3) we have
PPT Slide
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PPT Slide
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We can choose H 2 , H 3 and N such that H 3 N H 2 and (15)-(17) are satisfied. From Lemma 5, T has at least two fixed points in
PPT Slide
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respectively. The proof is complete. □
4. Examples
Some examples are given to illustrate our main results.
Example 1. Consider the following dynamic equations
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satisfying the boundary conditions
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where f ( t , v ) = v 3 , g ( t , u ) = u 2 , then conditions of Theorem 1 are satisfied. From Theorem 1, the BVP (18)-(19) has at least one positive solution.
Example 2. Consider the following dynamic equations
PPT Slide
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satisfying the boundary conditions
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where f ( t , v ) = v 2/3 , g ( t , u ) = u 3/4 , then conditions of Theorem 2 are satisfied. From Theorem 2, the BVP (20)-(21) has at least one positive solution.
Example 3. Consider the following system of boundary value problems
PPT Slide
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PPT Slide
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where
PPT Slide
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PPT Slide
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We can choose N = 100, then conditions of Theorem 3 are satisfied. From Theorem 3, the BVP (22)-(23) has at least two positive solutions.
BIO
A. K. Rao received M. Sc degree from Andhra University, Visakhapatnam. M. Phil degree from M. K. University, Madhurai and Ph. D from Andhra University under the esteemed guidence of Prof. K. R. Prasad. His research interest focuses on Boundary Value Problems.
Department of Mathematics, Gayatri Vidya Parishad College of Engineering for Women, Madhurawada, Visakhapatnam, 530 048, India.
e-mail: kamesh−1724@yahoo.com
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