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EXISTENCE AND NONEXISTENCE OF POSITIVE SOLUTIONS FOR A SYSTEM OF EVEN ORDER DYNAMIC EQUATION ON TIME SCALES
EXISTENCE AND NONEXISTENCE OF POSITIVE SOLUTIONS FOR A SYSTEM OF EVEN ORDER DYNAMIC EQUATION ON TIME SCALES
Journal of Applied Mathematics & Informatics. 2015. Sep, 33(5_6): 531-543
Copyright © 2015, Korean Society of Computational and Applied Mathematics
  • Received : September 23, 2014
  • Accepted : February 23, 2015
  • Published : September 30, 2015
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SABBAVARAPU NAGESWARA RAO

Abstract
We determine interval of two eigenvalues for which there existence and nonexistence of positive solution for a system of even-order dynamic equation on time scales subject to Sturm-Liouville boundary conditions. AMS Mathematics Subject Classification : 34B15, 39B10, 34B18.
Keywords
1. Introduction
The theory of time scales was introduced and developed by Hilger [9] to unify both continuous and discrete analysis. Time scales theory presents us with the tools necessary to understand and explain the mathematical structure underpinning the theories of discrete and continuous dynamic systems and allows us to connect them. The theory is widely applied to various situations like epidemic models, the stock market and mathematical modeling of physical and biological systems. Certain economically important phenomena contain processes that feature elements of both the continuous and discrete. The book on the subject of time scales by Bohner and Peterson [4 , 5] , summarizes and organizes much of the time scale calculus.
In recent years, the existence and nonexistence of positive solutions of the higher order boundary value problems (BVPs) on time scales have been studied extensively due to their striking applications to almost all area of science, engineering and technology, Anderson [2 , 3] , Chyan and Henderson [6] , Erbe and Peterson [7] , Kameswararao and Nageswararao [14] , Sun [16] .
We are concerned with determining values of λ and μ for which there exist and nonexist of positive solutions for the system of dynamic equations,
PPT Slide
Lager Image
with the Sturm-Liouville boundary conditions,
PPT Slide
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for 0 ≤ i n − 1, n ≥ 1 with a ∈ T kn , b ∈ T kn for a time scale T and σn ( a ) < ρn ( b ). Our interest in this paper is to investigate the existence and nonexistence of eigenvalues λ and μ that yields positive and no positive solutions to the associated boundary value problems, (1.1)-(1.2).
We assume that:
  • (A1)αj,βj,γj,δj≥ 0 anddj=γjβj+αjδj+αjγj(b−a) > 0;
  • (A2)f,g∈C([0,∞) × [0,∞), [0,∞));
  • (A3)p,q∈C([a,b], [0,∞)), and each does not vanish identically on any subinterval;
  • (A4) All of
PPT Slide
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The rest of this paper is organized as follows. In Section 2, we present some preliminaries and lemmas that will be used to prove our main results. In Section 3, we discuss the existence of a positive solution of the system (1.1)-(1.2). The intervals in which the parameters λ and μ can guarantee the existence of a solution are obtained. In Section 4, we will consider the conditions of the nonexistence of a positive solution.
2. Preliminary results
In this section, we state some lemmas that will be used to prove our results. Shortly we will be concerned with a completely continuous operator whose kernel is the Green’s function for the related homogeneous problem (−1) nu (Δ∇)n ( t ) = 0, t ∈ [ a , b ] satisfying boundary conditions (1.2). For 1 ≤ j n , let Gj ( t , s ) be the Green’s function for the boundary value problems,
PPT Slide
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PPT Slide
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First, we need few results on the related second order homogeneous boundary value problem (2.1)-(2.2).
Lemma 2.1. For 1 ≤ j n, let dj = γj βj + αj δj + αj γj ( b a ). The homogeneous boundary value problem (2.1)-(2.2) has only the trivial solution if and only if dj > 0.
Lemma 2.2. For 1 ≤ j n, the Green's function Gj ( t , s ) for the homogeneous boundary value problem (2.1)-(2.2), is given by
PPT Slide
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Lemma 2.3. Assume that condition ( A 1) is satisfied. Then, the Green's function Gj ( t , s ) satisfies the following inequality
PPT Slide
Lager Image
where
PPT Slide
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for 1 ≤ j n .
Proof . It is straightforward to see that
PPT Slide
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this expression yields both inequalities in (2.4) for gj as in (2.5). □
Lemma 2.4. Assume that the condition ( A 1) is satisfied, and Gj ( t , s ) as in (2.3). Let us define H 1 ( t , s ) = G 1 ( t , s ) , and recursively define
PPT Slide
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for 2 ≤ j n . Then Hn ( t , s ) is the Green's function for the corresponding homogeneous problem (1.1)-(1.2).
Let ξ and ω are chosen from T such that a < ξ < ω < b and also
PPT Slide
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for gj as in (2.5).
Let τ ∈ [ ξ , ω ] be defined by
PPT Slide
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Lemma 2.5. Assume that the condition ( A 1) holds. If we define
PPT Slide
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then the Green's function Hn ( t , s ) in Lemma 2.4 satisfies
PPT Slide
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and
PPT Slide
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where mn is given in (2.7),
PPT Slide
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Proof . We using mathematical induction on n it is straightforward. □
By using Green’s function, our problem (1.1)-(1.2) can be written equivalently as the following nonlinear system of integral equations
PPT Slide
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We consider the Banach space B = C [ a , b ] × C [ a , b ] with the norm
PPT Slide
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We define the cone P B by
PPT Slide
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For λ , μ > 0, we introduce the operators Qλ , Qμ : C [ a , b ] × C [ a , b ] → C [ a , b ] by
PPT Slide
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and an operator Q : C [ a , b ] × C [ a , b ] → C [ a , b ] × C [ a , b ] as
PPT Slide
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Then seeking solution to our BVP (1.1)-(1.2) is equivalent to looking for fixed points of the equation Q ( u , v ) = ( u , v ) in the Banach space B .
Lemma 2.6. Q : P P is completely continuous .
Proof . By using standard arguments, we can easily show that, under assumptions ( A 1) − ( A 2), the operator Q is completely continuous, we need only to prove Q ( P ) ⊆ P . Choose some ( u , v ) ∈ P . Then by Lemma 2.5 we have
PPT Slide
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and thus
PPT Slide
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which implies that Q ( P ) ⊆ P for every ( u , v ) ∈ P .
As Qλ and Qμ are integral operators, it is not difficult to see that using standard arguments we may conclude that both Qλ and Qμ are completely continuous, hence Q is completely continuous operator. □
3. Existence results
In this section, we apply Krasnosel’skii fixed point theorem [13] to obtain the solutions in a cone (that is, positive solution) of (1.1)-(1.2).
Theorem 3.1 ( Krasnosel'skii ). Let B be a Banach space, and let P B be a cone in B . Assume that Ω 1 and Ω 2 are open subsets of B with 0 ∈ Ω 1
PPT Slide
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⊂ Ω 2 , and let T : P ∩(
PPT Slide
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╲Ω 1 ) → P be 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 P ∩ (
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╲Ω 1 ).
For our first result, define positive numbers M 1 and M 2 by
PPT Slide
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Theorem 3.2. Assume that conditions ( A 1) − ( A 4) are satisfied. Then, for each λ , μ satisfying
PPT Slide
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there exists a pair ( u , v ) satisfying (1.1)-(1.2) such that u ( t ) > 0 and v ( t ) > 0 on ( a , b ).
Proof . Let λ , μ be as in (3.1). Let ϵ > 0 be chosen such that
PPT Slide
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Let Q be defined as in (2.10), then Q is a cone preserving, completely continuous operator. By the definitions of f 0 and g 0 , there exists H 1 > 0 such that f ( u , v ) ≤ ( f 0 + ϵ )( u + v ) for ( u , v ) ∈ P with 0 < ( u , v ) ≤ H 1 , and g ( u , v ) ≤ ( g 0 + ϵ )( u + v ) for ( u , v ) ∈ P with 0 < ( u , v ) ≤ H 1 . Set Ω 1 = {( u , v ) ∈ B :║( u , v )║< H 1 }. Now let ( u , v ) ∈ P ∩ ∂Ω 1 , i.e., let ( u , v ) ∈ P with║( u , v )║= H 1 . Then, in view of the inequality (2.8) and choice of ϵ , for a s b , we have
PPT Slide
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and so,
PPT Slide
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Similarly, we prove that║ Qμ ( u , v )║≤
PPT Slide
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║( u , v )║. Thus, for ( u , v ) ∈ P ∩ ∂Ω 1 it follows that
PPT Slide
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that is,
PPT Slide
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Due to the definition of f and g , there exists an
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> 0 such that f ( u , v ) ≥ ( f ϵ )( u + v ) for all u , v
PPT Slide
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and g ( u , v ) ≥ ( g ϵ )( u + v ) for all u , v
PPT Slide
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. Set H 2 =
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and define Ω 2 = {( u , v ) ∈ P :║( u , v )║< H 2 }. If ( u , v ) ∈ P with║( u , v )║= H 2 then, min t∈[ξ,ω] ( u + v )( t ) ≥
PPT Slide
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, by consequently, from (2.9) and choice of ϵ , for a s b , we have that
PPT Slide
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that is, Qλ ( u , v )( t ) ≥
PPT Slide
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║( u , v )║for all t τ and so, Qλ ( u , v )( t ) ≥
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║( u , v )║. Similarly, we find that Qμ ( u , v ) ≥
PPT Slide
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║( u , v )║. Thus, for ( u , v ) ∈ P ∩ ∂Ω 2 it follows that
PPT Slide
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that is,
PPT Slide
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Applying Theorem 3.1 to (3.2) and (3.3), we obtain that Q has a fixed point in P ∩ (
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╲Ω 1 ) such that H 1 ≤║( u , v )║≤ H 2 , and so (1.1)-(1.2) has a positive solution. The proof is complete. □
For our next result we define the positive numbers
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We are now ready to state and prove our main result.
Theorem 3.3. Assume that conditions ( A 1) − ( A 4) are satisfied. Then, for each λ , μ satisfying
PPT Slide
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there exists a pair ( u , v ) satisfying (1.1)-(1.2) such that u ( t ) > 0 and v ( t ) > 0 on ( a , b ).
Proof . Let λ , μ be as in (3.4) and choose a sufficiently small ϵ > 0 such that
PPT Slide
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By the definition of f 0 and g 0 , there exists an H 3 > 0 such that f ( u , v ) ≥ ( f 0 ϵ )( u + v ), for all ( u , v ) with 0 < ( u , v ) ≤ H 3 and g ( u , v ) ≥ ( g 0 ϵ )( u + v ), for all ( u , v ) with 0 < ( u , v ) ≤ H 3 . Set Ω 3 = {( u , v ) ∈ P :║( u , v )║< H 3 } and let ( u , v ) ∈ P ∩∂Ω 3 . Thus we have, from (2.9) and choice of ϵ , for a s b ,
PPT Slide
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that is,║ Qλ ( u , v )║≥
PPT Slide
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║( u , v )║. In a similar manner,║ Qμ ( u , v )║≥
PPT Slide
Lager Image
║( u , v )║. Thus, for an arbitrary ( u , v ) ∈ P ∩ ∂Ω 3 it follows that
PPT Slide
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and so,
PPT Slide
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Now let us define two functions f , g : [0,∞) → [0,∞) by
PPT Slide
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It follows that f ( u , v ) ≤ f ( t ) and g ( u , v ) ≤ g ( t ) for all ( u , v ) with 0 ≤ u + v t . It is clear that the function f and g are nondecreasing. Also, there is no difficulty to see that
PPT Slide
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In view of the definitions of f and g , there exists an
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such that
PPT Slide
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Set H 4 =
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, and Ω 4 = {( u , v ) : ( u , v ) ∈ P and║( u , v )║< H 4 }. Let ( u , v ) ∈ P ∩ ∂Ω 4 and observe that, by the definition of f , it follows that for any s ∈ [ a , b ], we have
PPT Slide
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In view of the observation and by the use of inequality (2.8),
PPT Slide
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which implies║ Qλ ( u , v )║≤
PPT Slide
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║( u , v )║. In a similar manner, we can prove that║ Qμ ( u , v )║≤
PPT Slide
Lager Image
║( u , v )║. Thus, for ( u , v ) ∈ P ∩ ∂Ω 4 , it follows that
PPT Slide
Lager Image
and so,
PPT Slide
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Applying Theorem 3.1 to (3.5) and (3.6), we obtain that Q has a fixed point in P ∩ (
PPT Slide
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╲Ω 3 ) such that H 3 ≤║( u , v )║≤ H 4 , and so (1.1)-(1.2) has a positive solution. The proof is complete. □
4. Nonexistence results
In this section, we give some sufficient conditions for the nonexistence of positive solutions to the BVP (1.1)-(1.2).
Theorem 4.1. Assume that ( A 1)−( A 4) hold. If f 0 , f , g 0 , g < ∞, then there exist positive constants λ 0 , μ 0 such that for every λ ∈ (0, λ 0 ) and μ ∈ (0, μ 0 ) , the boundary value problem (1.1)-(1.2) has no positive solution .
Proof . Since f 0 , f < ∞, we deduce that there exist
PPT Slide
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such that
PPT Slide
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We consider M 1 =
PPT Slide
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. Then, we obtain f ( u , v ) ≤ M 1 ( u + v ), ∀ u , v ≥ 0. Since g 0 , g < ∞, we deduce that there exist
PPT Slide
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such that
PPT Slide
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We consider M 2 =
PPT Slide
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Then, we obtain g ( u , v ) ≤ M 2 ( u + v ), ∀ u , v ≥ 0. We define
PPT Slide
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, where
PPT Slide
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. We shall show that for every λ ∈ (0, λ 0 ) and μ ∈ (0, μ 0 ), the problem (1.1)-(1.2) has no positive solution.
Let λ ∈ (0, λ 0 ) and μ ∈ (0, μ 0 ). We suppose that (1.1)-(1.2) has a positive solution ( u ( t ), v ( t )), t ∈ [ a , b ]. Then, we have
PPT Slide
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Therefore, we conclude
PPT Slide
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In a similar manner,
PPT Slide
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Therefore, we conclude
PPT Slide
Lager Image
Hence,║( u , v )║=║ u ║+║ v ║<
PPT Slide
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║( u , v )║+
PPT Slide
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║( u , v )║=║( u , v )║, which is a contradiction. So, the boundary value problem (1.1)-(1.2) has no positive solution. □
Theorem 4.2. Assume that ( A 1) − ( A 4) hold .
( i ) If f 0 , f > 0, then there exists a positive constant
PPT Slide
Lager Image
such that for every λ >
PPT Slide
Lager Image
and μ > 0, the boundary value problem (1.1)-(1.2) has no positive solution .
( ii ) If g 0 , g > 0, then there exists a positive constant
PPT Slide
Lager Image
such that for every μ >
PPT Slide
Lager Image
and λ > 0, the boundary value problem (1.1)-(1.2) has no positive solution .
( iii ) If f 0 , f , g 0 , g > 0, then there exist positive constants
PPT Slide
Lager Image
such that for every
PPT Slide
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, the boundary value problem (1.1)-(1.2) has no positive solution .
Proof . ( i ) Since f 0 , f > 0, we deduce that there exist
PPT Slide
Lager Image
such that
PPT Slide
Lager Image
We introduce m 1 =
PPT Slide
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. Then, we obtain f ( u , v ) ≥ m 1 ( u + v ), ∀ u , v ≥ 0. We define
PPT Slide
Lager Image
. We shall show that for every λ >
PPT Slide
Lager Image
and μ > 0 the problem (1.1)-(1.2) has no positive solution.
Let λ >
PPT Slide
Lager Image
and μ > 0. We suppose that (1.1)-(1.2) has a positive solution ( u ( t ), v ( t )), t ∈ [ a , b ]. Then, we obtain
PPT Slide
Lager Image
Therefore, we deduce
PPT Slide
Lager Image
and so,║( u , v )║=║ u ║+║ v ║≥║ u ║>║( u , v )║, which is a contradiction. Therefore, the boundary value problem (1.1)-(1.2) has no positive solution.
( ii ) Since g 0 , g > 0, we deduce that there exist
PPT Slide
Lager Image
such that
PPT Slide
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We introduce m 2 =
PPT Slide
Lager Image
. Then, we obtain g ( u , v ) ≥ m 2 ( u + v ), ∀ u , v ≥ 0. We define
PPT Slide
Lager Image
. We shall show that for every μ >
PPT Slide
Lager Image
and λ > 0 the problem (1.1)-(1.2) has no positive solution.
Let μ >
PPT Slide
Lager Image
and λ > 0. We suppose that (1.1)-(1.2) has a positive solution ( u ( t ), v ( t )), t ∈ [ a , b ]. Then, we obtain
PPT Slide
Lager Image
Therefore, we deduce
PPT Slide
Lager Image
and so,║( u , v )║=║ u ║+║ v ║≥║ v ║>║( u , v )║, which is a contradiction. Therefore, the boundary value problem (1.1)-(1.2) has no positive solution.
( iii ) Because f 0 , f , g 0 , g > 0, we deduce as above, that there exist m 1 , m 2 > 0 such that f ( u , v ) ≥ m 1 ( u + v ), g ( u , v ) ≥ m 2 ( u + v ), ∀ u , v ≥ 0. We define
PPT Slide
Lager Image
. Then for every λ >
PPT Slide
Lager Image
and μ >
PPT Slide
Lager Image
, the problem (1.1)-(1.2) has no positive solution.
Indeed, let λ >
PPT Slide
Lager Image
and μ >
PPT Slide
Lager Image
. We suppose that (1.1)-(1.2) has a positive solution ( u ( t ), v ( t )), t ∈ [ a , b ]. Then in a similar manner as above, we deduce
PPT Slide
Lager Image
and so,
PPT Slide
Lager Image
which is a contradiction. Therefore, the boundary value problem (1.1)-(1.2) has no positive solution. □
BIO
S.N. Rao received M.Sc. from Andhra University and Ph.D at Andhra University, India. Since 2014 he has been at Jazan University, Kingdom of Saudi Arabia. His research focuses on boundary value problems.
Department of Mathematics, Jazan University, Jazan, Kingdom of Saudi Arabia.
e-mail: snrao@jazanu.edu.sa
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