EXISTENCE AND NONEXISTENCE OF POSITIVE SOLUTIONS FOR A SYSTEM OF EVEN ORDER DYNAMIC EQUATION ON TIME SCALES

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.

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,

with the Sturm-Liouville boundary conditions,

for 0 ≤ i ≤ n − 1, n ≥ 1 with a ∈ Tkn, b ∈ Tkn 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:

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,

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 = γjβj+αjδj+αjγj(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

Lemma 2.3. Assume that condition (A1) is satisfied. Then, the Green's function Gj(t, s) satisfies the following inequality

where

for 1 ≤ j ≤ n.

Proof. It is straightforward to see that

this expression yields both inequalities in (2.4) for gj as in (2.5). □

Lemma 2.4. Assume that the condition (A1) is satisfied, and Gj(t, s) as in (2.3). Let us define H1(t, s) = G1(t, s), and recursively define

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

for gj as in (2.5).

Let τ ∈ [ξ, ω] be defined by

Lemma 2.5. Assume that the condition (A1) holds. If we define

then the Green's function Hn(t, s) in Lemma 2.4 satisfies

and

where mn is given in (2.7),

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

We consider the Banach space B = C[a, b] × C[a, b] with the norm

We define the cone P ⊂ B by

For λ, μ > 0, we introduce the operators Qλ,Qμ : C[a, b] × C[a, b] → C[a, b] by

and an operator Q : C[a, b] × C[a, b] → C[a, b] × C[a, b] as

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 (A1) − (A2), 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

and thus

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 Ω1 and Ω2 are open subsets of B with 0 ∈ Ω1 ⊂ ⊂ Ω2, and let T : P∩(╲Ω1) → P be a completely continuous operator such that either

Then, T has a fixed point in P ∩ (╲Ω1).

For our first result, define positive numbers M1 and M2 by

Theorem 3.2. Assume that conditions (A1) − (A4) are satisfied. Then, for each λ, μ satisfying

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

Let Q be defined as in (2.10), then Q is a cone preserving, completely continuous operator. By the definitions of f0 and g0, there exists H1 > 0 such that f(u, v) ≤ (f0 + ϵ)(u + v) for (u, v) ∈ P with 0 < (u, v) ≤ H1, and g(u, v) ≤ (g0 + ϵ)(u + v) for (u, v) ∈ P with 0 < (u, v) ≤ H1. Set Ω1 = {(u, v) ∈ B :║(u, v)║< H1}. Now let (u, v) ∈ P ∩ ∂Ω1, i.e., let (u, v) ∈ P with║(u, v)║= H1. Then, in view of the inequality (2.8) and choice of ϵ, for a ≤ s ≤ b, we have

and so,

Similarly, we prove that║Qμ(u, v)║≤║(u, v)║. Thus, for (u, v) ∈ P ∩ ∂Ω1 it follows that

that is,

Due to the definition of f∞ and g∞, there exists an > 0 such that f(u, v) ≥ (f∞ −ϵ)(u + v) for all u, v ≥ and g(u, v) ≥ (g∞ −ϵ)(u + v) for all u, v ≥ . Set H2 = and define Ω2 = {(u, v) ∈ P :║(u, v)║< H2}. If (u, v) ∈ P with║(u, v)║= H2 then, mint∈[ξ,ω](u + v)(t) ≥ , by consequently, from (2.9) and choice of ϵ, for a ≤ s ≤ b, we have that

that is, Qλ(u, v)(t) ≥║(u, v)║for all t ≥ τ and so, Qλ(u, v)(t) ≥║(u, v)║. Similarly, we find that Qμ(u, v) ≥║(u, v)║. Thus, for (u, v) ∈ P ∩ ∂Ω2 it follows that

that is,

Applying Theorem 3.1 to (3.2) and (3.3), we obtain that Q has a fixed point in P ∩ (╲Ω1) such that H1 ≤║(u, v)║≤ H2, and so (1.1)-(1.2) has a positive solution. The proof is complete. □

For our next result we define the positive numbers

We are now ready to state and prove our main result.

Theorem 3.3. Assume that conditions (A1) − (A4) are satisfied. Then, for each λ, μ satisfying

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

By the definition of f0 and g0, there exists an H3 > 0 such that f(u, v) ≥ (f0 − ϵ)(u + v), for all (u, v) with 0 < (u, v) ≤ H3 and g(u, v) ≥ (g0 − ϵ)(u + v), for all (u, v) with 0 < (u, v) ≤ H3. Set Ω3 = {(u, v) ∈ P :║(u, v)║< H3} and let (u, v) ∈ P ∩∂Ω3. Thus we have, from (2.9) and choice of ϵ, for a ≤ s ≤ b,

that is,║Qλ(u, v)║≥║(u, v)║. In a similar manner,║Qμ(u, v)║≥║(u, v)║. Thus, for an arbitrary (u, v) ∈ P ∩ ∂Ω3 it follows that

and so,

Now let us define two functions f∗, g∗ : [0,∞) → [0,∞) by

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

In view of the definitions of f∞ and g∞, there exists an such that

Set H4 = , and Ω4 = {(u, v) : (u, v) ∈ P and║(u, v)║< H4}. Let (u, v) ∈ P ∩ ∂Ω4 and observe that, by the definition of f∗, it follows that for any s ∈ [a, b], we have

In view of the observation and by the use of inequality (2.8),

which implies║Qλ(u, v)║≤║(u, v)║. In a similar manner, we can prove that║Qμ(u, v)║≤║(u, v)║. Thus, for (u, v) ∈ P ∩ ∂Ω4, it follows that

and so,

Applying Theorem 3.1 to (3.5) and (3.6), we obtain that Q has a fixed point in P ∩ (╲Ω3) such that H3 ≤║(u, v)║≤ H4, 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 (A1)−(A4) hold. If f0, f∞, g0, g∞ < ∞, then there exist positive constants λ0, μ0 such that for every λ ∈ (0, λ0) and μ ∈ (0, μ0), the boundary value problem (1.1)-(1.2) has no positive solution.

Proof. Since f0, f∞ < ∞, we deduce that there exist such that

We consider M1 = . Then, we obtain f(u, v) ≤ M1(u + v), ∀ u, v ≥ 0. Since g0, g∞ < ∞, we deduce that there exist such that

We consider M2 = Then, we obtain g(u, v) ≤ M2(u + v), ∀ u, v ≥ 0. We define , where . We shall show that for every λ ∈ (0, λ0) and μ ∈ (0, μ0), the problem (1.1)-(1.2) has no positive solution.

Let λ ∈ (0, λ0) and μ ∈ (0, μ0). We suppose that (1.1)-(1.2) has a positive solution (u(t), v(t)), t ∈ [a, b]. Then, we have

Therefore, we conclude

In a similar manner,

Therefore, we conclude

Hence,║(u, v)║=║u║+║v║<║(u, v)║+║(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 (A1) − (A4) hold.

(i) If f0, f∞ > 0, then there exists a positive constant such that for every λ > and μ > 0, the boundary value problem (1.1)-(1.2) has no positive solution.

(ii) If g0, g∞ > 0, then there exists a positive constant such that for every μ > and λ > 0, the boundary value problem (1.1)-(1.2) has no positive solution.

(iii)If f0, f∞, g0, g∞ > 0, then there exist positive constants such that for every , the boundary value problem (1.1)-(1.2) has no positive solution.

Proof. (i) Since f0, f∞ > 0, we deduce that there exist such that

We introduce m1 = . Then, we obtain f(u, v) ≥ m1(u + v), ∀ u, v ≥ 0. We define . We shall show that for every λ > and μ > 0 the problem (1.1)-(1.2) has no positive solution.

Let λ > and μ > 0. We suppose that (1.1)-(1.2) has a positive solution (u(t), v(t)), t ∈ [a, b]. Then, we obtain

Therefore, we deduce

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 g0, g∞ > 0, we deduce that there exist such that

We introduce m2 = . Then, we obtain g(u, v) ≥ m2(u + v), ∀ u, v ≥ 0. We define . We shall show that for every μ > and λ > 0 the problem (1.1)-(1.2) has no positive solution.

Let μ > and λ > 0. We suppose that (1.1)-(1.2) has a positive solution (u(t), v(t)), t ∈ [a, b]. Then, we obtain

Therefore, we deduce

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 f0, f∞, g0, g∞ > 0, we deduce as above, that there exist m1,m2 > 0 such that f(u, v) ≥ m1(u + v), g(u, v) ≥ m2(u + v), ∀ u, v ≥ 0. We define . Then for every λ > and μ > , the problem (1.1)-(1.2) has no positive solution.

Indeed, let λ > and μ > . 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

and so,

which is a contradiction. Therefore, the boundary value problem (1.1)-(1.2) has no positive solution. □