Research

# Positive solutions for a sixth-order boundary value problem with four parameters

Ravi P Agarwal1*, B Kovacs2 and D O’Regan3

Author Affiliations

1 Department of Mathematics, Texas A&M University-Kingsville, 700 University Blvd., Kingsville, 78363-8202, USA

2 Department of Analysis, University of Miskolc, Egyetemvaros, 3515, Hungary

3 School of Mathematics, Statistics and Applied Mathematics, National University of Ireland, Galway, Ireland

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Boundary Value Problems 2013, 2013:184  doi:10.1186/1687-2770-2013-184

 Received: 20 May 2013 Accepted: 30 July 2013 Published: 14 August 2013

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

### Abstract

This paper investigates the existence and multiplicity of positive solutions of a sixth-order differential system with four variable parameters using a monotone iterative technique and an operator spectral theorem.

MSC: 34B15, 34B18.

##### Keywords:
positive solutions; variable parameters; fixed point theorem; operator spectral theorem

### 1 Introduction

It is well known that boundary value problems for ordinary differential equations can be used to describe a large number of physical, biological and chemical phenomena. In recent years, boundary value problems for sixth-order ordinary differential equations, which arise naturally, for example, in sandwich beam deflection under transverse shear have been studied extensively, see [1-4] and the references therein. The deformation of the equilibrium state of an elastic circular ring segment with its two ends simply supported can be described by a boundary value problem involving a sixth-order ordinary differential equation

(1)

Liu and Li [5] studied the existence and nonexistence of positive solutions of the nonlinear fourth-order beam equation

(2)

They showed that there exists a such that the above boundary value problem has at least two, one, and no positive solutions for , and , respectively.

In this paper, we discuss the existence of positive solutions for the sixth-order boundary value problem

(3)

For this, we shall assume the following conditions throughout

(H1) is continuous;

(H2) , , , where , and , and with , and .

Let and .

Assumption (H2) involves a three-parameter nonresonance condition.

More recently Li [6] studied the existence and multiplicity of positive solutions for a sixth-order boundary value problem with three variable coefficients. The main difference between our work and [6] is that we consider boundary value problem not only with three variable coefficients, but also with two positive parameters λ and μ, and the existence of the positive solution depends on these parameters. In this paper, we shall apply the monotone iterative technique [7] to boundary value problem (3) and then obtain several new existence and multiplicity results. In the special case, in [8] by using the fixed point theorem and the operator spectral theorem, we establish a theorem on the existence of positive solutions for the sixth-order boundary value problem (3) with .

### 2 Preliminaries

Let and . It is well known that Y is a Banach space equipped with the norm . Set . For given and , we denote the norm by

We also need the space X, equipped with the norm

In [8], it is shown that X is complete with the norm and , and moreover , .

For , consider the linear boundary value problem

(4)

where a, b, c satisfy the assumption

(5)

and let . Inequality (5) follows immediately from the fact that is the first eigenvalue of the problem , , and is the first eigenfunction, i.e., . Since the line is the first eigenvalue line of the three-parameter boundary value problem , , if lies in , then by the Fredholm alternative, the existence of a solution of the boundary value problem (4) cannot be guaranteed.

Let , where , . It is easy to see that the equation has two real roots with . Let be a number such that . In this case, (4) satisfies the decomposition form

(6)

Suppose that () is the Green’s function associated with

(7)

We need the following lemmas.

Lemma 1[5,9]

Let, then () can be expressed as

(i) when,

(ii) when,

(iii) when,

Lemma 2[5]

() has the following properties

(i) , ;

(ii) , ;

(iii) , ,

where, , if; , , if; , , if.

In what follows, we let .

Lemma 3[10]

LetXbe a Banach space, Ka cone and Ω a bounded open subset ofX. Letandbe condensing. Suppose thatfor alland. Then.

Lemma 4[10]

LetXbe a Banach space, letKbe a cone ofX. Assume that (here, ) is a compact map such thatfor all. Iffor, then.

Now, since

(8)

the solution of boundary value problem (4) can be expressed as

(9)

Thus, for every given , the boundary value problem (4) has a unique solution , which is given by (9).

We now define a mapping by

(10)

Throughout this article, we shall denote the unique solution of the linear boundary value problem (4).

Lemma 5[8]

is linear completely continuous, where, and. Moreover, , if, then, and, .

We list the following conditions for convenience

(H3) is nondecreasing in u for ;

(H4) for all ;

(H5) uniformly for ;

(H6) for and , where is independent of ρ and u.

Suppose that is the Green’s function of the linear boundary value problem

(11)

Then, the boundary value problem

can be solved by using Green’s function, namely,

(12)

where . Thus, inserting (12) into the first equation in (3), yields

(13)

Let us consider the boundary value problem

(14)

Now, we consider the existence of a positive solution of (14). The function is a positive solution of (14), if , , and .

Let us rewrite equation (13) in the following form

(15)

For any , let

The operator is linear. By Lemmas 2 and 3 in [8], , , we have

where , , , , . Hence , and so . Also is a solution of (13) if satisfies , where , i.e.,

(16)

The operator maps X into X. From together with and the condition , and applying the operator spectral theorem, we find that exists and is bounded. Let , where , then the condition is fulfilled. Let , and let .

Let . Then (16) is equivalent to . By the Neumann expansion formula, H can be expressed by

(17)

The complete continuity of T with the continuity of guarantees that the operator is completely continuous.

Now , let , then , and , . Thus, we have

Hence

(18)

and so, , .

It is easy to see [11] that the following inequalities hold: ,

(19)

and, moreover,

(20)

Lemma 6[8]

is completely continuous, where, and, , , and, moreover, .

For any , define . From (H1), we have that is continuous. It is easy to see that , being a positive solution of (13), is equivalent to , being a nonzero solution of

(21)

Let us introduce the following notations

i.e., . Obviously, is completely continuous. We next show that the operator has a nonzero fixed point in .

Let , where . It is easy to see that P is a cone in Y, and now, we show .

Lemma 7andis completely continuous.

Proof It is clear that is completely continuous. Now , let , then . Using Lemma 6, i.e., , and by Lemma 2, for all , we have

Thus,

(22)

On the other hand, by Lemma 6 and (22), we have

Thus, . □

### 3 Main results

Lemma 8Letbe nondecreasing inuforandfor all, whereis a constant and. Then there existsandsuch that the operatorhas a fixed pointatwith.

Proof Set , where

It is easy to see that . Let and , where and , respectively. Then and , and from Lemma 6, we obtain

It is easy to see that

Indeed, for , let , then from (10), we have . Using equation (4) and (6), we obtain

(23)

and

(24)

Then by (23), we have for

because , and finally, from (24), we have

because and . From the equation

we have

(25)

i.e., for all . Finally, if and , then

i.e.,

(26)

and from (26), it follows that for , if then, we have

(27)

Set and ,  , . Then

where

□

Indeed, by Lemma 6, we have

Now, nondecreasing in u for , Lemma 2, and the Lebesgue convergence theorem guarantee that decreases to a fixed point of the operator .

Lemma 9Suppose that (H3)-(H5) hold, and. Set

wherefor some constants, . Then there exists a constantsuch thatfor all.

Proof Suppose, to the contrary, that there exists a sequence such that , where is a fixed point of the operator at (). Then

where .

Choose , so that

and such that

and , so that . Now,

and so,

Lemma 10Suppose that, (H3) and (H4) hold and that the operatorhas a positive fixed point inPatand. Then for everythere exists a functionsuch that.

Proof Let be a fixed point of the operator at . Then

where , . Hence

Set

and

and . Then

and

because is nondecreasing in u for and is also nondecreasing in u. Thus

(28)

where

□

Indeed, by Lemma 6, we have

Lemma 2 implies that decreases to a fixed point .

Lemma 11Suppose that, (H3)-(H5) hold. Let

Then Λ is bounded.

Proof Suppose, to the contrary, that there exists a fixed point sequence of at such that and . Then there are two cases to be considered: (i) there exists a subsequence such that , which is impossible by Lemma 9, so we only consider the next case: (ii) there exists a constant such that ,  . In view of (H3) and (H4), we can choose such that , and further, for . We know that

Let , i.e., . Then it follows that

(29)

Multiplying (29) by and integrating over , and then using integration by parts on the left side of (29), we have

Next, assume that (ii) holds. Then

and

lead to , which is a contradiction. The proof is complete. □

Lemma 12Suppose that, (H3)-(H4) hold. Let

and let. Then, where Λ is defined in Lemma 11.

Proof By Lemma 10, it follows that . We only need to prove . We may choose a distinct nondecreasing sequence such that . Set as a fixed point of at ,  , i.e., . By Lemma 9, is uniformly bounded, so it has a subsequence, denoted by , converging to . Note that

(30)

Taking the limit as on both sides of (30), and using the Lebesgue convergence theorem, we have

which shows that has a positive fixed point at . □

Theorem 1Suppose that (H3)-(H5) hold, and. For fixed, then there exists atsuch that (3) has at least two, one and has no positive solutions for, for, respectively.

Proof Suppose that (H3) and (H4) hold. Then there exists and such that has a fixed point at and . In view of Lemma 12, also has a fixed point , , and , , . For , there exists such that

for , . In this case, it is easy to see that

Indeed, we have

Similarly, it is easy to see that

Moreover, from (25), it follows that for we have

Finally, we have

By induction, it is easy to see that

(31)

Hence, using (31), we have

i.e.,

so that

Set . Then is completely continuous. Furthermore, for and . Indeed set . Then there exists such that and

By Lemma 3, .

Let k be such that

We know that uniformly for , so we may choose , so that

, so that

Set and . Then is completely continuous. It is easy to obtain

for and . Now , and so

In view of Lemma 4, . By the additivity of the fixed point index,

Thus has a fixed point in and has another fixed point in by choosing . □

Let us introduce the notation in the equation of (13), then we have

(32)

In this case, we can prove the following theorem, which is similar to Theorem 1.

Theorem 2Suppose that (H3)-(H5) hold, and. Then there exists atsuch that (32) has at least two, one and has no positive solutions for, for, respectively.

We follow exactly the same procedure, described in detail in the proof of Theorem 1 for .

Let us introduce the following notations for and

(33)

i.e., .

Lemma 13Suppose that (H3), (H4) and (H6) hold, and. Then for any, there exist real numberssuch that

where.

Proof For any from Lemma 6, we have

Note that for any , there exists an interval and a number such that for . In addition, by (H6), there exists and such that for . If , then ; if , then . Hence

where , , . □

Theorem 3Suppose that (H3), (H4) and (H6) hold, and. Then

(i) (32) has a unique positive solutionsatisfying

whereare constants.

(ii) For any, the sequence

() converges uniformly to the unique solution, and the rate of convergence is determined by

whereis a positive number.

Proof In view of (H3), (H4) and (H6), is a nondecreasing operator and satisfies for and . Indeed, let , , since is nondecreasing in u, then by using , for , it follows that

Moreover, from (25), it follows that for

(34)

Finally, since is nondecreasing in u, then by using form (34), , for , we have

i.e.,

By induction, it is easy to see that

(35)

Hence, using (35), we have

(36)

Now, we show that satisfies for and . Note that

Moreover, from (25), it follows that for ,

Finally, we have

i.e.,

By induction, it is easy to see that

(37)

Hence, using (35) and , , we have

(38)

By Lemma 13, there exists such that

Let

Pick and such that

(39)

and

(40)

Set , , , and ,  . From (36) and (38), we have

(41)

Indeed, from (39) , and , we have

and by induction

From (40), , and , we have

and by induction

Let . Then

(42)

In fact is clear. Assume that (42) holds with (k is a positive integer), i.e., . Then

By induction, it is easy to see that (42) holds. Furthermore, in view of (38), (41) and (42), we have

and

where z is a nonnegative integer. Thus, there exists such that

and is a fixed point of Q and satisfies

This means that , where .

Next we show that is the unique fixed point of Q in . Suppose, to the contrary, that there exists another such that . We can suppose that

Let . Then and . We assert . Otherwise, , and then

This means that , which is a contradiction of the definition of , because .

Let us introduce the following notations for

i.e., , where Q is given by (33). □

Theorem 4Suppose that (H3), (H4), (H6) andhold. Then (32) has a unique positive solutionfor any.

Proof Theorem 3 implies that for , the operator has a unique fixed point , that is . Then from Lemma 10, for every , there exists a function such that .

Thus, is a unique positive solution of (32) for every . □

### 4 Application

As an application of Theorem 1, consider the sixth-order boundary value problem

(43)

for a fixed , , and . In this case, , , and . We have , , , and . It is easy to see that , , and . Note also that , , , , and . Thus, if , then the conditions of Theorem 1 (note ) are fulfilled (in particular, (H3)-(H5) are satisfied). As a result, Theorem 1 can be applied to (43).

### Competing interests

The authors did not provide this information.

### Authors’ contributions

The authors did not provide this information.

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