Abstract
This paper investigates the existence and multiplicity of positive solutions of a sixthorder 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 theorem1 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 sixthorder ordinary differential equations, which arise naturally, for example, in sandwich beam deflection under transverse shear have been studied extensively, see [14] 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 sixthorder ordinary differential equation
Liu and Li [5] studied the existence and nonexistence of positive solutions of the nonlinear fourthorder beam equation
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 sixthorder boundary value problem
For this, we shall assume the following conditions throughout
(H2) , , , where , and , and with , and .
Assumption (H2) involves a threeparameter nonresonance condition.
More recently Li [6] studied the existence and multiplicity of positive solutions for a sixthorder 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 sixthorder 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
where a, b, c satisfy the assumption
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 threeparameter 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
Suppose that () is the Green’s function associated with
We need the following lemmas.
Let, then () can be expressed as
Lemma 2[5]
() has the following properties
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
the solution of boundary value problem (4) can be expressed as
Thus, for every given , the boundary value problem (4) has a unique solution , which is given by (9).
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 ;
(H6) for and , where is independent of ρ and u.
Suppose that is the Green’s function of the linear boundary value problem
Then, the boundary value problem
can be solved by using Green’s function, namely,
where . Thus, inserting (12) into the first equation in (3), yields
Let us consider the boundary value problem
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
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.,
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
The complete continuity of T with the continuity of guarantees that the operator is completely continuous.
Now , let , then , and , . Thus, we have
Hence
It is easy to see [11] that the following inequalities hold: ,
and, moreover,
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
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,
On the other hand, by Lemma 6 and (22), we have
3 Main results
Lemma 8Letbe nondecreasing inuforandfor all, whereis a constant and. Then there existsandsuch that the operatorhas a fixed pointatwith.
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
and
because , and finally, from (24), we have
because and . From the equation
we have
i.e., for all . Finally, if and , then
i.e.,
and from (26), it follows that for , if then, we have
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
and so,
which is a contradiction. □
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
Set
and
and
because is nondecreasing in u for and is also nondecreasing in u. Thus
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
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
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
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
Let k be such that
We know that uniformly for , so we may choose , so that
Set and . Then is completely continuous. It is easy to obtain
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
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
Lemma 13Suppose that (H3), (H4) and (H6) hold, and. Then for any, there exist real numberssuch that
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
Theorem 3Suppose that (H3), (H4) and (H6) hold, and. Then
(i) (32) has a unique positive solutionsatisfying
() converges uniformly to the unique solution, and the rate of convergence is determined by
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
Finally, since is nondecreasing in u, then by using form (34), , for , we have
i.e.,
By induction, it is easy to see that
Hence, using (35), we have
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
Hence, using (35) and , , we have
By Lemma 13, there exists such that
Let
and
Set , , , and , . From (36) and (38), we have
Indeed, from (39) , and , we have
and by induction
and by induction
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
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 .
4 Application
As an application of Theorem 1, consider the sixthorder boundary value problem
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
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Authors’ contributions
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