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
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
Liu and Li  studied the existence and nonexistence of positive solutions of the nonlinear fourth-order 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 sixth-order boundary value problem
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  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  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  to boundary value problem (3) and then obtain several new existence and multiplicity results. In the special case, in  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 .
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 , 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 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
Suppose that ( ) is the Green’s function associated with
We need the following lemmas.
Let , then ( ) can be expressed as
(i) when ,
(ii) when ,
(iii) when ,
( ) has the following properties
(i) , ;
(ii) , ;
(iii) , ,
where , , if ; , , if ; , , if .
In what follows, we let .
LetXbe a Banach space, Ka cone and Ω a bounded open subset ofX. Let and be condensing. Suppose that for all and . Then .
LetXbe a Banach space, letKbe a cone ofX. Assume that (here , ) is a compact map such that for all . If for , then .
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).
We now define a mapping by
Throughout this article, we shall denote the unique solution of the linear boundary value problem (4).
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
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
For any , let
The operator is linear. By Lemmas 2 and 3 in , , , 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
and so, , .
It is easy to see  that the following inequalities hold: ,
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 7 and is 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
On the other hand, by Lemma 6 and (22), we have
Thus, . □
3 Main results
Lemma 8Let be nondecreasing inufor and for all , where is a constant and . Then there exists and such that the operator has a fixed point at with .
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
Then by (23), we have for
because , and finally, from (24), we have
because and . From the equation
i.e., for all . Finally, if and , then
and from (26), it follows that for , if then, we have
Set and , , . Then
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
where for some constants , . Then there exists a constant such that for all .
Proof Suppose, to the contrary, that there exists a sequence such that , where is a fixed point of the operator at ( ). Then
Choose , so that
and such that
and , so that . Now,
which is a contradiction. □
Lemma 10Suppose that , (H3) and (H4) hold and that the operator has a positive fixed point inPat and . Then for every there exists a function such that .
Proof Let be a fixed point of the operator at . Then
where , . Hence
and . Then
because is nondecreasing in u for and is also nondecreasing in u. Thus
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
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 at such 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
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