We study the existence of positive solutions of the following fourthorder boundary value problem with integral boundary conditions, , , , , , , where is continuous, are nonnegative. The proof of our main result is based upon the KreinRutman theorem and the global bifurcation techniques.
1. Introduction
The deformations of an elastic beam in an equilibrium state, whose both ends are simple supported, can be described by the fourthorder boundary value problem
where is continuous; see Gupta [1, 2]. In the past twenty more years, the existence of solutions and positive solutions of these kinds of problems and the Lidstone problem has been extensively studied; see [3–9] and the references therein. In [3], Ma was concerned with the existence of positive solutions of (1.1) and (1.2) under the assumptions:
(H1) is continuous and there exist constants , with such that
uniformly for , and
uniformly for , where ;
(H2) for and ;
(H3)there exist constants satisfying and
Ma proved the following.
Theorem A (see [3, Theorem ]).
Let (H1), (H2), and (H3) hold. Assume that either
or
where denotes the first generalized eigenvalue of the generalized eigenvalue problem
Then (1.1) and (1.2) have at least one positive solution.
At the same time, we notice that a class of boundary value problems with integral boundary conditions appeared in heat conduction, chemical engineering underground water flow, thermoelasticity, and plasma physics. Such a kind of problems include twopoint, threepoint, multipoint and nonlocal boundary value problems as special cases and attracting the attention of a few readers; see [10–13] and the references therein. For example, In particular, Zhang and Ge [10] used GuoKrasnoselskii fixedpoint theorem to study existence and nonexistence of positive solutions of the following fourthorder boundary value problem with integral boundary conditions:
where may be singular at and (or) ; is continuous, and are nonnegative.
Motivated by [3, 10], in this paper, we consider the existence of positive solutions of the following fourthorder boundary value problem with integral boundary conditions:
under the assumption
(H4) are nonnegative, and . The main result of this paper is the following.
Theorem 1.1.
Let (H1), (H2), (H3), and (H4) hold. Assume that either
or
where
Then (1.9) has at least one positive solution.
Remark 1.2.
Theorem 1.1 generalizes [3, Theorem ] where the special case and was treated.
Remark 1.3.
Zhang and Ge [10] proved existence and nonexistence of positive solutions via GuoKrasnoselskii fixedpoint theorem under some conditions which do not involve the eigenvalues of (1.12)–(1.14). While our Theorem 1.1 is established under (1.10) or (1.11) which is related to the eigenvalues of (1.12)–(1.14). Moreover, (1.10) and (1.11) are optimal. Let us consider the problem
In this case, and the corresponding eigenfunction is . However, (1.15) and (1.16) has no positive solution. (In fact, suppose on the contrary that (1.15) and (1.16) has a positive solution . Multiplying (1.15) with and integrating from to , we get a desired contradiction!).
Suppose that is a real Banach space with norm . Let be a cone in . A nonlinear mapping is said to be positive if . It is said to be completely continuous if is continuous and maps bounded subsets of to precompact subset of . Finally, a positive linear operator on is said to be a linear minorant for if for . If is a continuous linear operator on , denote the spectral radius of . Define
The following lemma will play a very important role in the proof of our main results, which is essentially a consequence of Dancer [14, Theorem ].
Lemma 1.4.
Assume that
(i) has nonempty interior and ;
(ii) is completely continuous and positive, for , for and
where is a strongly positive linear compact operator on with the spectral radius , satisfies as locally uniformly in .
Then there exists an unbounded connected subset of
such that .
Moreover, if has a linear minorant and there exists a
such that and , then can be chosen in
Proof.
Since is a strongly positive compact endomorphism of and has nonempty interior, we have from Amann [15, Theorem ] that the set in [14, Theorem ] reduces to a single point . Now the desired result is a consequence of Dancer [14, Theorem ].
The rest of the paper is arranged as follows. In Section 2, we state and prove some preliminary results about the spectrum of (1.12)–(1.14). Finally, in Section 3, we proved our main result.
2. Generalized Eigenvalues
Lemma 2.1 (see [10]).
Assume that (H4) holds. Then for any , the boundary value problem
has a unique solution which is given by
where
Lemma 2.2 (see [10]).
Assume that (H4) holds. Then for any , the boundary value problem
has a unique solution which is given by
where
Lemma 2.3 (see [10]).
Assume that (H4) holds. Then one has
Let
(H5) be two given constants with .
Definition 2.4.
One says thatis a generalized eigenvalue of linear problem
if (2.8) and (2.9) have nontrivial solutions.
Let
Let with the norm . Let with the norm .
Let
For , from Lemma 2.1, it follows that
By simple calculations, we have
Combining this with (H4), we conclude that
This together with (2.12) and the fact that imply that
Since , we may define the norm of by
We claim that is a Banach space.
In fact, let be a Cauchy sequence, that is, as . From the definition of , it follows that
where is a normal in defined by . Thus,
By the completeness of , there exists , such that
From the fact that , we have that for arbitrary , there exists , such that
and subsequently,
Fixed and let , we get
This is,
Therefore, is a Banach space.
Let
Then the cone is normal and nonempty interior and .
In fact, for any , it follows from the definition of that
(1)there exist real number , such that
(2), .
From and (H4), we obtain that for some . Moreover,
and subsequently,
for some . We may take satisfying
Now, let us define
Then , and
Thus, . Obviously, .
Lemma 2.5.
Assume that holds. Then for any , one has
Proof.
In fact, for , we have that
From , we have that , and so , and accordingly .
We have from the fact that , , that
which implies that , and consequently .
For , define a linear operator by
Theorem 2.6.
Assume that (H4) and (H5) hold. Let be the spectral radius of . Then (2.8) and (2.9) has an algebraically simple eigenvalue, , with a positive eigenfunction . Moreover, there is no other eigenvalue with a positive eigenfunction.
Remark 2.7.
If , then can be explicitly given by
and the corresponding eigenfunction .
Proof of Theorem 2.6.
From Lemma 2.2, it is easy to check that (2.8) and (2.9) is equivalent to the integral equation
We claim that .
In fact, for ,we have that
Since
and for some constant , it concludes that
Hence,
where , it follows that , and accordingly .
If , then on , and accordingly
Thus , and accordingly .
Now, since , and is compactly embedded in , we have that is compact.
Next, we show that is positive.
For , if , from Lemma 2.3, we have
Combining this with (2.39), there exist such that
For , if , applying a similar proof process of (2.43), we have
Combining this with (2.39), there exist such that
This together with (2.9) and (H4) imply on .
Therefore, it follows from (2.43) and (2.45) that .
Now, by the KreinRutman theorem ([16, Theorem C]; [17, Theorem ]), has an algebraically simple eigenvalue with an eigenfunction . Moreover, there is no other eigenvalue with a positive eigenfunction. Correspondingly, with a positive eigenfunction of , is a simple eigenvalue of (2.8) and (2.9). Moreover, for (2.8) and (2.9), there is no other eigenvalue with a positive eigenfunction.
3. The Proof of the Main Result
Before proving Theorem 1.1, we denote by setting
where
It is easy to check that is compact.
Let be such that
Obviously,()impliesthat
Let
then is nondecreasing and
Let us consider
as a bifurcation problem from the trivial solution . It is to easy to check that (3.8) can be converted to the equivalent equation
From the proof process of Theorem 2.6, the operator ,
is compact and strongly positive. Define by
then we have from (3.4) and Lemma 2.5 that
locally uniformly in . Fromand Theorem 2.6 (with obvious changes), it follows that if is a nontrivial solution of (3.8) with , then . Combining this with Lemma 1.4, we conclude that there exists an unbounded connected subset of the set
such that .
Proof of Theorem 1.1.
It is clear that any solution of the form yields a solution of (1.9). We will show that crosses the hyperplane . To do this, it is enough to show that joins to . Let satisfy
we note that for all since is the only solution of (3.8) for and .
Case 1 ().
In this case, we show that
We divide the proof into two steps.
Step 1.
We show that if there exists a constant number such that
then joins to .
From (3.16), we have that . We divide the equation
by and set . Since is bounded in , choosing a subsequence and relabeling if necessary, we see that for some with . Moreover, we have from (3.7) and Lemma 2.5 that
Since Thus,
where , again choosing a subsequence and relabeling if necessary. Thus,
This together with Theorem 2.6 imply that . Therefore, joins to .
Step 2.
We show that there exists a constant be such that for all .
By Lemma 1.4, we only need to show that has a linear minorant and there exists a such that and .
By,there exist constants satisfying and
For , let
then is a linear minorant of . Moreover,
where
Combining this with (2.39), we conclude that , here, . Therefore, we have that from Lemma 1.4 that .
Case 2 ().
In this case, if such that
and then
and, moreover, .
Assume that there exists , such that for all ,
Applying a similar argument to that used in Step 1 of Case 1, after taking a subsequence and relabeling if necessary, it follows that
Again joins to and the result follows.
Acknowledgments
The authors are very grateful to the anonymous referees for their valuable suggestions. This paper was supported by the NSFC (no. 11061030), the Fundamental Research Funds for the Gansu Universities.
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