Existence of Positive Solutions of Fourth-Order Problems with Integral Boundary Conditions
© R. Ma and T. Chen. 2011
Received: 5 May 2010
Accepted: 7 July 2010
Published: 3 August 2010
We study the existence of positive solutions of the following fourth-order boundary value problem with integral boundary conditions, , , , , , , where is continuous, are nonnegative. The proof of our main result is based upon the Krein-Rutman theorem and the global bifurcation techniques.
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 , Ma was concerned with the existence of positive solutions of (1.1) and (1.2) under the assumptions:
uniformly for , where ;
(H2) for and ;
Ma proved the following.
Theorem A (see [3, Theorem ]).
Then (1.1) and (1.2) have at least one positive solution.
where may be singular at and (or) ; is continuous, and are nonnegative.
under the assumption
(H4) are nonnegative, and . The main result of this paper is the following.
Then (1.9) has at least one positive solution.
Theorem 1.1 generalizes [3, Theorem ] where the special case and was treated.
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!).
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 ].
has nonempty interior and ;
- (ii)is -completely continuous and positive, for , for and(1.18)
where is a strongly positive linear compact operator on with the spectral radius , satisfies as locally uniformly in .
such that .
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 ).
Lemma 2.2 (see ).
Lemma 2.3 (see ).
(H5) be two given constants with .
if (2.8) and (2.9) have nontrivial solutions.
Let with the norm . Let with the norm .
We claim that is a Banach space.
Therefore, is a Banach space.
Then the cone is normal and nonempty interior and .
In fact, for any , it follows from the definition of that
(2) , .
Thus, . Obviously, .
From , we have that , and so , and accordingly .
which implies that , and consequently .
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.
and the corresponding eigenfunction .
Proof of Theorem 2.6.
We claim that .
where , it follows that , and accordingly .
Thus , and accordingly .
Now, since , and is compactly embedded in , we have that is compact.
Next, we show that is positive.
This together with (2.9) and (H4) imply on .
Therefore, it follows from (2.43) and (2.45) that .
Now, by the Krein-Rutman 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
It is easy to check that is compact.
such that .
Proof of Theorem 1.1.
we note that for all since is the only solution of (3.8) for and .
Case 1 ( ).
We divide the proof into two steps.
then joins to .
This together with Theorem 2.6 imply that . Therefore, joins to .
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 .
Combining this with (2.39), we conclude that , here, . Therefore, we have that from Lemma 1.4 that .
Case 2 ( ).
and, moreover, .
Again joins to and the result follows.
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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