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On Linear and Nonlinear Fourth-Order Eigenvalue Problems with Nonlocal Boundary Condition
We determine the principal eigenvalue of the linear problem , , , where and . Moreover, we investigate the existence of positive solutions for the corresponding nonlinear problem. The proofs of our main results are based upon the Krein-Rutman theorem and fixed point index theory.
The deformations of an elastic beam can be described by the boundary value problems of the fourth-order ordinary differential equations. For example, an elastic beam in an equilibrium state whose both ends are simply supported can be described by the fourth-order boundary value problem of the form see Gupta [1, 2]. Owing to its significance in physics, it has been studied by many authors using nonlinear alternatives of Leray-Schauder, the fixed point index theory, and the method of lower and upper solutions; see, for example, [3–16].
Recently, Bai  investigated the existence of positive solutions for more general fourth-order nonlocal boundary value problem By using the Krasnoselskii fixed point theorem, the sufficient conditions for the existence of positive solutions of (2) are obtained. We can find that, however, those conditions do not involve the eigenvalues with respect to the relevant linear operator, and those sufficient conditions are not optimal.
The likely reason is that the spectrum structure of the linear eigenvalue problem is not clear.
It is the purpose of this paper to investigate the first eigenvalue of (3) by using the Krein-Rutman theorem [17, 18], and then we use this spectrum result to establish the existence of positive solutions of nonlinear nonlocal problem (2). The existence of positive solution is obtained by means of fixed point index theory under some conditions concerning the first eigenvalue with respect to the relevant linear operator. The obtained sufficient conditions in this paper are optimal. For the concepts and properties of fixed point index theory, we refer the reader to .
2. Spectrum of (3)
Let us assume that(A1), , , , , , .(A2).
To study the spectrum of (3), we need several preliminary results.
Lemma 2 (see ). Assume (A1) holds. Then one has(i), for ; , for ;(ii), for ,where , ; , .
Lemma 3 (see ). Assume (A1) holds. Then for any , solves the problem if and only if where
Let Then, there exist such that respectively.
In fact, from (ii) of Lemma 2, we can obtain where we can take and also we obtain where
Let for a certain , and let be the Banach space with the norm . For , we have Combining (16) and (9) with the fact that , we conclude that So, we may define the norm of by This norm is so called -norm.
It is easy to get the following lemmas.
Lemma 4. is a Banach space.
Proof. Let be Cauchy sequence of . Then, we have From (18) and (19), for all , there exists such that Hence, which implies that is a Cauchy sequence of . According to the completeness of , there exists such that Let in (20). Then, we have which implies that Hence, for Cauchy sequence , there exists satisfying Therefore, is a Banach space.
Lemma 5. is compactly embedded in .
Proof. Let be bounded. Then, according to the fact that is compactly embedded in , there exist and such that From (26), for all , there exists such that Since combining this fact with (27) we have which implies Therefore, is compactly embedded in .
Let Then, the cone is normal and has nonempty interior .
In the rest of this section, we will prove the existence of the first eigenvalue of (3). To wit, we get the following.
Theorem 6. Assume (A1) holds. Then (3) has an algebraically simple eigenvalue , with an eigenfunction , and there is no other eigenvalue with a positive eigenfunction.
Remark 7. If , then can be explicitly given by and the corresponding eigenfunction , .
Proof of Theorem 6. For , define a linear operator by
Then, by the definition of , we have
Therefore, we can obtain
We claim that . In fact, for , let . Then, from (18) and (17) we have, for , On the one hand, from (36)-(37) and (10) we have On the other hand, from (36)-(37) and (10) we have Therefore, . If , then , on . According to (36) and the fact that we have Furthermore, according to (41), the definition of , and the fact that we have Then , and accordingly, .
Now, since and is compactly embedded in , we have that is compact.
Next, we show that is strongly positive.
For , it is easy to check that there exist , such that for In fact, for every , it follows from Lemma 2 that where where
Therefore, it follows from (44) that .
Now, on the one hand, by the Krein-Rutman theorem ([17, Theorem ] and [18, Theorem 19.3]), has an algebraically simple eigenvalue with an eigenfunction . Moreover, there is no other eigenvalue with a positive eigenfunction. On the other hand, we have from the definition of and Lemma 3 that (3) is equivalent to the integral equation Therefore, with a positive eigenfunction is a simple eigenvalue of (3). Moreover, for (3), there is no other eigenvalue with a positive eigenfunction.
3. An Application of Theorem 6
For convenience, we introduce the following notations:
Theorem 8. Assume that (A1) and (A2) hold, and . Then, for each satisfying there exists at least one positive solution of (2).
Theorem 9. Assume that (A1) and (A2) hold, and . Then, for each satisfying there exists at least one positive solution of (2).
Remark 10. Bai  proved existence of positive solutions via Guo-Krasnoselskii fixed point theorem under some conditions which do not involve the eigenvalue of (3). While our Theorems 8 and 9 are established under (51) or (52) which are related to the eigenvalue of (3). Our Theorems 8 and 9 cover an undefined case in . Consider the following boundary value problem: In this case, , , , and According to Theorem 9, the above boundary value problem has at least one positive solution. For the above boundary value problem, however, we cannot obtain the above conclusion by [3, Theorem 3.1] since Moreover, (51) and (52) are optimal. In order to illustrate this point, consider the problem where In this case, , , , and However, (56) has no positive solution. In fact, suppose on the contrary that (56) has a positive solution . Multiplying the first equation of (56) with and integrating from 0 to 1, we get which is a contradiction.
Lemma 11. For every , there exist such that
Proof. By , there is , so
By , there is a such that , and so, for Hence , and accordingly,
We have from (18) that , which implies that and consequently where .
Let It is easy to show that is a completely continuous operator. In addition, we can verify that the nonzero fixed points of the operator are positive solutions of the problem (2).
Lemma 12 (see ). Let be Banach space, a cone in , and a bounded open set in . Suppose that is a completely continuous operator. If there exists such that then the fixed point index .
Lemma 13 (see ). Let be Banach space, a cone in , and a bounded open set in with . Suppose that is a completely continuous operator. If then the fixed point index .
Proof of Theorem 8. It follows from the first inequality of (51) that there exists , such that
Let be the positive eigenfunction of (see (33)) corresponding to . Thus .
Let . Then for every , we have from Lemma 11 that It follows from (70) that
We may suppose that has no fixed points on (otherwise, the proof is finished). Now we show that
Suppose the contrary, that there exists and such that . Hence and Put It is easy to see that and . We find from that Therefore, by (72), we have which contradicts the definition of . Hence (73) is true, and we have from Lemma 12 that
It follows from the second inequality of (51) that there exists and such that
Let , . Then is a bounded linear operator and .
Let where . It is clear that .
Let In the following, we prove that is bounded.
For any , set . Then, Thus, , for every . Since is the first eigenvalue of and , the first eigenvalue of , . Therefore, the inverse operator exists and It follows from that . So, we have , , and we conclude that is bounded.
Select . Then, from the invariance property of the fixed point index, we have By (78) and (84), we have that Then has at least one fixed point on . This means that the boundary value problem (2) has at least one positive solution.
Proof of Theorem 9. It follows from the second inequality of (52) that there exists , such that
Let . Then, for every , we have from Lemma 11 that
It follows from (86) that
Suppose there exists and satisfying . We may suppose that has no fixed points on (otherwise, the proof is finished). So . By (88), we have . By induction, we have , for all . Thus
By Gelfand’s formula, we have
which is a contradiction with . Hence
and we have from Lemma 13 that
It follows from the first inequality of (52) that there exist and