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ISRN Mathematical Analysis
Volume 2012 (2012), Article ID 354513, 13 pages
Bifurcation of Sign-Changing Solutions for -Point Boundary Value Problems
1Department of Mathematics, Shanghai Normal University, Shanghai 200235, China
2Department of Mathematics, Shanghai Institute of Technology, Shanghai 201418, China
Received 13 February 2012; Accepted 21 March 2012
Academic Editors: R. Avery and A. I. Hai
Copyright © 2012 Yulian An and Maoan Han. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
With the help of bifurcation techniques, some multiplicity results and global structure for sign-changing solutions of some -point boundary value problems are obtained when the nonlinear term is sublinear at 0.
1. Introduction and Main Results
In this paper, we consider the -point boundary value problems: where integer , is a positive parameter, and . We study the multiplicity and global structure of sign-changing solutions of (1.1) and (1.2) under the assumptions:(A1) for with ;(A2) satisfies for ;(A3);(A4).
Multipoint boundary value problems for ordinary differential equations arise in different areas of applied mathematics and physics. The existence of solutions of second-order multipoint boundary value problems has been extensively studied in the literature, see [1–4] and the references therein. Particularly, many authors have studied the existence of sign-changing solutions for various nonlinear boundary value problems, see for example [5–10].
Recently, the global structure of solutions of nonlinear multipoint boundary value problems has also been investigated by several authors using bifurcation methods, see [7–10]. These papers dealt with the case , and relatively little is known about the global structure of solutions when satisfying . The main reason is that the global bifurcation techniques cannot be used directly in this case. Very recently,  investigated the global structure of positive solutions for a class of boundary value problems with . However, to our knowledge there is no paper studying the global structure of sign-changing solutions for nonlinear multipoint boundary value problems under the assumption . The purpose of present paper is to fill this gap.
In this paper, we consider the global structure of nodal solutions of (1.1) and (1.2), a kind of sign-changing having a given number of zeros, when . We find that the discussion is more complicated, when sign-changing solutions are concerned. Eigenvalue theory and Sturm’s comparison theorem play important roles in our discussion.
Now, we introduce some notations as follows.
Let with the norm . Let , and equipped with the norm:
For any function , if and , then is called a simple zero of . For any integer and any , let be sets consisting of functions satisfying the following conditions:: (i) ; (ii) has only simple zeros in and has exactly zeros in .: (i) and ;(ii) has only simple zeros in and has exactly zeros in ;(iii) has a zero strictly between each two consecutive zeros of .
Remark 1.1. If , then or . The sets are open in and disjoint .
The main results of this paper are as follows.
Remark 1.5. Theorem 1.3 extends the result stated in . Meanwhile, Theorem 1.3 and Corollary 1.4 do not only obtain the multiplicity of nodal solutions of (1.1) and (1.2), but also describe the global structure of these solutions.
2. Preliminary Lemmas
The following definition and lemmas about superior limit and component are important to prove Theorem 1.3.
Definition 2.1 (See ). Let be a Banach space, and be a family of subsets of . Then, the superior limit of is defined by:
Lemma 2.2 (See ). Each connected subset of metric space is contained in a component, and each component of is closed.
Lemma 2.3 (See ). Let be a Banach space and a family of closed connected subsets of . Assume that:
(i) there exist , , and such that ;
(ii) , where ;
(iii) for all is a relative compact set of , where Then, there exists an unbounded connected component in such that .
Define a linear operator by: We consider the linear eigenvalues problem: Let be the th eigenvalue of (2.4), and an eigenfunction corresponding to . The following lemma or similar result can be found in [7–9].
Lemma 2.4. Let (A1) hold. Then, For each , algebraic multiplicity of is equal to 1, and the corresponding eigenfunction and is strictly positive on (0,1).
Define a map by: where
It is clear that is completely continuous provided that (A1) and (A2) hold.
3. Proof of the Main Results
We will construct a sequence of functions which is asymptotic linear at 0 and satisfies By means of some corresponding auxiliary equations, we can obtain a sequence of unbounded components via Rabinowitz’s global bifurcation theorem . Based on the sequence, we can find an unbounded component satisfying: and joining with . We do it as follows.
For each , define by: Then, with By (A3), it follows that Now let us consider the auxiliary family of problems: From Proposition 4.1 in , we obtain the following.
Lemma 3.1. Let (A1) and (A2) hold. If is a nontrivial solution of (3.6). Then, for some .
Let such that: Note that Let us consider as a bifurcation problem from the trivial solution .
Equation (3.9) can be converted to the equivalent form: Note that for near in . Applying Lemma 2.4, the global bifurcation result of Rabinowitz  for (3.9) can be stated as follows: for each integer , , there exists a continuum of solutions of (3.9) joining to infinity in . Moreover, .
For properties of , we give the following lemmas.
Lemma 3.2. Let (A1)–(A4) hold. Then for each fixed , joins to in (see Figure 1(b)).
Proof. We divide the proof into two steps.
Step 1. We show that . Assume on the contrary that . Let be such that: Similar to the argument of Lemma 2.5, we conclude that .
Since , we have Set . Then, , and Using , we can show that The proof is similar to that of Theorem 1 in , and therefore we omit it. Equations (3.13) and (3.14) imply that for some constant , independent of . Hence, has a convergent subsequence in . Without loss of generality, we assume that there exists with: such that Note that (3.12) is equivalent to Combining this with (3.16) and using (3.14) and the Lebesgue dominated convergence theorem, we have This contradicts (3.15). Therefore,
Step 2. We show that . On the contrary, assume that . Then, there exists a sequence such that From Remark 1.1, we can take a subsequences of , still denoted by , such that or . Without loss of generality, we suppose that . When is considered, the proof is similar. We omit it.
Note that satisfies the autonomous equation: Therefore, the graph of consists of a sequence of positive and negative bumps, together with a truncated bump at the right end of the interval , with the following properties (ignoring the truncated bump) (see ): all the positive (respectively, negative) bumps (i) have the same shape (the shapes of the positive and negative bumps may be different); (ii) attain the same maximum (minimum) value.
Let denote the zeros of in . Then, after taking a subsequence if necessary, . Clearly, . Set . We can choose at least one subinterval which is of length at least for some . Then, for this , if is large enough. Put .
Obviously, for the above given , and , have the same sign on for all . Without loss of generality, we assume Armed with the information on the shape of , it is easy to show that for the above given , .
Let be a constant with . Since is concave on , we have Then, there must exist constants with and such that On the other hand, note that Using the relation: and Sturm’s comparison theorem, we deduce that must change its sign on if is sufficiently large, contradicting (3.25). Therefore, Hence, joins to in .
Lemma 3.3. Let (A1)–(A4) hold. Then, there exists such that
Proof. The proof is similar to that of Lemma 4.3 in . We omit it.
Lemma 3.4. Let (A1)–(A4) hold, and let be as in Lemma 3.3. Then, there exist and such that for any and :
Proof. Suppose on the contrary that there exists such that
Now, the method used in the proof of Lemma 3.2, Step 2, is still valid. Let be a constant with . Taking subsequences again if necessary, still denoted by , such that . Without loss of generality, we can also derive an interval and such that It is easy to find an integer such that . This implies that Note that for , Combining these facts and the relation: and Sturm’s comparison theorem, we conclude that must change its sign on if is large enough. This contradicts (3.32), and the proof is done.
Lemma 3.5. Let (A1)–(A4) hold, and let be as in Lemma 3.4. Then, there exist and such that for any and :
Proof. Similar to the proof of Lemma 3.4, we can find a constant such that provided that being a solution of (1.1) and (1.2).
Let . We claim that there exists such that (3.36) holds. Suppose on the contrary that there exists satisfying We can discuss two cases.
Case 1. If . is compact in implies that there exists a subsequence, still denoted by , such that Obviously, is a solution of (1.1) and (1.2). It is impossible.Case 2. If . Taking subsequences again if necessary, still denoted by , such that . Using the same argument as Lemma 3.4, we can find a contradiction.
Proof of Theorem 1.3. We will prove that the superior limit of contains an unbounded component of solutions of (1.1) and (1.2), which joins to . For , introduce
Let and be as in Lemma 3.5. Firstly, for each given nonnegative integer , and with , we define the connected subset, , in satisfying (see Figure 2(a)):
(ii) joins with infinity in .By Lemmas 2.2 and 2.3, contains a component joining with infinity in (see Figure 2(b)).
It is easy to verify that if , then is a solution of (1.1) and (1.2), and .
Next, by using Lemma 2.3 and the method in  (see (4.22)–(4.30) in ), we can find a component in , which is unbounded both in and .
Finally, we show that joins with . This will be done by the following three steps.
Step 1. We show that for .
Suppose on the contrary that there exists with , and for some constant . Applying the method of proving Lemma 3.2, we can deduce a contradiction.Step 2. We show that . By a similar argument as Lemma 3.2, we can get the conclusion.Step 3. We show that for .
On the contrary, suppose that there exists with for some constant . The proof can be done by the same argument as Lemma 3.2.
This completes the proof of Theorem 1.3.
This paper is supported by NSFC (no. 10971139); China Postdoctoral Fund (no. 2011M500615); Scientific Innovation Projection of Shanghai Education Department (no. 11YZ225); SIT-YJ2009-16.
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