We consider the nonlinear eigenvalue problems , , , , where , and for with and satisfies for , and , where . We investigate the global structure of nodal solutions by using the Rabinowitz's global bifurcation theorem.
We study the global structure of nodal solutions of the problem
Here and for with is a positive parameter, and
In the case that the global structure of nodal solutions of nonlinear second-order -point eigenvalue problems (1.1), (1.2) have been extensively studied; see [1–5] and the references therein. However, relatively little is known about the global structure of solutions in the case that and few global results were found in the available literature when The likely reason is that the global bifurcation techniques cannot be used directly in the case. On the other hand, when -point boundary value condition (1.2) is concerned, the discussion is more difficult since the problem is nonsymmetric and the corresponding operator is disconjugate. In , we discussed the global structure of positive solutions of (1.1), (1.2) with However, to the best of our knowledge, there is no paper to discuss the global structure of nodal solutions of (1.1), (1.2) with
In this paper, we obtain a complete description of the global structure of nodal solutions of (1.1), (1.2) under the following assumptions:
Let with the norm
with the norm
respectively. Define by setting
For any function if then is a simple zero of if For any integer and any define sets consisting of functions satisfying the following conditions:
(ii) has only simple zeros in and has exactly zeros in ;
(ii) has only simple zeros in and has exactly zeros in
(iii) has a zero strictly between each two consecutive zeros of .
Obviously, if then or The sets are open in and disjoint.
The nodal properties of solutions of nonlinear Sturm-Liouville problems with separated boundary conditions are usually described in terms of sets similar to see . However, Rynne  stated that are more appropriate than when the multipoint boundary condition (1.2) is considered.
Next, we consider the eigenvalues of the linear problem
Let hold. The spectrum consists of a strictly increasing positive sequence of eigenvalues with corresponding eigenfunctions In addition,
(ii) for each and is strictly positive on
We can regard the inverse operator as an operator In this setting, each is a characteristic value of with algebraic multiplicity defined to be dim where denotes null-space and is the identity on
Let hold. For each the algebraic multiplicity of the characteristic value of is equal to 1.
Let under the product topology. As in , we add the points to our space Let Let denote the closure of set of those solutions of (1.1), (1.2) which belong to The main results of this paper are the following.
Let (A1)–(A4) hold.
(a)If then there exists a subcontinuum of with and
(b)If then there exists a subcontinuum of with
(c)If then there exists a subcontinuum of with is a bounded closed interval, and approaches as
Let (A1)–(A4) hold.
(a)If , then (1.1), (1.2) has at least one solution in for any
(b)If , then (1.1), (1.2) has at least one solution in for any
(c)If then there exists such that (1.1), (1.2) has at least two solutions in for any .
We will develop a bifurcation approach to treat the case . Crucial to this approach is to construct a sequence of functions which is asymptotic linear at and satisfies
By means of the corresponding auxiliary equations, we obtain a sequence of unbounded components via Rabinowitz's global bifurcation theorem , and this enables us to find unbounded components satisfying
The rest of the paper is organized as follows. Section 2 contains some preliminary propositions. In Section 3, we use the global bifurcation theorems to analyse the global behavior of the components of nodal solutions of (1.1), (1.2).
Definition 2.1 (see ).
Let be a Banach space and 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 connected component of is closed.
Lemma 2.3 (see ).
(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 and .
Define the map by
It is easy to verify that the following lemma holds.
Assume that (A1)-(A2) hold. Then is completely continuous.
For , let
Let (A1)-(A2) hold. If , then
The proof is similar to that of Lemma in ; we omit it.
Let (A1)-(A2) hold, and is a sequence of solutions of (1.1), (1.2). Assume that for some constant , and Then
From the relation we conclude that Then
which implies that is bounded whenever is bounded.
3. Proof of the Main Results
For each define by
By (A3), it follows that
Now let us consider the auxiliary family of the equations
Lemma 3.1 (see [1, Proposition ]).
Let (A1), (A2) hold. If is a nontrivial solution of (3.4), (3.5), then for some .
Let be such that
Let us consider
as a bifurcation problem from the trivial solution
Equation (3.8) can be converted to the equivalent equation
Further we note that for near in
The results of Rabinowitz  for (3.8) can be stated as follows. For each integer , there exists a continuum of solutions of (3.8) joining to infinity in Moreover,
Proof of Theorem 1.5.
Let us verify that satisfies all of the conditions of Lemma 2.3.
condition in Lemma 2.3 is satisfied with . Obviously
and accordingly, holds. can be deduced directly from the Arzela-Ascoli Theorem and the definition of . Therefore, the superior limit of , contains an unbounded connected component with .
From the condition (A2), applying Lemma 2.2 with in , we can show that the initial value problem
has a unique solution on for every and . Therefore, any nontrivial solution of (1.1), (1.2) has only simple zeros in and . Meanwhile, (A1) implies that [1, proposition 4.1]. Since , we conclude that . Moreover, by (1.1) and (1.2).
We divide the proof into three cases.
Case 1 ().
In this case, we show that .
Assume on the contrary that
then there exists a sequence such that
for some positive constant depending not on . From Lemma 2.6, we have
Set Then Now, choosing a subsequence and relabelling if necessary, it follows that there exists with
Since , we can show that
The proof is similar to that of the step 1 of Theorem in ; we omit it. So, we obtain
and subsequently, for . This contradicts (3.16). Therefore
Case 2 ().
In this case, we can show easily that joins with by using the same method used to prove Theorem in .
Case 3 ().
In this case, we show that joins with .
Let be such that
If is bounded, say, , for some depending not on , then we may assume that
Taking subsequences again if necessary, we still denote such that . If , all the following proofs are similar.
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 that
Moreover, we have
Combining this with the fact
and using the relation
we deduce that must change its sign on if is large enough. This is a contradiction. Hence is unbounded. From Lemma 2.6, we have that
Note that satisfies the autonomous equation
We see that 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, ):
(i)all the positive (resp., negative) bumps have the same shape (the shapes of the positive and negative bumps may be different);
(ii)each bump contains a single zero of , and there is exactly one zero of between consecutive zeros of ;
(iii)all the positive (negative) bumps attain the same maximum (minimum) value.
Armed with this information on the shape of it is easy to show that for the above given is an unbounded sequence. That is
Since is concave on , for any small enough,
This together with (3.31) implies that there exist constants with , such that
Hence, we have
Now, we show that .
Suppose on the contrary that, choosing a subsequence and relabeling if necessary, for some constant . This implies that
From (3.28) we obtain that must change its sign on if is large enough. This is a contradiction. Therefore .
Proof of Theorem 1.6.
and are immediate consequence of Theorem 1.5 and , respectively.
To prove , we rewrite (1.1), (1.2) to
By Lemma 2.5, for every and ,
Let be such that
Then for and ,
This means that
By Lemma 2.6 and Theorem 1.5, it follows that is also an unbounded component joining and in . Thus, (3.40) implies that for (1.1), (1.2) has at least two solutions in .
The author is very grateful to the anonymous referees for their valuable suggestions. This paper was supported by NSFC (no.10671158), 11YZ225, YJ2009-16 (no.A06/1020K096019).
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