Research Article | Open Access

Ruyun Ma, Jiemei Li, "Global Bifurcation for Second-Order Neumann Problem with a Set-Valued Term", *International Journal of Differential Equations*, vol. 2009, Article ID 373851, 16 pages, 2009. https://doi.org/10.1155/2009/373851

# Global Bifurcation for Second-Order Neumann Problem with a Set-Valued Term

**Academic Editor:**Qingkai Kong

#### Abstract

We study the global bifurcation of the differential inclusion of the form , where is a “set-valued representation” of a function with jump discontinuities along the line segment . The proof relies on a Sturm-Liouville version of Rabinowitz's bifurcation theorem and an approximation procedure.

#### 1. Introduction

We are concerned with the following differential inclusion which arises from a Budyko-North type energy balance climate models:

see [1–6] and the references therein. In particular, the set-valued right-hand side arises from a jump discontinuity of the albedo at the ice-edge in these models. By filling in such a gap, one arrives at the set-valued problem (1.1). As in [6], we are here interested in a considerably simplified version as compared to the situation from climate modeling; for example, a one-dimensional regular Sturm-Liouville differential operator substitutes for a two-dimensional Laplace-Beltrami operator or a singular Legendre-type operator, and the jump discontinuity is transformed to in a way, which resembles only locally the climatological problem.

Assume that

(H1), ;(H2), strictly increasing for , exists uniformly for , and on ,(H) satisfies that exists uniformly for ;(H3), , , .Let in (1.1) be given by and set

Throughout will be considered as subset of the Banach space under the norm

Let

Using a Sturm-Liouville version of Rabinowitz's bifurcation theorem and an approximation procedure, Hetzer [6] proved the following.

Theorem A (see [6, Theorem]). *Let (H1)–(H3) be fulfilled. Then there exist sequences of unbounded, closed, connected subsets of with and the property that has exactly zeroes, which are all simple, if . Moreover, is positive (negative) on an interval for some , if () and .*

It is easy to see from Theorem A that the effect of the discontinuity at zero is a solution branch which consists of infinitely many subbranches all meeting in . Two subbranches are distinguished by the number of zeroes of the respective solutions. However, Theorem A provides no any information about the asymptotic behavior of at infinity.

It is the purpose of this paper to study the asymptotic behavior of at infinity, and accordingly, to determine values of , for which there exist infinitely many *nodal solutions* of (1.1) (here and after, a function is a *nodal solution* of (1.1) if all of zeroes of are simple). To wit, we have the following.

Theorem 1.1. *Let (H1)–(H3) and (H) be fulfilled. Assume that *(H4)*
where
Then for each , joins with , joins with , where , , is the -th eigenvalue of the linear problem: *

Corollary 1.2. *Let (H1)–(H4) and (H) be fulfilled. Let be fixed. Then**(1) for each , (1.1) has infinitely many solutions:
**
which satisfies that has exactly simple zeroes and is positive on an interval for some , has exactly simple zeroes and is negative on an interval for some ;**(2) for each , (1.1) has infinitely many solutions:
**
which satisfies that has exactly simple zeroes, and is positive on an interval for some , has exactly simple zeroes, and is negative on an interval for some .*

#### 2. Notations and Preliminary Results

Recall Kuratowski's notion of lower and upper limits of sequences of sets.

*Definition 2.1 (see [7]). *Let be a metric space and let be a sequence of subsets of . The set
is called the upper limit of the sequence , whereas
is called the lower limit of the sequence .

*Definition 2.2 (see [7]). *A *component* of a set is meant a maximal connected subset of .

Lemma 2.3 (see [7]). *Suppose that is a compact metric space, and are nonintersecting closed subsets of , and no component of intersects both and . Then there exist two disjoint compact subsets and , such that , , .*

Using the above Whyburn Lemma, Ma and An [8] proved the following.

Lemma 2.4 (see [8, Lemma 2.1]). *Let be a Banach space and let be a family of closed connected subsets of . Assume that*(i)*there exist , and , such that ;*(ii)*;*(iii)*for every , is a relatively compact set of , where
**Then there exists an unbounded component in and .*

*Remark 2.5. *The limiting processes for sets go back at least to the work of Kuratowski [9]. Lemma 2.4 will play an important role in the proof of Theorem 1.1. It is a slight generalization of the following well-known results due to Whyburn [7].

Proposition 2.6 (Whyburn [7, page 12]). *Let be a Banach space and let be a family of closed connected subsets of . Let and is relatively compact. Then is nonempty, compact, and connected.*

Lemma 2.7. *Let . Let be such that
**
for some . Suppose that the sequence satisfies
**
with either
**
or
**
where with being a given closed subinterval of . Then
**
for some positive constant .*

*Proof. *We only deal with the case that for all sufficiently large. The other case can be treated by the similar way. We may assume that for all .

We divide the proof into three cases.*Case 1. *Let be a subinterval of satisfying(i);(ii);(iii) for all .

Let and be the unique solution of the problems:

respectively. Then it is easy to check is nondecreasing on , is nonincreasing on , and that Green's function of
is explicitly given by
Let and be the unique solution of the problems:
respectively. Then it is easy to check that is nondecreasing on and is nonincreasing on , and
Let and be the unique solution of the problems
respectively. Then, for

Since
it follows that for ,
where

Set
Then
By (2.5), we have that
which together with (2.15) and (2.20) imply that for ,
Therefore
*Case 2. *Let be a subinterval of satisfying(i);(ii);(iii) for all .

Let and be the unique solution of the problems:

respectively. Then it is easy to check that is nondecreasing on , is nonincreasing on , and Green's function of
is explicitly given by
By the similar method to prove Case 1, we may get the desired results.*Case 3. *Let be a subinterval of satisfying(i);(ii);(iii) for all .

Using the same method to prove Case 2, with obvious changes, we may show that (2.8) is true.*Case 4. *Let . We may assume that for all .

Let and be the unique solution of the problems
respectively. Then, it is easy to verify that is strictly increasing on and is strictly decreasing on . Using the same method to deal with Case 1, we may get the desired results.

#### 3. Proof of the Results

Recall the proof of Theorem A.

By [6, Remark 1], the hypotheses (H1)–(H3) imply that

Actually, such continua can be obtained as upper limits in the sense of Kuratowski of sequences of solution continua from associated continuous problems. To this end one sets

and selects an approximation sequence of satisfying

(A1) for and ;(A2) for and ; on ; on ;(A3) for and ; for and ;(A4) is nondecreasing in for ; is nonincreasing in for .*Remark 3.1. *Let
We may show that there exists a positive constant , independent of , such that for each ,
for some constant .

In fact, it is easy to see from the definition of that
for some positive constant , independent of .

Applying (H2) and (H), it concludes that
for some positive constant . Therefore, if we take
then (3.4) holds.

It is easy to see thanks to (H2) and (A1) that

falls into the scope of the Sturm-Liouville version of the celebrated Rabinowitz bifurcation theorem (cf. [10] for a more general, but somewhat different setting).

Indeed, denote the strictly increasing sequence of simple eigenvalues of

by and set

Then is a bifurcation point of the solution set of () for every , and for each , there exist two unbounded closed connected subsets of the solution set of () with the following.

(a). Moreover, is the only bifurcation point contained in .(b)If and , then possesses exactly simple zeroes (and no multiple zeroes) in and is positive on for some .(c)If and , then possesses exactly simple zeroes (and no multiple zeroes) in and is negative on for some .Combining the above with the fact

and utilizing Lemma 2.4, it concludes that there exists an unbounded component with

As an immediate consequence of [6, emma 4-6], we have the following

Lemma 3.2. *If , then is a solution of (1.1) and . Moreover, if with , has exactly simple zeroes in , and is positive on an interval for some ; if with , has exactly simple zeroes in , and is negative on an interval for some .*

Lemma 3.3. *Let (H1)–(H4), (H) and (A1)–(A4) be fulfilled. Then for each , the connected component joins with .*

*Proof. *Assume that for some fixed with
The case can be treated by the same way.

We divide the proof into two steps.*Step 1. *We show that if there exists a constant number such that
then joins with . In this case it follows that
Define
Then satisfies the problem:
Set
then is nondecreasing, and (H4) and (H) yields

Now, we divide () by and set . Since is bounded in , after taking a subsequence if necessary, we have that for some with Moreover, from the definition of and (3.19) and the fact that is nondecreasing, we have that
since
By standard limit procedure, we get
where again choosing a subsequence and relabeling if necessary. Moreover, the fact that , , has exactly simple zeroes in implies that has exactly simple zeroes in , too. Therefore .*Step 2. *We show that there exists a constant such that for all Suppose there is no such , choosing a subsequence and relabeling if necessary, it follows that

Let
denote the zeroes of , and set
Then, after taking a subsequence if necessary,
We claim that for all

Suppose on the contrary that there exists such that
Define a function by
Then by Remark 3.1, there exists , such that
Now we choose a closed interval with positive length, then we know from Lemma 2.7 that (after taking a subsequence if necessary) must change sign on . However, this contradicts the fact that for all sufficiently large, we have and
Therefore, (3.27) holds.

On the other hand, it follows
that
which contradicts (3.27).

Therefore
for some constant number , independent of .

Now we are in the position to prove Theorem 1.1.

*Proof of Theorem 1.1. *We only prove that has the desired property, the case of can be treated by the same way.

Assume that is a sequence with
We claim that

Assume on the contrary that (3.36) is not true. We divide the proof into two cases.*Case 1. *.In this case, we may take a subsequence of , denote it by again, with the property that there exists , such that for each ,

Since , it follows that for each , there exists a sequence , such that

Now let us consider the sequence . Obviously, we have that
?Equation (3.39) implies that
Noticing that in (3.30) is independent of and using Remark 3.1 and the method to prove Lemma 3.3 and with obvious changes, we may show that is bounded, and subsequently
However, this contradicts (3.37).*Case 2. *. In this case, after taking a subsequence of and relabeling if necessary, we may assume that
for some constant . Equation (3.35) together with (3.42) implies
Using the same notations as those in Case 1, we have from (3.43) that
Combining this with (3.40) and using Remark 3.1 and the similar method to prove Step 2 of Lemma 3.3 and noticing that in (3.30) is independent of , it concludes that is bounded. This is a contradiction.

*Remark 3.4. *It is easy to see from Theorem 1.1 and its proof that the “jumping” of at : does not affect the asymptotic behavior of at infinity. In other words, for any nonnegative function , the asymptotic behavior of at infinity is the same.

#### Acknowledgement

This work supported by the NSFC (no. 10671158), the NSF of Gansu Province (no. ZS051-A25-016), NWNU-KJCXGC-03-17, the Spring-sun program (no. Z2004-1-62033), SRFDP (no. 20060736001), and the SRF for ROCS, SEM (2006).

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#### Copyright

Copyright © 2009 Ruyun Ma and Jiemei Li. 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.