Boundary Value Problems
Volume 2011, Article ID 715836,12pages doi:10.1155/2011/715836
Research Article
Global Structure of Nodal Solutions for
Second-Order
m
-Point Boundary Value Problems
with Superlinear Nonlinearities
Yulian An
Department of Mathematics, Shanghai Institute of Technology, Shanghai 200235, China
Correspondence should be addressed to Yulian An,an [email protected]
Received 8 May 2010; Revised 1 August 2010; Accepted 23 September 2010
Academic Editor: Feliz Manuel Minh ´os
Copyrightq2011 Yulian An. 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.
We consider the nonlinear eigenvalue problemsuλfu 0, 0 < t < 1,u0 0,u1
m−2
i1 αiuηi, wherem ≥ 3,ηi ∈ 0,1,andαi > 0 fori 1, . . . , m−2,withim−12αi < 1,and
f ∈C1R\{0},R∩CR,Rsatisfiesfss > 0 for
s /0, andf0 ∞, wheref0lim|s| →0fs/s.
We investigate the global structure of nodal solutions by using the Rabinowitz’s global bifurcation theorem.
1. Introduction
We study the global structure of nodal solutions of the problem
uλfu 0, t∈0,1, 1.1
u0 0, u1
m−2
i1
αiu
ηi
. 1.2
Here m ≥ 3, ηi ∈ 0,1,and αi > 0 fori 1, . . . , m−2 with m−2
i1 αi < 1; λis a positive
parameter, andf∈C1R\ {0},R∩CR,R.
In the case thatf0∈0,∞,the global structure of nodal solutions of nonlinear
second-order m-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 thatf0∞,and few global results were found in the available literature
used directly in the case. On the other hand, whenm-point boundary value condition1.2
is concerned, the discussion is more difficult since the problem is nonsymmetric and the corresponding operator is disconjugate. In6, we discussed the global structure of positive solutions of1.1,1.2withf0 ∞.However, to the best of our knowledge, there is no paper
to discuss the global structure of nodal solutions of1.1,1.2withf0∞.
In this paper, we obtain a complete description of the global structure of nodal solutions of1.1,1.2under the following assumptions:
A1αi>0 fori1, . . . , m−2,with 0< m−2
i1 αi<1;
A2f ∈C1R\ {0},R∩CR,Rsatisfiesfss >0 fors /0;
A3f0:lim|s| →0fs/s∞; A4f∞:lim|s| → ∞fs/s∈0,∞.
LetY C0,1with the norm
u∞ max
t∈0,1|ut|. 1.3
Let
X
u∈C10,1|u0 0, u1
m−2
i1
αiu
ηi
,
E
u∈C20,1|u0 0, u1
m−2
i1
αiu
ηi
1.4
with the norm
uXmax
u∞, u ∞, umaxu∞, u ∞, u ∞, 1.5
respectively. DefineL:E → Y by setting
Lu:−u, u∈E. 1.6
ThenLhas a bounded inverseL−1 :Y → Eand the restriction ofL−1toX,that is,L−1:X → X
is a compact and continuous operator; see1,2,6.
For anyC1 function u, if ux0 0, then x0 is a simple zero of uif ux0/0.For
any integerk ≥ 1 and anyν ∈ {,−},define setsSν
k, Tkν ⊂ C20,1consisting of functions u∈C20,1satisfying the following conditions:
Sν
k:iu0 0, νu0>0,
iiuhas only simple zeros in0,1and has exactlyk−1 zeros in0,1;
Tν
k :iu0 0, νu0>0 andu1/0,
iiuhas only simple zeros in0,1and has exactlykzeros in0,1,
Remark 1.1. Obviously, ifu ∈ Tkν,then u ∈ Sνk oru ∈ Sνk1.The setsTkν are open inEand disjoint.
Remark 1.2. The nodal properties of solutions of nonlinear Sturm-Liouville problems with separated boundary conditions are usually described in terms of sets similar to Sνk; see
7. However, Rynne1stated thatTν
k are more appropriate than Sνk when the multipoint
boundary condition1.2is considered.
Next, we consider the eigenvalues of the linear problem
Luλu, u∈E. 1.7
We call the set of eigenvalues of1.7the spectrum ofLand denote it byσL. The following lemmas or similar results can be found in1–3.
Lemma 1.3. LetA1hold. The spectrumσLconsists of a strictly increasing positive sequence of eigenvaluesλk, k1,2, . . . ,with corresponding eigenfunctionsϕkx sin
λk x.In addition,
ilimk→ ∞λk∞;
iiϕk∈Tk,for eachk≥1,andϕ1is strictly positive on0,1.
We can regard the inverse operator L−1 : Y → E as an operator L−1 : Y → Y. In
this setting, eachλk, k 1,2, . . . ,is a characteristic value ofL−1,with algebraic multiplicity
defined to be dim∞j1NI−λkL−1j,whereNdenotes null-space andIis the identity on
Y.
Lemma 1.4. Let A1hold. For each k ≥ 1, the algebraic multiplicity of the characteristic value
λk, k1,2, . . . ,ofL−1:Y → Y is equal to 1.
LetER×Eunder the product topology. As in7, we add the points{λ,∞|λ∈R}
to our spaceE.LetΦνk R×Tkν.LetΣνkdenote the closure of set of those solutions of1.1,
1.2which belong toΦν
k.The main results of this paper are the following.
Theorem 1.5. Let (A1)–(A4) hold.
aIff∞0,then there exists a subcontinuumCνkofΣνkwith0,0∈ Cνkand
ProjRCνk 0,∞. 1.8
bIff∞∈0,∞,then there exists a subcontinuumCνkofΣνkwith
0,0∈ Cνk, ProjRCνk⊆
0, λ1
f∞
. 1.9
cIff∞ ∞,then there exists a subcontinuumCνkofΣkν with 0,0 ∈ Cνk, ProjRCνk is a
bounded closed interval, andCν
Theorem 1.6. Let (A1)–(A4) hold.
aIff∞0, then1.1,1.2has at least one solution inTkνfor anyλ∈0,∞.
bIff∞∈0,∞, then1.1,1.2has at least one solution inTkνfor anyλ∈0, λ1/f∞.
cIff∞ ∞,then there existsλ∗ >0such that1.1,1.2has at least two solutions inTkν
for anyλ∈0, λ∗.
We will develop a bifurcation approach to treat the case f0 ∞. Crucial to this
approach is to construct a sequence of functions{fn} which is asymptotic linear at 0 and
satisfies
fn−→f, fn
0−→ ∞. 1.10
By means of the corresponding auxiliary equations, we obtain a sequence of unbounded components{Ckνn}via Rabinowitz’s global bifurcation theorem8, and this enables us to find unbounded componentsCν
ksatisfying
0,0∈ Cνk⊂lim supCνkn. 1.11
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 of1.1,1.2.
2. Preliminaries
Definition 2.1see9. LetW be a Banach space and{Cn |n1,2, . . .}a family of subsets of W.Then the superior limitDof{Cn}is defined by
D:lim sup
n→ ∞
Cn{x∈W| ∃{ni} ⊂Nandxni ∈Cni, such thatxni −→x}. 2.1
Lemma 2.2see9. Each connected subset of metric spaceW is contained in a component, and each connected component ofWis closed.
Lemma 2.3see6. Assume that
ithere existzn∈Cn n1,2, . . .andz∗∈W,such thatzn → z∗;
iirn ∞, wherernsup{x |x∈Cn};
iiifor allR >0, ∞n1Cn∩BRis a relative compact set ofW, where
BR{x∈W| x ≤R}. 2.2
Define the mapTλ:Y → Eby
Tλut λ
1
0
Ht, sfusds, 2.3
where
Ht, s Gt, s
m−2 i1 αiG
ηi, s
1−m−i12αiηi
t, Gt, s
⎧ ⎨ ⎩
1−ts, 0≤s≤t≤1,
t1−s, 0≤t≤s≤1. 2.4
It is easy to verify that the following lemma holds.
Lemma 2.4. Assume that (A1)-(A2) hold. ThenTλ:Y → Eis completely continuous.
Forr >0, let
Ωr {u∈Y | u∞< r}. 2.5
Lemma 2.5. Let (A1)-(A2) hold. Ifu∈∂Ωr, r >0, then
Tλu∞≤λMr
1
m−2
i1 αi
1−m−i12αiηi
1
0
Gs, sds, 2.6
whereMr 1max0≤|s|≤r{|fs|}.
Proof. The proof is similar to that of Lemma 3.5 in6; we omit it.
Lemma 2.6. Let (A1)-(A2) hold, and {μl, yl} ⊂ Φνk is a sequence of solutions of 1.1,1.2.
Assume thatμl≤C0for some constantC0>0, andliml→ ∞yl∞.Then
lim
l→ ∞ yl ∞∞. 2.7
Proof. From the relation ylt μl 1
0Ht, sfylsds, we conclude that y1t
μl
1
0 Htt, sfylsds.Then
y
l ∞≤C0
1
m−2 i1 αi
1−m−i12αiηi
1
0
f
ylsds, 2.8
3. Proof of the Main Results
For eachn∈N,definefns:R → Rby
fns
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩
fs, s∈
1
n,∞
∪
−∞,−1
n , nf 1 n
s, s∈
−1 n, 1 n . 3.1
Thenfn∈CR,R∩C1R\ {±1/n},Rwith
fnss >0, ∀s /0, fn
0nf
1
n
. 3.2
ByA3, it follows that
lim
n→ ∞
fn
0∞. 3.3
Now let us consider the auxiliary family of the equations
uλfnu 0, t∈0,1, 3.4
u0 0, u1
m−2
i1
αiu
ηi
. 3.5
Lemma 3.1see1, Proposition 4.1. Let (A1), (A2) hold. Ifμ, u∈Eis a nontrivial solution of 3.4,3.5, thenu∈Tkνfor somek, ν.
Letζn∈CR,Rbe such that
fnu fn
0uζ
nu nf1
n
uζnu. 3.6
Note that
lim
|s| →0
ζns
s 0. 3.7
Let us consider
Lu−λfn
0uλζ
nu 3.8
Equation3.8can be converted to the equivalent equation
ut
1
0
Ht, sλfn
0us λζ
nusds
:λL−1fn
0u·
t λL−1ζnu·t.
3.9
Further we note thatL−1ζnuouforunear 0 inE.
The results of Rabinowitz8for3.8can be stated as follows. For each integerk ≥
1, ν ∈ {,−}, there exists a continuum{Cνkn}of solutions of3.8joiningλk/fn0,0to
infinity inE.Moreover,{Cνkn} \ {λk/fn 0,0} ⊂Φνk.
Proof of Theorem1.5. Let us verify that{Cνkn}satisfies all of the conditions of Lemma2.3. Since
lim
n→ ∞
λk
fn
0 lim
n→ ∞
λk
nf1/n0, 3.10
conditioniin Lemma2.3is satisfied withz∗ 0,0. Obviously
rnsup
λ y |λ, y∈ Ckνn ∞, 3.11
and accordingly,iiholds.iiican be deduced directly from the Arzela-Ascoli Theorem and the definition of fn. Therefore, the superior limit of {Ckνn}, Dνk, contains an unbounded connected componentCν
kwith0,0∈ Cνk.
From the conditionA2, applying Lemma2.2withp2 in10, we can show that the initial value problem
vλfv 0, t∈0,1,
vt0 0, vt0 β
3.12
has a unique solution on 0,1 for everyt0 ∈ 0,1 and β ∈ R. Therefore, any nontrivial
solutionuof1.1,1.2has only simple zeros in0,1andu0/0. Meanwhile,A1implies thatu1/0 1, proposition 4.1. SinceCkνn ⊂ Φνk, we conclude thatCνk ⊂ Φνk. Moreover,
Cν
k⊂Σνkby1.1and1.2.
We divide the proof into three cases.
Case 1f∞0. In this case, we show that ProjRCνk 0,∞.
Assume on the contrary that
then there exists a sequence{μl, yl} ⊂ Cνksuch that
lim
l→ ∞ yl ∞, μl≤C0, 3.14
for some positive constantC0depending not onl. From Lemma2.6, we have
lim
l→ ∞ yl ∞∞. 3.15
Setvlt ylt/yl∞.Thenvl∞1.Now, choosing a subsequence and relabelling
if necessary, it follows that there existsμ∗, v∗∈0, C0×Ewith
v∗∞1, 3.16
such that
lim
l→ ∞
μl, vl
μ∗, v∗, inR×E. 3.17
Since lim|u| → ∞fu/u0, we can show that
lim
l→ ∞ f
ylt
yl ∞
0. 3.18
The proof is similar to that of the step 1 of Theorem 1 in7; we omit it. So, we obtain
v∗t μ∗·00, t∈0,1, 3.19
v∗0 0, v∗1
m−2
i1
αiv∗ηi
, 3.20
and subsequently,v∗t≡0 fort∈0,1. This contradicts3.16. Therefore
supλ|λ, y∈ Cνk∞. 3.21
Case 2f∞ ∈ 0,∞. In this case, we can show easily thatCjoins0,0withλk/f∞,∞by
using the same method used to prove Theorem 5.1 in2.
Case 3f∞∞. In this case, we show thatCνkjoins0,0with0,∞.
Let{μl, yl} ⊂ Cνkbe such that
If{yl}is bounded, say,yl ≤ M1, for someM1 depending not onl, then we may
assume that
lim
l→ ∞μl∞. 3.23
Taking subsequences again if necessary, we still denote{μl, yl}such that{yl} ⊂Tkν∩Sνk. If {yl} ⊂Tkν∩Sνk1, all the following proofs are similar.
Let
0τl0< τl1 <· · ·< τlk−1 3.24
denote the zeros ofyl in0,1. Then, after taking a subsequence if necessary, liml→ ∞τlj :
τ∞j , j ∈ {0,1, . . . , k−1}. Clearly,τ∞0 0. Setτ∞k 1. We can choose at least one subinterval
τ∞j , τ∞j1 I∞j which is of length at least 1/kfor somej ∈ {0,1, . . . , k−1}. Then, for this j, τlj1−τlj>3/4kiflis large enough. Putτlj, τlj1Ilj.
Obviously, for the above given k, νand j, ylthave the same sign on Ilj for alll.
Without loss of generality, we assume that
ylt>0, t∈Ilj. 3.25
Moreover, we have
max
t∈Ij l
|ult| ≤M1. 3.26
Combining this with the fact
fylt
ylt ≥inf
!
fs
s |0< s≤M1
"
>0, t∈τlj, τlj1, 3.27
and using the relation
ylt μl
fylt
ylt ylt 0, t∈
τlj, τlj1, 3.28
we deduce thatylmust change its sign onτlj, τlj1iflis large enough. This is a contradiction.
Hence{yl}is unbounded. From Lemma2.6, we have that
lim
l→ ∞ yl ∞∞. 3.29
Note that{μl, yl}satisfies the autonomous equation
ylμlf
yl
We see that yl consists of a sequence of positive and negative bumps, together with a
truncated bump at the right end of the interval0,1,with the following propertiesignoring the truncated bump see,1:
iall the positive resp., negative bumps have the same shape the shapes of the positive and negative bumps may be different;
iieach bump contains a single zero ofyl, and there is exactly one zero ofylbetween
consecutive zeros ofyl;
iiiall the positivenegativebumps attain the same maximumminimumvalue. Armed with this information on the shape ofyl,it is easy to show that for the above
givenIlj, {ylIj
l,∞:maxI j lylt}
∞
l1is an unbounded sequence. That is
lim
l→ ∞ yl
Ilj,∞∞. 3.31
Sinceylis concave onIlj, for anyσ >0 small enough,
ylt≥σ yl Ij
l,∞, ∀t∈
τljσ, τlj1−σ. 3.32
This together with3.31implies that there exist constantsα, βwithα, β ⊂I∞j , such
that
lim
l→ ∞ylt ∞, uniformly fort∈ #
α, β$. 3.33
Hence, we have
lim
l→ ∞
fylt
ylt ∞, uniformly fort∈ #
α, β$. 3.34
Now, we show that liml→ ∞μl0.
Suppose on the contrary that, choosing a subsequence and relabeling if necessary,μl≥ b0for some constantb0>0. This implies that
lim
l→ ∞μl
fylt
ylt ∞, uniformly fort∈ #
α, β$. 3.35
From 3.28we obtain that yl must change its sign onα, βif l is large enough. This is a
contradiction. Therefore liml→ ∞μl0.
Proof of Theorem1.6. a and b are immediate consequence of Theorem 1.5a and b, respectively.
To provec, we rewrite1.1,1.2to
uλ
1
0
By Lemma2.5, for everyr >0 andu∈∂Ωr,
Tλu∞≤λMr
1
m−2
i1 αi
1−m−i12αiηi
1
0
Gs, sds, 3.37
whereMr 1max0≤|s|≤r{|fs|}.
Letλr >0 be such that
λrMr
1
m−2
i1 αi
1−m−i12αiηi
1
0
Gs, sdsr. 3.38
Then forλ∈0, λrandu∈∂Ωr,
Tλu∞<u∞. 3.39
This means that
Σν
k∩ {λ, u∈0,∞×E|0< λ < λr, u∈E:u∞r}∅. 3.40
By Lemma2.6and Theorem1.5, it follows thatCν
kis also an unbounded component joining
0,0and0,∞in0,∞×Y. Thus,3.40implies that forλ∈0, λr,1.1,1.2has at least
two solutions inTν k.
Acknowledgments
The author is very grateful to the anonymous referees for their valuable suggestions. This paper was supported by NSFCno.10671158, 11YZ225, YJ2009-16no.A06/1020K096019.
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