Volume 2009, Article ID 970135,20pages doi:10.1155/2009/970135
Research Article
Multiplicity Results Using Bifurcation
Techniques for a Class of Fourth-Order
m
-Point
Boundary Value Problems
Yansheng Liu
1and Donal O’Regan
21Department of Mathematics, Shandong Normal University, Jinan 250014, China
2Department of Mathematics, National University of Ireland, Galway, Ireland
Correspondence should be addressed to Yansheng Liu,[email protected]
Received 13 March 2009; Accepted 12 April 2009 Recommended by Juan J. Nieto
By using bifurcation techniques, this paper investigates the existence of nodal solutions for a class of fourth-orderm-point boundary value problems. Our results improve those in the literature. Copyrightq2009 Y. Liu and D. O’Regan. 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.
1. Introduction
Consider the following fourth orderm-point boundary value problemBVP, for short
u4t fut, ut, t∈0,1
u0 0, u1
m−2
i1
αiu
ηi
u0 0, u1
m−2
i1
αiu
ηi
,
1.1
wheref : R×R → Ris a given sign-changing continuous function,m ≥3,ηi ∈0,1,and
αi>0 fori1, . . . , m−2 with
m−2
i1
Multi-point boundary value problems for ordinary differential equations arise in different areas of applied mathematics and physics. The existence of solutions of the second order multi-point boundary value problems has been studied by many authors and the methods used are the nonlinear alternative of Leray-Schauder, coincidence degree theory, fixed point theorems in cones and global bifurcation techniquessee1–9, and the references therein. In5, Ma investigated the existence and multiplicity of nodal solutions for
ut fut 0, t∈0,1;
u0 0, u1
m−2
i1
αiu
ηi
1.3
when
ηi ∈Qi1,2, . . . , m−2 with 0< η1< η2 <· · ·< ηm−2<1, 1.4
andαi >0 fori1, . . . , m−2 satisfying1.2. He obtained some results on the spectrum of
the linear operator corresponding to1.1. It should be pointed out that the main tool used in
5is results on bifurcation coming from the trivial solutions and we note no use was made of global results on bifurcation from infinity.
Recently 10 Wei and Pang studied the existence and multiplicity of nontrivial solutions for the fourth orderm-point boundary value problems:
u4t fut, ut, t∈0,1
u0 0, u1
m−2
i1
αiu
ηi
u0 0, u1
m−2
i1
αiu
ηi
,
1.5
wheref : R×R → Ris a given sign-changing continuous function,m ≥3,ηi ∈0,1, and
αi>0 fori1, . . . , m−2 satisfies1.2.
Motivated by5,10, in this paper we consider the existence and multiplicity of nodal solutions for BVP1.1. The method used here is Rabinowitz’s global bifurcation theorem. To the best of our best knowledge, only10seems to have considered the existence of nontrivial or positive solutions of the nonlinear multi-point boundary value problems for fourth order differential equations. As in5,10we suppose1.2is satisfied throughout.
value ofL. The set of real characteristic values ofLwill be denoted byσL. The multiplicity ofμ∈σLis
dim ∞
j1
NI−μLj, 1.6
where NAdenotes the null space of A. Suppose that H : R×E → Eis compact and
Hλ, u ouatu0 uniformly on boundedλintervals. Then
uλLuHλ, u 1.7
possesses the line of trivial solutionsΘ {λ,0 |λ ∈R}. It is well known that ifμ ∈ R, a necessary condition forμ,0to be a bifurcation point of1.7with respect toΘis thatμ ∈ σL. Ifμis a simple characteristic value ofL, letvdenote the eigenvector ofLcorresponding toμnormalized sov 1. ByΣwe denote the closure of the set of nontrivial solutions of
1.7. A component ofΣis a maximal closed connected subset. It was shown inRabinowitz
11, Theorems 1.3, 1.25, 1.27, the following.
Lemma 1.1. Ifμ∈σLis simple, thenΣcontains a componentCμwhich can be decomposed into
two subcontinuaCμ,Cμ−such that for some neighborhoodBofμ,0,
λ, u∈Cμ
C−μ
∩B, λ, u/μ,0 1.8
impliesλ, u λ, αvwwhereα >0α <0and|λ−μ|o1,wo|α|atα0. Moreover, each ofCμ,Cμ−either
imeets infinity inΣ, or
iimeetsμ, 0whereμ /μ∈σL, or
iiicontains a pair of pointsλ, u,λ,−u,u /0.
The following are global results for1.7on bifurcation from infinitysee, Rabinowitz
9, Theorem 1.6 and Corollary 1.8.
Lemma 1.2. SupposeLis compact and linear,Hλ, uis continuous onR×E,Hλ, u ouat
u ∞uniformly on boundedλintervals, andu2Hλ, u/u2is compact. Ifμ∈σLis of odd multiplicity, thenΣpossesses an unbounded componentDμwhich meetsμ,∞. Moreover ifΛ⊂R
is an interval such thatΛ∩σL {μ}and℘is a neighborhood ofμ,∞whose projection onRlies inΛand whose projection onEis bounded away from0, then either
iDμ\℘is bounded inR×Ein which caseDμ\℘meetsΘ {λ,0|λ∈R}or
iiDμ\℘is unbounded.
If (ii) occurs andDμ \℘has a bounded projection onR, thenDμ\℘meets μ, ∞where
μ /μ∈σL.
Lemma 1.3. Suppose the assumptions ofLemma 1.2 hold. Ifμ ∈ σLis simple, then Dμ can be
decomposed into two subcontinuaDμ,D−μ and there exists a neighborhoodI⊂℘ofμ,∞such that
λ, u ∈ DμD−μ∩I and λ, u/ μ,∞implies λ, u λ, αvwwhere α > 0α < 0and
2. Preliminaries
LetX C0,1with the normu maxt∈0,1|ut|,Y {u ∈ C10,1 : u0 0, u1
m−2
i1 αiuηi}with the normu1 max{u,u},Z {u ∈ C20,1 : u0 0, u1
m−2
i1 αiuηi} with the norm u2 max{u,u,u
}. Then X, Y, and Z are Banach spaces.
For anyC1 function u, ifut
0 0, thent0 is a simple zero ofuif ut0/0. For any
integerk∈Nand anyν∈ {±}, as in6, define setsTkν⊂Zconsisting of the set of functions
u∈Zsatisfying the following conditions:
iu0 0, νu0>0 andu1/0;
iiuhas only simple zeros in0,1, and has exactlyk−1 such zeros;
iiiuhas a zero strictly between each two consecutive zeros ofu.
NoteTk−−Tkand letTkTk−∪Tk. It is easy to see that the setsTk−andTkare disjoint
and open inZ. Moreover, ifu∈Tν
k, thenuhas at leastk−2 zeros in0,1, and at mostk−1
zeros in0,1.
LetER×Yunder the product topology. As in12, we add the points{λ,∞:λ∈R}
to the spaceE. LetΦkR×Tk,Φk−R×Tk−, andΦkR×Tk.
We first convert BVP1.1into another form. Supposeutis a solution of BVP1.1. Letvt −ut. Notice that
ut vt 0, t∈I;
u0 0, u1
m−2
i1
αiu
ηi
.
2.1
Thusutcan be written as
ut Lvt, 2.2
where the operatorLis defined by
Lvt:
1
0
Ht, svsds, ∀v∈Y, 2.3
where
Ht, s Gt, s
m−2
i1 αiG
ηi, s
1− mi1−2αiηi
,
Gt, s
⎧ ⎨ ⎩
1−t, 0≤s≤t≤1;
1−s, 0≤t≤s≤1.
Therefore we obtain the following equivalent form of1.1
vt fLvt,−vt 0, t∈0,1;
v0 0, v1
m−2
i1
αiv
ηi
.
2.5
For the rest of this paper we always suppose that the initial value problem
vt fLvt,−vt 0, t∈0,1;
vt0 vt0 0
2.6
has the unique trivial solution v ≡ 0 on 0,1 for any t0 ∈ 0,1; in fact some suitable
conditions such as a Lipschitz assumption orf∈C1guarantee this.
Define two operators onYby
Avt: LFvt, Fvt:fLvt,−vt, t∈I, v∈Y. 2.7
Then it is easy to see the following lemma holds.
Lemma 2.1. The linear operatorLand operatorAare both completely continuous fromYtoYand
Lv1≤Mv ≤Mv1, ∀v∈Y, 2.8
whereMmax{1,1/81 im1−2αi/1− mi1−2αiηi}.
Moreover,u∈ C40,1is a solution of BVP1.1if and only ifv −u is a solution of the operator equationvAv.
Let the functionΓsbe defined by
Γs coss−
m−2
i1
αicosηis, s∈R. 2.9
Then we have the following lemma.
Lemma 2.2. iFor eachk≥1,Γshas exactly one zerosk∈Ik: k−1π, kπ, so
s1< s2<· · ·< sk−→∞ k−→∞; 2.10
iithe characteristic value ofLis exactly given byμks2k, k1,2, . . ., and the eigenfunction
φkcorresponding toμkisφkt cosskt;
iiithe algebraic multiplicity of each characteristic valueμkofLis1;
Proof. From5and by a similar analysis as in the proof of6, Lemma 3.3we obtainiand
ii.
Now we assertiiiholds. Suppose, on the contrary, there existsy∈Y such thatI− μkLyμk−1φk. Theny∈Zand
−y−s2
kycosskt. 2.11
Fromy0 0 we know the general solution of this differential equation is
yCcosskt−
1 2skt
sinskt. 2.12
Fromiandiiof this lemma,Ccossktsatisfies the boundary condition. Thus
cossk m−2
i1
αicosηisk, sinsk m−2
i1
αiηisinηisk. 2.13
Then, by1.2,
1
m−2
i1
αicosηisk
2
m−2
i1
αiηisinηisk
2
≤m−2
i,j1
αiαjcosηiskcosηjsksinηisksinηjsk
≤ m−2
i1
αi
2
<1,
2.14
a contradiction. Thus the algebraic multiplicity of each characteristic valueμkofLis 1.
Finally, fromsk∈k−1π, kπands1∈0, π/2, it is easy to see thativholds.
Lemma 2.3. Ford d1, d2∈R×R\ {0,0}, define a linear operator
Ldvt
d1L2d2L
vt, ∀t∈I, v∈Y, 2.15
whereLis defined as in2.3. Then the generalized eigenvalues ofLdare simple and are given by
0< λ1Ld< λ2Ld<· · ·< λkLd−→∞ k−→∞, 2.16
where
λkLd
μ2
k
d1d2μk.
The generalized eigenfunction corresponding toλkLdis
φkt cosskt, 2.18
whereμk, sk, φkare as inLemma 2.2.
Proof. Suppose there existλandv /0 such thatvλLdv. Setut Lvt. Then from2.2–
2.7and2.15it is easy to see thatu /0 and
u4t λd1ut−d2u
t, t∈0,1;
u0 0, u1
m−2
i1
αiu
ηi
;
u”’0 0, u1
m−2
i1
αiu
ηi
.
2.19
DenoteL−1u −u foru ∈Z. Then there exist two complex numbersr
1 andr2 such
that
u4t−λd1ut−d2u
tL−1−r2I
L−1−r1I
ut 0. 2.20
Now if there exists somerii1,2such that
L−1−riI
ut 0, 2.21
then byLemma 2.2we knowrisk2 μkfor somek∈N, and consequently
ut cosskt 2.22
is a nontrivial solution. Substituting2.22into2.19, we have
λ μ
2
k
d1d2μk.
2.23
On the other hand, suppose, for example,
L−1−r 1I
ut/0, L−1−r 2I
L−1−r 1I
Let wt : L−1 − r
1Iut. Then L−1 − r2Iwt 0. Reasoning as previously
mentioned, we haver2 s2k for somek ∈ N, and consequently wt acosskta /0is a
nontrivial solution. Therefore,
L−1−r1I
ut acosskt. 2.25
Ifr1s2k, then the general solution of the differential equation2.25, satisfyingu0
0, is
ut Ccosskt− a
2skt
sinskt, 2.26
which is similar to2.12. Reasoning as in the proof ofLemma 2.2we can get a contradiction. Thusr1/s2kand the general solution of2.25, satisfyingu0 0, is
ut ut acosskt s2
k−r1
, 2.27
where utis the general solution of homogeneous differential equation corresponding to
2.25
L−1−r1I
ut 0. 2.28
Notice the termacosskt/s2k−r1in2.27satisfies the boundary condition of1.1at
t1, soutalso satisfies
u0 0, u1
m−2
i1
αiu
ηi
. 2.29
Therefore, byLemma 2.2we knowut Ccossjtfor somej∈N, and consequently
r1s2j/s2k, ut Ccossjta
cosskt
s2
k−s2j
. 2.30
By substituting this into2.19, we have
aλd1d2μk
aμ2k, Cλd1d2μj
Cμ2j. 2.31
Sinceμj/μk, if there exists someλsuch that2.31holds, then
d1d2μk
d1d2μj
μ2
k
which implies
d1d2/0, d1 1 μk 1 μj
−d2, 2.33
a contradiction withd1>0 andd2>0.
Consequently,2.24does not hold. This together with2.20–2.23andLemma 2.2
guarantee that the generalized eigenvalues ofLdare given by
0< λ1Ld< λ2Ld<· · ·< λkLd−→∞ k−→∞, 2.34
whereλkLd μ2k/d1 d2μk. The generalized eigenfunction corresponding toλkLdis
φkt cosskt.
Now we are in a position to show the generalized eigenvalues ofLdare simple.
Clearly, from above we know forλk:λkLd,I−λkLdφk0 and dimNI−λkLd 1.
Suppose there exists anv∈C2such that
I−λkLdv
1
μkφkt.
2.35
This together with2.3and2.15guarantee thatv ∈ Y. If we letut Lvtas above, then we have
u4t−λk
d1ut−d2u
tcosskt, t∈0,1, 2.36
u0 0, u1
m−2
i1
αiu
ηi
; u”’0 0, u1
m−2
i1
αiu
ηi
. 2.37
Consider the following homogeneous equation corresponding to2.36:
u4t− μ
2
k
d1d2μk
d1ut−d2u
t0. 2.38
The characteristic equation associated with2.38is
λ4− μ
2
k
d1d2μk
d1−d2λ2
0. 2.39
Then there exists a real numberηsuch that
λ2μk
λ2−ηλ4− μ
2
k
d1d2μk
d1−d2λ2
0. 2.40
First we consider the cased1 >0. In this case the general solution of2.38is
c1e
√ηt
c2e−√ηtc3cossktc4sinskt. 2.41
After computation we obtain that the general solution of2.36is
ut c1e
√ηt
c2e−√ηtc3cossktc4sinsktatsinskt, 2.42
wherea −d1d2μk/2sk2d1μkd2μ2k. From boundary conditionu0 u”’0 0 in
2.37it follows that
ηc1−c2skc40;
ηηc1−c2−s3kc4 0.
2.43
Byη >0 andμk>0, we knowc1−c20 andc40. Then2.42can be rewritten as
ut c1
e√ηte−√ηtc3cossktatsinskt. 2.44
Notice that the termc3cossktsatisfies2.37. From the boundary condition
u1
m−2
i1
αiu
ηi
, u1
m−2
i1
αiu
ηi
,
tsinskt
2skcosskt−s2ktsinskt
2.45
we have
c1
e√ηe−√ηasinsk m−2
i1
αi
c1
e√ηηie−√ηηiaη
isinskηi
, 2.46
c1η
e√ηe−√η−as2ksinsk m−2
i1
αi
c1η
e√ηηie−√ηηi−aη
is2ksinskηi
. 2.47
Multiply2.46bys2kand then add to2.47to obtain
c1
ηs2ke√ηe−√ηc1
ηs2k
m−2
i1
αi
e√ηηi e−√ηηi. 2.48
On the other hand, from1.2it can be seen that
e√ηe−√η>
m−2
i1
αi
This together with2.48guarantee thatc1 0. Therefore,2.42reduces to
ut c3cossktatsinskt. 2.50
Similar to2.12, a contradiction can be derived.
Next consider the cased10. Thenη0 from above. In this case the general solution
of2.38is
c1c2tc3cossktc4sinskt. 2.51
By a similar process, one can easily get a contradiction.
To sum up, the generalized eigenvalues ofLdare simple, and the proof of this lemma
is complete.
3. Main Results
We now list the following hypotheses for convenience.
H1There existsa a1, a2∈R×R\ {0,0}such that
fx, ya1x−a2yox, y, asx, y−→0, 3.1
wherex, y∈R×R, and|x, y|:max{|x|,|y|}.
H2There existsb b1, b2∈R×R\ {0,0}such that
fx, yb1x−b2yox, y, asx, y−→ ∞. 3.2
H3There existsR >0 such that
f
x, y< R M, for
x, y∈x, y:|x| ≤MR,y≤R, 3.3
whereMis defined as inLemma 2.1.
H4There exist two constants r1 < 0 < r2 such thatfx,−r1 ≥ 0 and fx,−r2 ≤ 0
for x ∈ −Mr, Mr, and fx,−ysatisfies a Lipschitz condition iny forx, y ∈
−Mr, Mr×r1, r2, wherermax{|r1|, r2}.
Now we are ready to give our main results.
Theorem 3.1. Suppose (H1)-(H2) hold. Suppose there exists two integersi0≥0andk >0such that
either
μ2
i0k
a1a2μi0k
<1< μ
2
i01
b1b2μi01
or
μ2i0k b1b2μi0k
<1< μ
2
i01
a1a2μi01
3.5
holds. Then BVP1.1has at least2knontrivial solutions.
Theorem 3.2. Suppose (H1) and (H2) hold and one of (H3) and (H4) hold. Suppose there exists two
integersi0andj0such that
μ2
i0
a1a2μi0
<1, μ
2
j0
b1b2μj0
<1. 3.6
Then BVP1.1has at least2i0j0solutions.
To set it up we first consider global results for the equation
vλAv, 3.1λ
onY, whereλ∈R, and the operatorAis defined as in2.7. Under the conditionH1,3.1λ can be rewritten as
vλLavHaλ, v, 3.7
hereHaλ, v λAv−λLav,Lais defined as in2.12 replacingdwitha. Obviously, byH1
andLemma 2.1–2.3, it can be seen thatHaλ, visov1forvnear 0 uniformly on bounded
λintervals andLais a compact linear map onY. A solution of3.1λis a pairλ, v∈E. By
H1, the known curve of solutions{λ,0|λ∈R}will henceforth be referred to as the trivial solutions. The closure of the set on nontrivial solutions of3.1λwill be denoted byΣas in
Lemma 1.1.
IfHaλ, v≡0, then3.7becomes a linear system
vλLav. 3.8
ByLemma 2.3,3.8possesses an increasing sequence of simple eigenvalues
0< λ1< λ2<· · ·< λk<· · ·, with λk
μ2k a1a2μk
ask−→∞. 3.9
Any eigenfunctionφkcossktcorresponding toλkis inTk.
A similar analysis as in6, Proposition 4.1yields the following results.
Lemma 3.4. Assume that (H1) holds andλkis defined by3.9. Then for each integerk >0and each
ν , or−, there exists a continuaCν
kof solutions of 3.1λinΦνk∪ {λk,0}, which meets{λk,0}
and∞inΣ.
Proof. Consider 3.7 as a bifurcation problem from the trivial solution. From Lemma 1.1
and conditionH1it follows that for each positive integer k ∈ N,Σcontains a component
Ck ⊆E R×Y which can be decomposed into two subcontinuaCk,C−k such that for some
neighborhoodBofλk,0,
λ, v∈CkC−k∩B, λ, v/ λk,0 3.10
implyλ, v λ, αφkw, whereα >0α <0and|λ−λk|o1,w1o|α|atα0.
By3.7and the continuity of the operatorA:Y → Z, the setCkνlies inR×Zand the injectionCν
k → R×Zis continuous. Thus,Cνkis also a continuum inR×Z, and the above
properties hold inR×Z.
SinceTkis open inZandφk∈Tk, we know
v α φk
w α ∈T
k 3.11
for 0/αsufficiently small. Then there existsε0>0 such that forε∈0, ε0, we have
λ, v∈Φk,
Ck
{λk,0}
∩Bε⊂Φk, 3.12
whereBεis an open ball inR×Zof radiusεcentered atλk,0. It follows from the proof of
6, Proposition 4.1that
λ, v∈Ck∩R×∂Tk ⇒u0, 3.13
which meansCk\ {λk,0} ∩∂Φk∅. Consequently,Cklies inΦk∪ {λk,0}.
Similarly we have thatCνklies inΦνk∪ {λk,0} ν or−.
Next we show alternative ii of Lemma 1.1 is impossible. If not, without loss of generality, assume thatCk meetsλi,0with λk/λi ∈ σLa. Then there exists a sequence
ξm, zm∈Ckwithξm → λiandzm → 0 asm → ∞. Notice that
zmξmLazmHξm, zm. 3.14
Dividing this equation by zm1 and usingLemma 2.1andHξm, zm ozm1asm →
∞, we may assume without loss of generality thatzm/zm1 → zasm → ∞. Thus from
3.14it follows that
zλiLaz. 3.15
Ti−are open inZ. Therefore,zm ∈TiorTi−formsufficiently large, which is a contradiction
withzm∈Tkm≥1,i /k. Hence alternativeiiofLemma 1.1is not possible.
Finally it remains to show alternativeiiiofLemma 1.1is impossible. In fact, notice thatTk− −Tk, andTk−∩Tk ∅. Ifu ∈Tk, then−u ∈ Tk−. This guarantees thatCνkdoes not contain a pair of pointsλ, v,λ,−v,v /0.
Therefore alternativeiofLemma 1.1holds. This implies that for each integerk ∈N
and eachν , or−, there exists a continuaCνkof solutions of3.1λinΦνk∪ {λk,0}, which
meets{λk,0}and∞inΣ.
Under the conditionH2,3.1λcan be rewritten as
vλLbvHbλ, v, 3.16
hereHbλ, v λAv−λLbv,Lbis defined as in2.12 replacingdwithb.
Lethx, y: fx, y−b1xb2y. Then fromH2it follows that lim|x,y| → ∞hx, y/
|x, y|0. Define a function
hr:maxhx, y:x, y≤r. 3.17
Thenhris nondecreasing and
lim
r→ ∞
hr
r 0. 3.18
Obviously, by 3.18 and Lemma 2.1, it can be seen thatHbλ, vis ov1 for v near ∞
uniformly on boundedλintervals andLbis a compact linear map onY.
Similar to 3.8, by Lemma 2.3, Lb possesses an increasing sequence of simple
eigenvalues
0< λ1< λ2<· · ·< λk<· · ·, with λk
μ2k b1b2μk
ask−→∞. 3.19
Noteφkcossktis an eigenfunction corresponding toλk. Obviously, it is inTk.
Lemma 3.5. Assume that (H1)-(H2) holds. Then for each integerk >0and eachν , or−, there
exists a continuaDν
kofΣinΦνk∪ {λk,∞}coming from{λk,∞}, which meetsλk,0or has an
unbounded projection onR.
Proof. From2.7,3.16, and3.18 it follows thatHbλ, v is continuous onE,Hbλ, v
ov1atv∞uniformly on boundedλintervals. Moreover, as in the proof of12, Theorem
2.4, one can see thatv21Hbλ, v/v21is compact. FromLemma 2.3we knowλkis a simple
characteristic value ofLb for each integerk ∈ N. Thus by Lemmas1.2and1.3,Σcontains a
componentDkwhich can be decomposed into two subcontinuaDk,Dk−which meet{λk,∞}.
Now we show that for a smaller neighborhoodI ⊂℘ofλk,∞,λ, v∈DkDk−∩I
exists a neighborhood I ⊂ ℘ofλk,∞satisfying λ, v ∈ DkD−k∩Iand λ, v/ λk,∞
implyλ, v λ, αvkwwhereα >0α <0and|λ−λk|o1,w1 o|α|at|α|∞.
As in the proof ofLemma 3.4,Dνkis also a continuum inR×Z, and the above properties hold inR×Z. SinceTν
k is open inZandw/αis smaller compared toφk ∈Tknearα ∞,
φkw/αand thereforevαφkw∈Tkforαnear∞. Here and in the following the same
argument works ifis replaced by−.
Therefore,Dk∩I⊂R×Tk∪λk,∞. Now we have two cases to consider, that is,Dk\I
is bounded or unbounded. First supposeDk\Iis bounded. Then there existsλ, v∈ ∂Dk
with v ∈ ∂Tk. Ifv /0, by Lemma 3.3we know v ∈ Tjν for some positive integerj /k and
ν ∈ {,−}. As in the proof ofLemma 3.4, we get a contradiction, which meansv 0. Thus there exists a sequenceξm, zm∈Dkwithzm → v≡0 asm → ∞. This together withH1
guarantee thatξm, zmsatisfies3.14.
As in the proof of Lemma 3.4, we may assume without loss of generality that
zm/zm1 → zandξm → ξasm → ∞. Then we have
zξLaz. 3.20
Sincez /0,ξ /0 is an eigenvalue of operatorLa. From this,3.9, and3.19it follows that
ξ λj for some positive integerj. Then by Lemma 2.3,zbelongs toTj orTj−. Notice that
zm−z1 → 0 and sozm−z2 → 0 as in the proof ofLemma 3.4. Thuszm ∈TjorTj− for
msufficiently large sinceTj andTj−are open. This together withzm∈Tk m≥1guarantee
thatkj. This meansDkmeetsλk,0ifDk\Iis bounded.
Next supposeDk\Iis unbounded. In this case we showDk \Ihas an unbounded projection onR. If not, then there exists a sequence ζm, ym ∈ Dk \I with ζm → ζand
ym1 → ∞asm → ∞. Letxm:ym/ym1,m≥1. From the fact that
ymζmLbymHb
ζm, ym
3.21
it follows that
xmζmLbxm
Hb
ζm, ym
ym
1
. 3.22
Notice thatLb : Y → Y is completely continuous. We may assume that there existsw ∈Y
withw11 such thatxm−w1 → 0 asm → ∞.
Letting m → ∞in 3.22and noticingHbζm, ym/ym1 → 0 asm → ∞ one
obtains
wζLbw. 3.23
Sincew /0,ζ /0 is an eigenvalue of operator Lb, that is, ζ λk0 for some positive integer
k0/k. Then byLemma 2.3wbelongs toTk0orTk−0. Notice the fact thatxm−w1 → 0 and
soxm−w2 → 0 as in the proof ofLemma 3.4. Thusxm∈Tk0orTk−0 formsufficiently large
sinceTk0 andTk−0 are open. This is a contradiction withxm∈Tkm≥1. ThusDk\Ihas an
Proof ofTheorem 3.1. Suppose first that
μ2
i0k
a1a2μi0k
<1< μ
2
i01
b1b2μi01
. 3.24
Using the notation of3.9and3.19, this meansλi0k<1< λi01and so fromLemma 2.3we
know
λi01< λi02<· · ·< λi0k<1< λi01 < λi02<· · ·< λi0k. 3.25
Consider3.7as a bifurcation problem from the trivial solution. We need only show that
Cνi0j{1} ×Y/∅, j1,2, . . . , k; ν ,−. 3.26
Suppose, on the contrary and without loss of generality, that
Ci0i{1} ×Y ∅, for somei, 1≤i≤k. 3.27
ByLemma 3.4we know thatCi0ijoinsλi0i,0to infinity inΣandλ, v 0,0is the unique
solution of3.1λ in whichλ0inE. This together withλi0i<1 guarantee that there exists
a sequence{ζm, ym} ⊂ Ci0i such thatζm ∈ 0,1andym1 → ∞asm → ∞. We may
assume thatζm → ζ ∈ 0,1asm → ∞. Letxm : ym/ym1,m ≥ 1,then3.22holds.
Similarly, we may assume that there existsw ∈ Y withw1 1 such thatxm−w1 → 0
asm → ∞and 3.23holds. From the proof of Lemma 3.5one can see ζ λi0i, which
contradictsλi0i>1. Thus3.27is not true, which means3.26holds.
Next suppose that
μ2i0k b1b2μi0k
<1< μ
2
i01
a1a2μi01
. 3.28
This means
λi01< λi02<· · ·< λi0k<1< λi01 < λi02<· · ·< λi0k. 3.29
Consider3.16as a bifurcation problem from infinity. As above we need only to prove that
Dνi0j{1} ×Y/∅, j 1,2, . . . , k; ν ,−. 3.30
FromLemma 3.5, we know thatDν
i0jcomes from{λi0j,∞}, meetsλi0j,0or has an
unbounded projection onR. If it meetsλi0j,0, then the connectedness ofD
ν
guarantees that3.30is satisfied. On the other hand, ifDνi0jhas an unbounded projection on
R, notice thatλ, v 0,0is the unique solution of3.1λ in whichλ 0inE, so3.30 also holds.
Proof ofTheorem 3.2. First suppose thatH3holds. Then there existsε >0 such that
1εfx, y< R M, for
x, y∈x, y:|x| ≤MR, y≤R. 3.31
Letλ, vbe a solution of 3.1λsuch that 0 ≤ λ <1εandv1 ≤R. Then by2.7,
3.1λ,3.31andLemma 2.1it is easy to see
v1λAv1λLFv1 ≤λMFvMmax
t∈0,1λfLvt,−vt< M
R
M R. 3.32
Therefore,
Σ∩0,1ε×∂BR
∅. 3.33
This together with3.32and Lemmas3.4and3.5implies that
Cνk∩0,1ε×BR
⊂0,1ε×BR, k1,2, . . . , i0; 3.34
Dνj ∩0,1ε×∂BR
∅, j1,2, . . . , j0; 3.35
whereBR {v∈Y | v1 < R}andBR {v∈Y | v1 ≤R},CkνandDjνare obtained from
Lemmas3.4and3.5, respectively. SinceCν
kis a unbounded component of solutions of3.1λjoiningλk,0inE, it follows
from3.33and3.34thatCνkcrosses the hyperplane{1} ×Ywith1, vνsuch thatvν
1< R
ν or−,k1,2, . . . , i0. This means BVP2.5has 2i0nontrivial solutions{vνi} i0
1 in which
v1 andv1−are positive and negative solutions, respectively.
On the other hand, by3.33,3.35, andLemma 3.5one can obtain
Dνj ∩{1} ×Y\BR
/
∅, j 1,2, . . . , j0. 3.36
This means BVP2.5has 2j0nontrivial solutions{wiν} j0
1 in whichw1andw1−are positive and
negative solutions, respectively.
Now it remains to show this theorem holds when the conditionH4is satisfied. From the above we need only to prove that
iforλ, v∈Cνi ν or−,i1,2, . . . , i0,
iiforλ, v∈Dνj ν or−,j1,2, . . . , j0, we have that either
max
t∈0,1vt> r2, t∈0,1 3.38
or
min
t∈0,1vt< r1, t∈0,1. 3.39
In fact, like in13, suppose on the contrary that there existsλ, v∈Cνi Djνsuch that either
max{vt:t∈0,1}r2 3.40
or
min{vt:t∈0,1}r1 3.41
for somei, j.
First consider the case max{vt:t∈0,1}r2. Then there existst∈0,1such that
vt r2. Let
τ0 : inf
t∈0, t:vs≥ fors∈t, t,
τ1 : sup
t∈t,1:vs≥0 fors∈t, t.
3.42
Then
max{vt:t∈τ0, τ1}r2, 3.43
0≤vt≤r2, t∈τ0, τ1. 3.44
Therefore,vtis a solution of the following equation
−vt λfLvt,−vt, t∈τ0, τ1 3.45
withvτ0 vτ1 0 if 0< τ0< τ1<1 andvτ0 0 ifτ0 0.
By H4, there existsM ≥ 0 such that fx,−y My is strictly increasing iny for
x, y∈r1, r2×−Mr, Mr, wherermax{|r1|, r2}. Then
UsingH4andLemma 2.1again, we can obtain
−r2−vt
λMr2−vt
−λfLvt,−vt Mvt−Mr2
−λfLvt,−vt Mvt−fLvt,−r2 Mr2
−λfLvt,−r2
≥0, t∈τ0, τ1
3.47
and ifτ11, by1.2we knowv1< r2. Therefore,
r2−vτ0>0, r2−vτ1>0 if 0< τ0< τ1<1;
r2−vτ00 ifτ00;
r2−vτ1>0 ifτ11.
3.48
This together with3.47and the maximum principle14imply thatr2−vt> 0 in
τ0, τ1, which contradicts3.43.
The proof in the case min{vt:t∈0,1}r1is similar, so we omit it.
Acknowledgments
The Project Supported by NNSF of P. R. China10871120, the Key Project of Chinese Ministry of EducationNo: 209072, and the Science & Technology Development Funds of Shandong Education CommitteeJ08LI10.
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