Volume 2009, Article ID 572512,18pages doi:10.1155/2009/572512
Research Article
The Existence of Countably Many Positive
Solutions for Nonlinear
n
th-Order Three-Point
Boundary Value Problems
Yude Ji and Yanping Guo
College of Sciences, Hebei University of Science and Technology, Shijiazhuang, Hebei 050018, China
Correspondence should be addressed to Yude Ji,[email protected]
Received 5 July 2009; Revised 30 August 2009; Accepted 30 October 2009
Recommended by Kanishka Perera
We consider the existence of countably many positive solutions for nonlinearnth-order three-point boundary value problemuntatfut 0,t∈0,1,u0 αuη,u0 · · ·un−20 0,
u1 βuη, wheren≥ 2, α ≥0, β ≥0,0 < η < 1, α β−αηn−1 < 1,at ∈Lp0,1for some
p ≥ 1 and has countably many singularities in0,1/2. The associated Green’s function for the nth-order three-point boundary value problem is first given, and growth conditions are imposed on nonlinearityf which yield the existence of countably many positive solutions by using the Krasnosel’skii fixed point theorem and Leggett-Williams fixed point theorem for operators on a cone.
Copyrightq2009 Y. Ji and Y. Guo. 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
The existence of positive solutions for nonlinear second-order and higher-order multipoint boundary value problems has been studied by several authors, for example, see 1–12
and the references therein. However, there are a few papers dealing with the existence of positive solutions for thenth-order multipoint boundary value problems with infinitely many singularities. Hao et al.13discussed the existence and multiplicity of positive solutions for the followingnth-order nonlinear singular boundary value problems:
unt atft, u 0, t∈0,1,
u0 0, u0 · · ·un−20 0, u1 αuη,
1.1
iff is either superlinear or sublinear by applying the Krasnosel’skii-Guo theorem on cone expansion and compression.
In14, Kaufmann and Kosmatov showed that there exist countably many positive
solutions for the two-point boundary value problems with infinitely many singularities of following form:
−ut atfut, 0< t <1,
u0 0, u1 0, 1.2
whereat∈Lp0,1for somep≥1 and has countably many singularities in0,1/2.
In15, Ji and Guo proved the existence of countably many positive solutions for the
nth-order ordinary differential equation
unt atfut 0, t∈0,1, 1.3
with one of the followingm-point boundary conditions:
u0 m−2
i1
kiuξi, u0 · · ·un−20 0, u1 0,
u0 0, u0 · · ·un−20 0, u1
m−2
i1
kiuξi,
1.4
wheren≥ 2,ki > 0i1,2, . . . , m−2, 0< ξ1 < ξ2 < · · ·< ξm−2 <1,f ∈C0,∞,0,∞, at∈Lp0,1for somep≥1 and has countably many singularities in0,1/2.
Motivated by the result of 13–15, in this paper we are interested in the existence
of countably many positive solutions for nonlinear nth-order three-point boundary value problem
unt atfut 0, t∈0,1,
u0 αuη, u0 · · ·un−20 0, u1 βuη,
1.5
wheren≥2,α≥0,β≥0, 0< η <1,α β−αηn−1<1,f ∈C0,∞,0,∞,at∈Lp0,1
for somep ≥1 and has countably many singularities in0,1/2. We show that the problem 1.5has countably many solutions ifaandfsatisfy some suitable conditions. Our approach is based on the Krasnosel’skii fixed point theorem and Leggett-Williams fixed point theorem in cones.
Suppose that the following conditions are satisfied.
H1There exists a sequence{tk}∞k1such thattk1< tkk∈N,t1<1/2, limk→ ∞tkt∗≥
0, and limt→tkat ∞for allk1,2, . . . .
H2There existsm >0 such thatat≥mfor allt∈t∗,1−t∗.
The paper is organized as follows. In Section 2, we provide some necessary back-ground material such as the Krasnosel’skii fixed-point theorem and Leggett-Williams fixed point theorem in cones. InSection 3, the associated Green’s function for thenth-order three-point boundary value problem is first given and we also look at some properties of the Green’s function associated with problem 1.5. In Section 4, we prove the existence of countably many positive solutions for problem1.5under suitable conditions onaandf. InSection 5, we give two simple examples to illustrate the applications of obtained results.
2. Preliminary Results
Definition 2.1. LetEbe a Banach space overR. A nonempty convex closed setP ⊂Eis said to be a cone provided that
iau∈Pfor allu∈P and for alla≥0; iiu,−u∈Pimpliesu0.
Definition 2.2. The map α : P → 0,∞ is said to be a nonnegative continuous concave functional onPprovided thatαis continuous and
αtx 1−ty≥tαx 1−tαy, 2.1
for allx, y∈Pand 0≤t≤1.Similarly, we say that the mapγ :P → 0,∞is a nonnegative continuous convex functional onPprovided thatγis continuous and
γtx 1−ty≤tγx 1−tγy, 2.2
for allx, y∈Pand 0≤t≤1.
Definition 2.3. Let 0< a < bbe given and letαbe a nonnegative continuous concave functional onP. Define the convex setsPr andPα, a, bby
Pr {x∈P | x< r},
Pα, a, b {x∈P|a≤αx,x ≤b}. 2.3
The following Krasnosel’skii fixed point theorem and Leggett-Williams fixed point theorem play an important role in this paper.
Theorem 2.416, Krasnosel’skii fixed point theorem. LetEbe a Banach space and letP ⊂E
be a cone. Assume thatΩ1,Ω2are bounded open subsets ofEsuch that0 ∈Ω1 ⊂Ω1⊂Ω2. Suppose
that
T :PΩ2\Ω1
is a completely continuous operator such that, either
iTu ≤ u, u∈P∂Ω1, andTu ≥ u, u∈P
∂Ω2, or iiTu ≥ u, u∈P∂Ω1, andTu ≤ u, u∈P∂Ω2.
ThenT has a fixed point inPΩ2\Ω1.
Theorem 2.517, Leggett-Williams fixed point theorem. LetA : Pc → Pcbe a completely
continuous operator and letαbe a nonnegative continuous concave functional onP such thatαx≤
xfor allx∈Pc. Suppose there exist0< a < b < d≤csuch that
C1{x∈Pα, b, d|αx> b}/∅, andαAx> b forx∈Pα, b, d, C2Ax< a forx ≤a,
C3αAx> b forx∈Pα, b, c, withAx> d.
ThenAhas at least three fixed pointsx1, x2, andx3such that
x1< a, b < αx2, x3> a with αx3< b. 2.5
In order to establish some of the norm inequalities in Theorems2.4and 2.5we will need Holder’s inequality. We use standard notation ofLpa, bfor the space of measurable
functions such that
1
0 fsp
ds <∞, 2.6
where the integral is understood in the Lebesgue sense. The norm onLpa, b, · , is defined
by
fp 1
0
|fs|pds 1/p
. 2.7
Theorem 2.618, Holder’s inequality. Letf ∈ Lpa, band g ∈ Lqa, b, wherep > 1and
1/p1/q1. Thenfg∈L1a, band, moreover
1
0
fsgsds≤fpgq. 2.8
Letf∈L1a, bandg∈L∞a, b. Thenfg ∈L1a, band
1
0
3. Preliminary Lemmas
To prove the main results, we need the following lemmas.
Lemma 3.1see15. Foryt∈C0,1,the boundary value problem
unt yt 0, t∈0,1,
u0 0, u0 · · ·un−20 0, u1 0
3.1
has a unique solution
ut − t 0
t−sn−1
n−1! ysdstn−1 1
0
1−sn−1
n−1! ysds. 3.2
Lemma 3.2see15. The Green’s function for the boundary value problem
−unt 0, t∈0,1,
u0 0, u0 · · ·un−20 0, u1 0
3.3
is given by
gt, s 1 n−1!
⎧ ⎨ ⎩
tn−11−sn−1−t−sn−1, 0≤s≤t≤1,
tn−11−sn−1, 0≤t≤s≤1. 3.4
Lemma 3.3see15. The Green’s functiongt, sdefined by3.4satisfies that
igt, s≥0is continuous on0,1×0,1;
iigt, s ≤ gθ1s, s for allt, s ∈ 0,1and there exists a constantγτ > 0 for anyτ ∈
0,1/2such that
min
t∈τ,1−τgt, s≥γτgθ1s, s≥γτg
t, s, ∀t, s∈0,1, 3.5
where
γτ min
τ
θ1s n−1
, τ
1−θ1s
,
θ1s
s
1−1−sn−1/n−2 s < θ1s<1.
Lemma 3.4. Supposeα β−αηn−1/1,then foryt∈C0,1,the boundary value problem
unt yt 0, t∈0,1,
u0 αuη, u0 · · ·un−20 0, u1 βuη 3.7
has a unique solution
ut − t 0
t−sn−1 n−1! ysds
− α
1−α−β−αηn−1
η
0
η−sn−1
n−1! ysds
αηn−1 1−α−β−αηn−1
1
0
1−sn−1 n−1! ysds
1−αtn−1 1−α−β−αηn−1
1
0
1−sn−1
n−1! ysds−
β−αtn−1 1−α−β−αηn−1
η
0
η−sn−1 n−1! ysds.
3.8
Proof. The general solution ofunt yt 0 can be written as
ut − t 0
t−sn−1
n−1! ysdsAtn−1
n−2
i1
AitiB. 3.9
Sinceui0 0 fori1,2, . . . , n−2, we getA
i 0 fori 1,2, . . . , n−2. Now we solve for
A, Bbyu0 αuηandu1 βuη, it follows that
B−α
η
0
η−sn−1
n−1! ysdsαAηn−1αB
− 1
0
1−sn−1
n−1! ysdsAB−β
η
0
η−sn−1
n−1! ysdsβAηn−1βB.
3.10
By solving the above equations, we get
A 1
1−α−β−αηn−1
1−α 1 0
1−sn−1
n−1! ysds−
β−α
η
0
η−sn−1 n−1! ysds
,
B 1
1−α−β−αηn−1
−α
η
0
η−sn−1
n−1! ysdsαηn−1 1
0
1−sn−1 n−1! ysds
.
Therefore,3.7has a unique solution
ut − t 0
t−sn−1 n−1! ysds
− α
1−α−β−αηn−1
η
0
η−sn−1
n−1! ysds
αηn−1 1−α−β−αηn−1
1
0
1−sn−1 n−1! ysds
1−αtn−1 1−α−β−αηn−1
1
0
1−sn−1
n−1! ysds−
β−αtn−1 1−α−β−αηn−1
η
0
η−sn−1
n−1! ysds. 3.12
Lemma 3.5. Suppose0< α β−αηn−1<1, the Green’s function for the boundary value problem
unt yt 0, t∈0,1,
u0 αuη, u0 · · ·un−20 0, u1 βuη
3.13
is given by
Gt, s gt, s
β−αtn−1α 1−α−β−αηn−1g
η, s, 3.14
wheregt, sis defined by3.4.
We omit the proof as it is immediate fromLemma 3.4and3.4.
Lemma 3.6. Suppose0< α β−αηn−1 <1, the Green’s functionGt, sdefined by3.14satisfies
that
iGt, s≥0is continuous on0,1×0,1;
iiGt, s≤ Jsfor allt, s∈ 0,1and there exists a constantγτ >0for anyτ ∈0,1/2
such that
min
t∈τ,1−τGt, s≥γτJs≥γτG
t, s, ∀t, s∈0,1, 3.15
where
Js gθ1s, s
maxα, β 1−α−β−αηn−1g
η, s,
γτmin ⎧ ⎨ ⎩τn−1,
minβ−ατn−1,β−α1−τn−1α
maxα, β
≤min
⎧ ⎨ ⎩
τ θ1s
n−1
, τ
1−θ1s ,
minβ−ατn−1,β−α1−τn−1α
maxα, β
⎫ ⎬ ⎭
min
⎧ ⎨ ⎩γτ,
minβ−ατn−1,β−α1−τn−1α
maxα, β
⎫ ⎬ ⎭.
3.16
Proof. iFromLemma 3.3and3.14, we get
Gt, s≥0 is continuous on 0,1×0,1. 3.17
iiFromLemma 3.3and3.14, we have
Gt, s gt, s
β−αtn−1α 1−α−β−αηn−1g
η, s
≤gθ1s, s
maxα, β 1−α−β−αηn−1g
η, sJs.
3.18
Next, we prove that3.15holds.
FromLemma 3.3and3.14, fort∈τ,1−τ, we have
Gt, s gt, s
β−αtn−1α 1−α−β−αηn−1g
η, s
≥γτgθ1s, s
minβ−ατn−1,β−α1−τn−1α
1−α−β−αηn−1 g
η, s
γτgθ1s, s
minβ−ατn−1,β−α1−τn−1α
maxα, β ×
maxα, β 1−α−β−αηn−1g
η, s
≥γτ
gθ1s, s
maxα, β 1−α−β−αηn−1g
η, s
γτJs ≥γτG
t, s,
3.19
for allt∈0,1, whereγτ min{τn−1,min{β−ατn−1,β−α1−τn−1}α/max{α, β}},
We use inequality3.15to define our cones. LetEC0,1, thenEis a Banach space with the normumaxt∈0,1|ut|. For a fixedτ∈0,1/2, define the coneP ⊂Eby
P
u∈E|ut≥0 on0,1, and min
t∈τ,1−τut≥γτu
. 3.20
Define the operatorTby
Tut 1
0
Gt, sasfusds, 0≤t≤1. 3.21
Obviously,utis a solution of1.5if and only ifutis a fixed point of operatorT.
Theorems 2.4and 2.5require the operatorT to be completely continuous and cone preserving. IfTis continuous and compact, then it is completely continuous. The next lemma shows thatT :P → Pforτ∈0,1/2and thatT is continuous and compact.
Lemma 3.7. The operatorT is completely continuous andT :P → Pfor eachτ ∈0,1/2.
Proof. Fixτ ∈0,1/2. Sinceasfus≥0 for alls∈0,1,u∈P and sinceGt, s≥0 for allt, s∈0,1, thenTut≥0 for allt∈0,1, u∈P.
Letu∈P, by3.15and3.21we have
min
t∈τ,1−τut t∈minτ,1−τ 1
0
Gt, sasfusds
≥ 1
0 min
t∈τ,1−τGt, sasfusds
≥γτ
1
0
Gt, sasfusds
≥γτTu
t,
3.22
for allt∈0,1. Thus
min
t∈τ,1−τut≥γτTu. 3.23
Clearly operator3.21is continuous. By the Arzela-Ascoli theoremT is compact. Hence, the operatorT is completely continuous and the proof is complete.
4. Main Results
For convenience, we denote
Λ1 1
maxt∈0,1τ1−1τ1Gt, sds·m
, Λ2 1
Jq· ap. 4.1
Theorem 4.1. Suppose conditionsH1andH2hold, let{τk}∞k1be such thattk1 < τk < tk, k
1,2, . . . .Let{Rk}∞k1and{rk}∞k1be such that
Rk1< γτkrk< rk< Rk, Mrk< LRk, k1,2, . . . , 4.2
whereM∈Λ1,∞,L∈0,Λ2. Furthermore, for each natural numberk, assume thatf satisfies
the following two growth conditions:
H3fu≤LRk for allu∈0, Rk,
H4fu≥Mrk for allu∈γτkrk, rk.
Then problem1.5has countably many positive solutions{uk}∞k1such thatrk≤ uk ≤Rkfor each
k1,2, . . . .
Proof. Consider the sequences{Ω1,k}∞k1and{Ω2,k}∞k1of open subsets ofEdefined by
Ω1,k {u∈E| u< Rk},
Ω2,k {u∈E| u< rk}.
4.3
Let{τk}∞k1be as in the hypothesis and note thatt0 < tk1 < τk< tk< 1/2,for allk ∈N. For
eachk∈N, define the conePkby
Pk
u∈E|ut≥0 on0,1, and min
t∈τk,1−τk
ut≥γτku
. 4.4
Fixedkand letu∈Pk
∂Ω2,k. Fors∈τk,1−τk, we have
γτkrkγτku ≤ min s∈τk,1−τk
By conditionH4, we get
Tumax
t∈0,1 1
0
Gt, sasfusds
≥max
t∈0,1 1−τk
τk
Gt, sasfusds
≥max
t∈0,1 1−τk
τk
Gt, sasds·Mrk
≥mMrk·max t∈0,1
1−τ1
τ1
Gt, sds
≥rku.
4.6
Now letu∈Pk∂Ω1,k, thenus≤ uRkfor alls ∈0,1. By conditionH3, we get
Tumax
t∈0,1 1
0
Gt, sasfusds
≤ 1
0
Jsasds·LRk
≤ Jqap·LRk
≤Rku.
4.7
It is obvious that 0∈Ω2,k⊂Ω2,k ⊂Ω1,k. Therefore, byTheorem 2.4, the operatorThas
at least one fixed pointuk ∈ Pk
Ω1,k \Ω2,ksuch thatrk ≤ uk ≤ Rk. Sincek ∈ Nwas
arbitrary,Theorem 4.1is completed.
Let τk is defined by Theorem 4.1. We define the nonnegative continuous concave
functionalsαkuonPby
αku min t∈τk,1−τk
ut. 4.8
We observe here that, for eachu∈P,αu≤ u. For convenience, we denote
Λ Jq· ap, Γk min t∈τk,1−τk
1−τk
τk
Theorem 4.2. Suppose conditionsH1andH2hold, let{τk}∞k1be such thattk1 < τk < tk, k
1,2, . . . .Let{ak}∞k1,{bk}∞k1, and{ck}∞k1be such that
ck1 < ak< bk≤min
γτk,
M L
ck< ck, k1,2, . . . , 4.10
whereL ∈Λ,∞,M ∈0,Γk. Furthermore, for each natural numberk, assume thatf satisfies
the following growth conditions:
H5fu≤ck/L for allu∈0, ck,
H6fu< ak/L for allu∈0, ak,
H7fu≥bk/M for allu∈bk, bk/γτk.
Then problem1.5has three infinite families of solutions{u1k}∞k1,{u2k}∞k1, and{u3k}∞k1such that
u1k< ak, min t∈τk,1−τk
u2kt> bk, u3k> ak, with min t∈τk,1−τk
u3kt< bk, 4.11
for eachk1,2, . . . .
Proof. We note first that T : Pck → Pck is completely continuous operator. If u ∈ P, then
from properties of Gt, s, Tut ≥ 0, and by Lemma 3.7, mint∈τk,1−τkTut ≥ γτkTu.
Consequently,T:P → P.
Ifu∈Pck, thenu ≤ck, and by conditionH5, we have
Tu max
t∈0,1|Tut|
max
t∈0,1
1
0
Gt, sasfusds
≤ ck
L
1
0
Jsasds
≤ ck
L · Jq· ap≤ck.
4.12
Therefore,T :Pck → Pck. Standard applications of Arzela-Ascoli theorem imply thatT
is completely continuous operator.
In a completely analogous argument, conditionH6implies that conditionC2 of
Theorem 2.5is satisfied.
We now show that conditionC1ofTheorem 2.5is satisfied. Clearly,
u∈P
αk, bk,
bk
γτk
|αku> bk
/
Ifu∈Pαk, bk, bk/γτk, thenbk≤us≤bk/γτk, fors∈τk,1−τk. By conditionH7, we get
αkTu min t∈τk,1−τk
1
0
Gt, sasfusds
≥ min
t∈τk,1−τk 1−τk
τk
Gt, sasfusds
≥m· bk
Mt∈minτk,1−τk 1−τk
τk
Gt, sds≥bk.
4.14
Therefore, conditionC1ofTheorem 2.5is satisfied.
Finally, we show that conditionC3ofTheorem 2.5is also satisfied. Ifu∈Pαk, bk, ckandTu> bk/γτk, then
αkTu min t∈τk,1−τk
Tut≥γτkTu> bk. 4.15
Therefore, conditionC3is also satisfied. ByTheorem 2.5, There exist three infinite families of solutions{u1k}∞k1,{u2k}k∞1, and{u3k}∞k1for problem1.5such that
u1k< ak, min t∈τk,1−τk
u2kt> bk, u3k> ak, with min t∈τk,1−τk
u3kt< bk, 4.16
for eachk1,2, . . . .Thus,Theorem 4.2is completed.
5. Example
In this section, we cite an example see 15 to verify existence of at, and two simple examples are presented to illustrate the applications for obtained conclusion of Theorems
4.1and4.2.
Example 5.1. As an example of problem1.5, we mention the boundary value problem
u3t atfut 0, t∈0,1,
u0 1
2u
1 2
, u0 0, u1 u
1 2
,
whereatis defined by15, Example 6.1andε1/4, fu ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
32×10−4k2−1/2×10−4k1 1/625×10−8k4−10−8k1
u2−10−8k1
1 2 ×10−
4k1, u∈!10−4k1, 1
25 ×10− 4k2",
18×10−4k2×sinπ u−1/25×10−4k2 10−4k2−1/25×10−4k2
32×10−4k2, u∈
!
1 25×10
−4k2,10−4k2",
32×10−4k2−1/2×10−4k
10−8k4−10−8k
u2−10−8k
1 2 ×10−
4k, u∈#10−4k2,10−4k$k1,2, . . .,
1 2 ×10
−4, u∈#10−4,∞.
5.2
We notice thatn3,α1/2,β1,η1/2. If we taket0 5/16,tk t0−
%k−1
i0 1/i24,τk 1/2tktk1,k 1,2, . . . ,then tk1 < τk< tk, and 1/5< t∗< τk< τ11/4−1/2×34<1/4, γτkmin{τ
2
k,min{β−ατk2,β−
α1−τk2}α/max{α, β}}>1/25,k1,2, . . . .
It follows from a direct calculation that
1−τ1
τ1
Gt, sds > 1−1/4 1/4
Gt, sds
3/4 1/4
gt, sds4 3
1t2 3/4
1/4 g
1 2, s
ds
1 2
t
1/4 &
t21−s2−t−s2'ds 3/4
t
t21−s2ds
4 3
1t2( 1/2
1/4
1 41−s
2−
1 2 −s
2
ds
3/4
1/2 1 41−s
2ds )
1
576
−96t3122t2−18t 25 2
,
so
max
t∈0,1 1−τ1
τ1
Gt, sds≥ max
t∈1/4,1−1/4 1−1/4
1/4
Gt, sds 31×7 24×6×32 >
1 32,
J12 1
0 J12sds
1/2 ≤ 5
6, a2
* + +
,√2π2 3 −
9 4
.
5.4
In addition, if we takerk10−4k2,Rk10−4k,M32,L1/2,m 4/31/4, then
at≥
4 3
1/4
m, t∈t∗,1−t∗,
Rk110−4k1< 1 25×10
−4k2< γ
τk·rk< rk10
−4k2< R
k10−4k,
Mrk32×10−4k2< LRk 1
2 ×10
−4k, k1,2, . . . ,
Λ1
1
maxt∈0,1τ11−τ1G1t, sds·m
≤ 1
1/32×4/31/4 <32M,
Λ2 1
J12· a2
≥ 1
5/6×-√2π2/3−9/4
> L 1 2,
5.5
andfusatisfies the following growth conditions:
fu≤LRk 1
2 ×10
−4k, u∈&0,10−4k',
fu≥Mrk32×10−4k2, u∈ !
1 25×10
−4k2,10−4k2".
5.6
Then all the conditions of Theorem 4.1 are satisfied. Therefore, byTheorem 4.1 we know that problem5.1has countably many positive solutions{uk}k∞1such that 10−4k2 ≤ uk ≤10−4kfor eachk1,2, . . . .
Example 5.2. As another example of problem1.5, we mention the boundary value problem
u3t atfut 0, t∈0,1,
u0 1
2u
1 2
, u0 0, u1 u
1 2
whereatis defined by15, Example 6.1andε1/4, fu ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ u
2 u∈
#
10−4k1,10−4k3$,
1/2×10−4k3−45×10−4k2 10−8k6−10−8k4
u2−10−8k4
45×10−4k2, u∈#10−4k3,10−4k2$,
5×10−4k2×sinπ u−10−4k2 25×10−4k2−10−4k2
45×10−4k2, u∈#10−4k2,25×10−4k2$,
45×10−4k2−1/2×10−4k
625×10−8k4−10−8k
u2−10−8k
1 2 ×10−
4k, u∈#25×10−4k2,10−4k$,k1,2, . . .,
1 2×10−
4, u∈#10−4,∞.
5.8
We notice thatn3,α1/2,β1,η1/2.
If we taket0 5/16,tk t0−%ki−011/i24,τk 1/2tktk1,k1,2, . . . ,then tk1 < τk< tk, and 1/5< t∗< τk< τ11/4−1/2×34<1/4,γτkmin{τ
2
k,min{β−ατk2,β−
α1−τk2}α/max{α, β}}>1/25,k1,2, . . . .
It follows from a direct calculation that
Λ J2· a2 ≤ 5 6
* + +
,√2π2 3 − 9 4 , min
t∈τk,1−τk 1−τk
τk
Gt, sds≥ min
t∈1/5,1−1/5 1−1/4
1/4
Gt, sds 3253 3200×45 >
1 45.
5.9
In addition, if we takeak10−4k3,bk10−4k2,ck10−4k,M1/45,L2,m4/31/4, then
at≥
4 3
1/4
m, t∈t0,1−t0,
ck110−4k1< ak10−4k3< bk10−4k2
< 1 90×10
−4kmin
γτk,
M L
M 1 45 <
1 45×
4 3
1/4
< min
t∈τk,1−τk 1−τk
τk
Gt, sds·m Γk, k1,2, . . . ,
Λ J2· a2≤ 5 6
* + +
,√2π2 3 −
9 4
<2L,
5.10
andfusatisfies the following growth conditions:
fu≤ ck
L
10−4k
2 , u∈
&
0,10−4k',
fu< ak
L
10−4k3
2 , u∈
&
0,10−4k3',
fu≥ bk
M
10−4k2
1/45 45×10
−4k2, u∈&10−4k2,25×10−4k2'.
5.11
Then all the conditions of Theorem 4.2 are satisfied. Therefore, byTheorem 4.2 we know that problem5.7has countably many positive solutions{uk}∞k1such that
u1k<10−4k3, min t∈τk,1−τk
u2kt>10−4k2
u3k>10−4k3, with min t∈τk,1−τk
u3kt<10−4k2,
5.12
for eachk1,2, . . . .
Remark 5.3. In 8–12, the existence of solutions for local or nonlocal boundary value
problems of higher-order nonlinear ordinaryfractionaldifferential equations that has been treated did not discuss problems with singularities. In13, the singularity only allowed to
appear att0 and/ort1, the existence and multiplicity of positive solutions were asserted under suitable conditions onf. Although,14,15seem to have considered the existence of countably many positive solutions for the second-order and higher-order boundary value problems with infinitely many singularities in0,1/2. However, in15, only the boundary
conditionsu0 0 oru1 0 have been considered. It is clear that the boundary conditions of Examples 5.1 and 5.2 are u0/0 and u1/0. Hence, we generalize second-order and higher-order multipoint boundary value problem.
Acknowledgments
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