Volume 2010, Article ID 368169,15pages doi:10.1155/2010/368169
Research Article
Positive Solutions of Singular Complementary
Lidstone Boundary Value Problems
Ravi P. Agarwal,
1Donal O’Regan,
2and Svatoslav Stan ˇek
3 1Department of Mathematical Sciences, Florida Institute of Technology, Melbourne,FL 32901-6975, USA
2Department of Mathematics, National University of Ireland, Galway, Ireland
3Department of Mathematical Analysis, Faculty of Science, Palack´y University, Tˇr. 17. listopadu 12,
771 46 Olomouc, Czech Republic
Correspondence should be addressed to Ravi P. Agarwal,[email protected]
Received 7 October 2010; Accepted 21 November 2010
Academic Editor: Irena Rach ˚unkov´a
Copyrightq2010 Ravi P. Agarwal et al. 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 investigate the existence of positive solutions of singular problem−1mx2m1 ft, x, . . . ,
x2m,x0 0,x2i−10 x2i−1T 0, 1≤i≤m. Here,m≥1 and the Carath´eodory function
ft, x0, . . . , x2mmay be singular in all its space variablesx0, . . . , x2m. The results are proved by regularization and sequential techniques. In limit processes, the Vitali convergence theorem is used.
1. Introduction
LetTbe a positive constant,J 0, TandÊ− −∞,0,Ê 0,∞,Ê0 Ê\{0}. We consider
the singular complementary Lidstone boundary value problem
−1mx2m1t ft, xt, . . . , x2mt, m≥1, 1.1
x0 0, x2i−10 x2i−1T 0, 1≤i≤m, 1.2
wherefsatisfies the local Carath´eodory function onJ× Df ∈CarJ× Dwith
D
⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ Ê
2
×Ê0×Ê− ×Ê0×Ê × · · · ×Ê×Ê0
4k−1
ifm2k−1,
Ê
2
×Ê0×Ê− ×Ê0×Ê × · · · ×Ê−×Ê0
4k1
The functionft, x0, . . . , x2mis positive and may be singular at the value zero of all its space
variablesx0, . . . , x2m.
Leti∈ {0,1, . . . ,2m}. We say thatf is singular at the value zero of its space variablexiif
for a.e.t∈Jand allxj, 0≤j≤2m,j /isuch thatx0, . . . , xi, . . . , x2m∈ D, the relation
lim
xi→0ft, x0, . . . , xi, . . . , x2m ∞ 1.4
holds.
A functionx ∈ AC2mJ i.e.,xhas absolutely continuous 2mth derivative on Jis
a positive solution of problem 1.1, 1.2if xt > 0 for t ∈ 0, T, x satisfies the boundary conditions1.2and1.1holds a.e. onJ.
The regular complementary Lidstone problem
−1mx2m1t ht, xt, . . . , xqt, m≥1, qfixed, 0≤q≤2m,
x0 α0, x2i−10 αi, x2i−11 βi, 1≤i≤m
1.5
was discussed in1. Here,h:0,1×Êq1 → Êis continuous at least in the interior of the
domain of interest. Existence and uniqueness criteria for problem 1.5 are proved by the
complementary Lidstone interpolating polynomial of degree 2m. No contributions exist, as far as we know, concerning the existence of positive solutions of singular complementary Lidstone problems.
We observe that differential equations in complementary Lidstone problems as well as derivatives in boundary conditions are odd orders, in contrast to the Lidstone problem
−1mx2mt pt, xt, . . . , xrt, m≥1, r fixed, 0≤r ≤2m−1,
x2i0 a
i, x2i1 bi, 1≤i≤m−1,
1.6
where the differential equation and derivatives in the boundary conditions are even orders. Forai bi 0 1 ≤ i ≤ m−1, regular Lidstone problems were discussed in 2–9, while
singular ones in10–15.
The aim of this paper is to give the conditions on the functionf in1.1which gua-rantee that the singular problem1.1,1.2has a solution. The existence results are proved by regularization and sequential techniques, and in limit processes, the Vitali convergence theorem16,17is applied.
Throughout the paper,x∞ max{|xt| : t ∈ J}and xCn nk0xk∞,n ≥ 1
stands for the norm in C0J and CnJ, respectively. L1J denotes the set of functions
Lebesgueintegrable onJand measMthe Lebesgue measure ofM ⊂J.
We work with the following conditions on the functionfin1.1.
H1f ∈CarJ× Dand there existsa∈0,∞such that
a≤ft, x0, . . . , x2m, 1.7
H2For a.e.t∈Jand for allx0, . . . , x2m∈ D, the inequality
ft, x0, . . . , x2m≤h ⎛ ⎝t,2m
j0
xj⎞⎠2m j0
ωjxj 1.8
is fulfilled, whereh∈ CarJ×0,∞is positive and nondecreasing in the second variable,ωj:Ê → Êis nonincreasing, 0≤j ≤2m,
lim sup
v→ ∞ 1 v
T
0
ht, Kvdt <1, K
⎧ ⎪ ⎨ ⎪ ⎩
T2m1−1
T−1 if T /1,
2m1 if T 1,
1
0 ω2j
s2ds <∞,
1
0
ω2j1sds <∞ if 0≤j≤m−1,
1
0
ω2msds <∞.
1.9
The paper is organized as follows. In Section2, we construct a sequence of auxiliary regular differential equations associated with 1.1. Section 3 is devoted to the study of auxiliary regular complementary Lidstone problems. We show that the solvability of these problems is reduced to the existence of a fixed point of an operatorH. The existence of a fixed
point ofHis proved by a fixed point theorem of cone compression type according to
Guo-Krasnosel’skii18,19. The properties of solutions to auxiliary problems are also investigated here. In Section4, applying the results of Section3, the existence of a positive solution of the singular problem1.1,1.2is proved.
2. Regularization
Let mbe from1.1. For n ∈ Æ, define χn, ϕn, τn,m ∈ C
0
Ê, Ên ⊂ Ê, andDn ⊂ Ê
2m1 by
the formulas
χnu ⎧ ⎪ ⎨ ⎪ ⎩
u foru≥ 1
n, 1
n foru < 1 n,
ϕnu ⎧ ⎪ ⎨ ⎪ ⎩
−1n foru >−1 n,
u foru≤ −1
n,
τn,m ⎧ ⎨ ⎩
χn ifm2k−1,
ϕn ifm2k,
Ên
−∞,−1
n
∪
1 n,∞
,
DnÊ
2×
Ên×Ê×Ên×Ê× · · · ×Ê×Ên
2m1
.
Letf∈CarJ× D. Chosen∈Æand put
f∗
nt, x0, x1, x2, x3, x4, . . . , x2m−1, x2m
ft, χnx0, χnx1, x2, ϕnx3, x4, . . . , τn,mx2m−1, x2m
2.2
fort, x0, x1, x2, x3, x4, . . . , x2m−1, x2m∈J× Dn. Now, define an auxiliary functionfnby means
of the following recurrence formulas:
fn,0t, x0, x1, . . . , x2m fn∗t, x0, x1, . . . , x2m fort, x0, x1, . . . , x2m∈J× Dn,
fn,it, x0, x1, . . . , x2m
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
fn,i−1t, x0, x1, . . . , x2m if|x2i| ≥
1 n, n
2
fn,i−1
t, x0, . . . , x2i−1,n1, x2i1, . . . , x2m
x2i1n
−fn,i−1
t, x0, . . . , x2i−1,−1n, x2i1, . . . , x2m
x2i− 1n
if|x2i|< 1 n,
2.3
for 1≤i≤m, and
fnt, x0, x1, . . . , x2m fn,mt, x0, x1, . . . , x2m fort, x0, x1, . . . , x2m∈J×Ê
2m1. 2.4
Then, under conditionH1,fn∈CarJ×Ê
2m1and
a≤fnt, x0, x1, . . . , x2m for a.e. t∈J and allx0, x1, . . . , x2m∈Ê
2m1. 2.5
ConditionH2gives
fnt, x0, x1, . . . , x2m≤h ⎛
⎝t,2m12m j0
xj⎞⎠2m j0
ωjxjωj1,
for a.e. t∈J and allx0, x1, . . . , x2m∈Ê
2m1
0 ,
2.6
fnt, x0, x1, . . . , x2m≤h ⎛
⎝t,2m12m j0
xj⎞⎠2m j0
ωj
1 n
,
for a.e. t∈J and allx0, x1, . . . , x2m∈Ê
2m1.
2.7
We investigate the regular differential equation
−1mx2m1t f
n
t, xt, . . . , x2mt. 2.8
3. Auxiliary Regular Problems
Letj∈Æ and denote byGjt, sthe Green function of the problem
x2jt 0, x2i0 x2iT 0, 0≤i≤j−1. 3.1
Then,
G1t, s
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
s
Tt−T for 0≤s≤t≤T, t
Ts−T for 0≤t≤s≤T.
3.2
By2,3,20, the Green functionGjcan be expressed as
Gjt, s T
0
G1t, τGj−1τ, sdτ, j >1, 3.3
and it is known thatsee, e.g.,3,20
−1jGjt, s>0 fort, s∈0, T×0, T, j≥1. 3.4
Lemma 3.1see10, Lemmas 2.1 and 2.3. Fort, s∈J×Jandj∈Æ, the inequalities
−1jGjt, s≤T
2j−3
6j−1 sT−s, 3.5
−1jGjt, s≥ T
2j−5
30j−1tsT−tT−s 3.6
hold.
Letγ∈L1Jand letu∈AC2m−1Jbe a solution of the differential equation
−1mu2mt γt, 3.7
satisfying the Lidstone boundary conditions
u2i0 u2iT 0, 0≤i≤m−1. 3.8
It follows from the definition of the Green functionGjthat
−1ju2jt −1m−j
T
0
It is easy to check thatx∈AC2mJis a solution of problem2.8,1.2if and only ifx0 0, and its derivativexis a solution of a problem involving the functional differential equation
−1mu2mt fn
t,
t
0
usds, ut, . . . , u2m−1t
3.10
and the Lidstone boundary conditions3.8. From3.9 forj0, we see thatu∈AC2m−1J is a solution of problem3.10,3.8exactly if it is a solution of the equation
ut −1m
T
0
Gmt, sfn
s,
s
0
uτdτ, us, . . . , u2m−1sds, 3.11
in the setC2m−1J. Consequently,xis a solution of problem2.8,1.2if and only if it is a solution of the equation
xt −1m
t
0
T
0
Gms, τfn
τ, xτ, . . . , x2mτdτ
ds, 3.12
in the setC2mJ. It means thatxis a solution of problem2.8,1.2ifxis a fixed point of the
operatorH:C2mJ → C2mJdefined as
Hxt −1m
t
0
T
0
Gms, τfn
τ, xτ, . . . , x2mτdτ
ds. 3.13
We prove the existence of a fixed point of H by the following fixed point result of cone
compression type according to Guo-Krasnosel’skiisee, e.g.,18,19.
Lemma 3.2. LetX be a Banach space, and letP ⊂ X be a cone inX. LetΩ1,Ω2be bounded open balls ofX centered at the origin withΩ1 ⊂Ω2. Suppose thatF:P ∩Ω2\Ω1 → P is completely continuous operator such that
Fx ≥ x forx∈P∩∂Ω1, Fx ≤ x forx∈P∩∂Ω2 3.14
holds. Then,Fhas a fixed point inP∩Ω2\Ω1.
We are now in the position to prove that problem2.8,1.2has a solution.
Lemma 3.3. Let (H1) and (H2) hold. Then, problem2.8,1.2has a solution.
Proof. Let the operatorH:C2mJ → C2mJbe given in3.13, and let
P x∈C2mJ:xt≥0 for t∈J. 3.15
Then,P is a cone inC2mJand since−1mG
mt, s>0 fort, s∈0, T×0, Tby3.4and
operator follows fromfn∈CarJ×Ê
2m1, from Lebesgue dominated convergence theorem, and from the Arzel`a-Ascoli theorem.
Choosex∈Pand putyt Hxtfort∈J. Then,cf.2.5
−1my2m1t fn
t, xt, . . . , x2mt≥a >0 for a.e. t∈J. 3.16
Sincey0 0 andy2i−10 y2i−1T 0 for 1≤i≤m, the equalityyjξ
j 0 holds with
someξj∈Jfor 0≤j ≤2m. We now use the equalityy2mξ2m 0 and have
y2mt
t
ξ2m
y2m1sds
≥a|t−ξ2m| fort∈J. 3.17
Hence,y2m
∞≥aT/2, and so
HxC2m > aT
2 . 3.18
Next, we deduce from the relation
y2mt
t
ξ2m
fn
s, xs, . . . , x2msds
≤ T
0 fn
s, xs, . . . , x2msds 3.19
and from2.7that
y2mt≤T 0
hs,2m1xC2mdsT
2m
j0 ωj
1 n
fort∈J. 3.20
Therefore,
y2m
∞≤
T
0
hs,2m1xC2mdsV, 3.21
whereV T2jm0ωj1/n. Sinceyjξj 0 for 0≤j≤2m, we have
yj ∞≤T
2m−jy2m
∞, 0≤j≤2m. 3.22
The last inequality together with3.21gives
yC2m ≤Ky2m
∞≤K
T
0
hs,2m1xC2mdsV
, 3.23
x∈P, inequalities3.18and
HxC2m ≤K T
0
hs,2m1xC2mdsV
3.24
hold. ByH2, there existsC >0 such that
1 v
T
0
hs,2m1KvdsV
≤1 ∀v≥ C
K, 3.25
and therefore,
K
T
0
hs,2m1vdsV
≤v ∀v≥C. 3.26
Let
Ω1
x∈C2mJ:x
C2m < aT
2 , Ω2
x∈C2mJ:x
C2m < C . 3.27
Then, it follows from3.18,3.24, and3.26that
HxC2m ≥ xC2m forx∈P∩∂Ω1, HxC2m ≤ xC2m forx∈P∩∂Ω2. 3.28
The conclusion now follows from Lemma3.2forXC2mJandFH.
The properties of solutions to problem2.8,1.2are collected in the following lemma.
Lemma 3.4. Let (H1) and (H2) be satisfied. Letxnbe a solution of problem2.8,1.2. Then, for all
n∈Æ, the following assertions hold:
i −1jxn2j1t>0fort∈0, T,0≤j≤m−1, and−1mxn2m1t≥afor a.e.t∈J,
iixnis increasing onJ, and for0≤j ≤m−1,−1jxn2j2is decreasing onJ, and there is a uniqueξj,n∈0, Tsuch thatxn2j2ξj,n 0,
iiithere exists a positive constantAsuch that
x2m
n t≥A|t−ξm−1,n|,
x2j2
n t≥At−ξj,n2 if 0≤j≤m−2,
x2j1
n t≥AtT−t if0≤j≤m−1,
xnt≥At2,
3.29
fort∈J,
Proof. Let us choose an arbitraryn∈Æ. By2.5,
−1mxn2m1t fnt, xnt, . . . , xn2mt≥a for a.e. t∈J, 3.30
and it follows from the definition of the Green functionGjthat the equality
−1jxn2j1t −1m−j T
0
Gm−jt, sfn
s, xns, . . . , xn2ms
ds 3.31
holds fort ∈ J and 0 ≤ j ≤ m−1. Now, using 1.2,3.4,3.30, and 3.31, we see that assertioniis true. Hence,−1jxn2j2is decreasing onJfor 0≤j ≤m−1 andxnis increasing
on this interval. Due toxn2i−10 xn2i−1T 0 for 1≤i≤m, there exists a uniqueξj,n∈0, T
such thatun2j2ξj,n 0 for 0≤j≤m−1. Consequently, assertioniiholds.
Next, in view of2.5,3.6, and3.31,
x2j1
n t≥ T
2m−j−5a
30m−j−1 tT−t
T
0
sT−sds
T2m−j−2a
6·30m−j−1tT−t fort∈J, 0≤j ≤m−1.
3.32
Since
x2j2
n t t
ξj,n
x2j3
n sds 3.33
and, by13, Lemma 6.2,
t
ξj,n
sT−sds≥ T
6
t−ξj,n2, 3.34
we have
x2j2
n t≥ T
2m−j−3a
36·30m−j−2
t−ξj,n2 fort∈J, 0≤j≤m−2. 3.35
Furthermore,
x2m
n t
t
ξm−1,n
fn
s, xns, . . . , xn2ms
andcf.3.32forj0
xnt t
0 x
nsds≥ T
2m−2a
6·30m−1
t
0
sT−sds
T2m−2a
36·30m−1t
23T−2t≥ T2m−1a 36·30m−1t
2 fort∈J,
3.37
sincexn>0 on0, Tby assertionii. Let
Aa·min
!
1, A1, A2, T2m−1 36·30m−1
"
, 3.38
where
A1min
!
T2m−j−2
6·30m−j−1 : 0≤j≤m−1
"
,
A2min
!
T2m−j−3
36·30m−j−2 : 0≤j ≤m−2
"
.
3.39
Then estimate3.29follows from relations3.32–3.37.
It remains to prove the boundedness of the sequence{xn}inC2mJ. We use estimate
3.29, the properties ofωjgiven inH2, and the inequality
tT−t≥ ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
T
2t for 0< t≤
T 2, T
2T−t for T
2 < t < T
3.40
and have
T
0
ω2mxn2ms
ds≤ T
0
ω2mA|s−ξm−1,n|ds
A1
Aξm−1,n
0
ω2msds
AT−ξm−1,n
0
ω2msds
< A2
AT
0
ω2msds, T
0
ω2j2xn2j2s
ds≤ T
0 ω2j2
As−ξj,n2
ds
√1
A
√
AT−ξj,n
−√Aξj,n
ω2j2
< √2
A
√
AT
0
ω2j2
s2ds for 0≤j≤m−2,
T
0
ω2j1xn2j1s
ds≤ T
0
ω2j1AsT−sds
<
T/2
0 ω2j1
ATs 2 ds T T/2 ω2j1
ATT−s
2
ds
< AT4
AT2/4
0
ω2j1sds for 0≤j ≤m−1,
T
0
ω0|xns|ds≤ T
0
ω0As2ds √1
A
√
AT
0
ω0s2ds.
3.41
In particular,
T
0
ω2mxn2ms
ds < A2
AT
0
ω2msds,
T
0
ω2j2xn2j2s
ds < √2
A
√
AT
0
ω2j2
s2ds for 0≤j≤m−2,
T
0
ω2j1xn2j1s
ds < AT4
AT2/4
0
ω2j1sds for 0≤j≤m−1,
T
0
ω0|xns|ds≤ √1
A
√
AT
0
ω0s2ds,
3.42
for alln∈Æ. Now, from the above estimates, from2.6and fromx
2m
n ξm−1,n 0 for some
ξm−1,n∈0, T, which is proved inii, we get
x2m
n t
t
ξm−1,n
fn
s, xns, . . . , xn2ms ds ≤ T 0 fn
s, xns, . . . , xn2ms ds ≤ T 0 h ⎛
⎝s,2m12m j0
xnjs
⎞
⎠ds2m j0
T
0
ωjxnjs
ωj1 ds < T 0 h ⎛
⎝s,2m12m j0
xnj∞
⎞ ⎠ds Λ,
where
Λ A2
AT
0
ω2msds√2
A
m−2
j0
√
AT
0
ω2j2
s2ds
AT4 m−1
j0
AT2/4
0
ω2j1sds 1
√
A
√
AT
0
ω0s2dsT2m
j0 ωj1.
3.44
Notice thatΛ<∞byH2. Consequently,
x2m
n ∞< T
0 h
⎛
⎝s,2m12m j0
xnj∞
⎞
⎠ds Λ forn∈Æ. 3.45
Sincexnj∞ ≤T2m−jx2m
n ∞for 0≤j ≤2m, which follows from the fact thatxnjvanishes
inJby1.2and assertionii, inequality3.45yields
x2m
n ∞< T
0
hs,2m1Kxn2m∞
ds Λ forn∈Æ, 3.46
whereKis fromH2. Due to the condition
lim sup
v→ ∞ 1 v
T
0
ht, Kvdv <1 3.47
inH2, there exists a positive constantSsuch that for allv≥Sthe inequality
T
0
ht,2m1Kvdt Λ≤v 3.48
is fulfilled. The last inequality together with estimate3.46givesxn2m∞ < Sforn ∈ Æ.
Consequently,xnj∞< T2m−jSfor 0≤j≤2n,n∈
Æ, and assertionivfollows.
The following result gives the important property of {fnt, xnt, . . . , xn2mt} for
applying the Vitali convergent theorem in the proof of Theorem4.1.
Lemma 3.5. Let (H1) and (H2) hold. Letxnbe a solution of problem2.8,1.2. Then, the sequence
fn
t, xnt, . . . , xn2mt
⊂L1J 3.49
is uniformly integrable on J, that is, for each ε > 0, there exists δ > 0 such that ifM ⊂ J and measM< δ, then
Mfn
t, xnt, . . . , xn2mt
Proof. By Lemma3.4iv, there existsS > 0 such that forn∈ Æ, the inequalityxn C2m < S
holds. Now, we conclude from2.5and2.6, from the properties ofhandωjgiven inH2,
and finally from3.29that forj∈Jandn∈Æ, the estimate
a≤fn
t, xnt, . . . , xn2mt
≤ht,2m1S ω0At2
m−1
j0
ω2j1AtT−t
m−2
j0 ω2j2
At−ξj,n2
ω2mA|t−ξm−1,n|
m
j0 ωj1
3.51
is fulfilled, whereAis a positive constant. Since the functionsht,2m1S,ω0At2, and ω2j1AtT −t 0 ≤ j ≤ m−1 belong to the setL1J by assumption H2, in order to prove that{fnt, xnt, . . . , xn2mt}is uniformly integrable onJ, it suffices to show that the
sequences
{ω2mA|t−ξm−1,n|},
ω2j2
At−ξj,n2
, 0≤j≤m−2 3.52
are uniformly integrable onJ. Due to#01ω2msds <∞and#01ω2js2ds <∞for 1≤j≤m−1
byH2, this fact follows from13, Criterion 11.10withbAandr1,2.
4. The Main Result
The following theorem is the existence result for the singular problem1.1,1.2.
Theorem 4.1. Let (H1) and (H2) hold. Then, problem 1.1, 1.2 has a positive solution x∈AC2mJand
xt>0 fort∈0, T, −1jx2j1t>0 fort∈0, T, 0≤j≤m−1. 4.1
Proof. Lemma 3.3guarantees that problem 2.8, 1.2 has a solution xn. Consider the
se-quence{xn}. By Lemma3.4,{xn}is bounded inC2mJ,
−1jxn2j1t>0 fort∈0, T, 0≤j≤m−1, 4.2
andxnfulfils estimate3.29, whereAis a positive constant andξj,n ∈ 0, T. Furthermore,
the sequence {fnt, xnt, . . . , xn2mt} is uniformly integrable on J by Lemma 3.5, and
therefore, we deduce from the equality−1mxn2m1t fnt, xnt, . . . , xn2mtfor a.e.t∈J
that{xn2m} is equicontinuous onJ. Now, by the Arzel`a-Ascoli theorem and the Bolzano-Weierstrass theorem, we may assume without loss of generality that{xn}is convergent in C2mJand{ξ
0≤j≤m−1. Thenx∈C2mJandxsatisfies the boundary conditions1.2. Lettingn → ∞ in3.29and4.2, we getfort∈J
x2mt≥A|t−ξ
m−1|, x2j2t≥A
t−ξj2 if 0≤j≤m−2
−1jx2j1t≥AtT−t if 0≤j≤m−1, xt≥At2.
4.3
Keeping in mind the definition offn, we conclude from4.3that
lim
n→ ∞fn
t, xnt, . . . , xn2mt
ft, xt, . . . , x2mt for a.e. t∈J. 4.4
Then, by the Vitali theorem,ft, xt, . . . , x2mt∈L1Jand
lim
n→ ∞
t
0 fn
s, xns, . . . , xn2ms
ds
t
0
fs, xs, . . . , x2msds fort∈J. 4.5
Lettingn → ∞in the equality
x2m
n t xn2m0 t
0 fn
s, xns, . . . , xn2ms
ds, 4.6
we get
x2mt x2m0 t
0
fs, xs, . . . , x2msds fort∈J. 4.7
As a result,x∈AC2mJandxis a solution of1.1. Consequently,xis a positive solution of problem1.1,1.2and inequality4.1follows from4.3.
Example 4.2. Consider problem1.1,1.2with
ft, x0, . . . , x2m pt
2m
k0
akt|xk|αk bkt |xk|βk
4.8
onJ×D, wherep, ak∈L1J,bk∈L∞J that is,bkis essentially bounded and measurable on
Jare nonnegative,pt≥a >0 for a.e.t∈J. Ifαk∈0,1for 0≤k ≤2mandβ2k ∈0,1/2, β2m, β2k1∈0,1for 0≤k≤m−1, then, by Theorem4.1, the problem has a positive solution x∈AC2mJsatisfying inequality4.1.
Acknowledgment
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