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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 t

Volume 2010, Article ID 368169, 15 pages

doi:10.1155/2010/368169

Research Article

Positive Solutions of Singular Complementary

Lidstone Boundary Value Problems

Ravi P Agarwal,1 Donal O’Regan,2 and Svatoslav Stan ˇek3

1 Department of Mathematical Sciences, Florida Institute of Technology, Melbourne,

FL 32901-6975, USA

2 Department of Mathematics, National University of Ireland, Galway, Ireland

3 Department 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,agarwal@fit.edu

Received 7 October 2010; Accepted 21 November 2010

Academic Editor: Irena Rach ˚unkov´a

Copyrightq 2010 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−1m x 2m1  ft, x, ,

x 2m , x0  0, x 2i−1 0  x 2i−1 T  0, 1 ≤ i ≤ m Here, m ≥ 1 and the Carath´eodory function

ft, x0, , x2m  may be singular in all its space variables x0, , x2m The results are proved by regularization and sequential techniques In limit processes, the Vitali convergence theorem is used

1 Introduction

LetT be a positive constant, J  0, T andÊ −  −∞, 0,Ê   0, ∞,Ê 0 Ê\{0} We consider the singular complementary Lidstone boundary value problem

wheref satisfies the local Carath´eodory function on J × D f ∈ CarJ × D with

D 

⎧

⎪

⎨

⎪

⎩

Ê 2

 ×Ê 0×Ê − ×Ê 0×Ê  × · · · ×Ê ×Ê 0

4k−1

ifm  2k − 1,

Ê 2

 ×Ê 0×Ê − ×Ê 0×Ê  × · · · ×Ê −×Ê 0

4k1

The functionft, x0, , x2 m is positive and may be singular at the value zero of all its space variablesx0, , x2 m

Leti ∈ {0, 1, , 2m} We say that f is singular at the value zero of its space variable x iif for a.e.t ∈ J and all x j, 0≤ j ≤ 2m, j / i such that x0, , xi , , x2 m ∈ D, the relation

lim

holds

A functionx ∈ AC2m J i.e., x has absolutely continuous 2mth derivative on J is

a positive solution of problem 1.1, 1.2 if xt > 0 for t ∈ 0, T, x satisfies the boundary

conditions1.2 and 1.1 holds a.e on J.

The regular complementary Lidstone problem

−1m x 2m1 t  ht, xt, , x q t, m ≥ 1, q fixed, 0 ≤ q ≤ 2m,

x0  α0, x 2i−1 0  α i , x 2i−1 1  β i , 1 ≤ i ≤ m

1.5

was discussed in1 Here, h : 0, 1 ×Ê

q1 → 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

−1m x 2m t  pt, xt, , x r t, m ≥ 1, r fixed, 0 ≤ r ≤ 2m − 1,

x 2i 0  a i , x 2i 1  b i , 1 ≤ i ≤ m − 1,

1.6

where the differential equation and derivatives in the boundary conditions are even orders Fora i  b i  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 in 1.1 which gua-rantee that the singular problem1.1, 1.2 has a solution The existence results are proved

by regularization and sequential techniques, and in limit processes, the Vitali convergence theorem16,17 is applied

Throughout the paper,x∞  max{|xt| : t ∈ J} and x C n  n

k0 x k∞,n ≥ 1

stands for the norm in C0J and C n J, respectively L1J denotes the set of functions

Lebesgue integrable on J and meas M the Lebesgue measure of M ⊂ J.

We work with the following conditions on the functionf in 1.1

H1 f ∈ CarJ × D and there exists a ∈ 0, ∞ such that

for a.e.t ∈ J and each x0, , x2 m ∈ D

H2 For a.e t ∈ J and for all x0, , x2 m ∈ D, the inequality

ft, x0, , x2 m  ≤ h

⎛

⎝t,2m

j0

x j⎞⎠ 2m

j0

ω j x j 1.8

is fulfilled, whereh ∈ CarJ × 0, ∞ is positive and nondecreasing in the second

variable,ω j:Ê  → Ê is nonincreasing, 0≤ j ≤ 2m,

lim sup

v → ∞

1

v

T

0 ht, Kvdt < 1, K 

⎧

⎪

⎪

T2m1− 1

1

0

ω2 j



s2

ds < ∞,

1

0

ω2 j1 sds < ∞ if 0 ≤ j ≤ m − 1,

1

0

ω2 m sds < ∞.

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 ofH is 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, 1.2 is proved

2 Regularization

Let m be from 1.1 For n ∈ Æ, define χ n , ϕ n , τ n,m ∈ C0Ê, Ên ⊂ Ê, andDn ⊂ Ê

2m1 by the formulas

χ n u 

⎧

⎪

⎪

u for u ≥ 1n ,

1

n foru <

1

n ,

ϕ n u 

⎧

⎪

⎪

−1n foru > −1n ,

n ,

τ n,m

⎧

⎨

⎩

χ n ifm  2k − 1,

ϕ n ifm  2k,

Ên 



−∞, −1

n



∪

 1

n , ∞



,

DnÊ

2×Ên×Ê×Ên×Ê× · · · ×Ê×Ên

2m1

.

2.1

Letf ∈ CarJ × D Chose n ∈Æand put

f∗

n t, x0, x1, x2, x3, x4, , x2m−1 , x2 m

 ft, χ n x0, χ n x1, x2, ϕn x3, x4, , τn,m x2m−1 , x2m 2.2

fort, x0, x1, x2, x3, x4, , x2m−1 , x2 m  ∈ J × D n Now, define an auxiliary functionf nby means

of the following recurrence formulas:

f n,0 t, x0, x1, , x2m   f∗

n t, x0, x1, , x2m  for t, x0, x1, , x2m  ∈ J × D n ,

f n,i t, x0, x1, , x2m



⎧

⎪

⎪

⎪

⎪

⎪

⎪

f n,i−1 t, x0, x1, , x2m if|x2i| ≥ n1,

n

2



f n,i−1



t, x0, , x2 i−1 , n1, x2 i1 , , x2 m



x2 i1n



−f n,i−1



t, x0, , x2 i−1 , −1

n , x2 i1 , , x2 m



x2 i− 1

n



if|x2i | < n1,

2.3

for 1≤ i ≤ m, and

f n t, x0, x1, , x2m   f n,m t, x0, x1, , x2m  for t, x0, x1, , x2m  ∈ J ×Ê

2m1 2.4 Then, under conditionH1, f n ∈ CarJ ×Ê

2m1 and

a ≤ f n t, x0, x1, , x2m  for a.e t ∈ J and allx0, x1, , x2m ∈Ê

2m1 2.5 ConditionH2 gives

f n t, x0, x1, , x2m  ≤ h

⎛

⎝t, 2m  1 2m

j0

x j⎞⎠ 2m

j0



ω j x j   ω j1,

for a.e t ∈ J and all x0 , x1, , x2 m ∈Ê

2m1

2.6

f n t, x0, x1, , x2m  ≤ h

⎛

⎝t, 2m  1 2m

j0

x j⎞⎠ 2m

j0

ω j

 1

n



,

for a.e t ∈ J and all x0, x1, , x2 m ∈Ê

2m1

2.7

We investigate the regular differential equation

−1m x 2m1 t  f n



If a functionx ∈ AC2m J satisfies 2.8 for a.e t ∈ J, then x is called a solution of 2.8

3 Auxiliary Regular Problems

Letj ∈Æ and denote byG j t, s the Green function of the problem

x 2j t  0, x 2i 0  x 2i T  0, 0 ≤ i ≤ j − 1. 3.1 Then,

G1 t, s 

⎧

⎪

⎪

s

T t − T for 0 ≤ s ≤ t ≤ T, t

T s − T for 0 ≤ t ≤ s ≤ T.

3.2

By2,3,20, the Green function G jcan be expressed as

G j t, s 

T

0

G1 t, τG j−1 τ, sdτ, j > 1, 3.3 and it is known thatsee, e.g., 3,20

−1j G j t, s > 0 for t, s ∈ 0, T × 0, T, j ≥ 1. 3.4

Lemma 3.1 see 10, Lemmas 2.1 and 2.3 For t, s ∈ J × J and j ∈Æ, the inequalities

−1j G j t, s ≤ T2j−3

−1j G j t, s ≥ T2j−5

hold.

Letγ ∈ L1J and let u ∈ AC2m−1 J be a solution of the differential equation

−1m u 2m t  γt, 3.7 satisfying the Lidstone boundary conditions

u 2i 0  u 2i T  0, 0 ≤ i ≤ m − 1. 3.8

It follows from the definition of the Green functionG jthat

−1j u 2j t  −1 m−jT

0

G m−j t, sγsds for t ∈ J, 0 ≤ j ≤ m − 1. 3.9

It is easy to check thatx ∈ AC2m J is a solution of problem 2.8, 1.2 if and only if x0  0,

and its derivativexis a solution of a problem involving the functional differential equation

−1m u 2m t  f n



t,

t

0

usds, ut, , u 2m−1 t



3.10

and the Lidstone boundary conditions3.8 From 3.9 for j  0, we see that u ∈ AC2m−1 J

is a solution of problem3.10, 3.8 exactly if it is a solution of the equation

ut  −1 m

T

0

G m t, sf n



s,

s

0uτdτ, us, , u 2m−1 s



ds, 3.11

in the setC2m−1 J Consequently, x is a solution of problem 2.8, 1.2 if and only if it is a solution of the equation

xt  −1 m

t

0

T

0

G m s, τf n



τ, xτ, , x 2m τdτ



ds, 3.12

in the setC2m J It means that x is a solution of problem 2.8, 1.2 if x is a fixed point of the

operatorH : C2m J → C2m J defined as

Hxt  −1 mt

0

T

0

G m s, τf nτ, xτ, , x 2m τdτ



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 Let X be a Banach space, and let P ⊂ X be a cone in X Let Ω1, Ω2be bounded open balls of X centered at the origin with Ω1 ⊂ Ω2 Suppose thatF : P ∩ Ω2\Ω1 → P is completely

continuous operator such that

Fx ≥ x for x ∈ P ∩ ∂Ω1, Fx ≤ x for x ∈ P ∩ ∂Ω2 3.14

holds Then, F has a fixed point in P ∩ Ω2\ Ω1.

We are now in the position to prove that problem2.8, 1.2 has a solution

Lemma 3.3 Let (H1) and (H2 ) hold Then, problem2.8, 1.2 has a solution.

Proof Let the operator H : C2m J → C2m J be given in 3.13, and let

Then,P is a cone in C2m J and since −1 m G m t, s > 0 for t, s ∈ 0, T × 0, T by 3.4 and

f nsatisfies2.5, we see that H : C2m J → P The fact that H is a completely continuous

operator follows fromf n ∈ CarJ ×Ê

2m1, from Lebesgue dominated convergence theorem, and from the Arzel`a-Ascoli theorem

Choosex ∈ P and put yt  Hxt for t ∈ J Then, cf 2.5

−1m y 2m1 t  f n



t, xt, , x 2m t≥ a > 0 for a.e t ∈ J. 3.16

Sincey0  0 and y 2i−1 0  y 2i−1 T  0 for 1 ≤ i ≤ m, the equality y j ξ j  0 holds with someξ j ∈ J for 0 ≤ j ≤ 2m We now use the equality y 2m ξ2m  0 and have



y 2m t 



t

ξ2m

y 2m1 sds



 ≥ a|t − ξ2m | for t ∈ J. 3.17 Hence,y 2m∞≥ aT/2, and so

Hx C2m > aT

Next, we deduce from the relation



y 2m t 



t

ξ2m

f n



s, xs, , x 2m sds



 ≤

T

0

f n



s, xs, , x 2m sds 3.19

and from2.7 that



y 2m t ≤T

0

hs, 2m  1  x C2m ds  T2m

j0

ω j1 n



Therefore,

y 2m

T

whereV  T2m

j0 ω j 1/n Since y j ξ j   0 for 0 ≤ j ≤ 2m, we have

y j

∞≤ T2m−jy 2m

The last inequality together with3.21 gives

y C2m ≤ Ky 2m

T

0 hs, 2m  1  x C2m ds  V



, 3.23 whereK is from H2  Since x ∈ P is arbitrary, relations 3.18 and 3.21 imply that for all

x ∈ P, inequalities 3.18 and

Hx C2m ≤ K

T

0 hs, 2m  1  x C2m ds  V



3.24 hold ByH2, there exists C > 0 such that

1

v

T

0 hs, 2m  1  Kvds  V



≤ 1 ∀v ≥ K C , 3.25 and therefore,

K

T

0 hs, 2m  1  vds  V



≤ v ∀v ≥ C. 3.26 Let

Ω1

x ∈ C2m J : x C2m < aT

2 , Ω2 

x ∈ C2m J : x C2m < C 3.27 Then, it follows from3.18, 3.24, and 3.26 that

Hx C2m ≥ x C2m forx ∈ P ∩ ∂Ω1, Hx C2m ≤ x C2m forx ∈ P ∩ ∂Ω2. 3.28 The conclusion now follows from Lemma3.2for X  C2m J and F  H.

The properties of solutions to problem2.8, 1.2 are collected in the following lemma

Lemma 3.4 Let (H1) and (H2) be satisfied Letx n be a solution of problem2.8, 1.2 Then, for all

n ∈Æ, the following assertions hold:

i −1j x n 2j1 t > 0 for t ∈ 0, T, 0 ≤ j ≤ m − 1, and −1 m x 2m1 n t ≥ a for a.e t ∈ J,

ii x n is increasing on J, and for 0 ≤ j ≤ m − 1, −1 j x 2j2 n is decreasing on J, and there is a unique ξ j,n ∈ 0, T such that x n 2j2 ξ j,n   0,

iii there exists a positive constant A such that



x 2m n t ≥ A|t − ξ

m−1,n |,



x 2j2 n t ≥ A

t − ξ j,n2

if 0 ≤ j ≤ m − 2,



x 2j1 n t ≥ AtT − t if 0 ≤ j ≤ m − 1,

x n t ≥ At2,

3.29

for t ∈ J,

iv the sequence {x n } is bounded in C2m J.

Proof Let us choose an arbitrary n ∈Æ By2.5,

−1m x 2m1 n t  f nt, x n t, , x 2m n t≥ a for a.e t ∈ J, 3.30 and it follows from the definition of the Green functionG jthat the equality

−1j x 2j1 n t  −1 m−jT

0

G m−j t, sf ns, x n s, , x 2m n sds 3.31

holds fort ∈ J and 0 ≤ j ≤ m − 1 Now, using 1.2, 3.4, 3.30, and 3.31, we see that assertioni is true Hence, −1j x 2j2 n is decreasing onJ for 0 ≤ j ≤ m−1 and x nis increasing

on this interval Due tox 2i−1 n 0  x 2i−1 n T  0 for 1 ≤ i ≤ m, there exists a unique ξ j,n ∈ 0, T

such thatu 2j2 n ξ j,n   0 for 0 ≤ j ≤ m − 1 Consequently, assertion ii holds.

Next, in view of2.5, 3.6, and 3.31,



x n 2j1 t ≥ T2m−j−5a

30m−j−1 tT − t

T

0

sT − sds

 T2m−j−2a

6· 30m−j−1 tT − t for t ∈ J, 0 ≤ j ≤ m − 1.

3.32

Since

x n 2j2 t 

t

ξ j,n

and, by13, Lemma 6.2,







t

ξ j,n

sT − sds



 ≥

T

6



t − ξ j,n2

we have



x 2j2 n t ≥ T2m−j−3a

36· 30m−j−2



t − ξ j,n2

Furthermore,



x 2m n t 



t

ξ f n



s, x n s, , x 2m n sds



 ≥ a|t − ξ m−1,n |, t ∈ J, 3.36

andcf 3.32 for j  0

x n t 

t

0

x

n sds ≥ T2m−2 a

6· 30m−1

t

0sT − sds

 T2m−2 a

36· 30m−1 t23T − 2t ≥ T2m−1 a

36· 30m−1 t2 fort ∈ J,

3.37

sincex

n > 0 on 0, T by assertion ii Let

A  a · min

!

1, A1, A2, T2m−1

36· 30m−1

"

where

A1 min

!

T2m−j−2

6· 30m−j−1 : 0≤ j ≤ m − 1

"

, A2 min

!

T2m−j−3

36· 30m−j−2 : 0≤ j ≤ m − 2

"

.

3.39

Then estimate3.29 follows from relations 3.32–3.37

It remains to prove the boundedness of the sequence{x n } in C2m J We use estimate

3.29, the properties of ω jgiven 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

ω2 mx 2m

n s

ds ≤

T

0

ω2 m A|s − ξ m−1,n |ds

 A1

Aξ m−1,n

0

ω2 m sds 

AT−ξ m−1,n

0

ω2 m sds



< A2

AT

0

ω2 m sds,

T

0

ω2 j2x 2j2

n s

ds ≤

T

0

ω2 j2



As − ξ j,n2

ds

 √1

A

√

AT−ξ j,n



s2

ds

< √2

A

√

AT

0

ω2 j2s2

ds for 0 ≤ j ≤ m − 2,

T

0

ω2 j1x 2j1

n s

ds ≤

T

0

ω2 j1 AsT − sds

<

T/2

0

ω2 j1

ATs 2



ds 

T

T/2 ω2j1

2



ds

< AT4

AT2/4

0

ω2 j1 sds for 0 ≤ j ≤ m − 1,

T

0

ω0 |x n s|ds ≤

T

0

ω0As2

ds  √1

A

√

AT

0

ω0s2

ds.

3.41

In particular,

T

0

ω2 mx 2m

n s

ds < A2

AT

0

ω2 m sds,

T

0

ω2 j2x 2j2

n s

ds < √2

A

√

AT

0

ω2 j2s2

ds for 0 ≤ j ≤ m − 2,

T

0

ω2 j1x 2j1

n s

ds < AT4

AT2/4

0

ω2 j1 sds for 0 ≤ j ≤ m − 1,

T

0

ω0 |x n s|ds ≤ √1

A

√

AT

0

ω0s2

ds,

3.42

for alln ∈Æ Now, from the above estimates, from2.6 and from x 2m n ξ m−1,n  0 for some

ξ m−1,n ∈ 0, T, which is proved in ii, we get



x 2m n t 



t

ξ m−1,n

f ns, x n s, , x 2m n sds





≤

T

0

f n



s, x n s, , x 2m n sds

≤

T

0

h

⎛

⎝s, 2m  1 2m

j0



x j n s⎞⎠ds 2m

j0

T

0



ω jx j

n s

 ω j1ds

<

T

0

h

⎛

⎝s, 2m  1 2m

j0

x n j

∞

⎞

⎠ds  Λ,

3.43

2 Regularization