Hindawi Publishing Corporation
Boundary Value Problems
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 Staněk3
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, Palacký University, Tř. 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 Rachůnková
Copyright q 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 x2m1 f t, x, . . . ,
x2m , x0 0, x2i−1 0 x2i−1 T 0, 1 ≤ i ≤ m. Here, m ≥ 1 and the Carathéodory function
f t, 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
Let T be a positive constant, J 0, T and Ê− −∞, 0, Ê 0, ∞, Ê0 Ê \{0}. We consider
the singular complementary Lidstone boundary value problem
−1m x2m1 t f t, xt, . . . , x2m t ,
x0 0,
x2i−1 0 x2i−1 T 0,
m ≥ 1,
1 ≤ i ≤ m,
1.1
1.2
where f satisfies the local Carathéodory function on J × D f ∈ CarJ × D with
D
⎧ 2
Ê × Ê0 × Ê− × Ê0 × Ê × · · · × Ê × Ê0 if m 2k − 1,
⎪
⎪
⎪
⎨
4k−1
⎪
Ê2 × Ê0 × Ê− × Ê0 × Ê × · · · × Ê− × Ê0 if m 2k.
⎪
⎪
⎩ 4k1
1.3
2
Boundary Value Problems
The function ft, x0 , . . . , x2m is positive and may be singular at the value zero of all its space
variables x0 , . . . , x2m .
Let i ∈ {0, 1, . . . , 2m}. We say that f is singular at the value zero of its space variable xi if
for a.e. t ∈ J and all xj , 0 ≤ j ≤ 2m, j / i such that x0 , . . . , xi , . . . , x2m ∈ D, the relation
lim ft, x0 , . . . , xi , . . . , x2m ∞
1.4
xi → 0
holds.
A function x ∈ 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
conditions 1.2 and 1.1 holds a.e. on J.
The regular complementary Lidstone problem
−1m x2m1 t h t, xt, . . . , xq t ,
x0 α0 ,
x2i−1 0 αi ,
m ≥ 1, q fixed, 0 ≤ q ≤ 2m,
x2i−1 1 βi ,
1.5
1≤i≤m
was discussed in 1. Here, h : 0, 1 × Êq1 → Ê is 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 x2m t p t, xt, . . . , xr t ,
x2i 0 ai ,
m ≥ 1, r fixed, 0 ≤ r ≤ 2m − 1,
x2i 1 bi ,
1.6
1 ≤ i ≤ m − 1,
where the differential equation and derivatives in the boundary conditions are even orders.
For ai bi 0 1 ≤ i ≤ m − 1, regular Lidstone problems were discussed in 2–9, while
singular ones in 10–15.
The aim of this paper is to give the conditions on the function f in 1.1 which guarantee that the singular problem 1.1, 1.2 has a solution. The existence results are proved
by regularization and sequential techniques, and in limit processes, the Vitali convergence
theorem 16, 17 is applied.
Throughout the paper, x∞ max{|xt| : t ∈ J} and xCn nk0 xk ∞ , n ≥ 1
stands for the norm in C0 J and Cn J, respectively. L1 J 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 function f in 1.1.
H1 f ∈ CarJ × D and there exists a ∈ 0, ∞ such that
a ≤ ft, x0 , . . . , x2m ,
for a.e. t ∈ J and each x0 , . . . , x2m ∈ D.
1.7
Boundary Value Problems
3
H2 For a.e. t ∈ J and for all x0 , . . . , x2m ∈ D, the inequality
⎞
2m 2m
xj ⎠ ωj xj ft, x0 , . . . , x2m ≤ h⎝t,
⎛
j0
1.8
j0
is fulfilled, where h ∈ CarJ × 0, ∞ is positive and nondecreasing in the second
variable, ωj : Ê → Ê is nonincreasing, 0 ≤ j ≤ 2m,
lim sup
v→∞
1
1
v
T
ht, Kvdt < 1,
0
ω2j s2 ds < ∞,
0
1
1
⎧
2m1
⎪
−1
⎨T
T −1
K
⎪
⎩2m 1
if T / 1,
if T 1,
ω2j1 sds < ∞ if 0 ≤ j ≤ m − 1,
1.9
0
ω2m sds < ∞.
0
The paper is organized as follows. In Section 2, 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 operator H. The existence of a fixed
point of H is proved by a fixed point theorem of cone compression type according to GuoKrasnosel’skii 18, 19. The properties of solutions to auxiliary problems are also investigated
here. In Section 4, applying the results of Section 3, the existence of a positive solution of the
singular problem 1.1, 1.2 is proved.
2. Regularization
Let m be from 1.1. For n ∈
the formulas
χn u τn,m ⎧
⎪
⎨u
Æ,
define χn , ϕn , τn,m ∈ C0 Ê,
1
,
n
1
for u < ,
n
for u ≥
⎪
⎩1
n
⎧
⎨χn if m 2k − 1,
⎩ϕ
n
if m 2k,
⎧
1
⎪
⎨−
n
ϕn u ⎪
⎩u
Ên
−∞, −
Ên
Ê,
and Dn ⊂
Ê2m1
by
1
for u > − ,
n
1
for u ≤ − ,
n
1
1
∪
,∞ ,
n
n
Dn Ê2 × Ên × Ê × Ên × Ê × · · · × Ê × Ên .
2m1
⊂
2.1
4
Boundary Value Problems
Let f ∈ CarJ × D. Chose n ∈ Æ and put
fn∗ t, x0 , x1 , x2 , x3 , x4 , . . . , x2m−1 , x2m f t, χn x0 , χn x1 , x2 , ϕn x3 , x4 , . . . , τn,m x2m−1 , x2m
2.2
for t, x0 , x1 , x2 , x3 , x4 , . . . , x2m−1 , x2m ∈ J × Dn . Now, define an auxiliary function fn by means
of the following recurrence formulas:
fn,0 t, x0 , x1 , . . . , x2m fn∗ t, x0 , x1 , . . . , x2m for t, x0 , x1 , . . . , x2m ∈ J × Dn ,
fn,i t, x0 , x1 , . . . , x2m ⎧
⎪
⎪
⎪fn,i−1 t, x0 , x1 , . . . , x2m ⎪
⎪
⎪
⎪
⎪
1
1
⎨n
fn,i−1 t, x0 , . . . , x2i−1 , , x2i1 , . . . , x2m
x2i 2
n
n
⎪
⎪
⎪
⎪
1
1
⎪
⎪
⎪
−f
,
x
t,
x
x
,
.
.
.
,
x
,
−
,
.
.
.
,
x
−
n,i−1
0
2i−1
2i1
2m
2i
⎪
⎩
n
n
if |x2i | ≥
1
,
n
if |x2i | <
1
,
n
2.3
for 1 ≤ i ≤ m, and
fn t, x0 , x1 , . . . , x2m fn,m t, x0 , x1 , . . . , x2m for t, x0 , x1 , . . . , x2m ∈ J × Ê2m1 .
2.4
Then, under condition H1 , fn ∈ CarJ × Ê2m1 and
a ≤ fn t, x0 , x1 , . . . , x2m for a.e. t ∈ J and allx0 , x1 , . . . , x2m ∈ Ê2m1 .
2.5
Condition H2 gives
⎞
2m 2m ωj xj ωj 1 ,
fn t, x0 , x1 , . . . , x2m ≤ h⎝t, 2m 1 xj ⎠ ⎛
j0
j0
2.6
for a.e. t ∈ J and all x0 , x1 , . . . , x2m ∈ Ê2m1
,
0
⎛
⎞
2m 2m
1
⎝
⎠
fn t, x0 , x1 , . . . , x2m ≤ h t, 2m 1 xj
ωj
,
n
j0
j0
2.7
for a.e. t ∈ J and all x0 , x1 , . . . , x2m ∈ Ê2m1 .
We investigate the regular differential equation
−1m x2m1 t fn t, xt, . . . , x2m t .
If a function x ∈ AC2m J satisfies 2.8 for a.e. t ∈ J, then x is called a solution of 2.8.
2.8
Boundary Value Problems
5
3. Auxiliary Regular Problems
Let j ∈ Æ and denote by Gj t, s the Green function of the problem
x2j t 0,
x2i 0 x2i T 0,
0 ≤ i ≤ j − 1.
3.1
Then,
G1 t, s ⎧s
⎪
⎪
⎨ T t − T
for 0 ≤ s ≤ t ≤ T,
⎪
t
⎪
⎩ s − T
T
for 0 ≤ t ≤ s ≤ T.
3.2
By 2, 3, 20, the Green function Gj can be expressed as
Gj t, s T
G1 t, τGj−1 τ, sdτ,
j > 1,
3.3
0
and it is known that see, e.g., 3, 20
−1j Gj 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
T 2j−3
sT − s,
6j−1
3.5
T 2j−5
tsT − tT − s
30j−1
3.6
−1j Gj t, s ≤
−1j Gj t, s ≥
hold.
Let γ ∈ L1 J and let u ∈ AC2m−1 J be a solution of the differential equation
−1m u2m t γ t,
3.7
satisfying the Lidstone boundary conditions
u2i 0 u2i T 0,
0 ≤ i ≤ m − 1.
3.8
It follows from the definition of the Green function Gj that
j 2j
−1 u
m−j
t −1
T
0
Gm−j t, sγ sds
for t ∈ J, 0 ≤ j ≤ m − 1.
3.9
6
Boundary Value Problems
It is easy to check that x ∈ AC2m J is a solution of problem 2.8, 1.2 if and only if x0 0,
and its derivative x is a solution of a problem involving the functional differential equation
m 2m
−1 u
t
2m−1
t fn t, usds, ut, . . . , u
t
3.10
0
and the Lidstone boundary conditions 3.8. From 3.9 for j 0, we see that u ∈ AC2m−1 J
is a solution of problem 3.10, 3.8 exactly if it is a solution of the equation
ut −1m
s
Gm t, sfn s, uτdτ, us, . . . , u2m−1 s ds,
T
0
3.11
0
in the set C2m−1 J. Consequently, x is a solution of problem 2.8, 1.2 if and only if it is a
solution of the equation
m
xt −1
t T
0
Gm s, τfn τ, x τ, . . . , x
2m
τ dτ
ds,
3.12
0
in the set C2m J. It means that x is a solution of problem 2.8, 1.2 if x is a fixed point of the
operator H : C2m J → C2m J defined as
m
Hxt −1
t T
0
Gm s, τfn τ, x τ, . . . , x
2m
τ dτ ds.
3.13
0
We prove the existence of a fixed point of H by the following fixed point result of cone
compression type according to Guo-Krasnosel’skii see, e.g., 18, 19.
Lemma 3.2. Let X be a Banach space, and let P ⊂ X be a cone in X. Let Ω1 , Ω2 be bounded open
balls of X centered at the origin with Ω1 ⊂ Ω2 . Suppose that F : 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 problem 2.8, 1.2 has a solution.
Lemma 3.3. Let (H1 ) and (H2 ) hold. Then, problem 2.8, 1.2 has a solution.
Proof. Let the operator H : C2m J → C2m J be given in 3.13, and let
P x ∈ C2m J : xt ≥ 0 for t ∈ J .
3.15
Then, P is a cone in C2m J and since −1m Gm t, s > 0 for t, s ∈ 0, T × 0, T by 3.4 and
fn satisfies 2.5, we see that H : C2m J → P . The fact that H is a completely continuous
Boundary Value Problems
7
operator follows from fn ∈ CarJ × Ê2m1 , from Lebesgue dominated convergence theorem,
and from the Arzelà-Ascoli theorem.
Choose x ∈ P and put yt Hxt for t ∈ J. Then, cf. 2.5
−1m y2m1 t fn t, xt, . . . , x2m t ≥ a > 0
for a.e. t ∈ J.
3.16
Since y0 0 and y2i−1 0 y2i−1 T 0 for 1 ≤ i ≤ m, the equality yj ξj 0 holds with
some ξj ∈ J for 0 ≤ j ≤ 2m. We now use the equality y2m ξ2m 0 and have
t
2m 2m1
y
t sds ≥ a|t − ξ2m |
y
ξ2m
for t ∈ J.
3.17
Hence, y2m ∞ ≥ aT/2, and so
HxC2m >
aT
.
2
3.18
Next, we deduce from the relation
t
T 2m 2m
fn s, xs, . . . , x
fn s, xs, . . . , x2m s ds
t s ds ≤
y
ξ2m
0
3.19
and from 2.7 that
2m
T
1
2m hs, 2m 1 xC2m ds T
ωj
t ≤
y
n
0
j0
for t ∈ J.
3.20
Therefore,
T
2m hs, 2m 1 xC2m ds V,
y
≤
∞
where V T
2m
j0
3.21
0
ωj 1/n. Since yj ξj 0 for 0 ≤ j ≤ 2m, we have
j y ≤ T 2m−j y2m ,
∞
∞
0 ≤ j ≤ 2m.
3.22
The last inequality together with 3.21 gives
y C2m
T
2m ≤ K y
hs, 2m 1 xC2m ds V ,
≤K
∞
3.23
0
where K is from H2 . Since x ∈ P is arbitrary, relations 3.18 and 3.21 imply that for all
8
Boundary Value Problems
x ∈ P , inequalities 3.18 and
HxC2m ≤ K
T
0
hs, 2m 1 xC2m ds V
3.24
hold. By H2 , there exists C > 0 such that
1
v
T
hs, 2m 1 Kvds V
≤ 1 ∀v ≥
0
C
,
K
3.25
and therefore,
K
T
hs, 2m 1 vds V
≤v
∀v ≥ C.
3.26
0
Let
Ω1 x ∈ C2m J : xC2m <
aT
,
2
Ω2 x ∈ C2m J : xC2m < C .
3.27
Then, it follows from 3.18, 3.24, and 3.26 that
HxC2m ≥ xC2m
for x ∈ P ∩ ∂Ω1 ,
HxC2m ≤ xC2m
for x ∈ P ∩ ∂Ω2 .
3.28
The conclusion now follows from Lemma 3.2 for X C2m J and F H.
The properties of solutions to problem 2.8, 1.2 are collected in the following lemma.
Lemma 3.4. Let (H1 ) and (H2 ) be satisfied. Let xn be a solution of problem 2.8, 1.2. Then, for all
n ∈ Æ , the following assertions hold:
2j1
i −1j xn
2m1
t > 0 for t ∈ 0, T, 0 ≤ j ≤ m − 1, and −1m xn
2j2
−1j xn
ii xn is increasing on J, and for 0 ≤ j ≤ m − 1,
2j2
ξj,n 0,
unique ξj,n ∈ 0, T such that xn
t ≥ a for a.e. t ∈ J,
is decreasing on J, and there is a
iii there exists a positive constant A such that
2m xn t ≥ A|t − ξm−1,n |,
2
2j2 if 0 ≤ j ≤ m − 2,
t ≥ A t − ξj,n
xn
2j1 t ≥ AtT − t if 0 ≤ j ≤ m − 1,
xn
xn t ≥ At2 ,
for t ∈ J,
iv the sequence {xn } is bounded in C2m J.
3.29
Boundary Value Problems
9
Proof. Let us choose an arbitrary n ∈ Æ . By 2.5,
2m1
−1m xn
2m
t fn t, xn t, . . . , xn t ≥ a
for a.e. t ∈ J,
3.30
and it follows from the definition of the Green function Gj that the equality
2j1
−1j xn
t
m−j
−1
T
0
2m
Gm−j t, sfn s, xn s, . . . , xn s ds
3.31
holds for t ∈ J and 0 ≤ j ≤ m − 1. Now, using 1.2, 3.4, 3.30, and 3.31, we see that
2j2
assertion i is true. Hence, −1j xn
is decreasing on J for 0 ≤ j ≤ m−1 and xn is increasing
2i−1
2i−1
on this interval. Due to xn
0 xn
T 0 for 1 ≤ i ≤ m, there exists a unique ξj,n ∈ 0, T
2j2
such that un
ξj,n 0 for 0 ≤ j ≤ m − 1. Consequently, assertion ii holds.
Next, in view of 2.5, 3.6, and 3.31,
T
2m−j−5
a
2j1 T
tT − t
sT − sds
t ≥
xn
30m−j−1
0
3.32
T
a
tT − t for t ∈ J, 0 ≤ j ≤ m − 1.
6 · 30m−j−1
2m−j−2
Since
2j2
xn
t
t
ξj,n
2j3
xn
sds
3.33
and, by 13, Lemma 6.2,
T
t
2
sT − sds ≥
t − ξj,n ,
6
ξj,n
3.34
we have
2
T 2m−j−3 a 2j2 t − ξj,n
t ≥
xn
36 · 30m−j−2
for t ∈ J, 0 ≤ j ≤ m − 2.
3.35
Furthermore,
t
2m
2m f s, xn s, . . . , xn s ds ≥ a|t − ξm−1,n |,
xn t ξm−1,n n
t ∈ J,
3.36
10
Boundary Value Problems
and cf. 3.32 for j 0
xn t t
0
xn sds
T 2m−2 a
≥
6 · 30m−1
t
sT − sds
0
3.37
T 2m−2 a 2
T 2m−1 a 2
t
t
−
2t
≥
3T
36 · 30m−1
36 · 30m−1
for t ∈ J,
since xn > 0 on 0, T by assertion ii. Let
"
T 2m−1
,
A a · min 1, A1 , A2 ,
36 · 30m−1
!
3.38
where
!
"
T 2m−j−2
: 0 ≤j ≤ m−1 ,
A1 min
6 · 30m−j−1
!
"
T 2m−j−3
A2 min
:0≤j ≤m−2 .
36 · 30m−j−2
3.39
Then estimate 3.29 follows from relations 3.32–3.37.
It remains to prove the boundedness of the sequence {xn } in C2m J. We use estimate
3.29, the properties of ωj given in H2 , and the inequality
⎧
T
⎪
⎪
⎨2t
tT − t ≥
for 0 < t ≤
T
,
2
3.40
⎪
⎪
⎩ T T − t for T < t < T
2
2
and have
T
0
T
2m ω2m xn s ds ≤
ω2m A|s − ξm−1,n |ds
0
1
A
<
T
0
2j2 ω2j2 xn
s ds ≤
Aξ
2
A
T
m−1,n
ω2m sds AT −ξm−1,n 0
0
AT
ω2m sds,
0
2 ds
ω2j2 A s − ξj,n
0
1
√
A
√AT −ξj,n √
− Aξj,n
ω2j2 s2 ds
ω2m sds
Boundary Value Problems
11
√AT
2
<√
ω2j2 s2 ds
A 0
T
T
2j1 ω2j1 xn
ω2j1 AsT − sds
s ds ≤
0
for 0 ≤ j ≤ m − 2,
0
<
T/2
ω2j1
0
<
T
ω0 |xn s|ds ≤
4
AT
T
0
0
AT 2 /4
T
ATs
ATT − s
ds ds
ω2j1
2
2
T/2
ω2j1 sds for 0 ≤ j ≤ m − 1,
0
1
ω0 As ds √
A
2
√AT
ω0 s2 ds.
0
3.41
In particular,
T
0
2 AT
2m ω2m xn s ds <
ω2m sds,
A 0
√
AT
2j2 2
ω2j2 xn
ω2j2 s2 ds for 0 ≤ j ≤ m − 2,
s ds < √
A 0
0
T
2
2j1 4 AT /4
ω2j1 xn
ω2j1 sds for 0 ≤ j ≤ m − 1,
ds
<
s
AT 0
0
T
T
0
1
ω0 |xn s|ds ≤ √
A
√AT
3.42
ω0 s2 ds,
0
for all n ∈ Æ . Now, from the above estimates, from 2.6 and from xn
ξm−1,n ∈ 0, T, which is proved in ii, we get
2m
ξm−1,n 0 for some
t
2m 2m
f s, xn s, . . . , xn s ds
xn t ξm−1,n n
≤
T
0
≤
T
0
<
T
0
2m
fn s, xn s, . . . , xn s ds
⎛
⎞
2m 2m T j j
ωj xn s ωj 1 ds
h⎝s, 2m 1 xn s⎠ds j0
⎛
h⎝s, 2m 1 j0
⎞
2m j ⎠
ds Λ,
xn j0
∞
0
3.43
12
Boundary Value Problems
where
2
Λ
A
AT
0
2 m−2
ω2m sds √
A j0
4 m−1
AT j0
AT 2 /4
0
√AT
ω2j2 s2 ds
0
1
ω2j1 sds √
A
√AT
3.44
2m
ω0 s2 ds T ωj 1.
j0
0
Notice that Λ < ∞ by H2 . Consequently,
⎞
⎛
T
2m j
2m h⎝s, 2m 1 xn ⎠ds Λ
xn <
∞
j
j0
0
∞
for n ∈ Æ .
3.45
j
2m
Since xn ∞ ≤ T 2m−j xn ∞ for 0 ≤ j ≤ 2m, which follows from the fact that xn vanishes
in J by 1.2 and assertion ii, inequality 3.45 yields
T 2m 2m h s, 2m 1 K xn ds Λ
xn <
∞
∞
0
for n ∈ Æ ,
3.46
where K is from H2 . Due to the condition
1
lim sup
v
v→∞
T
3.47
ht, Kvdv < 1
0
in H2 , there exists a positive constant S such that for all v ≥ S the inequality
T
3.48
ht, 2m 1 Kvdt Λ ≤ v
0
2m
is fulfilled. The last inequality together with estimate 3.46 gives xn ∞ < S for n ∈
j
Consequently, xn ∞ < T 2m−j S for 0 ≤ j ≤ 2n, n ∈ Æ , and assertion iv follows.
2m
The following result gives the important property of {fn t, xn t, . . . , xn
applying the Vitali convergent theorem in the proof of Theorem 4.1.
Æ.
t} for
Lemma 3.5. Let (H1 ) and (H2 ) hold. Let xn be a solution of problem 2.8, 1.2. Then, the sequence
2m
fn t, xn t, . . . , xn t ⊂ L1 J
3.49
is uniformly integrable on J, that is, for each ε > 0, there exists δ > 0 such that if M ⊂ J and
meas M < δ, then
M
2m
fn t, xn t, . . . , xn t dt < ε
for n ∈ Æ .
3.50
Boundary Value Problems
13
Proof. By Lemma 3.4 iv, there exists S > 0 such that for n ∈ Æ , the inequality xn C2m < S
holds. Now, we conclude from 2.5 and 2.6, from the properties of h and ωj given in H2 ,
and finally from 3.29 that for j ∈ J and n ∈ Æ , the estimate
2m
a ≤ fn t, xn t, . . . , xn t
m−1
ω2j1 AtT − t
≤ ht, 2m 1 S ω0 At2 j0
3.51
m
2 ω2m A|t − ξm−1,n | ωj 1
ω2j2 A t − ξj,n
m−2
j0
j0
is fulfilled, where A is a positive constant. Since the functions ht, 2m 1 S, ω0 At2 , and
ω2j1 AtT − t 0 ≤ j ≤ m − 1 belong to the set L1 J by assumption H2 , in order to
2m
prove that {fn t, xn t, . . . , xn t} is uniformly integrable on J, it suffices to show that the
sequences
{ω2m A|t − ξm−1,n |},
2 ,
ω2j2 A t − ξj,n
0≤ j ≤ m−2
3.52
#1
#1
are uniformly integrable on J. Due to 0 ω2m sds < ∞ and 0 ω2j s2 ds < ∞ for 1 ≤ j ≤ m − 1
by H2 , this fact follows from 13, Criterion 11.10 with b A and r 1, 2.
4. The Main Result
The following theorem is the existence result for the singular problem 1.1, 1.2.
Theorem 4.1. Let (H1 ) and (H2 ) hold. Then, problem 1.1, 1.2 has a positive solution
x ∈ AC2m J and
xt > 0
for t ∈ 0, T,
−1j x2j1 t > 0
for t ∈ 0, T, 0 ≤ j ≤ m − 1.
4.1
Proof. Lemma 3.3 guarantees that problem 2.8, 1.2 has a solution xn . Consider the sequence {xn }. By Lemma 3.4, {xn } is bounded in C2m J,
2j1
−1j xn
t > 0
for t ∈ 0, T, 0 ≤ j ≤ m − 1,
4.2
and xn fulfils estimate 3.29, where A is a positive constant and ξj,n ∈ 0, T. Furthermore,
2m
the sequence {fn t, xn t, . . . , xn t} is uniformly integrable on J by Lemma 3.5, and
2m1
2m
therefore, we deduce from the equality −1m xn
t fn t, xn t, . . . , xn t for a.e. t ∈ J
2m
that {xn } is equicontinuous on J. Now, by the Arzelà-Ascoli theorem and the BolzanoWeierstrass theorem, we may assume without loss of generality that {xn } is convergent in
C2m J and {ξj,n } is convergent in Ê for 0 ≤ j ≤ m − 1. Let limn → ∞ xn x and limn → ∞ ξj,n ξj
14
Boundary Value Problems
0 ≤ j ≤ m−1. Then x ∈ C2m J and x satisfies the boundary conditions 1.2. Letting n → ∞
in 3.29 and 4.2, we get for t ∈ J
2m t ≥ A|t − ξm−1 |,
x
2
2j2 t ≥ A t − ξj
x
if 0 ≤ j ≤ m − 2
−1j x2j1 t ≥ AtT − t if 0 ≤ j ≤ m − 1, xt ≥ At2 .
4.3
Keeping in mind the definition of fn , we conclude from 4.3 that
2m
lim fn t, xn t, . . . , xn t f t, xt, . . . , x2m t
n→∞
for a.e. t ∈ J.
4.4
Then, by the Vitali theorem, ft, xt, . . . , x2m t ∈ L1 J and
t
lim
n→∞
0
t 2m
fn s, xn s, . . . , xn s ds f s, xs, . . . , x2m s ds for t ∈ J.
4.5
0
Letting n → ∞ in the equality
2m
xn
2m
t xn
0 t
0
2m
fn s, xn s, . . . , xn s ds,
4.6
we get
x
2m
t x
2m
t 0 f s, xs, . . . , x2m s ds
for t ∈ J.
4.7
0
As a result, x ∈ AC2m J and x is a solution of 1.1. Consequently, x is a positive solution of
problem 1.1, 1.2 and inequality 4.1 follows from 4.3.
Example 4.2. Consider problem 1.1, 1.2 with
ft, x0 , . . . , x2m pt 2m
k0
ak t|xk |
αk
bk t
|xk |βk
4.8
on J ×D, where p, ak ∈ L1 J, bk ∈ L∞ J that is, bk is essentially bounded and measurable on
J are nonnegative, pt ≥ a > 0 for a.e. t ∈ J. If αk ∈ 0, 1 for 0 ≤ k ≤ 2m and β2k ∈ 0, 1/2,
β2m , β2k1 ∈ 0, 1 for 0 ≤ k ≤ m − 1, then, by Theorem 4.1, the problem has a positive solution
x ∈ AC2m J satisfying inequality 4.1.
Acknowledgment
This work was supported by the Council of Czech Government MSM no. 6198959214.
Boundary Value Problems
15
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