Báo cáo hóa học: "EXISTENCE AND MULTIPLICITY OF SOLUTIONS FOR SOME THREE-POINT NONLINEAR BOUNDARY VALUE PROBLEMS" pptx

EXISTENCE AND MULTIPLICITY OF SOLUTIONS FOR SOME
THREE-POINT NONLINEAR BOUNDARY VALUE PROBLEMS
XU XIAN AND DONAL O’REGAN
Received 30 September 2004; Accepted 20 October 2004
We study the existence and multiplicity of solutions for the three-point nonlinear bound-
ary value problem u

(t)+λa(t) f (u) = 0, 0 <t<1; u(0) = 0 = u(1) − γu(η), where η ∈
(0,1), γ ∈ [0,1), a(t)and f (u) are assumed to be positive and have some singularities, and
λ is a positive parameter. Under certain conditions, we prove that there exists λ
∗
> 0such
that the three-point nonlinear boundary value problem has at least two positive solutions
for 0 <λ<λ
∗
, at least one solution for λ = λ
∗
, and no solution for λ>λ
∗
.
Copyright © 2006 X. Xian and D. O’Regan. This is an open access article distributed un-
der the Creative Commons Attribution License, which permits unrestricted use, dist ri-
bution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
In this paper, we consider the following second-order three-point boundary value prob-
lem (BVP)
u

(t)+λa(t) f (u) = 0, 0 <t<1,
u(0)
= 0 = u(1) − γu(η),

(1.1
λ
)
where η
∈ (0,1),γ ∈ [0, 1), a ∈ C((0,1),(0,+∞)), and f ∈ C(R
+
\{0},R
+
), here λ is a pos-
itive parameter and
R
+
= [0, +∞).
Now a(t) may have a singular ity at t
= 0andt = 1, f (u) may have a singularity at
u
= 0, so the BVP (1.1
λ
) is a singular problem. The BVP (1.1
λ
) in the case when γ = 0can
be reduced to the Dirichlet BVP
u

(t)+λa(t) f (u) = 0, 0 <t<1,
u(0)
= 0 = u(1).
(1.2
λ
)

The BVP (1.2
λ
) has been studied extensively in the literature, see [1, 2, 5, 9, 12] and the
references therein. Choi [1] studied the particular case where f (u)
= e
u
, a ∈ C
1
(0,1],
a>0 in (0,1), and a can be singular at t
= 0, but is at most O(1/t
2−δ
)ast → 0
+
for some
δ. Using the shooting method, he established the following result.
Hindawi Publishing Corporation
Journal of Inequalities and Applications
Volume 2006, Article ID 79653, Pages 1–17
DOI 10.1155/JIA/2006/79653
2 Solutions of three-point nonlinear BVPs
Theorem 1.1 (see [1]). There exists λ
0
> 0 such that the BVP (1.2
λ
)hasasolutionin
C
2
(0,1] ∩ C[0,1] for 0 <λ<λ
0

, while there is no solution for λ>λ
0
.
Wong [9] studied the more general BVP (1.2
λ
). Using also the shooting method, Wong
proved some existence results for positive solutions of the BVP (1.2
λ
). Recently, Dalmasso
[2]improvedTheorem 1.1 and the main results in [9]. Using the upper and lower solu-
tions technique and the fixed point index method, Dalmasso [2] proved the following
result.
Theorem 1.2 (see [2]). Let a and f satisfy the following assumptions:
(A
1
) a ∈ C((0, 1), [0, ∞)), a ≡ 0 in (0,1), and there exists α,β ∈ [0,1) such that

1
0
s
α
(1 − s)
β
a(s)ds < ∞; (1.1)
(A
2
) f ∈ C([0, ∞), (0, ∞)) is nondecreasing.
Then,
(i) there exists λ
0

> 0 such that the BVP (1.2
λ
) has at least one positive solution in
C
2
(0,1) ∩ C[0,1] for 0 <λ<λ
0
,
(ii) if in addition f satisfies the condition that
(A
3
) there exists d>0 such that f (u) ≥ du for u ≥ 0.
Then there exists λ
∗
> 0 such that the BVP (1.2
λ
) has at least one positive solution in C
2
(0,
1)
∩ C[0,1] for 0 <λ<λ
∗
while there is no such solution for λ>λ
∗
.
Ha and Lee [5]alsoconsideredtheBVP(1.2
λ
) in the case when f (u) ≥ e
u
.Theyproved

Theorems 1.3 and 1.4.
Theorem 1.3 (see [5]). Assume the following conditions hold
(B
1
) a>0 on (0,1);
(B
2
) a(t) is singular at t = 0 satisfying

1
0
sa(s)ds < ∞;
(B
3
) f (u) ≥ e
u
for all u ∈ R.
Then there exists λ
0
such that the BVP (1.2
λ
)hasnosolutionforλ>λ
0
and at least one
solution for 0 <λ<λ
0
.
Theorem 1.4 (see [5]). Consider (1.2
λ
), where a and f are continuous and satisfy (B

1
)–
(B
3
). Also assume that
(B
4
) f is nondecreasing.
Then the number λ
0
given by Theorem 1.3 is such that
(i) (1.2
λ
)hasnosolutionforλ>λ
0
;
(ii) (1.2
λ
) has at least one solution for λ = λ
0
;
(iii) (1.2
λ
) has at least two solutions for 0 <λ<λ
0
.
Xu and Ma [12] generalized the main results of [1, 2, 5, 9]toanoperatorequationin
arealBanachspaceE. In recent years, the multipoint BVP has been extensively studied
(see [3, 4, 6–8, 10, 11, 13] and the references therein). For example, Ma and Castaneda
[7] using the well-known fixed point theorem in cones established some results on the

existence of at least one positive solution for some m-point boundary value problems
if the nonlinearit y f is either superlinear or sublinear. The purpose of this paper is to
X. Xian and D. O’Regan 3
extend the main results of [1, 2, 5, 9] to the nonlinear three-point BVP (1.1
λ
). We will
consider the existence and multiplicity of positive solution for the nonlinear three-point
BVP (1.1
λ
). The results of this paper are improvements of the main results in [1, 2, 5, 9].
2. Several lemmas
Let us list some conditions to be used in this paper.
(H
1
) γ ∈ [0,1),a ∈ C((0,1),(0,∞)), and

1
0
s(1 − s)a(s)ds < ∞. (2.1)
(H
2
) f (u) = g(u)+h(u), where g :(0,∞) → (0,∞) is continuous and nonincreasing,
h :
R
+
→ R
+
is continuous, and
h(u)
≥ b

0
u
w
, u ∈ R
+
, (2.2)
for some b
0
> 0andw ≥ 1.
(H
3
) There exists M>0suchthat
h

u
2

−
h

u
1

≥−
M

u
2
− u
1


(2.3)
for all u
1
,u
2
∈ R
+
with u
2
≥ u
1
The main results of this paper are the following theorems.
Theorem 2.1. Assume that (H
1
)and(H
2
) hold. Then there exists λ
∗
> 0 such that the BVP
(1.1
λ
) has at least one positive solution for 0 <λ<λ
∗
and no solution for λ>λ
∗
.
Moreover, the B VP (1.1
λ
) has at least one positive solution if ω>1.

Theorem 2.2. Assume that (H
1
), (H
2
), and (H
3
)hold,ω>1, and there exists constant
c
≥ 0 such that g(u) = c for all u ∈ (0,+∞). Then there exists λ
∗
> 0 such that the BVP
(1.1
λ
) has at least two positive solutions for 0 <λ<λ
∗
, at least one solution for λ = λ
∗
,and
no solution for λ>λ
∗
.
Remark 2.3. Our theorems generalize Theorems 1.1–1.4 and the main results in [9]. In
fact, Theorems 1.1–1.4 are corollaries of our theorems. Moreover, the nonlinear term
f (u) may have singularity at u
= 0, therefore, even in the case when γ = 0, Theorem 2.1
cannot be obtained by Theorems 1.1–1.4 and the abstract results in [12].
Remark 2.4. The nonlinear term f was assumed to be nondecreasing in Theorems 1.2
and 1.4,butinTheorem 2.2 in this paper, we do not assume that the nonlinear term f is
nondecreasing. Thus, even in the case when γ
= 0, Theorem 2.2 cannot be obtained from

Theorem 1.4.
Let n
∈ N and let N be the natural numbers set. First, let us consider the BVP of the
form
u

(t)+λa(t)

g

u +
1
n

+ h(u)

=
0, 0 <t<1,
u(0)
= 0 = u(1) − γu(η).
(2.1
λ
n
)
4 Solutions of three-point nonlinear BVPs
Definit ion 2.5. α
∈ C([0, 1], R) ∩ C
2
((0,1),R) is called a lower solution of (2.1
λ

n
)if
α

(t)+λa(t)

g

α(t)+
1
n

+ h

α(t)


≥
0, t ∈ (0,1),
α(0)
≤ 0, α(1) − γα(η) ≤ 0.
(2.4)
β
∈ C([0,1],R) ∩ C
2
((0,1),R) is called an upper solution of (2.1
λ
n
)if
β


(t)+λa(t)

g

β(t)+
1
n

+ h

β(t)


≤
0, t ∈ (0,1),
β(0)
≥ 0, β(1) − γβ(η) ≥ 0.
(2.5)
According to [13, Lemma 4], we have the following lemma.
Lemma 2.6. Assume that (H
1
)holdsandτ ≥ 0. Then the initial value problems
u

(t) = τa(t)u(t), 0 ≤ α<t<1,
u(α)
= 0, u

(α) = 1,

u

(t) = τa(t)u(t), 0 <t<β≤ 1,
u(β)
= 0, u

(β) =−1
(2.6)
have unique positive solutions p
α,τ
(t) ∈ AC[α,1) ∩ C
1
[α,1) and q
β,τ
(t) ∈ AC(0,β] ∩
C
1
(0,β],respectively.Moreover,p
α,τ
and q
β,τ
are strictly c onvex. As a result,
t
− α ≤ p
α,τ
(t) ≤ p
α,τ
(a)
(t
− α)

(a − α)
, α
≤ t ≤ a ≤ 1,
β
− t ≤ q
β,τ
(t) ≤ q
β,τ
(b)
(β
− t)
(β − b)
,0
≤ b ≤ t ≤ β
(2.7)
for any a
∈ [α,1) and b ∈ [0, β).
When 0
≤ α<β≤ 1,fort ∈ [α, β],
W
(τ)
[α,β]
(t) =





q
β,τ

(t), p
α,τ
(t)
q

β,τ
(t), p

α,τ
(t)





=
q
β,τ
(α) = p
α,τ
(β). (2.8)
It is well known that C[0,1] is a Banach space w ith maximum norm
·.Forτ ≥ 0,
denote θ
τ
by
θ
τ
=
γ(1 − η)

p
0,τ
(η)+q
1,τ
(η)
min

p
0,τ
(η)
p
0,τ
(1) + p
0,τ
(η)
,
q
1,τ
(η)
q
1,τ
(0) + q
1,τ
(η)

. (2.9)
X. Xian and D. O’Regan 5
Let P
={x ∈ C[0, 1]|x(t) ≥ 0 for t ∈ [0,1]} and Q
τ

={x ∈ P|x(t) ≥ θ
τ
xt for t ∈ [0,1]}.
It is easy to s ee that P and Q
τ
are cones in C[0,1].Forτ ≥ 0 and each n ∈ N,defineoperators
L
τ
and F
n
: C[0,1] → C[0,1] by

L
τ
x

(t) =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
p
0,τ
(1)
p
0,τ
(1) − γp
0,τ
(η)

1
0
G
(τ)
[0,1]
(η,s)a(s)x(s)ds, t = η,


η
0
G
(τ)
[0,η]
(t,s)a(s)x(s)ds +(L
τ
x)(η)
p
0,τ
(t)
p
0,τ
(η)
, t
∈ [0,η],

1
η
G
(τ)
[η,1]
(t,s)a(s)x(s)ds +(L
τ
x)(η)
q
1,τ
(t)+γp
η,τ
(t)

q
1,τ
(η)
, t
∈ [η,1],
(2.10)
and (F
n
x)(t) = g(x(t)+1/n)+h(x(t)) for t ∈ [0,1],where
G
(τ)
[α,β]
(t,s):=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
q
β,τ
(t)
p
α,τ
(s)

p
α,τ
(β)
, α
≤ s ≤ t ≤ β,
p
α,τ
(t)
q
β,τ
(s)
q
β,τ
(α)
, α
≤ t ≤ s ≤ β.
(2.11)
From [ 13,Theorem5],wehaveLemmas2.7 and 2.9.
Lemma 2.7. Assume that (H
1
)holds,τ ≥ 0,andh ∈ C([0,1],R). Then w(t) is the solution
of the three-point BVP
−w

(t)+τa(t)w(t) = a(t)h(t), 0 ≤ α<t≤ 1,
w(α)
= 0 = w(1) − γw(η)
(2.12)
if and only if w
∈ C[0,1] is the solution of the integral equation

w(t)
=

L
τ
h

(t), t ∈ [0,1]. (2.13)
Remark 2.8. To e n su re th a t p
α,τ
(1) − γp
α,τ
(η) > 0, the following condition is assumed in
[13,Theorem5]:
τa(t) >
3γ
(1 − η)
2
. (2.14)
If 0
≤ γ<1, we have
p
α,τ
(1) − γp
α,τ
(η) >p
α,τ
(η)

1+


1
η
τa(s)q
1,τ
(s)ds− γ

> 0 . (2.15)
Thus, if 0
≤ γ<1, condition (2.14)canberemoved.
6 Solutions of three-point nonlinear BVPs
Lemma 2.9. Assume that (H
1
)holds,τ,α,ξ
∗
,η
∗
≥ 0, h ∈ C([0, 1], R
+
). Also suppose that
w
∈ C[α,1] satisfies
−w

(t)+τa(t)w(t) = a(t)h(t), α<t<1,
w(α)
= ξ
∗
, w(1) − γw(η) = η
∗

.
(2.16)
Then w(t)
≥ 0 for t ∈ [α,1].
Lemma 2.10. Assume that (H
1
)holdsandτ ≥ 0. Then L
τ
: P → Q
τ
is a completely continu-
ous and increasing operator.
Proof. From Lemma 2.6,wehaveforanyx
∈ P and t ∈ [0, 1],

L
τ
x

(t) ≥
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪

⎪
⎪
⎩

L
τ
x

(η)
p
0,τ
(t)
p
0,τ
(η)
, t
∈ [0, η],

L
τ
x

(η)
q
1,τ
(t)+γp
η,τ
(t)
q
1,τ

(η)
, t
∈ [η,1],
≥
⎧
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎩

L
τ
x

(η)
t
p
0,τ
(η)
, t
∈ [0, η],


L
τ
x

(η)
1
− t + γ(t − η)
q
1,τ
(η)
, t
∈ [η,1],
≥

L
τ
x

(η)
γ(1
− η)t
p
0,τ
(η)+q
1,τ
(η)
,
(2.17)

L

τ
x

(η) =
p
0,τ
(1)
p
0,τ
(1) − γp
0,τ
(η)


η
0
q
1,τ
(η)
p
0,τ
(s)
p
0,τ
(1)
a(s)x(s)ds
+

1
η

p
0,τ
(η)
q
1,τ
(s)
q
1,τ
(0)
a(s)x(s)ds

≥
q
1,τ
(η)
p
0,τ
(1) − γp
0,τ
(η)

η
0
p
0,τ
(s)a(s)x( s)ds,
(2.18)
(L
τ
x)(η) =

p
0,τ
(1)
p
0,τ
(1) − γp
0,τ
(η)


η
0
q
1,τ
(η)
p
0,τ
(s)
p
0,τ
(1)
a(s)x(s)ds
+

1
η
p
0,τ
(η)
q

1,τ
(s)
q
1,τ
(0)
a(s)x(s)ds

≥
p
0,τ
(η)
p
0,τ
(1) − γp
0,τ
(η)

1
η
q
1,τ
(s)a(s)x( s)ds.
(2.19)
X. Xian and D. O’Regan 7
By (2.18)andLemma 2.6,wehaveforanyt
∈ [0, η],

L
τ
x


(t) =

t
0
q
η,τ
(t)
p
0,τ
(s)
p
0,τ
(η)
a(s)x(s)ds
+

η
t
p
0,τ
(t)
q
η,τ
(s)
q
η,τ
(0)
a(s)x(s)ds +


L
τ
x

(η)
p
0,τ
(t)
p
0,τ
(η)
≤

t
0
q
η,τ
(0)
p
0,τ
(s)
p
0,τ
(η)
a(s)x(s)ds +

η
t
p
0,τ

(s)
q
η,τ
(0)
q
η,τ
(0)
a(s)x(s)ds +

L
τ
x

(η)
=

η
0
p
0,τ
(s)a(s)x( s)ds +

L
τ
x

(η)
≤
q
1,τ

(0) + q
1,τ
(η)
q
1,τ
(η)

L
τ
x

(η);
(2.20)
here we have used the facts that q
η,τ
(0) = p
0,τ
(η)andp
0,τ
(1) = q
1,τ
(0). From (2.19)and
Lemma 2.6,wehaveforanyt
∈ [η,1],

L
τ
x

(t)

≤

t
η
q
1,τ
(s)
p
η,τ
(1)
p
η,τ
(1)
a(s)x(s)ds
+

1
t
p
η,τ
(1)
q
1,τ
(s)
q
1,τ
(η)
a(s)x(s)ds+

L

τ
x

(η)
q
1,τ
(t)+γp
η,τ
(t)
q
1,τ
(η)
≤

1
η
q
1,τ
(s)a(s)x( s)ds+

L
τ
x

(η)
q
1,τ
(η)

(1 − t)/(1 − η)


+γp
η,τ
(1)

(t−η)/(1−η)

q
1,τ
(η)
≤

1
η
q
1,τ
(s)a(s)x( s)ds +

L
τ
x

(η)
≤
p
0,τ
(1) + p
0,τ
(η)
p

0,τ
(η)

L
τ
x

(η);
(2.21)
here we have used the fact p
η,τ
(1) = q
1,τ
(η). By (2.20)and(2.21), we have

L
τ

(η) ≥ min

q
1,τ
(η)
q
1,τ
(0) + q
1,τ
(η)
,
p

0,τ
(η)
p
0,τ
(1) + p
0,τ
(η)


L
τ
x. (2.22)
By (2.17)and(2.22), we have
(L
τ
x)(t) ≥ θ
τ
L
τ
xt. (2.23)
This implies that L
τ
: P → Q
τ
.
Now we will show that L
τ
: P → Q
τ
is completely continuous. It is easy to show that

L
τ
: P → Q
τ
is continuous and bounded. Let B ⊂ P be a bounded set such that x≤R
0
8 Solutions of three-point nonlinear BVPs
and
L
τ
x≤R
0
for some R
0
> 0. For any ε>0, by (H
1
) there exists δ
1
> 0suchthat
2R
0

δ
1
0
G
(τ)
[0,η]
(s,s)a(s)ds+2R
0


η
η
−δ
1
G
(τ)
[0,η]
(s,s)a(s)ds
≤ 2R
0
q
η,τ
(0)

δ
1
0
(η − s)s
η
2
a(s)ds+2R
0
p
0,τ
(η)

η
η
−δ

1
(η − s)s
η
2
a(s)ds <
ε
3
.
(2.24)
It is easy to see that there exists δ>0suchthatforanyt
1
,t
2
∈ [0, η], |t
1
− t
2
| <δ,
R
0

η−δ
1
δ
1



G
(τ)

[0,η]

t
1
,s

−
G
(τ)
[0,η]

t
2
,s




a(s)ds <
ε
3
,
R
0


p
0,τ

t

2

−
p
0,τ

t
1



p
0,τ
(η)
<
ε
3
.
(2.25)
By (2.24)–(2.25), we have for any x
∈ B and t
1
,t
2
∈ [0, η], |t
1
− t
2
| <δ,




L
τ
x

t
2

−

L
τ
x

t
1



≤

η
0



G
(τ)
[0,η]


t
2
,s

−
G
(τ)
[0,η]

t
1
,s




a(s)x(s)ds
+

L
τ
x

(η)


p
0,τ


t
2

−
p
0,τ

t
1



p
0,τ
(η)
≤ 2R
0

δ
1
0
G
(τ)
[0,η]
(s,s)a(s)ds
+2R
0

η
η

−δ
1
G
(τ)
[0,η]
(s,s)a(s)ds
+ R
0

η−δ
1
δ
1



G
(τ)
[0,η]

t
1
,s

−
G
(τ)
[0,η]

t

2
,s




a(s)ds
+ R
0


p
0,τ

t
2

−
p
0,τ

t
1



p
0,τ
(η)
<ε.

(2.26)
Thus, L
τ
(B)isequicontinuouson[0,η]. Similarly, L
τ
(B)isalsoequicontinuouson[η,1].
By the Arzela-Ascoli theorem, L
τ
(B) ⊂ C[0, 1] is a relatively compact set. Therefore, L
τ
:
P
→ Q
τ
is a completely continuous operator.
Finally, we show that L
τ
: P → Q
τ
is increasing. For any x
1
,x
2
∈ P, x
1
≤ x
2
∈ P,let
y
1

= L
τ
x
1
and y
2
= L
τ
x
2
, u = y
2
− y
1
.Then,byLemma 2.7,wehave
−u

(t)+τa(t)u(t) = a(t)

x
2
(t) − x
1
(t)

≥
0, t ∈ (0,1),
u(0)
= 0 = u(1) − γu(η).
(2.27)

Then Lemma 2.9 implies that u(t)
≥ 0fort ∈ [0,1], and so, y
2
≥ y
1
.Theproofiscom-
plete.

Lemma 2.11. Assume (H
1
)and(H
2
)hold.Letλ>0 be fixed. If there exists R
λ
> 0 such that
(2.1
λ
n
) has at least one positive solution x
n
with x
n
≤R
λ
for each positive integer n, then
there exist
¯
x
∈ C[0, 1] and a subsequence {x
n

k
}
+∞
k=1
of {x
n
}
+∞
n=1
such that x
n
k
→
¯
x as k
→ +∞.
Moreover,
¯
x is a positive solution of the BVP (1.1
λ
)
X. Xian and D. O’Regan 9
Proof. Let z
0
(t) = 1fort ∈ [0,1], and z
λ
(t) = λg(R
λ
+1)(L
τ

z
0
)(t)fort ∈ [0,1]. Since L
0
is
increasing and g is nonincreasing, then we have for any n
∈ N,
x
n
(t) = λ

L
0
F
n
x
n

(t) ≥ λg

R
λ
+1

L
0
z
0

(t) = z

λ
(t), t ∈ [0,1]. (2.28)
Let us define the function F by
F(t)
=

1
t
(1 − s)a(s)ds, t ∈ (0,1]. (2.29)
Obviously, F
∈ C(0, 1], F(1) = 0, and F is nonincreasing on (0,1]. For each n ∈ N, x
n
is a
concave function on [0,1]. Then there exists t
n
∈ (0,1) such that x

n
(t
n
) = 0. By (H
2
), we
have
−x

n
(t) ≤ λa(t)g

x

n
(t)


1+
¯
h

R
λ

g

R
λ
+1


, t ∈ (0,1), (2.30)
where
¯
h(R
λ
) = max
s∈[0,R
λ
]
h(s). Integrate (2.30)fromt
n
to t (t ∈ (t

n
,1)) to obtain
−x

n
(t)
g

x
n
(t)

≤
λ

1+
¯
h

R
λ

g

R
λ
+1




t
t
n
a(s)ds. (2.31)
Then integrate (2.31)fromt
n
to 1 to o btain

x
n
(t
n
)
x
n
(1)
ds
g(s)
≤ λ

1+
¯
h

R
λ

g

R

λ
+1



1
t
n
(1 − s)a(s)ds = λ

1+
¯
h

R
λ

g

R
λ
+1


F

t
n

. (2.32)

On the other hand, by (2.28), we have

x
n
(t
n
)
x
n
(1)
ds
g(s)
≥
x
n

t
n

−
x
n
(1)
g

x
n
(1)

≥

x
n
(η)(1 − γ)
g

x
n
(1)

≥
z
λ
(η)(1 − γ)
g

z
λ
(1)

. (2.33)
By (2.32)and(2.33), we have
F

t
n

≥

λ


1+
¯
h

R
λ

g

R
λ
+1


−1
z
λ
(η)(1 − γ)
g

z
λ
(1)

. (2.34)
Let β
0
∈ (0, 1] be such that
F


β
0

=

λ

1+
¯
h

R
λ

g

R
λ
+1


−1
z
λ
(η)(1 − γ)
g

z
λ
(1)


.
(2.35)
Then (2.34) implies that t
n
≤ β
0
. Similarly, we can show that there exists α
0
> 0suchthat
t
n
≥ α
0
for each n ∈ N. Let us define the function I : R
+
→ R
+
by I(x) =

x
0
ds/g(s)for
10 Solutions of three-point nonlinear BVPs
x
∈ R
+
.Foranyt
1
,t

2
∈ [β
0
,1], t
1
<t
2
,by(2.31), we have
I

x
n

t
1

−
I

x
n

t
2

=

x
n
(t

1
)
x
n
(t
2
)
ds
g(s)
=

t
2
t
1
−
x

n
(s)ds
g

x
n
(s)

≤
λ

1+

¯
h

R
λ

g

R
λ
+1



t
2
t
1
dt

t
0
a(s)ds
≤ λ

1+
¯
h

R

λ

g

R
λ
+1



t
2
t
1

t
2
− s

a(s)ds+

t
2
− t
1


t
1
0

a(s)ds

≤
λ

1+
¯
h

R
λ

g

R
λ
+1



t
2
t
1
(1 − s)a(s)ds+

t
2
− t
1



1−(t
2
−t
1
)
0
a(s)ds

.
(2.36)
This and the inequalities (2.21)in[11] imply that the set I(
{x
n
}
+∞
n=1
)isequicontinuous
on [β
0
,1]. It is easy to see that I
−1
, the inverse function of I, is uniformly continuous
on [0,I(R
λ
)]. Therefore, the set {x
n
}
+∞

n=1
is equcontinuous on [β
0
,1]. Similarly, {x
n
}
+∞
n=1
is
equcontinuous on [0,α
0
].
From ( 2.30), we have for any t
∈ [α
0
,β
0
],


x

n
(t)


≤
λ

g


min
t∈[α
0
,β
0
]
z
λ
(t)

+
¯
h

R
λ



β
0
α
0
a(s)ds. (2.37)
Thus,
{x
n
}
+∞

n=1
is equcontinuous on [α
0
,β
0
]. Then, by the Arzela-Ascoli theorem, we see
that
{x
n
}
+∞
n=1
⊂ C[0,1] is a relatively compact set. Thus, there exist
¯
x ∈ C[0,1] and a sub-
sequence
{x
n
k
}
+∞
k=1
of {x
n
}
+∞
n=1
such that x
n
k

→
¯
x. By a standard argument (see [11]), we
have that
¯
x is a positive solution of the BVP (1.1
λ
). The proof is complete. 
Lemma 2.12. Assume that (H
1
)and(H
2
) hold. Then for small enough λ>0,theBVP(1.1
λ
)
has at least one positive solution.
Proof. Let R
0
> 0andλ
0
be such that
0 <λ
0
<
1
2

R
0
γR

0
ds
g(s)


1
0
s(1 − s)a(s)ds

−1

1+
¯
h

R
0

g

R
0
+1


−1
. (2.38)
By Lemma 2.10, λ
0
L

0
F
n
: P → Q
0
is a completely continuous operator for each n ∈ N.
Now we will show that
μλ
0
L
0
F
n
u = u, μ ∈ [0,1], u ∈ ∂B

θ,R
0

, n ∈ N, (2.39)
where B(θ,R
0
) ={x ∈ Q
0
|x <R
0
}.Suppose(2.39) is not true. Then there exist μ
0
∈
[0,1], u
0

∈ ∂B(θ, R
0
), and n
0
∈ N such that μ
0
λ
0
L
0
F
n
0
u
0
= u
0
.Obviously,μ
0
> 0.
X. Xian and D. O’Regan 11
By Lemma 2.7,wehave
u

0
(t)+μ
0
λ
0
a(t)


g

u
0
+
1
n
0

+ h

u
0


=
0, 0 <t<1,
u
0
(0) = 0 = u
0
(1) − γu
0
(η).
(2.40)
Thus u
0
is a concave function on [0,1], and there exists t
0

∈ (0, 1) such that u

0
(t
0
) = 0.
A similar argument as in the proof of (2.32) guarantees that

u
0

t
0

u
0
(1)
ds
g(s)
≤ λ
0
μ
0

1+
¯
h

R
0


g

R
0
+1



1
t
0
(1 − s)a(s)ds
≤
λ
0
μ
0
t
0

1+
¯
h

R
0

g


R
0
+1



1
0
s(1 − s)a(s)ds,

u
0

t
0

u
0
(0)
ds
g(s)
≤
λ
0
μ
0
1 − t
0

1+

¯
h

R
0

g

R
0
+1



1
0
s(1 − s)a(s)ds.
(2.41)
Since u
0
(t
0
) = R
0
and u
0
(1) = γu
0
(η) ≤ γR
0

,by(2.41), we have
λ
0
≥
1
2

1+
¯
h

R
0

g

R
0
+1



1
0
s(1 − s)a(s)ds

−1

R
0

γR
0
ds
g(s)
, (2.42)
which contradicts (2.38). Therefore, (2.39) holds, and so
i

λ
0
L
0
F
n
,B

θ,R
0

,Q
0

=
1, n ∈ N. (2.43)
This means that for each n
∈ N,theoperatorλ
0
L
0
F

n
has at least one positive fixed point
x
n
such that x
n
≤R
0
.ByLemma 2.7,theBVP(2.1
λ
n
) has a positive solution x
n
such that
x
n
≤R
0
.ThenbyLemma 2.11,theBVP(1.1
λ
) has at least one positive solution. The
proof is complete.

Lemma 2.13. Let α(t) and β(t) beloweranduppersolutionsof(2.1
λ
n
)forsomen ∈ N and
λ>0, 0
≤ α(t) ≤ β(t).Then(2.1
λ

n
) has at least one positive solution u
n,λ
such that
α(t)
≤ u
n,λ
(t) ≤ β(t), t ∈ [0,1]. (2.44)
Proof. Let us define the function F
∗
n
by

F
∗
n
x

(t) =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪

⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
g

β(t)+
1
n

+ h

β(t)

, x ≥ β(t),
g

x +
1
n


+ h(x), α(t) <x<β(t),
g

α(t)+
1
n

+ h

α(t)

, α(t) <x,
(2.45)
12 Solutions of three-point nonlinear BVPs
for x
∈ P. Then there exists a constant C
n
> 0suchthat0≤ (F
∗
n
x)(t) ≤ C
n
for x ∈ P.Now
Lemma 2.10 and Schauder’s fixed point theorem guarantees that the operator λL
0
F
∗
n
has

at least one fixed point. Then the BVP
u

(t)+λa(t)

F
∗
n
u

(t) = 0, t ∈ (0,1),
u(0)
= 0 = u(1) − γu(η)
(2.46)
has at least one solution u
n,λ
(t). Now, we will show that α(t) ≤ u
n,λ
(t) ≤ β(t)fort ∈
[0,1]. Su ppose that ε
0
= max
t∈[0,1]
{u
n,λ
(t) − β(t)} > 0. Let y
n,λ
(t) = u
n,λ
(t) − β(t). Then,

y
n,λ
(t) ≤ ε
0
for t ∈ [0,1]. Let t
0
∈ [t
1
,t
2
] ⊂ [0,1] be such that
(a) y
n,λ
(t
0
) = ε
0
,
(b) y
n,λ
(t) > 0fort ∈ (t
1
,t
2
),
(c) [t
1
,t
2
] is the maximal interval which has the properties (a) and (b).

Then we have the following three cases.
(1) If t
0
∈ (0, 1 ), then t
0
∈ (t
1
,t
2
), y

n,λ
(t
0
) = 0. Also
−y

n,λ
(t) ≤ λa(t)

g

β(t)+
1
n

+ h

β(t)


−
g

β(t)+
1
n

−
h

β(t)


=
0 (2.47)
for t
∈ [t
1
,t
2
]. Then y

n,λ
(t) ≤ 0fort ∈ (t
1
,t
0
), and y

n,λ

(t) ≥ 0fort ∈ (t
0
,t
2
).
Since y
n,λ
(t
0
) = max
t∈[0,1]
y
n,λ
(t), then y
n,λ
(t) = ε
0
for t ∈ [t
1
,t
2
], contradicting
the proper ties (b) and (c).
(2) If t
0
= 1, then y
n,λ
(1) = u
n,λ
(1) − β(1) ≤ γ(u

n,λ
(η) − β(η)) = γy
n,λ
(η) ≤ γy
n,λ
(1),
and so y
n,λ
(1) = 0, a contradiction.
(3) If t
0
= 0, then y
n,λ
(0) = u
n,λ
(0) − β(0) < 0, a contradiction.
Therefore, β(t)
≥ u
n,λ
(t)fort ∈ [0,1]. Similarly, we can show that α(t) ≤ u
n,λ
(t)for
t
∈ [0, 1]. Thus, u
n,λ
(t) is a positive solution of (2.1
λ
n
). The proof is complete. 
3. Proof of the main results

Proof of Theorem 2.1. Let
Λ
={λ ∈ (0, +∞)|(1.1
λ
) has at least one positive solution}. (3.1)
By Lemma 2.12, Λ
=∅. Assume that λ
0
∈ Λ. Then we can show that
(1) λ

∈ Λ for any 0 <λ

≤ λ
0
,
(2)
λ
0
≤
p
0,0
(1) − γp
0,0
(η)
q
1,0
(η)



η
(1/2)η
s
2
a(s)ds

−1
max

1
b
0
θ
ω
0
,
1
g(2)

. (3.2)
Assume that (1.1
λ
) has a positive solution z
0
(t). It is easy to see that z
0
(t)and0areupper
and lower solutions of (2.1
n
λ


)foreachn ∈ N, respectively. By Lemma 2.13,foreachn ∈ N,
(2.1
n
λ

) has a positive solution x
n,λ

such that 0 ≤ x
n,λ

≤ z
0
.Thus,byLemma 2.11,there
exist
¯
x
λ

∈ C[0,1] and a subsequence {x
n
k
,λ

}
+∞
k=1
of {x
n,λ


}
+∞
n=1
such that x
n
k
,λ

→
¯
x
λ

as
k
→ +∞ and
¯
x
λ

is a positive solution of (1.1
λ

). Thus, λ

∈ Λ.
X. Xian and D. O’Regan 13
From Lemma 2.7,wehavex
n

k
,λ

= λ

L
0
F
n
k
x
n
k
,λ

.ThenbyLemma 2.10,
x
n
k
,λ

(t) ≥ θ
0


x
n
k
,λ




t, t ∈ [0,1]. (3.3)
If
x
n
k
,λ

≤1, then by (H
2
), we have
1
≥


x
n
k
,λ



≥
x
n
k
,λ

(η) ≥

g(2)λ

p
0,0
(1)
p
0,0
− γp
0,0
(η)

1
0
G
(0)
[0,1]
(η,s)a(s)ds
=
g(2)λ

p
0,0
(1)
p
0,0
(1) − γp
0,0
(η)



η
0
q
1,0
(η)
p
0,0
(s)
p
0,0
(1)
a(s)ds+

1
η
p
0,0
(η)
q
1,0
(s)
q
1,0
(0)
a(s)ds

≥
g(2)λ

q

1,0
(η)
p
0,0
(1) − γp
0,0
(η)

η
(1/2)η
sa(s)ds,
(3.4)
and so
λ

≤
p
0,0
(1) − γp
0,0
(η)
g(2)q
1,0
(η)


η
(1/2)η
sa(s)ds


−1
. (3.5)
If
x
n
k
,λ

≥1, then by (H
2
)and(3.3), we have


x
n
k
,λ



≥
x
n
k
,λ

(η)
≥
b
0

λ

p
0,0
(1)
p
0,0
(1) − γp
0,0
(η)

1
0
G
(0)
[0,1]
(η,s)a(s)

x
n
k
,λ


w
ds
≥
b
0
λ


q
1,0
(η)θ
ω
0
p
0,0
(1) − γp
0,0
(η)

η
(1/2)η
s
2
a(s)ds


x
n
k
,λ



w
≥
b
0

λ

q
1,0
(η)θ
ω
0
p
0,0
(1) − γp
0,0
(η)

η
(1/2)η
s
2
a(s)ds


x
n
k
,λ



,
(3.6)
and so

λ

≤
p
0,0
(1) − γp
0,0
(η)
b
0
θ
ω
0
q
1,0
(η)


η
(1/2)η
s
2
a(s)ds

−1
. (3.7)
Then, (3.2)followsfrom(3.5)and(3.7), and (3.2) implies that Λ is a bounded set. Let
λ
∗
= supΛ. Therefore, (1.1

λ
) has at least one positive solution for 0 <λ<λ
∗
.
Finally, we will show that λ
∗
∈ Λ if ω>1. Let {λ
n
}⊂Λ be an increasing number se-
quence such that λ
n
→ λ
∗
as n → +∞,andλ
n
≥ λ
∗
/2forn = 1,2, Assume that (1.1
λ
n
)
has positive solution z
n
for each n ∈ N.Thenz
n
is an upper solution of (2.1
k
λ
n
)and0is

a lower solution of (2.1
k
λ
n
)foreachk ∈ N.ByLemma 2.13,(2.1
k
λ
n
) has a positive solution
z
n,k
such that 0 ≤ z
n,k
≤ z
n
.Then,byLemma 2.7,
z
n,k
= λ
n
L
0
F
k
z
n,k
. (3.8)
14 Solutions of three-point nonlinear BVPs
Let k
∈ N be fixed. Now we will show that {z

n,k
}
+∞
n=1
is bounded. In fact, by (3.8)and
Lemmas 2.6 and 2.10,wehave


z
n,k


≥

λ
n
L
0
F
k
z
n,k

(η)
≥
λ
∗
p
0,0
(1)

2

p
0,0
(1) − γp
0,0
(η)


η
(1/2)η
q
1,0
(η)
p
0,0
(s)
p
0,0
(1)
a(s)h

z
n,k
(s)

ds
≥
λ
∗

b
0
q
1,0
(η)
2

p
0,0
(1) − γp
0,0
(η)


η
(1/2)η
p
0,0
(s)a(s)

z
n,k
(s)

w
ds
≥
λ
∗
b

0
θ
w
0
q
1,0
(η)
2

p
0,0
(1) − γp
0,0
(η)


η
(1/2)η
s
2
a(s)


z
n,k
(s)


w
ds,

(3.9)
and so


z
n,k


≤

2

p
0,0
(1) − γp
0,0
(η)

λ
∗
q
1,0
(η)b
0
θ
w
0


η

(1/2)η
s
2
a(s)ds

−1

1/(w−1)
. (3.10)
This means that
{z
n,k
}
+∞
n=1
is a bounded set. Using the fact that L
0
: P → Q
0
is a com-
pletely continuous operator and
{λ
n
}
+∞
n=1
is a bounded set, we see that {z
n,k
} is a rela-
tively compact set. Without loss of generalit y, we assume that z

n,k
→ z
0,k
as n → +∞.Now
the Lebesgue dominant convergence theorem guarantees that z
0,k
= λ
∗
L
0
F
k
z
0,k
.Then,by
Lemma 2.7, z
0,k
is a positive solution of (1.1
k
λ
∗
). By Lemma 2.11,(1.1
λ
∗
) has a positive
solution u
∗
. The proof is complete. 
Proof of Theorem 2.2. Let λ
∗

be defined as in Theorem 2.1 and let λ ∈ (0, λ
∗
)befixed.Let
us define the nonlinear operators F and T
λ
by
(Fx)(t)
= f

x(t)

+ Mx(t), t ∈ [0,1], x ∈ P, (3.11)
and (T
λ
x)(t) = (λL
λM
Fx)(t)forallx ∈ P and t ∈ [0,1]. It follows from Lemma 2.7 that to
show that (1.1
λ
) has at least two positive solutions, we only need to show that the operator
T
λ
has at least two fixed points.
Let z
0
(t)=1fort ∈ [0,1] and Ω
λ
={x ∈ Q
λM
|∃τ>0suchthatT

λ
x≤u
∗
−τ(L
λM
z
0
)(t)}.
Since u
∗
is a positive solution of (1.1
λ
∗
), then
−

u
∗


(t)+λMa(t)u
∗
(t) = λa(t)

Fu
∗

(t)+

λ

∗
− λ

a(t) f

u
∗
(t)

,0<t<1,
u
∗
(0) = 0, u
∗
(1) = γu
∗
(η).
(3.12)
By Lemma 2.7,wehaveu
∗
= T
λ
u
∗
+(λ
∗
− λ)L
λM
f (u
∗

). Since L
λM
is increasing and
f (u
∗
) ≥ c,thenwehave
T
λ
u
∗
≤ u
∗
− c

λ
∗
− λ

L
λM
z
0

(t). (3.13)
This means that u
∗
∈ Ω
λ
,andsoΩ
λ

=∅.
X. Xian and D. O’Regan 15
For any x
0
∈ Ω
λ
,byLemma 2.10,wehave


u
∗


≥

T
λ
x

(η) ≥
λp
0,λM
(1)
p
0,λM
(1) − γp
0,λM
(η)

η

(1/2)η
q
1,λM
(η)
p
0,λM
(s)
p
0,λM
(1)
a(s)h

x(s)

ds
≥
λq
1,λM
(η)
p
0,λM
(1) − γp
0,λM
(η)

η
(1/2)η
sa(s)b
0


x
0
(s)

w
ds
≥
b
0
λθ
ω
λM
q
1,λM
(η)
p
0,λM
(1) − γp
0,λM
(η)

η
(1/2)η
s
2
a(s)


x
0

(s)


w
ds,
(3.14)
and so


x
0


≤

p
0,λM
(1) − γp
0,λM
(η)
b
0
λθ
ω
λM
q
1,λM
(η)



η
(1/2)η
s
2
a(s)ds

−1


u
∗



1/w
=: R
0
. (3.15)
This means that Ω
λ
is a bounded set.
For any x
0
∈ Ω
λ
, there exists τ
0
> 0suchthatT
λ
x

0
≤ u
∗
− τ
0
(L
λM
z
0
)(t). For any x ∈
Q
λM
,byLemma 2.10,wehavefort ∈ [0,1],

T
λ
x

(t) −

T
λ
x
0

(t) =

λL
λM


Fx− Fx
0

(t) ≤ λ


Fx− Fx
0



L
λM
z
0

(t), (3.16)
and since F is continuous on Q
λM
, then there exists δ>0suchthat
λ
Fx− Fx
0
≤
τ
0
2
(3.17)
for any x
∈ Q

λM
with x − x
0
 <δ.
By (3.16)and(3.17), we have

T
λ
x

(t) ≤ T
λ
x
0
(t)+
τ
0
2

L
λM
z
0

(t) ≤ u
∗
(t) −
τ
0
2


L
λM
z
0

(t), t ∈ [0,1], (3.18)
for any x
∈ Q
λM
with x − x
0
 <δ. This implies that x ∈ Ω
λ
,andsoΩ
λ
is an open set.
Now we will show that
μT
λ
x = x, x ∈ ∂Ω
λ
, μ ∈ [0,1]. (3.19)
Suppose (3.19) is not true. Then there exist x
0
∈ ∂Ω
λ
, μ
0
∈ [0,1] such that μ

0
T
λ
x
0
= x
0
.
Obviously, T
λ
x
0
≤ u∗,andsox
0
= μ
0
T
λ
x
0
≤ u
∗
. Since T
λ
is increasing, we have
T
λ
x
0
≤ T

λ
u
∗
≤ u
∗
− c

λ
∗
− λ

L
λM
z
0

(t). (3.20)
This implies that x
0
∈ Ω
λ
, a contradiction. Thus, (3.19) holds, and so
i

T
λ
,Ω
λ
,Q
λM


=
i

θ,Ω
λ
,Q
λM

=
1. (3.21)
Let
R

0
=

p
0,λM
(1) − γp
0,λM
(η)
b
0
θ
w
M
λq
1,λM
(η)



η
(1/2)η
s
2
a(s)ds

−1

1/(w−1)
,
(3.22)
16 Solutions of three-point nonlinear BVPs
and R
1
> max{R
0
,R

0
}.Foranyx ∈ ∂(B(θ, R
1
) ∩ Q
λM
), we have


T
λ

x


≥
(λTx)(η) ≥
λp
0,λM
(1)
p
0,λM
(1) − γp
0,λM
(η)

η
η/2
q
1,λM
(η)
p
0,λM
(s)
p
0,λM
(1)
a(s)h

x(s)

ds

≥
λb
0
q
1,λM
(η)
p
0,λM
(1) − γp
0,λM
(η)

η
η/2
sa(s)

x(s)

w
ds
≥
θ
w
M
λb
0
q
1,λM
(η)
p

0,λM
(1) − γp
0,λM
(η)

η
η/2
s
2
a(s)


x(s)


w
ds>R
1
.
(3.23)
Then, we have
i

T
λ
,B

θ,R
1


∩
Q
λM
,Q
λM
) = 0. (3.24)
By (3.21)and(3.24), we have
i

λ,

B

θ,R
1

∩
Q
λM

\
¯
Ω
λ
,Q
λM

=
0 − 1 =−1. (3.25)
It follows from (3.21)and(3.25)thatT

λ
has at least two fixed points in (B(θ,R
1
) ∩
Q
λM
)\
¯
Ω
λ
and Ω
λ
, respectively. Thus (1.1
λ
) has at least two positive solutions for 0 <λ<
λ
∗
. 
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Xu Xian: Department of Mathematics, Xuzhou Normal University, Xuzhou, Jiangsu 221116, China
E-mail address:
Donal O’Regan: Department of Mathematics, National University of Ireland, Galway,
University Road, Galway, Ireland
E-mail address: