2 Preliminaries and lemmas
In this section, we shall state some necessary definitions and preliminaries.
Definition 2.1. Let E be a real Banach space. A nonempty closed convex set K ⊂ E is called a cone if it satisfies the following two conditions:
(1) x ∈ K, λ > 0 implies λx ∈ K;
(2) x ∈ K, -x ∈ K implies x = 0.
Definition 2.2. An operator T is called completely continuous if it is continuous and maps bounded sets into precompact sets.
For convenience, we list the following assumptions:
(H1) For i ∈ {1,2, ⋯, m - 2}, ξi ∈ (0, 1), 0 < ξ1 < ξ2 < ⋯ < ξm-2 < 1 and αi, βi ∈ [0, +∞) satisfying 0
<
∑
i
=
1
m
-
2
α
i
<
1
, 0
<
∑
i
=
1
m
-
2
β
i
<
1
and 0
<
∑
i
=
1
m
-
2
α
i
ξ
i
<
1
.
(H2) f ∈ C((0, 1) × R × R × R, R) and there exist functions p, q ∈ L1((0, 1), (0, +∞)), g ∈ C(R × R × R, (0, +∞)) such that
-
p
(
t
)
≤
f
(
t
,
x
1
,
x
2
,
x
3
)
≤
q
(
t
)
g
(
x
1
,
x
2
,
x
3
)
for
(
t
,
x
1
,
x
2
,
x
3
)
∈
(
0
,
1
)
×
R
×
R
×
R
.
(H3)
lim
(
|
x
1
|
+
|
x
2
|
+
|
x
3
|
)
→
+
∞
f
(
t
,
x
1
,
x
2
,
x
3
)
|
x
1
|
+
|
x
2
|
+
|
x
3
|
=
+
∞
for t uniformly on [0,1].
Remark 2.1. From (H2) we know that for given points t1, t2,..., tm on [0,1], the functions p, q = (0, 1)\{ti, i = 1, 2,..., m} → (0, +∞) are continuous and integrable, that is 0
<
∫
0
1
(
p
(
t
)
+
q
(
t
)
)
d
t
<
+
∞
. The condition (H2) also implies that f may have finitely singularities at t1, t2,..., tm on [0,1].
Lemma 2.1. Suppose that (H1) and (H2) hold. Then the problem (1.1) has a positive solution if and only if the following nonlinear second-order integro-differential equation
x
″
(
t
)
+
λ
f
t
,
∫
0
t
(
t
-
u
)
x
(
u
)
d
u
,
∫
0
t
x
(
u
)
d
u
,
x
(
t
)
=
0
,
0
<
t
<
1
,
x
′
(
0
)
=
∑
i
=
1
m
-
2
β
i
x
′
(
ξ
i
)
,
x
(
1
)
=
∑
i
=
1
m
-
2
α
i
x
(
ξ
i
)
has a positive solution.
Proof. Let y(t) be a positive solution of the problem (1.1) and let x(t) = y''(t). Then it follows from the problem (1.1) and combining with exchanging the integral sequence we know that
y
(
t
)
=
∫
0
t
(
t
-
u
)
x
(
u
)
d
u
,
y
′
(
t
)
=
∫
0
t
x
(
u
)
d
u
.
Thus x(t) = y''(t) is a positive solution of the second-order integro-differential equation multi-point boundary value problem (2.1).
Conversely, let x(t) be a positive solution of the problem (2.1), then y
(
t
)
=
∫
0
t
(
t
-
u
)
x
(
u
)
d
u
is a positive solution of the problem (1.1). In fact,
y
′
(
t
)
=
∫
0
t
x
(
u
)
d
u
,
y
″
(
t
)
=
x
(
t
)
, which implies that y(0) = 0, y'(0) = 0. The proof is complete. □
Now, let X = C[0,1]. Then X is a real Banach space with norm ∥
x
∥
=
max
t
∈
[
0
,
1
]
|
x
(
t
)
|
for x ∈ C[0, 1]. Let
C
+
[
0
,
1
]
=
{
x
∈
C
[
0
,
1
]
:
x
(
t
)
≥
0
,
t
∈
[
0
,
1
]
}
.
Lemma 2.2. Suppose that (H1) holds. In addition, assume that u(t) ∈ L1(0, 1) and u(t) ≥ 0. Then the following problem
x
″
(
t
)
+
u
(
t
)
=
0
,
0
<
t
<
1
,
x
′
(
0
)
=
∑
i
=
1
m
-
2
β
i
x
′
(
ξ
i
)
,
x
(
1
)
=
∑
i
=
1
m
-
2
α
i
x
(
ξ
i
)
has a unique positive solution
x
(
t
)
=
-
∫
0
t
(
t
-
s
)
u
(
s
)
d
s
+
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
t
+
1
1
-
∑
i
=
1
m
-
2
α
i
∫
0
1
(
1
-
s
)
u
(
s
)
d
s
-
1
1
-
∑
i
=
1
m
-
2
α
i
∑
i
=
1
m
-
2
α
i
∫
0
ξ
i
(
ξ
i
-
s
)
u
(
s
)
d
s
+
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
satisfies x(t) ≥ 0, t ∈ [0,1] and
min
t
∈
[
0
,
1
]
x
(
t
)
≥
w
∥
x
∥
,
where
ω
=
∑
i
=
1
m
-
2
α
i
(
1
-
ξ
i
)
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
.
Proof. From (2.2), we have x'' (t) = -u(t), 0 < t < 1. For t ∈ [0,1], integrating from 0 to t we get
x
′
(
t
)
=
x
′
(
0
)
-
∫
0
t
u
(
s
)
d
s
.
Thus
x
′
(
0
)
=
∑
i
=
1
m
-
2
β
i
x
′
(
ξ
i
)
=
∑
i
=
1
m
-
2
β
i
∑
i
=
1
m
-
2
β
i
-
1
∫
0
ξ
i
u
(
s
)
d
s
.
For t ∈ [0, 1], integrating (2.6) from t to 1 yields
x
(
1
)
-
x
(
t
)
=
∑
i
=
1
m
-
2
β
i
(
1
-
t
)
∑
i
=
1
m
-
2
β
i
-
1
∫
0
ξ
i
u
(
s
)
d
s
-
∫
0
t
(
s
-
t
)
u
(
s
)
d
s
-
∫
0
1
(
1
-
s
)
u
(
s
)
d
s
,
which means that
-
x
(
t
)
=
-
∑
i
=
1
m
-
2
α
i
x
(
ξ
i
)
+
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
-
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
t
+
∫
0
t
(
t
-
s
)
u
(
s
)
d
s
-
∫
0
1
(
1
-
s
)
u
(
s
)
d
s
.
From (2.9), we have
x
(
ξ
i
)
=
1
1
-
∑
i
=
1
m
-
2
α
i
∫
0
1
(
1
-
s
)
u
(
s
)
d
s
-
∑
i
=
1
m
-
2
β
i
(
1
-
ξ
i
)
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
-
∫
0
ξ
i
(
ξ
i
-
s
)
u
(
s
)
d
s
.
It follows from (2.9) and (2.10) that
x
(
t
)
=
-
∫
0
t
(
t
-
s
)
u
(
s
)
d
s
+
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
t
+
∫
0
1
(
1
-
s
)
u
(
s
)
d
s
+
1
1
-
∑
i
=
1
m
-
2
α
i
∑
i
=
1
m
-
2
α
i
∫
0
1
(
1
-
s
)
u
(
s
)
d
s
-
∑
i
=
1
m
-
2
α
i
∫
0
ξ
i
(
ξ
i
-
s
)
u
(
s
)
d
s
-
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
⋅
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
=
-
∫
0
t
(
t
-
s
)
u
(
s
)
d
s
+
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
t
+
1
1
-
∑
i
=
1
m
-
2
α
i
∫
0
1
(
1
-
s
)
u
(
s
)
d
s
-
1
1
-
∑
i
=
1
m
-
2
α
i
∑
i
=
1
m
-
2
α
i
∫
0
ξ
i
(
ξ
i
-
s
)
u
(
s
)
d
s
+
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
.
Combining (2.11) with (H1) we know that
x
(
0
)
=
1
1
-
∑
i
=
1
m
-
2
α
i
∫
0
1
(
1
-
s
)
u
(
s
)
d
s
-
∑
i
=
1
m
-
2
α
i
∫
0
ξ
i
(
ξ
i
-
s
)
u
(
s
)
d
s
-
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
≥
1
1
-
∑
i
=
1
m
-
2
α
i
∑
i
=
1
m
-
2
α
i
∫
0
1
(
1
-
s
)
u
(
s
)
d
s
-
∑
i
=
1
m
-
2
α
i
∫
0
ξ
i
(
ξ
i
-
s
)
u
(
s
)
d
s
-
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
∑
i
=
1
m
-
2
β
i
-
1
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
≥
∑
i
=
1
m
-
2
α
i
1
-
∑
i
=
1
m
-
2
α
i
∫
ξ
i
1
(
1
-
s
)
u
(
s
)
d
s
+
∑
i
=
1
m
-
2
β
i
∫
0
ξ
i
u
(
s
)
d
s
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
1
-
∑
i
=
1
m
-
2
β
i
1
-
∑
i
=
1
m
-
2
α
i
≥
0
.
From the fact that x'' (t) = -u(t) ≤ 0, we know that the graph of x(t) is concave on [0,1].
Thus
If x(1) ≥ 0, we know that x(t) ≥ 0 for all t ∈[0,1].
If x(1) < 0, from the concavity of x once again we know that
x
(
ξ
i
)
ξ
i
≥
x
(
1
)
1
for i ∈ {1, 2, ..., m - 2}. This implies
x
(
1
)
=
∑
i
=
1
m
-
2
α
i
x
(
ξ
i
)
≥
∑
i
=
1
m
-
2
α
i
ξ
i
x
(
1
)
,
which contracts with the fact 0
<
∑
i
=
1
m
-
2
α
i
ξ
i
<
1
. Thus we know that (2.4) holds.
Again from x''(t) = -u(t) ≤ 0, we see that x'(t) is non-increasing on (0, 1). Combining the condition 0
<
∑
i
=
1
m
-
2
β
i
<
1
we have x' (0) ≤ 0 and
x
′
(
t
)
=
x
′
(
0
)
-
∫
0
t
u
(
s
)
d
s
≤
0
for t ∈ (0, 1). Hence x(t) is non-increasing on (0, 1). By making use of the concavity of x(t) on (0, 1) we get ||x|| = x(0) and
min
t
∈
[
0
,
1
]
x
(
t
)
=
x
(
1
)
. Therefore, for all i = 1, 2, ⋯, m - 2, we obtain
∥
x
∥
=
x
(
0
)
≤
x
(
1
)
+
x
(
ξ
i
)
-
x
(
1
)
1
-
ξ
i
(
1
-
0
)
=
x
(
1
)
+
x
(
ξ
i
)
1
-
ξ
i
-
x
(
1
)
1
-
ξ
i
=
x
(
ξ
i
)
-
x
(
1
)
ξ
i
1
-
ξ
i
=
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
∑
i
=
1
m
-
2
α
i
(
1
-
ξ
i
)
x
(
1
)
,
which implies that
min
t
∈
[
0
,
1
]
x
(
t
)
≥
∑
i
=
1
m
-
2
α
i
(
1
-
ξ
i
)
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
∥
x
∥
=
ω
∥
x
∥
where ω is given by (2.5). This completes the proof. □
Lemma 2.3. Suppose that (H1) holds. In addition, assume that p ∈ L1((0, 1), (0, +∞))
.
Then the following boundary value problem
x
″
(
t
)
+
λ
p
(
t
)
=
0
,
0
<
t
<
1
,
x
′
(
0
)
=
∑
i
=
1
m
-
2
β
i
x
′
(
ξ
i
)
,
x
(
1
)
=
∑
i
=
1
m
-
2
α
i
x
(
ξ
i
)
has a unique positive solution z satisfying z(t) ≥ 0, t ∈ [0,1],
min
t
∈
[
0
,
1
]
z
(
t
)
≥
ω
∥
z
∥
and
z
(
t
)
≤
λ
G
ω
∫
0
1
p
(
s
)
d
s
,
where
G
=
1
1
-
∑
i
=
1
m
-
2
α
i
+
∑
i
=
1
m
-
2
β
i
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
1
-
∑
i
=
1
m
-
2
α
i
1
-
∑
i
=
1
m
-
2
β
i
ω
-
1
,
ω is given by (2.5).
Proof. From Lemma 2.2. we have z(t) ≥ 0 and
min
t
∈
[
0
,
1
]
z
(
t
)
≥
ω
∥
z
∥
,
t ∈ [0,1]. By making use of (2.3) we get
z
(
t
)
≤
λ
1
−
∑
i
=
1
m
−
2
α
i
(
∫
0
1
(
1
−
s
)
p
(
s
)
d
s
−
∑
i
=
1
m
−
2
α
i
∫
0
ξ
i
(
ξ
i
−
s
)
p
(
s
)
d
s
+
∑
i
=
1
m
−
2
β
i
∫
0
ξ
i
p
(
s
)
d
s
1
−
∑
i
=
1
m
−
2
β
i
)
⋅
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
≤
λ
1
−
∑
i
=
1
m
−
2
α
i
(
∫
0
1
p
(
s
)
d
s
+
∑
i
=
1
m
−
2
β
i
∫
0
1
p
(
s
)
d
s
1
−
∑
i
=
1
m
−
2
β
i
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
)
=
λ
∫
0
1
p
(
s
)
d
s
(
1
1
−
∑
i
=
1
m
−
2
α
i
+
∑
i
=
1
m
−
2
β
i
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
(
1
−
∑
i
=
1
m
−
2
α
i
)
(
1
−
∑
i
=
1
m
−
2
β
i
)
)
=
λ
G
ω
∫
0
1
p
(
s
)
d
s
.
The proof is complete. □
Let
[
v
(
t
)
]
*
=
v
(
t
)
,
v
(
t
)
≥
0
,
0
,
v
(
t
)
<
0
,
0
<
t
<
1
.
and
F
(
t
,
[
x
(
t
)
−
z
(
t
)
]
*
)
=
f
(
t
,
∫
0
t
(
t
−
u
)
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
∫
0
t
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
[
x
(
t
)
−
z
(
t
)
]
*
)
+
p
(
t
)
.
Lemma 2.4. Suppose that (H1) and (H2) hold. Then the following nonlinear second-order integro-differential equation boundary value problem
{
x
″
(
t
)
+
λ
[
f
(
t
,
∫
0
t
(
t
−
u
)
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
∫
0
t
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
[
x
(
t
)
−
z
(
t
)
]
*
)
+
p
(
t
)
]
=
0
,
0
<
t
<
1
,
x
′
(
0
)
=
∑
i
=
1
m
−
2
β
i
x
′
(
ξ
i
)
,
x
(
1
)
=
∑
i
=
1
m
−
2
α
i
x
(
ξ
i
)
has a positive solution x(t) with x(t) ≥ z(t) for t ∈ [0, 1] if and only if y(t) = x(t) - z(t) is a nonnegative solution (positive on (0, 1)) of the problem (2.1).
Proof. Assume that y(t) = x(t) - z(t) is a nonnegative solution (positive on (0,1)) of the problem (2.1). Then we know that x(t) ≥ z(t) and
y
″
(
t
)
+
λ
f
t
,
∫
0
t
(
t
-
u
)
y
(
u
)
d
u
,
∫
0
t
y
(
u
)
d
u
,
y
(
t
)
=
0
.
Noticing that z is a positive solution of the problem (2.13). Thus we get
−
x
″
(
t
)
=
λ
[
f
(
t
,
∫
0
t
(
t
−
u
)
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
∫
0
t
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
[
x
(
t
)
−
z
(
t
)
]
*
)
+
p
(
t
)
]
Therefore x(t) is a positive solution of the problem (2.15) with x(t) ≥ z(t) for t ∈ [0,1].
Conversely, we assume that x(t) and z(t) are positive solutions of the problem (2.15) and the problem (2.13), respectively, and it implies that the boundary conditions of the problem (2.13) are also satisfied. Thus y(t) = x(t) - z(t) is a nonnegative solution (positive on (0, 1)) of the problem (2.1). The proof is complete. □
Remark 2.2. Combining Lemma 2.4. with Lemma 2.1. we know that if the problem (2.15) has a positive solution, then the fourth-order multi-point boundary value problem (1.1) has a positive solution. So, we need only to study the problem (2.15).
Remark 2.3. For any fixed x ∈ C+[0,1], let L
=
max
t
∈
[
0
,
1
]
x
(
t
)
. Noticing that [x(u) - z(u)]* ≤ x(u) ≤ L and
∫
0
s
(
s
-
u
)
[
x
(
u
)
-
z
(
u
)
]
*
d
u
≤
∫
0
1
L
d
u
=
L
, by virtue of (H2), we obtain
∫
0
1
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
≤
∫
0
1
[
q
(
s
)
g
(
∫
0
s
(
s
−
u
)
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
∫
0
s
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
[
x
(
s
)
−
z
(
s
)
]
*
)
+
p
(
s
)
]
d
s
≤
∫
0
1
(
p
(
s
)
+
M
q
(
s
)
)
d
s
≤
M
∫
0
1
(
p
(
s
)
+
q
(
s
)
)
d
s
<
+
∞
,
where
M
=
max
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
∈
[
0
,
L
]
×
[
0
,
L
]
×
[
0
,
L
]
g
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
+
1
.
We introduce an integral mapping T : C+[0, 1] → C+[0, 1] defined by
(
T
x
)
(
t
)
=
−
λ
(
∫
0
t
(
t
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
−
t
∑
i
=
1
m
−
2
β
i
∫
0
ξ
i
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
1
−
∑
i
=
1
m
−
2
β
i
)
+
λ
1
−
∑
i
=
1
m
−
2
α
i
(
∫
0
1
(
1
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
−
∑
i
=
1
m
−
2
α
i
∫
0
ξ
i
(
ξ
i
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
+
∑
i
=
1
m
−
2
β
i
∫
0
ξ
i
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
1
−
∑
i
=
1
m
−
2
β
i
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
)
.
Denote
K
=
{
y
∈
C
+
[
0
,
1
]
:
min
0
≤
t
≤
1
y
(
t
)
≥
ω
|
|
y
|
|
}
,
where ω is given by the problem (2.5). It is obvious that K is a positive cone of C[0,1].
Lemma 2.5. Suppose that (H1) - (H3) hold. Then T : K →K is a completely continuous operator.
Proof. It follows from Lemma 2.2. we see T (K) ⊂ K. Combining (H1) with (2.18) we know that T (K) is equicontinuous and uniformly bounded. In fact, let D ⊂ C+[0,1] be a bounded set. Then there exists M0 > 0 such that ||x|| ≤ M0 for all x ∈ D. By virtue of (H2) we obtain
|
(
T
x
)
(
t
)
|
=
|
−
λ
∫
0
t
(
t
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
−
λ
∑
i
=
1
m
−
2
β
i
∫
0
ξ
i
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
1
−
∑
i
=
1
m
−
2
β
i
t
+
λ
1
−
∑
i
=
1
m
−
2
α
i
(
∫
0
1
(
1
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
−
∑
i
=
1
m
−
2
α
i
∫
0
ξ
i
(
ξ
i
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
−
∑
i
=
1
m
−
2
β
i
∫
0
ξ
i
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
∑
i
=
1
m
−
2
β
i
−
1
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
)
|
≤
|
λ
1
−
∑
i
=
1
m
−
2
α
i
(
∫
0
1
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
+
∑
i
=
1
m
−
2
β
i
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
1
−
∑
i
=
1
m
−
2
β
i
∫
0
1
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
)
|
=
λ
1
−
∑
i
=
1
m
−
2
α
i
(
1
+
∑
i
=
1
m
−
2
β
i
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
1
−
∑
i
=
1
m
−
2
β
i
)
∫
0
1
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
≤
λ
M
(
1
−
∑
i
=
1
m
−
2
β
i
∑
i
=
1
m
−
2
α
i
ξ
i
)
(
1
−
∑
i
=
1
m
−
2
α
i
)
(
1
−
∑
i
=
1
m
−
2
β
i
)
∫
0
1
(
q
(
s
)
+
p
(
s
)
)
d
s
<
+
∞
,
which implies that T (K) is uniformly bounded.
On the other hand, for all x ∈ D, once again from (H2) we have
0
≤
|
(
T
x
)
′
(
t
)
|
=
λ
|
−
∫
0
t
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
−
∑
i
=
1
m
−
2
β
i
∫
0
ξ
i
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
1
−
∑
i
=
1
m
−
2
β
i
|
≤
λ
|
∫
0
1
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
+
∑
i
=
1
m
−
2
β
i
1
−
∑
i
=
1
m
−
2
β
i
∫
0
1
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
|
=
λ
1
−
∑
i
=
1
m
−
2
β
i
∫
0
1
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
≤
λ
M
1
−
∑
i
=
1
m
−
2
β
i
∫
0
1
[
q
(
s
)
+
p
(
s
)
]
d
s
=
λ
M
M
*
1
−
∑
i
=
1
m
−
2
β
i
<
+
∞
,
here
M
*
=
∫
0
1
[
p
(
s
)
+
q
(
s
)
]
d
s
. So, for any 0 ≤ t1 < t2 ≤ 1, and for all x ∈ D, we get
|
(
T
x
)
(
t
1
)
-
(
T
x
)
(
t
2
)
|
=
∫
t
1
t
2
(
T
x
)
′
(
t
)
d
t
≤
λ
M
M
*
1
-
∑
i
=
1
m
-
2
β
i
|
t
2
-
t
1
|
.
By the absolutely continuous of integral, we know that T(K) is equicontinuous on [0,1]. Thus, an application of the Ascoli-Arzela theorem we know that T(K) is a relatively compact set.
Now we show that T is continuous. Let xn → x* (n → ∞), xn, x* ∈ C+[0,1]. It follows from the Lebesgue control convergence theorem that we obtain
∥
(
T
x
n
)
(
t
)
-
(
T
x
*
)
(
t
)
∥
→
0
(
n
→
∞
)
,
t
∈
[
0
,
1
]
.
Therefore T : K →K is a completely continuous operator. The proof is complete. □
Lemma 2.6. 17 Let X = (X, ||·||) be a Banach space and K ⊂ X be a cone. For r > 0 define Kr = {u ∈ K : ||u|| < r}. Assume that T
:
K
¯
r
→
K
is a completely continuous operator, such that Tu ≠ u for u ∈ ∂Kr = {u ∈ K : ||u|| = r}, and
(1) If ||Tu|| ≥ ||u|| for u ∈ ∂Kr, then i(T, Kr, K) = 0,
(2) If ||Tu|| ≤ ||u|| for u ∈ ∂Kr, then i(T, Kr, K) = 1.
3 Main results
In this section, we shall apply Lemma 2.6. to establish the existence of at least one positive solutions of the problem (1.1).
Theorem 3.1. Suppose that (H1)-(H3) hold. Then there exists sufficiently small λ* > 0 such that the problem (1.1) has at least one positive solution for any λ ∈ (0, λ*)
.
Proof. Let r > 1 and λ ∈ (0, λ*) be fixed, where
λ
*
=
min
r
M
1
∫
0
1
(
p
(
s
)
+
q
(
s
)
)
d
s
1
-
∑
i
=
1
m
-
2
β
i
∑
i
=
1
m
-
2
α
i
ξ
i
1
-
∑
i
=
1
m
-
2
α
i
1
-
∑
i
=
1
m
-
2
β
i
-
1
,
r
G
∫
0
1
p
(
s
)
d
s
,
1
here
M
1
=
max
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
∈
[
0
,
r
]
×
[
0
,
r
]
×
[
0
,
r
]
{
g
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
}
+
1
. Choose Ωr = {x ∈ C+ [0,1]: ||x|| < r}. If there is a fixed point on ∂Ωr, we complete the proof. Without loss of generality, we may assume that there is no fixed point on ∂Ωr. Thus, for any x ∈ K ∩∂Ωr, from (3.1) we get
(
T
x
)
(
t
)
≤
λ
1
−
∑
i
=
1
m
−
2
α
i
∫
0
1
(
1
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
+
λ
∑
i
=
1
m
−
2
β
i
∫
0
ξ
i
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
(
1
−
∑
i
=
1
m
−
2
α
i
)
(
1
−
∑
i
=
1
m
−
2
β
i
)
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
≤
λ
M
1
(
1
−
∑
i
=
1
m
−
2
β
i
∑
i
=
1
m
−
2
α
i
ξ
i
)
∫
0
1
(
p
(
s
)
+
q
(
s
)
)
d
s
(
1
−
∑
i
=
1
m
−
2
α
i
)
(
1
−
∑
i
=
1
m
−
2
β
i
)
<
r
=
∥
x
∥
.
So
∥
T
x
∥
<
|
|
x
∥
for all
x
∈
K
∩
∂
Ω
r
.
It follows from Lemma 2.6. we know that
i
(
T
,
Ω
r
,
Ω
)
=
1
.
Let d be a real number such that
1
2
λ
d
w
≥
r
.
Choose R > max{r, λdω + 1} such that if φ
>
1
r
R
ω
, then
F
(
t
,
φ
)
φ
≥
d
for
t
∈
[
0
,
1
]
and
1
-
λ
G
p
(
s
)
R
∫
0
1
p
(
s
)
d
s
≥
1
r
.
Let z
=
p
(
s
)
z
̃
, where
z
̃
is the unique solution of the problem (2.13). Denote
Ω
R
=
{
x
∈
C
+
[
0
,
1
]
:
∥
x
∥
<
R
}
.
From (2.14) we have
z
(
s
)
=
p
(
s
)
z
̃
(
s
)
≤
λ
p
(
s
)
G
w
∫
0
1
p
(
s
)
d
s
≤
λ
p
(
s
)
G
∫
0
1
p
(
s
)
d
s
x
(
s
)
∥
x
∥
,
∀
x
∈
K
∩
∂
Ω
R
.
Thus
x
(
s
)
-
z
(
s
)
≥
1
-
λ
G
p
(
s
)
∫
0
1
p
(
s
)
d
s
R
x
(
s
)
.
Combining (3.7) with (3.5) and by making use of Lemma 2.2. we know that
x
(
s
)
-
z
(
s
)
≥
1
r
x
(
s
)
≥
1
r
min
s
∈
[
0
,
1
]
x
(
s
)
≥
1
r
∥
x
∥
w
≥
1
r
R
w
,
s
∈
[
0
,
1
]
,
x
∈
K
∩
∂
Ω
R
.
From (3.8) together with (3.4), we see that
F
(
s
,
(
x
-
z
)
)
≥
d
(
x
-
z
)
≥
R
d
w
r
,
s
∈
[
0
,
1
]
.
Then
∫
0
s
(
s
−
u
)
[
x
(
u
)
−
z
(
u
)
]
*
d
u
+
∫
0
s
[
x
(
u
)
−
z
(
u
)
]
*
d
u
+
[
x
(
s
)
−
z
(
s
)
]
*
=
∫
0
s
(
s
−
u
)
[
x
(
u
)
−
z
(
u
)
]
d
u
+
∫
0
s
[
x
(
u
)
−
z
(
u
)
]
d
u
+
[
x
(
s
)
−
z
(
s
)
]
>
[
x
(
s
)
−
z
(
s
)
]
≥
1
r
R
w
,
s
∈
[
0
,
1
]
.
Thus, for any x ∈ K ∩ ∂ΩR, it follows from (H3) we know that
f
(
s
,
∫
0
s
(
s
−
u
)
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
∫
0
s
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
[
x
(
s
)
−
z
(
s
)
]
*
)
≥
d
(
∫
0
s
(
s
−
u
)
[
x
(
u
)
−
z
(
u
)
]
*
d
u
+
∫
0
s
[
x
(
u
)
−
z
(
u
)
]
*
d
u
+
[
x
(
s
)
−
z
(
s
)
]
*
)
+
p
(
s
)
≥
d
[
x
(
s
)
−
z
(
s
)
]
*
≥
d
[
x
(
s
)
−
z
(
s
)
]
≥
d
R
w
r
,
s
∈
[
0
,
1
]
.
Therefore, in view of (2.18) and (3.9) together with (3.3) we get
(
T
x
)
(
0
)
=
λ
1
−
∑
i
=
1
m
−
2
α
i
(
∫
0
1
(
1
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
−
∑
i
=
1
m
−
2
α
i
∫
0
ξ
i
(
ξ
i
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
−
∑
i
=
1
m
−
2
β
i
∫
0
ξ
i
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
∑
i
=
1
m
−
2
β
i
−
1
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
)
≥
λ
1
−
∑
i
=
1
m
−
2
α
i
[
∫
0
1
(
1
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
−
∑
i
=
1
m
−
2
α
i
∫
0
1
(
ξ
i
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
]
=
λ
∫
0
1
(
1
−
s
)
F
(
s
,
[
x
(
s
)
−
z
(
s
)
]
*
)
d
s
=
λ
∫
0
1
(
1
−
s
)
[
f
(
s
,
∫
0
s
(
s
−
u
)
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
∫
0
s
[
x
(
u
)
−
z
(
u
)
]
*
d
u
,
[
x
(
s
)
−
z
(
s
)
]
*
)
+
p
(
s
)
]
d
s
≥
λ
d
R
w
2
r
≥
R
which implies that ||Tx|| ≥ || x || for all x ∈K ∩ ∂ΩR. It follows from Lemma 2.6 that
i
(
T
,
K
∩
Ω
R
,
K
)
=
0
.
Combining (3.2) with (3.10) and the additivity of fixed point index, we know that
i
(
T
,
K
∩
(
Ω
¯
R
\
Ω
r
)
,
K
)
=
i
(
T
,
K
∩
Ω
¯
R
,
K
)
-
i
(
T
,
K
∩
Ω
r
,
K
)
=
-
1
.
As a result, there exists
x
*
∈
K
∩
(
Ω
¯
R
\
Ω
r
)
satisfying Tx* = x* and r ≤ ||x*|| ≤ R. From (3.1) we have
x
*
(
t
)
-
z
(
t
)
≥
w
∥
x
*
∥
-
λ
G
w
∫
0
1
p
(
s
)
d
s
≥
w
∥
x
*
∥
-
r
w
>
0
.
Let y(t) = x*(t) - z(t). Then y(t) is a positive solution of the problem (2.1). By virtue of Lemma 2.1. we know that y
(
t
)
=
∫
0
t
(
t
-
s
)
x
*
(
s
)
d
s
is a positive solution of the fourth-order multi-point boundary value problem (1.1). This completes the proof. □
Remark 3.1. In the case, when f = f(t, u) and f has lower bound i. e. f(t, u) + M ≥ 0 for some M > 0, we can study the second-order multi-point boundary value problem under suitable condition by making use of the similar method. In particular, if p(t) = M, the conclusion of Theorem 3.1. is still valid.
Remark 3.2. The constant λ in problem (1.1) is usually called the Thiele modulus, in ap-plications, one is interested in showing the existence of positive solutions for semipositone problems for small enough λ > 0.
4 Examples
Example 4.1. Consider the following singular fourth-order semipositone boundary value problem:
{
y
(
4
)
(
t
)
+
4
λ
3
(
1
−
t
)
2
3
[
sin
8
(
|
y
(
t
)
|
+
|
y
′
(
t
)
|
)
+
e
|
y
(
t
)
|
+
|
y
′
(
t
)
|
+
|
y
′
′
(
t
)
|
+
(
|
y
(
t
)
|
+
|
y
′
(
t
)
|
+
|
y
′
′
(
t
)
|
)
1
3
]
−
2
t
=
0
,
t
∈
(
0
,
1
)
y
(
0
)
=
y
′
(
0
)
=
0
,
y
′
′
(
1
)
=
1
3
y
′
′
(
1
2
)
,
y
′
′
′
(
0
)
=
1
4
y
′
′
′
(
1
2
)
Proof. Let
f
(
t
,
u
1
,
u
2
,
u
3
)
=
4
3
(
1
−
t
)
2
3
[
sin
8
(
|
u
1
|
+
|
u
2
|
)
+
e
|
u
1
|
+
|
u
2
|
+
|
u
3
|
+
(
|
u
1
|
+
|
u
2
|
+
|
u
3
|
)
1
3
]
−
2
t
Then
-
p
(
t
)
≤
f
(
t
,
u
1
,
u
2
,
u
3
)
≤
q
(
t
)
g
(
u
1
,
u
2
,
u
3
)
and
lim
(
|
u
1
|
+
|
u
2
|
+
|
u
3
|
)
→
+
∞
f
(
t
,
u
1
,
u
2
,
u
3
)
|
u
1
|
+
|
u
2
|
+
|
u
3
|
=
+
∞
where p
(
t
)
=
1
t
,
q
(
t
)
=
4
3
(
1
-
t
)
2
3
,
g
(
u
1
,
u
2
,
u
3
)
=
sin
8
(
|
u
1
|
+
|
u
2
|
)
+
e
|
u
1
|
+
|
u
2
|
+
|
u
3
|
+
(
|
u
1
|
+
|
u
2
|
+
|
u
3
|
)
1
3
, which implies that (H1)-(H3) hold. Since α
=
1
3
,
β
=
1
4
,
ξ
=
1
2
, then we know that
∫
0
1
(
p
(
s
)
+
q
(
s
)
)
d
s
=
∫
0
1
2
s
+
4
3
(
1
-
s
)
2
3
d
s
=
8
.
Take r = 2, then
M
1
=
max
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
∈
[
0
,
2
]
×
[
0
,
2
]
×
[
0
,
2
]
{
g
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
}
+
1
=
max
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
∈
[
0
,
2
]
×
[
0
,
2
]
×
[
0
,
2
]
{
[
sin
8
(
|
u
1
|
+
|
u
2
|
)
+
e
|
u
1
|
+
|
u
2
|
+
|
u
3
|
+
(
|
u
1
|
+
|
u
2
|
+
|
u
3
|
)
1
3
]
}
+
1
=
2
+
e
6
+
6
3
.
G
=
(
1
1
−
∑
i
=
1
m
−
2
α
i
+
∑
i
=
1
m
−
2
β
i
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
(
1
−
∑
i
=
1
m
−
2
α
i
)
(
1
−
∑
i
=
1
m
−
2
β
i
)
)
(
1
−
∑
i
=
1
m
−
2
α
i
ξ
i
)
∑
i
=
1
m
−
2
α
i
(
1
−
ξ
i
)
=
115
12
.
r
G
∫
0
1
p
(
s
)
d
s
=
2
G
∫
0
1
2
s
d
s
=
6
115
.
r
[
M
1
∫
0
1
(
p
(
s
)
+
q
(
s
)
)
d
s
(
1
−
∑
i
=
1
m
−
2
β
i
∑
i
=
1
m
−
2
α
i
ξ
i
)
(
1
−
∑
i
=
1
m
−
2
α
i
)
(
1
−
∑
i
=
1
m
−
2
β
i
)
]
−
1
=
3
23
(
2
+
e
6
+
6
3
)
.
It follows from Theorem 3.1 that the problem (4.1) has at least one positive solution for any
λ
*
∈
0
,
3
23
(
2
+
e
6
+
6
3
)
. □
Example 4.2. Consider the following singular fourth-order semipositone boundary value problem:
{
y
(
4
)
(
t
)
+
8
λ
π
t
(
1
−
t
)
[
sin
19
(
|
y
(
t
)
|
+
|
y
′
(
t
)
|
)
+
28
e
|
y
(
t
)
|
+
|
y
′
(
t
)
|
+
|
y
′
′
(
t
)
|
+
(
|
y
(
t
)
|
+
|
y
′
(
t
)
|
+
|
y
′
′
(
t
)
|
)
1
8
]
−
3
1
−
t
=
0
,
t
∈
(
0
,
1
)
y
(
0
)
=
y
′
(
0
)
=
0
,
y
′
′
(
1
)
=
1
3
y
′
′
(
1
2
)
+
1
6
y
″
(
2
3
)
+
1
10
y
′
′
(
5
6
)
,
y
′
′
′
(
0
)
=
1
6
y
′
′
′
(
1
2
)
+
1
4
y
′
′
′
(
2
3
)
+
1
10
y
′
′
′
(
5
6
)
Proof. Let
f
(
t
,
u
1
,
u
2
,
u
3
)
=
8
π
t
(
1
−
t
)
[
sin
19
(
|
u
1
|
+
|
u
2
|
)
+
28
e
|
u
1
|
+
|
u
2
|
+
|
u
3
|
+
(
|
u
1
|
+
|
u
2
|
+
|
u
3
|
)
1
8
]
−
3
1
−
t
Then
-
p
(
t
)
≤
f
(
t
,
u
1
,
u
2
,
u
3
)
≤
q
(
t
)
g
(
u
1
,
u
2
,
u
3
)
and
lim
(
|
u
1
|
+
|
u
2
|
+
|
u
3
|
)
→
+
∞
f
(
t
,
u
1
,
u
2
,
u
3
)
|
u
1
|
+
|
u
2
|
+
|
u
3
|
=
+
∞
where p
(
t
)
=
3
1
-
t
,
q
(
t
)
=
8
π
t
(
1
-
t
)
, g
(
u
1
,
u
2
,
u
3
)
=
sin
19
(
|
u
1
|
+
|
u
2
|
)
+
28
e
|
u
1
|
+
|
u
2
|
+
|
u
3
|
+
(
|
u
1
|
+
|
u
2
|
+
|
u
3
|
)
1
8
, which implies that (H1)-(H3) hold. Since
α
1
=
1
3
,
α
2
=
1
2
,
α
3
=
1
5
,
ξ
1
=
1
2
,
ξ
2
=
2
3
,
ξ
3
=
5
6
,
β
1
=
1
6
,
β
2
=
1
4
,
β
3
=
1
10
, then we know that
∫
0
1
(
p
(
s
)
+
q
(
s
)
)
d
s
=
∫
0
1
3
1
-
s
+
8
π
s
(
1
-
s
)
d
s
=
6
+
8
=
14
.
Take r = 3, then
M
1
=
max
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
∈
[
0
,
3
]
×
[
0
,
3
]
×
[
0
,
3
]
{
g
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
}
+
1
=
max
(
|
u
1
|
,
|
u
2
|
,
|
u
3
|
)
∈
[
0
,
3
]
×
[
0
,
3
]
×
[
0
,
3
]
{
[
sin
19
(
|
u
1
|
+
|
u
2
|
)
+
28
e
|
u
1
|
+
|
u
2
|
+
|
u
3
|
+
(
|
u
1
|
+
|
u
2
|
+
|
u
3
|
)
1
8
]
}
+
1
=
2
+
28
e
9
+
3
4
.
G
=
1
1
-
∑
i
=
1
m
-
2
α
i
+
∑
i
=
1
m
-
2
β
i
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
1
-
∑
i
=
1
m
-
2
α
i
1
-
∑
i
=
1
m
-
2
β
i
1
-
∑
i
=
1
m
-
2
α
i
ξ
i
∑
i
=
1
m
-
2
α
i
(
1
-
ξ
i
)
=
1
2
5
+
31
60
×
23
36
2
5
×
29
60
×
23
36
43
180
=
1010275
89784
.
r
G
∫
0
1
p
(
s
)
d
s
=
3
1010275
89784
×
6
=
44892
1010275
.
r
M
1
∫
0
1
(
p
(
s
)
+
q
(
s
)
)
d
s
1
-
∑
i
=
1
m
-
2
β
i
∑
i
=
1
m
-
2
α
i
ξ
i
1
-
∑
i
=
1
m
-
2
α
i
1
-
∑
i
=
1
m
-
2
β
i
-
1
=
3
×
(
2
+
28
e
9
+
3
4
)
×
14
2
5
×
1
-
31
60
×
13
36
29
60
-
1
=
3132
61495
(
2
+
28
e
9
+
3
4
)
.
It follows from Theorem 3.1 that the problem (4.2) has at least one positive solution for any
λ
*
∈
0
,
3132
61495
(
2
+
28
e
9
+
3
4
)
. □