# Nontrivial Solutions for a Boundary Value Problem ofnth-Order Impulsive Differential Equation

Authors: Jiafa Xu; Zhongli Wei
Journal: Chinese Journal of Mathematics
 (2014)

---

## Abstract

We study the existence of nontrivial solutions fornth-order boundary value problem with impulsive effects. We utilize Leray-Schauder degree theory to establish our main results. Furthermore, our nonlinear term f is allowed to grow superlinearly and sublinearly.


---

## Body

## 1. Introduction

In this paper, we consider the existence of nontrivial solutions for the followingnth-order boundary value problem involving impulsive effects:
(1)
u
(
n
)
(
t
)
+
f
(
t
,
u
(
t
)
)
=
0
,
t
∈
[
0,1
]
,
t
≠
t
k
,
-
Δ
u
(
n
-
1
)
∣
t
=
t
k
=
I
k
(
u
(
t
k
)
)
,
k
=
1,2
,
…
,
m
,
u
(
0
)
=
∫
0
1
‍
u
(
t
)
d
α
(
t
)
,
u
(
1
)
=
∫
0
1
‍
u
(
t
)
d
β
(
t
)
,
u
′
(
0
)
=
⋯
=
u
(
n
-
3
)
(
0
)
=
u
(
n
-
2
)
(
0
)
=
0
,
where f
∈
C
(
[
0,1
]
×
ℝ
,
ℝ
), I
k
∈
C
(
ℝ
,
ℝ
+
), ∫
0
1
‍
u
(
t
)
d
α
(
t
) and ∫
0
1
‍
u
(
t
)
d
β
(
t
) denote the Riemann-Stieltjes integral, and α and β are right continuous on [
0,1
), left continuous at t
=
1, and nondecreasing on [
0,1
], with α
(
0
)
=
β
(
0
)
=
0.The theory of impulsive differential equations describes processes which experience a sudden change of their state at certain moments. Therefore, many research workers pay their attention to boundary value problems for impulsive differential equations; for example, see [1–10] and the references therein.In [1], D. Guo utilized fixed point theory to establish some existence theorems of positive solutions for an infinite boundary value problem of nth-order nonlinear impulsive singular integrodifferential equations on the half-line in Banach spaces.In [2, 3], he dealt with multiple positive solutions for the following infinite boundary value problem of first order impulsive superlinear integrodifferential equations on the half-line:
(2)
u
′
(
t
)
=
f
(
t
,
u
(
t
)
,
(
T
u
)
(
t
)
,
(
S
u
)
(
t
)
)
,
∀
t
∈
J
+
′
,
Δ
u
∣
t
=
t
k
=
I
k
(
u
(
t
k
)
)
,
k
=
1,2
,
3
,
…
,
u
(
∞
)
=
β
u
(
0
)
.
In [2], the function f is continuous, that is, no singularity, but in [3] the author discussed the singular case, that is, f
→
∞ as t
→
0
+ and u
→
0
+. Compare with [2], the key point of [3] is apart from the singular point u
=
0 by some appropriate functions.Motivated by the works cited above, in particular [1–3], in this paper, we utilize the methods in [11] to study the existence of nontrivial solutions for (1). Nevertheless, our work here improves and extends the corresponding ones in [11]. We first note the impulsive effect as a perturbation to the corresponding problem of (1) without impulse, so we can construct an integral operator for the corresponding boundary value problem without the impulsive terms and find out its first eigenvalue and eigenfunction. Then we establish a special cone associated with the Green function of (1). Finally, by employing Leray-Schauder degree theory, two existence theorems of nontrivial solutions for (1) are obtained.


## 2. Preliminaries

We first offer some related preliminaries used in the ensuing section. In this work, we always assume that the following two conditions are satisfied:(H1)
f
∈
C
(
[
0,1
]
×
ℝ
,
ℝ
), I
k
∈
C
(
ℝ
,
ℝ
+
),
(H2)
κ
1
≥
0
,
κ
4
≥
0, κ
:
=
κ
1
κ
4
-
κ
2
κ
3
>
0, where
(3)
κ
1
:
=
1
-
∫
0
1
‍
(
1
-
t
n
-
1
)
d
α
(
t
)
,
κ
2
:
=
∫
0
1
‍
t
n
-
1
d
α
(
t
)
,
κ
3
:
=
∫
0
1
‍
(
1
-
t
n
-
1
)
d
β
(
t
)
,
κ
4
:
=
1
-
∫
0
1
‍
t
n
-
1
d
β
(
t
)
.For any g
∈
C
[
0,1
], the boundary value problem
(4)
u
(
n
)
(
t
)
+
g
(
t
)
=
0
,
t
∈
[
0,1
]
,
u
(
0
)
=
u
′
(
0
)
=
⋯
=
u
(
n
-
3
)
(
0
)
=
u
(
n
-
2
)
(
0
)
=
u
(
1
)
=
0
,
if and only if u can be expressed by u
(
t
)
=
∫
0
1
‍
H
(
t
,
s
)
g
(
s
)
d
s, where
(5)
H
(
t
,
s
)
=
1
(
n
-
1
)
!
{
(
1
-
s
)
n
-
1
t
n
-
1
-
(
t
-
s
)
n
-
1
,
0
≤
s
≤
t
≤
1
,
(
1
-
s
)
n
-
1
t
n
-
1
,
0
≤
t
≤
s
≤
1
.Lemma 1 (see [12], Lemma 2.1).
H
(
t
,
s
) has the following properties:(i)
0
≤
H
(
t
,
s
)
≤
K
(
s
), ∀
t
,
s
∈
[
0,1
], where K
(
s
)
:
=
s
(
1
-
s
)
n
-
1
/
(
n
-
2
)
!,
(ii)
H
(
t
,
s
)
≥
ζ
(
t
)
K
(
s
), ∀
t
,
s
∈
[
0,1
], where ζ
(
t
)
:
=
1
/
(
n
-
1
)
min
⁡
{
t
n
-
1
,
(
1
-
t
)
t
n
-
2
}.Let E
:
=
C
[
0,1
]
,
∥
u
∥
:
=
max
⁡
t
∈
[
0,1
]
|
u
(
t
)
|, and P
:
=
{
u
∈
E
:
u
(
t
)
≥
0
,
∀
t
∈
[
0,1
]
}. Then (
E
,
∥
·
∥
) is a real Banach space and P is a cone on E. Set J
∶
=
[
0,1
], J
′
=
[
0,1
]
∖
{
t
1
,
t
2
,
…
,
t
m
} and introduce the following space:
(6)
P
C
n
-
1
[
0,1
]
∶
=
{
u
∈
C
[
0,1
]
:
u
(
n
-
1
)
∣
(
t
k
,
t
k
+
1
)
∈
C
(
t
k
,
t
k
+
1
)
,
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
u
(
n
-
1
)
(
t
k
-
)
=
u
(
n
-
1
)
(
t
k
)
,
∃
u
(
n
-
1
)
(
t
k
+
)
}
,
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
k
=
1,2
,
…
,
m
,
with the norm ∥
u
∥
P
C
n
-
1
:
=
max
⁡
{
∥
u
∥
,
∥
u
′
∥
,
…
,
∥
u
(
n
-
1
)
∥
}. Clearly, (
P
C
n
-
1
[
0,1
]
,
∥
·
∥
P
C
n
-
1
) is also a real Banach space.Lemma 2 (see [9], Lemma 2.1).
Let (H1) and (H2) hold. Then (1) is equivalent to the following integral equation:
(7)
u
(
t
)
:
=
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
I
k
(
u
(
t
k
)
)
,
where
(8)
G
(
t
,
s
)
=
H
(
t
,
s
)
+
κ
-
1
(
κ
4
(
1
-
t
n
-
1
)
+
κ
3
t
n
-
1
)
∫
0
1
‍
H
(
t
,
s
)
d
α
(
t
)
+
κ
-
1
(
κ
2
(
1
-
t
n
-
1
)
+
κ
1
t
n
-
1
)
∫
0
1
‍
H
(
t
,
s
)
d
β
(
t
)
.
Letη
(
t
)
:
=
min
⁡
{
t
n
-
1
,
1
-
t
n
-
1
} and
(9)
γ
(
t
)
:
=
min
⁡
{
ζ
(
t
)
,
η
(
t
)
∫
0
1
‍
ζ
(
t
)
d
α
(
t
)
(
α
(
1
)
)
-
1
,
η
(
t
)
∫
0
1
‍
ζ
(
t
)
d
β
(
t
)
(
β
(
1
)
)
-
1
}
.
Here, ζ
(
t
) is determined by Lemma 1. Clearly, γ
(
t
) is nonnegative and continuous on [
0,1
].Lemma 3.
One hasG
(
t
,
s
)
≥
γ
(
t
)
G
(
τ
,
s
), ∀
t
,
s
,
τ
∈
[
0,1
].Proof .
By Lemma1 and (8), we obtain G
(
t
,
s
)
≤
K
(
s
)
+
κ
-
1
α
(
1
)
(
κ
4
+
κ
3
)
K
(
s
)
+
κ
-
1
β
(
1
)
(
κ
2
+
κ
1
)
K
(
s
)
:
=
ℱ
1
(
s
), and
(10)
G
(
t
,
s
)
≥
ζ
(
t
)
K
(
s
)
+
κ
-
1
(
κ
4
(
1
-
t
n
-
1
)
+
κ
3
t
n
-
1
)
ccccccccc
×
∫
0
1
‍
ζ
(
t
)
d
α
(
t
)
K
(
s
)
ccccccccc
+
κ
-
1
(
κ
2
(
1
-
t
n
-
1
)
+
κ
1
t
n
-
1
)
∫
0
1
‍
ζ
(
t
)
d
β
(
t
)
K
(
s
)
≥
γ
(
t
)
[
K
(
s
)
+
κ
-
1
α
(
1
)
(
κ
4
+
κ
3
)
K
(
s
)
ccccccccccc
+
κ
-
1
β
(
1
)
(
κ
2
+
κ
1
)
K
(
s
)
]
=
γ
(
t
)
ℱ
1
(
s
)
.
Combining these, we can find G
(
t
,
s
)
≥
γ
(
t
)
ℱ
1
(
s
)
≥
γ
(
t
)
G
(
τ
,
s
), ∀
t
,
s
,
τ
∈
[
0,1
]. This completes the proof.Let(11)
(
A
u
)
(
t
)
:
=
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
(
s
)
)
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
I
k
(
u
(
t
k
)
)
,
(
L
u
)
(
t
)
:
=
∫
0
1
‍
G
(
t
,
s
)
u
(
s
)
d
s
,
u
∈
E
.
Clearly, A
:
E
→
E is a completely continuous nonlinear operator, and the existence of the solutions of (1) is equivalent to that of fixed points of A; L
:
E
→
E is a completely continuous linear operator, satisfying L
(
P
)
⊂
P. Namely, L is a positive, completely continuous, linear operator. By the positivity of G
(
t
,
s
), ∀
t
,
s
∈
(
0,1
), the spectral radius of L, denoted by r
(
L
), is positive. The Krein-Rutman theorem [13] then asserts that there are p
∈
P
∖
{
0
} and w
∈
L
(
0,1
)
∖
{
0
} such that
(12)
∫
0
1
‍
G
(
t
,
s
)
p
(
s
)
d
s
=
r
(
L
)
p
(
t
)
,
∫
0
1
‍
G
(
t
,
s
)
w
(
t
)
d
t
=
r
(
L
)
w
(
s
)
,
∫
0
1
‍
w
(
t
)
d
t
=
1
.
Denote δ
:
=
∫
0
1
‍
γ
(
t
)
w
(
t
)
d
t
>
0 and P
0
:
=
{
u
∈
P
:
∫
0
1
‍
u
(
t
)
w
(
t
)
d
t
≥
δ
∥
u
∥
}.Lemma 4.
One hasL
(
P
)
⊂
P
0.Proof.
Suppose thatu
∈
P, then
(13)
∫
0
1
‍
(
L
u
)
(
t
)
w
(
t
)
d
t
=
∫
0
1
‍
∫
0
1
‍
G
(
t
,
s
)
u
(
s
)
d
s
w
(
t
)
d
t
≥
∫
0
1
‍
γ
(
t
)
w
(
t
)
d
t
∫
0
1
‍
G
(
τ
,
s
)
u
(
s
)
d
s
=
δ
(
L
u
)
(
τ
)
,
for all τ
∈
[
0,1
] and thus ∫
0
1
‍
(
L
u
)
(
t
)
w
(
t
)
d
t
≥
δ
∥
L
u
∥. This completes the proof.Lemma 5 (see [14]).
LetE be a real Banach space and Ω
⊂
E a bounded open set with 0
∈
Ω. Suppose that A
:
Ω
¯
→
E is a completely continuous operator. If there is y
0
∈
E
∖
{
0
} such that u
≠
A
u
+
λ
y
0, ∀
u
∈
∂
Ω, λ
≥
0, then deg
⁡
(
I
-
A
,
Ω
,
0
)
=
0, where deg
⁡ stands for the Leray-Schauder topological degree in E.


## 3. Main Results

Letλ
1
-
1
:
=
r
(
L
). We now list our assumptions on f.(H3)
liminf
⁡
u
→
+
∞
(
f
(
t
,
u
)
/
u
)
>
λ
1 and limsup
⁡
u
→
-
∞
(
f
(
t
,
u
)
/
u
)
<
λ
1 uniformly in t
∈
[
0,1
].
(H4)
There is a positive numbera
1
∈
(
0
,
δ
/
(
r
(
L
)
(
δ
+
∑
k
=
1
m
w
(
t
k
)
‍
)
)
) such that
(14)
limsup
⁡
u
→
0
|
f
(
t
,
u
)
|
|
u
|
<
a
1
,
limsup
⁡
u
→
0
I
k
(
u
)
|
u
|
<
a
1
ccccccccccccccccccccc
uniformly
in
t
∈
[
0,1
]
.
(H5)
liminf
⁡
u
→
0
+
(
f
(
t
,
u
)
/
u
)
>
λ
1 and limsup
⁡
u
→
0
-
(
f
(
t
,
u
)
/
u
)
<
λ
1 uniformly in t
∈
[
0,1
].
(H6)
Fora
1 in (H4), assume that
(15)
limsup
⁡
u
→
∞
|
f
(
t
,
u
)
|
|
u
|
<
a
1
,
limsup
⁡
u
→
∞
I
k
(
u
)
|
u
|
<
a
1
ccccccccccccccccccccc
uniformly
in
t
∈
[
0,1
]
.Two Examples. Let α
(
t
)
=
t, β
(
t
)
=
0, for all t
∈
[
0,1
]. Then κ
1
=
1
/
n, κ
2
=
1
/
n, κ
3
=
0, κ
4
=
1, and (H2) holds true.(1)
Letf
(
t
,
u
)
=
b
1
u
+
u
2
+
u
4, where b
1
∈
(
0
,
a
1
). Then f satisfies (H1). Moreover, we obtain
(16)
liminf
⁡
u
→
+
∞
f
(
t
,
u
)
u
=
liminf
⁡
u
→
+
∞
(
b
1
+
u
+
u
3
)
=
+
∞
>
λ
1
,
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
uniformly
in
t
∈
[
0,1
]
,
limsup
⁡
u
→
-
∞
f
(
t
,
u
)
u
=
limsup
⁡
u
→
-
∞
(
b
1
+
u
+
u
3
)
=
-
∞
<
λ
1
,
ccccccccccccccccccccccccccc
uniformly
in
t
∈
[
0,1
]
,
limsup
⁡
u
→
0
|
f
(
t
,
u
)
|
|
u
|
=
limsup
⁡
u
→
0
|
b
1
+
u
+
u
3
|
=
b
1
<
a
1
,
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
uniformly
in
t
∈
[
0,1
]
.
So, f satisfies (H3) and (H4). Also, let I
k
(
u
)
=
u
4. Then I
k satisfies (H1) and
(17)
limsup
⁡
u
→
0
I
k
(
u
)
|
u
|
=
limsup
⁡
u
→
0
|
u
|
3
=
0
<
a
1
.
Hence, I
k satisfies (H4). Consequently, (H1), (H3), and (H4) hold true.
(2)
Letf
(
t
,
u
)
=
b
2
u
+
2
u
2
/
3
+
3
u
4
/
5, I
k
(
u
)
=
|
u
|, where b
2
∈
(
0
,
a
1
). Then
(18)
liminf
⁡
u
→
0
+
f
(
t
,
u
)
u
=
liminf
⁡
u
→
0
+
(
b
2
+
2
u
-
1
/
3
+
3
u
-
1
/
5
)
=
+
∞
>
λ
1
cccccccccccccccccccccccccccccccccc
uniformly
in
t
∈
[
0,1
]
,
limsup
⁡
u
→
0
-
f
(
t
,
u
)
u
=
limsup
⁡
u
→
0
-
(
b
2
+
2
u
-
1
/
3
+
3
u
-
1
/
5
)
=
-
∞
<
λ
1
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
c
uniformly
in
t
∈
[
0,1
]
.Moreover,
(19)
limsup
⁡
u
→
∞
|
f
(
t
,
u
)
|
|
u
|
=
limsup
⁡
u
→
∞
|
b
2
+
2
u
-
1
/
3
+
3
u
-
1
/
5
|
=
b
2
<
a
1
ccccccccccccccccccccccccccccccccc
uniformly
in
t
∈
[
0,1
]
,
limsup
⁡
u
→
∞
I
k
(
u
)
|
u
|
=
limsup
⁡
u
→
∞
1
|
u
|
=
0
<
a
1
.
So, (H1), (H5), (H6) hold.Theorem 6.
Assume that (H1)–(H4) hold, and (1) has at least one nontrivial solution.Proof.
(H3) implies that there areɛ
∈
(
0
,
λ
1
) and c
1
>
0 such that f
(
t
,
u
)
≥
(
λ
1
+
ɛ
)
u
-
c
1
,
u
≥
0, t
∈
[
0,1
] and f
(
t
,
u
)
≥
(
λ
1
-
ɛ
)
u
-
c
1, u
≤
0
,
t
∈
[
0,1
]. Note that if (
t
,
u
)
∈
[
0,1
]
×
[
0
,
+
∞
), f
(
t
,
u
)
≥
(
λ
1
+
ɛ
)
u
-
c
1
≥
(
λ
1
-
ɛ
)
u
-
c
1, and if (
t
,
u
)
∈
[
0,1
]
×
(
-
∞
,
0
], f
(
t
,
u
)
≥
(
λ
1
-
ɛ
)
u
-
c
1
≥
(
λ
1
+
ɛ
)
u
-
c
1. That is, we have
(20)
f
(
t
,
u
)
≥
(
λ
1
+
ɛ
)
u
-
c
1
,
∀
u
∈
ℝ
,
t
∈
[
0,1
]
,
(21)
f
(
t
,
u
)
≥
(
λ
1
-
ɛ
)
u
-
c
1
,
∀
u
∈
ℝ
,
t
∈
[
0,1
]
.
Let ℳ
1
:
=
{
u
∈
E
:
u
=
A
u
+
λ
p
,
λ
≥
0
}, where p
∈
P
∖
{
0
} is given by (12). We claim that ℳ
1 is bounded in E. Indeed, if u
0
∈
ℳ
1, there is λ
0
≥
0 such that
(22)
u
0
(
t
)
=
(
A
u
0
)
(
t
)
+
λ
0
p
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
0
(
s
)
)
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
I
k
(
u
0
(
t
k
)
)
+
λ
0
p
(
t
)
≥
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
0
(
s
)
)
d
s
.
Combining this with (20) yields
(23)
u
0
(
t
)
≥
(
λ
1
+
ɛ
)
(
L
u
0
)
(
t
)
-
c
1
(
L
1
)
(
t
)
,
where 1 stands for the constant function 1
(
t
)
≡
1. Multiply (23) by w
(
t
) on both sides and integrate over [
0,1
] and use (12) to obtain ∫
0
1
‍
u
0
(
t
)
w
(
t
)
d
t
≥
(
1
+
ɛ
r
(
L
)
)
∫
0
1
‍
u
0
(
t
)
w
(
t
)
d
t
-
c
1
r
(
L
), and thus ∫
0
1
‍
u
0
(
t
)
w
(
t
)
d
t
≤
ɛ
-
1
c
1. Note that (21) holds. Define the Nemytskii operator F
:
E
→
E by (
F
u
)
(
t
)
:
=
f
(
t
,
u
(
t
)
). Now (22) is equivalent to
(24)
u
0
(
t
)
-
(
λ
1
-
ɛ
)
(
L
u
0
)
(
t
)
+
c
1
(
L
1
)
(
t
)
=
(
A
u
0
)
(
t
)
-
(
λ
1
-
ɛ
)
(
L
u
0
)
(
t
)
+
c
1
(
L
1
)
(
t
)
+
λ
0
p
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
[
f
(
s
,
u
0
(
s
)
)
-
(
λ
1
-
ɛ
)
u
0
(
s
)
+
c
1
]
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
I
k
(
u
0
(
t
k
)
)
+
λ
0
p
(
t
)
=
(
L
[
F
u
0
-
(
λ
1
-
ɛ
)
u
0
+
c
1
1
]
)
(
t
)
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
I
k
(
u
0
(
t
k
)
)
+
λ
0
p
(
t
)
.
Lemma 4 implies that u
0
-
(
λ
1
-
ɛ
)
L
u
0
+
c
1
L
1
∈
P
0. Therefore,
(25)
∥
u
0
-
(
λ
1
-
ɛ
)
L
u
0
+
c
1
L
1
∥
≤
δ
-
1
∫
0
1
‍
[
u
0
(
t
)
-
(
λ
1
-
ɛ
)
(
L
u
0
)
(
t
)
+
c
1
(
L
1
)
(
t
)
]
w
(
t
)
d
t
≤
ɛ
δ
-
1
r
(
L
)
∫
0
1
‍
u
0
(
t
)
w
(
t
)
d
t
+
c
1
δ
-
1
r
(
L
)
≤
2
c
1
δ
-
1
r
(
L
)
.
Since (
λ
1
-
ɛ
)
r
(
L
)
<
1, the operator I
-
(
λ
1
-
ɛ
)
L has the bounded inverse operator (
I
-
(
λ
1
-
ɛ
)
L
)
-
1. Therefore, there is 𝒩
1
>
0 such that ∥
u
∥
≤
𝒩
1, ∀
u
∈
ℳ
1. This proves the boundedness of ℳ
1. For each R
>
sup
⁡
u
∈
ℳ
1
∥
u
∥, we have u
≠
A
u
+
λ
p, ∀
u
∈
∂
B
R, λ
≥
0, which leads to
(26)
deg
⁡
(
I
-
A
,
B
R
,
0
)
=
0
.
On the other hand, (H4) implies that there is ar
>
0 such that |
f
(
t
,
u
)
|
≤
a
1
|
u
|, I
k
(
u
)
≤
a
1
|
u
|, ∀
|
u
|
≤
r, t
∈
[
0,1
]. Notice we may choose r
>
0 so that r
<
sup
⁡
u
∈
ℳ
1
∥
u
∥. Thus we have
(27)
|
(
A
u
)
(
t
)
|
≤
∫
0
1
‍
G
(
t
,
s
)
|
f
(
s
,
u
(
s
)
)
|
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
I
k
(
u
(
t
k
)
)
≤
a
1
[
∫
0
1
‍
G
(
t
,
s
)
|
u
(
s
)
|
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
|
u
(
t
k
)
|
]
for all u
∈
B
¯
r and t
∈
[
0,1
]. We claim that
(28)
u
≠
μ
A
u
,
∀
u
∈
∂
B
r
,
μ
∈
[
0,1
]
.
Suppose the contrary. Then there are u
0
∈
∂
B
r and μ
0
∈
[
0,1
] such that u
0
=
μ
0
A
u
0. Let v
(
t
)
=
|
u
0
(
t
)
|. Then v
∈
P and
(29)
v
(
t
)
≤
a
1
[
∫
0
1
‍
G
(
t
,
s
)
v
(
s
)
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
v
(
t
k
)
]
.
Multiply (29) by w
(
t
) on both sides and integrate over [
0,1
] and use (12) to obtain
(30)
∫
0
1
‍
v
(
t
)
w
(
t
)
d
t
≤
a
1
r
(
L
)
∫
0
1
‍
v
(
t
)
w
(
t
)
d
t
+
a
1
r
(
L
)
∑
k
=
1
m
‍
v
(
t
k
)
w
(
t
k
)
,
and then by Lemma 4,
(31)
δ
(
1
-
a
1
r
(
L
)
)
∥
v
∥
≤
(
1
-
a
1
r
(
L
)
)
∫
0
1
‍
v
(
t
)
w
(
t
)
d
t
≤
a
1
r
(
L
)
∑
k
=
1
m
‍
v
(
t
k
)
w
(
t
k
)
≤
a
1
r
(
L
)
∥
v
∥
∑
k
=
1
m
‍
w
(
t
k
)
,
That is, a
1
r
(
L
)
(
δ
+
∑
k
=
1
m
‍
w
(
t
k
)
)
≥
δ, which contracts to a
1
∈
(
0
,
δ
/
(
r
(
L
)
(
δ
+
∑
k
=
1
m
w
(
t
k
)
)
)
). Consequently, ∥
v
∥
=
0, and thus v
=
0
⇒
u
0
≡
0. This contradicts to u
0
∈
∂
B
r. As a result of this, (28) holds true. Hence I and I
-
A are homotopic on ∂
B
r. The homotopy invariance of topological degree implies that deg
⁡
(
I
-
A
,
B
r
,
0
)
=
1. This and (26) together imply that deg
⁡
(
I
-
A
,
B
R
∖
B
¯
r
,
0
)
=
deg
⁡
(
I
-
A
,
B
R
,
0
)
-
deg
⁡
(
I
-
A
,
B
r
,
0
)
=
-
1. Therefore the operator A has at least one point in B
R
∖
B
¯
r. Equivalently, (1) has at least one nontrivial solution. This completes the proof.Theorem 7.
Assume that (H1), (H2), (H5), and (H6) hold; (1) has at least one nontrivial solution.Proof.
(H5) implies there areɛ
∈
(
0
,
λ
1
) and r
>
0 such that f
(
t
,
u
)
≥
(
λ
1
+
ɛ
)
u, u
∈
[
0
,
r
], t
∈
[
0,1
], and f
(
t
,
u
)
≥
(
λ
1
-
ɛ
)
u, u
∈
[
-
r
,
0
], t
∈
[
0,1
]. Note that if (
t
,
u
)
∈
[
0,1
]
×
[
0
,
r
], f
(
t
,
u
)
≥
(
λ
1
+
ɛ
)
u
≥
(
λ
1
-
ɛ
)
u, and if (
t
,
u
)
∈
[
0,1
]
×
[
-
r
,
0
], f
(
t
,
u
)
≥
(
λ
1
-
ɛ
)
u
≥
(
λ
1
+
ɛ
)
u. That is, we have
(32)
f
(
t
,
u
)
≥
(
λ
1
+
ɛ
)
u
∀
u
∈
[
-
r
,
r
]
,
t
∈
[
0,1
]
,
(33)
f
(
t
,
u
)
≥
(
λ
1
-
ɛ
)
u
,
∀
u
∈
[
-
r
,
r
]
,
t
∈
[
0,1
]
.
Nowadays, we claim
(34)
u
≠
A
u
+
λ
p
,
∀
u
∈
∂
B
r
,
λ
≥
0
,
where p
∈
P
∖
{
0
} is given by (12). Indeed, if the claim is false, then there are u
0
∈
∂
B
r and λ
0
≥
0 such that
(35)
u
0
(
t
)
=
(
A
u
0
)
(
t
)
+
λ
0
p
(
t
)
=
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
0
(
s
)
)
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
I
k
(
u
0
(
t
k
)
)
+
λ
0
p
(
t
)
≥
∫
0
1
‍
G
(
t
,
s
)
f
(
s
,
u
0
(
s
)
)
d
s
.
Combining this with (32) yields
(36)
u
0
(
t
)
≥
(
λ
1
+
ɛ
)
(
L
u
0
)
(
t
)
.
Multiply (36) by w
(
t
) and integrate over [
0,1
] and use (12) to obtain ∫
0
1
‍
u
0
(
t
)
w
(
t
)
d
t
≥
(
1
+
ɛ
r
(
L
)
)
∫
0
1
‍
u
0
(
t
)
w
(
t
)
d
t, and thus ∫
0
1
‍
u
0
(
t
)
w
(
t
)
d
t
≤
0. Note that (33) holds, and then we find
(37)
u
0
(
t
)
-
(
λ
1
-
ɛ
)
(
L
u
0
)
(
t
)
=
(
A
u
0
)
(
t
)
-
(
λ
1
-
ɛ
)
(
L
u
0
)
(
t
)
+
λ
0
p
(
t
)
=
(
L
[
F
u
0
-
(
λ
1
-
ɛ
)
u
0
]
)
(
t
)
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
I
k
(
u
0
(
t
k
)
)
+
λ
0
p
(
t
)
.
Lemma 4 implies that u
0
-
(
λ
1
-
ɛ
)
L
u
0
∈
P
0. Therefore, ∥
u
0
-
(
λ
1
-
ɛ
)
L
u
0
∥
≤
δ
-
1
∫
0
1
‍
[
u
0
(
t
)
-
(
λ
1
-
ɛ
)
(
L
u
0
)
(
t
)
]
w
(
t
)
d
t
≤
ɛ
δ
-
1
r
(
L
)
∫
0
1
‍
u
0
(
t
)
w
(
t
)
d
t
≤
0. Notice that (
λ
1
-
ɛ
)
r
(
L
)
<
1. We have u
0
=
0, contradicting u
0
∈
∂
B
r. As a result of this, (34) is true. Invoking Lemma 5 gives
(38)
deg
⁡
(
I
-
A
,
B
r
,
0
)
=
0
.
On the other hand, (H6) implies that there isc
2
>
0 such that
(39)
|
f
(
t
,
u
)
|
≤
a
1
|
u
|
+
c
2
,
I
k
(
u
)
≤
a
1
|
u
|
+
c
2
,
ccccccccccccccccccccc
∀
u
∈
ℝ
,
t
∈
[
0,1
]
.
Letℳ
2
:
=
{
u
∈
E
:
u
=
λ
A
u
,
∀
λ
∈
[
0,1
]
}. We shall now show that ℳ
2 is bounded in E. Indeed, if u
0
∈
ℳ
2, then u
0
=
λ
0
A
u
0 for some λ
0
∈
[
0,1
]. In view of (39), let v
(
t
)
=
|
u
0
(
t
)
|, we have v
∈
P and
(40)
v
(
t
)
≤
a
1
[
∫
0
1
‍
G
(
t
,
s
)
v
(
s
)
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
v
(
t
k
)
]
+
c
2
[
∫
0
1
‍
G
(
t
,
s
)
d
s
+
∑
k
=
1
m
‍
G
(
t
,
t
k
)
]
.
Multiply (40) by w
(
t
) on both sides and integrate over [
0,1
] and use (12) to obtain
(41)
∫
0
1
‍
v
(
t
)
w
(
t
)
d
t
≤
a
1
r
(
L
)
∫
0
1
‍
v
(
t
)
w
(
t
)
d
t
+
a
1
r
(
L
)
∑
k
=
1
m
‍
v
(
t
k
)
w
(
t
k
)
+
c
2
r
(
L
)
(
1
+
∑
k
=
1
m
‍
w
(
t
k
)
)
,
and then by Lemma 4,
(42)
δ
(
1
-
a
1
r
(
L
)
)
∥
v
∥
≤
(
1
-
a
1
r
(
L
)
)
∫
0
1
‍
v
(
t
)
w
(
t
)
d
t
≤
a
1
r
(
L
)
∑
k
=
1
m
‍
v
(
t
k
)
w
(
t
k
)
+
c
2
r
(
L
)
(
1
+
∑
k
=
1
m
‍
w
(
t
k
)
)
≤
a
1
r
(
L
)
∥
v
∥
∑
k
=
1
m
‍
w
(
t
k
)
+
c
2
r
(
L
)
(
1
+
∑
k
=
1
m
‍
w
(
t
k
)
)
.
Hence, ∥
v
∥
≤
(
c
2
r
(
L
)
(
1
+
∑
k
=
1
m
‍
w
(
t
k
)
)
)
/
(
δ
(
1
-
a
1
r
(
L
)
)
-
a
1
r
(
L
)
∑
k
=
1
m
‍
w
(
t
k
)
). This proves the boundedness of ℳ
2. Choosing R
>
sup
⁡
u
∈
ℳ
2
∥
u
∥ and R
>
r, we have u
≠
λ
A
u, ∀
u
∈
∂
B
R, λ
∈
[
0,1
]. The homotopy invariance of Leray-Schauder topological degree implies that deg
⁡
(
I
-
A
,
B
R
,
0
)
=
1. Combining this with (38) we get deg
⁡
(
I
-
A
,
B
R
∖
B
¯
r
,
0
)
=
deg
⁡
(
I
-
A
,
B
R
,
0
)
-
deg
⁡
(
I
-
A
,
B
r
,
0
)
=
1. Therefore the operator A has at least one point in B
R
∖
B
¯
r. Equivalently, (1) has at least one nontrivial solution. This completes the proof.



---

*Source: 379761-2014-01-23.xml*