# Multiple Results to Some Biharmonic Problems

Authors: Xingdong Tang; Jihui Zhang
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We study a nonlinear elliptic problem defined in a bounded domain involving biharmonic operator together with an asymptotically linear term. We establish at least three nontrivial solutions using the topological degree theory and the critical groups.


---

## Body

## 1. Introduction

We consider the following biharmonic problem:(1)
Δ
2
u
=
f
(
x
,
u
)
in
Ω
,
u
=
Δ
u
=
0
,
on
⁡
∂
Ω
,
where Ω
∈
ℝ
N  (
N
⩾
5
) is a smooth bounded domain and f
:
Ω
×
ℝ
→
ℝ is of class C
1 with f
(
x
,
0
)
=
0.In the past decades, biharmonic operators have attracted much attention of many researchers and experts. Whilef
(
x
,
u
)
=
b
[
(
u
+
1
)
+
-
1
], the solutions of (1) characterized the travelling waves in a suspension bridge; see [1].In 1998, Micheletti and Pistoia [2] considered the following biharmonic problem:
(2)
Δ
2
u
+
a
2
Δ
u
=
b
[
(
u
+
1
)
+
-
1
]
,
in
Ω
,
u
=
Δ
u
=
0
,
on
⁡
∂
Ω
,
where a, b are constants and Ω
⊂
ℝ
n is a bounded smooth domain, and they established multiple results by using a minimax process.Three years later, Zhang [3] considered a more general condition; that is,
(3)
Δ
2
u
+
c
Δ
u
=
f
(
x
,
u
)
,
in
Ω
,
u
=
Δ
u
=
0
,
on
⁡
∂
Ω
,
where c
∈
ℝ and f satisfies the subcritical growth; at least one nontrivial solution was obtained.Since then, a lot of papers dealing with biharmonic problems by the critical point theory sprung up, and so forth [4, 5].At the same time, Leray-Schauder degree as a very wonderful tool was introduced to handle biharmonic problems; see [6–9]. To our best knowledge, there are few papers considered (1) by combining the critical point theory (especially Morse theory) with Leray-Schauder degree.Our argument was originally developed by Hofer [10] and Zhang [11]. Following Hofer [10] and Zhang [11], there are some papers dealing with second-order elliptic problems, and so forth [12].Zhang [11] first considered the following second-order elliptic problem:
(4)
-
Δ
u
=
g
(
x
,
u
)
,
in
Ω
,
B
u
=
0
,
on
⁡
∂
Ω
,
where B denotes Neumann operator or Dirichlet operator. Sub- and sup-solutions methods with critical point theory were used to obtain at least two distinct solutions. Also under subcritical growth condition, Chang [13] proved that if p
0 is an isolated critical point of J, then, for all q
∈
ℕ, C
q
(
J
~
,
p
0
)
=
C
q
(
J
,
p
0
) with integral coefficients, where J
~, J denote the energy functional under C
0
(
Ω
¯
)
∩
C
1
(
Ω
¯
) and H
0
1
(
Ω
), C
q means qth critical group corresponding to (4), which inspires us to consider (1).Bartsch et al. [12] considered (4) and obtained more results in this direction. Then, some other results in this direction were also obtained; see [14].As far as we know, there are few papers concerned with the biharmonic problem (1) using this method; only Qian and Li [5] considered
(5)
Δ
2
u
+
c
Δ
u
=
f
(
x
,
u
)
,
in
Ω
,
u
=
Δ
u
=
0
,
on
⁡
∂
Ω
;
And they proved that if u
0 is an isolated critical point of J, then, for all q
∈
ℕ, C
q
(
J
~
,
u
0
)
=
C
q
(
J
,
u
0
) with integral coefficients, where J
~, J denote the energy functional on the space C
0
(
Ω
¯
)
∩
C
1
(
Ω
¯
) and H
0
1
(
Ω
)
∩
H
2
(
Ω
), C
q means qth critical group. In our paper, the results in [5] are improved, and some new results are obtained.Let0
<
λ
1
<
λ
2
⩽
λ
3
⩽
⋯
⩽
λ
n
⩽
⋯ denote the eigenvalues of (
-
Δ
,
H
0
1
(
Ω
)
) (counting with their multiplicity) with corresponding eigenfunctions e
1
,
e
2
,
e
3
,
…
,
e
n
,
…. We may choose e
1
>
0 in Ω. Let μ
k
=
λ
k
2, k
=
1,2
,
…
,
n
,
…, then μ
1
<
μ
2
⩽
μ
3
⩽
⋯
⩽
μ
n
⩽
⋯ are eigenvalues of the following biharmonic problem [3] corresponding eigenfunctions e
1
,
e
2
,
e
3
,
…
,
e
n
,
…;
(6)
Δ
2
u
=
μ
u
,
in
Ω
,
u
=
Δ
u
=
0
,
on
⁡
∂
Ω
.In order to obtain nontrivial solutions, we now assume that the nonlinearityf satisfies the following conditions:(f1)
f
∈
C
1
(
Ω
¯
×
ℝ
,
ℝ
), f
(
x
,
u
)
u
⩾
0 for all (
x
,
u
)
∈
Ω
¯
×
ℝ, and there exist constant numbers C
>
0 and α with 1
<
α
<
(
N
+
4
)
/
(
N
-
4
), such that
(7)
|
f
u
′
(
x
,
u
)
|
⩽
c
(
1
+
|
u
|
α
-
1
)
;
(f2)
there existsi
∈
ℕ with μ
2
i
<
μ
2
i
+
1, such that f
u
′
(
x
,
0
)
=
μ
2
i, for all x
∈
Ω
¯;
(f3)
limsup
⁡
|
t
|
→
∞
f
(
x
,
u
)
/
u
<
μ
1 uniformly for x
∈
Ω
¯;
(f4)
there is somer
>
0 small, such that
(8)
μ
2
i
u
2
⩽
F
(
x
,
u
)
<
μ
2
i
+
1
u
2
,
u
∈
ℝ
,
|
u
|
⩽
r
,
a
.
e
.
x
∈
Ω
,
whereF
(
x
,
u
)
=
∫
0
u
‍
f
(
x
,
s
)
d
s.The main result of this paper is the followingTheorem 1.
Supposef satisfies (f1)–(f4). Then (1) has at least three solutions.


## 2. Preliminaries

In this section, we first recall some lemmas and preliminaries.LetC
0
(
Ω
¯
)
∩
C
k
(
Ω
¯
) denote the set of f
:
Ω
¯
→
ℝ which are k-times continuous differentiable in Ω
¯ and identically vanishing on ∂
Ω with the norm ∥
u
∥
k
=
∑
i
=
0
k
‍
∥
u
(
k
)
∥
0, where ∥
u
∥
0
=
max
⁡
x
∈
Ω
¯
u
(
x
), P
k
=
{
u
∈
C
0
(
Ω
¯
)
∩
C
k
(
Ω
¯
)
:
u
(
x
)
⩾
0
,
∀
x
∈
Ω
}, ∀
k
∈
ℕ.Lemma 2.
P
2 is a solid cone of C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
); that is, P
°
2
≠
∅.It is well known that the positive coneP
1 is a solid cone of C
0
(
Ω
¯
)
∩
C
1
(
Ω
¯
). Our proof depends on the fact above; what is more, the technique we used here is originated from [15, page 628].Proof.
SinceP
1 is a closed positive cone of (
C
0
(
Ω
¯
)
∩
C
1
(
Ω
¯
)
,
∥
·
∥
1
), by the definition of ∥
·
∥
k, k
=
0,1
,
2
,
…
,
n
,
…, for u
∈
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
), ∥
u
∥
1
⩽
∥
u
∥
2, thus the embedding  i
:
(
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
,
∥
·
∥
2
)
↪
(
C
0
(
Ω
¯
)
∩
C
1
(
Ω
¯
)
,
∥
·
∥
1
) is continuous. i
-
1
(
P
1
) is closed in (
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
,
∥
·
∥
2
) (in fact P
2
=
i
-
1
(
P
1
)). Obviously P
°
1
∩
(
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
,
∥
·
∥
2
)
≠
∅; thus P
2
=
i
-
1
(
P
1
) has nonempty interior. The proof is finished.Remark 3.
Using the method above, it is not difficult to know thatP
k is a solid cone in (
C
0
(
Ω
¯
)
∩
C
k
(
Ω
¯
)
,
∥
·
∥
k
), k
=
2,3
,
4
,
…
,
n
,
….Remark 4.
For anyu
∈
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
), if u is an interior point of P
1 in (
C
0
(
Ω
¯
)
∩
C
1
(
Ω
¯
)
,
∥
·
∥
1
), then u is an interior point of P
2 in (
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
,
∥
·
∥
2
).In what follows, we will use the Hilbert spaceV
=
H
0
1
(
Ω
)
∩
H
2
(
Ω
), and the norm on V is given by ∥
u
∥
V
=
∫
Ω
‍
|
Δ
u
|
2
d
x. It is well known that solutions of (1) are critical points of the functional
(9)
Ψ
(
u
)
=
1
2
∫
Ω
‍
|
Δ
u
|
2
d
x
-
∫
Ω
‍
F
(
x
,
u
)
d
x
,
where F
(
x
,
u
)
=
∫
0
u
‍
f
(
x
,
s
)
d
s. Since f
∈
C
1
(
Ω
¯
×
ℝ
,
ℝ
), it is easy to know that Ψ
∈
C
2
(
V
,
ℝ
), and
(10)
〈
Ψ
′
(
u
)
,
v
〉
=
∫
Ω
‍
[
Δ
u
Δ
v
-
f
(
x
,
u
)
v
]
d
x
,
〈
Ψ
′′
(
u
)
v
,
h
〉
=
∫
Ω
‍
[
Δ
v
Δ
h
-
f
′
(
x
,
u
)
v
h
]
d
x
.Corresponding to the eigenvaluesμ
j
′
s we have the splitting V
=
H
-
⊕
N
⊕
H
+ where
(11)
H
-
=
⨁
j
=
1
2
i
-
1
e
j
,
N
=
span
⁡
{
e
2
i
}
,
H
+
=
⨁
j
=
2
i
+
1
+
∞
e
j
¯
.Consider the problem(12)
Δ
2
u
=
h
,
in
Ω
,
u
=
Δ
u
=
0
,
on
⁡
∂
Ω
.For allr
∈
ℝ
+, denote B
(
0
,
r
)
≜
{
u
∈
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
:
∥
u
∥
2
<
r
}, U
(
0
,
r
)
≜
{
u
∈
V
:
∥
u
∥
V
<
r
}, P
2
r
=
P
2
∩
B
(
0
,
r
), ∂
P
2
r
=
P
2
∩
∂
B
(
0
,
r
).LetK denote the solution operator of (12), and (
f
u
)
(
x
)
≜
f
(
x
,
u
(
x
)
). Under condition (f1), it is easy to see that A
≜
K
f
:
V
→
V is of class C
1. Since f
:
V
→
V
* is completely continuous [3], then A
:
V
→
V is completely continuous.Lemma 5 (see [16]).
Supposeh
∈
L
q
(
Ω
), q
⩾
2; then the weak solution u
=
K
(
h
) of (12) satisfies ∥
u
∥
W
4
,
q
⩽
C
∥
h
∥
L
q; what is more, we have that
(13)
K
:
L
q
(
Ω
)
⟼
W
4
,
q
(
Ω
)
∩
W
0
1
,
q
(
Ω
)
is continuous.Remark 6.
Actually, for allh
∈
V
*, there exists a unique weak solution u
=
K
(
h
)
∈
V of (12). Since by Riesz representation theorem, for all h
∈
V
*, there exists a unique Θ
=
Θ
(
h
) such that 〈
h
,
v
〉
=
(
Θ
,
v
)  ∀
v
∈
V; thus Θ
=
K
(
h
) is the corresponding weak solution.Consider the Cauchy problem inV,
(14)
d
d
t
u
(
t
)
=
-
u
(
t
)
+
K
f
u
(
t
)
,
u
(
0
)
=
u
0
.Lemma 7 (see [13]).
LetH be a real Hilbert space, and let ψ
∈
C
2
(
H
,
ℝ
) satisfy the (
P
S
) condition. Assume that
(15)
ψ
′
(
v
)
=
v
-
A
v
,
v
∈
H
,
where A is a compact mapping, and that p
0 is an isolated critical point of f. Then we have
(16)
ind
(
ψ
′
,
p
0
)
=
∑
q
=
0
∞
‍
(
-
1
)
q
rank
⁡
C
q
(
ψ
,
p
0
)
.LetX be a retract of a real Banach space E, let U be a relatively open subset of X, and let A
:
U
¯
X
→
X be a completely continuous operator. Suppose that A has no fixed points on ∂
X
U and that the fixed point of A is bounded. The following lemma establishes the relationship of fixed point index and topological degree.Lemma 8 (see [17]).
If any fixed point of inU is an interior point of X, then there exists an open subset O of E with O
⊂
U such that O contains all fixed points of A in U and
(17)
deg
⁡
(
I
-
A
,
O
,
0
)
=
i
(
A
,
U
,
X
)
.Remark 9.
LetO be a bounded open subset of U, and let there be no zero points of I
-
A on ∂
O. Since C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
) can be compactly embedded into V, it follows from the bootstrap argument and the definition of Leray-Schauder degree that
(18)
de
g
V
(
I
-
A
,
O
,
0
)
=
de
g
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
(
I
-
A
,
O
,
0
)
.In what follows,de
g
C
0
2
(
Ω
¯
) is denoted simply by deg
⁡.Remark 10 (see [18]).
Remark9 implies that two topological degrees in both de
g
C
0
2
(
Ω
¯
) and de
g
V are the same. Combining with Lemma 7, we can obtain the connection between the topological degree and the critical group:
(19)
deg
⁡
(
I
-
A
,
O
,
0
)
=
∑
q
=
0
∞
‍
(
-
1
)
q
rank
⁡
C
q
(
J
,
p
0
)
.Lemma 11.
Letu
(
t
,
u
0
) be the unique solution of (14) with the maximal interval [
0
,
η
(
u
0
)
). We have the following conclusions.(i)
Ifu
0
∈
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
), then {
u
(
t
,
u
0
)
:
0
⩽
t
<
η
(
u
0
)
}
⊂
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
), and u
(
t
,
u
0
) is continuous as a function of t from [
0
,
η
(
u
0
)
) to C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
).
(ii)
Ifu
0, u
⋆
∈
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
), u
⋆
=
K
f
u
⋆, and ∥
u
(
t
,
u
0
)
-
u
⋆
∥
2
→
0 as t
→
η
(
u
0
), then ∥
u
(
t
,
u
0
)
-
u
⋆
∥
2
→
0 as t
→
η
(
u
0
).
(iii)
Ifu
0
∈
C
0
(
Ω
¯
)
∩
C
2
,
μ
(
Ω
¯
) for some μ
∈
(
0,1
) then {
u
(
t
,
u
0
)
:
0
⩽
t
<
η
(
u
0
)
}
⊂
C
0
(
Ω
¯
)
∩
C
2
μ
(
Ω
¯
) and is bounded in the C
0
2
,
μ norm.Lemma11 essentially comes from [19].Proof.
We only need to construct the embedding chains like (5) and (6) of [19]; the rest can be proved similar to [19, Lemma 2].
Without loss of generality,α can be assumed to satisfy max
⁡
{
8
/
(
N
-
4
)
,
1
}
<
α
<
(
N
+
4
)
/
(
N
-
4
). We can choose δ
>
0, such that
(20)
α
<
δ
+
(
N
+
4
)
(
1
-
δ
)
N
-
4
.
Let q
0
′
=
2
N
/
(
N
-
4
), and define q
i by
(21)
1
q
i
+
1
′
=
α
q
i
′
-
2
N
,
i
=
0,1
,
3
,
…
.
A direct computation shows that(22)
q
n
′
⩾
(
5
5
-
4
δ
)
n
q
0
′
.
Hence there exists a number n
⩾
3 such that
(23)
q
0
′
<
q
1
′
<
⋯
<
q
n
-
3
′
<
N
α
2
⩽
q
n
-
2
′
.
Let(24)
q
i
=
q
i
′
,
i
=
0,1
,
2
,
…
,
n
-
3
,
and choose q
n
-
2 and q
n
-
1 such that
(25)
q
n
-
3
<
q
n
-
2
<
N
α
2
,
q
n
-
1
=
α
N
.
Let(26)
p
i
=
q
i
α
,
i
=
0,1
,
2
,
…
,
n
-
1
.
Define
(27)
X
0
=
L
q
0
(
Ω
)
,
X
i
+
1
=
W
4
,
p
i
(
Ω
)
∩
W
0
1
,
p
i
,
Y
i
=
L
p
i
(
Ω
)
,
Z
i
=
L
q
i
(
Ω
)
,
i
=
0,1
,
…
,
n
-
1
.
Then we have the following imbedding chains:(28)
V
↓
X
n
⟶
X
n
-
1
⟶
X
n
-
2
⟶
⋯
⟶
X
1
⟶
X
0
↓
↓
↓
↓
↓
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
⟶
Z
n
-
1
⟶
Z
n
-
2
⟶
⋯
⟶
Z
1
⟶
Z
0
,
Y
n
-
1
⟶
Y
n
-
2
⟶
⋯
⟶
Y
1
⟶
Y
0
.What is more, we have the chains of bounded and continuous operators
(29)
Z
i
→
f
Y
i
→
K
X
i
+
1
,
i
=
0,1
,
2
,
…
,
n
-
1
.Lemma 12.
Suppose that (f1) and (f3) hold. ThenΨ satisfies the (
P
S
) condition.The proof of this lemma is similar to the proof of [5, Lemma 2.1]. We omit it here.SinceA
:
V
→
V is completely continuous, then by the above bootstrap iteration, A
:
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
→
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
) is completely continuous. For our application, sometimes we would consider the restriction Ψ
~ of Ψ on a smaller Banach space C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
). The functional may lose the (PS) condition. However following [20], the following two lemmas can be obtained.Lemma 13 (see [20]).
Suppose that (f1) and (f3) hold. ThenΨ
~ possesses the following properties. (i)
Ψ
~
(
K
) is a closed subset.
(ii)
For each paira
<
b, K
∩
Ψ
~
-
1
(
a
,
b
)
=
∅ implies that Ψ
~
a is a strong deformation retract of    Ψ
~
b
∖
K
b, where K denotes the critical set of   Ψ (and also Ψ
~).Lemma 14 (see [20]).
C
*
(
Ψ
~
,
p
0
)
=
C
*
(
Ψ
,
p
0
) with integral coefficients.Here and in what follows, we always assume thatΨ has only finitely many critical points.Lemma 15 (see [21]).
Let0 be an isolated critical point of   Ψ
∈
C
2
(
E
,
ℝ
), where N
=
ker
⁡
[
Ψ
′
(
0
)
]. Denote μ
=
dim
⁡
E
-
<
∞, ν
=
dim
⁡
N
<
∞, and assume that Ψ has a local linking at 0 with respect to a direct sum decomposition E
=
W
-
⊕
W
+, where W
-
=
E
-
⊕
N; that is, there exists r
>
0 small such that
(30)
Ψ
(
u
)
>
0
f
o
r
u
∈
W
+
,
0
<
∥
u
∥
V
⩽
r
,
Ψ
(
u
)
⩽
0
f
o
r
u
∈
W
-
,
∥
u
∥
V
⩽
r
.
Then
(31)
C
q
(
Ψ
,
0
)
=
δ
q
,
k
𝔽
f
o
r
k
=
μ
+
ν
.


## 3. Calculation of Degree

Lemma 16.
Suppose that (f1),  (f3), and (f4) hold. Then there existsr
l
k
>
0, such that, for all r
∈
(
0
,
r
l
k
],
(32)
deg
⁡
(
I
-
A
,
B
(
0
,
r
)
,
0
)
=
0
.Proof.
Since{
e
j
}
j
=
1
∞ is an orthogonal basis of V, for u
∈
V, there exist {
a
j
}
j
=
1
∞
⊂
ℝ, such that u
=
∑
j
=
1
∞
‍
a
j
e
j. Let
(33)
E
-
=
H
-
⊕
N
,
E
+
=
H
+
.
Since E
- is finite dimensional, we have that, for given r
>
0, there exists some ρ
>
0 such that if
(34)
u
∈
E
-
,
∥
u
∥
V
⩽
ρ
,
then
(35)
|
u
(
x
)
|
⩽
r
3
<
r
,
a
.
e
.
x
∈
Ω
.
By (f4), for u
∈
E
- with ∥
u
∥
⩽
ρ,
(36)
Ψ
(
u
)
=
1
2
∫
Ω
‍
(
|
Δ
u
|
2
)
d
x
-
∫
Ω
‍
F
(
x
,
u
)
d
x
⩽
1
2
∫
Ω
‍
(
|
Δ
u
|
2
-
μ
2
i
u
2
)
d
x
=
1
2
∫
Ω
‍
(
|
Δ
∑
j
=
1
2
i
‍
a
j
e
j
|
2
-
μ
2
i
∑
j
=
1
2
i
‍
a
j
e
j
)
d
x
⩽
1
2
∫
Ω
‍
(
|
Δ
u
|
2
-
m
u
2
i
u
2
)
d
x
=
1
2
∫
Ω
‍
(
|
∑
j
=
1
2
i
‍
μ
j
a
j
e
j
|
2
-
∑
j
=
1
2
i
‍
μ
2
i
a
j
e
j
)
d
x
⩽
0
.
For u
∈
E
+ with 0
<
∥
u
∥
⩽
ρ,
(37)
Ψ
(
u
)
=
1
2
∫
Ω
‍
(
|
Δ
u
|
2
)
d
x
-
∫
Ω
‍
F
(
x
,
u
)
d
x
>
1
2
∫
Ω
‍
(
|
Δ
u
|
2
-
μ
2
i
+
1
u
2
)
d
x
=
1
2
∫
Ω
‍
(
|
Δ
∑
j
=
2
i
+
1
∞
‍
a
j
e
j
|
2
-
m
u
2
i
+
1
∑
j
=
2
i
+
1
∞
‍
a
j
e
j
)
d
x
=
1
2
∫
Ω
‍
(
|
∑
j
=
2
i
∞
‍
μ
j
a
j
e
j
|
2
-
∑
j
=
2
i
+
1
∞
‍
μ
2
i
+
1
a
j
e
j
)
d
x
⩾
0
.
Thus Ψ possesses a local linking at the origin. By Lemma 15, the critical groups of Ψ at the origin satisfy
(38)
C
μ
2
i
(
Ψ
,
0
)
≠
0
.
Then there exists r
l
k
>
0 small such that there is no other critical point in B
(
0
,
r
l
k
) except 0, for all r
∈
(
0
,
r
l
k
], and the following can be obtained by Lemma 13 and Remark 10:
(39)
deg
⁡
(
I
-
A
,
B
(
0
,
r
)
,
0
)
=
1
.Lemma 17.
Suppose (f1) and (f2) hold. There existsr
1
∈
(
0
,
r
l
k
] (r
l
k is defined in Lemma 16), such that, for all r
∈
(
0
,
r
1
],
(40)
i
(
A
,
P
2
r
,
P
2
)
=
0
,
i
(
A
,
-
P
2
r
,
-
P
2
)
=
0
.Proof.
We only prove thati
(
A
,
P
2
r
,
P
2
)
=
0. By (f1),  A
(
P
2
)
⊂
P
2, it follows from the condition (f2) that there exist δ
1
>
0 and ρ
1
∈
(
0
,
r
l
k
] such that
(41)
f
(
x
,
t
)
⩾
μ
1
(
1
+
δ
1
)
t
,
(
x
,
t
)
∈
Ω
¯
×
[
0
,
ρ
1
]
.
Ifu
=
A
u
+
ν
e
1 for some ν
⩾
0 and u
∈
∂
P
2
r, where r
∈
(
0
,
ρ
1
] is a positive number, that is
(42)
Δ
2
u
=
f
(
x
,
u
)
+
ν
e
1
,
in
Ω
.
u
=
Δ
u
=
0
,
on
⁡
∂
Ω
,
then we have from (41) that
(43)
Δ
2
u
⩾
μ
1
(
1
+
δ
1
)
u
,
in
Ω
.
u
=
Δ
u
=
0
,
on
⁡
∂
Ω
.
Thus,
(44)
μ
1
∫
Ω
‍
u
(
x
)
e
1
(
x
)
d
x
⩾
μ
1
(
1
+
δ
1
)
∫
Ω
‍
u
(
x
)
e
1
(
x
)
d
x
,
and this is a contradiction. Therefore, according to the property of fixed point index, we get
(45)
i
(
A
,
P
2
r
,
P
2
)
=
0
.
Similarly, we can also show that there existsρ
2
∈
(
0
,
r
0
] such that i
(
A
,
-
P
2
r
,
-
P
2
)
=
0 for all r
∈
(
0
,
ρ
2
]. Let r
1
=
min
⁡
{
ρ
1
,
ρ
2
}. Then the conclusion holds.Lemma 18.
Suppose that (f1) and (f3) hold. Then there existsO
+
⊂
P
2
∖
{
0
}, O
-
⊂
(
-
P
2
∖
{
0
}
), such that
(46)
deg
⁡
(
I
-
A
,
O
+
,
0
)
=
1
,
deg
⁡
(
I
-
A
,
O
-
,
0
)
=
1
.Proof.
Sincelim
⁡
|
t
|
→
∞
f
(
x
,
t
)
/
t
<
μ
1 uniformly for x
∈
Ω
¯, there exist constants δ
∈
(
0,1
) and C
1
>
0 such that
(47)
f
(
x
,
t
)
⩽
μ
1
(
1
-
δ
)
t
+
C
1
,
(
x
,
t
)
∈
Ω
¯
×
[
0
,
∞
)
.
We will first show that any solution of (1) is bounded. Suppose u
0 is a solution; then u
0 satisfy
(48)
Δ
2
u
0
=
f
(
x
,
u
0
)
,
in
Ω
,
u
0
=
Δ
u
0
=
0
,
on
⁡
∂
Ω
.
Multiplying by u
0, we have
(49)
∥
u
0
∥
V
2
=
∫
Ω
‍
f
(
x
,
u
0
)
u
0
,
x
∈
Ω
;
by (47), it is easy to see
(50)
∥
u
0
∥
V
2
=
∫
Ω
‍
μ
1
(
1
-
δ
)
u
0
2
+
∫
Ω
‍
C
1
u
0
,
x
∈
Ω
.
Let ϵ
<
μ
1
δ
/
(
2
C
1
), and using Young inequality, there exists constant C
2
=
C
2
(
δ
,
C
1
) such that
(51)
∥
u
0
∥
V
2
⩽
∫
Ω
‍
μ
1
(
1
-
δ
2
)
u
0
2
+
C
2
|
Ω
|
1
/
2
.
By Poincaré inequality, there exists C
3
>
0 only dependent on Ω, such that
(52)
∥
u
0
∥
V
<
C
3
.
By a bootstrap argument, there exists R
1
>
0 such that ∥
u
0
∥
2
<
R
1.
LetR
>
R
1, and we will show that
(53)
A
u
≠
ν
u
,
∀
x
∈
P
2
∩
∂
B
(
0
,
R
)
,
ν
⩾
1
.
Suppose there exists ν
0
⩾
1,  ∥
u
0
∥
V
=
R, such that A
u
0
=
ν
0
u
0; then ν
0
>
1. By (f3), such that f
(
x
,
t
)
⩽
μ
1
(
1
-
δ
)
t
+
C
1, for all t
⩾
0,
(54)
ν
0
u
0
=
A
u
0
⩽
μ
1
(
1
-
δ
)
K
u
0
+
C
4
K
1
.
Then [
1
-
μ
1
(
1
-
δ
)
/
K
]
u
0
⩽
C
5, where C
5
=
C
1
∥
K
∥
L
(
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
,
C
0
(
Ω
¯
)
∩
C
2
(
Ω
¯
)
). Then u
0
⩽
C
6, where C
6
=
∥
(
I
-
K
)
-
1
∥
C
5. Let R
=
C
6; then, for all u
∈
∂
B
(
0
,
R
), we have
(55)
A
u
≠
ν
u
∀
x
∈
∂
P
2
R
,
ν
⩾
1
.
Then
(56)
i
(
A
,
P
2
R
,
P
2
)
=
1
.
From Lemma17, we have
(57)
i
(
A
,
P
2
R
∖
P
2
r
¯
,
P
2
)
=
1
.
By Lemma18 and the strong maximum principle of second order elliptic problem, it is easy to know that if u
∈
P
2
∖
{
0
} is a solution of (1), then u
∈
P
°
2, and thus, by Lemma 2, u
∈
P
°
2. Using Lemma 8, there is a bounded open O
1
⊂
P
2
∩
(
B
(
0
,
R
)
∖
B
(
0
,
r
)
), such that
(58)
deg
⁡
(
I
-
A
,
O
+
,
0
)
=
1
.
Similarly, there is a bounded open subset O
-
⊂
-
(
P
2
R
∖
P
¯
2
r
), such that
(59)
deg
⁡
(
I
-
A
,
O
-
,
0
)
=
1
.


## 4. Proof of Main Result

Proof.
By conditions (f1) and (f3), it is easy to know that(60)
ind
(
I
-
A
,
∞
)
=
ind
(
I
-
A
′
(
∞
)
,
0
)
=
(
-
1
)
0
=
1
;
that is, there exists R
¯
>
R large enough, such that
(61)
deg
⁡
(
I
-
A
,
B
(
0
,
R
¯
)
,
0
)
=
ind
(
I
-
A
,
∞
)
=
1
.
IfA has no fixed point in B
(
0
,
R
¯
)
∖
(
P
2
R
∪
(
-
P
2
R
)
), then the additivity property of degree implies
(62)
deg
⁡
⁡
(
I
-
A
,
B
(
0
,
R
¯
)
,
0
)
=
deg
⁡
⁡
(
I
-
A
,
O
+
,
0
)
+
deg
⁡
(
I
-
A
,
O
-
,
0
)
+
deg
⁡
(
I
-
A
,
B
(
0
,
r
2
)
,
0
)
.
It follows that 1
=
1
+
1
+
1. This is a contradiction. Thus (1) has at least a solution u
3 in B
(
0
,
R
¯
)
∖
(
P
2
R
∪
(
-
P
2
R
)
∪
B
(
0
,
(
r
/
2
)
)
).



---

*Source: 267052-2014-04-15.xml*