# Eigenvalues for a Neumann Boundary Problem Involving thep

Authors: Qing Miao
Journal: Advances in Mathematical Physics
 (2015)

---

## Abstract

We study the existence of weak solutions to the following Neumann problem involving thep
(
x
)-Laplacian operator:  -
Δ
p
(
x
)
u
+
e
(
x
)
|
u
|
p
(
x
)
-
2
u
=
λ
a
(
x
)
f
(
u
), in
Ω, ∂
u
/
∂
ν
=
0, on
∂
Ω. Under some appropriate conditions on the functions p,  e,  a, and  f, we prove that there exists λ
¯
>
0 such that any λ
∈
(
0
,
λ
¯
) is an eigenvalue of the above problem. Our analysis mainly relies on variational arguments based on Ekeland’s variational principle.


---

## Body

## 1. Introduction

This paper is considered with the existence of weak solutions to the following Neumann problem:(1)
-
Δ
p
x
u
+
e
(
x
)
u
p
x
-
2
u
=
λ
a
x
f
u
in
Ω
,
∂
u
∂
ν
=
0
on
⁡
∂
Ω
,where Ω
⊂
R
N
(
N
≥
1
) is a bounded domain with a smooth boundary, p
(
x
)
∈
C
(
Ω
¯
) with p
(
x
)
>
1 on Ω
¯, and λ
>
0 is a parameter. e
(
x
)
∈
C
(
Ω
¯
) is nonnegative, f
∈
C
(
R
), and a
(
x
)
∈
L
r
(
x
)
(
Ω
) for some r
∈
C
(
Ω
¯
).The operatorΔ
p
(
x
)
u
=
div
(
|
∇
u
|
p
(
x
)
-
2
∇
u
) is called a p
(
x
)-Laplacian. If p
(
x
)
≡
p is a constant, then operator is the well-known p-Laplacian, and (1) is the usual p-Laplacian equation. As a matter of fact, the p
(
x
)-Laplacian has more complicated nonlinearities than the p-Laplacian. For example, it is not homogeneous. The study of differential equations with p
(
x
)-growth conditions is an interesting and attractive topic and has been the object of considerable attention in recent years [1–6].In [3], by applying a variational principle due to B. Ricceri and the theory of the variable exponent Sobolev spaces, the author considered the Neumann problem of p
(
x
)-Laplacian: (2)
-
Δ
p
(
x
)
u
+
λ
(
x
)
u
p
(
x
)
-
2
u
=
f
(
x
,
u
)
+
g
(
x
,
u
)
in
Ω
,
∂
u
∂
ν
=
0
on
⁡
∂
Ω
.In [5], based on the three critical points theorem due to B. Ricceri, the author studied the problem(3)
-
Δ
p
x
u
+
e
x
u
p
x
-
2
u
=
λ
a
(
x
)
f
(
x
,
u
)
+
μ
g
x
,
u
in
Ω
,
∂
u
∂
ν
=
0
on
⁡
∂
Ω
.In [6], the author established the existence of at least three solutions of the problem(4)
-
Δ
p
x
u
+
u
p
x
-
2
u
=
λ
α
(
x
)
f
(
u
)
+
β
x
g
u
in
Ω
,
∂
u
∂
ν
=
0
on
⁡
∂
Ω
.However, in [3, 5, 6], the function p
(
x
) assumed that p
-
>
N. In this paper, by using Ekeland’s variational principle, we show that when 1
<
p
(
x
)
≤
N, there exists λ
¯
>
0 such that any λ
∈
(
0
,
λ
¯
); problem (1) has at least one nontrivial weak solution. For more applications of Ekeland’s variational principle to other problems, see, for example, [7–10]. Our result is partly motivated by these nice papers.This paper is organized as follows. In Section2, we recall some basic facts about the variable exponent Lebesgue and Sobolev spaces and also give some propositions that will be used later. In Section 3, we obtain the existence of weak solutions of problem (1).


## 2. Preliminaries

In order to deal with thep
(
x
)-Laplacian problem, we need some theories on spaces L
p
(
x
)
(
Ω
), W
1
,
p
(
x
)
(
Ω
) and properties of p
(
x
)-Laplacian which will be used later. Let(5)
C
+
Ω
¯
=
h
∣
h
∈
C
Ω
¯
,
h
x
>
1
for
x
on
⁡
Ω
¯
.Through this paper, for any h
∈
C
(
Ω
¯
), we use the notations(6)
h
+
∶
=
sup
⁡
x
∈
Ω
¯
h
x
,
h
-
∶
=
inf
⁡
x
∈
Ω
¯
h
x
;denote (7)
L
p
x
Ω
=
u
∣
u
is
a
measurable
real-valued
function
,
∫
Ω
‍
∫
Ω
‍
u
x
p
(
x
)
d
x
<
∞
.We can introduce the norm on L
p
(
x
)
(
Ω
) by (8)
u
p
(
x
)
=
inf
⁡
λ
>
0
∣
∫
Ω
‍
u
x
λ
p
x
d
x
≤
1
,and (
L
p
(
x
)
(
Ω
)
,
·
p
(
x
)
) becomes a Banach space; we call it a generalized Lebesgue space.The variable exponent Sobolev spaceW
1
,
p
(
x
)
(
Ω
) is defined by(9)
W
1
,
p
x
(
Ω
)
=
u
∈
L
p
x
(
Ω
)
∣
∇
u
∈
L
p
x
(
Ω
)
,and it can be equipped with the norm (10)
u
=
u
p
(
x
)
+
∇
u
p
(
x
)
,
∀
u
∈
W
1
,
p
(
x
)
(
Ω
)
.The spaces L
p
(
x
)
(
Ω
), W
1
,
p
(
x
)
(
Ω
) are separable, reflexive Banach spaces [11].From now on, we denoteX by the space W
1
,
p
(
x
)
(
Ω
). For any u
∈
X, define(11)
u
e
=
inf
⁡
λ
>
0
∣
∫
Ω
‍
∇
u
x
λ
p
x
+
e
x
u
x
λ
p
x
d
x
≤
1
;then, it is easy to see that u
e is a norm on X and equivalent to u. In the following, we will use u
e instead of u on X.Proposition 1 (see [11, 12]).
The conjugate space ofL
p
(
x
)
(
Ω
) is L
p
0
(
x
)
(
Ω
), where 1
/
p
(
x
)
+
1
/
p
0
(
x
)
=
1. For any u
∈
L
p
(
x
)
(
Ω
) and v
∈
L
p
0
(
x
)
(
Ω
), one has the following Hölder-type inequality: (12)
∫
Ω
‍
u
v
d
x
≤
1
p
-
+
1
p
0
-
u
p
(
x
)
v
p
0
(
x
)
≤
2
u
p
(
x
)
v
p
0
(
x
)
.Moreover, if h
i
∈
C
+
(
Ω
¯
) with 1
/
h
1
(
x
)
+
1
/
h
2
(
x
)
+
1
/
h
3
(
x
)
=
1, then, for u
∈
L
h
1
(
x
)
(
Ω
), v
∈
L
h
2
(
x
)
(
Ω
), w
∈
L
h
3
(
x
)
(
Ω
), one has(13)
∫
Ω
‍
u
v
w
d
x
≤
1
h
1
-
+
1
h
2
-
+
1
h
3
-
u
h
1
x
v
h
2
x
w
h
3
x
≤
3
u
h
1
(
x
)
v
h
2
(
x
)
w
h
3
(
x
)
.Proposition 2 (see [11]).
Putρ
(
u
)
=
∫
Ω
‍
|
u
|
p
(
x
)
d
x, ∀
u
∈
L
p
(
x
)
(
Ω
); then(i)
|
u
|
p
(
x
)
<
1
(
=
1
;
>
1
)
⇔
ρ
(
u
)
<
1
(
=
1
;
>
1
);
(ii)
|
u
|
p
(
x
)
>
1
⇒
|
u
|
p
(
x
)
p
-
≤
ρ
(
u
)
≤
|
u
|
p
(
x
)
p
+; |
u
|
p
(
x
)
<
1
⇒
|
u
|
p
(
x
)
p
+
≤
ρ
(
u
)
≤
|
u
|
p
(
x
)
p
-;
(iii)
|
u
|
p
(
x
)
→
0
⇔
ρ
(
u
)
→
0; |
u
|
p
(
x
)
→
∞
⇔
ρ
(
u
)
→
∞.From Proposition2, the following inequalities hold:(14)
u
e
p
-
≤
∫
Ω
‍
∇
u
x
p
x
+
e
x
u
x
p
x
d
x
≤
u
e
p
+
,
if
u
e
≥
1
;
(15)
u
e
p
+
≤
∫
Ω
‍
∇
u
x
p
x
+
e
x
u
x
p
x
d
x
≤
u
e
p
-
,
if
u
e
≤
1
.Proposition 3 (see [13]).
Letq
∈
L
∞
(
Ω
) be such that 1
≤
p
(
x
)
q
(
x
)
≤
∞ for a.e.  x
∈
Ω. Let u
∈
L
p
(
x
)
(
Ω
)
,
u
≠
0. Then, one has the following:(i)
if|
u
(
x
)
|
p
(
x
)
q
(
x
)
≤
1, then |
u
|
p
(
x
)
q
(
x
)
q
+
≤
u
q
x
p
(
x
)
≤
|
u
|
p
(
x
)
q
(
x
)
q
-;
(ii)
if|
u
(
x
)
|
p
(
x
)
q
(
x
)
≥
1, then |
u
|
p
(
x
)
q
(
x
)
q
-
≤
u
q
x
p
(
x
)
≤
|
u
|
p
(
x
)
q
(
x
)
q
+.For anyx
∈
Ω
¯, let(16)
p
*
(
x
)
=
N
p
(
x
)
N
-
p
(
x
)
,
if
p
(
x
)
<
N
,
∞
,
if
p
x
≥
N
.Proposition 4 (see [11, 14]).
Assume that1
≤
q
(
x
)
∈
C
(
Ω
¯
) satisfy q
(
x
)
<
p
*
(
x
) on Ω
¯; then the imbedding from W
1
,
p
(
x
)
(
Ω
) to L
q
(
x
)
(
Ω
) is compact and continuous.We defineΦ
:
X
→
R by(17)
Φ
(
u
)
=
∫
Ω
‍
1
p
(
x
)
∇
u
x
p
(
x
)
+
e
(
x
)
p
(
x
)
u
x
p
(
x
)
d
x
,
Ψ
(
u
)
=
∫
Ω
‍
a
(
x
)
F
(
u
)
d
x
,
F
(
t
)
=
∫
0
t
‍
f
(
s
)
d
s
,
I
λ
(
u
)
=
Φ
(
u
)
-
λ
Ψ
(
u
)
.Proposition 5 (see [15]).
(i)Φ is weakly lower semicontinuous, Φ
∈
C
1
(
X
,
R
), and(18)
Φ
′
u
,
v
=
∫
Ω
‍
∇
u
p
x
-
2
∇
u
∇
v
+
e
x
u
p
x
-
2
u
v
d
x
,
∀
u
,
v
∈
X
;
(ii)Φ
′ is a mapping of type (
S
+
); that is, if u
n
⇀
u and lim
inf
⁡
n
→
∞
(
Φ
′
(
u
n
)
,
u
n
-
u
)
≤
0, then u
n
→
u;
(iii)Φ
′
:
X
→
X
* is a homeomorphism.


## 3. Main Result

We say thatu
∈
X is a weak solution of problem (1) if(19)
∫
Ω
‍
∇
u
p
x
-
2
∇
u
∇
v
+
e
x
u
p
x
-
2
u
v
d
x
=
λ
∫
Ω
‍
a
x
f
u
v
d
x
,
∀
v
∈
X
.It follows that we can seek for weak solutions of problem (1). We need the following assumptions.(A) There existγ
1
≥
γ
2
>
0, 0
<
β
<
1, and q
1
,
q
2
,
r
∈
C
+
(
Ω
¯
) such that(20)
1
<
q
1
(
x
)
≤
q
2
(
x
)
<
p
(
x
)
≤
N
<
r
x
on
⁡
Ω
¯
,
(21)
0
≤
F
(
t
)
≤
γ
1
t
q
1
(
x
)
for
t
∈
R
,
(22)
F
(
t
)
≥
γ
2
t
q
2
(
x
)
for
t
∈
[
-
β
,
β
]
.(B) Considera
(
x
)
∈
L
r
(
x
)
(
Ω
) and there exists a subset Ω
1
⊂
Ω with meas
⁡
(
Ω
1
)
>
0 such that a
(
x
)
>
0 for x
∈
Ω
1, where meas
⁡
(
·
) denotes the Lebesgue measure.Remark 6.
Regarding condition (A), we have(23)
r
0
x
q
i
x
<
p
*
x
,
s
i
x
=
r
x
q
i
x
r
x
-
q
i
x
<
p
*
x
,
for
i
=
1,2
,where 1
/
r
(
x
)
+
1
/
r
0
(
x
)
=
1. Thus, by Proposition 4, the embeddings X
↪
L
r
0
(
x
)
q
i
(
x
)
(
Ω
) and X
↪
L
s
i
(
x
)
Ω, i
=
1,2 are continuous and compact.By a standard argument, we have thatΨ is weakly lower semicontinuous, Ψ
∈
C
1
(
X
,
(
R
)
), and(24)
Ψ
′
u
,
v
=
∫
Ω
‍
a
x
f
u
v
d
x
,
∀
u
,
v
∈
X
.Remark 7.
I
λ is weakly lower semicontinuous, I
λ
∈
C
1
(
X
,
R
), and(25)
I
λ
′
u
,
v
=
∫
Ω
‍
∇
u
p
x
-
2
∇
u
∇
v
+
e
x
u
p
x
-
2
u
v
d
x
-
λ
∫
Ω
‍
a
x
f
u
v
d
x
,
∀
v
∈
X
.Thus, u is weak solution of problem (1) if and only if u is a critical point of I
λ.Note from Remark6 that the embedding X
↪
L
r
0
(
x
)
q
1
(
x
)
(
Ω
) is continuous; then, there exists a positive constant d
>
1 such that(26)
u
r
0
(
x
)
q
1
(
x
)
≤
d
u
e
,
for
any
u
∈
X
.Lemma 8.
For anyα
∈
(
0,1
/
d
), there exist λ
¯
>
0 and γ
>
0 such that I
λ
(
u
)
≥
γ for any λ
∈
(
0
,
λ
¯
) and u
∈
X with u
e
=
α.Proof.
Letα
∈
(
0,1
/
d
) be fixed. Then, α
<
1, and, from (26), we know that(27)
u
r
0
x
q
1
x
<
1
,
for
any
u
∈
X
,
u
e
=
α
.From Propositions 1 and 3, (15), (21), and (26), it follows that (28)
I
λ
u
≥
1
p
+
∫
Ω
‍
∇
u
p
x
+
e
x
u
p
x
d
x
-
λ
∫
Ω
‍
a
x
F
u
d
x
≥
1
p
+
u
e
p
+
-
λ
γ
1
∫
Ω
‍
a
x
u
q
1
x
d
x
≥
1
p
+
u
e
p
+
-
2
λ
γ
1
a
x
r
x
u
q
1
x
r
0
x
≥
1
p
+
u
e
p
+
-
2
λ
γ
1
a
r
(
x
)
u
r
0
(
x
)
q
1
(
x
)
q
1
-
≥
1
p
+
u
e
p
+
-
2
λ
γ
1
d
q
1
-
a
r
(
x
)
u
e
q
1
-
=
1
p
+
α
p
+
-
2
λ
γ
1
d
q
1
-
a
r
(
x
)
α
q
1
-
.Hence, if we let(29)
λ
¯
=
2
α
p
+
-
q
1
-
5
p
+
γ
1
d
q
1
-
a
r
(
x
)
,then, for any λ
∈
(
0
,
λ
¯
) and u
∈
X with u
e
=
α, there exists γ
=
α
p
+
/
5
p
+
>
0 such that I
λ
(
u
)
≥
γ.Lemma 9.
There existsϕ
∈
X such that ϕ
≥
0, ϕ
≠
0, and I
λ
(
t
ϕ
)
<
0 for t
>
0 small enough.Proof.
From condition (B) and (20), there exists a subset Ω
1
⊂
Ω, and q
2
(
x
)
<
p
(
x
) on Ω
1. If we let q
^
2
=
min
⁡
x
∈
Ω
¯
1
q
2
(
x
) and p
^
=
min
⁡
x
∈
Ω
¯
1
p
(
x
), then there exists δ
>
0 such that q
^
2
+
δ
<
p
^. Moreover, since q
2
(
x
)
∈
C
(
Ω
1
), there exists an open set Ω
2
⊂
Ω
1, meas
⁡
(
Ω
2
)
>
0, and |
q
2
(
x
)
-
q
^
2
|
<
δ for x
∈
Ω
2. Thus, q
2
(
x
)
<
q
^
2
+
δ
<
p
^ in Ω
2.
Letϕ
∈
C
0
∞
(
Ω
) be nontrivial such that supp
(
ϕ
)
⊂
Ω
2
⊂
Ω
1, ϕ
≥
0 and ϕ
≠
0 in Ω
2. From (22), then for 0
<
t
<
min
⁡
{
1
,
β
/
(
max
⁡
x
∈
Ω
2
ϕ
x
)
}, we have (30)
I
λ
t
ϕ
=
∫
Ω
‍
1
p
x
∇
t
ϕ
p
x
+
e
x
t
ϕ
p
x
d
x
-
λ
∫
Ω
‍
a
x
F
t
ϕ
d
x
=
∫
Ω
2
‍
t
p
x
p
x
∇
ϕ
p
x
+
e
x
ϕ
p
x
d
x
-
λ
∫
Ω
2
‍
a
x
F
t
ϕ
d
x
≤
t
p
^
p
-
∫
Ω
2
‍
∇
ϕ
p
x
+
e
x
ϕ
p
x
d
x
-
λ
γ
2
∫
Ω
2
‍
a
x
t
ϕ
q
2
x
d
x
≤
t
p
^
p
-
∫
Ω
2
‍
∇
ϕ
p
x
+
e
x
ϕ
p
x
d
x
-
λ
γ
2
t
q
^
2
+
δ
∫
Ω
2
‍
a
(
x
)
ϕ
q
2
x
d
x
.Since ∫
Ω
2
‍
(
|
∇
(
ϕ
)
|
p
(
x
)
+
e
(
x
)
|
ϕ
|
p
(
x
)
)
d
x
>
0, in fact, if this is not true, ∫
Ω
2
‍
(
|
∇
(
ϕ
)
|
p
(
x
)
+
e
(
x
)
|
ϕ
|
p
(
x
)
)
d
x
=
0; by Proposition 2, we have ϕ
e
=
0 and so ϕ
≡
0 in Ω, which is a contradiction. Hence, I
λ
(
t
ϕ
)
<
0 for(31)
0
<
t
<
min
⁡
1
,
β
max
⁡
x
∈
Ω
2
⁡
ϕ
x
,
λ
γ
2
p
-
∫
Ω
2
‍
a
(
x
)
ϕ
q
2
x
∫
Ω
2
‍
∇
ϕ
p
x
+
e
x
ϕ
p
x
d
x
1
/
(
p
^
-
q
^
2
-
δ
)
<
min
⁡
λ
γ
2
p
-
∫
Ω
2
‍
a
(
x
)
ϕ
q
2
x
∫
Ω
2
‍
∇
ϕ
p
x
+
e
x
ϕ
p
x
d
x
1
/
(
p
^
-
q
^
2
-
δ
)
.In this paper, we study problem (1) by using Ekeland’s variation principle, which is recalled below.Lemma 10 (see [16]).
LetM be a complete metric space and let J
:
M
→
(
∞
,
∞
] be a lower semicontinuous functional, bounded from below, and not identically equal to ∞. Let ϵ
>
0 be given and z
∈
M be such that(32)
J
(
z
)
≤
inf
⁡
M
J
+
ϵ
.Then, there exists v
∈
M such that(33)
J
(
v
)
≤
J
(
z
)
≤
inf
⁡
M
J
+
ϵ
,
d
(
z
,
v
)
≤
1
,and for any u
≠
v in M,(34)
J
(
v
)
<
J
(
u
)
+
ϵ
d
(
v
,
u
)
,where d
(
·
,
·
) denotes the distance between two elements in M.We now state our main theorem.Theorem 11.
Assume that conditions (A) and (B) hold. Then, there existsλ
¯
>
0 such that any λ
∈
(
0
,
λ
¯
); problem (1) has at least one nontrivial weak solution.Proof.
Letλ
¯ be defined by (29), and λ
∈
(
0
,
(
¯
λ
)
). By Lemma 8, we have(35)
inf
⁡
∂
B
α
(
0
)
I
λ
>
0
,where B
α
(
0
) is the ball in X and ∂
B
α
(
0
) is the boundary of B
α
(
0
). For any u
∈
B
α
(
0
), by an argument similar to those used in Lemma 8, we can obtain that(36)
I
λ
u
≥
1
p
+
u
e
p
+
-
2
λ
γ
1
d
q
1
-
a
r
x
u
e
q
1
-
.Note from Lemma 9 that there exists ϕ
∈
X such that I
λ
(
t
ϕ
)
<
0 for t
>
0 small enough. Then, from (35) and (36),(37)
-
∞
<
c
_
λ
∶
=
inf
⁡
∂
B
α
¯
(
0
)
I
λ
<
0
.Let(38)
0
<
ϵ
<
inf
⁡
∂
B
α
(
0
)
I
λ
-
c
_
λ
.Applying Lemma 10, we see that there exists u
ϵ
∈
B
α
(
0
)
¯ such that(39)
c
_
λ
≤
I
λ
(
u
ϵ
)
≤
c
_
λ
+
ϵ
,
I
λ
(
u
ϵ
)
<
I
λ
(
u
)
+
ϵ
u
-
u
ϵ
e
for
u
≠
u
ϵ
.Since I
λ
(
u
ϵ
)
≤
c
_
λ
+
ϵ
<
inf
⁡
∂
B
α
(
0
)
I
λ, we have u
ϵ
∈
B
α
(
0
).
Now, define a functionalJ
λ
:
B
α
¯
(
0
)
→
R by(40)
J
λ
(
u
)
=
I
λ
(
u
)
+
ϵ
u
-
u
ϵ
e
.By (40), u
ϵ is a minimum point of J
λ, and for t
>
0 small enough and all v
∈
B
α
(
0
), we have(41)
J
λ
(
u
ϵ
+
t
v
)
-
J
λ
(
u
ϵ
)
t
≥
0
;then(42)
I
λ
(
u
ϵ
+
t
v
)
-
I
λ
(
u
ϵ
)
t
+
ϵ
v
e
≥
0
;when t
→
0, we have(43)
I
λ
′
u
ϵ
,
v
+
ϵ
v
e
≥
0
∀
v
∈
B
α
(
0
)
.Together with (39), there exists a sequence {
u
n
}
⊂
B
α
(
0
) such that(44)
I
λ
(
u
n
)
⟶
c
_
λ
,
I
λ
′
(
u
n
)
⟶
0
.By the reflexivity of X, there exists u
0
∈
X such that u
n
⇀
u
0. Note that(45)
I
λ
′
u
n
,
u
n
-
u
0
≤
I
λ
′
u
n
,
u
n
+
I
λ
′
u
n
,
u
0
≤
I
λ
′
(
u
n
)
u
n
+
I
λ
′
(
u
n
)
u
0
.Then, from (44), we have(46)
lim
n
→
∞
(
I
λ
′
(
u
n
)
,
u
n
-
u
0
)
=
0
.From (21) and Propositions 1 and 3, we have(47)
Ψ
′
u
n
,
u
n
-
u
0
≤
3
q
1
+
γ
1
a
r
x
max
⁡
⁡
u
n
q
1
x
q
1
+
-
1
,
u
n
q
1
x
q
1
-
-
1
u
n
-
u
0
s
1
(
x
)
,where s
1
(
x
) is defined in Remark 6; by the continuous and compact embedding of X
↪
L
q
1
(
x
)
(
Ω
) and X
↪
L
s
1
(
x
)
(
Ω
), we can get(48)
lim
n
→
∞
Ψ
′
u
n
,
u
n
-
u
0
=
0
.Now, we conclude that(49)
lim
n
→
∞
(
Φ
′
(
u
n
)
,
u
n
-
u
0
)
=
0
.Thus, by Proposition 5, we have u
n
→
u
0. Hence,(50)
I
λ
(
u
0
)
=
c
_
λ
<
0
,
I
λ
′
(
u
0
)
=
0
.Therefore, u
0 is a nontrivial weak solution of problem (1).



---

*Source: 632745-2015-03-11.xml*