Electronic Journal of Differential Equations, Vol. 2015 (2015), No. 158, pp. 1–8.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
EXISTENCE AND NON-EXISTENCE OF SOLUTIONS FOR A
p(x)-BIHARMONIC PROBLEM
GHASEM A. AFROUZI, MARYAM MIRZAPOUR, NGUYEN THANH CHUNG
Abstract. In this article, we study the following problem with Navier boundary conditions
∆(|∆u|p(x)−2 ∆u) + |u|p(x)−2 u = λ|u|q(x)−2 u + µ|u|γ(x)−2 u
u = ∆u = 0
in Ω,
on ∂Ω.
where Ω is a bounded domain in RN with smooth boundary ∂Ω, N ≥ 1.
p(x), q(x) and γ(x) are continuous functions on Ω, λ and µ are parameters.
Using variational methods, we establish some existence and non-existence results of solutions for this problem.
1. Introduction
In recent years, the study of differential equations and variational problems with
p(x)-growth conditions was an interesting topic, which arises from nonlinear electrorheological fluids and elastic mechanics. In that context we refer the reader to
Ruzicka [11], Zhikov [16] and the reference therein; see also [2, 4, 5, 7].
Fourth-order equations appears in many context. Some of theses problems come
from different areas of applied mathematics and physics such as Micro ElectroMechanical systems, surface diffusion on solids, flow in Hele-Shaw cells (see [8]). In
addition, this type of equations can describe the static from change of beam or the
sport of rigid body. El Amrouss et al [1] studied a class of p(x)-biharmonic of the
form
∆(|∆u|p(x)−2 ∆u) = λ|u|p(x)−2 u + f (x, u)
u = ∆u = 0
in Ω,
on ∂Ω,
where Ω is a bounded domain in RN , with smooth boundary ∂Ω, N ≥ 1, λ ≤ 0 and
some assumptions on the Carathéodory function f : Ω × R → R. They obtained
the existence and multiplicity of solutions.
In a recent article, Lin Li et al [9] considered the above problem and using
variational methods, by the assumptions on the Carathéodory function f , they
establish the existence of at least one solution and infinitely many solutions of the
problem.
2010 Mathematics Subject Classification. 35J60, 35B30, 35B40.
Key words and phrases. p(x)-Biharmonic; variable exponent; critical points;
minimum principle; fountain theorem; dual fountain theorem.
c
2015
Texas State University - San Marcos.
Submitted July 22, 2014. Published June 15, 2015.
1
2
G. A. AFROUZI, M. MIRZAPOUR, N. T. CHUNG
EJDE-2015/158
Inspired by the above references and the work of Jinghua Yao [13], the aim of this
article is to study the existence and multiplicity of weak solutions of the following
fourth-order elliptic equation with Navier boundary conditions
∆(|∆u|p(x)−2 ∆u) + |u|p(x)−2 u = λ|u|q(x)−2 u + µ|u|γ(x)−2 u in Ω,
u = ∆u = 0
on ∂Ω.
(1.1)
where Ω is a bounded domain in RN with smooth boundary ∂Ω, N ≥ 1, p(x), q(x)
and γ(x) are continuous functions on Ω with inf x∈Ω p(x) > 1, inf x∈Ω q(x) > 1,
inf x∈Ω γ(x) > 1 and λ and µ are parameters. Throughout the paper, we assume
that λ2 + µ2 6= 0.
2. Preliminaries
To study p(x)-Laplacian problems, we need some results on the spaces Lp(x) (Ω)
and W k,p(x) (Ω), and properties of p(x)-Laplacian, which we use later. Let Ω be a
bounded domain of RN , denote
C+ (Ω) = {h(x); h(x) ∈ C(Ω), h(x) > 1, ∀x ∈ Ω}.
For any h ∈ C+ (Ω), we define
h+ = max{h(x); x ∈ Ω},
h− = min{h(x); x ∈ Ω};
For any p ∈ C+ (Ω), we define the variable exponent Lebesgue space
n
Lp(x) (Ω) = u; u is a measurable real-valued function such that
Z
o
|u(x)|p(x) dx < ∞ ,
Ω
endowed with the so-called Luxemburg norm
Z
u(x) p(x)
|u|p(x) = inf µ > 0;
|
|
dx ≤ 1 .
µ
Ω
Then (Lp(x) (Ω), | · |p(x) ) becomes a Banach space.
Proposition 2.1 ([6]). The space (Lp(x) (Ω), |·|p(x) ) is separable, uniformly convex,
reflexive and its conjugate space is Lq(x) (Ω) where q(x) is the conjugate function of
p(x), i.e.,
1
1
+
= 1,
p(x) q(x)
for all x ∈ Ω. For u ∈ Lp(x) (Ω) and v ∈ Lq(x) (Ω), we have
Z
1
1 uvdx ≤ − + − |u|p(x) |v|q(x) ≤ 2|u|p(x) |v|q(x) .
p
q
Ω
The Sobolev space with variable exponent W k,p(x) (Ω) is defined as
W k,p(x) (Ω) = {u ∈ Lp(x) (Ω) : Dα u ∈ Lp(x) (Ω), |α| ≤ k},
|α|
where Dα u = ∂xα1 ∂x∂α2 ...∂xαN u, with α = (α1 , . . . , αN ) is a multi-index and |α| =
1
2
N
PN
k,p(x)
(Ω) equipped with the norm
i=1 αi . The space W
X
kukk,p(x) =
|Dα u|p(x) ,
|α|≤k
EJDE-2015/158
p(x)-BIHARMONIC PROBLEM
3
also becomes a separable and reflexive Banach space. For more details, we refer the
reader to [3, 6, 10, 13]. Denote
(
N p(x)
if kp(x) < N,
p∗k (x) = N −kp(x)
+∞
if kp(x) ≥ N
for any x ∈ Ω, k ≥ 1.
Proposition 2.2 ([6]). For p, r ∈ C+ (Ω) such that r(x) ≤ p∗k (x) for all x ∈ Ω,
there is a continuous embedding
W k,p(x) (Ω) ,→ Lr(x) (Ω).
If we replace ≤ with <, the embedding is compact.
k,p(x)
We denote by W0
(Ω) the closure of C0∞ (Ω) in W k,p(x) (Ω). Note that the
weak solutions of problem (1.1) are considered in the generalized Sobolev space
1,p(x)
X = W 2,p(x) (Ω) ∩ W0
(Ω)
equipped with the norm
Z o
n
u(x) p(x) ∆u(x) p(x)
+ λ
dx ≤ 1 .
kuk = inf µ > 0 :
µ
µ
Ω
Remark 2.3. According to [14], the norm k · k2,p(x) is equivalent to the norm
|∆ · |p(x) in the space X. Consequently, the norms k · k2,p(x) , k · k and |∆ · |p(x) are
equivalent.
R
Proposition 2.4 ([1]). If we denote ρ(u) = Ω (|∆u|p(x) + |u|p(x) )dx, then for
u, un ∈ X, we have
(1) kuk < 1 (respectively=1; > 1) ⇐⇒ ρ(u) < 1 (respectively = 1; > 1);
+
−
(2) kuk ≤ 1 ⇒ kukp ≤ ρ(u) ≤ kukp ;
−
+
(3) kuk ≥ 1 ⇒ kukp ≤ ρ(u) ≤ kukp ;
(4) kuk → 0 (respectively → ∞) ⇐⇒ ρ(u) → 0 (respectively → ∞).
It is clear that the energy functional associated with (1.1) is defined by
Z
Z
Z
1
1
1
p(x)
p(x)
q(x)
(|∆u|
+ |u|
)dx − λ
|u|
dx − µ
|u|γ(x) dx.
Iλ,µ (u) =
p(x)
q(x)
γ(x)
Ω
Ω
Ω
Let us define the functional
Z
J(u) =
Ω
1
(|∆u|p(x) + |u|p(x) )dx.
p(x)
It is well known that J is well defined, even and C 1 in X. Moreover, the operator
L = J 0 : X → X ∗ defined as
Z
hL(u), vi = (|∆u|p(x)−2 ∆u∆v + |u|p(x)−2 uv)dx
Ω
for all u, v ∈ X satisfies the following assertions.
Proposition 2.5 ([1]).
(1) L is continuous, bounded and strictly monotone.
(2) L is a mapping of (S+ ) type, namely: un * u, and lim supn→+∞ L(un )(un −
u) ≤ 0 implies un → u.
(3) L is a homeomorphism.
4
G. A. AFROUZI, M. MIRZAPOUR, N. T. CHUNG
EJDE-2015/158
3. Main results and proofs
In this section, we study the existence and non-existence of weak solutions for
problem (1.1). We use the letter ci in order to denote a positive constant.
Theorem 3.1. Assume that q(x), γ(x) ∈ C+ (Ω) and p+ < q − ≤ q(x) < p∗2 (x),
γ + < p− for any x ∈ Ω. Then we have
(i) For every λ > 0, µ ∈ R, (1.1) has a sequence of weak solutions (±uk ) such
that Iλ,µ (±uk ) → +∞ as k → +∞.
(ii) For every µ > 0, λ ∈ R, (1.1) has a sequence of weak solutions (±vk ) such
that Iλ,µ (±vk ) < 0 and Iλ,µ (±vk ) → 0 as k → +∞.
(iii) For every λ < 0, µ < 0, (1.1) has no nontrivial weak solution.
We will use the following Fountain theorem to prove (i) and the Dual of the
Fountain theorem to prove (ii).
Lemma 3.2 ([15]). Let X be a reflexive and separable Banach space, then there
exist {ej } ⊂ X and {e∗j } ⊂ X ∗ such that
X ∗ = span{e∗j : j = 1, 2, . . . },
X = span{ej : j = 1, 2, . . . },
and
(
hei , e∗j i =
1
0
if i = j,
if i =
6 j,
We define
Xj = span{ej },
Yk = ⊕kj=1 Xj ,
Zk = ⊕∞
j=k Xj .
(3.1)
Then we have the following Lemma.
Lemma 3.3 ([1]). If q(x), γ(x) ∈ C+ (Ω), q(x) < p∗2 (x), and γ(x) < p∗2 (x) for all
x ∈ Ω, denote
βk = sup{|u|q(x) ; kuk = 1, u ∈ Zk }
θk = sup{|u|γ(x) ; kuk = 1, u ∈ Zk },
then limk→∞ βk = 0, limk→∞ θk = 0.
Lemma 3.4 (Fountain Theorem [12]). Let
(A1) I ∈ C 1 (X, R) be an even functional, where (X, k · k) is a separable and
reflexive Banach space, the subspaces Xk , Yk and Zk are defined by (3.1).
If for each k ∈ N, there exist ρk > rk > 0 such that
(A2) inf{I(u) : u ∈ Zk , kuk = rk } → +∞ as k → +∞.
(A3) max{I(u) : u ∈ Yk , kuk = ρk } ≤ 0.
(A4) I satisfies the (PS) condition for every c > 0.
Then I has an unbounded sequence of critical points.
Lemma 3.5 (Dual Fountain Theorem [12]). Assume (A1) is satisfied and there is
k0 > 0 such that, for each k ≥ k0 , there exist ρk > rk > 0 such that
(B1) ak = inf{I(u) : u ∈ Zk , kuk = ρk } ≥ 0.
(B2) bk = max{I(u) : u ∈ Yk , kuk = rk } < 0.
(B3) dk = inf{I(u) : u ∈ Zk , kuk ≤ ρk } → 0 as k → +∞.
(B4) I satisfies the (P S)∗c condition for every c ∈ [dk0 , 0).
Then I has a sequence of negative critical values converging to 0.
EJDE-2015/158
p(x)-BIHARMONIC PROBLEM
5
Definition 3.6. We say that Iλ,µ satisfies the (P S)∗c condition (with respect to
(Yn )), if any sequence {unj } ⊂ X such that nj → +∞, unj ∈ Ynj , Iλ,µ (unj ) → c
and (Iλ,µ |Ynj )0 (unj ) → 0, contains a subsequence converging to a critical point of
Iλ,µ .
Proof of Theorem 3.1. (i) First we verify Iλ,µ satisfies the (PS) condition. Suppose that (un ) ⊂ X is (PS) sequence, i.e.,
|Iλ,µ (un )| ≤ c9 ,
0
Iλ,µ
(un ) → 0
as n → ∞.
By Propositions 2.2 and 2.1, we know that if we denote
Z
Z
1
1
φ(u) = −λ
|u|q(x) , dx, ψ(u) = −µ
|u|γ(x) , dx,
Ω q(x)
Ω γ(x)
then they are both weakly continuous and their derivative operators are compact.
0
By Proposition 2.5, we deduce that Iλ,µ
= L + φ0 + ψ 0 is also of type (S+ ). Thus it
is sufficient to verify that (un ) is bounded. Assume kun k > 1 for convenience. For
n large enough, we have
c9 + 1 + kun k
1 0
(un ), un i
≥ Iλ,µ (un ) − − hIλ,µ
q
Z
Z
i
hZ
1
1
1
(|∆un |p(x) + |un |p(x) )dx − λ
|un |q(x) dx − µ
|un |γ(x) dx
=
q(x)
Ω γ(x)
Ω p(x)
Z
ZΩ
Z
i
1 h
− −
(|∆un |p(x) + |un |p(x) )dx − λ
|un |q(x) dx − µ
|un |γ(x) dx
q
Ω
Ω
Ω
1
1 p−
γ+
≥ + − − ||un || − c10 kun k .
p
q
(3.2)
Since q − > p+ and p− > γ + , we know that {un } is bounded in X. In the following
we will prove that if k is large enough, then there exist ρk > rk > 0 such that (A2)
and (A3) hold.
(A2) For any u ∈ Zk , kuk = rk > 1 (rk will be specified below), we have
Z
Z
Z
1
1
1
Iλ,µ (u) =
(|∆u|p(x) + |u|p(x) )dx − λ
|u|q(x) dx − µ
|u|γ(x) dx
Ω p(x)
Ω q(x)
Ω γ(x)
Z
+
1
λ
c11 |µ|
p−
≥ + kuk − −
|u|q(x) dx −
kukγ .
−
p
q
γ
Ω
+
−
|µ|
γ
≤ 2p1+ kukp
Since p− > γ + , there exists r0 > 0 large enough such that c11
γ − kuk
R
−
as r = kuk ≥ r0 . If |u|q(x) ≤ 1 then Ω |u|q(x) dx ≤ |u|qq(x) ≤ 1. However, if
R
+
+
|u|q(x) > 1 then Ω |u|q(x) dx ≤ |u|qq(x) ≤ (βk kuk)q . So, we conclude that
(
1
p−
12
− λc
if |u|q(x) ≤ 1,
+ kuk
q−
Iλ,µ (u) ≥ 2p1
−
λ
p
q+
− q− (βk kuk)
if |u|q(x) > 1.
2p+ kuk
≥
−
+
1
λ
kukp − − (βk kuk)q − c13 ,
p+
q
6
choose rk =
G. A. AFROUZI, M. MIRZAPOUR, N. T. CHUNG
+
2λ + q
q − q βk
EJDE-2015/158
1
p− −q +
, we have
1 1
1 p−
Iλ,µ (u) =
−
r − c13 → ∞
2 p+
q+ k
ask → ∞,
because of p+ < q − ≤ q + and βk → 0.
(A3) Let u ∈ Yk such that kuk = ρk > rk > 1. Then
Z
Z
Z
1
1
1
(|∆u|p(x) + |u|p(x) )dx − λ
|u|q(x) dx − µ
|u|γ(x) dx
Iλ,µ (u) =
Ω q(x)
Ω γ(x)
Ω p(x)
Z
Z
1
λ
|µ|
p+
q(x)
≤ − kuk − +
|u|
|u|γ(x) dx.
dx + −
p
q
γ
Ω
Ω
Since dim Yk < ∞, all norms are equivalent in Yk , we obtain
Iλ,µ (u) ≤
+
−
+
λ
|µ|
1
kukp − + kukq + − kukγ .
−
p
q
γ
We get that: Iλ,µ (u) → −∞ as kuk → +∞ since q − > p+ and γ + < p− . So (A2)
holds. From the proof of (A2) and (A3), we can choose ρk > rk > 0. Obviously
Iλ,µ is even and the proof of (i) is complete.
(ii) We use the Dual Fountain theorem to prove conclusion (ii). Now we prove
that there exist ρk > rk > 0 such that if k is large enough (B1), (B2) and (B3) are
satisfied.
(B1) For any u ∈ Zk we have
Z
Z
Z
1
1
1
Iλ,µ (u) =
(|∆u|p(x) + |u|p(x) )d − λ
|u|q(x) dx − µ
|u|γ(x) dx
Ω p(x)
Ω q(x)
Ω γ(x)
Z
1
c14 |λ|
µ
p+
q−
≥ + kuk − − kuk − −
|u|γ(x) dx.
p
q
γ
Ω
−
|λ|
q
Since q − > p+ , there exists ρ0 > 0 small enough such that c14
≤
q − kuk
as 0 < ρ = kuk ≤ ρ0 . Then from the proof above, we have
(
1
p+
15
− µc
if |u|γ(x) ≤ 1,
+ kuk
γ−
Iλ,µ (u) ≥ 2p1
+
µ
p
γ+
− γ − (θk kuk)
if |u|γ(x) > 1.
2p+ kuk
Choose ρk =
γ
2p+ µθk
γ−
+
1
p+ −γ +
1
p+
2p+ kuk
(3.3)
, then
Iλ,µ (u) =
+
+
1
1
(ρk )p − + (ρk )p = 0.
+
2p
2p
Since p− > γ + , θk → 0, we know ρk → 0 as k → ∞.
(B2) For u ∈ Yk with kuk ≤ 1, we have
Z
Z
Z
1
1
1
p(x)
p(x)
q(x)
Iλ,µ (u) =
(|∆u|
+ |u|
)dx − λ
|u|
dx − µ
|u|γ(x) dx
p(x)
q(x)
γ(x)
Ω
Ω
Ω
Z
Z
−
1
|λ|
µ
≤ − kukp + −
|u|q(x) dx − +
|u|γ(x) dx.
p
q
γ
Ω
Ω
Since dimYk = k, conditions γ + < p− and p+ < q − imply that there exists a
rk ∈ (0, ρk ) such that Iλ,µ (un ) < 0 when kuk = rk . So we obtain
max
u∈Yk ,kuk=rk
Iλ,µ (u) < 0,
EJDE-2015/158
p(x)-BIHARMONIC PROBLEM
7
i.e., (B2) is satisfied.
(B3) Because Yk ∩ Zk 6= ∅ and rk < ρk , we have
dk =
inf
u∈Zk ,kuk≤ρk
Iλ,µ (u) ≤ bk =
max
u∈Yk ,kuk=rk
Iλ,µ (u) < 0.
From (3.3) , for u ∈ Zk , kuk ≤ ρk small enough we can write
Iλ,µ (u) ≥
+
+
+
λ +
λ +
1
kukp − − θkγ kukγ ≥ − − θkγ kukγ ,
2p+
γ
γ
Since θk → 0 and ρk → 0 as k → ∞, (B3) holds. Finally we verify the (P S)∗c
condition. Suppose {unj } ⊂ X such that
nj → +∞,
unj ∈ Ynj ,
Iλ,µ (unj ) → c16 ,
(Iλ,µ |Ynj )0 (unj ) → 0.
If λ ≥ 0, similar to (3.2), we can get the boundedness of kunj k. Assume kunj k ≥ 1
for convenience. If λ < 0, for n > 0 large enough, we have
1 0
c16 + 1 + kunj k ≥ Iλ,µ (unj ) − + hIλ,µ
(unj ), unj i
q
Z
Z
h
1
1
p(x)
p(x)
=
(|∆unj |
|unj |q(x) dx
+ |unj |
)dx − λ
p(x)
q(x)
Ω
Ω
Z
Z
i
1 h
1
|unj |γ(x) dx − +
(|∆unj |p(x) + |unj |p(x) )dx
−µ
γ(x)
q
Ω
ZΩ
Z
i
q(x)
γ(x)
−λ
|unj |
dx − µ
|unj |
dx
Ω
Ω
1
−
+
1 ≥ + − + kunj kp − c17 kunj kγ .
p
q
Since p− > γ + and q + > p+ , we know that {unj } is bounded in X. Hence there
exists u ∈ X such that unj → u in x. Observe now that X = ∪nj Ynj , then we can
find vnj ∈ Ynj such that vnj → u. We have
0
0
0
hIλ,µ
(unj ), unj − ui = hIλ,µ
(unj ), unj − vnj i + hIλ,µ
(unj ), vnj − ui.
Having in mind that (unj − vnj ) ∈ Ynj , it yields
0
0
hIλ,µ
(unj ), unj − ui = h(Iλ,µ |Ynj )0 (unj ), unj − vnj i + hIλ,µ
(unj ), vnj − ui → 0 (3.4)
0
as n → ∞. By Proposition 2.5, the operator Iλ,µ
is obviously of (S+ ) type. Using
0
0
this fact with (3.4), we deduce that unj → u in X, furthermore Iλ,µ
(unj ) → Iλ,µ
(u).
We claim now that u is in fact a critical point of Iλ,µ . Taking ωk ∈ Yk , notice
that when nj ≥ k we have
0
0
0
0
hIλ,µ
(u), ωk i = hIλ,µ
(u) − Iλ,µ
(unj ), ωk i + hIλ,µ
(unj ), ωk i
D
E
0
0
= hIλ,µ (u) − Iλ,µ (unj ), ωk i + (Iλ,µ |Ynj )0 (unj ), ωk .
Going to the limit on the right side of the above equation reaches
0
hIλ,µ
(u), ωk i = 0,
∀ωk ∈ Yk ,
0
so Iλ,µ
(u) = 0, this show that Iλ,µ satisfies the (P S)∗c condition for every c ∈ R.
8
G. A. AFROUZI, M. MIRZAPOUR, N. T. CHUNG
EJDE-2015/158
(iii) Assume for the sake of contradiction, u ∈ X\{0} is a weak solution of
problem (1.1). Then multiplying the equation in (1.1) by u, integrating by parts
we obtain
Z
Z
Z
p(x)
p(x)
q(x)
(|∆u|
+ |u|
)dx = λ
|u|
+µ
|u|γ(x) .
Ω
Ω
Ω
This leads to contradiction and the proof is complete.
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Ghasem A. Afrouzi
Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran
E-mail address: afrouzi@umz.ac.ir
Maryam Mirzapour
Department of Mathematics, Faculty of Mathematical Sciences, University of Mazandaran, Babolsar, Iran
E-mail address: mirzapour@stu.umz.ac.ir
Nguyen Thanh Chung
Department of Mathematics, Quang Binh University, 312 Ly Thuong Kiet, Dong Hoi,
Quang Binh, Vietnam
E-mail address: ntchung82@yahoo.com