Volume 2009, Article ID 191649, 12 pagesdoi:10.1155/2009/191649 Research Article Multiple Solutions for a Class of px-Laplacian Systems Yongqiang Fu and Xia Zhang Department of Mathemati
Trang 1Volume 2009, Article ID 191649, 12 pages
doi:10.1155/2009/191649
Research Article
Multiple Solutions for a Class of
px-Laplacian Systems
Yongqiang Fu and Xia Zhang
Department of Mathematics, Harbin Institute of Technology, Harbin 150001, China
Correspondence should be addressed to Xia Zhang,piecesummer1984@163.com
Received 19 November 2008; Accepted 11 February 2009
Recommended by Ondrej Dosly
We study the multiplicity of solutions for a class of Hamiltonian systems with the px-Laplacian.
Under suitable assumptions, we obtain a sequence of solutions associated with a sequence of positive energies going toward infinity
Copyrightq 2009 Y Fu and X Zhang This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited
1 Introduction and Main Results
Since the space L px and W 1,px were thoroughly studied by Kov´aˇcik and R´akosn´ık 1, variable exponent Sobolev spaces have been used in the last decades to model various phenomena In2, R ˚uˇziˇcka presented the mathematical theory for the application of variable exponent spaces in electro-rheological fluids
In recent years, the differential equations and variational problems with px-growth conditions have been studied extensively; see for example3 6 In 7, De Figueiredo and Ding discussed the multiple solutions for a kind of elliptic systems on a smooth bounded
domain Motivated by their work, we will consider the following sort of px-Laplacian
systems with “concave and convex nonlinearity”:
−div∇upx−2 ∇u |u| px−2 u H u x, u, v, x ∈ Ω,
−div∇vpx−2 ∇v |v| px−2 v −H v x, u, v, x ∈ Ω,
1.1
whereΩ ⊂ RN is a bounded domain, p is continuous on Ω and satisfies 1 < p−≤ px ≤ p <
Trang 2of Hamiltonians H such that
where 1 < α− ≤ αx ≤ px, px βx p∗x Here we denote
p sup
x∈Ω
and denote by px βx the fact that inf x∈Ω βx − px > 0 Throughout this paper, Fx, u, v satisfies the following conditions:
H1 F ∈ C1Ω × R2, R Writing z u, v, Fx, 0 ≡ 0, F z x, 0 ≡ 0;
H2 there exist px < q1x p∗x, 1 < q2−≤ q2x < px such that
F u x, u, v,F v x, u, v ≤ a0
1 |u| q1x−1 |v| q2x−1
where a0is positive constant;
H3 there exist μx, νx ∈ C1Ω with px μx p∗x, 1 < ν− ≤ νx ≤ px, and R0> 0 such that
1
μx F u x, u, vu 1
when|u, v| ≥ R0.
As8, Lemma 1.1, from assumption H3, there exist b0, b1> 0 such that
|u| μx |v| νx
for anyx, u, v ∈ Ω × R2 We can also get that there exists b2> 0 such that
1
μx F u x, u, vu 1
for anyx, u, v ∈ Ω × R2 In this paper, we will prove the following result.
Theorem 1.1 Assume that hypotheses (H1)–(H3) are fulfilled If Fx, z is even in z, then problem
1.1 has a sequence of solutions {z n } such that
I
z n
Ω
∇u npxu npx
as n → ∞.
Trang 32 Preliminaries
on these spaces, we refer to1,9 11
LetPΩ be the set of all Lebesgue measurable functions p : Ω → 1, ∞ and
|u| px inf λ > 0 :
Ω
u λpx dx ≤ 1
The variable exponent space L px Ω is the class of all functions u such that Ω|ux| px
dx <
∞ Under the assumption that p < ∞, L pxΩ is a Banach space equipped with the norm
2.1
L px Ω such that |∇u| ∈ L pxΩ and it can be equipped with the norm
u 1,px |u| px |∇u| px 2.2
For u ∈ W 1,px Ω, if we define
|||u||| inf λ > 0 :
Ω
|u| px |∇u| px
then|||u||| and u 1,px are equivalent norms on W 1,px Ω.
By W01,px Ω we denote the subspace of W 1,px Ω which is the closure of C∞
0 Ω with respect to the norm2.2 and denote the dual space of W 1,px
know that ifΩ ⊂ RN is a bounded domain,||u|| 1,px and |∇u| px are equivalent norms on
W01,px Ω.
Under the condition 1 < p− ≤ p < ∞, W01,pxΩ is a separable and reflexive Banach space, then there exist{e n}∞n1 ⊂ W 1,px
0 Ω and {f m}∞m1 ⊂ W −1,p xΩ such that
f m e n 1 if n m,
0 if n / m,
W01,pxΩ spane i : i 1, , n,
,
W −1,p xΩ spanf j : j 1, , m,
.
2.4
In the following, we will denote that E E1⊕ E2, where
E1 {0} × W 1,px
0 Ω, E2 W 1,px
Trang 4For any z ∈ E, define the norm ||z|| ||u, v|| |||u||| |||v||| For any n ∈ N, set e1
n
0, e n , e2
X n spane11, , e n1
⊕ E2, X n E1⊕ spane12, , e2n
denote the complement of X n in E by X n⊥ span{e2
n1 , e2
n2 , }.
Ω
∇u
0px−2 ∇u0∇u u0px−2
u0u − ∇v0px−2 ∇v0∇v
−v0px−2
x, u0, v0
x, u0, v0
3.1
In this section, we denote that V m span{e i : i 1, , m}, for any m ∈ N, and c i is
positive constant, for any i 0, 1, 2
Lemma 3.2 Any PS sequence {z n } ⊂ E, that is, |Iz n | ≤ c and I z n → 0, as n → ∞, is bounded.
infx∈Ω 1 s/νx − 1/px > 0, l3 supx∈Ω 1/αx − 1 s/μx > 0, l4 supx∈Ω1
s/νx − 1/ βx > 0.
Let{z n } ⊂ E be such that |Iz n | ≤ c and I z n → 0, as n → ∞ We get
I
z n
−
z n
,
1 s
1 s
Ω
1
μx
∇u
npxu npx
1 su n
μx2 ∇u npx−2 ∇u n ∇μ
1 s
px
∇v npxv npx
− 1 sv n
νx2 ∇v npx−2 ∇v n ∇ν
1 s
x, u n , v n
x, u n , v n
v n − Fx, u n , v n
1 s
αx
u
nαx
1 s
βx
v
nβx
dx
≥
Ω
l1∇u npx l2∇v npx sFx, u n , v n
− l3u nαx l4v nβx
1 su n
μx2 ∇u npx−2 ∇u n ∇μ − 1 sv n
νx2 ∇v npx−2 ∇v n ∇ν − 1 sb2
dx.
3.2
Trang 5As μx, νx ∈ C1Ω, by the Young inequality, we can get that for any ε1, ε2∈ 0, 1,
1 su n
μx2 ∇u npx−2 ∇u n ∇μ
≤ c0∇u npx−1u n
≤ c0
ε1px − 1
1−px
1
px u npx
≤ c0
ε1∇u npx ε1−p
1 u npx
,
1 sv n
νx2 ∇v npx−2 ∇v n ∇ν
≤ c1
ε2∇v npx ε1−p
2 v npx
.
3.3
Let ε1, ε2be sufficiently small such that
c0ε1≤ l1
2 , c1ε2≤ l2
then
I
z n
−
z n
,
1 s
1 s
≥
Ω
l1
2 ∇u npx l2
2 ∇v npx sb0u nμx b0v nνx − b1
−l3u nαx c0ε1−p1 u npx
l4v nβx − c1ε1−p2 v npx
− 1 sb2
dx.
3.5
Note that αx ≤ px μx, px βx, by the Young inequality, for any ε3, ε4, ε5∈ 0, 1,
we get
u nαx≤ ε3αxu nμx
αx/αx−μx
3
≤ ε3u nμx ε −α/μ−α−
u npx≤ ε4px
px/px−μx
4
≤ ε4u nμx ε −p/μ−p−
v npx≤ ε5px
px/ px−βx
5
≤ ε5v nβx ε −p/ β−p−
3.6
Trang 6Let ε3, ε4, ε5 be sufficiently small such that l3ε3 c0ε1−p1 ε4 ≤ sb0 and c1ε1−p2 ε5 ≤ l4, then we
get
I
z n
−
z n
,
1 s
1 s
≥
Ω
l1
2 ∇u npx l2
2 ∇v npx − c2
Note that
I
,
1 s
1 s
≤I z n ·1 s
1νx s v n
≤ c3I
z n ·∇1 s
px
∇1νx s v n
px
≤ c4I
z n ·∇u n
px∇v n
px
,
3.8
and for n ∈ N being large enough, we have
c4I
z n ≤ min l1
4 , l2
4
It is easy to know that if|∇u n|px ≥ 1 and |∇v n|px ≥ 1,
∇u n
px ≤
Ω∇u npx
px≤
Ω∇v npx
thus we get
I
z n
≥
Ω
l1
4 ∇u npx l2
4 ∇v npx − c2
get that|∇u n|px , |∇v n|pxare bounded It is immediate to get that{z n } is bounded in E.
Lemma 3.3 Any PS sequence contains a convergent subsequence.
As E is reflexive, passing to a subsequence, still denoted by {z n }, we may assume that there
Trang 7exists z ∈ E such that z n → z weakly in E Then we can get u n → u weakly in W 1,px
0 Ω Note that
− I z,u n − u, 0
Ω
∇u
npx−2 ∇u n−∇u px−2 ∇u∇u n − u
u
npx−2
u n − |u| px−2 u
−u
nαx−2
u n − |u| αx−2 u
−F u
x, u n , v n
− F u x, u, vu n − udx.
3.12
It is easy to get that
z n
− I z,u n − u, 0−→ 0,
and u n → u in L px Ω, u n → u in L αx Ω, as n → ∞ Then
Ω
u
npx−2
u n − |u| px−2 u
Ω
u
nαx−2
u n − |u| αx−2 u
3.14
as n → ∞ By condition H2, we obtain
Ω
F u
x, u n , v n
≤
Ωa0
1u nq1x−1v nq2x−1u
n − udx
≤ a1
u
n − u
1u
nq1x−1
q1x·u n − u
q1xv
nq2x−1
q2x·u n − u
q2x
.
3.15
It is immediate to get that |u n − u|1 → 0, ||u n|q1x−1|q1x , ||v n|q2x−1|q2x are bounded and
|u n − u| q1x → 0, |u n − u| q2x → 0, then we get
ΩF u
x, u n , v n
Ω
∇u
npx−2 ∇u n−∇u px−2 ∇u∇u n − udx −→ 0,
3.16
as n → ∞ Similar to 3,4, Theorem 3.1, we divide Ω into two parts:
Trang 8OnΩ1, we have
Ω 1
∇u n − ∇upx
dx
≤ c5
Ω 1
∇u
npx−2 ∇u n−∇u px−2 ∇u
∇u n − ∇upx/ 2
×∇u
npx∇u px2−px/2
dx
≤ c6∇u
npx−2 ∇u n−∇u px−2 ∇u
∇u n − ∇upx / 2
2/ px ,Ω1
×∇u
npx∇u px2−px/2
2/2−px,Ω1
,
3.18
then
Ω 1|∇u n − ∇u| px dx → 0 On Ω2, we have
Ω 2
|∇u n − ∇u| px dx ≤ c7
Ω 2
|∇u n|px−2 ∇u n − |∇u| px−2 ∇u∇u n − ∇udx −→ 0. 3.19
Ω|∇u n − ∇u| px dx → 0 Then u n → u in W 1,px
v n → v in W 1,px
0 Ω.
Lemma 3.4 There exists R m > 0 such that Iz ≤ 0 for all z ∈ X m with ||z|| ≥ R m
Iz ≤
Ω
∇u px |u| px
px − ∇v px |v| px
dx
≤
Ω
∇u px |u| px
p− − ∇v px |v| px
dx.
3.20
In the following, we will consider
Ω|∇u| px |u| px /p−− b0|u| μx dx.
i If |||u||| ≤ 1 We have
Ω
∇u px |u| px
ii If |||u||| > 1 Note that μ, p ∈ CΩ, px μx For any x ∈ Ω, there exists Qx
which is an open subset ofΩ such that
y∈Qx
Trang 9then{Qx} x∈Ωis an open covering ofΩ As Ω is compact, we can pick a finite subcovering {Qx} n
i1forΩ Thus there exists a sequence of open set {Ω i}n
i1such thatΩ n
i1Ωiand
p i sup
x∈Ω i
x∈Ω i
for i 1, , n Denote that r i |||u|||Ωi , then we have
Ω
∇u px |u| px
dx
n
i1
Ωi
∇u px |u| px
dx
r i >1
Ωi
∇u px |u| px
dx
r i≤1
Ωi
∇u px |u| px
dx
r i >1
|||u||| p i
Ωi
p− − b0k m i |||u||| μ i−
Ωi
3.24
where k m i infu∈V m|Ωi , |||u||| Ωi1
k m i > 0, for i 1, , n.
We denote by s i the maximum of polynomial t p i /p− − b0k m i t μ i− on 0, ∞, for i
1, , n Then there exists t0> 1 such that
t p i
for t > t0and i 1, , n, where c8 n
i1 s i n/p− b1measΩ.
Let R m max{2, 2pc8 1/ p−1 /p−, 2nt0} If ||z|| ≥ R m , we get |||u||| ≥ R m /2 or
|||v||| ≥ R m /2.
i If |||u||| ≥ R m /2, |||u||| ≥ nt0 > 1 It is easy to verify that there exists at least i0such that|||u|||Ωi0 ≥ t0> 1, thus
Ωi0
p− − b0k m i0 |||u||| μ i0−
Trang 10ii If |||v||| ≥ R m /2, |||v||| ≥ pc8 1/p−1/p− We obtain
p− − |||v||| p−
Now we get the result
Lemma 3.5 There exist r m > 0 and a m → ∞ m → ∞ such that Iz ≥ a m , for any z ∈ X m−1⊥
with ||z|| r m
Let||z|| ≥ 1, we get
Iz
Ω
∇u px |u| px
αx − Fx, u, 0
dx
≥
Ω
∇u px |u| px
dx
≥
Ω
∇u px |u| px
3.29
Denote that
u∈V m⊥
|||u|||≤1
thus
Let
p−
c10pq1θ m
1/q1−p−
,
2c11pq1
1/p−
Trang 11By5, Lemma 3.3, we get that θm → 0, as m → ∞, then
Iz ≥ r m p−
a m ,
3.33
when m is sufficiently large and ||z|| r m It is easy to get that a m → ∞, as m → ∞.
Lemma 3.6 I is bounded from above on any bounded set of X m
Iz ≤
Ω
∇u px |u| px
By conditionsH2 and H3, we know that if |u, v| ≥ R0, Fx, u, v ≥ 0 and if |u, v| <
Iz ≤
Ω
∇u px |u| px
and it is easy to get the result
complete the proof
Acknowledgments
This work is supported by Science Research Foundation in Harbin Institute of Technology
HITC200702 and The Natural Science Foundation of Heilongjiang Province A2007-04
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... two parts: < /p> Trang 8OnΩ1, we have< /p>
< /p>
Ω 1 < /p>
∇u... nβ x ε ? ?p / β? ?p − < /p>
3.6 < /p> Trang 6
Let... < /p>
y∈Q x < /p> Trang 9
then{Q x } x? ??Ωis an open covering of< i>Ω As Ω is compact,