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 orig
Trang 1Volume 2009, Article ID 563767, 17 pages
doi:10.1155/2009/563767
Research Article
Caisheng Chen,1 Lanfang Shi,1, 2and Hui Wang1, 3
1 Department of Mathematics, Hohai University, Nanjing 210098, Jiangsu, China
2 College of Mathematics and Physics, Nanjing University of Information Science and Technology, Nanjing 210044, Jiangsu, China
3 Department of Mathematics, Ili Normal University, Yining 835000, Xinjiang, China
Correspondence should be addressed to Caisheng Chen,cshengchen@hhu.edu.cn
Received 4 April 2009; Revised 9 July 2009; Accepted 24 July 2009
Recommended by Zhitao Zhang
We study the long-time behavior of solution for the m-Laplacian equation u t − div|∇u| m−2 ∇u
λ |u| m−2 u fx, u gx in R N × R, in which the nonlinear term f x, u is a function like fx, u
−hx|u| q−2 u with h x ≥ 0, 2 ≤ q < m, or fx, u ax|u| α−2 u − hx|u| β−2 u with a x ≥ hx ≥ 0 and α > β ≥ m We prove the existence of a global L2R N , L p R N -attractor for any p > m.
Copyrightq 2009 Caisheng Chen et al 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
In this paper we are interested in the existence of a globalL2R N , L p R N-attractor for the
m-Laplacian equation
with initial data condition
where the m-Laplacian operatorΔm u div|∇u| m−2∇u, 2 ≤ m < N, λ > 0.
For the case m 2, the existence of global L2R N , L2R N-attractor for 1.1-1.2 is
2 studied the existence of global L2R N , L m∗R N-attractor for 1.1-1.2 with m∗
mN/ N−m Yang et al in 3 investigated the global L2R N , L p R N ∩W 1,m R N-attractor
Trang 2Ap under the assumptions fx, uu ≥ a1|u| p − a2|u| m − a3x and f u x, u ≥ a4x with the constants a1, a2 > 0 and the functions a3, a4 ∈ L1R N ∩ L∞R N We note that the global
the existence of globalL2R N , L p R N-attractor for 1.1-1.2, which the term λ|u| m−2u is
For a typical example is fx, u ax|u| α−2u − hx|u| β−2u with a x ≥ hx ≥ 0, α > β ≥ 2,
h x ∈ L2R N ∩ L∞R N In 4, we assume that fx, u satisfies
u
0
f
x, η
dη Lx|u| ≤ k2
with some k2> 0 and L x ∈ L2R N ∩ L∞R N
satisfy the assumption1.3
In this paper, motivated by2 4, we are interested in the global L2R N , L p R N
order of polynomial for u on fx, u.
of solution of1.1-1.2 in L p R N , p ≥ 2, we derive L∞ estimate of solutions by Moser’s technique as in4,6,7 We will prove that the existence of the global attractor Ap in L p R N under weaker conditions
some lemmas for the solution of1.1-1.2 By the a priori estimates inSection 2, the existence
of globalL2R N , L p R N-attractor for 1.1-1.2 is established inSection 3
2 Preliminaries
We denote by L p and W 1,m the space L p R N and W 1,m R N, and the relevant norms by · p
and · 1,m , respectively It is well known that W 1,m R N W 1,m
0 R N In general, · Edenotes
the norm of the Banach space E.
For the proof of our results, we will use the following lemmas
Lemma 2.1 8 10 Gagliardo-Nirenberg Let β ≥ 0, 1 ≤ r ≤ q ≤ m1 βN/N − m when
N > m and 1 ≤ r ≤ q ≤ ∞ when N ≤ m Suppose u ∈ L r and |u| β u ∈ W 1,m Then there exists C0 such that
u q ≤ C 1/β10 u 1−θ
r ∇|u| β uθ/ β1
with θ 1 βr−1− q−1/N−1− m−1 1 βr−1, where C0is a constant independent of q, r, β, and θ if N / m and a constant depending on q/1 β if N m.
Lemma 2.2 7 Let yt be a nonnegative differentiable function on 0, T satisfying
Trang 3with A, θ > 0, λθ ≥ 1, B, C ≥ 0, k ≤ 1, and 0 ≤ δ < 1 Then one has
y t ≤ A −1/θ
2λ 2BT1−k1/θ
t −λ 2Cλ BT1−k−1
Lemma 2.3 11 Let yt be a nonnegative differential function on 0, ∞ satisfying
with A, μ > 0, B ≥ 0 Then one has
y t ≤BA−11/1μ
Aμt−1/μ
, t > 0. 2.5
First, the following assumptions are listed.
A 1 Let fx, u ∈ C1R N1, fx, 0 0 and there exist the nontrivial nonnegative functions
h x ∈ L q1∩ L∞and h1 x ∈ L1, such that F x, u ≤ k1fx, uu and
f x, u − fx, vu − v ≥ −k2
1 |u| q−2 |v| q−2
where F x, u u
0f x, sds, 2 ≤ q < m, q1 m/m − q and some constants k1, k2≥ 0.
A 2 Let fx, u ∈ C1R N1, fx, 0 0 and there exists the nontrivial nonnegative function
h1 x ∈ L1, such that F x, u ≤ k1fx, uu and
a1 |u| α − a2|u| m ≤ fx, uu ≤ b1|u| α b2|u| m h1x,
f x, u − fx, vu − v ≥ −k4
1 |u| α−2 |v| α−2
where a2< λ, m < α < m 2m/N, and a1, b1, b2> 0, k1, k2≥ 0.
A typical example is f x, u ax|u| α−2u −hx|u| β−2u with a x, hx ≥ 0, and α > β ≥ m.
The assumptionA 2 is similar to [ 3 , 1.3–1.7].
Remark 2.4 If f x, u −hx|u| q−2u, q > m, the problem1.1-1.2 has no nontrivial solution
for some hx ≥ 0, see 12
We first establish the following theorem
Theorem 2.5 Let g ∈ L m ∩ L∞and u0 ∈ L2 IfA 1 holds, then the problem 1.1-1.2 admits a
unique solution u t satisfying
u t ∈ X ≡ C0, ∞, L2
∩ L m
loc
0, ∞, W 1,m
∩ L∞ loc
0, ∞, L2
,
u t ∈ L m
loc
0, ∞, W −1,m
,
2.9
Trang 4and the following estimates:
ut 2
2≤ C0
gm
m h q1
q1
t u0 2
m λ ut m
m ≤ C0
gm
m h q1
q1 h1 1 t−1 u0 2
2, t > 0, 2.11
t
s
u t τ 2
2dτ ≤ C0
gm
m h q1
q1 h1 1 s−1 u0 2
with m m/m − 1 The constant C0 depends only on m, N, q, λ, and C1depends on h, g, u0, and T.
Proof For any T > 0, the existence and uniqueness of solution u t for 1.1-1.2 in the class
XT ≡ C0, T, L2
∩ L m
0, T, W 1,m
∩ L∞
0, T, L2
2.14
In the following, we will derive the estimates2.10–2.13 The solution is in fact given
estimates that the solutions under consideration are appropriately smooth We begin with the estimate of ut 2
1 2
d
dt ut 2
2 ∇ut m
m λ ut m
m
R N
Since
−
R N
f x, ututdx ≤
R N
h x|ut| q dx ≤ λ0 ut m
m C0 h q1
q1,
R N
g xutdx ≤ λ0 ut m
m C0gm
m
2.16
1 2
d
dt ut 2
2 ∇ut m
m 2λ0 ut m
m ≤ C0
gm
m h q1
q1
Integrating2.17 with respect to t, we obtain
1
2 ut 2
2
t
0
m 2λ0 uτ m
m
dτ ≤ C0
gm
m h q1
q1
t1
2 u0 2
Trang 5This implies2.10 and the existence of t∗∈ 0, t such that
∇ut∗ m
m 2λ0 ut∗ m
m ≤ C0
gm
m h q1
q1
t−1 u0 2
2, t > 0. 2.19
On the other hand, multiplying1.1 by u tand integrating ons, t × R N, we get
t
s
u t τ 2
2dτ 1
m λ
m
R N
F x, ut − gxutdx
m λ
m
R N
F x, us − gxusdx.
2.20
By2.6, we have Fx, u ≥ −hx|u| qand
−
R N
F x, utdx ≤
R N
h x|ut| q dx ≤ ε ut m
m C0 h q1
R N
g xut dx ≤ ε ut m
m C0gm
m ,
R N
g xuS dx ≤ us m
mgm
m ,
R N
F x, usdx ≤ k1
R N
h x|us| q h1xdx
≤ C0
us m
m h q1
q1 h1 1.
2.22
t
s
u t τ 2
2dτ 1
m λ
m ≤ C0
m M1
where
M1gm
m h q1
Further, we let s t∗in2.23 and obtain from 2.19 that
m λ ut m
m ≤ C0
M1 t−1 u0 2
2
, t > 0,
t
s
u t τ 2
2dτ ≤ C0
M1 s−1 u0 2
2
, 0 < s < t.
2.25
Trang 6Thus, the solution ut satisfies 2.10–2.12 We now derive 2.13 by Moser’s technique as in 5,6 In the sequel, we will write u p instead of |u| p−1u when p ≥ 1 Also,
Multiplying1.1 by |u| p−2u, p ≥ 2, we get
1
p
d
dt ut p
p C1p1−m∇u pm−2/mm
m λ ut p m−2
p m−2≤
R N
g x − fx, u|u| p−2u dx.
2.26
It follows from Young’s inequality that
R N
g x |u| p−1dx ≤ λ0 u p m−2
p m−2 λ 1−p/m−10 gα p
α p
−
R N
f x, u|u| p−2u dx ≤ λ0 u p m−2
p m−2 λ 2−p−q/m−q0 h β p
β p
2.27
1
p
d
dt ut p
p C1p1−m∇u pm−2/mm
m 2λ0 ut p m−2
p m−2
≤ λ 1−p/m−10 gα p
α p λ 2−p−q/m−q0 h β p
β p
2.28
Let R > m/2, p1 2, p n Rp n−1− m − 2, n 2, 3, Then, byLemma 2.1, we see
∇u p n m−2/mm
m ≥ C −m/θ n
0 u p n m−21−θ−1
n
p n−1 u p n m−2θ−1
n
where
θ n p n m − 2
m
1
p n−1− 1
p n
1
N− 1
mp n−1
−1
1− p n−1p n−1
Inserting2.29 into 2.28 p p n, we find
d
dt ut p n
p n C1C−m/θ n
0 p2−mn u p n r n
p n u m −2−r n
where r n p n m − 2θ−1
n − p nand
A n λ 2−p n −q/m−q
0 h μ n
μ n λ 1−p n /m−1
0 gλ n
with λ n p n m − 2/m − 1, μ n p n m − 2/m − q, n 1, 2,
Trang 7We claim that there exist the bounded sequences{ξ n } and {s n} such that
Indeed, by2.10, this holds for n 1 if we take s1 0, ξ1 M1T 1/2 u0 2 If2.33 is
true for n − 1, then we have from 2.31 that
where yt ut p n
p n , τ n s n p nand
θ r n p−1n , s n 1 s n−1r n − m 2r−1
n , A C1C −m/θ n
0 p n2−mξ m −2−r n
ξ n ξ n−1
C−11 C m/θ n
0 p n m−1s−1n 1/r n
2A n s−1n 1/p n
for n 2, 3,
bounded, see6 Then, 2.13 follows from 2.33 as n → ∞.
We now consider the uniqueness and continuity of the solution for1.1-1.2 in L2 Let
u1, u2be two solutions of1.1-1.2, which satisfy 2.10–2.13 Denote ut u1t − u2t Then ut solves
u t− Δm u1− Δm u2 λ|u1|m−2u1 − |u2|m−2u2
Multiplying2.37 by u, we get from 2.7 and 2.13 that
1
2
d
dt ut 2
2 γ0 ∇ut m
m γ1 ut m
m ≤ k2
R N
1 |u1|q−2 |u2|q−2
u2dx
≤ k2
R N
1 u1t q−2
∞ u2t q−2
∞
u2dx ≤ C0
1 t −s0q−2
ut 2 2
2.38
with some γ0, γ1 > 0 Since s0q − 2 < 1 and u0 0, 2.38 implies that ut 2 ≡ 0 in 0, T and u1t u2t in 0, T.
Further, let t > s≥ 0 Note that
ut − us 2
2
R N
s
u t τdτ
2
dx≤
t
s
u t τ 2
Theorem 2.5is completed
Trang 8Remark 2.6 By2.23, we know that if u0∈ W 1,m, then
t
0
u t τ 2
2dτ 1
m λ
m ≤ C0 u0 m
Theorem 2.7 Assume A1 and g ∈ L m ∩ L∞ Suppose also u0x ∈ W 1,m Then, the unique solution u t in Theorem 2.5 also satisfies
u t ∈ Y ≡ L∞
0, ∞, W 1,m
, u t ∈ L2
0, ∞, L2
and the estimate2.40.
Now consider the assumptionA 2 Since m < α < m 2m/N, one has s0α − 2 Nα −
2/2m m − 2N < 1 By a similar argument in the proof ofTheorem 2.5 , one can establish the following theorem.
Theorem 2.8 Assume A2 and g ∈ L m ∩ L∞, u0 ∈ L2 Then the problem 1.1-1.2 admits a
unique solution u t which satisfies
u t ∈ X ≡ C0, ∞, L2
∩ L m
loc
0, ∞, W 1,m
∩ L∞ loc
0, ∞, L2
,
u t ∈ L m
loc
0, ∞, W −1,m
,
2.42
and the following estimates:
ut 2
2≤ C0tgm
m u0 2
2, t ≥ 0,
m λ ut m
m ut α
α ≤ C0
gm
m h1 1 t−1 u0 2
2, t > 0,
t
s
u t τ 2
2dτ ≤ C0
gm
m h1 1 s−1 u0 2
2, 0 < s ≤ t, ut ∞≤ C1t−s0, s0 N2m m − 2N−1, 0 < t ≤ T.
2.43
Further, if u0∈ W 1,m , the unique solution u t∈ Y satisfies
t
0
u t τ 2
m ut m
m ut α
α ≤ C0
u0 m
1,m h1 1gm
m
where C0depends only on m, N, λ, α, and C1on the given data g, h1, u0, and T > 0.
properties:
Trang 91 St : L2 → L2for t ≥ 0, and S0u0 u0for u0 ∈ L2 or St : W 1,m → W 1,mfor
t ≥ 0, and S0u0 u0for u0∈ W 1,m;
2 St s StSs for t, s ≥ 0;
Lemma 2.9 Suppose A1 (or A2) and g ∈ L m ∩ L∞ LetB0be a bounded subset of L2 Then, there exists T0 T0B0 such that StB0⊂ D for every t ≥ T0, where
m ≤ M1
2.45
with M1 h q1
q1 h1 1 g m
m ifA 1 holds, and M1 h1 1 g m
m ifA 2 holds.
Now it is a position of Theorem 2.5 to establish some continuity of S t with respect to the
initial data u0, which will be needed in the proof for the existence of attractor.
Lemma 2.10 Assume that all the assumptions in Theorem 2.5 are satisfied Let S tφ n and S tφ
be the solutions of problem1.1-1.2 with the initial data φ n and φ, respectively If φ n → φ in
0, T as n → ∞.
Proof Let u n t Stφ n , ut Stφ, n 1, 2, Then, w n t u n t − ut solves
w nt− Δm u n− Δm u λ|u n|m−2u n − |u| m−2u
and w n x, 0 φ n x − φx.
Multiplying2.46 by |w n|p−2w n, we get from8, Chapter 1, Lemma 4.4 and 2.13 that
1
p
d
dt w n t p
p γ0
R N
|∇w n|m |w n|p−2dx λ w n t p m−2
p m−2
≤ k2
R N
1 |u| q−2t |u n|q−2t|w n t| p
p dx
≤ C0
1 u n t q−2
∞ ut q−2
∞
w n t p p
≤ C0
1 t −s0q−2 w n t p
2.47
w n t p ≤ w n0 pexp
C0
T1− s0q − 2−1T1−s0q−2
φ n − φ
pexp
C0
T1− s0
q− 2−1T1−s0q−2, 0≤ t ≤ T, 2.48
Trang 10with s0q − 2 Nq − 2m − 2N 2m−1 < 1 Letting n → ∞, we obtain the desired result
Lemma 2.11 Suppose that all the assumptions in Theorem 2.5 are satisfied Let u t be the solution
of 1.1-1.2 with u0 ∈ L2, u0 2 ≤ M0 Then,∃T0> 0, such that for any p > m, one has
where α0 m − 2 m2/N /p − m and A p , B p > 0, which depend only on p, N, m and the given data g α p , h β p , M0with α p p m − 2/m − 1, β p p m − 2/m − q.
Proof Multiplying1.1 by |u| p−2u, we have
1
p
d
dt ut p p γ p∇
|u| p−2/m um
m λ u p m−2
p m−2≤
R N
g x − fx, uu |u| p−2dx 2.50
with γ p m m p − 1m p − 2 −m Note that
R N
g x|u| p−2u dx ≤ ε u p m−2
p m−2 C pgα p
α p ,
−
R N
f x, uu|u| p−2dx≤
R N
h x|u| p q−2 dx ≤ ε u p m−2
p m−2 C p h β p
β p
2.51
with 0 < ε < λ/4 Then2.50 becomes
1
p
d
dt ut p
p γ p∇
|u| p−2/m um
mλ
2 u p m−2
p m−2 ≤ C p
h β p
β pgα p
α p
ByLemma 2.1, we get
∇|ut| τ u tm
m ≥ C0 ut m 1τ/θ1
with
τ p− 2
p
1
N τ
m
−1
, τ1 m1− θ−1
1
ByLemma 2.9,∃T0> 0, such that t ≥ T0, ut m ≤ M1 Therefore, we have from2.52 and2.53 that
1
p
d
dt ut p
p C0M τ1
1 ut p 1α0
h β p
β pgα p
α p
, t > T0 2.55
Trang 11p 1 α0 m 1 τ
p − m > 0. 2.56
ut p p≤AM −τ1
1 C−10 1/1α0
C0M τ1
1α0 t − T0−1/α0, t > T0. 2.57
ByLemma 2.11, we now establish
Lemma 2.12 Assume that all the assumptions in Theorem 2.5 are satisfied LetB0be a bounded set in
L2and u t be a solution of 1.1-1.2 with u0∈ B0 Then, for any η > 0 and p > m,∃r0 r0η, B0,
T1 T1η, B0, such that r ≥ r0, t≥ T1,
where B c
r {x ∈ R N | |x| ≥ r}.
Proof We choose a suitable cut-off function for the proof Let
φ0 s
⎧
⎪
⎨
⎪
⎩
n − k−1n s − 1 k − ks − 1 n
, 1 < s < 2;
2.59
in which n> k > m will be determined later It is easy to see that φ0s ∈ C10, ∞, 0 ≤
φ0s ≤ 1, 0 ≤ φ0s ≤ β0φ1−1/k0 s for s ≥ 0, where β0 kn/n − k 1/k For every r > 0,
∇x φ r, x ≤ β1
r φ
1−1
with β1 Nβ0
Multiplying1.1 by |u| p−2uφ, p > m, we obtain
1
p
d
dt
R N
|u| p φ dx
R N
|∇u| m−2∇u∇|u| p−2uφ
dx λ 2
R N
|u| p m−2 φ dx
≤ C p
h β p
β B c
r gα p
α B c
r,
2.61
Trang 12where and in the sequel, we let f p
pΩ Ω|fx| p dx Note that D1
R N |∇u| m−2∇u∇|u| p−2uφ
dxp− 1
with
D2
R N
|∇u| m−2∇u∇φ|u| p−2u dx
≤
R N
|∇u| m−1 ∇φ |u| p−1dx
r
R N
|∇u| m−1|u| p−1φ1−1/kdx
r
R N
|∇u| m |u| p−2φ |u| p m−2 φ1−m/k
dx.
2.63
Therefore, if r ≥ 2β1/p − 1,
D1 ≥ p− 1 2
R N |∇u| m |u| p−2φ dx−β1
r
R N |u| p m−2 φ1−m/kdx. 2.64 Further, we estimate the first term of the right-hand side in2.64 Since
∂
∂x i
uφ 1/p τ
uφ 1/p
τ 1|u| τ φ τ/p φ 1/p ∂u
∂x i u
p
∂φ
∂x i
φ 1/p−1
, i 1, 2, , N,
∇ uφ 1/p τ
uφ 1/p 2
τ 12|u| 2τ φ 2τ/p
|∇u|2φ 2/pu2
p2 ∇φ 2φ 2/p−22u
p φ 2/p−1 ∇u∇φ ,
2.65
we have
D3 ∇ uφ 1/p τ
uφ 1/p m
∇ uφ 1/p τ
uφ 1/p 2m/2
≤ λ0
|u| τm |∇u| m φ mτ2 |u| mτ0|∇φ| m φ m τ2 −1 |u| mτ m/2|∇u| ∇φ m/2
φ mτ2−m/2
,
2.66
of2.66 is
2.662≤ β m1
r m |u| p −2m φ1m−2/p−m/k≤ C1
r |u| p −2m φ1m−2/p−m/k, r ≥ 1, 2.67
... mτ2−m/2,
2.66
of 2.66 is
2.662≤ β m1
r