Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 615345, 27 pages
doi:10.1155/2012/615345
Research Article
On Existence, Uniform Decay Rates, and Blow-Up
for Solutions of a Nonlinear Wave Equation with
Dissipative and Source
Xiaopan Liu
Department of Mathematics, Southeast University, Nanjing 210018, China
Correspondence should be addressed to Xiaopan Liu, liuxiaopan112@126.com
Received 24 May 2012; Revised 17 July 2012; Accepted 17 July 2012
Academic Editor: Narcisa C. Apreutesei
Copyright q 2012 Xiaopan Liu. 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.
This paper studies the blow-up and existence, and asymptotic behaviors of the solution of a
nonlinear hyperbolic equation with dissipative and source terms. By using Galerkin procedure and
the perturbed energy method, the local and global existence of solution is established. In addition,
by the concave method, the blow-up of solutions can be obtained.
1. Introduction
In this paper, we investigate the following nonlinear wave equation:
|ut |ρ utt Δ2 u |ut |m ut γΔ2 ut u
∂u
0,
∂n
N
∂
σi uxi |u|p−1 u,
∂xi
i1
x, t ∈ ∂Ω × 0, T ,
ux, 0 u0 x, ut x, 0 u1 x,
x, t ∈ Ω × 0, T ,
1.1
x ∈ Ω,
where Ω is a bounded domain in Rn with smooth boundary ∂Ω, Δ is a Laplace operator,
and ∂u/∂n|∂Ω indicates derivative of u in outward normal direction of ∂Ω. In addition to, if
n > 3, 1 < p < n 2/n − 2 and if n 1, 2, then p > 1. γ ≥ 0 is a constant, σi s i 1, . . . , N
are given in A1 later.
2
Abstract and Applied Analysis
In 1968, Greenberg et al. 1 first suggested and studied the following equation:
utt − uxxt σux x .
1.2
Under the condition σ s > 0 and higher smooth conditions on σs and initial data, they
claimed the global existence of classical solutions for the initial boundary value problem of
1.2.
The multidimensional form of the following:
utt − Δut −
N
∂
ai x, t, uxi fx, t
∂xi
i1
1.3
was first studied by Clement 2, 3. Exploiting the monotone operator method, he obtained
the global existence of weak solutions for the initial boundary value problem of 1.2.
Our model comes from 4. In 4, Yang has studied the global existence, asymptotic
behavior, and blow-up of solutions for a nonlinear wave equations with dissipative term:
utt Δ2 u λut N
∂
σi uxi ,
∂x
i
i1
1.4
with the same initial and boundary conditions as that of 1.1. In our model, we add damping
and source terms which enhance the difficulty of proving the existence and decay of solution
of 1.2.
More related studies of the damped hyperbolic equation with dissipative term or
damping term can be found in papers 5–15.
The paper is organized as follows. In Section 2, we present some notations, and results
needed later and main results. Section 3 contains the statement and the proofs of the decay of
solutions. Section 4 gives the statement and the proofs of the blow-up of solutions.
2. Preliminaries
We first introduce the following abbreviations:
QT Ω × 0, T ,
H m W m,2 ,
Lp Lp Ω,
H0m W0m,2 ,
W m,p W m,p Ω,
·2 ·L2 ,
·1,2 ·H01 .
1,p
2.1
Let ·, · denote the L2 -inner product. We denote the dual of W0 by W −1,p , with p p/p−1,
and · −1,p · W −1,p . Now we make the following assumptions:
A1 σi ∈ C1 R, σi ss > 0 and
C1 |s|r ≤ |σi s| ≤ C2 |s|r ,
2.2
Abstract and Applied Analysis
3
for some r ≥ 1, if n 1, 2, else if n ≥ 3, then r ≤ n/n − 2, so that ∇u2r ≤ B1 Δu2 ,
i 1, 2, . . . , N, where B1 is the optimal embedding constant.
A2 The initial data
u0 ∈ H02 ,
2.3
u1 ∈ H01 .
A3 m < ρ 1; If n > 3, then 0 < ρ < m < 4/n − 2 and if n 1, 2, then 0 < ρ < m.
A4 If n > 3, 1 < p < n 2/n − 2 and if n 1, 2, then p > 1. Without loss of generality,
we assume that r < p.
And
B1 σi ∈ C1 R, σi ss < 0, and
C1 |s|r ≤ |σi s| ≤ C2 |s|r ,
2.4
for some r ≥ 1, if n 1, 2, else if n ≥ 3, then r ≤ n/n − 2, so that ∇u2r ≤ B1 Δu2 ,
i 1, 2, . . . , N. where B1 is the optimal embedding constant, and C2 < r 1/p 1.
B2 If n > 3, then m < r − 1 < p − 1 < 4/n − 2 and if n 1, 2, then 0 < m < r − 1 < p − 1.
B3 ρ 1 < r, and if n > 3, then ρ < 4/n − 2 and if n 1, 2, then 0 < ρ.
Throughout this paper, we use the embedding H02 Ω → Lq Ω which implies uq ≤
B2 Δu2 when
2≤q≤
2n
n−2
if n ≥ 3,
2.5
q ≥ 1 if n 1, 2,
where B2 is an optimal embedding constant. We introduce the following functionals:
Ai s s
σi η dη,
2.6
0
N
1
1
ρ2
Et ut ρ2 Δu22 ρ2
2
i1
p1
Ω
Ai uxi dx −
up1
p1
,
2.7
p1
up1
1
C2
Jt Δu22 −
,
∇ur1
r1 −
2
r1
p1
p1
It Δu22 − C2 ∇ur1
r1 − up1 .
2.8
2.9
Because r < p, we have
Jt p−r
It
r−1
p1
Δu22 up1 .
r 1 2r 1
p 1 r 1
2.10
4
Abstract and Applied Analysis
Theorem 2.1. Assume that (A1)–(A4) hold, then problem 1.1 has a unique solution u satisfying
u ∈ L∞ 0, T ; H02 ∩ W 1,∞ 0, T ; Lρ2 ;
ut ∈
L2 0, T ; H02
∩L
∞
2.11
0, T ; Lρ2 ,
where T < 1.
Theorem 2.2. Assume that (A1)–(A4) hold, u is the local solution of the problem 1.1. And
C0
2r 1
E0
r−1
r−1/2
2r 1
E0
r−1
p−1/2 < 1,
2.12
I0 > 0,
where C0 max{B1 C2 , B2 }, then u is a global bounded solution, moreover,
Et ≤
M 1 t1/m2 tm−ρ/m2 t1/2 ,
1t
∀t ∈ 0, ∞,
ρ2
lim Δu22 lim ut ρ2 0,
t→∞
t→∞
2.13
2.14
where M > 0 is a constant.
Theorem 2.3. Assume that (A1)–(A4) and
pB2 2r 1 p1/2
E0p−1/2 < 1
p1
r−1
2.15
hold, u is the local solution of the problem 2.7. Consider 2.12 are satisfied, then
Et ≤ Ke−κt ,
∀t ∈ 0, ∞,
2.16
where K, κ > 0 are constants.
Remark 2.4. When γ > 0, we will use perturbed energy method, which is different to the
method of the proof of Theorem 2.2, to prove Theorem 2.3.
Theorem 2.5. Assume that (B1)–(B3), (A4) hold, and there exist some u0 , u1 ∈ H 2 Ω × Lρ2 Ω,
then the solution of the problem 1.1 blows-up at the limited time T ∗ > 0.
Abstract and Applied Analysis
5
3. Decay of Solutions
In this section, we prove Theorems 2.1–2.3. First, we give the following Lemma.
Lemma 3.1 see 2. Let Ω be any bounded domain in RN , {ωk }∞
k1 be an orthogonal basis in
L2 Ω. Then for any ε > 0, there exists a positive number Nε such that
u ≤
N
ε
1/2
2
3.1
εu1,p ,
u, ωk k1
1,p
for all u ∈ W0
2 ≤ p < ∞.
Proof of Theorem 2.1. We look for approximate solutions un t of problem 1.1 of the form
un t :
n
Tjn tωj ,
3.2
j1
where {ωj }nj1 is an orthogonal basis in H02 , and also in L2 , and the coefficients {Tjn }nj1 satisfy
Tjn t un t, ωj with
n ρ n
u t u t, ωj Δ2 un t, ωj un m un t, ωj γ Δ2 un t, ωj
t
tt
t
t
t
∂ n
σi uxi t , ωj |un t|p−1 un t, ωj ,
∂xi
i
un 0 un0 ,
t > 0, j 1, . . . , n,
3.3
unt 0 un1 .
Since C0∞ is dense in H02 and L2 , we choose un0 , un1 ∈ C0∞ such that
un0 −→ u0
un1 −→ u1
in H02 ,
in H01
as n −→ ∞.
3.4
The above system of o.d.e. has a local solution un t defined in some interval 0, Tn . The
following will prove that the Tn can be substituted by some T > 0.
t and summing up about j, we get
Multiply 3.3 by Tjn
d
1 un ρ2 1 Δun 2 un m2 γ Δun 2
2
t ρ2
t m2
t 2
dt ρ 2
2
N σin uxi ,
i1
unxi t
n p−1 n
|u |
u
, unt
.
3.5
6
Abstract and Applied Analysis
A simple integration of 3.5 over 0, t leads to
t
1 n 2
un ρ2 1 Δun 2 γu
uns m2
s
2
t ρ2
m2
2 ds
ρ2
2
0
t
N t
n n
C3 σi uxi , uxi s ds |un |p−1 un , uns ds,
i1
0
3.6
0
ρ2
where C3 1/ρ 2un1 ρ2 1/2Δun0 22 > 0.
We now estimate the last two terms at the right-hand side of 3.6. Using Hölder
inequality, Young’s inequality, and the embedding theorem, we know there exist q, w > 0,
satisfying that
1 1 n−2
1,
q w
2n
3.7
where
r − 1q 2n
,
n−2
2n
,
p−1 q n−2
1
n 2 r − 1n − 2 n − 2
−
≥
,
w
2n
2n
2n
p − 1 n − 2 n − 2
1
n2
−
≥
.
w
2n
2n
2n
3.8
Consider the following:
t t
N
n n ≤
C
σ
ds
u
,
u
|∇un |r−1 |∇un ||∇uns |dx ds
2
xi s
i1 0 i xi
0 Ω
≤ C2
t
0
≤ C2 C
n
n
∇un r−1
r−1q ∇u w ∇us 2n/n−2 ds
t
0
C2 C
t
0
n
n
Δun r−1
2 Δu 2 Δus 2 ds
Δun r2 Δuns 2 ds
1 t
t
n 2r
n 2
≤ C2 C
Δu 2 ds Δus 2 ds
2 0
2 0
⎛ r/p
t
1
2p
n
≤ C2 C⎝
Δu 2 ds
2
0
1
2
t
p−r/p
1 ds
0
2
t
0
⎞
Δuns 22 ds⎠
Abstract and Applied Analysis
7
1 p
≤ C2 C
2 r
C2 C
2
t
0
t
0
2p
Δun 2 ds
Δuns 22 ds,
C2 C
1 p−r
2 p
∀ > 0,
3.9
assuming that t < 1.
Similarly,
t
p−1
|un |p−1 un , uns ds ≤ un p−1q un w uns 2n/n−2 ds
t
0
0
≤C
t
0
≤C
1
2
p−1
3.10
Δun 2 Δun 2 Δuns 2 ds
t
2p
Δun 2 ds C
2
0
t
0
Δuns 22 ds,
∀ > 0.
Using 3.6–3.10, we have
t
0
t
1
1 un ρ2
Δuns 22 ds Δun 22 t ρ2
2
ρ
2
0
t
t
2p
≤ C4 Cε Δun 2 ds C Δuns 22 .
2
0
0
uns m2
m2 ds
γ
3.11
Choosing 1/Cγ in 3.11, we have
t
0
1
1 un ρ2 γ
Δun 22 2
ρ 2 t ρ2 2
t
2p
≤ C4 C Δun 2 ds,
uns m2
m2 ds
t
0
Δuns 22 ds
3.12
0
where C4 C3 C2 C/2p − r/p. Assuming Yn t 2C4 C
p
Yn t ≤ 2CYn .
t
0
2p
Δun 2 , we have
3.13
A simple integration of 3.13 over 0, t leads to
1−p
−1/p−1
;
Yn t ≤ Yn 0 − 2C p − 1 t
3.14
8
Abstract and Applied Analysis
this implies that
Δun 22 −1/p−1
2 un ρ2 ≤ Yn1−p 0 − 2C p − 1 t
.
t ρ2
ρ2
3.15
Though Yn t may blow up, there exists 0 < T < min{1, Tn } satisfying
ρ2
Δun 22 unt ρ2 ≤ C,
∀t ∈ 0, T ,
3.16
where C is independent of n.
Moreover,
t
0
uns m2
m2 ds
t
0
Δuns 22 ds ≤ C.
3.17
By 3.17,
t
0
n
|us s|m uns sm2/m1 ds ≤
m2/m1
t
0
≤ C,
uns sm2
m2 ds
3.18
t ∈ 0, T .
By 3.16,
p1/p
p1
p1
n
p−1
≤ un sp1 ≤ Δun s2 ≤ C.
|u s| un s
p1/p
3.19
From 3.16–3.19, we have
un ∈ L∞ 0, T ; H02 ;
unt ∈ L2 0, T ; H02 ∩ L∞ 0, T ; Lρ2 .
3.20
So the solution un t of problem 3.3 exists on 0, T for each n. On the other hand, we can
extract a subsequence from un , still denoted by un , such that
weak∗ in L∞ 0, T ; H02 ,
3.21
weak∗ in L∞ 0, T ; Lρ2 ∩ L2 0, T ; H02 ,
3.22
un −→ u
unt −→ ut
Abstract and Applied Analysis
9
as n → ∞. By 3.21, the Sobolev embedding theorem and the continuity of σi s, for t ∈
0, T ,
∇un t −→ ∇ut strongly in L2 , a.e. on Ω,
σi unxi t −→ σi uxi t, a.e. on Ω, i 1, . . . N,
3.23
as n → ∞. By Lemma 3.1, 3.21-3.22, for any ε > 0, there exist positive constant N1ε and
N2ε independent of un and unt , respectively, such that, as n → ∞,
u t − ut ≤
n
N
1ε
1/2
u − u, ωk n
2
εun t − ut1,2 ≤ Mε,
k1
T
0
N T
2ε
2
n
n
2
u s − ut s ds ≤ 2
ut s − ut s, ωk ds
t
2
ε
3.24
0
k1
2
T
0
n
u s − ut s2 ds ≤ Mε2 .
t
1,2
By the arbitrariness of ε we get
un −→ u
strongly in L∞ 0, T ; L2 , a.e. on QT ,
unt −→ ut
strongly in L2 QT , a.e. on QT .
3.25
From the continuity of |ut |m ut and 3.25 we know that |unt |m unt → |ut |m ut a.e. on QT . With
the same methods used above we easily get |un |p−1 un → |u|p−1 u a.e. on QT . Integrating 3.3
over 0, t, t < T gets
t
t
t
nm n
1 n ρ1
ut t
, ωj Δun s, Δωj ds |us | us s, ωj ds γ Δuns s, Δωj ds
ρ1
0
0
0
t
t
∂ n
σi uxi s , ωj ds |un s|p−1 un s, ωj ds, t > 0, j 1, . . . , n.
∂x
i
0
0
i
3.26
Exploiting 3.16-3.17, we have
t
Δu s, Δωj ds ≤ C
n
0
0
n m n
u t u t, ωj ≤ un m1 ωj t
t
t
t
ρ2
Δun 22 ds ≤ CT,
ρ2/m1
ρ2 ρ2/m1
≤ unt ρ2 ωj ρ2/m1 ≤ C,
10
Abstract and Applied Analysis
t
0
|uns s|m uns s, ωj ds ≤
t
t
0
m2 uns m2
m2 ωj m2 ds ≤ CT,
|un s|p−1 un s, ωj ds ≤ CT.
0
3.27
Let n → ∞ in 3.26 and we deduce from 3.23, 3.27 and the Lebesgue-dominated
convergence theorem that
t
t
t
1 ρ1
m
Δus, Δωj ds |us t| us s, ωj ds γ Δus s, Δωj ds
|ut t| , ωj ρ1
0
0
0
t
t ∂
σi uxi s, ωj ds |us|p−1 us, ωj ds, t > 0, j 1, . . . , n.
0 ∂xi
0
i
3.28
This implies u ∈ L∞ 0, T ; H02 is a local weak solution of problem 1.1. The proof of
Theorem 2.1 is completed.
Secondly, we prove Theorem 2.2. First we give two lemmas. It is easy to prove what
follows.
Lemma 3.2. The modified energy functional satisfies, along solutions of 1.1,
2
E t −ut m2
m2 − γΔut 2 ≤ 0.
3.29
Lemma 3.3. Assume that (A1)–(A3) hold, satisfying
C0
2r 1
E0
r−1
r−1/2
2r 1
E0
r−1
p−1/2 < 1,
3.30
I0 > 0.
Then It > 0.
Proof. Since I0 > 0, then there exists by continuity T∗ ≤ T such that It > 0, for all t ∈
0, T∗ , this gives
Jt p−r
r−1
It
p1
Δu22 up1
r 1 2r 1
p 1 r 1
p−r
r−1
p1
≥
up1 .
Δu22 2r 1
p 1 r 1
3.31
Abstract and Applied Analysis
11
By using 2.8, 2.9, 3.31, and Lemma 3.2, we easily have
2r 1
2r 1
2r 1
Jt ≤
Et ≤
E0,
r−1
r−1
r−1
Δu22 ≤
3.32
∀t ∈ 0, T∗ .
We then exploit 2.12, and 3.32 to obtain
C2 ∇ur1
r1
≤
B1 C2 Δur1
2
p1
≤
p1
up1 ≤ B2 Δu2
2
B1 C2 Δur−1
2 Δu2
≤ B2
2r 1
E0
r−1
≤ B1 C2
p−1/2
2r 1
E0
r−1
Δu22 ,
r−1/2
Δu22 ,
∀t ∈ 0, T∗ .
3.33
Using 3.33, we have
C2 ∇ur1
r1
p1
up1
r−1/2
p−1/2
2r 1
2r 1
2
E0
E0
≤ B1 C2
Δu2 B2
Δu22
r−1
r−1
r−1/2 p−1/2 2r 1
2r 1
E0
E0
≤ C0
Δu22
r−1
r−1
< Δu22 ,
∀t ∈ 0, T∗ ,
3.34
where C0 max{B2 , B1 C2 }.
Therefore,
p1
It Δu22 − C2 ∇ur1
r1 − up1 > 0,
3.35
for all t ∈ 0, T∗ . By repeating this procedure, and using the fact that
lim C0
t → T∗
2r 1
Et
r−1
≤ lim C0
t → T∗
r−1/2
2r 1
E0
r−1
2r 1
Et
r−1
r−1/2
p−1/2 2r 1
E0
r−1
3.36
p−1/2
< 1,
the proof is completed.
Lemma 3.4. Assume that (A1)–(A4) hold, 2.12 satisfy. Then the solution is global existence. More,
exist positive constant M > 0 has
Δu22
p2
ut p2
≤ M;
t
0
us m2
m2 ds
≤ M;
t
0
Δus 22 ds ≤ M,
∀t ∈ 0, ∞.
3.37
12
Abstract and Applied Analysis
Proof. It suffices to show that
ρ2
Δu22 ut ρ2
3.38
is bounded independently of t. To achieve this, we use 2.9, 3.31, and Lemma 3.2 to get
E0 Et t
0
≥ Jt us m2
m2 ds
γ
1
ρ2
ut ρ2 ρ2
t
0
t
0
Δus 22 ds
us m2
m2 ds γ
t
0
Δus 22 ds
p−r
r−1
p1
Δu22 up1
2r 1
p 1 r 1
t
t
1
ρ2
ds
γ
Δus 22 ds
ut ρ2 us m2
m2
ρ2
0
0
t
t
1
r−1
ρ2
2
m2
≥
Δu2 ut ρ2 us m2 ds γ Δus 22 ds.
2r 1
ρ2
0
0
≥
3.39
Since It, Jt are positive. Therefore,
ρ2
Δu22 ut ρ2 ≤ CE0.
3.40
Moreover,
t
0
us m2
m2 ds
≤ E0;
t
γ
0
Δus 22 ds ≤ E0,
3.41
where C is a positive constant, which depends only on r.
Lemma 3.5. Assume that (A1)–(A4) hold, 2.12 satisfy. Then exist C > 0 has
t
0
ρ2
us ρ2 ds ≤ Ctm−ρ/m2 ;
t
0
Δu22 ds
≤C
t
Isds.
0
3.42
3.43
Abstract and Applied Analysis
13
Proof. Using Lemma 3.4, we have
t
0
ρ2
us ρ2 ds
≤
t
0
≤C
ρ2/m2 t
us m2
ρ2 ds
t
0
m−ρ/m2
1 ds
0
ρ2/m2
3.44
tm−ρ/m2
us m2
m2 ds
≤ Ctm−ρ/m2 .
Using 3.34, we have
C2 ∇ur1
r1
p1
up1
≤ BC0
:
2r 1
E0
r−1
r−1/2 2r 1
E0
r−1
p−1/2 Δu22
3.45
ηΔu22 .
So
1 − η Δu22 ≤ It;
3.46
the proof is complete.
Lemma 3.6. Assume that (A1)–(A4) hold, 2.12 satisfy. Then there exists a C > 0, having
t
Isds ≤ C t1/2 tm−ρ/m2 t1/m2 1 .
3.47
0
Proof. By multiplying the differential equation in 1.1 by u and integrating over Ω, using
integration by parts 2.9 and assumption A1, we obtain
u ρ2
t ρ2
d |ut |ρ ut
0
,u −
Δu22 |ut |m ut , u
dt ρ 1
ρ1
N
p1
γ Δ2 ut , u σi uxi , uxi − up1
i1
u ρ2
t ρ2
d |ut |ρ ut
,u −
Δu22 |ut |m ut , u
≥
dt ρ 1
ρ1
p1
γΔut , Δu − C2 ∇ur1
r1 − up1
u ρ2
t ρ2
d |ut |ρ ut
,u −
It
dt ρ 1
ρ1
ut m ut , u γΔut , Δu.
3.48
14
Abstract and Applied Analysis
So
t
0
ρ1
ρ2
Isds ≤ ut ρ2 uρ2 u1 ρ2 u0 ρ2 t
us m us , u ds γ
t
0
t u ρ2
s ρ2
0
ρ1
ds
3.49
Δus , Δuds.
0
Using A3, 3.42-3.43, we have
ρ1
ρ2 ρ1/ρ2
ut ρ2 uρ2 ≤ C ut ρ2
Δu2 ≤ C,
t 0
Ω
|us |m us u dx ds ≤
t
0
≤C
us m1
m2 um2 ds ≤ C
t
0
t
γ
Δus , Δuds ≤ γ
0
us m1
m2 ds
t
0
≤C
t
0
t
0
us m1
m2 Δu2 ds
m1/m2
us m2
m2
Δus 2 Δu2 ds ≤ C
t
0
t1/m2 ≤ Ct1/m2 ,
1/2
Δus 22 ds
t1/2 ≤ Ct1/2 .
3.50
Therefore,
t
Isds ≤ M 1 t1/m2 tm−ρ/m2 t1/2 .
3.51
0
Proof of Theorem 2.2. First, t → 1 tEt is also absolutely continuous, and we have
d
1 tEt ≤ Et.
dt
A simple integration of 3.52 over 0, t leads to
1 tEt ≤ E0 t
Esds
0
t
1 t
1
ρ2
us ρ2 ds Δu22 ds
ρ2 0
2 0
t
N t 1
p1
Ai uxi dx ds −
up1 ds
p
1
0
0
Ω
0
≤ E0 3.52
Abstract and Applied Analysis
15
t
1
1 t
ρ2
us ρ2 ds Δu22 ds
ρ2 0
2 0
t
N t 1
p1
Ai uxi dx ds up1 ds
p
1
0 Ω
0
0
≤ E0 t
1
1 t
ρ2
us ρ2 ds Δu22 ds
ρ2 0
2 0
t
1
C2 t
p1
ds
∇ur1
up1 ds.
r1
r1 0
p1 0
≤ E0 3.53
Using the upper inequality, 3.45, 3.46, and 3.52, we have
1 tEt ≤ E0 t
1
ρ2
0
t
1
≤ E0 ρ2
0
ρ2
us ρ2 ds ρ2
us ρ2 ds
1
2
C
t
0
Δu22 ds C
t
t
0
Δu22 ds
3.54
Isds.
0
Apply 3.42, 3.47, we can get 2.13.
Using 3.31 and Lemma 3.3, we get Jt ≥ 0, It > 0. 2.13 implies lim t → ∞Et ρ2
0. So when t → ∞, we have ut ρ2 → 0 and Jt → 0. It is that 2.14 is satisfied.
Theorem 2.2 is complete.
Following we will prove Theorem 2.3. For this purpose we set
Lt : Et εΨt,
3.55
where ε is a positive constant and
Ψt 1
ρ1
Ω
|ut |ρ ut u dx.
3.56
Lemma 3.7. Let ε be small enough. Then there exist two positive constants α1 and α2 such that
α1 Lt ≤ Et ≤ α2 Lt.
Proof. By Lemma 3.4 and Young’s inequality, a direct computation gives
ε
|ut | dx |u|ρ2 dx
ρ2 Ω
Ω
ε
εB2 2r 1E0 ρ/2
ρ2
≤ Et Δu22
ut ρ2 ρ2
ρ2
r−1
ε
Lt ≤ Et ρ2
ρ2
3.57
16
Abstract and Applied Analysis
εB2 2r 1 ρ2/2
E0ρ/2 Et
≤ Et εEt ρ2
r−1
≤
1
Et.
α1
3.58
Similarly, we have
ε
|ut | dx −
|u|ρ2 dx
ρ2 Ω
Ω
ε
εB2 2r 1E0 ρ/2
ρ2
≥ Et −
Δu22
ut ρ2 −
ρ2
ρ2
r−1
εB2 2r 1 ρ2/2
≥ Et − εEt −
E0ρ/2 Et
ρ2
r−1
ε
Lt ≥ Et −
ρ2
≥
ρ2
3.59
1
Et,
α2
provided that ε is small enough.
Lemma 3.8. Assume that the conditions of Theorem 2.3 hold, then the function
Ψt :
1
ρ1
Ω
|ut |ρ ut u dx
3.60
satisfies, along the solution of 1.1,
1
2
Ψ t ≤ − Et μut m2
m2 ωΔu2 .
4
3.61
Proof. Applying equations of 1.1, we see
Ψ t 1
ρ2
ut ρ2 ρ1
1
ρ2
ut ρ2 ρ1
Ω
|ut |ρ utt u dx
N
∂
p−1
2
−Δ u − |ut | ut σi uxi − γΔ ut |u| u u dx
∂xi
i1
m
2
Ω
N
1
ρ2
p1
ut ρ2 − Δu22 − |ut |m ut , u − σi uxi , uxi − γΔut , Δu up1
ρ1
i1
−
N
1
ρ2
ut ρ2 − Δu22 −
ρ2
i1
Ω
Ai uxi dx N i1
Ω
Ai uxi − σi uxi , uxi dx
Abstract and Applied Analysis
17
p1
− |ut |m ut , u − γΔut , Δu up1 2ρ 3
ρ2
ut ρ2
ρ2 ρ1
≤ − Et − |ut |m ut , u − γΔut , Δu 2ρ 3
ρ2
ut ρ2
ρ2 ρ1
N
p
p1
up1 p1
i1
Ω
Ai uxi − σi uxi uxi dx.
3.62
Exploiting the assumption A1 and Young’s inequality, we have
N Ω
i1
Ai uxi − σi uxi uxi dx ≤ 0.
p1 p1
pB2
pB2
p
p1
p1
up1 ≤
Δu2 ≤
p1
p1
p1
m
|ut | ut , u ≤
2r 1
r−1
p1/2
E0p−1/2 Et
δm2
ut m2
m2
m2
m 1
δB2 m2 2r 1E0 m/2
m2
≤
u
Δu22
t
m2
m2
r−1
m 2δm2/m1
1
2δm2/m1
1
≤
m 2δm2/m1
δB2 m2
m2
ut m2
m2 ut m2
m2
2r 1
r−1
m2/2
E0m/2 Et,
−1
γΔut , Δu ≤ γη Δut 2 γη Δu2 ,
2
2
2
2
∀δ > 0
∀η > 0.
3.63
Exploiting 3.63 and 3.62, we get
p2 Ψ t ≤ − 1 −
pB2
p1
2r 1
r−1
δB2 m2
−
m2
1
m 2δ
p1/2
2r 1
r−1
E0p−1/2
m2/2
ut m2
m2 m2/m1
E0
γη
−1
m/2
2r 1
Et
− γη
r−1
ρ/ρ2 2
2ρ 3
B2 Δu22 .
ρ 2 E0
ρ1 ρ2
3.64
18
Abstract and Applied Analysis
Choosing δ satisfies
δB2 m2
m2
2r 1E0
r−1
m/2
p2 pB2
2r 1 p1/2
1
p−1/2
1−
E0
4
p1
r−1
3.65
and η satisfies
p2 pB2
2r 1 p1/2
2r 1 1
p−1/2
,
1−
γη
E0
r−1
4
p1
r−1
3.66
at the above; the proof of 3.61 is completed.
Proof of Theorem 2.3. Using 3.61 and Lemma 3.7, we have
L t E t εΨ t
ε
− Et εμut m2
≤ −ut m2
−
γΔu
t
m2
m2 εωΔut 4
ε
− Et − γ − εω Δut − 1 − εμ ut m2
m2
4
εα1
Lt − γ − εω Δut − 1 − εμ ut m2
≤−
m2 .
4
3.67
Choosing ε satisfies ε ≤ min{γ/ω, 1/μ}. So we have
L t ≤ −
εα1
Lt,
4
∀t ≥ 0.
3.68
A simple integration of 3.68 over 0, t leads to
Lt ≤ L0e−εα1 /4t ,
∀t ≥ 0.
3.69
Exploiting Lemma 3.7 again, we have
Et ≤ α2 L0e−εα1 /4t : Ke−κt ,
∀t ≥ 0,
3.70
where K, κ > 0 are constants. The proof of Theorem 2.3 is complete.
4. Blow-Up of Solutions
Proof of Theorem 2.5. Assuming that the solution of 1.1 is global, we have
2
E t −ut m2
m2 − γΔut 2 ≤ 0.
4.1
Abstract and Applied Analysis
19
So
4.2
Δu22 ≤ C3 ;
we set
Qt : −
t
u20 dx,
4.3
u20 dx − E0.
4.4
Esds ot ω
0
Ω
where o, ω are constants and will be given later.
Consider
Q t −Et o
Ω
u20 dx ≥ o
Ω
Choosing o satisfies the following condition in 4.4:
o
Ω
u20 dx − E0 Q 0 > 0,
4.5
so we have
Q t ≥ Q 0 > 0,
∀t ∈ 0, T .
4.6
Moreover, we have
Q t − Q 0 E0 − Et −
t
E sds 0
t
0
2
γΔu
us m2
s
m2
2 ds.
4.7
Define
Kt : Q
1−γ
ε
t ρ1
t 0
Ω
|us |ρ us u dx ds,
4.8
where ε > 0 will be given later, and
r−ρ−1
p−m−1
r−m−1
0 < γ ≤ min .
,
,
ρ 2 r 1 p 1 m 1 m 1r 1
4.9
20
Abstract and Applied Analysis
Multiplying 1.1 by u and a direct computation yield
K t 1 − γ Q−γ Q t 1 − γ Q−γ Q t ε
t 0
Ω
−ε
0
ε
ε
ρ1
Ω
|u1 |ρ u1 u0 dx ds Ω
|u1 |ρ u1 u0 dx ds Ω
|u1 |ρ u1 u0 dx ds Ω
Δ uu dx ds − ε
2
t ε
ρ1
ε
ρ1
Ω
0
ε
ρ1
m
Ω
0
N t i1
ε
ρ1
t 0
Ω
0
Ω
t |us |ρ us u s dx ds
|us |ρ2 dx ds
|us |ρ uss u dx ds
1 − γ Q−γ Q t t ε
ρ1
|us | us u dx ds − εγ
∂
σi uxi u dx ds ε
∂xi
t 0
Ω
t 0
Ω
0
Ω
t |us |ρ2 dx ds
Δ2 us u dx ds
4.10
|u|p−1 uu dx ds
t ε
ε
ρ
1 − γ Q Q t |u1 | u1 u0 dx ds |us |ρ2 dx ds
ρ1 Ω
ρ1 0 Ω
t t t Δus Δu dx ds − ε
−ε
|Δu|2 dx ds − εγ
|us |m us u dx ds
0
−ε
−γ
Ω
N t
i1
0
0
Ω
Ω
σi uxi uxi dx ds ε
0
t
0
Ω
p1
up1 ds.
Exploiting 4.7 and B1, we have
t γ
0
γ
Δus Δu dx ds ≤ 2
η
Ω
t
0
Δus 22 ds
η γ
2
t
0
Δu22 ds
4.11
γ ≤ 2 Q t − Q 0 η2 γC3 T,
η
t 0
Ω
|us |m us u dx ds ≤
ςm2
m2
ςm2
≤
m2
−
t 0
Ω
t 0
Ω
t 0
Ω
|u|m2 dx ds |u|m2 dx ds m 1 −m1/m2
ς
m2
t 0
Ω
|us |m2
m2 dx ds
m 1 −m1/m2 ς
Q t − Q 0
m2
σi uxi uxi dx ds ≥ C4
t
0
∇ur1
r1 ds.
4.12
4.13
Abstract and Applied Analysis
21
Using 4.10–4.13, we have
K t ≥ 1 − γ Q−γ Q t εC4
t
0
ε
ρ1
∇ur1
r1 ds − ε
Ω
t
0
|u1 |ρ u1 u0 dx Δu22 ds ε
t
0
ε
ρ1
t 0
Ω
|us |ρ2 dx ds
p1
up1 ds − εη2 γTC3
m 1 −m1/m2 γ
ςm2 t
ς
−ε
2 Q t − ε
|u|m2 dx ds
m2
m2 0 Ω
η
m 1 −m1/m2 γ
ς
2 Q 0.
ε
m2
η
4.14
Choosing ς, η satisfies
1
M2 Q−γ t.
η2
ς−m2/m1 M1 Q−γ t,
4.15
Then we have
m1
ε
−γ K t ≥ 1 − γ − ε
M1 − εγM2 Q Q t |u1 |ρ u1 u0 dx ds
m2
ρ1 Ω
t m1
ε
M1 εγM2 Q−γ Q 0
|us |ρ2 dx ds ε
ρ1 0 Ω
m2
t
M−m−1 γm1 t
Q
−ε 1
|u|m2 dx ds − εM2−1 Qγ γTC3 − ε Δu22 ds
m2
0 Ω
0
t
t
p1
ε up1 ds εC4 ∇ur1
r1 ds.
0
4.16
0
A simple computing implies
t
γ
2
u0 dx
Q − Esds ot ω
γ
Ω
0
⎞
⎤γ
⎡ ⎛
t up1
p1
⎠ds oT ωu0 2 ⎦
C2 ∇ur1
≤⎣ ⎝
2
r1
p1
0
≤2
γ−1
γ
C5 oT ω
2γ
u0 2
t
0
p1
up1
∇ur1
r1
4.17
γ ds
.
22
Abstract and Applied Analysis
Using 4.17, B2, embedding theorem, and Hölder inequality, we can get
Q
γm1
≤ 2
γ−1
C5 oT ω
γ
2γ
u0 2
t
0
p1
up1
∇ur1
r1
γ m1
ds
⎡
γm1
t
p1
r1
γ−1m1−1
⎣
≤2
C6
up1 ∇ur1 ds
0
4.18
⎤
oT t α
0
m2
Ω
|u|
dx ds ≤ α
t Ω
0
|u|
p1
2γm1
⎦,
ωγm1 u0 2
m2/p1
ds|Ω|p−m−1/p1
dx
p−m−1/p1 p−m−1/p1
≤ α|Ω|
t
T
0
: C8
t
0
m2/p1
p1
up1 ds
4.19
m2/p1
p1
up1 ds
,
where C8 α|Ω|p−m−1/p1 T p−m−1/p1 .
Consider
1 − α
t 0
Ω
t
|u|m2 dx ds ≤ 1 − αB
0
≤ 1 − αB
ds
∇um2
2
t
0
≤ C7
t
0
m2/r1
tr−m−1/r1
∇ur1
2 ds
4.20
m2/r1
∇ur1
r1 ds
.
Using 4.18–4.20, B1, we get
Q
γm1
t 0
Ω
|u|m2 dx ds
γ−1m1−1
≤2
⎡
⎤
γm1
t
p1
2γm1
r1
γm1
⎦
C6 ⎣
oT ω
u0 2
up1 ∇ur1 ds
0
⎡
× ⎣C8
t
0
m2/p1
p1
up1 ds
C7
t
0
m2/r1 ⎤
⎦
∇ur1
r1 ds
Abstract and Applied Analysis
⎡ γ−1m1−1
≤2
23
t
C6 ⎣C8
p1
up1
0
∇ur1
r1
γm1m2/p1
ds
2γm1
oT ωγm1 u0 2
t
γm1m2/p1 ⎤
p1
⎦
× C8
up1 ∇ur1
r1 ds
0
⎡
γ−1m1−1
2
C6 ⎣C7
t
0
p1
up1
∇ur1
r1
γm1m2/r1
ds
2γm1
oT ωγm1 u0 2
t
γm1m2/r1 ⎤
p1
⎦,
× C7
up1 ∇ur1
r1 ds
0
4.21
Using 4.9, 4.17, and Young inequality, we have
Q
γm1
t 0
Ω
|u|
m2
dx ds ≤ C9 1 t
0
p1
up1
∇ur1
r1
ds .
4.22
Similarly, we have
T C3 Q ≤ C10 1 γ
t
0
p1
up1
∇ur1
r1
4.23
ds .
Additionally, we choose ε satisfying
ε≤
1−γ
.
m 1/m 2M1 γM2
4.24
Using 4.16, 4.21–4.24, we get
t ε
|u1 | u1 u0 dx ds |us |ρ2 dx ds
ρ
1
0 Ω
Ω
−m−1
t
M1
γ
2
C9 C10
− ε Δu2 ds − ε
m2
M2
0
ε
K t ≥
ρ1
×
t
ρ
0
p1
up1
∇ur1
r1
ε min{1, C4 }
t
0
−m−1
M1
γ
C9 C10
−ε
m2
M2
p1
∇ur1
r1 up1 ξQt
24
Abstract and Applied Analysis
ξ
N t uxi
i1
−ξ
Ω
0
t up1
p1
0
ξ
2
t
0
p1
Ω
ds ξ
ρ1
t
0
ρ2
us ρ2 ds
Δu22 ds − ξot ωu0 22 ,
t ux
0
σi τdτ dx ds
0
i
σi τdτ dx ds −
t up1
p1
0
0
p1
ds
t up1
C2 t
p1
r1
≥−
ds
∇ur1 ds −
r1 0
p
1
0
t
1
p1
≥ −
∇ur1
r1 up1 ds.
p1 0
4.25
Using 4.25, we get
t
ξ
K t ≥ ξQt −1
ε
Δu22 ds
2ε
0
0
t
−m−1
γ
1 ξ M1
p1
ε min{1, C4 } −
−
C9 − C10
up1 ∇ur1
r1
p1ε
m2
M2
0
t
M1−m−1
γ
1
ξ
ρ
2
ε
,
C9 − C10
|u1 | u1 u0 dx − oT ω u0 dx −
ρ1 Ω
ε
m2
M2
0
ε
ξ
ρ2 ρ1
t
ρ2
us ρ2 ds
4.26
choosing 2ε < ξ < p 1ε, u0 , u1 satisfying
1 ξ
> 0,
p1ε
t
1
ξ
|u1 |ρ u1 u0 dx − oT ω u20 dx > 0.
ρ1 Ω
ε
0
min{1, C4 } −
4.27
Then we choose M1 , M2 big enough, satisfying
γ
1 ξ M1−m−1
−
C9 − C10
>0
p1ε
m2
M2
t
M−m−1
γ
1
ξ
ρ
C9 − C10
> 0.
|u1 | u1 u0 dx − oT ω u20 dx − 1
ρ1 Ω
ε
m
2
M
2
0
min{1, C4 } −
4.28
Abstract and Applied Analysis
25
Using the two above inequalities, we have
K t ≥ Cε Qt t
0
ρ2
us ρ2 ds
t
0
Δu22 ds
t
∇ur1
r1
0
p1
up1
ds ν ,
4.29
where ν > 0. So
1−γ
Kt > K0 ≥ ω
u20 dx
> 0.
4.30
Ω
Using B1–B3, A4, and Hölder and Young inequalities, we have
t
1 t 1
ρ1
ρ
|u | u u dx ds ≤
u ∇ur1 ds
ρ 1 0 Ω s s
ρ 1 0 s ρ2
C
≤
ρ1
t
0
ρ1μ
us ρ2
4.31
∇uθr1 ds,
where 1/μ 1/θ 1.
So we have
1/1−γ
t
1/1−γ t
1/1−γ
1 t ρ1μ
ρ
θ
≤C
|u | u u dx ds
us ρ2 ds
∇ur1 ds
ρ 1 0 Ω s s
0
0
≤C
t
0
C
μρ1/1−γρ2
ρ1μ·ρ2/ρ1μ
ds
us ρ2
t
0
C
t
0
C
θ/1r1−γ
θ·1r/θ
ds
∇ur1
μρ1/1−γρ2
ρ2
us ρ2 ds
t
0
θ/1r1−γ
∇u1r
r1 ds
.
4.32
Using 4.9, 4.32 and choosing μ satisfy μρ 1/ρ 21 − γ 1, then θ/1 r1 − γ < 1.
26
Abstract and Applied Analysis
So
1/1−γ
t
t
1 t ρ2
ρ
1r
≤C
|u | u u dx ds
us ρ2 ds ∇ur1 ds β
ρ 1 0 Ω s s
0
0
≤C
t
ρ2
us ρ2 ds 0
t
0
Δu22 ds
t
4.33
∇u1r
r1 ds
0
t
0
p1
up1
β,
where β > 0 is a constant. Finally, we can easily get
⎛
K 1/1−γ
1/1−γ ⎞
1 t ⎠.
≤ 21/1−γ ⎝Qt ε1/1−γ |us |ρ us u dx ds
ρ 1 0 Ω
4.34
Combining 4.29 and 4.33-4.34, we have
K t ≥ CK 1/1−γ t,
4.35
t ≤ T,
for some constant C > 0. Integrating the above inequality in 0, t, we get
1
K 1/1−γ ≥ K −γ/1−γ 0
1/γ ,
− ct
∀t ≤ T.
4.36
The above inequality implies Kt blows-up on some time T ∗ . Since u exists globally, so we
have
ε
ρ1
t Ω
0
|us |ρ us u dx ds < ∞.
4.37
And we know that Kt → ∞, as t → T ∗ , so
Q1−γ t −→ ∞;
this implies Qt → ∞, that is to say −
−
t
0
t
4.38
Esds → ∞. Because
Esds ≤ −tEt,
4.39
0
we know
Et −→ −∞,
as t −→ T ∗ .
4.40
Abstract and Applied Analysis
27
This contradicts with the assumption that u is a global solution. So the solution of 1.1 blowsup on time T ∗ .
Acknowledgment
This work was supported by the National Natural Science Foundation of China 10771032.
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