Gen. Math. Notes, Vol. 8, No. 2, February 2012, pp.34-48
c
ISSN 2219-7184; Copyright ICSRS
Publication, 2012
www.i-csrs.org
Available free online at http://www.geman.in
Global Existence, Exponential Decay
and Blow-up of Solutions for Coupled a Class
of Nonlinear Higher-order Wave Equations
Erhan Pis.kin1 and Necat Polat2
1
Department of Mathematics, Dicle University, 21280, Diyarbakir, Turkey
E-mail: episkin@dicle.edu.tr
2
Department of Mathematics, Dicle University, 21280, Diyarbakir, Turkey
E-mail: npolat@dicle.edu.tr
(Received: 10-1-12/ Accepted: 14-2-12)
Abstract
In this paper we consider the decay rate and the blow up of solutions for the
initial and Dirichlet boundary value problem for coupled a class of nonlinear
higher order wave equations, in a bounded domain.
Keywords: decay, blow-up, higher-order wave equations, weak damping.
1
Introduction
In this paper we consider the following initial-boundary value problem

utt + P u + ut = f1 (u, v) ,
(x, t) ∈ Ω × (0, T ) ,




(x, t) ∈ Ω × (0, T ) ,
 vtt + P v + vt = f2 (u, v) ,
u (x, 0) = u0 (x) , ut (x, 0) = u1 (x) ,
x ∈ Ω,
(1.1)


v (x, 0) = v0 (x) , vt (x, 0) = v1 (x) ,
x ∈ Ω,


 α
α
D u (x, t) = D v (x, t) = 0, |α| ≤ m − 1, x ∈ ∂Ω,
where P = (−△)m , m ≥ 1 is a natural number, Ω is a bounded domain
n
P
αi , D α =
with smooth boundary ∂Ω in Rn ; α = (α1 , α2 , ..., αn ) , |α| =
n
Q
i=1
i=1
∂ αi
α ,
∂xi i
x = (x1 , x2 , ..., xn ) .
Global Existence, Exponential Decay...
35
The motivation of our work is due to some results regarding higher-order
wave equation. A single wave equation of the problem (1.1) becomes as following
utt + (−△)m u + g (ut ) = |u|p−1 u.
(1.2)
When m = 1 and g (ut ) = ut , the equation in (1.2) becomes a wave equation
utt − △u + ut = |u|p−1 u,
(1.3)
was considered by many authors, see [2, 4, 5]. In [4], Ikehata showed certain
decay rate for the total energy E (t) and L2 norm of a solution to equation
(1.3). Also, in [2, 5], the authors obtained a time-decay result.
When m = 2, the equation in (1.2) becomes a Petrovsky equation, studied
by many authors [8, 11], for g (ut ) = |ut |q−1 ut (q > 1).
Recently, Ye [12] investigated equation (1.2) for m ≥ 1 and g (ut ) =
|ut |q−1 ut (q > 1), and showed the existence of global solutions if the initial
energy is sufficiently small. Also, Zhou et. al. [13] extended the results of [12].
In this paper, under some restrictions on the initial data, we establish the
uniform decay rates. After that, we show blow up of solution with negative
and nonnegative initial energy, using the same techniques as in [7].
This paper is organized as follows. In section 2, we present some lemmas,
and the local existence theorem. In section 3, the global existence and the
decay of the solution are given. In section 4, we show the blow up properties
of solution.
2
Preliminaries
In this section, we shall give some assumptions and lemmas which will be used
throughout this work. Let k.k and k.kp denote the usual L2 (Ω) norm and
Lp (Ω) norm, respectively.
Concerning the functions f1 (u, v) and f2 (u, v) , we take
h
i
2(r+1)
r
r+2
f1 (u, v) = k |u + v|
(u + v) + l |u| u |v|
,
h
i
f2 (u, v) = k |u + v|2(r+1) (u + v) + l |u|r+2 |v|r v ,
where k, l > 0 are constants and r satisfies
−1 < r if n ≤ 2m,
3m−n
−1 < r ≤ n−2m
if n > 2m.
(2.1)
According to the above equalities can easily verify that
u f1 (u, v) + vf2 (u, v) = 2 (r + 2) F (u, v) , ∀ (u, v) ∈ R2 ,
(2.2)
36
Erhan Pis.kin et al.
where
F (u, v) =
i
h
1
k |u + v|2(r+2) + 2l |uv|r+2 .
2 (r + 2)
(2.3)
We have the following result.
Lemma 2.1 [9]. There exist two positive constants c0 and c1 such that
c0 |u|2(r+2) + |v|2(r+2) ≤ 2 (r + 2) F (u, v) ≤ c1 |u|2(r+2) + |v|2(r+2) (2.4)
is satisfied.
Let’s define
Z
1 12 2 21 2
J (t) =
P u + P v − F (u, v) dx,
2
Ω
Z
1 2 1 2
2 2 I (t) = P u + P v − 2 (r + 2) F (u, v) dx,
(2.5)
(2.6)
Ω
and also the energy function as follows
Z
1
1 12 2 21 2
2
2
E (t) =
kut k + kvt k +
P u + P v − F (u, v) dx. (2.7)
2
2
Ω
Lemma 2.2. E (t) is a nonincreasing function for t ≥ 0 and
E ′ (t) = − kut k2 + kvt k2 ≤ 0.
(2.8)
Proof. Multiplying first equation of (1.1) by ut and second equation by
vt , integrating over Ω using integrating by parts and summing up the product
results, we get
Z t
(2.9)
kuτ k2 + kvτ k2 dτ for t ≥ 0.
E (t) − E (0) = −
0
Moreover, the following energy inequality holds:
Z t
kuτ k2 + kvτ k2 dτ ≤ E (s) , for 0 ≤ s ≤ t < T.
E (t) +
(2.10)
s
Lemma 2.3 (Sobolev-Poincare inequality) [1]. If 2 ≤ p ≤
if n = 2q) , then
q
2
kukp ≤ C∗ (−△) u for u ∈ H0q (Ω)
2q
[n−2q]+
holds with some constant C∗ , where we put [a]+ = max {0, a} ,
[a]+ = 0.
(2 ≤ p < ∞
(2.11)
1
[a]+
= ∞ if
37
Global Existence, Exponential Decay...
The following integral inequality plays an important role in our proof of
the energy decay of the solutions to problem (1.1).
Lemma 2.4 [6]. Let h : [0, ∞) −→ [0, ∞) be a nonincreasing function and
assume that there exists a constant c > 0 such that
Z ∞
h (τ ) dτ ≤ ch (t) , ∀t ∈ [0, ∞) .
t
Then we have
−1
h (t) ≤ h (0) e1−tc , ∀t ≥ c.
Lemma 2.5 [7]. Let δ > 0 and B (t) ∈ C 2 (0, ∞) be a nonnegative function
satisfying
B ′′ (t) − 4 (δ + 1) B ′ (t) + 4 (δ + 1) B (t) ≥ 0.
If
B ′ (0) > r2 B (0) + K0 ,
p
with r2 = 2 (δ + 1) − 2 (δ + 1) δ, then B ′ (t) > K0 for t > 0, where K0 is a
constant.
Lemma 2.6 [7]. If H (t) is a nonincreasing function on [t0 , ∞) and satisfies
the differential inequality
1
2
[H ′ (t)] ≥ a + b [H (t)]2+ δ , for t ≥ t0 ,
where a > 0, b ∈ R, then there exists a finite time T ∗ such that
lim H (t) = 0.
t−→T ∗−
Upper bounds for T ∗ are estimated
p as follows:
(i) If b < 0 and H (t0 ) < min 1, − ab then
p a
−b
1
∗
T ≤ t0 + √ ln p a
.
− b − H (t0 )
−b
(ii) If b = 0, then
T ∗ ≤ t0 +
H (t0 )
.
H ′ (t0 )
(iii) If b > 0, then
i
h
1
3δ+1 δc
H (t0 )
− 2δ
∗
2δ
√ 1 − (1 + cH (t0 ))
,
or T ≤ t0 + 2
T ≤ √
a
a
∗
where c =
1
a 2+ δ
b
.
38
Erhan Pis.kin et al.
Next, we state the local existence theorem that can be established by combining arguments of [3, 8].
Theorem 2.1. (Local existence) Suppose that (2.1) holds, and further
(u0 , v0 ) ∈ H0m (Ω) × H0m (Ω) , (u1 , v1 ) ∈ L2 (Ω) × L2 (Ω) . Then problem (1.1)
has a unique local solution
u, v ∈ C ([0, T ) ; H0m (Ω)) ,
ut ∈ C [0, T ) ; L2 (Ω) ∩L2 (Ω × [0, T )) and vt ∈ C [0, T ) ; L2 (Ω) ∩L2 (Ω × [0, T )) .
Moreover, at least one of the following statements holds true:
i) T = ∞,
1 2 1 2
ii) kut k2 + kvt k2 + P 2 u + P 2 v −→ ∞ as t −→ T − .
Remark 2.1. We denote by C various positive constants which may be
different at different occurrences.
3
Global Existence and Energy Decay
In this section, we consider the global existence and energy decay of solutions
for problem (1.1).
Lemma 3.1 [9]. Suppose that (2.1) holds. Then there exists η > 0 such
that for any (u, v) ∈ H0m (Ω) × H0m (Ω) , the inequality
r+2
12 2 21 2
2(r+2)
r+2
ku + vk2(r+2) + 2 kuvkr+2 ≤ η P u + P v (3.1)
is satisfied.
To prove our result and for the sake of simplicity, we take k = l = 1 and
introduce
1
r+2
1
1
−
−
α2∗ ,
(3.2)
B = η 2(r+2) , α∗ = B r+1 , E1 =
2 2 (r + 2)
where η is the optimal constant in (3.1). Next, we will state and prove a lemma
which similar to the one introduced firstly by Vitillaro in [10] to study a class
of a single wave equation.
Lemma 3.2. Suppose that (2.1) holds. Let (u, v) be the solution of system
(1.1). Assume further that E (0) < E1 and
1
12 2 21 2 2
< α∗ .
(3.3)
P u0 + P v 0 Global Existence, Exponential Decay...
Then
for all t ∈ [0, T ) .
1
21 2 21 2 2
< α∗ ,
P u + P v 39
(3.4)
Proof. First have from (2.7), (2.11), (3.1) and the definition of B,
Z
1 21 2 12 2
E (t) ≥
P u + P v − F (u, v) dx
2
Ω
2 1 2 1
1 1
2 2 2(r+2)
r+2
ku
+
vk
+
2
kuvk
=
P
−
P
u
+
v
r+2
2(r+2)
2
2 (r + 2)
r+2
1 1
21 2 12 2
21 2 12 2
η P u + P v ≥
P u + P v −
2
2 (r + 2)
r+2
B 2(r+2) 1 21 2 12 2
12 2 21 2
≥
(3.5)
.
P u + P v −
P u + P v 2
2 (r + 2)
So we get
21 2 12 2
E (t) ≥ G P u + P v for t ≥ 0,
(3.6)
2(r+2)
where G (α) = 21 α2 − B2(r+2) α2(r+2) . Note that G (α) has the maximum at
1
α∗ = r+2
and maximum value is
B r+1
1
1
α2∗ .
(3.7)
−
E1 = G (α∗ ) =
2 2 (r + 2)
Now we establish (3.4) by contradiction. Suppose (3.4) does not hold, then
it follows from the continuity of (u (t) , v (t)) that there exists t0 ∈ (0, T ) such
that
2 1
2 21
12
2
= α∗ .
(3.8)
P u (t0 ) + P v (t0 )
By (3.5), we see that
#
"
2 1
2 21
1
= G (α∗ ) = E1 .
E (t0 ) ≥ G P 2 u (t0 ) + P 2 v (t0 )
(3.9)
This is impossible since E (t) ≤ E (0) < E1 , t ≥ 0. Hence (3.4) is established.
Theorem 3.1. (Global existence and energy decay) Assume that (2.1)
hold. If the initial data (u0 , u1 ) ∈ H0m (Ω)×L2 (Ω) , (v0 , v1 ) ∈ H0m (Ω)×L2 (Ω) ,
satisfy E (0) < E1 and
1
21 2 21 2 2
< α∗ ,
(3.10)
P u0 + P v 0 40
Erhan Pis.kin et al.
where the constants α∗ and E1 are defined in (3.2), then the corresponding
solution to system (1.1) globally exists, i.e., T = ∞.
Moreover, if
r+1
2 (r + 2)
E (0)
> 0,
(3.11)
1−η
r+1
then we have the following decay estimates
−1
E (t) ≤ E (0) e1−µC4
for every t ≥
C4
,
µ
t
(3.12)
where C4 is positive constant.
Proof. First, we prove that T = ∞. Since E (0) < E1 and
1
21 2 21 2 2
< α∗ ,
P u0 + P v 0 it follows from Lemma 3.2 that
1 2 1 2
1
2 P u + P 2 v < α2∗ = η − r+1
which implies that
Z
1 2 1 2
2 2 I (t) = P u + P v − 2 (r + 2) F (u, v) dx
Ω
2 1 2 r+2
1 2 1 2
1
≥0
≥ P 2 u + P 2 v − η P 2 u + P 2 v for t ∈ [0, T ) . Furthermore, (2.5) and (2.6), we get
1
r+1
12 2 21 2
J (t) =
I (t)
P u + P v +
2 (r + 2)
2 (r + 2)
r+1
12 2 21 2
≥
P u + P v .
2 (r + 2)
From (2.7) and E (t) ≤ E (0) , we deduce that
1 2 1 2 2 (r + 2)
2 (r + 2)
2 (r + 2)
2 J (t) ≤
E (t) ≤
E (0) (3.13)
P u + P 2 v ≤
r+1
r+1
r+1
for t ∈ [0, T ) . So it follows from (3.13) and Lemma 2.2
1
1
r+1
12 2 21 2
kut k2 + kvt k2 ≤ J (t) +
kut k2 + kvt k2
P u + P v +
2 (r + 2)
2
2
= E (t) ≤ E (0) < E1 , ∀t ∈ [0, T )
41
Global Existence, Exponential Decay...
which implies
1 2 1 2
kut k + kvt k + P 2 u + P 2 v < C ′ E1 ,
2
2
(3.14)
o
n
. Then, by Theorem 2.1, we have the global exiswhere C ′ = max 21 , 2(r+2)
r+1
tence result.
Next, we want to derive the decay rate of energy function for problem
(1.1). By multiplying first equation of system (1.1) by u and second equation
of system (1.1) by v, integrating them over Ω × [t1 , t2 ] (0 ≤ t1 ≤ t2 ) , using
integration by parts and summing up, we have
Z
Z t2
Z
Z t2
t2
2
t2
uut dx|t1 −
kut k dt +
vvt dx|t1 −
kvt k2 dt
Ω
t1
Ω
t1
Z
Z
Z t2 Z
Z t2 Z
t
t
2
2
12 2
21 2
+
uut dxdt +
uut dxdt
P v dt +
P u dt +
t1
t1
Ω
t1
t1
Ω
Z t2 Z
[uf1 (u, v) + vf2 (u, v)] dxdt.
=
Ω
t1
It follows from (2.7)
Z t2
Z t2 Z
2
E (t) dt − 2 (r + 1)
F (u, v) dxdt
t1
t1
Ω
Z t2
Z
Z
t2
2
2
= −
(uut + vvt ) dx|t1 + 2
kut k + kvt k dt −
Ω
t1
t2
t1
= A1 + A2 + A3 .
Z
(uut + uut ) dxdt
Ω
(3.15)
In what follows we will estimate A1 , A2 , A3 in (3.15). Firstly, by Hölder,
Young and Sobolev Poincare inequalities, we have
Z
1
1
1
1
(|uut | + |vvt |) dx ≤
ku (t)k2 + kut (t)k2 + kv (t)k2 + kvt (t)k2
2
2
2
2
Ω
2 1
2 1
1
1
C
C
2
2
2
2
≤
P u (t) + kut (t)k + P v (t) + kvt (t)k .
2
2
2
2
Then, by (3.13), we have
Z
A1 ≤
Ω
(|uut | + |vvt |) dx|tt21 ≤ 2C1 E (t1 ) .
(3.16)
For A2 in (3.15), applying kut k2 + kvt k2 ≤ −E ′ (t) from (2.8), we have
Z t2
A2 = 2
kut k2 + kvt k2 dt ≤ 2C2 E (t1 ) .
(3.17)
t1
42
Erhan Pis.kin et al.
We also have the following estimate
Z t2 Z
A3 =
(uut + vvt ) dxdt
t1
Ω
Z
Z
1 t2 d
1 t2 d
2
kuk dt +
kvk2 dt
=
2 t1 dt
2 t1 dt
1
1
ku (t2 )k2 − ku (t1 )k2 +
kv (t2 )k2 − kv (t1 )k2
=
2
2
2 (r + 2)
E (t1 ) = C3 E (t1 ) .
≤
r+1
(3.18)
Inserting estimates (3.16)-(3.18) into (3.15), we arrive at
2
Z
t2
t1
E (t) dt − 2 (r + 1)
Z
t2
t1
Z
Ω
F (u, v) dxdt ≤ C4 E (t1 ) ,
(3.19)
where C4 = 2C1 + 2C2 + C3 .
On the other hand, from (3.1) and (3.13), we have
Z
r+1
2(r+2)
r+2
2 (r + 1)
F (u, v) dx =
ku + vk2(r+2) + 2 kuvkr+2
r+2
Ω
r+2
r+1
21 2 12 2
η P u + P v ≤
r+2
r+1
2 (r + 2)
≤ 2η
E (0)
E (t)
r+1
which implies
2
Z
t2
t1
E (t) dt − 2 (r + 1)
≥ 2 1−η
Z
2 (r + 2)
E (0)
r+1
t2
Z
F (u, v) dxdt
t1
Ω
r+1 ! Z t2
E (t) dt.
t1
Noting that E (0) < E1 , we see
1−η
2 (r + 2)
E (0)
r+1
r+1
> 0.
Thus, combining (3.19) and (3.20), we have
r+1 ! Z t2
2 (r + 2)
E (0)
E (t) dt ≤ C4 E (t1 ) ,
2 1−η
r+1
t1
(3.20)
43
Global Existence, Exponential Decay...
that is
Z
t2
E (t) dt ≤ C4 E (t1 ) ,
r+1 2(r+2)
.
where µ = 2 1 − η r+1 E (0)
µ
(3.21)
t1
We rewrite (3.21) as
µ
Z
∞
t
E (t) dt ≤ C4 E (t)
(3.22)
for every t ∈ [0, ∞) .
Since µ > 0 from the assumption of conditions, by Lemma 2.4, we have
−1
for every t ≥
4
C4
.
µ
E (t) ≤ E (0) e1−µC4
t
The proof is completed.
Blow up of Solution
In this section, we deal with the blow up of solution of problem (1.1).
Definition 4.1. A solution (u, v) of (1.1) is called blow-up if there exists
a finite time T ∗ such that
Z
Z tZ
2
2
2
2
(4.1)
u + v dxds = ∞.
lim∗−
u + v dx +
t−→T
Let
a (t) =
0
Ω
Z
2
u +v
Ω
2
dx +
Z tZ
0
Ω
Ω
u2 + v 2 dxds, for t ≥ 0.
(4.2)
Lemma 4.1. Assume (2.1) holds and that 0 < δ ≤ 4r , then we have
Z
Z t
2
2
′′
ut + vt dx+(−4 − 8δ) E (0)+(4 + 8δ)
kut k2 + kvt k2 dt.
a (t) ≥ 4 (δ + 1)
Ω
0
Proof. From (4.2), we have
Z
′
a (t) = 2 (uut + vvt ) dx + kuk2 + kvk2 .
(4.3)
(4.4)
Ω
By (1.1) and Divergence theorem, we get
Z
Z
Z
2
2
′′
ut + vt dx + 2 (uutt + vvtt ) dx + 2 (uut + vvt ) dx
a (t) = 2
Ω
Ω
Ω
Z
1 2 1 2
u2t + vt2 dx − 2 P 2 u + P 2 v = 2
Ω
Z
+2 (r + 2) F (u, v) dx.
(4.5)
Ω
44
Erhan Pis.kin et al.
Then from (2.9) and (4.5), we have
Z
Z t
2
2
′′
ut + vt dx + (−4 − 8δ) E (0) + (4 + 8δ)
kut k2 + kvt k2 dt
a (t) ≥ 4 (δ + 1)
0
Ω Z
1 2 2
1
+4δ P 2 u + P 2 v + (2r − 8δ) F (u, v) dx.
Ω
Since 0 < δ ≤ 4r , 2r − 8δ ≥ 0, we obtain (4.3).
Lemma 4.2. Assume (2.1) holds and one of the following statements are
satisfied:
(i) E (0) < 0,
R
(ii) E (0) = 0, and Ω (u0 u1 + v0 v1 ) dx > 0,
(iii) E (0) > 0, and
K1
′
a (0) > r2 a (0) +
+ ku0 k2 + kv0 k2
4 (δ + 1)
(4.6)
holds.
Then a′ (t) > ku0 k2 + kv0 k2 for t > t∗ , where t0 = t∗ is given by (4.7) in
case (i) and t0 = 0 in cases (ii) and (iii).
Where K1 and t∗ are defined in (4.12) and (4.7), respectively.
Proof. (i) If E (0) < 0, then from (4.3), we have
a′ (t) ≥ a′ (0) − 4 (1 + 2δ) E (0) t, t ≥ 0.
Thus we get a′ (t) > ku0 k2 + kv0 k2 for t > t∗ , where
(
)
2
2
′
a
(0)
−
ku
k
+
kv
k
0
0
t∗ = max
,0 .
4 (1 + 2δ) E (0)
(4.7)
R
(ii) If E (0) = 0, and Ω (u0 u1 + v0 v1 ) dx > 0, then a′′ (t) ≥ 0 for t ≥ 0. We
have a′ (t) > ku0 k2 + kv0 k2 , t ≥ 0.
(iii) If E (0) > 0, we first note that
Z tZ
uut dxdt = kuk2 − ku0 k2 .
(4.8)
2
0
Ω
By Hölder inequality and Young inequality, we have from (4.8)
Z
Z t
2
2
2
kuk dt + kut k2 dt.
kuk ≤ ku0 k +
0
Ω
(4.9)
45
Global Existence, Exponential Decay...
Similarly,
2
2
kvk ≤ kv0 k +
Z
t
0
2
kvk dt +
Z
Ω
kvt k2 dt.
(4.10)
By Hölder inequality, Young inequality and inequalities (4.9) and (4.10), we
have
Z
Z t
2
2
2
2
′
kut k2 + kvt k2 dt. (4.11)
ut + vt dx +
a (t) ≤ a (t) + ku0 k + kv0 k +
0
Ω
Hence, by (4.3) and (4.11), we obtain
a′′ (t) − 4 (δ + 1) a′ (t) + 4 (δ + 1) a (t) + K1 ≥ 0,
where
K1 = (4 + 8δ) E (0) + 4 (δ + 1) ku0 k2 + kv0 k2 .
Let
b (t) = a (t) +
(4.12)
K1
, t > 0.
4 (δ + 1)
Then b (t) satisfies Lemma 2.5. Consequently, from (4.6), a′ (t) > ku0 k2 + kv0 k2 , t >
0, where r2 is given in Lemma 2.5.
Theorem 4.1. Assume (2.1) holds and one of the following statements
are satisfied ( for 0 < δ ≤ 4r ):
(i) E (0) < 0,
R
(ii) E (0) = 0, and Ω (u0 u1 + v0 v1 ) dx > 0,
2
(a′ (t0 )−(ku0 k2 +kv0 k2 ))
(iii) 0 < E (0) < 8 a(t )+(T −t ) ku k2 +kv k2 , and (4.6) holds.
[ 0
1
0 (
0
0 )]
Then the solution (u, v) blows up in finite time T ∗ in the sense of (4.1). In
case (i),
H (t0 )
.
(4.13)
T ∗ ≤ t0 − ′
H (t0 )
p Furthermore, if H (t0 ) < min 1, − ab , we have
p a
−b
1
,
T ∗ ≤ t0 + √ ln p a
− b − H (t0 )
−b
where
2
a = δ 2 H 2+ δ (t0 )
h
a′ (t0 ) − ku0 k2 − kv0 k2
2
b = 8δ 2 E (0) .
i
1
− 8E (0) H − δ (t0 ) > 0,
(4.14)
(4.15)
(4.16)
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Erhan Pis.kin et al.
In case (ii),
T ∗ ≤ t0 −
H (t0 )
.
H ′ (t0 )
(4.17)
In case (iii),
a 2+ 1δ δ
3δ+1
H (t0 )
√
or T ∗ ≤ t0 + 2 2δ
T∗ ≤ √
b
a
a
(
1− 1+
a 2+ 1δ
b
H (t0 )
− 2δ1 )
,
(4.18)
where a and b are given (4.15), (4.16).
Proof. Let
−δ
H (t) = a (t) + (T1 − t) ku0 k2 + kv0 k2
, for t ∈ [0, T1 ] ,
(4.19)
where T1 > 0 is a certain constant which will be specified later. Then we get
−δ−1 ′
a (t) − ku0 k2 + kv0 k2
H ′ (t) = −δ a (t) + (T1 − t) ku0 k2 + kv0 k2
1
= −δH 1+ δ (t) a′ (t) − ku0 k2 + kv0 k2 ,
(4.20)
2
H ′′ (t) = −δH 1+ δ (t) a′′ (t) a (t) + (T1 − t) ku0 k2 + kv0 k2
2
2
+δH 1+ δ (t) (1 + δ) a′ (t) − ku0 k2 + kv0 k2
2
= −δH 1+ δ (t) V (t) .
(4.21)
where
2
V (t) = a′′ (t) a (t) + (T1 − t) ku0 k2 + kv0 k2 −(1 + δ) a′ (t) − ku0 k2 + kv0 k2 .
(4.22)
For simplicity of calculation, we define
R
R
Rt
Rt
Pu = Ω u2 dx, Ru = Ω u2t dx, Qu = 0 kuk2 dt, Su = 0 kut k2 dt,
Rt
Rt
R
R
Pv = Ω v 2 dx, Rv = Ω vt2 dx, Qv = 0 kvk2 dt, Sv = 0 kvt k2 dt.
From (4.4), (4.8) and Hölder inequality, we get
Z
Z tZ
2
2
′
(uut + vvt ) dxdt
(4.23)
a (t) = 2 (uut + vvt ) dx + ku0 k + kv0 k + 2
0
Ω
Ω
p
p
p
p
Ru Pu + Qu Su + Rv Pv + Qv Sv + ku0 k2 + kv0 k2 .
≤ 2
If case (i) or (ii) holds, by (4.3) we have
a′′ (t) ≥ (−4 − 8δ) E (0) + 4 (1 + δ) (Ru + Su + Rv + Sv ) .
(4.24)
47
Global Existence, Exponential Decay...
Thus, from (4.19) and (4.22)- (4.24), we obtain
1
V (t) ≥ [(−4 − 8δ) E (0) + 4 (1 + δ) (Ru + Su + Rv + Sv )] H − δ (t)
p
2
p
p
p
−4 (1 + δ)
Ru Pu + Qu Su + Rv Pv + Qv Sv .
From (4.2)
a (t) =
Z
2
u +v
2
Ω
dx +
Z tZ
0
u2 + v 2 dxds = Pu + Pv + Qu + Qv .
Ω
From the above inequality and (4.19), we get
1
V (t) ≥ (−4 − 8δ) E (0) H − δ (t) + 4 (1 + δ)
(Ru + Su + Rv + Sv ) (T1 − t) ku0 k2 + kv0 k2 + Θ (t) ,
where
Θ (t) = (Ru + Su + Rv + Sv ) (Pu + Qu + Pv + Qv )
p
2
p
p
p
−
Ru Pu + Qu Su + Rv Pv + Qv Sv .
By the Schwarz inequality, and Θ (t) being nonnegative, we have
1
V (t) ≥ (−4 − 8δ) E (0) H − δ (t) , t ≥ t0 .
(4.25)
Therefore, by (4.21) and (4.25), we get
1
H ′′ (t) ≤ 4δ (1 + 2δ) E (0) H 1+ δ (t) , t ≥ t0 .
(4.26)
By Lemma 4.2, we know that H ′ (t) < 0 for t ≥ t0 . Multiplying (4.27) by H ′ (t)
and integrating it from t0 to t, we get
1
H ′2 (t) ≥ a + bH 2+ δ (t)
for t ≥ t0 , where a, b are defined in (4.15) and (4.16) respectively.
If case (iii) holds, by the steps of case (i), we get a > 0 if and only if
E (0) <
a′ (t0 ) − ku0 k2 + kv0 k2
2
.
8 a (t0 ) + (T1 − t0 ) ku0 k2 + kv0 k2
Then by Lemma 2.6, there exists a finite time T ∗ such that
∗
lim H (t) = 0
t−→T ∗−
and upper bound of T is estimated according to the sign of E (0) . This means
that (4.1) holds.
48
Erhan Pis.kin et al.
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