1 Introduction
In this article, we concerned with the problem
u
t
t
+
Δ
2
u
-
∫
0
t
g
(
t
-
s
)
Δ
2
u
(
s
)
d
s
=
u
p
u
,
x
∈
Ω
,
τ
>
0
u
(
x
,
t
)
=
∂
ν
u
(
x
,
t
)
=
0
,
x
∈
∂
Ω
,
t
≥
0
u
(
x
,
0
)
=
u
0
(
x
)
,
u
t
(
x
,
0
)
=
u
1
(
x
)
,
x
∈
Ω
where Ω ⊂ Rn is a bounded domain with smooth boundary ∂Ω in order that the divergence theorem can be applied. ν is the unit normal vector pointing toward the exterior of Ω and p > 0. Here, g represents the kernel of the memory term satisfying some conditions to be specified later.
In the absence of the viscoelastic term, i.e., (g = 0), we motivate our article by presenting some results related to initial-boundary value Petrovsky problem
u
t
t
+
Δ
2
u
=
f
(
u
,
u
t
)
,
x
∈
Ω
,
t
>
0
u
(
x
,
t
)
=
∂
ν
u
(
x
,
t
)
=
0
,
x
∈
∂
Ω
,
t
≥
0
u
(
x
,
0
)
=
u
0
(
x
)
,
u
t
(
x
,
0
)
=
u
1
(
x
)
,
x
∈
Ω
.
Research of global existence, blow-up and energy decay of solutions for the initial boundary value problem (1.2) has attracted a lot of articles (see 1234 and references there in).
Amroun and Benaissa 1 investigated (1.2) with f(u, ut) = b|u|p-2u-h(ut) and proved the global existence of solutions by means of the stable set method in
H
0
2
(
Ω
)
combined with the Faedo-Galerkin procedure. In 3, Messaoudi studied problem (1.2) with f(u, ut) = b|u|p-2u-a|ut|m-2ut. He proved the existence of a local weak solution and showed that this solution blows up in finite time with negative initial energy if p > m.
In the presence of the viscoelastic terms, Rivera et al. 5 considered the plate model:
u
t
t
+
Δ
2
u
-
∫
0
t
g
(
t
-
s
)
Δ
2
u
(
s
)
d
s
=
0
in a bounded domain Ω ⊂ RN and showed that the energy of solution decay exponentially provided the kernel function also decay exponentially. For more related results about the existence, finite time blow-up and asymptotic properties, we refer the reader to 5678910111213141516.
In the present article, we devote our study to problem (1.1). We will prove the existence of weak solutions under some appropriate assumptions on the function g and blow-up behavior of solutions. In order to obtain the existence of solutions, we use the Faedo-Galerkin method and to get the blow-up properties of solutions with non-positive and positive initial energy, we modify the method in 17. Estimates for the blow-up time T* are also given.
2 Preliminaries
We define the energy function related with problem (1.1) is given by
E
(
t
)
=
1
2
u
t
2
+
1
-
∫
0
t
g
(
s
)
d
s
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
-
1
p
+
2
u
p
+
2
p
+
2
,
where
(
g
⊙
v
)
(
t
)
=
∫
0
t
g
(
t
-
s
)
v
(
t
)
-
v
(
s
)
2
2
d
s
.
We denote by ∥.∥k, the Lk-norm over Ω. In particular, the L2-norm is denoted ∥.∥2. We use the familiar function spaces
H
0
2
,
H
4
and throughout this article we assume
u
0
∈
H
0
2
(
Ω
)
∩
H
4
(
Ω
)
and
u
1
∈
H
0
2
(
Ω
)
∩
L
2
(
Ω
)
.
In the sequel, we state some hypotheses and three well-known lemmas that will be needed later.
(A1) p satisfies
0
<
p
≤
∞
(
N
≤
4
)
,
0
<
p
≤
2
(
N
-
2
)
N
-
4
(
N
≥
5
)
.
(A2) g is a positive bounded C1 function satisfying g(0) > 0, and for all t > 0
1
-
∫
0
∞
g
(
t
)
d
s
=
l
>
0
,
also there exists positive constants L0, L1 such that
(A3)
-
L
0
≤
g
′
(
t
)
≤
0
,
0
≤
g
″
(
t
)
≤
L
1
.
Lemma 1 (Sobolev-Poincare's inequality). Let p be a number that satisfies (A1), then there is a constant C* = C(Ω, p) such that
u
p
≤
C
*
Δ
u
2
,
u
∈
H
0
2
(
Ω
)
Lemma 2 4. Let δ > 0 and B(t) ∈ C2(0, ∞) be a nonnegative function satisfying
B
″
(
t
)
-
4
(
δ
+
1
)
B
′
(
t
)
+
4
(
δ
+
1
)
B
(
t
)
≥
0
.
If
B
′
(
0
)
>
r
2
B
(
0
)
+
K
0
,
with
r
2
=
2
(
δ
+
1
)
-
2
δ
(
δ
+
1
)
, then B'(t) > K0 for t > 0, where K0 is a constant.
Lemma 3 4. If Y(t) is a non-increasing function on [t0, ∞) and satisfies the differential inequality
Y
′
(
t
)
2
≥
a
+
b
Y
(
t
)
2
+
δ
-
1
f
o
r
t
≥
t
0
≥
0
,
where a > 0, δ > 0 and b ∈ R, then there exists a finite time T* such that
lim
t
→
T
*
-
Y
(
t
)
=
0
.
Upper bounds for T* is estimated as follows:
(i) If b < 0, then
T
*
≤
t
0
+
1
-
b
l
n
-
a
b
-
a
b
-
Y
(
t
0
)
.
(ii) If b = 0, then
T
*
≤
t
0
+
Y
(
t
0
)
Y
′
(
t
0
)
.
(iii) If b > 0, then
T
*
≤
Y
(
t
0
)
a
,
or
T
*
≤
t
0
+
2
3
δ
+
1
2
δ
c
δ
a
1
-
[
1
+
c
Y
(
t
0
)
]
-
1
2
δ
,
where c
=
a
b
2
+
1
δ
.
3 Existence of solutions
In this section, we are going to obtain the existence of weak solutions to the problem (1.1) using Faedo-Galerkin's approximation.
Theorem 1 Let the assumptions (A1)-(A3) hold. Then there exists at least a solution u of (1.1) satisfying
u
∈
L
∞
(
0
,
∞
;
H
0
2
(
Ω
)
∩
H
4
(
Ω
)
)
,
u
′
∈
L
∞
(
0
,
∞
;
H
0
2
(
Ω
)
∩
L
2
(
Ω
)
)
,
u
″
∈
L
∞
(
0
,
∞
;
L
2
(
Ω
)
)
and
u
(
x
,
t
)
→
u
0
(
x
)
i
n
H
0
2
(
Ω
)
∩
H
4
(
Ω
)
u
′
(
x
,
t
)
→
u
1
(
x
)
i
n
H
0
2
(
Ω
)
∩
L
2
(
Ω
)
as t → 0.
Proof We choose a basis {ωk} (k = 1, 2, ...) in
H
0
2
(
Ω
)
∩
H
4
(
Ω
)
which is orthonormal in L2(Ω) and ωk being the eigenfunctions of biharmonic operator subject to the homogeneous Dirichlet boundary condition.
Let Vm be the subspace of H02(Ω) ∩ H4(Ω)
generated by the first m vectors. Define
u
m
(
t
)
=
∑
k
=
1
m
d
m
k
(
t
)
ω
k
,
where um(t) is the solution of the following Cauchy problem
u
m
″
(
t
)
,
ω
k
+
(
Δ
u
m
(
t
)
,
Δ
ω
k
)
-
∫
0
t
(
t
-
s
)
(
Δ
u
m
(
s
)
,
Δ
ω
k
)
d
s
-
u
m
(
t
)
p
u
m
(
t
)
,
ω
k
=
0
∀
k
=
1
,
m
.
with the initial conditions (when m → ∞)
u
m
(
0
)
=
∑
k
=
1
m
(
u
m
(
0
)
,
ω
k
)
ω
k
→
u
0
i
n
H
0
2
(
Ω
)
∩
H
4
(
Ω
)
u
m
′
(
0
)
=
∑
k
=
1
m
u
m
′
(
0
)
,
ω
k
ω
k
→
u
1
i
n
H
0
2
(
Ω
)
∩
L
2
(
Ω
)
The approximate systems (3.3) and (3.4) are the normal one of differential equations which has a solution in [0, Tm) for some Tm > 0. The solution can be extended to the [0, T] for any given T > 0 by the first estimate below.
First estimation. Substituting
u
m
′
(
t
)
instead of ωk in (3.3), we find
d
d
t
1
2
u
m
′
2
+
1
2
Δ
u
m
2
-
u
m
p
+
2
p
+
2
p
+
2
-
∫
0
t
g
(
t
-
s
)
(
Δ
u
m
(
s
)
,
Δ
u
m
′
(
t
)
)
d
s
=
0
.
Simple calculation similar to 11 yield
-
∫
0
t
g
(
t
-
s
)
(
Δ
u
m
(
s
)
,
Δ
u
m
′
(
t
)
)
d
s
=
-
∫
0
t
g
(
t
-
s
)
∫
Ω
Δ
u
m
(
t
)
Δ
u
m
′
(
t
)
d
x
d
s
-
∫
0
t
g
(
t
-
s
)
∫
Ω
(
Δ
u
m
(
s
)
-
Δ
u
m
(
t
)
)
Δ
u
m
′
(
t
)
d
x
d
s
=
1
2
∫
0
t
g
(
t
-
s
)
d
d
t
Δ
u
m
(
s
)
-
Δ
u
m
(
t
)
2
d
s
-
1
2
∫
0
t
g
(
t
-
s
)
d
d
t
Δ
u
m
(
t
)
2
d
s
=
1
2
d
d
t
(
g
⊙
Δ
u
m
)
(
t
)
-
1
2
(
g
′
⊙
Δ
u
m
)
(
t
)
-
1
2
d
d
t
∫
0
t
g
(
s
)
d
s
Δ
u
m
(
t
)
2
d
s
+
1
2
g
(
t
)
Δ
u
m
(
t
)
2
.
Combining (3.5) and (3.6), we find
d
d
t
1
2
u
m
′
2
+
1
2
1
-
∫
0
t
g
(
s
)
d
s
Δ
u
m
2
+
1
2
(
g
⊙
Δ
u
m
)
(
t
)
-
u
m
p
+
2
p
+
2
p
+
2
=
1
2
(
g
′
⊙
Δ
u
m
)
(
t
)
-
1
2
g
(
t
)
Δ
u
m
(
t
)
2
,
integrating (3.7) over (0, t) and using assumption (A3) we infer that
u
m
′
2
+
Δ
u
m
2
+
(
g
⊙
Δ
u
m
)
(
t
)
-
u
m
p
+
2
p
+
2
≤
C
1
,
where C1 is a positive constant depending only on ∥u0∥, ∥u1∥, p, and l. It follows from (3.8) that
{
u
m
}
i
s
u
n
i
f
o
r
m
l
y
b
o
u
n
d
e
d
i
n
L
∞
(
0
,
T
;
H
0
2
(
Ω
)
)
{
u
m
′
}
i
s
u
n
i
f
o
r
m
l
y
b
o
u
n
d
e
d
i
n
L
∞
(
0
,
T
;
L
2
(
Ω
)
)
Second estimation. Differentiating (3.3) with respect to t, we get
u
m
‴
(
t
)
,
ω
k
+
(
Δ
u
m
′
(
t
)
,
Δ
ω
k
)
-
∫
0
t
g
′
(
t
-
s
)
(
Δ
u
m
(
s
)
,
Δ
ω
k
)
d
s
-
g
(
0
)
(
Δ
u
m
(
t
)
,
Δ
ω
k
)
-
(
p
+
1
)
u
m
(
t
)
p
u
m
′
(
t
)
,
ω
k
=
0
.
If we substitute
u
m
″
(
t
)
instead of ωk in (3.10), it holds that
d
d
t
1
2
u
m
″
2
+
1
2
Δ
u
m
′
2
-
d
d
t
∫
0
t
g
′
(
t
-
s
)
Δ
u
m
(
s
)
,
Δ
u
m
′
(
t
)
d
s
+
∫
0
t
g
″
(
t
-
s
)
Δ
u
m
(
s
)
,
Δ
u
m
′
(
t
)
d
s
+
g
′
(
0
)
Δ
u
m
(
t
)
,
Δ
u
m
′
(
t
)
-
g
(
0
)
d
d
t
Δ
u
m
(
t
)
,
Δ
u
m
′
(
t
)
+
g
(
0
)
Δ
u
m
′
(
t
)
,
Δ
u
m
′
(
t
)
-
(
p
+
1
)
u
m
(
t
)
p
u
m
′
(
t
)
,
u
m
″
(
t
)
=
0
.
Since H2(Ω) ↪ L2p+2(Ω), using Lemma 2, Hölder and Young's inequalities and (3.8)
(
p
+
1
)
u
m
(
t
)
p
u
m
′
(
t
)
,
u
m
″
(
t
)
≤
(
p
+
1
)
u
m
(
t
)
2
p
+
2
p
.
u
m
′
(
t
)
2
p
+
2
.
u
m
″
(
t
)
2
≤
C
(
γ
)
Δ
u
m
′
(
t
)
2
+
γ
u
m
″
(
t
)
2
.
Combining the relations (3.11), (3.12) and integrating over (0, t) for all t ∈ [0, T] with arbitrary fixed T, we obtain
1
2
u
m
″
2
+
1
2
Δ
u
m
′
2
≤
1
2
u
m
″
(
0
)
2
+
∫
0
t
g
′
(
t
-
s
)
(
Δ
u
m
(
s
)
,
Δ
u
m
′
(
t
)
)
d
s
+
1
2
Δ
u
m
′
(
0
)
2
-
∫
0
t
∫
0
τ
g
″
(
τ
-
s
)
(
Δ
u
m
(
s
)
,
Δ
u
m
′
(
τ
)
)
d
s
d
τ
-
g
′
(
0
)
∫
0
t
Δ
u
m
(
s
)
,
Δ
u
m
′
(
s
)
+
g
(
0
)
Δ
u
m
(
t
)
,
Δ
u
m
′
(
t
)
-
g
(
0
)
Δ
u
m
(
0
)
,
Δ
u
m
′
(
0
)
-
g
(
0
)
∫
0
t
Δ
u
m
′
(
s
)
2
d
s
+
C
(
γ
)
∫
0
t
Δ
u
m
′
(
s
)
2
d
s
+
γ
∫
0
t
u
m
″
(
s
)
2
d
s
.
From (3.4) and (3.8), we deduce that
|
1
2
Δ
u
m
′
(
0
)
2
-
g
(
0
)
(
Δ
u
m
(
0
)
,
Δ
u
m
′
(
0
)
)
|
≤
L
2
,
where L2 is a positive constant independent of m. In the following, we find the upper bound for
u
m
″
(
0
)
2
. Again we substitute um″(t)
instead of ωk in (3.3), and choosing t = 0, we arrive at
u
m
″
(
0
)
,
u
m
″
(
0
)
+
Δ
u
m
(
0
)
,
Δ
u
m
″
(
0
)
-
u
m
(
0
)
p
u
m
(
0
)
,
u
m
″
(
0
)
=
0
,
which combined with the Green's formula imply
u
m
″
(
0
)
2
+
Δ
2
u
m
(
0
)
,
u
m
″
(
0
)
-
u
m
(
0
)
p
u
m
(
0
)
,
u
m
″
(
0
)
=
0
.
By using (A1), (3.4) and Young's inequality, we deduce that
u
m
″
(
0
)
≤
L
3
,
where L3 > 0 is a constant independent of m.
Owing to (3.8), (3.5) and Young's inequality with (A3), we deduce that
∫
0
t
g
′
(
t
-
s
)
(
Δ
u
m
(
s
)
,
Δ
u
m
′
(
t
)
)
d
s
)
=
Δ
u
m
′
(
t
)
,
∫
0
t
g
′
(
t
-
s
)
Δ
u
m
(
s
)
d
s
≤
γ
Δ
u
m
′
(
t
)
2
+
1
4
γ
∫
Ω
∫
0
t
g
′
(
t
-
s
)
Δ
u
m
(
s
)
d
s
2
d
x
≤
γ
Δ
u
m
′
(
t
)
2
+
L
0
2
4
γ
∫
0
t
Δ
u
m
(
s
)
2
d
s
≤
γ
Δ
u
m
′
(
t
)
2
+
L
4
(
T
)
,
-
∫
0
t
∫
0
τ
g
″
(
τ
-
s
)
(
Δ
u
m
(
s
)
,
Δ
u
m
′
(
τ
)
)
d
s
d
τ
=
∫
0
t
Δ
u
m
′
(
τ
)
,
∫
0
τ
g
″
(
τ
-
s
)
Δ
u
m
(
s
)
d
s
d
τ
≤
1
2
∫
0
t
Δ
u
m
′
(
s
)
2
d
s
+
1
2
∫
0
t
∫
Ω
∫
0
τ
g
″
(
τ
-
s
)
Δ
u
m
(
s
)
d
s
2
d
x
d
τ
≤
1
2
∫
0
t
Δ
u
m
′
(
s
)
2
d
s
+
T
L
1
2
2
∫
0
t
Δ
u
m
(
s
)
2
d
s
≤
1
2
∫
0
t
Δ
u
m
′
(
s
)
2
d
s
+
L
5
(
T
)
,
-
g
′
(
0
)
∫
0
t
(
Δ
u
m
(
s
)
,
Δ
u
m
′
(
s
)
)
d
s
≤
L
0
∫
0
t
Δ
u
m
′
(
s
)
2
d
s
+
L
6
(
T
)
,
and
g
(
0
)
(
Δ
u
m
(
t
)
,
Δ
u
m
′
(
t
)
)
≤
γ
Δ
u
m
′
(
t
)
2
+
L
7
(
γ
)
.
Now we choose γ > 0 small enough and combining (A3), (3.8), (3.13), (3.14), and (3.16)-(3.20), we get
1
2
u
m
″
2
+
1
2
Δ
u
m
′
2
≤
L
8
∫
0
t
u
m
″
(
s
)
2
d
s
+
∫
0
t
Δ
u
m
′
(
s
)
2
d
s
+
L
9
.
By using Gronwall's lemma we arrive at
1
2
u
m
″
2
+
1
2
Δ
u
m
′
2
≤
L
10
,
for all t ∈ [0, T], and L10 is a positive constant independent of m. Estimate (3.22) implies
{
u
m
″
}
i
s
u
n
i
f
o
r
m
l
y
b
o
u
n
d
e
d
i
n
L
∞
(
0
,
T
;
L
2
(
Ω
)
)
{
u
m
′
}
i
s
u
n
i
f
o
r
m
l
y
b
o
u
n
d
e
d
i
n
L
∞
(
0
,
T
;
H
0
2
(
Ω
)
)
By attention to (3.9) and (3.23), there exists a subsequence {ui} of {um} and a function u such that
u
i
⇀
u
w
e
a
k
l
y
s
t
a
r
i
n
L
∞
(
0
,
T
;
H
0
2
(
Ω
)
)
u
i
′
⇀
u
′
w
e
a
k
l
y
s
t
a
r
i
n
L
∞
(
0
,
T
;
H
0
2
(
Ω
)
)
u
i
″
⇀
u
″
w
e
a
k
l
y
s
t
a
r
i
n
L
∞
(
0
,
T
;
L
2
(
Ω
)
)
By Aubin-Lions compactness lemma 18, it follows from (3.24) that
u
i
→
u
s
t
r
o
n
g
l
y
i
n
C
(
[
0
,
T
]
;
H
0
2
(
Ω
)
)
u
i
′
→
u
s
t
r
o
n
g
l
y
i
n
C
(
[
0
,
T
]
)
;
L
2
(
Ω
)
)
In the sequel we will deal with the nonlinear term. By (3.9) and Sobolev embedding theorem, we obtain
u
m
p
u
m
i
s
u
n
i
f
o
r
m
l
y
b
o
u
n
d
e
d
i
n
L
∞
(
0
,
T
;
L
2
(
Ω
)
)
,
and therefore we can extract a subsequence {ui} of {um} such that
u
i
p
u
i
⇀
u
p
u
w
e
a
k
l
y
s
t
a
r
i
n
L
∞
(
0
,
T
;
L
2
(
Ω
)
)
.
Applying (3.24), (3.27) and letting i → ∞ in (3.3), we see that u satisfies the equation. For the initial conditions by using (3.4), (3.25) and the simple inequality
u
-
u
0
H
0
2
(
Ω
)
≤
u
-
u
i
H
0
2
(
Ω
)
+
u
i
-
u
i
(
0
)
H
0
2
(
Ω
)
+
u
i
(
0
)
-
u
0
H
0
2
(
Ω
)
,
we get the first initial condition immediately. In the similar way, we can show the second initial condition and the proof is complete.
4 Blow-up of solutions
In this section, we study blow-up property of solutions with non-positive initial energy as well as positive initial energy, and estimate the lifespan of solutions. For this purpose, we assume that g is positive and C1 function satisfying
(A4)
g
(
0
)
>
0
,
g
′
(
s
)
≤
0
,
1
-
∫
0
∞
g
(
s
)
d
s
=
l
>
0
,
and we make the following extra assumption on g
(A5)
∫
0
∞
g
(
s
)
d
s
<
p
1
+
p
.
From (2.1), (A4) and Lemma 1, we have
E
(
t
)
≥
1
2
1
-
∫
0
t
g
(
s
)
d
s
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
-
1
p
+
2
u
p
+
2
p
+
2
≥
1
2
l
Δ
u
2
+
g
⊙
Δ
u
)
(
t
)
-
C
1
p
+
2
l
p
+
2
2
p
+
2
Δ
u
p
+
2
≥
G
l
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
,
t
≥
0
,
where G
(
λ
)
=
1
2
λ
2
-
C
1
p
+
2
p
+
2
λ
p
+
2
,
C
1
=
C
*
l
. It is easy to verify that G(λ) has a maximum at
λ
1
=
C
1
-
p
+
2
p
and the maximum value is
E
1
=
p
2
p
+
4
C
1
-
2
p
+
4
p
.
Lemma 4 Let (A4) hold andu be a local solution of (1.1). Then E(t) is a non-increasing function on [0, T] and
d
d
t
E
(
t
)
=
1
2
(
g
′
⊙
Δ
u
)
(
t
)
-
1
2
g
(
t
)
Δ
u
2
≤
0
,
for almost every t ∈ [0, T].
Proof Multiplying (1.1) by ut, integrating over Ω, and finally integrating by parts, we obtain (4.2) for any regular solution. Then by density arguments, we have the result.
Lemma 5 Let (A4) hold and u be a local solution of (1.1) with initial data satisfying E(0) < E1 and
l
1
2
Δ
u
0
>
λ
1
. Then there exists λ2 > λ1 such that
l
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
≥
λ
2
2
.
Proof See Li and Tsai 11.
The choice of the functional is standard (see 19)
ψ
(
t
)
=
u
2
.
It is clear that
ψ
′
(
t
)
=
2
(
u
,
u
t
)
,
and from (1.1)
ψ
″
(
t
)
=
2
u
t
2
-
2
Δ
u
2
+
2
u
p
+
2
p
+
2
+
2
∫
0
t
g
(
t
-
s
)
(
Δ
u
(
t
)
,
Δ
u
(
s
)
)
d
s
.
Lemma 6 Let u be a solution of (1.1) and (A4), (A5) hold, then we have
ψ
″
(
t
)
-
(
4
+
p
)
∫
Ω
u
t
2
d
x
≥
m
(
l
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
)
-
(
4
+
2
p
)
E
(
0
)
,
where m
=
(
1
+
p
)
-
1
l
>
0
.
Proof Using the Hölder and Young's inequalities, we arrive at
∫
0
t
g
(
t
-
s
)
(
Δ
u
(
t
)
,
Δ
u
(
s
)
)
d
s
≥
-
1
2
(
g
⊙
Δ
u
)
(
t
)
+
1
2
∫
0
t
g
(
s
)
d
s
Δ
u
2
+
∫
0
t
g
(
s
)
d
s
Δ
u
2
,
therefore (4.6) becomes
ψ
″
(
t
)
-
(
4
+
p
)
u
t
2
≥
-
(
2
+
p
)
u
t
2
-
2
Δ
u
2
-
(
g
⊙
Δ
u
)
(
t
)
+
∫
0
t
g
(
s
)
d
s
Δ
u
2
+
2
u
p
+
2
p
+
2
.
Then, using (4.2), we obtain
ψ
″
(
t
)
-
(
4
+
p
)
u
t
2
≥
-
(
4
+
2
p
)
E
(
0
)
+
p
Δ
u
2
+
(
1
+
p
)
(
g
⊙
Δ
u
)
(
t
)
-
(
1
+
p
)
∫
0
t
g
(
s
)
d
s
Δ
u
2
-
(
2
+
p
)
∫
0
t
(
g
′
⊙
Δ
u
)
(
s
)
d
s
,
and so by (2.5) and (A5), we deduce
ψ
″
(
t
)
-
(
4
+
p
)
u
t
2
≥
-
(
4
+
2
p
)
E
(
0
)
+
(
p
-
(
1
+
p
)
(
1
-
l
)
)
Δ
u
2
+
(
1
+
p
)
(
g
⊙
Δ
u
)
(
t
)
,
if we set m
:
=
(
1
+
p
)
-
1
l
then inequality (4.8) yields the desired result.
Consequently, we have the following result.
Lemma 7 Assume that (A4) and (A5) hold. u be a local solution of (1.1) and that either one of the following four conditions is satisfied:
(i) E(0) < 0
(ii) E(0) = 0 and ψ'(0) > 0
(iii) 0
<
E
(
0
)
<
m
p
E
1
and
l
1
2
Δ
u
0
>
λ
1
(iv)
m
p
E
1
≤
E
(
0
)
and
ψ
′
(
0
)
>
r
2
ψ
(
0
)
+
(
4
+
2
p
)
E
(
0
)
4
+
p
.
Then ψ' (t) > 0 for t > t*, where
in case (i)
t
*
=
max
0
,
ψ
′
(
0
)
(
4
+
2
p
)
E
(
0
)
,
in cases (ii), (iv)
t
*
=
0
,
and in case (iii)
t
*
=
max
0
,
-
ψ
′
(
0
)
(
4
+
2
p
)
m
p
E
1
-
E
(
0
)
.
Proof Suppose that condition (i) is satisfied. Then from (4.5), we have
ψ
′
(
t
)
≥
ψ
′
(
0
)
-
(
4
+
2
p
)
E
(
0
)
t
,
t
≥
0
.
Thus ψ'(t) > 0 for t > t*, and it is easy to see that t* satisfies (4.9).
If E(0) = 0, then by using (4.3) we have ψ" (t) ≥ 0, and since ψ'(0) > 0 we arrive at
ψ
′
(
t
)
>
0
,
f
o
r
t
>
0
.
If 0 < E(0) < m p E1
and l 1 2 Δu0 > λ1
then by Lemma 4, we see that
m
(
l
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
)
-
(
4
+
2
p
)
E
(
0
)
≥
m
λ
2
2
-
(
4
+
2
p
)
E
(
0
)
>
m
4
+
2
p
p
E
1
-
(
4
+
2
p
)
E
(
0
)
=
(
4
+
2
p
)
m
p
E
1
-
E
(
0
)
.
Thus from (4.5), we have
ψ
″
(
t
)
≥
m
(
l
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
)
-
(
4
+
2
p
)
E
(
0
)
>
(
4
+
2
p
)
m
p
E
1
-
E
(
0
)
>
0
,
and integrating (4.12) from 0 to t gives
ψ
′
(
t
)
>
0
,
f
o
r
t
≥
t
*
,
where t* satisfies (4.11).
Let m p E1 ≤ E(0)
, this assumption causes that
ψ
″
(
t
)
-
(
4
+
p
)
u
t
2
+
(
4
+
2
p
)
E
(
0
)
≥
0
,
and by using Hölder and Young's inequalities, we get
u
t
2
≥
ψ
′
(
t
)
-
ψ
(
t
)
,
thus
ψ
″
(
t
)
-
(
4
+
p
)
ψ
′
(
t
)
+
(
4
+
p
)
ψ
(
t
)
+
(
4
+
2
p
)
E
(
0
)
≥
0
.
We see that the hypotheses of Lemma 2 are fulfilled with
δ
=
p
4
a
n
d
B
(
t
)
=
ψ
(
t
)
+
(
4
+
2
p
)
E
(
0
)
4
+
p
and the conclusion of Lemma 2.2 gives us
ψ
′
(
t
)
>
0
,
f
o
r
t
>
0
.
Therefore the proof is complete.
To estimate the life-span of ψ(t), we define the following functional
Y
(
t
)
=
ψ
(
t
)
-
p
4
,
f
o
r
t
≥
0
.
Then we have
Y
′
(
t
)
=
p
4
Y
(
t
)
1
+
4
p
ψ
′
(
t
)
,
Y
″
(
t
)
=
-
p
4
Y
(
t
)
1
+
8
p
ψ
″
(
t
)
ψ
(
t
)
-
1
+
p
4
(
ψ
′
(
t
)
)
2
.
Using (4.4)-(4.6) and exploiting Holder's inequality on ψ'(t), we get
ψ
″
(
t
)
ψ
(
t
)
-
1
+
p
4
(
ψ
′
(
t
)
)
2
≥
[
(
l
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
)
-
(
4
+
2
p
)
E
(
0
)
+
(
4
+
p
)
u
t
2
]
ψ
(
t
)
-
4
1
+
p
4
u
t
2
ψ
(
t
)
=
[
(
l
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
)
-
(
4
+
2
p
)
E
(
0
)
]
Y
(
t
)
-
4
p
.
Utilizing the last inequality into (4.16) yields
Y
″
(
t
)
≤
-
p
4
[
(
l
Δ
u
2
+
(
g
⊙
Δ
u
)
(
t
)
)
-
(
4
+
2
p
)
E
(
0
)
]
Y
(
t
)
1
+
4
p
.
Now we should assume different values for initial energy E(0).
(1) At first if E(0) ≤ 0 then from (4.17) we have
Y
″
(
t
)
≤
p
4
(
4
+
2
p
)
E
(
0
)
Y
(
t
)
1
+
4
p
,
on the other hand by Lemma 7, Y'(t) < 0 for t > t*. Multiplying (4.18) by Y'(t) and integrating from t* to t, we deduce that
Y
′
(
t
)
2
≥
α
+
β
Y
(
t
)
2
+
4
p
f
o
r
t
≥
t
*
,
where
α
=
p
2
16
Y
(
t
*
)
2
+
8
p
ψ
′
(
t
*
)
2
-
8
E
(
0
)
Y
(
t
*
)
-
4
p
>
0
,
and
β
=
p
2
2
E
(
0
)
.
Then the hypotheses of Lemma 3 are fulfilled with δ
=
p
4
,
t
0
=
t
*
and using the conclusion of Lemma 3, there exists a finite time T* such that
lim
t
→
T
*
-
Y
(
t
)
=
0
, i.e., in this case some solutions blow up in finite time T*.
(2) If 0 < E(0) < m p E1
, then from (4.17) and (4.12) we have
Y
″
(
t
)
≤
-
p
4
(
4
+
2
p
)
m
p
E
1
-
E
(
0
)
Y
(
t
)
1
+
4
p
.
Then using the same arguments as in (1), we get
Y
′
(
t
)
2
≥
α
1
+
β
1
Y
(
t
)
2
+
4
p
f
o
r
t
≥
t
*
,
where
α
1
=
p
2
16
Y
(
t
*
)
2
+
8
p
(
ψ
′
(
t
*
)
2
+
8
m
p
E
1
-
E
(
0
)
Y
(
t
*
)
-
4
p
)
>
0
,
and
β
1
=
p
2
2
E
(
0
)
-
m
p
E
1
.
Thus by Lemma 3, there exists a finite time T* such that
lim
t
→
T
*
-
ψ
(
t
)
=
∞
.
(3) m p E1 ≤ E(0)
. In this case, it is easy to see that by using (4.19) and (4.20) into discussion in part (1), we obtain
α
>
0
i
f
a
n
d
o
n
l
y
i
f
E
(
0
)
<
ψ
′
(
t
*
)
2
8
ψ
(
t
*
)
.
Hence, Lemma 3 yields the blow-up property in this case.
Therefore, we proved the following theorem.
Theorem 2 Assume that (A4) and (A5) hold. u be a local solution of (1.1) and that either one of the following four conditions is satisfied:
(i) E(0) < 0
(ii) E(0) = 0 and ψ'(0) > 0
(iii) 0 < E(0) < m p E1
and l 1 2 Δu0 > λ1
(iv) m p E1 ≤ E(0)
and ψ′(0) > r2 ψ(0) + (4 + 2p)E(0) 4 + p
holds.
Then the solution u blows up at finite time T*. Moreover, the upper bounds for T* can be estimated according to the sign of E(0):
in case (i)
T
*
≤
t
*
-
Y
(
t
*
)
Y
′
(
t
*
)
.
Furthermore, if Y
(
t
*
)
<
min
{
1
,
α
-
β
}
, then
T
*
≤
t
*
+
1
-
β
l
n
α
-
β
α
-
β
-
Y
(
t
*
)
in cases (ii)
T
*
≤
t
*
-
Y
(
t
*
)
Y
′
(
t
*
)
o
r
T
*
≤
t
*
+
Y
(
t
*
)
α
in case (iii)
T
*
≤
t
*
-
Y
(
t
*
)
Y
′
(
t
*
)
.
Furthermore, if Y(t*) <min{1, α
-β}
, then
T
*
≤
t
*
+
1
-
β
1
l
n
α
1
-
β
1
α
1
-
β
1
-
Y
(
t
*
)
and in case (iv)
T
*
≤
Y
(
t
*
)
α
o
r
T
*
≤
t
*
+
2
3
p
+
4
2
p
p
c
4
α
1
-
(
1
+
c
Y
(
t
*
)
)
-
2
p
,
where d
=
β
α
p
p
+
8
. Here α, β, α1, and β1 are given in (4.19)-(4.22), respectively. Note that each t* in the above cases satisfy the same case in Lemma 7.
Competing interests
The authors declare that they have no competing interests.
Authors' contributions
The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.