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C. R. Mecanique 346 (2018) 402–407Contents lists available at ScienceDirect
Comptes Rendus Mecanique
www.sciencedirect.com
Blow-up of solutions to a quasilinear wave equation for high 
initial energy
Fang Li a, Fang Liu b,∗
a School of Mathematics, Jilin University, Changchun 130012, PR China
b School of Basic Sciences, Changchun University of Technology, Changchun 130012, PR China
a r t i c l e i n f o a b s t r a c t
Article history:
Received 13 January 2018
Accepted after revision 1 March 2018
Available online 6 March 2018
Keywords:
Viscoelastic hyperbolic equation
Energy estimate method
High initial energy
This paper deals with blow-up solutions to a nonlinear hyperbolic equation with variable 
exponent of nonlinearities. By constructing a new control function and using energy 
inequalities, the authors obtain the lower bound estimate of the L2 norm of the solution. 
Furthermore, the concavity arguments are used to prove the nonexistence of solutions; at 
the same time, an estimate of the upper bound of blow-up time is also obtained. This 
result extends and improves those of [1,2].
© 2018 Published by Elsevier Masson SAS on behalf of Académie des sciences.
1. Introduction
In this paper, we consider the following quasilinear hyperbolic problem⎧⎪⎨
⎪⎩
utt − div(|∇u|p(x)−2∇u) − ut = |u|q(x)−2u, (x, t) ∈  × (0, T ) := Q T
u(x, t) = 0, (x, t) = ∂ × (0, T ) := T
u(x,0) = u0(x), ut(x,0) = u1(x), x ∈ 
(1.1)
where  ⊂ R N (N  1) is a bounded domain, ∂ is Lipschitz continuous. It will be assumed throughout the paper that the 
exponents p(x), q(x) satisfy the following conditions.
2  p−  p(x)  p+ < ∞, 1 < q−  q(x)  q+ < ∞
Problem (1.1) models many physical problems such as viscoelastic fluids, electro-rheological fluids, processes of filtration 
through a porous media, fluids with temperature-dependent viscosity, etc. The interested readers may refer to [3–5] and the 
references therein. When p, q are fixed constants, many authors discussed the existence of solutions, finite-time blow-up of 
solutions for low initial energy and arbitrarily high initial energy as well as some estimates of a lower bound for blow-up 
times – the interested readers may refer to [6–14]. When p, q are continuous functions, S.N. Antontsev in [1,15] discussed 
the blowing-up properties of solutions to the initial and homogeneous boundary value problem of quasilinear wave equa-
tions involving the p(x, t)-Laplacian operator and a negative initial energy. Guo–Gao of [2] proved that the solution blew 
up in finite time for positive initial energy. Later, Guo, in [16], applied the interpolation inequality and energy inequalities 
to obtain an estimate of the lower bound for the blow-up time when the source is super-linear. In addition, Messaoudi and 
* Corresponding author.
E-mail addresses: lf8583 @jlu .edu .cn (F. Li), fairfang @sina .com (F. Liu).https://doi.org/10.1016/j.crme.2018.03.002
1631-0721/© 2018 Published by Elsevier Masson SAS on behalf of Académie des sciences.
F. Li, F. Liu / C. R. Mecanique 346 (2018) 402–407 403Talahmeh in [17,18] discussed blow-up properties of solutions to the nonlinear wave equations with weak damping terms 
and a p(.)-Laplacian operator. However, there are few works that discuss blow-up properties of solutions for high initial 
energy. In fact, dealing with such problems, one has to face some difficulties:
• how can one give an estimate of the lower bound of the norm ‖∇u‖Lp(.)()? In fact, due to the presence of strong 
damping term ut , the technique used in [17] is not applicable, so, we have to find some new methods or techniques;
• how can one establish the quantitative relationship between the term ∫

|u|q(.)dx and the initial energy? Since the 
initial energy is arbitrary, we can not apply the technique used in [2] to give the quantitative relation between 
∫

|u|q(.)dx
and E(0).
In this paper, we construct a new control function and apply energy estimate inequalities to bypass the first difficulty 
above. Furthermore, by modifying the functional constructed in [2] and utilizing the quantitative relationship between the 
term 
∫

|u|2dx and the initial energy, we prove that the solution blows up in finite time for arbitrary positive initial energy. 
In particular, it is worth pointing out that our results extend and improve those of [1,2]. For the existence of solutions, we 
may refer to [1,2,15].
Define the energy functional as the following:
E(t) = 1
2
∫

|ut |2dx +
∫

1
p(x)
|∇u|p(x)dx −
∫

1
q(x)
|u|q(x)dx
First of all, due to pt = qt = 0 and a(x, t) = b(x, t) = 1, we follow the line of the proof of Lemma 2.1 in [1] or Lemma 1.1 
of [2] to obtain the energy functional E(t), which satisfies the following identity.
Lemma 1.1. Suppose that u ∈ Lq(x)(Q T ) ∩ L∞(0, T ; W 1,p(x)0 ()), ut ∈ L2(0, T ; H1()) is a solution to Problem (1.1), then E(t)
satisfies the following identity:
E(t) +
t∫
0
∫

|∇us|2dx ds = E(0), t  0 (1.2)
if the following conditions are fulfilled
⎧⎪⎪⎪⎪⎨
⎪⎪⎪⎪⎩
2  p−  p(x)  p+ < ∞, 1 < q−  q(x)  q+ < p−∗
p−∗ =
{
Np−
N−p− , if 1 < p
− < N
∞, if p−  N
u0 ∈ W 1,p(x)0 () ∩ Lq(x)(), u1 ∈ L2()
(1.3)
Let λ1 be the first eigenvalue of the following problem
{
−ψ = λψ, x ∈ 
ψ = 0, x ∈ ∂
and set B1 = min{λ1, p+(q−+2)2(q−−p+) }.
Lemma 1.2. If all the conditions of Lemma 1.1 remain true and q− > p+ , then the solution to Problem (1.1) satisfies the following 
inequality
∫

uutdx  eM0t
[∫

u0u1dx − q
−
M0
E(0) − ||
M0
]
+ q
−
M0
E(t) + ||
M0
, t > 0 (1.4)
where M0 = 4(q−−p+)q− B1+ − > 0.p (B1+4q )
404 F. Li, F. Liu / C. R. Mecanique 346 (2018) 402–407Proof. Define F (t) = ∫

uut dx. Then the definition of the weak solution shows that
F ′(t) =
∫

|ut |2dx +
∫

uuttdx
=
∫

|ut |2dx +
∫

[
− |∇u|p(x) − ∇u∇ut + |u|q(x)
]
dx

∫

|ut |2dx −
∫

|∇u|p(x)dx −
∫

∇u∇ut dx + q−
[∫

[ |ut |2
2
+ |∇u|
p(x)
p(x)
]
dx − E(t)
]
= q
− + 2
2
∫

|ut |2dx +
∫

q− − p(x)
p(x)
|∇u|p(x)dx − q−E(t) −
∫

∇u∇ut dx
(1.5)
The Cauchy–Schwarz inequality implies that
∣∣∣ ∫

∇u∇utdx
∣∣∣  (q− − p+)B1
p+(B1 + 4q−)
∫

|∇u|2dx + (B1 + 4q
−)p+
4(q− − p+)B1
∫

|∇ut |2dx (1.6)
Combining (1.5) and (1.6) with (1.2), we have
d
dt
[
F (t) − (B1 + 4q
−)p+
4(q− − p+)B1 E(t)
]
= F ′(t) − (B1 + 4q
−)p+
4(q− − p+)B1 E
′(t)
 q
− + 2
2
∫

|ut |2dx + q
− − p+
p+
∫

|∇u|p(x)dx − q−E(t) − (q
− − p+)B1
p+(B1 + 4q−)
∫

|∇u|2dx
(1.7)
Moreover, by virtue of p−  2 and the inequality 
∫

|v|2dx  1
λ1
∫

|∇v|2dx for v ∈ H10(), it is easy to verify that∫

|∇u|p(x)dx =
∫
{x∈:|∇u|1}
|∇u|p(x)dx +
∫
{x∈:|∇u|1}
|∇u|p(x)dx

∫
{x∈:|∇u|1}
|∇u|2dx 
∫

|∇u|2dx −
∫
{x∈:|∇u|1}
|∇u|2dx

∫

|∇u|2dx − ||  λ1
∫

|u|2dx − ||  B1
∫

|u|2dx − ||
(1.8)
So, we apply Inequalities (1.7) and (1.8) to obtain that
d
dt
[
F (t) − q
−
M0
E(t) − ||
M0
]
 q
− + 2
2
∫

|ut |2dx − || − q−E(t) + M0
B1
∫

|∇u|2dx
 q
− + 2
2
∫

|ut |2dx − || − q−E(t) + M0
∫

|u|2dx
 M0
[∫

|ut |2dx − q
−
M0
E(t) − ||
M0
+
∫

|u|2dx
]
 M0
[
F (t) − q
−
M0
E(t) − ||
M0
]
(1.9)
Inequality (1.9) indicates that
F (t)  eM0t
[
F (0) − q
−
M0
E(0) − ||
M0
]
+ q
−
M0
E(t) + ||
M0

Our main results are as follows.
F. Li, F. Liu / C. R. Mecanique 346 (2018) 402–407 405Theorem 1.1. Assume that all the conditions of Lemma 1.1 and the following conditions are satisfied,
(H1) q
− > p+ > p−  2
(H2)
∫

u0u1dx 
p+(B1 + 4q−)
4(q− − p+)B1
[
E(0) + ||
q−
]
> 0
(H3)min
{(
B1
∫

|u0|2dx
) p−
2
,
(
B1
∫

|u0|2dx
) p+
2
}
> p+(1 + ||)
p+(2−3p−)
p− E(0)
then the solution to Problem (1.1) blows up in finite time.
Proof. Case 1. For all t  0, we first assume that E(t)  0. We will divide the proof into three steps.
Step 1. According to (H2) and Inequality (1.4), we observe that the following facts remain true
d
dt
∫

|u|2dx = 2
∫

uutdx  0, t  0 (1.10)
which implies that the term 
∫

|u|2dx is nondecreasing with respect to the time variable.
Step 2. By Hölder inequality and Corollary 3.34 in [4], we have∫

|∇u|2dx  (1 + ||)3− 2p− ‖∇u‖2p(.) (1.11)
Further, using the inequality min{‖u‖p−p(.), ‖u‖p
+
p(.)} 
∫

|u|p(.)dx  max{‖u‖p−p(.), ‖u‖p
+
p(.)} and Inequality (1.11), we get∫

|∇u|p(.)dx  min{‖∇u‖p−p(.),‖∇u‖p
+
p(.)}
 min
{
(1 + ||)2−3p−‖∇u‖p−2 , (1 + ||)
2p+
p− −3p+‖∇u‖p+2
}
 (1 + ||)
p+(2−3p−)
p− min
{
‖∇u‖p−2 ,‖∇u‖p
+
2
}
 (1 + ||)
p+(2−3p−)
p− min
{
(
√
B1‖u‖2)p− , (
√
B1‖u‖2)p+
}
(1.12)
Obviously, applying ‖u(., t)‖2  ‖u0(.)‖2 and (1.12), it is not difficult to prove∫

|∇u|p(.)dx  (1 + ||)
p+(2−3p−)
p− min
{
(
√
B1‖u0‖2)p− , (
√
B1‖u0‖2)p+
}
(1.13)
Step 3. Define
L(t) = 1
2
∫

|u|2dx + 1
2
t∫
0
∫

|∇u|2dxdτ − t
2
‖∇u0‖22 + β(t + t0)2
where β > 0, t0 > max
{ ‖∇u0‖22
β
, ‖u1u0‖12β
}
will be determined later. A direct computation shows that
L′(t) =
∫

uutdx +
t∫
0
∫

∇u∇uτ dxdτ + 2β(t + t0)
L′′(t) =
∫

ut utdx +
∫

uuttdx +
∫

∇u∇ut dx + 2β
406 F. Li, F. Liu / C. R. Mecanique 346 (2018) 402–407According to the definition E(t), the expression of L′′(t), and Lemma 1.2, we have
L′′(t)  q
− + 2
2
‖ut‖22 +
q−
p+
∫

|∇u|p(x)dx + q−
t∫
0
∫

|∇uτ |2dxdτ − q−E(0) + 2β
 q
− + 2
2
[
‖ut‖22 +
t∫
0
∫

|∇uτ |2dxdτ + 4β
]
where
β = q
−(1 + ||)
p+(2−3p−)
p−
2(q− + 1)p+ min
{(
B1
∫

|u0|2dx
) p−
2
,
(
B1
∫

|u0|2dx
) p+
2
}
− q
−E(0)
2(q− + 1) > 0
Following the lines of the proof of Theorem 1.1 of [2], we have
L(t)L′′(t) − q
− + 2
4
(L′(t))2  0
where q− > 2, which implies
(L1−
q−+2
4 (t))′′  0, for t > 0
Noting that L1−
q−+2
4 (0) > 0, (L1−
q−+2
4 )′(0) < 0, then
L1−
q−+2
4 (T ∗) = 0, for some T ∗ ∈ (0, −L
1− q−+24 (0)
(L1−
q−+2
4 )′(0)
)
Case 2. There exists t1 > 0 such that E(t1) < 0. Noting that E(0) > 0 and considering the continuity of E(t), we know that 
there exists t2 ∈ (0, t1) such that E(t2) = 0. In addition, we apply the monotonicity of E(t) to obtain E(t)  0, 0 < t  t2. 
Similar as the proof of Case 1, we may prove that the solution to Problem (1.1) blows up before the time t1. 
It is worth pointing out that the principle significance of the condition (H3) is that it allows us to establish an explicit 
upper bound of the blow-up time. In fact, if this condition is removed, we also prove the nonexistence of solutions.
Theorem 1.2. Assume that the exponents p(x), q(x) and the initial data u0(x), u1(x) satisfy the conditions of Lemma 1.1 and the 
following conditions
(H4) q
− > p+ > p−  2
(H5)
∫

u0u1dx >
p+(B1 + 4q−)
4(q− − p+)B1
[
E(0) + ||
q−
]
> 0
then the solution to Problem (1.1) blows up in finite time.
Proof. Case 1. For all t  0, we first assume that E(t)  0. According to (H5) and to Inequality (1.4), it is easy to check that
d
dt
∫

|u|2dx = 2
∫

uutdx  2eM0t H(0), t  0 (1.14)
where H(0) = ∫

u0u1dx − p+(B1+4q−)4(q−−p+)B1
[
E(0) + ||q−
]
> 0. Assume by contradiction that the solution u is global. Then, it is 
easily seen that
‖u(., t)‖2 = ‖u0‖2 + 2
t∫
0
∫

u(., τ )uτ (., τ )dxdτ  ‖u0‖2 + 2
t∫
0
eM0τ H(0)dτ
= ‖u0‖2 + 2H(0)
(
eM0t − 1
) (1.15)M0
F. Li, F. Liu / C. R. Mecanique 346 (2018) 402–407 407On the other hand, by Lemma 1.1, Minkowski inequality and Hölder inequality, we have
‖u(., t)‖2  ‖u0‖2 + ‖u(., t) − u0‖2  ‖u0‖2 +
∥∥∥
t∫
0
uτ dτ
∥∥∥
2
 ‖u0‖2 +
t∫
0
‖uτ ‖2dτ
 ‖u0‖2 + 1√
B1
t∫
0
‖∇uτ ‖2dτ  ‖u0‖2 +
√
t√
B1
( t∫
0
∫

|∇uτ |2dxdτ
) 1
2
 ‖u0‖2 +
√
t√
B1
(
E(0) − E(t)
) 1
2  ‖u0‖2 +
√
t E(0)
B1
(1.16)
which contradicts Inequality (1.15).
Case 2. There exists t1 > 0 such that E(t1) < 0. This proof is similar to the argument of the second case of Theorem 1.1. 
We omit it here. 
Acknowledgements
The first author’s research was supported by NSF of China (11201170), and the second author’s one was supported by 
The Scientific and Technological Project of Jilin Provinces’s Education Department in Thirteenth-five-Year (JJKH20170538KJ).
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