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C. R. Mecanique 347 (2019) 710–715Contents lists available at ScienceDirect
Comptes Rendus Mecanique
www.sciencedirect.com
Non-global existence of solutions to pseudo-parabolic 
equations with variable exponents and positive initial energy
Menglan Liao a,b,∗
a School of Mathematics, Jilin University, Changchun, Jilin Province 130012, China
b Department of Mathematics, Michigan State University, East Lansing, MI 48824, USA
a r t i c l e i n f o a b s t r a c t
Article history:
Received 17 May 2018
Accepted after revision 21 September 2019
Available online 9 October 2019
Keywords:
Non-global existence
Variable exponents
Positive initial energy
Pseudo-parabolic equation
This paper deals with a pseudo-parabolic equation involving variable exponents under 
Dirichlet boundary value condition. The author proves that the solution is not global in 
time when the initial energy is positive. This result extends and improves a recent result 
obtained by Di et al. (2017) [1].
© 2019 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.
1. Introduction
In this paper, we study the following m(x)-Laplacian equation:⎧⎪⎨
⎪⎩
ut − ut − div(|∇u|m(x)−2∇u) = |u|p(x)−2u, in  × (0, T ),
u(x, t) = 0, on ∂ × (0, T ),
u(x,0) = u0(x), in 
(1.1)
where  ⊂ RN is a bounded domain with smooth boundary ∂, T > 0. The exponents m(x) and p(x) are two continuous 
functions satisfying
2 < m−  m(x)  m+ < p−  p(x)  p+ < ∞ (1.2)
and both satisfy the following condition:
|p(x) − p(y)|  A
log
(
1
|x−y|
) , for all x, y ∈  with |x − y| < δ (1.3)
where A > 0 and 0 < δ < 1. The upper bound for the blow-up time was obtained when the initial energy is nonpositive [1]. 
However, it is natural to pose an interesting problem: does the blow-up phenomena occur when the initial energy is 
positive? To the best of our knowledge, there exist only rare results about such problems.
* Correspondence to: School of Mathematics, Jilin University, Changchun, Jilin Province 130012, China.
E-mail address: liaoml14@mails.jlu.edu.cn.https://doi.org/10.1016/j.crme.2019.09.003
1631-0721/© 2019 Académie des sciences. Published by Elsevier Masson SAS. All rights reserved.
M. Liao / C. R. Mecanique 347 (2019) 710–715 711Pseudo-parabolic equations describe many important physical processes, such as the seepage of homogeneous fluids 
through a fissured rock, the aggregation of populations, etc. The interested readers may refer to [2,3] and the references 
therein. For the study of pseudo-parabolic equations, many results have been obtained [4–6]. Especially, Xu and Su [4]
studied a class of semilinear pseudo-parabolic equations (where m(x) = 2, and the source term |u|p(x)−2 is replaced by up). 
They proved global existence, non-existence, and asymptotic behavior of solutions with initial energy by introducing a family 
of potential wells and obtained finite-time blow-up with high initial energy by the comparison principle. Since there exist 
variable exponents, we cannot obtain the blow-up result with high initial energy by trying to consider the potential-well 
method for Problem (1.1). Variable exponents and the term ut cause some difficulties, thus we have to develop other 
techniques to overcome these difficulties. Before stating our result, let us recall some results on the existence and blow-up 
of the solution to Problem (1.1) with nonpositive initial energy.
Theorem 1.1. [1] Assume that (1.2) and (1.3) hold, then for any u0 ∈ W 1,m(·)0 () ∩ Lp(·)(), there exists a T0 ∈ (0, T ] such that the 
Problem (1.1) has a unique local solution u ∈ L∞([0, T0], W 1,m(·)() ∩ Lp(·)()) with ut ∈ L2([0, T0], W 1,2()) satisfying
(ut, v) + (∇ut,∇v) + (|∇u|m(x)−2∇u,∇v) = (|u|p(x)−2u, v), for v ∈ W 1,m(·)0 () ∩ Lp(·)() (1.4)
Define the energy functional by
E(t) =
∫

( 1
m(x)
|∇u|m(x) −
∫

1
p(x)
|u|p(x)
)
dx (1.5)
It obviously follows that
E ′(t) = −‖ut‖2H10  0 (1.6)
which implies
E(t) = E(0) −
t∫
0
‖us‖2H10 ds (1.7)
Theorem 1.2. [1] Assume that (1.2) and (1.3) hold. Let u0 ∈ W 1,m(·)0 () ∩ Lp(·)() such that ‖u0‖H10() > 0 and E(0)  0. Then the 
solution u(x, t) to Problem (1.1) blows up in a finite time T ∗. Moreover, an upper bound for the blow-up time is given by
T ∗ 
2(‖u0‖2H10 )
1− m−2
(m− − 2)β
where β > 0 is a constant.
Our main result is as follows:
Theorem 1.3. Assume that (1.2) and (1.3) hold. Let u0 ∈ W 1,m(·)0 () ∩ Lp(·)() be non-zero. If the following condition holds:
0 < E(0) < C (1.8)
then the solution u(x, t) to Problem (1.1) is non-global, where
C = p
− − m+
m+
1
p−
min
{ 2
m+
(m−||
2
)− m+−22 (m−
2
λ1
1 + λ1
)m+
2 ‖u0‖m+H10 ,
2
m−
(m−||
2
)− m−−22 (m−
2
λ1
1 + λ1
)m−
2
(m+ − 2
m+
m−
m− − 2
)− m−−22 ‖u0‖m−H10
}
Here λ1 is the first eigenvalue of the problem{
ω + λω = 0, in ,
ω = 0, on ∂
In comparison with Theorem 1.2, we add something substantial to E(0), in the sense that we obtain the non-global 
existence of solutions when the initial energy is positive. From this point of view, this improvement is in the spirit of the 
one performed by Merle and Zaag [7] for some parabolic equation derived from the semilinear heat equation; the interested 
readers can refer to Proposition 2.1 in [7].
712 M. Liao / C. R. Mecanique 347 (2019) 710–7152. Proof of Theorem 1.3
In order to prove our main result, we argue by contradiction. We assume that the solution is global in time. Next, we 
will divide this proof into two steps.
Step 1. It directly follows from [1] that
d
dt
‖u‖2
H10
= 2
∫

|u|p(x) dx − 2
∫

|∇u|m(x) dx
 2p−
∫

|u|p(x)
p(x)
dx − 2
∫

|∇u|m(x) dx
= 2p−
∫

|u|p(x)
p(x)
dx − 2p−
∫

|∇u|m(x)
m(x)
dx + 2p−
∫

|∇u|m(x)
m(x)
dx − 2
∫

|∇u|m(x) dx
 −2p−E(t) + 2(p
− − m+)
m+
∫

|∇u|m(x) dx
(2.1)
Step 2. In order to complete the proof, we consider the following two cases.
Case 1. We first consider the case E(t)  0, for all t > 0.
Firstly, we give an estimate of a lower bound for the solution in H10 norm. In fact, by virtue of (1.8), we choose an α
satisfying the following inequality
1 < α <
C
E(0)
(2.2)
Combining (1.7) with (2.1) (2.2) and E(t)  0, we get
d
dt
‖u‖2
H10
 −2p−E(t) + 2(p
− − m+)
m+
∫

|∇u|m(x) dx
= 2p−(α − 1)E(t) − 2p−αE(t) + 2(p
− − m+)
m+
∫

|∇u|m(x) dx
 −2p−αE(0) + 2p−α
t∫
0
‖us‖2H10 ds +
2(p− − m+)
m+
∫

|∇u|m(x) dx
(2.3)
Young’s inequality implies that
‖∇u‖22 
2σ
m−
∫

|∇u|m(x) dx + m
+ − 2
m+
||max
{
σ
−2
m+−2 ,σ
−2
m−−2
}
(2.4)
for all σ > 0. It follows from Poincaré’s inequality that ‖∇u‖22  λ1‖u‖22. Therefore, we get
‖∇u‖22 =
λ1
1 + λ1 ‖∇u‖
2
2 +
1
1 + λ1 ‖∇u‖
2
2 
λ1
1 + λ1 ‖u‖
2
H10
(2.5)
Combining (2.4) with (2.5), and replacing m
−
2σ
λ1
1+λ1 by κ , then∫

|∇u|m(x) dx  κ‖u‖2
H10
− m
+ − 2
m+
m−
2
||C1(κ) (2.6)
where C1(κ) = max
{(
m− λ1
) −m+
m+−2 κ
m+
m+−2 , 
(
m− λ1
) −m−
m−−2 κ
m−
m−−2
}
. Inserting (2.6) into (2.3), it directly follows that2 1+λ1 2 1+λ1
M. Liao / C. R. Mecanique 347 (2019) 710–715 713d
dt
‖u‖2
H10
 −2p−αE(0) + 2p−α
t∫
0
‖us‖2H10 ds +
2(p− − m+)κ
m+
‖u‖2
H10
− 2(p
− − m+)
m+
m+ − 2
m+
m−
2
||C1(κ)
(2.7)
Removing the second term on the right-hand side of the above inequality, and then solving the ordinary differential in-
equality, we obtain
‖u‖2
H10
 m
+
2(p− − m+)κ
(
2p−αE(0) + 2(p
− − m+)
m+
m+ − 2
m+
m−
2
||C1(κ)
)
×
(
1 − e 2(p
−−m+)κt
m+
)
+ ‖u0‖2H10 e
2(p−−m+)κt
m+
(2.8)
Secondly, we prove that the solution is not global in time. Define y(t) = ∫ t0 ‖u‖2H10 ds, then y′(t) = ‖u‖2H10 , y′′(t) =
d
dt ‖u‖2H10 . By inserting (2.8) into (2.7), we have
y′′(t)  2p−α
t∫
0
‖us‖2H10 ds +
(2(p− − m+)κ
m+
‖u0‖2H10
− 2(p
− − m+)
m+
m+ − 2
m+
m−
2
||C1(κ) − 2p−αE(0)
)
e
2(p−−m+)κt
m+
(2.9)
Due to (2.2), we can take ε > 0 small enough that
ε <
2(C − αE(0))
α‖u0‖2H10
(2.10)
And then we may choose sufficiently large c > 0 satisfying
c >
1
4
ε−2‖u0‖4H10 (2.11)
Denoting ϕ(t) = y2(t) + ε−1‖u0‖2H10 y(t) + c, we get
ϕ′(t) =
(
2y(t) + ε−1‖u0‖2H10
)
y′(t) (2.12)
ϕ′′(t) =
(
2y(t) + ε−1‖u0‖2H10
)
y′′(t) + 2(y′(t))2 (2.13)
Set δ = 4c − ε−2‖u0‖4H10 , then δ > 0 with the help of (2.11). From (2.12), we obtain that
(ϕ′(t))2 = (4ϕ(t) − δ)(y′(t))2
which implies
4ϕ(t)(y′(t))2 = (ϕ′(t))2 + δ(y′(t))2 (2.14)
(2.13) and (2.14) imply that
4ϕ(t)ϕ′′(t) = 4
(
2y(t) + ε−1‖u0‖2H10
)
y′′(t)ϕ(t) + 2
(
(ϕ′(t))2 + δ(y′(t))2
)
(2.15)
Noticing that
t∫
0
∫

(uus + ∇u∇us)dxds 
( t∫
0
∫

(u2 + |∇u|2)dx ds
) 1
2
( t∫
0
∫

(u2s + |∇us|2)dx ds
) 1
2
it is easy to verify that
714 M. Liao / C. R. Mecanique 347 (2019) 710–715(y′(t))2 = ‖u‖4
H10
=
(
‖u0‖2H10 + 2
t∫
0
∫

(uus + ∇u∇us)dx ds
)2
 ‖u0‖4H10 + 4‖u0‖
2
H10
(y(t))
1
2
( t∫
0
‖us‖2H10 ds
) 1
2 + 4y(t)
t∫
0
‖us‖2H10 ds
 ‖u0‖4H10 + 2ε‖u0‖
2
H10
y(t) + 2ε−1‖u0‖2H10
t∫
0
‖us‖2H10 ds + 4y(t)
t∫
0
‖us‖2H10 ds
(2.16)
for all ε > 0. Therefore, combining (2.16) with (2.15) and (2.9), we have
4ϕ′′(t)ϕ(t) − (2 + αp−)(ϕ′(t))2
= 4
(
2y(t) + ε−1‖u0‖2H10
)
y′′(t)ϕ(t) + 2
(
(ϕ′(t))2 + δ(y′(t))2
)
− (2 + αp−)(ϕ′(t))2
= 4
(
2y(t) + ε−1‖u0‖2H10
)
y′′(t)ϕ(t) + 2δ(y′(t))2 − αp−(ϕ′(t))2
= 4
(
2y(t) + ε−1‖u0‖2H10
)
y′′(t)ϕ(t) + 2δ(y′(t))2 − αp−(4ϕ(t) − δ)(y′(t))2
= 4
(
2y(t) + ε−1‖u0‖2H10
)
y′′(t)ϕ(t) + (2 + αp−)δ(y′(t))2 − 4αp−ϕ(t)(y′(t))2
 4
(
2y(t) + ε−1‖u0‖22
)
y′′(t)ϕ(t) − 4αp−ϕ(t)(y′(t))2
 4
(
2y(t) + ε−1‖u0‖22
)(
2p−α
t∫
0
‖us‖2H10 ds + G(κ)e
2(p−−m+)κt
m+
)
ϕ(t) − 4αp−
× ϕ(t)
(
‖u0‖4H10 + 2ε‖u0‖
2
H10
y(t) + 2ε−1‖u0‖2H10
t∫
0
‖us‖2H10 ds + 4y(t)
t∫
0
‖us‖2H10 ds
)
(2.17)
where
G(κ) = 2(p
− − m+)κ
m+
‖u0‖2H10 −
2(p− − m+)
m+
m+ − 2
m+
m−
2
||C1(κ) − 2p−αE(0)
Next we prove the non-negativity for the right-hand side of (2.17). Solving the equation G ′(κ) = 0, then we obtain one 
of κ1 and κ2, respectively
κ1 = ‖u0‖m+−2H10
(m−
2
||
)− m+−22 (m−
2
λ1
1 + λ1
)m+
2
κ2 = ‖u0‖m−−2H10
(m−
2
||
)− m−−22 (m−
2
λ1
1 + λ1
)m−
2
(m+ − 2
m+
m−
m− − 2
)− m−−22
Noticing that
G(κ1) = 2(p
− − m+)
m+
2
m+
(m−
2
||
)− m+−22 (m−
2
λ1
1 + λ1
)m+
2 ‖u0‖m+H10 − 2p
−αE(0)
G(κ2) = 2(p
− − m+)
m+
2
m−
(m−
2
||
)− m−−22 (m−
2
λ1
1 + λ1
)m−
2
(m+ − 2
m+
m−
m− − 2
)− m−−22 ‖u0‖m−H10
− 2p−αE(0)
we apply (2.10) to obtain
G(κmax)  min{G(κ1), G(κ2)} = 2p−(C − αE(0)) > εp−α‖u0‖2H10
κmax ∈ {κ1, κ2} is the maximum point of the function G . Utilizing the fact that e
2(p−−m+)κt
m+  1 and (2.10), (2.17), we obtain
M. Liao / C. R. Mecanique 347 (2019) 710–715 7154ϕ′′(t)ϕ(t) − (2 + αp−)(ϕ′(t))2
 4
(
2y(t) + ε−1‖u0‖2H10
)(
2p−α
t∫
0
‖us‖2H10 ds + εp
−α‖u0‖2H10
)
ϕ(t)
− 4αp−ϕ(t)
(
‖u0‖4H10 + 2ε‖u0‖
2
H10
y(t) + 2ε−1‖u0‖2H10
t∫
0
‖us‖2H10 ds + 4y(t)
t∫
0
‖us‖2H10 ds
)
= 0
which indicates
ϕ′′(t)ϕ(t) − 2 + αp
−
4
(ϕ′(t))2  0
Applying ϕ(0) > 0, ϕ′(0) > 0, and the concavity method [8–10], we get the maximal existence time of the solution Tmax =
4ϕ(0)
(αp−−2)ϕ′(0) , which shows that the solution u(x, t) is non-global. This is a contradiction with the assumption.
Case 2. There exists some t̃ > 0 such that E(t̃) < 0. Because the energy functional E(t) is a continuous function and E(0) > 0, 
we can assume that there exists a first time t0 > 0 such that E(t0) = 0 and E(t̃) < 0 for some t̃ > t0. Integrating (2.1) from 
t0 to t̃ , then ‖u(x, ̃t)‖H10() > 0. Now we choose u(x, ̃t) as a new initial datum, then it is clear from Theorem 1.2 that the 
solution is non-global. This is a contradiction with the assumption.
This completes the proof of Theorem 1.3.
Acknowledgements
The author wishes to express her gratitude to the China Scholarship Council for the scholarship and Professor Wenjie 
Gao for his support and constant encouragement.
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Non-global existence of solutions to pseudo-parabolic equations with variable exponents and positive initial energy

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