The Exponential Growth of Solution, Upper and Lower Bounds for the
Blow-Up Time for a Viscoelastic Wave Equation with Variable-
Exponent Nonlinearities
SOUFIANE BENKOUIDER
Laboratory of Pure and Applied Mathematics,
University Amar Telidji, Laghouat,
ALGERIA
ABITA RAHMOUNE
Department of Technical Sciences,
Laboratory of Pure and Applied Mathematics,
Laghouat University
ALGERIA
Abstract: - This paper aims to study the model of a nonlinear viscoelastic wave equation with damping and
source terms involving variable-exponent nonlinearities. First, we prove that the energy grows exponentially,
and thus in 𝐿 and 𝐿 norms. For the case 2≤𝑘(.)<𝑝(.), we reach the exponential growth result of a blow-
up in finite time with positive initial energy and get the upper bound for the blow-up time. For the case 𝑘(.)=
2, we use the concavity method to show a finite time blow-up result and get the upper bound for the blow-up
time. Furthermore, for the case 𝑘(.)≥2, under some conditions on the data, we give a lower bound for the
blow-up time when the blow-up occurs.
Key-Words: - Viscoelastic wave equations; Exponential Growth; Blow-up; Lower and upper bound; Sobolev
spaces with variable exponents; variable nonlinearity.
Received: October 29, 2022. Revised: May 2, 2023. Accepted: May 22, 2023. Published: June 2, 2023.
1 Introduction
In recent years, many researchers, in the different
cases of the values of the memory kernel,
specifically when 𝑔=0 or 𝑔>0, a nonlinear wave
equation with a memory, damping, and source terms
associated with the Laplace operator with Dirichlet
type condition has been considered
𝑢−Δ𝑢−𝛾Δ𝑢+
𝑔(𝑡−𝑠)Δ𝑢(𝑥,𝑠)ds
+𝑟|𝑢|𝑢(𝑥,𝑡)
=|𝑢|𝑢(𝑥,𝑡) in Ω × ℝ,
𝑢=0 on Γ × (0,+∞),
𝑢(𝑥,0)=𝑢(𝑥),𝑢(𝑥,0)=𝑢(𝑥),𝑥∈Ω,
(1.1)
where 𝑇>0,𝑝,𝑘≥2,𝑟,𝛾 are positive
constants, Δ stands for the Laplacian with respect to
the spatial variables, Ω is a bounded domain of ℝ
(𝑛≥1) with a smooth boundary 𝜕Ω. The relaxation
function 𝑔 is a positive and uniformly decaying
function, 𝑢,𝑎𝑛𝑑 𝑢 are given functions belonging
to suitable spaces.
Under suitable conditions on 𝑔,𝑝,𝑘,𝑢, and 𝑢,
using some known theorems in the mathematical
literature, the global existence in time, blow up in
finite time, the asymptotic behavior, and a lower
bound for the blow-up time of the unique weak
solution have been discussed.
From the physical point of view, this type of
problem arises usually in viscoelasticity. It has been
considered first by, [12], in 1970, where the general
decay was discussed. Related problems to (1.1) have
attracted a great deal of attention in the last two
decades, and many results have appeared on the
existence and long-time behavior of solutions. Look
at this area, in, [6], [12], [17], [11], [18], [16], [19],
and references therein.
In this work given positive measurable functions
𝑝(.),𝑘(.) On Ω satisfying:
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1 2
ΩΩ
1Ω
2Ω
2 inf sup ,
2 inf
sup ,
xx
x
x
p ess p x p ess p x
k ess k x k x
k ess k x
(1.2)
we consider the following semilinear generalized
hyperbolic boundary value problem governed by
partial differential equations that describe the
evolution of viscoelastic materials with
nonlinearities of variable exponent type under
Dirichlet type condition:
𝑢−Δ𝑢−𝛾Δ𝑢+∫
𝑔(𝑡−𝑠)Δ𝑢(𝑥,𝑠)d𝑠
+𝑟|𝑢|()𝑢(𝑥,𝑡)
=|𝑢|()𝑢(𝑥,𝑡) in Ω × ℝ, (1.3) 𝑢=0 on Γ × (0,+∞), (1.4)
𝑢(𝑥,0)=𝑢(𝑥),𝑢(𝑥,0)=𝑢(𝑥),𝑥∈Ω. (1.5)
Problems with variable exponents of nonlinearity
arise from many important mathematical models in
engineering and physical sciences. For example,
modeling of physical phenomena such as flows of
electro-rheological fluids or fluids with temperature-
dependent viscosity, thermoelasticity, nonlinear
viscoelasticity, filtration processes through porous
media, image processing, nuclear science, chemical
reactions, heat transfer, population dynamics,
biological sciences, etc., More details on these
problems can be found in, [10], [13], [1], [2], [4],
[8], [9], [3], [18], [5], [7], [15], and references
therein.
As far as we have known, there is little
information on the bounds for blow-up time to
problem (1.3)-(1.5) when the initial energy is
positive with 𝑝(.) and 𝑘(.) are not constants. So, it
is natural to analyze the problem (1.3)-(1.5) and
give further results on the behavior of solutions.
The contents of this paper are as follows. In
Section 2 we give Preliminaries. In Section 3 we
prove the exponential growth of the energy 𝐸(𝑡) of
a solution. In Section 4, we consider an upper bound
for the blow-up time in case 2≤𝑘≤𝑘<𝑝. An
upper bound for the blow-up time in case 𝑘(𝑥)=2,
∀𝑥, is proved in Section 5. Section 6, is devoted to
proving a lower bound for the blow-up time in case
𝑘≥2.
2 Preliminaries
In the following section, we introduce some
preliminaries and notations, which will be used
throughout this paper.
Given a function 𝑝:Ω→[𝑝,𝑝]⊂(2,∞), 𝑝,=
const, we define the set
𝐿(.)(Ω)
=𝑣:Ω→ℝ:𝑣 measurablefunctionsonΩ,
𝜚(𝑣)=𝜚(.)(𝑣)=
|𝑣(𝑥)|()d𝑥<∞.
The variable-exponent space 𝐿(.)(Ω) equipped with
the Luxemburg norm
‖𝑢‖(.)=inf𝜆>0,
𝑢𝜆()d𝑥≤1,
becomes a Banach space. In general, variable-
exponent Lebesgue spaces are similar to classical
Lebesgue spaces in many aspects, see the first
discussed the 𝐿() spaces and 𝑊,() spaces by
Kovàcik and Rákosnik in [15].
We also assume that 𝑝(𝑥), 𝑘(.) satisfies the
following Zhikov–Fan uniform local continuity
condition:
|𝑘(𝑥)−𝑘(𝑦)|+|𝑝(𝑥)−𝑝(𝑦)|≤𝑀
log|𝑥−𝑦|,
for all 𝑥,𝑦 in Ω with |𝑥−𝑦|<1
2,𝑀>0.
(2.1)
Let us list some properties of the spaces 𝐿(.)(Ω)
which will be used in the study of the problem (1.3)-
(1.5).
• If 𝑝(𝑥) is measurable and
1<𝑝≤𝑝(𝑥)≤𝑝<∞, in Ω,
then 𝐿(.)(Ω) is a reflexive and separable Banach
space, and 𝐶(Ω) is dense in 𝐿(.)(Ω).
• If condition (2.1) is fulfilled, and Ω has a finite
measure and 𝑝, 𝑞 are variable exponents so that
𝑝(𝑥)≤𝑞(𝑥) almost everywhere in Ω, the inclusion
𝐿(.)(Ω)⊂𝐿(.)(Ω)
is continuous and
∀𝑣∈𝐿(.)(Ω) ‖𝑢‖(.)≤𝐶‖𝑢‖(.),𝐶=𝐶|Ω|,𝑝,
(2.2)
• It follows directly from the definition of the norm
that min‖𝑢‖(.)
,‖𝑢‖(.)
≤𝜚(.)(𝑢)
≤max‖𝑢‖(.)
,‖𝑢‖(.)
,
(2.3)
and
min𝜚(.)(𝑢)
,𝜚(.)(𝑢)
≤‖𝑢‖(.)≤
max𝜚(.)(𝑢)
,𝜚(.)(𝑢)
. (2.4)
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• If 𝑝:Ω→[𝑝,𝑝]⊂[1,+∞) is a
measurable function and 𝑝∗>𝑒𝑠𝑠sup
{∈}𝑝(𝑥) with
𝑝∗≤
, then the embedding 𝐻(Ω)↪𝐿(.)(Ω) is
continuous and compact.
2.1 Mathematical Hypotheses
We begin this section by introducing some
hypotheses and our main result. Throughout this
paper, we use standard functional spaces, and
denote that (.,.),‖.‖ the inner products, and norms
in 𝐿(Ω) and 𝐻(Ω) represented and they are given
by:
(𝑢,𝑣)=𝑢(𝑥)𝑣(𝑥)d𝑥 and ‖𝑢‖()
=‖𝑢‖
=𝑢d𝑥;
‖𝑢‖()
=‖𝑢‖=|∇𝑢|d𝑥.
Next, we state the assumptions for the problem
(1.3)-(1.5).
Let 𝑘(.) and 𝑝(.) are given measurable functions
on Ω satisfying the following conditions
1 2
1 2
1 2
1 2
2
2
.
, 2
2
and 2 if 2,
2
2 , 2
2
and 2 if 2.
n
p p x p n
n
p p n
n
k k x k n
n
k k n
(2.5)
𝑔:ℝ→ℝ is a nonincreasing differentiable
function satisfying
1−∫
𝑔(𝑠)d𝑠=𝑙>0,∀𝑡∈ℝ . (2.6)
By Corollary 3.3.4 in [13], we know 𝐿(Ω)↪
𝐿(.)(Ω). So, it is a consequence of the embedding
𝐻(Ω)↪𝐿(Ω) and Poincaré
inequality that ‖𝑢‖(.)≤𝐵‖∇𝑢‖, (2.7)
where 𝐵 is the best constant of the embedding
𝐻(Ω)↪𝐿(.)(Ω) determined by
𝐵=inf|∇𝑢|:𝑢∈𝐻(Ω),‖𝑢‖(.)=1.
The following constants play a crucial role in the
proof of our results. Let 𝐵, 𝛼, 𝛼, 𝐸 be constants
satisfying
𝐵=max(1,𝐵),𝛼=𝑙
𝐵
,
𝛼=‖∇𝑢‖, 𝐸=𝑙1
2−1
𝑝𝛼.
(2.8)
3 Exponential Growth
In this section, will prove that the energy grows
exponentially, and thus so the 𝐿 and 𝐿 norms to
the problem (1.3)-(1.5) if the variable exponents
𝑝(.), and 𝑘(.) satisfy some conditions and the initial
data are large enough (in the energy viewpoint).
Firstly, we start with a local existence result for the
problem (1.3)-(1.5), which can be obtained by the
combination of the Faedo-Galerkin argument and
the compactness method together with the Banach
fixed point theorem. Hereafter, for simplicity, we
take 𝑎=1, and we have
Lemma 3.1 Let 2≤𝑝≤𝑝(𝑥)≤𝑝≤𝑞 and
𝑚𝑎𝑥2,
≤𝑘≤𝑘(𝑥)≤𝑘≤𝑞. Then
given (𝑢,𝑢)∈𝐻(𝛺)×𝐿(𝛺) there exists 𝑇>0
and a unique solution 𝑢 of the problem (1.3)-(1.5)
on (0,𝑇) such that
𝑢∈𝐶0,𝑇;𝐻(Ω)∩𝐶0,𝑇;𝐿(Ω),
𝑢∈𝐿0,𝑇;𝐻(Ω)∩𝐿(.)(0,𝑇)×Ω.
The first main result of this paper reads as follows
Theorem 3.2 Let 𝑘<𝑝≤𝑝(𝑥)≤𝑝 where 2≤
𝑝≤𝑝(𝑥)≤𝑝≤𝑞. Assume the initial value 𝑢 is
chosen to ensure that 𝐸(0)<𝐸 and 𝐵≥
‖∇𝑢‖>𝛼 hold. Then under the above
conditions, the solution of problem (1.3)-(1.5) will
grow exponentially in the 𝐿 and 𝐿 norms.
For this purpose, we start with the following lemma
defining the energy of the solution
Lemma 3.3 The corresponding energy to the
problem (1.3)-(1.5) is given by
𝐸(𝑡)=1
2(𝑔∘∇𝑢)+1
2‖𝑢‖
+1
21−
𝑔(𝑠)d𝑠‖∇𝑢‖
−1
𝑝(𝑥)|𝑢|()d𝑥,
(3.1)
furthermore, by easily verified formula
()
=
(𝑔∘∇𝑢)−
𝑔(𝑡)‖∇𝑢‖
−
𝛾∫|∇𝑢|d𝑥−𝑟∫|𝑢|()d𝑥≤0, (3.2)
the inequality 𝐸(𝑡)≤𝐸(0) is obtained, where
𝐸(0)=1
2‖𝑢‖
+1
2‖∇𝑢‖
−1
𝑝(𝑥)|𝑢|()d𝑥,
(3.3)
and
(𝑔∘∇𝑢)(𝑡)
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=
𝑔(𝑡−𝑠)∥∇𝑢(𝑡)−∇𝑢(𝑠)∥
d𝑠≥0.
We conclude from (2.3) and (3.1) that
𝐸(𝑡)≥1
2(𝑔∘∇𝑢)+1
21−
𝑔(𝑠)d𝑠‖∇𝑢‖
−1
𝑝max‖𝑢‖(.)
,‖𝑢‖(.)
≥1
21−
𝑔(𝑠)d𝑠‖∇𝑢‖
−1
𝑝max((𝐵‖∇𝑢‖),(𝐵‖∇𝑢‖))
≥1
2𝑙𝛼−1
𝑝max((𝐵𝛼),(𝐵𝛼)):=ℎ(𝛼)∀𝛼
∈[0,+∞),
(3.4)
where 𝛼=‖∇𝑢‖.
Lemma 3.4 Let 𝑓:[0,+∞)→ℝ be defined by
𝑓(𝛼)=
𝑙𝛼−
(𝐵𝛼). (3.5)
Then the following claims hold under the
hypotheses of Theorem 3.2: 𝑓 is increasing for
0<𝛼≤𝛼 and decreasing for 𝛼≥𝛼,
lim
→𝑓(𝛼)=−∞ and 𝑓(𝛼)=𝐸.
Proof. By the assumption that 𝐵>1 and 𝑝>2,
one can see that 𝑓(𝛼)=ℎ(𝛼), for 0<𝛼≤𝐵.
Furthermore, 𝑓(𝛼) is differentiable and continuous
in [0,+∞).
𝑓(𝛼)=𝛼𝑙−𝐵𝛼, 0≤𝛼<𝐵.
Then (i) follows. Since 𝑝−2>0, we have
lim
→𝑓(𝛼)=−∞. A usual computation yields
𝑓(𝛼)=𝐸. Then (ii) holds valid.
Lemma 3.5 Under the assumptions of Theorem 3.2,
there exists a positive constant 𝛼>𝛼 such that
‖∇𝑢‖≥𝛼,𝑡≥0, (3.6)
∫|𝑢(𝑥,𝑡)|()d𝑥≥(𝐵𝛼). (3.7)
Proof. Since 𝐸(0)<𝐸, it follows from Lemma 3.4
that there exists a positive constant 𝛼>𝛼, such
that 𝐸(0)=𝑓(𝛼). By (3.4), we have 𝑓(𝛼)=
ℎ(𝛼)≤𝐸(0)=𝑓(𝛼), it follows from Lemma
3.4(i) that 𝛼≥𝛼, so (3.6) holds for 𝑡=0. Now
we prove (3.6) by contradiction. Suppose that
‖∇𝑢(𝑡∗)‖ <𝛼 for some 𝑡∗>0. By the
continuity of ‖∇𝑢(.,𝑡)‖ and 𝛼>𝛼, we may take
𝑡∗ such that 𝛼>‖∇𝑢(𝑡∗)‖>𝛼, then it results
from (3.4) that
𝐸(0)=𝑓(𝛼)<𝑓(‖∇𝑢(𝑡∗)‖)≤𝐸(𝑡∗),
which contradicts Lemma 3.3, and (3.6) holds.
By (3.1) and (3.2), we obtain
1
𝑝|𝑢(𝑥,𝑡)|()d𝑥≥1
𝑝(𝑥)|𝑢(𝑥,𝑡)|()d𝑥
≥1
2𝑙‖∇𝑢‖
−𝐸(0)
≥1
2𝑙𝛼−𝐸(0)=1
2𝑙𝛼−𝑓(𝛼)=1
𝑝(𝐵𝛼),
(3.8)
and (3.7) follows.
Let 𝐻(𝑡)=𝐸−𝐸(𝑡) for 𝑡≥0, we have
the following lemma.
Lemma 3.6 Under the assumptions of Theorem 3.2,
the functional 𝐻(𝑡) illustrated above provides the
following estimates:
0<𝐻(0)≤𝐻(𝑡)≤1
𝑝(𝑥)|𝑢(𝑥,𝑡)|()d𝑥
≤1
𝑝𝜚(𝑢), 𝑡≥0.
(3.9)
Proof. By Lemma 3.3, 𝐻(𝑡) is nondecreasing in 𝑡.
Thus
𝐻(𝑡)≥𝐻(0)=𝐸−𝐸(0)>0, 𝑡≥0.
(3.10)
Combining (3.1), (2.8), (3.6) and 𝛼>𝛼, we have
𝐻(𝑡)−1
𝑝(𝑥)|𝑢(𝑥,𝑡)|()d𝑥
≤𝐸−
𝑙‖∇𝑢‖
≤𝑙
−
𝛼−
𝑙𝛼<0,𝑡≥0. (3.11)
and (3.9) follows from (3.10) and (3.11).
Based on the above three lemmas, we can give
proof of Theorem 3.2.
Proof of Theorem 3.2. For 𝜀>0 small to be chosen
later, we then define the auxiliary function
𝐿(𝑡)=𝐻(𝑡)+𝜀𝑢𝑢d𝑥+1
2𝜀𝛾|∇𝑢|d𝑥.
(3.12)
Let us observe that 𝐿 is a small perturbation of the
energy. By taking the time derivative of (3.12),
we obtain:
d𝐿(𝑡)
d𝑡 =𝛾‖∇𝑢‖
+𝜀‖𝑢‖
−1
2(𝑔∘∇𝑢)(𝑡)+1
2𝑔(𝑡)‖∇𝑢‖
+𝜀∫𝑢𝑢d𝑥 +𝜀𝛾∫∇𝑢∇𝑢d𝑥. (3.13)
Using problem (1.3)-(1.5), the equation (3.13)
becomes:
d𝐿(𝑡)
d𝑡 =𝛾|∇𝑢|d𝑥+𝑟|𝑢|()d𝑥+𝜀‖𝑢‖
−𝜀‖∇𝑢‖
−𝜀𝑟|𝑢|()𝑢𝑢d𝑥
+𝜀∫|𝑢(𝑡)|()d𝑥 (3.14)
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+𝜀∇𝑢(𝑡)
𝑔(𝑡−𝑠)∇𝑢(𝑠)d𝑠d𝑥.
To estimate the last term on the right-hand side of
the previous equality, let 𝛿>0 be determined later.
Young’s inequality drives to:
|𝑢|()𝑢𝑢d𝑥≤1
𝑘𝛿()|𝑢|()d𝑥
+𝑘−1
𝑘𝛿()
()|𝑢|()d𝑥.
This yields by substitution in (3.14):
()
≥𝛾∫|∇𝑢|d𝑥+𝑟∫|𝑢|()d𝑥+
𝜀‖𝑢‖
+𝜀∫|𝑢(𝑡)|()d𝑥
−𝜀|∇𝑢|d𝑥−𝜀𝑟1
𝑘𝛿()|𝑢|()d𝑥
−𝜀𝑟𝑘−1
𝑘𝛿()
()|𝑢|()d𝑥
+𝜀‖∇𝑢‖
𝑔(𝑠)d𝑠
+𝜀
𝑔(𝑡−𝑠)∇𝑢(𝑡)∇𝑢(𝑠)−∇𝑢(𝑡)d𝑥d𝑠
(3.15)
and for some positive number 𝜂 to be determined
later,
Ω 0
0 Ω
2
20
2
20
. d d
ds
1
1
4
t
t
t
t
u t g t s u s dsdx
g t s u t u t u s x s
u g s
u g s ds g u t
We want now to estimate the term involving
∫|𝑢|()d𝑥 in (3.15), we have
‖𝑢‖=‖𝑢‖
‖𝑢‖
≤𝐶‖𝑢‖(.)
‖∇𝑢‖
,
and ‖𝑢‖=‖𝑢‖
‖𝑢‖
≤𝐶‖𝑢‖(.)
‖∇𝑢‖
,
which operates for:
2𝑛
𝑛−2≥𝑘≥𝑘>2 𝑎𝑛𝑑 0<𝑛
2−𝑛
𝑘≤𝑛
2−𝑛
𝑘
≤𝑠<1,
Thus, we have the following inequality:
∫|𝑢|()d𝑥≤‖𝑢‖
+‖𝑢‖
≤𝐶max𝜚(.)(𝑢)()
,𝜚(.)(𝑢)()
‖∇𝑢‖
+𝐶max𝜚(.)(𝑢)()
,𝜚(.)(𝑢)()
‖∇𝑢‖
If 𝑠<𝑚𝑖𝑛
,
, using again Young’s
inequality, we get:
max𝜚(.)(𝑢)()
,𝜚(.)(𝑢)()
‖∇𝑢‖
≤𝐶max𝜚(.)(𝑢)()
,𝜚(.)(𝑢)()
+𝐶(‖∇𝑢‖
)
,
(3.16)
and
max𝜚(.)(𝑢)()
,𝜚(.)(𝑢)()
‖∇𝑢‖
≤𝐶max𝜚(.)(𝑢)()
,𝜚(.)(𝑢)()
+𝐶(‖∇𝑢‖
)
,
(3.17)
for 1/𝜇+1/𝜃=1. Here we choose (𝜃,𝜇)=
,
, and (𝜃,𝜇)=
,
in (3.16) and
(3.17), respectively. Therefore the previous
inequality becomes
|𝑢|()d𝑥≤𝐶max𝜚(𝑢)()
(),𝜚(𝑢)()
()
+𝐶max𝜚(𝑢)()
(),𝜚(𝑢)()
()+𝐶‖∇𝑢‖
.
(3.18)
Now, picking 𝑠 such that:
0<𝑠≤min
⎝
⎜
⎛
2(𝑝−𝑘)
𝑘(𝑝−2),2(𝑝−𝑘)
𝑘(𝑝−2)
,2(𝑝−𝑘)
𝑘(𝑝−2),2(𝑝−𝑘)
𝑘(𝑝−2)
⎠
⎟
⎞
<1
we get
0<𝑚𝑎𝑥()
(),()
(),
()
(),()
()≤1. (3.19)
When the inequality (3.19) is satisfied, we apply the
classical algebraic inequality:
𝑧≤(𝑧+1)≤1+1
𝜔(𝑧+𝜔),∀𝑧≥0,
0<𝑑≤1,𝜔≥0,
to get the following estimate:
max𝜚(𝑢)()
(),𝜚(𝑢)()
()
+max𝜚(𝑢)()
(),𝜚(𝑢)()
()
≤𝐶(1+𝐻(0))𝜚(𝑢)+𝐻(0)
≤𝐶𝜚(𝑢)+𝐻(𝑡) ∀𝑡≥0.
(3.20)
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Injecting the estimate (3.20) into (3.18) we obtain
the following inequality
|𝑢|()d𝑥≤𝐶(𝜚(𝑢)+2𝐻(𝑡)+𝑙‖∇𝑢‖
),
(3.21)
which yields finally:
∫|𝑢|()d𝑥≤𝐶𝜚(𝑢)+2𝐸−‖𝑢‖
+∫
()|𝑢|()d𝑥
≤𝐶2𝐸−‖𝑢‖
+1+2
𝑝|𝑢|()d𝑥.
(3.22)
Therefore by injecting the inequality (3.22) into the
inequality (3.15), we obtain:
()
≥𝛾‖∇𝑢‖
+𝜀1+
max𝛿,𝛿‖𝑢‖
−2𝐶
max𝛿,𝛿𝐸
−𝜀‖∇𝑢‖
+𝜀1−𝑟𝐶
𝑘max𝛿,𝛿1
+2
𝑝|𝑢(𝑡)|()d𝑥
+𝑟1
−𝜀(𝑘−1)
𝑘max𝛿
,𝛿
|𝑢|()dx
+𝜀1−1
4𝜂‖∇𝑢‖
𝑔(𝑠)d𝑠−𝜀𝜂(𝑔∘∇𝑢),
(3.23)
for some positive number 𝜂 to be determined later.
From the inequality
2𝐻(𝑡)=−‖𝑢‖
−2𝐸+(𝑔∘∇𝑢)
+1−
𝑔(𝑠)d𝑠‖∇𝑢‖
−2
𝑝(𝑥)|𝑢|()d𝑥,
we have
−𝑙‖∇𝑢‖
≥−1−
𝑔(𝑠)d𝑠‖∇𝑢‖
=2𝐻(𝑡)+‖𝑢‖
+(𝑔∘∇𝑢)
−2𝐸−2
𝑝(𝑥)|𝑢|()d𝑥
≥2𝐻(𝑡)−2𝐸+‖𝑢‖
+(𝑔∘∇𝑢)
−2
𝑝|𝑢|()d𝑥
(3.24)
Thus injecting it in (3.23), we get the following
inequality:
d𝐿(𝑡)
d𝑡 ≥𝛾‖∇𝑢‖
+𝜀+𝑟𝐶
𝑘max𝛿,𝛿‖𝑢‖
+𝜀1−2
𝑝−𝑟𝐶
𝑘max𝛿,𝛿1
+2
𝑝|𝑢|()d𝑥
+𝜀2𝐻(𝑡)−21+
max𝛿,𝛿𝐶𝐸
+𝑟1−
()
max𝛿
,𝛿
∫|𝑢|()d𝑥
+𝜀1−1
4𝜂‖∇𝑢‖
𝑔(𝑠)d𝑠
+(1−𝜀𝜂)(𝑔∘∇𝑢) (3.25)
Using the definition of 𝛼 and 𝐸 (see equation (2.8)
and the lemma 3.5), we have
−2𝐸−4
max𝛿,𝛿𝐶𝐸
−2𝐸(𝐵𝛼)
(𝐵𝛼)
−4𝑟
𝑘max𝛿,𝛿𝐶𝐸(𝐵𝛼)
(𝐵𝛼)
≥−2𝐸(𝐵𝛼)
−4𝐶𝑟
𝑘max𝛿,𝛿𝐸(𝐵𝛼)
|𝑢|()d𝑥
Finally, we obtain
d𝐿(𝑡)
d𝑡 ≥𝛾‖∇𝑢‖
+𝜀2+𝑟𝐶
𝑘max𝛿,𝛿‖𝑢‖
+𝜀
⎝
⎜
⎜
⎛
1−2
𝑝−2𝐸(𝐵𝛼)
−𝑟𝐶
𝑘max𝛿,𝛿 1+2
𝑝
+4𝐸(𝐵𝛼)
⎠
⎟
⎟
⎞
∫|𝑢|()d𝑥
+2𝜀𝐻(𝑡)+
max𝛿,𝛿𝐶𝐸
+𝜀1−1
4𝜂‖∇𝑢‖
𝑔(𝑠)d𝑠
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+(1−𝜀𝜂)(𝑔∘∇𝑢)
+𝑟1
−𝜀(𝑘−1)
𝑘max𝛿
,𝛿
|𝑢|()d𝑥,
(3.26)
we have
1−2
𝑝−2𝐸(𝐵𝛼)
>0
since 𝛼>𝐵
.
We choose now 𝛿 small enough such that
⎝
⎜
⎜
⎛
1−2
𝑝−2𝐸(𝐵𝛼)
−𝑟𝐶
𝑘max𝛿,𝛿 1+2
𝑝
+4𝐸(𝐵𝛼)
⎠
⎟
⎟
⎞
>0
and taking 𝜂>
. Once 𝛿 and 𝜂 are fixed, we choose
𝜀 small enough such that:
1−𝜀𝜂>0,1−𝜀(𝑘−1)
𝑘max𝛿
,𝛿
>0𝑎𝑛𝑑𝐿(0)>0
Therefore, the inequality (3.26) becomes
d𝐿(𝑡)
d𝑡 ≥𝜀𝜅𝐻(𝑡)+‖𝑢‖
+|𝑢|()d𝑥+𝐸
(3.27)
for some 𝜅>0.
Next, it is clear that, by Young’s inequality and
Poincaré’s inequality, we get
𝐿(𝑡)≤𝜆𝐻(𝑡)+‖𝑢‖
+‖∇𝑢‖
for some 𝜆>0. (3.28)
From (3.11), we have
‖∇𝑢‖
≤2𝑙𝐸+2
𝑙𝑝|𝑢(𝑥,𝑡)|()d𝑥, 𝑡≥0.
Thus, the inequality (3.28) becomes:
𝐿(𝑡)≤𝜁𝐻(𝑡)+‖𝑢‖
+|𝑢|()d𝑥
+𝐸 for some 𝜁>0.
(3.29)
From the two inequalities (3.27) and (3.29), we
finally obtain the differential inequality:
d𝐿(𝑡)
d𝑡 ≥𝜇𝐿(𝑡) for some 𝜇>0.
(3.30)
Integrating the previous differential inequality
(3.30) on (0,𝑡) gives the following estimate for the
function 𝐿:
𝐿(𝑡)≥𝐿(0)𝑒. (3.31)
On the other hand, from the definition of the
function 𝐿 (and for small values of the parameter 𝜀),
it results:
𝐿(0)𝑒≤𝐿(𝑡)≤1
𝑝|𝑢|()d𝑥
≤1
𝑝max|𝑢|d𝑥,|𝑢|d𝑥
(3.32)
From the two inequalities (3.31) and (3.32) we
deduce the exponential growth of the solution in the
𝐿 and 𝐿-norms.
Now, we state the blow-up results as follows.
4 An Upper Bound for the Blow-
Up Time: The Case 𝟐≤𝒌𝟏≤𝒌𝟐<𝒑𝟏
In this section, we prove the blow-up result under
the condition of 2≤𝑘≤𝑘<𝑝 with positive
initial energy and use 𝐶 to denote a generic positive
constant.
Theorem 4.1 For any fixed 𝛿<1, assume that 𝑢,
𝑢 satisfy
𝐼(0)<0, 𝐸(0)<𝛿𝐸.
(4.1)
and (2.6) holds. Suppose that
𝑔(𝑠)d𝑠
≤𝑝−2
𝑝−2+(1−𝛿)(𝑝−2)+2(1−𝛿)
(4.2)
where 𝛿=max{0,𝛿}. Under the condition of
Lemma 3.1, if
2≤𝑘≤𝑘(𝑥)≤𝑘<𝑝≤𝑝(𝑥)≤𝑝≤2𝑛
𝑛−2,
the solution of problem (1.3)-(1.5) blows up in a
finite time 𝑇, in the sense that
lim
→(‖𝑢(𝑡)‖
+∥∇𝑢(𝑡)∥
)=+∞. (4.3)
Furthermore, the upper bound for 𝑇 can be
estimated from above by
𝑇≤(1−𝑎)𝐶
𝜒𝜀𝑎(𝐿(0))
,
(4.4)
where the positive constants 𝑎, 𝐶, 𝜒, and 𝜀 to be
determined later.
Before the proof, we want to introduce some
materials and lemmas firstly we provide the
following functions:
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𝐸(𝑡)=1
2(𝑔∘∇𝑢)+1
2‖𝑢‖
+1
21−
𝑔(𝑠)d𝑠‖∇𝑢‖
−1
𝑝(𝑥)|𝑢|()d𝑥,
(4.5)
𝐼(𝑢(𝑡)=1−
𝑔(𝑠)d𝑠
∥∇𝑢(𝑡)∥
+(𝑔∘∇𝑢)(𝑡)
−|𝑢|()d𝑥,
(4.6)
𝐽𝑢(𝑡)=1
21−
𝑔(𝑠)d𝑠
∥∇𝑢(𝑡)∥
+(𝑔∘∇𝑢)(𝑡)
−1
𝑝(𝑥)|𝑢|()d𝑥,
(4.7)
𝐸(𝑡)=𝐸(𝑢(𝑡),𝑢(𝑡))=𝐽(𝑡)+1
2‖𝑢(𝑡)‖
(4.8)
Secondly, we need the following two lemmas.
Lemma 4.2 Under the same conditions as
in Theorem 4.1, one has 𝐼(𝑡)<0
and 𝐸<𝑝−2
2𝑝1−
𝑔(𝑠)d𝑠∥∇𝑢∥
+(𝑔∘∇𝑢)
<𝑝−2
2𝑝|𝑢|()d𝑥
for all 𝑡∈0,𝑇). (4.9)
Proof. By (3.2) and (4.1), we have 𝐸(𝑡)≤𝛿𝐸, for
all 𝑡∈0,𝑇). Besides, we can get 𝐼(𝑡)<0 for all
𝑡∈0,𝑇). If it is not true, then there exists some 𝑡∗∈
0,𝑇) such that 𝐼(𝑡∗)=0. So 𝐼(𝑡)<0 for all 0≤
𝑡<𝑡∗, i.e.
1−
𝑔(𝑠)d𝑠∥∇𝑢(𝑡)∥
+(𝑔∘∇𝑢)(𝑡)
<|𝑢|()d𝑥<‖𝑢‖(.)
,
0≤𝑡<𝑡∗
(4.10)
By the proof of Lemma 3.4, we have
𝐸=𝑝−2
2𝑝𝑙
1
𝐵
≤𝑝−2
2𝑝𝑙∥∇𝑢∥
‖𝑢‖(.)
<𝑝−2
2𝑝
⎝
⎜
⎜
⎜
⎛
1−∫
𝑔(𝑠)d𝑠∥∇𝑢∥
+(𝑔∘∇𝑢)(𝑡)
1−∫
𝑔(𝑠)d𝑠∥∇𝑢(𝑡)∥
+(𝑔∘∇𝑢)(𝑡)
⎠
⎟
⎟
⎟
⎞
=𝑝−2
2𝑝1−
𝑔(𝑠)d𝑠∥∇𝑢(𝑡)∥
+(𝑔∘∇𝑢)(𝑡),
0≤𝑡<𝑡∗
(4.11)
Joined with (4.10) and (4.11), we obtain
|𝑢|()d𝑥> 2𝑝
𝑝−2𝐸>0, 0≤𝑡<𝑡∗
By the continuity of 𝑡↦∫|𝑢(𝑡)|()d𝑥, we get
𝑢(𝑡∗)≠0. By (4.7), we get
𝐸≤𝑝−2
2𝑝|𝑢(𝑡∗)|()d𝑥≤𝐽𝑢(𝑡∗)
which contradicts 𝐽𝑢(𝑡∗)≤𝐸(𝑡∗)<𝐸. By
repeating the previous step, we obtain (4.9). This
completes the proof.
We set
𝐻(𝑡)=𝛿𝐸−𝐸(𝑡),
(4.12)
then under the condition of theorem 4.1, we obtain
2
Ω
2
2Ω Ω
1
d2
1d d 0
2
t
k x k x
t t
H t u x g u
g t u r u x r u x
(4.13)
and
0<𝐻(0)≤𝐻(𝑡)<𝛿𝐸+1
𝑝(𝑥)|𝑢|()d𝑥
<𝛿𝑝−2
2𝑝+1
𝑝|𝑢|()d𝑥.
(4.14)
It’s easy to examine the following lemma
Lemma 4.3 Under the assumptions of Theorem 4.1,
we have
‖𝑢(𝑡)‖
≤𝐶−𝐻(𝑡)−‖𝑢‖
−(𝑔∘∇𝑢)(𝑡)
+|𝑢|()d𝑥
(4.15)
for any 𝑢∈𝐻(Ω) and 2≤𝑠≤𝑝
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Proof of Theorem 4.1. Assume by contradiction that
(4.3) does not hold true. Then for all 𝑇∗<+∞ and
all 𝑡∈0,𝑇∗], we have
‖𝑢(𝑡)‖
+∥∇𝑢(𝑡)∥
≤𝐶, (4.16)
with 𝐶 is a positive constant. Motivated by [19], we
set the function
𝐿(𝑡)=𝐻(𝑡)+𝜀𝑢𝑢d𝑥+𝜀𝛾2|∇𝑢|d𝑥,𝑡
≥0,
(4.17)
where 𝜀>0 is a positive constant to be chosen
later, and
0<𝑎≤min
,
()<1, (4.18)
derivative the Eq (4.17) and using Eq. (1.3)-(1.5) we
obtain
𝐿(𝑡)=(1−𝑎)𝐻(𝑡)𝐻(𝑡)+𝜀‖𝑢(𝑡)‖
+𝜀|𝑢|()
d𝑥−𝜀𝛾‖∇𝑢(𝑡)‖
−𝜀𝑟|𝑢|()
𝑢𝑢d𝑥
+𝜀∇
𝑢(𝑡).𝑔
(𝑡−𝑠)∇𝑢(𝑠)d𝑠d𝑥.
(4.19)
Applying the relation
𝑝𝐻(𝑡)= 𝑝𝛿𝐸−
‖𝑢(𝑡)‖
−
1−∫
𝑔(𝑠)d𝑠∥∇𝑢(𝑡)∥
−
(𝑔∘∇𝑢)(𝑡)+𝑝∫
()|𝑢|()d𝑥
and Young’s inequality, we get from (4.19)
that 𝐿(𝑡)=(1−𝑎)𝐻(𝑡)𝐻(𝑡)
+𝜀1+𝑝
2‖𝑢(𝑡)‖
−𝜀𝑝𝛿𝐸+𝜀𝑝𝐻(𝑡)
+𝜀𝑝
21−𝑔
(𝑠)d𝑠−1‖∇𝑢(𝑡)‖
−𝜀𝑟|𝑢|()
𝑢𝑢d𝑥+𝜀𝑝
2(𝑔∘∇𝑢)(𝑡)
+𝜀|𝑢|()
d𝑥−𝑝1
𝑝(𝑥)
|𝑢|()d𝑥
+𝜀𝑔
(𝑡−𝑠)‖∇𝑢(𝑡)‖
d𝑠
+𝜀𝑔
(𝑡−𝑠)∇
𝑢(𝑡).∇𝑢(𝑠)−∇𝑢(𝑡)d𝑥d𝑠
≥𝑟(1−𝑎)𝐻(𝑡)|𝑢|()d𝑥
+𝜀1+𝑝
2‖𝑢(𝑡)‖
−𝜀𝑝𝛿𝐸+𝜀𝑝𝐻(𝑡)
+𝜀
⎣
⎢
⎢
⎡
𝑝
2−1
−𝑝
2−1+1
4𝜂
𝑔(𝑠)d𝑠
⎦
⎥
⎥
⎤
∥∇𝑢(𝑡)∥
+𝜀𝑝
2−𝜂(𝑔∘∇𝑢)(𝑡)−𝜀𝑟|𝑢|()𝑢𝑢d𝑥
(4.20)
By (4.9), estimate (4.20) becomes
𝐿(𝑡)≥ 𝑟(1−𝑎)𝐻(𝑡)|𝑢|()d𝑥
+𝜀1+𝑝
2‖𝑢(𝑡)‖
+𝜀𝑝𝐻(𝑡)
+𝜀(1−𝛿)𝑝
2−1
−1−𝛿𝑝
2−1
+1
4𝜂
𝑔(𝑠)d𝑠
⎭
⎪
⎪
⎬
⎪
⎪
⎫
∥∇𝑢(𝑡)∥
+𝜀1−𝛿𝑝
2−1
+(1−𝜂) (𝑔∘∇𝑢)(𝑡)
−𝜀𝑟|𝑢|()𝑢𝑢d𝑥
(4.21)
Now, by using Young’s inequality, we estimate the
last term in (4.21) as follows
|𝑢|()𝑢𝑢d𝑥
≤1
𝑘𝛿()|𝑢|()d𝑥
+𝑘−1
𝑘𝛿()
()|𝑢|()d𝑥,
∀𝛿>0
(4.22)
Therefore by taking 𝛿 so that
𝛿()
()=𝑀𝐻(𝑡)
for a large constant 𝑀 to be specified later, and
substituted in (4.22)
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|𝑢|()|𝑢|d𝑥
≤1
𝑘𝑀()|𝑢|()𝐻(())(𝑡)d𝑥
+𝑘−1
𝑘𝑀𝐻(𝑡)|𝑢|()d𝑥
Then estimate (4.21) takes the form
𝐿(𝑡)≥𝑟(1−𝑎)−𝜀𝑘−1
𝑘𝑀
𝐻(𝑡)]|𝑢|()d𝑥+𝜀1+𝑝
2‖𝑢(𝑡)‖
+𝜀𝑝𝐻(𝑡)−𝑀
𝑘𝐻()(𝑡)|𝑢|()d𝑥
+𝜀(1−𝛿)𝑝
2−1
−(1−𝛿)𝑝
2−1+1
4𝜂
𝑔(𝑠)d𝑠∥∇𝑢∥
+𝜀(1−𝛿)𝑝
2−1+(1−𝜂)(𝑔∘∇𝑢)(𝑡)
Applying (4.2) and picking
𝜂<(1−𝛿)𝑝
2−1+1,
we can get
𝐿(𝑡)≥𝑟[(1−𝑎)
−𝜀𝑘−1
𝑘𝑀𝐻(𝑡)|𝑢|()d𝑥
+𝜀1+𝑝
2‖𝑢(𝑡)‖
+𝜀𝑝𝐻(𝑡)−𝑀
𝑘𝐻()(𝑡)|𝑢|()d𝑥
+𝜀𝑀∥∇𝑢(𝑡)∥
+𝜀𝑀(𝑔∘∇𝑢)(𝑡),
(4.23)
where
𝑀=(1−𝛿)𝑝
2−1
−(1−𝛿)𝑝
2−1
+1
4𝜂
𝑔(𝑠)d𝑠>0,
𝑀=(1−𝛿)𝑝
2−1+(1−𝜂)>0.
Using
𝐻()(𝑡)|𝑢|()d𝑥
≤𝐶𝛿𝑝−2
2𝑝+1
𝑝()
∥𝑢(𝑡)∥(.)
()
hence (4.23) yields
𝐿(𝑡)≥𝑟(1−𝑎)−𝜀𝑘−1
𝑘𝑀
𝐻(𝑡)]|𝑢|()d𝑥++𝜀1+𝑝
2‖𝑢(𝑡)‖
+𝜀𝑝𝐻(𝑡)−𝑀
𝑘𝐶𝛿𝑝−2
2𝑝+1
𝑝()
∥𝑢(𝑡)∥(.)
()
+𝜀𝑀∥∇𝑢(𝑡)∥
+𝜀𝑀(𝑔∘∇𝑢)(𝑡).
(4.24)
We then use Lemma 4.3 and (4.18), for 𝑠=𝑘+
𝑝𝑎(𝑘−1)≤𝑝, to deduce from (4.24)
𝐿(𝑡)≥𝑟(1−𝑎)
−𝜀𝑘−1
𝑘𝑀𝐻|𝑢|()d𝑥
+𝜀1+𝑝
2‖𝑢(𝑡)‖
+𝜀𝑀∥∇𝑢(𝑡)∥
+𝜀𝑀(𝑔∘∇𝑢)(𝑡)
+𝜀[𝑝𝐻(𝑡)−𝐶𝑀
×−𝐻(𝑡)−‖𝑢(𝑡)‖
−(𝑔∘∇𝑢)(𝑡)
+𝜚(𝑢)
≥𝑟(1−𝑎)
−𝜀𝑘−1
𝑘𝑀𝐻(𝑡)|𝑢|()d𝑥
+𝜀1+𝑝
2+𝐶𝑀‖𝑢(𝑡)‖
+𝜀𝑀∥∇𝑢(𝑡)∥
+𝜀 𝑀
+𝐶𝑀(𝑔∘∇𝑢)(𝑡)
+𝜀𝑝+𝐶𝑀𝐻(𝑡)−𝜀𝐶𝑀𝜚(𝑢)
(4.25)
where 𝐶=
𝛿
+
(). By (3.1) and
(4.12), we obtain
𝐻(𝑡)≥1
𝑝𝜚(𝑢)−1
2‖𝑢(𝑡)‖
−1
2‖∇𝑢(𝑡)‖
−1
2(𝑔∘∇𝑢)(𝑡),
writing 𝑝=2𝑀+(𝑝−2𝑀), where 𝑀=
min{𝑀,𝑀}, the estimate (4.25) yields
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𝐿(𝑡)≥𝑟(1−𝑎)
−𝜀𝑘−1
𝑘𝑀𝐻|𝑢|()d𝑥
+𝜀1+𝑝
2+𝐶𝑀−𝑀‖𝑢(𝑡)‖
+𝜀(𝑀−𝑀)∥∇𝑢(𝑡)∥
+𝜀𝑀+𝐶𝑀−𝑀(𝑔∘∇𝑢)(𝑡)
+𝜀𝑝−2𝑀+𝐶𝑀𝐻(𝑡)
+𝜀2𝑀
𝑝−𝐶𝑀𝜚(𝑢).
(4.26)
We choose 𝑀 large enough, (4.26) becomes
𝐿(𝑡)≥ 𝑟(1−𝑎)−
𝜀𝑘−1
𝑘𝑀𝐻(𝑡)|𝑢|()d𝑥
+𝜒𝜀𝐻(𝑡)+‖𝑢(𝑡)‖
+𝜚(𝑢)
+(𝑔∘∇𝑢)(𝑡),
(4.27)
for some positive constant 𝜒. Once 𝑀 is fixed, we
choose 𝜀 small enough such that
(1−𝑎)−𝜀𝑘−1
𝑘𝑀>0,
and
𝐿(0)=𝐻(0)+𝜀𝑢𝑢d𝑥+𝜀𝛾
2‖∇𝑢‖
>0
Hence, we have
𝐿(𝑡)≥𝜒𝜀𝐻(𝑡)+‖𝑢(𝑡)‖
+𝜚(𝑢)
+(𝑔∘∇𝑢)(𝑡) (4.28)
On the other hand, we have
𝐿
(𝑡) =
⎝
⎜
⎛
𝐻(𝑡)
+𝜀𝑢(𝑡)𝑢(𝑡)d𝑥
+𝜀𝛾2∥∇𝑢(𝑡)∥
⎠
⎟
⎞
≤𝐶
⎝
⎜
⎛
𝐻(𝑡)+𝑢(𝑡)𝑢(𝑡)d𝑥
+∥∇𝑢(𝑡)∥
⎠
⎟
⎞
.
(4.29)
By Hölder’s and Young’s inequalities, (4.16), and
Lemma 4.3, we get
∫𝑢
(𝑡)𝑢(𝑡)d𝑥
≤𝐶(‖𝑢(𝑡)‖‖𝑢(𝑡)‖)
≤𝐶‖𝑢(𝑡)‖(.)
‖𝑢(𝑡)‖
≤‖𝑢(𝑡)‖(.)
+‖𝑢(𝑡)‖
≤𝐶𝐻(𝑡)+‖𝑢(𝑡)‖
+ϱ(𝑢)+(𝑔∘∇𝑢)(𝑡),
(4.30)
And ∥∇𝑢(𝑡)∥
≤𝐶
. (4.31)
By Poincaré’s inequality and (4.16), we have
∥𝑢(𝑡)∥(.)
≤𝐵∥∇𝑢(𝑡)∥
≤𝐵𝐶
(4.32)
By virtue of (4.14) and (4.32), we get 𝐻(𝑡) is
bounded. There exists a positive constant 𝐶 such
that
𝐻(𝑡)+𝐶
≤𝐶𝐻(𝑡)
(4.33)
Therefore, we obtain
𝐿
(𝑡)≤𝐶𝐻(𝑡)+‖𝑢(𝑡)‖
+𝜚(𝑢)
+(𝑔∘∇𝑢)(𝑡)
(4.34)
By joining (4.28) and (4.34), we reach that
𝐿(𝑡)≥𝜀𝜒
𝐶𝐿
(𝑡)
(4.35)
A simple integration of (4.35) over [0,𝑡], yields that
𝐿
(𝑡)≥ 1
𝐿
(0)−
()𝑡, ∀𝑡≥0
This shows that 𝐿(𝑡) blows up in a finite time 𝑇,
where
𝑇≤(1−𝑎)𝐶
𝜒𝜀𝑎[𝐿(0)]
If we choose 𝑇∗≥()
[()]
, then we obtain 𝑇≤
𝑇∗, which contradicts our assumption. This
completes the proof.
5 An Upper Bound for the Blow-
Up Time: The Case 𝒌(𝒙)=
𝟐,∀𝒙
In this section, we prove a finite time blow-up
result. We need the following lemma.
Lemma 5.1 ([14], Lemma1.1 and, [16],
Logarithmic convexity methods) Assume that 𝜑∈
𝐶([0,𝑇)) satisfying:
𝜑𝜑−(1+𝛼)(𝜑)≥0, 𝛼>0
and
𝜑(0)>0, 𝜑(0)>0,
then
𝜑→∞as𝑡→𝑡≤𝑡=𝜑(0)
𝛼𝜑(0).
Theorem 5.2 For any fixed 𝛿<1, suppose that
(2.6) holds and 𝑢, 𝑢 satisfy
𝐼(0)<0,𝐸(0)<𝛿𝐸. (5.1)
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Assume that
𝑔(𝑠)d𝑠
≤𝑝−2
𝑝−2+(1−𝛿)𝑝+2𝛿(1−𝛿)
(5.2)
where 𝛿=max{0,𝛿}, and suppose further that
∫𝑢𝑢d𝑥>0 for 0≤𝐸(0)<𝐸. Under the
assumption of Lemma 3.1, if
2=𝑘=𝑘(𝑥)=𝑘<𝑝≤𝑝(𝑥)≤𝑝≤2𝑛
𝑛−2,
the solution of problem (1.3)-(1.5) blows up in a
finite time 𝑇, in the sense that
lim
→
⎣
⎢
⎢
⎢
⎡
∥𝑢(𝑡)∥
+
∥∇𝑢(𝑠)∥
d𝑠
+
∥𝑢(𝑠)∥
d𝑠
⎦
⎥
⎥
⎥
⎤
=+∞
Further, the upper bound for 𝑇 can be estimated by
𝑇≤2(𝑝−2)∥𝑢∥
+8(𝛾‖∇𝑢‖
+𝑟‖𝑢‖
)
(𝑝−2)∫𝑢𝑢d𝑥 ,
with some 𝑡>0 and 𝜑 is defined in (5.1).
Proof. Assume by contradiction that the solution 𝑢
is global. Then for any 𝑇>0, we define the
functional 𝜑 as follows
𝜑(𝑡)=∥𝑢(𝑡)∥
+𝛾
∥∇𝑢(𝑠)∥
ds
+𝑟
∥𝑢(𝑠)∥
d𝑠
+(𝑇−𝑡)[𝛾‖∇𝑢‖
+𝑟‖𝑢‖
]
+(𝑡+𝑡), 𝑡<𝑇
(5.3)
where 𝑡,𝑇 and 𝛽 are positive constants to be
chosen later. Then using equation (1.3) and
integration by parts, to get
Ω
0 Ω
0
0 Ω
2 d
2 . d d
d d 2 .
t
t
s
t
s
t u t u t x
u s u s x s
u s u s x s t t
(5.4)
And
𝜑(𝑡)=2‖𝑢(𝑡)‖
+2∫𝑢(𝑡)𝑢(𝑡)d𝑥
−21−∫
𝑔(𝑠)d𝑠∥∇𝑢(𝑡)∥
+2𝛾∫∇𝑢(t).∇𝑢(𝑡)d𝑥
+2𝑟∫𝑢(𝑡)𝑢(𝑡)d𝑥+2.
(5.5)
Furthermore 𝜑(𝑡)≥2‖𝑢(𝑡)‖
−21−𝑔
(𝑠)d𝑠‖∇𝑢(𝑡)‖
−1𝜀𝑔
(𝑠)d𝑠‖∇𝑢(𝑡)‖
−𝜀(𝑔∘∇𝑢)(𝑡)+2
+𝑝(𝑔∘∇𝑢)+𝑝‖𝑢‖
+𝑝1−𝑔
(𝑠)d𝑠‖∇𝑢‖
−2𝑝𝐸(𝑡)
≥(𝑝+2)‖𝑢(𝑡)‖
+(𝑝−2)−𝑝−2+1𝜀𝑔
(𝑠)d𝑠‖∇𝑢‖
+(𝑝−𝜀)(𝑔∘∇𝑢)(𝑡)−2𝑝𝐸(0)+2,
(5.6)
where 𝑡,𝑇 are constants to be determined later.
Case 1: If 𝛿<0, then 𝐸(0)<0, we choose 𝜀=𝑝
in (5.6). Then, by (5.3), (5.4), (5.5), (5.2), and (5.6),
we have
⎩
⎪
⎪
⎨
⎪
⎪
⎧
2 2
2
0 0 0 0
2 2
Ω
2
0
0 d
0
u x x T u r u
t
;
𝜑(0)=2𝑢𝑢d𝑥+2𝑡>0;
𝜑(𝑡)≥2−2𝑝𝐸(0)>0∀𝑡≥0.
Therefore 𝜑 and 𝜑 are both positive. Thus, from
(5.3)-(5.6) and (5.8), the following inequality,
inferred for all (𝜁,𝜂)∈ ℝ
, which implies that
𝜑(𝑡)𝜑(𝑡)−𝑝+2
4𝜑(𝑡)≥0.
(5.7)
Case2: If 0≤𝛿<1, then
0≤𝐸(0)<𝛿𝐸<𝐸,
we choose 𝜀=(1−𝛿)𝑝+2𝛿 in (5.6), using (5.2)
Then, we have
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2
12
2
1 1 2
0
1
1 1
22
1 1 2
20
1
2
1 1 1
2
2
1
2 2 d
1 2
2 2 0 2
2 2 1 d
2 0 2
2 2 0 2 0.
t
t
t
t
t
t p u t
p p g s s u
p
p g u t p E
p u t p g s s u g u t
p E
p u t p E E
(5.8)
Then, by (5.3), (5.4), (5.5), (5.2), and (5.6), we have
⎩
⎪
⎪
⎨
⎪
⎪
⎧
𝜑(0)=𝑢(𝑥)d𝑥+𝑡
+𝑇[𝛾‖∇𝑢‖
+𝑟‖𝑢‖
]>0;
𝜑(0)=2𝑢𝑢d𝑥+2𝑡>0;
𝜑(𝑡)≥2−2𝑝𝐸(0)>0∀𝑡≥0.
Therefore 𝜑 and 𝜑 are both positive.
Then using Lemma5.1, to infer
𝜑(𝑡)→∞
as 𝑡→𝑇∗, where,
𝑇∗≤2∥𝑢∥
+2𝑇[𝛾‖∇𝑢‖
+𝑟‖𝑢‖
]+2𝑡
(𝑝−2)∫𝑢𝑢d𝑥+𝑡.
Now we go to choose appropriate 𝑡 and 𝑇. Let 𝑡
be any number that depends only on 𝑝, 𝐸−
𝐸(0) and ‖𝑢‖() as
𝑡≥2(𝛾‖∇𝑢‖
+𝑟‖𝑢‖
)
(𝑝−2)
Fix 𝑡, then 𝑇 can be picked as
𝑇=2∥𝑢∥
+2𝑇[𝛾‖∇𝑢‖
+𝑟‖𝑢‖
]+2𝑡
(𝑝−2)∫𝑢𝑢d𝑥+𝑡
so that
𝑇
=2∥𝑢∥
+2𝑡
(𝑝−2)𝑡+(𝑝−2)∫𝑢𝑢d𝑥−2𝛾‖∇𝑢‖
+𝑟‖𝑢‖
Therefore the lifespan of the solution 𝑢(𝑥,𝑡) is
bounded by
𝑇∗≤inf
2∥𝑢∥
+2𝑡
(𝑝−2)𝑡+(𝑝−2)∫𝑢𝑢d𝑥−
2(𝛾‖∇𝑢‖
+𝑟‖𝑢‖
)
=2(𝑝−2)∥𝑢∥
+8(𝛾‖∇𝑢‖
+𝑟‖𝑢‖
)
(𝑝−2)∫𝑢𝑢d𝑥 .
6 A Lower Bound for the Blow-Up
Time: The Case 𝒌𝟏≥𝟐.
In this section, by using a first-order differential
inequality technique for a suitably defined auxiliary
function and some Sobolev-type inequalities, we
give a lower bound for the blow-up time 𝑇 for the
solution 𝑢(𝑥,𝑡) of the problem (1.3)-(1.5) if
2≤𝑘≤𝑘(𝑥)≤𝑘<𝑝
≤𝑝(𝑥)≤𝑝≤
, (6.1)
holds.
Theorem 6.1 Under the condition of Lemma3.1,
assume that (6.1) holds, then the solution of
problem (1.3)-(1.5) will blow up in finite time 𝑇.
Moreover, the blow-up time 𝑇 can be estimated
from above by 𝑇, where 𝑇
=max
⎝
⎜
⎜
⎜
⎜
⎜
⎛
2ln2
𝑙(‖𝑢‖
+∥∇𝑢∥
)
+|Ω|
𝑝−2 ,
2ln2
𝑙(‖𝑢‖
+∥∇𝑢∥
)
+|Ω|
𝑝−2
⎠
⎟
⎟
⎟
⎟
⎟
⎞
(6.2)
and |Ω|=∫d𝑥.
Proof. We assume that 𝑢(𝑥,𝑡) blows up at time 𝑇
and define the auxiliary functional
𝜑(𝑡)=1
2‖𝑢(𝑡)‖
+1
21−
𝑔(𝑠)d𝑠
∥∇𝑢(𝑡)∥
+(𝑔∘∇𝑢)(𝑡)
(6.3)
Taking a derivative of 𝜑(𝑡), and using (1.3), we get
𝜑(𝑡)= 𝑢(𝑡)𝑢(𝑡)d𝑥
+1−
𝑔(𝑠)d𝑠∇𝑢(𝑡).∇𝑢(𝑡)d𝑥
−𝑔(𝑡)∥∇𝑢(𝑡)∥
+(𝑔∘∇𝑢)(𝑡)
+∇𝑢(𝑡)
𝑔(𝑡−𝑠)(∇𝑢(𝑡)−∇𝑢(𝑠))d𝑠d𝑥
=−𝛾‖∇𝑢(𝑡)‖
−𝑔(𝑡∥∇𝑢(𝑡)∥
+(𝑔∇𝑢)(𝑡)
+𝑢𝑢|𝑢|()d𝑥−𝑟|𝑢|()d𝑥
≤𝑢𝑢|𝑢|()d𝑥
(6.4)
Using Young’s inequality, we have
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∫|𝑢|()𝑢𝑢d𝑥
≤
∫𝑢d𝑥+
∫|𝑢|()d𝑥
≤
∫𝑢d𝑥
+
max∫|𝑢|d𝑥,∫|𝑢|d𝑥
≤
∫𝑢d𝑥
+
max𝐶∥∇𝑢∥
,𝐶∥∇𝑢∥
≤
𝜑+
max
𝜑,
𝜑
(6.5)
where
𝐶=|Ω|
2
, 𝐶=|Ω|
2
.
By (2.6), (6.4), and (6.5), we can obtain that
𝜑(𝑡)≤1
2𝜑+1
2max𝐶
𝑙𝜑,𝐶
𝑙𝜑
(6.6)
Integrating inequality (6.6), we have
𝜑(𝑡)
≤max
⎝
⎜
⎜
⎜
⎜
⎜
⎜
⎛
(𝜑(0))+𝐶
𝑙𝑒()
−𝐶
𝑙
,(𝜑(0))+𝐶
𝑙𝑒()
−𝐶
𝑙
⎠
⎟
⎟
⎟
⎟
⎟
⎟
⎞
Let
0<𝑇∗:=max2
𝑝−2ln𝑙
𝐶(𝜑(0))
+1, 2
𝑝−2ln𝑙
𝐶(𝜑(0))
+1<∞.
(6.7)
then 𝜑(𝑡) blows up at time 𝑇∗. Hence, 𝑢(𝑥,𝑡)
discontinues at some finite time 𝑇≤𝑇∗, that is to
means, 𝑢(𝑥,𝑡) blows up at a finite time 𝑇.
Next, we estimate 𝑇. By the values of 𝐶, 𝐶, we
have
(𝜑(0))+1≤
(‖‖
∥∇∥
)||
||
,
𝑙
𝐶(𝜑(0))+1
≤2
𝑙(‖𝑢‖
+∥∇𝑢∥
)+|Ω|
|Ω|
.
The above pair inequalities coupling (6.7) give 𝑇≤
𝑇∗≤𝑇, where 𝑇 is fixed in (6.2).
7 General Comments and Issues
This paper is devoted to studying a model of a
nonlinear viscoelastic wave equation with damping
and source terms involving variable-exponent
nonlinearities (1.3)-(1.5).
1. We prove that the energy grows exponentially,
and thus so the 𝐿 and 𝐿-norms. For the case
2≤k(.)<p(.), we reach the exponential growth result
in a blow-up in finite time with positive initial
energy and get the upper bound for the blow-up
time.
2. For the case k(.)=2, we use the concavity method
to show a finite time blow-up result and get the
upper bound for the blow-up time of the solutions.
3. Furthermore, for the case k(.)≥2, under some
conditions on the data, we give a lower bound for
the blow-up time when the blow-up occurs.
-The natural question that we can ask is whether the
obtained decay rate (3.32) is optimal.
-The second question is the extension of our results
to the case of other boundary conditions than (1,4),
especially the proof of the lack of exponential
stability.
-The last interesting question we note here is
proving the stability of (1.3)-(1.5) in the whole
space ℝ (𝑛≥1) (instead of Ω).
Acknowledgments:
The authors would like to sincerely thank the
anonymous referees and the handling editor for their
reading and relevant remarks/suggestions on several
points of the paper.
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Contribution of Individual Authors to the
Creation of a Scientific Article (Ghostwriting
Policy)
Soufiane Benkouider and Abita Rahmoune wrote
the main manuscript text. All authors reviewed the
manuscript.
Sources of Funding for Research Presented in a
Scientific Article or Scientific Article Itself
No funding was received for conducting this study.
Conflict of Interest
The authors have no conflict of interest to declare.
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(Attribution 4.0 International, CC BY 4.0)
This article is published under the terms of the
Creative Commons Attribution License 4.0
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WSEAS TRANSACTIONS on MATHEMATICS
DOI: 10.37394/23206.2023.22.51
Soufiane Benkouider, Abita Rahmoune
E-ISSN: 2224-2880
465
Volume 22, 2023
