International Journal of Analysis and Applications

Volume 17, Number 4 (2019), 620-629

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-17-2019-620

BLOW-UP, EXPONENTIAL GROUTH OF SOLUTION FOR A NONLINEAR

PARABOLIC EQUATION WITH p (x)− LAPLACIAN

AMAR OUAOUA∗ AND MESSAOUD MAOUNI

Laboratory of Applied Mathematics and History and Didactics of Mathematics (LAMAHIS)

University of 20 August 1955, Skikda, Algeria

∗Corresponding author: a.ouaoua@univ-skikda.dz

Abstract. In this paper, we consider the following equation

ut − div
(
|∇u|p(x)−2 ∇u

)
+ ω |u|m(x)−2 ut = b |u|r(x)−2 u.

We prove a finite time blowup result for the solutions in the case ω = 0 and exponential growth in the

case ω > 0, with the negative initial energy in the both case.

1. Introduction

We consider the following boundary problem:


ut − div
(
|∇u|p(x)−2 ∇u

)
+ ω |u|m(x)−2 ut = b |u|

r(x)−2
u in Ω × (0,T) ,

u (x,t) = 0, x ∈ ∂Ω, t ≥ 0,

u (x, 0) = u0 (x) in Ω.

(1.1)

where Ω is a bounded domain in Rn,n ≥ 1 with smooth boundary ∂Ω and b > 0, ω ≥ 0 are constants, p (.) ,

m (x) and r (.) are given measurable functions on Ω satisfying

2 ≤ m1 ≤ m (x) ≤ m2 < p1 ≤ p (x) ≤ p2 < r1 ≤ r (x) ≤ r2 ≤ p∗ (x) . (1.2)

Received 2019-04-06; accepted 2019-05-07; published 2019-07-01.

2010 Mathematics Subject Classification. 35K55; 35K61; 35K60 .

Key words and phrases. nonlinear parabolic equation; p (x)− Laplacian; blow-up, exponential grouth.
c©2019 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

620

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-17-2019-620


Int. J. Anal. Appl. 17 (4) (2019) 621

p1 : = ess
x∈Ω

inf(p (x)), p2 := ess
x∈Ω

sup(p (x)),

r1 : = ess
x∈Ω

inf(r (x)), r2 := ess
x∈Ω

sup(r (x)),

m1 : = ess
x∈Ω

inf(m (x)), m2 := ess
x∈Ω

sup(m (x)),

and

p∗ (x) =




np(x)
esssup

x∈Ω
(n−p(x)) if p2 < n

+∞ if p2 ≥ n
.

We also assume that p (.) , m (.) and r (.) satisfy the log-Hölder continuity condition:

|q (x) −q (y)| ≤−
A

log |x−y|
, for a.e. x, y ∈ Ω, with |x−y| < δ, (1.3)

A > 0, 0 < δ < 1.

Equation (1.1) can be viewed as a generalization of the evolutional p-Laplacian equation

ut − div
(
|∇u|p−2 ∇u

)
+ ω |u|m−2 ut = b |u|

r−2
u,

with the constant exponent of nonlinearity p, m, r ∈ (2, ∞) , which appears in various physical contexts.

In particular, this equation arises from the mathematical description of the reaction-diffusion/ diffusion,

heat transfer, population dynamics processus, and so on (see [11]) and references therein). Recently in [1],

in the case ω = 0, Agaki proved an existence and blow up result for the initial datum u0 ∈ Lr(). Ôtani

[17] studied the existence and the asymptotic behavior of solutions of (1.1) and overcome the difficulties

caused by the use of nonmonotone perturbation theory. The quasilinear case, with p 6= 2, requires a strong

restriction on the growth of the forcing term |u|r−2u, which is caused by the loss of the elliptic estimate for

the p−Laplacian operator defined by ∆pu = div(|∇u|p−2∇u) (see [2]).

Alaoui et al [12] considered the following nonlinear heat equation




ut − div
(
|∇u|p(x)−2 ∇u

)
= |u|r(x)−2 u + f, in Ω × (0,T) ,

u (x,t) = 0, x ∈ ∂Ω × (0,T) ,

u (x, 0) = u0 (x) in Ω.

(1.4)

Where Ω is a bounded domain in Rn with smooth boundary ∂Ω. Under suitable conditions on r and p

and for f = 0, they showed that any solution with nontrivial initial datum blows up in finite time. In the

absence of the diffusion term in equation (1.1) when p (x) = p and r (x) = r proved the existence and plow

up results have been established by many authors (See [1 − 3, 9, 14, 17]).



Int. J. Anal. Appl. 17 (4) (2019) 622

We should also point out that Polat [18] established a blow-up result for the solution with vanishing initial

energy of the following initial boundary value problem

ut −uxx + |u|
m−2

ut = |u|
p−2

u. (1.5)

Where m and p are real constants.

In recent years, mush attention has been paid to the study of mathematical models of electro-theological

fluids. This models inclode hyperbolic, parapolic or elliptic equations which are nonlinear with respect to

the gradient of the thought solution with variable exponents of nonlinearity, (see [4, 5, 10, 15]).

Our objective in this paper is to study: In the section 3, the blow up of the solutions of the problem (1.1)

in the case ω = 0, in the section 4, exponential growth of solution when ω > 0.

2. Preliminaries

We present in this section some Lemmas about the Lebesque and sobolev space with variables conponents

(See [6 − 8, 12, 13]). Let p : Ω → [1, + ∞] be a measurable function, where Ω is adomain of Rn.

We define the Lebesque space with a variale exponent p (.) by

Lp(.) (Ω) :=
{
v : Ω → R : measurable in Ω, Ap(.) (λv) < +∞, for some λ > 0

}
,

where Ap(.) (v) =
∫
Ω

|v (x)|p(x) dx.

The set Lp(.) (Ω) equipped with the norm ( Luxemburg’s norm)

‖v‖p(.) := inf


λ > 0 :

∫
Ω

∣∣∣∣v (x)λ
∣∣∣∣p(x) dx ≤ 1


 ,

Lp(.) (Ω) is a Banach space [13].

We next, define the variable-exponent Sobolev space W 1,p(.) (Ω) as follows:

W 1,p(.) (Ω) :=
{
v ∈ Lp(.) (Ω) such that ∇v exists and |∇v| ∈ Lp(.) (Ω)

}
.

This is a Banach space with respect to the norm ‖v‖W1,p(.)(Ω) = ‖v‖p(.) + ‖∇v‖p(.) .

Furthmore, we set W 1,p(.) (Ω) to be the closure of C∞0 (Ω) in the space W
1,p(.)
0 (Ω). Let us note that the

space W 1,p(.) (Ω) has a differenet definition in the case of variable exponents.

However, under condition (1.3) , both definitions are equivalent [13] . The space W−1,p
′
(.) (Ω) , dual of

W
1,p(.)
0 (Ω) , is defined in the same way as the classical Sobolev spaces, where

1
p(.)

+ 1
p
′
(.)

= 1.

Lemma 2.1. (Poincaré’s inequality) Let Ω ⊂ Rn be a bounded domain and suppose that p (.) satisfies (1.3) ,

then

‖v‖p(.) ≤ c‖∇v‖p(.) , for all v ∈ W
1,p(.)
0 (Ω) .



Int. J. Anal. Appl. 17 (4) (2019) 623

Where c > 0 is a constant which depends on p1, p2, and Ω only. In particular, ‖∇v‖p(.) define an equivalent

norm on W
1,p(.)
0 (Ω) .

Lemma 2.2. If p (.) ∈ C
(
Ω
)

and q : Ω → [1, + ∞) is a measurable function such that

ess inf
x∈Ω

(p∗ (x) −q (x)) > 0 with p∗ (x) =




np(x)
esssup

x∈Ω
(n−p(x)) if p2 < n

+∞ if p2 ≥ n.

Then the embedding W
1,p(.)
0 (Ω) ↪→ L

q(.) (Ω) is continuous and compact.

Lemma 2.3. ( Hölder’s Inequality) Suppose that p, q, s ≥ 1 are measurable functions defined on Ω such

that

1

s (y)
=

1

p (y)
+

1

q (y)
, for a.e. y ∈ Ω.

If u ∈ Lp(.) (Ω) and v ∈ Lq(.) (Ω) , then uv ∈ Ls(.) (Ω) , with

‖uv‖s(.) ≤ 2‖u‖p(.) ‖v‖q(.) .

Lemma 2.4. If p a measurable function on Ω satisfying (1.2) , then we have

min
{
‖u‖p1

p(.)
, ‖u‖p2

p(.)

}
≤ Ap(.) (u) ≤ max

{
‖u‖p1

p(.)
, ‖u‖p2

p(.)

}
,

for any u ∈ Lp(.) (Ω) .

3. Blow up

In this section, we prove that the solution of equation (1.1) blow up in finite time when ω = 0. we recall

that (1.1), becomes 


ut − div
(
|∇u|p(x)−2 ∇u

)
= b |u|r(x)−2 u in Ω × (0,T) ,

u (x,t) = 0, x ∈ ∂Ω, t ≥ 0,

u (x, 0) = u0 (x) in Ω.

(3.1)

We start with a local existence result for the problem (1.1), which is a direct result of the existence

theorem by Agaki and Ôtani [2].

Proposition 3.1. For all u0 ∈ W
1,p(.)
0 (Ω), there exists a number T0 ∈ (0,T] such that the problem (1.1) has

a solution u on [0,T0] satisfying:

u ∈ Cw([0,T0]; W
1,p(.)
0 (Ω)) ∩C([0,T0], L

r(.)(Ω)) ∩W 1,2(0,T0; L2(Ω)).

We define the energy functional associaeted of the problem (1.1)

E (t) =

∫
Ω

1

p (x)
|∇u|p(x) dx− b

∫
Ω

1

r (x)
|u|r(x) dx. (3.2)



Int. J. Anal. Appl. 17 (4) (2019) 624

Theorem 3.1. Let the assumptions of proposition 1, be satisfied and assume that

E (0) < 0. (3.3)

Then the solution of the problem (3.1) , blow up in finite time.

Now, we let

H (t) := −E (t) , (3.4)

and

L (t) =
1

2

∫
Ω

u2dx. (3.5)

To prove our result, we first establesh some Lemmas.

Lemma 3.1. Assume that (1.2) and (1.3) , hold and E (0) < 0. Then

Ap(.) (∇u) <
bp2
r1
Ar(.) (u) , (3.6)

and

r1
b
H (0) < Ar(.) (u) . (3.7)

Proof. We multiply the first equation of (3.1) by ut and integratying over the domain Ω, we get

d

dt


∫

Ω

1

p (x)
|∇u|p(x) dx− b

∫
Ω

1

r (x)
|u|r(x) dx


 = −‖ut‖22 ,

then

E
′
(t) = −‖ut‖

2
2 ≤ 0. (3.8)

Integrating (3.8) over (0, t) , we obtain

E (t) ≤ E (0) < 0. (3.9)

By (3.2) and (3.9) , we have

∫
Ω

1

p (x)
|∇u|p(x) dx < b

∫
Ω

1

r (x)
|u|r(x) dx,

so that ∫
Ω

1

p2
|∇u|p(x) dx <

∫
Ω

b

r1
|u|r(x) dx.

On the other hand, we have

H (t) = −
∫
Ω

1

p (x)
|∇u|p(x) dx + b

∫
Ω

1

r (x)
|u|r(x) dx

≤ b
∫
Ω

1

r (x)
|u|r(x) dx. (3.10)



Int. J. Anal. Appl. 17 (4) (2019) 625

Then, by (3.10) , (3.4) and (3.9) , we obtain

0 < H (0) < H (t) <
b

r1
Ar(.) (u) .

�

Lemma 3.2. [16] Assume that (1.2), (1.3) hold and E (0) < 0. Then the solution of (3.1) , satisfies for some

c > 0,

Ar(.) (u) ≥ c‖u‖
r1
r1
. (3.11)

Proof of theorem 1. We have

L
′
(t) =

∫
Ω

uutdx

=

∫
Ω

u
(

div
(
|∇u|p(x)−2 ∇u

)
+ b |u|r(x)−2 u

)
dx

= −Ap(.) (∇u) + bAr(.) (u) . (3.12)

Combining of (3.12) , (3.11) and (3.6) , leads to

L
′
(t) ≥ cb

(
1 −

p2
r1

)
‖u‖r1r1 . (3.13)

Now, we estimate L
r1
2 (t) , by the embedding of Lr1 (Ω) ↪→ L2 (Ω) , we get

L
r1
2 (t) ≤

(
1

2
‖u‖2r1

)r1
2

≤ c‖u‖r1r1 . (3.14)

By combining (3.14) and (3.13) , we obtain

L
′
(t) ≥ ξL

r1
2 (t) . (3.15)

A direct integration of (3.15) , then yields

L
r1
2
−1 (t) ≥

1

L1−
r1
2 (0) − ξt

.

Therefore, L blow up in a time t∗ ≤ 1
L

r1
2
−1

(0)
. �

4. Exponential growth

In this section, we prove that the solution of equation (1.1) exponential growth when ω > 0.

Lemma 4.1. Suppose that (1.2) holds and E (0) < 0. Then,∫
Ω

|u|m(x) dx ≤ c
(
‖u‖r1r1 + H (t)

)
. (4.1)



Int. J. Anal. Appl. 17 (4) (2019) 626

Proof. ∫
Ω

|u|m(x) dx =
∫
Ω−

|u|m(x) dx +
∫
Ω+

|u|m(x) dx,

where

Ω+ = {x ∈ Ω / |u (x, t)| ≥ 1} and Ω− = {x ∈ Ω / |u (x, t)| < 1} .

So, we get

∫
Ω

|u|m(x) dx ≤ c




∫

Ω−

|u|r1 dx




m1
r1

+


∫

Ω+

|u|r1 dx




m2
r1




≤ c
(
‖u‖m1r1 + ‖u‖

m2
r1

)
.

Exploiting the algebric inequality

zv ≤ (z + 1) ≤
(

1 +
1

a

)
(z + a) , ∀z > 0, 0 < v ≤ 1, a ≥ 0,

we have

‖u‖m1r1 ≤ c
(
‖u‖r1r1

)m1
r1 ≤ c

(
1 +

1

H (0)

)(
‖u‖r1r1 + H (0)

)
≤ c

(
‖u‖r1r1 + H (t)

)
.

Similarly,

‖u‖m2r1 ≤ c
(
‖u‖r1r1

)m2
r1 ≤ c

(
1 +

1

H (0)

)(
‖u‖r1r1 + H (0)

)
≤ c

(
‖u‖r1r1 + H (t)

)
.

This gives ∫
Ω

|u|m(x) dx ≤ c
(
‖u‖r1r1 + H (t)

)
.

�

Theorem 4.1. Let the assumptions of proposition 1, be satisfied and assume that (3.3) holds. Then the

solution of the problem (1.1) , grows exponentially.

Proof. By the same procedure of the proof the Lemma 5, we get

E
′
(t) = −‖ut‖

2
2 −ω

∫
Ω

|u|m(x)−2 u2t ≤ 0, (4.2)

then, we have

H
′
(t) = ‖ut‖

2
2 + ω

∫
Ω

|u|m(x)−2 u2tdx ≥ 0. (4.3)



Int. J. Anal. Appl. 17 (4) (2019) 627

We define

G (t) = H (t) + �L (t) . (4.4)

for � small to be chosen later.

The time derivative of (4.4) , we obtain

G
′
(t) = H

′
(t) + �

∫
Ω

uutdx.

By using (1.1) , we get

G
′
(t) = H

′
(t) − �Ap(.) (∇u) + �bAr(.) (u) − �ω

∫
Ω

|u|m(x)−2 utudx. (4.5)

To estimate the last term in the right hand side of (4.5) , by using the following Young’s Inequality

XY ≤ δX2 + δ−1Y 2, X, Y ≥ 0, δ > 0.

∫
Ω

|u|m(x)−2 utudx =
∫
Ω

|u|
m(x)−2

2 ut |u|
m(x)−2

2 udx

≤ δ
∫
Ω

|u|m(x)−2 u2tdx + δ
−1
∫
Ω

|u|m(x) dx.

We conclude

G
′
(t) ≥ (1 − �δ)

∫
Ω

|u|m(x)−2 u2tdx + ‖ut‖
2
2 − �Ap(.) (∇u)

+�bAr(.) (u) − �ωδ
−1
∫
Ω

|u|m(x) dx. (4.6)

Then

G
′
(t) ≥ (1 − �δ)

∫
Ω

|u|m(x)−2 u2tdx + ‖ut‖
2
2 − �ωδ

−1
∫
Ω

|u|m(x) dx

+� (1 −µ) r1H (t) + �bµAr(.) (u) + �
(

(1 −µ)
r1
p2
− 1
)
Ap(.) (∇u) ,

where µ is a constant such that 0 < µ ≤ 1 − p2
r1
.

Also, by using (3.6) , we obtain

G
′
(t) ≥ (1 − �δ)

∫
Ω

|u|m(x)−2 u2tdx + ‖ut‖
2
2 − �ωδ

−1
∫
Ω

|u|m(x) dx

+� (1 −µ) r1H (t) + �
(
bµ + 1 −µ−

p2
r1

)
Ar(.) (u) . (4.7)



Int. J. Anal. Appl. 17 (4) (2019) 628

Then, by Lemma 7 and (3.11) , (4.7) becomes

G
′
(t) ≥ (1 − �δ)

∫
Ω

|u|m(x)−2 u2tdx + ‖ut‖
2
2 − �c ωδ

−1 (‖u‖r1r1 + H (t))
+� (1 −µ) r1H (t) + �c

(
bµ + 1 −µ−

p2
r1

)
‖u‖r1r1 . (4.8)

So that

G
′
(t) ≥ (1 − �δ)

∫
Ω

|u|m(x)−2 u2tdx + ‖ut‖
2
2 + �

(
(1 −µ) r1 − c ωδ−1

)
H (t)

+�

(
c

(
bµ + 1 −µ−

p2
r1

)
− c ωδ−1

)
‖u‖p1p1 . (4.9)

So, we chosen δ large sufficient and � small enough for that we can find λ1, λ2 > 0, such that

G
′
(t) ≥ λ1H (t) + λ2 ‖u‖

r1
r1
≥ K1

(
H (t) + ‖u‖r1r1

)
, (4.10)

and

G (0) = H (0) + �L (0) > 0.

Similarly in (4.7) , we have

‖u‖22 ≤ c
(
H (t) + ‖u‖r1r1

)
. (4.11)

On the other hand, by (4.11) , we get

G (t) ≤ K2
(
H (t) + ‖u‖r1r1

)
. (4.12)

Combining with (4.12) and (4.10) , we arrive at

G
′
(t) ≥ ηG (t) . (4.13)

Finally, a simple integration of (4.13) gives

G (t) ≥ G (0) eηt, ∀t ≥ 0. (4.14)

Thus completes the proof. �

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	1. Introduction
	2. Preliminaries
	3. Blow up
	4. Exponential growth
	References