Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 70, pp. 1–13.

ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu

IMPROVED BLOWUP TIME ESTIMATES FOR FOURTH-ORDER

DAMPED WAVE EQUATIONS WITH STRAIN TERM AND

ARBITRARY POSITIVE INITIAL ENERGY

SHAOHUA CHEN, RUNZHANG XU, CHAO YANG

Abstract. We propose a new differential inequality that improve the upper

bound of the blowup time estimate for nonlinear fourth-order damped wave
equations with strain term and arbitrary positive initial energy. We also give

two new initial conditions to expand the range of the initial data leading to

the finite time blowup of solutions. We obtain a sharp result of finite time
blowup for the special case of the new differential inequality. We illustrate our

results with some simulations.

1. Introduction

We consider the initial boundary value problem (IBVP) for the nonlinear fourth-
order damped wave equation with strain term

utt + ∆2u+

n∑
i=1

∂

∂xi
σi(uxi) +mut = 0, in (x, t) ∈ Ω× (0, T ), (1.1)

u(x, 0) = u0(x), ut(x, 0) = u1(x), in x ∈ Ω, (1.2)

u =
∂u

∂ν
= 0 or u = ∆u = 0, on (x, t) ∈ ∂Ω× (0, T ), (1.3)

where m ≥ 0, u0(x) and u1(x) are the initial data, Ω is a bounded domain in Rn
(n ≥ 1) with smooth boundary, T is the maximum existence time of the solution,
and σi(s) satisfies

(H1) (i) σi(s) ∈ C1 and σi(0) = σ′i(0) = 0;
(ii) σi(s) is monotone for −∞ < s <∞ and is convex when for s > 0, and
concave for s < 0;
(iii) (p+ 1)Gi(s) ≤ sσi(s) and |σi(s)| ≤ a|s|p for some a > 0, with 1 < p <
∞ if n ≤ 2, and 1 < p < n

n−2 if n ≥ 3;

(iv) Gi(s) =
∫ s

0
σi(v) dv, 1 ≤ i ≤ n.

Model (1.1) was introduced to describe the longitudinal motion of an elasto-
plastic bar [1], and then this model with the corresponding conditions attracted a
lot of mathematicians’ attention. Initial boundary-value problem (1.1)-(1.3) was
considered in [3, 5, 14] by the potential well method proposed in [10, 11]. The

2020 Mathematics Subject Classification. 35L35, 35B44, 35L05, 35D30.
Key words and phrases. Finite time blowup; fourth-order wave equation; strain term;

arbitrary positive initial energy.
©2022. This work is licensed under a CC BY 4.0 license.

Submitted May 22, 2022. Published October 12, 2022.

1



2 S. H. CHEN, R. Z. XU, C. YANG EJDE-2022/70

finite time blowup of solutions with negative initial energy E(0) < 0 was treated
by Yang in [14]. Liu and Xu obtained global solutions, and finite time blowup of
solutions for positive initial energy E(0) ≤ d (d denotes the mountain pass level) in
[5]. Han et al. [3] showed the finite time blowup of solutions for arbitrarily positive
initial energy E(0) > 0. Relevant problems with damping and source terms were
considered in [2, 6, 9, 15]. Some representative works [8, 12, 13] studied the control
mechanism of the initial data on the dynamic behavior of solutions by the potential
well method.

In studying the relationship between the dynamic behaviors of solutions and ini-
tial data in the framework of the potential well method, one of the central problems
is investigating the finite time blowup of solutions. We often expect to utilize some
differential inequalities as tools to prove the finite blowup of solutions and estimate
blowup time. Korpusov proposed a differential inequality in [4] to estimate the
upper bound of blowup time of solutions to the IBVP for a generalized dissipative
high-order equation of Klein-Gordon type. This inequality was applied by Lin and
Luo in [7] to do the same estimate for problem (1.1)-(1.3). As we know, the smaller
the upper bound of blowup time we obtain, the better the estimate of blowup time
is. In this paper, we propose a new differential inequality to improve that in [4]
and obtain a better upper bound of blowup time. We also try to expand the range
of the initial data leading to finite time blowup by giving two different sets of ini-
tial data. In a special case, we obtain a sharp result for finite time blowup to the
new differential inequality. Some simulations are also conducted to verify the main
results of the present paper.

The rest of this paper is organized as follows: Section 2 presents and proves
theorems about the new differential inequality. Section 3 uses the theorems in
Section 2 to prove the finite time blowup and improve the upper bound estimate
of the blowup time of solutions.

2. Improved differential inequality and new initial conditions

In this section, we present an improved differential inequality and some new
initial conditions, which will be used in Section 3 to prove the finite time blowup
and improve the upper bound estimate of the blowup time of solutions.

Theorem 2.1 (Improved differential inequality). Suppose that Φ(t) ∈ C2 satisfies

ΦΦ′′ − α(Φ′)2 + γΦΦ′ + βΦ ≥ µΦ2, α > 1, β, γ, µ ≥ 0; (2.1)

Φ′(0) >
Φ(0)θ

α− 1
> 0, (2.2)

where

θ =
1

2

(
γ +

√
max{γ2 + 4(α− 1)(βΦ−1(0)− µ), 0}

)
. (2.3)

Then Φ(t) blows up in finite time T , where

T <


1

2θ−γ ln
(

(2θ−γ)Φ(0)
(α−1)Φ′(0)−θΦ(0) + 1

)
if 2θ > γ,

Φ(0)
(α−1)Φ′(0)−γΦ(0)/2 if 2θ ≤ γ.

(2.4)

Proof. Dividing both sides of (2.1) by Φ1+α yields( Φ′

Φα

)′
+ γ

Φ′

Φα
+

β

Φα
≥ µ

Φα−1
,



EJDE-2022/70 IMPROVED BLOWUP TIME ESTIMATES 3

that is
1

1− α
(
Φ1−α)′′ + γ

1− α
(
Φ1−α)′ + βΦ−α − µΦ1−α ≥ 0.

Let z(t) = Φ1−α(t). Then

z′′(t) + γz′(t)− β(α− 1)zα1(t) + µ(α− 1)z(t) ≤ 0, α1 =
α

α− 1
. (2.5)

We first assume γ2 + 4(α − 1)(βΦ−1(0) − µ) > 0. Let y(t) = eθtz(t), where θ is
defined in (2.3). It is easy to see that 2θ > γ. Then

z′(t) =
(
e−θty(t)

)′
= −θe−θty(t) + e−θty′(t),

z′′(t) = e−θty′′(t)− 2θe−θty′(t) + θ2e−θty(t) ,

z′′(t) + γz′(t)− β(α− 1)zα1(t) + µ(α− 1)z(t)

= e−θt
(
y′′(t)− 2θy′(t) + θ2y(t) + γy′(t)− θγy(t)

− β(α− 1)eθ(1−α1)tyα1(t) + µ(α− 1)y(t)
)
≤ 0,

i.e.,

y′′(t) ≤ (2θ−γ)y′(t)−y(t)
(
θ(θ − γ) + µ(α− 1)− β(α− 1)e−

θt
α−1 y

1
α−1 (t)

)
. (2.6)

Note that

y′(t) = θeθtz(t) + eθtz′(t)

= eθt
(
θΦ1−α(t) + (1− α)Φ−α(t)Φ′(t)

)
= eθtΦ−α(t) (θΦ(t)− (α− 1)Φ′(t)) .

By (2.2), we have

y′(0) = Φ−α(0) (θΦ(0)− (α− 1)Φ′(0)) < 0. (2.7)

We claim that y′(t) < 0 for all t < T . Arguing by contradiction, we suppose that
there exists a t0 ∈ (0, T ), such that

y′(t0) = 0, y′(t) < 0, 0 < y(t) < y(0) (2.8)

for t ∈ [0, t0). In fact, y(0) = Φ1−α(0) > 0. By the definition of the maximum
existence time T and the continuity of y(t) in t, the sign of y(t) cannot change over
[0, T ), because y(t∗) = 0 means that the solution Φ(t) blows up at the finite time
t∗. Hence we have y(t) > 0 over [0, T ). Integrating (2.6) from 0 to t, we find

y′(t) ≤ y′(0) +

∫ t

0

(
(2θ − γ)y′(τ)

− y(τ)
(
θ(θ − γ) + µ(α− 1)− β(α− 1)e−

θτ
α−1 y

1
α−1 (τ)

))
dτ.

(2.9)

Now

θ(θ − γ) + µ(α− 1)− β(α− 1)e−
θτ
α−1 y

1
α−1 (τ)

≥ θ(θ − γ) + µ(α− 1)− β(α− 1)y
1

α−1 (0)

= θ(θ − γ) + µ(α− 1)− β(α− 1)Φ−1(0).

(2.10)



4 S. H. CHEN, R. Z. XU, C. YANG EJDE-2022/70

It follows from (2.3) that

θ(θ − γ) =
(1

2

√
γ2 + 4(α− 1)

(
βΦ−1(0)− µ

)
+
γ

2

)
×
(1

2

√
γ2 + 4(α− 1)

(
βΦ−1(0)− µ

)
− γ

2

)
= (α− 1)

(
βΦ−1(0)− µ

)
.

(2.11)

Combining (2.10) and (2.11), we obtain

θ(θ − γ) + µ(α− 1)− β(α− 1)e−
θt
α−1 y

1
α−1 (t) ≥ 0 for t ≤ t0. (2.12)

From (2.9), we have

y′(t) ≤ y′(0) < 0, t ∈ [0, t0],

which is a contradiction with (2.8). Hence y′(t) ≤ y′(0) for all t < T and

y(t) ≤ y(0) + y′(0)t, (2.13)

which implies that Φ(t) blows up in finite time.
To estimate the blowup time T , we use (2.9) and (2.12) to obtain

y′(t) ≤ y′(0) +

∫ t

0

(2θ − γ)y′(τ) dτ = y′(0) + (2θ − γ)(y(t)− y(0)). (2.14)

For convenience, let σ = −y′(0) > 0. Rewrite (2.14) as

d

dt

(
e−(2θ−γ)ty(t)

)
≤ − (σ + (2θ − γ)y(0)) e−(2θ−γ)t. (2.15)

Integrating (2.15) from 0 to t, we obtain

e−(2θ−γ)ty(t) ≤ y(0)− σ + (2θ − γ)y(0)

2θ − γ

(
1− e−(2θ−γ)t

)
= y(0)e−(2θ−γ)t − σ

2θ − γ

(
1− e−(2θ−γ)t

)
,

i.e.,

0 < y(t) ≤ y(0)− σ

2θ − γ

(
e(2θ−γ)t − 1

)
, t ∈ [0, T ).

So we obtain
(2θ − γ)

σ
y(0) + 1 > e(2θ−γ)t, t ∈ [0, T ).

Hence

T <
1

2θ − γ
ln
( 2θ − γ
σΦα−1(0)

+ 1
)

=
1

2θ − γ
ln
( (2θ − γ)Φ(0)

(α− 1)Φ′(0)− θΦ(0)
+ 1
)
.

If γ2 + 4(α− 1)(βΦ−1(0)− µ) ≤ 0, then, setting θ = γ/2 in (2.6), we have

y′′(t) ≤ −y(t)
(−γ

4
+ µ(α− 1)− β(α− 1)e−

θt
α−1 y

1
α−1 (t)

)
< −y(t)

(
− γ

4
+ (α− 1)(µ− βy

1
α−1 (0))

)
≤ 0,

which implies that y(t) < y(0) + y′(0)t. Hence, Φ(t) blows up in finite time and
T < y(0)/|y′(0)|. �



EJDE-2022/70 IMPROVED BLOWUP TIME ESTIMATES 5

Remark 2.2. It is easy to verify that for µ = 0,

Φ′(0) >
γ

α− 1
Φ(0) +

√
2β

2α− 1
Φ(0) >

Φ(0)

2(α− 1)

(
γ +

√
γ2 + 4β(α− 1)Φ−1(0)

)
,

while Φ′(0) > γ
α−1Φ(0) +

√
2β

2α−1Φ(0) is required in [4]. Hence in Theorem 2.1, this

condition is relaxed to include a large range of initial data.

Remark 2.3. We consider some specific examples to show that by applying the
improved differential inequality (Theorem 2.1), we can obtain some better estimates
of the blowup time in some cases than the results derived by the method proposed
in [4]. We also show some numerical simulations in Figure 1.

Let α = 2, β = γ = 1, Φ(0) = 2, and denote T ∗ as the upper bound of blowup
time and consider the following three cases:

(i) If Φ′(0) = 2.74 and µ = 0, then (2.2) is satisfied, but the conditions of
[4] are not satisfied. So we can get T ∗ ≈ 3.51 by the improved differential
inequality, and T ∗ ≈ 1.71 by numerical computations when (2.1) is equality
(Figure 1, (a), the dotted line).

(ii) If Φ′(0) = 3.2 and µ = 0, then both (2.2) and the conditions of [4] are
satisfied. We can obtain T ∗ ≈ 6.12 based on [4], T ∗ ≈ 1.23 by the improved
differential inequality, and T ∗ ≈ 1.11 by numerical computations when (2.1)
is equality (Figure 1, (a), the solid line).

(iii) If we choose Φ′(0) = 2.4 and µ = 0.3, then (2.2) is satisfied and the solution
blows up with blowup time T ∗ < 2.37 based on (2.4). The numerical
blowup time T ∗ ≈ 1.8 when (2.1) is equality (Figure 1, (b), the left-hand
solid line). Also, numerical computations show that the solution will blow
up when Φ′(0) = 1.1 (Figure 1, (b), the dotted line), but the solution won’t
blow up when Φ′(0) = 1.0 (2.4) (Figure 1, (b), the right-hand solid line).

0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6

t

0

20

40

60

80

100

120

140

160

180

200

(0)=2.74, T
*
=1.71

(0)=3.20, T
*
=1.11

0 5 10 15 20 25

t

0

10

20

30

40

50

60

70

80

90

100

(0)=1.0, no blowup

(0)=1.1, T
*
=30.53

(0)=2.4, T
*
=1.80

(a) µ = 0 (b) µ = 0.3

Figure 1. Blowup solutions for ΦΦ′′ − α(Φ′)2 + γΦΦ′ + βΦ = µΦ2

Next, we present a theorem that allows us to estimate the upper bound of blowup
time for some new sets of initial data not included in Theorem 2.1.

Theorem 2.4 (New initial condition I). Suppose that Φ(t) ∈ C2 satisfies (2.1) with
µ = 0 and

Φ′(0) >
γΦ(0)

α− 1
+
β(α− 1)

γα
> 0. (2.16)



6 S. H. CHEN, R. Z. XU, C. YANG EJDE-2022/70

Then Φ(t) blows up in finite time T ≤ Γ, where Γ is the unique solution of

Φ(0) =
α− 1

αγ2

(
(α− 1)2β

(
1− e−

γΓ
α−1

)
+ ω(eγΓ − 1)

)
, (2.17)

and

ω = αγ
(

Φ′(0)− γΦ(0)

α− 1
− β(α− 1)

αγ

)
. (2.18)

Proof. Let θ = γ and µ = 0 in (2.6). Then (2.6) becomes

y′′(t) ≤ γy′(t) + β(α− 1)e−
γt
α−1 yα1(t), α1 =

α

α− 1
. (2.19)

Multiplying (2.19) on both sides by e
γt
α−1 yields

e
γt
α−1 y′′(t) ≤ γy′(t)e

γt
α−1 + β(α− 1)yα1(t),

i.e.,
d

dt

(
e
γt
α−1 y′(t)

)
≤
(
γ +

γ

α− 1

)
y′(t)e

γt
α−1 + (α− 1)βyα1(t). (2.20)

Integrating (2.20) from 0 to t, we obtain

e
γt
α−1 y′(t) ≤ y′(0) +

∫ t

0

( αγ

α− 1
y′(τ)e

γt
α−1 + (α− 1)βyα1(τ)

)
dτ. (2.21)

Note that, from (2.16) and (2.18),

αγ

α− 1
y′(0) + (α− 1)βyα1(0)

=
αγ

α− 1
Φ−α(0) (γΦ(0)− (α− 1)Φ′(0)) + (α− 1)βΦ−α(0)

= −αγΦ−α(0)
(

Φ′(0)− γΦ(0)

α− 1
− β(α− 1)

αγ

)
= −ωΦ−α(0) < 0.

(2.22)

In particular, y′(0) < 0. By the continuity of y′(t), there exist a t1 > 0 such that

αγ

α− 1
y′(t)e

γt
α−1 + (α− 1)βyα1(t) ≤ 0, (2.23)

for 0 < t ≤ t1, which implies that

e
γt
α−1 y′(t) ≤ y′(0), y′(t) < 0 and 0 < y(t) < y(0) (2.24)

for 0 < t ≤ t1. Substituting the first and third inequalities of (2.24) into (2.21) and
using (2.22), we find

e
γt
α−1 y′(t) ≤ y′(0) +

∫ t

0

( αγ

α− 1
y′(0) + (α− 1)βyα1(0)

)
dτ

≤ y′(0)− ω1t,

(2.25)

where ω1 = ωΦ−α(0). From (2.25), we can see that (2.23) is a strict inequality,
that is t1 = T . Substituting (2.25) into (2.21) and using (2.22), we have

e
γt
α−1 y′(t) ≤ y′(0) +

∫ t

0

( αγ

α− 1
(y′(0)− ω1τ) + (α− 1)βyα1(0)

)
dτ

≤ y′(0)− ω1t−
αγω1

α− 1

t2

2!
.



EJDE-2022/70 IMPROVED BLOWUP TIME ESTIMATES 7

After n steps we obtain

e
γt
α−1 y′(t) ≤ y′(0)− ω1

n∑
k=1

( αγ

α− 1

)k−1 tk

k!
. (2.26)

Then, substituting (2.26) into (2.21), we deduce

e
γt
α−1 y′(t)

≤ y′(0) +

∫ t

0

( αγ

α− 1

(
y′(0)− ω1

n∑
k=1

( αγ

α− 1

)k−1 τk

k!

)
+ (α− 1)βyα1(0)

)
dτ

≤ y′(0)− ω1t− ω1

n∑
k=1

( αγ

α− 1

)k tk+1

(k + 1)!

= y′(0)− ω1

n+1∑
k=1

( αγ

α− 1

)k−1 tk

k!
.

By mathematical induction, (2.26) is satisfied for all n ≥ 2. Let n→∞, we obtain

e
γt
α−1 y′(t) ≤ y′(0)− (α− 1)ω1

αγ

(
e
αγt
α−1 − 1

)
, (2.27)

that is

y′(t) ≤
(
y′(0) +

(α− 1)ω1

αγ

)
e−

γt
α−1 − (α− 1)ω1

αγ
eγt. (2.28)

Note that from (2.22) we have

y′(0) +
(α− 1)ω1

αγ
= − (α− 1)2

αγ
βyα1(0). (2.29)

Integrating (2.28) from 0 to t, we obtain

y(t) ≤ y(0)− (α− 1)3

αγ2
βyα1(0)

(
1− e−

γt
α−1
)
− (α− 1)ω1

αγ2
(eγt − 1).

Since y(t) > 0, we obtain

0 < Φ1−α(0)− (α− 1)3β

αγ2
Φ−α(0)

(
1− e−

γt
α−1
)
− (α− 1)ω

αγ2
Φ−α(0)(eγt − 1).

Hence t < Γ, where Γ is the unique solution of (2.17). �

Remark 2.5. If Φ(0) is small, condition (2.2) is better than condition (2.16).
If Φ(0) is not small, condition (2.16) is better than condition (2.2). Let us use
numerical simulations to show the above conclusion. Choose α = 2, β = γ = 1, µ =
0, and denote T ∗ as the upper bound of blowup time.

(i) Small initial data case. If Φ(0) = 0.4, Φ′(0) = 0.88, then (2.2) is satisfied
((2.16) is not satisfied), so the solution blows up. By (2.4), we can obtain
T ∗ ≈ 1.62, and the numerical T ∗ ≈ 0.82.

(ii) Large initial data case. If Φ(0) = 2, Φ′(0) = 2.52, then (2.16) is satisfied (
(2.2) is not satisfied), so the solution blows up. We can obtain Γ ≈ 4.34 by
numerical method, and the numerical blowup time T ∗ ≈ 2.72 when (2.1) is
equality.

Next, we show another theorem to give some new sets of initial data, whose
result is sharp based on numerical computations if µ is not small.



8 S. H. CHEN, R. Z. XU, C. YANG EJDE-2022/70

Theorem 2.6 (New initial condition II). Suppose that Φ(t) ∈ C2 satisfies (2.1)
and

Φ′(0) > 0, µ > max
{
βΦ−1(0),

4θ2 + γ2(α− 1)2

2(α+ 1)(α− 1)2

}
, (2.30)

where θ is defined in (2.3). Then Φ(t) blows up in finite time.

Proof. Multiplying (2.5) on both sides by eγt yields

d

dt

(
eγtz′(t)

)
≤ (α− 1) (βzα1(t)− µz(t)) eγt. (2.31)

Integrating (2.31) from 0 to t, we obtain

z′(t) ≤ e−γtz′(0) +

∫ t

0

(α− 1)z(τ)(βz1/(α−1)(τ)− µ)eγ(τ−t)dτ. (2.32)

Since z′(t) = (1 − α)Φ−α(t)Φ′(t) and z′(0) = (1 − α)Φ−α(0)Φ′(0) < 0, we have
z′(τ) < 0 or z(τ) < z(0) for small t > 0. Then

(α−1)z(τ)(βzα1(τ)−µ)eγ(τ−t) < (α−1)z(τ)(βz1/(α−1)(0)−µ)eγ(τ−t) < 0, (2.33)

which implies that z′(t) ≤ e−γtz′(0) from (2.32) for all t such that the solution
exists (also (2.33) is satisfied for such t).

To prove that Φ(t) blows up in finite time, we use a proof by contradiction.
Assume that the solution exists for all t > 0. From z′(t) < 0, we obtain Φ′(t) > 0
for all t > 0. We first claim that limt→∞Φ(t) = ∞. Otherwise, if limt→∞ Φ(t) =
a <∞, then by monotonicity, limt→∞Φ′(t) = limt→∞ Φ′′(t) = 0. Taking limits in
(2.1), we obtain βa − µa2 ≥ 0. Since a > Φ(0), we obtain a contradiction against
(2.30). Thus, limt→∞ Φ(t) =∞.

Now, we discuss two cases:
(i) There exist a t2 > 0, such that Φ′(t2) > θΦ(t2)/(α− 1);
(ii) Φ′(t) ≤ θΦ(t)/(α− 1) for all t > 0.

In case (i), we can use Theorem 2.1 with Φ(0) replaced by Φ(t2) to show that
Φ(t) blows up in finite time. In case (ii), we rewrite (2.1) as

ΦΦ′′ + γΦΦ′ + Φ′
2 ≥ (α+ 1)Φ′

2
+ µΦ2 − βΦ. (2.34)

We denote

M =
γ2(α− 1)2(α+ 1)

4θ2 + γ2(α− 1)2
. (2.35)

Then

α+ 1−M =
4(α+ 1)θ2

4θ2 + γ2(α− 1)2
> 0, (2.36)

and by Cauchy inequality, we obtain

0 < γΦΦ′ = 2
√
MΦ′

γ

2
√
M

Φ ≤MΦ′
2

+
γ2

4M
Φ2. (2.37)

Substituting (2.37) into (2.34) yields

d

dt
(ΦΦ′) = ΦΦ′′ + Φ′

2 ≥ (α+ 1−M)Φ′
2

+
(
µ− γ2

4M

)
Φ2 − βΦ. (2.38)

By the second inequality of (2.30), we can find ε > 0, such that

µ− 4θ2 + γ2(α− 1)2

2(α+ 1)(α− 1)2
− 2ε = 0. (2.39)



EJDE-2022/70 IMPROVED BLOWUP TIME ESTIMATES 9

Since limt→∞Φ(t) =∞, we can find a t3 > t2 such that

εΦ2(t)− βΦ(t) ≥ 0 for t ≥ t3.

Integrating (2.38) from t3 to t, we find

ΦΦ′ ≥ Φ(t3)Φ′(t3) +

∫ t

t3

(
(α+ 1−M)Φ′

2
(t) +

(
µ− γ2

4M
− ε
)
Φ2(τ)

)
dτ.

In case (ii), we have
θ

α− 1
Φ2 ≥ ΦΦ′.

Let

f(t) =
θ

α− 1
Φ2(t)− Φ(t3)Φ′(t3)

−
∫ t

t3

(
(α+ 1−M)Φ′

2
(τ) +

(
µ− γ2

4M
− ε
)
Φ2(τ)

)
dτ.

Then f(t) ≥ 0 for t ≥ t3, and

f ′(t) =
2θ

α− 1
ΦΦ′ − (α+ 1−M)Φ′

2 −
(
µ− γ2

4M
− ε
)
Φ2. (2.40)

Using Cauchy inequality again, we obtain

2θ

α− 1
ΦΦ′ = 2

√
α+ 1−MΦ′

θ

(α− 1)
√
α+ 1−M

Φ (2.41)

≤ (α+ 1−M)Φ′
2

+
θ2

(α− 1)2(α+ 1−M)
Φ2.

Note that from (2.36) and (2.35) we have

θ2

(α− 1)2(α+ 1−M)
=

4θ2 + γ2(α− 1)2

4(α+ 1)(α− 1)2
=

γ2

4M
,

which combined with (2.39) yields

µ− 2γ2

4M
− ε = ε.

Substituting (2.41) into (2.40) yields

f ′(t) ≤ −εΦ2,

which implies that f(t) ≥ 0 is not satisfied for all t > t3. This is a contradiction.
Hence Φ(t) blows up in a finite time. �

Remark 2.7. Note that Φ(t) ≡ β/µ is a constant solution to (2.1). Under the
condition µ > (4θ2 + γ2(α− 1)2)/(2(α− 1)2(α+ 1)), if Φ′(0) > 0 and Φ(0) > β/µ,
then the solution blows up in finite time. By the uniqueness, the solution of (2.1)
satisfies Φ(t) < β/µ if Φ′(0) < 0 and Φ(0) < β/µ when (2.1) is equality. Hence
the result of Theorem 2.6 is sharp. To simulate this result, we choose α = 3,
γ = 1, β = 1, µ = 0.26. If Φ′(0) = 0.01 and Φ(0) = 3.8463, then the conditions of
Theorem 2.6 are satisfied, and the solution blows up (Figure 2, the solid line). If
Φ′(0) = −0.01 and Φ(0) = 3.8462, numerical computations show that the solution
will go to zero at a finite time (Figure 2, the dotted line).



10 S. H. CHEN, R. Z. XU, C. YANG EJDE-2022/70

5 10 15 20 25 30

t

0

10

20

30

40

50

60

70

(0)=0.01, T
*
=34.6

(0)=-0.01, no blowup

Figure 2. Sharp result

3. Finite time blowup for Problem (1.1)-(1.3)

In this section, we apply our results established in Section 2 to prove the finite
time blowup and improve the upper bound estimate of the blowup time of solutions
to problem (1.1)-(1.3).

We denote the norm of L2(Ω) as ‖ · ‖, the inner product in L2(Ω) as (·, ·), and
the dual pairing between H(Ω) and H−2(Ω) as 〈·, ·〉, where

H(Ω) =

{
H2

0 (Ω), when u = ∂u
∂ν = 0 on ∂Ω;

H2(Ω) ∩H1
0 (Ω), when u = ∆u = 0 on ∂Ω,

(3.1)

and ν is the unit outer normal. Define the energy functional of Problem (1.1)-(1.3)
as follows

E(t) =
1

2
‖ut‖2 +

1

2
‖∆u‖2 −

n∑
i=1

∫
Ω

Gi(uxi) dx.

Note that problem(1.1)-(1.3) is a special case of that in Lian et al. [9], and the
existence of a local solution can be obtained from [9, Theorem 3.2]. We need the
following Lemma before presenting our main results.

Lemma 3.1. If u ∈ H(Ω), then λ1‖u‖ ≤ ‖∆u‖, and the equality is satisfied if and
only if u = cψ, where λ1 and ψ are the first eigenvalue and eigenfunction of −∆
with Dirichlet boundary conditions.

Proof. By Poincare Lemma, we have ‖u‖2 ≤ λ1
−1‖∇u‖2 and

‖∇u‖2 = −
∫

Ω

u∆udx ≤ ‖u‖‖∆u‖.

Hence, the conclusion follows. �

Theorem 3.2 (Blowup time estimate by Theorem 2.1). For Problem (1.1)-(1.3),
let u0 ∈ H(Ω), u1 ∈ L2(Ω), E(0) > 0 and σi satisfy rm (H1) in Section 1. Assume
that

ξ :=
√

max
{
m2 + 2(p2 − 1)E(0)‖u0‖−2 − λ1

2(p− 1)2, 0
}
, (3.2)

where p is defined in (H1), and

(u0, u1) >
1

p− 1
‖u0‖2 (m+ ξ) > 0. (3.3)



EJDE-2022/70 IMPROVED BLOWUP TIME ESTIMATES 11

Then the solution u blows up, and there exists a T such that

lim
t→T−

‖u‖2 = +∞

with

T <


1
ξ ln

( 2ξ‖u0‖2
(p−1)(u0,u1)−(m+ξ)‖u0‖2 + 1

)
if ξ > 0,

2‖u0‖2
(p−1)(u0,u1)−m‖u0‖2 if ξ ≤ 0.

(3.4)

Proof. We define Φ(t) := ‖u(t)‖2. Then

Φ′(t) = 2(u, ut), Φ′′(t) = 2〈utt, u〉+ 2‖ut‖2. (3.5)

We denote H(t) = ‖ut‖2. Multiplying (1.1) by u and integrating over Ω, we obtain

〈utt, u〉+m(ut, u) + ‖∆u‖2 −
n∑
i=1

∫
Ω

uxiσi(uxi) dx = 0. (3.6)

From (3.5) and (3.6), we have

1

2
Φ′′(t) +

m

2
Φ′(t)−H(t) + ‖∆u‖2 =

n∑
i=1

∫
Ω

uxiσi(uxi) dx. (3.7)

Multiplying (1.1) by ut and integrating over Ω, we obtain

d

dt

(1

2
H(t) +

1

2
‖∆u‖2

)
+m‖ut‖2 =

d

dt

( n∑
i=1

∫
Ω

Gi(uxi) dx
)
. (3.8)

Integrating (3.8) over (0, t) and using (H1), we deduce that

1

2
H(t) +

1

2
‖∆u‖2 +m

∫ t

0

‖uτ‖2 dτ − E(0) =

n∑
i=1

∫
Ω

Gi(uxi) dx

≤ 1

p+ 1

n∑
i=1

∫
Ω

uxiσi(uxi) dx.

(3.9)

Substituting (3.7) into (3.9) yields

1

2
H(t) +

1

2
‖∆u‖2 +m

∫ t

0

‖uτ‖2 dτ − E(0)

≤ 1

p+ 1

(1

2
Φ′′(t) +

m

2
Φ′(t)−H(t) + ‖∆u‖2

)
,

that is,

1

2
Φ′′(t) +

m

2
Φ′(t) + (p+ 1)E(0) ≥ p+ 3

2
H(t) +

p− 1

2
‖∆u‖2. (3.10)

By Cauchy-Schwarz inequality, we have

(Φ′(t))2 ≤ 4Φ(t)H(t). (3.11)

Multiplying both sides of (3.10) by 2Φ(t) and using (3.11) and Lemma 3.1,

Φ′′(t)Φ(t)− p+ 3

4
(Φ′(t))

2
+mΦ′(t)Φ(t) + 2(p+ 1)E(0)Φ(t)

≥ (p− 1)‖∆u‖2Φ(t)

≥ λ1
2(p− 1)Φ2(t),

(3.12)



12 S. H. CHEN, R. Z. XU, C. YANG EJDE-2022/70

then we set

α =
p+ 3

4
> 1, γ = m ≥ 0, β = 2(p+ 1)E(0) > 0, µ = λ1

2(p− 1) > 0.

Then it is easy to verify that (3.12), (3.2), and (3.3) satisfy the conditions in
Theorem 2.1. Hence, using Theorem 2.1, we obtain that the solution of (1.1)-(1.3)
blows up in finite time, and we obtain the upper bound estimate of blowup time as
(3.4). �

Theorem 3.3 (Blowup time estimate by Theorem 2.4). For problem (1.1)-(1.3),
let u0 ∈ H(Ω), u1 ∈ L2(Ω), E(0) > 0 and σi satisfy (H1). Assume that

(u0, u1) >
2m

p− 1
‖u0‖2 +

(p2 − 1)E(0)

m(p+ 3)
> 0, (3.13)

where p is defined in (H1). Then the solution to Problem (1.1)-(1.3) blows up in
finite time T ≤ Γ, where Γ is the unique solution of

‖u0‖2 =
p− 1

(p+ 3)m2

( (p− 1)2(p+ 1)E(0)

8
(1− e

4mΓ
1−p ) + w(emΓ − 1)

)
, (3.14)

and

w =
(p+ 3)m

4

(
2(u0, u1)− 4m

p− 1
‖u0‖2 −

2E(0)(p2 − 1)

(p+ 3)m

)
. (3.15)

Proof. By arguments similar to those in Theorem 3.2, we obtain (3.12), which
implies that

Φ′′(t)Φ(t)− p+ 3

4
(Φ′(t))

2
+mΦ′(t)Φ(t) + 2(p+ 1)E(0)Φ(t) ≥ 0 . (3.16)

We set

α =
p+ 3

4
> 1, γ = m ≥ 0, β = 2(p+ 1)E(0) > 0, µ = 0.

Then combining (3.13), (3.16), and Theorem 2.4, we obtain the proof. �

Theorem 3.4 (Blowup based on Theorem 2.6). For problem (1.1)-(1.3), let u0 ∈
H(Ω), u1 ∈ L2(Ω), E(0) > 0 and σi satisfy (H1). Assume that (u0, u1) > 0 and

λ2
1(p− 1) > max

{
2(p+ 1)E(0)‖u0‖−2,

32(m+ ξ)2 + 2m2(p− 1)2

(p+ 7)(p− 1)2

}
, (3.17)

where ξ and p are defined in (3.2) and (H1), respectively. Then the solution of
(1.1)-(1.3) blows up in finite time.

Proof. By arguments similar those in Theorem 3.2, combining (3.12), (3.17), and
Theorem 2.6, we obtain the proof. �

Acknowledgments. The authors would like to thank the referees their valuable
constructive and valuable comments, which led to the improvement of our paper.
Shaohua Chen was supported by NSERC Grant RGPIN-2019-05940. Runzhang Xu
was supported by the National Natural Science Foundation of China (11871017) and
by the Fundamental Research Funds for the Central Universities. Chao Yang was
supported by the Ph. D. Student Research and Innovation Fund of the Fundamental
Research Funds for the Central Universities (3072022GIP2403).



EJDE-2022/70 IMPROVED BLOWUP TIME ESTIMATES 13

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Shaohua Chen

School of Science and Technology, Cape Breton University, Sydney, Nova Scotia B1P
6L2, Canada

Email address: george chen@cbu.ca

Runzhang Xu

College of Mathematical Sciences, Harbin Engineering University, Harbin 150001, China

Email address: xurunzh@163.com; xurunzh@hrbeu.edu.cn

Chao Yang (corresponding author)
College of Mathematical Sciences, Harbin Engineering University, Harbin 150001, China.
Faculty of Applied Mathematics, AGH University of Science and Technology, Kraków
30-059, Poland

Email address: yangchao yc98@163.com; yangchao @hrbeu.edu.cn


	1. Introduction
	2. Improved differential inequality and new initial conditions
	3. Finite time blowup for Problem (??)-(??)
	Acknowledgments

	References