Electronic Journal of Differential Equations, Vol. 2022 (2022), No. 38, pp. 1–30.

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

GLOBAL SOLUTIONS AND BLOW-UP FOR A

KIRCHHOFF-TYPE PROBLEM ON A GEODESIC BALL OF THE

POINCARÉ BALL MODEL

HANG DING, JUN ZHOU

Abstract. This article concerns a Kirchhoff-type parabolic problem on a ge-

odesic ball of hyperbolic space. Firstly, we obtain conditions for finite time
blow-up, and for the existence of global solutions for J(u0) ≤ d, where J(u0)

denotes the initial energy and d denotes the depth of the potential well. Sec-

ondly, we estimate the upper and lower bounds of the blow-up time. In addi-
tion, we derive the growth rate of the blow-up solution and the decay rate of

the global solution. Thirdly, we establish a new finite time blow-up condition

which is independent of d and prove that the solution can blow up in finite
time with arbitrary high initial energy, by using this blow-up condition. Fi-

nally, we present some equivalent conditions for the solution existing globally

or blowing up in finite time.

1. Introduction

In this article, we consider the Kirchhoff-type parabolic problem

ut −
(
a+ b

∫
BR

|∇Hu(σ)|2 dµ
)

∆Hu = ξ|u|q−1u, σ ∈ BR, t > 0,

u(σ, t) = 0, σ ∈ ∂BR, t > 0,

u(σ, 0) = u0(σ), σ ∈ BR,

(1.1)

where ∆H is the Laplace-Beltrami operator on the Poincaré ball model B3, which
is a model of the hyperbolic space H3, BR ⊂ B3 denotes a geodesic ball centered in
zero with radius R, the initial value u0 ∈ H1

0 (BR), and the parameters a, b, ξ and
q satisfy

a ≥ 0, b > 0, ξ > 0, 3 < q < 5. (1.2)

We first recall the definitions of B3, ∆H , BR, H1
0 (BR) and ∇H , which can be

found in [2, 30].
(1) The Poincaré ball is

B3 := {σ = (x1, x2, x3) ∈ R3 : |σ| < 1}

2020 Mathematics Subject Classification. 35K55, 35B40, 35B44.
Key words and phrases. Parabolic problem of Kirchhoff type; hyperbolic space;

poincaré ball model; global solution; blow-up.
©2022. This work is licensed under a CC BY 4.0 license.

Submitted July 18, 2021 Published May 13, 2022.

1



2 H. DING, J. ZHOU EJDE-2022/38

endowed with the Riemannian metric

gij :=
4

(1− |σ|2)2
δij (σ ∈ B3; i, j = 1, 2, 3),

where δij and | · | denote the usual Kronecker delta and the Euclidean distance,
respectively.

(2) For i, j = 1, 2, 3, we define

gij := (gij)
−1 and g := det(gij).

In this setting the operator ∆H is locally defined by

∆H :=
1
√
g

3∑
i=1

∂

∂xi

(√
g

3∑
j=1

gij
∂

∂xj

)
.

As usual, let

dµ :=
√
gdx =

8

(1− |σ|2)3
dx

be the Riemannian volume element in B3, where dx is the standard Lebesgue mea-
sure in the Euclidean space R3. Therefore, if (note that |σ| < 1)

dH(σ, 0) := 2

∫ |σ|
0

1

1− t2
dt = log(

1 + |σ|
1− |σ|

)

denotes the geodesic distance of σ ∈ B3 from the origin, a direct calculation ensures
that the operator ∆H has the more convenient form

∆H =
1

4
(1− |σ|2)2

3∑
i=1

∂2

∂x2
i

+
1

2
(1− |σ|2)

3∑
i=1

xi
∂

∂xi
.

(3) The geodesic ball BR and its surface ∂BR are defined by

BR := {σ ∈ B3 : dH(σ, 0) < R}, ∂BR := {σ ∈ B3 : dH(σ, 0) = R},

where

dH(σ1, σ2) := cosh−1
(

1 +
2|σ2 − σ1|2

(1− |σ1|2)(1− |σ2|2)

)
, ∀σ1, σ2 ∈ B3

denotes the hyperbolic distance in the Poincaré ball model B3.
(4) For a geodesic ball BR ⊂ B3, we denote by H1

0 (BR) the completion of
C∞0 (BR) with respect to the Hilbertian norm

‖u‖ :=
(∫

BR

|∇Hu(σ)|2 dµ
)1/2

, (1.3)

where

∇H :=
(1− |σ|2

2

)2

∇,

denotes the hyperbolic gradient. Then we have∫
BR

(∆Hu)φdµ =

∫
BR

(∇Hu)(∇Hφ) dµ, ∀u, φ ∈ C∞0 (BR). (1.4)

By a denseness argument, H1
0 (BR) denotes the Sobolev space of the functions

u ∈ L2(BR) such that ∇Hu exists in the sense of distributions and |∇Hu| is in
L2(BR), endowed with the natural norm (1.3).



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 3

(5) From standard theory, we know that the embedding H1
0 (BR) ↪→ Lν(BR) is

continuous for any ν ∈ [1, 6], while it is compact whenever ν ∈ [1, 6). Therefore,
there is a positive constant Cν such that

‖u‖Lν(BR) ≤ Cν‖u‖ for all u ∈ H1
0 (BR) and ν ∈ [1, 6]. (1.5)

Below we introduce the research history of problem (1.1). Kirchhoff in 1883
proposed the model

ρ
∂2u

∂t2
−
(P0

h
+

E

2L

∫ L

0

∣∣∂u(x)

∂x

∣∣2dx)∂2u

∂x2
= 0,

which was as a generalization of the well-known D’Alembert wave equation for free
vibrations of elastic strings. The above parameters have the following definitions:
ρ denotes the mass density, P0 denotes the initial tension, h denotes the area of
the cross-section, E denotes the Young modulus of the material and L denotes the
length of the string.

Recently, Xiang et al. [34] considered the Kirchhoff-type parabolic problem in-
volving the fractional Laplacian,

ut +M([u]2s)LKu = |u|p−2u, in Ω× (0,+∞),

u(x, 0) = u0(x), in Ω,

u(x, t) = 0, in (Rn \ Ω)× (0,+∞).

(1.6)

Firstly, by using the classical Galerkin method, the authors showed the local exis-
tence of solutions. Secondly, they obtained the finite time blow-up of solutions with
negative initial energy. Finally, they also estimated the upper and lower bounds
of the blow-up time by some differential inequality techniques. For more recent
references on Kirchhoff-type problems, we refer to [6, 7, 8, 12, 13, 14, 26, 27, 28,
32, 33, 35, 36, 37, 39, 41].

It is worth mentioning that the potential well method was introduced by Sat-
tinger in [31] to study the global existence of solutions to the nonlinear hyperbolic
equations. From then on, many researchers applied this method to study the non-
linear evolution equations, see [5, 9, 10, 18, 21, 22, 23, 24, 25, 29, 38, 43]. Especially,
Payne and Sattinger [29] investigated the existence and finite time blow-up of so-
lutions to the initial boundary value problem of semilinear parabolic equations
and semilinear hyperbolic equations. Ikehata and Suzuki [18] studied the stable
and unstable sets for the parabolic equations and hyperbolic equations. Liu et
al. [22, 23, 24, 25, 38] treated the existence of the global solution for the double
dispersion equations, semilinear wave equations and parabolic equations.

In recent years, there has been a lot of work on evolution/steady-state problems
in the hyperbolic space, especially in the Poincaré ball model. The research contents
include existence, uniqueness, multiplicity, global existence and blow-up, see [1, 2,
4, 15, 16, 17, 30] and references therein. In particular, the reference [2] dealt with
the steady-state problem corresponding to problem (1.1), i.e.,

−
(
a+ b

∫
BR

|∇Hu(σ)|2 dµ
)

∆Hu = ξf(u), in BR,

u(σ, t) = 0, on ∂BR.

(1.7)

By using the topological and variational methods, the existence and multiplicity of
the weak solution to the above problem were studied.



4 H. DING, J. ZHOU EJDE-2022/38

Motivated by the above work, in this article, we consider the evolution problem
corresponding to (1.7) with f(u) = |u|q−1u, i.e., problem (1.1). By using the
potential well theory, we obtain

(1) conditions for the existence of global solutions and for finite time blow-up;
(2) growth rate of the blow-up solution and the decay rate of the global solution;
(3) upper and lower bounds of the blow-up time;
(4) necessary and sufficient conditions for the solution existing globally or blow-

ing up in finite time.

The remaining parts of this article are organized as follows. In Section 2, we give
the main results of this paper. In Section 3, we introduce some important lemmas.
In Section 4, we prove the main results.

2. Main results

To introduce the main results, we first introduce some notation. Throughout
this paper, the norm of the space L%(BR) for 1 ≤ % ≤ +∞ is denoted by ‖ · ‖%.
Namely, for any v ∈ L%(BR) ,

‖v‖% =

{( ∫
BR
|v(σ)|% dµ

)1/%
, if 1 ≤ % < +∞;

ess supσ∈BR |v(σ)|, if % = +∞.

Moreover, the inner product of the Hilbert space L2(BR) is defined by

(u, v) :=

∫
BR

uv dµ, ∀u, v ∈ L2(BR).

Secondly, the energy functional J and the Nehari functional I are defined by

J(u) :=
1

2

(
a‖u‖2 +

b

2
‖u‖4

)
− ξ

q + 1
‖u‖q+1

q+1, (2.1)

I(u) := 〈J ′(u), u〉 = a‖u‖2 + b‖u‖4 − ξ‖u‖q+1
q+1, (2.2)

where 〈·, ·〉 denotes the dual product between H−1(BR) and H1
0 (BR). By 3 < q < 5

and (1.5), we know that J and I are well defined in H1
0 (BR). Moreover, from (1.4),

we see that the critical points of J are weak solutions of the steady-state problem
corresponding to (1.1) (see [2]).

Obviously, from (2.1) and (2.2), one has

J(u) =
(q − 1)a

2(q + 1)
‖u‖2 +

(q − 3)b

4(q + 1)
‖u‖4 +

1

q + 1
I(u). (2.3)

The depth of the potential well is defined by

d := inf
u∈N

J(u), (2.4)

where N denotes the Nehari manifold and

N := {u ∈ H1
0 (BR) \ {0} : I(u) = 0}. (2.5)

From Lemma 3.3, we see that d is a positive constant and

d ≥M :=
2ar2

0(q − 1) + br4
0(q − 3)

4(q + 1)
, (2.6)

where r0 > 0 is the constant given in Lemma 3.2.
In addition, we set

N+ := {u ∈ H1
0 (BR) : I(u) > 0}, (2.7)



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 5

N− := {u ∈ H1
0 (BR) : I(u) < 0}. (2.8)

Thirdly, we define the potential well W and the outer space of the potential well
V as follows:

W := {u ∈ H1
0 (BR) : J(u) < d, I(u) > 0} ∪ {0}, (2.9)

V := {u ∈ H1
0 (BR) : J(u) < d, I(u) < 0}. (2.10)

To introduce the main results, we need the following three definitions.

Definition 2.1. Let u0 ∈ H1
0 (BR) and T > 0. A function u = u(t) in the space

L∞(0, T ;H1
0 (BR)) with ut ∈ L2(0, T ;L2(BR)) is said to be a weak solution of (1.1),

if ∫
BR

utφdµ+
(
a+ b

∫
BR

|∇Hu(σ)|2 dµ
)∫

BR

∇Hu∇Hφdµ

= ξ

∫
BR

|u|q−1uφ dµ,

(2.11)

for any φ ∈ H1
0 (BR). In addition, the following energy inequality holds

J(u(t)) +

∫ t

0

‖uτ‖22 dτ ≤ J(u0) (2.12)

for a.e. t ∈ (0, T ).

Definition 2.2. Assume u = u(t) is a weak solution of (1.1), then the maximal
existence time T of u is defined by:

(1) If there is a t0 ∈ (0,+∞) such that u exists for t ∈ [0, t0), but does not
exist at t = t0, then the maximal existence time T = t0;

(2) If u exists for all t ∈ [0,+∞), then the maximal existence time T = +∞.

Definition 2.3. Assume u = u(t) is a weak solution of (1.1). If the maximal
existence time T < +∞ and

lim
t→T−

∫ t

0

‖u‖22 dτ = +∞, (2.13)

then we say that u blows up in finite time.

Now, we introduce the main results of the present paper. Firstly, we give the
existence of global solutions.

Theorem 2.4. Assume (1.2) holds and u0 ∈ H1
0 (BR). If J(u0) < d and I(u0) > 0,

then (1.1) admits a global weak solution u(t) ∈ L∞(0,+∞;H1
0 (BR)) with ut ∈

L2(0,+∞;L2(BR)) and u(t) ∈ W for all t ∈ [0,+∞). In addition, if the weak
solution is bounded, then it is unique. Furthermore, if J(u0) < d0, then

‖u‖22 ≤
‖u0‖22

Dt‖u0‖22 + 1
,

where

D := 2λ2
1

[
b− ξCq+1

q+1

(4(q + 1)J(u0)

(q − 3)b

) q−3
4
]
> 0, d0 :=

(q − 3)b

4(q + 1)

( b

ξCq+1
q+1

) 4
q−3

.

Here, Cq+1 is defined in (1.5), and λ1 > 0 is the first eigenvalue of the eigenvalue
problem

−∆Hu = λu, in BR;

u = 0, on ∂BR,
(2.14)



6 H. DING, J. ZHOU EJDE-2022/38

which can be characterized as

λ1 = inf
u∈H1

0 (BR)\{0}

‖u‖2

‖u‖22
. (2.15)

Remark 2.5. Note that d0 ≤ d. Indeed, for any u ∈ N , it follows from (2.3) and
Lemma 3.2(3) that

J(u) =
(q − 1)a

2(q + 1)
‖u‖2 +

(q − 3)b

4(q + 1)
‖u‖4

≥ (q − 3)b

4(q + 1)
‖u‖4

≥ (q − 3)b

4(q + 1)
r4
0

=
(q − 3)b

4(q + 1)
(

b

ξCq+1
q+1

)
4
q−3 = d0.

From Theorem 2.4, we can obtain the following corollary.

Corollary 2.6. Assume (1.2) holds and u0 ∈ H1
0 (BR). If J(u0) ≤ d and I(u0) ≥ 0,

then (1.1) admits a global weak solution u(t) ∈ L∞(0,+∞;H1
0 (BR)) with ut ∈

L2(0,+∞;L2(BR)) and u(t) ∈W for all t ∈ [0,+∞).

Next, we introduce a result about finite time blow-up.

Theorem 2.7. Assume (1.2) holds and u0 ∈ H1
0 (BR). Let u = u(t) be a weak

solution of (1.1). If J(u0) ≤ d and I(u0) < 0, then u(t) blows up at some finite
time T . Furthermore,

(1) if J(u0) < d, then T can be estimated by

T ≤ 4q‖u0‖22
(q + 1)(q − 1)2(d− J(u0))

;

(2) if 3 < q < 11/3, then

T >
‖u0‖2−2γ

2

2Ĉ(γ − 1)
and ‖u‖2 >

(
2Ĉ(T − t)(γ − 1)

) 1
2(1−γ) ,

where

γ =
10− 2q

11− 3q
> 1, Ĉ =

(ξC̃q+1

b
3q−3

8

) 8
11−3q

.

Here, C̃ is the best constant in the inequality

‖u‖q+1 ≤ C̃‖u‖1−θ‖u‖θ2, (2.16)

and

θ =
5− q

2(q + 1)
∈ (0, 1). (2.17)

Remark 2.8. The constant C̃ in (2.16) is well-defined. In fact, by (1.5), we have

‖u‖6 ≤ C6‖u‖.

Since 3 < q < 5, by using the interpolation inequality (see [3, 11]), we obtain

‖u‖q+1 ≤ ‖u‖1−θ6 ‖u‖θ2,



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 7

where θ is given in (2.17). Combining the above two inequalities, we obtain

‖u‖q+1 ≤ C1−θ
6 ‖u‖1−θ‖u‖θ2.

So, C̃ is well-defined and C̃ ≤ C1−θ
6 .

The next theorem gives the growth rate of blow-up solutions.

Theorem 2.9. Assume (1.2) holds and u0 ∈ H1
0 (BR). Let u = u(t) be a weak

solution of (1.1). If J(u0) ≤ M and I(u0) < 0, then for any ε ∈ (0, 1/3], there is
a tε ∈ (0, T ) such that u satisfies

‖u‖22 ≥ Cε
(
t
(q+1)ε

2 − t
(q+1)ε

2 −1tε

) 2
2−(q+1)ε

for all t ∈ [tε, T ), where

Cε :=
[(

1− (q + 1)ε

2

)
F−

(q+1)ε
2 (tε)F

′(tε)
] 2

2−(q+1)ε

, F (t) :=

∫ t

0

‖u‖22 dτ.

Next, we give a new blow-up condition which is independent of d.

Theorem 2.10. Assume (1.2) holds and u0 ∈ H1
0 (BR). Let u = u(t) be a weak

solution of (1.1). If

J(u0) <
(q − 1)aλ1

2(q + 1)
‖u0‖22 +

(q − 3)bλ2
1

4(q + 1)
‖u0‖42, (2.18)

then u(t) blows up at some finite time T . Furthermore, T can be estimated by

T ≤ 16q‖u0‖22
(q − 1)2[2(q − 1)aλ1‖u0‖22 + (q − 3)bλ2

1‖u0‖42 − 4(q + 1)J(u0)]
.

In addition,

‖u‖22 ≥
2(q + 1)

S
J(u0) +

(
‖u0‖22 −

2(q + 1)

S
J(u0)

)
eSt,

where λ1 is defined in (2.15) and S =
2(q−1)aλ1+(q−3)bλ2

1‖u0‖22
2 .

Next, we give a finite time blow-up result with arbitrary high initial energy.

Theorem 2.11. For any constant P > d, there is a function uP ∈ H1
0 (BR), which

satisfies J(uP ) = P and (2.18). Then the weak solution u of problem (1.1) with the
initial value uP blows up in finite time.

Next, we give a result related to the asymptotic behavior of the energy functional.

Theorem 2.12. Let u = u(t) be a weak solution of (1.1) and T be the maximal
existence time of u. If J(u0) ≤ d, I(u0) < 0 or (2.18) holds, then

lim
t→T

J(u(t)) = −∞. (2.19)

The next theorem is about some equivalent conditions for the solution blowing
up in finite time or existing globally.

Theorem 2.13. Let u = u(t) be a weak solution of (1.1) and T be the maximum
existence time of u,

(1) if J(u0) < d and u0 ∈ H1
0 (BR) \ {0}, then it holds

(a) I(u0) < 0⇔ T < +∞⇔ there is a t0 ∈ [0, T ) such that J(u(t0)) < 0;
(b) I(u0) > 0⇔ T = +∞⇔ J(u(t)) > 0 for all t ∈ [0, T );



8 H. DING, J. ZHOU EJDE-2022/38

(2) if J(u0) = d and u0 ∈ H1
0 (BR) \ {N ∪ {0}}, then it holds

(a) I(u0) < 0⇔ T < +∞⇔ there is a t0 ∈ [0, T ) such that J(u(t0)) < 0;
(b) I(u0) > 0⇔ T = +∞⇔ J(u(t)) > 0 for all t ∈ [0, T ),

where N is defined in (2.5).

3. Preliminaries

Lemma 3.1 (see [19, 20]). Assume that 0 < T ≤ +∞ and ρ(t) ∈ C2[0, T ) is a
nonnegative function satisfying

ρ′′(t)ρ(t)− (1 + γ)(ρ′(t))2 ≥ 0,

where γ is a positive constant. If ρ(0) > 0 and ρ′(0) > 0, then T ≤ ρ(0)
γρ′(0) < +∞

and ρ(t)→ +∞ as t→ T .

Lemma 3.2. Let u ∈ H1
0 (BR) and (1.2) hold,

(1) if 0 < ‖u‖ < r0, then I(u) > 0;
(2) if I(u) < 0, then ‖u‖ > r0;
(3) if I(u) = 0, then ‖u‖ ≥ r0 or ‖u‖ = 0,

where r0 =
(

b

ξCq+1
q+1

) 1
q−3 > 0.

Proof. (1) It follows from 0 < ‖u‖ < r0 and (1.5) that

ξ‖u‖q+1
q+1 ≤ ξC

q+1
q+1‖u‖q+1 = ξCq+1

q+1‖u‖q−3‖u‖4 < b‖u‖4 ≤ a‖u‖2 + b‖u‖4,

which, together with the definition of I(u), implies I(u) > 0.
(2) Because I(u) < 0, we infer that ‖u‖ 6= 0. By (1.5), one has

a‖u‖2 + b‖u‖4 < ξ‖u‖q+1
q+1 ≤ ξC

q+1
q+1‖u‖q+1,

which yields

ξCq+1
q+1‖u‖q−1 > a+ b‖u‖2 ≥ b‖u‖2,

this gives ‖u‖ > r0.
(3) If ‖u‖ = 0, then we obtain I(u) = 0. If I(u) = 0 and ‖u‖ 6= 0, then we obtain

from (1.5) that

a‖u‖2 + b‖u‖4 = ξ‖u‖q+1
q+1 ≤ ξC

q+1
q+1‖u‖q+1,

which implies

ξCq+1
q+1‖u‖q−1 ≥ a+ b‖u‖2 ≥ b‖u‖2,

this gives ‖u‖ ≥ r0. �

Lemma 3.3. Let (1.2) hold. Then

d ≥ 2ar2
0(q − 1) + br4

0(q − 3)

4(q + 1)
, (3.1)

where d and r0 are defined in (2.4) and Lemma 3.2, respectively.

Proof. For all u ∈ N , we have I(u) = 0 and u ∈ H1
0 (BR) \ {0}. Then from (2.3)

and Lemma 3.2(3) we obtain

J(u) =
(q − 1)a

2(q + 1)
‖u‖2 +

(q − 3)b

4(q + 1)
‖u‖4 +

1

q + 1
I(u)

=
(q − 1)a

2(q + 1)
‖u‖2 +

(q − 3)b

4(q + 1)
‖u‖4



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 9

≥ 2ar2
0(q − 1) + br4

0(q − 3)

4(q + 1)
,

which implies (3.1). �

Lemma 3.4. Let (1.2) hold. If u ∈ H1
0 (BR) and I(u) < 0, then there is a r∗ ∈ (0, 1)

such that I(r∗u) = 0.

Proof. We divide the proof into two cases.

Case 1: a = 0. For r > 0, we set φ(r) := ξrq−3‖u‖q+1
q+1, then it is clear that

I(ru) = br4‖u‖4 − ξrq+1‖u‖q+1
q+1 = r4(b‖u‖4 − φ(r)). (3.2)

It follows from I(u) < 0, (3.2) and Lemma 3.2(2) that

φ(1) > b‖u‖4 > br4
0 > 0. (3.3)

Furthermore, according to the definition of φ(r), we reach

lim
r→0+

φ(r) = 0,

which, together with (3.3), implies that there is a r∗ ∈ (0, 1) such that φ(r∗) = b‖u‖4
and I(r∗u) = 0.

Case 2: a > 0. For r > 0, we set φ(r) := ξrq−1‖u‖q+1
q+1 − br2‖u‖4, then we have

I(ru) = ar2‖u‖2 + br4‖u‖4 − ξrq+1‖u‖q+1
q+1 = r2(a‖u‖2 − φ(r)). (3.4)

It follows from I(u) < 0, (3.4) and Lemma 3.2(2) that

φ(1) > a‖u‖2 > ar2
0 > 0. (3.5)

Furthermore, from the definition of φ(r), we have

lim
r→0+

φ(r) = 0,

which, together with (3.5), implies that there is a r∗ ∈ (0, 1) such that φ(r∗) =
a‖u‖2 and I(r∗u) = 0. �

Lemma 3.5. Let (1.2) hold. If u ∈ H1
0 (BR) and I(u) < 0, then

I(u) < (q + 1)(J(u)− d). (3.6)

Proof. By Lemma 3.4, we see that there is a r∗ ∈ (0, 1) such that I(r∗u) = 0. Let

f(r) := (q + 1)J(ru)− I(ru), r > 0,

then we have

f(r) =
a(q − 1)

2
r2‖u‖2 +

b(q − 3)

4
r4‖u‖4.

It follows from Lemma 3.2(2) that

f ′(r) = a(q − 1)r‖u‖2 + b(q − 3)r3‖u‖4 ≥ b(q − 3)r3‖u‖4 > b(q − 3)r3r4
0 > 0,

which implies that f(r) is strictly increasing for r > 0. Then we obtain from
0 < r∗ < 1 that f(1) > f(r∗), i.e.,

(q + 1)J(u)− I(u) > (q + 1)J(r∗u)− I(r∗u) = (q + 1)J(r∗u) ≥ (q + 1)d,

which means (3.6). �



10 H. DING, J. ZHOU EJDE-2022/38

Lemma 3.6. Assume (1.2) holds and u0 ∈ H1
0 (BR). Let u = u(t) be a weak

solution of (1.1). Then

1

2

d

dt
‖u‖22 = −I(u), ∀t ∈ [0, T ), (3.7)

where T is the maximum existence time of u.

Proof. Let φ = u(t) in (2.11), one has∫
BR

utu dµ+
(
a+ b

∫
BR

|∇Hu(σ)|2 dµ
)∫

BR

|∇Hu|2 dµ = ξ

∫
BR

|u|q+1 dµ,

i.e.,

1

2

d

dt
‖u‖22 = −a‖u‖2 − b‖u‖4 + ξ‖u‖q+1

q+1,

which, along with the definition of I(u), yields (3.7). �

Lemma 3.7. Let u = u(t) be a weak solution of (1.1) and T be the maximum
existence time of u. If J(u0) ≤ d, then the sets N− and N+ are both invariant for
u(t), namely, if u0 ∈ N− (resp. u0 ∈ N+), then u(t) ∈ N− (resp. u(t) ∈ N+) for
all t ∈ [0, T ).

Proof. Because the proof of the invariance of N+ and N− is similar, we only show
the invariance of N−. We divide the proof into two cases.

Case 1: J(u0) < d. By contradiction, if not, then there must exist a t0 ∈ (0, T )
such that I(u(t)) < 0 for t ∈ [0, t0) and I(u(t0)) = 0. Then it follows from Lemma
3.2(2) that ‖u‖ > r0 > 0 for t ∈ [0, t0), which means ‖u(t0)‖ ≥ r0 > 0. Thus, we
infer that u(t0) ∈ N and J(u(t0)) ≥ d, which contradicts that J(u(t0)) ≤ J(u0) < d
(see (2.12)).

Case 2: J(u0) = d. By contradiction, if not, then there must exist a t1 ∈ (0, T )
such that I(u(t)) < 0 for t ∈ [0, t1) and I(u(t1)) = 0. Then it follows from Lemma
3.2(2) that ‖u‖ > r0 > 0 for t ∈ [0, t1), which implies u(t1) 6= 0. Consequently,
we infer that u(t1) ∈ N and J(u(t1)) ≥ d. Furthermore, by Lemma 3.6, we know

(ut, u) = −I(u(t)) > 0 for t ∈ [0, t1), which means
∫ t1

0
‖uτ‖22 dτ > 0. Hence, we

obtain from (2.12) that

J(u(t1)) ≤ J(u0)−
∫ t1

0

‖uτ‖22 dτ < d,

which contradicts that J(u(t1)) ≥ d. �

4. Proofs of main results

Proof of Theorem 2.4. We divide the proof into three steps.

Step 1: Existence of the global weak solution. Let ωj , j = 1, 2, . . . be the
eigenfunction of the Laplace-Beltrami operator subject to the Dirichlet boundary
condition

−∆Hωj = λjωj , σ ∈ BR,
ωj = 0, σ ∈ ∂BR.



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 11

Furthermore, we normalize ωj such that ‖ωj‖2 = 1. Then we see that {ωj}∞j=1 is a

basis of H1
0 (BR). Constructing the following approximate solution um(t) of (1.1),

um =

m∑
j=1

gjm(t)ωj(σ), m = 1, 2 . . . (4.1)

which satisfies∫
BR

umtωj dµ+
(
a+ b

∫
BR

|∇um(σ)|2 dµ
)∫

BR

∇Hum∇Hωj dµ

= ξ

∫
BR

|um|q−1umωj dµ,

(um(0), ωj) = ζjm

(4.2)

for j = 1, 2, . . . ,m, where the constant ζjm satisfies

um(0) =

m∑
j=1

ζjmωj(σ)→ u0 in H1
0 (BR) as m→ +∞. (4.3)

According to the standard theory of ODEs, we infer that there is a T > 0
depending only on ζjm (j = 1, 2, . . . ,m) such that gjm ∈ C1[0, T ] and gjm(0) = ζjm.
Hence, um ∈ C1([0, T ];H1

0 (BR)).
Multiplying the first equation of (4.2) by g′jm(t) and summing for j = 1, 2, . . . ,m,

then integrating with respect to time from 0 to t, one has

J(um(t)) +

∫ t

0

‖umτ‖22 dτ = J(um(0)), t ∈ [0, T ].

It follows from (4.3) and gjm(0) = ζjm that

lim
m→+∞

J(um(0)) = J(u0) < d, lim
m→+∞

I(um(0)) = I(u0) > 0.

We conclude that I(um(0)) > 0 and

J(um(t)) +

∫ t

0

‖umτ‖22 dτ = J(um(0)) < d, t ∈ [0, T ] (4.4)

for sufficiently large m, which implies um(0) ∈W .
Now, for any t ∈ [0, T ] and sufficiently large m, we show that um(t) ∈ W . In

fact, if not, we infer that there is a t0 ∈ (0, T ] and a sufficiently large m such
that um(t0) ∈ ∂W , which implies um(t0) ∈ H1

0 (BR) \ {0} and I(um(t0)) = 0 or
J(um(t0)) = d. According to (4.4), we know that J(um(t0)) = d is impossible.
Hence, we obtain um(t0) ∈ N , then we infer that J(um(t0)) ≥ d, a contradiction.
Therefore, for any t ∈ [0, T ] and sufficiently large m, we have um(t) ∈W .

According to I(um(t)) > 0, (4.4), and

J(um(t)) =
(q − 1)a

2(q + 1)
‖um‖2 +

(q − 3)b

4(q + 1)
‖um‖4 +

1

q + 1
I(um(t)),

we readily obtain

(q − 1)a

2(q + 1)
‖um‖2 +

(q − 3)b

4(q + 1)
‖um‖4 +

∫ t

0

‖umτ‖22 dτ < d

for any t ∈ [0, T ] and sufficiently large m. Then∫ t

0

‖umτ‖22 dτ < d, ∀t ∈ [0, T ], (4.5)



12 H. DING, J. ZHOU EJDE-2022/38

‖um‖4 <
4(q + 1)d

(q − 3)b
, ∀t ∈ [0, T ]. (4.6)

Thus, T = +∞ and um(t) ∈W for t ∈ [0,+∞). It follows from (1.5) and (4.6) that∫
BR

∣∣|um|q−1um
∣∣ q+1
q dµ =

∫
BR

|um|q+1 dµ ≤ Cq+1
q+1‖um‖q+1

< Cd :=
(4(q + 1)dC4

q+1

(q − 3)b

) q+1
4

, ∀t ∈ [0,+∞).

(4.7)

It follows from (4.5), (4.6), and (4.7) that there exists a function u = u(t) ∈
L∞(0,+∞;H1

0 (BR)) with ut ∈ L2(0,+∞;L2(BR)) and a subsequence of {um}∞m=1

(still denoted by {um}∞m=1) such that for each T̃ > 0, as m→ +∞,

umt ⇀ ut weakly in L2(0, T̃ ;L2(BR)), (4.8)

um ⇀ u weakly star in L∞(0, T̃ ;H1
0 (BR)), (4.9)

um ⇀ u weakly in L2(0, T̃ ;H1
0 (BR)), (4.10)

|um|q−1um ⇀ |u|q−1u weakly star in L∞(0, T̃ ;L
q+1
q (BR)), (4.11)

|um|q−1um ⇀ |u|q−1u weakly in L2(0, T̃ ;L
q+1
q (BR)). (4.12)

In addition, it is clear from 3 < q < 5 that H1
0 (BR) ↪→ Lq+1(BR) compactly. Then

from [42], we conclude that{
u : u ∈ L2

(
0, T̃ ;H1

0 (BR)
)
, ut ∈ L2(0, T̃ ;L2(BR))

}
↪→ L2(0, T̃ ;Lq+1(BR))

compactly. Consequently,

um → u strongly in L2(0, T̃ ;Lq+1(BR)). (4.13)

Now, we choose a function h ∈ C1([0, T̃ ];H1
0 (BR)) and fix a integer s > 0 such that

h =

s∑
j=1

fj(t)ωj(σ), (4.14)

where {fj(t)}sj=1 are arbitrary given C1 functions. Taking m ≥ s in the first
equation of (4.2), and multiplying it by fj(t), summing for j = 1, 2, . . . , s, then

integrating with respect to t from 0 to T̃ , one has∫ T̃

0

∫
BR

umth dµ dt+

∫ T̃

0

(
a+ b

∫
BR

|∇Hum(σ)|2 dµ
)∫

BR

∇Hum∇Hh dµ dt

= ξ

∫ T̃

0

∫
BR

|um|q−1umh dµ dt.

Taking m → +∞ in the above equality, we obtain from (4.8), (4.10) and (4.12)
that ∫ T̃

0

∫
BR

uth dµ dt+

∫ T̃

0

(
a+ b

∫
BR

|∇Hu(σ)|2 dµ
)∫

BR

∇Hu∇Hh dµ dt

= ξ

∫ T̃

0

∫
BR

|u|q−1uh dµ dt.



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 13

Because the set of functions h is dense in L2(0, T̃ ;H1
0 (BR)), we know the above

equality holds for all h ∈ L2(0, T̃ ;H1
0 (BR)). From the arbitrariness of T̃ > 0, we

infer that∫
BR

utφdµ+
(
a+ b

∫
BR

|∇Hu(σ)|2 dµ
)∫

BR

∇Hu∇Hφdµ = ξ

∫
BR

|u|q−1uφ dµ

holds for any φ ∈ H1
0 (BR) and a.e. t ∈ (0,+∞).

According to (4.8) and (4.10), we know that um(0) ⇀ u(0) weakly in L2(BR).
Then it follows from (4.1), (4.3) and gjm(0) = ζjm that u(0) = u0 ∈ H1

0 (BR).
Now, we show that (2.12) holds for a.e. t ∈ (0,+∞). Indeed, for a.e. t ∈ (0,+∞),

we select T̃ > t. Then we obtain from (4.13) that um(t) → u(t) strongly in
Lq+1(BR). Thus, it follows from (4.7) and (4.12) that, as m→ +∞,

∣∣ ∫
BR

|um|q+1 dµ−
∫
BR

|u|q+1 dµ
∣∣

≤
∣∣ ∫
BR

u(um|um|q−1 − u|u|q−1) dµ
∣∣+
∣∣ ∫
BR

(um − u)um|um|q−1 dµ
∣∣

≤
∣∣ ∫
BR

u(um|um|q−1 − u|u|q−1) dµ
∣∣+

q+1
q

√
Cd‖um − u‖q+1 → 0,

(4.15)

which, along with (4.1), (4.3), (4.4), (4.8), (4.10), (4.13), and gjm(0) = ξjm, yields

∫ t

0

‖uτ‖22 dτ +
a

2
‖u‖2 +

b

4
‖u‖4

≤ lim inf
m→+∞

∫ t

0

‖umτ‖22 dτ +
a

2
lim inf
m→∞

‖um‖2 +
b

4
lim inf
m→+∞

‖um‖4

≤ lim inf
m→+∞

(∫ t

0

‖umτ‖22 dτ +
a

2
‖um‖2 +

b

4
‖um‖4

)
= lim inf
m→+∞

(∫ t

0

‖umτ‖22 dτ + J(um) +
ξ

q + 1
‖um‖q+1

q+1

)
= lim
m→+∞

(
ξ

q + 1
‖um‖q+1

q+1 + J(um(0)))

=
ξ

q + 1
‖u‖q+1

q+1 + J(u0),

which means (2.12) holds for a.e. t ∈ (0,+∞). In addition, similar to the proof of
um(t) ∈W , we can show that u(t) ∈W for all t ∈ [0,+∞).

Step 2: Uniqueness of the bounded global weak solution. Let u and v be
two bounded weak solutions of problem (1.1), then

(ut, φ) + a(∇Hu,∇Hφ) + b‖∇Hu‖22(∇Hu,∇Hφ) = ξ(|u|q−1u, φ),

(vt, φ) + a(∇Hv,∇Hφ) + b‖∇Hv‖22(∇Hv,∇Hφ) = ξ(|v|q−1v, φ)



14 H. DING, J. ZHOU EJDE-2022/38

for any φ ∈ H1
0 (BR). Subtracting the above two equalities, then letting φ = u−v ∈

H1
0 (BR) and integrating over (0, t) for any t > 0, one has∫ t

0

∫
BR

(u− v)τ (u− v) + a|∇H(u− v)|2

+ (b‖∇Hu‖22∇Hu− b‖∇Hv‖22∇Hv)∇H(u− v) dµ dτ

= ξ

∫ t

0

∫
BR

(|u|q−1u− |v|q−1v)(u− v) dµ dτ.

(4.16)

By means of the Cauchy-Schwarz inequality, we conclude that

a‖∇H(u− v)‖22 + b

∫
BR

(‖∇Hu‖22∇Hu− ‖∇Hv‖22∇Hv)(∇Hu−∇Hv) dµ

≥ b
(
‖∇Hu‖42 − ‖∇Hu‖22

∫
BR

∇Hu∇Hv dµ
)

− b
(
‖∇Hv‖22

∫
BR

∇Hu∇Hv dµ− ‖∇Hv‖42
)

≥ b
(
‖∇Hu‖42 − ‖∇Hu‖22 ·

‖∇Hu‖22 + ‖∇Hv‖22
2

)
− b
(
‖∇Hv‖22 ·

‖∇Hu‖22 + ‖∇Hv‖22
2

− ‖∇Hv‖42
)

=
b

2
(‖∇Hu‖22 − ‖∇Hv‖22)2 ≥ 0.

(4.17)

It follows from (4.16) and (4.17) that

ξ

∫ t

0

∫
BR

(|u|q−1u− |v|q−1v)(u− v) dµ dτ ≥
∫ t

0

∫
BR

(u− v)τ (u− v) dµ dτ.

Since (u− v)(σ, 0) = 0 and 3 < q < 5, we obtain from the boundedness of u and v
that

‖φ‖22 ≤ C
∫ t

0

‖φ‖22 dτ,

where the positive constant C depends only on q, ξ and the bounds of u, v. Then
by Gronwall’s inequality, we have ‖φ‖22 = 0. Hence, we obtain φ = 0 a.e. in
BR × (0,+∞) and the proof of this step is complete.

Step 3: Decay estimates. Because d0 ≤ d, it follows from J(u0) < d0, I(u0) > 0
and step 1 that I(u) ≥ 0 for t ∈ [0,+∞). By (2.3) and (2.12), we obtain

J(u0) ≥ J(u) =
(q − 1)a

2(q + 1)
‖u‖2 +

(q − 3)b

4(q + 1)
‖u‖4 +

1

q + 1
I(u) ≥ (q − 3)b

4(q + 1)
‖u‖4. (4.18)

Then it is clear from (1.5) that

‖u‖q+1 ≤ Cq+1‖u‖ ≤ Cq+1

(4(q + 1)J(u0)

(q − 3)b

)1/4

. (4.19)

From (1.5) and (4.19), we have

‖u‖q+1
q+1 = ‖u‖q−3

q+1‖u‖4q+1 ≤ C4
q+1‖u‖

q−3
q+1‖u‖4

≤ Cq+1
q+1 (

4(q + 1)J(u0)

(q − 3)b
)
q−3
4 ‖u‖4.

(4.20)



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 15

Moreover, from Lemma 3.6 we obtain

d

dt
‖u‖22 = −2I(u) = 2ξ‖u‖q+1

q+1 − 2a‖u‖2 − 2b‖u‖4. (4.21)

Then by (2.15) and (4.20), we obtain

d

dt
‖u‖22 ≤ −2b‖u‖4 + 2ξ‖u‖q+1

q+1

≤ −2b‖u‖4 + 2ξCq+1
q+1

(4(q + 1)J(u0)

(q − 3)b

) q−3
4 ‖u‖4

= −2
[
b− ξCq+1

q+1

(4(q + 1)J(u0)

(q − 3)b

) q−3
4
]
‖u‖4

≤ −2λ2
1

[
b− ξCq+1

q+1

(4(q + 1)J(u0)

(q − 3)b

) q−3
4
]
‖u‖42,

which implies

‖u‖22 ≤
‖u0‖22

Dt‖u0‖22 + 1
,

where

D = 2λ2
1

[
b− ξCq+1

q+1

(4(q + 1)J(u0)

(q − 3)b

) q−3
4
]
> 0.

�

Proof of Corollary 2.6. If u0 = 0, then the proof is complete because problem (1.1)
admits a global weak solution u(t) ≡ 0. Hence, in the following, we assume that
u0 ∈ H1

0 (BR) \ {0}, and then the proof is divided into three cases:

Case 1: J(u0) < d and I(u0) = 0. Because u0 ∈ N , then we infer that J(u0) ≥ d,
which is a contradiction and this case cannot happen.

Case 2: J(u0) < d and I(u0) > 0. By Theorem 2.4, we know that problem (1.1)
admits a global weak solution.

Case 3: J(u0) = d and I(u0) ≥ 0. In this case, we let hm = 1− 1
m (m = 2, 3, . . . ).

Considering the problem

ut −
(
a+ b

∫
BR

|∇Hu(σ)|2 dµ
)

∆Hu = ξ|u|q−1u, σ ∈ BR, t > 0,

u(σ, t) = 0, σ ∈ ∂BR, t > 0,

u(σ, 0) = u0m(σ) := hmu0, σ ∈ BR.

(4.22)

It follows from u0 ∈ H1
0 (BR) \ {0}, hm ∈ (0, 1) and I(u0) ≥ 0 that

I(u0m) = ah2
m‖u0‖2 + bh4

m‖u0‖4 − ξhq+1
m ‖u0‖q+1

q+1

= h2
m

(
a‖u0‖2 + bh2

m‖u0‖4 − ξhq−1
m ‖u0‖q+1

q+1

)
> 0.

Furthermore, from the definition of J(u), we obtain

d

dhm
J(hmu) = ahm‖u‖2 + bh3

m‖u‖4 − ξhqm‖u‖
q+1
q+1

=
1

hm
(ah2

m‖u‖2 + bh4
m‖u‖4 − ξhq+1

m ‖u‖q+1
q+1)

=
1

hm
I(hmu).



16 H. DING, J. ZHOU EJDE-2022/38

Then
d

dhm
J(hmu0) =

1

hm
I(hmu0) =

1

hm
I(u0m) > 0, (4.23)

which means

J(u0m) = J(hmu0) < J(u0) = d. (4.24)

By Theorem 2.4, for each m, we see that problem (4.22) admits a global weak so-
lution um(t) ∈ L∞(0,+∞;H1

0 (BR)) with umt ∈ L2(0,+∞;L2(BR)), which satisfies
um(t) ∈W for all t ∈ [0,+∞) and∫

BR

umtφdµ+
(
a+ b

∫
BR

|∇Hum(σ)|2 dµ
)∫

BR

∇Hum∇Hφdµ

= ξ

∫
BR

|um|q−1umφdµ

(4.25)

for a.e. t > 0 and any φ ∈ H1
0 (BR). In addition,

J(um(t)) +

∫ t

0

‖umτ‖22 dτ = J(u0m) < d. (4.26)

Since

J(um(t)) =
(q − 1)a

2(q + 1)
‖um‖2 +

(q − 3)b

4(q + 1)
‖um‖4 +

1

q + 1
I(um(t)),

it follows from (4.26) that

(q − 1)a

2(q + 1)
‖um‖2 +

(q − 3)b

4(q + 1)
‖um‖4 +

∫ t

0

‖umτ‖22 dτ < d.

Then the remainder of the proof is similar to that in the proof of Theorem 2.4. �

Proof of Theorem 2.7. We divide the proof into the following three steps.

Step 1: Blow-up in finite time. We divide this part of the proof into two cases.
Case 1: J(u0) < d. Let u = u(t), t ∈ [0, T ), be a weak solution of (1.1) and T be
the maximum existence time of u. Since J(u0) < d and I(u0) < 0, then it follows
from (2.12) and Lemma 3.7 that u(t) ∈ V . Next, we show that u(t) blows up at
some finite time T . By contradiction, we assume that the weak solution u exists
globally and T = +∞. Set

Q(t) :=

∫ t

0

‖u‖22 dτ, t ∈ [0, T ).

Then we obtain from Lemma 3.6 that

Q′(t) = ‖u‖22, Q′′(t) = 2(u, ut) = −2I(u). (4.27)

It follows from (2.3) and (2.12) that

(q − 1)a

2(q + 1)
‖u‖2 +

(q − 3)b

4(q + 1)
‖u‖4 +

1

q + 1
I(u) +

∫ t

0

‖uτ‖22 dτ ≤ J(u0),

which implies

−2I(u) ≥ (q − 1)a‖u‖2 +
(q − 3)b

2
‖u‖4 − 2(q + 1)J(u0) + 2(q + 1)

∫ t

0

‖uτ‖22 dτ,



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 17

then by (1.5), we reach

Q′′(t)

≥ (q − 1)a‖u‖2 +
(q − 3)b

2
‖u‖4 − 2(q + 1)J(u0) + 2(q + 1)

∫ t

0

‖uτ‖22 dτ

≥ (q − 3)b

2
‖u‖4 − 2(q + 1)J(u0) + 2(q + 1)

∫ t

0

‖uτ‖22 dτ

≥ (q − 3)b

2C4
2

‖u‖42 − 2(q + 1)J(u0) + 2(q + 1)

∫ t

0

‖uτ‖22 dτ

=
(q − 3)b

2C4
2

(Q′(t))2 − 2(q + 1)J(u0) + 2(q + 1)

∫ t

0

‖uτ‖22 dτ.

(4.28)

Furthermore, we have(∫ t

0

(uτ , u) dτ
)2

=
(1

2

∫ t

0

d

dτ
‖u‖22 dτ

)2

=
1

4
(‖u‖22 − ‖u0‖22)2

=
1

4
(‖u‖42 − 2‖u0‖22‖u‖22 + ‖u0‖42)

=
1

4
((Q′(t))2 − 2‖u0‖22Q′(t) + ‖u0‖42).

(4.29)

Then it follows from (4.28), (4.29), and Schwarz inequality that

Q(t)Q′′(t)− q + 1

2
(Q′(t))2

≥ (q − 3)b

2C4
2

Q(t)(Q′(t))2 − (q + 1)‖u0‖22Q′(t)− 2(q + 1)J(u0)Q(t) +
q + 1

2
‖u0‖42

+ 2(q + 1)

∫ t

0

‖uτ‖22 dτ
∫ t

0

‖u‖22 dτ − 2(q + 1)
(∫ t

0

(uτ , u) dτ
)2

≥ (q − 3)b

2C4
2

Q(t)(Q′(t))2 − (q + 1)‖u0‖22Q′(t)− 2(q + 1)J(u0)Q(t).

(4.30)

Furthermore, from Q′′(t) = −2I(u) > 0, we deduce that Q′(t) ≥ Q′(0) =
‖u0‖22 > 0. Then it is clear from (4.30) that

Q(t)Q′′(t)− q + 1

2
(Q′(t))2

≥ (q − 3)b‖u0‖22
2C4

2

Q(t)Q′(t)− (q + 1)‖u0‖22Q′(t)− 2(q + 1)J(u0)Q(t).
(4.31)

From (2.12), (4.27), and Lemma 3.5, we know that, for all t ∈ [0,+∞),

Q′′(t) = −2I(u) > 2(q + 1)(d− J(u)) ≥ 2(q + 1)(d− J(u0)) := C1 > 0. (4.32)

Then for all t ∈ [0,+∞), one has

Q′(t) ≥ Q′(0) + C1t = ‖u0‖22 + C1t > C1t, Q(t) > Q(0) +
C1

2
t2 =

C1

2
t2.

Hence,
lim

t→+∞
Q(t) = +∞, lim

t→+∞
Q′(t) = +∞.



18 H. DING, J. ZHOU EJDE-2022/38

Then there is a t0 ≥ 0 such that

(q − 3)b‖u0‖22
4C4

2

Q′(t) > 2(q + 1)J(u0), t0 ≤ t < +∞,

(q − 3)b‖u0‖22
4C4

2

Q(t) > (q + 1)‖u0‖22, t0 ≤ t < +∞,

which, together with (4.31), implies, for t ∈ [t0,+∞),

Q(t)Q′′(t)− q + 1

2
(Q′(t))2 ≥ (

(q − 3)b‖u0‖22
4C4

2

Q′(t)− 2(q + 1)J(u0))Q(t)

+ (
(q − 3)b‖u0‖22

4C4
2

Q(t)− (q + 1)‖u0‖22)Q′(t) > 0.

It follows from Lemma 3.1 that the maximum existence time T1 of Q(t) satisfies
T1 < +∞ and limt→T1

Q(t) = +∞, which is a contradiction.

Case 2: J(u0) = d. According to Lemma 3.7, we know I(u(t)) < 0 for all t ≥ 0,
then we infer from Lemma 3.6 that (ut, u) = −I(u(t)) > 0 for t ≥ 0, which implies
‖ut‖22 > 0 for t ≥ 0. Thus, it follows from (2.12) that there is a t1 > 0 such that

J(u(t1)) ≤ J(u0)−
∫ t1

0

‖uτ‖22 dτ < d.

Taking t1 as the initial time, then the remainder of the proof is similar to case 1.

Step 2: Upper bound estimate of the blow-up time. Let u = u(t), t ∈ [0, T ),
be a weak solution of (1.1) and T be the maximum existence time of u. Since
J(u0) < d and I(u0) < 0, then it follows from Step 1 that T < +∞. Furthermore,
by Lemma 3.7, we know that I(u) < 0 for t ∈ [0, T ). We set

a(t) :=
(∫ t

0

‖u‖22 dτ
)1/2

, b(t) :=
(∫ t

0

‖uτ‖22 dτ
)1/2

, ∀t ∈ [0, T ).

We define
H(t) := η(t+ κ)2 + a2(t) + (T − t)‖u0‖22, ∀t ∈ [0, T ), (4.33)

where κ, η > 0 are two constants which will be specified later. It follows from
Lemma 3.6 and I(u) < 0 that

H ′(t) = 2η(t+ κ) + ‖u‖22 − ‖u0‖22 ≥ 2η(t+ κ) > 0, t ∈ [0, T ). (4.34)

Then we have

H(t) ≥ H(0) = ηκ2 + T‖u0‖22 > 0, t ∈ [0, T ). (4.35)

In addition, from Lemma 3.5, Lemma 3.6, and (2.12), we conclude that

H ′′(t) = 2η − 2I(u) > 2η + 2(q + 1)(d− J(u))

≥ 2η + 2(q + 1)(d− J(u0)) + 2(q + 1)b2(t), t ∈ [0, T ).
(4.36)

Using the Cauchy-Schwarz inequality, one has

a(t)b(t) =
(∫ t

0

‖u‖22 dτ
)1/2(∫ t

0

‖uτ‖22 dτ
)1/2

≥
∫ t

0

‖u‖2‖uτ‖2 dτ ≥
∫ t

0

(u, uτ ) dτ

=
1

2

∫ t

0

d

dτ
‖u‖22 dτ, t ∈ [0, T ),



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 19

which, together with the definition of H(t), implies

(H(t)− (T − t)‖u0‖22)(b2(t) + η)

= (a2(t) + η(t+ κ)2)(b2(t) + η)

= a2(t)b2(t) + ηa2(t) + η(t+ κ)2b2(t) + η2(t+ κ)2

≥ a2(t)b2(t) + 2ηa(t)b(t)(t+ κ) + η2(t+ κ)2

≥
[
η(t+ κ) +

1

2

∫ t

0

d

dτ
‖u‖22 dτ

]2
, t ∈ [0, T ).

(4.37)

According to (4.34) and (4.37), we have

4H(t)(b2(t) + η) ≥ 4
(
η(t+ κ) +

1

2

∫ t

0

d

dτ
‖u‖22ds

)2

= (H ′(t))2, t ∈ [0, T ). (4.38)

Then it follows from (4.35), (4.36) and (4.38) that

H(t)H ′′(t)− q + 1

2
(H ′(t))2

> H(t)
[
2η + 2(q + 1)(d− J(u0)) + 2(q + 1)b2(t)− 2(q + 1)b2(t)− 2(q + 1)η

]
= H(t) [2(q + 1)(d− J(u0))− 2qη] , t ∈ [0, T ).

Choosing η small enough such that

0 < η ≤ (q + 1)(d− J(u0))

q
, (4.39)

we obtain

H(t)H ′′(t)− q + 1

2
(H ′(t))2 ≥ 0, t ∈ [0, T ).

By Lemma 3.1, one has

T ≤ H(0)

( q+1
2 − 1)H ′(0)

=
1

q − 1
(κ+

‖u0‖22
ηκ

T ). (4.40)

Choosing κ large enough such that

κ >
‖u0‖22

(q − 1)η
. (4.41)

Then by (4.40), we obtain

T ≤ ηκ2

(q − 1)ηκ− ‖u0‖22
. (4.42)

Consequently,

T ≤ inf
(δ,κ)∈Ψ

g(δ, κ), (4.43)

where δ := κη,

g(δ, κ) :=
δκ

(q − 1)δ − ‖u0‖22
, Ψ :=

{
(δ, κ) : δ >

‖u0‖22
q − 1

, κ ≥ qδ

(q + 1)(d− J(u0))

}
.

Because g(δ, κ) is increasing with respect to κ, we infer that

T ≤ inf
δ>
‖u0‖22
q−1

g
(
δ,

qδ

(q + 1)(d− J(u0))

)



20 H. DING, J. ZHOU EJDE-2022/38

= inf
δ>
‖u0‖22
q−1

qδ2

(q + 1)(d− J(u0))((q − 1)δ − ‖u0‖22)

=
qδ2

(q + 1)(d− J(u0))((q − 1)δ − ‖u0‖22)

∣∣∣
δ=

2‖u0‖22
q−1

=
4q‖u0‖22

(q + 1)(q − 1)2(d− J(u0))
.

Step 3: Lower bound estimate of the blow-up time. By Lemma 3.7, we see
that I(u) < 0 for t ∈ [0, T ). Now, we estimate the lower bound of the blow-up time
T and the blow-up rate. Set L(t) = 1

2‖u‖
2
2, then it is obvious that

L(T ) = +∞. (4.44)

It follows from Lemma 3.6 and the definition of I(u) that

1

2

d

dt
‖u‖22 = −I(u) = ξ‖u‖q+1

q+1 − a‖u‖2 − b‖u‖4. (4.45)

Then according to (2.16) and I(u) < 0, we conclude that

‖u‖q+1
q+1 ≤ C̃q+1‖u‖(1−θ)(q+1)‖u‖θ(q+1)

2

=
C̃q+1

b
(1−θ)(q+1)

4

(b‖u‖4)
(1−θ)(q+1)

4 ‖u‖θ(q+1)
2

≤ C̃q+1

b
(1−θ)(q+1)

4

(a‖u‖2 + b‖u‖4)
(1−θ)(q+1)

4 ‖u‖θ(q+1)
2

<
ξ

(1−θ)(q+1)
4 C̃q+1

b
(1−θ)(q+1)

4

(‖u‖q+1
q+1)

(1−θ)(q+1)
4 (‖u‖22)

θ(q+1)
2 .

(4.46)

From θ = 5−q
2(q+1) and 3 < q < 11

3 , we have

(1− θ)(q + 1)

4
=

3q − 3

8
< 1.

Hence, from (4.46) we obtain

‖u‖q+1
q+1 <

(ξ (1−θ)(q+1)
4 C̃q+1

b
(1−θ)(q+1)

4

) 4
4−(1−θ)(q+1)

(‖u‖22)
2θ(q+1)

4−(1−θ)(q+1) . (4.47)

By a simply calculation, one has

γ :=
2θ(q + 1)

4− (1− θ)(q + 1)
=

10− 2q

11− 3q
> 1.

Then it follows from (4.45) and (4.47) that

L′(t) = ξ‖u‖q+1
q+1 − a‖u‖2 − b‖u‖4 ≤ ξ‖u‖

q+1
q+1 < Ĉ(‖u‖22)γ = 2γĈ(L(t))γ , (4.48)

where

Ĉ =
( ξC̃q+1

b
(1−θ)(q+1)

4

) 4
4−(1−θ)(q+1)

=
(ξC̃q+1

b
3q−3

8

) 8
11−3q

.



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 21

Next, we claim that L(t) > 0 for t ∈ [0, T ). Indeed, if not, then there is a t1 ≥ 0
such that ‖u(t1)‖22 = 0, which conflicts with (4.47). Thus, we obtain from (4.48)
that

L′(t)

(L(t))γ
< 2γĈ. (4.49)

Integrating (4.49) from 0 to t yields

(L(0))1−γ − (L(t))1−γ < 2γĈ(γ − 1)t,

letting t→ T in the above inequality, we obtain from (4.44) that

T >
(L(0))1−γ

2γĈ(γ − 1)
=
‖u0‖2−2γ

2

2Ĉ(γ − 1)
.

Integrating (4.49) from t to T , it follows from (4.44) that

L(t) >
(

2γĈ(T − t)(γ − 1)
) 1

1−γ
,

i.e.,

‖u‖2 >
(

2Ĉ(T − t)(γ − 1)
) 1

2(1−γ)
.

�

Proof of Theorem 2.9. Let u = u(t), t ∈ [0, T ), be a weak solution of (1.1) and T
be the maximum existence time of u. Since J(u0) ≤ M and I(u0) < 0, then we
obtain I(u) < 0 for t ∈ [0, T ) from Lemma 3.7. Set

F (t) :=

∫ t

0

‖u‖22 dτ, t ∈ [0, T ),

then from Lemma 3.6 we obtain

F ′(t) = ‖u‖22, F ′′(t) = −2I(u) > 0. (4.50)

From (2.3), (2.6), (2.12), (4.50), and Lemma 3.2 it follows that

F ′′(t)

= (q − 1)a‖u‖2 +
(q − 3)b

2
‖u‖4 − 2(q + 1)J(u)

≥ (q − 1)a‖u‖2 +
(q − 3)b

2
‖u‖4 − 2(q + 1)J(u0) + 2(q + 1)

∫ t

0

‖uτ‖22 dτ

≥ 2(q + 1)

∫ t

0

‖uτ‖22 dτ + (q − 1)ar2
0 +

(q − 3)b

2
r4
0 − 2(q + 1)J(u0)

= 2(q + 1)

∫ t

0

‖uτ‖22 dτ + 2(q + 1)(M − J(u0)).

(4.51)

Moreover, from

4
(∫ t

0

(u, uτ ) dτ
)2

= (

∫ t

0

d

dτ
‖u‖22 dτ)2

= (F ′(t)− F ′(0))2

= (F ′(t))2 − 2F ′(t)F ′(0) + (F ′(0))2,



22 H. DING, J. ZHOU EJDE-2022/38

we readily obtain

(F ′(t))2 = 2‖u0‖22F ′(t)− ‖u0‖42 + 4
(∫ t

0

(u, uτ ) dτ
)2

, (4.52)

which, along with (4.51) and Cauchy-Schwarz inequality, yields

F (t)F ′′(t)− q + 1

2
(F ′(t))2

≥ 2(q + 1)(M − J(u0))F (t)− (q + 1)‖u0‖22F ′(t) +
q + 1

2
‖u0‖42

+ 2(q + 1)

∫ t

0

‖uτ‖22 dτ
∫ t

0

‖u‖22 dτ − 2(q + 1)
(∫ t

0

(u, uτ ) dτ
)2

≥ 2(q + 1)(M − J(u0))F (t)− (q + 1)‖u0‖22F ′(t)
≥ −(q + 1)‖u0‖22F ′(t).

Then it is clear that, for any ε ∈ (0, 1/3],

F (t)F ′′(t)− (q + 1)ε

2
(F ′(t))2 ≥ (q + 1)(1− ε)

2
(F ′(t))2− (q+ 1)‖u0‖22F ′(t). (4.53)

By Theorem 2.7, we see that u(t) blows up in finite time, then

lim
t→T−

F ′(t) = lim
t→T−

‖u‖22 = +∞.

Therefore, from (4.53) that there exists a tε ∈ (0, T ) such that

F (t)F ′′(t)− (q + 1)ε

2
(F ′(t))2 > 0, t ∈ [tε, T ). (4.54)

Since (
F 1− (q+1)ε

2 (t)
)′

=
(

1− (q + 1)ε

2

)
F−

(q+1)ε
2 (t)F ′(t),

then we obtain from (4.54) that(
F 1− (q+1)ε

2 (t)
)′′

=
(

1− (q + 1)ε

2

)
F−

(q+1)ε
2 −1(t)

×
[
F (t)F ′′(t)− (q + 1)ε

2
(F ′(t))2

]
> 0, t ∈ [tε, T ).

Because 2− (q + 1)ε ≥ 2− q+1
3 > 0 and F (tε) > 0, we conclude that

F (t) =
[ ∫ t

tε

(F 1− (q+1)ε
2 (τ))′ dτ + F 1− (q+1)ε

2 (tε)
] 2

2−(q+1)ε

≥
[
(t− tε)(F 1− (q+1)ε

2 (τ))′
∣∣
τ=tε

+ F 1− (q+1)ε
2 (tε)

] 2
2−(q+1)ε

≥
[(

1− (q + 1)ε

2

)
(t− tε)F−

(q+1)ε
2 (tε)F

′(tε)
] 2

2−(q+1)ε

= Cε(t− tε)
2

2−(q+1)ε , t ∈ [tε, T ),

(4.55)

where

Cε :=
[(

1− (q + 1)ε

2

)
F−

(q+1)ε
2 (tε)F

′(tε)
] 2

2−(q+1)ε

.

Furthermore, it follows from F ′′(t) > 0 for t ∈ [0, T ) that tF ′(t) ≥
∫ t

0
F ′(τ) dτ , i.e.,

t‖u‖22 ≥ F (t), t ∈ [0, T ). (4.56)



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 23

From (4.55) and (4.56), we reach, for all t ∈ [tε, T ) and any ε ∈ (0, 1/3],

‖u‖22 ≥ Cε
(
t
(q+1)ε

2 − t
(q+1)ε

2 −1tε

) 2
2−(q+1)ε

.

�

Proof of Theorem 2.10. We divide the proof into three steps.

Step 1: Blow-up in finite time. Let u = u(t), t ∈ [0, T ), be a weak solution of
(1.1) and T be the maximum existence time of u. According to (2.3) and (2.18),
we obtain

I(u0) = (q + 1)J(u0)− (q − 1)a

2
‖u0‖2 −

(q − 3)b

4
‖u0‖4

≤ (q + 1)J(u0)− (q − 1)aλ1

2
‖u0‖22 −

(q − 3)bλ2
1

4
‖u0‖42 < 0.

Now, we show that I(u(t)) < 0 for t ∈ [0, T ). Indeed, if not, then there exists
a t0 ∈ (0, T ) such that I(u(t)) < 0 for t ∈ [0, t0) and I(u(t0)) = 0. It follows from
Lemma 3.6 that ‖u‖22 is strictly increasing for t ∈ [0, t0). Thus,

J(u0) <
(q − 1)aλ1

2(q + 1)
‖u0‖22 +

(q − 3)bλ2
1

4(q + 1)
‖u0‖42

<
(q − 1)aλ1

2(q + 1)
‖u(t0)‖22 +

(q − 3)bλ2
1

4(q + 1)
‖u(t0)‖42.

(4.57)

Furthermore, it follows from (2.3) and (2.12) that

J(u0) ≥ J(u(t0))

≥ (q − 1)a

2(q + 1)
‖u(t0)‖2 +

(q − 3)b

4(q + 1)
‖u(t0)‖4

≥ (q − 1)aλ1

2(q + 1)
‖u(t0)‖22 +

(q − 3)bλ2
1

4(q + 1)
‖u(t0)‖42,

which conflicts with (4.57). Hence, we obtain I(u(t)) < 0 for all t ∈ [0, T ).
Next, we show that u(t) blows up in finite time. By contradiction, we choose

T̃ =
(4q‖u0‖22+1)2+1

%(q−1)2 , where % := (q−1)aλ1‖u0‖22 +
(q−3)bλ2

1

2 ‖u0‖42−2(q+1)J(u0) > 0,

and we assume that u(t) exists globally on [0, T̃ ]. Set

M(t) := ω(t+ δ)2 +

∫ t

0

‖u‖22 dτ + (T̃ − t)‖u0‖22, t ∈ [0, T̃ ],

where ω > 0 and δ > 0 are two constants which will be specified later.
It follows from Lemma 3.6 and I(u) < 0 for all t ∈ [0, T ) that ‖u‖22 is strictly

increasing for t ∈ [0, T ). Then we have

M ′(t) = 2ω(t+ δ) + ‖u‖22 − ‖u0‖22 ≥ 2ω(t+ δ) > 0, t ∈ [0, T̃ ]



24 H. DING, J. ZHOU EJDE-2022/38

and

M ′′(t) = 2ω − 2I(u)

= 2ω + (q − 1)a‖u‖2 +
(q − 3)b

2
‖u‖4 − 2(q + 1)J(u)

≥ 2ω + (q − 1)aλ1‖u‖22 +
(q − 3)bλ2

1

2
‖u‖42

− 2(q + 1)J(u0) + 2(q + 1)

∫ t

0

‖uτ‖22 dτ

≥ 2ω + (q − 1)aλ1‖u0‖22 +
(q − 3)bλ2

1

2
‖u0‖42

− 2(q + 1)J(u0) + 2(q + 1)

∫ t

0

‖uτ‖22 dτ > 0, t ∈ [0, T̃ ].

(4.58)

We define

α(t) :=
(∫ t

0

‖u‖22 dτ
)1/2

, β(t) :=
(∫ t

0

‖uτ‖22 dτ
)1/2

.

By Schwarz inequality and Hölder’s inequality, we obtain[
ω(t+ δ)2 +

∫ t

0

‖u‖22 dτ
][
ω +

∫ t

0

‖uτ‖22 dτ
]
−
[
ω(t+ δ) +

1

2
(‖u‖22 − ‖u0‖22)

]2
= [ω(t+ δ)2 + α2(t)][ω + β2(t)]−

[
ω(t+ δ) +

1

2

∫ t

0

d

dτ
‖u‖22 dτ

]2
≥ [ω(t+ δ)2 + α2(t)][ω + β2(t)]−

[
ω(t+ δ) +

∫ t

0

‖u‖2‖uτ‖2 dτ
]2

≥ [ω(t+ δ)2 + α2(t)][ω + β2(t)]− [ω(t+ δ) + α(t)β(t)]2

= [
√
ωα(t)]2 − 2ω(t+ δ)α(t)β(t) + [

√
ω(t+ δ)β(t)]2

= [
√
ωα(t)−

√
ω(t+ δ)β(t)]2 ≥ 0.

Then it is clear that

−(M ′(t))2 = −4
(
ω(t+ δ) +

1

2
(‖u‖22 − ‖u0‖22)

)2

= −4
(
M(t)− (T̃ − t)‖u0‖22

)(
ω +

∫ t

0

‖uτ‖22 dτ
)

+ 4
(
ω(t+ δ)2 +

∫ t

0

‖u‖22 dτ
)(
ω +

∫ t

0

‖uτ‖22 dτ
)

− 4
(
ω(t+ δ) +

1

2
(‖u‖22 − ‖u0‖22)

)2

≥ −4M(t)
(
ω +

∫ t

0

‖uτ‖22 dτ
)
.

It follows from (4.58) and the above inequality that

M(t)M ′′(t)− q + 1

2
(M ′(t))2

≥M(t)
(
M ′′(t)− 2(q + 1)

(
ω +

∫ t

0

‖uτ‖22 dτ
))



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 25

≥M(t)
(

(q − 1)aλ1‖u0‖22 +
(q − 3)bλ2

1

2
‖u0‖42 − 2(q + 1)J(u0)− 2qω

)
.

Taking ω = %/(4q), we obtain

M(t)M ′′(t)− q + 1

2
(M ′(t))2 ≥ 0.

It follows from Lemma 3.1 that

T ≤ 2M(0)

(q − 1)M ′(0)
=

δ

q − 1
+
‖u0‖22

ωδ(q − 1)
T̃ , lim

t→T
M(t) = +∞.

Taking δ =
4q‖u0‖22+1
%(q−1) , we obtain T < T̃ , a contradiction. Therefore, u(t) blows up

in finite time.

Step 2: Upper bound estimate of the blow-up time. For T1 ∈ (0, T ), we set

B(t) := ω(t+ δ)2 +

∫ t

0

‖u‖22 dτ + (T − t)‖u0‖22, t ∈ [0, T1],

where ω, δ > 0 are two constants to be determined later. Similar to Step 1, we infer
that

B(t)B′′(t)− q + 1

2
(B′(t))2

≥ B(t)
(

(q − 1)aλ1‖u0‖22 +
(q − 3)bλ2

1

2
‖u0‖42 − 2(q + 1)J(u0)− 2qω

)
.

Taking ω small enough such that

0 < ω ≤ %

2q
, (4.59)

we have

B(t)B′′(t)− q + 1

2
(B′(t))2 ≥ 0.

It follows from Lemma 3.1 that

T1 ≤
2B(0)

(q − 1)B′(0)
=

δ

q − 1
+
‖u0‖22

ωδ(q − 1)
T, ∀T1 ∈ [0, T ).

Letting T1 → T , we obtain

T ≤ δ

q − 1
+
‖u0‖22

ωδ(q − 1)
T. (4.60)

Taking δ large enough such that

δ >
‖u0‖22

(q − 1)ω
, (4.61)

then it is clear from (4.60) that

T ≤ ωδ2

ωδ(q − 1)− ‖u0‖22
.

According to (4.59) and (4.61), we define

Θ :=
{

(ω, δ) : 0 < ω ≤ %

2q
, δ >

‖u0‖22
(q − 1)ω

}
=
{

(ω, δ) :
‖u0‖22

(q − 1)δ
< ω ≤ %

2q
, δ >

2q‖u0‖22
(q − 1)%

}
,



26 H. DING, J. ZHOU EJDE-2022/38

then

T ≤ inf
(ω,δ)∈Θ

ωδ2

ωδ(q − 1)− ‖u0‖22
.

Let µ = ωδ and

χ(δ, µ) :=
µδ

µ(q − 1)− ‖u0‖22
.

Obviously, χ(δ, µ) is decreasing with respect to µ. Hence,

T ≤ inf
δ>

2q‖u0‖22
(q−1)%

χ
(
δ,
%δ

2q

)
= inf
δ>

2q‖u0‖22
(q−1)%

%δ2

%δ(q − 1)− 2q‖u0‖22

=
%δ2

%δ(q − 1)− 2q‖u0‖22

∣∣∣
δ=

4q‖u0‖22
(q−1)%

=
8q‖u0‖22
(q − 1)2%

.

Then it follows from the definition of % that

T ≤ 16q‖u0‖22
(q − 1)2[2(q − 1)aλ1‖u0‖22 + (q − 3)bλ2

1‖u0‖42 − 4(q + 1)J(u0)]
.

Step 3: Growth estimates.
By Step 1, we know that I(u) < 0 for all t ∈ [0, T ), then it follows from Lemma

3.6 that ‖u‖22 is strictly increasing for t ∈ [0, T ). Furthermore, we obtain from
Lemma 3.6 and (2.3) that

d

dt

(
‖u‖22 −

4(q + 1)

2(q − 1)aλ1 + (q − 3)bλ2
1‖u0‖22

J(u0)
)

= −2I(u) = (q − 1)a‖u‖2 +
(q − 3)b

2
‖u‖4 − 2(q + 1)J(u),

which, together with (2.12) and (2.15), implies

d

dt

(
‖u‖22 −

4(q + 1)

2(q − 1)aλ1 + (q − 3)bλ2
1‖u0‖22

J(u0)
)

≥ (q − 1)aλ1‖u‖22 +
(q − 3)bλ2

1

2
‖u‖42 − 2(q + 1)J(u0)

≥ (q − 1)aλ1‖u‖22 +
(q − 3)bλ2

1‖u0‖22
2

‖u‖22 − 2(q + 1)J(u0)

=
2(q − 1)aλ1 + (q − 3)bλ2

1‖u0‖22
2

(
‖u‖22 −

4(q + 1)

2(q − 1)aλ1 + (q − 3)bλ2
1‖u0‖22

J(u0)
)
,

this gives

‖u‖22 ≥
2(q + 1)

S
J(u0) +

(
‖u0‖22 −

2(q + 1)

S
J(u0)

)
eSt,

where S =
2(q−1)aλ1+(q−3)bλ2

1‖u0‖22
2 . �



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 27

Proof of Theorem 2.11. For all P > d, let BR1
and BR2

be two arbitrary disjoint
open subsets of BR. Furthermore, let ψ ∈ H1

0 (BR1) \ {0} be an arbitrary function
such that I(ψ) > 0. Next, we show that there must exist some sufficiently large
ς > 0 such that

(q − 1)aλ1

2(q + 1)
‖ςψ‖22 +

(q − 3)bλ2
1

4(q + 1)
‖ςψ‖42 > P, J(ςψ) ≤ 0. (4.62)

Indeed, when ς > ( 4P (q+1)
(q−3)bλ2

1‖ψ‖42
)1/4, we have

(q − 1)aλ1

2(q + 1)
‖ςψ‖22 +

(q − 3)bλ2
1

4(q + 1)
‖ςψ‖42 ≥

(q − 3)bλ2
1

4(q + 1)
‖ςψ‖42 > P.

On the other hand, from the definition of J(u) we have

J(ςψ) =
aς2

2
‖ψ‖2 +

bς4

4
‖ψ‖4 − ξςq+1

q + 1
‖ψ‖q+1

q+1

= ς2
[a

2
‖ψ‖2 + ς2

( b
4
‖ψ‖4 − ξςq−3

q + 1
‖ψ‖q+1

q+1

)]
.

(4.63)

Then we infer from (4.63) that there must exist some sufficiently large ς such
that J(ςψ) ≤ 0. Therefore, (4.62) holds for some sufficiently large ς > 0. For
such a ς, we pick a function ϕ ∈ H1

0 (BR2) such that J(ϕ) = P − J(ςψ). Then for
uP = ςψ + ϕ, we obtain

J(uP ) = J(ςψ) + J(ϕ) = P,

(q − 1)aλ1

2(q + 1)
‖uP ‖22 +

(q − 3)bλ2
1

4(q + 1)
‖uP ‖42 ≥

(q − 1)aλ1

2(q + 1)
‖ςψ‖22 +

(q − 3)bλ2
1

4(q + 1)
‖ςψ‖42

> J(uP ).

Taking uP as the initial time, then by Theorem 2.10, we see that the weak solution
u blows up in finite time. �

Proof of Theorem 2.12. Let u = u(t), t ∈ [0, T ), be a weak solution of (1.1) and T
be the maximum existence time of u. Since J(u0) ≤ d, I(u0) < 0 or (2.18) holds,
then it follows from Theorem 2.7 and Theorem 2.10 that

lim
t→T
‖u‖2 = +∞. (4.64)

Furthermore, by Lemma 3.7 and the proof of Theorem 2.10, we see that I(u) < 0
for all t ∈ [0, T ). Then we infer from Lemma 3.6 that ‖u‖22 is strictly increasing for
t ∈ [0, T ).

In addition, from [40, Proposition 3.3] we obtain

‖u‖2 − ‖u0‖2 ≤ ‖u(t)− u0‖2 =
∥∥∫ t

0

uτ dτ
∥∥

2
≤
∫ t

0

‖uτ‖2 dτ,

which, together with ‖u‖22 begin strictly increasing for t ∈ [0, T ), implies(∫ t

0

‖uτ‖2 dτ
)2

≥ (‖u‖2 − ‖u0‖2)2.

It follows from Hölder’s inequality that∫ t

0

‖uτ‖22 dτ ≥
1

t
(

∫ t

0

‖uτ‖2 dτ)2 ≥ 1

t
(‖u‖2 − ‖u0‖2)2.



28 H. DING, J. ZHOU EJDE-2022/38

According to (2.12), we know that

J(u) ≤ J(u0)−
∫ t

0

‖uτ‖22 dτ ≤ J(u0)− 1

t
(‖u‖2 − ‖u0‖2)2.

Then by (4.64), we readily obtain limt→T J(u(t)) = −∞. �

Proof of Theorem 2.13. Let u = u(t), t ∈ [0, T ), be a weak solution of (1.1) and T
be the maximum existence time of u. We divide the proof into the following two
cases.

Case 1: J(u0) < d and u0 ∈ H1
0 (BR) \ {0}. Firstly, we prove I(u0) 6= 0. If

I(u0) = 0, then it follows from u0 6= 0 and the definition of d that J(u0) ≥ d, a
contradiction.

(1) If I(u0) < 0, then we infer from J(u0) < d and Theorem 2.7 that u blows up
in finite time, so T < +∞. Next, we show that if T < +∞, then I(u0) < 0. In fact,
if not, then we have I(u0) > 0, which, together with J(u0) < d and Theorem 2.4,
implies T = +∞, a contradiction. In addition, if I(u0) < 0, then it follows from
J(u0) < d and Theorem 2.12 that there is a t0 ∈ [0, T ) such that J(u(t0)) < 0.
Next, we prove that

there is a t0 ∈ [0, T ) such that J(u(t0)) < 0 implies I(u0) < 0.

In fact, it follows from J(u(t0)) < 0 and (2.3) that I(u(t0)) < 0. Hence, one can
choose t0 as the initial time, and it follows from Theorem 2.7 that u blows up in
finite time. According to Theorem 2.4, we see that I(u0) > 0 is impossible, so we
obtain I(u0) < 0.

(2) If I(u0) > 0, then we obtain T = +∞ from J(u0) < d and Theorem 2.4.
Next, we show that if T = +∞, then I(u0) > 0. In fact, if I(u0) < 0, then it follows
from (1) that T < +∞, a contradiction. In addition, if I(u0) > 0, then we obtain
from J(u0) < d and Lemma 3.7 that I(u) > 0 for t ∈ [0,+∞). Thus, it follows
from (2.3) that J(u) > 0 for all t ∈ [0,+∞). Next, we prove that

J(u(t)) > 0 for all t ∈ [0, T )⇒ I(u0) > 0.

In fact, if I(u0) < 0, then it follows from J(u0) < d and Theorem 2.12 that
limt→T J(u) = −∞. Hence, we infer that there is a t0 such that J(u(t0)) < 0, a
contradiction.

Case 2: J(u0) = d and u0 ∈ H1
0 (BR) \ {N ∪ {0}}. Because u0 ∈ H1

0 (BR) \
{N ∪ {0}}, we know that I(u0) 6= 0.

(3) If I(u0) < 0, then we obtain from J(u0) = d and Theorem 2.7 that u blows
up in finite time, so T < +∞. Next, we show that if T < +∞, then I(u0) < 0. In
fact, if not, then we have I(u0) > 0, which, together with J(u0) = d and Corollary
2.6, implies T = +∞, a contradiction. In addition, if I(u0) < 0, then it follows
from J(u0) = d and Theorem 2.12 that there is a t0 ∈ [0, T ) such that J(u(t0)) < 0.
Next, we prove that

there is a t0 ∈ [0, T ) such that J(u(t0)) < 0 ⇒ I(u0) < 0.

In fact, it follows from J(u(t0)) < 0 and (2.3) that I(u(t0)) < 0. Hence, we can
choose t0 as the initial time, and it follows from Theorem 2.7 that u blows up in
finite time. From Corollary 2.6, we see that I(u0) > 0 is impossible, so we obtain
I(u0) < 0.

(4) If I(u0) > 0, then we obtain T = +∞ from J(u0) = d and Corollary 2.6.
Next, we show that if T = +∞, then I(u0) > 0. In fact, if I(u0) < 0, then it follows



EJDE-2022/38 GLOBAL SOLUTIONS AND BLOW-UP FOR A KIRCHHOFF PROBLEM 29

from (3) that T < +∞, a contradiction. In addition, if I(u0) > 0, then it is clear
from J(u0) = d and Lemma 3.7 that I(u) > 0 for all t ∈ [0,+∞). Therefore, it
follows from (2.3) that J(u) > 0 for all t ∈ [0,+∞). Next, we prove that

J(u(t)) > 0 for all t ∈ [0, T )⇒ I(u0) > 0.

In fact, if I(u0) < 0, then by J(u0) = d and Theorem 2.12, one has limt→T J(u(t)) =
−∞. Consequently, we infer that there is a t0 such that J(u(t0)) < 0, a contradic-
tion. �

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Hang Ding

School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
Email address: hding0527@163.com

Jun Zhou (corresponding author)

School of Mathematics and Statistics, Southwest University, Chongqing 400715, China
Email address: jzhouwm@163.com


	1. Introduction
	2. Main results
	3. Preliminaries
	4. Proofs of main results
	References