1. Introduction
We are concerned with the global existence and blowup phenomenon for a quasilinear parabolic equation with nonlinear inner absorption term
(1)
u
t
=
[
(

∇
u

p
+
1
)
u
,
i
]
,
i

f
(
u
)
,
(
x
,
t
)
∈
Ω
×
(
0
,
t
*
)
,
subjected to the nonlinear Neumann boundary and initial conditions
(2)
(

∇
u

p
+
1
)
∂
u
∂
ν
=
g
(
u
)
,
(
x
,
t
)
∈
∂
Ω
×
(
0
,
t
*
)
,
(3)
u
(
x
,
0
)
=
u
0
(
0
)
≥
0
,
x
∈
Ω
¯
,
where
Ω
is a bounded starshaped region of
R
N
(
N
≥
2
)
with smooth boundary
∂
Ω
,
ν
is the unit outward normal vector on
∂
Ω
,
p
≥
0
,
t
*
is the blowup time if blowup occurs, or else
t
*
=
+
∞
, the symbol,
i
denotes partial differentiation with respect to
x
i
,
i
=
1,2
,
…
,
N
, the repeated index indicates summation over the index, and
∇
is gradient operator.
Many physical phenomena and biological species theories, such as the concentration of diffusion of some nonNewton fluid through porous medium, the density of some biological species, and heat conduction phenomena, have been formulated as parabolic equation (1) (see [1–3]). The nonlinear Neumann boundary condition (2) can be physically interpreted as the nonlinear radial law (see [4, 5]).
In the past decades, there have been many works dealing with existence and nonexistence of global solutions, blowup of solutions, bounds of blowup time, blowup rates, blowup sets, and asymptotic behavior of solutions to nonlinear parabolic equations; see the books [6–8] and the survey papers [9–11]. Specially, we would like to know whether the solution blows up and at which time when blowup occurs. A variety of methods have been used to study the problem above (see [12]), and in many cases, these methods, used to show that solutions blow up, often provide an upper bound for the blowup time. However, lower bounds for blowup time may be harder to be determined. For the study of the initial boundary value problem of a parabolic equation with homogeneous Dirichlet boundary condition, see [13, 14]. Payne et al. [13] considered the following quasilinear parabolic equation:
(4)
u
t
=
div
(
ρ

∇
u

2
∇
u
)
+
f
(
u
)
,
(
x
,
t
)
∈
Ω
×
(
0
,
t
*
)
,
where
Ω
is a bounded domain in
R
3
with smooth boundary
∂
Ω
. To get the lower bound for the blowup time, the authors assumed that
ρ
is a positive
C
1
function which satisfies
(5)
ρ
(
s
)
+
s
ρ
′
(
s
)
>
0
,
s
>
0
.
The lower bound for the blowup time of solution to (4) with Robin boundary condition was obtained in [15], where
ρ
is also assumed to satisfy the condition (5). However, under this boundary condition, the best constant of Sobolev inequality used in [13] is no longer applicable. They imposed suitable conditions on
f
and
ρ
and determined a lower bound for the blowup time if blowup occurs and determined when blowup cannot occur. Marras and Vernier Piro [14] studied the nonlinear parabolic problem with time dependent coefficients
(6)
k
1
(
t
)
div
(
g
(

∇
u

2
∇
u
)
)
+
k
2
(
t
)
f
(
u
)
=
k
3
(
t
)
u
t
,
gdfgfdgdfgfdgfdgfdgfdgfdgf
(
x
,
t
)
∈
Ω
×
(
0
,
t
*
)
,
where
Ω
is a bounded domain in
R
N
with smooth boundary
∂
Ω
. Under some conditions on the data and geometry of the spatial domain, they obtained upper and lower bounds of the blowup time. Moreover, the sufficient conditions for global existence of the solution were derived.
For the study of the initial boundary value problem of a parabolic equation with Robin boundary condition, we refer to [15–19]. Li et al. [16] investigated the problem of the nonlinear parabolic equation
(7)
u
t
=
[
(

∇
u

p
+
1
)
u
,
i
]
,
i
+
f
(
u
)
,
(
x
,
t
)
∈
Ω
×
(
0
,
t
*
)
,
where
Ω
is a bounded domain in
R
3
with smooth boundary
∂
Ω
. They derived the lower bound for the blowup time when the blowup occurs. Clearly,

∇
u

p
+
1
does not satisfy the condition (5). Enache [17] discussed the quasilinear parabolic problem
(8)
u
t
=
(
g
(
u
)
u
,
i
)
,
i
+
f
(
u
)
,
where
Ω
is a bounded domain in
R
N
(
N
≥
2
)
with smooth boundary
∂
Ω
. By virtue of a firstorder differential inequality technique, they showed the sufficient conditions to guarantee that the solution
u
(
x
,
t
)
exists globally or blows up. In addition, a lower bound for the blowup time when blowup occurs was also obtained. Ding [18] studied the nonlinear parabolic problem
(9)
(
b
(
u
)
)
t
=
∇
·
(
g
(
u
)
∇
u
)
+
f
(
u
)
,
(
x
,
t
)
∈
Ω
×
(
0
,
t
*
)
,
where
Ω
is a bounded domain in
R
3
with smooth boundary
∂
Ω
. They derived conditions on the data which guarantee the blowup or the global existence of the solution. A lower bound on blowup time when blowup occurs was also obtained. For the problem of the nonlinear nonlocal porous medium equation, we read the paper of Liu [19].
Recently, for the problems with nonlinear Neumann boundary conditions, Payne et al. [20] studied the semilinear heat equation with inner absorption term
(10)
u
t
=
Δ
u

f
(
u
)
,
(
x
,
t
)
∈
Ω
×
(
0
,
t
*
)
.
They established conditions on nonlinearity to guarantee that the solution
u
(
x
,
t
)
exists for all time
t
>
0
or blows up at some finite time
t
*
. Moreover, an upper bound for
t
*
was derived. Under somewhat more restrictive conditions, a lower bound for
t
*
was derived. Thereafter, they considered the quasilinear parabolic equation
(11)
u
t
=
∇
·
(

∇
u

p
∇
u
)
,
(
x
,
t
)
∈
Ω
×
(
0
,
t
*
)
,
and they showed that blowup occurs at some finite time under certain conditions on the nonlinearities and the data; upper and lower bounds for the blowup time were derived when blowup occurs; see [21]. Liu et al. The authors [22, 23] studied the reaction diffusion problem with nonlocal source and inner absorption terms or with local source and gradient absorption terms. Very recently, Fang et al. [24] considered lower bounds estimate for the blowup time to nonlocal problem with homogeneous Dirichlet or Neumann boundary condition.
Motivated by the above work, we intend to study the global existence and the blowup phenomena of problem (1)–(3), and the results of the semilinear equations are extended to the quasilinear equations. Unfortunately, the techniques used for semilinear equation to analysis of blowup phenomena are no longer applicable to our problem. As a consequence, by using the suitable techniques of differential inequalities, we establish, respectively, the conditions on the nonlinearities
f
and
g
to guarantee that
u
(
x
,
t
)
exists globally or blows up at some finite time. If blowup occurs, we derive upper and lower bounds of the blowup time.
The rest of our paper is organized as follows. In Section 2, we establish conditions on the nonlinearities to guarantee that
u
(
x
,
t
)
exists globally. In Section 3, we show the conditions on data forcing the solution
u
(
x
,
t
)
to blow up at some finite time
t
*
and obtain an upper bound for
t
*
. A lower bound of blowup time under some assumptions is derived in Section 4.
2. The Global Existence
In this section, we establish the conditions on the nonlinearity
f
and nonlinearity
g
to guarantee that
u
(
x
,
t
)
exists globally. We state our result as follows.
Theorem 1.
Assume that the nonnegative functions
f
and
g
satisfy
(12)
f
(
ξ
)
≥
k
1
ξ
q
,
ξ
≥
0
,
g
(
ξ
)
≤
k
2
ξ
s
,
ξ
≥
0
,
where
k
1
>
0
,
k
2
≥
0
,
s
>
1
,
2
s
<
q
+
1
, and
s

1
<
p
<
q

1
. Then the (nonnegative) solution
u
(
x
,
t
)
of problem (1)(3) does not blow up; that is,
u
(
x
,
t
)
exists for all time
t
>
0
.
Proof.
Set
(13)
Ψ
(
t
)
=
∫
Ω
u
2
d
x
.
Similar to Theorem 2.1 in [20], we get
(14)
Ψ
′
(
t
)
≤
{
2
δ
2
∫
Ω
u
2
s
d
x

k
1
∫
Ω
u
q
+
1
d
x
}
+
{
2
k
2
N
ρ
0
∫
Ω
u
s
+
1
d
x
f
d
f
d
d

2
∫
Ω

∇
u

p
+
2
d
x

k
1
∫
Ω
u
q
+
1
d
x
2
k
2
N
ρ
0
}
=
I
1
+
I
2
,
where
δ
=
k
2
(
s
+
1
)
d
/
2
ρ
0
,
ρ
0
=
min
x
∈
∂
Ω
(
x
·
ν
)
,
d
=
max
x
∈
∂
Ω

x

, and
(15)
I
1
≤
(
∫
Ω
u
q
+
1
d
x
)
(
s
+
1
)
/
(
q
+
1
)
×
{
A
1
~

Ω

(
q

s
)
/
(
q
+
1
)

A
2
~
(
∫
Ω
u
q
+
1
d
x
)
(
q

s
)
/
(
q
+
1
)
}
,
where
A
1
~
=
2
δ
2
α
ε
(
α

1
)
/
α
,
A
2
~
=
k
1

2
δ
2
(
1

α
)
ɛ
,
α
=
(
q
+
1

2
s
)
/
(
q

s
)
<
1
,
ɛ
>
0
.
Next, we estimate
I
2
=
(
2
k
2
N
/
ρ
0
)
∫
Ω
u
s
+
1
d
x

2
∫
Ω

∇
u

p
+
2
d
x

k
1
∫
Ω
u
q
+
1
d
x
. Since
(16)

∇
u
(
p
/
2
)
+
1

2
=
(
p
2
+
1
)
2
u
p

∇
u

2
,
it follows from Hölder inequality that
(17)
∫
Ω

∇
u
(
p
/
2
)
+
1

2
d
x
≤
(
p
2
+
1
)
2
(
∫
Ω

∇
u

p
+
2
d
x
)
2
/
(
p
+
2
)
×
(
∫
Ω
u
p
+
2
d
x
)
p
/
(
p
+
2
)
.
Furthermore, we have
(18)
∫
Ω
u
p
+
2
d
x
≤
[
(
p
+
2
)
2
4
λ
1
]
(
p
/
2
)
+
1
∫
Ω

∇
u

p
+
2
d
x
,
which follows from (17) and membrane inequality
(19)
λ
1
∫
Ω
ω
2
d
x
≤
∫
Ω

∇
ω

2
d
x
,
where
λ
1
is the first eigenvalue in the fixed membrane problem
(20)
Δ
ω
+
λ
ω
=
0
,
ω
>
0
in
Ω
,
ω
=
0
on
∂
Ω
.
Combining
I
2
and (18), we have
(21)
I
2
≤
2
k
2
N
ρ
0
∫
Ω
u
s
+
1
d
x

2
[
4
λ
1
(
p
+
2
)
2
]
(
p
/
2
)
+
1
×
∫
Ω
u
p
+
2
d
x

k
1
∫
Ω
u
q
+
1
d
x
.
=
{
2
k
2
N
ρ
0
∫
Ω
u
s
+
1
d
x

3
[
4
λ
1
(
p
+
2
)
2
]
(
p
/
2
)
+
1
∫
Ω
u
p
+
2
d
x
}
+
{
[
4
λ
1
(
p
+
2
)
2
]
(
p
/
2
)
+
1
∫
Ω
u
p
+
2
d
x

k
1
∫
Ω
u
q
+
1
d
x
}
=
I
21
+
I
22
.
Making use of Hölder inequality, we obtain
(22)
∫
Ω
u
s
+
1
d
x
≤
(
∫
Ω
u
p
+
2
d
x
)
(
s
+
1
)
/
(
p
+
2
)

Ω

(
p

s
+
1
)
/
(
p
+
2
)
,
(23)
Ψ
(
t
)
=
∫
Ω
u
2
d
x
≤
(
∫
Ω
u
s
+
1
d
x
)
2
/
(
s
+
1
)

Ω

(
s

1
)
/
(
s
+
1
)
.
Combining (21), (22) with (23), we get
(24)
I
21
≤
(
∫
Ω
u
s
+
1
d
x
)
{
B
1

B
2
Ψ
(
p

s
+
1
)
/
2
}
,
with
(25)
B
1
=
2
k
2
N
ρ
0
,
B
2
=
3
[
4
λ
1
(
p
+
2
)
2
]
(
p
/
2
)
+
1

Ω


(
p

s
+
1
)
/
2
.
Applying Hölder inequality, we obtain
(26)
∫
Ω
u
p
+
2
d
x
≤
(
∫
Ω
u
q
+
1
d
x
)
(
p
+
2
)
/
(
q
+
1
)

Ω

(
q

p

1
)
/
(
q
+
1
)
,
Ψ
(
t
)
=
∫
Ω
u
2
d
x
≤
(
∫
Ω
u
q
+
1
d
x
)
2
/
(
q
+
1
)

Ω

(
q

1
)
/
(
q
+
1
)
.
It follows from (26) that
(27)
I
22
≤
(
∫
Ω
u
q
+
1
d
x
)
(
p
+
2
)
/
(
q
+
1
)
{
C
1

C
2
Ψ
(
q

p

1
)
/
2
}
,
where
(28)
C
1
=
[
4
λ
1
(
p
+
2
)
2
]
(
p
/
2
)
+
1

Ω

(
q

p

1
)
/
(
q
+
1
)
,
C
2
=
k
1

Ω

(
1

q
)
(
q

p

1
)
/
2
(
q
+
1
)
.
Combining (14), (15), (21), and (24) with (27), we obtain
(29)
Ψ
′
(
t
)
≤
(
∫
Ω
u
q
+
1
d
x
)
(
s
+
1
)
/
(
q
+
1
)
{
A
1

A
2
Ψ
(
t
)
(
q

s
)
/
2
}
+
(
∫
Ω
u
s
+
1
d
x
)
{
B
1

B
2
Ψ
(
p

s
+
1
)
/
2
}
+
(
∫
Ω
u
q
+
1
d
x
)
(
p
+
2
)
/
(
q
+
1
)
{
C
1

C
2
Ψ
(
q

p

1
)
/
2
}
,
with
(30)
A
1
=
A
1
~

Ω

(
q

s
)
/
(
q
+
1
)
,
A
2
=
A
2
~

Ω

(
1

q
)
(
q

s
)
/
2
(
q
+
1
)
.
We conclude from (29) that
Ψ
(
t
)
is decreasing in each time interval on which we obtain
(31)
Ψ
(
t
)
≥
max
{
(
A
1
A
2
)
2
/
(
q

s
)
,
(
B
1
B
2
)
2
/
(
p

s
+
1
)
,
(
C
1
C
2
)
2
/
(
q

p

1
)
}
,
so that
Ψ
(
t
)
remains bounded for all time under the conditions in Theorem 1. This completes the proof of Theorem 1.
4. Lower Bounds for <inlineformula>
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In this section, under the assumption that
Ω
is a star shaped domain in
R
3
, convex in two orthogonal directions, we seek a lower bound for the blowup time
t
*
. Now we state the result as follows.
Theorem 3.
Let
u
(
x
,
t
)
be the nonnegative solution of problem (1)(3) and
u
(
x
,
t
)
blows up at
t
*
; moreover, the nonnegative functions
f
and
g
satisfy
(46)
f
(
ξ
)
≥
k
1
ξ
q
,
ξ
≥
0
,
g
(
ξ
)
≤
k
2
ξ
s
,
ξ
≥
0
,
with
k
1
>
0
,
k
2
>
0
,
q
>
1
,
s
>
1
,
q
<
s
. Define
(47)
φ
(
t
)
=
∫
Ω
u
n
(
s

1
)
d
x
,
where
n
is a parameter restricted by the condition
(48)
n
>
max
{
4
,
2
s

1
}
.
Then
φ
(
t
)
satisfies inequality
(49)
φ
′
(
t
)
≤
Γ
(
φ
)
,
for some computable function
Γ
(
φ
)
. It follows that
t
*
is bounded from below. We have
(50)
t
*
≥
∫
φ
(
0
)
∞
d
η
Γ
(
η
)
d
η
.
Proof.
Differentiating (47) and making use of the boundary condition (2) together with the conditions (46), we have
(51)
φ
′
(
t
)
=
n
(
s

1
)
∫
Ω
u
n
(
s

1
)

1
u
t
d
x
=
n
(
s

1
)
∫
Ω
u
n
(
s

1
)

1
f
d
s
f
d
f
d
f
d
d
d
×
[
(
(

∇
u

p
+
1
)
u
,
i
)
,
i

f
(
u
)
]
d
x
=
n
(
s

1
)
∫
∂
Ω
u
n
(
s

1
)

1
(

∇
u

p
+
1
)
∂
u
∂
ν
d
S

n
(
s

1
)
[
n
(
s

1
)

1
]
×
∫
Ω
u
n
(
s

1
)

2

∇
u

p
+
2
d
x

n
(
s

1
)
[
n
(
s

1
)

1
]
×
∫
Ω
u
n
(
s

1
)

2

∇
u

2
d
x

n
(
s

1
)
∫
Ω
u
n
(
s

1
)

1
f
(
u
)
d
x
≤
k
2
n
(
s

1
)
∫
∂
Ω
u
(
n
+
1
)
(
s

1
)
d
S

n
(
s

1
)
[
n
(
s

1
)

1
]
×
∫
Ω
u
n
(
s

1
)

2

∇
u

p
+
2
d
x

n
(
s

1
)
[
n
(
s

1
)

1
]
×
∫
Ω
u
n
(
s

1
)

2

∇
u

2
d
x

k
1
n
(
s

1
)
∫
Ω
u
n
(
s

1
)
+
q

1
d
x
.
Applying inequality
(
2.7
)
in [20] to the first term on the right hand side of (51), we have
(52)
∫
∂
Ω
u
(
n
+
1
)
(
s

1
)
d
S
≤
3
ρ
0
∫
Ω
u
(
n
+
1
)
(
s

1
)
d
x
+
(
n
+
1
)
(
s

1
)
d
ρ
0
∫
Ω
u
(
n
+
1
)
(
s

1
)

1

∇
u

d
x
.
Substituting (52) into (51), we obtain
(53)
φ
′
(
t
)
≤
3
k
2
n
(
s

1
)
ρ
0
∫
Ω
u
(
n
+
1
)
(
s

1
)
d
x
+
k
2
n
(
n
+
1
)
(
s

1
)
2
d
ρ
0
∫
Ω
u
(
n
+
1
)
(
s

1
)

1

∇
u

d
x

n
(
s

1
)
[
n
(
s

1
)

1
]
∫
Ω
u
n
(
s

1
)

2

∇
u

p
+
2
d
x

n
(
s

1
)
[
n
(
s

1
)

1
]
∫
Ω
u
n
(
s

1
)

2

∇
u

2
d
x

k
1
n
(
s

1
)
∫
Ω
u
n
(
s

1
)
+
q

1
d
x
.
Making use of arithmeticgeometric mean inequality, we derive
(54)
∫
Ω
u
(
n
+
1
)
(
s

1
)

1

∇
u

d
x
≤
μ
2
∫
Ω
u
n
(
s

1
)

2

∇
u

2
d
x
+
1
2
μ
∫
Ω
u
(
n
+
2
)
(
s

1
)
d
x
,
for all
μ
>
0
. Choose
μ
>
0
such that
(55)
k
2
n
(
n
+
1
)
(
s

1
)
2
d
μ
2
ρ
0

n
(
s

1
)
[
n
(
s

1
)

1
]
=
0
.
We rewrite (53) as
(56)
φ
′
(
t
)
≤
3
k
2
n
(
s

1
)
ρ
0
∫
Ω
u
(
n
+
1
)
(
s

1
)
d
x
+
k
2
n
(
n
+
1
)
(
s

1
)
2
d
2
μ
ρ
0
∫
Ω
u
(
n
+
2
)
(
s

1
)
d
x

n
(
s

1
)
[
n
(
s

1
)

1
]
∫
Ω
u
n
(
s

1
)

2

∇
u

p
+
2
d
x

k
1
n
(
s

1
)
∫
Ω
u
n
(
s

1
)
+
q

1
d
x
.
Using Hölder inequality, we get
(57)
∫
Ω
u
n
(
s

1
)
d
x
≤
(
∫
Ω
u
n
(
s

1
)
+
q

1
d
x
)
n
(
s

1
)
/
(
n
(
s

1
)
+
q

1
)
×

Ω

(
q

1
)
/
(
n
(
s

1
)
+
q

1
)
.
Combining (56) with (57), we obtain
(58)
φ
′
(
t
)
≤
3
k
2
n
(
s

1
)
ρ
0
∫
Ω
u
(
n
+
1
)
(
s

1
)
d
x
+
k
2
n
(
n
+
1
)
(
s

1
)
2
d
2
μ
ρ
0
∫
Ω
u
(
n
+
2
)
(
s

1
)
d
x

n
(
s

1
)
[
n
(
s

1
)

1
]
×
∫
Ω
u
n
(
s

1
)

2

∇
u

p
+
2
d
x

k
1
n
(
s

1
)

Ω

(
1

q
)
/
n
(
s

1
)
φ
(
n
(
s

1
)
+
q

1
)
/
n
(
s

1
)
=
3
k
2
n
(
s

1
)
ρ
0
J
1
(
t
)
+
k
2
n
(
n
+
1
)
(
s

1
)
2
d
2
μ
ρ
0
J
2
(
t
)

n
(
s

1
)
[
n
(
s

1
)

1
]
ω
(
t
)

k
1
n
(
s

1
)

Ω

(
1

q
)
/
n
(
s

1
)
φ
(
n
(
s

1
)
+
q

1
)
/
n
(
s

1
)
,
where
(59)
J
1
(
t
)
=
∫
Ω
u
(
n
+
1
)
(
s

1
)
d
x
,
J
2
(
t
)
=
∫
Ω
u
(
n
+
2
)
(
s

1
)
d
x
,
ω
(
t
)
=
∫
Ω
u
n
(
s

1
)

2

∇
u

p
+
2
d
x
.
Using Sobolev type inequality
(
A
.
5
)
derived by Payne et al. [21], we obtain
(60)
J
1
(
t
)
=
∫
Ω
u
(
n
+
1
)
(
s

1
)
d
x
≤
{
3
ρ
0
∫
Ω
u
(
2
/
3
)
(
n
+
1
)
(
s

1
)
d
x
+
(
n
+
1
)
(
s

1
)
3
f
d
f
×
(
1
+
d
ρ
0
)
∫
Ω
u
(
2
/
3
)
(
n
+
1
)
(
s

1
)

1

∇
u

d
x
}
3
/
2
.
We now make use of Hölder inequality to bound the second integral on the right hand side of (60) as follows:
(61)
∫
Ω
u
(
2
/
3
)
(
n
+
1
)
(
s

1
)

1

∇
u

d
x
≤
(
∫
Ω
u
(
2
/
3
)
(
n
+
1
)
(
s

1
)
(
1

δ
1
)
d
x
)
(
p
+
1
)
/
(
p
+
2
)
ω
1
/
(
p
+
2
)
,
with
(62)
δ
1
=
(
n

2
)
(
s

1
)
+
3
p
2
(
n
+
1
)
(
s

1
)
(
p
+
1
)
.
We note that
δ
1
<
1
for
n
>
(
3
p

2
(
s

1
)
(
p
+
2
)
)
/
(
s

1
)
(
2
p
+
1
)
, an inequality satisfied in view of (48). Using again Hölder's inequality, we obtain
(63)
∫
Ω
u
(
2
/
3
)
(
n
+
1
)
(
s

1
)
(
1

δ
1
)
d
x
≤
φ
2
(
n
+
1
)
(
1

δ
1
)
/
3
n

Ω

1

(
2
(
n
+
1
)
(
1

δ
1
)
/
3
n
)
,
∫
Ω
u
(
2
/
3
)
(
n
+
1
)
(
s

1
)
d
x
≤
φ
2
(
n
+
1
)
/
3
n

Ω

1

(
2
(
n
+
1
)
/
3
n
)
,
where

Ω

=
∫
Ω
d
x
is the volume of
Ω
. Substituting (61) and (63) in (60), we obtain the following inequality:
(64)
J
1
(
t
)
≤
{
c
1
~
φ
2
(
n
+
1
)
/
3
n
+
c
2
~
φ
(
2
(
n
+
1
)
(
1

δ
1
)
/
3
n
)
(
(
p
+
1
)
/
(
p
+
2
)
)
f
d
s
×
ω
1
/
(
p
+
2
)
}
3
/
2
≤
c
1
φ
(
n
+
1
)
/
n
+
c
2
φ
(
(
n
+
1
)
(
1

δ
1
)
/
n
)
(
(
p
+
1
)
/
(
p
+
2
)
)
ω
3
/
2
(
p
+
2
)
,
where
c
1
,
c
2
are computable positive constants. Note that the last inequality in (64) follows from Hölder inequality under the particular form
(
a
+
b
)
3
/
2
≤
2
(
a
3
/
2
+
b
3
/
2
)
. Similarly, we can bound
J
2
and get
(65)
J
2
(
t
)
≤
c
3
φ
(
n
+
2
)
/
n
+
c
4
φ
(
(
n
+
2
)
(
1

δ
2
)
/
n
)
(
(
p
+
1
)
/
(
p
+
2
)
)
ω
3
/
2
(
p
+
2
)
,
where
c
3
,
c
4
are computable positive constants,
(66)
δ
2
=
(
n

4
)
(
s

1
)
+
3
p
2
(
n
+
1
)
(
s

1
)
(
p
+
1
)
.
We note that
δ
2
<
1
for
n
>
(
3
p

4
(
s

1
)
(
p
+
2
)
)
/
(
s

1
)
(
2
p
+
1
)
, an inequality satisfied in view of (48). Inserting (64) and (65) in (58), we arrive at
(67)
φ
′
(
t
)
≤
d
1
~
φ
(
n
+
1
)
/
n
+
d
2
~
φ
(
(
n
+
1
)
(
1

δ
1
)
/
n
)
λ
ω
3
/
2
(
p
+
2
)
+
d
3
φ
(
n
+
2
)
/
n
+
d
4
~
φ
(
(
n
+
2
)
(
1

δ
2
)
/
n
)
λ
ω
3
/
2
(
p
+
2
)

n
(
s

1
)
[
n
(
s

1
)

1
]
ω
(
t
)

k
1
n
(
s

1
)

Ω

(
1

q
)
/
n
(
s

1
)
φ
(
n
(
s

1
)
+
q

1
)
/
n
(
s

1
)
,
where
λ
=
(
p
+
1
)
/
(
p
+
2
)
,
d
3
and
d
j
~
(
j
=
1,2
,
4
)
are computable positive constants. Next, we want to eliminate the quantity
ω
(
t
)
in inequality (67). By using the following inequality:
(68)
φ
α
ω
β
=
(
γ
ω
)
β
{
φ
α
/
(
1

β
)
γ
β
/
(
1

β
)
}
1

β
≤
γ
β
ω
+
(
1

β
)
γ
β
/
(
β

1
)
+
(
1

β
)
γ
β
/
(
β

1
)
φ
α
/
(
1

β
)
,
valid for
0
<
β
<
1
, where
γ
is an arbitrary positive constant, then we have
(69)
d
2
~
φ
(
(
n
+
1
)
(
1

δ
1
)
/
n
)
λ
ω
3
/
2
(
p
+
2
)
≤
γ
1
ω
(
t
)
+
d
2
φ
(
2
(
n
+
1
)
(
1

δ
1
)
(
p
+
2
)
/
n
(
2
p
+
1
)
)
λ
,
d
4
~
φ
(
(
n
+
2
)
(
1

δ
2
)
/
n
)
λ
ω
3
/
2
(
p
+
2
)
≤
γ
2
ω
(
t
)
+
d
4
φ
(
2
(
n
+
2
)
(
1

δ
2
)
(
p
+
2
)
/
n
(
2
p
+
1
)
)
λ
,
with arbitrary positive constants
γ
1
,
γ
2
and computable positive constants
d
2
,
d
4
. Substitute (69) in (67) and choose the arbitrary (positive) constants
γ
1
,
γ
2
such that
γ
1
+
γ
2

n
(
s

1
)
[
n
(
s

1
)

1
]
=
0
. We obtain
(70)
φ
′
(
t
)
≤
d
1
~
φ
(
n
+
1
)
/
n
+
d
2
φ
(
2
(
n
+
1
)
(
1

δ
1
)
(
p
+
2
)
/
n
(
2
p
+
1
)
)
λ
+
d
3
φ
(
n
+
2
)
/
n
+
d
4
φ
(
2
(
n
+
2
)
(
1

δ
2
)
(
p
+
2
)
/
n
(
2
p
+
1
)
)
λ

k
1
n
(
s

1
)

Ω

(
1

q
)
/
n
(
s

1
)
φ
(
n
(
s

1
)
+
q

1
)
/
n
(
s

1
)
.
We eliminate the last term in (70), by using the following inequality:
(71)
φ
(
n
+
1
)
/
n
=
{
m
φ
(
n
(
s

1
)
+
q

1
)
/
n
(
s

1
)
}
(
2
n

1
)
(
s

1
)
/
(
(
2
n

1
)
(
s

1
)
+
s

q
)
×
{
m
(
2
n

1
)
(
1

s
)
/
(
s

q
)
φ
3
}
(
s

q
)
/
(
(
2
n

1
)
(
s

1
)
+
s

q
)
≤
(
2
n

1
)
(
s

1
)
(
2
n

1
)
(
s

1
)
+
s

q
m
φ
(
n
(
s

1
)
+
q

1
)
/
n
(
s

1
)
+
s

q
(
2
n

1
)
(
s

1
)
+
s

q
m
(
2
n

1
)
(
1

s
)
/
(
s

q
)
φ
3
,
valid for
q
<
s
and arbitrary
m
>
0
, and choose
m
such that
(72)
(
2
n

1
)
(
s

1
)
(
2
n

1
)
(
s

1
)
+
s

q
d
1
~
m

k
1
n
(
s

1
)

Ω

(
1

q
)
/
n
(
s

1
)
=
0
.
Then (70) can be rewritten as
(73)
φ
′
(
t
)
≤
d
1
φ
3
+
d
2
φ
(
2
(
n
+
1
)
(
1

δ
1
)
(
p
+
2
)
/
n
(
2
p
+
1
)
)
λ
+
d
3
φ
(
n
+
2
)
/
n
+
d
4
φ
(
2
(
n
+
2
)
(
1

δ
2
)
(
p
+
2
)
/
n
(
2
p
+
1
)
)
λ
.
Integrating (73) over
[
0
,
t
]
, we conclude
(74)
t
*
≥
∫
φ
(
0
)
∞
d
η
f
d
f
d
d
d
×
(
d
1
η
3
+
d
2
η
(
2
(
n
+
1
)
(
1

δ
1
)
(
p
+
2
)
/
n
(
2
p
+
1
)
)
λ
d
s
d
s
d
s
d
s
s
s
+
d
3
η
(
n
+
2
)
/
n
+
d
4
η
(
2
(
n
+
2
)
(
1

δ
2
)
(
p
+
2
)
/
n
(
2
p
+
1
)
)
λ
)

1
.
This completes the proof of Theorem 3.