# A Nonhomogeneous Dirichlet Problem for a Nonlinear Pseudoparabolic Equation Arising in the Flow of Second-Grade Fluid

Authors: Le Thi Phuong Ngoc; Truong Thi Nhan; Nguyen Thanh Long
Journal: Discrete Dynamics in Nature and Society
 (2016)

---

## Abstract

We study the following initial-boundary value problem{
u
t − μ
t
+
α
t
(
∂
/
∂
t
)
∂
2
u
/
∂
x
2
+
(
γ
/
x
)
(
∂
u
/
∂
x
) + f
u = f
1
x
,
t, 1
<
x
<
R, t
>
0; u
(
1
,
t
)
=
g
1
(
t
), u
(
R
,
t
)
=
g
R
(
t
); u
(
x
,
0
)
=
u
~
0
(
x
)
}, where γ
>
0
,
R
>
1 are given constants and f
,
f
1
,
g
1
,
g
R
,
u
~
0
,
α
, and μ are given functions. In Part 1, we use the Galerkin method and compactness method to prove the existence of a unique weak solution of the problem above on (
0
,
T
)
, for every T
>
0
. In Part 2, we investigate asymptotic behavior of the solution as t
→
+
∞
. In Part 3, we prove the existence and uniqueness of a weak solution of problem {
u
t − μ
t
+
α
t
(
∂
/
∂
t
)
∂
2
u
/
∂
x
2
+
(
γ
/
x
)
(
∂
u
/
∂
x
) + f
u = f
1
x
,
t, 1
<
x
<
R, t
>
0; u
(
1
,
t
)
=
g
1
(
t
), u
(
R
,
t
)
=
g
R
(
t
)
} associated with a “(
η
,
T
)-periodic condition” u
(
x
,
0
)
=
η
u
(
x
,
T
), where 0
<
η
≤
1 is given constant.


---

## Body

## 1. Introduction

In this paper, we consider the following nonlinear pseudoparabolic equation:(1)
u
t
-
μ
t
+
α
t
∂
∂
t
∂
2
u
∂
x
2
+
γ
x
∂
u
∂
x
+
f
u
=
f
1
x
,
t
,
1
<
x
<
R
,
t
>
0
,associated with the boundary conditions(2)
u
1
,
t
=
g
1
t
,
u
R
,
t
=
g
R
tand the initial condition(3)
u
x
,
0
=
u
~
0
x
,or the “(
η
,
T
)-periodic condition”(4)
u
x
,
0
=
η
u
x
,
T
,where γ
>
0, R
>
1, and 0
<
η
≤
1 are given constants and f, f
1, g
1, g
R, u
~
0, α, and μ are given functions satisfying conditions specified later.In the case ofγ
=
1, μ
(
t
)
=
μ
>
0, and α
(
t
)
=
α
>
0 being the constants, the initial-boundary value problems (1)–(3) are classical and have a long history of applications and mathematical development. We refer to the monographs of Al’shin et al. [1] and of Carroll and Showalter [2] for references and results on pseudoparabolic or Sobolev type equations. We also refer to [3] for asymptotic behavior and to[4] for nonlinear problems. Problems of this type arise in material science and physics, which have been extensively studied, and several results concerning existence, regularity, and asymptotic behavior have been established.Equation (1) arises within frameworks of mathematical models in engineering and physical sciences (see [5–11] for references therein and interesting results on second grade fluids or a fourth grade fluid or other unsteady flows). It is well known that fluid solid mixtures are generally considered as second-grade fluids and are modeled as fluids with variable physical parameters; thus, an analysis is performed for a second-grade fluid with space dependent viscosity, elasticity, and density.In [9], some unsteady flow problems of a second-grade fluid were considered. The flows are generated by the sudden application of a constant pressure gradient or by the impulsive motion of a boundary. Here, the velocities of the flows are described by the partial differential equations and exact analytic solutions of these differential equations are obtained. Suppose that the second-grade fluid is in a circular cylinder and is initially at rest, and the fluid starts suddenly due to the motion of the cylinder parallel to its length. The axis of the cylinder is chosen as the z-axis. Using cylindrical polar coordinates, the governing partial differential equation is(5)
∂
w
∂
t
=
ν
+
α
∂
∂
t
∂
2
∂
r
2
+
1
r
∂
∂
r
w
r
,
t
-
N
w
,
0
<
r
<
a
,
t
>
0
,
w
a
,
t
=
W
,
t
>
0
,
w
r
,
0
=
0
,
0
≤
r
<
a
,where w
(
r
,
t
) is the velocity along the z-axis, ν is the kinematic viscosity, α is the material parameter, and N is the imposed magnetic field. In the boundary and initial conditions, W is the constant velocity at r
=
a and a is the radius of the cylinder.In [6], two types of time-dependent flows were investigated. An eigenfunction expansion method was used to find the velocity distribution. The obtained solutions satisfy the boundary and initial conditions and the governing equation. Remarkably, some exact analytic solutions are possible for flows involving second-grade fluid with variable material properties in terms of trigonometric and Chebyshev functions.In [5], Mahmood et al. have considered the longitudinal oscillatory motion of second-grade fluid between two infinite coaxial circular cylinders, oscillating along their common axis with given constant angular frequencies Ω
1 and Ω
2. Velocity field and associated tangential stress of the motion were determined by using Laplace and Hankel transforms. In order to find exact analytic solutions for the flow of second-grade fluid between two longitudinally oscillating cylinders, the following problem was studied:(6)
∂
v
∂
t
=
μ
+
α
∂
∂
t
∂
2
∂
r
2
+
1
r
∂
∂
r
v
r
,
t
,
R
1
<
r
<
R
2
,
t
>
0
,
v
R
1
,
t
=
V
1
sin
⁡
Ω
1
t
,
v
R
2
,
t
=
V
2
sin
⁡
Ω
2
t
,
v
r
,
0
=
0
,
R
1
≤
r
≤
R
2
,where 0
<
R
1
<
R
2, μ, α, V
1, V
2, Ω
1, and Ω
2 are positive constants. The solutions obtained have been presented under series form in terms of Bessel functions J
0
(
x
), Y
0
(
x
), J
1
(
x
), Y
1
(
x
), J
2
(
x
), and Y
2
(
x
), satisfying the governing equation and all imposed initial and boundary conditions.The nonlinear parabolic problems of the form (1)–(3), with/without the term u
r
r
+
(
γ
/
r
)
u
r, were also studied in [12, 13] and references therein. In [12], by using the Galerkin and compactness method in appropriate Sobolev spaces with weight, the authors proved the existence of a unique weak solution of the following initial and boundary value problem for nonlinear parabolic equation:(7)
u
t
-
a
t
u
r
r
+
γ
r
u
r
+
F
r
,
u
=
f
r
,
t
,
0
<
r
<
1
,
0
<
t
<
T
,
lim
r
→
0
+
⁡
r
γ
/
2
u
r
r
,
t
<
+
∞
,
u
r
1
,
t
+
h
t
u
1
,
t
-
u
¯
0
=
0
,
u
r
,
0
=
u
0
r
.Furthermore, asymptotic behavior of the solution ast
→
+
∞ was studied. In [13], the following nonlinear heat equation associated with Dirichlet-Robin conditions was investigated:(8)
u
t
-
∂
∂
x
μ
x
,
t
u
x
+
f
u
=
f
1
x
,
t
,
x
,
t
∈
0,1
×
0
,
T
,
u
x
0
,
t
=
h
0
u
0
,
t
+
g
0
t
,
-
u
x
1
,
t
=
h
1
u
1
,
t
+
g
1
t
,
u
x
,
0
=
u
0
x
.Condition (4), which we call “(
η
,
T
)-periodic condition,” is known as a drifted periodic condition (see [14]). Indeed, if u
(
t
)
=
η
u
(
t
+
T
), ∀
t
≥
0, in the case of 0
<
η
≤
1, then we have(9)
u
t
+
T
=
1
η
u
t
=
u
t
+
1
η
-
1
u
t
,
∀
t
≥
0
,which means(10)
u
t
+
T
=
u
t
+
δ
t
,
∀
t
≥
0
,with δ
(
t
)
=
(
1
/
η
-
1
)
u
(
t
) satisfying the condition(11)
δ
t
=
η
δ
t
+
T
,
∀
t
≥
0
.Note that (11) holds by the fact that(12)
η
δ
t
+
T
=
η
1
η
-
1
u
t
+
T
=
1
η
-
1
u
t
=
δ
t
,
∀
t
≥
0
.Withη
=
1, (4) leads to T-periodic condition(13)
u
x
,
0
=
u
x
,
T
,and with η
=
-
1, we have the antiperiodic condition(14)
u
x
,
0
=
-
u
x
,
T
.The present paper is concerned with the second-grade fluid in a circular cylinder associated with the initial condition (3) or a drifted periodic condition (10). The extensive study of such flows is motivated by both their fundamental interest and their practical importance (see [9]).This paper is a continuation of paper [15] dealing with the nonlinear pseudoparabolic equation (1) associated with the mixed inhomogeneous condition, in the case of γ
=
1, μ
(
t
)
=
μ
>
0, α
(
t
)
=
α
>
0 being the constants. It consists of five sections. First, preliminaries are done in Section 2. Under appropriate conditions, the existence of a unique weak solution of problems (1)–(3) is proved in Section 3. Next, an asymptotic behavior of the solution of problems (1)–(3), as t
→
+
∞, is discussed in Section 4. Finally, Section 5 is devoted to the establishment, the existence, and uniqueness of a weak solution of problems (1), (2), and (4).Because of mathematical context, the results obtained here generalize relatively the ones in [12, 13, 15], by improving the techniques used as before and with appropriate modifications.


## 2. Preliminaries

PutΩ
=
(
1
,
R
), Q
T
=
Ω
×
(
0
,
T
), T
>
0
. We omit the definitions of the usual function spaces: C
m
Ω
¯, L
p
Ω, and W
m
,
p
Ω. We define W
m
,
p
=
W
m
,
p
Ω, L
p
=
W
0
,
p
Ω and H
m
=
W
m
,
2
Ω, 1
≤
p
≤
∞, m
=
0,1
,
…. The norm in L
2 is denoted by ·
. We also denote by (
·
,
·
) the scalar product in L
2. We denote by ·
X the norm of a Banach space X and by X
′ the dual space of X
. We denote by L
p
(
0
,
T
;
X
)
1
≤
p
≤
∞ for the Banach space of the real functions u
:
(
0
,
T
)
→
X measurable, such that (15)
u
L
p
0
,
T
;
X
=
∫
0
T
u
t
X
p
d
t
1
/
p
<
∞
for
1
≤
p
<
∞
,
u
L
∞
0
,
T
;
X
=
ess
sup
0
<
t
<
T
⁡
u
t
X
for
p
=
∞
.Letu
(
t
)
,
u
′
(
t
)
=
u
t
(
t
), u
′
′
(
t
)
=
u
t
t
(
t
), u
x
(
t
), and u
x
x
(
t
) denote u
(
x
,
t
), (
∂
u
/
∂
t
)
(
x
,
t
), (
∂
2
u
/
∂
t
2
)
(
x
,
t
), (
∂
u
/
∂
x
)
(
x
,
t
), (
∂
2
u
/
∂
x
2
)
(
x
,
t
), respectively.OnH
1, we shall use the following norm:(16)
v
H
1
=
v
2
+
v
x
2
1
/
2
.We put(17)
H
0
1
=
v
∈
H
1
Ω
:
v
1
=
v
R
=
0
.H
0
1 is a closed subspace of H
1 and on H
0
1, two norms v
H
1 and v
x are equivalent.Note thatL
2 and H
1 are also the Hilbert spaces with respect to the corresponding scalar products(18)
u
,
v
=
∫
1
R
x
γ
u
x
v
x
d
x
,
u
,
v
+
u
x
,
v
x
,respectively. The norms in L
2 and H
1 induced by the corresponding scalar products are denoted by ·
0 and ·
1, respectively. H
0
1 is continuously and densely embedded in L
2. Identifying L
2 with (
L
2
)
′ (the dual of L
2), we have H
0
1
↪
L
2
↪
(
H
0
1
)
′
=
H
-
1; on the other hand, the notation ·
,
· is used for the pairing between H
0
1 and H
-
1.We then have the following lemmas, the proofs of which can be found in [16].Lemma 1.
We have the following inequalities:(19)
i
v
≤
v
0
≤
R
γ
v
,
∀
v
∈
L
2
,
i
i
v
H
1
≤
v
1
≤
R
γ
v
H
1
,
∀
v
∈
H
1
.Lemma 2.
The imbeddingH
1
↪
C
0
(
Ω
¯
) is compact.Lemma 3.
The imbeddingH
0
1
↪
C
0
(
Ω
¯
) is compact and(20)
i
v
C
0
Ω
¯
≤
R
-
1
v
x
∀
v
∈
H
0
1
,
i
i
v
≤
R
-
1
2
v
x
∀
v
∈
H
0
1
,
i
i
i
v
0
≤
R
γ
2
R
-
1
v
x
0
∀
v
∈
H
0
1
.Remark 4.
OnL
2, two norms v
↦
v and v
↦
v
0 are equivalent. So there are two norms v
↦
v
H
1 and v
↦
v
1 on H
1 and four norms v
↦
v
H
1, v
↦
v
1, v
↦
v
x, and v
↦
v
x
0 on H
0
1
.Considera
(
·
,
·
) is the symmetric bilinear form on H
0
1
×
H
0
1 defined by(21)
a
u
,
w
=
u
x
,
w
x
,
∀
u
,
w
∈
H
0
1
.Then, the symmetric bilinear forma
(
·
,
·
) is continuous on H
0
1
×
H
0
1 and coercive on H
0
1
.We have also the following lemma.Lemma 5.
There exists the Hilbert orthonormal base{
w
j
} of L
2 consisting of the eigenfunctions w
j corresponding to the eigenvalue λ
¯
j such that(22)
0
<
λ
¯
1
≤
λ
¯
2
≤
⋯
≤
λ
¯
j
≤
λ
¯
j
+
1
≤
⋯
,
l
i
m
j
→
+
∞
λ
¯
j
=
+
∞
,
a
w
j
,
w
=
λ
¯
j
w
j
,
w
∀
w
∈
H
0
1
,
j
=
1,2
,
…
.
Furthermore, the sequence{
w
j
/
λ
¯
j
} is also the Hilbert orthonormal base of H
0
1 with respect to the scalar product a
(
·
,
·
)
.
On the other hand, we also havew
j satisfying the following boundary value problem:(23)
-
w
j
x
x
+
γ
x
w
j
x
=
λ
¯
j
w
j
,
in
1
,
R
,
w
j
1
=
w
j
R
=
0
,
w
j
∈
C
∞
1
,
R
.The proof of Lemma5 can be found in [17, p. 87, Theorem 7.7], with H
=
L
2 and V
=
H
0
1 and a
(
·
,
·
) as defined by (21).


## 3. The Existence and the Uniqueness

Now, we consider problems (1)–(3) in which γ is a positive constant and make the following assumptions:H
1
u
~
0
∈
H
1
.
H
2
g
1
,
g
R
∈
W
1,1
0
,
T, u
~
0
1
-
g
1
0
=
u
~
0
R
-
g
R
0
=
0
.
H
3
α
∈
W
1,1
0
,
T, α
t
≥
α
∗
>
0, ∀
t
∈
0
,
T
.
H
4
μ
∈
W
1,1
0
,
T, μ
t
≥
μ
∗
>
0, ∀
t
∈
0
,
T
.
H
5
f
1
∈
L
1
0
,
T
;
L
2
.
H
6
f
∈
C
0
R
;
R satisfies the condition that there exists positive constant δ such that y
-
z
f
y
-
f
z
≥
-
δ
y
-
z
2, for all y
,
z
∈
R.In caseg
1
≠
0 or g
R
≠
0, it is clearly that problems (1)–(3) reduce to a problem with homogeneous boundary conditions by the suitable transformation. Indeed, putting φ
(
x
,
t
)
=
(
x
-
1
/
R
-
1
)
g
R
(
t
)
+
(
(
R
-
x
)
/
(
R
-
1
)
)
g
1
(
t
), by the transformation v
(
x
,
t
)
=
u
(
x
,
t
)
-
φ
(
x
,
t
), problems (1)–(3) reduce to the following problem:(24)
v
t
-
μ
t
+
α
t
∂
∂
t
∂
2
v
∂
x
2
+
γ
x
∂
v
∂
x
+
f
v
+
φ
=
f
¯
1
x
,
t
,
1
<
x
<
R
,
t
>
0
,
v
1
,
t
=
v
R
,
t
=
0
,
v
x
,
0
=
v
~
0
x
,where(25)
f
¯
1
x
,
t
=
f
1
x
,
t
-
1
R
-
1
x
-
1
g
R
′
t
+
R
-
x
g
1
′
t
+
γ
R
-
1
x
μ
t
g
R
t
-
g
1
t
+
α
t
g
R
′
t
-
g
1
′
t
,
v
~
0
x
=
u
~
0
x
-
φ
x
,
0
,
v
~
0
∈
H
0
1and u
~
0, g
1, and g
R satisfy the condition u
~
0
(
1
)
-
g
1
(
0
)
=
u
~
0
(
R
)
-
g
R
(
0
)
=
0.The weak formulation of the initial-boundary value problem (24) can be given in the following manner: Find v
∈
L
∞
(
0
,
T
;
H
0
1
) with t
v
t
∈
L
2
(
0
,
T
;
H
0
1
), such that v satisfies the following variational equation:(26)
d
d
t
v
t
,
w
+
α
t
a
v
t
,
w
+
μ
t
-
α
′
t
a
v
t
,
w
+
f
v
t
+
φ
t
,
w
=
f
¯
1
t
,
w
,
∀
w
∈
H
0
1
,
a.e.
,
t
∈
0
,
T
,
v
0
=
v
~
0
,where a
(
·
,
·
) is the symmetric bilinear form on H
0
1
×
H
0
1 defined by (21).Then, we have the following theorem.Theorem 6.
LetT
>
0 and (
H
1
)
–
(
H
6
) hold. Then, problem (24) has a unique weak solution v such that(27)
v
∈
L
∞
0
,
T
;
H
0
1
,
t
v
t
∈
L
2
0
,
T
;
H
0
1
.
Moreover, if(
H
5
) is replaced by f
1
∈
L
2
(
Q
T
), then the solution v satisfies(28)
v
∈
L
∞
0
,
T
;
H
0
1
,
v
t
∈
L
2
0
,
T
;
H
0
1
.Proof.
The proof consists of several steps.
Step  1 (the Faedo-Galerkin approximation (introduced by Lions [18])). Consider the basis {
w
j
} for H
0
1 as in Lemma 5. We find the approximate solution of problem (24) in the form(29)
v
m
t
=
∑
j
=
1
m
c
m
j
t
w
j
,where the coefficients c
m
j satisfy the system of linear differential equations(30)
v
m
′
t
,
w
j
+
α
t
a
v
m
′
t
,
w
j
+
μ
t
a
v
m
t
,
w
j
+
f
v
m
t
+
φ
t
,
w
j
=
f
¯
1
t
,
w
j
,
1
≤
j
≤
m
,
v
m
0
=
v
0
m
,where(31)
v
0
m
=
∑
j
=
1
m
α
m
j
w
j
⟶
v
~
0
strongly  in
H
0
1
.
The system of (30) can be rewritten in the form(32)
c
m
j
′
t
+
λ
¯
j
μ
t
1
+
λ
¯
j
α
t
c
m
j
t
+
1
1
+
λ
¯
j
α
t
f
v
m
t
+
φ
t
,
w
j
=
1
1
+
λ
¯
j
α
t
f
¯
1
t
,
w
j
,
c
m
j
0
=
α
m
j
,
1
≤
j
≤
m
.
It is clear that for eachm there exists a solution v
m
(
t
) in the form of (29) which satisfies (30) almost everywhere on 0
≤
t
≤
T
~
m for some T
~
m, 0
<
T
~
m
≤
T
. The following estimates allow one to take T
~
m
=
T for all m.
Step  2 (a priori estimates)
(a) The First Estimate. Multiplying the jth equation of (30) by c
m
j
(
t
) and summing up with respect to j, afterwards, integrating by parts with respect to the time variable from 0 to t, we get after some rearrangements:(33)
v
m
t
0
2
+
α
t
v
m
x
t
0
2
=
v
0
m
0
2
+
α
0
v
0
m
x
0
2
-
∫
0
t
2
μ
s
-
α
′
s
v
m
x
s
0
2
d
s
-
2
∫
0
t
f
v
m
s
+
φ
s
,
v
m
s
d
s
+
2
∫
0
t
f
¯
1
s
,
v
m
s
d
s
.
Byv
0
m
→
v
~
0 strongly in H
0
1, we have(34)
v
0
m
0
2
+
α
0
v
0
m
x
0
2
≤
S
¯
0
,
∀
m
,where S
¯
0 always indicates a bound depending on v
~
0.
Put(35)
S
m
t
=
v
m
t
0
2
+
α
∗
v
m
x
t
0
2
.
By the assumptions(
H
3
)
–
(
H
6
), we estimate without difficulty the following terms in (33) as follows:(36)
-
∫
0
t
2
μ
s
-
α
′
s
v
m
x
s
0
2
d
s
≤
1
α
∗
∫
0
t
2
μ
s
-
α
′
s
S
m
s
d
s
;
-
2
∫
0
t
f
v
m
s
+
φ
s
,
v
m
s
d
s
≤
2
δ
∫
0
t
v
m
s
0
2
d
s
+
2
∫
0
t
f
φ
s
0
v
m
s
0
d
s
≤
2
δ
+
1
∫
0
t
v
m
s
0
2
d
s
+
∫
0
T
f
φ
s
0
2
d
s
;
2
∫
0
t
f
¯
1
s
,
v
m
s
d
s
≤
f
¯
1
L
1
0
,
T
;
L
2
+
∫
0
t
f
¯
1
s
0
v
m
s
0
2
d
s
≤
f
¯
1
L
1
0
,
T
;
L
2
+
∫
0
t
f
¯
1
s
0
S
m
s
d
s
.
Hence, it follows from (33), (34), and (36) that(37)
S
m
t
≤
C
T
1
+
∫
0
t
d
T
1
s
S
m
s
d
s
,where(38)
C
T
1
=
S
¯
0
+
f
¯
1
L
1
0
,
T
;
L
2
+
∫
0
T
f
φ
s
0
2
d
s
,
d
T
1
s
=
1
+
2
δ
+
f
¯
1
s
0
+
1
α
∗
2
μ
s
-
α
′
s
,
d
T
1
∈
L
1
0
,
T
.
By Gronwall’s lemma, we obtain from (37) that(39)
S
m
t
≤
C
T
1
exp
⁡
∫
0
t
d
T
1
s
d
s
≤
C
T
,for all m
∈
N, for all t, 0
≤
t
≤
T
~
m
≤
T; that is, T
~
m
=
T, where C
T always indicates a bound depending on T.
(b) The Second Estimate. Multiplying the jth equation of (30) by 2
t
2
c
m
j
′
(
t
) and summing up with respect to j, we have(40)
2
t
v
m
′
t
0
2
+
2
α
t
t
v
m
x
′
t
0
2
+
d
d
t
μ
t
t
v
m
x
t
0
2
=
t
2
μ
t
′
v
m
x
t
0
2
-
2
t
2
f
v
m
t
+
φ
t
,
v
m
′
t
+
2
t
2
f
¯
1
t
,
v
m
′
t
.
Integrating (40), we get(41)
2
∫
0
t
s
v
m
′
s
0
2
d
s
+
2
∫
0
t
α
s
s
v
m
x
′
s
0
2
d
s
+
μ
t
t
v
m
x
t
0
2
=
∫
0
t
s
2
μ
s
′
v
m
x
s
0
2
d
s
-
2
∫
0
t
s
f
v
m
s
+
φ
s
,
s
v
m
′
s
d
s
+
2
∫
0
t
s
f
¯
1
s
,
s
v
m
′
s
d
s
.
We shall estimate the terms of (41) as follows:(42)
∫
0
t
s
2
μ
s
′
v
m
x
s
0
2
d
s
≤
1
α
∗
∫
0
t
s
2
μ
s
′
S
m
s
d
s
≤
C
T
α
∗
∫
0
T
s
2
μ
s
′
d
s
;
(43)
2
∫
0
t
s
f
¯
1
s
,
s
v
m
′
s
d
s
≤
2
∫
0
T
s
f
¯
1
s
0
2
d
s
+
1
2
∫
0
t
s
v
m
′
s
0
2
d
s
.
On the other hand, we have(44)
v
m
x
,
s
+
φ
x
,
s
≤
v
m
s
C
0
Ω
¯
+
g
1
s
+
g
R
s
≤
R
-
1
C
T
α
∗
+
g
1
C
0
0
,
T
+
g
R
C
0
0
,
T
≡
C
¯
T
,and hence(45)
2
∫
0
t
s
f
v
m
s
+
φ
s
,
s
v
m
′
s
d
s
≤
2
∫
0
t
s
f
v
m
s
+
φ
s
0
2
d
s
+
1
2
∫
0
t
s
v
m
′
s
0
2
d
s
≤
2
∫
0
t
s
2
d
s
∫
1
R
x
γ
sup
z
≤
C
¯
T
⁡
f
2
z
d
x
+
1
2
∫
0
t
s
v
m
′
s
0
2
d
s
≤
2
T
2
R
γ
+
1
-
1
γ
+
1
sup
z
≤
C
¯
T
⁡
f
2
z
d
x
+
1
2
∫
0
t
s
v
m
′
s
0
2
d
s
.
It follows from (41)–(43) and (45) that(46)
∫
0
t
s
v
m
′
s
0
2
d
s
+
2
α
∗
∫
0
t
s
v
m
x
′
s
0
2
d
s
+
μ
∗
t
v
m
x
t
0
2
≤
C
T
α
∗
∫
0
T
s
2
μ
s
′
d
s
+
2
∫
0
T
s
f
¯
1
s
0
2
d
s
+
2
T
2
R
γ
+
1
-
1
γ
+
1
sup
z
≤
C
¯
T
⁡
f
2
z
d
x
≤
C
T
,for all m
∈
N, for all t
∈
[
0
,
T
], where C
T always indicates a bound depending on T
.
By(
t
v
m
x
)
′
=
t
v
m
x
′
+
v
m
x and (39) and (46), we deduce that(47)
t
v
m
x
′
L
2
Q
T
≤
t
v
m
x
′
L
2
Q
T
+
v
m
x
L
2
Q
T
≤
∫
0
T
s
v
m
x
′
s
0
2
d
s
+
T
v
m
L
2
0
,
T
;
H
0
1
≤
C
T
.
Step  3 (the limiting process). By (39), (46), and (47), we deduce that there exists a subsequence of {
v
m
}, still denoted by {
v
m
} such that(48)
v
m
⟶
v
in
L
∞
0
,
T
;
H
0
1
weakly
∗
,
t
v
m
′
⟶
t
v
′
in
L
2
0
,
T
;
H
0
1
weakly
.
Using a compactness lemma ([18], Lions, p. 57), applied to (48), we can extract from the sequence {
v
m
} a subsequence still denoted by {
v
m
}, such that(49)
t
v
m
⟶
t
v
strongly  in
L
2
Q
T
.By the Riesz-Fischer theorem, we can extract from {
v
m
} a subsequence still denoted by {
v
m
}, such that(50)
v
m
x
,
t
⟶
v
x
,
t
a.e.
x
,
t
in
Q
T
.
Becausef is continuous, it gives(51)
f
v
m
x
,
t
+
φ
x
,
t
⟶
f
v
x
,
t
+
φ
x
,
t
a.e.
x
,
t
in
Q
T
.
On the other hand, by(
H
6
), it follows from (44) that(52)
f
v
m
x
,
t
+
φ
x
,
t
≤
sup
z
≤
C
¯
T
⁡
f
z
≤
C
T
,where C
T is a constant independent of m
.
Using the dominated convergence theorem, (51) and (52) yield(53)
f
v
m
+
φ
⟶
f
v
+
φ
strongly  in
L
2
Q
T
.
Passing to the limit in (30) by (31), (48), and (53), we obtain(54)
d
d
t
v
t
,
w
+
α
t
a
v
t
,
w
+
μ
t
-
α
′
t
a
v
t
,
w
+
f
v
t
+
φ
t
,
w
=
f
¯
1
t
,
w
,
∀
w
∈
H
0
1
,
a.e.
,
t
∈
0
,
T
,
v
0
=
v
~
0
.
Step  4 (uniqueness of the solution). First, we shall need the following lemma.
Lemma  7. Let
v
be the weak solution of the following problem:
(55)
v
t
-
μ
t
+
α
t
∂
∂
t
∂
2
v
∂
x
2
+
γ
x
∂
v
∂
x
=
f
~
x
,
t
,
1
<
x
<
R
,
0
<
t
<
T
,
v
1
,
t
=
v
R
,
t
=
0
,
v
x
,
0
=
v
~
0
x
,
v
∈
L
∞
0
,
T
;
H
0
1
,
t
v
t
∈
L
2
0
,
T
;
H
0
1
,
μ
,
α
∈
W
1,1
0
,
T
.
Then,
(56)
v
t
0
2
+
α
t
v
x
t
0
2
+
∫
0
t
2
μ
s
-
α
′
s
v
x
s
0
2
d
s
≥
v
~
0
0
2
+
α
0
v
~
0
x
0
2
+
2
∫
0
t
f
~
s
,
v
s
d
s
.
Furthermore, if
v
~
0
=
0,then the equality in (56) holds.
Lemma  7 is a slight improvement of a lemma used in [12] (or it can be found in Lions’s book [18]).
Now, we will prove the uniqueness of the solutions.
Letv
1 and v
2 be two weak solutions of (24). Then, v
=
v
1
-
v
2 is a weak solution of (55) with the right-hand side function replaced by f
~
(
x
,
t
)
=
-
f
(
v
1
+
φ
)
+
f
(
v
2
+
φ
) and v
~
0
=
0
. Using Lemma7, we have equality(57)
σ
1
t
=
-
∫
0
t
2
μ
s
-
α
′
s
v
x
s
0
2
d
s
-
2
∫
0
t
f
v
1
+
φ
-
f
v
2
+
φ
,
v
s
d
s
,where(58)
σ
1
t
=
v
t
0
2
+
α
t
v
x
t
0
2
.
ByH
6, we obtain(59)
2
∫
0
t
f
v
1
+
φ
-
f
v
2
+
φ
,
v
s
d
s
≥
-
2
δ
∫
0
t
v
s
0
2
d
s
≥
-
2
δ
∫
0
t
σ
1
s
d
s
;
-
∫
0
t
2
μ
s
-
α
′
s
v
x
s
0
2
d
s
≤
1
α
∗
∫
0
t
2
μ
s
-
α
′
s
σ
1
s
d
s
.
It follows from (57)–(59) that(60)
σ
1
t
≤
∫
0
t
2
δ
+
1
α
∗
2
μ
s
-
α
′
s
σ
1
s
d
s
.
By Gronwall’s lemma,v
=
0
.
Assume now that(
H
5
) is replaced by f
1
∈
L
2
(
Q
T
); then we only have to show that {
v
m
′
} is bounded in L
2
(
0
,
T
;
H
0
1
)
.
Indeed, multiplying thejth equation of (30) by c
m
j
′
(
t
) and summing up with respect to j, afterwards, integrating with respect to the time variable from 0 to t, we get after some rearrangements(61)
X
m
t
=
X
m
0
+
∫
0
t
μ
′
s
v
m
x
s
0
2
d
s
+
2
∫
0
t
f
¯
1
s
,
v
m
′
s
d
s
-
2
∫
0
t
f
v
m
s
+
φ
s
,
v
m
′
s
d
s
,where(62)
X
m
t
=
2
∫
0
t
v
m
′
s
0
2
+
α
s
v
m
x
′
s
0
2
d
s
+
μ
t
v
m
x
t
0
2
.
By the same estimates as above, we obtain(63)
X
m
0
=
μ
0
v
0
m
x
0
2
≤
μ
0
α
∗
S
¯
0
;
∫
0
t
μ
′
s
v
m
x
s
0
2
d
s
≤
∫
0
t
μ
′
s
α
∗
S
m
s
d
s
≤
C
T
α
∗
∫
0
T
μ
′
s
d
s
;
2
∫
0
t
f
¯
1
s
,
v
m
′
s
d
s
≤
2
f
¯
1
L
2
Q
T
2
+
1
2
∫
0
t
v
m
′
s
0
2
d
s
≤
2
f
¯
1
L
2
Q
T
2
+
1
4
X
m
t
;
-
2
∫
0
t
f
v
m
s
+
φ
s
,
v
m
′
s
d
s
≤
2
T
R
γ
+
1
-
1
γ
+
1
sup
z
≤
C
¯
T
⁡
f
2
z
+
1
2
∫
0
t
v
m
′
s
0
2
d
s
≤
2
T
R
γ
+
1
-
1
γ
+
1
sup
z
≤
C
¯
T
⁡
f
2
z
+
1
4
X
m
t
.
This implies(64)
X
m
t
≤
2
α
∗
μ
0
S
¯
0
+
C
T
∫
0
T
μ
′
s
d
s
+
4
f
¯
1
L
2
Q
T
2
+
4
T
R
γ
+
1
-
1
γ
+
1
sup
z
≤
C
¯
T
⁡
f
2
z
=
C
¯
T
1
.
Then, the sequence{
v
m
′
} is bounded in L
2
(
0
,
T
;
H
0
1
).
Applying a similar argument used as above, the limitv of the sequence {
v
m
} in suitable function spaces is a unique weak solution of (24) satisfying (28).
Therefore, Theorem6 is proved.


## 4. Asymptotic Behavior of the Solution ast
→
+
∞

In this part, letT
>
0, (
H
1
)
–
(
H
6
) hold. Then, there exists a unique solution u
=
v
+
φ of problems (1)–(3) such that (65)
u
-
φ
=
v
∈
L
∞
0
,
T
;
H
0
1
,
t
u
t
-
φ
t
=
t
v
t
∈
L
2
0
,
T
;
H
0
1
.We shall study asymptotic behavior of the solutionu
(
t
) as t
→
+
∞.We make the following supplementary assumptions on the functionsg
1
(
t
), g
R
(
t
), α
(
t
), μ
(
t
), and f
1
x
,
t:H
2
′
g
1
,
g
R
∈
W
1,1
(
R
+
), u
~
0
(
1
)
-
g
1
(
0
)
=
u
~
0
(
R
)
-
g
R
(
0
)
=
0, and there exist the positive constants C
-
1, C
-
R, γ
-
1, and γ
-
R, such that(66)
g
i
t
+
g
i
′
t
≤
C
-
i
e
-
γ
-
i
t
,
∀
t
≥
0
,
i
∈
1
,
R
.
H
3
′
α
∈
W
1,1
(
R
+
), α
(
t
)
≥
α
∗
>
0, ∀
t
≥
0, α
(
0
)
≤
α
(
T
).
H
4
′
μ
∈
W
1,1
(
R
+
), and there exist the positive constants μ
∗, μ
-
∗, μ
∞, C
-
μ, and γ
-
μ, such that(67)
μ
t
≥
μ
∗
>
0
,
∀
t
≥
0
,
μ
t
-
μ
∞
≤
C
-
μ
e
-
γ
-
μ
t
,
∀
t
≥
0
,
2
μ
t
-
α
′
t
≥
2
μ
-
∗
>
0
,
∀
t
≥
0
.
H
5
′
f
1
∈
L
∞
(
0
,
∞
;
L
2
), and there exist the positive constants C
1, γ
1 and the function f
1
∞
∈
L
2, such that f
1
(
t
)
-
f
1
∞
0
≤
C
1
e
-
γ
1
t, ∀
t
≥
0.
H
6
′
f
∈
C
0
(
R
;
R
) and there exists a positive constant δ, with 0
<
δ
<
2
μ
∞
/
R
γ
(
R
-
1
)
2, such that (
y
-
z
)
f
(
y
)
-
f
(
z
)
≥
-
δ
y
-
z
2, for all y
,
z
∈
R
.First, we consider the following stationary problem:(68)
-
μ
∞
∂
2
u
∂
x
2
+
γ
x
∂
u
∂
x
+
f
u
=
f
1
∞
x
,
1
<
x
<
R
,
u
1
,
t
=
u
R
,
t
=
0
.The weak solution of problem (68) is obtained from the following variational problem. Find u
∞
∈
H
0
1 such that(69)
μ
∞
a
u
∞
,
w
+
f
u
∞
,
w
=
f
1
∞
,
w
,for all w
∈
H
0
1, where a
(
·
,
·
) is the symmetric bilinear form on H
0
1
×
H
0
1 defined by (21).We then have the following theorem.Theorem 8.
LetH
4
′
–
H
6
′ hold. Then, there exists a unique solution u
∞ of the variational problem (69) such that u
∞
∈
H
0
1
.Proof.
Consider the basis{
w
j
} for H
0
1 as in Lemma 5. Put(70)
y
m
=
∑
j
=
1
m
d
m
j
w
j
,where d
m
j satisfies the following nonlinear equation system:(71)
μ
∞
a
y
m
,
w
j
+
f
y
m
,
w
j
=
f
1
∞
,
w
j
,
1
≤
j
≤
m
.
By Brouwer’s lemma (see Lions [18], Lemma 4.3, p. 53), it follows from the hypotheses H
4
′
–
H
6
′ that systems (70) and (71) have a solution y
m
.
Multiplying thejth equation of system (71) by d
m
j, and then summing up with respect to j, we have(72)
μ
∞
a
y
m
,
y
m
+
f
y
m
,
y
m
=
f
1
∞
,
y
m
.
By usingH
6, we obtain(73)
f
y
m
,
y
m
=
∫
1
R
x
γ
f
y
m
x
-
f
0
y
m
x
d
x
+
∫
1
R
x
γ
f
0
y
m
x
d
x
≥
-
δ
∫
1
R
x
γ
y
m
2
x
d
x
+
∫
1
R
x
γ
f
0
y
m
x
d
x
≥
-
δ
y
m
0
2
-
ε
1
y
m
0
2
-
1
4
ε
1
∫
1
R
x
γ
f
2
0
d
x
=
-
δ
+
ε
1
y
m
0
2
-
1
4
ε
1
R
γ
+
1
-
1
γ
+
1
f
2
0
,
∀
ε
1
>
0
.
By using inequalities (20)(iii) and (73), we obtain from (72) that (74)
μ
∞
y
m
x
0
2
≤
δ
+
ε
1
y
m
0
2
+
1
4
ε
1
R
γ
+
1
-
1
γ
+
1
f
2
0
+
f
1
∞
0
y
m
0
≤
δ
+
ε
1
y
m
0
2
+
1
4
ε
1
R
γ
+
1
-
1
γ
+
1
f
2
0
+
1
4
ε
1
f
1
∞
0
2
+
ε
1
y
m
0
2
=
δ
+
2
ε
1
y
m
0
2
+
1
4
ε
1
R
γ
+
1
-
1
γ
+
1
f
2
0
+
f
1
∞
0
2
=
δ
+
2
ε
1
R
γ
2
R
-
1
2
y
m
x
0
2
+
1
4
ε
1
R
γ
+
1
-
1
γ
+
1
f
2
0
+
f
1
∞
0
2
.
By0
<
δ
<
2
μ
∞
/
R
γ
(
R
-
1
)
2, choose ε
1
>
0 such that δ
+
2
ε
1
(
R
γ
/
2
)
(
R
-
1
)
2
<
μ
∞
.
Hence, we deduce from (74) that(75)
y
m
x
0
≤
R
γ
+
1
-
1
/
γ
+
1
f
2
0
+
f
1
∞
0
2
4
ε
1
μ
∞
-
δ
+
2
ε
1
R
γ
/
2
R
-
1
2
=
D
~
1
,and D
~
1 is a constant independent of m
.
By means of (75) and Lemma 3, the sequence {
y
m
} has a subsequence still denoted by {
y
m
} such that(76)
y
m
⟶
u
∞
in
H
0
1
w
e
a
k
l
y
,
y
m
⟶
u
∞
in
C
0
1
,
R
strongly
.
On the other hand, by (76)2 and the continuity of f, we have(77)
f
y
m
⟶
f
u
∞
in
C
0
1
,
R
strongly
.
Passing to the limit in (71), we find without difficulty from (76) and (77) that u
∞ satisfies the equation(78)
μ
∞
a
u
∞
,
w
j
+
f
u
∞
,
w
j
=
f
1
∞
,
w
j
.
Equation (78) holds for every j
=
1,2
,
…; that is, (69) is true.
The solution of problem (69) is unique, which can be shown by the same arguments as in the proof of Theorem 6.Now we consider asymptotic behavior of the solutionu
(
t
) as t
→
+
∞
.We then have the following theorem.Theorem 9.
LetH
1, H
2
′
–
H
5
′, and H
6 hold. Let f satisfy the following condition, in addition,H
6
′
′
∀
M
>
0, ∃
k
M
>
0
:
f
(
y
)
-
f
(
z
)
≤
k
M
y
-
z, ∀
y
,
z
∈
[
-
M
,
M
]
.And let δ
>
0 in H
6 satisfy the following condition, in addition, (
H
6
′
′
′
)
0
<
δ
<
(
2
/
R
γ
(
R
-
1
)
2
)
m
i
n
{
μ
∞
,
μ
-
∗
}
.
Then we have(79)
u
t
-
u
∞
1
≤
C
¯
e
-
γ
¯
t
,
∀
t
≥
0
,where γ
¯
>
0, C
¯
>
0 are constants independent of t
.Proof.
PutZ
m
(
t
)
=
v
m
(
t
)
-
y
m
. Let us subtract (30)1 with (71) to obtain(80)
Z
m
′
t
,
w
j
+
α
t
a
Z
m
′
t
,
w
j
+
μ
t
a
Z
m
t
,
w
j
+
μ
t
-
μ
∞
a
y
m
,
w
j
+
f
v
m
t
+
φ
t
-
f
y
m
,
w
j
=
f
¯
1
t
-
f
1
∞
,
w
j
,
1
≤
j
≤
m
,
Z
m
0
=
v
0
m
-
y
m
.By multiplying (80)1 by c
m
j
(
t
)
-
d
m
j and summing up in j, we obtain(81)
d
d
t
Z
m
t
0
2
+
α
t
Z
m
x
t
0
2
+
2
μ
t
-
α
′
t
Z
m
x
t
0
2
+
2
μ
t
-
μ
∞
a
y
m
,
Z
m
t
+
2
f
v
m
t
+
φ
t
-
f
y
m
+
φ
t
,
Z
m
t
+
2
f
y
m
+
φ
t
-
f
y
m
,
Z
m
t
=
2
f
¯
1
t
-
f
1
t
,
Z
m
t
+
2
f
1
t
-
f
1
∞
,
Z
m
t
.
By the assumptionsH
2
′
–
H
5
′, H
6, H
6
′
′, and H
6
′
′
′ and using inequality (20)(iii), and with ε
1
>
0, we estimate without difficulty the following terms in (81) as follows:
(i)Estimate
2
μ
(
t
)
-
α
′
(
t
)
Z
m
x
(
t
)
0
2, as(82)
2
μ
t
-
α
′
t
Z
m
x
t
0
2
≥
2
μ
¯
∗
Z
m
x
t
0
2
;
(ii)Estimate
2
μ
t
-
μ
∞
a
y
m
,
Z
m
t, as(83)
2
μ
t
-
μ
∞
a
y
m
,
Z
m
t
≤
2
μ
t
-
μ
∞
y
m
x
0
Z
m
x
t
0
≤
2
C
¯
μ
e
-
γ
¯
μ
t
C
~
Z
m
x
t
0
≤
1
ε
1
C
~
2
C
¯
μ
2
e
-
2
γ
¯
μ
t
+
ε
1
Z
m
x
t
0
2
.
(iii)Estimate
2
〈
f
(
v
m
(
t
)
+
φ
(
t
)
)
-
f
(
y
m
+
φ
(
t
)
)
,
Z
m
(
t
)
〉, as(84)
2
f
v
m
t
+
φ
t
-
f
y
m
+
φ
t
,
Z
m
t
≥
-
2
δ
Z
m
t
0
2
≥
-
2
δ
R
γ
2
R
-
1
2
Z
m
x
t
0
2
.
(iv)Estimate
2
f
y
m
+
φ
t
-
f
y
m
,
Z
m
t. Note that, from the inequalities(85)
φ
x
,
t
≤
g
1
t
+
g
R
t
≤
C
¯
1
+
C
¯
R
,
y
m
C
0
Ω
¯
≤
R
-
1
y
m
x
≤
R
-
1
D
~
1
,
y
m
+
φ
C
0
Ω
¯
≤
R
-
1
D
~
1
+
C
¯
1
+
C
¯
R
=
M
1
,and H
6
′
′, we deduce that(86)
f
y
m
+
φ
t
-
f
y
m
≤
k
M
1
φ
x
,
t
≤
k
M
1
g
1
t
+
g
R
t
=
k
M
1
Ψ
t
,where(87)
Ψ
t
=
C
¯
1
e
-
γ
¯
1
t
+
C
¯
R
e
-
γ
¯
R
t
.
Hence,(88)
f
y
m
+
φ
t
-
f
y
m
0
2
≤
1
γ
+
1
R
γ
+
1
-
1
k
M
1
2
Ψ
2
t
.
Thus,(89)
2
f
y
m
+
φ
t
-
f
y
m
,
Z
m
t
≤
1
ε
1
f
y
m
+
φ
t
-
f
y
m
0
2
+
ε
1
Z
m
t
0
2
≤
1
ε
1
1
γ
+
1
R
γ
+
1
-
1
k
M
1
2
Ψ
2
t
+
ε
1
R
γ
2
R
-
1
2
Z
m
x
t
0
2
.
(v)Estimate
〈
f
¯
1
(
t
)
-
f
1
(
t
)
,
Z
m
(
t
)
〉. We have(90)
f
¯
1
x
,
t
-
f
1
x
,
t
=
-
1
R
-
1
x
-
1
g
R
′
t
+
R
-
x
g
1
′
t
+
γ
R
-
1
x
μ
t
g
R
t
-
g
1
t
+
α
t
g
R
′
t
-
g
1
′
t
.
Hence,(91)
f
¯
1
x
,
t
-
f
1
x
,
t
≤
g
1
′
t
+
g
R
′
t
+
γ
R
-
1
α
t
+
μ
t
g
1
t
+
g
1
′
t
+
g
R
t
+
g
R
′
t
≤
C
¯
1
e
-
γ
¯
1
t
+
C
¯
R
e
-
γ
¯
R
t
+
γ
R
-
1
α
L
∞
R
+
+
μ
L
∞
R
+
C
¯
1
e
-
γ
¯
1
t
+
C
¯
R
e
-
γ
¯
R
t
=
1
+
γ
R
-
1
α
L
∞
R
+
+
μ
L
∞
R
+
C
¯
1
e
-
γ
¯
1
t
+
C
¯
R
e
-
γ
¯
R
t
=
1
+
γ
R
-
1
α
L
∞
R
+
+
μ
L
∞
R
+
Ψ
t
≡
C
~
α
,
μ
Ψ
t
.
It follows that(92)
f
¯
1
t
-
f
1
t
0
2
≤
1
γ
+
1
R
γ
+
1
-
1
C
~
2
α
,
μ
Ψ
2
t
.
Thus,(93)
2
f
¯
1
t
-
f
1
t
,
Z
m
t
≤
1
ε
1
f
¯
1
t
-
f
1
t
0
2
+
ε
1
Z
m
t
0
2
≤
1
ε
1
1
γ
+
1
R
γ
+
1
-
1
C
~
2
α
,
μ
Ψ
2
t
+
ε
1
R
γ
2
R
-
1
2
Z
m
x
t
0
2
.
(vi)Estimate
〈
f
1
(
t
)
-
f
1
∞
,
Z
m
(
t
)
〉, as(94)
2
f
1
t
-
f
1
∞
,
Z
m
t
≤
1
ε
1
f
1
t
-
f
1
∞
0
2
+
ε
1
Z
m
t
0
2
≤
1
ε
1
C
1
2
e
-
2
γ
1
t
+
ε
1
R
γ
2
R
-
1
2
Z
m
x
t
0
2
.
It follows from (81)–(84), (89), (93), and (94) that(95)
d
d
t
Z
m
t
0
2
+
α
t
Z
m
x
t
0
2
+
2
μ
¯
∗
-
ε
1
-
2
δ
+
3
ε
1
R
γ
2
R
-
1
2
Z
m
x
t
0
2
≤
1
ε
1
C
~
2
C
¯
μ
2
e
-
2
γ
¯
μ
t
+
C
1
2
e
-
2
γ
1
t
+
1
ε
1
R
γ
+
1
-
1
γ
+
1
k
M
1
2
+
C
~
2
α
,
μ
Ψ
2
t
≡
ψ
~
t
.
By0
<
δ
<
2
/
R
γ
(
R
-
1
)
2
min
⁡
{
μ
∞
,
μ
¯
∗
}
≤
2
μ
¯
∗
/
R
γ
(
R
-
1
)
2, choose ε
1
>
0 such that 2
γ
~
=
2
μ
¯
∗
-
ε
1
-
2
δ
+
3
ε
1
(
R
γ
/
2
)
(
R
-
1
)
2
>
0.
Putγ
¯
0
=
m
i
n
{
γ
1
,
γ
¯
1
,
γ
¯
R
,
γ
¯
μ
}, and we have ψ
~
(
t
)
≤
C
¯
0
e
-
2
γ
¯
0
t for all t
≥
0, as(96)
d
d
t
Z
m
t
0
2
+
α
t
Z
m
x
t
0
2
+
2
γ
~
Z
m
x
t
0
2
≤
ψ
~
t
≤
C
¯
0
e
-
2
γ
¯
0
t
.
By(97)
Z
m
x
t
0
2
=
1
2
Z
m
x
t
0
2
+
1
2
Z
m
x
t
0
2
≥
1
2
1
α
t
α
t
Z
m
x
t
0
2
+
1
2
2
R
γ
R
-
1
2
Z
m
t
0
2
≥
1
2
1
α
L
∞
R
+
α
t
Z
m
x
t
0
2
+
1
2
2
R
γ
R
-
1
2
Z
m
t
0
2
≥
β
1
α
t
Z
m
x
t
0
2
+
Z
m
t
0
2
,where β
1
=
(
1
/
2
)
m
i
n
{
1
/
α
L
∞
(
R
+
)
,
2
/
R
γ
(
R
-
1
)
2
}, it follows from (96) and (97) that(98)
d
d
t
Z
m
t
0
2
+
α
t
Z
m
x
t
0
2
+
2
γ
~
β
1
Z
m
t
0
2
+
α
t
Z
m
x
t
0
2
≤
ψ
~
t
≤
C
¯
0
e
-
2
γ
¯
0
t
.
Chooseγ
¯
>
0 such that γ
¯
<
m
i
n
{
γ
¯
0
,
2
γ
~
β
1
}, and then we have from (98) that(99)
d
d
t
Z
m
t
0
2
+
α
t
Z
m
x
t
0
2
+
2
γ
¯
Z
m
t
0
2
+
α
t
Z
m
x
t
0
2
≤
C
¯
0
e
-
2
γ
¯
0
t
.
Hence, we obtain from (99) that(100)
Z
m
t
0
2
+
α
∗
Z
m
x
t
0
2
≤
Z
m
t
0
2
+
α
t
Z
m
x
t
0
2
≤
Z
m
0
0
2
+
α
0
Z
m
x
0
0
2
+
C
¯
0
2
γ
¯
0
-
γ
¯
e
-
2
γ
¯
t
.
Lettingm
→
+
∞ in (100), we obtain(101)
v
t
-
u
∞
0
2
+
α
∗
v
x
t
-
u
∞
x
0
2
≤
liminf
m
→
+
∞
⁡
v
m
t
-
y
m
0
2
+
α
∗
v
m
x
t
-
y
m
x
0
2
≤
v
~
0
-
u
∞
0
2
+
α
0
v
~
0
x
-
u
∞
x
0
2
+
C
¯
0
2
γ
¯
0
-
γ
¯
e
-
2
γ
¯
t
,
∀
t
≥
0
,or(102)
v
t
-
u
∞
1
≤
D
~
2
e
-
γ
¯
t
,
∀
t
≥
0
,where(103)
D
~
2
=
1
m
i
n
1
,
α
∗
v
~
0
-
u
∞
0
2
+
α
0
v
~
0
x
-
u
∞
x
0
2
+
C
¯
0
2
γ
¯
0
-
γ
¯
.
Note that(104)
φ
x
,
t
≤
g
R
t
+
g
1
t
≤
C
¯
1
e
-
γ
¯
1
t
+
C
¯
R
e
-
γ
¯
R
t
=
Ψ
t
;
φ
x
x
,
t
=
g
R
t
-
g
1
t
R
-
1
≤
1
R
-
1
Ψ
t
.
Hence,(105)
φ
t
1
2
=
φ
t
0
2
+
φ
x
t
0
2
≤
1
γ
+
1
R
γ
+
1
-
1
1
+
1
R
-
1
2
Ψ
2
t
≤
D
~
3
2
e
-
2
γ
¯
t
.
It follows from (102) and (105) that(106)
u
t
-
u
∞
1
≤
v
t
-
u
∞
1
+
φ
t
1
≤
D
~
2
+
D
~
3
e
-
γ
¯
t
,
∀
t
≥
0
.
This completes the proof of Theorem9.


## 5. The Existence and the Uniqueness of a(
η
,
T
)-Periodic Weak Solution

In this section, we shall consider problems (1), (2), and (4) with R
>
1, 0
<
η
≤
1 as given constants and μ, α, f, f
1, g
1, and g
R as given functions satisfying the following assumptions:H
-
2
g
1
,
g
R
∈
W
1,1
(
0
,
T
), and g
1 and g
R are (
η
,
T
)-periodic; that is, (107)
g
1
0
=
η
g
1
T
,
g
R
0
=
η
g
R
T
.
H
-
3
α
∈
W
1,1
(
0
,
T
), α
(
t
)
≥
α
∗
>
0, ∀
t
∈
[
0
,
T
], 0
<
α
(
0
)
≤
α
(
T
);
H
-
4
μ
∈
W
1,1
(
0
,
T
), μ
(
t
)
≥
μ
∗
>
0, ∀
t
∈
[
0
,
T
], 2
μ
t
-
α
′
t
≥
2
μ
-
∗
>
0, a.e., t
∈
[
0
,
T
];
H
-
5
f
1
,
f
1
′
∈
L
2
(
Q
T
), f
1 is (
η
,
T
)-periodic, f
1
(
0
)
=
η
f
1
(
T
);
H
-
6
f
∈
C
0
(
R
;
R
) and there exists a positive constant δ, with(108)
0
<
δ
<
2
μ
-
∗
R
γ
R
-
1
2
,
such that
y
-
z
f
y
-
f
z
≥
-
δ
y
-
z
2
,
∀
y
,
z
∈
R
.Remark 10.
An example of the functionsg
1
,
g
R satisfying (
H
¯
2
) is(109)
g
k
t
=
ζ
k
e
p
t
,where p
>
0, ζ
k, and k
∈
{
1
,
R
} are constants. It is obvious that (
H
¯
2
) holds, because (110)
g
k
t
=
e
-
p
T
ζ
k
e
p
t
+
T
=
η
g
k
t
+
T
,with η
=
e
-
p
T and (111)
g
k
0
=
η
g
k
T
,
k
∈
1
,
R
.
Similarly, by the transformationv
(
x
,
t
)
=
u
(
x
,
t
)
-
φ
(
x
,
t
), with φ
(
x
,
t
)
=
(
x
-
1
/
R
-
1
)
g
R
(
t
)
+
(
(
R
-
x
)
/
(
R
-
1
)
)
g
1
(
t
), φ
(
x
,
0
)
=
η
φ
(
x
,
T
), problems (1), (2), and (4) reduce to the following problem:(112)
v
t
-
μ
t
+
α
t
∂
∂
t
∂
2
v
∂
x
2
+
γ
x
∂
v
∂
x
+
f
v
+
φ
=
f
¯
1
x
,
t
,
1
<
x
<
R
,
0
<
t
<
T
,
v
1
,
t
=
v
R
,
t
=
0
,
v
x
,
0
=
η
v
x
,
T
,where f
¯
1
(
x
,
t
) is defined by (25)1
.
The weak formulation of problem (112) can be given in the following manner: Find v
∈
L
∞
(
0
,
T
;
H
0
1
) with v
t
∈
L
2
(
0
,
T
;
H
0
1
), such that v satisfies the following variational equation:(113)
∫
0
T
v
′
t
,
w
t
d
t
+
∫
0
T
α
t
a
v
′
t
,
w
t
d
t
+
∫
0
T
μ
t
a
v
t
,
w
t
d
t
+
∫
0
T
f
v
t
+
φ
t
,
w
t
d
t
=
∫
0
T
f
¯
1
t
,
w
t
d
t
,
∀
w
∈
L
2
0
,
T
;
H
0
1
,
v
0
=
η
v
T
,where a
(
·
,
·
) is the symmetric bilinear form on H
0
1
×
H
0
1 defined by (21).Then, we have the following theorem.Theorem 11.
LetT
>
0 and H
¯
2
–
H
¯
6 hold. Then, problem (112) has a unique weak solution v such that(114)
v
∈
L
∞
0
,
T
;
H
0
1
,
v
t
∈
L
2
0
,
T
;
H
0
1
.Proof.
The proof consists of several steps.
Step  1 (the Faedo-Galerkin approximation (introduced by Lions [18])). Consider the basis {
w
j
} for H
0
1 as in Lemma 5. Let V
m be the linear space generated by w
1
,
w
2
,
…
,
w
m
. We consider the following problem.
Find a functionv
m
(
t
) in the form (29) satisfying the nonlinear differential equation system (30)1 and the (
η
,
T
)
- periodic condition:(115)
v
m
0
=
η
v
m
T
.
We consider an initial value problem given by (30), where v
0
m is given in V
m
.
It is clear that, for eachm, there exists a solution v
m
(
t
) in the form (29) which satisfies (30) almost everywhere on 0
≤
t
≤
T
~
m for some T
~
m, 0
<
T
~
m
≤
T
. The following a priori estimates allow us to take T
~
m
=
T for all m.
Step  2 (a priori estimates). Multiplying the jth equation of (30)1 by c
m
j
(
t
) and summing up with respect to j, we get(116)
d
d
t
v
m
t
0
2
+
α
t
v
m
x
t
0
2
+
2
μ
t
-
α
′
t
v
m
x
t
0
2
+
2
f
v
m
t
+
φ
t
,
v
m
t
=
2
f
¯
1
t
,
v
m
t
.
By the same estimates as in Section3, and with ε
1
>
0, we obtain(117)
2
f
v
m
t
+
φ
t
,
v
m
t
=
2
f
v
m
t
+
φ
t
-
f
φ
t
,
v
m
t
+
2
f
φ
t
,
v
m
t
≥
-
2
δ
+
ε
1
R
γ
2
R
-
1
2
v
m
x
t
0
2
-
1
ε
1
f
φ
t
0
2
;
2
f
¯
1
t
,
v
m
t
≤
1
ε
1
f
¯
1
t
0
2
+
ε
1
v
m
t
0
2
≤
1
ε
1
f
¯
1
t
0
2
+
ε
1
R
γ
2
R
-
1
2
v
m
x
t
0
2
.
Hence, it follows from (116) and (117) that(118)
d
d
t
v
m
t
0
2
+
α
t
v
m
x
t
0
2
+
2
μ
¯
∗
-
δ
+
ε
1
R
γ
2
R
-
1
2
v
m
x
t
0
2
≤
1
ε
1
f
¯
1
t
0
2
+
f
φ
t
0
2
.
By0
<
δ
<
2
μ
¯
∗
/
R
γ
(
R
-
1
)
2, choose ε
1
>
0 such that γ
~
=
μ
¯
∗
-
δ
+
ε
1
(
R
γ
/
2
)
(
R
-
1
)
2
>
0.
It is similar to (97); we get(119)
v
m
x
t
0
2
≥
β
1
α
t
v
m
x
t
0
2
+
v
m
t
0
2
,where β
1
=
1
/
2
min
⁡
{
1
/
α
L
∞
(
0
,
T
)
,
2
/
R
γ
(
R
-
1
)
2
}
.
From (118) and (119), it leads to(120)
d
d
t
v
m
t
0
2
+
α
t
v
m
x
t
0
2
+
2
β
1
γ
~
v
m
t
0
2
+
α
t
v
m
x
t
0
2
≤
f
¯
t
,in which f
¯
(
t
)
=
(
1
/
ε
1
)
f
¯
1
(
t
)
0
2
+
f
(
φ
(
t
)
)
0
2
.
Integrating (120), we have(121)
v
m
t
0
2
+
α
t
v
m
x
t
0
2
≤
v
0
m
0
2
+
α
0
v
0
m
x
0
2
+
∫
0
t
e
2
β
1
γ
~
s
f
¯
s
d
s
e
-
2
β
1
γ
~
t
≤
ρ
2
+
v
0
m
0
2
+
α
0
v
0
m
x
0
2
-
ρ
2
e
-
2
β
1
γ
~
t
,where ρ
2
=
sup
0
≤
t
≤
T
⁡
ρ
1
(
t
), with(122)
ρ
1
t
=
1
e
2
β
1
γ
~
t
-
1
∫
0
t
e
2
β
1
γ
~
s
f
¯
s
d
s
,
0
<
t
≤
T
,
1
2
β
1
γ
~
f
¯
0
,
t
=
0
.
Therefore, if we choosev
0
m such that v
0
m
0
2
+
α
(
0
)
v
0
m
x
0
2
≤
ρ
2, we obtain from (121) that(123)
v
m
t
0
2
+
α
t
v
m
x
t
0
2
≤
ρ
2
,
i.e.
,
T
~
m
=
T
∀
m
.
LetB
¯
m
(
ρ
) be a closed ball in the space V
m with the norm v
0
m
↦
v
0
m
V
m
=
v
0
m
0
2
+
α
(
0
)
v
0
m
x
0
2
1
/
2
.
By0
<
α
(
0
)
≤
α
(
T
), we obtain(124)
v
m
T
V
m
2
=
v
m
T
0
2
+
α
0
v
m
x
T
0
2
≤
v
m
T
0
2
+
α
T
v
m
x
T
0
2
≤
ρ
2
.
Let us define(125)
F
m
:
B
¯
m
ρ
⟶
B
¯
m
ρ
v
0
m
⟼
F
m
v
0
m
=
η
v
m
T
.
We prove thatF
m is a contraction. Let v
0
m
,
v
¯
0
m
∈
B
¯
m
(
0
,
ρ
) and let y
m
(
t
)
=
v
m
(
t
)
-
v
¯
m
(
t
), where v
m
(
t
) and v
¯
m
(
t
) are solutions of system (30)1 on [
0
,
T
] satisfying the initial conditions v
m
(
0
)
=
v
0
m and v
¯
m
(
0
)
=
v
¯
0
m, respectively. Then, y
m
(
t
) satisfies the following differential equation system:(126)
y
m
′
t
,
w
j
+
α
t
a
y
m
′
t
,
w
j
+
μ
t
a
y
m
t
,
w
j
+
f
v
m
t
+
φ
t
-
f
v
¯
m
t
+
φ
t
,
w
j
=
0
,where 1
≤
j
≤
m, with initial condition(127)
y
m
0
=
v
0
m
-
v
¯
0
m
.
By using the same arguments as before, we can show that(128)
d
d
t
y
m
t
0
2
+
α
t
y
m
x
t
0
2
+
2
β
2
y
m
t
0
2
+
α
t
y
m
x
t
0
2
≤
0
,where β
2
=
β
1
μ
¯
∗
-
δ
R
γ
/
2
(
R
-
1
)
2
>
0, β
1
=
(
1
/
2
)
m
i
n
{
1
/
α
L
∞
(
0
,
T
)
,
2
/
R
γ
(
R
-
1
)
2
}
.
Integrating inequality (128), we obtain (129)
y
m
t
0
2
+
α
t
y
m
x
t
0
2
≤
e
-
2
β
2
t
y
m
0
0
2
+
α
0
y
m
x
0
0
2
=
e
-
2
β
2
t
y
m
0
V
m
2
,
∀
t
∈
0
,
T
.
By0
<
α
(
0
)
≤
α
(
T
), it follows that (130)
y
m
T
V
m
2
=
y
m
T
0
2
+
α
0
y
m
x
T
0
2
≤
y
m
T
0
2
+
α
T
y
m
x
T
0
2
≤
e
-
2
β
2
T
y
m
0
V
m
2
=
e
-
2
β
2
T
v
0
m
-
v
¯
0
m
V
m
2
,or(131)
F
m
v
0
m
-
F
m
v
¯
0
m
V
m
=
η
y
m
T
V
m
≤
η
e
-
β
2
T
v
0
m
-
v
¯
0
m
V
m
;that is, F
m is a contraction.
Therefore, there exists a unique functionv
0
m
∈
B
¯
m
(
ρ
) such that the solution of the initial value problem (30) is a solution of system (30)1, (115). This solution satisfies inequality (124) a.e., in [
0
,
T
]
.
On the other hand, multiplying thejth equation of (30)1 by c
m
j
′
(
t
) and summing up with respect to j, afterwards, integrating with respect to the time variable from 0 to T, we get after some rearrangements(132)
2
∫
0
T
v
m
′
t
0
2
+
α
t
v
m
x
′
t
0
2
d
t
+
∫
0
T
d
d
t
μ
t
v
m
x
t
0
2
d
t
+
2
∫
0
T
f
v
m
t
+
φ
t
,
v
m
′
t
d
t
=
∫
0
T
μ
′
t
v
m
x
t
0
2
d
t
+
2
∫
0
T
f
¯
1
t
,
v
m
′
t
d
t
.
We estimate without difficulty the following terms in (132):(133)
-
∫
0
T
d
d
t
μ
t
v
m
x
t
0
2
d
t
=
-
μ
T
v
m
x
T
0
2
+
μ
0
v
m
x
0
0
2
≤
μ
0
v
m
x
0
0
2
≤
μ
0
ρ
2
α
∗
.
Note that(134)
φ
t
C
0
1
,
R
≤
g
1
t
+
g
R
t
≤
s
u
p
0
≤
t
≤
T
g
1
t
+
g
R
t
,
(135)
v
m
t
+
φ
t
C
0
1
,
R
≤
v
m
t
C
0
1
,
R
+
φ
t
C
0
1
,
R
≤
R
-
1
ρ
α
∗
+
s
u
p
0
≤
t
≤
T
g
1
t
+
g
R
t
=
M
1
T
.
Hence,(136)
2
∫
0
T
f
v
m
t
+
φ
t
,
v
m
′
t
d
t
≤
2
∫
0
T
f
v
m
t
+
φ
t
0
2
d
t
+
1
2
∫
0
T
v
m
′
t
0
2
d
t
≤
M
2
T
+
1
2
∫
0
T
v
m
′
t
0
2
d
t
,where (137)
M
2
T
=
2
T
R
γ
+
1
-
1
γ
+
1
sup
z
≤
M
1
T
⁡
f
2
z
.
Moreover,(138)
2
∫
0
T
f
¯
1
t
,
v
m
′
t
d
t
≤
2
∫
0
T
f
¯
1
t
0
2
d
t
+
1
2
∫
0
T
v
m
′
t
0
2
d
t
;
∫
0
T
μ
′
t
v
m
x
t
0
2
d
t
≤
ρ
2
α
∗
∫
0
T
μ
′
t
d
t
.
It follows from (132), (133), (136), and (138) that(139)
∫
0
T
v
m
′
t
0
2
d
t
+
2
α
∗
∫
0
T
v
m
x
′
t
0
2
d
t
≤
μ
0
ρ
2
α
∗
+
M
2
T
+
2
∫
0
T
f
¯
1
t
0
2
d
t
+
ρ
2
α
∗
∫
0
T
μ
′
t
d
t
≤
C
T
,for all m
∈
N, for all t
∈
[
0
,
T
], where C
T always indicates a bound depending on T.
Step  3 (the limiting process). By (123) and (139), we deduce that there exists a subsequence of {
v
m
}, still denoted by {
v
m
} such that(140)
v
m
⟶
v
in
L
∞
0
,
T
;
H
0
1
weakly
∗
,
v
m
′
⟶
v
′
in
L
2
0
,
T
;
H
0
1
weakly
.
From (115), we obtain(141)
v
0
=
η
v
T
.
Indeed, we prove (141) as follows.
Byv
0
m
x
≤
(
1
/
α
∗
)
v
0
m
V
m
≤
ρ
/
α
∗, and by the imbedding H
0
1
↪
C
0
(
Ω
¯
) being compact, there exists a subsequence of {
v
0
m
}, still denoted by {
v
0
m
} such that(142)
v
0
m
⟶
v
~
0
in
H
0
1
weakly
,
v
0
m
⟶
v
~
0
in
C
0
1
,
R
strongly
.
Byv
m
(
t
)
=
v
m
(
0
)
+
∫
0
t
v
m
′
(
s
)
d
s, we deduce from (140) and (142) that(143)
v
t
=
v
~
0
+
∫
0
t
v
′
s
d
s
.
This implies(144)
v
0
=
v
~
0
,
(145)
v
0
m
⟶
v
0
in
H
0
1
weakly
,
v
0
m
⟶
v
0
in
C
0
1
,
R
strongly
.
From (115), we obtain(146)
v
m
0
,
w
j
=
η
v
m
T
,
w
j
=
η
v
m
0
,
w
j
+
∫
0
T
v
m
′
t
,
w
j
d
t
,
∀
j
∈
N
.
By (140), (145), and (146), it yields(147)
v
0
,
w
j
=
η
v
0
,
w
j
+
∫
0
T
v
′
t
,
w
j
d
t
=
η
v
T
,
w
j
,
∀
j
∈
N
.
Therefore,(148)
v
0
=
η
v
T
.
Therefore, (141) is proved.
Using a compactness lemma ([18], Lions, p. 57), applied to (140), we can extract from the sequence {
v
m
} a subsequence still denoted by {
v
m
}, such that(149)
v
m
⟶
v
strongly  in
L
2
Q
T
.
By the Riesz-Fischer theorem, we can extract from{
v
m
} a subsequence still denoted by {
v
m
}, such that (150)
v
m
x
,
t
⟶
v
x
,
t
a.e.
,
x
,
t
in
Q
T
,
Becausef is continuous, then(151)
f
v
m
x
,
t
+
φ
x
,
t
⟶
f
v
x
,
t
+
φ
x
,
t
a.e.
,
x
,
t
i
n
Q
T
.
On the other hand,(152)
f
v
m
x
,
t
+
φ
x
,
t
≤
s
u
p
z
≤
M
1
T
f
z
,
a.e.
,
x
,
t
in
Q
T
,where M
1
(
T
) is the constant defined by (135).
Using the dominated convergence theorem, (151) and (152) yield(153)
f
v
m
+
φ
⟶
f
v
+
φ
strongly  in
L
2
Q
T
.
Denote by{
ζ
i
,
i
=
1,2
,
…
} the orthonormal base in the real Hilbert space L
2
(
0
,
T
). The set {
ζ
i
w
j
,
i
,
j
=
1,2
,
…
} forms an orthonormal base in L
2
(
0
,
T
;
H
0
1
)
. From (30)1 we have(154)
∫
0
T
v
m
′
t
,
w
j
ζ
i
t
d
t
+
∫
0
T
α
t
a
v
m
′
t
,
w
j
ζ
i
t
d
t
+
∫
0
T
μ
t
a
v
m
t
,
w
j
ζ
i
t
d
t
+
∫
0
T
f
v
m
t
+
φ
t
,
w
j
ζ
i
t
d
t
=
∫
0
T
f
¯
1
t
,
w
j
ζ
i
t
d
t
,for all i
,
j, 1
≤
j
≤
m, i
∈
N
.
Fori
,
j fixed, passing to the limit in (154) by (140) and (153), we get(155)
∫
0
T
v
′
t
,
w
j
ζ
i
t
d
t
+
∫
0
T
α
t
a
v
′
t
,
w
j
ζ
i
t
d
t
+
∫
0
T
μ
t
a
v
t
,
w
j
ζ
i
t
d
t
+
∫
0
T
f
v
t
+
φ
t
,
w
j
ζ
i
t
d
t
=
∫
0
T
f
¯
1
t
,
w
j
ζ
i
t
d
t
.
Note that (155) holds for every i
,
j
∈
N; that is, the equality(156)
∫
0
T
v
′
t
,
w
t
d
t
+
∫
0
T
α
t
a
v
′
t
,
w
t
d
t
+
∫
0
T
μ
t
a
v
t
,
w
t
d
t
+
∫
0
T
f
v
t
+
φ
t
,
w
t
d
t
=
∫
0
T
f
¯
1
t
,
w
t
d
t
,
∀
w
∈
L
2
0
,
T
;
H
0
1
,is fulfilled.
Step  4 (uniqueness of the solutions). Let v
1 and v
2 be two solutions of (113). Then v
=
v
1
-
v
2 satisfies the following problem:(157)
∫
0
T
v
′
t
,
w
t
d
t
+
∫
0
T
α
t
a
v
′
t
,
w
t
d
t
+
∫
0
T
μ
t
a
v
t
,
w
t
d
t
+
∫
0
T
f
v
1
t
+
φ
t
-
f
v
2
t
+
φ
t
,
w
t
d
t
=
0
,
∀
w
∈
L
2
0
,
T
;
H
0
1
,
v
0
=
η
v
T
,
v
∈
L
∞
0
,
T
;
H
0
1
,
v
t
∈
L
2
0
,
T
;
H
0
1
.
Takingw
=
2
v in (157)1 and using (157)2, we get(158)
∫
0
T
2
μ
t
-
α
′
t
v
x
t
0
2
d
t
+
∫
0
T
d
d
t
v
t
0
2
d
t
+
∫
0
T
d
d
t
α
t
v
x
t
0
2
d
t
+
2
∫
0
T
f
v
1
t
+
φ
t
-
f
v
2
t
+
φ
t
,
v
t
d
t
=
0
.
By2
μ
(
t
)
-
α
′
(
t
)
≥
μ
¯
∗
>
0, we have(159)
∫
0
T
2
μ
t
-
α
′
t
v
x
t
0
2
d
t
≥
2
μ
¯
∗
∫
0
T
v
x
t
0
2
d
t
.
On the other hand,(160)
∫
0
T
d
d
t
v
t
0
2
d
t
=
v
T
0
2
-
v
0
0
2
=
1
-
η
2
v
T
0
2
≥
0
;
∫
0
T
d
d
t
α
t
v
x
t
0
2
d
t
=
α
t
v
x
T
0
2
-
α
0
v
x
0
0
2
≥
α
0
1
-
η
2
v
x
T
0
2
≥
0
.
Hence,(161)
2
μ
¯
∗
∫
0
T
v
x
t
0
2
d
t
≤
-
2
∫
0
T
f
v
1
t
+
φ
t
-
f
v
2
t
+
φ
t
,
v
t
d
t
≤
2
δ
∫
0
T
v
t
0
2
d
t
≤
2
δ
R
γ
2
R
-
1
2
∫
0
T
v
x
t
0
2
d
t
.
By0
<
δ
<
2
μ
¯
∗
/
R
γ
(
R
-
1
)
2, implying δ
(
R
γ
/
2
)
(
R
-
1
)
2
<
μ
¯
∗, we deduce from (161) that ∫
0
T
v
x
(
t
)
0
2
d
t
=
0; that is, v
=
v
1
-
v
2
=
0
.
This completes the proof of Theorem11.



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*Source: 3875324-2016-12-29.xml*