# Kawahara-Burgers Equation on a Strip

Authors: N. A. Larkin
Journal: Advances in Mathematical Physics
 (2015)

---

## Abstract

An initial-boundary value problem for the 2D Kawahara-Burgers equation posed on a channel-type strip was considered. The existence and uniqueness results for regular and weak solutions in weighted spaces as well as exponential decay of small solutions without restrictions on the width of a strip were proven both for regular solutions in an elevated norm and for weak solutions in theL
2-norm.


---

## Body

## 1. Introduction

We are concerned with an initial-boundary value problem (IBVP) for the two-dimensional Kawahara-Burgers (KB) equation(1)
u
t
+
u
x
-
u
x
x
+
u
u
x
+
u
x
x
x
+
u
x
y
y
-
∂
x
5
u
=
0posed on a strip modeling an infinite channel {
(
x
,
y
)
∈
R
2
:
x
∈
R
,
y
∈
(
0
,
B
)
,
B
>
0
}. This equation is a two-dimensional analog of the Kawahara type equation(2)
u
t
-
∂
x
5
u
+
F
u
,
u
x
,
u
x
x
,
u
x
x
x
=
0which includes dissipation and dispersion and has been studied intensively in the last years due to its applications in mechanics and physics [1–9]. The KB equation may be considered as a perturbation of the Zakharov-Kuznetsov-Burgers equation which describes the propagation of nonlinear ionic-sonic waves in a plasma submitted to a magnetic field directed along the x-axis [10, 11]. Equations (1) and (2) are typical examples of the so-called dispersive equations which attract considerable attention of both pure and applied mathematicians in the past decades. The theory of the Cauchy problem for (2) and other dispersive equations like the KdV equation has been extensively studied and is considerably advanced today [1, 3–9, 12–17]. Results on IBVPs for one-dimensional dispersive equations both in bounded and in unbounded domains may be found in [4, 6, 18–22]. It was shown in [18, 19, 23–26] that the KdV and Kawahara equations have an implicit internal dissipation. This allowed the proof of exponential decay of small solutions in bounded domains without adding any artificial damping term [21, 23]. Later, this effect has been proven for a wide class of dispersive equations of any odd order with one space variable [27]. The control and stability for the linear KdV equation with the transport term u
x may fail for critical domains, but it is possible to eliminate the term u
x by simple scaling when the KdV and Kawahara equations are posed on the whole line. The same is true also for (1) posed on a strip (
y
∈
(
0
,
B
)
,
x
∈
R
,
t
>
0
) [28]. Recently, interest in dispersive equations became to be extended to multidimensional models such as Kadomtsev-Petviashvili (KP), Zakharov-Kuznetsov (ZK) equations [11], and dispersive equations of higher orders [29]. As far as the ZK equation and its generalizations are concerned, the results on IVPs can be found in [30–34] and IBVPs were studied in [24, 35–39]. It was shown that IBVP for the ZK equation posed on a half-strip unbounded in x direction with the Dirichlet conditions on the boundaries possesses regular solutions which decay exponentially as t
→
∞ provided initial data are sufficiently small and the width of a half-strip is not too large [25, 38]. Similar result was established for the 2D Kawahara equation posed on a half-strip [24]. This means that a multidimensional dispersive equation may create an internal dissipative mechanism for some types of IBVPs. The goal of our paper is to prove that the KB equation on a strip also may create a dissipative effect without adding any artificial damping. We must mention that IBVP for the ZK equation on a strip (
x
∈
(
0,1
)
,
y
∈
R
) has been studied in [39, 40] and IBVPs on a strip (
y
∈
(
0
,
L
)
,
x
∈
R
) for the ZK equation and Zakharov-Kuznetsov-Burgers equation were considered in [28, 35] and for the ZK equation with some internal damping in [37]. In the domain (
y
∈
(
0
,
B
)
,
x
∈
R
,
t
>
0
), the term u
x in (1) can be scaled out by a simple change of variables. Nevertheless, it can not be safely ignored for problems posed both on finite and on semi-infinite intervals as well as on infinite in y direction bands without changes in the original domain [40, 41]. The main results of our paper are the existence and uniqueness of regular and weak global-in-time solutions for (1) posed on a strip with the Dirichlet boundary conditions and the exponential decay rate of these solutions as well as continuous dependence on initial data. To explore dissipativity of the term u
x
y
y, we used exponential weight e
2
b
x which was implied to define solutions of (1) as the product (3)
e
b
x
u
t
-
u
x
x
+
u
u
x
+
u
x
x
x
+
u
x
y
y
-
∂
x
5
u
=
0
in
L
2
S
.We must mention that this idea has been proposed earlier in [12].The paper has the following structure. Section1 is Introduction. Section 2 contains formulation of the problem. In Section 3, we prove global existence and uniqueness theorems for regular solutions in some weighted spaces as well as continuous dependence on initial data. In Section 4, we prove exponential decay of small regular solutions in an elevated norm. In Section 5, we prove the existence, uniqueness, and continuous dependence on initial data for weak solutions as well as the exponential decay rate of the L
2
(
S
)-norm for small solutions without limitations on the width of the strip.


## 2. Problem and Preliminaries

LetB
,
T
,
r be finite positive numbers. Define S
=
{
(
x
,
y
)
∈
R
2
:
x
∈
R
,
y
∈
(
0
,
B
)
};S
r
=
{
(
x
,
y
)
∈
R
2
:
x
∈
(
-
r
,
+
∞
)
,
y
∈
(
0
,
B
)
}; and S
T
=
S
×
(
0
,
T
)
.Hereafter subscriptsu
x, u
x
y, and so forth denote the partial derivatives, as well as ∂
x or ∂
x
y
2 when it is convenient. Operators ∇ and Δ are the gradient and Laplacian acting over S. By (
·
,
·
) and · we denote the inner product and the norm in L
2
(
S
), and ·
H
k stands for norms in the L
2-based Sobolev spaces. We will use also the spaces H
s
∩
L
b
2, where L
b
2
=
L
2
(
e
2
b
x
d
x
); see [12].Consider the following IBVP:(4)
L
u
≡
u
t
-
u
x
x
+
u
u
x
+
u
x
x
x
+
u
x
y
y
-
∂
x
5
u
=
0
,
in
S
T
;
(5)
u
x
,
0
,
t
=
u
x
,
B
,
t
=
0
,
x
∈
R
,
t
>
0
;
(6)
u
x
,
y
,
0
=
u
0
x
,
y
,
x
,
y
∈
S
.


## 3. Existence of Regular Solutions

Approximate Solutions. We will construct solutions to (4)–(6) by the Faedo-Galerkin method: the functions (7)
w
j
y
=
2
B
sin
⁡
λ
j
y
,where λ
j
=
j
2
π
2
/
B
2, are orthonormal in L
2
(
S
) eigenfunctions of the following Dirichlet problem:(8)
w
j
y
y
+
λ
j
w
j
=
0
,
y
∈
0
,
B
;
w
j
0
=
w
j
B
=
0
,
j
=
1,2
,
3
,
…
.Define approximate solutions of (4)–(6) as follows:(9)
u
N
x
,
y
,
t
=
∑
j
=
1
N
w
j
y
g
j
x
,
t
,where g
j
(
x
,
t
) are solutions to the following Cauchy problem for the system of N generalized Kawahara equations:(10)
∂
∂
t
g
j
x
,
t
+
∂
3
∂
x
3
g
j
x
,
t
-
∂
2
∂
x
2
g
j
x
,
t
-
λ
j
∂
∂
x
g
j
x
,
t
-
∂
5
∂
x
5
g
j
x
,
t
+
∫
0
B
u
N
x
,
y
,
t
u
x
N
x
,
y
,
t
w
j
y
d
y
=
0
,
(11)
g
j
x
,
0
=
∫
0
B
w
j
y
u
0
x
,
y
d
y
,
j
=
1
,
…
,
N
.It can be shown that, for g
j
(
x
,
0
)
∈
H
s
,
s
≥
5, the Cauchy problem (10) and (11) has a unique regular solution g
j
∈
L
∞
(
0
,
T
;
H
s
(
R
)
∩
L
b
2
(
R
)
)
∩
L
2
(
0
,
T
;
H
s
+
2
(
R
)
∩
L
b
2
(
R
)
) [1, 8, 12, 16]. To prove the existence of global solutions for (4)–(6), we will establish uniform in N global in t estimates of approximate solutions u
N
(
x
,
y
,
t
).Estimate 1.
Multiply thejth equation of (10) by g
j, sum up over j
=
1
,
…
,
N, and integrate the result with respect to x over R to obtain (12)
d
d
t
u
N
2
t
+
2
u
x
N
2
t
=
0
,which implies(13)
u
N
2
t
+
2
∫
0
t
u
x
N
2
s
d
s
=
u
N
2
0
∀
t
∈
0
,
T
.It follows from here that for N sufficiently large and ∀
t
>
0
(14)
u
N
2
t
+
2
∫
0
t
u
x
N
2
s
d
s
=
u
N
2
0
≤
2
u
0
2
.In our calculations we will drop the index N where it is not ambiguous.Estimate 2.
For some positiveb, multiply the jth equation of (10) by e
2
b
x
g
j, sum up over j
=
1
,
…
,
N, and integrate the result with respect to x over R. Dropping the index N, we get(15)
d
d
t
e
2
b
x
,
u
2
t
+
2
+
6
b
-
40
b
3
e
2
b
x
,
u
x
2
t
+
2
b
e
2
b
x
,
u
y
2
t
+
10
b
e
2
b
x
,
u
x
x
2
t
-
4
b
3
e
2
b
x
,
u
3
t
-
4
b
2
+
8
b
3
-
32
b
5
e
2
b
x
,
u
2
t
=
0
.
In our calculations, we will frequently use the following multiplicative inequalities [42].Proposition 1.
(i) For allu
∈
H
1
(
R
2
),(16)
u
L
4
R
2
2
≤
2
u
L
2
R
2
∇
u
L
2
R
2
.
(ii) For allu
∈
H
1
(
D
),(17)
u
L
4
D
2
≤
C
D
u
L
2
D
u
H
1
D
,where the constant C
D depends on a way of continuation of u
∈
H
1
(
D
) as u
~
(
R
2
) such that u
~
(
D
)
=
u
(
D
).Extendingu
N
(
x
,
y
,
t
) for a fixed t into the exterior of S by 0 and exploiting (16), we find(18)
4
b
3
e
2
b
x
u
3
t
≤
b
e
2
b
x
,
u
y
2
t
+
2
b
e
2
b
x
,
u
x
2
t
+
2
b
3
+
8
b
9
u
0
N
2
e
2
b
x
,
u
2
t
.Substituting this into (15) and making use of the inequality (19)
e
2
b
x
,
u
x
2
≤
ϵ
e
2
b
x
,
u
x
x
2
+
1
4
ϵ
+
2
b
2
e
2
b
x
,
u
2with an appropriate ϵ
>
0, when (
3
-
20
b
2
<
0
), we come to the inequality(20)
d
d
t
e
2
b
x
,
u
2
t
+
e
2
b
x
,
u
x
2
+
u
y
2
+
u
x
x
2
t
≤
C
b
1
+
u
0
2
e
2
b
x
,
u
2
t
.By the Gronwall lemma, (21)
e
2
b
x
,
u
2
t
≤
C
b
,
T
,
u
0
e
2
b
x
,
u
0
2
.Returning to (20) gives(22)
e
2
b
x
,
u
N
2
t
+
∫
0
t
e
2
b
x
,
∇
u
N
2
+
u
x
x
N
2
τ
d
τ
≤
C
b
,
T
,
u
0
e
2
b
x
,
u
0
2
∀
t
∈
0
,
T
.It follows from this estimate and (13) that uniformly in N and for any r
>
0 and t
∈
(
0
,
T
)
(23)
u
N
2
t
+
∫
0
t
∫
0
B
∫
-
r
+
∞
∇
u
N
2
+
u
x
x
N
2
d
x
d
y
d
s
≤
C
r
,
b
,
T
,
u
0
e
2
b
x
,
u
0
2
,where C does not depend on N.Estimates (22), (23) make it possible to prove the existence of a weak solution to (4)–(6) passing to the limit in (10) as N
→
∞. For details of passing to the limit in the nonlinear term, see [12].We will need the following lemma.Lemma 2.
Letu
(
x
,
y
)
:
S
→
R be such that (24)
∫
S
e
2
b
x
u
2
x
,
y
+
∇
u
x
,
y
2
+
u
x
y
2
x
,
y
d
x
d
y
<
∞and for all x
∈
R there is some y
0
∈
[
0
,
B
] such that u
(
x
,
y
0
)
=
0. Then(25)
sup
S
⁡
e
b
x
u
x
,
y
,
t
2
≤
δ
1
+
2
b
2
e
2
b
x
,
u
y
2
t
+
2
δ
e
2
b
x
,
u
x
y
2
t
+
2
δ
1
δ
e
2
b
x
,
u
x
2
t
+
1
δ
1
δ
1
+
2
δ
1
b
2
e
2
b
x
,
u
2
t
,where δ
,
δ
1 are arbitrary positive numbers.Proof.
Denotev
=
e
b
x
u. Then simple calculations give(26)
sup
S
⁡
v
2
x
,
y
,
t
≤
δ
v
y
2
t
+
v
x
y
2
t
+
1
δ
v
x
2
t
+
v
2
t
.Returning to the function u
(
x
,
y
,
t
), we prove Lemma 2.Estimate 3.
Multiplying thejth equation of (10) by -
(
e
2
b
x
g
j
x
)
x, and dropping the index N, we come to the equality(27)
d
d
t
e
2
b
x
,
u
x
2
t
+
2
+
6
b
-
40
b
3
e
2
b
x
,
u
x
x
2
t
+
2
b
e
2
b
x
,
u
x
y
2
t
+
10
b
e
2
b
x
,
u
x
x
x
2
t
-
4
b
2
+
8
b
3
-
32
b
5
e
2
b
x
,
u
x
2
t
+
e
2
b
x
,
u
x
3
t
-
2
b
e
2
b
x
,
u
u
x
2
t
=
0
.Making use of Proposition 1, we estimate(28)
I
1
=
e
2
b
x
,
u
x
3
t
≤
u
x
t
e
b
x
u
x
L
4
S
2
t
≤
2
u
x
t
e
b
x
u
x
t
∇
e
b
x
u
x
t
≤
δ
e
2
b
x
,
2
u
x
x
2
+
u
x
y
2
t
+
2
δ
b
2
+
u
x
2
t
2
δ
e
2
b
x
,
u
x
2
t
.Similarly,(29)
I
2
=
2
b
e
2
b
x
,
u
u
x
2
t
≤
δ
e
2
b
x
,
2
u
x
x
2
+
u
x
y
2
t
+
2
b
2
δ
+
4
b
2
δ
u
0
2
e
2
b
x
,
u
x
2
t
.Substituting I
1
,
I
2 into (27) with 2
δ
=
b, we obtain ∀
t
∈
(
0
,
T
)
(30)
e
2
b
x
,
u
x
N
2
t
+
∫
0
t
e
2
b
x
,
∇
u
x
N
2
+
u
x
x
x
N
2
s
d
s
≤
C
b
,
T
,
u
0
e
2
b
x
,
u
0
x
2
.Estimate 4.
Multiplying thejth equation of (10) by -
2
(
e
2
b
x
λ
j
g
j
), and dropping the index N, we come to the equality(31)
d
d
t
e
2
b
x
,
u
y
2
t
+
2
+
6
b
-
40
b
3
e
2
b
x
,
u
x
y
2
t
+
2
b
e
2
b
x
,
u
y
y
2
t
+
10
b
e
2
b
x
,
u
x
x
y
2
t
-
4
b
2
+
8
b
3
-
32
b
5
e
2
b
x
,
u
y
2
t
+
2
1
-
b
e
2
b
x
,
u
x
u
y
2
t
=
0
.Making use of Proposition 1, we estimate(32)
I
=
2
1
-
b
e
2
b
x
,
u
x
u
y
2
t
≤
2
C
D
1
+
b
u
x
t
e
b
x
u
y
t
e
b
x
u
y
H
1
S
t
≤
δ
e
2
b
x
,
2
u
x
y
2
+
u
y
y
2
t
+
2
δ
1
+
b
2
+
C
D
2
1
+
b
2
u
x
2
t
δ
e
2
b
x
,
u
y
2
t
.Taking δ
=
b, we transform (31) into the inequality(33)
d
d
t
e
2
b
x
,
u
y
2
t
+
e
2
b
x
,
u
x
y
2
+
u
y
y
2
+
u
x
x
y
2
t
≤
C
b
1
+
u
x
t
2
e
2
b
x
,
u
y
2
t
.Making use of (14) and the Gronwall lemma, we get ∀
t
∈
(
0
,
T
)
(34)
e
2
b
x
,
u
y
N
2
t
+
∫
0
t
e
2
b
x
,
u
y
y
N
2
+
u
x
y
N
2
+
u
x
x
y
N
2
s
d
s
≤
C
b
,
T
,
u
0
e
2
b
x
,
u
0
y
2
.
This and (30) give ∀
t
∈
(
0
,
T
)
(35)
e
2
b
x
,
∇
u
N
2
t
+
∫
0
t
e
2
b
x
,
∇
u
x
N
2
+
∇
u
x
x
N
2
+
u
y
y
N
2
s
d
s
≤
C
b
,
T
,
u
0
e
2
b
x
,
∇
u
0
2which imply that for all finite r
>
0 and all t
∈
(
0
,
T
)
(36)
u
N
t
H
1
S
r
≤
C
r
,
b
,
T
,
u
0
e
2
b
x
,
∇
u
0
2
.Estimate 5.
Multiplying thejth equation of (10) by (
e
2
b
x
g
j
x
x
)
x
x, and dropping the index N, we come to the equality(37)
d
d
t
e
2
b
x
,
u
x
x
2
t
+
2
+
6
b
-
40
b
3
e
2
b
x
,
u
x
x
x
2
t
+
2
b
e
2
b
x
,
u
x
x
y
2
t
+
10
b
e
2
b
x
,
u
x
x
x
x
2
t
-
4
b
2
+
8
b
3
-
32
b
5
e
2
b
x
,
u
x
x
2
t
-
2
b
e
2
b
x
,
u
u
x
x
2
t
+
5
e
2
b
x
u
x
,
u
x
x
2
t
=
0
.
Using (16), we estimate(38)
I
=
-
2
b
e
2
b
x
,
u
u
x
x
2
t
+
5
e
2
b
x
u
x
,
u
x
x
2
t
≤
2
δ
e
2
b
x
,
2
u
x
x
x
2
+
u
x
x
y
2
t
+
4
b
2
δ
+
25
δ
u
x
2
t
+
4
b
2
δ
u
2
t
e
2
b
x
,
u
x
x
2
t
.Taking 2
δ
=
b and substituting I into (37), we obtain(39)
d
d
t
e
2
b
x
,
u
x
x
2
t
+
e
2
b
x
,
u
x
x
x
2
+
u
x
x
y
2
+
u
x
x
x
x
2
t
≤
C
b
1
+
u
x
2
t
+
u
2
t
e
2
b
x
,
u
x
x
2
t
.Making use of (14), we find(40)
e
2
b
x
,
u
x
x
N
2
t
+
∫
0
t
e
2
b
x
,
∇
u
x
x
N
2
+
u
x
x
x
x
N
2
s
d
s
≤
C
b
,
T
,
u
0
e
2
b
x
,
u
0
x
x
2
∀
t
∈
0
,
T
.Estimate 6.
Differentiate (10) by t and multiply the result by e
2
b
x
g
j
t to obtain(41)
d
d
t
e
2
b
x
,
u
t
2
t
+
2
+
6
b
-
40
b
3
e
2
b
x
,
u
x
t
2
t
+
2
b
e
2
b
x
,
u
t
y
2
t
+
10
b
e
2
b
x
,
u
t
x
x
2
t
-
4
b
2
+
8
b
3
-
32
b
5
e
2
b
x
,
u
t
2
t
+
2
-
2
b
e
2
b
x
u
x
,
u
t
2
t
=
0
.Making use of (16), we estimate(42)
I
=
2
-
2
b
e
2
b
x
u
x
,
u
t
2
t
≤
2
2
+
2
b
u
x
t
e
b
x
u
t
t
∇
e
b
x
u
t
t
≤
δ
e
2
b
x
,
2
u
x
t
2
+
u
t
y
2
t
+
2
b
2
δ
+
2
+
2
b
2
u
x
t
2
δ
e
2
b
x
,
u
t
2
t
.Taking δ
=
b and substituting I into (41), we get(43)
d
d
t
e
2
b
x
,
u
t
2
t
+
e
2
b
x
,
u
x
t
2
+
u
t
y
2
+
u
t
x
x
2
t
≤
C
b
1
+
u
x
t
2
e
2
b
x
,
u
t
2
t
.
This implies∀
t
∈
(
0
,
T
)
(44)
e
2
b
x
,
u
t
N
2
t
+
∫
0
t
e
2
b
x
,
∇
u
s
N
2
+
u
s
x
x
N
2
s
d
s
≤
C
b
,
T
,
u
0
e
2
b
x
,
u
t
2
0
≤
C
b
,
T
,
u
0
J
0
,where (45)
J
0
=
u
0
2
+
e
2
b
x
,
u
0
2
+
∇
u
0
2
+
∇
u
0
x
2
+
u
0
2
u
0
x
2
+
Δ
u
0
x
2
+
∂
x
5
u
0
2
.Estimate 7.
Multiplying thejth equation of (10) by -
e
2
b
x
g
j
x and dropping the index N, we come to the equality(46)
e
2
b
x
,
u
x
y
2
+
u
x
x
x
2
t
=
e
2
b
x
,
u
u
x
2
t
+
e
2
b
x
u
t
,
u
x
t
+
8
b
2
-
1
e
2
b
x
,
u
x
x
2
t
+
b
+
2
b
2
-
8
b
4
e
2
b
x
,
u
x
2
t
.Using (16), we estimate(47)
I
=
e
2
b
x
,
u
u
x
2
t
≤
δ
e
2
b
x
,
2
u
x
x
2
+
u
x
y
2
t
+
2
b
2
δ
+
u
0
2
δ
e
2
b
x
,
u
x
2
t
.Taking 2
δ
=
1, using (30)–(44) and substituting I into (46), we get(48)
e
2
b
x
,
u
x
x
x
N
2
+
u
x
y
N
2
t
≤
C
b
,
T
,
u
0
J
0
∀
t
∈
0
,
T
.Estimate 8.
Multiplying thejth equation of (10) by e
2
b
x
g
j
x
x
x, we come, dropping the index N, to the equality(49)
e
2
b
x
,
u
x
x
y
2
+
u
x
x
x
x
2
t
=
-
e
2
b
x
u
t
-
u
x
x
,
u
x
x
x
t
-
e
2
b
x
u
u
x
,
u
x
x
x
t
+
2
b
2
e
2
b
x
,
u
x
y
2
t
+
2
b
2
-
1
e
2
b
x
,
u
x
x
x
2
t
.
Using Lemma2 and (14), we estimate(50)
I
=
e
2
b
x
u
u
x
,
u
x
x
x
t
≤
u
t
sup
S
⁡
e
b
x
u
x
x
,
y
,
t
e
b
x
u
x
x
x
t
≤
u
0
2
2
e
2
b
x
,
u
x
x
x
2
t
+
1
2
1
δ
1
+
2
b
2
e
2
b
x
,
u
x
2
t
+
2
δ
e
2
b
x
u
x
x
2
t
+
δ
1
+
2
b
2
e
2
b
x
,
u
x
y
2
t
+
2
δ
e
2
b
x
,
u
x
x
y
2
t
.
Takingδ sufficiently small and positive and substituting I into (49), we find(51)
e
2
b
x
,
∇
u
x
x
N
2
+
∂
x
4
u
N
2
t
≤
C
b
,
T
,
u
0
J
0
∀
t
∈
0
,
T
.
Consequently, it follows from the equalities(52)
-
e
2
b
x
u
t
N
-
u
x
x
N
+
u
x
x
x
N
+
u
x
y
y
N
+
u
N
u
x
N
-
∂
x
5
u
N
,
u
y
y
N
t
=
0
,
-
e
2
b
x
u
t
N
-
u
x
x
N
+
u
x
x
x
N
+
u
x
y
y
N
+
u
N
u
x
N
-
∂
x
5
u
N
,
∂
x
5
u
N
t
=
0
,
e
2
b
x
u
t
N
-
u
x
x
N
+
u
x
x
x
N
+
u
x
y
y
N
+
u
N
u
x
N
-
∂
x
5
u
N
,
u
x
y
y
N
t
=
0that(53)
e
2
b
x
,
u
y
y
N
2
+
u
x
y
y
N
2
+
∂
x
5
u
N
2
+
u
x
x
x
y
N
2
t
≤
C
b
,
T
,
u
0
J
0
∀
t
∈
0
,
T
.
Jointly, estimates (30), (34), (35), (48), (51), and (53) read(54)
e
2
b
x
,
u
N
2
+
∇
u
N
2
+
∇
u
x
N
2
+
∇
u
y
N
2
+
∇
u
x
x
N
2
+
Δ
u
x
N
2
+
∇
u
x
x
x
N
2
+
∂
x
5
u
N
2
t
≤
C
b
,
T
,
u
0
J
0
∀
t
∈
0
,
T
.In other words,(55)
e
b
x
u
N
,
e
b
x
u
x
N
∈
L
∞
0
,
T
;
H
2
S
,
e
b
x
∇
u
x
x
x
N
,
e
b
x
∂
x
5
u
N
∈
L
∞
0
,
T
;
L
2
Sand these inclusions are uniform in N.Estimate 9.
Multiplying thejth equation of (10) by e
2
b
x
λ
j
2
g
j, we come, dropping the index N, to the equality(56)
b
e
2
b
x
,
5
u
x
x
y
y
2
+
u
y
y
y
2
t
=
2
b
2
+
4
b
3
-
16
b
5
e
2
b
x
,
u
y
y
2
t
+
20
b
3
-
3
b
-
1
e
2
b
x
,
u
x
y
y
2
t
+
e
2
b
x
u
t
y
,
u
y
y
y
t
+
e
2
b
x
u
u
x
y
,
u
y
y
y
t
+
e
2
b
x
u
y
u
x
,
u
y
y
y
t
.We estimate (57)
I
1
=
-
e
2
b
x
,
u
t
y
,
u
y
y
y
t
≤
ϵ
2
e
2
b
x
,
u
y
y
y
2
t
+
1
2
ϵ
e
2
b
x
,
u
y
t
2
t
,
I
2
=
e
2
b
x
u
y
u
x
,
u
y
y
y
t
≤
u
x
t
e
b
x
u
x
y
y
y
t
sup
S
⁡
e
b
x
u
y
x
,
y
,
t
≤
ϵ
2
e
2
b
x
,
u
x
y
y
y
2
t
+
u
x
2
t
2
ϵ
1
+
2
b
2
e
2
b
x
,
u
y
2
t
+
2
e
2
b
x
,
u
x
y
2
t
+
1
+
2
b
2
e
2
b
x
,
u
y
y
2
t
+
2
e
2
b
x
,
u
x
y
y
2
t
,
I
3
=
e
2
b
x
u
u
x
y
,
u
y
y
y
t
≤
u
t
e
b
x
u
y
y
y
t
sup
S
⁡
e
b
x
u
x
y
x
,
y
,
t
≤
u
0
2
ϵ
1
2
e
2
b
x
,
u
y
y
y
2
t
+
1
2
ϵ
1
2
δ
e
2
b
x
,
u
x
x
y
y
2
t
+
2
δ
e
2
b
x
,
u
x
x
y
2
t
+
δ
1
+
2
b
2
e
2
b
x
,
u
x
y
y
2
t
+
1
δ
1
+
2
b
2
e
2
b
x
,
u
x
y
2
t
.Choosing ϵ
,
ϵ
1
,
δ sufficiently small and positive, after integration, we transform (56) into the form(58)
∫
0
T
e
2
b
x
,
u
x
x
y
y
N
2
+
u
y
y
y
N
2
t
d
t
≤
C
b
,
T
,
u
0
J
0
.Acting similarly, we get from the scalar product (59)
e
2
b
x
u
t
N
-
u
x
x
N
+
u
x
x
x
N
+
u
x
y
y
N
+
u
N
u
x
N
-
∂
x
5
u
N
,
u
x
y
y
y
y
N
t
=
0the estimate(60)
∫
0
T
e
2
b
x
,
u
x
y
y
y
N
2
+
∂
x
3
u
y
y
N
2
t
d
t
≤
C
b
,
T
,
u
0
J
0
.Estimate 10.
Differentiate thejth equation of (10) two times with respect to x and multiply the result by e
2
b
x
∂
x
7
g
j to get the equation (61)
e
2
b
x
u
x
x
t
N
-
∂
x
4
u
N
+
∂
x
5
u
N
+
∂
x
3
u
y
y
N
+
u
N
u
x
N
x
x
-
∂
x
7
u
N
,
∂
x
7
u
N
t
=
0
,which in turn, dropping the index N, can be rewritten in the form (62)
e
b
x
∂
x
7
u
t
≤
e
b
x
u
x
x
t
+
∂
x
5
u
-
∂
x
4
u
+
∂
x
3
u
y
y
+
3
u
x
u
x
x
+
u
∂
x
3
u
t
.
Estimates (54), (55), (58), and (60) guarantee that(63)
e
b
x
u
N
,
e
b
x
u
x
N
∈
L
∞
0
,
T
;
H
2
S
∩
L
2
0
,
T
;
H
3
S
,
e
b
x
∂
x
5
u
N
∈
L
∞
0
,
T
;
L
2
S
,
e
b
x
∂
x
3
u
y
y
N
∈
L
2
0
,
T
;
L
2
Sand inclusions do not depend on N. These estimates show the smoothing effect of the Kawahara operator first observed for the KdV equation in [12] and for dispersive equations of any finite order in [27]. Independence of (14), (63) of N allows us to pass to the limit in (10) and to prove the following result.Theorem 3.
Letu
0
(
x
,
y
)
:
R
2
→
R be such that u
0
(
x
,
0
)
=
u
0
(
x
,
B
)
=
0 and for some b
>
0 satisfying the following inequality: (64)
J
0
=
u
0
2
+
e
2
b
x
,
u
0
2
+
∇
u
0
2
+
∇
u
0
x
2
+
u
0
2
u
0
x
2
+
Δ
u
0
x
2
+
∂
x
5
u
0
2
<
∞
.Then there exists a regular solution to (4)–(6), u
(
x
,
y
,
t
):(65)
u
∈
L
∞
0
,
T
;
L
2
S
,
u
x
∈
L
2
0
,
T
;
L
2
S
;
e
b
x
u
,
e
b
x
u
x
∈
L
∞
0
,
T
;
H
2
S
∩
L
2
0
,
T
;
H
3
S
;
e
b
x
u
t
∈
L
∞
0
,
T
;
L
2
S
∩
L
2
0
,
T
;
H
1
S
;
e
b
x
∂
x
5
u
∈
L
∞
0
,
T
;
L
2
S
∩
L
2
0
,
T
;
H
1
S
;
e
b
x
u
x
x
t
∈
L
2
0
,
T
;
L
2
S
,which for a
.
e
.
t
∈
(
0
,
T
) satisfies the identity(66)
e
b
x
u
t
-
u
x
x
+
u
x
x
x
+
u
u
x
+
u
x
y
y
-
∂
x
5
u
,
ϕ
x
,
y
t
=
0
,where ϕ
(
x
,
y
) is an arbitrary function from L
2
(
S
)
.Proof.
Rewrite (10) in the form(67)
e
b
x
u
t
N
-
u
x
x
N
+
u
N
u
x
N
+
u
x
x
x
N
+
u
x
y
y
N
-
∂
x
5
u
N
,
Φ
N
y
Ψ
x
t
=
0
,where Φ
N
(
y
) is an arbitrary function from the set of linear combinations ∑
i
=
1
N
‍
α
i
w
i
(
y
) and Ψ
(
x
) is an arbitrary function from H
1
(
R
). Taking into account estimates (14), (63) and fixing Φ
N, we can easily pass to the limit as N
→
∞ in linear terms of (67). To pass to the limit in the nonlinear term, we must use (36) and repeat arguments of [12]. Since linear combinations [
∑
i
=
1
N
‍
α
i
w
i
(
y
)
]
Ψ
(
x
) are dense in L
2
(
S
), we come to (66). Rewriting it in the form (68)
e
b
x
∂
x
5
u
=
e
b
x
u
t
-
u
x
x
+
u
u
x
+
u
x
x
x
+
u
x
y
y
≡
F
x
,
y
,
tand making use of (63), one can see that F
∈
L
∞
(
0
,
T
;
L
2
(
S
)
)
∩
L
2
(
0
,
T
;
H
1
(
S
)
). The proof of Theorem 3 is complete.Remark 4.
Estimates (14), (63) are valid also for the limit function u
(
x
,
y
,
t
) and (14) obtains its sharp form:(69)
u
t
2
+
2
∫
0
t
u
x
2
s
d
s
=
u
0
2
∀
t
∈
0
,
T
.
Uniqueness of a Regular SolutionTheorem 5.
The regular solution from Theorem3 is uniquely defined.Proof.
Letu
1
,
u
2 be two distinct regular solutions of (4)–(6); then z
=
u
1
-
u
2 satisfies the following initial-boundary value problem:(70)
z
t
-
z
x
x
+
z
x
x
x
+
z
x
y
y
-
∂
x
5
z
+
1
2
u
1
2
-
u
2
2
x
=
0
in
S
T
,
(71)
z
x
,
0
,
t
=
z
x
,
B
,
t
=
0
,
x
∈
R
,
t
>
0
,
(72)
z
x
,
y
,
0
=
0
.
x
,
y
∈
S
.Multiplying (70) by 2
e
b
x
z, we get(73)
d
d
t
e
2
b
x
,
z
2
t
+
2
+
6
b
-
40
b
3
e
2
b
x
,
z
x
2
t
+
10
b
e
2
b
x
,
z
x
x
2
t
-
4
b
2
+
8
b
3
-
32
b
5
e
2
b
x
,
z
2
t
+
e
2
b
x
u
1
x
+
u
2
x
,
z
2
t
+
2
b
e
2
b
x
,
z
y
2
t
-
b
e
2
b
x
u
1
+
u
2
,
z
2
t
=
0
.
We estimate(74)
I
1
=
e
2
b
x
u
1
x
+
u
2
x
,
z
2
t
≤
u
1
x
+
u
2
x
t
e
b
x
z
L
4
S
2
t
≤
2
u
1
x
+
u
2
x
t
e
b
x
z
t
∇
e
b
x
z
t
≤
δ
e
2
b
x
,
2
z
x
2
+
z
y
2
t
+
2
b
2
δ
+
2
δ
u
1
x
2
t
+
u
2
x
2
t
e
2
b
x
,
z
2
t
,
I
2
=
b
e
2
b
x
u
1
+
u
2
,
z
2
t
≤
b
u
1
+
u
2
t
e
b
x
z
L
4
S
2
t
≤
2
b
u
1
+
u
2
t
e
b
x
z
t
∇
e
b
x
z
t
≤
δ
e
2
b
x
,
2
z
x
2
+
z
y
2
t
+
2
b
2
δ
+
2
b
2
δ
u
1
2
t
+
u
2
2
t
e
2
b
x
,
z
2
t
.Substituting I
1
,
I
2 into (73) and taking δ
>
0 sufficiently small, we find(75)
d
d
t
e
2
b
x
,
z
2
t
≤
C
b
1
+
u
1
2
t
+
u
2
2
t
+
u
1
x
2
t
+
u
2
x
2
t
e
2
b
x
,
z
2
t
.
Since(76)
u
i
∈
L
∞
0
,
T
;
L
2
S
,
u
i
x
∈
L
2
0
,
T
;
L
2
S
i
=
1,2
,then, by the Gronwall lemma, (77)
e
2
b
x
,
z
2
t
=
0
∀
t
∈
0
,
T
.Hence, u
1
=
u
2
a.e. in S
T
.Remark 6.
Changing initial condition (72) for z
(
x
,
y
,
0
)
=
z
0
(
x
,
y
)
≠
0, and repeating the proof of Theorem 5, we obtain from (75) that (78)
e
2
b
x
,
z
2
t
≤
C
b
,
T
,
u
0
e
2
b
x
,
z
0
2
∀
t
∈
0
,
T
.This means continuous dependence of regular solutions on initial data.


## 4. Decay of Regular Solutions

In this section we will prove exponential decay of regular solutions in an elevated weighted norm. We start with the following theorem which is crucial for the main result.Theorem 7.
Letb
∈
(
0
,
b
0
)
,
u
0
≤
3
π
/
8
B and u
(
x
,
y
,
t
) be a regular solution of (4)–(6). Then for all finite B
>
0 the following inequalities are true:(79)
e
b
x
u
2
t
≤
e
-
χ
t
e
b
x
u
0
2
,
(80)
∫
0
t
e
χ
s
e
2
b
x
,
∇
u
2
+
u
x
x
2
s
d
s
≤
C
b
,
u
0
1
+
t
e
2
b
x
,
u
0
2
,where (81)
χ
=
b
0
π
2
4
B
2
,
b
0
=
min
⁡
0,6
2
,
1
5
-
1
+
1
+
5
π
2
4
B
2
.Proof. Multiplying (4) by 2
e
2
b
x
u, we get the equality(82)
d
d
t
e
2
b
x
,
u
2
t
+
2
+
6
b
-
40
b
3
e
2
b
x
,
u
x
2
t
+
10
b
e
2
b
x
,
u
x
x
2
t
+
2
b
e
2
b
x
,
u
y
2
t
-
4
b
3
e
2
b
x
,
u
3
t
-
4
b
2
+
8
b
3
-
32
b
5
e
2
b
x
,
u
2
t
=
0
.Taking into account (16), we estimate(83)
I
=
4
b
3
e
2
b
x
,
u
3
t
≤
b
e
2
b
x
,
u
y
2
+
2
u
x
2
+
2
b
2
u
2
t
+
16
b
9
u
0
2
e
2
b
x
,
u
2
t
.The following proposition is principal for our proof.Proposition 8.
The following inequality is true:(84)
∫
R
∫
0
B
e
2
b
x
u
2
x
,
y
,
t
d
y
d
x
≤
B
2
π
2
∫
R
∫
0
B
e
2
b
x
u
y
2
x
,
y
,
t
d
y
d
x
.Proof.
Sinceu
(
x
,
0
,
t
)
=
u
(
x
,
B
,
t
)
=
0, fixing (
x
,
t
), we can use with respect to y the following Steklov inequality: if f
(
y
)
∈
H
0
1
(
0
,
π
), then (85)
∫
0
π
f
2
y
d
y
≤
∫
0
π
f
y
y
2
d
y
.After a corresponding process of scaling we prove Proposition 8.Making use of (84) and the fact that b
∈
(
0
,
b
0
) implies 2
b
0
(
3
-
20
b
0
2
)
≥
0 and substituting I into (82), we come to the following inequality:(86)
d
d
t
e
2
b
x
,
u
2
t
+
e
2
b
x
,
u
x
2
t
+
b
π
2
B
2
-
4
b
2
-
10
b
3
-
16
b
9
u
0
2
e
2
b
x
,
u
2
t
≤
0
,which can be rewritten as(87)
d
d
t
e
2
b
x
,
u
2
t
+
χ
e
2
b
x
,
u
2
t
≤
0
,where (88)
χ
=
b
π
2
B
2
-
4
b
-
10
b
2
-
16
u
0
2
9
.Since we need χ
>
0, define(89)
4
b
+
10
b
2
=
γ
π
2
B
2
,
16
u
0
2
9
=
1
-
γ
2
π
2
B
2
,where γ
∈
(
0,1
). It implies χ
=
b
A
γ
π
2
/
B
2 with A
(
γ
)
=
γ
(
1
-
γ
).It is easy to see that(90)
sup
γ
∈
0,1
⁡
A
γ
=
A
1
2
=
1
4
.Solving (89), we find (91)
b
=
1
5
-
1
+
1
+
5
π
2
4
B
2
,
u
0
≤
3
π
8
B
,
χ
=
b
π
2
4
B
2
,and from (87) we get (92)
e
2
b
x
,
u
2
t
≤
e
-
χ
t
e
2
b
x
,
u
0
2
.The last inequality implies (79).To prove (80), we return to (82) and multiply it by e
χ
t to obtain(93)
d
d
t
e
χ
t
e
2
b
x
,
u
2
t
+
e
χ
t
2
+
6
b
2
-
40
b
3
e
2
b
x
,
u
x
2
t
+
2
b
e
2
b
x
,
u
y
2
t
+
10
b
e
2
b
x
,
u
x
x
2
t
=
4
b
e
χ
t
3
e
2
b
x
,
u
3
t
+
e
χ
t
χ
+
4
b
2
+
8
b
3
-
32
b
5
e
2
b
x
,
u
2
t
.We find(94)
I
=
4
b
3
e
2
b
x
,
u
3
t
≤
b
e
2
b
x
,
2
u
x
2
+
u
y
2
t
+
10
b
e
2
b
x
,
u
x
x
2
t
+
2
b
2
+
16
b
u
0
2
9
e
2
b
x
,
u
2
t
.Substituting this into (93), we get(95)
d
d
t
e
χ
t
e
2
b
x
,
u
2
t
+
e
χ
t
e
2
b
x
,
∇
u
2
t
+
10
b
e
χ
t
e
2
b
x
,
u
x
x
2
t
≤
C
b
,
u
0
e
χ
t
e
2
b
x
,
u
2
t
.Integration and (79) imply(96)
e
χ
t
e
2
b
x
,
u
2
t
+
∫
0
t
e
χ
s
e
2
b
x
,
∇
u
2
+
u
x
x
2
s
d
s
≤
C
b
,
u
0
1
+
t
e
2
b
x
,
u
0
2
.The proof of Theorem 7 is complete.Observe that, differently from [24, 25, 38], we do not have any restrictions on the width of a strip B.The main result of this section is the following assertion.Theorem 9.
Let all the conditions of Theorem7 be fulfilled. Then regular solutions of (4)–(6) satisfy the following inequality:(97)
e
2
b
x
,
u
2
+
∇
u
2
+
u
x
x
2
t
≤
C
b
,
u
0
1
+
t
e
-
χ
t
e
2
b
x
,
u
0
2
+
∇
u
0
2
+
u
0
x
x
2or (98)
e
b
x
u
H
1
S
2
t
+
e
b
x
u
x
x
2
t
≤
C
b
,
u
0
1
+
t
e
-
χ
t
e
2
b
x
,
u
0
2
+
∇
u
0
2
+
u
0
x
x
2
.Proof. We start with the following lemma.Lemma 10.
Regular solutions of (4)–(6) satisfy the following equality:(99)
e
χ
t
e
2
b
x
,
∇
u
2
+
u
x
x
2
t
+
2
∫
0
t
e
χ
s
1
+
3
b
-
20
b
3
e
2
b
x
,
∇
u
x
2
+
u
x
x
x
2
s
+
b
e
2
b
x
,
∇
u
y
2
+
u
x
x
y
2
s
+
5
b
e
2
b
x
,
∇
u
x
x
2
+
∂
x
4
u
2
s
d
s
=
∫
0
t
e
χ
s
χ
+
4
b
2
+
8
b
3
-
32
b
5
e
2
b
x
,
∇
u
2
+
u
x
x
2
s
d
s
+
∫
0
t
e
χ
s
4
b
e
2
b
x
u
,
u
x
2
s
+
2
e
2
b
x
u
u
x
,
u
y
y
+
4
b
2
u
x
x
+
4
b
u
x
x
x
+
∂
x
4
u
s
-
e
2
b
x
u
2
,
u
x
x
x
+
2
b
u
x
x
s
d
s
+
e
2
b
x
,
∇
u
0
2
+
u
0
x
x
2
.Proof.
First we transform the scalar product(100)
-
2
e
2
b
x
u
t
-
u
x
x
+
u
x
x
x
+
u
x
y
y
+
u
u
x
-
∂
x
5
u
,
u
y
y
+
u
x
x
-
∂
x
4
u
t
=
0into the following equality:(101)
d
d
t
e
2
b
x
,
∇
u
2
+
u
x
x
2
t
+
2
1
+
3
b
-
20
b
3
e
2
b
x
,
∇
u
x
2
+
u
x
x
x
2
t
+
2
b
e
2
b
x
,
∇
u
y
2
+
u
x
x
y
2
t
+
10
b
e
2
b
x
,
∇
u
x
x
2
+
∂
x
4
u
2
t
=
4
b
2
1
+
2
b
-
8
b
3
e
2
b
x
,
∇
u
2
+
u
x
x
2
t
+
2
e
2
b
x
u
u
x
,
u
y
y
+
4
b
2
u
x
x
+
4
b
u
x
x
x
+
∂
x
4
u
t
-
e
2
b
x
u
2
,
u
x
x
x
+
2
b
u
x
x
t
+
4
b
e
2
b
x
,
u
u
x
2
t
.Multiplying this by e
χ
t and integrating, we prove (99).Making use of Lemma2, estimate separate terms in (99) as follows:(102)
I
1
=
2
e
2
b
x
u
u
x
,
u
y
y
+
4
b
2
u
x
x
+
4
b
u
x
x
x
+
∂
x
4
u
t
≤
2
u
t
sup
S
⁡
e
b
x
u
x
t
e
b
x
u
y
y
+
4
b
2
u
x
x
+
4
b
u
x
x
x
+
∂
x
4
u
t
≤
ϵ
1
+
u
0
2
e
2
b
x
,
u
y
y
2
+
u
x
x
2
+
u
x
x
x
2
+
∂
x
4
u
2
t
+
C
b
ϵ
2
δ
e
2
b
x
,
u
x
y
2
+
u
x
x
y
2
t
+
3
δ
e
2
b
x
,
u
x
x
2
+
u
x
2
t
,
I
2
=
4
b
e
2
b
x
,
u
u
x
2
s
≤
δ
e
2
b
x
,
2
u
x
x
2
+
u
x
y
2
s
+
2
b
2
δ
+
16
1
+
4
b
2
u
0
2
δ
e
2
b
x
,
u
x
2
s
,
I
3
=
e
2
b
x
u
2
,
u
x
x
x
+
2
b
u
x
x
t
≤
u
t
sup
S
⁡
e
b
x
u
t
e
b
x
u
x
x
x
+
2
b
u
x
x
t
≤
ϵ
2
e
2
b
x
,
u
x
x
x
2
+
u
x
x
2
t
+
C
b
u
0
2
δ
e
2
b
x
,
u
x
y
2
t
+
1
δ
2
e
2
b
x
,
∇
u
2
+
u
2
t
.Choosingϵ
,
δ sufficiently small, substituting I
1
–
I
3 into (99), and taking into account (79), we prove that(103)
e
χ
t
e
2
b
x
,
∇
u
2
+
u
x
x
2
t
≤
C
b
,
u
0
1
+
t
e
2
b
x
,
u
0
2
+
∇
u
0
2
+
u
0
x
x
2
.Adding (79), we complete the proof of Theorem 9.


## 5. Weak Solutions

Here we will prove the existence and uniqueness and continuous dependence on initial data as well as exponential decay results for weak solutions of (4)–(6) when the initial function u
0
∈
L
2
(
S
).Theorem 11.
Letu
0
∈
L
2
(
S
)
∩
L
b
2
(
S
). Then for all finite positive T and B there exists at least one function u
(
x
,
y
,
t
): (104)
u
∈
L
∞
0
,
T
;
L
2
S
,
u
x
∈
L
2
0
,
T
;
L
2
Ssuch that (105)
e
b
x
u
∈
L
∞
0
,
T
;
L
2
S
∩
L
2
0
,
T
;
H
1
S
,
e
b
x
u
x
x
∈
L
2
0
,
T
;
L
2
Sand the following integral identity takes place:(106)
e
b
x
u
,
v
T
+
∫
0
T
-
e
b
x
u
,
v
t
t
-
1
2
e
2
b
x
u
2
,
b
v
+
v
x
t
+
e
b
x
u
x
x
,
v
x
x
x
+
3
b
v
x
x
+
3
b
2
-
1
v
x
+
b
3
-
b
-
1
v
t
+
e
b
x
u
y
,
b
v
y
+
v
x
y
t
d
t
=
e
b
x
u
0
,
v
x
,
y
,
0
,where v
∈
C
∞
(
S
T
) is an arbitrary function.Proof.
In order to justify our calculations, we must operate with sufficiently smooth solutionsu
m
(
x
,
y
,
t
). With this purpose, we consider first initial functions u
0
m
(
x
,
y
), which satisfy conditions of Theorem 3, and obtain estimates (14), (36) for functions u
m
(
x
,
y
,
t
). This allows us to pass to the limit as m
→
∞ in the identity(107)
e
b
x
u
m
,
v
T
+
∫
0
T
-
e
b
x
u
m
,
v
t
t
-
1
2
e
2
b
x
u
m
2
,
b
v
+
v
x
t
+
e
b
x
u
x
x
m
,
v
x
x
x
+
3
b
v
x
x
+
3
b
2
-
1
v
x
+
b
3
-
b
-
1
v
t
+
e
b
x
u
y
m
,
b
v
y
+
v
x
y
t
d
t
=
e
b
x
u
0
m
,
v
x
,
y
,
0and come to (106).Uniqueness of a Weak SolutionTheorem 12.
A weak solution of Theorem11 is uniquely defined.Proof.
Actually, this proof is provided by Theorem5. It is sufficient to approximate the initial function u
0
∈
L
2
(
S
) by regular functions u
0
m in the form: (108)
lim
m
→
∞
⁡
u
0
m
-
u
0
=
0
,where u
0
m satisfies the conditions of Theorem 3. This guarantees the existence of the unique regular solution to (4)–(6) and allows us to repeat all the calculations which have been done during the proof of Theorem 5 and to come to the following inequality:(109)
d
d
t
e
2
b
x
,
z
m
2
t
+
e
2
b
x
,
∇
z
m
2
t
≤
C
b
1
+
u
1
m
2
t
+
u
2
m
2
t
+
u
1
x
m
2
t
+
u
2
x
m
2
t
e
2
b
x
,
z
m
2
t
.By the generalized Gronwall’s lemma,(110)
e
2
b
x
,
z
m
2
t
≤
exp
⁡
∫
0
t
C
b
1
+
u
1
m
2
s
+
u
2
m
2
s
+
u
1
x
m
2
s
+
u
2
x
m
2
s
d
s
e
2
b
x
,
z
0
m
2
t
.Functions u
1
m and u
2
m for m sufficiently large satisfy the estimate(111)
u
i
m
2
t
+
2
∫
0
t
u
i
m
x
2
s
d
s
=
u
0
m
2
≤
2
u
0
2
,
i
=
1,2
.
Hence,(112)
exp
⁡
∫
0
t
C
b
1
+
u
1
m
2
s
+
u
2
m
2
s
+
u
1
x
m
2
s
+
u
2
x
m
2
s
d
s
≤
C
b
,
T
,
u
0
.Since e
b
x
z
(
x
,
y
,
t
) is a weak limit of regular solutions {
e
b
x
z
m
(
x
,
y
,
t
)
}, then (113)
e
2
b
x
,
z
2
t
≤
e
2
b
x
,
z
m
2
t
=
0
.This implies u
1
≡
u
2
a.e. in S
T
. The proof of Theorem 12 is complete.Remark 13.
Changing initial conditionz
(
x
,
y
,
0
)
≡
0 for z
(
x
,
y
,
0
)
=
z
0
(
x
,
y
)
≠
0, and repeating the proof of Theorem 12, we obtain that (114)
e
2
b
x
,
z
2
t
≤
C
b
,
T
,
u
0
e
2
b
x
,
z
0
2
∀
t
∈
0
,
T
.This means continuous dependence of weak solutions on initial data.Decay of Weak SolutionsTheorem 14.
Letb
∈
(
0
,
b
0
)
,
u
0
≤
3
π
/
16
B, and u
(
x
,
y
,
t
) be a regular solution of (4)–(6). Then for all finite B
>
0 the following inequality is true: (115)
e
b
x
u
2
t
≤
e
-
χ
t
e
b
x
u
0
2
0
,where (116)
χ
=
b
0
π
2
4
B
2
,
b
0
=
min
⁡
0,6
2
,
1
5
-
1
+
1
+
5
π
2
4
B
2
.Proof.
Similar to the proof of the uniqueness result for a weak solution, we approximateu
0
∈
L
2
(
S
) by sufficiently smooth functions u
0
m in order to work with regular solutions. Acting in the same manner as by the proof of Theorem 7, we come to the following inequality:(117)
e
b
x
u
m
2
t
≤
e
-
χ
t
e
b
x
u
0
2
0
,where (118)
χ
=
b
0
π
2
4
B
2
.Since u
(
x
,
y
,
t
) is a weak limit of regular solutions {
u
m
(
x
,
y
,
t
)
}, then (119)
e
2
b
x
,
u
2
t
≤
e
2
b
x
,
u
m
2
t
≤
e
-
χ
t
e
2
b
x
,
u
0
2
.The proof of Theorem 14 is complete.We have in this theorem a more strict conditionu
0
≤
3
π
/
16
B instead of u
0
≤
3
π
/
8
B in the case of decay for regular solution because for weak solutions we do not have the sharp estimate (69), but only (14).Remark 15.
We have established our results on the existence and uniqueness of regular and weak solutions as well as their exponential decay without restrictions on the widthB due to absence of the travel term u
x. Of course, the presence of this term will cause restrictions on B as in [38].Remark 16.
The rate of exponential decay in Theorem7 and Theorem 14 is optimal for the fixed B
>
0: (120)
e
b
x
u
2
t
≤
e
-
χ
t
e
b
x
u
0
2
,where (121)
χ
=
b
0
π
2
4
B
2
b
0
=
min
⁡
0,6
2
,
1
5
-
1
+
1
+
5
π
2
4
B
2
.Remark 17.
The results established in Theorems7 and 14 show that a hot plasma simulated by system (4)–(6) can be stabilized by suitable boundary conditions provided perturbations of the initial state are sufficiently small.



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*Source: 269536-2015-12-10.xml*