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 the
j
th 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 positive
b
, multiply the
j
th 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 all
u
∈
H
1
(
R
2
)
,
(16)
u
L
4
R
2
2
≤
2
u
L
2
R
2
∇
u
L
2
R
2
.
(ii) For all
u
∈
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
)
.
Extending
u
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
2
with 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.
Let
u
(
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.
Denote
v
=
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 the
j
th 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 the
j
th 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
2
which 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 the
j
th 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 the
j
th 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 the
j
th 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 Lemma 2 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
.
Taking
δ
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
=
0
that
(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
S
and these inclusions are uniform in
N
.
Estimate 9.
Multiplying the
j
th 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
=
0
the 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 the
j
th 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
S
and 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.
Let
u
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
,
t
and 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 Solution
Theorem 5.
The regular solution from Theorem 3 is uniquely defined.
Proof.
Let
u
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.
Let
b
∈
(
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.
Since
u
(
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 Theorem 7 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
2
or
(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
=
0
into 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 Lemma 2, 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
.
Choosing
ϵ
,
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.
Let
u
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
S
such 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
S
and 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 solutions
u
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
,
0
and come to (106).
Uniqueness of a Weak Solution
Theorem 12.
A weak solution of Theorem 11 is uniquely defined.
Proof.
Actually, this proof is provided by Theorem 5. 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 condition
z
(
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 Solutions
Theorem 14.
Let
b
∈
(
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 approximate
u
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 condition
u
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 width
B
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 Theorem 7 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 Theorems 7 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.