14 / 17
Moreover, for
q
(
x
)
∈ D
(Ω)
we have:
(
u, q
)
e
H
= (
∇
u,
∇
q
)
L
2
(Ω)
= (
√
Lu,
√
Lq
)
L
2
(Ω)
=
=
−
∞
Z
0
λ
cos(
√
λt
)
d
(
E
(
λ
)
f, q
)
−
∞
Z
0
√
λ
sin(
√
λt
)
d
(
E
(
λ
)
g, q
) =
= 2
∞
Z
0
cos(
zt
)
z
3
m
1
(
z
2
)
dz
+ 2
∞
Z
0
z
2
sin(
zt
)
m
2
(
z
2
)
dz.
(47)
By (43) and (45) the following integrals are finite:
∞
Z
0
z
3
|
m
1
(
z
2
)
|
dz
+
∞
Z
0
z
2
|
m
2
(
z
2
)
|
dz
=
=
1
2
∞
Z
0
λ
|
m
1
(
λ
)
|
dλ
+
1
2
∞
Z
0
√
λ
|
m
2
(
λ
)
|
dλ <
∞
.
Therefore, applying the Riemann – Lebesgue Lemma to (47), we obtain
lim
t
→∞
(
u
(
t, x
)
, q
(
x
))
e
H
= 0
.
(48)
The space
D
(Ω)
is dense in
e
H
. After closure the relation (48) with respect
to
q
we obtain (48) for all
q
∈
e
H
. Let us prove that for any
R >
0
lim
t
→∞
k
u
t
(
t, x
)
k
L
2
(Ω
R
)
= 0
.
(49)
By (32) the set of functions
{
u
t
(
t, x
)
}
,
t >
0
is a compact set in
L
2
(Ω
R
)
.
Let
{
h
j,R
(
x
)
}
,
j
= 1
,
2
, . . .
, be an orthonormal basis in
L
2
(Ω
R
)
. By the
compactness criterion [21], for any
ε >
0
there exists
N >
0
such that
u
t
(
t, x
) =
N
P
j
=1
c
j,R
(
t
)
h
j,R
(
x
) +
∞
P
j
=
N
+1
c
j,R
(
t
)
h
j,R
(
x
)
for
t >
0
,
x
∈
Ω
R
and
k
∞
P
j
=
N
+1
c
j,R
h
j,R
k
L
2
(Ω
R
)
< ε
for all
t >
0
. Therefore,
k
u
t
k
2
L
2
(Ω
R
)
=
k
N
X
j
=1
c
j,R
h
j,R
k
2
L
2
(Ω
R
)
+
k
∞
X
j
=
N
+1
c
j,R
h
j,R
k
2
L
2
(Ω
R
)
<
<
N
X
j
=1
c
2
j,R
+
ε
2
=
N
X
j
=1
(
u
t
, h
j,R
)
2
L
2
(Ω)
+
ε
2
.
(50)
By the relation (46)
lim
t
→∞
(
u
t
, h
j,R
)
L
2
(Ω)
= 0
for
j
= 1
,
2
, . . . , N
. Apply (50)
16
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2015. № 3