15 / 17
we conclude that for any
ε >
0 lim sup
t
→∞
k
u
t
k
2
L
2
(Ω
R
)
≤
ε
2
.
It means that
lim
t
→∞
k
u
t
k
L
2
(Ω
R
)
= 0
.
(51)
Let us prove now that for any
R >
0 lim
t
→∞
k∇
u
k
L
2
(Ω
R
)
= 0
.
It now follows
from (32) that the set of functions
{
u
(
t, x
)
}
,
t >
0
, is a compact set in
the space
e
H
R
. Let
{
h
j,R
(
x
)
}
,
j
= 1
,
2
, . . .
, be an orthonormal basis in
e
H
R
. By the compactness criterion in the space
e
H
R
with basis
{
h
j,R
}
for
any
ε >
0
there exists
N >
0
such that
u
(
t, x
) =
N
P
j
=1
b
j,R
(
t
)
h
j,R
(
x
) +
+
∞
P
j
=
N
+1
b
j,R
(
t
)
h
j,R
(
x
)
,
and
k
∞
P
j
=
N
+1
b
j,R
h
j,R
k
e
H
R
< ε
for all
t >
0
.
Therefore,
k
u
k
2
e
H
R
=
k
N
X
j
=1
b
j,R
h
j,R
k
2
e
H
R
+
k
∞
X
j
=
N
+1
b
j,R
h
j,R
k
2
e
H
R
<
<
N
X
j
=1
b
2
j,R
+
ε
2
=
N
X
j
=1
(
u, h
j,R
)
2
e
H
R
+
ε
2
=
N
X
j
=1
u,
˜
h
j,R
2
e
H
+
ε
2
,
(52)
where the functions
˜
h
j,R
∈
e
H
satisfy the equality
v,
˜
h
j,R
e
H
= (
v, h
j,R
)
e
H
R
for all
v
∈
e
H
. By the relation (48) we obtain
lim
t
→∞
u,
˜
h
j,R
e
H
= 0
for
j
= 1
, . . . , N
. Apply the equality (52), we have for any
ε >
0
lim sup
t
→∞
k∇
u
k
2
L
2
(Ω
R
)
= lim sup
t
→∞
k
u
k
2
e
H
R
≤
N
X
j
=1
lim
t
→∞
u,
˜
h
j,R
2
e
H
+
ε
2
=
ε
2
.
In other words,
lim
t
→∞
k∇
u
k
L
2
(Ω
R
)
= 0
.
(53)
Now, for any bounded
Ω
0
⊂
Ω
we take
R
sufficiently large such that
Ω
0
⊂
Ω
R
. Finally, combining the relations (51) and (53), we get the equality
(41). Proof of the Theorem 6 is complete.
I
Conclusion.
The results explained in the previous sections show how
the basic information about spectrum of the Laplace operator allows us
to study the qualitative properties of solutions of the mixed problem to
hyperbolic equation. To obtain the rate of decay for local energy function
E
Ω
0
(
t
)
we need the estimates to resolvent function of the Laplace operator
in the complex domain [2, 11–13].
This work was supported by the RFBR Grant (no. 11-01-00989).
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2015. № 3
17