lim
t
→∞
1
t
t
Z
0
|
s
(
τ
)
|
2
dτ
= 0
.
(23)
Now we prove the main result of this section.
Theorem 5.
Let
σ
p
(
L
) =
∅
. Then for all
f
∈
o
H
1
(Ω)
,
g
∈
L
2
(Ω)
and
all bounded domains
Ω
0
⊂
Ω
lim
t
→∞
1
t
t
Z
0
E
Ω
0
(
τ
)
dτ
= 0
.
(24)
J
It is sufficient to prove the relation (24) for
f, g
∈ D
(Ω)
. For any
f
∈
o
H
1
(Ω)
,
g
∈
L
2
(Ω)
and arbitrary
ε >
0
there exist
˜
f ,
˜
g
∈ D
(Ω)
such
that
k
f
−
˜
f
k
H
1
(Ω)
< ε
,
k
g
−
˜
g
k
L
2
(Ω)
< ε
. Therefore, for
Ω
0
= Ω
R
= Ω
∩{|
x
|
<
< R
}
we have
E
Ω
R
(
t
) =
Z
Ω
R
u
2
t
+
|∇
u
|
2
dx
≤
2
Z
Ω
R
˜
u
2
t
+
|∇
˜
u
|
2
dx
+
+2
Z
Ω
((
u
−
˜
u
)
t
)
2
+
|∇
(
u
−
˜
u
)
|
2
dx
= 2
Z
Ω
R
˜
u
2
t
+
|∇
˜
u
|
2
dx
+
+2
k
f
−
˜
f
k
2
H
1
(Ω)
+
k
g
−
˜
g
k
2
L
2
(Ω)
<
2
Z
Ω
R
˜
u
2
t
+
|∇
˜
u
|
2
dx
+ 4
ε
2
for
t >
0
. Now, to prove (24) suppose that
f, g
∈ D
(Ω)
. Note that
f, g
∈ D
(Ω)
⊂
D
(
L
p
)
,
p
= 1
,
2
, . . .
, where
D
(
L
p
)
is a domain of the
p
-th
power of the Laplace operator
L
=
−
Δ
with Dirichlet boundary condition.
Thence [17, Ch. 9, Par. 1] for
f, g
∈ D
(Ω)
we have the inequalities
∞
Z
0
λ
2
p
d
(
E
(
λ
)
f, f
)
<
∞
;
∞
Z
0
λ
2
p
d
(
E
(
λ
)
g, g
)
<
∞
, p
= 0
,
1
,
2
, . . .
(25)
Now, by the operator calculus for self-adjoint operators [17, Ch. 9, Par. 1]
for
q
(
x
)
∈ D
(Ω)
and
t >
0
we obtain the relations
(
u
t
, q
)
L
2
(Ω)
=
−
∞
Z
0
sin(
√
λt
)
√
λd
(
E
(
λ
)
f, q
) +
∞
Z
0
cos(
√
λt
)
d
(
E
(
λ
)
g, q
) =
=
∞
Z
0
sin
ztdm
1
(
z
) +
∞
Z
0
cos
ztdm
2
(
z
);
(26)
10
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2015. № 3