13 / 17
Theorem 6.
Let
σ
(
L
) =
σ
ac
(
L
)
. Then for all
f
∈
o
H
1
(Ω)
,
g
∈
L
2
(Ω)
and all bounded domains
Ω
0
⊂
Ω
lim
t
→∞
E
Ω
0
(
t
) = 0
.
(41)
J
We can assume without loss of generality (as in the proof of
Theorem 5) that
f, g
∈ D
(Ω)
⊂
D
(
L
p
)
for all
p
= 1
,
2
, . . .
So, we can
prove (41) for this case and suppose that the inequalities (25) holds. Now,
for an arbitrary function
q
(
x
)
∈ D
(Ω)
we have the equality
(
u
t
, q
)
L
2
(Ω)
=
−
∞
Z
0
sin(
√
λt
)
√
λd
(
E
(
λ
)
f, q
)+
+
∞
Z
0
cos(
√
λt
)
d
(
E
(
λ
)
g, q
)
.
(42)
It follows from
σ
(
L
) =
σ
ac
(
L
)
and (25) that
d
(
E
(
λ
)
f, q
) =
m
1
(
λ
)
dλ
,
d
(
E
(
λ
)
g, q
) =
m
2
(
λ
)
dλ
where
∞
Z
0
λ
2
p
|
m
1
(
λ
)
|
dλ <
∞
;
∞
Z
0
λ
2
p
|
m
2
(
λ
)
|
dλ <
∞
(43)
for
p
= 0
,
1
,
2
, . . .
Therefore, by (42) we have
(
u
t
, q
)
L
2
(Ω)
=
−
∞
Z
0
√
λ
sin(
√
λt
)
m
1
(
λ
)
dλ
+
∞
Z
0
cos(
√
λt
)
m
2
(
λ
)
dλ
=
=
−
2
∞
Z
0
sin(
zt
)
z
2
m
1
(
z
2
)
dz
+ 2
∞
Z
0
cos(
zt
)
zm
2
(
z
2
)
dz.
(44)
It now follows from the inequality
√
λ
≤
λ
+ 1
2
(45)
and (43) that
∞
Z
0
z
2
|
m
1
(
z
2
)
|
dz
+
∞
Z
0
z
|
m
2
(
z
2
)
|
dz
=
=
1
2
∞
Z
0
√
λ
|
m
1
(
λ
)
|
dλ
+
1
2
∞
Z
0
|
m
2
(
λ
)
|
dλ <
∞
.
Now, applying the Riemann – Lebesgue Lemma to (44), we obtain
lim
t
→∞
(
u
t
, q
)
L
2
(Ω)
= 0
.
(46)
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2015. № 3
15