6 / 17
0 = lim
T
→∞
1
T
T
Z
0
Z
Ω
0
u
x
k
(
t, x
)
η
(
x
)
dx
sin(
p
λ
m
t
)
dt
=
=
1
2
Z
Ω
0
X
λ
j
=
λ
m
b
j
p
λ
j
(
v
j
)
x
k
(
x
)
η
(
x
)
dx
(17)
for
m
= 1
,
2
, . . .
The equalities (16), (17) mean that
X
λ
j
=
λ
m
a
j
(
v
j
)
x
k
(
x
) = 0;
X
λ
j
=
λ
m
b
j
p
λ
j
(
v
j
)
x
k
(
x
) = 0
(18)
for
x
∈
Ω
0
and
m
= 1
,
2
, . . .
Using the analyticity of eigenfunctions in
Ω
we obtain from (18) that
X
λ
j
=
λ
m
a
j
(
v
j
)
x
k
(
x
) = 0;
X
λ
j
=
λ
m
b
j
p
λ
j
(
v
j
)
x
k
(
x
) = 0
, k
= 1
, . . . , n,
for
x
∈
Ω
and
m
= 1
,
2
, . . .
It now follows from the boundary condition
(10) that
X
λ
j
=
λ
m
a
j
v
j
(
x
) = 0
,
X
λ
j
=
λ
m
b
j
p
λ
j
v
j
(
x
) = 0
for
x
∈
Ω
and
m
= 1
,
2
, . . .
By the orthogonality of eigenfunctions we have
a
j
=
b
j
= 0
for all
j
:
λ
j
=
λ
m
,
m
= 1
,
2
, . . .
Proof of Theorem 2 is complete.
I
In the case of unbounded domain we say that the energy scatters to
infinity if for any bounded
Ω
0
⊂
Ω
lim
t
→∞
E
Ω
0
(
t
) = 0
.
(19)
The following theorem means that in the case of
σ
p
(
L
)
6
=
∅
(the continuous
spectrum
σ
c
(
L
)
can be non-empty too) the relation (19) does not holds even
for smooth and finite initial functions.
Theorem 3.
Let
σ
p
(
L
)
6
=
∅
. Then there exist the functions
f, g
∈ D
(Ω)
and domain
Ω
0
is compact embedded to
Ω
such that
lim inf
t
→∞
E
Ω
0
(
t
)
>
0
.
(20)
J
Let
λ
∈
σ
p
(
L
)
,
v
(
x
)
∈
o
H
1
(Ω)
is a corresponding eigenfunction. It
is sufficient to consider
λ >
0
because
0
/
∈
σ
p
(
L
)
. Really, for
λ
= 0
an
eigenfunction is a harmonic function from
o
H
1
(Ω)
and vanishes in
Ω
[2,
Ch. 2, Par. 4, no. 4.7]. Consider the solution
u
(
t, x
) = cos(
√
λt
)
v
(
x
)
of
the problem (8)–(10) with
f
=
v
,
g
= 0
we obtain
E
Ω
0
(
t
) =
1
2
Z
Ω
0
|∇
v
|
2
+
λv
2
dx
+
cos(2
√
λt
)
2
Z
Ω
0
|∇
v
|
2
−
λv
2
dx.
(21)
8
ISSN 1812-3368. Вестник МГТУ им. Н.Э. Баумана. Сер. “Естественные науки”. 2015. № 3