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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

)

|

+

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

R

)

= 0

.

(49)

By (32) the set of functions

{

u

t

(

t, x

)

}

,

t >

0

is a compact set in

L

2

R

)

.

Let

{

h

j,R

(

x

)

}

,

j

= 1

,

2

, . . .

, be an orthonormal basis in

L

2

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

R

)

< ε

for all

t >

0

. Therefore,

k

u

t

k

2

L

2

R

)

=

k

N

X

j

=1

c

j,R

h

j,R

k

2

L

2

R

)

+

k

X

j

=

N

+1

c

j,R

h

j,R

k

2

L

2

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