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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

)

,

d

(

E

(

λ

)

g, q

) =

m

2

(

λ

)

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

)

+

Z

0

cos(

λt

)

m

2

(

λ

)

=

=

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

)

|

+

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