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Об одной обратной краевой задаче для гиперболического уравнения второго порядка с интегральным условием первого рода.

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F T ra n sf o
A B B Y Y.c
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F T ra n sf o
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- 517.95
w.
A B B Y Y.c
.
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,
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,
,
.
:
,
,
,
.
1.
,
.
.
,
[1],
[2]
[3],
.
2.
utt (x,t) uxx(x,t) a(t)u(x,t) f (x,t)
DT
{( x, t ) : 0
(1)
x 1, 0 t T }
u (x ,0)
(x ), ut (x ,0)
u x (0, t ) 0
(x ) (0 x 1) ,
(2)
(0 t T ) ,
(3)
1
u( x, t )dx
0
(0 t T )
0
u (0, t )
2
,
2
h(t ) (0 t
T),
(5)
0 ), f ( x, t ), ( x), ( x ), h(t ) -
(
a(t ) -
u (1, t )
(4)
u ( x, t )
,
.
.
{u ( x, t ), a(t )}
(1)-(5)
a(t ) ,
u ( x, t )
:
1)
u ( x, t )
2)
3)
a(t )
DT
,
(1);
[ 0, T ] ;
(1)-(5)
.
1.
1
( x) C[0,1],
1
( x)dx 0, ( x) C[0,1],
0
( x)dx 0 ,
0
1
h(t ) C 2 [0, T ], h (t )
0, f ( x, t ) C ( DT ),
f ( x, t ) dx
0 (0 t
T ),
0
( 0)
(1)
h(0),
( 0)
(1)
h (0).
(1)-(5)
u ( x, t )
a(t ) ,
1)
2)
(1)-(5),
(1),(2),
u x (1, t )
h (t ) ( u xx (0, t )
u xx (1, t ))
0 (0
a (t ) h (t )
t
T ),
f ( 0, t )
(6)
f (1, t ) (0
t
T).
(7)
om
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F T ra n sf o
A B B Y Y.c
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F T ra n sf o
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Y
u ( x , t ), a (t )
.
(1)
0
x,
1
w.
A B B Y Y.c
(1)-(5).
:
2 1
1
d
u ( x, t )dx u x (1, t ) u x (0, t )
d t2 0
1
a(t ) u ( x, t )dx
f ( x, t )dx (0 t
0
T)
.
0
(8)
1
f ( x, t )dx
,
0
(3)
0
(6).
2
h(t ) C [0, T ] ,
,
(5)
utt (0, t )
utt (1, t )
,
:
h (t ) (0
t
(1),
:
u tt (0, t )
u tt (1, t ) ( u xx (1, t )
u xx (1, t ))
a (t )( u (0, t )
u (1, t ))
f ( 0, t )
f (1, t ) 0 t T .
,
(5) (9),
u ( x , t ), a (t )
,
,
(3) (6),
T)
.
(9)
(10)
(7).
(1)-(3),(6),(7).
(8),
:
y (t ) a (t ) y (t )
0 (0 t T ) ,
(11)
1
y (t )
u ( x, t )dx
0
1,
1
y (0)
u ( x,0) dx
1
( x)dx 0
y (0)
(13)
0
( x )dx 0
.
0
y (t ) 0 (0 t T ) .
,
(10),
1
ut ( x,0) dx
,
0
(11),
(4).
, (7)
(12)
,
1
0
(0 t T ) .
(13)
,
(12),
:
2
d
( u (0, t )
dt 2
u (1, t ) h(t ))
a (t )( u (0, t )
(1)
( 0)
h (0) ,
( 0)
u (0,0)
(14)
h(0),
u (1,0) h(0)
(15),
T)
.
ut (0,0)
0,
u (1, t ) h (t )) (0 t
(1)
ut (1,0) h (0)
,
(14)
:
0
.
(15)
(5).
.
3.
u ( x, t )
{u ( x, t ), a (t )}
u ( x, t )
(1)-(3),(6),(7)
u k (t ) cos
k
(
x
k
:
k )
,
k 0
(16)
1
uk (t )
2 u ( x, t ) cos
k
xdx ( k
0,1,...)
.
0
,
,
uk (t )
(1)
2
k
uk (t )
u k (0)
(2)
:
Fk (t ; u , a) (0 t
k , u k (0 )
k
T; k
(k
0,1,...) ,
0,1,...),
1
Fk (t ; u , a )
f k (t ) a(t )uk (t ), f k (t )
2 f ( x, t ) cos
0
k
xdx
,
(17)
(18)
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A B B Y Y.c
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1
1
2
k
( x) cos
k
xdx,
2
k
0
( x) cos
k
xdx (k
A B B Y Y.c
0,1,...)
.
0
(17), (18),
w.
:
t
u 0 (t )
t
0
(t
0
) F0 ( ; u , a) d
,
0
u k (t )
k
1
cos k t
k
sin
1
t
k
k
(19)
t
Fk ( ; u, a) sin k (t
)d
(k 1,2,...)
.
k 0
(20)
,
t
u k (t )
k sin
k
kt
k cos
kt
Fk ( ; u, a) cos
k
(t
)d ( k
1,2,...)
, (21)
0
2
k
u k (t )
cos
k
k
t
k
k
sin
k
t
t
Fk ( ; u , a ) sin
k
k
(t
)d
Fk (t ; u , a ) ( k
1,2,...).
(22)
0
u k (t ) ( k
(1)-(3),(6),(7)
0,1,...)
u ( x, t )
(16),
:
t
u ( x, t )
t
0
(t
0
) F0 ( ; u , a)d
0
k
cos k t
1
k
k 1
,
sin
k
t
k
(7),
(16),
1
t
Fk ( ; u, a) sin
(t
)d
cos
x.
:
2
k
f (1, t ))
(
( 1) k )u k (t ) .
k 1
(24)
a (t )
,
u ( x, t ), a (t )
a(t )
k
(23)
h 1 (t ) h (t ) ( f (0, t )
a(t )
k
k 0
(1)-(3),(6),(7)
(20)
h 1 (t ){ h (t ) ( f (0, t )
2
k
f (1, t ))
(
(24):
( 1) k )(
k
cos k t
k 1
1
k sin kt
k
,
Fk ( ;u)sin k (t
k 0
)d .
(25)
(23), (25)
(1)-(3),(6),(7)
u ( x, t )
a (t ) .
{u ( x, t ), a (t )} -
2.
1t
(1)-(3),(6),(7) ,
1
uk (t )
2 u ( x, t ) cos
k
xdx ( k
0,1,...)
0
(19),(20).
2
.
,
(23), (25)
.
B2,T
[4]
.
(1)-(3),(6),(7)
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Y
F T ra n sf o
A B B Y Y.c
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k
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k
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C
w.
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w
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y
ABB
PD
re
to
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2.0
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y
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Y
F T ra n sf o
ABB
PD
er
Y
u ( x, t )
uk (t ) cos
kx
(
,
u k (t ) ( k
DT ,
u0 (t ) C[0,T ]
0,1,...)
u k (t ) C[0,T ]
k
[0,T ]
1
2
2
k 1
,
0.
u ( x, t )
B2 , T
ET
z ( x, t )
B2,T
,
A B B Y Y.c
k )
k
k 0
J (u )
w.
:
J (u )
B2 ,T
C[0, T ]
z ( x, t ) {u ( x, t ), a(t )}
–
u ( x, t )
ET
.
a (t ) C[ 0,T ]
B2 ,T
ET
.
.
ET3
,
(u , a ) {
1
(u, a) u~ ( x, t )
1
(u , a ),
u~k (t ) cos
k
2
(u , a )} ,
x,
2
k 0
u~0 (t ), u~k (t ) ( k 1,2,...) , a~ (t )
a~ (t ),
(u , a )
(19),(20),(25).
:
T
u~0 (t ) C [ 0,T ]
T
0
T 2 a(t ) C [ 0,T ] u0 (t ) C [ 0,T ]
f0 ( ) d
T T
0
1
2
2
0
3
k
u~k . (t ) C [0,T ]
1
2
2
2
3
k
2
k 1
,(26)
1
2
k 1
T
2
k
2 T
1
(
6
k 1
2
fk ( ) d
(
3
k
2T a(t ) C[ 0,T ]
u k (t ) C[ 0,T ]
2
k 1
h 1 (t ) C [ 0,T ] h" C 0,T
)
k
1
2
0 k 1
a~(t ) C [0,T ]
2
2
k
2
k
1
2
3
k
k
)2
f 0, t
1
2
(
k 1
2
k
k
)2
,
(27)
f (1, t ) C 0,T
1
2
T
T
(
2
k
f k ( ) )2 d
1
2
0 k 1
k 1
T a (t )
1
2
k
C [ 0 ,T ]
k
u k (t )
C [ 0 ,T ]
)
2
1
2
.
k 1
(28)
(1)-(3),(6),(7)
2
:
1.
( x) C [0,1],
( x)
L2 (0,1)
( 0)
(1)
0;
2.
( x) C 1[0,1],
( x)
L2 (0,1)
(0 )
(1)
0;
3. f ( x, t ), f x ( x, t )
f x (0, t )
f x (1, t )
C ( DT ), f xx ( x, t )
0 (0 t T ) ;
2
4. h(t ) C [0, T ], h(t ) 0 (0 t T ) .
L2 ( DT )
om
Y
F T ra n sf o
A B B Y Y.c
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C
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k
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k
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C
w.
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w
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rm
y
ABB
PD
re
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Y
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Y
F T ra n sf o
ABB
PD
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Y
(26)-(28)
:
u~0 (t ) C[ 0,T ]
( x)
T T f ( x, t )
3
k
u~k . (t ) C [ 0 ,T ]
1
2
2
T
L2 ( 0 ,1)
( x) L
T 2 a (t )
L2 ( DT )
2
( x)
k 1
C [ 0 ,T ]
u 0 (t )
( x)
C [ 0 ,T ]
(29)
L2 ( 0,1)
1
2 T ( f xx ( x, t )
a~(t ) C[ 0,T ]
1
(
6
)
L2 ( DT )
1
h (t )
( x)
2T a(t ) C [0,T ]
T f xx ( x , t )
L 2 ( 0 ,1 )
u k (t ) C [0,T ] )
2
2
k 1
h (t ) C[ 0,T ]
C [ 0,T ]
3
k
(
,
f (0, t )
(30)
f (1, t ) C[ 0,T ]
3
k
T
L 2 ( DT )
u k (t )
1
2
2
C 0 ,T
k 1
+
.
,
(29)
(30)
u~ ( x , t )
A1 (T )
2
( x)
B1 (T )
h (t )
A1 (T )
T (T
( x)
( x)
1
2 ) a (t )
T T f ( x, t )
L2 ( 0,1)
)
L2
a~(t ) C[ 0,T ]
f (0, t )
( x)
0 ,1
1
(
6
:
) h 1 (t )
u~( x, t )
B1 (T ), B (T ) T (T
L2 ( DT )
u ( x, t )
2)
,
B23,T
L2
,
(32)
L2 ( 0 ,1)
.
B23,T
,
(33)
f (1, t ) C [ 0,T ]
T f xx ( x, t )
0 ,1
C [ 0 ,T ]
B 23 ,T
( x)
L2 ( DT )
B1 (T ) B2 (T ) a(t ) C [0,T ] u ( x, t )
( x)
(33)
C [ 0 ,T ]
2 T f xx ( x, t )
L2 ( 0 ,1)
h (t ) C [ 0 ,T ]
C [ 0 ,T ]
B2 (T )
A(T )
2
a~ (t ) C[ 0,T ]
1
(
6
+
(32)
A1 (T )
B 23 ,T
L2 ( 0 ,1)
:
(31)
:
T
(31),
A B B Y Y.c
2 ( 0 ,1)
2
L2 ( 0 ,1)
w.
L2 ( DT )
,
T
.
A(T ) B(T ) a (t ) C [ 0,T ] u ( x, t )
B23,T
,
(34)
B2 (T ) .
.
1.
1-4
B(T )( A(T ) 2) 2
K
(1)-(3),(6),(7)
.
KR( z
ET3
(35)
A(T ) 2)
ET3
ET3
.
z
z {u, a} ,
(23), (25)
1.
i
(u , a )
(36)
(u , a )
K
(u , a)
.
z,
KR ( z
ET3
R
ET3 .
(34)
,
z , z1, z2
KR
:
A(T ) 2)
om
Y
F T ra n sf o
A B B Y Y.c
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y
ABB
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Y
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Y
F T ra n sf o
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Y
z
z1
(37)
z2
(35),
KR
K
.
B(T )R a1 (t) a2 (t) C[0,T ]
ET3
(38),
A(T ) B(T ) a(t ) C[ 0,T ] u ( x, t )
ET3
.
(38)
KR
.
(23), (25).
B23,T ,
u ( x, t ) ,
u x ( x, t ) , u xx ( x, t )
,
(21)
DT .
(22)
2
k
:
2
uk . (t ) C [ 0,T ]
1
2
2
( x)
k 1
k
uk . (t ) C[ 0,T ]
2
L2 ( 0,1)
2
3
3 T a(t )u xx ( x, t )
(39), (40)
,
( x)
f xx ( x, t )
KR
2 T a(t )u xx ( x, t )
L2 ( 0,1)
3
L2 ( 0 ,1)
L2 ( DT )
( x)
f xx ( x, t )
(1)
f x ( x, t ) C[ 0,T ]
, (39)
L2 ( 0,1)
.
(40)
DT .
(2), (3), (6),(7)
(1)-(3),(6),(7) .
.
L2 ( DT )
L2 ( 0,1)
3 a(t )u x ( x, t )
u t ( x , t ) , utt ( x, t )
,
{u ( x, t ), a (t )}
K
( x)
1
2
k 1
,
2,T
A B B Y Y.c
(37)
K
(36),
KR
u1 (x, t) u2 ( x, t) B3
w.
,
,
{u , a} ,
K
B23,T
.
2
.
1,
(1)-(5).
2.
1
1
1
( x) dx
0
0,
1
( x)dx
0, f ( x, t ) dx
0
( 0)
(1)-(5)
.
0 (0 t
T ),
0
(1)
K
h(0),
KR z
ET3
( 0)
A(T ) 2
(1)
h (0).
ET3
In this work an inverse problem for the hyperbolic equation of second order with periodical boundary conditions is
investigated. For this reason, first of all the initial problem reduces to the equivalent problem, for which the theorem of
existence and uniqueness proves. Then using these facts the existence and uniqueness of the classical solution of initial
problem is proved.
The key words: Inverse boundary problem, hyperbolic equation, method Fourier, classic solution.
1.
1980, .16,
2.
//
3.
, 1982, .18,
4.
.
11, .1925-1935.
.
.
, 1977, .13,
.
1, .72-81. .
.,
//
.
//
.
2, .294-304.
.
.
:
, 2010, 168 .
om
Y
F T ra n sf o
A B B Y Y.c
bu
to
re
he
C
lic
k
he
k
lic
C
w.
om
w
w
w
w
rm
y
ABB
PD
re
to
Y
2.0
2.0
bu
y
rm
er
Y
F T ra n sf o
ABB
PD
er
Y
.–
,
, yashar_aze@mail.ru
w.
A B B Y Y.c
om
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