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Один из случаев решения задачи Маркушевича в замкнутой форме.

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517.544.8
1
. .
2
, . .
-
.
-
.
,
b(t)
.
Ʉɥɸɱɟɜɵɟ ɫɥɨɜɚ: ɵɟ , ɚɱɚ , , .
,
[1].
-
[2–4],
.
L.
ψ + (t ) = a (t )ψ − (t ) + b(t )ψ + (t ) + f (t )
a (t ), b(t ), f (t ) ∈ H ( L) –
κ
κ = Ind L a (t ), a (t ) = a+ (t )t a− (t ) –
t∈L.
.
a (t )
(1)
φ+ (t ) = t κ φ− (t ) + b1 (t )φ+ (t ) + f 0 (t ),
φ± (t ) =
ψ ± (t )
a± (t )
, b1 (t ) = b(t )
) b1 (t ) + 1 ≠ 0, t ∈ L;
) b1 (t ) + 1
,
κ1 .
(2)
,
(2):
,
(2)
,
:
φ+ (t ) =
f1 (t ) =
κ
2b (t )
t
Re φ+ (t ) + f1 (t ) ,
φ− (t ) + 1
b1 (t ) + 1
b1 (t ) + 1
f (t )
.
a+ (t )(b1 (t ) + 1)
(3)
κ
,
(3)
Re φ+ (t )
t
= κ − κ1 , κ1 = Ind L (b1 (t ) + 1) .
b1 (t ) + 1
κ0 ≥ 0 ,
(3)
[5]
φ ( z ) = χ ( z )[ F ( z ) + Pκ 0 −1 ( z )] .
κ 0 = Ind L
F ( z) =
-
a+ (t )
f (t )
, f 0 (t ) =
.
a+ (t )
a+ (t )
b1 (t )
D−
(1)
, a (t ) ≠ 0,
.
(4)
º dτ
1 ª 2b1 (τ ) Re φ+ (τ )
+ f1 (τ ) »
, Pκ 0 −1 ( z ) –
«
³
2π i L ¬ (b1 (τ ) + 1)
¼τ − z
κ0 − 1,
1
2
$ *!"#% *!"# &' – ()+,, "-( )./% 00&"&, 12,)-3!4#"&% 5)#(# ,,6% ,& #&.
E-mail: patraleksej@yandex.ru
$ 7!, 8#&!4 , – ()+,, "-( 9&"!(,)% 00&"&, 12,)-3!4#"&% 5)#(# ,,6% ,& #&.
2013,
5,
1
63
­ 1, z ∈ D+ ,
°
χ ( z ) = ® b1 ( z ) + 1
, z ∈ D−
°
¯ zκ
–
.
κ 0 < 0,
:
ª 2b1 (τ ) Re φ+ (τ ) º k −1
k −1
»τ dτ − ³ f1 (τ )τ dτ = 0, k = 1,", −κ 0 .
τ
(
(
)
1)
+
b
1
¼
L
L
³ «¬
φ+ ( z ) ,
L
φ+ ( z ) ,
(3),
.
(5)
D+ ,
-
2 Re φ+ (t )
− f1 (t ) :
b1 (t ) + 1
º 1 ª 2 Re φ+ (τ )
º dτ
1 ª 2 Re φ+ (t )
− f1 (t ) » +
− f1 (τ ) »
= d + Pκ 0 −1 (t ) .
«
«
³
2 ¬ (b1 (t ) + 1)
¼ 2π i L ¬ (b1 (τ ) + 1)
¼τ − t
d=
1 Re φ+ (τ )
dτ − c0 ,
2π i ³L
τ
c0 –
(6)
.
d
º dτ
1 ª 2 Re φ+ (τ )
− f1 (τ ) »
= d + a0 ,
«
³
2π i L ¬ b1 (τ ) + 1
¼ τ
a0 = Pκ 0 −1 (0) –
.
κ 0 ≤ 0,
a0 = 0.
(6)
[6]
2 Re φ+ (t )
− f1 (t ) = d + Pκ 0 −1 (t ) − ϕ1− (t ) ,
b1 (t ) + 1
ϕ1− (t ) –
,
D−
.
1
f (t )
Re φ+ (t ) = (b1 (t ) + 1)[d + Pκ 0 −1 (t ) − ϕ1− (t )] +
2
2a+ (t )
f1 (t ) =
,
f (t )
.
a+ (t )(b1 (t ) + 1)
,
,
(7)
(7) –
-
­
f (t ) ½
Re[−i (b1 (t ) + 1)ϕ1− (t )] = Im ®(b1 (t ) + 1)[d + Pκ 0 −1 (t )] +
¾,
a+ (t ) ¿
¯
.
(8)
[7]
1
ª F ( z ) + F ( z ) + Q ( z ) + Q ( z* ) º .
−iϕ1− ( z ) =
0
1
κ1 −1
κ1 −1
¼»
b ( z ) + 1 ¬«
(8)
(9)
1
F0 ( z ) =
­ f (t ) ½ τ + z
1 c(τ )(τ + z )
1
Im ®
dτ , F1 ( z ) =
dτ , Qκ1 −1 ( z ) –
¾
³
³
2π i L (τ − z )τ
2π i L ¯ a+ (t ) ¿ (τ − z )τ
{
}
κ1 − 1 , c(t ) = Im (b1 (t ) + 1)[d + Pκ 0 −1 (t )] .
κ1 ≤ 0 ,
Qκ1 −1 ( z ) ≡ 0 .
ª c(τ )(τ + z )
º
­ f (t ) ½ τ + z
1
dτ + ³ Im ®
dτ »
«³
¾
2π i (b1 ( z ) + 1) «¬ L (τ − z )τ
¯ a+ (t ) ¿ (τ − z )τ »¼
L
64
.
«
.
(10)
.
»
..,
..
(8)
,
-
­ f (t ) ½ dτ
c(τ )
³ τ k +1 dτ = α k , α k = − ³ Im ®¯ a+ (t ) ¾¿ τ k +1 , k = 0, ", −κ1 .
(11)
L
L
(7), (9)
φ ( z)
Re φ+ (t ) ,
(3).
­
ª 1 g (τ )
º
a+ ( z ) «
dτ + G ( z ) + Pκ 0 −1 ( z ) » , z ∈ D+ ,
°
³
°°
¬« 2π i L τ − z
¼»
ψ ( z) = ®
º
° a− ( z )(b1 ( z ) + 1) ª 1 g (τ )
+
+
d
τ
G
(
z
)
P
(
z
)
«
» , z ∈ D− .
1
κ
−
°
³
0
zκ
¬« 2π i L τ − z
¼»
¯°
g (t ) = b1 (t ) ¬ª d + Pκ 0 −1 (t ) ¼º −
g1 (t ) =
ib1 (t )
ª F0− (t ) + Qκ −1 (t ) + Qκ −1 (t ) º ,
1
1
¬
¼
b1 (t ) + 1
(12)
1 g1 (τ )
dτ ,
2π i ³ τ − z
G( z) =
º
b1 (t ) ª f (t )
− iF1− (t ) » .
«
b1 (t ) + 1 ¬ a+ (t )
¼
f (t ) ≡ 0 ,
-
,
(12) ( G ( z ) ≡ 0 ),
ɨ
;
(12) ( G ( z ) ≡ 0 ), ( Pκ 0 −1 ( z ) ≡ 0 ),
ɨ-
ɟ
);
(12) ( G ( z ) ≡ 0 ,
, r –
,
2κ 0 − r
,
);
4)
κ1 ≤ 0, κ 0 < 0,
(5) ( f (t ) ≡ 0 ),
( G ( z ) ≡ 0 , Qκ1 −1 ( z ) ≡ 0 , Pκ 0 −1 ( z ) ≡ 0 ); ɜ
2.
f (t ) ∈ H ( L) , a (t ) ≠ 0 , t ∈ L , ɚ
L
,
ɨ
,ɜ
ɜ
:
1)
κ1 > 0, κ 0 ≥ 0
2κ
1
-
r1 = 2κ1 ,
r = 2κ 0 ,
(11) (
5,
ɸ
, r1 –
(5) (
3)
κ1 ≤ 0, κ 0 ≥ 0
Qκ1 −1 ( z ) ≡ 0 ),
(1) ( f (t ) ≡ 0 )
ɚD− ∪ L ,
ɟ-
b1 (t ) + 1
ɜ
2κ1 − r1
ɟ
G( z) ≡ 0 .
.
.
1.
a (t ), b(t ) ∈ H ( L), a (t ) ≠ 0, t ∈ L , κ = Ind L a (t ), ɚ
L,
,
,
,ɜ
κ1 , κ 0 = κ − κ1 .
(1) ( f (t ) ≡ 0 ) ɜ
,
κ1 > 0, κ 0 ≥ 0
,
1)
2κ 0 + 2κ1 = 2κ
κ1 > 0, κ 0 < 0
2)
2013,
-
,
b1 (t ) + 1
(11) ( f (t ) ≡ 0 )
(12)
,
.
a (t ), b(t ) ∈ H ( L) ,
b1 (t ) + 1
ɜ
D− ∪ L ,
.
-
ɨ-
,
,
-
(12),
;
65
κ1 > 0, κ 0 < 0
2)
(12) ( Pκ 0 −1 ( z ) ≡ 0) ,
−κ 0 − r1
(5)),
,
2κ1 − 2r1
);
( r1 –
(
κ1 ≤ 0, κ 0 ≥ 0
3)
ɟr1 = κ1 , ɟ-
(12) ( Qκ1 −1 ( z ) ≡ 0 ),
,
−κ1 + 1 − r
(11)),
(
r = κ0
4)
κ1 ≤ 0, κ 0 < 0
( Qκ1 −1 ( z ) ≡ 0, Pκ 0 −1 ( z ) ≡ 0 ),
(11)
-
,
(r –
ɩɨ-
2κ 0 − 2r
);
,
−κ1 + 1
,
−κ 0
(12)
-
(5).
1.
-
L:
ψ + (t ) = (2t + 1) 2ψ − (t ) + (3t + 1)ψ + (t ), t ∈ L .
(13)
(13)
,
.
(2t + 1)2
2(3t + 1)
ψ − (t ) +
Reψ + (t ) .
3t + 2
3t + 2
(2t + 1) 2
κ = Ind L (2t + 1)2 = 2, κ1 = Ind L (3t + 2) = 1 , κ 0 = Ind L
= 1.
3t + 2
ψ + (t ) =
(2t + 1)2
3t + 2
(14)
,
L,
,
D− .
,
L
:
3t + 1
1 (3τ + 1) Reψ + (τ ) dτ
Reψ + (t ) + ³
+α,
ψ + (t ) =
π i L (3τ + 2)(τ − t )
3t + 2
α –
-
(15)
.
,
ψ + (t ) = Reψ + (t ) +
c0 –
1 Reψ + (τ )dτ
1 Reψ + (τ )dτ
−
+ c0 ,
³
πi L
τ −t
τ
2π i ³L
.
(16)
(15), (16)
-
Reψ + (t ) 1 Reψ + (τ )dτ
1ª
1 Reψ + (τ )dτ º
+ ³
= 2d , d = «α + c0 +
».
π i L (3τ + 2)(τ − t )
τ
3t + 2
2 «¬
2π i ³L
»¼
(17)
Reψ + (t ) = (3t + 2)(d − ϕ − (t )) ,
−
ϕ ( z) –
D−
(18) –
(17)
(18)
,
.
,
:
Im ª¬ (3t + 2)Ψ − (t ) º¼ = 0 ,
Ψ − (t ) = d − ϕ − (t ) –
(19)
D− , Ψ − (∞) = d .
,
(19)
:
-
3
(3t + 2)Ψ − (t ) − ( + 2)Ψ − (t ) = 0 ,
t
66
.
«
.
.
»
..,
..
Ψ1+ (t ) =
­°Ψ − ( z*), | z |< 1,
(3t + 2)t −
Ψ1 (t ), Ψ1 ( z ) = ®
,
3 + 2t
°̄ Ψ − ( z ), | z |> 1.
(20)
­ z
, | z |< 1,
°°
χ1 ( z ) = ® 3 + 2 z
° 1 , | z |> 1,
¯° 3 z + 2
χ1 ( z*) = χ1 ( z )
,
(20).
,
Ψ − ( z) =
α 0 ,α1 , α 2 ∈ R .
-
:
1
(2α 0 + α1 ( z + z −1 ) + α 2i( z − z −1 )) ,
3z + 2
(18)
Reψ + (t ) = 2α 0 + α1 (t + t −1 ) + α 2i(t − t −1 ).
α + iα 2
α1 + iα 2 1
Ψ − (∞) = 1
,
= (α + c0 + 2α 0 ) ,
,
,
3
2
3
−1
−1
1 (2α 0 + α1 (τ + τ ) + α 2i(τ − τ ))dτ
1 Reψ + (τ )dτ
2α 0 =
.
=
³
2π i L
2π i ³L
τ
τ
(13)
­
°
2α 0 + 2(α1 + iα 2 ) z − c0 , | z |< 1,
°
°°
3 z + 2 § 4α 0 11(α1 + iα 2 )
ψ ( z) = ®
−
−
,
¨
3z + 2
(2 z + 1) 2 © 3z + 2
°
° α − iα
2(α1 + iα 2 )
·
2
°− 1
+
− c0 − 2α 0 ¸ , | z |> 1.
z
3
¹
¯°
α 0 ,α1 , α 2 , ic0 ∈ R.
1.
, . .
/ . .
//
. 4. – 10(227). – . 29–37.
, . .
». – 2011. –
2.
.
«
.
.
/
. .
– 2010. –
3.
. .
97.
4.
. .
.
5.
6.
. .
7.
.–
2013,
//
.
«
. 6. – 35(211). – . 4–12.
, . .
//
, . .
, . .
. .
. .
,
//
:
5,
1
/
4. – . 82–
. – 2010. –
//
.
.
/
. .
. – 2011.– . 11. –
. 2. – . 4–12.
. – .:
, 1963. – 640 .
/
. – 1966. – . 2,
, . .
-
,
».
. .
/
.
,
-
2. – . 533–544.
/
. .
-
- , 1977. – 301 .
67
A VARIANT OF THE SOLUTION OF MARKUSHEVICH BOUNDARY PROBLEM
1
A.A. Patrushev , E.V. Patrusheva
2
In the article an explicit method for the solution of Markushevich boundary value problem in the
class of piecewise analytic functions is suggested. Boundary condition of the problem is given on the
unit circle. The problem is found in a closed form under additional restriction on the coefficient b(t) of
the problem.
Keywords: boundary problems for analytic functions, Riemann boundary problem, Hilbert boundary problem, Markushevich boundary problem.
References
1. Patrushev A.A. Zadacha Markushevicha v klasse avtomorfnykh funkciij v sluchae proizvolnoj
okruzhnosti [The Markushevich problem in the lass of automorphic functions for arbitrary circle].
Vestnik YuUrGU. Seriya “Matematika. Mexanika. Fizika”. 2011. Issue 4. no. 10(227). pp. 29–37. (in
Russ.).
2. Adukov V.M., Patrushev A.A. Algoritm tochnogo resheniya chetyrekhlementnoj zadachi linejnogo sopryazheniya s racionalnymi koefficientami i ego programmnaya realizaciya [Algorithm of exact
solution of generalized four-element Riemann–Hilbert boundary problem with rational coefficients and
its programm realization]. Vestnik YuUrGU. Seriya “Matematicheskoe modelirovanie i programmirovanie”. 2010. Issue 6. no. 35(211). pp. 4–12. (in Russ.).
3. Patrushev A.A. Izvestiya Smolenskogo gosudarstvennogo universiteta. 2010. no. 4. pp. 82–98. (in
Russ.).
4. Adukov V.M., Patrushev A.A. Izvestiya Saratovskogo gosudarstvennogo universiteta. Novaya
seriya. Seriya Matematika. Mekhanika. Informatika. 2011. Vol. 11. Issue 2. pp. 4–12. (in Russ.).
5. Gaxov F.D. Kraevye zadachi (Boundary problems). Moscow: Fizmatgiz, 1963. 640 p. (in Russ.).
6. Gaxov F.D. Differencialnye uravneniya. 1966. Vol. 2, no. 2. pp. 533–544.
7. Chibrikova L.I. Osnovnye granichnye zadachi dlya analiticheskikh funkcij [Basic boundary problems for analytic functions]. Kazan: Izdatelstvo Kazanskogo universiteta, 1977. 301 p. (in Russ.).
18 2012 .
1
2
Patrushev Alexey Alexeevich is Associate Professor, Department of General Mathematics, South Ural State University.
E-mail: patraleksej@yandex.ru
Patrusheva Elena Vasilievna is Associate Professor, Applied Mathematics Department, South Ural State University.
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