On the uniqueness of a solution of the inverse problem for a simple-layer potential
We prove the uniqueness of solution of the inverse problem of single-layer potential for star-shaped smooth surfaces in the case of the metaharmonic equation Δv - K² v = 0. For the Laplace equation, a similar statement is not true.
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| Date: | 2008 |
|---|---|
| Main Authors: | , |
| Format: | Article |
| Language: | Russian English |
| Published: |
Institute of Mathematics, NAS of Ukraine
2008
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| Online Access: | https://umj.imath.kiev.ua/index.php/umj/article/view/3207 |
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| Journal Title: | Ukrains’kyi Matematychnyi Zhurnal |
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860509257828401152 |
|---|---|
| author | Kapanadze, D. V. Капанадзе, Д. В. Капанадзе, Д. В. |
| author_facet | Kapanadze, D. V. Капанадзе, Д. В. Капанадзе, Д. В. |
| author_sort | Kapanadze, D. V. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2020-03-18T19:48:23Z |
| description | We prove the uniqueness of solution of the inverse problem of single-layer potential
for star-shaped smooth surfaces in the case of the metaharmonic equation Δv - K² v = 0. For the Laplace equation, a similar statement is not true. |
| first_indexed | 2026-03-24T02:38:14Z |
| format | Article |
| fulltext |
UDK 517.956
D. V. Kapanadze (Tbyl. un-t, Hruzyq)
O EDYNSTVENNOSTY REÍENYQ OBRATNOJ ZADAÇY
DLQ POTENCYALA PROSTOHO SLOQ
We prove the uniqueness of solution of the inverse problem of single-layer potential for star-shaped
smooth surfaces in the case of the metaharmonic equation ∆ v – K 2
v = 0. For the Laplace equation, a
similar statement is not true.
Dovedeno [dynist\ rozv’qzku oberneno] zadaçi potencialu prostoho ßaru dlq zirkovyx hladkyx
poverxon\ u vypadku metaharmoniçnoho rivnqnnq ∆ v – K 2
v = 0. Analohiçne tverdΩennq u vy-
padku rivnqnnq Laplasa [ xybnym.
Reßenye obratnoj zadaçy teoryy potencyala ymeet teoretyçeskoe y prakty-
çeskoe znaçenye. Yzvestno, çto obratnaq zadaça teoryy potencyala qvlqetsq
matematyçeskoj model\g razvedky polezn¥x yskopaem¥x y yzuçenyq vnutren-
neho stroenyq Zemly. Dlq praktyky trebuetsq dal\nejßee razvytye teoryy.
Xarakternaq osobennost\ obratn¥x zadaç — yx nekorrektnost\ po Adamaru.
VaΩn¥m momentom pry teoretyçeskom yssledovanyy zadaçy, nekorrektnoj po
Adamaru, qvlqetsq dokazatel\stvo teorem¥ edynstvennosty. Vperv¥e edyn-
stvennost\ ee reßenyq v klasse zvezdn¥x oblastej postoqnnoj plotnosty v
sluçae obæemnoho potencyala b¥la dokazana P. S. Novykov¥m [1], rezul\tat¥
yssledovanyq kotoroho rasßyren¥ v rabotax [2 – 7]. V rabote Y. M. Rapoporta
[2] yssledovanyq provodqtsq dlq loharyfmyçeskoho potencyala prostoho
sloq. Ym, v çastnosty, dokazano, çto v otlyçye ot obæemnoho potencyala obrat-
naq zadaça dlq potencyala prostoho sloq v sluçae edynyçnoj plotnosty y v
klasse zvezdn¥x oblastej moΩet ymet\ ne edynstvennoe reßenye.
V nastoqwej stat\e rassmatryvaetsq metaharmonyçeskoe uravnenye y doka-
z¥vaetsq edynstvennost\ reßenyq obratnoj zadaçy, esly kusoçno-hladkaq zamk-
nutaq ohranyçennaq poverxnost\ Si , i = 1, 2, ne soderΩyt ploskug çast\.
Plotnost\ µ ∈ C ( R3
), µ ( x ) > 0, x ∈ R3
. Krome toho, edynstvennost\ reßenyq
obratnoj zadaçy ustanavlyvaetsq dlq zvezdn¥x poverxnostej Si ∈ C
(
2,
α
), i =
= 1, 2.
Rassmotrym metaharmonyçeskoe uravnenye ∆ u – K2
u = 0 (yly uravnenye
Klejna – Hordona), K > 0. Pust\
Γ ( x, y ) =
1
4π −
− −e
x y
k x y
, x ∈ R3, y ∈ R3,
— hlavnoe fundamental\noe reßenye [8, s. 46] uravnenyq ∆ U – K2
U = 0. Po-
tencyal prostoho sloq y obæemn¥j potencyal oboznaçym tak:
Uψ
( x ) = Γ
Ω
( ) ( )
∂
∫ x y y dSy, ψ , V
f
( x ) = Γ
Ω
( ) ( )∫ x y f y dSy, ,
hde Ω — kusoçno-hladkaq ohranyçennaq oblast\, ψ ∈ L1 ( ∂ Ω ), f ∈ L1 ( Ω ). V
dal\nejßem çerez Si , i = 1, 2, oboznaçaetsq kusoçno-hladkaq zamknutaq po-
verxnost\ yz prostranstva R 3
. Kusoçno-hladkug ohranyçennug odnosvqznug
oblast\, hranyca kotoroj est\ Si , i = 1, 2, oboznaçym çerez Ωi , i = 1, 2 ( Si =
= ∂ Ω1 ), svqznug komponentu dopolnenyq R3 – Q , kotoraq soderΩyt besko-
neçno udalennug toçku, — çerez Q∞, ∅ — pustoe mnoΩestvo. Podrazumevaet-
© D. V. KAPANADZE, 2008
892 ISSN 1027-3190. Ukr. mat. Ωurn., 2008, t. 60, # 7
O EDYNSTVENNOSTY REÍENYQ OBRATNOJ ZADAÇY … 893
sq, çto kaΩdaq hladkaq çast\ poverxnosty prynadleΩyt C
(
2,
α
)
.
Opredelenye 1. Pust\ S — kusoçno-hladkaq zamknutaq poverxnost\ ( S ⊂
⊂ R3
, yly oblast\ Ω, ∂Ω = S ). Budem hovoryt\, çto dlq hladkoj toçky x0 ∈ S
( x0 ∈ ∂ Ω ) kryvyzna ρ ( x0 ) ≠ 0, esly suwestvuet ploskost\ P0 , kotoraq pro-
xodyt çerez normal\ νx0
, y kryvyzna ploskoj kryvoj S ∩ P0 v toçke x0 ot-
lyçna ot nulq.
1. O edynstvennosty reßenyq obratnoj zadaçy dlq potencyala pros-
toho sloq v sluçae edynyçnoj plotnosty. Sformulyruem obratnug zadaçu
teoryy potencyala. Pust\ S1 , S2 — kusoçno-hladkye zamknut¥e poverxnosty
yz R3 ( Si = ∂ Ωi , i = 1, 2 ). Voznykaet vopros: pry kakyx uslovyqx yz ravenstva
potencyalov
Γ Γ( ) = ( )∫ ∫x y dS x y dSy
S
y
S
, ,
1 2
, x ∈ Ω∞, (1)
sleduet, çto S1 = S2 ( Ω = Ω1 ∪ Ω2 ) ?
Teorema 1. Pust\ Si , i = 1, 2, — kusoçno-hladkaq zamknutaq poverxnost\
yz R3 ( Si = ∂ Ωi , i = 1, 2 ). Esly Ω1 ⊂ Ω2 ( S1 ≠ S2 ), to potencyal¥ poverx-
nostej S1 , S2 ne sovpadagt na Ω∞
.
Dokazatel\stvo. Budem predpolahat\, çto ∂ Ωi = ∂Ωi, i = 1, 2. Dopustym
protyvnoe, t. e.
Γ Γ( ) = ( )∫ ∫x y dS x y dSy
S
y
S
, ,
1 2
, x ∈ Ω∞. (2)
V dannom sluçae oçevydno, çto Ω∞ = Ω2
∞
.
UmnoΩym pred¥duwee ravenstvo potencyalov na funkcyg | x | eK | x |
. Tohda
x e e
x y
dS
x e e
x y
dS
K x K x y
y
S
K x K x y
y
S
− − − −
−
=
−∫ ∫
1 2
, x ∈ Ω2
∞.
Perejdem k predelu pry x → ∞. V rezul\tate poluçym
| S1 | = | S2 | = | ∂ Ω1 | = | ∂ Ω2 |,
hde | Si | , i = 1, 2, — plowad\ poverxnosty Si , i = 1, 2. S druhoj storon¥, yz
ravenstva potencyalov (2) poluçaem ( Uγ
( x ) < 1, x ∈ Ω2 , U γ
( x ) = 1 kvazyvsgdu
na S2 )
U y dS U y dSy
S
y
S
γ γ( ) = ( )∫ ∫
1 2
,
hde γ — ravnovesnaq mera [9] kompakta S2 = ∂Ω2 .
PokaΩem, çto U γ
( x ) = 1 dlq kaΩdoj hladkoj toçky x = ∂Ω2 y U γ
( x ) < 1
pry x ∈ Ω2 . V samom dele, pust\ Ωn
1
, n = 1, 2, 3, … , — posledovatel\nost\
hladkyx odnosvqzn¥x oblastej, udovletvorqgwaq uslovyqm Ωn
1 ∈ C
(
2,
α
)
,
Ωn +1
1 ⊂ Ω n , Ωn
1
1
∞∩ = Ωn . Pust\, dalee, Uγn
— ravnovesn¥j potencyal [9,
s.K179] kompakta ∂Ωn
1
, Vn ( x ) — reßenye zadaçy Dyryxle v oblasty Ωn
1
, kohda
hranyçnoe znaçenye ϕn ( x ) = 1 pry x ∈ ∂Ωn
1
. Yzvestno, çto posledovatel\nost\
ISSN 1027-3190. Ukr. mat. Ωurn., 2008, t. 60, # 7
894 D. V. KAPANADZE
Vn ( x ) sxodytsq [10, s. 92 – 194] dlq toçky x ∈ Ω2 . Oboznaçym
lim
n
nV x
→∞
( ) = V ( x ), x ∈ Ω2 ( Vγn
( x ) = Vn ( x ), x ∈ Ωn
1
).
Poskol\ku hladkaq toçka x = ∂Ω2 est\ toçka ustojçyvosty zadaçy Dyryxle [10,
s. 194], to V ( x ) = 1. Suwestvuet podposledovatel\nost\ { γnk
}, slabo sxodq-
waqsq [9] k γ0 . Znaçyt,
lim
n
U xnk
→∞
( )γ
= Uγ0
( x ), x ∈ Ω2 ( Uγ0
( x ) = V ( x ), x ∈ Ω2 ).
Potencyal Uγ0
dlq hladkoj toçky x0 = ∂Ω2 udovletvorqet uslovyg [9,
s.K261]
lim
x
x x
U x
∈
→
( )
ν
γ
1
0
0 = Uγ0
( x0 ) = 1 ( Uγ0
( x ) = Vγ
( x ), x ∈ Ω2, γ0 = γ ),
hde ν1 — vnutrennqq yly vneßnqq normal\ v toçke x0 . V sylu pryncypa mak-
symuma metaharmonyçeskoho uravnenyq Uγ
( x ) < 1, x ∈ Ω2 ( Uγ
( x ) ≤ 1, x ∈ Ω2 ).
Takym obrazom,
U y dSy
S
γ ( )∫
1
< | S1 |, U x dSx
S
γ ( )∫
2
< | S2 |.
Poluçennoe protyvoreçye dokaz¥vaet teoremu.
Teorema 2. Pust\ S1 , S2 — kusoçno-hladkye ohranyçenn¥e zamknut¥e po-
verxnosty. PredpoloΩym, çto Si , i = 1, 2, ne soderΩyt ploskug çast\. Tohda
reßenye obratnoj zadaçy teoryy potencyala edynstvenno ( Si = ∂ Ωi , Ω = Ω1 ∪
∪ Ω2 ).
Dokazatel\stvo. Dopustym protyvnoe, t. e.
Γ Γ( ) = ( )∫ ∫x y dS x y dSy
S
y
S
, ,
1 2
, x ∈ Ω∞, (3)
no poverxnosty ne sovpadagt. Poskol\ku Si , i = 1, 2, ne soderΩyt çast\ plos-
kosty, suwestvuet hladkaq toçka x0 ∈ ∂Ω1 ∩ ∂ ∞Ω1 , x0 ∉ Ω2, ρ ( x0 ) ≠ 0 ( yly
x0 ∈ ∂Ω2 ∩ ∂Ω∞, x0 ∉ Ω1 ). V¥polnym lynejnoe preobrazovanye v prostranstve
R3
, posle kotoroho ploskost\ Px0
(sm. opredelenye 1) sovpadaet s ploskost\g
x1 O x3 y normal\ v toçke x0 parallel\na ploskosty x1 O x3
. Oboznaçym σ =
= { x : ( x – x0 ) < ε } ∩ ( Px0
∩ S1 ),
σ1 =
x x x x x: − <
( > )0 3 3
0
2
ε σ∩ ∩ , x0 = ( )x x x1
0
2
0
3
0, , .
Uravnenye kryvoj σ1 ymeet vyd x3 = τ ( x1 ), hde x0 – ε < x1 < x0
, 0 < ε1 ≤ ε / 2
(yly x1
0 < x1 < x1
0 + ε1 ). Qsno, çto kryvaq σ1 leΩyt na ploskosty x1 O x3
.
Podrazumevaetsq, çto S1 ∪ S2 ⊂ { x : x3 > 0 }.
Ravenstvo potencyalov (3) perepyßem sledugwym obrazom:
ISSN 1027-3190. Ukr. mat. Ωurn., 2008, t. 60, # 7
O EDYNSTVENNOSTY REÍENYQ OBRATNOJ ZADAÇY … 895
U y dS U y dSy
S
y
S
ψ ψ( ) = ( )∫ ∫
1 2
, ψ ∈ C ( ∂ Ω∞
).
Rassmotrym hladkug ohranyçennug odnosvqznug oblast\ Ω0 , kotoraq
udovletvorqet uslovyqm S1 ∪ S2 ⊂ Ω0, Ω0 ∈ C
(
2,
α
), σ ⊂ ∂Ω0 , σ ∩ S2 = ∅. Yz
pred¥duweho ravenstva poluçaem
V x dS V x dS V x dSx x
S
x
S
( ) = ( ) − ( )∫ ∫ ∫
−σ σ2 1
, (4)
hde V — harmonyçeskaq funkcyq yz C( )Ω0 , dlq kotoroj spravedlyvo pred-
stavlenye
V ( x ) = − ∂ ( )
∂
( )
∂
∫ G x y
y dS
y
y
,
ν
ϕ
Ω0
, ϕ ∈ C ( ∂ Ω0 ).
Rassmotrym sledugwee hranyçnoe znaçenye harmonyçeskoj funkcyy V na
kryvoj σ1 [11]:
ϕ = δx1
× δ0 × Y3 .
Zdes\ δx1
, δ0 — odnomern¥e mer¥ Dyraka, sosredotoçenn¥e sootvetstvenno v
toçkax ξ1 = x1 , x2 = 0. Raspredelenye ϕ dejstvuet sledugwym obrazom ( x1
0 –
– ε1 < x1 < x1
0
) :
( ϕ, f ) = δ δ τ τ
σ
x f y y y dS f x x x
1 0 1 2 3 1 1 1
20 1× ( )
= ( ) + [ ′( )]∫ ( ), , , , , .
Pust\, dalee, fn
1, n = 1, 2, 3, … , — posledovatel\nost\ neprer¥vn¥x funk-
cyj, dlq kotoroj v¥polnqetsq ravenstvo
lim
k
nf y y dy
→∞
( ) ( )∫ 1
1 1 1 1ψ = ψ1 ( x1 ),
ψ ∈ C x x[ − ]1
0
1 1
0ε , , x1 ∈ ( − )x x1
0
1 1
0ε , .
Rassmotrym posledovatel\nost\ hranyçn¥x znaçenyj
ϕn ( y1 , y2 , y3 ) = f y f yn
1
1 2 30( ) ( ) , n = 1, 2, 3, … .
Yz (4) poluçaem (predpolahaq, çto f2 ( 0 ) = 1)
f y y dS f y y y dyn n
x
x
1
1 3
1
1 1 1
2
11 1
1 1
0
1
1
0
( )⋅ ⋅ = ( ) ( ) + [ ′( )]∫ ∫
−σ ε
τ τ ,
lim
n
nf y y dS x x
→∞
( )⋅ ⋅ = ( ) + [ ′( )]∫ 1
1 3 1 1
21 1
1σ
τ τ , x1 ∈ ( − < < )x x x1
0
1 1 1
0ε .
Oçevydno, çto ( Vϕ ( x ) ≠ 0, x ∈ Ω0 )
ISSN 1027-3190. Ukr. mat. Ωurn., 2008, t. 60, # 7
896 D. V. KAPANADZE
lim
x x
x
1 1
0
1 1
2
→
+ [ ′( )]τ = ∞. (5)
S druhoj storon¥, tak kak
min
x S
y
x y
∈
∈
−
2
1σ
> 0, min
x S
y
x y
∈ −
∈
−
1
1
σ
σ
> 0,
poluçaem ( ϕ = δx1
× δ0 × Y3 )
Sup
x S
V x
∈
( )
2
ϕ < ∞, Sup
x S
V x
∈ −
( )
1 σ
ϕ < ∞. (6)
Zametym, çto funkcyq Hryna oblasty Ω0 udovletvorqet uslovyg [12]
G x y
G x y
x
C
x yk
( ) + ∂ ( )
∂
≤
−
,
, 1
2 , x ∈ Ω0 , y ∈ Ω0 , k = 1, 2, 3.
Yz (4) – (6) poluçaem protyvoreçye.
Teorema 2 dokazana.
2. O edynstvennosty reßenyq obratnoj zadaçy dlq zvezdn¥x oblastej.
Teorema 3. Pust\ Si , i = 1, 2, — hladkaq ohranyçennaq zamknutaq poverx-
nost\ yz klassa C
(
2,
α
)
. PredpoloΩym, çto pereseçenye ∂Ω1 ∩ ∂Ω2 ∩ ∂Ω∞
—
koneçnoe çyslo kryv¥x. Tohda potencyal¥ poverxnostej S1 y S2 ne sovpada-
gt na Ω∞ ( Ω = Ω1 ∪ Ω2
, ∂ Ωi = Si , i = 1, 2 ).
Dokazatel\stvo. Opredelym dyametr obæedynenyq S1 ∪ S2
:
d = Sup
x S
y S
x y z z
∈
∈
− = −
1
2
1 2 .
V sylu uslovyq teorem¥ lehko ubeΩdaemsq, çto v okrestnosty toçky z1
(yly z2 ) suwestvuet hladkaq toçka x0 ∈ S1 ∩ ∂Ω∞, x0 ∉ Ω2, ρ ( x0 ) ≠ 0 (sm.
opredelenye 1). Posle πtoho dostatoçno povtoryt\ rassuΩdenyq yz dokazatel\-
stva teorem¥ 2.
Teorema 4. Pust\ Si , i = 1, 2, — hladkye ohranyçenn¥e zamknut¥e poverx-
nosty yz C
(
2,
α
)
, zvezdn¥e otnosytel\no obwej toçky z0 ( z0 ∈ Ω1 ∩ Ω2 ), S1 =
= ∂ Ω1 , S2 = ∂ Ω2 . Tohda reßenye obratnoj zadaçy potencyala prostoho sloq
edynstvenno.
Dokazatel\stvo. Esly Ω1 ⊂ Ω2 (yly Ω2 ⊂ Ω1), to dokazatel\stvo sle-
duet yz teorem¥ 1.
Poskol\ku S1 ∈ C
(
2,
α
), S2 ⊂ C
(
2,
α
)
, vozmoΩn¥ sledugwye sluçay:
1) pereseçenye ∂Ω1 ∩ ∂Ω2 ∩ ∂Ω∞
est\ koneçnoe çyslo kryv¥x
yly
2) suwestvuet obwaq poverxnost\ σ ⊂ ∂Ω1 ∩ ∂Ω2 ∩ ∂Ω∞
.
V pervom sluçae dokazatel\stvo sleduet yz teorem¥ 3. Teper\ predpolo-
Ωym, çto pereseçenye soderΩyt obwug poverxnost\ σ. Oçevydno, çto
Γ Γ( ) = ( )
− −
∫ ∫x y dS x y dSy
S
y
S
, ,
1 2σ σ
, x ∈ Ω∞. (7)
ISSN 1027-3190. Ukr. mat. Ωurn., 2008, t. 60, # 7
O EDYNSTVENNOSTY REÍENYQ OBRATNOJ ZADAÇY … 897
Oboznaçym ω =
ωkk
N
=1∪ = ( )Ω Ω1 2∪ – ( )Ω Ω1 2∩ ω = ωk
N
1∪ , hde ωk , k = 1,
2, … , N, — odnosvqznaq oblast\. Yz (7) ymeem
U y dS U y dSy
S
y
S
ϕ
σ
ϕ
σ
( ) = ( )
− −
∫ ∫
1 2
, ϕ ∈ C(∂ )ω . (8)
Otmetym, çto R3
— ω -svqznoe mnoΩestvo. Lehko ubedyt\sq, çto mnoΩestvo
potencyalov { Uϕ, ϕ ∈ C(∂ )ω } plotno v prostranstve L2 ( ∂ω ). Znaçyt, su-
westvuet posledovatel\nost\ potencyalov, udovletvorqgwaq uslovyqm {∂ω =
= ∂( − )R3 ω }
lim
n
xU x dSn
→∞ ∂
( )( ) −∫ ϕ
ω
1 2
1
= 0,
ϕn ∈ C(∂ )ω , (9)
lim
n
xU x dSn
→∞ ∂
( )( ) −∫ ϕ
ω
1 2
0
= 0,
hde ω0 = ω – ω1 (predel\naq funkcyq V ( x ) = 1, x ∈ ∂ω1, V ( x ) = 0, x ∈ ∂ω0 ).
Yz (8) y (9) poluçaem
1
1
1⋅ = ∂
∂
∫ dSx
ω
ω = 0.
Pryßly k protyvoreçyg.
Teorema 4 dokazana.
Zametym, çto dokazannoe utverΩdenye v sluçae operatora Laplasa ( k = 0 )
neverno [2]. ∏to svqzano s tem, çto ravnovesn¥j potencyal kompakta S2 = ∂ Ω2
udovletvorqet uslovyg
Uγ
( x ) ≡ 1, x ∈ Ω2 .
3. O edynstvennosty reßenyq obratnoj zadaçy dlq plotnosty µµµµ ∈∈∈∈
∈∈∈∈ C (((( R3
)))), µµµµ (((( x )))) > 0, x ∈ R3
.
Teorema 5. Pust\ S1 , S2 — kusoçno-hladkye ohranyçenn¥e zamknut¥e po-
verxnosty na R3
. Esly Ω1 ⊂ Ω2 ( Ω1 ≠ Ω2
, ∂ Ωi = Si , i = 1, 2 ), to potencya-
l¥ poverxnostej S1 , S2 ne sovpadagt na Ω∞ = Ω2
∞
.
Dokazatel\stvo. Dopustym protyvnoe, t. e.
Γ Γ( ) ( ) = ( ) ( )∫ ∫x y y dS x y y dSy
S
y
S
, ,µ µ
1 2
, x ∈ Ω∞
, (10)
no S1 ≠ S2 .
Kak y pry dokazatel\stve teorem¥ 1, poluçaem
µ µ( ) = ( )∫ ∫y dS y dSy
S
y
S1 2
. (11)
Pust\ γ — ravnovesnaq mera kompakta S2 , Uγ
( x ) = 1 dlq hladkoj toçky
x ∈ S2 . Posle yntehryrovanyq po γ yz (10) ymeem
ISSN 1027-3190. Ukr. mat. Ωurn., 2008, t. 60, # 7
898 D. V. KAPANADZE
U x x dS U x x dS y dSx
S
x
S
y
S
γ γµ µ µ( ) ( ) = ( ) ( ) = ( )∫ ∫ ∫
1 2 2
. (12)
Poskol\ku Uγ
( x ) < 1, x ∈ Ω2 ( ∂ Ωi = Si ), to
U x y dS y dSx
S
y
S
γ µ µ( ) ( ) < ( )∫ ∫
1 2
. (13)
Yz (11) – (13) poluçaem protyvoreçye.
Teorema 5 dokazana.
Teorema 6. Pust\ Si , i = 1, 2, — kusoçno-hladkaq ohranyçennaq zamknutaq
poverxnost\. PredpoloΩym, çto S i , i = 1, 2, ne soderΩyt ploskug çast\.
Tohda reßenye obratnoj zadaçy v sluçae plotnosty µ edynstvenno.
Dokazatel\stvo. V sylu uslovyq teorem¥ suwestvuet hladkaq toçka x0 ∈
∈ S1 ∩ ∂Ω∞, x0 ∉ Ω2, ρ ( x0 ) ≠ 0. Pust\ σ = { x : | x – x0 | < ε } ∩ ( S1 ∩ Px0
), σ ∩
∩ Ω2 = ∅, σ1 = { x : | x – x0 | < ε / 2 } ∩ ( S1 ∩ P0 ). V¥polnym lynejnoe preobrazo-
vanye koordynatnoj system¥, posle kotoroho ploskost\ P0 ( P0 = Px0
) sovpadet
s ploskost\g x1 O x3 y normal\ νx0
budet parallel\na ploskosty x1 O x3 .
Dopustym, çto potencyal¥ sovpadagt na Ω∞
, t. e. ( Ω = Ω1 ∪ Ω2 )
Γ Γ( ) ( ) = ( ) ( )∫ ∫x y y dS x y y dSy
S
y
S
, ,µ µ
1 2
, x ∈ Ω∞
,
no poverxnosty S1 y S2 ne sovpadagt (podrazumevaetsq, çto S1 ∪ S2 ⊂ { x : x3 >
> 0 } ).
Pust\ ohranyçennaq odnosvqznaq oblast\ Ω0 udovletvorqet uslovyqm σ ⊂
⊂ ∂Ω0 , Ω0 ∈ C
(
2,
α
), S1 ∪ S2 ⊂ Ω0.
Yz pred¥duweho ravenstva potencyalov poluçaem
U x x dS U x x dSx
S
x
S
ψ ψµ µ( ) ( ) = ( ) ( )∫ ∫
1 2
, ψ ∈ C ( ∂ Ω 0 ) .
Otsgda ymeem
V x x dS V x x dSx
S
x
S
( ) ( ) = ( ) ( )∫ ∫µ µ
1 2
,
hde V — harmonyçeskaq funkcyq yz C( )Ω0 .
Oçevydno, çto (sm. teoremu 2)
V x x dS V x x dS V x x dSx x
S
x
S
( ) ( ) = ( ) ( ) − ( ) ( )∫ ∫ ∫
−
µ µ µ
σ σ2 1
. (14)
Na kryvoj σ1 rassmotrym raspredelenye [11]
ϕ = δx1
⋅ δ0 ⋅ Y3 , ( x1, 0, y3 ) ∈ σ1 .
Yz ravenstva (14) poluçaem
lim , , , ,
n
n yf y y y y dS x x x x
→∞
( ) ( ) = ( ) ( ) + [ ′( )]∫ [ ]1
1 3 1 3 1 1 1 1
20 0 1µ τ µ τ τ
σ
.
Zdes\ posledovatel\nost\ { }( )f xn
1
udovletvorqet uslovyg
ISSN 1027-3190. Ukr. mat. Ωurn., 2008, t. 60, # 7
O EDYNSTVENNOSTY REÍENYQ OBRATNOJ ZADAÇY … 899
lim
k
nf y y dy x
→∞
( ) ( ) = ( )∫ 1
1 1 1 1ψ ψ , x1
0 – ε1 < x1 < x1
0 , ψ ∈ C0 ( R1
).
Kak y pry dokazatel\stve teorem¥ 2, ymeem
lim , ,
x x
x x x x
1 1
0 1 1 1 1
20 1
→
( ) ( ) + [ ′( )][ ]τ µ τ τ = ∞,
mynymal\noe rasstoqnye
d ( σ1 , σ2 ) > 0, d ( σ1 , S1 – σ ) > 0.
Sledovatel\no,
Sup
x S
V x
∈
( )
2
< ∞, Sup
x S
V x
∈ −
( )
1 σ
< ∞.
Pryßly k protyvoreçyg.
Analohyçno dokaz¥vaetsq teorema 6 v sluçae plotnosty µ.
1. Novykov P. S. O edynstvennosty reßenyq obratnoj zadaçy teoryy potencyala // Dokl.
ANKSSSR. – 1938. – 18, # 3. – S. 165 – 168.
2. Rapoport Y. M. Ob odnoj zadaçe teoryy potencyala // Ukr. mat. Ωurn. – 1950. – 2, # 2. –
S. 50 – 58.
3. Sretenskyj L. N. O edynstvennosty opredelenyq form¥ prytqhyvagwehosq tela po zna-
çenyqm eho vneßneho potencyala // Dokl. AN SSSR. – 1954. – 99, # 1. – S. 20 – 22.
4. Yvanov V. K. Obratnaq zadaça potencyala dlq tela, blyzkoho k dannomu // Yzv. AN SSSR.
Ser. mat. – 1956. – 20. – S. 793 – 818.
5. Íaßkyn G. A. O edynstvennosty obratnoj zadaçy teoryy potencyala // Dokl. AN SSSR. –
1957. – 115, # 1. – S. 64 – 66.
6. Straxov V. N., Brodskyj M. A. O edynstvennosty v obratnoj zadaçe loharyfmyçeskoho po-
tencyala // Yzv. AN SSSR. Fyzyka Zemly. – 1985. – # 6. – S. 27 – 47.
7. Kapanadze D. V. O edynstvennosty reßenyq obratnoj zadaçy teoryy potencyala // Soobw.
AN Hruzyy. – 1992. – 145, # 1. – S. 78 – 80.
8. Bycadze A. V. Kraev¥e zadaçy dlq πllyptyçeskyx uravnenyj vtoroho porqdka. – M.: Nauka,
1966.
9. Landhof N. S. Osnov¥ sovremennoj teoryy potencyala. – M.: Nauka, 1966.
10. Keld¥ß M. V. Matematyka. – Yzbr. trud¥. – M.: Nauka, 1985.
11. Vladymyrov V. S. Obobwenn¥e funkcyy v matematyçeskoj fyzyke. – M.: Nauka, 1979.
12. ∏ydus\ D. M. Ocenky proyzvodn¥x funkcyy Hryna // Dokl. AN SSSR. – 1956. – 106, # 1. –
S. 207 – 209.
Poluçeno 13.06.06
ISSN 1027-3190. Ukr. mat. Ωurn., 2008, t. 60, # 7
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| id | umjimathkievua-article-3207 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | rus English |
| last_indexed | 2026-03-24T02:38:14Z |
| publishDate | 2008 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/b3/005646b3f14e9ea95093aa4ce7e016b3.pdf |
| spelling | umjimathkievua-article-32072020-03-18T19:48:23Z On the uniqueness of a solution of the inverse problem for a simple-layer potential O единственности решения обратной задачи для потенциала простого слоя Kapanadze, D. V. Капанадзе, Д. В. Капанадзе, Д. В. We prove the uniqueness of solution of the inverse problem of single-layer potential for star-shaped smooth surfaces in the case of the metaharmonic equation &Delta;v - K&sup2; v = 0. For the Laplace equation, a similar statement is not true. Доведено єдиність розв'язку оберненої задачі потенціалу простого шару для зіркових гладких поверхонь у випадку метагармонічного рівняння &Delta;v - K&sup2; v = 0. Аналогічне твердження у випадку рівняння Лапласа є хибним. Institute of Mathematics, NAS of Ukraine 2008-07-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/3207 Ukrains’kyi Matematychnyi Zhurnal; Vol. 60 No. 7 (2008); 892–899 Український математичний журнал; Том 60 № 7 (2008); 892–899 1027-3190 rus en https://umj.imath.kiev.ua/index.php/umj/article/view/3207/3160 https://umj.imath.kiev.ua/index.php/umj/article/view/3207/3161 Copyright (c) 2008 Kapanadze D. V. |
| spellingShingle | Kapanadze, D. V. Капанадзе, Д. В. Капанадзе, Д. В. On the uniqueness of a solution of the inverse problem for a simple-layer potential |
| title | On the uniqueness of a solution of the inverse problem for a simple-layer potential |
| title_alt | O единственности решения обратной задачи для потенциала простого слоя |
| title_full | On the uniqueness of a solution of the inverse problem for a simple-layer potential |
| title_fullStr | On the uniqueness of a solution of the inverse problem for a simple-layer potential |
| title_full_unstemmed | On the uniqueness of a solution of the inverse problem for a simple-layer potential |
| title_short | On the uniqueness of a solution of the inverse problem for a simple-layer potential |
| title_sort | on the uniqueness of a solution of the inverse problem for a simple-layer potential |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/3207 |
| work_keys_str_mv | AT kapanadzedv ontheuniquenessofasolutionoftheinverseproblemforasimplelayerpotential AT kapanadzedv ontheuniquenessofasolutionoftheinverseproblemforasimplelayerpotential AT kapanadzedv ontheuniquenessofasolutionoftheinverseproblemforasimplelayerpotential AT kapanadzedv oedinstvennostirešeniâobratnojzadačidlâpotencialaprostogosloâ AT kapanadzedv oedinstvennostirešeniâobratnojzadačidlâpotencialaprostogosloâ AT kapanadzedv oedinstvennostirešeniâobratnojzadačidlâpotencialaprostogosloâ |