On the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0
We prove the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0.
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| Дата: | 2006 |
|---|---|
| Автори: | , |
| Формат: | Стаття |
| Мова: | Російська Англійська |
| Опубліковано: |
Institute of Mathematics, NAS of Ukraine
2006
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| Онлайн доступ: | https://umj.imath.kiev.ua/index.php/umj/article/view/3499 |
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| Назва журналу: | Ukrains’kyi Matematychnyi Zhurnal |
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860509598453071872 |
|---|---|
| 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 | 2020-03-18T19:56:00Z |
| description | We prove the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0. |
| first_indexed | 2026-03-24T02:43:39Z |
| format | Article |
| fulltext |
UDK 517.956
D. V. Kapanadze (Tbylys. un-t, Hruzyq)
O EDYNSTVENNOSTY REÍENYQ ZADAÇY
S NAKLONNOJ PROYZVODNOJ DLQ URAVNENYQ ∆∆∆∆n
vvvv = 0
We prove the uniqueness of solution of the problem with directional derivative for the equation ∆n
v = 0.
Dovodyt\sq [dynist\ rozv’qzku zadaçi z poxylog poxidnog dlq rivnqnnq ∆n
v = 0.
∏llyptyçeskye zadaçy rassmatryvalys\ v rabotax [1 – 3], a zadaça s naklonnoj
proyzvodnoj — v [4 – 8]. Yzvestno, çto esly napravlenye dyfferencyrovanyq
xotq b¥ v odnoj toçke kasaetsq hranyc¥ oblasty, to zadaça s naklonnoj proyz-
vodnoj perestaet b¥t\ ne tol\ko fredhol\movoj, no daΩe neterovoj. Ona lybo
ne razreßyma normal\no, lybo ee yndeks beskoneçen, lybo odnovremenno ymeet
mesto y to, y druhoe.
V nastoqwej stat\e rassmatryvaetsq vopros o edynstvennosty reßenyq zada-
çy s naklonnoj proyzvodnoj dlq uravnenyq ∆
n
v = 0. Dlq prostot¥ yzloΩenyq
budem rassmatryvat\ trexmernoe prostranstvo R
3
( ∆ — operator Laplasa).
Vvedem nekotor¥e oboznaçenyq. Obæemn¥e potencyal¥ y potencyal¥ pros-
toho sloq oboznaçym sledugwym obrazom [9]:
V
µ
( x ) =
1
4π
µ
Ω
∫ −
( )y dy
x y
,
U
ψ
( x ) =
1
4π
ψ
∂
∫ −
Ω
( )y d Sy
x y
,
hde ∂ Ω — hranyca oblasty Ω ∈ C
(
2,α
)
, µ ∈ L1 ( Ω ), ψ ∈ L1 ( ∂ Ω ). Esly G —
funkcyq Hryna zadaçy Dyryxle dlq oblasty Ω, to reßenye zadaçy Dyryxle v
oblasty Ω ymeet vyd
v ( x ) = –
∂
∫ ∂
∂
Ω
G x y
y dSy
y
( , )
( )
ν
ϕ , x ∈ Ω, ϕ ∈ C ( ∂ Ω ).
Zdes\ ν — vneßnqq normal\ v toçke y ∈ ∂ Ω,
G ( x, y ) =
1
4
1
π
ε
x y
U yx
−
−
′ ( )
— funkcyq Hryna v oblasty Ω dlq zadaçy Dyryxle, U yx′ε ( ) — potencyal
prostoho sloq, ′εx — plotnost\ v¥metannoj [9, c. 255] mer¥ Dyraka εx yz Ω
na ∂ Ω. Çerez lx , x ∈ ∂ Ω, oboznaçym hladkoe napravlenye v toçke x ∈ ∂ Ω,
lx ∈ C
(
3,α
)
, | l | = 1. Plotnost\ v¥metannoj mer¥ dlq obæemnoj plotnosty
µ ∈ C Ω( ) opredelym sledugwym obrazom:
T µ ( y ) = µ ′ ( y ) = –
Ω
∫ ∂
∂
G x y
x d x
y
( , )
( )
ν
µ .
Yzvestno, çto spravedlyvo ravenstvo [9, c. 260]
V
µ
( x ) = U x′µ ( ) , x ∈ R
3 – Ω .
Çerez E oboznaçym mnoΩestvo toçek
© D. V. KAPANADZE, 2006
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 6 835
836 D. V. KAPANADZE
E = x l xx x: cos ,ν
Ò( ) = ∈∂
0 Ω ,
Γ ( x, y ) =
1
4π x y−
— qdro N\gtona. Symvol ∅ oboznaçaet pustoe mnoΩestvo.
V dal\nejßem ponadobytsq odno vspomohatel\noe utverΩdenye.
Lemma. Dlq funkcyy Hryna zadaçy Dyryxle spravedlyvo ravenstvo
∂
∂
G x y
ly
( , )
=
cos
( , )ν
νy y
y
l
G x yÒ( ) ∂
∂
, x ∈ Ω, y ∈ ∂ Ω – E.
Dokazatel\stvo. Rassmotrym potencyal (potencyal Hryna)
W ( x ) = –
Ω
∫ G x y f y dy( , ) ( ) , f ∈ C1 Ω( ).
Qsno, çto
∂
∂
W x
lx
( )
= –
Ω
∫ ∂
∂
G x y
l
f y dy
x
( , )
( ) , x ∈ ∂ Ω.
S druhoj storon¥,
W ( x ) = –
Ω
∫ G x y f y dy( , ) ( ) = –
Ω
Γ∫ ( , ) ( )x y f y dy +
∂Ω
Γ∫ ′( , ) ( )x y f y dSy .
Yzvestno, çto
Ω
Γ∫ ( , ) ( )x y f y dy =
∂Ω
Γ∫ ′( , ) ( )x y f y dSy , x ∈ R
3 – Ω .
Krome toho, esly cos νx xl
Ò( ) ≠ 0, x ∈ ∂ Ω, to dlq potencyala prostoho sloq
U
Ψ
( x ) =
∂Ω
Γ Ψ∫ ( , ) ( )x y y dSy , Ψ ∈ C
1
( ∂ Ω ),
spravedlyvo ravenstvo (predel yznutry y yzvne)
∂
∂
U x
l
i
x
Ψ ( )
=
1
2
Ψ( )cosx lx xν
Ò( ) +
∂Ω
Γ Ψ∫ ∂
∂
( , )
( )
x y
l
y dSy
x
.
Analohyçno,
∂
∂
U x
l
e
x
Ψ ( )
= –
1
2
Ψ( )cosx lx xν
Ò( ) +
∂Ω
Γ Ψ∫ ∂
∂
( , )
( )
x y
l
y dSy
x
.
Krome toho, oçevydno, çto
0 =
∂
∂
W x
l
e
x
( )
=
= –
Ω
Γ∫ ∂
∂
( , )
( )
x y
l
f y dy
x
–
1
2
′ ( )f x lx x( )cos ν
Ò
+
∂Ω
Γ∫ ∂
∂
′( , )
( )
x y
l
f y dSy
x
,
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 6
O EDYNSTVENNOSTY REÍENYQ ZADAÇY S NAKLONNOJ PROYZVODNOJ … 837
∂
∂
W x
l
i
x
( )
= –
Ω
Γ∫ ∂
∂
( , )
( )
x y
l
f y d y
x
+
1
2
′ ( )f x lx x( )cos ν
Ò
+
∂Ω
Γ∫ ∂
∂
′( , )
( )
x y
l
f y dSy
x
.
Yz raznosty
∂
∂
W x
l
i
x
( )
–
∂
∂
W x
l
e
x
( )
= –
Ω
∫ ∂
∂
G x y
l
f y d y
x
( , )
( )
poluçaem
Ω
∫ ∂
∂
G x y
l
f y d y
x
( , )
( ) =
′ ( )f x lx x( )cos ν
Ò
, f ′ = T f.
Sledovatel\no,
–
Ω
∫ ∂
∂
G x y
l
f y d y
x
( , )
( ) = –
cos
( , )
( )νx x
x
l
G x y
l
f y d y
Ò( ) ∂
∂∫
Ω
.
Takym obrazom,
∂
∂
G x y
lx
( , )
= cos
( , )ν
νx x
x
l
G x yÒ( ) ∂
∂
, x ∈ ∂ Ω – E, y ∈ Ω.
Yzvestno sledugwee utverΩdenye zadaçy s naklonnoj proyzvodnoj dlq
uravnenyq Laplasa ∆ u = 0.
Teorema (Û. Bulyhan). Pust\ v — harmonyçeskaq funkcyq yz prostran-
stva C1 Ω( ), Ω ∈ C
(
2,
α
)
. Esly v ( x ) = 0, x ∈ E, y
∂
∂
v( )y
ly
= 0, y ∈ ∂ Ω,
to v ( x ) = 0, x ∈ Ω.
V πtoj stat\e ustanavlyvaetsq teorema Û. Bulyhana dlq uravnenyq ∆
n
v = 0.
Spravedlyva sledugwaq teorema.
Teorema 1. Pust\ Ω — ohranyçennaq oblast\ yz klassa C
(
4,
α
)
, v — re-
ßenye uravnenyq ∆
3
v = 0 yz prostranstva C( , )4 α Ω( ) , lx ∈ C
(
3,
α
)
. Pust\, da-
lee, v ( x ) = 0, x ∈ ∂ Ω, ∆ v ( x ) = ∆
2
v ( x ) = 0 , x ∈ E. Esly
∂
∂
v( )x
lx
=
∂
∂
2
2
v( )x
lx
= 0, x ∈ ∂ Ω – E,
to v ( x ) = 0, x ∈ Ω .
Dokazatel\stvo. Dlq reßenyq uravnenyq ∆
3
v = 0, v ∈ C( , )4 α Ω( ) , spra-
vedlyvo sledugwee predstavlenye:
v ( x ) = H0 ( x ) –
Ω
∫ G x y H y d y( , ) ( )1 +
Ω Ω
∫ ∫G x y G y z H z dz d y( , ) ( , ) ( )2 , x ∈ Ω,
hde H 0 , H 1 , H 2 — harmonyçeskye funkcyy, dlq kotor¥x Hk ( x ) = ∆
k
v ( x ),
x ∈ ∂ Ω, k = 0, 1, 2.
Oboznaçym
H ( Ω, E ) = ω ω ω ω: , ( ) , , ( ) ,∈ ( ) = ∈ = ∈{ }C x x x x EΩ ∆ Ω0 0 .
Çerez H E( , )Ω oboznaçym popolnenye lynejnoho mnoΩestva H ( Ω, E ) po nor-
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 6
838 D. V. KAPANADZE
me L2 ( Ω ). Netrudno vydet\, çto spravedlyvo sledugwee razloΩenye pros-
transtva L2 ( Ω ):
L2 ( Ω ) = H E( , )Ω ⊕ H E⊥ ( , )Ω ,
hde H E⊥ ( , )Ω — ortohonal\noe dopolnenye dlq H E( , )Ω . Otsgda dlq
funkcyy
V xG
H2 ( ) =
Ω
∫ G x y H y d y( , ) ( )2
poluçaem
V xG
H2 ( ) = H3 ( x ) + Φ ( x ),
hde
H3 ∈ H E( , )Ω , Φ ∈ H E⊥ ( , )Ω
Ω
Φ∫ =
H x x d x3 0( ) ( ) .
Zdes\ H3 ∈ H E( , )Ω , Φ ∈ H E⊥ ( , )Ω . Znaçyt,
v ( x ) = –
Ω
∫ G x y H y d y( , ) ( )1 +
Ω Ω
∫ ∫G x y G y z H z dz d y( , ) ( , ) ( )2 , x ∈ Ω.
Sohlasno uslovyg teorem¥ na ∂ Ω – E ymeem
∂
∂
v( )x
lx
= 0 = –
Ω
∫ ∂
∂
G x y
l
H y d y
x
( , )
( )1 +
+
Ω
∫ ∂
∂
G x y
l
H y d y
x
( , )
( )3 +
Ω
Φ∫ ∂
∂
G x y
l
y d y
x
( , )
( ) .
Sledovatel\no,
–
Ω
∫ H y H y d y( ) ( )1 +
Ω
∫ H y H y d y( ) ( )3 +
Ω
Φ∫ H y y d y( ) ( ) = 0,
hde H ∈ H E( , )Ω . Poskol\ku
Ω
Φ∫ H x x dx( ) ( ) = 0, Φ ∈ H E⊥ ( , )Ω ,
to
Ω
∫ H x H x dx( ) ( )1 =
Ω
∫ H y H y dy( ) ( )3 , H ∈ H E( , )Ω .
Otsgda poluçaem H x1( ) = H x3( ) , x ∈ Ω. Teper\ predstavlenye reßenyq v
( ∆
3
v = 0 ) prynymaet vyd
v ( x ) =
Ω
Φ∫ G x y y d y( , ) ( ) .
V sylu uslovyq teorem¥
∂
∂
v( )x
lx
=
Ω
Φ∫ ∂
∂
G x y
l
y d y
x
( , )
( ) , x ∈ ∂ Ω – E.
Znaçyt, sohlasno lemme
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 6
O EDYNSTVENNOSTY REÍENYQ ZADAÇY S NAKLONNOJ PROYZVODNOJ … 839
Φ ′ ( x ) = T Φ ( x ) = 0, x ∈ ∂ Ω – E.
Krome toho, dlq reßenyq v ymeem
v ( x ) =
Ω
Φ∫ G x y y d y( , ) ( ) =
Ω
Γ Φ∫ ( , ) ( )x y y d y –
∂Ω
Γ Φ∫ ′( , ) ( )x y y dSy .
Sohlasno uslovyg
∂
∂
2
2
v( )x
lx
= 0, x ∈ ∂ Ω – E.
Takym obrazom,
∂
∂
2
2
V x
l
i
x
Φ( )
–
∂
∂
2
2
U
l
i
x
′Φ
= 0, x ∈ ∂ Ω – E.
Oçevydno takΩe, çto
∂
∂
2
2
V x
l
i
x
Φ( )
–
∂
∂
2
2
V x
l
e
x
Φ( )
+
∂
∂
2
2
V x
l
e
x
Φ( )
–
∂
∂
2
2
U
l
i
x
′Φ
= 0, x ∈ ∂ Ω – E.
Yz teoryy potencyala yzvestno, çto
V xΦ( ) = U x′Φ ( ) , x ∈ ∂ Ω.
Ytak,
∂
∂
2
2
V x
l
e
x
Φ( )
=
∂
∂
2
2
U x
l
e
x
′Φ ( )
= 0, x ∈ ∂ Ω.
Poskol\ku Φ ′ ( x ) = 0, x ∈ ∂ Ω – E, to
∂
∂
2
2
V x
l
e
x
Φ( )
–
∂
∂
2
2
U
l
i
x
′Φ
= 0, x ∈ ∂ Ω – E.
Krome toho, v sylu formul¥ yz [10, c. 115]
0 =
∂
∂
2
2
V x
l
i
x
Φ( )
–
∂
∂
2
2
V x
l
e
x
Φ( )
= – Φ ( x )cos ( , )2 νx xl , x ∈ ∂ Ω – E.
Otsgda poluçaem Φ ( x ) = 0, x ∈ ∂ Ω – E. Dalee, tak kak
V xG
H2 ( ) = H3 + Φ,
to Φ ( x ) = 0, x ∈ ∂ Ω. Otsgda sleduet, çto H3 ( x ) = 0, x ∈ ∂ Ω, y
V xG
H2 ( ) = Φ ( x ).
Poskol\ku Φ ′ ( x ) = 0, x ∈ ∂ Ω – E, to
Ω
∫ V x H x dxG
H2 ( ) ( ) = 0, H ∈ H E( , )Ω .
Ytak,
Ω
∫ V x H x dxG
H2
2( ) ( ) = 0.
Takym obrazom, πnerhyq [9, c. 120] plotnosty H2 ravna nulg. Sledovatel\no,
H2 ( x ) = 0, x ∈ Ω,
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 6
840 D. V. KAPANADZE
v ( x ) =
Ω
∫ G x y V x dxG
H( , ) ( )2 = 0, x ∈ Ω.
Teorema 1 dokazana.
Teper\ rassmotrym uravnenye ∆
n
v = 0.
Teorema 2. Pust\ ohranyçennaq oblast\ Ω prynadleΩyt klassu C
(
2
n
–
2,
α
)
,
a v — reßenye uravnenyq ∆
n
v = 0 yz prostranstva C n( , )2 2− ( )α Ω . Pust\,
dalee, ∆
k
v ( x ) = 0, x ∈ E, k = 1, 2, … , n – 1, l ∈ C
(
n
,
α
)
. Esly v ( x ) = 0, x ∈ ∂ Ω,
y
∂
∂
k
x
k
x
l
v( )
= 0, x ∈ ∂ Ω – E, k = 1, 2, … , n – 1,
to v ( x ) = 0, x ∈ ∂Ω .
Dokazatel\stvo. Dlq reßenyq v spravedlyvo predstavlenye
v ( x ) = –
Ω
∫ G x y H y dy( , ) ( )1 1 1 1 +
Ω Ω
∫ ∫G x y G y y H y dy dy( , ) ( , ) ( )1 1 2 2 2 2 1 +
+ ( ) ( , ) ( , ) ( , ) ( )− … …−
− − − −∫ ∫ ∫1 1
1 1 2 2 1 1 1 1 1
n
n n n nG x y G y y G y y H y dy dy
Ω Ω Ω
.
Oçevydno takΩe sledugwee predstavlenye:
v ( x ) = –
Ω
Φ∫ G y x y dy( ) ( , )1 1 1 1,
hde
Φ1 ( y1 ) = H1 ( y1 ) –
Ω
∫ G y y H y dy( , ) ( )1 2 2 2 2 + …
… + ( ) ( , ) ( , ) ( )− … …∫ ∫ − − − − −1 1 2 2 1 1 1 1 2
n
n n n n nG y y G y y H y dy dy
Ω Ω
.
Netrudno vydet\, çto ∆ v ( x ) = Φ1 ( x ), x ∈ Ω , y
0 =
∂
∂
v( )x
lx
= –
Ω
Φ∫ ∂
∂
G x y
l
y dy
x
( , )
( )1
1 1 1,
0 =
∂
∂
2
2
v( )x
lx
= Φ1 ( x )cos ( , )2 νx xl , x ∈ ∂ Ω – E.
V sylu uslovyq teorem¥ Φ1 ( x ) = 0, x ∈ ∂ Ω. Krome toho,
∆
n
–
1
( ∆ v ) = ∆
n
–
1
Φ1 ( x ) = 0, x ∈ Ω.
Ytak, Φ1 — reßenye uravnenyq
∆
n
–
1
Φ1 ( x ) = 0, x ∈ Ω.
Yspol\zuem metod matematyçeskoj yndukcyy. Teorema 2 dokazana v sluçae
n = 3 (zametym, çto teorema 2 lehko dokaz¥vaetsq takΩe dlq byharmonyçeskoho
uravnenyq ∆
2
v = 0 ). Dopustym, çto teorema 2 spravedlyva dlq uravnenyq
∆
n
–
1
v = 0. DokaΩem teoremu dlq uravnenyq ∆
n
v = 0. Poskol\ku Φ1 ( x ) = 0,
x ∈ ∂ Ω, dlq lgboho k, 1 ≤ k ≤ n – 2, poluçaem
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 6
O EDYNSTVENNOSTY REÍENYQ ZADAÇY S NAKLONNOJ PROYZVODNOJ … 841
0 =
∂
∫
Ω
Φ1 2
0
( )
( )
x
x
l
dS
k
x
x
∂ ϕ
∂
= ( )
( )
( )−
∂
∫1 1
0
k
k
x
k x
x
l
x dS
Ω
Φ∂
∂
ϕ ,
ϕ ∈ C Rn
0
2 3− ( ), x0 ∈ ∂ Ω – E.
Zdes\ Φ1 rassmatryvaetsq kak obobwennaq funkcyq. Yz pred¥duweho raven-
stva sleduet
∂
∂
k
x
k
x
l
Φ1( )
= 0, x ∈ ∂ Ω – E, k = 1, 2, … , n – 2.
Netrudno vydet\, çto ∆ ∆k xv( )( ) = ∆
k
Φ1 ( x ) = 0, x ∈ E, k = 1, 2, … , n – 2.
Poskol\ku teorema 2 spravedlyva dlq uravnenyq ∆
n
–
1
v = 0, poluçaem
Φ1 ( x ) = 0, x ∈ Ω. Yz ravenstva ∆ v ( x ) = Φ1 ( x ) = 0, x ∈ Ω, sleduet, çto v —
harmonyçeskaq funkcyq v oblasty Ω. V sylu hranyçnoho uslovyq v ( x ) = 0,
x ∈ Ω .
Teorema 2 dokazana.
1. Lopatynskyj Q. B. Ob odnom sposobe pryvedenyq hranyçn¥x zadaç dlq system¥ dyffe-
rencyal\n¥x uravnenyj πllyptyçeskoho typa k rehulqrn¥m uravnenyqm // Ukr. mat. Ωurn. –
1953. – 5, # 2. – S. 132 – 151.
2. Vyßyk M. Y. Lekcyy po v¥roΩdagwymsq πllyptyçeskym zadaçam // Sed\maq letnqq mat.
ßkola (Kacyvely, 1969). – Kyev: Yn-t matematyky AN USSR, 1970.
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Poluçeno 29.06.2005,
posle dorabotky — 21.11. 2005
ISSN 1027-3190. Ukr. mat. Ωurn., 2006, t. 58, # 6
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| id | umjimathkievua-article-3499 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | rus English |
| last_indexed | 2026-03-24T02:43:39Z |
| publishDate | 2006 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/ba/93a62b6d5d97e34ba8c6f9de68717bba.pdf |
| spelling | umjimathkievua-article-34992020-03-18T19:56:00Z On the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0 O единственности решения задачи с наклонной производной для уравнения Δ n v = 0 Kapanadze, D. V. Капанадзе, Д. В. Капанадзе, Д. В. We prove the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0. Доводиться єдиність розв'язку задачі з похилою похідною для рівняння Δ n v = 0. Institute of Mathematics, NAS of Ukraine 2006-06-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/3499 Ukrains’kyi Matematychnyi Zhurnal; Vol. 58 No. 6 (2006); 835–841 Український математичний журнал; Том 58 № 6 (2006); 835–841 1027-3190 rus en https://umj.imath.kiev.ua/index.php/umj/article/view/3499/3735 https://umj.imath.kiev.ua/index.php/umj/article/view/3499/3736 Copyright (c) 2006 Kapanadze D. V. |
| spellingShingle | Kapanadze, D. V. Капанадзе, Д. В. Капанадзе, Д. В. On the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0 |
| title | On the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0 |
| title_alt | O единственности решения задачи с наклонной производной для уравнения Δ n v = 0 |
| title_full | On the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0 |
| title_fullStr | On the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0 |
| title_full_unstemmed | On the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0 |
| title_short | On the uniqueness of a solution of the problem with oblique derivative for the equation Δ n v = 0 |
| title_sort | on the uniqueness of a solution of the problem with oblique derivative for the equation δ n v = 0 |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/3499 |
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