Задачи по векторному анализу. Михайлов В.К - 91 стр.

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91
ãäå H
1
, H
2
, H
3
— êîýôôèöèåíòû Ëàìý äàííîé ñèñòåìû êðèâîëè-
íåéíûõ êîîðäèíàò.  ÷àñòíîñòè, â öèëèíäðè÷åñêèõ êîîðäèíàòàõ
(q
1
=ρ
, q
2
=ϕ
, q
3
=z; H
1
= 1, H
2
=ρ
, H
3
= 1):
d
az
d
az
dz
az
ρ
ρϕ
ρϕ
ρϕ ρϕ
123
(,,) (,,) (,,)
==
;
â ñôåðè÷åñêèõ êîîðäèíàòàõ (q
1
=r, q
2
=θ
, q
3
=ϕ
; H
1
= 1, H
2
=r,
H
3
=rsin
θ
):
dr
ar
rd
ar
rd
ar
123
(, , ) (, , )
sin
(, , )
θϕ
θ
θϕ
θϕ
θϕ
==
.
Ïðèìåð 1. Íàéòè âåêòîðíûå ëèíèè ïîëÿ
ρρ ρ
ae e=+ =
ρϕ
ϕϕ
{, , }10
.
Ðåøåíèå. Äèôôåðåíöèàëüíûå óðàâíåíèÿ âåêòîðíûõ ëèíèé
dd
dz
ρρϕ
ϕ
10
==
.
Îòñþäà z=C
1
,
ρ
=C
2
ϕ
. Ýòî ñïèðàëè Àðõèìåäà â ïëîñêîñòÿõ, ïåð-
ïåíäèêóëÿðíûõ îñè z .
4.3.2. Ãðàäèåíò
Ïóñòü çàäàíî ñêàëÿðíîå ïîëå u (q
1
, q
2
, q
3
). Òîãäà åãî ãðàäèåíò
∇=
u
H
u
qH
u
qH
u
q
11 1
11 2 2 3 3
∂
∂
∂
∂
∂
∂
,,
. (4.6.)
 ÷àñòíîñòè, â öèëèíäðè÷åñêèõ êîîðäèíàòàõ:
=∇
z
u
,
u
æ
,
æ
u
u
∂
∂
∂ϕ
∂
∂
∂
1
;
â ñôåðè÷åñêèõ êîîðäèíàòàõ:
∇=
u
u
rr
u
r
u
∂
∂
∂
∂θ θ
∂
∂ϕ
,,
sin
11
.
ãäå H1, H2, H3 — êîýôôèöèåíòû Ëàìý äàííîé ñèñòåìû êðèâîëè-
íåéíûõ êîîðäèíàò.  ÷àñòíîñòè, â öèëèíäðè÷åñêèõ êîîðäèíàòàõ
(q1 = ρ, q2 = ϕ, q3 = z; H1 = 1, H2 = ρ, H3 = 1):
                     dρ             ρdϕ              dz
                               =               =
                 a1( ρ, ϕ , z ) a2 ( ρ, ϕ , z ) a3( ρ, ϕ , z ) ;
â ñôåðè÷åñêèõ êîîðäèíàòàõ (q1 = r, q2 = θ, q3 = ϕ; H1 = 1, H2 = r,
H3 = r sinθ):

                     dr           rdθ         r sin θdϕ
                              =             =
                 a1(r, θ , ϕ ) a2(r, θ , ϕ ) a3(r, θ , ϕ ) .
                                          ρ ρ       ρ
     Ïðèìåð 1. Íàéòè âåêòîðíûå ëèíèè ïîëÿ a = eρ + ϕeϕ = {1, ϕ ,0} .
     Ðåøåíèå. Äèôôåðåíöèàëüíûå óðàâíåíèÿ âåêòîðíûõ ëèíèé
                             dρ ρdϕ dz
                                =   =
                              1   ϕ   0 .
Îòñþäà z = C1, ρ = C2ϕ. Ýòî ñïèðàëè Àðõèìåäà â ïëîñêîñòÿõ, ïåð-
ïåíäèêóëÿðíûõ îñè z .

                            4.3.2. Ãðàäèåíò
     Ïóñòü çàäàíî ñêàëÿðíîå ïîëå u (q1, q2, q3). Òîãäà åãî ãðàäèåíò

                        1 ∂u 1 ∂u 1 ∂ u 
                  ∇u =        ,      ,       .                   (4.6.)
                        H1 ∂q1 H2 ∂q2 H3 ∂q3 
 ÷àñòíîñòè, â öèëèíäðè÷åñêèõ êîîðäèíàòàõ:
                              ∂u 1 ∂u ∂u 
                        ∇u =  ,      , ;
                             ∂æ æ ∂ϕ ∂z 
â ñôåðè÷åñêèõ êîîðäèíàòàõ:

                     ∇u = ∂u , 1 ∂u , 1 ∂u  .
                                                
                           ∂r  r ∂θ  r sin θ ∂ϕ 




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