Задачи по векторному анализу. Михайлов В.К - 102 стр.

                  ∂u = −a ∂u = −ρa ∂u
                  ∂ρ     ρ ,
                             ∂ϕ    ϕ ,    = −az ;                        (4.12)
                                       ∂z
â ñôåðè÷åñêèõ êîîðäèíàòàõ ýòà ñèñòåìà èìååò âèä:
               ∂u = −a ∂u = −ra ∂u = −r sin θ a
                      r,       θ , ∂ϕ          ϕ.                        (4.13)
               ∂r        ∂θ
     Ïðèìåð 3. Íàéòè ïîòåíöèàë ïîëÿ
           ρ                       ρ           ρ     ln ρ ρ
           a = − 1 arctg z + cos ϕ  eρ + sin ϕ eϕ =       e
                ρ                                   1 + z2 z .
     Ðåøåíèå. Ïðåæäå âñåãî óáåäèìñÿ â ïîòåíöèàëüíîñòè ýòîãî
                                                   ρ
ïîëÿ. Íåñëîæíûå âû÷èñëåíèÿ ïîêàçûâàþò, ÷òî rot a = 0 . Îäíàêî
                             ρ
îáëàñòü îïðåäåëåíèÿ ïîëÿ a íåîäíîñâÿçíà: îíî íåîïðåäåëåíî íà
îñè z, ò. å. ïðè ρ = 0. Ñëåäîâàòåëüíî, íåîáõîäèìî ïðîâåðèòü ðà-
                                   ρ
âåíñòâî íóëþ öèðêóëÿöèè ïîëÿ a ïî êàêîìó-ëèáî êîíòóðó, îõ-
âàòûâàþùåìó îñü z. Âûáåðåì â êà÷åñòâå òàêîãî êîíòóðà îêðóæ-
íîñòü ρ = R, z = 0. Ýòî äàåò:

             ρ    ρ                                   2π

           ∫ a ⋅ dl   = ∫ (aρ dρ + ρaϕ dϕ )ρ = R = R ∫ sin ϕ dϕ = 0 .
                                                      0

                        ρ
Ñëåäîâàòåëüíî, ïîëå a ÿâëÿåòñÿ ïîòåíöèàëüíûì. Èñêîìûé ïî-
òåíöèàë u (ρ, ϕ, z) ÿâëÿåòñÿ ðåøåíèåì ñèñòåìû (4.12):

                            ∂u = 1 arctg z + cos ϕ ,
                            ∂ρ ρ
                            ∂u
                            ∂ϕ = −ρ sin ϕ ,
                                                                       (4.14)
                            ∂u = ln ρ2 .
                            ∂z 1 + z

Èíòåãðèðîâàíèå ýòîé ñèñòåìû ìîæíî íà÷èíàòü ñ ëþáîãî óðàâ-
íåíèÿ, íî ïðîùå íà÷àòü ñ òðåòüåãî:
                          u = ln ρ ⋅ arctg z + C1( ρ, ϕ ) .
Ïîäñòàâëÿÿ ýòî âûðàæåíèå â ïåðâîå óðàâíåíèå (4.14), ïîëó÷àåì:

                 1          ∂C ( ρ, ϕ ) 1
                   arctg z + 1         = arctg z + cos ϕ ,
                 ρ            ∂ρ        ρ


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