Laplace s equation in Cylindrical Coordinates

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1 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 Lplce eqution in Cylindricl Coordinte 1- Circulr cylindricl coordinte The circulr cylindricl coordinte (, φ, z ) re relted to the rectngulr Crtein coordinte ( x, y, z ) by the forul (ee Fig.): Circulr cylindricl coordinte. x = co φ, y = in φ, <, φ π, < z < z = z. The invere reltion re: y = x y, tn φ =, z = z. x An infiniteil length d i ( φ ) d = d d dz An infiniteil volue eleent i: di = d dφ dz. The grdient, divergence, curl nd Lplcin becoe, in cylindricl coordinte re: Grdient V 1 V ˆ ˆ V V = φ zˆ z Divergent 1 1 i v = ( v ) ( vφ ) ( v z ) z Curl 1 v v z φ 1 v v z ˆ 1 v v = ˆ φ ( v ) zˆ φ z z 1

2 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 Lplcin 1 V 1 V V V = = z Lplce eqution in two dienion (Conult Jckon (pge 111) ) Exple: Solve Lplce' eqution by eprtion of vrible in cylindricl coordinte, uing there i no dependence on z (cylindricl yetry). Mke ure tht you find ll olution to the rdil eqution. Doe your reult ccoodte the ce of n infinite line chrge? Anwer: For yte with cylindricl yetry the electrottic potentil doe not depend on z. V Thi ieditely iplie tht =. Under thi uption Lplce' eqution red: z 1 V 1 V V = = Conider poible olution of V: V (, φ) = R Φ ( φ) Subtituting thi olution into Lplce' eqution we obtin Φ( φ) R R Φ( φ) = Multiplying ech ter in thi eqution by nd dividing by R( ) Φ ( φ) we obtin R 1 Φ( φ) = R Φ( φ) The firt ter in thi eqution depend only on while the econd ter in thi eqution depend only on φ. Thi eqution cn therefore be only vlid for every nd every φ if ech ter i equl to contnt. Thu we require tht: R = γ contnt (A) R nd 1 Φ( φ) = γ (B) Φ( φ) 1- conider the ce in which γ = >. The differentil eqution for Φ ( φ) cn be rewritten Φ ( φ) Φ ( φ) = The ot generl olution of thi differentil olution i φ φ Φ ( φ ) = Ce De However, in cylindricl coordinte the ngleφ ut be unique, nely, Φ ( φ π) =Φ ( φ) nd therefore the generl olution of the eqution d Φ Φ= i not tified for thi olution, nd dφ we conclude tht γ = >. The differentil eqution for Φ ( φ) cn be rewritten

3 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 Φ ( φ) ( φ) The ot generl olution of thi differentil olution i Φ φ = C co φ D in φ The condition tht Φ ( φ) =Φ ( φ π) require tht i n integer. Now conider the rdil function R ( ). We will firt conider the ce in which γ = >. - Conider the following olution for R ( ) : k R = A, A = contnt Subtituting thi olution into eqution (A) we obtin k A k k = = A Therefore, the contnt k cn tke on the following two vlue: k =, k = The ot generl olution for R( ) under the uption tht R = A Now conider the olution for R ( ) when R R B > i therefore =. In thi ce we require tht = = = contnt Thi eqution cn be rewritten R( ) = If = then the olution of thi differentil eqution i b contnt R = = If then the olution of thi differentil eqution i ( ) ln b R = Cobining the olution obtined for = with the olution obtined for > we conclude tht the ot generl olution for R( ) i given by B R = ln b A = 1 Therefore, the ot generl olution of Lplce' eqution for yte with cylindricl yetry i B V (, φ) = ln b A ( C co( φ) D in ( φ) ) = 1 3

4 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 Lplce eqution in three dienion Lplce' eqution in cylindricl coordinte tke the for: 1 V 1 V V V = = (1) z Conider poible olution of V: V (, φ, z) = R Φ ( φ) Z ( z) () Subtituting () into (1) we obtin 1 R( ) 1 Φ( φ) Z z V = z = (3) V R Φ( φ) Z ( z) Tking ( Z / z )/ Z to the right-hnd ide of the eqution we hve n expreion independent of z on the left, fro which we conclude tht either expreion (on the right or on the left) ut be contnt. Explicitly putting in the ign (which ut till be deterined fro the boundry condition) of the eprtion contnt, we hve: 1- Aziuthl direction: 1 Φ( φ) = γ. (3) Φ( φ) Firt conider the ce in which γ = >. The differentil eqution for Φ ( φ) cn be rewritten Φ ( φ) Φ ( φ) = The ot generl olution of thi differentil olution i φ ( φ ) C e D e φ Φ =. (3b) However, in cylindricl coordinte the ngleφ ut be unique, nely, Φ ( φ π) =Φ ( φ) nd therefore the generl olution of the eqution d Φ Φ= i not tified for thi olution, nd dφ we conclude tht γ = >. The differentil eqution for Φ ( φ) cn be rewritten Φ ( φ) Φ ( φ) = (4) The ot generl olution of (4) i Φ ( φ) = A in( φ) B co( φ) (5) The condition tht Φ ( φ) =Φ ( φ π) require tht i n integer. Next, conider the econd prt, i.e.: Z ( z) z = λ inh coh Z ( z) Z z = A λ z B λ z (4) Or, lterntively, 4

5 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 Z ( z) z = λ Z ( z) Z = A in ( λ z ) B co( λ z ) (5) For the choice (4), Eq (3) reduce to: R( ) Φ( φ) λ = = (6) R( ) Φ( φ) Finlly, conider the rdil function R ( ), in the for: ( ) R ( λ ) R = (7) i the Beel eqution, hving olution R = EJ( λ) FN ( λ) (8) where J nd N re Beel nd Neunn function of order. Hd we picked the negtive eprtion contnt in eqution (7), we would hve obtined for R ( ) the odified Beel eqution: R( ) ( λ ) R = (9) hving olution the odified Beel function I ( λ ) nd K ( λ ). R = EI ( λ) FK ( λ) (11) Note: K ( λ ) nd N ( λ ) diverge t r = nd re therefore excluded fro proble where the region of interet include r =, while J ( λ) nd I ( λ ) diverge r nd will therefore be excluded fro ny exterior olution. The coplete olution i then of the for: J ( λ) in ( φ) inh ( λz ) V (, φ, z) = λ, N ( λ) co( φ) coh ( λz ) (1) I ( λ) in ( φ) in ( λz ) λ, K ( λ) co( φ) co( λz ) where the brce { } tnd for the rbitrry liner cobintion of the two ter within. 5

6 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 Exple (6): A cylinder of rdiu nd height L, i plced prllel to the z xi. It bi t it fce t =. The bi t z = L i held t given potentil Vo (, ϕ ) ( given function). Find the potentil everywhere within the cylinder. z = i grounded, nd o i Solution: Let u firt conider the generl olution of the three eqution, with the boundry condition: () B. C. V (, φ,) = (1 ) V (, φ, L) = V (1 b) V (, φ, z) = () V (, φ, z) = finite (3) V = Vo (, ϕ) V = i- The ngleφ ut be unique, nely, Φ ( φ π) =Φ ( φ) nd therefore the generl olution of the d eqution Φ Φ = will be: dφ Φ ( φ) = A co( φ) B in( φ) with n integer. ii- In our ce, Z ut vnih t z =, but not t z = L, which en we hve the eqution d Z k Z = nd the Z function i of the for: dz Z ( z) = C inh( kz ) D coh( kz ), iii- Due to the bove ite i nd ii, R ut be the olution of the eqution d R 1 dr k R= nd tken to be of the for: d d R = EJ ( k ) FN ( k ) 6

7 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 (b) The generl olution of the Lplce eqution for the proble in cylindricl coordinte conit of u (uperpoition) of ter of the for: ( φ ) V,, z =R Φ( φ) Z ( z) [ EJ ( k ) FN ( k )] A in ( φ) B co( φ) C inh ( kz ) D coh ( kz ) = = I- B.C. 1 nd 1b iplie D = II- B.C. 3 iplie F = IIIx n B.C. iplie J( k n) = k n =, n = 1,,3, x n i the n th root of J( k n ). Reeber tht, Beel function h n infinite nuber of root, nd therefore κ tke n infinite nuber of dicrete vlue, ll of the re root of the th Beel function. Nely, k i the n th root of the th Beel function. n It follow tht the generl olution of our proble i n= 1 = { } V (, φ, z) = J ( k )inh( k z) A in φ B co φ We now ipoe the boundry condition t z = L: n= 1 = n n n n { } V (, φ) = J ( k )inh( k L) A in φ B co φ n n n n Thi i Fourier erie in φ nd Fourier-Beel erie in. We now ue thi property with the boundry condition t z ter of the given function Vo (, ϕ ). Firt, we ue (Fourier trick) the delt function of the trigonoetric function in the for: if n nπy πy coco dy = δn = if = n if = n = if n nπy πy inin dy = δn = if = n nπy πy inco dy = if n i even nπ y in dy = if n i odd nπ to obtin = L to deterine ll the coefficient in 7

8 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 1 An inh( k nl) J ( k n) = V (, ϕ)in ( ϕ) dϕ π π π 1 B n inh( k nl) J ( k n) = V (, ϕ)co ( ϕ) dϕ π Secondly, we ue the orthonorl property of the Beel function, which cn be written in the for J ( x n ' ) J ( x n ) d = J 1 ( x n ) ν ν nn ' ν ν δ ν ν. Then π An = dϕ J (, )in k n V φ ( φ) d, πj ( x )inh( k L) 1 n n π B = dϕ J ( k ) V (, φ)co φ d n n πj 1( x n)inh( k nl) which coplete the olution. For the pecil but iportnt ce of ziuthl yetry, for which V i independent of φ, i.e. =, we obtin: A n = 4δ, B n = J ( k n ) V ( ) d J ( x )inh( k L) 1 n n The reon we obtined dicrete vlue for k w the dend tht φ vnih t =. If we let, then k will be continuou vrible, nd inted of u over k, we will obtin n integrl. Thi i copletely nlogou to the trnition fro Fourier erie to Fourier trnfor, but we will not purue it further. 8

9 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 Exple: A hollow right circulr cylinder of rdiu h it xi coincident with the z xi nd it end t z = nd z = L. The potentil on the end fce i zero, while the potentil on the cylindricl urfce i given contnt V.Uing the pproprite eprtion of vrible in polr coordinte; () Write down the boundry condition (condition). (b) Ue the phyicl principl to write down the generl olution. (c) Ue the boundry condition in () to iplify the generl olution in the eprte coordinte. Write your reon for dropping ny ter or ter. (d) Find erie olution for the potentil nywhere inide the cylinder. Solution: () B. C. V (, φ,) = (1 ) V (, φ, L) = (1 b) V (, φ, z) = V () V (, φ, z) = finite (3) i- The ngle φ ut be unique, nely, Φ ( φ π ) =Φ ( φ) nd therefore the generl olution of the eqution d Φ ν Φ = will be: dφ B Φ ( φ) = Aco νφ in νφ with ν n integer. ii- In our ce, Z ut vnih t z = nd z = L, which en we hve the eqution d Z k Z = nd the Z function i of the for: dz Z ( z) = C in kz Dco kz, iii- Due to the bove ite i nd ii, R ut be the olution of the eqution d R 1 dr k R= nd tken to be of the for: d d R ( ) = EI ( k ) FK ( k ) n n n n (b) The generl olution of the Lplce eqution for the proble in cylindricl coordinte conit of u (uperpoition) of ter of the for: V, φ, z =R Φ( φ) Z ( z) [ E I ( k ) F K ( k ) ] A in( φ) B co( φ) C in( k ) D co( k ) = ν ν nν ν ν ν ν ν ν = For Z-direction: B1- Boundry condition (1) iplie D =. B- Boundry condition (1b) iplie 9

10 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 Z( L) = C in kl = kl = nπ nπ k n =, n = 1,,3, L n = give trivil olution. For Z-direction: Since we're looking for the potentil inide the cylinder nd there i no chrge t the origin, the olution ut be finite (B. C. 3), which require F =. Then the potentil expnion becoe t (4) ( φ ) = ( νφ) ( νφ) V,, z Anν in Bn ν co in k nz Iν( k n) = n = 1ν = (5) ( φ ) = = ( νφ) ( νφ) V,, z V Anν in Bn ν co in k nz Iν( k n) n = 1ν = Multiplying (5) by in ( k n ' z ) nd integrte, we find:. L ( k z ) co n ' V k n ' ( νφ ) ( νφ ) V in k n' z dz = Iν( k n) An ν in Bn ν co in k nz in k n' z dz n = 1ν = So: LV L co ( n' π ) 1 = Iν( k n' ) An' ν in ( νφ ) Bn' ν co( νφ ) n ' π (6) ν = n ' ( 1) Relbeling n = n', then for n odd, LV L = Iν( k n) An ν in( νφ ) Bn ν co( νφ ) nπ (7) Eqution (7) iplie A ν =, n And finlly, ν = ν = (No ter contin in ( νφ ) or (, φ, ) V z B L L δnn ' co νφ in the left hnd ide), then 4V 1 n = (8) nπ I( k n) = π n = 1,3,5, ( n ) 4V 1 in k z I ( k n) n I ( k ) n 1

11 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 Appendix (Repeted) Helholtz eqution in cylindricl coordinte In thi ppendix we will diue the generl olution of Helholtz eqution ψ K ψ =, in cylindricl coordinte. Strting with the 1 ψ 1 ψ ψ k ψ = (1) z z Expnd eqution (1) nd ultiply by, we hve: ψ ψ 1 ψ ψ K ψ = () z Uing the ethod of eprtion of vrible, ψ = R Φ( φ) Z( z) (3) eq. () will be reduced to, fter dividing by RΦ Z, 1 R Z R Φ K = (4) R R Φ Z z Since the left hnd ide of (4) i function of nd φ nd the right hnd ide i function of Z only, we y write 1 Z = λ (6) Z z with olution Z ( z) = A co( λ z ) B in ( λ z ) (6) where λ i contnt nd R R 1 Φ K λ (7) R R = Φ On eprting the vrible in (7), we obtin 1 Φ = (8) Φ with olution Φ ( φ) = C co( φ) Din( φ) (9) At thi tge, λ nd re unknown contnt. However, we will be intereted in proble in which the dependence on the ngle i uniquely defined, Φ ( φ π) =Φ ( φ) nd therefore = n, where n i n integer. Clerly we know the generl olution for the differentil eqution of Z ( z ) nd Φ ( φ). Wht bout the function R? The third eqution then red R R K λ (1) { } R= For K λ = α, where α = contnt, eqution (1) becoe 11

12 Prof. Dr. I. Ner Phy 571, T-131 -Oct-13 R R { α } P = (1) We now ke the following chnge of the dienionle vrible ξ : ξ = α So tht d d dξ d = = α, d dξ d dξ d d = α d dξ By ue of thi chnge of vrible, eq. (1) reduce to: R 1 R 1 R= ξ ξ ξ ξ (11) Which i the Beel eqution, nd it olution re clled Beel ( ) function (or cylindricl function) 1

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