數(shù)學(xué)物理方程第十章球函數(shù)ppt課件

1、0u0sin1)(sinsin1)(12222222ururrurrr偏微分方程常微分方程組分別變量 本征值問(wèn)題廣義傅立葉級(jí)數(shù)勒讓德多項(xiàng)式貝塞耳函數(shù)特殊函數(shù)特殊函數(shù)特殊函數(shù)勒讓德、埃米特、拉蓋爾等多項(xiàng)式；貝塞耳、虛宗量貝塞耳、球貝塞耳、超幾何，集合超幾何等函數(shù)。10.1 軸對(duì)稱球函數(shù)第十章 球函數(shù)0) 1(2)1 (222lldxdxdxdx一、勒讓德多項(xiàng)式1x有限,2, 1 ,0lkkakklklkkkllkka) 1)(2() 1)() 1)(2() 1() 1(2設(shè)最后一個(gè)不為零點(diǎn)系數(shù)有l(wèi)k llallla) 12(2) 1(22lk2) !(2)!2() 12(2) 1(llllll1

2、.代數(shù)表示那么對(duì)1)(0 xPcos)(1 xxP) 1cos3(41) 13(21)(222xxP適當(dāng)乘本征函數(shù)以常數(shù)使得)!2()!1(2)!22() 1(lllll)!2()!(2 !)!22() 1(2klklkklalkklklllkklxklklkklxP22/0)!2()!(2 !)!22() 1()(勒讓德多項(xiàng)式：2/l ：小于、等于l 的最大整數(shù)。0)0(12kP總有 x 。!2 !)!2() 1()0(22kkkPkkk獨(dú)一不含 x 的項(xiàng)kl2)cos33cos5(81)35(21)(33xxxP2(2 )!2 ( !)lllal-1-0.50.51-1-0.50.510.

3、511.522.53-1-0.50.51) 11(),(xxPl)0(),(coslP2. 微分表示羅德里格斯公式lllllxdxdlxP) 1(!21)(2證：證：)(202!)!(!) 1(!21) 1(!21kllkklllxkklllxlkllklkxklkklkl20) !2( !)!(2)!22() 1()(20!)!(21) 1(kllklkxkkl2()011(1)( 1)2 !2 ()! !llllkl kllllkddxxl dxdxlkk20(22 )(221)(21)( 1)2 ()! !lklklklklklkxlkk3.積分表示施列夫積分Clllllllldzxzz

4、lixdxdlxP122)() 1(!2121) 1(!21)(由科西公式C 繞 z=x 點(diǎn)。設(shè)半徑為12xC 上iexxz12dexidzi12dexiexexxlidzxzzliilililClll1)1(1)1(!2121)()1(!212120122212dexiexexexxxiiliii1121) 1(12212022222dexexexxxliii022222121) 1(12211) 1 (1PlP) 1() 1(1deexxlii02)(2111dxixlcos1120dxixxPllcos11)(20cosxdixPllcossincos1)(0dixPll0cossinc

5、os1)(dl02/222cossincos1dl02/22sincos1110d即1)(xPl二、 正交關(guān)系和模1)(x0)()(11dxxPxPlk1. 正交關(guān)系一個(gè)公式2. 模dxxPNll2112)() 1() 1()!21(2112112lllllllxdxddxdxdxddxl) 1() 1() 1() 1()!21(2211111121122lllllllllllllxdxddxdxdxddxxdxdxdxdl第一項(xiàng)為零，即) 1() 1() 1()!21(221111122llllllllxdxddxdxdxddxlN進(jìn)展 l 次分步積分后lllllllxdxdxdxlN)1(

6、)1()1()!21(22221122只需最高次冪才不為零，故lllllxxdxllN) 1() 1() 1() !(2)!2(11222再逐次進(jìn)展分步積分，得1222lNl即122lNl三、廣義傅立葉級(jí)數(shù)定義在區(qū)間 的函數(shù) 可以展開(kāi)為廣義傅立葉級(jí)數(shù) 1 , 1)(xf0),()(lllxPfxf展開(kāi)系數(shù)為dxxPxflfll)()(21211或區(qū)間 的函數(shù) 展開(kāi)為, 0)(f0),(cos)(lllPff系數(shù)為dPflfllsin)(cos)(2120例：例：在 ，中將 展開(kāi)為廣義傅立葉級(jí)數(shù)。 1 , 1432)(3xxxfkllklkklxklklkklxP22/0)!2()!(2 !)!

7、22() 1() 1()(解：解：比較展開(kāi)式最多含三階勒讓德多項(xiàng)式。1)(0 xPxxP)(1) 13(21)(22xxP)35(21)(33xxxP432)(3xxxf33221100PfPfPfPf)35 (213310 xxfxff3331025)23(xfxfff5215631f2253f40f543f32331ff310545214)(PPPxf例例2xxf)()(xfxdxxPxflfll)()(212110),()(lllxPfxf)()()(2121001dxxPxdxxPxlll)()()()(2121001dxxPxxdxPxllldxxPxPxlll)()(21210是奇

8、函數(shù)：)(12xPk012kfdxxPxkfkk)(22142102dxxdxdxkkkkkk22221022) 1() !2(214) 1() 1() !2(214221212101022121222dxxdxdxdxdxkkkkkkkkk1022222222) 1() !2(214kkkkxdxdkk1022202222210222222) 1() 1(nnkknnkkkkkkxCdxdxdxd因找出) 1(22kn項(xiàng)，它在 x=0 才不為零。10221212222) 1(kkkkkkxCdxd122)!22() 1(kkkCk121222)!22() 1() !2(214kkkkkCkk

9、kf10021xdxf)()!22() !2(214) 1()(2121222110 xPCkkkxPxkkkkkk例例32cos, 00rruu)(cos)(),(01llllllPrBrAru解：由軸對(duì)稱球內(nèi)含0r所以0lB)(cos),(0llllPrAru2cos0rru2000cos)(cos),(llllPrAru拉普拉斯方程的軸對(duì)稱問(wèn)題邊境條件與角 無(wú)關(guān)，可以推斷解也與角 無(wú)關(guān)。故0m邊境條件：1)(0 xP) 13(21)(22xxP20231)(32xPxP)(cos31)(cos32)(cos0200PPPrAllll310A20232rA )(cos32)(cos31),

10、(22020PrrPru32202rA例例 4cos0u0q, 0u)20(cos00uurr02u解：,cos/sin/,sin2ururuyuryruyury偶延拓：)2(cos00uurrcos0u)(cos),(0llllPrAruxuxPrArullll0000)(),()()!22() !2(214) 1()(2121222110 xPCkkkxPxkkkkkkxuxPrArullll0000)(),()()!22() !2(214) 1()(21212221100 xPCkkkxPukkkkkk)()!22() !2(214) 1(2),(220212220110 xPrrCkk

11、kuurukkkkkkkk例例 5均勻電場(chǎng)中放置介電常數(shù)的球，求介質(zhì)球內(nèi)、外的電場(chǎng)。0E0E解：無(wú)窮遠(yuǎn)處有邊境條件，球面處有銜接條件。取球坐標(biāo)，z-方向沿 。0Er軸對(duì)稱拉普拉斯問(wèn)題, 0u內(nèi)外分別討論，然后銜接起來(lái)。邊境條件：cos0rEure銜接條件：iuiueueuInternal:External:00rrerriuu電勢(shì)延續(xù)：電位移延續(xù)：0000rrerriruru)0( ru延續(xù)軸對(duì)稱拉普拉斯方程度解的普通方式：)(cos)(),(01llllllPrBrAru球內(nèi) 有限：), 0(ru)(cos),(0lllliPrAru球外無(wú)窮遠(yuǎn)邊值：cos)(cos),(00rEPrCrul

12、llle01EC10lCl)(coscos),(0100llllePrDrECru利用銜接條件：)(coscos)(cos01000000llllllllPrDrECPrA)(cos) 1(cos)(cos0200010llllllllPrDlEPrlA00CA 2010001rDrErA100llllrDrA301012rDEA2010) 1(llllrDlrlA00CA 解得00D0123EA030121ErD1 , 00lDAllcos23),(00rEAruicos121cos),(203000rErrEArue球內(nèi)電場(chǎng)強(qiáng)度：kEzArui0023),(kzujyuixuuEiiiii

13、023E0E10EEi四、母函數(shù)定義：0)(cos)(lllPzzF叫勒讓德多項(xiàng)式的母函數(shù)。dRr04 電荷在單位球的北極。求球內(nèi)任一點(diǎn)電勢(shì)。2cos2111rrd它又是拉普拉斯方程度內(nèi)解：)(coscos21102llllPrArr令000) 1 (11lllllllrAPrAr又011llrr所以1lA)(coscos21102lllPrrr即2cos211rr是勒讓德多項(xiàng)式的母函數(shù)。球外)(coscos211012llllPrBrr11r)(cos)1(cos1211012llllPrBrrr令00111111llrrrr所以)(cos1cos211012lllPrrr半徑 R 的球：)

14、(coscos210122llllPRrrrRR)(coscos210122llllPrRrrRR例6r1rd解：解：利用知結(jié)果。導(dǎo)體內(nèi)：等勢(shì)。導(dǎo)體外：無(wú)導(dǎo)體時(shí)20200cos2rrrrqu有導(dǎo)體時(shí)，設(shè)),(0rvuu0u0),(au接地0),(u00u20200cos2),(aarrqau0),(v0v2020cos2),(aarrqav0),(0uq0raq)(cos),(01llllPrBrv)(coscos2)(cos010202001llllllllPraqaarrqPaB1012lllraqB)(cos1cos2),(1010122020lllllPrraqaarrqru又2022

15、02010020cos2)()(cos1)(),(rrraraqraPrraqrarvlllluv是 處電荷 的電勢(shì)。這個(gè)電荷叫原電荷的鏡像。02ra0raq是原電荷的電勢(shì)與鏡像電荷的電勢(shì)的疊加。五、遞推公式)(21102xPrrrxlll兩邊求導(dǎo))()21 (012/32xPlrrrxrxlll或)()21 (210122xPlrrrxrrxrxlll)()21 ()()(0120 xPlrrrxxPrrxllllll兩邊同冪的系數(shù))() 1()(2)() 1()()(111xPkxxkPxPkxPxxPkkkkk0)()() 12()() 1(11xkPxxPkxPkkkk遞推公式)( )

16、( )() 12(11xPxPxPllll10.2連帶勒讓德函數(shù))()1 (22xyxmxyxmyxmm12222)1 ()1 (yxxmmyxmxyxmyxmmmm222212212222)1)(1()1 ()1 (2 )1 ( 01) 1()1(222xmlldxdxdxd01) 1(2 )1 (222xmllxx0)1(1) 1()1 (2)1 (2)1)(1()1 ()1 (2 )1 (222221222221222222122yxxmllyxxmyxxyxxmmyxmxyxmyxmmmmmmm0)1 (2)1)(1()1 ()1 (1)1)(1()1 (2)1 (2 )1 (2122

17、2122222222222222122yxxmyxxmmyxmyxxmyxllyxxxyxmyxmmmmmmmm1.函數(shù)設(shè)m是規(guī)定的0)1 (2)1)(1(1) 1(22 )1 (2122122212yxxmyxxmmmyyxmyllxymxyyx0)1 (11)1 () 1() 1( 2 )1 (2222212yxxmyxmxyllxymyx0)1() 1() 1( 2 )1 (12ymmllxymyx0) 1(2)1()(222mlllplldxdpxdxpdx) 1() 1()2(2)(222) 1(2)1 ()1(mlmlmlmlpmmmxppxdxpdx)() 1()(222mlml

18、mlmpxpdxdpx0) 1(22) 1(2)1 ()()() 1() 1() 1()2(2mlmlmlmlmlmlpllmpxppmmmxppx0) 1() 1() 1( 2)1 ()()() 1()2(2mlmlmlmlpllpmmxpmpx)()1 (22xPxPmlmml是 l 次多項(xiàng)式，求 l+1 次導(dǎo)數(shù)后變?yōu)榱?。lm, 2 , 1 , 0llPP 02. 微分表示lllllxdxdlP) 1(!212lmlmllmmlxdxdlxP) 1(!2)1 (222m情況：lmlmllmmlxdxdlxP) 1(!2)1 (22201) 1()1(222xmlldxdxdxd這也是勒讓德

19、方程滿足自然邊境條件的解。二階微分方程至少有兩個(gè)獨(dú)立解，但滿足特定邊境條件的解是獨(dú)一的，故這兩個(gè)解只相差一個(gè)常數(shù)。lmlmllmlmlmlmlmllmlmlmllmmlmlxdxdxdxdxxdxdlxxdxdlxPP) 1() 1()1 () 1(!2)1 () 1(!2)1 (222222222同項(xiàng)冪的比應(yīng)該就是這個(gè)常數(shù)。例如最高次冪：mklklkklklkmlmllmlmlxmlklkkllxklkldxdxdxd2202202) 1()!()!( !)!22( !) 1()!( !) 1(最高項(xiàng)：0kmlxmlkl)!()!22()!()!() 1()!()!22()!()!22()

20、1(2mlmlxmlklxmlklxPPmmlmlmmmlml3.積分表示CmlllmlmlmllmmldzxzzimllxxdxdlxP1222222)() 1(2)!(!2)1 () 1(!2)1 (4.正交關(guān)系0)()(11dxxPxPmlmk)(lk 5.模dxxPNmlml1122)()(dxxdxdxdxdlmlmllmlmllmlmllm112222) 1() 1() !(21)!()!() 1(多次分步積分：dxxdxdxdxdlmlmllllllllm112222) 1() 1() !(21)!()!() 1(dxxPxPmlmlNllml112)()()!()!()(122

21、)!()!(lmlml6.廣義傅立葉級(jí)數(shù))()(0 xPfxfmllldxxPxfmlmllfmll)()()!()!(21211m是規(guī)定的例1sin)1 ()(2/1211xxP)3()1 (2/12xx)13(21)1 ()(22/1212xxxP2sin23cossin3)(61)()!12()!12() 1()(1212112xPxPxP2sin41)13(21)1 ()(22/2222xxxP)1 (32x)2cos1 (23sin322sin)(xf)(cos3122P例2102)(dxxPldxxdxdlxllll) 1(!2)1 (222210dxxdxdxlllll) 1()

22、1 (!21222210) 1(2) 1()1(!2121110102112dxxdxdxxdxdxllllllll) 1() 1(2) 1()1(!2110211102102112llllllllllxdxdxdxdxxdxdxl1010211102112)(2) 1() 1()1(!21xxPxdxdxdxdxlllllllll1010211102112)(2) 1() 1()1(!21xxPxdxdxdxdxlllllllll02112112) 1() 1()1(!212xlllllllxdxdxdxdxl0)(2011)(20112) 1() 1()1(!212xkkllkklllkk

23、llkkllllxCdxdxCdxdxl01211212) 1()!1()!( 2) 1()!1()!( 2)1(!212xkklllkklkklllkklllklxClklxCxl01211212) 1()!1()!( 2) 1()!1()!( 2)1(!212xkklllkklkklllkklllklxClklxCxl第一項(xiàng)在 ，第二項(xiàng)在 不為零。12 kl12 kl) 1()!1()!21( 2) 1()!1()!21( 2!21221212121lllllllllCllCl) 1()!22(!2) 1()!2()!1( 2)!12(21211121212nnnnnnnnnCnnCn12

24、 nl)!22(!2)!2()!1( 2!)!1()!12()!12(21) 1(212nnnnnnnnnn)!22()!1(1!)!2(121) 1(22nnnnnn)!1()!22(1) 12() 1(21) 1(2212nnnnnn)!1()!22(1) 12() 1(21) 1(2)!12()!12(234212012nnnnmnumnnfnnn)()!1()!22(1) 12() 1(21) 1(2)!12()!12(234)(1202120 xPnnnnmnumnnxfmnnnn7.遞推公式由勒讓德多項(xiàng)式的遞推公式得之。10.3球函數(shù)1. 球函數(shù), 3 , 2 , 1, 2 , 1

25、 , 0cossin)(cos),(llmmmPYmlml2.正交關(guān)系klnmddYYnkSml, 0sin),(),(3.模2022112122)!()!(cossinsin)(cossin),(lmlmldmmdPddYmmlSml4.球面上的廣義傅立葉級(jí)數(shù))(cossincos),(0mlmlmmlmlPmBmAf 例1cos)(coscossin11P1, 1lmsin)(cossinsin11P例21cossin3),(22f留意：lmcos,cos122cos1sin3),(2f2cossin231sin23222cossin23) 1cos3(21222cos)(cos21)(cos2122PPl例3偶極矩的電場(chǎng)中的電勢(shì)qqrsqxyzq),(zyx解1)(14),(2222220zyxzysx