第三篇:特殊函數(shù)第二章勒讓德多項(xiàng)式主要內(nèi)容:◆勒讓德多項(xiàng)式(軸對(duì)稱問(wèn)題)及性質(zhì)◆連帶勒讓德函數(shù)(轉(zhuǎn)動(dòng)對(duì)稱問(wèn)題)◆球函數(shù)(一般問(wèn)題)在分離變量法一章中,我們已經(jīng)知道拉普拉斯方程10,2,Ou10102(sin0)+0Orarrsin8aesinaop在球坐標(biāo)系下分離變量后得到歐拉型常微分方程dr+2rx-l(+1)R=0和球諧函數(shù)方程SInsineae+(+1)y=0a0sin0a繼續(xù)分離變數(shù),令(,q)=6(0)(q),得到關(guān)于o的方程Φ"+=0(1)2=0時(shí),①(q)=C1p+C2=C2a(+2r)=o()(2)=m,p()=Acosmo+Bsinmpa()=Acosmo+Bsinmo,m=0,1,2,3,球函數(shù)Y(,p)=(Acosmp+Bsinmop)⊙(O,其中⊙(0)需從連帶勒讓德方程解出:(1⊙j⊙=0,x=cosm=0時(shí),成為階勒讓德方程1-x24-/4(+1)=0用常點(diǎn)鄰域令⊙=y的級(jí)數(shù)解法a1X=40()+y(x)={4(x,為偶數(shù)時(shí)ay(x),/為奇數(shù)同樣若記O=arccosxy(x)=Q(x)則上述方程也可寫為下列形式的l階勒讓德方程[(1-x2)]+l(1+1)y=0dxdx軸對(duì)稱球函數(shù)現(xiàn)在注意:m=O時(shí),①()=Acosmo+Bsinmop=A(常數(shù))u(r,0,)=R(r)o()()=AR(r)(0)=u(r,0)l與g無(wú)關(guān),只與r,θ有關(guān)。意味著當(dāng)r,θ一定時(shí),φ可任意改變,u不變即在以,O構(gòu)成的錐體上各點(diǎn)的值相同問(wèn)題關(guān)于極軸(z軸)對(duì)稱球函數(shù)Y(O,9)=A⊙(O)~稱為軸對(duì)稱球函數(shù)前面已學(xué):勒讓德方程在x=+1自然邊界條件:yx→有限,從而構(gòu)成本征值問(wèn)題,本征值是1(+1),l=0,1,2,3…在為整數(shù)條件下,勒讓德方程的兩個(gè)線性獨(dú)立特解y(x)=a00(x)+ay(x)之一退化為次多項(xiàng)式0(0)=y(r)為2k(偶數(shù))any(x)~x261為2k+1(奇數(shù):ay1(x)~x261將它們分別乘上適當(dāng)?shù)某?shù),叫做階勒讓德y多項(xiàng)式,記作P(P(x)→>軸對(duì)稱情況下的球函數(shù)。Y(,q)=A⊙()=Ay(x)→P§2·1勒讓德多項(xiàng)式·勒讓德方程的求解勒讓德多項(xiàng)式勒讓德多項(xiàng)式的性質(zhì)、母函數(shù)和遞推公式·勒讓德多項(xiàng)式的應(yīng)用(2)勒讓德多項(xiàng)式通常約定:用適當(dāng)?shù)某?shù)乘以本征函數(shù)使最高次冪項(xiàng)x的系數(shù)為:a1=2(!)2P(x)=∑ax2(相鄰兩項(xiàng)相差2次)利用系數(shù)遞推公式:a(k+2)(k+1)(k-D)(k+l+1)推算出其他系數(shù):ana12na14,a12,a2)(21-1)2(21-1)2(1!2(21A1)2(-)!(-1)(1-2(21-2)!(2l-2)!2(l-1)!(l-2)2(l-1)!(-2)21-3)2H2)-301-41(1-2)2!2(-2)(-4)!4=(1)22-532(-3)(-6)2(-n)l-2n2勒讓德多項(xiàng)式:y=P(x=2421-2n)n!2(-n)!-2n將指標(biāo)n→kk2(1=2,按降冪排列的次多項(xiàng)式。Pa=∑y(2-2k)勒讓德方程的解:我們知道:在自然邊界條件下,勒讓德方程的解P(x)為P(x)=∑(-1)2k)!2