工程數(shù)學(xué)2012 ch09本征函數(shù)法

§ ,2u(x,

a22u(x,t) 0x 0 utt(x,t)a2uxx(x,FT/其中a FT/u(x,u(x,

0 u(x,

初始條件u(x

0x u(x,t)X(x)T Xd

2d2 T或者寫(xiě)為 aX XT XT X(x)X(x)

(t)a2T(t) X(0)T(t) X(0) X(l)T(t) 即X(l) X(x)T(0) X(x)T(0) 由方 X(x)X(x) X(0)X(l) 構(gòu)成的問(wèn)題稱(chēng)為本征值問(wèn)題使本征值問(wèn)題有非平庸解(非恒零解)的值稱(chēng)為方程(4)在邊界條件(4a)下如果0,方程(4)X(x)A由(4a)得A

B0,X(x)0.即0如果0,方程(4)X(x)A xB l0.因此ln,n1,n 本征值n ,本征函數(shù)Xn(x) 將nT(t)a2T(t)nTn(t)Cn natDn nCcosnatDsinn un(x,t)Xn(x)TnsinnxCcosnatDsinn

t,

u(x, un(x, xCn tDn u(x,應(yīng)用初始條件u(x

0x (x)Cn

nasinnx 系數(shù) 2l()sinl l0

na 0u(x,t)CcosnatDsinn l

A

n

t

sinnx n x CnAncosn DnAnsinnAC2D21/2, arccotC/D 每個(gè)un(xt)對(duì)應(yīng)于一種駐波§ u(x, u(x,定解問(wèn)題

a22u(x,u(x,u(x,

0x 00,0u(x,t)

0x

l

tu(x,t)X(x)T T X"(x) X X"(x)X(x)0,T'(t)a2T(t) X'(0)X'(L) X"(x)X(x) X'(0)X'(l)0時(shí)，方程的通解X(x)A滿(mǎn)足邊界條件的解X0x0時(shí)，方程的通解X(x)A xB X(x)Acosn

n T'(t)a2T(t)00時(shí) T0(t)

,

ana2 Cne an u(x,t)C0 Cne

(x)u(x,t) C

Ccosn 利用系數(shù)

nnC0

1l()dl

Cn

lll

nd. an u(x,ll

C0Cne

C0

1l()dl

Cn

()cosnll ll當(dāng)t u(x,t)C0.u(x, 定解問(wèn)題u(x

a22u(x,u(x,u(x,

0x 00,0u(x,

(x), 0

u(x,t)X(x)T T X"(x) X X"(x)X(x)0,T'(t)a2T(t) X(0) X'(l) X"(x)X(x) X(0)0,X'(l)0X(x)ABx,滿(mǎn)足邊界條件X0x)0,00時(shí)，方程的通解X(x)A xB

l本征值ln1 n1

l 對(duì)應(yīng)的本征函數(shù)X(x)n

n1 T'(t)a2T(t)

n1n1,2,...時(shí),T(t)Ce

2lt2 2 l

u(x,t)Cen2

sinn1

2 (x)u(x,t)

C

n1 2 利用系數(shù) ()sinn2

l 2l()sinn1

22 2

l

u(x,t)

n

tsinn122(

2 C

l 2u(xu(x,當(dāng)tu(xt0.

§ f lxf(x)f(x2ml) n f(x)C0Cncos n

sin

n1 1 2l1

lCn lll1Dn 1

l lndl f 0x

(x)f lxff

(xf(x2ml) 1

f(x)

l2 lDnl

l

dl

f() l0lf(x)滿(mǎn)足第一類(lèi)邊界條件f(0)f(l) f 0x

(x)f lx(xf(x2ml)

偶函數(shù)展開(kāi)f(x)C0ll

l

12ll

1l 1l0

lCnl

l

2l 2l0

f(x)滿(mǎn)足第二類(lèi)邊界條件f

f

f 0

x延拓,令函數(shù) f(x)f(2l lxf (x)f 2lxff(xf(x4ml)內(nèi)其它區(qū)域式中,m12,.... f(x)

2

2lf() 01 0l f()sin2ldf()sin2ldl 變量代換2l或2l,利用f(f(2l f()

d

f(2l)

d(2l f()sin 其中m2n(n1,2,...)時(shí)取號(hào),m2n1時(shí)取號(hào).1 l因此Dm f()sin2ldf()sin2ldl 2 l l ll2lDm2l0

令m2n1Dm和f(x)ll2lDn2l0

f(x)

f(x)滿(mǎn)足第三類(lèi)邊界條件f(0)0,f

u 0x 0 |x0T

ux

xl

0,0u (x),0x t v(x,t)u(x,t)v 0x 0 |x0

vx

xl

0v (x)T,0x t結(jié) u(x,t)v(x,t)u 0x 0 u|xlT2,0u u

0x v(x,t)u(x,t)并使(0T1,(l)v 0x 0 v (x) 0x t u(x,t)v(x,t)例如可設(shè)(x)T2T1u a/(c 0x 0 |x0T0

ux

xl

Q/,0u T 0x t

v(x,t)u(x,t)

0 v 0x 0 v|x0 v|t0

vx|xl0, 0t 0xl v(x,t)Cen2lt

n1 2 Cnl()sinn2l 0由于(Q

Cn

2 1n2

u(x,t)

x

sinn12n0n1 2 定解問(wèn)題,例

0xa,0yu|x0T0 u|xaT0,0yu

u

T,0x令v(x,y)u(x,y) 或u(x,y)v(x,y)vxxvyy 0xa,0yv|x00,v

0,0y v

T

,0xv(xyX(x)Yy)XX Y X(0)X(a)n

a

Xn

(x)sinnaY(y)CchnyDshn a v(xy)yDnsh a v(x,y)Cnch yDnsh ysina

y00

再由v|ybT

T0TT0Dn

bsin Dnsh

ba

a(TT0) a04(TT

0 u(x,y)T0v(x,T4(TT0

nysinn0n1,3,...n

nb T04(TT0)

(2k (2k (2k k1(2k1)sh

0xa,0y u| f(y), u| f(y),0ybu g u| g( 令u(x,y)uIx,yuIIx,使uIx,y)uI 0xa,0y uI f( uI f(y),0y uI

uI

0,0xuIIx,y)uII 0xa,0y uII|x0 uII|xa0,0yuII g uII g(x),0x uIx,y)和uIIx,y)可分別用分離變量法求解§9.5按本征函數(shù)系展開(kāi)方非方程的解可用對(duì)應(yīng)方程的本征函數(shù)展開(kāi).例1求定解問(wèn)題 f(x, 0x 0v|xv|x0 v|xl 0v v 0,0x t t對(duì)應(yīng) 方程為

(x)sinnx,l

按本征函數(shù)展開(kāi)f(x

f(x,t)

fn(t) 2 fn(t)f(,t) 按本征函數(shù)展開(kāi)v(x

lv(x,t)Tn(t)l將f(xt)和v(xt) Tn(t)fn(t) n1

2

(t)

(t) 從v|t00vt|t0

(t)|t0 n設(shè)T(p) T(t)e 即T(t)0Tn(t)pTn(p)Tn(0)pTn(

(T(t)p2T(p)pT(0)T(0)p2T( fn(t)fn(

(t)

(t) 得 (p)

na2T(p)f(p) T(p) fn( p2

2 sinnp2

T(p) f(

f( sinna(t

T0 p2na0

lv(x,t)Tn(t)l na(t

n1

na

fn()

t2 na(t

nalf(,) n1 0 例2 a2w f(x, 0x 0 w w w w (x),0x t t w(x,t)u(x,t)v(x,其中u(xt)滿(mǎn)足弦振動(dòng)方程的第一類(lèi)邊值問(wèn)題(9.1 0x 0 u|x0 u|xl0,0u u

|t

(x),0xv(x,t)滿(mǎn)足非方程的定解問(wèn)題(本節(jié)例 f(x, 0x 0 v v v 0,0x t t§9.6正交曲線(xiàn)坐標(biāo)系中的度規(guī)系數(shù)和算符PDF"pdfFactoryPro"試用版本創(chuàng)建? 算符2

x

y

z2u

u

算符2

x

rsin

yrsinsinz rz2u

2u 2

u

1

r r

§9.7亥姆方程的分 vtta22v T

k u(r

T(t)a2k2T(t) vta22v T

k u(r T(t)a2k2T(t) 方程的一個(gè)特例(kx y

z 1

u

k2u

u(,,z)R()()Z(z),2 d

dR

RZd2

RdZ d

d

k d dR d 1d Rdd

2d Z d dR

d 1d 令Rdd

2d Z Z2Z ddR d2

k22Rd d

2d Rd

dRd

2)2

dd

m2 得 m2 d dR

m2 d

d

d dR

m2)R d

d

(k22 d dRm2Rdd 2d

d

mR稱(chēng)型方程.其m0 R0()C0D0lnm0 Rm()m,m0,1,2, , 一般解：u(,)CDlnC AcosmBsinm 漸近條件：u(,)E0cos

ln邊界條件：u(a,由漸近條件得 Const,D ,CE, 0, 0由邊界條件得：Aa2E, 0, 0,Const 000 u(,) ln

Ecos1a2 a

2Eu

a2 a2 E0cos12?E0sin12

2

d dR

m2)R d

d

22 2d2RdR

(k22)2

R d k2 x,R(k2

y(x)dR()dy(x)dxdy(x)xd d 2Rx2yx2yxy(x2m2)y

xyzz

rsinrsinr

sin

2u 2

u

1

2u2k2u

u(r,,)R(r)()(),dr2dR

d d

d r

sind

r2sin2d2kR1d2dR d d

1d或Rdr

drk sindsin

sin2d 1令R

2dRk2r2 sin d d sin sind d

1dsin2d

dR r得R(r)程 1rR

k2r2

dr2dRk

2Rr2 dr r2 d d 1d dsindsind d d 1d 令dsindsind 得 d

m2d 2

sin sin2d d 令xcosy(x)(dxsin

d

dydxdx

sind dy

1x2y 上式稱(chēng)為連帶勒讓德方程(AssociatedLegenderEquationm0時(shí),上式稱(chēng)為勒讓德方程.§9.8姆－本征PDF"pdfFactoryPro"試用版本創(chuàng)建?dkdx

dyq(x)yw(x)ydx

ax上式稱(chēng)為姆型方程k(x)1,q(x)0,w(x)

ym2k(x)x,q(x)

,w(x)xdxdx

dym2dx

yxy為m階貝賽耳方程(k22k(x)1x2,q(x)0,w(x)d ddx

dydx

y為勒讓德方程(222k(x)1x,q(x) 22

,w(x)d 1(1x2)dy yydx dx 1為連帶勒讓德方程(2k(x)x2,q(x)l(l