第二十三讲柱函数(二) §23.1 Bessel方程的本征值问题 现在从一个具体问题入手,讨论 Bessel方程的本征值问题 求四周固定的圆形薄膜的固有频率 注意,这个问题不同于过去讨论过的偏微分方程定解问题:现在并没有给出初始条件,所要 求的也不是描写圆形薄膜振动的位移如何随时间和空间而变化.现在要求的是固有频率,即求出 给定偏微分方程和边界条件下的所有各种振动模式的角频率也正是因为现在的问题中并没有给 出初始条件,所以也不能得出位移转动不变的结论 取平面极坐标系,坐标原点放置在圆形薄膜的中心这样,偏微分方程和边界条件就是 (23.1a) u=有界, u=a=0 (23.1b) au au u=0=1=2= (23.1c) = 现在要求的就是在边界条件(23.1b)和(3.1c)的限制下,到底许可哪些值,使得方程 (23.1a)有非零解 u(r,, t)=(, )eiut (23.2) 将此解式代入方程(23.1)及边界条件(23.1b)和23.1c),并令k=c,就可以得到 1 12v +20b2+k2v=0 0有界=a=0 av v=0==2 a中=00中=2n 再令v(,中)=R(r)(中),分离变量,就得到两个本征值问题 ()+m2()=0 (23.3a) (0)=重(2元),重(0)=重(2) (23.3b) 和 (23.4a) R(0)有界,R(a)=0 (23.4b)
Wu Chong-shi ✁✂✄☎ ✆ ✝ ✞ (✁ ) §23.1 Bessel ✟✠✡☛☞✌✍✎ ✏✑✒✓✔✕✖✗✘✙✚✛✜✢ Bessel ✣✤✥✦✧★✗✘✩ ✪✫✬ ✭✮✥ ✯✰✱✲✥ ✭✳✴✵ ✶✷✛✸✹ ✺✻✼ ✽✾✿❀❁❂✿❃❄❅❆❇❈❉❊ ✺✻❋●❍■❏❑▲ ▼◆❖P◗✛❘❙ ❚❃❯✼❱❲ ❳❨❩ ❬❭❪❫❃❴❵❛❜❝❞ ❡❢ ❣❡❤✐❥✩●❍❙❚❃❱ ❦❑❧♠✛♥❚ ▼ ▲❉❄❅❆❇❈❢♦♣P◗q❃❘❑rs❪❫t✉❃ ✈❧♠✩❯✇❱ ①②●❍❃ ✺✻ ③■❏❑▲ ▼◆❖P◗✛❘ ④❯✼⑤⑥ ▼❴❵⑦❫✼✐❃⑧❂✩ ⑨⑩❶❷❸❹❺✛❸❹❻❼❽❾✑ ✯✰✱✲✥ ❿➀✩➁➂✛➃➄➅✣✤➆➇➈➉➊➋➌ ∂ 2u ∂t2 − c 2 1 r ∂ ∂r r ∂u ∂r + 1 r 2 ∂ 2u ∂φ2 = 0, (23.1a) u r=0✳ ➈, u r=a = 0, (23.1b) u φ=0 = u φ=2π , ∂u ∂φ φ=0 = ∂u ∂φ φ=2π . (23.1c) ✏ ✑ ➍ ✪ ✥ ➋ ➌ ✑ ➇ ➈ ➉ ➊ (23.1b) ➆ (23.1c) ✥ ➎ ➏ ➐ ✛➑ ➒ ➓ ➔ → ➣ ω ★ ✛↔ ↕ ✣ ✤ (23.1a) ✳➙➛➜ u(r, φ, t) = v(r, φ)eiωt . (23.2) ➝➞➜➟➠✙ ✣✤ (23.1) ➡➇➈➉➊ (23.1b) ➆ (23.1c) ✛➢➤ k = ω/c ✛ ➋ ➔➥↕➑ 1 r ∂ ∂r r ∂v ∂r + 1 r 2 ∂ 2v ∂φ2 + k 2 v = 0, v r=0 ✳ ➈ v r=a = 0, v φ=0 = v φ=2π , ∂v ∂φ φ=0 = ∂v ∂φ φ=2π . ➦➤ v(r, φ) = R(r)Φ(φ) ✛➅➧➨➩✛➋ ↕➑➫✔✦✧★✗✘ Φ 00(φ) + m2Φ(φ) = 0, (23.3a) Φ(0) = Φ(2π), Φ 0 (0) = Φ 0 (2π) (23.3b) ➆ 1 r d dr r dR(r) dr + k 2 − m2 r 2 R(r) = 0, (23.4a) R(0)✳ ➈, R(a) = 0. (23.4b)
§231 Bessel方程的本征值问题 本征值问题(23.3)已经多次见到过,对应于它的本征值 m=0,1,2,3, 本征函数为 所以,在本征值问题(23.4)中,参数m2是已知的,而k2是本征值,待求 可以证明 PA((rdr=mR(G)r(T+dr ar, 所以,一定有本征值k2>0·通过作变换x=Mr,列(x)=B(r),就可以将微分方程(234)化为 Bessel方程,从而求得它的通解 R(r)=CJm(kr)+ DNm(kr) 考虑到边界条件(23.4b)的要求,R(O)有界,故D=0;又由于要求R(a)=0,就得到 将m阶Bsel函数Jn(x)的第i个正零点(由小到大排列记作r),i=1,2,3…,则本征值间 题(23.4)的解是 本征值 i=1,2,3, 237a) 本征函数Fmi(r)=Jm(kmr) 于是就求得了圆形薄膜的固有振动的角频率 其中m)是m阶Bes函数Jn(x)的第i个正零点 在上述求解过程中,实际上用到了有关J(x)零点的结论:当v>-1或为整数时,J(x)有 无穷多个零点,它们全部都是实数,对称地分布在实轴上
Wu Chong-shi §23.1 Bessel ➭➯➲➳➵➸➺➻ ➼ 2 ➽ ✦✧★✗✘ (23.3) ➾➚➪➶➹➑➘✛➴➷➬➮✥✦✧★ m2 , m = 0, 1, 2, 3, · · ·, ✦✧➱✃❐ Φm(φ) = cos mφ, sin mφ. ❒➥✛✑✦✧★✗✘ (23.4) ❿ ✛❮✃ m2 ➌ ➾❰✥✛Ï k 2 ➌✦✧★✛Ð✪✩ ➔➥Ñ Ò k 2 Z a 0 R(r)R ∗ (r)rdr = m2 Z a 0 R(r)R ∗ (r) dr r + Z a 0 dR(r) dr dR∗ (r) dr rdr, ❒➥✛✓✮✳✦✧★ k 2 > 0 ✩Ó➘Ô➨Õ x = kr ✛ y(x) = R(r) ✛ ➋ ➔➥➝➄➅✣✤ (23.4a) Ö❐ Bessel ✣✤✛✒Ï✪↕➮✥ Ó➜ R(r) = CJm(kr) + DNm(kr). (23.5) ×Ø➑ ➇➈➉➊ (23.4b) ✥ ➍✪✛ R(0) ✳ ➈ ✛Ù D = 0 ÚÛ Ü➬➍✪ R(a) = 0 ✛ ➋ ↕➑ Jm(ka) = 0. (23.6) ➝ m Ý Bessel ➱✃ Jm(x) ✥Þ i ✔ß➛❼ (Üà➑áâã) ä Ô µ (m) i ✛ i = 1, 2, 3, · · · ✛å ✦✧★✗ ✘ (23.4) ✥ ➜ ➌ ✦ ✧ ★ k 2 mi = µ (m) i a !2 , i = 1, 2, 3, · · · , (23.7a) ✦✧➱✃ Rmi(r) = Jm(kmir). (23.7b) ➬ ➌➋✪↕æ ✯✰✱✲✥ ✭✳çè✥é✴✵ ωmi = µ (m) i a c, (23.8) ê ❿ µ (m) i ➌ m Ý Bessel ➱✃ Jm(x) ✥Þ i ✔ß➛❼✩ ✑ëì✪➜➘ ✤ ❿ ✛íîëï➑æ✳ð Jν(x) ➛❼ ✥ñ✢❋ò ν > −1 ó❐ô✃õ✛ Jν(x) ✳ ö÷➪ ✔➛❼✛➮øùúû➌ í ✃ ✛➴üý➅þ✑íÿë✩
第二十三讲柱函数( 第3页 Zeros of the functions J,(a)& Nv(z) 1. Real zeros When v is real, the functions Jv(a)& Nv(z)each have an infinite number of zeros, all of which are simple with the possible exception of z=0. For non-negative v the sth positive zeros of these functions are denoted by ju,,s and nu, s respectively 3s2 3.831710.89358 7.015593.95768 312 38.6537310.173477.068058.59601 411.7915313.3236910.2223511.74915 514.9309216.4706313.3611014.89744 618.0710619.6158616.5009218.04340 721.2116422.7600819.6413121.18807 824.3524725.9036722.7820324.33194 927.4934829.0468325.9229627.47529 030.6346132.1896829.0640330.61829 2. McMahons expansions for large zeros -14(-1)(71-31)32(1-1)(8312-982+3779) 3(86) 64(-1)(89-1535415436272372-…,s>n,H=4D2 105(86 6= 3. Complex zeros of J,(a) When 12-l the zeros of J,(z) are all real. If v< -l and v is not an integer the number of complex zeros of J,(z) is twice the integer part of (-v; if the integer part of (-v)is odd two of these zeros lie on the imaginary axis 4. Complex zeros of Nv(a When v is real the pattern of the complex zeros of N,(z)depends on the non-integer part of Attention is confined here to the case v=n.a tive integer or zero
Wu Chong-shi ✁✂✄☎ ✆ ✝ ✞ (✁ ) ➼ 3 ➽ Zeros of the functions Jν(z) & Nν (z) 1. Real zeros When ν is real, the functions Jν(z) & Nν (z) each have an infinite number of zeros, all of which are simple with the possible exception of z = 0. For non-negative ν the sth positive zeros of these functions are denoted by jν,s and nν,s respectively. s j0,s j1,s n0,s n1,s 1 2.40483 3.83171 0.89358 2.19714 2 5.52008 7.01559 3.95768 5.42968 3 8.65373 10.17347 7.06805 8.59601 4 11.79153 13.32369 10.22235 11.74915 5 14.93092 16.47063 13.36110 14.89744 6 18.07106 19.61586 16.50092 18.04340 7 21.21164 22.76008 19.64131 21.18807 8 24.35247 25.90367 22.78203 24.33194 9 27.49348 29.04683 25.92296 27.47529 10 30.63461 32.18968 29.06403 30.61829 2. McMahon’s expansions for large zeros jν,s, nν,s ∼ β − µ − 1 8β − 4(µ − 1)(7µ − 31) 3(8β) 3 − 32(µ − 1)(83µ 2 − 982µ + 3779) 15(8β) 5 − 64(µ − 1)(6949µ 3 − 153855µ 2 + 1585743µ − 6277237) 105(8β) 7 − · · · · · · , s ν, µ = 4ν 2 , β = s + ν 2 − 1 4 π, for jν,s s + ν 2 − 3 4 π, for nν,s 3. Complex zeros of Jν(z) When ν ≥ −1 the zeros of Jν(z) are all real. If ν < −1 and ν is not an integer the number of complex zeros of Jν(z) is twice the integer part of (−ν); if the integer part of (−ν) is odd two of these zeros lie on the imaginary axis. 4. Complex zeros of Nν (z) When ν is real the pattern of the complex zeros of Nν(z) depends on the non-integer part of ν. Attention is confined here to the case ν = n, a positive integer or zero
231 Bessel方程的本征值问题 第4页 The figure 23. 1 shows the approximate distribution of the complex zeros of Nn(z) in the region I arg al<T. The figure is symmetrical about the real axis. The two curves on the left extend to infinit having the asymptotes Imz=±-ln3=±0.54931.. There are an infinite number of zeros near each of The two curves extending from z=-n to z=n and bounding an eye-shaped domain intersect the imaginary axis at the points ti(na+b), where Figure 23.1 Zeros of Nn(2) f-1=0.6274 and to=1.19968 is the positive root of cotht=t. There are n zeros near each of these curves. Complex zeros of No(z) Complex zeros of Ni(a) Real part Imaginary part Real part Imaginary part 2.403020.53988 0.78624 5.519880.54718 0.54841 7.01590
Wu Chong-shi §23.1 Bessel ➭➯➲➳➵➸➺➻ ➼ 4 ➽ Zeros of Nn(z) The figure 23.1 shows the approximate distribution of the complex zeros of Nn(z) in the region | arg z| ≤ π. The figure is symmetrical about the real axis. The two curves on the left extend to infinity, having the asymptotes Im z = ± 1 2 ln 3 = ±0.54931 . . .. . . There are an infinite number of zeros near each of these curves. The two curves extending from z = −n to z = n and bounding an eye-shaped domain intersect the imaginary axis at the points ±i(na + b), where Figure 23.1 Zeros of Nn(z) a = q t 2 0 − 1 = 0.66274 . . .. . . b = 1 2 q 1 − t −2 0 ln 2 = 0.19146 . . .. . . and t0 = 1.19968 . . .. . . is the positive root of coth t = t. There are n zeros near each of these curves. Complex zeros of N0(z) Complex zeros of N1(z) Real part Imaginary part Real part Imaginary part −2.40302 0.53988 −0.50274 0.78624 −5.51988 0.54718 −3.83353 0.56236 −8.65367 0.54841 −7.01590 0.55339
第二十三讲柱函数( 第5页 为了在分离变量法中的应用,自然要讨论上面得到的本征函数的正交归关系.下面,介绍 种略为不同的做法,可以同时得到本征函数的正交归一关系 首先,写出本征函数Rm1(r)=Jn(kmr)所满足的微分方程和边界条件, r2 Jm(kimi) (23.9a) Jn(0)有界,Jn(kma)=0. 同时,再写出函数R(r)=Jm(kr)所满足的微分方程和边界条件, 1d「dn(kr) dr Jm(O)有界 由于其中的k为任意实数,所以一般说来,不会有Jm(ka)=0 再用rJm(kr)和rJm(kmr)分别乘方程(239a)和(23.10a) d「dJn(kmin) m(kmir)Jm(er) Jm(mir)dram(kr2 72rJm(kmir)J m(kr)=0, 相减,并在区间0,]上积分,就得到 代入边界条件(23.9b)和(23.10b),可以将上面的结果化为 (kmi-k2)/Jm(kmir) Jm(kr)rdr=-kimiaJm(k a) 'm(kmia (2311) 我们对两个特殊情形感兴趣.第一种情形是k=km≠km这时就有Jm1(km (23.11)式的右端为0.但由于km≠km,所以 即对应于不同本征值的本征函数在区间0,a]上以权重r正交 另一种情形是k=km,这时(23.11)式的两端均为0.我们可以先将(23.11)式的两端同除以 k2,然后取极限k→km,这样就得到 Aim gJm(ka) J'm(kmia)=2 I'm(kimia 2313) 这正是本征函数Jm( kmir)的模方 如果将本征值问题(239)中r=a端的齐次边界条件(23.9b)改为第二类或第三类边界条件, 也可以类似地讨论 事实上,可以把这三种情形统一写成 1 d dR(n1+(k2- R(r)=0, (2314a)
Wu Chong-shi ✁✂✄☎ ✆ ✝ ✞ (✁ ) ➼ 5 ➽ ❐ æ✑➅➧➨➩✟ ❿✥➷ï✛ ✠✡➍✜✢ë❶↕➑✥✦✧➱✃✥ß☛☞✓ð❺✩ ➐ ❶✛✌✍ ✓✎✏❐✑✒✥✓ ✟✛➔➥✒õ ↕➑✦✧➱✃✥ß☛☞✓ð❺✩ ✔✕✛✖✗✦✧➱✃ Rmi(r) = Jm(kmir) ❒✘✙✥ ➄➅✣✤➆➇➈➉➊✛ 1 r d dr r dJm(kmir) dr + k 2 mi − m2 r 2 Jm(kmir) = 0, (23.9a) Jm(0)✳ ➈, Jm(kmia) = 0. (23.9b) ✒õ ✛➦✖✗➱✃ R(r) = Jm(kr) ❒✘✙✥ ➄➅✣✤➆➇➈➉➊✛ 1 r d dr r dJm(kr) dr + k 2 − m2 r 2 Jm(kr) = 0, (23.10a) Jm(0)✳ ➈. (23.10b) Ü ➬ê ❿✥ k ❐✚✛í ✃ ✛❒➥✓✜✢✣✛ ✑✤✳ Jm(ka) = 0 ✩ ➦ï rJm(kr) ➆ rJm(kmir) ➅✥✦✣✤ (23.9a) ➆ (23.10a) ✛ Jm(kr) d dr r dJm(kmir) dr + k 2 mi − m2 r 2 rJm(kmir)Jm(kr) = 0, Jm(kmir) d dr r dJm(kr) dr + k 2 − m2 r 2 rJm(kmir)Jm(kr) = 0, ✧★✛➢✑✩✪ [0, a] ë✫➅✛➋ ↕➑ k 2 mi − k 2 Z a 0 Jm(kmir)Jm(kr)rdr = r Jm(kmir) dJm(kr) dr − Jm(kr) dJm(kmir) dr r=a r=0 . ➠✙ ➇➈➉➊ (23.9b) ➆ (23.10b) ✛➔➥➝ë❶ ✥ñ✬Ö❐ k 2 mi − k 2 Z a 0 Jm(kmir)Jm(kr)rdr = −kmiaJm(ka)J0 m(kmia). (23.11) ✭ø➴➫✔✮✯✰✰✱✲✳✩ Þ ✓✎✰✰➌ k = kmj 6= kmi ✩➁õ➋✳ Jm(kmja) = 0 ✛✴➞ (23.11) ➟ ✥✵✶❐ 0 ✩✷ Ü ➬ kmj 6= kmi ✛❒➥ Z a 0 Jm(kmir)Jm(kmj r)rdr = 0, kmi 6= kmj , (23.12) ✸➴➷➬✑✒✦✧★✥✦✧➱✃✑✩✪ [0, a] ë➥✹✺ r ß☛✩ ✻✓✎✰✰➌ k = kmi ✛➁ õ (23.11) ➟ ✥ ➫ ✶✼❐ 0 ✩✭ø➔➥✕➝ (23.11) ➟ ✥ ➫ ✶✒✽➥ k 2 mi − k 2 ✛✡✾⑨❷➎ k → kmi ✛➁➂➋ ↕➑ Z a 0 J 2 m(kmir)rdr = − lim k→kmi kmia k 2 mi − k 2 Jm(ka)J0 m(kmia) = a 2 2 [J0 m (kmia)]2 . (23.13) ➁ß➌✦✧➱✃ Jm(kmir) ✥✿✣ ✩ ❀ ✬ ➝ ✦✧★✗✘ (23.9) ❿ r = a ✶✥❁➶➇➈➉➊ (23.9b) ❂❐Þ❃❄óÞ❅❄➇➈➉➊✛ ❆➔➥❄❇ý✜✢✩ ❈íë✛➔➥❉➁ ❅ ✎✰✰❊ ✓✖❋ 1 r d dr r dR(r) dr + k 2 − m2 r 2 R(r) = 0, (23.14a)