问题引入：
由分离变量法有:
Δu+λu=0Δu=0 }−−−−→ 令u=R(r)Φ(φ)Z(z)
Φ ′′ +n 2 Φ=0→Φ n (φ)=A n cosnφ+B n sinnφ
Z ′′ +μZ=0→Z(z)=c 1 e kz +d 2 e −kz (μ=−k 2 )
ρ 2 R ′′ +ρR ′ +(k 2 ρ 2 −n 2 )R=0→R(ρ)=?
x=kρ,R(ρ)=y(x),k 2 =λ−μ≥0
x 2 y ′′ (x)+xy ′ (x)+(x 2 −n 2 )y(x)=0→y(x)→?
§15.1Bessel函数
附：二阶线性常微分方程的级数解法2
对于:W ′′ (z)+p(z)W ′ (z)+q(z)W(z)=0(∗)
若z 0 是它的奇点,即在z 0 点其系数p(z)和q(z)之一或均不解析,且z 0 是p(z)不高于一阶的极点、q(z)不高于二阶的极点,则称z 0 是方程的正则奇点。
在正则奇点z=z 0 的邻域0<|z−z 0 |<R内,方程至少有形式为
W(z)=(z−z 0 ) ρ ∑ k=0 ∞ C k (z−z 0 ) k (∗∗)
的幂级数解.将形式解(∗∗)代入方程(∗)中通过比较方程两边最低次幂的系数,便可得到
一关于ρ的二次方程,称为指标方程。由此方程可求得两个指标：ρ 1 ,ρ 2 (设ρ 1 >ρ 2 ),
那么，方程
W ′′ (z)+p(z)W ′ (z)+q(z)W(z)=0(∗)
的两线性无关的根为:
W 1 (z)=(z−z 0 ) ρ 1 ∑ k=0 ∞ c k (z−z 0 ) k
W 2 (z)=⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (z−z 0 ) ρ 2 ∑ k=0 ∞ d k (z−z 0 ) k ,ρ 1 −ρ 2 ≠整数aw 1 (z)ln(z−z 0 )+(z−z 0 ) ρ 2 ∑ k=0 ∞ d ′ k (z−z 0 ) k ,ρ 1 −ρ 2 =整数
15.1.1Bessel方程的级数解
x 2 y ′′ (x)+xy ′ (x)+(x 2 −v 2 )y(x)=0(1)→y(x)=?
p(x)=1x ,q(x)=1−(vx ) 2 x=0−方程的正则奇点.
1.令y=∑ ∞ k=0 c k x k+ρ ,则
y ′ =∑ k=0 ∞ c k (k+ρ)x k+ρ−1 ,y ′′ =∑ k=0 ∞ c k (k+ρ)(k+ρ−1)x k+ρ−2
(1)→∑ k=0 ∞ [(k+ρ) 2 −v 2 ]c k x k+ρ +∑ k=0 ∞ c k x k+ρ+2 =0(2)
2.比较最低次幂x ρ 的系数:
(ρ 2 −v 2 )c 0 =0(c 0 ≠0)
→判定方程:ρ 2 −v 2 =0→ρ 1 =v,ρ 2 =−v(设v>0)
3.令y 1 =∑ k=0 ∞ c k x k+ρ 1 =∑ k=0 ∞ c k x k+v
(1)→∑ k=0 ∞ [(k+v) 2 −v 2 ]c k x k+v +∑ k=0 ∞ c k x k+v+2 =0
x v :(v 2 −v 2 )c 0 =0(设c 0 ≠0)
x v+1 :[(v+1) 2 −v 2 ]c 1 =0→c 1 =0
x v+k :[(v+k) 2 −v 2 ]c k +c k−2 =0→c k =−c k−2 k(2v+k) (3)
c 2 =−c 0 2⋅2(v+1) ,c 3 =−c 1 3⋅(3+2v) =0
c 4 =(−1) 2 c 0 2 4 ⋅2(v+2)(v+1) ,c 5 =0
c 2n =(−1) n c 0 2 2n n!(v+n)(v+n−1)⋯(v+1) ,c 2n+1 =0
→c 2n =(−1) n c 0 Γ(v+1)2 2n n!Γ(v+n+1) →y 1 (x)=∑ n=0 ∞ (−1) n c 0 Γ(v+1)2 2n n!Γ(v+n+1) x 2n+v
4.取ρ=ρ 2 =−v,类似可得
→y 2 (x)=∑ n=0 ∞ (−1) n c 0 Γ(−v+1)2 2n n!Γ(−v+n+1) x 2n−v
15.1.2解的敛散性
1.解的奇点少于或等于方程的奇点
W ′′ (z)+p(z)W ′ (z)+q(z)W(z)=0(∗)
2.贝塞尔方程的奇点为:0,∞
x 2 y ′′ (x)+xy ′ (x)+(x 2 −v 2 )y(x)=0(1)
令x=1t
t 2 y ′′ (t)+ty ′ (t)+(1t 2 −v 2 )y(t)=0(1)
3.y 1 (x),y 2 (x)均在0<|x|<∞中收敛.
y 1 (x)在|x|<∞中收敛,而y 2 (x)当x→0时发散.
15.1.3贝塞尔函数
1.定义:在y 1 (x)中,取c 0 =12 v Γ(v+1) ,并记y 1 (x)=J v (x)
−称之为v阶贝塞尔函数.
y 1 (x)=J v (x)=∑ k=0 ∞ (−1) k k!Γ(v+k+1) (x2 ) 2k+v (4)
在y 2 (x)中，取c 0 =12 −v Γ(−v+1) ,并记y 2 (x)=J −v (x)
y 2 (x)=J −v (x)=∑ ∞ k=0 (−1) k k!Γ(−v+k+1) (x2 ) 2k−v (5)
2.线性相关性
(1)当v≠n时,J v (x)与J −v (x)是线性无关的.
x→0:J v (x)≈1Γ(v+1) (x2 ) v ,J −v (x)≈1Γ(−v+1) (x2 ) −v
J v (x)J −v (x) ≈Γ(−v+1)Γ(v+1) (x2 ) 2v −随x的值而变.
通解:y c (x)=c v J v (x)+d v J −v (x)
(2)当v=n时,J −n (x)=(−1) n J n (x)(6)
J n (−x)=(−1) n J n (x)=J −n (x)
15.1.4本征值问题
{ρ 2 R ′′ (ρ)+ρR ′ (ρ)+(k 2 ρ 2 −n 2 )R(ρ)=0(7)[αR(ρ)+βR ′ (ρ)] ρ=a =0(8)
1.{ρ 2 R ′′ (ρ)+ρR ′ (ρ)+(k 2 ρ 2 −n 2 )R(ρ)=0(9)R(a)=0(10)
即:{x 2 y ′′ +xy ′ +(x 2 −n 2 )y=0(9) ′ y| x=ka =0(10) ′
(1)J n (x)是一振荡函数.
J 0 (x)=∑ k=0 ∞ (−1) k (k!) 2 (x2 ) 2k =1−(x2 ) 2 +12! 2 (x2 ) 4 −⋯
J 1 (x)=∑ k=0 ∞ (−1) k k!(k+1)! (x2 ) 2k+1 =x2 −11!2! (x2 ) 3 +⋯
(2)J n (x)与x轴有无穷个交点.
J n (x n m )=0(m=1,2,⋯)→x n m −称为J n (x)的第m个零点.
(3)本征值问题(9)(10)或(9) ′ (10) ′ 的
本征值:k n m =x n m a ,m=1,2,⋯
本征函数:y m (kρ)=J n (x n m a ρ),m=1,2,⋯,
2.{ρ 2 R ′′ (ρ)+ρR ′ (ρ)+(k 2 ρ 2 −n 2 )R(ρ)=0,ρ<a(11)R ′ (a)=0(12)
J ′ n (x ~ n m )=0(m=1,2,⋯)→x ~ n m −称为J ′ n (x)的第m个零点.
本征值:k ~ n m =x ~ n m a ,m=1,2,⋯
本征函数:R m (kρ)=J n (x ~ n m a ρ),m=1,2,⋯
15.1.5小结
1.x 2 y ′′ (x)+xy ′ (x)+(x 2 −v 2 )y(x)=0(1)
y 1 (x)=J v (x)=∑ k=0 ∞ (−1) k k!Γ(v+k+1) (x2 ) 2k+v (4)
y 2 (x)=J −v (x)=∑ k=0 ∞ (−1) k k!Γ(−v+k+1) (x2 ) 2k−v (5)
当v≠n时:y c (x)=c v J v (x)+d v J −v (x)
当v=n时:J −n (x)=(−1) n J n (x)(6)
2.{ρ 2 R ′′ (ρ)+ρR ′ (ρ)+(k 2 ρ 2 −n 2 )R(ρ)=0(9)R(a)=0(10)
J n (x n m )=0(m=1,2,⋯)→x n m −称为J n (x)第m个零点.
本征值:k n m =x n m a ,m=1,2,⋯
本征函数:R m (kρ)=J n (x n m a ρ),m=1,2,⋯
{J n (k n m ρ)}:J n (k n 1 ρ),J n (k n 2 ρ),J n (k n 3 ρ),⋯−n阶贝塞尔函数系
对于:W ′′ (z)+p(z)W ′ (z)+q(z)W(z)=0(∗)
在正则奇点z=z 0 的邻域0<|z−z 0 |<R内,方程至少有形式为
W(z)=(z−z 0 ) ρ ∑ k=0 ∞ C k (z−z 0 ) k (∗∗)
若ρ的指标的两个指标为:ρ 1 ,ρ 2 (设ρ 1 >ρ 2 ),那么,方程(∗)的两线性无关的根为:
W 1 (z)=(z−z 0 ) ρ 1 ∑ k=0 ∞ c k (z−z 0 ) k
W 2 (z)=⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ (z−z 0 ) ρ 2 ∑ k=0 ∞ d k (z−z 0 ) k ,ρ 1 −ρ 2 ≠整数aw 1 (z)ln(z−z 0 )+(z−z 0 ) ρ 2 ∑ k=0 ∞ d ′ k (z−z 0 ) k ,ρ 1 −ρ 2 =整数
§15.2贝塞尔函数的性质
15.2.1母函数关系式
e x2 (t−1t ) =∑ k=0 ∞ J n (x)t n (1)
e z =∑ k=0 ∞ 1k! z k ,|z|<∞
证明:∵e x2 t =∑ l=0 ∞ 1l! (x2 t) l ,|t|<∞
e −x2t =∑ m=0 ∞ 1m! (−x2t ) m ,|t|>0
e x2 (t−1t ) =e x2 t ⋅e −x2t =∑ l=0 ∞ 1l! (x2 t) l ⋅∑ m=0 ∞ 1m! (−x2t ) m
=∑ l=0 ∞ ∑ m=0 ∞ (−1) m l!m! (x2 ) l+m t l−m
令l−m=n,则l=m+n
→∑ l=0 ∞ →∑ m+n=0 ∞ →∑ n=−m ∞ →∑ n=−∞ ∞
e x2 (t−1t ) =∑ n=−∞ ∞ ∑ m=0 ∞ (−1) m (m+n)!m! (x2 ) 2m+n t n
=∑ n=−∞ ∞ J n (x)t n
J n (x)=∑ k=0 ∞ (−1) k k!(n+k)! (x2 ) 2k+n
f(z)=∑ k=−∞ ∞ c k (z−b) k ,c k =12πi ∮ l f(z)(z−b) k+1 dz
e x2 (t−1t ) =∑ n=−∞ ∞ J n (x)t n (1)
问:1.J n (x)的积分分？
J n (x)=12π ∫ π −π e i(xsinθ−nθ) dθ
或J n (x)=1π ∫ π 0 cos(xsinθ−nθ)dθ
2.J n (x)的微分式?
3.J v (x)(v≠n)有母函数关系吗?
J v (x)=∑ k=0 ∞ (−1) k k!Γ(v+k+1) (x2 ) 2k+v (4)
15.2.2递推公式
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ddx [x v J v (x)]=x v J v−1 (x)(2)ddx [x −v J v (x)]=−x −v J v+1 (x)(3)
用途:(1)可派生出其它递推公式
xJ ′ v (x)+vJ v (x)=xJ v−1 (x)(4)
xJ ′ v (x)−vJ v (x)=−xJ v+1 (x)(5)
2J ′ v (x)=J v−1 (x)−J v+1 (x)(6)
2vx J v (x)=J v−1 (x)+J v+1 (x)(7)
(2)只要查J 0 (x)和J 1 (x)表,可计算出任一J v (x)
用途:(3)可用来计算含J v (x)的积分
例1:∫ a 0 x 3 J 0 (x)dx=a 3 J 1 (a)−2a 2 J 2 (a)
例2:∫J 1 (x)dx=? −J 0 (x)+CJ 1 (x)=−J ′ 0 (x)
例3:∫J 3 (x)dx=−J 0 (x)−2J 2 (x)+c
15.2.3正交性
∫ a 0 ρJ n (k n m ρ)J n (k n l ρ)dρ=a 2 2 J 2 n+1 (k n l a)δ ml (8)
证明:∵ρ 2 R ′′ (ρ)+ρR ′ (ρ)+(k 2 ρ 2 −n 2 )R(ρ)=0
→ddρ (ρdRdρ )+(k 2 ρ−n 2 ρ )R=0
ddρ [ρdJ n (k n m ρ)dρ ]+[(k n m ) 2 ρ−n 2 ρ ]J n (k n m ρ)=0(9)
ddρ [ρdJ n (k n l ρ)dρ ]+[(k n l ) 2 ρ−n 2 ρ ]J n (k n l ρ)=0(10)
J n (k n m a)=0,m=1,2,⋯,l,⋯
∫ a 0 [(9)⋅J n (k n l ρ)−(10)⋅J n (k n m ρ)]dρ:
[(k n m ) 2 −(k n l ) 2 ]∫ a 0 ρJ n (k n m ρ)J n (k n l ρ)dρ
=∫ a 0 J n (k n m ρ)ddρ [ρdJ n (k n l ρ)dρ ]dρ−∫ a 0 J n (k n l ρ)ddρ [ρdJ n (k n m ρ)dρ ]dρ
=ρJ n (k n m ρ)dJ n (k n l ρ)dρ | a 0 −∫ a 0 ρdJ n (k n l ρ)dρ dJ n (k n m ρ)dρ dρ
−ρJ n (k n l ρ)dJ n (k n m ρ)dρ | a 0 +∫ a 0 ρdJ n (k n m ρ)dρ dJ n (k n l ρ)dρ dρ
[(k n m ) 2 −(k n l ) 2 ]∫ a 0 ρJ n (k n m ρ)J n (k n l ρ)dρ
=ρ[J n (k n m ρ)dJ n (k n l ρ)dρ −J n (k n l ρ)dJ n (k n m ρ)dρ ]| 1 0
=0(∴J n (k n m a)=0,m=1,2,⋯,l,⋯)
1.若m≠l:∫ a 0 ρJ n (k n m ρ)J n (k n l ρ)dρ=0
2.若m=l,令m→l
∫ a 0 ρJ n (k n m ρ)J n (k n l ρ)dρ=lim k n m →k n l aJ n (k n m a)J ′ n (k n l a)k n l (k n m ) 2 −(k n l ) 2
=lim k n m →k n l a 2 k n l J ′ n (k n m a)J ′ n (k n l a)2k n m =a 2 2 [J ′ n (k n l a)] 2 =a 2 2 J 2 n+1 (k n l a)
15.2.4广义傅氏展开
若f(ρ)在[0,a]上有连续的一阶导数,分段连续的二阶导数,且f(ρ)| ρ=0 →有界,f(ρ)| ρ=a =0,则
f(ρ)=∑ m=1 ∞ c m J n (k n m ρ)
c m =1a 2 2 J 2 n+1 (k n m a) ∫ a 0 ρf(ρ)J n (k n m ρ)dρ
例4:一半径为a,高为h的均匀圆柱体,其下底和侧面保持温度为零度,上端温度为u 0 ,求柱内的稳定温度分布.
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ Δu=0,0≤ρ≤a(1)u(a,z)=0(2)u(ρ,0)=0(3)u(ρ,h)=u 0 (4)
解:1.令u(ρ,z)=R(ρ)Z(z)
(1)→{Z ′′ +μZ=0(5)ρ 2 R ′′ +ρR ′ +(k 2 ρ 2 −0)R=0(6)
(2)→R(a)=0(7);(3)→Z(0)=0(8)
2.解本征值问题(6)(7)得
k 2 =−μ=(x 0 m a ) 2 ,R m (ρ)=J 0 (k 0 m ρ),m=1,2,⋯
3.解方程(5):
{Z ′′ +μZ=0(5)→Z m (z)=c m sinh(k 0 m z)Z(0)=0(8)
4.叠加,定系数:
u(ρ,z)=∑ m=1 ∞ c m sinh(k 0 m z)J 0 (k 0 m ρ)
→∑ m=1 ∞ c m sinh(k 0 m h)J 0 (k 0 m ρ)=u 0
c m =1a 2 2 J 2 1 (k 0 m a)sinh(k 0 m h) ∫ a 0 u 0 ρJ 0 (k 0 m ρ)dρ
令x=k 0 m ρ
∫ a 0 ρJ 0 (k 0 m ρ)dρ=1(k 0 m ) 2 ∫ (k 0 m a) 0 xJ 0 (x)dx=ak 0 m J 1 (k 0 m a)
问:J 1 (k 0 m a)=0?
c m =1a 2 2 J 2 1 (k 0 m a)sinh(k 0 m h) ∫ a 0 u 0 ρJ 0 (k 0 m ρ)dρ
=2u 0 (k 0 m a)sinh(k 0 m h)J 1 (k 0 m a)
u=∑ m=1 ∞ 2u 0 x 0 m sinh(k 0 m z)sinh(k 0 m h) J 0 (k 0 m ρ)J 1 (k 0 m a)
15.2.5小结
(1)母函数关系式
e x2 (t−1t ) =∑ n=−∞ ∞ J n (x)t n (1)
(2)递推公式:
⎧ ⎩ ⎨ ⎪ ⎪ ⎪ ⎪ ddx [x v J v (x)]=x v J v−1 (x)(2)ddx [x −v J v (x)]=−x −v J v+1 (x)(3)
(3)正交性
∫ a 0 ρJ n (k n m ρ)J n (k n l ρ)dρ=a 2 2 J 2 n+1 (k n l a)δ ml (8)
(4)广义傅氏展开
f(ρ)=∑ m=1 ∞ c m J n (k n m ρ)
c m =1a 2 2 J 2 n+1 (k n m a) ∫ a 0 ρf(ρ)J n (k n m ρ)dρ
