分离变量法习题课

n
令 ψ (x, t) = ϕ(x) f (t)
n2π 2h2 nπ 2 ϕ′′(x) + λϕ = 0#→ λn = ( )→ En = 2 ϕn (x) = Cn sin nπ (x + a) 解 8a µ 2a 2a

a
a
ϕ(−a) = ϕ(a) = 0
#
1 −i h2 t π ψ (x, t) = e sin (x + a) a a
∞
u( x,0) = 3sin x → ∑ An sin nx = 3sin x
n=1
∞
n=1
ut (x,0) = 0
Wuhan University
→ A = 3 , An = 0 (n ≠ 1) 1 → ∑Bnna sin nx = 0 → Bn = 0
n=1
u( x, t) = 3cos at sin x 16
二、齐次问题
√ 2、求下列高维波动问题的解： 解
utt = a2∆u , 0 < x < 1 , 0 < y < 1 , 0 < z < 1 u(x, y, z;0) = sin πx sin πy sin πz ut (x, y, z;0) = 0 u(0, y, z; t) = u(1, y, z; t) = 0 u(x,0, z; t) = u(x,1, z; t) = 0 u(x, y,0; t) = u(x, y,1; t) = 0
∆u = 0 , (ρ < a , α < ϕ < β ) u |ϕ=α = 0 , u |ϕ=β = 0 u |ρ =a = f (ϕ)
①令
<1> <2> <3>
u = R(ρ)Φ(ϕ)
则由式<1>，得：
Φ′′ + n2Φ = 0 2 ρ R′′ + ρR′ − n2 R = 0
Wuhan University
Φ(α) = 0 <2>* 由<2>式，得： Φ(β ) = 0 Φ′′(ϕ) + µΦ = 0 ② 解： 令： = ϕ −α θ Φ(α) = 0 , Φ(β ) = 0 nπ 2 Φ′′(θ ) + µΦ = 0 µ =[ ] , n = 1,2,L → β −α Φ |θ =0 = 0 , Φ |θ =β −α = 0 nπ nπ Φ(ϕ) = Φ(θ ) = an sin θ = an sin (ϕ −α) β −α β −α
Wuhan University
15
二、齐次问题
√
1、求解
u(x, t) = ∑( An cos nat + Bn sin nat) sin nx
∞
utt = a2uxx , 0 < x < π , t > 0 u(x,0) = 3sin x , 0 ≤ x ≤π ut (x,0) = 0 u(0, t) = u(π , t) = 0;
解 X (0) = 0 X (1) = 0 →
X m (x) = am sin mπx ,
α = −m2π 2 , m = 1,2,...
Yn ( y) = bn sin nπy ,
Y′′ − βY = 0 解 Y (0) = 0 → Y (1) = 0 Z′′ − γZ = 0 解 Z (0) = 0 → Z (1) = 0
∞
Wuhan University
nπ β −α
#
8
二、齐次问题 定解问题
utt = a2uxx , 0 < x < l u x=0 = 0, u x=l = 0 u = ϕ(x), u t =0 t t =0 = ψ( x)
∞
(1) (2) (3)
nπa nπa nπ u(x,t) = ∑(Ancos t + Bn sin t)sin x (11) l l l n=1 2 l nπ 2 l nπ An = ∫ ϕ(α ) sin αdα , Bn = ∫0ψ (α) sin l αdα (12) 0 l l nπa
7
nπ 2 µຫໍສະໝຸດ Baidu=[ ] ③求解<5> ρ R′′ + ρR′ − µR = 0 β −α nπ 因为ρ < a R (ρ) = bρ µ = b ρ β −α
2
C0 + D0 ln ρ ρ R′′ + ρR′ − n R = 0 → n −n 一、正交曲线坐标系中的分离变量 Cn ρ + Dn ρ
E
Wuhan University
21
h(t) − g(t ) −∫ P( x) dx w(x, t) = x + g(t) ∫ P( x)dxdx + c) Y (x) = e 三、非齐次边界条件的定解问题(∫ Q(x)e l
ut = Duxx 令 u(x, t) = v(x, t) + w(x, t) x √ u x=0 = ct, u x=l = 0 # w( x, t ) = ct (1− ) l ∞ u t =0 = 0 nπx 令 v(x, t) = ∑Tn (t) sin x l 2 n=1 2 vt − Dvxx = c( l −1) Tn′(t) + D n π Tn (t) = fn (t) → l2 → v x=0 = 0, v x=l = 0 Tn (0) = 0 v = 0 2 l α nπα 2c t =0 dα = − fn (t) = ∫ c( −1) sin l 0 l l nπ
∑=B y z sin = cos 3πat sin πx 0 → y sin πz u(lx,mnl,ω; t) mπx sin nπy sin lπz = sin πBmnl = 0 m,n, 1
Wuhan University
∞
→ A = 1 , Amnl = 0 (m ≠ 1, n ≠ 1, l ≠ 1) 111
<1> <2> <3> <4> <5> <6> <3>*
①令： u(r,θ ) = R(r)Θ(θ ) <1> <3>
Θ′′(θ ) + m2Θ = 0 2 2 2 r R′′ + rR − m R = 0 R(r2 ) = 0
3
Wuhan University
一、正交曲线坐标系中的分离变量
2
Y (0) = 0 X (0) = 0 < 5 >→ < 5 >′ < 4 >′; < 4 >→ Y (1) = 0 X (1) = 0 Z (0) = 0 < 6 >′ < 6 >→ Z (1) = 0 18 Wuhan University
二、齐次问题
② X ′′ −αX = 0
<1> <2> <3> <4> <5> <6>
Wuhan University
17
二、齐次问题 ① 令 u(x, y, z; t) = X (x)Y ( y)Z(z);T (t)
utt = a2∆u , <1> T′′ − a µT = 0 <7> u(x, y, z;0) = sin πx sin πy sin πz <8> X ′′ −αX = 0 #<1> → ut (x, y, z;0) = 0 <3> <9> u(0, y, z; t) = u(1, y, z; t) = 0 <4> Y′′ − βY = 0 Z′′ − γZ = 0 <10> u(x,0, z; t) = u(x,1, z; t) = 0 <5> u(x, y,0; t) = u(x, y,1; t) = 0 <6> 其中 µ = α + β + γ ) (
Mathematical Methods in Physics
武 汉 大 学 物理科学与技术学院
Wuhan University
第三章 分离变量法
The Method of Separation of Variables
习 题 课
*本章内容小结 *典型例题分析
一、正交曲线坐标系中的分离变量 二、齐次方程、齐次边界条件的定解问题 三、非齐次边界条件的定解问题
2 2 r β1 = 2 1 2 u(r,θ ) = r1 r − r2 ⋅ sin θ 2 2 r − r2 1 r r − r2 1
Wuhan University
5
Z′′ + µZ = 0 (−µ = k 2 ) ′′ + n2Φ = 0 一、正交曲线坐标系中的分离变量 Φ ρ 2 R′′ + ρR′ + (k 2 ρ 2 − n2 )R = 0 2、求解扇形区域中的狄氏问题： ：
Θ′′(θ ) + m2Θ = 0 ②解 Θ(θ + 2π ) = Θ(θ ) 得： Θ(θ ) = Am cos mθ + Bm sin mθ ：
C0 ln r + D0 ③解<6>： Rm (r) = ： m −m Cmr + Dmr
m = 0， 2， 1 ... ， m=0
m≠ 0
20
nπ 2 X ′′ − µX = 0 µ = −( ) , n = 1,2,... l 二、齐次问题 (0) = 0 X nπx X (l) = 0 X n (x) = Cn 3、求处于一维无限深势阱中的粒子状态。 sin l
E −i t df ih = E ⋅ f → f (t) = bne h dt ∂ψ (x, t) h2 ∂2 ih =− ψ (x, t) #→ 2 2 ∂t 2µ ∂x2 − h d ϕ = Eϕ ψ 2µ dx2 (−a, t) =ψ (a, t) = 0 2µE 1 π ϕ(−a) = ϕ(a) = 0 记 2 = λ (x,0) = sin (x + a) ψ h
Wuhan University
1
Wuhan University
2
Z′′ + µZ = 0 (−µ = k 2 ) ′′ + n2Φ = 0 一、正交曲线坐标系中的分离变量 → Φ ρ 2 R′′ + ρR′ + (k 2 ρ 2 − n2 )R = 0 √1、求圆环的狄氏问题
∆u = 0 , (r2 < r < r ) 1 u(r ,θ ) = sin θ 1 u(r ,θ ) = 0 2
④
u = ∑um (r,θ ) = ∑Rm (r)Θm (θ )
= α0 (ln r − ln r2 ) +∑
m=1 ∞
∞
∞
m
2
m=0
m=0
r
2m
− r2 m r
2m
[αm cos mθ + βm sin mθ ]
# 由u(r1,θ ) = sin θ 有： αn = 0; βm = 0, m ≠ 1
Wuhan University
n =? n ≠ 0,1,L
2
<4> <5>
6
nπ 2 X ′′ − µX = 0 µ = −( ) , n = 1,2,... l X (0) = 0 一、正交曲线坐标系中的分离变量 nπ X (l) = 0 <4>x X n (x) = Cn sin 则由式<1>，得： Φ′′ + µΦ = 0 l 2 <5> ρ R′′ + ρR′ − µR = 0
Wuhan University
β = −n2π 2 , n = 1,2,...
Zl ( y) = cl sin lπz ,
γ = −l 2π 2 , l = 1,2,...
19
二、齐次问题 ③ T ′′ + a2 (n2 + m2 + l 2 )π 2T = 0 ′ ′ Tm,n,l (t) = Amnl cosωt + Bmnl sin ω
2 2
n
n
nπ ④ u(ρ,ϕ) = ∑Cn ρ sin (ϕ −α) β −α n=1 nπ ∞ nπ β −α sin (ϕ −α) 由式<3>，得： f (ϕ) = ∑Cna β −α n=1 −nπ 2 β nπ β −a ∴ Cn = a ∫α f (ϕ) sin β −α (ϕ −α)dϕ β −α
④ u(x, y, z; t) =
∞
ω = a ∞n + m + l )π (
2 2 2 2 2
m,n,l =1
2
∑( A
mnl
cosωt + Bmnl sin ωt)
m,n,l =1
∑A
sin mπx sin nπy sin lπz
mnl
sin mπx sin nπy sin lπz = sin πx sin πy sin πz
Rm (r) = C0 (ln r − ln r2 ) + Cm (r − r2 r )
m
Wuhan University
<3>*
2m −m
4
Rm (r) = C0 (ln r − ln r2 ) 一、正交曲线坐标系中的分离变量 m − r 2mr −m ) Θ(θ ) = Am cos mθ + Bm sin mθ + C (r