2 2 2
x y" + 2xy' +λx y = 0
若 λ=0 :
2
x y" + 2xy' = 0
m
令 y = x 得
m(m-1) + 2m = 0
m = 0 or -1
-1
y = c1 + c2 x
由 y(0) = finite 得 c2 = 0
由 y'(1) = -2y(1) 知
-2 -1 ｜
-c2 x = -2c1 -2c2 x ｜x=1
｜c2=0
得 c1 = 0
故 y = 0 (無非零解)
若 λ≠0 :
2 2 2
x y" + 2xy' +λx y = 0
-1/2
令 y = z x
-1/2 1 -3/2
則 y' = z'x - --- z x
2
-1/2 1 -3/2 1 -3/2 3 -5/2
y" = z"x - --- z'x - --- z'x + --- z x
2 2 4
-1/2 -3/2 3 -5/2
= z"x - z'x + --- z x
4
帶回原式
2 -1/2 -3/2 3 -5/2 -1/2 1 -3/2 2 2 -1/2
x (z" x - z'x + --- z x ) + 2x(z'x - --- z x ) + λx (z x ) = 0
4 2
1/2
z的各次微分項合併，並同乘以 x
2 2 2 1
x z" + xz' + [λx - --- ]z = 0
4
將自變數換成ξ=λx，z原本是對x微分，改為對ξ微分:
d dξ d d
--- = --- --- = λ---
dx dx dξ dξ
2 2
d 2 d
--- = λ -----
2 2
dx dξ
則
2 2 2 2 1
λx z" + λxz' + [λx - ---]z = 0
4
2 2 1 2
ξz" + ξz' + [ ξ - (---) ]z = 0
2
此為Bessel function(自變數為ξ)
z = c1 J (ξ) + c2 J (ξ)
1/2 -1/2
-1/2 * sinλx * cosλx
y = z x = c1 -------- + c2 --------
x x
*
由 y(0)=finite 得 c2 = 0
由 y'(1) = -2y(1) 知
* λcosλx sinλx * sinλx ｜
c1 ( ---------- - -------- ) = c1 (-2 --------)｜
x 2 x ｜x=1
x
λcosλ - sinλ = -2 sinλ
λcosλ = -sinλ
tanλ = -λ
λ為 y=tanx 及 y=-x 交點之x座標
eigenvalues 為λ=λn
sin(λn x)
eigenfunctions 為 yn = bn ------------
x
--
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◆ From: 218.173.128.217
※ 編輯: youmehim 來自: 218.173.128.217 (11/07 13:18)