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 --

※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 218.173.128.217 ※ 編輯: youmehim 來自: 218.173.128.217 (11/07 13:18)