背景 1 d sin e d() +|入 sin 0 de e(6)=0 O(0)有界 e(m)有界 和 =0 (0)=p(27)(0)=y(2m)
Associated Legendre Functions Spheroidal Harmonics Eigenproblem of Associated Legendre Eqns Orthogonality of Associated Legendre Ftns µ 1 sin θ d dθ sin θ dΘ(θ) dθ + λ − µ sin2 θ Θ(θ) = 0 Θ(0)k. Θ(π)k. Ú Φ 00 + µΦ = 0 Φ(0) = Φ(2π) Φ 0 (0) = Φ 0 (2π) C. S. Wu 1lù ¥¼ê(n)
背景 p"+p=0 cos mo (0)=更(2) pn(口) (0)=(2m) Sin mo m=0.1.2.… 又出现连带 Legendre方程的本征值问题 de( sin b de 0)有界
Associated Legendre Functions Spheroidal Harmonics Eigenproblem of Associated Legendre Eqns Orthogonality of Associated Legendre Ftns µ Φ 00 + µΦ = 0 Φ(0) = Φ(2π) Φ 0 (0) = Φ 0 (2π) =⇒ µ = m2 Φm(φ) = ( cos mφ sin mφ m = 0, 1, 2, · · · qÑyëLegendre§ ¯K 1 sin θ d dθ sin θ dΘ(θ) dθ + λ − m2 sin2 θ Θ(θ) = 0 Θ(0)k. Θ(π)k. C. S. Wu 1lù ¥¼ê(n)
背景 p"+p=0 cos mo (0)=更(2) pn(口) (0)=(2m) Sin mo m=0.1.2.… 又出现连带 Legendre方程的本征值问题 1 d de(0 sin e +入 sin ede e(6)=0 sIn O(0)有界 e(m)有界
Associated Legendre Functions Spheroidal Harmonics Eigenproblem of Associated Legendre Eqns Orthogonality of Associated Legendre Ftns µ Φ 00 + µΦ = 0 Φ(0) = Φ(2π) Φ 0 (0) = Φ 0 (2π) =⇒ µ = m2 Φm(φ) = ( cos mφ sin mφ m = 0, 1, 2, · · · qÑyëLegendre§ ¯K 1 sin θ d dθ sin θ dΘ(θ) dθ + λ − m2 sin2 θ Θ(θ) = 0 Θ(0)k. Θ(π)k. C. S. Wu 1lù ¥¼ê(n)
连带 Legendre方程在有界条件下的本征值问题 1 d de(e sIn ode d+|入 (6)=0 sIn O(0)有界O()有界 通常作变换x=c0s.(x)=6(0),并且把待定 参数写成(+1),本征值问题就变为 (±1)有界
Associated Legendre Functions Spheroidal Harmonics Eigenproblem of Associated Legendre Eqns Orthogonality of Associated Legendre Ftns ëLegendre§3k.^e ¯K 1 sin θ d dθ sin θ dΘ(θ) dθ + λ − m2 sin2 θ Θ(θ) = 0 Θ(0)k. Θ(π)k. Ï~Cx = cos θ, y(x) = Θ(θ)§¿ r½ ëêλ¤ν(ν + 1)§ ¯KÒC d dx 1 − x 2 dy dx + ν(ν + 1) − m2 1 − x 2 y = 0 y(±1)k. C. S. Wu 1lù ¥¼ê(n)
连带 Legendre方程在有界条件下的本征值问题 1 d de(e sIn 0 de d+|入 (6)=0 sIn (0)有界6(丌)有界 通常作变换x=cosθ,y(x)=6(θ),并且把待定 参数λ写成v(+1),本征值问题就变为 d d d (v+1) y(±1)有界
Associated Legendre Functions Spheroidal Harmonics Eigenproblem of Associated Legendre Eqns Orthogonality of Associated Legendre Ftns ëLegendre§3k.^e ¯K 1 sin θ d dθ sin θ dΘ(θ) dθ + λ − m2 sin2 θ Θ(θ) = 0 Θ(0)k. Θ(π)k. Ï~Cx = cos θ, y(x) = Θ(θ)§¿ r½ ëêλ¤ν(ν + 1)§ ¯KÒC d dx 1 − x 2 dy dx + ν(ν + 1) − m2 1 − x 2 y = 0 y(±1)k. C. S. Wu 1lù ¥¼ê(n)