主题
Search

椭圆积分奇异值--k_1

第一类椭圆积分 K(k) 的第一个奇异值 k_1 ，对应于

K^'(k_1)=K(k_1),
(1)

由下式给出

k_1=1/(sqrt(2))
(2)
k_1^'=1/(sqrt(2)).
(3)

值 K(k_1) 由下式给出

K(1/(sqrt(2)))=int_0^1(dt)/(sqrt((1-t^2)(1-1/2t^2))),
(4)

可以变换为

K(1/(sqrt(2)))=sqrt(2)int_0^1(dt)/(sqrt(1-t^4)).
(5)

设

u=t^4
(6)
du=4t^3dt
(7)
=4u^(3/4)dt
(8)
dt=1/4u^(-3/4)du,
(9)

则

K(1/(sqrt(2)))=(sqrt(2))/4int_0^1u^(-3/4)(1-u)^(-1/2)du
(10)
=(sqrt(2))/4B(1/4,1/2)
(11)
=(Gamma(1/4)Gamma(1/2))/(Gamma(3/4))(sqrt(2))/4,
(12)

其中 B(a,b) 是贝塔函数，Gamma(z) 是伽玛函数。现在使用

Gamma(1/2)=sqrt(pi)
(13)

和

1/(Gamma(1-x))=(sin(pix))/piGamma(x),
(14)

所以

1/(Gamma(3/4))=1/(Gamma(1-1/4))=(sin(pi/4))/piGamma(1/4)=1/(pisqrt(2))Gamma(1/4).
(15)

因此，

K(1/(sqrt(2)))=(Gamma^2(1/4)sqrt(pi)sqrt(2))/(4pisqrt(2))=(Gamma^2(1/4))/(4sqrt(pi)).
(16)

现在考虑

E(1/(sqrt(2)))=int_0^1sqrt((1-1/2t^2)/(1-t^2))dt.
(17)

设

t^2=1-u^2
(18)
2tdt=-2udu
(19)
dt=-1/tudu
(20)
=u(1-u^2)^(-1/2)du,
(21)

所以

E(1/(sqrt(2)))=int_0^1sqrt((1-1/2(1-u^2))/(1-(1-u^2)))u(1-u^2)^(-1/2)du
(22)
=int_0^1(sqrt(1/2(1+u^2)))/uu(1-u^2)^(-1/2)du
(23)
=1/(sqrt(2))int_0^1sqrt((1+u^2)/(1-u^2))du.
(24)

现在注意到

(1/(sqrt(1-u^4))+(u^2)/(sqrt(1-u^4)))^2=(1+u^2)/(1-u^2),
(25)

所以

E(1/(sqrt(2)))=1/(sqrt(2))int_0^1sqrt((1+u^2)/(1-u^2))du
(26)
=1/(sqrt(2))int_0^1(1/(sqrt(1-u^4))+(u^2)/(sqrt(1-u^4)))du
(27)
=1/2K(1/(sqrt(2)))+1/(sqrt(2))int_0^1(u^2du)/(sqrt(1-u^4)).
(28)

现在设

t=u^4
(29)
dt=4u^3du,
(30)

所以

int_0^1(u^2du)/(sqrt(1-u^4))=1/4int_0^1t^(1/2)t^(-3/4)(1-t)^(-1/2)dt
(31)
=1/4int_0^1t^(-1/4)(1-t)^(-1/2)dt
(32)
=1/4B(3/4,1/2)=(Gamma(3/4)Gamma(1/2))/(4Gamma(5/4)).
(33)

但是

[Gamma(5/4)]^(-1)=[1/4Gamma(1/4)]^(-1)
(34)
Gamma(3/4)=pisqrt(2)[Gamma(1/4)]^(-1)
(35)
Gamma(1/2)=sqrt(pi),
(36)

所以

int_0^1(u^2du)/(sqrt(1-u^4))=1/4(pisqrt(2)·4sqrt(pi))/(Gamma^2(1/4))=(sqrt(2)pi^(3/2))/(Gamma^2(1/4))
(37)
E(1/(sqrt(2)))=1/2K+(pi^(3/2))/(Gamma^2(1/4))
(38)
=(Gamma^2(1/4))/(8sqrt(pi))+(pi^(3/2))/(Gamma^2(1/4))
(39)
=1/4sqrt(pi/2)[(Gamma(1/4))/(Gamma(3/4))+(Gamma(3/4))/(Gamma(5/4))].
(40)
(41)

总结 (◇) 和 (41) 给出

K(1/(sqrt(2)))=(Gamma^2(1/4))/(4sqrt(pi))
(42)
K^'(1/(sqrt(2)))=(Gamma^2(1/4))/(4sqrt(pi))
(43)
E(1/(sqrt(2)))=(Gamma^2(1/4))/(8sqrt(pi))+(pi^(3/2))/(Gamma^2(1/4))
(44)
E^'(1/(sqrt(2)))=(Gamma^2(1/4))/(8sqrt(pi))+(pi^(3/2))/(Gamma^2(1/4)).
(45)

使用 探索

请引用为

Weisstein, Eric W. "椭圆积分奇异值--k_1." 来自 Web 资源。 https://mathworld.net.cn/EllipticIntegralSingularValuek1.html

主题分类