Jacobi-Gauss-Lobatto积分点和积分权

这次介绍的是关于Jacobi正交多项式的零点计算问题，谷歌学术里面可以搜索到很多相关学术文章。由于在Galerkin-Spectral方法中经常使用Jacobi正交多项式，所以整理了一些相关知识点。

Jacobi正交多项式的递推公式：

$J_0^{alpha,eta}(x)=1$, $quad J_1^{alpha,eta}(x)=frac12 (alpha+eta+2)x+frac12(alpha-eta),$

$J_{n+1}^{alpha,eta}(x)=Big(a_n^{alpha,eta}x-b_n^{alpha,eta}Big)J_{n}^{alpha,eta}(x)-c_{n}^{alpha,eta}J_{n-1}^{alpha,eta}(x),quad ngeq 1.$

其中

$a_n^{alpha,eta}=frac{ig(  2n+alpha+eta+1 ig)ig(  2n+alpha+eta+2 ig)}{2ig(  n+1 ig)ig(  n+alpha+eta+1 ig)},$

$b_n^{alpha,eta}=frac{ig(  eta^2-alpha^2 ig)ig(  2n+alpha+eta+1 ig)}{2ig(  n+1 ig)ig(  n+alpha+eta+1 ig)ig(  2n+alpha+etaig)},$

$c_n^{alpha,eta}=frac{ig(  n+alpha ig)ig(  n+etaig)ig(  2n+alpha+eta+2 ig)}{ig(  n+1 ig)ig(  n+alpha+eta+1 ig)ig(  2n+alpha+etaig)}.$

Jacobi正交多项式的导函数

$partial_x J_{n}^{alpha,eta}(x)=frac12ig(   n+alpha+eta+1 ig)J_{n-1}^{alpha+1,eta+1}(x),$

$partial^k_x J_{n}^{alpha,eta}(x)=d_{n,k}^{alpha,eta}J_{n-k}^{alpha+k,eta+k}(x),quad ngeq k.$

$d_{n,k}^{alpha,eta}=frac{Gamma(n+k+alpha+eta+1)}{2^kGamma(n+alpha+eta+1)}.$

Jacobi-Gauss-Lobatto 积分点和积分权：

积分点：${  x_j }_{j-1}^{N-1}$是多项式$partial_x J_{N}^{alpha,eta}(x)$的零点，也即正交多项式$frac12ig(   N+alpha+eta+1 ig)J_{N-1}^{alpha+1,eta+1}(x)$的零点.

积分权：$w_0=frac{2^{alpha+eta+1}(eta+1)Gamma^2(eta+1)Gamma(N)Gamma(N+alpha+1)}{Gamma(N+eta+1)Gamma(N+alpha+eta+2)}=frac{2^{alpha+eta+1}Gamma(eta+2)N!(alpha+1)_NGamma(alpha+1)}{(eta+1)_N(alpha+eta+2)_NGamma(alpha+eta+2)},$

$w_N=frac{2^{alpha+eta+1}(alpha+1)Gamma^2(alpha+1)Gamma(N)Gamma(N+eta+1)}{Gamma(N+alpha+1)Gamma(N+alpha+eta+2)},$

$w_j=frac{1}{1-x_j^2}frac{G_{N-2}^{alpha+1,eta+1}}{J_{N-2}^{alpha+1,eta+1}partial_x J_{N-1}^{alpha+1,eta+1}(x_j)},quad 1leq jleq N-1,$

其中

$G_N^{alpha,eta}=frac{2^{alpha+eta}(2N+alpha+eta+2)Gamma(N+alpha+1)Gamma(N+eta+1)}{(N+1)!Gamma(N+alpha+eta+2)}.$