Skip to main content
1 of 5
ina
  • 23
  • 7

Ellipsoidal harmonics - A Series expansion for Lame functions of the second kind

In the theory of ellipsoidal harmonics, Lame functions of the second kind $F_n$ arise as the second linearly independent solution (the first being Lame functions of the first kind $E_n$) of Laplace's equation in ellipsoidal coordinates which is the so called Lame's equation. Lame functions of the second kind contain the improper elliptic integral $I_n(\rho)$ and for this reason they are obtainable through numerical approximations e.g. Gauss-Legrendre quadrature. A really comprehensive book on the theory of ellipsoidal harmonics is https://books.google.gr/books?isbn=0521113091 .

For "investigational" purposes my goal is to find an expansion in powers of 1/ρ (and its first 2 or 3 terms) of the elliptic integral: \begin{equation}I_n(\rho)=\int_\rho^{+\infty}\frac{dt}{(E_n(t))^2\sqrt{t^2-h_2^2}\sqrt{t^2-h_3^2}},\quad \rho \ge h_2\end{equation} where $E_n(t)$ is a polynomial in descending powers of n of the form: \begin{equation} E_n(t)=\sum_{k=0}^\infty a_kt^{n-2k}, \quad a_0\neq 0\end{equation} where for negative powers summation is terminated. A first step to solution is to set $t\rightarrow \frac{h_2}{t}$ and this leads to \begin{equation} I_n(\rho)=\int_0^\frac{h_2}{\rho}\frac{t^{2n}/h_2^{2n}}{(E_n(t))^2\sqrt{1-t^2}\sqrt{1-\frac{h_3^2}{h_2^2}t^2}}dt. \end{equation} Using binomial theorem \begin{align} (1-t^2)^{-1/2}&=\sum_{n=0}^{+\infty}(-1)^n\binom{-\frac{1}{2}}{n}t^{2n},\\ (1-\frac{h_3^2}{h_2^2}t^2)^{-1/2}&=\sum_{n=0}^{+\infty}(-1)^n\binom{-\frac{1}{2}}{n}\frac{h_3^{2n}}{h_2^{2n}}t^{2n}, \end{align} and multiplying gives the series expansion \begin{equation}\sum_{n=0}^{+\infty}\sum_{k=0}^n(-1)^n\binom{-\frac{1}{2}}{n-k}\binom{-\frac{1}{2}}{k}\frac{h_3^{2k}}{h_2^{2k+1}}t^{2n}\end{equation} so: \begin{equation}I_n(\rho)=\int_0^{h_2/\rho}\left[\frac{t^{2n}/h_2^{2n}}{(E_n(t))^2 }\sum_{n=0}^{+\infty}\sum_{k=0}^n(-1)^n\binom{-\frac{1}{2}}{n-k}\binom{-\frac{1}{2}}{k}\frac{h_3^{2k}}{h_2^{2k+1}}t^{2n}\right] dt .\end{equation}

And here starts the problem:

The reciprocal of the square of $E_n$ is: \begin{equation} \frac{1}{E_n[(t)]^2}=\sum_{n=0}^{+\infty}c_nt^n \end{equation} where: \begin{align} c_0&=\frac{1}{a_0}\\ c_n&=-\frac{1}{na_0^2}\sum_{k=0}^{n-1}(2n-k)a_{n-k}c_k, \quad n\ge 1 \end{align}

If everything up to now is correct then how do i proceed? Hope you like the subject and offer some help on the matter. Thank you very much :)

ina
  • 23
  • 7