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$\underline{Intro \;to \;skip}$

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. For anyone interested in ellipsoidal harmonics the main bibliographical references (as fas as I know) are Ellipsoidal Harmonics by Dassios, The Theory of Spherical and Ellipsoidal Harmonics by Hobson, An Elementary Treatise on Fourier's Series by Byerly.

$\underline{Question}$

My goal is to find an expansion in powers of 1/ρ (and its first 2 or 3 terms) of the quantity \begin{equation}\label{eq:1} F(\rho,\mu,\nu)=(2n+1)E_n(\rho)E_n(\mu)E_n(\nu)I_n(\rho),\quad \rho \ge h_2\end{equation} where \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}} \end{equation} $\rho,\mu,\nu$ are the ellipsoidal coordinates and \begin{equation} E_n(t)=t^{n-2r}\sum_{i=0}^ra_it^{2i}, \end{equation} where \begin{align} r&=n/2, \quad n \quad \text{even},\\ r&=(n-1)/2, \quad n \quad \text{odd}. \end{align}

$\underline{What\; have\;I\; tried}$

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}

$\underline{The\;problem}$

I don t know how to take it from here. The basic problem is that by trying to express everything as a sum in the quantity of F which is the brute force/obvious way, it gets so complicated to perform multiplication between these sums that I don't know if I could find my 2-3 first terms by the resulting series expansion.

Hope you like the subject and offer some help on the matter. Thank you very much :)

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  • $\begingroup$ Why don't you expand the integrand in $I_n(\rho)$ in powers of $1/t$ and integrate term by term? You are unlikely to come up with a closed form series, but if you want just two or three terms, this ought to work. $\endgroup$ – Michael Renardy Jul 15 '16 at 13:58
  • $\begingroup$ @MichaelRenardy Can you give me an example because I don't think I understand? $\endgroup$ – ina Jul 15 '16 at 17:08
  • $\begingroup$ There is also math.SE copy of this question: math.stackexchange.com/questions/1858009/… $\endgroup$ – Martin Sleziak Jul 19 '16 at 7:26
  • $\begingroup$ @Martin Sleziak Thank you for that. I ve deleted the other question $\endgroup$ – ina Jul 19 '16 at 9:01
  • $\begingroup$ Personally, I do not think it is necessary to delete it (but as it is your question, it's your call. However, in this specific case I guess you will not be able to delete it while there is still an active bounty.) My main point was that if you are posting on both MO and math.SE, you should clearly mention this in the question and add link to the other question. $\endgroup$ – Martin Sleziak Jul 19 '16 at 9:07
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just to follow up on Michael Renardy's suggestion: let's take $n$ even $>2$, then the integrand in $I_n(t)$ has the small-$t$ expansion [using $E_n(t)=a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)$, with $n$-dependent coefficients $a_p$]:

$$K(t)=\frac{(t/h_2)^{2n}}{E_n(t)^2\sqrt{1-t^2}\sqrt{1-h_3^2(t/h_2)^2}}=c_0 (t/h_2)^{2n}+c_1 (t/h_2)^{2n+2}+c_2 (t/h_2)^{2n+4} + {\cal O}(t^{2n+6})$$

with coefficients

$$c_0=\frac{1}{a_0^2},\;\;c_1=\frac{a_0 (h_2^2+h_3^2)-4 a_1 h_2^2}{2 a_0^3},$$

$$c_2=\frac{ h_2^4\left(3 a_0^2-8 a_0 a_1-16a_0 a_2+24 a_1^2\right)+3 a_0^2 h_3^4+2 a_0 h_2^2 h_3^2 (a_0-4 a_1)}{8 a_0^4}$$

integration $I_n(\rho)=\int_0^{h_2/\rho}K(t)dt$ gives you the desired first three terms of the expansion for $n$ even,

$$I_n(\rho)=\frac{c_0 h_2 }{2n+1}(1/\rho)^{2n+1}+\frac{c_1 h_2}{2n+3}(1/\rho)^{2n+3}+\frac{c_2 h_2}{2n+5}(1/\rho)^{2n+5}+{\cal O}(1/\rho)^{2n+7}$$

for $n$ odd $>3$ we have instead $E_n(t)=t\times[a_0+a_1 t^2+a_2 t^4+ {\cal O}(t^6)]$ so the expansion becomes

$$I_n(\rho)=\frac{c_0/ h_2 }{2n-1}(1/\rho)^{2n-1}+\frac{c_1/ h_2}{2n+1}(1/\rho)^{2n+1}+\frac{c_2/ h_2}{2n+3}(1/\rho)^{2n+3}+{\cal O}(1/\rho)^{2n+5}$$

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  • $\begingroup$ Thank you very much! Just to be sure: you found the coefficients $c_0, c_1,c_2,\dots$ by equating $\left[\frac{t^{2n}/h_2^{2n}}{(a_0+a_1t^2+a_2t^4)^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]=c_0(t/h_2)^{2n}+c_1(t/h_2)^{2n+2}+c_2(t/h_2)^{2n+4} $ ? $\endgroup$ – ina Jul 19 '16 at 8:58
  • $\begingroup$ no binomial expansion is needed; I just took $K(t)$, substituted the series expansion for $E_n(t)$ and made a Taylor expansion of the whole thing; the first three terms are the ones I listed. $\endgroup$ – Carlo Beenakker Jul 19 '16 at 11:09
  • $\begingroup$ How do i proceed with $(2n+1)E(\rho)E(\mu)E(\nu)I_n(\rho)$ ? Do I just take the first few terms and multiplicate with the $I_n(\rho)$ that i ve found? ie $(2n+1)(a_0\rho^n+a_1\rho^{n-2}+a_2\rho^{n-4}+\cdots)(a_0\mu^n+a_1 \mu^{n-2}+a_2\mu^{n-4}+\cdots)(a_0\nu^n+a_1\nu^{n-2}+a_2\nu^{n-4}+\cdots)$ $\endgroup$ – ina Jul 27 '16 at 9:30
  • $\begingroup$ You have assumed that $\rho$ is much larger than one, so the appropriate expansion of $E(\rho)$ is in powers of $1/\rho$, not in powers of $\rho$. $\endgroup$ – Carlo Beenakker Jul 28 '16 at 6:06

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