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I would like to find the relation between the solutions of a differential equation obtained by two different authors. The first solution is given in terms of the hypergeometric function $_2F_1$:

$$\left(1+\beta p^2\right)^{\frac{1}{4} (-1-2 n-2 \xi)} \, _2F_1\left(-n,-n-2 \xi ;-n-\xi +\frac{1}{2};\frac{1}{2}\left(1 + i p \sqrt{\beta}\right)\right),$$

where $p$ is a real variable, $\beta>0$, $n$ is a positive integer and $\xi= \sqrt{4 + \beta^2}/(2\beta)$.

The second one involves the Gegenbauer polynomials $C_n^\lambda(z)$

$$ \left(1 + \beta p^2\right)^{-\frac{1}{2} \left(\xi +\frac 12\right)} C_n^{\left(\xi +\frac{1}{2}\right)}\left(\frac{\sqrt{\beta } p}{\sqrt{\beta p^2+1}}\right).$$

I have omitted some normalization constants. Perhaps it is easy, but I do not see it. I know the relations between the hypergeometric function and the Gegenbauer polynomials but I can't apply them to this problem. In particular, I'm not sure how to transform the argument of $_2F_1$ to match the argument of $C_n^\lambda(z)$.

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  • $\begingroup$ @MartinNicholson $n$ is an arbitrary positive integer (and also zero): $n = 0,1,2,\ldots$. But if you have a relation that involves even or odd $n$ it's a good start. $\endgroup$ – Pincopallino Nov 24 '15 at 18:48
  • $\begingroup$ formulas 20-23 mathworld.wolfram.com/GegenbauerPolynomial.html $\endgroup$ – Nemo Nov 24 '15 at 19:46
  • $\begingroup$ @MartinNicholson I have already seen these formulas but I can't use them, or at least I can't find the proper variable change from $(1+i\sqrt{\beta}p)/2$ to $\sqrt{\beta} p / \sqrt{\beta p^2 +1}$ $\endgroup$ – Pincopallino Nov 25 '15 at 15:53
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I will consider the case when $n$ is even. The fact that these two solutions are equivalent when $n$ is odd can be proved in an analogous manner.

Let $n=2\nu$ be even nonnegative integer, $\lambda=\xi+\frac{1}{2}$ and let's absorb $\sqrt{\beta}$ into the definition of $p$. Then the ratio of the two expressions above can be written as $$ \displaystyle\frac{ \displaystyle\phantom{}_2F_1\left(-2\nu,-2\nu-2\lambda+1,-2\nu-\lambda+1;\frac{1+pi}{2}\right)}{ \displaystyle(1+p^2)^\nu C_{2\nu}^\lambda\left(\frac{p}{\sqrt{p^2+1}}\right)}. $$ By eq. (20): $$ C_{2\nu}^\lambda\left(\frac{p}{\sqrt{p^2+1}}\right)=\left({\textstyle 2\nu+2\lambda-1\atop \textstyle 2\nu}\right)\phantom{}_2F_1\left(-\nu,\nu+\lambda,\lambda+\frac{1}{2};\frac{1}{p^2+1}\right). $$ By eq. (15.8.2) this is equal, up to a constant coefficient: $$ \frac{1}{(p^2+1)^\nu}\cdot \phantom{}_2F_1\left(-\nu,-\nu-\lambda+\frac{1}{2},-2\nu-\lambda+1;p^2+1\right). $$ By Kummer's quadratic transformation, eq. 15.8.18: $$ \phantom{}_2F_1\left(-\nu,-\nu-\lambda+\frac{1}{2},-2\nu-\lambda+1;p^2+1\right)=\phantom{}_2F_1\left(-2\nu,-2\nu-2\lambda+1,-2\nu-\lambda+1;\frac{1\pm pi}{2}\right). $$ So we proved that the expression $$ \displaystyle\frac{ \displaystyle\phantom{}_2F_1\left(-2\nu,-2\nu-2\lambda+1,-2\nu-\lambda+1;\frac{1+pi}{2}\right)}{ \displaystyle(1+p^2)^\nu C_{2\nu}^\lambda\left(\frac{p}{\sqrt{p^2+1}}\right)}. $$ is constant, i.e. doesn't depend on $p$, as required.

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  • $\begingroup$ Thank you! But I think that you used 15.8.2 instead of 15.8.1 and 15.8.18 instead of 15.8.17, am I right? Otherwise I can't explain the passages. $\endgroup$ – Pincopallino Nov 30 '15 at 16:16
  • $\begingroup$ @Pincopallino You are right, thanks for mentioning that. $\endgroup$ – Nemo Nov 30 '15 at 16:37
  • $\begingroup$ While I was working out the normalization constant, I came up with an interesting thing. The normalization constant of the two functions should be a number close to unity. After applying Eq. (15.8.2) of the NIST DLMF I obtained a very large $\Gamma(2 \lambda)$ in the normalization. But by using the corresponding formula on the Abramowitz & Stegun, (15.3.7), which is slightly different, then everything is correct. I wonder why the two formulas are different and why the older one works for me and the other doesn't $\endgroup$ – Pincopallino Dec 1 '15 at 20:59
  • $\begingroup$ @Pincopallino try to use (1.8.6) instead of (1.8.2). $\endgroup$ – Nemo Dec 1 '15 at 21:03
  • $\begingroup$ Yes (1.8.6) is the right one to use, thank you. Still, I don't understand the problem with (15.8.2)... $\endgroup$ – Pincopallino Dec 2 '15 at 11:13

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