Appendix/Mathematics/Hypergeometric

Hypergeometric Differential Equation

According to §9.151 (p. 1045) of Gradshteyn & Ryzhik (1965), "… a hypergeometric series is one of the solutions of the differential equation,

$0$

$=$

$z(1-z)\frac{d^{2}u}{d{z}^2}+[\gamma -(\alpha +\beta +1)z]\frac{du}{dz}-\alpha \beta u,$

which is called the hypergeometric equation. And, according to §9.10 (p. 1039) of Gradshteyn & Ryzhik (1965), "A hypergeometric series is a series of the form,

$F(\alpha ,\beta ;\gamma ;z)$

$=$

$1+\left[\frac{\alpha \cdot \beta }{\gamma \cdot 1}\right]z+\left[\frac{\alpha (\alpha +1)\beta (\beta +1)}{\gamma (\gamma +1)\cdot 1\cdot 2}\right]z^2+\left[\frac{\alpha (\alpha +1)(\alpha +2)\beta (\beta +1)(\beta +2)}{\gamma (\gamma +1)(\gamma +2)\cdot 1\cdot 2\cdot 3}\right]z^3+\dots$

Among other attributes, Gradshteyn & Ryzhik (1965) note that this, "… series terminates if $\alpha$ or $\beta$ is equal to a negative integer or to zero."

LAWE

Familiar Foundation

Drawing from an accompanying discussion, we have the,

LAWE:   Linear Adiabatic Wave (or Radial Pulsation) Equation

$\frac{d^{2}x}{d{r}_0^2}+[\frac{4}{r_0}-(\frac{g_0{\rho }_0}{P_0}\left)\right]\frac{dx}{d{r}_0}+\left(\frac{{\rho }_0}{{\gamma }_{\mathrm{g}}{P}_0}\right)[{\omega }^2+(4-3{\gamma }_{\mathrm{g}})\frac{g_0}{r_0}]x=0$

where,

$g_0$

$=$

$-\frac{1}{{\rho }_0}\frac{d{P}_0}{d{r}_0}.$

Multiplying through by $R^2$, and making the variable substitutions,

$x$	$\to$	$f,$
$\frac{r_0}{R}$	$\to$	$x,$
$(4-3{\gamma }_g)$	$\to$	$-\alpha {\gamma }_g,$

the LAWE may be rewritten as,

$0$	$=$	$\frac{d^{2}f}{d{x}^2}+[\frac{4}{x}-(\frac{g_0{\rho }_{0}R}{P_0}\left)\right]\frac{df}{dx}+\left(\frac{{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0}\right)[{\omega }^2-\frac{\alpha {\gamma }_g{g}_0}{r_0}]f$
	$=$	$\frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-(\frac{g_0{\rho }_0{r}_0}{P_0}\left)\right]\frac{df}{dx}+\left[\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0})-\frac{\alpha {\gamma }_g{g}_0}{r_0}(\frac{{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0}\left)\right]f$
	$=$	$\frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-(\frac{g_0{\rho }_0{r}_0}{P_0}\left)\right]\frac{df}{dx}+\left[\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0})-\frac{\alpha }{x^2}(\frac{g_0{r}_0{\rho }_0}{P_0}\left)\right]f.$

If we furthermore adopt the variable definition,

$\mu$

$\equiv$

$\left(\frac{g_0{\rho }_0{r}_0}{P_0}\right)=-\frac{d\ln P_0}{d\ln r_0},$

we obtain what we will refer to as the,

Kopal (1948) LAWE

$0$	$=$	$\frac{d^{2}f}{d{x}^2}+\frac{(4-\mu )}{x}\cdot \frac{df}{dx}+\left[\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0})-\frac{\alpha \mu }{x^2}]f.$
📚 Kopal (1948), p. 378, Eq. (6) 📚 Van der Borght (1970), p. 325, Eq. (1)

Specifically for Polytropes

Let's look at the expression for the function, $\mu$, that arises in the context of polytropic spheres.

General Expression for the Function μ

First, we note that,

$r_0$	$=$	$a\xi ,$
${\rho }_0$	$=$	${\rho }_c{\theta }^n,$
$P_0$	$=$	$K[{\rho }_c{\theta }^n{]}^{(n+1)/n},$
$M_r$	$=$	$4\pi a^3{\rho }_c(-{\xi }^2\frac{d\theta }{d\xi }),$
$g_0\equiv \frac{G{M}_r}{r_0^2}$	$=$	$4\pi Ga{\rho }_c(-\frac{d\theta }{d\xi }),$

where,

$a$	$\equiv$	$[\frac{(n+1)K}{4\pi G}{]}^{1/2}{\rho }_c^{(1-n)/2n}.$
$\Rightarrow K$	$=$	$\left[\frac{4\pi G}{(n+1)}\right]a^2{\rho }_c^{(n-1)/n}.$

Hence,

$\mu =\left(\frac{g_0{\rho }_0{r}_0}{P_0}\right)$	$=$	$4\pi Ga{\rho }_c(-\frac{d\theta }{d\xi }){\rho }_c{\theta }^{n}a\xi \left[{\rho }_c{\theta }^n{]}^{-(n+1)/n}\right[\frac{(n+1)}{4\pi G}]a^{-2}{\rho }_c^{-(n-1)/n}$
	$=$	$(n+1)(-\frac{d\theta }{d\xi }){\theta }^n\xi {\theta }^{-(n+1)}{\rho }_c^{2-(n+1)/n-(n-1)/n}$
	$=$	$(n+1)(-\frac{\xi }{\theta }\cdot \frac{d\theta }{d\xi })=(n+1)(-\frac{d\ln \theta }{d\ln \xi }).$

Alternatively,

$\mu =-\frac{d\ln P_0}{d\ln r_0}$	$=$	$-\frac{r_0}{P_0}\cdot \frac{d{P}_0}{d{r}_0}$
	$=$	$-\xi {\theta }^{-(n+1)}\cdot \frac{d}{d\xi }\left[{\theta }^{(n+1)}\right]$
	$=$	$-(n+1)(\frac{\xi }{\theta }\cdot \frac{d\theta }{d\xi })=(n+1)(-\frac{d\ln \theta }{d\ln \xi }).$

Yes!

Trial Displacement Function

Now, building on an accompanying discussion, let's guess,

$f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}$	$=$	$\frac{3(n-1)}{2n}[1+(\frac{n-3}{n-1}\left)\right(\frac{1}{\xi {\theta }^n}\left)\frac{d\theta }{d\xi }\right]$
$\Rightarrow \left[\frac{2n}{3(n-1)}\right]f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}$	$=$	$1+\left(\frac{n-3}{n-1}\right){\xi }^{-1}{\theta }^{-n}[\frac{\theta }{\xi }\cdot \frac{d\ln \theta }{d\ln \xi }]$
	$=$	$1+\left(\frac{3-n}{n-1}\right){\xi }^{-2}{\theta }^{(1-n)}\cdot (-\frac{d\ln \theta }{d\ln \xi })$
	$=$	$1+\left[\frac{3-n}{(n+1)(n-1)}\right]{\xi }^{-2}{\theta }^{(1-n)}\cdot \mu$

Flipping it around, we have alternatively,

$\mu$

$=$

$\left[\frac{(n+1)(n-1)}{3-n}\right]\left\{\right[\frac{2n}{3(n-1)}]f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-1\}{\xi }^2{\theta }^{(n-1)}$

Plug into Kopal (1948) LAWE

Replace f_trial by μ

Plugging this trial function into the Kopal (1948) LAWE and recognizing that $x=\xi /{\xi }_1$, we find,

${\xi }_1^{-2}\cdot \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$\frac{d^2{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d{\xi }^2}+\frac{(4-\mu )}{\xi }\cdot \frac{d{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d\xi }+\left[\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0{\xi }_1^2})-\frac{\alpha \mu }{{\xi }^2}]f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}$
$\Rightarrow \left[\frac{2n}{3(n-1)}\right]{\xi }_1^{-2}\cdot \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$\frac{d^2}{d{\xi }^2}\{1+[\frac{3-n}{(n+1)(n-1)}]{\xi }^{-2}{\theta }^{(1-n)}\cdot \mu \}+\frac{(4-\mu )}{\xi }\cdot \frac{d}{d\xi }\{1+[\frac{3-n}{(n+1)(n-1)}]{\xi }^{-2}{\theta }^{(1-n)}\cdot \mu \}$
		$+\left[\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0{\xi }_1^2})-\frac{\alpha \mu }{{\xi }^2}]\{1+[\frac{3-n}{(n+1)(n-1)}]{\xi }^{-2}{\theta }^{(1-n)}\cdot \mu \}$
$\Rightarrow \left[\frac{2n(n+1)}{3}\right]{\xi }_1^{-2}\cdot \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$\frac{d^2}{d{\xi }^2}\{(3-n){\xi }^{-2}{\theta }^{(1-n)}\cdot \mu \}+\frac{(4-\mu )}{\xi }\cdot \frac{d}{d\xi }\{(3-n){\xi }^{-2}{\theta }^{(1-n)}\cdot \mu \}$
		$+\left[\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0{\xi }_1^2})-\frac{\alpha \mu }{{\xi }^2}]\{(n+1)(n-1)+(3-n){\xi }^{-2}{\theta }^{(1-n)}\cdot \mu \}$
$\Rightarrow \left[\frac{2n(n+1)}{3(3-n)}\right]{\xi }_1^{-2}\cdot \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$\frac{d^2}{d{\xi }^2}\{{\xi }^{-2}{\theta }^{(1-n)}\cdot \mu \}+\frac{(4-\mu )}{\xi }\cdot \frac{d}{d\xi }\{{\xi }^{-2}{\theta }^{(1-n)}\cdot \mu \}+\left[\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0{\xi }_1^2})-\frac{\alpha \mu }{{\xi }^2}]\{\frac{(n+1)(n-1)}{(3-n)}+{\xi }^{-2}{\theta }^{(1-n)}\cdot \mu \}.$

Noting that, $R/{\xi }_1=a$ and

$\frac{{\rho }_0}{P_0}$

$=$

${\rho }_c{\theta }^n{K}^{-1}{\rho }_c^{-(n+1)/n}{\theta }^{-(n+1)}=K^{-1}{\rho }_c^{-1/n}{\theta }^{-1},$

the frequency-squared term may be rewritten as,

$\frac{{\omega }^2}{{\gamma }_g}\left(\frac{{\rho }_0{a}^2}{P_0}\right)$

$=$

$\left(\frac{{\omega }^2}{{\gamma }_g}\right)K^{-1}{\rho }_c^{-1/n}{\theta }^{-1}\left[\frac{(n+1)K}{4\pi G}\right]{\rho }_c^{(1-n)/n}=\left(\frac{{\omega }^2}{{\gamma }_g}\right)\left[\frac{(n+1)}{4\pi {\rho }_{c}G}\right]{\theta }^{-1}.$

Replace μ by f_trial

Making instead the alternate substitution, namely,

$\mu$	$=$	$\left[\frac{(n+1)(n-1)}{3-n}\right]\left\{\right[\frac{2n}{3(n-1)}]f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-1\}{\xi }^2{\theta }^{(n-1)}$
	$=$	$\{\frac{2n(n+1)}{3(3-n)}\cdot f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-\frac{(n+1)(n-1)}{3-n}\}{\xi }^2{\theta }^{(n-1)}$
	$=$	$\frac{1}{3(3-n)}[\underset{A}{\underbrace{2n(n+1)}}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-\underset{B}{\underbrace{3(n+1)(n-1)}}]{\xi }^2{\theta }^{(n-1)}$

we have,

${\xi }_1^{-2}\cdot \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$\frac{d^2{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d{\xi }^2}+\left\{\frac{4}{\xi }\right\}\frac{d{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d\xi }-\left\{\frac{\mu }{\xi }\right\}\frac{d{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d\xi }+\left(\frac{{\omega }^2}{{\gamma }_g}\right)\left[\frac{(n+1)}{4\pi {\rho }_{c}G}\right]\frac{f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{\theta }-\left(\frac{\alpha }{{\xi }^2}\right)\mu f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}$
$\Rightarrow \left[\frac{3(3-n)}{{\xi }_1^2}\right]\mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$3(3-n)\frac{d^2{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d{\xi }^2}+\left\{\frac{12(3-n)}{\xi }\right\}\frac{d{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d\xi }-[A{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-B]\xi {\theta }^{(n-1)}\frac{d{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d\xi }$
		$+3(3-n)\left(\frac{{\omega }^2}{{\gamma }_g}\right)\left[\frac{(n+1)}{4\pi {\rho }_{c}G}\right]\frac{f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{\theta }-\alpha [A{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-B]{\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}.$

Noting that,

$\frac{d}{d\xi }\left[\xi {\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}\right]$	$=$	$\{{\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}+\xi f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}(n-1){\theta }^{(n-2)}\frac{d\theta }{d\xi }+\xi {\theta }^{(n-1)}\frac{d{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d\xi }\}$
$\Rightarrow \xi {\theta }^{(n-1)}\frac{d{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d\xi }$	$=$	$\frac{d}{d\xi }\left[\xi {\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}\right]-[{\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}+\xi f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}(n-1){\theta }^{(n-2)}\frac{d\theta }{d\xi }],$

we furthermore can write,

$\Rightarrow \left[\frac{3(3-n)}{{\xi }_1^2}\right]\mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}$	$=$	$3(3-n)\frac{d^2{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d{\xi }^2}+\left\{\frac{12(3-n)}{\xi }\right\}\frac{d{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d\xi }+3(3-n)\left(\frac{{\omega }^2}{{\gamma }_g}\right)\left[\frac{(n+1)}{4\pi {\rho }_{c}G}\right]\frac{f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{\theta }$
		$-\alpha [A{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-B]{\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-[A{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-B]\left\{\frac{d}{d\xi }\right[\xi {\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}]-[{\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}+\xi f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}(n-1){\theta }^{(n-2)}\frac{d\theta }{d\xi }\left]\right\}$
	$=$	$3(3-n)\frac{d^2{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d{\xi }^2}+\left\{\frac{12(3-n)}{\xi }\right\}\frac{d{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{d\xi }+3(3-n)\left(\frac{{\omega }^2}{{\gamma }_g}\right)\left[\frac{(n+1)}{4\pi {\rho }_{c}G}\right]\frac{f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}}{\theta }$
		$-[A{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}-B]\left\{\frac{d}{d\xi }\right[\xi {\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}]-[{\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}+\xi f_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}(n-1){\theta }^{(n-2)}\frac{d\theta }{d\xi }]+\alpha {\theta }^{(n-1)}{f}_{\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{a}\mathrm{l}}\}$

Seek Hypergeometric Form

Start with the standard LAWE, namely,

$0$	$=$	$\frac{d^{2}f}{d{\xi }^2}+\frac{(4-\mu )}{\xi }\cdot \frac{df}{d\xi }+\left[\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0})-\frac{\alpha \mu }{{\xi }^2}]f$
	$=$	$\frac{d^{2}f}{d{\xi }^2}+\frac{1}{\xi }[4+(n+1)\frac{\xi {\theta }^{\text{'}}}{\theta }]\cdot \frac{df}{d\xi }+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}})\frac{1}{\theta }+\frac{\alpha {\theta }^{\text{'}}}{\xi \theta }]f.$

Try switching the independent variable from $\xi$ to $z$ such that,

$z$ $=$ ${\xi }^{-1}{\theta }^{-n}({\theta }^{\text{'}})$ $\Rightarrow \frac{dz}{d\xi }$ $=$ $-{\xi }^{-2}{\theta }^{-n}({\theta }^{\text{'}})-n{\xi }^{-1}{\theta }^{-n-1}({\theta }^{\text{'}})+{\xi }^{-1}{\theta }^{-n}({\theta }^{{\text{'}}^{\prime }})$   $=$ $-[{\xi }^{-2}{\theta }^{-n}({\theta }^{\text{'}})+n{\xi }^{-1}{\theta }^{-n-1}({\theta }^{\text{'}})+{\xi }^{-1}{\theta }^{-n}({\theta }^n+\frac{2{\theta }^{\prime }}{\xi }\left)\right]$   $=$ $-{\theta }^{\text{'}}[3{\xi }^{-2}{\theta }^{-n}+n{\xi }^{-1}{\theta }^{-n-1}]-{\xi }^{-1}$ and, $\frac{d^{2}z}{d{\xi }^2}$ $=$ ${\xi }^{-2}+[{\theta }^n+\frac{2{\theta }^{\text{'}}}{\xi }][3{\xi }^{-2}{\theta }^{-n}+n{\xi }^{-1}{\theta }^{-n-1}]-{\theta }^{\text{'}}\cdot \frac{d}{d\xi }[3{\xi }^{-2}{\theta }^{-n}+n{\xi }^{-1}{\theta }^{-n-1}]$   $=$ ${\xi }^{-2}+\left[{\theta }^n\right][3{\xi }^{-2}{\theta }^{-n}+n{\xi }^{-1}{\theta }^{-n-1}]+\left[\frac{2{\theta }^{\text{'}}}{\xi }\right][3{\xi }^{-2}{\theta }^{-n}+n{\xi }^{-1}{\theta }^{-n-1}]$     $-{\theta }^{\text{'}}[-6{\xi }^{-3}{\theta }^{-n}-3n{\xi }^{-2}{\theta }^{-n-1}({\theta }^{\text{'}})-n{\xi }^{-2}{\theta }^{-n-1}-n(n+1){\xi }^{-1}{\theta }^{-n-2}({\theta }^{\text{'}})]$   $=$ ${\xi }^{-2}+[3{\xi }^{-2}+n{\xi }^{-1}{\theta }^{-1}]+2({\theta }^{\text{'}})[3{\xi }^{-3}{\theta }^{-n}+n{\xi }^{-2}{\theta }^{-n-1}]$     $-({\theta }^{\text{'}})[-6{\xi }^{-3}{\theta }^{-n}-n{\xi }^{-2}{\theta }^{-n-1}-3n{\xi }^{-2}{\theta }^{-n-1}({\theta }^{\text{'}})-n(n+1){\xi }^{-1}{\theta }^{-n-2}({\theta }^{\text{'}})]$

Example Density- and Pressure-Profiles

Properties of Analytically Defined, Spherically Symmetric, Equilibrium Structures
Model	$\rho (x)$	$P(x)$	$P^{\text{'}}(x)$	$\mu (x)$	$\frac{\rho (x)}{P(x)}$
Uniform-density	$1$	$1-x^2$	$-2x$	$\frac{2x^2}{(1-x^2)}$	$\frac{1}{(1-x^2)}$
Linear	$1-x$	$(1-x{)}^2(1+2x-\frac{9}{5}{x}^2)$	$-\frac{12}{5}x(1-x)(4-3x)$	$\frac{\frac{12}{5}{x}^2(4-3x)}{(1-x)(1+2x-\frac{9}{5}{x}^2)}$	$\frac{1}{(1-x)(1+2x-\frac{9}{5}{x}^2)}$
Parabolic	$1-x^2$	$(1-x^2{)}^2(1-\frac{1}{2}{x}^2)$	$-x(1-x^2)(5-3x^2)$	$\frac{x^2(5-3x^2)}{(1-x^2)(1-\frac{1}{2}{x}^2)}$	$\frac{1}{(1-x^2)(1-\frac{1}{2}{x}^2)}$
$n=1$ Polytrope	$\frac{\sin x}{x}$	$(\frac{\sin x}{x}{)}^2$	$\frac{2}{x}[\cos x-\frac{\sin x}{x}]\frac{\sin x}{x}$	$2(1-x\cot x)$	$\frac{x}{\sin x}$

Uniform Density

In the case of a uniform-density, incompressible configuration, the Kopal (1948) LAWE becomes,

$0$	$=$	$\frac{d^{2}f}{d{x}^2}+\frac{(4-\mu )}{x}\cdot \frac{df}{dx}+\left[\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0})-\frac{\alpha \mu }{x^2}]f$
	$=$	$\frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-\frac{2x^2}{(1-x^2)}]\frac{df}{dx}+\left[\right(\frac{{\omega }^2{\rho }_c{R}^2}{{\gamma }_{\mathrm{g}}{P}_c})\frac{1}{(1-x^2)}-(\frac{2\alpha }{1-x^2}\left)\right]f$
	$=$	$(1-x^2)\cdot \frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-6x^2]\frac{df}{dx}+\left[\right(\frac{{\omega }^2{\rho }_c{R}^2}{{\gamma }_{\mathrm{g}}{P}_c})-2\alpha ]f.$

Given that, in the equilibrium state,

$\frac{{\rho }_c{R}^2}{P_c}$

$=$

$\frac{6}{4\pi G{\rho }_c}$

we obtain the LAWE derived by 📚 T. E. Sterne (1937, MNRAS, Vol. 97, pp. 582 - 593) — see his equation (1.91) on p. 585 — namely,

$0$	$=$	$(1-x^2)\cdot \frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-6x^2]\frac{df}{dx}+\left[6\right(\frac{{\omega }^2}{4\pi {\gamma }_{\mathrm{g}}G{\rho }_c})-2\alpha ]f$
	$=$	$(1-x^2)\cdot \frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-6x^2]\frac{df}{dx}+\mathfrak{𝔉}f,$

where,

$\mathfrak{𝔉}$

$\equiv$

$\left[6\right(\frac{{\omega }^2}{4\pi {\gamma }_{\mathrm{g}}G{\rho }_c})-2\alpha ].$

This also matches, respectively, equations (8) and (9) of 📚 Z. Kopal (1948, Proc. NAS, Vol. 34, Issue 8, pp.377-384), aside from what, we presume, is a type-setting error that appears in the numerator of the second term on the RHS of his equation (8):  $(4-x^2)$ appears, whereas it should be $(4-6x^2)$.

In order to see if this differential equation is of the same form as the hypergeometric expression, we'll make the substitution,

$z$	$\equiv$	$x^2$
$\Rightarrow dz$	$=$	$2xdx$
$\Rightarrow \frac{df}{dx}$	$=$	$\frac{dz}{dx}\cdot \frac{df}{dz}=2x\cdot \frac{df}{dz}=2z^{1/2}\cdot \frac{df}{dz}$
$\Rightarrow \frac{d^{2}f}{d{x}^2}$	$=$	$2z^{1/2}\cdot \frac{d}{dz}[2z^{1/2}\cdot \frac{df}{dz}]=2z^{1/2}[z^{-1/2}\cdot \frac{df}{dz}+2z^{1/2}\cdot \frac{d^{2}f}{d{z}^2}]=[2\cdot \frac{df}{dz}+4z\cdot \frac{d^{2}f}{d{z}^2}],$

in which case the 📚 Sterne (1937) LAWE may be rewritten as,

$0$	$=$	$(1-z)[2\cdot \frac{df}{dz}+4z\cdot \frac{d^{2}f}{d{z}^2}]+\frac{1}{z^{1/2}}[4-6z]2z^{1/2}\frac{df}{dz}+\mathfrak{𝔉}f$
	$=$	$(1-z)[4z\cdot \frac{d^{2}f}{d{z}^2}]+(1-z)[2\cdot \frac{df}{dz}]+2[4-6z]\frac{df}{dz}+\mathfrak{𝔉}f$
	$=$	$4z(1-z)\cdot \frac{d^{2}f}{d{z}^2}+2[5-7z]\frac{df}{dz}+\mathfrak{𝔉}f.$

This is, indeed, of the hypergeometric form if we set $(\alpha ,\beta ;\gamma ;z)$

$\gamma$	$=$	$\frac{5}{2},$
$(\alpha +\beta +1)$	$=$	$\frac{7}{2},$
$\alpha \beta$	$=$	$-\frac{\mathfrak{𝔉}}{4}.$

Combining this last pair of expressions gives,

$0$	$=$	$-\frac{\mathfrak{𝔉}}{4}-\alpha [\frac{5}{2}-\alpha ]$
	$=$	${\alpha }^2-\left(\frac{5}{2}\right)\alpha -\frac{\mathfrak{𝔉}}{4}$
$\Rightarrow \alpha$	$=$	$\frac{1}{2}\{\frac{5}{2}\pm [(\frac{5}{2}{)}^2-\mathfrak{𝔉}{]}^{1/2}\}$
	$=$	$\frac{5}{4}\{1\pm [1-\left(\frac{4}{25}\right)\mathfrak{𝔉}{]}^{1/2}\};$

and,

$\beta$

$=$

$\frac{5}{2}-\frac{5}{4}\{1\pm [1-\left(\frac{4}{25}\right)\mathfrak{𝔉}{]}^{1/2}\}.$

Example α = -1

If we set $\alpha =-1$, then the eigenvector is,

$u_1=F(-1,\frac{7}{2};\frac{5}{2};x^2)$

$=$

$1-\left[\frac{\beta }{\gamma }\right]x^2=1-\left(\frac{7}{5}\right)x^2;$

and the corresponding eigenfrequency is obtained from the expression,

$-1$	$=$	$\frac{5}{4}\{1\pm [1-\left(\frac{4}{25}\right)\mathfrak{𝔉}{]}^{1/2}\}$
$\Rightarrow -\frac{9}{5}$	$=$	$\pm [1-(\frac{4}{25})\mathfrak{𝔉}{]}^{1/2}$
$\Rightarrow \frac{3^4}{5^2}$	$=$	$1-\left(\frac{4}{25}\right)\mathfrak{𝔉}$
$\Rightarrow \mathfrak{𝔉}$	$=$	$\left(\frac{5^2}{4}\right)[1-\frac{3^4}{5^2}]=\frac{1}{4}[5^2-3^4]=14.$

As we have reviewed in a separate discussion, this is identical to the eigenvector identified by 📚 Sterne (1937) as mode "$j=1$".

More Generally

More generally, in agreement with 📚 Sterne (1937), for any (positive integer) mode number, $0\le j\le \mathrm{\infty }$, we find,

${\alpha }_j$

$=$

$-j;$       ${\beta }_j=\frac{5}{2}+j;$       $\gamma =\frac{5}{2};$       $\mathfrak{𝔉}=2j(2j+5).$

And, in terms of the hypergeometric function series, the corresponding eigenfunction is,

$u_j=$

$=$

$F({\alpha }_j,{\beta }_j;\frac{5}{2};x^2).$

See Also

• In an article titled, "Radial Oscillations of a Stellar Model," 📚 C. Prasad (1949, MNRAS, Vol 109, pp. 103 - 107) investigated the properties of an equilibrium configuration with a prescribed density distribution given by the expression,

  $\rho (r)={\rho }_c[1-(\frac{r}{R}{)}^2],$

  where, ${\rho }_c$ is the central density and, $R$ is the radius of the star.

• MathProjects/EigenvalueProblemN1:   In the most general context, the LAWE takes the form,

  $\left[P\right]\frac{d^2{{𝒢}}_{\sigma }}{d{x}^2}+[\frac{4P}{x}+P^{\text{'}}]\frac{d{{𝒢}}_{\sigma }}{dx}+[{\sigma }^2\rho +\frac{\alpha P^{\text{'}}}{x}]{{𝒢}}_{\sigma }$

  $=$

  $0.$

  Properties of Analytically Defined Astrophysical Structures
  Model	$\rho (x)$	$P(x)$	$P^{\text{'}}(x)$
  Uniform-density	$1$	$1-x^2$	$-2x$
  Linear	$1-x$	$(1-x{)}^2(1+2x-\frac{9}{5}{x}^2)$	$-\frac{12}{5}x(1-x)(4-3x)$
  Parabolic	$1-x^2$	$(1-x^2{)}^2(1-\frac{1}{2}{x}^2)$	$-x(1-x^2)(5-3x^2)$
  $n=1$ Polytrope	$\frac{\sin x}{x}$	$[\frac{\sin x}{x}{]}^2$	$\frac{2}{x}[\cos x-\frac{\sin x}{x}]\frac{\sin x}{x}$