Appendix/Mathematics/Hypergeometric

Hypergeometric Differential Equation

Gradshteyn & Ryzhik

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."

Van der Borght

General Value for b

Comment by J. E. Tohline: In Van der Borght (1970), the fourth argument of the hypergeometric function is ax² whereas the more general power law form, ax^b, applies.

In association with his equation (3), 📚 R. Van der Borght (1970, Proc. Astr. Soc. Australia, Vol. 1, Issue 7, pp. 325 - 326) states that a displacement function of the form,

$\xi$

$=$

$x^{c}F(\alpha ,\beta ,\gamma ,a{x}^b),$

provides a solution to the following 2^nd-order ODE:

$0$

$=$

$x^2(a{x}^b-1)\cdot \frac{d^2\xi }{d{x}^2}+(ap{x}^b+\lambda )x\cdot \frac{d\xi }{dx}+(ar{x}^b+s)\cdot \xi .$

Is this ODE essentially the same as the above-defined hypergeometric equation?

A mapping between the two differential equations requires,

$a{x}^b$

$\leftrightarrow$

$z,$

and,

$\frac{\xi }{x^c}$

$\leftrightarrow$

$u,$

$\Rightarrow \frac{dx}{dz}$

$=$

$\frac{d}{dz}(\frac{z}{a}{)}^{1/b}=a^{-1/b}\cdot \frac{d{z}^{1/b}}{dz}=\frac{1}{b}\cdot a^{-1/b}{z}^{-1+1/b}=\frac{1}{bz}\cdot (\frac{z}{a}{)}^{1/b}=\frac{x}{bz}=\frac{x^{1-b}}{ab},$

in which case,

$0$	$=$	$z(1-z)\frac{d^{2}u}{d{z}^2}+[\gamma -(\alpha +\beta +1)z]\frac{du}{dz}-\alpha \beta u$
	$=$	$a{x}^b(1-a{x}^b)\{\frac{dx}{dz}\cdot \frac{d}{dx}[\frac{dx}{dz}\cdot \frac{d}{dx}\left(\xi x^{-c}\right)\left]\right\}+[\gamma -(\alpha +\beta +1)a{x}^b]\frac{dx}{dz}\cdot \frac{d}{dx}\cdot \left(\xi x^{-c}\right)-\alpha \beta \left(\xi x^{-c}\right)$
	$=$	$a{x}^b(1-a{x}^b)\{\frac{x^{1-b}}{ab}\cdot \frac{d}{dx}[\frac{x^{1-b}}{ab}\cdot (x^{-c}\frac{d\xi }{dx}-c\xi x^{-1-c})\left]\right\}+[\gamma -(\alpha +\beta +1)a{x}^b]\cdot \frac{x^{1-b}}{ab}\cdot (x^{-c}\frac{d\xi }{dx}-c\xi x^{-1-c})-\alpha \beta \left(\xi x^{-c}\right)$
	$=$	$a{x}^b(1-a{x}^b)\{\frac{x^{1-b}}{a^2{b}^2}\cdot \frac{d}{dx}[x^{1-b-c}\frac{d\xi }{dx}-c\xi x^{-b-c}\left]\right\}+\frac{[\gamma -(\alpha +\beta +1)a{x}^b]}{ab}\cdot (x^{1-b-c}\frac{d\xi }{dx}-c\xi x^{-b-c})-\alpha \beta \left(\xi x^{-c}\right)$
	$=$	$a{x}^b(1-a{x}^b)\cdot \frac{x^{1-b}}{a^2{b}^2}\cdot [x^{1-b-c}\frac{d^2\xi }{d{x}^2}+(1-b-c)x^{-b-c}\frac{d\xi }{dx}-c{x}^{-b-c}\frac{d\xi }{dx}-c(-b-c)\xi x^{-1-b-c}]$
		$+\frac{[\gamma -(\alpha +\beta +1)a{x}^b]}{ab}\cdot [x^{1-b-c}\frac{d\xi }{dx}-c\xi x^{-b-c}]-\left(\alpha \beta x^{-c}\right)\xi$
	$=$	$(1-a{x}^b)\cdot \frac{x}{a{b}^2}\cdot \left[x^{1-b-c}\frac{d^2\xi }{d{x}^2}\right]+(1-a{x}^b)\cdot \frac{x}{a{b}^2}\cdot [(1-b-c)x^{-b-c}\frac{d\xi }{dx}-c{x}^{-b-c}\frac{d\xi }{dx}]+(1-a{x}^b)\cdot \frac{x}{a{b}^2}\cdot [c(b+c)\xi x^{-1-b-c}]$
		$+\frac{[\gamma -(\alpha +\beta +1)a{x}^b]}{ab}\cdot \left[x^{1-b-c}\frac{d\xi }{dx}\right]-\frac{[\gamma -(\alpha +\beta +1)a{x}^b]}{ab}\cdot \left[c\xi x^{-b-c}\right]-\left(\alpha \beta x^{-c}\right)\xi$
	$=$	$\frac{(1-a{x}^b)}{a{b}^2}\cdot x^2\left[x^{-b-c}\frac{d^2\xi }{d{x}^2}\right]+\{\frac{(1-a{x}^b)}{a{b}^2}\cdot [(1-b-c)x^{1-b-c}-c{x}^{1-b-c}]+\frac{[\gamma -(\alpha +\beta +1)a{x}^b]}{ab}\cdot [x^{1-b-c}\left]\right\}\frac{d\xi }{dx}$
		$+\{-\frac{c[\gamma -(\alpha +\beta +1)a{x}^b]}{ab}\cdot [x^{-b-c}]-(\alpha \beta x^{-b-c})x^b+\frac{(1-a{x}^b)}{a{b}^2}\cdot [c(b+c)x^{-b-c}\left]\right\}\xi .$

Multiplying through by $(-a{b}^2{x}^{b+c})$ gives,

$(-a{b}^2{x}^{b+c})\times 0$	$=$	$x^2(a{x}^b-1)\frac{d^2\xi }{d{x}^2}+\{(a{x}^b-1)\cdot [(1-b-c)-c]-b[\gamma -(\alpha +\beta +1)a{x}^b]\}x\cdot \frac{d\xi }{dx}$
		$+\{bc[\gamma -(\alpha +\beta +1)a{x}^b]+(\alpha \beta a{b}^2)x^b-(1-a{x}^b)\cdot [c(b+c)\left]\right\}\xi$
	$=$	$x^2(a{x}^b-1)\frac{d^2\xi }{d{x}^2}+\{-(1+b\gamma -b-2c)+a{x}^b(1-b-2c)+b(\alpha +\beta +1)a{x}^b\}x\cdot \frac{d\xi }{dx}$
		$+\{bc\gamma -c(b+c)-bc(\alpha +\beta +1)a{x}^b+\alpha \beta a{b}^2{x}^b+c(b+c)a{x}^b\}\xi$
	$=$	$x^2(a{x}^b-1)\frac{d^2\xi }{d{x}^2}+(ap{x}^b+\lambda )x\cdot \frac{d\xi }{dx}+(ar{x}^b+s)\xi ,$

which matches equation (3) of 📚 Van der Borght (1970) if the expressions for the four new scalar coefficients are,

Required Mapping Expressions

1st: $p$ $=$ $(1-b-2c)+b(\alpha +\beta +1)=1-2c+b(\alpha +\beta ),$ 2nd: $\lambda$ $=$ $-(1+b\gamma -b-2c),$ 3rd: $s$ $=$ $bc\gamma -c(b+c)=c(b\gamma -b-c),$ 4th: $r$ $=$ $-bc(\alpha +\beta +1)+\alpha \beta b^2+c(b+c).$

If, for any given problem, we are given the values of these four scalar coefficients along with a choice of the exponent, $b$, that appears in the fourth argument of the hypergeometric series, we can determine values the other three arguments of the hypergeometric series — $\alpha ,\beta ,\gamma$ — and the exponent, $c$. In what follows we show how this is done.

Determining the Value of the Exponent, c

Equating $(b\gamma )$ in the 2^nd and 3^rd of the required mapping expressions, gives,

$-(1-b-2c+\lambda )$	$=$	$\frac{s}{c}+b+c$
$\Rightarrow (1+\lambda )$	$=$	$c-\frac{s}{c}$
$\Rightarrow 0$	$=$	$c^2-c(1+\lambda )-s.$

The pair of roots, $c_{\pm }$, of this quadratic equation are then obtained from the relation,

$2c_{\pm }$

$=$

$(1+\lambda )\pm [(1+\lambda {)}^2+4s{]}^{1/2}.$

Note for further use below that,

$2(c_{+}-c_{-})$	$=$	$\{(1+\lambda )+[(1+\lambda {)}^2+4s{]}^{1/2}\}-\{(1+\lambda )-[(1+\lambda {)}^2+4s{]}^{1/2}\}.$
	$=$	$2[(1+\lambda {)}^2+4s{]}^{1/2}.$

Consistent with our derivation, 📚 Van der Borght (1970) states, "… if $A_1$, $A_2$ are the solutions of $A^2-(\lambda +1)A-s=0$ … then $c=A_1$ …"

Determining the Value of the Coefficient, γ

Combining our 3^rd required mapping expression with the quadratic equation for $c_{\pm }$ in such a way as to eliminate $s$, we find,

$cb(\gamma -1)-c^2$	$=$	$c^2-c(1+\lambda )$
$\Rightarrow b(\gamma -1)$	$=$	$2c_{\pm }-(1+\lambda ).$

Adopting the superior sign, we find that,

$b(\gamma -1)$	$=$	$2c_{+}-(1+\lambda )=[(1+\lambda {)}^2+4s{]}^{1/2}$
	$=$	$(c_{+}-c_{-}),$

where, in order to make this last step we have drawn from the relation derived immediately above.

Consistent with this derivation, 📚 Van der Borght (1970) states, "… $(1-\gamma )b=A_2-A_1$ …"

Determining the Values of the Coefficients, α and β

Combining the 1^st and 4^th required mapping expressions in such a way as to cancel terms involving $bc(\alpha +\beta )$, we find,

$(1-p)c-2c^2+bc(\alpha +\beta )$	$=$	$bc(\alpha +\beta )+bc-\alpha \beta b^2-c(b+c)+r$
$\Rightarrow (1-p)c-c^2$	$=$	$-\alpha \beta b^2+r$
$\Rightarrow (b\alpha )(b\beta )$	$=$	$-(1-p)c+c^2+r.$

Also, from the 1^st required mapping expression alone we can write,

$(b\alpha )$

$=$

$(p-1)+2c-b\beta .$

Together, then, we have,

$(b\beta )[(p-1)+2c-b\beta ]$	$=$	$-(1-p)c+c^2+r$
$\Rightarrow 0$	$=$	$(b\beta {)}^2-(b\beta )\underset{(b\alpha +b\beta )}{\underbrace{[(p-1)+2c]}}+\stackrel{(b\alpha )\cdot (b\beta )}{\overbrace{[c^2+(p-1)c+r]}}.$

The pair of roots, $(b\beta {)}_{\pm }$, of this quadratic equation are then obtained from the relation,

$2(b\beta {)}_{\pm }$

$=$

$[(p-1)+2c]\pm \{[(p-1)+2c{]}^2-4[c^2+(p-1)c+r]{\}}^{1/2}.$

Finally, plugging this expression for $(b{\beta }_{\pm })$ into the 1^st required mapping expression gives $(b{\alpha }_{\pm }).$

Alternate Determination of α and β by Completing Squares

Again, let's draw upon the 📚 Van der Borght (1970) statement that, "… if $A_1$, $A_2$ are the solutions of $A^2-(\lambda +1)A-s=0$ … then $c=A_1$ …" 📚 Van der Borght (1970) also states that if, "… $B_1,B_2$ are solutions of $B^2-(p-1)B+r=0$, then … $b\alpha =A_1+B_1$ and $b\beta =A_1+B_2.$" Let's see if we draw these same conclusions.

First, Complete the square in the quadratic equation for $B^2$:

$B^2-(p-1)B$	$=$	$-r$
$\Rightarrow B^2-(p-1)B+[\frac{(p-1)}{2}{]}^2$	$=$	$[\frac{(p-1)}{2}{]}^2-r$
$\Rightarrow [B-\frac{(p-1)}{2}{]}^2$	$=$	$[\frac{(p-1)}{2}{]}^2-r$

Second, complete the square in the quadratic equation for $A^2$ — which also completes the square for $c^2$:

$A^2-(\lambda +1)A$	$=$	$s$
$\Rightarrow A^2-(\lambda +1)A+[\frac{(\lambda +1)}{2}{]}^2$	$=$	$s+[\frac{(\lambda +1)}{2}{]}^2$
$\Rightarrow [A-\frac{(\lambda +1)}{2}{]}^2$	$=$	$s+[\frac{(\lambda +1)}{2}{]}^2$

Third, complete the square in the quadratic equation for $(b\beta {)}^2$:

$(b\beta {)}^2-(b\beta )[(p-1)+2c]$	$=$	$-[c^2+(p-1)c+r]$
$\Rightarrow (b\beta {)}^2-(b\beta )[(p-1)+2c]+[\frac{(p-1)+2c}{2}{]}^2$	$=$	$[\frac{(p-1)+2c}{2}{]}^2-[c^2+(p-1)c+r]$
$\Rightarrow [(b\beta )-\frac{(p-1)+2c}{2}{]}^2$	$=$	$[\frac{(p-1)+2c}{2}{]}^2-[c^2+(p-1)c+\frac{(p-1{)}^2}{2^2}]-r+\frac{(p-1{)}^2}{2^2}$
	$=$	$[\frac{(p-1)}{2}+c{]}^2-[c+\frac{(p-1)}{2}{]}^2-r+\frac{(p-1{)}^2}{2^2}$
	$=$	$-r+\frac{(p-1{)}^2}{2^2}$
	$=$	$[B-\frac{(p-1)}{2}{]}^2.$

Taking the positive root of both sides of this expression, we find that,

$(b\beta )-\frac{(p-1)}{2}-c_{+}$	$=$	$B_{+}-\frac{(p-1)}{2}$
$\Rightarrow (b\beta )$	$=$	$B_{+}+c_{+}.$

But, $c_{\pm }=A_{\pm }$. So we conclude, as did 📚 Van der Borght (1970), that,

$(b\beta )$

$=$

$B_{+}+A_{+}.$

Alternatively, taking the negative root of the RHS of this expression, we find that,

$(b\beta )-\frac{(p-1)}{2}-c_{+}$	$=$	$-B_{-}+\frac{(p-1)}{2}$
$\Rightarrow (b\beta )$	$=$	$\frac{(p-1)}{2}+c_{+}-B_{-}+\frac{(p-1)}{2}$
	$=$	$c_{+}-B_{-}+(p-1).$

Also, given that,

$1-2c+b(\alpha +\beta )$	$=$	$p$
$\Rightarrow (b\alpha )$	$=$	$p-1+2c-(b\beta )$
	$=$	$(p-1)+2c-[c_{+}-B_{-}+(p-1)]$
	$=$	$2c-c_{+}+B_{-}.$

As long as we assume that $c=c_{+}$ in this expression, we also obtain the 📚 Van der Borght (1970) expression for $(b\alpha )$, namely,

$(b\alpha )$

$=$

$B_{-}+A_{+}.$

If b = 2

$0$	$=$	$z(1-z)\frac{d^{2}u}{d{z}^2}+[\gamma -(\alpha +\beta +1)z]\frac{du}{dz}-\alpha \beta u$
	$=$	$a{x}^2(1-a{x}^2)\{\frac{dx}{dz}\cdot \frac{d}{dx}[\frac{dx}{dz}\cdot \frac{d}{dx}\left(\xi x^{-c}\right)\left]\right\}+[\gamma -(\alpha +\beta +1)a{x}^2]\frac{dx}{dz}\cdot \frac{d}{dx}\cdot \left(\xi x^{-c}\right)-\alpha \beta \left(\xi x^{-c}\right)$
	$=$	$a{x}^2(1-a{x}^2)\{\frac{1}{2ax}\cdot \frac{d}{dx}[\frac{1}{2ax}\cdot (x^{-c}\frac{d\xi }{dx}-c\xi x^{-1-c})\left]\right\}+[\gamma -(\alpha +\beta +1)a{x}^2]\cdot \frac{1}{2ax}\cdot (x^{-c}\frac{d\xi }{dx}-c\xi x^{-1-c})-\alpha \beta \left(\xi x^{-c}\right)$
	$=$	$a{x}^2(1-a{x}^2)\{\frac{1}{4a^{2}x}\cdot \frac{d}{dx}[(x^{-1-c}\frac{d\xi }{dx}-c\xi x^{-2-c})\left]\right\}+[\gamma -(\alpha +\beta +1)a{x}^2]\cdot \frac{1}{2a}\cdot (x^{-1-c}\frac{d\xi }{dx}-c\xi x^{-2-c})-\alpha \beta \left(\xi x^{-c}\right)$
	$=$	$\frac{x(1-a{x}^2)}{4a}\{x^{-1-c}\frac{d^2\xi }{d{x}^2}-(1+c)x^{-2-c}\frac{d\xi }{dx}-c{x}^{-2-c}\frac{d\xi }{dx}+(2c+c^2)\xi x^{-3-c}\}$
		$+\left[\frac{\gamma -(\alpha +\beta +1)a{x}^2}{2a}\right]\cdot \left(x^{-1-c}\frac{d\xi }{dx}\right)-\left\{\right[\frac{\gamma -(\alpha +\beta +1)a{x}^2}{2a}]\cdot c{x}^{-2-c}+\alpha \beta x^{-c}\}\xi$
	$=$	$\frac{x(1-a{x}^2)}{4a}\left\{x^{-1-c}\frac{d^2\xi }{d{x}^2}\right\}+\left\{\frac{x(1-a{x}^2)}{4a}\right[-(1+c)x^{-2-c}-c{x}^{-2-c}]+[\frac{\gamma -(\alpha +\beta +1)a{x}^2}{2a}]\cdot x^{-1-c}\}\frac{d\xi }{dx}$
		$+\left\{\right[\frac{x(1-a{x}^2)}{4a}](2c+c^2)x^{-3-c}-[\frac{\gamma -(\alpha +\beta +1)a{x}^2}{2a}]\cdot c{x}^{-2-c}-\alpha \beta x^{-c}\}\xi .$

Multiplying through by a term proportional to $x^{2+c}$ gives us,

$[-4a{x}^{2+c}]\times \left[0\right]$	$=$	$x^2(a{x}^2-1)\left\{\frac{d^2\xi }{d{x}^2}\right\}+\{x(1-a{x}^2)[(1+c)+c]-2[\gamma -(\alpha +\beta +1)a{x}^2]\cdot x\}\frac{d\xi }{dx}$
		$+\{(a{x}^2-1)(2c+c^2)-2c[\gamma -(\alpha +\beta +1)a{x}^2]+4a\alpha \beta x^2\}\xi$
	$=$	$x^2(a{x}^2-1)\left\{\frac{d^2\xi }{d{x}^2}\right\}+\{(1+2c-2\gamma )-a{x}^2(1+2c)+2(\alpha +\beta +1)a{x}^2]\}x\cdot \frac{d\xi }{dx}$
		$+\{(a{x}^2-1)(2c+c^2)-2c[\gamma -(\alpha +\beta +1)a{x}^2]+4a\alpha \beta x^2\}\xi$

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,

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

Part I

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

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

and,

$\frac{d^{2}z}{d{\xi }^2}$	$=$	$-\{-{\xi }^{-2}(1+3z)+3{\xi }^{-1}\cdot \frac{dz}{d\xi }+n{\theta }^{n-1}{z}^2+n(n-1)\xi {\theta }^{n-2}({\theta }^{\text{'}})z^2+2n\xi {\theta }^{n-1}z\cdot \frac{dz}{d\xi }\}$
	$=$	$-\{-{\xi }^{-2}(1+3z)-3{\xi }^{-1}\cdot [{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2]+n{\theta }^{n-1}{z}^2+n(n-1)\xi {\theta }^{n-2}(\xi {\theta }^{n}z)z^2-2n\xi {\theta }^{n-1}z\cdot [{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2\left]\right\}$
	$=$	$\{{\xi }^{-2}(1+3z)+3{\xi }^{-1}\cdot [{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2]-n{\theta }^{n-1}{z}^2-n(n-1)\xi {\theta }^{n-2}(\xi {\theta }^{n}z)z^2+2n\xi {\theta }^{n-1}z\cdot [{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2\left]\right\}$
	$=$	$\{{\xi }^{-2}(1+3z)+[3{\xi }^{-2}(1+3z)+3n{\theta }^{n-1}{z}^2]-n{\theta }^{n-1}{z}^2-n(n-1){\xi }^2{\theta }^{2(n-1)}{z}^3+[2n{\theta }^{n-1}z(1+3z)+2n^2{\xi }^2{\theta }^{2(n-1)}{z}^3\left]\right\}$
	$=$	$4{\xi }^{-2}(1+3z)+2n{\theta }^{n-1}[z(1+3z)+z^2]+[2n^2-n(n-1)]{\xi }^2{\theta }^{2(n-1)}{z}^3$
	$=$	$4{\xi }^{-2}(1+3z)+2n{\theta }^{n-1}[1+4z]z+n(n+1){\xi }^2{\theta }^{2(n-1)}{z}^3.$

Part II

Part I Summary …

$({\theta }^{\text{'}})$ $=$ $\xi {\theta }^{n}z,$ $\frac{dz}{d\xi }$ $=$ $-[{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2],$ $\frac{d^{2}z}{d{\xi }^2}$ $=$ $4{\xi }^{-2}(1+3z)+2n{\theta }^{n-1}[1+4z]z+n(n+1){\xi }^2{\theta }^{2(n-1)}{z}^3.$

---

Also,

$\frac{df}{d\xi }$	$\to$	$\frac{dz}{d\xi }\cdot \frac{df}{dz};$
$\frac{d^{2}f}{d{\xi }^2}=\frac{d}{d\xi }[\frac{dz}{d\xi }\cdot \frac{df}{dz}]$	$\to$	$\frac{d^{2}z}{d{\xi }^2}\cdot \frac{df}{dz}+[\frac{dz}{d\xi }{]}^2\cdot \frac{d^{2}f}{d{z}^2}$

As a result,

LAWE	$=$	$\frac{d^{2}f}{d{\xi }^2}+\frac{1}{\xi }[4+(n+1)\frac{\xi {\theta }^{\text{'}}}{\theta }]\cdot \frac{df}{d\xi }+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}})\frac{1}{\theta }+\frac{\alpha {\theta }^{\text{'}}}{\xi \theta }]f$
	$=$	$\frac{d^{2}f}{d{\xi }^2}$
		$+\frac{1}{\xi }[4+(n+1){\xi }^2{\theta }^{n-1}z]\cdot \frac{df}{d\xi }$
		$+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}})\frac{1}{\theta }+\alpha {\theta }^{n-1}z]f$
	$=$	$[\frac{dz}{d\xi }{]}^2\cdot \frac{d^{2}f}{d{z}^2}+\frac{d^{2}z}{d{\xi }^2}\cdot \frac{df}{dz}$
		$+[4{\xi }^{-1}+(n+1)\xi {\theta }^{n-1}z]\frac{dz}{d\xi }\cdot \frac{df}{dz}$
		$+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}){\theta }^{-1}+\alpha {\theta }^{n-1}z]f$
	$=$	$\{\frac{dz}{d\xi }{\}}^2\cdot \frac{d^{2}f}{d{z}^2}$
		$+\{\frac{d^{2}z}{d{\xi }^2}+[4{\xi }^{-1}+(n+1)\xi {\theta }^{n-1}z\left]\frac{dz}{d\xi }\right\}\frac{df}{dz}$
		$+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}){\theta }^{-1}+\alpha {\theta }^{n-1}z]f$
	$=$	$\{{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2{\}}^2\cdot \frac{d^{2}f}{d{z}^2}+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-1}+\alpha {\theta }^{n-1}z]f$
		$+\left\{\right[4{\xi }^{-2}(1+3z)+2n{\theta }^{n-1}[1+4z]z+n(n+1){\xi }^2{\theta }^{2(n-1)}{z}^3]-[4{\xi }^{-1}+(n+1)\xi {\theta }^{n-1}z\left]\right[{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2\left]\right\}\frac{df}{dz}$
	$=$	$[{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2][{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2]\cdot \frac{d^{2}f}{d{z}^2}+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}){\theta }^{-1}+\alpha {\theta }^{n-1}z]f$
		$+[4{\xi }^{-2}(1+3z)+2n{\theta }^{n-1}[1+4z]z+n(n+1){\xi }^2{\theta }^{2(n-1)}{z}^3]\frac{df}{dz}$
		$-\left[4{\xi }^{-1}\right][{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2]\frac{df}{dz}-[(n+1)\xi {\theta }^{n-1}z][{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2]\frac{df}{dz}$
	$=$	$[{\xi }^{-2}(1+3z{)}^2+2n(1+3z)z^2{\theta }^{n-1}+n^2{\xi }^2{\theta }^{2(n-1)}{z}^4]\cdot \frac{d^{2}f}{d{z}^2}+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-1}+\alpha {\theta }^{n-1}z]f$
		$+[4{\xi }^{-2}(1+3z)+2n{\theta }^{n-1}[1+4z]z+n(n+1){\xi }^2{\theta }^{2(n-1)}{z}^3]\frac{df}{dz}$
		$-[4{\xi }^{-2}(1+3z)+4n{\theta }^{n-1}{z}^2]\frac{df}{dz}-[(n+1){\theta }^{n-1}(1+3z)z+n(n+1){\xi }^2{\theta }^{2(n-1)}{z}^3]\frac{df}{dz}$
	$=$	$[{\xi }^{-2}(1+3z{)}^2+2n(1+3z)z^2{\theta }^{n-1}+n^2{\xi }^2{\theta }^{2(n-1)}{z}^4]\cdot \frac{d^{2}f}{d{z}^2}+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-1}+\alpha {\theta }^{n-1}z]f$
		$+\left\{\right[2n{\theta }^{n-1}(1+4z)]-[4n{\theta }^{n-1}z]-[(n+1){\theta }^{n-1}(1+3z)\left]\right\}z\cdot \frac{df}{dz}$
	$=$	$[{\xi }^{-2}(1+3z{)}^2+2n(1+3z)z^2{\theta }^{n-1}+n^2{\xi }^2{\theta }^{2(n-1)}{z}^4]\cdot \frac{d^{2}f}{d{z}^2}+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-1}+\alpha {\theta }^{n-1}z]f$
		$+\left\{\right[2n+8nz]-[4nz]-[(n+1)+3(n+1)z\left]\right\}{\theta }^{n-1}z\cdot \frac{df}{dz}$
	$=$	$[{\xi }^{-2}(1+3z{)}^2+2n(1+3z)z^2{\theta }^{n-1}+n^2{\xi }^2{\theta }^{2(n-1)}{z}^4]\cdot \frac{d^{2}f}{d{z}^2}+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-1}+\alpha {\theta }^{n-1}z]f$
		$+\{(n-1)+(n-3)z\}{\theta }^{n-1}z\cdot \frac{df}{dz}.$

Part III

Now, suppose that $f=(a_0+b_{0}z)$. We have,

LAWE	$=$	$[{\xi }^{-2}(1+3z{)}^2+2n(1+3z)z^2{\theta }^{n-1}+n^2{\xi }^2{\theta }^{2(n-1)}{z}^4]\cdot {\xcancel{\frac{d^{2}f}{d{z}^2}}}^0+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-1}+\alpha {\theta }^{n-1}z](a_0+b_{0}z)+\{(n-1)+(n-3)z\}{\theta }^{n-1}z\cdot b_0$
	$=$	$(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\left){\theta }^{-1}\right](a_0+b_{0}z)+\{(n+1)\alpha (a_0+b_{0}z)+b_0(n-1)+b_0(n-3)z\}{\theta }^{n-1}z$
	$=$	$(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\left){\theta }^{-1}\right](a_0+b_{0}z)+\{(n+1)\alpha (a_0)+b_0(n-1)+[(n-3)+(n+1)\alpha \left]b_{0}z\right\}{\theta }^{n-1}z.$

Now, in order for the last term to be zero, we need,

$0$	$=$	$[(n-3)+(n+1)\alpha ]$
$\Rightarrow \alpha$	$=$	$\frac{3-n}{n+1}.$

This is precisely the relation that results from the definition of $\alpha \equiv (3-4/{\gamma }_g)$ if the model is evolved assuming ${\gamma }_g=(n+1)/n$. We simultaneously seek the relation,

$0$	$=$	$(n+1)\alpha (a_0)+b_0(n-1)$
$\Rightarrow b_0$	$=$	$\left[\frac{n+1}{1-n}\right]\alpha (a_0)$
	$=$	$a_0\left[\frac{3-n}{1-n}\right]$

It appears as though the leading coefficient, $a_0$, is arbitrary, so we will set it equal to unity. This means that the displacement function is,

$f$

$=$

$1+\left[\frac{3-n}{1-n}\right]z.$

This expression for the displacement function, $f$, is identical to the expression found inside the square brackets of our separately derived exact solution of the polytropic LAWE. Furthermore, given the notation, $({\sigma }_c^2/{\gamma }_g)=(\mathfrak{𝔉}-2\alpha )$, the first term on the RHS of the LAWE will go to zero when, $\mathfrak{𝔉}=2(3-n)/(n+1)$.

Part IV

If we divide through by $({\theta }^{n-1}z)$, the LAWE that was derived above in Part II assumes the following form,

$\left[{\theta }^{1-n}{z}^{-1}\right]\times$ LAWE	$=$	${\theta }^{1-n}{z}^{-1}[{\xi }^{-2}(1+3z{)}^2+2n(1+3z)z^2{\theta }^{n-1}+n^2{\xi }^2{\theta }^{2(n-1)}{z}^4]\cdot \frac{d^{2}f}{d{z}^2}+\{(n-1)+(n-3)z\}\cdot \frac{df}{dz}+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-n}{z}^{-1}+\alpha ]\cdot f$
	$=$	$[{\xi }^{-2}{\theta }^{1-n}{z}^{-1}(1+3z{)}^2+2n(1+3z)z+n^2{\xi }^2{\theta }^{(n-1)}{z}^3]\cdot \frac{d^{2}f}{d{z}^2}+\{(n-1)+(n-3)z\}\cdot \frac{df}{dz}+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-n}{z}^{-1}+\alpha ]\cdot f,$

which resembles the above-discussed hypergeometric differential equation, namely,

$0$

$=$

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

For the record we note that the coefficient (in square brackets) of the first term on the RHS of our LAWE expression is the square of the first derivative of $z$ with respect to $\xi$; that is,

${\theta }^{1-n}{z}^{-1}(\frac{dz}{d\xi }{)}^2$	$=$	${\theta }^{1-n}{z}^{-1}[{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2{]}^2$
	$=$	${\theta }^{1-n}{z}^{-1}[{\xi }^{-2}(1+3z{)}^2+2n(1+3z){\theta }^{n-1}{z}^2+n^2{\xi }^2{\theta }^{2(n-1)}{z}^4].$

Part V

Now suppose that, $f=(a_0+b_{0}z+c_0{z}^2)$, where again,

$z$

$=$

${\xi }^{-1}{\theta }^{-n}({\theta }^{\text{'}}).$

Recalling that,

Part I Summary …

$({\theta }^{\text{'}})$ $=$ $\xi {\theta }^{n}z,$ $\frac{dz}{d\xi }$ $=$ $-[{\xi }^{-1}(1+3z)+n\xi {\theta }^{n-1}{z}^2],$ $\frac{d^{2}z}{d{\xi }^2}$ $=$ $4{\xi }^{-2}(1+3z)+2n{\theta }^{n-1}[1+4z]z+n(n+1){\xi }^2{\theta }^{2(n-1)}{z}^3.$

it may prove useful to recognize that,

${\theta }^{\text{'}}=\frac{d\theta }{d\xi }$	$\to$	$\frac{dz}{d\xi }\cdot \frac{d\theta }{dz}$
$\Rightarrow \frac{d\theta }{dz}$	$=$	$(\frac{dz}{d\xi }{)}^{-1}\xi {\theta }^{n}z$
$\Rightarrow \frac{d\ln \theta }{d\ln z}$	$=$	$(\frac{dz}{d\xi }{)}^{-1}\xi {\theta }^{n-1}{z}^2$
$\Rightarrow \frac{dz}{d\xi }$	$=$	$(\frac{d\ln \theta }{d\ln z}{)}^{-1}\xi {\theta }^{n-1}{z}^2$

In this case we have,

LAWE	$=$	$[-(\frac{d\ln \theta }{d\ln z}{)}^{-1}\xi {\theta }^{n-1}{z}^2{]}^2\cdot (2c_0)+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}){\theta }^{-1}+\alpha {\theta }^{n-1}z](a_0+b_{0}z+c_0{z}^2)$
		$+\{(n-1)+(n-3)z\}{\theta }^{n-1}z\cdot (b_0+2c_{0}z)$

Useful ?????

Try again …

$\left[{\theta }^{1-n}{z}^{-1}\right]\times$ LAWE	$=$	${\theta }^{1-n}{z}^{-1}[\frac{dz}{d\xi }{]}^2\cdot \frac{d^2(a_0+b_{0}z+c_0{z}^2)}{d{z}^2}+\{(n-1)+(n-3)z\}\cdot \frac{d(a_0+b_{0}z+c_0{z}^2)}{dz}+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-n}{z}^{-1}+\alpha ]\cdot (a_0+b_{0}z+c_0{z}^2)$
	$=$	$[{\xi }^{-2}(1+3z{)}^2+2n(1+3z)z^2{\theta }^{n-1}+n^2{\xi }^2{\theta }^{2(n-1)}{z}^4]\cdot (2c_0)+\{(n-1)+(n-3)z\}\cdot (b_0+2c_{0}z)+(n+1)[\left(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\right){\theta }^{-n}{z}^{-1}+\alpha ]\cdot (a_0+b_{0}z+c_0{z}^2)$
	$=$	$[(2c_0){\xi }^{-2}(1+6z+9z^2)+4c_{0}n(z^2+3z^3){\theta }^{n-1}+2c_0{n}^2{\xi }^2{\theta }^{2(n-1)}{z}^4]+(n-1)(b_0+2c_{0}z)$
		$+(n-3)(b_{0}z+2c_0{z}^2)+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_{\mathrm{g}}}\left){\theta }^{-n}{z}^{-1}\right]\cdot (a_0+b_{0}z+c_0{z}^2)+(n+1)\alpha (a_0+b_{0}z+c_0{z}^2).$

Looks pretty hopeless!

Example Density- and Pressure-Profiles

Properties of Analytically Defined, Spherically Symmetric, Equilibrium Structures Note:  $x\equiv \frac{r_0}{R}$
Model	$\frac{{\rho }_0(x)}{{\rho }_c}$	$\frac{P_0(x)}{P_c}$	$\frac{R}{P_c}\cdot P_0^{\text{'}}(x)$	$\mu (x)=-[\frac{P_0(x)}{P_c}{]}^{-1}\cdot [\frac{R}{P_c}\cdot P_0^{\text{'}}(x)]x$	$\frac{P_c}{{\rho }_c}\cdot \frac{{\rho }_0(x)}{P_0(x)}$
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	$\frac{\sin x}{x}$	$(\frac{\sin x}{x}{)}^2$	$\frac{2}{x}[\cos x-\frac{\sin x}{x}]\frac{\sin x}{x}$	$2(1-x\cot x)$	$\frac{x}{\sin x}$

Parabolic Density Distribution

Relevant, Parabolic LAWE

In the case of a parabolic density distribution, we have found that the equilibrium configuration is defined by the relations:

$x$	$=$	$\frac{r_0}{R},$
$\frac{{\rho }_0}{{\rho }_c}$	$=$	$(1-x^2),$
$\frac{P_0}{P_c}$	$=$	$(1-x^2{)}^2(1-\frac{1}{2}{x}^2),$       where,       $P_c=\left(\frac{4\pi }{15}\right)G{\rho }_c^2{R}^2,$
$g_0$	$=$	$\left(\frac{4\pi }{3}\right)G{\rho }_{c}Rx(1-\frac{3}{5}{x}^2),$

in which case,

$\mu =\frac{g_0{\rho }_0{r}_0}{P_0}$	$=$	$\left(\frac{4\pi }{3}\right)G{\rho }_{c}Rx(1-\frac{3}{5}{x}^2){\rho }_c(1-x^2)Rx\left[\right(\frac{4\pi }{15})G{\rho }_c^2{R}^2(1-x^2{)}^2(1-\frac{1}{2}{x}^2){]}^{-1}$
	$=$	$x^2(5-3x^2)[(1-x^2)(1-\frac{1}{2}{x}^2){]}^{-1},$
$\left(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0}\right)$	$=$	$\left(\frac{{\omega }^2{R}^2}{{\gamma }_g}\right){\rho }_c(1-x^2)\left[\right(\frac{4\pi }{15})G{\rho }_c^2{R}^2(1-x^2{)}^2(1-\frac{1}{2}{x}^2){]}^{-1}$
	$=$	$\left(\frac{15{\omega }^2}{4\pi G{\rho }_c{\gamma }_g}\right)[(1-x^2)(1-\frac{1}{2}{x}^2){]}^{-1}.$

Hence, in the case of a parabolic density distribution, the Kopal (1948) LAWE becomes,

$0$	$=$	$\frac{d^{2}f}{d{x}^2}+\frac{(4-\mu )}{x}\cdot \frac{df}{dx}+\left\{\right(\frac{{\omega }^2{\rho }_0{R}^2}{{\gamma }_{\mathrm{g}}{P}_0})-\frac{\alpha \mu }{x^2}\}f$
	$=$	$\frac{d^{2}f}{d{x}^2}+\frac{1}{x}\{4-x^2(5-3x^2)[(1-x^2)(1-\frac{1}{2}{x}^2){]}^{-1}\}\cdot \frac{df}{dx}+\{\left(\frac{15{\omega }^2}{4\pi G{\rho }_c{\gamma }_g}\right)-\alpha (5-3x^2)\left\}\right[(1-x^2)(1-\frac{1}{2}{x}^2){]}^{-1}f.$

Multiplying through by $[(1-x^2)(1-\frac{1}{2}{x}^2)]$ gives,

$0$	$=$	$(1-x^2)(1-\frac{1}{2}{x}^2)\cdot \frac{d^{2}f}{d{x}^2}+\frac{1}{x}[(1-x^2)(4-2x^2)-x^2(5-3x^2)]\cdot \frac{df}{dx}+\left[\right(\frac{15{\omega }^2}{4\pi G{\rho }_c{\gamma }_g})-\alpha (5-3x^2)]\cdot f$
	$=$	$(1-x^2)(1-\frac{1}{2}{x}^2)\cdot \frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-11x^2+5x^4]\cdot \frac{df}{dx}+\left[\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha x^2]\cdot f$

where,

$\mathfrak{𝔉}$

$\equiv$

$2\left[\right(\frac{3{\omega }^2}{4\pi {\gamma }_{\mathrm{g}}G{\rho }_c})-\alpha ].$

Change of Variable

In an effort to shift this LAWE into a 2^nd-order ODE that has the form of an hypergeometric equation, let's try …

First Try

$z=x^2-1$	$\Rightarrow$	$x=(1+z{)}^{1/2};$       also,       $1-\frac{1}{2}{x}^2=\frac{1}{2}(1-z);$
$\Rightarrow \frac{dz}{dx}$	$=$	$2x=2(1+z{)}^{1/2};$
$\Rightarrow \frac{df}{dx}=\frac{dz}{dx}\cdot \frac{df}{dz}$	$=$	$2(1+z{)}^{1/2}\cdot \frac{df}{dz};$
$\Rightarrow \frac{d^{2}f}{d{x}^2}=\frac{dz}{dx}\cdot \frac{d}{dz}[2(1+z{)}^{1/2}\cdot \frac{df}{dz}]$	$=$	$2(1+z{)}^{1/2}[(1+z{)}^{-1/2}\cdot \frac{df}{dz}+2(1+z{)}^{1/2}\cdot \frac{d^{2}f}{d{z}^2}]$
	$=$	$[2\cdot \frac{df}{dz}+4(1+z)\cdot \frac{d^{2}f}{d{z}^2}].$

Hence, the parabolic LAWE takes the form,

$0$	$=$	$(1-x^2)(1-\frac{1}{2}{x}^2)\cdot \frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-11x^2+5x^4]\cdot \frac{df}{dx}+\left[\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha x^2]\cdot f$
	$=$	$(-z)(1-z)\cdot [\frac{df}{dz}+2(1+z)\cdot \frac{d^{2}f}{d{z}^2}]+(1+z{)}^{-1/2}[4-11(1+z)+5(1+z{)}^2]2(1+z{)}^{1/2}\cdot \frac{df}{dz}+[\left(\frac{5}{2}\right)\mathfrak{𝔉}+3\alpha (1+z)]\cdot f$
	$=$	$-z(1-z)\cdot [2(1+z)\cdot \frac{d^{2}f}{d{z}^2}]-z(1-z)\cdot \frac{df}{dz}+[8-22(1+z)+10(1+z{)}^2]\cdot \frac{df}{dz}+[(\frac{5}{2}\cdot \mathfrak{𝔉}+3\alpha )+3\alpha z]\cdot f$
	$=$	$-2z(1-z^2)\cdot \frac{d^{2}f}{d{z}^2}+[8-22-22z+10+20z+10z^2+z^2-z]\cdot \frac{df}{dz}+\left[\right(\frac{5}{2}\cdot \mathfrak{𝔉}+3\alpha )+3\alpha z]\cdot f$
	$=$	$-2z(1-z^2)\cdot \frac{d^{2}f}{d{z}^2}+[-4-3z+11z^2]\cdot \frac{df}{dz}+\left[\right(\frac{5}{2}\cdot \mathfrak{𝔉}+3\alpha )+3\alpha z]\cdot f$

Second Try

$z=F{x}^B-G$	$\Rightarrow$	$x=(\frac{G+z}{F}{)}^{1/B};$       also,       $1-\frac{1}{2}{x}^2=1-\frac{1}{2}(\frac{G+z}{F}{)}^{2/B};$
$\Rightarrow \frac{dz}{dx}$	$=$	$BF{x}^{B-1}=BF(\frac{G+z}{F}{)}^{(B-1)/B};$
$\Rightarrow \frac{df}{dx}=\frac{dz}{dx}\cdot \frac{df}{dz}$	$=$	$BF(\frac{G+z}{F}{)}^{(B-1)/B}\frac{df}{dz};$
$\Rightarrow \frac{d^{2}f}{d{x}^2}=\frac{dz}{dx}\cdot \frac{d}{dz}\left[BF\right(\frac{G+z}{F}{)}^{(B-1)/B}\frac{df}{dz}]$	$=$	$BF\left(\frac{G+z}{F}{)}^{(B-1)/B}\right\{BF\cdot F^{-1+1/B}\left[\frac{B-1}{B}\right](G+z{)}^{-1/B}\cdot \frac{df}{dz}+BF(\frac{G+z}{F}{)}^{(B-1)/B}\frac{d^{2}f}{d{z}^2}\}$
	$=$	$B{F}^{1/B}(G+z{)}^{1-1/B}\{F^{1/B}(B-1)(G+z{)}^{-1/B}\cdot \frac{df}{dz}+B{F}^{1/B}(G+z{)}^{1-1/B}\frac{d^{2}f}{d{z}^2}\}$
	$=$	$B{F}^{2/B}(G+z{)}^{1-2/B}[(B-1)\cdot \frac{df}{dz}+B(G+z)\cdot \frac{d^{2}f}{d{z}^2}].$

This should reduce to the "First Try" example by setting:  $(B,F,G)=(2,1,1)$. Let's see …

$\frac{dz}{dx}$	$=$	$BF(\frac{G+z}{F}{)}^{(B-1)/B}=2(1+z{)}^{1/2};$
$\frac{df}{dx}$	$=$	$BF(\frac{G+z}{F}{)}^{(B-1)/B}\frac{df}{dz}=2(1+z{)}^{1/2}\cdot \frac{df}{dz};$
$\frac{d^{2}f}{d{x}^2}$	$=$	$B{F}^{2/B}(G+z{)}^{1-2/B}[(B-1)\cdot \frac{df}{dz}+B(G+z)\cdot \frac{d^{2}f}{d{z}^2}]=2(1+z{)}^0[\frac{df}{dz}+2(1+z)\frac{d^{2}f}{d{z}^2}].$

Functional Expressions from "First Try" $z=x^2-1$ $\Rightarrow$ $x=(1+z{)}^{1/2};$       also,       $1-\frac{1}{2}{x}^2=\frac{1}{2}(1-z);$ $\frac{dz}{dx}$ $=$ $2(1+z{)}^{1/2};$ $\frac{df}{dx}$ $=$ $2(1+z{)}^{1/2}\cdot \frac{df}{dz};$ $\frac{d^{2}f}{d{x}^2}$ $=$ $[2\cdot \frac{df}{dz}+4(1+z)\cdot \frac{d^{2}f}{d{z}^2}].$

Hence, the parabolic LAWE takes the form,

$0$	$=$	$(1-x^2)(1-\frac{1}{2}{x}^2)\cdot \frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-11x^2+5x^4]\cdot \frac{df}{dx}+\left[\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha x^2]\cdot f$
	$=$	$[1-(\frac{G+z}{F}{)}^{2/B}\left]\right[1-\frac{1}{2}\left(\frac{G+z}{F}{)}^{2/B}\right]B{F}^{2/B}(G+z{)}^{1-2/B}[(B-1)\cdot \frac{df}{dz}+B(G+z)\cdot \frac{d^{2}f}{d{z}^2}]$
		$+\left(\frac{G+z}{F}{)}^{-1/B}\right[4-11x^2+5x^4]\cdot BF(\frac{G+z}{F}{)}^{1-1/B}\frac{df}{dz}$
		$+\left[\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha (\frac{G+z}{F}{)}^{2/B}]\cdot f$
	$=$	$[F^{2/B}-(G+z{)}^{2/B}\left]\right[F^{2/B}-\frac{1}{2}(G+z{)}^{2/B}]B{F}^{-2/B}(G+z{)}^{1-2/B}[B(G+z)\cdot \frac{d^{2}f}{d{z}^2}]$
		$+[F^{2/B}-(G+z{)}^{2/B}\left]\right[F^{2/B}-\frac{1}{2}(G+z{)}^{2/B}]B{F}^{-2/B}(G+z{)}^{1-2/B}[(B-1)\cdot \frac{df}{dz}]$
		$+[4F^{4/B}-11F^{2/B}(G+z{)}^{2/B}+5(G+z{)}^{4/B}]\cdot B{F}^{-2/B}(G+z{)}^{1-2/B}\cdot \frac{df}{dz}$
		$+\left[\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha F^{-2/B}(G+z{)}^{2/B}]\cdot f$

Dividing through by, $B{F}^{-2/B}(G+z{)}^{1-2/B}$, we have,

$0$	$=$	$\left\{\right[F^{2/B}-(G+z{)}^{2/B}][F^{2/B}-\frac{1}{2}(G+z{)}^{2/B}]B(G+z)\}\cdot \frac{d^{2}f}{d{z}^2}$
		$+\left\{\right[F^{2/B}-(G+z{)}^{2/B}][F^{2/B}-\frac{1}{2}(G+z{)}^{2/B}](B-1)+[4F^{4/B}-11F^{2/B}(G+z{)}^{2/B}+5(G+z{)}^{4/B}\left]\right\}\cdot \frac{df}{dz}$
		$+B^{-1}(G+z{)}^{-1+2/B}[F^{2/B}\left(\frac{5}{2}\right)\mathfrak{𝔉}+3\alpha (G+z{)}^{2/B}]\cdot f$
	$=$	$\frac{1}{2}[2F^{4/B}-3F^{2/B}(G+z{)}^{2/B}+(G+z{)}^{4/B}]B(G+z)\cdot \frac{d^{2}f}{d{z}^2}+B^{-1}(G+z{)}^{-1+2/B}[F^{2/B}\left(\frac{5}{2}\right)\mathfrak{𝔉}+3\alpha (G+z{)}^{2/B}]\cdot f$
		$+\frac{1}{2}\left\{\right[2F^{4/B}-3F^{2/B}(G+z{)}^{2/B}+(G+z{)}^{4/B}](B-1)+[8F^{4/B}-22F^{2/B}(G+z{)}^{2/B}+10(G+z{)}^{4/B}\left]\right\}\cdot \frac{df}{dz}$
	$=$	$\frac{1}{2}[2F^{4/B}-3F^{2/B}(G+z{)}^{2/B}+(G+z{)}^{4/B}]B(G+z)\cdot \frac{d^{2}f}{d{z}^2}+B^{-1}(G+z{)}^{-1+2/B}[F^{2/B}\left(\frac{5}{2}\right)\mathfrak{𝔉}+3\alpha (G+z{)}^{2/B}]\cdot f$
		$+\frac{1}{2}[(2B+6)F^{4/B}-(3B+19)F^{2/B}(G+z{)}^{2/B}+(B+9)(G+z{)}^{4/B}]\cdot \frac{df}{dz}$

Now, this should reduce to the "First Try" example by setting:  $(B,F,G)=(2,1,1)$. Let's see …

$0$	$=$	$\frac{1}{2}[2-3(1+z)+(1+z{)}^2]2(1+z)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{2}[10-25(1+z)+11(1+z{)}^2]\cdot \frac{df}{dz}+\frac{1}{2}\left[\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha (1+z\left)\right]\cdot f$
	$=$	$[2-3-3z+(1+2z+z^2\left)\right](1+z)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{2}[10-25-25z+11(1+2z+z^2\left)\right]\cdot \frac{df}{dz}+\frac{1}{2}\left[\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha (1+z\left)\right]\cdot f$
	$=$	$z(z-1)(1+z)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{2}[-4-3z+11z^2]\cdot \frac{df}{dz}+\frac{1}{2}\left[\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha (1+z\left)\right]\cdot f.$

Compare with LAWE from "First Try" $0$ $=$ $-2z(1-z^2)\cdot \frac{d^{2}f}{d{z}^2}+[-4-3z+11z^2]\cdot \frac{df}{dz}+\left[\right(\frac{5}{2}\cdot \mathfrak{𝔉}+3\alpha )+3\alpha z]\cdot f$

Next, let's try setting $B=2$ while leaving $F$ and $G$ arbitrary. The LAWE becomes,

$0$	$=$	$\frac{1}{2}[2F^2-3F(G+z)+(G+z{)}^2]2(G+z)\cdot \frac{d^{2}f}{d{z}^2}+2^{-1}(G+z{)}^0\left[F\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha (G+z\left)\right]\cdot f$
		$+\frac{1}{2}[10F^2-25F(G+z)+11(G+z{)}^2]\cdot \frac{df}{dz}$
	$=$	$[2F^2-3F(G+z)+(G+z{)}^2](G+z)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{2}[10F^2-25F(G+z)+11(G+z{)}^2]\cdot \frac{df}{dz}+\frac{1}{2}\left[F\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha (G+z\left)\right]\cdot f.$

---

Now, let's set $G=-1$ to obtain,

$0$	$=$	$[2F^2-3F(z-1)+(z-1{)}^2](z-1)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{2}[10F^2-25F(z-1)+11(z-1{)}^2]\cdot \frac{df}{dz}+\frac{1}{2}\left[F\right(\frac{5}{2})\mathfrak{𝔉}+3\alpha (z-1\left)\right]\cdot f$
	$=$	$[2F^2-3Fz+3F+z^2-2z+1](z-1)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{2}[10F^2+25F-25Fz+11z^2-22z+11]\cdot \frac{df}{dz}+\frac{1}{2}\left[F\right(\frac{5}{2})\mathfrak{𝔉}-3\alpha +3\alpha z]\cdot f$
	$=$	$[(2F^2+3F+1)-(3F+2)z+z^2](z-1)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{2}[(10F^2+25F+11)-(25F+22)z+11z^2]\cdot \frac{df}{dz}+\frac{1}{2}\left[F\right(\frac{5}{2})\mathfrak{𝔉}-3\alpha +3\alpha z]\cdot f.$

Notice that if $F=-\frac{2}{3}$, we have,

$0$	$=$	$[z^2-\frac{1}{9}](z-1)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{2}[\frac{40}{9}+\frac{50}{3}+11-(\frac{50}{3}+22)z+11z^2]\cdot \frac{df}{dz}+\frac{1}{2}\left[F\right(\frac{5}{2})\mathfrak{𝔉}-3\alpha +3\alpha z]\cdot f$
	$=$	$\frac{1}{9}[9z^2-1](z-1)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{18}[201-348z+99z^2]\cdot \frac{df}{dz}+\frac{1}{2}\left[F\right(\frac{5}{2})\mathfrak{𝔉}-3\alpha +3\alpha z]\cdot f$

---

Alternatively, let's set,

$0$	$=$	$z^2+G[2F^2-3F(G+z)+(G+z{)}^2]$
	$=$	$2G{F}^2-3GF(G+z)+G(G+z{)}^2+z^2$
	$=$	$2G[F^2-\frac{3F}{G}(G+z)+\frac{3^2}{2^2{G}^2}(G+z{)}^2]-\frac{3^2}{2G}(G+z{)}^2+G(G+z{)}^2+z^2$
	$=$	$2G[F-\frac{3}{2G}(G+z){]}^2-\frac{3^2}{2G}(G+z{)}^2+G(G+z{)}^2+z^2$

in which case,

$0$

$=$

$z^2(\frac{z}{G}-1)\cdot \frac{d^{2}f}{d{z}^2}+\frac{1}{2}[10F^2-25F(G+z)+11(G+z{)}^2]\cdot \frac{df}{dz}+\frac{1}{2}[F\left(\frac{5}{2}\right)\mathfrak{𝔉}+3\alpha (G+z)]\cdot f.$

The coefficient of the first-derivative term also may be rewritten ad,

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,

$\mathfrak{𝔉}$

$\equiv$

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

Summary for PowerPoint Slide
LAWE:    $0$ $=$ $(1-x^2)\cdot \frac{d^{2}f}{d{x}^2}+\frac{1}{x}[4-6x^2]\frac{df}{dx}+\mathfrak{𝔉}f,$ where, $\mathfrak{𝔉}$ $\equiv$ $\left[\right(\frac{{\omega }^2{\rho }_c{R}^2}{{\gamma }_{\mathrm{g}}{P}_c})-2\alpha ].$

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

• Our "Ramblings" chapter titled, Other Analytically Definable, Spherical Equilibrium Structures
• 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	$\frac{\sin x}{x}$	$[\frac{\sin x}{x}{]}^2$	$\frac{2}{x}[\cos x-\frac{\sin x}{x}]\frac{\sin x}{x}$