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,

$0$

$=$

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

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,

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

$=$

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

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,

$\xi$

$=$

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

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

$0$

$=$

$x^{2}(ax^{b}-1)\cdot {\frac {d^{2}\xi }{dx^{2}}}+(apx^{b}+\lambda )x\cdot {\frac {d\xi }{dx}}+(arx^{b}+s)\cdot \xi \,.$

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

A mapping between the two differential equations requires,

$ax^{b}$

$\leftrightarrow$

$z\,,$

and,

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

$\leftrightarrow$

$u\,,$

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

$=$

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

in which case,

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

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

$(-ab^{2}x^{b+c})\times 0$	$=$	$x^{2}(ax^{b}-1){\frac {d^{2}\xi }{dx^{2}}}+{\biggl \{}(ax^{b}-1)\cdot {\biggl [}(1-b-c)-c{\biggr ]}-b[\gamma -(\alpha +\beta +1)ax^{b}]{\biggr \}}x\cdot {\frac {d\xi }{dx}}$
		$+{\biggl \{}bc[\gamma -(\alpha +\beta +1)ax^{b}]+{\biggl (}\alpha \beta ab^{2}{\biggr )}x^{b}-(1-ax^{b})\cdot {\biggl [}c(b+c){\biggr ]}{\biggr \}}\xi$
	$=$	$x^{2}(ax^{b}-1){\frac {d^{2}\xi }{dx^{2}}}+{\biggl \{}-(1+b\gamma -b-2c)+ax^{b}(1-b-2c)+b(\alpha +\beta +1)ax^{b}{\biggr \}}x\cdot {\frac {d\xi }{dx}}$
		$+{\biggl \{}bc\gamma -c(b+c)-bc(\alpha +\beta +1)ax^{b}+\alpha \beta ab^{2}x^{b}+c(b+c)ax^{b}{\biggr \}}\xi$
	$=$	$x^{2}(ax^{b}-1){\frac {d^{2}\xi }{dx^{2}}}+(apx^{b}+\lambda )x\cdot {\frac {d\xi }{dx}}+(arx^{b}+s)\xi \,,$

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

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 c

i.) Determining the value of the exponent, $c$, from $\lambda$ and $s$
Equating $b\gamma$ in the 2nd and 3rd of these 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 γ

ii.) Determining the value of, $b(1-\gamma )$, from $\lambda$ and $s$
Combining our 3rd 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 in box "i." immediately above.
Consistent with this derivation, 📚 Van der Borght (1970) states, "… $(1-\gamma )b=A_{2}-A_{1}$ …"

Determining the Value of β

iii.) Determining $b\beta$
Combining our 1st and 4th 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 1st expression alone we can write, $(b\alpha )$ $=$ $(p-1)+2c-b\beta \,.$ Together, then, we have, $(b\beta ){\biggl [}(p-1)+2c-b\beta {\biggr ]}$ $=$ $-(1-p)c+c^{2}+r$ $\Rightarrow ~~~0$ $=$ $(b\beta )^{2}-(b\beta )\underbrace {[(p-1)+2c]} _{(b\alpha +b\beta )}+\overbrace {[c^{2}+(p-1)c+r]} ^{(b\alpha )\cdot (b\beta )}$ The pair of roots, $(b\beta )_{\pm }$, of this quadratic equation are then obtained from the relation, $2(b\beta )_{\pm }$ $=$ $[(p-1)+2c]\pm {\biggl \{}[(p-1)+2c]^{2}-4[c^{2}+(p-1)c+r]{\biggr \}}^{1/2}$
📚 Van der Borght (1970) states that if, "… $B_{1},B_{2}$ are solutions of $B^{2}-(p-1)B+r=0$, then … $b\beta =A_{1}+B_{2}.$" Let's see if our derivation leads to this same conclusion. First note that the roots, $B_{\pm }$, of this Van der Borght quadratic equation are, $2B_{\pm }$ $=$ $(p-1)\pm [(p-1)^{2}-4r]^{1/2}\,.$ If we assign the inferior root with Van der Borght's notation, $B_{2}$, then, $2(A_{1}+B_{2})=2(c_{+}+B_{-})$ $=$ $\{(1+\lambda )+[(1+\lambda )^{2}+4s]^{1/2}\}+\{(p-1)-[(p-1)^{2}-4r]^{1/2}\}$

Completing the Squares to Determine the Value of α

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

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

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

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

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

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

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,

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

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

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

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

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

LAWE

Familiar Foundation

Drawing from an accompanying discussion, we have the,

LAWE:   Linear Adiabatic Wave (or Radial Pulsation) Equation

${\frac {d^{2}x}{dr_{0}^{2}}}+{\biggl [}{\frac {4}{r_{0}}}-{\biggl (}{\frac {g_{0}\rho _{0}}{P_{0}}}{\biggr )}{\biggr ]}{\frac {dx}{dr_{0}}}+{\biggl (}{\frac {\rho _{0}}{\gamma _{\mathrm {g} }P_{0}}}{\biggr )}{\biggl [}\omega ^{2}+(4-3\gamma _{\mathrm {g} }){\frac {g_{0}}{r_{0}}}{\biggr ]}x=0$

where,

$g_{0}$

$=$

$-{\frac {1}{\rho _{0}}}{\frac {dP_{0}}{dr_{0}}}\,.$

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

$x$	$\rightarrow$	$f\,,$
${\frac {r_{0}}{R}}$	$\rightarrow$	$x\,,$
$(4-3\gamma _{g})$	$\rightarrow$	$-\alpha \gamma _{g}\,,$

the LAWE may be rewritten as,

$0$	$=$	${\frac {d^{2}f}{dx^{2}}}+{\biggl [}{\frac {4}{x}}-{\biggl (}{\frac {g_{0}\rho _{0}R}{P_{0}}}{\biggr )}{\biggr ]}{\frac {df}{dx}}+{\biggl (}{\frac {\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}}}{\biggr )}{\biggl [}\omega ^{2}-{\frac {\alpha \gamma _{g}g_{0}}{r_{0}}}{\biggr ]}f$
	$=$	${\frac {d^{2}f}{dx^{2}}}+{\frac {1}{x}}{\biggl [}4-{\biggl (}{\frac {g_{0}\rho _{0}r_{0}}{P_{0}}}{\biggr )}{\biggr ]}{\frac {df}{dx}}+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}}}{\biggr )}-{\frac {\alpha \gamma _{g}g_{0}}{r_{0}}}{\biggl (}{\frac {\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}}}{\biggr )}{\biggr ]}f$
	$=$	${\frac {d^{2}f}{dx^{2}}}+{\frac {1}{x}}{\biggl [}4-{\biggl (}{\frac {g_{0}\rho _{0}r_{0}}{P_{0}}}{\biggr )}{\biggr ]}{\frac {df}{dx}}+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}}}{\biggr )}-{\frac {\alpha }{x^{2}}}{\biggl (}{\frac {g_{0}r_{0}\rho _{0}}{P_{0}}}{\biggr )}{\biggr ]}f\,.$

If we furthermore adopt the variable definition,

$\mu$

$\equiv$

${\biggl (}{\frac {g_{0}\rho _{0}r_{0}}{P_{0}}}{\biggr )}=-{\frac {d\ln P_{0}}{d\ln r_{0}}}\,,$

we obtain what we will refer to as the,

Kopal (1948) LAWE

$0$	$=$	${\frac {d^{2}f}{dx^{2}}}+{\frac {(4-\mu )}{x}}\cdot {\frac {df}{dx}}+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}}}{\biggr )}-{\frac {\alpha \mu }{x^{2}}}{\biggr ]}f\,.$
📚 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,

$r_{0}$	$=$	$a\xi \,,$
$\rho _{0}$	$=$	$\rho _{c}\theta ^{n}\,,$
$P_{0}$	$=$	$K{\biggl [}\rho _{c}\theta ^{n}{\biggr ]}^{(n+1)/n}\,,$
$M_{r}$	$=$	$4\pi a^{3}\rho _{c}{\biggl (}-\xi ^{2}{\frac {d\theta }{d\xi }}{\biggr )}\,,$
$g_{0}\equiv {\frac {GM_{r}}{r_{0}^{2}}}$	$=$	$4\pi Ga\rho _{c}{\biggl (}-{\frac {d\theta }{d\xi }}{\biggr )}\,,$

where,

$a$	$\equiv$	${\biggl [}{\frac {(n+1)K}{4\pi G}}{\biggr ]}^{1/2}\rho _{c}^{(1-n)/2n}\,.$
$\Rightarrow ~~~K$	$=$	${\biggl [}{\frac {4\pi G}{(n+1)}}{\biggr ]}a^{2}\rho _{c}^{(n-1)/n}\,.$

Hence,

$\mu ={\biggl (}{\frac {g_{0}\rho _{0}r_{0}}{P_{0}}}{\biggr )}$	$=$	$4\pi Ga\rho _{c}{\biggl (}-{\frac {d\theta }{d\xi }}{\biggr )}\rho _{c}\theta ^{n}a\xi {\biggl [}\rho _{c}\theta ^{n}{\biggr ]}^{-(n+1)/n}{\biggl [}{\frac {(n+1)}{4\pi G}}{\biggr ]}a^{-2}\rho _{c}^{-(n-1)/n}$
	$=$	$(n+1){\biggl (}-{\frac {d\theta }{d\xi }}{\biggr )}\theta ^{n}\xi \theta ^{-(n+1)}\rho _{c}^{2-(n+1)/n-(n-1)/n}$
	$=$	$(n+1){\biggl (}-{\frac {\xi }{\theta }}\cdot {\frac {d\theta }{d\xi }}{\biggr )}=(n+1){\biggl (}-{\frac {d\ln \theta }{d\ln \xi }}{\biggr )}\,.$

Alternatively,

$\mu =-{\frac {d\ln P_{0}}{d\ln r_{0}}}$	$=$	$-{\frac {r_{0}}{P_{0}}}\cdot {\frac {dP_{0}}{dr_{0}}}$
	$=$	$-\xi \theta ^{-(n+1)}\cdot {\frac {d}{d\xi }}{\biggl [}\theta ^{(n+1)}{\biggr ]}$
	$=$	$-(n+1){\biggl (}{\frac {\xi }{\theta }}\cdot {\frac {d\theta }{d\xi }}{\biggr )}=(n+1){\biggl (}-{\frac {d\ln \theta }{d\ln \xi }}{\biggr )}\,.$

Yes!

Trial Displacement Function

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

$f_{\mathrm {trial} }$	$=$	${\frac {3(n-1)}{2n}}{\biggl [}1+{\biggl (}{\frac {n-3}{n-1}}{\biggr )}{\biggl (}{\frac {1}{\xi \theta ^{n}}}{\biggr )}{\frac {d\theta }{d\xi }}{\biggr ]}$
$\Rightarrow ~~~{\biggl [}{\frac {2n}{3(n-1)}}{\biggr ]}f_{\mathrm {trial} }$	$=$	$1+{\biggl (}{\frac {n-3}{n-1}}{\biggr )}\xi ^{-1}\theta ^{-n}{\biggl [}{\frac {\theta }{\xi }}\cdot {\frac {d\ln \theta }{d\ln \xi }}{\biggr ]}$
	$=$	$1+{\biggl (}{\frac {3-n}{n-1}}{\biggr )}\xi ^{-2}\theta ^{(1-n)}\cdot {\biggl (}-{\frac {d\ln \theta }{d\ln \xi }}{\biggr )}$
	$=$	$1+{\biggl [}{\frac {3-n}{(n+1)(n-1)}}{\biggr ]}\xi ^{-2}\theta ^{(1-n)}\cdot \mu$

Flipping it around, we have alternatively,

$\mu$

$=$

${\biggl [}{\frac {(n+1)(n-1)}{3-n}}{\biggr ]}{\biggl \{}{\biggl [}{\frac {2n}{3(n-1)}}{\biggr ]}f_{\mathrm {trial} }-1{\biggr \}}\xi ^{2}\theta ^{(n-1)}$

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,

$\xi _{1}^{-2}\cdot ~\mathrm {LAWE}$	$=$	${\frac {d^{2}f_{\mathrm {trial} }}{d\xi ^{2}}}+{\frac {(4-\mu )}{\xi }}\cdot {\frac {df_{\mathrm {trial} }}{d\xi }}+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}\xi _{1}^{2}}}{\biggr )}-{\frac {\alpha \mu }{\xi ^{2}}}{\biggr ]}f_{\mathrm {trial} }$
$\Rightarrow ~~~{\biggl [}{\frac {2n}{3(n-1)}}{\biggr ]}\xi _{1}^{-2}\cdot ~\mathrm {LAWE}$	$=$	${\frac {d^{2}}{d\xi ^{2}}}{\biggl \{}1+{\biggl [}{\frac {3-n}{(n+1)(n-1)}}{\biggr ]}\xi ^{-2}\theta ^{(1-n)}\cdot \mu {\biggr \}}+{\frac {(4-\mu )}{\xi }}\cdot {\frac {d}{d\xi }}{\biggl \{}1+{\biggl [}{\frac {3-n}{(n+1)(n-1)}}{\biggr ]}\xi ^{-2}\theta ^{(1-n)}\cdot \mu {\biggr \}}$
		$+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}\xi _{1}^{2}}}{\biggr )}-{\frac {\alpha \mu }{\xi ^{2}}}{\biggr ]}{\biggl \{}1+{\biggl [}{\frac {3-n}{(n+1)(n-1)}}{\biggr ]}\xi ^{-2}\theta ^{(1-n)}\cdot \mu {\biggr \}}$
$\Rightarrow ~~~{\biggl [}{\frac {2n(n+1)}{3}}{\biggr ]}\xi _{1}^{-2}\cdot ~\mathrm {LAWE}$	$=$	${\frac {d^{2}}{d\xi ^{2}}}{\biggl \{}(3-n)\xi ^{-2}\theta ^{(1-n)}\cdot \mu {\biggr \}}+{\frac {(4-\mu )}{\xi }}\cdot {\frac {d}{d\xi }}{\biggl \{}(3-n)\xi ^{-2}\theta ^{(1-n)}\cdot \mu {\biggr \}}$
		$+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}\xi _{1}^{2}}}{\biggr )}-{\frac {\alpha \mu }{\xi ^{2}}}{\biggr ]}{\biggl \{}(n+1)(n-1)+(3-n)\xi ^{-2}\theta ^{(1-n)}\cdot \mu {\biggr \}}$
$\Rightarrow ~~~{\biggl [}{\frac {2n(n+1)}{3(3-n)}}{\biggr ]}\xi _{1}^{-2}\cdot ~\mathrm {LAWE}$	$=$	${\frac {d^{2}}{d\xi ^{2}}}{\biggl \{}\xi ^{-2}\theta ^{(1-n)}\cdot \mu {\biggr \}}+{\frac {(4-\mu )}{\xi }}\cdot {\frac {d}{d\xi }}{\biggl \{}\xi ^{-2}\theta ^{(1-n)}\cdot \mu {\biggr \}}+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}\xi _{1}^{2}}}{\biggr )}-{\frac {\alpha \mu }{\xi ^{2}}}{\biggr ]}{\biggl \{}{\frac {(n+1)(n-1)}{(3-n)}}+\xi ^{-2}\theta ^{(1-n)}\cdot \mu {\biggr \}}\,.$

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,

${\frac {\omega ^{2}}{\gamma _{g}}}{\biggl (}{\frac {\rho _{0}a^{2}}{P_{0}}}{\biggr )}$

$=$

${\biggl (}{\frac {\omega ^{2}}{\gamma _{g}}}{\biggr )}K^{-1}\rho _{c}^{-1/n}\theta ^{-1}{\biggl [}{\frac {(n+1)K}{4\pi G}}{\biggr ]}\rho _{c}^{(1-n)/n}={\biggl (}{\frac {\omega ^{2}}{\gamma _{g}}}{\biggr )}{\biggl [}{\frac {(n+1)}{4\pi \rho _{c}G}}{\biggr ]}\theta ^{-1}\,.$

Replace μ by f_trial

Making instead the alternate substitution, namely,

$\mu$	$=$	${\biggl [}{\frac {(n+1)(n-1)}{3-n}}{\biggr ]}{\biggl \{}{\biggl [}{\frac {2n}{3(n-1)}}{\biggr ]}f_{\mathrm {trial} }-1{\biggr \}}\xi ^{2}\theta ^{(n-1)}$
	$=$	${\biggl \{}{\frac {2n(n+1)}{3(3-n)}}\cdot f_{\mathrm {trial} }-{\frac {(n+1)(n-1)}{3-n}}{\biggr \}}\xi ^{2}\theta ^{(n-1)}$
	$=$	${\frac {1}{3(3-n)}}{\biggl [}\underbrace {2n(n+1)} _{A}f_{\mathrm {trial} }-\underbrace {3(n+1)(n-1)} _{B}{\biggr ]}\xi ^{2}\theta ^{(n-1)}$

we have,

$\xi _{1}^{-2}\cdot ~\mathrm {LAWE}$	$=$	${\frac {d^{2}f_{\mathrm {trial} }}{d\xi ^{2}}}+{\biggl \{}{\frac {4}{\xi }}{\biggr \}}{\frac {df_{\mathrm {trial} }}{d\xi }}-{\biggl \{}{\frac {\mu }{\xi }}{\biggr \}}{\frac {df_{\mathrm {trial} }}{d\xi }}+{\biggl (}{\frac {\omega ^{2}}{\gamma _{g}}}{\biggr )}{\biggl [}{\frac {(n+1)}{4\pi \rho _{c}G}}{\biggr ]}{\frac {f_{\mathrm {trial} }}{\theta }}-{\biggl (}{\frac {\alpha }{\xi ^{2}}}{\biggr )}\mu f_{\mathrm {trial} }$
$\Rightarrow ~~~{\biggl [}{\frac {3(3-n)}{\xi _{1}^{2}}}{\biggr ]}~\mathrm {LAWE}$	$=$	$3(3-n){\frac {d^{2}f_{\mathrm {trial} }}{d\xi ^{2}}}+{\biggl \{}{\frac {12(3-n)}{\xi }}{\biggr \}}{\frac {df_{\mathrm {trial} }}{d\xi }}-{\biggl [}Af_{\mathrm {trial} }-B{\biggr ]}\xi \theta ^{(n-1)}{\frac {df_{\mathrm {trial} }}{d\xi }}$
		$+3(3-n){\biggl (}{\frac {\omega ^{2}}{\gamma _{g}}}{\biggr )}{\biggl [}{\frac {(n+1)}{4\pi \rho _{c}G}}{\biggr ]}{\frac {f_{\mathrm {trial} }}{\theta }}-\alpha {\biggl [}Af_{\mathrm {trial} }-B{\biggr ]}\theta ^{(n-1)}f_{\mathrm {trial} }\,.$

Noting that,

${\frac {d}{d\xi }}{\biggl [}\xi \theta ^{(n-1)}f_{\mathrm {trial} }{\biggr ]}$	$=$	${\biggl \{}\theta ^{(n-1)}f_{\mathrm {trial} }+\xi f_{\mathrm {trial} }(n-1)\theta ^{(n-2)}{\frac {d\theta }{d\xi }}+\xi \theta ^{(n-1)}{\frac {df_{\mathrm {trial} }}{d\xi }}{\biggr \}}$
$\Rightarrow ~~~\xi \theta ^{(n-1)}{\frac {df_{\mathrm {trial} }}{d\xi }}$	$=$	${\frac {d}{d\xi }}{\biggl [}\xi \theta ^{(n-1)}f_{\mathrm {trial} }{\biggr ]}-{\biggl [}\theta ^{(n-1)}f_{\mathrm {trial} }+\xi f_{\mathrm {trial} }(n-1)\theta ^{(n-2)}{\frac {d\theta }{d\xi }}{\biggr ]}\,,$

we furthermore can write,

$\Rightarrow ~~~{\biggl [}{\frac {3(3-n)}{\xi _{1}^{2}}}{\biggr ]}~\mathrm {LAWE}$	$=$	$3(3-n){\frac {d^{2}f_{\mathrm {trial} }}{d\xi ^{2}}}+{\biggl \{}{\frac {12(3-n)}{\xi }}{\biggr \}}{\frac {df_{\mathrm {trial} }}{d\xi }}+3(3-n){\biggl (}{\frac {\omega ^{2}}{\gamma _{g}}}{\biggr )}{\biggl [}{\frac {(n+1)}{4\pi \rho _{c}G}}{\biggr ]}{\frac {f_{\mathrm {trial} }}{\theta }}$
		$-\alpha {\biggl [}Af_{\mathrm {trial} }-B{\biggr ]}\theta ^{(n-1)}f_{\mathrm {trial} }-{\biggl [}Af_{\mathrm {trial} }-B{\biggr ]}{\biggl \{}{\frac {d}{d\xi }}{\biggl [}\xi \theta ^{(n-1)}f_{\mathrm {trial} }{\biggr ]}-{\biggl [}\theta ^{(n-1)}f_{\mathrm {trial} }+\xi f_{\mathrm {trial} }(n-1)\theta ^{(n-2)}{\frac {d\theta }{d\xi }}{\biggr ]}{\biggr \}}$
	$=$	$3(3-n){\frac {d^{2}f_{\mathrm {trial} }}{d\xi ^{2}}}+{\biggl \{}{\frac {12(3-n)}{\xi }}{\biggr \}}{\frac {df_{\mathrm {trial} }}{d\xi }}+3(3-n){\biggl (}{\frac {\omega ^{2}}{\gamma _{g}}}{\biggr )}{\biggl [}{\frac {(n+1)}{4\pi \rho _{c}G}}{\biggr ]}{\frac {f_{\mathrm {trial} }}{\theta }}$
		$-{\biggl [}Af_{\mathrm {trial} }-B{\biggr ]}{\biggl \{}{\frac {d}{d\xi }}{\biggl [}\xi \theta ^{(n-1)}f_{\mathrm {trial} }{\biggr ]}-{\biggl [}\theta ^{(n-1)}f_{\mathrm {trial} }+\xi f_{\mathrm {trial} }(n-1)\theta ^{(n-2)}{\frac {d\theta }{d\xi }}{\biggr ]}+\alpha \theta ^{(n-1)}f_{\mathrm {trial} }{\biggr \}}$

Seek Hypergeometric Form

Start with the standard LAWE, namely,

$0$	$=$	${\frac {d^{2}f}{d\xi ^{2}}}+{\frac {(4-\mu )}{\xi }}\cdot {\frac {df}{d\xi }}+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}}}{\biggr )}-{\frac {\alpha \mu }{\xi ^{2}}}{\biggr ]}f$
	$=$	${\frac {d^{2}f}{d\xi ^{2}}}+{\frac {1}{\xi }}{\biggl [}4+(n+1){\frac {\xi \theta ^{'}}{\theta }}{\biggr ]}\cdot {\frac {df}{d\xi }}+(n+1){\biggl [}{\biggl (}{\frac {\sigma _{c}^{2}}{6\gamma _{\mathrm {g} }}}{\biggr )}{\frac {1}{\theta }}+{\frac {\alpha \theta ^{'}}{\xi \theta }}{\biggr ]}f\,.$

Part I

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

$z$	$=$	$\xi ^{-1}\theta ^{-n}(\theta ^{'})~~~\Rightarrow ~~~(\theta ^{'})=\xi \theta ^{n}z$
$\Rightarrow ~~~{\frac {dz}{d\xi }}$	$=$	$-\xi ^{-2}\theta ^{-n}(\theta ^{'})-n\xi ^{-1}\theta ^{-n-1}(\theta ^{'})^{2}+\xi ^{-1}\theta ^{-n}(\theta ^{''})$
	$=$	$-{\biggl [}\xi ^{-2}\theta ^{-n}(\theta ^{'})+n\xi ^{-1}\theta ^{-n-1}(\theta ^{'})^{2}+\xi ^{-1}\theta ^{-n}{\biggl (}\theta ^{n}+{\frac {2\theta '}{\xi }}{\biggr )}{\biggr ]}$
	$=$	$-{\biggl [}\xi ^{-2}\theta ^{-n}(\xi \theta ^{n}z)+n\xi ^{-1}\theta ^{-n-1}(\xi \theta ^{n}z)^{2}+\xi ^{-1}\theta ^{-n}{\biggl (}\theta ^{n}{\biggr )}+\xi ^{-1}\theta ^{-n}{\biggl (}{\frac {2\theta '}{\xi }}{\biggr )}{\biggr ]}$
	$=$	$-{\biggl [}\xi ^{-1}z+n\xi \theta ^{n-1}z^{2}+\xi ^{-1}+2\xi ^{-1}z{\biggr ]}$
	$=$	$-{\biggl [}\xi ^{-1}(1+3z)+n\xi \theta ^{n-1}z^{2}{\biggr ]}\,;$

and,

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

Part II

Part I Summary …

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

---

Also,

${\frac {df}{d\xi }}$	$\rightarrow$	${\frac {dz}{d\xi }}\cdot {\frac {df}{dz}}\,;$
${\frac {d^{2}f}{d\xi ^{2}}}={\frac {d}{d\xi }}{\biggl [}{\frac {dz}{d\xi }}\cdot {\frac {df}{dz}}{\biggr ]}$	$\rightarrow$	${\frac {d^{2}z}{d\xi ^{2}}}\cdot {\frac {df}{dz}}+{\biggl [}{\frac {dz}{d\xi }}{\biggr ]}^{2}\cdot {\frac {d^{2}f}{dz^{2}}}$

As a result,

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

Part III

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

LAWE	$=$	${\biggl [}\xi ^{-2}(1+3z)^{2}+2n(1+3z)z^{2}\theta ^{n-1}+n^{2}\xi ^{2}\theta ^{2(n-1)}z^{4}{\biggr ]}\cdot {\cancelto {0}{\frac {d^{2}f}{dz^{2}}}}+(n+1){\biggl [}{\biggl (}{\frac {\sigma _{c}^{2}}{6\gamma _{\mathrm {g} }}}{\biggr )}\theta ^{-1}+\alpha \theta ^{n-1}z{\biggr ]}(a_{0}+b_{0}z)+{\biggl \{}(n-1)+(n-3)z{\biggr \}}\theta ^{n-1}z\cdot b_{0}$
	$=$	$(n+1){\biggl [}{\biggl (}{\frac {\sigma _{c}^{2}}{6\gamma _{\mathrm {g} }}}{\biggr )}\theta ^{-1}{\biggr ]}(a_{0}+b_{0}z)+{\biggl \{}(n+1)\alpha (a_{0}+b_{0}z)+b_{0}(n-1)+b_{0}(n-3)z{\biggr \}}\theta ^{n-1}z$
	$=$	$(n+1){\biggl [}{\biggl (}{\frac {\sigma _{c}^{2}}{6\gamma _{\mathrm {g} }}}{\biggr )}\theta ^{-1}{\biggr ]}(a_{0}+b_{0}z)+{\biggl \{}(n+1)\alpha (a_{0})+b_{0}(n-1)+{\biggl [}(n-3)+(n+1)\alpha {\biggr ]}b_{0}z{\biggr \}}\theta ^{n-1}z\,.$

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

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

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,

$0$	$=$	$(n+1)\alpha (a_{0})+b_{0}(n-1)$
$\Rightarrow ~~~b_{0}$	$=$	${\biggl [}{\frac {n+1}{1-n}}{\biggr ]}\alpha (a_{0})$
	$=$	$a_{0}{\biggl [}{\frac {3-n}{1-n}}{\biggr ]}$

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,

$f$

$=$

$1+{\biggl [}{\frac {3-n}{1-n}}{\biggr ]}z\,.$

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,

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

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

$0$

$=$

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

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,

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

Part V

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

$z$

$=$

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

Recalling that,

Part I Summary …

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

it may prove useful to recognize that,

$\theta ^{'}={\frac {d\theta }{d\xi }}$	$\rightarrow$	${\frac {dz}{d\xi }}\cdot {\frac {d\theta }{dz}}$
$\Rightarrow ~~~{\frac {d\theta }{dz}}$	$=$	${\biggl (}{\frac {dz}{d\xi }}{\biggr )}^{-1}\xi \theta ^{n}z$
$\Rightarrow ~~~{\frac {d\ln \theta }{d\ln z}}$	$=$	${\biggl (}{\frac {dz}{d\xi }}{\biggr )}^{-1}\xi \theta ^{n-1}z^{2}$
$\Rightarrow ~~~{\frac {dz}{d\xi }}$	$=$	${\biggl (}{\frac {d\ln \theta }{d\ln z}}{\biggr )}^{-1}\xi \theta ^{n-1}z^{2}$

In this case we have,

LAWE	$=$	${\biggl [}-{\biggl (}{\frac {d\ln \theta }{d\ln z}}{\biggr )}^{-1}\xi \theta ^{n-1}z^{2}{\biggr ]}^{2}\cdot (2c_{0})+(n+1){\biggl [}{\biggl (}{\frac {\sigma _{c}^{2}}{6\gamma _{\mathrm {g} }}}{\biggr )}\theta ^{-1}+\alpha \theta ^{n-1}z{\biggr ]}(a_{0}+b_{0}z+c_{0}z^{2})$
		$+{\biggl \{}(n-1)+(n-3)z{\biggr \}}\theta ^{n-1}z\cdot (b_{0}+2c_{0}z)$

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${\biggl [}\theta ^{1-n}z^{-1}{\biggr ]}\times$ LAWE	$=$	$\theta ^{1-n}z^{-1}{\biggl [}{\frac {dz}{d\xi }}{\biggr ]}^{2}\cdot {\frac {d^{2}(a_{0}+b_{0}z+c_{0}z^{2})}{dz^{2}}}+{\biggl \{}(n-1)+(n-3)z{\biggr \}}\cdot {\frac {d(a_{0}+b_{0}z+c_{0}z^{2})}{dz}}+(n+1){\biggl [}{\biggl (}{\frac {\sigma _{c}^{2}}{6\gamma _{\mathrm {g} }}}{\biggr )}\theta ^{-n}z^{-1}+\alpha {\biggr ]}\cdot (a_{0}+b_{0}z+c_{0}z^{2})$
	$=$	${\biggl [}\xi ^{-2}(1+3z)^{2}+2n(1+3z)z^{2}\theta ^{n-1}+n^{2}\xi ^{2}\theta ^{2(n-1)}z^{4}{\biggr ]}\cdot (2c_{0})+{\biggl \{}(n-1)+(n-3)z{\biggr \}}\cdot (b_{0}+2c_{0}z)+(n+1){\biggl [}{\biggl (}{\frac {\sigma _{c}^{2}}{6\gamma _{\mathrm {g} }}}{\biggr )}\theta ^{-n}z^{-1}+\alpha {\biggr ]}\cdot (a_{0}+b_{0}z+c_{0}z^{2})$
	$=$	${\biggl [}(2c_{0})\xi ^{-2}(1+6z+9z^{2})+4c_{0}n(z^{2}+3z^{3})\theta ^{n-1}+2c_{0}n^{2}\xi ^{2}\theta ^{2(n-1)}z^{4}{\biggr ]}+(n-1)(b_{0}+2c_{0}z)$
		$+(n-3)(b_{0}z+2c_{0}z^{2})+(n+1){\biggl [}{\biggl (}{\frac {\sigma _{c}^{2}}{6\gamma _{\mathrm {g} }}}{\biggr )}\theta ^{-n}z^{-1}{\biggr ]}\cdot (a_{0}+b_{0}z+c_{0}z^{2})+(n+1)\alpha (a_{0}+b_{0}z+c_{0}z^{2})\,.$

Looks pretty hopeless!

Example Density- and Pressure-Profiles

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

Uniform Density

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

$0$	$=$	${\frac {d^{2}f}{dx^{2}}}+{\frac {(4-\mu )}{x}}\cdot {\frac {df}{dx}}+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{0}R^{2}}{\gamma _{\mathrm {g} }P_{0}}}{\biggr )}-{\frac {\alpha \mu }{x^{2}}}{\biggr ]}f$
	$=$	${\frac {d^{2}f}{dx^{2}}}+{\frac {1}{x}}{\biggl [}4-{\frac {2x^{2}}{(1-x^{2})}}{\biggr ]}{\frac {df}{dx}}+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{c}R^{2}}{\gamma _{\mathrm {g} }P_{c}}}{\biggr )}{\frac {1}{(1-x^{2})}}-{\biggl (}{\frac {2\alpha }{1-x^{2}}}{\biggr )}{\biggr ]}f$
	$=$	$(1-x^{2})\cdot {\frac {d^{2}f}{dx^{2}}}+{\frac {1}{x}}{\biggl [}4-6x^{2}{\biggr ]}{\frac {df}{dx}}+{\biggl [}{\biggl (}{\frac {\omega ^{2}\rho _{c}R^{2}}{\gamma _{\mathrm {g} }P_{c}}}{\biggr )}-2\alpha {\biggr ]}f\,.$

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,

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

where,

${\mathfrak {F}}$

$\equiv$

${\biggl [}6{\biggl (}{\frac {\omega ^{2}}{4\pi \gamma _{\mathrm {g} }G\rho _{c}}}{\biggr )}-2\alpha {\biggr ]}\,.$

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,

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

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

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

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,

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

and,

$\beta$

$=$

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

Example α = -1

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

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

$=$

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

and the corresponding eigenfrequency is obtained from the expression,

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

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, $0\leq j\leq \infty$, 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,

$u_{j}=$

$=$

$F{\biggl (}\alpha _{j},\beta _{j};{\frac {5}{2}};x^{2}{\biggr )}\,.$

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,

  $\rho (r)=\rho _{c}{\biggl [}1-{\biggl (}{\frac {r}{R}}{\biggr )}^{2}{\biggr ]}\,,$

  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,

  $~{\biggl [}P{\biggr ]}{\frac {d^{2}{\mathcal {G}}_{\sigma }}{dx^{2}}}+{\biggl [}{\frac {4P}{x}}+P^{'}{\biggr ]}{\frac {d{\mathcal {G}}_{\sigma }}{dx}}+{\biggl [}\sigma ^{2}\rho +{\frac {\alpha P^{'}}{x}}{\biggr ]}{\mathcal {G}}_{\sigma }$

  $~=$

  $~0\,.$

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