SSC/Stability/n1PolytropeLAWE/Pt2

Radial Oscillations of n = 1 Polytropic Spheres (Pt 2)

Part I:   Search for Analytic Solutions

Part II:  New Ideas

III:  What About Bipolytropes?

New Idea Involving Logarithmic Derivatives

Simplistic Layout

Let's begin, again, with the relevant LAWE, as provided above. After dividing through by $x$, we have,

$(\sin \xi )\frac{{\xi }^2}{x}\cdot \frac{d^{2}x}{d{\xi }^2}+2[\sin \xi +\xi \cos \xi ]\frac{\xi }{x}\cdot \frac{dx}{d\xi }+[{\sigma }^2{\xi }^3-2\alpha (\sin \xi -\xi \cos \xi )]=0,$

where,

${\sigma }^2$	$\equiv$	$\frac{{\omega }^2}{2\pi G{\rho }_c{\gamma }_g},$
$\alpha$	$\equiv$	$3-\frac{4}{{\gamma }_g}.$

Now, in addition to recognizing that,

$\frac{\xi }{x}\cdot \frac{dx}{d\xi }$

$=$

$\frac{d\ln x}{d\ln \xi },$

in a separate context, we showed that, quite generally,

$\frac{{\xi }^2}{x}\cdot \frac{d^{2}x}{d{\xi }^2}$

$=$

$\frac{d}{d\ln \xi }\left[\frac{d\ln x}{d\ln \xi }\right]-[1-\frac{d\ln x}{d\ln \xi }]\cdot \frac{d\ln x}{d\ln \xi }.$

Hence, if we assume that the eigenfunction is a power-law of $\xi$, that is, assume that,

$x=a_0{\xi }^{c_0},$

then the logarithmic derivative of $x$ is a constant, namely,

$\frac{d\ln x}{d\ln \xi }=c_0,$

and the two key derivative terms will be,

$\frac{\xi }{x}\cdot \frac{dx}{d\xi }=c_0,$

and

$\frac{{\xi }^2}{x}\cdot \frac{d^{2}x}{d{\xi }^2}=c_0(c_0-1).$

In this case, the LAWE is no longer a differential equation but, instead, takes the form,

$-{\sigma }^2{\xi }^3$	$=$	$c_0(c_0-1)\sin \xi +2c_0[\sin \xi +\xi \cos \xi ]-2\alpha (\sin \xi -\xi \cos \xi )$
	$=$	$\sin \xi [c_0(c_0-1)+2c_0-2\alpha ]+\xi \cos \xi [2(c_0+\alpha )]$
	$=$	$\sin \xi [c_0^2+c_0-2\alpha ]+\xi \cos \xi [2(c_0+\alpha )].$

Now, the cosine term will go to zero if $c_0=-\alpha$; and the sine term will go to zero if,

$\alpha$	$=$	$3$
$\Rightarrow {\gamma }_g$	$=$	$\mathrm{\infty }.$

If these two — rather strange — conditions are met, then we have a marginally unstable configuration because, ${\sigma }^2=0$. This, in and of itself, is not very physically interesting. However, it may give us a clue regarding how to more generally search for a physically reasonable radial eigenfunction.

More general Assumption

Try,

$x$	$=$	${\xi }^{c_0}[a_0+b_0\sin \xi +d_0\xi \cos \xi ]$
$\Rightarrow \frac{dx}{d\xi }$	$=$	${\xi }^{c_0}\frac{d}{d\xi }[a_0+b_0\sin \xi +d_0\xi \cos \xi ]+c_0{\xi }^{c_0-1}[a_0+b_0\sin \xi +d_0\xi \cos \xi ]$
	$=$	${\xi }^{c_0}[b_0\cos \xi -d_0\xi \sin \xi +d_0\cos \xi ]+c_0{\xi }^{c_0-1}[a_0+b_0\sin \xi +d_0\xi \cos \xi ]$
$\Rightarrow \frac{d\ln x}{d\ln \xi }$	$=$	$\xi [b_0\cos \xi -d_0\xi \sin \xi +d_0\cos \xi ][a_0+b_0\sin \xi +d_0\xi \cos \xi {]}^{-1}+c_0$
	$=$	$[(b_0+d_0)\xi \cos \xi -d_0{\xi }^2\sin \xi ][a_0+b_0\sin \xi +d_0\xi \cos \xi {]}^{-1}+c_0$

Another Viewpoint

Development

Multiplying through the above LAWE by $(x{\xi }^{-3})$ gives,

$0$

$=$

$\frac{\sin \xi }{\xi }\cdot \frac{d^{2}x}{d{\xi }^2}+2\left[\frac{\sin \xi +\xi \cos \xi }{{\xi }^2}\right]\frac{dx}{d\xi }+[{\sigma }^2-2\alpha (\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right]x$

Notice that,

$\frac{d}{d\xi }\left[\frac{\sin \xi }{\xi }\right]$	$=$	$-\frac{\sin \xi }{{\xi }^2}+\frac{\cos \xi }{\xi }$
	$=$	$\left[\frac{\xi \cos \xi -\sin \xi }{{\xi }^2}\right].$

And, hence,

$\frac{d^2}{d{\xi }^2}\left[\frac{\sin \xi }{\xi }\right]$	$=$	$\frac{d}{d\xi }[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}]$
	$=$	$-\frac{\cos \xi }{{\xi }^2}-\frac{\sin \xi }{\xi }+\frac{2\sin \xi }{{\xi }^3}-\frac{\cos \xi }{{\xi }^2}$
	$=$	$-\frac{\sin \xi }{\xi }+2\left[\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\right].$

So, we can write,

$\frac{d^2}{d{\xi }^2}\left\{\right(\frac{\sin \xi }{\xi }\left)x\right\}$	$=$	$\frac{d}{d\xi }\left\{\right(\frac{\sin \xi }{\xi })\frac{dx}{d\xi }+x\frac{d}{d\xi }[\left(\frac{\sin \xi }{\xi }\right)\left]\right\}$
	$=$	$\frac{\sin \xi }{\xi }\cdot \frac{d^{2}x}{d{\xi }^2}+2\frac{dx}{d\xi }\cdot \left[\frac{d}{d\xi }\right(\frac{\sin \xi }{\xi }\left)\right]+x\cdot \frac{d^2}{d{\xi }^2}\left(\frac{\sin \xi }{\xi }\right)$
	$=$	$\frac{\sin \xi }{\xi }\cdot \frac{d^{2}x}{d{\xi }^2}+2\frac{dx}{d\xi }\cdot \left[\frac{\xi \cos \xi -\sin \xi }{{\xi }^2}\right]+x\cdot \{-\frac{\sin \xi }{\xi }+2[\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left]\right\}.$

This means that we can rewrite the LAWE as,

$0$	$=$	$\frac{d^2}{d{\xi }^2}\left\{\right(\frac{\sin \xi }{\xi }\left)x\right\}-2\frac{dx}{d\xi }\cdot \left[\frac{\xi \cos \xi -\sin \xi }{{\xi }^2}\right]-x\cdot \{-\frac{\sin \xi }{\xi }+2[\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left]\right\}+2\left[\frac{\sin \xi +\xi \cos \xi }{{\xi }^2}\right]\frac{dx}{d\xi }+[{\sigma }^2-2\alpha (\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right]x$
	$=$	$\frac{d^2}{d{\xi }^2}\left\{\right(\frac{\sin \xi }{\xi }\left)x\right\}+4\left[\frac{\sin \xi }{{\xi }^2}\right]\frac{dx}{d\xi }+\{\frac{\sin \xi }{\xi }+{\sigma }^2-2(1+\alpha )(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right\}\cdot x.$

We recognize, also, that,

$\frac{1}{\xi }\cdot \frac{d}{d\xi }\left[\right(\frac{\sin \xi }{\xi }\left)x\right]$	$=$	$\left[\frac{\xi \cos \xi -\sin \xi }{{\xi }^3}\right]x+\left(\frac{\sin \xi }{{\xi }^2}\right)\frac{dx}{d\xi }.$
$\Rightarrow 4\left(\frac{\sin \xi }{{\xi }^2}\right)\frac{dx}{d\xi }$	$=$	$\frac{4}{\xi }\cdot \frac{d}{d\xi }\left[\right(\frac{\sin \xi }{\xi }\left)x\right]+4\left[\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\right]x.$

So the LAWE becomes,

$0$	$=$	$\frac{d^2}{d{\xi }^2}\left\{\right(\frac{\sin \xi }{\xi }\left)x\right\}+\frac{4}{\xi }\cdot \frac{d}{d\xi }\left[\right(\frac{\sin \xi }{\xi }\left)x\right]+4\left[\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\right]x+\{\frac{\sin \xi }{\xi }+{\sigma }^2-2(1+\alpha )(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right\}\cdot x$
	$=$	$\frac{d^2}{d{\xi }^2}\left\{\right(\frac{\sin \xi }{\xi }\left)x\right\}+\frac{4}{\xi }\cdot \frac{d}{d\xi }\left[\right(\frac{\sin \xi }{\xi }\left)x\right]+\{\frac{\sin \xi }{\xi }+{\sigma }^2+[4-2(1+\alpha )](\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right\}\cdot x$
	$=$	$\frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+\frac{4}{\xi }\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }+\mathrm{{\rm Y}}+[{\sigma }^2+2(1-\alpha )(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right]\cdot x,$

where we have introduced the new, modified eigenfunction,

$\mathrm{{\rm Y}}\equiv \left(\frac{\sin \xi }{\xi }\right)x.$

Alternatively, the LAWE may be written as,

$0$

$=$

$\frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+\frac{4}{\xi }\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }+[{\sigma }^2+2(1-\alpha )(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3})+\frac{\sin \xi }{\xi }]\cdot x;$

or,

$0$	$=$	$\frac{{\xi }^2}{\mathrm{{\rm Y}}}\cdot \frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+\frac{4\xi }{\mathrm{{\rm Y}}}\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }+[{\sigma }^2+2(1-\alpha )(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3})+\frac{\sin \xi }{\xi }]\cdot \frac{{\xi }^3}{\sin \xi }$
	$=$	$\frac{{\xi }^2}{\mathrm{{\rm Y}}}\cdot \frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+\frac{4\xi }{\mathrm{{\rm Y}}}\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }+\left[{\sigma }^2\right(\frac{{\xi }^3}{\sin \xi })+2(1-\alpha )(1-\xi \cot \xi )+{\xi }^2]$

Now, if we adopt the homentropic convention that arises from setting, $\gamma =(n+1)/n$, then for our $n=1$ polytropic configuration, we should set, $\gamma =2$ and, hence, $\alpha =1$. This will mean that the lat term in this LAWE naturally goes to zero. Hence, we have,

$-{\sigma }^{2}x$

$=$

$\frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+\frac{4}{\xi }\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }+\mathrm{{\rm Y}};$

or,

$0$

$=$

$\frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+\frac{4}{\xi }\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }+[1+{\sigma }^2(\frac{\xi }{\sin \xi }\left)\right]\mathrm{{\rm Y}};$

or,

$0$

$=$

$\frac{{\xi }^2}{\mathrm{{\rm Y}}}\cdot \frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+\frac{4\xi }{\mathrm{{\rm Y}}}\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }+[{\xi }^2+{\sigma }^2(\frac{{\xi }^3}{\sin \xi }\left)\right].$

Does this help?

Check for Mistakes

Given the definition of $\mathrm{{\rm Y}}$, its first derivative is,

$\frac{d\mathrm{{\rm Y}}}{d\xi }$

$=$

$\left(\frac{\sin \xi }{\xi }\right)\frac{dx}{d\xi }+x[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}],$

and its second derivative is,

$\frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}$	$=$	$\frac{d}{d\xi }\left\{\right(\frac{\sin \xi }{\xi })\frac{dx}{d\xi }+x[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}\left]\right\}$
	$=$	$\left(\frac{\sin \xi }{\xi }\right)\frac{d^{2}x}{d{\xi }^2}+2[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}]\cdot \frac{dx}{d\xi }+x\cdot \frac{d}{d\xi }[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}]$
	$=$	$\left(\frac{\sin \xi }{\xi }\right)\frac{d^{2}x}{d{\xi }^2}+2[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}]\cdot \frac{dx}{d\xi }+x[-\frac{\sin \xi }{\xi }-\frac{2\cos \xi }{{\xi }^2}+\frac{2\sin \xi }{{\xi }^3}]$

Hence, the "upsilon" LAWE becomes,

$-{\sigma }^{2}x$	$=$	$\frac{d^2\mathrm{{\rm Y}}}{d{\xi }^2}+\frac{4}{\xi }\cdot \frac{d\mathrm{{\rm Y}}}{d\xi }+\mathrm{{\rm Y}}+[2(1-\alpha )(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right]\cdot x$
	$=$	$\left(\frac{\sin \xi }{\xi }\right)\frac{d^{2}x}{d{\xi }^2}+2[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}]\cdot \frac{dx}{d\xi }+x[-\frac{\sin \xi }{\xi }-\frac{2\cos \xi }{{\xi }^2}+\frac{2\sin \xi }{{\xi }^3}]+\frac{4}{\xi }\cdot \left\{\right(\frac{\sin \xi }{\xi })\frac{dx}{d\xi }+x[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}\left]\right\}+[\frac{\sin \xi }{\xi }+2(1-\alpha )(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right]\cdot x$
	$=$	$\left(\frac{\sin \xi }{\xi }\right)\frac{d^{2}x}{d{\xi }^2}+\left\{\right(\frac{4\sin \xi }{{\xi }^2})+2[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}\left]\right\}\cdot \frac{dx}{d\xi }+[-\frac{\sin \xi }{\xi }-\frac{2\cos \xi }{{\xi }^2}+\frac{2\sin \xi }{{\xi }^3}+\frac{4\cos \xi }{{\xi }^2}-\frac{4\sin \xi }{{\xi }^3}+\frac{\sin \xi }{\xi }+2(1-\alpha )(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right]\cdot x$
	$=$	$\left(\frac{\sin \xi }{\xi }\right)\frac{d^{2}x}{d{\xi }^2}+[\frac{2\cos \xi }{\xi }+\frac{2\sin \xi }{{\xi }^2}]\cdot \frac{dx}{d\xi }+[-2(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3})+(2-2\alpha )(\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right]\cdot x$
	$=$	$\left(\frac{\sin \xi }{\xi }\right)\frac{d^{2}x}{d{\xi }^2}+2[\frac{\sin \xi }{{\xi }^2}+\frac{\cos \xi }{\xi }]\cdot \frac{dx}{d\xi }+[-2\alpha (\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right]\cdot x.$

This should be compared with the first expression, above, namely,

$0$

$=$

$\frac{\sin \xi }{\xi }\cdot \frac{d^{2}x}{d{\xi }^2}+2\left[\frac{\sin \xi +\xi \cos \xi }{{\xi }^2}\right]\frac{dx}{d\xi }+[{\sigma }^2-2\alpha (\frac{\sin \xi -\xi \cos \xi }{{\xi }^3}\left)\right]x,$

and it matches! Q.E.D.

Motivated by Yabushita's Discovery

Initial Exploration

This subsection is being developed following our realization — see the accompanying overview — that the eigenfunction is known analytically for marginally unstable, pressure-truncated configurations having $3\le n\le \mathrm{\infty }$. Specifically, from the work of Yabushita (1975) we have the following,

Exact Solution to the Isothermal LAWE
${\sigma }_c^2=0$	and	$x=1-\left(\frac{1}{\xi e^{-\psi }}\right)\frac{d\psi }{d\xi }.$

And from our own recent work, we have discovered the following,

Precise Solution to the Polytropic LAWE
${\sigma }_c^2=0$	and	$x_P\equiv \frac{3(n-1)}{2n}[1+(\frac{n-3}{n-1}\left)\right(\frac{1}{\xi {\theta }^n}\left)\frac{d\theta }{d\xi }\right]$

if the adiabatic exponent is assigned the value, ${\gamma }_g=(n+1)/n$, in which case the parameter, $\alpha =(3-n)/(n+1)$. Using this polytropic displacement function as a guide, let's try for the case of $n=1$, an expression of the form,

$x$	$=$	$A-B\left[\right(\frac{1}{\xi \theta }\left)\frac{d\theta }{d\xi }\right]$
	$=$	$A-B\left[\right(\frac{1}{\sin \xi }\left)\frac{d}{d\xi }\right(\frac{\sin \xi }{\xi }\left)\right]$
	$=$	$A-B\left(\frac{1}{\sin \xi }\right)[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}]$
	$=$	$A+\frac{B}{{\xi }^2}(1-\frac{\xi \cos \xi }{\sin \xi }),$

in which case,

$\frac{dx}{d\xi }$	$=$	$-B\left\{\right(\frac{-\cos \xi }{{\sin }^2\xi }\left)\right[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}]+(\frac{1}{\sin \xi }\left)\right[-\frac{\sin \xi }{\xi }-\frac{\cos \xi }{{\xi }^2}-\frac{\cos \xi }{{\xi }^2}+\frac{2\sin \xi }{{\xi }^3}\left]\right\}$
	$=$	$-\frac{B}{{\xi }^3}\left\{\right(\frac{\cos \xi }{{\sin }^2\xi }\left)\right[-{\xi }^2\cos \xi +\xi \sin \xi ]+[2-{\xi }^2-\frac{2\xi \cos \xi }{\sin \xi }\left]\right\}$
	$=$	$-\frac{B}{{\xi }^3}\{2-{\xi }^2-\frac{\xi \cos \xi }{\sin \xi }-\frac{{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }\},$

What if, instead, we try the more generalized form, $x$ $=$ $A+\frac{B}{(\lambda \xi {)}^2}[1-\frac{\lambda \xi \cos (\lambda \xi )}{\sin (\lambda \xi )}].$ Then we have, $\frac{1}{\lambda B}\cdot \frac{dx}{d\xi }$ $=$ $-\frac{1}{(\lambda \xi {)}^3}\{2-(\lambda \xi {)}^2-\frac{\lambda \xi \cos (\lambda \xi )}{\sin (\lambda \xi )}-\frac{(\lambda \xi {)}^2{\cos }^2(\lambda \xi )}{{\sin }^2(\lambda \xi )}\},$ Probably this also means, $\frac{1}{{\lambda }^{2}B}\cdot \frac{dx}{d\xi }$ $=$ $\frac{1}{(\lambda \xi {)}^4}\{6-2(\lambda \xi {)}^2-\frac{2\lambda \xi \cos (\lambda \xi )}{\sin (\lambda \xi )}-\frac{2(\lambda \xi {)}^2{\cos }^2(\lambda \xi )}{{\sin }^2(\lambda \xi )}-\frac{2(\lambda \xi {)}^3\cos (\lambda \xi )}{\sin (\lambda \xi )}-\frac{2(\lambda \xi {)}^3{\cos }^3(\lambda \xi )}{{\sin }^3(\lambda \xi )}\}.$

Let's check against the more general derivation, which gives after recognizing that, $B\leftrightarrow (3-n)/(n-1)$,

$\frac{dx}{d\xi }$	$=$	$\left(\frac{3-n}{n-1}\right)\{\frac{1}{\xi }+\frac{n({\theta }^{\text{'}}{)}^2}{\xi {\theta }^{n+1}}+\frac{3{\theta }^{\text{'}}}{{\xi }^2{\theta }^n}\}$
	$=$	$\frac{B}{{\xi }^3}\{{\xi }^2+{\xi }^2(\frac{\xi }{\sin \xi }{)}^2[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}{]}^2+\frac{3{\xi }^2}{\sin \xi }[\frac{\cos \xi }{\xi }-\frac{\sin \xi }{{\xi }^2}\left]\right\}$
	$=$	$\frac{B}{{\xi }^3}\{{\xi }^2+3[\frac{\xi \cos \xi }{\sin \xi }-1]+[\frac{\xi \cos \xi }{\sin \xi }-1{]}^2\}$
	$=$	$\frac{B}{{\xi }^3}\{{\xi }^2+3[\frac{\xi \cos \xi }{\sin \xi }-1]+[(\frac{\xi \cos \xi }{\sin \xi }{)}^2-2(\frac{\xi \cos \xi }{\sin \xi })+1]\}$
	$=$	$\frac{B}{{\xi }^3}\{{\xi }^2+[\frac{\xi \cos \xi }{\sin \xi }-2]+(\frac{\xi \cos \xi }{\sin \xi }{)}^2\}.$

This matches the preceding, direct derivation.

Also,

$\frac{d^{2}x}{d{\xi }^2}$	$=$	$\frac{3B}{{\xi }^4}\left\{\right(\frac{\cos \xi }{{\sin }^2\xi }\left)\right[-{\xi }^2\cos \xi +\xi \sin \xi ]+[2-{\xi }^2-\frac{2\xi \cos \xi }{\sin \xi }\left]\right\}$
		$-\frac{B}{{\xi }^3}\left\{\right[-\frac{1}{\sin \xi }-\frac{2{\cos }^2\xi }{{\sin }^3\xi }\left]\right[-{\xi }^2\cos \xi +\xi \sin \xi ]+(\frac{\cos \xi }{{\sin }^2\xi }\left)\right[-2\xi \cos \xi +\sin \xi +{\xi }^2\sin \xi +\xi \cos \xi ]$
		$+[-2\xi -\frac{2\cos \xi }{\sin \xi }+\frac{2\xi \sin \xi }{\sin \xi }+\frac{2\xi {\cos }^2\xi }{{\sin }^2\xi }]\}$
	$=$	$\frac{B}{{\xi }^4}\left\{\right(\frac{3\cos \xi }{{\sin }^2\xi }\left)\right[-{\xi }^2\cos \xi +\xi \sin \xi ]+[6-3{\xi }^2-\frac{6\xi \cos \xi }{\sin \xi }]+[\frac{1}{\sin \xi }+\frac{2{\cos }^2\xi }{{\sin }^3\xi }\left]\right[-{\xi }^3\cos \xi +{\xi }^2\sin \xi ]$
		$+\left(\frac{\cos \xi }{{\sin }^2\xi }\right)[2{\xi }^2\cos \xi -\xi \sin \xi -{\xi }^3\sin \xi -{\xi }^2\cos \xi ]+[2{\xi }^2+\frac{2\xi \cos \xi }{\sin \xi }-\frac{2{\xi }^2\sin \xi }{\sin \xi }-\frac{2{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }]\}$
	$=$	$\frac{B}{{\xi }^4}\left\{\right[-\frac{3{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }+\frac{3\xi \cos \xi }{\sin \xi }]+[6-3{\xi }^2-\frac{6\xi \cos \xi }{\sin \xi }]+[-\frac{{\xi }^3\cos \xi }{\sin \xi }+{\xi }^2]+[-\frac{2{\xi }^3{\cos }^3\xi }{{\sin }^3\xi }+\frac{2{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }]$
		$+[\frac{2{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }-\frac{\xi \cos \xi }{\sin \xi }-\frac{{\xi }^3\cos \xi }{\sin \xi }-\frac{{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }]+[2{\xi }^2+\frac{2\xi \cos \xi }{\sin \xi }-\frac{2{\xi }^2\sin \xi }{\sin \xi }-\frac{2{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }]\}$
	$=$	$\frac{B}{{\xi }^4}\{6-2{\xi }^2-\frac{2\xi \cos \xi }{\sin \xi }-\frac{2{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }-\frac{2{\xi }^3\cos \xi }{\sin \xi }-\frac{2{\xi }^3{\cos }^3\xi }{{\sin }^3\xi }\}.$

Let's also check this against the more general derivation, which gives after again recognizing that, $B\leftrightarrow (3-n)/(n-1)$,

$\frac{d^{2}x}{d{\xi }^2}$	$=$	$\left(\frac{n-3}{n-1}\right)\{\frac{4}{{\xi }^2}+\frac{2n({\theta }^{\text{'}})}{\xi \theta }+\frac{12{\theta }^{\text{'}}}{{\xi }^3{\theta }^n}+\frac{8n({\theta }^{\text{'}}{)}^2}{{\xi }^2{\theta }^{n+1}}+(n+1)\frac{n({\theta }^{\text{'}}{)}^3}{\xi {\theta }^{n+2}}\}$
	$=$	$-B\{\frac{4}{{\xi }^2}+\frac{2}{\xi \theta }[\frac{\sin \xi }{{\xi }^2}(\frac{\xi \cos \xi }{\sin \xi }-1)]+\frac{12}{{\xi }^3\theta }[\frac{\sin \xi }{{\xi }^2}(\frac{\xi \cos \xi }{\sin \xi }-1)]+\frac{8}{{\xi }^2{\theta }^2}[\frac{\sin \xi }{{\xi }^2}(\frac{\xi \cos \xi }{\sin \xi }-1){]}^2+\frac{2}{\xi {\theta }^3}\left[\frac{\sin \xi }{{\xi }^2}\right(\frac{\xi \cos \xi }{\sin \xi }-1\left){]}^3\right\}$
	$=$	$-\frac{B}{{\xi }^4}\{4{\xi }^2+2{\xi }^2(\frac{\xi \cos \xi }{\sin \xi }-1)+12(\frac{\xi \cos \xi }{\sin \xi }-1)+8(\frac{\xi \cos \xi }{\sin \xi }-1{)}^2+2(\frac{\xi \cos \xi }{\sin \xi }-1{)}^3\}$
	$=$	$-\frac{2B}{{\xi }^4}\{2{\xi }^2+\frac{{\xi }^3\cos \xi }{\sin \xi }-{\xi }^2+\frac{6\xi \cos \xi }{\sin \xi }-6+\frac{4{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }-\frac{8\xi \cos \xi }{\sin \xi }+4+(\frac{{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }-\frac{2\xi \cos \xi }{\sin \xi }+1\left)\right(\frac{\xi \cos \xi }{\sin \xi }-1\left)\right\}$
	$=$	$-\frac{2B}{{\xi }^4}\{-2+{\xi }^2-\frac{2\xi \cos \xi }{\sin \xi }+\frac{{\xi }^3\cos \xi }{\sin \xi }+\frac{4{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }-(\frac{{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }-\frac{2\xi \cos \xi }{\sin \xi }+1)+\frac{{\xi }^3{\cos }^3\xi }{{\sin }^3\xi }-\frac{2{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }+\frac{\xi \cos \xi }{\sin \xi }\}$
	$=$	$-\frac{2B}{{\xi }^4}\{-3+{\xi }^2+\frac{\xi \cos \xi }{\sin \xi }+\frac{{\xi }^3\cos \xi }{\sin \xi }+\frac{{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }+\frac{{\xi }^3{\cos }^3\xi }{{\sin }^3\xi }\}$
	$=$	$\frac{B}{{\xi }^4}\{6-2{\xi }^2-\frac{2\xi \cos \xi }{\sin \xi }-\frac{2{\xi }^2{\cos }^2\xi }{{\sin }^2\xi }-\frac{2{\xi }^3\cos \xi }{\sin \xi }-\frac{2{\xi }^3{\cos }^3\xi }{{\sin }^3\xi }\}.$

A cross-check with the first attempt to derive this second derivative expression initially unveiled a couple of coefficient errors. These have now been corrected and both expressions agree.

Succinct Demonstration

Given that, for $n=1$, we should set ${\gamma }_{\mathrm{g}}=(n+1)/n=2\Rightarrow \alpha =(3-4/{\gamma }_{\mathrm{g}})=+1$, and,

$Q\equiv -\frac{d\ln \theta }{d\ln \xi }$

$=$

$-\frac{{\xi }^2}{\sin \xi }\cdot \frac{d}{d\xi }\left[\frac{\sin \xi }{\xi }\right]=1-\xi \cot \xi .$

If we then employ the displacement function,

$x$

$=$

$A+\frac{B}{{\xi }^2}[1-\xi \cot \xi ],$

the LAWE becomes,

LAWE	$=$	$\frac{d^{2}x}{d{\xi }^2}+[4-(n+1)Q]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_g})\frac{{\xi }^2}{\theta }-\alpha Q]\frac{x}{{\xi }^2}$
	$=$	$\frac{d^{2}x}{d{\xi }^2}+[4-2Q]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+\left[\right(\frac{{\sigma }_c^2}{6})\frac{{\xi }^3}{\sin \xi }-2Q]\frac{x}{{\xi }^2}$
	$=$	$\frac{d^{2}x}{d{\xi }^2}+[2+\frac{2\xi \cos \xi }{\sin \xi }]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+[-2+\frac{2\xi \cos \xi }{\sin \xi }]\frac{x}{{\xi }^2}+\left[\right(\frac{{\sigma }_c^2}{6}\left)\frac{\xi }{\sin \xi }\right]x$
	$=$	$\frac{2B}{{\xi }^4}\{3-{\xi }^2-\frac{\xi \cos \xi }{\sin \xi }-(\frac{\xi \cos \xi }{\sin \xi }{)}^2-\frac{{\xi }^3\cos \xi }{\sin \xi }-\left(\frac{\xi \cos \xi }{\sin \xi }{)}^3\right\}$
		$+\frac{2B}{{\xi }^4}[1+\frac{\xi \cos \xi }{\sin \xi }]\{{\xi }^2-2+\frac{\xi \cos \xi }{\sin \xi }+(\frac{\xi \cos \xi }{\sin \xi }{)}^2\}$
		$+[-2+\frac{2\xi \cos \xi }{\sin \xi }][\frac{A}{{\xi }^2}+\frac{B}{{\xi }^4}(1-\frac{\xi \cos \xi }{\sin \xi }\left)\right]+\left[\right(\frac{{\sigma }_c^2}{6}\left)\frac{\xi }{\sin \xi }\right]x$
	$=$	$\frac{2B}{{\xi }^4}\{3-{\xi }^2-\frac{\xi \cos \xi }{\sin \xi }-(\frac{\xi \cos \xi }{\sin \xi }{)}^2-\frac{{\xi }^3\cos \xi }{\sin \xi }-(\frac{\xi \cos \xi }{\sin \xi }{)}^3$
		$+{\xi }^2\left(\frac{\xi \cos \xi }{\sin \xi }\right)-2\left(\frac{\xi \cos \xi }{\sin \xi }\right)+(\frac{\xi \cos \xi }{\sin \xi }{)}^2+(\frac{\xi \cos \xi }{\sin \xi }{)}^3+{\xi }^2-2+\frac{\xi \cos \xi }{\sin \xi }+\left(\frac{\xi \cos \xi }{\sin \xi }{)}^2\right\}$
		$-\frac{2B}{{\xi }^4}[1-\frac{2\xi \cos \xi }{\sin \xi }+(\frac{\xi \cos \xi }{\sin \xi }{)}^2]+\frac{2A}{{\xi }^2}[\frac{\xi \cos \xi }{\sin \xi }-1]+[\left(\frac{{\sigma }_c^2}{6}\right)\frac{\xi }{\sin \xi }]x$
	$=$	$\frac{2A}{{\xi }^2}[\frac{\xi \cos \xi }{\sin \xi }-1]+\left[\right(\frac{{\sigma }_c^2}{6}\left)\frac{\xi }{\sin \xi }\right]x$

Pretty amazing degree of cancelation! So the above-hypothesized displacement function does satisfy the $n=1$, polytropic LAWE — for any value of the coefficient, $B$ — if we set $A=0$ and ${\sigma }_c^2=0$. If we set $B=3$, the function will be normalized such that it goes to unity at the center. In summary, then, we have,

$x_P{|}_{n=1}$

$=$

$\frac{3}{{\xi }^2}[1-\xi \cot \xi ].$

See Also