Appendix/Ramblings/51AnalyticStabilitySynopsis

More Focused Search for Analytic EigenVector of (5,1) Bipolytropes

The ideas that are captured in this chapter have arisen after a review of a previous hunt for the desired analytic eigenvector and as an extension of our accompanying "renormalization" of the Analytic51 bipolytrope.

Review of Attempt 4B

Structure

From a separate search that we labeled Attempt 4B, we draw the following information regarding the structure of the envelope.

$\varphi$

$=$

$a_0\left[\frac{\sin (\eta -b_0)}{\eta }\right],$

and,

$\frac{d\varphi }{d\eta }$

$=$

$\frac{a_0}{{\eta }^2}[\eta \cos (\eta -b_0)-\sin (\eta -b_0)],$

and,

$\frac{d^2\varphi }{d{\eta }^2}$

$=$

$-\frac{a_0}{\eta }\cdot \sin (\eta -b_0)-\frac{2a_0}{{\eta }^2}\cdot \cos (\eta -b_0)+\frac{2a_0}{{\eta }^3}\cdot \sin (\eta -b_0).$

This satisfies the Lane-Emden equation for any values of the parameter pair, $a_0$ and $b_0$. Note that,

$Q\equiv -\frac{d\ln \varphi }{d\ln \eta }$	$=$	$[1-\eta \cot (\eta -b_0)]$
$\Rightarrow \eta \cot (\eta -b_0)$	$=$	$(1-Q).$

LAWE

Now, guided by a separate parallel discussion we also showed in Attempt 4B that, in the case of a bipolytropic configuration for which $n_e=1$, the

Trial Displacement Function
${\sigma }_c^2=0$	and	$x_P$	$\equiv \frac{3c_0(n-1)}{2n}[1+(\frac{n-3}{n-1}\left)\right(\frac{1}{\eta {\varphi }^n}\left)\frac{d\varphi }{d\eta }\right]$
			$=-\left(\frac{3c_0}{\eta \varphi }\right)\frac{d\varphi }{d\eta }=\frac{3c_0}{{\eta }^2}\cdot Q,$

precisely satisfies the

Governing LAWE
$0$	$=$	$\frac{d^2{x}_P}{d{\eta }^2}+[4-2Q]\frac{1}{\eta }\cdot \frac{d{x}_P}{d\eta }-2Q\cdot \frac{x_P}{{\eta }^2}.$

Note for later use that, $\frac{d\ln x_P}{d\ln \eta }=\frac{\eta }{x_P}\cdot \frac{d}{d\eta }[\frac{3c_0}{{\eta }^2}\cdot Q]$ $=$ $3c_0\eta \left[\frac{{\eta }^2}{3c_0\cdot Q}\right]\cdot \frac{d}{d\eta }\left[\frac{Q}{{\eta }^2}\right]$   $=$ $\left[\frac{{\eta }^3}{Q}\right]\cdot [\frac{1}{{\eta }^2}\frac{dQ}{d\eta }-\frac{2Q}{{\eta }^3}]$   $=$ $[\frac{d\ln Q}{d\ln \eta }-2].$ Note as well that, $Q$ $=$ $[1-\frac{\eta \cdot \cos (\eta -b_0)}{\sin (\eta -b_0)}]$ $\Rightarrow \frac{dQ}{d\eta }$ $=$ $-\left[\frac{\cos (\eta -b_0)}{\sin (\eta -b_0)}\right]+\left[\frac{\eta \cdot \sin (\eta -b_0)}{\sin (\eta -b_0)}\right]+\left[\frac{\eta \cdot {\cos }^2(\eta -b_0)}{{\sin }^2(\eta -b_0)}\right]$   $=$ $\eta +\eta {\cot }^2(\eta -b_0)-\cot (\eta -b_0)$ $\Rightarrow \frac{d\ln Q}{d\ln \eta }$ $=$ $Q^{-1}[{\eta }^2+{\eta }^2{\cot }^2(\eta -b_0)-\eta \cot (\eta -b_0)]$   $=$ $Q^{-1}[{\eta }^2+(1-Q{)}^2+Q-1].$

While it is rather amazing that we have been able to identify this analytic solution to the LAWE, the solution seems troubling because it blows up at the surface, where, ${\eta }_s-b_0=\pi$. We will ignore this undesired behavior for the time being.

Transition at Interface

Here, as a numerical example, we will adopt the parameters that are relevant to Amodel2 from an associated discussion. For example, $({\mu }_e/{\mu }_c)=0.31$ and ${\xi }_i=9.0149598$.

Under "Attempt 1" of our accompanying discussion, we have shown that, at the core/envelope interface (note the following mappings:   $b\to 3c_0$ and $B\to b_0$),

${\eta }_i\cot ({\eta }_i-b_0)$	$=$	$1-\left(\frac{{\mu }_e}{{\mu }_c}\right)\left[\frac{3{\xi }_i^2}{3+{\xi }_i^2}\right]$
$\Rightarrow Q_i$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\left[\frac{3{\xi }_i^2}{3+{\xi }_i^2}\right]=0.8968919;$
${\eta }_i$	$=$	$3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\xi }_i[1+\frac{{\xi }^2}{3}{]}^{-1}=0.1723205$
	$=$	$3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right)\left[\frac{3{\xi }_i}{3+{\xi }^2}\right]=\frac{3^{1/2}{Q}_i}{{\xi }_i};$

and,

$3c_0$	$=$	$x_i{\eta }_i^2[1-{\eta }_i\cot ({\eta }_i-b_0){]}^{-1}$
	$=$	$\frac{3}{5}\left(\frac{{\mu }_e}{{\mu }_c}\right)\left[\frac{15-{\xi }_i^2}{3+{\xi }_i^2}\right];$
$b_0$	$=$	${\eta }_i-\frac{\pi }{2}+{\tan }^{-1}[\frac{1}{{\eta }_i}-\frac{{\xi }_i}{\sqrt{3}}]=-0.8592701.$

As viewed from the perspective of the envelope, then,

$[\frac{d\ln x_P}{d\ln \eta }{]}_i$	$=$	$[\frac{d\ln Q}{d\ln \eta }{]}_i-2$
	$=$	$Q_i^{-1}[{\eta }_i^2+(1-Q_i{)}^2+Q_i-1]-2$
	$=$	$-0.0700000-2=-2.0700000.$

As viewed from the perspective of the core, we have instead,

$[\frac{d\ln x_P}{d\ln \eta }{]}_i$	$=$	$3(\frac{{\gamma }_c}{{\gamma }_e}-1)+\frac{{\gamma }_c}{{\gamma }_e}[\frac{d\ln x}{d\ln \xi }{]}_i$
	$=$	$3(\frac{3}{5}-1)-\frac{3}{5}[\frac{2{\xi }_i^2}{15-{\xi }_i^2}{]}_i$
	$=$	$\frac{3}{5}\left\{\right[\frac{2{\xi }_i^2}{{\xi }_i^2-15}{]}_i-2\}=+0.2716182.$

Playing Around

Evidently, for our chosen example "Amodel2", $d\ln Q/d\ln \eta =-7/100$ exactly. How can this be?

See Also