SSC/Structure/BiPolytropes/Analytic51Renormalize

BiPolytrope with $n_c=5$ and $n_e=1$

This chapter very closely parallels our original analytic derivation — see also, 📚 P. P. Eggleton, J. Faulkner, & R. C. Cannon (1998, MNRAS, Vol. 298, issue 3, pp. 831 - 834) — of the structure of bipolytropes in which the core has an $n_c=5$ polytropic index and the envelope has an $n_e=1$ polytropic index. Our primary objective, here, is to renormalize the principal set of variables, replacing the central density with the configuration's total mass, so that the mass is held fixed along each model sequence.

From Table 1 of our original analytic derivation, we see that,

$(\frac{{\mu }_e}{{\mu }_c}{)}^2{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}(\frac{K_c}{G}{)}^{3/2}{\rho }_0^{-1/5}$
$\Rightarrow {\rho }_0$	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^5,$

where,

${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$

$\equiv$

$\left(\frac{2}{\pi }{)}^{1/2}{\theta }_i^{-1}\right(-{\eta }^2\frac{d\varphi }{d\eta }{)}_s=(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}.$

Steps 2 & 3

Based on the discussion presented elsewhere of the structure of an isolated $n=5$ polytrope, the core of this bipolytrope will have the following properties:

$\theta (\xi )=[1+\frac{1}{3}{\xi }^2{]}^{-1/2}\Rightarrow {\theta }_i=[1+\frac{1}{3}{\xi }_i^2{]}^{-1/2};$

$\frac{d\theta }{d\xi }=-\frac{\xi }{3}[1+\frac{1}{3}{\xi }^2{]}^{-3/2}\Rightarrow (\frac{d\theta }{d\xi }{)}_i=-\frac{{\xi }_i}{3}[1+\frac{1}{3}{\xi }_i^2{]}^{-3/2}.$

The first zero of the function $\theta (\xi )$ and, hence, the surface of the corresponding isolated $n=5$ polytrope is located at ${\xi }_s=\mathrm{\infty }$. Hence, the interface between the core and the envelope can be positioned anywhere within the range, $0<{\xi }_i<\mathrm{\infty }$.

Step 4: Throughout the core ($0\le \xi \le {\xi }_i$)

Specify: $K_c$ and ${\rho }_0\Rightarrow$
$\rho$	$=$	${\rho }_0{\theta }^{n_c}$	$=$	${\rho }_0(1+\frac{1}{3}{\xi }^2{)}^{-5/2}$
$P$	$=$	$K_c{\rho }_0^{1+1/n_c}{\theta }^{n_c+1}$	$=$	$K_c{\rho }_0^{6/5}(1+\frac{1}{3}{\xi }^2{)}^{-3}$
$r$	$=$	$[\frac{(n_c+1)K_c}{4\pi G}{]}^{1/2}{\rho }_0^{(1-n_c)/(2n_c)}\xi$	$=$	$\left[\frac{K_c}{G{\rho }_0^{4/5}}{]}^{1/2}\right(\frac{3}{2\pi }{)}^{1/2}\xi$
$M_r$	$=$	$4\pi \left[\frac{(n_c+1)K_c}{4\pi G}{]}^{3/2}{\rho }_0^{(3-n_c)/(2n_c)}\right(-{\xi }^2\frac{d\theta }{d\xi })$	$=$	$\left[\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left[{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]$

---

Specify: $K_c$ and $M_{\mathrm{t}\mathrm{o}\mathrm{t}}\Rightarrow$

$\rho$	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^5\right(1+\frac{1}{3}{\xi }^2{)}^{-5/2}$
	$=$	$\left(\frac{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^5\right(\frac{K_c}{G}{)}^{15/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}\right(1+\frac{1}{3}{\xi }^2{)}^{-5/2};$
$P$	$=$	$K_c\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^6\right(1+\frac{1}{3}{\xi }^2{)}^{-3}$
	$=$	$\left(\frac{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^6{K}_c^{10}{G}^{-9}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}(1+\frac{1}{3}{\xi }^2{)}^{-3};$
$r$	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^{-2}\right[\frac{K_c}{G}{]}^{1/2}(\frac{3}{2\pi }{)}^{1/2}\xi$
	$=$	$\left(\frac{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{)}^{-2}\right(\frac{K_c}{G}{)}^{-5/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^4\right(\frac{3}{2\pi }{)}^{1/2}\xi ;$
$M_r$	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^{-1}\right[\frac{K_c^3}{G^3}{]}^{1/2}\left(\frac{2\cdot 3}{\pi }{)}^{1/2}\right[{\xi }^3(1+\frac{1}{3}{\xi }^2{)}^{-3/2}]$
	$=$	$\left(\frac{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}}\right)\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left[{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}].$

New Normalization $\tilde{\rho }$ $\equiv$ $\rho \left[\right(\frac{K_c}{G}{)}^{3/2}\frac{1}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{-5};$ $\tilde{P}$ $\equiv$ $P\left[K_c^{-10}{G}^9{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^6\right];$ $\tilde{r}$ $\equiv$ $r\left[\right(\frac{K_c}{G}{)}^{5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-2}],$ ${\tilde{M}}_r$ $\equiv$ $\frac{M_r}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}};$ $\tilde{H}$ $\equiv$ $H\left[K_c^{-5/2}{G}^{3/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}\right].$

After applying this new normalization, we have throughout the core,

$\tilde{\rho }$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}\right(1+\frac{1}{3}{\xi }^2{)}^{-5/2};$
$\tilde{P}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}\right(1+\frac{1}{3}{\xi }^2{)}^{-3};$
$\tilde{r}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^4\right(\frac{3}{2\pi }{)}^{1/2}\xi ;$
${\tilde{M}}_r$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\right(\frac{2\cdot 3}{\pi }{)}^{1/2}\left[{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}].$

Step 8: Throughout the envelope (${\eta }_i\le \eta \le {\eta }_s$)

Given (from above) that,

${\rho }_0$

$=$

$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^5,$

we have throughout the envelope,

$\rho$	$=$	${\rho }_0\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\varphi$
	$=$	$\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^5\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^5\varphi$
	$=$	$\left\{\right(\frac{K_c}{G}{)}^{15/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-5}\left\}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-9}{\theta }_i^5\varphi ;$
$P$	$=$	$K_c{\rho }_0^{6/5}{\theta }_i^6{\varphi }^2$
	$=$	$K_c\left\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{K_c}{G}{)}^{3/2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^6{\theta }_i^6{\varphi }^2$
	$=$	$\left\{K_c^{10}{G}^{-9}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-6}\right\}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\theta }_i^6{\varphi }^2;$
$r$	$=$	$\left[\frac{K_c}{G{\rho }_0^{4/5}}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta$
	$=$	$\left[\frac{K_c}{G}{]}^{1/2}\right\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\left(\frac{K_c}{G}{)}^{3/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^{-2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta$
	$=$	$\left\{\right(\frac{K_c}{G}{)}^{-5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^2\left\}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^3{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta ;$
$M_r$	$=$	$\left[\frac{K_c^3}{G^3{\rho }_0^{2/5}}{]}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$
	$=$	$\left[\frac{K_c^3}{G^3}{]}^{1/2}\right\{{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\left(\frac{K_c}{G}{)}^{3/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-1}{\}}^{-1}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\right(\frac{2}{\pi }{)}^{1/2}(-{\eta }^2\frac{d\varphi }{d\eta })$
	$=$	$M_{\mathrm{t}\mathrm{o}\mathrm{t}}{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta }).$

Adopting the new normalization then gives,

$\tilde{\rho }$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5(\frac{{\mu }_e}{{\mu }_c}{)}^{-9}{\theta }_i^5\varphi ;$
$\tilde{P}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^6(\frac{{\mu }_e}{{\mu }_c}{)}^{-12}{\theta }_i^6{\varphi }^2;$
$\tilde{r}$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}(\frac{{\mu }_e}{{\mu }_c}{)}^3{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta ;$
${\tilde{M}}_r$	$=$	${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}{\theta }_i^{-1}\left(\frac{2}{\pi }{)}^{1/2}\right(-{\eta }^2\frac{d\varphi }{d\eta }).$

Behavior of Central Density Along Equilibrium Sequence

Each equilibrium sequence will be defined as a sequence of models having the same jump in the mean-molecular weight, ${\mu }_e/{\mu }_c$. Along a given sequence, we vary the location of the core/envelope interface, ${\xi }_i$. Our desire is to analyze the behavior of the central density, while holding the total mass fixed, as the location of the interface is varied.

The central density is given by the expression,

${\tilde{\rho }}_c$

$=$

${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}\right[(1+\frac{1}{3}{\xi }^2{)}^{-5/2}{]}_{\xi =0}={{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5(\frac{{\mu }_e}{{\mu }_c}{)}^{-10},$

where,

${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$

$=$

$(\frac{2}{\pi }{)}^{1/2}\frac{A{\eta }_s}{{\theta }_i}.$

In order to evaluate ${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$ for a given specification of the interface location, ${\xi }_i$, we need to know that,

${\theta }_i$	$=$	$(1+\frac{1}{3}{\xi }_i^2{)}^{-1/2},$
${\eta }_i$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}{\theta }_i^2{\xi }_i,$
${\mathrm{\Lambda }}_i$	$=$	$\frac{{\xi }_i}{\sqrt{3}}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{1}{{\theta }_i^2{\xi }_i^2}-1],$
$A$	$=$	${\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2},$
${\eta }_s$	$=$	$\frac{\pi }{2}+{\eta }_i+{\tan }^{-1}({\mathrm{\Lambda }}_i).$

Keep in mind, as well, that,

$\nu \equiv \frac{M_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}{)}^2\sqrt{3}\right[\frac{{\xi }_i^3{\theta }_i^4}{A{\eta }_s}],$
$q\equiv \frac{r_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{R}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}\left[\frac{{\xi }_i{\theta }_i^2}{{\eta }_s}\right].$

Bipolytropic (5, 1) Equilibrium Sequences

Model Pairings

Here we work in the context of the B-KB74 conjecture. We will stick with the sequence corresponding to ${\mu }_e/{\mu }_c=0.31$, and continue to examine the model pairings (B1 and B2) associated with the degenerate model (A) at ${\nu }_{\mathrm{m}\mathrm{a}\mathrm{x}}$. Specifically …

file = Dropbox/WorkFolder/Wiki edits/Bipolytrope/Stability/qAndNuMax.xlsx --- worksheet = B-KB74 thru MinuPreparation Bipolytrope with $(n_c,n_e)=(5,1)$ Selected Pairings along the ${\mu }_e/{\mu }_c=0.31$ Sequence
Pairing	${\xi }_i$	${\mathrm{\Lambda }}_i$	$\nu$	$q$
A	$9.014959766$	$0.59835053$	$0.3372170064$	$0.0755022550$
B1	$9.12744$	$0.60069262$	$0.3372001445$	$0.0746451491$
B2	$8.90394$	$0.59610192$	$0.33720014467$	$0.0763642133$

Bipolytropic (5, 1) Equilibrium Sequences

Bipolytropic (5, 1) Equilibrium Sequences

Core
$m_r$	B1		B2		$\frac{\delta r}{r}=\frac{{\tilde{r}}_{\mathrm{B}\mathrm{1}}-{\tilde{r}}_{\mathrm{B}\mathrm{2}}}{2({\tilde{r}}_{\mathrm{B}\mathrm{1}}+{\tilde{r}}_{\mathrm{B}\mathrm{2}})}$
	$\xi$	$\tilde{r}(m_r)$	$\xi$	$\tilde{r}(m_r)$
0.0	0.0	0.0	0.0	0.0	--
0.005	0.430797	0.0007299	0.430395	0.0007331	-0.00109
0.05	1.054468	0.0017865	1.053194	0.0017938	-0.00102
0.1	1.502081	0.0025449	1.499761	0.0025544	-0.00093
0.15	1.963871	0.0033273	1.959917	0.0033382	-0.00082
0.20	2.5329785	0.0042915	2.525985	0.0043023	-0.00063
0.25	3.366385	0.0057034	3.352268	0.0057097	-0.00028
0.30	5.000525	0.0084721	4.959790	0.0084477	+0.00072
0.33715	9.11445	0.015442	8.89185	0.0151449	+0.00486
0.3372001	9.12744	0.015464	8.90394	0.0151654	+0.00487

Envelope
$m_r$	B1		B2		$\frac{\delta r}{r}=\frac{{\tilde{r}}_{\mathrm{B}\mathrm{1}}-{\tilde{r}}_{\mathrm{B}\mathrm{2}}}{2({\tilde{r}}_{\mathrm{B}\mathrm{1}}+{\tilde{r}}_{\mathrm{B}\mathrm{2}})}$
	$\eta$	$\tilde{r}(m_r)$	$\eta$	$\tilde{r}(m_r)$
0.3372001	0.1703455	0.015464	0.1743134	0.0151654	+0.00487
0.35	0.3073375	0.0279002	0.309463	0.0269236	+0.00891
0.40	0.5753765	0.0522328	0.576515	0.0501574	+0.01013
0.45	0.748189	0.0679208	0.749101	0.0651726	+0.01032
0.50	0.8885645	0.0806641	0.8893695	0.0773761	+0.01040
0.55	1.0122575	0.091893	1.012999	0.088132	+0.01045
0.60	1.126297	0.1022455	1.1269968	0.0980499	+0.01047
0.65	1.2347644	0.1120922	1.2354345	0.1074841	+0.01049
0.70	1.3405518	0.1216956	1.3411998	0.1166858	+0.01051
0.75	1.4461523	0.131282	1.4467833	0.1258716	+0.01052
0.80	1.5542198	0.1410924	1.5548378	0.1352725	+0.01053
0.85	1.6683004	0.1514487	1.668908	0.1451967	+0.01054
0.90	1.794487	0.1629039	1.7950862	0.1561743	+0.01055
0.95	1.94764	0.1768072	1.9482325	0.1694982	+0.01055
1.00	2.2820704	0.2071669	2.282658	0.1985936	+0.01056

Attempt at Constructing Analytic Eigenfunction Expression

Background

In our accompanying discussion of eigenvectors associated with the radial oscillation of pressure-truncated polytropes, we derived the following,

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

Drawing on the definition of $\theta (\xi )$ for n = 5 polytropes, as given in an accompanying chapter, we deduce that,

$x_P{|}_{n=5}$	$=$	$\frac{6}{5}[1+\frac{1}{2}(\frac{1}{\xi {\theta }^5})\frac{d\theta }{d\xi }{]}_{n=5}$
	$=$	$\frac{6}{5}-\frac{3}{5\xi }(1+\frac{{\xi }^2}{3}{)}^{5/2}\frac{\xi }{3}(1+\frac{{\xi }^2}{3}{)}^{-3/2}$
	$=$	$\frac{6}{5}-\frac{1}{5}(1+\frac{{\xi }^2}{3})$
	$=$	$1-\frac{{\xi }^2}{15}.$

And, given that for n = 1 polytropes,

$\theta (\xi )=\frac{\sin \xi }{\xi },$

we also find,

$x_P{|}_{n=1}$	$=$	$-3\left[\right(\frac{1}{\xi \theta })\frac{d\theta }{d\xi }{]}_{n=1}$
	$=$	$\frac{3}{\xi }\left(\frac{\xi }{\sin \xi }\right)[\frac{\sin \xi }{{\xi }^2}-\frac{\cos \xi }{\xi }]$
	$=$	$\frac{3}{{\xi }^2}[1-\xi \cot \xi ].$

Core

Allowing for an overall leading scale factor, $\alpha$, a viable displacement function for the $(n=5)$ core of our bipolytropic configuration is,

$\frac{x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{\alpha }$

$=$

$[1-\frac{{\xi }^2}{15}].$

Throughout the core, the corresponding Lagrangian radial coordinate, $\tilde{r}$, is given by the expression,

${\tilde{r}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\left(\frac{{\mu }_e}{{\mu }_c}{)}^4\right(\frac{3}{2\pi }{)}^{1/2}\xi .$

For "model A" the range is,

$0\le \xi \le {\xi }_i=9.0149598.$

SLOPE:  What is the slope of the function, $x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(\tilde{r})$, at the interface? $\frac{d{x}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{d\xi }{|}_i$ $=$ $-\frac{2\alpha {\xi }_i}{15},$ $\Rightarrow \frac{d{x}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{d\tilde{r}}{|}_i$ $=$ $-\frac{2\alpha {\xi }_i}{15}\left[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^4(\frac{3}{2\pi }{)}^{1/2}{]}^{-1}=-707.53765\alpha ,$ where, for "model A," we have set $({\mu }_e/{\mu }_c)=0.31$ and ${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}=1.938127063$. Note as well that, $\frac{d\ln (x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}})}{d\ln \tilde{r}}{|}_i=\frac{d\ln (x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}})}{d\ln \xi }{|}_i$ $=$ $-\frac{2{\xi }_i^2}{15}[1-\frac{{\xi }_i^2}{15}{]}^{-1}=+2.452697.$

Envelope

As we have demonstrated in a separate structure discussion, the radial profile of the $(n=1)$ envelope of our bipolytropic configuration is governed by the modified sinc-function,

$\varphi (\eta )$	$=$	$A\left[\frac{\sin (\eta -B)}{\eta }\right],$
$\Rightarrow -\frac{d\varphi }{d\eta }$	$=$	$\frac{A}{{\eta }^2}[\sin (\eta -B)-\eta \cos (\eta -B)].$

where, for "model A," $A=0.200812422$ and $B=-0.859270052$.

Again allowing for an overall leading scale factor, $\beta$, a viable displacement function for the $(n=1)$ envelope of our bipolytropic configuration is,

$\frac{x_{\mathrm{e}\mathrm{n}\mathrm{v}}}{3\beta }$	$=$	$\frac{1}{\eta \varphi }(-\frac{d\varphi }{d\eta })$
	$=$	$\frac{A}{{\eta }^3}[\sin (\eta -B)-\eta \cos (\eta -B)]\cdot \left[\frac{\eta }{A\sin (\eta -B)}\right]$
	$=$	$\frac{1}{{\eta }^2}[1-\eta \cot (\eta -B)].$

Throughout the envelope, the corresponding Lagrangian radial coordinate is,

${\tilde{r}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}(\frac{{\mu }_e}{{\mu }_c}{)}^3{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta .$

For "model A" the range is,

$({\eta }_i=0.1723205)\le \eta \le ({\eta }_s=2.282322601).$

SLOPE:  As we have detailed elsewhere, the slope of the function, $x_{\mathrm{e}\mathrm{n}\mathrm{v}}(\tilde{r})$, is related to the slope of $x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}(\tilde{r})$ at the interface via the expression, $\{\frac{d\ln x}{d\ln r}{|}_i{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}=\{\frac{d\ln x}{d\ln \eta }{|}_i{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ $=$ $3(\frac{{\gamma }_c}{{\gamma }_e}-1)+\frac{{\gamma }_c}{{\gamma }_e}\{\frac{d\ln x}{d\ln \xi }{|}_i{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}.$ In our case, ${\gamma }_c=6/5$ and ${\gamma }_e=2\Rightarrow {\gamma }_c/{\gamma }_e=3/5$. Hence, from the point of view of the envelope displacement function, at the interface, $[\frac{\tilde{r}}{x_{\mathrm{e}\mathrm{n}\mathrm{v}}}\cdot \frac{d{x}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{d\tilde{r}}{]}_i$ $=$ $\frac{3}{5}\left\{\right[\frac{d\ln (x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}})}{d\ln \xi }{]}_i-2\}$   $=$ $\frac{3}{5}\left\{\right[2.452697-2\}=0.271618.$ Now, at the interface of any bipolytrope, the ratio $\tilde{r}/x$ should have the same numerical value whether it is viewed from the point of view of the core or the envelope. Given that, for our particular "model A", $[\frac{\tilde{r}}{x}{]}_i=\frac{0.015315}{0.00485976}=3.15139,$ we should expect the slope of the envelope's displacement function at the interface to be, $\frac{d{x}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{d\tilde{r}}{|}_i$ $=$ $0.08619.$

Trial Displacement Function

The blue curve in the following figure results from plotting $x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ versus ${\tilde{r}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$ after setting the leading coefficient, $\alpha =-0.0011$. The red-dotted curve results from plotting $(x_{\mathrm{e}\mathrm{n}\mathrm{v}}+x_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}})$ versus ${\tilde{r}}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ after setting the leading coefficient, $\beta =-0.000062$, and $x_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}}=+0.0105$.

ASSESSMENT:

• Our analytically specified displacement function, $x_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$, appears to be an excellent match to the displacement function obtained throughout the core by implementing the B-KB74 conjecture.
• At first glance, the plot of $(x_{\mathrm{e}\mathrm{n}\mathrm{v}}+x_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}})$ appears to provide a reasonably good fit to the approximate displacement function that we have obtained throughout the envelope by implementing the B-KB74 conjecture. But, in reality, there are two fatal flaws:
  1. We have presented the behavior of our analytically specified envelope displacement function only up to the radial coordinate, $\eta =2.19707({\tilde{r}}_{\mathrm{e}\mathrm{n}\mathrm{v}}=0.19526)$. Between this point and the surface, ${\eta }_s=2.2823226({\tilde{r}}_{\mathrm{e}\mathrm{n}\mathrm{v}}=0.2028415)$ — where the argument of the cotangent, $({\eta }_s-B)\to \pi$ — the analytic function dives steeply to negative infinity. This violently departs from the behavior derived via the B-KB74 conjecture.
  2. While our analytically specified displacement function, $x_{\mathrm{e}\mathrm{n}\mathrm{v}}$, satisfies the "n = 1" polytropic LAWE, this satisfaction is destroyed by adding $x_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}}$ to the displacement function.

Let's examine the slope of the displacement function at the interface. From the perspective of the core, our analytic prescription for the displacement function matches the K-BK74-derived displacement function very well. An analytic evaluation of the slope at the inferface — as derived above — gives,

$\frac{d{x}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{d\tilde{r}}{|}_i=-707.53765\alpha =+0.77829$.

The black-dashed line segment that appears in the following figure has this slope and goes through the point of intersection; it appears to be tangent to the analytic displacement function, as expected. Alternatively, the orange-dashed line segment that appears in this same figure, also goes through the point of intersection, but it has a slope that matches our expectation for the envelope's displacement function; that is, it has a slope as derived of,

$\frac{d{x}_{\mathrm{e}\mathrm{n}\mathrm{v}}}{d\tilde{r}}{|}_i=+0.08619$.

This orange-dashed line segment does not appear to lie tangent to the K-BK74-derived displacement function for the envelope.

Numerical Integration Through Envelope

In an effort to numerically determine the eigenfunction of the envelope, we will follow the procedure described in an accompanying stability analysis of pressure-truncated polytropes to integrate the $n=1$ envelope from the core/envelope interface to the surface. In a closely related chapter titled, Radial Oscillations of n = 1 Polytropic Spheres, we have tried to find analytic expressions for the eigenvector of marginally unstable configurations.

Setup

Continuous Form of LAWE

We begin by writing our generic version of the polytropic LAWE,

	$0=\frac{d^{2}x}{d{\xi }^2}+[4-(n+1)Q]\frac{1}{\xi }\cdot \frac{dx}{d\xi }+(n+1)\left[\right(\frac{{\sigma }_c^2}{6{\gamma }_g})\frac{{\xi }^2}{\theta }-\alpha Q]\frac{x}{{\xi }^2}$
where:    $Q(\xi )\equiv -\frac{d\ln \theta }{d\ln \xi },$    ${\sigma }_c^2\equiv \frac{3{\omega }^2}{2\pi G{\rho }_c},$     and,     $\alpha \equiv (3-\frac{4}{{\gamma }_{\mathrm{g}}})$

then focus on the $n=1$ case — setting ${\gamma }_g=1+1/n=2$ and $\alpha =+1$ — the relevant LAWE becomes,

$0$

$=$

$\frac{d^{2}x}{d{\eta }^2}+\{4-2Q\}\frac{1}{\eta }\cdot \frac{dx}{d\eta }+2\left\{\right(\frac{{\sigma }_c^2}{12})\frac{{\eta }^2}{\varphi }-Q\}\frac{x}{{\eta }^2},$

where, in the broadest context,

$\varphi (\eta )$	$=$	$A\left[\frac{\sin (\eta -B)}{\eta }\right],$
$-\frac{d\varphi }{d\eta }$	$=$	$\frac{A}{{\eta }^2}[\sin (\eta -B)-\eta \cos (\eta -B)],$
$\Rightarrow Q(\eta )\equiv -\frac{d\ln \varphi }{d\ln \eta }$	$=$	$\frac{A}{{\eta }^2}[\sin (\eta -B)-\eta \cos (\eta -B)]\cdot \frac{{\eta }^2}{A\sin (\eta -B)}$
	$=$	$[1-\eta \cot (\eta -B)].$

For the $n=1$ LAWE we therefore have,

$0$	$=$	$\frac{d^{2}x}{d{\eta }^2}+\{4-2[1-\eta \cot (\eta -B)\left]\right\}\frac{1}{\eta }\cdot \frac{dx}{d\eta }+2\left\{\right(\frac{{\sigma }_c^2}{12})\frac{{\eta }^3}{A\sin (\eta -B)}-[1-\eta \cot (\eta -B)\left]\right\}\frac{x}{{\eta }^2}$
	$=$	$\frac{d^{2}x}{d{\eta }^2}+2\{1+\eta \cot (\eta -B)\}\frac{1}{\eta }\cdot \frac{dx}{d\eta }+2\left\{\right(\frac{{\sigma }_c^2}{12})\frac{{\eta }^3}{A\sin (\eta -B)}-1+\eta \cot (\eta -B)\}\frac{x}{{\eta }^2}.$

Multiplying through by $\varphi$, we can write,

$\left[\frac{A\sin (\eta -B)}{\eta }\right]\frac{d^{2}x}{d{\eta }^2}$

$=$

$-\frac{2A}{\eta }\{\sin (\eta -B)+\eta \cos (\eta -B)\}\frac{1}{\eta }\cdot \frac{dx}{d\eta }+\frac{2A}{\eta }\{\sin (\eta -B)-\eta \cos (\eta -B)\}\frac{x}{{\eta }^2}-2\left(\frac{{\sigma }_c^2}{12}\right)x.$

Discrete Form of LAWE

In order to integrate this 2^nd-order ODE numerically, we will build from the more general expression for polytropes used in our separate development of a finite-difference scheme, namely,

${\theta }_i{x_i}^{″}$

$=$

$-[4{\theta }_i-(n+1){\xi }_i(-{\theta }^{\text{'}}{)}_i]\frac{{x_i}^{\prime }}{{\xi }_i}-(n+1)[\frac{{\sigma }_c^2}{6{\gamma }_g}-\frac{\alpha }{{\xi }_i}(-{\theta }^{\text{'}}{)}_i]x_i.$

Making the notation substitutions, $(\xi ,\theta )\to (\eta ,\varphi )$, we have instead,

${\varphi }_i{x_i}^{″}$

$=$

$-[4{\varphi }_i-(n+1){\eta }_i(-{\varphi }^{\text{'}}{)}_i]\frac{{x_i}^{\prime }}{{\eta }_i}-(n+1)[\frac{{\sigma }_c^2}{6{\gamma }_g}-\frac{\alpha }{{\eta }_i}(-{\varphi }^{\text{'}}{)}_i]x_i.$

Now, adopting the finite-difference expressions,

${x_i}^{\prime }$	$\approx$	$\frac{x_{+}-x_{-}}{2{\mathrm{\Delta }}_{\eta }},$    and,
${x_i}^{″}$	$\approx$	$\frac{x_{+}-2x_i+x_{-}}{{\mathrm{\Delta }}_{\eta }^2},$

the discrete form of the LAWE becomes,

$x_{+}\stackrel{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{1}}{\overbrace{[2{\varphi }_i+\frac{4{\mathrm{\Delta }}_{\eta }{\varphi }_i}{{\eta }_i}-{\mathrm{\Delta }}_{\eta }(n+1)(-{\varphi }^{\text{'}}{)}_i]}}$

$=$

$x_{-}\stackrel{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{2}}{\overbrace{[\frac{4{\mathrm{\Delta }}_{\eta }{\varphi }_i}{{\eta }_i}-{\mathrm{\Delta }}_{\eta }(n+1)(-{\varphi }^{\text{'}}{)}_i-2{\varphi }_i]}}+x_i\stackrel{\mathrm{T}\mathrm{E}\mathrm{R}\mathrm{M}\mathrm{3}}{\overbrace{\{4{\varphi }_i-\frac{{\mathrm{\Delta }}_{\eta }^2(n+1)}{3}[\frac{{\sigma }_c^2}{{\gamma }_g}-2\alpha (-\frac{3{\varphi }^{\text{'}}}{\eta }{)}_i]\}}}.$

When applied specifically to an $n=1$, polytropic configuration, we should insert the following specific expressions:

${\gamma }_g$	$=$	$1+\frac{1}{n}=2,$
$\alpha$	$=$	$3-\frac{4}{{\gamma }_g}=+1,$
${\varphi }_i$	$=$	$A\left[\frac{\sin ({\eta }_i-B)}{{\eta }_i}\right],$
$(-{\varphi }^{\prime }{)}_i$	$=$	$\frac{A}{{\eta }_i^2}[\sin ({\eta }_i-B)-{\eta }_i\cos ({\eta }_i-B)].$

Pressure-Truncated n = 1 Polytrope

In the case of an isolated, pressure-truncated $n=1$ polytrope, we must set $B=0$; in addition, it is customary to set $A=1$. The relevant LAWE is, then,

$0$

$=$

$\frac{d^{2}x}{d{\eta }^2}+2\{1+\eta \cot (\eta )\}\frac{1}{\eta }\cdot \frac{dx}{d\eta }+2\left\{\right(\frac{{\sigma }_c^2}{12})\frac{{\eta }^3}{\sin (\eta )}-1+\eta \cot (\eta )\}\frac{x}{{\eta }^2}.$

Review of Trial Analytic Eigenfunction

This is the same 2^nd-order ODE that we derived in a separate discussion; there it was accompanied by the surface boundary condition,

$-\frac{d\ln x}{d\ln \xi }{|}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$	=	$\left(\frac{3-n}{n+1}\right)+\frac{n{\sigma }_c^2}{6(n+1)}[\frac{\xi }{{\theta }^{\prime }}{]}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$
$\Rightarrow -\frac{d\ln x}{d\ln \eta }{|}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$	=	$1+\frac{{\sigma }_c^2}{12}[\frac{{\eta }^3}{(\eta \cos \eta -\sin \eta )}{]}_{\eta =\pi }=1-\frac{{\pi }^2{\sigma }_c^2}{12}.$

From, for example, a separate succinct demonstration, we appreciate that if the displacement function is assumed to be,

$x_P$

$=$

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

… that is, $x_P$ $=$ $\frac{3}{{\eta }^2}-\frac{3\cos \eta }{\eta \sin \eta },$ in which case, $\frac{d{x}_P}{d\eta }$ $=$ $-\frac{6}{{\eta }^3}+3[\frac{\cos \eta }{{\eta }^2\sin \eta }+\frac{1}{\eta }+\frac{{\cos }^2\eta }{\eta {\sin }^2\eta }],$ and, $\frac{d^2{x}_P}{d{\eta }^2}$ $=$ $+\frac{18}{{\eta }^4}+3\frac{d}{d\eta }\left[\frac{\cos \eta }{{\eta }^2\sin \eta }\right]+3\frac{d}{d\eta }\left[\frac{1}{\eta }\right]+3\frac{d}{d\eta }\left[\frac{{\cos }^2\eta }{\eta {\sin }^2\eta }\right]$   $=$ $+\frac{18}{{\eta }^4}-3[\frac{2\cos \eta }{{\eta }^3\sin \eta }+\frac{{\cos }^2\eta }{{\eta }^2{\sin }^2\eta }+\frac{\sin \eta }{{\eta }^2\sin \eta }]-\left[\frac{3}{{\eta }^2}\right]-3[\frac{{\cos }^2\eta }{{\eta }^2{\sin }^2\eta }+\frac{2{\cos }^3\eta }{\eta {\sin }^3\eta }+\frac{2\cos \eta }{\eta \sin \eta }].$ Hence, LAWE $=$ $\{\frac{2}{\eta }+\frac{2\cos \eta }{\sin \eta }\}\{-\frac{6}{{\eta }^3}+[\frac{3\cos \eta }{{\eta }^2\sin \eta }+\frac{3}{\eta }+\frac{3{\cos }^2\eta }{\eta {\sin }^2\eta }\left]\right\}+2\left\{\right(\frac{{\sigma }_c^2}{12})\frac{{\eta }^3}{\sin (\eta )}-1+\frac{\eta \cos \eta }{\sin \eta }\}[\frac{3}{{\eta }^4}-\frac{3\cos \eta }{{\eta }^3\sin \eta }]$     $+\frac{18}{{\eta }^4}-3[\frac{2\cos \eta }{{\eta }^3\sin \eta }+\frac{{\cos }^2\eta }{{\eta }^2{\sin }^2\eta }+\frac{\sin \eta }{{\eta }^2\sin \eta }]-\left[\frac{3}{{\eta }^2}\right]-3[\frac{{\cos }^2\eta }{{\eta }^2{\sin }^2\eta }+\frac{2{\cos }^3\eta }{\eta {\sin }^3\eta }+\frac{2\cos \eta }{\eta \sin \eta }]$   $=$ $[\frac{2}{\eta }+2\cot \eta ][-\frac{6}{{\eta }^3}+\frac{3\cot \eta }{{\eta }^2}+\frac{3}{\eta }+\frac{3{\cot }^2\eta }{\eta }]+\left\{\right(\frac{{\sigma }_c^2}{6}\left)\frac{{\eta }^3}{\sin \eta }\right\}[\frac{3}{{\eta }^4}-\frac{3\cot \eta }{{\eta }^3}]+[-2+2\eta \cot \eta ][\frac{3}{{\eta }^4}-\frac{3\cot \eta }{{\eta }^3}]$     $+\frac{18}{{\eta }^4}-[\frac{6\cot \eta }{{\eta }^3}+\frac{3{\cot }^2\eta }{{\eta }^2}+\frac{3}{{\eta }^2}]-\left[\frac{3}{{\eta }^2}\right]-[\frac{3{\cot }^2\eta }{{\eta }^2}+\frac{6{\cot }^3\eta }{\eta }+\frac{6\cot \eta }{\eta }]$   $=$ $\left\{\right(\frac{{\sigma }_c^2}{2}\left)\frac{1}{\eta \sin \eta }\right\}[1-\eta \cot \eta ]+[-\frac{12}{{\eta }^4}+\frac{6}{{\eta }^2}]+[2\cot \eta ][\frac{3}{{\eta }^3}+\frac{3\cot \eta }{{\eta }^2}]+[2\cot \eta ][-\frac{6}{{\eta }^3}+\frac{3\cot \eta }{{\eta }^2}+\frac{3}{\eta }+\frac{3{\cot }^2\eta }{\eta }]$     $+[-\frac{6}{{\eta }^4}+\frac{6\cot \eta }{{\eta }^3}]+[2\cot \eta ][\frac{3}{{\eta }^3}-\frac{3\cot \eta }{{\eta }^2}]+\frac{18}{{\eta }^4}-\frac{6}{{\eta }^2}+[2\cot \eta ][-\frac{3{\cot }^2\eta }{\eta }-\frac{3\cot \eta }{{\eta }^2}-\frac{3}{\eta }-\frac{3}{{\eta }^3}]$   $=$ $\left\{\right(\frac{{\sigma }_c^2}{2}\left)\frac{1}{\eta \sin \eta }\right\}[1-\eta \cot \eta ]+[2\cot \eta ][-\frac{3}{{\eta }^3}+\frac{6\cot \eta }{{\eta }^2}+\frac{3}{\eta }+\frac{3{\cot }^2\eta }{\eta }]$     $+[2\cot \eta ][\frac{3}{{\eta }^3}-\frac{3{\cot }^2\eta }{\eta }-\frac{6\cot \eta }{{\eta }^2}-\frac{3}{\eta }]$   $=$ $\left\{\right(\frac{{\sigma }_c^2}{2}\left)\frac{1}{\eta \sin \eta }\right\}[1-\eta \cot \eta ].$

the $n=1$ LAWE reduces to …

LAWE

$=$

$\left\{\right(\frac{{\sigma }_c^2}{2}\left)\frac{1}{\eta \sin \eta }\right\}[1-\eta \cot \eta ].$

ASSESSMENT: 

1. If we set ${\sigma }_c^2=0$, the right-hand-side of this expression goes to zero — and, hence, the $n=1$ LAWE is satisfied — for any chosen truncation radius in the range, $0<{\eta }_i<\pi$. (We have not included the isolated $n=1$ polytrope because $x_P$ blows up at its surface, ${\eta }_i=\pi$.)
2. At the surface, ${\eta }_i$, the slope of this trial eigenfunction is,

   $\frac{1}{3}\cdot \frac{d{x}_P}{d\eta }{|}_i$

   $=$

   $[\frac{\cot {\eta }_i}{{\eta }_i^2}+\frac{1}{{\eta }_i}+\frac{{\cot }^2{\eta }_i}{{\eta }_i}]-\frac{2}{{\eta }_i^3}.$

   By contrast, as stated above, the eigenvalue problem will be properly solved only if the surface slope is,

   $-\frac{d\ln x}{d\ln \eta }{|}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$	=	$1+\frac{{\xcancel{{\sigma }_c^2}}^0}{12}[\frac{{\eta }^3}{(\eta \cos \eta -\sin \eta )}{]}_{\eta =\pi }=1$
   $\Rightarrow \frac{1}{3}\cdot \frac{d{x}_P}{d\eta }{|}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$	=	$-\frac{x_P}{3\eta }{|}_i$
   	=	$-\frac{1}{{\eta }_i^3}[1-{\eta }_i\cot {\eta }_i]=\frac{\cot {\eta }_i}{{\eta }_i^2}-\frac{1}{{\eta }_i^3}.$

   These two slopes do not appear to be the same, for any allowed choice of ${\eta }_i$. We conclude, therefore, that no model along the sequence of pressure-truncated $n=1$ polytropes is marginally unstable.

Determining Discrete Representation of Eigenfunction

Let's numerically integrate the discrete form of the $n=1$ LAWE over the radial coordinate range, $0\le {\eta }_i\le {\eta }_s$. Following our discussion of the more general polytropic case, we will kickstart integration from the center, outward, via the expression,

$x_2$

$=$

$x_1[1-\frac{(n+1)\mathfrak{𝔉}{\mathrm{\Delta }}_{\eta }^2}{60}],$

where,

$\mathfrak{𝔉}$

$\equiv$

$[\frac{{\sigma }_c^2}{{\gamma }_g}-2\alpha ].$

Here, we will restrict our investigation to the case where ${\gamma }_g=(n+1)/n=2$, in which case, $\alpha =(3-4/{\gamma }_g)=+1$, $\mathfrak{𝔉}=({\sigma }_c^2-4)/2$, and

$x_2$

$=$

$x_1[1-\frac{({\sigma }_c^2-4){\mathrm{\Delta }}_{\eta }^2}{60}].$

Isolated n = 1 Polytrope

If we integrate all the way out to the natural, zero-pressure surface of our $n=1$ polytrope, then ${\eta }_s=\pi$ and — as derived in our discussion of the equilibrium structure of n = 1 polytropes — $({\rho }_c/\overline{\rho })={\pi }^2/3$. In line with our discussion of Schwarzschild's model of oscillations in $n=3$ polytropes, we therefore expect the boundary condition at the surface of our $n=1$ configurations to be given by the expression,

$-\frac{d\ln x}{d\ln \eta }{|}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}$	$=$	$-\frac{1}{2}\left\{\right[\mathfrak{𝔉}+2\alpha \left]\right(\frac{{\rho }_c}{\overline{\rho }})-2\alpha \}$
	$=$	$1-\left(\frac{{\sigma }_c^2{\pi }^2}{12}\right),$

as reviewed immediately above.

Pressure-Truncated n = 1 Polytrope

Drawing from an accompanying discussion, if the polytropic configuration is truncated by the pressure, $P_e$, of a hot, tenuous external medium, then the solution to the LAWE is subject to the outer boundary condition,

$-\frac{d\ln x}{d\ln \eta }$

$=$

$3$     at     $\eta =\tilde{\eta }$.

Bipolytropic Envelope

For the $n=1$ envelope of a $(n_c,n_e)=(5,1)$ bipolytrope, the relevant LAWE is,

LAWE	$=$	$\frac{d^{2}x}{d{\eta }^2}$
		$+2\{1+\frac{\eta \cos (\eta -B)}{\sin (\eta -B)}\}\frac{1}{\eta }\cdot \frac{dx}{d\eta }$
		$+2\left\{\right(\frac{{\sigma }_c^2}{12})\frac{{\eta }^3}{A\sin (\eta -B)}-1+\frac{\eta \cos (\eta -B)}{\sin (\eta -B)}\}\frac{x}{{\eta }^2}.$

Three terms in this expression blow up at the surface, where $(\eta -B)\to \pi$ and, hence, $\sin (\eta -B)\to 0$. We can improve the behavior of this LAWE expression by assuming that the eigenfunction is of the form,

$x$

$=$

$f(\eta )[\sin (\eta -B){]}^m,$

in which case,

$\frac{dx}{d\eta }$	$=$	$[\sin (\eta -B){]}^m\frac{df}{d\eta }+mf(\eta )[\sin (\eta -B){]}^{m-1}\cos (\eta -B);$     and,
$\frac{d^{2}x}{d{\eta }^2}$	$=$	$[\sin (\eta -B){]}^m\frac{d^{2}f}{d{\eta }^2}+2m[\sin (\eta -B){]}^{m-1}\cos (\eta -B)\cdot \frac{df}{d\eta }+m(m-1)f(\eta )[\sin (\eta -B){]}^{m-2}{\cos }^{}(\eta -B)-mf(\eta )[\sin (\eta -B){]}^{m-1}\sin (\eta -B).$

This gives,

LAWE	$=$	$[\sin (\eta -B){]}^m\frac{d^{2}f}{d{\eta }^2}+2m[\sin (\eta -B){]}^{m-1}\cos (\eta -B)\cdot \frac{df}{d\eta }+m(m-1)f(\eta )[\sin (\eta -B){]}^{m-2}{\cos }^{}(\eta -B)-mf(\eta )[\sin (\eta -B){]}^{m-1}\sin (\eta -B)$
		$+2\{1+\frac{\eta \cos (\eta -B)}{\sin (\eta -B)}\}\frac{1}{\eta }\cdot \left\{\right[\sin (\eta -B){]}^m\frac{df}{d\eta }+mf(\eta )[\sin (\eta -B){]}^{m-1}\cos (\eta -B)\}$
		$+2\left\{\right(\frac{{\sigma }_c^2}{12})\frac{\eta }{A\sin (\eta -B)}-\frac{1}{{\eta }^2}+\frac{\cos (\eta -B)}{\eta \sin (\eta -B)}\}f(\eta )[\sin (\eta -B){]}^m$
	$=$	$[\sin (\eta -B){]}^m\frac{d^{2}f}{d{\eta }^2}+2\{\frac{1}{\eta }+\frac{\cos (\eta -B)}{\sin (\eta -B)}\left\}\right[\sin (\eta -B){]}^m\frac{df}{d\eta }+2m[\sin (\eta -B){]}^{m-1}\cos (\eta -B)\cdot \frac{df}{d\eta }$
		$+m(m-1)[\sin (\eta -B){]}^{m-2}{\cos }^{}(\eta -B)\cdot f(\eta )-m[\sin (\eta -B){]}^{m-1}\sin (\eta -B)\cdot f(\eta )$
		$+2m\{\frac{1}{\eta }+\frac{\cos (\eta -B)}{\sin (\eta -B)}\}[\sin (\eta -B){]}^{m-1}\cos (\eta -B)\cdot f(\eta )+2\{\left(\frac{{\sigma }_c^2}{12}\right)\frac{\eta }{A\sin (\eta -B)}-\frac{1}{{\eta }^2}+\frac{\cos (\eta -B)}{\eta \sin (\eta -B)}\left\}\right[\sin (\eta -B){]}^{m}f(\eta )$
	$=$	$[\sin (\eta -B){]}^m\frac{d^{2}f}{d{\eta }^2}+2\{\frac{1}{\eta }[\sin (\eta -B)]+(m+1)\cos (\eta -B)\left\}\right[\sin (\eta -B){]}^{m-1}\frac{df}{d\eta }$
		$+\{m(m-1){\cos }^2(\eta -B)-m[\sin (\eta -B){]}^2\left\}\right[\sin (\eta -B){]}^{m-2}\cdot f(\eta )$
		$+\left\{\frac{2m}{\eta }\right[\sin (\eta -B)]\cos (\eta -B)+2m{\cos }^{}(\eta -B)+[\left(\frac{{\sigma }_c^2}{6}\right)\frac{\eta }{A}\left]\right[\sin (\eta -B)]-\frac{2}{{\eta }^2}[\sin (\eta -B){]}^2+\frac{2\cos (\eta -B)}{\eta }[\sin (\eta -B)]\left\}\right[\sin (\eta -B){]}^{m-2}\cdot f(\eta ).$

Try m = 1 and m = 2

If we set $m=1$, there are still terms in the LAWE expression that blow up at the surface, where $(\eta -B)\to \pi$ and, hence, $\sin (\eta -B)\to 0$. Instead, let's try $m=2$:

$\mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}{|}_{m=2}$	$=$	$[\sin (\eta -B){]}^2\frac{d^{2}f}{d{\eta }^2}+2\{\frac{1}{\eta }[\sin (\eta -B)]+3\cos (\eta -B)\left\}\right[\sin (\eta -B)]\frac{df}{d\eta }$
		$+\{2{\cos }^2(\eta -B)-2[\sin (\eta -B){]}^2\}\cdot f(\eta )$
		$+\left\{\frac{4}{\eta }\right[\sin (\eta -B)]\cos (\eta -B)+4{\cos }^{}(\eta -B)+[\left(\frac{{\sigma }_c^2}{6}\right)\frac{\eta }{A}\left]\right[\sin (\eta -B)]-\frac{2}{{\eta }^2}[\sin (\eta -B){]}^2+\frac{2\cos (\eta -B)}{\eta }[\sin (\eta -B)]\}\cdot f(\eta ),$

which, at the surface $\eta \to {\eta }_s=(\pi +B)$, reduces to …

$\{\mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}{|}_{m=2}{\}}_{{\eta }_s}$

$=$

$6f({\eta }_s).$

Hence, this LAWE will be satisfied for any function, $f(\eta )$, that goes to zero at the surface.

Try m = 3

Setting $m=3$, we obtain,

$\mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}{|}_{m=3}$	$=$	$[\sin (\eta -B){]}^3\frac{d^{2}f}{d{\eta }^2}+2\{\frac{1}{\eta }[\sin (\eta -B)]+4\cos (\eta -B)\left\}\right[\sin (\eta -B){]}^2\frac{df}{d\eta }$
		$+\{6{\cos }^2(\eta -B)-3[\sin (\eta -B){]}^2\left\}\right[\sin (\eta -B)]\cdot f(\eta )$
		$+\left\{\frac{6}{\eta }\right[\sin (\eta -B)]\cos (\eta -B)+6{\cos }^{}(\eta -B)+[\left(\frac{{\sigma }_c^2}{6}\right)\frac{\eta }{A}\left]\right[\sin (\eta -B)]-\frac{2}{{\eta }^2}[\sin (\eta -B){]}^2+\frac{2\cos (\eta -B)}{\eta }[\sin (\eta -B)]\left\}\right[\sin (\eta -B)]\cdot f(\eta ).$

which trivially reduces to zero at the surface because, $\eta \to {\eta }_s=(\pi +B)\Rightarrow \sin (\eta -B)\to 0$. For all other relevant radial positions in the envelope, ${\eta }_i\le \eta <{\eta }_s$, we can divide through by ${\sin }^3(\eta -B)$ to obtain,

$[\sin (\eta -B){]}^{-3}\times \mathrm{L}\mathrm{A}\mathrm{W}\mathrm{E}{|}_{m=3}$

$=$

$\frac{d^{2}f}{d{\eta }^2}+2\{\frac{1}{\eta }+4\cot (\eta -B)\}\cdot \frac{df}{d\eta }+\{12{\cot }^{}(\eta -B)-3+\frac{8}{\eta }[\cot (\eta -B)]+[\left(\frac{{\sigma }_c^2}{6}\right)\frac{\eta }{A\sin (\eta -B)}]-\frac{2}{{\eta }^2}\}\cdot f(\eta ),$

Boundary Condition

In addition, there is a (boundary condition) constraint on the slope of the eigenfunction at the surface. So, let's examine …

$\frac{d\ln x}{d\ln \eta }{|}_{m=2}\equiv [\frac{\eta }{x}\cdot \frac{dx}{d\eta }{]}_{m=2}$	$=$	$\frac{\eta }{f(\eta ){\sin }^2(\eta -B)}\{{\sin }^2(\eta -B)\frac{df}{d\eta }+2f(\eta )\sin (\eta -B)\cos (\eta -B)\}$
	$=$	$\frac{d\ln f}{d\ln \eta }+2\eta \cot (\eta -B)$

Now, from above, we appreciate that when $\varphi =A\sin (\eta -B)/\eta$,

$\eta \cot (\eta -B)$	$=$	$1+\frac{d\ln \varphi }{d\ln \xi }$
$\Rightarrow \frac{d\ln x}{d\ln \eta }{|}_{m=2}$	$=$	$\frac{d\ln f}{d\ln \eta }+2[1+\frac{d\ln \varphi }{d\ln \xi }].$

It therefore appears as though we should adopt the function relation,

$\frac{d\ln f}{d\ln \eta }$

$=$

$-2\frac{d\ln \varphi }{d\ln \eta }$

$x$

$=$

$f(\eta )[\sin (\eta -B){]}^m,$

---

Let's now examine "model A" from above, for which, $A=0.200812422$ and $B=-0.859270052$. If we set ${\sigma }_c^2=0$, this LAWE becomes,

$0$

$=$

$\frac{d^{2}x}{d{\eta }^2}+[4-2Q]\frac{1}{\eta }\cdot \frac{dx}{d\eta }-2\left[\alpha Q\right]\frac{x}{{\eta }^2}.$