# Continue Search for Marginally Unstable (5,1) Bipolytropes

This Ramblings Appendix chapter — see also, various trials — provides some detailed trial derivations in support of the accompanying, thorough discussion of this topic.

## Key Differential Equation

In an accompanying discussion, we derived the so-called,

 $~ \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x = 0$

whose solution gives eigenfunctions that describe various radial modes of oscillation in spherically symmetric, self-gravitating fluid configurations. After adopting an appropriate set of variable normalizations — as detailed here — this becomes,

 $~0$ $~=$ $~ \frac{d^2x}{dr*^2} + \biggl\{ 4 -\biggl(\frac{\rho^*}{P^*}\biggr)\frac{ M_r^*}{(r^*)}\biggr\}\frac{1}{r^*} \frac{dx}{dr*} + \biggl(\frac{\rho^*}{ P^* } \biggr)\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}} ~-~\frac{\alpha_\mathrm{g} M_r^*}{(r^*)^3}\biggr\} x \, ,$

where, $~\alpha_g \equiv (3 - 4/\gamma_g)$. Alternatively — see, for example, our introductory discussion — for polytropic configurations we may write,

 $~0$ $~=$ $~\frac{d^2x}{d\xi^2} + \biggl[4 - (n+1) \biggl(- \frac{d\ln \theta}{d\ln \xi} \biggr)\biggr] \frac{1}{\xi} \frac{dx}{d\xi} + \biggl\{ \frac{(n+1)}{\theta} \biggl[ \frac{\sigma_c^2}{6\gamma_g}\biggr] - (n+1) \biggl(- \frac{d\ln \theta}{d\ln \xi} \biggr) \frac{\alpha_g}{\xi^2 } \biggr\} x \, .$

### Applied to the Core

As we have already summarized in an accompanying discussion, throughout the core we have,

 $~r^*$ $~=$ $~\biggl( \frac{3}{2\pi} \biggr)^{1/2} \xi \, ;$ $~\frac{\rho^*}{P^*}$ $~=$ $~\biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2} \, ;$ $~\frac{M_r^*}{r^*}$ $~=$ $~ 2 \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3/2} \, .$

So the relevant core LAWE becomes,

 $~0$ $~=$ $~ \biggl( \frac{2\pi}{3} \biggr) \frac{d^2x}{d\xi^2} + \biggl( \frac{2\pi}{3} \biggr) \biggl\{ 4 - \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2} \biggl[ 2 \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr]\biggr\}\frac{1}{\xi} \frac{dx}{d\xi} + \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2}\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}} ~-~\biggl( \frac{2\pi}{3} \biggr)\frac{\alpha_\mathrm{g} }{\xi^2} \biggl[ 2 \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr] \biggr\} x$ $~\Rightarrow ~~~ \biggl( \frac{3}{4\pi} \biggr) \cdot 0$ $~=$ $~ \frac{1}{2}\cdot \frac{d^2x}{d\xi^2} + \biggl[ 2 - \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr] \frac{1}{\xi} \frac{dx}{d\xi} + \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2}\biggl[ \frac{\sigma_c^2}{2\gamma_\mathrm{g}} ~-~\alpha_\mathrm{g} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr] x \, .$

Now, following our separate discussion of an analytic solution to this LAWE, we try,

 $~x_P\biggr|_\mathrm{core}$ $~\equiv$ $~1 - \frac{\xi^2}{15}$ $~\Rightarrow~~~\frac{dx_P}{d\xi}\biggr|_\mathrm{core}$ $~\equiv$ $~- \frac{2\xi}{15}$ $~\Rightarrow~~~\frac{d\ln x_P}{d\ln \xi}\biggr|_\mathrm{core}$ $~\equiv$ $~- \frac{2\xi^2}{15} \biggl[ \frac{(15 - \xi^2)}{15} \biggr]^{-1} = - \frac{2\xi^2}{(15 - \xi^2)} \, .$

Plugging this trial function into the relevant LAWE gives,

 LAWE $~=$ $~ \frac{1}{2} \biggl( -\frac{2}{3\cdot 5}\biggr) + \biggl( -\frac{2}{3\cdot 5}\biggr)\biggl[ 2 - \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr] + \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2}\biggl[ \frac{\sigma_c^2}{2\gamma_\mathrm{g}} ~-~\alpha_\mathrm{g} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr] \biggl[1 - \frac{\xi^2}{15}\biggr]$ $~=$ $~ - \frac{1}{3} + \biggl( \frac{2}{3\cdot 5}\biggr)\biggl[ \xi^2 \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-1} \biggr] + \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{1 / 2}\biggl[ \frac{\sigma_c^2}{2\gamma_\mathrm{g}} ~-~\alpha_\mathrm{g} \biggl( 1 + \frac{1}{3}\xi^2 \biggr)^{-3 / 2} \biggr] \biggl[1 - \frac{\xi^2}{15}\biggr]$

Now, if we set $~\sigma_c^2 = 0$ and $~\gamma_g = \gamma_c = \tfrac{6}{5} ~~\Rightarrow ~~ \alpha_g = -1/3$, we find that the terms on the RHS sum to zero. It therefore appears that we have identified a dimensionless displacement function that satisfies the core LAWE.

### Applied to the Envelope

And as we have also summarized in the same accompanying discussion, throughout the envelope we have,

 $~r^*$ $~=$ $~\biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2}\eta \, ;$ $~\frac{\rho^*}{P^*}$ $~=$ $~ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi(\eta)^{-1} \, ;$ $~\frac{M_r^*}{r^*}$ $~=$ $~ 2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i \biggl(-\eta \frac{d\phi}{d\eta} \biggr) \, .$

So the relevant envelope LAWE becomes,

 $~\mathrm{LAWE}$ $~=$ $~ \frac{d^2x}{dr*^2} + \biggl\{ 4 -\biggl(\frac{\rho^*}{P^*}\biggr)\frac{ M_r^*}{r^*}\biggr\}\frac{1}{r^*} \frac{dx}{dr*} + \biggl(\frac{\rho^*}{ P^* } \biggr)\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}} ~-~\frac{\alpha_\mathrm{g} M_r^*}{(r^*)^3}\biggr\} x$ $~=$ $~ \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2} \biggr]^{-2}\frac{d^2x}{d\eta^2} + \biggl\{ 4 - \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi^{-1}\biggr] \biggl[ 2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i \biggl(-\eta \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\}\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2} \biggr]^{-2}\frac{1}{\eta} \frac{dx}{d\eta}$ $~ + \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi^{-1}\biggr]\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}} ~-~\frac{\alpha_\mathrm{g}}{\eta^2} \biggl[ 2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i \biggl(-\eta \frac{d\phi}{d\eta} \biggr) \biggr] \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-2}_i (2\pi)^{-1/2} \biggr]^{-2}\biggr\} x$ $~\Rightarrow ~~~ \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{2} \theta^{4}_i (2\pi) \biggr]^{-1} \cdot~ \mathrm{LAWE}$ $~=$ $~ \frac{d^2x}{d\eta^2} + \biggl\{ 4 - \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi^{-1}\biggr] \biggl[ 2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i \biggl(-\eta \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\} \frac{1}{\eta} \frac{dx}{d\eta}$ $~ + \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr) \theta^{-1}_i \phi^{-1}\biggr]\biggl\{ \frac{2\pi \sigma_c^2}{3\gamma_\mathrm{g}}\biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{2} \theta^{4}_i (2\pi) \biggr]^{-1} ~-~\frac{\alpha_\mathrm{g}}{\eta^2} \biggl[ 2 \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta_i \biggl(-\eta \frac{d\phi}{d\eta} \biggr) \biggr] \biggr\} x$ $~=$ $~ \frac{d^2x}{d\eta^2} + \biggl\{ 4 - \biggl[ 2 \biggl(-\frac{d\ln \phi}{d\ln \eta} \biggr) \biggr] \biggr\} \frac{1}{\eta} \frac{dx}{d\eta} + \biggl\{ \frac{\sigma_c^2}{3\gamma_\mathrm{g}} \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-5}_i \phi^{-1}\biggr] ~-~\frac{\alpha_\mathrm{g}}{\eta^2} \biggl[ 2 \biggl(- \frac{d\ln \phi}{d\ln \eta} \biggr) \biggr] \biggr\} x$ $~=$ $~ \frac{d^2x}{d\eta^2} + \biggl\{ 4 - 2Q_\eta \biggr\} \frac{1}{\eta} \frac{dx}{d\eta} + \biggl\{ \frac{\sigma_c^2}{3\gamma_\mathrm{g}} \biggl[ \biggl( \frac{\mu_e}{\mu_c} \biggr)^{-1} \theta^{-5}_i \phi^{-1}\biggr] ~-~(2Q_\eta)\frac{\alpha_\mathrm{g}}{\eta^2} \biggr\} x$

where,

 $~\phi(\eta)$ $~=$ $~\frac{A\sin(\eta - B)}{\eta}$ and $~Q_\eta$ $~\equiv$ $~- \frac{d\ln \phi}{d\ln\eta} = \biggl[1 - \eta\cot(\eta-B) \biggr] \, .$

If we set $~\sigma_c^2 = 0$ and $~\gamma_g = \gamma_e = 2 ~~\Rightarrow ~~ \alpha_g = +1$, the envelope LAWE simplifies to the form,

 $~ \biggl(\frac{r^*}{\eta}\biggr)^2 \cdot~ \mathrm{LAWE}$ $~=$ $~ \frac{d^2x}{d\eta^2} + \biggl\{ 4 - 2Q_\eta \biggr\} \frac{1}{\eta} \frac{dx}{d\eta} - \biggl\{ \frac{2Q_\eta}{\eta^2} \biggr\} x \, .$

In yet another Ramblings Appendix derivation we have explored a trial dimensionless displacement for the envelope of the form,

 $~x_P\biggr|_\mathrm{env}$ $~= \frac{3c_0}{\eta^2} \cdot Q_\eta \, .$

In this case,

 $~\frac{1}{3c_0}\cdot \frac{dx_P}{d\eta}$ $~=$ $~\frac{1}{\eta^2} \frac{dQ_\eta}{d\eta} - \frac{2Q_\eta}{\eta^3}$ $~=$ $~ \frac{1}{\eta^2}\biggl[\eta -\cot(\eta - B) +\eta\cot^2(\eta - B) \biggr] - \frac{2Q_\eta}{\eta^3}$ $~=$ $~ \frac{1}{\eta^3 \sin^2(\eta-B)} \biggl[\eta^2 + \eta\sin(\eta-B)\cos(\eta-B) -2\sin^2(\eta-B)\biggr]$ $~=$ $~ \frac{1}{\eta\sin^2(\eta - B)} + \frac{\cot(\eta-B)}{\eta^2} - \frac{2}{\eta^3} \, ;$ $~\frac{1}{3c_0}\cdot \frac{d^2x_P}{d\eta^2}$ $~=$ $~ \frac{d}{d\eta}\biggl[\frac{1}{\eta\sin^2(\eta - B)} + \frac{\cot(\eta-B)}{\eta^2} - \frac{2}{\eta^3} \biggr]$ $~=$ $~ \frac{6}{\eta^4} - \frac{2\cot(\eta-B)}{\eta^3} - \frac{2}{\eta^2\sin^2(\eta-B)} - \frac{2\cos(\eta-B)}{\eta \sin^3(\eta-B)} \, ,$

and it can be shown that the simplified envelope LAWE is perfectly satisfied. Notice that, with this adopted segment of the eigenfunction for the envelope, we have,

 $~\frac{d\ln x_P}{d\ln\eta}\biggr|_\mathrm{env} = \frac{\eta^3}{3c_0 Q_\eta}\cdot \frac{dx_P}{d\eta}$ $~=$ $~ \frac{\eta^3}{Q_\eta} \biggl\{ \frac{1}{\eta^2}\biggl[\eta -\cot(\eta - B) +\eta\cot^2(\eta - B) \biggr] - \frac{2Q_\eta}{\eta^3} \biggr\}$ $~=$ $~ \frac{1}{Q_\eta} \biggl[\eta^2 - \eta \cot(\eta - B) +\eta^2 \cot^2(\eta - B) \biggr] - 2$ $~=$ $~ \frac{[\eta^2 - 2 + \eta \cot(\eta - B)+\eta^2 \cot^2(\eta - B) ] }{[1 - \eta\cot(\eta-B)]}$ $~=$ $~ \frac{[\eta^2 - 2\sin^2(\eta-B) + \eta \sin(\eta-B) \cos(\eta - B) ] }{[\sin^2(\eta-B) - \eta \sin(\eta-B) \cos(\eta-B)]} \, .$

## Interface Matching

According to our accompanying discussion of the interface matching condition — as we presently understand it — the proper eigenfunction will exhibit a discontinuity in the slope of the dimensionless displacement function such that,

 $~\frac{d\ln x_\mathrm{env}}{d\ln \eta} \biggr|_{\eta=\eta_i}$ $~=$ $~3\biggl(\frac{\gamma_c}{\gamma_e} -1\biggr) + \frac{\gamma_c}{\gamma_e} \biggl( \frac{d\ln x_\mathrm{core}}{d\ln \xi} \biggr)_{\xi=\xi_i}$ $~=$ $~\frac{3}{5}\biggl[ \biggl( \frac{d\ln x_\mathrm{core}}{d\ln \xi} \biggr)_{\xi=\xi_i} -2 \biggr] \, .$