Appendix/Ramblings/51BiPolytropeStability/BetterInterface

Better Interface for 51BiPolytrope Stability Study

Content Pointing to Previous Work

Tilded Menu Pointers

1. Murphy & Fiedler (1985b): SSC/Stability/MurphyFiedler85
   1. Interface Conditions as promoted by Ledoux & Walraven (1958)
   2. Numerical Integration
      1. General Approach
      2. Special Handling at the Center
      3. Special Handling at the Interface
   3. Reconcile Approaches
2. Excellent Foundation (no pointer from Tiled Menu): SSC/Stability/Biipolytropes
3. Our Broader Analysis: SSC/Stability/BiPolytropes/HeadScratching
4. Succinct Discussion: SSC/Stability/BiPolytropes/SuccinctDiscussion

Ramblings: Analyzing Five-One Bipolytropes

1. Assessing the Stability of Spherical, BiPolytropic Configurations
2. Searching for Analytic EigenVector for (5,1) Bipolytropes
3. See (below) Discussing Patrick Motl's 2019 BiPolytrope Simulations
4. Continue Search
5. Renormalize Structure
6. Renormalize Structure (Part 2)
7. More Carefully Exam Step Function Behavior
8. More Focused Search for Analytic EigenVector if (5,1) Bipolytropes
9. Do Not Confine Search to Analytic Eigenvector
10. Clean, Methodical Examination
11. Rethink Handling of n = 1 Envelope
12. Improved Treatment of Core-Envelope Interface

Solid Foundation

Here we pull primarily from the chapters labeled II and III, above.

Entire Configuration

Beginning with the familiar,

Adiabatic Wave (or Radial Pulsation) Equation

$\frac{d^{2}x}{d{r}_0^2}+[\frac{4}{r_0}-(\frac{g_0{\rho }_0}{P_0}\left)\right]\frac{dx}{d{r}_0}+\left(\frac{{\rho }_0}{{\gamma }_{\mathrm{g}}{P}_0}\right)[{\omega }^2+(4-3{\gamma }_{\mathrm{g}})\frac{g_0}{r_0}]x=0$

where,

$g_0$

$=$

$\frac{G{M}_r^{*}}{(r^{*}{)}^2}\left[{\rho }_c^{3/5}\right(\frac{K_c}{G}{)}^{1/2}],$

if we adopt the variable normalizations,

${\rho }^{*}$	$\equiv$	$\frac{{\rho }_0}{{\rho }_c}$	;	$r^{*}$	$\equiv$	$\frac{r_0}{[K_c^{1/2}/(G^{1/2}{\rho }_c^{2/5})]}$
$P^{*}$	$\equiv$	$\frac{P_0}{K_c{\rho }_c^{6/5}}$	;	$M_r^{*}$	$\equiv$	$\frac{M_r}{[K_c^{3/2}/(G^{3/2}{\rho }_c^{1/5})]}$

the LAWE takes the form,

$0$

$=$

$\frac{d^{2}x}{dr{*}^2}+\frac{{\mathscr{H}}}{r^{*}}\frac{dx}{dr*}+\left[\right(\frac{{\sigma }_c^2}{{\gamma }_g}){{𝒦}}_1-{\alpha }_g{{𝒦}}_2]x,$

where,

${\mathscr{H}}$	$\equiv$	$\{4-(\frac{{\rho }^{*}}{P^{*}}\left)\frac{M_r^{*}}{(r^{*})}\right\}$	,	${{𝒦}}_1$	$\equiv$	$\frac{2\pi }{3}\left(\frac{{\rho }^{*}}{P^{*}}\right)$	and	${{𝒦}}_2$	$\equiv$	$\left(\frac{{\rho }^{*}}{P^{*}}\right)\frac{M_r^{*}}{(r^{*}{)}^3},$
${\sigma }_c^2$	$\equiv$	$\frac{3{\omega }^2}{2\pi G{\rho }_c}$	,	${\alpha }_g$	$\equiv$	$(3-\frac{4}{{\gamma }_g}).$

Core

Given that, in the core, ${\gamma }_g=6/5$ and,

$r^{*}$

$=$

$(\frac{3}{2\pi }{)}^{1/2}\xi ,$

we can rewrite the LAWE to read,

$0$

$=$

$\frac{d^{2}x}{d{\xi }^2}+\frac{{\mathscr{H}}}{\xi }\frac{dx}{d\xi }+\left(\frac{1}{4\pi }\right)[5{\sigma }_c^2{{𝒦}}_1+2{{𝒦}}_2]x,$

where,

${{𝒦}}_1$	$=$	$\frac{2\pi }{3}(1+\frac{1}{3}{\xi }^2{)}^{1/2},$
${\mathscr{H}}$	$=$	$4-2{\xi }^2(1+\frac{1}{3}{\xi }^2{)}^{-1},$
${{𝒦}}_2$	$=$	$\left(\frac{4\pi }{3}\right)(1+\frac{1}{3}{\xi }^2{)}^{-1}.$

Structure at the Interface

Once ${\mu }_e/{\mu }_c$ and ${\xi }_i$ have been specified, other parameter values at the interface are:

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

Linearized Perturbation at the Interface

At all radial locations throughout the equilibrium configuration, the three spatially dependent quantities — $p\equiv \delta p/P^{*},d\equiv \delta \rho /{\rho }^{*},$ and $x\equiv \delta r/r^{*}$ — are related to one another via the set of linearized governing relations, namely,

Linearized Equation of Continuity $r_0\frac{dx}{d{r}_0}=-3x-d,$ Linearized Euler + Poisson Equations $\frac{P_0}{{\rho }_0}\frac{dp}{d{r}_0}=(4x+p)g_0+{\omega }^2{r}_{0}x,$ Linearized Adiabatic Form of the First Law of Thermodynamics $p={\gamma }_{\mathrm{g}}d.$

Combining the 2^nd and 3^rd equations, we find,

$(4x+{\gamma }_{g}d)g_0+{\omega }^2{r}_{0}x$

$=$

$\frac{P_0}{{\rho }_0}\frac{d({\gamma }_{g}d)}{d{r}_0}$

---

At the interface, presumably the dimensional structural variables, $P^{*}$ and $r^{*}$ have the same values, whether viewed from the perspective of the core or from the perspective of the envelope. But ${\rho }^{*}$ has a different value, depending on the point of view. Specifically,

${\rho }^{*}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$\left(\frac{{\mu }_e}{{\mu }_c}\right){\rho }^{*}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}.$

Hence, from the perspective of the core, the linearized equation of continuity may be written as,

$[r_0\cdot \frac{dx}{d{r}_0}+3x{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$-\frac{\delta \rho }{[{\rho }^{*}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}};$

while, from the perspective of the envelope, the linearized equation of continuity may be written as,

$[r_0\cdot \frac{dx}{d{r}_0}+3x{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$-\frac{\delta \rho }{[{\rho }^{*}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}$

---

Try again

From here, we know …

$\frac{P}{{\rho }^{{\gamma }_g}}$

$=$

$\exp \left[\frac{s({\gamma }_g-1)}{\mathfrak{\Re }/\overline{\mu }}\right].$

And, from my discussions with Patrick Motl, we find …

CORE:   Throughout the core,

$P^{*}$

$=$

$(1+\frac{{\xi }^2}{3}{)}^{-3}$

and

${\rho }^{*}$

$=$

$(1+\frac{{\xi }^2}{3}{)}^{-5/2}$

and

${\gamma }_g$

$=$

$\frac{6}{5}.$

Hence, independent of the radial location, $\xi$, throughout the core,

$\frac{s}{\mathrm{\Re }/\overline{\mu }}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$5\ln (5).$

ENVELOPE:   Throughout the envelope,

$P^{*}$

$=$

${\theta }_i^6[\varphi (\eta ){]}^2$

and

${\rho }^{*}$

$=$

$\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5[\varphi (\eta )]$

and

${\gamma }_g$

$=$

$2.$

Hence, independent of the radial location, $\eta$, throughout the envelope,

$\frac{s}{\mathrm{\Re }/\overline{\mu }}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$\ln \left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}]$

$=$

$\ln \left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}(1+\frac{{\xi }_i^2}{3}{)}^2].$

Envelope

Given that, throughout the envelope ${\gamma }_g=2$ and,

$r^{*}$

$=$

$(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}(2\pi {)}^{-1/2}\eta ,$

we can rewrite the LAWE to read,

$0$

$=$

$\frac{d^{2}x}{d{\eta }^2}+\frac{{\mathscr{H}}}{\eta }\frac{dx}{d\eta }+\frac{1}{2\pi {\theta }_i^4}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\right[\left(\frac{{\sigma }_c^2}{2}\right){{𝒦}}_1-{{𝒦}}_2]x,$

where,

${{𝒦}}_1$	$=$	$\frac{2\pi }{3}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^{-1}\left[\frac{\eta }{A\sin (\eta -B)}\right],$
${\mathscr{H}}$	$=$	$2[1+\frac{\eta }{\tan (\eta -B)}],$
${{𝒦}}_2$	$=$	$4\pi \left(\frac{{\mu }_e}{{\mu }_c}{)}^2{\theta }_i^4\right[1-\frac{\eta }{\tan (\eta -B)}]\frac{1}{{\eta }^2}.$

Entropy as a Step Function

Useful Chapters:

• Appendix/Mathematics/Stepfunction
• Step-Function Behavior of Specific Entropy

Review

The unit — or, Heaviside — step function, $H(x)$, is defined such that,

$H(x)=\{\begin{array}{ll}0;& x<0\\ 1;& x>0\end{array}$ [MF53], Part I, §2.1 (p. 123), Eq. (2.1.6)

In evaluating this function at $x=0$, we will adopt the half-maximum convention and set $H(0)=\frac{1}{2}$. As has been pointed out in, for example, a relevant Wikipedia discussion, the derivative of the unit step function is,

$\frac{dH(x)}{dx}$

$=$

$\delta (x),$

where, $\delta (x)$ is the Dirac Delta function.

Perturbed Density

Let,

$\zeta$	$\equiv$	$\frac{r}{r_i}-1$
$\Rightarrow [{\rho }^{*}{]}_{\mathrm{e}\mathrm{q}}$	$\equiv$	$H(\zeta )\cdot {\rho }_{\mathrm{e}\mathrm{n}\mathrm{v}}^{*}+H(-\zeta )\cdot {\rho }_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}^{*};$

and, more generally after a perturbation, $\delta \rho (\zeta )$,

${\rho }^{*}=[{\rho }^{*}{]}_{\mathrm{e}\mathrm{q}}+\delta \rho$

$=$

$[{\rho }^{*}{]}_{\mathrm{e}\mathrm{q}}\{1+\frac{\delta \rho }{[{\rho }^{*}{]}_{\mathrm{e}\mathrm{q}}}\}=[{\rho }^{*}{]}_{\mathrm{e}\mathrm{q}}\{1+d{e}^{i\omega t}\}.$

Hence, in the linearized version of the continuity equation, we recognize that,

$d{e}^{i\omega t}$

$=$

$\frac{\delta \rho }{[{\rho }^{*}{]}_{\mathrm{e}\mathrm{q}}}.$

CORE:

${\rho }_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}^{*}$

$=$

$(1+\frac{{\xi }^2}{3}{)}^{-5/2}$

and

${\gamma }_g$

$=$

$\frac{6}{5}$

and

${{𝒮}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\equiv \frac{s({\gamma }_g-1)}{\mathrm{\Re }/\overline{\mu }}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$\ln (5).$

ENVELOPE:

${\rho }_{\mathrm{e}\mathrm{n}\mathrm{v}}^{*}$

$=$

$\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5[\varphi (\eta )]$

and

${\gamma }_g$

$=$

$2$

and

${{𝒮}}_{\mathrm{e}\mathrm{n}\mathrm{v}}\equiv \frac{s({\gamma }_g-1)}{\mathrm{\Re }/\overline{\mu }}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$

$=$

$\ln \left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}(1+\frac{{\xi }_i^2}{3}{)}^2].$

Perturbed Pressure

$P^{*}$

$=$

$({\rho }^{*}{)}^{{\gamma }_g}\cdot \exp [\frac{s({\gamma }_g-1)}{\mathrm{\Re }/\overline{\mu }}]$

$[P^{*}{]}_{\mathrm{e}\mathrm{q}}$

$\equiv$

$H(\zeta )\cdot P_{\mathrm{e}\mathrm{n}\mathrm{v}}^{*}+H(-\zeta )\cdot P_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}^{*};$

and, more generally after a perturbation, $\delta P(\zeta )$,

$P^{*}=[P^{*}{]}_{\mathrm{e}\mathrm{q}}+\delta P$

$=$

$[P^{*}{]}_{\mathrm{e}\mathrm{q}}\{1+\frac{\delta P}{[P^{*}{]}_{\mathrm{e}\mathrm{q}}}\}=[P^{*}{]}_{\mathrm{e}\mathrm{q}}\{1+p{e}^{i\omega t}\}.$

Hence, in the linearized version of the first law of thermodynamics, we recognize that,

$p{e}^{i\omega t}$

$=$

$\frac{\delta P}{[P^{*}{]}_{\mathrm{e}\mathrm{q}}}.$

Obtaining Perturbed Density from Perturbed Pressure

Given that, quite generally,

$\frac{P}{{\rho }^{{\gamma }_g}}$

$=$

$\exp \left[\frac{s({\gamma }_g-1)}{\mathfrak{\Re }/\overline{\mu }}\right],$

let's define the density-like quantity,

${{𝒬}}^{*}$

$\equiv$

$e^{-{𝒮}/{{\gamma }}_{{ℊ}}}[P^{*}{]}^{1/{\gamma }_g},$

in which case,

${\rho }_{\mathrm{e}\mathrm{q}}^{*}$

$\equiv$

$H(\zeta )\cdot [{{𝒬}}_{\mathrm{e}\mathrm{q}}^{*}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\cdot [{{𝒬}}_{\mathrm{e}\mathrm{q}}^{*}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}.$

What happens if we perturb the pressure? In either region (core or envelope),

${{𝒬}}^{*}$

$=$

$e^{-{𝒮}/{{\gamma }}_{{ℊ}}}[P_{\mathrm{e}\mathrm{q}}^{*}{]}^{1/{\gamma }_g}[1+\frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}{]}^{1/{\gamma }_g}\approx e^{-{𝒮}/{{\gamma }}_{{ℊ}}}[P_{\mathrm{e}\mathrm{q}}^{*}{]}^{1/{\gamma }_g}[1+\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}]=[Q_{\mathrm{e}\mathrm{q}}^{*}][1+\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}].$

As a result,

${\rho }^{*}={\rho }_{\mathrm{e}\mathrm{q}}^{*}+\delta \rho$	$\equiv$	$H(\zeta )\cdot [{{𝒬}}^{*}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\cdot [{{𝒬}}^{*}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$
	$=$	$H(\zeta )\cdot \{[Q_{\mathrm{e}\mathrm{q}}^{*}][1+\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}]{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\cdot \{[Q_{\mathrm{e}\mathrm{q}}^{*}][1+\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}]{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$
	$=$	$H(\zeta )\cdot [Q_{\mathrm{e}\mathrm{q}}^{*}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\cdot [Q_{\mathrm{e}\mathrm{q}}^{*}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$
		$+H(\zeta )\cdot \{[Q_{\mathrm{e}\mathrm{q}}^{*}][\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}]{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\cdot \{[Q_{\mathrm{e}\mathrm{q}}^{*}][\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}]{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$
	$=$	${\rho }_{\mathrm{e}\mathrm{q}}^{*}+H(\zeta )\cdot \{[Q_{\mathrm{e}\mathrm{q}}^{*}][\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}]{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\cdot \{[Q_{\mathrm{e}\mathrm{q}}^{*}][\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}]{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$
$\Rightarrow \frac{\delta \rho }{{\rho }_{\mathrm{e}\mathrm{q}}^{*}}$	$=$	$\frac{H(\zeta )}{{\rho }_{\mathrm{e}\mathrm{q}}^{*}}\cdot \{[Q_{\mathrm{e}\mathrm{q}}^{*}][\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}]{\}}_{\mathrm{e}\mathrm{n}\mathrm{v}}+\frac{H(-\zeta )}{{\rho }_{\mathrm{e}\mathrm{q}}^{*}}\cdot \{[Q_{\mathrm{e}\mathrm{q}}^{*}][\frac{1}{{\gamma }_g}\cdot \frac{\delta P}{P_{\mathrm{e}\mathrm{q}}^{*}}]{\}}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

Set of Linearized Equations

Borrowing from an accompanying discussion, we have …

Linearized Equation of Continuity $r_0\frac{dx}{d{r}_0}=-3x-d,$ Linearized Euler + Poisson Equations $\frac{P_0}{{\rho }_0}\frac{dp}{d{r}_0}=(4x+p)g_0+{\omega }^2{r}_{0}x,$ Linearized Adiabatic Form of the First Law of Thermodynamics $p={\gamma }_{\mathrm{g}}d.$

Rearranging terms in the "Linearized Euler + Poisson Equations" as follows …

$\frac{1}{{\rho }_0}\cdot \frac{dp}{d{r}_0}$

$=$

$\frac{1}{P_0}[(4x+p)g_0+{\omega }^2{r}_{0}x]$

we realize that the expression on the RHS has the same value at the interface, whether you're viewing the equation from the point of view of the core or the envelope; and we recognize as well that ${\rho }_0$ is a simple step function at the interface. Hence, letting a prime indicate differentiation with respect to $r_0$, we can write,

$H(\zeta )\cdot (p^{\prime }{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\cdot (p^{\prime }{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$\frac{[{\rho }_0{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{P_0}[(4x+p)g_0+{\omega }^2{r}_{0}x]\{H(\zeta )\cdot [\frac{{\mu }_e}{{\mu }_c}]+H(-\zeta )\}.$

Analogously, the "Linearized Equation of Continuity" can be rewritten as,

$H(\zeta )\cdot (x^{\prime }{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\cdot (x^{\prime }{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

$=$

$-\frac{3x}{r_0}-\frac{1}{r_0}\{H(\zeta )\cdot [\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right\}.$

Now, given that $\zeta =(r_0/r_i-1)$, we see that,

$\frac{dH(\zeta )}{d{r}_0}$

$=$

$\frac{\partial H(\zeta )}{\partial \zeta }\cdot \frac{d\zeta }{d{r}_0}=r_o^{-1}\delta (\zeta );$

and,

$\frac{dH(-\zeta )}{d{r}_0}$

$=$

$-\frac{\partial H(\zeta )}{\partial \zeta }\cdot \frac{d\zeta }{d{r}_0}=-r_o^{-1}\delta (\zeta ).$

Hence, differentiation of the "Linearized Equation of Continuity" gives,

$\frac{d}{d{r}_0}[H(\zeta )\cdot (x^{\prime }{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}]+\frac{d}{d{r}_0}[H(-\zeta )\cdot (x^{\prime }{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}]$	$=$	$-[\frac{3}{r_0}\cdot x^{\prime }-\frac{3x}{r_0^2}]-\frac{1}{r_0}\frac{d}{d{r}_0}\{H(\zeta )\cdot [\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right\}+\frac{1}{r_0^2}\{H(\zeta )\cdot [\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right\}.$
$\Rightarrow (x^{\prime }{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}\frac{d}{d{r}_0}[H(\zeta )]+H(\zeta )\cdot (x^{″}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}+(x^{\prime }{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\frac{d}{d{r}_0}[H(-\zeta )]+H(-\zeta )\cdot (x^{″}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$-[\frac{3}{r_0}\cdot x^{\prime }-\frac{3x}{r_0^2}]+\frac{1}{r_0^2}\{H(\zeta )\cdot [\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right\}.$
		$-\frac{1}{r_0}\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\frac{d}{d{r}_0}\right\{H(\zeta )\}-\frac{1}{r_0}\{H(\zeta )\cdot \left[\frac{p^{\prime }}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right\}-\frac{1}{r_0}\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\frac{d}{d{r}_0}\right\{H(-\zeta )\}-\frac{1}{r_0}\{H(-\zeta )\left[\frac{p^{\prime }}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right\}$
$\Rightarrow (x^{\prime }{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}[\frac{\delta (\zeta )}{r_0}]+H(\zeta )\cdot (x^{″}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}-(x^{\prime }{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}[\frac{\delta (\zeta )}{r_0}]+H(-\zeta )\cdot (x^{″}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$-[\frac{3}{r_0}\cdot x^{\prime }-\frac{3x}{r_0^2}]+\frac{1}{r_0^2}\{H(\zeta )\cdot [\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}+H(-\zeta )\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right\}.$
		$-\frac{1}{r_0}\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right\{\frac{\delta (\zeta )}{r_0}\}-\frac{1}{r_0}\{H(\zeta )\cdot \left[\frac{p^{\prime }}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right\}+\frac{1}{r_0}\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right\{\frac{\delta (\zeta )}{r_0}\}-\frac{1}{r_0}\{H(-\zeta )\left[\frac{p^{\prime }}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right\}$

From the Perspective of the Core

When $r_0/r_i\le 1$ — that is, from the perspective of the core while including the interface,

$-(x^{\prime }{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}[\frac{\delta (\zeta )}{r_0}]+H(-\zeta )\cdot (x^{″}{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$-[\frac{3}{r_0}\cdot x^{\prime }-\frac{3x}{r_0^2}]+\frac{1}{r_0^2}\{H(-\zeta )[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\}.$
		$-\frac{1}{r_0}\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right\{\frac{\delta (\zeta )}{r_0}\}+\frac{1}{r_0}[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\left\{\frac{\delta (\zeta )}{r_0}\right\}-\frac{1}{r_0}\{H(-\zeta )[\frac{p^{\prime }}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\}$

And examining only the interface, where $\delta (\zeta )=1$ while $H(-\zeta )=0$,

$-(x^{\prime }{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}[\frac{\delta (\zeta )}{r_0}]$	$=$	$-[\frac{3}{r_0}\cdot x^{\prime }-\frac{3x}{r_0^2}]-\frac{1}{r_0}\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right\{\frac{\delta (\zeta )}{r_0}\}+\frac{1}{r_0}[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\left\{\frac{\delta (\zeta )}{r_0}\right\}$
$\Rightarrow -(x^{\prime }{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$-[3x^{\prime }-\frac{3x}{r_0}]-\frac{1}{r_0}[\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}+\frac{1}{r_0}[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$

From the Perspective of the Envelope

When $r_0/r_i\ge 1$ — that is, from the perspective of the envelope while including the interface,

$(x^{\prime }{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}[\frac{\delta (\zeta )}{r_0}]+H(\zeta )\cdot (x^{″}{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$-[\frac{3}{r_0}\cdot x^{\prime }-\frac{3x}{r_0^2}]+\frac{1}{r_0^2}\{H(\zeta )\cdot [\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\}.$
		$-\frac{1}{r_0}\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right\{\frac{\delta (\zeta )}{r_0}\}-\frac{1}{r_0}\{H(\zeta )\cdot \left[\frac{p^{\prime }}{{\gamma }_g}{]}_{\mathrm{e}\mathrm{n}\mathrm{v}}\right\}+\frac{1}{r_0}\left[\frac{p}{{\gamma }_g}{]}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\right\{\frac{\delta (\zeta )}{r_0}\}$

Focus on Nonlinear Continuity Equation

A spherical shell of core density, ${\rho }_0$, where the inner radius of the shell is $(r_0-\mathrm{\Delta }r)$ and its outer radius is $(r_0+\mathrm{\Delta }r)$ has a shell mass given by the expression,

$(\mathrm{\Delta }m{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$\frac{4\pi }{3}\cdot {\rho }_0[(r_0+\mathrm{\Delta }r{)}^3-(r_0-\mathrm{\Delta }r{)}^3]$
	$=$	$\frac{4\pi }{3}\cdot {\rho }_0{r}_0^3\left[\right(1+\frac{\mathrm{\Delta }r}{r_0}{)}^3-(1-\frac{\mathrm{\Delta }r}{r_0}{)}^3]$
$\Rightarrow \left[\frac{3}{4\pi {\rho }_0{r}_0^3}\right](\mathrm{\Delta }m{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$[1+\frac{2\mathrm{\Delta }r}{r_0}+(\frac{\mathrm{\Delta }r}{r_0}{)}^2\left]\right(1+\frac{\mathrm{\Delta }r}{r_0})-[1-\frac{2\mathrm{\Delta }r}{r_0}+\left(\frac{\mathrm{\Delta }r}{r_0}{)}^2\right](1-\frac{\mathrm{\Delta }r}{r_0})$
	$=$	$[1+\frac{3\mathrm{\Delta }r}{r_0}+3(\frac{\mathrm{\Delta }r}{r_0}{)}^2+\left(\frac{\mathrm{\Delta }r}{r_0}{)}^3\right]-[1-\frac{3\mathrm{\Delta }r}{r_0}+3(\frac{\mathrm{\Delta }r}{r_0}{)}^2-\left(\frac{\mathrm{\Delta }r}{r_0}{)}^3\right]$
	$=$	$[\frac{6\mathrm{\Delta }r}{r_0}+2(\frac{\mathrm{\Delta }r}{r_0}{)}^3]$
$\Rightarrow (\mathrm{\Delta }m{)}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$	$=$	$\left[\frac{4\pi {\rho }_0{r}_0^3}{3}\right][\frac{6\mathrm{\Delta }r}{r_0}+2(\frac{\mathrm{\Delta }r}{r_0}{)}^3].$

Similarly, as viewed from the perspective of the envelope, it has a shell mass of,

$(\mathrm{\Delta }m{)}_{\mathrm{e}\mathrm{n}\mathrm{v}}$	$=$	$\frac{4\pi }{3}\cdot \left(\frac{{\mu }_e}{{\mu }_c}\right){\rho }_0[(r_0+\mathrm{\Delta }r{)}^3-(r_0-\mathrm{\Delta }r{)}^3]$
	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\left[\frac{4\pi {\rho }_0{r}_0^3}{3}\right][\frac{6\mathrm{\Delta }r}{r_0}+2(\frac{\mathrm{\Delta }r}{r_0}{)}^3].$

Better yet, pick the two edges of the shell, $r_{+}$ and $r_{-}$, and let $r_0\equiv (r_{+}+r_{-})/2$ and $\mathrm{\Delta }r=(r_{+}-r_{-})/2$. Given the value of ${\rho }_0$, the unperturbed mass in the shell is given by the above expression, that is,

$(\mathrm{\Delta }m{)}_0$

$=$

$\left[\frac{4\pi {\rho }_0{r}_0^3}{3}\right][\frac{6\mathrm{\Delta }r}{r_0}+2(\frac{\mathrm{\Delta }r}{r_0}{)}^3].$

Now, let $r_{+}\to (r_{+}{)}_0+(\delta r_{+}),r_{-}\to (r_{-}{)}_0+(\delta r_{-}),$ and, ${\rho }_0\to ({\rho }_0{)}_0+(\delta \rho )$. We then have,

$2r_0$	$=$	$[(r_{+}{)}_0+(\delta r_{+})]+[(r_{-}{)}_0+(\delta r_{-})]=2(r_0{)}_0+[\delta r_{+}+\delta r_{-}];$
$2\mathrm{\Delta }r$	$=$	$[(r_{+}{)}_0+(\delta r_{+})]-[(r_{-}{)}_0+(\delta r_{-})]=2(\mathrm{\Delta }r{)}_0+[\delta r_{+}-\delta r_{-}];$
$\Rightarrow \frac{\mathrm{\Delta }r}{r_0}$	$=$	$\frac{2(\mathrm{\Delta }r{)}_0+[\delta r_{+}-\delta r_{-}]}{2(r_0{)}_0+[\delta r_{+}+\delta r_{-}]}=\frac{(\mathrm{\Delta }r{)}_0}{(r_0{)}_0}\times \{1+[\frac{\delta r_{+}-\delta r_{-}}{2(\mathrm{\Delta }r{)}_0}\left]\right\}\{1+[\frac{\delta r_{+}+\delta r_{-}}{2(r_0{)}_0}]{\}}^{-1}$
	$\approx$	$\frac{(\mathrm{\Delta }r{)}_0}{(r_0{)}_0}\times \{1+[\frac{\delta r_{+}-\delta r_{-}}{2(\mathrm{\Delta }r{)}_0}\left]\right\}\{1-[\frac{\delta r_{+}+\delta r_{-}}{2(r_0{)}_0}\left]\right\}$
	$\approx$	$\frac{(\mathrm{\Delta }r{)}_0}{(r_0{)}_0}\times \{1+[\frac{\delta r_{+}-\delta r_{-}}{2(\mathrm{\Delta }r{)}_0}]-[\frac{\delta r_{+}+\delta r_{-}}{2(r_0{)}_0}\left]\right\}$

In order for the shell to have the same $\mathrm{\Delta }m$ in both the unperturbed and perturbed cases, the following relation must hold:

$\left[\frac{8\pi ({\rho }_0{)}_0(r_0{)}_0^3}{3}\right]\{\frac{3(\mathrm{\Delta }r{)}_0}{(r_0{)}_0}+[\frac{(\mathrm{\Delta }r{)}_0}{(r_0{)}_0}{]}^3\}$	$=$	$\left\{\frac{\pi [({\rho }_0{)}_0+\delta \rho ][2r_0{]}^3}{2\cdot 3}\right\}$
		$\times 2\{\frac{3\mathrm{\Delta }r}{r_0}+(\frac{\mathrm{\Delta }r}{r_0}{)}^3\}$

Work With Pair of First-Order Linearized Equations

Equilibrium Structures Using Preferred Normalizations

Working from our earlier "new" normalization — which was done in the context of our examination of the B-KB74 conjecture — that is, by setting,

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

and after adopting the notation,

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

$\equiv$

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

(see definitions of $A$, ${\eta }_s$, and ${\theta }_i$ given below) we have throughout the core,

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

For later use, note that, $\frac{\tilde{\rho }}{\tilde{P}}\cdot \frac{{\tilde{M}}_r}{{\tilde{r}}^2}$ $=$ $\left[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}(1+\frac{1}{3}{\xi }^2{)}^{-5/2}]\left[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-6}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{12}(1+\frac{1}{3}{\xi }^2{)}^3]$     $\times \left[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\right(\frac{{\mu }_e}{{\mu }_c}{)}^2\left(\frac{2\cdot 3}{\pi }{)}^{1/2}{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}\left]\right[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^4\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-8}\right(\frac{3}{2\pi }{)}^{-1}{\xi }^{-2}]$   $=$ ${{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^2\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-4}\right(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi (\frac{2^3\pi }{3}{)}^{1/2}$   $=$ $2\left[\frac{\xi }{\tilde{r}}\right](1+\frac{1}{3}{\xi }^2{)}^{-1}\xi$ Note as well that, $\frac{{\tilde{r}}^3}{{\tilde{M}}_r}$ $=$ $\left[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-1}\right(\frac{{\mu }_e}{{\mu }_c}{)}^2\left(\frac{2\cdot 3}{\pi }{)}^{1/2}{\xi }^3\right(1+\frac{1}{3}{\xi }^2{)}^{-3/2}{]}^{-1}\left[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^4(\frac{3}{2\pi }{)}^{1/2}\xi {]}^3$   $=$ $\left[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\left(\frac{4\cdot 3}{2\pi }{)}^{-1/2}{\xi }^{-3}\right(1+\frac{1}{3}{\xi }^2{)}^{3/2}\left]\right[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-6}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{12}\right(\frac{3}{2\pi }{)}^{3/2}{\xi }^3]$   $=$ $\left[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-5}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{10}\left(\frac{3}{4\pi }\right)(1+\frac{1}{3}{\xi }^2{)}^{3/2}]$

Similarly, throughout the envelope, we find,

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

where,

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

and,

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

Keep in mind, as well, that,

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

Linearized Equations With Preferred Normalizations

Review and Elaborate

Linearized Equation of Continuity $r_0\frac{dx}{d{r}_0}=-3x-d,$ Linearized Euler + Poisson Equations $\frac{P_0}{{\rho }_0}\frac{dp}{d{r}_0}=(4x+p)g_0+{\omega }^2{r}_{0}x,$ Linearized Adiabatic Form of the First Law of Thermodynamics $p={\gamma }_{\mathrm{g}}d.$

The LHS of the "linearized Euler + Poisson" equation is rewritten as,

$\mathrm{L}\mathrm{H}\mathrm{S}=\frac{P_0}{{\rho }_0}\frac{dp}{d{r}_0}$	$=$	$\left[K_c^{-10}{G}^9{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^6{]}^{-1}\tilde{P}\right[\left(\frac{K_c}{G}{)}^{3/2}\frac{1}{M_{\mathrm{t}\mathrm{o}\mathrm{t}}}{]}^{-5}{\tilde{\rho }}^{-1}\right[\left(\frac{K_c}{G}{)}^{5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-2}\right]\frac{dp}{d\tilde{r}}$
	$=$	$\frac{\tilde{P}}{\tilde{\rho }}\frac{dp}{d\tilde{r}}\left[K_c^{10}{G}^{-9}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-6}\right]\left[G^{15/2}{K}_c^{-15/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^5\right]\left[K_c^{5/2}{G}^{-5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-2}\right]$
	$=$	$\frac{\tilde{P}}{\tilde{\rho }}\frac{dp}{d\tilde{r}}\left[K_c^5{G}^{-4}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-3}\right];$

and,

$g_0=\frac{G{M}_r}{r_0^2}$	$=$	$G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}{\tilde{M}}_r\left[\right(\frac{K_c}{G}{)}^{5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-2}{]}^2{\tilde{r}}^{-2}$
	$=$	$\frac{{\tilde{M}}_r}{{\tilde{r}}^2}\left[\right(\frac{K_c}{G}{)}^{5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-2}{]}^{2}G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}$
	$=$	$\frac{{\tilde{M}}_r}{{\tilde{r}}^2}\left[K_c^5{G}^{-4}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-3}\right].$

Therefore, multiplying the full equation through by $[K_c^{-5}{G}^4{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^3]$ gives,

$\frac{\tilde{P}}{\tilde{\rho }}\frac{dp}{d\tilde{r}}$	$=$	$(4x+p)\frac{{\tilde{M}}_r}{{\tilde{r}}^2}+\left[K_c^{-5}{G}^4{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^3\right]\left[\right(\frac{K_c}{G}{)}^{5/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^{-2}{]}^{-1}{\omega }^2\tilde{r}x$
	$=$	$(4x+p)\frac{{\tilde{M}}_r}{{\tilde{r}}^2}+{\tau }_c^2{\omega }^2\tilde{r}x,$

where the square of the characteristic timescale,

${\tau }_c^2$

$\equiv$

$\left[K_c^{-15/2}{G}^{13/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^5\right].$

ASIDE:  Building on an associated discussion, the square of the dimensionless frequency also can be represented by the expression, ${\sigma }_c^2\equiv \frac{3{\omega }^2}{2\pi G{\rho }_c}$ $=$ $\frac{3{\omega }^2}{2\pi {\tilde{\rho }}_c}\left[\right(\frac{G}{K_c}{)}^{15/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^5]G^{-1}=\frac{3{\tau }_c^2{\omega }^2}{2\pi {\tilde{\rho }}_c},$ where, ${\tilde{\rho }}_c$ $=$ $m_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}.$

Hence we can write,

$\frac{dp}{d\tilde{r}}$	$=$	$\frac{\tilde{\rho }}{\tilde{P}}\cdot \frac{{\tilde{M}}_r}{{\tilde{r}}^2}[(4x+p)+{\tau }_c^2{\omega }^2(\frac{{\tilde{r}}^3}{{\tilde{M}}_r}\left)x\right]$
	$=$	$\frac{\tilde{\rho }}{\tilde{P}}\cdot \frac{{\tilde{M}}_r}{{\tilde{r}}^2}[(4x+p)+{\sigma }_c^2(\frac{2\pi }{3}\cdot \frac{{\tilde{\rho }}_c{\tilde{r}}^3}{{\tilde{M}}_r}\left)x\right].$

---

Focusing on the core …

As demonstrated earlier, the leading term on the RHS of this expression can be rewritten to give,

$\frac{dp}{d\tilde{r}}$	$=$	$2\left[\frac{\xi }{\tilde{r}}\right](1+\frac{1}{3}{\xi }^2{)}^{-1}\xi [(4x+p)+{\sigma }_c^2(\frac{2\pi }{3}\cdot \frac{{\tilde{\rho }}_c{\tilde{r}}^3}{{\tilde{M}}_r})x]$
$\Rightarrow \frac{dp}{d\xi }$	$=$	$2(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi [(4x+p)+{\sigma }_c^2(\frac{2\pi }{3}\cdot \frac{{\tilde{\rho }}_c{\tilde{r}}^3}{{\tilde{M}}_r})x].$

Noting as well that,

${\tilde{\rho }}_c\left(\frac{{\tilde{r}}^3}{{\tilde{M}}_r}\right)$

$=$

$m_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^5\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-10}\right[{{𝓂}}_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^{-5}\left(\frac{{\mu }_e}{{\mu }_c}{)}^{10}\right(\frac{3}{4\pi }\left)\right(1+\frac{1}{3}{\xi }^2{)}^{3/2}]=\frac{3}{4\pi }(1+\frac{1}{3}{\xi }^2{)}^{3/2},$

we have,

$\frac{dp}{d\xi }$

$=$

$2(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi [(4x+p)+\frac{{\sigma }_c^2}{2}(1+\frac{1}{3}{\xi }^2{)}^{3/2}x].$

At the Center

All σ²

According to our discussion in an appendix chapter, starting from the center of the equilibrium configuration, the displacement function can be represented by a power-series expression of the form,

$x$

$\approx$

$1-\left[\frac{(n+1)\mathfrak{𝔉}}{60}\right]{\xi }^2,$

where, $\mathfrak{𝔉}\equiv ({\sigma }_c^2/\gamma -2\alpha )$, and (see, for example, here),

$\xi$

$=$

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

Note that, at the center of our $(n_c,n_e)=(5,1)$ bipolytrope, ${\gamma }_g=6/5$, so $\alpha =-1/3$. Hence, for this particular investigation, the central boundary condition is,

$x$	$\approx$	$1-[\frac{{\sigma }_c^2}{12}+\frac{1}{15}]{\xi }^2$
	$\approx$	$1-[\frac{{\sigma }_c^2}{12}+\frac{1}{15}]\left[m_{\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{f}}^4\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-8}\left(\frac{2\pi }{3}\right)]{\tilde{r}}^2.$

Also, the derivative of this displacement function is,

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

$\approx$

$-2[\frac{{\sigma }_c^2}{12}+\frac{1}{15}]\xi .$

Hence,

$p=-{\gamma }_c[3x+\xi \cdot \frac{dx}{d\xi }]$	$\approx$	$-\frac{6}{5}\left\{3\right[1-(\frac{{\sigma }_c^2}{12}+\frac{1}{15}){\xi }^2]-2(\frac{{\sigma }_c^2}{12}+\frac{1}{15}\left){\xi }^2\right\}$
	$=$	$-\frac{18}{5}[1-\frac{5}{3}(\frac{{\sigma }_c^2}{12}+\frac{1}{15}\left){\xi }^2\right].$

Furthermore, we find,

$(4x+p)$	$=$	$4[1-(\frac{{\sigma }_c^2}{12}+\frac{1}{15}\left){\xi }^2\right]-\frac{18}{5}[1-\frac{5}{3}(\frac{{\sigma }_c^2}{12}+\frac{1}{15}\left){\xi }^2\right].$
	$=$	$[\frac{20}{5}-4(\frac{{\sigma }_c^2}{12}+\frac{1}{15}\left){\xi }^2\right]+[-\frac{18}{5}+6(\frac{{\sigma }_c^2}{12}+\frac{1}{15}\left){\xi }^2\right]$
	$=$	$\frac{2}{5}+2(\frac{{\sigma }_c^2}{12}+\frac{1}{15}){\xi }^2.$

Hence we have,

$\frac{dp}{d\xi }$	$=$	$2(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi \{\frac{2}{5}+2(\frac{{\sigma }_c^2}{12}+\frac{1}{15}){\xi }^2+\frac{{\sigma }_c^2}{2}(1+\frac{1}{3}{\xi }^2{)}^{3/2}[1-(\frac{{\sigma }_c^2}{12}+\frac{1}{15}){\xi }^2\left]\right\}$
	$=$	$2(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi \{24+2(5{\sigma }_c^2+4){\xi }^2+\frac{{\sigma }_c^2}{2}(1+\frac{1}{3}{\xi }^2{)}^{3/2}[60-(5{\sigma }_c^2+4){\xi }^2\left]\right\}\frac{1}{60}$
	$=$	$\frac{1}{30}(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi \{24+8{\xi }^2+10{\sigma }_c^2{\xi }^2+30{\sigma }_c^2(1+\frac{1}{3}{\xi }^2{)}^{3/2}-\frac{{\sigma }_c^2}{2}(1+\frac{1}{3}{\xi }^2{)}^{3/2}(4+5{\sigma }_c^2){\xi }^2\}$
	$=$	$\frac{1}{30}(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi \{[24+8{\xi }^2+10{\sigma }_c^2{\xi }^2]+(1+\frac{1}{3}{\xi }^2{)}^{3/2}[30{\sigma }_c^2-2{\sigma }_c^2{\xi }^2-\frac{5}{2}{\sigma }_c^4{\xi }^2\left]\right\}$
	$=$	$\frac{1}{15}(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi [12+4{\xi }^2+5{\sigma }_c^2{\xi }^2]+\frac{{\sigma }_c^2\xi }{30}(1+\frac{1}{3}{\xi }^2{)}^{1/2}[30-2{\xi }^2-\frac{5}{2}{\sigma }_c^2{\xi }^2]$

Just σ² = 0

If we set ${\sigma }_c^2=0$, we have,

$p$	$=$	$-\frac{18}{5}+\frac{2}{5}{\xi }^2$
$\Rightarrow \frac{dp}{d\xi }$	$=$	$\frac{4}{5}\xi .$

Alternatively, from immediately above,

$\frac{dp}{d\xi }{|}_{{\sigma }_c^2=0}$	$=$	$\frac{1}{15}(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi [12+4{\xi }^2+{\xcancel{5{\sigma }_c^2{\xi }^2}}^0]+{\xcancel{\frac{{\sigma }_c^2\xi }{30}}}^0(1+\frac{1}{3}{\xi }^2{)}^{1/2}[30-2{\xi }^2-\frac{5}{2}{\sigma }_c^2{\xi }^2]$
	$=$	$\frac{12}{15}(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi [1+\frac{1}{3}{\xi }^2]$
	$=$	$\frac{4}{5}\xi .$

Yes! It matches!

Summary

Moving from the center, outward thorough the core — that is, interior to the interface — we can assign values of $x(\xi )$ and $p(\xi )$ using the following approximate (exact if ${\sigma }_c^2=0$) relations:

$x$	$\approx$	$1-[\frac{{\sigma }_c^2}{12}+\frac{1}{15}]{\xi }^2;$
$p$	$\approx$	$-\frac{18}{5}[1-\frac{5}{3}(\frac{{\sigma }_c^2}{12}+\frac{1}{15}\left){\xi }^2\right].$

For all radial shells throughout the entire bipolytropic configuration, the pair of first derivatives can be evaluated using the following relations:

$\frac{dx}{d\tilde{r}}$	$=$	$-\frac{1}{\tilde{r}}[3x+\frac{p}{{\gamma }_g}];$
$\frac{dp}{d\tilde{r}}$	$=$	$\frac{\tilde{\rho }}{\tilde{P}}\cdot \frac{{\tilde{M}}_r}{{\tilde{r}}^2}[(4x+p)+{\sigma }_c^2(\frac{2\pi }{3}\cdot \frac{{\tilde{\rho }}_c{\tilde{r}}^3}{{\tilde{M}}_r}\left)x\right].$

Near the center, this pressure-derivative expression can be checked against the relation,

$\frac{dp}{d\xi }$

$=$

$\frac{1}{15}(1+\frac{1}{3}{\xi }^2{)}^{-1}\xi [12+4{\xi }^2+5{\sigma }_c^2{\xi }^2]+\frac{{\sigma }_c^2\xi }{30}(1+\frac{1}{3}{\xi }^2{)}^{1/2}[30-2{\xi }^2-\frac{5}{2}{\sigma }_c^2{\xi }^2];$

notice that, in order to make this comparison, you need to multiply this last expression through by the ratio,

$\frac{\xi }{\tilde{r}}$

$=$

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

The comparison should be especially accurate in the case of ${\sigma }_c^2=0$.

At the Interface

See below.

At the Surface

Drawing from a separate discussion, the surface boundary condition is,

$r_0\frac{dx}{d{r}_0}$

$=$

$(4-3{\gamma }_g+\frac{{\omega }^2{R}^3}{G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}})\frac{x}{{\gamma }_g}$        at         $r_0=R,$

that is,

$\frac{d\ln x}{d\ln \tilde{r}}{|}_s$

$=$

$-\alpha +\frac{{\omega }^2{R}^3}{{\gamma }_{g}G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}},$

where (see also, here),

$\alpha$

$\equiv$

$3-\frac{4}{{\gamma }_g}.$

Note that since,

$R^3$

$=$

${\tilde{r}}_s^3\left[G^{15/2}{K}_c^{-15/2}{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}^6\right],$

in terms of our adopted normalizations, the frequency-squared term should be rewritten as,

$\frac{{\omega }^2{R}^3}{{\gamma }_{g}G{M}_{\mathrm{t}\mathrm{o}\mathrm{t}}}$

$=$

${\tilde{r}}_s^3\left[\frac{{\omega }^2{\tau }_c^2}{{\gamma }_g}\right].$

Note as well that, at the surface of our $(n_c,n_e)=(5,1)$ bipolytrope, ${\gamma }_g=2$, so $\alpha =+1$. Hence, for this particular investigation, the surface boundary condition is,

$\frac{d\ln x}{d\ln \tilde{r}}{|}_s$	$=$	$\frac{1}{2}\left[{\tilde{r}}_s^3{\omega }^2{\tau }_c^2\right]-1$
	$=$	$\frac{{\tilde{r}}_s^3}{6}\left[2\pi {\tilde{\rho }}_c{\sigma }_c^2\right]-1$
	$=$	$\frac{{\tilde{r}}_s^3}{6}\left[2\pi {\tilde{\rho }}_c{\sigma }_c^2\right]\frac{3}{4\pi {\tilde{r}}_s^3{\tilde{\rho }}_{\mathrm{m}\mathrm{e}\mathrm{a}\mathrm{n}}}-1$
	$=$	$\frac{1}{4}\left(\frac{{\tilde{\rho }}_c}{{\tilde{\rho }}_{\mathrm{m}\mathrm{e}\mathrm{a}\mathrm{n}}}\right){\sigma }_c^2-1.$

This result should be compared with our separate discussion of eigenfunction details.

Discretize for Numerical Integration

General Discretization

First Approximation

Now, let's set up a grid associated with a uniformly spaced spherical radius, where the subscript $J$ denotes the grid zone at which all terms in the finite-difference representation of the governing relations will be evaluated. More specifically,

${\tilde{r}}_{J-1}$

$=$

${\tilde{r}}_J-\mathrm{\Delta }\tilde{r}$

and

${\tilde{r}}_{J+1}$

$=$

${\tilde{r}}_J+\mathrm{\Delta }\tilde{r};$

also,

$(\frac{dx}{d\tilde{r}}{)}_J$

$\approx$

$\frac{(x_{J+1}-x_{J-1})}{2\mathrm{\Delta }\tilde{r}}$

and

$(\frac{dp}{d\tilde{r}}{)}_J$

$\approx$

$\frac{(p_{J+1}-p_{J-1})}{2\mathrm{\Delta }\tilde{r}}.$

And at each grid location, the governing relations establish the local evaluation of the derivatives, that is,

$(\frac{dx}{d\tilde{r}}{)}_J$

$=$

$-\frac{1}{{\tilde{r}}_J}[3x+\frac{p}{{\gamma }_g}{]}_J,$

and

$(\frac{dp}{d\tilde{r}}{)}_J$

$=$

$\frac{{\tilde{\rho }}_J}{{\tilde{P}}_J}[(4x+p)\frac{{\tilde{M}}_r}{{\tilde{r}}^2}+{\tau }_c^2{\omega }^2\tilde{r}x{]}_J.$

So, integrating step-by-step from the center of the configuration, outward, once all the variable values are known at grid locations $J$ and $(J-1)$, the values of $x$ and $p$ at $(J+1)$ are given by the expressions,

$x_{J+1}$

$=$

$x_{J-1}-2\mathrm{\Delta }\tilde{r}\left\{\frac{1}{\tilde{r}}\right[3x+\frac{p}{{\gamma }_g}]{\}}_J,$

and

$p_{J+1}$

$=$

$p_{J-1}+2\mathrm{\Delta }\tilde{r}\{\frac{\tilde{\rho }}{\tilde{P}}\cdot \frac{{\tilde{M}}_r}{{\tilde{r}}^2}[(4x+p)+{\sigma }_c^2(\frac{2\pi }{3}\cdot \frac{{\tilde{\rho }}_c{\tilde{r}}^3}{{\tilde{M}}_r})x]{\}}_J.$

Then we will obtain the "$x_J$" and "$p_J$" values via the average expressions,

$x_J$

$=$

$\frac{1}{2}(x_{J-1}+x_{J+1}),$

and

$p_J$

$=$

$\frac{1}{2}(p_{J-1}+p_{J+1}).$

Consider implementing a more implicit finite-difference analysis. Wherever a "$J$" index appears in the source term, replace it with the average expressions. The general form of the source term expressions is, $x_{J+1}$ $=$ $x_{J-1}+2\mathrm{\Delta }\tilde{r}\{{𝒜}{x}_J+{ℬ}{p}_J\}$   $=$ $x_{J-1}+\mathrm{\Delta }\tilde{r}\left\{{𝒜}\right[x_{J+1}+x_{J-1}]+{ℬ}[p_{J+1}+p_{J-1}\left]\right\}$ $\Rightarrow x_{J+1}[1-\mathrm{\Delta }\tilde{r}{𝒜}]$ $=$ $x_{J-1}[1+\mathrm{\Delta }\tilde{r}{𝒜}]+\mathrm{\Delta }\tilde{r}{ℬ}[p_{J+1}+p_{J-1}],$ where, ${𝒜}$ $\equiv$ $-\frac{3}{\tilde{r}},$       and       ${ℬ}$ $\equiv$ $-\frac{1}{{\gamma }_g\tilde{r}};$ and, $p_{J+1}$ $=$ $p_{J-1}+2\mathrm{\Delta }\tilde{r}\{{𝒞}{x}_J+{𝒟}{p}_J\}$   $=$ $p_{J-1}+\mathrm{\Delta }\tilde{r}\left\{{𝒞}\right[x_{J+1}+x_{J-1}]+{𝒟}[p_{J+1}+p_{J-1}\left]\right\}$ $\Rightarrow p_{J+1}[1-\mathrm{\Delta }\tilde{r}{𝒟}]$ $=$ $p_{J-1}[1+\mathrm{\Delta }\tilde{r}{𝒟}]+\mathrm{\Delta }\tilde{r}{𝒞}[x_{J+1}+x_{J-1}],$ where, ${𝒞}$ $\equiv$ $\left\{{𝒟}\right[4+{\sigma }_c^2(\frac{2\pi }{3}\cdot \frac{{\tilde{\rho }}_c{\tilde{r}}^3}{{\tilde{M}}_r})]{\}}_J,$       and       ${𝒟}$ $\equiv$ $\{\frac{\tilde{\rho }}{\tilde{P}}\cdot \frac{{\tilde{M}}_r}{{\tilde{r}}^2}{\}}_J.$ In both cases, the two unknowns are $x_{J+1}$ and $p_{J+1}$. Combining this pair of equations gives, $x_{J+1}[1-\mathrm{\Delta }\tilde{r}{𝒜}]$ $=$ $x_{J-1}[1+\mathrm{\Delta }\tilde{r}{𝒜}]+\mathrm{\Delta }\tilde{r}{ℬ}\left[p_{J-1}\right]+\mathrm{\Delta }\tilde{r}{ℬ}[1-\mathrm{\Delta }\tilde{r}{𝒟}{]}^{-1}\{p_{J-1}[1+\mathrm{\Delta }\tilde{r}{𝒟}]+\mathrm{\Delta }\tilde{r}{𝒞}[x_{J+1}+x_{J-1}]\}$ $\Rightarrow x_{J+1}\{1-\mathrm{\Delta }\tilde{r}{𝒜}-\mathrm{\Delta }\tilde{r}{ℬ}[1-\mathrm{\Delta }\tilde{r}{𝒟}{]}^{-1}\left[\mathrm{\Delta }\tilde{r}{𝒞}\right]\}$ $=$ $x_{J-1}[1+\mathrm{\Delta }\tilde{r}{𝒜}]+\mathrm{\Delta }\tilde{r}{ℬ}\left[p_{J-1}\right]+\mathrm{\Delta }\tilde{r}{ℬ}[1-\mathrm{\Delta }\tilde{r}{𝒟}{]}^{-1}\{p_{J-1}[1+\mathrm{\Delta }\tilde{r}{𝒟}]+\mathrm{\Delta }\tilde{r}{𝒞}\left[x_{J-1}\right]\},$ which determines $x_{J+1}$, which then allows the straightforward determination of $p_{J+1}$. Via the average expressions, we can also then determine — and record — the self-consistent values of $x_J$ and $p_J$. Dropping terms ${𝒪}[(\mathrm{\Delta }\tilde{r}{)}^2]$ and higher gives, $x_{J+1}[1-\mathrm{\Delta }\tilde{r}{𝒜}{]}_J$ $\approx$ $x_{J-1}[1+\mathrm{\Delta }\tilde{r}{𝒜}{]}_J+\mathrm{\Delta }\tilde{r}{{ℬ}}_J[p_{J-1}],$ and, in turn, $p_{J+1}[1-\mathrm{\Delta }\tilde{r}{𝒟}{]}_J$ $\approx$ $p_{J-1}[1+\mathrm{\Delta }\tilde{r}{𝒟}{]}_J+\mathrm{\Delta }\tilde{r}{{𝒞}}_J[x_{J-1}]+\mathrm{\Delta }\tilde{r}{{𝒞}}_J[x_{J+1}].$

Second Approximation

Let's assume that we know the three quantities, $x_{J-1},x_J$, and $(x_J{)}^{\text{'}}\equiv (dx/d\tilde{r}{)}_J$ and want to project forward to determine, $x_{J+1}$. We should assume that, locally, the displacement function $x$ is quadratic in $\tilde{r}$, that is,

$x$	$=$	$a+b\tilde{r}+c{\tilde{r}}^2$
$\Rightarrow \frac{dx}{d\tilde{r}}$	$=$	$b+2c\tilde{r},$

where we have three unknowns, $a,b,c$. These can be determined by appropriately combining the three relations,

$(x_J{)}^{\text{'}}$	$=$	$b+2c{\tilde{r}}_J,$
$x_J$	$=$	$a+b{\tilde{r}}_J+c{\tilde{r}}_J^2,$
$x_{J-1}$	$=$	$a+b({\tilde{r}}_J-\mathrm{\Delta }\tilde{r})+c({\tilde{r}}_J-\mathrm{\Delta }\tilde{r}{)}^2.$

We have,

$b$	$=$	$(x_J{)}^{\text{'}}-2c{\tilde{r}}_J,$
$-a$	$=$	$-x_J+[(x_J{)}^{\text{'}}-2c{\tilde{r}}_J]{\tilde{r}}_J+c{\tilde{r}}_J^2,$
$-a$	$=$	$-x_{J-1}+[(x_J{)}^{\text{'}}-2c{\tilde{r}}_J]({\tilde{r}}_J-\mathrm{\Delta }\tilde{r})+c({\tilde{r}}_J-\mathrm{\Delta }\tilde{r}{)}^2.$

Combining the last two expressions gives,

$-x_J+[(x_J{)}^{\text{'}}-2c{\tilde{r}}_J]{\tilde{r}}_J+c{\tilde{r}}_J^2$	$=$	$-x_{J-1}+[(x_J{)}^{\text{'}}-2c{\tilde{r}}_J]({\tilde{r}}_J-\mathrm{\Delta }\tilde{r})+c({\tilde{r}}_J-\mathrm{\Delta }\tilde{r}{)}^2$
$\Rightarrow -x_J+(x_J{)}^{\text{'}}{\tilde{r}}_J-c{\tilde{r}}_J^2$	$=$	$-x_{J-1}+(x_J{)}^{\text{'}}({\tilde{r}}_J-\mathrm{\Delta }\tilde{r})-2c{\tilde{r}}_J({\tilde{r}}_J-\mathrm{\Delta }\tilde{r})+c[{\tilde{r}}_J^2-2r_J\mathrm{\Delta }\tilde{r}+(\mathrm{\Delta }\tilde{r}{)}^2]$
$\Rightarrow -x_J-c{\tilde{r}}_J^2$	$=$	$-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}+c[{\tilde{r}}_J^2-2r_J\mathrm{\Delta }\tilde{r}+(\mathrm{\Delta }\tilde{r}{)}^2-2{\tilde{r}}_J^2+2{\tilde{r}}_J(\mathrm{\Delta }\tilde{r})]$
$\Rightarrow c(\mathrm{\Delta }\tilde{r}{)}^2$	$=$	$x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}.$

Therefore, also,

$b$	$=$	$(x_J{)}^{\text{'}}-[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\frac{2{\tilde{r}}_J}{(\mathrm{\Delta }\tilde{r}{)}^2},$
$a$	$=$	$x_J-(x_J{)}^{\text{'}}{\tilde{r}}_J+[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\frac{{\tilde{r}}_J^2}{(\mathrm{\Delta }\tilde{r}{)}^2}$

Hence,

$x_{J+1}$	$=$	$a+b({\tilde{r}}_J+\mathrm{\Delta }\tilde{r})+c({\tilde{r}}_J+\mathrm{\Delta }\tilde{r}{)}^2$
	$=$	$x_J-(x_J{)}^{\text{'}}{\tilde{r}}_J+[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\frac{{\tilde{r}}_J^2}{(\mathrm{\Delta }\tilde{r}{)}^2}+({\tilde{r}}_J+\mathrm{\Delta }\tilde{r})\{(x_J{)}^{\text{'}}-[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}\left]\frac{2{\tilde{r}}_J}{(\mathrm{\Delta }\tilde{r}{)}^2}\right\}$
		$+({\tilde{r}}_J+\mathrm{\Delta }\tilde{r}{)}^2[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\frac{1}{(\mathrm{\Delta }\tilde{r}{)}^2}$
	$=$	$x_J-(x_J{)}^{\text{'}}{\tilde{r}}_J+\{{\tilde{r}}_J^2+2{\tilde{r}}_J({\tilde{r}}_J+\mathrm{\Delta }\tilde{r})+({\tilde{r}}_J+\mathrm{\Delta }\tilde{r}{)}^2\}[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\frac{1}{(\mathrm{\Delta }\tilde{r}{)}^2}$
	$=$	$x_J-(x_J{)}^{\text{'}}{\tilde{r}}_J+[2{\tilde{r}}_J+\mathrm{\Delta }\tilde{r}{]}^2[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\frac{1}{(\mathrm{\Delta }\tilde{r}{)}^2}.$

WRONG!! Try again …

Third Approximation

Let's assume that we know the three quantities, $x_{J-1},x_J$, and $(x_J{)}^{\text{'}}\equiv (dx/d\tilde{r}{)}_J$ and want to project forward to determine, $x_{J+1}$. We should assume that, locally, the displacement function $x$ is quadratic in $\tilde{r}$, that is,

$x$	$=$	$a+b\tilde{r}+c{\tilde{r}}^2$
$\Rightarrow \frac{dx}{d\tilde{r}}$	$=$	$b+2c\tilde{r},$

where we have three unknowns, $a,b,c$. These can be determined by appropriately combining the three relations,

$(x_J{)}^{\text{'}}$	$=$	$b+2c{\tilde{r}}_J,$
$x_J$	$=$	$a+b{\tilde{r}}_J+c{\tilde{r}}_J^2,$
$x_{J-1}$	$=$	$a+b({\tilde{r}}_J-\mathrm{\Delta }\tilde{r})+c({\tilde{r}}_J-\mathrm{\Delta }\tilde{r}{)}^2.$

The difference between the last two expressions gives,

$x_J-x_{J-1}$	$=$	$[b{\tilde{r}}_J+c{\tilde{r}}_J^2]-[b({\tilde{r}}_J-\mathrm{\Delta }\tilde{r})+c({\tilde{r}}_J-\mathrm{\Delta }\tilde{r}{)}^2]$
	$=$	$b\mathrm{\Delta }\tilde{r}+c(2{\tilde{r}}_J\mathrm{\Delta }\tilde{r}-\mathrm{\Delta }{\tilde{r}}^2).$

Combining this with the first of the three expressions then gives,

$\mathrm{\Delta }\tilde{r}[(x_J{)}^{\text{'}}-2c{\tilde{r}}_J]+c(2{\tilde{r}}_J\mathrm{\Delta }\tilde{r}-\mathrm{\Delta }{\tilde{r}}^2)$	$=$	$x_J-x_{J-1}$
$\Rightarrow c(2{\tilde{r}}_J\mathrm{\Delta }\tilde{r}-\mathrm{\Delta }{\tilde{r}}^2)-c\left[2{\tilde{r}}_J\mathrm{\Delta }\tilde{r}\right]$	$=$	$x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}$
$\Rightarrow -c\mathrm{\Delta }{\tilde{r}}^2$	$=$	$x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}$
$\Rightarrow c$	$=$	$\frac{1}{\mathrm{\Delta }{\tilde{r}}^2}[-x_J+x_{J-1}+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}].$

Hence,

$b$	$=$	$(x_J{)}^{\text{'}}-2c{\tilde{r}}_J$
	$=$	$(x_J{)}^{\text{'}}+\frac{2{\tilde{r}}_J}{\mathrm{\Delta }{\tilde{r}}^2}[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}];$
$a$	$=$	$x_J-b{\tilde{r}}_J-c{\tilde{r}}_J^2$
	$=$	$x_J-\{(x_J{)}^{\text{'}}+\frac{2{\tilde{r}}_J}{\mathrm{\Delta }{\tilde{r}}^2}[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}\left]\right\}{\tilde{r}}_J-\left\{\frac{1}{\mathrm{\Delta }{\tilde{r}}^2}\right[-x_J+x_{J-1}+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\}{\tilde{r}}_J^2$
	$=$	$x_J-(x_J{)}^{\text{'}}{\tilde{r}}_J-\{\frac{2{\tilde{r}}_J^2}{\mathrm{\Delta }{\tilde{r}}^2}[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}\left]\right\}+\left\{\frac{{\tilde{r}}_J^2}{\mathrm{\Delta }{\tilde{r}}^2}\right[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\}$
	$=$	$x_J-(x_J{)}^{\text{'}}{\tilde{r}}_J-\frac{{\tilde{r}}_J^2}{\mathrm{\Delta }{\tilde{r}}^2}[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}].$

As a result,

$x_{J+1}$	$=$	$\left\{a\right\}+({\tilde{r}}_J+\mathrm{\Delta }\tilde{r})\left\{b\right\}+({\tilde{r}}_J+\mathrm{\Delta }\tilde{r}{)}^2\{c\}$
	$=$	$\{x_J-(x_J{)}^{\text{'}}{\tilde{r}}_J-\frac{{\tilde{r}}_J^2}{\mathrm{\Delta }{\tilde{r}}^2}[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}\left]\right\}+({\tilde{r}}_J+\mathrm{\Delta }\tilde{r})\{(x_J{)}^{\text{'}}+\frac{2{\tilde{r}}_J}{\mathrm{\Delta }{\tilde{r}}^2}[x_J-x_{J-1}-(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}\left]\right\}$
		$+({\tilde{r}}_J^2+2{\tilde{r}}_J\mathrm{\Delta }\tilde{r}+\mathrm{\Delta }{\tilde{r}}^2)\left\{\frac{1}{\mathrm{\Delta }{\tilde{r}}^2}\right[-x_J+x_{J-1}+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\}$
	$=$	$x_J-(x_J{)}^{\text{'}}{\tilde{r}}_J+{\tilde{r}}_J^2\{\frac{1}{\mathrm{\Delta }{\tilde{r}}^2}[-x_J+x_{J-1}+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}\left]\right\}+(x_J{)}^{\text{'}}({\tilde{r}}_J+\mathrm{\Delta }\tilde{r})-2{\tilde{r}}_J({\tilde{r}}_J+\mathrm{\Delta }\tilde{r})\{\frac{1}{\mathrm{\Delta }{\tilde{r}}^2}[-x_J+x_{J-1}+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}\left]\right\}$
		$+({\tilde{r}}_J^2+2{\tilde{r}}_J\mathrm{\Delta }\tilde{r}+\mathrm{\Delta }{\tilde{r}}^2)\left\{\frac{1}{\mathrm{\Delta }{\tilde{r}}^2}\right[-x_J+x_{J-1}+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]\}$
	$=$	$x_J+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}+[{\tilde{r}}_J^2-2{\tilde{r}}_J({\tilde{r}}_J+\mathrm{\Delta }\tilde{r})+({\tilde{r}}_J^2+2{\tilde{r}}_J\mathrm{\Delta }\tilde{r}+\mathrm{\Delta }{\tilde{r}}^2)\left]\right\{\frac{1}{\mathrm{\Delta }{\tilde{r}}^2}[-x_J+x_{J-1}+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}\left]\right\}$
	$=$	$x_J+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}+[-x_J+x_{J-1}+(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}]$
	$=$	$x_{J-1}+2(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}.$

GOOD! This is the same as our first approximation expression stated above.

This is test ... $\mathrm{\Delta }\tilde{r}$ $x_J$ $x_{J-1}$ $(x_J{)}^{\text{'}}$ $(x_{J-1}{)}^{\text{'}}$ 0.001936393 -4.695376 -4.547832 -116.0119 -76.19513

$x_{J+1}$ $=$ $x_{J-1}+2(x_J{)}^{\text{'}}\mathrm{\Delta }\tilde{r}=-4.997121.$

Part 2

For a continuation (part 2) of this discussion, go here.

See Also