Appendix/Ramblings/51BiPolytropeStability/RethinkEnvelope

Rethink Handling of n = 1 Envelope

Solution Steps

Drawing from an accompanying discussion …

• Step 1: Choose $n_c$ and $n_e$.
• Step 2: Adopt boundary conditions at the center of the core ($\theta =1$ and $d\theta /d\xi =0$ at $\xi =0$), then solve the Lane-Emden equation to obtain the solution, $\theta (\xi )$, and its first derivative, $d\theta /d\xi$ throughout the core; the radial location, $\xi ={\xi }_s$, at which $\theta (\xi )$ first goes to zero identifies the natural surface of an isolated polytrope that has a polytropic index $n_c$.
• Step 3 Choose the desired location, $0<{\xi }_i<{\xi }_s$, of the outer edge of the core.
• Step 4: Specify $K_c$ and ${\rho }_0$; the structural profile of, for example, $\rho (r)$, $P(r)$, and $M_r(r)$ is then obtained throughout the core — over the radial range, $0\le \xi \le {\xi }_i$ and $0\le r\le r_i$ — via the relations shown in the $2^{\mathrm{n}\mathrm{d}}$ column of Table 1.
• Step 5: Specify the ratio ${\mu }_e/{\mu }_c$ and adopt the boundary condition, ${\varphi }_i=1$; then use the interface conditions as rearranged and presented in Table 3 to determine, respectively:
  • The gas density at the base of the envelope, ${\rho }_e$;
  • The polytropic constant of the envelope, $K_e$, relative to the polytropic constant of the core, $K_c$;
  • The ratio of the two dimensionless radial parameters at the interface, ${\eta }_i/{\xi }_i$;
  • The radial derivative of the envelope solution at the interface, $(d\varphi /d\eta {)}_i$.
• Step 6: The last sub-step of solution step 5 provides the boundary condition that is needed — in addition to our earlier specification that ${\varphi }_i=1$ — to derive the desired particular solution, $\varphi (\eta )$, of the Lane-Emden equation that is relevant throughout the envelope; knowing $\varphi (\eta )$ also provides the relevant structural first derivative, $d\varphi /d\eta$, throughout the envelope.
• Step 7: The surface of the bipolytrope will be located at the radial location, $\eta ={\eta }_s$ and $r=R$, at which $\varphi (\eta )$ first drops to zero.
• Step 8: The structural profile of, for example, $\rho (r)$, $P(r)$, and $M_r(r)$ is then obtained throughout the envelope — over the radial range, ${\eta }_i\le \eta \le {\eta }_s$ and $r_i\le r\le R$ — via the relations provided in the $3^{\mathrm{r}\mathrm{d}}$ column of Table 1.

Setup

Drawing from the accompanying Table 1, we have …

Core	Envelope
$n_c=5$	$n_e=1$
$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d\theta }{d\xi }\right)=-{\theta }^5$ sol'n: $\theta (\xi )$	$\frac{1}{{\eta }^2}\frac{d}{d\eta }\left({\eta }^2\frac{d\varphi }{d\eta }\right)=-\varphi$ sol'n: $\varphi (\eta )$
Specify: $K_c$ and ${\rho }_0\Rightarrow$ $\rho$ $=$ ${\rho }_0{\theta }^5$ $P$ $=$ $K_c{\rho }_0^{6/5}{\theta }^6$ $r$ $=$ $[\frac{3K_c}{2\pi G}{]}^{1/2}{\rho }_0^{-2/5}\xi$ $M_r$ $=$ $4\pi \left[\frac{3K_c}{2\pi G}{]}^{3/2}{\rho }_0^{-1/5}\right(-{\xi }^2\frac{d\theta }{d\xi })$	Knowing: $K_e$ and ${\rho }_e\Rightarrow$ $\rho$ $=$ ${\rho }_e\varphi$ $P$ $=$ $K_e{\rho }_e^2{\varphi }^2$ $r$ $=$ $[\frac{K_e}{2\pi G}{]}^{1/2}\eta$ $M_r$ $=$ $4\pi \left[\frac{K_e}{2\pi G}{]}^{3/2}{\rho }_e\right(-{\eta }^2\frac{d\varphi }{d\eta })$

From an accompanying discussion of $(n_c,n_e)=(5,1)$ bipolytropes, we know that the solution to the pair of Lane-Emden equations is …

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

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

and,

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

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

Let's shift from the envelope's standard radial coordinate, $\eta$, to $\mathrm{\Delta }$ $\equiv$ $\eta -B$ $\Rightarrow \varphi$ $=$ $A\left[\frac{\sin \mathrm{\Delta }}{\mathrm{\Delta }+B}\right]$ The solid blue curve in the following plot shows how $\varphi$ varies with $\mathrm{\Delta }$ when $(A,B)=(0.608404,-0.265127)$. Notice that $\varphi$ is zero when $\mathrm{\Delta }=\pi ,2\pi ,3\pi ,4\pi ,5\pi$; more generally, it crosses zero when $\mathrm{\Delta }=\pm m\pi$, for all positive values of the integer, $m$. Notice as well that when $\mathrm{\Delta }=-B$ … (that is, when $\eta =0$) … $\varphi \to \pm \mathrm{\infty }$. phi vs Delta The solid blue curve exhibits an extremum when, $\frac{d\varphi }{d\mathrm{\Delta }}$ $=$ $\frac{A}{(\mathrm{\Delta }+B{)}^2}[(\mathrm{\Delta }+B)\cos \mathrm{\Delta }-\sin \mathrm{\Delta }]\to 0.$ This occurs when the quantity, $[\varphi -A\cos \mathrm{\Delta }]$ (the dotted grey curve) goes to zero and/or when the quantity, $[\tan \mathrm{\Delta }-(\mathrm{\Delta }+B)]$ (the dotted orange curve) goes to zero. NOTE:  A very similar expression arises in our accompanying discussion of bipolytropes with $(n_c,n_e)=(1,5)$. Specifically, $\frac{{\xi }_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}}}{\tan ({\xi }_{\mathrm{t}\mathrm{r}\mathrm{a}\mathrm{n}\mathrm{s}})}$ $=$ $1-\frac{3}{2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}.$ I'm not sure whether this is relevant information or not! The solid black, vertical line segments in this plot bracket the regime $2\pi <\mathrm{\Delta }<3\pi$; the function, $\varphi$ is positive everywhere across this interval. In this interval, one extremum arises at $\mathrm{\Delta }={\mathrm{\Delta }}_{\mathrm{e}\mathrm{x}\mathrm{t}}\approx 7.72832$; it is identified in the plot by the dashed red, vertical line segment. The function, $\varphi (\mathrm{\Delta })$, can be used to construct a physically viable envelope over the interval, ${\mathrm{\Delta }}_{\mathrm{e}\mathrm{x}\mathrm{t}}\le \mathrm{\Delta }\le 3\pi$, because, across this interval, $\varphi$ is everywhere positive (if not zero) and $d\varphi /d\mathrm{\Delta }$ is everywhere negative (if not zero). The blue curve in the following plot is identical to the one depicted in the previous plot, except the function, $\varphi$, is plotted versus $\eta$ rather than versus $\mathrm{\Delta }$. This is analogous to the blue curve shown in Figure 3 of our accompanying discussion of Shrivastava's ${\theta }_{5F}$ Function. phi vs eta

Adopting the same normalizations as before, namely,

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

we have,

Core	Envelope
${\rho }^{*}\equiv \frac{\rho }{{\rho }_0}$ $=$ ${\theta }^5$ $P^{*}\equiv \frac{P}{K_c{\rho }_0^{6/5}}$ $=$ ${\theta }^6$ $r^{*}\equiv r\left[\frac{G^{1/2}{\rho }_0^{2/5}}{K_c^{1/2}}\right]$ $=$ $[\frac{3}{2\pi }{]}^{1/2}\xi$ $M_r^{*}\equiv M_r\left[\frac{G^{3/2}{\rho }_0^{1/5}}{K_c^{3/2}}\right]$ $=$ $4\pi \left[\frac{3}{2\pi }{]}^{3/2}\right(-{\xi }^2\frac{d\theta }{d\xi })$	${\rho }^{*}$ $=$ $\left(\frac{{\rho }_e}{{\rho }_0}\right)\varphi$ $P^{*}$ $=$ $\left[\frac{K_e{\rho }_e^2}{K_c{\rho }_0^{6/5}}\right]{\varphi }^2$ $r^{*}$ $=$ ${\rho }_0^{2/5}[\frac{K_e}{2\pi K_c}{]}^{1/2}\eta$ $M_r^{*}$ $=$ $4\pi \left[{\rho }_e{\rho }_0^{1/5}\right]\left[\frac{K_e}{2\pi K_c}{]}^{3/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$

Interface Conditions

Drawing from Table 2 in our accompanying discussion, we see that the interface conditions give,

$\left(\frac{{\rho }_0}{{\mu }_c}\right){\theta }_i^5$	$=$	$\left(\frac{{\rho }_e}{{\mu }_e}\right){\varphi }_i$
$\Rightarrow \frac{{\rho }_e}{{\rho }_0}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5{\varphi }_i^{-1},$

and,

$K_c{\rho }_0^{6/5}{\theta }_i^6$	$=$	$K_e{\rho }_e^2{\varphi }_i^2$
$\Rightarrow (\frac{K_e}{K_c}{)}^{1/2}$	$=$	$\frac{{\rho }_0^{3/5}}{{\rho }_e}\cdot \frac{{\theta }_i^3}{{\varphi }_i}.$

As a result, throughout the envelope,

$P^{*}$	$=$	$\left[\frac{K_e{\rho }_e^2}{K_c{\rho }_0^{6/5}}\right]{\varphi }^2=\left(\frac{{\theta }_i^6}{{\varphi }_i^2}\right){\varphi }^2$
$r^{*}$	$=$	${\rho }_0^{2/5}[\frac{K_e}{2\pi K_c}{]}^{1/2}\eta =(2\pi {)}^{-1/2}{\rho }_0^{2/5}[\frac{{\rho }_0^{3/5}}{{\rho }_e}\cdot \frac{{\theta }_i^3}{{\varphi }_i}]\eta =\left[\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^5{\varphi }_i^{-1}{]}^{-1}(2\pi {)}^{-1/2}\left[\frac{{\theta }_i^3}{{\varphi }_i}\right]\eta =(2\pi {)}^{-1/2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}\eta$
$M_r^{*}$	$=$	$4\pi \left[{\rho }_e{\rho }_0^{1/5}\right]\left[\frac{K_e}{2\pi K_c}{]}^{3/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })=2(2\pi {)}^{-1/2}\left[{\rho }_e{\rho }_0^{1/5}\right][\frac{{\rho }_0^{3/5}}{{\rho }_e}\cdot \frac{{\theta }_i^3}{{\varphi }_i}{]}^3(-{\eta }^2\frac{d\varphi }{d\eta })=2(2\pi {)}^{-1/2}\left[\frac{{\rho }_e}{{\rho }_0}{]}^{-2}\right[\frac{{\theta }_i^3}{{\varphi }_i}{]}^3(-{\eta }^2\frac{d\varphi }{d\eta })$
	$=$	$2(2\pi {)}^{-1/2}[\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5{\varphi }_i^{-1}{]}^{-2}\left[\frac{{\theta }_i^9}{{\varphi }_i^3}\right](-{\eta }^2\frac{d\varphi }{d\eta })=\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}({\theta }_i{\varphi }_i{)}^{-1}(-{\eta }^2\frac{d\varphi }{d\eta }).$

In summary, then,

Core

Envelope

${\rho }^{*}$ $=$ ${\theta }^5=[1+\frac{1}{3}{\xi }^2{]}^{-5/2}$ $P^{*}$ $=$ ${\theta }^6$ $r^{*}$ $=$ $[\frac{3}{2\pi }{]}^{1/2}\xi$ $M_r^{*}$ $=$ $4\pi \left[\frac{3}{2\pi }{]}^{3/2}\right(-{\xi }^2\frac{d\theta }{d\xi })$   $=$ $\left[\frac{2^4\cdot 3^3{\pi }^2}{2^3{\pi }^3}{]}^{1/2}\right\{\frac{{\xi }^3}{3}[1+\frac{1}{3}{\xi }^2{]}^{-3/2}\}$   $=$ $\left(\frac{6}{\pi }{)}^{1/2}{\xi }^3\right[1+\frac{1}{3}{\xi }^2{]}^{-3/2}$

${\rho }^{*}$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5{\varphi }_i^{-1}\varphi =\frac{A}{{\varphi }_i}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\left[\frac{\sin (\eta -B)}{\eta }\right]$ $P^{*}$ $=$ $\left(\frac{{\theta }_i^6}{{\varphi }_i^2}\right){\varphi }^2={\theta }_i^6\left(\frac{A}{{\varphi }_i}{)}^2\right[\frac{\sin (\eta -B)}{\eta }{]}^2.$ $r^{*}$ $=$ $(2\pi {)}^{-1/2}[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}\right]\eta$ $M_r^{*}$ $=$ $\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}({\theta }_i{\varphi }_i{)}^{-1}(-{\eta }^2\frac{d\varphi }{d\eta })$   $=$ $\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\left\{\frac{A}{{\varphi }_i}\right[\sin (\eta -B)-\eta \cos (\eta -B)\left]\right\}$

Notice that by setting the pressure to be the same at the interface, we have the relation,

${\theta }_i^6$	$=$	${\theta }_i^6\left(\frac{A}{{\varphi }_i}{)}^2\right[\frac{\sin ({\eta }_i-B)}{{\eta }_i}{]}^2$
$\Rightarrow \frac{A}{{\varphi }_i}$	$=$	$\frac{{\eta }_i}{\sin ({\eta }_i-B)}.$

Pulling from Table 3 in our accompanying discussion, two other constraints come from making sure that the radius of the configuration and the enclosed mass match at the interface, whether you examine it from the point of view of the core or of the envelope. In principle, these constraints can provide expressions for the two unknown constants, $A$ and $B$. Let's do the radius first.

$(2\pi {)}^{-1/2}(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}{\eta }_i$	$=$	$[\frac{3}{2\pi }{]}^{1/2}{\xi }_i$
$\Rightarrow \frac{{\eta }_i}{{\xi }_i}$	$=$	$3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2.$

Now, from the enclosed mass constraint,

$\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}({\theta }_i{\varphi }_i{)}^{-1}(-{\eta }^2\frac{d\varphi }{d\eta })$	$=$	$4\pi \left[\frac{3}{2\pi }{]}^{3/2}\right(-{\xi }^2\frac{d\theta }{d\xi })$
$\Rightarrow (\frac{d\varphi }{d\eta }{)}_i$	$=$	$4\pi \left[\frac{3^3}{2^3{\pi }^3}\right(\frac{\pi }{2}\left){]}^{1/2}\right(\frac{{\xi }_i}{{\eta }_i}{)}^2\times (\frac{{\mu }_e}{{\mu }_c}{)}^2({\theta }_i{\varphi }_i)(\frac{d\theta }{d\xi }{)}_i$
	$=$	$3^{3/2}\left[3^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2{]}^{-2}\times (\frac{{\mu }_e}{{\mu }_c}{)}^2({\theta }_i{\varphi }_i)(\frac{d\theta }{d\xi }{)}_i$
	$=$	$3^{1/2}({\theta }_i^{-3}{\varphi }_i)(\frac{d\theta }{d\xi }{)}_i$
	$=$	$-\frac{{\xi }_i}{\sqrt{3}}.$

Alternatively, the ratio of these two expressions gives,

$\frac{{\eta }_i}{{\xi }_i}\left\{\right(\frac{d\varphi }{d\eta }{)}_i{\}}^{-1}$	$=$	$3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^2\{3^{1/2}({\theta }_i^{-3}{\varphi }_i)(\frac{d\theta }{d\xi }{)}_i{\}}^{-1}$
$\Rightarrow \frac{{\eta }_i{\varphi }_i}{(d\varphi /d\eta {)}_i}$	$=$	$\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{{\xi }_i{\theta }_i^5}{(d\theta /d\xi {)}_i};$

and their product gives,

$\frac{{\eta }_i}{{\xi }_i}(\frac{d\varphi }{d\eta }{)}_i$	$=$	$3^{1/2}({\theta }_i^{-3}{\varphi }_i)(\frac{d\theta }{d\xi }{)}_i\times 3^{1/2}(\frac{{\mu }_e}{{\mu }_c}){\theta }_i^2$
$\Rightarrow \frac{{\eta }_i}{{\varphi }_i}(\frac{d\varphi }{d\eta }{)}_i$	$=$	$\frac{3{\xi }_i}{{\theta }_i}\left(\frac{d\theta }{d\xi }{)}_i\right(\frac{{\mu }_e}{{\mu }_c}).$

Shift from η to Δ

Again, let's shift from the envelope's standard radial coordinate, $\eta$, to

$\mathrm{\Delta }$	$\equiv$	$\eta -B$
$\Rightarrow \varphi$	$=$	$A\left[\frac{\sin \mathrm{\Delta }}{\mathrm{\Delta }+B}\right];$

and,

$\frac{d\varphi }{d\mathrm{\Delta }}$	$=$	$\frac{A}{(\mathrm{\Delta }+B{)}^2}[(\mathrm{\Delta }+B)\cos \mathrm{\Delta }-\sin \mathrm{\Delta }]$
$\Rightarrow \frac{1}{\varphi }\cdot \frac{d\varphi }{d\mathrm{\Delta }}$	$=$	$\left[\frac{\mathrm{\Delta }+B}{\sin \mathrm{\Delta }}\right]\frac{1}{(\mathrm{\Delta }+B{)}^2}[(\mathrm{\Delta }+B)\cos \mathrm{\Delta }-\sin \mathrm{\Delta }]$
	$=$	$\frac{1}{(\mathrm{\Delta }+B)}[(\mathrm{\Delta }+B)\cot \mathrm{\Delta }-1]$
	$=$	$\cot \mathrm{\Delta }-\frac{1}{(\mathrm{\Delta }+B)}.$

The pair of constraints obtained from matching the radius and the enclosed mass, respectively, are,

${\eta }_i$	$=$	$3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\xi }_i{\theta }_i^2$
$\Rightarrow {\mathrm{\Delta }}_i+B$	$=$	$3^{1/2}\left(\frac{{\mu }_e}{{\mu }_c}\right){\xi }_i{\theta }_i^2;$

and,

$3^{1/2}({\theta }_i^{-3})(\frac{d\theta }{d\xi }{)}_i$	$=$	$\left[\frac{1}{\varphi }\right(\frac{d\varphi }{d\mathrm{\Delta }}){]}_i$
$\Rightarrow -\frac{{\xi }_i}{\sqrt{3}}$	$=$	$\cot {\mathrm{\Delta }}_i-\frac{1}{({\mathrm{\Delta }}_i+B)}$
$\Rightarrow \cot {\mathrm{\Delta }}_i$	$=$	$\frac{1}{{\eta }_i}-\frac{{\xi }_i}{\sqrt{3}}.$

Cross-checking against our earlier tabulation of parameter values — specifically the parameter, ${\mathrm{\Lambda }}_i$ — we recognize that,

${\mathrm{\Lambda }}_i$	$=$	$\frac{1}{{\eta }_i}-\frac{{\xi }_i}{\sqrt{3}}$
$\Rightarrow \cot {\mathrm{\Delta }}_i$	$=$	${\mathrm{\Lambda }}_i$
$\Rightarrow {\mathrm{\Delta }}_i$	$=$	${\tan }^{-1}\left(\frac{1}{{\mathrm{\Lambda }}_i}\right)+m\pi .$

For the record: ${\mathrm{\Lambda }}_i=\frac{1}{{\eta }_i}-\frac{{\xi }_i}{\sqrt{3}}$ $=$ $\frac{\sqrt{3}-{\eta }_i{\xi }_i}{\sqrt{3}{\eta }_i}$   $=$ $\frac{\sqrt{3}-\sqrt{3}({\mu }_e/{\mu }_c){\theta }_i^2{\xi }_i^2}{3({\mu }_e/{\mu }_c){\theta }_i^2{\xi }_i}$   $=$ $\frac{1-({\mu }_e/{\mu }_c){\theta }_i^2{\xi }_i^2}{\sqrt{3}({\mu }_e/{\mu }_c){\theta }_i^2{\xi }_i}$   $=$ $\frac{(1+{\xi }_i^2/3)-({\mu }_e/{\mu }_c){\xi }_i^2}{\sqrt{3}({\mu }_e/{\mu }_c){\xi }_i}$   $=$ $\frac{3+{\xi }_i^2[1-3({\mu }_e/{\mu }_c)]}{3^{3/2}({\mu }_e/{\mu }_c){\xi }_i}$ $\Rightarrow \frac{1}{{\mathrm{\Lambda }}_i}$ $=$ $\frac{3^{3/2}({\mu }_e/{\mu }_c){\xi }_i}{3+{\xi }_i^2[1-3({\mu }_e/{\mu }_c)]}.$

NOTE: By adding the additional term, $m\pi$, we are able to take advantage of the oscillatory nature of the density function, $\varphi$. As a result, we see that,

$B$

$=$

${\eta }_i-{\tan }^{-1}\left(\frac{1}{{\mathrm{\Lambda }}_i}\right)-m\pi ;$

and, given that,

$\sin {\mathrm{\Delta }}_i$	$=$	$\sin \left\{{\tan }^{}\right(\frac{1}{{\mathrm{\Lambda }}_i})+m\pi \}$
	$=$	$\sin \left[{\tan }^{}\right(\frac{1}{{\mathrm{\Lambda }}_i}\left)\right]\cdot \cos (m\pi )+\cos \left[{\tan }^{}\right(\frac{1}{{\mathrm{\Lambda }}_i}\left)\right]\cdot {\xcancel{\sin (m\pi )}}^0$
	$=$	$\left[\frac{1/{\mathrm{\Lambda }}_i}{(1+1/{\mathrm{\Lambda }}_i^2{)}^{1/2}}\right]\cdot \cos (m\pi )$
	$=$	$\cos (m\pi )\cdot [1+{\mathrm{\Lambda }}_i^2{]}^{-1/2},$

the other constant is,

$\frac{A}{{\varphi }_i}$

$=$

$\frac{{\mathrm{\Delta }}_i+B}{\sin {\mathrm{\Delta }}_i}=\frac{{\eta }_i}{\sin {\mathrm{\Delta }}_i}=\frac{{\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2}}{\cos (m\pi )}.$

As in the earlier case depicted below, let's draw from the accompanying B2 model for which, ${\mu }_e/{\mu }_c=0.25$ and ${\xi }_i=2.4782510$ and …

${\theta }_i$	${\eta }_i$	${\mathrm{\Lambda }}_i$	$A$	$B$	${\eta }_s$		$Q_{\rho }$	$Q_m$	$Q_r$
0.572857	0.352159	1.408807	0.608404	-0.265127	$2.876465$		0.00938349	13.558308	7.0373055

${\theta }_i$	$(\frac{d\theta }{d\xi }{)}_i$	${\eta }_i$	$b_i$	$(y_i{)}_{+}$	$(y_i{)}_{-}$	$A$	$B$	${\eta }_s$		$Q_{\rho }$	$Q_m$	$Q_r$
0.572857	-0.1552971	0.352159	-1.430819	0.672019	-0.987752	0.608404	-0.265127	$2.876465$		0.00938349	13.558308	7.0373055

Useful?

This matches our earlier derivation. Remember, as well, that ${\varphi }_i=1$, that is to say,

$A$

$=$

$\left[\frac{{\eta }_i}{\sin ({\eta }_i-B)}\right].$

Suppose we use $\eta$ as the primary abscissa. Throughout the envelope, for various values of $\eta$, we set

${\rho }^{*}=Q_{\rho }\left[\frac{\sin (\eta -B)}{\eta }\right],M^{*}=Q_m[\sin (\eta -B)-\eta \cos (\eta -B)],\xi =Q_r\eta$

where,

$Q_{\rho }$	$\equiv$	$A\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5$
$Q_m$	$\equiv$	$A\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}$
$Q_r$	$\equiv$	$3^{-1/2}\left[\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}]$

Note as well that,

$q$	$\equiv$	$\frac{{\xi }_i}{{\xi }_s}=\frac{{\xi }_i}{Q_r{\eta }_s};$
$\nu$	$\equiv$	$\frac{M_r^{*}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_r^{*}{|}_{\mathrm{t}\mathrm{o}\mathrm{t}}}=\frac{\sin ({\eta }_i-B)-{\eta }_i\cos ({\eta }_i-B)}{{\eta }_s}.$

Earlier Example

In our earlier analysis, we determined that the following relations hold in an equilibrium bipolytrope.

Keep in mind that, once ${\mu }_e/{\mu }_c$ and ${\xi }_i$ have been specified, other parameter values at the interface are: ${\theta }_i$ $=$ $(1+\frac{1}{3}{\xi }_i^2{)}^{-1/2},$ ${\eta }_i$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}{\theta }_i^2{\xi }_i,$ ${\mathrm{\Lambda }}_i$ $=$ $\frac{1}{{\eta }_i}+(\frac{d\varphi }{d\eta }{)}_i=(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{1}{\sqrt{3}{\xi }_i{\theta }_i^2}-\frac{{\xi }_i}{\sqrt{3}},$ $A$ $=$ ${\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2},$ $B$ $=$ ${\eta }_i-\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i),$ ${\eta }_s$ $=$ $B+\pi .$

As a test case, let's draw from the accompanying B2 model for which, ${\mu }_e/{\mu }_c=0.25$ and ${\xi }_i=2.4782510$ and …

${\theta }_i$	${\eta }_i$	${\mathrm{\Lambda }}_i$	$A$	$B$	${\eta }_s$		$Q_{\rho }$	$Q_m$	$Q_r$
0.572857	0.352159	1.408807	0.608404	-0.265127	$2.876465$		0.00938349	13.558308	7.0373055

The following pair of plots show how the normalized density, ${\rho }^{*}$, and normalized integrated mass, $M_r^{*}$, varies over the radial-coordinate range, $0\le \eta \le 3$, for both the core description and the envelope description for Model B2. Both plots present the same four curves except, in the "first plot", the density has been magnified by a factor of 35 to aid in visualizing the shapes of the curves. In the "first plot" the maximum ordinate value is 40, which comfortably accommodates the maximum value of both mass curves. In the "second plot" the maximum ordinate value is 0.09, which permits us to zoom in on the behavior of the (unmagnified) density curves in the vicinity of the core-envelope interface.

More specifically, here are the expressions that were used to generate each of the four curves (in both plots).

Grey dotted curve: After setting $\xi =Q_r\eta$ for each value of $\eta$ over the specified range …

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

$=$

$[1+\frac{{\xi }^2}{3}{]}^{-5/2}.$

Orange curve: After setting $\xi =Q_r\eta$ for each value of $\eta$ over the specified range …

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

$=$

$\left(\frac{6}{\pi }{)}^{1/2}{\xi }^3\right[1+\frac{1}{3}{\xi }^2{]}^{-3/2}.$

Dark-blue dotted curve: Acknowledging that $B=-0.265127$ for each value of $\eta$ over the specified range …

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

$=$

$Q_{\rho }\left[\frac{\sin (\eta -B)}{\eta }\right].$

Red curve: Acknowledging that $B=-0.265127$ for each value of $\eta$ over the specified range …

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

$=$

$Q_m[\sin (\eta -B)-\eta \cos (\eta -B)].$

Model B2 — first plot

Model B2 — second plot

Things to notice: Because $B\ne 0$ and ${\rho }^{*}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ is proportional to ${\eta }^{-1}$, the envelope-density (dark-blue dotted) curve shoots up to infinity as $\eta \to 0$. Nevertheless, as the red curve in the "first plot" shows, the integrated envelope mass, $M_r^{*}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$, is well behaved; it goes to $Q_m\sin (-B)=3.5527$ as $\eta \to 0$. As seen in the "second plot," the envelope-density (dark-blue dotted curve) first goes to zero when $\eta \to {\eta }_s=\pi +B=2.876465$. As the red curve in the "first plot" shows, this is also where $M_r^{*}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ reaches its maximum value, $Q_m{\eta }_s=39.00000$. The gray-dotted curve in the "first plot" shows how the "core density" varies over the entire examined range. At the center — where $\eta \to 0$ and, hence, $\xi \to 0$ — the core density is unity; as $\eta$ climbs, the core density drops smoothly toward zero, but always remains positive. As the orange curve in the "first plot" shows, the integrated core mass is zero at $\eta =0$; as $\eta$ increases, the integrated core mass smoothly increases, heading toward a limiting value of $M_r^{*}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}\to 3^{3/2}(6/\pi {)}^{1/2}=7.18096$ as $\eta \to \mathrm{\infty }$ and, hence, $\xi \to \mathrm{\infty }$. As the "first plot" shows, the (red) curve representing the envelope mass intersects the (orange) curve representing the envelope mass twice. Moving from the center, outward, the first intersection occurs at the Model B2 core-envelope interface, where $\eta ={\eta }_i=0.352159$ and $\xi ={\xi }_i=2.47825$. As can be seen in the "second plot," the two "density" curves do not intersect at the interface. However, by design and construction, at the core-envelope interface the value of ${\rho }^{*}{|}_{\mathrm{e}\mathrm{n}\mathrm{v}}$ is precisely a factor of ${\mu }_e/{\mu }_c=0.25$ smaller than ${\rho }^{*}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}$; in the "second plot," the vertical red line-segment highlights this discontinuous drop in the density at the interface.

Model B2 — third plot	Model B2 — fourth plot
Things to notice:

Obtain ξ from η

Again, let's set ${\mu }_e/{\mu }_c=0.25$ but this time specify the value of ${\eta }_i$ and work backwards — through the definition of ${\theta }_i$ — to determine ${\xi }_i$. Specifically, we find that,

${\theta }_i^2{\xi }_i$	$=$	$\left(\frac{3{\xi }_i}{3+{\xi }_i^2}\right)=(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{3}^{-1/2}{\eta }_i$
$\Rightarrow 0$	$=$	$3-3c_0{\xi }_i+{\xi }_i^2$

where, $c_0\equiv ({\mu }_e/{\mu }_c)3^{1/2}{\eta }_i^{-1}$. That is,

$2{\xi }_i$	$=$	$3c_0\pm [3^2{c}_0^2-12{]}^{1/2}$
NOTE: Real root implies, ${\eta }_i\le \frac{3}{2}\left(\frac{{\mu }_e}{{\mu }_c}\right)$; and, at this limit, $({\xi }_i{)}_{\pm }=\sqrt{3}.$

TEST:   As in Model B2, set ${\mu }_e/{\mu }_c=0.25$ and set ${\eta }_i=0.352159$. Then, $c_0=1.22959405$ and, $({\xi }_i{)}_{+}=2.47825101$ while $({\xi }_i{)}_{-}=1.210531136$. The value for $({\xi }_i{)}_{+}$ matches the value for Model B2.

Keep in mind that, once ${\mu }_e/{\mu }_c$ and ${\xi }_i$ have been specified, other parameter values at the interface are: ${\theta }_i$ $=$ $(1+\frac{1}{3}{\xi }_i^2{)}^{-1/2},$ ${\eta }_i$ $=$ $\left(\frac{{\mu }_e}{{\mu }_c}\right)\sqrt{3}{\theta }_i^2{\xi }_i,$ ${\mathrm{\Lambda }}_i$ $=$ $\frac{1}{{\eta }_i}+(\frac{d\varphi }{d\eta }{)}_i=(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}\frac{1}{\sqrt{3}{\xi }_i{\theta }_i^2}-\frac{{\xi }_i}{\sqrt{3}},$ $A$ $=$ ${\eta }_i(1+{\mathrm{\Lambda }}_i^2{)}^{1/2},$ $B$ $=$ ${\eta }_i-\frac{\pi }{2}+{\tan }^{-1}({\mathrm{\Lambda }}_i),$ ${\eta }_s$ $=$ $B+\pi .$

$\frac{{\mu }_e}{{\mu }_c}=0.25$
${\eta }_i$	$c_0$	${\xi }_i$		${\theta }_i$	${\mathrm{\Lambda }}_i$	$A$	$B$	${\eta }_s$		$Q_{\rho }$	$Q_m$	$Q_r$		$q\equiv \frac{{\xi }_i}{Q_r{\eta }_s}$	$\nu \equiv \frac{M_r^{*}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_r^{*}{|}_{\mathrm{t}\mathrm{o}\mathrm{t}}}$
0.352159	1.229594	$(+)$ B2	2.478253	0.572857	1.408807	0.608404	- 0.265128	2.875465		0.00938347	13.55831	7.037311		0.122470	0.101429
		$(-)$	1.210530	0.819655	2.140726	0.832073	-0.084850	3.056743		0.0769585	12.95956	3.437456		0.115207	0.034078

Note as well that,

$q$	$\equiv$	$\frac{{\xi }_i}{{\xi }_s}=\frac{{\xi }_i}{Q_r{\eta }_s};$
$\nu$	$\equiv$	$\frac{M_r^{*}{|}_{\mathrm{c}\mathrm{o}\mathrm{r}\mathrm{e}}}{M_r^{*}{|}_{\mathrm{t}\mathrm{o}\mathrm{t}}}=\frac{\sin ({\eta }_i-B)-{\eta }_i\cos ({\eta }_i-B)}{{\eta }_s}.$

The figure here, on the right, is intended to illustrate that we can reproduce the results displayed in Figure 2 of our accompanying discussion — see also here. The displayed sequences are, as labeled, for ${\mu }_e/{\mu }_c=0.25$ and for ${\mu }_e/{\mu }_c=0.309$.

Consider Larger Interface-Value for Function φ

Core	Envelope
${\rho }^{*}\equiv \frac{\rho }{{\rho }_0}$ $=$ ${\theta }^5$ $P^{*}\equiv \frac{P}{K_c{\rho }_0^{6/5}}$ $=$ ${\theta }^6$ $r^{*}\equiv r\left[\frac{G^{1/2}{\rho }_0^{2/5}}{K_c^{1/2}}\right]$ $=$ $[\frac{3}{2\pi }{]}^{1/2}\xi$ $M_r^{*}\equiv M_r\left[\frac{G^{3/2}{\rho }_0^{1/5}}{K_c^{3/2}}\right]$ $=$ $4\pi \left[\frac{3}{2\pi }{]}^{3/2}\right(-{\xi }^2\frac{d\theta }{d\xi })$	${\rho }^{*}$ $=$ $\left(\frac{{\rho }_e}{{\rho }_0}\right)\varphi$ $P^{*}$ $=$ $\left[\frac{K_e{\rho }_e^2}{K_c{\rho }_0^{6/5}}\right]{\varphi }^2$ $r^{*}$ $=$ ${\rho }_0^{2/5}[\frac{K_e}{2\pi K_c}{]}^{1/2}\eta$ $M_r^{*}$ $=$ $4\pi \left[{\rho }_e{\rho }_0^{1/5}\right]\left[\frac{K_e}{2\pi K_c}{]}^{3/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$

Redo Interface Conditions

Now, at the core-envelope interface …

• ${\rho }^{*}{|}_c={\theta }_i^5$
• ${\rho }^{*}{|}_e=({\rho }_e/{\rho }_0){\varphi }_i$
• Leave specification of ${\varphi }_i$ arbitrary
• ${\rho }^{*}{|}_e{\mu }_c={\rho }^{*}{|}_c{\mu }_e$

Hence,

$\frac{{\rho }_e}{{\rho }_0}$

$=$

$\frac{{\rho }^{*}{|}_e}{{\varphi }_i}=\frac{{\rho }^{*}{|}_c}{{\varphi }_i}\left(\frac{{\mu }_e}{{\mu }_c}\right)=\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{{\theta }_i^5}{{\varphi }_i}.$

Also, setting the value of $P^{*}$ equal across the boundary gives us,

${\theta }_i^6$	$=$	$\left[\frac{K_e{\rho }_e^2}{K_c{\rho }_0^{6/5}}\right]{\varphi }_i^2={\rho }_0^{4/5}\left(\frac{{\rho }_e}{{\rho }_0}{)}^2\right(\frac{K_e}{K_c}){\varphi }_i^2={\rho }_0^{4/5}(\frac{K_e}{K_c}\left)\right(\frac{{\mu }_e}{{\mu }_c}{)}^2{\theta }_i^{10}$
$\Rightarrow {\rho }_0^{4/5}\left(\frac{K_e}{K_c}\right)$	$=$	$(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}$

As a result, throughout the envelope,

$P^{*}$	$=$	$\left[{\rho }_0^{4/5}\right(\frac{K_e}{K_c}\left)\right](\frac{{\rho }_e}{{\rho }_0}{)}^2{\varphi }^2=[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}\right]\left[\right(\frac{{\mu }_e}{{\mu }_c})\frac{{\theta }_i^5}{{\varphi }_i}{]}^2{\varphi }^2=[\frac{{\theta }_i^6}{{\varphi }_i^2}]{\varphi }^2;$
$r^{*}$	$=$	$(2\pi {)}^{-1/2}[{\rho }_0^{4/5}\cdot \frac{K_e}{K_c}{]}^{1/2}\eta =(2\pi {)}^{-1/2}[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}\right]\eta ;$
$M_r^{*}$	$=$	$2(2\pi {)}^{-1/2}[\frac{{\rho }_e}{{\rho }_0}\left]\right({\rho }_0^{4/5}\cdot \frac{K_e}{K_c}{)}^{3/2}(-{\eta }^2\frac{d\varphi }{d\eta })=2(2\pi {)}^{-1/2}[\left(\frac{{\mu }_e}{{\mu }_c}\right)\frac{{\theta }_i^5}{{\varphi }_i}\left]\right[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-4}{]}^{3/2}\right(-{\eta }^2\frac{d\varphi }{d\eta })$
	$=$	$\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\frac{1}{{\theta }_i{\varphi }_i}(-{\eta }^2\frac{d\varphi }{d\eta }).$

In summary, then,

Core

Envelope

${\rho }^{*}$ $=$ ${\theta }^5=[1+\frac{1}{3}{\xi }^2{]}^{-5/2}$ $P^{*}$ $=$ ${\theta }^6$ $r^{*}$ $=$ $[\frac{3}{2\pi }{]}^{1/2}\xi$ $M_r^{*}$ $=$ $4\pi \left[\frac{3}{2\pi }{]}^{3/2}\right(-{\xi }^2\frac{d\theta }{d\xi })$   $=$ $\left[\frac{2^4\cdot 3^3{\pi }^2}{2^3{\pi }^3}{]}^{1/2}\right\{\frac{{\xi }^3}{3}[1+\frac{1}{3}{\xi }^2{]}^{-3/2}\}$   $=$ $\left(\frac{6}{\pi }{)}^{1/2}{\xi }^3\right[1+\frac{1}{3}{\xi }^2{]}^{-3/2}$

${\rho }^{*}$ $=$ $\left[\right(\frac{{\mu }_e}{{\mu }_c}\left)\frac{{\theta }_i^5}{{\varphi }_i}\right]\varphi =\frac{A}{{\varphi }_i}\left(\frac{{\mu }_e}{{\mu }_c}\right){\theta }_i^5\left[\frac{\sin (\eta -B)}{\eta }\right]$ $P^{*}$ $=$ $\left(\frac{{\theta }_i^6}{{\varphi }_i^2}\right){\varphi }^2$ $r^{*}$ $=$ $(2\pi {)}^{-1/2}[\left(\frac{{\mu }_e}{{\mu }_c}{)}^{-1}{\theta }_i^{-2}\right]\eta$ $M_r^{*}$ $=$ $\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}\frac{1}{{\theta }_i{\varphi }_i}(-{\eta }^2\frac{d\varphi }{d\eta })$   $=$ $\left(\frac{2}{\pi }{)}^{1/2}\right(\frac{{\mu }_e}{{\mu }_c}{)}^{-2}{\theta }_i^{-1}\left\{\frac{A}{{\varphi }_i}\right[\sin (\eta -B)-\eta \cos (\eta -B)\left]\right\}$

Eureka (NOT!)

Here is an approach that has been mapped out with some success today (23 July 2023).

Lane-Emden Equation

Drawing from our accompanying general description of the Lane-Emden equation, we need to solve the following second-order ODE relating the two unknown functions, $\rho$ and $H$:

$\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dH}{dr}\right)=-4\pi G\rho$ .

It is customary to replace $H$ and $\rho$ in this equation by a dimensionless polytropic enthalpy, ${\mathrm{\Theta }}_H$, such that,

${\mathrm{\Theta }}_H\equiv \frac{H}{H_c}=(\frac{\rho }{{\rho }_c}{)}^{1/n},$

where the mathematical relationship between $H/H_c$ and $\rho /{\rho }_c$ comes from the adopted barotropic (polytropic) relation. To accomplish this, we replace $H$ with $H_c{\mathrm{\Theta }}_H$ on the left-hand-side of the governing differential equation and we replace $\rho$ with ${\rho }_c{\mathrm{\Theta }}_H^n$ on the right-hand-side, then gather the constant coefficients together on the left. The resulting ODE is,

$\left[\frac{1}{4\pi G}\right(\frac{H_c}{{\rho }_c}\left)\right]\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{d{\mathrm{\Theta }}_H}{dr}\right)=-{\mathrm{\Theta }}_H^n$ ,

or,

$\frac{a_n^2}{r^2}\cdot \frac{d}{d(r/a_n)}\left[\right(\frac{r}{a_n}{)}^2\frac{d{\mathrm{\Theta }}_H}{dr}]=-{\mathrm{\Theta }}_H^n$ ,

where,

$a_{\mathrm{n}}\equiv \left[\frac{1}{4\pi G}\right(\frac{H_c}{{\rho }_c}){]}^{1/2}=[\frac{(n+1)K_n}{4\pi G}\cdot {\rho }_c^{(1-n)/n}{]}^{1/2}.$

If we follow tradition and define a normalized radius, $\xi \equiv r/a_n$, then our governing ODE becomes what is referred to in the astrophysics literature as the,

Lane-Emden Equation

$\frac{1}{{\xi }^2}\frac{d}{d\xi }\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right)=-{\mathrm{\Theta }}_H^n$

Our task is to solve this ODE to determine the behavior of the function ${\mathrm{\Theta }}_H(\xi )$ — and, from it in turn, determine the radial distribution of various dimensional physical variables — for various values of the polytropic index, $n$.

ASIDE:  Following along the lines of Chapter IV of [C67] — see also our accompanying derivation — the expression for the enclosed mass is,

$M_r$	$=$	$\underset{0}{\overset{r}{\int }}4\pi \rho r^{2}dr=4\pi {\rho }_c{a}_n^3\underset{0}{\overset{\xi }{\int }}{\mathrm{\Theta }}_H^n{\xi }^{2}d\xi$
	$=$	$-4\pi {\rho }_c{a}_n^3\underset{0}{\overset{\xi }{\int }}\left[\frac{1}{{\xi }^2}\frac{d}{d\xi }\right({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\left)\right]{\xi }^{2}d\xi$
	$=$	$-4\pi {\rho }_c{a}_n^3\left({\xi }^2\frac{d{\mathrm{\Theta }}_H}{d\xi }\right).$
§IV.5.b of [C67], p. 97, Eq. (67)

Modified Approach for n = 1

Modified Governing ODE

Here, we deviate from tradition and adopt the following expression for the dimensionless radius,

$\eta -C$

$\equiv$

$\frac{r}{a_n}$

and, adopt the above notation for the (n_e = 1) envelope of our bipolytrope — namely, ${\mathrm{\Theta }}_H\to \varphi$ — the governing ODE becomes,

$-\varphi$

$=$

$\frac{1}{(\eta -C{)}^2}\frac{d}{d\eta }[(\eta -C{)}^2\frac{d\varphi }{d\eta }].$

Modified Expression for Integrated Mass

The modified expression for the integrated mass in our $n=1$ envelope is,

$M_r$	$=$	$\underset{r=0}{\overset{r=R}{\int }}4\pi \rho r^{2}dr=4\pi {\rho }_c{a}_n^3\underset{\eta =C}{\overset{\eta =C+R/a_n}{\int }}\varphi (\eta -C{)}^{2}d\eta$
	$=$	$-4\pi {\rho }_c{a}_n^3\underset{\eta =C}{\overset{\eta =C+R/a_n}{\int }}\left\{\frac{1}{(\eta -C{)}^2}\frac{d}{d\eta }\right[(\eta -C{)}^2\frac{d\varphi }{d\eta }]\}(\eta -C{)}^{2}d\eta$
	$=$	$[-4\pi {\rho }_c{a}_n^3(\eta -C{)}^2\frac{d\varphi }{d\eta }{]}_{\eta =C}^{\eta =C+R/a_n}.$

Guess Solution

Let's guess a solution of the form,

$\varphi$

$=$

$A\cdot \frac{\sin (\eta -B)}{(\eta -C)},$

in which case,

$\frac{1}{A}\cdot \frac{d\varphi }{d\eta }$

$=$

$\frac{\cos (\eta -B)}{(\eta -C)}-\frac{\sin (\eta -B)}{(\eta -C{)}^2};$

and,

$\frac{1}{A}\cdot \frac{d^2\varphi }{d{\eta }^2}$

$=$

$-\frac{\sin (\eta -B)}{(\eta -C)}-\frac{2\cos (\eta -B)}{(\eta -C{)}^2}+\frac{2\sin (\eta -B)}{(\eta -C{)}^3}.$

Hence, the RHS of the governing ODE becomes,

RHS	$=$	$\frac{1}{(\eta -C{)}^2}\frac{d}{d\eta }[(\eta -C{)}^2\frac{d\varphi }{d\eta }]=\frac{d^2\varphi }{d{\eta }^2}+\frac{2}{(\eta -C)}\cdot \frac{d\varphi }{d\eta }$
	$=$	$A[-\frac{\sin (\eta -B)}{(\eta -C)}-\frac{2\cos (\eta -B)}{(\eta -C{)}^2}+\frac{2\sin (\eta -B)}{(\eta -C{)}^3}]+A[\frac{2\cos (\eta -B)}{(\eta -C{)}^2}-\frac{2\sin (\eta -B)}{(\eta -C{)}^3}]$
	$=$	$A[-\frac{\sin (\eta -B)}{(\eta -C)}]=-\varphi ,$

which is precisely the expression for the LHS of the governing ODE.

See Also