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===Is There an Analytic Expression for the Eigenfunction?=== After noticing that, in Figure 6, the ''envelope'' segments of all of the marginally unstable eigenfunctions merge into the same curve, we began to wonder whether a single expression — and, even better, an ''analytically defined'' expression — would perfectly describe the eigenfunction. We had reason to believe that this might actually be possible because, in [[SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|pressure-truncated polytropic configurations, we have derived analytic expressions for the marginally unstable, fundamental-mode eigenfunctions]] of both <math>~n = 5</math> and <math>~n=1</math> systems. Very quickly, we convinced ourselves that a parabolic function does indeed perfectly match the "core" segment of each displayed eigenfunction. Specifically, throughout the core <math>~(0 \le \xi \le \xi_i)</math>, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_P\biggr|_\mathrm{core}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~1 - \frac{\xi^2}{15}</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~\frac{dx_P}{d\xi}\biggr|_\mathrm{core}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~- \frac{2\xi}{15} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~\frac{d\ln x_P}{d\ln \xi}\biggr|_\mathrm{core}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~- \frac{2\xi^2}{15} \biggl[ \frac{(15 - \xi^2)}{15} \biggr]^{-1} = - \frac{2\xi^2}{(15 - \xi^2)} \, .</math> </td> </tr> </table> The envelope segment posed a much greater challenge. =====Attempt 1===== Building on our [[SSC/Stability/InstabilityOnsetOverview#Configurations_Having_an_Index_Less_Than_Three|accompanying discussion of ''Pressure-Truncated Configurations Having a Polytropic Index less than Three'']] — see, for example, a relevant [[SSC/Stability/n1PolytropeLAWE#Succinct_Demonstration|''succinct demonstration'']] — a promising analytic expression is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_P\biggr|_\mathrm{env}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{b}{\eta^2} \biggl[1 - \eta\cot(\eta - C) \biggr] \, , </math> </td> </tr> </table> where the values of the pair of coefficients, <math>~b</math> and <math>~C</math>, is to be determined. Most likely, we should set <math>~C = B</math>. Focusing on the case of <math>~\mu_e/\mu_c = 1</math>, <math>~\gamma_c = 6/5</math>, and <math>~\gamma_e = 2</math>, here are some parameters that we ''think'' we know. <table border="1" align="center" cellpadding="10"> <tr> <th align="center" colspan="5">Relevant Parameters for<br />Marginally Unstable Model <math>~(\Omega^2 = 0)</math> <br />with <math>~\mu_e/\mu_c = 1</math> and <math>~(\gamma_c, \gamma_e) = (\tfrac{6}{5}, 2)</math></th> </tr> <tr> <td align="right"><math>~\xi_i</math></td> <td align="center"> </td> <td align="center">numerically<br />determined</td> <td align="center"><math>~\approx</math></td> <td align="left">1.6686460157</td> </tr> <tr> <td align="right"><math>~x_i</math></td> <td align="center"><math>~\equiv</math></td> <td align="left"><math>~1 - \frac{\xi_i^2}{15} </math></td> <td align="center"><math>~=</math></td> <td align="left">0.814374699</td> </tr> <tr> <td align="right"><math>~\biggl[\frac{d\ln x}{d\ln \xi}\biggr]_i</math></td> <td align="center"><math>~\equiv</math></td> <td align="left"><math>~- \frac{2\xi_i^2}{(15 - \xi_i^2)}</math></td> <td align="center"><math>~=</math></td> <td align="left">-0.455871976</td> </tr> <tr> <td align="right"><math>~\eta_i</math></td> <td align="center"><math>~\equiv</math></td> <td align="left"><math>~\biggl(\frac{\mu_e}{\mu_c}\biggr) \sqrt{3} \xi_i \biggl[1 + \frac{\xi_i^2}{3}\biggr]^{-1 } </math></td> <td align="center"><math>~=</math></td> <td align="left">1.498957494</td> </tr> <tr> <td align="right"><math>~B</math></td> <td align="center"><math>~\equiv</math></td> <td align="left"><math>~\eta_i - \frac{\pi}{2} + \tan^{-1}\biggl[ \frac{1}{\eta_i} - \frac{\xi_i}{\sqrt{3}} \biggr] </math></td> <td align="center"><math>~=</math></td> <td align="left">- 0.359863579</td> </tr> <tr> <td align="right"><math>~\eta_s</math></td> <td align="center"><math>~\equiv</math></td> <td align="left"><math>~\pi + B </math></td> <td align="center"><math>~=</math></td> <td align="left">2.781729075</td> </tr> <tr> <td align="right"><math>~\biggl[\frac{d\ln x}{d\ln \eta}\biggr]_i</math></td> <td align="center"><math>~\equiv</math></td> <td align="left"><math>~3\biggl( \frac{\gamma_c}{\gamma_e}-1\biggr) + \frac{\gamma_c}{\gamma_e} \biggl[\frac{d\ln x}{d\ln \xi}\biggr]_i</math></td> <td align="center"><math>~=</math></td> <td align="left">-1.473523186</td> </tr> <tr> <td align="right"><math>~\alpha_e</math></td> <td align="center"><math>~\equiv</math></td> <td align="left"><math>~3 - \frac{4}{\gamma_e} </math></td> <td align="center"><math>~=</math></td> <td align="left">+ 1</td> </tr> <tr> <td align="right"><math>~\biggl[\frac{d\ln x}{d\ln \eta}\biggr]_\mathrm{surf}</math></td> <td align="center"><math>~\equiv</math></td> <td align="left"><math>~\frac{\cancelto{0}{\Omega^2}}{\gamma_e} - \alpha_e </math></td> <td align="center"><math>~=</math></td> <td align="left">- 1</td> </tr> <tr> <td align="right"><math>~b</math></td> <td align="center"><math>~\equiv</math></td> <td align="left"><math>~\frac{3}{5}\biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{15-\xi_i^2}{3+\xi_i^2}\biggr]</math></td> <td align="center"><math>~=</math></td> <td align="left">1.26097406</td> </tr> </table> Notice that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\cot(\eta_i - B)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan\biggl[ \frac{\pi}{2} -(\eta_i - B ) \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ \frac{1}{\eta_i} - \frac{\xi_i}{\sqrt{3}} \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \eta_i \cot(\eta_i - B)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left" colspan="3"> <math>~\biggl[ 1 - \frac{\xi_i \eta_i}{\sqrt{3}} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left" colspan="3"> <math>~1 - \xi_i^2 \biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[1 + \frac{\xi_i^2}{3}\biggr]^{-1 } </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left" colspan="3"> <math>~1 - \biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{3\xi_i^2 }{3 + \xi_i^2}\biggr] \, .</math> </td> </tr> </table> So if we adopt the expression for <math>~x_P</math> as given above (with C = B), then we can evaluate the leading ''b'' factor by examining the function at the interface, that is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~b</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ x_i \eta_i^2 \biggl[1 - \eta_i \cot(\eta_i - B) \biggr]^{-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ x_i \eta_i^2 \biggl[\frac{\xi_i \eta_i }{\sqrt{3}} \biggr]^{-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \eta_i \biggl[\frac{\sqrt{3}}{\xi_i} \biggr]x_i </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl(\frac{\mu_e}{\mu_c}\biggr) \sqrt{3} \xi_i \biggl[1 + \frac{\xi_i^2}{3}\biggr]^{-1 } \biggl[\frac{\sqrt{3}}{\xi_i} \biggr] \biggl[1 - \frac{\xi_i^2}{15} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 3\biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{3}{3+\xi_i^2}\biggr]\biggl[\frac{15-\xi_i^2}{15} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3}{5}\biggl(\frac{\mu_e}{\mu_c}\biggr) \biggl[\frac{15-\xi_i^2}{3+\xi_i^2}\biggr] \, . </math> </td> </tr> </table> <table border="0" align="right"> <tr> <th align="center">Figure 7: Analytic Trial</th> </tr> <tr><td align="center">[[File:AnalyticAttempt1.png|400px|First Analytic Trial]]</td></tr> </table> An even clearer way of looking at this is to realize that, quite generally, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(\eta - B)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\eta - \eta_i + \frac{\pi}{2} - \tan^{-1} f \, ,</math> </td> </tr> </table> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~f</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\frac{1}{\eta_i} - \frac{\xi_i}{\sqrt{3}} \, .</math> </td> </tr> </table> Hence, we can write, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\cot(\eta - B)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan\biggl[ \frac{\pi}{2} - (\eta - B) \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan\biggl[ (\eta_i - \eta) + \tan^{-1}f \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\tan(\eta_i - \eta) + f}{1 - f\tan(\eta_i - \eta)} \, .</math> </td> </tr> </table> As a result, we can rewrite the expression for our ''guess'' of the envelope segment of the eigenfunction in the form, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_P\biggr|_\mathrm{env}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{b}{\eta^2} \biggl\{ 1 - \eta \biggl[ \frac{\tan(\eta_i - \eta) + f}{1 - f\tan(\eta_i - \eta)} \biggr] \biggr\} \, . </math> </td> </tr> </table> This blows up when <math>\eta \rightarrow \eta_s</math> because, as it turns out, <math>~f = 1/\tan(\eta_i - \eta_s)</math>. We should point out, as well, that the expression for ''b'' can be rewritten in the form, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~b</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\eta_i^2 (x_P)_i}{(1 - \eta_i f)} \, . </math> </td> </tr> </table> In Figure7 we have reprinted the numerically determined fundamental-mode eigenfunction that was first displayed in [[#Eigenfunction_Details|Figure 5, above]]. We have added to this plot the eigenfunction segments that are defined by our trial analytic functions: The ''core'' segment, <math>~x_P|_\mathrm{core}</math>, matches the numerically determined segment with sufficient precision that the two curve segments are indiscernible from one another. However, our analytically defined "env" segment, <math>~x_P|_\mathrm{env}</math> — identified in Figure 7 by the solid black, small circular markers — does not match the numerically determined envelope segment at all. We therefore have more work to do! =====Attempt 2===== Using an Excel spreadsheet as a sandbox, we employed crude, brute force iterations in an effort to fit the numerically constructed envelope eigenfunction. Here is a trial function that works pretty well. Using <math>~\eta_\mathrm{F}</math> to represent the envelope's dimensionless radial coordinate, over the range, <div align="center"> <math>~\eta_i \le \eta_\mathrm{F} \le \eta_s \, ,</math> </div> and defining the parameter, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~g_\mathrm{F}</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \frac{\pi}{8(\eta_s - \eta_i)} \, ,</math> </td> </tr> </table> <table border="1" cellpadding="5" align="right"> <tr> <th align="center" colspan="4">Limiting Parameter Values</th> </tr> <tr> <td align="center"> </td> <td align="center">min</td> <td align="center">max</td> <td align="center"><math>~\alpha = \alpha_s</math> </tr> <tr> <td align="center"><math>~\eta_\mathrm{F}</math></td> <td align="center"><math>~\eta_i</math></td> <td align="center"><math>~\eta_s</math></td> <td align="center"><math>~\frac{8}{\pi} ( \eta_s - \eta_i )^2 + 2\eta_s - \eta_i</math></td> </tr> <tr> <td align="center"><math>~\alpha</math></td> <td align="center"><math>~-\frac{\pi}{2}</math></td> <td align="center"><math>~-\frac{5\pi}{8}</math></td> <td align="center"><math>~\eta_i - \eta_s - \frac{3\pi}{4}</math></td> </tr> <tr> <td align="center"><math>~\Lambda</math></td> <td align="center"><math>~\eta_i - \frac{\pi}{4}</math></td> <td align="center"><math>~\eta_i - \frac{\pi}{8}</math></td> <td align="center"><math>~\eta_s</math></td> </tr> </table> we propose, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_\mathrm{trial}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \biggl[ \frac{\tan(\alpha) + f_\alpha}{1 - f_\alpha \cdot \tan(\alpha)} \biggr] \biggr\} - a_0 \, , </math> </td> </tr> </table> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{f_\alpha} = \tan(\alpha_s)</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \tan[ - (\eta_s - \eta_i + \tfrac{3\pi}{4}) ] \, ,</math> </td> </tr> <tr> <td align="right"> <math>~\alpha(\eta_\mathrm{F})</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~g_\mathrm{F} \biggl[ 5\eta_i - 4\eta_s - \eta_\mathrm{F}\biggr] \, ,</math> </td> </tr> <tr> <td align="right"> <math>~\Lambda(\eta_\mathrm{F})</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~ \eta_i + g_\mathrm{F} \biggl[ \eta_i - 2\eta_s + \eta_\mathrm{F} \biggr] \, .</math> </td> </tr> </table> <table border="0" align="right"> <tr> <th align="center">Figure 8: Another Analytic Trial</th> </tr> <tr><td align="center">[[File:AnalyticAttempt2.png|400px|Second Analytic Trial]]</td></tr> </table> This function, <math>~x_\mathrm{trial}</math>, is displayed as the black-dotted curve segment in Figure 8 with the tuning/scaling parameters set to the values, <math>~(a_0, b_0) = (0.31, 0.96)</math>. We should point out that, when plotting this curve segment in Figure 8, the dimensionless radial coordinate has been defined by the relation, <math>~r^*/R^* = \eta_F/\eta_s</math>. <font color="red">'''[16 February 2019: Comment by Tohline]'''</font> When assessed visually, this ''trial'' function appears to match pretty well the numerically derived eigenfunction for the envelope. We have not yet critically assessed whether or not the function satisfies the LAWE or whether it satisfies either one (or both) of the required boundary values. This work is still to be done. A couple of days after inserting this <font color="red">''Comment''</font>, we recognized for the first time that, quite generally, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\alpha</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\eta_i - \biggl(\Lambda + \frac{3\pi}{4} \biggr) \, .</math> </td> </tr> </table> Hence, the parameter, <math>~\alpha</math>, can be straightforwardly removed from the expression for the trial eigenfunction to give, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_\mathrm{trial}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{b_0}{\Lambda^2} \biggl\{ 1 - \Lambda \biggl[ \frac{\tan(\eta_i - \Lambda - 3\pi/4) + f_\alpha}{1 - f_\alpha \cdot \tan(\eta_i - \Lambda - 3\pi/4)} \biggr] \biggr\} - a_0 \, . </math> </td> </tr> </table> Drawing from our [[SSC/Stability/InstabilityOnsetOverview#Polytropic_Stability|accompanying discussion of pressure-truncated polytropes]], we need the eigenfunction to satisfy the, <div align="center"> <font color="maroon"><b>Polytropic LAWE (linear adiabatic wave equation)</b></font><br /> {{ Math/EQ_RadialPulsation02 }} </div> (Note that, in order to bring the notation of this ''Key Equation'' in line with the notation used elsewhere in this chapter, we will hereafter adopt the variable mapping <math>~\xi \rightarrow \eta</math> and <math>~\theta \rightarrow \phi</math>.) Here we are especially focused on finding a solution in the case where <math>~\sigma_c^2 = 0</math> and <math>~n = 1</math>, that is — see also our [[#Envelope:|above discussion]] — the relevant envelope LAWE is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{d^2x}{d\eta^2} + \biggl[ 2 - Q \biggr] \frac{2}{\eta}\cdot \frac{dx}{d\eta} -2\alpha_g Q \cdot \frac{x}{\eta^2} \, , </math> </td> </tr> </table> where, drawing from our [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|discussion of the n = 1 envelope's equilibrium structure]], <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~Q</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ - \frac{d\ln\phi}{d\ln\eta} \biggr]_\mathrm{n=1} = - \frac{\eta}{\phi}\biggl[ \frac{d\phi}{d\eta} \biggr]_\mathrm{n=1}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \biggl[ \frac{1}{\sin(\eta - B)} \biggr]\biggl[ \eta\cos(\eta-B) - \sin(\eta-B) \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[1 - \eta\cot(\eta-B) \biggr] \, .</math> </td> </tr> </table> In an [[SSC/Stability/n1PolytropeLAWE#Succinct_Demonstration|accompanying discussion]] — see also a [[SSC/Stability/InstabilityOnsetOverview#Configurations_Having_an_Index_Less_Than_Three|short summary of the same]] — we have shown that an analytically specified displacement function that precisely satisfies this LAWE for pressure-truncated configurations (i.e., when B = 0) is, <div align="center"> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_P\biggr|_{n=1}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3}{\eta^2}\biggl[ 1- \eta \cot\eta \biggr] \, . </math> </td> </tr> </table> </div> In still [[SSC/Stability/n1PolytropeLAWE#What_About_Bipolytropes.3F|another related discussion]], we have attempted to construct an analytic eigenfunction expression that satisfies the LAWE when <math>~B \ne 0</math>. =====Attempt 3===== ======Straightforward Trial====== Let's adopt a trial eigenfunction of the form <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_\mathrm{trial}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{b_0}{\Lambda^2} \biggl[ 1 - \Lambda \cot(\Lambda - E) \biggr] - a_0 \, , </math> </td> </tr> </table> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Lambda</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Lambda_0 + g_\mathrm{F} \eta \, .</math> </td> </tr> </table> <table border="1" width="60%" align="center" cellpadding="8"><tr><td align="left"> <font color="red">'''NOTE:'''</font> We can retrieve the empirical expression for <math>~x_\mathrm{trial}</math> obtained above in '''Attempt 2''' if we eventually set, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Lambda_0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\eta_i + g_\mathrm{F}(\eta_i - 2\eta_s)</math> </td> </tr> <tr> <td align="right"> <math>~g_\mathrm{F}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\pi}{8(\eta_s - \eta_i)} \, ,</math> and, </td> </tr> <tr> <td align="right"> <math>~E</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\eta_i - \frac{5\pi}{4} + \tan^{-1} f_\alpha \, .</math> </td> </tr> </table> The last of these expressions arises because, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\cot(\Lambda - E)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan \biggl[ \frac{\pi}{2} - (\Lambda - E)\biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan\biggl[ \frac{\pi}{2} - \Lambda + \eta_i - \frac{5\pi}{4} + \tan^{-1} f_\alpha \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan\biggl[ (\eta_i - \Lambda - \tfrac{3\pi}{4}) + \tan^{-1}f_\alpha \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{\tan(\eta_i - \Lambda - \tfrac{3\pi}{4}) + f_\alpha }{1 - f_\alpha \tan(\eta_i - \Lambda - \tfrac{3\pi}{4} )} \, .</math> </td> </tr> </table> </td></tr></table> Because the LAWE requires derivatives of <math>~x_\mathrm{trial}</math> with respect to <math>~\eta</math>, we will often need to recognize that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d}{d\eta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{d\Lambda}{d\eta} \cdot \frac{d}{d\Lambda} = g_\mathrm{F} \cdot \frac{d}{d\Lambda} \, .</math> </td> </tr> </table> Hence, in particular, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d}{d\eta}\biggl[\cot(\Lambda - E) \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- g_\mathrm{F} [1 + \cot^2(\Lambda - E) ] \, .</math> </td> </tr> </table> The first derivative gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d}{d\eta}\biggl[ x_\mathrm{trial} \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{d}{d\eta} \biggl\{ \frac{b_0}{\Lambda^2} \biggl[ 1 - \Lambda \cot(\Lambda - E) \biggr] - a_0 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -~\frac{2b_0 g_\mathrm{F}}{\Lambda^3} \biggl[1 - \Lambda \cot(\Lambda - E)\biggr] + \frac{b_0}{\Lambda^2} \biggl[- g_\mathrm{F} \cot(\Lambda - E)\biggr] + \frac{b_0}{\Lambda^2} \biggl[g_\mathrm{F} \Lambda [1 + \cot^2(\Lambda - E) ] \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ g_\mathrm{F} b_0 \biggl\{ -~\frac{2}{\Lambda^3} \biggl[1 - \Lambda \cot(\Lambda - E)\biggr] - \frac{1}{\Lambda^2} \biggl[\cot(\Lambda - E)\biggr] + \frac{1}{\Lambda} \biggl[1 + \cot^2(\Lambda - E) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ g_\mathrm{F} b_0 \biggl\{ \frac{1}{\Lambda} -\frac{2}{\Lambda^3} +\frac{1}{\Lambda^2} \biggl[\cot(\Lambda - E)\biggr] + \frac{1}{\Lambda} \biggl[\cot^2(\Lambda - E) \biggr] \biggr\} \, . </math> </td> </tr> </table> The second derivative gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{g_\mathrm{F}b_0} \cdot \frac{d^2}{d\eta^2}\biggl[ x_\mathrm{trial} \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{d}{d\eta}\biggl\{ \frac{1}{\Lambda} -\frac{2}{\Lambda^3} +\frac{1}{\Lambda^2} \biggl[\cot(\Lambda - E)\biggr] + \frac{1}{\Lambda} \biggl[\cot^2(\Lambda - E) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\frac{g_\mathrm{F}}{\Lambda^2} +\frac{6 g_\mathrm{F}}{\Lambda^4} -\frac{2g_\mathrm{F}}{\Lambda^3} \biggl[\cot(\Lambda - E)\biggr] - \frac{g_\mathrm{F}}{\Lambda^2} \biggl[1 + \cot^2(\Lambda - E)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \frac{g_\mathrm{F}}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr] - \frac{2g_\mathrm{F}}{\Lambda} \biggl[\cot(\Lambda - E) \biggr] \biggl[1 + \cot^2(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \frac{1}{g^2_\mathrm{F}b_0} \cdot \frac{d^2}{d\eta^2}\biggl[ x_\mathrm{trial} \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\frac{1}{\Lambda^2} +\frac{6 }{\Lambda^4} -\frac{2}{\Lambda^3} \biggl[\cot(\Lambda - E)\biggr] - \frac{1}{\Lambda^2} - \frac{1}{\Lambda^2} \biggl[\cot^2(\Lambda - E)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \frac{1}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr] - \frac{2}{\Lambda} \biggl[\cot(\Lambda - E) + \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\frac{1}{\Lambda^2} - \frac{1}{\Lambda^2} +\frac{6 }{\Lambda^4} -\frac{2}{\Lambda^3} \biggl[\cot(\Lambda - E)\biggr] - \frac{2}{\Lambda} \biggl[\cot(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \frac{1}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr] - \frac{1}{\Lambda^2} \biggl[\cot^2(\Lambda - E)\biggr] - \frac{2}{\Lambda} \biggl[ \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\frac{2}{\Lambda^2} +\frac{6 }{\Lambda^4} - \biggl[ \frac{2}{\Lambda^3} + \frac{2}{\Lambda}\biggr]\biggl[\cot(\Lambda - E)\biggr] - \frac{2}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr] - \frac{2}{\Lambda} \biggl[ \cot^3(\Lambda - E) \biggr] \, . </math> </td> </tr> </table> So, appreciating that, <math>~\eta = (\Lambda - \Lambda_0 )/g_\mathrm{F}</math>, and dividing the relevant LAWE through by <math>~b_0 g_\mathrm{F}^2</math>, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{b_0 g_\mathrm{F}^2} \cdot \frac{d^2x_\mathrm{trial} }{d\eta^2} + \frac{1}{b_0 g_\mathrm{F}^2} \biggl[ 2 - Q \biggr] \frac{2g_\mathrm{F}}{(\Lambda - \Lambda_0 )}\cdot \frac{dx_\mathrm{trial}}{d\eta} - \frac{2Q}{b_0 g_\mathrm{F}^2} \cdot \frac{g^2_\mathrm{F} x_\mathrm{trial}}{(\Lambda - \Lambda_0 )^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{(\Lambda - \Lambda_0 )^2} \biggl\{ \frac{(\Lambda - \Lambda_0 )^2}{b_0 g_\mathrm{F}^2} \cdot \frac{d^2x}{d\eta^2} + \frac{ (\Lambda - \Lambda_0 )}{b_0 g_\mathrm{F}} \biggl[ 2 - Q \biggr] \frac{dx}{d\eta} - \frac{2Q}{b_0 } \cdot x \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \frac{ (\Lambda - \Lambda_0 )^2}{b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (\Lambda - \Lambda_0 )^2 \biggl\{ \frac{1}{b_0 g_\mathrm{F}^2} \cdot \frac{d^2x}{d\eta^2} \biggr\} + (\Lambda - \Lambda_0 ) \biggl[ 4 - 2Q \biggr] \biggl\{ \frac{1}{b_0 g_\mathrm{F}} \frac{dx}{d\eta} \biggr\} - 2Q \biggl\{ \frac{x}{b_0} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (\Lambda - \Lambda_0 )^2 \biggl\{ -\frac{2}{\Lambda^2} +\frac{6 }{\Lambda^4} - \biggl[ \frac{2}{\Lambda^3} + \frac{2}{\Lambda}\biggr]\biggl[\cot(\Lambda - E)\biggr] - \frac{2}{\Lambda^2} \biggl[\cot^2(\Lambda - E) \biggr] - \frac{2}{\Lambda} \biggl[ \cot^3(\Lambda - E) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (\Lambda - \Lambda_0 ) \biggl[ 4 - 2Q \biggr] \biggl\{ \frac{1}{\Lambda} -\frac{2}{\Lambda^3} +\frac{1}{\Lambda^2} \biggl[\cot(\Lambda - E)\biggr] + \frac{1}{\Lambda} \biggl[\cot^2(\Lambda - E) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - 2Q \biggl\{ \frac{1}{\Lambda^2} \biggl[1 - \Lambda \cot(\Lambda - E) \biggr] - \frac{ a_0 }{b_0} \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (\Lambda - \Lambda_0 )^2 \biggl\{ (6-2\Lambda^2) - (2\Lambda + 2\Lambda^3 ) \cot(\Lambda - E) - 2\Lambda^2 \cot^2(\Lambda - E) - 2\Lambda^3 \cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (\Lambda - \Lambda_0 ) (4 - 2Q ) \biggl\{ \Lambda^3 -2\Lambda +\Lambda^2 \cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - 2Q \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (\Lambda - \Lambda_0 ) \biggl\{ (3-\Lambda^2)(\Lambda - \Lambda_0 ) - (\Lambda + \Lambda^3 )(\Lambda - \Lambda_0 ) \cot(\Lambda - E) - \Lambda^2 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E) - \Lambda^3 (\Lambda - \Lambda_0 )\cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^3 -4\Lambda +2\Lambda^2 \cot(\Lambda - E) + 2\Lambda^3 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -Q \biggl\{ \Lambda^3(\Lambda - \Lambda_0 ) -2\Lambda (\Lambda - \Lambda_0 ) +\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E) + \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - Q \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} \, . </math> </td> </tr> </table> ---- The right-hand-side of this expression should simplify considerably if we let <math>~E \rightarrow B</math>, if we set <math>~(g_\mathrm{F}, \Lambda_0) = (1, 0) ~\Rightarrow~ \Lambda = \eta</math>, and if <math>~(a_0, b_0) = (0, 3)</math>. Let's see. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> RHS </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Lambda \biggl\{ (3-\Lambda^2)\Lambda - (\Lambda + \Lambda^3 )\Lambda \cot(\Lambda - E) - \Lambda^3 \cot^2(\Lambda - E) - \Lambda^4 \cot^3(\Lambda - E) +2\Lambda^3 -4\Lambda +2\Lambda^2 \cot(\Lambda - E) + 2\Lambda^3 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -Q \biggl\{ \Lambda^4 -2\Lambda^2 +\Lambda^3 \cot(\Lambda - E) + \Lambda^4 \cot^2(\Lambda - E) +\Lambda^2 - \Lambda^3 \cot(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Lambda \biggl\{ (-\Lambda +\Lambda^3) + (\Lambda^2 - \Lambda^4) \cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) - \Lambda^4 \cot^3(\Lambda - E) -\Lambda^3 + \Lambda - \Lambda^3 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +\Lambda\cot(\lambda-E) \biggl\{ \Lambda^4 -2\Lambda^2 +\Lambda^3 \cot(\Lambda - E) + \Lambda^4 \cot^2(\Lambda - E) +\Lambda^2 - \Lambda^3 \cot(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Lambda \biggl\{ (\Lambda^2 - \Lambda^4) \cot(\Lambda - E) - \Lambda^4 \cot^3(\Lambda - E) \biggr\} +\Lambda\cot(\lambda-E) \biggl\{ \Lambda^4 -\Lambda^2 + \Lambda^4 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 0 \, . </math> </td> </tr> </table> EXCELLENT !! ---- Let's return to the more general expression and see if it can be simplified. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Lambda (\Lambda - \Lambda_0 ) \biggl\{ (3-\Lambda^2) - (\Lambda + \Lambda^3 ) \cot(\Lambda - E) - \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \Lambda_0 (\Lambda - \Lambda_0 ) \biggl\{ (3-\Lambda^2) - (\Lambda + \Lambda^3 ) \cot(\Lambda - E) - \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \Lambda (\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^2 -4 +2\Lambda \cot(\Lambda - E) + 2\Lambda^2 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -Q(\Lambda - \Lambda_0 ) \biggl\{ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr\} - Q \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Lambda (\Lambda - \Lambda_0 ) \biggl\{ (\Lambda^2-1) + (\Lambda - \Lambda^3 ) \cot(\Lambda - E) + \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr\} - \Lambda (\Lambda - \Lambda_0 ) \biggl\{ \Lambda^2 -2 +\Lambda \cot(\Lambda - E) + \Lambda^2 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \Lambda_0 (\Lambda - \Lambda_0 ) \biggl\{ (3-\Lambda^2) - (\Lambda + \Lambda^3 ) \cot(\Lambda - E) - \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr\} - \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta\cot(\eta-B) (\Lambda - \Lambda_0 ) \biggl\{ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr\} + \eta\cot(\eta-B) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Lambda (\Lambda - \Lambda_0 ) \biggl\{ 1 - \Lambda^3 \cot(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \Lambda_0 (\Lambda - \Lambda_0 ) \biggl\{ (3-\Lambda^2) - (\Lambda + \Lambda^3 ) \cot(\Lambda - E) - \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr\} - \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta\cot(\eta-B) (\Lambda - \Lambda_0 ) \biggl\{ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr\} + \eta\cot(\eta-B) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} </math> </td> </tr> </table> Simplifying again produces, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> RHS </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Lambda^2 \biggl\{ 1 - \Lambda^3 \cot(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr\} - \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \Lambda^2 \cot(\Lambda-E) \biggl\{ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr\} + \Lambda \cot(\Lambda-E) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \Lambda^3 \cot(\Lambda - E) - \Lambda^5 \cot(\Lambda - E) - \Lambda^5 \cot^3(\Lambda - E) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ +\Lambda^5 \cot(\Lambda-E) - 2\Lambda^3 \cot(\Lambda-E) +\Lambda^4\cot^2(\Lambda - E) + \Lambda^5 \cot^3(\Lambda - E) + \Lambda^3 \cot(\Lambda-E) - \Lambda^4 \cot^2(\Lambda - E) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 0\, . </math> </td> </tr> </table> EXCELLENT !! Keep trying to simplify the more general expression … <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ \Lambda^2 - \Lambda^5 \cot(\Lambda - E) - \Lambda^5 \cot^3(\Lambda - E) - \Lambda^2 + \Lambda^3 \cot(\Lambda - E) + \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} - \Lambda_0 \biggl\{ \Lambda - \Lambda^4 \cot(\Lambda - E) - \Lambda^4 \cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \Lambda_0 (\Lambda - \Lambda_0 ) \biggl\{ (3-\Lambda^2) - (\Lambda + \Lambda^3 ) \cot(\Lambda - E) - \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta\cot(\eta-B) (\Lambda ) \biggl\{ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr\} + \eta\cot(\eta-B) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \eta\cot(\eta-B) (\Lambda_0) \biggl\{ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \Lambda^2 - \Lambda^5 \cot(\Lambda - E) - \Lambda^5 \cot^3(\Lambda - E) - \Lambda^2 + \Lambda^3 \cot(\Lambda - E) + \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta\cot(\eta-B) \biggl[ \Lambda^4 -2\Lambda^2 +\Lambda^3 \cot(\Lambda - E) + \Lambda^4 \cot^2(\Lambda - E) + \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \Lambda_0 \biggl\{ (\Lambda - \Lambda_0 ) \biggl[ (3-\Lambda^2) - (\Lambda + \Lambda^3 ) \cot(\Lambda - E) - \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr] + \biggl[ \Lambda - \Lambda^4 \cot(\Lambda - E) - \Lambda^4 \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta\cot(\eta-B) \biggl[ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 + (\Lambda^3 - \Lambda^5) \cot(\Lambda - E) - \Lambda^5 \cot^3(\Lambda - E) \biggr] + \eta\cot(\eta-B) \biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4 - \Lambda^2 + \Lambda^4 \cot^2(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \Lambda_0 \biggl\{ (\Lambda - \Lambda_0 ) \biggl[ (3-\Lambda^2) - (\Lambda + \Lambda^3 ) \cot(\Lambda - E) - \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr] + \biggl[ \Lambda - \Lambda^4 \cot(\Lambda - E) - \Lambda^4 \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta\cot(\eta-B) \biggl[ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr] \biggr\} \, . </math> </td> </tr> </table> ---- Simplify again … <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> RHS </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 + (\Lambda^3 - \Lambda^5) \cot(\Lambda - E) - \Lambda^5 \cot^3(\Lambda - E) \biggr] + \eta\cot(\eta-B) \biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4 - \Lambda^2 + \Lambda^4 \cot^2(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 + (\Lambda^3 - \Lambda^5) \cot(\Lambda - E) - \Lambda^5 \cot^3(\Lambda - E) \biggr] + \biggl[ (\Lambda^5 - \Lambda^3) \cot(\Lambda-E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^5\cot(\Lambda-E) + \Lambda^5 \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl( \frac{ a_0 }{b_0}\biggr) \biggl[ \Lambda^4 - \Lambda^5\cot(\Lambda-E)\biggr] \, . </math> </td> </tr> </table> ======First Argument Relationships Guess====== Let's try the relationship, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~(\eta - B)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~m(\Lambda - E) \, .</math> </td> </tr> </table> Then, for example, if <math>~m = 2</math>, we can make the replacement, <div align="center"> <math>~\cot(\eta - B) = \cot[2(\Lambda - E)] = \frac{\cot^2(\Lambda - E) - 1}{2\cot(\Lambda - E)} \, .</math> </div> And, alternatively, if <math>~m = 3</math>, we can make the replacement, <div align="center"> <math>~\cot(\eta - B) = \cot[3(\Lambda - E)] = \frac{\cot^3(\Lambda - E) - 3\cot(\Lambda - E)}{3\cot^2(\Lambda - E)-1} \, .</math> </div> <table border="1" width="60%" align="center" cellpadding="8"><tr><td align="left"> <div align="center"><font color="red">'''IMPLICATIONS'''</font></div> Given that we also are assuming the relationship, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\eta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{g_\mathrm{F}} (\Lambda - \Lambda_0) \, ,</math> </td> </tr> </table> we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~m\Lambda - \eta </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~mE - B</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ m\Lambda - \frac{1}{g_\mathrm{F}} (\Lambda - \Lambda_0)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~mE - B</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \Lambda \biggl[m - \frac{1}{g_\mathrm{F}} \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~mE - B - \frac{\Lambda_0}{g_\mathrm{F}} \, . </math> </td> </tr> </table> In order for this statement to be true for all <math>~\Lambda</math>, the RHS and the LHS must independently be zero. Hence, we require, <div align="center"> <math>~g_\mathrm{F} = \frac{1}{m}</math> and <math>~\Lambda_0 = E - \frac{B}{m} \, .</math> </div> </td></tr></table> Let's try <math>~m=2</math>: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ 2\cot(\Lambda - E) \cdot \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\cot(\Lambda - E)\biggl[ \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 + (\Lambda^3 - \Lambda^5) \cot(\Lambda - E) - \Lambda^5 \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + 2(\Lambda - \Lambda_0) [\cot^2(\Lambda - E) - 1]\biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4 - \Lambda^2 + \Lambda^4 \cot^2(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \Lambda_0 \biggl\{ 2(\Lambda - \Lambda_0 )\cot(\Lambda - E) \biggl[ (3-\Lambda^2) - (\Lambda + \Lambda^3 ) \cot(\Lambda - E) - \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + 2(\Lambda - \Lambda_0)[\cot^2(\Lambda - E) - 1] \biggl[ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + 2\cot(\Lambda - E)\biggl[ \Lambda - \Lambda^4 \cot(\Lambda - E) - \Lambda^4 \cot^3(\Lambda - E) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>~ \Rightarrow ~~~\cot(\Lambda - E) \cdot \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \Lambda_0 (\Lambda - \Lambda_0) \biggl[ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - (\Lambda - \Lambda_0) \biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4 - \Lambda^2 + \Lambda^4 \cot^2(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ \cot(\Lambda - E)\biggl[ \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 + (\Lambda^3 - \Lambda^5) \cot(\Lambda - E) - \Lambda^5 \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -\Lambda_0 (\Lambda - \Lambda_0 )\cot(\Lambda - E) \biggl[ (3-\Lambda^2) - (\Lambda + \Lambda^3 ) \cot(\Lambda - E) - \Lambda^2 \cot^2(\Lambda - E) - \Lambda^3 \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \Lambda_0 \cot(\Lambda - E)\biggl[ \Lambda - \Lambda^4 \cot(\Lambda - E) - \Lambda^4 \cot^3(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (\Lambda - \Lambda_0) \cot^2(\Lambda - E) \biggl[ \biggl(1- \frac{ a_0 }{b_0}\biggr)\Lambda^4 - \Lambda^2 + \Lambda^4 \cot^2(\Lambda - E) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - \Lambda_0 (\Lambda - \Lambda_0)\cot^2(\Lambda - E) \biggl[ \Lambda^3 -2\Lambda +\Lambda^2\cot(\Lambda - E) + \Lambda^3 \cot^2(\Lambda - E) \biggr] </math> </td> </tr> </table> ======Second Argument Relationships Guess====== Let's go back up to the general expression, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\Rightarrow~~~ \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (\Lambda - \Lambda_0 ) \biggl\{ (3-\Lambda^2)(\Lambda - \Lambda_0 ) - (\Lambda + \Lambda^3 )(\Lambda - \Lambda_0 ) \cot(\Lambda - E) - \Lambda^2 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E) - \Lambda^3 (\Lambda - \Lambda_0 )\cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^3 -4\Lambda +2\Lambda^2 \cot(\Lambda - E) + 2\Lambda^3 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -Q \biggl\{ \Lambda^3(\Lambda - \Lambda_0 ) -2\Lambda (\Lambda - \Lambda_0 ) +\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E) + \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - Q \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} \, , </math> </td> </tr> </table> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~Q</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ 1 - \eta \cot(\eta - B) \biggr] \, .</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 - \biggl[\frac{(\Lambda - \Lambda_0)}{g_\mathrm{F}}\biggr] \cot[ (\Lambda - \Lambda_0)/g_\mathrm{F} - B] \, . </math> </td> </tr> </table> Then, let's try setting, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\cot[ (\Lambda - \Lambda_0)/g_\mathrm{F} - B]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan(\Lambda - E)</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \cot\biggl\{ \frac{\pi}{2} - \biggl[ \frac{\pi}{2} + B - (\Lambda - \Lambda_0)/g_\mathrm{F} \biggr] \biggr\}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\tan(\Lambda - E)</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \frac{\pi}{2} + B - \frac{ (\Lambda - \Lambda_0) }{g_\mathrm{F}} </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Lambda - E \, .</math> </td> </tr> </table> This will only work for all <math>~\Lambda</math> if, <math>~g_\mathrm{F} = -1</math>; in which case, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{\pi}{2} + B + (\Lambda - \Lambda_0) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Lambda - E </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ E </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\Lambda_0 - \frac{\pi}{2} - B \, . </math> </td> </tr> </table> Hence, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~Q</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 + (\Lambda - \Lambda_0)\tan(\Lambda - E) </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ Q \cot(\Lambda-E)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\cot(\Lambda-E) + (\Lambda - \Lambda_0) \, .</math> </td> </tr> </table> This means that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\cot(\Lambda-E) \cdot \frac{ \Lambda^4(\Lambda - \Lambda_0 )^2}{2b_0 g_\mathrm{F}^2} \cdot 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \cot(\Lambda-E)(\Lambda - \Lambda_0 ) \biggl\{ (3-\Lambda^2)(\Lambda - \Lambda_0 ) - (\Lambda + \Lambda^3 )(\Lambda - \Lambda_0 ) \cot(\Lambda - E) - \Lambda^2 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E) - \Lambda^3 (\Lambda - \Lambda_0 )\cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \cot(\Lambda-E)(\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^3 -4\Lambda +2\Lambda^2 \cot(\Lambda - E) + 2\Lambda^3 \cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -[\cot(\Lambda-E) + (\Lambda - \Lambda_0)] \biggl\{ \Lambda^3(\Lambda - \Lambda_0 ) -2\Lambda (\Lambda - \Lambda_0 ) +\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E) + \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - [\cot(\Lambda-E) + (\Lambda - \Lambda_0)] \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ (\Lambda - \Lambda_0 ) \biggl\{ (3-\Lambda^2)(\Lambda - \Lambda_0 )\cot(\Lambda-E) - (\Lambda + \Lambda^3 )(\Lambda - \Lambda_0 ) \cot^2(\Lambda - E) - \Lambda^2 (\Lambda - \Lambda_0 )\cot^3(\Lambda - E) - \Lambda^3 (\Lambda - \Lambda_0 )\cot^4(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + (\Lambda - \Lambda_0 ) \biggl\{ 2\Lambda^3\cot(\Lambda-E) -4\Lambda \cot(\Lambda-E) +2\Lambda^2 \cot^2(\Lambda - E) + 2\Lambda^3 \cot^3(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -(\Lambda - \Lambda_0) \biggl\{ \Lambda^3(\Lambda - \Lambda_0 ) -2\Lambda (\Lambda - \Lambda_0 ) +\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E) + \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - (\Lambda - \Lambda_0) \biggl\{ \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -[\cot(\Lambda-E) ] \biggl\{ \Lambda^3(\Lambda - \Lambda_0 ) -2\Lambda (\Lambda - \Lambda_0 ) +\Lambda^2 (\Lambda - \Lambda_0 )\cot(\Lambda - E) + \Lambda^3 (\Lambda - \Lambda_0 )\cot^2(\Lambda - E) + \Lambda^2 - \Lambda^3 \cot(\Lambda - E) - \biggl( \frac{ a_0 }{b_0}\biggr)\Lambda^4 \biggr\} \, . </math> </td> </tr> </table> ====Attempt 4A==== Try, the ''structural'' function, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\theta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{\xi} \biggl[A\sin\xi - B\cos\xi \biggr] </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{a_0}{\xi} \cdot \sin(\xi - b_0) \, ,</math> </td> </tr> </table> where, recognizing that, <math>~f \equiv B/A</math>, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~a_0 \equiv A \sqrt{1 + f^2}</math> </td> <td align="center"> and </td> <td align="left"> <math>~b_0 \equiv \tan^{-1}f \, .</math> </td> </tr> </table> Does this ''structural'' function satisfy the Lane-Emden equation, namely … <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\xi^2 \frac{d^2\theta}{d\xi^2} + 2\xi \frac{d\theta}{d\xi} + \xi^2\theta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, .</math> </td> </tr> </table> First, recognize that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d\theta}{d\xi}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{a_0}{\xi} \cdot \cos(\xi - b_0) - \frac{a_0}{\xi^2} \cdot \sin(\xi - b_0) \, , </math> </td> </tr> </table> and, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d^2\theta}{d\xi^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{a_0}{\xi} \cdot \sin(\xi - b_0) - \frac{a_0}{\xi^2} \cdot \cos(\xi - b_0) - \frac{a_0}{\xi^2} \cdot \cos(\xi - b_0) + \frac{2a_0}{\xi^3} \cdot \sin(\xi - b_0) \, . </math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~ \xi^2 \frac{d^2\theta}{d\xi^2} + 2\xi \frac{d\theta}{d\xi} + \xi^2\theta </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \xi^2 \biggl\{ - \frac{a_0}{\xi} \cdot \sin(\xi - b_0) - \frac{a_0}{\xi^2} \cdot \cos(\xi - b_0) - \frac{a_0}{\xi^2} \cdot \cos(\xi - b_0) + \frac{2a_0}{\xi^3} \cdot \sin(\xi - b_0) \biggr\} + 2\xi \biggl\{ \frac{a_0}{\xi} \cdot \cos(\xi - b_0) - \frac{a_0}{\xi^2} \cdot \sin(\xi - b_0) \biggr\} + \xi^2\biggl\{ \frac{a_0}{\xi}\cdot \sin(\xi - b_0) \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl\{ -a_0 \xi \cdot \sin(\xi - b_0) - 2a_0 \cdot \cos(\xi - b_0) + \frac{2a_0}{\xi} \cdot \sin(\xi - b_0) \biggr\} + \biggl\{ 2a_0\cdot \cos(\xi - b_0) - \frac{2a_0}{\xi} \cdot \sin(\xi - b_0) \biggr\} + a_0 \xi\cdot \sin(\xi - b_0) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 0 \, . </math> </td> </tr> </table> Good! ====Attempt 4B==== Let's continue with the same '''4A''' approach, but shift to the variable notation that we have used in the [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|accompanying description of how to build an equilibrium, (n<sub>c</sub>, n<sub>e</sub>) = (5, 1) bipolytrope]]. For example, the envelope's ''structural function'' will be referred to as <math>~\phi(\eta)</math> instead of <math>~\theta(\xi)</math>. Specifically, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\phi</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~a_0 \biggl[ \frac{\sin(\eta - b_0)}{\eta} \biggr] \, ,</math> </td> </tr> </table> and, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d\phi}{d\eta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{a_0}{\eta^2} \biggl[ \eta \cos(\eta - b_0) - \sin(\eta - b_0) \biggr] \, ,</math> </td> </tr> </table> and, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d^2\phi}{d\eta^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ - \frac{a_0}{\eta} \cdot \sin(\eta - b_0) - \frac{2a_0}{\eta^2} \cdot \cos(\eta - b_0) + \frac{2a_0}{\eta^3} \cdot \sin(\eta - b_0) \, . </math> </td> </tr> </table> This satisfies the Lane-Emden equation for any values of the parameter pair, <math>~a_0</math> and <math>~b_0</math>. Note that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~Q \equiv - \frac{d\ln \phi}{d\ln\eta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{a_0}{\eta^2} \biggl[\sin(\eta - b_0) - \eta \cos(\eta - b_0) \biggr] \cdot \frac{\eta}{\phi}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[1 - \eta \cot(\eta - b_0) \biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~ \eta \cot(\eta - b_0) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~(1 - Q ) \, .</math> </td> </tr> </table> <table border="1" align="center" cellpadding="10" width="80%"><tr><td align="left"> Note that for the <math>~\mu_e/\mu_c = 1</math> model having a core-envelope interface of <math>~\xi_i = 1.668646016</math> (and an adopted normalization, φ<sub>i</sub> = 1), we also have, <table border="0" cellpadding="8" align="center"> <tr> <td align="right"><math>~\theta_i =</math></td> <td align="right">0.720165</td> </tr> <tr> <td align="right"><math>~-\biggl(\frac{d\theta}{d\xi}\biggr)_i =</math></td> <td align="right">0.207749</td> </tr> <tr> <td align="right"><math>~\eta_i =</math></td> <td align="right">1.498956</td> </tr> <tr> <td align="right"><math>~-\biggl(\frac{d\phi}{d\eta}\biggr)_i =</math></td> <td align="right">0.963393</td> </tr> <tr> <td align="right"><math>~\Lambda_i =</math></td> <td align="right">- 0.296262</td> </tr> <tr> <td align="right"><math>~b_0 =</math></td> <td align="right">- 0.359864 (or +2.78173)</td> </tr> <tr> <td align="right"><math>~a_0 =</math></td> <td align="right">1.563355</td> </tr> <tr> <td align="right"><math>~\eta_s = \pi + b_0 =</math></td> <td align="right">+2.78173 (or - 0.359864 = 5.92332)</td> </tr> </table> Some of this is also detailed [[#Attempt_1|above in the Attempt 1 subsection]]. </td></tr></table> Now, guided by a [[SSC/Stability/n1PolytropeLAWE#Succinct_Demonstration|separate parallel discussion]], we want to see whether, in the case of a bipolytropic configuration for which <math>n_e=1</math>, the <table border="0" cellpadding="5" align="center"> <tr> <td align="center" colspan="4"><font color="maroon"><b>Precise Solution to the Polytropic LAWE</b></font></td> </tr> <tr> <td align="right"> <math>~\sigma_c^2 = 0</math> </td> <td align="center"> and </td> <td align="left"> <math>~x_P </math> </td> <td align="left"> <math>~ \equiv \frac{3c_0 (n-1)}{2n}\biggl[1 + \biggl(\frac{n-3}{n-1}\biggr) \biggl( \frac{1}{\eta \phi^{n}}\biggr) \frac{d\phi}{d\eta}\biggr] </math> </td> </tr> <tr> <td align="right" colspan="3"> </td> <td align="left"> <math>~= -\biggl( \frac{3c_0}{\eta \phi}\biggr) \frac{d\phi}{d\eta} = \frac{3c_0}{\eta^2} \cdot Q \, , </math> </td> </tr> </table> satisfies the governing LAWE, namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{d^2x_P}{d\eta^2} + \biggl[4 - 2Q\biggr]\frac{1}{\eta}\cdot \frac{dx_P}{d\eta} - 2Q\cdot \frac{x_P}{\eta^2} \, . </math> </td> </tr> </table> Well … recognizing that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d}{d\eta} \biggl[\cot(\eta - b_0)\biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \biggl[ 1 + \cot^2(\eta - b_0)\biggr] \, ,</math> </td> </tr> </table> we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{dQ}{d\eta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \eta -\cot(\eta - b_0) +\eta\cot^2(\eta - b_0) \, , </math> </td> </tr> </table> and, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d^2Q}{d\eta^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{d}{d\eta}\biggl[ \eta -\cot(\eta - b_0) +\eta\cot^2(\eta - b_0) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 1 + \biggl[ 1 + \cot^2(\eta - b_0)\biggr] +\cot^2(\eta - b_0) -2\eta \cot(\eta - b_0)\biggl[ 1 + \cot^2(\eta - b_0)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2 -2\eta \cot(\eta - b_0) + 2\cot^2(\eta - b_0) -2\eta \cot^3(\eta - b_0) \, . </math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{3c_0}\cdot \frac{dx_P}{d\eta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{1}{\eta^2} \frac{dQ}{d\eta} - \frac{2Q}{\eta^3} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\eta^2}\biggl[\eta -\cot(\eta - b_0) +\eta\cot^2(\eta - b_0) \biggr] - \frac{2Q}{\eta^3} \, , </math> </td> </tr> </table> and, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{1}{3c_0}\cdot \frac{d^2x_P}{d\eta^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\eta^2} \frac{d^2Q}{d\eta^2} - \frac{2}{\eta^3} \frac{dQ}{d\eta} + \frac{6Q}{\eta^4} - \frac{2}{\eta^3} \frac{dQ}{d\eta} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{1}{\eta^2} \biggl[2 -2\eta \cot(\eta - b_0) + 2\cot^2(\eta - b_0) -2\eta \cot^3(\eta - b_0) \biggr] - \frac{4}{\eta^3} \biggl[ \eta -\cot(\eta - b_0) +\eta\cot^2(\eta - b_0) \biggr] + \frac{6Q}{\eta^4} \, . </math> </td> </tr> </table> <span id="proof">So the relevant LAWE becomes,</span> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl( \frac{\eta^4}{3c_0} \biggr) \cdot</math> LAWE </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \eta^2 \biggl[2 -2\eta \cot(\eta - b_0) + 2\cot^2(\eta - b_0) -2\eta \cot^3(\eta - b_0) \biggr] - 4\eta \biggl[ \eta -\cot(\eta - b_0) +\eta\cot^2(\eta - b_0) \biggr] + 6Q </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta^3 \biggl[4 - 2Q\biggr]\biggl\{ \frac{1}{\eta^2}\biggl[\eta -\cot(\eta - b_0) +\eta\cot^2(\eta - b_0) \biggr] - \frac{2Q}{\eta^3} \biggr\} - 2Q^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\eta^2 \biggl[1 - \eta \cot(\eta - b_0) + \cot^2(\eta - b_0) - \eta \cot^3(\eta - b_0) \biggr] - 4 \biggl[ \eta^2 - \eta\cot(\eta - b_0) +\eta^2\cot^2(\eta - b_0) \biggr] + 6Q </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + 4\biggl[\eta^2 - \eta\cot(\eta - b_0) + \eta^2\cot^2(\eta - b_0) - 2Q \biggr] - 2Q \biggl[\eta^2 - \eta\cot(\eta - b_0) + \eta^2\cot^2(\eta - b_0) - 2Q \biggr] - 2Q^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 2\eta^2 \biggl[Q + \cot^2(\eta - b_0) - \eta \cot^3(\eta - b_0) \biggr] - 2Q - 2Q \biggl[\eta^2 - \eta\cot(\eta - b_0) + \eta^2\cot^2(\eta - b_0) - 2Q \biggr] - 2Q^2 </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow~~~\biggl( \frac{\eta^4}{6c_0} \biggr) \cdot</math> LAWE </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\eta^2 \cot^2(\eta - b_0) - \eta^3 \cot^3(\eta - b_0) \biggr] + \biggl[ 1-\eta\cot(\eta - b_0)\biggr] \biggl[\eta\cot(\eta - b_0) - \eta^2\cot^2(\eta - b_0) \biggr] - Q(1-Q) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\eta\cot(\eta - b_0) - \eta^2\cot^2(\eta - b_0) \biggr] - Q(1-Q) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 0\, . </math> </td> </tr> </table> ====Attempt 5==== =====New Strategy===== In the vast majority of our prior attempts to derive an analytic expression for the envelope's eigenvector, we have started with the presumption — as voiced by [http://adsabs.harvard.edu/abs/1988Ap%26SS.147..219B Beech (1988)] and repeated in [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|our accompanying derivation of the structure of the <math>~(n_c, n_e) = (5, 1)</math> bipolytrope's structure]] — that the most general solution to the n = 1 Lane-Emden equation can be written in the form, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\phi</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>A \biggl[ \frac{\sin(\eta - B)}{\eta} \biggr] \, .</math> </td> </tr> </table> But, as we have [[SSC/Structure/Polytropes#Primary_E-Type_Solution|emphasized in a separate context]], another expression that satisfies the relevant Lane-Emden equation has the form, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\phi</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- A \biggl[ \frac{\cos(\eta - B)}{\eta} \biggr] \, .</math> </td> </tr> </table> <table border="1" width="80%" align="center" cellpadding="10"><tr><td align="left"> We fully appreciate that for appropriately chosen and ''different'' values of the parameter, <math>~B</math>, these two functions can be made equal to one another. But, for now, let's work through our analysis pretending that they are different functions. </td></tr></table> This alternate "cosine" expression was not the solution of choice when we were seeking a mathematical description of the structure of an ''isolated'', n = 1 polytrope because it does not satisfy the relevant central boundary conditions. However, it occurs to us that this alternate expression might work in the context we are considering now <font color="red">[20 April 2019]</font>, which deals with properties of the envelope of a bipolytrope. In this case, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~Q \equiv - \frac{d\ln\phi}{d\ln\eta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~A \biggl[\frac{\cos(\eta-B)}{\eta^2} + \frac{\sin(\eta-B)}{\eta} \biggr]\cdot \frac{\eta^2}{A\cos(\eta-B)}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[\cos(\eta-B) + \eta\sin(\eta-B) \biggr]\cdot \frac{1}{\cos(\eta-B)}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~1 + \eta\tan(\eta-B) \, ;</math> </td> </tr> </table> and, following along the lines of our earlier '''[[#Attempt_4B|Attempt 4B]]''' discussion, a reasonable ''guess'' for the dimensionless displacement function is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_P</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3c_1 Q}{\eta^2}</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3c_1 }{\eta^2}\biggl[1 + \eta\tan(\eta-B)\biggr] \, .</math> </td> </tr> </table> What are the first and second derivatives of this trial eigenfunction? <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{dx_P}{d\eta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3c_1}{\eta^2}\biggl[\tan(\eta - B) + \frac{\eta}{\cos^2(\eta-B)}\biggr] - \frac{6c_1}{\eta^3} \biggl[1 + \eta\tan(\eta-B)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3c_1}{\eta^3} \biggl[\frac{\eta^2}{\cos^2(\eta-B)} -2 - \eta\tan(\eta-B)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr]\, . </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \frac{d^2x_P}{d\eta^2}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] 3c_1 \frac{d}{d\eta}\biggl[\eta^{-3}\cos^{-2}(\eta-B) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \frac{3c_1}{\eta^3\cos^2(\eta-B)}\cdot \frac{d}{d\eta}\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] 3c_1 \biggl[2\eta^{-3}\sin(\eta-B)\cos^{-3}(\eta-B) -3\eta^{-4}\cos^{-2}(\eta-B) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[2\eta + 4\sin(\eta-B)\cos(\eta-B) - \sin(\eta-B)\cos(\eta-B)- \eta \cos^2(\eta-B)+ \eta\sin^2(\eta-B)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] \cdot \biggl[2\eta\sin(\eta-B) -3\cos(\eta-B) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[2\eta^2\cos(\eta-B) + 4\eta \sin(\eta-B)\cos^2(\eta-B) - \eta \sin(\eta-B)\cos^2(\eta-B)- \eta^2 \cos^3(\eta-B)+ \eta^2\sin^2(\eta-B)\cos(\eta-B)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl\{ 2\eta^3 \sin(\eta-B) - 4\eta\sin(\eta-B)\cos^2(\eta-B) - 2\eta^2 \sin^2(\eta-B)\cos(\eta-B) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -3\eta^2 \cos(\eta-B) + 6\cos^3(\eta-B) +3 \eta\sin(\eta-B)\cos^2(\eta-B) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta \biggl[ 4\sin(\eta-B)\cos^2(\eta-B) - \sin(\eta-B)\cos^2(\eta-B)\biggr] + \eta^2 \biggl[ 2\cos(\eta-B) - \cos^3(\eta-B)+ \sin^2(\eta-B)\cos(\eta-B)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl\{6\cos^3(\eta-B) + \eta^3 \biggl[2\sin(\eta-B)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta \biggl[ 4\sin(\eta-B)\cos^2(\eta-B) - \sin(\eta-B)\cos^2(\eta-B) - 4\sin(\eta-B)\cos^2(\eta-B) +3 \sin(\eta-B)\cos^2(\eta-B) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \eta^2 \biggl[ 2\cos(\eta-B) - \cos^3(\eta-B)+ \sin^2(\eta-B)\cos(\eta-B) - 2 \sin^2(\eta-B)\cos(\eta-B) -3 \cos(\eta-B) \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[ 6\cos^3(\eta-B) + 2\eta \sin(\eta-B)\cos^2(\eta-B) - 2 \eta^2 \cos(\eta-B) + 2\eta^3 \sin(\eta-B) \biggr] \, . </math> </td> </tr> </table> =====Illustration===== <table border="1" align="center" width="710px" cellpadding="8"> <tr><td align="center"> [[File:Frame06good.png|700px|Attempt5 Trial Eigenfunction]] </td></tr> <tr><td align="left"> ''Above & Below:'' The solid, light-blue circular markers trace how the function, <math>~x_P(\eta)</math>, varies over the "radial" range, <math>~- \pi \le \eta \le \pi</math>, when <math>~c_1 = 1</math>, for various values of the parameter, <math>~B</math>; in the above panel as well as in each frame of the animation, the chosen value of <math>~B</math> is recorded in the upper-right corner of the image. (These values are also recorded in the table immediately below the animation.) The red vertical dashed line segment identifies the value of <math>~\eta</math> at which the argument of the tangent function goes to <math>~\tfrac{\pi}{2}</math> and, hence, where the function <math>~x_P(\eta)</math> flips discontinuously from plus- to minus-infinity. As explained further, below, the solid purple curve shows how the x<sub>P</sub>-''intercept'' function varies with <math>~\eta</math>; as defined, this curve is independent of the parameter, <math>~B</math>, so it is unchanging in the animation sequence. The single larger yellow-circular marker (with a black border) shows where this "intercept" curve intersects the <math>~x_P</math> function and, therefore, where along this trial eigenfunction <math>~d\ln x_P/d\ln\eta = -1</math>; the coordinates (abscissa & ordinate) of this yellow marker are recorded in the accompanying table, for each illustrative value of <math>~B</math>. </td></tr> <tr><td align="center"> [[File:TangentTrialAnalytic51movie.gif|700px|Animation of Attempt5 Trial Eigenfunction]] </td></tr> <tr><td align="center"> <table border="0" cellpadding="8"> <tr> <th align="center" colspan="2"> </th> <th align="center" colspan="2"><math>~\frac{d\ln x_P}{d\ln\eta} = -1</math><br /><hr /></th> </tr> <tr> <th align="center">Animation<br />Frame</th> <th align="center"><math>~B</math></th> <th align="center">Abscissa<br /><math>~\eta</math></th> <th align="center">Ordinate<br /><math>~x_P(\eta)</math></th> </tr> <tr> <td align="center"><math>~1</math></td> <td align="center"><math>~1.02 \times \tfrac{\pi}{2} = 1.6022</math></td> <td align="center"><math>~0.014</math></td> <td align="center"><math>~27,165</math></td> </tr> <tr> <td align="center"><math>~2</math></td> <td align="center"><math>~1.2 \times \tfrac{\pi}{2} = 1.8850</math></td> <td align="center"><math>~0.157</math></td> <td align="center"><math>~242.7</math></td> </tr> <tr> <td align="center"><math>~3</math></td> <td align="center"><math>~1.4 \times \tfrac{\pi}{2} = 2.1991</math></td> <td align="center"><math>~0.311</math></td> <td align="center"><math>~60.31</math></td> </tr> <tr> <td align="center"><math>~4</math></td> <td align="center"><math>~1.6 \times \tfrac{\pi}{2} = 2.5133</math></td> <td align="center"><math>~0.462</math></td> <td align="center"><math>~26.52</math></td> </tr> <tr> <td align="center"><math>~5</math></td> <td align="center"><math>~1.8 \times \tfrac{\pi}{2} = 2.8274</math></td> <td align="center"><math>~0.606</math></td> <td align="center"><math>~14.68</math></td> </tr> <tr> <td align="center"><math>~6</math></td> <td align="center"><math>~2 \times \tfrac{\pi}{2} = 3.1416</math></td> <td align="center"><math>~0.739</math></td> <td align="center"><math>~9.187</math></td> </tr> <tr> <td align="center"><math>~7</math></td> <td align="center"><math>~0.2 \times \tfrac{\pi}{2} = 0.3142</math></td> <td align="center"><math>~0.856</math></td> <td align="center"><math>~6.200</math></td> </tr> <tr> <td align="center"><math>~8</math></td> <td align="center"><math>~0.4 \times \tfrac{\pi}{2} = 0.6283</math></td> <td align="center"><math>~0.949</math></td> <td align="center"><math>~4.381</math></td> </tr> <tr> <td align="center"><math>~9</math></td> <td align="center"><math>~0.5 \times \tfrac{\pi}{2} = 0.7854</math></td> <td align="center"><math>~0.981</math></td> <td align="center"><math>~3.724</math></td> </tr> <tr> <td align="center"><math>~10</math></td> <td align="center"><math>~0.5 \times \tfrac{\pi}{2} = 0.9425</math></td> <td align="center"><math>~0.998</math></td> <td align="center"><math>~3.178</math></td> </tr> <tr> <td align="center"><math>~11</math></td> <td align="center"><math>~0.8 \times \tfrac{\pi}{2} = 1.2566</math></td> <td align="center"><math>~0.955</math></td> <td align="center"><math>~2.313</math></td> </tr> <tr> <td align="center"><math>~12</math></td> <td align="center"><math>~0.9 \times \tfrac{\pi}{2} = 1.4137</math></td> <td align="center"><math>~0.840</math></td> <td align="center"><math>~1.944</math></td> </tr> <tr> <td align="center"><math>~13</math></td> <td align="center"><math>~0.98 \times \tfrac{\pi}{2} = 1.5394</math></td> <td align="center"><math>~0.545</math></td> <td align="center"><math>~1.632</math></td> </tr> </table> </td></tr> </table> What is the expression for the logarithmic derivative of this eigenfunction guess? <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d\ln x_P}{d\ln\eta} = \frac{\eta}{x_P} \cdot \frac{dx_P}{d\eta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\eta^3}{3c_1} \biggl[1 + \eta\tan(\eta-B)\biggr]^{-1} \biggl\{ \frac{3c_1}{\eta^2}\biggl[\tan(\eta - B) + \frac{\eta}{\cos^2(\eta-B)}\biggr] - \frac{6c_1}{\eta^3} \biggl[1 + \eta\tan(\eta-B)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[\frac{\cos (\eta-B)}{\cos(\eta-B) + \eta\sin(\eta-B)}\biggr] \biggl\{ \frac{\eta}{\cos^2(\eta-B)} \biggl[\sin(\eta - B)\cos(\eta-B) +\eta\biggr] - \frac{2}{\cos^2(\eta-B)} \biggl[\cos^2(\eta-B) + \eta\sin(\eta-B)\cos(\eta-B)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{\eta^2 - \eta\sin(\eta - B)\cos(\eta-B) - 2\cos^2(\eta-B)}{\cos^2(\eta-B) + \eta\sin(\eta-B) \cos(\eta-B)} \, . </math> </td> </tr> </table> At the surface of the bipolytropic configuration — that is, presumably when <math>~\eta = \eta_s</math> — we must find that this logarithmic derivative is negative one. So, for a given value of the parameter, <math>~B</math>, what is the value of <math>~\eta_s</math>? Well … <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\frac{d\ln x_P}{d\ln\eta}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~-1</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ -\cos^2(\eta-B) - \eta\sin(\eta-B) \cos(\eta-B)</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\eta^2 - \eta\sin(\eta - B)\cos(\eta-B) - 2\cos^2(\eta-B)</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ 0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\eta^2 - \cos^2(\eta-B)</math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~ \eta</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\pm \cos(\eta-B)\, .</math> </td> </tr> </table> Now, given that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\tan(\eta-B) </math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\pm \biggl[ \frac{\sqrt{1-\cos^2(\eta-B)}}{\cos(\eta-B)} \biggr] \, ,</math> </td> </tr> </table> we see that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~x_P\biggr|_\mathrm{intercept}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \eta \biggl[ \frac{\sqrt{1-\cos^2(\eta-B)}}{\cos(\eta-B)} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \eta \biggl[ \frac{\sqrt{1-\cos^2(\eta-B)}}{\cos(\eta-B)} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \eta \biggl[ \frac{\sqrt{1-\eta^2}}{\eta} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\frac{3c_1 }{\eta^2}\biggl\{ 1 \pm \sqrt{1-\eta^2}\biggr\} \, .</math> </td> </tr> </table> The solid-purple, x<sub>P</sub>-''intercept'' curve that appears in the above figure/animation is defined by this function. Notice that this function never exceeds unity. This presumably means that if the tangent-based <math>~x_P</math> eigenfunction is the correct solution to the envelope's LAWE, then the dimensionless radius, <math>~\eta_s</math>, of the bipolytrope must never exceed unity. The following diagram — the original of which appears in our [[SSC/Structure/BiPolytropes/Analytic51#Parameter_Values|accompanying discussion of the equilibrium properties of bipolytropic configurations having]] <math>~(n_c, n_e) = (5, 1)</math> — shows how <math>~\eta_i</math> (purple curve) and <math>~\eta_s</math> (green curve) vary with the interface location <math>~\xi_i</math> (ordinate). The solid yellow circular markers (with black edges) identify where the logarithmic derivative of the dimensionless displacement function, <math>~d\ln x_P/d\ln\eta</math>, equals negative one. If <math>~x_P</math> is the correct eigenfunction for the marginally unstable bipolytropic configuration, one of these yellow circular markers should coincide with the green curve, that is, it should be associated with the configuration's surface. Since the curve identified by the yellow circular markers does not appear to intersect the green curve, we conclude that we have not yet identified the correct eigenfunction. <div align="center"> [[File:Bipolytrope51Boundaries03b.png|500px|Bipolytrope Properties]] </div> =====Is This Compatible With LAWE===== In an effort to track the two <math>~Q(\eta)</math> functions separately, we will add a subscript zero to the one that applies to the ''structural'' properties of the underlying equilibrium configuration. Again, we will be focused on finding a solution in the case where <math>~\sigma_c^2 = 0</math> and <math>~n = 1</math>, that is — see also our [[#Envelope:|above discussion]] — the relevant envelope LAWE is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{d^2x_P}{d\eta^2} + \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{dx_P}{d\eta} -2\alpha_g Q_0 \cdot \frac{x_P}{\eta^2} \, , </math> </td> </tr> </table> where, drawing from our [[SSC/Structure/BiPolytropes/Analytic51#Step_6:__Envelope_Solution|discussion of the n = 1 envelope's equilibrium structure]], <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~Q_0</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[ - \frac{d\ln\phi}{d\ln\eta} \biggr]_\mathrm{n=1} = - \frac{\eta}{\phi}\biggl[ \frac{d\phi}{d\eta} \biggr]_\mathrm{n=1}</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~- \biggl[ \frac{1}{\sin(\eta - B_0)} \biggr]\biggl[ \eta\cos(\eta-B_0) - \sin(\eta-B_0) \biggr]</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~\biggl[1 - \eta\cot(\eta-B_0) \biggr] \, .</math> </td> </tr> </table> Hence, after recognizing that for this specific case, <math>~\alpha_g = +1</math>, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> LAWE </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{d^2x_P}{d\eta^2} + \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{dx_P}{d\eta} -2 Q_0 \cdot \frac{x_P}{\eta^2} </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~</math> LAWE </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \frac{3c_1}{\eta^4\cos^3(\eta-B)} \biggl[ 6\cos^3(\eta-B) + 2\eta \sin(\eta-B)\cos^2(\eta-B) - 2 \eta^2 \cos(\eta-B) + 2\eta^3 \sin(\eta-B) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl[ 2 - Q_0 \biggr] \frac{2}{\eta}\cdot \frac{3c_1}{\eta^3\cos^2(\eta-B)} \biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] - \frac{2 Q_0}{\eta^2} \cdot \frac{3c_1 }{\eta^2}\biggl[1 + \eta\tan(\eta-B)\biggr] </math> </td> </tr> <tr> <td align="right"> <math>~\Rightarrow ~~~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math> LAWE </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ \biggl[ 3\cos^3(\eta-B) + \eta \sin(\eta-B)\cos^2(\eta-B) - \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl[ 2 - Q_0 \biggr]\cos(\eta - B)\biggl[\eta^2 -2\cos^2(\eta-B) - \eta\sin(\eta-B)\cos(\eta-B)\biggr] - Q_0 \cos^3(\eta - B) \biggl[1 + \eta\tan(\eta-B)\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 3\cos^3(\eta-B) + \eta \sin(\eta-B)\cos^2(\eta-B) - \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ -4\cos^3(\eta-B) - 2\eta\sin(\eta-B)\cos^2(\eta-B) + 2\eta^2\cos(\eta - B) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ - Q_0\biggl\{ \biggl[\eta^2\cos(\eta - B) -2\cos^3(\eta-B) - \eta\sin(\eta-B)\cos^2(\eta-B)\biggr] + \biggl[\cos^3(\eta - B) + \eta\sin(\eta-B)\cos^2(\eta - B)\biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ -\cos^3(\eta-B) - \eta \sin(\eta-B)\cos^2(\eta-B) + \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B) + Q_0\biggl[ \cos^3(\eta-B) - \eta^2\cos(\eta - B) \biggr] </math> </td> </tr> </table> <table border="1" width="85%" cellpadding="10" align="center"><tr><td align="left"> <font color="red">'''Quick Check:'''</font> Now, if we set <math>~Q_0 = 1 + \eta\tan(\eta-B)</math>, these RHS terms should sum to zero. Let's check. <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math> LAWE </td> <td align="center"> <math>~\rightarrow</math> </td> <td align="left"> <math>~ -\cos^3(\eta-B) - \eta \sin(\eta-B)\cos^2(\eta-B) + \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl[\cos(\eta - B)+ \eta\sin(\eta-B) \biggr] \cdot \biggl[ \cos^2(\eta-B) - \eta^2 \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~ 0 \, . </math> </td> </tr> </table> Excellent! </td></tr></table> Now, let's plug in the expression for the ''structural'' <math>~Q_0</math>. Specifically, we want, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~Q_0</math> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\biggl[ 1 - \eta\cot(\eta-B_0) \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>~\equiv</math> </td> <td align="left"> <math>~\biggl[ 1 + \eta\tan(\eta-B_0 - \tfrac{\pi}{2} \pm m\pi) \biggr] \, ,</math> </td> </tr> </table> in which case we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[ \frac{\eta^4\cos^3(\eta-B)}{6c_1} \biggr] \cdot</math> LAWE </td> <td align="center"> <math>~\rightarrow</math> </td> <td align="left"> <math>~ -\cos^3(\eta-B) - \eta \sin(\eta-B)\cos^2(\eta-B) + \eta^2 \cos(\eta-B) + \eta^3 \sin(\eta-B) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math>~ + \biggl[ 1 + \eta\tan(\eta-B_0 - \tfrac{\pi}{2} \pm m\pi) \biggr] \cdot \biggl[ \cos^3(\eta-B) - \eta^2\cos(\eta-B) \biggr] \, . </math> </td> </tr> </table> As we have just shown, above, in the context of a "<font color="red">'''Quick Check'''</font>", the expression on the RHS will go to zero if we adopt the transformation, <math>~[B_0 + \tfrac{\pi}{2} \mp m\pi] \rightarrow B</math>. Does this help shift the coordinate, <math>~\eta</math>?
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