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__FORCETOC__ <!-- __NOTOC__ will force TOC off --> =Hypergeometric Differential Equation= ==Gradshteyn & Ryzhik== According to §9.151 (p. 1045) of [http://www.mathtable.com/gr/ Gradshteyn & Ryzhik (1965)], <font color="darkgreen">"… a hypergeometric series is one of the solutions of the differential equation, <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> z(1-z) \frac{d^2u}{dz^2} + [\gamma - (\alpha + \beta + 1)z] \frac{du}{dz} - \alpha \beta u \, ,</math> </td> </tr> </table> which is called the ''hypergeometric equation.''</font> And, according to §9.10 (p. 1039) of [http://www.mathtable.com/gr/ Gradshteyn & Ryzhik (1965)], <font color="darkgreen">"A ''hypergeometric series'' is a series of the form,</font> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>F(\alpha, \beta; \gamma; z)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 1 + \biggl[\frac{\alpha \cdot \beta}{\gamma \cdot 1} \biggr]z + \biggl[\frac{ \alpha(\alpha+1)\beta(\beta+1) }{ \gamma(\gamma+1)\cdot 1\cdot 2 }\biggr]z^2 + \biggl[\frac{ \alpha(\alpha+1)(\alpha+2)\beta(\beta+1)(\beta+2) }{ \gamma(\gamma+1)(\gamma+2)\cdot 1\cdot 2 \cdot 3}\biggr]z^3 + \dots </math> </td> </tr> </table> Among other attributes, [http://www.mathtable.com/gr/ Gradshteyn & Ryzhik (1965)] note that this, <font color="darkgreen">"… series terminates if <math>\alpha</math> or <math>\beta</math> is equal to a negative integer or to zero."</font> ==Van der Borght== ===General Value for b=== [[File:CommentButton02.png|right|100px|Comment by J. E. Tohline: In Van der Borght (1970), the fourth argument of the hypergeometric function is ax<sup>2</sup> whereas the more general power law form, ax<sup>b</sup>, applies.]]In association with his equation (3), {{ VdBorght70full }} states that a displacement function of the form, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\xi</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> x^c F(\alpha, \beta, \gamma, ax^b) \, , </math> </td> </tr> </table> provides a solution to the following 2<sup>nd</sup>-order ODE: <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> x^2(ax^b - 1) \cdot \frac{d^2\xi}{dx^2} + (apx^b + \lambda)x \cdot \frac{d\xi}{dx} + (ar x^b + s)\cdot \xi \, .</math> </td> </tr> </table> Is this ODE essentially the same as the above-defined ''hypergeometric equation''? A mapping between the two differential equations requires, <!-- <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>ax^2</math> </td> <td align="center"> <math>\leftrightarrow</math> </td> <td align="left"> <math> z \, , </math> </td> <td align="center"> and, <td align="right"> <math>\frac{\xi}{x^c}</math> </td> <td align="center"> <math>\leftrightarrow</math> </td> <td align="left"> <math> u \, ,</math> </td> </tr> </table> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\Rightarrow~~~ \frac{dx}{dz}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d}{dz}\biggl(\frac{z}{a}\biggr)^{1 / 2} = a^{-1 / 2} \cdot \frac{dz^{1 / 2}}{dz} = \frac{1}{2} a^{-1 / 2} z^{-1 / 2} = \frac{1}{2ax} \, ,</math> </td> </tr> </table> --> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>ax^b</math> </td> <td align="center"> <math>\leftrightarrow</math> </td> <td align="left"> <math> z \, , </math> </td> <td align="center"> and, <td align="right"> <math>\frac{\xi}{x^c}</math> </td> <td align="center"> <math>\leftrightarrow</math> </td> <td align="left"> <math> u \, ,</math> </td> </tr> </table> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\Rightarrow~~~ \frac{dx}{dz}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d}{dz}\biggl(\frac{z}{a}\biggr)^{1 / b} = a^{-1 / b} \cdot \frac{dz^{1 / b}}{dz} = \frac{1}{b} \cdot a^{-1 / b} z^{-1 + 1/b} = \frac{1}{bz} \cdot \biggl(\frac{z}{a}\biggr)^{1 / b} = \frac{x}{bz} = \frac{x^{1-b}}{ab} \, ,</math> </td> </tr> </table> in which case, <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> z(1-z) \frac{d^2u}{dz^2} + [\gamma - (\alpha + \beta + 1)z] \frac{du}{dz} - \alpha \beta u </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> ax^b(1-ax^b)\biggl\{ \frac{dx}{dz} \cdot \frac{d}{dx}\biggl[\frac{dx}{dz} \cdot \frac{d}{dx}\biggl( \xi x^{-c} \biggr) \biggr]\biggr\} + [\gamma - (\alpha + \beta + 1)ax^b] \frac{dx}{dz} \cdot \frac{d}{dx}\cdot\biggl( \xi x^{-c} \biggr) - \alpha \beta \biggl( \xi x^{-c} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> ax^b(1-ax^b)\biggl\{ \frac{x^{1-b}}{ab}\cdot \frac{d}{dx}\biggl[\frac{x^{1-b}}{ab} \cdot \biggl(x^{-c} \frac{d\xi}{dx} - c \xi x^{-1-c} \biggr) \biggr]\biggr\} + [\gamma - (\alpha + \beta + 1)ax^b] \cdot \frac{x^{1-b}}{ab}\cdot \biggl(x^{-c} \frac{d\xi}{dx} - c \xi x^{-1-c} \biggr) - \alpha \beta \biggl( \xi x^{-c} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> ax^b(1-ax^b)\biggl\{ \frac{x^{1-b}}{a^2b^2}\cdot \frac{d}{dx}\biggl[ x^{1-b-c} \frac{d\xi}{dx} - c \xi x^{-b-c} \biggr]\biggr\} + \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl(x^{1-b-c} \frac{d\xi}{dx} - c \xi x^{-b-c} \biggr) - \alpha \beta \biggl( \xi x^{-c} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> ax^b(1-ax^b) \cdot \frac{x^{1-b}}{a^2b^2}\cdot \biggl[ x^{1-b-c} \frac{d^2\xi}{dx^2} + (1-b-c)x^{-b-c} \frac{d\xi}{dx} - c x^{-b-c}\frac{d\xi}{dx} - c(-b-c) \xi x^{-1-b-c} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ x^{1-b-c} \frac{d\xi}{dx} - c \xi x^{-b-c} \biggr] - \biggl( \alpha \beta x^{-c} \biggr)\xi </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-ax^b) \cdot \frac{x}{ab^2}\cdot \biggl[ x^{1-b-c} \frac{d^2\xi}{dx^2} \biggr] + (1-ax^b) \cdot \frac{x}{ab^2}\cdot \biggl[ (1-b-c)x^{-b-c} \frac{d\xi}{dx} - c x^{-b-c}\frac{d\xi}{dx} \biggr] + (1-ax^b) \cdot \frac{x}{ab^2}\cdot \biggl[ c(b+c) \xi x^{-1-b-c}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ x^{1-b-c} \frac{d\xi}{dx} \biggr] - \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ c \xi x^{-b-c} \biggr] - \biggl( \alpha \beta x^{-c} \biggr)\xi </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{(1-ax^b)}{ab^2} \cdot x^2 \biggl[ x^{-b-c} \frac{d^2\xi}{dx^2} \biggr] +\biggl\{ \frac{(1-ax^b) }{ab^2}\cdot \biggl[ (1-b-c)x^{1-b-c} - c x^{1-b-c}\biggr] + \frac{[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ x^{1-b-c} \biggr] \biggr\} \frac{d\xi}{dx} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> +\biggl\{ - \frac{c[\gamma - (\alpha + \beta + 1)ax^b]}{ab} \cdot \biggl[ x^{-b-c} \biggr] - \biggl( \alpha \beta x^{-b-c} \biggr)x^b + \frac{(1-ax^b)}{ab^2} \cdot \biggl[ c(b+c) x^{-b-c}\biggr] \biggr\}\xi \, . </math> </td> </tr> </table> Multiplying through by <math>(-ab^2 x^{b+c})</math> gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>(-ab^2 x^{b+c}) \times 0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> x^2(ax^b-1) \frac{d^2\xi}{dx^2} +\biggl\{ (ax^b-1) \cdot \biggl[ (1-b-c) - c \biggr] - b[\gamma - (\alpha + \beta + 1)ax^b] \biggr\} x \cdot \frac{d\xi}{dx} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> +\biggl\{ bc[\gamma - (\alpha + \beta + 1)ax^b] + \biggl( \alpha \beta a b^2 \biggr)x^b - (1-ax^b) \cdot \biggl[ c(b+c) \biggr] \biggr\}\xi </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> x^2(ax^b-1) \frac{d^2\xi}{dx^2} +\biggl\{ -(1 + b\gamma -b-2c) + ax^b (1-b-2c) + b(\alpha + \beta + 1)ax^b \biggr\} x \cdot \frac{d\xi}{dx} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> +\biggl\{ bc\gamma - c(b+c) - bc(\alpha + \beta + 1)ax^b + \alpha \beta a b^2 x^b + c(b+c)ax^b \biggr\}\xi </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> x^2(ax^b-1) \frac{d^2\xi}{dx^2} + (apx^b + \lambda) x \cdot \frac{d\xi}{dx} + (arx^b + s) \xi \, , </math> </td> </tr> </table> which matches equation (3) of {{ VdBorght70 }} if the expressions for the four new scalar coefficients are, <table border="1" align="center" width="60%" cellpadding="10"> <tr><td align="center">'''Required Mapping Expressions'''</td></tr> <tr><td align="left"> <table border="0" cellpadding="5" align="center"> <tr> <td align="left" width="25%"> 1<sup>st</sup>: </td> <td align="right"> <math>p</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-b-2c) + b(\alpha + \beta + 1) = 1 - 2c + b(\alpha + \beta) \, , </math> </td> </tr> <tr> <td align="left" width="25%"> 2<sup>nd</sup>: </td> <td align="right"> <math>\lambda</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -(1 + b\gamma -b-2c) \, , </math> </td> </tr> <tr> <td align="left" width="25%"> 3<sup>rd</sup>: </td> <td align="right"> <math>s</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> bc\gamma - c(b+c) = c(b\gamma - b - c)\, , </math> </td> </tr> <tr> <td align="left" width="25%"> 4<sup>th</sup>: </td> <td align="right"> <math>r</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - bc(\alpha + \beta + 1) + \alpha \beta b^2 + c(b+c) \, . </math> </td> </tr> </table> </td></tr></table> If, for any given problem, we are given the values of these four scalar coefficients along with a choice of the exponent, <math>b</math>, that appears in the fourth argument of the hypergeometric series, we can determine values the other three arguments of the hypergeometric series — <math>\alpha, \beta, \gamma</math> — and the exponent, <math>c</math>. In what follows we show how this is done. ====Determining the Value of the Exponent, <i>c</i>==== Equating <math>(b\gamma)</math> in the 2<sup>nd</sup> and 3<sup>rd</sup> of the ''required mapping expressions'', gives, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>-(1 -b-2c +\lambda)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{s}{c} + b + c </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ (1+\lambda) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> c - \frac{s}{c} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ 0 </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> c^2 - c(1+\lambda)- s \, . </math> </td> </tr> </table> The pair of roots, <math>c_\pm</math>, of this quadratic equation are then obtained from the relation, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>2c_\pm </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1+\lambda) \pm [ (1+\lambda)^2 + 4s ]^{1 / 2} \, . </math> </td> </tr> </table> <span id="cplusminus">Note for further use below that,</span> <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>2(c_+ - c_-) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \{(1+\lambda) + [(1+\lambda)^2 + 4s]^{1 / 2} \} - \{(1+\lambda) - [(1+\lambda)^2 + 4s]^{1 / 2} \} \, . </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2[(1+\lambda)^2 + 4s]^{1 / 2} \, . </math> </td> </tr> </table> </td></tr> <tr> <td align="left"> Consistent with our derivation, {{ VdBorght70 }} states, <font color="darkgreen">"… if <math>A_1</math>, <math>A_2</math> are the solutions of <math>A^2 - (\lambda + 1)A-s = 0</math> … then <math>c = A_1</math> …"</font> </td> </tr> ====Determining the Value of the Coefficient, <i>γ</i>==== <tr><td align="left"> Combining our 3<sup>rd</sup> ''required mapping expression'' with the quadratic equation for <math>c_\pm</math> in such a way as to eliminate <math>s</math>, we find, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>cb(\gamma - 1) -c^2</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> c^2 - c(1+\lambda) </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ b(\gamma - 1) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2c_\pm - (1+\lambda) \, . </math> </td> </tr> </table> Adopting the ''superior'' sign, we find that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>b(\gamma - 1) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2c_+ - (1+\lambda) = [(1+\lambda)^2 + 4s]^{1 / 2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (c_+ - c_-) \, , </math> </td> </tr> </table> where, in order to make this last step we have drawn from the relation derived [[#cplusminus|immediately above]]. </td></tr> <tr> <td align="left"> Consistent with this derivation, {{ VdBorght70 }} states, <font color="darkgreen">"… <math>(1-\gamma)b = A_2 - A_1</math> …"</font> </td> </tr> ====Determining the Values of the Coefficients, <i>α</i> and <i>β</i>==== Combining the 1<sup>st</sup> and 4<sup>th</sup> ''required mapping expressions'' in such a way as to cancel terms involving <math>bc(\alpha + \beta)</math>, we find, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>(1 - p)c - 2c^2 + bc(\alpha + \beta)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> bc(\alpha + \beta) + bc - \alpha \beta b^2 - c(b+c) + r </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ (1 - p)c - c^2 </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \alpha \beta b^2 + r </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ (b\alpha)(b\beta ) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - (1 - p)c + c^2 + r \, . </math> </td> </tr> </table> Also, from the 1<sup>st</sup> ''required mapping expression'' alone we can write, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math> (b\alpha) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>(p-1) + 2c - b\beta \, .</math> </td> </tr> </table> Together, then, we have, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>(b\beta ) \biggl[ (p-1) + 2c - b\beta \biggr] </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - (1 - p)c + c^2 + r </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ 0 </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (b\beta )^2 - (b\beta)\underbrace{[ (p-1) + 2c ]}_{(b\alpha + b\beta)} + \overbrace{[c^2 + (p - 1 )c + r]}^{(b\alpha)\cdot(b\beta) } \, . </math> </td> </tr> </table> The pair of roots, <math>(b\beta)_\pm</math>, of this quadratic equation are then obtained from the relation, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>2(b\beta)_\pm </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> [ (p-1) + 2c ] \pm \biggl\{ [ (p-1) + 2c ]^2 - 4[c^2 + (p - 1 )c + r] \biggr\}^{1 / 2} \, . </math> </td> </tr> </table> Finally, plugging this expression for <math>(b\beta_\pm)</math> into the 1<sup>st</sup> ''required mapping expression'' gives <math>(b\alpha_\pm).</math> <!-- {{ VdBorght70 }} states that if, <font color="darkgreen">"… <math>B_1, B_2</math> are solutions of <math>B^2 - (p-1)B + r = 0</math>, then … <math>b\beta = A_1 + B_2.</math>"</font> Let's see if our derivation leads to this same conclusion. First note that the roots, <math>B_\pm</math>, of ''this'' Van der Borght quadratic equation are, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>2B_\pm </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (p-1) \pm [(p-1)^2 -4r]^{1 / 2} \, . </math> </td> </tr> </table> If we assign the ''inferior'' root with Van der Borght's notation, <math>B_2</math>, then, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>2(A_1 + B_2) = 2(c_+ + B_-) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \{ (1+\lambda) + [(1+\lambda)^2 + 4s]^{1 / 2} \} + \{(p-1) - [(p-1)^2 -4r]^{1 / 2} \} </math> </td> </tr> </table> --> ====Alternate Determination of <i>α</i> and <i>β</i> by Completing Squares==== <table border="1" cellpadding="10" width="80%" align="center"> <tr><td align="left"> Again, let's draw upon the {{ VdBorght70 }} statement that, <font color="darkgreen">"… if <math>A_1</math>, <math>A_2</math> are the solutions of <math>A^2 - (\lambda + 1)A-s = 0</math> … then <math>c = A_1</math> …"</font> {{ VdBorght70 }} also states that if, <font color="darkgreen">"… <math>B_1, B_2</math> are solutions of <math>B^2 - (p-1)B + r = 0</math>, then … <math>b\alpha = A_1 + B_1</math> and <math>b\beta = A_1 + B_2.</math>"</font> Let's see if we draw these same conclusions. </td></tr> </table> First, ''Complete the square'' in the quadratic equation for <math>B^2</math>: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>B^2 - (p-1)B </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -r </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ B^2 - (p-1)B + \biggl[\frac{(p-1)}{2}\biggr]^2 </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[\frac{(p-1)}{2}\biggr]^2 - r </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ B - \frac{(p-1)}{2}\biggr]^2 </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[\frac{(p-1)}{2}\biggr]^2 - r </math> </td> </tr> </table> Second, ''complete the square'' in the quadratic equation for <math>A^2</math> — which also completes the square for <math>c^2</math>: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>A^2 - (\lambda + 1)A </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> s </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ A^2 - (\lambda + 1)A + \biggl[\frac{(\lambda + 1)}{2}\biggr]^2</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> s + \biggl[\frac{(\lambda + 1)}{2}\biggr]^2 </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ A - \frac{(\lambda + 1)}{2} \biggr]^2</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> s + \biggl[\frac{(\lambda + 1)}{2}\biggr]^2 </math> </td> </tr> </table> Third, ''complete the square'' in the quadratic equation for <math>(b\beta)^2</math>: <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math> (b\beta )^2 - (b\beta)[ (p-1) + 2c ] </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - [c^2 + (p - 1 )c + r] </math> </td> </tr> <tr> <td align="right"> <math> \Rightarrow ~~~ (b\beta )^2 - (b\beta)[ (p-1) + 2c ] + \biggl[ \frac{(p-1)+2c}{2} \biggr]^2 </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ \frac{(p-1)+2c}{2} \biggr]^2 - [c^2 + (p - 1 )c + r] </math> </td> </tr> <tr> <td align="right"> <math> \Rightarrow ~~~ \biggl[ (b\beta ) - \frac{(p-1)+2c}{2} \biggr]^2 </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ \frac{(p-1)+2c}{2} \biggr]^2 - \biggl[c^2 + (p - 1 )c + \frac{(p-1)^2}{2^2}\biggr] - r + \frac{(p-1)^2}{2^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ \frac{(p-1)}{2}+c \biggr]^2 - \biggl[c + \frac{(p-1)}{2}\biggr]^2 - r + \frac{(p-1)^2}{2^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - r + \frac{(p-1)^2}{2^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ B - \frac{(p-1)}{2}\biggr]^2 \, . </math> </td> </tr> </table> Taking the ''positive'' root of both sides of this expression, we find that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math> (b\beta ) - \frac{(p-1)}{2} - c_+ </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> B_+ - \frac{(p-1)}{2} </math> </td> </tr> <tr> <td align="right"> <math> \Rightarrow ~~~ (b\beta ) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> B_+ + c_+ \, . </math> </td> </tr> </table> But, <math>c_\pm = A_\pm</math>. So we conclude, as did {{ VdBorght70 }}, that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math> (b\beta ) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> B_+ + A_+ \, . </math> </td> </tr> </table> Alternatively, taking the ''negative'' root of the RHS of this expression, we find that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math> (b\beta ) - \frac{(p-1)}{2} - c_+ </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - B_- + \frac{(p-1)}{2} </math> </td> </tr> <tr> <td align="right"> <math> \Rightarrow ~~~ (b\beta ) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{(p-1)}{2} + c_+ - B_- + \frac{(p-1)}{2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> c_+ - B_- + (p-1) \, . </math> </td> </tr> </table> Also, given that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math> 1 - 2c + b(\alpha + \beta) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>p</math> </td> </tr> <tr> <td align="right"> <math> \Rightarrow ~~~ (b\alpha) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>p - 1 + 2c - (b\beta)</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>(p - 1) + 2c - \biggl[ c_+ - B_- + (p-1) \biggr]</math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>2c - c_+ + B_- \, .</math> </td> </tr> </table> As long as we assume that <math>c = c_+</math> in this expression, we also obtain the {{ VdBorght70 }} expression for <math>(b\alpha)</math>, namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math> (b\alpha ) </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> B_- + A_+ \, . </math> </td> </tr> </table> ===If b = 2=== <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> z(1-z) \frac{d^2u}{dz^2} + [\gamma - (\alpha + \beta + 1)z] \frac{du}{dz} - \alpha \beta u </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> ax^2(1-ax^2)\biggl\{ \frac{dx}{dz} \cdot \frac{d}{dx}\biggl[\frac{dx}{dz} \cdot \frac{d}{dx}\biggl( \xi x^{-c} \biggr) \biggr]\biggr\} + [\gamma - (\alpha + \beta + 1)ax^2] \frac{dx}{dz} \cdot \frac{d}{dx}\cdot\biggl( \xi x^{-c} \biggr) - \alpha \beta \biggl( \xi x^{-c} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> ax^2(1-ax^2)\biggl\{ \frac{1}{2ax}\cdot \frac{d}{dx}\biggl[\frac{1}{2ax} \cdot \biggl(x^{-c} \frac{d\xi}{dx} - c \xi x^{-1-c} \biggr) \biggr]\biggr\} + [\gamma - (\alpha + \beta + 1)ax^2] \cdot \frac{1}{2ax}\cdot \biggl(x^{-c} \frac{d\xi}{dx} - c \xi x^{-1-c} \biggr) - \alpha \beta \biggl( \xi x^{-c} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> ax^2(1-ax^2)\biggl\{ \frac{1}{4a^2 x}\cdot \frac{d}{dx}\biggl[ \biggl(x^{-1-c} \frac{d\xi}{dx} - c \xi x^{-2-c} \biggr) \biggr]\biggr\} + [\gamma - (\alpha + \beta + 1)ax^2] \cdot \frac{1}{2a}\cdot \biggl(x^{-1-c} \frac{d\xi}{dx} - c \xi x^{-2-c} \biggr) - \alpha \beta \biggl( \xi x^{-c} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{x(1-ax^2)}{4a}\biggl\{ x^{-1-c}\frac{d^2\xi}{dx^2} -(1+c)x^{-2-c} \frac{d\xi}{dx} -cx^{-2-c}\frac{d\xi}{dx} + (2c+c^2)\xi x^{-3-c} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl[\frac{\gamma - (\alpha + \beta + 1)ax^2}{2a} \biggr] \cdot \biggl(x^{-1-c} \frac{d\xi}{dx} \biggr) - \biggl\{ \biggl[\frac{\gamma - (\alpha + \beta + 1)ax^2}{2a} \biggr] \cdot c x^{-2-c} + \alpha \beta x^{-c} \biggr\}\xi </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{x(1-ax^2)}{4a}\biggl\{ x^{-1-c}\frac{d^2\xi}{dx^2} \biggr\} + \biggl\{ \frac{x(1-ax^2)}{4a}\biggl[ - (1+c)x^{-2-c} - cx^{-2-c} \biggr] + \biggl[\frac{\gamma - (\alpha + \beta + 1)ax^2}{2a} \biggr] \cdot x^{-1-c} \biggr\}\frac{d\xi}{dx} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ \biggl[ \frac{x(1-ax^2)}{4a}\biggr] (2c+c^2) x^{-3-c} - \biggl[\frac{\gamma - (\alpha + \beta + 1)ax^2}{2a} \biggr] \cdot c x^{-2-c} - \alpha \beta x^{-c} \biggr\}\xi \, . </math> </td> </tr> </table> Multiplying through by a term proportional to <math>x^{2+c}</math> gives us, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\biggl[ -4a x^{2+c}\biggr] \times \biggl[0\biggr]</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> x^2(ax^2-1)\biggl\{ \frac{d^2\xi}{dx^2} \biggr\} + \biggl\{ x(1-ax^2)\biggl[ (1+c) + c \biggr] - 2[\gamma - (\alpha + \beta + 1)ax^2] \cdot x \biggr\}\frac{d\xi}{dx} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ (ax^2-1)(2c+c^2) - 2c[\gamma - (\alpha + \beta + 1)ax^2] + 4a \alpha \beta x^{2} \biggr\}\xi </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> x^2(ax^2-1)\biggl\{ \frac{d^2\xi}{dx^2} \biggr\} + \biggl\{ (1+2c- 2\gamma) -ax^2(1+2c) + 2(\alpha + \beta + 1)ax^2] \biggr\} x \cdot \frac{d\xi}{dx} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ (ax^2-1)(2c+c^2) - 2c[\gamma - (\alpha + \beta + 1)ax^2] + 4a \alpha \beta x^{2} \biggr\}\xi </math> </td> </tr> </table> =LAWE= ==Familiar Foundation== Drawing from an [[SSC/Perturbations#2ndOrderODE|accompanying discussion]], we have the, <div align="center" id="2ndOrderODE"> <font color="#770000">'''LAWE: Linear Adiabatic Wave''' (or ''Radial Pulsation'') '''Equation'''</font><br /> {{Math/EQ_RadialPulsation01}} </div> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>g_0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>- \frac{1}{\rho_0} \frac{dP_0}{dr_0} \, .</math> </td> </tr> </table> Multiplying through by <math>R^2</math>, and making the variable substitutions, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>x</math> </td> <td align="center"> <math>\rightarrow</math> </td> <td align="left"> <math>f \, ,</math> </td> </tr> <tr> <td align="right"> <math>\frac{r_0}{R}</math> </td> <td align="center"> <math>\rightarrow</math> </td> <td align="left"> <math>x \, ,</math> </td> </tr> <tr> <td align="right"> <math>(4 - 3\gamma_g)</math> </td> <td align="center"> <math>\rightarrow</math> </td> <td align="left"> <math>-\alpha \gamma_g \, ,</math> </td> </tr> </table> the LAWE may be rewritten as, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{dx^2} + \biggl[\frac{4}{x} - \biggl(\frac{g_0 \rho_0 R}{P_0}\biggr) \biggr] \frac{d f}{dx} + \biggl(\frac{\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 - \frac{\alpha \gamma_g g_0}{r_0} \biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[4 - \biggl(\frac{g_0 \rho_0 r_0}{P_0}\biggr) \biggr] \frac{d f}{dx} + \biggl[\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr) - \frac{\alpha \gamma_g g_0}{r_0}\biggl(\frac{\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr) \biggr] f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[4 - \biggl(\frac{g_0 \rho_0 r_0}{P_0}\biggr) \biggr] \frac{d f}{dx} + \biggl[\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr) - \frac{\alpha }{x^2}\biggl(\frac{ g_0 r_0\rho_0 }{ P_0} \biggr) \biggr] f \, . </math> </td> </tr> </table> If we furthermore adopt the variable definition, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\mu</math> </td> <td align="center"> <math>\equiv</math> </td> <td align="left"> <math>\biggl(\frac{g_0 \rho_0 r_0}{P_0}\biggr) = - \frac{d\ln P_0}{d\ln r_0} \, ,</math> </td> </tr> </table> we obtain what we will refer to as the, <div align="center" id="Kopal48Expression"> <font color="#770000">'''Kopal (1948) LAWE'''</font><br /> <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{dx^2} + \frac{(4-\mu)}{x} \cdot \frac{df}{dx} + \biggl[\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr) - \frac{\alpha \mu}{x^2} \biggr]f \, . </math> </td> </tr> <tr> <td align="center" colspan="3"> {{ Kopal48 }}, p. 378, Eq. (6)<br /> {{ VdBorght70 }}, p. 325, Eq. (1) </td> </tr> </table> </div> ==Specifically for Polytropes== Let's look at the expression for the function, <math>\mu</math>, that arises in the context of polytropic spheres. ===General Expression for the Function μ=== First, we note that, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>r_0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>a \xi \, ,</math> </td> </tr> <tr> <td align="right"> <math>\rho_0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\rho_c \theta^n \, ,</math> </td> </tr> <tr> <td align="right"> <math>P_0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>K\biggl[\rho_c \theta^n\biggr]^{(n+1)/n} \, ,</math> </td> </tr> <tr> <td align="right"> <math>M_r</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>4\pi a^3\rho_c \biggl( - \xi^2 \frac{d\theta}{d\xi}\biggr) \, ,</math> </td> </tr> <tr> <td align="right"> <math>g_0 \equiv \frac{GM_r}{r_0^2}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>4\pi G a\rho_c \biggl( - \frac{d\theta}{d\xi}\biggr) \, ,</math> </td> </tr> </table> where, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>a</math> </td> <td align="center"> <math>\equiv</math> </td> <td align="left"> <math>\biggl[\frac{(n+1)K}{4\pi G}\biggr]^{1 / 2} \rho_c^{(1-n)/2n} \, .</math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ K</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \frac{4\pi G}{(n+1)} \biggr] a^2 \rho_c^{(n-1)/n} \, .</math> </td> </tr> </table> Hence, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\mu = \biggl(\frac{g_0 \rho_0 r_0}{P_0}\biggr)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 4\pi G a\rho_c \biggl( - \frac{d\theta}{d\xi}\biggr) \rho_c \theta^n a\xi \biggl[\rho_c \theta^n\biggr]^{-(n+1)/n}\biggl[ \frac{(n+1)}{4\pi G} \biggr] a^{-2} \rho_c^{-(n-1)/n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (n+1)\biggl( - \frac{d\theta}{d\xi}\biggr) \theta^n \xi \theta^{-(n+1)} \rho_c^{2-(n+1)/n-(n-1)/n} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (n+1)\biggl( - \frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi}\biggr) = (n+1)\biggl( - \frac{d\ln \theta}{d\ln \xi}\biggr) \, . </math> </td> </tr> </table> Alternatively, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\mu= - \frac{d\ln P_0}{d\ln r_0}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -\frac{r_0}{P_0} \cdot \frac{dP_0}{d r_0} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -\xi \theta^{-(n+1)}\cdot \frac{d}{d\xi} \biggl[\theta^{(n+1)}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -(n+1) \biggl(\frac{\xi}{\theta}\cdot \frac{d\theta}{d\xi} \biggr) = (n+1)\biggl( - \frac{d\ln \theta}{d\ln \xi}\biggr) \, . </math> </td> </tr> </table> Yes! ===Trial Displacement Function=== Now, building on an [[SSC/Stability/InstabilityOnsetOverview#Analyses_of_Radial_Oscillations|accompanying discussion]], let's guess, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>f_\mathrm{trial}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{3(n-1)}{2n} \biggl[ 1 + \biggl( \frac{n-3}{n-1} \biggr)\biggl(\frac{1}{\xi \theta^n}\biggr)\frac{d\theta}{d\xi} \biggr] </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ \frac{2n}{3(n-1)} \biggr] f_\mathrm{trial}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 1 + \biggl( \frac{n-3}{n-1} \biggr)\xi^{-1} \theta^{-n}\biggl[ \frac{\theta}{\xi} \cdot \frac{d\ln\theta}{d\ln\xi} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 1 + \biggl( \frac{3-n}{n-1} \biggr)\xi^{-2} \theta^{(1-n)}\cdot \biggl( - \frac{d\ln\theta}{d\ln\xi} \biggr) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 1 + \biggl[ \frac{3-n}{(n+1)(n-1)} \biggr]\xi^{-2} \theta^{(1-n)}\cdot \mu </math> </td> </tr> </table> Flipping it around, we have alternatively, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\mu</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ \frac{(n+1)(n-1)}{3-n} \biggr]\biggl\{ \biggl[ \frac{2n}{3(n-1)} \biggr] f_\mathrm{trial} - 1 \biggr\} \xi^{2} \theta^{(n-1)} </math> </td> </tr> </table> ===Plug into Kopal (1948) LAWE=== ====Replace f<sub>trial</sub> by μ==== Plugging this trial function into the Kopal (1948) LAWE and recognizing that <math>x = \xi/\xi_1</math>, we find, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\xi_1^{-2} \cdot ~\mathrm{LAWE}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f_\mathrm{trial}}{d\xi^2} + \frac{(4-\mu)}{\xi} \cdot \frac{df_\mathrm{trial}}{d\xi} + \biggl[\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0 \xi_1^2} \biggr) - \frac{\alpha \mu}{\xi^2} \biggr]f_\mathrm{trial} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ \frac{2n}{3(n-1)} \biggr] \xi_1^{-2} \cdot ~\mathrm{LAWE}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 }{d\xi^2} \biggl\{ 1 + \biggl[ \frac{3-n}{(n+1)(n-1)} \biggr]\xi^{-2} \theta^{(1-n)}\cdot \mu \biggr\} + \frac{(4-\mu)}{\xi} \cdot \frac{d}{d\xi} \biggl\{ 1 + \biggl[ \frac{3-n}{(n+1)(n-1)} \biggr]\xi^{-2} \theta^{(1-n)}\cdot \mu \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl[\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0 \xi_1^2} \biggr) - \frac{\alpha \mu}{\xi^2} \biggr]\biggl\{ 1 + \biggl[ \frac{3-n}{(n+1)(n-1)} \biggr]\xi^{-2} \theta^{(1-n)}\cdot \mu \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ \frac{2n(n+1)}{3} \biggr] \xi_1^{-2} \cdot ~\mathrm{LAWE}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 }{d\xi^2} \biggl\{ (3-n) \xi^{-2} \theta^{(1-n)}\cdot \mu \biggr\} + \frac{(4-\mu)}{\xi} \cdot \frac{d}{d\xi} \biggl\{ (3-n) \xi^{-2} \theta^{(1-n)}\cdot \mu \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl[\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0 \xi_1^2} \biggr) - \frac{\alpha \mu}{\xi^2} \biggr]\biggl\{ (n+1)(n-1) + (3-n)\xi^{-2} \theta^{(1-n)}\cdot \mu \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ \frac{2n(n+1)}{3(3-n)} \biggr] \xi_1^{-2} \cdot ~\mathrm{LAWE}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 }{d\xi^2} \biggl\{ \xi^{-2} \theta^{(1-n)}\cdot \mu \biggr\} + \frac{(4-\mu)}{\xi} \cdot \frac{d}{d\xi} \biggl\{ \xi^{-2} \theta^{(1-n)}\cdot \mu \biggr\} + \biggl[\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0 \xi_1^2} \biggr) - \frac{\alpha \mu}{\xi^2} \biggr]\biggl\{ \frac{(n+1)(n-1)}{(3-n)} + \xi^{-2} \theta^{(1-n)}\cdot \mu \biggr\} \, . </math> </td> </tr> </table> Noting that, <math>R/\xi_1 = a</math> and <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\frac{\rho_0}{P_0}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \rho_c \theta^n K^{-1} \rho_c^{-(n+1)/n}\theta^{-(n+1)} = K^{-1}\rho_c^{-1/n} \theta^{-1} \, , </math> </td> </tr> </table> the frequency-squared term may be rewritten as, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\frac{\omega^2}{\gamma_g}\biggl(\frac{\rho_0 a^2}{ P_0 } \biggr)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl( \frac{\omega^2}{\gamma_g} \biggr) K^{-1}\rho_c^{-1/n} \theta^{-1} \biggl[\frac{(n+1)K}{4\pi G}\biggr] \rho_c^{(1-n)/n} = \biggl( \frac{\omega^2}{\gamma_g} \biggr) \biggl[\frac{(n+1)}{4\pi \rho_c G}\biggr]\theta^{-1} \, . </math> </td> </tr> </table> ====Replace μ by f<sub>trial</sub>==== Making instead the alternate substitution, namely, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\mu</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ \frac{(n+1)(n-1)}{3-n} \biggr]\biggl\{ \biggl[ \frac{2n}{3(n-1)} \biggr] f_\mathrm{trial} - 1 \biggr\} \xi^{2} \theta^{(n-1)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl\{ \frac{2n(n+1)}{3(3-n)} \cdot f_\mathrm{trial} - \frac{(n+1)(n-1)}{3-n} \biggr\} \xi^{2} \theta^{(n-1)} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{3(3-n)} \biggl[ \underbrace{2n(n+1)}_{A} f_\mathrm{trial} - \underbrace{3(n+1)(n-1)}_{B} \biggr]\xi^{2} \theta^{(n-1)} </math> </td> </tr> </table> we have, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\xi_1^{-2} \cdot ~\mathrm{LAWE}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f_\mathrm{trial}}{d\xi^2} + \biggl\{\frac{4}{\xi} \biggr\} \frac{df_\mathrm{trial}}{d\xi} - \biggl\{ \frac{\mu}{\xi} \biggr\} \frac{df_\mathrm{trial}}{d\xi} + \biggl( \frac{\omega^2}{\gamma_g} \biggr)\biggl[\frac{(n+1)}{4\pi \rho_c G}\biggr]\frac{f_\mathrm{trial}}{\theta} - \biggl(\frac{\alpha}{\xi^2} \biggr) \mu f_\mathrm{trial} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ \frac{3(3-n) }{\xi_1^2} \biggr] ~\mathrm{LAWE}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 3(3-n)\frac{d^2 f_\mathrm{trial}}{d\xi^2} + \biggl\{\frac{12(3-n)}{\xi} \biggr\} \frac{df_\mathrm{trial}}{d\xi} - \biggl[ A f_\mathrm{trial} - B \biggr]\xi \theta^{(n-1)}\frac{df_\mathrm{trial}}{d\xi} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + 3(3-n)\biggl( \frac{\omega^2}{\gamma_g} \biggr)\biggl[\frac{(n+1)}{4\pi \rho_c G}\biggr]\frac{f_\mathrm{trial}}{\theta} - \alpha \biggl[ A f_\mathrm{trial} - B \biggr]\theta^{(n-1)} f_\mathrm{trial} \, . </math> </td> </tr> </table> Noting that, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\frac{d}{d\xi} \biggl[ \xi \theta^{(n-1)}f_\mathrm{trial} \biggr]</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl\{ \theta^{(n-1)}f_\mathrm{trial} + \xi f_\mathrm{trial} (n-1)\theta^{(n-2)}\frac{d\theta}{d\xi} + \xi\theta^{(n-1)}\frac{df_\mathrm{trial}}{d\xi} \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \xi\theta^{(n-1)}\frac{df_\mathrm{trial}}{d\xi} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d}{d\xi} \biggl[ \xi \theta^{(n-1)}f_\mathrm{trial} \biggr] - \biggl[ \theta^{(n-1)}f_\mathrm{trial} + \xi f_\mathrm{trial} (n-1)\theta^{(n-2)}\frac{d\theta}{d\xi} \biggr] \, , </math> </td> </tr> </table> we furthermore can write, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\Rightarrow ~~~ \biggl[ \frac{3(3-n) }{\xi_1^2} \biggr] ~\mathrm{LAWE}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 3(3-n)\frac{d^2 f_\mathrm{trial}}{d\xi^2} + \biggl\{\frac{12(3-n)}{\xi} \biggr\} \frac{df_\mathrm{trial}}{d\xi} + 3(3-n)\biggl( \frac{\omega^2}{\gamma_g} \biggr)\biggl[\frac{(n+1)}{4\pi \rho_c G}\biggr]\frac{f_\mathrm{trial}}{\theta} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> - \alpha \biggl[ A f_\mathrm{trial} - B \biggr]\theta^{(n-1)} f_\mathrm{trial} - \biggl[ A f_\mathrm{trial} - B \biggr] \biggl\{ \frac{d}{d\xi} \biggl[ \xi \theta^{(n-1)}f_\mathrm{trial} \biggr] - \biggl[ \theta^{(n-1)}f_\mathrm{trial} + \xi f_\mathrm{trial} (n-1)\theta^{(n-2)}\frac{d\theta}{d\xi} \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 3(3-n)\frac{d^2 f_\mathrm{trial}}{d\xi^2} + \biggl\{\frac{12(3-n)}{\xi} \biggr\} \frac{df_\mathrm{trial}}{d\xi} + 3(3-n)\biggl( \frac{\omega^2}{\gamma_g} \biggr)\biggl[\frac{(n+1)}{4\pi \rho_c G}\biggr]\frac{f_\mathrm{trial}}{\theta} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> - \biggl[ A f_\mathrm{trial} - B \biggr] \biggl\{ \frac{d}{d\xi} \biggl[ \xi \theta^{(n-1)}f_\mathrm{trial} \biggr] - \biggl[ \theta^{(n-1)}f_\mathrm{trial} + \xi f_\mathrm{trial} (n-1)\theta^{(n-2)}\frac{d\theta}{d\xi} \biggr] + \alpha \theta^{(n-1)} f_\mathrm{trial} \biggr\} </math> </td> </tr> </table> =Seek Hypergeometric Form= Start with the standard LAWE, namely, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{d\xi^2} + \frac{(4-\mu)}{\xi} \cdot \frac{df}{d\xi} + \biggl[\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr) - \frac{\alpha \mu}{\xi^2} \biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{d\xi^2} + \frac{1}{\xi} \biggl[4 + (n+1)\frac{\xi \theta^'}{\theta}\biggr] \cdot \frac{df}{d\xi} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \frac{1}{\theta} + \frac{\alpha \theta^'}{\xi\theta}\biggr]f \, . </math> </td> </tr> </table> ==Part I== Try switching the independent variable from <math>\xi</math> to <math>z</math> such that, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>z</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \xi^{-1}\theta^{-n} (\theta^') ~~~\Rightarrow ~~~ (\theta^') = \xi \theta^n z </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{dz}{d\xi}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -\xi^{-2}\theta^{-n} (\theta^') -n \xi^{-1}\theta^{-n-1} (\theta^')^2 + \xi^{-1}\theta^{-n} (\theta^{''}) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \biggl[ \xi^{-2}\theta^{-n} (\theta^') +n \xi^{-1}\theta^{-n-1} (\theta^')^2 + \xi^{-1}\theta^{-n} \biggl( \theta^n + \frac{2\theta'}{\xi} \biggr) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \biggl[ \xi^{-2}\theta^{-n} (\xi \theta^n z) +n \xi^{-1}\theta^{-n-1} (\xi \theta^n z)^2 + \xi^{-1}\theta^{-n} \biggl( \theta^n \biggr) + \xi^{-1}\theta^{-n} \biggl( \frac{2\theta'}{\xi} \biggr) \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \biggl[ \xi^{-1}z +n \xi \theta^{n-1} z^2 + \xi^{-1} + 2\xi^{-1}z \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \biggl[\xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] \, ; </math> </td> </tr> </table> and, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\frac{d^2z}{d\xi^2}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \biggl\{ -\xi^{-2}(1 + 3z) + 3\xi^{-1}\cdot \frac{dz}{d\xi} + n \theta^{n-1} z^2 + n (n-1)\xi \theta^{n-2} (\theta^') z^2 + 2n \xi \theta^{n-1} z \cdot \frac{dz}{d\xi} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \biggl\{ -\xi^{-2}(1 + 3z) - 3\xi^{-1}\cdot \biggl[\xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] + n \theta^{n-1} z^2 + n (n-1)\xi \theta^{n-2} (\xi \theta^n z) z^2 - 2n \xi \theta^{n-1} z \cdot \biggl[\xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl\{ \xi^{-2}(1 + 3z) + 3\xi^{-1}\cdot \biggl[\xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] - n \theta^{n-1} z^2 - n (n-1)\xi \theta^{n-2} (\xi \theta^n z) z^2 + 2n \xi \theta^{n-1} z \cdot \biggl[\xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl\{ \xi^{-2}(1 + 3z) + \biggl[3\xi^{-2}(1 + 3z) + 3n \theta^{n-1} z^2 \biggr] - n \theta^{n-1} z^2 - n (n-1)\xi^2 \theta^{2(n-1)} z^3 + \biggl[2n \theta^{n-1} z (1 + 3z) + 2n^2 \xi^2 \theta^{2(n-1)} z^3 \biggr] \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 4\xi^{-2}(1 + 3z) + 2n \theta^{n-1} [z (1 + 3z) + z^2] + [2n^2 - n (n-1)]\xi^2 \theta^{2(n-1)} z^3 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 4\xi^{-2}(1 + 3z) + 2n \theta^{n-1} [1 + 4z]z + n(n+1)\xi^2 \theta^{2(n-1)} z^3 \, . </math> </td> </tr> </table> ==Part II== <table border="1" align="center" cellpadding="8" width="60%"> <tr> <td align="center"><b>Part I Summary …</b></td> </tr> <tr><td align="left"> <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>(\theta^')</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \xi\theta^{n} z \, , </math> </td> </tr> <tr> <td align="right"> <math>\frac{dz}{d\xi}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \biggl[\xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] \, , </math> </td> </tr> <tr> <td align="right"> <math>\frac{d^2z}{d\xi^2}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 4\xi^{-2}(1 + 3z) + 2n \theta^{n-1} [1 + 4z]z + n(n+1)\xi^2 \theta^{2(n-1)} z^3 \, . </math> </td> </tr> </table> </td></tr></table> ---- Also, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\frac{df}{d\xi}</math> </td> <td align="center"> <math>\rightarrow</math> </td> <td align="left"> <math> \frac{dz}{d\xi} \cdot \frac{df}{dz} \, ; </math> </td> </tr> <tr> <td align="right"> <math>\frac{d^2f}{d\xi^2} = \frac{d}{d\xi} \biggl[\frac{dz}{d\xi} \cdot \frac{df}{dz}\biggr]</math> </td> <td align="center"> <math>\rightarrow</math> </td> <td align="left"> <math> \frac{d^2z}{d\xi^2} \cdot \frac{df}{dz} + \biggl[ \frac{dz}{d\xi} \biggr]^2 \cdot \frac{d^2f}{dz^2} </math> </td> </tr> </table> As a result, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> LAWE </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{d\xi^2} + \frac{1}{\xi} \biggl[4 + (n+1)\frac{\xi \theta^'}{\theta}\biggr] \cdot \frac{df}{d\xi} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \frac{1}{\theta} + \frac{\alpha \theta^'}{\xi\theta}\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{d\xi^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \frac{1}{\xi} \biggl[4 + (n+1)\xi^2 \theta^{n-1} z\biggr] \cdot \frac{df}{d\xi} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \frac{1}{\theta} + \alpha \theta^{n-1} z\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ \frac{dz}{d\xi} \biggr]^2 \cdot \frac{d^2f}{dz^2} + \frac{d^2z}{d\xi^2} \cdot \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl[4\xi^{-1} + (n+1)\xi \theta^{n-1} z\biggr] \frac{dz}{d\xi} \cdot \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl\{ \frac{dz}{d\xi} \biggr\}^2 \cdot \frac{d^2f}{dz^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ \frac{d^2z}{d\xi^2} + \biggl[4\xi^{-1} + (n+1)\xi \theta^{n-1} z\biggr] \frac{dz}{d\xi} \biggr\} \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl\{ \xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr\}^2 \cdot \frac{d^2f}{dz^2} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ \biggl[ 4\xi^{-2}(1 + 3z) + 2n \theta^{n-1} [1 + 4z]z + n(n+1)\xi^2 \theta^{2(n-1)} z^3 \biggr] - \biggl[4\xi^{-1} + (n+1)\xi \theta^{n-1} z\biggr] \biggl[ \xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] \biggr\} \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ \xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr]\biggl[ \xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] \cdot \frac{d^2f}{dz^2} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl[ 4\xi^{-2}(1 + 3z) + 2n \theta^{n-1} [1 + 4z]z + n(n+1)\xi^2 \theta^{2(n-1)} z^3 \biggr] \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> - \biggl[4\xi^{-1} \biggr] \biggl[ \xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] \frac{df}{dz} - \biggl[(n+1)\xi \theta^{n-1} z\biggr] \biggl[ \xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \xi^{-2}(1+3z)^2 + 2n(1+3z)z^2 \theta^{n-1} + n^2 \xi^2 \theta^{2(n-1)}z^4 \biggr] \cdot \frac{d^2f}{dz^2} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl[ 4\xi^{-2}(1 + 3z) + 2n \theta^{n-1} [1 + 4z]z + n(n+1)\xi^2 \theta^{2(n-1)} z^3 \biggr] \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> - \biggl[ 4\xi^{-2}(1 + 3z) + 4n \theta^{n-1} z^2 \biggr] \frac{df}{dz} - \biggl[ (n+1)\theta^{n-1} (1 + 3z)z + n(n+1) \xi^2 \theta^{2(n-1)} z^3 \biggr] \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \xi^{-2}(1+3z)^2 + 2n(1+3z)z^2 \theta^{n-1} + n^2 \xi^2 \theta^{2(n-1)}z^4 \biggr] \cdot \frac{d^2f}{dz^2} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ \biggl[ 2n \theta^{n-1} (1 + 4z) \biggr] - \biggl[ 4n \theta^{n-1} z \biggr] - \biggl[ (n+1)\theta^{n-1} (1 + 3z) \biggr] \biggr\} z \cdot \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \xi^{-2}(1+3z)^2 + 2n(1+3z)z^2 \theta^{n-1} + n^2 \xi^2 \theta^{2(n-1)}z^4 \biggr] \cdot \frac{d^2f}{dz^2} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ \biggl[ 2n + 8n z \biggr] - \biggl[ 4n z \biggr] - \biggl[(n+1) + 3(n+1)z \biggr] \biggr\} \theta^{n-1} z \cdot \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \xi^{-2}(1+3z)^2 + 2n(1+3z)z^2 \theta^{n-1} + n^2 \xi^2 \theta^{2(n-1)}z^4 \biggr] \cdot \frac{d^2f}{dz^2} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr]f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ (n-1) + (n - 3)z \biggr\} \theta^{n-1} z \cdot \frac{df}{dz} \, . </math> </td> </tr> </table> ==Part III== Now, suppose that <math>f = (a_0 + b_0z)</math>. We have, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> LAWE </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \xi^{-2}(1+3z)^2 + 2n(1+3z)z^2 \theta^{n-1} + n^2 \xi^2 \theta^{2(n-1)}z^4 \biggr] \cdot \cancelto{0}{\frac{d^2f}{dz^2}} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr](a_0 + b_0z) + \biggl\{ (n-1) + (n - 3)z \biggr\} \theta^{n-1} z \cdot b_0 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} \biggr](a_0 + b_0z) + \biggl\{ (n+1)\alpha (a_0 + b_0z) + b_0(n-1) + b_0(n - 3)z \biggr\} \theta^{n-1} z </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} \biggr](a_0 + b_0z) + \biggl\{ (n+1)\alpha (a_0 ) + b_0(n-1) + \biggl[(n - 3) + (n+1)\alpha \biggr]b_0z \biggr\} \theta^{n-1} z \, . </math> </td> </tr> </table> Now, in order for the last term to be zero, we need, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[(n - 3) + (n+1)\alpha \biggr] </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \alpha</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{3-n}{n+1} \, . </math> </td> </tr> </table> This is precisely the relation that results from the definition of <math>\alpha \equiv (3 - 4/\gamma_g)</math> if the model is evolved assuming <math>\gamma_g = (n+1)/n</math>. We simultaneously seek the relation, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (n+1)\alpha (a_0 ) + b_0(n-1) </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ b_0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ \frac{n+1}{1-n} \biggr] \alpha (a_0 ) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> a_0 \biggl[ \frac{3-n}{1-n} \biggr] </math> </td> </tr> </table> It appears as though the leading coefficient, <math>a_0</math>, is arbitrary, so we will set it equal to unity. This means that the displacement function is, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>f</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 1 + \biggl[ \frac{3-n}{1-n} \biggr]z \, . </math> </td> </tr> </table> This expression for the displacement function, <math>f</math>, is identical to the [[SSC/Stability/InstabilityOnsetOverview#ExactPolytropicSolution|expression found inside the square brackets of our separately derived exact solution of the polytropic LAWE]]. Furthermore, given the notation, <math>(\sigma_c^2/\gamma_g) = (\mathfrak{F}-2\alpha)</math>, the first term on the RHS of the LAWE will go to zero when, <math>\mathfrak{F} = 2(3-n)/(n+1)</math>. ==Part IV== If we divide through by <math>(\theta^{n-1}z)</math>, the LAWE that was derived above in Part II assumes the following form, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\biggl[ \theta^{1-n}z^{-1}\biggr] \times</math> LAWE </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\theta^{1-n}z^{-1}\biggl[ \xi^{-2}(1+3z)^2 + 2n(1+3z)z^2 \theta^{n-1} + n^2 \xi^2 \theta^{2(n-1)}z^4 \biggr] \cdot \frac{d^2f}{dz^2} + \biggl\{ (n-1) + (n - 3)z \biggr\} \cdot \frac{df}{dz} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-n} z^{-1} + \alpha \biggr]\cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \xi^{-2}\theta^{1-n}z^{-1}(1+3z)^2 + 2n(1+3z)z + n^2 \xi^2 \theta^{(n-1)}z^3 \biggr] \cdot \frac{d^2f}{dz^2} + \biggl\{ (n-1) + (n - 3)z \biggr\} \cdot \frac{df}{dz} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-n} z^{-1} + \alpha \biggr]\cdot f \, , </math> </td> </tr> </table> which resembles the [[#Hypergeometric_Differential_Equation|above-discussed ''hypergeometric differential equation'']], 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> z(1-z) \frac{d^2u}{dz^2} + [\gamma - (\alpha + \beta + 1)z] \frac{du}{dz} - \alpha \beta u \, .</math> </td> </tr> </table> For the record we note that the coefficient (in square brackets) of the first term on the RHS of our LAWE expression is the square of the first derivative of <math>z</math> with respect to <math>\xi</math>; that is, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\theta^{1-n}z^{-1}\biggl(\frac{dz}{d\xi} \biggr)^2</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \theta^{1-n}z^{-1} \biggl[\xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr]^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \theta^{1-n}z^{-1} \biggl[ \xi^{-2}(1 + 3z)^2 + 2n(1 + 3z) \theta^{n-1} z^2 + n^2 \xi^2 \theta^{2(n-1)} z^4 \biggr] \, . </math> </td> </tr> </table> ==Part V== Now suppose that, <math>f = (a_0 + b_0z + c_0z^2)</math>, where again, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>z</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \xi^{-1} \theta^{-n}(\theta^') \, . </math> </td> </tr> </table> Recalling that, <table border="1" align="center" cellpadding="8" width="60%"> <tr> <td align="center"><b>Part I Summary …</b></td> </tr> <tr><td align="left"> <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>(\theta^')</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \xi\theta^{n} z \, , </math> </td> </tr> <tr> <td align="right"> <math>\frac{dz}{d\xi}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> - \biggl[\xi^{-1}(1 + 3z) + n \xi \theta^{n-1} z^2 \biggr] \, , </math> </td> </tr> <tr> <td align="right"> <math>\frac{d^2z}{d\xi^2}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 4\xi^{-2}(1 + 3z) + 2n \theta^{n-1} [1 + 4z]z + n(n+1)\xi^2 \theta^{2(n-1)} z^3 \, . </math> </td> </tr> </table> </td></tr></table> it may prove useful to recognize that, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\theta^' = \frac{d\theta}{d\xi}</math> </td> <td align="center"> <math>\rightarrow</math> </td> <td align="left"> <math> \frac{dz}{d\xi} \cdot \frac{d\theta}{dz} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{d\theta}{dz}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl(\frac{dz}{d\xi}\biggr)^{-1} \xi \theta^n z </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{d \ln\theta}{d\ln z}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl(\frac{dz}{d\xi}\biggr)^{-1} \xi \theta^{n-1} z^2 </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{dz}{d\xi}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl(\frac{d\ln \theta}{d\ln z}\biggr)^{-1} \xi \theta^{n-1} z^2 </math> </td> </tr> </table> In this case we have, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> LAWE </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ - \biggl(\frac{d\ln \theta}{d\ln z}\biggr)^{-1} \xi \theta^{n-1} z^2 \biggr]^{2} \cdot (2c_0) + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-1} + \alpha \theta^{n-1} z\biggr](a_0 + b_0z + c_0z^2) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ (n-1) + (n - 3)z \biggr\} \theta^{n-1} z \cdot (b_0 + 2c_0z) </math> </td> </tr> </table> <font color="red"><b>Useful ?????</b></font> Try again … <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\biggl[ \theta^{1-n}z^{-1}\biggr] \times</math> LAWE </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\theta^{1-n}z^{-1}\biggl[ \frac{dz}{d\xi} \biggr]^2 \cdot \frac{d^2(a_0 + b_0z + c_0z^2)}{dz^2} + \biggl\{ (n-1) + (n - 3)z \biggr\} \cdot \frac{d(a_0 + b_0z + c_0z^2)}{dz} + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-n} z^{-1} + \alpha \biggr]\cdot (a_0 + b_0z + c_0z^2) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ \xi^{-2}(1+3z)^2 + 2n(1+3z)z^2 \theta^{n-1} + n^2 \xi^2 \theta^{2(n-1)}z^4 \biggr] \cdot (2c_0) + \biggl\{ (n-1) + (n - 3)z \biggr\} \cdot (b_0 + 2c_0z) + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-n} z^{-1} + \alpha \biggr]\cdot (a_0 + b_0z + c_0z^2) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>\biggl[ (2c_0)\xi^{-2}(1+6z + 9z^2) + 4c_0n(z^2+3z^3) \theta^{n-1} + 2 c_0 n^2 \xi^2 \theta^{2(n-1)}z^4 \biggr] + (n-1)(b_0 + 2c_0z) </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + (n - 3)(b_0z + 2c_0z^2) + (n+1)\biggl[\biggl(\frac{\sigma_c^2}{6\gamma_\mathrm{g} } \biggr) \theta^{-n} z^{-1} \biggr]\cdot (a_0 + b_0z + c_0z^2) + (n+1)\alpha (a_0 + b_0z + c_0z^2) \, . </math> </td> </tr> </table> <font color="red"><b>Looks pretty hopeless!</b></font> =Example Density- and Pressure-Profiles= <table border="1" cellpadding="5" align="center" width="90%"> <tr> <th align="center" colspan="6">Properties of Analytically Defined, Spherically Symmetric, Equilibrium Structures Note: <math>x \equiv \frac{r_0}{R}</math> </th> </tr> <tr> <td align="center" width="10%">Model</td> <td align="center"><math>\frac{\rho_0(x)}{\rho_c}</math></td> <td align="center"><math>\frac{P_0(x)}{P_c}</math></td> <td align="center"><math>\frac{R}{P_c} \cdot P_0^'(x)</math></td> <td align="center"> <math> \mu(x) = - \biggl[ \frac{P_0(x)}{P_c} \biggr]^{-1} \cdot \biggl[ \frac{R}{P_c} \cdot P_0^'(x) \biggr]x </math> </td> <td align="center"><math>\frac{P_c}{\rho_c} \cdot \frac{\rho_0(x)}{P_0(x)}</math></td> </tr> <tr> <td align="center">[[SSC/Stability/UniformDensity#The_Stability_of_Uniform-Density_Spheres|Uniform-density]]</td> <td align="center"><math>1</math></td> <td align="center"><math>1 - x^2</math></td> <td align="center"><math>-2x</math></td> <td align="center"><math>\frac{2x^2}{ (1 - x^2)}</math></td> <td align="center"><math>\frac{1}{ (1 - x^2)}</math></td> </tr> <tr> <td align="center">[[SSC/Structure/OtherAnalyticModels#Linear_Density_Distribution|Linear]]</td> <td align="center"><math>1-x</math></td> <td align="center"><math>(1-x)^2(1 + 2x - \tfrac{9}{5}x^2)</math></td> <td align="center"><math>-\tfrac{12}{5}x(1-x)(4-3x)</math></td> <td align="center"><math>\frac{\tfrac{12}{5}x^2(4-3x)}{(1-x)(1 + 2x - \tfrac{9}{5}x^2)}</math></td> <td align="center"><math>\frac{1}{(1-x)(1 + 2x - \tfrac{9}{5}x^2)}</math></td> </tr> <tr> <td align="center">[[SSC/Structure/OtherAnalyticModels#Parabolic_Density_Distribution|Parabolic]]</td> <td align="center"><math>1-x^2</math></td> <td align="center"><math>(1-x^2)^2(1 - \tfrac{1}{2} x^2)</math></td> <td align="center"><math>-x(1-x^2)(5-3x^2)</math></td> <td align="center"><math>\frac{x^2(5-3x^2)}{ (1-x^2)(1 - \tfrac{1}{2} x^2) }</math></td> <td align="center"><math>\frac{1}{(1-x^2)(1 - \tfrac{1}{2} x^2)} </math></td> </tr> <tr> <td align="center">[[SSC/Stability/Polytropes#n_.3D_1_Polytrope|<math>~n=1</math> Polytrope]]</td> <td align="center"><math>\frac{\sin x }{ x}</math></td> <td align="center"><math>\biggl(\frac{\sin x}{x}\biggr)^2</math></td> <td align="center"><math>\frac{2}{x} \biggl[ \cos x - \frac{\sin x}{x} \biggr]\frac{\sin x}{x}</math></td> <td align="center"><math>2 (1 - x \cot x )</math> <td align="center"><math>\frac{x}{\sin x}</math> </tr> </table> ==Parabolic Density Distribution== ===Relevant, Parabolic LAWE=== In the case of a parabolic density distribution, we have found that the equilibrium configuration is defined by the relations: <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>x</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{r_0}{R} \, , </math> </td> </tr> <tr> <td align="right"> <math>\frac{\rho_0}{\rho_c}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2) \, , </math> </td> </tr> <tr> <td align="right"> <math>\frac{P_0}{P_c}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2)^2 (1 - \tfrac{1}{2} x^2) \, ,</math> where, <math>P_c = \biggl(\frac{4\pi}{15}\biggr) G\rho_c^2 R^2 \, ,</math> </td> </tr> <tr> <td align="right"> <math>g_0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl(\frac{4\pi}{3}\biggr) G\rho_c R ~x(1- \tfrac{3}{5}x^2) \, , </math> </td> </tr> </table> in which case, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\mu = \frac{g_0 \rho_0 r_0}{P_0}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl(\frac{4\pi}{3}\biggr) G\rho_c R ~x(1- \tfrac{3}{5}x^2) \rho_c (1-x^2) Rx \biggl[ \biggl(\frac{4\pi}{15}\biggr) G\rho_c^2 R^2 (1-x^2)^2 (1 - \tfrac{1}{2} x^2)\biggr]^{-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> x^2(5 - 3x^2) \biggl[(1-x^2) (1 - \tfrac{1}{2} x^2) \biggr]^{-1} \, , </math> </td> </tr> <tr> <td align="right"> <math>\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl(\frac{\omega^2 R^2}{\gamma_g}\biggr) \rho_c (1-x^2) \biggl[ \biggl(\frac{4\pi}{15}\biggr) G\rho_c^2 R^2 (1-x^2)^2 (1 - \tfrac{1}{2} x^2)\biggr]^{-1} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl(\frac{15 \omega^2 }{4\pi G \rho_c\gamma_g}\biggr) \biggl[ (1-x^2) (1 - \tfrac{1}{2} x^2)\biggr]^{-1} \, . </math> </td> </tr> </table> Hence, in the case of a parabolic density distribution, the [[#Kopal48Expression|Kopal (1948) LAWE]] becomes, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{dx^2} + \frac{(4-\mu)}{x} \cdot \frac{d f}{dx} + \biggl\{ \biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr) - \frac{\alpha \mu}{x^2} \biggr\} f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl\{ 4 - x^2(5 - 3x^2) \biggl[(1-x^2) (1 - \tfrac{1}{2} x^2) \biggr]^{-1} \biggr\}\cdot \frac{d f}{dx} + \biggl\{ \biggl(\frac{15 \omega^2 }{4\pi G \rho_c\gamma_g}\biggr) - \alpha (5 - 3x^2) \biggr\}\biggl[(1-x^2) (1 - \tfrac{1}{2} x^2) \biggr]^{-1} f \, . </math> </td> </tr> </table> Multiplying through by <math>[(1-x^2) (1 - \tfrac{1}{2} x^2)]</math> gives, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2) (1 - \tfrac{1}{2} x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ (1-x^2) (4 - 2 x^2) - x^2(5 - 3x^2) \biggr] \cdot \frac{d f}{dx} + \biggl[ \biggl(\frac{15 \omega^2 }{4\pi G \rho_c\gamma_g}\biggr) - \alpha (5 - 3x^2) \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2) (1 - \tfrac{1}{2} x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 11x^2 + 5 x^4 \biggr] \cdot \frac{d f}{dx} + \biggl[ \biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha x^2 \biggr] \cdot f </math> </td> </tr> </table> where, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\mathfrak{F}</math> </td> <td align="center"> <math>\equiv</math> </td> <td align="left"> <math> 2\biggl[\biggl(\frac{3\omega^2}{4\pi \gamma_\mathrm{g} G \rho_c} \biggr) - \alpha \biggr] \, . </math> </td> </tr> </table> ===Change of Variable=== In an effort to shift this LAWE into a 2<sup>nd</sup>-order ODE that has the form of an hypergeometric equation, let's try … ====First Try==== <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>z = x^2 - 1</math> </td> <td align="center"> <math>\Rightarrow</math> </td> <td align="left"> <math> x = (1+z)^{1 / 2} \, ; </math> also, <math>1 - \tfrac{1}{2} x^2 = \tfrac{1}{2}(1-z) \, ;</math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{dz}{dx}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2x = 2(1+z)^{1 / 2} \, ; </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{df}{dx} = \frac{dz}{dx} \cdot \frac{df}{dz}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2(1+z)^{1 / 2} \cdot \frac{df}{dz} \, ; </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{d^2f}{dx^2} = \frac{dz}{dx} \cdot \frac{d}{dz} \biggl[ 2(1+z)^{1 / 2} \cdot \frac{df}{dz} \biggr] </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2(1+z)^{1 / 2} \biggl[ (1+z)^{-1 / 2} \cdot \frac{df}{dz} + 2(1+z)^{1 / 2} \cdot \frac{d^2 f}{dz^2}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ 2 \cdot \frac{df}{dz} + 4(1+z) \cdot \frac{d^2 f}{dz^2}\biggr] \, . </math> </td> </tr> </table> Hence, the ''parabolic'' LAWE takes the form, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2) (1 - \tfrac{1}{2} x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 11x^2 + 5 x^4 \biggr] \cdot \frac{d f}{dx} + \biggl[ \biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha x^2 \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (-z) (1 - z) \cdot \biggl[ \frac{df}{dz} + 2(1+z) \cdot \frac{d^2 f}{dz^2}\biggr] + (1+z)^{-1 / 2}\biggl[ 4 - 11(1+z) + 5 (1+z)^2 \biggr] 2(1+z)^{1 / 2} \cdot \frac{df}{dz} + \biggl[ \biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha (1+z) \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -z(1 - z) \cdot \biggl[ 2(1+z) \cdot \frac{d^2 f}{dz^2}\biggr] -z(1 - z) \cdot \frac{df}{dz} + \biggl[ 8 - 22(1+z) + 10 (1+z)^2 \biggr] \cdot \frac{df}{dz} + \biggl[ \biggl( \frac{5}{2} \cdot \mathfrak{F} + 3\alpha\biggr) + 3\alpha z \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -2z(1 - z^2) \cdot \frac{d^2 f}{dz^2} + \biggl[ 8 - 22 - 22z + 10 + 20z + 10z^2 + z^2-z \biggr] \cdot \frac{df}{dz} + \biggl[ \biggl( \frac{5}{2} \cdot \mathfrak{F} + 3\alpha\biggr) + 3\alpha z \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -2z(1 - z^2) \cdot \frac{d^2 f}{dz^2} + \biggl[ -4 - 3z + 11z^2 \biggr] \cdot \frac{df}{dz} + \biggl[ \biggl( \frac{5}{2} \cdot \mathfrak{F} + 3\alpha\biggr) + 3\alpha z \biggr] \cdot f </math> </td> </tr> </table> ====Second Try==== <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>z = Fx^B - G</math> </td> <td align="center"> <math>\Rightarrow</math> </td> <td align="left"> <math> x = \biggl( \frac{G + z}{F} \biggr)^{1 / B} \, ; </math> also, <math>1 - \tfrac{1}{2} x^2 = 1 - \frac{1}{2}\biggl( \frac{G + z}{F} \biggr)^{2 / B} \, ;</math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{dz}{dx}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> BFx^{B-1} = BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B} \, ; </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{df}{dx} = \frac{dz}{dx} \cdot \frac{df}{dz}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\frac{df}{dz} \, ; </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{d^2f}{dx^2} = \frac{dz}{dx} \cdot \frac{d}{dz}\biggl[ BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\frac{df}{dz} \biggr]</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\biggl\{ BF \cdot F^{-1 + 1/B} \biggl[ \frac{B-1}{B}\biggr] \biggl( G + z \biggr)^{-1 / B} \cdot \frac{df}{dz} + BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\frac{d^2f}{dz^2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> BF^{1/B}\biggl( G+z \biggr)^{1 - 1/B}\biggl\{ F^{1/B} (B-1) \biggl( G + z \biggr)^{-1 / B} \cdot \frac{df}{dz} + BF^{1/B}\biggl( G+z \biggr)^{1 - 1/B}\frac{d^2f}{dz^2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> BF^{2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[ (B-1) \cdot \frac{df}{dz} + B( G + z ) \cdot \frac{d^2f}{dz^2} \biggr] \, . </math> </td> </tr> </table> This should reduce to the "First Try" example by setting: <math>(B, F, G) = (2, 1, 1)</math>. Let's see … <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\frac{dz}{dx}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B} = 2(1+z)^{1 / 2} \, ; </math> </td> </tr> <tr> <td align="right"> <math>\frac{df}{dx} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> BF\biggl( \frac{G + z}{F} \biggr)^{(B-1) / B}\frac{df}{dz} = 2(1+z)^{1 / 2} \cdot \frac{df}{dz} \, ; </math> </td> </tr> <tr> <td align="right"> <math> \frac{d^2f}{dx^2} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> BF^{2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[ (B-1) \cdot \frac{df}{dz} + B( G + z ) \cdot \frac{d^2f}{dz^2} \biggr] = 2(1+z)^{0}\biggl[\frac{df}{dz} + 2(1+z) \frac{d^2f}{dz^2} \biggr] \, . </math> </td> </tr> </table> <table border="1" width="80%" cellpadding="10" align="center"><tr><td align="left"> <div align="center"><b>Functional Expressions from "First Try"<b></div> <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>z = x^2 - 1</math> </td> <td align="center"> <math>\Rightarrow</math> </td> <td align="left"> <math> x = (1+z)^{1 / 2} \, ; </math> also, <math>1 - \tfrac{1}{2} x^2 = \tfrac{1}{2}(1-z) \, ;</math> </td> </tr> <tr> <td align="right"> <math>\frac{dz}{dx}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2(1+z)^{1 / 2} \, ; </math> </td> </tr> <tr> <td align="right"> <math>\frac{df}{dx}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2(1+z)^{1 / 2} \cdot \frac{df}{dz} \, ; </math> </td> </tr> <tr> <td align="right"> <math>\frac{d^2f}{dx^2} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ 2 \cdot \frac{df}{dz} + 4(1+z) \cdot \frac{d^2 f}{dz^2}\biggr] \, . </math> </td> </tr> </table> </td></tr></table> Hence, the ''parabolic'' LAWE takes the form, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2) (1 - \tfrac{1}{2} x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 11x^2 + 5 x^4 \biggr] \cdot \frac{d f}{dx} + \biggl[ \biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha x^2 \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ 1- \biggl( \frac{G+z}{F} \biggr)^{2/B} \biggr] \biggl[ 1 - \frac{1}{2} \biggl( \frac{G+z}{F} \biggr)^{2/B}\biggr] BF^{2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[ (B-1) \cdot \frac{df}{dz} + B( G + z ) \cdot \frac{d^2f}{dz^2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl( \frac{G+z}{F} \biggr)^{-1/B}\biggl[ 4 - 11x^2 + 5 x^4 \biggr] \cdot BF\biggl( \frac{G + z}{F} \biggr)^{1 - 1/B}\frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl[ \biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( \frac{G+z}{F} \biggr)^{2/B} \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ F^{2/B} - \biggl( G+z \biggr)^{2/B} \biggr] \biggl[ F^{2/B} - \frac{1}{2} \biggl( G+z \biggr)^{2/B}\biggr] BF^{-2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[ B( G + z ) \cdot \frac{d^2f}{dz^2} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> +\biggl[ F^{2/B} - \biggl( G+z \biggr)^{2/B} \biggr] \biggl[ F^{2/B} - \frac{1}{2} \biggl( G+z \biggr)^{2/B}\biggr] BF^{-2/B}\biggl( G+z \biggr)^{1 - 2/B}\biggl[ (B-1) \cdot \frac{df}{dz} \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl[ 4F^{4/B} - 11 F^{2/B}\biggl( G+z \biggr)^{2/B} + 5 \biggl( G+z \biggr)^{4/B} \biggr] \cdot BF^{-2/B}\biggl( G + z \biggr)^{1 - 2/B} \cdot \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl[ \biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha F^{-2/B} \biggl( G+z \biggr)^{2/B} \biggr] \cdot f </math> </td> </tr> </table> Dividing through by, <math>BF^{-2/B}( G+z )^{1 - 2/B}</math>, we have, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl\{ \biggl[ F^{2/B} - \biggl( G+z \biggr)^{2/B} \biggr] \biggl[ F^{2/B} - \frac{1}{2} \biggl( G+z \biggr)^{2/B}\biggr] B( G + z ) \biggr\} \cdot \frac{d^2f}{dz^2} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \biggl\{ \biggl[ F^{2/B} - \biggl( G+z \biggr)^{2/B} \biggr] \biggl[ F^{2/B} - \frac{1}{2} \biggl( G+z \biggr)^{2/B}\biggr] (B-1) + \biggl[ 4F^{4/B} - 11 F^{2/B}\biggl( G+z \biggr)^{2/B} + 5 \biggl( G+z \biggr)^{4/B} \biggr] \biggr\} \cdot \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + B^{-1}( G+z )^{-1 + 2/B}\biggl[ F^{2/B}\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr)^{2/B} \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{2}\biggl[ 2F^{4/B} - 3 F^{2/B} \biggl( G+z \biggr)^{2/B} + \biggl( G+z \biggr)^{4/B}\biggr] B( G + z ) \cdot \frac{d^2f}{dz^2} + B^{-1}( G+z )^{-1 + 2/B}\biggl[ F^{2/B}\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr)^{2/B} \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \frac{1}{2} \biggl\{ \biggl[ 2F^{4/B} - 3 F^{2/B} \biggl( G+z \biggr)^{2/B} + \biggl( G+z \biggr)^{4/B}\biggr] (B-1) + \biggl[ 8F^{4/B} - 22 F^{2/B}\biggl( G+z \biggr)^{2/B} + 10 \biggl( G+z \biggr)^{4/B} \biggr] \biggr\} \cdot \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{2}\biggl[ 2F^{4/B} - 3 F^{2/B} \biggl( G+z \biggr)^{2/B} + \biggl( G+z \biggr)^{4/B}\biggr] B( G + z ) \cdot \frac{d^2f}{dz^2} + B^{-1}( G+z )^{-1 + 2/B}\biggl[ F^{2/B}\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr)^{2/B} \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \frac{1}{2} \biggl[ (2B+6)F^{4/B} - ( 3B + 19) F^{2/B} \biggl( G+z \biggr)^{2/B} + (B+9)\biggl( G+z \biggr)^{4/B}\biggr] \cdot \frac{df}{dz} </math> </td> </tr> </table> Now, this should reduce to the "First Try" example by setting: <math>(B, F, G) = (2, 1, 1)</math>. Let's see … <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{2}\biggl[ 2 - 3 \biggl( 1+z \biggr) + \biggl( 1+z \biggr)^{2}\biggr] 2( 1 + z ) \cdot \frac{d^2f}{dz^2} + \frac{1}{2} \biggl[ 10 - 25 \biggl( 1+z \biggr) + 11\biggl( 1+z \biggr)^{2}\biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ \biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( 1+z \biggr) \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ 2 - 3 -3z + \biggl( 1+2z +z^2\biggr)\biggr] ( 1 + z ) \cdot \frac{d^2f}{dz^2} + \frac{1}{2} \biggl[ 10 - 25 - 25z + 11\biggl( 1+ 2z + z^2 \biggr)\biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ \biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( 1+z \biggr) \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> z(z-1) ( 1 + z )\cdot \frac{d^2f}{dz^2} + \frac{1}{2} \biggl[ -4 - 3z + 11 z^2 \biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ \biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( 1+z \biggr) \biggr] \cdot f \, . </math> </td> </tr> </table> <table border="1" width="80%" cellpadding="10" align="center"><tr><td align="left"> <div align="center"><b>Compare with LAWE from "First Try"<b></div> <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -2z(1 - z^2) \cdot \frac{d^2 f}{dz^2} + \biggl[ -4 - 3z + 11z^2 \biggr] \cdot \frac{df}{dz} + \biggl[ \biggl( \frac{5}{2} \cdot \mathfrak{F} + 3\alpha\biggr) + 3\alpha z \biggr] \cdot f </math> </td> </tr> </table> </td></tr></table> Next, let's try setting <math>B = 2</math> while leaving <math>F</math> and <math>G</math> arbitrary. The LAWE becomes, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{2}\biggl[ 2F^{2} - 3 F \biggl( G+z \biggr) + \biggl( G+z \biggr)^{2}\biggr] 2( G + z ) \cdot \frac{d^2f}{dz^2} + 2^{-1}( G+z )^{0}\biggl[ F\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr) \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> </td> <td align="left"> <math> + \frac{1}{2} \biggl[ 10F^{2} - 25 F \biggl( G+z \biggr) + 11\biggl( G+z \biggr)^{2}\biggr] \cdot \frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ 2F^{2} - 3 F \biggl( G+z \biggr) + \biggl( G+z \biggr)^{2}\biggr] ( G + z ) \cdot \frac{d^2f}{dz^2} + \frac{1}{2} \biggl[ 10F^{2} - 25 F \biggl( G+z \biggr) + 11\biggl( G+z \biggr)^{2}\biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ F\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr) \biggr] \cdot f \, . </math> </td> </tr> </table> ---- Now, let's set <math>G = -1</math> to obtain, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ 2F^{2} - 3 F \biggl( z - 1 \biggr) + \biggl( z - 1 \biggr)^{2}\biggr] ( z - 1 ) \cdot \frac{d^2f}{dz^2} + \frac{1}{2} \biggl[ 10F^{2} - 25 F \biggl( z - 1 \biggr) + 11\biggl( z - 1 \biggr)^{2}\biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ F\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( z - 1 \biggr) \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ 2F^{2} - 3 F z + 3F + z^2 - 2z + 1 \biggr] ( z - 1 ) \cdot \frac{d^2f}{dz^2} + \frac{1}{2} \biggl[ 10F^{2} + 25F - 25 F z + 11 z^2 - 22z + 11 \biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ F\biggl( \frac{5}{2} \biggr) \mathfrak{F} -3\alpha + 3\alpha z \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ (2F^{2} + 3F + 1 ) - (3 F +2) z + z^2 \biggr] ( z - 1 ) \cdot \frac{d^2f}{dz^2} + \frac{1}{2} \biggl[ ( 10F^{2} + 25F + 11) - (25 F + 22) z + 11 z^2 \biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ F\biggl( \frac{5}{2} \biggr) \mathfrak{F} -3\alpha + 3\alpha z \biggr] \cdot f \, . </math> </td> </tr> </table> Notice that if <math>F = -\tfrac{2}{3}</math>, we have, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl[ z^2 - \frac{1}{9} \biggr] ( z - 1 ) \cdot \frac{d^2f}{dz^2} + \frac{1}{2} \biggl[ \frac{40}{9} + \frac{50}{3} + 11 - \biggl(\frac{50}{3} + 22 \biggr) z + 11 z^2 \biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ F\biggl( \frac{5}{2} \biggr) \mathfrak{F} -3\alpha + 3\alpha z \biggr] \cdot f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{9}\biggl[ 9z^2 - 1 \biggr] ( z - 1 ) \cdot \frac{d^2f}{dz^2} + \frac{1}{18} \biggl[ 201 - 348 z + 99 z^2 \biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ F\biggl( \frac{5}{2} \biggr) \mathfrak{F} -3\alpha + 3\alpha z \biggr] \cdot f </math> </td> </tr> </table> ---- Alternatively, let's set, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> z^2 + G\biggl[ 2F^{2} - 3 F \biggl( G+z \biggr) + \biggl( G+z \biggr)^{2}\biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2GF^{2} - 3 GF \biggl( G+z \biggr) + G\biggl( G+z \biggr)^{2} + z^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2G \biggl[ F^{2} - \frac{3F}{G} \biggl( G+z \biggr) + \frac{3^2}{2^2G^2} \biggl( G+z \biggr)^2\biggr] - \frac{3^2}{2G} \biggl( G+z \biggr)^2 + G\biggl( G+z \biggr)^{2} + z^2 </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2G \biggl[ F - \frac{3}{2G} \biggl( G+z \biggr)\biggr]^2 - \frac{3^2}{2G} \biggl( G+z \biggr)^2 + G\biggl( G+z \biggr)^{2} + z^2 </math> </td> </tr> </table> in which case, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> z^2\biggl( \frac{z}{G} - 1 \biggr) \cdot \frac{d^2f}{dz^2} + \frac{1}{2} \biggl[ 10F^{2} - 25 F \biggl( G+z \biggr) + 11\biggl( G+z \biggr)^{2}\biggr] \cdot \frac{df}{dz} + \frac{1}{2}\biggl[ F\biggl( \frac{5}{2} \biggr) \mathfrak{F} + 3\alpha \biggl( G+z \biggr) \biggr] \cdot f \, . </math> </td> </tr> </table> The coefficient of the first-derivative term also may be rewritten ad, ==Uniform Density== In the case of a uniform-density, incompressible configuration, the [[#Kopal48Expression|Kopal (1948) LAWE]] becomes, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{dx^2} + \frac{(4-\mu)}{x} \cdot \frac{d f}{dx} + \biggl[\biggl(\frac{\omega^2\rho_0 R^2}{\gamma_\mathrm{g} P_0} \biggr) - \frac{\alpha \mu}{x^2} \biggr] f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - \frac{2x^2}{(1-x^2)} \biggr] \frac{d f}{dx} + \biggl[\biggl(\frac{\omega^2\rho_c R^2}{\gamma_\mathrm{g} P_c} \biggr) \frac{1}{(1-x^2)} - \biggl(\frac{2\alpha }{1-x^2}\biggr) \biggr] f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 6x^2 \biggr] \frac{d f}{dx} + \biggl[\biggl(\frac{\omega^2\rho_c R^2}{\gamma_\mathrm{g} P_c} \biggr) - 2\alpha \biggr] f \, . </math> </td> </tr> </table> Given that, in the [[SSC/Structure/UniformDensity#Isolated_Uniform-Density_Sphere|equilibrium state]], <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\frac{\rho_c R^2}{P_c}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{6}{4\pi G \rho_c} </math> </td> </tr> </table> we obtain the LAWE derived by {{ Sterne37full }} — see his equation (1.91) on p. 585 — namely, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 6x^2 \biggr] \frac{d f}{dx} + \biggl[6\biggl(\frac{\omega^2}{4\pi \gamma_\mathrm{g} G \rho_c} \biggr) - 2\alpha \biggr] f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 6x^2 \biggr] \frac{d f}{dx} + \mathfrak{F} f \, , </math> </td> </tr> </table> where, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\mathfrak{F}</math> </td> <td align="center"> <math>\equiv</math> </td> <td align="left"> <math> \biggl[6\biggl(\frac{\omega^2}{4\pi \gamma_\mathrm{g} G \rho_c} \biggr) - 2\alpha \biggr] \, . </math> </td> </tr> </table> <table border="1" width="80%" cellpadding="8" align="center"> <tr> <th align="center">Summary for PowerPoint Slide</th> </tr> <tr> <td> <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <b>LAWE:</b> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-x^2) \cdot \frac{d^2 f}{dx^2} + \frac{1}{x}\biggl[ 4 - 6x^2 \biggr] \frac{d f}{dx} + \mathfrak{F} f \, , </math> </td> </tr> </table> where, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\mathfrak{F}</math> </td> <td align="center"> <math>\equiv</math> </td> <td align="left"> <math> \biggl[\biggl(\frac{\omega^2\rho_c R^2}{\gamma_\mathrm{g} P_c} \biggr) - 2\alpha \biggr] \, . </math> </td> </tr> </table> </td> </tr> </table> This also matches, respectively, equations (8) and (9) of {{ Kopal48full }}, aside from what, we presume, is a type-setting error that appears in the numerator of the second term on the RHS of his equation (8): <math>(4 - x^2)</math> appears, whereas it should be <math>(4 - 6x^2)</math>. In order to see if this differential equation is of the same form as the ''hypergeometric expression'', we'll make the substitution, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>z</math> </td> <td align="center"> <math>\equiv</math> </td> <td align="left"> <math> x^2 </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ dz</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2x dx </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{df}{dx}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{dz}{dx} \cdot\frac{df}{dz} = 2x \cdot\frac{df}{dz} = 2z^{1 / 2} \cdot\frac{df}{dz} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \frac{d^2f}{dx^2}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 2z^{1 / 2} \cdot \frac{d}{dz}\biggl[ 2z^{1 / 2} \cdot\frac{df}{dz} \biggr] = 2z^{1 / 2} \biggl[ z^{-1 / 2} \cdot\frac{df}{dz} + 2z^{1 / 2} \cdot\frac{d^2f}{dz^2}\biggr] = \biggl[ 2\cdot\frac{df}{dz} + 4z \cdot\frac{d^2f}{dz^2}\biggr] \, , </math> </td> </tr> </table> in which case the {{ Sterne37 }} LAWE may be rewritten as, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-z) \biggl[ 2\cdot\frac{df}{dz} + 4z \cdot\frac{d^2f}{dz^2}\biggr] + \frac{1}{z^{1 / 2}}\biggl[ 4 - 6z \biggr]2 z^{1 /2} \frac{d f}{dz} + \mathfrak{F} f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> (1-z) \biggl[ 4z \cdot\frac{d^2f}{dz^2}\biggr] + (1-z) \biggl[ 2\cdot\frac{df}{dz} \biggr] + 2\biggl[ 4 - 6z \biggr] \frac{d f}{dz} + \mathfrak{F} f </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 4z(1-z) \cdot\frac{d^2f}{dz^2} + 2\biggl[ 5 - 7z \biggr] \frac{d f}{dz} + \mathfrak{F} f \, . </math> </td> </tr> </table> This is, indeed, of the hypergeometric form if we set <math>(\alpha, \beta; \gamma ; z)</math> <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\gamma</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{5}{2} \, , </math> </td> </tr> <tr> <td align="right"> <math>(\alpha + \beta +1)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{7}{2} \, , </math> </td> </tr> <tr> <td align="right"> <math>\alpha\beta</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -\frac{\mathfrak{F}}{4} \, . </math> </td> </tr> </table> Combining this last pair of expressions gives, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>0</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> -\frac{\mathfrak{F}}{4} - \alpha\biggl[\frac{5}{2}-\alpha \biggr] </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \alpha^2 - \biggl(\frac{5}{2}\biggr)\alpha -\frac{\mathfrak{F}}{4} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~ \alpha</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{1}{2}\biggl\{ \frac{5}{2} \pm \biggl[\biggl(\frac{5}{2}\biggr)^2 - \mathfrak{F} \biggr]^{1 / 2} \biggr\} </math> </td> </tr> <tr> <td align="right"> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{5}{4}\biggl\{ 1 \pm \biggl[1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} \biggr]^{1 / 2} \biggr\} \, ; </math> </td> </tr> </table> and, <table border=0 cellpadding=2 align="center"> <tr> <td align="right"> <math>\beta</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{5}{2} - \frac{5}{4}\biggl\{ 1 \pm \biggl[1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} \biggr]^{1 / 2} \biggr\} \, . </math> </td> </tr> </table> ===Example α = -1=== If we set <math>\alpha = -1</math>, then the eigenvector is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>u_{1} = F\biggl (-1, \frac{7}{2}; \frac{5}{2}; x^2 \biggr)</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 1 - \biggl[\frac{ \beta}{\gamma} \biggr]x^2 = 1 - \biggl(\frac{ 7}{5} \biggr)x^2 \, ; </math> </td> </tr> </table> and the corresponding eigenfrequency is obtained from the expression, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>-1</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \frac{5}{4}\biggl\{ 1 \pm \biggl[1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} \biggr]^{1 / 2} \biggr\} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~-\frac{9}{5}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \pm \biggl[1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} \biggr]^{1 / 2} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~\frac{3^4}{5^2}</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> 1 - \biggl(\frac{4}{25}\biggr)\mathfrak{F} </math> </td> </tr> <tr> <td align="right"> <math>\Rightarrow ~~~\mathfrak{F} </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> \biggl(\frac{5^2}{4}\biggr)\biggl[ 1 - \frac{3^4}{5^2} \biggr] = \frac{1}{4}\biggl[ 5^2 - 3^4 \biggr] = 14 \, . </math> </td> </tr> </table> As we have reviewed in a [[SSC/Stability/UniformDensity#Sterne's_Presentation|separate discussion]], this is identical to the eigenvector identified by {{ Sterne37 }} as mode "<math>j=1</math>". ===More Generally=== More generally, in agreement with {{ Sterne37 }}, for any (positive integer) mode number, <math>0 \le j \le \infty</math>, we find, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>\alpha_j</math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math>- j \, ;</math> <math>\beta_j = \frac{5}{2} + j \, ;</math> <math>\gamma = \frac{5}{2} \, ;</math> <math>\mathfrak{F} = 2j(2j+5) \, . </math> </td> </tr> </table> And, in terms of the hypergeometric function ''series'', the corresponding eigenfunction is, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>u_{j} = </math> </td> <td align="center"> <math>=</math> </td> <td align="left"> <math> F\biggl (\alpha_j, \beta_j; \frac{5}{2}; x^2 \biggr) \, . </math> </td> </tr> </table> =See Also= <ul> <li> Our "Ramblings" chapter titled, [[SSC/Structure/OtherAnalyticModels|Other Analytically Definable, Spherical Equilibrium Structures]] </li> <li> In an article titled, "Radial Oscillations of a Stellar Model," {{ Prasad49full }} investigated the properties of an equilibrium configuration with a prescribed density distribution given by the expression, <div align="center"> <math>\rho(r) = \rho_c\biggl[ 1 - \biggl(\frac{r}{R} \biggr)^2 \biggr] \, ,</math> </div> where, <math>~\rho_c</math> is the central density and, <math>~R</math> is the radius of the star. </li> <li> [[MathProjects/EigenvalueProblemN1|MathProjects/EigenvalueProblemN1]]: In the most general context, the LAWE takes the form, <table border="0" cellpadding="5" align="center"> <tr> <td align="right"> <math>~\biggl[P \biggr]\frac{d^2\mathcal{G}_\sigma}{dx^2} + \biggl[\frac{4P}{x} + P^' \biggr]\frac{d\mathcal{G}_\sigma}{dx} + \biggl[ \sigma^2 \rho + \frac{\alpha P^'}{x} \biggr]\mathcal{G}_\sigma</math> </td> <td align="center"> <math>~=</math> </td> <td align="left"> <math>~0 \, .</math> </td> </tr> </table> <table border="1" cellpadding="5" align="center" width="90%"> <tr> <th align="center" colspan="4">Properties of Analytically Defined Astrophysical Structures</th> </tr> <tr> <td align="center" width="10%">Model</td> <td align="center"><math>~\rho(x)</math> <td align="center"><math>~P(x)</math> <td align="center"><math>~P^'(x)</math> </tr> <tr> <td align="center">[[SSC/Stability/UniformDensity#The_Stability_of_Uniform-Density_Spheres|Uniform-density]]</td> <td align="center"><math>~1</math> <td align="center"><math>~1 - x^2</math> <td align="center"><math>~-2x</math> </tr> <tr> <td align="center">[[SSC/Structure/OtherAnalyticModels#Linear_Density_Distribution|Linear]]</td> <td align="center"><math>~1-x</math> <td align="center"><math>~(1-x)^2(1 + 2x - \tfrac{9}{5}x^2)</math> <td align="center"><math>~-\tfrac{12}{5}x(1-x)(4-3x)</math> </tr> <tr> <td align="center">[[SSC/Structure/OtherAnalyticModels#Parabolic_Density_Distribution|Parabolic]]</td> <td align="center"><math>~1-x^2</math> <td align="center"><math>~(1-x^2)^2(1 - \tfrac{1}{2} x^2)</math> <td align="center"><math>~-x(1-x^2)(5-3x^2)</math> </tr> <tr> <td align="center">[[SSC/Stability/Polytropes#n_.3D_1_Polytrope|<math>~n=1</math> Polytrope]]</td> <td align="center"><math>~\frac{\sin x }{ x}</math> <td align="center"><math>~\biggl[\frac{\sin x}{x}\biggr]^2</math> <td align="center"><math>~\frac{2}{x} \biggl[ \cos x - \frac{\sin x}{x} \biggr] \frac{\sin x}{x}</math> </tr> </table> </li> </ul> {{ SGFfooter }}
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