SSC/Stability/n1PolytropeLAWE

Radial Oscillations of n = 1 Polytropic Spheres

Groundwork

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

Adiabatic Wave (or Radial Pulsation) Equation

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

whose solution gives eigenfunctions that describe various radial modes of oscillation in spherically symmetric, self-gravitating fluid configurations. Because this widely used form of the radial pulsation equation is not dimensionless but, rather, has units of inverse length-squared, we have found it useful to also recast it in the following dimensionless form:

$\frac{d^{2}x}{d{\chi }_0^2}+[\frac{4}{{\chi }_0}-(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{P_0}{P_c}{)}^{-1}\left(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\right)]\frac{dx}{d{\chi }_0}+(\frac{{\rho }_0}{{\rho }_c}\left)\right(\frac{P_0}{P_c}{)}^{-1}\left(\frac{1}{{\gamma }_{\mathrm{g}}}\right)[{\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}^2{\omega }^2+(4-3{\gamma }_{\mathrm{g}})(\frac{g_0}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}\left)\frac{1}{{\chi }_0}\right]x=0,$

where,

$g_{\mathrm{S}\mathrm{S}\mathrm{C}}\equiv \frac{P_c}{R{\rho }_c},$       and       ${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}\equiv [\frac{R^2{\rho }_c}{P_c}{]}^{1/2}.$

In a separate discussion, we showed that specifically for isolated, polytropic configurations, this linear adiabatic wave equation (LAWE) can be rewritten as,

$0$	$=$	$\frac{d^{2}x}{d{\xi }^2}+\left[\frac{4-(n+1)V(\xi )}{\xi }\right]\frac{dx}{d\xi }+\left[\frac{{\omega }^2}{{\gamma }_g\theta }\right(\frac{n+1}{4\pi G{\rho }_c})-(3-\frac{4}{{\gamma }_g})\cdot \frac{(n+1)V(x)}{{\xi }^2}]x$
	$=$	$\frac{d^{2}x}{d{\xi }^2}+[\frac{4}{\xi }-\frac{(n+1)}{\theta }(-\frac{d\theta }{d\xi }\left)\right]\frac{dx}{d\xi }+\frac{(n+1)}{\theta }[\frac{{\sigma }_c^2}{6{\gamma }_g}-\frac{\alpha }{\xi }(-\frac{d\theta }{d\xi }\left)\right]x,$

where we have adopted the dimensionless frequency notation,

${\sigma }_c^2$

$\equiv$

$\frac{3{\omega }^2}{2\pi G{\rho }_c}.$

Here we focus on an analysis of the specific case of isolated, $n=1$ polytropic configurations, whose unperturbed equilibrium structure can be prescribed in terms of analytic functions. Our hope — as yet unfulfilled — is that we can discover an analytically prescribed eigenvector solution to the governing LAWE.

Search for Analytic Solutions to the LAWE

Setup

From our derived structure of an n = 1 polytrope, in terms of the configuration's radius $R$ and mass $M$, the central pressure and density are, respectively,

$P_c=\frac{\pi G}{8}\left(\frac{M^2}{R^4}\right)$ ,

and

${\rho }_c=\frac{\pi M}{4R^3}$ .

Hence the characteristic time and acceleration are, respectively,

${\tau }_{\mathrm{S}\mathrm{S}\mathrm{C}}=[\frac{R^2{\rho }_c}{P_c}{]}^{1/2}=[\frac{2R^3}{GM}{]}^{1/2}=[\frac{\pi }{2G{\rho }_c}{]}^{1/2},$

and,

$g_{\mathrm{S}\mathrm{S}\mathrm{C}}=\frac{P_c}{R{\rho }_c}=\left(\frac{GM}{2R^2}\right).$

The required functions are,

• Density:

$\frac{{\rho }_0({\chi }_0)}{{\rho }_c}=\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}$ ;

• Pressure:

$\frac{P_0({\chi }_0)}{P_c}=[\frac{\sin (\pi {\chi }_0)}{\pi {\chi }_0}{]}^2$ ;

• Gravitational acceleration:

$\frac{g_0(r_0)}{g_{\mathrm{S}\mathrm{S}\mathrm{C}}}=\frac{2}{{\chi }_0^2}\left[\frac{M_r({\chi }_0)}{M}\right]=\frac{2}{\pi {\chi }_0^2}[\sin (\pi {\chi }_0)-\pi {\chi }_0\cos (\pi {\chi }_0)].$

So our desired Eigenvalues and Eigenvectors will be solutions to the following ODE:

$\frac{d^{2}x}{d{\chi }_0^2}+\frac{2}{{\chi }_0}[1+\pi {\chi }_0\cot (\pi {\chi }_0)]\frac{dx}{d{\chi }_0}+\frac{1}{{\gamma }_{\mathrm{g}}}\left\{\frac{\pi {\chi }_0}{\sin (\pi {\chi }_0)}\right[\frac{\pi {\omega }^2}{2G{\rho }_c}]+\frac{2}{{\chi }_0^2}(4-3{\gamma }_{\mathrm{g}})[1-\pi {\chi }_0\cot (\pi {\chi }_0)\left]\right\}x=0,$

or, replacing ${\chi }_0$ with $\xi \equiv \pi {\chi }_0$ and dividing the entire expression by ${\pi }^2$, we have,

$\frac{d^{2}x}{d{\xi }^2}+\frac{2}{\xi }[1+\xi \cot \xi ]\frac{dx}{d\xi }+\frac{1}{{\gamma }_{\mathrm{g}}}\left\{\frac{\xi }{\sin \xi }\right[\frac{{\omega }^2}{2\pi G{\rho }_c}]+\frac{2}{{\xi }^2}(4-3{\gamma }_{\mathrm{g}})[1-\xi \cot \xi \left]\right\}x=0.$

This is identical to the formulation of the wave equation that is relevant to the (n = 1) core of the composite polytrope studied by J. O. Murphy & R. Fiedler (1985b); for comparison, their expression is displayed, here, in the following boxed-in image.

n = 1 Polytropic Formulation of Wave Equation as Presented by Murphy & Fiedler (1985b)
Murphy & Fiedler (1985b) Murphy & Fiedler (1985b)

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Attempt at Deriving an Analytic Eigenvector Solution

Multiplying the last expression through by ${\xi }^2\sin \xi$ gives,

$({\xi }^2\sin \xi )\frac{d^{2}x}{d{\xi }^2}+2[\xi \sin \xi +{\xi }^2\cos \xi ]\frac{dx}{d\xi }+[{\sigma }^2{\xi }^3-2\alpha (\sin \xi -\xi \cos \xi )]x=0,$

where,

${\sigma }^2$	$\equiv$	$\frac{{\omega }^2}{2\pi G{\rho }_c{\gamma }_g},$
$\alpha$	$\equiv$	$3-\frac{4}{{\gamma }_g}.$

The first two terms can be folded together to give,

$\frac{1}{{\xi }^2{\sin }^2\xi }\cdot \frac{d}{d\xi }\left[{\xi }^2{\sin }^2\xi \frac{dx}{d\xi }\right]$	$=$	$\frac{1}{{\xi }^2\sin \xi }[2\alpha (\sin \xi -\xi \cos \xi )-{\sigma }^2{\xi }^3]x$
	$=$	$-\left[\frac{2\alpha }{{\xi }^2}\right(\frac{\xi \cos \xi }{\sin \xi }-1)+{\sigma }^2(\frac{\xi }{\sin \xi }\left)\right]x$
	$=$	$-[\frac{2\alpha }{{\xi }^2}\frac{{\xi }^2}{\sin \xi }\cdot \frac{d}{d\xi }(\frac{\sin \xi }{\xi })+{\sigma }^2(\frac{\xi }{\sin \xi }\left)\right]x$
	$=$	$-\left[\frac{2\alpha }{\xi }\frac{d}{d\xi }\right(\frac{\sin \xi }{\xi })+{\sigma }^2]\left(\frac{\xi }{\sin \xi }\right)x,$

where, in order to make this next-to-last step, we have recognized that,

$\frac{d}{d\xi }\left(\frac{\sin \xi }{\xi }\right)$

$=$

$\frac{\sin \xi }{{\xi }^2}[\frac{\xi \cos \xi }{\sin \xi }-1].$

It would seem that the eigenfunction, $x(\xi )$, should be expressible in terms of trigonometric functions and powers of $\xi$; indeed, it appears as though the expression governing this eigenfunction would simplify considerably if $x\propto \sin \xi /\xi$. With this in mind, we have made some attempts to guess the exact form of the eigenfunction. Here is one such attempt.

First Guess (n1)

Let's try,

$x=\frac{\sin \xi }{\xi },$

which means,

$x^{\text{'}}\equiv \frac{dx}{d\xi }$

$=$

$\frac{\sin \xi }{{\xi }^2}[\frac{\xi \cos \xi }{\sin \xi }-1].$

Does this satisfy the governing expression? Let's see. The right-and-side (RHS) gives:

RHS

$=$

$-\left[\frac{2\alpha }{\xi }\frac{d}{d\xi }\right(\frac{\sin \xi }{\xi })+{\sigma }^2]\left(\frac{\xi }{\sin \xi }\right)x=-[\frac{2\alpha x^{\text{'}}}{\xi }+{\sigma }^2].$

At the same time, the left-hand-side (LHS) may, quite generically, be written as:

LHS	$=$	$\frac{x^{\text{'}}}{\xi }\{\frac{\xi }{({\xi }^2{\sin }^2\xi )x^{\text{'}}}\cdot \frac{d[({\xi }^2{\sin }^2\xi )x^{\text{'}}]}{d\xi }\}$
	$=$	$\frac{x^{\text{'}}}{\xi }\left[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }\right].$

Putting the two sides together therefore gives,

$\frac{x^{\text{'}}}{\xi }[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }+2\alpha ]$	$=$	$-{\sigma }^2$
$\Rightarrow [\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}{]}^{1/(2\alpha )}}{d\ln \xi }+1]$	$=$	$-\frac{{\sigma }^2}{2\alpha }\left(\frac{\xi }{x^{\text{'}}}\right)$
$\Rightarrow \frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}{]}^{-1/(2\alpha )}}{d\ln \xi }$	$=$	$1+\frac{{\sigma }^2}{2\alpha }\left(\frac{\xi }{x^{\text{'}}}\right).$

[Comment from J. E. Tohline on 6 April 2015: I'm not sure what else to make of this.]

Second Guess (n1)

Adopting the generic rewriting of the LHS, and leaving the RHS fully generic as well, we have,

$\frac{x^{\text{'}}}{\xi }\left[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }\right]$	$=$	$-\left[\frac{2\alpha }{\xi }\frac{d}{d\xi }\right(\frac{\sin \xi }{\xi })+{\sigma }^2]\left(\frac{\xi }{\sin \xi }\right)x$
$\Rightarrow \frac{x^{\text{'}}}{x}\left[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }\right]$	$=$	$-2\alpha \left(\frac{\xi }{\sin \xi }\right)\frac{d}{d\xi }\left(\frac{\sin \xi }{\xi }\right)-{\sigma }^2\left(\frac{{\xi }^2}{\sin \xi }\right)$
$\Rightarrow \frac{d\ln (x)}{d\ln \xi }\left[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }\right]$	$=$	$-2\alpha \left[\frac{d\ln (\sin \xi /\xi )}{d\ln \xi }\right]-{\sigma }^2\left(\frac{{\xi }^3}{\sin \xi }\right).$
$\Rightarrow -{\sigma }^2$	$=$	$\left(\frac{\sin \xi }{{\xi }^3}\right)\left\{\frac{d\ln (x)}{d\ln \xi }\right[\frac{d\ln [({\xi }^2{\sin }^{}\xi )x^{\text{'}}]}{d\ln \xi }]+2\alpha [\frac{d\ln (\sin \xi /\xi )}{d\ln \xi }\left]\right\}.$

[Comment from J. E. Tohline on 6 April 2015: I'm not sure what else to make of this.]

Third Guess (n1)

Let's rewrite the polytropic (n = 1) wave equation as follows:

$\sin \xi [{\xi }^2{x}^{{\text{'}}^{\prime }}+2\xi x^{\text{'}}-2\alpha x]+\cos \xi [2{\xi }^2{x}^{\text{'}}+2\alpha \xi x]+{\sigma }^2{\xi }^{3}x=0.$

It is difficult to determine what term in the adiabatic wave equation will cancel the term involving ${\sigma }^2$ because its leading coefficient is ${\xi }^3$ and no other term contains a power of $\xi$ that is higher than two. After thinking through various trial eigenvector expressions, $x(\xi )$, I have determined that a function of the following form has a chance of working because the second derivative of the function generates a leading factor of ${\xi }^3$ while the function itself does not introduce any additional factors of $\xi$ into the term that contains ${\sigma }^2$:

$x$	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})][A\sin \xi +B\cos \xi ]$
$\Rightarrow x^{\text{'}}$	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+\frac{d[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]}{d\xi }\cdot [A\sin \xi +B\cos \xi ]$
	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+[A\sin \xi +B\cos \xi ]\{5a{\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+b[\frac{5}{2}{\xi }^{3/2}{\cos }^{}({\xi }^{5/2})-\frac{5}{2}{\xi }^{3/2}{\sin }^{}({\xi }^{5/2})]-5c{\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})\}$
	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+[A\sin \xi +B\cos \xi ]\{5(a-c){\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+\frac{5b}{2}{\xi }^{3/2}[1-2{\sin }^{}({\xi }^{5/2})\left]\right\}$
$\Rightarrow x^{{\text{'}}^{\prime }}$	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d^2[A\sin \xi +B\cos \xi ]}{d^2\xi }$
		$+\{5(a-c){\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+\frac{5b}{2}{\xi }^{3/2}[1-2{\sin }^{}({\xi }^{5/2})\left]\right\}\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+[A\sin \xi +B\cos \xi ]\{\frac{15}{2}(a-c){\xi }^{1/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+\frac{25}{2}(a-c){\xi }^3{\cos }^{}({\xi }^{5/2})-\frac{25}{2}(a-c){\xi }^3{\sin }^{}({\xi }^{5/2})$
		$+\frac{15b}{4}{\xi }^{1/2}[1-2{\sin }^2({\xi }^{5/2})]-25b{\xi }^3\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})\}$
	$=$	$[a{\sin }^2({\xi }^{5/2})+b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+c{\cos }^{}({\xi }^{5/2})]\cdot \frac{d^2[A\sin \xi +B\cos \xi ]}{d^2\xi }$
		$+\{5(a-c){\xi }^{3/2}\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+\frac{5b}{2}{\xi }^{3/2}[1-2{\sin }^{}({\xi }^{5/2})\left]\right\}\cdot \frac{d[A\sin \xi +B\cos \xi ]}{d\xi }$
		$+[A\sin \xi +B\cos \xi ]\left\{\frac{15}{4}{\xi }^{1/2}\right[2(a-c)\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+b(1-2{\sin }^{}({\xi }^{5/2}))]$
		$+\frac{25}{2}{\xi }^3[-2b\sin ({\xi }^{5/2})\cos ({\xi }^{5/2})+(a-c)(1-2{\sin }^{}({\xi }^{5/2})\left)\right]\}$

[Comment from J. E. Tohline on 9 April 2015: I'm not sure what else to make of this.]

[Additional comment from J. E. Tohline on 15 April 2015: It is perhaps worth mentioning that there is a similarity between the argument of the trigonometric function being used in this "third guess" and the Lane-Emden function derived by Srivastava for $n=5$ polytropes; and also a similarity between Srivastava's function and the functional form of the LHS that we constructed, above, in connection with our "second guess."]

Fourth Guess (n1)

Again, working with the polytropic (n = 1) wave equation written in the following form,

$\sin \xi [{\xi }^2{x}^{{\text{'}}^{\prime }}+2\xi x^{\text{'}}-2\alpha x]+\cos \xi [2{\xi }^2{x}^{\text{'}}+2\alpha \xi x]+{\sigma }^2{\xi }^{3}x=0.$

Now, let's try:

$x=a_0+b_1\xi \sin \xi +c_2{\xi }^2\cos \xi ,$

which means,

$x^{\text{'}}$	$=$	$b_1\sin \xi +b_1\xi \cos \xi +2c_2\xi \cos \xi -c_2{\xi }^2\sin \xi$
	$=$	$(b_1-c_2{\xi }^2)\sin \xi +(b_1+2c_2)\xi \cos \xi ,$
$x^{{\text{'}}^{\prime }}$	$=$	$(-2c_2\xi )\sin \xi +(b_1-c_2{\xi }^2)\cos \xi +(b_1+2c_2)\cos \xi -(b_1+2c_2)\xi \sin \xi$
	$=$	$-(2c_2+b_1+2c_2)\xi \sin \xi +(2b_1+2c_2-c_2{\xi }^2)\cos \xi .$

The LHS of the wave equation then becomes,

LHS	$=$	$\sin \xi \left\{{\xi }^2\right[-(2c_2+b_1+2c_2)\xi \sin \xi +(2b_1+2c_2-c_2{\xi }^2)\cos \xi ]+2\xi [(b_1-c_2{\xi }^2)\sin \xi +(b_1+2c_2)\xi \cos \xi ]-2\alpha [a_0+(b_1\xi )\sin \xi +(c_2{\xi }^2)\cos \xi \left]\right\}$
		$+\cos \xi \left\{2{\xi }^2\right[(b_1-c_2{\xi }^2)\sin \xi +(b_1+2c_2)\xi \cos \xi ]+2\alpha \xi [a_0+(b_1\xi )\sin \xi +(c_2{\xi }^2)\cos \xi \left]\right\}+{\sigma }^2{\xi }^3[a_0+b_1\xi \sin \xi +c_2{\xi }^2\cos \xi ]$
	$=$	$\sin \xi \left\{\right[-(2c_2+b_1+2c_2){\xi }^3\sin \xi +(2b_1+2c_2-c_2{\xi }^2){\xi }^2\cos \xi ]+[2(b_1-c_2{\xi }^2)\xi \sin \xi +2(b_1+2c_2){\xi }^2\cos \xi ]-2\alpha [a_0+(b_1\xi )\sin \xi +(c_2{\xi }^2)\cos \xi \left]\right\}$
		$+\cos \xi \left\{\right[2(b_1-c_2{\xi }^2){\xi }^2\sin \xi +2(b_1+2c_2){\xi }^3\cos \xi ]+[2a_0\alpha \xi +2b_1\alpha {\xi }^2\sin \xi +2c_2\alpha {\xi }^3\cos \xi \left]\right\}+{\sigma }^2[a_0{\xi }^3+b_1{\xi }^4\sin \xi +c_2{\xi }^5\cos \xi ]$
	$=$	$\sin \xi \{-2\alpha a_0+[-(2c_2+b_1+2c_2){\xi }^3+2(b_1-c_2{\xi }^2)\xi -2\alpha (b_1\xi )]\sin \xi +[(2b_1+2c_2-c_2{\xi }^2){\xi }^2+2(b_1+2c_2){\xi }^2-2\alpha (c_2{\xi }^2)]\cos \xi \}$
		$+\cos \xi \{+2a_0\alpha \xi +[2(b_1-c_2{\xi }^2){\xi }^2+2b_1\alpha {\xi }^2]\sin \xi +[2(b_1+2c_2){\xi }^3+2c_2\alpha {\xi }^3]\cos \xi \}+{\sigma }^2[a_0{\xi }^3+b_1{\xi }^4\sin \xi +c_2{\xi }^5\cos \xi ]$
	$=$	${\sigma }^2{a}_0{\xi }^3+[-(2c_2+b_1+2c_2){\xi }^3+2(b_1-c_2{\xi }^2)\xi -2\alpha (b_1\xi )]{\sin }^2\xi +[2(b_1+2c_2){\xi }^3+2c_2\alpha {\xi }^3](1-{\sin }^2\xi )$
		$+[(2b_1+2c_2-c_2{\xi }^2){\xi }^2+2(b_1+2c_2){\xi }^2-2\alpha (c_2{\xi }^2)]\sin \xi \cos \xi +[2(b_1-c_2{\xi }^2){\xi }^2+2b_1\alpha {\xi }^2]\sin \xi \cos \xi$
		$+{\sigma }^2[b_1{\xi }^4\sin \xi +c_2{\xi }^5\cos \xi ]+2a_0\alpha \xi \cos \xi -2\alpha a_0\sin \xi$
	$=$	$[{\sigma }^2{a}_0+2(b_1+2c_2)+2c_2\alpha ]{\xi }^3+\{+2(b_1)\xi -2\alpha (b_1\xi )+[-2c_2-2(b_1+2c_2)-2c_2\alpha -(2c_2+b_1+2c_2)]{\xi }^3\}{\sin }^2\xi$
		$+\{[(2b_1+2c_2)+2(b_1+2c_2)-2\alpha (c_2)+2(b_1)+2b_1\alpha ]{\xi }^2-3c_2{\xi }^4\}\sin \xi \cos \xi$
		$+\sin \xi [{\sigma }^2{b}_1{\xi }^4-2\alpha a_0]+\xi \cos \xi [{\sigma }^2{c}_2{\xi }^4+2a_0\alpha ]$
	$=$	$[{\sigma }^2{a}_0+2(b_1+2c_2)+2c_2\alpha ]{\xi }^3+\{2b_1(1-\alpha )-[2c_2(5+\alpha )+3b_1]{\xi }^2\}\xi {\sin }^2\xi$
		$+\{2(3-\alpha )(b_1+c_2)-3c_2{\xi }^2\}{\xi }^2\sin \xi \cos \xi +\sin \xi [{\sigma }^2{b}_1{\xi }^4-2\alpha a_0]+\xi \cos \xi [{\sigma }^2{c}_2{\xi }^4+2a_0\alpha ].$

See Also