SSC/Stability/MoreGeneralApproach

More General Approach to the Parabolic Eigenvalue Problem

The material presented in this chapter is an extension of the chapter titled, Other Analytic Models and could also be considered to be a subsection of the associated chapter titled, Other Analytic Ramblings. More specifically, in the following "Introduction," we repeat a manipulation of the LAWE that was originally developed in the subsection of that chapter titled, "Consider Parabolic Case".

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Introduction

In the case of a parabolic density distribution, the LAWE may be written in the form,

$\frac{2}{(1-x^2)(2-x^2)}\left[\right(\alpha +\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }})(5-3x^2)-{\sigma }^2]$

$=$

$\left(\frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}\right)+\frac{4}{x^2}\cdot \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}$

Let's try,

${{𝒢}}_{\sigma }$

$=$

$(a_0+a_2{x}^2{)}^n\cdot (b_0+b_2{x}^2{)}^m,$

which implies,

${{𝒢}}_{\sigma }^{\text{'}}$	$=$	$n(a_0+a_2{x}^2{)}^{n-1}(2a_{2}x)\cdot (b_0+b_2{x}^2{)}^m+m(a_0+a_2{x}^2{)}^n\cdot (b_0+b_2{x}^2{)}^{m-1}(2b_{2}x)$
$\Rightarrow \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}$	$=$	$n(a_0+a_2{x}^2{)}^{-1}(2a_2{x}^2)+m(b_0+b_2{x}^2{)}^{-1}(2b_2{x}^2)$
	$=$	$\frac{2x^2}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}[n{a}_2(b_0+b_2{x}^2)+m{b}_2(a_0+a_2{x}^2)]$
	$=$	$\frac{2x^2}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}[(n{a}_2{b}_0+m{b}_2{a}_0)+(n{a}_2{b}_2+m{b}_2{a}_2)x^2],$

and,

${{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}$	$=$	$nm(a_0+a_2{x}^2{)}^{n-1}(2a_{2}x)\cdot (b_0+b_2{x}^2{)}^{m-1}(2b_{2}x)+n(a_0+a_2{x}^2{)}^{n-1}(2a_2)\cdot (b_0+b_2{x}^2{)}^m+n(n-1)(a_0+a_2{x}^2{)}^{n-2}(2a_{2}x{)}^2\cdot (b_0+b_2{x}^2{)}^m$
		$+mn(a_0+a_2{x}^2{)}^{n-1}(2a_{2}x)\cdot (b_0+b_2{x}^2{)}^{m-1}(2b_{2}x)+m(a_0+a_2{x}^2{)}^n\cdot (b_0+b_2{x}^2{)}^{m-1}(2b_2)+m(m-1)(a_0+a_2{x}^2{)}^n\cdot (b_0+b_2{x}^2{)}^{m-2}(2b_{2}x{)}^2$
$\Rightarrow \frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}$	$=$	$8nm{a}_2{b}_2{x}^2(a_0+a_2{x}^2{)}^{-1}\cdot (b_0+b_2{x}^2{)}^{-1}+n2a_2(a_0+a_2{x}^2{)}^{-1}+n(n-1)4a_2^2{x}^2(a_0+a_2{x}^2{)}^{-2}+m2b_2(b_0+b_2{x}^2{)}^{-1}+m(m-1)4b_2^2{x}^2(b_0+b_2{x}^2{)}^{-2}$
	$=$	$\frac{2n{a}_2(b_0+b_2{x}^2)+2m{b}_2(a_0+a_2{x}^2)}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}+[\frac{4n(n-1)a_2^2}{(a_0+a_2{x}^2{)}^2}+\frac{8nm{a}_2{b}_2}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}+\frac{4m(m-1)b_2^2}{(b_0+b_2{x}^2{)}^2}]x^2$

So, we have for the LAWE:

LHS	$=$	$\frac{2}{(1-x^2)(2-x^2)}\left[\right(\alpha +\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }})(5-3x^2)-{\sigma }^2]$
	$=$	$\frac{2}{(1-x^2)(2-x^2)(a_0+a_2{x}^2)(b_0+b_2{x}^2)}\left\{\right[\alpha (a_0+a_2{x}^2)(b_0+b_2{x}^2)+2x^2(n{a}_2{b}_0+m{b}_2{a}_0)+2x^4(n{a}_2{b}_2+m{b}_2{a}_2)](5-3x^2)$
		$-{\sigma }^2(a_0+a_2{x}^2)(b_0+b_2{x}^2)\};$
RHS	$=$	$\left(\frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}\right)+\frac{4}{x^2}\cdot \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}$
	$=$	$\frac{2n{a}_2(b_0+b_2{x}^2)+2m{b}_2(a_0+a_2{x}^2)}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}+[\frac{4n(n-1)a_2^2}{(a_0+a_2{x}^2{)}^2}+\frac{8nm{a}_2{b}_2}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}+\frac{4m(m-1)b_2^2}{(b_0+b_2{x}^2{)}^2}]x^2$
		$+\frac{8}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}[(n{a}_2{b}_0+m{b}_2{a}_0)+(n{a}_2{b}_2+m{b}_2{a}_2)x^2]$
	$=$	$\frac{1}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}\{2n{a}_2(b_0+b_2{x}^2)+2m{b}_2(a_0+a_2{x}^2)+8(n{a}_2{b}_0+m{b}_2{a}_0)+8(n{a}_2{b}_2+m{b}_2{a}_2)x^2$
		$+[8nm{a}_2{b}_2+\frac{4n(n-1)a_2^2(b_0+b_2{x}^2)}{(a_0+a_2{x}^2)}+\frac{4m(m-1)b_2^2(a_0+a_2{x}^2)}{(b_0+b_2{x}^2)}]x^2\}.$

Putting these together gives,

$0$	$=$	$[\alpha (a_0+a_2{x}^2)(b_0+b_2{x}^2)+2x^2(n{a}_2{b}_0+m{b}_2{a}_0)+2x^4(n{a}_2{b}_2+m{b}_2{a}_2)](5-3x^2)-{\sigma }^2(a_0+a_2{x}^2)(b_0+b_2{x}^2)$
		$-[n{a}_2(b_0+b_2{x}^2)+m{b}_2(a_0+a_2{x}^2)+4(n{a}_2{b}_0+m{b}_2{a}_0)+4(n{a}_2{b}_2+m{b}_2{a}_2)x^2+4nm{a}_2{b}_2{x}^2](1-x^2)(2-x^2)$
		$-\frac{(1-x^2)(2-x^2)}{(a_0+a_2{x}^2)(b_0+b_2{x}^2)}[2n(n-1)a_2^2(b_0+b_2{x}^2{)}^2+2m(m-1)b_2^2(a_0+a_2{x}^2{)}^2]x^2.$

Additional Setup

Benefitting from our earlier exploration of this problem, let's divide through by the product, $(a_0{b}_0)$, and introduce the new variable notations,

$\lambda \equiv \frac{a_2}{a_0},$       and       $\eta \equiv \frac{b_2}{b_0}.$

The LAWE becomes,

$0$	$=$	$[\alpha (1+\lambda x^2)(1+\eta x^2)+2x^2(n\lambda +m\eta )+2x^4(n\lambda \eta +m\eta \lambda )](5-3x^2)-{\sigma }^2(1+\lambda x^2)(1+\eta x^2)$
		$-[n\lambda (1+\eta x^2)+m\eta (1+\lambda x^2)+4(n\lambda +m\eta )+4(n\lambda \eta +m\eta \lambda )x^2+4nm\lambda \eta x^2](1-x^2)(2-x^2)$
		$-\frac{(1-x^2)(2-x^2)}{(1+\lambda x^2)(1+\eta x^2)}[2n(n-1){\lambda }^2(1+\eta x^2{)}^2+2m(m-1){\eta }^2(1+\lambda x^2{)}^2]x^2.$

Multiplying through by the denominator of the last term(s) — that is, multiplying through by $(1+\lambda x^2)(1+\eta x^2)$ — will give us a polynomial with coefficient expressions for 6 terms $(x^0,x^2,x^4,x^6,x^8,x^{10})$ expressed in terms of 5 unknowns $({\sigma }^2,n,m,\lambda ,\eta )$.

Wouldn't a better strategy be to insert yet another quadratic factor — specifically, $(1+\beta x^2{)}^{\ell }$ — which will introduce two additional unknowns but only add one more term into the polynomial expression? This would bring the total number of coefficient expressions to 7 while simultaneously raising the number of unknowns to 7. It will be tedious and messy, but worth the try.

Expanding from Two to Three Quadratic Terms

Here we rearrange terms in the "parabolic" LAWE to construct the governing ODE as,

${\sigma }^2$

$=$

$5(1-\frac{3}{5}{x}^2)[\alpha +\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}]-(1-x^2)(1-\frac{1}{2}{x}^2)\left[\right(\frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }})+\frac{4}{x^2}\cdot \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}]$

Let's try,

${{𝒢}}_{\sigma }$

$=$

$(1+\lambda x^2{)}^n\cdot (1+\eta x^2{)}^m\cdot (1+\beta x^2{)}^{\ell },$

or, in an effort to permit writing more compact expressions,

${{𝒢}}_{\sigma }$

$=$

$N^n\cdot M^m\cdot L^{\ell },$

where,

$N\equiv (1+\lambda x^2);$       $M\equiv (1+\eta x^2);$       and       $L\equiv (1+\beta x^2).$      

This implies (after some whiteboard derivations),

$\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}$	$=$	$\frac{2x^2}{N\cdot M\cdot L}[\ell \beta M\cdot N+m\eta L\cdot N+n\lambda L\cdot M],$
$\frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }}$	$=$	$\frac{4x^2}{N\cdot M\cdot L}[\ell \beta (m\eta N+n\lambda M)+m\eta (\ell \beta N+n\lambda L)+n\lambda (\ell \beta M+m\eta L)]$
		$+\frac{2\ell \beta }{L^2}[1+x^2\beta (2\ell -1)]+\frac{2m\eta }{M^2}[1+x^2\eta (2m-1)]+\frac{2n\lambda }{N^2}[1+x^2\lambda (2n-1)].$

Specific Values of Quadratic Coefficients

Now, if we assume that,

$\lambda =-1;$       $\eta =-\frac{1}{2};$       and       $\beta =-\frac{3}{5}.$      

the "parabolic" LAWE becomes,

$L\cdot {\sigma }^2$

$=$

$5L^2[\alpha +\frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}]-N\cdot M\cdot L\left[\right(\frac{{{𝒢}}_{\sigma }^{{\text{'}}^{\prime }}}{{{𝒢}}_{\sigma }})+\frac{4}{x^2}\cdot \frac{x{{𝒢}}_{\sigma }^{\text{'}}}{{{𝒢}}_{\sigma }}].$

Then, plugging in the expressions for ${{𝒢}}_{\sigma }$ and its derivatives, we have,

$L\cdot {\sigma }^2$	$=$	$5L^2\{\alpha +\frac{2x^2}{N\cdot M\cdot L}[\ell \beta M\cdot N+m\eta L\cdot N+n\lambda L\cdot M\left]\right\}$
		$-\left\{4x^2\right[\ell \beta (m\eta N+n\lambda M)+m\eta (\ell \beta N+n\lambda L)+n\lambda (\ell \beta M+m\eta L)]+8[\ell \beta M\cdot N+m\eta L\cdot N+n\lambda L\cdot M\left]\right\}$
		$-N\cdot M\cdot L\left\{\frac{2\ell \beta }{L^2}\right[1+x^2\beta (2\ell -1)]+\frac{2m\eta }{M^2}[1+x^2\eta (2m-1)]+\frac{2n\lambda }{N^2}[1+x^2\lambda (2n-1)\left]\right\}$
	$=$	$5L^2\{\alpha -\frac{2x^2}{N\cdot M\cdot L}[\frac{3}{5}\ell M\cdot N+\frac{1}{2}mL\cdot N+nL\cdot M\left]\right\}$
		$+\{-4x^2[\frac{3}{5}\ell (\frac{1}{2}mN+nM)+\frac{1}{2}m(\frac{3}{5}\ell N+nL)+n(\frac{3}{5}\ell M+\frac{1}{2}mL)]+8[\frac{3}{5}\ell M\cdot N+\frac{1}{2}mL\cdot N+nL\cdot M\left]\right\}$
		$+\left\{\frac{6\ell N\cdot M}{5L}\right[1-\frac{3}{5}(2\ell -1)x^2]+\frac{mN\cdot L}{M}[1-\frac{1}{2}(2m-1)x^2]+\frac{2nM\cdot L}{N}[1-(2n-1)x^2\left]\right\}$
$\Rightarrow N\cdot M\cdot L^2{\sigma }^2$	$=$	$5L^2\{N\cdot M\cdot L\cdot \alpha -2x^2[\frac{3}{5}\ell M\cdot N+\frac{1}{2}mL\cdot N+nL\cdot M]\}$
		$+N\cdot M\cdot L\{-4x^2[\frac{3}{5}\ell (\frac{1}{2}mN+nM)+\frac{1}{2}m(\frac{3}{5}\ell N+nL)+n(\frac{3}{5}\ell M+\frac{1}{2}mL)]+8[\frac{3}{5}\ell M\cdot N+\frac{1}{2}mL\cdot N+nL\cdot M]\}$
		$+\{\frac{6}{5}\ell N^2\cdot M^2[1-\frac{3}{5}(2\ell -1)x^2]+m{N}^2\cdot L^2[1-\frac{1}{2}(2m-1)x^2]+2n{M}^2\cdot L^2[1-(2n-1)x^2]\}$
$\Rightarrow {\sigma }^2$	$=$	$5L\{\alpha -2x^2[\frac{3}{5}\ell \left(\frac{1}{L}\right)+\frac{1}{2}m\left(\frac{1}{M}\right)+n\left(\frac{1}{N}\right)\left]\right\}$
		$+\frac{1}{L}\{-4x^2[\frac{3}{5}\ell (\frac{1}{2}mN+nM)+\frac{1}{2}m(\frac{3}{5}\ell N+nL)+n(\frac{3}{5}\ell M+\frac{1}{2}mL)]+8[\frac{3}{5}\ell M\cdot N+\frac{1}{2}mL\cdot N+nL\cdot M]\}$
		$+\frac{1}{L}\left\{\frac{6\ell }{5}\right[\frac{N\cdot M}{L}\left]\right[1-\frac{3}{5}(2\ell -1)x^2]+m[\frac{N\cdot L}{M}\left]\right[1-\frac{1}{2}(2m-1)x^2]+2n[\frac{M\cdot L}{N}\left]\right[1-(2n-1)x^2\left]\right\}$
	$=$	$5L\{\alpha +2[(L-1)\ell \left(\frac{1}{L}\right)+(M-1)m\left(\frac{1}{M}\right)+(N-1)n\left(\frac{1}{N}\right)\left]\right\}$
		$+\frac{4}{L}\{\frac{6}{5}\ell M\cdot N+mL\cdot N+2nL\cdot M+(L-1)\ell (\frac{1}{2}mN+nM)+(M-1)m(\frac{3}{5}\ell N+nL)+(N-1)n(\frac{3}{5}\ell M+\frac{1}{2}mL)\}$
		$+\frac{1}{L}\left\{\frac{6\ell }{5}\right[\frac{N\cdot M}{L}\left]\right[1+(L-1)(2\ell -1)]+m[\frac{N\cdot L}{M}\left]\right[1+(M-1)(2m-1)]+2n[\frac{M\cdot L}{N}\left]\right[1+(N-1)(2n-1)\left]\right\}$
	$=$	$5L\{\alpha +2[(1-\frac{1}{L})\ell +(1-\frac{1}{M})m+(1-\frac{1}{N})n\left]\right\}$
		$+\frac{4}{L}\{\frac{6}{5}\ell M\cdot N+mL\cdot N+2nL\cdot M+\ell (\frac{1}{2}mN\cdot L+nM\cdot L)-\ell (\frac{1}{2}mN+nM)+m(\frac{3}{5}\ell N\cdot M+nL\cdot M)-m(\frac{3}{5}\ell N+nL)+n(\frac{3}{5}\ell M\cdot N+\frac{1}{2}mL\cdot N)-n(\frac{3}{5}\ell M+\frac{1}{2}mL)\}$
		$+\frac{1}{L}\left\{\frac{6\ell }{5}\right[\frac{N\cdot M}{L}\left]\right[2(1-\ell )+L(2\ell -1)]+m[\frac{N\cdot L}{M}\left]\right[2(1-m)+M(2m-1)]+2n[\frac{M\cdot L}{N}\left]\right[2(1-n)+N(2n-1)\left]\right\}$
	$=$	$5L\{\alpha +2(\ell +m+n)-2[\left(\frac{\ell }{L}\right)+\left(\frac{m}{M}\right)+\left(\frac{n}{N}\right)\left]\right\}$
		$+\frac{2}{L}\{\frac{6\ell }{5}N\cdot M[2+m+n]+mL\cdot N[2+\ell +n]+2nL\cdot M[2+\ell +m\left]\right\}\}+\frac{1}{L}\{\frac{6\ell }{5}N\cdot M[(2\ell -1)]+mL\cdot N[(2m-1)]+2nL\cdot M[(2n-1)]\}$
		$+\frac{1}{L}\{\frac{6\ell }{5}N\cdot M[\frac{2}{L}(1-\ell )]+mL\cdot N[\frac{2}{M}(1-m)]+2nL\cdot M[\frac{2}{N}(1-n)\left]\right\}-\frac{4}{L}\left\{\right[\ell mN(\frac{1}{2}+\frac{3}{5})+\ell nM(1+\frac{3}{5})+mnL(1+\frac{1}{2})]$
	$=$	$5L\{\alpha +2(\ell +m+n)-2[\left(\frac{\ell }{L}\right)+\left(\frac{m}{M}\right)+\left(\frac{n}{N}\right)\left]\right\}+\frac{(3+2m+2n+2\ell )}{L}[\frac{6\ell }{5}N\cdot M+mL\cdot N+2nL\cdot M]$
		$-\frac{4}{L}\left\{\frac{3}{5}\right[\frac{N\cdot M}{L}]\ell (\ell -1)+\frac{1}{2}[\frac{L\cdot N}{M}]m(m-1)+[\frac{L\cdot M}{N}]n(n-1)\}-\frac{4}{L}\left\{\right[\ell mN(\frac{1}{2}+\frac{3}{5})+\ell nM(1+\frac{3}{5})+mnL(1+\frac{1}{2})\left]\right\}$

See Also