Appendix/Mathematics/Hypergeometric

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Hypergeometric Differential Equation

According to §9.151 (p. 1045) of Gradshteyn & Ryzhik (1965), "… a hypergeometric series is one of the solutions of the differential equation,

0

=

z(1−z)d2udz2+[γ−(α+β+1)z]dudz−αβu,

which is called the hypergeometric equation. And, according to §9.10 (p. 1039) of Gradshteyn & Ryzhik (1965), "A hypergeometric series is a series of the form,

F(α,β;γ;z)

=

1+[α⋅βγ⋅1]z+[α(α+1)β(β+1)γ(γ+1)⋅1⋅2]z2+[α(α+1)(α+2)β(β+1)(β+2)γ(γ+1)(γ+2)⋅1⋅2⋅3]z3+…

Among other attributes, Gradshteyn & Ryzhik (1965) note that this, "… series terminates if α or β is equal to a negative integer or to zero."

LAWE

Drawing from an accompanying discussion, we have the,

LAWE:   Linear Adiabatic Wave (or Radial Pulsation) Equation

d2xdr02+[4r0−(g0ρ0P0)]dxdr0+(ρ0γgP0)[ω2+(4−3γg)g0r0]x=0

where,

g0

=

−1ρ0dP0dr0.

Multiplying through by R2, and making the variable substitutions,

x

→

f,

r0R

→

x,

(4−3γg)

→

−αγg,

the LAWE may be rewritten as,

0

=

d2fdx2+[4x−(g0ρ0RP0)]dfdx+(ρ0R2γgP0)[ω2−αγgg0r0]f

 

=

d2fdx2+1x[4−(g0ρ0r0P0)]dfdx+[(ω2ρ0R2γgP0)−αγgg0r0(ρ0R2γgP0)]f

 

=

d2fdx2+1x[4−(g0ρ0r0P0)]dfdx+[(ω2ρ0R2γgP0)−αx2(g0r0ρ0P0)]f.

If we furthermore adopt the variable definition,

μ

≡

(g0ρ0r0P0)=−dln⁡P0dln⁡r0,

we obtain equation (1) of R. Van der Borght (1970), namely,

Borght's LAWE

0

=

d2fdx2+(4−μ)x⋅dfdx+[(ω2ρ0R2γgP0)−αμx2]f.

Example Density- and Pressure-Profiles

Properties of Analytically Defined, Spherically Symmetric, Equilibrium Structures
Model ρ(x) P(x) P'(x) μ(x) ρ(x)P(x)
Uniform-density 1 1−x2 −2x 2x2(1−x2) 1(1−x2)
Linear 1−x (1−x)2(1+2x−95x2) −125x(1−x)(4−3x) 125x2(4−3x)(1−x)(1+2x−95x2) 1(1−x)(1+2x−95x2)
Parabolic 1−x2 (1−x2)2(1−12x2) −x(1−x2)(5−3x2) x2(5−3x2)(1−x2)(1−12x2) 1(1−x2)(1−12x2)
n=1 Polytrope sin⁡xx (sin⁡xx)2 2x[cos⁡x−sin⁡xx]sin⁡xx 2(1−xcot⁡x) xsin⁡x

Uniform Density

In the case of a uniform-density, incompressible configuration, Borght's LAWE becomes,

0

=

d2ξdx2+(4−μ)x⋅dξdx+[(ω2ρ0R2γgP0)−αμx2]ξ

 

=

d2ξdx2+1x[4−2x2(1−x2)]dξdx+[(ω2ρcR2γgPc)1(1−x2)−(2α1−x2)]ξ

 

=

(1−x2)⋅d2ξdx2+1x[4−6x2]dξdx+[(ω2ρcR2γgPc)−2α]ξ.

Given that, in the equilibrium state,

ρcR2Pc

=

64πGρc

See Also


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