Hypergeometric Equation And Hypergeometric Function

Hypergeometric Equation And Hypergeometric Function :

        The linear equation

     x(1-x)y" + [γ-(1+α+β)x]y'-αβy = 0, .....(1)

     Where α,β and γ are constants which can take various real or complex values is called the Hypergeometric equation or the Gauss equation .

    The equation (1) can be put to the form 

  y" + [γ-(1+α+β)x]y'/x(1-x) - αβy/x(1-x) = 0

or      y"+(1+x+x²+...)[γ-(1+α+β)x]y'/x

               -αβ(1+x+x²+...)y/x = 0

or    y" + p(x)y' + q(x)y = 0,

  Since xp(x) and x²q(x) can be expressed as power series , x=0 is a regular singularity . Hence the above equation can be solved by Frobenius method .

 We write
            ∞
      y = Σ cₙ xⁿ⁺ʳ , where c₀≠0,
           n=0

so that  y' = Σcₙ(n+r)xⁿ⁺ʳ⁻¹ and

              y" = Σcₙ(n+r)(n+r-1)xⁿ⁺ʳ⁻²

   Substituting these expressions into the equation(1), we have

  Σcₙ(n+r)(n+r-1+γ)xⁿ⁺ʳ⁻¹

              - Σcₙ [(n+r)(n+r+α+β)-αβ]xⁿ⁺ʳ = 0

or   ΣAₙxⁿ⁺ʳ⁻¹ - ΣBₙxⁿ⁺ʳ = 0 .........(2)

where Aₙ = cₙ(n+r)(n+r-1+γ)

and      Bₙ = cₙ[(n+r)(n+r+α+β)-αβ]
                     ∞
or , A₀xʳ⁻¹+ Σ (Aₙ₊₁-Bₙ)xⁿ⁺ʳ = 0
                   n=0

Equating the coefficient of xʳ⁻¹ to zero, we get

         A₀ = 0

⇒      c₀ r(r-1+γ) = 0
⇒      r=0 or r= 1-γ

 Equating the coefficient of xⁿ⁺ʳ to zero , we obtain the recurrence relation

        Aₙ₊₁ = Bₙ

i.e. cₙ₊₁(n+r+1)(n+r+γ) =

                 cₙ[(n+r)(n+r+α+β)-αβ]

or   cₙ₊₁ = (n+r+α)(n+r+β)cₙ/(n+r+1)(n+r+γ)

                                               n= 0,1,2...  ......(4)

Let us put n=0,1,2 we get

          c₁ = (r+α)(r+β)c₀/(r+1)(r+γ)
         

          c₂ = (r+α+1)(r+β+1)c₁/(r+2)(r+γ+1)

              = (r+α)(r+α+1)(r+β)(r+β+1)c₀ /
                        (r+1)(r+2)(r+γ)(r+γ+1)

and so on .

 Thus

y = c₀xʳ[1+ (r+α)(r+β)x / (r+1)(r+γ) +

                (r+α)(r+α+1)(r+β)(r+β+1)x² / (r+1)                  (r+2)(r+γ)(r+γ+1) + ......].....(5)

If we set c₀ = 1 and r = 0 , we get from (5)

y₁ = 1+ αβx/1.γ + α(α+1)β(β+1)x² /  1.2.γ(γ+1)
                                +........ .......(6)

When r = 1-γ , with c₀ = 1 , we get

y₂ = x^(1-γ)[1+ (α-γ+1)(β-γ+1)x/1.(2-γ)

          + (α-γ+1)(α-γ+2)(β-γ+1)(β-γ+2)x²/

                              1.2.(2-γ)(3-γ) + .......] .....(7)

If we introduce abbreviation

(λ)₀ = 1,(λ)ₙ= λ(λ+1)....(λ+n-1), n=1,2,..... ...(8)

the solutions (6) and (7) can be put to the form
        ∞
y₁ = Σ (α)ₙ(β)ₙxⁿ/n!(γ)ₙ = F(α,β;γ;x), γ≠0,-1,-2...
      n=0

                                                          ........(9)
                           ∞
and y₂= x^(1-γ)Σ (1-γ+α)ₙ(1-γ+β)ₙxⁿ/n!(2-γ)ₙ
                          n=0

                                            , γ≠2,3,4......

           = x^(1-γ)F(1-γ+α,1-γ+β;2-γ;x) ......(10)

The series(9) is known as the hypergeometric series and is absolutely convergent for |x|<1(can be tested by ratio test). We shall refer the function as given in(9) as a hypergeometric function . If γ≠0,1,2,....the two solutions (9) and (10) exist simultaneously and are linearly independent . The general solution in this case is   y = Ay₁ + By₂,

Where A and B are constants .    ......(11)

Note 1. 

          If α= 1, β=γ, y₁ reduces to a geometric series . y₁ being the generalisation of the geometric series is called hypergeometric series.

Note 2.

        It is possible to express many elementary functions as hypergeometric series . As for example :

(i) F(α,β;β;x) = 1+ α.βx/1.β +                                                      α(α+1)β(β+1)x²/1.2.β(β+1)+.....

                      = (1-x)^(-α)

(ii) F(1,1;2;x) = 1+ 1.1x/1.2 +                                                           1.2.1.2.x²/1.2.2.3+........

                      = 1/x [x+x²/2 + x³/3 +.....]

                      = -1/x log(1-x).