Solution of a Class of the Riemann-Papperitz Equation with Two Singular Points Milan Batista University of Ljubljana Faculty of Maritime Studies and Transport Portoroz, Slovenia, EU
[email protected]
(Aug 10, 2005)
Abstract
This paper provides the solution of the Riemann-Paperitz equation with singular points at z = ±iˆ . This solution is obtained by mapping the singular points into points 0, ∞ . The solution is then obtained in terms of the Gauss hypergeometric function.
1 Introduction The object is the following equation
(1 + z 2 )
2
d2y dy + 2az (1 + z 2 ) + 4 ( b + cz ) y = 0 2 dz dz
(1)
where y ( z ) is an unknown function, z ∈ ^ is a variable and a, b, c ∈ ^ are constants. (Factors 2 and 4 in (1) are introduced for convenience). The equation is a special case of the Riemann-Papperitz equation ([3],[4]) with only two regular singular points at
z = ±iˆ . The equation of type (1) appears in the dynamics of a disk (having finite thickness) rolling on a rough plane.
1
Note. The solution of the equation of type (1 + z 2 )
2
d2y dy + az (1 + z 2 ) + by = 0 can be 2 dz dz
found in [2] (equation 2.368)
The equation will be solved in the usual way ([3],[4]). First the equation’s singular points will be mapped by bilinear transformations into the points 0, ∞ . Then the transformed equation will be reduced to the Gauss hypergeometric equation ([3],[4]). This equation has the form
t (1 − t )
d 2Y dY + ⎡⎣γ − (α + β + 1) t ⎤⎦ − αβ Y = 0 2 dt dt
(2)
and has solutions including
Y = C1 F (α , β , γ , t ) + C2t1−γ F (α − γ + 1, β − γ + 1, 2 − γ , t )
(3)
where F is the Gauss hypergeometric function with α , β , γ as parameters and t as the argument, and C1 , C2 are arbitrary constants.
2 Solution The function t=
z − iˆ z + iˆ
(4)
where iˆ = −1 , maps singular points z = −iˆ, iˆ to points t = 0, ∞ . Also it maps the interior of the unit circle t ≤ 1 to the upper half space ℑz ≥ 0 . The inverse transformation of (4) is 1+ t z = iˆ 1− t
2
(5)
Using (5), equation (1) is transformed to
t 2 (1 − t )
(
) (
)
d2y dy ˆ t − b + ic ˆ ⎤y=0 + t ⎡⎣ a − ( 2 − a ) t ⎤⎦ + ⎡ b − ic 2 ⎦ dt dt ⎣
(6)
where y ( z (t ) ) = y ( t ) . The solution of (6) is assumed to be of the form ([3])
y (t ) = t λ Y (t )
(7)
where λ is constant and Y ( t ) the function which are both to be determined. Substituting (7) into (6) yields the hypergeometric equation
(
)
d 2Y dY ⎡ 2 ˆ ⎤Y = 0 − λ + (1 − a ) λ − b − ic t (1 − t ) 2 + ⎡⎣( 2λ + a ) − ( 2 + 2λ − a ) t ⎤⎦ ⎦ dt dt ⎣
(8)
where λ is any one of the solutions of the quadratic equation ˆ )=0 λ 2 − (1 − a ) λ − ( b + ic
(9)
The solution Y ( t ) of (8) has the form (3), where, by comparing (2) and (8), the parameters α , β , γ are
γ = 2λ + a
ˆ ) αβ = λ 2 + (1 − a ) λ − ( b − ic
α + β = 1 + 2λ − a
It is seen from (10) that α and β
(10)
are solutions of the quadratic equation
z 2 − (α + β ) z + αβ = 0 . If the solution of (9) is selected to be
λ=
1− a + ∆ , 2
∆≡
3
(1 − a )
2
(
+ 4 b + iˆc
)
(11)
then by (10), it follows that
α = 1− a +
∆ + ∆* , 2
β = 1− a +
∆ − ∆* , 2
(
ˆ ∆ ≡ (1 − a ) + 4 b − ic *
2
γ = 1+ ∆
)
(12)
The solution of (1), using (7), (11) and (12), can now be written in the form
y = C1t λ F (α , β , γ , t ) + C2t1+λ −γ F (α − γ + 1, β − γ + 1, 2 − γ , t )
(13)
or, reverting from t to the variable z, using (4) , of λ
⎛ z − iˆ ⎞ ⎛ z − iˆ ⎞ y = C1 ⎜ F ⎜α , β , γ , ⎟ ⎟ z + iˆ ⎠ ⎝ z + iˆ ⎠ ⎝ 1+ λ −γ
⎛ z − iˆ ⎞ + C2 ⎜ ⎟ ⎝ z + iˆ ⎠
⎛ z − iˆ ⎞ F ⎜ α − γ + 1, β − γ + 1, 2 − γ , ⎟ z + iˆ ⎠ ⎝
(14)
Solution (14) can be put in various forms by using the transforming formulas for F functions ([1][3][4]).
3 Conclusion
A solution of the equation
(1 + z )
2 2
d2y dy + 2az (1 + z 2 ) + 4 ( b + cz ) y = 0 2 dz dz
(15)
is λ
⎛ z − iˆ ⎞ ⎛ z − iˆ ⎞ α β γ , , , y = C1 ⎜ F ⎟ ⎜ ⎟ z + iˆ ⎠ ⎝ z + iˆ ⎠ ⎝ 1+ λ −γ
⎛ z − iˆ ⎞ + C2 ⎜ ⎟ ⎝ z + iˆ ⎠
⎛ z − iˆ ⎞ F ⎜ α − γ + 1, β − γ + 1, 2 − γ , ⎟ z + iˆ ⎠ ⎝
where
4
(16)
λ=
1− a + ∆ , 2
α = 1− a +
and ∆≡
(1 − a )
2
∆ + ∆* , 2
(
ˆ + 4 b + ic
)
β = 1− a +
∆* ≡
(1 − a )
2
∆ − ∆* , 2
(
+ 4 b − iˆc
γ = 1+ ∆
)
(17)
(18)
References
[1] M.Abramowitz, I.A.Stegun. Handbook of Mathematical Functions, With Formulas, Graphs, and Mathematical Tables. Dover Publications (1974) [2] E.Kamke. Differentialgleichungen, Leipzig 1959 [3] N.M.Temme. Special Functions : An Introduction to the Classical Functions of Mathematical Physics, Wiley-Interscience (1996) [4] E. T. Whittaker, G. N. Watson. A Course of Modern Analysis. Cambridge University Press; 4 edition (1978)
5
