MATH 543: FUCHSIAN DIFFERENTIAL EQUATIONS HYPERGEOMETRIC FUNCTION

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1 SET 7 MATH 543: FUCHSIAN DIFFERENTIAL EQUATIONS HYERGEOMETRIC FUNCTION References: DK and Sadri Hassan. Historical Notes: lease read the book Linear Differential Equations and the Group Theory by Jeremy J. Gray, Birkhouser, 000 for the contributions of Euler, faff, Gauss, Riemann, Kummer, Jacobi and others on the symboll and on the Hypergeometric equation and hyoergeometric function. Definition. Linear ordinary differential equations having only regular singular points are called Fuchsian Differential Equations (FDE). FDE with two regular singular points Definition. (Regular singular point at ). If the transformed DE by z = has regular singular point at t = 0 then the original DE has a regular t singular point (in the extended complex plane) at z =. roposition. Second order FDE having only two singular points is equivalent to a DE with constant coefficients, hence solvable in terms of the elementary functions sin z,cos z and polynomial in z. FDE with three regular singular points: The Riemann equation and Riemann Symbol. Riemann has put a second order linear FDE with three regular singular points z = z, z = z and z = z 3 into the form u + ( + α + α + β + β + + γ + ) γ z z z z z z 3 (z z )(z z 3 )ββ z z + (z 3 z )(z 3 z )γ γ z z 3 ) + ( (z z )(z z 3 )α α z z + u (z z )(z z )(z z 3 ) = 0, ()

2 where α, α, β, β, γ, γ are constant satisfying the constraint α + α + β + β + γ + γ = The solution of () is denoted as the Riemann -symbol u(z) = The columns in the symbol indicate the location of the regular singular points and the corresponding roots of the indical equation, i.e.,. z = z, r = α and r = α.. z = z, r = β and r = β. 3. z = z 3, r = γ and r = γ. One can reduce the 9 number of parameters in the Riemann equation (or in the Riemann symbol) into three parameters by using the following type of transformation (i) u(z) = (z z ) r (z z ) s (z z 3 ) t v(z), r + s + t = 0 (ii) The Mobius transformation: z = (z z ) (z 3 z ) (z z ) (z 3 z ) we obtain first (from (i) above) and the using (ii) we obtain () 0 a 0 z ( ) z z α ( ) z z3 γ = z z z z (3) ( ) z z α ( ) z z3 γ = z z z z 0 0 a 0 z (4),

3 where z = (z z )(z 3 z ), a = α + β + γ, (5) (z z )(z 3 z ) b = α + β + γ, c = + α α (6) The symboll in the right hand side of (4) is the hypergeomtric function, F (a, b; c; z ) roblems F (a, b; c; z) = 0 0 a 0 z. rove that the hypergeometric function defined above (7) satisfies the differential equation (hypergeometric DE) (7) z( z) u + [c ( + a + b) z) u ab u = 0 (8). Find (Kummer) transformations leaving the Reiemann equation form invariant. 3. rove that where a n = F = Γ(a + n), b n = Γ(a) 4. rove the following roposition. roposition. n=0 a n b n c n z n n! Γ(b + n), c n = Γ(b) Γ(c + n) Γ(c) z Solutions of the hypergeometric function bout its regular singular points z = 0, z = and z =, provided c, b a, and c a b are not integers, are respectively given by 3

4 u(z) = A F (a, b; c; z) + B z c F (b c +, a c + ; c; z), (9) u(z) = A z a F (a, a c + ; a b + ; z ) + B z b F (b, b c + ; b a + ; ), (0) z u(z) = A 3 F (a, b; a + b + c; z) + B 3 z c a b F (c b, c a; + c a b; z), () where A i, B i, (i =,, 3) are arbitrary constants. 5. Find the solutions (0) and () about the regular singular points z = and z = by the use of Kummer s transformations mentioned in the r.. 6. rove the following: The Jacobi function (α,β) λ satisfying the Jacobi equation equation ( z )u + [β α (α + β + )z]u + λ(λ + α + β + )u = 0 can be written in terms of the hypergeometric function (α,β) λ = Γ(n + α + ) z F ( λ, λ + α + β; α + ; Γ(n + )Γ(α + ) ) When λ = n a non-negative integer the the solution becomes Jacobi polynomials containing the Legendre, and Tchbechev polynomials. 7. rove that when the regular singular points z = z and z 3 of the Riemann equation are pushed out then the resulting function function is the confluent hypergeometric function Φ(a, c; z) in 8. rove that Φ(a, c, z) = lim b F (a, b; c; z b ) 9. rove that the confluent hypergeometric function satisfies the DE z u + (c z) u a u = 0 () 0. rove that the point z = of the confluent hypergeometric equation () is an irregular singular point. 4

5 . Find solutions of the confluent hypergeometric equation about all it regular singular points. The form of the equation is of the form u(z) = A Φ(a, c; z) + B z c Φ(a, c, b z) where A and B are arbitrary constants. Find the constants a, c and b. rove that Bessel s function J v (z) satisfying the Bessel equation u + z u + ( ν z )u = 0 is given by 3. rove that d n dz Use induction. J v (z) = Γ(ν + ) (z/)ν Φ(ν + /, ν + ; iz) n F (a, b;, c; z) = Γ(a + n)γ(b + n)γ(c) Γ(a)Γ(b)Γ(c + n) 4. rove the following: (i) F ( a, b; b; z) = ( + z) a, (ii) F (, ; ; z) = ln( + z), z (iii) F (, ; 3; z ) = z sin z, (iv) F (, ; ; z ) = π (v) e z = lim b F ((, b,, z b ) (vi) The error function Erf(z) = π 0 0 F (a + n, b + n; c + n; z) dθ z sin θ, e t dt 5. rove that the Hermite-Weber differential equation u + (ν + 4 z )u = 0 can be converted to the confluent hypergeometric equation by with u(z) = e z 4 v(ξ), ξ = z v(ξ) = Φ( ν/, /; ξ) 5