Integral Solutions To Heun’s Di Erential EquationVia Some Rational Transformation

A. Anjorin

Abstract-The present work determines the integral form of solutions obtained from the transformation of Heun’s equation tohypergeometric equation by rational substitution. All relevant solutions are provided. AMS Subject Classification: 33CXX Key Words : Hypergeometric functions, Heun’s functions.

γ

t+

δ

t− 1+ εt− d

)du

dt+

αβt− qt(t− 1)(t− d)

u = 0, (1.1)

where {α, β, γ, δ, ε, d, q} (d 6 = 0, 1) are parameters, generally complex and arbitrary, linked by the Fuschian constraint α + β + 1 = γ + δ + ε . This equation has four regular singular points at {0, 1,a,∞}, with the exponents of these singularities being respectively, {0, 1 − γ}, {0, 1 − δ}, {0, 1 − ε}, and {α, β}. The equation (1.1) is called Heun’s equation[3]The hypergeometric equation[3]

z(1− z) d2ydz2 + [c− (a+ b+ 1)z]

dydz− aby = 0, (1.2)

The hypergeometric equation has three regular singular points. Heun’s equation has four singular points. The prob-lem of conversion from Heun’s equation to hypergeometric equation has been teated in the works of K. Kuiken [1]. The purpose of this work is to derive some integrated forms of solutions to the Heun’s equation via some rational transformation as stated in [1]. The steps taken shall be the conversion of Heun’s functions to the hypergeometricfunctions then taken the integration, and through a push and pull back process we arrive back to a new Heun’s functions di erent from the original Heun’s function.Every homogenous linear second order di erential equation with four regular singularities can be transformed into[3]

d2u dt2

+ (

has three regular singular points. In the works of [1], it has been shown that these two equations above can be trans

formed to one another via six rational polynomials z = R(t), where R(t) = t2, 1 − t2, (t − 1)2, 2t − t2, (2t − 1)2, 4t(1

− t). The following parameter relations were deduced[1].For the polynomial R(t) = t2

• α + β = 2(a + b), αβ = 4ab, γ = −1 + 2c, δ = 1 + a + b − c, ε = δ , q = 0 and d = −1 .

For the polynomial R(t) = 1− t2

• α + β = 2(a + b), αβ = 4ab, γ = 1 − 2c + 2a + 2b, δ = c, ε = δ = c, q = 0 and

d = −1.

Department of Mathematics, Lagos State University(LASU) PMB 1087 Apapa Lagos Nigeria.Email and Phone number: anjomaths@yahoo.com, +2348033809538 and +2348025818386

Global Journal of Science Frontier Research Vol.10 Issue7(Ver1.0),November 2010 P a g e | 47

I INTRODUCTION

The hypergeometric equation has three regular singular points. Heun’s equation has four singular points. The problem of conversion from Heun’s equation to hypergeometric equation has been teated in the works of K. Kuiken [1]. The purpose of this work is to derive some integrated forms of solutions to the Heun’s equation via some rational transformation as stated in [1]. The steps takenshall be the conversion of Heun’s functions to the hypergeometric functions then taken the integ-ration, and through a push and pull back process we arrive back to a new Heun’s functions dierent from the original Heun’s function.Every homogenous linear second order di erential equationwith four regular singularities can be transformed into[3

GJSFR Classification - F (FOR) 230107p

For the polynomial R(t) = 2t− t2

• α+ β = 2(a+ b), αβ = 4ab, γ = c, δ = 1− 2c+ 2a+ 2b, ε = δ = c, q = 4ab andd = 2.

For the polynomial R(t) = (2t− 1)2

• α+ β = 2(a+ b), αβ = 4ab, γ = −1 + a+ b− c, δ = γ, ε = δ = −1 + 2c, q = 2aband d = 1/2.

For the polynomial R(t) = 4t(1− t)

• α + β = 2(a + b), αβ = 4ab, γ = c, δ = γ, ε = 1 − 2c + 2a + 2b, q = 2ab andd = 1/2.

Assuming H(d, q;α, β, γ, δ, ε; t) and 2F1(a, b; c; z = R(t)) are representative forms of thesolutions of (1.1) and (1.2) respectively, toghether with the parametres above relationscan de extablished between these two forms via the polynomials data given above. Weprovide an answer to this in this paper. In deed, we provide that the integral of thesolution of GHE can be expressed in terms of another GHE solution.

Main Results

2 Integral solutions

In this section we shall apply the relations above in deriving the integral form of solu-tions via these polynomial transformations. Let I =

∫C be a integral operator defined

over a compact interval C. Since (a)n−1 = (a−1)na−1 , we have I2F1(a, b; c; z = R(t)) =

R∗(t)(c−1)(a−1)(b−1) 2F1(a− 1, b− 1; c− 1; z = R(t)), where R∗(t) is a polynomial factor derivedfrom the integrand and through a push and pull-back processes we have the followingpossible solutions;

1. For polynomial R(t) = t2

(a) Using c = (γ + 1)/2, we obtain

IH(−1, 0;α, β, γ, δ, ε; t) =2(γ − 1)t3

3(α− 2)(β − 2)2F1(β − 2

2,α− 2

2;γ − 1

2;R(t) = t2)|C

=2(γ − 1)t3

3(α− 2)(β − 2)H(−1, 0;α− 2, β − 2, γ − 2,

α+ β − γ − 12

,α+ β − γ − 1

2; t))|C .

(2.1)

(b) Using c = 1− δ + a+ b, we get

IH(−1, 0;α, β, γ, δ, ε; t)

P a g e |48 Vol. 10 Issue7(Ver1.0),November 2010 Global Journal of Science Frontier Research

IV IVII

=4(α+ β − 2δ)t3

3(α− 2)(β − 2)2F1(β − 2

2,α− 2

2;α+ β − 2δ;R(t) = t2)|C

=4(α+ β − 2δ)t3

3(α− 2)(β − 2)

×H(−1, 0;α− 2, β − 2, 2(α+ β − 2δ)− 1,4δ − (α+ β)− 2

2,4δ − (α+ β)− 2

2; t))|C .

(2.2)

2. For polynomial R(t) = 1− t2

(a) Using c = δ, we have

IH(−1, 0;α, β, γ, δ, ε; t)

=4(δ − 1)(3t− t3)3(α− 2)(β − 2) 2F1(

β − 22

,α− 2

2; δ − 1;R(t) = 1− t2)|C

=4(δ − 1)(3t− t3)3(α− 2)(β − 2)

H(−1, 0;α− 2, β − 2, α+ β − 2δ − 1, δ − 1, δ − 1; t))|C .

(2.3)

(b) Using c = ε, we have

IH(−1, 0;α, β, γ, δ, ε; t)

=4(ε− 1)(3t− t3)3(α− 2)(β − 2) 2F1(

β − 22

,α− 2

2; ε− 1;R(t) = 1− t2)|C

=4(ε− 1)(3t− t3)3(α− 2)(β − 2)

H(−1, 0;α− 2, β − 2, α+ β − 2ε− 1, ε− 1, ε− 1; t))|C .

(2.4)

(c) Using c = (1− γ + 2a+ 2b)/2, we arrive at

IH(−1, 0;α, β, γ, δ, ε; t)

=2(α+ β − γ − 1)(3t− t3)

3(α− 2)(β − 2) 2F1(β − 2

2,α− 2

2;α+ β − γ − 1

2;R(t) = 1− t2)|C

=2(α+ β − γ − 1)(3t− t3)

3(α− 2)(β − 2)

×H(−1, 0;α− 2, β − 2, γ − 2,α+ β − γ − 1

2,α+ β − γ − 1

2; t)|C .

(2.5)

3. For polynomial R(t) = 2t− t2

(a) Using c = (δ + 1)/2, we obtain

IH(2, αβ; α, β, γ, δ, ε; t)

=2(δ − 1)t2(3− t2)3(α− 2)(β − 2) 2F1(

β − 22

,α− 2

2;δ − 1

2;R(t) = 2t− t2)|C

=2(δ − 1)t2(3− t2)3(α− 2)(β − 2)

×H(2, (α− 2)(β − 2);α− 2, β − 2,α+ β − δ − 1

2, δ − 2,

α+ β − δ − 12

; t)|C .(2.6)

Global Journal of Science Frontier Research Vol.10 Issue7(Ver1.0),November 2010 P a g e | 49

(b) Using c = 1 + a+ b− γ, we get

IH(2, αβ; β , α, γ, δ, ε; t)=2(α+ β − 2γ)t2(3− t2)

3(α− 2)(β − 2) 2F1(β − 2

2,α− 2

2; γ + 1;R(t) = 2t− t2)|C

=2(α+ β − 2γ)t2(3− t2)

3(α− 2)(β − 2)

×H(2, (α− 2)(β − 2);α− 2, β − 2,γ − 2

2, α+ β − γ − 1,

γ − 22

; t))|C .(2.7)

4. For polynomial R(t) = (t− 1)2

(a) Using c = (1− δ + 2a+ 2b)/2, we get

IH(2, αβ, α, β, γ, δ, ε; t)

=2(α+ β − δ − 1)(t− 1)3

3(α− 2)(β − 2) 2F1(β − 2

2,α− 2

2;α+ β − δ − 1

2;R(t) = (t− 1)2)|C

=2(α+ β − δ − 1)(t− 1)3

3(α− 2)(β − 2)

×H(2, (α− 2)(β − 2);α− 2, β − 2,α+ β − δ − 1

2,α+ β − δ − 1

2,α+ β − δ − 1

2; t))|C .

(2.8)

(b) Using c = γ , we have

IH(2, αβ;α, β, γ, δ, ε; t)

=2(γ − 1)(t− 1)3

3(α− 2)(β − 2) 2F1(β − 2

2,α− 2

2; γ − 1;R(t) = (t− 1)2)|C

=2(γ − 1)

(α− 2)(β − 2)×H(2, (α− 2)(β − 2);α− 2, β − 2, γ − 1, α+ β − 2γ − 1, α+ β − 2γ − 1; t))|C .

(2.9)

(c) By changing γ to ε in (2.9), similar relation can be obtained.

5. For polynomial R(t) = (2t− 1)2

(a) Using c = (ε+ 1)/2 = (δ + 1)/2

IH(1/2, αβ/2;α, β, γ, δ, ε; t)

=2(ε− 1)(2t− 1)3

6(α− 2)(β − 2) 2F1(β − 2

2,α− 2

2;ε− 1

2;R(t) = (2t− 1)2)|C

=2(ε− 1)(2t− 1)3

6(α− 2)(β − 2)

×H(1/2,(α− 2)(β − 2)

2;α− 2, β − 2,

α+ β − ε− 52

,α+ β − ε− 5

2, ε− 2; t)|C .

(2.10)

By changing ε to δ a similar expression can be obtained.

P a g e |50 Vol. 10 Issue7(Ver1.0),November 2010 Global Journal of Science Frontier Research

(b) Using c = −1 + a+ b− γ, we obtain

IH(1/2, αβ/2;α, β, γ, δ, ε; t)

=2(α+ β − 2(γ + 2))(2t− 1)3

6(α− 2)(β − 2) 2F1(β − 2

2,α− 2

2;α+ β − 2γ − 4

2;R(t) = (2t− 1)2)|C

=2(α+ β − 2(γ + 2))(2t− 1)3

6(α− 2)(β − 2)

×H(1/2,(α+ 2)(β + 2)

2;α− 2, β − 2, γ − 1, γ − 1, α+ β − 2γ − 5; t)|C .

(2.11)

6. For polynomial R(t) = 4t(1− t)

(a) Using c = γ, we have

IH(1/2, αβ/2;β, α, γ, δ, ε, ; t)

=4(γ − 1)2t2(3− 2t)

3(α− 2)(β − 2) 2F1(β − 2

2,α− 2

2; γ − 1;R(t) = 4t(1− t))|C

=4(γ − 1)2t2(3− 2t)

3(α− 2)(β − 2)

×H(1/2,(α− 2)(β − 2)

2;α− 2, β − 2, γ − 1, γ − 1, α+ β − 2γ − 1; t)|C .

(2.12)

(b)

IH(1/2, αβ/2;β, α, γ, δ, ε, ; t)

=4(δ − 1)2t2(3− 2t)

3(α− 2)(β − 2) 2F1(β − 2

2,α− 2

2; δ − 1;R(t) = 4t(1− t))|C

=4(δ − 1)2t2(3− 2t)

3(α− 2)(β − 2)

×H(1/2,(α− 2)(β − 2)

2;α− 2, β − 2, δ − 1, δ − 1, α+ β − 2δ − 1; t)|C .

(2.13)

(c) Using c = (1− ε+ 2a+ 2b)/2

IH(1/2, αβ/2;α, β, γ, δ, ε, ; t)

=2(α+ β − ε− 1)2t2(3− 2t)

3(α− 2)(β − 2) 2F1(β − 2

2,α− 2

2;α+ β − ε− 1

2;R(t) = 4t(1− t))|C

=2(α+ β − ε− 1)2t2(3− 2t)

3(α− 2)(β − 2)

×H(1/2,(β − 2)(α− 2)

2;α− 2, β − 2,

α+ β − ε− 12

,α+ β − ε− 1

2, ε− 2; t)|C .

(2.14)

Global Journal of Science Frontier Research Vol.10 Issue7(Ver1.0),November 2010 P a g e | 51

III. CONCLUDING REMARKS AND SUGGESTIONS

In this paper, we have shown that the parameterrelations obtained in the works of K. Kuiken[1] leadsto some integral forms of solutions to the generalHeun’s equation. The multiple choise of close formsolutions arises from the papameter relations. Forexample, consider the quadratic equation arising fromthe relations + = 2(a+b) and = 4ab leads to theparameter choise a = /2 and b = /2 or a = /2 and b =/2. The first leads to all the relations above while thelater repeates all the relations described above bychanging to . This method has being extended in theworks of [2] to the work of Robert Maier [11] pp 15.

IV REFERENCES

1) [1]K. Kuiken, Heun’s equations and theHypergeometric equations, S. I. A. M. J. Math. Anal.10(3), (1979), 655-657. 2) [2]M. N. Hounkonnou, A. Ronveaux and A.Anjorin, Derivative of Heun’s equation from someproperties of hypergeometric ;;,kfunction. Proceedingof International Workshop on geometry and physicsMarseille France (In press, 2007). 3) [3]A. Ronveaux, Heun Di erential equation(Oxford University press, Oxford, 1995). 4) [4]A. O. Smirinov, C.R.M. Proceedings andLecture notes 32, (2002), 287. 5) [5]N. H. Christ and T. D. Lee, Phys. Rev. D12, (1975), 1606; 6) D. P. Jatker, C. N. Kumar, and A. Khare,Phys. Lett. A 142, (1989), 200; A. Khare and B. P. Mandal, Phys. Lett. A 239,(1998), 197. 7) [6]R. K. Bhadari, A. Khare, J. Law, M. V. N. Murthy and D. Sen, J. Phys. A: Math. Gen. 30,(1997), 2557. 8) [7]M. Suzuki, E. Takasugi and H. Umetsu,Prog. Theor. Phys. 100, (1998), 491. 9) [8]K. Takemura, Commun. Math. Phys. 235,(2003), 467; J. Nonlinear Math. Phys. 11, (2004), 21. 10) [9]P. Dorey, J. Suzuki and R. Tateo, J. Phys.A 37, (2004), 2047. 11) [10]A. Ambramowitz and J. A. Stegun,Handbook of mathematical functions (New york,Dover, 1965). 12) [11]R. S. Maier, Heun-to-hypergeometrictransformations, contribution to the conference ofFoundations of Computational Mathematics 02(2002); downloadable fromhttp://www.math.umn.edu/Ȉfocm/c−/Maier.pdf 13) [12]E. S. Cheb-Terrab, New closed formsolutions in terms of pFq for families of the general,Confluent and Bi-Confluent Heun di erentialequations, J. Phys. A: Math Gen. 37, (2004), 9923

14) [12]E. S. Cheb-Terrab, New closed formsolutions in terms of pFq for families of the general,Confluent and Bi-Confluent Heun di erentialequations, J. Phys. A: Math Gen. 37, (2004), 9923. 15) [13]A. Ronveaux, Factorisation of the Heundi erential operator, J. App. Math. Comp. 141,(2003), 177-184. 16) [14]H. Exton, A new solution of thebiconfluent Heun equation, Rendiconti diMathematica Serie VII 18, (1998), 615. 17) [15]A. Decarreau, M. C. I. Dumont-Lepage, P. Maroni, A. Robert, and A. Ronveaux, Formescanoniques des ´equations confluentes de l’quation deHeun, Ann. Soc. Sci. Bruxelles T. 92 I-II , (1978),151-189. 18) [16]A. Ishkhanyan and K-A. Souminen,New solutions of Heun general equation, J. Phys. A:Math. Gen 36, (2003), L81-L85. 19) [17]G. Vallent, International Conference onDi erential Equations, Special Functions andApplications, Munich, (in press) 2005

P a g e |52 Vol. 10 Issue7(Ver1.0),November 2010 Global Journal of Science Frontier Research