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Solution of Inhomogeneous and Homogeneous Heun’s Differential Equations in Nonstandard Analysis

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12 November 2024

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12 November 2024

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Abstract
Discussions are presented by Morita and Sato on the problem of obtaining the particular solution of an inhomogeneous differential equation with polynomial coefficients in terms of the Green’s function. In succeeding papers, Morita gave discussions of this problem on the basis of nonstandard analysis. It was applied to the hypergeometric, the Hermite, a simple ordinary and a fractional differential equation. In the present paper, this method is applied to the solutions of inhomogeneous and homogeneous Heun’s differential equations.
Keywords: 
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1. Introduction

In a series of papers, Morita and Sato [1,2] and Morita [3,4,5] studied the problem of obtaining the particular solutions of differential equations by using the Green’s function and nonstandard analysis.
In paper [1], this problem is studied in the framework of distribution theory, where the method is applied to Kummer’s and the hypergeometric differential equation. In paper [2], this problem is studied in the framework of nonstandard analysis, where a recipe of solution of the present problem is presented, and it is applied to a simple fractional and a first-order ordinary differential equation. In paper [3], a compact recipe based on nonstandard analysis is obtained by revising the one given in [2], and is applied to Kummer’s differential equation.
In [4], we adopt a recipe without the Green’s function, and is applied to the hypergeometric differential equation, the differential equations treated in [2] and the Hermite differential equation.
In [5], we study the same differential equations as in [4], but the solutions are expressed in terms of the Green’s function.
It is the purpose of the present paper to give solutions of inhomogeneous Heun’s differential equation, by using the method presented in [5].
The presentation in this paper follows those in [1,2,3], in Introduction and in many descriptions in the following sections.
In the present paper, we use Riemann-Liouville fractional integrals and derivatives, whose definition is given in [6,7], and also in [3,4,5]. The property which we use is presened in Section 1.1. The properties which we use in nonstandard analysis, are presented in Section 1.2, following papers [3,4,5], and then contents of the following sections are given in Section 1.3,

1.1. Riemann-Liouville Fractional Integrals and Derivatives

We give here some notations to be used. Z is the set of all integers, R and C are the sets of all real numbers and all complex numbers, respectively, and Z > a = { n Z n > a } , Z < b = { n Z n < b } and Z [ a , b ] = { n Z a n b } for a , b Z satisfying a < b . We also use R > a = { x R x > a } for a R , and C + = { z C Re z > 0 } .
We use the step function H ( t ) for t R , which is equal to 1 if t > 0 , and to 0 if t 0 , and h k , which denotes h k = 1 if k Z > 1 , and h k = 0 if k Z < 0 .
We use the Riemann-Liouville fractional integral and derivative R R D t ρ for ρ C , which is defined in the following remark, that is given in [3,4,5].
Remark 1.
Let g ν ( t ) = 1 Γ ( ν ) t ν 1 H ( t ) for ν C . Then g ν ( t ) = 0 if ν Z < 1 , and if ν Z < 1 ,
R R D t ρ g ν ( t ) = R R D t ρ 1 Γ ( ν ) t ν 1 H ( t ) = 1 Γ ( ν ρ ) t ν ρ 1 H ( t ) = g ν ρ ( t ) .
As a consequence, we have R R D t ν + n g ν ( t ) = g n ( t ) = 0 for n Z > 1 .
In distribution theory [1,8,9,10], we use distribution H ˜ ( t ) , which corresponds to function H ( t ) , differential operator D and distribution δ ( t ) = D H ˜ ( t ) , which is called Dirac’s delta function.

1.2. Preliminaries on Nonstandard Analysis

In nonstandard analysis [11], infinitesimal numbers appear. We denote the set of all infinitesimal real numbers by R 0 . We also use R > 0 0 = { ϵ R 0 ϵ > 0 } , which is such that if ϵ R > 0 0 , there exists N Z > 0 satisfying ϵ < 1 N . We use R n s , which has subsets R and R 0 . If x R n s and x R , x is expressed as x 1 + ϵ by x 1 R and ϵ R 0 , where x 1 may be 0 R . Equation x y for x R n s and y R n s , is used, when x y R 0 . We denote the set of all infinitesimal complex numbers by C 0 , which is the set of complex numbers z which satisfy Re z + Im z R 0 . We use C n s , which has subsets C and C 0 . If z C n s and z C , z is expressed as z 1 + ϵ by z 1 C and ϵ C 0 , where z 1 may be 0 C .
In place of (1), we now use
R R D t ρ g ν + ϵ ( t ) = R R D t ρ 1 Γ ( ν + ϵ ) t ν 1 + ϵ H ( t ) = g ν ρ + ϵ ( t ) = 1 Γ ( ν ρ + ϵ ) t ν ρ 1 + ϵ H ( t ) ,
for all ρ C and ν C , where ϵ R > 0 0 .
Lemma 1.
Let ρ 1 C , ρ 2 C , ν C , ϵ R > 0 0 and g ν + ϵ ( t ) = 1 Γ ( ν + ϵ ) t ν + ϵ 1 H ( t ) . Then the index law:
R R D t ρ 1 R R D t ρ 2 g ν + ϵ ( t ) = R R D t ρ 1 + ρ 2 g ν + ϵ ( t ) = g ν ρ 1 ρ 2 + ϵ ( t ) ,
always holds.
In the present study in nonstandard analysis, in place of H ˜ ( t ) and δ ( t ) in distribution theory, H ϵ ( t ) and δ ϵ ( t ) are used, which are given by
H ϵ ( t ) = R R D t ϵ H ( t ) = g 1 + ϵ ( t ) = 1 Γ ( ϵ + 1 ) t ϵ H ( t ) , δ ϵ ( t ) = d d t H ϵ ( t )
for ϵ R > 0 0 . We note that they tend to H ( t ) and 0, respectively, in the limit of ϵ 0 .
Lemma 2.
In the notation in Remark 1, H ϵ ( t ) = g 1 + ϵ ( t ) , δ ϵ ( t ) = g ϵ ( t ) , and
R R D t ϵ H ϵ ( t ) = R R D t ϵ g 1 + ϵ ( t ) = g 1 ( t ) = H ( t ) , R R D t ϵ δ ϵ ( t ) = R R D t ϵ g ϵ ( t ) = g 0 ( t ) = 0 .

1.3. Contents of the Following Sections

In Section 2, we present Heun’s differential equation. In Section 2.1, transformed differential equations of Heun’s differential equation are presented, which are used to obtain the particular and complementary solutions of Heun’s differential equation, in Section 2.2, Section 3 and Section 4. Section 5 is for Conclusion.

2. Heun’s Differential Equation

Before writing Heun’s differential equation, we present a related differential equation given by
p ( R D t , t ) u ( t ) : = { ( t t 3 ) ( t t 1 ) ( t t 2 ) d 2 d t 2 + [ γ 3 ( t t 1 ) ( t t 2 ) + γ 1 ( t t 2 ) ( t t 3 ) + γ 2 ( t t 3 ) ( t t 1 ) ] d d t + ( α 1 β 1 t α 1 β 1 q 0 ) } u ( t ) + D 0 · R R D t 1 u ( t ) = f ( t ) ,
where t 1 , t 2 , t 3 , γ 1 , γ 2 , γ 3 , α 1 , β 1 , q 0 and D 0 are constants. We express this equation as follows:
p ( R D t , t ) u ( t ) = [ ( A 0 + A 1 t + A 2 t 2 + A 3 t 3 ) d 2 d t 2 + ( B 0 + B 1 t + B 2 t 2 ) d d t + ( C 0 + C 1 t ) ] u ( t ) + D 0 · R R D t 1 u ( t ) = f ( t ) ,
where
A 0 = t 1 t 2 t 3 , A 1 = t 1 t 2 + t 2 t 3 + t 3 t 1 , A 2 = t 1 t 2 t 3 , A 3 = 1 , B 0 = γ 1 t 2 t 3 + γ 2 t 3 t 1 + γ 3 t 1 t 2 , B 1 = γ 1 ( t 2 + t 3 ) γ 2 ( t 3 + t 1 ) γ 3 ( t 1 + t 2 ) , B 2 = γ 1 + γ 2 + γ 3 , C 0 = α 1 β 1 q 0 , C 1 = α 1 β 1 , D 0 = 0 .
Heun’s equation is given by
p H e ( t , R R D t ) u ( t ) : = { t ( t 1 ) ( t t 2 ) d 2 d t 2 + [ γ 3 t 2 [ α 1 + β 1 + 1 γ 1 + ( γ 1 + γ 3 ) t 2 ] t + ( α 1 + β 1 + 1 ) t 2 ] d d t α 1 β 1 q 0 + α 1 β 1 t } u ( t ) = f ( t ) .
This equation is a special one of Equations (6), in which t 1 = 1 , t 3 = 0 , γ 2 = α 1 + β 1 + 1 γ 1 γ 3 and D 0 = 0 . As a consequence, we have the following lemma.
Lemma 3.
Heun’s equation (9) is expressed by the equation which is obtained from Equation (7), by replacing p ( t , R R D t ) by p H e ( t , R R D t ) , and adopting
A 0 = 0 , A 1 = t 2 , A 2 = ( 1 + t 2 ) , A 3 = 1 , B 0 = γ 3 t 2 , B 2 = γ 1 + γ 2 + γ 3 = α 1 + β 1 + 1 , B 1 = [ α 1 + β 1 + 1 γ 1 + ( γ 1 + γ 3 ) t 2 ] = [ γ 1 t 2 + γ 2 + γ 3 ( t 2 + 1 ) ] ,
C 0 = α 1 β 1 q 0 , C 1 = α 1 β 1 , D 0 = 0 ,
in place of Equation (8).

2.1. Transformed Equations of Equation (7)

We now present a transformed equation of Equation (7), which is satisfied by w ˜ ( t ) = R R D t β u ˜ ( t ) = R R D t β R R D t ϵ u ( t ) = R R D t ρ u ( t ) , for β C , ϵ R > 0 o and ρ = β + ϵ , when u ( t ) satifies Equation (7).
Lemma 4.
Let u ( t ) be a solution of Equation (7), and w ˜ ( t ) be given by w ˜ ( t ) = R R D t ρ u ( t ) . Then we see that
p ρ ( R D t , t ) w ˜ ( t ) : = R R D t ρ p ( R D t , t ) R R D t ρ w ˜ ( t ) = { A 0 d 2 d t 2 + [ A 1 t d 2 d t 2 + B ˜ 0 ( ρ ) d d t ] + [ A 2 t 2 d 2 d t 2 + B ˜ 1 ( ρ ) t d d t + C ˜ 0 ( ρ ) ] + [ A 3 t 3 d 2 d t 2 + B ˜ 2 ( ρ ) t 2 d d t + C ˜ 1 ( ρ ) t ] + D ˜ 0 ( ρ ) · R R D t 1 } w ˜ ( t ) = f ˜ β ( t ) : = R R D t ρ f ( t ) ,
where
B ˜ 0 ( ρ ) = B 0 A 1 ρ , B ˜ 2 ( ρ ) = B 2 A 3 · 3 ρ , B ˜ 1 ( ρ ) = B 1 A 2 · 2 ρ , C ˜ 0 ( ρ ) = C 0 B 1 ρ + A 2 · ρ ( ρ + 1 ) + A 3 · 3 ρ ( ρ + 1 ) , C ˜ 1 ( ρ ) = C 1 B 2 · 2 ρ + A 3 · 3 ρ ( ρ + 1 ) , D ˜ 0 ( ρ ) = D 0 C 1 ρ + B 2 · ρ ( ρ + 1 ) A 3 · ρ ( ρ + 1 ) ( ρ + 2 ) .
is a transformed equation of Equation (7). When Equation (8) with Equation (13) is adopted, Equation (12) is a transformed equation of Equation (6).
Proof. 
Remark 9 in [3] shows that when ν C , n Z > 1 , u ˜ ( t ) = t ν + ϵ Γ ( ν + ϵ + 1 ) and u ˜ n ( t ) = d n d t n u ˜ ( t ) , we have
R R D t ρ [ t u ˜ n ( t ) ] = t · R R D t ρ u ˜ n ( t ) ρ · R R D t ρ 1 u ˜ n ( t ) , R R D t ρ [ t 2 u ˜ n ( t ) ] = t · R R D t ρ [ t u ˜ n ( t ) ] ρ · R R D t ρ 1 [ t u ˜ n ( t ) ] = t 2 · R R D t ρ u ˜ n ( t ) 2 ρ t · R R D t ρ 1 u ˜ n ( t ) + ρ ( ρ + 1 ) · R R D t ρ 2 u ˜ n ( t ) , R R D t ρ [ t 3 u ˜ n ( t ) ] = t 2 · R R D t ρ [ t u ˜ n ( t ) ] 2 ρ t · R R D t ρ 1 [ t u ˜ n ( t ) ] + ρ ( ρ + 1 ) · R R D t ρ 2 [ t u ˜ n ( t ) ] = t 3 · R R D t ρ u ˜ n ( t ) 3 ρ t 2 · R R D t ρ 1 u ˜ n ( t ) + 3 ρ ( ρ + 1 ) t · R R D t ρ 2 u ˜ n ( t ) ρ ( ρ + 1 ) ( ρ + 2 ) · R R D t ρ 3 u ˜ n ( t ) .
By using these relations in Equation (7), we obtain the following equation:
p ρ ( R D t , t ) w ˜ ( t ) : = R R D t ρ p ( R D t , t ) u ( t ) = { ( A 0 + A 1 t + A 2 t 2 + A 3 t 3 ) d 2 d t 2 + ( B 0 + B 1 t + B 2 t 2 A 1 ρ A 2 · 2 ρ t A 3 · 3 ρ t 2 ) d d t + C 0 + C 1 t B 1 ρ B 2 · 2 ρ t + A 2 · ρ ( ρ + 1 ) + A 3 · 3 ρ ( ρ + 1 ) t + [ D 0 C 1 ρ + B 2 · ρ ( ρ + 1 ) A 3 · ρ ( ρ + 1 ) ( ρ + 2 ) ] R R D t 1 } w ˜ ( t ) = f ˜ β ( t ) : = R R D t ρ f ( t ) .
By using this equation, we obtain Equation (12). □
As a corollary of this lemma, we have the following lemma.
Lemma 5.
Let u ( t ) be a solution of Equation (7), and u ˜ ( t ) be given by u ˜ ( t ) = R R D t ϵ u ( t ) . Then we have p ϵ ( R D t , t ) u ˜ ( t ) which is obtained from Equation (12), by replacing ρ by ϵ and w ˜ ( t ) by u ˜ ( t ) , and the following equation:
p ϵ ( R D t , t ) u ˜ ( t ) : = R R D t ϵ p ( R D t , t ) R R D t ϵ u ˜ ( t ) = f ˜ ( t ) : = R R D t ϵ f ( t ) ,
which is the transformed equation of Equation (7), when Equations (8) and (13) are adopted.

2.1.1. Transformed Equations of Heun’s Differential Equation

We denote the transformed equations of Heun’s equation (9), which correspond to Equations (12) and (16), by Equations (12-He) and (16-He), respectively.
Lemma 6.
Lemmas 4 and 3 show that Equation (12-He) is obtained from Equation (12) by replacing p ρ ( R D t , t ) by p ρ , H e ( R D t , t ) , and p ( R D t , t ) by p H e ( R D t , t ) , and using Equations (10) and (11) in place of Equations (8). In this replacement, Equation (13) is replaced by
B ˜ 0 ( ρ ) = ( γ 3 ρ ) t 2 , , B ˜ 2 ( ρ ) = γ 1 + γ 2 + γ 3 3 ρ = α 1 + β 1 + 1 3 ρ , B ˜ 1 ( ρ ) = [ α 1 + β 1 + 1 γ 1 + ( γ 1 + γ 3 ) t 2 ] + ( 1 + t 2 ) · 2 ρ = B 1 + ( 1 + t 2 ) · 2 ρ , C ˜ 0 ( ρ ) = α 1 β 1 q 0 B 1 ρ + ( 2 t 2 ) · ρ ( ρ + 1 ) , C ˜ 1 ( ρ ) = α 1 β 1 ( α 1 + β 1 + 1 ) · 2 ρ + 3 ρ ( ρ + 1 ) = ( α 1 2 ρ ) ( β 1 2 ρ ) ρ 2 + ρ , D ˜ 0 ( ρ ) = α 1 β 1 ρ + ( α 1 + β 1 + 1 ) · ρ ( ρ + 1 ) ρ ( ρ + 1 ) ( ρ + 2 ) = ( α 1 ρ 1 ) ( β 1 ρ 1 ) ρ .
Lemmas 4 and 5 show that Equation (16-He) is obtained from Equation (12-He) by replacing ρ by ϵ, and w ˜ ( t ) by u ˜ ( t ) .
Lemma 7.
Lemma 6 shows that when we put ρ = 0 and replace w ˜ ( t ) by u ( t ) , Equation (12-He) is Heun’s equation (9).

2.1.2. Summary of Section 2.2, Section 3 and Section 4

Lemma 6 shows that Equations (12-He) and (16-He) are transformed equations of Heun’s equation (9).
In Section 2.2 and Section 3, we solve them. We first obtain the solution w ˜ ( t ) of Equation (12-He) for the inhomogeneous term f ˜ β ( t ) = δ ϵ ( t ) = g ϵ ( t ) , and then we obtain u ˜ ( t ) and u ( t ) , given by u ˜ ( t ) = R R D t β w ˜ ( t ) and u ( t ) = R R D t β + ϵ w ˜ ( t ) . They are the solutions of Equation (16-He) for the inhomogeneous term given by f ˜ ( t ) = R R D t β δ ϵ ( t ) = g ϵ β ( t ) and, Heun’s differential equation (9) for f ( t ) = R R D t β δ ϵ ( t ) = g β ( t ) for β Z > 1 , respectively. When β = 0 , f ( t ) = 0 and hence the solution of Heun’s equation is a complementary solution, which is studied in Section 4, and we do not consider the case of β = n Z > 0 , for which f ( t ) = g n ( t ) = 0 .

2.2. Solutions of Heun’s Differential Equation

We now use w ˜ ( t ) and f ˜ β ( t ) expressed by
w ˜ ( t ) = k = 0 p k 1 Γ ( α + k + 1 ) t α + k H ( t ) = k = 0 p k g α + k + 1 ( t ) ,
f ˜ β ( t ) = k = 0 c k 1 Γ ( ϵ + k ) t ϵ + k 1 H ( t ) ,
where p 0 0 , α = ν + ϵ or α = ν , ν C Z < 0 and ϵ R > 0 0 . We then prepare the following equations:
d d t w ˜ ( t ) = k = 0 p k g α + k ( t ) , t d 2 d t 2 w ˜ ( t ) = k = 0 p k ( α + k 1 ) g α + k ( t ) , w ˜ ( t ) = k = 1 p k 1 · g α + k ( t ) , t n d n d t n w ˜ ( t ) = k = 1 p k 1 ( α + k n ) n · g α + k ( t ) , n Z [ 1 , 2 ] , t n d n 1 d t n 1 w ˜ ( t ) = k = 2 p k 2 ( α + k n ) n · g α + k ( t ) , n Z [ 1 , 3 ] ; R R D t 1 w ˜ ( t ) = k = 2 p k 2 · g α + k ( t ) . R
By using Equation (20), f ˜ β ( t ) given by Equation (19), and Equations (10) and (17), Equation (12-He) is expressed as follows:
p ρ , H e ( R D t , t ) w ˜ ( t ) : = R R D t ρ p H e ( R D t , t ) R R D t ρ w ˜ ( t ) = k = 0 { p k [ A 1 ( α + k 1 ) + B ˜ 0 ( ρ ) ] h k 1 p k 1 Q k ( α , ρ ) + h k 2 p k 2 R k ( α , ρ ) } 1 Γ ( α + k ) t α + k 1 H ( t ) = f ˜ β ( t ) ,
where h k l = 1 if k l Z > 1 , h k l = 0 if k l Z < 0 , and
A 1 ( α + k 1 ) + B ˜ 0 ( ρ ) = t 2 ( α + k 1 + γ 3 ρ ) ,
Q k ( α , ρ ) = [ A 2 ( α + k 2 ) + B ˜ 1 ( ρ ) ] ( α + k 1 ) C ˜ 0 ( ρ ) , k Z > 0 ,
R k ( α , ρ ) = [ [ A 3 ( α + k 3 ) + B ˜ 2 ( ρ ) ] ( α + k 2 ) + C ˜ 1 ( ρ ) ] ( α + k 1 ) + D ˜ 0 ( ρ ) , k Z > 1 .
Lemma 8.
Let f ˜ β ( t ) be given by Equation (19), A 1 ( α + k 1 ) + B ˜ 0 ( ρ ) , Q k ( α , ρ ) and R k ( α , ρ ) be given by Equations (22), (23) and (24), and p k and α be so determined that
p 0 t 2 ( α 1 + γ 3 ρ ) 1 Γ ( α ) t α 1 H ( t ) = c 0 1 Γ ( ϵ ) t ϵ 1 H ( t ) , p 1 t 2 ( α + γ 3 ρ ) p 0 Q 1 ( α , ρ ) ] 1 Γ ( α + 1 ) t α H ( t ) = c 1 1 Γ ( ϵ + 1 ) t ϵ H ( t ) , p k t 2 ( α + k 1 + γ 3 ρ ) p k 1 Q k ( α , ρ ) + p k 2 R k ( α , ρ ) ] 1 Γ ( α + k ) t α + k 1 H ( t )
= c k 1 Γ ( ϵ + k ) t ϵ + k 1 H ( t ) , k Z > 1 .
Then w ˜ ( t ) given by Equation (18) is a solution of Equation (21).
Lemma 9.
When c 0 = 1 . Equation (25) is satisfied by α = ϵ and p 0 = 1 t 2 ( ϵ 1 + γ 3 ρ ) .
Lemma 10.
When c 0 = 0 or ϵ = 0 , the righthand side of Equation (25) is 0. In this case, Equation (25) is satisfied by α = 0 or α = 1 γ 3 + ρ , and by any value of p 0 .
Lemma 11.
When c k = 0 for k Z > 0 , we use p ˜ k in place of p k p 0 for k Z > 1 , Then the equations in Equation (26) are expressed as p ˜ 0 = 1 and
p ˜ k = 1 t 2 ( k 1 + α + γ 3 ρ ) [ p ˜ k 1 Q k ( α , ρ ) h k 2 p ˜ k 2 R k ( α , ρ ) ] , k Z > 0 .
We also use the coefficients P k in place of p ˜ k . They are defined by p ˜ 0 = P 0 = 1 and
p ˜ k = 1 t 2 k ( α + γ 3 ρ ) k P k , i . e . P k = t 2 k ( α + γ 3 ρ ) k p ˜ k , k Z > 1 .
Now in place of Equation (27), we have P 0 = 1 and
P k = P k 1 Q k ( α , ρ ) t 2 ( k 2 + α + γ 3 ρ ) h k 2 P k 2 R k ( α , ρ ) , k Z > 0 .

3. Particular Solutions

In the present section, we consider the solution w ˜ ( t ) of Equation (21) in the form of Equation (18), assuming that f ˜ β ( t ) = δ ϵ ( t ) , c 0 = 1 , c k = 0 for k Z > 0 and α = ϵ , in Equations (25) and (26).
Theorem 1. (i) In the above condition, Lemmas 9 and 11 show that the coefficints p 0 and p ˜ k = p k p k for k Z > 1 are given by
p 0 = 1 t 2 ( 1 + γ 3 β ) , p ˜ 0 = 1 ,
p ˜ k = 1 t 2 ( k 1 + γ 3 β ) [ p ˜ k 1 Q k ( ϵ , ρ ) h k 2 p ˜ k 2 R k ( ϵ , ρ ) ] , k Z > 0 ,
and the solution of Equation (21) is expressed by
w ˜ ( t ) = k = 0 p k 1 Γ ( ϵ + k + 1 ) t ϵ + k H ( t ) = p 0 k = 0 p ˜ k 1 Γ ( ϵ + k + 1 ) t ϵ + k H ( t ) .
Section 2.1.2 shows that by using Equation (32), we obtain the solution u ˜ ( t ) = R R D t β w ˜ ( t ) of Equation (16-He) for f ˜ ( t ) = R R D t β δ ϵ ( t ) and β Z > 0 , as follows:
u ˜ ( t ) = p 0 k = 0 p ˜ k 1 Γ ( ϵ + k + 1 β ) t ϵ + k β H ( t ) = p 0 1 Γ ( ϵ + 1 β ) k = 0 p ˜ k 1 ( ϵ + 1 β ) k t ϵ + k β H ( t ) ,
and u ( t ) = R R D t ϵ u ˜ ( t ) is obtained from Equation (33).
(ii) We note that if we replace Q k ( ϵ , ρ ) and R k ( ϵ , ρ ) in Equation (31) by Q k ( 0 , β ) and R k ( 0 , β ) , respectively, so that p ˜ 0 = 1 , and
p ˜ k 1 t 2 ( k 1 + γ 3 β ) [ p ˜ k 1 Q k ( 0 , β ) h k 2 p ˜ k 2 R k ( 0 , β ) ] , k Z > 0 ,
w ˜ ( t ) and u ˜ ( t ) given by Equations (32) and (33), respectively, are deviated by a contribution of O ( ϵ ) , which can be neglected, and hence we can adopt it.
By Equations (23), (24) and (17), Q k ( 0 , β ) and R k ( 0 , β ) are given by
Q 1 ( 0 , β ) = α 1 β 1 q 0 + B 1 β + ( 2 t 2 ) · β ( β + 1 ) , Q k ( 0 , β ) = [ ( 1 + t 2 ) ( k 2 2 β ) B 1 ] ( k 1 ) + α 1 β 1 q 0 + B 1 β + ( 2 t 2 ) · β ( β + 1 ) , R k ( 0 , β ) = [ ( k 2 + α 1 3 2 β ) ( k 2 + β 1 3 2 β ) ( α 1 + β 1 ) 1 2 β + 3 4 β 2 + β ] ( k 1 ) [ ( α 1 β 1 ) ( β 1 β 1 ) β 1 ] β , k Z > 1 .
Remark 2.
Following Lemma 2.4 in [5], we denote the solution expressed by w ˜ ( t ) by G H e u n , β , ϵ ( t , 0 ) . Remark 10 shows that when we put ϵ = 0 in this solution, the obtained G H e u n , β , 0 ( t , 0 ) = R R D t ϵ G H e u n , β , ϵ ( t , 0 ) is a complementary solution of Equation (21) for ϵ = 0 .
The solutions u ˜ ( t ) and u ( t ) are expressed by R R D t β G H e u n , β , ϵ ( t , 0 ) and R R D t β + ϵ G H e u n , β , ϵ ( t , 0 ) , respectively. When β = 0 , these solutions are expressed by G H e u n , ϵ ( t , 0 ) and G H e u n , 0 ( t , 0 ) = R R D t ϵ G H e u n , ϵ ( t , 0 ) , respectively.
Corollary 1. (i) When β = 0 , u ˜ ( t ) given by Equation (33), in which β = 0 , is a particular solution of Equation (16-He) for f ˜ ( t ) = δ ϵ ( t ) . In this case, in place of Equation (31), we have the equations which are obtained those in it by replacing β by 0 and ρ by ϵ.
(ii) Followig Theorem 1(ii), we may use Q k ( 0 ) and R k ( 0 ) in place of Q k ( 0 , β ) and R k ( 0 , β ) in Equation (34), so that p ˜ 0 = 1 and
p ˜ k 1 t 2 ( k 1 + γ 3 ) [ p ˜ k 1 Q k ( 0 ) h k 2 p ˜ k 2 R k ( 0 ) ] , k Z > 0 ,
where Q k ( 0 ) and R k ( 0 ) are given by
Q 1 ( 0 ) : = Q 1 ( 0 , 0 ) = α 1 β 1 q 0 , Q k ( 0 ) : = Q k ( 0 , 0 ) = [ ( 1 + t 2 ) ( k 2 ) B 1 ] ( k 1 ) + α 1 β 1 q 0 , k Z > 0 , R k ( 0 ) : = R k ( 0 , 0 ) = ( k 2 + α 1 ) ( k 2 + β 1 ) ( k 1 ) , k Z > 1 .
Remark 3.
In Remark 2, the solution w ˜ ( t ) which appears in Theorems 1 is called G H e u n , β , ϵ ( t , 0 ) .

3.1. Use of Coefficients P k

Theorem 2. (i) In Theorem 1(i), we have the particular solution of Equation (16-He), given by Equation (33). We now define P k by Equation (28) for α = ϵ , that is
p ˜ k = 1 t 2 k ( γ 3 β ) k P k , i . e . P k = t 2 k ( γ 3 β ) k p ˜ k , k Z > 1 .
By using Equation (29) for α = ϵ , we obtain P 0 = 1 and
P k = P k 1 Q k ( ϵ , ρ ) t 2 ( k 2 + γ 3 β ) h k 2 P k 2 R k ( ϵ , ρ ) , k Z > 0 .
and then the particular solution of Equation (16-He), given by Equation (33), is expressed by
u ˜ ( t ) = p 0 k = 0 P k 1 ( γ 3 β ) k Γ ( k + 1 β + ϵ ) t 2 k t k β + ϵ H ( t ) .
and u ( t ) = R R D t ϵ u ˜ ( t ) is obtained from Equation (40).
(ii) In Theorem 1(ii), it is proposed to use Equation (34) in place of Equation (31). We now propose to use the following eqution in place of Equation (39) :
P k P k 1 Q k ( 0 , β ) t 2 ( k 2 + γ 3 β ) h k 2 P k 2 R k ( 0 , β ) , k Z > 0 .
Corollary 2. (i) When β = 0 , u ˜ ( t ) given by Equation (40) is a particular solution of Equation (16-He) for f ˜ ( t ) = δ ϵ ( t ) , where Equation (39) for β = 0 is used.
(ii) Corresponding to Theorem 2(ii), we propose to use Equation (41), by replacing Q k ( 0 , β ) and R k ( 0 , β ) by Q k ( 0 ) and R k ( 0 ) , respectively, where Q k ( 0 ) and R k ( 0 ) are given in Equation (37).

3.2. Use of Coefficients a k

Theorem 3. (i) In Theorem 1, we have a particular solution u ˜ ( t ) of Equation (16-He) for the inhomogeneous term f ˜ ( t ) = R R D t β δ ϵ ( t ) = g ϵ β ( t ) . We now define a k by
p ˜ k = ( ϵ β + 1 ) k · a k , i . e . a k = 1 ( ϵ β + 1 ) k p ˜ k , k Z > 1 ,
and then we see that the solution u ˜ ( t ) of Equation (16-He), given by Equation (33), is expressed by
u ˜ ( t ) = p 0 k = 0 p ˜ k 1 Γ ( ϵ β + k + 1 ) t ϵ β + k H ( t ) = p 0 1 Γ ( ϵ β + 1 ) k = 0 a k t ϵ β + k H ( t ) ,
where a k satisfy a 0 = 1 and
a k = 1 ( ϵ β + k ) 1 t 2 ( k 1 + γ 3 β ) [ a k 1 Q k ( ϵ , ρ ) 1 ϵ β + k 1 h k 2 a k 2 R k ( ϵ , ρ ) ] , k Z > 0 .
Section 2.1.2 shows that u ( t ) = R R D t ϵ u ˜ ( t ) is obtained from Equation (43).
(ii) Following Theorems 1(ii) and 2(ii), we now propose to use the following equtions in place of Equation (44) :
a k 1 ( ϵ β + k ) 1 t 2 ( k 1 + γ 3 β ) [ a k 1 Q k ( 0 , β ) 1 ϵ β + k 1 h k 2 a k 2 R k ( 0 , β ) ] , k Z > 0 .
Proof. 
By using the first equation of Equation (42) in Equation (31), we obtain
( ϵ β + 1 ) k a k = 1 t 2 ( k 1 + γ 3 β ) [ ( ϵ β + 1 ) k 1 a k 1 Q k ( ϵ , ρ ) ( ϵ β + 1 ) k 2 h k 2 a k 2 R k ( ϵ , ρ ) ] ,
This gives Equation (44). □
Corollary 3. (i) When β = 0 , u ˜ ( t ) given by Equation (43) for β = 0 , is a particular solution of Equation (16-He) for f ˜ ( t ) = δ ϵ ( t ) , where Equation (44) for β = 0 is used.
(ii) Corresponding to Theorem 3(ii), we propose to use Equation (45) for β = 0 , by replacing Q k ( 0 , β ) and R k ( 0 , β ) by Q k ( 0 ) and R k ( 0 ) , respectively,
Remark 4.
In Remark 2, the solutions u ˜ ( t ) and u ( t ) which appear in Theorems 1, 2 and 3 are called R R D t β G H e u n , β , ϵ ( t , 0 ) and R R D t β + ϵ G H e u n , β , ϵ ( t , 0 ) , respectively.
Remark 5.
In Remark 2, u ˜ ( t ) which appear in Corollaries 1, 2 and 3, are called G H e u n , ϵ ( t , 0 ) .

4. Complementary Solutions

In the present section, we apply the results in Section 2.2, to the cases of β = 0 and f ( t ) = f ˜ β ( t ) = 0 . Lemma 10 shows two choices. We first consider the case of α = ϵ = 0 .
Remark 6.
In Corollaries 1(i), 2(i) and 3(i), the solutions u ˜ ( t ) for the cases of β = 0 , f ( t ) = f ˜ β ( t ) = δ ϵ ( t ) and α = ϵ are given. The solutions u ( t ) in the present section, are obtained from them by u ( t ) = R R D t ϵ u ˜ ( t ) or by replacing u ˜ ( t ) by u ( t ) , ϵ by 0, and a value of p 0 by an arbitrary number.
Theorem 4.
In the case stated above, Lemmas 10 and 11 show that by using Equations (19) and (27) for α = ϵ = 0 and ρ = 0 , a complementary solution u ( t ) of Equation (9) is given by
u ( t ) = k = 0 p k 1 k ! t k H ( t ) = p 0 k = 0 p ˜ k 1 k ! t k H ( t ) ,
where p 0 is any number, p k = p 0 p ˜ k for k Z > 1 , and p ˜ k for k Z > 1 satisfy p ˜ 0 = 1 and
p ˜ k = 1 t 2 ( k 1 + γ 3 ) [ p ˜ k 1 Q k ( 0 ) h k 2 p ˜ k 2 R k ( 0 ) ] , k Z > 0 ,
where Q k ( 0 ) and R k ( 0 ) are given in Equation (37).
Note here that Equation (48) is obtained from Equation (36), by replacing ≃ by =.
Theorem 5.
Lemmas 10 and 11 show that by using Equations (28) and (29) for α = ϵ = 0 and ρ = 0 , P k is defined by p ˜ k = 1 t 2 k ( γ 3 ) k P k , and the solution u ( t ) of Equation (9), given by Equation (47), is expressed as follows:
u ( t ) = p 0 k = 0 P k 1 ( γ 3 ) k k ! ( t t 2 ) k H ( t ) .
Here p 0 is any number, and P k for k Z > 1 are given by P 0 = 1 and
P k = P k 1 Q k ( 0 ) t 2 ( k 2 + γ 3 ) h k 2 P k 2 R k ( 0 ) , k Z > 0 ,
Theorem 6.
The complementary solution of Heun’s differential equation (9), given by Equation (47), is also expressed as follows:
u ( t ) = p 0 k = 0 a k t k ,
where p 0 is any number, and a k are related with p ˜ k by
p ˜ k = a k k ! , i . e . a k = p ˜ k 1 k ! , k Z > 1 .
Then we cofirm that a k satisfy a 0 = 1 and
a k = 1 ( γ 3 + k 1 ) k t 2 [ a k 1 Q k ( 0 ) 1 k 1 h k 2 a k 2 R k ( 0 ) ] = 1 ( γ 3 + k 1 ) k t 2 { [ a k 1 [ ( 1 + t 2 ) ( k 2 + γ 3 ) + γ 2 + γ 1 t 2 ] ( k 1 ) + α 1 β 1 q 0 ] h k 2 a k 2 ( k 2 + α 1 ) ( k 2 + β 1 ) } , k Z > 0 .
Proof. 
By using the first equation of Equation (52) in Equation (48), we obtain
k ! · a k = 1 t 2 ( γ 3 + k 1 ) [ ( k 1 ) ! · a k 1 Q k ( 0 ) ( k 2 ) ! · h k 2 a k 2 R k ( 0 ) ] .
This gives the first equality in Equation (53). □
This result is given in Section 3.3 in [12] and in Section 8.2 in [13].
Remark 7.
Remark 6 states that when p 0 is given by Equation (30) for β = 0 , the solutions u ( t ) in Equations (47), (48) and (51) are obtained from the solutions u ˜ ( t ) given in Corollaries 1(i), 2(i) and 3(i). In Remark 2, the solutions u ( t ) are called G H e u n , 0 ( t , 0 ) .

4.1. Complementary Solution, II

In Theorems 4∼6, we studied the case of f ˜ ( t ) = 0 , β = 0 and α = ϵ = 0 . We now study the case of α = 1 γ 3 and ϵ = 0 in place of α = ϵ = 0 .
Theorem 7.
Lemmas 10 and 11 show that by using Equations (19) and (27) for α = 1 γ 3 and ρ = 0 , we obtain the complementary solution of Equation (9), given by
u ( t ) = p 0 k = 0 p ˜ k 1 Γ ( 2 γ 3 + k ) t 1 γ 3 + k H ( t ) = 1 Γ ( 2 γ 3 ) t 1 γ 3 p 0 k = 0 p ˜ k 1 ( 2 γ 3 ) k t k H ( t ) ,
where p 0 is any number, p ˜ 0 = 1 and
p ˜ k = 1 t 2 k [ p ˜ k 1 Q k ( 1 γ 3 ) h k 2 p ˜ k 2 R k ( 1 γ 3 ) ] , k Z > 0 ,
Q k ( 1 γ 3 ) : = Q k ( 1 γ 3 , 0 ) = [ ( 1 + t 2 ) ( k 1 γ 3 ) B 1 ] ( k γ 3 ) + α 1 β 1 q 0 , k Z > 0 , R k ( 1 γ 3 ) : = R k ( 1 γ 3 , 0 ) = [ [ ( k 2 γ 3 ) + α 1 + β 1 + 1 ] ( k 1 γ 3 ) + α 1 β 1 ] ( k γ 3 ) = ( k 1 γ 3 + α 1 ) ( k 1 γ 3 + β 1 ) ( k γ 3 ) , k Z > 1 .
Theorem 8.
In Theorem 7, p 0 is any number, and p ˜ k satisfiy Equation (55). By using Equation (28) for α = 1 γ 3 and ρ = 0 , we define P k by
p ˜ k = 1 t 2 k k ! P k , i . e . P k = t 2 k k ! · p ˜ k , k Z > 1 .
Then P k satisfy P 0 = 1 , and
P k = P k 1 Q k ( 1 γ 3 ) t 2 ( k 1 ) h k 2 P k 2 R k ( 1 γ 3 ) , k Z > 0 .
By using Equation (57) in Equation (54), the complementary solution of Equation (9) is expressed by
u ( t ) = p 0 k = 0 P k 1 t 2 k k ! · Γ ( 1 γ 3 + k + 1 ) t 1 γ 3 + k H ( t ) .
Proof. 
Using the first equation of Equation (57) in Equation (55), we obtain
1 t 2 k k ! P k = 1 t 2 k [ 1 t 2 k 1 ( k 1 ) ! P k 1 Q k ( 1 γ 3 ) 1 t 2 k 2 ( k 2 ) ! h k 2 P k 2 R k ( 1 γ 3 ) ] ,
which gives Equation (58). □
Theorem 9.
The complementary solution of Heun’s equation, given by Equation (54), is also expressed as follows:
u ( t ) = p 0 1 Γ ( 2 γ 3 ) t 1 γ 3 k = 0 a k t k H ( t ) ,
where p 0 is any number, and a k are defined by
p ˜ k = ( 2 γ 3 ) k a k , i . e . a k = 1 ( 2 γ 3 ) k p ˜ k , k Z > 1 .
By using the first equation of Equation (62) in Equation (55), we obtain a 0 = p ˜ 0 = 1 , and
a k = 1 ( 2 γ 3 ) k p k = 1 ( 2 γ 3 ) k t 2 k [ p ˜ k 1 Q k ( 1 γ 3 ) h k 2 p ˜ k 2 R k ( 1 γ 3 ) ] = 1 ( 2 γ 3 + k 1 ) t 2 k [ a k 1 Q k ( 1 γ 3 ) 1 2 γ 3 + k 2 h k 2 a k 2 R k ( 1 γ 3 ) ] = 1 t 2 k ( 1 γ 3 + k ) { a k 1 [ [ ( 1 + t 2 ) ( k 1 ) + γ 2 + γ 1 t 2 ] ( k γ 3 ) + α 1 β 1 q 0 ] h k 2 a k 2 ( k 1 γ 3 + α 1 ) ( k 1 γ 3 + β 1 ) } , k Z > 0 .

5. Conclusion

In a preceding paper [5] of the present author, the particular solutions of Kummer’s and the hypergeometric differential equation are obtained for the inhomogeneous term given by f ( t ) = g β ( t ) = 1 Γ ( β ) t 1 β for β C Z > 1 . When the desired solution of Kummer’s equation is u ( t ) , we construct a transformed differntial equation of Kummer’s equation, which is satisfied u ˜ ( t ) = R R D t ϵ u ( t ) , and obtain its solutioon u ˜ ( t ) and the desired soltion by u ( t ) = R R D t ϵ u ˜ ( t ) .
In the present paper, we present the solution of the same problem for the case of Heun’s equation. The solutions obtained are given in three formats.
In Section 4, we obtain two complementary solutions of Heun’s equation. They are expressed in three formats. One of the complementary solutions in one format is in agreement with a solution presented in the past, given in Section 3.3 in [12] and in Section 8.2 in [13].

Acknowledgments

The author is indebted to Ken-ichi Sato, who collaborated in writing preceding papers and suggested the present study, but left from mathematics because of his illness.

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