Acta Polytechnica https://doi.org/10.14311/AP.2022.62.0165 Acta Polytechnica 62(1):165–189, 2022 © 2022 The Author(s). Licensed under a CC-BY 4.0 licence Published by the Czech Technical University in Prague ON GENERALIZED HEUN EQUATION WITH SOME MATHEMATICAL PROPERTIES Nasser Saad University of Prince Edward Island, Department of Mathematics and Statistics, 550 University Avenue, Charlottetown, PEI, Canada C1A 4P3. correspondence: nsaad@upei.ca Abstract. We study the analytic solutions of the generalized Heun equation, ( α0 + α1 r + α2 r2 + α3 r 3 ) y′′ + ( β0 + β1 r + β2 r2 ) y′ + ( ε0 + ε1 r ) y = 0, where |α3| + |β2| ≠ 0, and {αi}3i=0, {βi} 2 i=0, {εi}1i=0 are real parameters. The existence conditions for the polynomial solutions are given. A simple procedure based on a recurrence relation is introduced to evaluate these polynomial solutions explicitly. For α0 = 0, α1 ̸= 0, we prove that the polynomial solutions of the corresponding differential equation are sources of finite sequences of orthogonal polynomials. Several mathematical properties, such as the recurrence relation, Christoffel-Darboux formulas and the norms of these polynomials, are discussed. We shall also show that they exhibit a factorization property that permits the construction of other infinite sequences of orthogonal polynomials. Keywords: Heun equation, confluent forms of Heun’s equation, polynomial solutions, sequences of orthogonal polynomials. 1. Introduction It seems as a simple question to ask: Under what conditions does the differential equation π3(r) y′′ + π2(r) y′ + π1(r) y = (λn + µnπ0(r)) y, where λn and µn are constants and πj (r), j = 0, 1, 2, 3 are polynomials of unknown degree to be found, has n-degree monic polynomial solutions yn = ∑n k=0 ck r k, c0 ̸= 0, ck = 1? A simple approach to deduce the possible degrees of πj , j = 0, 1, 2, 3, is to examine the degrees for the (possible) polynomial solutions yn: For n = 0, y0(r) = 1, we must have π1(r) = λ0 + µ0π0(r) and the degree of the polynomial π1(r) must have the same degree as that of π0(r), so we may combine the same degree polynomial coefficients of y and write the equation as π3(r)y′′ + π2(r)y′ + π1(r)y = 0. Next, for a polynomial solution of degree one, say y1(r) = r + α, the differential equation reduces to π2(r) + π1(r)(r + α) = 0 and the degree of π2 should be the degree of π1(r) plus one. Similarly, for a second-order polynomial solution, say y(r) = r2 + α r + β, it follows by substitution that π3(r) + π2(r)(2r + α) + π1(r)(r2 + αr + β) = 0 which indicates that the degree of π3(r) should be the degree of π2 plus one, which, in turn, is a polynomial of π1 degree plus one. This simple argument shows that for the polynomial solutions of the linear second-order differential equation with polynomial coefficients, the degree of the polynomial coefficients πj (r), j = 3, 2, 1 must be of degree n, n − 1 165 https://doi.org/10.14311/AP.2022.62.0165 https://creativecommons.org/licenses/by/4.0/ https://www.cvut.cz/en Nasser Saad Acta Polytechnica and n − 2, respectively. So, without the loss of generality, we may direct out attention to the following: Under what conditions on the equation parameters αk, βk, and εk, for k = 0, 1, . . . , n, does the differential equation( n∑ k=0 αk r k ) y′′(r) + ( n−1∑ k=0 βk r k ) y′(r) + ( n−2∑ k=0 εk r k ) y(r) = 0, n ≥ 2 (1) has polynomial solutions y = ∑m k=0 Cj r j ? A logical approach is to examine the differential equation using the series solution  y = ∞∑ j=0 Cj rj , y′ = ∞∑ j=0 j Cj rj−1, y′′ = ∞∑ j=0 j (j − 1)Cj rj−2 in (1) and enforce the coefficients Cj = 0 for all j ≥ m + 1, m = 0, 1, 2, · · · to find the condition so that Cm ≠ 0. This approach leads to a conclusion that for equation (1) to have m degree polynomial solution, it is necessary that εn−2 = −m (m − 1) αn − m βn−1, n = 2, 3, · · · . (2) Did this answer the question? Indeed, no. Consider, for example, this simple equation r3y′′ + 2 r2y′ + (−2 r + 5)y = 0. Clearly, the necessary condition (2) is satisfied for m = 1 and one expects the existence of a first degree polynomial solution, say y = r + b, for an arbitrary value of b ∈ R, however, 2 r2 + (−2r + 5)(r + b) ̸= 0 for any real value of b. Therefore, for n ≥ 3, the condition (2) is necessary but not sufficient for the existence of polynomial solutions of the differential equation (1). Note, for n = 2, equation (1) is the classical hypergeometric-type differential equation [1–4] (α2 r2 + α1 r + α0) y′′ + (β1 r + β0) y′ + ε0 y = 0 (3) with the necessary and sufficient condition [2] for the polynomial solutions ε0 = −m (m − 1) α2 − m β1, m = 0, 1, 2, · · · . For n = 3, the differential equation (1) assumes the form p3(r) y′′ + p2(r) y′ + p1(r) y = 0,  p3(r) = 3∑ j=0 αj r j , p2(r) = 2∑ j=0 βj r j , p1(r) = 1∑ j=0 εj r j , αj , βj , εj ∈ R , (4) which includes as a special case or with elementary substitutions, the classical Heun differential equation [5, 6] y′′ + ( γ r + δ r − 1 + ε r − a ) y′ + α β r − q r(r − 1)(r − a) y = 0, (5) subject to the regularity (at infinity) condition α + β + 1 = γ + δ + ε, and its four confluent forms (Confluent, Doubly-Confluent, Biconfluent and Triconfluent Heun Equations). These equations are indispensable from the point of view of a mathematical analysis [5–11] and for its valuable applications in many areas of theoretical physics [5, 6, 12–20]. In the present work 166 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . • From equation (4), we will extract the possible differential equations that can be solved using two-term recurrence formulas. • From equation (4), we will extract all the differential equations whose series solutions can be evaluated with a three-term recurrence formula. • For a0 ̸= 0, we shall devise a procedure based on the Asymptotic Iteration Method [21] to find the series and polynomial solutions of the differential equation (4). • In the neighbordhoud of a singular point r = 0, i.e., with a0 = 0, we will prove that the series solution can be written as y(r) = ∞∑ k=0 (−1)k Pk;s(ε0) αk1 ( β0 α1 + s ) k (1 + s)k rk+s, where s is a root of the indicial equation. Also, we show that {Pk;s(ε0)}∞k=0 is an infinite sequence of orthogonal polynomials with several interesting properties. • By imposing the termination conditions, we study the mathematical properties of the finite sequences of the orthogonal polynomials {Pk;s(ε0)}nk=0 and explore the factorization property associated with these polynomials. 2. Elementary observations The classical approach to study the analytical solutions of equation (4) relies on the nature of the singular points of the leading polynomial coefficients L ≡ α0 + α1 r + α2 r2 + α3 r3 in addition to the point r = ∞ in the extended plane. For real coefficients and α0 ̸= 0, the odd-degree polynomial L is factored into either a product of a linear polynomial and an irreducible quadratic polynomial or a product of three linear factors. In the first case, the polynomial L can be written as L = α3(r − ξ)(r2 + br + c) where r2 + br + c is an irreducible polynomial. In this case, ξ is regular, real, singular point and ∞ is irregular for otherwise, the differential equation can be solved in terms of elementary functions according to the classical theory of ordinary differential equations. In this case, the differential equation can be written as d2y dr2 + ( µ1 r − ξ + µ2 r2 + br + c ) dy dr + ε1 r + ε0 α3(r − ξ)(r2 + b r + c) y = 0. (6) The second case, the polynomial L can be written as L = α3(r − ξ1)(r − ξ2)(r − ξ3) where ξj , j = 1, 2, 3 and ∞ are all regular singular points, i.e., the differential equation of Fushsian type, d2y dr2 +   3∑ j=1 µj r − ξj   dy dr + ε1 r + ε0 α3(r − ξ1)(r − ξ2)(r − ξ3) y = 0. (7) where µj are constants depending on the differential equation parameters. One can then study the series solutions of equations (6) and (7) using the classical Frobenius method. Another approach, recently adopted, to study (4), depends on the possible combination of the parameters αj , j = 0, 1, 2, 3 such that the polynomial L does not vanish identically. There are fifteen possible combinations in total. These fifteen combinations can be classified into two main classes: the first class is characterized by α0 ≠ 0, which has eight equations in total, the second class characterized by α0 = 0 includes the remaining seven equations. Each of these two classes will be studied in the next sections. First, we consider some elementary observations regarding the differential equation (4). 167 Nasser Saad Acta Polytechnica We assume no common factor among the polynomial coefficients pj (r), j = 1, 2, 3, we start our study of equation (4) by asking the following simple question: Under what conditions the series solutions of the differential equation (4) can be evaluated using a two-term recurrence relation [22]? For, in this case, the two linearly independent series solutions can be found explicitly. Theorem 2.1. The necessary and sufficient conditions for the linear differential equation p2(r) u′′(r) + p1(r) u′(r) + p0(r) u(r) = 0, (8) to have a two-term recurrence relationship that relates the successive coefficients in its series solution is that in the neighbourhood of the singular regular point r0 (where p2(r0) = 0), the equation (8) can be written as: [q2,0 + q2,h (r − r0)h]︸ ︷︷ ︸ q2(r) (r − r0)2−m u′′(r) + [q1,0 + q1,h (r − r0)h]︸ ︷︷ ︸ q1(r) r1−m u′(r) + [q0,0 + q0,h (r − r0)h]︸ ︷︷ ︸ q0(r) (r − r0)−m u(r) = 0, (9) where, for m ∈ Z , h ∈ Z+, j = 0, 1, 2, qj (r) ≡ ∞∑ k=0 qj,k(r − r0)k = pj (r) (r − r0)m−j , (10) when at least one of qj,0, j = 0, 1, 2 and qj,h, j = 0, 1, 2, is different from zero. In this case, the two-term recurrence formula is given by ck ck−h = − (k + λ − h)[q2,h (k + λ − h − 1) + q1,h ] + q0,h (k + λ)[q2,0 (k + λ − 1) + q1,0] + q0,0 , (11) where c0 ̸= 0, and λ = λ1, λ2 are the roots of the indicial equation q2,0 λ (λ − 1) + q1,0 λ + q0,0 = 0. (12) The closed form of the series solution generated by (11) can be written in terms of the generalized hypergeometric function as u(r; λ) =zλ ∞∑ k=0 chkr hk = rλ3F2 ( 1, 2λ−12h + q1,h 2 h q2,h − √ (q1,h−q2,h)2−4q0,hq2,h 2 h q2,h , 2λ−1 2h + q1,h 2 h q2,h + √ (q1,h−q2,h)2−4q0,hq2,h 2 h q2,h ; 1 + 2λ−12 h + q1,0 2 h q2,0 − √ (q1,0−q2,h)2−4q0,0q2,0 2 h q2,0 , 1 + 2λ−12h + q1,0 2 h q2,0 + √ (q1,0−q2,h)2−4q0,0q2,0 2 h q2,0 ; − q2,h q2,0 rh ) . (13) Applying this theorem, equation (4) generates the following solvable equations: • Differential equation: r2 (α2 + α3 r) u′′(r) + r (β1 + β2 r) u′(r) + (ε0 + ε1 r) u(r) = 0, ε0 ̸= 0, (14) Recurrence relation: For k = 1, 2, · · · , and c0 = 1, ck ck−1 = − (k + λ − 1)[α3 (k + λ − 2) + β2] + ε1 (k + λ)[α2 (k + λ − 1) + β1 ] + ε0 , (15) where λ = λ+, λ− are the roots of the indical equation α2 λ (λ − 1) + β1 λ + ε0 = 0, namely λ± = α2 − β1 ± √ (α2 − β1)2 − 4α2ε0 2α2 . 168 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . The two linearly independent solutions generated by (15), in terms of the Gauss hypergeometric functions, are: u± =r α2 −β1 ± √ (α2 −β1 )2 −4α2 ε0 2α2 2F1 ( β2 2α3 − β12α2 ± √ (α2−β1)2−4α2ε0 2α2 − √ (α3−β2)2−4α3ε1 2α3 , β2 2α3 − β12α2 ± √ (α2−β1)2−4α2ε0 2α2 + √ (α3−β2)2−4α3c1 2α3 ; α2± √ (α2−β1)2−4α2ε0 α2 ; − α3 α2 r ) . (16) • Differential equation: (α1 r + α3 r3) u′′(r) + (β0 + β2 r2) u′(r) + ε1 r u(r) = 0, (β0 ̸= 0), (17) Recurrence relation: ck ck−2 = − (k + λ − 2)[α3 (k + λ − 3) + β2] + ε1 (k + λ)[α1(k + λ − 1) + β0] , (c0, c1 ̸= 0, k = 2, 3, · · · ), (18) where λ = λ+, λ− are the roots of the indicial equation α1 λ(λ − 1) + β0 λ = 0, i.e λ+ = 0, λ− = 1 − β0/α1. The two linearly independent series solutions generated by (18) are: u+(r) = 2F1 ( β2 4α3 − 1 4 − √ (α3 − β2)2 − 4α3ε1 4α3 , β2 4α3 − 1 4 + √ (α3 − β2)2 − 4α3ε1 4α3 ; 1 2 + β0 2α1 ; − α3 α1 r2 ) , (19) and u−(r) =r 1− β0 α1 × 2F1 ( 1 4 + β2 4α3 − β0 2α1 − √ (α3 − β2)2 − 4α3ε1 4α3 , 1 4 + β2 4α3 − β0 2α1 + √ (α3 − β2)2 − 4α3ε1 4α3 ; 3 2 − β0 2α1 ; − α3 α1 r2 ) . (20) • Differential equation (α0 + α3 r3) u′′(r) + β2 r2 u′(r) + ε1r u(r) = 0, α0 ̸= 0, (21) Recurrence relation: ck ck−3 = − (k + λ − 3)[α3(k + λ − 4) + β2] + ε1 α0 (k + λ)(k + λ − 1) , (c0 ̸= 0), (22) where λ = λ1, λ2 are the roots of the indicial equation α0 λ (λ − 1) = 0, namely, λ1 = 0, λ2 = 1. The two linearly independent series solutions are: u1(r) = 2F1 ( − α3 − β2 + √ (α3 − β2)2 − 4α3ε1 6α3 , −α3 + β2 + √ (α3 − β2)2 − 4α3ε1 6α3 ; 2 3 ; − α3 α0 r3 ) , (23) and u2(r) = r 2F1 ( α3 + β2 − √ (α3 − β2)2 − 4α3ε1 6α3 , α3 + β2 + √ (α3 − β2)2 − 4α3ε1 6α3 ; 4 3 ; − α3 α0 r3 ) . (24) Out of the three generic equations (14), (17) and (21), five exactly solvable differential equations (Cases 1, 4, 5, 8, and 10) of the type (4) follows and other five (Cases 2, 3, 6, 7, 9) that can be derived directly from them by taking the limits of the equation parameters. For direct use, the ten equations are listed in Table 1. 169 Nasser Saad Acta Polytechnica DEs and their linearly independent solutions 1 α2 r2u′′ + (β1r + β2r2) u′ + (ε0 + ε1r) u = 0 u = r 1 2 − β1 2α2 + 12α2 √ (α2−β1)2−4α2ε0 1F1 ( 1 2 − β1 2α2 + ε1 β2 + √ (α2−β1)2−4α2ε0 2α2 ; 1 + √ (α2−β1)2−4α2ε0 α2 ; − β2 α2 r ) , u = r − 12 + β1 2α2 − 12α2 √ (α2−β1)2−4α2ε0 1F1 ( 1 2 − β1 2α2 + ε1 β2 − √ (α2−β1)2−4α2ε0 2α2 ; 1 − √ (α2−β1)2−4α2ε0 α2 ; − β2 α2 r ) . 2 α2 r2u′′ + β1r u′ + (ε0 + ε1r) u = 0 u = r 1 2 − β1 2α2 + 12α2 √ (α2−β1)2−4α2ε0 0F1 ( −; 1 + √ (α2−β1)2−4α2ε0 α2 ; − ε1 α2 r ) , u = r− 1 2 + β1 2α2 − 12α2 √ (α2−β1)2−4α2ε0 0F1 ( −; 1 − √ (α2−β1)2−4α2ε0 α2 ; − β2 α2 r ) . 3 α2 r2u′′ + β2r2u′ + (ε0 + ε1r) u = 0 u = r 1 2 − √ α2 −4ε0 2√α2 1F1 ( 1 2 + ε1 β2 − √ α2−4ε0 2 √ α2 ; 1 − √ α2−4ε0√ α2 ; − β2 α2 r ) , u = r 1 2 + √ α2 −4ε0 2√α2 1F1 ( 1 2 + ε1 β2 + √ α2−4ε0 2 √ α2 ; 1 + √ α2−4ε0√ α2 ; − β2 α2 r ) . 4 (α2 r2 + α3 r3)u′′ + β1 r u′ + (ε0 + ε1r) u = 0 u = r 1 2 − β1 2α2 + 12α2 √ (α2−β1)2−4α2ε0 2F1 ( 1 2 √ α3−4ε1 α3 − β12α2 + √ (α2−β1)2−4α2ε0 2α2 , − 12 √ α3−4ε1 α3 − β12α2 + √ (α2−β1)2−4α2ε0 2α2 ; 1 + √ (α2−β1)2−4α2ε0 α2 ; − α3 α2 r ) , u = r 1 2 − β1 2α2 − 12α2 √ (α2−β1)2−4α2ε0 2F1 ( 1 2 √ α3−4ε1 α3 − β12α2 − √ (α2−β1)2−4α2ε0 2α2 , − 12 √ α3−4ε1 α3 − β12α2 − √ (α2−β1)2−4α2ε0 2α2 ; 1 − √ (α2−β1)2−4α2ε0 α2 ; − α3 α2 r ) . 5 (α2 r2 + α3 r3)u′′ + β2r2 u′ + (ε0 + ε1r) u = 0 u = r 1 2 − 1 2√α2 √ α2−4ε0 2F1 ( β2 2α3 − 12 √ α2−4ε0 α2 + √ (α3−β2)2−4α3ε1 2α3 , β2 2α3 − 12 √ α2−4ε0 α2 − √ (α3−β2)2−4α3ε1 2α3 ; 1 − √ α2−4ε0√ α2 ; − α3 α2 r ) , u = r 1 2 + 1 2√α2 √ α2−4ε0 2F1 ( β2 2α3 + 12 √ α2−4ε0 α2 − √ (α3−β2)2−4α3ε1 2α3 , β2 2α3 + 12 √ α2−4ε0 α2 + √ (α3−β2)2−4α3ε1 2α3 ; 1 + √ α2−4ε0√ α2 ; − α3 α2 r ) . 6 (α2 r2 + α3 r3)u′′ + (ε0 + ε1 r) u = 0 u = r 1 2 − 1 2√α2 √ α2−4ε0 2F1 ( − 12 √ α2−4ε0 α2 − √ α3−4ε1 2 √ α3 , − 12 √ α2−4ε0 α2 − √ α3−4ε1 2 √ α3 ; 1 − √ α2−4ε0√ α2 ; − α3 α2 r ) , u = r 1 2 + 1 2√α2 √ α2−4ε0 2F1 ( 1 2 √ α2−4ε0 α2 − √ α3−4ε1 2 √ α3 , 12 √ α2−4ε0 α2 + √ α3−4ε1 2 √ α3 ; 1 + √ α2−4ε0√ α2 ; − α3 α2 r ) . 7 α2 r2 u′′ + (ε0 + ε1 r) u = 0 u = r 1 2 + 1 2 √ 1− 4ε0 α2 0F1 ( ; 1 + √ 1 + 4ε0 α2 ; − ε1 α2 r ) , u = r 1 2 − 1 2 √ 1− 4ε0 α2 0F1 ( ; 1 − √ 1 − 4ε0 α2 ; − ε1 α2 r ) 8 α1 r u′′ + (β0 + β2r2) u′ + ε1 r u = 0, u = 1F1 ( ε1 2β2 ; 12 + β0 2α1 ; − β22α1 r 2 ) , u = r1− β0 α1 1F1 ( 1 2 − β0 2α1 + ε12β2 ; 3 2 − β0 2α1 ; − β22α1 r 2 ) . 9 α1 r u′′ + β0 u′ + ε1 r u = 0, u = 0F1 ( −; 12 + β0 2α1 ; − ε14α1 z 2 ) , u = r1− β0 α1 0F1 ( −; 32 − β0 2α1 ; − ε14α1 r 2 ) . 10 α0 u′′ + β2 r2u′ + ε1 r u = 0, u = 1F1 ( ε1 3β2 ; 23 ; − β2 3α0 r3 ) , u = r 1F1 ( 1 3 + ε1 3β2 ; 43 ; − β2 3α0 r3 ) . Table 1. Ten solvable equations of the type (4) that follows from the generic equations (14), (17), and (21). 170 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . 3. The solutions in the neighbourhood of an ordinary point 3.1. Series solutions In the case of α0 ̸= 0, r = 0, there is an ordinary point for the differential equations (4). The classical theory of differential equation ensure that the (4) has two linearly independent power series solutions in the neighbourhood of r = 0 and valid to the nearest real singular point of the leading polynomial coefficient L ≡ α0 + α1 r + α2 r2 + α3 r3 = 0. Indeed, the polynomial L = 0 has the discriminate [23]: ∆ = 18 α3 α2 α1 α0 − 4 α32 α0 + α 2 2 α 2 1 − 4 α3 α 3 1 − 27 α 2 3 α 2 0. (25) The nature of the L roots as given by (25) along with the corresponding eight differential equations are summarized in Table 2. For these differential equations, the following theorem, that can be easily proved using Frobenius method, holds. Theorem 3.1. (Formal series solutions) In the neighbourhood of the ordinary point r = 0, the coefficients of the series solution y(r) = ∑∞ k=0 Ck r k to the differential equation (4) satisfy the four-term recurrence relation ((k − 1) ((k − 2) α3 + β2) + ε1) Ck−1 + (k((k − 1) α2 + β1) + ε0) Ck + (k + 1)(kα1 + β0) Ck+1 + (k + 2) (k + 1) α0 Ck+2 = 0, (26) where k = 0, 1, 2, · · · , with C−1 = 0 and arbitrary nonzero constants C0 and C1. The radius of convergence of these series solutions is extended from r = 0 to the nearest singular point of the leading polynomial coefficient L = 0. The first few terms of the series solution are given explicitly by C2 = − ε02α0 C0 − β0 2α0 C1, C3 = (α1+β0) ε0−α0 ε1 6α20 C0 + β0(α1+β0)−α0(β1+ε0) 6 α20 C1, · · · . For α0 ̸= 0, using (26), we can extract the following differential equations with series solution from (4) using a three-term recurrence relation: • Differential equation: ( α0 + α1 r + α3 r3 ) u′′(r) + ( β0 + β2 r2 ) u′(r) + ε1 r u(r) = 0. (27) Recurrence formula: Ck+2 = − (k + 1)(k α1 + β0) (k + 1) (k + 2) α0 Ck+1 − (k − 1)((k − 2) α3 + β2) + ε1 (k + 1) (k + 2) α0 Ck−1. (28) • Differential equation: ( α0 + α2r2 + α3r3 ) u′′(r) + (β1 r + β2 r2)u′(r) + ε1 r u(r) = 0. (29) Recurrence formula: Ck+2 = − k(k − 1)α2 + k β1 (k + 1) (k + 2) α0 Ck − (k − 1)(k − 2)α3 + (k − 1)β2 + ε1 (k + 1)(k + 2) α0 Ck−1. (30) • Differential equation: ( α0 + α2 r2 ) u′′(r) + (β1 r + β2 r2) u′(r) + ε1 r u(r) = 0. (31) Recurrence formula: Ck+2 = − k (k − 1)α2 + k β1 (k + 1) (k + 2) α0 Ck − (k − 1)β2 + ε1 (k + 1) (k + 2) α0 Ck−1. (32) 171 Nasser Saad Acta Polytechnica DE α3 α2 α1 α0 Discriminant Roots of L Domain definition I ∆3 > 0 ξ1 ̸= ξ2 ̸= ξ3 |r| < mini=1,2,3 ξi α3 α2 α1 α0 ∆3 = 0 ξ1 = ξ2 = ξ3 = ξ |r| < ξ ∆3 < 0 ξ ∈ R |r| < ξ Differential equation: (α0 + α1 r + α2 r2 + α3 r3) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Discriminant: ∆3 = 18 α3 α2 α1 α0 − 4 α23 α0 + α22α12 − 4 α3 α13 − 27 α32 α02 II ∆3 > 0 ξ1 ̸= ξ2 |r| < mini=1,2 ξi 0 α2 α1 α0 ∆3 = 0 ξ1 = ξ2 = ξ |r| < ξ ∆3 < 0 None |r| < ∞ Differential Equation: (α0 + α1 r + α2 r2) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Discriminant: ∆3 = α22(−4 α2 α0 + α12) III 0 0 α1 α0 α1α0 > 0 r = −α0/α1 −∞ < r < −α0/α1 α1α0 < 0 r = −α0/α1 −α0/α1 < r < ∞ Differential Equation: (α0 + α1 r) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Discriminant: ∆3 = 0 IV 0 0 0 α0 None None −∞ < r < ∞ Differential Equation: α0 y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Discriminant: ∆3 = 0 V ∆3 > 0 ξ1 ̸= ξ2 ̸= ξ3 |r| < mini=1,2,3 ξi α3 0 α1 α0 ∆3 = 0 ξ1 = ξ2 = ξ3 = ξ |r| < ξ ∆3 < 0 ξ ∈ R |r| < ξ Differential Equation: (α0 + α1 r + α3 r3) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Discriminant: ∆3 = −4 α3 α13 − 27 α32 α02 VI ∆3 > 0 ξ1 ̸= ξ2 ̸= ξ3 |r| < mini=1,2,3 ξi α3 α2 0 α0 ∆3 = 0 ξ1 = ξ2 = ξ3 = ξ |r| < ξ ∆3 < 0 ξ ∈ R |r| < ξ Differential Equation: (α0 + α2 r2 + α3 r3) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Discriminant: ∆3 = −4 α23 α0 − 27 α32 α02 VII α3 0 0 α0 α0 α3 < 0 or α0 α3 > 0 ξ = 3 √ −α0/α3 |r| < ξ Differential Equation: (α0 + α3 r3) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Discriminant: ∆3 = −27 α32 α02 VIII 0 α2 0 α0 α2α0 < 0 r = ± √ − α0 α2 − √ − α0 α2 < r <√ − α0 α2 α2α0 > 0 None −∞ < r < ∞ Differential Equation: (α0 + α2 r2) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Discriminant: ∆3 = −4 α23 α0 Table 2. Tabulating the eight different types of differential equations, which apply to Theorem 3.1. 172 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . • Differential equation: ( α0 + α3 r3 ) u′′(r) + (β1 r + β2 r2) u′(r) + r ε1 u(r) = 0. (33) Recurrence formula: Ck+2 = − k β1 (k + 1) (k + 2) α0 Ck − (k − 1)(k − 2)α3 + (k − 1)β2 + ε1 (k + 1) (k + 2) α0 Ck−1. (34) • Differential equation: α0 u ′′(r) + (β1 r + β2 r2) u′(r) + ε1 r u(r) = 0. (35) Recurrence formula: Ck+2 = − k β1 (k + 1) (k + 2) α0 Ck − (k − 1) β2 + ε1 (k + 1) (k + 2) α0 Ck−1. (36) • Differential equation: ( α0 + α2 r2 + α3 r3 ) u′′(r) + β1 r u′(r) + ε1 r u(r) = 0. (37) Recurrence formula: Ck+2 = − k (k − 1)α2 + k β1 (k + 1) (k + 2) α0 Ck − (k − 2)(k − 1)α3 + ε1 (k + 1) (k + 2) α0 Ck−1. (38) • Differential equation: ( α0 + α3 r3 ) u′′(r) + β1 r u′(r) + ε1 r u(r) = 0. (39) Recurrence formula: Ck+2 = − k β1 (k + 1)(k + 2) α0 Ck − (k − 2) (k − 1) α3 + ε1 (k + 1)(k + 2) α0 Ck−1. (40) • Differential equation: ( α0 + α2 r2 ) u′′(r) + β1 r u′(r) + ε1 r u(r) = 0. (41) Recurrence formula: Ck+2 = − k(k − 1)α2 + k β1 (k + 1)(k + 2)α0 Ck − ε1 (k + 1)(k + 2) α0 Ck−1. (42) • Differential equation: ( α0 + α2 r2 ) u′′(r) + (β1 + β2 r2)u′(r) + ε1 r u(r) = 0. (43) Recurrence formula: Ck+2 = − k(k − 1)α2 + β1 k) (k + 1) (k + 2) α0 Ck − (k − 1)β2 + ε1 (k + 1) (k + 2) α0 Ck−1. (44) • Differential equation: u′′(r) + β1 r u′(r) + ε1 r u(r) = 0, (45) Recurrence formula: Ck+2 = − k β1 (k + 1) (k + 2) Ck − ε1 (k + 1) (k + 2) Ck−1. (46) 3.2. Polynomial solutions The series solution y(r) = ∑∞ k=0 Ck r k terminates to an nth-degree polynomial if Cn ̸= 0 and Cj = 0 for all j ≥ n + 1. It is not difficult to show by direct substitution that for polynomial solutions of Pn(r) = ∑n k=0 Ck r k, it is necessary that ε1 = −n (n − 1) α3 − n β2, n = 0, 1, 2, · · · . (47) 173 Nasser Saad Acta Polytechnica Furthermore, the polynomial solution coefficients {Ck}nk=0 satisfy a four-term recurrence relation, see (26),( (k − 1) ( (k − 2)α3 + β2 ) + ε1;n ) Ck−1 + ( k ( (k − 1)α2 + β1 ) + ε0;n ) Ck + (k + 1)(kα1 + β0) Ck+1 + (k + 1)(k + 2)α0 Ck+2 = 0 , k = 0, 1, . . . , n + 1 , (48) that generates a system of (n + 2) linear equations in {Ck}nk=0: n−equations︷ ︸︸ ︷︸ ︷︷ ︸ (n+2)−equations The first n equations are  k = 0, → ε0C0 + β0C1 + 2 α0 C2 = 0 k = 1, → ε1C0 + (β1 + ε0)C1 + 2(α1 + β0)C2 + 6 α0 C3 = 0 k = 2, → (β2 + ε1)C1 + (2 α2 + 2 β1 + ε0)C2 + 3(2 α1 + β0)C3 + 12 α0 C4 = 0 k = 3, → (2α3 + 2 β2 + ε1)C2 + (6 α2 + 3 β1 + ε0)C3 + 4(3 α1 + β0)C4 + 20 α0 C5 = 0 ... k = n − 1, → ( (n − 2) ( (n − 3) α3 + β2 ) + ε1;n ) Cn−2 + ( (n − 1) ( (n − 2) α2 + β1 ) + ε0;n ) Cn−1 +n ( (n − 1) α1 + β0 ) Cn = 0. (49) These equations permit the evaluation, using say Cramer’s rule, of the coefficients {Ck}nk=1 of the polynomial solution in terms of the non-zero constant C0. The (n + 1)th equation( (n − 1) (n − 2) α3 + (n − 1) β2 + ε1 ) Cn−1 + ( n(n − 1) α2 + n β1 + ε0 ) Cn = 0 , (50) gives our sufficient condition that relates ε0 ≡ ε0;n to the remaining parameters of the differential equation. Finally, the (n + 2)th equation ε1;n = −n (n − 1) α3 − n β2 , n = 0, 1, · · · , (51) re-establishes the necessary condition (ε1 ≡ ε1;n) for the existence of the n-degree polynomial solution, see (47). For a non-zero solution, the n + 1 linear equations generated by the recurrence relation (48) require the vanishing of the (n + 1) × (n + 1)-determinant (with four main diagonals and all other entries being zeros) ∆n+1 = ∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ S0 T1 η1 γ1 S1 T2 η2 γ2 S2 T3 η3 . . . . . . . . . . . . γn−2 Sn−2 Tn−1 ηn−1 γn−1 Sn−1 Tn γn Sn ∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ , where Sk = ε0;n + k ( (k − 1)α2 + β1 ) , Tk = k ( (k − 1)α1 + β0 ) , γk = ε1;n + (k − 1) ( (k − 2)α3 + β2 ) , ηk = k(k + 1)α0 , and for fixed n , ε1;n = −n (n − 1) α3 − n β2 . (52) 174 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . A simple relation to evaluate this determinant in terms of lower-degree determinants is given by ∆k+1 = Sk ∆k − γk Tk ∆k−1 + γk γk−1 ηk−1 ∆k−2 , (∆−2 = ∆−1 = 0, ∆0 = 1, k = 0, 1, . . . , n). (53) Although there is a classical theorem [24] that guarantees the simple distinct real roots of the three diagonal matrix, to the best of our knowledge, there is no such theorem available for the matrix-type (52). However, we shall assume, in the following example, that the matrix entries allow for the distinct real roots of the resulting polynomial of ε0;n. Illustrative example: • For the zero-degree polynomial solution P0 (r) = 1, i.e., n = 0, the coefficients Cj = 0 for all j ≥ 1 and the recurrence relation (28) for k = 0, 1 gives, respectively, the necessary and sufficient conditions ε1;0 = 0 , ε0;0 = 0. (54) • For a first-degree polynomial solution, n = 1, the coefficients Cj = 0 for all j ≥ 2 where k = 0, 1, 2 give the following three equations   ε0;1 C0 + β0 C1 = 0, ε1;1 C0 + (β1 + ε0;1) C1 = 0, (β2 + ε1;1) C1 = 0. (55) So, for C0 = 1, it is necessary that ε1;1 = −β2 and therefore, C1 = −ε0;1/β0 where ε0;1 are now the roots of the quadratic equation β0 β2 + β1ε0;1 + ε20;1 = 0. Let εℓ0;1, ℓ = 1, 2, denote, if any, the two distinct real roots ε00;1 ̸= ε10;1 of this quadratic equation. Then, for the two (distinct) differential equations( α0 + α1 r + α2 r2 + α3 r3 ) P′′1;ℓ (r) + ( β0 + β1 r + β2 r2 ) P′1;ℓ (r) + ( εℓ0;1 − β2 r ) P1;ℓ (r) = 0 , ℓ = 1, 2 , (56) the first-order polynomial solutions are P1;ℓ (r) = 1 − εℓ0;1 β0 r, β0 β2 + β1 εℓ0;1 + (εℓ0;1)2 = 0, ℓ = 1, 2 . (57) • For a second-degree polynomial solution, n = 2, the coefficients Cj = 0 for all j ≥ 3 where k = 0, 1, 2, 3 give the four linear equations   ε0;2 C0 + β0 C1 + 2 α0 C2 = 0, ε1;2 C0 + (β1 + ε0;2) C1 + 2(α1 + β0) C2 = 0, (β2 + ε1;2) C1 + (2 α2 + 2 β1 + ε0;2) C2 = 0, (2α3 + 2 β2 + ε1;2)C2 = 0. (58) The very last equation of (58), correspondent to k = 3, gives the necessary condition ε1;2 = −2 α3 − 2 β2 , (59) and for k = 0, 1, the coefficients of the polynomial solution y(r) = 1 + C1 r + C2 r2 read  C1 = ∣∣∣∣∣ −ε0;2 2α02 α3 + 2 β2 2α1 + 2β0, ∣∣∣∣∣∣∣∣∣∣ β0 2α0β1 + ε0;2 2α1 + 2β0 ∣∣∣∣∣ , C2 = ∣∣∣∣∣ β0 −ε0;2β1 + ε0;2 2 α3 + 2 β2 ∣∣∣∣∣∣∣∣∣∣ β0 2α0β1 + ε0;2 2α1 + 2β0 ∣∣∣∣∣ . (60) 175 Nasser Saad Acta Polytechnica The equation corresponding to k = 2 and n = 2 establishes the sufficient condition∣∣∣∣∣∣∣∣ εℓ0;2 β0 2α0 −2 α3 − 2 β2 β1 + εℓ0;2 2α1 + 2β0 0 β2 − 2 α3 − 2 β2 2 α2 + 2 β1 + εℓ0;2 ∣∣∣∣∣∣∣∣ = 0, (61) where ℓ = 1, 2, 3 refers to the three distinct simple roots εℓ0;2, ℓ = 1, 2, 3, if any, of the polynomial generated by the determinant (61). Hence, for each index ℓ = 1, 2, 3, the differential equation( α0 + α1 r + α2 r2 + α3 r3 ) P′′2;ℓ (r) + ( β0 + β1 r + β2 r2 ) P′2;ℓ (r) + ( εℓ0;2 − (2α3 + 2β2) r ) P2;ℓ (r) = 0 , (62) has the polynomial solution (for ℓ = 1, 2, 3 .) P2;ℓ (r) = 1 + ∣∣∣∣∣∣ −εℓ0;2 2α0 2 α3 + 2 β2 2α1 + 2β0 ∣∣∣∣∣∣∣∣∣∣∣∣ β0 2α0 β1 + εℓ0;2 2α1 + 2β0 ∣∣∣∣∣∣ r + ∣∣∣∣∣∣ β0 −εℓ0;2 β1 + εℓ0;2 2 α3 + 2 β2 ∣∣∣∣∣∣∣∣∣∣∣∣ β0 2α0 β1 + εℓ0;2 2α1 + 2β0 ∣∣∣∣∣∣ r2, (63) The above constructive approach can be continued to generate higher-order polynomial solutions of an arbitrary degree. Theorem 3.2. Suppose the polynomial in εℓ0;n generated by the determinant (52) has n + 1 distinct real roots arranged in ascending order ε00;n < ε10;n < ε20;n < · · · < εn0;n, then, the eigenvalue problem ( α0 + α1 r + α2 r2 + α3 r3 ) d2Pn;ℓ dr2 + ( β0 + β1 r + β2 r2 ) dPn;ℓ dr − n ( (n − 1) α3 + β2 ) r Pn;ℓ = −εℓ0;n Pn;ℓ, (64) has a polynomial solution of the degree n, for ℓ = 1, 2, . . . , n + 1. This theorem is illustrated by Figure 1, for n = 0, 1, 2, 3, n = 0 ε10;0 P0;1(r) n = 1 ε10;1 P1;1(r) ε20;1 P1;2(r) n = 2 ε10;2 P2;1(r) ε20;2 P2;2(r) ε30;2 P2;3(r) n = 3 ε10;3 P3;1(r) ε20;3 P3;2(r) ε30;3 P3;3(r) ε40;3 P3;4(r) · · · · · · · · · Figure 1. A graphical representation of Theorem 3.2 Open problem: It is an open question to establish the condition(s) on the parameters so that the polynomial generated by the determinant (32) has simple and real distinct roots. 4. The solutions in the neighbourhood of a singular point 4.1. Series Solution and infinite sequence of orthogonal polynomials {Pk(ε0)}∞k=0 As mentioned earlier, if α0 = 0, there are seven subclasses characterized by the equation r ( α1 + α2 r + α3r2 ) y′′ + ( β0 + β1 r + β2 r2 ) y′ + ( ε0 + ε1 r ) y = 0 . (65) 176 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . The classification of these seven equations along with their singularities and the associated domains are summarized in Table 3. From this Table, it is noted that if α1 ̸= 0, there are four subclasses where the point r = 0 is a regular singular point, while if α1 = 0, the condition β0 = 0 is necessary to ensure that r = 0 is a regular singular point for two additional subclasses and the last equation is a class where r = 0 is irregular singular point unless we reduce to Euler’s type (α1 = α2 = β1 = β0 = ε0 = 0). In the neighbourhood of the regular singular point r = 0, the formal series solution y(r) = rs ∑∞ k=0 Ckr k is then valid within the interval (0, ζ) where ζ is the nearest singular point obtained via the roots of the quadratic equation α1 + α2 r + α3r2 = 0. Here, s are the roots of the indicial equation α1 s(s − 1) + β0 s = 0, i.e. s1 = 0 and s = 1 − β0/α1. Using Frobenius method, it is straightforward to show that the coefficients {Ck}∞k=0 satisfy the three-term recurrence relation (k + s + 1) ( α1(k + s) + β0 ) Ck+1 + ( (k + s)[α2 (k + s − 1) + β1] + ε0 ) Ck + ( (k + s − 1)[α3(k + s − 2) + β2] + ε1 ) Ck−1 = 0 , (66) where k = 1, 2, . . . . For   C−1 = 0, C0 = 1, C1 = − s(α2 (s − 1) + β1) + ε0 (α1 s + β0)(s + 1) = − P1;s(ε0) α1 ( s + β0 α1 ) (s + 1) , this equation can be written as Ck+2 = λ0(k) Ck+1 + s0(k) Ck, where   λ0(k) = − (α2 (k + s) + β1) (k + s + 1) + ε0 (α1(k + s + 1) + β0) (k + s + 2) , s0(k) = − (α3(k + s − 1) + β2) (k + s) + ε1 (α1(k + s + 1) + β0) (k + s + 2) , From this equation, we note that Ck+3 = λ1(k) Ck+1 + s1(k) Ck,   λ1(k) = λ0(k + 1) λ0(k) + s0(k + 1)s1(k) = λ0(k + 1) s0(k), Ck+4 = λ2(k) Ck+1 + s2(k) Ck,   λ2(k) = λ1(k + 1)λ0(k) + s1(k + 1)s2(k) = λ1(k + 1)s0(k), Ck+5 = λ3(k) Ck+1 + s3(k) Ck,   λ3(k) = λ2(k + 1)λ0(k) + s2(k + 1)s3(k) = λ2(k + 1)s0(k), and in general Ck+m = λm−2(k) Ck+1 + sm−2(k) Ck,   λm(k) = λm−1(k + 1)λ0(k) + sm−1(k + 1)sm(k) = λm−1(k + 1)s0(k), and therefore C2 = (s + 1)(α2 s + β1) + ε0 (α1(s + 1) + β0)(s + 2) ( s(α2 (s − 1) + β1) + ε0 (α1 s + β0)(s + 1) ) − s(α3(s − 1) + β2) + ε1 (α1(s + 1) + β0)(s + 2) = P2;s(ε0) α21 ( s + β0 α1 ) 2 (s + 1)2 . (67) 177 Nasser Saad Acta Polytechnica DE α3 α2 α1 Condition Roots of LPC Domain definition I α3 α2 α1 a22 − 4a1a3 > 0 r = 0, ξ+ ̸= ξ− r ∈ (0, min ξ±) if ξ± > 0 r ∈ (max ξ±, 0) if ξ± < 0 r ∈ (0, ξ+) if ξ− < 0 < ξ+, |ξ−| > ξ+ r ∈ (ξ−, 0) if ξ− < 0 < ξ+, |ξ−| < ξ+ a22 − 4a1a3 = 0 r = 0, ξ+ = ξ− = ξ r ∈ (0, ξ) Differential Equation: r (ξ1 − r)(ξ2 − r) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Roots: r = 0; r = ξ± ≡ (−α2 ± √ α22 − 4α1α3)/(2α3) Singularity: r = 0, ξ±, ∞: Regular Differential Equation: r (ξ − r)2 y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Roots: r = 0; r = ξ ≡ −α2/(2α3) Singularity: r = 0, ξ: Regular; r = ∞: Irregular II 0 α2 α1 r = 0, r = −α1/α2 r ∈ (0, −α1/α2) if α1/α2 < 0 r ∈ (−α1/α2, 0) if α1/α2 > 0 Differential Equation: r(α1 + α2 r) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Singularity: r = 0, −α1/α2: Regular; r = ∞: Irregular III α3 0 α1 a1a3 < 0 r = 0, ± √ −α1/α3 r ∈ (0, √ −α1/α3) α1α3 > 0 r = 0, r ∈ (0, ∞) Differential Equation: r (α3r2 + α1) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0, α3α1 < 0 Singularity: r = 0, ± √ −α1/α3, ∞: Regular Differential Equation: r (α3r2 + α1) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0, α3α1 > 0 Singularity: r = 0: Regular; r = ∞: Irregular IV α3 α2 0 β0 = 0 r = 0, −α2/α3 r ∈ (0, −α2/α3) if α2/α3 < 0 r ∈ (−α2/α3, 0) if α2/α3 > 0 Differential Equation: r2(α3r + α2) y′′ + r(β1 + β2 r) y′ + (ε0 + ε1 r) y = 0 Singularity: r = 0, −α2/α3: Regular; r = ∞: Irregular V 0 0 α1 r = 0 r ∈ (0, ∞) Differential Equation: α1 r y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Singularity: r = 0: Regular; r = ∞: Irregular VI 0 α2 0 β0 = 0 r = 0 r ∈ (0, ∞) Differential Equation: α2 r2 y′′ + r(β1 + β2 r) y′ + (ε0 + ε1 r) y = 0 Singularity: r = 0 : Regular; r = ∞: Irregular VII α3 0 0 r = 0 r ∈ (0, ∞) Differential Equation: α3 r3 y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0 Singulaity: r = 0, ∞: Irregular Table 3. Tabulating the seven different types of differential equations, which apply to Theorem 4.1. 178 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . initiated with P2,s(ε) = ((s + 1)(α2 s + β1) + ε0)P1;s(ε0) − (α1 s + β0)(s + 1)(s(α3(s − 1) + β2) + ε1) Continuing with this process, it is straightforward to conclude that the series solution can be written as y(r) = rs ∞∑ k=0 Ck rk = ∞∑ k=0 (−1)k Pk;s(ε0) αk1 ( β0 α1 + s ) k (1 + s)k rk+s, (68) where the k-degree polynomials of the parameter ε0, namely {Pk;s(ε0)}∞k=0, satisfy the following three-term recurrence relation: Pk+1;s(ε0) = ( (k + s) [ (k + s − 1)α2 + β1 ] + ε0 ) Pk;s(ε0) − (k + s) ( (k + s − 1)α1 + β0 )( (k + s − 1) × [ (k + s − 2)α3 + β2 ] + ε1 ) Pk−1;s(ε0), (69) initiated with P−1;s(ε0) = 0 and P0;s(ε0) = 1. For the classes I-IV in Table 3, including, of course, the classical Heun equation, r = 0 is a regular singular point with one of the exponents of singularities being s = 0, in which case, the coefficients {Ck}∞k=0 of the series solution y(r) = ∑∞ k=0 Ckr k satisfy the three-term recurrence relation( (k + 1)(k α1 + β0) ) Ck+1 + ( k ( (k − 1)α2 + β1 ) + ε0 ) Ck + ( (k − 1) ( (k − 2)α3 + β2 ) + ε1 ) Ck−1 = 0, (70) and we have the following general result concerning the series solutions of the equation (65): Theorem 4.1. In the neighbourhood of the regular singular point r = 0, the series solution y(r) = ∑∞ k=0 Ckr k of the differential equation (65), with α1 ̸= 0, is explicitly given by y(r) = ∞∑ k=0 (−1)k Pk(ε0) k! αk1 ( β0 α1 ) k rk , (71) where the infinite sequence {Pk(ε0)}∞k=0 is evaluated using the three-term recurrence relation Pk+1(ε0) = ( k(k − 1)α2 + kβ1 + ε0 ) Pk(ε0) − k ( (k − 1)α1 + β0 ) × ( (k − 1)(k − 2)α3 + (k − 1)β2 + ε1 ) Pk−1(ε0), (72) where P−1(ε0) = 0, and P0(ε0) = 1. Here, (α)n refers to the Pochhammer symbol (α)n = α(α + 1) · · · (α − n + 1) = Γ(α + n)/Γ(α) which is defined in terms of Gamma functions and satisfies the identity (−n)k = 0 for any positive integers k ≥ n + 1. Equation (72) in Theorem follows directly by substituting the coefficients of (71) in the recurrence relation (65) and eliminates the common terms. Corollary 4.2. In the neighbourhood of the regular singular point r = 0, the series solution y(r) = ∑∞ k=0 Ckr k of the differential equation r (α1 + α3r2) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0, (73) is given, explicitly, by y(r) = ∞∑ k=0 (−1)k Pk(ε0) k! αk1 ( β0 α1 ) k rk, (74) where Pk+1(ε0) = (k β1 + ε0) Pk(ε0) − k ((k − 1)α1 + β0) × ((k − 1)(k − 2)α3 + (k − 1)β2 + ε1) Pk−1(ε0), (75) initiated with P−1(ε0) = 0, P0(ε0) = 1. 179 Nasser Saad Acta Polytechnica Corollary 4.3. In the neighbourhood of the regular singular point r = 0, the series solution y(r) = ∑∞ k=0 Ckr k of the differential equation r (α1 + α2 r) y′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0, (76) is given, explicitly, by y(r) = ∞∑ k=0 (−1)k Pk(ε0) k! αk1 ( β0 α1 ) k rk, (77) where Pk+1(ε0) = (k(k − 1)α2 + kβ1 + ε0)Pk(ε0) − k((k − 1)α1 + β0)((k − 1)β2 + ε1)Pk−1(ε0), (78) initiated with P−1(ε0) = 0, P0(ε0) = 1. Corollary 4.4. In the neighbourhood of the regular singular point r = 0, the series solution y(x) = ∑∞ k=0 Ckr k of the differential equation α1 r y ′′ + (β0 + β1 r + β2 r2) y′ + (ε0 + ε1 r) y = 0, (79) is given, explicitly, by y(r) = ∞∑ k=0 (−1)k Pk(ε0) k! αk1 ( β0 α1 ) k rk, (80) where Pk+1(ε0) = (kβ1 + ε0)Pk(ε0) − k ((k − 1) α1 + β0)((k − 1)β2 + ε1)Pk−1(ε0), (81) initiated with P−1(ε0) = 0, P0(ε0) = 0. Corollary 4.5. In the neighbourhood of the regular singular point x = 0, the series solution y(x) = ∑∞ k=0 Ckx k of the differential equation α1r y ′′ + (β0 + β2 r2) y′ + (ε0 + ε1 r) y = 0, (82) is given, explicitly, by y(r) = ∞∑ k=0 (−1)k Pk(ε0) k! αk1 ( β0 α1 ) k rk, (83) where Pk+1(ε0) = ε0Pk(ε0) − k((k − 1)α1 + β0)((k − 1)β2 + ε1)Pk−1(ε0), (84) initiated with P−1(ε0) = 0, P1(ε0) = 1. Remark 4.6. If, in addition to α0 = 0, we also have α1 = 0, then r = 0 is a regular singular point only if β0 = 0, in which case the differential equation reduces to an equation that resembles Euler’s equation, namely r2 ( α2 + α3r ) y′′ + r ( β1 + β2 r ) y′ + ( ε0 + ε1 r ) y = 0. (85) The exponents of the singularity r = 0 are s± = ( α2 − β1 ± √ (α2 − β1)2 − 4α2ε0 ) /(2α2). From the relation (66), the coefficients of the formal series solution y(r) = rs ∑∞ k=0 Ck r k satisfy the two-term recurrence relation (k = 1, 2, . . . , C0 = 1), Ck = − (k−1+s±)(k−2+s±)α3+(k−1+s±)β2+ε1 (k+s±)(k−1+s±)α2+(k+s±)β1+ε0 Ck−1, = k∏ j=1 (−1)j (j−1+s±)(j−2+s±)α3+(j−1+s±)β2+ε1(j+s±)(j−1+s±)α2+(j+s±)β1+ε0 , (86) that allows to obtain a closed form of the series solution of (71) in terms of the generalized hypergeometric function as y(r) =rs± 3F2 ( 1, s± − 12 + β2 2α3 − √ (α3 −β2 )2 −4α3 ε1 2α3 , s± + 12 + β2 2α3 − √ (α3 −β2 )2 −4α3 ε1 2α3 ; s± + 12 + β1 2α2 − √ (α2 −β1 )2 −4α2 ε0 2α2 , s± + 12 + β1 2α2 + √ (α2 −β1 )2 −4α2 ε0 2α2 ; − α3 α2 r ) . (87) 180 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . 4.2. Polynomial Solution and finite sequence of orthogonal polynomials Theorem 4.7. The necessary condition for the second-order linear differential equation (65) to have an nth- degree polynomial solution yn(r) = ∑n k=0 Ck r k , n = 0, 1, 2, . . ., in the neighbourhood of the regular singular point r = 0 with one of the indicial equation exponents s = 0, is ε1;n = −n (n − 1) α3 − n β2 , n = 0, 1, 2, . . . , (88) along with the sufficient condition, relating the remaining coefficients, given by the vanishing of the tridiagonal (n + 1) × (n + 1)-determinant ∆n+1 ≡ 0 given by ∆n+1 = ∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ S0 T1 γ1 S1 T2 γ2 S2 T3 . . . . . . . . . γn−2 Sn−2 Tn−1 γn−1 Sn−1 Tn γn Sn ∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣ , (89) where, for fixed n : ε1;n = −n (n − 1) α3 − n β2,  Sk = ε0;n + k ( (k − 1)α2 + β1 ) , Tk = −k ( (k − 1)α1 + β0 ) , γk = −ε1;n − (k − 1) ( (k − 2)α3 + β2 ) , and all other entries are zeros. In this case, the polynomial solutions are given explicitly by yn(r) = n∑ k=0 (−1)k P nk (ε0;n) k! αk1 ( β0 α1 ) k rk , (90) where the finite orthogonal sequences {P nk (ε0;n)} n k=0 are evaluated using the three-term recurrence relation P nk+1(ε0;n) = ( Sk + ε0;n ) P nk (ε0;n) − γkTkP n k−1(ε0;n) , or, more explicitly, P nk+1(ε0;n) = ( k(k − 1)α2 + kβ1 + ε0;n ) P nk (ε0;n) + k(n − k + 1) ( (k − 1)α1 + β0 ) × ( β2 + α3(k + n − 2) ) P nk−1(ε0;n) , (91) where P n−1(ε0;n) = 0, and P n0 (ε0;n) = 1 for the non-negative integer n. Expanding ∆k+1 with respect to the last column, it is clear that the determinant (89) satisfies a three-term recurrence relation { ∆k+1 = (Sk + ε0;n) ∆k − γk Tk ∆k−1, ∆0 = 1, ∆−1 = 0, k = 0, 1, . . . , n , (92) that allow to compute the determinant ∆k recursively in terms of lower-order determinants. We now show, by induction on k, that ∆k+1 = Pk+1(ε0;n). (93) For k = 0, we find by (89) that ∆1 = (S0 + ε0;n) where the right hand side equals to P n1 (ε0;n) using (91). Next, suppose that ∆j = Pj (ε0;n), for j = 0, 1, 2, · · · , k, then from (91) P nk+1(ε0;n) = ( Sk + ε0;n ) P nk (ε0;n) − γk Tk P n k−1(ε0;n) = ( Sk + ε0;n ) ∆k − γk Tk ∆k−1 = ∆k+1 and the induction step is reached. These results can be represented by the graphical representation (Figure 2). Some of the mathematical properties of the finite sequence of polynomials {P nk (ε0;n)} n k=0 will be explored in later sections. 181 Nasser Saad Acta Polytechnica ∆k+1 = det S0 T1 0 ... 0 γ1 S1 T2 ... 0 0 γ2 S2 ... 0 · · · · · · · · · . . . ... 0 0 0 · · · Sn     P n1 P n2 P n3 P nn Figure 2. A demonstration of how the polynomials {P nk (ε0;n)} n k=0 may be obtained from the (k + 1)-determinant ∆k+1 for k = 0, 1, 2, . . . , n . Remark 4.8. For α3 + α2 + α1 = 0, the canonical form of Heun’s equation can be deduced from (65) by means of the following substitutions: y′′(r) +   β0+β1+β3α3−α1 r − 1 + β0 α1 r + α23β0+α1α3β1+α 2 1β2 α1α3(α1−α3) (r − α1 α3 )   y′(r) + ε1α3 r + ε0α3 r (r − 1) ( r − α1 α3 ) y(r) = 0. (94) or, simply in the standard form as y′′(r) + ( γ r + δ r − 1 + ε r − b ) y′(r) + α β r − q r(r − 1)(r − b) y(r) = 0, (95) where γ δ ε α β q b ⇓ ⇓ ⇓ ⇓ ⇓ ⇓ β0 α1 β2 + β1 + β0 α3 − α1 β2α 2 1 + β1α1α3 + β0α23 α3α1(α1 − α3) β2 + (n − 1)α3 α3 −n − ε0 α3 α1 α3 ⇑ ⇑ ⇑ ⇑ ⇑ ⇑ γ δ ε β α q b where, in either case, it follows γ + δ + ε = α + β + 1 that ensures the regularity of the singular point ∞. With these parameters, the Sturm-Liouville form of the differential equation (65) is − d dr ( rγ (r − 1)δ (r − b)ε dy dr ) + α β rγ (r − 1)δ−1(r − b)ε−1y = q rγ−1(r − 1)δ−1(r − b)ε−1 y (96) where, for b ≥ 1, γ ≥ 0, δ ≥ 1, r ∈ (0, 1) . Corollary 4.9. The second-order linear differential equation r2(α3 r + α2)y′′(r) + r (β2 r + β1) y′ + (−(n(n − 1) α3 + n β2) r + ε0) y = 0, (97) where r ∈ (−α2/α3, 0) if α2α3 > 0 or r ∈ (0, α2/α3) if α2α3 < 0, has a polynomial solution of degree n subject to   n∏ k=0 (ε0 + k((k − 1)α2 + β1) = 0 =⇒ ε0 = −n (n − 1) α2 − n β1, n = 0, 1, 2, · · · . (98) 182 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . In particular, the differential equation r2(α3r + α2) y′′(r) + r(β2 r + β1) y′(r) − ( (n (n − 1) α3 + n β2) r + n (n − 1) α2 + n β1 ) y(r) = 0, (99) has the polynomial solutions yn(r) = rn, n = 0, 1, 2, . . . . (100) Proof. Follows immediately from Theorem 4.7 with α1 = β0 = 0. 5. Mathematical properties of the orthogonal polynomials {Pk(ε0)}∞k=0 As pointed out by Theorem 4.1, in the neighbourhood of the singular point r = 0 with an indicial exponent root zero, the series solution of the differential equation with four singular points, see (65), r ( α1 + α2 r + α3r2 ) y′′ + ( β0 + β1 r + β2 r2 ) y′ + ( ε0 + ε1 r ) y = 0 . can be written as y(r) = ∞∑ k=0 (−1)k Pk(ε0) k! αk1 ( β0 α1 ) k rk , (101) where the infinite sequence of polynomials {Pk(ε0)}∞k=0 in the real variable ε0 satisfies the three-term recurrence relation Pk+1(ε0) = (ε0 − Ak)Pk(ε0) − BkPk−1(ε0), (102) initiated with P−1(ε0) = 0, P0(ε0) = 1, k = 1, 2, 3, · · · . where Ak = −k(k − 1)α2 − kβ1, Bk = k ( (k − 1)α1 + β0 )( (k − 1)((k − 2)α3 + β2) + ε1 ) . For Ak, Bk ∈ R and if Bk > 0, then according to Favard Theorem [25], see also [26, Theorem 2.14], there exists a positive Borel measure µ such that {Pk}∞k=0 is orthogonal with respect to the inner product ⟨Pk, Pk′ ⟩ = ∫ R Pk(ε0)Pp′ (ε0)dµ (103) such that ∫ R Pk(ε0)Pk′ (ε0)dµ = pkpk′ δkk′ , ∫ R dµ = 1, (104) where δkk′ is the Kronecker symbol. In particular,∫ R εk0 Pk′ (ε0)dµ = 0 f or all 0 < k < k ′. (105) The norm pk can be found using the recurrence relations (102) by multiplying with εk−10 and taking the integral over ε0 with respect to µ that yields ∫ R εk0 Pk(ε0)dµ = Bk ∫ R εk−10 Pk−1(ε0)dµ = Bk Bk−1 ∫ R εk−20 Pk−2(ε0)dµ = · · · =   k∏ j=2 Bj  ∫ R dµ (106) 183 Nasser Saad Acta Polytechnica and ∫ R Pk(ε0)Pk′ (ε0) dµ = k! (α1 α3)k β0ε1 ( β0 α1 ) k × ( −α3 + β2 − √ (α3 − β2)2 − 4α3ε1 2α3 ) k × ( −α3 + β2 + √ (α3 − β2)2 − 4α3ε1 2α3 ) k δkk′ . (107) Using the recurrence relation (102), it also follows that ∫ R ε0[Pk(ε0)]2dµ = − k((k − 1)α2 + β1)k! (α1 α3)k β0ε1 ( β0 α1 ) k ( −α3 + β2 − √ (α3 − β2)2 − 4α3ε1 2α3 ) k × ( −α3 + β2 + √ (α3 − β2)2 − 4α3ε1 2α3 ) k . (108) Further, for k = 0, 1, 2, · · · ,∫ R ε0Pk+1(ε0)Pk(ε0)dµ = (k + 1)! (α1 α3)k+1 β0ε1 × ( β0 α1 ) k+1 ( −α3 + β2 − √ (α3 − β2)2 − 4α3ε1 2α3 ) k+1 ( −α3 + β2 + √ (α3 − β2)2 − 4α3ε1 2α3 ) k+1 . (109) Other integrals can be evaluated similarly, for example ∫ R[ε0Pk(ε0)] 2dµ can be evaluated by multiplying (102) by ε0Pk(ε0) and integrate with respect to the measure µ using (107), (108), and (109) and we continue similarly for ∫ R εm0 [Pk(ε0)] 2dµ, m = 0, 1, 2, · · · . The recurrence relations (102) for x = ε0 and y = ε′0 read Pk+1(x) = ( x − Ak ) P nk (x) − BkP n k−1(x) , Pk+1(y) = ( y − Ak ) P nk (y) − BkP n k−1(y), respectively. By multiplying the first by Pk(y) and the second by Pk(x) and subtracting, the resulting equation becomes (x − y)Pk(y)Pk(x) = Qk+1(x, y) − Bk Qk(x, y) (110) where Qk+1(x, y) = Pk+1(x)Pk(y) − Pk(x)Pk+1(y). Thus, recursively over k, we have (x − y)Pk(x)Pk(y) = Qk+1(x, y) − Bk Qk(x, y) (x − y)Pk−1(x)Pk−1(y) = Qk(x, y) − Bk−1 Qk−1(x, y) ... (x − y)P n0 (x)P n 0 (y) = Q1(x, y) , from which it is straightforward to obtain (x − y) [ Pk(x)Pk(y) + BkPk−1(x)Pk−1(y) + Bk Bk−1Pk−2(x)Pk−2(y) + Bk Bk−1Bk−2Pk−3(x)Pk−3(y) + · · · + λk+1λkλk−1λk−2 . . . λ2P0(x)P0(y) ] = Qk+1(ε0, y) . Dividing both sides by (x − y)Bk Bk−1Bk−2 . . . B2 and summing over k results in k∑ j=0 Pj (x)Pj (y) Bj Bj−1Bj−2 . . . B2 = (Bk Bk−1Bk−2 . . . B2)−1 × Pk+1(x)Pk(y) − P nk (x)Pk+1(y) x − y . 184 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . (101) then follows using Bk Bk−1Bk−2 . . . B2 = k∏ i=2 Bi = k! (α1 α3)k β0ε1 ( β0 α1 ) k × ( −α3 + β2 − √ (α3 − β2)2 − 4α3ε1 2α3 ) k × ( −α3 + β2 + √ (α3 − β2)2 − 4α3ε1 2α3 ) k and finally, we have, for k ≥ 0, Christoffel-Darboux identities: k∑ j=0 Pj (x)Pj (y) j! (α1 α3)j ( β0 α1 ) j (ξ+)j (ξ−)j = Pk+1(x)Pk(y) − Pk(x)Pk+1(y) k! (α1 α3)k ( β0 α1 ) k (ξ+)k (ξ−)k (x − y) , (111) where ξ± = −α3 + β2 ± √ (α3 − β2)2 − 4α3ε1 2α3 and by evaluating the limit of both sides as y → x, its confluent form k∑ j=0 [Pj (x)]2 j! (α1 α3)j (ξ+)j (ξ−)j = P ′k+1(x)Pk(x) − P ′ k(x)Pk+1(x) k! (α1 α3)k ( β0 α1 ) k (ξ+)k (ξ−)k (112) follows. Here, the prime refers to the derivative with respect to the variable x. As a direct consequence of the Christoffel-Darboux formula (112), all the zeros of the n-degree polynomial Pn(ε) are simple. To prove that they are also real, we note that the recurrence relation (102) can be written in a matrix form as x   P0(x) P1(x) P2(x) ... Pk−1(x)   =   A0 1 0 · · · 0 0 B1 A1 1 · · · 0 0 0 B2 A2 · · · 0 0 ... ... ... . . . ... ... 0 0 0 · · · Bk−1 Ak−1     P0(x) P1(x) P2(x) ... Pk−1(x)   + Pk(x)   0 0 0 ... 1   (113) Thus, if xi is a zero of Pk(x), it is an eigenvalue of the given tridiagonal matrix. Since, by the hypothesis of (102), Bk > 0 for all k ≥ 1, the results of Arscott [24] confirm that (i) the zeros of Pk−1(x) and Pk(x) interlace – that is, between two consecutive zeros of either polynomial lies precisely one zero of the other (ii) at the zeros of Pk(x) the values of Pk−1(x) are alternately positive and negative, (iii) all the zeros of Pk(x) – i.e. all the eigenvalues of tridiagonal matrix are real and different. 6. Mathematical properties of the finite orthogonal polynomials {P nk (ε0)}nk=0 In this section, we shall study some of the mathematical properties of the orthogonal polynomials {P nk (ε0;n)} n k=0. First, the zeros of the polynomial generated by the aforementioned determinant are all simple. This fact can be confirmed by establishing the Christoffel-Darboux formula. Denote x = ε0;k and y = ε0;k′ , where k ̸= k′ and k, k′ = 0, 1, 2, · · · , n − 1: For x ̸= y k∑ j=0 P nj (x)P n j (y) j!(α1α3)j (−n)j ( β0 α1 ) j ( β2 α3 + n − 1 ) j = P nk+1(x)P n k (y) − P n k (x)P n k+1(y) k!(α1α3)k(−n)k ( β0 α1 ) k ( β2 α3 + n − 1 ) k (x − y) , (114) while, for the limit y → x, k∑ j=0 ( P nj (x) )2 j!(α1α3)j (−n)j ( β0 α1 ) j ( β2 α3 + n − 1 ) j = [P nk+1(x)] ′P nk (x) − [P n k (x)] ′P nk+1(x) k!(α1α3)k(−n)k ( β0 α1 ) k ( β2 α3 + n − 1 ) k . (115) 185 Nasser Saad Acta Polytechnica Here, the prime refers to the derivative with respect to the variable x. If x = xk is a zero of the polynomial P nk (x) with multiplicity > 1, then P n k (xk) = 0 and (115) yields the contradiction 0 < k−1∑ j=0 ( P nj (xi) )2 j!(α1α3)j (−n)j ( β0 α1 ) j ( β2 α3 + n − 1 ) j = 0, (116) and the zeros of the polynomial P nk (x), k = 1, 2, · · · n are distinct. 6.1. Norms of the orthogonal polynomials Denote ε0;n = x, the general theory of orthogonal polynomials [27] guarantees that the finite sequence of polynomials {Pk(x)}nk=0 form a set of orthogonal polynomials for each n. This implies the existence of a certain weight function, W(x), which can be normalized as∫ dW = 1 , (117) for which ∫ Pk(x)Pk′ (x)dW = pk pk′ δkk′ , 0 ≤ k, k′ ≤ n , (118) where pk denotes the norms of polynomials Pk(x). These norms can be found from the recurrence relations (36) by multiplying with xk−1W(x) and taking the integral over x yields the recurrence formula∫ x k P nk (x) W(x) dx = −k(n − k + 1) ( (k − 1)α1 + β0 ) × ( β2 + α3(k + n − 2) )∫ x k−1P nk−1(x)W(x)dx , (119) and thus ∫ P nk (x) x k W(x) dx = k! (α1α3)k (−n)k ( β0 α1 ) k ( β2 α3 + n − 1 ) k . (120) From which it follows∫ [P nk (x)] 2W(x)dx = p2k = k! (α1α3) k (−n)k ( β0 α1 ) k ( β2 α3 + n − 1 ) k (121) for all 0 ≤ k ≤ n. Because of the Pochhammer identity (−n)k = 0 for k > n, it follows from (71) that the norms of all polynomials P nk (x) with k ≥ n + 1 vanish. Thus pk = 0 , k ≥ n + 1 . (122) We may also note, using the recurrence relation, that∫ x[Pk(x)]2W(x)dx = −k((k − 1)α2 + β1) k! (α1α3)k (−n)k ( β0 α1 ) k ( β2 α3 + n − 1 ) k . (123) 6.2. The zeros of the polynomials {P nk (ε0;n)} n k=0 One of the important properties of the polynomials P nn+1(ε0;n) concerns their zeros. An argument provided by Arscott [24] proves that if the product (γk · Tk) > 0 for all k = 1, 2, . . . , n, then the polynomials that satisfy the tri-diagonal determinant (67) are real and simple. Let us denote that the roots of the polynomials P nn+1(ε0;n) = 0 by εℓ0;n, ℓ = 0, 1, . . . , n such that P nn+1(ε ℓ 0;n) = 0 , (124) where ε00;n < ε 1 0;n < · · · < ε n 0;n . 186 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . In particular, since P nn+1(ε0;n) is of degree n + 1 and all the roots are simple and different, it follows that P nn+1(ε0;n) = n∏ ℓ=0 (ε0;n − εℓ0;n) . (125) The ‘discrete’ weight function W can be computed numerically [28] using (118), (119) and (125) for the given n. Denote Pℓ(ε0) = P nℓ (ε0), and let the roots of P n n+1(ε0;n) = 0 be ε j 0;n arranged in ascending order for j = 0, 1, 2, · · · , n. The weights Wj , j = 0, 1, · · · , n, for the orthogonal polynomials {P nℓ (ε0;n)} n k=0 can be computed by solving the linear system n∑ j=0 Wj P nℓ (ε j 0;n) = 0 (126) for ℓ = 0, 1, · · · , n. 6.3. Factorization property Another interesting property of the polynomials {P nk (x)} n k=0, aside from being an orthogonal sequence, is that when the parameter n takes positive integer values, the polynomials exhibit a factorization property. Clearly, the factorization occurs because the third term in the recursion relation (36) vanishes when k = n + 1, so that all subsequent polynomials have a common factor P nn+1(ζ) called a critical polynomial. Indeed, all the polynomials P nk+n+1(x), beyond the critical polynomial P n n+1(x) are factored into the product P nk+n+1(x) = Q n k (x) P n n+1(x), k = 0, 1, . . . , (127) where the sequence {Qnk (x)} are polynomials of degree k = 0, 1, . . . . Interestingly, the quotient polynomials {Qnk (x)} ∞ k=0 form an infinite sequence of orthogonal polynomials. To prove this claim, we substitute (128) into (36) and re-index the polynomials to eliminate the common factor P nn+1(ζ) from both sides. The recurrence relation (36) then reduces to a three-term recurrence relation for the polynomials {Qnk (ζ)}k≥0 that reads Qnk (x) = ( (k + n)(k + n − 1)α2 + (k + n)β1 + x ) Qnk−1(x) − (k + n)(k − 1) ( (k + n − 1)α1 + β0 ) × ( β2 + α3(k + 2n − 2) ) Qnk−2(x) , (128) where Qn−1(ζ) = 0, and Qn0 (ζ) = 1. Hence, the quotient polynomials Qnk (ζ) also form a new sequence of orthogonal polynomials for each value of n. For example, if n = 2, the critical polynomial is P 23 (x) = x 3 + (2α2 + 3β1)x2 + 2 ( (3α3 + 2β2)β0 + β1(α2 + β1) + α1(2α3 + β2) ) x + 4β0(α2 + β1)(α3 + β2) . (129) and P 24 (x) = ( x + 6α2 + 3β1 ) P 23 (x) , P 25 (x) = ( x 2 + (18α2 + 7β1)x − 4 ( (3α1 + β0)(4α3 + β2) − 3(2α2 + β1)(3α2 + β1) )) P 23 (x) , P 26 (x) = ( x 3 + (38α2 + 12β1)x2 + ( 432α22 + 290α2β1 + 47β 2 1 − 2(124α1α3 + 33α3β0 + 26α1β2 + 7β0β2) ) x − 10 ( β1(84α1α3 + 23α3β0 − 6β21 + 18α1β2 + 5β0β2) + 2α2 ( 31α3β0 − 27β21 + 7β0β2 + 12α1(9α3 + 2β2) ) − 144α32 − 156α 2 2β1 )) P 23 (x) , ... from which we have Q0(x) = 1 , Q1(x) = x + 6α2 + 3β1 , Q2(x) = x2 + (18α2 + 7β1)x − 4 ( (3α1 + β0)(4α3 + β2) − 3(2α2 + β1)(3α2 + β1) ) , Q3(x) = x3 + (38α2 + 12β1)x2 + ( 432α22 + 290α2β1 + 47β 2 1 − 2(124α1α3 + 33α3β0 + 26α1β2 + 7β0β2) ) x − 10 ( β1(84α1α3 + 23α3β0 − 6β21 + 18α1β2 + 5β0β2) + 2α2 ( 31α3β0 − 27β21 + 7β0β2 + 12α1(9α3 + 2β2) ) − 144α32 − 156α 2 2β1 ) , ... 187 Nasser Saad Acta Polytechnica and so on. The Christoffel-Darboux formula for this infinite sequence of orthogonal polynomials reads k∑ j=0 Qnj (x)Qnj (y) j! (α1α3)j (n + 2)j ( β0 α1 + n + 1 ) j ( β2 α3 + 2n ) j = Qnk+1(x)Q n k (y) − Q n k (x)Q n k+1(y) k! (α1α3)k(n + 2)k ( β0 α1 + n + 1 ) k ( β2 α3 + 2n ) k (x − y) , (130) and as y → x k∑ j=0 ( Qnj (x) )2 j! (α1α3)j (n + 2)j ( β0 α1 + n + 1 ) j ( β2 α3 + 2n ) j = [Qnk+1(x)] ′Qnk (x) − [Q n k (x)] ′Qnk+1(x) k! (α1α3)k(n + 2)k ( β0 α1 + n + 1 ) k ( β2 α3 + 2n ) k . (131) Theorem 6.1. The norms of all polynomials Qnk (ξ) are given by GQk = k! (α1α3) k(n + 2)k ( β0 α1 + n + 1 ) k ( β2 α3 + 2n ) k . (132) Proof. The proof follows by multiplying the recurrence relation (128) by xk−2ρ(x), with the normalized weight function ∫ ρ(x)dx = 1, and integrating over x. This procedure yields a two-term recurrence relation GQk = k (k + n + 1) ((k + n)α1 + β0) (β2 + α3(k + 2n − 1))) G Q k−1 , where GQk = ∫ |Qnk (x)| 2ρ(z)dz = ∫ xkQnk (x)ρ(x)dx with a solution given by (132). We see that, in general, the norm of the polynomials Qnk (x) does not vanish. Acknowledgements Partial financial support of this work under grant number GP249507 from the Natural Sciences and Engineering Research Council of Canada is gratefully acknowledged. References [1] A. F. Nikiforov, V. B. Uvarov. Special functions of mathematical physics. Birkhäuser Verlag, Basel, 1988. https://doi.org/10.1007/978-1-4757-1595-8. [2] E. J. Routh. On some properties of certain solutions of a differential equation of the second order. Proceedings of the London Mathematical Society 16:245–262, 1884/85. https://doi.org/10.1112/plms/s1-16.1.245. [3] N. Saad, R. L. Hall, H. Ciftci. Criterion for polynomial solutions to a class of linear differential equations of second order. Journal of Physics A: Mathematical and General 39(43):13445–13454, 2006. https://doi.org/10.1088/0305-4470/39/43/004. [4] N. Saad, R. L. Hall, V. A. Trenton. Polynomial solutions for a class of second-order linear differential equations. Applied Mathematics and Computation 226:615–634, 2014. https://doi.org/10.1016/j.amc.2013.10.056. [5] A. Ronveaux (ed.). Heun’s differential equations. Oxford University Press, New York, 1995. [6] S. Y. Slavyanov, W. Lay. Special functions: a unified theory based on singularities. Oxford University Press, Oxford, 2000. ISBN 0-19-850573-6. [7] A. Decarreau, M.-C. Dumont-Lepage, P. Maroni, et al. Formes canoniques des équations confluentes de l’équation de Heun. Annales de la Societé Scientifique de Bruxelles Série I Sciences Mathématiques, Astronomiques et Physiques 92(1-2):53–78, 1978. [8] A. Decarreau, P. Maroni, A. Robert. Sur les équations confluentes de l’équation de Heun. Annales de la Societé Scientifique de Bruxelles Série I Sciences Mathématiques, Astronomiques et Physiques 92(3):151–189, 1978. [9] K. Heun. Zur Theorie der Riemann’schen Functionen zweiter Ordnung mit vier Verzweigungspunkten. Mathematische Annalen 33(2):161–179, 1888. https://doi.org/10.1007/BF01443849. [10] F. Beukers, A. van der Waall. Lamé equations with algebraic solutions. Journal of Differential Equations 197(1):1–25, 2004. https://doi.org/10.1016/j.jde.2003.10.017. [11] A. Turbiner. On polynomial solutions of differential equations. Journal of Mathematical Physics 33(12):3989–3993, 1992. https://doi.org/10.1063/1.529848. [12] R. V. Craster, V. H. Hoàng. Applications of Fuchsian differential equations to free boundary problems. Proceedings of the Royal Society A Mathematical, Physical and Engineering Sciences 454(1972):1241–1252, 1998. https://doi.org/10.1098/rspa.1998.0204. 188 https://doi.org/10.1007/978-1-4757-1595-8 https://doi.org/10.1112/plms/s1-16.1.245 https://doi.org/10.1088/0305-4470/39/43/004 https://doi.org/10.1016/j.amc.2013.10.056 https://doi.org/10.1007/BF01443849 https://doi.org/10.1016/j.jde.2003.10.017 https://doi.org/10.1063/1.529848 https://doi.org/10.1098/rspa.1998.0204 vol. 62 no. 1/2022 On Generalized Heun equation with some mathematical . . . [13] P. P. Fiziev. The Heun functions as a modern powerful tool for research in different scientific domains, 2015. arXiv:1512.04025v1. [14] Mathematical physics. In U. Camcı, I. Semiz (eds.), Proceedings of the 13th Regional Conference held in Antalya, October 27–31, 2010, pp. 23–39. World Scientific Publishing Co. Pte. Ltd., Hackensack, NJ, 2013. https://doi.org/10.1142/8566. [15] H. Ciftci, R. L. Hall, N. Saad, E. Dogu. Physical applications of second-order linear differential equations that admit polynomial solutions. Journal of Physics A: Mathematical and Theoretical 43(41):415206, 14, 2010. https://doi.org/10.1088/1751-8113/43/41/415206. [16] Y.-Z. Zhang. Exact polynomial solutions of second order differential equations and their applications. Journal of Physics A: Mathematical and Theoretical 45(6):065206, 2012. https://doi.org/10.1088/1751-8113/45/6/065206. [17] B.-H. Chen, Y. Wu, Q.-T. Xie. Heun functions and quasi-exactly solvable double-well potentials. Journal of Physics A: Mathematical and Theoretical 46(3):035301, 2013. https://doi.org/10.1088/1751-8113/46/3/035301. [18] F. Caruso, J. Martins, V. Oguri. Solving a two-electron quantum dot model in terms of polynomial solutions of a Biconfluent Heun equation. Annals of Physics 347:130–140, 2014. https://doi.org/10.1016/j.aop.2014.04.023. [19] A. V. Turbiner. One-dimensional quasi-exactly solvable Schrödinger equations. Physics Reports A Review Section of Physics Letters 642:1–71, 2016. https://doi.org/10.1016/j.physrep.2016.06.002. [20] H. Karayer, D. Demirhan, F. Büyükkılıç. Solution of Schrödinger equation for two different potentials using extended Nikiforov-Uvarov method and polynomial solutions of biconfluent Heun equation. Journal of Mathematical Physics 59(5):053501, 2018. https://doi.org/10.1063/1.5022008. [21] H. Ciftci, R. L. Hall, N. Saad. Asymptotic iteration method for eigenvalue problems. Journal of Physics A: Mathematical and General 36(47):11807–11816, 2003. https://doi.org/10.1088/0305-4470/36/47/008. [22] H. Scheffé. Linear differential equations with two-term recurrence formulas. Journal of Mathematics and Physics 21(1-4):240–249, 1942. https://doi.org/10.1002/sapm1942211240. [23] R. S. Irving. Integers, polynomials, and rings. Springer-Verlag, New York, 2004. https://doi.org/10.1007/b97633. [24] F. M. Arscott. Latent roots of tri-diagonal matrices. Edinburgh Mathematical Notes 44:5–7, 1961. https://doi.org/10.1017/S095018430000330X. [25] J. Favard. Sur les polynômes de Tchebicheff. Comptes Rendus Hebdomadaires des Séances de l’Académie des Sciences, Paris 200:2052–2053, 1935. [26] F. Marcellán, R. Álvarez Nodarse. On the “Favard theorem” and its extensions. Journal of Computational and Applied Mathematics 127(1-2):231–254, 2001. https://doi.org/10.1016/S0377-0427(00)00497-0. [27] M. E. H. Ismail. Classical and quantum orthogonal polynomials in one variable. Cambridge University Press, Cambridge, 2009. [28] A. Krajewska, A. Ushveridze, Z. Walczak. Bender-Dunne orthogonal polynomials general theory. Modern Physics Letters A 12(16):1131–1144, 1997. https://doi.org/10.1142/S0217732397001163. 189 http://arxiv.org/abs/1512.04025v1 https://doi.org/10.1142/8566 https://doi.org/10.1088/1751-8113/43/41/415206 https://doi.org/10.1088/1751-8113/45/6/065206 https://doi.org/10.1088/1751-8113/46/3/035301 https://doi.org/10.1016/j.aop.2014.04.023 https://doi.org/10.1016/j.physrep.2016.06.002 https://doi.org/10.1063/1.5022008 https://doi.org/10.1088/0305-4470/36/47/008 https://doi.org/10.1002/sapm1942211240 https://doi.org/10.1007/b97633 https://doi.org/10.1017/S095018430000330X https://doi.org/10.1016/S0377-0427(00)00497-0 https://doi.org/10.1142/S0217732397001163 Acta Polytechnica 62(1):165–189, 2022 1 Introduction 2 Elementary observations 3 The solutions in the neighbourhood of an ordinary point 3.1 Series solutions 3.2 Polynomial solutions 4 The solutions in the neighbourhood of a singular point 4.1 Series Solution and infinite sequence of orthogonal polynomials {Pk (0)}k=0 4.2 Polynomial Solution and finite sequence of orthogonal polynomials 5 Mathematical properties of the orthogonal polynomials {Pk (0)}k=0 6 Mathematical properties of the finite orthogonal polynomials {Pkn(0)}k=0n 6.1 Norms of the orthogonal polynomials 6.2 The zeros of the polynomials {Pkn(0;n)}k=0n 6.3 Factorization property Acknowledgements References