1 Superelliptic affine Lie algebras

Let R be a ring of the form $\mathbb {C}[t,t^{-1}, u]$ , where $u^m\in \mathbb {C}[t]$ and $m\geq 2$ . Thus, R has a basis consisting of $t^i$ , $t^iu$ , $t^iu^2\dots ,t^iu^{m-1}$ for $i\in \mathbb {Z}$ . We will assume that $P(t)=\sum ^{D}_{i=0}a_it^i$ , where $a_D=1$ and $a_0,a_1$ are not both 0 and that $P(t)$ has no multiple roots. The equation $u^m=P(t)$ defines a superelliptic curve. Set $R^i$ for $\mathbb {C}[t,t^{-1}]u^i$ . Then $R=R^0\oplus R^1\oplus \cdots \oplus R^{m-1}$ is a $\mathbb {Z}/m\mathbb {Z}$ -grading.

Let $\mathfrak {g}$ be a simple finite-dimensional complex Lie algebra. The Lie algebra ${\mathcal {G}}=\mathfrak {g} \otimes R$ is an example of superelliptic loop algebras. The $\mathbb {Z}/m\mathbb {Z}$ -grading induces the structure of a $\mathbb {Z}/m\mathbb {Z}$ -graded Lie algebra on ${\mathcal {G}}$ by setting ${\mathcal {G}}^i=\mathfrak {g}\otimes R^i$ ( $i=0,1,2,\dots , m-1$ ).

Let $\hat {\mathcal {G}}={\mathcal {G}}\oplus C$ be the universal central extension of ${\mathcal {G}}$ , where C is the center of $\hat {\mathcal {G}}$ . By [Reference Kassel14], the center C is linearly isomorphic to $\Omega _R^1/dR$ , the space of Kähler differentials of R modulo the exact differentials. Our goal is to determine a basis for $\Omega _R^1/dR$ .

Let $F=R\otimes R$ be the left R-module with action $f(g\otimes h)=fg\otimes h$ for $f, g, h\in R$ . Let K be the submodule generated by the elements $1\otimes fg-f\otimes g-g\otimes f$ . Then $\Omega _R^1=F/K$ is the module of Kähler differentials. We denote the element $f\otimes g+K$ of $\Omega _R^1$ by $fdg$ . We define a map $d:R\rightarrow \Omega _R^1$ by $d(f)=df=1\otimes f+K$ , and we denote the coset of $fdg$ modulo $dR$ by $\overline {fdg}$ . We have

(1.1) $$ \begin{align} [x\otimes f, y\otimes g]=[xy]\otimes fg+(x,y)\overline{fdg},&& [x\otimes f,\omega]=0,&& [\omega, \omega']=0, \end{align} $$

where $x,y\in \mathfrak {g}$ , $f,g\in R$ and $\omega , \omega ' \in \Omega _R^1/dR$ ; here, $(x,y)$ denotes the Killing form on $\mathfrak {g}$ .

The elements $t^iu^k\otimes t^ju^l$ , with $i,j\in \mathbb {Z}$ and $k,l\in \{0,1,\dots , m-1\}$ form a basis of $R\otimes R$ . The following result is fundamental to the description of the universal central extension for $R=\mathbb {C}[t,t^{-1},u|\,\, u^m=P(t)]$ .

Theorem 1.1 ([Reference Albino dos Santos2], Theorem 2.4).

The following elements form a basis of $\Omega _R^1/dR$ :

$$ \begin{align*}\overline{t^{-1}dt}, \, \overline{t^{-1}u^ldt}, \, \dots, \, \overline{t^{-D}u^ldt},\end{align*} $$

where $l\in \{1,2,\dots ,m-1\}$ and we omit $\overline {t^{-D}u^ldt}$ if $a_0=0$ .

We have the following relation between the basis elements:

Proposition 1.2 ([Reference Cox and Futorny7], Lemma 2.0.2).

If $u^m=P(t)$ and $R=[t,t^{-1},u|\,\, u^m=P(t)\,\, ]$ , then in $\Omega _R^1/dR$ , one has

(1.2) $$ \begin{align} ((m+1)n+im)t^{n+i-1}udt\equiv\sum_{j=0}^{n-1}((m+1)j+mi)a_jt^{i+j-1}udt\quad \mod{dR}. \end{align} $$

This allows to describe the universal central extension of the superelliptic algebra.

Let us define the sequence of polynomials in $D+3$ parameters $P_{m,n,l}(a_{0},a_{1},\dots ,a_{D-1}):=P_{n}$ for $n\geq -D$ , $m, l\in \mathbb {Z}_+$ and $a_{0},a_{1},\dots ,a_{D-1}\in \mathbb {C}$ as follows:

(1.3) $$ \begin{align} (Dl+(1+n)m)P_n=\sum_{k=0}^{D-1}\left(-a_k\left(kl+(-D+1+n+k)m\right) \right) P_{-D+n+k}, \end{align} $$

with initial condition

(1.4) $$ \begin{align} P_{-D}=t^{-D}u^ldt, \, P_{-D+1}=t^{-D+1}u^ldt, \, \dots, \, P_{-1}=t^{-1}u^ldt. \end{align} $$

Then $P_n= t^{n}u^ldt$ for all $n\geq 0$ .

If $a_0\neq 0$ , we define the sequence of polynomials in $D+3$ parameters $Q_{m,n,l}(a_{0},a_{1},\dots ,a_{D-1}):=Q_{n}$ for $n\leq -D-1$ , $m, l\in \mathbb {Z}_+$ and $a_{0},a_{1},\dots ,a_{D-1}\in \mathbb {C}$ as follows:

(1.5) $$ \begin{align} (a_0(1+n)m)Q_n=\sum_{k=1}^{D}\left(-a_k\left(kl+(1+n+k)m\right) \right) Q_{n+k} \end{align} $$

with initial conditions

(1.6) $$ \begin{align} Q_{-D}=t^{-D}u^ldt, \, Q_{-D+1}=t^{-D+1}u^ldt, \, \dots, \, Q_{-1}=t^{-1}u^ldt. \end{align} $$

Then $Q_n=t^{n}u^ldt$ for $n\leq -D-1$ .

If $a_0=0$ , we define the sequence of polynomials in $D+3$ parameters $R_{m,n,l}(a_{0},a_{1},\dots ,a_{D-1}):=R_{n}$ for $n\leq -D-1$ , $m, l\in \mathbb {Z}_+$ and $a_{0},a_{1},\dots ,a_{D-1}\in \mathbb {C}$ as follows:

(1.7) $$ \begin{align} a_1(l+(1+n+1)m)R_n=\sum_{k=2}^{D}\left(-a_k\left(kl+(1+n+k)m\right) \right) R_{n+k} \end{align} $$

with initial condition

(1.8) $$ \begin{align} R_{-D+1}=t^{-D+1}u^ldt, \, R_{-D+2}=t^{-D+2}u^ldt, \, \dots, \, R_{-1}=t^{-1}u^ldt. \end{align} $$

We see that $R_n=t^{n}u^ldt$ , $n\leq -D-1$ .

Set

(1.9) $$ \begin{align} \psi_{i,j}= \begin{cases} P_{i+j-1} \textrm{ if } i+j\geq -D+1;\\ Q_{i+j-1} \textrm{ if } i+j\leq -D \textrm{ and } a_0\neq0;\\ R_{i+j-1} \textrm{ if } i+j\leq -D\textrm{ and } a_0=0. \end{cases} \end{align} $$

Then $ t^{i}u^ld(t^j) = j \psi _{i,j}$ [[Reference Albino dos Santos2], Proposition 3.4].

We have the following description of the Lie algebra structure on $\hat {\mathcal {G}}$ . Set

(1.10) $$ \begin{align} \omega_0=\overline{t^{-1}dt} \textrm{ and } \omega_{ij}=\overline{t^{i}u^jdt} \textrm{ for } j\neq 0. \end{align} $$

Theorem 1.3 ([Reference Albino dos Santos2], Theorem 3.5).

The superelliptic affine Lie algebra $\hat {\mathcal {G}}$ has a $\mathbb {Z}/m\mathbb {Z}$ -grading in which

$$ \begin{align*} \hat{\mathcal{G}}^0=\mathfrak{g}\otimes\mathbb{C}[t,t^{-1}]\oplus \mathbb{C}\omega_0,&& \hat{\mathcal{G}}^l=\mathfrak{g}\otimes\mathbb{C}[t,t^{-1}]u^l\bigoplus_{n=1}^D \mathbb{C}\omega_{-n,l}. \end{align*} $$

The subalgebra $\hat {\mathcal {G}}^0$ is an untwisted affine Kac-Moody Lie algebra with commutation relations

$$ \begin{align*} [x\otimes t^i,y\otimes t^j]=[x,y]\otimes t^{i+j}\oplus \delta_{i+j,0}(x,y)j\omega_0. \end{align*} $$

The commutation relations in $\mathcal {\hat {G}}$ are

$$ \begin{align*} [x\otimes t^iu^{l_1},y\otimes t^ju^{l_2}]=[x,y]\otimes (t^{i+j}u^{l_1+l_2})+(x,y)\left(\frac{jl_1-il_2}{l_1+l_2}\right) \omega_{i+j-1,l_1+l_2}, \end{align*} $$

if $l_1+l_2\leq m-1$ . When $l_1+l_2>m-1$ ,

$$ \begin{align*} {}[x\otimes t^iu^{l_1},y\otimes t^ju^{l_2}]=[x,y]\otimes \left(\sum_{k=0}^D a_kt^{i+j+k}u^{l_1+l_2-m}\right)+(x,y)\left(\frac{jl_1-il_2}{l_1+l_2}\right) \omega_{i+j-1,l_1+l_2}. \end{align*} $$

The subspace $\mathcal {\hat {G}}^l$ is a $\mathcal {\hat {G}}^0$ -module with

(1.11) $$ \begin{align} [x\otimes t^iu^n,y\otimes t^j]=&[x,y]\otimes (t^{i+j}u^{n+1})+(x,y)j\psi_{i,j}. \end{align} $$

In this paper, we will consider the superelliptic affine Lie algebras with the polynomial relation

$$ \begin{align*}u^m=1-2 c t^2+t^4.\end{align*} $$

In this case, letting $k=i+3$ , the recursion relation (1.2) becomes

$$ \begin{align*} (4+m+k m) \overline{t^k \text{udt}}=-(k-3) m \overline{t^{-4+k} \text{udt}}+2 c (2+(k-1) m) \overline{t^{-2+k} \text{udt}}. \end{align*} $$

Consider a family $P_k:=P_k(c)$ of polynomials in c satisfying the recursion relation

(1.12) $$ \begin{align} (k m+m+4) P_k(c)=2 c ((k-1) m+2) P_{k-2}(c)-(k-3) m P_{k-4}(c) \end{align} $$

for $k\geq 0$ . Set

$$ \begin{align*} P(c,z)=\sum _{k=-4}^{\infty } z^{k+4} P_k(c)=\sum _{k=0}^{\infty } z^{k} P_{k-4}(c). \end{align*} $$

After a straightforward rearrangement of terms, we have

$$ \begin{align*} 0&=\sum _{k=0}^{\infty } (k m+m+4) z^k P_k(c)-(2 c) \sum _{k=0}^{\infty } (k m-m+2) z^k P_{k-2}(c) \\& \quad +\sum _{k=0}^{\infty } (k-3) m z^k P_{k-4}(c)\\&= ((4 - 3 m)z^{-4} + (2 c (-2 + 3 m)) z^{-2} - 3 m) P(c,z) + m \left(z^{-3}-2cz^{-1}+z \right)\frac{d}{\textrm{d}z} P(c,z) \\& \quad + \left(2 c (2-3 m) z^{-2}+(3 m-4)z^{-4}\right)P_{-4}(c) + 2 \left(-2 c (m-1) z^{-1}+(m-2)z^{-3}\right)P_{-3}(c) \\& \quad + (m-4) P_{-2}(c)z^{-2}-4 P_{-1}(c) z^{-1}. \end{align*} $$

Hence, $P(c, z)$ satisfies the differential equation

(1.13) $$ \begin{align} &\frac{d}{\textrm{d}z}P(c,z)+\left(\frac{6 c m z^2-4 c z^2-3 m \left(z^4+1\right)+4}{m z \left(-2 c z^2+z^4+1\right)}\right)P(c,z) \nonumber \\& \quad =\frac{(4 z^3) P_{-1} + ((-m+4)z^2) P_{-2}+ (2\left(2 c (m-1) z^2-m+2\right)z) P_{-3} +({ \left(-2 c (2-3 m) z^2-3 m+4\right)})P_{-4}}{m \left(-2 c z^3+z^4+z\right)}. \end{align} $$

It has an integrating factor

$$ \begin{align*} \mu(c,z)&=\exp{\left\{\int_{\cdot}^z \left(\frac{6 c m w^2-4 c w^2-3 m \left(w^4+1\right)+4}{m w \left(-2 c w^2+w^4+1\right)}\right) \textrm{d}w\right\}} \\ &= \frac{1}{z^{3-\frac{4}{m}}(-2 c z^2+z^4+1)^{1/m}}. \end{align*} $$

Now, we consider different cases depending on the initial conditions.

1.1 Superelliptic case 1

Assume that $P_{-3}=P_{-2}=P_{-1}=0$ and $P_{-4}=1$ , and denote the generating function in this case by $ P_{-4}(c,z)$ . Then

$$ \begin{align*} P_{-4}(c,z)=\sum _{k=-4}^{\infty } P_{-4,k}(c)z^{k+4} =\sum _{k=0}^{\infty }P_{-4,k-4}(c) z^{k} \end{align*} $$

can be written in terms of a superelliptic integral

(1.14) $$ \begin{align} P_{-4}(c,z)&=z^{3-4/m} (1-2 c z^2+z^4)^{1/m}\nonumber\\&\quad \times\lim_{\epsilon\to 0}\left[\int_{\epsilon}^{z} \frac{(4-3 m+2 c(-2+3 m) w^2)}{mw^{4-4/m} (1-2 c w^2+w^4)^{1+1/m}} \textrm{d}w+\epsilon^{4/m-3}+2c(3-2/m)\phi_m(\epsilon)\right], \end{align} $$

where

$$\begin{align*}\phi_m(\epsilon)= \left\{ \begin{array}{cc} \frac{\epsilon^{4/m-1}}{4/m-1} & m\neq 4\\ \log{\epsilon} & m=4. \end{array} \right. \end{align*}$$

Remark 1.15. The correction term inside of the limit $\epsilon \to 0$ in the formula (1.14) is necessary to ensure finiteness of the limit.

1.2 Superelliptic case 2

Suppose now that $P_{-3}=P_{-4}=P_{-1}=0$ and $P_{-2}=1$ . Then we arrive to the generating function

$$ \begin{align*} P_{-2}(c,z)=\sum _{k=-4}^{\infty } P_{-2,k}(c)z^{k+4} =\sum _{k=0}^{\infty }P_{-2,k-4}(c) z^{k}, \end{align*} $$

which is defined in terms of a superelliptic integral

$$ \begin{align*} P_{-2}(c,z)=z^{3-4/m} (1-2 c z^2+z^4)^{1/m}\lim_{\epsilon\to 0}\left[\int_{\epsilon}^{z} \frac{(4 - m) }{m w^{2-4/m} (1-2 c w^2+w^4)^{1+1/m}} \textrm{d}w+\epsilon^{4/m-1}\right]. \end{align*} $$

1.3 Gegenbauer case 3

If we take $P_{-1}(c) = 1$ , and $P_{-2}(c) = P_{-3}(c) = P_{-4}(c) = 0$ and set $P_{-1}(c,z)=\sum _{n\geq 0}P_{-1,n-4}z^n$ , then

$$ \begin{align*} P_{-1}(c,z)=\frac{\left(4 z^{3-\frac{4}{m}} \left(-2 c z^2+z^4+1\right)^{1/m}\right)}{m} \sum_{n=0}^{\infty} \frac{z^{\frac{4}{m}+2n} Q_n^{\left(1+\frac{1}{m}\right)}(c)}{\frac{4}{m}+2 n}, \end{align*} $$

where $Q_n^{\left (1+\frac {1}{m}\right )}(c)$ is the n-th Gegenbauer polynomial.

1.4 Gegenbauer case 4

If we take $P_{-3}(c) = 1$ , and $P_{-1}(c) = P_{-2}(c) = P_{-4}(c) = 0$ and set $P_{-3}(c,z)=\sum _{n\geq 0}P_{-1,n-4}z^n$ , then

$$ \begin{align*} P_{-3}(c,z)=\frac{\left(z^{3-\frac{4}{m}} \left(-2 c z^2+z^4+1\right)^{1/m}\right)}{2} \sum _{n=0}^{\infty } z^{\frac{4}{m}+2 n} Q_n^{\left(1+\frac{1}{m}\right)}(c) \left(\frac{m-2}{z^2 (m (n-1)+2)}-\frac{2 c (m-1)}{m n+2}\right). \end{align*} $$

We will see in the last section that the family of polynomials described in the superelliptic case 1 is an example of associated ultraspherical polynomials.

2 Orthogonal polynomials in superelliptic cases

2.1 Superelliptic type 1

Let us reindex the polynomials $P_{-4,n}$ as follows:

$$ \begin{align*} P_{-4}(c,z)&=z^{3-4/m} (1-2 c z^2+z^4)^{1/m}\\& \quad \times\lim_{\epsilon\to 0}\left[\int_{\epsilon}^{z} \frac{(4-3 m+2 c(-2+3 m) w^2)}{mw^{4-4/m} (1-2 c w^2+w^4)^{1+1/m}} \textrm{d}w+\epsilon^{4/m-3}+2c(3-2/m)\phi_m(\epsilon)\right]\\&= \sum _{k=0}^{\infty }P_{-4,k-4}(c) z^{k}\\&= 1+\frac{3 m }{m+4}z^4+\frac{6 c m (m+2) }{(m+4) (3 m+4)}z^6 +\frac{3 m \left(4 c^2 (m+2) (3 m+2)-m (3 m+4)\right)}{(m+4) (3 m+4) (5 m+4)}z^8\\& \quad +\frac{12 c m (3 m+2) \left(2 c^2 (m+2) (5 m+2)-m (5 m+8)\right)}{(m+4) (3 m+4) (5 m+4) (7 m+4)}z^{10}\\& \quad +3 m\left(\frac{ (16 c^4 (m+2) (3 m+2) (5 m+2) (7 m+2))}{(m+4) (3 m+4) (5 m+4) (7 m+4) (9 m+4)}\right. \\& \quad +\left.\frac{-12 c^2 m (3 m+2) (5 m+2) (7 m+12)+5 m^2 (3 m+4) (7 m+4)}{(m+4) (3 m+4) (5 m+4) (7 m+4) (9 m+4)}\right)z^{12} + O(z^{14}). \end{align*} $$

The first nonzero polynomials are

$$ \begin{align*} P_{-4,0}(c) &= 1,\\P_{-4,4}(c)&=\frac{3 m }{m+4}, \\P_{-4,6}(c)&=\frac{6 c m (m+2) }{(m+4) (3 m+4)}, \\P_{-4,8}(c)&=\frac{3 m \left(4 c^2 (m+2) (3 m+2)-m (3 m+4)\right)}{(m+4) (3 m+4) (5 m+4)}, \\P_{-4,10}(c)&=\frac{12 c m (3 m+2) \left(2 c^2 (m+2) (5 m+2)-m (5 m+8)\right)}{(m+4) (3 m+4) (5 m+4) (7 m+4)}, \end{align*} $$

and $P_n(c)=P_{-4,n}(c)$ satisfies the recursion

(2.1) $$ \begin{align} 0=(2 c ((k-1) m+2)) P_{k+2}(c)-(k m+m+4) P_{k+4}(c)-((k-3) m) P_k(c). \end{align} $$

Our main result is the following theorem which generalizes [[Reference Cox, Futorny and Tirao9], Theorem 3.1.1].

Theorem 2.1. The polynomials $P_n=P_{-4,n}$ satisfy the following fourth-order linear differential equation:

(2.2) $$ \begin{align} 0&=16 \left(c^2-1\right)^2 m^2 P^ {(iv)}_n +160 c \left(c^2-1\right) m^2 P^{\prime\prime\prime}_n \nonumber\\& \quad +\left(8 m (m ((n-6) n-10)+4 (n-6))\right.\nonumber\\& \quad -\left.8 c^2 \left(m^2 (n-10) (n+4)+4 m (n-4)+8\right)\right) P^{\prime\prime}_n \nonumber\\& \quad -24 c \left(m^2 (n-6) n+4 m (n-4)+8\right) P^{\prime}_n \nonumber\\& \quad +(n-4) n (m (n-6)+4) (m (n-2)+4) P_n. \end{align} $$

Proof. See Appendix.

Denote $v:=\ln z$ and, consequently, $\frac {\partial }{\partial v}=z\frac {\partial }{\partial z}$ .

Corollary 2.2. The generating function $P_{-4}=P_{-4}(c,z)$ satisfies the following linear PDE of fourth order:

(2.3) $$ \begin{align} 0&= m^2\left[4(1-c^2)\frac{\partial^2 }{\partial c^2}-12 c\frac{\partial }{\partial c}+\frac{\partial^2 }{\partial v^2} +2 (\frac{2}{m}-3)\frac{\partial }{\partial v}+4(1-\frac{2}{m})\right]^2 P_{-4}\notag\\&\quad +16(m^2-8m)\frac{\partial^2 P_{-4}}{\partial c^2}-16(2-m)^2(c\frac{\partial }{\partial c}+1)^2P_{-4}, \end{align} $$

with $(c,z)\in (-1,1)\times (0,1)$ and with the following boundary conditions:

(2.4) $$ \begin{align} P_{-4}(c,0)=1, \frac{\partial P_{-4}}{\partial z}(c,0)=\frac{\partial^2 P_{-4}}{\partial z^2}(c,0)=\frac{\partial^3 P_{-4}}{\partial z^3}(c,0)=0. \end{align} $$

Proof. It follows from (1.14) that $P_{-4}$ is an analytic function represented by the series $P_{-4}(c,z)=\sum \limits _{n=0}^{\infty } P_{-4,n}(c)z^n$ (in some ball of small radius). Hence, by Theorem 2.1 and some elementary algebraic calculations, $P_{-4}$ satisfies the linear PDE (2.3) in the ball. By the uniqueness of analytic continuation, it satisfies the PDE (2.3) in the domain $(-1,1)\times (0,1)$ . The boundary conditions (2.4) immediately follow from the definition of $P_{-4}$ .

Remark 2.5. Boundary conditions for $P_{-4}$ on other parts of the boundary of the domain $[-1,1]\times [0,1]$ can be explicitly calculated using formula (1.14).

2.2 Superelliptic type 2

Now we consider the second family of polynomials $P_{-2,n}$ . We have

$$ \begin{align*} P_{-2}(c,z)&=\sum _{k=0}^{\infty }P_{-2,k-4}(c) z^{k} \\&=z^{3-\frac{4}{m}} \left(-2 c z^2+z^4+1\right)^{1/m} \lim_{\epsilon\to 0}\left[\int_{\epsilon}^{z} \frac{(4-m) w^{\frac{4}{m}-2}}{m \left(-2 c w^2+w^4+1\right)^{\frac{1}{m}+1}} \, dw +\epsilon^{4/m-1}\right] \\&=z^2-\frac{2 c (m-2)}{m+4}z^4+\frac{z^6 \left(m (m+4)-4 c^2 \left(m^2-4\right)\right)}{(m+4) (3 m+4)}\\& \quad -\frac{4 z^8 \left(c (m+2) \left(2 c^2 (m-2) (3 m+2)-3 m^2\right)\right)}{(m+4) (3 m+4) (5 m+4)}+\textrm{O}(z^9), \end{align*} $$

and the polynomials $P_{-2,n}(c)$ satisfy the following recursion:

(2.6) $$ \begin{align} 0=2 c (km-m+2) P_{-2,k+2}(c)-(k m+m+4) P_{-2,k+4}(c)-(km-3m) P_{-2,k}(c). \end{align} $$

In particular, the first nonzero polynomials of the sequence are

$$ \begin{align*} P_{-2,2}(c)&=1, \\P_{-2,4}(c)&=-\frac{2 c (m-2)}{m+4}, \\P_{-2,6}(c)&=\frac{m (m+4)-4 c^2 \left(m^2-4\right)}{(m+4) (3 m+4)}, \\P_{-2,8}(c)&=-\frac{4 c (m+2) \left(2 c^2 (m-2) (3 m+2)-3 m^2\right)}{(m+4) (3 m+4) (5 m+4)}. \end{align*} $$

Theorem 2.3. The polynomials $P_n=P_{-2,n}(c)$ satisfy the following fourth-order linear differential equation:

(2.7) $$ \begin{align} 0&=16 \left(c^2-1\right)^2 m^2 P_n{}^{(iv)}+160 c \left(c^2-1\right) m^2 P^{\prime\prime\prime}_n\nonumber\\& \quad +\Big[8 m (m (n^2-6n-2)+4 (n-8))\nonumber\\& \quad -\left.8 c^2 \left(m^2 ((n-6) n-32)+4 m (n-6)+8\right)\right] P^{\prime\prime}_n\nonumber\\& \quad -24 c \left(m^2 (n-4) (n-2)+4 m (n-6)+8\right) P^{\prime}_n\nonumber\\& \quad +(n^2-4) (m n-8m+4) (m n-4m+4) P_n. \end{align} $$

Corollary 2.4. Generating function $P_{-2}=P_{-2}(c,z)$ satisfies the following linear PDE of fourth order:

(2.8) $$ \begin{align} 0&=m^2\left[4(1-c^2)\frac{\partial^2 }{\partial c^2}-12 c\frac{\partial }{\partial c}+\frac{\partial^2 }{\partial v^2} +2 (\frac{2}{m}-3)\frac{\partial }{\partial v}-4\right]^2 P_{-2}\notag\\&\quad +144 m^2\frac{\partial^2 P_{-2}}{\partial c^2}-16(3m-2)^2(c\frac{\partial }{\partial c}+1)^2 P_{-2}, \end{align} $$

(where $v=\ln z$ , $\frac {\partial }{\partial v}=z\frac {\partial }{\partial z}$ ) with $(c,z)\in (-1,1)\times (0,1)$ and with the following boundary conditions:

(2.9) $$ \begin{align} \frac{\partial^2 P_{-2}}{\partial z^2}(c,0)=1, P_{-2}(c,0)=\frac{\partial P_{-2}}{\partial z}(c,0)=\frac{\partial^3 P_{-2}}{\partial z^3}(c,0)=0. \end{align} $$

Proof. The proof is analogous to the proof of Corollary 2.2.

Remark 2.10. The linear PDEs (2.3) and (2.8), as long as authors aware, does not seem to fall in any known category of PDEs with explicit solutions.

3 Uniqueness of solutions

Let us check now if the differential equations (2.2) and (2.7) have other polynomial solutions. The argument below was suggested by A. Shyndyapin.

Clearly, if $\{P_n\}$ is a family of polynomials satisfying (2.2) (respectively, (2.7)), then for any choice of nonzero scalars $\lambda _n\in \mathbb {C}$ , the family $\{\lambda _n P_n\}$ also satisfies (2.2) (respectively, (2.7)). We will call such families of polynomials equivalent. Also, we will ignore the families of polynomials obtained by zeroing some members of the family.

Consider first the equation (2.2) and assume that

$$ \begin{align*}Q_n=\sum\limits_{i=0}^{r(n)} \alpha_i(n) c^i\end{align*} $$

is its polynomial solution. We will simply write r for $r(n)$ and $\alpha _i$ for $\alpha _i(n)$ . After the substitution in (2.2), we obtain

$$ \begin{align*} \sum _{i=0}^r &((2 i-n+4) (2 i+n) (m (2 i-n+6)-4) (m (2 i+n-2)+4))\alpha_ic^{i}\\ &+(-8 (i-1) i m \left(m \left(4 i^2-(n-6) n-6\right)-4 (n-6)\right))\alpha_ic^{i-2}\\ &+(16 (i-3) (i-2) (i-1) i m^2)\alpha_ic^{i-4}=0, \end{align*} $$

which is equivalent to the following system of linear equations:

(3.1) $$ \begin{align} &[(2 i-n+4) (2 i+n) (m (2 i-n+6)-4) (m (2 i+n-2)+4)]\alpha_i\nonumber\\& \quad +[-8 (i+1) (i+2) m (m (4 i (i+4)-(n-6) n+10)-4 (n-6))]\alpha_{i+2}\\& \quad +[16 (i+1) (i+2) (i+3) (i+4) m^2]]\alpha_{i+4}=0,\,\,\,\,\, i=0,1, \ldots, r. \nonumber \end{align} $$

Here $\alpha _s$ is assumed to be zero if $s>r$ . The coefficient matrix of the system is upper triangular with

$$ \begin{align*}(2 i-n+4) (2 i+n) (m (2 i-n+6)-4) (m (2 i+n-2)+4), \,\, i=0, \ldots, r\end{align*} $$

on the diagonal. Hence, for each n, the solution of the system is unique and trivial if $(2 i-n+4) (m (2 i-n+6)-4)\neq 0$ for all $i=0,1, \ldots , r$ . Suppose $(2 i-n+4) (m (2 i-n+6)-4)= 0$ . First consider the case $m>4$ . Then we must have $2i-n+4=0$ . From (3.1), we get that $Q_{2k+1}=0$ for all $k>0$ , and $Q_{2k}$ is defined uniquely up to a scalar for each $k>1$ . The degree of $Q_{2k}$ is $\geq k-2$ . Now, if $m=4$ , then either $i=\frac {n}{2}-2$ or $i=\frac {n+1}{2}-3$ . Therefore, $Q_{2k}$ is defined uniquely up to a scalar as a polynomial of degree $\geq k-2$ , $Q_1=Q_3=0$ and $Q_{2k+1}$ is defined uniquely up to a scalar as a polynomial of degree $\geq k-2$ for each $k\geq 2$ .

Consider now the equation (2.7). Arguing as above, we come a linear system on the coefficients of the polynomial solution $Q_n$ . The matrix of this system is again upper triangular with diagonal entries

$$ \begin{align*}\left(4 (i+1)^2-n^2\right) (m (2 i-n+8)-4) (m (2 i+n-4)+4),\,\, i=0, \ldots, r.\end{align*} $$

Assume that $\left (4 (i+1)^2-n^2\right ) (m (2 i-n+8)-4) (m (2 i+n-4)+4)=0$ . If $m>4$ , then $Q_{2k+1}=0$ for all $k> 0$ , and $Q_{2k}$ is defined uniquely up to a scalar for each $k> 1$ , while $Q_2=0$ . The degree of $Q_{2k}$ is $\geq k-2$ . Suppose now that $m=4$ and $n>1$ . Then either $i=\frac {n}{2}-1$ or $i=\frac {n+1}{2}-3$ . Therefore, $Q_{2k}$ is defined uniquely up to a scalar as a polynomial of degree $\geq k-2$ , $Q_3=Q_5=0$ , and $Q_{2k+1}$ is defined uniquely up to a scalar as a polynomial of degree $\geq k-2$ for each $k\geq 3$ . When $m=4$ , $n=1$ and $i=1$ , we also get $0$ . Hence, $Q_1$ is defined uniquely up to a scalar.

We proved the following theorem.

4 Associated ultraspherical polynomials

After shifting the indices back by $4$ , we obtain that both families of the polynomials $P_{-4,n}(c)$ and $P_{-2,n}(c)$ satisfy the recurrence relation

(4.1) $$ \begin{align} (k m+m+4) P_k(c)=2 c (km-m+2)P_{k-2}(c)-(k-3) m P_{k-4}(c), \end{align} $$

where $P_k(c)\in \{P_{-4,k-4}(c), P_{-2,k-4}(c)\}$ . Note that all odd polynomials are zero. Set $k=2(n+1)$ and $q_s=q_s(c):=P_{2s}(c)$ . Then we have

(4.2) $$ \begin{align} 2 c (2 m n+m+2) q_n=(2m n+3m+4) q_{n+1}+m (2 n-1) q_{n-1}. \end{align} $$

For example, we have: $q_{0} =1, q_1=\frac {6 c m (m+2) }{(m+4) (3 m+4)}$ ,

$$ \begin{align*} q_2=\frac{3 m (3 m+2) \left(4 c^2 (m+2) -2m \right)}{(m+4) (3 m+4) (5 m+4)}, \,\,\, q_3=\frac{12cm (3 m+2)(5 m+2) \left(2 c^2 (m+2) -6m \right)}{(m+4) (3 m+4) (5 m+4) (7 m+4)}.\end{align*} $$

We will show that $\{q_n\}$ is a special case of the associated ultraspherical polynomials.

Let c and $\nu $ be two complex numbers. Let $C_n^{(\nu )}(x;c)$ denote the family of associated ultraspherical polynomials with initial conditions $C_{-1}^{(\nu )}(x;c)=0$ and $C_{0}^{(\nu )}(x;c)=1$ (cf. [Reference Bustoz and Ismail5], p. 729). Then they satisfy the following difference equation:

(4.3) $$ \begin{align} 2x(n+\nu+c)C_{n}^{(\nu)}(x;c)=(n+c+1)C_{n+1}^{(\nu)}(x;c)+(2\nu+n+c-1)C_{n-1}^{(\nu)}(x;c). \end{align} $$

Choosing $c=\frac {1}{2}+\frac {2}{m}$ and $\nu =-\frac {1}{m}$ , this equation becomes the recurrence relation (4.2). Hence, we conclude that polynomials $P_{-4,n}$ is a special case of associated ultraspherical polynomials. We immediately have the following:

5 Orthogonality of $P_{-2,n}$

Polynomials $P_{-2,n}$ satisfy the same recurrence relation as polynomials $P_{-4,n}$ but have different initial conditions. Hence, we do not know immediately whether they are associated ultraspherical polynomials and whether they are orthogonal. We will give an independent proof of the orthogonality of these polynomials. Set $q_s(c)=P_{-2,2s}(c)$ . In particular, we have

$$ \begin{align*} q_{1}=1\,\,\, q_{2}=-\frac{2 c (m-2)}{m+4}, \,\,\, q_{3}= \frac{m (m+4)-4 c^2 \left(m^2-4\right)}{(m+4) (3 m+4)}. \end{align*} $$

Now we set $ \overline {q}_n:=q_{n+1}$ , $n\geq -1$ . Then polynomials $\overline {q}_{n}$ have degree n, and (4.2) becomes

(5.1) $$ \begin{align} 2 (2m n+3m+2)c \overline{q}_{n}=(2 m n+m) \overline{q}_{n-1}+(2m n+5m+4) \overline{q}_{n+1}. \end{align} $$

We recall the following result.

Theorem 5.1 ([Reference Cox, Futorny and Tirao9], Theorem 5.0.4).

Let $\{p_n, n \geq 0\}$ be a sequence of polynomials, where $p_n$ has degree n for any n, satisfying the following recursion

(5.2) $$ \begin{align} \chi p_n = a_{n+1}p_{n+1}+b_np_n+c_{n-1}p_{n-1}, && n=0,1,2,\dots \end{align} $$

for some complex numbers $a_{n+1}$ , $b_n$ , $c_{n-1}$ , $n=0, 1, 2, \dots $ and $p_{-1}=0$ . Then $\{p_n, n\geq 0\}$ is a sequence of orthogonal polynomials with respect to some (unique) weight measure if and only if $b_n\in \mathbb {R}$ and $c_n=\overline {a}_n+1\neq 0$ for all $n\geq 0.$

We apply the theorem in the case when

$$ \begin{align*}a_{n+1}=\frac{(2m n+5m+4) }{2 (2m n+3m+2)}, \,\,\, b_{n}=0, \,\,\, c_{n-1}=\frac{(2 m n+m)}{2 (2m n+3m+2)}.\end{align*} $$

Proof. It is sufficient to check that there exists a family of orthonormal polynomials $f_n$ and constants $\lambda _n$ such that $q_n=\lambda _n f_n$ for all n.

We have the following recursion for $f_n$ :

$$ \begin{align*} f_n=\frac{(2 m n+m)\lambda_{n-1}}{2 (2m n+3m+2)c \lambda_{n} } f_{n-1}+\frac{(2m n+5m+4) \lambda_{n+1}}{2 (2m n+3m+2)c \lambda_{n} }f_{n+1}, \end{align*} $$

for $n\geq 1$ .

Set

$$ \begin{align*} A_{n+1} = \frac{(2m n+5m+4) \lambda_{n+1}}{2 (2m n+3m+2) \lambda_{n} }& & C_{n-1}=\frac{(2 m n+m)\lambda_{n-1}}{2 (2m n+3m+2) \lambda_{n} }. \end{align*} $$

Then $A_{n}=C_{n-1}$ if and only if

$$ \begin{align*} \lambda_n^2=\frac{(2m n+m+2) (2 m n+m)}{(2m n+3m+4)(2m n+3m+2) }\lambda_{n-1}^2. \end{align*} $$

Taking $\lambda _0 = 1$ , we can find a family of constants $\lambda _n$ satisfying this relation.

Appendix

Proof of Theorem 2.1.

The proof generalizes the proof of Theorem 3.1.1 in [Reference Cox, Futorny and Tirao9] in the case $m=2$ . We start off with the generating function

$$ \begin{align*} P_{-4}(c,z)&=z^{3-4/m} (1-2 c z^2+z^4)^{1/m}\\&\quad \times\lim_{\epsilon\to 0}\left[\int_{\epsilon}^{z}\frac{4 c \left(w^2 \left(\left(\frac{3 m}{2}-1\right) w^{\frac{4}{m}-2}\right)\right)-(3 m-4) w^{\frac{4}{m}-2}}{m w^2 \left(-2 c w^2+w^4+1\right)^{\frac{1}{m}+1}} \textrm{d}w+\epsilon^{4/m-3}+2c(3-2/m)\phi_m(\epsilon)\right] \\&=z^{3-4/m} (1-2 c z^2+z^4)^{1/m} \left(-\sum _{n=0}^{\infty } \frac{z^{\frac{4}{m}+2 n-3} \left((3 m-4) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right)}{m (2 n-3)+4}\right.\\& \quad +\left. \sum _{n=0}^{\infty } \frac{z^{\frac{4}{m}+2 n-1} \left(c (6 m-4) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right)}{m (2 n-1)+4}\right), \end{align*} $$

where $Q_n^\lambda (c)$ is the nth-Gegenbauer polynomial. The polynomials $Q_n^\lambda (c)$ satisfy the second-order linear differential equation

$$ \begin{align*} \left(1-c^2\right) y"-c (2 \lambda +1) y'+n y (2 \lambda +n)=0, \end{align*} $$

where the derivatives are with respect to c. Thus, for $\lambda = 1+\frac {1}{m}$ , we get

(5.3) $$ \begin{align} \left(1-c^2\right) (Q_n^{\left(1+\frac{1}{m}\right)})"(c)-c \left(\frac{2}{m}+3\right) (Q_n^{\left(1+\frac{1}{m}\right)})'(c)+n \left(\frac{2}{m}+n+2\right) Q_n^{\left(1+\frac{1}{m}\right)}(c)=0. \end{align} $$

Rewriting the expansion formula for $P_{-4}(c,z)$ , we get

(5.4) $$ \begin{align} z^{-3+4/m} (1-2 c z^2+z^4)^{-1/m}P_{-4}(c,z)&= \sum _{n=0}^{\infty } \frac{z^{\frac{4}{m}+2 n-1} \left(c (6 m-4) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right)}{m (2 n-1)+4} \nonumber\\& \quad - \sum _{n=0}^{\infty } \frac{z^{\frac{4}{m}+2 n-3} \left((3 m-4) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right)}{m (2 n-3)+4}. \end{align} $$

Now we apply the differential operator $L:=(1-c^2)\frac {d^2}{dc^2}-(c(3+\frac {2}{m}))\frac {d}{dc}$ to the right-hand side and use the identity (formula 4.7.27 in [Reference Szegö20])

(5.5)

Then we get

$$ \begin{align*} L \left(c(-\frac{4}{m}+6) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right)&=\left( (1-c^2)\frac{d^2}{dc^2}-(c(3+2/m))\frac{d}{dc} \right)\left(c(-4/m+6) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right)\\&=-\frac{2 (3 m-2)}{m} \left(2 m (n+1) Q_{n+1}^{\left(1+\frac{1}{m}\right)}(c)\right. \\& \quad +\left. c (n-1) (m n+m+2) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right). \end{align*} $$

From here, we deduce

$$ \begin{align*} &L \left(\sum _{n=0}^{\infty } \frac{z^{\frac{4}{m}+2 n-1} \left(c (6 m-4) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right)}{m (2 n-1)+4}\right)\\&\quad =\sum _{n=0}^{\infty } -\frac{z^{\frac{4}{m}+2 n-1} \left((2 (3 m-2)) \left(2 m (n+1) c_{n+1}^{\left(1+\frac{1}{m}\right)}(c)+c (n-1) (m n+m+2) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right)\right)}{m (m (2 n-1)+4)}\\&\quad =\sum _{n=0}^{\infty } -\frac{z^{\frac{4}{m}+2 n-1} \left((2 c (3 m-2) (n-1) (m n+m+2)) Q_n^{\left(1+\frac{1}{m}\right)}(c)\right)}{m (m (2 n-1)+4)}\\&\qquad +\sum _{n=0}^{\infty } -\frac{z^{\frac{4}{m}+2 n-1} \left((4 (3 m-2) (n+1)) c_{n+1}^{\left(1+\frac{1}{m}\right)}(c)\right)}{m (2 n-1)+4}. \end{align*} $$

Since

and

$$ \begin{align*} &\sum _{n=0}^{\infty } -\frac{z^{\frac{4}{m}+2 n-1} \left((4 (3 m-2) (n+1)) Q_{n+1}^{\left(1+\frac{1}{m}\right)}(c)\right)}{m (2 n-1)+4}\\&\quad =-\frac{(2 (3 m-4) (3 m-2)) \int_{0}^z w^{\frac{4}{m}-4} \left(\left(-2 c w^2+w^4+1\right)^{-\frac{1}{m}-1}-1\right) \, dw}{m^2}\\&\qquad -\frac{(4 (3 m-2)) \left(z^{\frac{4}{m}-3} \left(\left(-2 c z^2+z^4+1\right)^{-\frac{1}{m}-1}-1\right)\right)}{2 m}, \end{align*} $$

we get

In addition, we have

Applying the operator L to the left-hand side of (5.4), we get

$$ \begin{align*} &L \left(z^{\frac{4}{m}-3} \left(-2 c z^2+z^4+1\right)^{-1/m} P_{-4}(c,z)\right)\\&=\frac{z^{\frac{4}{m}-1} \left(-2 c z^2+z^4+1\right)^{-\frac{1}{m}-2} \left(4 c^2 (2 m+1) z^2-2 c (3 m+2) \left(z^4+1\right)+4 (m+1) z^2\right) P_{-4}(c,z)}{m^2}\\&\quad +\frac{z^{\frac{4}{m}-3} \left(-2 c z^2+z^4+1\right)^{-\frac{1}{m}-1} \left(6 c^2 m z^2-c (3 m+2) \left(z^4+1\right)+4 z^2\right) P^{\prime}_{-4}(c,z)}{m}\\&\quad -\left(c^2-1\right) z^{\frac{4}{m}-3} \left(-2 c z^2+z^4+1\right)^{-1/m} P^{\prime\prime}_{n-4}(c,z). \end{align*} $$

Hence, we obtain

$$ \begin{align*} &-\frac{3 (m+4) z^{\frac{4}{m}-3} \left(-2 c z^2+z^4+1\right)^{-\frac{1}{m}-1}}{4 m}-\frac{c (3 m-2) z^{\frac{4}{m}-1} \left(-2 c z^2+z^4+1\right)^{-\frac{1}{m}-1}}{2 m}\\&+\frac{(m+1) (3 m-4) \left(c-z^2\right) z^{\frac{4}{m}-1} \left(-2 c z^2+z^4+1\right)^{-\frac{1}{m}-2}}{m^2}\\&-\frac{2 c (m+1) (3 m-2) \left(c-z^2\right) z^{\frac{m+4}{m}} \left(-2 c z^2+z^4+1\right)^{-\frac{1}{m}-2}}{m^2}\\&+\frac{3 (m+4)}{4 m}\left(z^{\frac{4}{m}-3}\left(-2 c z^2+z^4+1\right)^{-1/m}P_{-4}(c,z)\right)\\&=\frac{z^{\frac{4}{m}-1} \left(-2 c z^2+z^4+1\right)^{-\frac{1}{m}-2} \left(4 c^2 (2 m+1) z^2-2 c (3 m+2) \left(z^4+1\right)+4 (m+1) z^2\right) P_{-4}(c,z)}{m^2}\\&+\frac{z^{\frac{4}{m}-3} \left(-2 c z^2+z^4+1\right)^{-\frac{1}{m}-1} \left(6 c^2 m z^2-c (3 m+2) \left(z^4+1\right)+4 z^2\right) P^{\prime}_{-4}(c,z)}{m}\\&-\left(c^2-1\right) z^{\frac{4}{m}-3} \left(-2 c z^2+z^4+1\right)^{-1/m} P^{\prime\prime}_{n-4}(c,z), \end{align*} $$

which gives us

$$ \begin{align*} &\frac{1}{4} \left(-z^4 \left(4 c^2 (m+2) (3 m-2)+(3 m+4) (5 m-4)\right)+2 c \left(9 m^2+6 m-8\right) z^6\right. \\& \qquad + \left. 4 c (3 m (m+2)-4) z^2-3 m (m+4)\right)\\& \quad =-\left(m \left(4 c^2 z^4-6 c \left(z^6+z^2\right)+3 z^8+2 z^4+3\right)+4 z^2 \left(-\left(c^2+1\right) z^2+c z^4+c\right)\right.\\& \qquad + \left. \frac{3}{4} m^2 \left(-2 c z^2+z^4+1\right)^2\right) P_{-4}(c,z)\\& \qquad +m \left(-2 c z^2+z^4+1\right) \left(6 c^2 m z^2-c (3 m+2) \left(z^4+1\right)+4 z^2\right) P^{\prime}_{-4}(c,z)\\& \qquad -\left(c^2-1\right) m^2 \left(-2 c z^2+z^4+1\right)^2 P^{\prime\prime}_{-4}(c,z). \end{align*} $$

Expanding this out and writing $P_{-4,k}(c)$ as $P_k$ , we will have the following equality:

(5.6) $$ \begin{align} 0&=\left(c^2 (-(m+2)) (3 m-2)-\frac{1}{2} m (3 m+4)+4\right) P_{n-4} +(c (3 m (m+2)-4)) P_{n-2} \nonumber\\&\quad -\frac{1}{4} (3 m (m+4)) P_n +\frac{1}{4} (-3 m (m+4)) P_{n-8} +(c (3 m (m+2)-4)) P_{n-6} \nonumber\\&\quad -(c m (3 m+2)) P^{\prime}_n +\left(4 \left(c^2 m (3 m+1)+m\right)\right) P^{\prime}_{n-2} \nonumber\\&\quad -\left(6 c m \left(2 c^2 m+m+2\right)\right) P^{\prime}_{n-4} +\left(4 \left(c^2 m (3 m+1)+m\right)\right) P^{\prime}_{n-6} \nonumber\\&\quad +(-c m (3 m+2)) P^{\prime}_{n-8} -\left(\left(c^2-1\right) m^2\right) P^{\prime\prime}_n +\left(4 c \left(c^2-1\right) m^2\right) P^{\prime\prime}_{n-2} \nonumber\\&\quad -\left(\left(c^2-1\right) m^2\right) P^{\prime\prime}_{n-8} +\left(4 c \left(c^2-1\right) m^2\right) P^{\prime\prime}_{n-6} \nonumber\\&\quad +\left(2 \left(-2 c^4+c^2+1\right) m^2\right) P^{\prime\prime}_{n-4}. \end{align} $$

We now differentiate twice with respect to c the recursion 2.1:

(5.7) $$ \begin{align} 0&=2 ((k-1) m+2) P_{k+2} +2 c ((k-1) m+2) P^{\prime}_{k+2} \nonumber\\& \quad -(k-3) m P^{\prime}_k -(k m+m+4) P^{\prime}_{k+4} \end{align} $$

(5.8) $$ \begin{align} 0&=(4 (k-1) m+8) P^{\prime}_{k+2} +2 c ((k-1) m+2) P^{\prime\prime}_{k+2} \nonumber\\&\quad -(k m+m+4) P^{\prime\prime}_{k+4} -(k-3) m P^{\prime\prime}_k. \end{align} $$

Substituting $k=n-8$ in the last equation, we obtain

(5.9) $$ \begin{align} 0&=4 (m (n-9)+2) P^{\prime}_{n-6} -m (n-11) P^{\prime\prime}_{n-8} \nonumber\\& \quad +2 c (m (n-9)+2) P^{\prime\prime}_{n-6} -(m (n-7)+4) P^{\prime\prime}_{n-4}. \end{align} $$

Multiplying (5.6) by $-11+n$ and adding it to the above equation multiplied by $(1-c^2) m$ gives us

(5.10) $$ \begin{align} 0&=4 c \left(c^2-1\right) m^2 (n-11) P^{\prime\prime}_{n-2} -\left(c^2-1\right) m^2 (n-11) P^{\prime\prime}_n \nonumber\\&\quad +4 m \left(c^2 (2 m (n-12)+n-13)+(m+1) (n-9)\right) P^{\prime}_{n-6} \nonumber\\&\quad -6 c m (n-11) \left(2 c^2 m+m+2\right) P^{\prime}_{n-4} \nonumber\\&\quad +2 c \left(c^2-1\right) m (m (n-13)-2) P^{\prime\prime}_{n-6} \nonumber\\&\quad -\left(c^2-1\right) m \left(m \left(4 c^2 (n-11)+n-15\right)-4\right) P^{\prime\prime}_{n-4} \nonumber\\&\quad +(n-11) \left(c^2 (-(m+2)) (3 m-2)-\frac{1}{2} m (3 m+4)+4\right) P_{n-4} \nonumber\\&\quad +4 (n-11) \left(c^2 m (3 m+1)+m\right) P^{\prime}_{n-2} -\frac{3}{4} (m+4) m (n-11) P_{n-8} \nonumber\\&\quad -\frac{3}{4} (m+4) m (n-11) P_n -c (3 m+2) m (n-11) P^{\prime}_{n-8} \nonumber\\&\quad -c (3 m+2) m (n-11) P^{\prime}_n +c (3 m (m+2)-4) (n-11) P_{n-6} \nonumber\\&\quad +c (3 m (m+2)-4) (n-11) P_{n-2}. \end{align} $$

Now substitute $k=n-8$ in (5.7) to get

(5.11) $$ \begin{align} 0&=2 (m (n-9)+2) P_{n-6} -m (n-11) P^{\prime}_{n-8} +2 c (m (n-9)+2) P^{\prime}_{n-6} \nonumber\\&\quad -(m (n-7)+4) P^{\prime}_{n-4}. \end{align} $$

Multiplying this equation by $-c (2+3 m)$ and adding it to the previous equation, we obtain

$$ \begin{align*} 0&=4 c \left(c^2-1\right) m^2 (n-11) P^{\prime\prime}_{n-2} -\left(c^2-1\right) m^2 (n-11) P^{\prime\prime}_n \\&\quad +2 c \left(c^2-1\right) m (m (n-13)-2) P^{\prime\prime}_{n-6} \\&\quad +\left(1-c^2\right) m \left(m \left(4 c^2 (n-11)+n-15\right)-4\right) P^{\prime\prime}_{n-4} \\&\quad +(n-11) \left(c^2 (-(m+2)) (3 m-2)-\frac{1}{2} m (3 m+4)+4\right) P_{n-4} \\&\quad +\left(2 c^2 (m (m (n-21)-14)-4)+4 m (m+1) (n-9)\right) P^{\prime}_{n-6} \\&\quad +\left(c (3 m+2) (m (n-7)+4)-6 c m (n-11) \left(2 c^2 m+m+2\right)\right) P^{\prime}_{n-4} \\&\quad +4 (n-11) \left(c^2 m (3 m+1)+m\right) P^{\prime}_{n-2} \\&\quad -\frac{3}{4} (m+4) m (n-11) P_{n-8} -\frac{3}{4} (m+4) m (n-11) P_n \\&\quad -c (3 m+2) m (n-11) P^{\prime}_n -c (m (3 m (n-7)-2 n+42)+4 (n-9)) P_{n-6} \\&\quad +c (3 m (m+2)-4) (n-11) P_{n-2}. \end{align*} $$

Finally, substitute $k=n-8$ in (2.1), multiply it by $-3 (4+m)$ and add to the previous equation multiplied by $4$ :

$$ \begin{align*} 0&=-2 c (m (m (9 n-69)+8 n-18)+8 (n-6)) P_{n-6} \\&\quad +\left(-4 c^2 (m+2) (3 m-2) (n-11)+16 (m+n-8)+m (4 n-3 m (n-15))\right) P_{n-4} \\&\quad +4 c (3 m (m+2)-4) (n-11) P_{n-2} -3 m (m+4) (n-11) P_n \\&\quad +\left(8 c^2 (m (m (n-21)-14)-4)+16 m (m+1) (n-9)\right) P^{\prime}_{n-6} \\&\quad +4 \left(c (3 m+2) (m (n-7)+4)-6 c m (n-11) \left(2 c^2 m+m+2\right)\right) P^{\prime}_{n-4} \\&\quad +16 (n-11) \left(c^2 m (3 m+1)+m\right) P^{\prime}_{n-2} -4 c m (3 m+2) (n-11) P^{\prime}_n \\&\quad +8 c \left(c^2-1\right) m (m (n-13)-2) P^{\prime\prime}_{n-6} \\&\quad +4 \left(1-c^2\right) m \left(m \left(4 c^2 (n-11)+n-15\right)-4\right) P^{\prime\prime}_{n-4} \\&\quad +16 c \left(c^2-1\right) m^2 (n-11) P^{\prime\prime}_{n-2} -4 \left(c^2-1\right) m^2 (n-11) P^{\prime\prime}_n. \end{align*} $$

We obtained the equation without terms with index $n-8$ .

The next step is to eliminate $P_{n-6}$ . After setting $k=n-6$ in (5.8), (5.7) and in the recursion, we obtain the following equations:

(5.12) $$ \begin{align} 0&=4 (m (n-7)+2) P^{\prime}_{n-4} -m (n-9) P^{\prime\prime}_{n-6} \nonumber\\&\quad +2 c (m (n-7)+2) P^{\prime\prime}_{n-4} -(m (n-5)+4) P^{\prime\prime}_{n-2}, \end{align} $$

(5.13) $$ \begin{align} 0&=2 (m (n-7)+2) P_{n-4} -m (n-9) P^{\prime}_{n-6} \nonumber\\&\quad +2 c (m (n-7)+2) P^{\prime}_{n-4} -(m (n-5)+4) P^{\prime}_{n-2}, \end{align} $$

$$ \begin{align*} \kern65pt 0=-m (n-9) P_{n-6} +2 c (m (n-7)+2) P_{n-4} -(m (n-5)+4) P_{n-2}. \end{align*} $$

Proceeding as in the previous case and using these three equations, we can eliminate all terms with index $n-6$ :

$$ \begin{align*} 0&=\left(-64 c^3 (m+1) (2 m+1) (m (n-4)+2)-4 c m (m (m (n (3 n-104)+629)\right.\\&\quad + 2 (n-54) n+714)-24 n+184)\Big) P^{\prime}_{n-4} \\&\quad -4 \left(c^2-1\right) m^3 (n-11) (n-9) P^{\prime\prime}_n \\&\quad +\left(m (n-9) \left(m^2 (29 n-179)+36 m (n-4)+16 (n-4)\right)\right.\\&\quad -\left. 16 c^2 (m+1) (2 m+1) (n-4) (m (n-6)+4)\right) P_{n-4} \\&\quad +m \left((m (n-5)+4) \left(8 c^2 \left(-m (n-21)+\frac{4}{m}+14\right)-16 (m+1) (n-9)\right)\right.\\&\quad+ 16 (n-11) (n-9) \left(c^2 m (3 m+1)+m\right)\bigg) P^{\prime}_{n-2} \\&\quad -4 \left(c^2-1\right) m \left(16 c^2 (m+1) (2 m+1)+m (n-9) (m (n-15)-4)\right) P^{\prime\prime}_{n-4} \\&\quad +8 c \left(c^2-1\right) m (m (m ((n-22) n+133)-2 n+42)+8) P^{\prime\prime}_{n-2} \\&\quad -3 (m+4) m^2 (n-11) (n-9) P_n -4 c (3 m+2) m^2 (n-11) (n-9) P^{\prime}_n \\&\quad +(4 c m (3 m (m+2)-4) (n-11) (n-9)\\&\quad +2 c (m (n-5)+4) (m (m (9 n-69)+8 n-18)+8 (n-6))) P_{n-2}. \end{align*} $$

The next step is to eliminate terms with index $n-4$ . Substituting $k=n-6$ in (5.8), (5.7) and in the recursion, we obtain the following equations:

$$ \begin{align*} 0&=4 (m (n-5)+2) P^{\prime}_{n-2} -m (n-7) P^{\prime\prime}_{n-4} \\&\quad +2 c (m (n-5)+2) P^{\prime\prime}_{n-2} -(m (n-3)+4) P^{\prime\prime}_n,\\[4pt]0&=2 (m (n-5)+2) P_{n-2} -m (n-7) P^{\prime}_{n-4} \\&\quad +2 c (m (n-5)+2) P^{\prime}_{n-2} -(m (n-3)+4) P^{\prime}_n,\\[4pt]0&=-m (n-7) P_{n-4} +2 c (m (n-5)+2) P_{n-2} -(m (n-3)+4) P_n. \end{align*} $$

Using these equations, we can eliminate terms with index $n-4$ :

(5.14) $$ \begin{align} 0&=2 c (n-2) (m (n-4)+4) \left(c^2 (m (n-5)+2)-m (n-8)\right) P_{n-2} \nonumber\\&\quad -(n-4) (m (n-6)+4) \left(c^2 (m (n-3)+4)-m (n-9)\right) P_n \nonumber\\&\quad -4 c \left(c^2 (m (n-4)+2) (m (n-3)+4)+m (m (6-(n-7) n)-6 n+46)\right) P^{\prime}_n \nonumber\\&\quad +8 c \left(c^2-1\right) m \left(c^2 (m (n-5)+2)-m (n-8)\right) P^{\prime\prime}_{n-2} \nonumber\\&\quad -4 \left(c^2-1\right) m \left(c^2 (m (n-3)+4)-m (n-9)\right) P^{\prime\prime}_n \nonumber\\&\quad +4 \left(2 c^4 (m (n-5)+2) (m (n-2)+2)+c^2 m (m ((28-3 n) n-47)-8 n+52)\right.\nonumber\\&\quad +\left. m^2 (n-9) (n-5)\right) P^{\prime}_{n-2}. \end{align} $$

Finally, we will eliminate all terms with index $n-2$ . Substituting $k=n-2$ in (5.8), (5.7) and in the recursion, we obtain

$$ \begin{align*} 0&=4 (m (n-3)+2) P^{\prime}_n -m (n-5) P^{\prime\prime}_{n-2} \\& \quad +2 c (m (n-3)+2) P^{\prime\prime}_n +(m (-n)+m-4) P^{\prime\prime}_{n+2}, \\0&=2 (m (n-3)+2) P_n -m (n-5) P^{\prime}_{n-2} +2 c (m (n-3)+2) P^{\prime}_n \\& \quad +(m (-n)+m-4) P^{\prime}_{n+2}, \\0&=-m (n-5) P_{n-2} +2 c (m (n-3)+2) P_n +(m (-n)+m-4) P_{n+2}. \end{align*} $$

These equations substitution $n\rightarrow n-2$ lead to the following equation:

(5.15) $$ \begin{align} 0&=2 c (n-4) (m (n-6)+4) (-m (n-3)-4) \left(c^2 (m (n-7)+2)-m (n-10)\right) P_n \nonumber \\&\quad +8 c \left(c^2-1\right) m (-m (n-3)-4) \left(c^2 (m (n-7)+2)-m (n-10)\right) P^{\prime\prime}_n \nonumber \\&\quad +4 c \left(4 c^4 (m (n-7)+2) (m (n-5)+2) (m (n-2)+2)\right.\nonumber\\&\quad +\left.c^2 m \left(m \left(-m (n-5) (7 (n-11) n+136)-34 n^2+404 n-1026\right)-40 n+296\right)\right.\nonumber\\&\quad +\left.m^2 (m (n (n (3 n-56)+303)-454)+2 n (5 n-78)+554)\right) P^{\prime}_{n-2} \nonumber \\&\quad +(n-2) (m (n-4)+4) \left(4 c^4 (m (n-7)+2) (m (n-5)+2)\right.\nonumber\\&\quad +\left.c^2 m (-m (n-5) (5 n-47)-12 (n-9))+m^2 (n-11) (n-7)\right) P_{n-2} \nonumber \\&\quad +4 \left(c^2-1\right) m \left(4 c^4 (m (n-7)+2) (m (n-5)+2)\right.\nonumber\\&\quad +\left.c^2 m (-m (n-5) (5 n-47)-12 (n-9))+m^2 (n-11) (n-7)\right) P^{\prime\prime}_{n-2} \nonumber \\&\quad +m (-m (n-3)-4) \left(\frac{8 c^4 (m (n-7)+2) (m (n-4)+2)}{m}\right.\nonumber\\&\quad -\left.4 c^2 (m (n (3 n-40)+115)+8 n-68)+4 m (n-11) (n-7)\right) P^{\prime}_n. \end{align} $$

Combining this with (5.14), we obtain

(5.16) $$ \begin{align} 0&=2 c (n-2) (m (n-4)+4) \left(c^2 (m (n-5)+2)-m (n-8)\right) P_{n-2} \nonumber\\&\quad -(n-4) (m (n-6)+4) \left(c^2 (m (n-3)+4)-m (n-9)\right) P_n \nonumber\\&\quad -4 c \left(c^2 (m (n-4)+2) (m (n-3)+4)+m (m (6-(n-7) n)-6 n+46)\right) P^{\prime}_n \nonumber\\&\quad +8 c \left(c^2-1\right) m \left(c^2 (m (n-5)+2)-m (n-8)\right) P^{\prime\prime}_{n-2} \nonumber\\&\quad -4 \left(c^2-1\right) m \left(c^2 (m (n-3)+4)-m (n-9)\right) P^{\prime\prime}_n \nonumber\\&\quad +4 \left(2 c^4 (m (n-5)+2) (m (n-2)+2)+c^2 m (m ((28-3 n) n-47)-8 n+52)\right.\nonumber\\&\quad +\left. m^2 (n-9) (n-5)\right) P^{\prime}_{n-2}. \end{align} $$

Now we use (5.16) and (5.15) to simplify the coefficient by $P^{\prime \prime }_{n-2}$ . Combining them, multiplying (5.16) by $-2 c (mn-7m+2)$ and adding it to (5.15) gives

$$ \begin{align*} 0=&-4 \left(c^2-1\right) m^2 (n-11) \left(c^2 (m (n-1)+4)-m (n-7)\right) P^{\prime\prime}_{n-2} \\&+8 c \left(c^2-1\right) (3 m+2) m^2 (n-11) P^{\prime\prime}_n \\&+4 m (n-11) \left(c^2 \left(m^2 ((n-10) n+39)+4 m (n-1)+8\right)-m (n-7) (m (n-3)+4)\right) P^{\prime}_n \\&-4 c m (n-11) \left(c^2 (m (n-2)+2) (m (n-1)+4)+m (m (16-(n-3) n)-6 n+34)\right) P^{\prime}_{n-2} \\&-m (n-11) (n-2) (m (n-4)+4) \left(c^2 (m (n-1)+4)-m (n-7)\right) P_{n-2} \\&+2 c (3 m+2) m (n-11) (n-4) (m (n-6)+4) P_n. \end{align*} $$

From now on, we assume that $n> 11$ . Then the equation simplifies to

(5.17) $$ \begin{align} 0&=\left(4 m (n-7) (m (n-3)+4)-4 c^2 \left(m^2 ((n-10) n+39)+4 m (n-1)+8\right)\right) P^{\prime}_n \nonumber\\&\quad +(n-2) (m (n-4)+4) \left(c^2 (m (n-1)+4)-m (n-7)\right) P_{n-2} \nonumber\\&\quad +4 c \left(c^2 (m (n-2)+2) (m (n-1)+4)+m (m (16-(n-3) n)-6 n+34)\right) P^{\prime}_{n-2} \nonumber\\&\quad +4 \left(c^2-1\right) m \left(c^2 (m (n-1)+4)-m (n-7)\right) P^{\prime\prime}_{n-2} \nonumber\\&\quad -8 c \left(c^2-1\right) m (3 m+2) P^{\prime\prime}_n -2 c (3 m+2) (n-4) (m (n-6)+4) P_n. \end{align} $$

Multiplying it by $-2 c (m n-5m+2)$ and adding it to (5.14) multiplied by $4+mn-m $ , we get

(5.18) $$ \begin{align} 0&=\left(4 c^2 \left(m^2 ((n-6) n+23)+4 m (n+1)+8\right)-4 m (n-5) (m (n-1)+4)\right) P^{\prime}_{n-2} \nonumber\\&\quad +(n-4) (m (n-6)+4) \left(c^2 m (n-7)-m n+m-4\right) P_n \nonumber\\&\quad -4 c \left(m \left(c^2 (n-7) (m (n-6)+2)-m ((n-13) n+24)-2 n+38\right)+8\right) P^{\prime}_n \nonumber\\&\quad +8 c \left(c^2-1\right) m (3 m+2) P^{\prime\prime}_{n-2} \nonumber\\&\quad +4 \left(c^2-1\right) m \left(c^2 m (n-7)-m n+m-4\right) P^{\prime\prime}_n \nonumber\\&\quad +2 c (3 m+2) (n-2) (m (n-4)+4) P_{n-2}. \end{align} $$

Multiplying this equation by $c (-m n+m-4)$ and adding it to (5.17) multiplied by $6 m+4$ , we get

(5.19) $$ \begin{align} 0&=c (n-4) (m (n-6)+4) \left(c^2 (m (n-1)+4)-m (n+5)-8\right) P_n \nonumber\\&\quad +4 c \left(m \left(c^2 (n-5) (m (n-1)+4)+m (13-(n-6) n)-4 n+44\right)+8\right) P^{\prime}_{n-2} \nonumber\\&\quad +8 \left(c^2-1\right) m (3 m+2) P^{\prime\prime}_{n-2} \nonumber\\&\quad +4 c \left(c^2-1\right) m \left(c^2 (m (n-1)+4)-m (n+5)-8\right) P^{\prime\prime}_n \nonumber\\&\quad +\left(-4 c^4 (m (n-6)+2) (m (n-1)+4)+4 c^2 (m (n-6)+2) (m (n+5)+8)\right.\nonumber\\&\quad - 8 (3 m+2) (m (n-3)+4)\Big) P^{\prime}_n .\nonumber\\&\quad +2 (3 m+2) (n-2) (m (n-4)+4) P_{n-2}. \end{align} $$

Multiplying it by $-c$ and adding to (5.18), we obtain the following equation:

(5.20) $$ \begin{align} 0&=4 \left(c^2-1\right) m P^{\prime\prime}_n +(n-4) (m (n-6)+4) P_n \nonumber \\&\quad +4 m (n-5) P^{\prime}_{n-2} -4 c (m (n-6)+2) P^{\prime}_n. \end{align} $$

Differentiate the last equation with respect to c to get

(5.21) $$ \begin{align} 0&=4 \left(c^2-1\right) m P^{\prime\prime\prime}_n +(n-6) (m (n-8)+4) P^{\prime}_n \nonumber \\&\quad +4 m (n-5) P^{\prime\prime}_{n-2} -4 c (m (n-8)+2) P^{\prime\prime}_n. \end{align} $$

Multiplying it by $-2 \left (c^2-1\right ) (3 m+2)$ and adding to (5.19) multiplied by $n-5$ gives us the following equation:

$$ \begin{align*} 0=&-8 \left(c^2-1\right)^2 m (3 m+2) P^{\prime\prime\prime}_n \\&+4 c \left(c^2-1\right) \left(m \left(c^2 (n-5) (m (n-1)+4)-m ((n-6) n+23)-4 (n-5)\right)+8\right) P^{\prime\prime}_n \\&+c (n-5) (n-4) (m (n-6)+4) \left(c^2 (m (n-1)+4)-m (n+5)-8\right) P_n \\&+4 c (n-5) \left(m \left(c^2 (n-5) (m (n-1)+4)+m (13-(n-6) n)-4 n+44\right)+8\right) P^{\prime}_{n-2} \\&+\left((n-5) \left(-4 c^4 (m (n-6)+2) (m (n-1)+4)+4 c^2 (m (n-6)+2) (m (n+5)+8)\right.\right.\\& -\left.\left. 8 (3 m+2) (m (n-3)+4)\right)-2 \left(c^2-1\right) (3 m+2) (n-6) (m (n-8)+4)\right) P^{\prime}_n \\&+2 (3 m+2) (n-5) (n-2) (m (n-4)+4) P_{n-2}. \end{align*} $$

Multiplying now (5.20) by $c \left (m \left (c^2 (n-5) (m (n-1)+4)+m (13-(n-6) n)-4 n+44\right )+8\right )$ and adding it to the equation above multiplied by $-m$ gives the following equation:

$$ \begin{align*} 0&=-24 c \left(c^2-1\right) m^2 P^{\prime\prime}_n -4 \left(c^2-1\right)^2 m^2 P^{\prime\prime\prime}_n \\&\quad +\left(c^2 (m (3 n (m (n-6)+4)-40)+16)\right.\\& \quad + m (-3 m ((n-6) n+4)-12 n+56)\Big) P^{\prime}_n \\&\quad +m (n-5) (n-2) (m (n-4)+4) P_{n-2} \\&\quad -c (n-4) (m (n-6)+4) (m (n-2)+2) P_n. \end{align*} $$

If we differentiate this equation, multiply by $-4$ and add to (5.20) multiplied by $(n-2) (m n-4m+4)$ , we will come to the desired linear differential equation of order $4$ . This proves the theorem.