ABSTRACT

A quasi-exactly solvable model refers to any second order differential equation with polynomial coefficients of the form $A\left(x\right)y^{″}\left(x\right)+B\left(x\right)y^{\prime }\left(x\right)+C\left(x\right)y\left(x\right)=0$ where a pair of exact polynomials $\left\{y\left(x\right),C\left(x\right)\right\}$ with respective degrees $\left\{\mathrm{deg}\left[y\right]=n,\mathrm{deg}\left[C\right]=p\right\}$ are to be found simultaneously in terms of the coefficients of two given polynomials $\left\{A\left(x\right),B\left(x\right)\right\}$. The existing methods for solving quasi-exactly solvable models require the solution of a system of nonlinear algebraic equations of which the dimensions depend on n, the degree of the exact polynomial solution $y\left(x\right)$. In this paper, a new method employing a set of polynomials, called canonical polynomials, is proposed. This method requires solving a system of nonlinear algebraic equations of which the dimensions depend only on p, the degree of $C\left(x\right)$, and do not vary with n. Several examples are implemented to testify the efficiency of the proposed method.

Keywords:

Polynomial Solutions, Canonical Polynomials, Tau Method, Quasi-Exactly Solvable Models

1. Introduction

Let D be a linear differential operator of degree $\nu \ge 1$ with polynomial coefficients, and let $y\left(x\right)$ be an exact solution of the differential equation:

$\left(Dy\right)\left(x\right)\equiv {\displaystyle \underset{i=0}{\overset{\nu }{\sum }}}{A}_i\left(x\right)\frac{{\text{d}}^{i}y}{\text{d}{x}^i}=A_{\nu +1}\left(x\right)$ (1)

where $\left\{A_i\left(x\right);i=0,1,\cdots ,\nu +1\right\}$ are polynomials in x for $i=0,1,\cdots ,\nu +1$.

Among the very efficient techniques to solve equations of the form (1) is the Tau Method. This method was invented by Lanczos in [1] to solve simple equations and it was extended later by Ortiz [2] to treat more sophisticated problems with applications in a wide range of scientific fields. The basic idea of the Tau Method consists of finding polynomial solutions to Equation (1) in terms of a special set of polynomials associated with D called canonical polynomials, denoted by $\left\{Q_k\left(x\right)\mathrm{<mo>\; ;\; </mo>}k\ge 0\right\}$. These are defined as the pseudo inverse image of the set of the power functions $\left\{x^k\mathrm{<mo>\; ;\; </mo>}k\ge 0\right\}$ by D.

In this paper, we use the canonical polynomials to develop a new method to solve quasi-exactly solvable (QES) second order ordinary differential equations. The QES problem, considered in this paper, can be stated as follows: Let p and n be two nonnegative integers, and suppose that $\left\{A\left(x\right)\mathrm{,}B\left(x\right)\mathrm{,}C\left(x\right)\right\}$ are polynomials with expressions

$A\left(x\right)={\displaystyle \underset{i=0}{\overset{p+2}{\sum }}}{A}_i{x}^i,B\left(x\right)={\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{B}_i{x}^i,\text{and}C\left(x\right)={\displaystyle \underset{i=0}{\overset{p}{\sum }}}{C}_i{x}^i.$

The coefficients $\left\{C_i;i=0,1,\cdots ,p\right\}$ are found in terms of $\left\{A_i;i=0,1,\cdots ,p+2\right\}$ and $\left\{B_i;i=0,1,\cdots ,p+1\right\}$ so that the function $y\left(x\right)$ defined implicitly by the differential equation

$\left(Dy\right)\left(x\right)=A\left(x\right)\frac{{\text{d}}^{2}y}{\text{d}{x}^2}+B\left(x\right)\frac{\text{d}y}{\text{d}x}+C\left(x\right)y\left(x\right)=0$ (2)

is an exact polynomial solution of degree n and written as $y\left(x\right)={\displaystyle {\sum }_{i=0}^n}{y}_i{x}^i$, where $\left\{y_i;i=0,1,2,\cdots ,n\right\}$ are determined in terms of $\left\{A_i\mathrm{,}{B}_i\mathrm{,}{C}_i\mathrm{<mo>\; ;\; </mo>}i\ge 0\right\}$.

Quasi-exactly solvable problems have applications in engineering, chemistry and physics among many other fields. This includes a wide range of mathematical settings that involve Schrödinger equations describing problems in quantum mechanics, for example, anharmonic singular potentials, coulombically repelling electrons on a multidimensional sphere, Planer Dirac electron in magnetic fields and Kink stability analysis among many other problems (see [3] [4] [5] [6] and the references given therein).

QES problems are usually solved by means of the Functional Bethe Ansatz method or by a constraint polynomial approach (see [7] ). With the latter, we derive a $n\times n$ system of algebraic equations, which gives the n roots of $y\left(x\right)$, and afterward the p coefficients of $C\left(x\right)$ are calculated. If one wishes to increase the order n of polynomial $y\left(x\right)$, then a new nonlinear algebraic system of higher dimensions has to be resolved. With our proposed method that employs the canonical polynomials, we solve first a $p\times p$ system of algebraic equations giving the coefficients of $C\left(x\right)$ and then the coefficients of $y\left(x\right)$ are computed. With this method, a polynomial solution $y\left(x\right)$ of higher order n can be obtained without increasing dimensions $p\times p$ of the algebraic system.

Section 2 introduces the canonical polynomials in the context that fits the problem under consideration. In Section 3 the main results of this paper are given and discussed. Sections 4 and 5 illustrate the implementation of the proposed method.

2. The Canonical Polynomials

In this section we recall the main features of the canonical polynomials associated with D (see Ortiz [2] ). We begin with this definition.

Definition 1. For any integer $k\ge 0$,

1) $P_k\left(x\right)$ is called the kth generating polynomial of D if $P_k\left(x\right)=D{x}^k$.

2) $Q_k^{\mathrm{<mo>\; *\; </mo>}}\left(x\right)$ is called a kth canonical function of D if $Q_k^{\mathrm{<mo>\; *\; </mo>}}\left(x\right)$ satisfies the differential equation $D{Q}_k^{*}=x^k$.

In [2] , Ortiz presented an algorithm in the form of a self-starting recursive formula for computing the $Q_k^{\mathrm{<mo>\; *\; </mo>}}$ 's associated with linear differential operators of arbitrary degree $\nu \ge 1$. The following theorem gives a variation of this algorithm for second order differential operators of the form (2) in which this paper is concerned:

Theorem 1. The canonical functions associated with the differential operator (2) can be generated by the following recursion:

$Q_{k+p}^{*}=\frac{1}{f_p^{\left(k\right)}}\left\{x^k-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_{i-2}^{\left(k\right)}{Q}_{k+i-2}^{*}\right\},k\ge 0$ (3)

where $f_{i-2}^{\left(k\right)}:=k\left(k-1\right)A_i+k{B}_{i-1}+C_{i-2}$ for any $i\ge 0$ with the convention that $A_i=B_i=C_i=0$ for $i<0$.

Proof. For any $k\ge 0$, let us expand the kth generating polynomial $P_k\left(x\right)$ in terms of $\left\{x^i;i=0,1,2,\cdots \right\}$ :

$\begin{array}{l}{P}_k\left(x\right):=D{x}^k=A\left(x\right)k\left(k-1\right)x^{k-2}+B\left(x\right)k{x}^{k-1}+C\left(x\right)x^k\\ =k\left(k-1\right){\displaystyle \underset{i=0}{\overset{p+2}{\sum }}}{A}_i{x}^{k+i-2}+k{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{B}_i{x}^{k+i-1}+{\displaystyle \underset{i=0}{\overset{p}{\sum }}}{C}_i{x}^{k+i}\\ =k\left(k-1\right)A_{p+2}{x}^{k+p}+k{B}_{p+1}{x}^{k+p}+C_p{x}^{k+p}\\ +k\left(k-1\right){\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{A}_i{x}^{k+i-2}+k{\displaystyle \underset{i=0}{\overset{p}{\sum }}}{B}_i{x}^{k+i-1}+{\displaystyle \underset{i=0}{\overset{p-1}{\sum }}}{C}_i{x}^{k+i}\end{array}$

$\begin{array}{l}=\left[k\left(k-1\right)A_{p+2}+k{B}_{p+1}+C_p\right]x^{k+p}+k\left(k-1\right){\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{A}_i{x}^{k+i-2}\\ +k{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{B}_{i-1}{x}^{k+i-2}+{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{C}_{i-2}{x}^{k+i-2}\\ =\left[k\left(k-1\right)A_{p+2}+k{B}_{p+1}+C_p\right]x^{k+p}\\ +{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}\left[k\left(k-1\right)A_i+k{B}_{i-1}+C_{i-2}\right]x^{k+i-2}\\ \equiv f_p^{\left(k\right)}{x}^{k+p}+{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_{i-2}^{\left(k\right)}{x}^{k+i-2}.\end{array}$ (4)

Since, by Definition 1, $D{Q}_{k+i-2}^{*}=x^{k+i-2}$, and since D is a linear operator, (4) becomes:

$D{x}^k=f_p^{\left(k\right)}{x}^{k+p}+D\left\{{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_{i-2}^{\left(k\right)}{Q}_{k+i-2}^{*}\right\},$ (5)

which implies that

$D\left\{x^k-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_{i-2}^{\left(k\right)}{Q}_{k+i-2}^{*}\right\}=f_p^{\left(k\right)}{x}^{k+p},$

from which follows the desired recursion (3):

$Q_{k+p}^{*}=\frac{1}{f_p^{\left(k\right)}}\left\{x^k-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_{i-2}^{\left(k\right)}{Q}_{k+i-2}^{*}\right\},\left(k\ge 0\right)$ $\square$

For instance, when $k=0$, Equation (3) gives

$\begin{array}{c}{Q}_p^{*}=\frac{1}{C_p}\left\{1-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{C}_{i-2}{Q}_{i-2}^{*}\right\}=\frac{1}{C_p}\left\{1-{\displaystyle \underset{i=0}{\overset{p-1}{\sum }}}{C}_i{Q}_i^{*}\right\}\\ =\frac{1}{C_p}-\frac{1}{C_p}\left(C_0{Q}_0^{*}+C_1{Q}_1^{*}+\cdots +C_{p-1}{Q}_{p-1}^{*}\right)\equiv Q_p\left(x\right)+R_p,\end{array}$

where

$Q_p\left(x\right):=\frac{1}{C_p}\text{and}{R}_p:=-\frac{1}{C_p}\left(C_0{Q}_0^{*}+C_1{Q}_1^{*}+\cdots +C_{p-1}{Q}_{p-1}^{*}\right).$ (6)

That is, $Q_p^{\mathrm{<mo>\; *\; </mo>}}$ has a polynomial component $Q_p\left(x\right)$ of degree 0 and a residual component $R_p\in \text{span}\left\{Q_i^{*};i=0,1,\cdots ,p-1\right\}$, which is the subspace generated by $\left\{Q_0^{\mathrm{<mo>\; *\; </mo>}}\mathrm{,}{Q}_1^{\mathrm{<mo>\; *\; </mo>}}\mathrm{,}\cdots \mathrm{,}{Q}_{p-1}^{\mathrm{<mo>\; *\; </mo>}}\right\}$.

For $k=1$, we have

$\begin{array}{c}{Q}_{p+1}^{*}=\frac{1}{B_{p+1}+C_p}\left\{x-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}\left(B_{i-1}+C_{i-2}\right)Q_{i-1}^{*}\right\}\\ =\frac{1}{B_{p+1}+C_p}\left\{x-\left(B_p+C_{p-1}\right)Q_p^{*}-{\displaystyle \underset{i=0}{\overset{p}{\sum }}}\left(B_{i-1}+C_{i-2}\right)Q_{i-1}^{*}\right\}\\ =\frac{1}{B_{p+1}+C_p}\left\{x-\left(B_p+C_{p-1}\right)\left(\frac{1}{C_p}+R_p\right)-{\displaystyle \underset{i=0}{\overset{p}{\sum }}}\left(B_{i-1}+C_{i-2}\right)Q_{i-1}^{*}\right\}\\ =\frac{1}{B_{p+1}+C_p}\left\{x-\frac{B_p+C_{p-1}}{C_p}\right\}+\frac{1}{B_{p+1}+C_p}\left\{-R_p-{\displaystyle \underset{i=0}{\overset{p}{\sum }}}\left(B_{i-1}+C_{i-2}\right)Q_{i-1}^{*}\right\}\\ =Q_{p+1}+R_{p+1},\end{array}$

where

$\begin{array}{l}{Q}_{p+1}:=\frac{1}{B_{p+1}+C_p}\left\{x-\frac{B_p+C_{p-1}}{C_p}\right\}\\ \text{and}{R}_{p+1}:=\frac{1}{B_{p+1}+C_p}\left\{-R_p-{\displaystyle \underset{i=0}{\overset{p}{\sum }}}\left(B_{i-1}+C_{i-2}\right)Q_{i-1}^{*}\right\}.\end{array}$ (7)

$Q_{p+1}\left(x\right)$ is a polynomial of degree 1 and the residual $R_{p+1}\in \text{span}\left\{Q_i^{*};i=0,1,\cdots ,p-1\right\}$.

Equations (6) and (7) suggest that any canonical function is the sum of a polynomial $\in \text{span}\left\{x^i\mathrm{<mo>\; ;\; </mo>}i\ge 0\right\}$, and a residual $\in \text{span}\left\{Q_i^{*};i=0,1,\cdots ,p-1\right\}$. This is formulated in the following theorem:

Theorem 2. For all $k\ge 0$, $Q_{k+p}^{\mathrm{<mo>\; *\; </mo>}}$ can be written as

$Q_{k+p}^{*}=Q_{k+p}+R_{k+p}$

where $\left\{Q_k\mathrm{<mo>\; ;\; </mo>}k\ge 0\right\}$ are called polynomials given by the recursion:

$Q_{k+p}=\frac{1}{f_p^{\left(k\right)}}\left\{x^k-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_{i-2}^{\left(k\right)}{Q}_{k+i-2}\right\},\left(k\ge 0\right)$ (8)

with $Q_j=0$ for all $j=0,1,\cdots ,p-1$. And

$R_{k+p}={\displaystyle \underset{j=0}{\overset{p-1}{\sum }}}{r}_{k+p}^{\left(j\right)}{Q}_j^{*}$

where $\left\{r_k^{\left(j\right)}\mathrm{<mo>\; ;\; </mo>}k\ge 0\right\}$ is a sequence of constants given by the recursion

$r_{k+p}^{\left(j\right)}=-\frac{1}{f_p^{\left(k\right)}}{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_{i-2}^{\left(k\right)}{r}_{k+i-2}^{\left(j\right)},\left(k\ge 0\text{and}j=0,1,\cdots ,p-1\right)$ (9)

with $r_i^{\left(j\right)}={\delta }_{ij}$ for all i and $j=0,1,\cdots ,p-1$.

Proof. Equations (6)-(7) show that this theorem holds for $k=0$ and $k=1$. The use of an induction argument allows to prove it for all k. To this end let us assume that

$Q_{k+j}^{*}=Q_{k+j}+R_{k+j},\left(j=0,1,2,\cdots ,p-1\right)$ (10)

with

$R_j={\displaystyle \underset{i=0}{\overset{p-1}{\sum }}}{r}_j^{\left(i\right)}{Q}_i^{*}$

and let us take $j=p$. Combining (10) with (3) gives:

$\begin{array}{c}{Q}_{k+p}^{*}=\frac{1}{f_p^{\left(k\right)}}\left\{x^k-{\displaystyle \underset{j=0}{\overset{p+1}{\sum }}}{f}_{j-2}^{\left(k\right)}\left(Q_{k+j-2}+R_{k+j-2}\right)\right\}\\ =\frac{1}{f_p^{\left(k\right)}}\left\{x^k-{\displaystyle \underset{j=0}{\overset{p+1}{\sum }}}{f}_{j-2}^{\left(k\right)}{Q}_{k+j-2}\right\}+\frac{1}{f_p^{\left(k\right)}}\left\{-{\displaystyle \underset{j=0}{\overset{p+1}{\sum }}}{f}_{j-2}^{\left(k\right)}{R}_{k+j-2}\right\}\\ =\frac{1}{f_p^{\left(k\right)}}\left\{x^k-{\displaystyle \underset{j=0}{\overset{p+1}{\sum }}}{f}_{j-2}^{\left(k\right)}{Q}_{k+j-2}\right\}+\frac{1}{f_p^{\left(k\right)}}\left\{-{\displaystyle \underset{j=0}{\overset{p+1}{\sum }}}{f}_{j-2}^{\left(k\right)}\left({\displaystyle \underset{i=0}{\overset{p-1}{\sum }}}{r}_{k+j-2}^{\left(i\right)}{Q}_i^{*}\right)\right\}\\ =\frac{1}{f_p^{\left(k\right)}}\left\{x^k-{\displaystyle \underset{j=0}{\overset{p+1}{\sum }}}{f}_{j-2}^{\left(k\right)}{Q}_{k+j-2}\right\}+\frac{1}{f_p^{\left(k\right)}}\left\{-{\displaystyle \underset{i=0}{\overset{p-1}{\sum }}}\left[{\displaystyle \underset{j=0}{\overset{p+1}{\sum }}}{f}_{j-2}^{\left(k\right)}\left(r_{k+j-2}^{\left(i\right)}\right)\right]Q_i^{*}\right\}\\ \equiv Q_{k+p}+R_{k+p}\end{array}$

as required.

Corollary 1. Suppose that the above assumptions and notation hold true.

1) If $p=1$, then $\left\{Q_k\mathrm{<mo>\; ;\; </mo>}k\ge 0\right\}$ are generated by the recursion:

$Q_0=0,Q_{k+1}=\frac{1}{f_1^{\left(k\right)}}\left\{x^k-f_0^{\left(k\right)}{Q}_k\right\},\left(k\ge 0\right).$

And

$R_{k+1}=r_{k+1}^{\left(0\right)}{Q}_0^{*},\left(k\ge 0\right)$

where

$r_0^{\left(0\right)}=1,r_{k+1}^{\left(0\right)}=-\frac{f_0^{\left(k\right)}}{f_1^{(k)}}{r}_k^{\left(0\right)},\left(k\ge 0\right)$

2) If $p=2$, then $\left\{Q_k\mathrm{<mo>\; ;\; </mo>}k\ge 0\right\}$ are generated by the recursion:

$Q_0=Q_1=0,Q_{k+2}=\frac{1}{f_2^{\left(k\right)}}\left\{x^k-f_0^{\left(k\right)}{Q}_k-f_1^{\left(k\right)}{Q}_{k+1}\right\},\left(k\ge 0\right).$

And

$R_{k+2}=r_{k+2}^{\left(0\right)}{Q}_0^{*}+r_{k+2}^{\left(1\right)}{Q}_1^{*}$

where $\left\{r_k^{\left(0\right)}\mathrm{,}{r}_k^{\left(1\right)}\mathrm{<mo>\; ;\; </mo>}k\ge 0\right\}$ are constants given by the recursion

$\begin{array}{l}{r}_0^{\left(0\right)}=r_1^{\left(1\right)}=1,r_0^{\left(1\right)}=r_1^{\left(0\right)}=0,\\ r_{k+2}^{\left(j\right)}=-\frac{1}{f_1^{(k)}}\left(f_0^{\left(k\right)}{r}_k^{\left(j\right)}+f_0^{\left(k\right)}{r}_{k+1}^{\left(j\right)}\right)\left(k\ge 0\text{and}j=0,1\right).\end{array}$

3. Solutions of Quasi-Exactly Solvable Models in Terms of {Q_n}

Having obtained Algorithm (8)-(9) that generates the canonical polynomials that fit to our context, we can now formulate the main result of this paper:

Theorem 3. Let $A\left(x\right)={\displaystyle \underset{i=0}{\overset{p+2}{\sum }}{A}_i{x}^i}$ and $B\left(x\right)={\displaystyle \underset{i=0}{\overset{p+1}{\sum }}{B}_i{x}^i}$ be two given polynomials of degree $p+2$ and $p+1$ respectively, and suppose that $C\left(x\right)={\displaystyle \underset{i=0}{\overset{p}{\sum }}{C}_i{x}^i}$ is an unknown polynomial of degree $p\ge 0$. Let $n\ge 0$. If $\left\{C_0\mathrm{,}{C}_1\mathrm{,}\cdots \mathrm{,}{C}_p\right\}$ satisfy the following system of algebraic equations

$C_p=-n\left(n-1\right)A_{p+2}-n{B}_{p+1}$ (11)

${\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}\left(n\left(n-1\right)A_i+n{B}_{i-1}+C_{i-2}\right)r_{n+i-2}^{\left(l\right)}=0,\left(l=0,1,2,\cdots ,p-1\right)$ (12)

where $\left\{r_i^{\left(l\right)}\right\}$ are given by (9), then the polynomial

$y\left(x\right)=x^n-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}\left(n\left(n-1\right)A_i+n{B}_{i-1}+C_{i-2}\right)Q_{n+i-2}$ (13)

is an exact solution for the differential Equation (2), where $\left\{Q_k\right\}$ are given by (8).

Proof. Let $n\ge 0$. If (11) holds true, then $f_p^{\left(n\right)}:=n\left(n-1\right)A_{p+2}+n{B}_{p+1}+C_p=0$ and therefore setting $k=n$ in (5) gives

$D{x}^n=f_p^{\left(n\right)}{x}^{n+p}+D\left\{{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_{i-2}^{\left(n\right)}{Q}_{n+i-2}^{*}\right\}=D\left\{{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_{i-2}^{\left(n\right)}{Q}_{n+i-2}^{*}\right\}.$

This, in turn, implies that

$D\left\{x^n-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}\left[n\left(n-1\right)A_i+n{B}_{i-1}+C_{i-2}\right]Q_{n+i-2}^{*}\right\}=0.$

That is,

$y\left(x\right)=x^n-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_i^{\left(n\right)}{Q}_{n+i-2}^{*}\equiv x^n-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_i^{\left(n\right)}\left(Q_{n+i-2}+R_{n+i-2}\right)$ (14)

is an exact solution for the differential Equation (2). For Expression (14) to be a polynomial, it must be independent of the residual terms. To this end, we proceed as follows:

$\begin{array}{c}y\left(x\right)=x^n-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_i^{\left(n\right)}\left(Q_{n+i-2}+R_{n+i-2}\right)\\ =x^n-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_i^{\left(n\right)}\left(Q_{n+i-2}+{\displaystyle \underset{l=0}{\overset{p-1}{\sum }}}{r}_{n+i-2}^{\left(l\right)}{Q}_l^{*}\right)\\ =x^n-{\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_i^{\left(n\right)}{Q}_{n+i-2}+{\displaystyle \underset{l=0}{\overset{p-1}{\sum }}}\left({\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_i^{\left(n\right)}{r}_{n+i-2}^{\left(l\right)}\right)Q_l^{*}\end{array}$

Now, for y(x) to be independent of the p undefined elements $\left\{Q_0^{\mathrm{<mo>\; *\; </mo>}}\mathrm{,}{Q}_1^{\mathrm{<mo>\; *\; </mo>}}\mathrm{,}\cdots \mathrm{,}{Q}_{p-1}^{\mathrm{<mo>\; *\; </mo>}}\right\}$, the coefficients of the latter must be set equal to zero leading to the following algebraic system that consists of p equations:

${\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}{f}_i^{\left(n\right)}{r}_{n+i-2}^{\left(l\right)}=0,\left(l=0,1,2,\cdots ,p-1\right).$

Explicitly,

${\displaystyle \underset{i=0}{\overset{p+1}{\sum }}}\left[n\left(n-1\right)A_i+n{B}_{i-1}+C_{i-2}\right]r_{n+i-2}^{\left(l\right)}=0,\left(l=0,1,2,\cdots ,p-1\right)$

with the p unknowns $\left\{C_0\mathrm{,}{C}_1\mathrm{,}{C}_2\mathrm{,}\cdots \mathrm{,}{C}_{p-1}\right\}$. This completes the proof of the theorem. $\square$

For illustration let us consider particular cases:

Case $p=1$. Then we have

$A\left(x\right)={\displaystyle \underset{i=0}{\overset{3}{\sum }}}{A}_i{x}^i,B\left(x\right)={\displaystyle \underset{i=0}{\overset{2}{\sum }}}{B}_i{x}^i,C\left(x\right)={\displaystyle \underset{i=0}{\overset{1}{\sum }}}{C}_i{x}^i$

where $\left\{C_0\mathrm{,}{C}_1\right\}$ satisfy the algebraic equation

$C_1=n\left(1-n\right)A_3-n{B}_2$

$n\left(n-1\right)\left(A_0{r}_{n-2}^{\left(0\right)}+A_1{r}_{n-1}^{\left(0\right)}+A_2{r}_n^{\left(0\right)}\right)+n\left(B_0{r}_{n-1}^{\left(0\right)}+n{B}_2{r}_n^{\left(0\right)}\right)+C_0{r}_n^{\left(0\right)}=0$

where $\left\{r_i^{\left(0\right)}\right\}$ are given by,

$\begin{array}{l}\left(k\left(k-1\right)A_3+k{B}_2+C_1\right)r_{k+1}^{\left(0\right)}\\ =-k\left(k-1\right)A_0{r}_{k-2}^{\left(0\right)}-\left(k\left(k-1\right)A_1+k{B}_0\right)r_{k-1}^{\left(0\right)}-\left(k\left(k-1\right)A_2+k{B}_1+C_0\right)r_k^{(0)}\end{array}$

with $r_0^{\left(0\right)}=1$. And

$y\left(x\right)=x^n-{\displaystyle \underset{i=0}{\overset{2}{\sum }}}\left(n\left(n-1\right)A_i+n{B}_{i-1}+C_{i-2}\right)Q_{n+i-2}$

where $\left\{Q_k;0\le k\le n\right\}$ are given by the recursion:

$\begin{array}{l}\left(k\left(k-1\right)A_3+k{B}_2+C_1\right)Q_{k+1}\\ =x^k-k\left(k-1\right)A_0{Q}_{k-2}-\left(k\left(k-1\right)A_1+k{B}_0\right)Q_{k-1}-\left(k\left(k-1\right)A_2+k{B}_1+C_0\right)Q_k\end{array}$

for $k=0,1,2,\cdots$, with $Q_0=0$.

Case $p=2$. In this case,

$A\left(x\right)={\displaystyle \underset{i=0}{\overset{4}{\sum }}}{A}_i{x}^i,B\left(x\right)={\displaystyle \underset{i=0}{\overset{3}{\sum }}}{B}_i{x}^i,C\left(x\right)={\displaystyle \underset{i=0}{\overset{2}{\sum }}}{C}_i{x}^i$

where $C_2=n\left(1-n\right)A_4-n{B}_3$ and $\left\{C_0\mathrm{,}{C}_1\right\}$ satisfy the $2\times 2$ algebraic system

$\begin{array}{l}{\displaystyle \underset{i=0}{\overset{3}{\sum }}}n\left(n-1\right)A_i{r}_{n+i-2}^{\left(0\right)}+{\displaystyle \underset{i=1}{\overset{3}{\sum }}}n{B}_{i-1}{r}_{n+i-2}^{\left(0\right)}+{\displaystyle \underset{i=2}{\overset{3}{\sum }}}{C}_{i-2}{r}_{n+i-2}^{\left(0\right)}=0,\\ {\displaystyle \underset{i=0}{\overset{3}{\sum }}}n\left(n-1\right)A_i{r}_{n+i-2}^{\left(1\right)}+{\displaystyle \underset{i=1}{\overset{3}{\sum }}}n{B}_{i-1}{r}_{n+i-2}^{\left(1\right)}+{\displaystyle \underset{i=2}{\overset{3}{\sum }}}{C}_{i-2}{r}_{n+i-2}^{\left(1\right)}=0,\end{array}$

where $\left\{r_k^{\left(0\right)}\mathrm{,}{r}_k^{\left(1\right)}\mathrm{<mo>\; ;\; </mo>}k\ge 0\right\}$ are given by the recursion

$\begin{array}{l}\left(k\left(k-1\right)A_4+k{B}_3+C_2\right)r_{k+2}^{\left(j\right)}\\ =-\left(C_1+k{B}_2+k\left(k-1\right)A_3\right)r_{k+1}^{\left(j\right)}-\left(C_0+k{B}_1+k\left(k-1\right)A_2\right)r_k^{\left(j\right)}\\ -\left(k{B}_0+k\left(k-1\right)A_1\right)r_{k-1}^{\left(j\right)}-k\left(k-1\right)A_0{r}_{k-2}^{\left(j\right)},j=0,1\end{array}$

And the exact solution $y\left(x\right)$ is given as:

$\begin{array}{c}y\left(x\right)=x^n-n\left(n-1\right)A_0{Q}_{n-2}-\left[n\left(n-1\right)A_1+n{B}_0\right]Q_{n-1}\\ -\left[n\left(n-1\right)A_2+n{B}_1+C_0\right]Q_n-\left[n\left(n-1\right)A_3+n{B}_2+C_1\right]Q_{n+1}\end{array}$

where $\left\{Q_k\mathrm{<mo>\; ;\; </mo>0}\le k\le n+1\right\}$ are given by the recursion:

$\begin{array}{l}\left(k\left(k-1\right)A_4+k{B}_3+C_2\right)Q_{k+2}\\ =x^k-\left[C_1+k{B}_2+k\left(k-1\right)A_3\right]Q_{k+1}-\left[C_0+k{B}_1+k\left(k-1\right)A_2\right]Q_k\\ -\left[k{B}_0+k\left(k-1\right)A_1\right]Q_{k-1}-k\left(k-1\right)a_0{Q}_{k-2}\end{array}$

4. Two Coulombically Repelling Electrons on a Sphere

This is a system governed by the ODE

$\left(\frac{s^2}{4R^2}-1\right)\frac{{\text{d}}^{2}U}{\text{d}{s}^2}+\left(\frac{\delta s}{4R^2}-\frac{1}{\gamma s}\right)\frac{\text{d}U}{\text{d}s}+\left(\frac{1}{s}-E\right)U\left(s\right)=0$ (15)

where s denotes the inter-electronic distance between two electrons interacting via a Coulomb potential constrained to remain on the surface of a D-dimensional sphere ( $D\ge 2$ ) of radius R, $\delta =2D-1$, $\gamma =\frac{1}{D-1}$, E is the unknown energy, and $U\left(s\right)$ stands for the unknown inter-electronic wave function (see [8] [9] ).

Setting $x=\frac{s}{2R}$ and $U\left(s\right)=U\left(2Rx\right)\equiv Y\left(x\right)$, then Equation (15) can be written in the form (2) with $p=1$ :

$\left(x^3-x\right)\frac{{\text{d}}^{2}Y}{\text{d}{x}^2}+\left(\delta x^2-\frac{1}{\gamma }\right)\frac{\text{d}Y}{\text{d}x}+\left(C_0+C_{1}x\right)Y\left(x\right)=0$ (16)

Let us apply the Algorithm (11)-(12) developed in the previous section to this example. Here

$\begin{array}{l}A\left(x\right)=x^3-x,A_0=A_2=0,A_1=-A_3=-1,\\ B\left(x\right)=\delta x^2-\frac{1}{\gamma },B_0=-\frac{1}{\gamma },B_1=0,B_2=\delta ,\\ C\left(x\right)=C_0+C_{1}x,C_0=2R,C_1=-4R^{2}E.\end{array}$ (17)

Let us determine $C_0$ and $C_1$, and thereafter obtain R and E. The implementation of Algorithm (11)-(12) yields the following algebraic system that consists of two nonlinear equations with unknowns $\left\{C_0\mathrm{,}{C}_1\right\}$ :

$C_1=n\left(1-n\right)-n\delta$ (18)

$-n\left(n-1\right)r_{n-1}^{\left(0\right)}+n\left(\left(-1/\gamma \right)r_{n-1}^{\left(0\right)}+n\delta r_n^{\left(0\right)}\right)+C_0{r}_n^{\left(0\right)}=0$ (19)

where $\left\{r_i^{\left(0\right)}\right\}$ are obtained by,

$\left(k\left(k-1\right)+k\delta +C_1\right)r_{k+1}^{\left(0\right)}=\left(k\left(k-1\right)+k/\gamma \right)r_{k-1}^{\left(0\right)}-C_0{r}_k^{(0)}$

which is equivalent to the explicit recursion

$r_{k+1}^{\left(0\right)}=\frac{\left(k\left(k-1\right)+k/\gamma \right)r_{k-1}^{\left(0\right)}-C_0{r}_k^{\left(0\right)}}{\left(k\left(k-1\right)+k\delta +C_1\right)}\mathrm{,}\text{with}{r}_0^{\left(0\right)}=1$

Then the desired solution $Y\left(x\right)$ follows from Equation (13):

$Y\left(x\right)=x^n-{\displaystyle \underset{i=0}{\overset{2}{\sum }}}\left(n\left(n-1\right)A_i+n{B}_{i-1}+C_{i-2}\right)Q_{n+i-2}$

where $\left\{Q_k\mathrm{<mo>\; ;\; </mo>0}\le k\le n\right\}$ are given by the recursion:

$\left(k\left(k-1\right)+k\delta +C_1\right)Q_{k+1}=x^k+\left(k\left(k-1\right)+k/\gamma \right)Q_{k-1}-C_0{Q}_k$

or, equivalently,

$Q_{k+1}=\frac{x^k+\left(k\left(k-1\right)+k/\gamma \right)Q_{k-1}-C_0{Q}_k}{\left(k\left(k-1\right)+k\delta +C_1\right)},\text{with}{Q}_0=0.$

To illustrate let us consider different values for n.

Case $n=2$ : In this case $C_1=-4D$ and the values of $C_0$ are the roots of the equation:

$-c_0\left(c_0^2+2\left(1-4D\right)D\right)=0$

which are $\left\{\mathrm{0,}\pm \sqrt{2D\left(4D-1\right)}\right\}$.

When $C_0=\sqrt{2D\left(4D-1\right)}$ we obtain:

$E=\frac{2}{4D-1}$

$R=\sqrt{\frac{4D^2-D}{2}}$

$C\left(x\right)=\sqrt{2D\left(4D-1\right)}-4Dx$

$Y\left(x\right)=\frac{D-1}{2D+1}+\frac{\sqrt{2D\left(4D-1\right)}}{2D+1}x+x^2$

$\Psi \left(s\right)=Y\left(\frac{s}{2R}\right)=\frac{D-1}{2D+1}+\frac{1}{2D+1}s-\frac{1}{2D-8D^2}{s}^2.$

The value $C_0=0$ implies that $R=0$ and if $C_0=-\sqrt{2D\left(4D-1\right)}$ lead to $R<0$. For these reasons, they are discarded.

Case $n=3$ : In this case $C_1=-3-6D$ and the values of $C_0$ are the roots of the equations

$-c_0^4+c_0^2\left(20D^2+20D+6\right)+9\left(-4D^4-8D^3+D^2+8D+3=0\right)$

which are

$\left\{\pm \sqrt{10D^2+10D+3\pm \Delta }\right\}$ where $\Delta =\sqrt{64D^4+128D^3+169D^2+132D+36}$

When $C_0=\sqrt{10D^2+10D+3\pm \Delta }$ we obtain:

$E=\frac{1}{2}\sqrt{10D^2+10D+3\pm \Delta }$

$R=\frac{6D+3}{10D^2+10D+3\pm \Delta }$

$C\left(x\right)=\left(\sqrt{10D^2+10D+3\pm \Delta }\right)-\left(6D+3\right)x$

$\begin{array}{c}Y\left(x\right)=\left[\frac{\left(-4D^2\pm \Delta -13D-6\right)\sqrt{10D^2\pm \Delta +10D+3}}{12\left(D+1\right)\left(2D+1\right)\left(2D+3\right)}\right]\\ +\left[\frac{4D^2\pm \Delta -5D-6}{4\left(D+1\right)\left(2D+3\right)}\right]x+\frac{\sqrt{10D^2\pm \Delta +10D+3}}{2D+3}{x}^2+x^3\end{array}$

$\Psi \left(s\right)=Y\left(\frac{s}{2R}\right).$

The values $C_0=-\sqrt{10D^2+10D+3\pm \Delta }$ lead to $R<0$ and therefore they are discarded.

Case $n=4$ : In this case $C_1=-8-8D$ and the values of $C_0$ are the roots $\xi$ of the equation

$-\xi \left({\xi }^4-2{\xi }^2\left(20D^2+45D+28\right)+8\left(32D^4+144D^3+185D^2+18D-64\right)\right)=0$

which are

$\left\{0,\pm \sqrt{20D^2+45D+28\pm 3\Delta }\right\}$ where