Analytical Computation of Hyper-Ellipsoidal Harmonics ^†

Abstract

The four-dimensional ellipsoid of an anisotropic hyper-structure corresponds to the four-dimensional sphere of an isotropic hyper-structure. In three dimensions, both theories for spherical and ellipsoidal harmonics have been developed by Laplace and Lamé, respectively. Nevertheless, in four dimensions, only the theory of hyper-spherical harmonics is hitherto known. This void in the literature is expected to be filled up by the present work. In fact, it is well known that the spectral decomposition of the Laplace equation in three-dimensional ellipsoidal geometry leads to the Lamé equation. This Lamé equation governs each one of the spectral functions corresponding to the three ellipsoidal coordinates, which, however, live in non-overlapping intervals. The analysis of the Lamé equation leads to four classes of Lamé functions, giving a total of 2n + 1 functions of degree n. In four dimensions, a much more elaborate procedure leads to similar results for the hyper-ellipsoidal structure. Actually, we demonstrate here that there are eight classes of the spectral hyper-Lamé equation and we provide a complete analysis for each one of them. The number of hyper-Lamé functions of degree n is (n + 1)²; that is, n² more functions than the three-dimensional case. However, the main difficulty in the four-dimensional analysis concerns the evaluation of the three separation constants appearing during the separation process. One of them can be extracted from the corresponding theory of the hyper-sphero-conal system, but the other two constants are obtained via a much more complicated procedure than the three-dimensional case. In fact, the solution process exhibits specific nonlinearities of polynomial type, itemized for every class and every degree. An example of this procedure is demonstrated in detail in order to make the process clear. Finally, the hyper-ellipsoidal harmonics are given as the product of four identical hyper-Lamé functions, each one defined in its own domain, which are explicitly calculated and tabulated for every degree less than five.

1. Introduction

The celebrated theory of ellipsoidal harmonics was introduced by Gabriel Lamé (1795–1870) in a series of papers [1,2,3,4,5,6] that finally led him to introduce the systems of curvilinear coordinates [3]. This theory, a complete exposition of which is included in [7], has been well-developed during the last two centuries [8,9,10,11,12]. However, the main difficulty with relevant applications during the first decades of the 20th century is due to the fact that in order to obtain the Lamé functions of degree greater than seven, we need to find the roots of irreducible polynomials of degree greater than four [9], a task for which no formulae are possible. On the other hand, there has been a growing interest in the use of ellipsoidal harmonics during the last few decades [13,14,15,16,17,18,19,20]. Obviously, this is connected to both the construction of very efficient numerical techniques [17] and the extraordinary improvement of computational power. Due to this high level of computational abilities, the next reasonable step would be a corresponding analysis of the ellipsoidal environment in four dimensions. This is the plan of the present effort. However, this extension is a task by no means trivial since the amount of work needed demands long, careful, and tedious calculations, as well as answers to some problems that are not easily extended from three to four dimensions. For possible applications of the introduced four-dimensional ellipsoidal harmonics, we refer the reader to [21] and the references therein, which involve applications of the corresponding degenerate case of hyper-spherical geometry.

Hence, in the present paper, we focus on the analysis of the hyper-Lamé equation and the classification of its solutions into eight classes. Once this is done, the corresponding ellipsoidal harmonics are obtained trivially as the product of four spectral functions. For each one of the eight classes, we provide a systematic way to obtain closed-form solutions for every degree (n). The totality of these solutions of degree n belonging to all classes is (n + 1)². Therefore, we obtain n² solutions more than the 2n + 1 ones which correspond to the case of three dimensions. In addition, the spectral decomposition of the 4D ellipsoidal Laplacian [22] leaves us with three independent separation constants. One of them can be obtained via the corresponding analysis of the hyper-sphero-conal system that we introduce and the corresponding 4D spherical system into which it degenerates. The calculation of the other two separation constants is by no means an easy task. In fact, we show that for every class and degree, the calculation of the other two separation constants demands the solution of a $2\times 2$ polynomial system of successively higher and higher degrees. All these are described in every detail for each one of the eight classes.

This work is organized as follows. Section 2 recapitulates the basic elements of the hyper-ellipsoidal system introduced by the first author in [22]. Section 3 is devoted to the spectral decomposition of the hyper-Laplacian, which leads to the hyper-Lamé equation. This equation governs each one of the four spectral functions, which are defined in different intervals. The study of the hyper-Lamé equation, which is included in Section 4, leads to eight classes of different solutions of the first kind, denoted by [${\rm K}j$], j = 1, 2, …, 8. Next, a detailed analysis of the [K1] class is given in Section 5, while the analysis of the classes [K2], [K3], and [K4], which are endowed with a symmetric structure, is given in Section 6. Similarly, because of relative symmetry, the classes [K5], [K6], and [K7] are analyzed in Section 7. Finally, Section 8 contains the analysis of the [K8] class, and Section 9 deals with some theoretical aspects concerning the separation constants.

At this point, the investigation of the hyper-Lamé classes is completed and a few Appendices are in order. Appendix A provides some complicated identities among the six principal semi-focal distances of the system, which are very useful for performing the calculations appearing in this work. Appendix B recapitulates the basic formulae of the 4D spherical harmonics, and Appendix C introduces the 4D sphero-conal geometry, as well as the corresponding Laplacian, from which we extract the value of one of the three separation constants. The last Appendix D contains all the solutions of the hyper-Lamé equation of degrees 0, 1, 2, 3, and 4 for all classes.

2. The Hyper-Ellipsoidal System

A complete analysis of second-degree surfaces that led to the definition of ellipsoidal coordinates in four dimensions has been demonstrated in [22]. This system is defined as

$$x_1={\displaystyle \frac{{\xi }_1{\xi }_2{\xi }_3{\xi }_4}{h_2{h}_3{h}_4}},$$

(1)

$$x_2={\displaystyle \frac{\sqrt{h_3^2-{\xi }_1^2}\sqrt{{\xi }_2^2-h_3^2}\sqrt{{\xi }_3^2-h_3^2}\sqrt{{\xi }_4^2-h_3^2}}{h_1{h}_3{h}_5}},$$

(2)

$$x_3={\displaystyle \frac{\sqrt{h_2^2-{\xi }_1^2}\sqrt{h_2^2-{\xi }_2^2}\sqrt{{\xi }_3^2-h_2^2}\sqrt{{\xi }_4^2-h_2^2}}{h_1{h}_2{h}_6}},$$

(3)

$$x_4={\displaystyle \frac{\sqrt{h_4^2-{\xi }_1^2}\sqrt{h_4^2-{\xi }_2^2}\sqrt{h_4^2-{\xi }_3^2}\sqrt{{\xi }_4^2-h_4^2}}{h_4{h}_5{h}_6}},$$

(4)

where ${\xi }_i, i=1,2,3,4$ are the ellipsoidal coordinates and $h_i, i=1,2,\dots ,6$ are the principal semi-focal distances, which are ordered in the following way:

$$0\le {\xi }_1^2\le h_3^2\le {\xi }_2^2\le h_2^2\le {\xi }_3^2\le h_4^2\le {\xi }_4^2\le +\infty .$$

(5)

In fact, the six semi-focal distances, which are defined as

$$\left.\begin{array}{c}{h}_1^2=a_2^2-a_3^2, h_2^2=a_1^2-a_3^2, h_3^2=a_1^2-a_2^2\\ h_4^2=a_1^2-a_4^2, h_5^2=a_2^2-a_4^2, h_6^2=a_3^2-a_4^2\end{array}\right\},$$

(6)

are connected via

$$h_1^2-h_2^2+h_3^2=0,$$

(7)

$$h_3^2-h_4^2+h_5^2=0,$$

(8)

$$h_2^2-h_4^2+h_6^2=0$$

(9)

and since $h_3$, $h_2$ are the corresponding independent parameters in the 3D ellipsoidal system, we choose $h_3$, $h_2$, $h_4$ as the three independent semi-focal distances for the 4D system.

The hyper-ellipsoid

$${\displaystyle \frac{x_1^2}{a_1^2}}+{\displaystyle \frac{x_2^2}{a_2^2}}+{\displaystyle \frac{x_3^2}{a_3^2}}+{\displaystyle \frac{x_4^2}{a_4^2}}=1$$

(10)

is the reference ellipsoid, which plays the role of the unit sphere in spherical coordinates, where we assume that

$$0<a_4<a_3<a_2<a_1<+\infty .$$

(11)

The hyper-ellipsoidal system is orthogonal and has the metric coefficients [22]:

$$h_{{\xi }_1}^2={\displaystyle \frac{\left({\xi }_1^2-{\xi }_2^2\right)\left({\xi }_1^2-{\xi }_3^2\right)\left({\xi }_1^2-{\xi }_4^2\right)}{\left({\xi }_1^2-h_3^2\right)\left({\xi }_1^2-h_2^2\right)\left({\xi }_1^2-h_4^2\right)}},$$

(12)

$$h_{{\xi }_2}^2={\displaystyle \frac{\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_2^2-{\xi }_3^2\right)\left({\xi }_2^2-{\xi }_4^2\right)}{\left({\xi }_2^2-h_3^2\right)\left({\xi }_2^2-h_2^2\right)\left({\xi }_2^2-h_4^2\right)}},$$

(13)

$$h_{{\xi }_3}^2={\displaystyle \frac{\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_2^2\right)\left({\xi }_3^2-{\xi }_4^2\right)}{\left({\xi }_3^2-h_3^2\right)\left({\xi }_3^2-h_2^2\right)\left({\xi }_3^2-h_4^2\right)}},$$

(14)

$$h_{{\xi }_4}^2={\displaystyle \frac{\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_3^2\right)}{\left({\xi }_4^2-h_3^2\right)\left({\xi }_4^2-h_2^2\right)\left({\xi }_4^2-h_4^2\right)}},$$

(15)

which define the diagonal metric dyadic

$$\tilde{g}={\displaystyle \sum _{k=1}^4{h}_{{\xi }_k}^2{\widehat{x}}_k\otimes {\widehat{x}}_k}.$$

(16)

Finally, the Laplace operator assumes the form

$$\begin{array}{c}\hfill \Delta ={\displaystyle \frac{1}{h_{{\xi }_1}^2}}\left[{\displaystyle \frac{{\partial }^2}{\partial {\xi }_1^2}}+\left({\displaystyle \frac{{\xi }_1}{{\xi }_1^2-h_3^2}}+{\displaystyle \frac{{\xi }_1}{{\xi }_1^2-h_2^2}}+{\displaystyle \frac{{\xi }_1}{{\xi }_1^2-h_4^2}}\right){\displaystyle \frac{\partial }{\partial {\xi }_1}}\right]\\ \hfill +{\displaystyle \frac{1}{h_{{\xi }_2}^2}}\left[{\displaystyle \frac{{\partial }^2}{\partial {\xi }_2^2}}+\left({\displaystyle \frac{{\xi }_2}{{\xi }_2^2-h_3^2}}+{\displaystyle \frac{{\xi }_2}{{\xi }_2^2-h_2^2}}+{\displaystyle \frac{{\xi }_2}{{\xi }_2^2-h_4^2}}\right){\displaystyle \frac{\partial }{\partial {\xi }_2}}\right]\\ \hfill +{\displaystyle \frac{1}{h_{{\xi }_3}^2}}\left[{\displaystyle \frac{{\partial }^2}{\partial {\xi }_3^2}}+\left({\displaystyle \frac{{\xi }_3}{{\xi }_3^2-h_3^2}}+{\displaystyle \frac{{\xi }_3}{{\xi }_3^2-h_2^2}}+{\displaystyle \frac{{\xi }_3}{{\xi }_3^2-h_4^2}}\right){\displaystyle \frac{\partial }{\partial {\xi }_3}}\right]\\ \hfill +{\displaystyle \frac{1}{h_{{\xi }_4}^2}}\left[{\displaystyle \frac{{\partial }^2}{\partial {\xi }_4^2}}+\left({\displaystyle \frac{{\xi }_4}{{\xi }_4^2-h_3^2}}+{\displaystyle \frac{{\xi }_4}{{\xi }_4^2-h_2^2}}+{\displaystyle \frac{{\xi }_4}{{\xi }_4^2-h_4^2}}\right){\displaystyle \frac{\partial }{\partial {\xi }_4}}\right],\end{array}$$

(17)

while, in terms of the generalized thermometric variables,

$${\tau }_1\left({\xi }_1\right)={\displaystyle \underset{0}{\overset{{\xi }_1}{\int }}\frac{dt}{\sqrt{h_3^2-t^2}\sqrt{h_2^2-t^2}\sqrt{h_4^2-t^2}}},$$

(18)

$${\tau }_2\left({\xi }_2\right)={\displaystyle \underset{h_3}{\overset{{\xi }_2}{\int }}\frac{dt}{\sqrt{t^2-h_3^2}\sqrt{h_2^2-t^2}\sqrt{h_4^2-t^2}}},$$

(19)

$${\tau }_3\left({\xi }_3\right)={\displaystyle \underset{h_2}{\overset{{\xi }_3}{\int }}\frac{dt}{\sqrt{t^2-h_3^2}\sqrt{t^2-h_2^2}\sqrt{h_4^2-t^2}}},$$

(20)

$${\tau }_4\left({\xi }_4\right)={\displaystyle \underset{h_4}{\overset{{\xi }_4}{\int }}\frac{dt}{\sqrt{t^2-h_3^2}\sqrt{t^2-h_2^2}\sqrt{t^2-h_4^2}}},$$

(21)

it is written as

$$\begin{array}{c}\Delta = {\displaystyle \frac{-1}{\left({\xi }_1^2-{\xi }_2^2\right)\left({\xi }_1^2-{\xi }_3^2\right)\left({\xi }_1^2-{\xi }_4^2\right)}}{\displaystyle \frac{{\partial }^2}{\partial {\tau }_1^2}}+{\displaystyle \frac{1}{\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_2^2-{\xi }_3^2\right)\left({\xi }_2^2-{\xi }_4^2\right)}}{\displaystyle \frac{{\partial }^2}{\partial {\tau }_2^2}}\\ +{\displaystyle \frac{-1}{\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_2^2\right)\left({\xi }_3^2-{\xi }_4^2\right)}}{\displaystyle \frac{{\partial }^2}{\partial {\tau }_3^2}}+{\displaystyle \frac{1}{\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_3^2\right)}}{\displaystyle \frac{{\partial }^2}{\partial {\tau }_4^2}},\end{array}$$

(22)

where the ${\xi }_k$ derivatives are connected to the ${\tau }_k$ derivatives by the relation

$$\begin{array}{c}{\displaystyle \frac{{\partial }^2}{\partial {\tau }_k^2}}={\left(-1\right)}^k\left({\xi }_k^2-h_3^2\right)\left({\xi }_k^2-h_2^2\right)\left({\xi }_k^2-h_4^2\right)\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\cdot \left[{\displaystyle \frac{{\partial }^2}{\partial {\xi }_k^2}}+\left({\displaystyle \frac{{\xi }_k}{{\xi }_k^2-h_3^2}}+{\displaystyle \frac{{\xi }_k}{{\xi }_k^2-h_2^2}}+{\displaystyle \frac{{\xi }_k}{{\xi }_k^2-h_4^2}}\right){\displaystyle \frac{\partial }{\partial {\xi }_k}}\right],\end{array}$$

(23)

for $k=1,2,3,4$.

3. Decomposing the Laplacian

Taking into consideration the analytic form of the Laplace equation:

$$\begin{gathered} \begin{array}{l}\Delta u= {\displaystyle \frac{1}{\left({\xi }_1^2-{\xi }_2^2\right)\left({\xi }_1^2-{\xi }_3^2\right)\left({\xi }_1^2-{\xi }_4^2\right)}}\left[\left({\xi }_1^2-h_3^2\right)\right.\left({\xi }_1^2-h_2^2\right)\left({\xi }_1^2-h_4^2\right){\displaystyle \frac{{\partial }^{2}u}{\partial {\xi }_1^2}}\\ \left.+{\xi }_1\left[\left({\xi }_1^2-h_2^2\right)\left({\xi }_1^2-h_4^2\right)+\left({\xi }_1^2-h_3^2\right)\left({\xi }_1^2-h_4^2\right)+\left({\xi }_1^2-h_3^2\right)\left({\xi }_1^2-h_2^2\right)\right]{\displaystyle \frac{\partial u}{\partial {\xi }_1}}\right]\end{array} \\ \begin{array}{l} +{\displaystyle \frac{1}{\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_2^2-{\xi }_3^2\right)\left({\xi }_2^2-{\xi }_4^2\right)}}\left[\left({\xi }_2^2-h_3^2\right)\right.\left({\xi }_2^2-h_2^2\right)\left({\xi }_2^2-h_4^2\right){\displaystyle \frac{{\partial }^{2}u}{\partial {\xi }_2^2}}\\ \left. +{\xi }_2\left[\left({\xi }_2^2-h_2^2\right)\left({\xi }_2^2-h_4^2\right)+\left({\xi }_2^2-h_3^2\right)\left({\xi }_2^2-h_4^2\right)+\left({\xi }_2^2-h_3^2\right)\left({\xi }_2^2-h_2^2\right)\right]{\displaystyle \frac{\partial u}{\partial {\xi }_2}}\right]\end{array} \\ \begin{array}{l} +{\displaystyle \frac{1}{\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_2^2\right)\left({\xi }_3^2-{\xi }_4^2\right)}}\left[\left({\xi }_3^2-h_3^2\right)\right.\left({\xi }_3^2-h_2^2\right)\left({\xi }_3^2-h_4^2\right){\displaystyle \frac{{\partial }^{2}u}{\partial {\xi }_3^2}}\\ \left. +{\xi }_3\left[\left({\xi }_3^2-h_2^2\right)\left({\xi }_3^2-h_4^2\right)+\left({\xi }_3^2-h_3^2\right)\left({\xi }_3^2-h_4^2\right)+\left({\xi }_3^2-h_3^2\right)\left({\xi }_3^2-h_2^2\right)\right]{\displaystyle \frac{\partial u}{\partial {\xi }_3}}\right]\end{array} \\ \begin{array}{l} +{\displaystyle \frac{1}{\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_3^2\right)}}\left[\left({\xi }_4^2-h_3^2\right)\right.\left({\xi }_4^2-h_2^2\right)\left({\xi }_4^2-h_4^2\right){\displaystyle \frac{{\partial }^{2}u}{\partial {\xi }_4^2}}\\ \left. +{\xi }_4\left[\left({\xi }_4^2-h_2^2\right)\left({\xi }_4^2-h_4^2\right)+\left({\xi }_4^2-h_3^2\right)\left({\xi }_4^2-h_4^2\right)+\left({\xi }_4^2-h_3^2\right)\left({\xi }_4^2-h_2^2\right)\right]{\displaystyle \frac{\partial u}{\partial {\xi }_4}}\right]\\ =0\end{array} \end{gathered}$$

(24)

and assuming the separation,

$$u\left({\xi }_1,{\xi }_2,{\xi }_3,{\xi }_4\right)=F_1\left({\xi }_1\right)F_2\left({\xi }_2\right)F_3\left({\xi }_3\right)F_4\left({\xi }_4\right),$$

(25)

we observe that Equation (24) is satisfied if

$$\begin{array}{l}\left({\xi }_k^2-h_3^2\right)\left({\xi }_k^2-h_2^2\right)\left({\xi }_k^2-h_4^2\right){\displaystyle \frac{d^2}{d{\xi }_k^2}}{F}_k\left({\xi }_k\right)\\ +\left[\left({\xi }_k^2-h_2^2\right)\left({\xi }_k^2-h_4^2\right)+\left({\xi }_k^2-h_4^2\right)\left({\xi }_k^2-h_3^2\right)+\left({\xi }_k^2-h_3^2\right)\left({\xi }_k^2-h_2^2\right)\right]{\xi }_k{\displaystyle \frac{d}{d{\xi }_k}}{F}_k\left({\xi }_k\right)=0,\end{array}$$

(26)

for $k=1,2,3,4$. It should be mentioned that the four ellipsoidal coordinates ${\xi }_1$, ${\xi }_2$, ${\xi }_3$, ${\xi }_4$ are of the same dimensional character, hence all the functions $F_k\left({\xi }_k\right)$, $k=1,2,3,4$ solve the same Equation (26). What differs is that the domain of ${\xi }_1^2$ is $\left(0,h_3^2\right)$, the domain of ${\xi }_2^2$ is the interval $\left(h_3^2,h_2^2\right)$, that of ${\xi }_3^2$ is $\left(h_2^2,h_4^2\right)$, while ${\xi }_4^2$ lives in $\left(h_4^2,+\infty \right)$. Nevertheless, the above separated Equation (26) does not involve the three independent separation constants, which are necessary for our 4D structure. One way to overcome this drawback would be to generalize Lamé’s relative argument, introduced in the 3D theory of ellipsoidal harmonics, which is described as follows.

If we multiply expression (22) by the product

$$\left({\xi }_1^2-{\xi }_2^2\right)\left({\xi }_1^2-{\xi }_3^2\right)\left({\xi }_1^2-{\xi }_4^2\right)\left({\xi }_2^2-{\xi }_3^2\right)\left({\xi }_2^2-{\xi }_4^2\right)\left({\xi }_3^2-{\xi }_4^2\right)$$

then Laplace’s equation takes the form

$$\begin{array}{l}\left({\xi }_3^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_3^2\right){\displaystyle \frac{{\partial }^{2}u}{\partial {\tau }_1^2}}+\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_3^2\right){\displaystyle \frac{{\partial }^{2}u}{\partial {\tau }_2^2}}\\ +\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_2^2\right){\displaystyle \frac{{\partial }^{2}u}{\partial {\tau }_3^2}}+\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_2^2\right){\displaystyle \frac{{\partial }^{2}u}{\partial {\tau }_4^2}}=0.\end{array}$$

(27)

Next, taking into account the form of the coefficients in (27), we can show that the following identities hold

$$\begin{array}{l}\left({\xi }_3^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_3^2\right)-\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_3^2\right)\\ +\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_2^2\right)-\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_2^2\right)=0,\end{array}$$

(28)

$$\begin{array}{l}{\xi }_1^2\left({\xi }_3^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_3^2\right)-{\xi }_2^2\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_3^2\right)\\ +{\xi }_3^2\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_2^2\right)-{\xi }_4^2\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_2^2\right)=0\end{array}$$

(29)

and

$$\begin{array}{l}{\xi }_1^4\left({\xi }_3^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_3^2\right)-{\xi }_2^4\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_3^2\right)\\ +{\xi }_3^4\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_2^2\right)-{\xi }_4^4\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_2^2\right)=0,\end{array}$$

(30)

where the negative sign in the second and fourth terms of (28)–(30) make up for the respective negative sign in (23). Consequently, multiplying (28) by $-Cu$, (29) by $-Bu$, (30) by $-Au$ and adding the resulting equations, we arrive at an expression that still vanishes.

Then, adding the above identity to Equation (27), we are led to the equation

$$\begin{array}{l}\left({\xi }_3^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_2^2\right)\left({\xi }_4^2-{\xi }_3^2\right)\left[{\displaystyle \frac{{\partial }^{2}u}{\partial {\tau }_1^2}}-A{\xi }_1^{4}u-B{\xi }_1^{2}u-Cu\right]\\ +\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_3^2\right)\left[{\displaystyle \frac{{\partial }^{2}u}{\partial {\tau }_2^2}}+A{\xi }_2^{4}u+B{\xi }_2^{2}u+Cu\right]\\ +\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_1^2\right)\left({\xi }_4^2-{\xi }_2^2\right)\left[{\displaystyle \frac{{\partial }^{2}u}{\partial {\tau }_3^2}}-A{\xi }_3^{4}u-B{\xi }_3^{2}u-Cu\right]\\ +\left({\xi }_2^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_1^2\right)\left({\xi }_3^2-{\xi }_2^2\right)\left[{\displaystyle \frac{{\partial }^{2}u}{\partial {\tau }_4^2}}+A{\xi }_4^{4}u+B{\xi }_4^{2}u+Cu\right]=0,\end{array}$$

(31)

which, in view of (23), yields the following identical hyper-Lamé equations

$$\begin{array}{l}\left({\xi }_i^2-h_3^2\right)\left({\xi }_i^2-h_2^2\right)\left({\xi }_i^2-h_4^2\right){\displaystyle \frac{d^2{F}_i\left({\xi }_i\right)}{d{\xi }_i^2}}+{\xi }_i\left[\left({\xi }_i^2-h_2^2\right)\left({\xi }_i^2-h_4^2\right)\right.\\ \left.+\left({\xi }_i^2-h_3^2\right)\left({\xi }_i^2-h_4^2\right)+\left({\xi }_i^2-h_3^2\right)\left({\xi }_i^2-h_2^2\right)\right]{\displaystyle \frac{d{F}_i\left({\xi }_i\right)}{d{\xi }_i}}+\left(A{\xi }_i^4+B{\xi }_i^2+C\right)F_i\left({\xi }_i\right)=0,\end{array}$$

(32)

with $i=1,2,3,4$, which now include the three separation constants $A,B$, and $C$.

As we will demonstrate in the following, the above Equation (32) is not amenable to straightforward analysis, despite being linear, mainly because of the three singularities at the points $h_3^2,h_2^2,h_4^2$, corresponding to the transition points connecting successively the domains of the $F_k$’s.

4. The Hyper-Lamé Equation

Our first goal in analyzing Equation (32) is to develop a systematic way to calculate the separation constants $A$, $B$, and $C$. One of these constants can be determined by reducing the 4D ellipsoidal system to the 4D sphero-conal one and then comparing the shero-conal with the 4D spherical system, just as Lamé did in three dimensions. This procedure leads to the value

$$A=-n\left(n+2\right)$$

(33)

as it is demonstrated in Appendix C. As we will show in the sequel, the constants $B$ and $C$ are calculated in a more complicated way.

A straightforward extension of Lamé arguments from the classical Lamé equation to the hyper-Lamé equation separates the totality of the hyper-Lamé functions into eight distinct classes, given in

Table 1. Classification of hyper-Lamé functions of degree $n$. The $P_k$’s are polynomials of degree $k$.

Generating Factors	Class	Form of Solutions
$1$	[K1]	$P_n$
$x_2$	[K2]	$\sqrt{\left|x^2-h_3^2\right|}{P}_{n-1}$
$x_3$	[K3]	$\sqrt{\left|x^2-h_2^2\right|}{P}_{n-1}$
$x_4$	[K4]	$\sqrt{\left|x^2-h_4^2\right|}{P}_{n-1}$
$x_2{x}_3$	[K5]	$\sqrt{\left|x^2-h_3^2\right|}\sqrt{\left|x^2-h_2^2\right|}{P}_{n-2}$
$x_2{x}_4$	[K6]	$\sqrt{\left|x^2-h_2^2\right|}\sqrt{\left|x^2-h_4^2\right|}{P}_{n-2}$
$x_3{x}_4$	[K7]	$\sqrt{\left|x^2-h_3^2\right|}\sqrt{\left|x^2-h_4^2\right|}{P}_{n-2}$
$x_2{x}_3{x}_4$	[K8]	$\sqrt{\left|x^2-h_3^2\right|}\sqrt{\left|x^2-h_2^2\right|}\sqrt{\left|x^2-h_4^2\right|}{P}_{n-3}$

.

The hyper-Lamé equation can be written as

$$\begin{array}{l}\left(x^6-H_1{x}^4+H_2{x}^2-H_3\right)F_n^{\prime \prime }\left(x\right)+x\left(3x^4-2H_1{x}^2+H_2\right)F_n^{\prime }\left(x\right)\\ +\left[H_{2}Q+H_{1}P x^2-n\left(n+2\right) x^4\right]F_n\left(x\right)=0,\end{array}$$

(34)

where

$$H_1=h_3^2+h_2^2+h_4^2,$$

(35)

$$H_2=h_3^2{h}_2^2+h_2^2{h}_4^2+h_4^2{h}_3^2,$$

(36)

$$H_3=h_3^2{h}_2^2{h}_4^2$$

(37)

and

$$B=H_{1}P ,$$

(38)

$$C=H_{2}Q ,$$

(39)

so that the two separation constants $B$ and $C$ are represented by the dimensionless parameters $P$ and $Q$, respectively. Note that the introduced factors $H_1$ and $H_2$ in (38) and (39) involve a symmetric representation of the geometric quantities $h_3$, $h_2$ and $h_4$.

Next, if we denote the hyper-Lamé function of the first kind of degree $n$ and order $m$ by $E_{n,4}^m$, then the corresponding interior hyper-ellipsoidal harmonic is defined as the product

$${\mathbb{E}}_{n,4}^m\left({\xi }_1,{\xi }_2,{\xi }_3,{\xi }_4\right)=E_{n,4}^m\left({\xi }_1\right)E_{n,4}^m\left({\xi }_2\right)E_{n,4}^m\left({\xi }_3\right)E_{n,4}^m\left({\xi }_4\right).$$

(40)

Finally, regarding the number of solutions for a given degree $n$, it should be noted that the ellipsoid is nothing more than a smooth and symmetric deformation of the sphere and, as such, the subspaces of solutions of the same degree for spherical and ellipsoidal geometries have the same cardinality. Hence, as shown in Appendix B, Formula (A40), we conclude that there are in total ${\left(n+1\right)}^2$ hyper-ellipsoidal harmonics as well as ${\left(n+1\right)}^2$ hyper-Lamé functions of degree $n$.

5. Class [K1]

Since the hyper-Lamé Equation (34) can be transformed into an equation in $x^2$, it follows that we look for solutions of the form

$$K_n^1\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{a}_k{x}^{n-2k}}.$$

(41)

Inserting (41) in (34), rearranging the indices, and setting equal to zero every coefficient, we arrive at the recurrence relation of the third order

$$\begin{array}{l}4\left(k+1\right)\left(k-n\right)a_{k+1}+H_1\left[P-\left(n-2k\right)\left(n-2k+1\right)\right] a_k\\ +H_2\left[Q+{\left(n-2k+2\right)}^2\right]a_{k-1}-H_3\left(n-2k+4\right)\left(n-2k+3\right)a_{k-2}=0 ,\end{array}$$

(42)

with $k$ being non-negative integer, while $a_k=0$ for $k<0$.

Specifically, for $k=0,1,2,\dots ,r+2$, with

$$r=\left\{\begin{array}{ll}\frac{n}{2},& \mathrm{for} n \mathrm{even} \\ \frac{n-1}{2},& \mathrm{for} n \mathrm{odd}, \end{array}\right.$$

(43)

we obtain the relations

$$4n{a}_1=H_1\left[P-n\left(n+1\right)\right] a_0,$$

(44)

$$8\left(n-1\right)a_2=H_1\left[P-\left(n-2\right)\left(n-1\right)\right] a_1+H_2\left[Q+n^2\right]a_0,$$

(45)

$$12\left(n-2\right)a_3=H_1\left[P-\left(n-4\right)\left(n-3\right)\right] a_2+H_2\left[Q+{\left(n-2\right)}^2\right]a_1-H_{3}n\left(n-1\right)a_0,$$

(46)

$$\begin{array}{l}16\left(n-3\right)a_4=H_1\left[P-\left(n-6\right)\left(n-5\right)\right] a_3+H_2\left[Q+{\left(n-4\right)}^2\right]a_2-H_3\left(n-2\right)\left(n-3\right)a_1,\\ ⋮\end{array}$$

(47)

$$\begin{array}{cc}\hfill 4r\left(n-r+1\right)a_r& =H_1\left[P-\left(n-2r+2\right)\left(n-2r+3\right)\right] a_{r-1}\hfill \\ & +H_2\left[Q+{\left(n-2r+4\right)}^2\right]a_{r-2}-H_3\left(n-2r+6\right)\left(n-2r+5\right)a_{r-3},\hfill \end{array}$$

(48)

$$\begin{array}{cc}\hfill 4\left(r+1\right)\left(n-r\right)a_{r+1}& =H_1\left[P-\left(n-2r\right)\left(n-2r+1\right)\right] a_r\hfill \\ & +H_2\left[Q+{\left(n-2r+2\right)}^2\right]a_{r-1}-H_3\left(n-2r+4\right)\left(n-2r+3\right)a_{r-2}\hfill \end{array}$$

(49)

as well as

$$\begin{array}{cc}\hfill 4\left(r+2\right)\left(n-r-1\right)a_{r+2}& =H_1\left[P-\left(n-2r-2\right)\left(n-2r-1\right)\right] a_{r+1}\hfill \\ & +H_2\left[Q+{\left(n-2r\right)}^2\right]a_r-H_3\left(n-2r+2\right)\left(n-2r+1\right)a_{r-1},\hfill \end{array}$$

(50)

$$\begin{array}{cc}\hfill 4\left(r+3\right)\left(n-r-2\right)a_{r+3}& =H_1\left[P-\left(n-2r-4\right)\left(n-2r-3\right)\right] a_{r+2}\hfill \\ & +H_2\left[Q+{\left(n-2r-2\right)}^2\right]a_{r+1}-H_3\left(n-2r\right)\left(n-2r-1\right)a_r.\hfill \end{array}$$

(51)

At this point, we explain how to deal with the system (44)–(51), which involves $r+3$ equations, in order to secure polynomial solutions for the hyper-Lamé Equation (34). This system involves the two unknown constants $P$ and $Q$, as well as the unknown coefficients $a_0,a_1,\dots ,a_{r+3}$. That is $r+6$ unknowns.

First, we observe that the last term in (51) is equal to zero for every $n$, due to the allowed values of $r$ (43). In fact, the vanishing of this term is the criterion for truncating the system at the particular Equation (51). Next, we can choose the dimensionless parameters $P$ and $Q$ to be solutions of the system of equations

$$a_{r+1}=0$$

(52)

and

$$a_{r+2}=0,$$

(53)

which reduces the number of unknowns to $r+4$. Then, (52) and (53) imply, via (51), that

$$a_{r+3}=0$$

(54)

and we are left with the system (44)–(50), which has $r+3$ unknowns and $r+2$ equations, while (41) takes the form

$$\begin{array}{cc}\hfill K_n^1\left(x\right)& ={\displaystyle \sum _{k=0}^r{a}_k{x}^{n-2k}}\hfill \\ & =a_0{x}^n+a_1{x}^{n-2}+a_2{x}^{n-4}+ \cdots +a_r{x}^{n-2r}.\hfill \end{array}$$

(55)

Note that the last term of the polynomial (55) is a constant for $n=\mathrm{even}$, or a constant times $x$ for $n=\mathrm{odd}$. Hence, $K_n^1\left(x\right)$ has $r+1$ unknown coefficients, i.e., $\left(n+2\right)/2$ for $n=\mathrm{even}$, or $\left(n+1\right)/2$ for $n=\mathrm{odd}$.

From the first $r+1$ Equations (44)–(49) we obtain a homogeneous system of linear equations in variables $a_0,a_1,a_2,\dots ,a_r$. Here we state the obvious condition that we can take, i.e., $a_0=1$, which further reduces the total number of unknowns to $r+2$. Hence, we are left with the condition of the vanishing determinant, which secures the existence of nontrivial solutions for the above homogeneous system, plus the remaining Equation (50). These two relations lead to a $2\times 2$ polynomial system in $P$ and $Q$, where each one of its solution pairs $\left(P,Q\right)$ provides a nontrivial solution of the linear system (44)–(49). Note that, as we have proved in Section 9 and demonstrated in Appendix D, these solution pairs $\left(P,Q\right)$ are all real.

In particular, we arrive at the $\left(r+1\right)\times \left(r+1\right)$ linear system

$$\begin{array}{c}\left[\begin{array}{cccccccccc}{M}_{11}^1& M_{12}^1& 0& 0& ·& 0& 0& 0& 0& 0\\ M_{21}^1& M_{22}^1& M_{23}^1& 0& ·& 0& 0& 0& 0& 0\\ M_{31}^1& M_{32}^1& M_{33}^1& M_{34}^1& ·& 0& 0& 0& 0& 0\\ ·& ·& ·& ·& \stackrel{·}{·}& ·& ·& ·& ·& ·\\ 0& 0& 0& 0& ·& M_{(r-1)(r-3)}^1& M_{(r-1)(r-2)}^1& M_{(r-1)(r-1)}^1& M_{(r-1)r}^1& 0\\ 0& 0& 0& 0& ·& 0& M_{r(r-2)}^1& M_{r(r-1)}^1& M_{rr}^1& M_{r(r+1)}^1\\ 0& 0& 0& 0& ·& 0& 0& M_{(r+1)(r-1)}^1& M_{(r+1)r}^1& M_{(r+1)(r+1)}^1\end{array}\right]\left[\begin{array}{c}{a}_0\\ a_1\\ a_2\\ ·\\ ·\\ a_{r-2}\\ a_{r-1}\\ a_r\end{array}\right]\hfill \\ =0\hfill \end{array}$$

(56)

with

$$M_{jj}^1=H_1\left[P-\left(n-2j+2\right)\left(n-2j+3\right)\right],\hspace{1em}j=1,2,\dots ,r+1,$$

(57)

$$M_{j\left(j+1\right)}^1=4j\left(j-n-1\right),\hspace{1em}j=1,2,\dots ,r,$$

(58)

$$M_{\left(j+1\right)j}^1=H_2\left[Q+{\left(n-2j+2\right)}^2\right],\hspace{1em}j=1,2,\dots ,r,$$

(59)

$$M_{\left(j+2\right)j}^1=-H_3\left(n-2j+2\right)\left(n-2j+1\right),\hspace{1em}j=1,2,\dots ,r-1.$$

(60)

Herein, nontrivial solutions exist if and only if we require that the corresponding determinant is equal to zero. Obviously, this leads to a polynomial equation of degree $r+1$ in $P$ and of a lower degree in $Q$. Besides the aforementioned condition, we also have the following one, which is directly derived from (50),

$$H_2\left[Q+{\left(n-2r\right)}^2\right]a_r-H_3\left(n-2r+2\right)\left(n-2r+1\right)a_{r-1}=0,$$

(61)

where the expressions of $a_r$ and $a_{r-1}$ are obtained from the solution of (56), assuming that $a_0=1$. So, utilizing the first polynomial equation in $P$ and $Q$ (vanishing determinant), we can express, e.g., $Q$ as a function of $P$ and then (61) will become a polynomial equation only in $P$.

In order to make this procedure clear, we provide the following example.

Example 1. 

Consider the case$n=4.$ Find the values of $$and$$that allow for solutions of the form

$$K_4^1\left(x\right)=a_0{x}^4+a_1{x}^2+a_2$$

(62)

and then calculate the coefficients$a_0$,$a_1$,$a_2$.

In this case, the Equations (56) and (61) are respectively written

$$H_1\left(P-20\right)a_0-16a_1=0,$$

(63)

$$H_2\left(Q+16\right)a_0+H_1\left(P-6\right)a_1-24a_2=0,$$

(64)

$$-12H_3{a}_0+H_2\left(Q+4\right)a_1+H_{1}P{a}_2=0$$

(65)

and

$$-2H_3{a}_1+H_{2}Q{a}_2=0.$$

(66)

Then, we pick up the system (63)–(65) and impose the condition

$$\left|\begin{array}{ccc}{H}_1\left(P-20\right)& -16& 0\\ H_2\left(Q+16\right)& H_1\left(P-6\right)& -24\\ -12H_3& H_2\left(Q+4\right)& H_{1}P\end{array}\right|=0$$

(67)

or

$$\begin{array}{l}{H}_1^{3}P\left(P-6\right)\left(P-20\right)-4608H_3\\ +16H_1{H}_{2}P\left(Q+16\right)+24H_1{H}_2\left(P-20\right)\left(Q+4\right)=0.\end{array}$$

(68)

Eliminating the coefficients $a_1$ and $a_2$ between Equations (63)–(66) and putting $a_0=1$, we arrive at the second polynomial equation in $P$ and $Q$

$$2H_1^2{H}_{3}P\left(P-20\right)-192H_2{H}_{3}Q+H_1{H}_2^2\left(P-20\right)Q\left(Q+4\right)=0.$$

(69)

The two polynomial Equations (68) and (69) are solved by pairs $\left(P,Q\right)$. In fact, eliminating the parameter Q, we end up with the 6th-degree polynomial equation in $P$

$$\begin{array}{l}{H}_1^2{H}_2\left(P-6\right)\left(P-20\right){\Pi }_1\left(P\right){\Pi }_3\left(P\right)+16H_2^2\left({\Pi }_3^2\left(P\right)+16{\Pi }_1\left(P\right){\Pi }_3\left(P\right)\right)\\ -48H_1{H}_3\left(P-20\right){\Pi }_1^2\left(P\right)=0,\end{array}$$

(70)

where

$${\Pi }_1\left(P\right)=40H_1{H}_2\left(P-12\right),$$

(71)

$${\Pi }_3\left(P\right)=-H_1^3{P}^3+26H_1^3{P}^2-\left(352H_1{H}_2+120H_1^3\right)P+1920H_1{H}_2+4608H_3$$

(72)

and

$$Q_j={\displaystyle \frac{{\Pi }_3\left(P_j\right)}{{\Pi }_1\left(P_j\right)}}, j=1,2,\dots ,6.$$

(73)

Hence, we obtain six pairs of simple real roots $\left(P_j,Q_j\right)$, $j=1,2,\dots ,6$ which give the six hyper-Lamé functions of degree 4

$$E_{4,4}^j\left(x\right)=x^4+{\displaystyle \frac{H_1}{16}}\left(P_j-20\right)x^2+{\displaystyle \frac{H_1{H}_3}{8H_2}}{\displaystyle \frac{\left(P_j-20\right)}{Q_j}}, j=1,2,\dots ,6.$$

(74)

6. Classes [K2]–[K4]

We remind that the classes [K2], [K3], and [K4] involve solutions of the form

$$K_n^2\left(x\right)=\sqrt{\left|x^2-h_3^2\right|}{L}_{n-1}^2\left(x\right),$$

(75)

$$K_n^3\left(x\right)=\sqrt{\left|x^2-h_2^2\right|}{L}_{n-1}^3\left(x\right),$$

(76)

$$K_n^4\left(x\right)=\sqrt{\left|x^2-h_4^2\right|}{L}_{n-1}^4\left(x\right),$$

(77)

respectively. Therefore, we should expect that the expressions of $L_{n-1}^2\left(x\right)$, $L_{n-1}^3\left(x\right)$ and $L_{n-1}^4\left(x\right)$ will differ only in the constants $h_3^2$, $h_2^2$ and $h_4^2$.

For [K2], we consider the function

$$L_{n-1}^2\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{b}_k{x}^{n-2k-1}}.$$

(78)

Inserting (75) in (34), we find that (78) has to satisfy the equation

$$\begin{array}{l}\left[x^6-H_1{x}^4+H_2{x}^2-H_3\right]{L_{n-1}^2}^{\prime \prime }\left(x\right)+x\left[5x^4-2\left(2H_1-h_3^2\right)x^2+H_2+2H_3{h}_3^{-2}\right]{L_{n-1}^2}^{\prime }\left(x\right)\\ +\left\{H_{2}Q+H_3{h}_3^{-2}+\left[H_1\left(P-2\right)+2h_3^2\right] x^2+\left[3-n\left(n+2\right)\right] x^4\right\}{L}_{n-1}^2\left(x\right)=0,\end{array}$$

(79)

from which we obtain

$$\begin{array}{l}{\displaystyle \sum \left\{4k\left(k-n-1\right) x^{n-2k+3}{b}_k+\left[H_1\left[P -\left(n-2k+1\right)\left(n-2k\right)\right]+2h_3^2\left(n-2k\right)\right]x^{n-2k+1}{b}_k\right.}\\ \left.+\left[H_2\left[Q+{\left(n-2k-1\right)}^2\right]+H_3{h}_3^{-2}\left(2n-4k-1\right)\right]x^{n-2k-1}{b}_k-H_3\left(n-2k-1\right)\left(n-2k-2\right)x^{n-2k-3}{b}_k\right\}\\ =0.\end{array}$$

(80)

Hence, the recurrence relation of the third order we obtain is

$$\begin{array}{l}4\left(k+1\right)\left(k-n\right) b_{k+1}+\left\{H_1\left[P -\left(n-2k+1\right)\left(n-2k\right)\right]+2h_3^2\left(n-2k\right)\right\}{b}_k\\ +\left\{H_2\left[Q+{\left(n-2k+1\right)}^2\right]+H_3{h}_3^{-2}\left(2n-4k+3\right)\right\}{b}_{k-1}-H_3\left(n-2k+3\right)\left(n-2k+2\right)b_{k-2}=0,\end{array}$$

(81)

with $k=0,1,2,\dots$.

We now define

$$r\equiv r_L=\left\{\begin{array}{ll}\frac{n}{2},& \mathrm{for} n \mathrm{even} \\ \frac{n+1}{2},& \mathrm{for} n \mathrm{odd}, \end{array}\right.$$

(82)

in terms of which we write down the relations

$$4n b_1=\left\{H_1\left[P -n\left(n+1\right)\right]+2h_3^{2}n\right\}{b}_0,$$

(83)

$$\begin{array}{cc}\hfill 8\left(n-1\right) b_2& =\left\{H_1\left[P -\left(n-1\right)\left(n-2\right)\right]+2h_3^2\left(n-2\right)\right\}{b}_1\hfill \\ & +\left\{H_2\left[Q+{\left(n-1\right)}^2\right]+H_3{h}_3^{-2}\left(2n-1\right)\right\}{b}_0,\hfill \end{array}$$

(84)

$$\begin{array}{cc}\hfill 12\left(n-2\right) b_3& =\left\{H_1\left[P -\left(n-3\right)\left(n-4\right)\right]+2h_3^2\left(n-4\right)\right\}{b}_2\hfill \\ & +\left\{H_2\left[Q+{\left(n-3\right)}^2\right]+H_3{h}_3^{-2}\left(2n-5\right)\right\}{b}_1-H_3\left(n-1\right)\left(n-2\right)b_0,\hfill \end{array}$$

(85)

$$\begin{array}{cc}\hfill 16\left(n-3\right) b_4& =\left\{H_1\left[P -\left(n-5\right)\left(n-6\right)\right]+2h_3^2\left(n-6\right)\right\}{b}_3\hfill \\ & +\left\{H_2\left[Q+{\left(n-5\right)}^2\right]+H_3{h}_3^{-2}\left(2n-9\right)\right\}{b}_2-H_3\left(n-3\right)\left(n-4\right)b_1,\hfill \\ & \hfill ⋮\hfill \end{array}$$

(86)

$$\begin{array}{l}4\left(r-1\right)\left(n-r+2\right) b_{r-1}=\left\{H_1\left[P -\left(n-2r+5\right)\left(n-2r+4\right)\right]+2h_3^2\left(n-2r+4\right)\right\}{b}_{r-2}\\ +\left\{H_2\left[Q+{\left(n-2r+5\right)}^2\right]+H_3{h}_3^{-2}\left(2n-4r+11\right)\right\}{b}_{r-3}-H_3\left(n-2r+7\right)\left(n-2r+6\right)b_{r-4},\end{array}$$

(87)

$$\begin{array}{l}4r\left(n-r+1\right) b_r=\left\{H_1\left[P -\left(n-2r+3\right)\left(n-2r+2\right)\right]+2h_3^2\left(n-2r+2\right)\right\}{b}_{r-1}\\ +\left\{H_2\left[Q+{\left(n-2r+3\right)}^2\right]+H_3{h}_3^{-2}\left(2n-4r+7\right)\right\}{b}_{r-2}-H_3\left(n-2r+5\right)\left(n-2r+4\right)b_{r-3},\end{array}$$

(88)

and

$$\begin{array}{l}4\left(r+1\right)\left(n-r\right) b_{r+1}=\left\{H_1\left[P -\left(n-2r+1\right)\left(n-2r\right)\right]+2h_3^2\left(n-2r\right)\right\}{b}_r\\ +\left\{H_2\left[Q+{\left(n-2r+1\right)}^2\right]+H_3{h}_3^{-2}\left(2n-4r+3\right)\right\}{b}_{r-1}-H_3\left(n-2r+3\right)\left(n-2r+2\right)b_{r-2},\end{array}$$

(89)

$$\begin{array}{l}4\left(r+2\right)\left(n-r-1\right) b_{r+2}=\left\{H_1\left[P -\left(n-2r-1\right)\left(n-2r-2\right)\right]+2h_3^2\left(n-2r-2\right)\right\}{b}_{r+1}\\ +\left\{H_2\left[Q+{\left(n-2r-1\right)}^2\right]+H_3{h}_3^{-2}\left(2n-4r-1\right)\right\}{b}_r-H_3\left(n-2r+1\right)\left(n-2r\right)b_{r-1} .\end{array}$$

(90)

As before, the last term of (90) vanishes due to (82), and we demand that $P$ and $Q$ solve the polynomial system of equations

$$b_r=0$$

(91)

and

$$b_{r+1}=0.$$

(92)

Then, we have

$$b_{r+2}=0$$

(93)

and, therefore,

$$b_k=0,\hspace{1em}k>r+2.$$

(94)

Hence, we end up with the $r\times r$ system

$$\begin{array}{c}\left[\begin{array}{cccccccccc}{M}_{11}^2& M_{12}^2& 0& 0& .& 0& 0& 0& 0& 0\\ M_{21}^2& M_{22}^2& M_{23}^2& 0& ·& 0& 0& 0& 0& 0\\ M_{31}^2& M_{32}^2& M_{33}^2& M_{34}^2& ·& 0& 0& 0& 0& 0\\ ·& ·& ·& ·& \stackrel{·}{·}& ·& ·& ·& ·& ·\\ 0& 0& 0& 0& ·& M_{(r-2)(r-4)}^2& M_{(r-2)(r-3)}^2& M_{(r-2)(r-2)}^2& M_{(r-2)(r-1)}^2& 0\\ 0& 0& 0& 0& ·& 0& M_{(r-1)(r-3)}^2& M_{(r-1)(r-2)}^2& M_{(r-1)(r-1)}^2& M_{(r-1)r}^2\\ 0& 0& 0& 0& ·& 0& 0& M_{r(r-2)}^2& M_{r(r-1)}^2& M_{rr}^2\end{array}\right]\left[\begin{array}{c}{b}_0\\ b_1\\ b_2\\ ·\\ ·\\ b_{r-3}\\ b_{r-2}\\ b_{r-1}\end{array}\right]\hfill \\ =0,\hfill \end{array}$$

(95)

with

$$M_{jj}^2=H_1\left[P -\left(n-2j+3\right)\left(n-2j+2\right)\right]+2h_3^2\left(n-2j+2\right),\hspace{1em}j=1,2,\dots ,r,$$

(96)

$$M_{j\left(j+1\right)}^2=4j\left(j-n-1\right),\hspace{1em}j=1,2,\dots ,r-1,$$

(97)

$$M_{\left(j+1\right)j}^2=H_2\left[Q+{\left(n-2j+1\right)}^2\right]+H_3{h}_3^{-2}\left(2n-4j+3\right),\hspace{1em}j=1,2,\dots ,r-1,$$

(98)

$$M_{\left(j+2\right)j}^2=-H_3\left(n-2j+1\right)\left(n-2j\right),\hspace{1em}j=1,2,\dots ,r-2,$$

(99)

for which nontrivial solutions exist if we require that the determinant is equal to zero. This leads to a polynomial equation of degree $r$ with respect to $P$ and a lower degree with respect to $Q$. As a result, if we express the constant Q as a function of $P$ and substitute it together with the calculated coefficients $b_{r-2}\left(P\right)$ and $b_{r-1}\left(P\right)$ into the additional Equation (89), which, due to (91) and (92), is written as

$$\left\{H_2\left[Q+{\left(n-2r+1\right)}^2\right]+H_3{h}_3^{-2}\left(2n-4r+3\right)\right\}{b}_{r-1}-H_3\left(n-2r+3\right)\left(n-2r+2\right)b_{r-2}=0,$$

(100)

we can then derive the appropriate values of $P$ and subsequently obtain the sought-out values of the coefficients $b_k$, $k=0,1,\dots ,r-1$, where we can assume that, e.g., $b_0=1$.

Next, following the exact same procedure as with the above [K2] class and assuming that the functions $L_{n-1}^3$ and $L_{n-1}^4$, embodied in (76) and (77), respectively, have the same form as (78), we arrive at similar results for the classes [K3] and [K4]. These can be directly derived from the corresponding relations of [K2], e.g., (79), (96)–(99), and (100), substituting the semi-focal distance $h_3$ with $h_2$ for [K3] or $h_3$ with $h_4$ for [K4] class.

In fact, for the [K3] class, considering the function

$$L_{n-1}^3\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{b}_k{x}^{n-2k-1}},$$

(101)

where $L_{n-1}^3$ satisfies

$$\begin{array}{l}\left[x^6-H_1{x}^4+H_2{x}^2-H_3\right]{L_{n-1}^3}^{\prime \prime }\left(x\right)+x\left[5x^4-2\left(2H_1-h_2^2\right)x^2+H_2+2H_3{h}_2^{-2}\right]{L_{n-1}^3}^{\prime }\left(x\right)\\ +\left\{H_{2}Q+H_3{h}_2^{-2}+\left[H_1\left(P-2\right)+2h_2^2\right] x^2+\left[3-n\left(n+2\right)\right] x^4\right\}{L}_{n-1}^3\left(x\right)=0 ,\end{array}$$

(102)

we are led to a $r\times r$ linear homogeneous system, specified by the non-zero coefficient matrix entries

$$M_{jj}^3=H_1\left[P -\left(n-2j+3\right)\left(n-2j+2\right)\right]+2h_2^2\left(n-2j+2\right),\hspace{1em}j=1,2,\dots ,r,$$

(103)

$$M_{j\left(j+1\right)}^3=4j\left(j-n-1\right),\hspace{1em}j=1,2,\dots ,r-1,$$

(104)

$$M_{\left(j+1\right)j}^3=H_2\left[Q+{\left(n-2j+1\right)}^2\right]+H_3{h}_2^{-2}\left(2n-4j+3\right),\hspace{1em}j=1,2,\dots ,r-1,$$

(105)

$$M_{\left(j+2\right)j}^3=-H_3\left(n-2j+1\right)\left(n-2j\right),\hspace{1em}j=1,2,\dots ,r-2$$

(106)

and we also obtain the following equation, corresponding to (100),

$$\left\{H_2\left[Q+{\left(n-2r+1\right)}^2\right]+H_3{h}_2^{-2}\left(2n-4r+3\right)\right\}{b}_{r-1}-H_3\left(n-2r+3\right)\left(n-2r+2\right)b_{r-2}=0.$$

(107)

Finally, for the class [K4] we take

$$L_{n-1}^4\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{b}_k{x}^{n-2k-1}},$$

(108)

where $L_{n-1}^4$ satisfies

$$\begin{array}{l}\left[x^6-H_1{x}^4+H_2{x}^2-H_3\right]{L_{n-1}^4}^{\prime \prime }\left(x\right)+x\left[5x^4-2\left(2H_1-h_4^2\right)x^2+H_2+2H_3{h}_4^{-2}\right]{L_{n-1}^4}^{\prime }\left(x\right)\\ +\left\{H_{2}Q+H_3{h}_4^{-2}+\left[H_1\left(P-2\right)+2h_4^2\right] x^2+\left[3-n\left(n+2\right)\right] x^4\right\}{L}_{n-1}^4\left(x\right)=0 \end{array}$$

(109)

and we now obtain the $r\times r$ homogeneous system with matrix elements

$$M_{jj}^4=H_1\left[P -\left(n-2j+3\right)\left(n-2j+2\right)\right]+2h_4^2\left(n-2j+2\right),\hspace{1em}j=1,2,\dots ,r,$$

(110)

$$M_{j\left(j+1\right)}^4=4j\left(j-n-1\right),\hspace{1em}j=1,2,\dots ,r-1,$$

(111)

$$M_{\left(j+1\right)j}^4=H_2\left[Q+{\left(n-2j+1\right)}^2\right]+H_3{h}_4^{-2}\left(2n-4j+3\right),\hspace{1em}j=1,2,\dots ,r-1,$$

(112)

$$M_{\left(j+2\right)j}^4=-H_3\left(n-2j+1\right)\left(n-2j\right),\hspace{1em}j=1,2,\dots ,r-2,$$

(113)

while the complementary relation (100) is now written as

$$\left\{H_2\left[Q+{\left(n-2r+1\right)}^2\right]+H_3{h}_4^{-2}\left(2n-4r+3\right)\right\}{b}_{r-1}-H_3\left(n-2r+3\right)\left(n-2r+2\right)b_{r-2}=0.$$

(114)

The particular expressions of $K_{n-1}^2$, $K_{n-1}^3$, $K_{n-1}^4$ for $n=0,1,2,3,4$ are given in Appendix D.

7. Classes [K5]–[K7]

The next classes [K5], [K6] and [K7] have solutions of the form

$$K_n^5\left(x\right)=\sqrt{\left|x^2-h_3^2\right|}\sqrt{\left|x^2-h_2^2\right|}{L}_{n-2}^5\left(x\right),$$

(115)

$$K_n^6\left(x\right)=\sqrt{\left|x^2-h_2^2\right|}\sqrt{\left|x^2-h_4^2\right|}{L}_{n-2}^6\left(x\right),$$

(116)

$$K_n^7\left(x\right)=\sqrt{\left|x^2-h_4^2\right|}\sqrt{\left|x^2-h_3^2\right|}{L}_{n-2}^7\left(x\right),$$

(117)

respectively.

In the [K5] class, the polynomial $L_{n-2}^5$ has to solve the equation

$$\begin{array}{l}\left[x^6-H_1{x}^4+H_2{x}^2-H_3\right]{L_{n-2}^5}^{\prime \prime }\left(x\right)+x\left[7x^4-2\left(3H_1-h_3^2-h_2^2\right)x^2+3H_2-2h_3^2{h}_2^2\right]{L_{n-2}^5}^{\prime }\left(x\right)\\ +\left\{H_2\left(Q+1\right)-h_3^2{h}_2^2+\left[H_1\left(P-6\right)+4\left(h_3^2+h_2^2\right)\right] x^2+\left[8-n\left(n+2\right)\right] x^4\right\}{L}_{n-2}^5\left(x\right)=0\end{array}$$

(118)

and if we insert the function

$$L_{n-2}^5\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{c}_k{x}^{n-2k-2}},$$

(119)

we arrive at the third-order recurrence relation

$$\begin{array}{l}4\left(k+1\right)\left(k-n\right) c_{k+1}+\left\{H_1\left[P-\left(n-2k\right)\left(n-2k+1\right)\right]+2\left(h_3^2+h_2^2\right)\left(n-2k\right)\right\}{c}_k\\ +\left\{H_2\left[Q+{\left(n-2k+1\right)}^2\right]-h_3^2{h}_2^2\left(2n-4k+1\right)\right\}{c}_{k-1}-H_3\left(n-2k+2\right)\left(n-2k+1\right)c_{k-2}=0,\end{array}$$

(120)

with

$$r\equiv r_M=\left\{\begin{array}{ll}\frac{n+2}{2},& \mathrm{for} n \mathrm{even} \\ \frac{n+1}{2},& \mathrm{for} n \mathrm{odd}. \end{array}\right.$$

(121)

Analyzing relation (120), we obtain

$$4n c_1=\left\{H_1\left[P-n\left(n+1\right)\right]+2\left(h_3^2+h_2^2\right)n\right\}{c}_0,$$

(122)

$$\begin{array}{cc}\hfill 8\left(n-1\right) c_2& =\left\{H_1\left[P-\left(n-2\right)\left(n-1\right)\right]+2\left(h_3^2+h_2^2\right)\left(n-2\right)\right\}{c}_1\hfill \\ & +\left\{H_2\left[Q+{\left(n-1\right)}^2\right]-h_3^2{h}_2^2\left(2n-3\right)\right\}{c}_0,\hfill \end{array}$$

(123)

$$\begin{array}{cc}\hfill 12\left(n-2\right) c_3& =\left\{H_1\left[P-\left(n-4\right)\left(n-3\right)\right]+2\left(h_3^2+h_2^2\right)\left(n-4\right)\right\}{c}_2\hfill \\ & +\left\{H_2\left[Q+{\left(n-3\right)}^2\right]-h_3^2{h}_2^2\left(2n-7\right)\right\}{c}_1-H_3\left(n-2\right)\left(n-3\right)c_0, \hfill \end{array}$$

(124)

$$\begin{array}{cc}\hfill 16\left(n-3\right) c_4& =\left\{H_1\left[P-\left(n-6\right)\left(n-5\right)\right]+2\left(h_3^2+h_2^2\right)\left(n-6\right)\right\}{c}_3\hfill \\ & +\left\{H_2\left[Q+{\left(n-5\right)}^2\right]-h_3^2{h}_2^2\left(2n-11\right)\right\}{c}_2-H_3\left(n-4\right)\left(n-5\right)c_1,\hfill \\ & \hfill ⋮\hfill \end{array}$$

(125)

$$\begin{array}{l}4\left(r-2\right)\left(n-r+3\right) c_{r-2}=\left\{H_1\left[P-\left(n-2r+6\right)\left(n-2r+7\right)\right]+2\left(h_3^2+h_2^2\right)\left(n-2r+6\right)\right\}{c}_{r-3}\\ +\left\{H_2\left[Q+{\left(n-2r+7\right)}^2\right]-h_3^2{h}_2^2\left(2n-4r+13\right)\right\}{c}_{r-4}-H_3\left(n-2r+8\right)\left(n-2r+7\right)c_{r-5},\end{array}$$

(126)

$$\begin{array}{l}4\left(r-1\right)\left(n-r+2\right) c_{r-1}=\left\{H_1\left[P-\left(n-2r+4\right)\left(n-2r+5\right)\right]+2\left(h_3^2+h_2^2\right)\left(n-2r+4\right)\right\}{c}_{r-2}\\ +\left\{H_2\left[Q+{\left(n-2r+5\right)}^2\right]-h_3^2{h}_2^2\left(2n-4r+9\right)\right\}{c}_{r-3}-H_3\left(n-2r+6\right)\left(n-2r+5\right)c_{r-4},\end{array}$$

(127)

and

$$\begin{array}{l}4r\left(n-r+1\right) c_r=\left\{H_1\left[P-\left(n-2r+2\right)\left(n-2r+3\right)\right]+2\left(h_3^2+h_2^2\right)\left(n-2r+2\right)\right\}{c}_{r-1}\\ +\left\{H_2\left[Q+{\left(n-2r+3\right)}^2\right]-h_3^2{h}_2^2\left(2n-4r+5\right)\right\}{c}_{r-2}-H_3\left(n-2r+4\right)\left(n-2r+3\right)c_{r-3},\end{array}$$

(128)

$$\begin{array}{l}4\left(r+1\right)\left(n-r\right) c_{r+1}=\left\{H_1\left[P-\left(n-2r\right)\left(n-2r+1\right)\right]+2\left(h_3^2+h_2^2\right)\left(n-2r\right)\right\}{c}_r\\ +\left\{H_2\left[Q+{\left(n-2r+1\right)}^2\right]-h_3^2{h}_2^2\left(2n-4r+1\right)\right\}{c}_{r-1}-H_3\left(n-2r+2\right)\left(n-2r+1\right)c_{r-2} .\end{array}$$

(129)

As in the previous classes, the last term of (129) vanishes by the particular choice of $r$ (121), and we choose $P$ and $Q$ so that

$$c_{r-1}=0$$

(130)

and

$$c_r=0.$$

(131)

Then, we obtain

$$c_{r+1}=0$$

(132)

and, therefore,

$$c_k=0,\hspace{1em}k>r+1.$$

(133)

These conditions imply that the function $L_{n-2}^5\left(x\right)$, defined in (119), degenerates to a polynomial of degree $n-2$. The coefficients $c_k$, $k=0,1,\dots ,r-2$ are obtained as the solution of the $\left(r-1\right)\times \left(r-1\right)$ system

$$\begin{array}{c}\left[\begin{array}{cccccccccc}{M}_{11}^5& M_{12}^5& 0& 0& ·& 0& 0& 0& 0& 0\\ M_{21}^5& M_{22}^5& M_{23}^5& 0& ·& 0& 0& 0& 0& 0\\ M_{31}^5& M_{32}^5& M_{33}^5& M_{34}^5& ·& 0& 0& 0& 0& 0\\ ·& ·& ·& ·& \stackrel{·}{·}& ·& ·& ·& ·& ·\\ 0& 0& 0& 0& ·& M_{(r-3)(r-5)}^5& M_{(r-3)(r-4)}^5& M_{(r-3)(r-3)}^5& M_{(r-3)(r-2)}^5& 0\\ 0& 0& 0& 0& ·& 0& M_{(r-2)(r-4)}^5& M_{(r-2)(r-3)}^5& M_{(r-2)(r-2)}^5& M_{(r-2)(r-1)}^5\\ 0& 0& 0& 0& ·& 0& 0& M_{(r-1)(r-3)}^5& M_{(r-1)(r-2)}^5& M_{(r-1)(r-1)}^5\end{array}\right]\left[\begin{array}{c}{c}_0\\ c_1\\ c_2\\ ·\\ ·\\ c_{r-4}\\ c_{r-3}\\ c_{r-2}\end{array}\right]\hfill \\ =0,\hfill \end{array}$$

(134)

with

$$M_{jj}^5=H_1\left[P-\left(n-2j+2\right)\left(n-2j+3\right)\right]+2\left(h_3^2+h_2^2\right)\left(n-2j+2\right),\hspace{1em}j=1,2,\dots ,r-1,$$

(135)

$$M_{j\left(j+1\right)}^5=4j\left(j-n-1\right),\hspace{1em}j=1,2,\dots ,r-2,$$

(136)

$$M_{\left(j+1\right)j}^5=H_2\left[Q+{\left(n-2j+1\right)}^2\right]-h_3^2{h}_2^2\left(2n-4j+1\right),\hspace{1em}j=1,2,\dots ,r-2,$$

(137)

$$M_{\left(j+2\right)j}^5=-H_3\left(n-2j\right)\left(n-2j-1\right),\hspace{1em}j=1,2,\dots ,r-3.$$

(138)

Again, for the above linear system we demand that its determinant vanishes, which leads to a polynomial equation of degree $r-1$ with respect to $P$. In addition, Equation (128), in view of (130) and (131), yields

$$\left\{H_2\left[Q+{\left(n-2r+3\right)}^2\right]-h_3^2{h}_2^2\left(2n-4r+5\right)\right\}{c}_{r-2}-H_3\left(n-2r+4\right)\left(n-2r+3\right)c_{r-3}=0,$$

(139)

where the coefficients $c_{r-3}$ and $c_{r-2}$ are functions of $P$ (or $Q$).

Following exactly the same steps as with the [K5] class, we arrive at almost identical equations for the classes [K6] and [K7]. These can be derived from the respective relations of [K5], substituting $h_3$ with $h_4$ for [K6] class or $h_2$ with $h_4$ for [K7].

Indeed, for [K6], if we assume that

$$L_{n-2}^6\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{c}_k{x}^{n-2k-2}},$$

(140)

we conclude that it has to satisfy the equation

$$\begin{array}{l}\left[x^6-H_1{x}^4+H_2{x}^2-H_3\right]{L_{n-2}^6}^{\prime \prime }\left(x\right)+x\left[7x^4-2\left(3H_1-h_4^2-h_2^2\right)x^2+3H_2-2h_4^2{h}_2^2\right]{L_{n-2}^6}^{\prime }\left(x\right)\\ +\left\{H_2\left(Q+1\right)-h_4^2{h}_2^2+\left[H_1\left(P-6\right)+4\left(h_4^2+h_2^2\right)\right] x^2+\left[8-n\left(n+2\right)\right] x^4\right\}{L}_{n-2}^6\left(x\right)=0 .\end{array}$$

(141)

Therefore, we are finally led to a $\left(r-1\right)\times \left(r-1\right)$ homogeneous system, whose matrix $(M_{ij}^6)$ contains the non-zero elements

$$M_{jj}^6=H_1\left[P-\left(n-2j+2\right)\left(n-2j+3\right)\right]+2\left(h_4^2+h_2^2\right)\left(n-2j+2\right),\hspace{1em}j=1,2,\dots ,r-1,$$

(142)

$$M_{j\left(j+1\right)}^6=4j\left(j-n-1\right),\hspace{1em}j=1,2,\dots ,r-2,$$

(143)

$$M_{\left(j+1\right)j}^6=H_2\left[Q+{\left(n-2j+1\right)}^2\right]-h_4^2{h}_2^2\left(2n-4j+1\right),\hspace{1em}j=1,2,\dots ,r-2,$$

(144)

$$M_{\left(j+2\right)j}^6=-H_3\left(n-2j\right)\left(n-2j-1\right),\hspace{1em}j=1,2,\dots ,r-3,$$

(145)

while the complementary equation corresponding to (139) is given by

$$\left\{H_2\left[Q+{\left(n-2r+3\right)}^2\right]-h_4^2{h}_2^2\left(2n-4r+5\right)\right\}{c}_{r-2}-H_3\left(n-2r+4\right)\left(n-2r+3\right)c_{r-3}=0.$$

(146)

In the same way, for the class [K7], we have the function

$$L_{n-2}^7\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{c}_k{x}^{n-2k-2}},$$

(147)

which has to satisfy the equation

$$\begin{array}{l}\left[x^6-H_1{x}^4+H_2{x}^2-H_3\right]{L_{n-2}^7}^{\prime \prime }\left(x\right)+x\left[7x^4-2\left(3H_1-h_3^2-h_4^2\right)x^2+3H_2-2h_3^2{h}_4^2\right]{L_{n-2}^7}^{\prime }\left(x\right)\\ +\left\{H_2\left(Q+1\right)-h_3^2{h}_4^2+\left[H_1\left(P-6\right)+4\left(h_3^2+h_4^2\right)\right] x^2+\left[8-n\left(n+2\right)\right] x^4\right\}{L}_{n-2}^7\left(x\right)=0,\end{array}$$

(148)

where the coefficients $c_k$, $k=0,1,\dots ,r-2$ satisfy a homogeneous system with corresponding matrix entries

$$M_{jj}^7=H_1\left[P-\left(n-2j+2\right)\left(n-2j+3\right)\right]+2\left(h_3^2+h_4^2\right)\left(n-2j+2\right),\hspace{1em}j=1,2,\dots ,r-1,$$

(149)

$$M_{j\left(j+1\right)}^7=4j\left(j-n-1\right),\hspace{1em}j=1,2,\dots ,r-2,$$

(150)

$$M_{\left(j+1\right)j}^7=H_2\left[Q+{\left(n-2j+1\right)}^2\right]-h_3^2{h}_4^2\left(2n-4j+1\right),\hspace{1em}j=1,2,\dots ,r-2,$$

(151)

$$M_{\left(j+2\right)j}^7=-H_3\left(n-2j\right)\left(n-2j-1\right),\hspace{1em}j=1,2,\dots ,r-3,$$

(152)

while the complementary equation reads

$$\left\{H_2\left[Q+{\left(n-2r+3\right)}^2\right]-h_3^2{h}_4^2\left(2n-4r+5\right)\right\}{c}_{r-2}-H_3\left(n-2r+4\right)\left(n-2r+3\right)c_{r-3}=0.$$

(153)

8. Class [K8]

Finally, in the [K8] class, where all three square roots are present, we have

$$K_n^8\left(x\right)=\sqrt{\left|x^2-h_3^2\right|}\sqrt{\left|x^2-h_2^2\right|}\sqrt{\left|x^2-h_4^2\right|}{L}_{n-3}^8\left(x\right)$$

(154)

and

$$L_{n-3}^8\left(x\right)={\displaystyle \sum _{k=0}^{\infty }{d}_k{x}^{n-2k-3}}.$$

(155)

The function $L_{n-3}^8$ has to satisfy the equation

$$\begin{array}{l}\left[x^6-H_1{x}^4+H_2{x}^2-H_3\right]{L_{n-3}^8}^{\prime \prime }\left(x\right)+3x\left[3x^4-2H_1{x}^2+H_2\right]{L_{n-3}^8}^{\prime }\left(x\right)\\ +\left[H_2\left(Q+1\right)+H_1\left(P-6\right) x^2+\left(15-n\left(n+2\right)\right) x^4\right]L_{n-3}^8\left(x\right)=0,\end{array}$$

(156)

which leads to the third-order recurrence relation

$$\begin{array}{l}4\left(k+1\right)\left(k-n\right) d_{k+1}+H_1\left[P-\left(n-2k\right)\left(n-2k-1\right)\right]d_k\\ +H_2\left[Q+{\left(n-2k\right)}^2\right]d_{k-1}-H_3\left(n-2k+1\right)\left(n-2k\right)d_{k-2}=0,\end{array}$$

(157)

with

$$r\equiv r_N=\left\{\begin{array}{ll}\frac{n+2}{2},& \mathrm{for} n \mathrm{even} \\ \frac{n+3}{2},& \mathrm{for} n \mathrm{odd}. \end{array}\right.$$

(158)

If we expand (157), we obtain

$$4n d_1=H_1\left[P-n\left(n-1\right)\right]d_0,$$

(159)

$$8\left(n-1\right) d_2=H_1\left[P-\left(n-2\right)\left(n-3\right)\right]d_1+H_2\left[Q+{\left(n-2\right)}^2\right]d_0,$$

(160)

$$12\left(n-2\right) d_3=H_1\left[P-\left(n-4\right)\left(n-5\right)\right]d_2+H_2\left[Q+{\left(n-4\right)}^2\right]d_1-H_3\left(n-3\right)\left(n-4\right)d_0,$$

(161)

$$\begin{array}{l}16\left(n-3\right) d_4=H_1\left[P-\left(n-6\right)\left(n-7\right)\right]d_3+H_2\left[Q+{\left(n-6\right)}^2\right]d_2-H_3\left(n-5\right)\left(n-6\right)d_1,\\ \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}⋮\end{array}$$

(162)

$$\begin{array}{cc}\hfill 4\left(r-3\right)\left(n-r+4\right) d_{r-3}& =H_1\left[P-\left(n-2r+8\right)\left(n-2r+7\right)\right]d_{r-4}\hfill \\ & +H_2\left[Q+{\left(n-2r+8\right)}^2\right]d_{r-5}-H_3\left(n-2r+9\right)\left(n-2r+8\right)d_{r-6},\hfill \end{array}$$

(163)

$$\begin{array}{cc}\hfill 4\left(r-2\right)\left(n-r+3\right) d_{r-2}& =H_1\left[P-\left(n-2r+6\right)\left(n-2r+5\right)\right]d_{r-3}\hfill \\ & +H_2\left[Q+{\left(n-2r+6\right)}^2\right]d_{r-4}-H_3\left(n-2r+7\right)\left(n-2r+6\right)d_{r-5},\hfill \end{array}$$

(164)

and

$$\begin{array}{cc}\hfill 4\left(r-1\right)\left(n-r+2\right) d_{r-1}& =H_1\left[P-\left(n-2r+4\right)\left(n-2r+3\right)\right]d_{r-2}\hfill \\ & +H_2\left[Q+{\left(n-2r+4\right)}^2\right]d_{r-3}-H_3\left(n-2r+5\right)\left(n-2r+4\right)d_{r-4},\hfill \end{array}$$

(165)

$$\begin{array}{cc}\hfill 4r\left(n-r+1\right) d_r& =H_1\left[P-\left(n-2r+2\right)\left(n-2r+1\right)\right]d_{r-1}\hfill \\ & +H_2\left[Q+{\left(n-2r+2\right)}^2\right]d_{r-2}-H_3\left(n-2r+3\right)\left(n-2r+2\right)d_{r-3} .\hfill \end{array}$$

(166)

Due to the allowed values of $r$, as they are given in (158), the last term in (166) equals zero, and if we demand that the constants $P$ and $Q$ are solutions of the system

$$d_{r-2}=0,$$

(167)

$$d_{r-1}=0,$$

(168)

then we have

$$d_r=0.$$

(169)

Hence, $d_k=0$, for $k>r$, while the sought-out function (155) turns into a polynomial of degree $n-3$, i.e.,

$$L_{n-3}^8\left(x\right)={\displaystyle \sum _{k=0}^{r-3}{d}_k{x}^{n-2k-3}}.$$

(170)

Herein, again, we obtain one equation more than the unknown coefficients $d_k$, i.e., relation (165), which is written as

$$H_2\left[Q+{\left(n-2r+4\right)}^2\right]d_{r-3}-H_3\left(n-2r+5\right)\left(n-2r+4\right)d_{r-4}=0.$$

(171)

We can ignore this equation and solve the rest of them, i.e., (159)–(164), to find all the coefficients. That gives the following $\left(r-2\right)\times \left(r-2\right)$ linear system in $d_k$, $k=0,1,\dots ,r-3,$

$$\begin{array}{c}\left[\begin{array}{cccccccccc}{M}_{11}^8& M_{12}^8& 0& 0& ·& 0& 0& 0& 0& 0\\ M_{21}^8& M_{22}^8& M_{23}^8& 0& ·& 0& 0& 0& 0& 0\\ M_{31}^8& M_{32}^8& M_{33}^8& M_{34}^8& ·& 0& 0& 0& 0& 0\\ ·& ·& ·& ·& \stackrel{·}{·}& ·& ·& ·& ·& ·\\ 0& 0& 0& 0& ·& M_{(r-4)(r-6)}^8& M_{(r-4)(r-5)}^8& M_{(r-4)(r-4)}^8& M_{(r-4)(r-3)}^8& 0\\ 0& 0& 0& 0& ·& 0& M_{(r-3)(r-5)}^8& M_{(r-3)(r-4)}^8& M_{(r-3)(r-3)}^8& M_{(r-3)(r-2)}^8\\ 0& 0& 0& 0& ·& 0& 0& M_{(r-2)(r-4)}^8& M_{(r-2)(r-3)}^8& M_{(r-2)(r-2)}^8\end{array}\right]\left[\begin{array}{c}{d}_0\\ d_1\\ d_2\\ ·\\ ·\\ d_{r-4}\\ d_{r-5}\\ d_{r-3}\end{array}\right]\hfill \\ =0,\hfill \end{array}$$

(172)

with

$$M_{jj}^8=H_1\left[P-\left(n-2j+2\right)\left(n-2j+1\right)\right], j=1,2,\dots ,r-2,$$

(173)

$$M_{j\left(j+1\right)}^8=4j\left(j-n-1\right),\hspace{1em}j=1,2,\dots ,r-3,$$

(174)

$$M_{\left(j+1\right)j}^8=H_2\left[Q+{\left(n-2j\right)}^2\right],\hspace{1em}j=1,2,\dots ,r-3,$$

(175)

$$M_{\left(j+2\right)j}^8=-H_3\left(n-2j-1\right)\left(n-2j-2\right),\hspace{1em}j=1,2,\dots ,r-4.$$

(176)

In a similar fashion, the generated linear system also involves the two separation constants $P$ and $Q$. Since this system is homogeneous, in order to obtain nontrivial solutions, we need to require that its determinant equals zero, yielding a non-linear relation among $P$ and $Q$. In addition, if we insert the calculated coefficients in the ignored Equation (171), we obtain a second non-linear relation among the parameters $P$ and $Q$, leading to a $2\times 2$ polynomial system for their determination. Finally, each pair of $\left(P,Q\right)$ is inserted in the expressions of the coefficients, and we arrive at the polynomial (170), which is part of the hyper-Lamé solution (154).

9. Theoretical Aspects

In the present section, we investigate the nature of the separation constants $P$ and $Q$. We mainly analyze the first class since the analysis for the rest of the classes is exactly the same.

Lemma 1. 

Let$E_n^j\left(x\right)$be a hyper-Lamé function of class [K1] which corresponds to the pair of separation constants$\left(P,Q.\right)$ Then, both$$and$$are real.

Proof. 

For ${\xi }_3\in \left(h_2,h_4\right)$, the function $E_n^j\left({\xi }_3\right)$ satisfies the equation

$$-R_1{\displaystyle \frac{d}{d{\xi }_3}}{R}_1{\displaystyle \frac{d}{d{\xi }_3}}{E}_n^j\left({\xi }_3\right)+\left[H_2{Q}_j+H_1{P}_j {\xi }_3^2-n\left(n+2\right) {\xi }_3^4\right]E_n^j\left({\xi }_3\right)=0,$$

(177)

where

$$R_1=\sqrt{\left({\xi }_3^2-h_3^2\right)\left({\xi }_3^2-h_2^2\right)\left(h_4^2-{\xi }_3^2\right)}.$$

(178)

On the other hand, $E_n^j\left({\xi }_2\right)$, for ${\xi }_2\in \left(h_3,h_2\right)$, satisfies the equation

$$R_2{\displaystyle \frac{d}{d{\xi }_2}}{R}_2{\displaystyle \frac{d}{d{\xi }_2}}{E}_n^j\left({\xi }_2\right)+\left[H_2{Q}_j+H_1{P}_j {\xi }_2^2-n\left(n+2\right) {\xi }_2^4\right]E_n^j\left({\xi }_2\right)=0,$$

(179)

where

$$R_2=\sqrt{\left({\xi }_2^2-h_3^2\right)\left(h_2^2-{\xi }_2^2\right)\left(h_4^2-{\xi }_2^2\right)}.$$

(180)

Next, we consider the function

$$F_n^j\left({\xi }_3,{\xi }_2\right)=E_n^j\left({\xi }_3\right)E_n^j\left({\xi }_2\right).$$

(181)

Multiplying (177) by $E_n^j\left({\xi }_2\right)$ and (179) by $E_n^j\left({\xi }_3\right)$ we obtain

$$-R_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{R}_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{F}_n^j\left({\xi }_3,{\xi }_2\right)+\left[H_2{Q}_j+H_1{P}_j {\xi }_3^2-n\left(n+2\right) {\xi }_3^4\right]F_n^j\left({\xi }_3,{\xi }_2\right)=0$$

(182)

and

$$R_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{R}_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{F}_n^j\left({\xi }_3,{\xi }_2\right)+\left[H_2{Q}_j+H_1{P}_j {\xi }_2^2-n\left(n+2\right) {\xi }_2^4\right]F_n^j\left({\xi }_3,{\xi }_2\right)=0.$$

(183)

Subtracting (183) from (182), we arrive at the relation

$$\begin{array}{l}-R_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{R}_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{F}_n^j-R_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{R}_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{F}_n^j\\ +\left[H_1{P}_j \left({\xi }_3^2-{\xi }_2^2\right)-n\left(n+2\right) \left({\xi }_3^4-{\xi }_2^4\right)\right]F_n^j=0.\end{array}$$

(184)

If $P_j$ is complex, then the conjugate value ${\overline{P}}_j$ generates the conjugate function ${\overline{F}}_n^j$, and we have the following conjugate to (184) equation

$$\begin{array}{l}-R_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{R}_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{\overline{F}}_n^j-R_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{R}_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{\overline{F}}_n^j\\ +\left[H_1{\overline{P}}_j \left({\xi }_3^2-{\xi }_2^2\right)-n\left(n+2\right) \left({\xi }_3^4-{\xi }_2^4\right)\right]{\overline{F}}_n^j=0.\end{array}$$

(185)

Multiplying (184) by ${\overline{F}}_n^j$, (185) by $F_n^j$ and subtracting the resulting equations we obtain

$$\begin{array}{l}{\overline{F}}_n^j\left[-R_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{R}_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{F}_n^j-R_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{R}_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{F}_n^j\right]-F_n^j\left[-R_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{R}_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{\overline{F}}_n^j-R_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{R}_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{\overline{F}}_n^j\right]\\ +\left(P_j-{\overline{P}}_j \right)H_1\left({\xi }_3^2-{\xi }_2^2\right)F_n^j{\overline{F}}_n^j=0,\end{array}$$

(186)

which is also written as

$$\begin{array}{l}{\displaystyle \frac{\left(P_j-{\overline{P}}_j \right)H_1\left({\xi }_3^2-{\xi }_2^2\right)F_n^j{\overline{F}}_n^j}{R{ }_1{R}_2}}\\ ={\displaystyle \frac{1}{R_2}}\left[{\overline{F}}_n^j\left({\displaystyle \frac{\partial }{\partial {\xi }_3}}{R}_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{F}_n^j\right)-F_n^j\left({\displaystyle \frac{\partial }{\partial {\xi }_3}}{R}_1{\displaystyle \frac{\partial }{\partial {\xi }_3}}{\overline{F}}_n^j\right)\right]\\ +{\displaystyle \frac{1}{R_1}}\left[{\overline{F}}_n^j\left({\displaystyle \frac{\partial }{\partial {\xi }_2}}{R}_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{F}_n^j\right)-F_n^j\left({\displaystyle \frac{\partial }{\partial {\xi }_2}}{R}_2{\displaystyle \frac{\partial }{\partial {\xi }_2}}{\overline{F}}_n^j\right)\right] .\end{array}$$

(187)

Then, integrating the expression (187) over the rectangle $\left(h_2,h_4\right)\times \left(h_3,h_2\right)$ we obtain

$$\begin{array}{c}\left(P_j-{\overline{P}}_j\right){\int }_{l_3}^{h_2}{\int }_{l_2}^{h_4}{\displaystyle \frac{H_1\left({\xi }_3^2-{\xi }_2^2\right)F_n^j{\overline{F}}_n^j}{R_1{R}_2}}d{\xi }_{3}d{\xi }_2\hfill \\ ={\int }_{l_3}^{h_2}{\displaystyle \frac{1}{R_2}}{\int }_{l_2}^{h_4}{\displaystyle \frac{\partial }{\partial {\xi }_3}}\left[R_1\left({\overline{F}}_n^j{\displaystyle \frac{\partial }{\partial {\xi }_3}}{F}_n^j-F_n^j{\displaystyle \frac{\partial }{\partial {\xi }_3}}{\overline{F}}_n^j\right)\right]d{\xi }_{3}d{\xi }_2\hfill \\ +{\int }_{l_2}^{h_4}{\displaystyle \frac{1}{R_1}}{\int }_{l_3}^{h_2}{\displaystyle \frac{\partial }{\partial {\xi }_2}}\left[R_2\left({\overline{F}}_n^j{\displaystyle \frac{\partial }{\partial {\xi }_2}}{F}_n^j-F_n^j{\displaystyle \frac{\partial }{\partial {\xi }_2}}{\overline{F}}_n^j\right)\right]d{\xi }_{2}d{\xi }_3,\hfill \end{array}$$

(188)

from which, taking into account the boundary conditions, we end up with

$$\begin{array}{l}\left(P_j-{\overline{P}}_j \right){\displaystyle {\int }_{h_3}^{h_2}{\displaystyle {\int }_{h_2}^{h_4}\frac{H_1\left({\xi }_3^2-{\xi }_2^2\right)E_n^j\left({\xi }_3\right){\overline{E}}_n^j\left({\xi }_3\right)E_n^j\left({\xi }_2\right){\overline{E}}_n^j\left({\xi }_2\right)}{R_1 R_2}d{\xi }_{3}d{\xi }_2}}=0\\ \end{array}$$

(189)

or

$$\begin{array}{l}{H}_1\left(P_j-{\overline{P}}_j \right){\displaystyle {\int }_{h_3}^{h_2}{\displaystyle {\int }_{h_2}^{h_4}\frac{{\left|E_n^j\left({\xi }_3\right)\right|}^2{\left|E_n^j\left({\xi }_2\right)\right|}^2\left({\xi }_3^2-{\xi }_2^2\right)}{R_1 R_2}d{\xi }_{3}d{\xi }_2}}=0 .\\ \end{array}$$

(190)

Hence, since $H_1$ as well as the value of the integrand are positive, it follows that

$$P_j={\overline{P}}_j ,$$

(191)

which implies that $P_j$ is real.

Next, we assume that $Q$ is complex. Then, besides the Equation (177), we also have its conjugate

$$-R_1{\displaystyle \frac{d}{d{\xi }_3}}{R}_1{\displaystyle \frac{d}{d{\xi }_3}}{\overline{E}}_n^j\left({\xi }_3\right)+\left[H_2{\overline{Q}}_j+H_1{P}_j {\xi }_3^2-n\left(n+2\right) {\xi }_3^4\right]{\overline{E}}_n^j\left({\xi }_3\right)=0.$$

(192)

Multiplying (177) by ${\overline{E}}_n^j\left({\xi }_3\right)$, (192) by $E_n^j\left({\xi }_3\right)$, and subtracting the resulting equations, we obtain

$$\begin{array}{l}\left({\displaystyle \frac{d}{d{\xi }_3}}{R}_1{\displaystyle \frac{d}{d{\xi }_3}}{E}_n^j\left({\xi }_3\right)\right){\overline{E}}_n^j\left({\xi }_3\right)-\left({\displaystyle \frac{d}{d{\xi }_3}}{R}_1{\displaystyle \frac{d}{d{\xi }_3}}{\overline{E}}_n^j\left({\xi }_3\right)\right)E_n^j\left({\xi }_3\right)\\ +{\displaystyle \frac{H_2}{R_1}}\left({\overline{Q}}_j-Q_j\right)E_n^j\left({\xi }_3\right){\overline{E}}_n^j\left({\xi }_3\right)=0 ,\end{array}$$

(193)

which is also written as

$${\displaystyle \frac{d}{d{\xi }_3}}{R}_1\left[E_n^j\left({\xi }_3\right){\displaystyle \frac{d}{d{\xi }_3}}{\overline{E}}_n^j\left({\xi }_3\right)-{\overline{E}}_n^j\left({\xi }_3\right){\displaystyle \frac{d}{d{\xi }_3}}{E}_n^j\left({\xi }_3\right)\right]+{\displaystyle \frac{H_2}{R_1}}\left(Q_j-{\overline{Q}}_j\right){\left|E_n^j\left({\xi }_3\right)\right|}^2=0.$$

(194)

Consequently, if we integrate (194) in the interval $\left(h_2,h_4\right)$ and use the form $R_1$ as it is given in (178) in order to kill the boundary terms at ${\xi }_3=h_2$ and ${\xi }_3=h_4$, we finally obtain

$$H_2\left(Q_j-{\overline{Q}}_j \right){\displaystyle {\int }_{h_2}^{h_4}\frac{{\left|E_n^j\left({\xi }_3\right)\right|}^2}{R_1 }d{\xi }_3}=0 .$$

(195)

Since $H_2$ and the integral are positive numbers, it follows that

$$Q_j={\overline{Q}}_j,$$

(196)

that is, $Q_j$ is also real.

That proves Lemma 1. □

In the sequel, we will demonstrate that all pairs $\left(P,Q\right)$ of the hyper-Lamé equation that lead to a solution are simple roots of the corresponding polynomial system. In fact, we will show this by proving that every such pair generates a unique linearly independent solution of the hyper-Lamé equation.

Lemma 2. 

The Wroskian of any two hyper-Lamé functions of the same class is always a polynomial.

Proof. 

This is obvious for the class [K1].

For the classes [K2], [K3], and [K4], the proofs are similar. In particular, for [K2] we have

$$\begin{gathered} \begin{array}{l}\sqrt{x^2-h_3^2}{E}_{n-1}^1{\displaystyle \frac{d}{dx}}\sqrt{x^2-h_3^2}{E}_{n-1}^2-\sqrt{x^2-h_3^2}{E}_{n-1}^2{\displaystyle \frac{d}{dx}}\sqrt{x^2-h_3^2}{E}_{n-1}^1\\ =\sqrt{x^2-h_3^2}{E}_{n-1}^1\left[{\displaystyle \frac{x}{\sqrt{x^2-h_3^2}}}{E}_{n-1}^2+\sqrt{x^2-h_3^2}{\displaystyle \frac{d}{dx}}{E}_{n-1}^2\right]\end{array} \\ \begin{array}{l}-\sqrt{x^2-h_3^2}{E}_{n-1}^2\left[{\displaystyle \frac{x}{\sqrt{x^2-h_3^2}}}{E}_{n-1}^1+\sqrt{x^2-h_3^2}{\displaystyle \frac{d}{dx}}{E}_{n-1}^1\right]\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em} \\ =\left(x^2-h_3^2\right)\left[E_{n-1}^1{\displaystyle \frac{d}{dx}}{E}_{n-1}^2-E_{n-1}^2{\displaystyle \frac{d}{dx}}{E}_{n-1}^1\right]\end{array} \end{gathered}$$

(197)

and in exactly the same way we prove the respective result when $\sqrt{x^2-h_3^2}$ is replaced with $\sqrt{x^2-h_2^2}$ and $\sqrt{x^2-h_4^2}$.

The classes [K5], [K6], and [K7] have also similar proofs. For [K5], we have

$$\begin{gathered} \begin{array}{l}\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}{E}_{n-2}^1{\displaystyle \frac{d}{dx}}\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}{E}_{n-2}^2\\ -\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}{E}_{n-2}^2{\displaystyle \frac{d}{dx}}\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}{E}_{n-2}^1\\ =\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}{E}_{n-2}^1\left[{\displaystyle \frac{x}{\sqrt{x^2-h_3^2}}}\sqrt{x^2-h_2^2}{E}_{n-2}^2\right. \end{array} \\ \begin{array}{l}\left.+\sqrt{x^2-h_3^2}{\displaystyle \frac{x}{\sqrt{x^2-h_2^2}}}{E}_{n-2}^2+\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}{\displaystyle \frac{d}{dx}}{E}_{n-2}^2\right]\\ - \cdots \\ =\left(x^2-h_3^2\right)\left(x^2-h_2^2\right)\left[E_{n-2}^1{\displaystyle \frac{d}{dx}}{E}_{n-2}^2-E_{n-2}^2{\displaystyle \frac{d}{dx}}{E}_{n-2}^1\right]\end{array} \end{gathered}$$

(198)

and it is obvious what happens with [K6] and [K7].

Finally, for class [K8], we have

$$\begin{gathered} \begin{array}{l}\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}\sqrt{x^2-h_4^2}{E}_{n-3}^1{\displaystyle \frac{d}{dx}}\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}\sqrt{x^2-h_4^2}{E}_{n-3}^2\\ - \cdots \\ =\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}\sqrt{x^2-h_4^2}{E}_{n-3}^1\left[{\displaystyle \frac{x}{\sqrt{x^2-h_3^2}}}\sqrt{x^2-h_2^2}\sqrt{x^2-h_4^2}{E}_{n-3}^2\right. \end{array} \\ \begin{array}{l}\left.+\sqrt{x^2-h_3^2}{\displaystyle \frac{x}{\sqrt{x^2-h_2^2}}}\sqrt{x^2-h_4^2}{E}_{n-3}^2+\sqrt{x^2-h_3^2}\sqrt{x^2-h_2^2}\sqrt{x^2-h_4^2}{\displaystyle \frac{d}{dx}}{E}_{n-3}^2\right]\\ - \cdots \\ =\left(x^2-h_3^2\right)\left(x^2-h_2^2\right)\left(x^2-h_4^2\right)\left[E_{n-3}^1{\displaystyle \frac{d}{dx}}{E}_{n-3}^2-E_{n-3}^2{\displaystyle \frac{d}{dx}}{E}_{n-3}^1\right] .\end{array} \end{gathered}$$

(199)

Hence, Lemma 2 is proved. □

Lemma 3. 

Each separation pair$\left(P,Q\right)$corresponds to a unique independent solution of the hyper-Lamé equation.

Proof. 

Suppose that $E_1$ and $E_2$ are two distinct hyper-Lamé solutions of the same degree and the same class corresponding to the pair $\left(P,Q\right)$. Then, rewriting (194) with the functions $E_n^j$ and ${\overline{E}}_n^j$ replaced by $E_1$ and $E_2$ respectively, we obtain

$$R_1\left(E_1\left({\xi }_3\right){\displaystyle \frac{d}{d{\xi }_3}}{E}_2\left({\xi }_3\right)-E_2\left({\xi }_3\right){\displaystyle \frac{d}{d{\xi }_3}}{E}_1\left({\xi }_3\right)\right)=\mathrm{constant}.$$

(200)

Note that the formula inside the above parenthesis is actually the Wroskian of the functions $E_1$, $E_2$.

From Lemma 2, we use the fact that the Wroskian of any two hyper-Lamé functions belonging to any one of the eight Lamé classes is always a polynomial. This indicates that the above product of an irrational by a rational function cannot be equal to a constant, and, therefore, the constant has to be zero. Hence, the Worskian of $E_1$, $E_2$ is zero, so these solutions are linearly dependent. Therefore, we cannot have more than one independent hyper-Lamé function for the pair $\left(P,Q\right)$.

This completes the proof of Lemma 3. □