Ellipsoidal Waves in Closed Form

Abstract

In the present work we propose an analytic method that leads to ellipsoidal waves in closed form. This is achieved by reducing 4D ellipsoidal harmonics to 3D ellipsoidal waves via appropriately replacing one of the 4D ellipsoidal coordinates with the time variable. The method is demonstrated in the case of a low-degree harmonic.

1. Introduction

In 1837 [1] Lamé published this celebrated paper on the isothermal surfaces of an ellipsoid, where he introduced the Theory of Ellipsoidal (3D) Harmonics. Then, in a series of fundamental papers, published within the next 23 years, he introduced the Theory of Curvilinear Coordinates [2,3,4,5,6]. The Ellipsoidal Harmonics was an active analytical subject for the whole 19th century and the early decades of the 20th Century. Due to the incomparable analytic and algebraic difficulties of the Lamé theory, this topic was not in the first line of research after the first quarter of the 20th century [7]. The situation changed during the last quarter of the 20th century due to the development of computers, where all the mathematical impossibilities of the theory found approximate numerical approaches.

However, an important problem of the theory remained intact. This is the problem of determination of closed-form solutions of the ellipsoidal wave equation. To the authors’ knowledge, every attempt to create ellipsoidal waves ended up either with a series or with an integral representation.

In the last couple of years, the authors published two papers, which extended Lamé’s work from a 3D to a 4D space [8,9]. This was a very long, tedious and extremely careful procedure that led to a complete analysis of the eight classes of Hyper-Lamé functions that resulted from the spectral analysis of the Laplace equation in 4D ellipsoidal coordinates. The idea behind this effort is to manage to construct 3D ellipsoidal waves in closed form by reducing the 4D Laplacian to the 3D d’Alembertian via the substitution $x_4=ict$. Then, via the same procedure, applied to 4D ellipsoidal harmonic solutions obtained in [9], to arrive at closed-form ellipsoidal waves. This is indeed the case, although as expected, it is not a trivial algorithm.

In the present work we discuss the analytic steps of this algorithm and apply them to a simple case. The result is a function of the ellipsoidal coordinates and time, which is expressed in closed form and satisfies the wave equation in 3D.

The paper is organized as follows. In Section 2 we demonstrate the elements of the 4D theory of Hyper-Ellipsoidal Harmonics. Section 3 includes the analytical steps needed to reduce a 4D hyper-ellipsoidal harmonic to a 3D ellipsoidal wave. Finally, and most important, in Section 4 we apply our procedure to a second-degree solution of the first kind and obtain the first, to the best of our knowledge, ellipsoidal wave in closed form.

It seems that only low-degree solutions can be calculated without the use of a computing machine. Related work can be found in references [10,11,12,13,14,15,16,17,18,19,20,21,22] as well as in [7] and the references therein.

2. The Hyper-Ellipsoidal System

For a detailed exhibition of this 4D system we refer to [8]. Starting with the reference ellipsoid

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

(1)

with semi-axes

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

(2)

we introduce the four dimensional ellipsoidal system:

$$\begin{array}{cc}\hfill x_1^2& ={\displaystyle \frac{{\xi }_1^2{\xi }_2^2{\xi }_3^2{\xi }_4^2}{h_2^2{h}_3^2{h}_4^2}},\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill x_2^2& ={\displaystyle \frac{\left(h_3^2-{\xi }_1^2\right)\left({\xi }_2^2-h_3^2\right)\left({\xi }_3^2-h_3^2\right)\left({\xi }_4^2-h_3^2\right)}{h_1^2{h}_3^2{h}_5^2}},\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill x_3^2& ={\displaystyle \frac{\left(h_2^2-{\xi }_1^2\right)\left(h_2^2-{\xi }_2^2\right)\left({\xi }_3^2-h_2^2\right)\left({\xi }_4^2-h_2^2\right)}{h_1^2{h}_2^2{h}_6^2}},\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill x_4^2& ={\displaystyle \frac{\left(h_4^2-{\xi }_1^2\right)\left(h_4^2-{\xi }_2^2\right)\left(h_4^2-{\xi }_3^2\right)\left({\xi }_4^2-h_4^2\right)}{h_4^2{h}_5^2{h}_6^2}},\hfill \end{array}$$

(6)

where the six semi-focal distances $h_i,i=1,2,3,4,5,6$, of the six principal ellipses appearing on the intersections of (1) with the coordinate planes, are given by

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

(7)

and they are connected via

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

(8)

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

(9)

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

(10)

Hence, only three of the above focal distances are independent and we decide to keep the distances

$$h_3<h_2<h_4$$

(11)

as the independent ones.

The ellipsoidal coordinate variables ${\xi }_i,i=1,2,3,4$ live in the following intervals

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

(12)

The reason why we write the relations (3)–(6) and (12) with the square of the variables ${\xi }_i$ has been chosen in order to avoid specifying every time in which one of the 16 compartments of the 4D system, the point $(x_1,x_2,x_3,x_4)\approx ({\xi }_1,{\xi }_2,{\xi }_3,{\xi }_4)$ is located.

Expressing the position vector in the generic form

$$\begin{array}{cc}\hfill \mathbf{r}& ={\displaystyle \frac{{\xi }_1{\xi }_2{\xi }_3{\xi }_4}{h_2{h}_3{h}_4}}{\widehat{\mathbf{x}}}_1+{\displaystyle \frac{\sqrt{h_3^2-{\xi }_1^2}\sqrt{{\xi }_2^2-h_3^2}\sqrt{{\xi }_3^2-h_3^2}\sqrt{{\xi }_4^2-h_3^2}}{h_1{h}_3{h}_5}}{\widehat{\mathbf{x}}}_2\hfill \\ \hfill & +{\displaystyle \frac{\sqrt{h_2^2-{\xi }_1^2}\sqrt{h_2^2-{\xi }_2^2}\sqrt{{\xi }_3^2-h_2^2}\sqrt{{\xi }_4^2-h_2^2}}{h_1{h}_2{h}_6}}{\widehat{\mathbf{x}}}_3\hfill \\ \hfill & +{\displaystyle \frac{\sqrt{h_4^2-{\xi }_1^2}\sqrt{h_4^2-{\xi }_2^2}\sqrt{h_4^2-{\xi }_3^2}\sqrt{{\xi }_4^2-h_4^2}}{h_4{h}_5{h}_6}}{\widehat{\mathbf{x}}}_4,\hfill \end{array}$$

(13)

we can calculate the non-vanishing diagonal elements of the first fundamental form

$$h_{{\xi }_k}^2={\displaystyle \frac{\partial \mathbf{r}}{\partial {\xi }_k}}·{\displaystyle \frac{\partial \mathbf{r}}{\partial {\xi }_k}},k=1,2,3,4,$$

(14)

which, after some tedious algebra assume the expressions

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

(15)

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

(16)

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

(17)

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

(18)

The off-diagonal elements of the first fundamental form are zero, securing the orthogonality of the system. Once the fundamental tensor

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

(19)

has been obtained we utilize the formula

$$\Delta ={\displaystyle \frac{1}{\sqrt{\det \tilde{\mathbf{g}}}}}{\displaystyle \frac{\partial }{\partial {\xi }_i}}\left(\sqrt{\det \tilde{\mathbf{g}}}{g}^{ij}{\displaystyle \frac{\partial }{\partial {\xi }_j}}\right),$$

(20)

where $g^{ij}$ are the elements of the inverse tensor ${\tilde{g}}^{-1}$ [23], to obtain the Laplacian

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

(21)

Lamé used an ingenious method to separate variables for the Laplace equation on the 3D ellipsoidal system. Following the same idea and using much harder identities between the focal distances, we managed to separate the 4D Laplace’s equation as well [8]. The separation procedure gave four identical ordinary differential equations for the four variables ${\xi }_i,i=1,2,3,4$. These spectral equations are

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

(22)

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

In [9] we proved that

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

(23)

and the hyper-Lamé Equation (22) is written as

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

(24)

where

$$\begin{array}{cc}\hfill H_1& =h_3^2+h_2^2+h_4^2,\hfill \end{array}$$

(25)

$$\begin{array}{cc}\hfill H_2& =h_3^2{h}_2^2+h_2^2{h}_4^2+h_4^2{h}_3^2,\hfill \end{array}$$

(26)

$$\begin{array}{cc}\hfill H_3& =h_3^2{h}_2^2{h}_4^2\hfill \end{array}$$

(27)

and

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

(28)

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

(29)

so that the two separation constants B and C are represented by the dimensionless parameters P and Q, respectively. Note that the expressions $H_1,H_2,H_3$ remain invariant under any cyclic permutation of the independent semi-focal distances $h_3^2,h_2^2,h_4^2$.

Equation (24) accepts eight classes of solutions, which are products of polynomials times all possible square roots of the quantities $|x^2-h_j^2|,j=2,3,4$, as shown in Table 1. A detailed analysis of these solutions is included in [9]. We note that the only mathematical difficulty that one finds in solving the hyper-Lamé equation for each one of the above classes is the exact determination of the remaining separations constants B and C, or equivalently of the dimensionless constants P and Q, which satisfy nonlinear algebraic systems. Hence, just as in the case of 3D ellipsoidal harmonics [7], in 4D we may also have to apply some numerical procedures to specify the values of P and Q.

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

Generating Factors	Class	Form of Solutions
1	${\mathrm{K}}_1$	$P_n$
$x_2$	${\mathrm{K}}_2$	$\sqrt{\left|x^2-h_3^2\right|}{P}_{n-1}$
$x_3$	${\mathrm{K}}_3$	$\sqrt{\left|x^2-h_2^2\right|}{P}_{n-1}$
$x_4$	${\mathrm{K}}_4$	$\sqrt{\left|x^2-h_4^2\right|}{P}_{n-1}$
$x_2{x}_3$	${\mathrm{K}}_5$	$\sqrt{\left|x^2-h_3^2\right|}\sqrt{\left|x^2-h_2^2\right|}{P}_{n-2}$
$x_2{x}_4$	${\mathrm{K}}_6$	$\sqrt{\left|x^2-h_2^2\right|}\sqrt{\left|x^2-h_4^2\right|}{P}_{n-2}$
$x_3{x}_4$	${\mathrm{K}}_7$	$\sqrt{\left|x^2-h_3^2\right|}\sqrt{\left|x^2-h_4^2\right|}{P}_{n-2}$
$x_2{x}_3{x}_4$	${\mathrm{K}}_8$	$\sqrt{\left|x^2-h_3^2\right|}\sqrt{\left|x^2-h_2^2\right|}\sqrt{\left|x^2-h_4^2\right|}{P}_{n-3}$

3. Reduction of a 4D Harmonic to a 3D Wave

It is well known that there are exactly eleven coordinate systems with coordinate surfaces of the second degree that separate variables for the Laplace equation [24]. Among these systems, the Cartesian is the simplest, and the ellipsoidal is the most complicated. On the other hand, these two systems are the only systems with coordinates that represent only lengths. As a consequence, the spectral equations that result from the decomposition are all of the same form.

The same structure holds true in the case of the four-dimensional ellipsoidal system and the four generated identical spectral equations. Therefore, if we denote by $F_n\left({\xi }_i\right)$, $i=1,2,3,4$ the hyper-Lamé functions in the interval where ${\xi }_i$ lives, then the corresponding hyper-ellipsoidal harmonic will be

$$u_n\left({\xi }_1,{\xi }_2,{\xi }_3,{\xi }_4\right)=F_n\left({\xi }_1\right)F_n\left({\xi }_2\right)F_n\left({\xi }_3\right)F_n\left({\xi }_4\right),$$

(30)

where n denotes the degree of the function $F_n$. Consequently, the only difference between the four functional factors in the right-hand side of (30) lies in the interval that each ${\xi }_i$ varies, as it is given in (12).

We are now ready to enumerate the series of steps we need in order to produce an ellipsoidal wave.

• STEP 1. Solve the hyper-Lamé Equation (24) for a particular class (e.g., ${\mathrm{K}}_1$) and a particular degree n. That will give us the function $F_n$.
• STEP 2. Write down the hyper-ellipsoidal harmonic as in (30).
• STEP 3. Perform the calculations indicated in (30) and write $u_n$ as a polynomial with terms

  $${\xi }_1^{n_1}{\xi }_2^{n_2}{\xi }_3^{n_3}{\xi }_4^{n_4},$$

  (31)

  where

  $$0⩽n_1+n_2+n_3+n_4⩽4n.$$

  (32)

  That may demand some simple transformation in order to secure complete symmetry of the resulting equation on the chosen variable, for example, ${\sigma }_i,i=1,2,3,4$.
• STEP 4. Express the new variables ${\sigma }_i$ in terms of the Cartesian variables $x_1,x_2,x_3,x_4$ given in (3)–(6).
• STEP 5. Introduce, in the Cartesian form of the solution obtained in Step 4, the transformation

  $$x_4=ict$$

  (33)

  for the variable $x_4$ and the classical 3D ellipsoidal coordinates $(\rho ,\mu ,\nu )$ for $(x_1,x_2,x_3)$ via the Lamé definition [7]

  $$\begin{array}{cc}\hfill x_1^2& ={\displaystyle \frac{{\rho }^2{\mu }^2{\nu }^2}{h_2^2{h}_3^2}},\hfill \end{array}$$

  (34)

  $$\begin{array}{cc}\hfill x_2^2& ={\displaystyle \frac{\left({\rho }^2-h_3^2\right)\left({\mu }^2-h_3^2\right)\left(h_3^2-{\nu }^2\right)}{h_1^2{h}_3^2}},\hfill \end{array}$$

  (35)

  $$\begin{array}{cc}\hfill x_3^2& ={\displaystyle \frac{\left({\rho }^2-h_2^2\right)\left(h_2^2-{\mu }^2\right)\left(h_2^2-{\nu }^2\right)}{h_1^2{h}_2^2}},\hfill \end{array}$$

  (36)

  where $h_1^2=h_2^2-h_3^2$. This will give us the function

  $$u_n=u_n\left(\rho ,\mu ,\nu ,t\right).$$

  (37)
• STEP 6. Since this program involves a lot of algebraic calculations, it is necessary to justify at the end that (37) indeed solves the wave equation

  $$\Delta u_n={\displaystyle \frac{1}{c^2}}{\displaystyle \frac{{\partial }^2}{\partial t^2}}{u}_n.$$

  (38)

  The function $u_n$ is then an ellipsoidal wave in closed form.

The fact that for almost two centuries this remained an open problem is reflected in the extended calculations that are needed to obtain even a low-degree ellipsoidal wave. This can be appreciated in the next Section, where the proposed procedure is applied to a very simple case, generating the first ellipsoidal wave in closed form.

4. The Realization

Let us consider the very simple case where we seek a second-degree polynomial solution of class ${\mathrm{K}}_1$ for the hyper-Lamé Equation (24), which for $n=2$ reads as

$$\begin{array}{cc}\hfill & \left(x^6-H_1{x}^4+H_2{x}^2-H_3\right){E_2}^{″}\left(x\right)\hfill \\ \hfill & +\left(3x^4-2H_1{x}^2+H_2\right)x{E_2}^{\prime }\left(x\right)\hfill \\ \hfill & +\left(H_{2}Q+H_{1}P{x}^2-8x^4\right)E_2\left(x\right)=0.\hfill \end{array}$$

(39)

From the theory of hyper-Lamé functions [9] we know that we actually look for a solution of the form

$$E_2\left(x\right)=x^2+a_0,$$

(40)

where $a_0$ is an unknown coefficient. Inserting (40) in (39) we obtain

$$\begin{array}{cc}\hfill & \left(x^6-H_1{x}^4+H_2{x}^2-H_3\right)2\hfill \\ \hfill & +\left(3x^4-2H_1{x}^2+H_2\right)2x^2\hfill \\ \hfill & +\left(H_{2}Q+H_{1}P{x}^2-8x^4\right)\left(x^2+a_0\right)=0.\hfill \end{array}$$

(41)

An appropriate value of the constant $a_0$ is

$$a_0={\displaystyle \frac{2H_3}{H_{2}Q}}.$$

(42)

Then, in order for the Equation (41) to be satisfied, we need to choose the separation constants P and Q such that

$$\begin{array}{cc}\hfill \overline{P}\overline{Q}-6H_1\overline{Q}-16H_3& =0,\hfill \end{array}$$

(43)

$$\begin{array}{cc}\hfill {\overline{Q}}^2+4H_2\overline{Q}+2H_3\overline{P}& =0,\hfill \end{array}$$

(44)

where

$$\overline{P}=H_{1}P,$$

(45)

$$\overline{Q}=H_{2}Q.$$

(46)

Eliminating $\overline{P}$ between (43) and (44) we arrive at the cubic equation

$${\overline{Q}}^3+4H_2{\overline{Q}}^2+12H_1{H}_3\overline{Q}+32H_3^2=0$$

(47)

for the constant $\overline{Q}$.

It is obvious from (42) that for every solution of (47) we have a constant $a_0$ and via (40) a different solution of Equation (39). Each one of these three distinct solutions ([9], Lemma 3) gives the hyper-ellipsoidal harmonic

$$\begin{array}{cc}\hfill & {\mathbb{E}}_2\left({\xi }_1,{\xi }_2,{\xi }_3,{\xi }_4\right)\hfill \\ \hfill & =\left({\xi }_1^2+a_0\right)\left({\xi }_2^2+a_0\right)\left({\xi }_3^2+a_0\right)\left({\xi }_4^2+a_0\right),\hfill \end{array}$$

(48)

which, if we set

$${\xi }_i^2={\sigma }_i+h_4^2,i=1,2,3,4,$$

(49)

is written as

$$\begin{array}{cc}\hfill & {\mathbb{E}}_2\left({\sigma }_1,{\sigma }_2,{\sigma }_3,{\sigma }_4\right)\hfill \\ \hfill & =\left({\sigma }_1+a_0+h_4^2\right)\left({\sigma }_2+a_0+h_4^2\right)\left({\sigma }_3+a_0+h_4^2\right)\left({\sigma }_4+a_0+h_4^2\right)\hfill \\ \hfill & ={\sigma }_1{\sigma }_2{\sigma }_3{\sigma }_4\hfill \\ \hfill & +\overline{\mathrm{A}}\left({\sigma }_1{\sigma }_2{\sigma }_3+{\sigma }_1{\sigma }_2{\sigma }_4+{\sigma }_1{\sigma }_3{\sigma }_4+{\sigma }_2{\sigma }_3{\sigma }_4\right)\hfill \\ \hfill & +{\overline{\mathrm{A}}}^2\left({\sigma }_1{\sigma }_2+{\sigma }_1{\sigma }_3+{\sigma }_1{\sigma }_4+{\sigma }_2{\sigma }_3+{\sigma }_2{\sigma }_4+{\sigma }_3{\sigma }_4\right)\hfill \\ \hfill & +{\overline{\mathrm{A}}}^3\left({\sigma }_1+{\sigma }_2+{\sigma }_3+{\sigma }_4\right)\hfill \\ \hfill & +{\overline{\mathrm{A}}}^4,\hfill \end{array}$$

(50)

where

$$\overline{\mathrm{A}}=a_0+h_4^2.$$

(51)

Then, by setting

$$\begin{array}{cc}\hfill {\Sigma }_1& ={\sigma }_1+{\sigma }_2+{\sigma }_3+{\sigma }_4,\hfill \end{array}$$

(52)

$$\begin{array}{cc}\hfill {\Sigma }_2& ={\sigma }_1{\sigma }_2+{\sigma }_1{\sigma }_3+{\sigma }_1{\sigma }_4+{\sigma }_2{\sigma }_3+{\sigma }_2{\sigma }_4+{\sigma }_3{\sigma }_4,\hfill \end{array}$$

(53)

$$\begin{array}{cc}\hfill {\Sigma }_3& ={\sigma }_1{\sigma }_2{\sigma }_3+{\sigma }_1{\sigma }_2{\sigma }_4+{\sigma }_1{\sigma }_3{\sigma }_4+{\sigma }_2{\sigma }_3{\sigma }_4,\hfill \end{array}$$

(54)

we have

$$\begin{array}{cc}\hfill & {\mathbb{E}}_2\left({\sigma }_1,{\sigma }_2,{\sigma }_3,{\sigma }_4\right)\hfill \\ \hfill & ={\sigma }_1{\sigma }_2{\sigma }_3{\sigma }_4+\overline{\mathrm{A}}{\Sigma }_3+{\overline{\mathrm{A}}}^2{\Sigma }_2+{\overline{\mathrm{A}}}^3{\Sigma }_1+{\overline{\mathrm{A}}}^4.\hfill \end{array}$$

(55)

Our next task is to express the quantities ${\Sigma }_1,{\Sigma }_2,{\Sigma }_3$ in terms of the Cartesian coordinates $x_1,x_2,x_3,x_4$. To this end, we introduce the new variables

$${\sigma }_i={\xi }_i^2-h_4^2,i=1,2,3,4,$$

(56)

in (3)–(6) and obtain

$$\begin{array}{cc}\hfill x_1^2& ={\displaystyle \frac{\left({\sigma }_1+h_4^2\right)\left({\sigma }_2+h_4^2\right)\left({\sigma }_3+h_4^2\right)\left({\sigma }_4+h_4^2\right)}{h_2^2{h}_3^2{h}_4^2}},\hfill \end{array}$$

(57)

$$\begin{array}{cc}\hfill x_2^2& ={\displaystyle \frac{-\left({\sigma }_1+h_5^2\right)\left({\sigma }_2+h_5^2\right)\left({\sigma }_3+h_5^2\right)\left({\sigma }_4+h_5^2\right)}{h_1^2{h}_3^2{h}_5^2}},\hfill \end{array}$$

(58)

$$\begin{array}{cc}\hfill x_3^2& ={\displaystyle \frac{\left({\sigma }_1+h_6^2\right)\left({\sigma }_2+h_6^2\right)\left({\sigma }_3+h_6^2\right)\left({\sigma }_4+h_6^2\right)}{h_1^2{h}_2^2{h}_6^2}},\hfill \end{array}$$

(59)

$$\begin{array}{cc}\hfill x_4^2& ={\displaystyle \frac{-{\sigma }_1{\sigma }_2{\sigma }_3{\sigma }_4}{h_4^2{h}_5^2{h}_6^2}}.\hfill \end{array}$$

(60)

If we perform the multiplications in (57)–(59), substitute the expression

$${\sigma }_1{\sigma }_2{\sigma }_3{\sigma }_4=-h_4^2{h}_5^2{h}_6^2{x}_4^2$$

(61)

and take into account relations (52)–(54), we can write (57)–(59) in the form

$$\begin{array}{cc}\hfill h_2^2{h}_3^2{h}_4^2{x}_1^2& =-h_4^2{h}_5^2{h}_6^2{x}_4^2+h_4^8+h_4^2{\Sigma }_3+h_4^4{\Sigma }_2+h_4^6{\Sigma }_1,\hfill \end{array}$$

(62)

$$\begin{array}{cc}\hfill h_1^2{h}_3^2{h}_5^2{x}_2^2& =h_4^2{h}_5^2{h}_6^2{x}_4^2-h_5^8-h_5^2{\Sigma }_3-h_5^4{\Sigma }_2-h_5^6{\Sigma }_1,\hfill \end{array}$$

(63)

$$\begin{array}{cc}\hfill h_1^2{h}_2^2{h}_6^2{x}_3^2& =-h_4^2{h}_5^2{h}_6^2{x}_4^2+h_6^8+h_6^2{\Sigma }_3+h_6^4{\Sigma }_2+h_6^6{\Sigma }_1\hfill \end{array}$$

(64)

or by defining

$$\begin{array}{cc}\hfill A^{\prime }& =h_2^2{h}_3^2{h}_4^2{x}_1^2+h_4^2{h}_5^2{h}_6^2{x}_4^2-h_4^8,\hfill \end{array}$$

(65)

$$\begin{array}{cc}\hfill B^{\prime }& =h_4^2{h}_5^2{h}_6^2{x}_4^2-h_1^2{h}_3^2{h}_5^2{x}_2^2-h_5^8,\hfill \end{array}$$

(66)

$$\begin{array}{cc}\hfill C^{\prime }& =h_1^2{h}_2^2{h}_6^2{x}_3^2+h_4^2{h}_5^2{h}_6^2{x}_4^2-h_6^8,\hfill \end{array}$$

(67)

we obtain the system

$$\begin{array}{cc}\hfill h_4^6{\Sigma }_1+h_4^4{\Sigma }_2+h_4^2{\Sigma }_3& =A^{\prime },\hfill \end{array}$$

(68)

$$\begin{array}{cc}\hfill h_5^6{\Sigma }_1+h_5^4{\Sigma }_2+h_5^2{\Sigma }_3& =B^{\prime },\hfill \end{array}$$

(69)

$$\begin{array}{cc}\hfill h_6^6{\Sigma }_1+h_6^4{\Sigma }_2+h_6^2{\Sigma }_3& =C^{\prime }.\hfill \end{array}$$

(70)

Solving the system (68)–(70) with respect to ${\Sigma }_1,{\Sigma }_2,{\Sigma }_3$ we get

$$\begin{array}{cc}\hfill {\Sigma }_1& ={\displaystyle \frac{A^{\prime }}{h_2^2{h}_3^2{h}_4^2}}-{\displaystyle \frac{B^{\prime }}{h_1^2{h}_3^2{h}_5^2}}+{\displaystyle \frac{C^{\prime }}{h_1^2{h}_2^2{h}_6^2}},\hfill \end{array}$$

(71)

$$\begin{array}{cc}\hfill {\Sigma }_2& =-{\displaystyle \frac{A^{\prime }}{h_2^2{h}_3^2{h}_4^2}}\left(h_5^2+h_6^2\right)+{\displaystyle \frac{B^{\prime }}{h_1^2{h}_3^2{h}_5^2}}\left(h_4^2+h_6^2\right)-{\displaystyle \frac{C^{\prime }}{h_1^2{h}_2^2{h}_6^2}}\left(h_4^2+h_5^2\right),\hfill \end{array}$$

(72)

$$\begin{array}{cc}\hfill {\Sigma }_3& ={\displaystyle \frac{A^{\prime }}{h_2^2{h}_3^2{h}_4^2}}{h}_5^2{h}_6^2-{\displaystyle \frac{B^{\prime }}{h_1^2{h}_3^2{h}_5^2}}{h}_4^2{h}_6^2+{\displaystyle \frac{C^{\prime }}{h_1^2{h}_2^2{h}_6^2}}{h}_4^2{h}_5^2.\hfill \end{array}$$

(73)

Next, inserting the values of $A^{\prime },B^{\prime },C^{\prime }$ from (65)–(67) in (71)–(73) we arrive at

$$\begin{array}{c}{\Sigma }_1=x_1^2+x_2^2+x_3^2+x_4^2+C_1,\hfill \\ {\Sigma }_2=-\left(h_5^2+h_6^2\right)x_1^2-\left(h_4^2+h_6^2\right)x_2^2-\left(h_4^2+h_5^2\right)x_3^2\hfill \end{array}$$

(74)

$$-{\displaystyle \frac{x_4^2}{h_1^2{h}_2^2{h}_3^2}}\left(h_1^2{h}_4^6-h_2^2{h}_5^6+h_3^2{h}_6^6\right)+C_2$$

(75)

and

$$\begin{array}{c}{\Sigma }_3=h_5^2{h}_6^2{x}_1^2+h_4^2{h}_6^2{x}_2^2+h_4^2{h}_5^2{x}_3^2\hfill \\ +{\displaystyle \frac{x_4^2}{h_1^2{h}_2^2{h}_3^2}}\left(h_1^2{h}_5^4{h}_6^4-h_2^2{h}_4^4{h}_6^4+h_3^2{h}_4^4{h}_5^4\right)+C_3,\hfill \end{array}$$

(76)

where

$$C_1=-\left(h_4^2+h_5^2+h_6^2\right),C_2=h_4^2{h}_5^2+h_5^2{h}_6^2+h_4^2{h}_6^2,C_3=-h_4^2{h}_5^2{h}_6^2.$$

(77)

At this stage we use the transformation

$$x_4^2=-c^2{t}^2$$

(78)

and inserting (61) and (74)–(76) in Equation (55) we obtain

$$\begin{array}{cc}\hfill & {\mathbb{E}}_2\left(x_1,x_2,x_3,t\right)\hfill \\ \hfill & =h_4^2{h}_5^2{h}_6^2{c}^2{t}^2+\overline{\mathrm{A}}\left(h_5^2{h}_6^2{x}_1^2+h_4^2{h}_6^2{x}_2^2+h_4^2{h}_5^2{x}_3^2\right)\hfill \\ \hfill & -{\displaystyle \frac{\overline{\mathrm{A}}}{h_1^2{h}_2^2{h}_3^2}}\left(h_1^2{h}_5^4{h}_6^4-h_2^2{h}_4^4{h}_6^4+h_3^2{h}_4^4{h}_5^4\right)c^2{t}^2\hfill \\ \hfill & -{\overline{\mathrm{A}}}^2\left[\left(h_5^2+h_6^2\right)x_1^2+\left(h_4^2+h_6^2\right)x_2^2+\left(h_4^2+h_5^2\right)x_3^2\right]\hfill \\ \hfill & +{\displaystyle \frac{{\overline{\mathrm{A}}}^2}{h_1^2{h}_2^2{h}_3^2}}\left(h_1^2{h}_4^6-h_2^2{h}_5^6+h_3^2{h}_6^6\right)c^2{t}^2+{\overline{\mathrm{A}}}^3\left(x_1^2+x_2^2+x_3^2-c^2{t}^2\right),\hfill \end{array}$$

(79)

where all the constants have been omitted, since the d’Alembert operator will eliminate them.

Moreover, we introduce the quantities

$$\begin{array}{cc}\hfill X& =3\left(h_4^2+h_5^2+h_6^2\right),\hfill \end{array}$$

(80)

$$\begin{array}{cc}\hfill Y& =2\left(h_4^2{h}_5^2+h_5^2{h}_6^2+h_6^2{h}_4^2\right)\hfill \end{array}$$

(81)

and verify the following relations

$$\begin{array}{cc}\hfill 4h_4^4-h_4^{2}X+Y-h_5^2{h}_6^2& =h_2^2{h}_3^2,\hfill \end{array}$$

(82)

$$\begin{array}{cc}\hfill 24h_4^4-4h_4^{2}X+2Y& =4H_2,\hfill \end{array}$$

(83)

$$\begin{array}{cc}\hfill h_3^2{h}_2^2{h}_4^2\left(48h_4^2-4X\right)& =12H_1{H}_3.\hfill \end{array}$$

(84)

The proofs of (82), (83) and (84) demand a series of long and tedious algebraic calculations. Hence, in view of (82)–(84), as well as some further algebra, we arrive at the expression

$$\begin{array}{cc}\hfill & {\mathbb{E}}_2\left(x_1,x_2,x_3,t\right)\hfill \\ \hfill & =\overline{\mathrm{A}}{h}_4^2{h}_5^2{h}_6^2\left({\displaystyle \frac{x_1^2}{h_4^2}}+{\displaystyle \frac{x_2^2}{h_5^2}}+{\displaystyle \frac{x_3^2}{h_6^2}}\right)+{\overline{\mathrm{A}}}^2\left(h_4^2{x}_1^2+h_5^2{x}_2^2+h_6^2{x}_3^2\right)\hfill \\ \hfill & -{\overline{\mathrm{A}}}^2\left(h_4^2+h_5^2+h_6^2-\overline{\mathrm{A}}\right)\left(x_1^2+x_2^2+x_3^2\right)\hfill \\ \hfill & +\left[{\overline{\mathrm{A}}}^3-{\overline{\mathrm{A}}}^2\left(h_4^2+h_5^2+h_6^2\right)-h_4^2{h}_5^2{h}_6^2\right.+\left.\overline{\mathrm{A}}\left(h_4^2{h}_5^2+h_5^2{h}_6^2+h_6^2{h}_4^2\right)\right]\left(-c^2{t}^2\right)\hfill \end{array}$$

(85)

for the solution ${\mathbb{E}}_2$. Remember that all constants in the above are absent.

The last step is to insert the 3D ellipsoidal coordinates in (85) via the defining equations [7]

$$\begin{array}{cc}\hfill x_1^2& ={\displaystyle \frac{{\rho }^2{\mu }^2{\nu }^2}{h_2^2{h}_3^2}},\hfill \end{array}$$

(86)

$$\begin{array}{cc}\hfill x_2^2& ={\displaystyle \frac{\left({\rho }^2-h_3^2\right)\left({\mu }^2-h_3^2\right)\left(h_3^2-{\nu }^2\right)}{h_1^2{h}_3^2}},\hfill \end{array}$$

(87)

$$\begin{array}{cc}\hfill x_3^2& ={\displaystyle \frac{\left({\rho }^2-h_2^2\right)\left(h_2^2-{\mu }^2\right)\left(h_2^2-{\nu }^2\right)}{h_1^2{h}_2^2}},\hfill \end{array}$$

(88)

using

$$h_1^2=h_5^2-h_6^2,$$

(89)

$$h_2^2=h_4^2-h_6^2,$$

(90)

$$h_3^2=h_4^2-h_5^2.$$

(91)

We know ([7], p. 50) that

$$x_1^2+x_2^2+x_3^2={\rho }^2+{\mu }^2+{\nu }^2-h_3^2-h_2^2,$$

(92)

thus we only have to express the quantities $h_4^2{x}_1^2+h_5^2{x}_2^2+h_6^2{x}_3^2$ and $h_5^2{h}_6^2{x}_1^2+h_4^2{h}_6^2{x}_2^2+h_4^2{h}_5^2{x}_3^2$ in the coordinates $\rho ,\mu ,\nu$. Some more algebra shows that

$$\begin{array}{c}{h}_4^2{x}_1^2+h_5^2{x}_2^2+h_6^2{x}_3^2\hfill \\ =\left({\rho }^2{\mu }^2+{\mu }^2{\nu }^2+{\nu }^2{\rho }^2\right)\hfill \\ +\left({\rho }^2+{\mu }^2+{\nu }^2\right)\left(-h_4^2+h_5^2+h_6^2\right)\hfill \\ +{\left(-h_4^2+h_5^2+h_6^2\right)}^2-h_5^2{h}_6^2\hfill \end{array}$$

(93)

and

$$\begin{array}{c}{h}_5^2{h}_6^2{x}_1^2+h_4^2{h}_6^2{x}_2^2+h_4^2{h}_5^2{x}_3^2\hfill \\ ={\rho }^2{\mu }^2{\nu }^2-h_4^2\left({\rho }^2{\mu }^2+{\mu }^2{\nu }^2+{\nu }^2{\rho }^2\right)\hfill \\ +h_4^4\left({\rho }^2+{\mu }^2+{\nu }^2\right)+h_4^2{h}_5^2{h}_6^2-h_4^6.\hfill \end{array}$$

(94)

Finally, inserting (92)–(94) in (85) we find, after the elimination of all constants, that

$$\begin{array}{c}{\mathbb{E}}_2\left(\rho ,\mu ,\nu ,t\right)\hfill \\ =\overline{\mathrm{A}}{\rho }^2{\mu }^2{\nu }^2+\left({\overline{\mathrm{A}}}^2-h_4^2\overline{\mathrm{A}}\right)\left({\rho }^2{\mu }^2+{\mu }^2{\nu }^2+{\nu }^2{\rho }^2\right)\hfill \\ +\left({\overline{\mathrm{A}}}^3-2h_4^2{\overline{\mathrm{A}}}^2+h_4^4\overline{\mathrm{A}}\right)\left({\rho }^2+{\mu }^2+{\nu }^2\right)\hfill \\ -\left[{\overline{\mathrm{A}}}^3-{\overline{\mathrm{A}}}^2\left(h_4^2+h_5^2+h_6^2\right)\right.+\overline{\mathrm{A}}\left(h_4^2{h}_5^2+h_5^2{h}_6^2+h_6^2{h}_4^2\right)-\left.h_4^2{h}_5^2{h}_6^2\right]c^2{t}^2,\hfill \end{array}$$

(95)

where

$$\overline{\mathrm{A}}={\displaystyle \frac{2H_3}{\overline{Q}}}+h_4^2$$

(96)

and $H_3,\overline{Q}$ are given in (27), (46) respectively.

CLAIM. 

${\mathbb{E}}_2\left(\rho ,\mu ,\nu ,t\right)$, as it is given in (95), solves the wave equation

$$\Delta {\mathbb{E}}_2={\displaystyle \frac{1}{c^2}}{\displaystyle \frac{{\partial }^2}{\partial t^2}}{\mathbb{E}}_2$$

(97)

in 3D ellipsoidal coordinates $(\rho ,\mu ,\nu )$.

Proof of the Claim. 

First we have to act on ${\mathbb{E}}_2$ with the ellipsoidal form [7] of the Laplace operator

$$\begin{array}{cc}\hfill & \Delta ={\displaystyle \frac{1}{\left({\rho }^2-{\mu }^2\right)\left({\rho }^2-{\nu }^2\right)\left({\mu }^2-{\nu }^2\right)}}·\hfill \\ \hfill & ·\left[\left({\mu }^2-{\nu }^2\right)\left({\rho }^2-h_3^2\right)\left({\rho }^2-h_2^2\right){\partial }_{\rho \rho }\right.+\left({\mu }^2-{\nu }^2\right)\rho \left(2{\rho }^2-h_3^2-h_2^2\right){\partial }_{\rho }\hfill \\ \hfill & -\left({\rho }^2-{\nu }^2\right)\left({\mu }^2-h_3^2\right)\left({\mu }^2-h_2^2\right){\partial }_{\mu \mu }-\left({\rho }^2-{\nu }^2\right)\mu \left(2{\mu }^2-h_3^2-h_2^2\right){\partial }_{\mu }\hfill \\ \hfill & +\left({\rho }^2-{\mu }^2\right)\left({\nu }^2-h_3^2\right)\left({\nu }^2-h_2^2\right){\partial }_{\nu \nu }+\left.\left({\rho }^2-{\mu }^2\right)\nu \left(2{\nu }^2-h_3^2-h_2^2\right){\partial }_{\nu }\right].\hfill \end{array}$$

(98)

Set

$$\begin{array}{cc}\hfill & {\Gamma }_1=\overline{\mathrm{A}},\hfill \end{array}$$

(99)

$$\begin{array}{cc}\hfill & {\Gamma }_2=\overline{\mathrm{A}}\left(\overline{\mathrm{A}}-h_4^2\right),\hfill \end{array}$$

(100)

$$\begin{array}{cc}\hfill & {\Gamma }_3=\overline{\mathrm{A}}\left({\overline{\mathrm{A}}}^2-2h_4^2\overline{\mathrm{A}}+h_4^4\right),\hfill \\ \hfill & {\Gamma }_4={\overline{\mathrm{A}}}^3-{\overline{\mathrm{A}}}^2\left(h_4^2+h_5^2+h_6^2\right)\hfill \end{array}$$

(101)

$$\begin{array}{cc}\hfill & +\overline{\mathrm{A}}\left(h_4^2{h}_5^2+h_5^2{h}_6^2+h_6^2{h}_4^2\right)-h_4^2{h}_5^2{h}_6^2\hfill \end{array}$$

(102)

and (95) is written as

$$\begin{array}{cc}\hfill & {\mathbb{E}}_2\left(\rho ,\mu ,\nu ,t\right)\hfill \\ \hfill & ={\Gamma }_1{\rho }^2{\mu }^2{\nu }^2+{\Gamma }_2\left({\rho }^2{\mu }^2+{\mu }^2{\nu }^2+{\nu }^2{\rho }^2\right)\hfill \\ \hfill & +{\Gamma }_3\left({\rho }^2+{\mu }^2+{\nu }^2\right)-{\Gamma }_4{c}^2{t}^2.\hfill \end{array}$$

(103)

This procedure also demands very long algebraic calculations, which, after all the necessary reductions with a set of non-linear identities connecting the six focal distances [8], leads to the expression

$$\Delta {\mathbb{E}}_2=2{\Gamma }_1{h}_3^2{h}_2^2+4{\Gamma }_2\left(h_3^2+h_2^2\right)+6{\Gamma }_3.$$

(104)

It is much simpler to find that

$${\displaystyle \frac{1}{c^2}}{\displaystyle \frac{{\partial }^2}{\partial t^2}}{\mathbb{E}}_2=-2{\Gamma }_4.$$

(105)

Therefore, ${\mathbb{E}}_2$ belongs to the kernel of the d’Alembert operator

$$\square =\Delta -{\displaystyle \frac{1}{c^2}}{\displaystyle \frac{{\partial }^2}{\partial t^2}},$$

(106)

if the following equation is satisfied

$${\Gamma }_1{h}_3^2{h}_2^2+2{\Gamma }_2\left(h_3^2+h_2^2\right)+3{\Gamma }_3=-{\Gamma }_4.$$

(107)

Inserting relations (99)–(102) for the $\Gamma$’s in (107), we end up with the equivalent equation

$$\begin{array}{cc}\hfill & 4{\overline{\mathrm{A}}}^3+3{\overline{\mathrm{A}}}^2\left(H_1-4h_4^2\right)\hfill \\ \hfill & +2\overline{\mathrm{A}}\left(3h_2^2{h}_3^2+3h_4^4-2H_2\right)\hfill \\ \hfill & +\left(H_2{h}_4^2-2H_3-h_4^6\right)\hfill \\ \hfill & =0,\hfill \end{array}$$

(108)

which, as the quantities $\overline{\mathrm{A}}$ and $\overline{Q}$ are connected via (96), is finally written

$${\overline{Q}}^3+4H_2{\overline{Q}}^2+12H_1{H}_3\overline{Q}+32H_3^2=0.$$

(109)

Equation (109) holds true, since it is actually the polynomial Equation (47), which determines the separation constants for the hyper-Lamé Equation (39). Consequently, solution (95) is indeed a closed-form ellipsoidal wave. □

5. Closing Notes

The present work, which ends up by demonstrating the first closed-form ellipsoidal wave and describes how one can generate any number of such waves, is actually founded on a very extended series of calculations, which were by no means straightforward. They demanded a selective order of calculations to take advantage of the symmetries of the 4D ellipsoidal system; some of the involved complicated identities can be found at the appendices of [8,9]. In concluding, we may have a way to generate ellipsoidal waves at this point, but this is not a simple exercise. Anyone that tries to fill in all computational gaps, even only for the special case of Section 4, will understand our point. However, work is under way for extending the presented methods, and we hope that some colleagues, too, will improve our work and make the whole process easier.