# Explicit solutions to the 3D incompressible Euler equations in Lagrangian
formulation

Tomi Saleva
Department of Physics and Mathematics
University of Eastern Finland
P.O. box 111, FI-80101 Joensuu, Finland
<EMAIL_ADDRESS>
&Jukka Tuomela
Department of Physics and Mathematics
University of Eastern Finland
P.O. box 111, FI-80101 Joensuu, Finland
<EMAIL_ADDRESS>

###### Abstract

We introduce many families of explicit solutions to the three dimensional
incompressible Euler equations for nonviscous fluid flows using the Lagrangian
framework. Almost no exact Lagrangian solutions exist in the literature prior
to this study. We search for solutions where the time component and the
spatial component are separated, applying the same ideas we used previously in
the two dimensional case. We show a general method to derive separate
constraint equations for the spatial component and the time component. Using
this provides us with a plenty of solution sets exhibiting several different
types of fluid behaviour, but since they are computationally heavy to analyze,
we have to restrict deeper analysis to the most interesting cases only. It is
also possible and perhaps even probable that there exist more solutions of the
separation of variables type beyond what we have found.

_M_ athematics Subject Classification (2020) 35A09; 35Q31; 76B99

_K_ eywords Explicit solutions, Euler equations, Lagrangian formulation, Fluid
mechanics

The first author was supported by Finnish Cultural Foundation.

##  1 Introduction

This is the third paper, following [13, 15], in our research of finding
explicit solutions to the incompressible Euler equations in the Lagrangian
framework. The Lagrangian framework is one of the two main approaches to
describing fluid motion. In the Lagrangian representation the flow is
described by giving the trajectories of individual particles. The other main
approach is the Eulerian framework which considers the velocity field or the
vorticity field of the flow. For general background on the Euler equations we
refer to [8], and for the Lagrangian formulation in particular we refer to
[5]. Lagrangian formulation has been applied to flow problems with group
theory in [4].

In our two previous articles we studied the two-dimensional case, finding
several solution sets, and in the present paper we apply similar ideas to the
three-dimensional case. The goal is to find solutions of the separation of
variables type, where space and time coordinates are treated separately. The
first explicit solutions of this type were already found in 19th century by
Gerstner and Kirchhoff in the 2D case [10, 11]. In the recent decades these
solutions have been generalized using harmonic maps, see for example [1, 2, 3,
7]. All these solutions were of the form where the time component and the
spatial component could be separated, and finally in [15] we systematically
found probably all possible solutions of the separation of variables type to
the 2D Euler equations in Lagrangian formulation, including ones that could
not be found with harmonic maps.

However, it seems that there are surprisingly few articles dealing with the 3D
case, the only exception we found is [18]. After all, many scientists have
studied this problem for a long time in the two-dimensional context and it
would seem natural to try to find similar solutions also in three dimensions
which is of course physically the most relevant case. Perhaps one reason is
that since classically complex functions were used to analyze the two-
dimensional case, the generalization to the three-dimensional case appeared
difficult.

In [18] the problem is approached in a slightly nonconventional variant of the
Lagrangian formulation, not describing the system for the particle trajectory
map $\varphi$ but rather for its differential $d\varphi$. For the spatial part
they in fact used the complex formulation for two of the components and a real
function for the third component. To find exact solutions, they considered
cases where the time component and the spatial component are separated for
$d\varphi$. A simple integration argument shows that then $\varphi$ is also of
the separation of variables form and thus these solutions can also be found
with the technique presented in this paper.

Anyway in [13, 15] we showed that even in two dimensions the situation is best
analyzed using real functions, and thus a priori it seemed natural to believe
that similar techniques should be applicable also in three dimensional case.
In the present paper we show that this is indeed the case: even though certain
classes of solutions can equivalently be analyzed with complex formulation,
one can find many more large classes of solutions operating with real
functions. In fact it is not possible to describe all solutions: there are
simply so many possibilities that one cannot treat them all in a short article
like this. Also it is not possible to describe all computations in detail.
Some intermediate formulas and expressions are simply too big to be written
down. However, this is not a problem in the sense that the verification of the
final results is a straightforward computation. Of course this computation
would be very tedious if done by hand.

Perhaps one should also clarify what we mean by the word ”explicit” in the
title. In some cases the solution set cannot really be described by completely
explicit formulas. In these cases one may think that the word ”explicit”
refers to the explicit description of the structure of the solutions. However,
we think that these structural descriptions can also be very useful in
applications: combining numerical and symbolic computation one can use them to
”build” specific solutions with the desired properties. Also one can then
analyze what kind of properties are actually possible for a given family of
solutions.

In Section 2 we introduce some useful notions and results which are needed
later. In Section 3 we formulate our framework precisely. Then in Sections 4
to 8 we present our solutions. We start with simple cases which can be
described more thoroughly; these cases show the reader the main ideas in the
computations. In later sections many details must be suppressed, but the
computations are really very similar to simpler cases. One does not need any
new methods to deal with the complicated cases, the difficulty comes just from
the sheer size of the problem. Unlike in the 2D case we studied in [15], here
we will not consider all possible families of solutions due to their vast
number. Instead, we concentrate on those cases which to us seem to be the most
promising from the physical point of view. However, we also indicate some
systems which we know lead to nontrivial explicit solutions but which are not
analyzed in the present paper. It is also highly possible that there are cases
that we did not discover altogether.

##  2 Preliminaries and notation

###  2.1 Euler parameters

We need to represent rotations in $\mathbb{R}^{3}$. There are several ways to
do this but Euler parameters are most convenient for our purposes. Let
$a=(a_{0},a_{1},a_{2},a_{3})$ and let us define the following matrices:

$\displaystyle\widetilde{H}_{a}$
$\displaystyle=\begin{pmatrix}-a_{1}&a_{0}&-a_{3}&a_{2}\\\
-a_{2}&a_{3}&a_{0}&-a_{1}\\\ -a_{3}&-a_{2}&a_{1}&a_{0}\end{pmatrix}\quad,\quad
H_{a}=\begin{pmatrix}-a_{1}&a_{0}&a_{3}&-a_{2}\\\ -a_{2}&-a_{3}&a_{0}&a_{1}\\\
-a_{3}&a_{2}&-a_{1}&a_{0}\end{pmatrix}$ $\displaystyle\tilde{K}_{a}$
$\displaystyle=\begin{pmatrix}a_{0}&a_{1}&a_{2}&a_{3}\\\
-a_{1}&a_{0}&-a_{3}&a_{2}\\\ -a_{2}&a_{3}&a_{0}&-a_{1}\\\
-a_{3}&-a_{2}&a_{1}&a_{0}\end{pmatrix}\quad,\quad
K_{a}=\begin{pmatrix}a_{0}&a_{1}&a_{2}&a_{3}\\\ -a_{1}&a_{0}&a_{3}&-a_{2}\\\
-a_{2}&-a_{3}&a_{0}&a_{1}\\\ -a_{3}&a_{2}&-a_{1}&a_{0}\end{pmatrix}$
$\displaystyle R_{a}=\widetilde{H}_{a}H^{T}_{a}$
$\displaystyle=\begin{pmatrix}a_{0}^{2}+a_{1}^{2}-a_{2}^{2}-a_{3}^{2}&2a_{1}a_{2}-2a_{0}a_{3}&2a_{1}a_{3}+2a_{0}a_{2}\\\
2a_{1}a_{2}+2a_{0}a_{3}&a_{0}^{2}-a_{1}^{2}+a_{2}^{2}-a_{3}^{2}&2a_{2}a_{3}-2a_{0}a_{1}\\\
2a_{1}a_{3}-2a_{0}a_{2}&2a_{2}a_{3}+2a_{0}a_{1}&a_{0}^{2}-a_{1}^{2}-a_{2}^{2}+a_{3}^{2}\end{pmatrix}\
.$

If $|a|=1$ then $R_{a}\in\mathbb{SO}(3)$ and $\tilde{K}_{a}$,
$K_{a}\in\mathbb{SO}(4)$. Since $R_{a}=R_{-a}$ the sphere $S^{3}$ is the
double cover of $\mathbb{SO}(3)$.

###  2.2 PDE

Let $u$ and $v$ be functions of two variables; there are two natural
conventions regarding the sign in the Cauchy–Riemann equations:

$\begin{cases}u_{x}-v_{y}=0\\\
u_{y}+v_{x}=0\end{cases}\quad,\quad\begin{cases}u_{x}+v_{y}=0\\\
u_{y}-v_{x}=0\end{cases}\ .$

It turns out that for us it is more convenient to use the second system. We
will call this the anti CR system.

For general PDE systems it is more convenient to use multiindices to denote
derivatives. Hence if $u=(u^{1},\dots,u^{m})$ is some map the derivatives of
its components are

$u_{\nu}^{j}=\partial^{\nu}u^{j}=\frac{\partial^{|\nu|}u^{j}}{\partial
x_{1}^{\nu_{1}}\cdots\partial x_{n}^{\nu_{n}}}\ .$

We will say several times below that a solution to a certain PDE system exists
by standard theorems. This means that the necessary details can easily be
found in [9]. We will also need some elementary notions of differential
geometry in the analysis; for details we refer to [12].

We will analyze below many rather complicated ODE and PDE systems. Since the
analysis often cannot be given very explicitly it is perhaps helpful to
discuss the general nature of systems and their solutions. For a general
treatment of these issues we refer to [16] and the many references therein.

Now all our equations are differential polynomials so that some kind of
algorithmic treatment is possible at least in principle. A fundamental theorem
in differential algebra says that

  * _any radical differential ideal is a finite intersection of prime differential ideals._

One may interpret this by saying that prime differential ideals correspond to
certain families of solutions, and hence the theorem says that there are in
fact only finitely many essentially different families of solutions.

Another complication is that our systems contain parameters, and in general
one must make case distinctions, depending on the values of parameters. The
differential elimination algorithm rifsimp [14] which is implemented in Maple
and which we will use at times follows the strategy that the default is to
look for the largest family. However, it must be kept in mind that there may
be also other solutions, not contained in the family given.

In the solutions given below we must at times make case distinctions so that
there are several families of solutions, i.e. the differential ideal is not
prime. However, we do not need to take into account all cases because we can
discard the cases which are irrelevant in the present context. We will explain
below more precisely what this means.

###  2.3 Minors

Let $A\in\mathbb{R}^{3\times m}$; the columns of $A$ are denoted by $A_{j}$.
Let us further set $p_{ijk}=\det(A_{i},A_{j},A_{k})$. The following formulas
will be useful in computations.

######  Lemma 2.1

$\displaystyle p_{123}p_{145}-p_{124}p_{135}+p_{125}p_{134}=0\,,$
$\displaystyle p_{123}p_{456}-p_{124}p_{356}+p_{125}p_{346}-p_{126}p_{345}=0$
$\displaystyle p_{234}A_{1}-p_{134}A_{2}+p_{124}A_{3}-p_{123}A_{4}=0$
$\displaystyle
p_{123}^{2}p_{456}+\det{\begin{pmatrix}p_{234}&p_{235}&p_{236}\\\
p_{134}&p_{135}&p_{136}\\\ p_{124}&p_{125}&p_{126}\end{pmatrix}}=0\,.$

The first two equations in Lemma 2.1 are examples of Plücker relations.

##  3 Euler equations

Let us consider the incompressible Euler equations

$\displaystyle u_{t}+u\nabla u+\nabla p=0$ (3.1) $\displaystyle\nabla\cdot
u=0$

in some domain $\Omega\subset\mathbb{R}^{n}$. This is called the Eulerian
description of the flow and the coordinates of $\Omega$, denoted $x$, are the
Eulerian coordinates. Below we will consider another description which is
almost the Lagrangian description of the flow.

Let $D\subset\mathbb{R}^{n}$ be another domain and let us consider a family of
diffeomorphisms $\varphi^{t}\,:\,D\to\Omega_{t}=\varphi^{t}(D)$. The
coordinates in $D$ are denoted by $z$. We can also define

$\varphi\,:\,D\times\mathbb{R}\to\mathbb{R}^{n}\quad,\quad\varphi(z,t)=\varphi^{t}(z)\,.$

Now given such $\varphi$ we can define the associated vector field $u$ by the
formula

$\frac{\partial}{\partial t}\varphi(z,t)=u(\varphi(z,t),t)\,.$ (3.2)

Our goal is to find maps $\varphi$ such that $u$ solves the Euler equations in
the three dimensional case.

To express the incompressibility condition $\nabla\cdot u=0$ we introduce the
volume forms $\mathsf{vol}_{x}=dx^{1}\wedge dx^{2}\wedge dx^{3}$ and
$\mathsf{vol}_{z}=dz^{1}\wedge dz^{2}\wedge dz^{3}$. Then
$\varphi^{\ast}\mathsf{vol}_{x}=\alpha\,\mathsf{vol}_{z}$ for some $\alpha$.
In the Lagrangian case we should require that $\alpha=1$ but here we only
require that it is independent of time and nonzero in $D$.

It will be convenient to interpret the first equation of (3.1) as well as the
equation (3.2) in terms of covectors instead of vectors. Then differentiating
the equation (3.2) we obtain

$\varphi^{\prime\prime}+dp=0\,.$

Then the condition of existence of local solutions is simply
$d\varphi^{\prime\prime}=0$. The exterior derivative cannot be directly
computed in $x$ coordinates so we have to pull back this condition to $z$
coordinates. To this end let us introduce

$h=d\,\varphi^{\ast}\,\varphi^{\prime}=d\varphi_{1}^{\prime}\wedge
d\varphi_{1}+d\varphi_{2}^{\prime}\wedge
d\varphi_{2}+d\varphi_{3}^{\prime}\wedge d\varphi_{3}\,.$

Note that if we interpret the vorticity $\zeta=du$ as a two-form then $h$ is
the vorticity in $z$ coordinates. A fundamental fact is that denoting by
$\mathcal{L}$ the Lie derivative we have [6, 17]

$\zeta_{t}+\mathcal{L}_{u}\zeta=0\,.$

We will often write for simplicity

$h=h^{3}dz^{1}\wedge dz^{2}-h^{2}dz^{1}\wedge dz^{3}+h^{1}dz^{2}\wedge
dz^{3}=\big{(}h^{1},h^{2},h^{3}\big{)}\ .$

The components $h^{j}$ are classically known as _Cauchy invariants_ [6], and
as the name suggests they should not depend on time. Accordingly we see that
$d\varphi^{\prime\prime}=0$ is equivalent to $h^{\prime}=0$. Let us summarize
this discussion with

######  Lemma 3.1

$\varphi$ provides a solution to Euler equations if and only if both $h$ and
$\alpha$ are independent of time.

Note that here is an essential difference to the 2D case: the condition for
$h$ is related to two-forms while the condition for $\alpha$ is formulated in
terms of three-forms. In 2D case both conditions are evidently formulated with
two-forms.

It is also interesting to express the vorticity $\zeta$ as a vector in $x$
coordinates in terms of $h$. To this end we express first $h$ as a vector
using the Hodge star and raising the index; in components this gives

$\tilde{h}=\sharp\ \ast
h=\tilde{h}^{1}\partial_{z_{1}}+\tilde{h}^{2}\partial_{z_{2}}+\tilde{h}^{3}\partial_{z_{3}}=\big{(}\tilde{h}^{1},\tilde{h}^{2},\tilde{h}^{3}\big{)}=\frac{1}{\alpha}\,\big{(}h^{1},h^{2},h^{3}\big{)}\
.$

From this we get _Cauchy’s vorticity formula_ [6]:

$\zeta=d\varphi\,\tilde{h}\,.$

This can be interpreted either as a pushforward or how the components of
vector field change under the change of coordinates.

Let us then consider the maps of the following form

$\varphi(z,t)=A(t)v(z)\,,$ (3.3)

where $A(t)\in\mathbb{R}^{3\times m}$, $v\,:\,D\to\mathbb{R}^{m}$ and
$D\subset\mathbb{R}^{3}$ is some coordinate domain. Our next task is to
compute convenient formulas for $h$ and $\alpha$ in this case.

Since all the analysis is local, the precise nature of $D$ is not important in
our context. We will try to find maps $\varphi$ such that the corresponding
vector field $u$ defined by the formula (3.2) is a solution to the Euler
equations. Hence we should find $A$ and $v$ such that the conditions in Lemma
3.1 are satisfied. Since we want that $\det(d\varphi)\neq 0$, this necessarily
implies that $\mathsf{rank}(A)=\mathsf{rank}(dv)=3$. Of course we can always
assume that $\det(d\varphi)>0$.

Let us denote the coordinates of the intermediate space $\mathbb{R}^{m}$ by
$y$. Then we can interpret $A$ as a map $\mathbb{R}^{m}\to\mathbb{R}^{3}$ and
we can write

$A^{\ast}\mathsf{vol}_{x}=\sum_{1\leq i<j<k\leq m}p_{ijk}dy^{i}\wedge
dy^{j}\wedge dy^{k}=\sum_{1\leq i<j<k\leq
m}\det(A_{i},A_{j},A_{k})dy^{i}\wedge dy^{j}\wedge dy^{k}\ ,$

where $A_{i}$ denote the columns of $A$. Similarly we set

$v^{\ast}dy^{i}\wedge dy^{j}\wedge dy^{k}=g_{ijk}\mathsf{vol}_{z}=\det(\nabla
v^{i},\nabla v^{j},\nabla v^{k})\mathsf{vol}_{z}\ .$

But from this we immediately obtain

$\alpha=\sum_{1\leq i<j<k\leq m}p_{ijk}g_{ijk}\ .$ (3.4)

This is known as the Cauchy–Binet formula. Then let us introduce

$\beta=\sum_{1\leq i<j\leq m}Q_{ij}dy^{i}\wedge dy^{j}\quad\mathrm{where}\quad
Q_{ij}=\langle A_{i}^{\prime},A_{j}\rangle-\langle
A_{j}^{\prime},A_{i}\rangle\ .$

Then putting $G_{ij}=v^{\ast}dy^{i}\wedge dy^{j}$ we can write

$h=v^{\ast}\beta=\sum_{1\leq i<j\leq m}Q_{ij}G_{ij}=\sum_{1\leq i<j\leq
m}Q_{ij}dv^{i}\wedge dv^{j}\ .$ (3.5)

To solve the Euler equations we need (3.4) and (3.5) to be constant in time by
Lemma 3.1. The crucial observation is that when we fix any $t$ in (3.4) and
(3.5), we see that for the time derivatives of $\alpha$ and $h$ to vanish the
spatial component only needs to satisfy constraints of the forms

$\sum\gamma_{ijk}g_{ijk}=0\textrm{ and }\sum\gamma_{ij}G_{ij}=0\,,$ (3.6)

where $\gamma_{ijk}$ and $\gamma_{ij}$ are constants. Furthermore, when we
substitute these equations back to (3.4) and (3.5), we obtain the constraints
that $p_{ijk}$ and $Q_{ij}$ have to satisfy.

There is the following curious connection between $p_{ijk}$ and $Q_{ij}$.

######  Lemma 3.2

$\Omega=A^{\ast}\mathsf{vol}_{x}\wedge\beta=0\ .$

Proof.  Let first $m=5$ so that $\Omega=\Omega_{12345}dy^{1}\wedge
dy^{2}\wedge dy^{3}\wedge dy^{4}\wedge dy^{5}$. Let us denote the rows of $A$
by $\hat{A}_{i}$. Then computing we see that

$\Omega_{12345}=\sum_{j=1}^{3}\det(\hat{A}_{1},\hat{A}_{2},\hat{A}_{3},\hat{A}_{j}^{\prime},\hat{A}_{j})=0\
.$

If $m>5$ then similar formulas are valid for each group of five indices.
$\square$

We have now our formulas for $\alpha$ and $h$ but we can still simplify them.
First it is important to remember that the domain $D$ is simply some parameter
domain which has no physical significance. Hence one can look for the
”simplest” possible parameter domain. For future reference let us record this
observation as

######  Lemma 3.3

Let $\psi\,:\,\hat{D}\to D$ be an arbitrary diffeomorphism and let
$\tilde{\varphi}^{t}=\varphi^{t}\circ\psi$. Then $\tilde{\varphi}$ provides
solutions to the Euler equations via formula (3.2) if and only if $\varphi$
does.

Proof.  This follows from standard properties of pullback. $\square$

In particular, if $g_{123}\neq 0$ then we can suppose that $v$ is of the form

$v(z)=\big{(}z_{1},z_{2},z_{3},f^{1}(z),\dots,f^{m-3}(z)\big{)}\,.$ (3.7)

In each case in which we find the general solution to $v$, the indices of $v$
are chosen such that $g_{123}=0$ would imply $\det(d\varphi)=0$. Thus we will
always write $v$ as in (3.7).

We can also write the time component in a more convenient way; recall that by
QR decomposition we can always write $A=RB$, where $R\in\mathbb{SO}(3)$ and
some $3\times 3$ submatrix $(B_{i},B_{j},B_{k})$ of $B$ is an upper triangular
matrix.

To see the effect of the QR decomposition on $h$ we first compute

$\langle A_{i}^{\prime},A_{j}\rangle=\langle
R^{T}R^{\prime}B_{i},B_{j}\rangle+\langle B_{i}^{\prime},B_{j}\rangle\ .$

Using the Euler parameters to represent the rotations we note that

$(R^{\prime})^{T}R=2H_{a^{\prime}}H_{a}^{T}=-2H_{a}H_{a^{\prime}}^{T}\,.$

This implies that

$Q_{ij}=4\langle H_{a}a^{\prime},B_{i}\times B_{j}\rangle+\langle
B_{i}^{\prime},B_{j}\rangle-\langle B_{i},B_{j}^{\prime}\rangle\ .$ (3.8)

But now we can in a sense eliminate $a$. Let us set $w=4\,H_{a}a^{\prime}$ and
$\hat{w}=(0,w_{1},w_{2},w_{3})$. Then we can write

$4\,K_{a}a^{\prime}=\hat{w}\quad\Longrightarrow\quad
a^{\prime}=\tfrac{1}{4}\,K_{a}^{T}\hat{w}=\tfrac{1}{4}\,\tilde{K}_{\hat{w}}^{T}a\
.$

Note that $\tilde{K}_{\hat{w}}$ is antisymmetric which guarantees that $|a|$
stays constant. $\tilde{K}_{\hat{w}}$ is also a Hamiltonian matrix so that
$a^{\prime}=\tfrac{1}{4}\,\tilde{K}_{\hat{w}}^{T}a$ is in fact a Hamiltonian
system with respect to standard symplectic form

$\omega=da_{0}\wedge da_{1}+da_{2}\wedge da_{3}\ .$

We can use $w$ as a kind of extra variables in our equations and we can thus
write

$Q_{ij}=\langle w,B_{i}\times B_{j}\rangle+\langle
B_{i}^{\prime},B_{j}\rangle-\langle B_{i},B_{j}^{\prime}\rangle\ .$ (3.9)

In what follows we will often have conditions where $p_{ijk}$ or $Q_{ij}$ is
some constant; in such cases these constants are always denoted by
$p_{ijk}=e_{ijk}$ and $Q_{ij}=c_{ij}$.

Note that the presentation of $\varphi=A(t)v(z)$ is not unique.

######  Lemma 3.4

If $\varphi=A(t)v(z)$ is a solution to the Euler equations and
$\tilde{v}(z)=H^{-1}v$, where $H$ is a regular $m\times m$ matrix, then there
is a matrix $\tilde{A}(t)$, namely $\tilde{A}=AH$, such that
$\varphi=Av=\tilde{A}\tilde{v}$.

Although the statement of the Lemma is really trivial, it is actually quite
useful in concrete computations and we will use it repeatedly to simplify the
problems below without losing any generality.

Then we note the following simple fact which allows us to discard
uninteresting cases.

######  Lemma 3.5

Suppose that in the representation (3.3)

  1. 1.

$A_{j}$ are linearly dependent over $\mathbb{R}$ or

  2. 2.

$dv^{j}$ are linearly dependent over $\mathbb{R}$.

Then we can write $\varphi=\hat{A}\hat{v}$ where
$\hat{A}(t)\in\mathbb{R}^{3\times k}$ and $k<m$.

The simple proof can be found in [15]. The problems we study typically have
several families of solutions. Some of them are reducible to a case of lower
$m$ by Lemma 3.5 and thus do not need to be considered in the same context. In
the solutions given below we have verified that all the reducible solution
families have been discarded and that they contain all the relevant solution
families.

In the following sections we find several solution classes for cases $m=3$,
$m=4$, $m=5$, and $m=6$. For cases $m=4$ and $m=5$ there are subcases that we
do not consider here since for each $m$ we concentrate on the solutions for
which the spatial component has the most freedom. In case $m=6$ the systems
for $A$ will always be overdetermined and we are lucky to find out that there
are at least some solution classes in that case, too.

Let us note that the explicit solutions given in [18] are particular cases of
either $m=3$ or $m=4$.

##  4 Cases $m=3$ and $m=4$

If $A\in\mathbb{R}^{3\times 3}$ and $v\in\mathbb{R}^{3}$, any constraint of
the form (3.6) would imply $\alpha=0$ and thus we may as well assume that
$v=z$ by Lemma 3.3. This case might be called the Kirchhoff case because it is
analogous to the Kirchhoff solution in 2D case.

In this case for (3.4) and (3.5) to be independent of time we need
$p_{123}=\det(A)$, as well as $Q_{12}$, $Q_{13}$, and $Q_{23}$ to be constant.
By scaling we may assume that $A\in\mathbb{SL}(3)$ and by elementary counting
arguments we suspect that there are 5 arbitrary functions in the general
solution. Writing $A=RB$, where $R\in\mathbb{SO}(3)$ and $B$ is an upper
triangular matrix, the condition $\alpha=1$ gives $b_{33}=1/(b_{11}b_{22})$.
Then solving $Q_{ij}=c_{ij}$ where $Q_{ij}$ are as in (3.9) we obtain

$\displaystyle b_{12}^{\prime}=$
$\displaystyle\frac{b_{12}b_{11}^{\prime}-w_{3}b_{11}b_{22}+c_{12}}{b_{11}}$
$\displaystyle b_{13}^{\prime}=$
$\displaystyle\frac{b_{13}b_{11}^{\prime}+w_{2}b_{11}b_{33}-w_{3}b_{11}b_{23}+c_{13}}{b_{11}}$
$\displaystyle b_{23}^{\prime}=$
$\displaystyle\frac{b_{11}b_{23}b_{22}^{\prime}-w_{1}+b_{11}c_{23}-b_{12}c_{13}+b_{13}c_{12}}{b_{22}b_{11}}\
.$

Hence there are 5 functions which can be chosen arbitrarily: $b_{11}$,
$b_{22}$, $w_{1}$, $w_{2}$ and $w_{3}$. Moreover, to demonstrate the
usefulness of Lemma 3.4 we present the following simplification.

######  Lemma 4.1

Without loss of generality $c_{12}$ and $c_{13}$ can be chosen to be zero.

Proof.  If some $c_{ij}$ is nonzero, by symmetry we may assume $c_{23}\neq 0$.
Then letting $\tilde{A}=AH$ and $\tilde{v}=H^{-1}v$, where

$H=\begin{pmatrix}1&0&0\\\ -c_{13}/c_{23}&1&0\\\
c_{12}/c_{23}&0&1\end{pmatrix}\,,$

we have $\varphi=Av=\tilde{A}\tilde{v}$ and $\tilde{c}_{12}=\tilde{c}_{13}=0$
for $\tilde{A}$. $\square$

Note that in this case $h$ is also constant in space and that the solution in
Euler coordinates is $u=A^{\prime}A^{-1}x$.

Then let $A\in\mathbb{R}^{3\times 4}$. Let us start with the constraint that
$\alpha$ in (3.4) must be constant. Using $v=\big{(}z,f(z)\big{)}$ we obtain

$\alpha=p_{123}+p_{124}f_{001}-p_{134}f_{010}+p_{234}f_{100}\ .$

Note that we cannot require that all $p_{ijk}$ are constant. Since
$p_{234}A_{1}-p_{134}A_{2}+p_{124}A_{3}-p_{123}A_{4}=0$ we must by Lemma 3.5
have a constraint of the form

$\sum\gamma_{ijk}g_{ijk}=0\ ,$

where $\gamma_{ijk}$ are constants.

######  Lemma 4.2

We may assume this constraint to be $g_{234}=0$.

Proof.  We may assume that $\gamma_{234}=1$. Then the transformation
$\tilde{v}=Hv$, where

$H=\begin{pmatrix}1&0&0&0\\\ -\gamma_{134}&1&0&0\\\ \gamma_{124}&0&1&0\\\
-\gamma_{123}&0&0&1\end{pmatrix},$

gives $\tilde{g}_{234}=0$. $\square$

The solution of the equation $g_{234}=0$ is
$v=(z_{1},z_{2},z_{3},f(z_{2},z_{3}))$ for an arbitrary function $f$. Then we
must have

$\displaystyle p_{123}=$ $\displaystyle b_{11}b_{22}b_{33}=e_{123}$
$\displaystyle p_{124}=$ $\displaystyle b_{11}b_{22}b_{34}=e_{124}$
$\displaystyle p_{134}=$ $\displaystyle
b_{11}\left(b_{23}b_{34}-b_{24}b_{33}\right)=e_{134}$

as well as $Q_{ij}=c_{ij}$ for each $i<j$. We can assume that $e_{123}\neq 0$
and by scaling we may further assume that $e_{123}=1$.

######  Lemma 4.3

Without loss of generality we may assume that $e_{124}=e_{134}=0$.

Proof.  We apply Lemma 3.4 with

$H=\begin{pmatrix}1&0&0&0\\\ 0&1&0&e_{134}\\\ 0&0&1&-e_{124}\\\
0&0&0&1\end{pmatrix}\ .$

$\square$

But if $e_{124}=e_{134}=0$ then we have $\alpha=1$ and we must also have
$b_{24}=b_{34}=0$ which gives

$B=\begin{pmatrix}b_{11}&b_{12}&b_{13}&b_{14}\\\ 0&b_{22}&b_{23}&0\\\
0&0&1/(b_{11}b_{22})&0\end{pmatrix}\ .$

This leads to

######  Theorem 4.4

In case $m=4$ the general solution of the time component can be written in the
following form:

$A=\hat{R}\begin{pmatrix}1&0&0\\\ 0&\cos(\theta)&-\sin(\theta)\\\
0&\sin(\theta)&\cos(\theta)\end{pmatrix}\begin{pmatrix}b_{11}&b_{12}&b_{13}&b_{14}\\\
0&b_{22}&b_{23}&0\\\ 0&0&1/(b_{11}b_{22})&0\end{pmatrix},$ (4.1)

where $b_{12}$, $b_{13}$, $b_{14}$ and $b_{23}$ satisfy the equations given in
the proof below. Functions $\theta$, $b_{11}$ and $b_{22}$ are arbitrary and
$\hat{R}$ is a constant rotation matrix.

Proof.  Substituting the above $B$ to the equations $Q_{ij}=c_{ij}$ we first
note that $Q_{14}=b_{14}b_{11}^{\prime}-b_{11}b_{14}^{\prime}=c_{14}\neq 0$
because otherwise the solution reduces. Then we compute that

$b_{12}=\frac{c_{24}b_{11}+c_{12}b_{14}}{c_{14}}\quad,\quad
b_{13}=\frac{c_{34}b_{11}+c_{13}b_{14}}{c_{14}}\ .$

Substituting these back to the equations we note that $w_{2}=w_{3}=0$ and

$b_{23}^{\prime}=\frac{b_{11}b_{23}b_{22}^{\prime}-w_{1}}{b_{22}b_{11}}+\Big{(}c_{23}+\frac{c_{12}c_{34}-c_{13}c_{24}}{c_{14}}\Big{)}\frac{1}{b_{22}}\
.$

Then since $w_{1}$ is arbitrary we may as well define
$\theta^{\prime}=w_{1}/2$ in which case one can easily write down the solution
of $a^{\prime}=\tfrac{1}{4}\tilde{K}_{\hat{w}}^{T}a$. By applying the constant
rotation the solution can then be written as indicated. $\square$

Note that if $\varphi$ is a solution and $\hat{R}$ is a constant rotation then
$\hat{R}\varphi$ is also a solution. In the sequel we will not anymore write
explicitly the rotation in this kind of situation.

The components of vorticity or the Cauchy invariants of the solutions of this
form are given by

$h=\Big{(}c_{24}f_{001}-c_{34}f_{010}+c_{23},-c_{14}f_{001}-c_{13},c_{14}f_{010}+c_{12}\Big{)}\
.$

Then one could ask how ”general” vorticity can be constructed with solutions
of this form.

######  Lemma 4.5

$\hat{h}=\big{(}h^{1},h^{2},h^{3}\big{)}$ is a vorticity for some solution
given in Theorem 4.1, if $h^{k}$ do not depend on $z_{1}$,
$h^{2}_{010}+h^{3}_{001}=0$ and $h^{1}=e_{2}h^{2}+e_{3}h^{3}$ for some
constants $e_{j}$.

Proof.  Eliminating $f$ from equations $h=\hat{h}$ gives readily the result.
$\square$

Note that since $dh=0$ there is always an equation
$h^{1}_{100}+h^{2}_{010}+h^{3}_{001}=0$ for $\hat{h}$.

If we suppose that $f$ depends only on one variable one could compute other
families of solutions. However, these solutions are less interesting
physically so we do not analyze them further.

##  5 Case $m=5$

In the case $m=5$ it turns out that there are numerous subcases providing
solutions, and it is impossible for us to consider all of them here. Hence, we
will concentrate on the cases for which the spatial component has the greatest
freedom, which happens when we have two equations of the form
$\sum\gamma_{ijk}g_{ijk}=0$ as the spatial constraints. The systems obtained
for the time component are then overdetermined and usually do not give any
solutions if the spatial constraints are chosen arbitrarily. However, we found
three cases that do provide solutions:

$\begin{cases}g_{134}+g_{235}=0\\\ g_{135}-g_{234}=0\end{cases}\ ,\
\begin{cases}g_{134}=0\\\ g_{235}=0\end{cases}\ \mathrm{and}\
\begin{cases}g_{134}+g_{235}=0\\\ g_{135}=0\end{cases}\ .$ (5.1)

These are analogous to the three cases we found for the 2D case in [15, sec.
4]. As we briefly noted in [15], the three 2D cases could be called elliptic,
hyperbolic, and parabolic, based on the second-order differential equation
that the spatial component functions satisfy in each case. We adopt this
terminology here and refer to the three cases of (5.1) as the _elliptic_ ,
_hyperbolic_ , and _parabolic_ case, respectively. In the current section we
present the solution for the spatial component and the time component for
these three cases. As examples of the cases we will not consider any further
but which give nontrivial families of solutions we may cite

$\begin{cases}G_{14}+G_{25}=0\\\
G_{15}-G_{24}=0\end{cases},\,\begin{cases}G_{15}=0\\\
G_{24}=0\end{cases},\,\begin{cases}G_{14}+G_{25}=0\\\ G_{15}=0\end{cases}.$

###  5.1 Elliptic case

Let the spatial constraints be

$g_{134}+g_{235}=g_{135}-g_{234}=0\,.$

The solution to this system is
$v=(z_{1},z_{2},z_{3},f^{1}(z_{1},z_{2},z_{3}),f^{2}(z_{1},z_{2},z_{3}))$,
where $f^{1}$ and $f^{2}$ are anti-CR with respect to $z_{1}$ and $z_{2}$. For
the time component,

$p_{123},p_{345},p_{134}-p_{235},p_{135}+p_{234},p_{124},p_{125},p_{145},p_{245}\,,$

as well as each $Q_{ij}$ has to be constant. Let us write $A=RB$ as before.
Since

$p_{245}A_{1}-p_{145}A_{2}+p_{125}A_{4}-p_{124}A_{5}=0$

by Lemma 2.1, we have $p_{124}=p_{125}=p_{145}=p_{245}=0$ by Lemma 3.5. This
immediately implies that $b_{34}=b_{35}=0$.

As for the rest of the determinant constants, we may assume without loss of
generality that $p_{123}=p_{345}=1$ and $p_{134}-p_{235}=p_{135}+p_{234}=0$.
The proof of this is essentially the same as in the 2D case found in [15,
Lemmas 5.3 and 5.4]. We simply replace 3 and 4 by 4 and 5 in all the
subscripts, respectively, and add a 3 to each of them, which does not affect
the calculations. Then we compute that

$\det(d\varphi)=1-\big{(}f^{1}_{100}\big{)}^{2}-\big{(}f^{1}_{010}\big{)}^{2}\neq
0\ .$

Then we compute that $b_{24}^{2}+b_{25}^{2}=b_{22}^{2}$ and thus we can write
$b_{24}=\sin(\theta)b_{22}$ and $b_{25}=\cos(\theta)b_{22}$. But then
$b_{14}=\cos(\theta)b_{11}+\sin(\theta)b_{12}$ and
$b_{15}=-\sin(\theta)b_{11}+\cos(\theta)b_{12}$ so that

$B=\begin{pmatrix}b_{11}&b_{12}&b_{13}&\cos(\theta)b_{11}+\sin(\theta)b_{12}&-\sin(\theta)b_{11}+\cos(\theta)b_{12}\\\
0&b_{22}&b_{23}&\sin(\theta)b_{22}&\cos(\theta)b_{22}\\\
0&0&1/(b_{11}b_{22})&0&0\end{pmatrix}\ .$

Then preliminary computations show that in any case we must have

$c_{45}=-c_{12}\neq 0\ ,\ c_{24}=c_{15}\ \mathrm{and}\ c_{25}=-c_{14}\ .$

######  Lemma 5.1

Without loss of generality we may assume that $c_{14}=c_{15}=0$.

Proof.  Eliminating $w_{j}$ from the equations $Q_{ij}=c_{ij}$ we eventually
find the following system:

$\frac{c_{12}}{b_{11}^{2}+b_{12}^{2}+b_{22}^{2}}\begin{pmatrix}\cos(\theta)&\sin(\theta)\\\
-\sin(\theta)&\cos(\theta)\end{pmatrix}\begin{pmatrix}2b_{11}b_{12}\\\
-b_{11}^{2}+b_{12}^{2}+b_{22}^{2}\end{pmatrix}=\begin{pmatrix}c_{14}\\\
c_{15}\end{pmatrix}\ .$

This implies that $c_{12}^{2}-c_{15}^{2}-c_{14}^{2}>0$, and hence we can
choose constants $d$ and $s$ such that

$\displaystyle c_{15}$ $\displaystyle=c_{12}\cos(d)\tanh(2s)$ $\displaystyle
c_{14}$ $\displaystyle=-c_{12}\sin(d)\tanh(2s)\,.$

Let the transformation matrix $H$ be

$H=\begin{pmatrix}\cosh(s)&0&0&\sinh(s)\cos(d)&\sinh(s)\sin(d)\\\
0&\cosh(s)&0&\sinh(s)\sin(d)&-\sinh(s)\cos(d)\\\ 0&0&1&0&0\\\
\sinh(s)\cos(d)&\sinh(s)\sin(d)&0&\cosh(s)&0\\\
\sinh(s)\sin(d)&-\sinh(s)\cos(d)&0&0&\cosh(s)\end{pmatrix}\,,$ (5.2)

and let $A=\tilde{A}H$, $\tilde{v}=Hv$. This transformation preserves the
spatial constraints as well as the determinant constants we fixed earlier. If
$d$ and $s$ are chosen as above, then $\tilde{c}_{14}=\tilde{c}_{15}=0$.
$\square$

######  Theorem 5.2

The solution to the time component is

$\theta^{\prime}=c_{12}/b_{11}^{2}=-w_{3}\,,\,w_{1}=w_{2}=0\,,\,b_{12}=b_{13}=b_{23}=0\,,\,b_{22}=b_{11}\,.$

Note that $c_{12}$ must be nonzero.

Proof.  Solving for $w_{j}$ and $\theta^{\prime}$, and using the previous
Lemma, we note that $b_{12}=0$ which then readily yields that $b_{22}=b_{11}$
and

$b_{13}=\big{(}k_{0}+k_{2}\cos(\theta)+k_{3}\sin(\theta)\big{)}b_{11}\,,\,b_{23}=\big{(}k_{1}-k_{3}\cos(\theta)+k_{2}\sin(\theta)\big{)}b_{11}\,.$

By Lemma 3.4 we can add a multiple of any $A_{j}$ to $A_{3}$ without loss of
generality and thus achieve $b_{13}=b_{23}=0$. Substituting these to the
formulas for $w_{j}$ and $\theta^{\prime}$ gives the rest. $\square$

The corresponding vorticity is then

$h=c_{12}\begin{pmatrix}-f^{1}_{100}f^{1}_{001}-f^{1}_{010}f^{2}_{001}\\\\[2.84526pt]
f^{1}_{100}f^{2}_{001}-f^{1}_{010}f^{1}_{001}\\\\[2.84526pt]
(f^{1}_{100})^{2}+(f^{1}_{010})^{2}+1\end{pmatrix}\,.$

###  5.2 Hyperbolic case

Here we consider the spatial constraints

$g_{134}=g_{235}=0\ ,$

which easily lead to
$v=\big{(}z_{1},z_{2},z_{3},f^{1}(z_{1},z_{3}),f^{2}(z_{2},z_{3})\big{)}$.
Then for the time dependent part each $Q_{ij}$ and the following minors have
to be constant:

$p_{123},p_{124},p_{125},p_{135},p_{145},p_{234},p_{245},p_{345}\,.$

Similarly to the previous case, we must have
$p_{124}=p_{125}=p_{145}=p_{245}=0$ which again implies that
$b_{34}=b_{35}=0$. For the rest of the determinant constants we can assume
that $p_{123}=1$, $p_{345}=-1$ and $p_{135}=p_{234}=0$. The proof is again
essentially the same as that of the corresponding 2D case [15, Lemma 5.6].
From this we obtain

$\det(d\varphi)=1-f^{1}_{100}f^{2}_{010}\neq 0\ .$

Then putting $b_{15}=\ell b_{11}$ we can now write

$B=\begin{pmatrix}b_{11}&b_{12}&b_{13}&b_{12}/\ell&\ell b_{11}\\\
0&b_{22}&b_{23}&b_{22}/\ell&0\\\ 0&0&1/(b_{11}b_{22})&0&0\end{pmatrix}\ .$

This gives

######  Theorem 5.3

We have $w=0$ and

$\ell^{\prime}=-1/b_{11}^{2}\,,\,b_{22}=\ell
b_{11}\,,\,b_{12}=b_{13}=b_{23}=0\,.$

Proof.  We may assume $c_{15}=1$ by scaling. From the equation $Q_{15}=1$ one
immediately gets the equation for $\ell$ and otherwise we solve $w_{j}$ from
equations $Q_{12}=c_{12}$, $Q_{13}=c_{13}$ and $Q_{23}=c_{23}$. Substituting
everything back we note that

$b_{12}=k_{1}\ell b_{11}\,,\,b_{22}=k_{2}\ell
b_{11}\,,\,b_{13}=(k_{3}\ell+k_{4})b_{11}\,,\,b_{23}=(k_{5}\ell+k_{6})b_{11}\,.$

Without loss of generality we may add a multiple of $A_{5}$ to $A_{2}$ and a
multiple of $A_{1}$ to $A_{4}$ in order to obtain $b_{12}=0$, after which we
add a suitable linear combination of the columns of $A$ to $A_{3}$ to obtain
$b_{13}=b_{23}=0$. By scaling we may also assume $k_{2}=1$. Finally we note
that this form of solutions forces $w=0$. $\square$

In this case the vorticity can be written as

$h=\Big{(}-f^{1}_{001}\,,\,-f^{2}_{001}\,,\,f^{1}_{100}+f^{2}_{010}\Big{)}\ .$

###  5.3 Parabolic case

Here we suppose that

$g_{134}+g_{235}=g_{135}=0\ ,$

which implies that $v$ is of the form

$v=\big{(}z,f^{1}+z_{2}f^{2}_{100},f^{2}\big{)}$

and $f^{j}=f^{j}(z_{1},z_{3})$. Then for the time component

$p_{123},p_{124},p_{125},p_{145},p_{234},p_{245},p_{345},p_{235}-p_{134}$

and each $Q_{ij}$ are constant. Hence by Lemma 2.1 and Lemma 3.5 we have
$e_{124}=e_{125}=e_{145}=e_{245}=0$. Again, the reduction of the other
determinant constants goes as in the corresponding 2D case so by [15, Lemma
5.8] we may assume that $e_{234}=1$ and $e_{123}=e_{345}=p_{235}-p_{134}=0$.
From this it follows that

$\det(d\varphi)=f_{100}^{1}+z_{2}f_{200}^{2}\neq 0\ .$

Thus we have

$A_{1}=\ell A_{2}\ \mathrm{and}\ A_{5}=\ell A_{4}\ ,$

where $\ell=p_{134}$. Now we have the $B$ matrix in the QR decomposition in
the following form

$B=\begin{pmatrix}\ell b_{12}&b_{12}&b_{13}&b_{14}&\ell b_{14}\\\
0&0&b_{23}&b_{24}&\ell b_{24}\\\
0&0&0&1/b_{12}b_{23}&\ell/b_{12}b_{23}\end{pmatrix}\ .$

Substituting to the equations $Q_{ij}=c_{ij}$ we see that $c_{12}$ and
$c_{45}$ must be nonzero and that $c_{15}=c_{24}=0$ and $c_{25}=-c_{14}$.
Moreover we have

######  Lemma 5.4

We may assume that $c_{14}=c_{13}=c_{23}=c_{34}=c_{35}=0$.

Proof.  Use the transformation $\tilde{A}_{2}=A_{2}+c_{14}/c_{45}A_{4}$,
$\tilde{A}_{1}=A_{1}+c_{14}/c_{45}A_{5}$ to obtain $c_{14}=0$ and thus
$c_{25}=0$. After that, write
$\tilde{A}_{3}=A_{3}+c_{23}/c_{12}A_{1}-c_{13}/c_{12}A_{2}-c_{35}/c_{45}A_{4}+c_{34}/c_{45}A_{5}$.
Since $c_{14}=c_{15}=c_{24}=c_{25}=0$, this transformation gives
$c_{13}=c_{23}=c_{34}=c_{35}=0$. $\square$

######  Theorem 5.5

The solution to the time constraints is

$A=\begin{pmatrix}\ell b_{12}&b_{12}&0&0&0\\\ 0&0&b_{12}^{-2}&0&0\\\
0&0&0&b_{12}&\ell b_{12}\end{pmatrix}\,,$

where $\ell^{\prime}=c_{12}/b_{12}^{2}$.

Proof.  Since so many constants are zero one immediately computes that
$\ell^{\prime}=c_{12}/b_{12}^{2}$ and $b_{13}=b_{14}=w_{2}=w_{3}=0$. Using
these we obtain that also $w_{1}=b_{24}=0$ and then we are left with the
equation $c_{45}b_{23}^{2}b_{12}^{4}+c_{12}=0$. Scaling gives then the
solution above. $\square$

The vorticity is now

$h=c_{12}\begin{pmatrix}-f^{2}_{100}f^{2}_{001}\\\\[2.84526pt]
f^{1}_{100}f^{2}_{001}-f^{2}_{100}f^{1}_{001}+z_{2}\big{(}f^{2}_{200}f^{2}_{001}-f^{2}_{100}f^{2}_{101}\big{)}\\\\[2.84526pt]
\left(f^{2}_{100}\right)^{2}+1\end{pmatrix}\ .$

##  6 $m=6$: hyperbolic case

In these final three sections we study cases where $m=6$. When $m=6$ the
computations really become much more involved than for smaller $m$. Also,
coming up with spatial constraints that yield solutions to both the spatial
and the time component is difficult. We managed to find three sets of spatial
constraints that provide solutions, which are again analogous to the three
cases we considered in [15, sec. 4] and to the ones considered in the previous
section. But then we also found that by restricting the time component of
these solutions we obtain more general solutions to the spatial component
which we could not have come up with otherwise.

In this section we consider the spatial constraints

$G_{14}=G_{25}=G_{36}=0\,.$

The solution of the spatial component is

$v=(z_{1},z_{2},z_{3},f_{1}(z_{1}),f_{2}(z_{2}),f_{3}(z_{3}))\,.$

Incidentally one can check that with this type of $v$ one can find nontrivial
solutions in any dimension.

The conditions for the time dependent part give that the following functions
are constant:

$\displaystyle
p_{123},p_{126},p_{135},p_{156},p_{234},p_{246},p_{345},p_{456},$
$\displaystyle
Q_{12},Q_{13},Q_{15},Q_{16},Q_{23},Q_{24},Q_{26},Q_{34},Q_{35},Q_{45},Q_{46},Q_{56}\,.$

Despite the fact that there are 20 equations to be satisfied by 18 functions,
there are nontrivial solutions for the time component; the simplest example is

$A=\begin{pmatrix}e^{c_{1}t}&0&0&0&0&e^{-c_{1}t}\\\
0&e^{c_{2}t}&0&e^{-c_{2}t}&0&0\\\
0&0&e^{-(c_{1}+c_{2})t}&0&e^{(c_{1}+c_{2})t}&0\end{pmatrix}\,,$

where $c_{j}$ are constants.

######  Lemma 6.1

We may assume that

  1. 1.

$e_{126}=e_{135}=e_{156}=e_{234}=e_{246}=e_{345}=0$ , or

  2. 2.

$e_{126}=e_{135}=e_{234}=e_{246}=e_{345}=e_{456}=0$ .

Proof.  Due to symmetry we may assume that $e_{123}\neq 0$. Add a multiple of
$A_{1}$ to $A_{4}$ to obtain $e_{234}=0$. Now, if also
$e_{246}=e_{345}=e_{456}=0$, we obtain case (2) by further adding a multiple
of $A_{2}$ to $A_{5}$ and a multiple of $A_{3}$ to $A_{6}$.

If $e_{246}$, $e_{345}$, and $e_{456}$ are not all zero, we may assume that
$e_{456}\neq 0$ by adding a multiple of $A_{2}$ to $A_{5}$ or a multiple of
$A_{3}$ to $A_{6}$ if necessary. In this case we show that we may assume
$e_{135}=e_{345}=e_{126}=e_{246}=0$. Apply the transformation $\tilde{A}=AH$,
where

$H=\begin{pmatrix}1&0&0&0&0&0\\\ 0&1&0&0&c_{1}&0\\\ 0&0&1&0&0&c_{2}\\\
0&0&0&1&0&0\\\ 0&c_{3}&0&0&1&0\\\ 0&0&c_{4}&0&0&1\end{pmatrix}\,.$

Writing

$\hat{H}=\begin{pmatrix}e_{123}&e_{156}\\\ e_{234}&e_{456}\end{pmatrix}\,,$

we have

$\begin{pmatrix}-\tilde{e}_{135}&\tilde{e}_{126}\\\
\tilde{e}_{345}&-\tilde{e}_{246}\end{pmatrix}=\begin{pmatrix}-e_{135}&e_{126}\\\
e_{345}&-e_{246}\end{pmatrix}+\hat{H}\begin{pmatrix}c_{1}&c_{2}\\\
c_{4}&c_{3}\end{pmatrix}\,.$

Since $\det(\hat{H})\neq 0$, we may choose the constants $c_{j}$ such that
$\tilde{e}_{135}=\tilde{e}_{126}=\tilde{e}_{345}=\tilde{e}_{246}=0$.

Finally, we may replace $A_{1}$ and $A_{4}$ by suitable linear combinations of
$A_{1}$ and $A_{4}$ to obtain $e_{234}=e_{156}=0$, yielding case (1).
$\square$

######  Lemma 6.2

In case (2) of Lemma 6.1 the solutions of $A$ reduce to a case of lower $m$.

Proof.  We may assume that $e_{123}=e_{156}=1$. To satisfy the determinant
conditions we need to have

$\displaystyle A_{4}$ $\displaystyle=\ell_{1}A_{2}+\ell_{2}A_{3}$
$\displaystyle A_{5}$ $\displaystyle=\ell_{3}A_{3}\quad,\quad
A_{6}=(1/\ell_{3})A_{2}\,.$

We may assume $c_{23}=c_{56}=0$. Then the computations show that

$A_{4}=\frac{c_{46}}{c_{26}}A_{2}+\frac{c_{45}}{c_{35}}A_{3}+\frac{c_{34}}{c_{35}}A_{5}+\frac{c_{24}}{c_{26}}A_{6}\,.$

Since $c_{26}$ and $c_{35}$ have to be nonzero, the columns $A_{2}$, $A_{3}$,
$A_{4}$, $A_{5}$, and $A_{6}$ are always linearly dependent and the statement
follows from Lemma 3.5. $\square$

In case (1) of Lemma 6.1 we obtain two families of solutions. To describe the
computations in detail is not really possible so that we merely present the
main ideas which lead to the solutions. However, it is easy to verify that the
claimed solutions are actually solutions since one can simply substitute
everything to the relevant equations and check that the conditions are
satisfied. Of course the computations by hand would be extremely tedious, but
with a convenient computer algebra system the verification is easy.

From the conditions of the case (1) it readily follows that there are
functions $\ell_{j}$ such that we can write either

$\displaystyle A_{6}=$ $\displaystyle\ell_{1}A_{1},\quad
A_{4}=\ell_{2}A_{2},\quad A_{5}=1/(\ell_{1}\ell_{2})A_{3}\quad\mathrm{or}$
$\displaystyle A_{5}=$ $\displaystyle\ell_{1}A_{1},\quad
A_{6}=\ell_{2}A_{2},\quad A_{4}=1/(\ell_{1}\ell_{2})A_{3}\,.$

By symmetry we may choose the first option. We can thus write $A=RB$ where

$B=\begin{pmatrix}b_{11}&b_{12}&b_{13}&\ell_{2}b_{12}&b_{13}/\ell_{1}\ell_{2}&\ell_{1}b_{11}\\\
0&b_{22}&b_{23}&\ell_{2}b_{22}&b_{23}/\ell_{1}\ell_{2}&0\\\
0&0&1/b_{11}b_{22}&0&1/\ell_{1}\ell_{2}b_{11}b_{22}&0\end{pmatrix}\ .$

and

$\det(d\varphi)=1+f^{1}_{100}f^{2}_{010}f^{3}_{001}\neq 0\ .$

Then the computations show that in all cases we must have $w=0$ so there is no
rotational component in the solutions and $A=B$.

Then we always have $\ell_{1}^{\prime}=-c_{16}/b_{11}^{2}$ and in particular
$c_{16}\neq 0$. Using this and other equations we notice that

$(\hat{m}_{2}\ell_{1}^{2}+\hat{m}_{1}\ell_{1}+\hat{m}_{0})\ell_{2}+\hat{k}_{1}\ell_{1}+\hat{k}_{0}=0$

for some constants $\hat{k}_{j}$ and $\hat{m}_{j}$. Now we have two cases:

  * (i)

it is possible to choose constants $c_{ij}$ such that
$\hat{k}_{j}=\hat{m}_{j}=0$; in this way we avoid a direct algebraic relation
between $\ell_{1}$ and $\ell_{2}$.

  * (ii)

if there is a direct algebraic relation then it turns out that the constants
must be chosen so that we can write

$\ell_{2}=\frac{1}{m_{1}\ell_{1}+m_{0}}\ .$

######  Theorem 6.3

The case (i) leads to the solution

$\displaystyle\ell_{1}^{\prime}=-\frac{c_{16}}{b_{11}^{2}}\quad,\quad\ell_{2}^{\prime}=-\frac{c_{24}}{b_{22}^{2}}$
$\displaystyle
c_{35}\ell_{1}^{2}\ell_{2}^{2}b_{11}^{4}b_{22}^{4}+c_{16}b_{22}^{2}\ell_{2}+c_{24}b_{11}^{2}\ell_{1}=0$
$\displaystyle b_{12}=b_{13}=b_{23}=0\ .$

Hence one function, for example $b_{11}$, can be freely chosen.

Proof.  Avoiding algebraic relation between $\ell_{1}$ and $\ell_{2}$ forces
all constants $c_{ij}$ to be zero except $c_{16}$, $c_{24}$ and $c_{35}$.
After this the solution given above follows. $\square$

Note that if we write $\ell_{j}=\exp(\lambda_{j})$ then we obtain

$\lambda_{1}^{\prime}\lambda_{2}^{\prime}\big{(}\lambda_{1}^{\prime}+\lambda_{2}^{\prime}\big{)}=c_{16}c_{24}c_{35}\
.$

Here the vorticity is

$h=-\Big{(}c_{35}f_{2}^{\prime}\,,\,c_{16}f_{3}^{\prime}\,,\,c_{24}f_{1}^{\prime}\Big{)}\,.$

######  Theorem 6.4

The case (ii) gives the following solution:

$\displaystyle(\ell_{1}^{\prime})^{3}=$ $\displaystyle
k_{0}\,\ell_{1}^{2}(m_{1}\ell_{1}+m_{0})^{2}\quad,\quad\ell_{2}=\frac{1}{m_{1}\ell_{1}+m_{0}}$
$\displaystyle b_{11}^{3}=$
$\displaystyle\frac{k_{1}}{\ell_{1}(m_{1}\ell_{1}+m_{0})}\quad,\quad
b_{22}=k_{2}b_{11}(m_{1}\ell_{1}+m_{0})$ $\displaystyle b_{12}=$
$\displaystyle k_{3}b_{11}(m_{1}\ell_{1}+m_{0})\quad,\quad
b_{13}=k_{4}b_{11}\ell_{1}\quad,\quad b_{23}=k_{5}b_{11}\ell_{1}\ .$

Proof.  Starting from $\ell_{1}^{\prime}=-c_{16}/b_{11}^{2}$ and
$\ell_{2}=1/(m_{1}\ell_{1}+m_{0})$ we find that
$b_{22}=k_{2}b_{11}(m_{1}\ell_{1}+m_{0})$ for some $k_{2}$. After this the
rest follows easily. $\square$

Evidently $k_{0}$, $k_{1}$, $k_{2}$, $m_{0}$ and $m_{1}$ are nonzero. The
constants $k_{3}$, $k_{4}$ and $k_{5}$ are arbitrary, except that choosing
$k_{3}=k_{4}=k_{5}=0$ is incompatible with the hypothesis
$\ell_{2}=1/(m_{1}\ell_{1}+m_{0})$. Apparently there is no explicit solution
to the differential equation for $\ell_{1}$. However, there is the following
discrete symmetry: if $\mu$ is a solution then
$\hat{\mu}(t)=m_{0}^{2}/(m_{1}^{2}\mu(-t))$ is also a solution.

In this case the vorticity has the form

$h=\begin{pmatrix}e_{0}+e_{1}f_{2}^{\prime}+e_{2}f_{3}^{\prime}+e_{3}f_{2}^{\prime}f_{3}^{\prime}\\\
e_{4}+e_{5}f_{1}^{\prime}+e_{6}f_{3}^{\prime}+e_{7}f_{1}^{\prime}f_{3}^{\prime}\\\
e_{8}+e_{9}f_{1}^{\prime}+e_{10}f_{2}^{\prime}+e_{11}f_{1}^{\prime}f_{2}^{\prime}\end{pmatrix}\
.$

where $e_{j}$ are some constants.

###  6.1 Extensions for the spatial component

Let us assume that $A$ is of the form

$A=\begin{pmatrix}b_{11}&0&0&0&0&b_{16}\\\ 0&b_{22}&0&b_{24}&0&0\\\
0&0&1/b_{11}b_{22}&0&1/b_{16}b_{24}&0\end{pmatrix}\,,$

where the functions $b_{ij}$ are like in Theorem 6.3. As we have seen in this
case the spatial part is of the form
$v=\big{(}z_{1},z_{2},z_{3},f^{1}(z_{1}),f^{2}(z_{2}),f^{3}(z_{3})\big{)}$.
However, if we a priori put some restrictions on $b_{ij}$ it is possible to
have more general solutions in the spatial part. There are several
possibilities to choose these restrictions, for example:

$\displaystyle\begin{cases}b_{11}=b_{22}\\\
b_{16}=b_{24}\end{cases}\quad\mathrm{or}\quad\frac{b_{11}}{b_{16}}+\frac{b_{16}}{b_{11}}+\frac{b_{22}}{b_{24}}=0$
$\displaystyle\mathrm{or}\quad\frac{b_{16}}{b_{11}}+\frac{b_{24}}{b_{22}}+\frac{b_{16}b_{24}}{b_{11}b_{22}}=0\,.$

All these choices lead to nontrivial solutions but we only consider the first
case as an example to give a flavor of the idea.

Now if $b_{11}=b_{22}$ and $b_{16}=b_{24}$ then it is easy to compute that
actually $b_{11}=e^{ct}$ and $b_{16}=e^{-ct}$. It follows that each $Q_{ij}$
is constant so $h$ is automatically independent of time, and for the
determinant condition the spatial part only needs to satisfy

$g_{145}+g_{256}=g_{236}-g_{134}=g_{125}=g_{346}=0\,.$

If $v=(z_{1},z_{2},z_{3},f^{1},f^{2},f^{3})$, then we obtain the following
PDE:

$\displaystyle f^{2}_{001}=f^{3}_{100}+f^{1}_{010}=0$ $\displaystyle
f^{1}_{100}f^{3}_{010}-f^{1}_{010}f^{3}_{100}=0$ $\displaystyle
f^{2}_{100}f^{3}_{001}-f^{2}_{010}f^{1}_{001}=0\,.$

For simplicity let us assume that

$\displaystyle f^{1}$ $\displaystyle=g^{1}(z_{1},z_{2})q(z_{3})$
$\displaystyle f^{2}$ $\displaystyle=g^{2}(z_{1},z_{2})$ $\displaystyle f^{3}$
$\displaystyle=g^{3}(z_{1},z_{2})q(z_{3})\ .$

Substituting this we obtain

$\displaystyle g^{1}_{10}g^{3}_{01}-g^{1}_{01}g^{3}_{10}=0$ $\displaystyle
g^{1}_{01}+g^{3}_{10}=0$ $\displaystyle g^{1}g^{2}_{01}+g^{3}g^{2}_{10}=0\ .$

We can ”solve” this as follows. From the first equation we have $\nabla
g^{3}=g\,\nabla g^{1}$ for some $g$. But then the second equation implies that
$g^{1}_{01}+g\,g^{1}_{10}=0$. But clearly we must in addition require
$g_{01}+g\,g_{10}=0$ which leads to the following triangular system:

$\displaystyle g_{01}+g\,g_{10}=0\quad,\quad g^{1}_{01}+g\,g^{1}_{10}=0$
$\displaystyle\nabla g^{3}=g\,\nabla g^{1}\quad,\quad
g^{1}g^{2}_{01}+g^{3}g^{2}_{10}=0\,.$

Existence of local solutions now follows from standard theorems. For example
if $g=(z_{1}-c_{1})/(z_{2}-c_{2})$ then

$g^{1}=q_{1}\Big{(}\frac{z_{2}-c_{2}}{z_{1}-c_{1}}\Big{)}\quad\mathrm{and}\quad
g^{3}=q_{2}\Big{(}\frac{z_{2}-c_{2}}{z_{1}-c_{1}}\Big{)}$

where $q_{2}^{\prime}(s)=q_{1}^{\prime}(s)/s$.

##  7 $m=6$: elliptic case

Let the spatial constraints be

$G_{34}=G_{15}+G_{26}=G_{16}-G_{25}=0\,.$

Doing the computations we find that $v$ is of the form

$v=(z_{1},z_{2},z_{3},f^{1}(z_{3}),f^{2}(z_{1},z_{2}),f^{3}(z_{1},z_{2}))\,,$

where $f^{1}$ is arbitrary, and $f^{2}$ and $f^{3}$ are anti-CR.

The simplest solution set to the time constraints is

$A=\begin{pmatrix}\cos(\theta_{0}t)&-\sin(\theta_{0}t)&\cos(\theta_{0}t)&\sin(\theta_{0}t)&0&0\\\
\sin(\theta_{0}t)&\cos(\theta_{0}t)&-\sin(\theta_{0}t)&\cos(\theta_{0}t)&0&0\\\
0&0&0&0&\cos(2\theta_{0}t)&\sin(2\theta_{0}t)\end{pmatrix}\ ,$ (7.1)

where $\theta_{0}\neq 0$ is a constant.

The determinant condition gives the following equations:

$\displaystyle p_{123}=e_{123}\ ,\
p_{124}=e_{124},p_{356}=e_{356},p_{456}=e_{456},$ $\displaystyle
p_{135}-p_{236}=e_{0},p_{136}+p_{235}=e_{1},p_{145}-p_{246}=e_{2},p_{146}+p_{245}=e_{3}\,.$

Let us first reduce as many constants to zero as possible.

######  Lemma 7.1

The determinant constants can be reduced to one of the following three cases:

  1. 1.

$p_{135}-p_{236}=p_{136}+p_{235}=p_{145}-p_{246}=p_{146}+p_{245}=0$.

  2. 2.

$p_{356}=p_{456}=p_{136}+p_{235}=p_{146}+p_{245}=0$.

  3. 3.

$p_{123}=p_{356}=p_{124}=p_{456}=0$.

In the proof of this Lemma we in fact use complex number formulation for
conciseness.

Proof.  Let $H$ be of the form

$H=\begin{pmatrix}h_{11}&-h_{21}&0&0&h_{15}&h_{25}\\\
h_{21}&h_{11}&0&0&h_{25}&-h_{15}\\\ 0&0&1&0&0&0\\\ 0&0&0&1&0&0\\\
h_{51}&h_{61}&0&0&h_{55}&-h_{65}\\\
h_{61}&-h_{51}&0&0&h_{65}&h_{55}\end{pmatrix}\,,$

and $\tilde{A}=AH$, $v=H\tilde{v}$. $H$ preserves the spatial constraints. Let
$h_{1}=h_{11}+ih_{21}$, $h_{2}=h_{51}+ih_{61}$, $h_{3}=h_{15}+ih_{25}$,
$h_{4}=h_{55}+ih_{65}$.

If $p_{123}=p_{356}=p_{124}=p_{456}=0$, we have case (3). Otherwise we may
assume that $p_{123}=1$. Now choosing $h_{1}=h_{4}=1$, $h_{2}=0$,
$2h_{3}=(p_{135}-p_{236})+i(p_{136}+p_{235})$, we may assume that
$p_{135}-p_{236}=p_{136}+p_{235}=0$. Then, since

$p_{123}p_{356}-p_{135}p_{236}+p_{136}p_{235}=p_{356}-p_{135}^{2}-p_{136}^{2}=0\,,$

we have $p_{356}\geq 0$. By scaling $A_{5}$ and $A_{6}$ we may assume that
$p_{356}=1$ or $p_{356}=0$. By choosing $h_{1}=1$, $h_{2}=h_{3}=0$ and letting
$h_{4}=e^{i\theta}$, where $\theta$ is suitable, we achieve
$p_{146}+p_{245}=0$.

If $p_{356}=0$, then also $p_{135}=p_{136}=p_{235}=p_{236}=0$. It follows that
$A_{5}\times A_{6}=0$, so $p_{456}=0$. Now the only nonzero constants are
$p_{123}$, $p_{124}$, and $p_{145}-p_{246}$, so we have reached case (2).

Then assume that $p_{356}=1$. By transformation
$\tilde{A}_{4}=A_{4}-p_{456}A_{3}$ we may assume that $p_{456}=0$. At this
point the nonzero constants are $p_{123}=p_{356}=p_{145}-p_{246}=1$ and
$p_{124}$. If $|p_{124}|>1$, let $k=p_{124}+\sqrt{p_{124}^{2}-1}$ and choose
$h_{1}=1$, $h_{2}=ik$, $h_{3}=k$, $h_{4}=i$. After this we have an instance of
case (1). If $|p_{124}|=1$, let $h_{1}=1$, $h_{2}=0$, $h_{3}=i$, $h_{4}=1$,
and we obtain case (2). Finally, if $|p_{124}|<1$, let $\sin{\mu}=p_{124}$.
Then let $h_{1}=1$, $h_{2}=e^{i\mu}$, $h_{3}=e^{i(\pi-\mu)}$, $h_{4}=1$. Then
$\tilde{A}$ belongs to case (3). For each transformation it can be checked
that $\det(H)\neq 0$. $\square$

Let us inspect the first case.

######  Lemma 7.2

In case (1) of Lemma 7.1 all solutions for $A$ reduce.

Proof.  We may assume that $p_{123}=1$ and $p_{124}=0$. Computations show that
we must have $p_{456}=0$ and $p_{356}\neq 0$, and we may assume by scaling
that $p_{356}=1$. Then we can write

$\displaystyle A_{4}$ $\displaystyle=\ell_{1}A_{1}+\ell_{2}A_{2}$
$\displaystyle A_{5}$ $\displaystyle=\cos{\theta}A_{1}-\sin{\theta}A_{2}$
$\displaystyle A_{6}$ $\displaystyle=\sin{\theta}A_{1}+\cos{\theta}A_{2}\,.$

Let us write $Q_{15}-Q_{26}=c_{1}$, $Q_{16}+Q_{25}=c_{2}$. Similarly to Lemma
5.1, we have $(c_{56}-c_{12})^{2}-c_{1}^{2}-c_{2}^{2}>0$ so we can write

$\displaystyle c_{2}$ $\displaystyle=(c_{12}-c_{56})\cos(d)\tanh(2s)$
$\displaystyle c_{1}$ $\displaystyle=(c_{56}-c_{12})\sin(d)\tanh(2s)\,.$

Thus using

$H=\begin{pmatrix}H_{1}&0&H_{2}\\\ 0&I&0\\\ H_{3}&0&H_{4}\end{pmatrix}\ ,$

where the blocks $H_{j}$ are as in the corners of (5.2), we can bring $c_{1}$
and $c_{2}$ to zero. But then we find that

$c_{24}A_{1}-c_{14}A_{2}+((c_{12}-c_{56})/2)A_{4}-c_{46}A_{5}+c_{45}A_{6}=0\,.$

Here $c_{12}-c_{56}=2\theta^{\prime}b_{11}$ must be nonzero so the columns of
$A$ are linearly dependent. Thus the solutions reduce. $\square$

The solutions of the second case also reduce.

######  Lemma 7.3

In case (2) of Lemma 7.1 all solutions for $A$ reduce to a case of lower $m$.

Proof.  Again, we may assume $p_{123}=1$, $p_{124}=0$. If $p_{145}-p_{246}=0$,
we can reduce the situation to case (1). Thus we may assume that
$p_{145}-p_{246}\neq 0$, and, after adding a multiple of $A_{4}$ to $A_{3}$,
that $p_{135}-p_{236}=0$. With these determinant constants we may write

$A_{4}=p_{234}A_{1}-p_{134}A_{2},\quad A_{5}=p_{125}A_{3},\quad
A_{6}=p_{126}A_{3}\,.$

The conditions for $Q_{35}$ and $Q_{36}$ yield

$c_{1}A_{3}+c_{36}A_{5}-c_{35}A_{6}=0$

for some constant $c_{1}$. Here $c_{35}$ and $c_{36}$ cannot be zero, so this
proves there are no irreducible solutions. $\square$

The third case of Lemma 7.1 does provide solutions. By replacing $A_{3}$ and
$A_{4}$ by suitable linear combinations of $A_{3}$ and $A_{4}$, we can further
without loss of generality assume that

$p_{135}-p_{236}=p_{146}+p_{245}=0,\quad
p_{136}+p_{235}=-p_{145}+p_{246}=1\,,$

and this leads to

$\det(d\varphi)=f^{2}_{100}+f^{2}_{010}f^{1}_{001}\neq 0\ .$

From this it follows that either $A_{1}$ and $A_{2}$ are parallel or $A_{5}$
and $A_{6}$ are parallel. Choosing the latter option leads to

$B=\begin{pmatrix}b_{11}&b_{12}&\ell_{1}b_{11}+\ell_{2}b_{12}&-\ell_{2}b_{11}+\ell_{1}b_{12}&\ell_{1}b_{15}&-\ell_{2}b_{15}\\\
0&b_{22}&\ell_{2}b_{22}&\ell_{1}b_{22}&\ell_{1}b_{25}&-\ell_{2}b_{25}\\\
0&0&0&0&\ell_{1}b_{35}&-\ell_{2}b_{35}\end{pmatrix}$

and in addition

$b_{11}b_{22}b_{35}\big{(}\ell_{1}^{2}+\ell_{2}^{2}\big{)}+1=0\ .$

Then we have

######  Lemma 7.4

There is a function $\theta$ and constants $k_{j}$, $\gamma$ such that

$\displaystyle\ell_{1}=k_{1}+k_{0}\cos(\theta),\quad\ell_{2}=k_{2}+k_{0}\sin(\theta)$
$\displaystyle\theta^{\prime}b_{11}b_{22}=\gamma\ .$

Proof.  Let $r_{ij}=\langle B_{i},B_{j}\rangle$ as before. Four of the
constraints can be written as follows:

$\displaystyle Q_{13}$
$\displaystyle=-r_{11}\ell_{1}^{\prime}-r_{12}\ell_{2}^{\prime}+c_{12}\ell_{2}=c_{13}$
$\displaystyle Q_{14}$
$\displaystyle=r_{11}\ell_{2}^{\prime}-r_{12}\ell_{1}^{\prime}+c_{12}\ell_{1}=c_{14}$
$\displaystyle Q_{23}$
$\displaystyle=-r_{12}\ell_{1}^{\prime}-r_{22}\ell_{2}^{\prime}-c_{12}\ell_{1}=c_{23}$
$\displaystyle Q_{24}$
$\displaystyle=r_{12}\ell_{2}^{\prime}-r_{22}\ell_{1}^{\prime}+c_{12}\ell_{2}=c_{24}\
.$

Solving $r_{ij}$ from the first three equations and substituting we obtain
from the fourth that $(\ell_{1},\ell_{2})$ traces a circle. This gives the
first formula and we have also $c_{23}=-2c_{12}k_{1}+c_{14}$ and
$c_{24}=2c_{12}k_{2}-c_{13}$. Using these $\ell_{j}$ we can compute the
expressions for $r_{ij}$. Since $b_{11}^{2}b_{22}^{2}=r_{11}r_{22}-r_{12}^{2}$
we get the second formula where

$\gamma^{2}=c_{12}^{2}k_{0}^{2}-\big{(}c_{12}k_{1}-c_{14}\big{)}^{2}-\big{(}c_{12}k_{2}-c_{13}\big{)}^{2}\
.$

$\square$

Hence to have nontrivial solutions we must have $k_{0}\neq 0$, $c_{12}\neq 0$
and $\gamma\neq 0$. Without loss of generality we can also assume that
$k_{1}=k\geq 0$, $k_{2}=0$ and $k_{0}=1$. It turns out that we have two
families of solutions.

######  Lemma 7.5

We have $b_{15}=b_{25}=0$, if $k\neq 1$. If $k=1$ then

$\displaystyle b_{15}=$
$\displaystyle\frac{c_{36}\sin\\!\left(\theta\right)-c_{46}(1-\cos\\!\left(\theta\right))}{2\gamma\sin\\!\left(\theta\right)}\,b_{22}$
$\displaystyle b_{25}=$
$\displaystyle\frac{m_{1}\sin\\!\left(\theta\right)+m_{0}\left(1+\cos\\!\left(\theta\right)\right)}{\gamma^{2}\left(1+\cos\\!\left(\theta\right)\right)}\,b_{22}\
,$

where the constants $m_{j}$ satisfy

$\left(c_{36}^{2}+c_{46}^{2}\right)\gamma^{2}+4c_{12}\big{(}c_{46}m_{0}+c_{36}m_{1}\big{)}+4m_{0}^{2}+4m_{1}^{2}=0\
.$

Hence both $b_{15}$ and $b_{25}$ must be nonzero, and also
$c_{46}m_{0}+c_{36}m_{1}\neq 0$.

Proof.  Substituting what we found already to the equations we notice that
there are four equations which can be written as

$M\begin{pmatrix}b_{11}\\\ b_{15}\\\ b_{25}\end{pmatrix}=0\ .$

Here the elements of $M$ are polynomials in $\cos(\theta)$, $\sin(\theta)$ and
various constants. To have nontrivial solutions all $3\times 3$ minors of $M$
have to be zero. This is possible; a necessary condition in all cases is
$c_{35}=c_{45}=0$. But when $k\neq 1$ then the first column must be zero and
the equations do not contain $b_{11}$. However, the second and third columns
are linearly independent for all acceptable choices of constants which implies
that $b_{15}=b_{25}=0$.

If $k=1$ then the solution is obtained by simple computation. $\square$

######  Theorem 7.6

The solution can be written as follows when $k\neq 1$:

$\displaystyle(\theta^{\prime})^{3}=$
$\displaystyle\frac{\gamma^{2}c_{56}\big{(}k^{2}+2k\cos(\theta)+1\big{)}^{2}}{k\cos(\theta)+1}$
$\displaystyle b_{35}=$
$\displaystyle-\frac{\theta^{\prime}}{\gamma\left(k^{2}+2k\cos\\!\left(\theta\right)+1\right)}$
$\displaystyle b_{11}^{2}=$
$\displaystyle\frac{\left(c_{14}-c_{12}k\right)\cos\\!\left(\theta\right)+c_{13}\sin\\!\left(\theta\right)-c_{12}}{\theta^{\prime}}$
$\displaystyle b_{12}=$
$\displaystyle\frac{\left(c_{14}-c_{12}k\right)\sin\\!\left(\theta\right)-c_{13}\cos\\!\left(\theta\right)}{b_{11}\theta^{\prime}}$
$\displaystyle b_{22}=$
$\displaystyle\frac{\gamma}{b_{11}\theta^{\prime}}\quad,\quad
w_{1}=w_{2}=0\quad,\quad w_{3}=-\frac{\gamma}{b_{11}^{2}}\ .$

Note that to have global solutions we must assume that $k<1$ and that
$|c_{12}|$ is sufficiently big.

Proof.  Substituting $b_{15}=b_{25}=0$ to the original equations we see that
also $c_{36}=c_{46}=0$. This implies that $w_{1}=w_{2}=0$ and we also get the
formula for $b_{12}$ as above. For $w_{3}$ we first get

$w_{3}=\frac{b_{11}b_{12}^{\prime}-b_{11}^{\prime}b_{12}+c_{12}}{b_{11}b_{22}}\
.$

This can be computed since
$b_{11}b_{12}^{\prime}-b_{11}^{\prime}b_{12}=r_{11}(r_{12}/r_{11})^{\prime}$
where $r_{ij}$ are as in Lemma 7.4. We have then only three equations
involving $\theta$, $b_{11}$ and $b_{35}$ left. Two of them give

$\displaystyle
Q_{56}=b_{35}^{2}\left(k\cos\\!\left(\theta\right)+1\right)\theta^{\prime}=c_{56}$
$\displaystyle
b_{11}b_{22}b_{35}\big{(}\ell_{1}^{2}+\ell_{2}^{2}\big{)}+1=\frac{\theta^{\prime}+\gamma
b_{35}\big{(}k^{2}+2\,k\,\cos\\!\left(\theta\right)+1\big{)}}{\theta^{\prime}}=0\
.$

This yields the first two equations and the expression for $b_{11}$ follow
from these. $\square$

The vorticity is given by

$h=\begin{pmatrix}-c_{13}f^{1}_{001}-2c_{12}k+c_{14}\\\\[2.84526pt]
-c_{14}f^{1}_{001}-c_{13}\\\\[2.84526pt] -c_{56}|\nabla
f^{2}|^{2}-\left(c_{16}+c_{25}\right)f^{2}_{100}+\left(c_{15}-c_{26}\right)f^{2}_{010}+c_{12}\end{pmatrix}\
.$

######  Theorem 7.7

If $k=1$ the solution can be written as follows: $b_{15}$ and $b_{25}$ are as
in Lemma 7.5 and for others we obtain

$\displaystyle(\theta^{\prime})^{3}=$ $\displaystyle
4\big{(}c_{56}\gamma^{2}-c_{36}m_{1}-c_{46}m_{0}\big{)}(1+\cos(\theta))$
$\displaystyle b_{35}=$
$\displaystyle-\frac{\theta^{\prime}}{2\gamma\left(1+\cos\\!\left(\theta\right)\right)}$
$\displaystyle b_{11}^{2}=$
$\displaystyle\frac{(m_{2}-m_{3})\cos(\theta)-2m_{4}\sin(\theta)+m_{2}+m_{3}}{4\theta^{\prime}}$
$\displaystyle b_{12}=$
$\displaystyle\frac{(m_{2}-m_{3})\sin(\theta)+2m_{4}\cos(\theta)}{4b_{11}\theta^{\prime}}$
$\displaystyle b_{22}=$
$\displaystyle\frac{\gamma}{b_{11}\theta^{\prime}}\quad,\quad
w_{1}=w_{2}=0\quad,\quad w_{3}=-\frac{\gamma}{b_{11}^{2}}\ ,$

where

$m_{2}=\frac{\gamma^{2}c_{36}^{2}+4m_{0}^{2}}{c_{36}m_{1}+c_{46}m_{0}}\quad,\quad
m_{3}=\frac{\gamma^{2}c_{46}^{2}+4m_{1}^{2}}{c_{36}m_{1}+c_{46}m_{0}}\quad,\quad
m_{4}=\frac{\gamma^{2}c_{36}c_{46}-4m_{0}m_{1}}{c_{36}m_{1}+c_{46}m_{0}}\ .$

Proof.  Substituting $b_{15}$, $b_{25}$ and various constants computed in
Lemma 7.5 to our equations we readily get $w_{1}=w_{2}=0$,
$w_{3}=-\gamma/b_{11}^{2}$ and $b_{12}$ as given above. Substituting
everything back to equations leaves us with three equations containing
$\theta$, $b_{35}$ and $b_{11}$. The expressions for these equations are too
big to be written down. But eliminating $b_{11}$ and $b_{35}$ from the
equations gives the surprisingly simple equation for $\theta$ given above and
then the formulas for $b_{11}$ and $b_{35}$ follow easily. $\square$

In the solutions $\theta$ reaches $\pi$ in finite time and the solution blows
up.

###  7.1 Extension for the spatial component

We can again look for other solutions by restricting the time component and
trying to find more general solutions to the spatial component. Let us
consider the matrix already mentioned in (7.1),

$A=\begin{pmatrix}\cos(\theta_{0}t)&-\sin(\theta_{0}t)&\cos(\theta_{0}t)&\sin(\theta_{0}t)&0&0\\\
\sin(\theta_{0}t)&\cos(\theta_{0}t)&-\sin(\theta_{0}t)&\cos(\theta_{0}t)&0&0\\\
0&0&0&0&\cos(2\theta_{0}t)&\sin(2\theta_{0}t)\end{pmatrix}\,,$

which is one particular form of the elliptic case. Now the conditions for the
spatial component are

$\displaystyle g_{125}+g_{345}=g_{126}+g_{346}=0$ $\displaystyle
g_{135}-g_{236}+g_{245}+g_{146}=0$ $\displaystyle
g_{136}+g_{246}-g_{145}+g_{235}=0\,.$

Writing $v=\big{(}z,f(z)\big{)}$ as before and supposing that $f^{2}$ and
$f^{3}$ are not anti CR with respect to $(z_{1},z_{2})$ we obtain a system of
the form

$f^{k}_{001}=g^{k}\quad,\quad 1\leq k\leq 3\quad,\quad f^{3}_{010}=g^{4}\ ,$

where the (real analytic) functions $g^{k}$ do not contain derivatives with
respect to $z_{3}$. The analysis of this system turns out to be surprisingly
tricky, so we simply give some indications of the necessary steps. The terms
in italics are defined and explained in [16].

If we had only the equations $f^{k}_{001}=g^{k}$ for $1\leq k\leq 3$ then the
local solution with initial conditions
$f^{k}(z_{1},z_{2},0)=h^{k}(z_{1},z_{2})$ would exist by Cauchy–Kovalevskaia
Theorem. Now one could hope that if the functions $h^{k}$ satisfied the final
equation then this would lead to a solution. Indeed there is a generalization
of Cauchy–Kovalevskaia Theorem, namely _Cartan–Kähler Theorem_ , which can be
applied to the overdetermined case. However, this theorem requires that the
system is _involutive_ ; in this case one can write the system in the _Cartan
normal form_ from which one can see what kind of initial conditions are
appropriate.

Unfortunately it happens that our system is not involutive. Now any system can
in principle be transformed to an involutive form by _prolongations_ and
_projections_. This is sometimes called the _Cartan–Kuranishi algorithm_.
Typically the concrete computations are difficult, even for a computer. We
checked that the _symbol_ of the system becomes involutive after two
prolongations, and then there is one _integrability condition_ of third order.
We did not check if adding this integrability condition produces an involutive
system or if there are further integrability conditions of higher order.
Anyway in this way one can in principle compute the involutive form after
which the existence of the local solution follows from the Cartan–Kähler
theorem.

##  8 $m=6$: parabolic case

This case has the spatial constraints

$G_{15}+G_{26}=G_{16}=G_{34}=0\,.$

The solution to this is

$v=(z_{1},z_{2},z_{3},f_{1}(z_{3}),z_{2}f_{3}^{\prime}(z_{1})+f_{2}(z_{1}),f_{3}(z_{1}))\,.$

For the time component the following quantities must be constant:

$\displaystyle p_{123},p_{356},p_{235},p_{135}-p_{236},$ $\displaystyle
p_{124},p_{456},p_{245},p_{145}-p_{246},$ $\displaystyle
Q_{12},Q_{13},Q_{14},Q_{15}-Q_{26},Q_{25},$ $\displaystyle
Q_{23},Q_{24},Q_{35},Q_{36},Q_{45},Q_{46},Q_{56}\,.$

The determinant constants can be reduced to several different cases. It is
best to omit the consideration of the ones that do not give solutions as they
just contain long calculations similar to what we have seen. The only reduced
case for the determinant constants that provides irreducible solutions is
where

$\displaystyle p_{123}=0,p_{356}=0,p_{235}=1,p_{135}-p_{236}=0,$
$\displaystyle p_{124}=0,p_{456}=0,p_{245}=0,p_{145}-p_{246}=1.$

It turns out that we must have either $p_{125}=p_{345}=0$ or
$p_{234}=p_{256}=0$ and we may assume $p_{125}=p_{345}=0$ by exchanging
$A_{1}$ with $A_{6}$ and $A_{2}$ with $A_{5}$ if needed. The determinant
conditions are thus satisfied if

$\displaystyle A_{1}$ $\displaystyle=\ell_{1}A_{2}\quad,\quad
A_{4}=\ell_{2}A_{5}$ $\displaystyle A_{6}$
$\displaystyle=\frac{1}{\ell_{2}}A_{3}+\ell_{1}A_{5}$

for some functions $\ell_{1}$ and $\ell_{2}$ and in this case we have

$\det(d\varphi)=f^{2}_{100}+z_{2}f^{3}_{200}-f^{1}_{001}f^{3}_{100}\neq 0\ .$

Now using the QR decomposition with respect to columns 2, 3 and 5 we can thus
write

$B=\begin{pmatrix}\ell_{1}b_{12}&b_{12}&b_{13}&\ell_{2}b_{15}&b_{15}&\ell_{1}b_{15}+b_{13}/\ell_{2}\\\
0&0&b_{23}&\ell_{2}b_{25}&b_{25}&\ell_{1}b_{25}+b_{23}/\ell_{2}\\\
0&0&-1/b_{12}b_{25}&0&0&-1/(\ell_{2}b_{12}b_{25})\end{pmatrix}\,.$

The analysis then is surprisingly similar to the hyperbolic case in Section 6.
Again it is not possible to give the computations in detail, but the basic
idea is as follows. First the preliminary computations show that we must have
$w=0$ so that there is no rotational component. Then computing further we
obtain

$(\hat{m}_{1}\ell_{2}+\hat{m}_{0})\ell_{1}+\hat{k}_{2}\ell_{2}^{2}+\hat{k}_{1}\ell_{2}+\hat{k}_{0}=0$

for some constants $\hat{k}$ and $\hat{m}$. At first it is possible to choose
constants $c_{ij}$ such that $\hat{k}_{j}=\hat{m}_{j}=0$ and thus apparently
to avoid the direct algebraic relation between $\ell_{1}$ and $\ell_{2}$.
However, the system is such that later we get this type of relation anyway.
Moreover it turns out that $\hat{m}_{0}=0$ in any case. Hence to get
nontrivial solution we must always have

$\ell_{1}=k_{2}\ell_{2}+k_{1}+k_{0}/\ell_{2}\ .$ (8.1)

We thus get three families of solutions: (1) $k_{2}k_{0}\neq 0$, (2) $k_{0}=0$
and (3) $k_{2}=0$.

######  Lemma 8.1

Suppose that $k_{2}k_{0}\neq 0$; then

$\displaystyle\left(\ell_{2}^{\prime}\right)^{3}=$
$\displaystyle\frac{k_{3}\ell_{2}^{4}}{k_{0}-k_{2}\ell_{2}^{2}}$
$\displaystyle l_{1}=$ $\displaystyle k_{2}\ell_{2}+k_{1}+k_{0}/\ell_{2}$
$\displaystyle b_{25}^{6}=$
$\displaystyle\frac{c_{45}^{3}(k_{0}-k_{2}\ell_{2}^{2})}{k_{3}\ell_{2}^{4}}$
$\displaystyle b_{13}=$ $\displaystyle\,b_{15}=0$ $\displaystyle b_{23}=$
$\displaystyle(k_{4}\ell_{2}-k_{0})b_{25}$ $\displaystyle b_{12}^{2}=$
$\displaystyle\frac{c_{12}b_{25}^{2}\ell_{2}^{2}}{c_{45}(k_{0}-k_{2}\ell_{2}^{2})}$

where $k_{j}$ are some constants.

Proof.  Starting from the equation (8.1) we first get some differential
equations for $b_{13}$ and $b_{15}$. However, substituting these back to the
system we notice that in fact $b_{13}=b_{15}=0$ and
$\ell_{2}^{\prime}=c_{45}/b_{25}^{2}$. Using these the rest follows easily.
$\square$

Note that the differential equation for $\ell_{2}$ has a discrete symmetry: if
$\mu$ is a solution then also $k_{0}/(k_{2}\mu)$ is a solution. Apparently one
cannot explicitly solve the equation for $\ell_{2}$.

The vorticity is now

$h=\begin{pmatrix}-f^{3}_{100}\left(c_{45}f^{1}_{001}+c_{35}\right)\\\\[2.84526pt]
z_{2}\left(c_{45}f^{1}_{001}+c_{35}\right)f^{3}_{200}+\left(c_{45}f^{2}_{100}+r_{1}f^{3}_{100}\right)f^{1}_{001}+c_{35}f^{2}_{100}-r_{0}f^{3}_{100}\\\\[2.84526pt]
c_{45}k_{2}\left(f^{3}_{100}\right)^{2}+\left(c_{15}-c_{26}\right)f^{3}_{100}+c_{12}\end{pmatrix}\
,$

where $r_{j}$ are some constants.

######  Lemma 8.2

Let us assume that $k_{0}=0$; then

$\displaystyle l_{1}=$ $\displaystyle k_{2}k_{4}t^{3}+k_{1}$ $\displaystyle
l_{2}=$ $\displaystyle k_{4}t^{3}$ $\displaystyle b_{12}=$ $\displaystyle
k_{5}/t$ $\displaystyle b_{13}=$ $\displaystyle k_{6}t^{2}$ $\displaystyle
b_{15}=$ $\displaystyle 0$ $\displaystyle b_{23}=$ $\displaystyle k_{7}t^{2}$
$\displaystyle b_{25}=$ $\displaystyle k_{3}/t$

where $k_{j}$ are constants.

Proof.  Substituting $w=0$ and $\ell_{1}=k_{2}\ell_{1}+k_{1}$ we can solve
$b_{13}$, $b_{15}$, $b_{23}$ and $b_{25}$ in terms of $\ell_{2}$ and $b_{12}$.
Substituting this back we obtain a system for $\ell_{2}$ and $b_{12}$ whose
solution is as above. Then the rest follows easily. $\square$

######  Lemma 8.3

Let us assume that $k_{2}=0$; then

$\displaystyle l_{1}=$ $\displaystyle k_{1}+k_{0}t^{3}/k_{4}$ $\displaystyle
l_{2}=$ $\displaystyle k_{4}/t^{3}$ $\displaystyle b_{15}=$ $\displaystyle
k_{6}t^{2}$ $\displaystyle b_{25}=$ $\displaystyle k_{5}t^{2}$ $\displaystyle
b_{12}=$ $\displaystyle k_{3}/t$ $\displaystyle b_{13}=$ $\displaystyle-
k_{0}k_{6}t^{2}+k_{7}/t$ $\displaystyle b_{23}=$ $\displaystyle-
k_{0}k_{5}t^{2}+k_{8}/t$

where $k_{j}$ are constants.

Proof.  Substituting $w=0$ and $\ell_{1}=k_{1}+k_{0}/\ell_{2}$ we can solve
$b_{12}$, $b_{13}$ and $b_{23}$ in terms of $\ell_{2}$ and $b_{25}$, and in
particular $b_{12}=\hat{k}b_{25}\ell_{2}$ for some $\hat{k}$. Substituting
these back to the equations we are left with equations for $\ell_{2}$ and
$b_{25}$ whose solution is given above. After this the rest follows easily.
$\square$

Note that $b_{12}$ is the same in both Lemma 8.2 and Lemma 8.3.

###  8.1 Extension for the spatial component

In both Lemmas 8.2 and 8.3 the entries of $A$ are constant linear combinations
of $t^{2}$ and $1/t$, so by linear transformations we can bring $A$ to the
form

$A=\begin{pmatrix}t^{2}&1/t&0&0&0&0\\\ 0&0&t^{2}&1/t&0&0\\\
0&0&0&0&t^{2}&1/t\end{pmatrix}\,.$

Let us present the constraints for $v$ when $A$ is as above. The equations $v$
needs to satisfy are

$g_{135}=g_{246}=g_{235}+g_{136}+g_{145}=0\,.$

Writing $v=(z,f(z))$ this gives

$\displaystyle f^{2}_{010}=0$ $\displaystyle
f^{1}_{100}f^{3}_{001}-f^{3}_{100}f^{1}_{001}=0$ $\displaystyle
f^{2}_{100}-f^{3}_{010}+f^{1}_{010}f^{2}_{001}=0\,.$

Since $f^{2}$ does not depend on $z_{2}$, the last equation gives

$f^{3}=z_{2}f^{2}_{100}+f^{1}f^{2}_{001}+g(z_{1},z_{3})\,.$

Substituting this to $f^{1}_{100}f^{3}_{001}-f^{3}_{100}f^{1}_{001}=0$ gives a
standard first order PDE for $f^{1}$ so the local solution exists and $f^{2}$
and $g$ are arbitrary. Note that here $f^{1}$ and $f^{3}$ depend nontrivially
on $z_{2}$.

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