# On global regularity of some bi-rotational Euler flows in $\mathbb{R}^{4}$

Kyudong Choi Department of Mathematical Sciences, Ulsan National Institute of
Science and Technology, 50 UNIST-gil, Eonyang-eup, Ulju-gun, Ulsan 44919,
Republic of Korea. Email<EMAIL_ADDRESS>In-Jee Jeong Department of
Mathematical Sciences and RIM, Seoul National University, 1 Gwanak-ro, Gwanak-
gu, Seoul 08826, Republic of Korea. Email<EMAIL_ADDRESS>Deokwoo Lim
The Research Institute of Basic Sciences, Seoul National University, 1 Gwanak-
ro, Gwanak-gu, Seoul 08826, Republic of Korea. Email<EMAIL_ADDRESS>
###### Abstract

In this paper, we consider incompressible Euler flows in $\mathbb{R}^{4}$
under bi-rotational symmetry, namely solutions that are invariant under
rotations in $\mathbb{R}^{4}$ fixing either the first two or last two axes.
With the additional swirl-free assumption, our first main result gives local
wellposedness of Yudovich-type solutions, extending the work of Danchin
[Uspekhi Mat. Nauk 62(2007), no.3, 73–94] for axisymmetric flows in
$\mathbb{R}^{3}$. The second main result establishes global wellposedness
under additional decay conditions near the axes and at infinity. This in
particular gives global regularity of $C^{\infty}$ smooth and decaying Euler
flows in $\mathbb{R}^{4}$ subject to bi-rotational symmetry without swirl.

††footnotetext: _2020 AMS Mathematics Subject Classification:_ 76B47,
35Q35††footnotetext: _Key words : Incompressible Euler equations; bi-
rotational symmetry; vorticity; local regularity; Biot–Savart law; global
regularity; BKM criterion_

##  1 Introduction

We consider the Cauchy problem for the incompressible Euler equations in
$\mathbb{R}^{n}$:

$\left\\{\begin{aligned} \partial_{t}u+(u\cdot\nabla)u+\nabla p&=0,\\\
\nabla\cdot u&=0,\\\ u|_{t=0}&=u_{0},\end{aligned}\right.$ (1.1)

where $u=(u^{1},\cdots,u^{n}):[0,\infty)\times\mathbb{R}^{n}\to\mathbb{R}^{n}$
and $p:[0,\infty)\times\mathbb{R}^{n}\to\mathbb{R}$. The system (1.1)
describes the motion of volume-preserving and inviscid fluid flows, and has
been studied extensively in the two- and three-dimensional cases, namely when
$n=2,3$. For $n\geq 3$, global regularity or finite time singularity for
$C^{\infty}$ smooth and decaying solutions to (1.1) is an outstanding open
problem.

The goal of this paper is to study the four-dimensional case $(n=4)$ under the
bi-rotational symmetry. The main motivation is to uncover new geometric
scenarios that could contribute to the understanding of global regularity or
the formation of finite-time singularities for the Euler equations.

###  1.1 The system for bi-rotationally symmetric solutions

The three-dimensional Euler equations have been extensively studied under the
axisymmetry assumption, which means that the solution is invariant under all
rotations fixing one axis in $\mathbb{R}^{3}$. Since the Euler equations
preserve rotational symmetries of the solution, it suffices to assume
axisymmetry for the initial data. The key simplification comes from the fact
that the number of independent spatial variables reduces from three to two.

While higher dimensional Euler equations ($n\geq 4$) has been studied under
the axisymmetry assumption as well, in this paper we shall investigate the bi-
rotational symmetry, or $O(2)\times O(2)$ symmetry, which is specific to
dimensions 4 or higher. As the name suggests, this symmetry refers to
solutions which are invariant under all rotations fixing either the first two
or the last two variables. To handle such solutions, it is natural to consider
the bi-polar coordinates in $\mathbb{R}^{4}$: the coordinate system
$(r,\theta,s,\phi)$ with $x_{1}=r\cos(\theta),\quad x_{2}=r\sin(\theta),\quad
x_{3}=s\cos(\theta)$ and $x_{4}=s\sin(\phi).$

Under the bi-rotational symmetry, the velocity vector can be written as

$\begin{split}u=u^{r}e^{r}+u^{\theta}e^{\theta}+u^{s}e^{s}+u^{\phi}e^{\phi},\end{split}$
(1.2)

where the functions $u^{*}$ depend only on $(t,r,s)$ and the unit vectors
$e^{*}$ are defined by

$\begin{split}e^{r}=\begin{pmatrix}\cos\theta\\\ \sin\theta\\\ 0\\\
0\end{pmatrix},\quad e^{\theta}=\begin{pmatrix}-\sin\theta\\\ \cos\theta\\\
0\\\ 0\end{pmatrix},\quad e^{s}=\begin{pmatrix}0\\\ 0\\\ \cos\phi\\\
\sin\phi\end{pmatrix},\quad e^{\phi}=\begin{pmatrix}0\\\ 0\\\ -\sin\phi\\\
\cos\phi\end{pmatrix}.\end{split}$

Similarly, the pressure is given by $p=q(t,r,s)$ for some scalar-valued
function $q$. We then compute that (1.1) turns into

$\left\\{\begin{aligned}
&\partial_{t}u^{r}+(u^{r}\partial_{r}+u^{s}\partial_{s})u^{r}=\frac{(u^{\theta})^{2}}{r}-\partial_{r}q\\\
&\partial_{t}u^{s}+(u^{r}\partial_{r}+u^{s}\partial_{s})u^{s}=\frac{(u^{\phi})^{2}}{s}-\partial_{s}q\end{aligned}\right.$
(1.3)

and

$\left\\{\begin{aligned}
&\partial_{t}u^{\theta}+(u^{r}\partial_{r}+u^{s}\partial_{s})u^{\theta}=-\frac{u^{r}}{r}u^{\theta}\\\
&\partial_{t}u^{\phi}+(u^{r}\partial_{r}+u^{s}\partial_{s})u^{\phi}=-\frac{u^{s}}{s}u^{\phi}\end{aligned}\right.$
(1.4)

supplemented with the divergence-free condition

$\begin{split}\partial_{r}u^{r}+\frac{u^{r}}{r}+\partial_{s}u^{s}+\frac{u^{s}}{s}=0.\end{split}$
(1.5)

The system (1.3)–(1.5) describes the 4D Euler equations under the bi-
rotational symmetry. We shall mainly work with its vorticity form (to avoid
dealing with the pressure), obtained by introducing the scalar “vorticity”

$\begin{split}w:=\partial_{r}u^{s}-\partial_{s}u^{r}.\end{split}$ (1.6)

From (1.3), we have that $w$ satisfies

$\begin{split}\partial_{t}w+(u^{r}\partial_{r}+u^{s}\partial_{s})w=\left(\frac{u^{r}}{r}+\frac{u^{s}}{s}\right)w+2\left(\frac{u^{\phi}}{s}\partial_{r}u^{\phi}-\frac{u^{\theta}}{r}\partial_{s}u^{\theta}\right).\end{split}$
(1.7)

We may introduce $\psi$ satisfying

$\begin{split}w=\partial_{r}\left(\frac{1}{r}\partial_{r}(r\psi)\right)+\partial_{s}\left(\frac{1}{s}\partial_{s}(s\psi)\right)\end{split}$
(1.8)

subject to the boundary condition $\psi(0,\cdot)=\psi(\cdot,0)=0$. The
velocity $(u^{r},u^{s})$ is obtained from $\psi$ by

$\displaystyle u^{r}$
$\displaystyle=-\frac{1}{s}\partial_{s}(s\psi)=-\frac{\psi}{s}-\partial_{s}\psi,\quad
u^{s}=\frac{1}{r}\partial_{r}(r\psi)=\frac{\psi}{r}+\partial_{r}\psi.$ (1.9)

The above system contains a few sub-systems; for instance, if one takes
$u^{\phi}=0$ (or similarly $u^{\theta}=0$), then it reduces to a system of two
scalar quantities which resembles the 3D axisymmetric Euler equations.
Moreover, if one takes

$u^{\theta}=u^{\phi}=0,$ (1.10)

then the system becomes

$\begin{split}D_{t}\left(\frac{w}{rs}\right)&=0,\qquad
D_{t}=\partial_{t}+u^{r}\partial_{r}+u^{s}\partial_{s},\end{split}$ (1.11)

which shall be referred to as the bi-rotational Euler equation without swirl.
This is equivalent with

$\begin{split}\partial_{t}w+(u^{r}\partial_{r}+u^{s}\partial_{s})w&=w\bigg{(}\frac{u^{r}}{r}+\frac{u^{s}}{s}\bigg{)},\end{split}$
(1.12)

which is simply (1.7) with the no-swirl condition $u^{\theta}=u^{\phi}\equiv
0$.

###  1.2 Main results

In this paper, we prove local and global regularity results for the Cauchy
problem to the bi-rotational Euler equations (1.12) without swirl. Before
stating the theorems, let us provide a quick review on the well-posedness
theory of the incompressible Euler equations. More refined statements which
naturally motivate the results in this work will be given in the following
section.

The incompressible Euler equations (1.1) in $\mathbb{R}^{n}$ are locally well-
posed in Hölder spaces $C^{k,\alpha}$ with $k\geq 1,0<\alpha<1$ (together with
some decay assumption at infinity) and Sobolev spaces $H^{s}$ with
$s>\frac{n}{2}+1$, in terms of $u$. While not much is known in general about
long-time behavior of the solutions, the Beale–Kato–Majda criterion ([2, 21])
states that these local solutions lose regularity at some time $T^{*}>0$ if
and only if

$\begin{split}\lim_{t\rightarrow T^{*}}\int_{0}^{t}\|(\nabla\times
u)(\tau)\|_{L^{\infty}}d\tau=+\infty.\end{split}$

In particular, the bi-rotational Euler equations (being a special case of the
four-dimensional Euler equations) are locally well-posed in such function
spaces. We are interested in utilizing the special nature of the bi-rotational
system to obtain well-posedness (existence and uniqueness) in the largest
possible borderline spaces, which are characterized by having a scale-
invariant norm: $\|\cdot\|$ satisfying $\|(\nabla\times
u)(\lambda\cdot)\|=\|\nabla\times u\|$ for any $\lambda>0$. In this direction,
we have a local well-posedness for Yudovich type initial data, extending
earlier works of Ukhovsksii–Yudovich [32] and Danchin [9] in the three-
dimensional axisymmetric Euler:

######  Theorem 1.1.

Assume that $w_{0}$ satisfy $w_{0}\in(L^{4,1}\cap L^{\infty})(\mathbb{R}^{4})$
and

${(1+r+s)\frac{w_{0}}{rs}\in L^{4,1}(\mathbb{R}^{4}).}$

Then, there exist $T>0$ and a unique solution to (1.1) satisfying $w\in
L^{\infty}(0,T;(L^{4,1}\cap L^{\infty})(\mathbb{R}^{4}))$ and

$(1+r+s)\frac{w}{rs}\in L^{\infty}(0,T;L^{4,1}(\mathbb{R}^{4})).$

In the statement above, $L^{p,q}$ refers to Lorentz spaces which are defined
in the following section. It is important that the uniqueness statement is not
merely among the solutions satisfying bi-rotational symmetry but is for all
possible solutions of the original Euler equations (1.1).

######  Remark 1.2.

The local-in-time solution $w$ from Theorem 1.1 loses its regularity at finite
time $T^{\ast}>0$ if and only if we have

$\int_{0}^{T_{\ast}}\bigg{[}\bigg{\|}\frac{w(t)}{r}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})}+\bigg{\|}\frac{w(t)}{s}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})}\bigg{]}dt=+\infty.$
(1.13)

Furthermore, under additional assumptions on the initial data, one can
guarantee global in time well-posedness.

######  Theorem 1.3.

Let $w_{0}$ from Theorem 1.1 additionally satisfy

$(1+r+s)\frac{w_{0}}{rs}\in
L^{\infty}(\mathbb{R}^{4})\qquad\mbox{and}\qquad{(1+r^{2}+s^{2})\frac{w_{0}}{rs}\in
L^{1}(\mathbb{R}^{4}).}$

Then the corresponding solution $w$ of (1.12) is global in time and satisfies
the estimate

$\displaystyle\|w(t)\|_{L^{\infty}(\mathbb{R}^{4})}\leq C\exp(Ct),$ (1.14)
$\displaystyle\bigg{\|}\frac{w(t)}{r}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})}\leq
C\exp(Ct),$ (1.15)
$\displaystyle\bigg{\|}\frac{w(t)}{s}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})}\leq
C\exp(Ct),$ (1.16)

for some constant $C>0$ depending only on

$\|w_{0}\|_{L^{\infty}(\mathbb{R}^{4})},\quad{\bigg{\|}(1+r+s)\frac{w_{0}}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})},\quad\text{and}\quad\bigg{\|}(1+r^{2}+s^{2})\frac{w_{0}}{rs}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}.}$

######  Remark 1.4 (Global well-posedness for smooth and decaying data).

If $u_{0}\in H^{s}(\mathbb{R}^{4})$ for some $s>5$ satisfies (1.2) and (1.10),
then the condition ${(1+r+s)}w_{0}/(rs)\in L^{\infty}(\mathbb{R}^{4})$ from
Theorem 1.3 is automatically satisfied. We will prove this in Proposition A.1
from the Appendix A. The other condition $(1+r^{2}+s^{2})w_{0}/(rs)\in
L^{1}(\mathbb{R}^{4})$ is satisfied when $\nabla u_{0}$ has some decay at
infinity. Therefore, we conclude the following: sufficiently smooth and
decaying initial data of (1.1) have globally unique smooth solutions under bi-
rotational symmetry without swirl.

###  1.3 Key ideas

Support growth of a compactly supported vorticity. For the moment, let us
consider an easier setting where the initial vorticity $w_{0}$ is compactly
supported away from the axes $\\{r=0\\}$ and $\\{s=0\\}$. Then using the
conservation of $L^{p}$-norm ($p\in[1,\infty]$) of $w/(rs)$ in time, we can
estimate the $L^{\infty}$-norm of the corresponding local-in-time solution $w$
from Theorem 1.1 as

$\|w(t)\|_{L^{\infty}(\mathbb{R}^{4})}\leq
R(t)S(t)\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}\lesssim[R(t)+S(t)]^{2}\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})},\quad
t\in[0,T_{\text{max}}),$

where $R(t)$ and $S(t)$ are given as

$R(t)=R(0)+\int_{0}^{t}\|u^{r}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}d\tau,\quad
S(t)=S(0)+\int_{0}^{t}\|u^{s}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}d\tau.$

These represent the highest distance where a point contained in the support of
$w_{0}$ can move along the flow in $r$ and $s$-direction, respectively. This
narrows our goal down to obtaining estimates of
$\|u^{r}(t)\|_{L^{\infty}(\mathbb{R}^{4})}$ and
$\|u^{s}(t)\|_{L^{\infty}(\mathbb{R}^{4})}$. This can be done by the following
time-independent estimate (Proposition 4.4):

$\|u^{r}\|_{L^{\infty}(\mathbb{R}^{4})},\,\|u^{s}\|_{L^{\infty}(\mathbb{R}^{4})}\lesssim\bigg{\|}(r^{2}+s^{2})\frac{w}{rs}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}^{1/2}\cdot\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}^{1/2}.$
(1.17)

Such an estimate is obtained by analyzing Biot–Savart kernels of $u^{r}$ and
$u^{s}$, which have symmetry relation. In addition, to prove the above
estimate, due to the scale-invariance of the Biot–Savart formula with respect
to the scaling $u\mapsto u(\lambda\cdot)$ and $w\mapsto\lambda
w(\lambda\cdot)$ for any $\lambda>0$, it is sufficient enough to show

$\sup_{\begin{subarray}{c}r,s\geq 0,\\\
r^{2}+s^{2}=1\end{subarray}}|u^{r}(r,s)|,\sup_{\begin{subarray}{c}r,s\geq
0,\\\
r^{2}+s^{2}=1\end{subarray}}|u^{s}(r,s)|\lesssim\bigg{\|}(r^{2}+s^{2})\frac{w}{rs}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}^{1/2}\cdot\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}^{1/2}.$

This idea using scaling was originally from Feng–Sverak [14] in the 3D
axisymmetric Euler without swirl. Once the estimate (1.17) is shown, then we
can get

$\begin{split}\frac{d}{dt}[R(t)+S(t)]&=\|u^{r}(t)\|_{L^{\infty}(\mathbb{R}^{4})}+\|u^{s}(t)\|_{L^{\infty}(\mathbb{R}^{4})}\lesssim_{w_{0}}[R(t)^{2}+S(t)^{2}]^{1/2}\leq
R(t)+S(t).\end{split}$

This gives the term $R(t)+S(t)$ an exponential upper bound with a constant
term added.

Here, let us emphasize an additional difficulty arising in the proof of
(1.17), compared to the axisymmetric case. In that situation, the velocity is
a function of $r,z$, and using scaling and translational invariance, it
suffices to prove the bound (1.17) only in the special case $(r,z)=(1,0)$. In
the current setup of bi-rotational symmetry, there are serious technical
difficulties arising not only from $(r,s)=(1,0)$ or $(0,1)$ but also from the
whole circle $r^{2}+s^{2}=1$. This is the reason why we need the control of
$(r^{2}+s^{2})w/(rs)\in L^{1}$ (and not just $rsw\in L^{1}$).

Furthermore, in the axisymmetric case, the Biot–Savart kernel can be
simplified into a composition between a single-variable function and some
scalar function with other simple terms multiplied. This idea was originally
from Feng–Sverak [14] and it makes the estimate of the Biot–Savart kernel much
easier as the process narrows down to estimating this single-variable
function. However, in our bi-rotational symmetry setting, we cannot simplify
the Biot–Savart kernel via some one variable function, so we had to directly
estimate the Biot–Savart kernel to achieve our goal.

Removing the compact support assumption. The case where the vorticity $w$ is
not necessarily compactly supported can be handled in a similar way as the
above easier case. The support size $R(t)+S(t)$ is replaced by the following
“length” function $L$:

$L(t)=L(0)+\int_{0}^{t}[\|u^{r}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}+\|u^{s}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}]d\tau.$

Then instead of compactly supportedness, if we assume decay assumptions near
rotational axes and at infinity, then we can obtain estimates of
$\|w\|_{L^{\infty}(\mathbb{R}^{4})}$ and the first term on the right-hand side
of (1.17):

$\|w(t)\|_{L^{\infty}(\mathbb{R}^{4})}\lesssim_{w_{0}}L(t)^{2},\qquad\bigg{\|}(r^{2}+s^{2})\frac{w}{rs}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}\lesssim_{w_{0}}L(t)^{2}.$

Using these and (1.17), we can get an exponential bound of $L(t)$ by the
following inequality,

$\frac{d}{dt}L(t)=\|u^{r}(t)\|_{L^{\infty}(\mathbb{R}^{4})}+\|u^{s}(t)\|_{L^{\infty}(\mathbb{R}^{4})}\lesssim_{w_{0}}L(t),$

which immediately implies that $\|w\|_{L^{\infty}(\mathbb{R}^{4})}$ is
exponentially bounded in time.

###  1.4 Literature review

To put our results in context, we review relevant literature, starting from
the axisymmetric Euler equations.

#### Axisymmetric solutions in three dimensions

In this section, we briefly review the theory of axisymmetric solutions to the
3D Euler equations, which naturally motivates the results in the current
paper. We fix the domain to be $\mathbb{R}^{3}$ and use cylindrical
coordinates $(r,\theta,z)$. Under axisymmery, 3D Euler equations reduce to the
system involving the axial components of the velocity and its curl, denoted by
$u^{\theta}$ and $\omega^{\theta}=\partial_{z}u^{r}-\partial_{r}u^{z}$ ([28]):

$\left\\{\begin{aligned}
&(\partial_{t}+u^{r}\partial_{r}+u^{z}\partial_{z})(ru^{\theta})=0,\\\
&(\partial_{t}+u^{r}\partial_{r}+u^{z}\partial_{z})(\frac{\omega^{\theta}}{r})=\frac{1}{r^{4}}\partial_{z}[(ru^{\theta})^{2}],\end{aligned}\right.$
(1.18)

supplemented with the divergence-free condition
$\partial_{r}u^{r}+\frac{u^{r}}{r}+\partial_{z}u^{z}=0.$

Global well-posedness in the no-swirl case. While the problem of singularity
formation for smooth solutions to (1.18) is still open, global regularity in
the no-swirl case $u^{\theta}=0$ is well-known. In this case, the system is
simply

$\begin{split}(\partial_{t}+u^{r}\partial_{r}+u^{z}\partial_{z})(\frac{\omega^{\theta}}{r})=0,\end{split}$
(1.19)

so that any $L^{p,q}$-norm (with respect to the Lebesgue measure on
$\mathbb{R}^{3}$) of $\frac{\omega^{\theta}}{r}$ is conserved in time. This
equation is analogous to the 2D vorticity equation, which is simply the
transport equation for the vorticity. In the 2D case, Yudovich theory [35]
gives global well-posedness for vorticities belonging to
$L^{\infty}(0,\infty;L^{\infty}\cap L^{1}(\mathbb{R}^{2}))$. Ideally, one
would like to prove well-posedness for (1.19) with merely bounded and decaying
$\omega^{\theta}$ as well, but there are subtle differences due to the factor
$r^{-1}$.

Regarding (1.19), existence of global solutions was first proved in
Ukhovsksii–Yudovich [32] under the assumptions
$\omega^{\theta}_{0},\frac{\omega^{\theta}_{0}}{r}\in L^{2}\cap L^{\infty}$.
Later, global well-posedness in the case $\omega_{0}^{\theta}\in H^{s}$ with
$s>\frac{3}{2}$ was obtained in [27, 30, 31]. Then, Danchin proved global
regularity under the assumption $\omega^{\theta}_{0}\in L^{\infty}\cap
L^{3,1}$, $\frac{\omega^{\theta}_{0}}{r}\in L^{3,1}$ in [9], and this seems to
be optimal.

To briefly explain the difference with the 2D Euler equation, let us return to
the equation for $\omega^{\theta}$ and observe the estimate

$\begin{split}\frac{d}{dt}\|\omega^{\theta}\|_{L^{\infty}}\leq\|r^{-1}u^{r}\|_{L^{\infty}}\|\omega^{\theta}\|_{L^{\infty}}.\end{split}$

However, having $\omega^{\theta}$ in $L^{\infty}\cap L^{1}$ does not guarantee
boundedness of $r^{-1}u^{r}$. On the other hand, estimate due to
Shirota–Yanagisawa ([31]) and Danchin ([9, Lemma 2]) gives

$\begin{split}\frac{|u^{r}(r,z)|}{r}\leq
C\int_{\mathbb{R}^{3}}\frac{1}{|x-\bar{x}|^{2}}\frac{|\omega^{\theta}(\bar{x})|}{\bar{r}}d\bar{x}\leq
C\|r^{-1}\omega^{\theta}\|_{L^{3,1}}=C\|r^{-1}\omega^{\theta}_{0}\|_{L^{3,1}},\end{split}$

which then allows us to get

$\begin{split}\frac{d}{dt}\|\omega^{\theta}\|_{L^{\infty}}\leq
C\|r^{-1}\omega^{\theta}_{0}\|_{L^{3,1}}\|\omega^{\theta}\|_{L^{\infty}}.\end{split}$

This guarantees that the vorticity remains uniformly bounded until any finite
time, and this gives not only uniqueness but also propagates higher Hölder or
Sobolev regularity of the vorticity.

We note that the condition $\frac{\omega^{\theta}_{0}}{r}\in
L^{3,1}(\mathbb{R}^{3})$ is automatic (via Sobolev embedding) when the initial
vorticity belongs to either $H^{s}(\mathbb{R}^{3})$ with $s>\frac{3}{2}$ or
$C^{\alpha}(\mathbb{R}^{3})$ with $\alpha>\frac{1}{3}$. We note that global
well-posedness for smooth and axisymmetric vortex patch solutions to (1.19)
was proved in [19] for initial data satisfying
$|\omega_{0}^{\theta}(r,z)|\lesssim|r|^{\beta}$ for $\beta>\frac{1}{3}$.
Moreover, global well-posedness of (1.19) in certain critical Besov spaces has
been established in [1, 8, 33].

On the other hand, optimality of $\frac{\omega^{\theta}_{0}}{r}\in
L^{3,1}(\mathbb{R}^{3})$ can be seen by directly using the kernel expression
for $u^{r}/r$: for any $\epsilon>0$, it is possible to find
$\omega^{\theta}_{0}\in L^{1}\cap L^{\infty}(\mathbb{R}^{3})$ such that
$\frac{\omega^{\theta}_{0}}{r}\in L^{3,1+\epsilon}(\mathbb{R}^{3})$ but the
corresponding velocity satisfies $u^{r}_{0}/r\notin
L^{\infty}(\mathbb{R}^{3})$. Indeed, strong illposedness of (1.19) was
obtained in critical Lorentz spaces in [23] based on this observation. Earlier
work of Bourgain–Li [3] obtained strong illposedness for (1.19) in the
critical $H^{3/2}(\mathbb{R}^{3})$ space.

Finite-time singularity formation for $C^{\alpha}$ data. The aforementioned
works may give the impression that the axisymmetric no-swirl equation is
globally well-posed in any function space in which the equation is locally
well-posed. Essentially111Strictly speaking, it is not clear if the BKM
criterion is valid for all spaces in which the Euler equations are well-posed
(see [9])., Danchin’s result shows that this is correct once
$r^{-1}\omega_{0}^{\theta}\in L^{3,1}$. However, recall that for
$\omega_{0}^{\theta}\in C^{\alpha}$, $\omega^{\theta}_{0}\in L^{3,1}$ is
guaranteed only when $\alpha>\frac{1}{3}$. Remarkably, recent work of Elgindi
[12] (see also [13]) proved that finite-time blow up occurs for the no-swirl
equation with initial data $\omega_{0}^{\theta}\in C^{\alpha}$ with $0<\alpha$
small. Although singularity formation for the axisymmetric no-swirl system is
limited to vorticities less regular than $C^{\frac{1}{3}}$, the situation
could be different in other geometries. It would be interesting to investigate
Elgindi type of singularity formation for the bi-rotational geometry as well.
However, in the current work, the initial data considered are regular enough
near the axis (e.g. $w\in L^{4,1}(\mathbb{R}^{4})$) which rules out this kind
of singularity formation. Inspired by this work, Chen–Hou [5] showed finite-
time singularity formation of some solution of 3D axisymmetric Euler equations
with swirl in a bounded cylinder with $C^{\alpha}$-initial vorticity.

#### Axisymmetric Euler equations in higher dimensions

As we mentioned earlier, axisymmetric solutions can be considered for the
Euler equations defined in higher dimensions as well. Interestingly, it turns
out that as dimension becomes larger, there seems to be a higher chance of
singularity formation, even for smooth and decaying initial data.

Drivas–Elgindi [11] considered pressureless Euler equations in each dimension.
They constructed a sequence of solutions for each dimension that is initially
smooth and blows up in finite time. Then, they sent the dimension to infinity
to show that there is a finite-time blowup in the “infinite dimension”, in
some sense. This lead them to propose an open question about whether there
would be a finite-time blowup of initially smooth, axisymmetric, and swirl-
free solution of the Euler equations in some high dimension. Miller [29] also
raised the same open question. In addition, Hou–Zhang [17, 18] investigated
potential finite-time singularity of the axisymmetric Euler equations without
swirl where the initial vorticity is in $C^{\alpha}$ for a large range of
$\alpha$ in both 3D and higher dimensions.

On the other hand, there are several works regarding global existence of
locally well-posed smooth solutions ([29]) with additional decay assumptions
at the origin and at infinity. In our previous work [7], it was shown that
such decay assumptions are sufficient for global regularity to hold in
$\mathbb{R}^{4}$. However, the decay assumption near the symmetry axis used in
[7] is not automatically satisfied for smooth data in $\mathbb{R}^{4}$, in
stark contrast to our global wellposedness result. In a more recent work by
the third author [26], it was shown that for any $d\geq 3$, a single-signed
vorticity with compact support and fast decay at the origin is globally
regular. Indeed, an explicit estimate of the size of vorticity confinement can
be obtained.

#### The lake equation and the vortex membrane equation

There is another equation that is closely related to the bi-rotational Euler
equations, namely, the lake equation in $\Omega\subset\mathbb{R}^{2}$;

$\begin{split}\partial_{t}{\bigg{(}\frac{\omega}{b}\bigg{)}}+U\cdot\nabla{\bigg{(}\frac{\omega}{b}\bigg{)}}&=0\quad\text{in}\quad\Omega,\\\
\nabla\cdot(bU)&=0\quad\text{in}\quad\Omega,\\\ U\cdot
n&=0\quad\text{in}\quad\partial\Omega,\\\
\omega&=-\partial_{2}U^{1}+\partial_{1}U^{2},\end{split}$ (1.20)

with a nonnegative depth function $b:\Omega\to[0,\infty)$. Note that when the
domain $\Omega$ is the half plane $\\{(r,z):r\geq 0,z\in\mathbb{R}\\}$ with
the depth function chosen as $b(r,z)=r$, then this corresponds to the 3D
axisymmetric Euler equations without swirl. Additionally, if we take $\Omega$
as the first quadrant $\\{(r,s):r,s\geq 0\\}$ and $b$ as $b(r,s)=rs$, then
this is exactly the bi-rotational Euler equations that we are considering in
this paper. There are several results about global existence of Yudovich-type
solutions in bounded domains ([25, 24]).

Also, works of Khesin–Yang [22] and Yang [34] presented that for integers
$m,l\in\mathbb{N}$, the Euler equations in $\mathbb{R}^{m+l+2}$ with sphere
product $\mathbb{S}^{m}\times\mathbb{S}^{l}$-symmetry (that is, when the
velocity $u$ and the pressure $p$ depend only on distances $|x|,|y|$ with
$x\in\mathbb{R}^{m+1},y\in\mathbb{R}^{l+1}$ to the origin) can be written as
the lake equation (1.20) in the first quadrant $\\{(|x|,|y|)\\}$ with
$b(|x|,|y|)=|x|^{m}|y|^{l}$. We can consider the Euler equations with such
symmetry as the bi-rotational Euler equations in higher dimensions
$\mathbb{R}^{{m+l+2}}$, with the equation

$\partial_{t}\bigg{(}\frac{w}{r^{m}s^{l}}\bigg{)}+(u^{r}\partial_{r}+u^{s}\partial_{s})\bigg{(}\frac{w}{r^{m}s^{l}}\bigg{)}=0,\quad
w=-\partial_{s}u^{r}+\partial_{r}u^{s}.$ (1.21)

In addition, their works show that the motion of a point vortex of a point
$(a,b)$ in the first quadrant for the equation (1.21) can be regarded as a
sphere product vortex membrane under skew-mean-curvature flow, and they
provided explicit point-vortex type solutions which exist for all time or
shows finite-time blowup with dependence on $m,l$. Here, a sphere product
vortex membrane is a singular vorticity supported on the sphere product
$\mathbb{S}^{m}(a)\times\mathbb{S}^{l}(b)$ with radii $a,b\geq 0$, and the
skew-mean-curvature flow is described by the following equation:

$\partial_{t}q=-J(H(q)),\quad q\in\Sigma,$

where $\Sigma^{n}\subset\mathbb{R}^{n+2}$ is a codimension 2 compact oriented
submanifold(membrane) in $\mathbb{R}^{n+2}$, $H(q)$ is the mean curvature
vector at the point $q\in\Sigma$, and $J$ is the operator of positive $\pi/2$
rotation in the 2-dimensional normal space $N_{q}\Sigma$ to $\Sigma$ at $q$.
This is a natural generalization from the case $n=1$, which is the binormal
equation:

$\partial_{t}\gamma=\kappa\textbf{b},$ (1.22)

where $\gamma$ is a curve in $\mathbb{R}^{3}$, $\kappa$ is the curvature of
$\gamma$ at some point, and b is the binormal vector at that point. Indeed,
when $n=1$, the mean curvature vector of $\gamma$ at a point is given as
$H=\kappa\textbf{n}$, with n being the outward normal vector at that point, so
we have (1.22):
$\partial_{t}\gamma=-J(\kappa\textbf{n})=\kappa\textbf{t}\times\textbf{n}=\kappa\textbf{b}$
(t is the tangent vector). However, it is not clear whether this dynamics of
point vortices can be well-approximated by that of smooth solutions.

#### Comparison with 2D and 3D Euler equations

Figure 1: Overall flow pattern for signed vorticity: $\mathbb{R}^{2}$ under
odd-odd symmetry (left), $\mathbb{R}^{3}$ under axisymmetry and odd symmetry
in $z$ (center), $\mathbb{R}^{4}$ under bi-rotational symmetry (right).

The bi-rotational Euler system can be considered in the first quadrant
$\\{(r,s):r,s\geq 0\\}$ with $r=0$ and $s=0$ as rotational axes. This is
similar with 2D Euler with odd-odd symmetry and 3D axisymmetric, swirl-free
Euler with odd symmetry in $z$, because under such symmetry assumptions, the
domain again reduces to the first quadrant. Furthermore, under the additional
assumption that the vorticity is signed, then in all three cases, it can be
shown that in an averaged sense, the overall fluid flow is directed southeast;
see Figure 1. In 2D Euler with odd-odd symmetry, Iftimie–Sideris–Gamblin [20]
proved that the horizontal center of mass grows linearly in time, using
single-signed vorticity. Similarly, in 3D axisymmetric Euler without swirl and
with $z$-odd symmetry, the first and the second author [6] showed sublinear
growth of impulse of a single-signed vorticity in time. This was enhanced by
Gustafson–Miller–Tsai [16]. One of the main difference from axisymmetric Euler
in 4D is that there seems to be no time-conserved quantity corresponding to
angular momentum.

We would like to point out an additional difference of bi-rotational symmetry
to axisymmetry: the variables $r$ and $s$ are symmetric, as in the case of 2D
Euler. Therefore, one may consider solutions which satisfy an additional odd
symmetry assumption across the diagonal line $\\{r=s\\}$. If one furthermore
considers signed vorticity (namely, $w\geq 0$ in the region $\\{s>r\\}$), then
the corresponding flow pattern looks like the one depicted in Figure 2.

Figure 2: Overall flow pattern for signed vorticity under an additional odd
symmetry

###  1.5 Outline of the paper

The rest of the paper is organized as follows. In section 2, we provide basic
definition and properties of the Lorentz space. Then, we introduce the
vorticity tensor in the bi-rotational geometry and derive the Biot–Savart law
of the bi-rotational Euler flow. In section 3, we conduct estimates of the
Biot–Savart kernel that was obtained in section 2 to show Theorem 1.1, the
local regularity of $w$. In section 4, the last section of this paper, we
prove Theorem 1.3, the global regularity of $w$, by doing finer estimate of
the Biot–Savart kernel.

##  2 Preliminaries

In this section, after giving a brief review of Lorentz spaces, we derive the
Biot–Savart law for the bi-rotational Euler equations.

###  2.1 Lorentz spaces

We briefly review the definition of Lorentz spaces and a few basic properties;
a classical reference is [15].

######  Definition 2.1.

Given a function $f:\mathbb{R}^{n}\rightarrow\mathbb{R}$, we define
$f^{*}:\mathbb{R}_{+}\rightarrow\mathbb{R}_{+}$ by
$f^{*}(t)=\inf\\{t>0:|\\{x\in\mathbb{R}^{n}:|f(x)|>\lambda\\}|\leq t\\}.$ We
then say $f\in L^{p,q}$ for $1\leq p,q\leq\infty$ if the quantity

$\begin{split}\|f\|_{L^{p,q}}=\begin{cases}\left(\int_{0}^{\infty}(t^{\frac{1}{p}}f^{*}(t))^{q}\frac{dt}{t}\right)^{\frac{1}{q}},\quad&q<\infty,\\\
\sup_{t>0}\,t^{\frac{1}{p}}f^{*}(t),\quad&q=\infty.\end{cases}\end{split}$

is finite. We write $L^{p,q}$ as the set of all measurable functions with
$\|f\|_{L^{p,q}}<\infty$.

Although the spaces can be defined even when either $p<1$ or $q<1$, we shall
not be concerned with such cases. In the following we shall list a few
properties which will be useful. The proofs can be found in [15, Section 1.4]
and the references therein.

######  Proposition 2.2.

The spaces $L^{p,q}$ satisfy the following properties:

  * •

The space $L^{p,p}$ coincides with the usual $L^{p}$ space.

  * •

The functional $\|\cdot\|_{L^{p,q}}$ is a quasi-norm; that is,
$\|f+g\|_{L^{p,q}}\leq C_{p,q}\left(\|f\|_{L^{p,q}}+\|g\|_{L^{p,q}}\right).$

  * •

With $1/p+1/p^{*}=1$ and $1/q+1/q^{*}=1$, we have
$(L^{p,q})^{*}=L^{p^{*},q^{*}}$ for any $p>1$ and $1\leq q<+\infty$.

  * •

We have the Hölder’s inequality $\|fg\|_{L^{r,s}}\leq
C_{p,q,s_{1},s_{2}}\|f\|_{L^{p,s_{1}}}\|g\|_{L^{q,s_{2}}}$ for any $1\leq
p,q,r,s_{1},s_{2}\leq+\infty$ satisfying
$\frac{1}{p}+\frac{1}{q}=\frac{1}{r},\,\frac{1}{s_{1}}+\frac{1}{s_{2}}=\frac{1}{s}.$

###  2.2 Vorticity formulation of the Euler equations

Introducing the vorticity tensor $\omega=(\omega^{i,j})_{1\leq i,j\leq n}$
where $\omega^{i,j}=\partial_{x_{j}}u^{i}-\partial_{x_{i}}u^{j},$ (1.1)
becomes

$\begin{split}\partial_{t}\omega^{i,j}+u\cdot\nabla\omega^{i,j}=-\sum_{k=1}^{n}\left(\omega^{k,j}\partial_{x_{k}}u^{i}-\omega^{k,i}\partial_{x_{k}}u^{j}\right).\end{split}$
(2.1)

The components of $\omega$ are subject to the following consistency relations:

$\begin{split}\partial_{x_{k}}\omega^{i,j}+\partial_{x_{i}}\omega^{j,k}+\partial_{x_{j}}\omega^{k,i}=0.\end{split}$
(2.2)

To close the system, we introduce $\Psi^{i,j}=\Delta^{-1}\omega^{i,j}$: then,
one can see that $u^{i}=\sum_{k=1}^{n}\partial_{x_{k}}\Psi^{i,k}$ and

$\begin{split}\partial_{x_{j}}u^{i}=-\sum_{k=1}^{n}R_{jk}\omega^{i,k}\end{split}$
(2.3)

where $R_{jk}=\partial_{x_{j}}\partial_{x_{k}}(-\Delta)^{-1}$. The system
(2.1) supplemented with (2.2) and (2.3) is the vorticity form of the
incompressible Euler equations ([4]).

###  2.3 Vorticity tensor in the bi-rotational geometry and derivation of the
Biot–Savart law

In the bi-rotational case, the vorticity tensor can be written as a linear
combination of $w$ and the derivatives of $u^{\theta}$ and $u^{\phi}$, with
coefficients depending on $\theta$ and $\phi$. The formulas are recorded in
the Appendix A. In the no-swirl case, that is, when $u^{\theta}=u^{\phi}=0$,
we have that $\omega^{1,2}=\omega^{3,4}=0$ and

$\begin{split}\omega^{1,3}=-\cos\theta\cos\phi
w,\quad\omega^{1,4}=-\cos\theta\sin\phi
w,\quad\omega^{2,3}=-\sin\theta\cos\phi
w,\quad\omega^{2,4}=-\sin\theta\sin\phi w.\end{split}$

We are now ready to obtain the kernel expressions for the “radial” velocities
$u^{r}$ and $u^{s}$ in terms of $w$, which will be our Biot-Savart law.
(Observe that $u^{r}$ and $u^{s}$ do not depend on $u^{\theta}$ and
$u^{\phi}$; we shall therefore assume that $u^{\theta}=u^{\phi}=0$.) Our
starting point is

$\begin{split}u^{i}=\sum_{j=1}^{4}\partial_{x_{j}}\Delta^{-1}\omega^{i,j}=\sum_{j=1}^{4}\Delta^{-1}\partial_{x_{j}}\omega^{i,j}.\end{split}$

Let us recall that $\Delta^{-1}$ is defined in $\mathbb{R}^{4}$ by convolution
against the kernel

$\begin{split}G(x)=-\frac{1}{4\pi^{2}}|x|^{-2}.\end{split}$

Hence the kernel for $\partial_{x_{j}}\Delta^{-1}$ is given by
$K=(K_{j})_{1\leq j\leq 4}$ with

$\begin{split}K_{j}(x)=\frac{1}{2\pi^{2}}\frac{x_{j}}{|x|^{4}}.\end{split}$

We now compute, using the formulas in the previous section,

$\begin{split}\partial_{x_{3}}\omega^{2,3}&=-\sin\theta\cos^{2}\phi\partial_{s}w-\sin\theta\sin^{2}\phi\frac{w}{s},\qquad\partial_{x_{4}}\omega^{2,4}=-\sin\theta
w\sin^{2}\phi\partial_{s}w-\sin\theta\cos^{2}\phi\frac{w}{s}.\end{split}$

In particular,

$\begin{split}\frac{u^{2}}{x_{2}}=-\frac{1}{x_{2}}\Delta^{-1}\left(\sin\theta(\partial_{s}w+\frac{w}{s})\right)=-\frac{1}{r}\widetilde{\Delta}^{-1}\left(\partial_{s}w+\frac{w}{s}\right)=\frac{u^{1}}{x_{1}}.\end{split}$

Here,

$\begin{split}\widetilde{\Delta}=\partial_{r}^{2}+\frac{\partial_{r}}{r}-\frac{1}{r^{2}}+\partial_{s}^{2}+\frac{\partial_{s}}{s}-\frac{1}{s^{2}}\end{split}$

Similarly, we obtain

$\begin{split}\frac{u^{3}}{x_{3}}=\frac{1}{x_{3}}\Delta^{-1}\left(\cos\theta(\partial_{r}w+\frac{w}{r})\right)=\frac{1}{s}\widetilde{\Delta}^{-1}\left(\partial_{r}w+\frac{w}{r}\right)=\frac{u^{4}}{x_{4}}.\end{split}$

Note that, under the no-swirl assumption,

$\begin{split}\frac{u^{2}}{x_{2}}=\frac{u^{1}}{x_{1}}=\frac{u^{r}}{r},\quad\frac{u^{3}}{x_{3}}=\frac{u^{4}}{x_{4}}=\frac{u^{s}}{s}.\end{split}$

It follows that

$\begin{split}\frac{u^{r}}{r}+\frac{u^{s}}{s}=-\frac{1}{r}\widetilde{\Delta}^{-1}\left(\partial_{s}w+\frac{w}{s}\right)+\frac{1}{s}\widetilde{\Delta}^{-1}\left(\partial_{r}w+\frac{w}{r}\right)\end{split}$

Based on the computations above, we shall derive the following expression for
the kernel of $u^{r}$ and $u^{s}$. We denote
$\Pi:=\\{(r,s)\in\mathbb{R}^{2}:r,s\geq 0\\}$.

######  Lemma 2.3.

We have

$u^{r}(r,s)=-\iint_{\Pi}F^{r}(r,s,\overline{r},\overline{s})w(\overline{r},\overline{s})d\overline{s}d\overline{r},$
(2.4)

with

$\displaystyle F^{r}(r,s,\overline{r},\overline{s})$
$\displaystyle=\frac{2}{\pi^{2}}\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}\overline{s}\cos\overline{\theta}(\overline{s}-s\cos\overline{\phi})}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})+2s\overline{s}(1-\cos\overline{\phi})]^{2}}d\overline{\phi}d\overline{\theta},$
(2.5)

and

$u^{s}(r,s)=\iint_{\Pi}F^{s}(r,s,\overline{r},\overline{s})w(\overline{r},\overline{s})d\overline{s}d\overline{r},$
(2.6)

with
$F^{s}(s,r,\overline{s},\overline{r})=F^{r}(r,s,\overline{r},\overline{s}).$

###### Proof of Lemma 2.3.

Let us briefly comment on the notations: $x,\bar{x}\in\mathbb{R}^{4}$ will be
expressed by $(r,\theta,s,\phi)$ and
$(\bar{r},\bar{\theta},\bar{s},\bar{\phi})$ in the bi-polar coordinates,
respectively. Recall from above that, in the rectangular coordinate system,

$\begin{split}(\frac{u^{r}}{r})(x)=-\frac{1}{4\pi^{2}x_{1}}\int_{\mathbb{R}^{4}}\frac{\cos\bar{\theta}}{|x-\bar{x}|^{2}}\left(\partial_{s}w+\frac{w}{\bar{s}}\right)d\bar{x}.\end{split}$

Rewriting in bi-polar coordinates and integrating by parts give that

$\begin{split}\frac{1}{x_{1}}\int_{\mathbb{R}^{4}}\frac{\cos\bar{\theta}}{|x-\bar{x}|^{2}}\left(\partial_{s}w+\frac{w}{\bar{s}}\right)d\bar{x}&=\frac{1}{r\cos\theta}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{2\pi}\cos\bar{\theta}\left(-\partial_{\bar{s}}\left(\frac{\bar{r}\bar{s}}{|x-\bar{x}|^{2}}\right)+\frac{\bar{r}}{|x-\bar{x}|^{2}}\right)w(\bar{r},\bar{s})d\bar{\theta}d\bar{\phi}d\bar{r}d\bar{s}\\\
&=\frac{1}{r\cos\theta}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{2\pi}2\bar{r}\bar{s}\cos\bar{\theta}\frac{\bar{s}-s\cos(\bar{\phi}-\phi)}{|x-\bar{x}|^{4}}w(\bar{r},\bar{s})d\bar{\theta}d\bar{\phi}d\bar{r}d\bar{s}.\end{split}$

Note that
$|x-\bar{x}|^{2}=r^{2}+\bar{r}^{2}-2r\bar{r}\cos(\bar{\theta}-\theta)+s^{2}+\bar{s}^{2}-2s\bar{s}\cos(\bar{\phi}-\phi)$
and a change of variables gives that

$\begin{split}(\frac{u^{r}}{r})(x)&=\frac{1}{4\pi^{2}r\cos\theta}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{2\pi}\int_{0}^{2\pi}2\bar{r}\bar{s}\frac{(\cos\bar{\theta}\cos\theta-\sin\bar{\theta}\sin\theta)(\bar{s}-s\cos\bar{\phi})}{(r^{2}+\bar{r}^{2}-2r\bar{r}\cos\bar{\theta}+s^{2}+\bar{s}^{2}-2s\bar{s}\cos\bar{\phi})^{2}}w(\bar{r},\bar{s})d\bar{\theta}d\bar{\phi}d\bar{r}d\bar{s}\\\
&=\frac{2}{\pi^{2}r}\int_{0}^{\infty}\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{\pi}\frac{2\bar{r}\bar{s}\cos\bar{\theta}(\bar{s}-s\cos\bar{\phi})}{(r^{2}+\bar{r}^{2}-2r\bar{r}\cos\bar{\theta}+s^{2}+\bar{s}^{2}-2s\bar{s}\cos\bar{\phi})^{2}}w(\bar{r},\bar{s})d\bar{\theta}d\bar{\phi}d\bar{r}d\bar{s}.\end{split}$

We have used that the integral term involving $\sin\theta$ vanishes after
integrating in $\bar{\theta}$. After normalization, this gives the desired
expression in (2.5). The proof for (2.6) is parallel. ∎

##  3 Local regularity of $w$

In this section, we estimate supremums of terms related to the velocity field
$u$ using the Biot–Savart kernels obtained from Lemma 2.3 to prove Theorem
1.1.

###  3.1 Estimates based on the Biot-Savart law

First, we present the following lemma, which says that supremums of $u$,
$r^{-1}u^{r}$, and $s^{-1}u^{s}$ are bounded by $L^{4,1}-$norms of $w$,
$r^{-1}w$, and $s^{-1}w$, respectively.

######  Lemma 3.1.

We have the estimates

$\begin{split}\|u^{r}\|_{L^{\infty}},\|u^{s}\|_{L^{\infty}}\leq
C\|w\|_{L^{4,1}}.\end{split}$ (3.1)
$\begin{split}\|r^{-1}u^{r}\|_{L^{\infty}}\leq
C\|r^{-1}w\|_{L^{4,1}},\quad\|s^{-1}u^{s}\|_{L^{\infty}}\leq
C\|s^{-1}w\|_{L^{4,1}}.\end{split}$ (3.2)

###### Proof.

Let us prove the estimate of $\|u^{r}\|_{L^{\infty}}$ and
$\|r^{-1}u^{r}\|_{L^{\infty}}$. We consider the kernel $F^{r}$:

$\begin{split}-2\pi^{2}F^{r}=\overline{r}\overline{s}\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{\cos\overline{\theta}(\overline{s}-s\cos\overline{\phi})}{|x-\bar{x}|^{4}}d\overline{\theta}d\overline{\phi}.\end{split}$

To prove (3.1), we just note that

$\begin{split}|F^{r}|\leq
C\int_{0}^{2\pi}\int_{0}^{2\pi}\frac{\overline{r}\overline{s}}{|x-\bar{x}|^{3}}d\overline{\theta}d\overline{\phi}.\end{split}$

which follows from $\overline{s}-s\cos\overline{\phi}\leq|x-\bar{x}|$. Then

$\begin{split}|u^{r}|\leq
C\int_{\mathbb{R}^{4}}\frac{1}{|x-\bar{x}|^{3}}|w|d\bar{x}\leq
C\|w\|_{L^{4,1}}\end{split}$

since $\bar{x}\mapsto\frac{1}{|x-\bar{x}|^{3}}$ belongs to
$L^{\frac{4}{3},\infty}$ for any $x$.

To prove (3.2), we return to the expression for $F^{r}$ and split the
$\overline{\theta}$-integral into $[-\frac{\pi}{2},\frac{\pi}{2})$ and
$[\frac{\pi}{2},\frac{3\pi}{2})$, and then make a change of variable
$\overline{\theta}\to\overline{\theta}-\pi$ in the second region. Defining
$\hat{x}$ to be the point corresponding to $(r,\pi,s,0)$, we have that

$\begin{split}-2\pi^{2}F^{r}=2\overline{r}\overline{s}\int_{0}^{2\pi}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\frac{r\overline{r}\cos^{2}\overline{\theta}(\overline{s}-s\cos\overline{\phi})}{|x-\bar{x}|^{4}|\hat{x}-\bar{x}|^{4}}\left(|x-\bar{x}|^{2}+|\hat{x}-\bar{x}|^{2}\right)d\overline{\theta}d\overline{\phi}.\end{split}$

This gives

$\begin{split}u^{r}(r,s)=c\int_{\mathbb{R}^{4}\cap\\{-\frac{\pi}{2}\leq\theta<\frac{\pi}{2}\\}}\frac{r\overline{r}\cos^{2}\overline{\theta}(\overline{s}-s\cos\overline{\phi})}{|x-\bar{x}|^{4}|\hat{x}-\bar{x}|^{4}}\left(|x-\bar{x}|^{2}+|\hat{x}-\bar{x}|^{2}\right)w(\overline{r},\overline{s})d\bar{x}.\end{split}$

First, noting that $|\overline{s}-s\cos\overline{\phi}|\leq|x-\bar{x}|$ and
$|x-\bar{x}|\leq|\hat{x}-\bar{x}|$,

$\begin{split}\frac{|u^{r}(r,s)|}{r}\leq
C\int_{\mathbb{R}^{4}\cap\\{-\frac{\pi}{2}\leq\theta<\frac{\pi}{2}\\}}\frac{\overline{r}}{|x-\bar{x}|^{3}|\hat{x}-\bar{x}|^{2}}|w(\overline{r},\overline{s})|d\bar{x}.\end{split}$

We now consider two cases: (i) $r\leq|x-\bar{x}|$ and (ii) $r>|x-\bar{x}|$. In
the case (i), we have

$\begin{split}\overline{r}=\overline{r}-r+r\leq 2|x-\bar{x}|\leq
2|\hat{x}-\bar{x}|\end{split}$

and hence

$\begin{split}\frac{\overline{r}}{|x-\bar{x}|^{3}|\hat{x}-\bar{x}|^{2}}\leq\frac{1}{|x-\bar{x}|^{3}\overline{r}}.\end{split}$

In the case (ii), we have $r^{2}\geq
r^{2}+\overline{r}^{2}-2r\overline{r}\cos\overline{\theta},$ which gives
$2r\cos\overline{\theta}\geq\overline{r}$ and hence

$\overline{r}\leq\sqrt{r^{2}+\overline{r}^{2}+2r\overline{r}\cos\overline{\theta}}\leq|\hat{x}-\bar{x}|,$

which gives the same bound. Therefore,

$\begin{split}\frac{|u^{r}(r,s)|}{r}\leq
C\int_{\mathbb{R}^{4}}\frac{1}{|x-\bar{x}|^{3}}\frac{w}{\overline{r}}d\bar{x}\leq
C\|r^{-1}w\|_{L^{4,1}}\end{split}$

since the function $\bar{x}\mapsto\frac{1}{|x-\bar{x}|^{3}}$ belongs to
$L^{\frac{4}{3},\infty}$ for any $x$. ∎

###  3.2 Local well-posedness in the no-swirl case

In this section, we prove a well-posedness result for the no-swirl system
(1.12):

$\begin{split}\partial_{t}w+(u^{r}\partial_{r}+u^{s}\partial_{s})w=\left(\frac{u^{r}}{r}+\frac{u^{s}}{s}\right)w,\end{split}$

supplemented with (1.5) and (1.6). In the remainder of this section, we shall
write $u=(u^{r},u^{s})$ and
$u\cdot\nabla=u^{r}\partial_{r}+u^{s}\partial_{s}$. Under the assumption $w\in
L^{4,1}(\mathbb{R}^{4})$ and $w\in L^{\infty}(\mathbb{R}^{4})$,
$u\in\mathrm{logLip}$ and therefore the flow map is well-defined as a measure-
preserving homeomorphism of $\mathbb{R}^{4}$. We borrow a proposition from [9]
which works in this setting:

######  Proposition 3.2 ([9, Prop. 2]).

Let $u$ be a divergence-free vector field in $\mathbb{R}^{4}$ and $h$ be a
solution to

$\begin{split}\partial_{t}h+u\cdot\nabla h=f\end{split}$

on some time interval $[0,T]$. Then for any $1\leq p,q\leq+\infty$,

$\begin{split}\|h(t)\|_{L^{p,q}}\leq\|h_{0}\|_{L^{p,q}}+\int_{0}^{t}\|f(\tau)\|_{L^{p,q}}d\tau,\end{split}$

for any $0<t\leq T$.

We now give the proof of Theorem 1.1.

###### Proof of Theorem 1.1.

As we are concerned with solutions having uniformly bounded vorticity and
velocity, uniqueness of the solution within such class is guaranteed in [9,
Section 3]. Moreover, once a priori estimates are given, existence of a
solution can be proved using standard smoothing procedure. Therefore let us
just prove the necessary a priori estimates. From the equation for $w$ and
$w/(rs)$, we have

$\begin{split}\frac{d}{dt}\|w\|_{L^{\infty}}\leq(\|r^{-1}u^{r}\|_{L^{\infty}}+\|s^{-1}u^{s}\|_{L^{\infty}})\|w\|_{L^{\infty}}\leq
C(\|r^{-1}w\|_{L^{4,1}}+\|s^{-1}w\|_{L^{4,1}})\|w\|_{L^{\infty}},\quad\bigg{\|}\frac{w(t)}{rs}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})}=\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})},\end{split}$
(3.3)

and similarly

$\begin{split}\frac{d}{dt}\|w\|_{L^{4,1}}\leq
C(\|r^{-1}w\|_{L^{4,1}}+\|s^{-1}w\|_{L^{4,1}})\|w\|_{L^{4,1}}\end{split}$
(3.4)

This shows that formally the norms $\|w\|_{L^{\infty}}$ and $\|w\|_{L^{4,1}}$
remain finite as long as $\|r^{-1}w\|_{L^{4,1}}+\|s^{-1}w\|_{L^{4,1}}$ does
not blow up. Now consider the equation for $\frac{w}{r}$ and $\frac{w}{s}$:

$\displaystyle\partial_{t}\frac{w}{r}+(u^{r}\partial_{r}+u^{s}\partial_{s})\frac{w}{r}$
$\displaystyle=\frac{u^{s}}{s}\frac{w}{r},$ (3.5)
$\displaystyle\partial_{t}\frac{w}{s}+(u^{r}\partial_{r}+u^{s}\partial_{s})\frac{w}{s}$
$\displaystyle=\frac{u^{r}}{r}\frac{w}{s}.$ (3.6)

Estimating similarly as in the above, we have

$\begin{split}\frac{d}{dt}\|r^{-1}w\|_{L^{4,1}}&\leq
C\|s^{-1}w\|_{L^{4,1}}\|r^{-1}w\|_{L^{4,1}},\qquad\frac{d}{dt}\|s^{-1}w\|_{L^{4,1}}\leq
C\|r^{-1}w\|_{L^{4,1}}\|s^{-1}w\|_{L^{4,1}}.\end{split}$

This gives

$\begin{split}\frac{d}{dt}(\|r^{-1}w\|_{L^{4,1}}+\|s^{-1}w\|_{L^{4,1}})\leq
C(\|r^{-1}w\|_{L^{4,1}}+\|s^{-1}w\|_{L^{4,1}})^{2}.\end{split}$

Therefore, for some interval of time $[0,T]$ with $T$ depending only on
$\|r^{-1}w_{0}\|_{L^{4,1}}+\|s^{-1}w_{0}\|_{L^{4,1}}$,
$\|r^{-1}w\|_{L^{4,1}}+\|s^{-1}w\|_{L^{4,1}}$ remains bounded. Finally, we
note that on the same time interval, $\|u\|_{L^{\infty}}\leq C\|w\|_{L^{4,1}}$
remains finite as well. ∎

######  Remark 3.3.

We point out that the solution is global in time once $\mathbb{R}^{4}$ is
replaced with a bounded domain $\Omega$ satisfying bi-rotational symmetry, for
example the $4$-ball defined by $\\{x\in\mathbb{R}^{4}:|x|\leq 1\\}$.
Actually, as we shall see below it is only necessary to be bounded either in
$r$ or $s$. One needs to add the condition that $\frac{w_{0}}{rs}\in
L^{4,1}(\mathbb{R}^{4})$. Then, there exists a unique global-in-time solution
$w(t)$ to (1.12). To see this, we estimate $r^{-1}w$ as follows:

$\begin{split}\frac{d}{dt}\left\|\frac{w(t)}{r}\right\|_{L^{4,1}}&\leq\left\|\frac{u^{s}(t)}{s}\frac{w(t)}{r}\right\|_{L^{4,1}},\end{split}$

and we bound

$\begin{split}\left\|\frac{u^{s}(t)}{s}\frac{w(t)}{r}\right\|_{L^{4,1}}&=\left\|u^{s}(t)\frac{w(t)}{rs}\right\|_{L^{4,1}}\leq
C\|u(t)\|_{L^{\infty}}\left\|\frac{w_{0}}{rs}\right\|_{L^{4,1}}.\end{split}$

On the other hand,

$\begin{split}\|u(t)\|_{L^{\infty}}\leq\|w(t)\|_{L^{4,1}}\leq
CR\left\|\frac{w(t)}{r}\right\|_{L^{4,1}}\end{split}$

where $R$ is the radius of $\Omega$ in the $r$-direction. Hence

$\begin{split}\left\|\frac{w(t)}{r}\right\|_{L^{4,1}}\leq\left\|\frac{w_{0}}{r}\right\|_{L^{4,1}}\exp\left(CR\left\|\frac{w_{0}}{rs}\right\|_{L^{4,1}}\right).\end{split}$
(3.7)

Then, it is easy to see that $\|s^{-1}w\|_{L^{4,1}}$ and therefore
$\|w\|_{L^{\infty}}$ remains bounded as well. Of course, the estimates for the
velocity needs to be re-proved since the domain is not $\mathbb{R}^{4}$
anymore. This can be done following [9, 10] and we omit the details.

##  4 Global regularity of $w$

In this section, we present key lemmas and estimates to prove Theorem 1.3.

###  4.1 Some technical lemma

Firstly, we introduce the following lemma which is frequently used in the
proof of Theorem 1.3. We denote

$f_{a}(\tau):=\int_{0}^{\pi}\frac{1}{[2(1-\cos\theta)+\tau]^{a}}d\theta,\quad\tau>0,\
a>\frac{1}{2}.$ (4.1)

######  Lemma 4.1.

$f_{a}$ from (4.1) satisfies

$f_{a}(\tau)\lesssim_{a}\min\bigg{\\{}\frac{1}{\tau^{a-1/2}},\frac{1}{\tau^{a}}\bigg{\\}},\quad\tau>0.$
(4.2)

###### Proof.

The upper bound $f_{a}(\tau)\lesssim\tau^{-a}$ immediately follows from the
form of $f_{a}$, so it suffices to show

$\tau^{a-1/2}f_{a}(\tau)\lesssim_{a}1,\quad 0<\tau<1.$

Before proving this, let us take a sufficiently small $\varepsilon_{0}>0$ that
satisfies

$2(1-\cos\theta)\geq\frac{\theta^{2}}{2},\quad\theta\in[0,\varepsilon_{0}].$

We fix $\tau\in(0,1)$ and split $\tau^{a-1/2}f_{a}(\tau)$ into

$\tau^{a-1/2}f_{a}(\tau)=\underbrace{\int_{0}^{\varepsilon_{0}}\frac{\tau^{a-1/2}}{[2(1-\cos\theta)+\tau]^{a}}d\theta}_{=(A)}+\underbrace{\int_{\varepsilon_{0}}^{\pi}\frac{\tau^{a-1/2}}{[2(1-\cos\theta)+\tau]^{a}}d\theta}_{=(B)}.$

For $(A)$, we use the change of variables $\alpha=\frac{\theta}{\sqrt{2\tau}}$
to get

$(A)\simeq\int_{0}^{\frac{\varepsilon_{0}}{\sqrt{2\tau}}}\frac{1}{(\alpha^{2}+1)^{a}}d\alpha<\infty.$

For $(B)$, we use the fact that $2(1-\cos\theta)\geq C$ for some $C>0$ when
$\theta\in[\varepsilon_{0},\pi]$ to get

$(B)\leq\int_{\varepsilon_{0}}^{\pi}\frac{\tau}{C^{a}}d\theta<\infty.$

Thus, we get $f_{a}(\tau)\lesssim_{a}1/\tau^{a-1/2},$ which finishes the
proof. ∎

###  4.2 Estimates of $\|w(t)\|_{L^{\infty}(\mathbb{R}^{4})}$,
$\|rs^{-1}w(t)\|_{L^{1}(\mathbb{R}^{4})}$, and
$\|r^{-1}sw(t)\|_{L^{1}(\mathbb{R}^{4})}$

In this subsection, we will show estimates of terms
$\|w(t)\|_{L^{\infty}(\mathbb{R}^{4})}$,
$\|rs^{-1}w(t)\|_{L^{1}(\mathbb{R}^{4})}$, and
$\|r^{-1}sw(t)\|_{L^{1}(\mathbb{R}^{4})}$ in time. The following lemma says
that if $w_{0}$ has decay near the axes $r=0$ and $s=0$, then the supremum of
$w(t)$ is bounded by the “length” function $L$ defined in (4.4).

######  Lemma 4.2.

Let $w_{0}$ satisfy $(1+r+s+rs)w_{0}/(rs)\in L^{\infty}(\mathbb{R}^{4})$. Then
we have

$\|w(t)\|_{L^{\infty}(\mathbb{R}^{4})}\lesssim\bigg{[}\|w_{0}\|_{L^{\infty}(\mathbb{R}^{4})}+\bigg{\|}\frac{w_{0}}{r}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}+\bigg{\|}\frac{w_{0}}{s}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}+\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}\bigg{]}\cdot\bigg{[}1+\int_{0}^{t}(\|u^{r}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}+\|u^{s}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})})d\tau\bigg{]}^{2}.$
(4.3)

###### Proof.

We fix $t\geq 0$ and define

$L(t):=1+\int_{0}^{t}[\|u^{r}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}+\|u^{s}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}]d\tau.$
(4.4)

In addition, we denote $\Phi_{t}$ as the flow map that solves the ODE

$\frac{d}{dt}\Phi_{t}(x)=u(t,\Phi_{t}(x)),\quad\Phi_{0}(x)=x.$

Also, we use the notation $r_{x}:=\sqrt{x_{1}^{2}+x_{1}^{2}}$,
$s_{x}:=\sqrt{x_{3}^{2}+x_{4}^{2}}$, and denote $\Phi_{t}^{r}$ and
$\Phi_{t}^{s}$ as the $r$ and $s$-component of $\Phi_{t}$, respectively.

We take $y=\Phi_{t}^{-1}(x)$, where $\Phi_{t}^{-1}$ is the inverse of
$\Phi_{t}$. Then note that we have

$x=\Phi_{t}(y)=\Phi_{0}(y)+\int_{0}^{t}u(\tau,\Phi_{\tau}(y))d\tau=y+\int_{0}^{t}u(\tau,\Phi_{\tau}(y))d\tau,$

which gives us

$\begin{split}r_{x}&=r_{y}+\int_{0}^{t}u^{r}(\tau,\Phi_{\tau}(y))d\tau,\quad
s_{x}=s_{y}+\int_{0}^{t}u^{s}(\tau,\Phi_{\tau}(y))d\tau.\end{split}$

Also, we use the conservation of the relative vorticity $\frac{w}{rs}$ in time
along the flow $\Phi_{t}$:

$\frac{w(t,x)}{r_{x}s_{x}}=\frac{w(t,\Phi_{t}(y))}{\Phi_{t}^{r}(y)\Phi_{t}^{s}(y)}=\frac{w_{0}(y)}{r_{y}s_{y}}.$

Then we have

$\begin{split}|w(t,x)|&=r_{x}s_{x}\cdot\frac{|w(t,x)|}{r_{x}s_{x}}=\frac{r_{x}s_{x}}{r_{y}s_{y}}\cdot|w_{0}(y)|=\bigg{(}1+\frac{1}{r_{y}}\int_{0}^{t}u^{r}(\tau,\Phi_{\tau}(y))d\tau\bigg{)}\bigg{(}1+\frac{1}{s_{y}}\int_{0}^{t}u^{s}(\tau,\Phi_{\tau}(y))d\tau\bigg{)}\cdot|w_{0}(y)|\\\
&\leq\bigg{(}1+\frac{1}{r_{y}}\int_{0}^{t}\|u^{r}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}d\tau\bigg{)}\bigg{(}1+\frac{1}{s_{y}}\int_{0}^{t}\|u^{s}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}d\tau\bigg{)}\cdot|w_{0}(y)|\\\
&\leq\|w_{0}\|_{L^{\infty}(\mathbb{R}^{4})}+\bigg{(}\bigg{\|}\frac{w_{0}}{r}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}+\bigg{\|}\frac{w_{0}}{s}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}\bigg{)}\cdot
L(t)+\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}\cdot
L(t)^{2}.\end{split}$

∎

Next, the lemma below says that if the relative vorticity $w_{0}/(rs)$ has a
weak decay at infinity, then the $L^{1}-$norms of $rs^{-1}w(t)$ and
$r^{-1}sw(t)$ are controlled by the length function $L$ from (4.4).

######  Lemma 4.3.

Let $w_{0}$ satisfy $(1+r^{2}+s^{2})w_{0}/(rs)\in L^{1}(\mathbb{R}^{4})$. Then
for any $t\geq 0$, we have

$\displaystyle\bigg{\|}\frac{rw(t)}{s}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}$
$\displaystyle\lesssim\bigg{[}\bigg{\|}\frac{rw_{0}}{s}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}+\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}\bigg{]}\cdot\bigg{[}1+\int_{0}^{t}(\|u^{r}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}+\|u^{s}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})})d\tau\bigg{]}^{2},$
(4.5) $\displaystyle\bigg{\|}\frac{sw(t)}{r}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}$
$\displaystyle\lesssim\bigg{[}\bigg{\|}\frac{sw_{0}}{r}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}+\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}\bigg{]}\cdot\bigg{[}1+\int_{0}^{t}(\|u^{r}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})}+\|u^{s}(\tau)\|_{L^{\infty}(\mathbb{R}^{4})})d\tau\bigg{]}^{2}.$
(4.6)

###### Proof of Lemma 4.3.

We use the same notations from the proof of the previous lemma. To prove
(4.5), let us consider the case when $r_{x}\leq L(t)$ or when $r_{x}>L(t)$.
For the first case, we have

$\begin{split}\bigg{\|}\frac{rw(t)}{s}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}&=\int_{\mathbb{R}^{4}}\frac{r_{x}|w(t,x)|}{s_{x}}dx=\int_{\mathbb{R}^{4}}r_{x}^{2}\cdot\frac{|w(t,x)|}{r_{x}s_{x}}dx=\int_{\mathbb{R}^{4}}[\Phi_{t}^{r}(y)]^{2}\cdot\frac{|w(t,\Phi_{t}(y))|}{\Phi_{t}^{r}(y)\Phi_{t}^{s}(y)}dy\\\
&=\int_{\mathbb{R}^{4}}[\Phi_{t}^{r}(y)]^{2}\cdot\frac{|w_{0}(y)|}{r_{y}s_{y}}dy\leq\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}\cdot
L(t)^{2}.\end{split}$

For the other case, we get $r_{y}>1$, which gives us

$\begin{split}\bigg{\|}\frac{rw(t)}{s}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}&=\int_{\mathbb{R}^{4}}[\Phi_{t}^{r}(y)]^{2}\cdot\frac{|w_{0}(y)|}{r_{y}s_{y}}dy\leq\int_{\mathbb{R}^{4}}\bigg{[}1+\frac{L(t)}{r_{y}}\bigg{]}^{2}\cdot\frac{r_{y}|w_{0}(y)|}{s_{y}}dy\lesssim\bigg{\|}\frac{rw_{0}}{s}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}\cdot
L(t)^{2}.\end{split}$

Combining the above two leads to (4.5). The proof of (4.6) is done in the
exact same way. ∎

###  4.3 Estimate of $u^{r}$ and $u^{s}$

Now let us present the key proposition of this section, which is about a time-
independent estimate of the supremum of the velocity field $u$.

######  Proposition 4.4.

We have

$\max\\{\|u^{r}\|_{L^{\infty}(\mathbb{R}^{4})},\|u^{s}\|_{L^{\infty}(\mathbb{R}^{4})}\\}\lesssim\bigg{[}\bigg{\|}\frac{rw}{s}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}+\bigg{\|}\frac{sw}{r}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}\bigg{]}^{1/2}\cdot\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}^{1/2}.$
(4.7)

###### Proof.

For any $\lambda>0$, let us consider $\widetilde{u}^{r},\widetilde{u}^{s}$,
and $\widetilde{w}$ defined as

$\widetilde{u}^{r}(r,s):=u^{r}(\lambda r,\lambda
s),\quad\widetilde{u}^{s}(r,s):=u^{s}(\lambda r,\lambda
s),\quad\widetilde{w}(r,s):=\lambda w(\lambda r,\lambda
s),\quad(r,s)\in\Pi=[0,\infty)^{2},$

where $\widetilde{u}^{r}$ (resp. $\widetilde{u}^{s}$) and $\widetilde{w}$
satisfy the Biot–Savart law (2.4) (resp. (2.6)) as well. Then to show (4.7),
it suffices to prove that for any $(r,s)\in\Pi$ satisfying $r^{2}+s^{2}=1$, we
have

$\max\\{|u^{r}(r,s)|,|u^{s}(r,s)|\\}\lesssim[\|r^{2}w\|_{L^{1}(\Pi)}+\|s^{2}w\|_{L^{1}(\Pi)}]^{1/2}\cdot\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(\Pi)}^{1/2}.$
(4.8)

Indeed, if (4.8) is true, then the same estimate holds for
$\widetilde{u}^{r}$, $\widetilde{u}^{s}$, and $\widetilde{w}$ as well:

$\begin{split}|\widetilde{u}^{r}(r,s)|=|u^{r}(\lambda r,\lambda
s)|,&\quad|\widetilde{u}^{s}(r,s)|=|u^{s}(\lambda r,\lambda s)|,\\\
[\|r^{2}\widetilde{w}\|_{L^{1}(\Pi)}+\|s^{2}\widetilde{w}\|_{L^{1}(\Pi)}]^{1/2}\cdot\bigg{\|}\frac{\widetilde{w}}{rs}\bigg{\|}_{L^{\infty}(\Pi)}^{1/2}&=[\lambda^{-3}\|r^{2}w\|_{L^{1}(\Pi)}+\lambda^{-3}\|s^{2}w\|_{L^{1}(\Pi)}]^{1/2}\cdot\lambda^{3/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(\Pi)}^{1/2}\\\
&=[\|r^{2}w\|_{L^{1}(\Pi)}+\|s^{2}w\|_{L^{1}(\Pi)}]^{1/2}\cdot\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(\Pi)}^{1/2}.\end{split}$

The proof is completed once we take supremum with respect to $\lambda>0$ on
the left-hand side.

Throughout the proof, let us use the following notations

$\begin{split}X_{--}&=X_{--}(r,s,\overline{r},\overline{s},\overline{\theta},\overline{\phi}):=(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})+2s\overline{s}(1-\cos\overline{\phi}),\\\
X_{+-}&=X_{+-}(r,s,\overline{r},\overline{s},\overline{\theta},\overline{\phi}):=(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1+\cos\overline{\theta})+2s\overline{s}(1-\cos\overline{\phi}),\\\
X_{-+}&=X_{-+}(r,s,\overline{r},\overline{s},\overline{\theta},\overline{\phi}):=(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})+2s\overline{s}(1+\cos\overline{\phi}),\\\
X_{++}&=X_{++}(r,s,\overline{r},\overline{s},\overline{\theta},\overline{\phi}):=(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1+\cos\overline{\theta})+2s\overline{s}(1+\cos\overline{\phi}).\end{split}$

We shall frequently use the inequalities

$X_{--}\leq\min\\{X_{+-},X_{-+}\\},\quad\max\\{X_{+-},X_{-+}\\}\leq X_{++},$

and the simple estimates

$\begin{split}|s-\overline{s}|,\quad|r-\overline{r}|\lesssim
X_{--}^{\frac{1}{2}},\qquad r\overline{r}(1-\cos\overline{\theta}),\quad
s\overline{s}(1-\cos\overline{\phi})\lesssim X_{--}.\end{split}$

This holds since each of the four terms in
$X_{--}=(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})+2s\overline{s}(1-\cos\overline{\phi})$
is nonnegative. Moreover, we shall split $F^{r}$ defined in (2.5) into

$\begin{split}F^{r}(r,s,\overline{r},\overline{s})&=(I)+(II),\\\
(I)&:=-\frac{2}{\pi^{2}}\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}\overline{s}\cos\overline{\theta}(\overline{s}-s)}{X_{--}^{2}}d\overline{\phi}d\overline{\theta},\quad(II):=-\frac{2}{\pi^{2}}\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}s\overline{s}\cos\overline{\theta}(1-\cos\overline{\phi})}{X_{--}^{2}}d\overline{\phi}d\overline{\theta}.\end{split}$

We shall frequently use Lemma 4.1 to estimate $(I)$ and $(II)$. More
precisely, we have for instance

$\begin{split}\int_{0}^{\pi}\frac{1}{X_{--}}d\overline{\theta}\leq\frac{1}{2r\bar{r}}\int_{0}^{\pi}\frac{1}{\frac{(r-\overline{r})^{2}+(s-\overline{s})^{2}}{2r\bar{r}}+(1-\cos\overline{\theta})}d\overline{\theta}\lesssim\frac{1}{(r\bar{r})^{\frac{1}{2}}}\frac{1}{((r-\overline{r})^{2}+(s-\overline{s})^{2})^{\frac{1}{2}}}.\end{split}$

Case 1. $1/4\leq r\leq 3/4$.

We split the domain $\Pi$ into

$\begin{split}A_{1}&:=\\{(\overline{r},\overline{s})\in\Pi:r-\frac{1}{8}\leq\overline{r}\leq
r+2,s-\frac{1}{8}\leq\overline{s}\leq s+2\\},\quad A_{2}:=\Pi\setminus
A_{1},\\\
A_{21}&:=A_{2}\cap\\{(\overline{r},\overline{s})\in\Pi:\overline{r}<r+2,\overline{s}<s+2\\},\quad
A_{22}:=\\{(\overline{r},\overline{s})\in\Pi:\overline{r}>r+2,\overline{s}<s+2\\}\subset
A_{2},\\\
A_{23}&:=\\{(\overline{r},\overline{s})\in\Pi:\overline{r}<r+2,\overline{s}>s+2\\}\subset
A_{2},\quad
A_{24}:=\\{(\overline{r},\overline{s})\in\Pi:\overline{r}>r+2,\overline{s}>s+2\\}\subset
A_{2}.\end{split}$

(i) $(\overline{r},\overline{s})\in A_{1}$.

For $(I)$, we have

$\begin{split}|(I)|&\lesssim\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}\overline{s}}{X_{--}^{3/2}}d\overline{\phi}d\overline{\theta}\lesssim\frac{\overline{r}\overline{s}^{1/2}}{s^{1/2}}\int_{0}^{\pi}\frac{1}{(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})}d\overline{\theta}\\\
&\lesssim\bigg{(}\frac{\overline{r}\overline{s}}{rs}\bigg{)}^{1/2}\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}\lesssim\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}.\end{split}$

For $(II)$, we get

$\begin{split}|(II)|&\lesssim\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}[s\overline{s}(1-\cos\overline{\phi})]^{1/2}}{X_{--}^{3/2}}d\overline{\phi}d\overline{\theta}\lesssim\int_{0}^{\pi}\frac{\overline{r}}{(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})}d\overline{\theta}\\\
&\lesssim\frac{\overline{r}^{1/2}}{r^{1/2}}\cdot\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}\lesssim\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}.\end{split}$

Thus, we have

$\begin{split}\bigg{|}\iint_{A_{1}}F^{r}(r,s,\overline{r},\overline{s})w(\overline{r},\overline{s})d\overline{s}d\overline{r}\bigg{|}&\lesssim\iint_{A_{1}}\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}|w(\overline{r},\overline{s})|d\overline{s}d\overline{r}\\\
&\leq\|w\|_{L^{1}(A_{1})}^{1/2}\|w\|_{L^{\infty}(A_{1})}^{1/2}\lesssim\|rsw\|_{L^{1}(A_{1})}^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{1})}^{1/2}{\leq[\|r^{2}w\|_{L^{1}(A_{1})}+\|s^{2}w\|_{L^{1}(A_{1})}]^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{1})}^{1/2}.}\end{split}$

(ii) $(\overline{r},\overline{s})\in A_{2}$.

Note that in this region, the term
$[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}$ has a constant lower
bound: $[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}\gtrsim 1.$ Then
using the triangular inequality, we have

$(\overline{r}^{2}+\overline{s}^{2})^{1/2}\leq[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}+(\underbrace{r^{2}+s^{2}}_{=1})^{1/2}\lesssim[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}.$

Keeping this bound in mind, let us analyze the kernel $F^{r}$. We shall use
the following notations:

$\begin{split}Y_{-}&:=\frac{X_{--}+X_{+-}}{2}=(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}+2s\overline{s}(1-\cos\overline{\phi}),\\\
Y_{+}&:=\frac{X_{-+}+X_{++}}{2}=(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}+2s\overline{s}(1+\cos\overline{\phi}).\end{split}$

Observe that $F^{r}$ can be rewritten as

$\begin{split}F^{r}(r,s,\overline{r},\overline{s})&\simeq\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}\overline{s}\cos\overline{\theta}(\overline{s}-s\cos\overline{\phi})}{X_{--}^{2}}d\overline{\phi}d\overline{\theta}\simeq\int_{0}^{\frac{\pi}{2}}\int_{0}^{\pi}\frac{\overline{r}^{2}\overline{s}\cos^{2}\overline{\theta}(\overline{s}-s\cos\overline{\phi})Y_{-}}{X_{--}^{2}X_{+-}^{2}}d\overline{\phi}d\overline{\theta}\\\
&=\int_{0}^{\frac{\pi}{2}}\int_{0}^{\frac{\pi}{2}}\overline{r}^{2}\overline{s}\cos^{2}\overline{\theta}\bigg{[}\frac{(\overline{s}-s\cos\overline{\phi})Y_{-}}{X_{--}^{2}X_{+-}^{2}}+\frac{(\overline{s}+s\cos\overline{\phi})Y_{+}}{X_{-+}^{2}X_{++}^{2}}\bigg{]}d\overline{\phi}d\overline{\theta}.\end{split}$

When $(\overline{r},\overline{s})\in A_{21}$, we have

$\begin{split}|F^{r}(r,s,\overline{r},\overline{s})|&\lesssim\frac{1}{X_{--}^{2}X_{+-}}+\frac{1}{X_{-+}^{2}X_{++}}\leq\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{3}}.\end{split}$

For the case $(\overline{r},\overline{s})\in A_{22}$, we get

$\begin{split}|F^{r}(r,s,\overline{r},\overline{s})|&\lesssim\frac{\overline{r}^{2}}{X_{--}^{2}X_{+-}}+\frac{\overline{r}^{2}}{X_{-+}^{2}X_{++}}\leq\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{2}}.\end{split}$

If $(\overline{r},\overline{s})\in A_{23}$, then we have

$\begin{split}|F^{r}(r,s,\overline{r},\overline{s})|&\lesssim\frac{\overline{s}^{2}}{X_{--}^{2}X_{+-}}+\frac{\overline{s}^{2}}{X_{-+}^{2}X_{++}}\leq\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{2}}.\end{split}$

Lastly, when $(\overline{r},\overline{s})\in A_{24}$, we get

$\begin{split}|F^{r}(r,s,\overline{r},\overline{s})|&\lesssim\frac{\overline{r}^{2}\overline{s}^{2}}{X_{--}^{2}X_{+-}}+\frac{\overline{r}^{2}\overline{s}^{2}}{X_{-+}^{2}X_{++}}\leq\frac{1}{(r-\overline{r})^{2}+(s-\overline{s})^{2}}.\end{split}$

In sum, in the region $A_{2}$, we obtain

$\begin{split}|F^{r}(r,s,\overline{r},\overline{s})|&\lesssim\frac{1}{(r-\overline{r})^{2}+(s-\overline{s})^{2}}.\end{split}$
(4.9)

Using this, we have by Hölder’s inequality that

$\begin{split}\bigg{|}\iint_{A_{2}}F^{r}(r,s,\overline{r},\overline{s})w(\overline{r},\overline{s})d\overline{s}d\overline{r}\bigg{|}&\lesssim\iint_{A_{2}}\frac{1}{(\overline{r}-1)^{2}+(\overline{s}-1)^{2}}\cdot(\overline{r}\overline{s})^{1/2}\cdot|w(\overline{r},\overline{s})|^{1/2}\cdot\frac{|w(\overline{r},\overline{s})|^{1/2}}{(\overline{r}\overline{s})^{1/2}}d\overline{s}d\overline{r}\\\
&\lesssim\|rsw\|_{L^{1}(A_{2})}^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{2})}^{1/2}{\leq[\|r^{2}w\|_{L^{1}(A_{2})}+\|s^{2}w\|_{L^{1}(A_{2})}]^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{2})}^{1/2}.}\end{split}$
(4.10)

Case 2. $0\leq r<1/4$.

Assume for a moment that $0<r<1/4$. In this case, we split the domain $\Pi$
into

$\begin{split}A_{1}^{\prime}&:=\\{(\overline{r},\overline{s})\in\Pi:0\leq\overline{r}\leq
r+\frac{1}{4},s-\frac{1}{4}\leq\overline{s}\leq s+2\\},\quad
A_{2}^{\prime}:=\Pi\setminus A_{1}^{\prime},\\\
A_{21}^{\prime}&:=\\{(\overline{r},\overline{s})\in\Pi:0\leq\overline{r}\leq
r+\frac{1}{4},0\leq\overline{s}<s-\frac{1}{4}\\}\subset A_{2}^{\prime},\quad
A_{22}^{\prime}:=\\{(\overline{r},\overline{s})\in\Pi:\overline{r}>r+\frac{1}{4},\overline{s}\leq
s+2\\}\subset A_{2}^{\prime},\\\
A_{23}^{\prime}&:=\\{(\overline{r},\overline{s})\in\Pi:0\leq\overline{r}\leq
r+\frac{1}{4},\overline{s}>s+2\\}\subset A_{2}^{\prime},\quad
A_{24}^{\prime}:\\{(\overline{r},\overline{s})\in\Pi:\overline{r}>r+\frac{1}{4},\overline{s}>s+2\\}\subset
A_{2}^{\prime}.\end{split}$

(i) $(\overline{r},\overline{s})\in A_{1}^{\prime}$.

We note that

$\begin{split}\bar{r}\leq\bar{r}^{1/2}(|r-\bar{r}|+r)^{1/2}\leq\bar{r}^{1/2}|r-\bar{r}|^{1/2}+(\bar{r}r)^{1/2}\leq\frac{1}{2}\bar{r}+2|r-\bar{r}|+(\bar{r}r)^{1/2},\end{split}$

which gives $\bar{r}\lesssim|r-\bar{r}|+(\bar{r}r)^{1/2}$. Then for $(I)$, we
have after applying Lemma 4.1 to integral in $\bar{\phi}$,

$\begin{split}|(I)|&\lesssim\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}\overline{s}}{X_{--}^{3/2}}d\overline{\phi}d\overline{\theta}\lesssim\frac{\overline{s}^{1/2}}{s^{1/2}}\int_{0}^{\pi}\frac{|r-\bar{r}|+(\bar{r}r)^{1/2}}{(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})}d\overline{\theta}.\end{split}$

Then, for the first term, we simply use
$|r-\bar{r}|\leq((r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta}))^{1/2}$
and for the second term, Lemma 4.1. This gives

$\begin{split}|(I)|\lesssim\bigg{(}\frac{\overline{s}}{s}\bigg{)}^{1/2}\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}\lesssim\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}.\end{split}$

For $(II)$, we get after applying
$\bar{r}\lesssim|r-\bar{r}|+(\bar{r}r)^{1/2}$,

$\begin{split}|(II)|&\lesssim\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}[s\overline{s}(1-\cos\overline{\phi})]^{1/2}}{X_{--}^{3/2}}d\overline{\phi}d\overline{\theta}\lesssim\int_{0}^{\pi}\frac{|r-\bar{r}|+(\bar{r}r)^{1/2}}{(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})}d\overline{\theta}\lesssim\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}\end{split}$

similarly as in the case of $(I)$. Hence, we obtain

$\begin{split}\bigg{|}\iint_{A_{1}^{\prime}}F^{r}(r,s,\overline{r},\overline{s})w(\overline{r},\overline{s})d\overline{s}d\overline{r}\bigg{|}&\lesssim\iint_{A_{1}^{\prime}}\frac{|w(\overline{r},\overline{s})|}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}d\overline{s}d\overline{r}\leq\|w\|_{L^{1}(A_{1}^{\prime})}^{1/2}\|w\|_{L^{\infty}(A_{1}^{\prime})}^{1/2}\lesssim\|s^{2}w\|_{L^{1}(A_{1}^{\prime})}^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{1}^{\prime})}^{1/2}.\end{split}$
(4.11)

(ii) $(\overline{r},\overline{s})\in A_{2}^{\prime}$.

Note that the term $[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}$ has a
constant lower bound:
$[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}\gtrsim 1.$ Then from the
triangular inequality, we have

$(\overline{r}^{2}+\overline{s}^{2})^{1/2}\leq[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}+(\underbrace{r^{2}+s^{2}}_{=1})^{1/2}\lesssim[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}.$

Using this, we repeat the process from Case 1-(ii) to get

$|F^{r}(r,s,\overline{r},\overline{s})|\lesssim\begin{cases}[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{-3}&\
\text{in}\ A_{21}^{\prime},\\\
[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{-2}&\ \text{in}\
A_{22}^{\prime}\cup A_{23}^{\prime},\\\
[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{-1}&\ \text{in}\
A_{24}^{\prime},\\\ \end{cases}$

which gives us

$\begin{split}\bigg{|}\iint_{A_{2}^{\prime}}F^{r}(r,s,\overline{r},\overline{s})w(\overline{r},\overline{s})d\overline{s}d\overline{r}\bigg{|}&\lesssim\iint_{A_{2}^{\prime}}\frac{1}{(\overline{r}-1)^{2}+(\overline{s}-1)^{2}}\cdot(\overline{r}\overline{s})^{1/2}\cdot|w(\overline{r},\overline{s})|^{1/2}\cdot\frac{|w(\overline{r},\overline{s})|^{1/2}}{(\overline{r}\overline{s})^{1/2}}d\overline{s}d\overline{r}\\\
&\lesssim\|rsw\|_{L^{1}(A_{2}^{\prime})}^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{2}^{\prime})}^{1/2}{\leq[\|r^{2}w\|_{L^{1}(A_{2}^{\prime})}+\|s^{2}w\|_{L^{1}(A_{2}^{\prime})}]^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{2}^{\prime})}^{1/2}.}\end{split}$

Now when $r=0$, for any $(\overline{r},\overline{s})\in\Pi$, we simply have

$\begin{split}F^{r}(0,1,\overline{r},\overline{s})=-\frac{1}{\pi^{2}}\int_{0}^{\pi}\frac{\overline{r}\overline{s}(\overline{s}-\cos\overline{\phi})}{[\overline{r}^{2}+(\overline{s}-1)^{2}+2\overline{s}(1-\cos\overline{\phi})]^{2}}\bigg{(}\int_{0}^{\pi}\cos\overline{\theta}d\overline{\theta}\bigg{)}d\overline{\phi}=0,\end{split}$

which gives us $u^{r}(0,1)=0.$

Case 3. $3/4<r\leq 1$.

In this case, we split the domain $\Pi$ into

$\begin{split}A_{1}^{\prime\prime}&:=\\{(\overline{r},\overline{s})\in\Pi:r-\frac{1}{4}\leq\overline{r}\leq
r+2,0\leq\overline{s}\leq s+\frac{1}{4}\\},\quad
A_{2}^{\prime\prime}:=\Pi\setminus A_{1}^{\prime\prime},\\\
A_{21}^{\prime\prime}&:=\\{(\overline{r},\overline{s})\in\Pi:0\leq\overline{r}<r-\frac{1}{4},0\leq\overline{s}\leq
s+\frac{1}{4}\\}\subset A_{2}^{\prime\prime},\quad
A_{22}^{\prime\prime}:=\\{(\overline{r},\overline{s})\in\Pi:\overline{r}>r+2,0\leq\overline{s}\leq
s+\frac{1}{4}\\}\subset A_{2}^{\prime\prime},\\\
A_{23}^{\prime\prime}&:=\\{(\overline{r},\overline{s})\in\Pi:0\leq\overline{r}\leq
r+2,\overline{s}>s+\frac{1}{4}\\}\subset A_{2}^{\prime\prime},\quad
A_{24}^{\prime}:=\\{(\overline{r},\overline{s})\in\Pi:\overline{r}>r+2,\overline{s}>s+\frac{1}{4}\\}\subset
A_{2}^{\prime\prime}.\end{split}$

(i) $(\overline{r},\overline{s})\in A_{1}^{\prime\prime}$.

We shall use the inequality $\bar{s}\lesssim(s\bar{s})^{1/2}+|s-\bar{s}|.$ For
$(I)$, we have

$\begin{split}|(I)|&\lesssim\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}\overline{s}}{X_{--}^{3/2}}d\overline{\phi}d\overline{\theta}\leq\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}((s\bar{s})^{1/2}+|s-\bar{s}|)}{X_{--}^{3/2}}d\overline{\phi}d\overline{\theta}\\\
&\lesssim\int_{0}^{\pi}\frac{\overline{r}}{(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})}d\overline{\theta}\lesssim\bigg{(}\frac{\overline{r}}{r}\bigg{)}^{1/2}\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}\lesssim\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}.\end{split}$

For $(II)$, we get

$\begin{split}|(II)|&\lesssim\int_{0}^{\pi}\int_{0}^{\pi}\frac{\overline{r}[s\overline{s}(1-\cos\overline{\phi})]^{1/2}}{X_{--}^{3/2}}d\overline{\phi}d\overline{\theta}\lesssim\int_{0}^{\pi}\frac{\overline{r}}{(r-\overline{r})^{2}+(s-\overline{s})^{2}+2r\overline{r}(1-\cos\overline{\theta})}d\overline{\theta}\\\
&\lesssim\bigg{(}\frac{\overline{r}}{r}\bigg{)}^{1/2}\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}\lesssim\frac{1}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}.\end{split}$

Hence, we obtain

$\begin{split}\bigg{|}\iint_{A_{1}^{\prime\prime}}F^{r}(r,s,\overline{r},\overline{s})w(\overline{r},\overline{s})d\overline{s}d\overline{r}\bigg{|}&\lesssim\iint_{A_{1}^{\prime\prime}}\frac{|w(\overline{r},\overline{s})|}{[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}}d\overline{s}d\overline{r}\leq\|w\|_{L^{1}(A_{1}^{\prime\prime})}^{1/2}\|w\|_{L^{\infty}(A_{1}^{\prime\prime})}^{1/2}\lesssim\|r^{2}w\|_{L^{1}(A_{1}^{\prime\prime})}^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{1}^{\prime\prime})}^{1/2}.\end{split}$
(4.12)

(ii) $(\overline{r},\overline{s})\in A_{2}^{\prime\prime}$.

Note that the term $[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}$ has a
constant lower bound:
$[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}\gtrsim 1.$ Then from the
triangular inequality, we have

$(\overline{r}^{2}+\overline{s}^{2})^{1/2}\leq[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}+(\underbrace{r^{2}+s^{2}}_{=1})^{1/2}\lesssim[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{1/2}.$

Using this, we repeat the process from Case 1-(ii) to get

$|F^{r}(r,s,\overline{r},\overline{s})|\lesssim\begin{cases}[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{-3}&\
\text{in}\ A_{21}^{\prime\prime},\\\
[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{-2}&\ \text{in}\
A_{22}^{\prime\prime}\cup A_{23}^{\prime\prime},\\\
[(r-\overline{r})^{2}+(s-\overline{s})^{2}]^{-1}&\ \text{in}\
A_{24}^{\prime\prime},\\\ \end{cases}$

which gives us

$\begin{split}\bigg{|}\iint_{A_{2}^{\prime\prime}}F^{r}(r,s,\overline{r},\overline{s})w(\overline{r},\overline{s})d\overline{s}d\overline{r}\bigg{|}&\lesssim\iint_{A_{2}^{\prime\prime}}\frac{1}{(\overline{r}-1)^{2}+(\overline{s}-1)^{2}}\cdot(\overline{r}\overline{s})^{1/2}\cdot|w(\overline{r},\overline{s})|^{1/2}\cdot\frac{|w(\overline{r},\overline{s})|^{1/2}}{(\overline{r}\overline{s})^{1/2}}d\overline{s}d\overline{r}\\\
&\lesssim\|rsw\|_{L^{1}(A_{2}^{\prime\prime})}^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{2}^{\prime\prime})}^{1/2}{\leq[\|r^{2}w\|_{L^{1}(A_{2}^{\prime\prime})}+\|s^{2}w\|_{L^{1}(A_{2}^{\prime\prime})}]^{1/2}\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(A_{2}^{\prime\prime})}^{1/2}.}\end{split}$
(4.13)

Note that when $r=1$ (which implies $s=0$), the same argument goes through
($(II)=0$ in this case).

Gathering Case 1–Case 3, we have shown that

$\sup_{\begin{subarray}{c}(r,s)\in\Pi\\\
r^{2}+s^{2}=1\end{subarray}}|u^{r}(r,s)|\lesssim[\|r^{2}w\|_{L^{1}(\Pi)}+\|s^{2}w\|_{L^{1}(\Pi)}]^{1/2}\cdot\bigg{\|}\frac{w}{rs}\bigg{\|}_{L^{\infty}(\Pi)}^{1/2}.$

For $u^{s}$, since the kernel $F^{s}$ of $u^{s}$ has the similar form to the
kernel $F^{r}$ of $u^{r}$ (i.e.,
$F^{s}(r,s,\overline{r},\overline{s})=F^{r}(s,r,\overline{s},\overline{r})$),
we can repeat the same procedure to obtain the claimed bound for $u^{s}$. This
finishes the proof. ∎

###  4.4 Proof of global regularity

Finally, we finish the proof of Theorem 1.3.

###### Proof of Theorem 1.3.

We let $t\geq 0$ and assume that $w_{0}$ satisfies conditions from Theorem 1.1
and

$(1+r+s)\frac{w_{0}}{rs}\in
L^{\infty}(\mathbb{R}^{4}),\quad(1+r^{2}+s^{2})\frac{w_{0}}{rs}\in
L^{1}(\mathbb{R}^{4}).$

First, let us prove the estimate (1.14). To prove this, by Lemma 4.2, it
suffices to show that the function $L(t)$ from (4.4) satisfies

$L(t)\lesssim_{w_{0}}e^{ct},$ (4.14)

for some constant $c>0$ having dependence on the initial data $w_{0}$.

To prove this, note that using Proposition 4.4, its time derivative is
estimated as

$\begin{split}L^{\prime}(t)&=\|u^{r}(t)\|_{L^{\infty}(\mathbb{R}^{4})}+\|u^{s}(t)\|_{L^{\infty}(\mathbb{R}^{4})}\lesssim\bigg{[}\bigg{\|}\frac{rw(t)}{s}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}+\bigg{\|}\frac{sw(t)}{r}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}\bigg{]}^{1/2}\cdot\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}^{1/2},\end{split}$

where we used the conservation of
$\|\frac{w}{rs}\|_{L^{\infty}(\mathbb{R}^{4})}$ in time. Then using estimates
from Lemma 4.3, we obtain

$\begin{split}L^{\prime}(t)&\lesssim\bigg{[}\bigg{\|}\frac{sw_{0}}{r}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}+\bigg{\|}\frac{rw_{0}}{s}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}+\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{1}(\mathbb{R}^{4})}\bigg{]}^{1/2}\cdot
L(t)\cdot\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{\infty}(\mathbb{R}^{4})}^{1/2}\lesssim_{w_{0}}L(t).\end{split}$
(4.15)

Thus, we have shown (4.14). Then combined with the estimate (4.3) from Lemma
4.2, we have the estimate of the term $\|w(t)\|_{L^{\infty}(\mathbb{R}^{4})}$:

$\|w(t)\|_{L^{\infty}(\mathbb{R}^{4})}\lesssim_{w_{0}}e^{ct}.$ (4.16)

Now it is left to show that the $L^{4,1}$-norm of the term
$(1+r+s+rs)w(t)/(rs)$ is bounded for any $t\geq 0$. To begin with, the
boundedness of the term $\|w(t)/(rs)\|_{L^{4,1}(\mathbb{R}^{4})}$ is immediate
from the conservation of $w(t)/(rs)$ along the flow map:

$\bigg{\|}\frac{w(t)}{rs}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})}=\bigg{\|}\frac{w_{0}}{rs}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})}.$
(4.17)

Next, note that by the estimate (4.14) above, estimates (4.5) and (4.6) from
Lemma 4.3, and the estimate (4.7) from Proposition 4.4, we have exponential
estimates of terms $\|u^{r}(t)\|_{L^{\infty}(\mathbb{R}^{4})}$ and
$\|u^{s}(t)\|_{L^{\infty}(\mathbb{R}^{4})}$ as well:

$\|u^{r}(t)\|_{L^{\infty}(\mathbb{R}^{4})},\|u^{s}(t)\|_{L^{\infty}(\mathbb{R}^{4})}\lesssim_{w_{0}}e^{ct}.$
(4.18)

Now observe that the right-hand sides of equations (3.5) and (3.6) for $w/r$
and $w/s$ are $u^{s}w/(rs)$ and $u^{r}w/(rs)$, respectively. Hence, by
combining the estimate (4.18) and the conservation (4.17), we have exponential
bounds of terms $\|w(t)/r\|_{L^{4,1}(\mathbb{R}^{4})}$ and
$\|w(t)/s\|_{L^{4,1}(\mathbb{R}^{4})}$:

$\bigg{\|}\frac{w(t)}{r}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})},\bigg{\|}\frac{w(t)}{s}\bigg{\|}_{L^{4,1}(\mathbb{R}^{4})}\lesssim_{w_{0}}e^{ct}.$
(4.19)

Finally, plugging in the exponential estimate (4.19) in the inequality (3.4),
we obtain an double-exponential estimate of the term
$\|w(t)\|_{L^{4,1}(\mathbb{R}^{4})}$:

$\|w(t)\|_{L^{4,1}(\mathbb{R}^{4})}\lesssim_{w_{0}}e^{e^{ct}}.$ (4.20)

This finishes the proof. ∎

### Acknowledgments

KC has been supported by the National Research Foundation of Korea(NRF) grant
funded by the Korea government(MSIT)(grant No. 2022R1A4A1032094,
RS-2023-00274499). IJ and DL have been supported by the Samsung Science and
Technology Foundation under Project Number SSTF-BA2002-04.

## Appendix

##  Appendix A Vorticity computation

In this section, we assume that the solution of the 4D Euler equations satisfy
the bi-rotational symmetry, and compute the vorticity tensor $\omega$ in terms
of $w,u^{\theta},$ and $u^{\phi}$. The following formulas will be used: we
have

$\displaystyle\partial_{x_{1}}$
$\displaystyle=\cos\theta\partial_{r}-\sin\theta\frac{\partial_{\theta}}{r},\quad\partial_{x_{2}}=\sin\theta\partial_{r}+\cos\theta\frac{\partial_{\theta}}{r},\quad\partial_{x_{3}}=\cos\phi\partial_{s}-\sin\phi\frac{\partial_{\phi}}{s},\quad\partial_{x_{4}}=\sin\phi\partial_{s}+\cos\phi\frac{\partial_{\phi}}{s},$
(A.1)

and

$\begin{split}u^{1}&=\cos\theta u^{r}-\sin\theta u^{\theta},\quad
u^{2}=\sin\theta u^{r}+\cos\theta u^{\theta},\quad u^{3}=\cos\phi
u^{s}-\sin\phi u^{\phi},\quad u^{4}=\sin\phi u^{s}+\cos\phi
u^{\phi}.\end{split}$ (A.2)

Then, we can compute that

$\begin{split}\omega^{1,2}&=(\sin\theta\partial_{r}+\cos\theta\frac{\partial_{\theta}}{r})(\cos\theta
u^{r}-\sin\theta
u^{\theta})-(\cos\theta\partial_{r}-\sin\theta\frac{\partial_{\theta}}{r})(\sin\theta
u^{r}+\cos\theta
u^{\theta})=-\partial_{r}u^{\theta}-\frac{u^{\theta}}{r}\end{split}$

and similarly

$\begin{split}\omega^{3,4}=-\partial_{s}u^{\phi}-\frac{u^{\phi}}{s}.\end{split}$

Next,

$\begin{split}\omega^{1,3}&=-\cos\theta\cos\phi
w-\sin\theta\cos\theta\partial_{s}u^{\theta}+\cos\theta\sin\phi\partial_{r}u^{\phi},\\\
\omega^{1,4}&=-\cos\theta\sin\phi
w-\sin\theta\sin\phi\partial_{s}u^{\theta}-\cos\theta\cos\phi\partial_{r}u^{\phi},\\\
\omega^{2,3}&=-\sin\theta\cos\phi
w+\cos\theta\cos\phi\partial_{s}u^{\theta}+\sin\theta\sin\phi\partial_{r}u^{\phi},\\\
\omega^{2,4}&=-\sin\theta\sin\phi
w+\cos\theta\sin\phi\partial_{s}u^{\theta}-\sin\theta\cos\phi\partial_{r}u^{\phi}.\end{split}$
(A.3)

One sees that in the no-swirl case, every component of the vorticity tensor is
a multiple of $w$ with coefficient depending only on $\theta$ and $\phi$. In
particular when $w\in L^{p}$ (or in $L^{p,q}$), we have automatically
$\omega\in L^{p}$ (in $L^{p,q}$, respectively).

Now we provide a proof of the fact that quantities $w/r$, $w/s$, and $w/(rs)$
are uniformly bounded in $\mathbb{R}^{4}$ if their corresponding velocity $u$
with bi-rotational symmetry without swirl is sufficiently regular. Recall that
the initial data $u_{0}$ mentioned in Remark 1.4 is in $H^{s}(\mathbb{R}^{4})$
for some $s>5$. Then by the Sobolev embedding theorem, $u_{0}$ is in
$C^{3}(\mathbb{R}^{4})$, that is, its derivatives are continuous and uniformly
bounded in $\mathbb{R}^{4}$ up to order 3. This is the type of velocity that
is considered in the proposition below.

######  Proposition A.1.

Let $u\in C^{3}(\mathbb{R}^{4})$ satisfy the bi-rotational symmetry (1.2) and
the no-swirl condition (1.10). Then we have

$(1+r+s)\frac{w}{rs}\in L^{\infty}(\mathbb{R}^{4}).$ (A.4)

###### Proof.

The uniformly boundedness of $w/r$ and $w/s$ are immediate from the regularity
and the bi-rotational symmetry of $u$, so it suffices to prove $w/(rs)\in
L^{\infty}(\mathbb{R}^{4})$.

First, note from (A.3) that the bi-rotational flow without swirl satisfies

$\omega^{1,3}(x_{1},x_{2},x_{3},x_{4})=-\cos\theta\cos\phi w(r,s),\quad
x\in\mathbb{R}^{4}.$ (A.5)

Then note that if $x_{1}=0$, then we have either $\cos\theta=0$ or $r=0$,
where $r=0$ implies $w(r,s)=0$ because of the bi-rotational symmetry of $u$
and the regularity of $w$. Thus, we have

$\omega^{1,3}(0,x_{2},x_{3},x_{4})=0,\quad x_{2},x_{3},x_{4}\in\mathbb{R}.$

Likewise, we also have

$\omega^{1,3}(x_{1},x_{2},0,x_{4})=0,\quad x_{1},x_{2},x_{4}\in\mathbb{R}.$

In addition, note that from $u\in C^{3}(\mathbb{R}^{4})$, we get
$\omega^{1,3}\in C^{2}(\mathbb{R}^{4})$. Then for each
$x_{2},x_{4}\in\mathbb{R}$, the term

$\begin{split}\frac{\omega^{1,3}(x_{1},x_{2},x_{3},x_{4})}{x_{1}x_{3}}&=\frac{1}{x_{3}}\cdot\bigg{(}\frac{\omega^{1,3}(x_{1},x_{2},x_{3},x_{4})-\omega^{1,3}(0,x_{2},x_{3},x_{4})}{x_{1}}-\frac{\omega^{1,3}(x_{1},x_{2},0,x_{4})-\omega^{1,3}(0,x_{2},0,x_{4})}{x_{1}}\bigg{)}\\\
&=\frac{1}{x_{1}}\cdot\bigg{(}\frac{\omega^{1,3}(x_{1},x_{2},x_{3},x_{4})-\omega^{1,3}(x_{1},x_{2},0,x_{4})}{x_{3}}-\frac{\omega^{1,3}(0,x_{2},x_{3},x_{4})-\omega^{1,3}(0,x_{2},0,x_{4})}{x_{3}}\bigg{)}\end{split}$

converges to
$\partial_{x_{3}x_{1}}\omega^{1,3}(0,x_{2},0,x_{4})=\partial_{x_{1}x_{3}}\omega^{1,3}(0,x_{2},0,x_{4})<\infty$
as $x_{1}\to 0$ and $x_{3}\to 0$. From this, we have
$\omega^{1,3}/(x_{1}x_{3})\in L^{\infty}(\mathbb{R}^{4})$. Finally, by the
relation

$\frac{\omega^{1,3}(x_{1},x_{2},x_{3},x_{4})}{x_{1}x_{3}}=-\frac{w(r,s)}{rs},\quad
x_{1},x_{3}\neq 0,$

we obtain $w/(rs)\in L^{\infty}(\mathbb{R}^{4})$. ∎

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