# Remarks on stationary and uniformly-rotating vortex sheets: Rigidity results

Javier Gómez-Serrano, Jaemin Park, Jia Shi and Yao Yao

###### Abstract.

In this paper, we show that the only solution of the vortex sheet equation,
either stationary or uniformly rotating with negative angular velocity
$\Omega$, such that it has positive vorticity and is concentrated in a finite
disjoint union of smooth curves with finite length is the trivial one:
constant vorticity amplitude supported on a union of nested, concentric
circles. The proof follows a desingularization argument and a calculus of
variations flavor.

Keywords: incompressible, vortex sheet, stationary, desingularization

##  1\. Introduction

A vortex sheet is a weak solution of the 2D Euler equations:

$v_{t}+v\cdot\nabla v=-\nabla p,\quad\nabla\cdot v=0,$ (1.1)

whose vorticity $\omega=\text{curl}(v)$ is a delta function supported on a
curve or a finite number of curves $\Gamma_{i}=z_{i}(\alpha,t)$, i.e.

$\omega(x,t)=\sum_{i}\varpi_{i}(\alpha,t)\delta(x-z_{i}(\alpha,t)).$ (1.2)

Here $\varpi_{i}(\alpha,t)$ is the vorticity strength on $\Gamma_{i}$ with
respect to the parametrization $z_{i}$, and the above equation is understood
in the sense that

$\int_{\mathbb{R}^{2}}\varphi(x)d\omega(x,t)=\sum_{i}\int\varphi(z_{i}(\alpha,t))\varpi_{i}(\alpha,t)d\alpha$

for all test functions $\varphi(x)\in C_{0}^{\infty}(\mathbb{R}^{2})$.

The motivation of the study of the equation (1.1) with vortex sheet initial
data comes from the fact that in fluids with small viscosity, flows separate
from rigid walls and corners [24, 32]. To model it, one may think of a
solution to (1.1) with one incompressible fluid where the velocity changes
sign in a discontinuous (tangential) way across a streamline $z$. This
discontinuity induces vorticity in $z$.

The equations of motion of $\varpi_{i}$ and $z_{i}$ can be derived by means of
the Birkhoff-Rott operator ([7, 21, 24, 35]), namely:

$BR(z,\varpi)(x,t)=\frac{1}{2\pi}PV\int\frac{(x-z(\beta,t))^{\bot}}{|x-z(\beta,t)|^{2}}\varpi(\beta,t)d\beta,$
(1.3)

yielding

$\displaystyle\partial_{t}z_{i}(\alpha,t)$
$\displaystyle=\sum_{j}BR(z_{j},\varpi_{i})(z_{i}(\alpha,t))+c_{i}(\alpha,t)\partial_{\alpha}z_{i}(\alpha,t)$
(1.4) $\displaystyle\partial_{t}\varpi_{i}(\alpha,t)$
$\displaystyle=\partial_{\alpha}(c_{i}(\alpha,t)\varpi_{i}(\alpha,t)),$ (1.5)

where the term $c_{i}(\alpha,t)$ accounts for the reparametrization freedom of
the curves. See the paper [19] by Izosimov–Khesin where they propose geodesic,
group-theoretic, and Hamiltonian frameworks for their description.

The main goal of this paper is to establish radial symmetry properties of
stationary/uniformly-rotating vortex sheets to (1.1). To do so, we first
define what we mean by a stationary vortex sheet. Assume the initial data
$\omega_{0}$ of (1.2) is supported on a finite number of curves parametrized
by $z_{i}(\alpha)$, with strength $\varpi_{i}(\alpha)$ (with respect to the
parametrization $z_{i}$) respectively. If there exists some reparametrization
choice $c_{i}(\alpha)$ such that the right hand sides of (1.4)–(1.5) are both
identically zero for every $i$, it gives that $\omega(\cdot,t)$ is invariant
in time, and we say $\omega(\cdot,t)=\omega_{0}$ is a _stationary vortex
sheet_.

For any $x\in\mathbb{R}^{2}$ and $\Omega\in\mathbb{R}$, let $R_{\Omega t}x$
denote the rotation of $x$ counter-clockwise by angle $\Omega t$ about the
origin. We say $\omega(x,t)=\omega_{0}(R_{\Omega t}x)$ is a _uniformly-
rotating vortex sheet_ with angular velocity $\Omega$ if $\omega_{0}$ is
stationary in the rotating frame with angular velocity $\Omega$. (Note that in
the special case $\Omega=0$, the uniformly-rotating sheet is in fact
stationary.) In Lemma 2.1, we will derive the equations satisfied by a
stationary/rotating vortex sheet.

It is easy to see that if the $z_{i}$’s are concentric circles with constant
$\varpi_{i}$ (with respect to the constant-speed parametrization) for every
$i$, the solution is stationary, and it is also uniformly-rotating with any
$\Omega\in\mathbb{R}$. We would like to understand the reverse implication,
namely:

######  Question 1.

Under what conditions must a stationary/uniformly-rotating vortex sheet be
radially symmetric?

This type of rigidity question has been very lately understood for different
equations and different settings such as in the papers by
Koch–Nadirashvili–Sverak [20] for Navier-Stokes, Hamel–Nadirashvili [17, 16,
18] for the 2D Euler equation on a strip, punctured disk or the full plane,
Gómez-Serrano–Park–Shi–Yao [14] for the 2D Euler and modified SQG in the full
plane and Constantin–Drivas–Ginsberg [8] for the 2D and 3D Euler, as well as
the 2D Boussinesq and the 3D Magnetohydrostatic (MHS) equations.

The next theorem is the main result of the paper, solving it for the vortex
sheet equations:

######  Theorem 1.1.

Let $\omega(x,t)=\omega_{0}(R_{\Omega t}x)$ be a stationary/uniformly-rotating
vortex sheet with angular velocity $\Omega$. Assume that $\omega_{0}$ is
concentrated on $\Gamma$, which is a finite union of smooth curves, and
$\omega_{0}$ has positive vorticity strength on $\Gamma$. (See (H1)–(H3) in
Section 2 for the precise regularity and positivity assumptions.)

If $\Omega\leq 0$, $\Gamma$ must be a union of concentric circles, and
$\omega_{0}$ must have constant strength along each circle (with respect to
the constant-speed parametrization). In addition, if $\Omega<0$, all circles
must be centered at the origin.

We now go first over the history of the equations (1.4)–(1.5), focusing later
on the case of steady solutions. The study of those solutions is important due
to the ill-posedness of the vortex sheet equation, thus they represent
(unstable) structures for which there is global existence.

###  1.1. Brief history of the dynamical problem

The existence of solutions to (1.4)–(1.5) has been widely studied. The seminal
paper of Delort [9] proved global existence of weak solutions of (1.1) for an
initial velocity in $L^{2}_{loc}$ and a vorticity a positive Radon measure.
Majda [23] provided a simpler proof. See also the works by Schochet [33, 34]
and Evans–Muller [13]. All of them use the hypothesis that the vorticity has a
definite sign.

If the vorticity does not have a sign, Lopes Filho–Nussenzveig Lopes–Xin
proved existence in [22], in the case where the system enjoys reflection
symmetry. For the setting in which the curve $z_{i}$ is not closed and
represented as a graph, Sulem–Sulem–Bardos–Frisch [35] proved local existence
in the case of analytic initial data.

The first sign of singularities with analytic initial data goes back to Moore
[26], where he demonstrated that the curvature may blow up in finite time.
Ebin [11] showed ill-posedness in Sobolev spaces when $\gamma$ has a
distinguished sign, and Duchon–Robert [10] proved global existence for a class
of initial data in the unbounded setting. Caflisch–Orellana [6] also showed
global existence for a class of initial data, as well as ill-posedness in
$H^{s}$ for $s>\frac{3}{2}$ and simplified the analysis of Moore [5]. We also
mention here the work of Wu [38], in which she proved the existence of
solutions to (1.4)–(1.5) in spaces which are less regular than $H^{s}$.
Székelyhidi [36] (resp. Mengual–Székelyhidi [25]) constructed infinitely many
admissible weak solutions to (1.1) for vortex sheet initial data with (resp.
without necessarily) a distinguished sign.

###  1.2. Stationary and rotating solutions

Relative equilibria are an important family of solutions of fluid equations
since their structures persist for long times. This is specially important
when the equations of motion are ill-posed. In the particular case of
(1.4)–(1.5), our knowledge is very small and only very few explicit cases are
known: the circle and the straight line (with constant $\gamma$), which are
stationary, and the segment of length $2a$ and density

$\gamma(x)=\Omega\sqrt{a^{2}-x^{2}},\qquad x\in[-a,a],$ (1.6)

which is a rotating solution with angular velocity $\Omega$ [2]. Protas–Sakajo
[31] generalized this solution and proved the existence of several others made
out of segments rotating about a common center of rotation with endpoints at
the vertices of a regular polygon by solving a Riemann-Hilbert problem, even
finding some of them analytically.

In the paper [15] we prove the existence of a family of vortex sheet rotating
solutions with non-constant vorticity density supported on a non-radial curve,
bifurcating from the circle with constant density.

Numerically, some solutions have been computed before. O’Neil [27, 28] used
point vortices to approximate the vortex sheet and compute uniformly rotating
solutions and Elling [12] constructed numerically self-similar vortex sheets
forming cusps. O’Neil [29, 30] also found numerically steady solutions which
are combinations of point vortices and vortex sheets.

###  1.3. Structure of the proof

The proof is inspired by our recent rigidity result in the paper [14] on
stationary and rotating solutions of the 2D Euler equations both in the smooth
and vortex patch settings. To prove it, we constructed an appropriate
functional and showed, on one hand, that any stationary solution had to be a
critical point, and on the other, for any curve which is not a circle there
existed a vector field along which the first variation was non-zero. This
vector field is defined in terms of an elliptic equation in the interior of
the patch. In the case of the vortex sheet, this is not possible anymore.
Instead, we desingularize the problem by considering patches of thickness
$\sim\varepsilon$ which are tubular neighborhoods of the sheet. The drawback
is that we lose the property that any stationary solution has to be a critical
point if $\varepsilon>0$ and very careful, quantitative estimates need to be
done to show that indeed the first variation of a stationary solution tends to
0 as $\varepsilon\to 0$. This setup is also reminiscent of the numerical work
by Baker–Shelley [1], where they approximate the motion of a vortex sheet by a
vortex patch of very small width. In [3], Benedetto–Pulvirenti proved the
stability (for short time) of vortex sheet solutions with respect to solutions
to 2D Euler with a thin strip of vorticity around a curve. See also the work
by Caflisch–Lombardo–Sammartino [4] for more stability results with a
different desingularization.

###  1.4. Organization of the paper

In Section 2 the equations for the stationary/rotating vortex sheet are
derived, and in Section 3 we perform the desingularization procedure. Section
4 is devoted to construct the aforementioned divergence free vector-field
along which the first variation is non-zero. Finally in Section 5 we conclude
the quantitative estimates and prove the symmetry result from Theorem 1.1.

###  1.5. Notations

For a bounded domain $D\subset\mathbb{R}^{2}$, we denote $|D|$ by its area
(i.e. its Lebesgue measure). For $x\in\mathbb{R}^{2}$ and $r>0$, denote by
$B(x,r)$ or $B_{r}(x)$ the open ball centered at $x$ with radius $r$.

Through Section 3-5 of this paper, we will desingularize the vortex sheet into
a vortex layer with width $\sim\epsilon$, and obtain various quantitative
estimates. In all these estimates, we say a term $f$ is $O(g(\epsilon))$ if
$|f|\leq Cg(\epsilon)$ for some constant $C$ independent of $\epsilon$.

For a domain $U\subset\mathbb{R}^{2}$, in the boundary integral
$\int_{\partial U}\vec{f}\cdot nd\sigma$, $n$ denotes the outer normal of the
domain $U$.

##  2\. Equations for a stationary/rotating vortex sheet

Let $\omega(\cdot,t)=\omega_{0}(R_{\Omega t})$ be a stationary/rotating vortex
sheet solution to the incompressible 2D Euler equation, where
$\omega_{0}\in\mathcal{M}(\mathbb{R}^{2})\cap H^{-1}(\mathbb{R}^{2})$ is a
Radon measure. Here $\Omega=0$ corresponds to a stationary solution, and
$\Omega\neq 0$ corresponds to a rotating solution. Assume $\omega_{0}$ is
concentrated on $\Gamma$, which is a finite disjoint union of curves.
Throughout this paper we assume $\Gamma$ satisfies the following:

(H1) Each connected component of $\Gamma$ is smooth and with finite length,
and it is either a simple closed curve (denote them by
$\Gamma_{1},\dots,\Gamma_{n})$, or a non-self-intersecting curve with two
endpoints (denote them by $\Gamma_{n+1},\dots,\Gamma_{n+m}$). Here we require
$n+m\geq 1$, but allow either $n$ or $m$ to be 0.

Let us denote

$d_{\Gamma}:=\min_{k\neq i}\text{dist}(\Gamma_{i},\Gamma_{k}),$ (2.1)

which is strictly positive since we assume the curves
$\\{\Gamma_{i}\\}_{i=1}^{n+m}$ are disjoint. For $i=1,\dots,n+m$, denote by
$L_{i}$ the length of $\Gamma_{i}$. Let $z_{i}:S_{i}\to\Gamma_{i}$ denote a
constant-speed parameterization of $\Gamma_{i}$ (in counter-clockwise
direction if $\Gamma_{i}$ is a closed curve), where the parameter domain
$S_{i}$ is given by

$S_{i}:=\begin{cases}\mathbb{R}/\mathbb{Z}&\text{for $i=1,\dots,n$},\\\
[0,1]&\text{for $i=n+1,\dots,n+m$}.\end{cases}$

Note that this gives $|z_{i}^{\prime}|\equiv L_{i}$, and the arc-chord
constant

$F_{\Gamma}:=\max_{i=1,\dots,n+m}\sup_{\alpha\neq\beta\in
S_{i}}\frac{|\alpha-\beta|}{|z_{i}(\alpha)-z_{i}(\beta)|}$ (2.2)

is finite, since $\Gamma$ is non-self-intersecting. Let
$\mathbf{s}:\Gamma\to\mathbb{R}^{2}$ be the unit tangential vector on
$\Gamma$, given by
$\mathbf{s}(z_{i}(\alpha)):=\frac{z_{i}^{\prime}(\alpha)}{|z_{i}^{\prime}(\alpha)|}=\frac{z_{i}^{\prime}(\alpha)}{L_{i}}$,
and $\mathbf{n}:\Gamma\to\mathbb{R}^{2}$ be the unit normal vector, given by
$\mathbf{n}=\mathbf{s}^{\perp}$. See Figure 1 for an illustration.

For $i=1,\dots,n+m$, let us denote by $\gamma_{i}(\alpha)$ the vorticity
strength at $z_{i}(\alpha)$ with respect to the arclength parametrization,
which is related to $\varpi_{i}(\alpha)$ by

$\gamma_{i}(\alpha)=\frac{\varpi_{i}(\alpha)}{|z_{i}^{\prime}(\alpha)|}\quad\text{
for }\alpha\in S_{i}.$ (2.3)

Throughout this paper we will be working with $\gamma_{i}$, instead of
$\varpi_{i}$. We impose the following regularity and positivity assumptions on
$\gamma_{i}$:

(H2) Assume that $\gamma_{i}\in C^{2}(S_{i})$ for $i=1,\dots,n$ and
$\gamma_{i}\in C^{b}(S_{i})\cap C^{1}(S_{i}^{\circ})$ for some $b\in(0,1)$ for
$i=n+1,\dots,n+m$.111For an open curve $i=n+1,\dots,n+m$, note that (H2) does
not require $\gamma_{i}$ to be $C^{1}$ up to the boundary of $S_{i}$, and its
derivative is allowed to blow up at the endpoints. This is motivated by the
fact that in the explicit uniformly-rotating solution (1.6), its strength
$\gamma$ is Hölder continuous in $[-a,a]$ and smooth in the interior, but its
derivative blows up at the endpoints.

(H3) For $i=1,\dots,n$, assume $\gamma_{i}>0$ in $S_{i}$. And for
$i=n+1,\dots,n+m$, assume $\gamma_{i}>0$ in $S_{i}^{\circ}$, and
$\gamma_{i}(0)=\gamma_{i}(1)=0$.

Note that for a closed curve, (H3) implies that $\gamma_{i}$ is uniformly
positive; whereas for an open curve, $\gamma_{i}$ is positive in the interior
of $S_{i}$ but vanishes at its endpoints. This is because any
stationary/rotating vortex sheet with continuous $\gamma_{i}$ must have it
vanishing at the two endpoints of any open curve: if not, one can easily check
that $|BR(z_{i}(\alpha))\cdot\mathbf{n}(z_{i}(\alpha))|\to\infty$ as $\alpha$
approaches the endpoint, thus such a vortex sheet cannot be stationary in the
rotating frame.

With the above notations of $z_{i}$ and $\gamma_{i}$, the Birkhoff-Rott
integral (1.3) along the sheet can now be expressed as

$BR(z_{i}(\alpha))=\sum_{k=1}^{n+m}BR_{k}(z_{i}(\alpha)):=\sum_{k=1}^{n+m}PV\int_{S_{k}}K_{2}(z_{i}(\alpha)-z_{k}(\alpha^{\prime}))\,\gamma_{k}(\alpha^{\prime})|z_{k}^{\prime}(\alpha^{\prime})|\,d\alpha^{\prime},$
(2.4)

with the kernel $K_{2}$ given by

$K_{2}(x):=(2\pi)^{-1}\nabla^{\perp}\log|x|=\frac{x^{\perp}}{2\pi|x|^{2}},$
(2.5)

and the principal value in (2.4) is only needed for the integral with $k=i$.

Let $\mathbf{v}:\mathbb{R}^{2}\to\mathbb{R}^{2}$ be the velocity field
generated by $\omega_{0}$, given by
$\mathbf{v}:=\nabla^{\perp}(\omega_{0}*\mathcal{N})$. Note that $\mathbf{v}\in
C^{\infty}(\mathbb{R}^{2}\setminus\Gamma)$, but $\mathbf{v}$ is discontinuous
across $\Gamma$. Let $\mathbf{v}^{+},\mathbf{v}^{-}:\Gamma\to\mathbb{R}^{2}$
denote the two limits of $\mathbf{v}$ on the two sides of $\Gamma$ (with
$\mathbf{v}^{+}$ being the limit on the side that $\mathbf{n}$ points into –
see Figure 1 for an illustration), and
$[\mathbf{v}]:=\mathbf{v}^{-}-\mathbf{v}^{+}$ the jump in $\mathbf{v}$ across
the sheets. $[\mathbf{v}]$ is related to the vortex-sheet strength $\gamma$ as
follows (see [24, Eq. (9.8)] for a derivation):
$[\mathbf{v}]\cdot\mathbf{n}=0$, and

$[\mathbf{v}]\times\mathbf{n}=[\mathbf{v}]\cdot\mathbf{s}=\gamma.$

In addition, the Birkhoff-Rott integral (2.4) is the the average of
$\mathbf{v}^{+}$ and $\mathbf{v}^{-}$, namely

$BR(z_{i}(\alpha))=\frac{1}{2}(\mathbf{v}^{+}(z_{i}(\alpha))+\mathbf{v}^{-}(z_{i}(\alpha)))\quad\text{
for all }\alpha\in S_{i},i=1,\dots,n+m.$

Figure 1. Illustration of the closed curves $\Gamma_{1},\dots,\Gamma_{n}$ and
the open curves $\Gamma_{n+1},\dots,\Gamma_{n+m}$, and the definitions of
$\mathbf{n}$, $\mathbf{s}$, $\mathbf{v}^{+}$ and $\mathbf{v}^{-}$.

In the following lemma, we derive the equation that the Birkhoff-Rott integral
satisfies for a stationary/rotating vortex sheet.

######  Lemma 2.1.

Assume $\omega(\cdot,t)=\omega_{0}(R_{\Omega t}x)$ is a stationary/uniformly-
rotating vortex sheet with angular velocity $\Omega\in\mathbb{R}$, and
$\omega_{0}$ is concentrated on $\cup_{i=1}^{n+m}\Gamma_{i}$, with $z_{i}$ and
$\gamma_{i}$ defined as above. Then the Birkhoff-Rott integral $BR$ (2.4) and
the strength $\gamma_{i}$ satisfy the following two equations:

$(BR-\Omega
x^{\perp})\cdot\mathbf{n}=\mathbf{v}^{+}\cdot\mathbf{n}=\mathbf{v}^{-}\cdot\mathbf{n}=0\quad\text{
on }\Gamma,$ (2.6)

and

$(BR(z_{i}(\alpha))-\Omega
z_{i}^{\perp}(\alpha))\cdot\mathbf{s}(z_{i}(\alpha))\,\gamma_{i}(\alpha)=\begin{cases}C_{i}&\text{
on }S_{i}\text{ for }i=1,\dots,n,\\\ 0&\text{ on }S_{i}\text{ for
}i=n+1,\dots,n+m.\end{cases}$ (2.7)

In particular, the above two equations imply that $BR(z_{i}(\alpha))-\Omega
z_{i}^{\perp}(\alpha)\equiv\mathbf{0}$ for $i=n+1,\dots,n+m$.

###### Proof.

By definition of the stationary/uniformly-rotating solutions, $\omega_{0}$ is
a stationary vortex sheet in the rotating frame with angular velocity
$\Omega$. In this rotating frame, an extra velocity $-\Omega z_{i}^{\perp}$
should be added to the right hand side of (1.4). Therefore the evolution
equations (1.4)–(1.5) become the following in the rotating frame (where we
also use (2.4)):

$\displaystyle\partial_{t}z_{i}(\alpha,t)$
$\displaystyle=BR(z_{i}(\alpha,t))-\Omega
z_{i}^{\perp}(\alpha,t)+c_{i}(\alpha,t)\partial_{\alpha}z_{i}(\alpha,t)$ (2.8)
$\displaystyle\partial_{t}\varpi_{i}(\alpha,t)$
$\displaystyle=\partial_{\alpha}(c_{i}(\alpha,t)\varpi_{i}(\alpha,t)),$ (2.9)

where the term $c_{i}(\alpha,t)$ accounts for the reparametrization freedom of
the curves. Since $\omega_{0}$ is stationary in the rotating frame,
$z_{i}(\cdot,t)$ parametrizes the same curve as $z_{i}(\cdot,0)$. Therefore
$\partial_{t}z_{i}(\alpha,t)$ is tangent to the curve $\Gamma_{i}$, and
multiplying $\mathbf{n}(z_{i}(\alpha,t))$ to (2.8) gives

$0=\partial_{t}z_{i}(\alpha,t)\cdot\mathbf{n}(z_{i}(\alpha,t))=(BR(z_{i}(\alpha,t))-\Omega
z_{i}^{\perp}(\alpha,t))\cdot\mathbf{n}(z_{i}(\alpha,t)),$ (2.10)

where we use that
$\mathbf{n}(z_{i}(\alpha,t))\cdot\partial_{\alpha}z_{i}(\alpha,t)=0$. This
proves (2.6).

Now we prove (2.7). Towards this end, let us choose

$c_{i}(\alpha,t):=-\frac{(BR(z_{i}(\alpha,t))-\Omega
z_{i}^{\perp}(\alpha,t))\cdot\mathbf{s}(z_{i}(\alpha,t))}{|\partial_{\alpha}z_{i}(\alpha,t)|},$

so that multiplying $\mathbf{s}(z_{i}(\alpha,t))$ to (2.8) gives
$\partial_{t}z_{i}(\alpha,t)\cdot\mathbf{s}(z_{i}(\alpha,t))=0$, and combining
it with (2.10) gives $\partial_{t}z_{i}(\alpha,t)=0.$ In other words, with
such choice of $c_{i}$, the parametrization $z_{i}(\alpha,t)$ remains fixed in
time. Since $\omega_{0}$ is stationary in the rotating frame, we know that
with a fixed parametrization $z_{i}(\alpha,t)=z_{i}(\alpha,0)$, the strength
$\varpi_{i}(\alpha,t)$ must also remain invariant in time. Thus (2.9) becomes

$c_{i}(\alpha,t)\varpi_{i}(\alpha,t)\equiv C_{i}.$

Plugging the definition of $c_{i}$ into the equation above and using the fact
that $z_{i}$ is invariant in $t$, we have

$\frac{(BR(z_{i}(\alpha))-\Omega
z_{i}^{\perp}(\alpha))\cdot\mathbf{s}(z_{i}(\alpha))\varpi_{i}(\alpha)}{|\partial_{\alpha}z_{i}(\alpha)|}\equiv-
C_{i}\quad\text{ for all }\alpha\in S_{i},$

and finally the relationship between $\gamma_{i}$ and $\varpi_{i}$ in (2.3)
yields (2.7) for $i=1,\dots,n$.

And for the open curves $i=n+1,\dots,n+m$, note that we do not have any
reparametrization freedom at the two endpoints $\alpha=0,1$, therefore the
endpoint velocity $BR(z_{i}(0,t))-\Omega z_{i}^{\perp}(0,t)$ must be 0 to
ensure that $\omega_{0}$ is stationary in the rotating frame. This immediately
leads to $C_{i}=0$ for $i=n+1,\dots,n+m$, finishing the proof of (2.7). ∎

##  3\. Approximation by a thin vortex layer

Our aim in this section is to desingularize the vortex sheet $\omega_{0}$.
Namely, for $0<\epsilon\ll 1$, we will construct a vorticity
$\omega^{\epsilon}\in L^{\infty}(\mathbb{R}^{2})\cap L^{1}(\mathbb{R}^{2})$
that only takes values $0$ and $\epsilon^{-1}$, and is supported in an
$O(\epsilon)$ neighborhood of $\Gamma$, such that $\omega^{\epsilon}$ weakly
converges to $\omega_{0}$ as $\epsilon\to 0^{+}$.

For each $i=1,\dots,n+m$, we will describe a neighborhood of $\Gamma_{i}$
using the following change of coordinates: let
$R^{\epsilon}_{i}:S_{i}\times\mathbb{R}\to\mathbb{R}^{2}$ be given by

$R^{\epsilon}_{i}(\alpha,\eta):=z_{i}(\alpha)+\epsilon\gamma_{i}(\alpha)\mathbf{n}(z_{i}(\alpha))\eta,$
(3.1)

and let

$D^{\epsilon}_{i}:=\left\\{R^{\epsilon}_{i}(\alpha,\eta):\alpha\in
S_{i}^{\circ},\eta\in(-1,0)\right\\}.$

Note that each $D^{\epsilon}_{i}$ is a connected open set, and for all
$\epsilon>0$ sufficiently small, the sets $(D^{\epsilon}_{i})_{i=1}^{n+m}$ are
disjoint. For $i=1,\dots,n$, the domains $D_{i}^{\epsilon}$ are doubly-
connected with smooth boundary, and its inner boundary coincides with
$\Gamma_{i}$; see the left of Figure 2 for an illustration. And for
$i=n+1,\dots,n+m$, the domains $D_{i}^{\epsilon}$ are simply-connected, and
its boundary is smooth except at at most two points; see the right of Figure 2
for an illustration.

Figure 2. Illustration of the definitions of $R_{i}^{\epsilon}$ and
$D_{i}^{\epsilon}$ for a closed curve (left) and an open curve (right).

In addition, for $\epsilon>0$ that is sufficiently small, one can check that
$R^{\epsilon}_{i}:S_{i}^{\circ}\times(-1,0)\to D_{i}^{\epsilon}$ is a
diffeomorphism. Since $\gamma_{i}\in C^{1}(S_{i})$ and $z_{i}\in
C^{2}(S_{i})$, we only need to show
$R^{\epsilon}_{i}:S_{i}^{\circ}\times(-1,0)\to D_{i}^{\epsilon}$ is injective.
Below we prove this fact in a stronger quantitative version, which will be
used later.

######  Lemma 3.1.

For any $i=1,\dots,n+m$, assume $\Gamma_{i}$ and $\gamma_{i}$ satisfy
(H1)–(H2). Then the map $R^{\epsilon}_{i}:S_{i}^{\circ}\times(-1,0)\to
D^{\epsilon}_{i}$ given by (3.1) is injective. In addition, there exist some
$c_{0},\epsilon_{0}>0$ depending on $\|z_{i}\|_{C^{2}(S_{i})},$
$\rVert\gamma_{i}\rVert_{L^{\infty}(S_{i})}$ and $F_{\Gamma}$, such that for
all $\epsilon\in(0,\epsilon_{0})$ we have

$|R^{\epsilon}_{i}(\alpha^{\prime},\eta^{\prime})-R^{\epsilon}_{i}(\alpha,\eta)|\geq
c_{0}\big{(}|\alpha^{\prime}-\alpha|+\epsilon|\gamma_{i}(\alpha)\eta-\gamma_{i}(\alpha^{\prime})\eta^{\prime}|\big{)},$
(3.2)

for all $\alpha,\alpha^{\prime}\in
S_{i}^{\circ},~{}\eta,\eta^{\prime}\in(-1,0)$.222In fact, (3.2) also holds
(with a slightly smaller $\epsilon_{0}$ and $c_{0}$) for
$\eta,\eta^{\prime}\in(-2,2)$, even though such $R_{i}^{\epsilon}$ may not
belong to $D_{i}^{\epsilon}$. We will use this fact later in the proof of
Lemma 3.5.

###### Proof.

To begin with, note that (3.2) immediately implies that
$R^{\epsilon}_{i}:S_{i}^{\circ}\times(-1,0)\to D^{\epsilon}_{i}$ is injective,
where we used the positivity assumption $\gamma_{i}>0$ in $S_{i}^{\circ}$ in
(H2). Thus it suffices to prove (3.2). Throughout the proof, we fix any
$i\in\\{1,\dots,n+m\\}$, and we will omit the subscript $i$ for notational
simplicity. Using the definition (3.1), let us break
$R^{\epsilon}(\alpha^{\prime},\eta^{\prime})-R^{\epsilon}(\alpha,\eta)$ into

$R^{\epsilon}(\alpha^{\prime},\eta^{\prime})-R^{\epsilon}(\alpha,\eta)=\underbrace{z(\alpha^{\prime})-z(\alpha)}_{=:T_{1}}+\underbrace{\epsilon\left(\gamma(\alpha^{\prime})\eta^{\prime}-\gamma(\alpha)\eta\right)\mathbf{n}(z(\alpha^{\prime}))}_{=:T_{2}}+\underbrace{\epsilon\gamma(\alpha)\eta\left(\mathbf{n}(z(\alpha^{\prime}))-\mathbf{n}(z(\alpha))\right)}_{=:T_{3}}.$
(3.3)

For $T_{1}$ and $T_{3}$, we have

$\begin{split}\left|T_{1}-z^{\prime}(\alpha^{\prime})(\alpha^{\prime}-\alpha)\right|&\leq\rVert
z\rVert_{C^{2}(S)}|\alpha-\alpha^{\prime}|^{2},\\\
\left|T_{3}\right|&\leq\epsilon\gamma(\alpha)\rVert
z\rVert_{C^{2}(S)}|\alpha-\alpha^{\prime}|.\end{split}$ (3.4)

Also, using that $z^{\prime}(\alpha^{\prime})=L\mathbf{s}(z(\alpha^{\prime}))$
is perpendicular to $\mathbf{n}(z(\alpha^{\prime}))$, we have

$\begin{split}|z^{\prime}(\alpha^{\prime})(\alpha^{\prime}-\alpha)+T_{2}|&=\left|L(\alpha^{\prime}-\alpha)\mathbf{s}(z(\alpha^{\prime}))+\epsilon\left(\gamma(\alpha^{\prime})\eta^{\prime}-\gamma(\alpha)\eta\right)\mathbf{n}(z(\alpha^{\prime}))\right|\\\
&\geq\frac{1}{2}L|\alpha^{\prime}-\alpha|+\frac{1}{2}\epsilon\left|\gamma(\alpha^{\prime})\eta^{\prime}-\gamma(\alpha)\eta\right|,\end{split}$

where we use that $\sqrt{x^{2}+y^{2}}\geq\frac{1}{2}(|x|+|y|)$. Combining this
with (3.4) gives

$\displaystyle|T_{1}+T_{2}+T_{3}|$
$\displaystyle\geq|\alpha-\alpha^{\prime}|\left(\frac{L}{2}-\rVert
z\rVert_{C^{2}(S)}\left(|\alpha-\alpha^{\prime}|+\epsilon\gamma(\alpha)\right)\right)+\frac{1}{2}\epsilon|\gamma(\alpha^{\prime})\eta^{\prime}-\gamma(\alpha)\eta|,$

thus

$|R^{\epsilon}(\alpha^{\prime},\eta^{\prime})-R^{\epsilon}(\alpha,\eta)|\geq\frac{L}{4}|\alpha-\alpha^{\prime}|+\frac{1}{2}\epsilon|\gamma(\alpha^{\prime})\eta^{\prime}-\gamma(\alpha)\eta|$
(3.5)

for all $0<\epsilon<L(8\rVert
z\rVert_{C^{2}}\rVert\gamma\rVert_{L^{\infty}})^{-1}$ and
$|\alpha-\alpha^{\prime}|\leq\frac{L}{8\rVert z\rVert_{C^{2}}}$.

For $|\alpha-\alpha^{\prime}|>\frac{L}{8\rVert z\rVert_{C^{2}}}$, recall that
the definition of $F_{\Gamma}$ in (2.2) gives
$|z(\alpha^{\prime})-z(\alpha)|\geq F_{\Gamma}^{-1}|\alpha^{\prime}-\alpha|$.
Thus a crude estimate gives

$|R^{\epsilon}(\alpha^{\prime},\eta^{\prime})-R^{\epsilon}(\alpha,\eta)|\geq|z(\alpha^{\prime})-z(\alpha)|-2\epsilon\|\gamma\|_{L^{\infty}(S)}\geq\frac{1}{2F_{\Gamma}}|\alpha^{\prime}-\alpha|+\epsilon|\gamma(\alpha^{\prime})\eta^{\prime}-\gamma(\alpha)\eta|$
(3.6)

for $0<\epsilon<L(64F_{\Gamma}\|z\|_{C^{2}}\|\gamma\|_{L^{\infty}})^{-1}$.
(Note that for such $\epsilon$ we have
$4\epsilon\|\gamma\|_{L^{\infty}}\leq\frac{1}{2F_{\Gamma}}|\alpha^{\prime}-\alpha|$
due to our assumption that $|\alpha-\alpha^{\prime}|>\frac{L}{8\rVert
z\rVert_{C^{2}}}$).

Finally, combining (3.5) and (3.6), it follows that (3.2) holds for
$c_{0}=\min\\{\frac{L}{4},\frac{1}{2F_{\Gamma}},\frac{1}{2}\\}$ and
$\epsilon_{0}=\min\\{L(8\rVert
z\rVert_{C^{2}}\rVert\gamma\rVert_{L^{\infty}})^{-1},L(64F_{\Gamma}\|z\|_{C^{2}}\|\gamma\|_{L^{\infty}})^{-1}\\}$.
This finishes the proof. ∎

In the next lemma we compute the partial derivatives and Jacobian of
$R^{\epsilon}_{i}(\alpha,\eta)$, which will be useful later.

######  Lemma 3.2.

For any $i=1,\dots,n+m$, let $z_{i}$ be a constant-speed parameterization of
the curve $\Gamma_{i}$ (with length $L_{i}$), and let $R^{\epsilon}_{i}$ be
given by (3.1). Then its partial derivatives are

$\begin{split}\partial_{\alpha}R^{\epsilon}_{i}(\alpha,\eta)&=z_{i}^{\prime}(\alpha)+\epsilon\left(\gamma_{i}^{\prime}(\alpha)\frac{z_{i}^{\prime}(\alpha)^{\perp}}{L_{i}}\eta+\gamma_{i}(\alpha)\frac{z_{i}^{\prime\prime}(\alpha)^{\perp}}{L_{i}}\eta\right),\\\
\partial_{\eta}R^{\epsilon}_{i}(\alpha,\eta)&=\epsilon\gamma_{i}(\alpha)\frac{z_{i}^{\prime}(\alpha)^{\perp}}{L_{i}}.\end{split}$
(3.7)

Moreover, its Jacobian is given by

$\begin{split}\det(\nabla_{\alpha,\eta}R^{\epsilon}_{i})&=\epsilon
L_{i}\gamma_{i}(\alpha)-\epsilon^{2}L_{i}\gamma_{i}^{2}(\alpha)\kappa_{i}(\alpha)\eta,\end{split}$
(3.8)

where $\kappa_{i}(\alpha)$ denotes the signed curvature of $\Gamma_{i}$ at
$z_{i}(\alpha)$.

###### Proof.

Since $z_{i}$ is the constant-speed parameterization of $\Gamma_{i}$ (which
has length $L_{i}$), we have $|z_{i}^{\prime}|\equiv L_{i}$ and
$\mathbf{n}(z_{i}(\alpha))=z_{i}^{\prime}(\alpha)^{\perp}/L_{i}$. Taking the
$\alpha$ and $\eta$ partial derivatives of (3.1) directly yields (3.7).

Putting the two partial derivatives into columns of a $2\times 2$ matrix and
computing the determinant, we have

$\begin{split}\det(\nabla_{\alpha,\eta}R_{i}^{\epsilon})&=\epsilon\gamma_{i}(\alpha)\frac{|z_{i}^{\prime}(\alpha)|^{2}}{L_{i}}+\epsilon^{2}\gamma_{i}^{2}(\alpha)\frac{z_{i}^{\prime\prime}(\alpha)^{\perp}\cdot
z_{i}^{\prime}(\alpha)}{L_{i}^{2}}\eta\\\ &=\epsilon
L_{i}\gamma_{i}(\alpha)-\epsilon^{2}L_{i}\gamma_{i}^{2}(\alpha)\kappa_{i}(\alpha)\eta,\end{split}$

where in the second equality we used that
$z_{i}^{\prime\prime}(\alpha)=\kappa_{i}(\alpha)\mathbf{n}(z_{i}(\alpha))L_{i}^{2}$
(recall that $z_{i}$ has constant speed $L_{i}$). This finishes the proof. ∎

######  Remark 3.3.

We point out that for each $i=1,\dots,n+m$, the determinant formula (3.8)
immediately gives the following approximation of $|D^{\epsilon}_{i}|$, which
will be helpful in the proofs later:

$\displaystyle\frac{|D^{\epsilon}_{i}|}{\epsilon}=\frac{1}{\epsilon}\int_{D^{\epsilon}_{i}}1dx=\frac{1}{\epsilon}\int_{S_{i}}\int_{-1}^{0}\det(\nabla_{\alpha,\eta}R^{\epsilon}_{i}(\alpha,\eta))\,d\eta
d\alpha=L_{i}\int_{S_{i}}\gamma_{i}(\alpha)d\alpha+O(\epsilon),$ (3.9)

where the $O(\epsilon)$ error term has its absolute value bounded by
$C\epsilon$, with $C$ only depending on $\|z_{i}\|_{C^{2}(S_{i})}$ and
$\|\gamma_{i}\|_{L^{\infty}(S_{i})}$.

Finally, let $D^{\epsilon}:=\cup_{i=1}^{n+m}D_{i}^{\epsilon}$, and
$\omega^{\epsilon}:\mathbb{R}^{2}\to\mathbb{R}$ is defined as

$\omega^{\epsilon}(x):=\epsilon^{-1}1_{D^{\epsilon}}(x)=\epsilon^{-1}\sum_{i=1}^{n+m}1_{D_{i}^{\epsilon}}(x),$

and let

$\mathbf{v}^{\epsilon}=\nabla^{\perp}(\omega^{\epsilon}*\mathcal{N})$ (3.10)

be the velocity field generated by $\omega^{\epsilon}$.

In the next lemma we aim to obtain some fine estimate of
$\mathbf{v}^{\epsilon}$ in the thin vortex layer $D^{\epsilon}$. Our goal is
to show that along each cross section of the thin layer (i.e. fix $i$ and
$\alpha$, and let $\eta$ vary in $[-1,0]$), the function
$\eta\mapsto\mathbf{v}^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))$ is almost a
linear function in $\eta$, with the endpoint values (at $\eta=-1$ and $0$)
being almost $\mathbf{v}^{-}(z_{i}(\alpha))$ and
$\mathbf{v}^{+}(z_{i}(\alpha))$ respectively.

######  Lemma 3.4.

For $i=1,\dots,n+m$, assume $\Gamma_{i}$ and $\gamma_{i}$ satisfy (H1)–(H3).
Let

$g_{i}(\alpha,\eta):=BR(z_{i}(\alpha))-\Big{(}\eta+\frac{1}{2}\Big{)}[\mathbf{v}](z_{i}(\alpha))\quad\text{
for }\alpha\in S_{i},$

and note that $g_{i}(\alpha,0)=\mathbf{v}^{+}(z_{i}(\alpha))$ and
$g_{i}(\alpha,-1)=\mathbf{v}^{-}(z_{i}(\alpha))$ (see Figure 3 for an
illustration of $g_{i}(\alpha,\eta)$). Then for all sufficiently small
$\epsilon>0$, for all $i=1,\dots,n+m$ we have

$|\mathbf{v}^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))-g_{i}(\alpha,\eta)|\leq
C\epsilon^{b}|\log\epsilon|\quad\text{ for all }\alpha\in
S_{i},\eta\in[-1,0],$ (3.11)

where $b\in(0,1)$ is as in (H2), and $C$ depends on $b$,
$\max_{i}\|z_{i}\|_{C^{2}(S_{i})}$, $\max_{i}\|\gamma_{i}\|_{C^{b}(S_{i})}$,
$d_{\Gamma}$ and $F_{\Gamma}$.

Figure 3. Illustration of the definition of $g_{i}(\alpha,\cdot)$ (the orange
arrows).

###### Proof.

Let $i$ be any fixed index in $1,\dots,n+m$. We begin with breaking
$\mathbf{v}^{\epsilon}$ into contributions from different components
$\\{D_{k}^{\epsilon}\\}_{k=1}^{n+m}$, namely

$\mathbf{v}^{\epsilon}(x)=\sum_{k=1}^{n+m}\mathbf{v}_{k}^{\epsilon}(x):=\sum_{k=1}^{n+m}\epsilon^{-1}\int_{D_{i}^{\epsilon}}K_{2}(x-y)dy,$

where the kernel $K_{2}$ is given by (2.5). Similarly, we can break
$BR(z_{i}(\alpha))$ into $\displaystyle
BR(z_{i}(\alpha))=\sum_{k=1}^{n+m}BR_{k}(z_{i}(\alpha)),$ where $BR_{k}$ is
the contribution from the $k$-th integral in (2.4), and note that the PV
symbol is only needed for $k=i$.

$\bullet$ _Estimates for $k\neq i$ terms._ For any $k\neq i$, we aim to show
that

$|\mathbf{v}_{k}^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))-BR_{k}(z_{i}(\alpha))|\leq
C\epsilon,$ (3.12)

where $C$ depends on $d_{\Gamma},\max_{k}\|z_{k}\|_{C^{2}}$ and
$\max_{k}\|\gamma_{k}\|_{L^{\infty}}$. Applying a change of variable
$y=R_{k}^{\epsilon}(\alpha^{\prime},\eta^{\prime})$, we can rewrite
$\mathbf{v}_{k}^{\epsilon}$ as

$\begin{split}\mathbf{v}_{k}^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))&=\epsilon^{-1}\int_{D_{k}^{\epsilon}}K_{2}(R_{i}^{\epsilon}(\alpha,\eta)-y)\,dy\\\
&=\int_{S_{k}}\int_{-1}^{0}\underbrace{K_{2}(R_{i}^{\epsilon}(\alpha,\eta)-R_{k}^{\epsilon}(\alpha^{\prime},\eta^{\prime}))}_{=:T_{1}}\underbrace{\epsilon^{-1}\det(\nabla_{\alpha^{\prime},\eta^{\prime}}R_{k}^{\epsilon}(\alpha^{\prime},\eta^{\prime}))}_{=:T_{2}}\,d\eta^{\prime}d\alpha^{\prime}.\end{split}$
(3.13)

Using the facts that
$R_{i}^{\epsilon}(\alpha,\eta)-R_{k}^{\epsilon}(\alpha^{\prime},\eta^{\prime})=z_{i}(\alpha)-z_{k}(\alpha^{\prime})+O(\epsilon)$
as well as $|z_{i}(\alpha)-z_{k}(\alpha^{\prime})|\geq d_{\Gamma}>0$ (recall
that $d_{\Gamma}$ is as given in (2.1)), for all sufficiently small
$\epsilon>0$ we have
$T_{1}=K_{2}(z_{i}(\alpha)-z_{k}(\alpha^{\prime}))+O(\epsilon).$ For $T_{2}$,
the explicit formula (3.8) for the determinant gives
$T_{2}=L_{k}\gamma_{k}(\alpha^{\prime})+O(\epsilon).$ Plugging these into the
above integral yields

$\mathbf{v}_{k}^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))=\int_{S_{k}}K_{2}(z_{i}(\alpha)-z_{k}(\alpha^{\prime}))L_{k}\gamma_{k}(\alpha^{\prime})\,d\alpha^{\prime}+O(\epsilon)=BR_{k}(z_{i}(\alpha))+O(\epsilon),$

finishing the proof of (3.12).

$\bullet$ _Estimates for the $k=i$ term._ It will be more involved to control
the $k=i$ term, and our goal is to show that

$\left|\mathbf{v}_{i}^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))-BR_{i}(z_{i}(\alpha))+\Big{(}\eta+\frac{1}{2}\Big{)}[\mathbf{v}](z_{i}(\alpha))\right|\leq
C\epsilon^{b}|\log\epsilon|.$ (3.14)

To begin with, we again rewrite $\mathbf{v}_{i}^{\epsilon}$ as in (3.13) with
$k=i$, and plug in the formula (3.8) for the determinant. This leads to

$\begin{split}\mathbf{v}_{i}^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))&=\int_{S_{k}}\int_{-1}^{0}K_{2}(R_{i}^{\epsilon}(\alpha,\eta)-R_{i}^{\epsilon}(\alpha^{\prime},\eta^{\prime}))\left(L_{i}\gamma_{i}(\alpha^{\prime})-\epsilon
L_{i}\gamma_{i}^{2}(\alpha^{\prime})\kappa_{i}(\alpha^{\prime})\eta^{\prime}\right)\,d\eta^{\prime}d\alpha^{\prime}\\\
&=:I_{1}+I_{2},\end{split}$

where $I_{1},I_{2}$ are the contributions from the two terms in the last
parenthesis respectively. Let us control $I_{2}$ first, and we claim that

$|I_{2}|\leq C\epsilon|\log\epsilon|.$ (3.15)

Using (3.2) of Lemma 3.1 and the fact that $|K_{2}(x)|\leq|x|^{-1}$, we can
bound $I_{2}$ as

$\begin{split}|I_{2}|&=\left|\int_{S_{k}}\int_{-1}^{0}K_{2}(R_{i}^{\epsilon}(\alpha,\eta)-R_{i}^{\epsilon}(\alpha^{\prime},\eta^{\prime}))\,\epsilon
L_{i}\gamma_{i}^{2}(\alpha^{\prime})\kappa_{i}(\alpha^{\prime})\eta^{\prime}\,d\eta^{\prime}d\alpha^{\prime}\right|\\\
&\leq
C\epsilon\int_{S_{k}}\int_{-1}^{0}\frac{\gamma_{i}(\alpha^{\prime})}{|\alpha^{\prime}-\alpha|+\epsilon|\gamma_{i}(\alpha^{\prime})\eta^{\prime}-\gamma_{i}(\alpha)\eta|}\,d\eta^{\prime}d\alpha^{\prime}\\\
&\leq
C\epsilon\int_{S_{k}}\int_{-\|\gamma_{i}\|_{\infty}}^{\|\gamma_{i}\|_{\infty}}\frac{1}{|\alpha^{\prime}-\alpha|+\epsilon|\theta^{\prime}|}\,d\theta^{\prime}d\alpha^{\prime}\quad(\theta^{\prime}:=\gamma_{i}(\alpha^{\prime})\eta^{\prime}-\gamma_{i}(\alpha)\eta)\\\
&\leq
C\epsilon\int_{-1/\epsilon}^{1/\epsilon}\int_{-\|\gamma_{i}\|_{\infty}}^{\|\gamma_{i}\|_{\infty}}\frac{1}{|\beta^{\prime}|+|\theta^{\prime}|}\,d\theta^{\prime}d\beta^{\prime}\quad(\beta^{\prime}:=\epsilon^{-1}(\alpha^{\prime}-\alpha))\\\
&\leq C\epsilon|\log\epsilon|\end{split}$ (3.16)

where $C$ depends on $\|z_{i}\|_{C^{2}}$ and $\|\gamma_{i}\|_{L^{\infty}}$.

In the rest of the proof we focus on estimating
$I_{1}=\displaystyle\int_{S_{k}}\int_{-1}^{0}K_{2}(R_{i}^{\epsilon}(\alpha,\eta)-R_{i}^{\epsilon}(\alpha^{\prime},\eta^{\prime}))L_{i}\gamma_{i}(\alpha^{\prime})\,d\eta^{\prime}d\alpha^{\prime}.$
For $t\in[0,1]$, let us define

$\displaystyle
f(\alpha,\alpha^{\prime},\eta,\eta^{\prime};t):=R_{i}^{\epsilon}(\alpha,\eta-t\eta^{\prime})-R_{i}^{\epsilon}(\alpha^{\prime},\eta^{\prime}-t\eta^{\prime}),$
$\displaystyle
J(t):=\int_{S_{k}}\int_{-1}^{0}K_{2}(f(\alpha,\alpha^{\prime},\eta,\eta^{\prime};t))L_{i}\gamma_{i}(\alpha^{\prime})\,d\eta^{\prime}d\alpha^{\prime}.$
(3.17)

Note that in the definition of $f$, the argument $\eta-t\eta^{\prime}$ of
$R_{i}^{\epsilon}$ belongs to $[-1,1]$, instead of $[-1,0]$ as in the original
definition of (3.1). Here $R_{i}^{\epsilon}(\alpha,\eta-t\eta^{\prime})$ is
defined as in the formula (3.1), even though it might not belong to
$D^{\epsilon}_{i}$. Clearly, $J(0)=I_{1}$. The motivation for us to define
such $f$ and $J(t)$ is that at $t=1$, we have

$\begin{split}J(1)&=\int_{S_{k}}\int_{-1}^{0}K_{2}(R_{i}^{\epsilon}(\alpha,\eta-\eta^{\prime})-z_{i}(\alpha^{\prime}))L_{i}\gamma_{i}(\alpha^{\prime})\,d\eta^{\prime}d\alpha^{\prime}=\int_{-1}^{0}\mathbf{v}_{i}(R_{i}^{\epsilon}(\alpha,\eta-\eta^{\prime}))\,d\eta^{\prime},\end{split}$
(3.18)

where $\mathbf{v}_{i}$ is the velocity field generated by the sheet
$\Gamma_{i}$. Recall that $\mathbf{v}_{i}$ has a jump across $\Gamma_{i}$,
where we denote its limits on two sides by $\mathbf{v}_{i}^{\pm}$. Using Lemma
3.5, which we will prove momentarily, we have

$\mathbf{v}_{i}(R_{i}^{\epsilon}(\alpha,\eta-\eta^{\prime}))=\begin{cases}\mathbf{v}_{i}^{+}(z_{i}(\alpha))+O(\epsilon^{b}|\log\epsilon|)&\text{
if }\eta-\eta^{\prime}\in(0,2),\\\
\mathbf{v}_{i}^{-}(z_{i}(\alpha))+O(\epsilon^{b}|\log\epsilon|)&\text{ if
}\eta-\eta^{\prime}\in(-2,0).\end{cases}$ (3.19)

We can then split the integration domain on the right hand side of (3.18) into
$\eta^{\prime}\in(-1,\eta)$ and $\eta^{\prime}\in(\eta,0)$, and use (3.19) to
approximate the integrand in each interval. This gives

$\begin{split}J(1)&=(\eta+1)\mathbf{v}_{i}^{+}(z_{i}(\alpha))-\eta\mathbf{v}_{i}^{-}(z_{i}(\alpha))+O(\epsilon^{b}|\log\epsilon|)\\\
&=BR_{i}(z_{i}(\alpha))-\Big{(}\eta+\frac{1}{2}\Big{)}[\mathbf{v}](z_{i}(\alpha))+O(\epsilon^{b}|\log\epsilon|),\end{split}$
(3.20)

where in the last step we used that
$[\mathbf{v}](z_{i}(\alpha))=[\mathbf{v}_{i}](z_{i}(\alpha))$, since all other
$\mathbf{v}_{k}$ with $k\neq i$ are continuous across $\Gamma_{i}$.

Finally, it remains to control $|J(0)-J(1)|$. Note that by (3.2), we have

$f(\alpha,\alpha^{\prime},\eta,\eta^{\prime};t)\geq
c_{0}\big{(}|\alpha-\alpha^{\prime}|+\epsilon|\gamma_{i}(\alpha^{\prime})\eta^{\prime}-\gamma_{i}(\alpha)\eta|\big{)}.$

In addition, we have

$\left|\frac{\partial}{\partial
t}f(\alpha,\alpha^{\prime},\eta,\eta^{\prime};t)\right|=\left|\epsilon\big{(}\gamma_{i}(\alpha)\mathbf{n}(z_{i}(\alpha))-\gamma_{i}(\alpha^{\prime})\mathbf{n}(z_{i}(\alpha^{\prime}))\big{)}\eta^{\prime}\right|\leq
C\epsilon|\alpha-\alpha^{\prime}|^{b},$

where the last inequality follows from (H2) and the fact that
$\mathbf{n}(z_{i}(\alpha))\in C^{1}(S_{i})$. Therefore, for any $t\in(0,1)$,
taking the $t$ derivative of (3.17) and using that $|\nabla
K_{2}(x)|\leq|x|^{-2}$, we have

$\begin{split}|J^{\prime}(t)|&\leq
C\int_{S_{k}}\int_{-1}^{0}\frac{\epsilon|\alpha-\alpha^{\prime}|^{b}\gamma_{i}(\alpha^{\prime})}{\big{(}|\alpha-\alpha^{\prime}|+\epsilon|\gamma_{i}(\alpha^{\prime})\eta^{\prime}-\gamma_{i}(\alpha)\eta|\big{)}^{2}}d\eta^{\prime}d\alpha^{\prime}\\\
&\leq
C\epsilon\int_{S_{k}}\int_{-1}^{0}\frac{\gamma_{i}(\alpha^{\prime})}{|\alpha-\alpha^{\prime}|^{1-b}\left(|\alpha-\alpha^{\prime}|+\epsilon|\gamma_{i}(\alpha^{\prime})\eta^{\prime}-\gamma_{i}(\alpha)\eta|\right)}d\eta^{\prime}d\alpha^{\prime}\\\
&\leq
C\epsilon^{b}\int_{-1/\epsilon}^{1/\epsilon}\int_{-\|\gamma_{i}\|_{\infty}}^{\|\gamma_{i}\|_{\infty}}\frac{1}{|\beta^{\prime}|^{1-b}(|\beta^{\prime}|+|\theta^{\prime}|)}\,d\theta^{\prime}d\beta^{\prime}\quad(\theta^{\prime}:=\gamma_{i}(\alpha^{\prime})\eta^{\prime}-\gamma_{i}(\alpha)\eta,\,\beta^{\prime}:=\epsilon^{-1}(\alpha^{\prime}-\alpha))\\\
&\leq
C\epsilon^{b}\int_{-1/\epsilon}^{1/\epsilon}|\beta^{\prime}|^{b-1}\log\left(1+\frac{\|\gamma_{i}\|_{L^{\infty}}}{|\beta^{\prime}|}\right)d\beta^{\prime}\\\
&\leq C\epsilon^{b},\end{split}$

where $C$ depends on $b$, $\|\gamma_{i}\|_{C^{b}(S_{i})}$,
$\|z_{i}\|_{C^{2}(S_{i})}$ and $F_{\Gamma}$. This leads to

$|J(1)-I_{1}|=|J(1)-J(0)|\leq C\epsilon^{b}|\log\epsilon|.$

Finally, combining this with (3.20) and (3.15) yields (3.14), finishing the
proof of the $k=i$ case. We can then conclude the proof by taking the sum of
this estimate with all the $k\neq i$ estimates in (3.12). ∎

The following lemma proves (3.19). Let $\mathbf{v}_{i}$ be the velocity field
generated by the sheet $\Gamma_{i}$, which is smooth in
$\mathbb{R}^{2}\setminus\Gamma_{i}$, and has a discontinuity across
$\Gamma_{i}$. It is known that $\mathbf{v}_{i}$ converges to
$\mathbf{v}_{i}^{\pm}$ respectively on the two sides of $\Gamma_{i}$ [24].
However, we were unable to find a quantitative convergence rate (in terms of
the distance from the point to $\Gamma_{i}$) in the literature, especially
under the assumption that $\gamma_{i}$ is only in $C^{b}(S_{i})$ for the open
curves. Below we prove such an estimate.

######  Lemma 3.5.

For $i=1,\ldots,n+m$, let $\mathbf{v}_{i}$ be the velocity field generated by
the sheet $\Gamma_{i}$, given by

$\mathbf{v}_{i}(x):=\int_{S_{i}}K_{2}(x-z_{i}(\alpha^{\prime}))\,\gamma_{i}(\alpha^{\prime})|z_{i}^{\prime}(\alpha^{\prime})|\,d\alpha^{\prime}\quad\text{
for }x\in\mathbb{R}^{2}\setminus\Gamma_{i}.$

Then there exist constants $C,\epsilon_{0}>0$ depending on on $b$ (as in
(H2)), $\|z_{i}\|_{C^{2}(S_{i})}$, $\|\gamma_{i}\|_{C^{b}(S_{i})}$ and
$F_{\Gamma}$, such that for all $\epsilon\in(0,\epsilon_{0})$ and
$\eta\in(-2,2)$ we have

$\displaystyle\left|\mathbf{v}_{i}(R^{\epsilon}_{i}(\alpha,\eta))-\mathbf{v}_{i}^{+}(z_{i}(\alpha))\right|\leq
C\epsilon^{b}|\log\epsilon|\quad\text{ if }\eta\in(0,2),$ (3.21)
$\displaystyle\left|\mathbf{v}_{i}(R^{\epsilon}_{i}(\alpha,\eta))-\mathbf{v}_{i}^{-}(z_{i}(\alpha))\right|\leq
C\epsilon^{b}|\log\epsilon|\quad\text{ if }\eta\in(-2,0),$ (3.22)

where

$\mathbf{v}_{i}^{+}=BR_{i}(z_{i}(\alpha))+\frac{\mathbf{n}(z_{i}(\alpha))^{\perp}\gamma_{i}(\alpha)}{2},\quad\mathbf{v}_{i}^{-}=BR_{i}(z_{i}(\alpha))-\frac{\mathbf{n}(z_{i}(\alpha))^{\perp}\gamma_{i}(\alpha)}{2},$

and $BR_{i}$ is the contribution from the $i$-th integral in (2.4).

###### Proof.

We will show (3.21) only since (3.22) can be treated in the same way. From the
definition of $R^{\epsilon}_{i}$ in (3.1), we have

$\displaystyle\mathbf{v}_{i}(R^{\epsilon}_{i}(\alpha,\eta))$
$\displaystyle=\frac{L_{i}}{2\pi}\int_{S_{i}}\frac{\left(z_{i}(\alpha)-z_{i}(\alpha^{\prime})\right)^{\perp}\gamma_{i}(\alpha^{\prime})}{\left|z_{i}(\alpha)-z_{i}(\alpha^{\prime})+\epsilon\eta\mathbf{n}(z_{i}(\alpha))\gamma_{i}(\alpha)\right|^{2}}\,d\alpha^{\prime}+\frac{L_{i}}{2\pi}\int_{S_{i}}\frac{\epsilon\eta\mathbf{n}(z_{i}(\alpha))^{\perp}\gamma_{i}(\alpha)\gamma_{i}(\alpha^{\prime})}{\left|z_{i}(\alpha)-z_{i}(\alpha^{\prime})+\epsilon\eta\mathbf{n}(z_{i}(\alpha))\gamma_{i}(\alpha)\right|^{2}}\,d\alpha^{\prime}$
$\displaystyle=:A_{1}+A_{2}.$

We claim that for all $\epsilon>0$ sufficiently small and $\eta\in[0,2)$, we
have

$\displaystyle|A_{1}-BR_{i}(z(\alpha))|\leq C\epsilon^{b}|\log\epsilon|,$
(3.23)
$\displaystyle\left|A_{2}-\frac{\mathbf{n}(z(\alpha))^{\perp}\gamma(\alpha)}{2}\right|\leq
C\epsilon^{b},$ (3.24)

and note that these two claims immediately yield (3.21). From now on, let us
fix $i\in\left\\{1,\ldots,n+m\right\\}$ and omit it in the notation for
simplicity. Throughout this proof, let us denote

$\mathbf{y}(\alpha,\alpha^{\prime}):=z(\alpha)-z(\alpha^{\prime})\quad\text{
and
}\quad\mathbf{c}(\alpha):=\epsilon\eta\mathbf{n}(z(\alpha))\gamma(\alpha),$

so that

$A_{1}=\frac{L}{2\pi}\int_{S}\frac{\mathbf{y}^{\perp}(\alpha,\alpha^{\prime})\gamma(\alpha^{\prime})}{|\mathbf{y}(\alpha,\alpha^{\prime})+\mathbf{c}(\alpha)|^{2}}d\alpha^{\prime},\quad
A_{2}=\frac{L}{2\pi}\int_{S}\frac{\mathbf{c}^{\perp}(\alpha)\gamma(\alpha^{\prime})}{|\mathbf{y}(\alpha,\alpha^{\prime})+\mathbf{c}(\alpha)|^{2}}d\alpha^{\prime}.$

Note that

$F_{\Gamma}^{-1}|\alpha-\alpha^{\prime}|\leq|\mathbf{y}(\alpha,\alpha^{\prime})|\leq\|z\|_{C^{1}}|\alpha-\alpha^{\prime}|.$
(3.25)

For the closed curves with $i=1,\dots,n$, since $z$ has period 1, we can
always set $\alpha-\alpha^{\prime}\in[-\frac{1}{2},\frac{1}{2})$ in this
proof.

Applying (3.2) (with $\eta^{\prime}=0$), we have

$|\mathbf{y}(\alpha,\alpha^{\prime})+\mathbf{c}(\alpha)|^{2}\geq
c_{0}(|\alpha-\alpha^{\prime}|^{2}+\epsilon^{2}\eta^{2}\gamma^{2}(\alpha))=c_{0}(|\alpha-\alpha^{\prime}|^{2}+|\mathbf{c}(\alpha)|^{2}).$
(3.26)

Since $z^{\prime}(\alpha)=L\mathbf{s}(z(\alpha))$, let us define

$\tilde{\mathbf{y}}(\alpha,\alpha^{\prime}):=L\mathbf{s}(z(\alpha))(\alpha-\alpha^{\prime}),$

which is a close approximation of $\mathbf{y}$ in the sense that

$|\mathbf{y}(\alpha,\alpha^{\prime})-\tilde{\mathbf{y}}(\alpha,\alpha^{\prime})|\leq\|z\|_{C^{2}}(\alpha-\alpha^{\prime})^{2}.$
(3.27)

Using $\mathbf{s}(z(\alpha))\perp\mathbf{n}(z(\alpha))$, we have

$|\tilde{\mathbf{y}}(\alpha,\alpha^{\prime})+\mathbf{c}(\alpha)|^{2}=L^{2}|\alpha-\alpha^{\prime}|^{2}+\epsilon^{2}\eta^{2}\gamma^{2}(\alpha)=L^{2}|\alpha-\alpha^{\prime}|^{2}+|\mathbf{c}(\alpha)|^{2}.$
(3.28)

From now on, for notational simplicity, we compress the dependence of
$\mathbf{y}(\alpha,\alpha^{\prime}),\tilde{\mathbf{y}}(\alpha,\alpha^{\prime}),\mathbf{c}(\alpha)$
on $\alpha$ and $\alpha^{\prime}$ in the rest of the proof.

$\bullet$ _Estimate ( 3.23)._ Note that $BR_{i}(z(\alpha))$ can also be
written using the above notations as

$BR_{i}(z(\alpha))=\frac{L}{2\pi}PV\int_{S}\frac{\mathbf{y}^{\perp}}{|\mathbf{y}|^{2}}\gamma(\alpha^{\prime})d\alpha,$

thus $A_{1}-BR_{i}(z(\alpha))$ can be written as follows:

$\begin{split}A_{1}-BR_{i}(z(\alpha))&=\frac{L}{2\pi}PV\int_{S}\underbrace{\left(\frac{\mathbf{y}^{\perp}}{|\mathbf{y}+\mathbf{c}|^{2}}-\frac{\mathbf{y}^{\perp}}{|\mathbf{y}|^{2}}\right)}_{=:\mathbf{f}(\mathbf{y},\mathbf{c})}\gamma(\alpha^{\prime})d\alpha^{\prime}\\\
&=\frac{L}{2\pi}\int_{S}\mathbf{f}(\mathbf{y},\mathbf{c})(\gamma(\alpha^{\prime})-\gamma(\alpha))d\alpha^{\prime}+\frac{L\gamma(\alpha)}{2\pi}PV\int_{S}\mathbf{f}(\mathbf{y},\mathbf{c})d\alpha^{\prime}\\\
&=:A_{11}+A_{12}.\end{split}$

A direct computation gives

$\mathbf{f}(\mathbf{y},\mathbf{c})=-\frac{\mathbf{y}^{\perp}}{|\mathbf{y}|^{2}}\,\frac{2\mathbf{y}\cdot\mathbf{c}+|\mathbf{c}|^{2}}{|\mathbf{y}+\mathbf{c}|^{2}}.$
(3.29)

Since
$\mathbf{y}\cdot\mathbf{c}=(\mathbf{y}-\tilde{\mathbf{y}})\cdot\mathbf{c}\leq
C|\alpha-\alpha^{\prime}|^{2}|\mathbf{c}|,$ (where we use
$\tilde{\mathbf{y}}\perp\mathbf{n}(z(\alpha))$ and (3.27)), combining this
with (3.25) and (3.26) gives a crude bound

$|\mathbf{f}(\mathbf{y},\mathbf{c})|\lesssim\frac{|\alpha-\alpha^{\prime}|^{2}|\mathbf{c}|+|\mathbf{c}|^{2}}{|\alpha-\alpha^{\prime}|(|\alpha-\alpha^{\prime}|^{2}+|\mathbf{c}|^{2})}.$

Plugging this into $A_{11}$ and using the Hölder continuity of $\gamma$, we
have

$\begin{split}|A_{11}|&\lesssim\int_{S}\frac{|\alpha-\alpha^{\prime}|^{2}|\mathbf{c}|+|\mathbf{c}|^{2}}{|\alpha-\alpha^{\prime}|(|\alpha-\alpha^{\prime}|^{2}+|\mathbf{c}|^{2})}|\alpha-\alpha^{\prime}|^{b}d\alpha^{\prime}\\\
&\lesssim\int_{|\theta|<|\mathbf{c}|}(|\theta|^{1+b}|\mathbf{c}|^{-1}+|\theta|^{b-1})d\theta+\int_{|\mathbf{c}|\leq|\theta|\leq
1}(|\mathbf{c}||\theta|^{b-1}+|\mathbf{c}|^{2}|\theta|^{b-3})d\theta\quad(\theta:=\alpha^{\prime}-\alpha)\\\
&\lesssim|\mathbf{c}|^{b}\leq C\epsilon^{b},\end{split}$

where the last step follows from the fact that $|\mathbf{c}|\leq
2\epsilon\|\gamma\|_{\infty}$. Now let us turn to $A_{12}$, which requires a
more delicate estimate of $\mathbf{f}(\mathbf{y},\mathbf{c})$. Let us break
$A_{12}$ as

$\begin{split}A_{12}&=\frac{L\gamma(\alpha)}{2\pi}\int_{S}(\mathbf{f}(\mathbf{y},\mathbf{c})-\mathbf{f}(\tilde{\mathbf{y}},\mathbf{c}))d\alpha^{\prime}+\frac{L\gamma(\alpha)}{2\pi}PV\int_{S}\mathbf{f}(\tilde{\mathbf{y}},\mathbf{c})d\alpha^{\prime}=:B_{1}+B_{2}.\end{split}$

For $B_{1}$, let us take the gradient of $\mathbf{f}(\mathbf{y},\mathbf{c})$
(as in (3.29)) in the first variable. An elementary computation yields that

$|\nabla_{\mathbf{x}}\mathbf{f}(\mathbf{x},\mathbf{c})|\leq
C|\mathbf{x}|^{-2}\min\Big{\\{}1,\frac{|\mathbf{c}|}{|\mathbf{x}|}\Big{\\}}$
(3.30)

as long as $\mathbf{x}$ satisfies

$|\mathbf{x}+\mathbf{c}|^{2}\geq c_{0}(|\mathbf{x}|^{2}+|\mathbf{c}|^{2}).$
(3.31)

We point out that $\mathbf{x}=\xi\mathbf{y}+(1-\xi)\tilde{\mathbf{y}}$ indeed
satisfies (3.31) for all $\xi\in[0,1]$: to see this, in the proof of Lemma
3.1, if we replace $T_{1}$ in (3.3) by
$\xi\mathbf{y}+(1-\xi)\tilde{\mathbf{y}}$, one can easily check the proof
still goes through for $\xi\in[0,1]$. In addition, for any $\xi\in[0,1]$ we
also have

$|\xi\mathbf{y}+(1-\xi)\tilde{\mathbf{y}}|\geq c_{0}|\alpha-\alpha^{\prime}|.$
(3.32)

Thus the gradient estimate (3.30) together with (3.27) and (3.32) yields

$|f(\mathbf{y},\mathbf{c})-f(\tilde{\mathbf{y}},\mathbf{c})|\lesssim\min\\{1,|\mathbf{c}||\alpha-\alpha^{\prime}|^{-1}\\}\lesssim\min\\{1,\epsilon|\alpha-\alpha^{\prime}|^{-1}\\},$

and plugging this into $B_{1}$ gives

$\begin{split}|B_{1}|&\lesssim\epsilon+\int_{\epsilon<|\alpha-\alpha^{\prime}|<1}\epsilon|\alpha-\alpha^{\prime}|^{-1}d\alpha^{\prime}\lesssim\epsilon|\log\epsilon|.\end{split}$

As for $B_{2}$, using the definition of $\tilde{\mathbf{y}}$, the identity
(3.28) and the fact that $\tilde{\mathbf{y}}\cdot\mathbf{c}=0$, we have

$\begin{split}B_{2}&=\frac{L\gamma(\alpha)}{2\pi}PV\int_{S}-\frac{\tilde{\mathbf{y}}^{\perp}}{|\tilde{\mathbf{y}}|^{2}}\,\frac{|\mathbf{c}|^{2}}{|\tilde{\mathbf{y}}+\mathbf{c}|^{2}}d\alpha^{\prime}\\\
&=\frac{L\gamma(\alpha)|\mathbf{c}|^{2}\mathbf{n}(z(\alpha))}{2\pi
L}PV\int_{S}\frac{\alpha^{\prime}-\alpha}{|\alpha^{\prime}-\alpha|^{2}(L^{2}|\alpha^{\prime}-\alpha|^{2}+|\mathbf{c}|^{2})}d\alpha^{\prime}.\end{split}$

For the closed curves $i=1,\dots,n$, we immediately have $B_{2}=0$ since
$\alpha-\alpha^{\prime}\in[-\frac{1}{2},\frac{1}{2})$, and the integrand is an
odd function of $\alpha^{\prime}-\alpha$.

For the open curves $i=n+1,\dots,n+m$, the above integral becomes

$\begin{split}B_{2}&=\frac{L\gamma(\alpha)|\mathbf{c}|^{2}\mathbf{n}(z(\alpha))}{2\pi
L}PV\int_{-\alpha}^{1-\alpha}\frac{\theta}{|\theta|^{2}(L^{2}|\theta|^{2}+|\mathbf{c}|^{2})}d\theta\quad(\theta:=\alpha^{\prime}-\alpha)\\\
&=\frac{L\gamma(\alpha)|\mathbf{c}|^{2}\mathbf{n}(z(\alpha))}{2\pi
L}\int_{\alpha}^{1-\alpha}\frac{\theta}{\theta^{2}(L^{2}\theta^{2}+|\mathbf{c}|^{2})}d\theta,\end{split}$

where in the second inequality we used that the integral in $[-\alpha,\alpha]$
gives zero contribution to the principal value, since the integrand is odd.

Next we discuss two cases. If $\alpha>|\mathbf{c}|$, we bound the integrand by
$C\theta^{-3}$, which gives

$|B_{2}|\leq C\gamma(\alpha)|\mathbf{c}|^{2}\alpha^{-2}\leq
C|\mathbf{c}|^{2}\alpha^{b-2}\leq C|\mathbf{c}|^{b}\leq C\epsilon^{b}.$

where the second inequality follows from the assumption $\gamma(0)=0$ for an
open curve in (H3), as well as the Hölder continuity of $\gamma$. And if
$0<\alpha\leq|\mathbf{c}|$, the integrand can be bounded above by
$\theta^{-1}|\mathbf{c}|^{-2}$, which immediately leads to

$|B_{2}|\leq C\gamma(\alpha)|\log\alpha|\leq
C|\mathbf{c}|^{b}|\log|\mathbf{c}||\leq C\epsilon^{b}|\log\epsilon|.$

In both cases we have $|B_{2}|\leq C\epsilon^{b}|\log\epsilon|,$ and combining
it with the $B_{1}$ and $A_{11}$ estimates gives (3.23).

$\bullet$ _Estimate ( 3.24)._ We break $A_{2}$ into

$\displaystyle A_{2}$
$\displaystyle=\frac{L\mathbf{c}^{\perp}}{2\pi}\int_{S}\frac{\gamma(\alpha^{\prime})-\gamma(\alpha)}{|\mathbf{y}+\mathbf{c}|^{2}}d\alpha^{\prime}+\frac{L\mathbf{c}^{\perp}\gamma(\alpha)}{2\pi}\int_{S}\left(\frac{1}{|\mathbf{y}+\mathbf{c}|^{2}}-\frac{1}{|\tilde{\mathbf{y}}+\mathbf{c}|^{2}}\right)d\alpha^{\prime}+\frac{L\mathbf{c}^{\perp}\gamma(\alpha)}{2\pi}\int_{S}\frac{1}{|\tilde{\mathbf{y}}+\mathbf{c}|^{2}}d\alpha^{\prime}$
$\displaystyle=:A_{21}+A_{22}+A_{23}.$

For $A_{21}$, (3.26) and the Hölder continuity of $\gamma$ immediately lead to

$|A_{21}|\leq
C|\mathbf{c}|\int_{S}\frac{|\alpha-\alpha^{\prime}|^{b}}{|\alpha-\alpha^{\prime}|^{2}+|\mathbf{c}|^{2}}d\alpha^{\prime}\leq|\mathbf{c}|^{b}\leq
C\epsilon^{b}.$ (3.33)

For $A_{22}$, its integrand can be controlled as

$\left|\frac{1}{|\mathbf{y}+\mathbf{c}|^{2}}-\frac{1}{|\tilde{\mathbf{y}}+\mathbf{c}|^{2}}\right|\leq\frac{|\mathbf{y}-\mathbf{\tilde{y}}|(|\mathbf{y}+\mathbf{c}|+|\tilde{\mathbf{y}}+\mathbf{c}|)}{|\mathbf{y}+\mathbf{c}|^{2}|\tilde{\mathbf{y}}+\mathbf{c}|^{2}}\leq\frac{C|\alpha-\alpha^{\prime}|^{2}}{(|\alpha-\alpha^{\prime}|^{2}+|\mathbf{c}|^{2})^{3/2}},$

where the last step follows from (3.26), (3.27) and (3.28). This allows us to
control $A_{22}$ as

$|A_{22}|\leq
C|\mathbf{c}|\int_{-1}^{1}\frac{\theta^{2}}{(\theta^{2}+|\mathbf{c}|^{2})^{3/2}}d\theta\leq
C|\mathbf{c}|\,\big{|}\log|\mathbf{c}|\big{|}\leq C\epsilon|\log\epsilon|.$
(3.34)

Finally, for the $A_{23}$ term, (3.28) gives

$A_{23}=\frac{L\mathbf{c}^{\perp}\gamma(\alpha)}{2\pi}\int_{S}\frac{1}{L^{2}|\alpha^{\prime}-\alpha|^{2}+|\mathbf{c}|^{2}}d\alpha^{\prime}=\frac{\mathbf{n}^{\perp}(\alpha)\gamma(\alpha)}{2\pi}\int_{I}\frac{1}{\theta^{2}+1}d\theta\quad(\text{set
}\theta:=\frac{L(\alpha^{\prime}-\alpha)}{|\mathbf{c}|}),$

where the integration interval
$I=(-\frac{L}{2|\mathbf{c}|},\frac{L}{2|\mathbf{c}|})$ for $i=1,\dots,n$, and
$I=(-\frac{L\alpha}{|\mathbf{c}|},\frac{L(1-\alpha)}{|\mathbf{c}|})$ for
$i=n+1,\dots,n+m$, and in the last equality we also used that
$\frac{\mathbf{c}^{\perp}}{|\mathbf{c}|}=\mathbf{n}^{\perp}$. For
$i=1,\dots,n$, one can easily check that

$\left|\int_{I}\frac{1}{\theta^{2}+1}d\theta-\pi\right|=2\int_{\frac{L}{2|\mathbf{c}|}}^{\infty}\frac{1}{\theta^{2}+1}d\theta\leq
C|\mathbf{c}|\leq C\epsilon,$

which immediately leads to

$\left|A_{23}-\frac{\mathbf{n}(z(\alpha))^{\perp}\gamma(\alpha)}{2}\right|=\left|\frac{\mathbf{n}^{\perp}(\alpha)\gamma(\alpha)}{2\pi}\left(\int_{I}\frac{1}{\theta^{2}+1}d\theta-\pi\right)\right|\leq
C\epsilon$

for $i=1,\dots,n$. Next we turn to the open curves $i=n+1,\dots,n+m$, and let
us assume $\alpha\in[0,\frac{1}{2}]$ without loss of generality. In this case
we have

$\left|\int_{I}\frac{1}{\theta^{2}+1}d\theta-\pi\right|=\int^{-\frac{L\alpha}{|\mathbf{c}|}}_{-\infty}\frac{1}{\theta^{2}+1}d\theta+\int_{\frac{L(1-\alpha)}{|\mathbf{c}|}}^{\infty}\frac{1}{\theta^{2}+1}d\theta\leq\min\left\\{C\frac{|\mathbf{c}|}{\alpha},\frac{\pi}{2}\right\\}+C\epsilon.$

where we used $1-\alpha>\frac{1}{2}$ to control the second integral by
$C\epsilon$. Using the above inequality as well as the fact that
$\gamma(\alpha)\leq C\alpha^{b}$ due to (H3), we have

$\left|A_{23}-\frac{\mathbf{n}(z(\alpha))^{\perp}\gamma(\alpha)}{2}\right|=\frac{\gamma(\alpha)}{2\pi}\left|\int_{I}\frac{1}{\theta^{2}+1}d\theta-\pi\right|\leq
C\alpha^{b}\min\left\\{\frac{|\mathbf{c}|}{\alpha},1\right\\}+C\epsilon\leq
C(|\mathbf{c}|^{b}+\epsilon)\leq C\epsilon^{b}$

for $i=n+1,\dots,n+m$. Finally, combining the $A_{23}$ estimates together with
(3.33) and (3.34) yields (3.24). ∎

##  4\. Constructing a divergence-free perturbation

In this section, we aim to construct a divergence-free velocity field
$\mathbf{u}^{\epsilon}:D^{\epsilon}\to\mathbb{R}^{2}$, such that
$-\mathbf{u}^{\epsilon}$ tends to make each $D_{i}^{\epsilon}$ “more
symmetric”. Let $\mathbf{u}^{\epsilon}:D^{\epsilon}\to\mathbb{R}^{2}$ be given
by

$\mathbf{u}^{\epsilon}:=x+\nabla p^{\epsilon}\quad\text{ in }D^{\epsilon},$
(4.1)

where the function $p^{\epsilon}:\overline{D^{\epsilon}}\to\mathbb{R}$ is
chosen such that

$\nabla\cdot\mathbf{u}^{\epsilon}=0\quad\text{ in }D^{\epsilon},$ (4.2)

and on each connected component $l$ of $\partial D^{\epsilon}$, $u^{\epsilon}$
satisfies

$\int_{l}\mathbf{u}^{\epsilon}\cdot n\,d\sigma=0,$ (4.3)

where $n$ is the unit normal of $l$ pointing outwards of $D^{\epsilon}$. Note
that $\partial D^{\epsilon}$ has a total of $2n+m$ connected components:
$D_{i}^{\epsilon}$ is doubly-connected for $i=1,\dots,n$ (denote its outer and
inner boundaries by $\partial D_{i,\text{out}}^{\epsilon}$ and $\partial
D_{i,\text{in}}^{\epsilon}$; note that $\partial D_{i,\text{in}}^{\epsilon}$
coincides with $\Gamma_{i}$), whereas it is simply-connected for
$i=n+1,\dots,n+m$ (denote its boundary by $\partial D_{i}^{\epsilon}$).

Next we show that there indeed exists a function $p^{\epsilon}$ so that
$\mathbf{u}^{\epsilon}$ satisfies (4.2)–(4.3). Clearly, (4.2) requires that
$p^{\epsilon}$ satisfies

$\Delta p^{\epsilon}=-2\quad\text{ in }D^{\epsilon}.$ (4.4)

As for the boundary conditions, we let

$p^{\epsilon}|_{\partial D_{i}^{\epsilon}}=0\quad\text{for }i=n+1,\dots,n+m,$
(4.5)

so the divergence theorem yields that (4.3) is satisfied for each $l=\partial
D_{i}^{\epsilon}$ for $i=n+1,\dots,n+m$. As for $i=1,\dots,n$, we define

$p^{\epsilon}=\begin{cases}0&\text{ on }\partial
D_{i,\text{out}}^{\epsilon}\\\ c_{i}^{\epsilon}&\text{ on }\partial
D_{i,\text{in}}^{\epsilon}=\Gamma_{i}\end{cases}\quad\text{ for }i=1,\dots,n,$
(4.6)

where $c_{i}^{\epsilon}>0$ is the unique constant such that

$\int_{\partial U_{i}}\nabla p^{\epsilon}\cdot nd\sigma=-2|U_{i}|\quad\text{
for }i=1,\dots,n,$ (4.7)

where $U_{i}$ is the domain enclosed by $\partial
D_{i,\text{in}}^{\epsilon}=\Gamma_{i}$ (thus $U_{i}$ is independent of
$\epsilon$), and $n$ is the outer normal of $U_{i}$ (thus the inner normal of
$D_{i}^{\epsilon}$). The existence of $c_{i}^{\epsilon}$ is guaranteed by [14,
Lemma 2.5]. One can then check that $\int_{\partial
U_{i}}\mathbf{u}^{\epsilon}\cdot nd\sigma=0$. Applying the divergence theorem
in $D^{\epsilon}_{i}$ then gives us that $\int_{\partial
D_{i,\text{out}}^{\epsilon}}\mathbf{u}^{\epsilon}\cdot nd\sigma=0$ as well.

In [14] we proved a rearrangement inequality for such $p^{\epsilon}$ in a
similar spirit of Talenti’s rearrangement inequality for elliptic equations
[37], which we state below.

######  Lemma 4.1 ([14, Proposition 2.6]).

The function $p^{\epsilon}:\overline{D^{\epsilon}}\to\mathbb{R}$ defined in
(4.4)–(4.7) satisfies the following in each $D_{i}^{\epsilon}$ for
$i=1,\dots,n+m$:

$\sup_{D_{i}^{\epsilon}}p^{\epsilon}\leq\frac{|D_{i}^{\epsilon}|}{2\pi},$
(4.8)

and

$\int_{D_{i}^{\epsilon}}p^{\epsilon}(x)dx\leq\frac{|D_{i}^{\epsilon}|^{2}}{4\pi}.$
(4.9)

Moreover, each inequality above achieves equality if and only
$D_{i}^{\epsilon}$ is either a disk or an annulus.

Note that the inequalities (4.8)–(4.9) hold for any domain with $C^{1,\alpha}$
boundary. Even though the inequalities are strict when $D_{i}^{\epsilon}$ is
non-radial, they are not strong enough to rule out non-radial vortex sheets,
as we need quantitative versions of strict inequalities that are still valid
in the $\epsilon\to 0^{+}$ limit. As we will see in the proof of Proposition
5.2, the key step is to show that if some $\Gamma_{i}$ is either not a circle
or does not have a constant $\gamma_{i}$, then the following quantitative
version of (4.9) holds:
$\epsilon^{-2}\left(\frac{|D_{i}^{\epsilon}|^{2}}{4\pi}-\int_{D_{i}^{\epsilon}}p^{\epsilon}(x)dx\right)\geq
c_{0}>0$, where $c_{0}$ is independent of $\epsilon$.

In order to upgrade (4.9) into a quantitative version, we need to obtain some
fine estimates for $p^{\epsilon}$ that take into account the shape of the thin
domains $D^{\epsilon}_{i}$. For $i=n+1,\dots,n+m$, since $p^{\epsilon}=0$ on
$\partial D_{i}^{\epsilon}$, and the domain $D_{i}^{\epsilon}$ is a thin
simply-connected domain with width $\epsilon\ll 1$, intuitively one would
expect that $|p^{\epsilon}|\leq C\epsilon^{2}$. The next proposition shows
that this crude estimate is indeed true, and its proof is postponed to Section
4.1.

######  Proposition 4.2.

For any $i=n+1,\dots,n+m$, let
$p^{\epsilon}:\overline{D_{i}^{\epsilon}}\to\mathbb{R}$ be given by
(4.4)–(4.5). Then there exist $\epsilon_{1}$ and $C$ only depending on $\rVert
z_{i}\rVert_{C^{2}(S_{i})},\rVert\gamma_{i}\rVert_{L^{\infty}(S_{i})}$ and
$F_{\Gamma}$, such that

$|p^{\epsilon}|\leq C\epsilon^{2}\quad\text{ in }D_{i}^{\epsilon}$

for all $\epsilon\in(0,\epsilon_{1})$.

For $i=1,\dots,n$, the estimate is more involved, since $p^{\epsilon}$ takes
different values $c^{\epsilon}_{i}$ and $0$ on the inner and outer boundaries
of $D^{\epsilon}_{i}$. Heuristically speaking, since $D^{\epsilon}_{i}$ is a
doubly-connected thin tubular domain with width $\sim\epsilon$, we would
expect that $p^{\epsilon}_{i}$ (in $\alpha,\eta$ coordinate) changes almost
linearly from $0$ to $c^{\epsilon}_{i}$ as $\eta$ goes from $-1$ (outer
boundary) to $0$ (inner boundary). Next we will show that the error between
$p^{\epsilon}(R^{\epsilon}_{i}(\alpha,\eta))$ and the linear-in-$\eta$
function $c^{\epsilon}_{i}(1+\eta)$ is indeed controlled by $O(\epsilon^{2})$.
We will also obtain fine estimates of the gradient of the function
$c^{\epsilon}_{i}(1+\eta)$, as well as the boundary value $c^{\epsilon}_{i}$.
Again, its proof is postponed to Section 4.1.

######  Proposition 4.3.

For any $i=1,\dots,n$, let
$p^{\epsilon}:\overline{D_{i}^{\epsilon}}\to\mathbb{R}$ and
$c^{\epsilon}_{i}\in\mathbb{R}$ be given by (4.4) and (4.6)–(4.7). For such
$p^{\epsilon}$, let us define
$\tilde{p}^{\epsilon}$,$q^{\epsilon}:\overline{D_{i}^{\epsilon}}\mapsto\mathbb{R}$
as follows:

$\displaystyle\tilde{p}^{\epsilon}(R^{\epsilon}_{i}(\alpha,\eta))$
$\displaystyle:=c^{\epsilon}_{i}(1+\eta)\quad\text{ for }\alpha\in
S_{i},\eta\in[0,-1],$ (4.10) $\displaystyle q^{\epsilon}$
$\displaystyle:=p^{\epsilon}-\tilde{p}^{\epsilon}\quad\quad\text{ in
}\overline{D^{\epsilon}_{i}}.$

Also let

$\beta_{i}:=\frac{2|U_{i}|}{L_{i}\int_{S_{i}}\gamma_{i}^{-1}(\alpha)d\alpha}.$
(4.11)

Then there exist $\epsilon_{1}$ and $C$ only depending on $\rVert
z_{i}\rVert_{C^{3}(S_{i})},\rVert\gamma_{i}\rVert_{C^{2}(S_{i})}$ and
$F_{\Gamma}$, such that for all $\epsilon\in(0,\epsilon_{1})$ we have the
following:

$\displaystyle\begin{cases}|q^{\epsilon}|\leq C\epsilon^{2}&\text{ in
}D^{\epsilon}_{i},\\\ q^{\epsilon}=0&\text{ on }\partial
D^{\epsilon}_{i},\end{cases}$ (4.12)
$\displaystyle\left|\frac{c^{\epsilon}_{i}}{\epsilon}-\beta_{i}\right|\leq
C\epsilon,$ (4.13)
$\displaystyle\left|\nabla\tilde{p}^{\epsilon}(R^{\epsilon}_{i}(\alpha,\eta))-\frac{\beta_{i}}{\gamma_{i}(\alpha)}\mathbf{n}(z_{i}(\alpha))\right|\leq
C\epsilon\quad\text{ for }\alpha\in S_{i},\eta\in[0,-1].$ (4.14)

###  4.1. Proof of the quantitative lemmas for $p^{\epsilon}$

In this subsection we aim to prove Propositions 4.2 and 4.3. We start with a
technical lemma on estimating the solution of Poisson’s equation (with zero
boundary condition) in the domain $D^{\epsilon}_{i}$.

######  Lemma 4.4.

For any $i=1,\dots,n+m$, assume $\Gamma_{i}$ and $\gamma_{i}$ satisfy
(H1)–(H3). Let $v^{\epsilon}\in C^{2}(D^{\epsilon}_{i})\cap
C(\overline{D^{\epsilon}_{i}})$ solve the Poisson’s equation with zero
boundary condition:

$\begin{cases}\Delta v^{\epsilon}=-1&\text{ in }D^{\epsilon}_{i},\\\
v^{\epsilon}=0&\text{ on }\partial D^{\epsilon}_{i}.\end{cases}$ (4.15)

Then there exist positive constants $\epsilon_{0}=C(\rVert
z_{i}\rVert_{C^{2}(S_{i})},\rVert\gamma_{i}\rVert_{L^{\infty}(S_{i})},F_{\Gamma})$
and $C_{1},C_{2}=C(\|\gamma_{i}\|_{L^{\infty}(S_{i})})$, such that for all
$\epsilon\in(0,\epsilon_{0})$ we have

$0\leq v^{\epsilon}\leq C_{1}\epsilon^{2}\quad\text{ in }D^{\epsilon}_{i}$
(4.16)

and

$\rVert\nabla v^{\epsilon}\rVert_{L^{\infty}(\Gamma_{i})}\leq
C_{2}\epsilon\quad\text{ for }i=1,\ldots,n.$ (4.17)

###### Proof.

Throughout the proof, let $i\in\\{1,\dots,n+m\\}$ be fixed. For notational
simplicity, in the rest of the proof we omit the subscript $i$ in
$R^{\epsilon}_{i}$, $D^{\epsilon}_{i}$, $S_{i}$, $z_{i}$ and $\gamma_{i}$.

Step 1. We start with a simple geometric result that $D^{\epsilon}$ is “flat”
in a small neighborhood of any $z(\alpha)$. For any $\alpha\in S$, let
$V^{\epsilon}(\alpha):=D^{\epsilon}\cap
B_{6\epsilon\|\gamma\|_{\infty}}(z(\alpha))$, where $\|\cdot\|_{\infty}$
denotes $\|\cdot\|_{L^{\infty}(S)}$. We will show that any $y\in
V^{\epsilon}(\alpha)$ satisfies

$\big{|}(z(\alpha)-y)\cdot\mathbf{n}(z(\alpha))\big{|}\leq
2\epsilon\|\gamma\|_{\infty}$ (4.18)

for all sufficiently small $\epsilon>0$ (to be quantified in (4.23)). See
Figure 4(a) for an illustration.

Since $y\in V^{\epsilon}(\alpha)\subset D^{\epsilon}$, there exist $\beta\in
S$ and $\eta\in(-1,0)$ such that
$y=R^{\epsilon}(\beta,\eta)=z(\beta)+\epsilon\gamma(\beta)\mathbf{n}(z(\beta))\eta$.
It follows that

$\begin{split}\big{|}(z(\alpha)-y)\cdot\mathbf{n}(z(\alpha))\big{|}&\leq|(z(\alpha)-z(\beta))\cdot\mathbf{n}(z(\alpha))|+\epsilon\rVert\gamma\rVert_{\infty}\\\
&\leq\rVert
z^{\prime\prime}\rVert_{\infty}(\alpha-\beta)^{2}+\epsilon\rVert\gamma\rVert_{\infty},\end{split}$
(4.19)

where in the second inequality we used

$|(z(\alpha)-z(\beta))-z^{\prime}(\alpha)(\alpha-\beta)|\leq\|z^{\prime\prime}\|_{\infty}(\alpha-\beta)^{2}$
(4.20)

and $z^{\prime}(\alpha)\cdot\mathbf{n}(z(\alpha))=0$. To bound $\alpha-\beta$
on the right hand side of (4.19), the fact that $y\in
B_{6\epsilon\|\gamma\|_{\infty}}(z(\alpha))$ gives

$6\epsilon\rVert\gamma\rVert_{\infty}\geq|z(\alpha)-y|\geq|z(\alpha)-z(\beta)|-\epsilon\gamma(\beta),$
(4.21)

which implies $|z(\alpha)-z(\beta)|\leq 7\epsilon\rVert\gamma\rVert_{\infty}$.
Since the arc-chord constant $F_{\Gamma}$ given in (2.2) is finite, this
implies

$|\alpha-\beta|\leq 7F_{\Gamma}\rVert\gamma\rVert_{\infty}\epsilon.$ (4.22)

Plugging this into the right hand side of (4.19), we know (4.18) holds for all

$0<\epsilon\leq(49\|z^{\prime\prime}\|_{\infty}F_{\Gamma}^{2}\|\gamma\|_{\infty})^{-1}.$
(4.23)

(a) (b)

Figure 4. (a) In Step 1, $V^{\epsilon}(\alpha)$ (the yellow set) must lie
between the two dashed lines for small $\epsilon$. (b) In Step 2, $\partial
V^{\epsilon}(\alpha_{0})$ is decomposed into $\partial
V_{1}^{\epsilon}(\alpha_{0})$ (in dark green) and $\partial
V_{2}^{\epsilon}(\alpha_{0})$ (in purple).

Step 2. Next we prove (4.16). Note that $v^{\epsilon}$ is superharmonic in
$D^{\epsilon}$ and vanishes on the boundary, thus it follows from the maximum
principle that $v^{\epsilon}\geq 0$ in $D^{\epsilon}$. Denote $M:=\max_{x\in
D^{\epsilon}}v(x)$, and pick $x_{0}=R(\alpha_{0},\eta_{0})\in D^{\epsilon}$
such that $v(x_{0})=M$. Without loss of generality, we can assume that
$z(\alpha_{0})=(0,0)$ and $\mathbf{s}(z(\alpha_{0}))=\mathbf{e}_{1}:=(1,0)$,
so that $\mathbf{n}(z(\alpha_{0}))=(0,1)$ and
$x_{0}=(0,\epsilon\gamma(\alpha_{0})\eta_{0})$. Let us consider a barrier
function $b_{1}:\mathbb{R}^{2}\mapsto\mathbb{R}$ given by

$b_{1}(x_{1},x_{2})=x_{2}^{2}-\frac{x_{1}^{2}}{2}.$

Clearly $\Delta b_{1}=1$, so $v^{\epsilon}+b_{1}$ is harmonic in
$D^{\epsilon}$. It then follows from the maximum principle that
$\max_{\overline{V^{\epsilon}(\alpha_{0})}}(v^{\epsilon}+b_{1})$ is achieved
at some boundary point $\tilde{x}_{0}\in\partial V^{\epsilon}(\alpha_{0})$.
Let us break $\partial V^{\epsilon}(\alpha_{0})$ into $\partial
V^{\epsilon}_{1}(\alpha_{0})\cup\partial V^{\epsilon}_{2}(\alpha_{0})$ (see
Figure 4(b) for an illustration), given by

$\partial V^{\epsilon}_{1}(\alpha_{0}):=\partial D^{\epsilon}\cap
B_{6\epsilon\rVert\gamma\rVert_{\infty}}(z(\alpha_{0})),\quad\partial
V^{\epsilon}_{2}(\alpha_{0}):=\overline{D^{\epsilon}}\cap\partial
B_{6\epsilon\rVert\gamma\rVert_{\infty}}(z(\alpha_{0})).$ (4.24)

We claim that $\tilde{x}_{0}\in\partial V^{\epsilon}_{1}(\alpha_{0})$. To see
this, note that any $y=(y_{1},y_{2})\in\partial V^{\epsilon}_{2}(\alpha_{0})$
satisfies $|y|=6\epsilon\rVert\gamma\rVert_{\infty}$ and $|y_{2}|\leq
2\epsilon\rVert\gamma\rVert_{\infty}$, where the latter follows from (4.18)
and our assumption that $\mathbf{s}(z(\alpha_{0}))=\mathbf{e}_{1}$. This
implies that $|y_{1}|\geq 4\epsilon\rVert\gamma\rVert_{\infty}>|y_{2}|$, thus
$b_{1}(y)<0$. Using that $v^{\epsilon}(x_{0})=M\geq v^{\epsilon}(y)$ and
$b_{1}(x_{0})=b_{1}(0,\epsilon\gamma(\alpha_{0})\eta_{0})\geq 0$, we have
$(v^{\epsilon}+b_{1})(y)<(v^{\epsilon}+b_{1})(x_{0}).$ This shows that
$\max_{\overline{V^{\epsilon}(\alpha_{0})}}(v^{\epsilon}+b_{1})$ cannot be
achieved on $\partial V^{\epsilon}_{2}(\alpha_{0})$, finishing the proof of
the claim.

Since $\tilde{x}_{0}\in\partial V^{\epsilon}_{1}(\alpha_{0})\subset\partial
D^{\epsilon}$, the boundary condition in (4.15) yields that
$v^{\epsilon}(\tilde{x}_{0})=0$. Thus

$\displaystyle M+b_{1}(x_{0})=v^{\epsilon}(x_{0})+b_{1}(x_{0})\leq
v^{\epsilon}(\tilde{x}_{0})+b_{1}(\tilde{x}_{0})=b_{1}(\tilde{x}_{0}).$

Using $b_{1}(x_{0})=b_{1}(0,\epsilon\gamma(\alpha_{0})\eta_{0})\geq 0$, the
above inequality becomes

$\displaystyle M\leq b_{1}(\tilde{x}_{0})\leq|\tilde{x}_{0}|^{2}\leq
36\rVert\gamma\rVert_{\infty}^{2}\epsilon^{2},$ (4.25)

where the second inequality follows from the definition of $b_{1}$. This
proves (4.16) for $C_{1}=36\rVert\gamma\rVert_{\infty}^{2}$.

Step 3. It remains to prove (4.17). First note that for $i\in\\{1,\dots,n\\}$,
the assumptions (H1)–(H3) yield that $D_{i}^{\epsilon}$ has $C^{2}$ boundary,
therefore $v^{\epsilon}\in C^{2}(D_{i}^{\epsilon})\cap
C^{1}(\overline{D_{i}^{\epsilon}})$. Let us fix
$i\in\left\\{1,\ldots,n\right\\}$ and any $\alpha\in S$, and we aim to show
that $|\nabla v^{\epsilon}(z(\alpha))|\leq C_{2}\epsilon$. Again, without loss
of generality we can assume that $z(\alpha)=(0,0)$ and
$\mathbf{s}(z(\alpha))=\mathbf{e}_{1}$. Let us consider a new barrier function
$b_{2}:\mathbb{R}^{2}\to\mathbb{R}$

$b_{2}(x_{1},x_{2}):=x_{2}^{2}+4\epsilon\rVert\gamma\rVert_{\infty}x_{2}-\frac{x_{1}^{2}}{2},$
(4.26)

which satisfies $b_{2}(0,0)=0$, and one can easily check that its zero level
set has horizontal tangent at $(0,0)$ (thus tangent to $\partial D^{\epsilon}$
at $z(\alpha)$).

Again, let us decompose $\partial V^{\epsilon}(\alpha)$ as $\partial
V^{\epsilon}_{1}(\alpha)\cup\partial V^{\epsilon}_{2}(\alpha)$ as in (4.24)
(except that $\alpha_{0}$ now becomes $\alpha$). We claim that for all
sufficiently small $\epsilon>0$, the new barrier function $b_{2}$ satisfies

$\displaystyle\Delta b_{2}=1~{}~{}~{}\,\text{ in }V^{\epsilon}(\alpha),$
(4.27) $\displaystyle b_{2}\leq 0\quad~{}~{}~{}\text{ on }\partial
V^{\epsilon}_{1},$ (4.28) $\displaystyle b_{2}\leq-\epsilon^{2}\quad\text{ on
}\partial V^{\epsilon}_{2}.$ (4.29)

Let us assume for a moment that (4.27)–(4.29) are true. Then it follows that

$v^{\epsilon}+C_{2}b_{2}\leq 0\text{ in }V^{\epsilon}(\alpha),$ (4.30)

where $C_{2}:=\max\left\\{1,C_{1}\right\\}$ and $C_{1}$ is as in (4.16) (in
the end of step 2 we have $C_{1}=36\rVert\gamma\rVert_{\infty}^{2}$). To show
(4.30), note that $v^{\epsilon}+C_{2}b_{2}$ is subharmonic in
$V^{\epsilon}(\alpha)$ due to (4.27) and the definition of $C_{2}$, thus its
maximum is attained on its boundary. The boundary conditions in (4.15) and
(4.28) yield that $v^{\epsilon}+C_{2}b_{2}\leq 0$ on $\partial
V^{\epsilon}_{1}(\alpha)$; whereas (4.16), (4.29) and the definition of
$C_{2}$ yield that $v^{\epsilon}+C_{2}b_{2}\leq 0$ on $\partial
V^{\epsilon}_{2}(\alpha)$. Thus $v^{\epsilon}+C_{2}b_{2}\leq 0$ on $\partial
V^{\epsilon}_{1}(\alpha)\cup\partial V^{\epsilon}_{2}(\alpha)$, implying
(4.30).

However, $v^{\epsilon}+C_{2}b_{2}$ is actually zero at $z(\alpha)\in\partial
V^{\epsilon}(\alpha)$, therefore Hopf’s Lemma implies that
$\nabla\left(v^{\epsilon}+C_{2}b_{2}\right)(z(\alpha))\cdot\vec{n}(z(\alpha))>0$,
where $\vec{n}(z(\alpha))$ is the outer normal of $\partial D^{\epsilon}$ at
$z(\alpha)$. Hence

$\displaystyle|\nabla v^{\epsilon}(z(\alpha))|=-\nabla
v^{\epsilon}(z(\alpha))\cdot\vec{n}(z(\alpha))<C_{2}\nabla
b_{2}(z(\alpha))\cdot\vec{n}(z(\alpha))=4C_{2}\rVert\gamma\rVert_{\infty}\epsilon,$
(4.31)

where the first equality follows from the fact that $v^{\epsilon}$ is
superharmonic in $D^{\epsilon}$ and constant on $\partial D^{\epsilon}$, and
the second equality is a direct computation of $\nabla b_{2}$. Thus (4.31)
proves (4.17).

To complete the proof, we only need to prove (4.27)–(4.29) for small
$\epsilon>0$. Note that (4.27) follows immediately from computing the
Laplacian of $b_{2}$. For (4.28), let us pick $y\in\partial
V^{\epsilon}_{1}(\alpha)$, and we aim to show that $b_{2}(y)\leq 0$. Note that
$y=R^{\epsilon}(\beta,0)$ or $R^{\epsilon}(\beta,-1)$ for some $\beta\in S$.
We first deal with the first case.

Let us denote $y=(y_{1},y_{2})$. Rewriting (4.20) into two inequalities for
the two components, and using that $z(\alpha)=(0,0)$ and
$z^{\prime}(\alpha)=L\mathbf{e}_{1}$ ($L$ is the length of the curve
$\Gamma_{i}$), we have

$\displaystyle|0-y_{1}-L(\alpha-\beta)|\leq\|z^{\prime\prime}\|_{\infty}(\alpha-\beta)^{2}$
(4.32)
$\displaystyle|y_{2}|=|0-y_{2}|\leq\|z^{\prime\prime}\|_{\infty}(\alpha-\beta)^{2}.$
(4.33)

Also, (4.22) gives $|\alpha-\beta|\leq
7F_{\Gamma}\|\gamma\|_{\infty}\epsilon$. Applying it to (4.32), for all
$\epsilon>0$ sufficiently small we have that

$|y_{1}|\geq\frac{L}{2}|\alpha-\beta|.$ (4.34)

Plugging (4.34) and (4.33) into
$b_{2}(y)=-\frac{1}{2}y_{1}^{2}+y_{2}^{2}+4\epsilon\|\gamma\|_{\infty}y_{2}$,
we have

$\begin{split}b_{2}(y)&\leq-\frac{L^{2}}{8}(\alpha-\beta)^{2}+\|z^{\prime\prime}\|_{\infty}^{2}(\alpha-\beta)^{4}+4\epsilon\|\gamma\|_{\infty}\|z^{\prime\prime}\|_{\infty}(\alpha-\beta)^{2}\\\
&\leq\left(-\frac{L^{2}}{8}+C\epsilon^{2}+C\epsilon\right)(\alpha-\beta)^{2}\leq
0,\end{split}$

for all $\epsilon>0$ sufficiently small, where the second inequality follows
from (4.22). This finishes the proof of (4.28) for the case
$y=R^{\epsilon}(\beta,0)$.

Before we deal with the case $y=R^{\epsilon}(\beta,-1)$, let us prove (4.29)
first. For any $y=(y_{1},y_{2})\in\partial V^{\epsilon}_{2}(\alpha)$, (4.18)
gives $|y_{2}|\leq 2\epsilon\|\gamma\|_{\infty}$. Combining this with
$|y|=6\epsilon\|\gamma\|_{\infty}$ yields
$|y_{1}|\geq\sqrt{32}\epsilon\rVert\gamma\rVert_{\infty}$. Thus

$b_{2}(y)\leq(2\epsilon\|\gamma\|_{\infty})^{2}+4\epsilon\rVert\gamma\rVert_{\infty}(2\epsilon\|\gamma\|_{\infty})-\frac{(\sqrt{32}\epsilon\rVert\gamma\rVert_{\infty})^{2}}{2}\leq-4\epsilon^{2}\|\gamma\|_{\infty}^{2}.$

Finally we turn to the proof of (4.28) for the case
$y=R^{\epsilon}(\beta,-1)$. Note that the curve
$\\{R^{\epsilon}(\beta,-1):\beta\in S\\}\cap
B_{6\epsilon\|\gamma\|_{\infty}}(z(\alpha))$ lies in the interior of the
region bounded by $\Gamma\cap B_{6\epsilon\|\gamma\|_{\infty}}(z(\alpha))$ on
the top, $\partial B_{6\epsilon\|\gamma\|_{\infty}}(z(\alpha))$ on the sides,
and $y_{2}=-2\epsilon\|\gamma\|_{\infty}$ on the bottom. (The last one follows
from (4.18) and our assumption that $\mathbf{s}(z(\alpha))=\mathbf{e}_{1}$).
We have already shown $b_{2}\leq 0$ on $\Gamma\cap
B_{6\epsilon\|\gamma\|_{\infty}}(z(\alpha))$ and the lateral boundaries, and
it is easy to check that $b_{2}\leq 0$ on
$y_{2}=-2\epsilon\|\gamma\|_{\infty}$. Since the set $\\{b_{2}\leq 0\\}$ is
simply-connected, it implies that $b_{2}\leq 0$ in the interior of this
region, finishing the proof. ∎

Note that (4.16) of Lemma 4.4 immediately implies Proposition 4.2. (The only
difference is that $\Delta v^{\epsilon}=-1$ in Lemma 4.4 whereas $\Delta
p^{\epsilon}=-2$ in Proposition 4.2, so the constant $C$ in Proposition 4.2 is
twice of that in (4.16)). The lemma also implies the following corollary,
which will be helpful in the proof of Proposition 4.3.

######  Corollary 4.5.

For any $i=1,\dots,n+m$, assume $\Gamma_{i}$ and $\gamma_{i}$ satisfy
(H1)–(H3). Assume $v^{\epsilon}\in C^{2}(D^{\epsilon}_{i})\cap
C(\overline{D^{\epsilon}_{i}})$ satisfies that

$\begin{cases}\left|\Delta v^{\epsilon}\right|\leq C_{0}&\text{ in
}D^{\epsilon}_{i},\\\ v^{\epsilon}=0&\text{ on }\partial
D^{\epsilon}_{i},\end{cases}$

for some constant $C_{0}>0$. Then for the same constants
$\epsilon_{0},C_{1},C_{2}$ as in Lemma 4.4, the following holds for all
$\epsilon\in(0,\epsilon_{0})$:

$|v^{\epsilon}|\leq C_{0}C_{1}\epsilon^{2}\quad\text{ in }D^{\epsilon}_{i},$
(4.35)

and if $v^{\epsilon}\in C^{2}(D^{\epsilon}_{i})\cap
C^{1}(\overline{D^{\epsilon}_{i}})$, we have

$\rVert\nabla v^{\epsilon}\rVert_{L^{\infty}(\Gamma_{i})}\leq
C_{0}C_{2}\epsilon\quad\text{ for }\quad i=1,\ldots,n.$ (4.36)

###### Proof.

Let $\tilde{v}$ be a solution to

$\begin{cases}\Delta\tilde{v}=-C_{0}&\text{ in }D^{\epsilon}_{i},\\\
\tilde{v}=0&\text{ on }\partial D^{\epsilon}_{i}.\end{cases}$

It is clear that $v^{\epsilon}+\tilde{v}$ is super-harmonic and
$v^{\epsilon}-\tilde{v}$ is sub-harmonic in $D^{\epsilon}_{i}$, and they both
vanish on the boundary. Thus the maximum principle implies that

$-\tilde{v}\leq v^{\epsilon}\leq\tilde{v}\quad\text{ in }D^{\epsilon}_{i}.$
(4.37)

Applying (4.16) of Lemma 4.4 to $\frac{\tilde{v}}{C_{0}}$, we obtain
$0\leq\tilde{v}\leq C_{0}C_{1}\epsilon^{2}$ in $D^{\epsilon}_{i}$ for all
$\epsilon\in(0,\epsilon_{0})$, leading to (4.35). Furthermore, (4.37) and the
fact that $v^{\epsilon}$ and $v$ both have zero boundary condition imply that

$|\nabla v^{\epsilon}|\leq|\nabla\tilde{v}|\quad\text{ on }\partial
D^{\epsilon}_{i}.$

We then apply (4.17) of Lemma 4.4 to $\frac{\tilde{v}}{C_{0}}$ and obtain
$\rVert\nabla v^{\epsilon}\rVert_{L^{\infty}(\Gamma_{i})}\leq
C_{0}C_{2}\epsilon,$ which proves (4.36). ∎

Now we are ready to prove Proposition 4.3.

######  Proof of Proposition 4.3.

Throughout the proof, let $i\in\\{1,\dots,n\\}$ be fixed. For notational
simplicity, in the rest of the proof we omit the subscript $i$ from all terms.

We claim that

$\displaystyle\left|\nabla\tilde{p}^{\epsilon}(R^{\epsilon}(\alpha,\eta))-\frac{c^{\epsilon}}{\epsilon\gamma(\alpha)}\mathbf{n}(z(\alpha))\right|\leq
C\epsilon\quad\text{ for all }\alpha\in S,\eta\in[0,-1],$ (4.38)
$\displaystyle\rVert\Delta q^{\epsilon}\rVert_{L^{\infty}(D^{\epsilon})}\leq
C$ (4.39)

for some constant $C>0$ only depending on $\rVert
z_{i}\rVert_{C^{3}(S_{i})},\rVert\gamma_{i}\rVert_{C^{2}(S_{i})}$ and
$F_{\Gamma}$. Assuming these are true, let us explain how they lead to
(4.12)–(4.14). By (4.6) and (4.10), $p^{\epsilon}$ and $\tilde{p}^{\epsilon}$
have the same boundary condition, thus $q^{\epsilon}=0$ on $\partial
D^{\epsilon}$. This and (4.39) allow us to apply Corollary 4.5 to
$q^{\epsilon}$ to obtain the estimate (4.35), implying (4.12).

Due to (4.36) of Corollary 4.5, we also have

$\rVert\nabla q^{\epsilon}\rVert_{L^{\infty}(\Gamma)}\leq C\epsilon.$ (4.40)

Using (4.7) and $p^{\epsilon}=\tilde{p}^{\epsilon}+q^{\epsilon}$, we have

$\displaystyle-2|U|$ $\displaystyle=\int_{\partial
U}\nabla\tilde{p}^{\epsilon}\cdot{n}d\sigma+\int_{\partial U}\nabla
q^{\epsilon}\cdot nd\sigma$
$\displaystyle=-\frac{c^{\epsilon}L}{\epsilon}\int_{S}\gamma^{-1}(\alpha)d\alpha+O(\epsilon),$

where the second equality follows from (4.38) for $\eta=0$,
$n(z(\alpha))=-\mathbf{n}(z(\alpha))$ and $d\sigma=Ld\alpha$, as well as
(4.40). Rearranging the terms and using the definition of $\beta$ in (4.11)
yields (4.13).

Finally, note that (4.13) and (4.38) directly lead to (4.14), where we are
using the fact that $\gamma_{i}$ is uniformly positive for $i=1,\dots,n$, due
to (H3).

The rest of the proof is devoted to proving the claims (4.38) and (4.39). For
(4.38), we compute the gradient of $\tilde{p}^{\epsilon}$. Differentiating
(4.10) with respect to $\alpha$ and $\eta$, we obtain

$\displaystyle(\nabla_{\alpha,\eta}R^{\epsilon}(\alpha,\eta))^{t}\nabla\tilde{p}(R^{\epsilon}(\alpha,\eta))=\begin{pmatrix}0\\\
c^{\epsilon}\end{pmatrix},$ (4.41)

where $(\nabla_{\alpha,\eta}R^{\epsilon})^{t}$ denotes the transpose of the
Jacobian matrix of $R^{\epsilon}$. Since
$\nabla_{\alpha,\eta}R^{\epsilon}=(\partial_{\alpha}R^{\epsilon},\partial_{\eta}R^{\epsilon})$,
using the formula for inverses of $2\times 2$ matrices, we have

$\left(\left(\nabla_{\alpha,\eta}R^{\epsilon}\right)^{t}\right)^{-1}=\frac{1}{J(\alpha,\eta)}\big{(}\\!-\\!(\partial_{\eta}R^{\epsilon})^{\perp},(\partial_{\alpha}R^{\epsilon})^{\perp}\big{)}.$
(4.42)

where $J(\alpha,\eta):=\text{det}(\nabla_{\alpha,\eta}R^{\epsilon})$.
Multiplying the inverse matrix on both sides of (4.41), we have

$\nabla\tilde{p}^{\epsilon}(R^{\epsilon}(\alpha,\eta))=\frac{1}{J}\big{(}\\!-\\!(\partial_{\eta}R^{\epsilon})^{\perp},(\partial_{\alpha}R^{\epsilon})^{\perp}\big{)}\begin{pmatrix}0\\\
c^{\epsilon}\end{pmatrix}=\frac{c^{\epsilon}}{J}(\partial_{\alpha}R^{\epsilon})^{\perp}.$
(4.43)

Recall that Lemma 3.2 gives
$(\partial_{\alpha}R^{\epsilon})^{\perp}=z^{\prime}(\alpha)^{\perp}+O(\epsilon)=L\mathbf{n}(z(\alpha))+O(\epsilon)$,
and $J=\epsilon L\gamma+O(\epsilon^{2})$. Plugging these into (4.43) gives

$\nabla\tilde{p}^{\epsilon}(R(\alpha,\eta))=\frac{c^{\epsilon}}{\epsilon}\left(\frac{\mathbf{n}(z(\alpha))}{\gamma}+O(\epsilon)\right).$
(4.44)

Note that it follows from (4.8) that
$c^{\epsilon}\leq\frac{|D^{\epsilon}|}{2\pi}$, where $|D^{\epsilon}|\leq
C\epsilon$ due to (3.9). These imply

$\frac{c_{\epsilon}}{\epsilon}\leq C,$ (4.45)

and applying it to (4.44) yields (4.38).

To prove (4.39), since $q^{\epsilon}=p^{\epsilon}-\tilde{p}^{\epsilon}$ and
$\Delta p^{\epsilon}=-2$ in $D^{\epsilon}$, it suffices to show that

$\left|\Delta\tilde{p}^{\epsilon}\right|\leq C\quad\text{ in }D^{\epsilon},$
(4.46)

and we will begin with an explicit computation of
$\partial_{x_{1}x_{1}}\tilde{p}^{\epsilon}$ and
$\partial_{x_{2}x_{2}}\tilde{p}^{\epsilon}$. Let us denote
$R^{\epsilon}=:(R^{1},R^{2})$. For notational simplicity, in the rest of the
proof we will use subscripts on $R^{\epsilon}$, $R^{1}$, $R^{2}$ and $J$ to
denote their partial derivative, e.g.
$R^{1}_{\alpha}:=\partial_{\alpha}R^{1}$.

From (4.43), it follows that

$\partial_{x_{1}}\tilde{p}^{\epsilon}(R^{\epsilon}(\alpha,\eta))=-\frac{c^{\epsilon}}{J}R^{2}_{\alpha}.$

Differentiating in $\alpha$ and $\eta$, we get

$\begin{split}\nabla\left(\partial_{x_{1}}\tilde{p}^{\epsilon}\right)(R^{\epsilon}(\alpha,\eta))&=\left(\left(\nabla_{\alpha,\eta}R^{\epsilon}\right)^{t}\right)^{-1}\nabla_{\alpha,\eta}\left(-\frac{c^{\epsilon}}{J}R^{2}_{\alpha}\right)\\\
&=\frac{c^{\epsilon}}{J}\begin{pmatrix}R_{\eta}^{2}&-R_{\alpha}^{2}\\\\[2.84544pt]
-R_{\eta}^{1}&R_{\alpha}^{1}\end{pmatrix}\begin{pmatrix}\frac{J_{\alpha}}{J^{2}}R^{2}_{\alpha}-\frac{1}{J}R^{2}_{\alpha\alpha}\\\\[2.84544pt]
\frac{J_{\eta}}{J^{2}}R_{\alpha}^{2}-\frac{1}{J}R^{2}_{\alpha\eta}\end{pmatrix},\end{split}$

thus

$\partial_{x_{1}x_{1}}\tilde{p}^{\epsilon}(R(\alpha,\eta))=\frac{c^{\epsilon}}{J}\left(\frac{J_{\alpha}}{J^{2}}R^{2}_{\eta}R^{2}_{\alpha}-\frac{1}{J}R^{2}_{\eta}R^{2}_{\alpha\alpha}-\frac{J_{\eta}}{J^{2}}(R_{\alpha}^{2})^{2}+\frac{1}{J}R^{2}_{\alpha}R^{2}_{\alpha\eta}\right).$

Likewise, $\partial_{x_{2}x_{2}}\tilde{p}(R(\alpha,\eta))$ takes the same
expression except every $R^{2}$ is changed into $R^{1}$. Adding them together
gives

$\Delta\tilde{p}^{\epsilon}(R(\alpha,\eta))=\frac{c^{\epsilon}}{J}\left(\frac{J_{\alpha}}{J^{2}}R^{\epsilon}_{\eta}\cdot
R^{\epsilon}_{\alpha}-\frac{1}{J}R_{\eta}^{\epsilon}\cdot
R^{\epsilon}_{\alpha\alpha}-\frac{J_{\eta}}{J^{2}}R^{\epsilon}_{\alpha}\cdot
R^{\epsilon}_{\alpha}+\frac{1}{J}R^{\epsilon}_{\alpha}\cdot
R^{\epsilon}_{\alpha\eta}\right).$ (4.47)

Using the explicit formulae of $R_{\alpha},R_{\eta}$ and $J$ in Lemma 3.2, we
directly obtain $|R^{\epsilon}_{\alpha}|,|R^{\epsilon}_{\alpha\alpha}|\leq C$;
$|R^{\epsilon}_{\eta}|,|R^{\epsilon}_{\alpha\eta}|,|J_{\alpha}|\leq
C\epsilon$; $|J_{\eta}|\leq C\epsilon^{2}$; and $J^{-1}\leq C\epsilon^{-1}$
when $\epsilon$ is sufficiently small, where $C$ depends on $\rVert
z_{i}\rVert_{C^{3}(S_{i})}$ and $\rVert\gamma_{i}\rVert_{C^{2}(S_{i})}$. As a
result, all the four terms in the parenthesis of (4.47) are bounded by some
constant $C$ independent of $\epsilon$. Finally, (4.45) yields
$\frac{c_{\epsilon}}{J}\leq C$ as well, thus $|\Delta\tilde{p}^{\epsilon}|\leq
C$, and this proves the second claim (4.39). ∎

##  5\. Proof of the symmetry result

In this section we prove that a stationary vortex sheet with positive
vorticity must be radially symmetric up to a translation, and a rotating
vortex sheet with positive vorticity and angular velocity $\Omega<0$ must be
radially symmetric. The key idea of the proof is to define the integral

$\begin{split}I^{\epsilon}&:=\int_{D^{\epsilon}}\epsilon^{-1}\mathbf{u}^{\epsilon}\cdot\nabla\left(\omega^{\epsilon}*\mathcal{N}-\frac{\Omega}{2}|x|^{2}\right)dx\\\
&=\int_{D^{\epsilon}}\epsilon^{-1}(x+\nabla
p^{\epsilon})\cdot\nabla\left(\omega^{\epsilon}*\mathcal{N}-\frac{\Omega}{2}|x|^{2}\right)dx,\end{split}$
(5.1)

and compute it in two different ways. The motivation of the definition is as
follows. As discussed in [14, Section 2.1], $I^{\epsilon}$ can be thought of
as a first variation of an “energy functional”

$\mathcal{E}[\omega^{\epsilon}]:=\int\frac{1}{2}\omega^{\epsilon}(\omega^{\epsilon}*\mathcal{N})-\frac{\Omega}{2}\omega^{\epsilon}|x|^{2}\,dx$

when we perturb $\omega^{\epsilon}$ by a divergence free vector
$\mathbf{u}^{\epsilon}$ in $D^{\epsilon}$. (This functional $\mathcal{E}$ only
serves as our motivation, and will not appear in the proof.) On the one hand,
using that $\omega_{0}$ is stationary in the rotating frame with angular
velocity $\Omega$ and $\omega^{\epsilon}$ is a close approximation of
$\omega_{0}$, we will show in Proposition 5.1 that $I^{\epsilon}$ is of order
$O(\epsilon|\log\epsilon|)$, thus goes to zero as $\epsilon\to 0$. On the
other hand, using the particular $\mathbf{u}^{\epsilon}$ that we constructed
in Section 4, we will prove in Proposition 5.2 that if $\Omega=0$,
$I^{\epsilon}$ is strictly positive independently of $\epsilon$ unless all the
vortex sheets are nested circles with constant density; and also prove a
similar result in Corollary 5.3 for $\Omega<0$.

######  Proposition 5.1.

Assume $\omega(\cdot,t)=\omega_{0}(R_{\Omega t}\cdot)$ is a
stationary/uniformly-rotating vortex sheet with angular velocity
$\Omega\in\mathbb{R}$, where $\omega_{0}$ satisfies (H1)–(H3). Then there
exists some $C>0$ only depending on $b$ (as in (H2)),
$\max_{i}\|z_{i}\|_{C^{3}(S_{i})}$, $\max_{i\leq
n}\|\gamma_{i}\|_{C^{2}(S_{i})}$, $\max_{i>n}\|\gamma_{i}\|_{C^{b}(S_{i})}$,
$d_{\Gamma}$ and $F_{\Gamma}$, such that
$|I^{\epsilon}|<C\epsilon^{b}|\log\epsilon|$ for all sufficiently small
$\epsilon>0$.

###### Proof.

Let us decompose $I^{\epsilon}=:\sum_{i=1}^{n+m}I^{\epsilon}_{i}$, where
$I_{i}^{\epsilon}:=\int_{D^{\epsilon}_{i}}\epsilon^{-1}(x+\nabla
p^{\epsilon})\cdot\nabla(\omega^{\epsilon}*\mathcal{N}-\frac{\Omega}{2}|x|^{2})dx.$

We start with showing that $|I^{\epsilon}_{i}|\leq
C\epsilon^{b}|\log\epsilon|$ for $i=n+1,\dots,n+m$. For such $i$,
$p^{\epsilon}=0$ on $\partial D^{\epsilon}_{i}$, thus the divergence theorem
(and the fact that $\omega^{\epsilon}=\epsilon^{-1}$ in $D^{\epsilon}_{i}$)
gives

$I^{\epsilon}_{i}=\underbrace{\int_{D^{\epsilon}_{i}}\epsilon^{-1}x\cdot\nabla\left(\omega^{\epsilon}*\mathcal{N}-\frac{\Omega}{2}|x|^{2}\right)dx}_{=:T_{i}^{\epsilon}}-\int_{D^{\epsilon}_{i}}\epsilon^{-1}(\epsilon^{-1}-2\Omega)p^{\epsilon}(x)dx.$

Using the estimate $|p^{\epsilon}|\leq C\epsilon^{2}$ in Proposition 4.2 and
the fact that $|D^{\epsilon}_{i}|\leq C\epsilon$ from (3.9), we easily bound
the second integral by $C\epsilon$. To control the first integral
$T_{i}^{\epsilon}$, we rewrite it using the change of variables
$x=R^{\epsilon}_{i}(\alpha,\eta)$ and the definition
$\mathbf{v}^{\epsilon}:=\nabla^{\perp}(\omega^{\epsilon}*\mathcal{N})$ in
(3.10): (also note that on the right hand side we group $\epsilon^{-1}$ with
the determinant)

$\begin{split}T_{i}^{\epsilon}=\int_{S_{i}}\int_{-1}^{0}R_{i}^{\epsilon}(\alpha,\eta)\cdot\underbrace{\Big{(}\\!-\\!(\mathbf{v}^{\epsilon})^{\perp}(R_{i}^{\epsilon}(\alpha,\eta))-\Omega
R_{i}^{\epsilon}(\alpha,\eta)\Big{)}}_{=:J_{i}^{\epsilon}}\underbrace{\epsilon^{-1}\det(\nabla_{\alpha,\eta}R_{i}^{\epsilon}(\alpha,\eta))}_{=:K_{i}^{\epsilon}}\,d\eta
d\alpha.\end{split}$

Let us take a closer look at the integrand, which is a product of 3 terms.
Clearly, the definition of $R_{i}^{\epsilon}$ gives
$R_{i}^{\epsilon}(\alpha,\eta)=z_{i}(\alpha)+O(\epsilon).$ As for the middle
term $J_{i}^{\epsilon}$, Lemma 3.4 yields

$J_{i}^{\epsilon}(\alpha,\eta)=-BR^{\perp}(z_{i}(\alpha))+\left(\eta+\frac{1}{2}\right)[\mathbf{v}]^{\perp}(z_{i}(\alpha))-\Omega
z_{i}(\alpha)+O(\epsilon^{b}|\log\epsilon|).$ (5.2)

Using the fact that $BR(z_{i}(\alpha))=\Omega z_{i}^{\perp}(\alpha)$ for
$i=n+1,\dots,n+m$ (which follows from (2.6) and (2.7)), it becomes

$J_{i}^{\epsilon}(\alpha,\eta)=\left(\eta+\frac{1}{2}\right)[\mathbf{v}]^{\perp}(z_{i}(\alpha))+O(\epsilon^{b}|\log\epsilon|).$
(5.3)

Also it follows from (3.8) that
$K_{i}^{\epsilon}(\alpha,\eta)=L_{i}\gamma_{i}(\alpha)+O(\epsilon).$ Plugging
these three estimates into the above integral gives

$T_{i}^{\epsilon}=\int_{S_{i}}\int_{-1}^{0}z_{i}(\alpha)\cdot\left(\eta+\frac{1}{2}\right)[\mathbf{v}]^{\perp}(z_{i}(\alpha))L_{i}\gamma_{i}(\alpha)d\eta
d\alpha+O(\epsilon^{b}|\log\epsilon|)=O(\epsilon^{b}|\log\epsilon|),$

where the last step follows from the fact that
$\int_{-1}^{0}(\eta+\frac{1}{2})d\eta=0$. This finishes the proof that
$|I^{\epsilon}_{i}|\leq C\epsilon^{b}|\log\epsilon|$ for $i=n+1,\dots,n+m$,
where $C$ depends on $b$, $\max_{i}\|z_{i}\|_{C^{2}(S_{i})}$,
$\max_{i}\|\gamma_{i}\|_{C^{b}(S_{i})}$, $d_{\Gamma}$ and $F_{\Gamma}$.

In the rest of the proof we aim to show $|I^{\epsilon}_{i}|\leq
C\epsilon^{b}|\log\epsilon|$ for $i=1,\dots,n$, which is slightly more
involved. Recall that in Proposition 4.3 we defined $\tilde{p}^{\epsilon}$ and
$q^{\epsilon}$ in $D_{i}^{\epsilon}$ for $i=1,\dots,n$, where they satisfy
$p^{\epsilon}=\tilde{p}^{\epsilon}+q^{\epsilon}$ in $D_{i}^{\epsilon}$, and
$q^{\epsilon}=0$ on $\partial D_{i}^{\epsilon}$. This allows us to apply the
divergence theorem (to the $q^{\epsilon}$ term only) and decompose
$I_{i}^{\epsilon}$ as

$I^{\epsilon}_{i}=\int_{D^{\epsilon}_{i}}\epsilon^{-1}(x+\nabla\tilde{p}_{\epsilon})\cdot\nabla\left(\omega^{\epsilon}*\mathcal{N}-\frac{\Omega}{2}|x|^{2}\right)dx-\int_{D^{\epsilon}_{i}}\epsilon^{-1}(\epsilon^{-1}-2\Omega)q^{\epsilon}(x)dx=:I^{\epsilon}_{i,1}+I^{\epsilon}_{i,2}.$

We can easily show that $I^{\epsilon}_{i,2}=O(\epsilon)$: (4.12) of
Proposition 4.3 gives $|q^{\epsilon}|\leq C\epsilon^{2}$, and combining it
with $|D^{\epsilon}_{i}|\leq C\epsilon$ in (3.9) immediately yields the
desired estimate.

Next we turn to $I^{\epsilon}_{i,1}$. Again, the change of variables
$x=R^{\epsilon}_{i}(\alpha,\eta)$ and the definition
$\mathbf{v}^{\epsilon}:=\nabla^{\perp}(\omega^{\epsilon}*\mathcal{N})$ gives

$I_{i,1}^{\epsilon}=\int_{S_{i}}\int_{-1}^{0}\big{(}R_{i}^{\epsilon}(\alpha,\eta)+\nabla\tilde{p}^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))\big{)}\cdot\underbrace{\Big{(}\\!-\\!(\mathbf{v}^{\epsilon})^{\perp}(R_{i}^{\epsilon}(\alpha,\eta))-\Omega
R_{i}^{\epsilon}(\alpha,\eta)\Big{)}}_{=:J_{i}^{\epsilon}}\underbrace{\epsilon^{-1}\det(\nabla_{\alpha,\eta}R_{i}^{\epsilon}(\alpha,\eta))}_{=:K_{i}^{\epsilon}}\,d\eta
d\alpha.$

For the three terms in the product of the integrand, we will approximate the
first term using the definition of $R_{i}^{\epsilon}$ and (4.14) of
Proposition 4.3:

$R_{i}^{\epsilon}(\alpha,\eta)+\nabla\tilde{p}^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))=z_{i}(\alpha)+\frac{\beta_{i}}{\gamma_{i}(\alpha)}\mathbf{n}(z_{i}(\alpha))+O(\epsilon),$

where
$\beta_{i}:=\frac{2|U_{i}|}{L_{i}\int_{S_{i}}\gamma_{i}^{-1}(\alpha)d\alpha}$
is given by (4.11). Lemma 3.4 allows us to approximate the middle term
$J_{i}^{\epsilon}$ as (5.2), however (5.3) no longer holds since for
$i=1,\dots,n$ we do not have $BR(z_{i}(\alpha))=\Omega z_{i}^{\perp}(\alpha)$.
As for $K_{i}^{\epsilon}$, we again use (3.8) to approximate it by
$K_{i}^{\epsilon}(\alpha,\eta)=L_{i}\gamma_{i}(\alpha)+O(\epsilon).$ Plugging
these three estimates into the integrand of $I_{i,1}^{\epsilon}$ gives

$I_{i,1}^{\epsilon}=\int_{S_{i}}\left(z_{i}(\alpha)+\frac{\beta_{i}}{\gamma_{i}(\alpha)}\mathbf{n}(z_{i}(\alpha))\right)\cdot\Big{(}-\\!\\!BR^{\perp}(z_{i}(\alpha))-\Omega
z_{i}(\alpha)\Big{)}L_{i}\gamma_{i}(\alpha)d\alpha+O(\epsilon^{b}|\log\epsilon|),$

where we again use the fact that the $(\eta+\frac{1}{2})$ term gives zero
contribution since $\int_{-1}^{0}(\eta+\frac{1}{2})d\eta=0$. Next we will show
the integral on the right hand side is in fact 0. Since $\omega$ is a rotating
solution with angular velocity $\Omega$, the conditions (2.6) and (2.7) yield
that

$-BR^{\perp}(z_{i}(\alpha))-\Omega
z_{i}(\alpha)=C_{i}\gamma_{i}^{-1}(\alpha)\mathbf{n}(z_{i}(\alpha)),$

for some constant $C_{i}$. Plugging this into the above integral gives

$\begin{split}I_{i,1}^{\epsilon}&=C_{i}L_{i}\int_{S_{i}}\left(z_{i}(\alpha)\cdot\mathbf{n}(z_{i}(\alpha))+\frac{\beta_{i}}{\gamma_{i}(\alpha)}\right)d\alpha+O(\epsilon^{b}|\log\epsilon|)\\\
&=C_{i}L_{i}\left(\int_{S_{i}}z_{i}(\alpha)\cdot\mathbf{n}(z_{i}(\alpha))d\alpha+\frac{2|U_{i}|}{L_{i}}\right)+O(\epsilon^{b}|\log\epsilon|),\end{split}$

where the second step follows from the definition of $\beta_{i}$ in (4.11).
Let us compute the integral on the right hand side by changing to arclength
parametrization and applying the divergence theorem:

$\int_{S_{i}}z_{i}(\alpha)\cdot\mathbf{n}(z_{i}(\alpha))d\alpha=-\frac{1}{L_{i}}\int_{\partial
U_{i}}x\cdot nd\sigma=-\frac{2|U_{i}|}{L_{i}},$

which yields $I_{i,1}^{\epsilon}=O(\epsilon^{b}|\log\epsilon|)$, and finishes
the proof that $|I_{i}^{\epsilon}|\leq C\epsilon^{b}|\log\epsilon|$ for
$i=1,\dots,n$, where $C$ depends on $b$, $\rVert
z_{i}\rVert_{C^{3}(S_{i})},\rVert\gamma_{i}\rVert_{C^{2}(S_{i})}$,
$d_{\Gamma}$ and $F_{\Gamma}$.

Finally, summing the $I_{i}^{\epsilon}$ estimates for $i=1,\dots,n+m$ gives
$|I^{\epsilon}|\leq C\epsilon^{b}|\log\epsilon|$ for all sufficiently small
$\epsilon>0$, thus we can conclude. ∎

Now we will use a different way to compute $I^{\epsilon}$. Let us first define
a new integral $\tilde{I}^{\epsilon}$ that is the same as $I^{\epsilon}$
except with $\Omega$ set to zero:

$\begin{split}\tilde{I}^{\epsilon}&:=\int_{D^{\epsilon}}\epsilon^{-1}(x+\nabla
p^{\epsilon})\cdot\nabla\left(\omega^{\epsilon}*\mathcal{N}\right)dx.\end{split}$
(5.4)

Next we will prove that $\tilde{I}^{\epsilon}$ is strictly positive
independently of $\epsilon$ unless all the vortex sheets are nested circles
with constant density. As we will see in the proof, the key step is to show
that if some $\Gamma_{i}$ is either not a circle or does not have a constant
$\gamma_{i}$, then the estimates on $p^{\epsilon}$ in Propositions 4.2–4.3
lead to the following quantitative version of (4.9):
$\epsilon^{-2}\left(\frac{|D_{i}^{\epsilon}|^{2}}{4\pi}-\int_{D_{i}^{\epsilon}}p^{\epsilon}(x)dx\right)\geq
c_{0}>0$, where $c_{0}$ is independent of $\epsilon$.

######  Proposition 5.2.

Let $\tilde{I}^{\epsilon}$ be defined as in (5.4). Assume that $\Gamma_{i}$
and $\gamma_{i}$ satisfy (H1)–(H3) for $i=1,\dots,n+m$. Then we have
$\tilde{I}^{\epsilon}\geq 0$ for all sufficiently small $\epsilon>0$.

In addition, if $\Gamma$ is not a union of nested circles with constant
$\gamma_{i}$’s on each connected component, there exists some $c_{0}>0$
independent of $\epsilon$, such that $\tilde{I}^{\epsilon}>c_{0}>0$ for all
sufficiently small $\epsilon>0$.

###### Proof.

We start by decomposing $\tilde{I}^{\epsilon}$ as

$\tilde{I}^{\epsilon}=\int_{D^{\epsilon}}\epsilon^{-1}x\cdot\nabla(\omega^{\epsilon}*\mathcal{N})dx+\int_{D^{\epsilon}}\epsilon^{-1}\nabla
p^{\epsilon}\cdot\nabla(\omega^{\epsilon}*\mathcal{N})dx=:I^{\epsilon}_{1}+I^{\epsilon}_{2}.$

$I^{\epsilon}_{1}$ can be easily computed as

$\begin{split}I^{\epsilon}_{1}&=\frac{1}{2\pi\epsilon^{2}}\int_{D^{\epsilon}}\int_{D^{\epsilon}}\frac{x\cdot(x-y)}{|x-y|^{2}}dxdy=\frac{|D^{\epsilon}|^{2}}{4\pi\epsilon^{2}}=\frac{1}{4\pi\epsilon^{2}}\left(\sum_{i=1}^{n+m}|D^{\epsilon}_{i}|\right)^{2}\end{split}$
(5.5)

where the second equality is obtained by exchanging $x$ with $y$ and taking
the average with the original integral. As for $I^{\epsilon}_{2}$, we have

$\begin{split}I^{\epsilon}_{2}&=\frac{1}{\epsilon}\int_{\partial
D^{\epsilon}}p^{\epsilon}\nabla(\omega^{\epsilon}*\mathcal{N})\cdot
nd\sigma-\frac{1}{\epsilon}\int_{D^{\epsilon}}p^{\epsilon}\omega^{\epsilon}dx\\\
&=-\frac{1}{\epsilon}\sum_{i=1}^{n}c^{\epsilon}_{i}\int_{\partial
U_{i}}\nabla(\omega^{\epsilon}*\mathcal{N})\cdot
nd\sigma-\frac{1}{\epsilon^{2}}\int_{D^{\epsilon}}p^{\epsilon}dx\\\
&\geq-\frac{1}{\epsilon^{2}}\sum_{i=1}^{n}\sum_{j=1}^{n+m}\frac{|D^{\epsilon}_{i}|}{2\pi}\int_{U_{i}}1_{D^{\epsilon}_{j}}dx-\frac{1}{\epsilon^{2}}\sum_{i=1}^{n+m}\int_{D^{\epsilon}_{i}}p^{\epsilon}dx,\end{split}$
(5.6)

where the first equality follows from the divergence theorem, the second
equality follows from the boundary conditions (4.5) and (4.6) for
$p^{\epsilon}$ (as well as the fact that $\partial U_{i}$ and $\partial
D^{\epsilon}_{i}$ have opposite outer normals), and the last inequality
follows from the divergence theorem as well as the inequality
$c_{i}^{\epsilon}\leq\sup_{D^{\epsilon}_{i}}p\leq\frac{|D^{\epsilon}_{i}|}{2\pi}$
due to (4.8).

Let us denote $j\prec i$ if
$i\in\left\\{1,\ldots,n\right\\},j\in\\{1,\dots,n+m\\}$, $j\neq i$ and
$\Gamma_{j}$ lies in the interior of the domain enclosed by $\Gamma_{i}$ (that
is, $\Gamma_{j}\subset U_{i}$). If not, we denote $j\nprec i$. Note that for
sufficiently small $\epsilon>0$, we have

$\int_{U_{i}}1_{D^{\epsilon}_{j}}dx=\begin{cases}|D^{\epsilon}_{j}|&\text{ if
}j\prec i,\\\ 0&\text{ otherwise.}\end{cases}$ (5.7)

Applying this to (5.6) yields

$\begin{split}I^{\epsilon}_{2}&\geq-\frac{1}{2\pi\epsilon^{2}}\sum_{i,j=1}^{n+m}\mathbbm{1}_{j\prec
i}\,|D^{\epsilon}_{i}||D^{\epsilon}_{j}|-\frac{1}{\epsilon^{2}}\sum_{i=1}^{n+m}\int_{D^{\epsilon}_{i}}p^{\epsilon}_{i}dx\\\
&=-\frac{1}{4\pi\epsilon^{2}}\sum_{i,j=1}^{n+m}(\mathbbm{1}_{j\prec
i}+\mathbbm{1}_{i\prec
j})\,|D^{\epsilon}_{i}||D^{\epsilon}_{j}|-\frac{1}{\epsilon^{2}}\sum_{i=1}^{n+m}\int_{D^{\epsilon}_{i}}p^{\epsilon}_{i}dx\end{split}$
(5.8)

where in the first step we used that the $i=n+1,\dots,n+m$ terms have zero
contribution in the first sum, due to the definition of $j\prec i$.

Adding (5.5) and (5.8) together, we obtain

$\tilde{I}^{\epsilon}\geq\sum_{i=1}^{n+m}\underbrace{\frac{1}{\epsilon^{2}}\left(\frac{|D^{\epsilon}_{i}|^{2}}{4\pi}-\int_{D^{\epsilon}_{i}}p^{\epsilon}_{i}dx\right)}_{=:A_{i}^{\epsilon}}+\sum_{i,j=1}^{n+m}\underbrace{\frac{1}{\epsilon^{2}}\left(\mathbbm{1}_{i\neq
j}-\left(\mathbbm{1}_{j\prec i}+\mathbbm{1}_{i\prec
j}\right)\right)\frac{|D^{\epsilon}_{i}||D^{\epsilon}_{j}|}{4\pi}}_{=:B_{i,j}^{\epsilon}},$
(5.9)

From (4.9), it follows that $A_{i}^{\epsilon}\geq 0$ for all $i=1,\dots,n+m$,
with equality achieved if and only if each $D^{\epsilon}_{i}$ is a disk or an
annulus. Note that $B_{i,j}^{\epsilon}\geq 0$ as well for all $i$ and $j$,
since for any $i\neq j$, at most one of $i\prec j$ and $j\prec i$ can hold.
Putting these together yields that $\tilde{I}^{\epsilon}\geq 0$ for any
sufficiently small $\epsilon>0$.

In the rest of the proof, we assume $\Gamma$ is not a union of nested circles
with constant $\gamma_{i}$’s on each connected component. Therefore at least
one of the following 3 cases must be true. In each case we aim to show that
$\tilde{I}_{\epsilon}\geq c_{0}>0$, where $c_{0}$ is independent of $\epsilon$
for all sufficiently small $\epsilon>0$.

Case 1. There exists some open curve $\Gamma_{i}$ that is not a loop. In this
case $D^{\epsilon}_{i}$ is simply-connected, and $p^{\epsilon}=0$ on $\partial
D^{\epsilon}_{i}$ by (4.5). Applying Proposition 4.2 to $p^{\epsilon}$ in
$D^{\epsilon}_{i}$, we have $\sup_{D^{\epsilon}_{i}}p^{\epsilon}\leq
C\epsilon^{2}$, where $C$ is independent of $\epsilon$. This leads to
$\int_{D^{\epsilon}_{i}}p^{\epsilon}dx\leq C\epsilon^{3}$, since
$|D^{\epsilon}_{i}|=O(\epsilon)$ by (3.9). As a result, for the index $i$ we
have

$A_{i}^{\epsilon}=\frac{|D^{\epsilon}_{i}|^{2}}{4\pi\epsilon^{2}}-\epsilon^{-2}\int_{D^{\epsilon}_{i}}p^{\epsilon}_{i}dx\geq\frac{L_{i}^{2}}{4\pi}\left(\int_{S_{i}}\gamma_{i}(\alpha)d\alpha\right)^{2}-C\epsilon,$

where we again used (3.9) in the second inequality. This gives that
$A_{i}^{\epsilon}\geq\frac{L_{i}^{2}}{8\pi}(\int_{S_{i}}\gamma_{i}(\alpha)d\alpha)^{2}>0$
for all sufficiently small $\epsilon>0$.

Case 2. There exists some closed curve $\Gamma_{i}$ that is either not a
circle, or $\gamma_{i}$ is not a constant. In this case we aim to show that
$A_{i}^{\epsilon}\geq c_{0}>0$, and this will be done by finding good
approximations (independent of $\epsilon$) for both terms in
$A_{i}^{\epsilon}$. For the first term
$\frac{|D^{\epsilon}_{i}|^{2}}{4\pi\epsilon^{2}}$, using (3.9) we again have

$\frac{|D^{\epsilon}_{i}|^{2}}{4\pi\epsilon^{2}}\geq\frac{L_{i}^{2}}{4\pi}\left(\int_{S_{i}}\gamma_{i}(\alpha)d\alpha\right)^{2}-C\epsilon=:J_{i}-C\epsilon,$
(5.10)

where $J_{i}>0$ is independent of $\epsilon$. For the second term
$\epsilon^{-2}\int_{D_{i}^{\epsilon}}p_{i}^{\epsilon}dx$, rewriting the
integral using the change of variables $x=R_{i}^{\epsilon}(a,\eta)$ gives

$\epsilon^{-2}\int_{D_{i}^{\epsilon}}p_{i}^{\epsilon}dx=\int_{S_{i}}\int_{-1}^{0}\frac{p^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))}{\epsilon}\frac{\det(\nabla_{\alpha,\eta}R_{i}^{\epsilon})}{\epsilon}d\eta
d\alpha.$

Recall that in Proposition 4.3 we defined
$\tilde{p}^{\epsilon}(R^{\epsilon}_{i}(\alpha,\eta)):=c^{\epsilon}_{i}(1+\eta)$
and $q_{\epsilon}$ such that $p^{\epsilon}-\tilde{p}_{\epsilon}=q_{\epsilon}$.
By (4.12) and (4.13), for all $\alpha\in S_{i}$ and $\eta\in(-1,0)$ we have

$\left|\frac{p^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))}{\epsilon}-\beta_{i}(1+\eta)\right|\leq\left|\frac{p^{\epsilon}(R_{i}^{\epsilon}(\alpha,\eta))}{\epsilon}-\frac{c_{i}^{\epsilon}}{\epsilon}(1+\eta)\right|+\left|\frac{c_{i}^{\epsilon}}{\epsilon}-\beta_{i}\right|\leq
C\epsilon,$

where
$\beta_{i}:=\frac{2|U_{i}|}{L_{i}\int_{S_{i}}\gamma_{i}^{-1}(\alpha)d\alpha}$
is defined in (4.11). Combining this with the expression of the determinant in
(3.8), we have

$\begin{split}\epsilon^{-2}\int_{D_{i}^{\epsilon}}p_{i}^{\epsilon}dx&=\int_{S_{i}}\int_{-1}^{0}(\beta_{i}(1+\eta)+O(\epsilon))(L_{i}\gamma_{i}(\alpha)+O(\epsilon))d\eta
d\alpha\\\
&\leq\frac{|U_{i}|}{\int_{S_{i}}\gamma_{i}^{-1}(\alpha)d\alpha}\int_{S_{i}}\gamma_{i}(\alpha)d\alpha+C\epsilon=:K_{i}+C\epsilon,\end{split}$

where $K_{i}$ is independent of $\epsilon$. Putting this together with (5.10)
yields the following:

$\begin{split}A_{i}^{\epsilon}&\geq J_{i}-K_{i}-C\epsilon\\\
&=\frac{L_{i}^{2}}{4\pi}\frac{\int_{S_{i}}\gamma_{i}(\alpha)d\alpha}{\int_{S_{i}}\gamma_{i}^{-1}(\alpha)d\alpha}\left(\int_{S_{i}}\gamma_{i}^{-1}(\alpha)d\alpha\int_{S_{i}}\gamma_{i}(\alpha)d\alpha-\frac{4\pi|U_{i}|}{L_{i}^{2}}\right)-C\epsilon.\end{split}$
(5.11)

Let us take a closer look at the two terms inside the parenthesis. For the
first term, Cauchy-Schwarz inequality gives

$\int_{S_{i}}\gamma_{i}^{-1}(\alpha)d\alpha\int_{S_{i}}\gamma_{i}(\alpha)d\alpha\geq
1,$

with equality achieved if and only if $\gamma_{i}$ is a constant. For the
second term, the isoperimetric inequality yields

$\frac{4\pi|U_{i}|}{L_{i}^{2}}\leq 1,$

(recall that $L_{i}=|\partial U_{i}|$), with equality achieved if and only
$U_{i}$ is a disk. By the assumption of Case 2, at least one of the
inequalities must be strict, thus the parenthesis on the right hand side of
(5.11) is strictly positive (and independent of $\epsilon$). Therefore there
exists some constant $c_{0}>0$ such that $\tilde{I}^{\epsilon}\geq
A_{i}^{\epsilon}\geq c_{0}$ for all sufficiently small $\epsilon$.

Case 3. There exist $i\neq j$ such that $i\nprec j$ and $j\nprec i$. Then it
is clear that for such $i,j$, $B_{i,j}^{\epsilon}$ in (5.9) is given by
$B_{i,j}^{\epsilon}=\frac{|D^{\epsilon}_{i}||D^{\epsilon}_{j}|}{4\pi\epsilon^{2}}$.
Hence (3.9) gives

$B_{i,j}^{\epsilon}\geq
L_{i}L_{j}\bigg{(}\int_{S_{i}}\gamma_{i}(\alpha)d\alpha\bigg{)}\bigg{(}\int_{S_{j}}\gamma_{j}(\alpha)d\alpha\bigg{)}-C\epsilon,$

which yields
$\tilde{I}^{\epsilon}\geq\frac{1}{2}L_{i}L_{j}(\int_{S_{i}}\gamma_{i}d\alpha)(\int_{S_{j}}\gamma_{j}d\alpha)>0$
for all sufficiently small $\epsilon>0$.

This finishes our discussion on all 3 cases. To conclude, since $\Gamma$ is
not a union of nested circles with constant $\gamma_{i}$’s on each connected
component, at least one of the 3 cases must hold, and all of them lead to
$\tilde{I}^{\epsilon}\geq c_{0}>0$. ∎

The above proposition immediately leads to the following corollary for the
$\Omega<0$ case.

######  Corollary 5.3.

Assume that $\Gamma_{i}$ and $\gamma_{i}$ satisfy (H1)–(H3) for
$i=1,\dots,n+m$. Let $I^{\epsilon}$ be defined as in (5.1), and assume
$\Omega<0$. Then we have $I^{\epsilon}\geq 0$ for all sufficiently small
$\epsilon>0$. In addition, if $\Gamma$ is not a union of concentric circles
all centered at the origin with constant $\gamma_{i}$’s, there exists some
$c_{0}>0$ independent of $\epsilon$, such that $I^{\epsilon}>c_{0}>0$ for all
sufficiently small $\epsilon>0$.

###### Proof.

Let us decompose $I^{\epsilon}$ as follows (recall the definition of
$\tilde{I}^{\epsilon}$ in (5.4))

$I^{\epsilon}=\tilde{I}^{\epsilon}+(-\Omega)\left(\epsilon^{-1}\int_{D^{\epsilon}}(|x|^{2}+\nabla
p^{\epsilon}\cdot
x)dx\right)=:\tilde{I}^{\epsilon}+\underbrace{(-\Omega)}_{>0}J^{\epsilon}.$
(5.12)

Recall that Proposition 5.2 gives $\tilde{I}_{\epsilon}\geq c_{0}>0$ as long
as $\Gamma$ is not a union of nested circles with constant $\gamma_{i}$’s. By
[14, Lemma 2.11], we have

$\int_{D_{i}^{\epsilon}}(|x|^{2}+\nabla p^{\epsilon}\cdot x)dx\geq
0\quad\text{ for any $i=1,\dots,n+m$,}$

thus $J^{\epsilon}\geq 0$. Putting them together, and using the fact that
$\Omega<0$, we know $I^{\epsilon}\geq c_{0}>0$ if $\Gamma$ is not a union of
nested circles with constant $\gamma_{i}$’s.

To finish the proof, we only need to focus on the case that the $\Gamma_{i}$’s
are nested circles with constant $\gamma_{i}$’s, but not all of them are
centered at the origin. Assume that there exists $k\in\\{1,\dots,n\\}$ such
that $\Gamma_{k}$ is a circle with radius $r_{k}$ centered at $x_{k}\neq 0$.
Since $\gamma_{k}$ is a constant, $D_{k}^{\epsilon}$ is an annulus given by
$B(x_{k},r_{k}+\epsilon\gamma_{k})\setminus B(x_{k},r_{k})$. The symmetry of
$D_{k}^{\epsilon}$ about $x_{k}$ immediately leads to
$p^{\epsilon}|_{D_{k}^{\epsilon}}=-\frac{1}{2}|x-x_{k}|^{2}+\frac{1}{2}(r_{k}+\epsilon\gamma_{k})^{2}.$
An elementary computation gives

$\epsilon^{-1}\int_{D^{\epsilon}_{k}}(|x|^{2}+\nabla p^{\epsilon}\cdot
x)dx=\epsilon^{-1}\int_{D^{\epsilon}_{k}}|x|^{2}-(x-x_{k})\cdot
xdx=\epsilon^{-1}|x_{k}|^{2}|D_{k}^{\epsilon}|\geq 2\pi
r_{k}\gamma_{k}|x_{k}|^{2}>0,$

where in the second-to-last step we used that $|D_{k}^{\epsilon}|=2\pi\epsilon
r_{k}\gamma_{k}+\pi\epsilon^{2}\gamma_{k}^{2}$. Setting $c_{0}:=2\pi
r_{k}\gamma_{k}|x_{k}|^{2}$ gives $I^{\epsilon}\geq c_{0}>0$, thus we can
conclude.∎

Now we are ready to prove Theorem 1.1. Note that for $\Omega<0$, the symmetry
result immediately follows from Proposition 5.1 and Corollary 5.3. For
$\Omega=0$, Proposition 5.1–5.2 already imply that a stationary vortex sheet
with positive strength must be a union of nested circles with constant
strength on each of them. To finish the proof, we only need to show that these
nested circles must be concentric.

######  Proof of Theorem 1.1.

For a uniformly-rotating vortex sheet with $\Omega<0$, the symmetry result for
$\Omega<0$ is a direct consequence of Proposition 5.1 and Corollary 5.3. Next
we focus on the stationary (i.e. $\Omega=0$) case.

Combining Propsitions 5.1–5.2, we obtain that $\Gamma$ is a union of nested
circles, and $\gamma_{i}$ is constant on $\Gamma_{i}$ for all $i=1\ldots,n$.
It remains to show that all $\Gamma_{i}$’s are concentric. Let us denote by
$\mathbf{v}_{i}$ the contribution to the velocity field by $\Gamma_{i}$. Since
$\Gamma_{i}$ is a circle with constant strength $\gamma_{i}$, a quick
application of the divergence theorem yields that $\mathbf{v}_{i}\equiv 0$ in
the open disk enclosed by $\Gamma_{i}$, whereas
$\mathbf{v}_{i}(x)=\dfrac{\gamma_{i}L_{i}(x-x^{0}_{i})^{\perp}}{2\pi|x-x^{0}_{i}|^{2}}$
in the open set outside $\Gamma_{i}$, where $x^{0}_{i}$ is the center of the
circle $\Gamma_{i}$.

Without loss of generality, let us reorder the indices such that $\Gamma_{i}$
is nested inside $\Gamma_{j}$ for $i<j$. Towards a contradiction, let $k>1$ be
such that $\Gamma_{k}$ is the first circle that is not concentric with
$\Gamma_{1}$. From the above discussion, we know that $\mathbf{v}_{i}=0$ on
$\Gamma_{k}$ for $i=k+1,\dots,n$ (since $\Gamma_{k}$ is nested inside
$\Gamma_{i}$), whereas for $i=1,\dots,k-1$ we have
$\mathbf{v}_{i}=\dfrac{\gamma_{i}L_{i}(x-x^{0}_{1})^{\perp}}{2\pi|x-x^{0}_{1}|^{2}}$
on $\Gamma_{k}$, since all these $\Gamma_{i}$’s have the same center
$x^{0}_{1}$ and are nested inside $\Gamma_{k}$. Summing them up (and also
using the fact that $\Gamma_{k}$ contributes zero normal velocity on itself,
since it is a circle with constant strength), we have

$BR(x)\cdot\mathbf{n}=\sum_{i=1}^{n}\mathbf{v}_{i}(x)\cdot\mathbf{n}=\left(\sum_{i=1}^{k-1}\gamma_{i}L_{i}\right)\frac{(x-x^{0}_{1})^{\perp}\cdot\mathbf{n}}{2\pi|x-x^{0}_{1}|^{2}}\quad\text{
on }\Gamma_{k},$

where the right hand side is not a zero function since $\Gamma_{k}$ has a
different center from $x_{1}^{0}$. This causes a contradiction with the fact
that $\omega=\omega_{0}$ is stationary. As a result, all
$\Gamma_{1},\dots,\Gamma_{n}$ must be concentric circles, finishing the proof.
∎

## Acknowledgements

JGS was partially supported by the European Research Council through ERC-
StG-852741-CAPA. JP was partially supported by NSF through Grants NSF
DMS-1715418, and NSF CAREER Grant DMS-1846745. JS was partially supported by
NSF through Grant NSF DMS-1700180. YY was partially supported by NSF through
Grants NSF DMS-1715418, NSF CAREER Grant DMS-1846745, and Sloan Research
Fellowship. JGS would like to thank Toan Nguyen for useful discussions.

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Javier Gómez-Serrano
---
Department of Mathematics
Brown University
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and
Departament de Matem$\grave{a}$tiques i Inform$\grave{a}$tica
Universitat de Barcelona
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Jaemin Park
School of Mathematics, Georgia Tech
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Jia Shi
Department of Mathematics
Princeton University
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Yao Yao
School of Mathematics, Georgia Tech
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