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

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.


Introduction
A vortex sheet is a weak solution of the 2D Euler equations: v t + v · ∇v = −∇p, ∇ · v = 0, (1.1) whose vorticity ω = curl(v) is a delta function supported on a curve or a finite number of curves Γ i = z i (α, t), i.e.
ω(x, t) = i i (α, t)δ(x − z i (α, t)). (1.2) Here i (α, t) is the vorticity strength on Γ i with respect to the parametrization z i , and the above equation is understood in the sense that for all test functions ϕ(x) ∈ C ∞ 0 (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 main goal of this paper is to establish radial symmetry properties of stationary/uniformlyrotating vortex sheets to (1.1). To do so, we first define what we mean by a stationary vortex sheet. Assume the initial data ω 0 of (1.2) is supported on a finite number of curves parametrized by z i (α), with strength i (α) (with respect to the parametrization z i ) respectively. If there exists some reparametrization choice c i (α) such that the right hand sides of (1.4)-(1.5) are both identically zero for every i, it gives that ω(·, t) is invariant in time, and we say ω(·, t) = ω 0 is a stationary vortex sheet.
For any x ∈ R 2 and Ω ∈ R, let R Ωt x denote the rotation of x counter-clockwise by angle Ωt about the origin. We say ω(x, t) = ω 0 (R Ωt x) is a uniformly-rotating vortex sheet with angular velocity Ω if ω 0 is stationary in the rotating frame with angular velocity Ω. (Note that in the special case Ω = 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 i (with respect to the constant-speed parametrization) for every i, the solution is stationary, and it is also uniformlyrotating with any Ω ∈ R. We would like to understand the reverse implication, namely: 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 ω(x, t) = ω 0 (R Ωt x) be a stationary/uniformly-rotating vortex sheet with angular velocity Ω. Assume that ω 0 is concentrated on Γ, which is a finite union of smooth curves, and ω 0 has positive vorticity strength on Γ. (See (H1)-(H3) in Section 2 for the precise regularity and positivity assumptions.) If Ω ≤ 0, Γ must be a union of concentric circles, and ω 0 must have constant strength along each circle (with respect to the constant-speed parametrization). In addition, if Ω < 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 γ 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 > 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 γ), which are stationary, and the segment of length 2a and density which is a rotating solution with angular velocity Ω [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 ∼ ε 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 ε > 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 ε → 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. 3 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 ⊂ R 2 , we denote |D| by its area (i.e. its Lebesgue measure). For x ∈ 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 ∼ , and obtain various quantitative estimates. In all these estimates, we say a term For a domain U ⊂ R 2 , in the boundary integral ∂U f · ndσ, n denotes the outer normal of the domain U .

Equations for a stationary/rotating vortex sheet
Let ω(·, t) = ω 0 (R Ωt ) be a stationary/rotating vortex sheet solution to the incompressible 2D Euler equation, where ω 0 ∈ M(R 2 ) ∩ H −1 (R 2 ) is a Radon measure. Here Ω = 0 corresponds to a stationary solution, and Ω = 0 corresponds to a rotating solution. Assume ω 0 is concentrated on Γ, which is a finite disjoint union of curves. Throughout this paper we assume Γ satisfies the following: (H1) Each connected component of Γ is smooth and with finite length, and it is either a simple closed curve (denote them by Γ 1 , . . . , Γ n ), or a non-self-intersecting curve with two endpoints (denote them by Γ n+1 , . . . , Γ n+m ). Here we require n + m ≥ 1, but allow either n or m to be 0.

Let us denote
which is strictly positive since we assume the curves {Γ i } n+m i=1 are disjoint. For i = 1, . . . , n + m, denote by L i the length of Γ i . Let z i : S i → Γ i denote a constant-speed parameterization of Γ i (in counter-clockwise direction if Γ i is a closed curve), where the parameter domain S i is given by Note that this gives |z i | ≡ L i , and the arc-chord constant is finite, since Γ is non-self-intersecting. Let s : Γ → R 2 be the unit tangential vector on Γ, given by s(z i (α)) := L i , and n : Γ → R 2 be the unit normal vector, given by n = s ⊥ . See Figure 1 for an illustration.
Note that for a closed curve, (H3) implies that γ i is uniformly positive; whereas for an open curve, γ i is positive in the interior of S i but vanishes at its endpoints. This is because any stationary/rotating vortex sheet with continuous γ i must have it vanishing at the two endpoints of any open curve: if not, one can easily check that |BR(z i (α)) · n(z i (α))| → ∞ as α approaches the endpoint, thus such a vortex sheet cannot be stationary in the rotating frame.
With the above notations of z i and γ i , the Birkhoff-Rott integral (1.3) along the sheet can now be expressed as with the kernel K 2 given by 5) and the principal value in (2.4) is only needed for the integral with k = i.
Let v : R 2 → R 2 be the velocity field generated by ω 0 , given by v := ∇ ⊥ (ω 0 * N ). Note that v ∈ C ∞ (R 2 \ Γ), but v is discontinuous across Γ. Let v + , v − : Γ → R 2 denote the two limits of v on the two sides of Γ (with v + being the limit on the side that n points into -see Figure 1 for an illustration), and [v] := v − − v + the jump in v across the sheets.
In the following lemma, we derive the equation that the Birkhoff-Rott integral satisfies for a stationary/rotating vortex sheet.
Lemma 2.1. Assume ω(·, t) = ω 0 (R Ωt x) is a stationary/uniformly-rotating vortex sheet with angular velocity Ω ∈ R, and ω 0 is concentrated on ∪ n+m i=1 Γ i , with z i and γ i defined as above. Then the Birkhoff-Rott integral BR (2.4) and the strength γ i satisfy the following two equations: and Figure 1. Illustration of the closed curves Γ 1 , . . . , Γ n and the open curves Γ n+1 , . . . , Γ n+m , and the definitions of n, s, v + and v − .
Proof. By definition of the stationary/uniformly-rotating solutions, ω 0 is a stationary vortex sheet in the rotating frame with angular velocity Ω. In this rotating frame, an extra velocity −Ωz ⊥ i 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)): where the term c i (α, t) accounts for the reparametrization freedom of the curves. Since ω 0 is stationary in the rotating frame, z i (·, t) parametrizes the same curve as z i (·, 0). Therefore ∂ t z i (α, t) is tangent to the curve Γ i , and multiplying n(z i (α, t)) to (2.8) gives where we use that n(z i (α, t)) · ∂ α z i (α, t) = 0. This proves (2.6). Now we prove (2.7). Towards this end, let us choose so that multiplying s(z i (α, t)) to (2.8) gives ∂ t z i (α, t) · s(z i (α, t)) = 0, and combining it with (2.10) gives ∂ t z i (α, t) = 0. In other words, with such choice of c i , the parametrization z i (α, t) remains fixed in time. Since ω 0 is stationary in the rotating frame, we know that with a fixed parametrization z i (α, t) = z i (α, 0), the strength i (α, t) must also remain invariant in time. Thus (2.9) becomes c i (α, t) i (α, t) ≡ C i . Plugging the definition of c i into the equation above and using the fact that z i is invariant in t, and finally the relationship between γ i and i in (2.3) yields (2.7) for i = 1, . . . , n.
And for the open curves i = n + 1, . . . , n + m, note that we do not have any reparametrization freedom at the two endpoints α = 0, 1, therefore the endpoint velocity BR(z i (0, t)) − Ωz ⊥ i (0, t) must be 0 to ensure that ω 0 is stationary in the rotating frame. This immediately leads to C i = 0 for i = n + 1, . . . , n + m, finishing the proof of (2.7).

Approximation by a thin vortex layer
Our aim in this section is to desingularize the vortex sheet ω 0 . Namely, for 0 < 1, we will construct a vorticity ω ∈ L ∞ (R 2 ) ∩ L 1 (R 2 ) that only takes values 0 and −1 , and is supported in an O( ) neighborhood of Γ, such that ω weakly converges to ω 0 as → 0 + .
For each i = 1, . . . , n + m, we will describe a neighborhood of Γ i using the following change of coordinates: let R i : S i × R → R 2 be given by and let  Figure 2 for an illustration. And for i = n + 1, . . . , n + m, the domains D i are simply-connected, and its boundary is smooth except at at most two points; see the right of Figure 2 for an illustration. In addition, for > 0 that is sufficiently small, one can check that Below we prove this fact in a stronger quantitative version, which will be used later.

Thus a crude estimate gives
. Finally, combining (3.5) and (3.6), it follows that (3.2) holds for In the next lemma we compute the partial derivatives and Jacobian of R i (α, η), which will be useful later.
Lemma 3.2. For any i = 1, . . . , n + m, let z i be a constant-speed parameterization of the curve Γ i (with length L i ), and let R i be given by (3.1). Then its partial derivatives are Moreover, its Jacobian is given by where κ i (α) denotes the signed curvature of Γ i at z i (α). 8 Proof. Since z i is the constant-speed parameterization of Γ i (which has length L i ), we have |z i | ≡ L i and n(z i (α)) = z i (α) ⊥ /L i . Taking the α and η partial derivatives of (3.1) directly yields (3.7).
Putting the two partial derivatives into columns of a 2 × 2 matrix and computing the determinant, we have where in the second equality we used that z i (α) = κ i (α)n(z i (α))L 2 i (recall that z i has constant speed L i ). This finishes the proof. Remark 3.3. We point out that for each i = 1, . . . , n + m, the determinant formula (3.8) immediately gives the following approximation of |D i |, which will be helpful in the proofs later: where the O( ) error term has its absolute value bounded by C , with C only depending on be the velocity field generated by ω .
In the next lemma we aim to obtain some fine estimate of v in the thin vortex layer D . Our goal is to show that along each cross section of the thin layer (i.e. fix i and α, and let η vary in [−1, 0]), the function η → v (R i (α, η)) is almost a linear function in η, with the endpoint values (at η = −1 and 0) being almost v − (z i (α)) and v + (z i (α)) respectively. Lemma 3.4. For i = 1, . . . , n + m, assume Γ i and γ i satisfy (H1)-(H3). Let Figure 3 for an illustration of g i (α, η)). Then for all sufficiently small > 0, for all i = 1, . . . , n + m we have Proof. Let i be any fixed index in 1, . . . , n + m. We begin with breaking v into contributions Figure 3. Illustration of the definition of g i (α, ·) (the orange arrows).
where the kernel K 2 is given by (2.5). Similarly, we can break 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.
• Estimates for k = i terms. For any k = i, we aim to show that finishing the proof of (3.12).
• Estimates for the k = i term. It will be more involved to control the k = i term, and our goal is to show that To begin with, we again rewrite v i as in (3.13) with k = i, and plug in the formula (3.8) for the determinant. This leads to 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 Using (3.2) of Lemma 3.1 and the fact that |K 2 (x)| ≤ |x| −1 , we can bound I 2 as where C depends on z i C 2 and γ i L ∞ .
In the rest of the proof we focus on estimating The motivation for us to define such f and J(t) is that at t = 1, we have where v i is the velocity field generated by the sheet Γ i . Recall that v i has a jump across Γ i , where we denote its limits on two sides by v ± i . Using Lemma 3.5, which we will prove momentarily, we have We can then split the integration domain on the right hand side of (3.18) into η ∈ (−1, η) and η ∈ (η, 0), and use (3.19) to approximate the integrand in each interval. This gives where in the last step we used that In addition, we have where the last inequality follows from (H2) and the fact that n(z i (α)) ∈ C 1 (S i ). Therefore, for any t ∈ (0, 1), taking the t derivative of (3.17) and using that |∇K 2 (x)| ≤ |x| −2 , we have 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 = i estimates in (3.12).
The following lemma proves (3.19). Let v i be the velocity field generated by the sheet Γ i , which is smooth in R 2 \ Γ i , and has a discontinuity across Γ i . It is known that v i converges to v ± i respectively on the two sides of Γ i [24]. However, we were unable to find a quantitative convergence rate (in terms of the distance from the point to Γ i ) in the literature, especially under the assumption that γ i is only in C b (S i ) for the open curves. Below we prove such an estimate.
Then there exist constants C, 0 > 0 depending on on b (as in (H2)), z i C 2 (S i ) , γ i C b (S i ) and F Γ , such that for all ∈ (0, 0 ) and η ∈ (−2, 2) we have where and BR i is the contribution from the i-th integral in (2.4).
Plugging this into A 11 and using the Hölder continuity of γ, we have where the last step follows from the fact that |c| ≤ 2 γ ∞ . Now let us turn to A 12 , which requires a more delicate estimate of f (y, c). Let us break A 12 as For B 1 , let us take the gradient of f (y, c) (as in (3.29)) in the first variable. An elementary computation yields that as long as x satisfies |x + c| 2 ≥ c 0 (|x| 2 + |c| 2 ).
As for B 2 , using the definition ofỹ, the identity (3.28) and the fact thatỹ · c = 0, we have dα .
For the open curves i = n + 1, . . . , n + m, the above integral becomes where in the second inequality we used that the integral in [−α, α] gives zero contribution to the principal value, since the integrand is odd.
Next we discuss two cases. If α > |c|, we bound the integrand by Cθ −3 , which gives where the second inequality follows from the assumption γ(0) = 0 for an open curve in (H3), as well as the Hölder continuity of γ. And if 0 < α ≤ |c|, the integrand can be bounded above by θ −1 |c| −2 , which immediately leads to In both cases we have |B 2 | ≤ C b | log |, and combining it with the B 1 and A 11 estimates gives (3.23).

Constructing a divergence-free perturbation
In this section, we aim to construct a divergence-free velocity field u : D → R 2 , such that −u tends to make each D i "more symmetric". Let u : D → R 2 be given by where the function p : D → R is chosen such that where n is the unit normal of l pointing outwards of D . Note that ∂D has a total of 2n + m connected components: D i is doubly-connected for i = 1, . . . , n (denote its outer and inner boundaries by ∂D i,out and ∂D i,in ; note that ∂D i,in coincides with Γ i ), whereas it is simply-connected for i = n + 1, . . . , n + m (denote its boundary by ∂D i ).
Next we show that there indeed exists a function p so that u satisfies (4.2)-(4.3). Clearly, (4.2) requires that p satisfies ∆p = −2 in D .

(4.4)
As for the boundary conditions, we let p | ∂D i = 0 for i = n + 1, . . . , n + m, (4.5) so the divergence theorem yields that (4.3) is satisfied for each l = ∂D i for i = n + 1, . . . , n + m. As for i = 1, . . . , n, we define where c i > 0 is the unique constant such that where U i is the domain enclosed by ∂D i,in = Γ i (thus U i is independent of ), and n is the outer normal of U i (thus the inner normal of D i ). The existence of c i is guaranteed by [14, Lemma 2.5].
One can then check that ∂U i u · ndσ = 0. Applying the divergence theorem in D i then gives us that ∂D i,out u · ndσ = 0 as well.
In [14] we proved a rearrangement inequality for such p in a similar spirit of Talenti's rearrangement inequality for elliptic equations [37], which we state below.
Moreover, each inequality above achieves equality if and only D i is either a disk or an annulus.
Note that the inequalities (4.8)-(4.9) hold for any domain with C 1,α boundary. Even though the inequalities are strict when D i is non-radial, they are not strong enough to rule out nonradial vortex sheets, as we need quantitative versions of strict inequalities that are still valid in the → 0 + limit. As we will see in the proof of Proposition 5.2, the key step is to show that if some Γ i is either not a circle or does not have a constant γ i , then the following quantitative version of (4.9) holds: In order to upgrade (4.9) into a quantitative version, we need to obtain some fine estimates for p that take into account the shape of the thin domains D i . For i = n + 1, . . . , n + m, since p = 0 on ∂D i , and the domain D i is a thin simply-connected domain with width 1, intuitively one would expect that |p | ≤ C 2 . The next proposition shows that this crude estimate is indeed true, and its proof is postponed to Section 4.1. For i = 1, . . . , n, the estimate is more involved, since p takes different values c i and 0 on the inner and outer boundaries of D i . Heuristically speaking, since D i is a doubly-connected thin tubular domain with width ∼ , we would expect that p i (in α, η coordinate) changes almost linearly from 0 to c i as η goes from −1 (outer boundary) to 0 (inner boundary). Next we will show that the error between p (R i (α, η)) and the linear-in-η function c i (1 + η) is indeed controlled by O( 2 ). We will also obtain fine estimates of the gradient of the function c i (1 + η), as well as the boundary value c i . Again, its proof is postponed to Section 4.1. . For such p , let us definep ,q : D i → R as follows:

Also let
(4.11) 17 Then there exist 1 and C only depending on z i C 3 (S i ) , γ i C 2 (S i ) and F Γ , such that for all ∈ (0, 1 ) we have the following: Then there exist positive constants and Proof. Throughout the proof, let i ∈ {1, . . . , n + m} be fixed. For notational simplicity, in the rest of the proof we omit the subscript i in R i , D i , S i , z i and γ i .
To complete the proof, we only need to prove (4.27)-(4.29) for small > 0. Note that (4.27) follows immediately from computing the Laplacian of b 2 . For (4.28), let us pick y ∈ ∂V 1 (α), and we aim to show that b 2 (y) ≤ 0. Note that y = R (β, 0) or R (β, −1) for some β ∈ S. We first deal with the first case.
Now we are ready to prove Proposition 4.3.

21
Proof of Proposition 4.3. Throughout the proof, let i ∈ {1, . . . , n} be fixed. For notational simplicity, in the rest of the proof we omit the subscript i from all terms.
We claim that for some constant C > 0 only depending on z i C 3 (S i ) , γ i C 2 (S i ) and F Γ . Assuming these are true, let us explain how they lead to (4.12)-(4.14). By (4.6) and (4.10), p andp have the same boundary condition, thus q = 0 on ∂D . This and (4.39) allow us to apply Corollary 4.5 to q to obtain the estimate (4.35), implying (4.12).

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 Ω < 0 must be radially symmetric. The key idea of the proof is to define the integral and compute it in two different ways. The motivation of the definition is as follows. As discussed in [14, Section 2.1], I can be thought of as a first variation of an "energy functional" when we perturb ω by a divergence free vector u in D . (This functional E only serves as our motivation, and will not appear in the proof.) On the one hand, using that ω 0 is stationary in the rotating frame with angular velocity Ω and ω is a close approximation of ω 0 , we will show in Proposition 5.1 that I is of order O( | log |), thus goes to zero as → 0. On the other hand, using the particular u that we constructed in Section 4, we will prove in Proposition 5.2 that if Ω = 0, I is strictly positive independently of unless all the vortex sheets are nested circles with constant density; and also prove a similar result in Corollary 5.3 for Ω < 0.
Proposition 5.1. Assume ω(·, t) = ω 0 (R Ωt ·) is a stationary/uniformly-rotating vortex sheet with angular velocity Ω ∈ R, where ω 0 satisfies (H1)-(H3). Then there exists some C > 0 only depending on b (as in (H2)), Proof. Let us decompose I =: We start with showing that |I i | ≤ C b | log | for i = n + 1, . . . , n + m. For such i, p = 0 on ∂D i , thus the divergence theorem (and the fact that ω = −1 in D i ) gives Using the estimate |p | ≤ C 2 in Proposition 4.2 and the fact that |D i | ≤ C from (3.9), we easily bound the second integral by C . To control the first integral T i , we rewrite it using the change of variables x = R i (α, η) and the definition v := ∇ ⊥ (ω * N ) in (3.10): (also note that on the right hand side we group −1 with the determinant) Let us take a closer look at the integrand, which is a product of 3 terms. Clearly, the definition of R i gives R i (α, η) = z i (α) + O( ). As for the middle term J i , Lemma 3.4 yields Using the fact that BR(z i (α)) = Ωz ⊥ i (α) for i = n + 1, . . . , n + m (which follows from (2.6) and (2.7)), it becomes Also it follows from (3.8) that K i (α, η) = L i γ i (α) + O( ). Plugging these three estimates into the above integral gives where the last step follows from the fact that 0 −1 (η + 1 2 )dη = 0. This finishes the proof that In the rest of the proof we aim to show |I i | ≤ C b | log | for i = 1, . . . , n, which is slightly more involved. Recall that in Proposition 4.3 we definedp and q in D i for i = 1, . . . , n, where they satisfy p =p + q in D i , and q = 0 on ∂D i . This allows us to apply the divergence theorem (to the q term only) and decompose I i as We can easily show that I i,2 = O( ): (4.12) of Proposition 4.3 gives |q | ≤ C 2 , and combining it with |D i | ≤ C in (3.9) immediately yields the desired estimate.
Next we turn to I i,1 . Again, the change of variables x = R i (α, η) and the definition v := ∇ ⊥ (ω * N ) gives For the three terms in the product of the integrand, we will approximate the first term using the definition of R i and (4.14) of Proposition 4.3: i (α)dα is given by (4.11). Lemma 3.4 allows us to approximate the middle term J i as (5.2), however (5.3) no longer holds since for i = 1, . . . , n we do not have BR(z i (α)) = Ωz ⊥ i (α). As for K i , we again use (3.8) to approximate it by K i (α, η) = L i γ i (α) + O( ). Plugging these three estimates into the integrand of I i,1 gives where we again use the fact that the (η + 1 2 ) term gives zero contribution since 0 −1 (η + 1 2 )dη = 0. Next we will show the integral on the right hand side is in fact 0. Since ω is a rotating solution with angular velocity Ω, the conditions (2.6) and (2.7) yield that ), for some constant C i . Plugging this into the above integral gives where the second step follows from the definition of β i in (4.11). Let us compute the integral on the right hand side by changing to arclength parametrization and applying the divergence theorem: Finally, summing the I i estimates for i = 1, . . . , n + m gives |I | ≤ C b | log | for all sufficiently small > 0, thus we can conclude. Now we will use a different way to compute I . Let us first define a new integralĨ that is the same as I except with Ω set to zero: Next we will prove thatĨ is strictly positive independently of 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 Γ i is either not a circle or does not have a constant γ i , then the estimates on p in Propositions 4.2-4.3 lead to the following quantitative version of (4.9): Proposition 5.2. LetĨ be defined as in (5.4). Assume that Γ i and γ i satisfy (H1)-(H3) for i = 1, . . . , n + m. Then we haveĨ ≥ 0 for all sufficiently small > 0.
In addition, if Γ is not a union of nested circles with constant γ i 's on each connected component, there exists some c 0 > 0 independent of , such thatĨ > c 0 > 0 for all sufficiently small > 0.
Proof. We start by decomposingĨ as I 1 can be easily computed as where the second equality is obtained by exchanging x with y and taking the average with the original integral. As for I 2 , we have where the first equality follows from the divergence theorem, the second equality follows from the boundary conditions (4.5) and (4.6) for p (as well as the fact that ∂U i and ∂D i have opposite outer normals), and the last inequality follows from the divergence theorem as well as the inequality 2π due to (4.8). Let us denote j ≺ i if i ∈ {1, . . . , n} , j ∈ {1, . . . , n + m}, j = i and Γ j lies in the interior of the domain enclosed by Γ i (that is, Γ j ⊂ U i ). If not, we denote j ⊀ i. Note that for sufficiently small > 0, we have Applying this to (5.6) yields where in the first step we used that the i = n + 1, . . . , n + m terms have zero contribution in the first sum, due to the definition of j ≺ i. 26 Adding (5.5) and (5.8) together, we obtaiñ From (4.9), it follows that A i ≥ 0 for all i = 1, . . . , n + m, with equality achieved if and only if each D i is a disk or an annulus. Note that B i,j ≥ 0 as well for all i and j, since for any i = j, at most one of i ≺ j and j ≺ i can hold. Putting these together yields thatĨ ≥ 0 for any sufficiently small > 0.
In the rest of the proof, we assume Γ is not a union of nested circles with constant γ 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 thatĨ ≥ c 0 > 0, where c 0 is independent of for all sufficiently small > 0.

Case 1.
There exists some open curve Γ i that is not a loop. In this case D i is simply-connected, and p = 0 on ∂D i by (4.5). Applying Proposition 4.2 to p in D i , we have sup D i p ≤ C 2 , where C is independent of . This leads to D i p dx ≤ C 3 , since |D i | = O( ) by (3.9). As a result, for the index i we have where we again used (3.9) in the second inequality. This gives that A i ≥ Case 2. There exists some closed curve Γ i that is either not a circle, or γ i is not a constant. In this case we aim to show that A i ≥ c 0 > 0, and this will be done by finding good approximations (independent of ) for both terms in A i . For the first term |D i | 2 4π 2 , using (3.9) we again have where J i > 0 is independent of . For the second term −2 D i p i dx, rewriting the integral using the change of variables x = R i (a, η) gives Recall that in Proposition 4.3 we definedp (R i (α, η)) := c i (1 + η) and q such that p −p = q . By (4.12) and (4.13), for all α ∈ S i and η ∈ (−1, 0) we have where β i := 2|U i | L i S i γ −1 i (α)dα is defined in (4.11). Combining this with the expression of the determinant in (3.8), we have where K i is independent of . Putting this together with (5.10) yields the following: Let us take a closer look at the two terms inside the parenthesis. For the first term, Cauchy-Schwarz inequality gives with equality achieved if and only if γ i is a constant. For the second term, the isoperimetric inequality yields 4π|U i | L 2 i ≤ 1, (recall that L i = |∂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 ). Therefore there exists some constant c 0 > 0 such thatĨ ≥ A i ≥ c 0 for all sufficiently small .

Case 3.
There exist i = j such that i ⊀ j and j ⊀ i. Then it is clear that for such i, j, B i,j in (5.9) is given by B i,j = |D i ||D j | 4π 2 . Hence (3.9) gives which yieldsĨ ≥ 1 2 L i L j ( S i γ i dα)( S j γ j dα) > 0 for all sufficiently small > 0. This finishes our discussion on all 3 cases. To conclude, since Γ is not a union of nested circles with constant γ i 's on each connected component, at least one of the 3 cases must hold, and all of them lead toĨ ≥ c 0 > 0.
The above proposition immediately leads to the following corollary for the Ω < 0 case. Corollary 5.3. Assume that Γ i and γ i satisfy (H1)-(H3) for i = 1, . . . , n + m. Let I be defined as in (5.1), and assume Ω < 0. Then we have I ≥ 0 for all sufficiently small > 0. In addition, if Γ is not a union of concentric circles all centered at the origin with constant γ i 's, there exists some c 0 > 0 independent of , such that I > c 0 > 0 for all sufficiently small > 0. (|x| 2 + ∇p · x)dx ≥ 0 for any i = 1, . . . , n + m, thus J ≥ 0. Putting them together, and using the fact that Ω < 0, we know I ≥ c 0 > 0 if Γ is not a union of nested circles with constant γ i 's.
To finish the proof, we only need to focus on the case that the Γ i 's are nested circles with constant γ i 's, but not all of them are centered at the origin. Assume that there exists k ∈ {1, . . . , n} such that Γ k is a circle with radius r k centered at x k = 0. Since γ k is a constant, D k is an annulus given by B(x k , r k + γ k ) \ B(x k , r k ). The symmetry of D k about x k immediately leads to p | D k = − 1 2 |x − x k | 2 + 1 2 (r k + γ k ) 2 . An elementary computation gives where in the second-to-last step we used that |D k | = 2π r k γ k + π 2 γ 2 k . Setting c 0 := 2πr k γ k |x k | 2 gives I ≥ c 0 > 0, thus we can conclude. Now we are ready to prove Theorem 1.1. Note that for Ω < 0, the symmetry result immediately follows from Proposition 5.1 and Corollary 5.3. For Ω = 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 Ω < 0, the symmetry result for Ω < 0 is a direct consequence of Proposition 5.1 and Corollary 5.3. Next we focus on the stationary (i.e. Ω = 0) case.
Combining Propsitions 5.1-5.2, we obtain that Γ is a union of nested circles, and γ i is constant on Γ i for all i = 1 . . . , n. It remains to show that all Γ i 's are concentric. Let us denote by v i the contribution to the velocity field by Γ i . Since Γ i is a circle with constant strength γ i , a quick application of the divergence theorem yields that v i ≡ 0 in the open disk enclosed by Γ i , whereas i is the center of the circle Γ i .
Without loss of generality, let us reorder the indices such that Γ i is nested inside Γ j for i < j. Towards a contradiction, let k > 1 be such that Γ k is the first circle that is not concentric with Γ 1 . From the above discussion, we know that v i = 0 on Γ k for i = k + 1, . . . , n (since Γ k is nested inside Γ i ), whereas for i = 1, . . . , k − 1 we have v i = γ i L i (x − x 0 1 ) ⊥ 2π|x − x 0 1 | 2 on Γ k , since all these Γ i 's have the same center x 0 1 and are nested inside Γ k . Summing them up (and also using the fact that Γ k contributes zero normal velocity on itself, since it is a circle with constant strength), we have where the right hand side is not a zero function since Γ k has a different center from x 0 1 . This causes a contradiction with the fact that ω = ω 0 is stationary. As a result, all Γ 1 , . . . , Γ n must be concentric circles, finishing the proof.