Advances in Mathematics 416 (2023) 108917 Contents lists available at ScienceDirect Advances in Mathematics journal homepage: www.elsevier.com/locate/aim The global regularity of vortex patches revisited Joan Verdera Departament de Matemàtiques, Universitat Autònoma de Barcelona, Centre de Recerca Matemàtica, Barcelona, Catalonia, Spain a r t i c l e i n f o a b s t r a c t Article history: We prove persistence of the regularity of the boundary of Received 11 May 2022 vortex patches for a large class of transport equations in the Received in revised form 2 February plane. The velocity field is given by convolution of the vorticity 2023 with an odd kernel, homogeneous of degree −1 and of class Accepted 2 February 2023 Available online 17 February 2023 C 2 off the origin. Communicated by C. Fefferman © 2023 The Author. Published by Elsevier Inc. This is an open access article under the CC BY license (http:// MSC: creativecommons.org/licenses/by/4.0/). primary 35Q31, 35Q35 secondary 35Q49, 42B20 Keywords: Transport equation Vortex patch Calderón-Zygmund singular integral 1. Introduction The vorticity form of the Euler equation for incompressible ideal fluids is ∂t ω + v · ∇ω = 0, v(·, t) = ω(·, t)  ∇⊥ N, (1) ω(z, 0) = ω0 (z), E-mail address: joan.verdera@uab.cat. https://doi.org/10.1016/j.aim.2023.108917 0001-8708/© 2023 The Author. Published by Elsevier Inc. This is an open access article under the CC BY license (http://creativecommons.org/licenses/by/4.0/). 2 J. Verdera / Advances in Mathematics 416 (2023) 108917 where ω(z, t) is vorticity at the point z at time t, z = x + iy ∈ C = R2 , t ∈ R, and N (z) = 2π 1 log |z| is the fundamental solution of the laplacian in the plane. Since the kernel in the Biot-Savart law v(·, t) = ω(·, t) ∇⊥ N is an orthogonal gradient, the velocity field v has zero divergence, in accordance with the fact that the fluid is incompressible. It is a deep result of Yudovich [10] that incompressibility yields that (1) is well posed in L∞ (R2 ) ∩ L1 (R2 ), in the sense that there exists a unique weak solution of (1) for each given initial condition in L∞ (R2 ) ∩ L1 (R2 ). When one takes as initial condition the characteristic function ρ0 = χD0 of a bounded domain D0 , then ω(·, t) = χDt (·), where Dt is a bounded domain. This follows from the fact that (1) is a transport equation and one refers to such a solution as a vortex patch. In the eighties the problem of deciding if boundary smoothness persisted for all times was an open challenging question. Chemin proved in 1993 [4] that if D0 is a bounded simply connected domain with boundary of class C 1+γ , 0 < γ < 1, then the weak solution of (1) with initial condition χD0 is of the form ω(z, t) = χDt (z) with Dt a simply connected domain of class C 1+γ for all times t ∈ R. Indeed in [4] one proves a more general statement and the proof depends on para-differential calculus. In [1] a shorter proof, based on classical analysis methods, was devised. See also [8]. In this paper we consider the regularity problem for vortex patches for transport equations of the form ∂t ω + v · ∇ω = 0, v(·, t) = ω(·, t)  k, (2) ω(z, 0) = χD0 (z), where k : R2 \ {0} → R2 is an odd kernel, homogeneous of degree −1, of class C 2 off the origin, and D0 is a simply connected domain with boundary of class C 1+γ . These kernels produce in general velocity fields v with non-zero divergence. For instance, for k(z) = ∇N (z) one has div v(z, t) = χDt (z) and the divergence is non-zero but still bounded. This kernel appears in models for aggregation and chemotaxis phenomena in 1 biology (see [2] and the references given there). If k(z) = 2πz , the complex conjugate of ∇N (z), then   1 x2 − y 2 div v(z, t) = χDt (z)  div k(z) = − p. v.  χDt (z), z ∈ C, (3) π |z|4 which is a second order Riesz transform applied to a characteristic function. The second order Riesz transforms are basic examples of convolution principal value Calderón- Zygmund singular integrals, which do not preserve L∞ (R2 ). Thus, in general, we should expect the divergence to be an unbounded function in BMO(R2 ). Indeed, if the domain Dt is smoothly bounded, then an even Calderón-Zygmund singular integral with kernel of class C 1 off the origin sends χDt to a bounded function, an elementary fact that guaranties that v(z, t) is a Lipschitz field. Thus in dealing with equation (2) in the class J. Verdera / Advances in Mathematics 416 (2023) 108917 3 of smooth vortex patches one has still bounded divergence, but formula (3) anticipates serious difficulties. Our main result reads as follows. Theorem. If D0 is a bounded simply connected domain with boundary of class C 1+γ , 0 < γ < 1, then there exists a weak solution of (2) with initial condition χD0 of the form ω(z, t) = χDt (z) with Dt a simply connected domain of class C 1+γ for all times t ∈ R. This solution is unique in the class of characteristic functions of domains of class C 1+γ . This has been proved simultaneously in any dimension in [3] via an intricate construc- tion of appropriate defining functions. Our proof avoids the use of defining functions. We proceed to state the definition of weak solution to (2), forward in time. A function ω(z, t) ∈ L∞ ([0, T ), L1 (R2 ) ∩ L∞ (R2 )), T > 0, is a weak solution of (2) in R2 × [0, T ) if for each 0 < T0 < T and each compactly supported ϕ ∈ C 1 (R2 × [0, T0 ]) one has  T0      ϕ(z, T0 )ω(z, T0 ) − ϕ(z, 0)χD0 (z) dz = (∂t ϕ + v · ∇ϕ) ω + ϕ ω div(v) dzdt. R2 0 R2   If ω(z, t) ∈ C R2 × [0, T ) is a classical solution of (2), then it is a weak solution; to 1   show this one just applies the transport formula. Conversely, if ω(z, t) ∈ C 1 R2 × [0, T ) is a weak solution, then it is a classical solution. See [5, Chapter 8] for a discussion of weak solutions for the Euler vorticity equation. We now briefly outline the proof of the theorem. As in [1], first we take for granted that a local in time solution for the CDE (contour dynamics equation (17)) exists (see Chapter 8 of [5]) and then we continue by proving an a priori inequality for the relevant quantities giving the smoothness of Dt . In the appendix we give indications, for the sake of the reader, on how to solve locally in time the CDE in our case (the argument is basically routine). The core of the proof is the second step, in which we show the a priori inequality for the Hölder seminorm of order γ of the gradient of the local in time solution X(·, t) of the contour dynamics equation. This is achieved by bringing in a commutator, which appears after an application of Whitney’s extension theorem. Is at this moment when we have to leave the boundary of Dt and work in the ambient space. The third step consists in proving a logarithmic inequality which estimates ∇v(·, t)∞ in terms of ∇X(·, t)γ,∂D0 and the Lipschitz constant of the inverse of X(·, t). Again the estimate is performed by resorting to objects that live in the plane but off the boundary of Dt . The hypothesis on the kernel k are the more general one can reasonably think of. The fact that the kernel is odd and homogeneous of degree −1 gives that a first order derivative of the kernel is a constant multiple of the delta function plus an even Calderón- Zygmund singular integral. The kernels of these even singular integrals are C 1 off the 4 J. Verdera / Advances in Mathematics 416 (2023) 108917 origin since k is assumed to be C 2 off the origin, and thus send characteristic functions of smoothly bounded domains into bounded functions. 2. The a priori estimate Parametrize the boundary ∂D0 by a mapping h : T → ∂D0 , of class C 1+γ (T , R2 ) satisfying the bilipschitz condition c−1 0 |u − v| ≤ |h(u) − h(v)| ≤ c0 |u − v|, u, v ∈ T for some positive constant c0 . The mapping h induces a norm in the vector space of C 1+γ functions on the boundary of D0 with values in the plane defined on X ∈ C 1+γ (∂D0 , R2 ) by X1+γ = X∞,∂D0 + ∇Xγ,∂D0 , (4) where X∞,∂D0 = sup{|X(α)| : α ∈ ∂D0 } and | du d X(h(u)) − du d X(h(v))| ∇Xγ,∂D0 = sup{ : u, v ∈ T , u = v}. |u − v|γ We have used the notation (which is an abuse of language we adopt for the sake of conciseness) d d f (u) = f (h(eiθ )), u = h(eiθ ), du dθ for each differentiable function f defined on T . The contour dynamics equation is an ordinary differential equation defined in the open set Ω of C 1+γ (∂D0 , R2 ) consisting of those mappings in C 1+γ (∂D0 , R2 ) satisfying the bilipschitz condition |X(α) − X(β)| 0 < b = b(X) = inf{ : α, β ∈ ∂D0 , α = β}. (5) |α − β| For more details on the contour dynamics equation (CDE) in this context the reader may consult the last section. The CDE has a unique solution X(·, t) ∈ Ω for t ∈ (−T, T ) satisfying the initial condition X(·, 0) = I, where I stands for the identity map on ∂D0 . Then t → X(α, t) is the trajectory of the particle that at time 0 is at the point α ∈ ∂D0 and X(∂D0 , t) is a Jordan curve of class C 1+γ which encloses a simply connected domain Dt . The function J. Verdera / Advances in Mathematics 416 (2023) 108917 5 ω(z, t) = χDt (z) is a weak solution of equation (2) (forward in time) in R2 × [0, T ). This can be shown as in [5, proposition 8.6, p.332]. Thus the flow associated with the Lipschitz velocity field v(·, t) = χDt (·, t)  k, namely the solution of d X(α, t) = v(X(α, t), t), dt (6) X(α, t) = α ∈ R , 2 coincides with the solution of the CDE for α ∈ ∂D0 . What remains to be proven is that the maximal time of existence of the solution of the CDE is T = ∞ and for that we need an a priori inequality. By (6) t X(h(u), t) = h(u) + v(X(h(u), s), s) ds, u ∈ T. 0 Hence t   d dh d X(h(u), t) = (u) + ∇v(X(h(u), s), s) X(h(u), s) ds. (7) du du du 0 d Clearly du X(h(u), s) is a tangent vector to ∂Ds at the point X(h(u), s). We would like to show that   d ∇v(X(h(u), s), s) X(h(u), s) du is a commutator. We need three lemmas. The first works in Rn . The notion of jet is in [9, p. 176], although the term jet is not used there. Given a subset S of ⊂ Rn and a vector field F : S → Rn set |F (x) − F (y)| F γ,S = sup{ : x, y ∈ S, x = y}. |x − y|γ Lemma 1. Let Γ be a hypersurface of dimension n − 1 and of class C 1+γ in Rn . Let N (x) = (N1 (x), . . . , Nn (x)) be a normal field on Γ of class C γ (Γ). Then (0, N (x)) is a jet of class C 1+γ (Γ) with constant AN γ,Γ , A a constant depending only on γ. In other words, sup |(y − x) · N (x)| ≤ AN γ,Γ |y − x|1+γ . x,y∈Γ, x=y Proof. Assume, without loss of generality, that x = 0 and N (0) = (0, . . . , 0, Nn (0)) with N γ,Γ Nn (0) > 0. Set δ −γ = 2 |N (0)| . 6 J. Verdera / Advances in Mathematics 416 (2023) 108917 If |y| ≥ δ, then |y · N (0)| ≤ |y||N (0)| = |y|2N γ,Γ δ γ ≤ 2N γ,Γ |y|1+γ , as required. If y ∈ B(0, δ) ∩ Γ, then |N (0)| |N (y) − N (0)| ≤ N γ,Γ δ γ = . 2 Hence the tangent hyperplane to Γ at y forms an angle of less than 30◦ with the hor- izontal hyperplane. Consequently B(0, δ) ∩ Γ is the graph of a function yn = ϕ(y  ), y  = (y1 , . . . , yn−1 ), of class C 1+γ , defined on the projection U of B(0, δ) ∩ Γ on {y ∈ Rn : yn = 0}. Note that U is open in Rn−1 and that, for y ∈ B(0, δ) ∩ Γ, the segment with extremes 0 and y  is contained in U (by the implicit function theorem). It is also clear that |∇ϕ(y)| ≤ 1, y  ∈ U , because of the inclination of tangent hyperplanes with respect to the horizontal hyperplane. If y ∈ B(0, δ) ∩ Γ we have |y · N (0)| ≤ |ϕ(y  )||N (0)| ≤ sup |∇ϕ(z  )| |y  ||N (0)| ≤ ∇ϕγ,Γ |y  |1+γ |N (0)|. |z  |≤|y|, z  ∈U Set M (y  ) = (−∂1 ϕ(y  ), . . . , −∂n−1 ϕ(y  ), 1), y  ∈ U, which is a normal vector to Γ at y = (y  , ϕ(y  )). Hence N (y) M (y  ) = , y ∈ B(0, δ) ∩ Γ, Nn (y) and N (y) ∇ϕγ,U = M γ,U = . Nn (y) γ,U Take y, z ∈ B(0, δ) ∩ Γ. Then √ |y − z| = (|y  − z  |2 + |ϕ(y  ) − ϕ(z  )|2 )1/2 ≤ 2|y  − z  |. One has N (y) N (z) 1 |Nn (z) − Nn (y)| − ≤ |N (y) − N (z)| + |N (z)| . (8) Nn (y) Nn (z) |Nn (y)| |Nn (y)||Nn (z)| J. Verdera / Advances in Mathematics 416 (2023) 108917 7 By the definition of δ |N (0)| |Nn (y)| ≥ |Nn (0)| − |Nn (y) − Nn (0)| ≥ |N (0)| − N γ,Γ δ γ = . 2 Thus the right hand side of (8) is not greater than N γ,Γ  23+γ/2 |y − z  |γ . |N (0)| Therefore |y · N (0)| ≤ ∇ϕγ,U |N (0)| ≤ 23+γ/2 N γ,Γ |y|1+γ and the lemma follows with A = 23+γ/2 .  Lemma 2. Let Γ be a Jordan curve of class C 1+γ in the plane and τ (x) = (τ1 (x), τ2 (x)), x ∈ Γ, a tangent vector to Γ at x with τ ∈ C γ (Γ, R2 ). Then there exists g(x) = (g1 (x), g2 (x)) ∈ C γ (R2 , R2 ) such that g = τ on Γ, gγ,R2 ≤ Cτ γ,Γ for a positive constant C depending only on γ, and div g = 0 in R2 . Proof. The field (−τ2 (x), τ1 (x)) is normal to Γ at each x ∈ Γ. By the previous lemma (0, −τ2 (x), τ1 (x)) is a C 1+γ jet on Γ with constant Cγ τ γ,Γ and by Whitney’s extension theorem there is a function ϕ ∈ C 1+γ (R2 , R) such that ϕ = 0, ∂1 ϕ = −τ2 , ∂2 ϕ = τ1 , on Γ and ∇ϕγ,R2 ≤ C0 Cγ τ γ,Γ , with C0 an absolute constant [9, Theorem 4, p.177]. Therefore τ (x) = (∂2 ϕ(x), −∂1 ϕ(x)), x ∈ Γ. Set g = (∂2 ϕ, −∂1 ϕ), so that g has no divergence in the plane and gγ,R2 ≤ C τ γ,Γ , where C = C0 Cγ depends only on γ.  Lemma 3. Let τ (x) = (τ1 (x), τ2 (x)) ∈ C γ (Γ) be a tangent field on Γ = ∂Ds . Then ∇v(x, s)(τ (x)), x ∈ Γ, (9) is the restriction to Γ of the commutator  ∇v(x − y, s)(g(x) − g(y)) dy, x ∈ R2 , Ds where g is the field given by Lemma 2. 8 J. Verdera / Advances in Mathematics 416 (2023) 108917 Proof. The first component of (9) is ∂1 v1 (x, s)τ1 (x) + ∂2 v1 (x, s)τ2 (x) = (∂1 k1 ∗ χ)(x)τ1 (x) + (∂2 k1 ∗ χ)(x)τ2 (x) (10) with χ = χDs . In the sense of distributions (p.v. stands for principal value)  ∂1 k1 = p. v. ∂1 k1 + c1 δ0 , c1 = k1 (x)x1 dσ(x) |x|=1 and  ∂2 k1 = p. v. ∂2 k1 + c2 δ0 , c2 = k1 (x)x2 dσ(x). |x|=1 Let g be the field given by Lemma 2. Since (∂1 k1 ∗ χ)g1 = (p.v. ∂1 k1 ∗ χ)g1 + c1 χg1 and ∂1 k1 ∗ (χg1 ) = p.v. ∂1 k1 ∗ (χg1 ) + c1 χg1 , we get (∂1 k1 ∗ χ)g1 = (p.v. ∂1 k1 ∗ χ)g1 − p.v. ∂1 k1 ∗ (χg1 ) + ∂1 k1 ∗ (χg1 ). Similarly (∂2 k1 ∗ χ)g2 = (p.v. ∂2 k1 ∗ χ)g2 − p.v. ∂2 k1 ∗ (χg2 ) + ∂2 k1 ∗ (χg2 ). Note that ∂i k1 ∗ (χgi ) = k1 ∗ ∂i (χgi ) = k1 ∗ (gi ∂i χ + χ∂i gi ) and so, recalling that g has vanishing divergence, ∂1 k1 ∗ (χg1 ) + ∂2 k1 ∗ (χg2 ) = k1 ∗ (g · ∇χ + χ div g) = k1 ∗ (g · ∇χ) ≡ 0 since g|Γ = τ is tangent to Γ and ∇χ normal to Γ at each point of Γ. Thus if x ∈ Γ = ∂Ds  (∂1 k1 ∗ χ)(x)τ1 (x) + (∂2 k1 ∗ χ)(x)τ2 (x) = ∂1 k1 (x − y)(g1 (x) − g1 (y) dy Ds  + ∂2 k1 (x − y)(g2 (x) − g2 (y)) dy Ds is indeed a sum of two scalar commutators. One has a similar formula for the second component of ∇v(x, s)(τ (x)) and the lemma follows.  J. Verdera / Advances in Mathematics 416 (2023) 108917 9 Therefore, by the usual commutator estimate in the Hölder seminorm [1, Lemma, p.26] or [2, Lemma 3.2, p.3799], ∇v(x, s)(τ (x))γ,Γ ≤ C (1 + ∇v(·, s)∞ ) gγ,R2 ≤ C (1 + ∇v(·, s)∞ ) τ γ,Γ . From (7) we see that t d d d X(h(u), t) ≤ h(u) +C (1 + ∇v(·, s)∞ ) X(h(u), s) ds du γ,T du γ,T du γ,T 0 and then, by Gronwall’s lemma, we get the a priori inequality ⎛ ⎞ t d d X(h(u), t) ≤ h(u) exp ⎝C (1 + ∇v(·, s)∞ ) ds⎠ . du γ,T du γ,T 0 We also have an a priori inequality for X(α, t) − X(β, t) 1 b(t) ≡ inf = α=β |α − β| sup |α−β| |X(α,t)−X(β,t)| α=β ⎛ ⎞ t 1 = ≥ exp ⎝− ∇v(·, s)∞ ds⎠ . ∇X −1 (·, t)∞ 0 Setting d du X(h(u), t) γ,T q(t) = , −T < t < T, b(t)1+γ we finally obtain ⎛ ⎞ t q(t) ≤ C exp ⎝C (1 + ∇v(·, s)∞ ) ds⎠ , 0 < t < T. (11) 0 3. The logarithmic inequality Let K ∈ C 1 (R2 \{0}) be a real function, homogeneous of degree −2 and even. Set  T f (x) = p.v. K(x − y)f (y) dy, R2 so that T is a convolution, smooth, homogeneous, even Calderón–Zygmund operator. 10 J. Verdera / Advances in Mathematics 416 (2023) 108917 Let X ∈ C 1+γ (∂D0 , R2 ), D0 a domain of class C 1+γ . Assume that X is bilipschitz onto the image and let b stand for the inverse of Lipschitz norm of the inverse mapping, as in (5). Since X(∂D0 ) is a Jordan curve, the simply connected domain D enclosed by the curve satisfies ∂D = X(∂D0 ). The maximal singular integral of a function f is  ∗ T f (x) = sup K(x − y)f (y) dy, x ∈ R2 . ε>0 |y−x|>ε Lemma 4. We have    ∗ d 1 T (χD )(x) ≤ A 1 + log + |D|  [X(h(u))]γ,T 1+γ 1/2 du b    = A 1 + log+ |D|1/2 q(t) , x ∈ R2 , where A is a constant depending only on γ and the kernel K. Proof. By an elementary argument [6, p.409-410] it is enough to prove the inequality for points x ∈ ∂D. Without loss of generality one can assume that x = 0 = X(0) and that the unitary normal vectors to ∂D0 and ∂D at 0 are (0, 1). Define δ by 1 d δ −γ =  [X(h(u))]γ,T . b1+γ du Then for ε > 0    K(y)χD (y) dy = K(y)χD (y) dy + K(y)χD (y) dy |y|>ε δ>|y|>ε |D|1/2 >|y|>δ  (12) + K(y)χD (y) dy. |y|>|D|1/2 The third term is estimated by  1 χD (y) dy ≤ 1 |y|2 |y|>|D|1/2 and the second by     |D|1/2 d 1 2π sup |K(y)| log ≤ A log + |D| 1/2  [X(h(u))]γ,T 1+γ . |y|=1 δ du b J. Verdera / Advances in Mathematics 416 (2023) 108917 11 It remains to estimate the first term in the right hand side of (12). The argument that follows is similar to that of [1] in the proof of Proposition 1, p.23–24. Incidentally, Proposition 1 of [1] was rediscovered in Lemma 5, p. 1142 of [7]. Set Dε,δ = {y ∈ D : ε < |y| < δ}, H − = {y = (y1 , y2 ) ∈ R2 : y2 < 0} and Sρ = {y ∈ R2 , |y| = ρ}, ρ > 0. Since an even kernel with vanishing integral on the unit circumference has also vanishing integral on half a circumference, we get ⎛ ⎞  δ  ⎜ ⎟ dρ K(y) dy = ⎝ K(θ) dσ(θ)⎠ ρ Dε,δ ε {θ∈S 1 :ρθ∈D} ⎛ ⎞ δ   ⎜ ⎟ dρ = ⎝ K(θ) dσ(θ) − K(θ) dσ(θ)⎠ ρ ε {θ∈S 1 :ρθ∈D} S 1 ∩H − ⎛ ⎞ δ  ⎜ ⎟ dρ = ⎝ K(θ) dσ(θ)⎠ ρ ε {θ∈S 1 :ρθ∈(D\H − )∪(H − \D)} and  δ dρ K(y) dy ≤ sup |K(θ)| σ{θ ∈ S 1 : ρθ ∈ (D\H − ) ∪ (H − \D)} . |θ|=1 ρ Dε,δ 0 Since the domains D\H − and H − \D are “tangential” 1 σ{θ ∈ S 1 : ρθ ∈ (D\H − ) ∪ (H − \D)} ≤ A sup{|X2 (α)| : α ∈ ∂D0 and |X(α)| = ρ}. ρ Take |X(α)| = ρ < δ. If α = h(u) and 0 = h(u0 ), then d |X2 (α)| = |X2 (h(u)) − X2 (h(u0 ))| ≤  [X2 (h(u))]γ,T |u − u0 |1+γ du d 1 ≤ [X(h(u))]γ,T 1+γ |X(α)|1+γ du b d 1 =  [X(h(u))]γ,T 1+γ ρ1+γ . du b 12 J. Verdera / Advances in Mathematics 416 (2023) 108917 Thus  δ d 1 K(y) dy ≤ A [X(h(u))]γ,T 1+γ ργ−1 dρ du b Dε,δ 0 d 1 δγ = A [X(h(u))]γ,T 1+γ = A.  du b γ 4. End of the proof Having at our disposition the a priori estimate of section 2 and the logarithmic esti- mate it is a standard matter to complete the proof. We remark that ∇v(·, t) is a 2 × 2 matrix with entries which are either constant multiples of χDt or even homogeneous smooth Calderón -Zygmund operators applied to χDt . Inserting the a priori estimate (11) into the logarithmic inequality we obtain t ∇v(·, t)∞ ≤ C + C (1 + ∇v(·, s)∞ ) ds, t ∈ (0, T ). 0 The factor |D|1/2 in the inequality of Lemma 4 causes no trouble because of the standard estimate t ∇X(·, t)∞ ≤ exp ∇v(·, s)∞ ds, t ∈ (0, T ). 0 Gronwall’s lemma then yields ∇v(·, t)∞ ≤ C exp(Ct), t ∈ (0, T ). Hence, for t ∈ (0, T ), d  [X(h(u), t)]γ,T ≤ C exp(C exp(Ct)), du and b(t) ≥ C −1 exp(−C exp(Ct)) ≥ C −1 exp(−C exp(CT )). Consequently T = ∞. J. Verdera / Advances in Mathematics 416 (2023) 108917 13 5. Appendix: local in time existence for the CDE Our first remark is that if k : R2 \ {0} → R2 is an odd kernel, homogeneous of degree −1, differentiable off the origin, then k = ∂1 (x1 k) + ∂2 (x2 k), x = (x1 , x2 ) ∈ R2 \ {0}. (13) This follows straightforwardly from Euler’s theorem on homogeneous functions. Assume that ω(z, t) = χDt (z) is a weak solution of the general equation (2). The velocity field is ∂ ∂ v(·, t) = χDt  k = χDt  ( (xk) + (yk)) ∂x ∂y ∂ ∂ = χD  (xk) + χD  (yk). ∂x t ∂y t Since ∇χDt = −ndσ = idz, where dσ is the arc-length measure on the curve ∂Dt and dz = dz∂Dt , we have ∂ ∂ χD = −n1 dσ = −dy and χD = −n2 dσ = dx. ∂x t ∂y t Setting z = x + iy and w = u + iv we get   v(z, t) = − (x − u)k(z − w) dv + (y − v)k(z − w) du ∂Dt ∂Dt  (14) = k(z − w)−i(z − w) · dw, ∂Dt where ·, · denotes scalar product in the plane. The flow mapping is the solution of the ODE d X(α, t) = v(X(α, t), t), dt (15) X(α, t) = α ∈ R .2 Substituting the expression (14) for the velocity field in (15) and making the change of variables w = X(β, t) we get  d X(α, t) = v(X(α, t), t) = k(X(α, t) − w)−i(X(α, t) − w), dw, dt ∂Dt  = k(X(α, t) − X(β, t))−i(X(α, t) − X(β, t)), ∇X(β, t)(dβ). ∂D0 14 J. Verdera / Advances in Mathematics 416 (2023) 108917 In the last line β is a generic point of the curve ∂D0 and dβ the standard complex differential form associated with the curve. Then   ∂X1 ∂X1 ∂X2 ∂X2 ∇X(β, t)(dβ) = (β, t) dβ1 + (β, t) dβ2 , (β, t) dβ1 + (β, t) dβ2 . ∂β1 ∂β2 ∂β1 ∂β2 In fact ∇X(β, t) can be understood more intrinsecally as the differential DX(β, t) of the mapping X(·, t) as a differentiable mapping from the differentiable curve ∂D0 onto the differentiable curve ∂Dt . This mapping takes a tangent vector to ∂D0 at the point β into a tangent vector to ∂Dt at the point X(β, t). The point is that this operation involves only first derivatives of X(·, t) on ∂D0 . Consider the Banach space C 1+γ (∂D0 , R2 ) endowed with the norm (4) and the open set Ω consisting of those mappings in C 1+γ (∂D0 , R2 ) satisfying the bilipschitz condition (5). For each X ∈ Ω define  F (X)(α) = k(X(α) − X(β))−i(X(α) − X(β)), ∇X(β)(dβ), α ∈ ∂D0 . (16) ∂D0 The CDE is the autonomous ODE in Ω dX = F (X), X(·, 0) = I, (17) dt where I is the identity mapping on ∂D0 . Of course one has to check that F (X) belongs to C 1+γ (∂D0 , R2 ) for each X ∈ Ω. After that one wants to apply the existence and uniqueness theorem of Picard, and for that one needs to check that F is locally Lipschitz in Ω. All these facts are verified routinely and depend essentially on the fact that the kernel appearing in (16) is of class C 2 (R2 \ {0}, R2 ) and homogeneous of degree 0. The reader may consult [5, Chapter 8] for the case of the vorticity equation. For instance, when you take a derivative in α in (16) to prove that F (X) is in C 1+γ (∂D0 , R2 ), then you get an expression of the form  H(X(α) − X(β))L(∇X(β), dβ), (18) ∂D0 where H(·) is a kernel of homogeneity −1 of class C 1 (R2 \ {0}, R2 ) and L(∇X(β), dβ) a linear expression in ∂Xj /∂βk and dβj . Hence (18) may be understood as a (non- convolution) singular integral on the curve ∂D0 applied to a linear combination of the ∂Xj /∂βk , which are functions satisfying a Hölder condition of order γ. Then the result is a function of α in the same space. An analogous situation appears in checking that F is locally Lipschitz in Ω. The reader may also consult [2], where full details are provided in a similar context. J. Verdera / Advances in Mathematics 416 (2023) 108917 15 Declaration of competing interest The author has no conflict of interests to declare. Data availability Data sharing not applicable to this article as no datasets were generated or analysed during the current study. 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