In two dimensions, ω can be identified with a scalar, and the equations ..... obtained from the vorticity by a d × d-matrix of singular integral transforms for d = 2, 3. .... We provide a simple proof of the strong illposedness of the Euler equation in the ...... 1We have used the notation Ω − z = {x − z : x ∈ Ω} for Ω ⊂ R2 and z ∈ R2.
Dynamics of the incompressible Euler equations at critical regularity
In-Jee Jeong
A Dissertation Presented to the Faculty of Princeton University in Candidacy for the Degree of Doctor of Philosophy
Recommended for Acceptance by the Department of Mathematics Adviser: Yakov G. Sinai
November 2017
c Copyright by In-Jee Jeong, 2017.
All Rights Reserved
Abstract The purpose of this work is to study the incompressible Euler equations in scaling critical spaces, the function spaces which are left invariant under the natural scaling transformations for the Euler equations. We demonstrate that under appropriate symmetry assumptions, which depends on the dimension of the physical space, the Euler equations are well-posed in certain critical spaces. It turns out that such critical spaces contain solutions which are radially homogeneous, whose dynamics is necessarily described by a system of one less dimension. In the case of 2D Euler equation, we obtain from this procedure a new 1D fluid model, and it turns out that the long-time dynamics of this model can be analyzed under some mild assumptions on the initial data. We proceed to show well-posedness results of “hybrid type”, by which we mean that if the initial data consists of a radially homogeneous piece and a smooth piece vanishing at the origin, the solution continues to have this decomposition, with the homogeneous part solving the lower dimensional system. As an immediate consequence, we obtain a conditional blow-up result, which states that if there is a finite time singularity for the 2D model arising from the 3D Euler equations, then there is a finite time blow-up for a Lipschitz continuous solution of the 3D Euler equation with compact support and hence of finite energy. As an application of the critical spaces introduced in this work, we obtain a global well-posedness result for a certain class of singular vortex patches in two dimensions.
iii
Acknowledgements First of all, I would like to sincerely thank Prof. Yakov Sinai for all the guidance and support that he provided me during my years at Princeton. He has suggested me several interesting problems from various fields such as number theory, dynamical systems, fluid dynamics, and probability theory. I plan to work on some of these problems in the near future. I am indebted to him for opening up many doors of research – I was able to meet many very strong mathematicians through him and even collaborate with them in some cases. Almost all of these collaborations was either based on or motivated by his previous work. I would like to thank Dr. Jonathan Fickenscher as well, for organizing a wonderful seminar together with my advisor during the past few years. I take this opportunity to express my sincere gratitude to Prof. Tarek Elgindi. He have been extremely generous with his time and also in sharing valuable insights. During the years I have learned so many lessons from him, which cannot be learned from anywhere else. His influence on me is evident from the fact that this thesis is almost entirely based on our joint works. I am indebted to professors Dong Li, Sasha Sodin, and Benoit Pausader for successful collaborations. In these works they have provided me with key ideas for solving the respective problem. I am very thankful for professors Alex Ionescu, Peter Constantin, and Vlad Vicol for their wonderful lectures regarding partial differential equations, ranging from introductory materials to more advanced topics. These courses really helped me a lot. I also thank professor Javier GomezSerrano for several very helpful discussions. Moreover, I want to thank Jill Leclair for her tremendous help regarding all sorts of issues related to my Ph.D studies. Last but not least, I need to thank my friends from the mathematics department, Masoud, Artem, Yuchen, Seokhyeong, Joonhyun, Donghun, Junho, and Ross. Financial support from the Samsung scholarship foundation is greatly appreciated.
iv
To my parents.
v
Contents Abstract . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iii
Acknowledgements . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
iv
1 Introduction 1.1
1
Central problems in the mathematical study of the Euler equations . . . . . . . . . .
2
1.1.1
Behavior of smooth solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.1.2
Strong versus weak solutions . . . . . . . . . . . . . . . . . . . . . . . . . . .
3
1.2
Why critical spaces? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
5
1.3
A few themes in this work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6
2 Ill-posedness of the Euler equations in critical spaces 2.1
2.2
Ill-posedness in critical Sobolev spaces . . . . . . . . . . . . . . . . . . . . . . . . . .
13
2.1.1
Proof of H 1 -norm inflation for Cc∞ data . . . . . . . . . . . . . . . . . . . . .
16
2.1.2
Proof of non-existence in H 1 . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
Ill-posedness for singular vortex patches . . . . . . . . . . . . . . . . . . . . . . . . .
30
2.2.1
Ill-posedness for a single corner . . . . . . . . . . . . . . . . . . . . . . . . . .
30
2.2.2
Ill-posedness under the odd-odd symmetry
37
. . . . . . . . . . . . . . . . . . .
3 Well-posedness of the Euler equations in critical spaces under symmetry 3.1
3.2
12
47
Background material . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
3.1.1
Yudovich theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
3.1.2
Solutions with non-decaying vorticity and the symmetry condition . . . . . .
49
3.1.3
Local well-posedness in a critical space and bounds on ∇u . . . . . . . . . . .
51
3.1.4
Scale-invariant solutions for 2D Euler . . . . . . . . . . . . . . . . . . . . . .
52
Existence and uniqueness for the 2D Euler . . . . . . . . . . . . . . . . . . . . . . . .
54
3.2.1
55
Explicit solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
vi
3.3
3.4
3.2.2
Proof of the main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
57
3.2.3
Propagation of the angular regularity . . . . . . . . . . . . . . . . . . . . . .
68
Radially homogeneous solutions to the 2D Euler . . . . . . . . . . . . . . . . . . . .
72
3.3.1
The 1D system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
73
3.3.2
Trend to equilibrium under odd symmetry and positivity . . . . . . . . . . .
77
3.3.3
Measure-valued data and quasi-periodic solutions . . . . . . . . . . . . . . . .
84
Case of the 3D Euler equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
87
3.4.1
The symmetry condition and the key estimate . . . . . . . . . . . . . . . . .
88
3.4.2
Local well-posedness in critical spaces . . . . . . . . . . . . . . . . . . . . . .
95
3.4.3
A conditional blow-up result . . . . . . . . . . . . . . . . . . . . . . . . . . .
97
4 Global well-posedness of rotationally symmetric vortex patches with corners
101
4.1
Smooth vortex patches: an approach by Bertozzi-Constantin . . . . . . . . . . . . . 101
4.2
Local well-posedness for symmetric patches . . . . . . . . . . . . . . . . . . . . . . . 104
4.3
Global well-posedness for symmetric patches in an intermediate space . . . . . . . . 115
4.4
Global well-posedness for symmetric C 1,α -patches with corners . . . . . . . . . . . . 124 4.4.1
The geometric setup and the main statement . . . . . . . . . . . . . . . . . . 124
4.4.2
Local C 1,α -estimate near the corner . . . . . . . . . . . . . . . . . . . . . . . 128
4.4.3
Proof of the main result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137
4.4.4
Multiple corners and cusp formation in infinite time . . . . . . . . . . . . . . 140
4.4.5
Extensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146
vii
Chapter 1
Introduction The contents of this thesis are mostly based on collaborations with Tarek Elgindi. There are three main results in this thesis, which is in one-to-one correspondence with chapters following the introduction. In Chapter 2, we present some ill-posedness results for the 2D Euler equations. In the first section, we obtain ill-posedness in critical Sobolev spaces, and this material is from a joint work with Tarek Elgindi [38]. In the second section, we collect a few ill-posedness results for vortex patches which have smooth boundaries except for a few corners. This will be a part of a forthcoming paper with Tarek Elgindi [35]. In Chapter 3, we obtain well-posedness for the Euler equations in new critical spaces contained in L∞ for vorticity, under an appropriate symmetry assumption. The case of the 2D Euler equations have appeared in a joint work with Tarek Elgindi [37], and it relies upon his previous work [34]. In Chapter 4, we return to the problem of vortex patches with corners, and show well-posedness under the m-fold rotational symmetry assumption. The contents in this chapter will make up the main part of a forthcoming paper with Tarek Elgindi [35]. In the remainder of this introduction, we provide some background material which would not only motivate but also put our results in context.
1
1.1
Central problems in the mathematical study of the Euler equations
The incompressible Euler equations have the form
∂t u + (u · ∇)u + ∇p = 0,
∇ · u = 0,
(1.1)
where u(t, x) : [0, ∞) × Ω → Rd and p(t, x) : [0, ∞) × Ω → R denote the velocity and the pressure of the fluid at some time t and point x ∈ Ω ⊂ Rd . In the case when Ω has a boundary, it is most physical to impose u · n|∂Ω = 0, that is, just the tangential component of u to vanish on the boundary. It is also traditional to set |u| → 0 at infinity, when the domain is unbounded. An important physical quantity to introduce is the vorticity ω = ∇ × u, which measures the infinitesimal rotation of the fluid element at each point. One can re-formulate the Euler equations as
∂t ω + (u · ∇)ω = (∇u)ω,
u = ∇ × (−∆)−1 ω.
(1.2)
In two dimensions, ω can be identified with a scalar, and the equations simplifies into
u = ∇⊥ (−∆)−1 ω.
∂t ω + (u · ∇)ω = 0,
(1.3)
For PDEs describing evolution of a physical system, the most important question is that of wellposedness, that is, whether there exists (at least locally in time) a unique solution of the equations given initial data. Otherwise, the PDE is said to be ill-posed. This notion of well/ill-posedness highly depends on the choice of a function space, which roughly corresponds to deciding how much irregularities would be allowed on the initial data. Once the PDE is well-posed, the next important question is whether the every solution can be continued globally in time or there exists one which collapses in finite time. In the former case, the natural question is what are the possible asymptotic behavior of the solutions. If there are solutions collapsing in finite time, the interesting problem is to determine the exact form of blow-up. Returning to the specific case of the Euler equations, the main difficulty in answering these questions is that the systems are both non-linear and non-local ; the non-locality in (1.1) comes from the pressure, which corresponds to the fact that each “fluid particle” should communicate
2
with every other particle to ensure incompressibility. In vorticity form (1.2), (1.3), the non-locality manifests itself in the relation (the Biot-Savart law) recovering the velocity from the vorticity.
1.1.1
Behavior of smooth solutions
Although the above set of questions consists of the minimal requirements for the understanding of an evolutionary PDE, not much have been answered regarding the Euler and the Navier-Stokes equations, beyond well-posedness. The situation is different in two and three (or higher) spatial dimensions – in the case of 2D, the global well-posedness in standard function spaces is known, but in 3D, it is unknown whether the Euler and the Navier-Stokes solutions can blow up in finite time. Strikingly, for the 3D Euler equations, a single example of a (nontrivial, truly three-dimensional) global smooth solution is not known! Question 1.1.1. Are there smooth (C ∞ with sufficient decay at spatial infinity) solutions to Euler or Navier-Stokes equations in R3 which becomes singular in finite time? Since smooth solutions are known to exist globally in time in two dimensions, one may ask what happens to them in the limit as time goes to infinity. Surprisingly little is known rigorously in this direction, in the case of the Euler equations. (However, there are predictions based on statistical mechanics in the physics literature [60], [63], [70].) Question 1.1.2. What is the asymptotic behavior for “typical” smooth solution of the 2D Euler equations?
1.1.2
Strong versus weak solutions
Let me begin by explaining the notion of strong 1 solutions as well as (a version of) the precise definition of well-posedness. In the context of the Euler equations, a solution is strong provided that at each moment of time, it belongs to a function space X in which the Euler equations are well-posed in the classical sense of Hadamard: given an initial data u0 ∈ X, we require that • (Existence) For some T = T (u0 ) > 0, there is a solution in u(t, x) ∈ C([0, T ); X).2 • (Uniqueness) In the class C([0, T ); X), the solution is unique. • (Continuous dependence on data) The solution operator u0 7→ u(t, ·) is a continuous map from X to C([0, T ); X). 1 This 2 In
term is sometimes used in an interchangeable way with smooth. certain cases, one should work with a weaker topology on X to ensure continuity.
3
∃
∃! s − d/p
L1 , M L log L
Lp
C 0 , L∞
Cα d/2+ H
BM O H d/2
Regularity of vorticity
X = X s,p
Figure 1.1: Well-posedness theory for the 2D Euler equations. Two vertical lines represent critical scaling for unique existence and mere existence. (One may argue that these requirements are too strong, but they are natural from the dynamical systems point of view and the inviscid transport nature of the Euler equations.) From the mathematical viewpoint, strong solutions are almost as good as C ∞ solutions – indeed, one can show that if one has uniform control over any strong norm of a solution of the Euler equations during some time interval [0, T ], then the solution at time T is as smooth as the initial data. The diagram in Figure 1.1 suggests that if a space X consists of functions more regular (“supercritical”) than a critical threshold, the Euler equations are well-posed in the sense above. (Still, there are some counter-examples to this statement [47], [20] when the space does not have a structure to make up for the loss of one derivative present in the nonlinear term u · ∇u.) Exactly at the critical regularity, the well-posedness issue becomes very delicate and it really depends on the nature of the space. The remarkable theorem of Yudovich in 1963 [76] says that the 2D Euler equations are well-posed at the critical regularity ω0 ∈ L∞ , while the Lipschitz bound on velocity ∇u0 ∈ L∞ is not propagated by the Euler equations [13], [39]. Although there are results showing well-posedness in the “slightly” sub-critical regime ([33], [77], [74], [9], [8]), strictly below L∞ , well-posedness is expected to fail, and actually there are explicit examples showing loss of sub-critical Sobolev regularity [31], [35] and negative H¨ older regularity [5]. Going further down the regularity scale, non-uniqueness is 2 known – actually there are L∞ t Lx -solutions to Euler equations with compact support in space-time
(this goes by the name of Scheffer-Shnirelman paradox [65], [69]). The critical space L∞ in terms of the vorticity is actually a very natural setting to study the Euler equations. Not only it is the strongest conservation law in the case of 2D, the Euler equations enjoy the following scaling symmetry which leaves L∞ invariant:
ω λ (t, x) := ω(t, λx).
(1.4)
Note that if ω(t, x) is a solution on Rd , then for any λ > 0, ω λ is another solution to the Euler equations. Moreover, the space L∞ allows for vorticities which are characteristic functions of measurable
4
sets. Such a configuration of vorticity is referred as a vortex patch solution, and it serves as a good model for certain physical situations. The general question of long-time dynamics can be stated in this set-up of L∞ solutions, and again, little is known in towards this direction. Indeed, it is natural to pose the question of long-time dynamics in L∞ , as there is a priori no reason to expect smooth solutions to retain any higher regularity than L∞ uniformly up to t = +∞. On the other hand, in sub-critical spaces, one has to utilize smooth test functions to make sense of the equations (1.2) and (1.3), and solutions defined in this “averaged sense” are commonly referred as weak solutions.3 From the purely theoretical point of view, the study of weak solutions can be viewed as an extension to the well-posedness theory. However, weak solutions such as vortex sheets (that is, vorticity is a measure supported on a curve with smooth density) are of great physical and engineering interest, as these not-so-smooth structures appear in various practical situations. Numerical simulations for such weak solutions is notoriously difficult to carry out, and −1 hence developing mathematical theory would be meaningful. In the regularity class ω ∈ L∞ t (Hloc ∩
M)x (which is a natural functional set-up to cover vortex sheets), existence of a local solution is not known in general and uniqueness is expected to fail (see Pullin [62]). When the vorticity is a signed measure, then the celebrated Delort theorem [27] gives existence of a global in time vortex sheet solution. Majda [57] then showed that such a solution can be realized as a weak limit of a sequence Navier-Stokes solutions in the vanishing viscosity limit ν → 0. Question 1.1.3. Given an initial data ω0 ∈ H −1 ∩ M(R2 ), does there exist a solution in the same class? Is it unique? What happens when the initial data is a vortex sheet? More generally, in 2D, what is the “largest” space in which the Euler equations is well-posed? What if we require mere existence? On the other hand, what is the “smallest” space that uniquness of the Euler solutions fail?
1.2
Why critical spaces?
It should be mentioned, although it could become gradually clear from the forthcoming chapters, why studying the Euler equations at critical regularity is of some value. 1. From a purely theoretical point of view, a well-posedness result is more general if it works in a broader space. (See Diagram 1.1.) 3 It is customary to regard vorticity in L∞ as weak solutions, but at this critical regularity, the vorticity equation (1.3) in 2D can be re-formulated using the flow maps to avoid the use of test functions. In any case, the well-posedness theory in L∞ is not significantly worse than that for smooth solutions.
5
2. Critical norms give a tight control on the classical solutions. As a simple illustration, consider the estimate Z ku(t)kX .X ku0 kX exp
0
t
k∇u(s)kL∞ ds
where X is either H¨ older spaces C k,α or Sobolev spaces H s , or the celebrated Beale-KatoMajda criterion [6]. In several situations, most notably in the set-up of Kiselev and Sverak [50], it was shown that these estimates are sharp. 3. In 2D, ω ∈ L∞ is the strongest conservation law, and in principle there is no reason to expect the solutions to retain any higher regularity than L∞ in the long-time limit. 4. Related to the above item, interesting physical vorticity structures belong to critical spaces and not better. It is possible that the long time behavior of them dictate the long time dynamics of smooth solutions that are nearby.
1.3
A few themes in this work
Let us discuss some recurring themes in the present thesis, which are intertwined with each other. The main point of all of this is to go beyond the standard a priori estimates and to enable refined analysis on the solutions, to access problems such as the long-time behavior, dynamics of weak solutions, and the inviscid limit. Bootstrap arguments through Lagrangian approach There are two complementary approaches to the study of hydrodynamic PDEs, namely Lagrangian (based on the flow map and particle trajectory) and Eulerian (Fourier and harmonic analysis). Although the Eulerian-based methods are more traditional, it is becoming clear that to extract detailed information about the Euler solution, one has to work with the flow maps and Lagrangian trajectories. The flow map Φ(t, x) : [0, T ) × Ω → Ω is defined by solving the ODE d Φ(t, x) = u(t, Φ(t, x)), dt
6
Φ(0, x) = x
(1.5)
for each fixed x ∈ Ω. For each fixed time t ≥ 0, the map Φt (x) := Φ(t, x) is invertible, and we denote the inverse as Φ−1 t . Then the flow map recovers the solution via
ω(t, x) = ω0 (Φ−1 t (x))
(1.6)
−1 ω(t, x) = ∇Φt (Φ−1 t (x))ω0 (Φt (x))
(1.7)
in 2D and
in 3D. In several of my works [38], [37], [35] several bootstrap-type arguments were presented, which have the following general form: first, using simple information on the solution, such as symmetries, positivity, a priori estimates, and so on, one obtains some crude bounds on the flow maps, which in turn gives more information on the vorticity via (1.6), (1.7). Then, using the Biot-Savart law and (1.5), one may obtain a more refined information on the flow, with a careful analysis of the ODE system (1.5). In principle, the information so obtained can be feed back into (1.6), (1.7) to extract even more control on the solution. Recently such arguments were used in a few different contexts: (i) to show existence of Euler solutions with norm growth [50], [28], [29], [75], [79], (ii) to obtain blow-up/regularity for 1D model equations for Euler [23], [32], [22], (iii) to study vortex patch dynamics [35], [43]. This type of argument is most effective when used in conjunction with a specific scenario, which we now discuss, as it not only enables the very first step of the bootstrap argument, but also provides sharp and conditional estimates throughout the whole argument. Use of scenarios By a scenario, one is usually referring to a simple qualitative character of the fluid configurations, including but not limited to, symmetries, sign of the vorticity, flows in certain domains, and hyperbolic/rotational flow-lines. In mathematical fluid dynamics, the strong use of symmetries as a way to extract more detailed information on the solution is quite recent. Being a physical system, the Euler and Navier-Stokes equations respect various types of symmetries, in the sense that if the initial data satisfies certain symmetry properties, the solution will continue to enjoy the same properties. (This holds under the assumption that the solution is unique, and it is widely believed that the non-uniqueness in Euler
7
can come from symmetry breaking.) Two examples are: vorticity odd with respect to a line (then the velocity will be even), and m-fold rotationally symmetric vorticity around a point (the velocity is vectorially m-fold rotationally symmetric). If one has a sign (either non-negative or non-positive) on the initial vorticity, the solution to the Euler equations keeps the same sign for all times. Together with the odd symmetry, one is able to obtain flows which are hyperbolic at the origin for all time: just take some non-negative vorticity on the positive quadrant, and extend it as a function on the entire plane as an odd function in both coordinates. This particular scenario was utilized in the works [50],[79],[38],[75],[29]. As a yet another example of a scenario, a new maximum principle was proved in the setting of m-fold rotational symmetry [37], together with an appropriate odd and non-negativity assumptions on the initial data. This was then used to show that, under this particular situation, the solutions to the 2D Euler equations converge to a specific stationary solution in the long time limit. Moreover, recent investigations have shown that the shape of the physical domain may have a significant impact on the dynamics; to just provide a few examples, (i) fluid in domains with boundary are more prone to generate small scales [78],[50], (ii) singularity of the boundary can cause finite time blow-up even in 2D [51], and on the other hand (iii) in certain domains the flow is “more stable” in the long time limit [46], [45], [37]. Sharp estimates on singular integral operators A singular integral operator is a transformation of the form Z T f (x) = lim
→0
y∈Rd ,|x−y|>
K(x − y)f (y)dy,
where K(·) : Rd → R is (typically) given by a homogeneous kernel of degree −d with zero mean on disks centered at the origin. In the context of the fluid dynamics, the gradient of velocity is obtained from the vorticity by a d × d-matrix of singular integral transforms for d = 2, 3. Although the classical theory [14] gives estimates
k∇ukLp ≤ Cp kωkLp ,
1 0, there exists an initial data ω0 ∈ C ∞ (T2 ) and a time moment 0 < t < such that
kω0 kH 1 ∩L∞ < ,
supp(ω0 ) ⊂ B0 ()
and
kω(t, ·)kH 1 > −1 ,
where B0 () is the ball of radius around the origin. Next, we show that a localized solution which is initially small in L∞ ∩ H 1 can immediately escape H 1 for t > 0, which have also appeared in [12]: Theorem 2.1.2. For any s, p such that sp = 2 and 1 ≤ s < 6/5, there is ω0 ∈ L∞ (T2 ) ∩ W s,p (T2 ) which is C ∞ away from the origin that for any 0 < t0 ≤ 1, ess-sup0 0. In the work of Bourgain-Li [12], the existence of localized initial data which escapes H 1 was (n)
∞ obtained by carefully “patching” together an infinite sequence {ω0 }∞ data whose support n=1 of C
becomes smaller but grows in H 1 with a larger rate in a shorter period of time as n → ∞. It is possible that the C ∞ solutions that we construct in Theorem 3.3.7 can be patched together to obtain the desired statement as well. However, this seems to require a rather involved analysis, and we have chosen to establish Theorem 2.1.2 via exhibiting a simple explicit initial data in L∞ ∩ H 1 , see (2.13). The problem of well-posedness will not be an issue in the above statements as there is a unique, global-in-time solution of the extended system (2.1)–(2.3) in L∞ ([0, ∞) × L∞ (T2 )) (so-called Yudovich solutions) for ω0 just in L∞ (T2 ), even though in this case, u(t, ·) is only log-Lipschitz in general. A simple proof of this fact may be found in [59], for instance. The space H 1 is called critical since we barely cannot close the standard energy estimate d k∇ωkL2 (T2 ) ≤ k∇ukL∞ (T2 ) k∇ωkL2 (T2 ) , dt as ω ∈ H 1 (T2 ) or even ω ∈ H 1 (T2 ) ∩ L∞ (T2 ) does not guarantee that u is Lipschitz. This failure of Lipschitz regularity is at the heart of the possibility of rapid growth of vorticity gradient. It is explicit in the Bahouri-Chemin example [4]: Take ω(x) = 1 on [0, 1]2 and extend it to [−1, 1]2 as an odd function in both variables. This defines a stationary solution of 2D Euler in the sense of equations (2.1)–(2.3), and the flow near the origin is “hyperbolic” in the following specific sense: for 0 < x1 < x2 small, it can be computed that for some absolute constant c > 0 (see Denissov [28]) u(x1 , x2 ) = c −x1
1 1 ln + r1 (x1 , x2 ) , x2 ln + r2 (x1 , x2 ) x2 x2
(2.4)
with some smooth functions r1 , r2 . Certain perturbations of this stationary solution were utilized in the works of Denissov [28], ˇ ak [50], and Zlatoˇs [79] (in chronological order) to obtain growth of vorticity gradient Kiselev-Sver´ in the maximum norm L∞ . The growth rates of k∇ωkL∞ (T2 ) obtained in [28] and [79] were double exponential for arbitary long but finite time and exponential for all time, respectively. The ground-
14
breaking work [50] settled the possibility of double exponential growth of k∇ωkL∞ (D) for all time, ˇ ak (which was also utilized in when the domain is a disc. The “Key Lemma” of Kiselev and Sver´ [79]) is an essential tool in our arguments as well (see below Lemma 2.1.3). While our basic strategy to obtain growth of ω in H 1 (T2 ) is similar to that of the aforementioned works, there are a number of notable differences in our setting. First, while the idea of “linearizing” around the Bahouri-Chemin stationary solution makes sense when considering only bounded vorticities, this solution does not belong to H 1 (T2 ). Hence, we needed to consider a different type of “background” vorticity, and our choice was to take a suitably localized version of the following function:
α
ω0 (x1 , x2 ) = ∆ (x1 x2 |ln |x|| ) with 0 < α < 1/2. The advantage of this initial vorticity is that it belongs to H 1 ∩ L∞ and the corresponding velocity u0 satisfies ∇u0 ∈ / L∞ . Second, since we want localized solutions, it is not clear if the specific hyperbolic picture of the type (2.4) near the origin will be sustained, even for a very short periodic of time. In view of this, our strategy is to take an initial vorticity which extends over two different length scales N −1 and N −1/2 , and to show that vorticity outside the O(N −1 )-region, in the special time scale of ln ln N/ ln N , is sufficient to generate a hyperbolic flow which stretches the vorticity gradient on the O(N −1 )-region. Here, a caveat is that we could not exclude the possibility of our initial vorticity chunk getting “squeezed” in the angular direction even earlier than the scale ln ln N/ ln N , in which case we do not have a good lower bound on |∇u|. Hence our actual proof is based on a contradiction argument. This difficulty vanishes when the domain has a boundary: see Remark 1. Closing the introduction, let us mention that similar ill-posedness statements were recently established for the integer based C k spaces with k ≥ 1 of the velocity field u, independently in the works of Elgindi-Masmoudi [39] and Bourgain-Li [13]. Notation. Let us use the notation |f |p = kf kLp (T2 ) for p ∈ [1, ∞] for simplicity. We use letters C, C1 , c, · · · to denote various absolute positive constants, and their values may change from line to line. When a constant depends on some parameters, we explicitly indicate dependence as subscripts. We use superscripts to refer to components of a vector: for example, u = (u1 , u2 ) and Φ = (Φ1 , Φ2 ).
15
2.1.1
Proof of H 1 -norm inflation for Cc∞ data
Our initial vorticity ω0 will be odd both in the variables x1 and x2 . Since this symmetry persists for all time, we may view ω(t, ·) as defined just on [0, 1]2 . Pick a large integer N , and let us define the initial vorticity on [0, 1]2 as follows:
ω0 (r, θ) := χ(r)ψ(θ),
(2.5)
where χ and ψ are smooth bump functions. More specifically, they satisfy
χ(r) :=
1 0
for r ∈ N −1 , N −1/2
and
ψ(θ) :=
for r ∈ / N −1 /2, 2N −1/2 ,
1 0
for θ ∈ [π/4, π/3] for θ ∈ / [π/6, 5π/12] .
Since |∇ω0 |22 =
Z Z
r|∂r ω0 |2 drdθ +
Z Z
1 |∂θ ω0 |2 drdθ, r
the main contribution of |∇ω0 |2 comes from the angular variation: |∇ω0 |2 ≈ c(ln N )1/2 as N → ∞. As mentioned in the introduction, we need to work with a special time scale. Given τ ∗ > 0 and N , we set t∗ (τ, N ) = τ ∗ ln ln N/ ln N and we shall track the evolution of initial data (2.5) on the time interval [0, t∗ ]. To get an idea of how this scale appears, recall that the main idea is to stretch vorticity in the O(N −1 ) region using the chunk of vorticity “behind”. Since initially |∇ω0 |L2 (O(N −1 )) = O(1) while |∇ω0 |2 ≈ c(ln N )1/2 , we need to stretch the H 1 -norm in the local region by a factor of (ln N )1/2+ to obtain norm inflation. In view of |∇u0 |∞ ≈ c ln N , we achieve this goal once we sustain this lower bound on the velocity gradient during an interval of time [0, t∗ ]. It is important that in this time scale, fluid particles can move only up to a factor of ln N , see (2.8) and (2.11) below. ˇ ak Our main technical tool is the following expression for the velocity due to Kiselev and Sver´ [50]; we use a version by Zlatoˇs [79, Lemma 2.1] which works in the case of the torus T2 = [−1, 1)2 . Lemma 2.1.3 (Key Lemma). Let ω(t, ·) be odd in x1 and x2 . Then for x ∈ [0, 1/2)2 , we have ui (t, x) 4 = (−1)i xi π
Z [2x1 ,1)×[2x2 ,1)
y1 y2 ω(t, y)dy + Bi (t, x) |y|4
(2.6)
with |Bi | ≤ C|ω|∞ (1 + ln(1 + x3−i /xi )) for i ∈ {1, 2}. There are several striking features of this lemma, which we would like to emphasize. First,
16
the expression (2.6) essentially gives a pointwise control over the velocity gradient, just under the assumption that ω(t, ·) ∈ L∞ . It is surprising that such a control is available, especially because the formula is applicable even in situations where ∇u is unbounded. Next, the integral in (2.6) is monotone in ω(t, ·), so that for the purpose of obtaining a lower bound on the velocity gradient, it suffices to find a region in space where vorticity is uniformly bounded from below. On the other hand, one should note that Lemma 2.1.3 is applicable only when the integral term in (2.6) dominates the remainder term Bi . The following estimates are standard (cf. [58, 59]) and will play a complementary rˆole of the previous lemma. Lemma 2.1.4. Let (ω, u, Φ) to be the solution triple for the 2D Euler equations in T2 with initial data ω0 . The velocity is log-Lipschitz
|u(t, x) − u(t, y)| ≤ C|ω0 |∞ |x − y| (1 + ln(4/|x − y|)) ,
(2.7)
and the flow maps Φ(t, ·) : T2 → T2 for 0 ≤ t ≤ (C|ω0 |∞ )−1 satisfy quasi-Lipschitz estimates of the form c|x − y|exp(ct|ω|∞ ) ≤ |Φ(t, x) − Φ(t, y)| ≤ C|x − y|exp(−Ct|ω|∞ ) . Note that the argument of the logarithm in (2.7) is always greater than 1 since |x − y| ≤
(2.8) √
2 in
our torus [−1, 1)2 . Proof. Although these estimates are well-known, we provide a proof of (2.8) (assuming the bound in (2.7)), as it appears throughout the arguments given below. For simplicity, we set d(t) := |Φ(t, x) − Φ(t, y)|, and from the definition of flow we have d (Φ(t, x) − Φ(t, y)) = u(t, Φ(t, x)) − u(t, Φ(t, y)), dt and applying the estimate (2.7) gives a bound d d(t) ≤ C|ω0 |∞ d(t) 1 + ln 4 , dt d(t) which implies d ln 4 ≤ C|ω0 |∞ 1 + ln 4 . dt d(t) d(t) 17
Φ(t, V ) 1/2
1/2
V
1 1 R(t)
N −1
N −1/2
N −5/6 N −4/6
(N )
Figure 2.1: The figure on the left describes the initial data ω0 , where the two rectangles represent (N ) the region where ω0 ≡ 1 (inner) and V (outer). The right figure shows possible evolution of the set V under the flow. The shaded region represents R(t). Denoting f (t) and g(t) as the unique solution of the respective ODE system 4 4 d ln = C|ω0 |∞ 1 + ln , dt f (t) f (t)
d 4 4 ln = −C|ω0 |∞ 1 + ln dt g(t) g(t)
on the time interval [0, (C|ω0 |∞ )−1 ] with initial data f (0) = g(0) = d(0) = |x − y|, we obtain the desired estimates as g(t) ≤ d(t) ≤ f (t). Given lemmas above, we present the proof of Theorem 3.3.7. Proof of Theorem 3.3.7. We will instead show the following statement: Claim. For any M > 0, there exists some N0 , τ ∗ > 0 depending only on M such that for all N ≥ N0 , the solution associated with the initial data as in (2.5) satisfies, with an absolute constant C, |∇ω(tN , ·)|2 ≥ CM 1/2 |∇ω0 |2
for some
0 < tN ≤ τ ∗
ln ln N . ln N
(2.9)
Once it is established, we simply use the scaling symmetry of the Euler equation: given a solution ω(t, x) and λ > 0, ω λ (t, x) := λω(λt, x) is another solution with initial data λω0 , and we can pick λ = (ln N )−1/2 M −1/4 to achieve the statements of the theorem. Given M > 0, we fix τ ∗ = αM , where αM > 0 is a constant depending only on M to be defined below. In several places of the following argument, it is implicitly assumed that N is sufficiently large with respect to M and some absolute constants appearing in the proof.
18
Consider the annulus A = {r : N −5/6 ≤ r ≤ N −4/6 }. During the time interval [0, t∗ ], particles starting from the arc {(r, θ) : r = N −1 , π/4 ≤ θ ≤ π/3} remain in the region {r < N −5/6 } under the flow Φ(t, ·). Similarly, particles from {(r, θ) : r = N −1/2 , π/4 ≤ θ ≤ π/3} cannot escape {r > N −4/6 }. Both statements follow from (2.8) applied with y = 0 and |x| = N −m (where 1/2 ≤ m ≤ 1): we have c|x|exp(ct|ω|∞ ) ≤ |Φ(t, x)| ≤ C|x|exp(−Ct|ω|∞ ) ,
(2.10)
and since t = τ ln ln N/ ln N for some 0 ≤ τ ≤ τ ∗ , we obtain c(ln N )cτ ≤
|Φ(t, x)| ≤ C(ln N )Cτ |x|
(2.11)
with constants c, C > 0 uniform over 1/2 ≤ m ≤ 1. In particular, it implies that any line segment {(r, θ0 ) : N −1 ≤ r ≤ N −1/2 } should evolve in a way that it intersects each circle {r = r0 } for N −5/6 ≤ r0 ≤ N −4/6 . Take the domain V := {(r, θ) : ω0 (r, θ) ≥ 1/2} and consider the region R(t) := Φ(t, V ) ∩ A which is a curvilinear rectangle whose two opposite edges are bounded by A (see Figure 2.1). Note that ω(t, ·) ≥ 1/2 on R(t). For each r0 ∈ [N −5/6 , N −4/6 ], consider the closed set I(t, r0 ) := {0 ≤ θ ≤ π/2 : (r0 , θ) ∈ R(t)}, and let us denote its Lebesgue measure by |I(t, r0 )|. To show that the H 1 -norm grows, we are led to consider two different scenarios. Case I. Assume that there exists a time moment 0 < tcr ≤ t∗ such that for more than half (with respect to the Haar measure r−1 dr) of r0 ∈ [N −5/6 , N −4/6 ], we have |I(tcr , r0 )| ≤ M −1 . If r0 ∈ [N −5/6 , N −4/6 ] is such that |I(tcr , r0 )| ≤ M −1 , then we can definitely pick some θ0 = θ0 (r0 ) such that the points (r0 , θ0 ) and (r0 , θ0 + δ) with 0 < δ ≤ M −1 satisfy ω(tcr , r0 , θ0 ) = 1 and ω(tcr , r0 , θ0 + δ) = 1/2. This implies a lower bound
M
−1/2
Z I(tcr ,r0 )
!1/2 2
|∂θ ω(tcr , r0 , θ)| dθ
19
≥
Z
θ0 +δ
∂θ ω(tcr , r0 , θ)dθ = 1/2 θ0
which in turn gives that
|∇ω(tcr , ·)|22 ≥
Z
N −4/6
N −5/6
Z
|∂θ ω(tcr , r, θ)|2 dθ
I(tcr ,r)
dr ≥ CM ln N, r
with some absolute constant C > 0. We have established the Claim in this case, recalling that |∇ω0 |22 ≤ C(ln N ). Case II. For all t ∈ [0, t∗ ], for at least half (again with respect to the measure r−1 dr) of r0 ∈ [N −5/6 , N −4/6 ], we have |I(t, r0 )| ≥ M −1 . In this scenario, we will track the evolution of the following segment ∗
S = {(h, h) : N −1 ≤ h ≤ (ln N )Kτ N −1 } for the time interval [0, t∗ ], where K > 0 is an absolute constant to be determined below. Since ω(t, Φ(t, S)) ≡ 1 for all t ≥ 0, to show growth of the H 1 -norm of ω, it is enough to demonstrate that Φ(t∗ , S) is close enough to the vertical segment (where ω vanishes). In the remaining part of the proof, we will always assume that x ∈ S and t ∈ [0, t∗ ]. As a first step, we collect simple bounds on the trajectory of x = (h, h) ∈ S, which will in particular guarantee the applicability of Lemma 2.1.3. To begin with, applying (2.8) with y = (0, 0) gives Φ2 (t, x) ≤ |Φ(t, x)| ≤ h(ln N )Cτ
∗
(recall that t∗ and τ ∗ are related by t∗ = τ ∗ ln ln N/ ln N ). Next, to obtain a lower bound on Φ1 (t, x), we use the log-Lipschitz estimate: |u1 (t, Φ(t, x))| = |u1 (t, Φ(t, x)) − u1 (t, 0, Φ2 (t, x))| ≤ CΦ1 (t, x) 1 + ln
4 1 Φ (t, x)
and since d 1 Φ (t, x) = u1 (t, Φ(t, x)), dt proceeding exactly as in the proof of the estimate (2.8) of Lemma 2.1.4 gives that
Φ1 (t, x) ≥
h . (ln N )Cτ ∗
20
Hence, for x ∈ S, we have
∗ Φ2 (t, x) ≤ C(ln N )2Cτ 1 Φ (t, x)
for 0 < t < t∗ . On the other hand, with our assumption on |I(t, r)|, we estimate the integral appearing in Lemma 2.1.3 at the point x ˆ = (N −7/8 , N −7/8 ): Z Q(t, x ˆ) := [2N −7/8 ,1)2
1 y1 y2 ω(t, y)dy ≥ |y|4 2
Z
N −4/6
Z
N −5/6
I(t,r)
sin θ cos θ dθdr r
and upon setting Z cM :=
min
|I|=1/2M I⊂[0,π/2]
sin θ cos θdθ, I
we obtain Q(t, x ˆ) ≥ c1 cM ln N
(2.12)
for some c1 > 0, whenever 0 ≤ t ≤ t∗ . Therefore, we conclude that the Bi (t, x)-term can be neglected in Lemma 2.1.3 (by possibly adjusting the value of c1 in (2.12)) as long as we apply it to the trajectory of S. That is, −u1 (t, Φ(t, x)) ≥ CQ(t, x ˆ) ≥ CM ln N, Φ1 (t, x) for x ∈ S and 0 ≤ t ≤ t∗ for some constant CM > 0 depending only on M . Similarly, we deduce that u2 (t, Φ(t, x)) ≥ 0 on the same time interval for x ∈ S. From these bounds, it follows that the curve Sˆ := Φ(t∗ , S) is contained in the region ∗
{(y1 , y2 ) : y2 /y1 ≥ (ln N )CM τ }. ∗ The flow estimate (2.11) further gives that Sˆ intersects the circles {r = (ln N )Cτ /N } and {r = ∗
(ln N )(K−c)τ /N }. We could have taken K so that K − c > C (where c, C are constants from the ∗
∗
estimate (2.11)). Then, for each (ln N )Cτ /N ≤ r ≤ (ln N )(K−c)τ /N , we may find a point (in ∗
polar coordinates) of the form (r, θ∗ (r)) on Sˆ such that π/2 − θ∗ (r) ≤ C(ln N )−CM τ . Therefore, we deduce that ∗ 1 (ln N )CM τ ≤ C
Z 0≤θ≤π/2
∗
∗
|∂θ ω(t∗ , r, θ)|2 dθ
and integrating over (ln N )Cτ /N ≤ r ≤ (ln N )(K−c)τ /N against r−1 dr with the choice τ ∗ := 1/CM 21
gives 1 (ln N ) ln C
K −c−C ln N CM
≤
Z Z
1 |∂θ ω(t∗ , r, θ)|2 dθdr ≤ |∇ω(t∗ )|22 r
which gives the desired lower bound in (2.9). Remark 1. This construction carries over to the setting of the whole domain and a bounded open set, with minor modifications. In the case when the fluid domain is a disc (or more generally, a bounded open set with an axis of symmetry), we can utilize the boundary to achieve Theorem 3.3.7 without relying on a contradiction argument. To be more specific, assume for simplicity that our domain is the upper half-plane {(r, θ) : 0 ≤ θ ≤ π}. Take ω0 which is odd in x1 and equals a smoothed out version of the indicator function on the polar rectangle [N −1 , N −1/2 ] × [0, π/4] in the positive quadrant. Then it can be shown that for the time interval that we consider, we do not run out of angles; i.e. Case I does not happen. The same can be said for the proof of Theorem 2.1.2, and actually one can even show continuous-in-time loss of regularity of the solution. We expand on this point in our forthcoming work [38]. Remark 2. Inspecting the proof, one can check that CM = CM −2 works, and so that we may choose N ≥ CM exp(CM 2 ) as M → ∞. In other words, the initial data in (2.5) grows at least by a multiple of (ln N )1/2− (in both scenarios). Again, when we have a boundary available, it is not necessary to introduce M and we obtain growth by a factor of (ln N )K for any K > 0 as long as N is sufficiently large.
2.1.2
Proof of non-existence in H 1
This time, we consider an odd initial vorticity defined on [0, 1)2 by −α 1 ω0 (r, θ) = ln ψ(θ)ξ(r), r
(2.13)
where ψ(·) is the same angular bump function as in (2.5) and ξ(r) is a smooth bump function which identically equals 1 for 0 ≤ r ≤ /2 and vanishes for r ≥ 2/3. Clearly, ω0 is a bounded continuous function and by choosing > 0 small enough, we may assume that |ω0 |∞ , |∇ω0 |2 ≤ 1. Given s and p satisfying sp = 2 and 1 ≤ s < 6/5, we can find a value of 1/2 < α < 3/5 so that kω0 kW s,p < +∞:
22
note that 1 |∇| ω0 (r, θ) ≈ s r
−α 1 ln ψ 0 (θ)ξ(r), r
ω0 ∈ W s,p (T2 )
if and only if
s
r 1,
so that given sp = 2,
αp > 1.
It can be shown that the solution associated with the initial data (2.13) remains C ∞ -smooth away from the origin for all time (see Proposition 2.1.5 below). Hence, if we denote the solution by ω(t, ·), its H 1 -norm can be unambiguously defined by Z lim
δ→0+
|y|>δ
|∇ω(t, y)|2 dy,
which can take the value +∞. We will show that there exists a sequence of positive time moments {tM }M ≥1 and a sequence of radii {rM }M ≥1 , such that tM → 0+ , rM → 0+ , and for a fixed absolute constant c > 0, Z |y|>rM
For each fixed r > 0, the function
|∇ω(tM , y)|2 dy > cM 1/2 .
R |y|>r
|∇ω(t, y)|2 dy is continuous in time and provides a lower
bound for |∇ω(t, ·)|22 . Therefore, the existence of sequences satisfying above gives the statement in Theorem 2.1.2. The proof we present is strictly analogous to that of Theorem 3.3.7, as ω0 in (2.13) can be viewed as a “continuum” version of data from our previous proof. To be more specific, pick some large number N and radially truncate the function (2.13) at length scales N −1 and N −1/2 . Then (N )
this is essentially a scalar multiple of the smooth initial data ω0
from the previous section, and
recalling the scaling symmetry of the Euler equation, it follows that this truncated initial data grows in H 1 by a factor which diverges with N at some time moment 0 < t(N ) which converges to 0 as N → +∞. Therefore it is intuitively clear that the data (2.13) would escape H 1 immediately. Proof of Theorem 2.1.2. Given M > 0, we consider the time moment t∗ = τ ∗
ln ln N (ln N )1−5α/3
where τ ∗ = τ ∗ (M ) is to be determined later. It will be implicitly assumed that N is sufficiently 23
large with respect to M and a few absolute constants. In particular, as 1 − 5α/3 > 0, it guarantees that t∗ 1. Throughout the proof, it will be always assumed that the variable t take values in the interval [0, t∗ ]. The outline of the argument is as follows: we identify a “bulk” region which initially extends over length scales N −1/2 and O(1), and a “local” region near N −1 . In the special time interval that we consider, if there is too much angular squeezing of the bulk, then we are done. Otherwise, the bulk region has enough mass which stretches vorticity in the local region. We note in advance that, compared to the situation of Theorem 3.3.7, we have less precise information on the local particle trajectories, so we should apply Lemma 2.1.3 in a very careful manner. To begin with, using the basic estimate (2.10) (recall that |ω|∞ ≤ 1), we take 0 < a0 < /2 such that the fluid particles starting from |x| > a0 at t = 0 cannot cross the circle |x| = a0 /2 within [0, t∗ ], as t∗ can be taken to be much smaller than a few absolute constants. Similarly using the same estimate, we can ensure that the particles starting on the circle |x| = N −5/10 cannot escape the annulus {N −6/10 < |x| < N −4/10 } in the same time interval. Indeed, taking the logarithms of (2.10) (assuming |x|, |Φ(t, x)| small enough),
ect ln
1 1 1 − c1 ≥ ln ≥ e−Ct ln − C1 |x| |Φ(t, x)| |x|
so that in the time interval that we consider, ln(1/|Φ(t, x)|) is equivalent to ln(1/|x|) up to absolute constants which can be assumed arbitrarily close to 1, uniformly in t and |x|. Given these bounds, take the polar rectangle V := {(r, θ) : N −1/2 ≤ r ≤ a0 , π/4 ≤ θ ≤ 3π/8} and consider intersections of the form R(t) := Φ(t, V ) ∩ {(r, θ) : N −4/10 ≤ r ≤ a0 /2}. Note that on the “angular” sides of V , ω0 takes the values (ln r−1 )−α and β(ln r−1 )−α respectively, for some 0 < β < 1. For each t ∈ [0, t∗ ] and r ∈ [N −1/4 , a0 /2], we consider the (non-empty) set of
24
angles I(t, r) := {0 ≤ θ ≤ π/2 : (r, θ) ∈ R(t)}. We again consider two cases; introducing the set of radii with “enough” angles
A(t) := {r ∈ [N
−1/4
, a0 /2] : |I(t, r)| ≥ M
−1
−α/3 1 ln } r
(note the power −α/3) and first, assume that there exists some 0 < tcr < t∗ such that
Z
ln r∈A(tcr )
1 r
−5α/3
dr 1 ≤ r 2
Z
ln r∈[N −1/4 ,a
0 /2]
1 r
−5α/3
dr 1−5α/3 ≤ C (ln N ) . r
In this case, we argue exactly as Case I of the previous proof: whenever r ∈ / A(tcr ), we integrate over angle to get π/2
Z
−5α/3 1 |∂θ ω(tcr , r, θ)| dθ ≥ CM ln , r 2
0
where we have used that when r = |Φ(t, y)|, ω(t, Φ(t, y)) = ω0 (y) = (ln |y|−1 )−α for y having the form (r, π/4) in polar coordinates and similarly ω(t, Φ(t, y)) = β(ln |y|−1 )−α when y = (r, 3π/8), and that ln(1/|y|) and ln(1/|Φ(t, y)|) are equivalent up to some absolute constants arbitrarily close to 1 (relative to the difference between 1 and β). Integrating the above lower bound over r ∈ / A(tcr ) gives the desired estimate Z |y|≥N −4/10
|∇ω(tcr , y)|2 dy ≥ CM
Z
ln [N −4/10 ,a
0 /2]\A(tcr )
1 r
−5α/3
dr ≥ cM (ln N )1−5α/3 . r
Therefore, we may assume that for all t ∈ [0, t∗ ], we have a lower bound
Z
ln r∈A(t)
1 r
−5α/3
dr ≥ c(ln N )1−5α/3 . r
Under this hypothesis, we shall track the evolution of the diagonal segment in the “local” region: S := {(h, h) : N −1 ≤ h ≤ N −7/10 }. We may assume that the trajectories of the two endpoints of S are trapped in the annuli {N −11/10
0), we deduce Φ1 (t∗ , x) ≤ x1 (ln N )−1 . Combined with Φ2 (t∗ , x) ≥ x2 = x1 , Φ2 (t∗ , x) ≥ ln N. Φ1 (t∗ , x) This is a contradiction, so there must exist 0 ≤ t0 ≤ t∗ for which Φ2 (t0 , x) ≥ ln N. Φ1 (t0 , x) However, observe that for any point on the line y2 = (ln N )y1 , we have |B1 | ≤ ln ln N , so that the trajectory of x for t ≥ t0 cannot escape the region {y2 ≥ (ln N )y1 } unless |Φ(t, x)| becomes larger than |ˆ x|, which is impossible during the time interval [0, t∗ ]. This finishes the proof. Remark 3. One does not face the restriction α < 3/5 in the presence of a boundary. For the convenience of the reader, we give a proof that ω(t, ·) in the case of Theorem 2.1.2 actually stays C ∞ away from the origin. Proposition 2.1.5. Consider ω0 ∈ L∞ (T2 ) which is C ∞ away from a closed set A ∈ T2 . Then, the unique solution ω(t, ·) ∈ L∞ (T2 ) of the 2D Euler equation stays C ∞ away from Φ(t, A) for all t > 0. Proof. We may assume that kω0 kL∞ = 1. Once we show that ω(t, ·) is smooth away from Φ(t, A) for t ∈ [0, T ] with some absolute constant T > 0 then we may iterate the argument to extend the statement to any finite time moment. Take an open set O which is separated from A. It suffices to show that for t ∈ [0, T ], there exists some α > 0 that ω(t, ·) is uniformly C k,α in Φ(t, Uk ), for any integer k ≥ 0 and some open set Uk ⊃ O. We deduce this by inducting on k. For the base case of k = 0, take some open set U0 ⊃ cl(O) which is still separated away from A. Then we simply write
kω(t, ·)kC α (Φ(t,U0 )) = sup
x,x0 ∈U0
|ω(t, Φ(t, x)) − ω(t, Φ(t, x0 ))| |x − x0 | ≤ kω0 kC 1 (U0 ) sup 0 α 0 α |Φ(t, x) − Φ(t, x )| x,x0 ∈U0 |Φ(t, x) − Φ(t, x )|
which is bounded by an absolute constant via the H¨older estimate (2.8) once we choose α ≤ e−cT where c is the constant from (2.8). Now we assume that ω(t, ·) is C k,α in Φ(t, Uk ) with some k ≥ 0, where Uk ⊃ cl(O) and d(Uk , A) >
28
0. We first pick some open set Uk+1 which satisfies
Uk+1 ⊃ cl(O),
Uk ⊃ cl(Uk+1 ).
In particular, Uk+1 is separated away from A. For each 0 ≤ t ≤ T , take a smooth cutoff function 0 ≤ χt ≤ 1 which equals 1 on Φ(t, Uk+1 ) and vanishes outside of Φ(t, Uk ). Then, for x ∈ Φ(t, Uk+1 ), with the Biot-Savart kernel K, we write
u(t, x) = (K ∗ ωt )(x) = K ∗ (χt ωt )(x) + K ∗ ((1 − χt )ωt )(x). Regarding the first term, a classical singular integral estimate gives
kK ∗ (χt ωt )kC k+1,α (T2 ) ≤ Ck∇k (χt ωt )kC α (Φ(t,Uk )) . The second term is indeed C ∞ in Φ(t, Uk+1 ) simply because K(·) is C ∞ away from the origin. Hence we deduce that u(t, ·) is uniformly C k+1,α in Φ(t, Uk+1 ). At this point we may extend u(t, ·) to be C k+1,α on the entire domain T2 to obtain u ˜(t, ·). Then solving d ˜ ˜ x)), Φ(t, x) = u ˜(t, Φ(t, dt ˜ x) is a C k+1,α flow, which coincides with Φ(t, x) whenever x ∈ Φ(t, Uk+1 ) and gives that Φ(t, ˜ ·)kC k+1,α and then argue 0 ≤ t ≤ T . This can be done by obtaining an a priori estimate for kΦ(t, along a (smooth) sequence of approximate solutions. Note that
˜ x)| ≥ exp(− |∇Φ(t,
Z 0
t
k∇˜ u(τ, ·)kL∞ dτ )
k+1,α ˜ ·) is invertible and the inverse function theorem gives that Φ ˜ −1 so Φ(t, diffeomort (·) is also a C
phism of the domain. From
˜ −1 ω(t, z) = ω0 (Φ−1 t (z)) = ω0 (Φt (z)),
z ∈ Φ(t, Uk+1 ),
differentiating both sides k + 1 times, on the right hand side we obtain terms which contains up −1 to the k + 1th derivatives of ω0 (composed with Φ−1 also up t ) multiplied with some factors of Φt
to the k + 1th derivatives. Since each such factor is C α , we conclude that ω(t, ·) is C k+1,α . This 29
finishes the proof.
2.2
Ill-posedness for singular vortex patches
The classical result says that if the boundary of a vortex patch is C 1,α -regular with 0 < α < 1, then the 2D Euler equation propagates this regularity for all times. In this section, we present a few ill-posedness results when this regularity assumption fails in the simplest way possible.
2.2.1
Ill-posedness for a single corner
In this section, we show that if a patch is locally given by two C 1,α curves meeting at some nontrivial angle, then the solution will instantaneously lose the C 1,α -regularity up to the corner. In the case of an acute (or similarly for an obtuse) angle, we can show the following stronger result: Theorem 2.2.1. Assume that there exists some r > 0 such that Ω0 ∩ [r, r]2 is given by the region between two C 1 -curves with an angle θ0 ∈ (0, π/2): Ω0 ∩ [−r, r]2 = {(x1 , x2 ) : 0 < x1 < r, g0 (x1 ) < x2 < f0 (x1 )}, where
f0 , g0 ∈ C 1 [0, 2r],
g0 < f0 ,
0 < tan−1 (f00 (0)) − tan−1 (g00 (0))
0 and
R(t)(Ωt ) ∩ Bδ (Φt (0)) − Φt (0) = {(x1 , x2 ) : gt (x1 ) < x2 < ft (x1 ), 0 < x1 < δ} for some C 1 functions ft and gt defined on [0, 2δ], for each t ∈ [0, T ].1 Then, there cannot exist a modulus of continuity ϕ : [0, ∞) → [0, ∞) satisfying |ft0 (x) − ft0 (y)| + |gt0 (x) − gt0 (y)| ≤ ϕ(|x − y|) 1 We
have used the notation Ω − z = {x − z : x ∈ Ω} for Ω ⊂ R2 and z ∈ R2 .
30
uniformly for all x, y ∈ [0, δ] and t ∈ [0, T ]. That is, the boundary curves instantaneously lose all moduli of continuity as soon as t > 0. In particular, if initially f0 and g0 are C 1,α for some 0 < α ≤ 1, the C 1,α -norm of the boundary curves (although well-defined for all time away from the corner) blows up at the corner immediately for t > 0. We shall need the following lemma, which simply says that the velocity from a C 1 -cusp is strictly weaker than that from a corner. Lemma 2.2.2. Assume that a domain U is given locally by an angle θ with C 1 boundary: for some r > 0,
U ∩ [−r, r]2 = {(x1 , x2 ) : 0 < x1 < r, g(x1 ) < x2 < f (x1 )} with f, g ∈ C 1 [0, r] and f 0 (0) > g 0 (0) = 0, so that tan θ = f 0 (0). Then, the associated velocity u = K ∗ (χU ) satisfies sin θ u(x1 , x2 ) − u(0, 0) = − sin θ log |x| cos θ
cos θ x1 + v(x1 , x2 ) − sin θ x2
(2.15)
with v(x) |x| log |x| ≤ 1,
n r o |x| < min 1, , Cϕ 10
(2.16)
and v(x) |x| log |x| → 0,
|x| → 0
(2.17)
where Cϕ > 0 is a constant depending only on the modulus of continuity ϕ for f 0 and g 0 . Moreover, the rate of convergence in (2.17) depends only on ϕ once the value of r > 0 is fixed. Proof. We only consider for simplicity the case when θ = 0 and therefore two C 1 functions f and g are both tangent to the horizontal axis at x1 = 0. Let us further assume that g ≡ 0 and f ≥ 0. In the case of a (locally) exact sector of angle θ, the expression (2.15) is well-known; for instance see Bertozzi [11, Lemma 3.10] or Carrillo-Soler [15], of course with v being a Lipschitz remainder in this case. 31
In our setting, it suffices to show that |u(x) − u(0)| →0 |x| log |x| as x → 0, with a rate depending only on the modulus of f 0 . We shall restrict ourselves to points on the boundary x = (x1 , f (x1 )) (and similarly (x1 , 0)) since this is all we need in the proof of Theorem 2.2.1. (The other points can be treated similarly.) An explicit computation gives (for simplicity of notation, let us allow to use x in place of x1 )
1
1
u (x, f (x)) − u (0, 0) =
Z
r
log
0
(x − y)2 + f (x)2 (x − y)2 + (f (x) − f (y))2
− log
y2 y 2 + f (y)2
dy + Lipschitz,
where the Lipschitz term comes from the part of patch that is outside the box [−r, r]2 . We shall neglect this term from now on. Recall that x < r/10, and we split the integral into the regions 0 ≤ y < 2|x| and 2|x| < y < r. In the first region, it suffices to bound the first term; 2|x|
Z
log
0
(x − y)2 + f (x)2 (x − y)2 + (f (x) − f (y))2
dy
We change variables z = x − y and note that (x − y)2 + (f (x) − f (y))2 = z 2
1 1 + f 0 (x) + z
Z
x+z
x
f 0 (w) − f 0 (x)dw
2 !
and Z x+z 0 0 0 f (x) + 1 f (w) − f (x)dw ≤ 2ϕ(x), z x so that taking Cϕ to be ϕ(x)/10 and then for x < Cϕ , the integral in absolute value can be bounded by
Cϕ0 x
Z
x
+C
log 1 +
−x
f (x) z
2 ! dz ≤
Cϕ0 x
tan−1 (x/f (x)) f (x)2 1+ + log 1 + , x/f (x) x2
which satisfies the desired bounds in the lemma since f (x) → 0 as x → 0.
32
Turning to the outer integral, we combine logarithms to obtain Z
r
log
2x
(x − y)2 + f (x)2 y 2 + f (y)2 · y2 (x − y)2 + (f (x) − f (y))2
dy.
This time, we make a change variables z = y/x: Z
r/x
log
x 2
(1 − z
−1 2
) + (f (x)x
−1 −1 2
z
)
·
1 + (f (zx)z −1 x−1 )2 (1 − z −1 )2 + (f (x)x−1 z −1 − f (zx)x−1 z −1 )2
dz,
and note that we have simple bounds Z f (x) 1 x ≤ ϕ(w)dw ≤ ϕ(x) x x 0 and f (xz) xz ≤ ϕ(r). Using the above bounds and explicitly integrating out the log shows that the above integral is bounded by x log x, adjusting the value of the constant Cϕ if necessary. Moreover ϕ(x) → 0 as x → 0, so that the above integral divided by x log x converges to zero. The argument for the second component of the velocity is strictly analogous. This finishes the proof. Proof of Theorem 2.2.1. Given the assumptions of the theorem, we first proceed to show that the angle θt (which is well-defined since the boundary is given by the graphs of two C 1 curves) is continuous on [0, T ]. Since the initial angle θ0 is positive and less than 90 degrees, we may assume that 0 < f00 (0) < 1 and −1 < g00 (0) < 0. Translating the plane continuously in time if necessary, we shall assume without loss of generality that the origin is a fixed point for all time. Assume for the sake of contradiction that the angle generated by the flow of segment {(x1 , f0 (x1 )) : 0 ≤ x1 ≤ δ} with the positive horizontal axis R+ ≥0 × {0} is not continuous in time. Then, we may assume that there is a sequence of time moments tk → 0+ , so that ft0k (0) ≥ 1. (The other possibilities are that the angle can drop, or the segment {(x1 , f0 (x1 )) : 0 ≤ x1 ≤ δ} crosses the vertical axis that a rotation R(t) is necessary to realize the boundary of Ωt as a graph of a function – these cases can be treated in a similar way.)
33
From the assumptions above, we have
f0 (x) =
f00 (0)x
x
Z
f00 (y) − f00 (0)dy,
+ 0
and
ftk (x) =
ft0k (0)x
x
Z + 0
ft0k (y) − ft0k (0)dy
for x ∈ [0, δ], and therefore, |f0 (x) −
f00 (0)x|
|ftk (x) −
ft0k (0)x|
≤
x
Z
ϕ(y)dy, 0
and
≤
Z
x
ϕ(y)dy. 0
Let us set q = 1 − f00 (0) > 0 and then pick δ ≥ δ 0 > 0 so that
Rx 0
ϕ(y)dy < qx/10 for all x < δ 0 .
Then, inside the ball of radius δ 0 /2, the “fluid particles” with coordinates (x1 , f0 (x1 )) lie inside the sector
{(x1 , x2 ) : 0 < x2 < (1 − q/2)x1 }. On the other hand, at time tk , the flow of those particles must lie inside the sector
{(x1 , x2 ) : (1 − q/2)x1 < x2 }, and otherwise, they must escape the ball B0 (δ 0 /2). These sectors do not intersect at all, which is a contradiction since each particle trajectory is continuous in time. A completely analogous argument shows that the “lower piece” of the boundary also moves continuously in time, finishing the proof that θt is a continuous function of time. From now, for the convenience of applying Lemma 2.2.2, let us assume that initially the lower boundary curve g0 is tangent to the horizontal axis; in particular g00 (0) = 0 and 0 < f00 (0) = tan θ0 < +∞. From continuity, it will be assumed that |gt (x1 )| ≤ x1 and |ft (x1 ) − tan θ0 x1 | ≤ x1 , taking T = T () smaller if necessary. 34
Applying Lemma 2.2.2, we have (recall that we are assuming that the origin is fixed for all time)
sin θt ut (x) = − sin θt log |x| cos θt
cos θt x1 + vt (x1 , x2 ) − sin θt x2
(2.18)
for t ∈ [0, T ] and |x| < δ/10, and we may pick δ/10 ≥ δ 0 = δ 0 (1 ) > 0 so that vt (x) |x| log(1 + |x|−1 ) ≤ 1
(2.19)
for |x| < δ 0 uniformly over t ∈ [0, T ]. (Strictly speaking, for t > 0 the lower boundary of the patch is not exactly tangent to the horizontal axis, but the deviation is a sector with an arbitrary small angle in [0, T ], whose contribution to the velocity comes with a prefactor involving the sine of the angle, so that the corresponding term can be absorbed in the vt -term of (2.18), by possibly shrinking the interval of time, depending on 1 of (2.19).) Now take two “particles” x = (x1 , f0 (x1 )) and x = (x1 , g0 (x1 )) with x1 small enough that the trajectories of these two points stay in the ball B0 (δ 0 ) for [0, T ]. To see the evolution of the angle that the triangle with vertices (0, 0), Φt (x), and Φt (x) form at the origin, we compute 1 d Φ2t (x) ˙ 2 (x)Φ1 (x) − Φ˙ 1 (x)Φ2 (x) = Φ t t t t 2 dt Φ1t (x) (Φ1t (x)) for x ∈ {x, x}. For simplicity, let us set ζ t :=
Φ2t (x) , Φ1t (x)
ζ t :=
Φ2t (x) . Φ1t (x)
We compute that d 1 2 ζ t = sin θt log cos θt (1 − ζ t ) − 2 sin θt ζ t dt |Φt (x)| vt1 |Φt (x)| 2 vt2 |Φt (x)| − · · ζ + · . t |Φt (x)| sin θt log |Φt (x)|−1 Φ1t (x) |Φt (x)| sin θt log |Φt (x)|−1 Φ1t (x) Note that the terms involving vt1 and vt2 can be taken to be arbitrarily small uniformly for |x1 | < δ 0 and [0, T ], by taking 1 , T > 0 smaller if necessary depending on θ0 . Moreover, recall that
ect log
1 1 1 ≥ log ≥ e−ct log |x| |Φt (x)| |x|
35
for t > 0 with some constant c > 0. Therefore, we have d 1 2 ζ t ≤ (1 + 2 ) sin θt log cos θt (1 − ζ t ) − 2 sin θt ζ t + E 1 (t, x1 ) dt x1 for any 2 > 0 by taking T, δ 0 , 1 smaller if necessary depending on 2 , and similarly 1 d ζ t ≥ (1 − 2 ) sin θt log cos θt (1 − ζ 2t ) − 2 sin θt ζ t − E 2 (t, x1 ) dt x1 (where E 1 (t, x1 ) and E 2 (t, x1 ) are uniformly small) so that d 1 1 ζ t − ζ t ≤ − sin θt log ζ t − ζ t cos θt (ζ t + ζ t ) + sin θt + E 3 (t, x1 ) log < 0, dt x1 x1 where E 3 (t, x1 ) is uniformly small. At this point, we deduce that 1 d ζ t − ζ t ≤ −C(θ0 ) log ζt − ζt dt x1 where C(θ0 ) is some strictly positive constant depending only on θ0 , since the angle is continuous in time and π/2 > θ0 > 0. This inequality is valid for all points x1 < δ 0 /10 and time t ∈ [0, T ] for some small but fixed δ 0 , T > 0. This implies
1 ζ T − ζ T ≤ exp T C(θ0 ) log x1
.
Taking the limit x1 → 0+ , the above inequality shows that θT = 0. This is a contradiction. Remark 4. The corresponding statement for an initial obtuse angle π/2 < θ0 < π can be proved in a completely analogous manner. In this case, the angle would “open up” as soon as t > 0. Remark 5. A result of Carrillo and Soler [15] says that, if Ω0 is given by an acute angle with C 1 boundary, there exists some time interval during which the angle does not increase: θ(t) ≤ θ0 . It is not assumed that the boundaries remain C 1 curves, and therefore the angle should be understood in the extended sense as in (2.20). This essentially follows from the fact that at the initial time, the velocity field is explicitly log-Lipschitz and the particle trajectories are smooth in time and space away from the origin. Remark 6. Our result is consistent with the fact that all the numerically reported V -states with corners have corner angle exactly equal to π/2. 36
2.2.2
Ill-posedness under the odd-odd symmetry
In this section, we investigate the evolution of singular vortex patches under the odd-odd symmetry. To be more precise, we consider patches on the plane satisfying
ω(x1 , x2 ) = −ω(−x1 , x2 ) = −ω(x1 , −x2 ) = ω(−x1 , −x2 ),
x1 , x2 ∈ R
and having a corner at the origin (see Figure 2.3). It is well-known, after the work of Kiselev and Sverak [50], that vorticities that are non-negative on the positive quadrant R2+ with odd-odd symmetry is able to generate “small scales” near the origin. The study of such flows is facilitated by the Key Lemma (see below 2.2.6), which essentially gives a lower bound on the velocity gradient.2 Since the odd-odd symmetry is preserved for all times, we shall view the patch as defined only on the positive quadrant. Next, to state our results, let us formally define the “angle” 0 ≤ θ ≤ π/2 of a domain U ⊂ R2+ at the origin as follows: sup
cos θU := lim
δ→0+ x,x0 ∈U ∩[0,δ]2
x · x0 . |x||x0 |
(2.20)
In the definition above, it is assumed that U intersects any small square [0, δ]2 . Theorem 2.2.3. Consider an odd-odd vortex patch supported on Ω0 ⊂ R2 such that Ω0 ∩ {(x1 , x2 ) : 0 ≤ x1 , x2 ≤ 1/2} = {(x1 , x2 ) : 0 ≤ x1 ≤ x2 ≤ 1/2} . Then, there exist universal constants T ∗ , δ > 0, such that in the ball [0, δ]2 , we have bounds
2z1 (t, x)β(t) ≥ z2 (t, x) ≥
1 z1 (t, x)α(t) , 2
(2.21)
for some strictly decreasing positive and continuous functions defined on [0, T ∗ ] with α(0) = β(0) = 1. In particular, the angle of the patch at the origin becomes immediately π/2. On the other hand, if one considers the backwards in time evolution, there exists some time interval [−T 0 , 0) with T 0 > 0, during which the angle of the patch at the origin is zero. It is not hard to slightly modify the initial patch so that its angle at the origin starts at zero and becomes π/2 immediately. To state it precisely, we have the following 2 We note that very recently, the key lemma was utilized to obtain double exponential rate of growth in time for the curvature for smooth vortex patches touching the axis – see [54].
37
−1
+1
−1
+1
+1
−1
+1
−1
Figure 2.3: Evolution of a corner with odd-odd symmetry. Corollary 2.2.4. There exists an odd-odd vortex patch Ω0 which has zero angle at the origin but the angle of Φ(t, Ω0 ) at the origin is π/2 for any t > 0. To get a sense of the behavior described in Theorem 2.2.3, one may consider the following explicitly solvable example: Example 2.2.5. On the plane R2 , consider the flow given by the divergence-free vector field
v1 (x1 , x2 ) = −x1 ln
1 , |x1 |
v2 (x1 , x2 ) = x2 ln
1 . |x1 |
Then, denoting the trajectory of (x, x) with x > 0 by (z1 (t), z2 (t)), we may explicitly compute that
z1 (t) = xexp(t) ,
z2 (t) = x2−exp(t) .
In particular, for 0 < t < ln 2, the image of the line {(x, |x|) : x ∈ R} under the flow map is H¨ older continuous with exponent 2 exp(−t) − 1 and no better, viewed as the graph of a function over the x1 -axis. It can be checked with a direct computation that with Ω0 as in Theorem 2.2.3, the initial velocity vector field u0 has the form given in the above example. For t > 0, the Key Lemma guarantees that the velocity retains the same form. Remark 7. We may consider these data on the upper half-plane, and in this case, the boundary of the initial patch ∂Ω0 is given as the graph of a C 0,1 and C 1,0 -function, respectively, near the origin. This property is not maintained for any small time t > 0. On the other hand, if initially one considers odd-odd patch given by the region below the graph of a C 1,α -function whose derivative vanish at the origin (see Figure 2.4), it can be shown using the ideas of previous sections that the solution continues to satisfy these properties, at least for some time interval.3 Therefore, in a certain 3 See
also the recent work of Kiselev, Ryzhik, Yao, and Zlatoˇs [49] where global in time well-posedness is shown for
38
−1 +1
+1 −1
Figure 2.4: A C 1,α -cusp with odd-odd symmetry. sense, our result says that the vortex patch dynamics is ill-posed in the class of C 0,1 or C 1 boundaries. Remark 8. The second part of the theorem was established in an earlier work by Hoff and Perepelitsa [43], with a similar patch initial data but having just one odd symmetry with respect to the x1 -axis. It is expected that the dynamics in that case is equivalent to the odd-odd symmetry case up to a translation of the corner point. Here we offer a simplified proof using the Key Lemma. Before we begin the proof, let us recall here for convenience the key lemma (see Zlatos [79]): Lemma 2.2.6. Let ω(t, ·) ∈ L∞ (R2 ) be odd-odd. For x1 , x2 ∈ (0, 1/2], we have (−1)j
uj (t, x) 4 = xj π
Z Q(2x)
y1 y2 ω(t, y)dy + Bj (t, x) |y|4
(2.22)
where Q(2x) := [2x1 , 1] × [2x2 , 1] and |Bj (t, x)| ≤ Ckω(t, ·)k
L∞
x3−j 1 + ln 1 + xj
for j ∈ {1, 2}. For simplicity of notation, we will denote the integral in (2.22) as 4 I(t, x) := π
Z Q(2x)
y1 y2 ω(t, y)dy . |y|4
On the other hand, we have the well-known log-Lipschitz bound for velocity: for any x, x0 with |x − x0 | < 1/2, |u(t, x) − u(t, x0 )| ≤ Ckω(t, ·)kL∞ |x − x0 | ln
c . |x − x0 |
Proof of Theorem 2.2.3. We consider the case t ≥ 0. We shall work within a short time interval [0, T ∗ ] for some T ∗ > 0, and in several places, the value of T ∗ will be taken to be sufficiently small for the arguments to work. cusps touching the boundary.
39
We begin with a simple observation. Note that on the diagonal x = (x0 , x0 ) with 0 < x0 , we have u1 (t, x) I(t, x) + B1 (t, x) =− u2 (t, x) I(t, x) + B2 (t, x) and |Bj (t, x)| ≤ C for all time. Clearly, one can find some small δ1 > 0 and T ∗ > 0 such that I(t, (δ1 , δ1 )) ≥ 10C for all 0 ≤ t ≤ T ∗ , simply because I(t, (δ1 , δ1 )) is continuous in t, δ1 and I(0, (δ1 , δ1 )) → +∞ as δ1 → 0+ . Therefore, we take δ1 ≥ δ2 > 0 such that the triangle {0 ≤ x1 ≤ x2 ≤ δ2 } is contained in Φ(t, Ω0 ) for all 0 < t < T ∗ . Consider a point on the diagonal (x, x) with 0 < x < δ δ2 (the value of δ > 0 will be specified later) and denote its trajectory by z(t, x) = (z1 (t, x), z2 (t, x)) := Φ(t, (x, x)). From the basic log-Lipschitz estimate on u2 , d c z2 (t, x) ≤ Cz2 (t, x) ln , dt z2 (t, x) (since u2 (t, (z1 (t, x), 0)) = 0 by odd symmetry) and upon integration, we deduce that z2 (t, x) ≤ cxexp(−Ct) . Proceeding analogously for z1 (t), we obtain z1 (t, x) ≥ cxexp(Ct) this time. Inserting these crude bounds,
1 |B1 (t, z(t, x))| ≤ C 1 + exp(2CT ) ln x ∗
(2.23)
for 0 ≤ t ≤ T ∗ . Moreover, with β(t) := exp(−2Ct), we obtain z1 (t, x)β(t) ≥
1 z2 (t, x), 2
for 0 ≤ t ≤ T ∗ by choosing T ∗ smaller if necessary. A lower bound for the integral I(t, z(t, x)) comes from the fact that Φ(t, Ω0 ) contains a triangle. We could have chosen δ > 0 small so that for all x < δ, its trajectory satisfies the bound z2 (t) ≤ δ2 . In particular, the region Q(z(t, x)) contains the triangle with vertices (z2 (t, x), z2 (t, x)), (δ2 , z2 (t, x)), and (δ2 , δ2 ). Hence
I(t, z(t, x)) ≥ c ln
δ2 δ0 ≥ c exp(−Ct) ln 2 z2 (t, x) x
and comparing this with (2.23), we could have chosen δ, T ∗ > 0 smaller so that for 0 < t < T ∗ and
40
0 < x < δ,
I(t, z(t, x)) ≥
1 |B1 (t, z(t, x))| . 10
Therefore, we may neglect the B1 -term in (2.22) at the cost of changing the multiplicative constant, and deduce
−
u1 (t, z(t, x)) δ0 ≥ c ln 2 . z1 (t, x) x
In turn, this ensures that
z1 (t, x) ≤ c0 x1+ct with c0 → 1 as T ∗ → 0+ . From the trivial bound z2 (t, x) ≥ x, we obtain z2 (t, x) ≥
1 z1 (t, x)α(t) , 2
with α(t) := (1 + ct)−1 . This finishes the proof of the first part. We now consider the backwards in time dynamics. Instead of reversing time, we revert the sign of vorticity, so that now initially the direction of velocity is southeast on the diagonal segment. As in the above, we set z(t, x) := Φ(t, (x, x)), and restrict our attention to 0 < x < δ and 0 ≤ t ≤ T 0 , for small δ, T 0 > 0 to be chosen below. Similarly as before, using either the Key Lemma or the log-Lipschitz estimate on velocity gives
z2 (t, x) ≤ z1 (t, x) ≤ 2z2 (t, x)γ(t) ≤ 2xγ(t) for all 0 < x < δ and 0 ≤ t ≤ T 0 with some sufficiently small δ, T 0 > 0. Here γ(t) > 0 is some continuous monotonically decreasing function with γ(0) = 1 and γ(t) < 1 − ct. This gives a lower bound on the integral η(t)
I(t, z(t, x)) ≥ I(t, (z1 (t, x), z1 (t, x)) ≥ c
δ η(t) − z1 η(t)
≥ c0
δ˜ct − xct ct
where η(t) = 2(1/γ(t) − 1) & t satisfies η(0) = 0 and is monotonically increasing with t. Applying
41
the Key Lemma to each of z1 (t, x) and z2 (t, x), we see that d dt
z1 (t, x) z2 (t, x)
z1 (t, x) ≥ z2 (t, x)
z1 (t, x) 2I(t, z(t, x)) − C ln 1 + , z2 (t, x)
or equivalently, ct ˜ct d z1 (t, x) z1 (t, x) 0δ − x ln 1 + ≥c − C ln 1 + . dt z2 (t, x) ct z2 (t, x) Fix some small x > 0 and consider the ODE δ˜ct − xct d (x) f (t) = − Cf (x) (t), dt ct
f (0) = ln 2.
It is straightforward to show that, for all sufficiently small 0 < t ≤ T 0 , we have lim
f (x) (t) = +∞. x
lim
z1 (t, x) = +∞ z2 (t, x)
x→0+
Then, this implies that
x→0+
for all 0 < t ≤ T 0 , which shows that the angle of the patch is zero in the same time interval. Remark 9. From the rough bounds (2.21), one may extract more detailed information about the evolution of the boundary. These bounds imply that, as soon as t > 0, the support of vorticity contains most of the box [0, δ]2 . From this observation, it is not difficult to show the following upper and lower bounds for x ≤ δ 0 δ : 2 δ 2 L ln − C ≤ I(t, z(t, x)) ≤ ln , π z2 (t, x) π z2 (t, x) for some absolute constant L. This suggests that, for small time, we have asymptotics
2
z2 (t, x) ≈ const · xexp(− π t+o(t)) .
42
Proof of Corollary 2.2.4. Take a domain Ω0 which satisfies ( cl(Ω0 ) ∩ [0, 1/2]2 =
−1 ) 1 (x1 , x2 ) : 0 ≤ x2 ≤ x1 ln ln ∩ [0, 1/2]2 . x1
In view of Theorem 2.2.3, it suffices to show the following: Claim. For any small 0 < T 1, there exists some R = R(T ) such that cl(Φ(t, Ω0 )) ∩ [0, R]2 ⊃ {0 ≤ x1 ≤ x2 ≤ R} . Theorem 2.2.3 then immediately implies that at time 2T (say), the angle at the origin is π/2. We first make a simple observation on the direction of the velocity for x = (x1 , x2 ) lying on
x2 =
x1 . ln ln x11
At t = 0, 4 π
Z
y1 y2 ω (y)dy 4 0 Q(2x) |y| 1 Z 1/10 Z Z 1/10 10 ln ln 1 θ 1 r ≥C dθdr ≥ C 2 dr r 2x1 2x1 0 r ln ln 1r
I(0, x) =
and 1 |Bi (0, x)| ≤ C 1 + ln 1 + ln ln 2x1
i ∈ {1, 2} .
Noting that Z
1/10
2x1
1 r ln ln
dr 1 2 r
ln ln ln
1 x1
x1 1
(using for instance L’Hˆ opital’s rule), and by continuity of the integral I(t, x) in time, we can choose some small T ∗ , δ > 0 such that 1 |Bi (t, x)| , I(t, x) ≥ 10
∗
0≤t≤T ,
In particular, on the boundary region {x2 =
x=
x1 ln ln x1
1
43
x1 x1 , ln ln x11
! ,
0 ≤ x1 ≤ δ .
} the velocity is directed toward the northwest,
so that (
δ x1 0 ≤ x2 ≤ 1 , 0 ≤ x1 ≤ 2 ln ln x1
cl(Φ(t, Ω0 )) ⊃
) ,
0 ≤ t ≤ T∗ ,
(2.24)
by taking T ∗ > 0 smaller if necessary, since then “fluid particles” carrying zero vorticity cannot cross the line segment {x1 = δ/2, 0 ≤ x2 ≤ δ/2(ln ln(δ/2)−1 )−1 } during an arbitrarily short time interval. Now let T > 0 be an arbitrarily small number, which we may assume to be less than T ∗ . Take R0 = R0 (T ) δ/2 small so that for |x| ≤ R0 , supt∈[0,T ] |Φ(t, x)| ≤ δ/100. Let us show that for any x = (x1 , x2 ) with 0 ≤ x1 ≤ R0 ,
x2 =
x1 , ln ln x11
(2.25)
its trajectory satisfies Φ1 (T, x2 ) ≤ Φ2 (T, x1 ), i.e. the particle initially located at x crosses the diagonal at some time in the interval [0, T ]. Once this is shown, we deduce the statement of the Claim with R = R0 / ln ln(R0 )−1 . Assume towards a contradiction that there is some x as in (2.25), so that for all 0 ≤ t ≤ T , we have Φ1 (t, x) > Φ2 (t, x). From (2.24), we note that
I(t, Φ(t, x)) ≥ C
Z
δ/2
2Φ1 (t,x)
Z
1 10 ln ln 1 r
0
θ dθdr ≥ C r
Z
δ/2
2Φ1 (t,x)
Setting Z
δ/2
F (z) = 2z
1 r ln ln 1r
2 dr
for 0 < z < δ/2, we have −u1 (t, Φ(t, x)) ≥ CF (Φ1 (t, x)) , Φ1 (t, x) and therefore d 1 dt Φ (t, x) Φ1 (t, x)
≤ −CF (Φ1 (t, x)) .
44
1 r ln ln 1r
2 dr .
In particular Φ1 (t, x) is decreasing in time and since F (·) is a decreasing function,
Φ1 (t, x) ≤ x1 exp (−CF (x1 )t) . Regarding the second component, we simply note that Φ2 (t, x) ≥ x2 as u2 (t, ·) is nonnegative in the region {0 ≤ x1 ≤ x2 ≤ δ/2}. Therefore, at time t = T , −1 ln ln x11
2
Φ (T, x) ≥ , Φ1 (T, x) exp(−CF (x1 )) and L’Hˆ opital’s rule guarantees that the ratio lim +
x1 →0
ln ln x11
−1
exp(−CF (x1 ))
= lim + x1 →0
ln x11 · exp(CF (x1 )) exp(CF (x1 )) = lim = +∞ . 2 x1 →0+ ln ln x11 ln ln x11
Hence we could have taken R0 smaller so that
inf
x1 ≤R0
Φ2 (T, x) ≥1. Φ1 (T, x)
This is a contradiction, and therefore for any such x there exists a moment of time 0 ≤ t ≤ T for which Φ2 (t, x) ≥ Φ1 (t, x). However, note that on the diagonal segment {x1 = x2 , 0 ≤ x1 ≤ δ/2}, the velocity is directed towards the northeast for [0, T ] since |Bi | ≤ C for i = 1, 2. Hence we conclude Φ2 (T, x) ≥ Φ1 (T, x). As a yet another corollary (more precisely, corollary of the proof), we show that an initially L∞ ∩ W 1,p -vorticity may continuous lose integrability in time, when we consider the Euler dynamics on the upper half-plane. Let us set H = R × R+ . Corollary 2.2.7. There exists an initial data ω0 ∈ L∞ (H) ∩ (∩p 0. This allows us to solve the ODE (3.1.1) uniquely. These consider-
48
ations lead to an existence and uniqueness theory for L1 ∩ L∞ vorticities which was first established by Yudovich in 1963 [76]. Unfortunately, the fact that the L∞ bound on ω does not necessarily lead to Lipschitz control on the velocity field can be problematic since a non-Lipschitz velocity field can lead to a flow map Φ(t, ·) which loses its regularity in time. Since ω = ω0 ◦ Φ−1 , a bound on the Lipschitz norm of u is crucial to propagate fine-scale structures which may be present in the initial data. Unfortunately, as has been established in the works [39, 13], propagating a bound on the Lipschitz norm of u is, in general, not possible if u0 is only assumed to be Lipschitz continuous. In order to actually propagate a Lipschitz bound on u, the initial data must be taken to be smoother than just Lipschitz. Taking velocity in C 1,α or H s with s > 2 would do, but one can also propagate Lipschitz bounds on u using anisotropic regularity such as is the case with smooth vortex patches (see [17]). Propagating the Lipschitz bound on u for solutions which have corner like discontinuities was left open in previous works and will be returned to later on in this work (see the subsection on the propagation of angular regularity).
3.1.2
Solutions with non-decaying vorticity and the symmetry condition
The usual assumption that ω ∈ L1 ∩ L∞ may be argued physically unsatisfactory, since the vorticity and the velocity have to decay at infinity. Indeed, the well-posedness of the Euler equations with merely bounded velocity and vorticity has been achieved in the works of Serfati [67, 68], and they have been further expanded and generalized in [1, 48, 72, 73]. The main result states that (see [48] for details) given a pair of bounded functions (u0 , ω0 ) satisfying ∇ × u0 = ø0 and ∇ · u0 = 0, and an arbitrary continuous function U∞ (t) : R2 → R2 satisfying U∞ (0) = 0, there is a unique global-in-time solution to the 2D Euler equations where for each time, ut , ωt are bounded functions and satisfies the “renormalized Biot-Savart law”
ut (x) − u0 (x) = U∞ (t) + lim (aR K) ∗ (ωt − ω0 )(x) , R→∞
(3.1)
where K(·) denotes the Biot-Savart kernel in (2.1) and aR is some cutoff whose support increases to infinity with R. Here, the “behavior at infinity” U∞ (t) can be removed with the following change of variables Z u ¯(t, x) := u(t, x + 0
t
U∞ (s)ds) − U∞ (t) ,
Z p¯(t, x) := p(t, x + 0
49
t 0 U∞ (s)ds) + U∞ (t) · x .
With ø just in L∞ , the Biot-Savart law clearly does not converge. The key observation of Serfati was that, using the Euler equations and integration by parts, for smooth solutions one has the following Serfati identity: ut − u0 = U∞ (t) + (aK) ∗ (øt − ø0 ) + ((1 − a)K) ∗ (øt − ø0 ) Z t ∇∇⊥ [(1 − a)K] ∗ (u ⊗ u)(s)ds , = U∞ (t) + (aK) ∗ (øt − ø0 ) −
(3.2)
0
with a being some compactly supported cut-off function, and surprisingly the last expression makes sense with u ∈ L∞ , since ∇∇⊥ [(1 − a)K] has decay |x|−3 , which is integrable. The identities (3.1) and (3.2) are indeed equivalent, see [48]. We keep the assumption ω ∈ L∞ but replace the assumption u ∈ L∞ with ω being m-fold rotationally symmetric around the origin, for some integer m ≥ 3. Our main result shows that one can uniquely solve the 2D Euler equation in this symmetry class. Somewhat analogously to the Serfati case, the key fact we utilize is that under the symmetry assumption, the Biot-Savart kernel actually gains integrable decay at infinity (this fact was obtained in a very recent work of the first author [34]), which in particular enables us to recover the velocity from the vorticity. Indeed, using the symmetry of ω, one may rewrite m
u(x) = K ∗ ø(x) =
1 X m i=1
Z R2
i K(x − O2π/m y)ω(y)dy ,
with O2π/m being the counterclockwise rotation matrix by the angle 2π/m, and the point is that m X i=1
i K(x − O2π/m y) ≈ c
|x|m−1 |y|m
in the regime |y| ≥ C|x|. This is integrable for m ≥ 3, and barely fails to be so for m = 2. In this situation, the origin is a fixed point for all time, and the velocity and stream function have bounds
|Ψ(x)| ≤ CkωkL∞ |x|2 ,
|u(x)| ≤ CkωkL∞ |x| ,
(3.3)
which are natural in view of physical dimensions. This in particular removes the non-uniqueness issue arising from U∞ (t) in the Serfati case. More importantly, it says that the logarithmic correction ln |x| (which we usually expect) is absent at the origin, and this is a key ingredient in our proof. In fact, as we will show, the symmetry condition actually allows us to propagate global Lipschitz bounds on u even though ω may have a ”corner” jump discontinuity at the origin which, without
50
the symmetry, was shown to lead to unbounded Lipschitz norm in [39]. Moreover, the symmetry condition gives that for ω0 which is Lipschitz, its gradient can grow at most exponentially at the origin. This type of result was recently established by Itoh, Miura, and Yoneda [46, 45], which we discuss after giving the proof of our main result. Previous results Yudovich’s L1 ∩ L∞ result was improved mainly in two directions, one which weakens the L1 assumption1 , and the other which allows the vorticity to be (slightly) unbounded. Regarding the latter, we just refer the interested reader to works [8, 9, 77, 33, 74]. In the other direction, Benedetto, Marchioro, and Pulvirenti have shown in [7] that if the initial data (u0 , ω0 ) satisfy ω0 ∈ Lp ∩ L∞ (R2 ) and |u0 (x)| ≤ C(1 + |x|α ) for some α < 1 with αp < 2, there is a unique solution to the 2D Euler equation. Note that by imposing some restriction on the growth of the velocity at infinity, one can relax L1 up to any Lp with p < ∞, as α → 0+ . The authors also ask what happens for just L∞ vorticity. Comparing this to Theorem 3.2.1 below, we can treat velocities growing linearly in space, at the cost of imposing m-fold symmetry for m ≥ 3. Indeed, by imposing just 2-fold symmetry, one can obtain existence and uniqueness for Lp ∩ L∞ -vorticity (for any p < ∞), without restricting the growth of velocity.
3.1.3
Local well-posedness in a critical space and bounds on ∇u
A natural question is whether one can actually prove bounds on the Lipschitz norm of u even if ω has a non-smooth jump discontinuity at the origin. In fact, one of the basic themes of this thesis is to study the question of whether L∞ estimates can be established for singular integrals without extra regularity assumptions in the sense of scaling using only symmetry conditions. To this end, ˚0,α by: we define a scale of spaces C
kf kC˚0,α = kf kL∞ + k| · |α f kC˙ α . These spaces have the same scaling as L∞ but, under a symmetry condition, we prove boundedness of the singular integrals arising from the operator which sends ω to ∇u via the Biot-Savart law and more general problems. We prove: ˚0,α (R2 ) be m−fold symmetric for some m ≥ 3. Then, ∇2 (−∆)−1 f ∈ C ˚0,α (R2 ). Theorem. Let f ∈ C 1 It
is straightforward to see that the Yudovich theorem holds for vorticity in Lp ∩ L∞ for any p < 2.
51
˚0,α (R2 ) with compact support and which is 2-fold symmetric for Remark 10. There exists f ∈ C which ∇2 (−∆)−1 f 6∈ L∞ (B1 (0)). These bounds are crucial to prove existence and uniqueness for the SQG equation in a class of merely Lipschitz continuous initial data and allow us to propagate Lipschitz bounds on the velocity field in the 2D Euler equation even when the initial velocity field has a jump discontinuity. In fact, one can prove the following theorem: ˚0,α (R2 ) be m−fold symmetric for some m ≥ 3. Then there exists a unique Theorem. Let ω0 ∈ C ˚0,α for all t > 0. Moreover, global solution to the 2D Euler equation with initial data ω0 belonging to C k∇ukL∞ remains bounded for all finite times.
3.1.4
Scale-invariant solutions for 2D Euler
This extension to the Yudovich theory contains some interesting classes of solutions. A distinguished class is the case of bounded and radially homogeneous vorticity, i.e. ω satisfying ω(x) = h(x/|x|). Homogeneity is propagated by the Euler dynamics, and, by uniqueness, the system reduces to a simple but nontrivial 1D equation on the unit circle. In particular, h satisfies the following 1D active scalar equation:
∂t h(θ, t) + 2H(θ, t)∂θ h(θ, t) = 0 , 4H + H 00 = h . This 1D system is even more regular than the 2D Euler equations, in the sense that the advecting velocity field is two derivatives more regular than the advected quantity. As a result, ∂θ h can grow at most exponentially in time, in contrast with the double exponential rate for the case of 2D Euler. We show that this exponential rate is sharp when we have a boundary available, and rule it out in the absence of boundaries, under some extra assumptions. These are shown via establishing that there is a trend to equilibrium as time goes to infinity. We note that these solutions, while being infinite energy, can be placed into a natural uniqueness class, which is the class of bounded vorticities satisfying the symmetry assumption. As a consequence, they can also be used to show that solutions with finite energy experience growth of angular derivatives as t → ∞. In fact, using our analysis of the 1D model, we can prove that there exist solutions to the 2D Euler equation with Lipschitz velocity field and which are smooth in the angular variable for all times for which the angular derivative of ω, ∂θ ω experience (at least) linear-in-time L∞ growth. When a boundary 52
is present, this growth can actually be (at least) exponential in time. This follows simply from analyzing the behavior of the 1D system. Hence, we emphasize, while the solutions of the 1D system are infinite energy, any growth of ∂θ h implies the existence of compact support ω with ∂θ ω growing at least as fast. Previous examples of infinite energy solutions The consideration of radially homogenous vorticity is comparable to the well-known stagnation-point similitude ansatz, which in the case of 2D takes the form
u(t, x, y) = (f (t, x), −y∂x f (t, x))
(3.4)
(note that u satisfies the divergence free condition), and when inserted in the 2D Euler equation, one obtains the so-called Proudman-Johnson equation [61]
∂t ∂x2 f = ∂x f · ∂x2 f − f ∂x3 f .
(3.5)
To the best of our knowledge, in the context of the Euler equations, (3.5) was first studied by Stuart [71] who showed formation of singularities in finite-time. This is one of the unfortunate sides of the ansatz (3.4) because clearly the fact that these solutions form singularities in finite time has no bearing on singularity formation in the 2D Euler equation (since singularities cannot form in finite time for the 2D Euler equation). In this sense, the singularity is ”coming from infinity” and is a consequence of the solution being of infinite energy. The ansatz (3.4) can be inserted in many other nonlinear evolution equations to define a 1D system. For an example, inserting it to the SQG (surface quasi-geostrophic) equation [16], one obtains a particular case of the De Gregorio model. See also a recent work of Sarria and Wu [64] where they study the 2D Boussinesq model with (3.4). In higher dimensions, one can similarly put
u(t, x0 , xn ) = (f (t, x0 ), −xn ∇ · f (t, x0 )) ,
x0 = (x1 , · · · , xn−1 ) .
This was suggested in Gibbon, Fokas, and Doering [42] in the context of the 3D Euler equations, and was shown to blow-up in finite time in the papers of Childress, Ierley, Spiegel, and Young [21] and Constantin [24]. Notice that in all these cases, the vorticity is never a bounded function; indeed,
53
from (3.4), one sees that ø(x, y) = −yf 00 (x) grows linearly in space. Our radial homogeneity ansatz is special in this regard, and it seems that having bounded vorticity is the key to having uniqueness.
3.2
Existence and uniqueness for the 2D Euler
Let us state the main result of this section. Assume that the initial vorticity ω0 is just bounded and m-fold rotationally symmetric with some m ≥ 3. Then, it shows that under this setting, the velocity can be uniquely recovered from the vorticity (so that the 2D Euler system can be formulated in terms of the vorticity alone), and that this vorticity formulation is globally well-posed. Theorem 3.2.1. Assume that ω0 is a bounded and m-fold rotationally symmetric function on the plane, for some m ≥ 3. Then, there is a unique global solution to the 2D Euler equation ∂t ω + u · ∇ω = 0 , with ω ∈ L∞ ([0, ∞); L∞ (R2 )) and m-fold rotationally symmetric. Here, u = u(ω) is the unique solution to the system
∇×u=ω ,
∇·u=0 ,
under the assumptions |u(x)| ≤ C|x| and m-fold rotationally symmetric.2 It satisfies the principal value version of the Biot-Savart law: 1 R→∞ 2π
Z
u(t, x) = lim
|y|≤R
(x − y)⊥ ω(t, y)dy . |x − y|2
Given the L∞ theorem, it is straightforward to prove the global well-posedness in H¨older spaces, without the usual Lp assumption on the vorticity. One may proceed as in the proof of Theorem 3.2.15 using the flow maps. Corollary 3.2.2. Assume that ω0 ∈ C α (R2 ) with 0 ≤ α ≤ 1, and it is m-fold rotationally symmetric for some m ≥ 3. Then, there is a unique global solution to the 2D Euler equation with α 2 ω ∈ L∞ loc ([0, ∞); C (R )) and m-fold rotationally symmetric.
2 We say that a vector-valued function v : R2 → R2 is m-fold symmetric when v(O 2π/m x) = O2π/m (v(x)) for all x ∈ R2 .
54
3.2.1
Explicit solutions
Before going into the proof of Theorem 3.2.1, we give a few classes of very simple solutions, whose uniqueness is covered by our extension of the Yudovich theory. Some further examples will be provided in Section 3.3. Example 3.2.3 (Radial eddies). Radial and stationary solutions in 2D are widely known, see for instance the book of Bertozzi and Majda [58, Chap. 2]. Take some vorticity function which depends only on the radius; i.e. ω(x) = f (|x|) for some f smooth and compactly supported. Then, it defines a stationary solution with the velocity obtained via the following radial Biot-Savart law:
u(x) =
x⊥ |x|2
Z
|x|
sf (s)ds .
(3.6)
0
In the simplest case when f is the characteristic function on [0, R], one sees that u(x) = x⊥ /2 up to |x| ≤ R and then decays as |x|−1 . One may consider the limiting case when R → ∞: our result implies that, u(x) := x⊥ /2 on R2 is the unique solution to the Euler equation, as long as we require m-fold symmetry for some m ≥ 3 and the growth condition |u(x)| ≤ C|x|. More generally, the expression (3.6) defines the unique solution, with f just in L∞ (R). We now turn to some other stationary solutions, which does not seem to be well known. Example 3.2.4 (Stationary solutions with odd symmetry). Take the disk with some radius R > 0. For each integer m ≥ 1, consider the following vorticity configuration (in polar coordinates):
ω (m) (r, θ) =
+1 −1
if θ ∈ ∪m−1 j=0 [2jπ/m, (2j + 1)π/m) , otherwise .
These solutions are stationary for any m ≥ 1; indeed, the Euler equation preserve odd symmetry of the vorticity along a line, and for each line of discontinuity of ω (m) , it is odd so that the fluid particles cannot cross such a line. Now we take the limit R → ∞: our main result implies that, for m ≥ 3, the stationary one is the unique solution to the Euler equation. In the special case m = 2, these solutions for finite R are often called the Bahouri-Chemin solutions [4], and the limit R → ∞ is not covered by our analysis. It serves as an explicit counterexample for us, as the associated velocity vector field near the origin satisfies |u(x)| ≤ C|x| ln |x| (see also (3.18)). Actually, in [38] we utilized a suitably smoothed-out version of the Bahouri-Chemin solution to recover ill-posedness of the 2D Euler equation in Sobolev spaces scale as kωkL∞ , a fact established 55
first by Bourgain and Li [12]. Our main result confirms that, in a sense, the Bahouri-Chemin scenario is the “only option” for showing such an ill-posedness result, at least near the origin. Lastly, when m = 1 (i.e. odd vorticity), the velocity does not vanish at the origin. Actually, it is easy to check that |u(0, 0)| ≈ CR as R → ∞. There cannot exist a limiting solution. In an interesting recent work [55, 56], the authors have successfully classified all stationary solutions to the 2D Euler equation, which has the form
u = ∇⊥ (rλ Ψ(θ)) for some λ ∈ R, and with some regularity assumption on Ψ. The above stationary solutions correspond to the case λ = 2 but they have escaped the classification in [55, 56]; our understanding is that the authors work under the situation where u ∈ C 1 on the unit circle (which is natural to do in their framework), while our solutions only satisfy u ∈ C 0,1 . Taking any linear combination of the above two examples, we get another class of interesting solutions, which simply rotates around the origin with constant angular speed. Example 3.2.5 (Rotating solutions). For m ≥ 3, take any two constants c1 and c2 , and define using polar coordinates
ω(r, θ) =
c1 c2
m−1 if θ ∈ ∪j=0 [2jπ/m, (2j + 1)π/m) ,
otherwise .
on R2 . Then, the resulting unique solution simply rotates. This is based on the previous example and the following simple fact: if (u, ω) be a solution of (1.1), then for any constant c ∈ R, −1 ω ˜ (t, x) := ω(t, Oct x) + 2c
(3.7)
−1 u ˜(t, x) := Oct u(t, Oct x) + cx⊥
is also a solution, where Oθ denotes the matrix of counterclockwise rotation by angle θ. Remark 11. In all of the above examples, the solutions are well-defined in terms of the Yudovich theory on the disk B(0, R) for any R > 0. In this setting, it is an easy matter to show that the particle trajectories of these Yudovich solutions converge to those of ours in the limit R → ∞ uniformly on compact sets.
56
3.2.2
Proof of the main result
The proof proceeds as follows. First, we deal with the issue of uniquely solving for the stream function Ψ in −∆Ψ = ω in our setup, which is more or less equivalent with uniquely recovering the velocity from the vorticity. The existence is then easily shown thanks to the Biot-Savart formula. Once this is done, we have a well-defined vorticity formulation of the 2D Euler equation, and we first show uniqueness, using the Osgood type uniqueness for ODEs. Then the existence can be actually shown along the similar lines. These arguments are indeed an adaptation of an elegant proof of the Yudovich theorem given in the book [59, Chap. 2]. Let us actually prove a slightly more general version of the uniqueness result than we need, which hopefully helps to clarify the situation. Lemma 3.2.6 (Uniqueness of the Poisson problem). Assume that a function Ψ : R2 → R in ˙ 2,p (R2 ) for some 1 < p ≤ ∞ satisfies W • ∆Ψ = 0, • for all x ∈ R2 , |Ψ(x)| ≤ C|x|2 (1 + |x|1− ) for some , and • for all sufficiently large R > 0, Z
π
Z
π
Ψ(Rθ) exp(±iθ)dθ = 0 = −π
Ψ(Rθ) exp(±2iθ)dθ . −π
Then Ψ is identically zero. In particular, this implies that the Poisson problem
∆Ψ = ω
(3.8)
is uniquely solvable in the class of functions that grows at most quadratically and has m-fold rotational symmetry, up to a constant. Proof. For each R > 0 (which is assumed to be sufficiently large), we have the representation formula R2 − |x|2 Ψ(x) = 2πR
Z ∂B(0,R)
1 Ψ(y)dS(y) . |y − x|2
We will keep subtracting zeroes to gain more and more decay in y in the above formula.
57
First, from Ψ(0) = 0, we may subtract an appropriate multiple of Ψ(0) to obtain R2 − |x|2 Ψ(x) = 2πR
Z ∂B(0,R)
|y|2 − |y − x|2 Ψ(y)dS(y) . |y − x|2 |y|2
Note that the kernel now decays as R−3 for a fixed x. Next, from the vanishing condition Z ∂B(0,R)
y Ψ(y)dS(y) = 0 , |y|4
we further rewrite R2 − |x|2 2πR
|x|2 − 2x · y 2x · y − Ψ(y)dS(y) , 2 2 |y|4 ∂B(0,R) |y − x| |y| 2 2 Z |x| |y| + 2|x|2 x · y − 4(x · y)2 R2 − |x|2 Ψ(y)dS(y) . = 2πR |y − x|2 |y|4 ∂B(0,R)
Ψ(x) =
Z
Finally, from the vanishing of second Fourier modes, Z ∂B(0,R)
2y1 y2 Ψ(y)dS(y) = |y|6
Z ∂B(0,R)
y12 − y22 Ψ(y)dS(y) = 0 . |y|6
Hence, we can subtract appropriate multiples of the above to modify the kernel |x|2 |y|2 + 2|x|2 x · y − 4(x · y)2 2|x|2 (y 2 − y22 )(x21 − x22 ) 2x1 x2 y1 y2 + + 1 − . 2 4 4 |y − x| |y| |y| |y|6 |y|6 A direct computation shows that this expression decays as R−5 . Therefore, we bound
|Ψ(x)| ≤ 10CR
Z ∂B(0,R)
|x|3 3− 1 R dS(y) ≤ C 0 |x|3 , 5 R R
since |Ψ(y)| ≤ CR3− on ∂B(0, R). Fixing x and taking R → +∞ finishes the proof. Lemma 3.2.7. Assume that ω ∈ L∞ (R2 ) and satisfies Z
π
Z
π
ω(Rθ) exp(±iθ)dθ = 0 = −π
ω(Rθ) exp(±2iθ)dθ . −π
58
Then, the principal value 1 u(x) = lim R→∞ 2π
Z |y|≤R
(x − y)⊥ ω(y)dy |x − y|2
(3.9)
is pointwise well-defined, with a bound
(3.10)
|u(x)| ≤ CkωkL∞ |x| .
Proof. We consider two regions: when y satisfies |y| ≤ 2|x| and |y| > 2|x|. In the first region, a brute force bound gives
|u(x)| ≤ CkωkL∞
Z |y|≤2|x|
1 dy ≤ C|x|kωkL∞ . |x − y|
(3.11)
In the latter region, we proceed exactly as in the previous lemma; from the vanishing of first Fourier modes, we rewrite it as 1 p.v. 2π
Z |y|>2|x|
(x − y)⊥ y⊥ + 2 ω(y)dy , |x − y|2 |y|
(3.12)
and note that the first component equals 1 p.v. 2π
Z |y|>2|x|
y2 (x21 + x22 ) − x2 (y12 − y22 ) + 2x1 y1 y2 ω(y)dy , |x − y|2 |y|2
so that the principal value makes sense once we have vanishing of the second Fourier modes. A similar conclusion holds for the second component of the velocity. Subtracting appropriate quantities and directly integrating in y gives the desired bound |u(x)| ≤ CkωkL∞ |x|. Lemma 3.2.8. Under the same assumptions on ω, we have a log-Lipschitz estimate of the form
0
0
|u(x) − u(x )| ≤ CkωkL∞ |x − x | ln
c max(|x|, |x0 |) |x − x0 |
.
(3.13)
Proof. We start with the expression 1 u(x) − u(x ) = p.v. 2π 0
Z R2
(x0 − y)⊥ (x − y)⊥ − 0 ω(y)dy , |x − y|2 |x − y|2
59
(3.14)
and assume |x| ≥ |x0 |. We split the integration into 3 domains, A = {y : |x − y| ≤ 2|x − x0 |}, C = {y : |y| > 3|x|}, and B = R2 \(A ∪ C). In the region A, a brute force bound on the kernel gives a contribution
CkωkL∞ |x − x0 | ,
(3.15)
and in the annulus-shape region B, due to the singular nature of the kernel ∇K we obtain a logLipschitz contribution
0
CkωkL∞ |x − x | ln
c|x| |x − x0 |
.
(3.16)
Finally, in the region C, we re-write each of K(x, y) and K(x0 , y) as done in the previous lemma to gain decay in y, and combine the resulting expressions to obtain a bound of the form
C|x − x0 |
|x| |y|3
(3.17)
in the region C. Integrating in y completes the proof. In particular, when ω is bounded and m-fold rotationally symmetric for some m ≥ 3, there exists a velocity vector field which grows at most linearly and log-Lipschitz continuous. From the convergence of the Biot-Savart law, it follows that ∇ · u = 0 and ∇ × u = ω in the sense of distributions. Integrating this velocity vector field in space, one obtains the existence to the Poisson problem (3.8), which actually satisfies the growth condition |Ψ(x)| ≤ C|x|2 . Remark 12. The situation is different when we do not have vanishing of the second Fourier modes (equivalently, in the case of 2-fold rotational symmetry). Explicitly, for
ω(y1 , y2 ) = α1
y1 y2 y12 − y22 + α , 2 |y|2 |y|2
(which is bounded) we have the following solutions to the Poisson problem
Ψ(y1 , y2 ) = α1 y1 y2 ln |y| + β1 y1 y2 + α2 (y12 − y22 ) ln |y| + β2 (y12 − y22 ) . These examples were presented in [34]. 60
(3.18)
Remark 13. We have seen that in the frozen-time case, the necessary condition is vanishing of the first two Fourier modes. Unfortunately, this information does not propagate under the Euler dynamics, in general. Therefore we really have to stick to the m-fold rotationally symmetric assumption, with some m ≥ 3. For an explicit example, take the following vorticity configuration:
ω0 (r, θ) =
1 0
if θ ∈ [−a, a] ∪ [−a + π/2, a + π/2] , otherwise .
This satisfies the vanishing assumptions, while u0 · ∇ω0 does not. Note that this example was also presented in [34]. We have seen that the vanishing of the first two Fourier modes gives the decay |x|−3 of the Biot-Savart kernel. Indeed, upon assuming further vanishing conditions, one continues to gain extra decay by subtracting appropriate quantities. Of course, it is more efficient to prove it by explicitly symmetrizing the kernel when we assume m-fold symmetry. More formally, one has: Lemma 3.2.9 (see [34]). Let K denote the Biot-Savart kernel 1 (x − y)⊥ . 2π |x − y|2
K(x, y) :=
Then for each m ≥ 1, the m-fold symmetrization in y gives the decay |y|−m : m
1 X Pm (x, y) i K(x, O2π/m y) = Qm i 2 m i=1 i=1 |x − O2π/m y| where Pm (x, y) is a vector of homogeneous polynomials in the components of x and y of degree 2m − 1, which contains powers of y1 and y2 only up to the degree m. Proof. Fix some y0 6= 0. Then Pm (x, y0 ) =
∇⊥ x
m Y i=1
! |x −
i O2π/m y0 |2
,
where the function
Fy0 (x) :=
m Y i=1
i |x − O2π/m y0 |2
is a m-fold symmetric polynomial in x1 and x2 . Hence it only consists of terms which have degree 61
0, m, and 2m in x1 , x2 . The statement follows. As a corollary, we have Corollary 3.2.10. For m ≥ 3, define the kernel m
K (m) (x, y) =
1 X i K(x, O2π/m y). m i=1
Then, for |y| ≥ 2|x|, we have |K (m) (x, y)| ≤ Cm
|x|m−1 |y|m
for some constant Cm > 0. Moreover, for points y, y 0 satisfying |y|, |y 0 | ≥ 2|x| and |y − y 0 | ≥ |x|, |x|m−1 (m) K (x, y) − K (m) (x, y 0 ) ≤ Cm |y − y 0 | m+1 . |y| We now turn to the task of obtaining uniqueness of the Euler solution. We will reduce it to the following Osgood uniqueness condition for a suitable quantity. Lemma 3.2.11. If a continuous function f : [0, ) → R≥0 satisfy f (0) = 0 and f (t) ≤ C
Z
t
f (s) ln(1 + 0
1 )ds f (s)
(3.19)
then f ≡ 0. Proof. See for example the book of Marchioro and Pulvirenti [59, p. 68]. For simplicity, let us set ρ(a) := a ln(1 + 1/a) for a ≥ 0. Lemma 3.2.12. Given ω0 ∈ L∞ (R2 ) with 4-fold rotational symmetry, there is at most one solution ∞ of the 2D Euler equation (in vorticity formulation) with ω ∈ L∞ t Lx with 4-fold rotational symmetry.
The assumption that ω is 4-fold symmetric is just for concreteness and simplicity of notation; a similar argument goes through for any m ≥ 3. ˜ A simple observation is that Proof. Assume there exist two solution triples, (ω, u, Φ) and (˜ ω, u ˜, Φ). we have a linear bound
|Φ(t, x)| ≤ |x| +
Z 0
t
|u(s, Φ(s, x))|ds ≤ |x| + Ckω0 kL∞ 62
Z 0
t
|Φ(s, x)|ds .
(3.20)
Proceeding similarly for the difference |Φ(t, x) − x| and using Gronwall’s inequality, we get that 1−≤
˜ x) − x| |Φ(t, x) − x| |Φ(t, , ≤1+ |x| |x|
(3.21)
and hence ˜ t (x)| |Φt (x) − Φ ≤ 2 |x|
(3.22)
for all 0 ≤ t ≤ T = T (), uniformly in x. From now on, we restrict ourselves to such a time interval, with a choice = 1/10 (say). We want to close an estimate of the form in Lemma 3.2.11 in terms of the quantity Z f (t) := R2
˜ t (x)| |ω0 (x)| |Φt (x) − Φ · dx . |x| 1 + |x|3
(3.23)
Note that |ω0 (x)|/(1 + |x|3 )dx is a finite measure, so f is finite. We then write ˜ t (x) = Φt (x) − Φ
Z tZ 0
R2
h i ˜ s (y)) ω0 (y)dyds K(Φs (x) − Φs (y)) − K(Φs (x) − Φ
Z
t
+ 0
(3.24)
˜ s (x))ds , u ˜s (Φs (x)) − u ˜s (Φ
and integrate against 1 |ω0 (x)| dx . |x| 1 + |x|3 ˜ s (x)| ≈ |x|, we bound the second term in terms From the log-Lipschitz bound on u ˜, and |Φs (x)| ≈ |Φ of Z tZ 0
Z t ˜ s (x)| |Φs (x) − Φ c|x| |ω0 (x)| ln dxds ≤ Cρ(f (s))ds , ˜ s (x)| 1 + |x|3 |x| |Φs (x) − Φ 0
where we have used the Jensen’s inequality with respect to the finite measure precisely the bound we want.
63
|ω0 (x)| 1+|x|3 dx.
This is
Turning to the first term, we have Z
t
ds 0
Z Z h y
x
˜ s (y)) K(Φs (x) − Φs (y)) − K(Φs (x) − Φ
i 1 |ω (x)| 0 dx ω0 (y)dy . |x| 1 + |x|3
(3.25)
For each fixed y, once we obtain an estimate of the form Z h i 1 |ω (x)| 0 ˜ K(Φs (x) − Φs (y)) − K(Φs (x) − Φs (y)) dx ω0 (y) 3 |x| 1 + |x| x ˜ y)| |Φ(t, y) − Φ(t, |ω0 (y)| c|y| ≤C , ln ˜ |y| |Φ(t, y) − Φ(t, y)| 1 + |y|3 we obain a bound C
Rt 0
(3.26)
ρ(f (s))ds, and this completes the proof.
We consider 4 regions. First, A = {x : |x| ≤ |y|/10}, D = {x : |x| > 10|y|}. Then set ˜ y)|} and C = R2 \(A ∪ B ∪ D). Regions A, B, D do not overlap B = {x : |x − y| ≤ 3|Φ(t, y) − Φ(t, and C is an annulus-shape domain. (i) Region A We symmetrize each of the kernels in y. Then, combining two fractions together, we may pull ˜ y)|. Then note that each factor in the denominator is bounded below out a factor of |Φ(t, y) − Φ(t, by a constant multiple of |x − y|, which is in turn bounded below by a multiple of |y| in this region. Therefore, we obtain a bound of the form (see Lemma 3.2.9) Z 14 3 11 1 |ω0 (x)| ˜ y)| |x| + |x| |y| C|Φ(t, y) − Φ(t, dx |ω0 (y)| . |y|16 |x| 1 + |x|3 A
(3.27)
This is integrable in x, and integrating gives a bound
C
˜ y)| |ω0 (y)| |Φ(t, y) − Φ(t, . |y| 1 + |y|3
(3.28)
(ii) Region D This time, we symmtrize the kernel in x. Proceeding similarly as in the region A, we obtain a bound Z D
˜ y)| |Φ(t, y) − Φ(t,
|y|14 + |y|2 |x|12 1 |ω0 (x)| dx |ω0 (y)| . |x|16 |x| 1 + |x|3
64
(3.29)
This is integrable in x, and clearly we have Z D
|y|14 + |y|2 |x|12 |y| 1 + |y|3 dx ≤ C . |x|16 |x| 1 + |x|3
This results in the same bound as in the region A. (iii) Regions B and C In this case, we have |x| ≈ |y|, so we can freely interchange the powers of |x| with |y| in the denominator. We then estimate the kernel exactly as in the proof of the log-Lipschitz bound of u(x). Collecting the bounds, we obtain (3.26). Hence, we have obtained
f (t) ≤ C
t
Z
ρ(f (s))ds , 0
on some time interval t ∈ [0, T ] with T depending only on m and kω0 kL∞ . Lemma 3.2.11 guarantees that f ≡ 0 on [0, T ], and since kω0 kL∞ = kωT kL∞ , this argument can be extended to any finite time. ˜ x) coincide on the support Hence, we have shown that the particle trajectories Φ(t, x) and Φ(t, ˜ x) everywhere). of ω0 , for all time. This trivially implies that ω ≡ ω ˜ (and therefore, Φ(t, x) ≡ Φ(t, The proof is now complete. Proof of Theorem 3.2.1. It only remains to establish the existence. A particularly nice feature of the proof in [59], which we have adopted here, is that the existence is shown in a completely parallel manner as the uniqueness. We only recall the main steps. (i) Construction of the approximate sequence Starting with the initial value ω (0) (t, x) := ω0 (x), we inductively define (n)
(n−1)
ut (x) := p.v. (K ∗ ωt )(x) , d (n) (n) (n) Φ (x) := ut ◦ Φt (x) , dt t (n) (n) ωt (x) := ω0 ◦ (Φt )−1 (x) .
(3.30)
Our previous lemmas guarantee that each of these definitions makes sense. It is important that the symmetry property remains valid at each step of the iteration. (ii) Convergence of the sequence 65
Define for n ≥ 1, δ n (t) :=
(n)
(n−1)
|Φt (x) − Φt |x|
Z R2
(x)|
·
|ω0 (x)| dx . 1 + |x|3
The same argument shows that this quantity is finite and satisfies the inequality
δ n (t) ≤ C
Z
t
ρ(δ n (s))ds + C
0
Z
t
ρ(δ n−1 (s))ds .
0
Upon introducing
δ¯N (t) := sup δ n (t) , n>N
we obtain
δ¯N (t) ≤ C
Z
t
ρ(δ¯N −1 (s))ds .
0
This is sufficient to show that
lim δ¯N (t) → 0 ,
N →∞
uniformly in some short time interval [0, T ]. This length of the time interval depends only on kω0 kL∞ and m, so that this argument extends to any finite time. (iii) Properties of the limit We have thus shown that, for each fixed time t ∈ R+ , there exists a map Φt defined on the support of ω0 , satisfying Z R2
(n)
|Φt (x) − Φt (x)| |ω0 (x)| · dx −→ 0 , |x| 1 + |x|3
n→∞.
Then set
ωt (x) =
ω0 (Φ−1 (x)) t
0
if x = Φt (y) for some y ∈ supp(ω0 ) , otherwise .
This vorticity is m-fold rotationally symmetric and bounded. Therefore, we may define ut and then
66
Φt on the entire plane. It is direct to show that this map is a measure preserving homeomorphism for each time moment. The triple (ωt , ut , Φt ) solves the 2D Euler equation and satisfies our assumptions. This finishes the proof. Example 3.2.13. In this example, we collect several situations where Theorem 3.2.1 applies. 1. (Torus) Consider the 2D torus T2 = [−π, π)2 and assume that the initial vorticity ω0 ∈ L∞ (T2 ) satisfies the symmetry ω0 (x1 , x2 ) = ω0 (−x2 , x1 ) = ω0 (−x1 , −x2 ) = ω0 (x2 , −x1 ). Then, we may identify such an initial data with a vorticity defined on R2 which has 4-fold rotational symmetry around the point (0, 0). 2. (Square and equilateral triangle) Consider the 2D Euler equation on the unit square = [0, 1]2 with slip boundary conditions, i.e., u · n = 0 on the boundary where n is the unit normal vector. Assume that we are given ω0 ∈ L∞ () which is odd across the diagonal, that is, ω0 (x1 , x2 ) = −ω0 (x2 , x1 ) for x1 , x2 ∈ [0, 1]. One may then extend it as an odd function with respect to all the sides of the square to obtain a 4-fold symmetric initial vorticity on R2 . A similar procedure can be done for the case of an equilateral triangle. Here, we obtain a 6-fold symmetric vorticity on the plane. It is interesting to note that in these cases, assuming that the mean of vorticity is zero on the periodic domain, the velocity is actually bounded on R2 and therefore our solutions coincide with Serfati’s. 3. (Rational sectors) This time, consider the 2D Euler equation on the sector Sπ/m = {(r, θ) : 0 ≤ r < ∞, 0 ≤ θ ≤ π/m} for m ≥ 3 with slip boundary conditions. Given ω0 ∈ L∞ (Sπ/m ), we extend it as an odd function onto the whole plane across the boundaries; i.e. ω0 (r, θ) = −ω0 (r, θ + π/m) for all r, θ. Then, we obtain an m-fold symmetric vorticity. Remark 14. In all of the above examples, it is not hard to show that if we have additional regularity of the initial data, e.g. ω0 ∈ C k,α (D) for D ∈ {T2 , , Sπ/m }, then this regularity propagates by the Euler equation, even though the odd extension onto R2 could be discontinuous across the symmetry axis. It is well known that if initially ω0 ∈ C 0,1 (R2 ) then the maximum of the gradient can grow at most double exponentially in time. This double exponential bound can be excluded, at least at the
67
origin, in all of the above situations; indeed, this is a direct consequence of the estimate
|u(x)| ≤ Ckω0 kL∞ |x| which says that no fluid particle cannot approach the origin faster than the exponential rate. This recovers some of the very recent results of Itoh, Miura, and Yoneda [46, 45]. Corollary 3.2.14 (see [46, 45]). Assume that we are in one of the above domains and the initial vorticity ω0 satisfies the required symmetry assumptions. If in addition ω0 is Lipschitz, then we have the following exponential bound on the gradient at the origin for all time:
sup x6=0
3.2.3
|ω(t, x) − ω(t, 0)| ≤ k∇ω0 kL∞ exp(ckω0 kL∞ t) . |x|
Propagation of the angular regularity
In this subsection, we show well-posedness of the 2D Euler equation in certain scaling invariant spaces, which encodes regularity in the angle direction. Let us use the notation
kωkC˚α (R2 ) := kωkL∞ (R2 ) + k|x|α ωkC˙ α (R2 )
(3.31)
for 0 < α < 1 and the endpoint case is simply
kωkC˚0,1 (R2 ) := kωkL∞ (R2 ) +
sup
(|x||∇ω(x)|) .
x∈R2 \{0}
(3.32)
˚k,α can be defined for k ≥ 1: first when 0 < α < 1, Higher order norms C kωkC˚k,α (R2 ) := kωkC˚k−1,1 (R2 ) + k|x|k+α ∇k ωkC α (R2 )
(3.33)
and
kωkC˚k,1 (R2 ) := kωkL∞ (R2 ) +
sup x∈R2 \{0}
|x|k+1 |∇k+1 ω(x)| .
(3.34)
Here ∇d ω(x) is just a vector which consists of all expressions of the form ∂i1 · · · ∂id ω(x) for (i1 , · · · , id ) ∈ {1, 2}d . It is clear that these spaces deal with angular regularity. Indeed, in the ideal case when ω
68
depends only on the angle, i.e. ω(r, θ) = h(θ) for some profile h defined on the unit circle,
kωkC˚k,α (R2 ) ≈ khkC k,α (S 1 ) . ˚k,α (R2 ) then ω0 is actually C k,α away from the origin. Note that if we have initial data ω0 ∈ C This information propagates in time; the solution constructed in the previous subsection remains C k,α away from the origin for all time. Of course it needs to be proved that ω(t, ·) actually belongs ˚k,α (R2 ). Not surprisingly, the bound turns out to be double exponential in time. to C ˚k,α is (m + k)-fold rotationally symmetric with some m ≥ 3, Theorem 3.2.15. Assume that ω0 ∈ C with k ≥ 0 and 0 < α ≤ 1. Then, the unique solution ω(t, x) ∈ L∞ ([0, ∞); L∞ (R2 )) belongs to ˚k,α with a bound L∞ loc C
kω(t)kC˚k,α ≤ C exp(c1 exp(c2 t))
(3.35)
with constants depending only on k, α, and kω0 kC˚k,α . In the proof, we shall need the following simple calculation: Lemma 3.2.16. For any 0 < α ≤ 1, we have a bound k∇uk
L∞
≤ Cα kωk
L∞
kωkC˚α 1 + ln 1 + cα , kωkL∞
(3.36)
with some constants cα , Cα depending only on 0 < α ≤ 1. Proof. We recall that each entry of the matrix ∇u(x) has an explicit representation (see [10] for instance) involving a linear combination of the expressions Z p.v. R2
(x1 − y1 )(x2 − y2 ) ω(y)dy , |x − y|4
Z p.v. R2
(x1 − y1 )2 − (x2 − y2 )2 ω(y)dy |x − y|4
and a constant multiple of ω(x). Let us first deal with the first expression, restricting ourselves to the Lipschitz case α = 1. We split R2 into the regions (i)|x − y| ≤ l|x|, (ii) l|x| < |x − y| ≤ 2|x|, and (iii) 2|x| < |x − y|, where l ≤ 1/2 is a number to be chosen below. In the third region, we use symmetry of ω to gain integrable
69
decay of the kernel, which results in a bound of the form CkωkL∞ . In the first region, we may rewrite Z p.v. R2
(x1 − y1 )(x2 − y2 ) (ω(y) − ω(x))dy , |x − y|4
and the given Lipschitz bound implies
sup
Cl|x|
y:|x−y|≤l|x|
|∇ω(y)| ≤ Cl sup |z||∇ω(z)| ,
which together gives a bound Z p.v. |x−y|≤l|x|
C k| · |∇ω(·)kL∞ dy ≤ Clk| · |∇ω(·)kL∞ . |x||x − y|
Lastly, in the second region we directly integrate to obtain a bound
CkωkL∞ ln
c l
.
Optimizing l := min
kωkL∞ 1 , 2 sup |z||∇ω(z)|
establishes the claimed bound. The other expression can be treated in a similar fashion. Then the C α bound may be obtained in a parallel manner: one just use the H¨older assumption on the region (ii) to remove the singularity, and then optimize in l accordingly. Proof of Theorem 3.2.15. We first deal with the case k = 0. To establish the double exponential growth rate in time, it is most efficient to pass to the Lagrangian formulation directly (see the introduction of [50] for instance). Starting with d Φ(t, x) = u(t, Φ(t, x)) , dt
70
we obtain d (Φ(t, x) − Φ(t, x0 )) ≤ k∇u(t)kL∞ |Φ(t, x) − Φ(t, x0 )| dt ≤ 1 + ln(1 + kω(t)kC˚0,α ) |Φ(t, x) − Φ(t, x0 )| , assuming kωkL∞ = 1 for simplicity. Integrating, e−
Rt 0
1+ln(1+kω(s)kC ˚0,α )dt
R Φ(t, x) − Φ(t, x0 ) ≤ e 0t 1+ln(1+kω(s)kC˚0,α )dt , ≤ 0 x−x
and it is clear that the same upper and lower bounds are available for the inverse of the flow map Φ−1 t . Given the above bound, we estimate |ω(t, x) − ω(t, x0 )| , |x − x0 |α and we will assume for simplicity of the argument that |x| and |x0 | are comparable; |x0 |/2 ≤ |x| ≤ 2|x0 |. Then, from the transport formula ω(t, x) = ω0 (Φ−1 t (x)), |ω(t, x) − ω(t, x0 )| ≤ kω0 kC˚0,α
−1 0 α |Φ−1 t (x) − Φt (x )| . −1 0 α min(|Φt (x)|α , |Φ−1 t (x )| )
Ct Recall that ckω0 kL∞ e−Ct ≤ |Φ−1 and similar for x0 . Therefore |Φ−1 t (x)| ≤ ckω0 kL∞ e t (x)| and 0 Ct |Φ−1 t (x )| are comparable up to a factor of e , and from the above bound for inverse particle
trajectories, we obtain
|x|α |ω(t, x) − ω(t, x0 )| ≤ Cα |x − x0 |α kω0 kC˚0,α ecα
Rt 0
1+ln(1+kω(s)kC ˚0,α )dt
.
At this point it is not hard to show that the desired double exponential bound holds. Now we set k = 1. It suffices to establish the corresponding estimate from Lemma 3.2.16 with ∇u, ω replaced by ∇2 u, ∇ω respectively. We argue as in the proof of Lemma 3.2.16 but ∇ω is not rotationally symmetric anymore. Instead, we recall that to gain integrable decay in the kernel, it is sufficient for ∇ω to be orthogonal with respect to y1 y2 and y12 − y22 on large circles. To this end, we
71
simply compute Z 1 y1 y2 y1 y2 ∇ω(y)dy = − y1 y2 ω(y)dy = 0 ω(y)dy + |y| y B0 (R) B0 (R) ∂B0 (R) y1 2
Z
Z
for all R > 0, once we assume that ω(y) is rotationally symmetric for some m ≥ 4. The case k > 1 can be treated in a strictly analogous manner.
3.3
Radially homogeneous solutions to the 2D Euler
In this section, we investigate the evolution of the class of radially homogeneous solutions to the 2D Euler equation. More concretely, we are interested in solutions which satisfy
ω(r, θ) = ω(1, θ) ,
u(r, θ) = ru(1, θ) ,
Ψ(r, θ) = r2 Ψ(1, θ)
(3.37)
for all time, again with some m-fold rotational symmetry where m is greater than or equal to 3. A simple scaling analysis shows that the above factors of r are the only possible set of degrees of homogeneity which may be propagated by Euler. Indeed, observe that if the velocity is radially homogeneous with degree 1, it maps any line through the origin to another such line, and this keeps the degree 0 homogeneity assumption for the vorticity. In turn, this ensures that the velocity remains homogeneous with degree 1, explicitly by the Biot-Savart law. Therefore, one sees that the 2D Euler equation reduces to a 1D dynamical system defined on the unit circle, which is not volume (length) preserving in general, and hence non-trivial. This 1D system is easily shown to be well-posed (in L∞ and C α for instance), and the resulting solution provides the unique solution to Euler via (3.37). A crucial property of this model is that the corresponding angular velocity is smoother than the advected scalar by degree 2, which is striking in view of the relation ∇ × u = ω. Indeed, when we impose the radial homogeneity assumption on 2D Euler, the velocity vector field decomposes into the angular part and the radial part, and the latter is indeed only one degree smoother than the vorticity, but of course the 1D model is not affected by the radial velocity at all; see the expression in (3.43). In the following, we first introduce the system and collect a few simple a priori bounds. Then we proceed to check that its solution actually gives a radially homogeneous solution to the 2D Euler equation. 72
3.3.1
The 1D system
Consider the following transport system defined on the circle S 1 = {θ : −π ≤ θ < π}: ∂t h(t, θ) + 2H(t, θ)∂θ h(t, θ) h(0, θ)
=0,
(3.38)
= h0 (θ) ,
with h0 ∈ L∞ (S 1 ) and m-fold rotationally symmetric for some m ≥ 3. Here, H(t, ·) is the unique solution of 1 2π
h(t, θ) = 4H(t, θ) + H 00 (t, θ) ,
Z
π
H(θ) exp(±2iθ)dθ = 0 .
(3.39)
−π
Hence H has the same rotational symmetry as h, and therefore a solution to (3.38) stays m-fold rotationally symmetric for all t > 0. Here and in the following, given an integrable function on S 1 = [−π, π), we use the convention that its Fourier coefficients are given by the formula 1 fˆk = 2π
π
Z
f (θ) exp(ikθ)dθ . −π
Before we state and prove a few a priori inequalities, let us demonstrate that the transformation in (3.39) which sends h to H is given by a simple and explicit kernel. Lemma 3.3.1 (Biot-Savart law). We have
H(θ) =
1 2π
Z
π
−π
¯ θ)dθ ¯ KS 1 (θ − θ)h( ,
(3.40)
where
KS 1 (θ) :=
θ 1 1 π sin(2θ) − sin(2θ)θ − cos(2θ) . 2 |θ| 2 8
(3.41)
Proof. It suffices to check that the function KS 1 has Fourier coefficients 1/(4 − k 2 ) for |k| = 6 2, and 0 when |k| = 2. This is a simple computation. Indeed, one can easily arrive at this expression by observing that 1 1 = 2 4−k 4
1 1 + 2−k 2+k 73
1.0
0.5
-3
-2
-1
1
2
3
Figure 3.1: The graph of KS 1 and that πsign(θ) − θ has Fourier coefficients 1/(ik). Remark 15. The expression (3.40) is truly the Biot-Savart law for the 2D Euler equation restricted to radially homogeneous data. Actually, it is possible to derive the expression (3.41) by starting with the usual Biot-Savart law (where x ∈ R2 corresponds to (r, θ)): H(θ) =
1 1 − sin θ 1 1 u(x) · p.v. = 2 r 4π r cos θ
Z R2
(x − y)⊥ − sin θ ω(y)dy · , 2 |x − y| cos θ
re-writing in polar coordinates, and then explicitly integrating out the radial variable. Lemma 3.3.2 (a priori inequalities). Assume that h(t, θ) is a smooth solution to (3.38). For each fixed time, we have
kHkW n+2,∞ ≤ Cn khkW n,∞ ,
kHkW n+1,∞ ≤ Cn khkW n,1 .
Next, d kh(t)kL∞ = 0 , dt
d kh(t)kL1 ≤ Ckh0 kL∞ kh(t)kL1 , dt
d kh(t)kL1 ≤ ckh(t)k2 1 . L dt
Moreover, if initially the data is nonnegative, then kh(t)kL1 = kh0 kL1 . Lastly, the gradient may grow at most exponentially in time; d k∂θ h(t)kL∞ ≤ Ckh0 kL∞ k∂θ h(t)kL∞ . dt 74
Proof. We check the first statement for n = 0. The case n > 0 can be treated similarly. From (3.40), it is clear that
kH 0 kL∞ ≤ πkKS 1 kLip khkL1 . Since we have trivially kHkL∞ ≤ CkhkL1 ≤ C 0 khkL∞ , kH 00 kL∞ ≤ khkL∞ + 4kHkL∞ ≤ CkhkL∞ . We define the flow map on S 1 , by solving d φ(t, θ) = 2H(t, φ(t, θ)) , dt
φ(0, θ) = θ .
We therefore have d h(t, φ(t, θ)) = 0 , dt and the L∞ norm of h is conserved in time. To derive statements regarding its L1 norm, we multiply both sides of (3.38) by sign(h) and integrate in space to obtain d dt
Z S1
|h(t, θ)|dθ = −
Z S1
2H∂θ |h|dθ =
Z S1
2H 0 |h|dθ ,
and the inequalities follow from our previous bounds. If h0 ≥ 0, then non-negativity is preserved by the flow, and integrating (3.38) in space gives that d dt
Z
Z h(t, θ)dθ =
S1
0
Z
2H hdθ = S1
0
0
00
Z
8H H + 2H H = S1
S1
0 4(H 2 )0 − (H 0 )2 dθ = 0 ,
which shows that the L1 norm is conserved. This also states that we may assume that the mean of h0 is zero, without loss of generality. The dynamics in the general case may be recovered by means of the transformation (3.7). Regarding the last statement, by differentiating (3.38), we obtain
(∂t + 2H(θ)∂θ ) h0 (θ) = −2H 0 (θ)h0 (θ) .
75
(3.42)
Composing with the flow map and taking the L∞ -norm of both sides gives d 0 kh (t, ·)kL∞ ≤ 2kH 0 (t, ·)kL∞ kh0 (t, ·)kL∞ . dt This finishes the proof. Remark 16. The flow homeomorphisms Φ(t, ·) : R2 → R2 for all t ≥ 0 are well-defined as biLipschitz maps in this setting. Note also that the 1D flow φ(t, ·) is “volume preserving” if and only if 0 = H 0 (θ), and hence the Lp -norms for h will not be conserved for p < ∞ in general. Proposition 3.3.3. The system (3.38) is well-posed in h ∈ L∞ ([0, ∞), L∞ ) and in L∞ ([0, ∞), C k,α ) for any k ≥ 0 and 0 ≤ α ≤ 1. The solution gives the unique solution to 2D Euler by setting ω(t, r, θ) = h(t, θ) ,
r cos θ −r sin θ 0 u(t, r, θ) = 2H(t, θ) , − H (t, θ) r sin θ r cos θ
(3.43)
Ψ(t, r, θ) = r2 H(θ) . Proof. It is not difficult to show that the system is globally well-posed, first in the case h0 ∈ L∞ . Indeed, h ∈ L∞ guarantees that H is Lipschitz (indeed, even H 0 is Lipschitz), which allows us to solve for the flow map. A standard iteration scheme will give existence. The resulting solution is global thanks to the conservation of h in L∞ . Uniqueness can be shown along similar lines. Given the L∞ well-posedness, the corresponding statement in C k,α is direct to verify. We now check that it provides a solution to Euler. Indeed, from (3.43), we see that the equation
∂t ω + (u · ∇)ω = 0 simply reduces to (3.38). Moreover, a direct computation gives
∇ × u = 4H + H 00 = ω ,
∇·u=0 ,
u = ∇⊥ Ψ .
(3.44)
This establishes the statement, as Theorem 3.2.1 states that the 2D Euler solution with such a data is unique. Example 3.3.4. We begin by noting that the examples in Section 3.2 defines either a stationary or
76
a purely rotating solution to the system (3.38). The conservation of L1 for non-negative data gives more examples of rotating solutions. Take some interval of length L less than 2π/m, and place them m-fold rotationally symmetric around the circle. Then, these patch solutions simply rotates. In general, if we add more intervals, then they may “exchange” lengths.
3.3.2
Trend to equilibrium under odd symmetry and positivity
Let us now study the evolution of the 1D system (3.38) in more detail. For concreteness, we will assume from now on that the vorticity is 4-fold rotationally symmetric:
h0 (θ) = h0 (θ + π/2) = h0 (θ + π) = h0 (θ + 3π/2) .
Therefore, we may view the solution as defined on [−π/4, π/4] as a periodic function. It turns out that once we impose odd symmetry and positivity, we can get a fairly satisfactory picture of the long-time dynamics. That is, we assume
h0 (θ) = −h0 (−θ) ,
h0 ≥ 0 on [0, π/4] .
First, it is easy to check from (3.38) and (3.39) that the odd symmetry is going to be preserved, and since the endpoints of the interval [0, π/4] are fixed, the solution remains non-negative. One motivation for imposing odd symmetry as well as positivity is that this scenario is expected to exhibit the fastest possible rate of gradient growth. Indeed, a very important consequence of the above extra assumptions is that there is a sign for the velocity as well. This drives all the fluid particles from (−π/4, π/4) towards the fixed point θ∗ = 0, which stretches the gradient at that point. In more detail, for fast gradient growth, it is necessary to have a lower bound on the velocity gradient, and H 0 measured in the maximum norm scales as the L1 -norm of h, and the effect of oscillations in the sign of h would be to simply reduce the magnitude of H 0 . We show that for such initial data h0 which is supported near the fixed point θ∗ = 0, the supnorm of the gradient h0 (t, ·) grows without bound as time goes to infinity, while the solution itself converges to the rest state in all Lp (S 1 ) with p < ∞. This excludes the exponential growth rate, which is the optimal possible one. To be clear, it does not follow that the exponential growth rate is impossible for any initial data.
77
1.5
0.4 1.0
0.3
0.5
0.2 0.1 0.5
1.0
1.5
0.5
-0.5
1.0
1.5
-0.1 -1.0
-0.2
-1.5
-0.3
Figure 3.2: Left: an example of “odd and positive” h0 , drawn on [0, π/2]. Right: associated graphs of H0 (solid line) and H00 (dotted line), drawn on the same interval. Lemma 3.3.5. If h is odd and non-negative on [0, π/4], we have H(t, ·) < 0 and H 00 (t, ·) > 0 on (0, π/4). In particular, H 0 is strictly increasing on (0, π/4) with H 0 (0) < 0 < H 0 (π/4). Proof. This immediately follows from a simple manipulation on the kernel (3.41). First, using the 4-fold rotational symmetry, we may consider the symmetrized kernel 3
1X π KS 1 (θ + jπ/2) = |sin(2θ)| . 4 j=0 8
KS1 1 (θ) =
Then, integrating against an odd function h, 4 H(θ) = 2π
Z
π/4
−π/4
KS1 1 (θ
8 − θ )h(θ )dθ = 2π 0
0
π/4
Z 0
π (|sin(2θ − 2θ0 )| − |sin(2θ + 2θ0 )|) h(θ0 )dθ0 8
and then it is sufficient to note that the function |sin(2θ − 2θ0 )| − |sin(2θ + 2θ0 )| is strictly negative on (θ, θ0 ) ∈ (0, π/4)2 . Indeed, on [0, π/4], K(θ, θ0 ) := |sin(2θ − 2θ0 )| − |sin(2θ + 2θ0 )| equals − cos(2θ) sin(2θ0 ) for θ ≥ θ0 and K(θ, θ0 ) = K(θ0 , θ). Hence, H 00 = h − 4H > 0 on (0, π/4). As an immediate corollary, we have the following classification of stationary solutions: Corollary 3.3.6. An odd, non-negative h0 ∈ L∞ (S 1 ) defines a stationary solution of (3.38) if and only if it equals a constant on (0, π/4). That is, there are no nontrivial such stationary solutions. Proof. The function h0 ∈ L∞ (S 1 ) defines a stationary solution if and only if H0 h00 = 0 in the sense of distributions. Assuming that h0 is not identically zero and non-negative on (0, π/4), we have H0 < 0 on (0, π/4) by Lemma 3.3.5, which means that h0 equals a constant almost everywhere on this interval. Theorem 3.3.7. Let 0 6= h0 ∈ W 1,∞ (S 1 ) be odd, non-negative on [0, π/4], and vanishes on [π/10, π/4]. Then the solution converges to the rest state, while the gradient grows with rates t−1 78
and t respectively: c C ≤ kh(t, ·)kL1 (S 1 ) ≤ , t t
ct ≤ kh0 (t, ·)kL∞ (S 1 ) ≤ Ct,
t≥1,
where c, C > 0 are constants which may depend only on h0 . Proof. We first note that the equivalence of norms
c1 kH 0 kL∞ (S 1 ) ≤ khkL1 (S 1 ) ≤ C1 kH 0 kL∞ (S 1 ) . holds. It suffices to establish the second inequality; since H 0 is increasing on (0, π/4),
kH 0 kL∞ = max (−H 0 (0), H 0 (π/4)) . Then
khkL1 < 4
π/4
Z
h − 4Hdθ = 4(H 0 (π/4) − H 0 (0)) ≤ 8kH 0 kL∞ .
0
Integrating the system (3.38) on (0, π/4) then gives d dt
Z
π/4
π/4
Z
−2H(t, θ)h0 (t, θ)dθ =
h(t, θ)dθ = 0
0 0
0
2
Z
π/4
2H 0 (t)(4H(t) + H 00 (t))dθ
0
2
= −(H (t, 0)) + (H (t, π/4)) . Under the given assumption that the support of h0 inside [0, π/4] is contained in [0, π/10], we have
0
−H (t, 0) =
Z 0
π/4
4 cos(2θ)h(t, θ)dθ ≥ 3
Z
π/4
sin(2θ)h(t, θ)dθ = 0
4 0 H (t, π/4) , 3
where we have used the formula
0
Z
H (θ) = sin(2θ) 0
θ 0
0
0
h(θ ) sin(2θ )dθ − cos(2θ)
Z
π/4
h(θ0 ) cos(2θ0 )dθ0
θ
which follows directly from differentiating the kernel expression for H(θ). Therefore,
−c|H 0 (t, 0)|2 ≥
d kh(t)kL1 ≥ −C|H 0 (t, 0)|2 , dt
79
and since H 0 (0) = kH 0 kL∞ is bounded below and above up to absolute constants from khkL1 , integrating in time gives 1 1 ≥ kh(t)kL1 ≥ −1 . ct + kh0 k−1 Ct + kh 0 kL1 L1 Then, to obtain the desired upper bound for kh0 kL∞ , we note that for all x ∈ (0, π/4) with h00 (x) 6= 0 we have from (3.42) that d ln (h0 ◦ φ(t, x)) ≤ 2kH 0 kL∞ ≤ C dt t for all large t > 0 uniformly in x. On the other hand, to get the lower bound we simply observe the inequality
kh0 (t)kL∞ ≥
kh0 k2L∞ ≥ ct . kh(t)kL1
To see this, one may take a point of maximum θmax of h(t) on [0, π/4] and since
h(t, θ) ≥ kh0 kL∞ − kh0 (t)kL∞ |θ − θmax | holds, integrating h(t) over the interval (θmax − kh0 kL∞ /kh0 (t)kL∞ , θmax + kh0 kL∞ /kh0 (t)kL∞ ) gives the lower bound for h in L1 . Remark 17. It is likely that a more careful analysis can establish the above statements for all initial data which is odd and non-negative on (0, π/4). We do not dwell on this issue here. Sharp gradient growth in the presence of a boundary Inspired by the work of Kiselev and Sverak [50], let us demonstrate that when our domain has a boundary, we can achieve the sharp growth rate. The growth in our case actually occur away from the boundary but its role is to keep enough L1 -mass for all time, which in turn guarantees the uniform rate of growth. We consider the problem (3.38) on the compact interval Q := {−π/4 ≤ θ ≤ π/4}, where the
80
endpoints are not identified with each other. If we again consider the class of initial data on Q with odd symmetry around θ∗ = 0, then the system on Q is well-posed in h(t, ·) ∈ L∞ (Q) simply because it exactly corresponds to solving (3.38) on the whole circle with the rotationally extended initial data
˜ 0 (θ) = h0 (θ + kπ/2) , h
k∈Z.
Similarly, it can be shown that if h0 ∈ W n,∞ (Q) then we have (h, H) ∈ W n,∞ (Q) × W n+2,∞ (Q) for all time. Together with the odd symmetry assumption, this provides a unique solution to the 2D Euler equation on the sector
˜ := {(r, θ) ∈ R2 : 0 ≤ θ ≤ π/2} , Q which satisfy the slip boundary conditions. Our uniqueness proof can be easily adapted to this setting, and note that the rotational symmetry assumption is even hidden in this case. Let us mention that the Yudovich theorem has been successfully extended to domains with polygonal corners, very recently; see [53, 30, 52]. Theorem 3.3.8. Let h0 ∈ W 1,∞ (Q) be odd with respect to θ∗ = 0 and non-negative on [0, π/4]. If h0 (π/4) > 0, then h(t, ·) converges to the odd stationary solution which equals h0 (π/4) on (0, π/4): kh(t, ·) − h0 (0)kL1 (0,π/4) → 0 ,
t→∞.
In addition, the solution exhibits the sharp growth rate of the gradient;
kh0 (t)kL∞ ≥ C exp(ct) , with some constant c, C > 0 depending only on the norm kh0 kW 1,∞ . Proof. The fact that we have initially h0 (π/4) > 0 gives a global-in-time lower bound on the L1 norm; indeed, one can find some triangle near the fixed point π/4 which lies below the graph of h(t) for all t ≥ 0. Indeed using this argument one has for all small δ > 0 that kh(t, ·)kL1 (δ,π/4−δ) ≥ (δ) > 0 .
81
Since the kernel KS1 1 for H is strictly negative on (δ, π/4 − δ) × (δ, π/4 − δ), we obtain a global in time lower bound for −H(t, θ) for θ ∈ [δ, π/4 − δ]. The velocity is simply 2H and therefore the particle starting from π/4 − δ reaches the point δ after some finite time. Since δ > 0 is arbitrary and h(t) is continuous in space and uniformly bounded in L∞ , convergence in L1 to the stationary solution which equals h0 (π/4) is established. To obtain the exponential growth statement, take any x ∈ (0, π/4) for which h00 (x) 6= 0 and recall that d 0 h (t, φ(t, x)) = −2H 0 (t, φ(t, x))h0 (t, φ(t, x)) . dt We know that for any δ > 0 there is T (δ) > 0 such that for all t > T (δ), 0 < φ(t, x) < δ. Since the convergence in L1 to the stationary solution implies
0
−H (t, 0) =
Z
π/4
cos(2θ)h(t, θ)dθ ≥ ch0 (π/4)
0
for all sufficiently large t (and taking into account that |H 00 (t, θ)| is uniformly bounded in t, θ), we deduce that h0 (t, φ(t, x)) grows exponentially in time. Remark 18. The above result shows that the system is not globally stable in L1 (S 1 ) (and similarly (1)
(2)
in all Lp with p < ∞). That is, even if h0 , h0 ∈ C 0 are arbitrarily close in L1 , it is clear that (1) (2) kh(1) (t) − h(2) (t)kL1 ≥ C h0 (π/4) − h0 (π/4) for t sufficiently large, where C > 0 is an absolute constant. Remark 19. We close this section by noting that there is really nothing special about the 4-fold symmetry assumption, and analogous results can be obtained for m-fold symmetric data with any m ≥ 3. Let us only point out that when m ≥ 3, we have m−1 X j=0
KS 1 (θ + j2π/m) = c1m |sin(mθ/2)| + c2m .
with some constants c1m , c2m depending only on m.
82
Growth of compactly supported solutions to 2D Euler As we have discussed in the introduction, the fact that we have solutions to the 1D system whose gradient grows almost immediately implies that there are compactly supported solutions to the 2D Euler equation whose gradient grows at least as fast as the 1D solutions do. ˚0,1 (R2 ) whose unique solution Corollary 3.3.9. There exists 4-fold symmetric initial data ω0 ∈ C to the 2D Euler equation grows at least linearly in time:
kω(t)kC˚0,1 (R2 ) ≥ ct . If we consider the 2D Euler equation on the sector Q = {(r, θ) : 0 ≤ θ ≤ π/2} with slip boundary ˚0,1 (Q) which grows exponentially: conditions, then there is initial data ω0 ∈ C kω(t)kC˚0,1 (Q) ≥ c exp(ct) . In both cases, the solution can be compactly supported and the velocity is Lipschitz everywhere in space. Proof. In both cases, take an initial data h0 ∈ W 1,∞ (S 1 ) from the 1D system (3.38) which exhibits the desired growth rate of the gradient, and consider an initial data to the 2D Euler equation of the form
˚0,1 ω0 = ω02D + h0 (θ) ∈ C with any ω02D ∈ C 0,1 (R2 ), which enjoys the same set of symmetries with h0 . Here ω02D may be chosen that ω0 is compactly supported. Then, there is a unique global-in-time solution to 2D Euler ˚0,1 , and it is indeed straightforward to show that in the space C
ω 2D (t) := ω(t) − h(t) remains in C 0,1 (R2 ) for all time (See for instance the proof of a conditional blow-up result from [37] where we establish this type of statement for the SQG equation). Then, for any large t > 0, we have
83
a pointwise inequality
||x|∇ω(t, x)| ≥ ||x|∇h(t, θ)| − |x||∇ω 2D (t, x)| and since |∇ω 2D (t, x)| ≤ C(t), taking the limit |x| → 0 guarantees that sup x∈R2 \{0}
||x|∇ω(t, x)| ≥
1 k∂θ h(t)kL∞ (S 1 ) 2
holds.
3.3.3
Measure-valued data and quasi-periodic solutions
As it was noted in the previous sections, in the 1D system (3.38), the active scalar just in L1 was sufficient to guarantee that the velocity is Lipschitz. This implies that we can actually solve the equation with L1 initial data, or even with finite signed measures M. If we restrict to the class of atomic measures, i.e. measures supported on a finite set, then we obtain a well-posed dynamical system of point vortices. The resulting measure-valued solutions give vortex-sheet solutions to the 2D Euler equation. The associated velocity on 2D is not Lipschitz but only locally bounded in the radial direction, and it is unclear whether this is the unique solution in the class of measure-valued vorticity. We describe the 1D system of point vortices. For simplicity, we keep the assumption that the vorticity is 4-fold symmetric, and describe the data only in an interval of length π/2 in S 1 . Proposition 3.3.10. Consider initial data
h0 (θ) =
N X
aj δθj , 0
j=1
where aj ∈ R are weights and θ0j ∈ [0, π/2). The unique global in time solution of the following ODE system N d j 2X θ (t) = al sin(|2θl (t) − 2θj (t)|) , dt π l=1
84
θj (0) = θ0j ,
j = 1, · · · , N
(3.45)
gives the unique solution of (3.38) by setting
h(t, θ) =
N X
aj δθj (t) .
j=1
We note that the point vortices cannot collide with each other since the conservation of the norm kh(t, ·)kM(S 1 ) implies that the velocity is Lipschitz for all time. This also shows that point vortices can approach each other at an exponential rate. It will be convenient to assume that the points are distinct and ordered;
θ01 < θ02 < · · · < θ0N ,
|θ0N − θ01 |
0 small.
3.4.1
The symmetry condition and the key estimate
It is well known that there are only a few finite symmetry groups of the sphere not fixing great circle. Among those, we shall make of the Octahedral symmetry group, denoted by O, which is generated by the following three rotations: P1 (x1 , x2 , x3 ) := (x1 , −x3 , x2 ) P2 (x1 , x2 , x3 ) := (−x3 , x2 , x1 ) P3 (x1 , x2 , x3 ) := (−x2 , x1 , x3 ). This group is isomorphic to the symmetry group of 4 elements. Definition 3.4.1. We say that a vector quantity f = (f1 , f2 , f3 )T : R3 → R3 is symmetric with respect to O if for each element O ∈ O, f (Ox) = O(f (x))
(3.47)
holds for all x ∈ R3 . Notation. We follow the usual convention that u, ω are denoted by 3 × 1 column vectors. The gradient matrices ∇u, ∇ω are 3 × 3 matrices with columns ∂xi u, ∂xi ω, respectively. Moreover, throughout and H¨ older spaces – see Bardos-Titi [5].
88
the section, the kernel K will denote the three-dimensional Biot-Savart kernel on R3 which has the explicit form
1 K(x) = 4π|x|3
0 −x 3 x2
x3 0 −x1
−x2 x1 . 0
With this kernel K, the relation u = K ∗ ω holds. We write ∇K ∗ ω (defined by the principal value) to denote the matrix whose i-th column is (∂xi K) ∗ ω for i = 1, 2, 3. Then, (see [44]) we have
−ω3 (x) ω2 (x) 0 1 ∇u(x) = ∇K ∗ ω(x) + ω3 (x) 0 −ω1 (x) . 3 −ω2 (x) ω1 (x) 0
(3.48)
The Euler equations, in any dimension, respects rotational symmetry. This is obvious from the viewpoint of physics but we supply a proof. Proposition 3.4.2. Assume that ω0 is symmetric with respect to O and belongs to a well-posedness class4 to the 3D Euler equations. Then, the unique local-in-time solution ω(t, ·) stays symmetric with respect to O. Proof. Take any rotation matrix O, and note that the relation
O−1 K(Ox)O = K(x) holds for all x ∈ R3 . This implies that u = K ∗ ω is symmetric with respect to O whenever ω is. Next, given a vector f symmetric with respect to O, the gradient matrix ∇f is symmetric in the sense that
O−1 ∇f (Ox)O = ∇f (x) holds for all O ∈ O. In turn, this implies that when ω is symmetric, ∇u is symmetric in view of the formula (3.48). Therefore, both u · ∇ω and ω · ∇u are symmetric as well. This shows that the 3D Euler equations respect rotational symmetries. In particular, symmetry breaking can only occur from non-uniqueness. This finishes the proof. 4 This
means that at least for some non-empty time interval, the solution exists uniquely in the given class.
89
The key estimate is the following bound on the velocity gradient in terms of the critical norm of the vorticity, which holds only under the symmetry assumption. Lemma 3.4.3. Assume that the vorticity ω is symmetric with respect to O. Then, ∇u satisfies the bound
k∇ukL∞ (R3 ) ≤ Cα kωkC˚α (R3 ) ,
0 < α < 1.
Proof. Let us only consider the partial derivative ∂x1 u. The other partial derivatives can be treated in the same way. Explicit computations give that the matrix ∂x1 K has the form 1 4π|x|5
0 3x x 1 3 −3x1 x2
−3x1 x3
3x1 x2 −(x21
0 (x21 − x22 ) + (x21 − x23 )
−
x22 )
−
0
(x21
−
x23 ) .
Note that each entry is a harmonic polynomial of degree 2. We claim that each component of the vorticity is orthogonal to such harmonic polynomials. We consider the second component ω2 . Symmetry with respect to the rotation P1 gives ω2 (x) = ω3 (x1 , −x3 , x2 ). Applying the matrix P1 once again, we have ω2 (x) = −ω2 (−x1 , −x2 , x3 ). Similarly, using P32 , it follows that ω2 (x) = −ω2 (x1 , −x2 , −x3 ). Moreover, from P2 ω(x) = ω(P2 x), we have ω2 (x) = ω2 (−x3 , x2 , x1 ) = ω2 (−x1 , x2 , −x3 ). Given these relations, consider Z x1 x2 ω2 (x)dσ(x). |x|=1
Then, from the change of variables (x1 , x2 , x3 ) 7→ (−z1 , −z2 , z3 ) which changes the sign of ω2 but keeps the value of the harmonic polynomial, we see that the above integral vanishes. For the same reason, Z x3 x2 ω2 (x)dσ(x) |x|=1
vanishes as well, and Z x3 x1 ω2 (x)dσ(x) |x|=1
vanishes from the invariance of ω2 under the change of variables (x1 , x2 , x3 ) 7→ (−z3 , z2 , z1 ). It 90
similarly follows that Z |x|=1
(x21 − x22 )ω2 (x)dσ(x) =
Z |x|=1
(x22 − x23 )ω2 (x)dσ(x) = 0.
Now returning to showing boundedness of ∇u, we need to consider the expression Z p.v. R3
Qi (x − y)(ω(x) − ω(y))dy,
where Qi = ∂xi K. We split the space into three domains, (i) |x − y| ≤ |x|/2, (ii) |y| ≥ 2|x|, and (iii) the remainder. In the region (i), we use the H¨older assumption to bound the integral as Z C |x−y|≤|y|/2
kωkC˚α 1 dy ≤ Cα kωkC˚α . |x − y|3−α |x|α
For (iii), the remainder region is contained in the set |x|/2 < |x − y| ≤ c|x|, and a direct estimate gives Z |x|/2 0. Proof. It suffices to consider the case ω = 0, and let Ψ be a solution to ∆Ψ = 0, i.e., a harmonic function satisfying the assumptions in the lemma. We shall show that Ψ ≡ 0. We recall that the symmetry assumption gives Z
p(x)Ψi (x)dσ(x) = 0,
i ∈ {1, 2, 3}
∂B(0,R)
for any homogeneous polynomial p(x) of degree two. This implies that following integral Z
p(x)Ψi (x)dx = 0
B(0,R)
vanishes as well. We claim now that the integrals against linear functions Z
xj Ψi (x)dσ(x) = 0,
∂B(0,R)
i, j ∈ {1, 2, 3}
vanish as well. For i 6= j, it just follows from the symmetry condition with respect to Pj . For i = j, Z
xi Ψi (x)dx =
B(0,R)
1 2
Z ∂B(0,R)
x2i Ψi (x)dσ(x) −
1 2
Z
x2i ∂xi Ψi (x)dx.
B(0,R)
The first term vanishes, and using the divergence free condition, the second term equals 1 2
Z
x21 ∂x2 Ψ2 (x) + ∂x3 Ψ3 (x) dx,
B(0,R)
assuming i = 1 for simplicity. Using an integration by parts again, we see that it vanishes as well. Now we recall the representation formula for harmonic functions on a ball:
Ψ(x) =
R2 − |x|2 4πR
Z ∂B(0,R)
1 Ψ(y)dσ(y). |x − y|3
Using the fact that Ψ(0) = 0, and the orthogonality with respect to linear and quadratic polynomials, 94
one can show that the kernel actually decays as |y|−6 and this finishes the proof with the growth assumption on Ψ. (This part of the proof is analogous with the proof of Lemma 3.2.6.) Similarly in the case of 2D, one can show the estimate |u(x)| ≤ CkωkL∞ , |x| but unlike the case of 2D, this estimate will not play a role in the following as we do not have an a priori bound on the L∞ -norm of vorticity.
3.4.2
Local well-posedness in critical spaces
We are now ready to state and prove the local well-posedness result in critical spaces. ˚α (R3 ) Theorem 3.4.7. Assume that the initial vorticity ω0 is divergence free, belongs to the space C for some 0 < α < 1 and symmetric with respect to O. Then, for some T = T (kω0 kC˚α (R3 ) ) > 0, there ˚α (R3 )) symmetric with respect is a unique solution to the 3D Euler equations with ω ∈ L∞ ([0, T ); C to O. Moreover, the solution blows up at time T if and only if Z 0
T
kω(t)kL∞ dt = +∞.
˚α (R3 )-norm on the vorticity. Proof. We restrict ourselves to obtaining an a priori bound on the C Then the solution can be constructed as the limit of an approximation sequence using the transport nature of the system and the Lipschitz bound on the velocity. We begin with the formula d ω(t, Φ(t, x)) = ω(t, Φ(t, x)) · ∇u(t, Φ(t, x)). dt For simplicity, we write ω(t, Φ(t, x)) = ωt ◦ Φt (x) or even ω ◦ Φ(x). Taking two points x 6= x0 , d [ωt ◦ Φt (x) − ωt ◦ Φt (x0 )] = ωt ◦ Φt (x) · ∇ut ◦ Φt (x) − ωt ◦ Φt (x0 ) · ∇ut ◦ Φt (x0 ). dt
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We then compute d ωt ◦ Φt (x) − ωt ◦ Φt (x0 ) ωt ◦ Φt (x) · ∇ut ◦ Φt (x) − ωt ◦ Φt (x0 ) · ∇ut ◦ Φt (x0 ) = dt |Φt (x) − Φt (x0 )|α |Φt (x) − Φt (x0 )|α 0 ωt ◦ Φt (x) − ωt ◦ Φt (x ) (Φt (x) − Φt (x0 )) · (ut ◦ Φt (x) − ut ◦ Φt (x0 )) . −α |Φt (x) − Φt (x0 )|α |Φt (x) − Φt (x0 )|α+1 Lastly, we compute d |Φt (x)|α ωt ◦ Φt (x) − |Φt (x0 )|α ωt ◦ Φt (x0 ) dt |Φt (x) − Φt (x0 )|α Φt (x) · ut ◦ Φt (x) ωt ◦ Φt (x) Φt (x0 ) · ut ◦ Φt (x0 ) ωt ◦ Φt (x0 ) =α − α |Φt (x)|2−α |Φt (x) − Φt (x0 )|α |Φt (x0 )|2−α |Φt (x) − Φt (x0 )|α α 0 α 0 |Φt (x)| ωt ◦ Φt (x) · ∇ut ◦ Φt (x) − |Φt (x )| ωt ◦ Φt (x ) · ∇ut ◦ Φt (x0 ) + |Φt (x) − Φt (x0 )|α |Φt (x)|α ωt ◦ Φt (x) − |Φt (x0 )|α ωt ◦ Φt (x0 ) (Φt (x) − Φt (x0 )) · (ut ◦ Φt (x) − ut ◦ Φt (x0 )) −α |Φt (x) − Φt (x0 )|α |Φt (x) − Φt (x0 )|2 = I + II + III. We bound each terms on the right hand side. For simplicity, we fix some t and set z = Φt (x) and z 0 = Φt (x0 ). We start with z 0 · u(z 0 ) |z 0 |α ω(z 0 ) z · u(z) |z|α ω(z) − α |z|2 |z − z 0 |α |z 0 |2 |z − z 0 |α α 0 α z · u(z) |z| ω(z) − |z | ω(z 0 ) z · u(z) z 0 · u(z 0 ) |z 0 |α ω(z 0 ) =α +α − , |z|2 |z − z 0 |α |z|2 |z 0 |2 |z − z 0 |α
I=α
and the latter term can be further rewritten as α
z · (u(z) − u(z 0 )) + (z − z 0 ) · u(z 0 ) z 0 · u(z 0 ) z 0 · u(z 0 ) + − |z|2 |z|2 |z 0 |2
|z 0 |α ω(z 0 ) . |z − z 0 |α
Then, we may bound 0 α 0 0 0 0 u(z) |z 0 |α kωkL∞ kωk ˚α + k∇ukL∞ |z − z | |z | |ω|L∞ + |z − z |(|z| + |z |) |z 0 |2 u(z ) |I| . C 0 α 2 0 2 0 z L∞ |z| |z − z | |z| |z | z L∞ |z − z 0 |α and we could have assumed that |z| ≥ |z 0 |, |z| ≥ |z − z 0 |/2 (or switch the role of z and z 0 otherwise). This gives
|I| . kωkC˚α k∇ukL∞ .
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Next, we deal with
II =
|z|α ω(z)∇u(z) − |z 0 |α ω(z 0 )∇u(z) |z 0 |α ω(z 0 )(∇u(z) − ∇u(z 0 )) + . |z − z 0 |α |z − z 0 |α
The latter term can be rearranged to be
ω(z 0 )
0 α α |z|α ∇u(z) − |z 0 |α ∇u(z 0 ) 0 0 |z | − |z| + ω(z )∇u(z ) . |z − z 0 |α |z − z 0 |α
Taking absolute values,
|II| . kωkC˚α k∇ukL∞ + kωkL∞ k∇ukC˚α . Regarding the term III, we note that (z − z 0 ) · (u(z) − u(z 0 )) ≤ k∇ukL∞ . |z − z 0 |2 Hence, we bound
|III| . k∇ωkC˚α k∇ukL∞ . Collecting the bounds, we obtain d |z|α ω(z) − |z 0 |ω(z 0 ) ≤ Ck∇ukL∞ kωk ˚α . C dt |z − z 0 |α Taking the supremum over z 6= z 0 gives d kωkC˚α ≤ Ck∇ukL∞ kωkC˚α . dt The bound k∇ukL∞ ≤ CkωkC˚α gives an a priori estimate for k∇ωkC˚α . Using the logarithmic estimate for the velocity gradient, one can show the blow-up criterion stated, as it is the case for the 3D Euler equations in usual C α -spaces.
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3.4.3
A conditional blow-up result
In this section, we prove a local well-posedness of the hybrid type, which implies a conditional blow-up result. Throughout the section, we assume that the vorticity is symmetric with respect to O. If one considers initial vorticity of the form
ω0 (x) = h0 (x/|x|),
h0 ∈ C α (S 2 ),
i.e. radially homogeneous, then the local in time solution must retain the homogeneity by uniqueness. A system for the evolution of the profile h defined on the unit sphere can be written down. With some velocity vector field v = v(h) again defined on the sphere, the system takes the form
∂t h + (v · ∇)h = R(h)h,
(3.49)
where R is a matrix of singular integral operators acting on the components of h. ˚α (R3 ) Theorem 3.4.8 (Local well-posedness and conditional blow-up). Consider initial data ω0 ∈ C which is symmetric with respect to O, and has a decomposition ω0 (x) = ω03D (x) + ω02D (x), where ω03D ∈ C α (R3 ) and ω02D (x) = h0 (x/|x|) with h0 ∈ C 1,α (S 2 ) are again symmetric with respect to O. Then the following statements hold. 1. (Local well-posedness) There exists some time T = T (kω03D kC α (R3 ) , kh0 kC 1,α (S 2 ) ) > 0 such that there is a unique solution to the 3D Euler equations satisfying
ω(t) = ω 3D (t) + ω 2D (t), with ω 3D ∈ C([0, T ); C α ) and ω 2D (t, x) = h(t, x/|x|) where h is the unique local solution to the 2D system (3.49). This solution could be continued past some time T ∗ > 0 if and only if Z 0
T∗
kω 3D (t)kL∞ + kh(t)kL∞ dt < +∞.
2. (Conditional blow-up) Assume that there exists an initial data h0 ∈ C 1,α (S 2 ) whose unique local 98
solution to the 2D system (3.49) blows up at some finite time T ∗ . Then, for any initial data ω03D ∈ C α (R3 ), the unique solution given in the above to the initial data ω0 (x) = h0 (x/|x|) + ˚α (R3 ). In particular, the initial data ω0 ω03D (x) blows up at some finite time 0 < T ≤ T ∗ in C can be compactly supported. Proof. We work within in the time interval [0, T1 ] for which the 2D system (3.49) has a solution with initial data h0 ∈ C 1,α (S 2 ). Moreover, by taking smaller T1 > 0 if necessary, we may assume ˚α (R3 ) with initial data ω0 for [0, T1 ]. that the solution ω(t) exists in C We then define ω 3D := ω − ω 2D on [0, T1 ]. We need to propagate a C α -bound on ω 3D . To begin with, it is straightforward to check that it solves the system
∂t ω 3D + (u2D + u3D ) · ∇ω 3D + u3D · ∇ω 2D = 0.
(3.50)
We make a few preliminary remarks. First, we have the bound k∇u3D kC α ≤ Ckω 3D kC α . This follows from the classical singular integral bound, together with the symmetry assumption used to gain integrable decay of the kernel (otherwise an Lp -assumption has to be imposed on ω 3D ). Next, a Taylor expansion at the origin gives that u(x) does not have any linear terms in x (this follows from the symmetry and the divergence-free assumptions – see [37, Theorem 7] for a strictly analogous argument done for the SQG equation), which in particular shows that |u(x)| ≤ C|x|1+α . Similarly, |∇u(x)| ≤ C|x|α . Next, we note that kf gkC α ≤ Ckf kC˚α kgkC α , ˚α . once we assume that g(0) = 0. This inequality follows directly from the definition of the space C Returning to the system (3.50), upon composition with the flow generated by u2D + u3D , (but let us suppress from writing out the composition) d 3D ω = −u3D · ∇ω 2D , dt and from the radial homogeneous nature of ω 2D , we bound 3D u (x) d 3D khkC 1 ≤ Ck∇u3D kL∞ khkC 1 . kω kL∞ ≤ C dt |x| L∞ This gives the L∞ bound. To obtain a C α -estimate, neglecting issues with the flow maps (which
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can be dealt with following the lines of the proof of Theorem 3.4.7), we write d 3D u3D (x) u3D (x) kω kC α ≤ Ck hkC α + Ck · |x|∇hkC α dt |x| |x| and then using the above product rule in C α , d 3D kω kC α ≤ Ck∇u3D kC α khkC 1,α (S 2 ) ≤ Ckω 3D kC α khkC 1,α (S 2 ) . dt This gives an a priori estimate. We omit the proof of the blow-up criterion, which is straightforward. Now let us show the conditional blow-up statement. For the sake of contradiction, assume that the solution ω = ω 2D + ω 3D stays smooth up to time T ∗ , which implies in particular that
sup kω(t)kL∞ ≤ C < +∞.
t∈[0,T ∗ ]
On the other hand, we know that there is a sequence of time moments tk → T ∗ for which kh(tk )kL∞ → +∞, from the assumption of blow-up. However, fixing time t < T ∗ and taking the supremum in space,
lim sup |ωt (x)| = lim sup |ωt2D (x)| x→0
x→0
simply because |ωt3D (x)| → 0 as |x| → 0 for each t < T ∗ . Then, from C ≥ kω(t)kL∞ ≥ lim sup |ω 2D (x)| ≥ kh(t)kL∞ , x→0
we obtain a contradiction by taking the limit along the time sequence tk .
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Chapter 4
Global well-posedness of rotationally symmetric vortex patches with corners In this chapter, we establish global in time propagation of piecewise smoothness up to the corner for rotationally symmetric “petals” as described in Figure 4.1. This could be of some surprise in view of the ill-posedness result for corners described earlier. However, this is consistent with the existence of reported V -states of the corresponding form.
4.1
Smooth vortex patches: an approach by Bertozzi-Constantin
In this section, let us provide a brief outline of the elegant proof of Bertozzi and Constantin [10] on global regularity of smooth vortex patches. We restrict ourselves to domains (bounded open set in R2 ) Ω which has a level set φ : R2 → R such that: • We have φ(x) > 0 if and only if x ∈ Ω (Hence φ vanishes precisely on ∂Ω). • The tangent vector field of φ satisfies ∇⊥ φ ∈ C α (R2 ). • The function φ is non-degenerate near ∂Ω, i.e., k∇⊥ φkinf(∂Ω) := inf x∈∂Ω |∇⊥ φ| ≥ c > 0.
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Figure 4.1: A 3-fold rotationally symmetric vortex patch with corners Then we say that the patch Ω is C 1,α -regular, or a C 1,α -patch. Given such a φ, we associate the following characteristic quantity:
Γ=
k∇⊥ φkC∗α (R2 ) k∇⊥ φkinf(∂Ω)
!1/α ,
which quantifies the C 1,α -regularity of Ω. Note that it has units of inverse length, so that Γ−1 provides a C 1,α -characteristic length scale for Ω. An alternative way of defining C 1,α patches is to require that, for any point x ∈ ∂Ω, there exists a ball Bx (r) with some radius r > 0 uniform over x such that the intersection Bx (r) ∩ ∂U is given by the graph of a C 1,α function, rotating the plane if necessary. Indeed, one may take r to be 1/(10Γ) for instance. Taking the initial vorticity to be the characteristic function ω0 = χΩ , we may denote its unique solution by χΩt . Since the vorticity is simply being transported by the flow, once we define the evolution of φ via
(4.1)
∂t φ + (u · ∇)φ = 0, then it follows that
φ(t, x) > 0
if and only if
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x ∈ Ωt .
To show that Ωt stays as a C 1,α -patch, it suffices to establish an a priori bound on Γt . In [10], the authors have provided a proof that Γt remains bounded for all time, based on the following two “frozen-time” lemmas: Lemma 4.1.1 (L∞ -bound on ∇u). Consider the velocity u(x) = K ∗χΩ (x), where Ω is a C 1,α -patch with a level set φ. Then, we have a bound
k∇ukL∞ (R2 ) ≤ C
k∇⊥ φkC α (R2 ) 1 + log 1 + k∇⊥ φkinf(∂Ω)
!! (4.2)
Lemma 4.1.2 (Directional C α -bound on ∇u). We have a pointwise identity 1 ∇u∇ φ(x) = 2π ⊥
Z Ω
∇K(x − y) ∇⊥ φ(x) − ∇⊥ φ(y) dy,
(4.3)
and in particular, this gives a bound
k∇u∇⊥ φkC α (R2 ) ≤ Ck∇ukL∞ (R2 ) k∇⊥ φkC α (R2 ) .
(4.4)
The point of (4.4) is that we do not need to take the C α -norm of the velocity gradient. Given these lemmas, one can finish the global well-posedness proof with a simple Gronwall estimate (details of this argument can be found in [10]). We differentiate (4.1) to obtain
∂t ∇⊥ φ + (u · ∇)∇⊥ φ = ∇u∇⊥ φ. Working on the Lagrangian coordinates, and using the bound (4.4) and then the logarithmic estimate (4.2) allows one to close the estimates in terms of k∇⊥ φkC α to show the bound k∇⊥ φ(t)kC α (R2 ) ≤ C exp(C exp(Ct)) as well as
k∇⊥ φ(t)kinf(∂Ω) ≥ c exp(−ct) with positive constants depending only on the initial data ∇⊥ φ0 (and 0 < α < 1). We would like to point out that, although it was not necessary in the above global well-posedness
103
argument, the velocity gradient is indeed uniformly C α inside the patch, up to the boundary. There are a number of ways to obtain this piece of information. One approach, due to Serfati [66], is that from the directional H¨ older regularity (∇⊥ φ · ∇)u ∈ C α that we already have, one can “invert” this using ∇ · u = 0 and ∇ × u = 1 (inside the patch) to recover u ∈ C α . We adopt this idea in our proof of local well posedness (see Lemma 4.2.6). Alternatively, Friedman and Velazquez [41] has shown, directly working with the Biot-Savart kernel, the following estimate: Lemma 4.1.3 (Friedman and Velazquez [41]). Assume that a C 1,α -patch Ω is tangent to the horizontal axis at the origin, and that near the origin, ∂Ω is described as the graph of a C 1,α -function:
∂Ω ∩ [−δ, δ]2 = {(x1 , x2 ) : x2 = f (x1 )},
f ∈ C α ([−δ, δ]),
sup |f 0 | ≤ 1.
[−δ,δ]
Then, the velocity u = K ∗ χΩ is C 1,α along this portion of the boundary:
k∇u(x1 , f (x1 ))kCxα
1
[−δ/10,δ/10]
1 ≤ Ckf kC 1,α [−δ,δ] log 1 + δ
.
With elliptic regularity, the above lemma immediately implies that for C 1,α -patches, the velocity gradient is uniformly C α up to the boundary. The above lemma of Friedman and Velazquez actually gives C 1,α -regularity for velocity coming from a C 1,α -cusp: consider the domain Ω satisfying
Ω ∩ [−δ, δ]2 = {(x1 , x2 ) ⊂ [0, δ] × [−δ, δ] : g(x1 ) < x2 < f (x1 )} where g < f are C 1,α [0, δ]-functions with g(0) = f (0) = 0 and g 0 (0) = f 0 (0) = 0. Then, applying the lemma first with a C 1,α domain obtained by taking (x1 , f (x1 )) and the semi-axis {(x1 , 0) : x1 ≤ 0} as a portion of its boundary, and then using the lemma another time with a domain using (x1 , g(x1 )) instead of f establishes that ∇u is uniformly C α in [0, δ/10] × [−δ, δ] ∩ Ω. We shall use this bound a few times in our arguments.
4.2
Local well-posedness for symmetric patches
The local well-posedness results for smooth vortex patches is usually obtained via an iteration scheme, using the contour dynamics equation (see for instance [11], [58]). An alternative approach which works directly with the flow maps restricted to the patch was described in an illuminating
104
work of Huang [44].1 This method originates from a previous work of Friedman and Huang [40], and it seems to be applicable for a wide variety of situations. We shall adopt this approach to show local well-posedness (as well as continuation criteria) in the setting of Section 4.3, i.e., patches admitting ˚α . a level set function φ with ∇⊥ φ ∈ C The starting point of this method is to write the 2D Euler equation purely in terms of the flow maps: Z tZ Φ(x, t) = x + 0
R2
Z tZ =x+ 0
Ω0
K(Φ(x, s) − y)ω0 (Φ−1 (y, s))dyds,
(4.5)
K(Φ(x, s) − Φ(z, s))dzds.
At this point, note that we only need to know Φ(·, t) on Ω0 to determine the velocity of the Euler equation everywhere in R2 . It is easy to show that the above formulation is equivalent to the (usual) weak formulation of the 2D Euler equations under ω ∈ L∞ ∩ L1 , and the Yudovich theorem gives that there is a unique solution Φ satisfying (4.5). The formulation 4.5 suggests one to build an iteration scheme; all that is necessary to appropriately define the space of functions. Following [44], we consider ( B(M, T ) =
)
Φ(x, t) ∈ X : Φ(x, 0) = x, Φ(0, t) = 0, kΦkX ≤ M,
sup Ω0 ×[0,T ]
|∇Φ(x, t) − I| ≤ 1/2 (4.6)
where the space X is defined for functions Φ : Ω0 × [0, T ] → R2 with det(∇x Φ) ≡ 1 by the norm kΦkX = sup
t∈[0,T ]
k∇x ΦkC˚α (Ω0 ) + k∂t ΦkL∞ (Ω0 ) .
That is, we have simply replaced the assumption in [44] that ∇Φ(·, t) is uniformly C α (up to ˚α . The extra assumption that Φ(0, t) = 0 holds will be guaranteed by the boundary of Ω0 ) by C symmetry. Under the assumption |∇Φ(x, t)−I| ≤ 1/2, it follows that the inverse map Φ−1 t : Ωt → Ω 0 −1 is Lipschitz with |∇Φ−1 t | ≤ 2. Moreover, it is elementary to verify that for Φ ∈ B(M, T ), ∇Φt
˚α (Ωt ) with norm depending only on M (see below Lemma 4.2.5). belongs to C 1 The main result of this work is that C 1,α -patches in 3D is locally well-posed under the Euler equations. In the three-dimensional case, the vorticity does not remain a constant inside the patch even if initially so, and therefore the contour dynamics approach is not available.
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Then, we define a mapping F , Z tZ F (Φ)(x, t) = x + 0
Ω0
K(Φ(x, s) − Φ(z, s))dzds,
so that a fixed point of F provides a solution to the 2D Euler equation on [0, T ] with initial data ω0 = χΩ0 . We need to propagate the regularity of φ in time, where the level set function φ0 is given together with the initial data Ω0 . We observe that, as long as Φ ∈ B(M, T ), by defining φ(x, t) := φ0 (Φ−1 t (x)),
x ∈ Ωt ,
we have
Γt := sup sup ˚ t∈[0,T ]
t∈[0,T ]
k∇⊥ φt (·)kC˚α (Ωt )
!1/α
k∇⊥ φt (·)kinf(∂Ωt )
≤ C(M ),
(4.7)
where, here and in the following, we use the notation C(M ) to denote a positive and increasing function of M > 0 depending on ˚ Γ0 . This function may change from a line to another. We are now in a position to formally state the local well-posedness results: Proposition 4.2.1. Assume that Ω0 is m-fold symmetric for some m ≥ 3 admitting a level set φ0 satisfying Definition 4.3.1. Then there exists some T > 0, depending only on ˚ Γ0 , such that there is a unique local solution Φ ∈ X of (4.5). In particular, we can extend the solution beyond some T ∗ as long as supt∈[0,T ∗ ) ˚ Γt < +∞. In the case of C 1,α -patch with symmetric corners, we have: Proposition 4.2.2. Assume that Ω0 is a C 1,α -patch with symmetric corners satisfying Definition 4.4.1. Then there exists some T > 0, depending only on its initial C 1,α -characteristic Γ0 , such that supt∈[0,T ] Γt < +∞: that is, the associated flow map Φ ∈ X on the time interval [0, T ] provided by Proposition 4.2.1 satisfies Φt (x, f0 (x)), Φt (x, g0 (x)) ∈ Cx1,α [0, δ0 ] uniformly in t ∈ [0, T ]. In particular, we can extend the solution beyond some T ∗ > 0 as long as supt∈[0,T ∗ ) Γt < +∞. The proof is a direct consequence of the following estimates: Lemma 4.2.3. For any initial data Ω0 satisfying Definition 4.3.1, there exists some M, T > 0 depending only on ˚ Γ0 so that F maps the space B(M, T ) to itself.
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Lemma 4.2.4. Assume that we are in the situation where Lemma 4.2.3 holds. Then, there exists ˜ ∈ B(M, T ), some 0 < T1 ≤ T , depending only on M and ˚ Γ0 , so that for any Φ, Φ ˜ kF (Φ)(t) − F (Φ)(t)k L∞ (Ω0 ) Z t ˜ sk ∞ ˜ sk ∞ kΦs − Φ ≤ C(M ) 1 + log 1 + kΦ − Φ ds s L (Ω0 ) L (Ω0 ) 0
and ˜ k∇F (Φ)(t) − ∇F (Φ)(t)k L∞ (Ω0 ) Z t ˜ ˜ sk ∞ k∇Φs − ∇Φ ≤ C(M ) ds L (Ω0 ) 1 + log 1 + k∇Φs − ∇Φs kL∞ (Ω0 ) 0
hold for any t ∈ [0, T1 ]. Assuming the statements of Lemmas 4.2.3 and 4.2.4, let us just provide a sketch of the proof, as the argument is parallel to [44, Proof of Theorem 4.1]. Proof of Proposition 4.2.1. Take M and T1 such that the map F sends B(M, T1 ) to itself, and ˜ ∈ B(M, T1 ), moreover, for any Φ, Φ ˜ kF (Φ)(t) − F (Φ)(t)k W 1,∞ (Ω0 ) Z t ˜ ˜ ≤ C(M ) 1 + log 1 + kΦ(s) − Φ(s)k ds kΦ(s) − Φ(s)k 1,∞ 1,∞ W (Ω0 ) W (Ω0 ) 0
for t ∈ [0, T1 ]. Here, M and T1 depends only on ˚ Γ0 . Define a sequence {Φn }n≥0 in B(M, T1 ) by Φ0 (t, x) = x,
Φn+1 (x, t) = F (Φn )(x, t),
n ≥ 0.
It is straightforward to see that at each step of the iteration, the flow is m-fold symmetric around the origin and therefore Φn (0, t) = 0 for all n ≥ 0. Setting ρn (t) := kΦn+1 (t) − Φn (t)kW 1,∞ (Ω0 ) , we have
ρn (t) ≤ C(M )
Z
t
ρn−1 (s) (1 + log (1 + ρn−1 (s))) ds. 0
107
This is sufficient to deduce that, taking a smaller value of T1 depending only on M if necessary (see [59, Chapter 2] for instance), there exists a function Φ : Ω0 × [0, T1 ] → R2 such that kΦn − ΦkL∞ ([0,T1 ];W 1,∞ (Ω0 )) → 0. At this point, it is easy to see that Φ actually belongs to B(M, T1 ) and F (Φ) = Φ. Therefore, we have constructed a solution, which belongs to the desired class, to the 2D Euler equation with initial data Ω0 on the time interval [0, T1 ]. We briefly comment on the issue of continuing the solution past T1 . (All the details can be found in [44].) Take ΩT1 as the new initial data, which has associated level set function φT1 with its characteristic ˚ ΓT1 . Going through the exact same iteration scheme again with this new data, one obtains a unique solution on some time interval [0, T2 ], with T2 = T2 (˚ ΓT1 ) > 0. Then, by putting ˚1,α -level set, this solution together with the previous one, we obtain a patch solution, admitting a C to the 2D Euler equation on the time interval [0, T1 + T2 ] with initial data Ω0 . This procedure can go on as long as we have a bound on ˚ Γt . This finishes the proof. Proof of Proposition 4.2.2. The assumptions given in Definition 4.4.1 are strictly stronger than the ones in Definition 4.3.1, so we may work inside the time interval within which we have available the iterates Φn and the limit Φ belonging to the class X, defined in the above proof of Proposition 4.2.1. It suffices to carry the information that, by shrinking T if necessary in a way only depending on Γ0 , for some time interval [0, T ], each of Φn satisfies following the H¨older estimate uniformly in n:
kΦn ((x, f0 (x)), t)kC 1,α [0,δ0 ] + kΦn ((x, g0 (x)), t)kC 1,α [0,δ0 ] ≤ C(Γ0 ) < ∞. This follows directly from the a priori estimates given in the proof of Theorem 4.4.3. It is not difficult to see that Φ inherits the same H¨ older estimate. The lemmas 4.2.3, 4.2.4, and the bound (4.7) are direct consequences of the following simple lemmas. The first one provides substitutes for the usual calculus inequalities on C α -spaces. ˚α functions on some domain Ω ⊂ R2 . Then we have Lemma 4.2.5. Let f and g be C kf gkC˚α ≤ C kf kC˚α · kgkL∞ + kf kL∞ · kgkC˚α
108
(4.8)
and if we assume further that |f | > 0 on Ω, k1/f kC˚α ≤ C(kf kinf(Ω) )kf kC˚α .
(4.9)
Moreover, if Ψ is a Lipschitz diffeomorphism of R2 with Ψ(0) = 0, then
kf ◦ ΨkC˚α ≤ C (k∇ΨkL∞ , k∇Ψkinf ) kf kC˚α .
(4.10)
Proof. Let us note first that for two points at comparable distance, i.e. if x 6= x0 satisfy c1 |x0 | ≤ |x| ≤ c2 |x0 |, kf kC˚α |f (x) − f (x0 )| ≤C |x − x0 |α |x|α with C depending on c1 , c2 . We begin with (4.8). First, we have an L∞ -bound kf gkL∞ ≤ kf kL∞ · kgkL∞ . Now take two points x 6= x0 ∈ Ω and assume without loss of generality that |x| ≥ |x0 |. Consider two cases, (i) |x − x0 | ≤ |x|/2 and (ii) |x − x0 | > |x|/2. In the latter case, |x|α f (x)g(x) − |x0 |α f (x0 )g(x0 ) |x|α (f (x)g(x) − f (x0 )g(x0 )) + (|x|α − |x0 |α ) f (x0 )g(x0 ) = |x − x0 |α |x − x0 |α ≤ Ckf kL∞ · kgkL∞ . Next, when (i) holds, we rewrite |x|α (f (x)g(x) − f (x0 )g(x0 )) + (|x|α − |x0 |α ) f (x0 )g(x0 ) |x − x0 |α α |x| (f (x) − f (x0 ))g(x) |x|α f (x0 )(g(x) − g(x0 )) |x|α − |x0 |α = + + f (x0 )g(x0 ), |x − x0 |α |x − x0 |α |x − x0 |α which is bounded in absolute value by the right hand side of (4.8), noting that
|f (x) − f (x0 )| ≤ Ckf kC˚α
|x − x0 |α |x|α
whenever |x − x0 | ≤ |x|/2. The proof of (4.9) is strictly analogous, so let us omit it. To show the last statement (4.10), it suffices to treat the case when |x0 | ≤ |x| and |x − x0 | ≤ |x|/2.
109
Moreover, it suffices to bound the quantity
|x|α
0 0 α |f (Ψ(x)) − f (Ψ(x0 ))| α |f (Ψ(x)) − f (Ψ(x ))| |Ψ(x) − Ψ(x )| = |x| · . |x − x0 |α |Ψ(x) − Ψ(x0 )|α |x − x0 |α
Note that since Ψ(0) = 0,
k∇Ψkinf ≤
|Ψ(z)| ≤ k∇ΨkL∞ |z|
for any z, and since we have |x0 | ≤ |x| ≤ 2|x0 |, there exists some constants c1 , c2 > 0 so that c1 |Ψ(x0 )| ≤ |Ψ(x)| ≤ c2 |Ψ(x0 )|. This allows us to bound
|x|α
|x|α |f (Ψ(x)) − f (Ψ(x0 ))| |Ψ(x) − Ψ(x0 )|α α · · ≤ Ckf k · (k∇ΨkL∞ ) . α ˚ C |Ψ(x) − Ψ(x0 )|α |x − x0 |α |Ψ(x)|α
This finishes the proof. Next, we shall need the piece of information that in the setting of Proposition 4.2.1, for each ˚α (Ωt ). In the case of C 1,α -patches, this fixed time t, the velocity gradient ∇ut actually belongs to C is a direct consequence of velocity being C 1,α on the boundary, since then ∆ut = 0 in Ω and hence an elliptic regularity statement applies. It is likely that such an argument could be used here, but let us adopt the approach of Serfati [66] (see also recent papers by Bae and Kelliher [3], [2]): ˚α (Ω). Assume further Lemma 4.2.6. Let W be a vector field on a domain Ω with components in C that |W | ≥ c0 > 0 on Ω. Then, for ω = χΩ , the associated velocity satisfies k∇ukC˚α (Ω) ≤ C(c0 )kW · ∇ukC˚α (Ω) . Proof. With W = (W1 , W2 ) and u = (u1 , u2 ), one computes that 1 W1 ∂1 u1 = |W |2 W ∂2 u1 2
−W2 W1 ∂1 u1 + W2 ∂2 u1 W1 W1 ∂2 u1 − W2 ∂1 u1
110
and note that using ∂1 u1 + ∂2 u2 = 0 as well as ∂1 u2 − ∂2 u1 = ω ≡ constant, W1 ∂2 u1 − W2 ∂1 u1 = W · ∇u2 − W1 ω, ˚α . It follows that ∇u2 ∈ C ˚α as well. so that using (4.8) and (4.9), we conclude that ∇u1 ∈ C Remark 20. To apply the above lemma to the setting of Proposition 4.2.1, taking W0 := ∇⊥ φ0 is strictly speaking not allowed since it may vanish at some points in the interior of the initial patch Ω0 . This can be simply fixed as follows (see [3, Section 10]). First, we know that for points x ∈ Ω with d(x, ∂Ω0 ) < δ|x|, |∇⊥ φ0 | is bounded from below with a constant uniform in |x|, where δ can be ˜ 0 which does not vanish for taken as 1/(10˚ Γ0 ), for instance. Then it suffices to take a vector field W ˜ 0 vanishes on ∂Ω0 and points x ∈ Ω0 with d(x, ∂Ω0 ) ≥ δ|x|. It is easy to require in addition that W ˜0 ∈ C ˚α (R2 ).2 Then, we evolve the vector field by ∇·W ˜ (Φ(x, t), t) := (W ˜ 0 (x) · ∇)Φ(x, t), W which is consistent with the evolution of vector fields having the form ∇⊥ φ for some scalar function φ advected by the flow. Proof of Lemma 4.2.3. Given an initial vortex patch Ω0 satisfying conditions of Proposition 4.2.1, ˜ 0 described in the remark following Lemma 4.2.6, as well as the level set φ0 . we fix a vector field W ˜ 0 in a region Then, one may fix a vector field W0 which coincides with ∇⊥ φ0 near ∂Ω0 and with W ˚α . where ∇⊥ φ0 vanishes. We have ∇ · W0 ∈ C We have Z F (Φ)(x, t) = x +
t
u(Φ(x, s), s)ds 0
as well as Z (∇F (Φ))(x, t) = I + 0
t
∇u(Φ(x, s), s)∇Φ(x, s)ds.
We claim that the push-forward of the vector field W0 (recall that W (Φ(x, t), t) := (W0 (x)·∇)Φ(x, t)) 2 To construct such a vector field, one first considers the family of annuli A = {x ∈ R2 : 2−n−1 < |x| < 2−n+1 }. n By rescaling the region An ∩ Ω to a domain of size O(1), we obtain a region with boundary in C 1,α . Then in this rescaled subset of the annulus one constructs easily a vector field in C α with desired properties. Rescaling it back, ˜ 0. and patching all the vector fields together finishes the construction of W
111
satisfies supt∈[0,T ] kWt kC˚α (Ωt ) ≤ C(M ) as well as inf t∈[0,T ] kWt kinf(Ωt ) ≥ (C(M ))−1 > 0 (see [3], [2] for complete details of this proof in the context of C α vector fields – the proof can be adapted to our setting with straightforward modifications). It then follows from Lemma 4.2.6 that
k∇ukC˚α (Ωt ) ≤ C(M ). Then, using the inequalities from 4.2.5 we immediately obtain
k∇F (Φ)kC˚α ≤ C(M )T and also
sup Ω0 ×[0,T ]
|∇F (Φ) − I| ≤ C(M )T.
Taking T sufficiently small, we see that F (Φ) ∈ B(M, T ). Finally, we give a sketch of the proof of Lemma 4.2.4. Proof of Lemma 4.2.4. Fix some x ∈ Ω0 and t ∈ [0, T1 ], and let us first obtain a bound on ˜ |F (Φ)(x, t) − F (Φ)(x, t)|. We need to estimate Z Ω0
˜ ˜ s)) dz s) − Φ(z, K(Φ(x, s) − Φ(z, s)) − K(Φ(x,
(4.11)
for each s ∈ [0, t]. We split the integral: when |z − x| > , we have Z
˜ ˜ s)) dz s) − Φ(z, K(Φ(x, s) − Φ(z, s)) − K(Φ(x, Ω0 \B (x) Z 1 ˜ ˜ dz ≤ C(M )kΦ(s) − Φ(s)k ≤ C(M ) kΦ(s) − Φ(s)k L∞ · L∞ (1 + | log()|) , |x − z|2 Ω0 \B (x)
whereas Z Ω0 ∩B (x)
Z ˜ ˜ s)) dz ≤ C(M ) s) − Φ(z, K(Φ(x, s) − Φ(z, s)) − K(Φ(x,
Ω0 ∩B (x)
We have used the following elementary inequality: |K(a) − K(b)| ≤ C|a − b| 112
1 1 + 2 |a|2 |b|
.
1 dz ≤ C(M ). |x − z|
˜ Choosing = kΦ(s) − Φ(s)k L∞ establishes the desired inequality (assuming that the latter quantity is non-zero – otherwise the result is trivial). Turning to the next inequality, one sees that the key is to obtain a bound on the following integral: Z Ω0
˜ ˜ s)) dz, s) − Φ(z, ∇K(Φ(x, s) − Φ(z, s)) − ∇K(Φ(x,
˜ L∞ . modulo the terms which are trivially bounded by C(M )k∇Φ − ∇Φk To begin with, take some constant 0 > 0 (depending only on Ω0 ) with the property that, for any x ∈ ∂Ω0 , there is an open ball of radius 40 |x| contained in Ω0 and whose boundary contains x. Now let us take some < 0 , whose value will be determined later. We shall consider two cases: (i) d(x, ∂Ω0 ) > 2|x|, (ii) d(x, ∂Ω0 ) ≤ 2|x|. When (i) holds, let us split the integral as Z
Z
˜ ˜ s)) dz, s) − Φ(z, ∇K(Φ(x, s) − Φ(z, s)) − ∇K(Φ(x,
+ Ω0 \B|x| (x)
Ω0 ∩B|x| (x)
and in the former region, we further decompose into regions where |x| < |z − x| ≤ 10|x| and 10|x| < |z − x|. Then, in the case |x| < |z − x| ≤ 10|x|, using the mean value theorem with the decay of ∇∇K gives a bound ˜ L∞ (1 + | log()|). C(M )k∇Φ − ∇Φk Then, when 10|x| < |z − x| holds, one first symmetrizes the kernel to gain extra decay and then use the mean value theorem to obtain
˜ L∞ . C(M )k∇Φ − ∇Φk In the latter region, the integral is bounded by Z Σ
|∇K(z)dz| ≤ C
113
Z Σ
|z|−2 dz,
where ˜ |x| (x), s) − Φ(x, ˜ Σ = (Φ(B|x| (x), s) − Φ(x, s))∆(Φ(B s)) ˜ ˜ |x| (x), s)}. := {y − Φ(x, s) : y ∈ Φ(B|x| (x), s)}∆{y − Φ(x, s) : y ∈ Φ(B For any unit vector ω, define
r1 (ω) = min{r > 0 : rω ∈ Σ},
r2 (ω) = max{r > 0 : rω ∈ Σ}.
Then, the claim of Huang [44, (4.20) on p. 531] translates in our setting to give that (after the usual scaling argument in |x|) ˜ L∞ . r2 (ω) ≤ C(M )|x| α + k∇Φ − ∇Φk
r1 (ω) ≥ (C(M ))−1 |x|, Using these bounds, we integrate Z Σ
|∇K(z)dz| ≤ C(M ) ≤ C(M )
Z ∂B1 (0)
Z ∂B1 (0)
Z
r2 (ω)
r1 (ω)
1 drdω ≤ C(M ) r
r2 (ω) − r1 (ω) log 1 + dω r1 (ω) ∂B1 (0)
Z
r2 (ω) − r1 (ω) ˜ L∞ ). dω ≤ C(M )(α + k∇Φ − ∇Φk r1 (ω)
We have established the desired bound on (4.11), and it follows immediately that
˜ ˜ L∞ (1 + | log()|)) |∇u(Φ(x, s), s) − ∇˜ u(Φ(x, s), s)| ≤ C(M )(α + k∇Φ − ∇Φk when (i) holds, and with < 0 . Now, when (ii) holds for x ∈ Ω0 , we can select (by the assumption on 0 ) a point y ∈ Ω0 , such that d(y, ∂Ω0 ) ≥ 2|x| and |x − y| ≤ 2|x|. Then, ˜ |∇u(Φ(x, s), s) − ∇˜ u(Φ(x, s), s)| ≤ |∇u(Φ(x, s), s) − ∇u(Φ(y, s), s)| ˜ s), s)| + |∇u(Φ(y, s), s) − ∇˜ u(Φ(y, ˜ s), s) − ∇˜ ˜ + |∇˜ u(Φ(y, u(Φ(x, s), s)| ˜ L∞ (1 + | log()|)) + C(M )α , ≤ C(M )(α + k∇Φ − ∇Φk
114
˚α : where we have used that ∇u, ∇˜ u∈C |∇u(Φ(x, s), s) − ∇u(Φ(y, s), s)| ≤ C(M )
|Φ(x, s) − Φ(y, s)|α |x − y|α α ≤ C(M )k∇Φk ≤ C(M )α , ∞ · L |Φ(x, s)|α |x|α
and similarly for the other term. At this point, observe that d ˜ t) ≤ C(M ), ∇Φ(x, t) − ∇Φ(x, dt so that
˜ t)kL∞ ≤ C(M )t, k∇Φ(·, t) − ∇Φ(·, and therefore by taking T1 sufficiently small, relative to M and Ω0 , it can be assumed that ˜ L∞ ≤ sup k∇Φ − ∇Φk
t∈[0,T1 ]
1 α . 10 0
˜ s)kL∞ for each s ∈ [0, T1 ] (or just a sufficiently small Now we may take α = k∇Φ(·, s) − ∇Φ(·, constant when the latter is zero). This finishes the proof.
4.3
Global well-posedness for symmetric patches in an intermediate space
In this section, we show that if a vortex patch admits a level set whose gradient is, roughly speaking, C α in the angle and non-degenerate, then the corresponding Yudovich solution retains this property for all time. As a consequence, we shall have that the velocity, and hence the flow map and its inverse, are Lipschitz functions in space for all finite. In this setup, it is necessary to impose that the patch is m-fold rotationally symmetric for some m ≥ 3. ˚1,α -patch, if it admits a level set φ : R2 → R Definition 4.3.1. Let us say that a domain Ω is a C such that: • We have φ(x) > 0 if and only if x ∈ Ω. ˚α (R2 ) (In particular φ is Lipschitz). • The tangent vector field of φ satisfies ∇⊥ φ ∈ C 115
• The function φ is non-degenerate near ∂Ω, i.e., k∇⊥ φkinf(∂Ω) := inf x∈∂Ω |∇⊥ φ| ≥ c > 0. We are ready to state our main result of this section. Theorem 4.3.2. Assume that the initial patch Ω0 is m-fold symmetric for some m ≥ 3 and admits a level set φ0 described in Definition 4.3.1. Then, the Yudovich solution Ωt continues to have this property; more specifically, by defining φ(t) as the solution of (4.1), we have a global-in-time bounds
k∇⊥ φ(t)kC˚α (R2 ) ≤ C exp(C exp(Ct)),
(4.12)
k∇⊥ φ(t)kinf(∂Ωt ) ≥ c exp(−ct),
(4.13)
k∇u(t)kL∞ (R2 ) ≤ C exp(Ct),
(4.14)
and
with constants C, c > 0 depending only on ∇⊥ φ0 and 0 < α < 1. Remark 21. Note that, in the above theorem, we do not require the initial patch Ω0 to have compact ˚α -norm uniformly bounded on all support. However, we do require that the gradient ∇⊥ φ0 to have C of R2 . ˚α Recall from the previous section that the 2D Euler equation is globally well-posed with ω0 ∈ C under symmetry. Therefore, the global well-posedness of the patch admitting a level set (under ˚α is a natural analogue of the classical global the same symmetry assumption) with ∇⊥ φ0 ∈ C well-posedness result of C 1,α -patches. As an immediate consequence of the above theorem, we have that, Corollary 4.3.3. Under the assumptions of Theorem 4.3.2, the flow map Φt is a Lipschitz bijection of the plane with a Lipschitz inverse for all times t ≥ 0. Before we proceed to the proof, let us describe a few classes of vortex patches satisfying the requirements of Definition 4.3.1. Examples and Remarks. Theorem 4.3.2 establishes global well-posedness for each of the following classes of example, under the assumption of m-fold rotational symmetry with some m ≥ 3. (i) Sectors: Assume that for some ball B0 (r), the intersection Ω0 ∩ B0 (r) is a union of sectors meeting at the origin (see Figures 4.3, 4.5 for symmetric examples). In addition, assume that 116
∂Ω0 is C 1,α -smooth in the complement of B0 (r). Then, one may take a level set locally by φ0 (x) = rh0 (θ) in polar coordinates with some h0 (·) ∈ C 1,α (S 1 ), where h0 can be appropriately chosen that φ0 satisfies Definition 4.3.1. Moreover, the same holds for the image Ψ(Ω0 ) of such a patch Ω0 under a global C 1,α -diffeomorphism of the plane Ψ satisfying |Ψ(x)| ≤ C|x|1+α for some C > 0. These facts are proved in Lemma 4.4.4 of the next section, where we study in detail the evolution of such vortex patches, under the assumption of m-fold symmetry. This class of vortex patches (which are locally the C 1,α -diffeomorphic image of a union of sectors meeting at the origin) are studied in great detail in Section 4.4. Unfortunately, the fact ˚α for all time is not sufficient to conclude that the evolved patch is still that ∇⊥ φ stays in C given by the image of some C 1,α -diffeomorphism. Therefore, a careful local analysis should be supplemented to recover this information (see Subsection 4.4.2). (ii) Logarithmic spirals: Take some indicator function χI where I is some interval of S 1 = [0, 2π) and consider a patch Ω0 which is locally given by
ω0 (r, θ) = χI ◦ (−c log(r) + θ) ,
r < 1/2
where c > 0 is some constant. Taking h0 ∈ C 1,α (S 1 ) vanishing precisely on the endpoints of the interval I with non-zero derivatives, and then by setting φ0 = rh0 (−c log(r) + θ), one may check that this function satisfies the requirements of Definition 4.3.1 (assuming for instance Ω0 is a C 1,α patch in r ≥ 1/2). This boils down to checking that, for a given function ζ ∈ C α (S 1 ) ˚α (R2 ). For simplicity, take the case α = 1, and then with 0 < α ≤ 1, ζ ◦ (−c ln r + θ) ∈ C 1 1 ∂θ ζ = ζ 0 , r r
c ∂r ζ = − ζ 0 , r
so that switching to rectangular coordinates, |x||∇ζ(x)| ∈ L∞ (R2 ), or equivalently ζ(x) ∈ ˚1 (R2 ). Similarly as in the case of (i), one can treat patches which are given as the image of C an exact spiral by a C 1,α -diffeomorphism of the plane fixing the origin. In the special case when the initial vorticity is given exactly by ω0 = h0 (−c log r + θ), then a 1D evolution equation satisfied by h(·, t) can be derived, so that ω(x, t) := h(−c log r + θ, t) solves the 2D Euler equations. This remark is due to Julien Guillod (private communication). It is interesting question to see if one can start with a patch which locally looks like a union of sectors (as in the case (i)) and converges to a logarithmic spiral when t → +∞. 117
The patch corresponding to the case c = 5 and I = [0, 5π/24], with 3-fold symmetrization, is given in Figure 4.2. (iii) Cusps: Consider the (infinite) region bounded by two tangent C 1,α -functions f0 , g0 : [0, ∞) → R:
Ω0 = {(x1 , x2 ) : g0 (x1 ) < x2 < f0 (x1 )},
f00 (0) = g00 (0) = 0,
g0 < f0 on (0, ∞).
Here, we require that f0 and g0 are uniformly C 1,α in all of R. A model case is provided by taking f0 (x1 ) = x1+α and g0 (x1 ) = −x1+α (locally for x1 near 0). One may take a number of 1 1 such cusps (possibly with different boundary profiles for each of them) and rotate each of them around the origin to make them disjoint. In particular, the resulting union of cusps can be m-fold symmetric for any m ≥ 3. In this setting, it is convenient to consider the complement ˚α R2 \Ω, which is more-or-less a union of corners. Then one may take some φ0 with ∇⊥ φ0 ∈ C defined on R2 \Ω0 . It can be taken to be C 1,α smooth when one “crosses” each of the cusps (see Figure 4.8). We discuss them in some detail in Subsection 4.4.5. Danchin has shown in [26] that the cusp-like singularities in a smooth vortex patch propagates globally in time. It is likely that the following alternative argument for the global well-posedness would go through: first apply Theorem 4.3.2 to obtain global propagation in the intermediate ˚α , and supply an additional local argument to recover C 1,α -regularity up to the point of class C singularity. (iv) Bubbles accumulating at the origin: Take a sequence of smooth C 1,α -patches {Un }n≥0 , which for simplicity are assumed to have comparable diameters (say less than 1/2) and C 1,α -characteristic ˜n , and place it inside scales. Now rescale the n-th patch Un by a factor of 2−n , denote it by U the annulus An = {x : 2−n < |x| < 2−n+1 }. Then define Ω0 as the union of rescaled patches ˜n . It can be easily arranged that, by placing several disjoint patches in each annulus ∪n≥0 U region, the entire set Ω0 is m-fold symmetric for some m ≥ 3. Assuming m-fold symmetry, Theorem 4.3.2 applies to show that the evolution of the (rescaled) ˜n has boundary in C 1,α with its characteristic satisfying n-th patch U
c(T )2n ≤ Γn (t) ≤ C(T )2n ˜n back to a patch of diameter for any T > 0 and t ∈ [0, T ]. In particular, by rescaling each of U 118
Figure 4.2: A 3-fold symmetric logarithmic spiral. O(1), we have that their C 1,α -characteristics are uniformly bounded from above and below. ˜n stays in C 1,α for all Even without the symmetry, it can be shown that the boundary of each U time. However, a uniform bound (after rescaling) cannot hold in general. Indeed, such a nonuniform growth was utilized in the work of Bourgain-Li [12] (see also [38]), after smoothing out the patches appropriately, to produce examples of ω0 ∈ H 1 (R2 ) which escapes H 1 (R2 ) instantaneously for t > 0. The proof of Theorem 4.3.2 is parallel to the one given in [10] and based on two “frozen time” estimates, except that the m-fold rotational symmetry gets involved in the current setup. We first observe that in this setting, an identity of the form (4.3) still holds:
∇u∇⊥ φ(x) =
Z Ω
∇K(x − y) ∇⊥ φ(x) − ∇⊥ φ(y) dy,
(4.15)
since all that was necessary to establish the above formula is to have the vector field ∇⊥ φ divergence free and tangent to the boundary of the patch. Given the identity (4.15), we can prove the following estimate: Lemma 4.3.4. Assume that a domain Ω admits a level set φ satisfying Definition 4.3.1. Then, we have a bound
k∇u∇⊥ φkC˚α (R2 ) ≤ C 1 + k∇ukL∞ (R2 ) k∇⊥ φkC˚α (R2 ) .
119
˚α , which works in the setting This is just a particular case of a general estimate about the space C of convolution against classical Calderon-Zygmund kernels. It is worth noting that the symmetry is not necessary for this particular lemma. Next, Lemma 4.3.5. Under the assumptions of Lemma 4.3.4, we have the following logarithmic bound:
k∇ukL∞ (R2 ) ≤ Cα
1 + log 1 +
k∇⊥ φkC˚α (R2 )
!!
k∇⊥ φkinf(∂Ω)
(4.16)
˚α The symmetry assumption is essential here; basically, the information that ∇⊥ φ belongs to C gives an effective C 1,α bound on ∂Ω only in a region of O(|x|) at a given point x, and the procedure of “zooming out” it to a region of size O(1) will in general bring the logarithmic loss, unless the m-fold rotational symmetry for some m ≥ 3 is imposed on the set Ω. Given these lemmas, let us give a sketch of the proof. ˚α -characteristic for ∇⊥ φt Proof of Theorem 4.3.2. We have shown already that as long as the C remains finite, the solution can be extended further. It suffices to obtain a global-in-time a priori estimate for the characteristic quantity3
˚ Γt =
k∇⊥ φt kC˚α (R2 )
!1/α
k∇⊥ φt kinf(∂Ωt )
.
As we have mentioned earlier, this proof is completely parallel to the arguments of Bertozzi and Constantin [10]. We start with W := ∇⊥ φ, which satisfies ∂t W + (u · ∇)W = ∇uW. Then, solving this equation along the flow, d W (Φ(x, t), t) = ∇u(Φ(x, t), t)W (Φ(x, t), t). dt Integrating in time and then changing variables gives
W (x, t) = W0 (Φ−1 t (x)) +
Z
t
(∇uW )(Φ−1 t−s (x), s)ds.
0 3 Note that, unlike the C 1,α -characteristic quantity that appeared earlier in the case of smooth patches, this quantity is non-dimensional. We use the notation ˚ Γt to emphasize this fact from now on.
120
Using the bound on ∇Φ−1 in terms of the velocity gradient, this implies, for points x 6= x0 satisfying t |x0 | ≤ |x| and |x − x0 | ≤ |x|/2, Z t |x − x0 |α |W (x, t) − W (x0 , t)| ≤ kW0 kC˚α exp c k∇us kL∞ ds · |x|α 0 Z t Z t |x − x0 |α . + k∇us Ws kC˚α exp c k∇us0 kL∞ ds0 ds · |x|α 0 s Introducing Q(s) = k∇us kL∞ and using Lemma 4.3.4, kWt kC˚α
Z t Z t Z t 0 0 Q(s )ds ds. Q(s)kWs kC˚α exp c Q(s)ds + C ≤ kW0 kC˚α exp c s
0
0
(For a pair of points x 6= x0 and |x0 | ≤ |x| not satisfying |x − x0 | ≤ |x|/2, we can simply use the L∞ -bound ∂t kWt kL∞ ≤ Qt kWt kL∞ .) Then, writing Z t Q(s)ds , G(t) := kWt kC˚α exp −c 0
we have, after a little bit of manipulation,
G(t) ≤ kW0 kC˚α + C
Z
t
Q(s)G(s)ds, 0
so that by Gronwall’s Lemma, kWt kC˚α ≤ kW0 kC˚α exp (C + c)
t
Z
k∇us k
L∞
0
ds .
On the other hand, we have trivially
kWt kinf(∂Ω)
Z t ∞ ≥ kW0 kinf(∂Ω) exp − k∇us kL ds . 0
Combining these estimates, and then applying Lemma 4.3.5 finishes the proof. Proof of Lemma 4.3.4. Let us set
G(x) = ∇u∇⊥ φ(x). Then, we have trivially an L∞ bound: |G(x)| ≤ k∇ukL∞ k∇⊥ φkL∞ . Now the proof of the C α -
121
estimate for |x|α G(x) is strictly analogous to the proof of (4.4) given in [10, Proof of Corollary 1]. To see this, fix some x, h and consider the difference
|x|α G(x) − |x + h|α G(x + h). First, in the case |h| > |x|/2, after a rewriting the above expression is bounded in absolute value by ||x|α (G(x) − G(x + h)) + G(x + h)(|x|α − |x + h|α )| ≤ C|h|α kGkL∞ + |h|α kGkL∞ . Therefore, we may assume that |h| ≤ |x|/2. Then, we write with f := ∇⊥ φ |x|α G(x) − |x + h|α G(x + h) Z Z = |x|α ∇K(x − y)(f (x) − f (y))dy − |x + h|α ∇K(x + h − y)(f (x) − f (y))dy Ω Ω Z = |x|α ∇K(x − y)(f (x) − f (y))dy {|x−y|δ Z 1 1 + 0 dy ≤ Cδ −α . ≤ C|x − x0 |1−α 3 |x − y|3 |y|>δ |x − y|
1 |x − x0 |α
Now, let us separately consider two integrals (x − y1 )2 − (h(x) − y2 )2
Z I1 (x) := Ω∩[−δ,δ]2
2 dy
((x − y1 )2 + (h(x) − y2 )2 )
and
I2 (x) := f 0 (x)
2(x − y1 )(h(x) − y2 )
Z Ω∩[−δ,δ]2
((x − y1 )2 + (h(x) − y2 )2 )
2 dy.
We shall only consider I1 , and just briefly comment on the other term I2 below. One can further write:
I1 (x) =
4 X
I1j (x),
j=1
where
I1j (x) =
Z Ωj ∩[−δ,δ]2
(x − y1 )2 − (h(x) − y2 )2
((x − y1 )2 + (h(x) − y2 )2 )
2 dy
(recall that Ωj := Rπ(j−1)/2 (Ω1 )). We have, after integrating in y2 , I11 (x)
Z
Z
= 0≤y1 ≤δ
Z
δ
= 0
g(y1 )≤y2 ≤f (y1 )
(x − y1 )2 − (h(x) − y2 )2
2 dy2 dy1
((x − y1 )2 + (h(x) − y2 )2 )
h(x) − f (z) h(x) − g(z) − dz (x − z)2 + (h(x) − f (z))2 (x − z)2 + (h(x) − g(z))2
(we have renamed y1 by z for simplicity). Similarly, I12 (x)
Z
(x − y1 )2 − (h(x) − y2 )2
Z
= 0≤y2 ≤δ
Z = 0
δ
−f (y2 )≤y1 ≤−g(y2 )
2 dy1 dy2
((x − y1 )2 + (h(x) − y2 )2 )
x + f (z) x + g(z) − dz (x + f (z))2 + (h(x) − z)2 (x + g(z))2 + (h(x) − z)2
tations below.
131
Claim. The integral Z 0
δ
h(x) − f (z) x + f (z) + dz (x − z)2 + (h(x) − f (z))2 (x + f (z))2 + (h(x) − z)2
(4.19)
defines a C α -function of 0 ≤ x ≤ δ/10 with C α -norm bounded by the right hand side of (4.18). Once we show the Claim (together with the upper bound stated in (4.18)) for h = f and h = g, this concludes the proof that I1 (x) belongs to C α , since each of I11 + I12 and I13 + I14 belongs to C α , by symmetry. A similar argument can be given for the other term I2 (x): we write it as
I2 (x) = f 0 (x)
4 X
I2j (x),
j=1
where
I2j (x) =
Z Ωj ∩[−δ,δ]2
2(x − y1 )(h(x) − y2 ) dy. ((x − y1 )2 + (h(x) − y2 )2 )2
Then, we can integrate each of I2j once with respect to either y1 or y2 , resulting in similar expressions as above. Let us consider the case h(z) ≡ f (z), which is actually the most difficult case. In this specific case, we rewrite the integrand in (4.19) as
f 0 (x) 1 f (x) − f (z) − · (x − z)2 + (f (x) − f (z))2 1 + (f 0 (x))2 x − z f 0 (x) F f 0 (x) 1 F x + f (z) − · + + + (x + f (z))2 + (f (x) − z)2 1 + (f 0 (x))2 x + f (z) 1 + (f 0 (x))2 x − z x + f (z)
where F := f 0 (0). Let us first estimate in C α the last term, which we further rewrite as: f 0 (x) 1 + (f 0 (x))2
1 F + x−z x + Fz
+F ·
F z − f (z) . (x + f (z))(x + F z)
Since f 0 ∈ C α , it suffices to estimate in C α the integrals of two terms in large brackets. Regarding the first term, one just explicitly evaluate that Z 0
δ
F 1 + dz = log x−z x + Fz
Fδ + x δ−x
,
which is clearly bounded in C α by the right hand side of (4.19). Note that the logarithmically
132
divergent terms (as x → 0+ ) present in each of the integrals cancel each other exactly. Regarding the second term, we first note that it is uniformly bounded: Z Z δ δ z F F z − f (z) dz ≤ C . dz ≤ CF 2 + (F z)2 0 (x + f (z))(x + F z) x 1 + F2 0 To bound the C α -norm, we need to estimate for 0 ≤ x < x0 Z 0
δ
1 1 1 dz, · |F z − f (z)| · − 0 0 α 0 |x − x | (x + f (z))(x + F z) (x + f (z))(x + F z)
and simply using that |F z − f (z)| ≤ kf kC 1,α |z|1+α , we bound the above by δ
|x − x0 |1−α z 1+α (x + x0 + F z) dz 0 0 0 (x + f (z))(x + F z)(x + f (z))(x + F z) Z ∞ (|x − x0 |2 + z 2 )(x + x0 + F z) ≤ Ckf kC 1,α · dz ≤ Ckf kC 1,α , (x2 + (F z)2 )(x02 + (F z)2 ) 0 Z
Ckf kC 1,α ·
where we have used an elementary inequality
A1+α B 1−α ≤ C(A2 + B 2 ). It remains to estimate δ
Z T1 (x) = 0
f (x) − f (z) f 0 (x) 1 − · dz (x − z)2 + (f (x) − f (z))2 1 + (f 0 (x))2 x − z
and δ
Z T2 (x) = 0
x + f (z) f 0 (x) F − · dz. (x + f (z))2 + (f (x) − z)2 1 + (f 0 (x))2 x + f (z)
We begin with T1 (x). After a bit of re-arranging, we have Z
δ
T1 (x) = 0
Z = 0
1 x−z
1+
δ
1 x−z
f (x)−f (z) x−z
0
f (x)−f (z) x−z
2 −
1
1+
f (x) dz 1 + (f 0 (x))2
f (x)−f (z) x−z
1 + 1 + (f 0 (x))2
133
2 −
1 f (x) − f (z) 1 + (f 0 (x))2 x−z
f (x) − f (z) 0 − f (x) dz. x−z
Let us first estimate the latter term: dropping the multiplicative factor which belongs to C α , Z
δ
0
1 x−z
f (x) − f (z) 0 − f (x) dz. x−z
Take two points 0 ≤ x < x0 < δ/10, and let us further assume that |x| ≥ |x0 − x| (the other case is simpler). We need to take δ
1 f (x) − f (z) 1 f (x0 ) − f (z) 0 0 − f 0 (x) − 0 − f (x ) dz x−z x −z x0 − z 0 x−z # "Z Z x+|x−x0 |/2 Z x0 +|x−x0 |/2 Z δ x−|x−x0 |/2 1 · · · dz + + + = |x − x0 |α 0 x−|x−x0 |/2 x0 −|x−x0 |/2 x0 +|x−x0 |/2
1 |x − x0 |α
Z
(4.20)
To begin with, we treat the second integral of (4.20): we simply use the bound
|f (x) − f (z) − f 0 (x)(x − z)| ≤ kf kC 1,α |x − z|1+α (and similarly for x replaced by x0 ) to bound it in absolute value by kf kC 1,α |x − x0 |α
Z
x+|x−x0 |/2
x−|x−x0 |/2
α−1
|x − z|α−1 + (|x − x0 | + |x − z|)
dz ≤ Ckf kC 1,α .
The third integral from (4.20) can be treated in a parallel way. Turning to the first integral, we rewrite as 1 |x − x0 |α
Z 0
x−|x−x0 |/2
(f (x) − f (z) − f 0 (x)(x − z)) − (f (x0 ) − f (z) − f 0 (x0 )(x0 − z) (x − z)2 1 1 + − 0 (f (x0 ) − f (z) − f 0 (x0 )(x0 − z))dz (x − z)2 (x − z)2
Note that the numerator of the first term equals
(f (x) − f (x0 ) − f 0 (x)(x − x0 )) + ((x0 − x) + (x − z)) (f 0 (x0 ) − f 0 (x)), and simply using the bounds
|f (x) − f (x0 ) − f 0 (x)(x − x0 )| ≤ Ckf kC 1,α |x − x0 |1+α ,
134
|f 0 (x) − f 0 (x0 )| ≤ Ckf kC 1,α |x − x0 |α ,
we bound the first term by Ckf kC 1,α . The second one can be bounded by kf kC 1,α |x − x0 |1−α
Z
x−|x−x0 |/2
0
|x − z| + |x − x0 | 0 |x − z|α−1 dz (x − z)2
and after a change of variable v := (x − z)/|x0 − x|, ≤ Ckf k
C 1,α
0 1−α
|x − x |
0 α−1
· |x − x |
Z
∞
1/2
(1 + v)α dv ≤ Ckf kC 1,α . v2
Now the last integral of (4.20) can be treated in an analogous fashion. To finish the estimate of T1 , we still need to consider the expression δ
Z 0
f (x) − f (z) 1 · · x−z x−z 1 =− 1 + (f 0 (x))2
Z 0
δ
1 1+
f (x)−f (z) x−z
2 −
1 dz 1 + (f 0 (x))2
1 f (x) − f (z) · · x−z x−z
(z) − f 0 (x) · f (x)−f + f 0 (x) x−z dz, 2 ˜ f˜(z) 1 + F + f (x)− x−z
f (x)−f (z) x−z
where F = f 0 (0) and f˜(x) = f (x) − F x. Consider the expansion 1
1+ F +
f˜(x)−f˜(z) x−z
2 =
1
1 + F2 + F +
f˜(x)−f˜(z) x−z
2
− F2
m !2 m X ˜(x) − f˜(z) 1 f 1 · (−1)m · F + − F 2 , = 1 + F2 1 + F2 x−z m≥0
which is convergent simply because kf˜0 kL∞ [0,δ] ≤ 1/10 from our choice of δ. Inspecting the terms, it suffices to estimate in C α the following integrals: Z 0
δ
1 x−z
f˜(x) − f˜(z) x−z
!m
f (x) − f (z) − f 0 (x) dz. x−z
We have already treated the case m = 0 in the above. For any m ≥ 1, the same proof carries over, resulting in a bound (using that kf˜0 kL∞ [0,δ] ≤ 1/10)
Z
δ 1
0 x−z
f˜(x) − f˜(z) x−z
!m
f (x) − f (z)
0 − f (x) dz
x−z
C α [0,δ/10]
≤
Cm2m kf kC 1,α [0,δ] , 10m
which is clearly summable in m. This concludes the argument for T1 (x). The other term T2 (x) can
135
be treated similarly, and it is simpler since the corresponding integral is less singular than that of T1 . We now sketch a proof that Claim holds in the case h(x) ≡ g(x). In this case, the arguments are simpler since we have a gap |h(x) − g(x)| & |x|. It suffices to show that the differences δ
g(x) − f (z) Gx − F z − dz, (x − z)2 + (g(x) − f (z))2 (x − z)2 + (Gx − F z)2
δ
x + f (z) x + Fz − dz, (x + f (z))2 + (g(x) − z)2 (x + F z)2 + (Gx − z)2
Z
δ
Z 0
Z 0
and
0
Gx − F z x + Fz + dz (x − z)2 + (Gx − F z)2 (x + F z)2 + (Gx − z)2
belong to C α with appropriate bounds, where G := g 0 (0) < 0. To begin with, the last integral can be evaluated directly, F (−F + G)x 1 −1 + log (1 + G2 )x2 + (1 + F 2 )z 2 − 2xz(1 + F G) tan − 1 + F2 −(1 + F G)x + (1 + F 2 )z 2 δ x(1 + F G) F 2 2 2 2 + tan−1 − log (1 + G )x + 2(F − G)xz + (1 + F )z , (F − G)x + (1 + F 2 )z 2 0 which gives a C α -function of x: two logarithmic terms cancel each other, and a simple computation shows that
−1
tan
Ax
≤ C(A, B)δ −α Bx + δ C α [0,δ/10]
for nonzero constants A and B. Now we turn to the first integral, which equals Z 0
δ
(x − z)2 ((Gx − g(x)) − (F z − f (z))) ((x − z)2 + (g(x) − f (z))2 ) ((x − z)2 + (Gx − F z)2 ) (g(x) − f (z))(Gx − F z) ((g(x) − f (z)) − (Gx − F z)) + dz ((x − z)2 + (g(x) − f (z))2 ) ((x − z)2 + (Gx − F z)2 )
(4.21)
Here, the key points are: • On the numerator, we gain an extra power of |x|α or |z|α , from H¨older continuity of f 0 and g 0 . • The denominator is uniformly bounded from above and below by constant multiples of x2 + z 2 . We sketch the proof of C α -continuity for the first term only, since the second one can be treated 136
similarly. We need to estimate 1 |x − x0 |α
Z 0
δ
(x − z)2 ((Gx − g(x)) − (F z − f (z))) ((x − z)2 + (g(x) − f (z))2 ) ((x − z)2 + (Gx − F z)2 ) (x0 − z)2 ((Gx0 − g(x0 )) − (F z − f (z))) − 0 dz ((x − z)2 + (g(x0 ) − f (z))2 ) ((x0 − z)2 + (Gx0 − F z)2 )
(4.22)
and we may assume |x − x0 | ≤ |x|. Let us even further assume that the denominators in (4.22) are the same, as they are roughly of the same size (and bounded uniformly from below by a constant multiple of x2 + z 2 and x02 + z 2 , respectively). Then, the resulting difference is bounded by:
C(kf kC 1,α + kgkC 1,α )
1 |x − x0 |α
Z
δ
0
|x − x0 | (|x| + |x − x0 | + |z|) |x|1+α + |z|1+α dz (x2 + z 2 )2
and at this point, the C α -bound simply follows from rescaling the variable z = xv. The actual proof can be done for instance by expanding one of the denominators in (4.22) around the other denominator in a power series as we have done earlier. The argument for the other component
d dx u1
is completely analogous. We just note that along
a curve (x, h(x)), it has the form: Z 1 d −(h(x) − y2 ) d u1 (x, h(x)) = dy dx 2π dx Ω (x − y1 )2 + (h(x) − y2 )2 Z −h0 (x) (x − y1 )2 + (h(x) − y2 )2 + 2(h(x) − y2 ) ((x − y1 ) + h0 (x)(h(x) − y2 )) 1 = dy. 2 2π Ω ((x − y1 )2 + (h(x) − y2 )2 ) This finishes the proof.
4.4.3
Proof of the main result
We are now in a position to complete the proof of Theorem 4.4.3. Let us recall that as a consequence of Theorem 4.3.2, for any T > 0, we have L∞ -bounds
sup t∈[0,T ]
k∇Φt kL∞ + k∇Φ−1 t kL∞ + k∇ukL∞ ≤ C(T ),
(4.23)
and moreover, for any r > 0, sup k∇ut kC α (R2 \B0 (r)) ≤ C(T )r−α .
t∈[0,T ]
137
(4.24)
From the local well-posedness result (Proposition 4.2.2), we know that at least for some short time interval [0, T1 ], each piece of the boundary of Ω1 (t) remains uniformly C 1,α up to the origin. Proof of Theorem 4.4.3. We just need to show an a priori estimate which is sufficient to guarantee that for all t > 0, the boundary of Ω1 (t) is given by two C 1,α -curves ft and gt (after rotating the plane if necessary) on some nonempty interval [0, δt ]. Let us consider only the evolution of the upper boundary (x, f0 (x)), since the other part can be treated similarly. We shall fix some T > 0, and use the bounds (4.23) and (4.24). Since ft itself does not obey a simple evolution equation, let us work directly with the particle trajectories
η 1 (t, x) := Φ1 (t, (x, f (x))),
η 2 (t, x) := Φ2 (t, (x, f (x))),
which is well-defined on x ∈ [0, δ0 ]. Then, since η(t, ˙ x) = u(t, η(t, x)), we have upon differentiating ∂ ∂t
∂ ∂ η(t, x) = ∇u(t, η(t, x)) η(t, x) . ∂x ∂x
First, from (4.23) we have
sup |∂x ηt | ≤ C(T ) < +∞,
x∈[0,δ0 ]
inf
x∈[0,δ0 ]
|∂x ηt | ≥ c(T ) > 0.
Then, from Lemma 4.4.5 and (4.24), we have a bound
k∇ut ◦ ηkC α [0,δ0 ] ≤ C(T ) k∂x ηkC α [0,δ0 ] + δt−α for some δt > 0, as long as the boundary of the Ω1 (t) is given by the graph of two C 1,α curves ft and gt defined on [0, cδt ] for some constant c > 0. However, from the inverse function theorem, we can say that7
−1/α
c(T )k∂x ηkC α [0,δ0 ] ≤ δt 7 Strictly speaking, to apply the inverse function theorem, we need to estimate the C 1,α -norm of the lower piece of the boundary at the same time.
138
as well as
kft kC 1,α [0,δt ] + kgt kC 1,α [0,δt ] ≤ C(T )k∂x ηt kC α [0,δ0 ] for all t ∈ [0, T ] with some constants c(T ), C(T ) > 0 depending only on T . Then simply from the algebra property of the space C α , we deduce an a priori bound d k∂x η(t)kC α [0,δ0 ] ≤ k∇ut kL∞ k∂x η(t)kC α [0,δ0 ] + k∇ut ◦ ηt kC α [0,δ0 ] k∂x η(t)kL∞ dt ≤ C(T )k∂x η(t)kC α [0,δ0 ] . This shows that for all t > 0, k∂x ηkC α as well as δt remains finite and non-zero, respectively. It remains to show that the angle of each corner stays the same for all times. For this purpose, let us decompose
ω = ω homog + ω cusp + ω f ar , where ω homog is the 0-homogeneous vorticity which is the characteristic function of the m-fold symmetrization of the infinite sector
{(x1 , x2 ) : 0 < x1 , Gt x1 < x2 < Ft x1 },
Ft = ft0 (0),
Gt = gt0 (0),
and ω cusp is simply χB0 (δt ) · χΩt − ω homog . To be concrete, modulo m-fold symmetry,
ω cusp (x1 , x2 ) =
+1
if
−1
if
Ft x1 < x2 < ft (x1 )
or
Ft x1 > x2 > ft (x1 )
or
gt (x1 ) < x2 < Gt x1
,
gt (x1 ) > x2 > Gt x1
inside the ball B0 (δt ). Then ω f ar is defined as ω − ω homog − ω cusp , and one note that it is supported outside the ball B0 (δt ). Then, accordingly, we obtain a decomposition of the velocity
u = uhomog + ucusp + uf ar ,
and we claim that each of them induces the same rotation speed on the boundary curves (x1 , ft (x1 )) and (x1 , gt (x1 )) for all times. To begin with, the radially homogeneous component uhomog induces the same rotation speed on
139
the tangent lines (x1 , Ft x1 ) and (x1 , Gt x1 ). However, since ∇u is bounded for all time, the angle between (x1 , Ft x1 ) and (x1 , ft (x1 )), and also between (x1 , Gt x1 ) and (x1 , gt (x1 )) stays zero. Next, we know that uf ar is C 1,α (indeed, C ∞ ) inside B0 (δt /2). Therefore, the associated stream function ψ f ar is C 2,α in the ball.8 Taylor expansion gives
ψ f ar (x1 , x2 ) = A + Bx1 + Cx2 + D(x21 + x22 ) + Ex1 x2 + O(|x|2+α ). However, A = 0 by assumption and B = C = E = 0 is forced under the m-fold rotational symmetry. In particular,
uf ar (x1 , x2 ) = ∇⊥ ψ f ar (x1 , x2 ) = 2Dx⊥ + O(|x|1+α ) which is a uniform rotation on all of R2 as |x| → 0. Lastly, it is known that within each connected component of the complement of the (closure of the) support of ω cusp , the associated velocity ucusp is uniformly C 1,α up to the boundary. It follows from our computations in Subsection 4.4.2 but also directly from the arguments of Friedman and Vel´azquez [41] (see the statement of their Lemma in the previous section). Then, an identical argument as in the case of uf ar shows that, this time, ucusp is a uniform rotation up to a term of order |x|1+α in the complement of the support of ω cusp . The proof is now complete.
4.4.4
Multiple corners and cusp formation in infinite time
In the above main result, we have only dealt with the case when there is a single corner in a sector of angle 2π/m (which serves as a fundamental domain for rotations by multiples of 2π/m). In this case, we have seen that the angle of the patch is preserved for all time. However, one may consider the case when there are several corners (separated from each other by some angle; see Figure 4.5) in each fundamental domain, and then some interesting dynamics for the angles can be observed. We just note that an essentially identical proof carries over to this case to establish global wellposedness of such patches, and also the fact that the angles evolve exactly as in the case of infinite sectors, up to a constant overall rotation. We just modify the last item from the Definition 4.4.1 to allow such patches: • (multiple C 1,α corners) There exists a C 1,α diffeomorphism Ψ : R2 → R2 of the plane with 8 Here, although ω f ar may have non-compact support, the Poisson problem ∆ψ f ar = ω f ar has a unique solution with ψ f ar ∈ W 2,∞ (R2 ) under m-fold rotational symmetry with m ≥ 3 and ψ f ar (0) = 0; see [37, Lemma 2.6].
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Figure 4.5: A 3-fold symmetric patch with multiple corners. Ψ(0) = 0 and ∇Ψ|x=0 = I, such that for some δ > 0, the image Ψ(Ω1 ) is a union of exact sectors with total angle less than 2π/m:
Ψ(Ω1 ) ∩ B0 (δ) = ∪kj=1 Sβk (ζk ) ∩ B0 (δ)
(4.25)
with some 0 < ζj and −π ≤ βj < π satisfying βj + ζj < βj+1
and βj+1 + ζj+1 − β1 < 2π/m for all
j = 1, · · · , k − 1,
(the ordering is well-defined on the interval [−π, π], assuming without loss of generality that −π ≤ β1 < 0). Then as before, we define Ω = ∪m−1 i=0 R2πi/m (Ω1 ). Alternatively, we may describe the patch locally as a union of approximate sectors with angles ζ1 , · · · , ζk , in counter-clockwise order, with gaps between them γ1+1/2 , · · · , γk−1+1/2 where γj+1/2 := βj+1 − βj − ζj . Note that given some value of m ≥ 3, the values ζ1 , · · · , ζk together with γ1+1/2 , · · · , γk−1/2 determine the local shape of the patch, up to a rotation of the plane. Corollary 4.4.6 (Dynamics of the angles). Assume that Ω0 is a symmetric C 1,α -patch with multiple corners as defined in the above, with corner angles ζ1 (0), · · · , ζk (0) with separation angles γ1+1/2 (0), · · · , γk−1/2 (0). Then, the angles evolve according to the following system of ordinary
141
differential equations for all t ∈ R: "j−1 m X m m dζj (t) = Cm sin ζj sin ζl cos (2(βj − βl ) + (ζj − ζl )) dt 4 4 4 l=1
−
k X l=j+1
m m ζl cos (2(βj − βl ) + (ζj − ζl )) sin 4 4
(4.26)
and " j m X m m dγj+1/2 (t) = Cm sin γj+1/2 ζl cos ((βj+1 − βl ) + (βj − βl ) + (ζj − ζl )) sin dt 4 4 4 l=1
−
k X l=j+1
m m sin ζj cos ((βj+1 − βl ) + (βj − βl ) + (ζj − ζl )) 4 4 (4.27)
with βj − βl = γj−1/2 + · · · + γl+1/2 + (ζj−1 + · · · + ζl ) , βl − βj = γl−1/2 + · · · + γj−1/2 + (ζj−1 + · · · + ζl−1 ) ,
j>l l≥j
for some constant Cm > 0 depending only on m. Proof. It suffices to recall the 1D system describing the evolution of 0-homogeneous vorticities. On the unit circle, we are given initial vorticity
h0 (θ) =
m−1 k XX
χ(βj +2πi/m,βj +ζj +2πi/m) .
i=0 j=1
Moreover, given h, the corresponding angular velocity (counter-clockwise rotation) on the circle is defined explicitly by Z
π/m
v(θ) =
c1m sin
−π/m
m 2
|θ − θ0 | − c2m h(θ0 )dθ0
for some constants c1m > 0 and c2m depending only on m ≥ 3. Since the integral of h over the circle is conserved in time, one may redefine the angular velocity to be
v˜(θ) = c1m
Z
π/m
sin
m
−π/m
142
2
|θ − θ0 | h(θ0 )dθ0
up to an overall rotation. Therefore, d ζj (t) = v˜(βj + ζj ) − v˜(βj ) dt Z βl +ζl h j−1 m i X m (βj + ζj − θ) − sin (βj − θ) dθ = c1m sin 2 2 βl l=1 Z k βl +ζl h m i m X + (θ − βj − ζj ) − sin (θ − βj ) dθ c1m sin 2 2 βl l=j+1
=
j−1 X
c0m sin
l=1
−
k X
m m m ζj sin ζl cos (2(βj − βl ) + (ζj − ζl )) 4 4 4 c0m sin
l=j+1
m m m ζj sin ζl cos (2(βj − βl ) + (ζj − ζl )) 4 4 4
(note that the contribution from the j-th sector cancels out) and the relations βj − βl = γj−1/2 + · · · + γl+1/2 + (ζj−1 + · · · + ζl ) , βl − βj = γl−1/2 + · · · + γj−1/2 + (ζj−1 + · · · + ζl−1 ) ,
j>l l≥j
enables us to express the right hand side in terms of γ’s and ζ’s. Similarly, d γj+1/2 (t) = v˜(βj+1 ) − v˜(βj + ζj ) dt j m m m X γj+1/2 sin ζl cos ((βj+1 − βl ) + (βj − βl ) + (ζj − ζl )) = c0m sin 4 4 4 l=1
−
k X
c0m sin
l=j+1
m m γj+1/2 sin ζl cos ((βj+1 − βl ) + (βj − βl ) + (ζj − ζl )) 4 4 4
m
This finishes the proof. A case study: 8 sectors with 4-fold symmetry To demonstrate that there is some non-trivial dynamics of sectors, we consider the case when there are two sectors in a fundamental domain, assuming 4-fold rotational symmetry. Let us write Ω1 = Ω11 ∪ Ω21 , and in this case, up to a rotation of R2 , we only need to specify three angles; angles of each Ωj1 and the angle in between. We denote them by ζ1 , ζ2 , and γ, respectively (see Figure 4.6). Then, the systems of equations (4.26), (4.27) reduce to, up to a multiplicative constant which we neglect,
ζ˙1 = − sin(ζ1 ) sin(ζ2 ) cos(2γ + ζ1 + ζ2 ), 143
(4.28)
Ω21
ζ2 γ
Ω11 ζ1
Figure 4.6: Evolution of two angles ζ˙2 = sin(ζ2 ) sin(ζ1 ) cos(2γ + ζ1 + ζ2 ),
(4.29)
γ˙ = sin(γ) sin(ζ1 − ζ2 ) cos(γ + ζ1 + ζ2 )
(4.30)
and
(Observe that ζ1 + ζ2 is an invariant of motion, as it should be). It can be easily shown that for exact infinite sectors, the whole patch Ω defines a purely rotating state if and only if ζ2 = ζ1 and γ = π/4 − ζ1 (in which case Ω is indeed 8-fold symmetric) or one of the sectors is degenerate, that is, either ζ2 = 0 or ζ1 = 0. In this restricted setting, it can be established that any state converges as t → +∞ to such a purely rotating space, and generically to a state where one of the angles become zero. Indeed, fix ζ1 (0) + ζ2 (0) = π/4, and moreover γ(0) = ζ2 (0). Then using the system (4.28), (4.29), (4.30) one sees that γ = ζ2 for all time. Alternatively, assuming such an initial data, subtracting −1/2 from vorticity everywhere in the plane gives an odd configuration with respect to the line separating ζ2 and γ (see Figure 4.6). Since the Euler equations preserve odd symmetries of vorticity, it follows that γ = ζ2 for all time. Therefore, the one-dimensional system we get is:
γ˙ = sin(γ) sin(π/4 − 2γ) cos(γ + π/4).
(4.31)
On the other hand, if we keep the assumption ζ1 (0) + ζ2 (0) = π/4 but now take γ(0) = γ1 (0), we
144
0.8
0.6
0.4
0.2
0.0 0.0
0.2
0.4
0.6
0.8
Figure 4.7: The phase portrait of the system (4.28), (4.29), (4.30) under the assumption ζ1 = π/4−ζ2 . The axis correspond to variables ζ2 and γ taking values in [0, π/4], which play symmetric roles. instead obtain
γ˙ = − sin(γ) sin(π/4 − 2γ) cos(γ + π/4).
(4.32)
Then, one sees that for any initial value 0 < γ(0) < π/4, it always converges to π/8 for t → +∞ in the former case, and either 0 to pi/4 in the latter. Actually, under the constraint ζ1 + ζ2 = π/4, the former case is the only situation where the forward asymptotic state is 8-fold symmetric, not just 4. See a plot of the phase portrait in Figure 4.7. Appealing to Corollary 4.4.6, we have shown the following Corollary 4.4.7. Assume that Ω0 is a 4-fold symmetric C 1,α -patch with two angles 0 < ζ1 (0) and 0 < ζ2 (0) separated by an angle 0 < γ, in a fundamental domain. Further assume that ζ1 (0)+ζ2 (0) = π/4 and γ(0) 6= ζ2 (0). Then, depending on whether ζ2 (0) > γ(0) or γ(0) > ζ2 (0) holds, we have ζ1 (t) → 0 or ζ2 (t) → 0, respectively, as t → +∞. That is, one of the two components of Ωt in each fundamental domain cusps in infinite time. The speed of cusp formation is exponential in time. Proof. We only need to show that the angle collapses with an exponential rate. Without loss of generality, assume that ζ1 → 0 as t → +∞. Then, we have γ → 0 as well, and the equation (4.28) asymptotically becomes
ζ˙1 ≈ −Cζ1 for some positive constant C > 0. This finishes the proof.
145
It is likely that for any finite number of sectors, with constant vorticity, there is asymptotic convergence of a purely rotating state.
4.4.5
Extensions
Generality of Serfati and Chemin The results of Serfati [66], [67] and Chemin [17], [19], [18] demonstrates that propagation of boundary regularity for smooth patches is just a special instance – the Euler equations indeed propagates “striated” regularity of vorticity. Here we present the version given by Bae and Kelliher [3], [2] To formally state the general result, assume that a family of C α (R2 ) vector fields {Y0λ }λ∈Λ is given, and satisfies the following properties: inf2 sup Y0λ (x) ≥ c0 > 0
x∈R
λ
and
sup kY0λ kC α + k∇ · Y0λ kC α < +∞. λ
Moreover, assume that the initial vorticity satisfies
ω0 ∈ L1 ∩ L∞ (R2 ),
sup k(Y0λ · ∇)ω0 kC α−1 < +∞. λ
The latter condition says that ω0 is C α -regular in the direction of Y0λ . The negative index H¨ older spaces may be defined in terms of the Littlewood-Paley decomposition, but it can be avoided as the above condition is equivalent to K ∗ ((Y0λ · ∇)ω0 ) ∈ C α (see [3]), where K is the usual Biot-Savart kernel. We evolve the family of vector fields by
Ytλ (Φ(x, t)) := (Y0λ (x) · ∇)Φ(x, t). Theorem (See Theorem 8.1 of [3]). In the above setting, the Yudovich solution ωt and the vector fields Ytλ satisfy the global-in-time bounds
sup k(Ytλ · ∇)ωt kC α−1 ≤ C exp(exp(ct)) λ
146
(4.33)
and
sup kYtλ kC α + k∇ · Ytλ kC α ≤ C exp(exp(ct)).
(4.34)
λ
The associated velocity is Lipschitz in space and indeed uniformly C 1,α after being corrected by a smooth multiple of the vorticity. That is, there is a matrix At with kAt kC α ≤ C exp(exp(ct)) such that k∇ut kL∞ ≤ C exp(ct),
(4.35)
k∇ut − ωt At kC α ≤ C exp(exp(ct)) holds. Here, the constants C, c > 0 depend only on 0 < α < 1 and the initial data. Example 4.4.8. Let us present two examples from [3, Section 10]. (i) C 1,α Patches with C α vortex profile: Take some C 1,α -domain Ω0 and C α -function f0 , and then define ω0 = χΩ0 f0 . Then, we can take Y01 := ∇⊥ φ0 where φ0 is a C 1,α level set function for Ω0 . In addition, we may take some vector field Y02 so that {Y01 , Y02 } satisfy all the requirements described in the above (most importantly, Y02 should be non-vanishing whenever Y01 vanishes). We recover the usual vortex patch when the profile f0 is a constant function. This results show that the vorticity can actually have a C α -profile on the patch. This particular statement also follows directly from the main result of Huang [44]. (ii) Vorticity smooth along leaves of a C 1,α -foliation: Consider φ0 ∈ C 1,α with |∇⊥ φ0 | ≥ c > 0 on R2 , such that each level curve of φ0 crosses any vertical line exactly once. Under these assumptions, we define ξx1 (x2 ) so that φ0 (x1 , ξx1 (x2 )) = φ0 (0, x2 ). Take some bounded measurable function W : R → R supported on some bounded interval [c, d]. Then, fix some L > 0 and define
ω0 (x1 , x2 ) := χ[−L,L] (x1 )W (ξx1 (x2 )). The above theorem applies to this case, simply with Y0 = ∇⊥ φ0 . It follows that for all time, all the level curves of ω remain (uniformly in R2 ) C 1,α . In the words of Bae and Kelliher, “extreme lack of regularity of ω0 transversal to Y0 does not disrupt the regularity of the flow lines.”
147
The generalization described in the above theorem can be easily adapted to our setting. Let us only described the necessary modifications in the assumptions. To begin with, we require that ω0 ∈ L1 ∩ L∞ is m-fold symmetric for some m ≥ 3, as usual. We need in addition that there is a distinguished vector field in the family, say Y0c , which is m-fold symmetric and satisfies
inf
x∈B0 (r0 )
|Y0c (x)| ≥ c0 > 0,
for some r0 > 0,
and
kY0c kC˚α (R2 ) + k∇ · Y0c kC˚α (R2 ) + kK ∗ ((Y0c · ∇)ω0 )kC˚α (R2 ) < +∞. ˚α instead of C α when λ = c. Then, we claim that the bounds (4.33), and (4.34) hold, with C ˚α after Moreover, the velocity will be Lipschitz in space for all time, and its gradient will belong to C ˚α -matrix multiple of the vorticity. being corrected by a C Symmetric Cusps Consider a m-fold symmetric set Ω0 which is a union of C 1,α -cusps for some m ≥ 3 in some ball B0 (r0 ) and has C 1,α boundary outside B0 (r0 ). It is possible to show that, using the methods developed here (and the generalization described in the above), the boundaries of the cusp remain as C 1,α (uniformly up to the origin) curves for all time. This can be done as a two-step procedure – the same strategy we have utilized to prove the propagation of C 1,α -corners. Note that the complementary region B0 (r0 )\Ω0 is a disjoint union of regions, each of which can be given as the image of an exact sector under a C 1,α -diffeomorphism of the plane fixing the origin. Therefore, in each of these regions, we can place a (divergence-free) vector field Y0c just as in the case of C 1,α -corners (see Figure 4.8). This vector field can be extended to the interior of the ˚α , and finally ∇ · Y c ∈ C ˚α .9 After that, one cusp, so that Y0c is non-vanishing in B0 (r0 ), Y0c ∈ C 0 takes a complementary vector field Y0b which is C α (R2 ), tangent to the boundary of the patch, with divergence in C α (R2 ) and supported outside the ball B0 (r0 /2). This construction of vector fields ˚α -regularity of the patch. Moreover, the velocity {Y0c , Y0b } gives global-in-time propagation of the C is Lipschitz in space for all time. 9 To see this, consider the simple case of the C 1,1 -cusp given by the region {−x2 ≤ x ≤ x2 , x ≥ 0}. Then, 2 1 1 1 define Y (x1 , x2 ) = (1, 2x2 /x1 ) in the interior of the cusp. Then, ∂x2 Y = (0, 2/x1 ) and ∂x1 Y = −2x2 /x21 so that ˚1 . Finally, ∇ · Y = 2/x1 and hence ∇(∇ · Y ) = (−2/x2 , 0), ||x|∇Y (x)| ∈ L∞ , which is equivalent to saying that Y ∈ C 1 and since |x2 | ≤ x21 , |x||∇(∇ · Y )| ∈ L∞ as well.
148
Figure 4.8: Vector field associated with a symmetric union of cusps After that, to recover the extra information that the boundary of the cusp stays in C 1,α , one performs a local analysis which is parallel to the one given in 4.4.2. Indeed, the velocity generated by the cusps is uniformly C 1,α in the interior of the patch, up to the boundary. This finishes the argument.
149
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