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Long time confinement of vorticity around a stable
stationary point vortex in a bounded planar domain
Martin Donati and Dragos
,Iftimie
Abstract
In this paper we consider the incompressible Euler equation in a simply-connected
bounded planar domain. We study the confinement of the vorticity around a stationary
point vortex. We show that the power law confinement around the center of the unit
disk obtained in [2] remains true in the case of a stationary point vortex in a simply-
connected bounded domain. The domain and the stationary point vortex must satisfy
a condition expressed in terms of the conformal mapping from the domain to the unit
disk. Explicit examples are discussed at the end.
1 Introduction and main result
To study the behavior of an incompressible inviscid fluid, we consider the planar Euler equa-
tions in a bounded domain Ω⊂R2:
∂tu(x, t) + u(x, t)· ∇u(x, t) = −∇p(x, t),∀(x, t)∈Ω×R∗
+
u(x, 0) = u0(x),∀x∈Ω
∇ · u(x, t) = 0,∀(x, t)∈Ω×R+
u(x, t)·~n = 0,∀(x, t)∈∂Ω×R+
(1.1)
where udenotes the velocity of the fluid, pits internal pressure, and nis the exterior normal
to ∂Ω. If Ω = R2the boundary condition should be changed into a vanishing condition at
infinity. We define the fluid’s vorticity by ω=∂1u2−∂2u1, which satisfies the equation:
∂tω(x, t) + u(x, t)· ∇ω(x, t) = 0.(1.2)
If the domain Ωis smooth, then we have a unique global smooth solution of (1.1), see
[17]. In addition, if ∂Ω∈C1,1we have the following result due to Yudovitch (see [18]): for
any ω0∈L1∩L∞, there exists a unique solution to (1.1), with u∈L∞(R+, W 1,p), for every
1<p<∞, and ω∈L∞(R+, L1∩L∞). The result is true in more general domains, in
particular in domains with finite number of corners with angle strictly less than πand when
the vorticity is compactly supported in Ω, see [13] and [6]. In particular, one could take Ω
to be a convex polygon.
1
Let us denote by GΩthe Green’s function of Ω. Since the domain is supposed to be
simply-connected, the velocity can be recovered from the vorticity through the following
Biot-Savart law
u(x, t) = ZΩ∇⊥
xGΩ(x, y)ω(y, t) dy, (1.3)
where x⊥= (−x2, x1).
The point vortex system is a simplified version of the Euler equations where the vorticity is
assumed to be a finite sum of Dirac masses ω0=PN
i=1 aiδzi. It was introduced by Helmholtz
in [8], see also [15]. Since (1.2) is a transport equation, one expects the vorticity to remain
a sum of Dirac masses at some points zi(t). The Biot-Savart law (1.3) reads in this case
u(x, t) =
N
X
i=1
ai∇⊥
xGΩ(x, zi(t)).
However, if xis one of the points zi(t), then this velocity is not defined as GΩ(x, y)is singular
at y=x. But as xapproaches zi(t), the singular part of the velocity defined above is given
by fast rotation around that point. More precisely, since the map GΩ−GR2is harmonic in
both its variable on Ω, the function γΩ: Ω ×Ω→R,γΩ=GΩ−GR2=GΩ−1
2πln |x−y|is
smooth. So the singular part of ∇⊥
xGΩ(x, zi(t)) is given by (x−zi)⊥
2π|x−zi|2. The point vortex system
consists in ignoring this singular part which should have no influence on the motion of zi
itself. Denoting by eγΩ(x) = γ(x, x)the Robin function of the domain Ωwe obtain then the
following point vortex dynamic:
∀1≤i≤N, dzi(t)
dt=
N
X
j=1
j6=i
aj∇⊥
xGΩ(zi(t), zj(t)) + ai
1
2∇⊥eγΩ(zi(t)),(1.4)
where the (ai)1≤i≤N∈R\ {0}are the masses of the point vortices (zi(t))1≤i≤N. Equations
(1.4) are called Kirchhoff-Routh equations.
Due to the singularities of the Green’s function, these equations are valid only while the
points zi(t)stay distinct and do not leave the domain Ω. There exist configurations leading
to collapse of the point vortices, but they are exceptional, see [15] for the case of R2, and [14]
for the case of the unit disk (we will discuss this more in detail later).
An important question in fluid dynamics is whether the point vortex system is a good
approximation of the Euler equations. There are convergence results in both ways.
Let us first mention that the point vortex system was used as a numerical approximation
of the Euler system. More precisely, consider a smooth solution of the Euler equations and
construct an initial discrete vorticity which is a sum of Dirac masses located on a grid (hj)j∈Z2
where h∈Ris the length of the grid, with masses h2ω0(hj). Solve then the point vortex
system with this initial vorticity. In [4], the authors proved that this point vortex method is
consistent, stable, and converges to the smooth solution of the Euler equation.
The convergence from the Euler system to the point vortex system was also proved in
[16]. More precisely, consider a smooth initial vorticity that is sharply concentrated around
some initial point vortices zi: the support of ω0is included in the union of the disks of radius
εaround the points zi, with ε > 0being small. The authors of [16] proved that for any
2
time τ, and for any δ > 0, if ε=ε(τ, δ)>0is small enough then the solution stays sharply
concentrated within disks of radius δfrom the points zi(t)up to time τ. This statement can
be seen as a fixed time confinement result.
In this paper we are interested in the so-called long time confinement problem, that is we
want to know for how long confinement around point vortices remains true. More precisely,
we want to understand how ε,δand τare linked together and we wish to obtain a confinement
time τas large as possible. We already know from [16] that τgoes to infinity when εgoes
to 0, but we would like to obtain an explicit rate as good as possible.
This problem was already studied by Buttà and Marchioro [2]. These authors assumed
that δ=εβ, with β < 1/2, and made the following assumptions on the initial vorticity.
Assume that ω0∈L1∩L∞and there exists νsuch that
|ω0| ≤ ε−ν
ω0=
N
X
i=1
ω0,i,supp ω0,i ⊂D(zi, ε)
ω0,i has a definite sign
ZΩ
ω0,i dx=ai.
(1.5)
Let ω(x, t)the solution of (1.2). We denote by τε,β the exit time of the vorticity from the
disks of radius εβ:
τε,β = sup (t≥0,∀s∈[0, t],supp ω(·, s)⊂
N
[
i=1
D(zi(s), εβ)).(1.6)
For any N-tuple of distinct points (zi)∈Ω, there exists εsmall enough, such that the disks
D(zi(0), εβ)are disjoints, and therefore this exit time is well defined and strictly positive.
The aim is to obtain a lower bound on τε,β depending explicitly on ε. Two results have been
obtained in [2]. The first is a logarithmic confinement for the whole plane.
Theorem 1.1 ([2]).Assume that Ω = R2, that the initial vorticity satisfies (1.5) and that
the point vortex system with initial data Pn
i=1 aiδzihas a global solution. Then for every
β < 1/2there exists ε0>0and C > 0such that
∀ε<ε0, τε,β > C|ln(ε)|.
The second result is more restrictive, it holds true for the unit disk and for a single point
vortex located at the center, but the conclusion is much stronger since it gives a power-law
confinement.
Theorem 1.2 ([2]).Let Ω = Dand ω0satisfying (1.5) with N= 1 and z1= 0, so that it is
compactly supported within the disk D(0, ε). Then for every β < 1/2there exists ε0>0and
α > 0such that:
∀ε<ε0, τε,β > ε−α.
3
The aim of this paper is to extend Theorem 1.2 to more general domains. We also observe
that Theorem 1.1 can also be extended to bounded domains; we will discuss this problem in
a forthcoming paper.
We consider a single point vortex in a simply-connected bounded domain. We assume for
simplicity that the mass of the point vortex is 1, but the results below hold true for a general
mass. The first question that arises is the location of the point vortex. We will show that
there are special points that allow us to obtain the power-law lower bound while for others
the logarithmic bound is probably optimal. The dynamic of a single point vortex of mass a
reduces to the following ODE
d
dtz(t) = a1
2∇⊥eγΩ(z(t)).
It is obvious from this ODE that a point vortex is stationary if and only if it is a critical
point of the Robin function eγ. The Robin function has been studied, see [5], and we know
that such critical points always exist in a bounded domain. Let x0be a critical point of the
Robin function which is fixed for the rest of this paper. From the Riemann mapping theorem
we know that there exists a biholomorphic map Tfrom Ωto the unit disk D. We can chose
Tsuch that it maps x0to 0: T(x0) = 0. We shall see below that x0is a critical point of
the Robin function if and only if T00(x0)=0(see Proposition 2.4). This condition therefore
characterizes the fact that x0is a stationary point for the point vortex system. We call such
points stationary points.
Our main result is the following.
Theorem 1.3. Let Ωbe a simply connected bounded domain of R2with C1,1boundary. Let
x0be a stationary point such that T000(x0) = 0 where Tis a biholomorphism from Ωto the
unit disk, mapping x0to 0. Assume that ω0satisfies (1.5) with N= 1 and z1=x0. Then
for every β < 1/2and for any α < min(β , 2−4β), there exists ε0>0such that
∀ε<ε0, τε,β > ε−α.
This extends Theorem 1.2 to more general bounded domains. Indeed, in the case of the
unit disk we can choose T(z) = zso the hypothesis given above is verified for the center of
the disk. Let us also observe that the hypothesis that x0is stationary induces no restriction
on the domain Ω. Indeed, we recall that Gustafsson [5] proved that every simply-connected
smooth domain has at least a stationary point. However, the hypothesis that T000(x0)=0is
a condition that not all domains satisfy. We will comment on this in the last section. We
will see in particular that any domain which is invariant by some rotation of angle θ∈(0, π)
around x0satisfies the condition T000(x0) = 0.
In order to understand better the significance of the condition T000(x0)=0, one could
assume that the vorticity ωitself is a point vortex. We study in detail this perturbation
problem in Section 3. We will prove there that if |T000(x0)|<2|T0(x0)|3then τε,β =∞if εis
small enough while if |T000(x0)|>2|T0(x0)|3then τε,β is in general not better than C|ln ε|, see
Theorem 3.1. In other words, in this particular case we have long time confinement better
than C|ln ε|if and only if |T000(x0)|<2|T0(x0)|3. However, when ωis smooth we require the
stronger assumption T000(x0)=0.
4
The plan of the paper is the following. In Section 2 we introduce some notation and
discuss some facts about the Green’s function and the point vortex system. In section 3 we
consider the particular case when ωis a point vortex itself. In Section 4 we prove Theorem
1.3. The last section contains some final remarks and some examples of domains for which
our theorem applies.
2 Preliminary tools
List of notation:
•Ωis a C1,1bounded and simply connected domain of R2;
•D(x0, r)is the disk of center x0and of radius rand D=D(0,1);
•uis the velocity of the fluid and pits pressure, satisfying equations (1.1);
•ω=∂1u2−∂2u1is the vorticity of the fluid;
•supp fis the support of the function f, namely the closure of the set {x∈Ω, f (x)6= 0};
•δzis the Dirac mass in z;
•GΩor Gis the Green’s function of the domain Ω;
•γΩor γis the regular part of GΩ, see relation (2.1);
•eγΩ(x) = γΩ(x, x)is the Robin function;
•C, C1, C2, . . . ;K, K1, K2, . . . , L, are strictly positive constants that may vary from one line to
another, when their value is not important to the result;
•a·bis the scalar product of vectors in R2;
• ∇f,D2fand ∇ · gare respectively the gradient of f, its Hessian matrix, and the divergence
of g.
2.1 Green’s Function
We recall that the Green’s function of a domain Ωis the solution of
∆xGΩ(x, y) = δ(x−y)
vanishing at the boundary, and at infinity if Ωis unbounded. It is a symmetric function on
Ω2that satisfies for x6=y
∆x(GΩ(x, y)−GR2(x, y)) = 0,
which means that GΩ−GR2is a function, denoted by γΩ, which is harmonic in both of its
variable. Therefore, we have:
GΩ(x, y) = 1
2πln |x−y|+γΩ(x, y)(2.1)
5
where γΩis symmetric and smooth.
In the particular case of Ω = D, we have that
GD(x, y) = ln |x−y|
2π−ln |x−y∗||y|
2π(2.2)
where y∗=y
|y|2is the inverse relative to the unit circle. In particular,
γD(x, y) = −ln |x−y∗||y|
2π.(2.3)
Let x0∈Ω. From the Riemann Mapping Theorem (see for instance [1], chapter 6) and
recalling that Ωis simply-connected, we know that there exists a biholomorphic map Tfrom
Ωto the unit disk D. The map Tis unique up to compositions with the biholomorphisms of
the unit disk which are given by
φz0,λ(z) = λz0−z
1−z0z,
with z0∈Dand |λ|= 1 arbitrary. Choosing z0and λconveniently, we can assume without
loss of generality that T(x0) = 0 and that T0(x0)is a strictly positive real number. These two
conditions insure the uniqueness of the conformal map T. Let us observe that in Theorem
1.3 the mapping Tis not unique since we only assume that it maps x0to 0. However, the
condition T000(x0) = 0 does not depend on the choice of T(once we assumed that it maps
x0to 0). Indeed, if T1and T2are two biholomorphisms from Ωto Dmapping x0to 0,
then T1◦T−1
2is a biholomorphism from Dto Dmapping 0 to 0. So it must be a rotation:
there exists some λof modulus 1 such that T1◦T−1
2(z) = λz. So T1=λT2and therefore
T000
1(x0) = λT 000
2(x0). Then T000
1(x0) = 0 if and only if T000
2(x0)=0.
We will assume from now on that T(x0) = 0. The assumption that T0(x0)>0can be
made but is not necessary. In the following, T0(x0)is a complex number.
The properties of the conformal map Timply a precise description of the Green’s function
of Ω. Indeed, a Green’s function composed with a conformal mapping is another Green’s
function, see for example [1] chapter 6. Therefore the formula (2.2) yields the following
proposition.
Proposition 2.1. Let Tbe the biholomorphic mapping introduced above. Then
GΩ(x, y) = GD(T(x), T (y)) = ln |T(x)−T(y)|
2π−ln |T(x)−T(y)∗||T(y)∗|
2π.
In the following, we will have to use both characterizations of the map T, as a C→Cmap,
and as a R2→R2map. In particular, we recall that T0(x) = ∂1T(x) = ∂1T1(x) + i∂1T2(x),
and that ∂1T1=∂2T2and ∂2T1=−∂1T2. So for any map f∈C1(R2,R), we have that
∇(f◦T)(x) = ∂1T1(x)∂1f(T(x)) + ∂1T2(x)∂2f(T(x))
−∂1T2(x)∂1f(T(x)) + ∂1T1(x)∂2f(T(x)),
6
or as complex numbers:
∂1(f◦T)(x) + i∂2(f◦T)(x) = Re(T0(x))∂1f(T(x)) + Im(T0(x))∂2f(T(x))
+i[−Im(T0(x))∂1f(T(x)) + Re(T0(x))∂2f(T(x))]
=T0(x)(∂1f(T(x)) + i∂2f(T(x))).
Identifying ∇f=∂1f+i∂2f, we have:
∇(f◦T)(x) = T0(x)∇f(T(x)),(2.4)
where the product above must be understood as the product of two complex numbers. We
will frequently use this property in the following.
From Proposition 2.1 and from relation (2.1) applied for Ωand for Dwe get that
γΩ(x, y) = γD(T(x), T (y)) + 1
2πln |T(x)−T(y)|
|x−y|,
and letting y→xwe obtain at the level of Robin functions
∀x∈Ω,eγΩ(x) = eγD(T(x)) + 1
2πln |T0(x)|.(2.5)
Using the explicit expression of γD, see relation (2.3), we can compute the gradient of γΩ:
∇xγΩ(x, y) = T0(x)(T(x)−T(y))
2π|T(x)−T(y)|2−T0(x)(T(x)−T(y)∗)
2π|T(x)−T(y)∗|2−(x−y)
2π|x−y|2
=T0(x)
2π(T(x)−T(y)) −T0(x)
2π(T(x)−T(y)∗)−1
2π(x−y)
(2.6)
where we used relation (2.4).
The mapping eγΩ: Ω →R2, that will be named only eγwhen there is no ambiguity, has
been studied extensively in [5]. In particular, it was shown in that paper that −eγΩis a super-
harmonic function, that is that ∆eγΩ>0, and that it goes to infinity near the boundary like
−1
2πln(d(x, ∂Ω)). This implies that we have the following proposition (see [5]).
Proposition 2.2 ([5]).For every bounded and simply connected open set Ω, there exists at
least one point x0where eγreaches its minimum.
Critical points of eγΩwill be of special interest in the following. The proposition above
implies the existence of a critical point of the Robin function. Moreover, if Ωis convex, then
the critical point is also unique (see [5]). Though we will not use this result here, it may be
interesting to keep this in mind, especially when looking for explicit examples.
2.2 Point vortex system
The point vortex dynamic, which is described by equations (1.4), can exhibit finite time
blow-up of solutions. One scenario of blow-up is when two point vortices hit each other
in finite time, meaning that there exists i6=jand t < ∞such that zi(t) = zj(t). This
7
phenomena can happen, see for instance [15] or [10] for an example of finite-time collapse of
a self similar evolution of point vortices. Another scenario for blow-up is when a point vortex
hits the boundary. However, the finite time blow-up is exceptional in the case of the whole
plane, see [15], and for the unit disk, see [14]. In those cases, the N-dimensional Lebesgue
measure of the set of initial positions that lead to a collapse is 0. The case of a more general
bounded domain is an ongoing work.
Let us recall the convergence theorem obtained in [16], for configurations of point vortices
that do not lead to finite-time blow-up:
Theorem 2.3 ([16]).Let (ai, zi(t)) be a global solution of the point vortex system (1.4). For
every δ > 0and for every time τ, there exists ε > 0such that if ω0satisfies (1.5) for some
0< ν < 8/3, then the vorticity stays confined up to the time τin disks of radius δcentered
on zi(t).
In addition, the authors of [3] proved that τε,β →+∞as εgoes to 0 for any β < 1/3
(recall that τε,β was defined in relation (1.6)). However, these theorems don’t say anything
about how δdepends on the time τ, or conversely, for how long this confinement remains
true, depending on ε.
For some explicit examples of point vortices we refer to [9].
As mentioned in the introduction, the stationary point vortices are the critical points of
the Robin function. Indeed, the relation (1.4) with N= 1 reduces to
z0(t) = a
2∇⊥eγΩ(z(t)).
We refer to such a critical point x0by saying that it is a stationary point vortex.
Those points can also be characterized in terms of the conformal mapping Tas zeroes of
T00.
Proposition 2.4. The following conditions are equivalent:
(i) A single point vortex placed in x0is stationary,
(ii) ∇eγΩ(x0) = 0,
(iii) T00(x0) = 0.
Proof. This proposition was already proved in [5]. We recall the proof for the convenience of
the reader.
We already observed that (i)and (ii)are equivalent. We only need to prove that (ii)is
equivalent to (iii). From relations (2.5) and (2.4), and recalling that T0can’t vanish in Ω,
we deduce that
∇eγΩ(x) = ∇eγD(T(x))T0(x) + 1
2π
T0(x)T00(x)
|T0(x)|2.
One can easily check that 0is a stationary point for the unit disk, so ∇eγD(0) = 0. Thus
∇eγΩ(x0) = 1
2πT00(x0)
T0(x0).
Clearly ∇eγΩ(x0) = 0 is equivalent to T00(x0) = 0.
8
3 Confinement for Dirac mass around a stationary vortex
The aim of this section is to study the case where the vorticity ωitself is a Dirac mass. The
reason we consider this simpler case is because it is easier to have a complete description of
what happens. And this in turn gives us an indication of what to expect in the smooth case
considered in Theorem 1.3.
We consider a single point vortex located at z(t)which is close to the stationary point
vortex x0. Rescaling time if needed, we can assume without loss of generality that the mass
of the point-vortex is 1. The question of knowing if z(t)remains close to x0is closely related
to the notion of stability of x0that we discuss in what follows.
3.1 Stability of a stationary point vortex
Since the mass of the point vortex z(t)is 1, its equation of motion is the following
z0(t) = 1
2∇⊥eγΩ(z(t)).(3.1)
We have that d
dteγΩ(z(t)) = z0(t)· ∇eγΩ(z(t)) = 0
which means that the point vortex is evolving on the level set eγ(x) = cst. We know from [5]
that the Robin function eγ(x)goes to infinity as xapproaches the boundary of Ω. Therefore,
the point vortex z(t)can never reach the boundary so (3.1) has a global solution.
Since −eγΩis super-harmonic, the eigenvalues of the real symmetric matrix D2eγΩ(x0)have
positive sum, meaning that at least one of them is positive. So two main cases can occur:
either both eigenvalues are positive, either one is positive and one is negative. We skip the
study of the degenerate case when one eigenvalue is 0.
So let λ+>0and λ−be the two eigenvalues of D2eγΩ(x0). If λ−>0, the Morse Lemma
implies that there exists a change of variables y=φ(x)in the neighborhood of x0such that
in this neighborhood of x0,
eγΩ(x) = eγΩ(φ−1(y)) = eγΩ(x0)+(y−x0)2
1+ (y−x0)2
2.
In particular, in these new local coordinates, the level sets of eγΩare circles, so in the real
coordinates, they are diffeomorphic to circles. More precisely, the level set eγΩ(φ−1(y)) =
eγΩ(x0) + ris a circle of radius √r, provided that r > 0is small enough. Because φis a
C∞diffeomorphism, and φ(x0) = x0, there exist constants k, K > 0such that k|y−x0|<
|x−x0|< K|y−x0|and thus the level set eγΩ(x) = eγΩ(x0) + ris contained in the annulus
{k√r < |x−x0|< K√r}.
We conclude from these observations that if |z(0) −x0| ≤ εwith εsmall enough, then
we have that |z(t)−x0| ≤ εK
kfor every time t≥0. This a stability property: if εis small
enough and if we assume that the point vortex starts at distance εof x0, then it remains at
distance of order εfor any time.
Assume now that λ−<0. We have in this case that
eγΩ(x) = eγΩ(φ−1(y)) = eγΩ(x0)−(y−x0)2
1+ (y−x0)2
2,
9
so the level sets in a neighborhood of x0are in the local coordinates yhyperbolas, with the
special level set eγΩ(x) = eγΩ(x0)being a union of two line segments. In particular, we will see
later that one line segment is repulsive, meaning that when a point vortex evolves on that
segment, it moves away from the point x0. We call this situation unstable. In that case,
no matter how close the point vortex starts from the critical point x0, it goes away from x0
exponentially fast, as we will see in section 3.3.
Let us summarize what we observed above. If the eigenvalues of the matrix D2eγΩ(x0)
are both positive, x0is said to be stable, and a point vortex close enough to x0remains
indefinitely close to it. If one of the eigenvalues is negative, then x0is said to be unstable,
because there exists a point vortex evolving on a level set going away from x0. This notion
of stability is the same as the one introduced in [15], Chapter 3. Figure 1 shows examples of
the two situations.
Figure 1: Trajectories of a single point vortex in a domain with two stable points and an unstable
one.
3.2 Taylor expansion near a stationary vortex
The aim is now to express the stability of a stationary point vortex in terms of the conformal
map T. We already know from Proposition 2.4 that x0is stationary if and only if T00(x0)=0.
We will now make a Taylor expansion of Taround x0in order to have a better understanding
of the point vortex dynamic near a critical point. We therefore need to compute the Hessian
matrix D2eγΩin that point. By relation (2.5) we have that
1
2D2eγΩ(x0) = 1
2D2(eγD◦T)(x0) + 1
4πD2(ln |T0|)(x0).
This expression is actually explicit in terms of the conformal map T; indeed, we know
that eγD(x) = −1
2πln(1 − |x|2). We recall that T00(x0)=0, so ∂1T0(x0) = ∂2T0(x0)=0. We
recall also that for an holomorphic function ϕwe have that ∂1ϕ=ϕ0and ∂2ϕ=iϕ0. Some
direct computations give that
D2(eγD◦T)(x0) = 1
π|T0(x0)|20
0|T0(x0)|2
10
and
∂i∂j(ln |T0|)(x0) = ∂i∂jT0(x0)·T0(x0)
|T0(x0)|2
so that
D2(ln |T0|)(x0) = 1
|T0(x0)|2T000(x0)·T0(x0) (T000 (x0))⊥·T0(x0)
(T000(x0))⊥·T0(x0)−T000(x0)·T0(x0).
We conclude that 1
2D2eγΩ(x0) = 1
4π2µ2+p q
q2µ2−p,(3.2)
where
µ2=|T0(x0)|2
p=1
µ2T000(x0)·T0(x0)
q=1
µ2(T000(x0))⊥·T0(x0).
We compute the characteristic polynomial of the matrix 1
2D2eγΩ(x0):
det(1
2D2eγΩ(x0)−X I2) = X2−µ2
πX+1
16π24µ4−p2−q2
and the eigenvalues are
λ±=2µ2±pp2+q2
4π.
Furthermore, we notice that p2+q2=|T000(x0)|2
|T0(x0)|2. We deduce from the expression of the
determinant that x0is stable when λ−>0, that is when 2|T0(x0)|3>|T000(x0)|and unstable
when 2|T0(x0)|3<|T000(x0)|.
Another interesting thing to notice is that when T000(x0) = 0, the matrix D2eγΩ(x0) =
µ2
2πI2, with I2the identity matrix, and thus the trajectories of the linearized system z0(t) =
(D2eγΩ(x0)(z(t)−x0))⊥are circles of center x0.
3.3 Exit time around an unstable stationary point
We now consider the unstable case, meaning λ+λ−<0. We now go back to (3.1). Using
relation (3.2) and the subsequent calculations, one can check that the eigenvalues of the
Jacobian matrix of 1
2∇⊥eγΩin x0are ξ=p−λ+λ−and −ξ.
We thus know from [7] Theorem 6.1 of Chapter 9, and the corollary and remarks associated
to this theorem that there exist two local invariant manifolds for the equation, and thus there
exist two special solutions to equation (3.1) associated with each eigenvalue ±ξ. We denote
by z1the solution that goes away from x0, corresponding to the eigenvalue ξ. It satisfies that
∀0< c < ξ, lim
t→−∞ |z1(t)−x0|e−ct = 0.(3.3)
11
We denote by t1the first time when |z1(t1)−x0|=εβand by t2the last time when
|z1(t2)−x0|=ε. Provided εis small enough, we have that 0> t1> t2. We define
˜z(t) = z1(t+t2). This is a solution of equation (3.1) such that |˜z(0) −x0|=εand thus we
can set
τε,β = max t≥0,∀s∈[0, t],|˜z(t)−x0| ≤ εβ
which is the exit time as defined in relation 1.6 associated to a point vortex starting at
distance εof x0and evolving along the same trajectory as z1. By construction, it satisfies
that τε,β =t1−t2≤ −t2.
We also know from relation (3.3) that there exist a constant Mand a constant 0< c < ξ,
such that for every t < 0, we have that |z1(t)−x0|e−ct ≤M. Therefore,
ε=|z1(t2)−x0| ≤ Mect2.
Thus, ln ε≤ln M+ct2which means that τε,β ≤ −t2≤2
c|ln ε|for εsmall enough.
This result empowers the conjecture that one cannot expect any general confinement
result, as in Theorem 1.1, better than C|ln ε|, since we have an example of such an exit time
in the case where the vorticity is itself a point vortex.
We can summarize the results of this section in the following theorem:
Theorem 3.1. Assume that T(x0) = T00 (x0) = 0, so that x0is a stationary point. If
|T000(x0)|<2|T0(x0)|3then any single point vortex starting close to x0will remain indefinitely
close to it. More precisely, there exists some small constant ε0>0such that if ε≤ε0and
z(0) ∈D(x0, ε)then z(t)∈D(x0, Cε)for all times tand τε,β = +∞.
Conversely, the condition |T000(x0)|>2|T0(x0)|3insures that for every ε > 0, there exists
a starting position z(0) such that |z(0) −x0| ≤ ε, but the point vortex exits the disk D(x0, εβ)
with β < 1in finite time τε,β ≤C|ln ε|.
4 Power-law confinement near a stationary vortex
The present section is devoted to the proof of Theorem 1.3. To simplify the notation, we
denote from now on γ=γΩ,eγ=eγΩand G=GΩ.
Recall that Ωis a simply connected and bounded domain, that Tis a biholomorphism
from Ωto Dsuch that T(x0) = 0. We have that x0is a stationary point which, according to
Proposition 2.4, means that T00(x0)=0.
We assume that ω0satisfies (1.5) with N= 1 and z1=x0. Without loss of generality we
can assume that the mass a1is 1 (changing the mass means rescaling the time). We therefore
have that ω0is non negative, is supported in D(x0, ε)and kωkL1= 1.
Let us introduce
F(x, t) = Z∇⊥
xγ(x, y)ω(y, t) dy, (4.1)
which is the influence of the blob over itself through the boundary. The key point of the
proof of Theorem 1.2 from [2] in the case of the disk is the following property:
∀x, z ∈D(0, δ),∀t≥0,|F(x, t)−F(z, t)| ≤ C|x−z|δ2.(4.2)
12
To prove that the result remains true for more general domains, we will need a similar
inequality. It is important to notice that unlike the case when ωis a point vortex itself, we
don’t necessarily have that F(x0, t) = 0 a priori. According to (2.6) we have that
2π∇xγ(x0, y) = −T0(x0)
T(y)+T0(x0)
T(y)∗−1
x0−y.
Recalling that T(y)∗= 1/T (y)and making the Taylor expansion
T(y) = T0(x0)(y−x0) + T000 (x0)
6(y−x0)3+O(|y−x0|4)
we get that
2π∇xγ(x0, y) = −1
y−x0
1
1 + T000 (x0)
6T0(x0)(y−x0)2+O(|y−x0|3)+T0(x0)T(y)−1
x0−y
=−1
y−x01−T000(x0)
6T0(x0)(y−x0)2+O(|y−x0|3)
+T0(x0)T0(x0)(y−x0) + O(|y−x0|2)−1
x0−y.
We conclude that
2π∇xγ(x0, y) = T000 (x0)
6T0(x0)(y−x0) + |T0(x0)|2(y−x0) + O(|y−x0|2).(4.3)
4.1 Estimate of the influence of the boundary
The first step of the proof is a technical lemma giving a Lipschitz inequality for F. This is
the counterpart in Ωof the relation (4.2) valid for the unit disk.
Lemma 4.1. We have the following estimate, for δsufficiently small and for every x, y, z ∈
D(x0, δ)
∇xγ(x, y)− ∇xγ(z, y) = (x−z)T000(x0)
6πT 0(x0)+O(δ),(4.4)
where the term O(δ)is bounded by Cδ with Ca constant depending only on Ωand x0. In
particular, if we assume that T000(x0) = 0, there exists a constant K1=K1(Ω, x0)such that
∀x, y, z ∈D(x0, δ),|∇xγ(x, y)− ∇xγ(z, y)| ≤ K1|x−z|δ. (4.5)
Furthermore, if we assume that supp ω⊂D(x0, δ)we have that
∀x, z ∈D(x0, δ),|F(x, t)−F(z, t)| ≤ K1|x−z|δ. (4.6)
Proof. Let us define
R(x, y, z) = 2π(∇xγ(x, y)− ∇xγ(z, y)).
13
Using (2.6) we can write
R(x, y, z) = T0(x)
T(x)−T(y)−T0(x)
T(x)−T(y)∗−1
x−y−T0(z)
T(z)−T(y)−T0(z)
T(z)−T(y)∗−1
z−y
so
R(x, y, z) = T0(x)(T(z)−T(y)) −T0(z)(T(x)−T(y))
(T(x)−T(y))(T(z)−T(y)) +x−z
(x−y)(z−y)
−T0(x)(T(z)−T(y)∗)−T0(z)(T(x)−T(y)∗)
(T(x)−T(y)∗)(T(z)−T(y)∗).
We decompose Rinto two parts:
R=R1−R2
with
R1(x, y, z) = T0(x)(T(z)−T(y)) −T0(z)(T(x)−T(y))
(T(x)−T(y))(T(z)−T(y)) +x−z
(x−y)(z−y)
and
R2(x, y, z) = T0(x)(T(z)−T(y)∗)−T0(z)(T(x)−T(y)∗)
(T(x)−T(y)∗)(T(z)−T(y)∗)
=T0(x)T(z)−T0(z)T(x)
(T(x)−T(y)∗)(T(z)−T(y)∗)+T(y)∗(T0(z)−T0(x))
(T(x)−T(y)∗)(T(z)−T(y)∗)
≡R21 +R22.
Since Tis smooth and T(x0) = 0, we clearly have that |T(x)| ≤ Cδ and |T(y)| ≤ Cδ. So
we can estimate
|T(x)−T(y)∗| ≥ |T(y)∗|−|T(x)|=1
|T(y)|− |T(x)| ≥ 1
Cδ −Cδ ≥1
2Cδ
if δis sufficiently small. We can therefore bound R21 as follows:
|R21| ≤ 4C2δ2|T0(x)T(z)−T0(z)T(x)| ≤ Cδ2|x−z|.
Similarly
|T(x)−T(y)∗| ≥ |T(y)∗|−|T(x)|=1
|T(y)|− |T(x)| ≥ 1
2|T(y)|
so |T(y)∗|
|T(x)−T(y)∗|≤2|T(y)||T(y)∗|= 2.
Then
|R22|=|T(y)∗|
|T(x)−T(y)∗||T0(z)−T0(x)|
|T(z)−T(y)∗|≤4Cδ|T0(z)−T0(x)| ≤ Cδ|x−z|2≤C δ2|x−z|.
14
We conclude from the estimates above that
|R2(x, y, z)| ≤ Cδ2|x−z|.(4.7)
To estimate R1, we write it under the form
R1(x, y, z) = N(x, y, z)
(T(x)−T(y))(T(z)−T(y))(x−y)(z−y)
with
N(x, y, z)=[T0(x)(T(z)−T(y)) −T0(z)(T(x)−T(y))](x−y)(z−y)
+ (x−z)(T(x)−T(y))(T(z)−T(y)).(4.8)
As Tis holomorphic we observe that Nis holomorphic in the variables x,yet z. One
can notice that Nis 0 if x=z, so it can be factorized by x−z. Let us also recall that γis
smooth everywhere, so Ris smooth too. We proved above that R2is bounded in D(x0, δ)3,
so R1=R+R2is also bounded in this set. But the denominator of R1has a factor
(x−y)2(y−z)2, so it follows that N(x, y, z)can be factorized by (x−y)2(y−z)2. But we
observed that Ncan also be factorized by x−z. This implies that there exists a holomorphic
function N1(x, y, z)such that:
N(x, y, z) = (x−y)2(z−y)2(x−z)N1(x, y, z).(4.9)
Therefore R1(x, y, z)
x−z=(x−y)(z−y)N1(x, y, z)
(T(x)−T(y))(T(z)−T(y)).(4.10)
We need now to compute N1(x0, x0, x0). To do that, we will differentiate five times
relation (4.9) and evaluate it in (x0, x0, x0). It is clear that derivatives up to order 4 of
(x−y)2(z−y)2(x−z)all vanish at (x0, x0, x0). We can therefore write that
∂3
x∂2
zN(x0, x0, x0) = ∂3
x∂2
z(x−y)2(z−y)2(x−z)(x0, x0, x0)N1(x0, x0, x0) = 12N1(x0, x0, x0).
Differentiating relation (4.8) allows to find after some computations that
∂3
x∂2
zN(x0, x0, x0) = 4T0(x0)T000(x0).
Therefore,
N1(x0, x0, x0) = T0(x0)T000(x0)
3
so
N1(x, y, z) = N1(x0, x0, x0) + O(δ) = T0(x0)T000(x0)
3+O(δ).
We now go back to (4.10). We observe that
x−y
T(x)−T(y)
15
is smooth on D(x0, δ)3with value 1/T 0(x)when x=y. Therefore
x−y
T(x)−T(y)=1
T0(x0)+O(δ).
Similarly z−y
T(z)−T(y)=1
T0(x0)+O(δ).
Combining the previous relations results in
R1(x, y, z)
x−z=1
T0(x0)+O(δ)2T0(x0)T000(x0)
3+O(δ)=T000(x0)
3T0(x0)+O(δ)
so
R1(x, y, z)=(x−z)T000(x0)
3T0(x0)+O(δ).
Recalling (4.7) finally implies that
R(x, y, z) = (x−z)T000(x0)
3T0(x0)+O(δ),
which proves (4.4). Clearly (4.5) follows from (4.4) if T000(x0) = 0. Finally, relation (4.6)
follows from (4.5) after integrating and recalling that the mass of ωis 1.
Comparing (4.6) and (4.2), we see that we lose the factor δ2. But in the case T000 (x0)=0,
we still get a factor δand this is enough to make our argument work. Actually the proof of
[2] would still be correct assuming only that |F(x, t)−F(z, t)| ≤ C|x−z|δ. In our proof, a
factor δ2would improve the restriction over the power αin Theorem 1.3. However, please
notice that if T000(x0)6= 0 we lose the factor δand our proof does not work anymore.
4.2 Estimates of the trajectories
From now on we assume that T000(x0) = 0.
Let us introduce the center of vorticity:
B(t) = ZΩ
xω(x, t) dx, (4.11)
and the moment of inertia:
I(t) = ZΩ|x−B|2ω(x, t) dx.
For future needs, let us compute the time derivative of B. Recall that ωsatisfies the
16
equation (1.2) in the sense of distributions and that it is compactly supported. We have that
d
dtB(t) = Zx∂tω(x, t) dx
=−Zxu(x, t)· ∇ω(x, t) dx
=Z(u(x, t)· ∇)x ω(x, t) dx
=Zu(x, t)ω(x, t) dx
=ZZ (x−y)⊥
2π|x−y|2+∇⊥
xγ(x, y)ω(y, t)ω(x, t) dxdy
where we used (1.3) and (2.1).
Observing that (x−y)⊥
2π|x−y|2is antisymmetric when exchanging xand yand recalling the
definition of Fgiven in (4.1), we infer that
d
dtB(t) = ZF(x, t)ω(x, t) dx. (4.12)
Let us define
Rt= max{|x−B(t)|;x∈supp ω(t)}
and choose some X(t)∈supp ω(t)such that |X(t)−B(t)|=Rt. We denote by s7→ Xt(s)
the trajectory passing through X(t)at time tso that Xt(t) = X(t).
We have the following lemma which allows us to control the time evolution of Rt:
Lemma 4.2. For any t≤τε,β we have that
d
ds|Xt(s)−B(s)|s=t≤2K1εβRt+5
πR3
t
I(t) + K2ε−νZ|x−B|>Rt/2
ω(x, t) dx1/2
where νis the constant from relation (1.5),K1is the constant from Lemma 4.1 and K2is a
universal constant.
This lemma shows that in order to obtain upper bounds for the growth of the support
of ω, one needs estimates for I(t),B(t)and for the mass of vorticity far from the center of
mass.
Proof. We have that for any s≥0and t≥0,X0
t(s) = u(Xt(s), s), so
d
ds|Xt(s)−B(s)|= (u(Xt(s), s)−B0(s)) ·Xt(s)−B(s)
|Xt(s)−B(s)|.
We fix now the time t≥0, we take s=t, and we write Xinstead of Xt(t). By the Biot-Savart
law (1.3), the relation (4.1) and recalling that the vorticity is assumed to be of integral 1, we
have that
u(X, t) = F(X, t) + Z(X−y)⊥
2π|X−y|2ω(y, t) dy=ZF(X, t) + (X−y)⊥
2π|X−y|2ω(y, t) dy.
17
Relation (4.12) now implies that
d
ds|Xt(s)−B(s)|s=t=ZF(X, t) + (X−y)⊥
2π|X−y|2−F(y, t)ω(y, t) dy·X−B(t)
|X−B(t)|
=H1+H2
where
H1=Z(F(X, t)−F(y, t)) ω(y, t) dy·X−B(t)
|X−B(t)|,
and
H2=Z(X−y)⊥
2π|X−y|2ω(y, t) dy·X−B(t)
|X−B(t)|,
Thanks to Lemma 4.1 and recalling that t≤τε,β, we have that
|H1|=Z(F(X, t)−F(y, t)) ω(y, t) dy≤K1εβZ|X−y|ω(y, t) dy
≤K1εβ2Rt
where we used that supp ω⊂D(B, R). The second term H2is the same as the left hand side
of relation (2.28) in [2], and its estimate is the same:
|H2| ≤ 5
πR3
t
I(t) + 1
πε−νZ|x−B|>Rt/2
ω(x, t) dx1/2
The lemma is now proved.
4.3 Estimates of the moments of the vorticity
We have the following lemma:
Lemma 4.3. For every t < min(τε,β, ε−β), we have:
I(t)≤K3ε2.
and:
|B(t)−x0| ≤ K3ε,
where K3is a positive constant depending only on Ωand x0.
Proof. We differentiate I(t):
I0(t) = Z|x−B|2∂tω(x, t)−2B0(t)·(x−B)ω(x, t)dx=Z|x−B|2∂tω(x, t) dx
18
where we used relation (4.11). Next, we use the equation of ωgiven in (1.2) and write
I0(t) = Z−|x−B|2u(x, t)· ∇ω(x, t)dx
=Z2(x−B)·u(x, t)ω(x, t) dx
=ZZ 2(x−B)· ∇⊥
xG(x, y)ω(x, t)ω(y, t) dxdy
=ZZ 2(x−B)·∇⊥
xγ(x, y) + (x−y)⊥
2π|x−y|2ω(x, t)ω(y, t) dxdy
=ZZ 2(x−B)· ∇⊥
xγ(x, y)ω(x, t)ω(y, t) dxdy
+ZZ −x·y⊥−B·(x−y)⊥
π|x−y|2ω(x, t)ω(y, t) dxdy
Exchanging xand yshows that the last term above vanishes. So
I0(t) = ZZ 2(x−B)· ∇⊥
xγ(x, y)ω(x, t)ω(y, t) dxdy
=ZZ 2(x−B)·[∇⊥
xγ(x, y)− ∇⊥
xγ(B, y)]ω(x, t)ω(y, t) dxdy
≤2K1εβZZ |x−B|2ω(x, t)ω(y, t) dxdy
= 2K1εβI(t)
where we used (4.5) with δ=εβbecause (x, y, z)∈B(x0, εβ)(see the definition (1.6) of τε,β).
The Gronwall lemma now implies that I(t)≤I(0) exp(2K1εβt). Since at the initial time we
have that supp ω0and B(0) are in the disk D(x0, ε)we infer that I(0) ≤4ε2. We assumed
that t<ε−βso we finally obtain that
∀t≤min(τε,β, ε−β), I(t)≤4e2K1ε2.
We estimate now the center of vorticity. By relations (4.12) and (4.1), we have that
d
dt|B(t)−x0|2= 2B0(t)·(B(t)−x0)
= 2 ZZ ∇⊥
xγ(x, y)ω(y, t)ω(x, t) dxdy·(B(t)−x0).
To estimate this, we use Lemma 4.1 and relation (4.3) and we recall that we assume that
T000(x0) = 0 and supp ω∈D(x0, εβ)to obtain
2π∇⊥
xγ(x, y) = 2π(∇⊥
xγ(x, y)− ∇⊥
xγ(x0, y) + ∇⊥
xγ(x0, y))
= 2π(∇⊥
xγ(x, y)− ∇⊥
xγ(x0, y)) + |T0(x0)|2(y−x0)⊥+O|y−x0|2
=|T0(x0)|2(y−x0)⊥+εβ(|x−x0|+|y−x0|)O(1).
19
Putting the estimates above together we have that
d
dt|B(t)−x0|2= 2 ZZ |T0|2(x0)(y−x0)⊥
2π+εβ(|x−x0|+|y−x0|)O(1)
ω(y, t)ω(x, t) dxdy·(B(t)−x0).
But we have the following cancellation:
ZZ |T0|2(x0)(y−x0)⊥ω(y, t)ω(x, t) dxdy·(B(t)−x0)
=|T0|2(x0)ZZ (y−x0)⊥ω(y, t)ω(x, t) dxdy·(B(t)−x0)
=|T0|2(x0)(B(t)−x0)⊥·(B(t)−x0)
= 0.
Therefore, d
dt|B(t)−x0|2≤C|B(t)−x0|εβZ|x−x0|ω(x, t) dx.
We notice now that
Z|x−x0|2ω(x, t) dx=Z|x−B(t)|2ω(x, t) dx
−Z(x0−B(t)) ·(x−x0+x−B(t))ω(x, t) dx
=Z|x−B(t)|2ω(x, t) dx−(x0−B(t)) ·(B(t)−x0+B(t)−B(t))
=I(t) + |x0−B(t)|2.
By the Cauchy-Schwarz inequality
Z|x−x0|ω(x, t) dx≤Z|x−x0|2ω(x, t) dx1
2= (I+|B(t)−x0|2)1
2,
so d
dt|B(t)−x0| ≤ Cεβ(I1
2+|B(t)−x0|).
By the Gronwall lemma we infer that
|B(t)−x0| ≤ exp Cεβt|B(0) −x0|+CεβZt
0
I(s)1
2ds.
We already know that I(s)≤Cε2, and |B(0) −x0| ≤ ε, so
|B(t)−x0| ≤ exp(Cεβt)(ε+Cεβtε).
As we assumed that εβt≤1we conclude that
|B(t)−x0| ≤ Cε.
This completes the proof of the lemma.
20
We have just shown that up to the time min(τε,β, ε−β)the center of mass Bstays close
to x0and the moment of inertia Iremains small. We need a last technical lemma, which is
inspired by the appendix of [11].
Lemma 4.4. For every k≥1and t≤min(τε,β, ε−β)there exists a small constant ε0=ε0(k),
a large constant C(k)and a constant K4which depends only on Ωand x0, such that if
ε≤ε0and r4≥K4ε2(1 + kt ln(2 + t)) ,
then Z|x−B|>r
ω(x, t) dx≤C(k)εk/2
rk.
Proof. Let us introduce the moment of vorticity of order 4n
mn(t) = ZΩ|x−B(t)|4nω(x, t) dx.
One can differentiate to obtain, by using relations (1.3), (4.12) and recalling that ∇ · u= 0
m0
n(t) = −Z|x−B(t)|4nu(x, t)· ∇ω(x, t) dx
−4nZB0(t)·(x−B(t))|x−B(t)|4n−2ω(x, t) dx
= 4nZu(x, t)·(x−B(t))|x−B(t)|4n−2ω(x, t) dx
−4nB0(t)·Z(x−B(t))|x−B(t)|4n−2ω(x, t) dx
= 4nZZ ∇⊥
xG(x, y)·(x−B(t))|x−B(t)|4n−2ω(x, t)ω(y, t) dxdy
−4nZZ ∇⊥
xγ(z, y)ω(z, t)ω(y, t) dzdy·Z(x−B(t))|x−B(t)|4n−2ω(x, t) dx.
Recalling that G(x, y) = ln |x−y|
2π+γ(x, y)and that eω(x0, t) = ω(x0+B(t), t)), we can further
decompose
m0
n(t) = an(t) + bn(t)−cn(t)
where
an(t)=4nZZ (x0−y0)⊥
2π|x0−y0|2·x0|x0|4n−2eω(x0, t)eω(y0, t) dy0dx0,
bn(t) = 4nZZ ∇⊥
xγ(x, y)·(x−B(t))|x−B(t)|4n−2ω(x, t)ω(y, t) dxdy,
cn(t)=4nZZZ ∇⊥
xγ(z, y)·(x−B(t))|x−B(t)|4n−2ω(x, t)ω(y, t)ω(z, t) dxdydz.
We observe that anis exactly the same quantity that appears in [11] on the last line of
page 1726. Observing that the center of mass of eω(x, t)is in 0, we deduce that the estimates
21
given in [11] are true for an. More precisely, ansatisfies the estimate given on the line 5, page
1729 of [11]:
|an(t)| ≤ Cn2I(t)mn−1(t).
Applying Lemma 4.3 we obtain that
|an(t)| ≤ Cn2ε2mn−1(t).(4.13)
The terms bnand cnare similar. We decompose
∇⊥
xγ(x, y) = ∇⊥
xγ(x, y)− ∇⊥
xγ(x0, y) + ∇⊥
xγ(x0, y).
We apply Lemma 4.1 with δ=εβto deduce that
|∇⊥
xγ(x, y)− ∇⊥
xγ(x0, y)| ≤ K1|x−x0|εβ≤K1ε2β
for all x, y ∈supp ω(t). Moreover, we also have that |x−B(t)| ≤ Cεβat least for εsmall
enough, since |x−x0| ≤ εβand |B(t)−x0| ≤ K3ε. Using these estimates we infer that
|bn(t)| ≤ Cnε2βZ|x−B(t)|4n−1ω(x, t) dx+dn≤Cnε5βmn−1(t) + dn
where
dn= 4nZZ ∇⊥
xγ(x0, y)·(x−B(t))|x−B(t)|4n−2ω(x, t)ω(y, t) dxdy.
Decomposing ∇⊥
xγ(z, y) = ∇⊥
xγ(z, y)−∇⊥
xγ(x0, y) + ∇⊥
xγ(x0, y)we obtain that the same
estimate holds true for cn:
|cn(t)| ≤ Cnε5βmn−1(t) + dn.
We estimate now dn. From relation (4.3) we know that there exists a bounded function
c(y)such that ∇⊥
xγ(x0, y)=[a(y−x0) + c(y)|y−x0|2]⊥where a=|T0(x0)|2
2π. We thus have
that
dn= 4nZZ[a(y−x0) + c(y)|y−x0|2]⊥·(x−B(t))|x−B(t)|4n−2ω(x, t)ω(y, t) dxdy
≤4na Z(B(t)−x0)⊥·(x−B(t))|x−B(t)|4n−2ω(x, t) dx
+Cn ZZ |y−x0|2|x−B(t)|4n−1ω(x, t)ω(y, t) dxdy
≤4na Z(B(t)−x0)⊥·(x−B(t))|x−B(t)|4n−2ω(x, t)) dx+Cnε5βmn−1(t)
≤Cn(|B(t)−x0|ε3β+ε5β)mn−1(t).
Let us recall that Lemma 4.3 gives that |B(t)−x0| ≤ K3εand therefore
|bn(t)|+|cn(t)| ≤ Cnε5βmn−1(t)+2dn≤Cn(ε1+3β+ε5β)mn−1(t).
Together with relation (4.13) and recalling that m0
n(t) = an(t)+bn(t)−cn(t), the last estimate
yields that
|m0
n(t)| ≤ Cn2(ε2+ε1+3β+ε5β)mn−1(t).
22
Let us observe now that it suffices to assume that β > 2/5. Indeed, assume that Theorem
1.3 is proved for any β∈]2/5,1/2[. Let β0≤2/5and α < min(β0,2−4β0). Since α < 2/5,
one can easily check that there exists β∈]2/5,1/2[ such that α < min(β, 2−4β)(one can
choose βclose to 2/5). Since β0< β, we also have that τε,β ≤τε,β0so τε,β0> ε−α.
We assume in the sequel that β > 2/5. Due to this additional assumption, we have that
|m0
n(t)| ≤ Cn2ε2mn−1(t).(4.14)
Using Hölder’s inequality on f(x) = ω1/n(x)and g(x) = |x−B(t)|4n−4ω1−1/n(x)with p=n
and q=n
n−1, we have that
mn−1(t) = Z|x−B(t)|4n−4ω(x, t) dx
≤Zω(x, t) dx1/n Z|x−B(t)|4nω(t, x) dx(n−1)/n
=mn(t)(n−1)/n.
This last inequality combined with relation (4.14) gives that
m0
n(t)≤Cn2ε2mn(t)(n−1)/n.
We integrate to obtain that
mn(t)≤mn(0)1/n +Cnε2tn.
Clearly, mn(0) ≤(2ε)4nso, assuming that ε≤1,
mn(t)≤Lε2(1 + nt)n
for some constant Lwhich depends only on Ωand x0. Let us choose any k≥1,rand nsuch
that
r4≥2Lε21 + kln(2 + t)
ln 2 t
and
kln(2 + t)
ln 2 −1< n ≤kln(2 + t)
ln 2 .
This defines n≥1since k≥1and ln(2+t)
ln 2 ≥1. It also implies that 2n+1 >(2 + t)k. Thus we
have that
Z|x−B|>r
ω(x, t) dx=Z|x−B|>r
ω(x, t)|x−B(t)|4n
|x−B(t)|4ndx
≤mn(t)
r4n
≤(Lε2(1 + nt))n
rkr4n−k
23
≤1
rkLε2(1 + kln(2+t)
ln 2 t)n
2Lε2(1 + kln(2+t)
ln 2 t)n−k/4
=1
rkLε2(1 + kln(2+t)
ln 2 t)k/4
2n−k/4
=2k/4+1
rkεk/2L(1 + kln(2+t)
ln 2 t)k/4
2n+1
≤2k/4+1
rkεk/2L(1 + kln(2+t)
ln 2 t)k/4
(2 + t)k.
The function t7→ L(1 + kln(2+t)
ln 2 t)k/4
(2 + t)kis bounded on R+for every k, so there exists a
constant C(k)such that for εsmall enough:
Z|x−B|>r
ω(x, t) dx≤C(k)εk/2
rk.
This completes the proof of the lemma.
4.4 End of the proof of Theorem 1.3
We can now finish the proof of Theorem 1.3.
We recall that, according to Lemma 4.2, we have that for each particle such that |X(t)−
B(t)|=Rt,
d
ds|Xt(s)−B(s)|s=t≤2K1εβRt+5
πR3
t
I(t) + K2ε−νZ|x−B|>Rt/2
ω(x, t) dx1/2
.
Due to the estimates obtained in Lemma 4.3, taking t < min(τε,β, ε−β), we infer that
d
ds|Xt(s)−B(s)|s=t≤2K1εβRt+5K3ε2
πR3
t
+K2ε−νZ|x−B|>Rt/2
ω(x, t) dx1/2
.(4.15)
Let us introduce fthe solution of the ODE:
f0(t)=4K1εβf(t) + 4 max 5K3ε2
πf 3(t), K2ε−νZ|x−B|>f (t)/2
ω(x, t) dx1/2!
f(0) = 4ε.
(4.16)
We want to show that for every t∈[0,min(τε,β, ε−β)],Rt< f (t). Assume that this assertion
is false, and let t2be the first time when it breaks down. Since f(0) = 4εand R0≤2ε, we
24
infer that t2>0. Let s7→ Xt2(s)be a trajectory such that |Xt2(t2)−B(t2)|=Rt2=f(t2).
From (4.15) and (4.16), we see that
d
ds|Xt2(s)−B(s)|s=t2< f0(t2).(4.17)
However, we have that for every 0< h < t2,
|Xt2(t2−h)−B(t2−h)| ≤ Rt2−h< f(t2−h)
which implies that
|Xt2(t2−h)−B(t2−h)|−|Xt2(t2)−B(t2)|
−h>f(t2−h)−f(t2)
−h
since |Xt2(t2)−B(t2)|=Rt2=f(t2). Taking the limit as h→0, we get a contradiction with
(4.17).
We choose now αand k > 6such that
0< α < min(β, 2−4β)
and
k(1/2−β)+6β−4−ν > 0.
We define
t2= inf{t > 0, f(t) = εβ}
and
t1= sup{t<t2, f (t) = εβ/2}
so that t1< t2and
f(t1) = εβ
2, f(t2) = εβand εβ
2≤f(t)≤εβ∀t∈[t1, t2].
If t2≥ε−αthen Rt< f(t)≤εβfor all t∈[0,min(τε,β , ε−β, ε−α)] = [0,min(τε,β , ε−α)]. By
definition of τε,β we know that Rτε,β =εβ. So necessarily τε,β ≥ε−αwhich completes the
proof of Theorem 1.3.
We assume from now on that t2< ε−α.
We have the following inequality:
∀t∈[t1, t2],f(t)
24
≥K4ε2(1 + kt ln(2 + t)) (4.18)
which implies that Lemma 4.4 can be applied with r=f(t)/2for t∈[t1, t2]. Indeed, relation
(4.18) is true since
K4ε2(1 + kt ln(2 + t)) ≤K4ε2(1 + kε−αln(2 + ε−α)) ≤εβ
44
≤f(t)
24
25
for εsmall enough, as we chose α < 2−4β. Lemma 4.4 yields that
ε−νZ|x−B|>f(t)/2
ω(x, t) dx1/2
≤C(k)εk/2−ν
(f(t)/2)k1/2
∀t∈[t1, t2].(4.19)
Since we chose ksuch that k(1/2−β)+6β−4−ν > 0, for εsmall enough we have that
ε−(k−6)βεk/2−4−ν=εk(1/2−β)+6β−4−ν≤1
C(k)22k−6.
Recalling that f(t)≥εβ/2for every t∈[t1, t2], we infer that
fk−6(t)≥εβ
2k−6≥C(k)2kεk/2−4−ν
which in turns gives that
C(k)εk/2−ν
(f(t)/2)k≤ε4
f6(t).(4.20)
Using relations (4.19) and (4.20) in (4.16) yields that
f0(t)≤Cεβf(t) + Cε2
f3(t).
which implies that
(f4)0(t)≤C1εβf4(t) + C1ε2
for some constant C1. The end of the argument is now straightforward. We use the Gronwall
lemma to obtain that
f4(t2)≤f4(t1)eC1εβ(t2−t1)+ε2−β(eC1εβ(t2−t1)−1).
Since t2−t1≤ε−αwe have that C1εβ(t2−t1)≤C1εβ−α<1for εsmall enough. Using the
inequality ex≤1 + 2xfor 0≤x≤1and recalling that f(t1) = εβ/2and f(t2) = εβwe have
that
ε4β≤(εβ/2)4eC1εβ−α+ 2C1ε2−α.
which implies that
1≤eC1εβ−α
16 + 2C1ε2−4β−α.
Since α < min(β, 2−4β), the right hand side of this inequality goes to 1/16 as ε→0. So
we obtain a contradiction if εis small enough. We thus proved that t2≥ε−αif εis small
enough.
In conclusion, for any β < 1/2, and for any α < min(β, 2−4β), there exists ε0small
enough such that for every ε∈(0, ε0), we have that τε,β > ε−α. This completes the proof of
Theorem 1.3.
26
5 Final remarks
A natural question is the signification of the hypothesis T000(x0)=0. As we already discussed,
the condition T00(x0)=0means that x0is a critical point of the map eγ. Such critical
points always exist in a bounded simply connected domain Ω. But to have a strong result
of confinement as we proved, we need more than just a critical point of eγ. Indeed, we
observed in Section 3 that we should not expect a confinement time better than |ln ε|around
unstable points. So we need the stationary point x0to be at least stable. But the hypothesis
T000(x0) = 0 is stronger than the stability. Indeed, the stability is characterized by the
condition |T000(x0)|<2|T0(x0)|3which is significantly weaker than T000(x0) = 0.
So the hypothesis T000(x0) = 0 is more than just stability. Because we obtained the
explicit value of D2eγ(x0), see relation (3.2), we see that this condition is equivalent to the
fact D2eγ(x0)is a multiple of the identity. This means that the orbit of a single point vortex
in the neighborhood of x0is almost a circle. We don’t know whether this condition is indeed
necessary to have a strong confinement as proved in Theorem 1.3. In our proof we use some
crucial cancellations to prove the estimate of the moment of inertia from Lemma 4.3 that we
can’t reproduce without the hypothesis T000(x0)=0.
One can wonder about the existence of domains satisfying the condition T000(x0) = 0 in a
stationary point x0. We call these domains valid domains. Let us notice that if T(x0) = 0 and
T00(x0) = 0, then the condition T000 (x0) = 0 is equivalent to (T−1)000(0) = 0. This means that
the image Ω = f(D(0,1)) by any biholomorphic map of the form f(z) = x0+a1z+P∞
k=4 akzk
of the unit disk is a valid domain: there exists a biholomorphic map T: Ω →D(0,1) such
that T(x0) = T00(x0) = T000 (x0)=0(one can choose T=f−1). However, checking that a
map fof the form given above is indeed biholomorphic may not be an easy task.
Let us observe now that regular convex polygons are valid domains. Indeed, by the
Schwarz-Christoffel formula (see [1]), there exists a conformal map ffrom the unit disk to
the nsided regular polygon with vertices at ωn=ei2π/n, and its derivative has the form
f0(z) = c
n
Y
k=1 1−zω−k
n1−2
n.
We can therefore obtain an explicit value for the second and third derivatives, and check that
they vanish at 0. So regular convex polygons are valid domains.
On the other hand, an ellipse which is not a circle is not a valid domain. Indeed, we know
from [12] that a conformal mapping from the disk to an ellipse of foci ±1mapping 0 to 0 has
a Taylor expansion f(z) = z+A3z3+. . . near 0 with A3>0.
Another method to obtain valid domains is to analyze the effect of rotational invariance.
Assume that the domain Ωis invariant by rotation of angle θ∈(0,2π)around the point
x0= 0. This means that eγΩ(x) = eγΩ(eiθx)for every x∈Ω. Differentiating this relation and
using relation (2.4) implies that
∇eγΩ(eiθ x) = cos θsin θ
−sin θcos θ∇eγΩ(x)
and thus ∇eγΩ(0) = 0 implying that 0is a stationary point. Similarly, differentiating again
27
yields that
∂2
1eγΩ(x) = ∂2
1[eγΩ(eiθx)] = cos2θ∂2
1eγΩ(eiθ x) + sin2θ∂2
2eγΩ(eiθ x) + 2 cos θsin θ∂1∂2eγΩ(eiθ x)
∂2
2eγΩ(x) = ∂2
2[eγΩ(eiθ x)] = sin2θ∂2
1eγΩ(eiθ x) + cos2θ∂2
2eγΩ(eiθ x)−2 cos θsin θ∂1∂2eγΩ(eiθ x)
∂1∂2eγΩ(x) = ∂1∂2[eγΩ(eiθ x)] = (cos2θ−sin2θ)∂1∂2eγΩ(eiθ x)
+ cos θsin θ(∂2
2eγΩ(eiθ x)−∂2
1eγΩ(eiθ x)).
We set x= 0 and we subtract the first equation above from the second equation. We get
(1 −cos2θ+ sin2θ)(∂2
2eγΩ(0) −∂2
1eγΩ(0)) = −4 cos θsin θ∂1∂2eγΩ(0).
Using this in the third equation yields after some calculations that
θ=πor ∂1∂2eγΩ(0) = 0.
If ∂1∂2eγΩ(0) = 0 and θ6=πwe observe that ∂2
1eγΩ(0) = ∂2
2eγΩ(0). Therefore we have that
θ=πor ∃λ, D2eγΩ(0) = λI2.
From (3.2) we know that
D2eγΩ(0) = µ2
πI2+1
2πp q
q−p.
Clearly D2eγΩ(0) is a multiple of the identity if an only if p=q= 0. Therefore we have that
either θ=πor p=q= 0. From the definition of pand qgiven after relation (3.2) we see that
for any conformal map Tmapping 0 to 0 the condition p=q= 0 is equivalent to T000(0) = 0.
So if θ6=π, then T000(0) = 0 and the domain is therefore valid. This is another proof of the
fact that regular polygons are valid domains, and it also gives us many other valid domains.
We used Mathematica to plot the boundary of some valid domains. We consider different
functions f(z) = a1z+PN
k=4 akzk, with N≥4and |a1|>PN
k=4 k|ak|in order to ensure
that fis injective on the unit disk. We obtain a large class of valid domains, with the unit
disk of course and small variations of it (figure 2), but also larger perturbation of the disk
(figure 3), and even some very erratic domains (figure 4). Notice that these domains don’t
necessarily have symmetry properties. Finally, by using the rotational invariance property,
we can plot more complicated boundaries without knowing the biholomorphic map, like in
figure 5. We plotted images of the interval [0,1] by maps of the form b(x) = r(x)ei2π(x+θ(x)),
with r(x)>0. We choose θand rto be 1/p-periodic functions with p > 2an integer. This
way, bplots a closed curve in Cthat is invariant by rotation of angle 2π/p ∈(0, π). If this
curve does not self intersect and is smooth, then its interior is a valid domain.
28
Figure 2: Plot for f(z) = z(left), and f(z) = 40z+z4(right).
Figure 3: Plot for f(z) = 40z+z7(left), and f(z) = 50z+ (i+ 1)z4+z23 (right).
Figure 4: Plot for f(z) = 20z+ (2i+ 1)z4+z7(left), and f(z) = 19z+iz7+z10 (right).
29
Figure 5: Plot for b(x) = (2 + cos(8πx))ei2πx (left), and b(x) = exp 2iπ(x+1
8cos(6πx)
3
2+1
4cos(6πx)(right),
with x∈[0,1].
Acknowledgments. Dragos
,Iftimie has been partially funded by the LABEX MILYON
(ANR-10-LABX-0070) of Université de Lyon, within the program “Investissements d’Avenir”
(ANR-11-IDEX-0007) operated by the French National Research Agency (ANR). Martin
Donati thanks the Dipartimento di Matematica, SAPIENZA Università di Roma for its
hospitality. The authors wish to acknowledge helpful discussions with Paolo Buttà and
Carlo Marchioro.
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Martin Donati: Université de Lyon, CNRS, Université Lyon 1, Institut Camille Jordan, 43
bd. du 11 novembre, Villeurbanne Cedex F-69622, France.
Email: donati@math.univ-lyon1.fr
Dragos
,Iftimie: Université de Lyon, CNRS, Université Lyon 1, Institut Camille Jordan, 43
bd. du 11 novembre, Villeurbanne Cedex F-69622, France.
Email: iftimie@math.univ-lyon1.fr
Web page: http://math.univ-lyon1.fr/˜iftimie
31
