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Spatial asymptotic expansion of the Euler equation
by Saif Sultan
BE Electrical Engineering, NUST PakistanMS Mathematics, Northeastern University
A dissertation submitted to
The Faculty ofthe College of Science ofNortheastern University
in partial fulfillment of the requirementsfor the degree of Doctor of Philosophy
December 6, 2019
Dissertation directed by
Prof. Peter TopalovProfessor of Mathematics
1
Dedication
Dedicated to my parents Sher Sultan and Tahira Batool and my siblings for all their love,
support and patience.
2
Acknowledgments
I want to sincerely thank my advisor Prof. Peter Topalov for the continuous support of my
Ph.D. study and for his patience, immense knowledge and for being an amazing teacher. His
guidance helped me throughout my research and during writing this thesis. I could not have
imagined having a better advisor and mentor for my Ph.D. study.
3
Abstract of Dissertation
In this dissertation we study the Euler equation in dimension 2 and higher. We show that
the Euler equation is locally well-posed in the space of spatially asymptotic functions. Fur-
thermore, we prove that the 2d Euler equation is globally well-posed in the corresponding
asymptotic space. We discuss how the Euler equation preserves certain function spaces that
have asymptotic expansions. These asymptotic function spaces are known to have asymp-
totic terms that include logarithms. However, we show that structurally simpler spaces
without log terms are also preserved by the Euler equation. In 2 dimensional fluid flow the
complex structure can be used to find such smaller function spaces without log terms.
4
Table of Contents
Dedication 2
Acknowledgments 3
Abstract of Dissertation 4
Table of Contents 5
Introduction 7
Chapter 1 Asymptotic expansion in two dimensions 10
1.1 Asymptotic spaces and diffeomorphisms ZDm,pN . . . . . . . . . . . . . . . . 12
1.2 Cauchy operator in asymptotic spaces . . . . . . . . . . . . . . . . . . . . . . 20
1.3 The Euler equation on ZDm,pN . . . . . . . . . . . . . . . . . . . . . . . . . . 33
1.4 Global solutions of the 2d Euler equation on ZDm,pN . . . . . . . . . . . . . . 40
1.5 Proof of Lemma 1.1.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
Chapter 2 Asymptotic expansion in higher dimensions 61
2.1 Asymptotic spaces and diffeomorphism groups . . . . . . . . . . . . . . . . . 64
2.2 Laplace operator in asymptotic spaces . . . . . . . . . . . . . . . . . . . . . 72
2.3 Euler vector field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
2.4 Smoothness of Euler Vector Field . . . . . . . . . . . . . . . . . . . . . . . . 78
2.5 Proof of Theorem 2.0.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
2.6 Volume preserving asymptotic diffeomorphism . . . . . . . . . . . . . . . . . 81
5
2.7 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84
2.8 Remarks on solutions with constant asymptotic term . . . . . . . . . . . . . 86
2.9 Asymptotics preserved by Euler equation . . . . . . . . . . . . . . . . . . . . 86
Appendices 89
References 91
6
Introduction
Consider the incompressible Euler equation on Rd with d ≥ 2, vt + v · ∇v = −∇p, div v = 0,
v|t=0 = v0,(1)
where v(x, t) is the velocity of the fluid and p(x, t) is the scalar pressure. The divergence div
and the covariant derivative ∇ are computed with respect to the Euclidean metric on Rd. It
was proven in [14] that the Euler equation (1) is locally well-posed in the class of asymptotic
vector fields Am,pN ;0. By definition, a vector field v on Rd belongs to the class of asymptotic
vector fields Am,pN ;0 with integer regularity exponent m ≥ 0, a real 1 < p <∞, and an integer
parameter N ≥ 0, if it can be written in the form
v(x) = a0(θ)+a0
1(θ) + a11(θ) log r
r+...+
a0N(θ) + ...+ aNN(θ)(log r)N
rN+f(x), | x | ≥ 1/2, (2)
where r ≡ r(x) := | x |, θ ≡ θ(x) := x /| x | is a point on the unit sphere Sd−1, the coefficients
ajk for 0 ≤ k ≤ N and 0 ≤ j ≤ k are continuous vector-valued functions on the sphere
ajk : Sd−1 → Rd that have some additional regularity, and f : Rd → Rd is the remainder that
belongs to a weighted Sobolev space on Rd so that it is continuous, and f(x) = o(1/rN) as
r → ∞. We refer to [14, Section 2] for detail. In this sense, any vector field in Am,pN ;0 has a
partial asymptotic expansion at infinity of order N .
There are two points that have to be mentioned: The first is that the classAm,pN ;0 of asymptotic
vector fields is natural for the Euler equation for the following reason: even if we take a generic
smooth initial data with compactly supported components vj0 ∈ C∞c (Rd) , the corresponding
7
local solution of the Euler equation will develop non-trivial asymptotics at infinity of the
form (2) with non-vanishing leading asymptotic term ad+1(θ)/rd+1. The second point is that
if we take a generic initial data of the form (2) with vanishing coefficients in front of the log
terms, then the local solution of the Euler equation will develop non-trivial asymptotics with
non-vanishing log terms. The reason for this is that the space Am,pN ;0 with vanishing log terms
requires no restrictions on the spherical Fourier modes of the coefficients ajk, 0 ≤ k ≤ N and
0 ≤ j ≤ k. We refer to [14, Appendix B] for detail.
The goal of this work is to study well-posedness of the Euler equation in simpler asymptotic
spaces which, for example, do not involve log term. To this end we will be working with
asymptotic spaces Am,pn,N as defined below (cf. [13, Section 3])
Am,pn,N :={u∣∣u = χ(r)
∑n≤k≤N
ak(θ)
rk+ f such that f ∈ Wm,p
γNand ak ∈ Hm+1+N−k,p(Sd−1)
},
(3)
where χ : R → R is a C∞-smooth cut-off function with bounded derivatives such that
χ(ρ) = 1 for ρ ≥ 2 and χ(ρ) = 0 for 0 ≤ ρ ≤ 1, equipped with the norm
‖u‖Am,pn,N:=
∑n≤k≤N
‖ak‖Hm+1+N−k(Sd−1) + ‖f‖Wm,pγN
. (4)
Here Wm,pδ denotes weighted Sobolev space with weight δ ∈ R and smoothness m ≥ 0 defined
as
Wm,pδ :=
{f ∈ Hm,p
loc (C,C)∣∣ 〈x〉δ+|α|∂αf ∈ Lp for |α| ≤ m
},
supplied with the norm ‖f‖Wm,pδ
:=∑|α|≤m
∥∥〈x〉δ+|α|∂αf∥∥Lp
where 〈x〉 = (1 + | x |2)1/2, and
Hm,ploc is space of functions that are locally in Sobolev spaces. We will denote the weight
γN := N +γ0 where 0 < γ0 + 2p< 1 is given a priori and does not depend on the choice of N .
This choice of weights γN ensures that the remainder terms f ∈ Wm,pγN
satisfy f = o(1/rN) as
lim r →∞. Furthermore, we will generally denote Am,pN := Am,p0,N . For m > d/p these spaces
are Banach algebras.
V. Arnold in [2] viewed the fluid flow as a solution to an ODE on an infinite dimensional
8
group of diffeomorphisms of the underlying space. This approach was used in [7] and [3] to
show the well-posedness of Euler equation. The well-posedness of the incompressible Euler
equation in the weighted Sobolev spaces Wm,pδ was then studied [5] while the well-posedness
in asymptotics spaces Am,pN ;0 is established in [14].
9
Chapter 1
Asymptotic expansion in two
dimensions
In this chapter we restrict our attention to the two dimensional case (d = 2) and answer the
following questions:
1) Can one find a subspace of the asymptotic space Am,pN ;0 with non-trivial asymptotic part
and such that it is invariant with respect to the Euler equation and does not have log
terms in its asymptotic part?
2) Are the solutions of the Euler equation in such a space global in time?
We will show that the answer to the both questions is positive. An important feature of our
analysis is that we identify R2 with the complex plane by setting z = x+ iy for (x, y) ∈ R2,
i =√−1 and rewrite the Euler equation (1) in complex form as ut + uuz + uuz = −2∂zp, div u = uz + uz = 0,
u|t=0 = u0,(1.1)
where u : C→ C is the holomorphic component of the fluid velocity (i.e. u(z, z) = v1(z, z) +
v2(z, z) where v1, v2 are components of the velocity vector v) and the subscripts z and z denote
10
the partial differentiations with respect to the Cauchy operators ∂z and ∂z respectively. Also
we will inter-changeably denote r = |z| depending on the context. In particular, this allows
us to define the following asymptotic space for integer m > 2/p and N ≥ 0,
Zm,pN :={u∣∣u = χ(r)
∑0≤k+l≤N
aklzkzl
+ f such that f ∈ Wm,pγN
and akl ∈ C}, (1.2)
where k, l ≥ 0 and χ : R → R is a C∞-smooth cut-off function with bounded derivatives
such that χ(ρ) = 1 for ρ ≥ 2 and χ(ρ) = 0 for 0 ≤ ρ ≤ 1, equipped with the norm
‖u‖Zm,pN:=
∑0≤k+l≤N
|akl|+ ‖f‖Wm,pγN
. (1.3)
Here Wm,pδ denotes Weighted Sobolev space with weight δ ∈ R and smoothness m ≥ 0
defined as
Wm,pδ :=
{f ∈ Hm,p
loc (C,C)∣∣ 〈z〉δ+|α|∂αf ∈ Lp for |α| ≤ m
}, ∂α ≡ ∂α1
z ∂α2z ,
supplied with the norm ‖f‖Wm,pδ
:=∑|α|≤m
∥∥〈z〉δ+|α|∂αf∥∥Lp
where 〈z〉 = (1 + |z|2)1/2 and
Hm,ploc is space of functions that are locally in Sobolev spaces. We will denote the weight
γN := N +γ0 where 0 < γ0 + 2p< 1 is given a priori and does not depend on the choice of N .
This choice of weights γN ensures that the remainder terms f ∈ Wm,pγN
satisfy f = o(1/rN)
as lim r →∞. For 0 ≤ n ≤ N we will define following subspace of Zm,pn,N as
Zm,pn,N :={u∣∣u = χ(r)
∑n≤k+l≤N
aklzkzl
+ f such that f ∈ Wm,pγN
and akl ∈ C}
(1.4)
where we omit the summation term if n = N + 1 and set Zm,pN+1,N ≡ Wm,pγN
. Take ρ > 0 and
denote by BZm,pN(ρ) ∈ Zm,pN the open ball of radius ρ centered at zero . The main result of
this chapter is the following Theorem proved in Section 1.4.
Theorem 1.0.1. Assume m > 3+2/p where 1 < p <∞. Then for any u0 ∈ Zm,pN the Euler
equation (1.1) has a unique global solution in time u ∈ C([0,∞),Zm,pN
)∩C1
([0,∞),Zm−1,p
N
)such that the pressure p(t) lies in Zm+1,p
1,N for any t ≥ 0. The solution depends continuously
11
on the initial data u0 ∈ Zm,pN in the sense that for any given T > 0 the data-to-solution map
u0 7→ u, Zm,pN → C([0, T ],Zm,pN
)∩ C1
([0, T ],Zm−1,p
N
)is continuous.
1.1 Asymptotic spaces and diffeomorphisms ZDm,pN
We can define the space of asymptotic diffeomorphisms associated with the asymptotic space
Zm,pN as
ZDm,pN :={ϕ ∈ Diff1
+(R2)∣∣∣ ϕ = id +u, u ∈ Zm,pN
}, (1.5)
where Diff1+(R2)1 denotes the group of orientation preserving C1-diffeomorphisms of R2 and
id : R2 → R2 is the identity map. Note that the asymptotic space Zm,pN is a closed subspace
in Am,pN and therefore ZDm,pN is a subset of the asymptotic diffeomorphisms ADm,pN considered
in (cf. [14, Section 2])
ADm,pN :={ϕ ∈ Diff1
+(R2)∣∣∣ ϕ = id +u, u ∈ Am,pN
}.
In this section we show that ZDm,pN is a topological subgroup of ADm,pN (see Theorem 1.1.12).
For this we use the fact that ZDm,pN is a submanifold in ADm,pN and we will need to show
that:
1. ZDm,pN is closed under composition of diffeomorphisms.
2. ZDm,pN is closed under inversion of diffeomorphisms.
The following two propositions summarize the properties of the asymptotic spaces Zm,pN and
their reminder spaces Wm,pδ . For the proof we refer to Proposition 1.5.1, Proposition 1.5.3,
and Remark 15 in Section 1.5 (cf. [13, § 2]).
1We will denote the identity diffeomorphism on Rd by id. I will denote the identity matrix on Rd.
12
Proposition 1.1.1. (i) For any 0 ≤ m1 ≤ m2 and for any real weights δ1 ≤ δ2 we have
that the inclusion map Wm2,pδ2→ Wm1,p
δ1is bounded.
(ii) For any m ≥ 0, δ ∈ R, and for any k ∈ Z the map f 7→ f · χzk
, Wm,pδ → Wm,p
δ+k is
bounded. The same holds with z replaced by z. Here χ(z, z) ≡ χ(|z|) is a C∞-smooth
cut-off function such that χ : R→ R and χ|(−R,R) ≡ 1 for some R > 0.
(iii) For any regularity exponent m ≥ 0, δ ∈ R, and a multi-index α ∈ Z2≥0 such that
|α| ≤ m the map ∂α : Wm,pδ → W
m−|α|,pδ+|α| is bounded.
(iv) For m > 2/p and for any weights δ1, δ2 ∈ R the map (f, g) 7→ fg, Wm,pδ1×Wm,p
δ2→
Wm,pδ1+δ2+(2/p) is bounded. In particular, for δ ≥ 0 the space Wm,p
δ is a Banach algebra.
Proposition 1.1.1 implies
Proposition 1.1.2. Assume that m > 2/p. Then
(i) For any 0 ≤ n1 ≤ n2 ≤ N1 ≤ N2 we have that the inclusion map Zm,pn2,N2→ Zm,pn1,N1
is
bounded.
(ii) For any 0 ≤ n ≤ N , an integer k ≥ 0, a regularity exponent m > k + (2/p), and a
multi-index |α| ≤ k the map ∂α : Zm,pn,N → Zm−|α|,pn+|α|,N+|α| is bounded.
(iii) For m > 2/p and for any 0 ≤ n1 ≤ N1 and 0 ≤ n2 ≤ N2 the map
(f, g) 7→ fg, Zm,pn1,N1×Zm,pn2,N2
→ Zm,pn,N ,
where n := n1 + n2 and N := min(n1 +N2, n2 +N1) is bounded. In particular, Zm,pN is
a Banach algebra for any N ≥ 0.
For a given R > 0, we will also consider the space Zm,pN (BcR) that is the image of the
restriction map of u ∈ Zm,pN to BcR =
{x ∈ R2
∣∣ | x | > R}
. An equivalent norm we will use
with Zm,pN (BcR) is
|u|Zm,pN (BcR) =∑
0≤k+l≤N
|akl|Rk+l
+ ‖f‖Wm,pγN
(BcR) .
13
Remark 1. This norm is equivalent to the norm it inherits from Zm,pN . This follows from
the fact that any two norms (see (1.3)) on RM for M ≥ 1 are equivalent. We will later show
that Zm,pN (BcR) is a Banach algebra with this norm and more importantly there is a constant
CZ independent of the radius R such that |a · b|Zm,pN (BcR) ≤ CZ |a|Zm,pN (BcR)|b|Zm,pN (BcR).
Following Lemma allows us to work between the two spaces Zm,pN and Zm,pN (BcR).
Lemma 1.1.3. Assume that R > 0. Then, for u ∈ Hm,ploc (C,C) we have that u ∈ Zm,pN if and
only if u ∈ Hm,p(BR+1) and u ∈ Zm,pN (BcR). The norms ‖·‖Zm,pN
and ‖·‖Hm,p(BR+1)+|·|Zm,pN (BcR)
on Zm,pN are equivalent.
We begin with following technical lemma. Its proof involves several technical steps and is
defered to Section 1.5. The constant CZ appearing in below is independent of radius R > 0
of the ball BR.
Lemma 1.1.4. Let m > 2/p then there exists a constant CZ independent of radius R > 0
such that for u1, u2 ∈ Zm,pN (BcR)
|u1 · u2|Zm,pN (BcR) ≤ CZ |u1|Zm,pN (BcR) |u2|Zm,pN (BcR) .
Lemma 1.1.5. Let m > 2/p and let u ∈ Zm,pN such that |1 + u| > 0 and u has no constant
asymptotics (a00 = 0), then 11+u∈ Zm,pN and the map u 7→ 1
1+uis analytic at such u.
Proof. Let us denote C∗ = min(1/2, C−1Z /2). We first show that 1
1+u∈ Zm,pN . Note that by
Proposition 1.5.1 u is a continuous function. Also by Corollary 1.5.5, we can take R > 2
large enough such that, |u|Zm,pN (BcR) < C∗ so that supz∈BcR |u| ≤ |u|Zm,pN (BcR) by Corollary 1.5.5.
By expanding in geometric series, we see that
(1 + u)−1 =∑k≥0
uk
converges both pointwise and also in Zm,pN (BcR). Continuity of u shows that |1+u| is bounded
below in BR+1 by a positive number and so (1 +u)−1 is bounded. By directly differentiating
14
(1 + u)−1 we see that (1 + u)−1 ∈ Hm,p(BR+1). Hence 11+u∈ Zm,pN by Lemma 1.1.3.
Now let ψ = 1 + u, by continuity of multiplication in Zm,pN and the embedding Zm,pN ↪→ L∞
given by Corollary 1.5.5, we can find a open neighborhood V of 0 in Zm,pN such that for v ∈ V∥∥∥∥v · 1
ψ
∥∥∥∥Zm,pN
< C∗ and
∥∥∥∥v · 1
ψ
∥∥∥∥L∞
< C∗.
Then we can expand as
1
ψ + v=
1
ψ
1
1 + vψ
=1
ψ
∞∑j=0
(−vψ
)jwhich shows that the map is analytic at u.
Remark 2. The proof of this Lemma shows that this is also true for Zm,pN (BcR) for R > 2
instead of Zm,pN . That is if u ∈ Zm,pN (BcR) satisfies conditions of the above Lemma then
(1 + u)−1 ∈ Zm,pN (BcR).
We will need following space when working with the Cauchy operator in the next section.
(See Remark 10)
Zm,pN :={u|u =
χ
z2
∑0≤k+l≤N−2
aklzkzl
+ f such that f ∈ Wm,pγN
and akl ∈ C}
(1.6)
where χ ≡ χ(|z|) and we set that Zm,p1 ≡ Wm,pγ1
. Note that Zm,pN is a closed subspace in Zm,pN .
Proposition 1.1.6. Let m > 1 + 2/p and ϕ ∈ ZDm,pN . Then for any u ∈ Zm,pN we have
u ◦ ϕ ∈ Zm,pN . Similarly, for u ∈ Zm,pN we have u ◦ ϕ ∈ Zm,pN .
Proof. We can take ϕ = z + v such that v ∈ Zm,pN . First we show that if u = χ(z,z)z
then
u ◦ ϕ ∈ Zm,pN . We can write
u ◦ ϕ =χ ◦ ϕ(z, z)
z + v(z, z)=χ ◦ ϕ(z, z)
z
1(1 + v
z
) . (1.7)
For large enough R > 2 we have χ ◦ϕ ≡ 1 in BcR since v is bounded by Proposition 1.5.1(ii).
For |z| ≥ R, v/z ∈ Zm,pN (BcR) and since v/z has zero constant term therefore by Remark 2
15
following Lemma 1.1.5, we have (1 + v/z)−1 ∈ Zm,pN (BcR). Since ϕ is a C1-diffeomorphism of
the complex plane C = R2, the expression ϕ = z+v does not vanish on the support of χ◦ϕ.
This shows that u ◦ϕ = χ ◦ϕ/(z+ v) belongs to Hm,p(BR+1) and so u ◦ϕ ∈ Zm,pN by Lemma
1.1.3. Similarly it holds that if u = χ(z,z)z
that u ◦ ϕ ∈ Zm,pN and since Zm,pN is an algebra so
this holds for any asymptotic part. Finally we show that if f ∈ Wm,pγN
then f ◦ ϕ ∈ Wm,pγN
.
This follows directly from Corollary 6.1 in [13]. The case when u belongs to the auxiliary
space Zm,pN follows easily by the same arguments.
Remark 3. Formula (1.7) implies that for u = χ/z and for any ϕ ∈ ZDm,pN we have that
u ◦ ϕ =χ
z
(1 + f
), f ∈ Zm,p1,N+1, (1.8)
and a similar formula for u = χ/z. This together with Proposition 1.1.2 and Lemma 6.5 in
[13] then shows that the following stronger statement holds: for any ϕ ∈ ZDm,pN and for any
u ∈ Zm,pn,N , 0 ≤ n ≤ N + 1 we have that u ◦ ϕ ∈ Zm,pn,N .
Following Proposition uses the previous result and the fact that ADm,pN is a topological group
with same smoothness property as below.
Proposition 1.1.7. Let m > 1 + 2/p then ZDm,pN is closed under composition of diffeomor-
phisms. Also the composition map (ϕ, ψ) 7→ ϕ ◦ψ, ZDm,pN ×ZDm,pN → ZDm,pN , is continuous
and the associated map
(ϕ, ψ) 7→ ϕ ◦ ψ, ZDm+1,pN ×ZDm,pN → ZDm,pN ,
is C1-smooth.
Remark 4. Similarly, for m > 1 + (2/p) the composition map (u, ψ) 7→ u ◦ ψ, Zm,pN ×
ZDm,pN → Zm,pN , is continuous and the map
(u, ψ) 7→ u ◦ ψ, Zm+1,pN ×ZDm,pN → Zm,pN ,
is C1-smooth.
16
Proof. Let ϕ = z+v and ψ = z+ω and ϕ, ψ ∈ ZDm,pN then ϕ◦ψ = ψ+v◦ψ = z+ω+v◦ψ =
z + u where u = w + v ◦ ψ and last result shows that u is in Zm,pN . Since ϕ ◦ ψ ∈ Diff1+(R2)
we then conclude that ϕ ◦ ψ ∈ Zm,pN .
The last statement of the proposition then follows from the analogous result for the asymp-
totic group ADm,pN (see [13, Proposition 5.1]) and the fact that ZDm,pN is a closed submanifold
in ADm,pN .
In rest of this section we show that ZDm,pN is closed under inversion.This uses results that
follow directly from related results in [13] and are added here. This and Proposition 1.1.7
allow us to concldue that ZDm,pN is a topological subgroup ofADm,pN . (Theorem 1.1.12 below).
Proposition 1.1.8. Assume m > 2 + 2/p then for any ε > 0 small enough there exists a
Uε ε-neighborhood of id in ZDm,pN such that for any ϕ ∈ Uε we have ϕ−1 ∈ ZDm,pN .
Proposition 1.1.9. Assume m > 1 + 2/p, then any ϕ ∈ ZDm,pN can be connected by a
continuous path γ : [0, 1] → ZDm,pN to a ϕ0 := γ(0) ∈ ZDm,pN such that ϕ0 − id has compact
support and such that γ(1) = ϕ.
Proposition 1.1.10. Assume m > 2 + 2/p and let ϕ ∈ ZDm,pN then there exists a neigh-
borhood U of id ∈ ZDm−1,pN such that the left translation ψ 7→ Lϕ(ψ) := ϕ ◦ ψ is a C1-
diffeomorphism U → Lϕ(U) ⊂ ZDm−1,pN .
Proposition 1.1.11. Let m > 3 + 2/p then for any ϕ ∈ ZDm,pN we have that ϕ−1 ∈ ZDm,pN
and the map ϕ 7→ ϕ−1, ZDm,pN → ZDm,pN , is continuous and the map ϕ 7→ ϕ−1, ZDm,pN →
ZDm−1,pN is a C1-map.
Proof. Let ϕ ∈ ZDm,pN . By Proposition 1.1.9, ϕ is connected by a path γ : [0, 1] → ZDm,pN
to γ(0) = ϕ0 = id +f such that f has compact support and γ(1) = ϕ. Note that γ
is also continuous into ZDm−1,pN . In particular, we see that f ∈ Hm,p(C,C), and hence
ϕ0 ∈ Dm,p(R2) where Dm,p(R2) is the group of Sobolev type diffeomorphisms of R2 con-
sidered in [8]. Hence, ϕ−10 ∈ Dm,p(R2). Since, f has compact support we obtain that
17
ϕ−10 = id +g where g ∈ Hm,p(C,C) has compact support. This implies that ϕ−1
0 ∈ ZDm,pN
with vanishing asymptotic part.
Along this path for each ϕt, by Proposition 1.1.8, there exists Vt ⊂ ZDm−1,pN an open neigh-
borhood of id ∈ ZDm−1,pN in which every diffeomorphism is invertible within ZDm−1,p
N and
such that by Proposition 1.1.10, Ut = Lϕt(Vt) is open neighborhood of ϕt in ZDm−1,pN . Ev-
ery ϕ∗ in U0 is invertible in ZDm−1,pN since we can write ϕ∗ = ϕ0 ◦ ψ with ψ ∈ Vt, so
ϕ−1∗ = ψ−1 ◦ ϕ−1
0 ∈ ZDm−1,pN . Now we can cover the path connecting ϕ0 to ϕ = ϕ1 by
open sets U0, Ut1 , Ut2 , ..., U1(ordered by sequence in which they intersect consecutively). By
induction on k, we show that any ϕ∗ ∈ Utk is invertible in ZDm−1,pN . For k = 1 choose
ϕ ∈ U0 ∩Ut1 then ϕ = ϕt1 ◦ ψ for some ψ ∈ Vt1 , so we can write ϕ−1t1 = ψ−1 ◦ ϕ−1 hence it is
invertible in ZDm−1,pN . Continuing in this way we see that ϕ−1 ∈ ZDm−1,p
N .
Now we use a bootstrapping argument to show that ϕ−1 ∈ ZDm,pN . Let ϕ−1 = id +v and
ϕ = id +u, then we can write the differentials of ϕ−1 ◦ϕ = id as (Dv ◦ϕ)(I+Du) = −Du. If
det(I +Du)−1 ∈ Zm−1,pN then by writing Du ◦ϕ = −Du · (I +Du)−1 and by ϕ−1 ∈ ZDm−1,p
N
it follows that Dv ∈ ZDm−1,pN and so ϕ−1 ∈ ZDm,pN . Now det(I + Du) = 1 + w, w ∈ Zm,pN ,
then by Lemma 1.1.5 we get that det(I +Du)−1 ∈ Zm−1,pN .
The last statement of the proposition follows from the analogous results for the asymptotic
group ADm,pN (see [13, Proposition 5.2], [16]) and the fact that Zm,pN is a closed submanifold
in ADm,pN .
Theorem 1.1.12. For integers m > 3 + 2/p and N ≥ 0, ZDm,pN is a topological group under
composition. Further the composition (ψ, ϕ) 7→ ψ ◦ ϕ, ZDm,pN × ZDm−1,pN → ZDm−1,p
N and
the inverse map ϕ 7→ ϕ−1, ZDm,pN → ZDm−1,pN are C1-maps.
Remark 5. Similarly the map (u, ϕ) 7→ u ◦ ϕ, Zm,pN ×ZDm−1,pN → ZDm−1,p
N is a C1-map.
Lemma 1.1.13. Let u ∈ C ([0, T ],Zm,pN ) for some T > 0, N ≥ 0 and m > 2 + 2/p. Then
18
there exists a unique solution ϕ ∈ C1 ([0, T ],ZDm,pN ) of the equation
ϕ = u ◦ ϕ, ϕ∣∣t=0
= id . (1.9)
Proof. Let us denote F (t, ϕ) = u(t) ◦ ϕ, [0, T ] × ZDm−1,pN → Zm−1,p
N . By Theorem 1.1.12,
composition is C1-map and so the second partial derivative
D2F : [0, T ]×ZDm−1,pN → L(Zm−1,p
N ,Zm−1,pN )
is continuous. Hence F is locally Lipschitz in ϕ for any t ∈ [0, T ], therefore for any t0 ∈ [0, T ],
there exists ε0 > 0 such that there is a unique solution ϕ ∈ C1([t0 − ε0, t0 + ε0],ZDm−1,p
N
)of the equation
ϕ = u ◦ ϕ, ϕ∣∣t=t0
= id .
Now if ψ0 ∈ ZDm−1,pN , then since right translation with fixed ψ0, ϕ 7→ ϕ◦ψ0 is C∞ smooth so
for any t0 ∈ [0, T ], we have same ε0 > 0 such that ψ = ϕ◦ψ0 ∈ C1([t0 − ε0, t0 + ε0],ZDm−1,p
N
)is the unique solution of
ψ = u ◦ ψ, ψ∣∣t=t0
= ψ0.
By compactness of [0, T ] in [0, T ] × ZDm−1,pN we have a solution ϕ ∈ C1
([0, T ],Zm−1,p
N
)of
equation (1.9). It remains to show that ϕ(t) ∈ ZDm,pN . Applying differential to equation
(1.9) we get following equation.
(dϕ) = du ◦ ϕ · dϕ.
Consider the system Y = A(t)Y , Y∣∣t=0
= I, where2 A(t) = du(t)◦ϕ. Since A(t) is continuous
by remark following Proposition 1.1.7, this linear system has a unique solution Y (t) ∈ Zm−1,pN+1 .
Hence ϕz ∈ Zm−1,pN+1 . Now if we write ϕ = z + w,w ∈ Zm−1,p
N then ϕz = 1 + ∂zw gives
∂zw ∈ Zm−1,pN+1 . A priori from ϕ ∈ ZDm−1,p
N we have ∂zw ∈ Zm−2,p
N+1 . Therefore ∂zw ∈ Zm−1,p
N+1 .
Now by Theorem 1.2.7 w ∈ Zm,pN hence ϕ ∈ C1([0, T ],Zm,pN ).
2I denotes the d× d identity matrix.
19
1.2 Cauchy operator in asymptotic spaces
In [11] behavior of the Laplace operator on the weighted Sobolev spaces Wm,pδ is studied.
Depending on the weight, δ, the Laplace operator is shown to be injective or surjective
with well described kernel and closed image. This essentially depends on the form of the
fundamental solution kernel of the Laplacian and how it relates to the weights δ. This
same approach can be applied to the Cauchy operator to obtain similar properties. To
show injectivity of Cauchy operator, we will construct a bounded inverse that agrees with
fundamental solution kernel on its image. In particular, we show that as in the case of
the Laplace operator (cf. [11]), the Cauchy operator is an injective or surjective Fredholm
operator for all but a discrete values of the weight δ ∈ R. In the Fredholm case we describe
explicitly the kernel and the (closed) image of ∂z. Note that the fundamental solution of the
Cauchy operator ∂z is K := 1/πz. The following estimate will play an important role in our
analysis.
Lemma 1.2.1. Let Ψ be a C∞-smooth complex valued function such that Ψ(z, z) = 1/z for
|z| ≥ 2 and∣∣Ψ(z, z)
∣∣ ≤ 1 for z ∈ C. For z, w ∈ C and any given l ∈ Z≥0 define,
K(z, w) :=1
z − w, Kl(z, z;w, w) := Ψ(z, z)
l∑k=0
wkΨ(z, z)k.
Then, there exists a constant C ≡ Cl > 0 such that for any z, w ∈ C,
∣∣K(z, w)− Kl(z, z;w, w)∣∣ ≤ Cl
〈w〉l+1
|z − w|〈z〉l+1(1.10)
where 〈w〉 = (1 + |w|2)1/2
.
Proof of Lemma 1.2.1. We will prove (1.10) with 〈z〉l+1 and 〈w〉l+1 replaced by 1+ |z|l+1 and
1 + |w|l+1 respectively. These are equivalent by the equivalence of norms on Rd+1. Assume
20
that∣∣wz
∣∣ ≥ 12. If |z| ≥ 2 then
∣∣∣(K − Kl
)(z − w)(1 + |z|l+1)
∣∣∣ = (1 + |z|l+1)∣∣∣1− (z − w)Ψ(z, z)
l∑k=0
wkΨ(z, z)k∣∣∣ (1.11)
= (1 + |z|l+1)
∣∣∣∣∣1− (1− w
z
) l∑k=0
wk
zk
∣∣∣∣∣= (1 + |z|l+1)
∣∣∣wz
∣∣∣l+1
≤(
1 +1
|z|l+1
)(1 + |w|l+1)
≤ 2(1 + |w|l+1
)which proves (1.10) in the considered case. If |z| ≤ 2 then the right hand side of (1.11) is
estimated by(1+2l+1
)(1+3|w|
∑lk=0 |wk|
)since
∣∣Ψ(z, z)∣∣ ≤ 1 and |z−w| ≤ |z|+ |w| ≤ 3|w|.
This together with the estimate
|w|l∑
k=0
|w|k = |w| 1 + |w|l+1
1 + |w|≤ 1 + |w|l+1
then implies that the right hand side of (1.11) is bounded by 4(1 + 2l+1
)(1 + |w|l+1
). This
proves (1.10) in the case when∣∣wz
∣∣ ≥ 12.
Now, assume that∣∣wz
∣∣ ≤ 12. Then, K(z, w) = 1
z
∑∞k=0( w
z)k. If |z| ≥ 2 we have
∣∣∣K(z, w)− Kl(z, z;w, w)∣∣∣ =
∣∣∣∣∣1z∞∑
k=l+1
(wz
)k∣∣∣∣∣ ≤ 2
|z||w|l+1
|z|l+1
≤ 3
2
2
|z − w||w|l+1
|z|l+1≤ 3
1 + |w|l+1
|z − w|(1 + |z|l+1)
where we used that
|z − w| ≤ |z|+ |w| ≤ 3
2|z| and
|w|l+1
|z|l+1≤ 1 + |w|l+1
1 + |z|l+1. (1.12)
The second inequality in (1.12) follows from the estimate (in the case when∣∣wz
∣∣ ≤ 12)
|w|l+1(1 + |z|l+1
)− |z|l+1
(1 + |w|l+1
)= |w|l+1 − |z|l+1 =
( 1
2l+1− 1)|z|l+1 ≤ 0.
21
If |z| ≤ 2 then we obtain from∣∣Ψ(z, z)
∣∣ ≤ 1 and∣∣wz
∣∣ ≤ 12
that
∣∣∣K(z, w)− Kl(z, z;w, w)∣∣∣ =
∣∣∣∣∣1z∞∑
k=l+1
(wz
)k+
1
z
l∑k=0
(wz
)k(1− zk+1 Ψ(z, z)k+1
)∣∣∣∣∣≤ 1
|z|
∞∑k=l+1
∣∣∣wz
∣∣∣k +1
|z|
l∑k=0
(1
2
)k(1 + 2k+1
)≤ 2
|z||w|l+1
|z|l+1+
1
|z|(4 + 2l)
≤ C(l)(1 + |w|l+1)
|z − w|(1 + |z|l+1)
where, in order to conclude the final estimate, we used (1.12) and the fact that
1 ≤(1 + 2l+1
) 1 + |w|l+1
1 + |z|l+1
for any |z| ≤ 2 and for any w ∈ C. This completes the proof of the lemma.
Now we can directly use this estimate to construct a bounded map that inverts the Cauchy
operator on its image. Also the image is identified as the null space of certain functionals on
the range Lpδ+1. We will denote by p′ the exponent conjugate to p that is 1 = 1p
+ 1p′
. In the
following (· , ·) : Lpδ × Lp′
−δ → C is the functional pairing
(u, v) =
∫R2
u(x)v(x)d x .
Using this pairing, we can extend ∂z to Lpδ into the dual (W 1,p′
−δ−1)∗ by defining
∂zu(v) = −(u, ∂zv)
when v ∈ W 1,p′
−δ−1 and u ∈ Lpδ . By the weighted Holder Inequality (Lemma 1.5.4) we can
write | − (u, ∂zv)| ≤ ‖u‖Lpδ ‖∂zv‖Lp′−δ ≤ C ‖u‖Lpδ ‖v‖W 1,p′−δ−1
which shows that ∂zu ∈ (W 1,p′
−δ−1)∗.
Following lemma shows that this map extends ∂z.
Lemma 1.2.2. Assume that δ ∈ R, 1 < p < ∞ and p′ be exponent conjugate to p. If
u ∈ W 1,pδ and v ∈ W 1,p′
−δ−1 then
(∂zu, v) = −(u, ∂zv). (1.13)
22
Proof. When u, v ∈ C∞c we can use integration by parts to see these bilinear forms on
W 1,pδ ×W
1,p′
−δ−1 agree on a dense subset.
Lemma 1.2.3. If δ + 2p> 0 then ∂z : W 1,p
δ → Lpδ+1 is injective.
Remark 6. In fact, one can easily see by using the Fourier transform that the kernel of
the Cauchy operator ∂z : S ′ → S ′ consists of all polynomials of z with complex coefficients,
ker ∂z = C[z]. Then, Lemma 1.2.3 follows since for δ + (2/p) > 0 we have that zk /∈ Lpδ for
any k ≥ 0.
Proof. Assume that u ∈ Lpδ and ∂zu = 0. Take an arbitrary test function ϕ ∈ C∞c . Since
δ + (2/p) > 0 we conclude that (δ + 2)p′ > 2, p′ is conjugate exponent to p. This together
with the estimate K ∗ ϕ = O(1/〈z〉
)implies that K ∗ ϕ ∈ Lp
′
−δ−1. By using that K is the
fundamental solution of ∂z we obtain from (1.13) that
(ϕ, u
)=(∂z(K ∗ ϕ)
)(u) = −
(K ∗ ϕ, ∂zu
)= 0
since ∂zu = 0. Since this holds for any ϕ ∈ C∞c we conclude that u = 0.
Lemma 1.2.4. For any weight δ ∈ R we have that u ∈ Lpδ and ∂zu ∈ Lpδ+1 imply that
u ∈ W 1,pδ .
Remark 7. For α ∈ R, if u ∈ Lpδ then 〈z〉αu ∈ Lpδ−α follows directly from definition. Also
∂z (〈z〉αu) = z〈z〉〈z〉
α−1u+ 〈z〉α∂zu. Then as z〈z〉 is bounded we get that 〈z〉αu ∈ Lpδ−α+1.
Proof of Lemma 1.2.4. By above Remark 7 above we can reduce to the case when 0 <
δ + 2/p < 1. Therefore assume < δ + 2/p < 1 and let u ∈ Lpδ and f := ∂zu ∈ Lpδ+1. We can
take a sequence with compact support, (fk)k≥1 in C∞c such that fk → f in Lpδ+1 as k →∞.
Denote by uk the C∞ functions defined as
uk := K ∗ fk . (1.14)
23
Since K ∈ S ′ is the fundamental solution of the Cauchy operator ∂z we have that for any
k ≥ 1,
∂zuk = fk and ∂zuk = (∂zK) ∗ fk (1.15)
in distributional sense in S ′. We will first show by a direct computation that
∂zK = −p.v.1
πz2∈ S ′ (1.16)
where p.v. denotes the Cauchy principal value. To see this let ϕ ∈ C∞c be any test function
and let R > 0 sufficiently large, then we have
(∂zK
)(ϕ) = −(K, ∂zϕ) =
1
2πi
∫C
1
zϕz dz ∧ dz =
1
2πilimr→0+
∫r≤|z|≤R
1
zϕz dz ∧ dz
=1
2πilimr→0+
∫r≤|z|≤R
∂z
(1
zϕ)dz ∧ dz +
1
2πilimr→0+
∫r≤|z|≤R
1
z2ϕdz ∧ dz
=1
2ip.v.
∫C
1
πz2ϕdz ∧ dz.
Here we used Stokes’ theorem to obtain
limr→0+
∫r≤|z|≤R
∂z
(1
zϕ)dz ∧ dz = − lim
r→0+
∫r≤|z|≤R
d(1
zϕ dz
)= lim
r→0+
∮|z|=r
1
zϕ dz
= i limr→0+
∫ 2π
0
e2iθϕ(reiθ) dθ = 0.
By Lemma 1 in [11] (cf. Theorem B* in [20] and Lemma 2.4 in [10]) for 0 < δ + (2/p) < 1
the convolution with the fundamental solution K of the Cauchy operator ∂z extends to a
bounded linear map Lpδ+1 → Lpδ . Moreover, by Theorem 3 in [21, Ch. II, §4], Theorem 1 in
[19], and (1.16), for 0 < δ+ (2/p) < 2 the convolution with ∂zK extends to a bounded linear
map Lpδ+1 → Lpδ+1. (Note that the cancellation condition required in Theorem 3 in [21, Ch.
II,§4] holds since 1/z2 = e2iθ/r2 and∫ 2π
0e2iθ dθ = 0.) Therefore in (1.14) since fk → f in
Lpδ+1 we can pass to the limit to get
u := K ∗ f = limk→∞K ∗ fk ∈ Lpδ ,
24
which is to say that uk → u in Lpδ and hence in S ′ (see Remark 6) where the limit exists in
Lpδ . Similarly, by passing to the limit in S ′ in (1.15) we obtain that
∂zu = f ∈ Lpδ+1 and ∂zu = (∂zK) ∗ f ∈ Lpδ+1.
This implies that u ∈ W 1,pδ . Since ∂zu = f = ∂zu we obtain from the Lemma 1.2.3 that
u = u. Hence, u ∈ W 1,pδ . This completes the proof of the lemma.
Remark 8. From the proof it also follows that ∂zu := f ∈ H1,pδ+1 ⊂ Lpδ+1 then ∂z∂
−1z f =
∂−1z ∂zf that is ∂z, ∂
−1z commute.
Proposition 1.2.5. For 0 < δ + (2/p) < 1 the map ∂z : W 1,pδ → Lpδ+1 is an isomorphism of
Banach spaces.
Proof of Proposition 1.2.5. By Lemma 1.2.4 this map is injective so it only remains to show
that it is surjective. For this let us take any f ∈ Lpδ+1 and define u := K ∗ f . Since for
0 < δ + (2/p) < 1 the convolution with K extends to a bounded map Lpδ+1 → Lpδ we obtain
that u ∈ Lpδ . This follows by Lemma 1 in [11] (cf. Theorem B* in [20] and Lemma 2.4
in [10]). Furthermore since K is the fundamental solution of ∂z, we have ∂zu = f ∈ Lpδ+1.
To see this let (fk)k≥1 ∈ C∞c be a sequence such that fk → f in Lpδ+1 then uk := K ∗ fk
gives ∂zuk = fk, finally since fk → f implies uk → u and as u 7→ ∂zu is a bounded map
so ∂zuk → ∂zu. But also ∂zu := fk → f . Hence ∂zu = f and so we have that the map is
surjective and therefore by open mapping theorem it is also an isomorphism.
Proposition 1.2.6. For l+ 1 < δ + (2/p) < l+ 2 and l ∈ Z≥0 the map ∂z : W 1,pδ → Lpδ+1 is
injective with closed image in Lpδ+1 given by Rl :={f ∈ Lpδ+1
∣∣ (f, zk) = 0 for 0 ≤ k ≤ l}
.
Rl has (complex) co-dimension l + 1.
Proof of Proposition 1.2.6. Let us fix a given l ∈ Z≥0. For l ≥ 0 injectivity of this map
follows from Lemma 1.2.3. Next For z, w ∈ C denote
Fl(z, z;w, w) := K(z, w)− Kl(z, z;w, w). (1.17)
25
By Lemma 1.2.1 we have that
∣∣Fl(z, z;w, w)∣∣ ≤ Cl
〈w〉l+1
|z − w|〈z〉l+1. (1.18)
We will show that the following map
Fl : C∞c → C∞, Fl(u)(z, z) := − 1
2πi
∫CFl(z, z;w, w)u(w, w) dw ∧ dw, (1.19)
associated to the kernel Fl/π extends to a bounded linear map Lpδ+1 → Lpδ . In order to see
this note that (1.19) can be decomposed as
Fl = (M−l−1) ◦ F ◦Ml+1 (1.20)
where for any given weight µ ∈ R,
Ml+1 : Lpµ → Lpµ−(l+1), f 7→ 〈z〉l+1f,
is a linear isomorphism and F : C∞c → C∞ is of the form (1.19) with kernel F (z, z;w, w) :=
F−1(z, z;w, w) satisfying, by the estimate (1.18), the inequality
∣∣F (z, z;w, w)∣∣ ≤ C
|z − w|
with a constant C > 0 depending only on l ≥ 0. Since δ satisfies
0 <[δ − (l + 1)
]+ (2/p) < 1
then the above estimate for∣∣F (z, z;w, w)
∣∣ implies that the map F : C∞c → C∞ extends
to a bounded linear map Lp[δ−(l+1)]+1 → Lpδ−(l+1) as discussed in the proof of Lemma 1.2.4.
Therefore from the decomposition (1.20) it follows that the map (1.19) extends to a bounded
linear map
Fl : Lpδ+1 → Lpδ . (1.21)
Now for any f ∈ Lpδ+1 we can write from equation (1.17) that
K ∗ f = Kl(f) + Fl(f), (1.22)
26
where Kl : Lpδ+1 → C∞ is the integral operator with kernel Kl/π,
Kl(f) :=Ψ
π
l∑k=0
(f, zk
)Ψk. (1.23)
Note that for l+ 1 < δ+ (2/p) < l+ 2 we have that (δ+ 1− l)p′ > 2 where 1p
+ 1p′
= 1. This
implies that
zk ∈ Lp′
−(δ+1)+(l−k) ⊆ Lp′
−(δ+1) (1.24)
for any 0 ≤ k ≤ l. Thus for any 0 ≤ k ≤ l, the map u 7→(u, zk
), Lpδ+1 → C appearing in
(1.23) is well defined and continuous map.
Now we are ready to show that ∂z(W1,pδ := Rl ⊂ Lpδ . Take f ∈ Lpδ+1 ∩ Rl. By (1.23) and
the fact that f ∈ Rl we obtain that K(f) = 0. This, together with (1.22) and (1.21) then
implies that
K ∗ f = Fl(f) ∈ Lpδ .
Since K is the fundamental solution of ∂z we also have that ∂z(K ∗ f
)= f ∈ Lpδ+1. Then,
by Lemma 1.2.4 we obtain that K ∗ f ∈ W 1,pδ . This proves that Rl is contained in the image
of ∂z : W 1,pδ → Lpδ+1. On the other side, it follows from (1.24) that for u ∈ W 1,p
δ and for any
0 ≤ k ≤ l, (∂zu, z
k)
= −(u, ∂z(z
k))
= 0.
Hence, Rl is the image of ∂z : W 1,pδ → Lpδ+1. This completes the proof of the proposition.
Theorem 1.2.7. Assume that m ∈ Z≥0. Then
(i) For 0 < δ + (2/p) < 1 the map ∂z : Wm+1,pδ → Wm,p
δ+1 is an isomorphism of Banach
spaces.
(ii) For l + 1 < δ + (2/p) < l + 2, l ∈ Z≥0, and any regularity exponent m ≥ 0 the map
∂z : Wm+1,pδ → Wm,p
δ+1 is injective with closed image in Wm,pδ+1 Rl :=
{f ∈ Wm,p
δ+1
∣∣ (f, zk) =
0 for 0 ≤ k ≤ l}
. Rl has (complex) co-dimension l + 1.
27
Remark 9. Theorem 1.2.7 also holds with ∂z replaced by ∂z and Rl replaced by
Rl :={f ∈ Wm,p
δ+1
∣∣ (f, zk) = 0 for 0 ≤ k ≤ l}.
Proof Theorem 1.2.7. We will prove the theorem by induction in the regularity exponent
m ≥ 0. For m = 0 the statement follows from Proposition 1.2.5. Now, take m ≥ 1 and
assume that the statement holds for m replaced by m− 1 ≥ 0. Take f ∈ Wm,pδ+1 ⊆ Wm−1,p
δ+1 ⊆
Lpδ+1. By the induction hypothesis there exists u ∈ Wm,pδ such that ∂zu = f . Then for any
multi-index α ∈ Z2≥0 with |α| ≤ m we have that ∂αf ∈ Lpδ+|α|+1. Hence, ∂αu ∈ Lpδ+|α| and
∂z(∂αu
)= ∂α
(∂zu)
= ∂αz f ∈ Lpδ+|α|+1.
Then, we apply Lemma 1.2.4 to conclude that that ∂αu ∈ W 1,pδ+|α| for any |α| ≤ m. This
implies that u ∈ Wm+1,pδ . Hence, ∂z : Wm+1,p
δ → Wm,pδ+1 is onto. The injectivity of this map
follows from Lemma 1.2.3. The first statement of the theorem then follows from the open
mapping theorem. The same arguments together with Proposition 1.2.6 prove the second
statement of theorem.
Recall that for m,N ≥ 0 the weight γN of the remainder space Wm,pγN
of the asymptotic space
Zm,pγN(cf. (1.2)) are taken of the form γN = N + γ0 where 0 < γ0 + (2/p) < 1. In particular,
we see that
N < γN + (2/p) < N + 1. (1.25)
Then, by Theorem 1.2.7 above, the map ∂z : Wm+1,pγN
→ Wm,pγN+1 is injective with closed image
in Wm,pγN+1
∂z(Wm+1,pγN
)={f ∈ Wm,p
γN+1
∣∣ (f, zk) = 0 for 0 ≤ k ≤ N − 1}
where we assume that for N = 0 the set on the right hand side is equal to Wm,pγN+1, and hence,
the map is an isomorphism when N = 0. In fact, our method allows to construct a right
inverse of the map ∂z : Wm+1,pγN
→ Wm,pγN+1 that is defined on the whole of Wm,p
γN+1 and takes
values in the asymptotic space Zm+1,pN . More specifically, we have the following important
proposition.
28
Proposition 1.2.8. For any integers m,N ≥ 0 there exists a bounded map L : Wm,pγN+1 →
Zm+1,pN such that for any f ∈ Wm,p
γN+1 we have that ∂z(L(f)
)= f and
L(f) =χ
π
N∑k=1
(f, zk−1
) 1
zk+R(f)
where R : Wm,pγN+1 → Wm+1,p
γNis a bounded linear map and for N = 0 we omit the summation
term in the formula.
Proof of Proposition 1.2.8. The case when N = 0 is trivial since the map ∂z : Wm+1,pγN
→
Wm,pγN+1 is an isomorphism. For N ≥ 1 consider the bounded linear map P : Wm,p
γN+1 → Wm,pγN+1,
P(f) := f − 1
π
N∑k=1
(f, zk−1
) χzzk, (1.26)
where χz ≡ ∂zχ ∈ C∞c . This maps Wm,pγN+1
into RN ⊂ Wm,pγN+1
, the image of Wm+1,pγN
under ∂z.
To see this note that for any k ∈ Z we have by Stokes’ theorem that(χz, 1/z
k)
= − 1
2i
∫|z|≤R
d( χzkdz)
= − 1
2i
∮|z|=R
dz
zk= πδk1 (1.27)
where R > 0 is chosen sufficiently large. This, together with (1.26) then implies that(P(f), zk
)= 0 for any 0 ≤ k ≤ N − 1. Hence since P(f) = f for f ∈ RN := ∂z
(Wm+1,pγN
)by
Theorem 1.2.7, we get
P(Wm,pγN+1
)= ∂z
(Wm+1,pγN
).
Also by Theorem 1.2.7, the map ∂z : Wm+1,pγN
→ Wm,pγN+1 is injective with closed image we
obtain from the open mapping theorem that there is a bounded linear map ı : ∂z(Wm+1,pγN
)→
Wm+1,pγN
such that ∂z(ı(f)
)= f for any f ∈ ∂z
(Wm+1,pγN
). In particular, we have that
∂z((ı ◦ P
)(f))
= P(f) for any f ∈ Wm,pγN+1. This, together with (1.26) then gives that
∂z
((ı ◦ P
)(f) +
χ
π
N∑k=1
(f, zk−1
) 1
zk
)= f
for any f ∈ Wm,pγN+1. By setting R(f) :=
(ı ◦ P
)(f) and
L(f) :=χ
π
N∑k=1
(f, zk−1
) 1
zk+(ı ◦ P
)(f), f ∈ Wm,p
γN+1,
we conclude the proof of the proposition.
29
Recall that (see (1.6), (1.4))
Zm,pN ≡{u|u =
χ
z2
∑0≤k+l≤N−2
aklzkzl
+ f such that, f ∈ Wm,pγN
and akl ∈ C}
and
Zm,pn,N ≡{χ∑
n≤k+l≤N
aklzkzl
+ f∣∣∣ f ∈ Wm,p
γNand akl ∈ C
}(1.28)
where 0 ≤ n ≤ N + 1 and we set that Zm,pN+1,N ≡ Wm,pγN
and Zm,p1 ≡ Wm,pγ1
. We will next
study the properties of ∂z acting on scale of spaces Zm,pN .
Theorem 1.2.9. For any m > 1 + (2/p) and N ≥ 0 we have that ∂z : Zm+1,p1,N → Zm,pN+1 is
an isomorphism of Banach spaces.
Remark 10. Note that we have ∂z = τ ◦∂z ◦ τ where τ : u 7→ u is the operation of taking the
complex conjugate of a function. Using this we can obtain similar statement for the Cauchy
operator ∂z if we replace Zm,pN+1 by the space
(Zm,pN+1
)={ χz2
∑0≤k+l≤N−1
aklzkzl
+ f∣∣∣ f ∈ Wm,p
γN+1 and akl ∈ C}.
Furthermore we have that ∂−1z = τ ◦ ∂−1
z ◦ τ , and similarly ∂−1z = τ ◦ ∂−1
z ◦ τ . Also note that
∂−1z and ∂−1
z commute since the Cauchy operators ∂z and ∂z commute.
Proof of Theorem 1.2.9. First we show that ∂z maps Zm+1,p1,N into Zm,pN+1. For any k ≥ 1 and
l ≥ 0 we have
∂z
( χ
zkzl
)= −k χ
z2
1
zk−1zl+ g (1.29)
where g := χz/zkzl ∈ C∞c ⊆ Wm,p
γN+1. Similarly, ∂z(χ/zl
)= χz/z
l ∈ C∞c ⊆ Wm,pγN+1. Also since
∂z : Wm+1,pγN
→ Wm,pγN+1, we have ∂z
(Zm+1,p
1,N
)⊆ Zm,pN+1. Hence the map
∂z : Zm+1,p1,N → Zm,pN+1 (1.30)
is well defined. Further, note that Zm+1,p1,N ⊆ W 1,p
γ0where γ0 + (2/p) > 0. Hence, the map
(1.30) is injective by Lemma 1.2.3. Let us now prove that the map is onto. Take u ∈ Zm,pN+1,
30
then we can write u in general form as
u =χ
z2
∑0≤k+l≤N−1
aklzkzl
+ f, f ∈ Wm,pγN+1.
From (1.29) we obtain that
u = ∂z
( ∑0≤k+l≤N−1
(− aklk + 1
) χ
zk+1zl
)+(f − g
)= ∂z
( ∑0≤k+l≤N−1
(− aklk + 1
) χ
zk+1zl+ L
(f − g
))
where g ∈ C∞c ⊆ Wm,pγN+1 and L : Wm,p
γN+1 → Zm+1,p1,N is the map in Proposition 1.2.8. Note
that∑
0≤k+l≤N−1
(− akl
k+1
)χ
zk+1zl∈ Zm+1,p
1,N . This shows that (1.30) is a bijective map onto
its image. The statement of the theorem then follows from the open mapping theorem.
Proposition 1.2.10. For m > 2 + 2/p, the map
(ϕ, v) 7→ Rϕ ◦ ∂z ◦Rϕ−1(v), ZDm,pN ×Zm,pN → Zm−1,p
N+1 (1.31)
is real analytic.
Proof. Let v ∈ Zm,pN then we can write, with ψ = ϕ−1. First observe that Rϕ ◦∂z ◦Rϕ−1(v) ∈
Zm−1,p
N+1 follows from Theorem 1.2.9 and Lemma 1.1.6. By a direct computation we show that
this operator does not involve composition with ϕ.
∂z ◦Rϕ−1(v) = vz ◦ ψ · ψz + vz ◦ ψ · ψz
Rϕ ◦ ∂z ◦Rϕ−1(v) = vz · ψz ◦ ϕ+ vz · ψz ◦ ϕ
By differentiating with z and z the expression ψ ◦ ϕ = z we get
ψz ◦ ϕ · ϕz + ψz ◦ ϕ · ϕz = 1
ψz ◦ ϕ · ϕz + ψz ◦ ϕ · ϕz = 0.
31
Solving this system we get
ψz ◦ ϕ =ϕzD
ψz ◦ ϕ = − ϕzD
where D = ϕzϕz − ϕzϕz. This gives
Rϕ ◦ ∂z ◦Rϕ−1(v) =vzϕz − vzϕz
D.
We can write D = 1 + uzuz − uzuz + uz + uz := 1 + w, w ∈ Zm−1,pN . Since |D| > 0 and w
has no constant term in asymptotic expansion so by Lemma 1.1.5, ϕ 7→ 1D
, ZDm,pN → Zm,pN
is real analytic in ϕ. Since v 7→ vz is analytic so Rϕ ◦ ∂z ◦Rϕ−1(v) is analytic in (ϕ, v).
Proposition 1.2.11. For m > 3 + (2/p) the map
(ϕ, v) 7→(Rϕ ◦ ∂−1
z ◦Rϕ−1
)(v), ZDm,pN × Zm−1,p
N+1 → Zm,p1,N , (1.32)
is real analytic. Here ∂−1z denotes the inverse of the Cauchy operator given be Theorem 1.2.9.
Proof of Proposition 1.2.11. First, note that by Lemma 1.1.6 the map v 7→(Rϕ ◦ ∂z ◦
Rϕ−1
)(v), Zm,p1,N → Zm−1,p
N+1 , is a bounded linear map. By Theorem 1.2.9, this map is a
linear isomorphism with inverse
(Rϕ ◦ ∂z ◦Rϕ−1
)−1= Rϕ ◦ ∂−1
z ◦Rϕ−1 . (1.33)
In particular, Rϕ ◦ ∂z ◦ Rϕ−1 ∈ GL(Zm,p1,N , Z
m−1,pN+1
)where GL(X, Y ) denotes the Banach
manifold of linear isomorphisms between two Banach spaces X and Y . On the other side,
since the map (1.31) in Lemma 1.2.10 is real analytic, we obtain that its first partial derivative
G ≡ D2F with respect to the second argument
G : ZDm,pN → GL(Zm,p1,N , Z
m−1,pN+1
), ϕ 7→ Rϕ ◦ ∂z ◦Rϕ−1 , (1.34)
is also real analytic. Note in addition that the map
ı : GL(Zm,p1,N , Z
m−1,pN+1
)→ GL
(Zm−1,pN+1 ,Zm,p1,N
), L 7→ L−1,
32
is analytic. It then follows from (1.33) and (1.34) that the map
ı ◦G : ϕ 7→ Rϕ ◦ ∂−1z ◦Rϕ−1 , ZDm,pN → GL
(Zm−1,pN+1 ,Zm,p1,N
),
is real analytic as it is a composition of real analytic maps. This implies that the map (1.32)
is also real analytic.
In next section we will see that another operator Q : Zm,pN → Zm−1,p
N+1 defined by Q(a) =
(az)2 +azaz plays important role in studying the well-posedness of Euler Equation in asymp-
totic spaces Zm,pN .
Proposition 1.2.12. For m > 2+2/p, the map (ϕ, v) 7→ Rϕ◦Q◦Rϕ−1(v), ZDm,pN ×Zm,pN →
Zm−1,p
N+1 is real analytic.
Proof. First we show that Q maps into Zm−1,p
N+1 . However we have Zm−1,p
N+1 ⊂ Zm−1,pN+1 is an
ideal and Zm−1,pN+1 is closed under conjugation so az and az ∈ Z
m−1,p
N+1 hence Q(a) ∈ Zm−1,p
N+1 .
To show analyticity, as in previous proposition we see that first term of Rϕ ◦Q ◦Rϕ−1(v) is
Rϕ
(∂z(v ◦ ϕ−1)
)2= (Rϕ ◦ ∂z ◦Rϕ−1(v))2 =
(vzϕz − vzϕz)2
D2.
and this is real analytic by last proposition. Similarly the second term is
Rϕ
((∂z(v ◦ ϕ−1)
) (∂z(v ◦ ϕ−1)
))= (Rϕ∂zRϕ−1(v)) (Rϕ∂zRϕ−1(v))
which is again real analytic by last proposition.
1.3 The Euler equation on ZDm,pN
In this section we convert the Euler equation from Zm,pN to a dynamical system on the smooth
Banach manifold ZDm,pN × Zm,pN . In Appendix 2.9 it is shown that the Euler’s equation in
complex plane can be written using Cauchy operator as
u+ (uuz + uuz) = −2∂zp, div u = uz + uz = 0 (1.35)
u∣∣t=0
= u0
33
where p : R2 → R is the scalar pressure. For uniqueness of solutions of (1.35) we will require
that p ∈ Zm+1,p1,N . The operator ∂−1
z in equation (1.36) below is the operator defined in
Theorem 1.2.9. This operator is defined on specific spaces, namely Zm−1,pN+1 . Therefore to see
that in (1.36) below the right hand side is well defined note that for u ∈ Zm,pN we have that
uz, uz ∈ Zm−1,pN+1 and uz ∈ Zm−1,p
N+1 . Since Zm−1,pN+1 is an ideal in the Banach algebra Zm−1,p
N+1 we
conclude that (see Proposition 1.2.12)
Q(u) ≡ uzuz + uzuz ∈ Zm−1,pN+1 .
Proposition 1.3.1. Assume that m > 2 + 2/p and u ∈ C ([0, T ],Zm,pN )∩C1([0, T ],Zm−1,p
N
).
Then u satisfies (1.35) for some real-valued p(t) ∈ Zm+1,p1,N , t ∈ [0, T ], if and only if u satisfies
u+ (uuz + uuz) = ∂−1z
((uz)
2 + |uz|2)
(1.36)
with divergence free initial condition u0 = u∣∣t=0
, div u0 = 0.
Proof. Assume u satisfies equation (1.35) for some real-valued p ∈ Zm+1,p1,N . Apply divergence
to equation (1.35) and use div u = 0 and (since m > 2 + 2/p we can use div u = (div u))
div (−2∂zp) = −2(∂z∂zp + ∂z∂zp) = −4∂z∂zp.
Combine this with expression
div (uuz + uuz) = 2(uz)2 + 2|uz|2 = 2Q(u) (1.37)
(see equation (47)) we get equation
−2∂z(pz) = (uz)2 + |uz|2.
Since p ∈ Zm+1,p1,N we have that pz ∈ Zm,p2,N+1 ⊂ Z
m,p1,N and therefore by Theorem 1.2.9 we get
that −2pz = ∂−1z
((uz)
2 + |uz|2)
showing that u satisfies equation (1.36).
For the converse statement assume that u ∈ C([0, T ],Zm,pN
)∩ C1
([0, T ],Zm−1,p
N
)satisfies
34
equation (1.36) with divergence free initial data div u0 = 0. We apply divergence to equation
(1.36) and after simplification we get
(div u) + u ∂z(div u) + u ∂z(div u) = 0 (1.38)
Let ϕ be the solution of equation (1.9), which exists by Lemma 1.1.13 since m > 2 + 2/p.
Since ϕ ∈ C1 ([0, T ],ZDm,pN ) and div u ∈ C([0, T ],Zm−1,p
N
)∩ C1
([0, T ],Zm−2,p
N
)we conclude
that div u ∈ C1([0, T ]× C,R
)and ϕ ∈ C1([0, T ]× C,C). This together with (1.38) implies
that for any given (x, y) ∈ R2 and for any t ∈ [0, T ],
((div u) ◦ ϕ
)= 0, (1.39)
Hence for any t ∈ [0, T ], div u(t) = div u0 = 0 for any t ∈ [0, T ]. By Lemma 1.3.2 below
there exists p ∈ C([0, T ],Zm+1,p
1,N
)such that −2pz = ∂−1
z
((uz)
2 + |uz|2)
for any t ∈ [0, T ].
This completes the proof of the proposition.
Remark 11. Since u is divergence free we have that div u = uz + uz = 0 and by formula
(48) in Appendix 2.9,
ω := i curl~v ≡ 1
2
(uz − uz
)= uz. (1.40)
By applying the Cauchy operator ∂z to (1.36) we obtain that.ω = −
(uuz + uuz
)z
+((uz)
2 +
uzuz)
= −(uωz + u ωz
)that is the 2d Euler equation in vorticity form,
.ω + uωz + u ωz = 0. (1.41)
This and the arguments used to prove the analogous formula (1.39) then implies that ∂t(ω(t)◦
ϕ(t))
= 0 for any t ∈ [0, T ]. In particular, for any t ∈ [0, T ] we have that
ω(t) = ω0 ◦ ψ(t) (1.42)
where ψ(t) := ϕ(t)−1, ϕ ∈ C1([0, T ],ZDm,pN
), and ω0 = ∂zu0 ∈ Zm−1,p
N+1 . We will use this
preservation of vorticity in next section to prove global existence of solutions.
35
In the proof of Proposition 1.3.1 we used the following lemma.
Lemma 1.3.2. Assume that m > 2 + (2/p). For any divergence free u ∈ Zm,pN there
exists a real valued p ∈ Zm+1,p1,N such that −2pzz = (uz)
2 + |uz|2. Moreover, we have that
−2pz = ∂−1z
((uz)
2 + |uz|2)
and −2p = ∂−1z ∂−1
z
((uz)
2 + |uz|2).
Proof of Lemma 1.3.2. Take u ∈ Zm,pN such that div u = uz + uz = 0. Then we have
Q(u) ≡ (uz)2 + uzuz = −uz(uz) + uz(uz). (1.43)
We will first consider the case when N ≥ 1. Since uz, uz ∈ Zm−1,pN+1 we obtain from Proposition
1.1.1, Proposition 1.1.2 and (1.43) that Q(u) ∈ Zm−1,p4,N+3 and
Q(u) =χ
z2z2
∑0≤k+l≤N−1
aklzkzl
+ f, f ∈ Wm−1,pγN+3 . (1.44)
In particular, we see that Q(u) ∈ Wm−1,pγ3
. Then, by Proposition 1.2.8, ∂−1z Q(u) ∈ Zm,p1,2 and
∂−1z Q(u) =
1
π
[(Q(u), 1
) χz
+(Q(u), z
) χz2
]+ g, g ∈ Wm,p
γ2.
Since(Q(u), 1
)=(Q(u), z
)= 0 by Lemma 1.3.3 below we conclude that
∂−1z Q(u) ∈ Wm,p
γ2. (1.45)
Note that when considering ∂−1Q(u), we have to view Q(u) ∈ Wm−1,pγ3
instead of being in
Zm,p1,N+2, since in computing the momenta(Q(u), 1
)and
(Q(u), z
)the function Q(u) has to
be in the remainder spaces Wm,pγN
(see Theorem 1.2.9). On the other side, by Theorem 1.2.9
and the fact that Q(u) ∈ Zm−1,p4,N+3 we have that ∂−1
z Q(u) ∈ Zm,p1,N+2. Moreover, it follows from
(1.44) and Proposition 1.2.8 (cf. the proof of Theorem 1.2.9) that
∂−1z Q(u) = χ
b1
z+ χ
b2
z2+
(χ
zz2
∑0≤k+l≤N−1
bklzkzl
+χ
z3
∑3≤k≤N+2
bkzk−3
)+ h, h ∈ Wm,p
γN+2
where bk := 1π
(f, zk−1
). By comparing this with (1.45) we obtain that b1 = b2 = 0, and
hence
∂−1z Q(u) =
(χ
zz2
∑0≤k+l≤N−1
bklzkzl
+χ
z3
∑3≤k≤N+2
bkzk−3
)+ h, h ∈ Wm,p
γN+2. (1.46)
36
In particular, this implies that
∂−1z Q(u) ∈
(Zm,pN+2
).
Hence, by Theorem 1.2.9 and Remark 10, we obtain that
p := −1
2∂−1z ∂−1
z Q(u) ∈ Zm+1,p1,N+1 ⊆ Z
m+1,p1,N .
Let us now consider the case when N = 0. Then, u ∈ Zm,p0 and u = a00 + f , f ∈ Wm,pγ0
. This
implies that uz, uz ∈ Wm−1,pγ1
and hence by Proposition 1.1.1 and (1.43) we obtain that
Q(u) ∈ Wm−1,p2γ1+(2/p) ⊆ Wm−1,p
γ2
where we used that 1 < γ1 + (2/p) < 2. This, together with Proposition 1.2.8 and Lemma
1.3.3 below then implies that
∂−1z Q(u) ∈ Wm,p
γ1.
Finally, by applying Theorem 1.2.7 (i) we obtain that
p := −1
2∂−1z ∂−1
z Q(u) ∈ Wm+1,pγ0
≡ Zm+1,p1,0 .
Let us now prove that p is real valued. Recall that τ : u 7→ u is the operation of taking the
complex conjugate of a function. By Remark 10 we then have
−2p = τ ◦(∂−1z ∂−1
z Q(u))
=(τ ◦ ∂−1
z ◦ τ)◦(τ ◦ ∂−1
z τ)◦ τQ(u)
= ∂−1z ∂−1
z Q(u) = −2p
where we used that Q(u) is real valued by (1.43) and that ∂−1z = τ ◦ ∂−1
z ◦ τ and similarly
∂−1z = τ ◦ ∂−1
z ◦ τ . This completes the proof of the proposition.
Lemma 1.3.3. Assume that m > 2 + (2/p) and N ≥ 1. If u ∈ Zm,pN is divergence free then(Q(u), 1
)= 0 and
(Q(u), z
)= 0.
Proof of Lemma 1.3.3. Since Q(u) does not involve the constant asymptotic term of u ∈
Zm,pN we will assume without loss of generality that u ∈ Zm,p1,N . Then, the lemma follows
37
easily from the Stokes’ theorem and the relation (1.37). In fact, with f := uuz + uuz we
obtain from (1.37) that
2(Q(u), 1
)= − 1
2i
∫C
div f dz ∧ dz = − Im
(∫Cfz dz ∧ dz
)= − Im
(limR→∞
∮|z|=R
f dz
)= 0
where we used that f = O(1/|z|3
)and Q(u) = O
(1/|z|4
)as |z| → ∞. Similarly,
2(Q(u), z
)= − 1
2i
∫Cz div f dz ∧ dz = − 1
2i
∫Cz(fz + fz) dz ∧ dz =
1
2i
∫Cf dz ∧ dz (1.47)
where we used as above the Stokes’ theorem to conclude that∫Czfz dz ∧ dz =
∫C
(zf)zdz ∧ dz = 0
and ∫Czfz dz ∧ dz =
∫C
((zf)z − f
)dz ∧ dz = −
∫Cf dz ∧ dz.
On the other side, since div u = uz + uu = 0, we have
f = uuz + uuz =1
2(u2)z + (uu)z − uuz =
1
2(u2)z + (uu)z + uuz = (u2)z + (uu)z.
Since u2, uu = O(1/|z|2
)as |z| → ∞ we then conclude again from Stokes’ theorem that∫
C f dz ∧ dz = 0. This together with (1.47) then implies that(Q(u), z
)= 0.
Proposition 1.3.4. Let m > 2 + 2/p and u0 ∈ Zm,pN . There is a bijection between solutions
of the dynamical system (.ϕ,
.v) =
(v,(Rϕ ◦ ∂−1
z ◦Q ◦Rϕ−1
)(v))
(ϕ, v)|t=0 = (id, u0)(1.48)
on ZDm,pN ×Zm,pN and solutions of (1.36). More specifically, (ϕ, v) ∈ C1([0, T ],ZDm,pN ×Z
m,pN )
is a solution of (1.48) if and only if u = v ◦ ϕ−1 ∈ C ([0, T ],Zm,pN ) ∩ C1([0, T ],Zm−1,p
N
)is a
solution of (1.36).
Proof. Assume that (ϕ, v) is a solution of the dynamical system (1.48). By Theorem 1.1.12,
ϕ 7→ ϕ−1,ZDm,pN → ZDm,pN is continuous and since v(t) and ϕ(t) is continuous so u =
38
v ◦ ϕ−1 ∈ C([0, T ],Zm,pN ). Similarly by Theorem 1.1.12, since ϕ 7→ ϕ−1,ZDm,pN → ZDm−1,pN
is C1-map, and v(t) and ϕ(t) are C1-maps so by Theorem 1.1.12, u ∈ C1([0, T ],Zm−1,pN ). A
direct computation shows that with v = u ◦ ϕ
v = (u ◦ ϕ) = u ◦ ϕ+ uz ◦ ϕϕ+ uz ◦ ϕ ˙ϕ = (u+ uzu+ uzu) ◦ ϕ. (1.49)
Now from (1.49)
u+ uzu+ uzu = Rϕ−1(v) = Rϕ−1
(Rϕ ◦ ∂−1
z ◦Q ◦Rϕ−1(v))
= ∂−1z Q(u).
Conversely if u is a solution of (1.36) then by Proposition 1.1.13, we have a solution ϕ of
equation (1.9). By Sobolev embedding Zm−1,pN ⊂ C1 so u ∈ C1
([0, T ] × C,C
)and ϕ ∈
C1(C,C). Then pointwise we can write, with v := u ◦ ϕ,
v = (u+ uzu+ uzu) ◦ ϕ = Rϕ(∂−1z Q(u)) = Rϕ ◦ ∂−1
z ◦Q ◦Rϕ−1(v) = E2(ϕ, v).
By pointwise integration gives
v(t, (z, z)) = u0(z, z) +
∫ t
0
E2(ϕ, v)∣∣(s,(z,z))
ds .
Now as E2 is real analytic map, s 7→ E2(ϕ(s), v(s)), [0, T ] → Zm,pN+1 is continuous, so this
integral converges in Zm,pN . This shows that v ∈ C1([0, T ],Zm,pN
)satisfies
.v = E2(ϕ, v) where
ϕ ∈ C1([0, T ],ZDm,pN
)satisfies
.ϕ = u ◦ ϕ = v, ϕ|t=0 = id. This shows that the curve
(ϕ, v) ∈ C1([0, T ],ZDm,pN ×Zm,pN
)satisfies (1.48).
For any ρ > 0 denote by BZm,pN(ρ) the ball of radius ρ in Zm,pN centered at zero.
Theorem 1.3.5. Assume m > 3 + 2/p and 1 < p < ∞ and N ≥ 0. Then for any ρ > 0,
there exists T > 0 such that for any u0 ∈ BZm,pN(ρ) the Euler equation (1.35) has a unique
solution u ∈ C([0, T ],Zm,pN ) ∩ C1([0, T ],Zm−1,pN ) such that the pressure p(t) ∈ Zm+1,p
1,N for
t ∈ [0, T ]. This solution depends continuously on the initial data u0 ∈ BZm,pN(ρ) in the sense
that the data-to-solution map
u0 7→ u, BZm,pN(ρ)→ C([0, T ],Zm,pN ) ∩ C1([0, T ],Zm−1,p
N )
is continuous.
39
Proof. By Proposition 1.3.1 and Proposition 1.3.4 such solutions of the Euler equation are
given by solutions of the dynamical system (1.48). By the existence and uniqueness theorem
of solutions of an ODE in a Banach space (cf. [9]) there exist ρ0 > 0 and T0 > 0 such that
for any u0 ∈ BZm,pN(ρ0) there exists a unique solution (ϕ, v) ∈ C1([0, T ],ZDm,pN × Zm,pN ) and
therefore by Proposition 1.3.4 u = v ◦ ϕ−1 ∈ C([0, T ],Zm,pN ) ∩ C1([0, T ],Zm−1,pN ). Now by
scaling by a constant c > 0, uc = cu(ct), one can check that uc is a solution of (1.35) with
initial condition uc∣∣t=0
= cu0. Now for any ρ > 0, we can take T = ρ0ρT0.
1.4 Global solutions of the 2d Euler equation on ZDm,pN
In this section we will show that the solutions of the Euler equation given by Theorem 1.3.5
exist for all time t ≥ 0 within asymptotic space Zm,pN . For this we will use the known global
existence of solutions in the little Holder spaces cm,γb and the corresponding diffeomorphism
groups discussed below. The little Holder space cm,γb is the closure of C∞b in the Holder
spaces Cm,γb ≡ Cm,γ
b (R2,R2), 0 ≤ γ ≤ 1. Recall that the Holder space Cm,γb (R2,R) consists
of functions in f ∈ Cmb (R2,R) with finite Holder semi-norms [Dαf ]γ for any multi-index
α ∈ Z2≥0 such that |α| = m. The Holder semi-norm is defined as
[f ]γ := supx 6=y
∣∣f(x)− f(y)∣∣
|x− y|γ
and the norm in Cm,γb (R2,R) is |f |m,γ := |f |m + max|α|=m[Dαf ]γ where |f |m is the norm in
the space Cmb (R2,R) of Cm-smooth functions on R2 with bounded derivatives of order ≤ m,
that is |f |m =∑|α|≤m sup |Dαf |. For γ = 0 we have that Cm
b ≡ Cm,0b = cm,0b . Consider the
group of diffeomorphisms of R2 of class cm,γb defined as,
diffm,γ(R2) :={ϕ = id +u
∣∣u ∈ cm,γb and ∃ ε > 0 such that det(I + du) > ε}.
For m ≥ 1 and 0 < γ < 1, by Theorem 3.1 in [22] (cf. [15]) diffm,γ(R2) is a topological group
with regularity properties of the composition and the inversion of diffeomorphisms similar
40
to the ones in Theorem 1.1.12. We will show that if there is a global solution of the Euler
equation in cm,γb with initial velocity u0 ∈ Zm,pN |d−2 then in fact the solution stays in the space
Zm,pN |d−2 for all time t ≥ 0. We begin with the following.
Lemma 1.4.1. Assume that m > 1 + 2/p. Then for any u ∈ Zm,pN+2 and for any ϕ ∈ ZDm,pN
we have that u◦ϕ ∈ Zm,pN+2. Also this composition map (u, ϕ) 7→ u◦ϕ, Zm,pN+2×ZDm,pN → Zm,pN+2,
is continuous. If, in addition, u ∈ Zm,pN+2 then u ◦ ϕ ∈ Zm,pN+2.
The first and last statement of the lemma follow from Remark 3 (see formula (1.8)) while
continuity of the map follows from Proposition 1.1.2 (iii), and Lemma 6.5 in [13]. As a
consequence of Lemma 1.4.1 we obtain the following proposition.
Proposition 1.4.2. Assume that m > 3+(2/p). Let u ∈ C([0, T ],Zm,pN
)∩C1
([0, T ],Zm−1,p
N
)for T > 0 be a solution of the 2d Euler equation (1.35) with u|t=0 ∈ Zm,pN+1. Then, u ∈
C([0, T ],Zm,pN+1
)∩ C1
([0, T ],Zm−1,p
N+1
).
Proof of Proposition 1.4.2. Assume that u ∈ C([0, T ],Zm,pN
)∩C1
([0, T ],Zm−1,p
N
)is a solution
of the Euler equation (1.35) with initial velocity u0 := u|t=0 ∈ Zm,pN+1. Then, for any t ∈ [0, T ]
we have by the preservation of vorticity (see (1.41)) that ω(t) = ∂zu(t) is such that
ω(t) = ω0 ◦ ψ(t), ω0 ∈ Zm−1,pN+2 , (1.50)
where ψ(t) := ϕ(t)−1 and ϕ ∈ C1([0, T ],ZDm,pN
))is the Lagrangian coordinate of the solution
u. By Theorem 1.1.12, ψ ∈ C([0, T ],ZDm,pN
)∩ C1
([0, T ],ZDm−1,p
N
). This, together with
(1.50) and Lemma 1.4.1 above, implies that for any t ∈ [0, T ] we have that
∂zu(t) = ω(t) = ω0 ◦ ψ(t) ∈ Zm−1,pN+2 and ∂zu ∈ C
([0, T ], Zm−1,p
N+2
).
Hence, we conclude from Theorem 1.2.9 that u ∈ C([0, T ],Zm,pN+1
). This implies that the local
solution of the 2d Euler equation in Zm,pN+1 with initial data u0 ∈ Zm,pN+1 given by Theorem
1.3.5 has a finite Zm,pN+1-norm on the intersection of its maximal interval of existence with
41
[0, T ]. Thus, the maximal interval of existence of this solution contains [0, T ] otherwise by
local existence of solutions we can extend beyond the maximal interval of existence hence
u ∈ C([0, T ],Zm,pN+1
)∩ C1
([0, T ],Zm−1,p
N+1
)completing the proof of the proposition.
Let us recall that for m ≥ 0, 1 < p <∞, and any weight δ ∈ R,
Hm,pδ (R2,R) :=
{f ∈ Lploc(R
2,R)∣∣ 〈x〉δDαf ∈ Lp for |α| ≤ m
}, Dα ≡ ∂α1
x1∂α2x2.
These spaces appear in Proposition 1.4.4 below. In the proposition we start with initial
vorticity ω0 ∈ Z4,p1 and show that since the vorticity ω is preserved, that is, ω = ω0 ◦ ϕ−1
where ϕ ∈ diff3,γ(R2) therefore ω ∈ Z4,p1 as well. To this end we differentiate ω four times
and show that every time we differentiate the decay also increases. But since ϕ ∈ diff3,γ(R2),
ϕ − id does not necessarily decay at infinity. So apriori we only get that ω ∈ H4,pγ0+1. Note
that for m > 2/p and and real weights δ the space Hm,pδ (R2,R) is a Banach algebra with
respect to pointwise multiplication of functions (see Proposition 2.1 in [13]). We have the
following lemma.
Lemma 1.4.3. Assume that 1 ≤ p ≤ ∞. Then for 0 ≤ γ < 1 and any weight δ ∈ R we
have that the composition map
(u, ϕ) 7→ u ◦ ϕ, Lpδ × diff1,γ(R2)→ Lpδ ,
is continuous. More generally, for any integers m ≥ 1 and 0 ≤ k ≤ m we have that the
composition map (u, ϕ) 7→ u ◦ ϕ, Hk,pδ × diffm,γ(R2)→ Hk,p
δ , is continuous.
Proof of Lemma 1.4.3. It is enough to prove the lemma only for the case when γ = 0. Then
the general statement follows since diffm,γ(R2) is continuously embedded in diffm(R2) ≡
diffm,0(R2). Also the second statement of the lemma follows from the first one and the fact
that ϕ has bounded derivatives. So we will prove the first statement. Assume that 0 ≤ γ < 1,
1 ≤ p < ∞ and δ ∈ R and take u1, u2 ∈ Lpδ ≡ Lpδ(R2,R) and ϕ ∈ diff1(R2). By change of
variables y = ϕ(x) in the corresponding integral below we obtain that
‖u2 ◦ ϕ− u1 ◦ ϕ‖Lpδ = ‖(u2 − u1) ◦ ϕ‖Lpδ ≤ C ‖u2 − u1‖Lpδ (1.51)
42
where C ≡ C(ϕ) > 0 depends continuously on |ϕ−1|1. Since ϕ ∈ diff1(R2) is a topological
group (see Theorem 3.1 in [22]) so the map ϕ 7→ ϕ−1 is continuous and therefore the constant
C > 0 in (1.51) can be chosen locally uniformly in ϕ ∈ diff1(R2), i.e. the same constant C
can be chosen for any ψ in a small neighborhood U(ϕ) of ϕ such that (1.51) holds for ψ in
place of ϕ. Now for any u ∈ C∞c and ϕ1, ϕ2 ∈ diff1(R2) we can obtain from the mean-value
theorem that for any x ∈ R2,
(u ◦ ϕ2)(x)− (u ◦ ϕ1)(x) =(∫ 1
0
(∇u)|ζt(x) dt)·(ϕ2(x)− ϕ1(x)
)where ζt(x) := ϕ1(x) + t(ϕ2(x) − ϕ1(x)). By applying first the Cauchy-Schwartz inequality
and then Jensen’s inequality we obtain∣∣(u ◦ ϕ2)(x)− (u ◦ ϕ1)(x)∣∣p ≤ C
∣∣ϕ2 − ϕ1
∣∣p1
∫ 1
0
∣∣∣∇u(ζt(x))∣∣∣p dt
where the constant C > 0 depends only on the choice of 1 ≤ p < ∞. By multiplying this
inequality by 〈x〉pδ and then by integrating it over x ∈ R2 we obtain that∥∥u ◦ ϕ2 − u ◦ ϕ1
∥∥pLpδ≤ C
∣∣ϕ2 − ϕ1
∣∣p1
∫ 1
0
∥∥|∇u| ◦ ζt∥∥pLpδ dt (1.52)
with a constant C > 0 that depends only on the choice of 1 ≤ p <∞. It follows from (1.51)
that for ϕ2 chosen in an open neighborhood U(ϕ1) of ϕ1 in diff1(R2) we have that∥∥∇u ◦ ζt∥∥Lpδ ≤ C ‖∇u‖Lpδ .
This gives us that for any u ∈ H1,pδ and for any ϕ1 ∈ diff1(R2) and any ϕ2 in a small
neighborhood U(ϕ1) we have that∥∥u ◦ ϕ2 − u ◦ ϕ1
∥∥Lpδ≤ C(p)
∣∣ϕ2 − ϕ1
∣∣1
∥∥∇u∥∥Lpδ≤ C(p) ‖u‖H1,p
δ
∣∣ϕ2 − ϕ1
∣∣1. (1.53)
Finally, it follows from (1.51) and (1.53) that there exist an open neighborhood U(ϕ1) of
ϕ1 ∈ diff1(R2) and constants C1 > 0 and C2 > 0 such that for any ϕ2 ∈ U(ϕ1) and for any
u1, u2 ∈ Lpδ and u ∈ C∞c we have∥∥u2 ◦ ϕ2 − u1 ◦ ϕ1
∥∥Lpδ≤
∥∥u2 ◦ ϕ2 − u ◦ ϕ2
∥∥Lpδ
+∥∥u ◦ ϕ2 − u ◦ ϕ1
∥∥Lpδ
+∥∥u ◦ ϕ1 − u1 ◦ ϕ1
∥∥Lpδ
≤ C1‖u2 − u‖Lpδ + C2 ‖u‖H1,pδ
∣∣ϕ2 − ϕ1
∣∣1
+ C1‖u− u1‖Lpδ .
43
Now to show continuity at point (u1, ϕ1) ∈ H1,pδ × diff1(R2) we take ε > 0. For any u2 ∈ Lpδ
and u ∈ C∞c , u 6≡ 0, inside the open ball in Lpδ of radius ρ > 0 centered at u1 ∈ Lpδ the
inequality above then implies that
∥∥u2 ◦ ϕ2 − u1 ◦ ϕ1
∥∥Lpδ≤ 3C1ρ+ C2‖u‖H1,p
δ
∣∣ϕ2 − ϕ1
∣∣1.
Finally, by choosing 0 < ρ < ε/(6C1) and ϕ2 ∈ U(ϕ1) such that |ϕ2−ϕ1
∣∣1< ε/
(2C2 ‖u‖H1,p
δ
)we conclude that
∥∥u2 ◦ ϕ2 − u1 ◦ ϕ1
∥∥Lpδ
< ε. Finally the case p = ∞ follows similarly
completing the proof of the lemma.
Remark 12. By Proposition 1.5.1 (iii) in section 1.5 for m > 2/p, δ + (2/p) > 0, and
0 < γ < 1 chosen sufficiently small, we have the continuous embedding Wm,pδ ⊆ Ck,γ
b for any
integer 0 ≤ k < m− 2/p. Since the elements of Wm,pδ are approximated by functions in C∞c ,
we then conclude that Wm,pδ is continuously embedded in the little Holder space cm,γb . The
same arguments applied to the asymptotic coefficients and the definition of the norm in the
asymptotic space Zm,pN show that Zm,pN is continuously embedded in ck,γb . Since 1 < p < ∞
we conclude that 0 < 2/p < 2, and hence for m ≥ 3 we have the continuous embedding
Zm,pN ⊆ cm−2,γb . In what follows we will assume that 0 < γ < 1 is chosen so that these
embeddings hold.
In order to show global persistence of solutions to the Euler equation we will use the fact
that global solutions exist in spaces cm,γb . Since, for m ≥ 1, 0 < γ < 1, diffm,γ(R2) is a
topological group modeled on Banach space cm,γb we will say that ϕ ∈ C1 ([0, T ], diffm,γ(R2))
is a solution of the 2d Euler equation if ϕ ∈ C2 ([0, T ], diffm,γ(R2)) and it satisfies following
second order ODE (.ϕ,
.v) =
(v,(Rϕ ◦ ∂−1
z ◦Q ◦Rϕ−1
)(v))
(ϕ, v)|t=0 = (id, u0)(1.54)
Here the Cauchy operator is defined as acting on the tempered distributions S ′ and ∂−1z is the
extension of the operator in Proposition 1.2.5 to the space S ′. Furthermore since diffm,γ(R2)
44
is a topological group (cf. [22, Theorem 3.1]) therefore we have that
u := ϕ ◦ ϕ−1 ∈ C ([0, T ], cm,γb ) ∩ C1([0, T ], cm−1,γ
b
)(1.55)
Now as discussed in Proposition 1.3.4 if ϕ ∈ C1 ([0, T ], diffm,γ(R2)) is a solution of Euler
equation in Lagrangian coordinates then u := ϕ ◦ϕ−1 satisfies the 2d Euler equation (1.36).
Also it follows by similar arguments that u and ω = ∂zu satisfies (1.41) and therefore we
have that
ω(t) = ω0 ◦ ϕ−1(t). (1.56)
In the following proposition we show that if the initial velocity u0 is in asymptotic space
Z5,p0 then it evolves in the same space. We use that the vorticity ω(t) is preserved (see above
equation) and show in particular that it follows from (1.56) that ω ∈ Z4,p1 .
Proposition 1.4.4. Assume that for u0 ∈ Z5,p0 ⊆ c3,γ
b there exists a unique solution of the
2d Euler equation in Lagrangian coordinates ϕ ∈ C1([0, T ], diff3,γ(R2)
)for some T > 0 with
˙ϕ|t=0 = u0. Then the solution of the 2d Euler equation with u|t=0 = u0 given by Theorem
1.3.5 extends to the whole interval [0, T ] and u ∈ C([0, T ],Z5,p
0
)∩ C1
([0, T ],Z4,p
0
).
Proof of Proposition 1.4.4. Let us take u0 ∈ Z5,p0 so that we can write u0 = c + f0 where
c is a constant and f0 ∈ W 5,pγ0
. Assume that for T > 0, ϕ ∈ C1([0, T ], diff3,γ(R2)
)is the
solution of the 2d Euler equation in Lagrangian coordinates (see (1.54)). Assume that the
local solution given by Theorem 1.3.5 does not extend to all of the interval [0, T ] then
there exists 0 < τ < T such that [0, τ) is the maximal interval of existence of the solution
u ∈ C([0, τ),Z5,p
0
)∩ C1
([0, τ),Z4,p
0
)given by Theorem 1.3.5. Then, it follows easily from
Theorem 1.3.5 that
‖u(t)‖Z5,p0→∞ as t→ τ − 0, (1.57)
since otherwise we can again apply Theorem 1.3.5 to extend the solution beyond t = τ . We
will show that (1.57) cannot happen. We start by considering the curve
ω(t) := ω0 ◦ ψ(t), t ∈ [0, T ], (1.58)
45
where ψ(t) := ϕ(t)−1 and ω0 := ∂zu0 ∈ W 4,pγ0+1 ⊆ H4,p
γ0+1. Also as diff3,γ(R2) is a topological
group we get that ψ ∈ C ([0, T ], diff3,γ(R2)). Since in particular ω0 ∈ H3,pγ0+1, it follows from
Lemma 1.4.3 that
ω ∈ C([0, T ], H3,p
γ0+1
). (1.59)
Note that for t ∈ [0, τ) we have by the conservation of the vorticity ω ≡ ∂zu on [0, τ) (see
(1.42) in Remark 11) that
ω(t) = ω(t) and u(t) = c+ ∂−1z ω(t), t ∈ [0, τ). (1.60)
Let us denote r := ∂−1z ω(t), t ∈ [0, T ]. Note that by (1.59) we have that
∂αω ∈ C([0, T ], Lpγ0+1
), |α| ≤ 3. (1.61)
Since γ0 satisfies 0 < γ0 + (2/p) < 1 we obtain from Theorem 1.2.7 (i) and Remark 10 that
for any |α| ≤ 3 we have that ∂αr ∈ C([0, T ],W 1,p
γ0
), |α| ≤ 3 (See Remark 8). Therefore if we
define the curve
u := c+ r (1.62)
then (1.61) implies that
u− c ∈ C([0, T ],W 1,p
γ0
)and du ∈ C
([0, T ], H3,p
γ0+1
). (1.63)
Note that by (1.55) it follows that ϕ ∈ C1([0, T ], diff3,γ(R2)
)satisfies the equation
.ϕ = u ◦ ϕ, ϕ|t=0 = id .
This implies that Y := dϕ = I + df , where ϕ = id +f and f ∈ c3,γb , satisfies the equation
.Y = A(t)Y , Y |t=0 = I, where A := du ◦ ϕ ∈ C
([0, T ], H3,p
γ0+1
)by Lemma 1.4.3. We can
rearrange this equation in terms of df as
(df). = A(t)(df) + A(t), df |t=0 = 0, (1.64)
and by the Banach algebra property in H3,pγ0+1 we can consider it as a linear ODE on df in the
space of 2× 2 matrices with elements in H3,pγ0+1. Therefore by the existence and uniqueness
46
of solutions of ODEs in Banach spaces, we obtain that dϕ− I ∈ C([0, T ], H3,p
γ0+1
). Now since
ψ = ϕ−1 so we can write
dψ =[
Adj(dϕ)/ det(dϕ)]◦ ψ, (1.65)
where Adj(dϕ) is the transpose of the cofactor matrix of dϕ, we conclude from the Banach
algebra property in H3,pγ0+1, the arguments used to prove Lemma 1.1.5, and Lemma 1.4.3,
that
dψ − I ∈ C([0, T ], H3,p
γ0+1
). (1.66)
Remark 13. In particular, this implies that ∂z(ψ−z
)∈ H3,p
γ0+1 ⊆ Lpγ0+1. Hence, by Theorem
1.2.7 (i), we obtain that ψ−z−c0 ∈ W 1,pγ0
for some constant c0 ∈ C, and therefore ψ−z−c0 ∈
H4,pγ0
. Similarly for ϕ and ψ we get that
ϕ, ψ ∈ C([0, T ],AD4,p
0
)(1.67)
where ADm,pN with N ≥ 0 denotes the group of orientation preserving C1-smooth diffeomor-
phisms associated to the weighted Sobolev space H4,pγN
and defined in a similar fashion as
ZDm,pN in (1.5) (see Definition 5.1 in [13, Section 5] for more detail). By Corollary 6.1 in
[13], the composition map
(f, ν)→ f ◦ ν, H4,pγ0× AD4,p
0 → H4,pγ0,
is continuous.
This remark allows us to conclude that ω ∈ C([0, T ], H4,p
γ0
). A simple approximation argu-
ment involving Remark 13 also shows that we can differentiate the equality (1.58) four times
47
to obtain the following expressions for the differentials of ω of higher order,
dω =[(dω0) ◦ ψ
](dψ),
d2ω =[(d2ω0) ◦ ψ
](dψ, dψ) +
[(dω0) ◦ ψ
](d2ψ),
d3ω =[(d3ω0) ◦ ψ
](dψ, dψ, dψ) + 3
[(d2ω0) ◦ ψ
](d2ψ, dψ) +
[(dω0) ◦ ψ
](d3ψ), (1.68)
d4ω =[(d4ω0) ◦ ψ
](dψ, dψ, dψ, dψ) + 6
[(d3ω0) ◦ ψ
](d2ψ, dψ, dψ) + 4
[(d2ω0) ◦ ψ
](d3ψ, dψ)
+ 3[(d2ω0) ◦ ψ
](d2ψ, d2ψ) +
[(dω0) ◦ ψ
](d4ψ).
These formulas together with the fact that ω0 ∈ W 4,pγ0+1, ψ ∈ C
([0, T ], c3,γ
b
), as well as formula
(1.66), Lemma 1.4.3, and Proposition 1.5.1 (ii) imply that
ω ∈ C([0, T ],W 2,p
γ0+1
)and d3ω, d4ω ∈ C
([0, T ], Lpγ0+3
). (1.69)
To see this note that first dkψ is a bounded function for 1 ≤ k ≤ 2. Furthermore dkω0 ◦ ψ ∈
Lpγ0+1+k for 0 ≤ k ≤ 4. The only non-trivial part is to show that[(dω0) ◦ ψ
](dkψ) ∈ Lpγ0+3,
k ≥ 2. This will follow from (1.66) and the observation that |(dω0)◦ψ(x)∣∣ = O(1/〈x〉γ0+2+2/p)
and using 0 < γ0 + 2/p < 1. Now by Lemma 2.1 (d) in [13] and (1.67) we have that |ϕ− id |
is bounded and therefore for some constant Cϕ we have that 〈ϕ(x)〉 ≤ Cϕ〈x〉. Now as
dω0 ∈ W 3,pγ0+2 then by Proposition 1.5.1 (ii), with δ = γ0 + 2 and y = ψ(x) we have that
supx∈R2
(〈x〉γ0+2+2/p|dω0 ◦ ψ(x)|
)= sup
y∈R2
〈ϕ(y)〉γ0+2+2/p|dω0(y)| (1.70)
≤ Cγ0+2+2/pϕ sup
y∈R2
〈 y〉γ0+2+2/p|dω0(y)| ≤ C ‖dω0‖W 3,pγ0+2
By combining (1.69) with Theorem 1.2.7 (i), Remark 10, and (1.62) we obtain that
u− c ∈ C([0, T ],W 3,p
γ0
)and d4u, d5u ∈ C
([0, T ], Lpγ0+3
). (1.71)
It follows from (1.71) that du ∈ W 3,pγ0
and from following equations that A = du ◦ ϕ ∈ W 2,pγ0
.
dA =[(d2u) ◦ ϕ
](dϕ),
d2A =[(d3u) ◦ ϕ
](dϕ, dϕ) +
[(d2u) ◦ ϕ
](d2ϕ). (1.72)
48
Here we use that d2ϕ ∈ Lpγ0+1, d2u ∈ W 2,pγ0+1 and similarly to (1.70)
[(d2u)◦ϕ
]= O(1/〈x〉γ0+1+2/p).
Since W 2,pγ0
is a Banach algebra therefore from (1.64) it follows that dϕ − I ∈ W 2,pγ0
. Now
Proposition 2.2 in [13] with d2ϕ ∈ W 1,pγ0+1 and
[(d2u) ◦ ϕ
]∈ W 1,p
γ0+2 gives that d2A ∈ Lpγ0+3.
This together with (1.71) gives that A ∈ C([0, T ],W 2,pγ0+1) so that from (1.64) we have
dϕ − I ∈ W 2,pγ0+1. This along with (1.65) gives that d3ψ ∈ Lpγ0+2. Now from (1.68) we
get that ω ∈ C([0, T ],W 3,p
γ0+1
)and so u− c ∈ C
([0, T ],W 4,p
γ0
). By differentiating A ≡ du ◦ ϕ
we obtain following equation
d3A =[(d4u) ◦ ϕ
](dϕ, dϕ, dϕ) + 3
[(d3u) ◦ ϕ
](d2ϕ, dϕ) +
[(d2u) ◦ ϕ
](d3ϕ) (1.73)
which shows that d3A ∈ C([0, T ], Lpγ0+4). Here we use that Proposition 2.2 in [13] and[(d2u)◦ ϕ
]= O(1/〈x〉γ0+2+2/p). Again A ∈ C([0, T ],W 3,p
γ0+1) so that from (1.64) we have dϕ−
I ∈ W 3,pγ0+1. Finally using (1.65) the last equation in (1.68) gives that ω ∈ C
([0, T ],W 4,p
γ0+1
)and so u − c ∈ C
([0, T ],W 5,p
γ0
). By (1.60), this contradicts the blow-up of the Z5,p
0 -norm in
(1.57). Therefore, τ > T . This completes the proof of the theorem.
Now we can finally prove the main result about the global existence of solutions to the Euler
equation (see introduction). The above proposition considers the case when m ≥ 5. However
one case occurs when m = 4 and p > 2 so that we still have m > 3 + 2/p. The above proof
can directly be adopted to this special case.
Proof of Theorem 1.0.1. We will consider the case when m ≥ 5. The special case when
m = 4 and p > 2 is similar. Assume therefore that m > 5 and let uo ∈ Zm,pN ⊂ Z5,p0 .
Now Z5,p0 is continuously embedded into the little Holder space cm,γb for some 0 < γ <
1. Therefore from the results in [17, Theorem 2] and [15] it follows that there exists a
solution ϕ ∈ C ([0,∞), diff3,γ(R2)) of the Euler equation in Lagrangian coordinates with
initial velocity ϕ|t=0 = u0. By Proposition 1.4.4 we conclude that the local solution of the
Euler equation in Z5,p0 given by Theorem 1.3.5 extends to a global solution
u ∈ C([0,∞),Z5,p
0
)∩ C1
([0,∞),Z4,p
0
), ϕ ∈ C1
([0,∞),ZD5,p
0
).
49
Inductively we can conclude that in fact
u ∈ C ([0,∞),Zm,p0 ) ∩ C1([0,∞),Zm−1,p
0
). (1.74)
To see this let m = 6. Since u0 ∈ Z6,p0 so ω0 = ∂zu0 ∈ Z5,p
1 so by Lemma 1.4.1 we have
ω(t) = ω0 ◦ϕ−1 ∈ C([0,∞),Z5,p
1
). And therefore we have that u = ∂−1
z ω ∈ C([0,∞),Z6,p
0
).
This shows that the Z6,p0 -norm of the curve t 7→ u(t) is bounded and therefore by the local
existence theorem 1.3.5 we see that u ∈ C([0,∞),Z6,p
0
)∩ C1
([0,∞),Z5,p
0
). Using this
inductive argument we show that infact (1.74) holds for any m > 5.
New we can increase the asymptotic expansion of u from 0 to N using Proposition 1.4.2
directly. Finally the solution is unique and depends continuously on initial data by local
existence theorem 1.3.5.
1.5 Proof of Lemma 1.1.4
In this section we prove several technical results used in the main body of this chapter. First
we study properties of auxiliary space Zm,pN (BcR). To this end we study the remainder space
Wm,pδ (Bc
R) where BcR =
{x ∈ Rd
∣∣ |x| > R}
.
The key point in following proposition is to show that the constant C is independent of the
ball radius R (Lemma 1.1.4). We will need this later to show that composition is closed in
ZDm,pN . For a non-negative real number β ∈ R denote by bβc its integer part i.e. the largest
integer number that is less or equal than β. Similarly, let {β} := β − bβc be the fractional
part of β ≥ 0. We have the following weighted version of the Sobolev embedding theorem.
Proposition 1.5.1. Assume that 1 < p <∞, δ ∈ R, and R ≥ 1 then
(i) If 0 ≤ m < d/p then for any q ∈[p, dp
d−mp
]one has Wm,p
δ− dp
(BcR) ⊂ Lq
δ− dp
(BcR). Moreover,
there exists a constant C ≡ C(d, p, q,m, δ) > 0 independent of the choice of R ≥ 1 such
50
that for any f ∈ Wm,p
δ− dp
(BcR)
‖f‖Lqδ− dp
(BcR) ≤ C ‖f‖Wm,p
δ− dp(BcR) .
If mp = d the above statement holds for q ∈ [p,∞).
(ii) If mp > d, the for any integer 0 ≤ k < m− dp
one has that Wm,p
δ− dp
(BcR) ⊂ Ck(Bc
R) and
for any multi-index α such that |α| ≤ k,
supx∈BcR
(〈x〉δ+α |Dα f(x)|
)≤ C ‖f‖Wm,p
δ− dp(BcR) , Dα ≡ ∂α1
x1· · · ∂αdxd ,
with a constant C ≡ C(d, p, q,m, δ) > 0 independent of the choice of R ≥ 1.
(iii) Assume that m > d/p, 0 ≤ k < m − (d/p), and δ + (d/p) > 0. Choose a real number
γ such that 0 < γ <{m − k − (d/p)
}for
{m − k − (d/p)
}6= 0 or 0 < γ < 1 for{
m− k − (d/p)}
= 0. Then,
Wm,pδ (Rd) ⊆ Ck,γ
b (Rd)
and the embedding is bounded.
Remark 14. The statement of this lemma also holds with BcR replaced by Rd (see Lemma
1.2 in [13]). An important technical part of this proposition is that the constant C > 0 is
independent of the radius R ≥ 1.
Proof. Let us start with (i). Assume 0 ≤ m < d/p. For a given open set U ⊂ Rd denote
|||f |||pm,p,δ;U =∑
0≤α≤m
∫U
∣∣∣|x|δ+|α|Dαf∣∣∣p dx
For R ≥ 1 consider the annulus AR = B2R \ BR where BR denotes the open ball of radius
R in Rd centered at zero and BR is its closure. By changing variables in the corresponding
integrals one easily sees that for any f ∈ C∞c (AR),
|||fR|||m,p,δ;A1= R−δ−
dp |||f |||m,p,δ;AR , (1.75)
51
where fR(x) = f(R x) for x ∈ Rd. By using that R ≤ | x | ≤ 2R on AR
∑0≤α≤m
∫AR
(| x |δ−
dp
+|α|Rdp |Dαf |
)pdx ≤ |||f |||pm,p,δ;AR
≤∑
0≤α≤m
∫AR
(| x |δ−
dp
+|α|(2R)dp |Dαf |
)pdx,
which impleis
Rdp |||f |||m,p,δ− d
p;AR≤ |||f |||m,p,δ;AR ≤ (2R)
dp |||f |||m,p,δ− d
p;AR
. (1.76)
This together with (1.75) shows that for any 0 ≤ m < d/p there exist constants 0 < M1 < M2
independent of R ≥ 1 such that for any f ∈ C∞c (AR),
M1R−δ|||f |||m,p,δ− d
p;AR≤ |||fR|||m,p,δ;A1
≤M2R−δ|||f |||m,p,δ− d
p;AR
. (1.77)
By the Sobolev embedding theorem for any given q ∈[p, dp
d−mp
]there exists a constant
C ≡ C(d, q) such that for any g ∈ Wm,p(A1),
‖g‖Lq(A1) ≤ C ‖g‖Wm,p(A1) , (1.78)
where we have used following notation
‖g‖qq;A1=
∫A1
|g(x)|q dx
‖g‖m,p;A1=
∑0≤α≤m
‖Dαg‖Lp(A1) .
By applying (1.78) to | x |δg(x) with g ∈ C∞c (A1) and then using∣∣Dα| x |δ
∣∣ ≤ Cδ| x |δ−|α| we
obtain
‖g‖Lqδ(A1) =∥∥∥|x|δ g∥∥∥
Lq(A1)≤ C
∑|α|≤m
∥∥Dα(| x |δg(x))∥∥Lp(A1)
≤ C1
∑|α|≤m
∑β≤α
∥∥Dα−β| x |δDβg∥∥Lp(A1)
≤ C2
∑|α|≤m
∑β≤α
∥∥| x |δ−|α|+|β|Dβg∥∥Lp(A1)
≤ C2
∑|α|≤m
∑|β|≤|α|
∥∥| x |δ+|β|Dβg∥∥Lp(A1)
≤ C3
∑|α|≤m
∥∥| x |δ+|α|Dαg∥∥Lp(A1)
≤ C4|||g|||m,p,δ;A1.
52
By taking g = fR with f ∈ C∞c (AR) we get |||fR|||q,δ;A1≤ C4|||fR|||p,m,δ;A1
which together with
(1.77) and the fact that |||·|||Lqδ(U) ≡ |||·|||0,q,δ;U for any δ ∈ R and an open set U ⊂ Rd implies,
R−δ|||f |||q,δ− dq
;AR ≤ |||fR|||q,δ;A1≤ C0|||fR|||m,p,δ;A1
≤ CC0R−δ|||f |||m,p,δ− d
p;AR
hence for any f ∈ C∞c (AR),
|||f |||Lqδ− dq
(AR) ≤ C|||f |||m,p,δ− dp
;AR., (1.79)
with C > 0 independent of R ≥ 1. Finally we write BcR = AR ∪ A2R ∪ ... ∪ A2kR ∪ ... and
using (1.79) obtain that for any f ∈ C∞c (BcR),
‖f‖Lqδ− dq
(BcR) =
(∞∑k=0
∫A
2kR
(| x |δ−
dq |f |
)q) 1q
=
(∞∑k=0
|||f |||qq,δ− d
q;A
2kR
) 1q
≤ C
(∞∑k=0
|||f |||qm,p,δ− d
p;A
2kR
) 1q
≤ C
(∞∑k=0
|||f |||pm,p,δ− d
p;A
2kR
) 1p
= C |||f |||m,p,δ− dp
;BcR.
Here we used Lemma 1.5.2 below. The case when m = d/p is treated the same way. This
proves item (i).
To prove (ii) i.e. mp > d, applying Sobolev inequality to A1 and using∣∣Dβ| x |δ+|α|
∣∣ ≤(δ + |α|)| x |δ+|α|−|β| gives
supx∈A1
(| x |δ+|α| |Dαg|
)≤ C1
∑|β|≤m
∥∥Dβ(| x |δ+|α|Dαg)∥∥Lp(A1)
≤ C2
∑|β|≤m
∑|γ|≤|β|
∥∥| x |δ+|α|+|γ|−|β|Dα+γg∥∥Lp(A1)
(1.80)
≤ C2
∑|β|≤m
∑|γ|≤|β|
∥∥| x |δ+|α+γ|Dα+γg∥∥Lp(A1)
≤ C3|||g|||m,p,δ;A1
Now with g = fR and DαfR = R|α|(Dαf)R we get using (1.77)
R−δ supy∈AR
(|y|δ+|α| |Dαf(y)|
)≤ C3|||fR|||m,p,δ;A1
≤ C4R−δ|||f |||m,p,δ− d
p;AR
53
giving
supy∈AR
|y|δ+|α||Dαf(y)| ≤ C4|||f |||m,p,δ− dp
;AR ≤ C4|||f |||m,p,δ− dp
and taking sup over BcR gives the result.
Let us now prove item (iii). Assume that m > d/p, 0 ≤ k + (d/p) < m, δ + (d/p) > 0, and
choose γ ∈ R as described in the proposition. It follows from item (ii) that Wm,pδ (Bc
1) ⊆
Ck(Bc1) and that for any |α| ≤ k and f ∈ Wm,p
δ (Bc1) we have
supx∈Bc1
(〈x〉(δ+(d/p))+|α| |Dαf(x)|
)≤ C ‖f‖Wm,p
δ (Bc1) <∞.
Since δ + (d/p) > 0 we then obtain that Wm,pδ (Rd) ⊆ Ck
b (Rd). Let us now estimate the
Holder semi-norm of Dαf with |α| ≤ k. For R ≥ 1 consider the open annulus AR in Rd
defined above and let | · |0,γ;AR and [·]γ;AR be the Holder norm and semi-norm in the closure
AR of AR in Rd. For any δ ∈ R and for any g ∈ Ck(A1) we have
[g]γ;A1=[|x|−δ
(|x|δg
)]γ;A1≤ 2|δ|
[|x|δg
]γ;A1
+Kδ
∣∣|x|δg∣∣L∞(A1)
≤ Cδ∣∣|x|δg∣∣
γ;A1(1.81)
with constants depending only on the choice of δ ∈ R. Here we used that for any non-empty
U ⊆ Rd we have that [fg]γ ≤ |f |L∞(U)[g]γ;U + |g|L∞(U)[f ]γ;U for any f, g ∈ C0,γ(U). By the
Sobolev embedding theorem in the domain A1 and by the assumptions on the regularity
exponents m, k, and γ we obtain that for any |α| ≤ k there exists a constants C > 0 as well
as constants C1, C2, C3 > 0, such that for any g ∈ Hm,p(A1) we have that
∣∣|x|δDαg∣∣γ;A1≤ C
∑|β|≤m−|α|
∥∥Dβ(|x|δDαg
)∥∥Lp(A1)
≤ C1
∑|β|≤m−|α|
∑γ≤β
∥∥|x|δ+|γ|−|β|Dα+γg∥∥Lp(A1)
≤ C2
∑|α+γ|≤m
∥∥|x|δ+|α+γ|Dα+γg∥∥Lp(A1)
≤ C3|||g|||m,p,δ;A1(1.82)
where we argued as in (1.80). It follows from (1.81) and (1.82) that there exists a constant
C > 0 such that for any g ∈ Hm,p(A1) and for any |α| ≤ k we have that
[Dαg]γ;A1≤ C |||g|||m,p,δ;A1
. (1.83)
54
Now, take R ≥ 1 and let f ∈ Hm,p(AR). Then, we have
[DαfR]γ;A1= sup
x,y∈A1;x 6=y
∣∣(DαfR)(x)− (DαfR)(y)∣∣
|x− y|γ
= R|α| supx,y∈A1;x 6=y
∣∣(Dαf)(Rx)− (Dαf)(Ry)∣∣
|x− y|γ
= R|α|+γ[Dαf ]γ;AR . (1.84)
By combining (1.75) and (1.84) with (1.83) we obtain that there exists a constant C > 0
such that for any R ≥ 1, |α| ≤ k, and for any f ∈ Hm,p(AR),
[Dαf ]γ;AR ≤C
Rµ‖f‖Wm,p
δ (AR) (1.85)
where µ :=(δ + (d/p)
)+ γ + |α| > 0. In addition, by the Sobolev embedding theorem in
the unit ball B1 we also have that there exists a constant, denoted again by C > 0, such
that for any f ∈ Hm,p(B1),
[Dαf ]γ;B1≤ C ‖f‖Wm,p(B1) (1.86)
where B1 denotes the closure of B1 in Rd. Note that
Rd = B1 ∪( ⋃k≥0
A2k
).
Now, take f ∈ C∞b (Rd) and two points x, y ∈ Rd such that x 6= y. The closed segment
[x, y] :={
x + s(y − x)∣∣ 0 ≤ s ≤ 1
}intersects any given annulus A2k with k ≥ 0 as well as
the closed ball B1 at no more than two straight segments
A2k ∩ [x, y] = I ′k ∪ I ′′k , B1 ∩ [x, y] = I−1
where I ′k and I ′′k with k ≥ 0, and I−1, can be empty. If such a segment is not empty we set
I ′k = [x′k, y′k], I ′′k = [x′k, y
′k], I−1 = [x−1, y−1].
Then,
[x, y] = I−1 ∪⋃k≥0
(I ′k ∪ I ′′k
).
55
This, together with (1.85) and (1.86) then implies that for any f ∈ Wm,pδ (Rd) we have that∣∣(Dαf)(x)− (Dαf)(y)
∣∣|x− y|γ
≤∣∣(Dαf)(x−1)− (Dαf)(y−1)
∣∣|x−1 − y−1|γ
+∑k≥0
∣∣(Dαf)(x′k)− (Dαf)(y′k)∣∣
|x′k − y′k|γ
+∑k≥0
∣∣(Dαf)(x′′k)− (Dαf)(y′′k)∣∣
|x′′k − y′′k|γ≤ [Dαf ]γ;B1
+ 2∑k≥0
[Dαf ]γ;A2k
≤ C(
1 + 2∑k≥0
1
2µk
)‖f‖Wm,p
δ
where we omit the terms corresponding to empty intervals in the second estimate above and
use that for any k ≥ 0 we have that ‖f‖Wm,pδ (A
2k) ≤ ‖f‖Wm,p
δ (Rd) and a similar inequality
for the ball B1. Hence, there exists a positive constant C ≡ Cµ < ∞ such that for any
f ∈ Wm,pδ (Rd) and for any |α| ≤ k we have that [Dαf ]γ ≤ C ‖f‖Wm,p
δ. Since we already
proved that Wm,pδ (Rd) ⊆ Ck
b (Rd), this completes the proof of item (iii).
In the above Proposition 1.5.1 we used following simple Lemma about norms on lp, p ≥ 1
spaces. lp is a Banach space of complex-valued sequences a = (ak)k≥1 with finite lp norm
||a||lp =(∑
k≥1 |ak|p)1/p
.
Lemma 1.5.2. For 1 ≤ p ≤ q < ∞ one has that lp ⊂ lq so that for any a ∈ lp we have
||a||lq ≤ ||a||lp.
Proof. Without loss of generality we can take ||a||lp = 1. Then in particular |ak| ≤ 1 and
therefore since q ≥ p we have
||a||qlq =∑k≥1
|ak|q ≤∑k≥1
|ak|p = ||a||plp = 1.
Next Proposition give the multiplicative property of the weighted Sobolev spaces.
Proposition 1.5.3. For any real δ1, δ2 ∈ R and integers 0 ≤ k ≤ l ≤ m with m+ l−k > d/p
there exists a constant C ≡ C(d, p, l, k,m, δ1, δ2) > 0 independent of the choice of R ≥ 1 such
56
that for any f ∈ Wm,p
δ1− dp(Bc
R) and for any g ∈ W l,p
δ2− dp(Bc
R) we have that fg ∈ W k,p
δ1+δ2− dp(Bc
R)
and
‖fg‖Wk,p
δ1+δ2−dp
(BcR) ≤ C ‖f‖Wm,p
δ1−dp
(BcR) ‖g‖W l,p
δ2−dp
(BcR) . (1.87)
In particular, for m > d/p and δ ∈ R the weighted Sobolev space Wm,pδ (Bc
R) is a Banach
algebra.
Remark 15. Note that inequality (1.87) as well as items (i) and (ii) of Proposition 1.5.1
hold with BcR replaced by Rd (see Lemma 1.2 in [13]). This easily follows from Proposition
1.5.1 (i), (ii), Proposition 1.5.3, and Lemma 2.1.6.
Proof. By the generalized Holder inequality with weights (see Lemma below) we can write
‖fg‖Lpδ− dp
(BcR) ≤ ‖f‖Lq1δ1−
dq1
(BcR) ‖g‖Lq2δ2−
dq2
(BcR) (1.88)
where δ = δ1 + δ2 and 1p
= 1q1
+ 1q2
. First consider the case when 0 ≤ m ≤ d/p. Then we
have by Proposition 1.5.1 (i) that Wm,p
δ− dp
⊂ Lqδ− d
q
for p ≤ q ≤ dpd−mp is continuous embedding.
We need to find q1, q2 such that 1p
= 1q1
+ 1q2
then from (1.88) above
‖fg‖Lpδ− dp
(BcR) ≤ ‖f‖Wm,p
δ1−dp
(BcR) ‖g‖Lm,pδ2−
dp
(BcR) . (1.89)
We can find such q1, q2 if
d−mpdp
+d− lpdp
≤ 1
p
which reduces to dp≤ m + l. If mp = d then we can again select q1, q2 in range p ≤ q < ∞
to satisfy 1p
= 1q1
+ 1q2
and we obtain (1.89) in the same way. Finally if mp > d take q1 =∞
and q2 = p to get
‖fg‖Lpδ− dp
(BcR) ≤ ‖f‖L∞δ1 (BcR) ‖g‖Lpδ2−
dp
(BcR) . (1.90)
Now apply Proposition 1.5.1 (ii) to f and (i) or (ii) to g depending if l ≤ d/p or l > d/p we
obtain (1.89) again.
57
To complete the proof we want to show that | x ||α|Dα(fg) ∈ Lpδ1+δ2−|α|− dp
(BcR) for all |α| ≤
k ≤ l ≤ m.
| x ||α|Dα(fg) =∑β≤α
(α
β
)(| x ||β|Dβf
) (| x ||α−β|Dα−βg
)with γ = α − β we have | x ||β|Dβf ∈ W
m−|β|,pδ1− dp
and | x ||γ|Dγg ∈ Wl−|γ|,pδ2− dp
so the product
belongs to Lpδ1+δ2− dp
(BcR) if (m− |β|+ l − |γ|)p = (m+ l − |α|)p ≥ (m+ l − k)p > d.
In the Proposition above we used the following generalized weighted Holder inequality
Lemma 1.5.4. For any real δ1, δ2 ∈ R, 1 < p ≤ q1, q2 ≤ ∞ with 1p
= 1q1
+ 1q2
and R > 0 we
have that
‖fg‖Lpδ− dp
(BcR) ≤ ‖f‖Lq1δ1−
dq1
(BcR) ‖g‖Lq2δ2−
dq2
(BcR)
for any f ∈ Lq1δ1− d
q1
(BcR) and g ∈ Lq2
δ2− dq2
(BcR).
Proof. First consider case < p ≤ q1, q2 <∞. Then 1p
= 1q1
+ 1q2
implies that 1 = 1(q1/p)
+ 1(q2/p)
.
Then by using (standard) Holder inequaltiy we get
||fg||pLpδ1+δ2−
dp
=
∫BcR
∣∣∣fg| x |δ1+δ2− dp∣∣∣p dx =
∫BcR
∣∣∣f | x |δ1− dq1
∣∣∣p ∣∣∣g| x |δ2− dq2
∣∣∣p dx
=
(∫BcR
∣∣∣f | x |δ1− dq1
∣∣∣p dx
)p/q1 (∫BcR
∣∣∣g| x |δ2− dq2
∣∣∣p dx
)p/q2
.
This gives the weighted Holder inequality in this case. When q1 = p and q2 = ∞ it is east
to see this inequality holds.
Corollary 1.5.5. Assume m > 2/p then for any u ∈ Zm,pN we have
(i) |u|Zm,pN (BcR) → a00 as R→∞.
(ii) For R ≥ 2 we have that supz∈BcR |u| ≤ |u|Zm,pN (BcR).
(iii) There is a constant Cχ depending on the cut-off function used in asymptotic expansion
of u ∈ Zm,pN such that ||u||L∞ ≤ Cχ ‖u‖Zm,pN.
58
Proof. Take u ∈ Zm,pN such that u =∑
0≤k+l≤Naklχzk zl
+ f . By Proposition 1.5.1 it follows that
‖f‖Wm,pγN
(BcR) → 0 proving 1. To prove 2 we can write
supz∈BcR
|u| ≤∑
0≤k+l≤N
|akl|Rk+l
+ supz∈BcR
|f | ≤∑
0≤k+l≤N
|akl|Rk+l
+ ‖f‖Wm,pγN
(BcR) = |u|Zm,pN (BcR).
Finally since mp > 2, u is continuous so ||u||L∞ = sup |u| so with Cχ = max(1, supχ)
||u||L∞ = sup |u| ≤ Cχ
( ∑0≤i+j≤N
|a|ij + sup |f |
).
Finally use sup |f | = limR→0 supz∈BcR |f | ≤ limR→0 ‖f‖Wm,pγN
(BcR) = ‖f‖Wm,pγN
.
Proof of Lemma 1.1.4. Assume that m > 2/p and that R ≥ 1. In the first case if u1, u2 ∈
Wm,pγN
(BcR) then by Proposition 1.5.3
|u1u2|Zm,pN (BcR) ≤ C|u1|Zm,pN (BcR)|u2|Zm,pN (BcR).
If u1 = azk
and u2 ∈ Wm,pγN
(BcR) then u1.u2 ∈ Wm,p
γN(Bc
R) so that
|u1u2|pZm,pN (BcR)=
∑0≤|α|≤m
∫BcR
∣∣∣|z|γN+|α|Dα( azku2
)∣∣∣p dx dy
≤∑|α|
|a|p∫BcR
|z|γN+|α|∑
0≤|β|≤|α|
(α
β
) ∣∣∣∣Dα−β(1
zk)
∣∣∣∣ ∣∣Dβu2
∣∣p
dx dy
≤∑α
|a|p∫BcR
(|z|γN+|α|
∑β≤α
Cm,N1
|z|k+|α|−|β|
∣∣Dβu2
∣∣)p
dx dy
≤(|a|Rk
)pC0
∑α
∫BcR
(∑β≤α
|z|γN+|β| ∣∣Dβu2
∣∣)p
dx dy
≤ Cp1
(|a|Rk
)p|u2|pZm,pN (BcR)
.
We have obtained
|u1u2|Zm,pN (BcR) ≤ C1 |u1|Zm,pN (BcR) |u2|Zm,pN (BcR) . (1.91)
59
Finally we have the trivial relation when 0 ≤ (k1 + l1) + (k2 + l2) ≤ N∣∣∣∣ a
zk1 zl1b
zk2 zl2
∣∣∣∣Zm,pN (BcR)
=∣∣∣ a
zk1 zl1
∣∣∣Zm,pN (BcR)
∣∣∣∣ b
zk2 zl2
∣∣∣∣Zm,pN (BcR)
.
Also when (k1 + l1) + (k2 + l2) ≥ N + 1 we obtain (1.91) by direct integration with C1
independent of R.
60
Chapter 2
Asymptotic expansion in higher
dimensions
In the last chapter we considered in the 2-dimensional case the smaller spaces of the asymp-
totic Am,pn,N that are preserved by Euler equation. In particular to obtain these spaces we used
the complex structure of R2. Furthermore as we saw the Fourier modes in the numerator of
the asymptotic terms were limited to finitely many. In dimensions d ≥ 3 we do not have a
similar complex structure in Rd therefore we consider limiting the number of Fourier modes
in the asymptotics again. We will show by carefully selecting certain Fourier modes we can
obtain a Banach subalgebra of Am,pn,N such that the Euler equation is preserved. We will also
see that the method of selecting/limiting the Fourier modes in the asymptotic terms used in
this chapter works for any dimension d ≥ 2 however the case d = 2 presents a special case
in this approach and there the method used in chapter is infact needed.
Consider again the incompressible Euler equation on Rd for d ≥ 3, ut + u · ∇u = −∇p, div u = 0,
u|t=0 = u0,(2.1)
where u(x, t) is the velocity of the fluid and p(x, t) is the scalar pressure. As discussed in the
main introduction it is proven in [14] that the Euler equation (2.1) is locally well-posed in the
61
class of asymptotic vector fields Am,pN ;0 (see the introduction for definitions) with log-terms as
defined in (2). Spaces with asymptotic terms are natural spaces for studying Euler equation
since even with smooth initial velocity u0 ∈ C∞c (Rd,Rd), the solution develops asymptotics.
However the log-term asymptotics appear since certain Fourier modes in a0k(θ) give log-terms
when inverting the Laplacian.
In this chapter we find smaller spaces with restrictions on the Fourier modes of these asymp-
totic coefficients ajk(θ) such that the log-terms do not appear anymore. In particular through-
out this chapter we will work with the asymptotic spaces Am,pn,N (cf. [13, Section 2]) and it’s
properties discusses in Section 2 to work with its Banach subalgebra Zm,pN |l . Since we will
be working with restrictions on Fourier modes of the coefficients, we will use homogeneous
functions in Rd for functions on Sd−1. Let us first define the space of homogeneous functions
Zk|l =⊕l′≥l
rl′Hk−l′ (2.2)
where Hk is space of harmonic polynomials of homogeneous degree k in Rd. Note that this
sum is finite since Hk<0 = {0}. Thus Zk|l is the space of all homogeneous functions in Rd
of degree k whose restriction to unit sphere lies in the linear span of spherical harmonics of
degree at most k − l. By Mk we will denote homogeneous polynomials of degree k and we
have following representation (cf. [18, Theorem 22.2]).
Mk = Hk
⊕r2Hk−2 · · ·
⊕r2lHk−2l
⊕. . . (2.3)
Remark 16. We will generally denote elements in Zk|l as Jk and polynomials in Mk as
Pk. By above grading we can write Jk = Pk + rPk−1. Furthermore given Jk ∈ Zk|l we can
write Jk = rlJk−l where Jk−l ∈ Zk−l|0. For ease of notation, in the last factorization, we will
sometimes denote Jk := Jk−l and therefore we will have Jk ∈ Zk−l|0 with k, l depending on
the context. Also we have the relations Zk|0 = Mk
⊕rMk−1 and Zk|l = rlZk−l|0.
Now we can define the subspace of Am,pn,N with restricted Fourier modes in the asymptotic
62
coefficients. For integers m > d/p and N, l ≥ 0 we define
Zm,pN |l ={u∣∣∣u =
∑0≤k≤N
χ(r)
rkJk(r, θ)
rk+ f, Jk ∈ Zk|l and f ∈ Wm,p
γN
}, (2.4)
where χ : R → R is a C∞-smooth cut-off function with bounded derivatives such that
χ(r) ≡ 1 for r ≥ 2. Different choices of χ give same space Zm,pN |l with equivalent norms (cf.
[13, Remark 3.2]).
Remark 17. The first non-zero asymptotic coefficient in Zm,pN |l that occurs is of the order 1/rl
with constant coefficient (only constant Fourier mode). For next term 1/rl+1 the numerator
has constant and linear Fourier modes. For example in the interesting case of d = 3 and
l = d− 2 = 1 (See Theorem 2.0.1 below) we have following first two terms
u = χ(r)
(a0
r+
1
r2
(ax+ by + cz) + dr
r2+ · · ·+ a0
N(θ)
rN
)+ f
where x, y, z are coordinates on R3 and a0, a, b, c, d ∈ R are constants. Therefore J2(x, y, z) =
ax+ by + cz + dr.
Note that this space Zm,pN |l does not contain log-terms in the asymptotic expansion and
that the corresponding coefficients ak(θ) = Jk(r,θ)rk
have only the first k − l Fourier modes.
Furthermore we will show that Zm,pN |d−2 is preserved by the incompressible Euler equation in
three and higher dimensions (d ≥ 3). This space is equipped with following norm
‖u‖Zm,pN|l
=∑
0≤k≤N
sup |Jk(1, θ)|+ ‖f‖Wm,pγN
which is equivalent to the norm it inherits from Am,pn,N . This is true since all norms on the
finite dimensional subspace of asymptotic terms in Zm,pN |l are equivalent. Let us also define
the space of all divergence free vector fields in Zm,pN |l as Zm,pN |l . We will denote BZm,pN|l
(ρ) to be
the open ball in Zm,pN |l of radius ρ > 0.
The layout of this chapter is as follows. In section 2 we will study the properties of asymptotic
spaces Zm,pN |l . In section 3 we prove our main result.
63
Theorem 2.0.1. Let d ≥ 3 and m > 3 + d/p where 1 < p < ∞. Then for any ρ > 0 there
exists T > 0 such that for any divergence-free vector field u0 ∈ BZm,pN|d−2
(ρ) there exists a
unique solution u ∈ C0([0, T ], Zm,pN |d−2) ∩ C1([0, T ], Zm−1,pN |d−2) of Euler equation (2.1) such that
pressure p(t) ∈ Zm+1,N |d−2. Furthermore the solution depends continuously on initial data in the
sense that the data-to-solution map
u0 7→ u, BZm,pN|d−2
(ρ) ∩ Zm,pN |d−2 → C0([0, T ], Zm,pN |d−2) ∩ C1([0, T ], Zm−1,pN |d−2)
is continuous.
In section 4 we give examples of initial vector fields without log terms in Am,pN ;0 that develop
log terms in it’s asymptotic expansion. In the last section we discuss uniqueness of solutions
of Euler equation when we allow constant asymptotic terms.
2.1 Asymptotic spaces and diffeomorphism groups
Let us define the space of asymptotic diffeomorphism ZDm,pN |l associated with the space Zm,pN |l
as
ZDm,pN |l :={ϕ ∈ Diff1
+(Rd)∣∣∣ ϕ = id +u, u ∈ Zm,pN |l
}, (2.5)
which consists of diffeomorphisms that form a closed subgroup of the group of asymptotic
diffeomorphisms ADm,pn,N (cf. [13, Section 5]). Here id is the identity map on Rd and Diff1+(Rd)
denotes the group of orientation preserving C1-diffeomorphisms of Rd. We will show ZDm,pN |l
is topological subgroup of ADm,pn,N when m > 2 + d/p (see Theorem 2.1.16). We begin
with some properties of the asymptotic spaces Zm,pN |l . It is already clear that Zm,pN |l ⊂ ADm,pn,N
(see introduction). We begin by stating properties of remainder spaces Wm,pδ and asymptotic
spacesADm,pn,N that follow from [13, § 2, Lemma 3.1, 3.3] and [13, Proposition 3.1] respectively.
Proposition 2.1.1. (i) For any 0 ≤ m1 ≤ m2 and for any real weights δ1 ≤ δ2 we have
that the inclusion map Wm2,pδ2→ Wm1,p
δ1is bounded.
64
(ii) For any m ≥ 0, δ ∈ R, and for any k ∈ Z the map f 7→ f · χa(θ)rk
, Wm,pδ → Wm,p
δ+k is
bounded. Here χ(x) ≡ χ(| x |) is a C∞-smooth cut-off function such that χ : R→ R as
in (2.4) and a(θ) ∈ C∞(Sd−1,C), Sd−1 ⊂ Rd is the unit sphere.
(iii) For any regularity exponent m ≥ 0, δ ∈ R, and a multi-index α ∈ Zd≥0 such that
|α| ≤ m the map ∂α : Wm,pδ → W
m−|α|,pδ+|α| is bounded.
(iv) For m > d/p and for any weights δ1, δ2 ∈ R the map (f, g) 7→ fg, Wm,pδ1×Wm,p
δ2→
Wm,pδ1+δ2+(d/p) is bounded. In particular, for δ ≥ 0 the space Wm,p
δ is a Banach algebra.
Proposition 2.1.2. Assume that m > d/p. Then
(i) For any integers 0 ≤ n1 ≤ n2 and 0 ≤ N1 ≤ N2 we have that the inclusion map
Am,pn1,N1→ Am,pn2,N2
is bounded.
(ii) If m > 1 + d/pthen for 1 ≤ j ≤ d the map u 7→ ∂u/∂xj, Am,pn,N → Am−1,pn+1,N+1 is bounded.
(iii) Let m > d/p. For any 0 ≤ ni ≤ Ni, with i = 1, 2 and n = n1 + n2, N = min(N1 +
n2, N2 + n1) the map
(f, g) 7→ fg, Am,pn1,N1×Am,pn2,N2
→ Am,pn,N
,
is bounded. In particular, Am,pn,N is a Banach algebra for any 0 ≤ n ≤ N .
Now we can show that in fact Zm,pN |l is a Banach subalgebra of Am,pn,N .
Lemma 2.1.3. If m > d/p and u ∈ Zm,pN1|l1 , v ∈ Zm,pN2|l2 then uv ∈ Zm,pN1+N2|l1+l2
.
Proof. By Proposition 2.1.2 (iii) we see that uv ∈ Am,pN1+N2. It remains to show that the
product of two asymptotic terms is again in Zm,pN |l . Let Jkr2k∈ Zm,pN1|l1 and
Jk′
r2k′∈ Zm,pN2|l2 .
Then we need to show that Jk · Jk′ ∈ Zk+k′|l1+l2 . This follows from rl′1Hk−l′1 · r
l′2Hk′−l′2 ⊂
rl′1+l′2Mk+k′−(l′1+l′2) ⊂ Zk+k′|l1+l2 . Here we used that l′s ≥ ls for s = 1, 2 and the grading
(2.3).
65
Remark 18. The above proof shows in particular that the product of two asymptotic terms
in Zm,pN |l is again in Zm,pN |l .
In the special case we get the following Corollary.
Corollary 2.1.4. If m > d/p and u, v ∈ Zm,pN |l then uv ∈ Zm,pN |2l ⊂ Zm,pN |l .
Since Zm,pN |l consists of the remainder space Wm,pγN
and finitely many asymptotic terms we see
that Zm,pN |l is a closed vector subspace of Am,pN and therefore we get by Lemma 2.1.3
Corollary 2.1.5. Let m > d/p then Zm,pN |l is a Banach subalgebra of Am,pN .
For any R > 2 we will define the following spaces. Here let BR = {x || x | ≤ R} be the closed
ball of radius R > 0 in Rd and BcR be it’s complement.
Zm,pN |l (BcR) =
{u∣∣∣u =
∑0≤k≤N
1
rkJk(r, θ)
rk+ f, Jk ∈ Zk|l and f ∈ Wm,p
γN(Bc
R)}. (2.6)
Here Wm,pγN
(BcR) is defined similarly with Bc
R instead of Rd. We have Zm,pN |l (BcR) ⊂ Hm,p
loc (BcR).
This space has a norm equivalent to one restricted from Zm,pN |l and is given by
|u|Zm,pN|l (BcR) =
∑0≤k≤N
1
Rksup |Jk(1, θ)|+ ‖f‖Wm,p
γN(BcR) .
This space is defined analogously to Am,pn,N(BcR) [13, Section 3]. Following Lemma allows us
to work with Zm,pN |l (BcR) (cf. [14, Lemma A.4]).
Lemma 2.1.6. Assume that R > 0. Then, for u ∈ Hm,ploc (C,C) we have that u ∈ Zm,pN |l if and
only if u ∈ Hm,p(BR+1) and u ∈ Zm,pN |l (BcR). The norms ‖·‖Zm,p
N|land ‖·‖Hm,p(BR+1)+|·|Zm,p
N|l (BcR)
on Zm,pN |l are equivalent.
Lemma 2.1.7. Assume m > d/p and R > 2. If u ∈ Zm,pN |l (BcR) then
xjr2· u ∈ Zm,pN |l (Bc
R).
Proof. First assume that u is in the remainder space i.e. u ∈ Wm,pγN
(BcR). By Proposition
2.1.1 (ii) we have thatxjr2· u ∈ Wm,p
γN(Bc
R) ⊂ Zm,pN |l (BcR). Next if u is one of the asymptotic
terms in Zm,pN |l (BcR) so that u = 1
rkJk−l(r,θ)rk−l
where Jk−l ∈ Zk−l|0 (See Remark 16) then
xjr2· u =
1
rk+1
xjJk−l(r, θ)
rk−l+1.
66
Also by Remark 16 we have xjJk−l ∈ Zk−l+1|0 and thereforexjr2· u ∈ Zm,pN |l (Bc
R) completing
the proof.
Remark 19. The functionxjr
is bounded in BcR and when multiplied with remainder terms
u ∈ Wm,pγN
(BcR) the product is again in Wm,p
γN(Bc
R). However when multiplied with asymptotic
terms it introduces new harmonics in the numerator thusxjr· u ∈ Zm,pN |l−1.
Lemma 2.1.8. Assume m > 1 + d/p and N ≥ 0. Let ϕ ∈ ZDm,pN |l then for large enough
R > 2 and r ≥ R we have
1
r◦ ϕ =
1
|ϕ(x)|=
1
r+u
r(2.7)
where u ∈ Zm,pN |l (BcR).
Proof. Let us write ϕ = id +v with v ∈ Zm,pN |l . Then for x ∈ BcR(Rd) we can write,
|ϕ(x)|2 = r2 + |v|2 + 2d∑1
xivi = r2
(1 +|v|2
r2+ 2
d∑1
xivir2
)= r2(1 + v)
and we have by Lemma 2.1.7 that v ∈ Zm,pN |l (BcR). Since v ∈ Zm,pN |l is bounded and v has no
constant term so by choosing R > 2 large enough we have that 1 + v ≥ ε for some ε > 0.
Therefore by Lemma 6.1 in [13] (1 + v)−1/2 ∈ Am,pN (BcR). Also as in the proof of Lemma 6.1
in [13] the Taylor expansion (1 + v)−1/2 = 1 +∑N
j=1 cj vj +RN(v) where RN(v) ∈ Wm,p
γN(Bc
R)
shows that by Remark 18 following Lemma 2.1.3 we have (1 + v)−1/2 = 1 + u such that
u ∈ Zm,pN |l (BcR) so we can write
1
r◦ ϕ =
1
|ϕ(x)|=
1
r
1√1 + v
=1
r+u
r. (2.8)
Lemma 2.1.9. Assume m > 1 + d/p and N ≥ 0. Let ϕ ∈ ZDm,pN |l and θj =xjr
then for large
enough R > 2 and r ≥ R we have
θj ◦ ϕ = θj +xjru+ w (2.9)
where u, w ∈ Zm,pN |l (BcR).
67
Proof. We can write ϕ in the form ϕ = id +v with v ∈ Zm,pN |l . Then for x ∈ BcR(Rd) we can
write by Lemma 2.1.8
θj ◦ ϕ = (xj + vj)
(1
r◦ ϕ)
= (xj + vj)
(1
r+u
r
)= θj +
xjru+ w (2.10)
where w =vj+vj u
r∈ Zm,pN |l (Bc
R).
Lemma 2.1.10. Assume m > 1 + d/p and let ϕ ∈ ZDm,pN |l and u ∈ Zm,pN |l then u ◦ ϕ ∈ Zm,pN |l .
Proof. By Proposition 5.1 in [13] we know that if ϕ ∈ ZDm,pN |l and u ∈ Zm,pN |l then u◦ϕ ∈ Am,pN .
Therefore it only remains to show that u ◦ ϕ ∈ Zm,pN |l for the asymptotic terms in u. Assume
ϕ = id +v as before and u = χ(r) 1rkJk−lrk−l
such that Jk ∈ Mk−l is a homogeneous polynomial
of degree k− l (The case when Jk ∈Mk−l−1 is similar. See Remark 16). Now by Proposition
3.1 in [13] sup|x|≥R |v(x)| is bounded as R → ∞ therefore we can take R > 2 large enough
so that χ ◦ ϕ ≡ 1. For given ϕ we also take R even larger if necessary so that conditions
Lemma 2.1.8 and Lemma 2.1.9 are satisfied. Now since k ≥ k − l, each monomial in u has
factors of the formxjr2
and therefore for r ≥ R we can write by Lemma 2.1.8 and Lemma
2.1.9 that
xjr2◦ ϕ =
(1
r+u
r
)(xjr
+xjru+ w
)=xjr2
+ 2xjr2u+ w =
xjr2
+ wj (2.11)
where w ∈ Zm,pN |l (BcR) and by Lemma 2.1.7
xjr2u ∈ Zm,pN |l (Bc
R) and therefore wj ∈ Zm,pN |l (BcR).
Therefore for r ≥ R applying this to each monomial and again applying Lemma 2.1.7 we get
that 1rk−l
Jk−lrk−l◦ ϕ = 1
rk−lJk−lrk−l
+ w for some w ∈ Zm,pN |l (BcR) and therefore by Lemma 2.1.8 and
Lemma 2.1.7
u ◦ ϕ =
(1
r+u
r
)l(1
rk−lJk−lrk−l
+ w
)= u+ h
where h ∈ Zm,pN |l (BcR). Finally applying Lemma 2.1.6 completes the proof.
Lemma 2.1.11. Assume m > 1 + d/p, then for ρ > 0 small enough if ϕ ∈ Zm+1,pN |l with
‖ϕ− id‖Zm+1,pN|l
< ρ, then ϕ−1 ∈ Zm+1,pN |l .
68
Proof. The composition is a C1-map Am,pn,N ×Am,pn,N → A
m,pn,N by Proposition 5.1 in [13]. Since
ZDm,pN |l is a submanifold of ADm,pn,N closed under composition by Corollary 2.1.5 and Lemma
2.1.10 therefore the map F (ϕ, φ) = ϕ◦φ, ZDm+1,pN |l ×ZD
m,pN |l → ZD
m,pN |l is C1-map. The partial
derivative of F with respect to the first variable is D1F (ϕ, φ)|h = h ◦φ. At the identity map
we have F (id, id) = id and D1F (id, id) = idZm,pN|l
. Therefore by inverse mapping theorem,
there exists ρ > 0 and a C1-map inv : B(ρ)Zm,pN|l→ B(ρ)Zm,p
N|lsuch that F (ϕ, inv(ϕ)) = id.
Therefore ϕ−1 ∈ ZDm,pN |l . However by Proposition 5.2 in [13] ϕ−1 ∈ ADm+1,pN |l and so the
remainder term of ϕ−1 has smoothness m+ 1 while since ϕ−1 ∈ ZDm,pN |l , the asymptotics are
C∞-smooth. Therefore ϕ−1 ∈ ZDm,pN |l .
The following two propositions are used in the proof of Proposition 2.1.14. They follow
directly from similar statements for Am,pn,N as in [13, Lemmas 7.2, 7.3, 7.4].
Proposition 2.1.12. Assume m > 1+d/p, then for any ϕ ∈ ZDm,pN |l there exists a continuous
path γ : [0, 1]→ ZDm,pN |l such that γ(1) = ϕ and γ(0) = id +f where f has compact support.
Proposition 2.1.13. Assume m > 2 + d/p and let ϕ∗ ∈ ZDm,pN |l be fixed then there exists a
neighborhood U of identity id ∈ ZDm−1,pN |l such that the left translation map
ψ 7→ Lϕ∗(ψ) := ϕ∗ ◦ ψ, U → Lϕ∗(U) ⊂ ZDm−1,pN |l
is a C1-diffeomorphism.
Proposition 2.1.14. Let m > 2 + d/p then for any ϕ ∈ ZDm,pN |l we have that ϕ−1 ∈ ZDm,pN |l .
Furthermore the map ϕ 7→ ϕ−1,ZDm,pN |l → ZDm,pN |l is continuous and the associated map
ϕ 7→ ϕ−1,ZDm,pN |l → ZDm−1,pN |l is a C1-map.
Proof. Let ϕ ∈ ZDm,pN |l . By Proposition 2.1.12 ϕ is connected by a path γ : [0, 1] → ZDm,pN |l
to a ϕ0 := γ(0) ∈ ZDm,pN |l so that γ(1) = ϕ and ϕ0− id has compact support. By Proposition
5.1 in [13] ϕ−10 ∈ A
m,pN . Also ϕ−1
0 − id has compact support and in particular ϕ−10 ∈ Z
m,pN |l .
Let us denote ϕt := γ(t). Now by Lemma 2.1.11, for each ϕt along this path there exists
69
Ut ⊂ ZDm−1,pN |l open neighborhood of id in which every diffeomorphism is invertible within
ZDm−1,pN |l and such that by Proposition 2.1.13, Vt = Lϕt(Ut) is open neighborhood of ϕt.
Every ϕ∗ in V0 is invertible in ZDm−1,pN |l since we can write ϕ∗ = ϕ0 ◦ ψ with ψ ∈ U0,
so ϕ−1∗ = ψ−1 ◦ ϕ−1
0 ∈ ZDm−1,pN |l . Since both ψ−1, ϕ−1
0 ∈ ZDm−1,pN |l and ZDm−1,p
N |l is closed
under composition. Now we can cover the path connecting ϕ0 to ϕ = ϕ1 by open sets
V0, Vt1 , Vt2 , ..., V1(ordered by sequence in which they intersect consecutively). By induction
on k, we show that any ϕ∗ ∈ Vtk is invertible in ZDm−1,pN |l . By last argument it is enough
to show that ϕtk is invertible. For k = 1 choose ϕ ∈ V0 ∩ Vt1 then ϕ = ϕt1 ◦ ψ for some
ψ ∈ Ut1 , so we can write ϕ−1t1 = ψ ◦ ϕ−1 hence it is invertible in ZDm−1,p
N |l . Note that ϕ is
invertible in ZDm−1,pN |l by induction and that ϕ ∈ V0 ∩ Vt1 . Continuing in this way we see
that ϕ−1 ∈ ZDm−1,pN |l .
By Proposition 2.4 in [13] we infact have that ϕ ∈ Am,pN . Therefore the remainder term of
ϕ−1 − id is in Wm,pγN
while the asymptotic coefficients ak(θ) have only finitely many Fourier
modes (since ϕ−1 ∈ ZDm−1,pN |l ) so in fact ak ∈ C∞(Sd), hence ϕ−1 ∈ ZDm,pN |l .
Finally the last statement follows from Proposition 5.2 in [13] abd the fact that ZDm,pN |l is a
closed submanifold of ADm,pN .
Corollary 2.1.15. For m > 2 + d/p ZDm,pN |l is a smooth subgroup of ADm,pN .
Theorem 2.1.16. Let m > 1 + d/p then the composition and inverse
(u, ϕ) 7→ u ◦ ϕ, Zm,pN |l ×ZDm,pN |l → Z
m,pN |l
ϕ 7→ ϕ−1, ZDm,pN |l → ZDm,pN |l
are continuous maps. Furthermore the following
(u, ϕ) 7→ u ◦ ϕ, Zm+1,pN |l ×ZDm,pN |l → Z
m,pN |l
ϕ 7→ ϕ−1, ZDm+1,pN |l → ZDm,pN |l
are C1-mappings.
70
Lemma 2.1.17. Let m > 2 + d/p and u ∈ C([0, T ],Zm,pN |l ) for some T > 0 then there exists
a unique solution ϕ ∈ C1([0, T ],ZDm,pN |l ) of the equation
ϕ = u ◦ ϕ, ϕ|t=0 = id . (2.12)
Proof. By Theorem 2.1.16 the map F (t, ϕ) = u(t) ◦ ϕ, F : [0, T ] × ZDm−1,pN |l → Zm−1,p
N |l
is a continuous map and it’s partial derivative with respect to the second argument is a
continuous map D2F (t, ϕ) : [0, T ]×ZDm−1,pN |l → L(Zm−1,p
N |l ,Zm−1,pN |l ), the space of all bounded
linear maps. Thus for any t0 ∈ [0, T ], F is locally Lipschitz in ϕ. Therefore by existence and
uniqueness of solutions to ODEs in Banach spaces there exists a neighborhood [to−ε0, to+ε0]
and a unique solution ϕ ∈ C1([0, T ],ZDm−1,pN |l ) of (2.12). Since the equation ϕ = u ◦ ϕ is
invariant under translation and since right translation by any fixed ϕ0 ∈ ZDm−1,pN |l is a C∞-
map therefore ϕ = u ◦ϕ has a unique solution ϕ ∈ C1([t0− ε0, t0 + ε0],ZDm−1,pN |l ) with initial
condition ϕ|t=t0 = ϕ0. Here ε0 depends on t0. Since [0, T ] is compact we see that (2.12) has
unique solution ϕ ∈ C1([0, T ],ZDm−1,pN |l ).
It remains to show that ϕ ∈ C1([0, T ],ZDm,pN |l ). By Proposition 2.1 in [14] there is a unique
solution ϕ ∈ C1([0, T ],Am,pN ;0). Since Zm,pN |l is a subalgebra of Am,pN ;0 by uniqueness of solutions
of ODEs ϕ = ϕ ∈ C1([0, T ],ZDm−1,pN |l ). So for any t ∈ [0, T ], ϕ(t) ∈ ZDm−1,p
N |l . Since
the projection map Am,pN ;0 → Zm,pN |l is continuous again by uniqueness of solution of ODEs
ϕ = ϕ ∈ C1([0, T ],Zm,pN |l ).
Before end of this section let us define a subgroup of ZDm,pN |l . This is the subgroup of volume
preserving diffeomorphism. These arise as consequence of incompressibility of Euler equation
(div u = 0). We will later show that it is a smooth submanifold of ZDm,pN |l preserved by the
Euler equation. It is defined as
ZDm,pN |l ={ϕ ∈ ZDm,pN |l
∣∣∣ det[dϕ] ≡ 1}. (2.13)
Lemma 2.1.18. Let m > 2 + d/p and u ∈ C([0, T ], Zm,pN |l ) for some T > 0 then there exists
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a unique solution ϕ ∈ C1([0, T ], ZDm,pN |l ) of the equation
ϕ = u ◦ ϕ, ϕ|t=0 = id . (2.14)
Proof. Let u ∈ C([0, T ], Zm,pN |l ). First we apply ∂xj , for each 1 ≤ j ≤ d, to equation (2.12) to
get following ODE
[dϕ]· = [du] ◦ ϕ · [dϕ], [dϕ]t=0 = I . (2.15)
Let us write A := [du] ◦ ϕ then A ∈ C([0, T ],Md×d) where Md×d is space of d× d matrices.
The linear system X = A ·X,X|t=0 = I has the solution X = [dϕ]. By Wronskian identity
we get
det[ϕ](t, x) = e∫ t0 (div u)(s,ϕ(s,x)ds. (2.16)
Since u(t) ∈ Zm,pN |l so we have div u(t, x) ≡ 0 and from the above equation we see that
det[ϕ](t, x) = 1 therefore ϕ(t) ∈ ZDm,pN |l completing the proof.
2.2 Laplace operator in asymptotic spaces
The Laplace operator on the scale of weighted Sobolev spaces Wm,pδ with weight δ ∈ R,
except for discrete values of δ, is injective operator with closed image (cf. [11]). In fact for
δ + d/p > 0 the Laplace operator is injective and it is possible to define an isomorphism by
extending the spaces Wm,pγN
to the asymptotic spaces Am,pN as done in [12]. In Theorem 2.2.2
below we show that this isomorphism can be obtained for the smaller asymptotic spaces
Zm,pN |l . Let us start with following property of asymptotic spaces.
Lemma 2.2.1. If m > 1 + d/p then the map u 7→ ∂u∂xj
, Zm,pN |l → Zm−1,pN+1|l is bounded.
Proof. By Proposition 2.1.2 this map is bounded on Am,pN → Am−1,pN+1 therefore by Corollary
2.1.5 it is enough to show that the asymptotics still have the same form. For an asymptotic
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term we have
∂
∂xj
Jkr2k
=1
r2(k+1)(−2kxjJk + r2∂jJk).
First we show that J 7→ xjJ maps Zk|l → Zk+1|l by the grading (2.3). Now xj(Hk−l′) ⊂
Mk+1−l′ =⊕
µ≥0 r2µHk+1−l′−2µ. Therefore
xj(Zk|l) =⊕l′≥l
rl′xj(Hk−l′) ⊂
⊕l′≥l
rl′⊕µ≥0
r2µHk+1−l′−2µ ⊂⊕l′≥l
rl′Hk+1−l′ = Zk+1|l.
Similarly we show that J 7→ r2∂jJ maps Zk|l → Zk+1|l. This follows directly from r2∂j(rl′Hk−l′) ⊂
rl′xj(Hk−l′) + rl
′+2∂j(Hk−l′) ⊂ rl′Mk+1−l′ + rl
′+2Hk−1−l′ ⊂ Zk+1|l.
Remark 20. The spherical harmonics of degree l will be denoted as Y jl (θ) where 0 ≤ j ≤
nl − 1 where nl is the number of spherical harmonics of degree l. The Y jl ’s are the eigen-
vectors of the spherical Lapalcian −4S
−4SYjl (θ) = l(l + d− 2)Y j
l (θ).
We will denote by Yl an element in the linear span of spherical harmonics of degree l ≥ 0.
Also note that there is a one to one correspondence between spherical harmonics of degree l
and harmonic polynomials of degree l in Rd given by Yl = hl|Sd−1 where hl ∈ Hl is a harmonic
polynomial. Therefore nl is the number of linearly independent harmonic polynomials in Rd
of degree l.
Theorem 2.2.2. Assume m > 2 + d/p and N ≥ 0 then 4 : Zm,pN |d−2 → Zm−2,pN+2|d+1 is an
isomorphism.
Proof. First we show that4(Zm,pN |d−2) ⊂ Zm−2,pN |d+1 and we only need to show this for asymptotic
terms of u ∈ Zm,pN |d−2 since for the remainder term f ∈ Wm,pγN
, 4f ∈ Wm−2,pγN+2 ⊂ Z
m−2,pN+2|d+1.
Now let u = bk−2(θ)
rk−2 then we can write bk−2(θ) =∑k−2−(d−2)
l=0 Yl(θ) where Yl(θ) is a linear
combination of spherical harmonics of degree l. By direct computation we see that
∆
(bk−2(θ)
rk−2
)=
1
rk[∆Sbk−2(θ) + (k − 2)(k − d)bk−2(θ)] =
ak(θ)
rk. (2.17)
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Since Yl are spherical harmonics so −4SYl = l(l + d − 2)Yl and therefore we can write
ak(θ) =∑k−2−(d−2)
l=0 ck,lYl(θ). As k − 2 − (d − 2) = k − d, from here we conclude that
4u ∈ Zm−2,pN+2|d as ak(θ) contains only first k−d spherical harmonics. However writing k−d = λ
we see that ak(θ) = [∆Sbk−2(θ) + λ(λ+ d− 2)bk−2(θ)], which shows that ak does not contain
the (k−d)th spherical harmonics. Hence infact 4u ∈ Zm−2,pN+2|d+1 as desired. Also the operator
4 is injective since there are no harmonic functions decaying asymptotically in Rd.
To see4 is onto first let u ∈ Wm−2,pγN+2 be in the remainder space. By part (b) of Lemma A.3 in
[14] 4−1u ∈ Zm,pN |d−2 since when inverting the Laplacian on the remainder space Wm−2,pγN+2 the
only asymptotics that we get are of the form Yl(θ)rd−2+l such that Yl(θ)
rlis harmonic in Rd − {0}.
And these asymptotics are already in Zm,pN |d−2 Finally for asymptotic terms of form ak(θ)/rk
we can solve (2.17) linearly to find corresponding bk − 2(θ). This completes the proof.
Remark 21. The above isomorphism gives following commutative diagram
Zm,pN |d−2 Zm−2,pN+2|d+1
Zm−1,pN+1|d−2 Zm−3,p
N+3|d+1
4
∂j ∂j
4
from which it follows that 4−1 ◦ ∂j = ∂j ◦ 4−1 on Zm−2,pN+2|d+1.
Lemma 2.2.3. Assume m > 3 + d/p and N ≥ 0 then for any ϕ ∈ ZDm,pN |d−2, the right
translation map Rϕ : v 7→ v ◦ ϕ is an isomorphism
Rϕ : Zm−2,pN+2|d+1 → Z
m−2,pN+2|d+1.
Proof. We need to show this for asymptotic terms only. Let ϕ = id +v, v ∈ Zm,pN |d−2 and
u ∈ Zm−2,pN+2|d+1 such that
u =1
rk+d+1
Jk(x)
rk
where Jk(x) is a homogeneous polynomial upto a factor of r i.e. we can write Jk(x) =
Pk(x) + rPk−1(x) where Pj are homogeneous polynomials of degree j (we consider the case
74
when Jk = Pk here, the other case Jk = rPk−1 follows similarly. Also see Remark 16). Then
for large enough R > 2 as in (2.11), for r ≥ R we have
Pk(x)
r2k◦ ϕ =
Pk(x)
r2k+ w, w ∈ Zm−2,p
N |d−2 .
and by Lemma 2.1.9 1rd+1 ◦ ϕ = 1
rd+1 (1 + u), u ∈ Zm,pN |d−2 therefore(1
rk+d+1
Pk(x)
rk
)◦ ϕ =
1
rk+d+1
Pk(x)
rk+
1
rd+1
(w + wu+
Pk(x)
r2ku
)= u+
h
rd+1.
By Lemma 2.1.7 h ∈ Zm,pN |d−2(BcR) and so h
rd+1 ∈ Zm,pN |2d−1(BcR) ∈ Zm,pN |d+1(Bc
R) for d ≥ 2. This
completes the proof.
From Theorem 2.2.2 and Lemma 2.2.3 we directly get that
Corollary 2.2.4. Assume m > 3 + d/p then for any ϕ ∈ ZDm,pN |d−2, the map 4ϕ = Rϕ ◦4◦
Rϕ−1 is an isomorphism
4ϕ : Zm,pN |d−2 → Zm−2,pN+2|d+1
with inverse map (4ϕ)−1 = Rϕ ◦ 4−1 ◦ Rϕ−1 where 4−1 is inverse of the map in Theorem
2.2.2.
2.3 Euler vector field
We can eliminate the pressure term from the Euler equation (2.1) by applying divergence to
both sides to get
−4p = div(u · ∇u) = tr([du]2) + u · ∇(div u) = tr([du]2).
Here we used that div u = 0. Let us define Q(u) := tr([du]2). Note that in the above
equation the second order derivative term that was zero is u·∇(div u). We write u·∇(div u) =
∇((div u)u)−(div u)2. From this we will define Q(u) = Q(u)−(div u)2 = tr([du]2)−tr([du])2.
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Therefore we can write from above for the Euler equation−4p = Q(u) = Q(u). Substituting
into (2.1) we get ut + u · ∇u = ∇ ◦4−1 ◦ Q(u),
u|t=0 = u0,(2.18)
This modification allows us to remove certain terms from Q(u) that introduce log-terms
when inverting 4. Following lemma combined with Theorem 2.2.2 shows that the Euler
equation preserves the space Zm,pN |d−2.
Lemma 2.3.1. Assume m > 1 + d/p and N ≥ 0. Let u ∈ Zm,pN |d−2 then Q(u) ∈ Zm−1,pN+2|2d−2
(⊂ Zm−1,pN+2|d+1 if d ≥ 3).
Proof. Since Zm,pN |d−2 ⊂ Am,p1,N , so Q(u) ∈ Am−1,p
2,N+2. Therefore we only need to show that the
asymptotic part lies in the space Zm−1,pN+2|2d−2. If we write the vector field as (see Remark 16)
u = χ(r)∑
0≤k≤N
1
rk
~Jkrk
+ f = χ(r)∑
0≤k≤N
rd−2
r2k
~Jk + f
such that Jk,j ∈ Zk|d−2 and f ∈ Wm,pγN
. We have written Jk,j, the jth component of the vector
~Jk, with a factor of rd−2 so that Jk,j ∈ Zk−d+2|0. When differentiating below we will only
keep track of asymptotic terms and therefore ignore the derivatives of χ(r) since they do not
contribute to the asymptotic part. Then by a direct computation we see that the asymptotic
terms in div u can be written as
div u '∑
0≤k≤N
rd−2
r2k+2
(−λk ~Jk · ~r + r2 div ~Jk
)(2.19)
such that λk = 2k− d+ 2 and ~r is the position vector in Rd. And therefore we can write the
asymptotic part of (div u)2 as
(div u)2 '∑
0≤k+l+2≤N+1
r2(d−2)
r2(k+l+2)
(λkλl(
~Jk · ~r)( ~Jl · ~r) + r2Ak+l+4−2d
)(2.20)
so that As ∈ Zs|0. Similarly when computing Q(u) = tr([du]2) we write
d
[rd−2
r2k
~Jk
]α,β
=rd−2
r2k+2
(−λkxβJk,α + r2∂βJk,α
).
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Then the asymptotics part of Q(u) can be written as follows. Note that in the summation
below we have 0 ≤ k + l + 2 ≤ N + 1, k, l ≥ 0 and 1 ≤ α, γ ≤ d. (d is the dimension in Rd)
Q(u) '∑k,l
tr
(d
[rd−2
r2k
~Jk
]· d[rd−2
r2l
~Jl
])(2.21)
=∑k,l;γ,α
r2(d−2)
r2k+2l+4(−λkxγJk,α + r2∂γJk,α)(−λlxαJl,γ + r2∂αJl,γ)
=∑k,l;γ,α
r2(d−2)
r2k+2l+4
(λkλlxγxαJk,αJl,γ − 2r2(kxγJk,α∂βJl,γ + lxαJl,γ∂γJk,α) + r4(∂γJk,α∂βJl,γ)
)=∑k,l
r2(d−2)
r2(k+l+2)
(λkλl(
~Jk · ~r)( ~Jl · ~r) + r2Bk+l+4−2d
)(2.22)
where Bs ∈ Zs|0. From (2.20) and (2.22) we see that asymptotic part of Q(u) has numerator
r2d−2(Bk+l+4−2d − Ak+l+4−2d) in Zk|2d−2 hence u ∈ Zm−1,pN+1|2d−2.
Lemma 2.3.2. Assume m > 3 + d/p and let l = d − 2.The curve u ∈ C([0, T ],Zm,pN |l ) ∩
C1([0, T ],Zm−1,pN |l ) is a solution of equation (2.18) with initial data u0 ∈ Zm,pN |l if and only if
u ∈ C([0, T ], Zm,pN |l ) ∩C1([0, T ], Zm−1,pN |l ) and it is a solution of Euler equation (2.1) for some
−p ≡ p(t) ∈ Zm+1,pN |d−2 , t ∈ [0, T ].
Proof. Let u ∈ C([0, T ],Zm,pN |l ) ∩ C1([0, T ],Zm−1,pN |l ) be a solution of (2.18). Applying diver-
gence to both sides of (2.18) we get
(div u)t + u · ∇(div u) = −(div u)2. (2.23)
Let ϕ(t) be solution of (2.12). Since div u ∈ C([0, T ],Zm−1,p
N |l)∩ C1
([0, T ],Zm−2,p
N |l)
and
ϕ ∈ C1([0, T ],ZDm,pN |l
)with m > 3 + (d/p) we conclude that div u ∈ C1
([0, T ]× Rd,R
)and
ϕ ∈ C1(Rd,Rd). This together with (2.23) implies that ω = (div u) ◦ ϕ satisfies ωt = −ω2.
Solving this differential equation we can write
ω(t) =div u0
1 + t div u0
. (2.24)
Since we assume that initial data is divergence free, i.e. div u0 = 0. This implies that for any
t, div u(t) = 0. Thus u ∈ C([0, T ], Zm,pN |l )∩C1([0, T ], Zm−1,pN |l ). Now Q(u) ∈ T(ZDm−1,p
N+2|d+1) so
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by Theorem 2.2.2 we can define p = 4−1Q(u) so u ∈ Zm+1,pN |d−2
Conversely if u ∈ C([0, T ], Zm,pN |d−2) ∩ C1([0, T ], Zm−1,pN |d−2) is a solution of Euler equation (2.1)
then taking divergence on both sides of (2.1) we get −4p = div (u · ∇u) = tr([du]2) + u ·
∇ div u = Q(u) ∈ Zm−1,pN+2|2d−2 by Lemma 2.3.1. Now by Theorem 2.2.2 −p = 4−1Q(u) ∈
Zm+1,pN |d−2 . Substituting this in equation (2.1) shows that u satisfies equation (2.18) with
div u0 = 0.
Following Lemma 7.1 in [14] we write the Euler equation (2.1) in it’s equivalent form as a
dynamical system on the tangent bundle ZDm,pN |d−2 ×Zm,pN |d−2 as follows
(ϕ, v) = (v, E2(ϕ, v)) ≡ E(ϕ, v) (2.25)
(ϕ, v)|t=0 = (id, u0), u0 ∈ Zm,pN |d−2.
Here E2(ϕ, v) = (Rϕ ◦∆−1 ◦ Rϕ−1) ◦ (Rϕ ◦ ∇ ◦ Q ◦ Rϕ−1)(v) is the second component of the
Euler vector field E and Rϕ(u) = u ◦ ϕ is the composition operator.
Lemma 2.3.3. Assume m > 3 + d/p and N ≥ 0. Let (ϕ, v) ∈ T(ZDm,pN |d−2) := ZDm,pN |d−2 ×
Zm,pN |d−2 then E2(ϕ, v) ∈ Zm,pN |d−2. So that E2 defines a map E2 : ZDm,pN |d−2 ×Zm,pN |d−2 → Z
m,pN |d−2.
Proof. By Lemma 2.3.1 Q ◦ Rϕ−1(v) ∈ Zm−1,pN+2|d+1. We can write by Theorem 2.2.2 that
E2(ϕ, v) = Rϕ◦∇◦∆−1◦Q◦Rϕ−1(v). And by Theorem 2.2.2 now ∆−1◦Q◦Rϕ−1(v) ∈ Zm+1,pN |d−2
and finally Lemma 2.2.1 gives the result.
2.4 Smoothness of Euler Vector Field
In this section we will show that the Euler vector field E , defined as in (2.25) is a real-analytic
vector field on the tangent bundle of ZDm,pN |d−2. Let us start with following.
Lemma 2.4.1. For m > 3 + d/p the following map
(ϕ, v) 7→ 4ϕ(v), ZDm,pN |d−2 ×Zm,pN |d−2 → Z
m−2,pN+2|d+1
is real-analytic. Here 4ϕ = Rϕ ◦ 4 ◦Rϕ−1 as in Corollary 2.2.4.
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Proof. We can write following directly
Rϕ ◦ ∇ ◦Rϕ−1(v) = [dv] · [dϕ]−1 (2.26)
Rϕ ◦ div ◦Rϕ−1(v) = tr([dv] · [dϕ]−1). (2.27)
By Lemma 5.2 in [14] the above maps are analytic. Finally composing these maps we see
that from Corollary 2.2.4 we see that (ϕ, v) 7→ 4ϕ(v) is real-analytic.
Now consider the map
D : ZDm,pN |d−2 ×Zm−2,pN+2|d+1 → Z
m,pN |d−2, D(ϕ, v) = Rϕ ◦ 4−1 ◦Rϕ−1(v). (2.28)
Proposition 2.4.2. Assume m > 2 + d/p. Then the map (2.28) is real-analytic.
Proof. Let E = Zm,pN |d−2 and F = Zm−2,pN+2|d+1. Then the set of all invertible linear maps
GL(E,F ) is a open group inside L(E,F ), the Banach space of all linear bounded maps from
E to F . Furthermore the inverse map
inv : GL(E,F )→ GL(F,E), G 7→ G−1 (2.29)
is real-analytic using Neumann series expansion.
By Corollary 2.2.4 the map
ϕ 7→ 4ϕ, ZDm,pN |d−2 → GL(E,F ) (2.30)
is well-defined. Also 4ϕ = (D2C)(ϕ, 0) where C(ϕ, v) = 4ϕ(v) and D2 denote partial
derivative with respect to second argument. By Lemma 2.4.1 the map (2.30) is real-analytic.
Composing this with map (2.29) we see that (2.28) is real-analytic.
Theorem 2.4.3. The Euler vector field E : ZDm,pN |d−2 ×Zm,pN |d−2 → Z
m,pN |d−2 is real-analytic.
Proof. We only need to show that E2 is real-analytic. We can write
E2 =[(Rϕ ◦∆−1 ◦Rϕ−1) ◦ (Rϕ ◦ ∇ ◦Q ◦Rϕ−1)− (Rϕ ◦ div ◦Rϕ−1)2] (v).
By Theorem 6.1 in [14] we see that first term is real-analytic. Also from (2.27) (ϕ, v) 7→
(Rϕ ◦ div ◦Rϕ−1)2 (v) is real-analytic.
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2.5 Proof of Theorem 2.0.1
Lemma 2.5.1. Assume that m > 3 + d/p. Then the map
(ϕ, v) 7→ v ◦ ϕ−1, C1([0, T ],T(ZDm,pN |d−2))→ C([0, T ],Zm,pN |d−2) ∩ C1([0, T ],Zm−1,pN |d−2)
gives a bijection between the solutions of the dynamical system (2.25) and the solutions of
equation (2.18) in C([0, T ],Zm,pN |d−2) ∩ C1([0, T ],Zm−1,pN |d−2).
Proof. Let u ∈ C([0, T ],Zm,pN |d−2) ∩ C1([0, T ],Zm−1,pN |d−2) be a solution of (2.18). By Lemma
2.1.17, there exists a ϕ ∈ C1([0, T ],ZDm,pN |d−2) satisfying (2.12). Let v = u ◦ ϕ then (ϕ, v) ∈
C([0, T ],T(ZDm,pN |d−2)) by Theorem 2.1.16. We will first show that this map is C1 and sat-
isfies the dynamical system (2.25). By Sobolev embedding Zm−1,pN |d−2 ⊂ C1 therefore u, ϕ ∈
C1([0, T ]× Rd,Rd). We can differentiate point-wise to get
vt = ut ◦ ϕ+ [du] ◦ ϕ · ϕ = ut ◦ ϕ+ [du] ◦ ϕ · u ◦ ϕ = (ut + u · ∇u) ◦ ϕ.
From equation (2.25) and (2.18) we can write
vt = (Rϕ ◦∆−1 ◦ ∇ ◦ Q ◦Rϕ−1)(v) = E2(ϕ, v).
Integrating this equation point-wise we get
v(t, x) = u0(x) +
∫ t
0
E2(ϕ(s, x), v(s, x))ds.
As v ∈ C([0, T ],Zm,pN |d−2) and ϕ ∈ C([0, T ],ZDm,pN |d−2) and E2 is analytic so the integral
converges in Zm,pN |d−2. Therefore v ∈ C1([0, T ],Zm,pN |d−2) and v = E2(ϕ, v). Hence (ϕ, v) ∈
C1([0, T ],T(ZDm,pN |d−2)) and satisfies the dynamical system (2.25).
Conversely if (ϕ, v) ∈ C1([0, T ],T(ZDm,pN |d−2)) then u = v◦ϕ−1 ∈ C([0, T ],Zm,pN |d−2)∩C1([0, T ],Zm−1,pN |d−2)
by Theorem 2.1.16. Differentiating v = u ◦ ϕ we see that u satisfies equation (2.18).
Combining Lemma 2.5.1 and Lemma 2.3.2 we easily get following.
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Lemma 2.5.2. Assume that m > 3 + d/p. Then the map
(ϕ, v) 7→ v ◦ ϕ−1, C1([0, T ],T(ZDm,pN |d−2))→ C([0, T ], Zm,pN |d−2) ∩ C1([0, T ], Zm−1,pN |d−2)
gives a bijection between the solutions of the dynamical system (2.25) with u0 ∈ Zm,pN |d−2 and
the solutions of the Euler equation (2.1) in C([0, T ], Zm,pN |d−2) ∩ C1([0, T ], Zm−1,pN |d−2).
Proof of Theorem 2.0.1. By Theorem 2.4.3 the Euler vector field (2.25) is real-analytic so
the existence and uniqueness of solutions of ODEs on Banach spaces (see [9]) there exists
ρ0 > 0 and T0 > 0 such that for any u0 ∈ BZm,pN|d−2
(ρ0) there is a unique solution (ϕ, v) ∈
C1([0, T0],T(ZDm,pN |d−2)). By Lemma 2.5.2 this gives a solution u ∈ C0([0, T0],Zm,pN |d−2) ∩
C1([0, T0],Zm−1,pN |d−2). Now for any ρ > 0 we use symmetry of Euler equation u 7→ uc where
uc(t) = cu(ct) and take c = ρ0ρ
and T = cT0 so that for any u0 ∈ BZm,pN|d−2
(ρ) we have a
solution u ∈ C([0, T ],Zm,pN |d−2) ∩ C1([0, T ],Zm−1,pN |d−2).
2.6 Volume preserving asymptotic diffeomorphism
The Euler vector field is defined on the tangent bundle T(ZDm,pN |d−2) thus the exponential
map ExpE of the spray E can be defined as follows. As E is a real-analytic vector field we
see that by existence and uniqueness of solutions of an analytic ODE in a Banach space (see
[6]), that there exists an open neighborhood U of zero in Zm,pN |d−2 such that for any initial
data (ϕ, v)|t=0 = (id, u0), with u0 ∈ U , the ODE on T(ZDm,pN |d−2)
(ϕ, v) = E(ϕ, v) = (v, E2(ϕ, v)) (2.31)
has a unique real-analytic solution (−2, 2) → ZDm,pN |d−2 × Zm,pN |d−2, t 7→ (ϕ(t;u0), v(t;u0)), so
that the map
(−2, 2)× U → ZDm,pN |d−2 ×Zm,pN |d−2 (2.32)
is real-analytic. Therefore the exponential map of E ,
ExpE : U → ZDm,pN |d−2, u0 7→ ϕ(1;u0) (2.33)
81
is also real-analytic. Furthermore the differential at (id, 0) is d0 ExpE = idZm,pN|d−2
, the identity
map on Zm,pN |d−2. Therefore by inverse function theorem on Banach spaces we get the following.
Proposition 2.6.1. Assume that m > 3 + d/p. Then there exists an open neighborhood U
of zero in Zm,pN |d−2 and an open neighborhood V of id in ZDm,pN |d−2 so that ExpE : U → V is a
real-analytic diffeomorphism.
Next we prove the following. Here U ,V are the open sets as in Proposition 2.6.1 above.
Proposition 2.6.2. Assume that m > 3 + d/p, then for any u0 ∈ U ∩ Zm,pN |d−2, we have that
ExpE(u0) ∈ ZDm,pN |d−2. Furthermore the exponential map
ExpE : U ∩ Zm,pN |d−2 → V ∩ ZDm,pN |d−2 (2.34)
is a real-analytic diffeomorphism.
Proof. Assume that u0 ∈ U ∩ Zm,pN |d−2 and let (ϕ, v) ∈ C1([0, 1],T(ZDm,pN |d−2)) be a solu-
tion of the dynamical system (2.25). By Lemma 2.5.2 u := v ◦ ϕ−1 ∈ C([0, 1], Zm,pN |d−2) ∩
C1([0, 1], Zm−1,pN |d−2) is solution of Euler equation (2.1) hence in particular div u ≡ 0. By (2.25)
we also see that ϕ = v = u ◦ ϕ and ϕ|t=0 = id, so by Lemma 2.1.18 uniqueness of solutions
to (2.14) give ϕ(t) ∈ ZDm,pN |d−2. Hence ExpE(u0) ∈ V ∩ ZDm,pN |d−2.
Conversely we will show that if u0 ∈ U r Zm,pN |d−2 then ExpE(u0) ∈ V r ZDm,pN |d−2. This to-
gether with the fact that ExpE : U → V is a diffeomorphism will show that the map (2.34)
is bijective and therefore real-analytic diffeomorphism as U ∩ Zm,pN |d−2 ⊂ U is a real-analytic
submanifold. We let u0 ∈ U r Zm,pN |d−2 and let (ϕ, v) ∈ C1((−2, 2),T(ZDm,pN |d−2)) be a solution
of (2.25). By Lemma 2.5.1 u = v ◦ϕ−1 is solution of (2.18) so by equation (2.24) we get that∫ t
0
ω(s)ds =
∫ t
0
(div u) ◦ ϕ ds =
∫ t
0
div u0
1 + s div u0
ds = ln (1 + t div u0).
Since ϕ = v = u ◦ ϕ, and ϕ|t=0 = id therefore by equation (2.16)
det[ϕ](t, x) = 1 + t div u0(x). (2.35)
82
If the initial divergence is non-zero at some point x0, then (det[dϕ](t, x0) − 1) will remain
non-zero. And therefore ExpE(u0) = ϕ(1;u0) /∈ ZDm,pN |d−2.
Theorem 2.6.3. (a) ZDm,pN |d−2 is a real-analytic submanifold of ZDm,pN |d−2.
(b) For any ψ ∈ ZDm,pN |d−2 the tangent space Tψ(ZDm,pN |d−2) to ZDm,pN |d−2 at the point ψ
is(ψ,Rψ
(Zm,pN |d−2
)), where Rψ
(Zm,pN |d−2
)is the image of the right-translation map
Rψ : Zm,pN |d−2 → Zm,pN |d−2.
(c) The Euler vector field E is tangent to the submanifold T (ZDm,pN |d−2) in T (ZDm,pN |d−2).
Proof. (a) Since U∩Zm,pN |d−2 ⊂ U is a real-analytic submanifold by Proposition 2.6.1, its image
is real-analytic submanifold of V . By Proposition 2.6.2 the image is V∩ZDm,pN |d−2. So ZDm,pN |d−2
is a submanifold in open neighborhood of id in ZDm,pN |d−2. In general, for any fixed ψ ∈
ZDm,pN |d−2 the right translation Rψ : ZDm,pN |d−2 → ZDm,pN |d−2 is a real analytic diffeomorphism
so that Uψ = Rψ(U) is a neighborhood of ψ. Furthermore Uψ∩ZDm,pN |d−2 = Rψ(U ∩ZDm,pN |d−2).
(b) At the identity id we have Tid(ZDm,pN |d−2) = Zm,pN |d−2. Again general case follows since right
translation Rψ is a real analytic diffeomorphism on ZDm,pN |d−2 by part (a).
(c) Let u0 ∈ Zm,pN |d−2 and let (ϕ, v) ∈ C1([0, T ],T(ZDm,pN |d−2)) be a solution of (2.25). Then by
Lemma 2.5.2 and Lemma 2.1.18 we see that in fact ϕ ∈ C1([0, T ], ZDm,pN |d−2) and therefore v =
ϕ ∈ Tϕ(ZDm,pN |d−2) = Rϕ(Zm,pN |d−2). Since we already know (ϕ, v) ∈ C1(
[0, T ], T (ZDm,pN |d−2))
we have shown that (ϕ, v) ∈ C1(
[0, T ], T (ZDm,pN |d−2))
. In particular we see that E(id, u0) =
ddt
∣∣t=0
(ϕ, v) is tangent submanifold T (ZDm,pN |d−2). Now note that for any (ϕ, v) ∈ T (ZDm,pN |d−2)
and any fixed ψ ∈ ZDm,pN |d−2, E(Rψϕ,Rψv) = Rψ(E(ϕ, v)) where Rψ(v, w) = (Rψv,Rψw).
This is because Rψ maps solutions of (2.25) to solutions of (2.25) with right translated
initial conditions. Now take an arbitrary (ϕ0, v0) ∈ T (ZDm,pN |d−2). By part (b) there we have
v0 = u0◦ϕ0 for some u0 ∈ Zm,pN |d−2. We have E(ϕ0, v0) = Rϕ0(E(id, u0)). Finally Rϕ0(E(id, u0))
is tangent to T (ZDm,pN |d−2) at (ϕ0, v0) as E(id, u0) is tangent to T (ZDm,pN |d−2) at (id, u0) and
Rϕ0 : (ϕ, v) 7→ (ϕ ◦ ϕ0, v ◦ ϕ0) is a C∞ diffeomorphism.
83
2.7 Examples
Example 1. We construct a divergence-free initial vector field u that is asymptotically in
Am,pN and such that u contains log terms. This can be done using so called stream functions
in 3 dimensions. Such divergence-free vector fields are given by
u = ∇ψ ×∇φ = ∇× (ψ∇φ) . (2.36)
(ψ and φ are the stream functions.) We make specific choices of stream functions to get
a counter example. One such choice is by first fixing ψ(x, y, z) = x2
2+ z2
2. Now we take
φ(x, y, z) = −13
1r3− 1
21r2
. Then we get
∇ψ =
x
0
z
, ∇φ =
(1
r4+
1
r5
)~r (2.37)
u =
(1
r4+
1
r5
)−yz
0
xy
. (2.38)
This is the vector field away from the origin. Near the origin we use a cut-off function
that adds a compactly supported vector field to u. Note that in u the terms such as −yzr4
are asymptotics with highest harmonic in numerator. This we know is necessary to get a
counter example. Now we compute the differential as follows.
du =
4xyzr6
+ 5xyzr7, − z
r4− z
r5+ 4y2z
r6+ 5y2z
r7, − y
r4− y
r5+ 4yz2
r6+ 5yz2
r7
0, 0, 0
yr4
+ yr5− 4x2y
r6− 5x2y
r7, x
r4+ x
r5− 4xy2
r6− 5xy2
r7, −4xyz
r6− 5xyz
r7
. (2.39)
In tr ([du]2) we look at the numerator, T , of 1r11
that comes from products of form 1r5× 1
r6
and 1r4× 1
r7.
T = 2(5x2y2 + 5y2z2 + 4x2y2 + 4y2z2) = 18y2(x2 + z2
).
84
It is easly to check that T can be written as a non-zero harmonic polynomial H of degree 4
modulo r2 as follows.
T = H +18
7r2
(3r2
5+ y2
).
Therefore we can write
T
r11=
1
r7
T
r4.
We see that Tr4
contains a (k − d)th non-zero spherical harmonic (Here k = 7 and d = 3).
Therefore when computing ∆−1 tr ([du]2), the term Tr11
will give rise to log terms.
Example 2. Here we will provide another example of a initial velocity vector field that
develops log-term. In this example we show that u0 ∈ Zm,pN |1 is not enough to avoid log-term
and that in fact the divergence-free conditions are needed to avoid it. Such an example is
easy to construct since there is no divergence-free condition here. Consider the velocity field
given asymptotically as
u =
1r
+ 1r3xr
0
0
. (2.40)
The differential is easily computed as
du =
− xr3
+ −4x2+r2
r6, − y
r3− 4xy
r6, − z
r3− 4xz
r6
0, 0, 0
0, 0, 0
. (2.41)
So that tr([du]2) =(− xr3
+ −4x2+r2
r6
)2
contains the term −2x(−4x2+r2)r9
= 1r6
8x3−2xr2
r3. Now if the
numerator (8x3−2xr2) contains a harmonic polynomial modulo r2 of degree k−d = 6−3 = 3
then inverting the Laplacian we will get asymptotics with log-terms. A direct computation
shows that we can write x3 = h + 35xr2, such that h = x3 − 3
5xr2 is harmonic polynomial.
However when computing Q(u) we see that no log-terms appear.
85
2.8 Remarks on solutions with constant asymptotic
term
From the Euler equation we can see apriori that for any spatially constant vector field ~c(t),
the transformation p 7→ p + ~c · ~x transforms the Euler equation to
ut + u · ∇u = −∇p− ~c(t). (2.42)
Now comparing the asymptotics we see that ~a0(t) satisfies ~a0(t) = −~c(t). Note that ~c(t) was
arbitrary and this gives non-uniqueness. However uniqueness can be obtained by restricting
to solutions such that p(x) = o(|x|). However it is not obvious that solutions to (2.42) exist.
We can follow same approach and construct a corresponding dynamical system that gives
solutions to (2.42).
We can append constants to Zm,pN |d−2 to obtain Zm,pN |d−2 := R⊕Zm,pN |d−2. This space is still a Ba-
nach algebra in Am,p0;N and therefore we can similarly define ZDm,pN |d−2 and obtain a dynamical
system analogous to (2.25) on ZDm,pN |d−2 × Zm,pN |d−2.
(ϕ, v) = (v, E2(ϕ, v)) ≡ E(ϕ, v) (2.43)
(ϕ, v)|t=0 = (id, u0) u0 ∈ Zm,pN |d−2
where E2(ϕ, v) = E2(ϕ, v) − ~c(t). This vector is field is analytic and therefore has a unique
solution, depending on ~c(t), which gives a solution uc of the Euler equation.
2.9 Asymptotics preserved by Euler equation
Combined with the divergence-free condition on the asymptotics we will see that certain
asymptotic terms are constant in Euler equation. This is true for dimensions greater than
3. Let us write
u = χ(r)∑
0≤k≤N
1
rk
~Jkrk
+ f = χ(r)∑
0≤k≤N
rd−2
r2k
~Jk + f
86
then divergence is given by
div u = χ(r)∑
0≤k≤N
rd−2
r2k+2(−λk ~Jk · ~r + r2 div ~Jk) + f.
And so on the asymptotic terms the condition div u ≡ 0 gives from (2.19)
λk ~Jk · ~r = r2 div ~Jk
here λk = 2k − d+ 2. Using this we first write
Q(u) =∑k,l
tr
(d
[rd−2
r2k
~Jk
]· d[rd−2
r2l
~Jl
])(2.44)
=∑k,l;γ,α
r2(d−2)
r2k+2l+4(−λkxγJk,α + r2∂γJk,α)(−λlxαJl,γ + r2∂αJl,γ)
=∑k,l;γ,α
r2(d−2)
r2k+2l+4
(λkλlxγxαJk,αJl,γ − 2r2(kxγJk,α∂βJl,γ + lxαJl,γ∂γJk,α) + r4(∂γJk,α∂βJl,γ)
)=∑k,l
r2(d−2)
r2(k+l+2)
(λkλl(
~Jk · ~r)( ~JJl · ~r) + r2Bk+l+4−2d
)(2.45)
=∑k,l
r2(d−2)
r2(k+l+2)
(r4 div ~Jk div ~Jl + r2Bk+l+4−2d
). (2.46)
So that Q(u) has numerators in Zk+l+2|2d−2. So 4−1(Q(u)) has numerators in Zk+l|2d−4
expect top harmonic terms are added that come from the non-local part of the operator.
Now writing u · ∇u we get for the asymptotic terms
[u · ∇u]α =∑k,l
rd−2Jl,βr2l
∂βrd−2Jk,αr2k
=∑k,l
r2(d−2)
r2(k+l+1)(−λkJl,βxβJk,α + r2Jl,β∂βJk,α)
=∑k,l
r2(d−2)
r2(k+l+1)(−λk( ~Jl · ~x)Jk,α + r2 ~Jl · ∇Jk,α)
=∑k,l
r2(d−2)
r2(k+l+1)(r2λkλ
−1l div ~JlJk,α + r2 ~Jl · ∇Jk,α)
And so the numerator above lies in Zk+l+1|2d−2. Therefore in ut = −u · ∇u+∇◦4−1 ◦ Q(u)
the modes between d− 2 and 2d− 4. So there are d− 3 modes preserved.
87
88
Appendices
89
Eulers equation in complex
coordinates
Using the complex structure of C = R2 we rewrite the Euler equation in terms of the
holomorphic projection of the vector field v. The vector field v = A(x, y)∂x + B(x, y)∂y in
R2 can be written as
v = a(z, z)∂z + b(z, z)∂z,
where a = A+ iB, b = a. Here we use following transformations
∂x = ∂z + ∂z, ∂y = i(∂z − ∂z).
The real vector field v is now completely determined by one complex function a(z, z) and we
call it the complex function associated to the field v. We will use the above expression to
write divergence and Lapalce operators in terms of the Cauchy operator ∂z and the complex
function a.
The divergence and the gradient operators. Direct computation for divergence gives following
expression:
div v = ∂xA+ ∂yB = (∂z + ∂z)A+ i(∂z − ∂z)B
= ∂z(A+ iB) + ∂z(A− iB) = ∂za+ ∂zb
= 2Re(∂za)
89
Similar computation gives
v · ∇v = (aaz + aaz)∂z + (aaz + aaz)∂z
and applying divergence div a = 2 Re az to this, we get
div (u · ∇u) = ∂z(aaz + aaz) + ∂z(aaz + aaz) (47)
= (az)2 + (az)
2 + 2azaz = 2(az)2 + 2|az|2
Now if p : R2 → R is a scalar function such as the pressure term in Euler equation then we
can write its gradient field as
∇p = (∂xp)∂x + (∂yp)∂y = px(∂z + ∂z) + py(i(∂z − ∂z))
= (px + ipy)∂z + (px − ipy)∂z
= 2pz∂z + 2pz∂z
Finally let us compute the curl of v as
curl v =1
2i
(az − az
). (48)
Note that for a = a(z, z) divergence free we obtain that curl v = −i ∂za.
Lemma .0.1. The Euler equation in R2 can be written in terms of complex function a :
C→ C and real pressure function p : R2 → R as
at + (aaz + aaz) = −2∂zp, div a = az + az = 0. (49)
Proof. This follows directly by rewriting the vectors ~u, ~u · ∇~u, and ∇p in the basis ∂z and
∂z and then comparing coefficients.
90
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