# Strong illposedness for SQG in critical Sobolev spaces

In-Jee Jeong Department of Mathematics and RIM, Seoul National University.
E-mail<EMAIL_ADDRESS>Junha Kim Department of Mathematics, Chung-ang
University. E-mail<EMAIL_ADDRESS>
###### Abstract

We prove that the inviscid surface quasi-geostrophic (SQG) equations are
strongly ill-posed in critical Sobolev spaces: there exists an initial data
$H^{2}(\mathbb{T}^{2})$ without any solutions in $L^{\infty}_{t}H^{2}$.
Moreover, we prove strong critical norm inflation for $C^{\infty}$–smooth
data. Our proof is robust and extends to give similar ill-posedness results
for the family of modified SQG equations which interpolate the SQG with two-
dimensional incompressible Euler equations.

##  1 Introduction

###  1.1 Main results

In this paper, we are concerned with the Cauchy problem for the inviscid
surface quasi-geostrophic (SQG) equations on
$\mathbb{T}^{2}=(\mathbb{R}/\mathbb{Z})^{2}$

$\left\\{\begin{aligned} \partial_{t}\theta+u\cdot\nabla\theta=0,\\\
u=\nabla^{\perp}(-\Delta)^{-\frac{1}{2}}\theta.\end{aligned}\right.$ (SQG)

Our first main result shows that strong norm inflation occurs for the solution
map of (SQG) in $H^{2}(\mathbb{T}^{2})$ with $C^{\infty}$–smooth solutions.

######  Theorem A (Strong norm inflation).

For any $\epsilon,\delta,A>0$, there exists $\theta_{0}\in
C^{\infty}(\mathbb{T}^{2})$ satisfying $\|\theta_{0}\|_{H^{2}\cap
W^{1,\infty}}<\epsilon$ such that the unique local-in-time smooth solution
$\theta$ to (SQG) with initial data $\theta_{0}$ exists on $[0,\delta^{*}]$
for some $0<\delta^{*}\leq\delta$ and satisfies

$\sup_{t\in[0,\delta^{*}]}\|\theta(t,\cdot)\|_{H^{2}}>A.$

The above result implies that the solution operator defined from $H^{2}\cap
C^{\infty}$ to $H^{2}$ by $\theta_{0}\mapsto\theta(t)$ for any $t>0$ cannot be
continuous at the trivial solution. On the other hand, the following result
shows that actually it is impossible to define the solution operator from
$H^{2}$ to $L^{\infty}_{t}H^{2}$.

######  Theorem B (Nonexistence).

For any $\epsilon>0$, there exists $\theta_{0}\in H^{2}\cap
W^{1,\infty}(\mathbb{T}^{2})$ satisfying

$\begin{split}\|\theta_{0}\|_{H^{2}\cap W^{1,\infty}}<\epsilon\end{split}$

such that there is _no_ solution to (SQG) with initial data $\theta_{0}$
belonging to $L^{\infty}([0,\delta];H^{2}(\mathbb{T}^{2}))$ with any
$\delta>0$.

######  Remark 1.1.

We give a few remarks relevant to the statements above.

  * •

With a rather straightforward modification of our proof, the space $H^{2}$ in
Theorems A and B can be replaced with $W^{1+\frac{2}{p},p}$ with any $p>1$.
Later, we shall sketch the proof in the endpoint case $p=\infty$. Moreover,
the domain $\mathbb{T}^{2}$ can be replaced with $\mathbb{R}^{2}$ or bounded
domains having symmetry axes.

  * •

The initial data for which non-existence occur can be given explicitly; see
(4.4) below.

  * •

The arguments we present can be adapted to prove ill-posedness for the case of
modified (and logarithmically regularized) SQG equations; see Subsection 1.3
below.

###  1.2 Well-posedness theory for SQG

To put the above ill-posedness results into context, let us briefly recall the
well-posedness theory for the SQG equation. Depending on the regularity of the
solutions considered, one has the following categories:

  * •

Strong solutions: local existence and uniqueness. Using the Kato–Ponce
commutator estimate, one obtains the following a priori estimate

$\begin{split}\frac{d}{dt}\|\theta\|_{H^{s}}\leq C\|\nabla
u\|_{L^{\infty}}\|\theta\|_{H^{s}}\end{split}$

for a solution of (SQG), which allows one to close
$\|\theta(t)\|_{H^{s}}\lesssim\|\theta_{0}\|_{H^{s}}$ for
$t\lesssim\|\theta_{0}\|_{H^{s}}^{-1}$ once $s>2$, using that $\|\nabla
u\|_{L^{\infty}}\lesssim\|\theta\|_{H^{s}}$. Similarly, $H^{s}$ can be
replaced with $W^{s,p}$, as long as $s>1+\frac{2}{p}$. Based on this a priori
estimate, one can prove local existence and uniqueness of a strong solution in
the class $L^{\infty}_{t}W^{s,p}$ with $s>1+\frac{2}{p}$. On the other hand,
note that the borderline inequality $\|\nabla
u\|_{L^{\infty}}\lesssim\|\theta\|_{H^{2}}$ fails; this makes the Sobolev
space $H^{2}$ (and similarly $W^{1+2/p,p}$) critical for local well-posedness.
This space is also scaling-critical: the critical norm is left invariant under
the transformation

$\begin{split}\theta(t,x)\mapsto\lambda^{-1}\theta(t,\lambda x),\qquad
u(t,x)\mapsto\lambda^{-1}u(t,\lambda x).\end{split}$ (1.1)

While not much is known for long-time dynamics of (SQG), see a recent
breakthrough of He–Kiselev [27] (and references therein) for a construction of
smooth initial data with Sobolev norms growing at least exponentially for all
times. Moreover, existence of traveling-wave solutions ([36, 7]) and rotating
solutions ([26, 13, 8]) are known.

  * •

Weak solutions: global existence. Global existence of $L^{p}$–weak solutions
is known, thanks to the works [41, 38, 2]. While such solutions are in general
expected to be non-unique, see [12] for a uniqueness result for patches. On
the other hand, for “very” weak solutions, non-uniqueness has been
established–see [6, 11]. Note the gap of regularity between week and strong
solutions.

  * •

Ill-posedness in $W^{1,\infty}$. To the best of knowledge, the only critical
space ill-posedness result concerning (SQG) is the one given in [23] for
$W^{1,\infty},$ where a powerful general method for proving ill-posedness of
active scalar systems in $L^{\infty}$–type spaces is developed. To be precise,
the authors show that ([23, Section 9.2]) there exist smooth steady states
$\bar{\theta}$ and a sequence of perturbations
$\tilde{\theta}_{0}^{(\epsilon)}$ ($\epsilon\to 0^{+}$) so that the associated
(SQG) solution $\theta^{(\epsilon)}$ with data
$\bar{\theta}+\tilde{\theta}_{0}^{(\epsilon)}$ satisfies

$\begin{split}\|\theta^{(\epsilon)}(0,\cdot)-\bar{\theta}\|_{W^{1,\infty}}<\epsilon,\qquad\sup_{0<t<\epsilon}\|\theta^{(\epsilon)}(t,\cdot)-\bar{\theta}\|_{W^{1,\infty}}>c\end{split}$

where $c>0$ depends only on $\bar{\theta}$. It is very interesting to note
that the authors use well-posedness in critical Besov spaces with summability
index 1. Such Besov well-posedness theory goes back to the pioneering work
[42]. Our result (which applies in the $W^{1,\infty}$ case as well) basically
says that one can take $\bar{\theta}\equiv 0$ and replace $c$ by
$\epsilon^{-1}$. On the other hand, one can restore well-posedness in
$W^{1,\infty}$ by assuming some rotational symmetry and anisotrophic Hölder
regularity [22].

The current work settles the issue of strong ill-posedness of (SQG) at
critical Sobolev spaces, and we believe that this could be a first step in
understanding the dynamics of “slightly” supercritical and subcritical
solutions (e.g. evolution of $H^{s}$–data with $|s-2|\ll 1$), thereby bridging
the gap between the theory of weak and strong solutions. Indeed, in a very
recent work of Elgindi on singularity formation for the 3D Euler equations
[18], one of the key steps was to understand precisely the mechanism of
$C^{1}$–illposedness. Closing this section, let us mention some interesting
works which seem contradictory to our main results:

  * •

Miura [40] proved that the fractionally dissipative SQG system

$\left\\{\begin{aligned}
\partial_{t}\theta+u\cdot\nabla\theta+(-\Delta)^{\beta}\theta=0,\\\
u=\nabla^{\perp}(-\Delta)^{-\frac{1}{2}}\theta,\end{aligned}\right.$ (1.2)

is actually well-posed in the critical Sobolev space $H^{2-2\beta}$ for all
$\beta>0$ (for data of any size), and this seems to suggest $H^{2}$ well-
posedness of the inviscid system by taking $\beta\to 0$! See [35, 30, 31] for
related recent advances.

  * •

An invariant measure defined on $H^{2}(\mathbb{T}^{2})$ which guarantees
global well-posedness in $L^{\infty}_{t}H^{2}$ for any initial data in the
support of the measure was constructed in [24]. The data in Theorem B
certainly does not belong to the support of such a measure.

###  1.3 Generalized SQG equations

In the recent years, there has been significant interest in the study of so-
called generalized SQG equations, given by

$\left\\{\begin{aligned} \partial_{t}\theta+u\cdot\nabla\theta=0,\\\
u=\nabla^{\perp}P(\Lambda)\theta,\end{aligned}\right.$ (1.3)

where $P(\Lambda)$ is some Fourier multiplier, with
$\Lambda=(-\Delta)^{\frac{1}{2}}$. Two distinguished cases are
$P(\Lambda)=\Lambda^{-1}$ (SQG) and $P(\Lambda)=\Lambda^{-2}$ (2D
incompressible Euler). Of particular interest is the case of $\alpha$–SQG
systems given by $P(\Lambda)=\Lambda^{-\alpha}$ with $1\leq\alpha\leq 2$,
which interpolates the SQG and 2D Euler equations. The $L^{2}$–based critical
Sobolev space is then given by $H^{3-\alpha}$, and let us point out that the
methods developed in the current work can handle the entire range
$1\leq\alpha\leq 2$ without any essential change in the proof. One could
consider more general symbols such as
$P(\Lambda)=\Lambda^{-\alpha}\log^{-\gamma}(10+\Lambda)$ with $\gamma>0$,
which give rise to the so-called logarithmically regularized systems ([10, 9,
17]). It is known that if the power of the logarithm is sufficiently large,
then one can restore well-posedness in $H^{3-\alpha}$ ([10]), but at this
point it is more appropriate to regard a logarithmically singularized Sobolev
space to be critical. Indeed, one can see from our proof that there is a
“logarithmic” room111strictly speaking, some power of the logarithm in the
arguments and therefore the same proof can cover same ill-posedness results in
the slightly logarithmically regularized systems. We shall not dwell on this
issue any further.

###  1.4 Critical space illposedness for Euler

It should be emphasized that the strong Sobolev ill-posedness statements,
Theorems A and B, were first established in a groundbreaking work of Bourgain
and Li [4, 5], for the case of 2D and 3D Euler equations, respectively.
Further developments, including the current work, seem to have been strongly
inspired by these papers. Recently, Kwon [34] settled the problem of strong
ill-posedness in $H^{1}$ for logarithmically regularized (strictly speaking,
powers of the log less than $1/2$) 2D Euler equations, nicely complementing
previous $H^{1}$ well-posedness from [10]. On the other hand, much simpler
proofs of $H^{1}$ ill-posedness for 2D Euler, which also shows continuous-in-
time degeneration of the solution in Sobolev spaces, have appeared in [20,
29]. Some details of these simplified arguments will be given in the next
section.

##  2 Ingredients of the proof

The purpose of this section is to sketch the main ingredients of the proof.
Several key ideas have already appeared in earlier works establishing ill-
posedness in the Euler case; we briefly review those in Subsection 2.1.
Additional difficulties arising in the (generalized) SQG case and new ideas
are covered then in Section 2.2.

###  2.1 Strategy in the Euler case

In this section, let us give an overview of the ill-posedness proof in the 2D
Euler case. We recall that in $\mathbb{T}^{2}$, the Euler equations are given
by

$\left\\{\begin{aligned} \partial_{t}\omega+u\cdot\nabla\omega=0,\\\
u=\nabla^{\perp}(-\Delta)^{-1}\omega.\end{aligned}\right.$ (Euler)

In terms of $\omega$, the critical $L^{2}$–based Sobolev space is
$H^{1}(\mathbb{T}^{2})$; indeed, $\omega\in H^{1}$ barely fails to guarantee
$\nabla u\in L^{\infty}$, which is necessary to close the a priori estimate in
$H^{1}$.

Choice of data for Euler. As a starting point of discussion, we present an
interesting identity observed by T. Elgindi:

$\begin{split}\frac{d}{dt}(\|\partial_{2}\omega\|_{L^{2}}^{2}-\|\partial_{1}\omega\|_{L^{2}}^{2})=\frac{1}{2}\int_{\mathbb{T}^{2}}\partial_{1}u_{1}((\partial_{2}\omega)^{2}+(\partial_{1}\omega)^{2})+\omega\partial_{1}\omega\partial_{2}\omega\,\mathrm{d}x.\end{split}$
(2.1)

For $\omega_{0}\in L^{\infty}$, Yudovich theory provides a unique global
solution in $L^{\infty}([0,\infty)\times L^{\infty}$, and note that the last
term in (2.1) cannot contribute to a large growth of the $H^{1}$–norm in a
small time interval. Therefore, to prove existence of an $H^{1}\cap
L^{\infty}$–initial data $\omega_{0}$ which “escapes” $H^{1}$ instantaneously,
the goal would be to find $\omega_{0}\in H^{1}\cap L^{\infty}$ such that

$\begin{split}\int_{0}^{t}\int_{\mathbb{T}^{2}}\partial_{1}u_{1}(\partial_{2}\omega)^{2}\,\mathrm{d}x=+\infty\end{split}$
(2.2)

for any $t>0$, where $\omega$ is the Yudovich solution with data $\omega_{0}$
and $u=\nabla^{\perp}(-\Delta)^{-1}\omega.$ In particular, it is necessary
that at the initial time, we have

$\begin{split}\int_{\mathbb{T}^{2}}\partial_{1}u_{0,1}(\partial_{2}\omega_{0})^{2}\,\mathrm{d}x=+\infty.\end{split}$
(2.3)

The choice in [20] was

$\begin{split}\omega_{0}(x)\simeq\frac{x_{1}x_{2}}{|x|^{2}}|\ln|x||^{-\alpha},\qquad|x|\ll
1\end{split}$ (2.4)

since then ([3, 16])

$\begin{split}\partial_{1}u_{0,1}(x)\simeq|\ln|x||^{1-\alpha},\qquad|x|\ll
1\end{split}$ (2.5)

which in particular guarantees (2.3) for a range of $\alpha>0$.

Hyperbolic flow. Geometrically, vorticity which is positive on the first
quadrant and odd with respect to both axes (as in (2.4)) induces velocity
which is stretching in the $x_{1}$-direction and contracting in the other,
which leads to squeezing of the vorticity near the $x_{1}$-axis and growth of
$H^{1}$–norm. This so-called “hyperbolic flow scenario” has been used to
produce Euler solutions with gradient growth–see [33, 46, 43, 14, 16, 15, 21].
Flattening of the vorticity level sets in such a flow configuration was
studied in detail in [47, 28].

Regularization effect. The main task is then to ensure that the velocity
field, for a small time interval, retains its logarithmic divergence near the
origin: indeed, instantaneous blow-up of the $H^{1}$–norm is not too difficult
to see for the passive transport equation

$\begin{split}\partial_{t}\omega+u_{0}\cdot\nabla\omega=0,\end{split}$

by solving the equation along the flow generated by $u_{0}$. When one tries to
replace $u_{0}$ by $u$, a fundamental difficulty arises: anisotropic
stretching of the vorticity regularizes the velocity. Indeed, rather involved
computations in [19] suggests the asymptotics $\|\nabla
u(t)\|_{L^{\infty}}\lesssim t^{-1}$, which is barely non-integrable in time;
this indicates that it could be a very delicate problem to verify (2.2). This
upper bound of $t^{-1}$ can be seen for instance by solving the passive
problem above and re-calculating the associated velocity at later times.

Key Lemma and Lagrangian approach. Towards the goal of obtaining a lower bound
on the velocity gradient $|\nabla u(t)|\gtrsim t^{-1}$, one needs to have a
robust way of estimating the velocity gradient and prove some “stability” of
the initial data. Regarding the former, the celebrated Key Lemma of
Kiselev–Sverak asserts that (stated roughly)

$\begin{split}\frac{u_{1}(x)}{x_{1}}\simeq\int_{[x_{1},1]\times[x_{2},1]}\frac{y_{1}y_{2}}{|y|^{4}}\omega(y)\,\mathrm{d}y,\end{split}$
(2.6)

for $\omega\in L^{\infty}$ with odd-odd symmetry. Note that $u_{1}=0$ for
$x_{1}=0$ by symmetry, so that the left hand side is an approximation of
$\partial_{1}u_{1}$. The lower bound of the form (2.6) has proven to be
extremely powerful in establishing growth of the vorticity ([33, 46, 43, 32,
25, 18, 27]). It is interesting to note that Bourgain–Li independently derived
similar lower bounds in [4]. Next, regarding the issue of showing stability of
the data, the key observation is the hierarchy of vortex dynamics expressed in
(2.6): the vorticity around a point $x$ is being affected mainly by the
vorticity supported in $|y|\geq 2|x|$. This suggests that the chunk of
vorticity supported far away from the origin is more stable, thereby
contributing to the right hand side of (2.6) for a longer time interval, to
squeeze the vorticity closer to the origin. The proof of such stability and
squeezing phenomena should be done in the Lagrangian variable, using the
transport formulas

$\begin{split}\omega(t,x)=\omega_{0}(\Phi_{t}^{-1}x),\qquad\nabla\omega(t,x)=\nabla\omega_{0}(\Phi_{t}^{-1}(x))\nabla\Phi_{t}^{-1}(x)\end{split}$

where $\Phi_{t}$ is the flow map at time $t$. In the actual ill-posedness
proofs, Lagrangian versions of the formula (2.1) are used.

###  2.2 Difficulties in the SQG case

Overall, the strategy of the ill-posedness proof in the SQG case is similar to
that explained in the above for 2D Euler. Roughly speaking, the initial data
is now modified to be

$\begin{split}\theta_{0}\simeq\frac{x_{1}x_{2}}{|x|}|\ln|x||^{-\alpha},\qquad|x|\ll
1,\end{split}$

which is odd-odd and nonnegative in the first quadrant. The associated SQG
velocity then satisfies the asymptotics (2.5) with strong hyperbolicity near
the origin, which should stretch $\theta$ near the $x_{1}$-axis. The issue is
whether such a stretching effect is sufficiently strong to remove $\theta$
from the critical Sobolev space it started from. Let us now explain some main
differences with the Euler case and new ideas employed to handle those.

While the equation for $\theta$ in (SQG) is simply the transport equation
exactly as in the 2D Euler case, probably the most significant difference is
that while the $L^{\infty}$–norm is the common strongest conservation law, it
is critical for 2D Euler but one order weaker for SQG. Furthermore, there is
global well-posedness for 2D Euler with $\omega_{0}\in L^{\infty}$ (Yudovich
[44]), and the associated sharp estimates given by Yudovich theory have been
very useful in understanding the dynamics.222Even in the 3D Euler case,
Bourgain–Li [5] actually carefully identifies a class of initial data for
which $\omega\in L^{\infty}$ propagates locally in time. Then, one can prove
and utilize estimates similar to Yudovich’s in 3D. On the other hand, the
corresponding quantity in the SQG case, $\|\nabla\theta\|_{L^{\infty}}$, blows
up together with $\|\theta\|_{H^{2}}$.

It seems that the only way to handle this issue is to rely entirely on a
contradiction argument–we assume that there is an
$L^{\infty}([0,T];H^{2})$–solution, and then prove that for any $t>0$, the
$H^{2}$–norm of the solution must be actually infinite. The whole point in
this contradiction argument is that we can use the hypothetical $H^{2}$–bound
to control the solution, an idea originated in [4]. Again, the difficulty in
the SQG case is that this hypothetical $H^{2}$ control is the only useful
bound, whereas in the Euler case one has both $H^{1}$ and $L^{\infty}$
control. Fortunately, it turns out that having an $H^{2}$–bound guarantees
that the velocity is log-Lipschitz, which implies in particular uniqueness in
the class $L^{\infty}_{t}H^{2}$ (this guarantees propagation in time of odd-
odd symmetry and non-negativity) and existence of the flow map. That is, an
$L^{\infty}_{t}H^{2}$–solution is Lagrangian, and therefore we can apply
transport formulas to understand the dynamics.

Under the contradiction hypothesis, the main part of the argument is to derive
and apply a version of the Key Lemma adapted to the SQG case. Series technical
difficulties appear; to begin with, in the remainder estimate of the Key Lemma
(see estimates (3.1) and (3.2) in Lemma 3.2) we are only allowed to use
$\theta\in H^{2}$. As a consequence, the remainder term blows up super-
logarithmically (the power 3/2 in (3.2)) as the point $x$ approaches the axes,
whereas only logarithmic errors are allowed in the nonexistence proof. It
seems that the only way to overcome this issue is to track carefully the
geometry of the support of $\theta$ in time so that the problematic remainder
term disappears. To achieve this, we replace $\theta_{0}$ with a disjoint
union of dyadic “bubbles” satisfying the same asymptotics as $|x|\to 0$ (see
(4.4) below) and obtain detailed information on the location of these bubbles
for an interval of time inductively, starting from the largest one. Such
refined information appear in technical Claims I, II, III in the proof. In the
context of controlling bubbles, another significant difference with Euler is
that the “self-interaction” of a bubble is not a bounded term anymore. To
overcome this issue we need to track the location of the “top point” of each
bubble, which is the slowest point but does not suffer from self-interactions.

Closing this section, we remark that the versions of the Key Lemma derived in
this work should be useful in improving previous growth results for the active
scalar equations, as we handle the remainder term only with the critical
quantity.

###  2.3 Organization of the paper

The rest of this paper is organized as follows. The main technical tool, which
we shall refer to as Key Lemma, is stated and proved in Section 3. After that,
the proofs of Theorems B and A are given in Sections 4 and 5.

##  3 The Key Lemma

To begin with, we recall the famous Hardy’s inequality.

######  Lemma 3.1 (Hardy’s inequality).

Let $f$ be a smooth function defined on the interval $(0,l)$ with $f(0)=0$.
Then we have

$\begin{split}\|x^{-1}f(x)\|_{L^{2}(0,l)}\leq 2\|\partial
f(x)\|_{L^{2}(0,l)}\end{split}$

and

$\begin{split}\|x^{-2}f(x)\|_{L^{2}(0,l)}^{2}\leq
2\|\partial^{2}f(x)\|_{L^{2}(0,l)}^{2}\end{split}$

for all $l\in[0,1]$.

###### Proof.

By the fundamental theorem of calculus and the assumption for $f$, we see for
a.e. $x_{2}\in[0,z_{2}]$ that

$\displaystyle\int_{0}^{l}\frac{f(x)^{2}}{x^{2}}\,\mathrm{d}x$
$\displaystyle=-\frac{1}{l}f(l)^{2}+2\int_{0}^{l}\frac{f(x)}{x}\partial
f(x)\,\mathrm{d}x$ $\displaystyle\leq 2\int_{0}^{l}\frac{f(x)}{x}\partial
f(x)\mathrm{d}x.$

Using Hölder’s inequality and Young’s inequality, we have

$\int_{0}^{l}\frac{f(x)^{2}}{x^{2}}\,\mathrm{d}x\leq 4\int_{0}^{l}|\partial
f(x)|^{2}\,\mathrm{d}x.$

Similarly, we have for a.e. $x_{2}\in[0,z_{2}]$ that

$\displaystyle\int_{0}^{l}\frac{f(x)^{2}}{x^{4}}\,\mathrm{d}x$
$\displaystyle=-\frac{1}{3l^{3}}f(l)^{2}+\frac{2}{3}\int_{0}^{l}\frac{f(x)}{x^{3}}\partial
f(x)\,\mathrm{d}x\leq\frac{2}{3}\int_{0}^{l}\frac{f(x)}{x^{2}}\frac{\partial
f(x)}{x}\mathrm{d}x\leq\frac{1}{2}\int_{0}^{l}\frac{\partial
f(x)^{2}}{x^{2}}\,\mathrm{d}x.$

Applying the above estimate, we obtain

$\int_{0}^{l}\frac{f(x)^{2}}{x^{4}}\,\mathrm{d}x\leq
2\int_{0}^{l}|\partial^{2}f(x)|^{2}\,\mathrm{d}x.$

This completes the proof. ∎

We shall now state and prove the Key Lemma. For convenience, we shall
normalize the SQG Biot–Savart law in a way that

$u(t,x)=\sum_{n\in\mathbb{Z}^{2}}\int_{[-1,1)^{2}}\frac{(x-(y+2n))^{\perp}}{|x-(y+2n)|^{3}}\theta(t,y)\,\mathrm{d}y$

holds.

######  Lemma 3.2.

We impose the following assumptions on $\theta\in H^{2}$:

  * •

$\theta$ is odd with respect to both axes, i.e.,
$\theta(x)=-\theta(\bar{x})=\theta(-x)=-\theta(\tilde{x})$ where
$\bar{x}:=(x_{1},-x_{2})$ and $\tilde{x}:=(-x_{1},x_{2})$;

  * •

$\theta$ vanishes near the axis; to be precise, for any $x\neq(0,0)$
satisfying either $x_{1}=0$ or $x_{2}=0$, there exists an open neighborhood of
$x$ such that $\theta$ vanishes.

Then, for any $x$ satisfying $|x|<1/4$ and $x_{1}>x_{2}>0$, we have

$\left|\frac{u_{1}(x)}{x_{1}}-12\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq
B_{1}(x)$ (3.1)

and

$\left|\frac{u_{2}(x)}{x_{2}}+12\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq\left(1+\log\frac{x_{1}}{x_{2}}\right)B_{2}(x)+\left(1+\log\frac{x_{1}}{x_{2}}\right)^{\frac{3}{2}}B_{3}(x),$
(3.2)

where $Q(x):=[2x_{1},1]\times[0,1]$ and $B_{1},B_{2},B_{3}$ satisfy

$|B_{1}(x)|+|B_{2}(x)|\leq
C(\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}+\|\theta\|_{L^{\infty}([0,1]^{2})})$

and

$|B_{3}(x)|\leq
C(\|\nabla\theta\|_{L^{2}(R(x))}+\|y_{2}^{-1}\partial_{1}\theta(y)\|_{L^{2}(R(x))}),\qquad
R(x):=[x_{1}/2,2x_{1}]\times[2x_{2},1].$

######  Remark 3.3.

We clearly have that

$\|y_{2}^{-1}\partial_{1}\theta(y)\|_{L^{2}(R(x))}\leq
2\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

###### Proof.

We fix a point $x=(x_{1},x_{2})$ satisfying the assumptions of the lemma.
After a symmetrization, we have

$u(x)=\sum_{n\in\mathbb{Z}^{2}}\int_{[0,1]^{2}}\bigg{(}\frac{(x-(y+2n))^{\perp}}{|x-(y+2n)|^{3}}-\frac{(x-(\tilde{y}+2n))^{\perp}}{|x-(\tilde{y}+2n)|^{3}}\\\
+\frac{(x-(-y+2n))^{\perp}}{|x-(-y+2n)|^{3}}-\frac{(x-(\bar{y}+2n))^{\perp}}{|x-(\bar{y}+2n)|^{3}}\bigg{)}\theta(y)\,\mathrm{d}y.$
(3.3)

We estimate $u_{1}$ first. Consider

$\displaystyle I_{1}(n)$
$\displaystyle:=-\int_{[0,1]^{2}}\left(\frac{x_{2}-(y_{2}+2n_{2})}{|x-(y+2n)|^{3}}-\frac{x_{2}-(y_{2}+2n_{2})}{|x-(\tilde{y}+2n)|^{3}}\right)\theta(y)\,\mathrm{d}y,$
$\displaystyle I_{2}(n)$
$\displaystyle:=-\int_{[0,1]^{2}}\left(\frac{x_{2}-(-y_{2}+2n_{2})}{|x-(-y+2n)|^{3}}-\frac{x_{2}-(-y_{2}+2n_{2})}{|x-(\bar{y}+2n)|^{3}}\right)\theta(y)\,\mathrm{d}y$

so that from (3.3)

$u_{1}(x)=\sum_{n\in\mathbb{Z}^{2}}(I_{1}(n)+I_{2}(n)).$

We think of the cases $n=0$ and $n\neq 0$ separately. For $n\neq 0$, we see
that

$|I_{1}(n)+I_{1}(\tilde{n})|\leq
O(|n|^{-4})\|\theta\|_{L^{\infty}([0,1]^{2})}x_{1}$

and

$|I_{2}(n)+I_{2}(\bar{n})|\leq
O(|n|^{-4})\|\theta\|_{L^{\infty}([0,1]^{2})}x_{1}.$

Therefore,

$\Bigg{|}\sum_{n\in\mathbb{Z}^{2}\setminus\\{0\\}}(I_{1}(n)+I_{2}(n))\Bigg{|}\leq
Cx_{1}\|\theta\|_{L^{\infty}([0,1]^{2})}.$ (3.4)

We now estimate the case of $n=0$. Using

$\frac{1}{A^{3}}-\frac{1}{B^{3}}=\frac{(B^{2}-A^{2})(A^{2}+AB+B^{2})}{A^{3}B^{3}(A+B)},$
(3.5)

we have

$I_{1}(0)=-4x_{1}\int_{[0,1]^{2}}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y.$

Noting that
$[0,1]^{2}=Q(x)\cup[0,2x_{1}]\times[2x_{1},1]\cup[0,2x_{1}]\times[0,2x_{1}]$,
we estimate the integral for each set.

(i) $Q(x)$

In the case of $y\in Q(x)$, we can show that

$\begin{split}\frac{1}{4}|y|\leq|x-y|\leq|y|,\qquad\frac{1}{2}|y|\leq|x-\tilde{y}|\leq
2|y|\end{split}$ (3.6)

because the first inequality comes from

$|x-y|^{2}\geq|x_{1}-y_{1}|^{2}\geq\frac{1}{4}y_{1}^{2}\geq\frac{1}{8}|y|^{2},\qquad
y_{1}\geq y_{2}$

and

$|x-y|^{2}=|x_{1}-y_{1}|^{2}+|x_{2}-y_{2}|^{2}\geq\frac{1}{4}y_{1}^{2}+\frac{1}{4}y_{2}^{2},\qquad
y_{1}\leq y_{2}.$

The goal is to prove that

$\begin{split}-\int_{Q(x)}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y=:J\end{split}$

satisfies

$\begin{split}\left|J-\frac{3}{2}\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}}.\end{split}$ (3.7)

We separate

$J=J_{1}+J_{2}$

where

$\begin{split}J_{1}:=\int_{Q(x)}\frac{y_{1}y_{2}(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\mathrm{d}y\\\
\end{split}$

and

$\begin{split}J_{2}:=-\int_{Q(x)}\frac{y_{1}x_{2}(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\mathrm{d}y.\end{split}$

Using (3.6), we may estimate

$\begin{split}|J_{2}|\leq
C|x|\int_{Q(x)}\frac{1}{|y|^{2}}\frac{|\theta(y)|}{|y|^{2}}\,\mathrm{d}y.\end{split}$

Note that by Hölder’s inequality,

$|x|\int_{Q(x)}\frac{1}{|y|^{2}}\frac{|\theta(y)|}{|y|^{2}}\,\mathrm{d}y\leq|x|\left(\int_{2x_{1}}^{\infty}\frac{1}{r^{3}}\,\mathrm{d}r\right)^{\frac{1}{2}}\||y|^{-2}\theta(y)\|_{L^{2}(Q(x))}\leq
C\||y|^{-2}\theta(y)\|_{L^{2}([0,1]^{2})}.$

Then with the Hardy’s inequality we have

$|J_{2}|\leq C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

On the other hand, regarding $J_{1}$, we shall show that

$\begin{split}\left|J_{1}-\frac{3}{2}\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.\end{split}$

For this purpose we simply write $J_{1}=J_{11}+J_{12}+J_{13}$ where

$\begin{split}J_{11}&=\int_{Q(x)}\frac{y_{1}y_{2}|x-y|^{2}}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y,\\\
J_{12}&=\int_{Q(x)}\frac{y_{1}y_{2}|x-y||x-\tilde{y}|}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y,\\\
J_{13}&=\int_{Q(x)}\frac{y_{1}y_{2}|x-\tilde{y}|^{2}}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y\end{split}$

and show that

$\begin{split}\left|J_{1k}-\frac{1}{2}\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}\end{split}$

for each $k=1,2,3$. We supply the proof only for the case $k=1$, since the
others can be treated similarly. We directly compute

$\begin{split}J_{11}-\frac{1}{2}\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y&=\int_{Q(x)}y_{1}y_{2}\frac{2|y|^{5}-|x-y||x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}{2|y|^{5}|x-y||x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y.\end{split}$

We rewrite the numerator as

$\displaystyle 2|y|^{5}-|x-y||x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)$
$\displaystyle=(|y|^{2}-|x-y|^{2})|y|^{3}+|x-y|^{2}(|y|^{3}-|x-\tilde{y}|^{3})+(|y|-|x-y|)|y|^{4}+|x-y|(|y|^{4}-|x-\tilde{y}|^{4})$

and further rewriting

$|y|-|x-y|=\frac{|y|^{2}-|x-y|^{2}}{|y|+|x-y|},\qquad|y|^{3}-|x-\tilde{y}|^{3}=\frac{(|y|^{2}-|x-\tilde{y}|^{2})(|y|^{2}+|y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|y|+|x-\tilde{y}|},$

we see using (3.6) that

$\left|2|y|^{5}-|x-y||x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)\right|\leq
C|x||y|^{4}.$

Then, we can infer that

$\left|J_{11}-\frac{1}{2}\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq
C|x|\int_{Q(x)}\frac{1}{|y|^{2}}\frac{|\theta(y)|}{|y|^{2}}\,\mathrm{d}y\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

Collecting the estimates for $J_{1}$ and $J_{2}$ gives (3.7).

(ii) $[0,2x_{1}]\times[2x_{1},1]$

In the case of $y\in[0,2x_{1}]\times[2x_{1},1]$, using $y_{1}\leq y_{2}$, we
can see that

$\frac{1}{2}{y_{2}}\leq|x-y|\leq
2y_{2},\qquad\frac{1}{2}{y_{2}}\leq|x-\tilde{y}|\leq 2y_{2}.$

Thus, Hölder’s inequality and Hardy’s inequality imply that

$\displaystyle\left|-\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y\right|$
(3.8) $\displaystyle\hphantom{\qquad\qquad\quad}\leq
C\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{1}{y_{2}}\frac{\theta(y)}{y_{2}^{2}}\,\mathrm{d}y$
$\displaystyle\hphantom{\qquad\qquad\quad}\leq
C\left(\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{1}{y_{2}^{2}}\,\mathrm{d}y\right)^{\frac{1}{2}}\|y_{2}^{-2}\theta(y)\|_{L^{2}([0,1]^{2})}$
$\displaystyle\hphantom{\qquad\qquad\quad}\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

(iii) $[0,2x_{1}]^{2}$

Due to $\theta(y_{1},0)=0$, using integration by parts gives

$\displaystyle-\int_{[0,2x_{1}]^{2}}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y$
$\displaystyle\hphantom{\qquad\qquad}=\frac{1}{x_{1}}\int_{[0,2x_{1}]^{2}}\left(\frac{1}{|x-y|}-\frac{1}{|x-\tilde{y}|}\right)\partial_{2}\theta(y)\,\mathrm{d}y$
$\displaystyle\hphantom{\qquad\qquad\qquad\qquad}-\frac{1}{x_{1}}\int_{0}^{2x_{1}}\left(\frac{1}{|(x_{1}-y_{1},x_{2}-2x_{1})|}-\frac{1}{|(x_{1}+y_{1},x_{2}-2x_{1})|}\right)\theta(y_{1},2x_{1})\,\mathrm{d}y_{1}.$

By Hölder’s inequality we estimate the second integral as

$\left|-\frac{1}{x_{1}}\int_{0}^{2x_{1}}\left(\frac{1}{|(x_{1}-y_{1},x_{2}-2x_{1})|}-\frac{1}{|(x_{1}+y_{1},x_{2}-2x_{1})|}\right)\theta(y_{1},2x_{1})\,\mathrm{d}y_{1}\right|\leq
Cx_{1}^{-1}\|\theta\|_{L^{\infty}([0,2x_{1}]^{2})}.$

We notice that since $\theta$ vanishes near the axis it follows

$\displaystyle\|\theta\|_{L^{\infty}([0,2x_{1}]^{2})}$
$\displaystyle\leq\sup_{y_{1}\in[0,2x_{1}]}\int_{0}^{2x_{1}}|\partial_{2}\theta(y_{1},y_{2})|\,\mathrm{d}y_{2}$
(3.9)
$\displaystyle\leq(2x_{1})^{\frac{1}{2}}\bigg{\|}\sup_{y_{1}\in[0,2x_{1}]}|\partial_{2}\theta(y_{1},\cdot)|\bigg{\|}_{L^{2}(0,2x_{1})}$
$\displaystyle\leq
2x_{1}\|\partial_{1}\partial_{2}\theta\|_{L^{2}([0,2x_{1}]^{2})}.$

Thus, we have

$\left|-\frac{1}{x_{1}}\int_{0}^{2x_{1}}\left(\frac{1}{|(x_{1}-y_{1},x_{2}-2x_{1})|}-\frac{1}{|(x_{1}+y_{1},x_{2}-2x_{1})|}\right)\theta(y_{1},2x_{1})\,\mathrm{d}y_{1}\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

Calculating the first integral with Hölder’s inequality, we see that

$\displaystyle\left|\frac{1}{x_{1}}\int_{[0,2x_{1}]^{2}}\left(\frac{1}{|x-y|}-\frac{1}{|x-\tilde{y}|}\right)\partial_{2}\theta(y)\,\mathrm{d}y\right|$
$\displaystyle\leq\frac{2}{x_{1}}\int_{[0,2x_{1}]^{2}}\frac{|\partial_{2}\theta(y)|}{|x-y|}\,\mathrm{d}y$
$\displaystyle\leq\frac{2}{x_{1}}\left(\int_{0}^{4x_{1}}r^{-\frac{1}{3}}\,\mathrm{d}r\right)^{\frac{3}{4}}\|\partial_{2}\theta\|_{L^{4}([0,2x_{1}]^{2})}$
$\displaystyle\leq
Cx_{1}^{-\frac{1}{2}}\|\nabla\theta\|_{L^{4}([0,2x_{1}]^{2})}.$

Gagliardo-Nirenberg interpolation inequality implies

$x_{1}^{-\frac{1}{2}}\|\nabla\theta\|_{L^{4}([0,2x_{1}]^{2})}\leq
Cx_{1}^{-\frac{1}{2}}\|\nabla^{2}\theta\|_{L^{2}([0,2x_{1}]^{2})}^{\frac{3}{4}}\|\theta\|_{L^{2}([0,2x_{1}]^{2})}^{\frac{1}{4}}+Cx_{1}^{-2}\|\theta\|_{L^{2}([0,2x_{1}]^{2})},$

where the constant $C>0$ is independent of $x_{1}$. Applying Hardy’s
inequality to it, we have

$\displaystyle
x_{1}^{-\frac{1}{2}}\|\partial_{2}\theta\|_{L^{4}([0,2x_{1}]^{2})}$
$\displaystyle\leq C\|\nabla^{2}\theta\|_{L^{2}([0,2x_{1}]^{2})},$

and hence it follows that

$\left|\frac{1}{x_{1}}\int_{[0,2x_{1}]^{2}}\left(\frac{1}{|x-y|}-\frac{1}{|x-\tilde{y}|}\right)\partial_{2}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

Combining the above estimates, we obtain that

$\left|-\int_{[0,2x_{1}]^{2}}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

We collect the estimates for each region and deduce that

$\displaystyle\left|\frac{I_{1}(0)}{x_{1}}-6\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

We can estimate

$I_{2}(0)=4x_{1}\int_{[0,1]^{2}}\frac{y_{1}(x_{2}+y_{2})(|x+y|^{2}+|x+y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x+y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\,\mathrm{d}y$

similarly with $I_{1}(0)$, resulting in the bound

$\displaystyle\left|\frac{I_{2}(0)}{x_{2}}-6\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

We omit the details. Thus we have (3.1).

Now we estimate $u_{2}$. Note that

$u_{2}(x)=\sum_{n\in\mathbb{Z}^{2}}(I_{3}(n)+I_{4}(n)),$

where

$\displaystyle I_{3}(n)$
$\displaystyle:=\int_{[0,1]^{2}}\left(\frac{x_{1}-(y_{1}+2n_{1})}{|x-(y+2n)|^{3}}-\frac{x_{1}-(y_{1}+2n_{1})}{|x-(\bar{y}+2n)|^{3}}\right)\theta(y)\,\mathrm{d}y,$
$\displaystyle I_{4}(n)$
$\displaystyle:=\int_{[0,1]^{2}}\left(\frac{x_{1}-(-y_{1}+2n_{1})}{|x-(-y+2n)|^{3}}-\frac{x_{1}-(-y_{1}+2n_{1})}{|x-(\tilde{y}+2n)|^{3}}\right)\theta(y)\,\mathrm{d}y.$

Since we can similarly see that

$\Bigg{|}\sum_{n\in\mathbb{Z}^{2}\setminus\\{0\\}}(I_{3}(n)+I_{4}(n))\Bigg{|}\leq
Cx_{2}\|\theta\|_{L^{\infty}([0,1]^{2})},$

it suffices to estimate for $n=0$. Using (3.5), we have

$I_{3}(0)=4x_{2}\int_{[0,1]^{2}}\frac{y_{2}(x_{1}-y_{1})(|x-y|^{2}+|x-y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x-y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\,\mathrm{d}y.$

We divide the domain into four regions as
$[0,1]^{2}=Q(x)\cup[0,x_{1}/2]\times[0,2x_{1}]\cup[x_{1}/2,2x_{1}]\times[0,2x_{1}]\cup[0,2x_{1}]\times[2x_{1},1]$
and estimate the integral in each region.

(i) $Q(x)$

In the case of $y\in Q(x)$, recalling that (3.6) holds, we can prove

$\begin{gathered}\left|\int_{Q(x)}\frac{y_{2}(x_{1}-y_{1})(|x-y|^{2}+|x-y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x-y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\,\mathrm{d}y+\frac{3}{2}\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\\\
\leq C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.\end{gathered}$

Since the proof is parallel to the one for (3.7), we omit it.

(ii) $[0,2x_{1}]\times[2x_{1},1]$

For $y\in[0,2x_{1}]\times[2x_{1},1]$, it follow that

$\frac{1}{2}{y_{2}}\leq|x-y|\leq 2y_{2},\qquad{y_{2}}\leq|x-\bar{y}|\leq
2y_{2}.$

Hence, we can show

$\left|\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{y_{2}(x_{1}-y_{1})(|x-y|^{2}+|x-y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x-y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}$

in a similar way with (3.8).

(iii) $[0,x_{1}/2]\times[0,2x_{1}]$

Since $y\in[0,x_{1}/2]\times[0,2x_{1}]$ implies

$\frac{1}{2}x_{1}\leq|x-y|\leq
4x_{1},\qquad\frac{1}{2}x_{1}\leq|x-\bar{y}|\leq 4x_{1},$

with $y_{2}\leq 2x_{1}$ we have

$\displaystyle\left|\int_{[0,x_{1}/2]\times[0,2x_{1}]}\frac{y_{2}(x_{1}-y_{1})(|x-y|^{2}+|x-y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x-y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\,\mathrm{d}y\right|$
$\displaystyle\hphantom{\qquad\qquad\quad}\leq
Cx_{1}^{-1}\int_{[0,x_{1}/2]\times[0,2x_{1}]}\frac{\theta(y)}{y_{2}^{2}}\,\mathrm{d}y$
$\displaystyle\hphantom{\qquad\qquad\quad}\leq
Cx_{1}^{-1}\left(\int_{[0,x_{1}/2]\times[0,2x_{1}]}1\,\mathrm{d}y\right)^{\frac{1}{2}}\|y_{2}^{-2}\theta(y)\|_{L^{2}([0,1]^{2})}$
$\displaystyle\hphantom{\qquad\qquad\quad}\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

(iv) $[x_{1}/2,2x_{1}]\times[0,2x_{1}]$

We claim that

$\int_{[x_{1}/2,2x_{1}]\times[0,2x_{1}]}\frac{y_{2}(x_{1}-y_{1})(|x-y|^{2}+|x-y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x-y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\,\mathrm{d}y=:K$

satisfies

$|K|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}\left(1+\log\frac{x_{1}}{x_{2}}\right)+C(\|\nabla^{2}\theta\|_{L^{2}(R(x))}+\|y_{2}^{-1}\partial_{1}\theta(y)\|_{L^{2}(R(x))})\left(1+\log\frac{x_{1}}{x_{2}}\right)^{\frac{3}{2}}.$
(3.10)

Using integration by parts, we have

$K=K_{1}+K_{2}+K_{3},$

where

$\displaystyle K_{1}$
$\displaystyle:=-\frac{1}{x_{2}}\int_{[x_{1}/2,2x_{1}]\times[0,2x_{1}]}\left(\frac{1}{|x-y|}-\frac{1}{|x-\bar{y}|}\right)\partial_{1}\theta(y)\,\mathrm{d}y,$
$\displaystyle K_{2}$
$\displaystyle:=\frac{1}{x_{2}}\int_{0}^{2x_{1}}\left(\frac{1}{|(x_{1},x_{2}-y_{2})|}-\frac{1}{|(x_{1},x_{2}+y_{2})|}\right)\theta(2x_{1},y_{2})\,\mathrm{d}y_{2},$
$\displaystyle K_{3}$
$\displaystyle:=-\frac{1}{x_{2}}\int_{0}^{2x_{1}}\left(\frac{1}{|(x_{1}/2,x_{2}-y_{2})|}-\frac{1}{|(x_{1}/2,x_{2}+y_{2})|}\right)\theta(x_{1}/2,y_{2})\,\mathrm{d}y_{2}.$

With (3.9) we may estimate $K_{2}$ as follows

$\displaystyle|K_{2}|$
$\displaystyle=\left|\int_{0}^{2x_{1}}\frac{4y_{2}\theta(2x_{1},y_{2})}{|(x_{1},x_{2}-y_{2})||(x_{1},x_{2}+y_{2})|(|(x_{1},x_{2}-y_{2})|+|(x_{1},x_{2}+y_{2})|)}\,\mathrm{d}y_{2}\right|$
(3.11) $\displaystyle\leq
Cx_{1}^{-2}\int_{0}^{2x_{1}}\theta(2x_{1},y_{2})\,\mathrm{d}y_{2}$
$\displaystyle\leq Cx_{1}^{-1}\|\theta\|_{L^{\infty}([0,2x_{1}]^{2})}$
$\displaystyle\leq C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

And similarly, it is obtained

$|K_{3}|\leq C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

Noting that

$K_{1}=\int_{[x_{1}/2,2x_{1}]\times[0,2x_{1}]}\frac{-4y_{2}\partial_{1}\theta(y)}{|x-y||x-\bar{y}|(|x-y|+|x-\bar{y}|)}\,\mathrm{d}y,$

we set

$K_{1}=K_{11}+K_{12},$

where

$\displaystyle K_{11}$
$\displaystyle:=\int_{[x_{1}/2,2x_{1}]\times[0,2x_{2}]}\frac{-4y_{2}\partial_{1}\theta(y)}{|x-y||x-\bar{y}|(|x-y|+|x-\bar{y}|)}\,\mathrm{d}y,$
$\displaystyle K_{12}$
$\displaystyle:=\int_{[x_{1}/2,2x_{1}]\times[2x_{2},2x_{1}]}\frac{-4y_{2}\partial_{1}\theta(y)}{|x-y||x-\bar{y}|(|x-y|+|x-\bar{y}|)}\,\mathrm{d}y.$

From $|x-\bar{y}|\geq x_{2}+y_{2}\geq y_{2}$ we have

$|K_{11}|\leq
C\int_{0}^{2x_{2}}\frac{\sup_{y_{1}\in[x_{1}/2,2x_{1}]}|\partial_{1}\theta(y_{1},y_{2})|}{x_{2}+y_{2}}\int_{0}^{2x_{2}}\frac{1}{|x-y|}\,\mathrm{d}y_{1}\,\mathrm{d}y_{2}$

and

$|K_{12}|\leq
C\int_{2x_{2}}^{2x_{1}}\frac{\sup_{y_{1}\in[x_{1}/2,2x_{1}]}|\partial_{1}\theta(y_{1},y_{2})|}{x_{2}+y_{2}}\int_{0}^{2x_{2}}\frac{1}{|x-y|}\,\mathrm{d}y_{1}\,\mathrm{d}y_{2}.$

By Gagliardo-Nirenberg interpolation inequality with $y_{2}\leq 2x_{1}$, we
can see that

$\displaystyle
y_{2}^{-\frac{1}{2}}{\sup_{y_{1}\in[x_{1}/2,2x_{1}]}|\partial_{1}\theta(y_{1},y_{2})|}$
$\displaystyle\leq
C(\|\partial_{1}^{2}\theta(\cdot,y_{2})\|_{L^{2}(x_{1}/2,2x_{1})}+y_{2}^{-1}\|\partial_{1}\theta(\cdot,y_{2})\|_{L^{2}(x_{1}/2,2x_{1})}),$

where the constant $C>0$ does not depend on $x_{1}$. On the other hand,

$\displaystyle\int_{0}^{2x_{1}}\frac{1}{|x-y|}\,\mathrm{d}y_{1}$
$\displaystyle=2\int_{0}^{x_{1}}\frac{1}{\sqrt{\tau^{2}+(x_{2}-y_{2})^{2}}}\,\mathrm{d}\tau$
(3.12)
$\displaystyle=2\log\left(x_{1}+\sqrt{x_{1}^{2}+(x_{2}-y_{2})^{2}}\right)-2\log|x_{2}-y_{2}|$
$\displaystyle\leq C\log\left(1+\frac{x_{1}}{|x_{2}-y_{2}|}\right).$

Hence, with $y_{2}\leq 2x_{1}$ and Hölder’s inequality, we infer that

$\displaystyle|K_{11}|$ $\displaystyle\leq
C(\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}+\|y_{2}^{-1}\theta(y)\|_{L^{2}([0,1]^{2})})\left\\{\int_{0}^{2x_{2}}\frac{1}{x_{2}+y_{2}}\left|\log\left(1+\frac{x_{1}}{|x_{2}-y_{2}|}\right)\right|^{2}\mathrm{d}y_{2}\right\\}^{\frac{1}{2}}$

and

$\displaystyle|K_{12}|$ $\displaystyle\leq
C(\|\nabla^{2}\theta\|_{L^{2}(R(x))}+\|y_{2}^{-1}\theta(y)\|_{L^{2}(R(x))})\left\\{\int_{0}^{2x_{2}}\frac{1}{x_{2}+y_{2}}\left|\log\left(1+\frac{x_{1}}{|x_{2}-y_{2}|}\right)\right|^{2}\mathrm{d}y_{2}\right\\}^{\frac{1}{2}}.$

Using Hardy’s inequality and that

$\displaystyle\int_{0}^{2x_{2}}\frac{1}{x_{2}+y_{2}}\left|\log\left(1+\frac{x_{1}}{|x_{2}-y_{2}|}\right)\right|^{2}\mathrm{d}y_{2}$
$\displaystyle\leq\frac{1}{x_{2}}\int_{0}^{2x_{2}}\left(\log\frac{2x_{1}}{|x_{2}-y_{2}|}\right)^{2}\mathrm{d}y_{2}$
$\displaystyle=\frac{2}{x_{2}}\int_{0}^{x_{2}}\left(\log\frac{2x_{1}}{x_{2}-y_{2}}\right)^{2}\mathrm{d}y_{2}$
$\displaystyle=4+4\log\frac{2x_{1}}{x_{2}}+2\left(\log\frac{2x_{1}}{x_{2}}\right)^{2}$
$\displaystyle\leq C\left(1+\log\frac{x_{1}}{x_{2}}\right)^{2},$

we obtain

$|K_{11}|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}\left(1+\log\frac{x_{1}}{x_{2}}\right).$

By the following estimate

$\displaystyle\int_{2x_{2}}^{2x_{1}}\frac{1}{x_{2}+y_{2}}\left|\log\left(1+\frac{x_{1}}{|x_{2}-y_{2}|}\right)\right|^{2}\mathrm{d}y_{2}$
$\displaystyle\leq\int_{2x_{2}}^{2x_{1}}\frac{1}{y_{2}-x_{2}}\left(\log\frac{3x_{1}}{y_{2}-x_{2}}\right)^{2}\mathrm{d}y_{2}$
$\displaystyle=\frac{1}{3}\left(\log\frac{3x_{1}}{x_{2}}\right)^{3}-\frac{1}{3}\left(\log\frac{3x_{1}}{2x_{1}-x_{2}}\right)^{3}$
$\displaystyle\leq C\left(1+\log\frac{x_{1}}{x_{2}}\right)^{3},$

we have

$|K_{12}|\leq
C(\|\nabla^{2}\theta\|_{L^{2}(R(x))}+\|y_{2}^{-1}\theta(y)\|_{L^{2}(R(x))})\left(1+\log\frac{x_{1}}{x_{2}}\right)^{\frac{3}{2}}.$

This implies

$|K_{1}|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}\left(1+\log\frac{x_{1}}{x_{2}}\right)+C(\|\nabla^{2}\theta\|_{L^{2}(R(x))}+\|y_{2}^{-1}\partial_{1}\theta(y)\|_{L^{2}(R(x))})\left(1+\log\frac{x_{1}}{x_{2}}\right)^{\frac{3}{2}},$

and collecting the estimates for $K_{1}$, $K_{2}$, and $K_{3}$, we obtain
(3.10). Therefore, we arrive at

$\displaystyle\left|\frac{I_{3}(0)}{x_{2}}+6\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\mathrm{d}y\right|$
$\displaystyle\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}\left(1+\log\frac{x_{1}}{x_{2}}\right)+C(\|\nabla^{2}\theta\|_{L^{2}(R(x))}+\|y_{2}^{-1}\partial_{1}\theta(y)\|_{L^{2}(R(x))})\left(1+\log\frac{x_{1}}{x_{2}}\right)^{\frac{3}{2}}.$

Using (3.5), we can estimate

$I_{4}(0)=-4x_{2}\int_{[0,1]^{2}}\frac{y_{2}(x_{1}+y_{1})(|x+y|^{2}+|x+y||x-\tilde{y}|+|x-\bar{y}|^{2})}{|x+y|^{3}|x-\tilde{y}|^{3}(|x+y|+|x-\tilde{y}|)}\theta(y)\mathrm{d}y$

similarly with $I_{3}(0)$. Hence we have (3.2), and this completes the proof.
∎

######  Lemma 3.4.

Moreover, we have

$\left|\frac{u_{1}(x)}{x_{1}}-12\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\mathrm{d}y\right|\leq
B_{4}(x)$ (3.13)

and

$\left|\frac{u_{2}(x)}{x_{2}}+12\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\mathrm{d}y\right|\leq\left(1+\log\frac{x_{1}}{x_{2}}\right)B_{5}(x)+\left(1+\log\frac{x_{1}}{x_{2}}\right)^{2}B_{6}(x),$
(3.14)

where $B_{4}$, $B_{5}$, $B_{6}$ satisfy

$|B_{4}(x)|+|B_{5}(x)|\leq
C(\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}+\|\theta\|_{L^{\infty}([0,1]^{2})})$

and

$|B_{6}(x)|\leq C\|\nabla\theta\|_{L^{\infty}(R(x))}.$

###### Proof.

We follow the proof of Lemma 3.2. To obtain (3.13), we recall (3.4) and have
that

$|u_{1}(x)|=\Bigg{|}\sum_{n\in\mathbb{Z}^{2}}(I_{1}(n)+I_{2}(n))\Bigg{|}\leq
Cx_{1}\|\theta\|_{L^{\infty}([0,1]^{2})}+I_{1}(0)+I_{2}(0).$

We estimate

$I_{1}(0)=-4x_{1}\int_{[0,1]^{2}}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y$

for each set $R(2x)$ and $[0,2x_{1}]\times[0,1]$.

(i) $Q(x)$

Using the notation $J_{1}$ and $J_{2}$ in Lemma 3.2, it sufficies to obtain
that

$\left|J_{1}-\frac{3}{2}\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|+|J_{2}|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$ (3.15)

We already showed that

$|J_{2}|\leq
C|x|\int_{Q(x)}\frac{1}{|y|^{3}}\frac{|\theta(y)|}{|y|}\,\mathrm{d}y.$

By (3.6) and Hölder’s inequaltiy, we have

$|x|\int_{Q(x)}\frac{1}{|y|^{3}}\frac{|\theta(y)|}{|y|}\,\mathrm{d}y\leq|x|\left(\int_{|x|}^{\infty}\frac{1}{r^{2}}\,\mathrm{d}r\right)\||y|^{-1}\theta(y)\|_{L^{\infty}(Q(x))}\leq\||y|^{-1}\theta(y)\|_{L^{\infty}([0,1]^{2})}.$

Due to that $\theta$ vanishes near the axis, it follows

$|J_{2}|\leq C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$

Separating $J_{1}=J_{11}+J_{12}+J_{13}$ as in the proof of Lemma 3.2, we can
prove that

$\begin{split}\left|J_{1k}-\frac{1}{2}\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq
C|x|\int_{Q(x)}\frac{1}{|y|^{3}}\frac{|\theta(y)|}{|y|}\,\mathrm{d}y\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}\end{split}$

for each $k=1,2,3$. Therefore, (3.15) is obtained.

(ii) $[0,2x_{1}]\times[2x_{1},1]$

In (3.8) we observed that

$\displaystyle\left|-\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y\right|$
$\displaystyle\leq
C\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{1}{y_{2}^{2}}\frac{\theta(y)}{y_{2}}\,\mathrm{d}y.$

Since

$\displaystyle\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{1}{y_{2}^{2}}\frac{\theta(y)}{y_{2}}\,\mathrm{d}y\leq
C\left(\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{1}{y_{2}^{2}}\,\mathrm{d}y\right)\|y_{2}^{-1}\theta(y)\|_{L^{\infty}([0,1]^{2})}\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})},$ (3.16)

we have

$\left|-\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$

(iii) $[0,2x_{1}]^{2}$

We recall that

$\displaystyle-\int_{[0,2x_{1}]^{2}}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y$
$\displaystyle\hphantom{\qquad\qquad}=\frac{1}{x_{1}}\int_{[0,2x_{1}]^{2}}\left(\frac{1}{|x-y|}-\frac{1}{|x-\tilde{y}|}\right)\partial_{2}\theta(y)\,\mathrm{d}y$
$\displaystyle\hphantom{\qquad\qquad\qquad\qquad}-\frac{1}{x_{1}}\int_{0}^{2x_{1}}\left(\frac{1}{|(x_{1}-y_{1},x_{2}-2x_{1})|}-\frac{1}{|(x_{1}+y_{1},x_{2}-2x_{1})|}\right)\theta(y_{1},2x_{1})\,\mathrm{d}y_{1}.$

Using Hölder’s inequality, we have

$\displaystyle\left|\frac{1}{x_{1}}\int_{[0,2x_{1}]^{2}}\left(\frac{1}{|x-y|}-\frac{1}{|x-\tilde{y}|}\right)\partial_{2}\theta(y)\,\mathrm{d}y\right|$
$\displaystyle\leq\frac{2}{x_{1}}\left(\int_{[0,2x_{1}]^{2}}\frac{1}{|x-y|}\,\mathrm{d}y\right)\|\partial_{2}\theta\|_{L^{\infty}([0,1]^{2})}$
$\displaystyle\leq C\|\nabla\theta\|_{L^{\infty}([0,2x_{1}]^{2})}.$

From Hölder’s inequality and the mean value theroem, it follows

$\left|-\frac{1}{x_{1}}\int_{0}^{2x_{1}}\left(\frac{1}{|(x_{1}-y_{1},x_{2}-2x_{1})|}-\frac{1}{|(x_{1}+y_{1},x_{2}-2x_{1})|}\right)\theta(y_{1},2x_{1})\,\mathrm{d}y_{1}\right|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$

Therefore, we obtain that

$\left|-\int_{[0,2x_{1}]^{2}}\frac{y_{1}(x_{2}-y_{2})(|x-y|^{2}+|x-y||x-\tilde{y}|+|x-\tilde{y}|^{2})}{|x-y|^{3}|x-\tilde{y}|^{3}(|x-y|+|x-\tilde{y}|)}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla^{2}\theta\|_{L^{2}([0,1]^{2})}.$

Combining the above estimates, it follows that

$\displaystyle\left|\frac{I_{1}(0)}{x_{1}}-6\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$

In a similar way, we can show that

$I_{2}(0)=4x_{1}\int_{[0,1]^{2}}\frac{y_{1}(x_{2}+y_{2})(|x+y|^{2}+|x+y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x+y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\mathrm{d}y$

satisfies

$\displaystyle\left|\frac{I_{2}(0)}{x_{2}}-6\int_{R(2x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\mathrm{d}y\right|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$

We omit the details. Thus we have (3.13).

Now we estimate $u_{2}$. We already know that

$|u_{2}(x)|=\Bigg{|}\sum_{n\in\mathbb{Z}^{2}}(I_{3}(n)+I_{4}(n))\Bigg{|}\leq
Cx_{1}\|\theta\|_{L^{\infty}([0,1]^{2})}+I_{3}(0)+I_{4}(0).$

To estimate

$I_{3}(0)=4x_{2}\int_{[0,1]^{2}}\frac{y_{2}(x_{1}-y_{1})(|x-y|^{2}+|x-y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x-y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\mathrm{d}y,$

we divide the domain into four regions as
$[0,1]^{2}=Q(x)\cup[0,x_{1}/2]\times[0,2x_{1}]\cup[x_{1}/2,2x_{1}]\times[0,2x_{1}]\cup[0,2x_{1}]\times[2x_{1},1]$
and estimate the integral in each region.

(i) $Q(x)$

In the case of $y\in Q(x)$, recalling that (3.6) holds, we can prove

$\begin{gathered}\left|\int_{Q(x)}\frac{y_{2}(x_{1}-y_{1})(|x-y|^{2}+|x-y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x-y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\,\mathrm{d}y+\frac{3}{2}\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\,\mathrm{d}y\right|\\\
\leq C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.\end{gathered}$ (3.17)

We shall not give a proof as it requires little adjustment from (3.15).

(ii) $[0,2x_{1}]\times[2x_{1},1]$

Since $y\in[0,2x_{1}]\times[2x_{1},1]$ satisfies

$\frac{1}{2}{y_{2}}\leq|x-y|\leq 2y_{2},\qquad{y_{2}}\leq|x-\bar{y}|\leq
2y_{2},$

with (3.16) we can show that

$\left|\int_{[0,2x_{1}]\times[2x_{1},1]}\frac{y_{2}(x_{1}-y_{1})(|x-y|^{2}+|x-y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x-y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\,\mathrm{d}y\right|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$

(iii) $[0,x_{1}/2]\times[0,2x_{1}]$

From that $y\in[0,x_{1}/2]\times[0,2x_{1}]$ implies

$\frac{1}{2}x_{1}\leq|x-y|\leq
4x_{1},\qquad\frac{1}{2}x_{1}\leq|x-\bar{y}|\leq 4x_{1},$

we can see that

$\left|\int_{[0,x_{1}/2]\times[0,2x_{1}]}\frac{y_{2}(x_{1}-y_{1})(|x-y|^{2}+|x-y||x-\bar{y}|+|x-\bar{y}|^{2})}{|x-y|^{3}|x-\bar{y}|^{3}(|x-y|+|x-\bar{y}|)}\theta(y)\,\mathrm{d}y\right|\\\
\leq C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$

(iv) $[x_{1}/2,2x_{1}]\times[0,2x_{1}]$

Recalling the notation $K_{1}$, $K_{2}$, and $K_{3}$ in the proof of Lemma
3.2, we claim

$|K_{1}|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}\left(1+\log\frac{x_{1}}{x_{2}}\right)+C\|\nabla\theta\|_{L^{\infty}(R(x))}\left(1+\log\frac{x_{1}}{x_{2}}\right)^{2}$
(3.18)

and

$|K_{2}|+|K_{3}|\leq C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$ (3.19)

As in (3.11), we have with the mean value theorem that

$\displaystyle|K_{2}|\leq
Cx_{1}^{-1}\|\theta\|_{L^{\infty}([0,2x_{1}]^{2})}\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})},$

and similarly,

$|K_{3}|\leq C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}.$

Hence, (3.19) follows. We reacll

$K_{1}=K_{11}+K_{12}$

where

$\displaystyle K_{11}$
$\displaystyle=\int_{[x_{1}/2,2x_{1}]\times[0,2x_{2}]}\frac{-4y_{2}\partial_{1}\theta(y)}{|x-y||x-\bar{y}|(|x-y|+|x-\bar{y}|)}\,\mathrm{d}y,$
$\displaystyle K_{12}$
$\displaystyle=\int_{[x_{1}/2,2x_{1}]\times[2x_{2},2x_{1}]}\frac{-4y_{2}\partial_{1}\theta(y)}{|x-y||x-\bar{y}|(|x-y|+|x-\bar{y}|)}\,\mathrm{d}y.$

From Hölder’s inequality and (3.12), we can deduce that

$|K_{11}|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}\int_{0}^{2x_{2}}\frac{1}{x_{2}+y_{2}}\log\left(1+\frac{x_{1}}{|x_{2}-y_{2}|}\right)\mathrm{d}y_{2}$

and

$|K_{12}|\leq
C\|\nabla\theta\|_{L^{\infty}(R(x))}\int_{2x_{2}}^{2x_{1}}\frac{1}{x_{2}+y_{2}}\log\left(1+\frac{x_{1}}{|x_{2}-y_{2}|}\right)\mathrm{d}y_{2}.$

Since

$\displaystyle\int_{0}^{2x_{2}}\frac{1}{x_{2}+y_{2}}\log\left(1+\frac{x_{1}}{|x_{2}-y_{2}|}\right)\mathrm{d}y_{2}$
$\displaystyle\leq\frac{1}{x_{2}}\int_{0}^{2x_{2}}\log\frac{2x_{1}}{|x_{2}-y_{2}|}\,\mathrm{d}y_{2}$
$\displaystyle=2\log\frac{2x_{1}}{x_{2}}-2$ $\displaystyle\leq
C\left(1+\log\frac{x_{1}}{x_{2}}\right)$

and

$\displaystyle\int_{2x_{2}}^{2x_{1}}\frac{1}{x_{2}+y_{2}}\log\left(1+\frac{x_{1}}{|x_{2}-y_{2}|}\right)\mathrm{d}y_{2}$
$\displaystyle\leq\int_{2x_{2}}^{2x_{1}}\frac{1}{y_{2}-x_{2}}\log\frac{3x_{1}}{y_{2}-x_{2}}\,\mathrm{d}y_{2}$
$\displaystyle=\frac{1}{2}\left|\log\frac{3x_{1}}{x_{2}}\right|^{2}-\frac{1}{2}\left|\log\frac{3x_{1}}{2x_{1}-x_{2}}\right|^{2}$
$\displaystyle\leq C\left(1+\log\frac{x_{1}}{x_{2}}\right)^{2},$

it follows

$|K_{11}|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}\left(1+\log\frac{x_{1}}{x_{2}}\right)$

and

$|K_{12}|\leq
C\|\nabla\theta\|_{L^{\infty}(R(x))}\left(1+\log\frac{x_{1}}{x_{2}}\right)^{2}.$

This shows (3.18). Combining the estimates, we arrive at

$\displaystyle\left|\frac{I_{3}(0)}{x_{2}}+6\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(y)\mathrm{d}y\right|\leq
C\|\nabla\theta\|_{L^{\infty}([0,1]^{2})}\left(1+\log\frac{x_{1}}{x_{2}}\right)+C\|\nabla\theta\|_{L^{\infty}(R(x))}\left(1+\log\frac{x_{1}}{x_{2}}\right)^{2}.$

Using (3.5), we can estimate

$I_{4}(0)=-4x_{2}\int_{[0,1]^{2}}\frac{y_{2}(x_{1}+y_{1})(|x+y|^{2}+|x+y||x-\tilde{y}|+|x-\bar{y}|^{2})}{|x+y|^{3}|x-\tilde{y}|^{3}(|x+y|+|x-\tilde{y}|)}\theta(y)\mathrm{d}y$

similarly with $I_{3}(0)$. Hence we have (3.14), and this completes the proof.
∎

##  4 Nonexistence

In this section, we prove Theorem B. We begin with a simple uniqueness result
which in particular guarantees that the hypothetical solution in
$L^{\infty}_{t}H^{2}$ satisfies the same symmetries with the initial data.

######  Proposition 4.1.

Given $\theta_{0}\in H^{2}$ and $T>0,$ there exists at most one solution
belonging to $L^{\infty}([0,T];H^{2})$ to (SQG) with initial data
$\theta_{0}$.

###### Proof.

The proof can be given by simply adapting Yudovich’s inequalities derived in
([44, 45]). This statement could be found in [1] as well. ∎

###### Proof of Theorem B.

For convenience, we shall divide the proof into several parts.

1\. Velocity and flow map: an $L^{\infty}_{t}H^{2}$–solution is Lagrangian.
Assume that we are given a solution to (SQG) satisfying

$\sup_{t\in[0,T]}\|\theta(t,\cdot)\|_{H^{2}}\leq M.$

Then, by the Sobolev embedding,
$u=\nabla^{\perp}(-\Delta)^{-\frac{1}{2}}\theta$ satisfies

$\begin{split}\sup_{t\in[0,T]}(\|\nabla
u(t,\cdot)\|_{BMO}+\|u(t,\cdot)\|_{W^{1,1}})\leq
C\sup_{t\in[0,T]}\|u(t,\cdot)\|_{H^{2}}\leq CM\end{split}$

with some absolute constant $C>0$. In particular, $u$ is log-Lipschitz: for
any $x,y\in\mathbb{T}^{2}$, we have

$\begin{split}\sup_{t\in[0,T]}|u(t,x)-u(t,y)|\leq
CM|x-y|\ln\left(10+\frac{1}{|x-y|}\right).\end{split}$ (4.1)

On the time interval $[0,T]$, we consider the flow map
$\Phi(t,\cdot):\mathbb{T}^{2}\rightarrow\mathbb{T}^{2}$ defined by

$\left\\{\begin{aligned} &\frac{d}{dt}\Phi(t,x)=u(t,\Phi(t,x)),\\\
&\Phi(0,x)=x.\end{aligned}\right.$ (4.2)

It is well-known that under the estimate (4.1), there is a unique solution to
the ODE (4.2) for any $x\in\mathbb{T}^{2}$ ([37, 39]). The solution $\Phi$
satisfies the estimate

$\begin{split}|x-y|^{\exp(CMt)}\leq|\Phi(t,x)-\Phi(t,y)|\leq|x-y|^{\exp(-CMt)}\end{split}$
(4.3)

for some absolute constant $C>0$, uniformly in $x,y\in\mathbb{T}^{2}$
satisfying $|x-y|<\frac{1}{2}$ and $t\in[0,T]$. We have the representation

$\begin{split}\theta(t,\Phi(t,x))=\theta_{0}(x).\end{split}$

The estimate (4.3) shows that for each $t\in[0,T]$, $\Phi(t,\cdot)$ is a
Hölder continuous homeomorphism $\mathbb{T}^{2}\rightarrow\mathbb{T}^{2}$, and
we denote the inverse map by $\Phi_{t}^{-1}$. Then, with this notation, we
have

$\begin{split}\theta(t,x)=\theta_{0}(\Phi_{t}^{-1}(x)).\end{split}$

The inverse map $\Phi_{t}^{-1}$ is again Hölder continuous. As an immediate
consequence, we have that if $\theta_{0}$ is an odd function with respect to
both axes and satisfies

$\begin{split}{\mathrm{supp}}\,(\theta_{0})\cup\\{x:x_{1}=0\mbox{ or
}x_{2}=0\\}\subset\\{(0,0)\\},\end{split}$

then the same properties are satisfied by $\theta(t,\cdot)$, as long as
$\theta\in L^{\infty}([0,t];H^{2})$. Indeed, the uniqueness assertion from
Proposition 4.1 guarantees that $\theta(t,\cdot)$ is odd with respect to both
axes. Furthermore, Hölder continuity of the flow map and its inverse ensures
that $\theta(t,\cdot)$ vanishes near the axes, possibly except at the origin.
Therefore, the last assumption in Lemma 3.2 is satisfied.

2\. Choice of initial data. We fix some smooth bump function
$\phi:\mathbb{R}^{2}\rightarrow\mathbb{R}_{\geq 0}$ satisfying the following
properties:

  * •

$\phi$ is $C^{\infty}$-smooth and radial.

  * •

$\phi$ is supported in $B_{0}(\frac{1}{8})$ and $\phi=1$ in
$B_{0}(\frac{1}{32})$.

Then, we define

$\begin{split}\theta_{0}:=\sum_{n=n_{0}}^{\infty}n^{-\alpha}\theta^{(n)}_{0,loc}\end{split}$
(4.4)

for some $1/2<\alpha<3/4$, where

$\begin{split}\theta^{(n)}_{0,loc}(x):=4^{-n}\phi(4^{n}(x_{1}-4^{-n-1},x_{2}-2^{-1}4^{-n-1}))\end{split}$

for $x\in[0,1]^{2}$. The precise value of $\alpha$ will be determined later,
but for now let us just mention that it will be taken slightly larger than
$\frac{1}{2}$. Next, let us extend each of $\theta^{(n)}_{0,loc}$ (and
similarly $\theta_{0}$) to $\mathbb{T}^{2}$ as an odd function with respect to
both axes. Note that by taking $n_{0}\geq 1$ sufficiently large in a way
depending only on $\epsilon>0$, we can guarantee that

$\|\theta_{0}\|_{H^{2}\cap W^{1,\infty}(\mathbb{T}^{2})}<\epsilon.$

Towards a contradiction, we shall assume that there exists $M>0$ and $T>0$
such that

$\begin{split}\sup_{t\in[0,T]}\|\theta(t)\|_{\dot{H}^{2}(\mathbb{T}^{2})}\leq
M.\end{split}$ (4.5)

For simplicity, we shall assume that $M\geq 1$. Observe that the assumptions
in the Key Lemma 3.2 are satisfied by $\theta_{0}$. Recalling the discussion
above, we have that $\theta(t,\cdot)$ is odd with respect to both axes and
vanishes near the axes, except at the origin.

3\. Preliminary bounds on the solution. Let us remark in advance that in the
following proof, we shall take $T>0$ to be smaller, if necessary, to satisfy
$T\leq c/M$ for some small absolute constant $c>0$. We shall begin with a
simple result:

######  Lemma 4.2.

Assume that $\theta$ is a solution satisfying (4.5) with initial data (4.4).
Then, by redefining $T$ to satisfy $T\leq c/M$ if necessary, we have

$\begin{split}\theta(t,y)=0,\qquad 0\leq y_{1}\leq y_{2},\qquad 0\leq t\leq
T.\end{split}$

###### Proof.

Since $\theta(t,\Phi(t,x))=\theta_{0}(x)$, to prove the claim, it suffices to
show that for $x\in{\mathrm{supp}}\,(\theta_{0})\backslash\\{(0,0)\\}$,
$\Phi_{2}(t,x)\leq\Phi_{1}(t,x)$ for $0\leq t\leq T$. Let us fix some
$x\in{\mathrm{supp}}\,(\theta_{0})\backslash\\{(0,0)\\}$. Then, from the
choice of initial data, we have $2x_{2}\leq x_{1}$. From continuity in time of
the flow map, there exists some $0<T^{*}\leq T$ such that
$\Phi_{2}(t,x)<\Phi_{1}(t,x)$ for $0\leq t<T^{*}$. Then, on this time
interval, Key Lemma is applicable for $\Phi(t,x)$ and we compute

$\begin{split}\frac{d}{dt}\left(\frac{\Phi_{2}(t,x)}{\Phi_{1}(t,x)}\right)&=\frac{\Phi_{2}(t,x)}{\Phi_{1}(t,x)}\left(\frac{u_{2}(t,\Phi(t,x))}{\Phi_{2}(t,x)}-\frac{u_{1}(t,\Phi(t,x))}{\Phi_{1}(t,x)}\right)\\\
&\leq
C\frac{\Phi_{2}(t,x)}{\Phi_{1}(t,x)}\left(|B_{1}(\Phi(t,x))|+|B_{2}(\Phi(t,x))|+|B_{3}(\Phi(t,x))|\right)\\\
&\leq CM\frac{\Phi_{2}(t,x)}{\Phi_{1}(t,x)}.\end{split}$

Therefore, we actually obtain

$\begin{split}\frac{\Phi_{2}(t,x)}{\Phi_{1}(t,x)}\leq\frac{1}{2}\exp\left(CMt\right)<\frac{3}{4}\end{split}$

on $t\in[0,T^{*}]$, as long as $T^{*}\leq c/M$ for $c>0$ depending only on
$C$. This bootstrap procedure allows us to get $\Phi_{2}/\Phi_{1}<3/4$
uniformly in $x\in{\mathrm{supp}}\,(\theta_{0})\backslash\\{(0,0)\\}$ by the
time $\min\\{T,c/M\\}=T$. ∎

The above lemma guarantees that on $[0,T]$, Key Lemma is applicable to points
in ${\mathrm{supp}}\,(\theta(t,\cdot))$. Next, let us prove that by reducing
$c>0$ if necessary, the bubbles are “well-ordered” with respect to the
$x_{1}$-axis for $t\in[0,T]$ with $T\leq c/M$.

Claim I. We have that

$\begin{split}\sup_{x\in\Omega_{n}}\Phi_{1}(t,x)\leq
2\inf_{x\in\Omega_{n}}\Phi_{1}(t,x)\qquad\mbox{and}\qquad
2\sup_{x\in\Omega_{n+1}}\Phi_{1}(t,x)\leq\inf_{x\in\Omega_{n}}\Phi_{1}(t,x)\end{split}$
(4.6)

uniformly for all $n\geq n_{0}$ and $t\in[0,T]$, by reducing $T$ to satisfy
$T\leq c/(1+M)$ for some small absolute constant $c>0$.

For simplicity we let $\Omega_{n}:=\mathrm{supp}\,(\theta^{(n)}_{0,loc})$ and

$\widehat{\Phi}_{j}^{n}(t):=\sup_{x\in\Omega_{n}}\Phi_{j}(t,x)$

for $j=1,2$. We can prove the Claim I inductively in $n$, using the Key Lemma
which gives

$\left|\frac{u_{1}(t,x)}{x_{1}}-12\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(t,y)\mathrm{d}y\right|\leq
CM.$

In the proof, we shall take $T>0$ smaller several times, but in a way which is
independent of $n$. To begin with, for $x\in\Omega_{n_{0}}$ we have

$\left|\frac{\mathrm{d}}{\mathrm{d}t}\log\Phi_{1}(t,x)\right|\leq CM,$

thus,

$\log\Phi_{1}(t,x)-\log x_{1}\geq-CMt.$

We also have

$\frac{\mathrm{d}}{\mathrm{d}t}\log\widehat{\Phi}^{n_{0}}_{1}(t)-\frac{\mathrm{d}}{\mathrm{d}t}\log\Phi_{1}(t,x)\leq
2CM,$

and thus,

$\log\widehat{\Phi}^{n_{0}}_{1}(t)-\log\Phi_{1}(t,y)\leq
2CMt+(\log\widehat{x}^{n_{0}}_{1}-\log x_{1}).$

Since $\widehat{x}^{n_{0}}_{1}/x_{1}<3/2$, we can take $T>0$ sufficiently
small such that

$2CMT+(\log\widehat{x}^{n_{0}}_{1}-\log x_{1})\leq\log 2,$

which implies that

$\widehat{\Phi}_{1}^{n_{0}}(t)\leq 2\inf_{x\in\Omega_{n_{0}}}\Phi_{1}(t,x)$

for all $t\in[0,T]$. Indeed, it suffices to take $T=c/(1+M)$ with a small
absolute constant $c>0$. Note that for $x\in\Omega_{n_{0}}$,

$\frac{\mathrm{d}}{\mathrm{d}t}\log\Phi_{1}(t,x)-\frac{\mathrm{d}}{\mathrm{d}t}\log\widehat{\Phi}^{n_{0}+1}_{1}(t)\geq-12\int_{\Omega_{n_{0}}}\frac{\Phi_{1}(t,y)\Phi_{2}(t,y)}{|\Phi(t,y)|^{5}}\theta_{0}(y)\mathrm{d}y-2CM.$

With the above estimates, we have

$\displaystyle\int_{\Omega_{n_{0}}}\frac{\Phi_{1}(t,y)\Phi_{2}(t,y)}{|\Phi(t,y)|^{5}}\theta_{0}(y)\mathrm{d}y$
$\displaystyle\leq\Big{(}\sup_{x\in\Omega_{n_{0}}}\frac{x_{1}}{\Phi_{1}(t,x)}\Big{)}^{3}\int_{\Omega_{n_{0}}}\frac{\theta_{0}(y)}{y_{1}^{3}}\mathrm{d}y\leq
C_{0}e^{3CMT}.$

Using it, we obtain

$\frac{\mathrm{d}}{\mathrm{d}t}\log\Phi_{1}(t,x)-\frac{\mathrm{d}}{\mathrm{d}t}\log\widehat{\Phi}^{n_{0}+1}_{1}(t)\geq-12C_{0}e^{3CMT}-2CM$

and

$\log\Phi_{1}(t,x)-\log\widehat{\Phi}^{n_{0}+1}_{1}(t)\geq-12C_{0}e^{3CMT}t-2CMt+(\log
x_{1}-\log\widehat{x}^{n_{0}+1}_{1}).$

Since $x_{1}/\widehat{x}^{n_{0}+1}_{1}>2$, we can take $T>0$ sufficiently
small such that

$-12C_{0}e^{3CMT}T-2CMT+(\log x_{1}-\log\widehat{x}^{n_{0}+1}_{1})\geq\log 2,$

which implies that

$2\widehat{\Phi}^{n_{0}+1}_{1}(t)\leq\inf_{x\in\Omega_{1}}\Phi_{1}(t,x)$ (4.7)

for all $t\in[0,T]$. Now let $x\in\Omega_{n_{0}+1}$. Then we have

$\frac{\mathrm{d}}{\mathrm{d}t}\log\Phi_{1}(t,x)\geq-CM,$

thus,

$\log\Phi_{1}(t,x)-\log x_{1}\geq-CMt.$

Since $\widehat{x}^{n_{0}+1}_{1}/x_{1}$ takes the same value as in the
previous case, we see that

$2CMT+(\log\widehat{x}^{n_{0}+1}_{1}-\log x_{1})\leq\log 2,$

and therefore, we have

$\widehat{\Phi}_{1}^{n_{0}+1}(t)\leq
2\inf_{x\in\Omega_{n_{0}+1}}\Phi_{1}(t,x)$

for all $t\in[0,T]$. Note that by (4.7),

$\frac{\mathrm{d}}{\mathrm{d}t}\log\Phi_{1}(t,x)-\frac{\mathrm{d}}{\mathrm{d}t}\log\widehat{\Phi}^{n_{0}+2}_{1}(t)\geq-12\int_{\Omega_{n_{0}+1}}\frac{\Phi_{1}(t,y)\Phi_{2}(t,y)}{|\Phi(t,y)|^{5}}\theta_{0}(y)\mathrm{d}y-2CM.$

With the above estimates, we have

$\displaystyle\int_{\Omega_{n_{0}+1}}\frac{\Phi_{1}(t,y)\Phi_{2}(t,y)}{|\Phi(t,y)|^{5}}\theta_{0}(y)\mathrm{d}y$
$\displaystyle\leq\Big{(}\sup_{x\in\Omega_{n_{0}+1}}\frac{x_{1}}{\Phi_{1}(t,x)}\Big{)}^{n_{0}+2}\int_{\Omega_{n_{0}+1}}\frac{\theta_{0}(y)}{y_{1}^{n_{0}+2}}\mathrm{d}y\leq
C_{0}e^{3CMT}.$

Using it, we obtain

$\frac{\mathrm{d}}{\mathrm{d}t}\log\Phi_{1}(t,x)-\frac{\mathrm{d}}{\mathrm{d}t}\log\widehat{\Phi}^{n_{0}+2}_{1}(t)\geq-12C_{0}e^{3CMT}-2CM$

and

$\log\Phi_{1}(t,x)-\log\widehat{\Phi}^{n_{0}+2}_{1}(t)\geq-12C_{0}e^{3CMT}t-2CMt+(\log
x_{1}-\log\widehat{x}^{n_{0}+2}_{1}).$

Since $x_{1}/\widehat{x}^{n_{0}+2}_{1}>2$ is the same value as in the previous
case, it follows

$-12C_{0}e^{3CMT}T-2CMT+(\log x_{1}-\log\widehat{x}^{n_{0}+2}_{1})\geq\log 2$

and

$2\widehat{\Phi}^{n_{0}+2}_{1}(t)\leq\inf_{x\in\Omega_{n_{0}+1}}\Phi_{1}(t,x)$

for all $t\in[0,T]$. Repeating this argument, one can finish the proof of
Claim I. ∎

Claim II. There exists $T>0$ and $C>0$ such that

$\log\widehat{\Phi}_{2}^{n}(T)\leq\log\widehat{x}_{2}^{n}-10\sum_{n_{0}\leq
j\leq
n-1}\int_{0}^{T}\int_{\Omega_{j}}\frac{\Phi_{1}(t,y)\Phi_{2}(t,y)}{|\Phi(t,y)|^{5}}\theta_{0}(y)\mathrm{d}y\mathrm{d}t+CM$

uniformly for all $n>n_{0}$.

Recall that

$\frac{u_{2}(x)}{x_{2}}\leq-12\int_{Q(x)}\frac{y_{1}y_{2}}{|y|^{5}}\theta(t,y)\mathrm{d}y+CM\bigg{(}\log\frac{x_{1}}{x_{2}}\bigg{)},$

if $\theta(y)=0$ for $y$ satisfying $x_{1}/2\leq y_{1}\leq 2x_{1}$ and
$2x_{2}\leq y_{2}\leq 1$. According to the order of the bubbles, we have for
$x\in\Omega_{n}$ that

$\int_{R(2\Phi(t,x))}\frac{y_{1}y_{2}}{|y|^{5}}\theta(t,y)\mathrm{d}y=\sum_{n_{0}\leq
j\leq
n-1}\int_{\Omega_{j}}\frac{\Phi_{1}(t,y)\Phi_{2}(t,y)}{|\Phi(t,y)|^{5}}\theta_{0}(y)\mathrm{d}y.$

And note that

$\sup_{2\Phi_{2}(t,x)\geq\widehat{\Phi}_{2}^{n}(t)}CM\bigg{(}\log\frac{\Phi_{1}(t,x)}{\Phi_{2}(t,x)}\bigg{)}\leq
CM\bigg{(}\log\frac{2\widehat{\Phi}_{1}^{n}(t)}{\widehat{\Phi}_{2}^{n}(t)}\bigg{)}.$

Thus, we can see that

$\displaystyle\frac{\mathrm{d}}{\mathrm{d}t}\log\widehat{\Phi}_{2}^{n}(t)$
$\displaystyle\leq-12\sum_{n_{0}\leq j\leq
n-1}\int_{\Omega_{j}}\frac{\Phi_{1}(t,y)\Phi_{2}(t,y)}{|\Phi(t,y)|^{5}}\theta_{0}(y)\mathrm{d}y+CM\bigg{(}\log\frac{2\widehat{\Phi}_{1}^{n}(t)}{\widehat{\Phi}_{2}^{n}(t)}\bigg{)}$
$\displaystyle\leq-12\sum_{n_{0}\leq j\leq
n-1}\int_{\Omega_{j}}\frac{\Phi_{1}(t,y)\Phi_{2}(t,y)}{|\Phi(t,y)|^{5}}\theta_{0}(y)\mathrm{d}y+CM\log(2\widehat{\Phi}_{1}^{n}(t))-CM\log\widehat{\Phi}_{2}^{n}(t)$

and

$\displaystyle\log\widehat{\Phi}_{2}^{n}(t)$
$\displaystyle\leq\log\widehat{x}_{2}^{n}-12\sum_{n_{0}\leq j\leq
n-1}\int_{0}^{t}\int_{\Omega_{j}}\frac{\Phi_{1}(\tau,y)\Phi_{2}(\tau,y)}{|\Phi(\tau,y)|^{5}}\theta_{0}(y)\mathrm{d}y\mathrm{d}\tau$
$\displaystyle\hphantom{\qquad\qquad}+CM\int_{0}^{t}\log(2\widehat{\Phi}_{1}^{n}(\tau))\mathrm{d}\tau-
CM\int_{0}^{t}\log\widehat{\Phi}_{2}^{n}(\tau)\mathrm{d}\tau.$

Note that

$\displaystyle\log(2\widehat{\Phi}_{1}^{n}(t))$ $\displaystyle=\log
2+\log\widehat{\Phi}_{1}^{n}(t)$ $\displaystyle\leq\log
2+\log\widehat{x}_{1}^{n}+12\sum_{n_{0}\leq j\leq
n-1}\int_{0}^{t}\int_{\Omega_{j}}\frac{\Phi_{1}(\tau,y)\Phi_{2}(\tau,y)}{|\Phi(\tau,y)|^{5}}\theta_{0}(y)\mathrm{d}y\mathrm{d}\tau+CMt,$

hence,

$\int_{0}^{t}\log(2\widehat{\Phi}_{1}^{n}(\tau))\mathrm{d}\tau\leq t\log
2+t\log\widehat{x}_{1}^{n}+12t\sum_{n_{0}\leq j\leq
n-1}\int_{0}^{t}\int_{\Omega_{j}}\frac{\Phi_{1}(\tau,y)\Phi_{2}(\tau,y)}{|\Phi(\tau,y)|^{5}}\theta_{0}(y)\mathrm{d}y\mathrm{d}\tau+\frac{C}{2}Mt^{2}.$

Therefore, with a sufficiently small $T>0$ (independent of $n$), we can
complete the proof of Claim II. ∎

4\. Almost invariant timescales. We shall write
$\Phi(t,\Omega_{n})\sim\Omega_{n}$ if

$\begin{split}{\mathrm{supp}}\,(\Phi(t,\Omega_{n}))\subset
B((4^{-n-1},2^{-1}4^{-n-1}),4^{-n-1}).\end{split}$

Here, $B((4^{-n-1},2^{-1}4^{-n-1}),4^{-n-1})$ denotes the ball of radius
$4^{-n-1}$ centered at $(4^{-n-1},2^{-1}4^{-n-1})$. Recall from the definition
of initial data that

$\begin{split}\Omega_{n}=B((4^{-n-1},2^{-1}4^{-n-1}),2^{-1}4^{-n-1}).\end{split}$

An immediate consequence of $\Phi(t,\Omega_{n})\sim\Omega_{n}$ is that once we
define

$\begin{split}I_{n}(t)=\int_{\Omega_{n}}\frac{\Phi_{1}(t,y)\Phi_{2}(t,y)}{|\Phi(t,y)|^{5}}\theta_{0}(y)\mathrm{d}y,\end{split}$

we have

$\begin{split}I_{n}(t)\geq a_{0}I_{n}(0)\end{split}$

for some absolute constant $a_{0}$. The following claim gives the sharp bound
on the “almost invariant” timescale for each bubble.

Claim III. For all $n\geq n_{0}$, we have

$\begin{split}\Phi(t,\Omega_{n})\sim\Omega_{n},\qquad\mbox{for}\qquad 0\leq
t\leq\min\\{T,\frac{c}{M+\sum_{j=n_{0}}^{n-1}j^{-\alpha}}\\}=:T_{n}\end{split}$

with some constants $c,C>0$ independent of $n$.

This claim can be proved easily with an induction in $n$. In the base case
$n=n_{0}$, we simply note that for $x\in\Phi(t,\Omega_{n_{0}})$,

$\begin{split}\left|\frac{u_{j}(t,x)}{x_{j}}\right|\leq CM\end{split}$

from which the claim follows in this case. Assume that Claim III holds for all
$n<n_{0}+k$ for some $k\geq 1$. Note that using Claim I and the induction
hypothesis, we have for $x\in\Phi(t,\Omega_{n_{0}+k})$ that

$\begin{split}\left|\frac{u_{j}(t,x)}{x_{j}}\right|\leq
C\left(M+\sum_{\ell=0}^{k-1}I_{n_{0}+\ell}\right),\qquad 0\leq t\leq
T_{n_{0}+k-1}.\end{split}$

A simple application of Gronwall’s inequality gives Claim III. ∎

We have proven that the $n$-th bubble remains almost invariant for $T_{n}$,
which is bounded from below by

$\begin{split}T_{n}\geq\frac{c_{0}}{\sum_{j=n_{0}}^{n-1}j^{-\alpha}}\geq\frac{(1-\alpha)c_{0}}{n^{1-\alpha}}\end{split}$

for all $n\geq N$ with some large $N$ depending only on $M,T$. Now, we observe
that

$\begin{split}\int_{0}^{T_{n}}I_{n}(t)\,\mathrm{d}t\gtrsim
T_{n}I_{n}(0)\gtrsim\frac{1}{n},\end{split}$

with constants independent of $n$, recalling that $I_{n}(0)\gtrsim
n^{-\alpha}$. (We shall take $\alpha$ close to $\frac{1}{2}$.) Hence,
summation gives

$\begin{split}\sum_{k=\ell}^{n}I_{k}(0)T_{k}\geq
c_{0}\left(\frac{1}{\ell}+\cdots+\frac{1}{n}\right)\geq\log\left(\frac{n}{\ell}\right)^{c_{0}}\end{split}$
(4.8)

for some absolute constant $c_{0}>0$, as long as $\ell>N$.

5\. Norm inflation and conclusion the proof. We are now in a position to
complete the proof. For each $\ell>N$ and $n\gg\ell$ (so that
$\log\left(\frac{n}{\ell}\right)^{c_{0}}\gg M$), we can bound for
$x\in\Omega_{n}$

$\begin{split}\log\frac{\widehat{\Phi}^{n}_{2}(T_{\ell})}{\widehat{x}^{n}_{2}}\leq
CM-10\sum_{k=\ell}^{n}I_{k}(0)T_{k}\leq\log\left(\frac{n}{\ell}\right)^{-c_{0}}.\end{split}$

In other words, we have the growth

$\begin{split}\frac{\widehat{x}_{2}^{n}}{\widehat{\Phi}^{n}_{2}(T_{\ell})}\geq\left(\frac{n}{\ell}\right)^{c_{0}}.\end{split}$
(4.9)

Now, we can write the solution in the form

$\begin{split}\theta=\sum_{n=n_{0}}^{\infty}n^{-\alpha}\theta^{(n)},\qquad\theta^{(n)}(t,\Phi(t,x))=\theta_{0,loc}^{(n)}(x),\end{split}$

so that the support of $\theta^{(n)}$ is disjoint from each other. We now take
$t=T_{\ell}$. Since $\theta^{(n)}(T_{\ell},\cdot)=1$ in a region of area
$\gtrsim 4^{-2n}$ and $\theta^{(n^{\prime})}=0$ for $n^{\prime}\neq n$ in that
region, with Hardy’s inequality and (4.9), we obtain that

$\begin{split}\|\theta^{(n)}(T_{\ell})\|_{\dot{H}^{2}}^{2}\gtrsim\left(\frac{n}{\ell}\right)^{2c_{0}}.\end{split}$

This estimate holds for all sufficiently large $n$. Then

$\begin{split}\|\theta(T_{\ell})\|_{H^{2}}^{2}\geq\sum_{n\geq
n_{0}}n^{-2\alpha}\|\theta^{(n)}(T_{\ell})\|_{\dot{H}^{2}}^{2}\gtrsim\ell^{-2c_{0}}\sum_{n\gg\ell}n^{2c_{0}-2\alpha}.\end{split}$

In the last inequality, since $c_{0}>0$ is an absolute constant, and we could
have chosen $\alpha=\frac{1}{2}(1+c_{0})$. This gives a contradiction to
$\|\theta(T_{\ell})\|_{H^{2}}<\infty$ since
$\sum_{n\gg\ell}n^{-1+c_{0}}=\infty$. ∎

######  Remark 4.3.

The nonexistence of the solution in $W^{1,\infty}$ follows directly. Repeating
the above process with Lemma 3.4 instead of Lemma 3.2, we obtain (4.9). Since
we clearly have that

$\begin{split}\|\theta^{(n)}(T_{\ell})\|_{\dot{W}^{1,\infty}}\gtrsim\left(\frac{n}{\ell}\right)^{c_{0}},\end{split}$

it follows

$\begin{split}\|\theta(T_{\ell})\|_{W^{1,\infty}}\geq\sum_{n\geq
n_{0}}n^{-\alpha}\|\theta^{(n)}(T_{\ell})\|_{\dot{W}^{1,\infty}}\gtrsim\ell^{-c_{0}}\sum_{n\gg\ell}n^{c_{0}-\alpha}.\end{split}$

Therefore, for the same $\alpha=\frac{1}{2}(1+c_{0})$, we complete the proof.

##  5 Norm inflation for smooth data

We establish Theorem A in this section, by proving a quantitative norm
inflation result for data obtained by truncating the data used in the proof of
Theorem B.

######  Proposition 5.1 (Quantitative norm inflation).

We consider the $C^{\infty}$–smooth initial data

$\begin{split}\theta_{0}^{(N)}:=\sum_{n=n_{0}}^{N}n^{-\alpha}\theta^{(n)}_{0,loc}\end{split}$
(5.1)

where $\phi,\alpha,n_{0}$ are the same as in (4.4). Then, there exists
$N_{0}\geq 1$ depending only on $\phi,n_{0}$ such that for all $N\geq N_{0}$,
the unique local in time $C^{\infty}$–solution $\theta^{(N)}$ to (SQG) with
initial data $\theta_{0}^{(N)}$ exists on the time interval $[0,T^{*}]$ for
some $0<T^{*}\leq T_{N}$ and satisfies

$\begin{split}\|\theta_{0}^{(N)}\|_{H^{2}\cap
W^{1,\infty}}\leq\epsilon,\qquad\sup_{t\in[0,T^{*}]}\|\theta^{(N)}(t)\|_{H^{2}}>M_{N},\end{split}$
(5.2)

where

$\begin{split}M_{N}:=\frac{c_{0}}{2}\ln N,\qquad T_{N}:=\frac{1}{M_{N}\ln
M_{N}}\end{split}$ (5.3)

with $c_{0}>0$ from (4.8).

###### Proof.

We shall establish the proposition with a contradiction argument: let
$0<T^{*}\leq+\infty$ be the lifespan of the smooth solution associated with
the initial data $\theta_{0}^{(N)}$ and assume that

$\begin{split}\|\theta^{(N)}\|_{L^{\infty}([0,\min\\{T^{*},T_{N}\\}];H^{2})}\leq
M_{N}.\end{split}$

Under this contradiction hypothesis, we can actually prove that $T^{*}>T_{N}$,
so that

$\begin{split}\|\theta^{(N)}\|_{L^{\infty}([0,T_{N}];H^{2})}\leq
M_{N}.\end{split}$ (5.4)

This is simply because the $H^{2}$-norm gives a blow-up criterion. To
illustrate this point, we estimate the $H^{3}$-norm of $\theta:=\theta^{(N)}$
on $[0,T_{N}]$:333For simplicity, from now on we shall suppress from writing
out the dependence of the solution $\theta$ in $N$. from the equation for
$\Delta\theta$

$\begin{split}\partial_{t}\Delta\theta+u\cdot\nabla\Delta\theta+\Delta
u\cdot\nabla\theta+2\sum_{i=1,2}\partial_{i}u\cdot\nabla\partial_{i}\theta=0,\end{split}$

we estimate for $j=1,2$

$\begin{split}\frac{1}{2}\frac{\mathrm{d}}{\mathrm{d}t}\|\partial_{j}\Delta\theta\|_{L^{2}}^{2}\leq
C(\|\nabla
u\|_{L^{\infty}}+\|\nabla\theta\|_{L^{\infty}})\|\partial_{j}\Delta\theta\|_{L^{2}}^{2}+C\|\theta\|_{H^{2}}\|\theta\|_{H^{3}}^{2}.\end{split}$

Here, we have used $L^{4}$ boundedness of the Riesz operator $\theta\mapsto u$
to bound

$\begin{split}\|\nabla^{2}u\|_{L^{4}}+\|\nabla^{2}\theta\|_{L^{4}}\leq
C\|\theta\|_{H^{2}}^{\frac{1}{2}}\|\theta\|_{H^{3}}^{\frac{1}{2}}.\end{split}$

Next, we use the logarithmic Sobolev inequality

$\begin{split}\|\nabla\theta\|_{L^{\infty}}\leq
C\|\theta\|_{H^{2}}\log\left(10+\frac{\|\theta\|_{H^{3}}}{\|\theta\|_{H^{2}}}\right)\end{split}$

and

$\begin{split}\|\nabla u\|_{L^{\infty}}\leq
C\|u\|_{H^{2}}\log\left(10+\frac{\|u\|_{H^{3}}}{\|u\|_{H^{2}}}\right)\leq
C\|\theta\|_{H^{2}}\log\left(10+\frac{\|\theta\|_{H^{3}}}{\|\theta\|_{H^{2}}}\right)\end{split}$

(we have used the lower bound $\|u\|_{H^{2}}\geq C\|\theta\|_{H^{2}}$).
Lastly, using $\|\theta\|_{H^{2}}\leq M_{N}$, we may deduce the a priori
estimate

$\begin{split}\frac{\mathrm{d}}{\mathrm{d}t}\|\theta\|_{H^{3}}^{2}\leq
CM_{N}\log\left(10+\|\theta\|_{H^{3}}\right)\|\theta\|_{H^{3}}^{2}\end{split}$

which shows that the $H^{3}$–norm of $\theta$ must remain finite up to
$t=T_{N}$. Higher norms of $\theta$ can be similarly controlled, so that the
solution $\theta$ remains $C^{\infty}$–smooth up to $t=T_{N}$.

In the following argument, $N_{0}\gg n_{0}$ will be taken to be sufficiently
large (but in a way depending only on a few absolute constants) whenever it
becomes necessary. Recall that we are assuming $N\geq N_{0}$. The following
argument is mainly a repetition of the proof of Theorem B above. For
convenience, let us fix

$\begin{split}\ell_{N}:=M_{N}^{3}.\end{split}$ (5.5)

Then, note from the definition of $M_{N}$ in (5.3) that $n_{0}\ll\ell_{N}\ll
N$. Here and in the following, we write $A\ll B$ if $A/B\rightarrow 0$ as
$N\to\infty$, where $A$ and $B$ are some positive expressions involving $N$.

Observe that the solution $\theta$ defined on $[0,T_{N}]$ satisfies the
properties stated in Lemma 4.2 and Claim I, II, III on the entire time
interval $[0,T_{N}]$ (by taking $N_{0}$ larger if necessary), simply because
we have

$\begin{split}T_{N}\ll\frac{1}{M_{N}}\end{split}$

from our choice of $T_{N}$ in (5.3). As in the above, we write the solution in
the form

$\begin{split}\theta=\sum_{n=n_{0}}^{N}n^{-\alpha}\theta^{(n)},\qquad\theta^{(n)}(t,\Phi(t,x))=\theta_{0,loc}^{(n)}(x),\end{split}$

and $\theta^{(n)}$ will be referred to as the $n$-th bubble. Then, for any
$\ell_{N}\leq k\leq N$, we have that the invariant timescale $T_{k}$ for the
$k$-th bubble satisfies

$\begin{split}T_{k}\leq T_{N}\qquad\mbox{and}\qquad
T_{k}\gtrsim\frac{1}{M_{N}+\sum_{j=n_{0}}^{k-1}j^{-\alpha}}\gtrsim
k^{\alpha-1}.\end{split}$

We have used that $\alpha$ is close to $\frac{1}{2}$. Now we consider the
values of $n$ satisfying

$\begin{split}n\geq C\ell_{N}\exp(c_{0}^{-1}M_{N})\end{split}$ (5.6)

for a sufficiently large absolute constant $C>0$. Then, at $t=T_{\ell_{N}}$,
we obtain similarly as before

$\begin{split}\|\theta^{(n)}(T_{\ell_{N}})\|_{\dot{H}^{2}}^{2}\gtrsim\left(\frac{n}{\ell_{N}}\right)^{2c_{0}}\end{split}$

whenever $n\leq N$ satisfies (5.6). Hence

$\begin{split}\|\theta(T_{\ell_{N}})\|_{H^{2}}^{2}\gtrsim\sum_{n=1+\lfloor
C\ell_{N}\exp(c_{0}^{-1}M_{N})\rfloor}^{N}n^{-2\alpha}\left(\frac{n}{\ell_{N}}\right)^{2c_{0}}\gtrsim\ell_{N}^{-2c_{0}}N^{1-2\alpha+2c_{0}}\gg
M_{N}^{2},\end{split}$

recalling the definitions of $M_{N}$ and $\ell_{N}$. We have used that
$N\gg\ell_{N}\exp(c_{0}^{-1}M_{N})$ to derive the last inequality. In
particular, for all sufficiently large $N$, we obtain

$\begin{split}\|\theta(T_{\ell_{N}})\|_{H^{2}}>M_{N},\end{split}$

which is a contradiction. This finishes the proof. ∎

## Acknowledgement

IJ has been supported by the Samsung Science and Technology Foundation under
Project Number SSTF-BA2002-04 and the New Faculty Startup Fund from Seoul
National University.

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