# Self-similar solutions for the Muskat equation

Eduardo García-Juárez† ,  Javier Gómez-Serrano‡ ,  Huy Q. Nguyen∗ and  Benoît
Pausader§ †,‡ Departament de Matemàtiques i Informàtica, Universitat de
Barcelona, Gran Via de les Corts Catalanes 585, 08007 Barcelona, Spain.
<EMAIL_ADDRESS><EMAIL_ADDRESS>‡,∗,§ Department of Mathematics,
Brown University, Kassar House, 151 Thayer Street, Providence, RI 02912, USA.
<EMAIL_ADDRESS><EMAIL_ADDRESS><EMAIL_ADDRESS>
###### Abstract.

We show the existence of self-similar solutions for the Muskat equation. These
solutions are parameterized by $0<s\ll 1$; they are exact corners of slope $s$
at $t=0$ and become smooth in $x$ for $t>0$.

##  1\. Introduction

###  1.1. The Muskat problem

The Muskat problem describes the evolution of the free boundary between
immiscible and incompressible fluids permeating a porous medium in a gravity
field. Each fluid is assumed to have constant physical properties and their
velocities are governed by Darcy’s law,

$\frac{\mu}{\kappa}u(z,t)=-\nabla p(z,t)-\rho g(0,1),\qquad
z\in\mathbb{R}^{2},\quad t\in\mathbb{R}_{+},$

where $p$, $u$ denote the pressure and velocity of the fluids, $\rho$, $\mu$
are their density and viscosity, $\kappa$ is the permeability constant of the
medium, and $g$ the gravitational constant. With or without surface tension
effects, it has long been known that the problem can be reduced to an
evolution equation for the free interface [23, 32, 45]. The case of a two-
fluid graph interface

$\Gamma(t)=\\{(x,f(x,t)):x\in\mathbb{R}\\}$

with only gravity effects admits a particularly compact form [23], now called
the Muskat equation:

$\begin{split}\partial_{t}f=\frac{1}{\pi}\int_{\mathbb{R}}\frac{\partial_{x}\Delta_{\alpha}f}{1+(\Delta_{\alpha}f)^{2}}d\alpha,\qquad\Delta_{\alpha}f(x)=\frac{f(x)-f(x-\alpha)}{\alpha},\qquad
x\in\mathbb{R}.\end{split}$ (1.1)

Above, all physical constants have been normalized for notational simplicity.
The Muskat equation is well-posed locally in time for sufficiently smooth
initial data, and globally in time if the initial interface is sufficiently
flat [8, 20, 22, 23, 24, 45, 46, 47]. Most notably, an initially smooth
interface can turn [13] and later lose regularity in finite time [14].
Furthermore, many other behaviors are possible, with interfaces that turn and
then go back to the graph scenario [25, 26]. Thus, finding criteria for global
existence became one of the main questions for the Muskat equation. Since
equation (1.1) has a natural scaling given by

$f(x,t)\rightarrow\lambda^{-1}f(\lambda x,\lambda t),\qquad\lambda>0,$

these criteria are stated in terms of critical regularity, i.e., spaces that
scale like $\dot{W}^{1,\infty}$. In this sense, having the product of the
maximal and minimal slopes strictly less than $1$ is sufficient for global
existence [11]. See also [28]. Medium-size initial data in critical spaces but
with uniformly continuous slope guarantees global wellposedness [20]. If the
initial data is sufficiently small in $\dot{H}^{\frac{3}{2}}$, then the slope
can be arbitrarily large [27] and even unbounded [5, 7]. The result [5] also
shows local existence and uniqueness in $H^{\frac{3}{2}}$. This is currently
the best (lowest) regularity result in terms of the space of the initial data,
which is a problem that has garnered a lot of attention recently (e.g. [3, 6,
17, 18, 21, 42, 43, 44]).

###  1.2. Main result

In this paper, we show the existence of self-similar solutions for the Muskat
problem. These solutions correspond to the global-in-time evolution of
initially exact corners, and thus they do not fit into the aforementioned
results111The results in [12] allow for merely medium size bounded slopes but
require sublinear growth of the profile, while our solutions grow linearly in
space..

We can rewrite (1.1) in terms of a closed system for the slope
$h=\partial_{x}f$:

$\begin{split}0=\partial_{t}h-\frac{1}{\pi}\int_{\mathbb{R}}\frac{\Delta_{\alpha}\partial_{x}h}{1+(\fint_{\alpha}h)^{2}}d\alpha+\frac{2}{\pi}\int_{\mathbb{R}}(\Delta_{\alpha}h)^{2}\cdot\frac{\fint_{\alpha}h}{(1+(\fint_{\alpha}h)^{2})^{2}}d\alpha,\qquad\fint_{\alpha}h(x)=\frac{1}{\alpha}\int_{x-\alpha}^{x}h(z)dz.\end{split}$
(1.2)

Plugging the ansatz $h(x,t)=k(x/t)$ in (1.2), we arrive at the equation

$\begin{split}0=Sk+\frac{1}{\pi}\int_{\mathbb{R}}\frac{\Delta_{\alpha}\partial_{y}k}{1+(\fint_{\alpha}k)^{2}}d\alpha-\frac{2}{\pi}\int_{\mathbb{R}}(\Delta_{\alpha}k)^{2}\cdot\frac{\fint_{\alpha}k}{(1+(\fint_{\alpha}k)^{2})^{2}}d\alpha,\qquad
S:=y\partial_{y}.\end{split}$ (1.3)

for which we construct a local curve of solutions:

######  Theorem 1.1.

There exists $s_{\ast}>0$ such that for all $|s|<s_{\ast}$, there exists a
self-similar solution of (1.2) $h_{s}(x,t)=k_{s}(x/t)$ satisfying
$\lim_{y\to+\infty}k_{s}(y)=s$. In addition, we have that

$\begin{split}\|k_{s}(y)-s\frac{2}{\pi}\arctan(y)\|_{H^{1}}+|s|\|\partial_{s}k_{s}(y)-\frac{2}{\pi}\arctan(y)\|_{H^{1}}&\lesssim|s|^{3}.\end{split}$

######  Remark 1.2.

$(i)$ In particular, we see that $k_{s}\in
L^{\infty}\setminus\dot{H}^{\frac{1}{2}}(\mathbb{R})$. $(ii)$ In fact, we will
compare $k_{s}$ and $(2s/\pi)\arctan((1+s^{2}/3)y)$, but one can readily see
that the difference between the two ansätze is $O(s^{3})$. $(iii)$ Due to the
symmetry we can find solutions with negative $s$ by setting
$k_{-s}(y)=-k_{s}(y)$. From now on, we assume $s\geq 0$.

Despite the numerous works on the Muskat equation and the mathematically
equivalent vertical Hele-Shaw problem, self-similar solutions were only known
for the simplified thin film Muskat [30, 33, 38]. See also [31] where the
authors find traveling solutions for the Muskat problem with surface tension
effects included.

From our numerical results, it might look surprising that, no matter how big
the slope is, the initial corner instantly smooths out, as opposed to the
known “waiting time” phenomenon in the Hele-Shaw problem [1, 9, 10, 16, 19,
32, 35, 36, 37]. We must note however that those works correspond to a
horizontal Hele-Shaw cell (hence without gravity) and some include fluid
injection. Moreover, some of these works are in a one-phase setting, which
even for Muskat significantly changes the possible behaviors [2, 4, 15, 29,
34].

###  1.3. Outline of the paper

The rest of the paper is structured as follows. In Section 2 we summarize the
notation that will be used along the paper. Next, in Section 3, we first
extract the quasilinear structure of (1.3) and rewrite it as a fixed point
equation. This section contains the proof of the main Theorem 1.1 via
Proposition 3.3. Section 4 contains the analysis of all the terms involved in
the equation. We will use key cancellations provided by some “elementary
bricks” that we will be able to extract through the symmetrization of the
nonlinear terms. Finally in Section 5 we illustrate our main Theorem by
numerically computing part of the branch of self-similar solutions.

##  2\. Notations

###  2.1. General notations

In the following, we fix $\varphi\in C^{\infty}_{c}(-4/3,4/3)$, a nonnegative
even function such that $\varphi\equiv 1$ on $[-3/4,3/4]$. For simplicity of
notation, we let $\fint_{\pm\alpha,0}f$ be an arbitrary function among

$\fint_{\pm\alpha,0}f\in\\{f,\fint_{\alpha}f,\fint_{-\alpha}f\\}.$

We will work mostly on the Fourier side. We define the Fourier transform as

$\begin{split}\mathcal{F}(f)(\xi)=\widehat{f}(\xi):=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}f(y)e^{-iy\xi}dy.\end{split}$

We note that the Fourier transform of a real odd function is an odd function
taking purely imaginary values. We define the Fourier multiplier $|\nabla|$ by

$\mathcal{F}\left\\{|\nabla|f\right\\}(\xi)=|\xi|\widehat{f}(\xi).$

Given an operator $T$, we let

$\widehat{T}[f]=\mathcal{F}^{-1}T[\widehat{f}]$ (2.1)

be its conjugation by the Fourier transform. To avoid functional analytic
considerations, we say that a functional $g\mapsto\mathcal{N}(g)$ is analytic
around a function $L_{s}$ if for any choice of $g_{1},g_{2}$ in the
appropriate space (here $H^{1}$) the restricted function

$\begin{split}N_{g_{1},g_{2}}(t_{1},t_{2}):=\mathcal{N}(L_{s}+t_{1}g_{1}+t_{2}g_{2})\end{split}$

is analytic. In this case, we denote

$\begin{split}\mathcal{N}_{1}[g]:=\frac{d}{dt_{1}}N_{g,0}(0,0),\qquad\mathcal{N}_{\geq
2}[g]:=\mathcal{N}(L_{s}+g)-\mathcal{N}(L_{s})-\mathcal{N}_{1}[g],\qquad\mathcal{N}_{\geq
1}[g]:=\mathcal{N}_{1}[g]+\mathcal{N}_{\geq 2}[g],\end{split}$ (2.2)

and we observe that

$\begin{split}\mathcal{N}_{\geq 2}(L_{s}+g_{2})-\mathcal{N}_{\geq
2}(L_{s}+g_{1})&=\int_{t_{1}=0}^{1}\frac{d}{dt_{1}}\frac{d}{dt_{2}}N_{g_{1},g_{2}-g_{1}}(t_{1},0)dt_{1}+\int_{\theta=0}^{1}(1-\theta)\frac{d^{2}}{d\theta^{2}}N_{g_{1},g_{2}-g_{1}}(1,\theta)d\theta.\end{split}$
(2.3)

In these notations, the “center” $L_{s}$ is implicit, but since we will always
consider functionals around $L_{s}$ defined in (3.11), there should be no
ambiguity.

###  2.2. Algebra of operators

We will use operators of the form

$\begin{split}\mathcal{N}(f)(y)&:=\int_{\alpha=0}^{\infty}m(f;\alpha,y)\cdot
G(f,\fint_{\alpha}f,\fint_{-\alpha}f)\frac{d\alpha}{\alpha}\end{split}$

associated to some multilinear function $f\mapsto m(f;\alpha,y)$ (i.e.
multilinear in $f$ for each fixed $\alpha$, $y$) and some numerical analytic
function $G$. For such operators, we compute that

$\begin{split}\mathcal{N}_{1}[g]&=\int_{\alpha=0}^{\infty}m_{1}(g,L_{s},\dots,L_{s};\alpha,y)\cdot
G(L_{s},\fint_{\alpha}L_{s},\fint_{-\alpha}L_{s})\frac{d\alpha}{\alpha}\\\
&\quad+\int_{\alpha=0}^{\infty}m(L_{s};\alpha,y)\cdot{\bf v}_{g}\cdot\nabla
G(L_{s},\fint_{\alpha}L_{s},\fint_{-\alpha}L_{s})\frac{d\alpha}{\alpha},\qquad{\bf
v}_{g}:=(g,\fint_{\alpha}g,\fint_{-\alpha}g),\end{split}$ (2.4)

and

$\begin{split}\frac{d}{dt_{1}}\frac{d}{dt_{2}}N_{g_{1},g_{2}}(t_{1},t_{2})&=\int_{\alpha=0}^{\infty}m_{2}(g_{1},g_{2},L_{t_{1},t_{2}};\alpha,y)\cdot
G(L_{t_{1},t_{2}},\fint_{\alpha}L_{t_{1},t_{2}},\fint_{-\alpha}L_{t_{1},t_{2}})\frac{d\alpha}{\alpha}\\\
&\quad+\int_{\alpha=0}^{\infty}m_{1}(g_{1},L_{t_{1},t_{2}};\alpha,y)\cdot{\bf
v}_{g_{2}}\cdot\nabla
G(L_{t_{1},t_{2}},\fint_{\alpha}L_{t_{1},t_{2}},\fint_{-\alpha}L_{t_{1},t_{2}})\frac{d\alpha}{\alpha}\\\
&\quad+\int_{\alpha=0}^{\infty}m_{1}(g_{2},L_{t_{1},t_{2}};\alpha,y)\cdot{\bf
v}_{g_{1}}\cdot\nabla
G(L_{t_{1},t_{2}},\fint_{\alpha}L_{t_{1},t_{2}},\fint_{-\alpha}L_{t_{1},t_{2}})\frac{d\alpha}{\alpha}\\\
&\quad+\int_{\alpha=0}^{\infty}m(L_{t_{1},t_{2}};\alpha,y)\cdot\nabla^{2}G(L_{t_{1},t_{2}},\fint_{\alpha}L_{t_{1},t_{2}},\fint_{-\alpha}L_{t_{1},t_{2}})[{\bf
v}_{g_{1}},{\bf v}_{g_{2}}]\frac{d\alpha}{\alpha},\end{split}$ (2.5)

where $m_{1}(f,g,\dots g)=d_{g}m_{1}\cdot f$ and similarly for $m_{j}$, and
$L_{t_{1},t_{2}}:=L_{s}+t_{1}g_{1}+t_{2}g_{2}$. Similarly

$\begin{split}\frac{d^{2}}{d\theta^{2}}N_{g_{1},g_{2}-g_{1}}(1,\theta)&=\int_{\alpha=0}^{\infty}m_{2}(g_{2}-g_{1},g_{2}-g_{1},L_{\theta};\alpha,y)\cdot
G(L_{\theta},\fint_{\alpha}L_{\theta},\fint_{-\alpha}L_{\theta})\frac{d\alpha}{\alpha}\\\
&\quad+2\int_{\alpha=0}^{\infty}m_{1}(g_{1}-g_{1},L_{\theta};\alpha,y)\cdot{\bf
v}_{g_{2}-g_{1}}\cdot\nabla
G(L_{\theta},\fint_{\alpha}L_{\theta},\fint_{-\alpha}L_{\theta})\frac{d\alpha}{\alpha}\\\
&\quad+\int_{\alpha=0}^{\infty}m(L_{\theta};\alpha,y)\cdot\nabla^{2}G(L_{\theta},\fint_{\alpha}L_{\theta},\fint_{-\alpha}L_{\theta})[{\bf
v}_{g_{2}-g_{1}},{\bf v}_{g_{2}-g_{1}}]\frac{d\alpha}{\alpha}.\end{split}$
(2.6)

##  3\. Reduction to a fixed point estimate

###  3.1. Analysis of the quasilinear structure

We can extract the quasilinear part from (1.3). This will be defined in terms
of two main terms. We define the function $F$ and the operator $W$ as follows

$\begin{split}F(t):=\left[1+t^{2}\right]^{-1},\qquad
W[g](y):=\frac{1}{\pi}\int_{\mathbb{R}}\frac{(\fint_{\alpha}g(y))^{2}-g^{2}(y)}{1+(\fint_{\alpha}g(y))^{2}}\frac{d\alpha}{\alpha},\end{split}$
(3.1)

and we obtain the following expression:

######  Lemma 3.1.

The self-similar profile $k$ satisfies

$\begin{split}|\nabla|k-S\left[k+\frac{k^{3}}{3}\right]+W[k]\partial_{y}k&=R[k]+\mathcal{T}[k],\end{split}$
(3.2)

where the semilinear terms are defined as

$\begin{split}R[h]&:=\frac{1}{\pi}\int_{\mathbb{R}}\partial_{\alpha}\left\\{h(y-\alpha)\right\\}\cdot\frac{(\fint_{\alpha}h)^{2}-h^{2}}{1+(\fint_{\alpha}h)^{2}}\frac{d\alpha}{\alpha},\\\
\mathcal{T}[h]&:=(1+h^{2})T[h];\qquad
T[h]:=\frac{1}{\pi}\int_{\mathbb{R}}(\Delta_{\alpha}h)^{2}\cdot
F^{\prime}(\fint_{\alpha}h)d\alpha.\end{split}$ (3.3)

###### Proof of Lemma 3.1.

Dividing by $1+k^{2}$, it suffices to show that (1.3) can be rewritten as

$\begin{split}\left[F(k)|\nabla|-S\right]k+V[k]\partial_{x}k&=F(k)R[k]+T[k],\\\
V[g](y)&:=\frac{1}{\pi}\int_{\mathbb{R}}\left\\{F(g(y))-F(\fint_{\alpha}g(y))\right\\}\frac{d\alpha}{\alpha}.\end{split}$
(3.4)

The only nontrivial part is the decomposition of the second term in (1.3). We
can expand

$\begin{split}\frac{1}{\pi}\int_{\mathbb{R}}\partial_{y}(\Delta_{\alpha}g)\cdot
F(\fint_{\alpha}g)d\alpha&=-F(g)\cdot|\nabla|g+\frac{1}{\pi}\int_{\mathbb{R}}\partial_{y}(\Delta_{\alpha}g)\cdot\left\\{F(\fint_{\alpha}g)-F(g)\right\\}d\alpha\\\
&=-F(g)\cdot|\nabla|g+\partial_{y}g(y)\frac{1}{\pi}\int_{\mathbb{R}}\left\\{F(\fint_{\alpha}g)-F(g)\right\\}\frac{d\alpha}{\alpha}\\\
&\quad-\frac{1}{\pi}\int_{\mathbb{R}}\partial_{y}g(y-\alpha)\cdot\left\\{F(\fint_{\alpha}g)-F(g)\right\\}\frac{d\alpha}{\alpha},\end{split}$

and rearranging the terms, we arrive at (3.4).

∎

###  3.2. Study of the linear equation: Duhamel formula

Given a constant $\kappa>0$, we now consider the linear adjusted equation from
(3.4)

$\begin{split}\left(\kappa|\nabla|-S\right)k&=p_{1}+\partial_{y}p_{2}\end{split}$
(3.5)

for an odd function $k$. Taking the Fourier transform and using Duhamel’s
formula, we obtain the ODE

$\begin{split}\partial_{\xi}\left\\{\xi
e^{\kappa|\xi|}\widehat{k}\right\\}&=e^{\kappa|\xi|}\widehat{p}_{1}+i\xi
e^{\kappa|\xi|}\widehat{p}_{2}.\end{split}$

If we assume that $\xi\widehat{k}(\xi)$ is continuous at the origin, and we
integrate from $0$, we find that

$\begin{split}\widehat{k}(\xi)&=C\frac{e^{-\kappa|\xi|}}{\xi}+{\bf
T}_{-1}[\widehat{p}_{1}]+{\bf T}_{0}[\widehat{p}_{2}],\end{split}$ (3.6)

for some constant $C$, where the linear operators are given by

$\begin{split}{\bf
T}_{-1}:g&\mapsto\frac{1}{\xi}\int_{\eta=0}^{\xi}e^{\kappa(|\eta|-|\xi|)}g(\eta)d\eta,\qquad{\bf
T}_{0}:g\mapsto\frac{i}{\xi}\int_{\eta=0}^{\xi}\eta
e^{\kappa(|\eta|-|\xi|)}g(\eta)d\eta.\end{split}$ (3.7)

In particular, under our assumptions, the odd solutions to the free equation
$\left(\kappa|\nabla|-S\right)k=0$ with the condition
$\lim_{y\to+\infty}k(y)=s$ are given by

$\begin{split}L_{s,\kappa}(y)&:=\frac{2s}{\pi}\arctan(y/\kappa).\end{split}$
(3.8)

From now on, we shall restrict ourselves to the study of odd solutions.

####  3.2.1. Operator estimates

It remains to estimate the solution operators from (3.7).

######  Lemma 3.2.

The operators

$\begin{split}{\bf
T}_{-1}:g&\mapsto\frac{1}{\xi}\int_{\eta=0}^{\xi}e^{\kappa(\eta-\xi)}g(\eta)d\eta,\qquad{\bf
T}_{0}:g\mapsto\frac{i}{\xi}\int_{\eta=0}^{\xi}\eta
e^{\kappa(\eta-\xi)}g(\eta)d\eta,\\\ \end{split}$

defined for functions on $L^{2}_{loc}(0,\infty)$ satisfy the boundedness
properties

$\begin{split}\|\xi{\bf T}_{-1}\|_{L^{2}\to L^{2}}+\|{\bf T}_{-1}\|_{L^{3}\to
L^{2}}&\lesssim 1,\\\ \|{\bf T}_{0}\|_{L^{2}\to L^{2}}+\|{\bf
T}_{0}\|_{L^{2}(\xi^{2}d\xi)\to L^{2}(\xi^{2}d\xi)}&\lesssim 1.\\\
\end{split}$

###### Proof of Lemma 3.2.

To control ${\bf T}_{-1}$, we first use Hölder’s inequality to bound

$\begin{split}|{\bf
T}_{-1}[g](\xi)|&\leq\xi^{-\frac{1}{p}}\left(\frac{1-e^{-\kappa\xi}}{\kappa\xi}\right)^{\frac{1}{p^{\prime}}}\left(\int_{\eta=0}^{\xi}e^{\kappa(\eta-\xi)}|g(\eta)|^{p}d\eta\right)^{\frac{1}{p}}\lesssim\|g\|_{L^{p}}\cdot\min\\{\xi^{-\frac{1}{p}},\xi^{-1}\\},\end{split}$

which shows the second bound. For the first, we compute by duality that

$\begin{split}\langle\xi{\bf T}_{-1}[g],h\rangle&=\iint
K(\xi,\eta)\overline{g}(\eta)h(\xi)d\xi d\eta,\qquad
K(\xi,\eta):=\mathfrak{1}_{\\{0\leq\eta\leq\xi\\}}e^{\kappa(\eta-\xi)},\end{split}$

and Schur’s test allows to conclude. This also gives the second estimate on
${\bf T}_{0}$, and the first one follows a similar proof with kernel
$K^{\prime}(\xi,\eta)=(\eta/\xi)K(\xi,\eta)$.

∎

###  3.3. Fixed point formulation

####  3.3.1. Linearization at self-similar profile

The key observation we will use is that for the solutions of the linearized
problem (3.8), the quasilinear part simplifies significantly since
$L_{s,\kappa}^{2}$ is almost constant away from a neighborhood of the origin:

$\begin{split}L_{s,\kappa}^{2}(y)=s^{2}+p(y),\qquad
p(y)=O(\langle\kappa/y\rangle^{-1}),\qquad
L_{s,\kappa}^{3}=s^{2}L_{s,\kappa}+(L_{s,\kappa}^{2}-s^{2})L_{s,\kappa}.\end{split}$

We can now linearize (3.2) at the constant $s$ to get an equation of the form
(3.5) with

$\kappa:=\left[1+s^{2}/3\right]^{-1}.$ (3.9)

More precisely, we obtain

$\begin{split}|\nabla|k-\left[1+s^{2}/3\right]Sk=\frac{1}{3}S\left[(k^{2}-s^{2})k\right]-W[k]\partial_{y}k+R[k]+\mathcal{T}[k].\end{split}$

which we prefer to rewrite via a normal form as

$\begin{split}\left[\kappa|\nabla|-S\right]h&=\frac{\kappa^{2}}{3}|\nabla|\left[(k^{2}-s^{2})k\right]-\kappa
W[k]\partial_{y}k+\kappa R[k]+\kappa\mathcal{T}[k],\\\
h&:=k+\frac{\kappa}{3}(k^{2}-s^{2})k.\end{split}$ (3.10)

####  3.3.2. Fixed point formulation

We can now seek solutions of (3.10) as perturbations of (3.8) when $s,\kappa$
are related as in (3.9). We seek solutions of the form

$\begin{split}k(y)=L_{s}(y)+g(y),\qquad
L_{s}(y):=\frac{2s}{\pi}\arctan(y/\kappa)=s\frac{2}{\pi}\arctan((1+s^{2}/3)y).\end{split}$
(3.11)

We note for later use that $L_{s}$ is a smooth function of $s$ and

$\begin{split}\partial_{s}L_{s}&=\frac{2}{\pi}\arctan(y/\kappa)+\frac{4}{3\pi}\frac{s^{2}y}{1+(y/\kappa)^{2}},\\\
\partial_{s}(L_{s}^{2}-s^{2})&=2s\left(s^{-2}(L_{s}^{2}-s^{2})+\frac{2}{\pi}\arctan(y/\kappa)\frac{4}{3\pi}\frac{s^{2}y}{1+(y/\kappa)^{2}}\right).\end{split}$
(3.12)

We define $\pi_{s}$ by

$\pi_{s}:=(L_{s}^{2}-s^{2})L_{s}.$

Then, plugging (3.11) into (3.10), we obtain

$\begin{split}h&=L_{s}+(\kappa/3)\pi_{s}+(1+\kappa(L_{s}^{2}-s^{2}/3))g+\kappa
L_{s}g^{2}+(\kappa/3)g^{3},\\\
\left[\kappa|\nabla|-S\right]h&=\kappa^{2}|\nabla|\left[(1/3)\pi_{s}+(L_{s}^{2}-s^{2}/3)g+L_{s}g^{2}+g^{3}/3\right]\\\
&\quad-\kappa
W[L_{s}]\partial_{y}L_{s}-\kappa\left(W[L_{s}+g]-W[L_{s}]\right)\partial_{y}\left(L_{s}+g\right)-\kappa\partial_{y}\left(W[L_{s}]g\right)+\kappa
g\partial_{y}W[L_{s}]\\\ &\quad+\kappa(R[L_{s}]+R_{1}[g]+R_{\geq
2}[g]+\mathcal{T}[L_{s}]+\mathcal{T}_{1}[g]+\mathcal{T}_{\geq
2}[g]).\end{split}$

Thus, using (3.6) and the notation (2.1), we obtain two equations for $h$
(after identifying the element in the kernel for the second equation from the
limit at $\infty$):

$\begin{split}h&=L_{s}+(\kappa/3)\pi_{s}+\left\\{3(1+s^{2})/(3+s^{2})+\kappa(L_{s}^{2}-s^{2})\right\\}g+\kappa
L_{s}g^{2}+(\kappa/3)g^{3}\\\ &=L_{s}+\kappa\widehat{{\bf
T}}_{-1}\left\\{(\kappa/3)|\nabla|\pi_{s}-W[L_{s}]\partial_{y}L_{s}+R[L_{s}]+\mathcal{T}[L_{s}]\right\\}\\\
&\quad+\kappa\widehat{{\bf
T}}_{-1}\\{\kappa|\nabla|(L_{s}^{2}-s^{2}/3)g-\partial_{y}\left(W[L_{s}]g\right)+g\partial_{y}W[L_{s}]-W_{1}[g]\partial_{y}L_{s}+R_{1}[g]+\mathcal{T}_{1}[g]\\\
&\quad+\kappa\widehat{{\bf
T}}_{-1}\\{\kappa|\nabla|(L_{s}g^{2}+g^{3}/3)-W_{\geq
2}[g]\partial_{y}L_{s}-W^{hi}_{\geq
1}[g]\partial_{y}g-\partial_{y}(gW^{lo}_{\geq 1}[g])+g\partial_{y}W^{lo}[g]\\\
&\quad\,+R_{\geq 2}[g]+\mathcal{T}_{\geq 2}[g]\\},\end{split}$

where $R_{1},\mathcal{T}_{1}$, $R_{\geq 2}$, $\mathcal{T}_{\geq 2}$ follow the
convention in (2.2) and $W=W^{lo}+W^{hi}$ is a decomposition later introduced
in (4.18). Combining the two equations for $h$, we arrive at the fixed-point
formulation:

$\begin{split}\Pi&=Ag+\mathcal{N}(g),\\\ \end{split}$ (3.13)

with forcing term

$\begin{split}\Pi&:=(\kappa/3)\pi_{s}-(\kappa^{2}/3)\widehat{{\bf
T}}_{-1}[|\nabla|\pi_{s}]+\kappa\widehat{{\bf
T}}_{-1}\left[W[L_{s}]\partial_{y}L_{s}-R[L_{s}]-\mathcal{T}[L_{s}]\right],\\\
\end{split}$ (3.14)

linear operator

$\begin{split}Ag&:=-(1+\frac{2s^{2}}{3+s^{2}}+\kappa(L_{s}^{2}-s^{2}))g+\kappa^{2}\widehat{{\bf
T}}_{-1}\left[|\nabla|(L_{s}^{2}-s^{2}/3)g\right]\\\
&\quad+\kappa\widehat{{\bf
T}}_{-1}\left[g\partial_{y}W[L_{s}]-W_{1}[g]\partial_{y}L_{s}+R_{1}[g]\right]-\kappa\widehat{{\bf
T}}_{0}[W[L_{s}]g]+\kappa\widehat{{\bf
T}}_{-1}[\mathcal{T}_{1}[g]],\end{split}$ (3.15)

and nonlinearity

$\begin{split}\mathcal{N}(f)&:=-\kappa(L_{s}+g/3)g^{2}+\kappa^{2}\widehat{{\bf
T}}_{-1}\left[|\nabla|(L_{s}g^{2}+g^{3}/3)\right]\\\
&\quad\,\,+\kappa\widehat{{\bf T}}_{-1}\left[-W_{\geq
2}[g]\partial_{y}L_{s}-W^{hi}_{\geq
1}[g]\partial_{y}g-g\partial_{y}W^{lo}_{\geq 1}[g]\right]-\kappa\widehat{{\bf
T}}_{0}[gW^{lo}_{\geq 1}[g]]\\\ &\quad\,\,+\kappa\widehat{{\bf
T}}_{-1}\left[R_{\geq 2}[g]\right]+\kappa\widehat{{\bf
T}}_{-1}\left[\mathcal{T}_{\geq 2}[g]\right],\end{split}$ (3.16)

where we used that $\widehat{{\bf T}}_{-1}[\partial_{y}f]=\widehat{{\bf
T}}_{0}[f]$. In view of this, Theorem 1.1 follows from the following existence
result.

######  Proposition 3.3.

There exists $s_{\ast}$ such that for all $0<s<s_{\ast}$, there exists exactly
one solution ${\bf g}_{s}$ of (3.13) with (3.14)-(3.16) in a ball in $H^{1}$,
$B_{H^{1}}(0,s_{\ast})$. In addition the mapping $s\mapsto{\bf g}_{s}$ is
$C^{1}$ in $H^{1}$ and

$\begin{split}\|\partial_{s}{\bf g}_{s}\|_{H^{1}}\lesssim s^{2}.\end{split}$
(3.17)

This proposition will be an easy consequence of the following quantitative
estimates proved in the next section.

######  Lemma 3.4.

There holds that

$\begin{split}\|\Pi\|_{H^{1}}&\lesssim
s^{3},\qquad\|\partial_{s}\Pi\|_{H^{1}}\lesssim s^{2}.\end{split}$

######  Lemma 3.5.

There holds that

$\|A+Id\|_{H^{1}\to H^{1}}\lesssim s^{2},\qquad\|\partial_{s}A\|_{H^{1}\to
H^{1}}\lesssim s.$

In particular, there exists $s_{1}>0$ such that $A$ is invertible in $H^{1}$
for all $0\leq s\leq s_{1}$.

######  Lemma 3.6.

There holds that $\mathcal{N}(0)=0$ and whenever
$\|g_{1}\|_{H^{1}}+\|g_{2}\|_{H^{1}}\leq 1$,

$\begin{split}\|\mathcal{N}(g_{1})-\mathcal{N}(g_{2})\|_{H^{1}}&\lesssim\|g_{1}-g_{2}\|_{H^{1}}\cdot\left[s^{2}+\|g_{1}\|_{H^{1}}^{2}+\|g_{2}\|_{H^{1}}^{2}\right].\end{split}$

###### Proof of Proposition 3.3.

We are now ready to prove Proposition 3.3 via a fixed point formulation. For
$\varepsilon>0$, we consider $X:=\\{g\in
H^{1}:\|g\|_{H^{1}}\leq\varepsilon\\}$ and we want to show that (3.13) has a
unique solution in $X$, provided that $0<s\leq s_{\ast}$ is small enough. We
define the mapping

$\begin{split}\Phi:g\mapsto A^{-1}\left[\Pi-\mathcal{N}(g)\right],\end{split}$

which is well defined on $H^{1}$ since $A$ is invertible for $0<s\ll 1$ by
Lemma 3.5. Using Lemma 3.5 and Lemma 3.6, we see that $\Phi:X\to X$ provided
$0<\varepsilon\leq\varepsilon_{\ast}$ is small enough. Finally, decreasing
$\varepsilon_{\ast}$ and using Lemma 3.6 again, we see that $\Phi$ is a
contraction on $X$. By the Banach fixed point theorem it has a unique fixed
point in $X$.

In addition, we can study the smoothness of $s\mapsto{\bf g}_{s}$. Deriving
(3.13), we find that

$\begin{split}(A+d_{g_{s}}\mathcal{N})[\partial_{s}g_{s}]=\partial_{s}\Pi_{s}-\left(\partial_{s}A\right)g_{s},\end{split}$

and using Lemma 3.4, Lemma 3.5, and Lemma 3.6 again, we deduce that
$s\mapsto{\bf g}_{s}$ is $C^{1}$ and we have (3.17).

∎

It remains to prove Lemma 3.4, Lemma 3.5 and Lemma 3.6. This will be done in
the next section.

##  4\. Quantitative analysis

###  4.1. Analysis of the elementary bricks

We need to control many terms. Fortunately, in many of them, the key
cancellation is provided by simple “elementary bricks” which can be analyzed
separately. We define the quadratic expressions for $\alpha\geq 0$:

$\begin{split}\delta_{\alpha}[f](y)&:=(\fint_{\alpha}f(y))^{2}-(\fint_{-\alpha}f(y))^{2},\qquad
J_{\alpha}[h](y):=\left(\fint_{\alpha}h(y)\right)^{2}-2h^{2}(y)+\left(\fint_{-\alpha}h(y)\right)^{2},\\\
\end{split}$ (4.1)

and let $\delta_{\alpha}[f,g]$ and $J_{\alpha}[f,g]$ denote the bilinear
expression obtained by polarization. We see that $\delta_{\alpha}[f]$ is odd
if $f$ is odd, while $J_{\alpha}[f]$ is even if $f$ is either even or odd.

We start by estimating the key cancellation.

######  Lemma 4.1.

Given $\delta_{\alpha}[f,g]$ defined in (4.1), $y\geq 0$ and $\alpha\geq 0$,
there holds that

$\begin{split}s^{-2}|\delta_{\alpha}[L_{s}](y)|&\lesssim\mathfrak{1}_{\\{0\leq\alpha\leq
y\\}}\alpha\ln\langle\alpha\rangle\cdot\min\\{y,y^{-2}\\}+\mathfrak{1}_{\\{0\leq
y\leq\alpha\\}}y\min\\{\alpha,\alpha^{-1}\\},\\\
s^{-2}|\partial_{y}\delta_{\alpha}[L_{s}](y)|&\lesssim\mathfrak{1}_{\\{0\leq\alpha\leq
y\\}}\alpha\ln\langle\alpha\rangle\cdot\langle y-\alpha\rangle^{-1}\langle
y\rangle^{-2}+\mathfrak{1}_{\\{0\leq
y\leq\alpha\\}}\min\\{\alpha,\alpha^{-1}\\},\end{split}$ (4.2)

and

$\begin{split}|\delta_{\alpha}[L_{s},g]|&\lesssim
s\|g\|_{H^{1}}\cdot\min\\{\alpha^{\frac{1}{2}},\alpha^{-\frac{1}{2}}\\},\\\
|\delta_{\alpha}[g,h]|&\lesssim\|g\|_{H^{1}}\|h\|_{H^{1}}\min\\{\alpha^{\frac{1}{2}},\alpha^{-1}\\}.\end{split}$
(4.3)

In addition,

$\begin{split}|\delta_{\alpha}[L_{s},g]|&\lesssim
s\int_{\\{|t|\leq\alpha\\}}\left[|g^{\prime}(y+t)|+\langle\alpha\rangle^{-1}|g(y+t)|\right]dt,\\\
|\delta_{\alpha}[g_{1},g_{2}]|&\lesssim\|g_{2}\|_{H^{1}}\int_{\\{|t|\leq\alpha\\}}\min\\{\alpha^{-\frac{1}{2}},\alpha^{-\frac{3}{2}}\\}|g_{1}(y+t)|dt,\\\
&\quad+\|g_{1}\|_{H^{1}}\int_{\\{|t|\leq\alpha\\}}\min\\{\alpha^{-\frac{1}{2}},\alpha^{-\frac{3}{2}}\\}|g_{2}(y+t)|dt\end{split}$
(4.4)

and a derivative brings powers of $\alpha^{-1}$,

$\begin{split}|\alpha\partial_{y}\delta_{\alpha}[L_{s},g]|(y)&\lesssim
s\Big{(}|g(y+\alpha)|+|g(y)|+|g(y-\alpha)|+\frac{1}{\alpha}\int_{t=-\alpha}^{\alpha}|g(y+t)|dt\Big{)}.\end{split}$
(4.5)

###### Proof of Lemma 4.1.

The key observation is that $2\delta_{\alpha}[f,g]=\delta[f,g]+\delta[g,f]$
where

$\begin{split}\delta[f,g]&:=-\frac{1}{\alpha}\int_{-\alpha}^{\alpha}f(y+t)dt\cdot\frac{1}{\alpha}\int_{t=0}^{\alpha}\left\\{g(y+t)-g(y-t)\right\\}dt.\\\
\end{split}$ (4.6)

We see that the first term in $\delta$ vanishes to first order for odd $f$ if
$\alpha\gg 1$, while the second term contains a difference and vanishes to
first order for $\alpha\ll 1$ when $g$ is smooth. The bounds in (4.2) follow
directly from the formula (4.6) and the bounds

$\begin{split}s^{-1}\left|\frac{1}{\alpha}\int_{-\alpha}^{\alpha}L_{s}(y+t)dt\right|&\lesssim
y/(\langle y\rangle+\langle\alpha\rangle),\\\
s^{-1}\left|\frac{1}{\alpha}\int_{t=0}^{\alpha}\left\\{L_{s}(y+t)-L_{s}(y-t)\right\\}dt\right|&\lesssim\mathfrak{1}_{\\{0\leq\alpha\leq
y\\}}\alpha\ln\langle\alpha\rangle\cdot\langle
y\rangle^{-2}+\mathfrak{1}_{\\{0\leq
y\leq\alpha\\}}\min\\{\alpha,1\\},\end{split}$ (4.7)

and

$\begin{split}s^{-1}\left|\frac{1}{\alpha}\int_{-\alpha}^{\alpha}L_{s}^{\prime}(y+t)dt\right|&\lesssim\langle
y\rangle^{-1}\langle y-\alpha\rangle^{-1}\mathfrak{1}_{\\{0\leq\alpha\leq
y\\}}+\langle\alpha\rangle^{-1}\mathfrak{1}_{\\{0\leq y\leq\alpha\\}},\\\
s^{-1}\left|\frac{1}{\alpha}\int_{t=0}^{\alpha}\left\\{L^{\prime}_{s}(y+t)-L^{\prime}_{s}(y-t)\right\\}dt\right|&\lesssim\alpha\min\\{y,\langle
y-\alpha\rangle^{-1}\langle y\rangle^{-2}\\}\mathfrak{1}_{\\{0\leq\alpha\leq
y\\}}+\min\\{\alpha^{2},\alpha^{-1}\\}\mathfrak{1}_{\\{0\leq
y\leq\alpha\\}}.\end{split}$

For general odd functions, we obtain that

$\begin{split}\left|\frac{1}{\alpha}\int_{-\alpha}^{\alpha}g(y+t)dt\right|&\lesssim\min\\{\sqrt{y+\alpha},\langle\alpha\rangle^{-\frac{1}{2}}\\}\|g\|_{H^{1}},\\\
\left|\frac{1}{\alpha}\int_{t=0}^{\alpha}\left\\{g(y+t)-g(y-t)\right\\}dt\right|&\lesssim\int_{\\{|t|\leq\alpha\\}}|g^{\prime}(y+t)|dt,\\\
\left|\frac{1}{\alpha}\int_{t=0}^{\alpha}\left\\{g(y+t)-g(y-t)\right\\}dt\right|&\lesssim\min\\{\alpha^{\frac{1}{2}},\alpha^{-\frac{1}{2}}\\}\|g\|_{H^{1}},\end{split}$
(4.8)

from which we deduce (4.3) and (4.4). In addition, we observe that

$\begin{split}\delta[\partial_{x}g_{1},g_{2}]&=\frac{g_{1}(y-\alpha)-g_{1}(y+\alpha)}{\alpha}\cdot\frac{1}{\alpha}\int_{t=0}^{\alpha}\left\\{g_{2}(y+t)-g_{2}(y-t)\right\\}dt,\\\
\delta[g_{1},\partial_{x}g_{2}]&=-\frac{1}{\alpha}\int_{-\alpha}^{\alpha}g_{1}(y+t)dt\cdot\frac{g_{2}(y+\alpha)-2g_{2}(y)+g_{2}(y-\alpha)}{\alpha},\\\
\end{split}$

and using (4.7), we easily arrive at (4.5).

∎

######  Lemma 4.2.

Let $J_{\alpha}$ be defined as in (4.1). For $1\leq p\leq\infty$, there holds
that

$\begin{split}\|\fint_{\alpha}h\|_{L^{\infty}_{\alpha}L^{p}_{y}}+\|\alpha\partial_{\alpha}\left(\fint_{\pm\alpha}h\right)\|_{L^{\infty}_{\alpha}L^{p}_{y}}&\lesssim\|h\|_{L^{p}},\\\
\|J_{\alpha}[h]\|_{L^{\infty}}+\|\alpha\partial_{\alpha}J_{\alpha}[h]\|_{L^{\infty}}&\lesssim\|h\|_{L^{\infty}}^{2},\end{split}$
(4.9)

and for $L_{s}$ we have the more precise bound,

$\begin{split}s^{-2}|J_{\alpha}[L_{s}](y)|&\lesssim\alpha\ln\langle\alpha\rangle\cdot\langle
y\rangle^{-2}\mathfrak{1}_{\\{0\leq\alpha\leq
y\\}}+\min\\{\alpha^{2},1\\}\mathfrak{1}_{\\{0\leq
y\leq\alpha\\}}\lesssim\min\\{\alpha,1\\},\end{split}$ (4.10)

while

$\begin{split}s^{-1}|J_{\alpha}[h,L_{s}](y)|&\lesssim\int_{|t|\leq\alpha}\left[|h(y+t)|+|h^{\prime}(y+t)|\right]dt,\\\
|J_{\alpha}[h_{1},h_{2}](y)|&\lesssim\sum_{\\{a,b\\}=\\{1,2\\}}\|h_{a}\|_{L^{\infty}}\int_{|t|\leq\alpha}\left[|h_{b}(y+t)|+|h_{b}^{\prime}(y+t)|\right]dt.\end{split}$
(4.11)

###### Proof of Lemma 4.2.

For (4.9), the estimates in $\fint_{\alpha}h$ follow from direct computations.
They directly imply the estimates on $J_{\alpha}$. To analyze $J_{\alpha}$, we
can rewrite $J_{\alpha}[h]=J[h,h]$ where222Note that since $J$ is not
symmetric, $J_{\alpha}[g,h]=J[g,h]+J[h,g]$.

$\begin{split}J[g,h]&=\frac{1}{2\alpha}\int_{t=0}^{\alpha}(g(y+t)-2g(y)+g(y-t))dt\cdot\frac{1}{\alpha}\int_{t=0}^{\alpha}(h(y+t)+2h(y)+h(y-t))dt\\\
&\quad+\frac{1}{2}\left(\frac{1}{\alpha}\int_{t=0}^{\alpha}(g(y+t)-g(y-t))dt\right)\cdot\left(\frac{1}{\alpha}\int_{t=0}^{\alpha}(h(y+t)-h(y-t))dt\right).\end{split}$
(4.12)

For (4.10), the second term can be estimated using (4.7), while for the first
term, we rewrite

$\begin{split}\frac{1}{\alpha}\int_{t=0}^{\alpha}(h(y+t)-2h(y)+h(y-t))dt&=\frac{1}{\alpha}\int_{u=-\alpha}^{\alpha}(\alpha-|u|)^{2}h^{\prime\prime}(y+u)du.\end{split}$
(4.13)

It remains to check (4.11). For the linear term, this follows from (4.12) and
(4.7) for the second term, while for the first term, we use (4.13) in case
$g=L_{s}$ and

$\begin{split}\left|\frac{1}{\alpha}\int_{t=0}^{\alpha}(h(y+t)-2h(y)+h(y-t))dt\right|&\leq\int_{\\{|t|\leq\alpha\\}}|h^{\prime}(y+t)|dt\end{split}$

else. For the quadratic estimate, we proceed similarly for the first term in
(4.12) and we use (4.8) for the second term.

∎

###  4.2. Control on the nonlinear operators

In this section, we will upgrade the bounds on the basic bricks in Section 4.1
to bounds on more complicated operators. Since they will not be multilinear,
we will only operate under the assumption that

$\|g_{1}\|_{H^{1}}+\|g_{2}\|_{H^{1}}+\|g_{3}\|_{H^{1}}\leq 1.$ (4.14)

####  4.2.1. Symmetrization of the nonlinear operators

We can symmetrize to obtain a key cancellation in the operators involved in
(3.2). The main observation is that one can always extract $\delta$ defined in
(4.6). Using that

$\begin{split}\int_{\mathbb{R}}H(\alpha)\frac{d\alpha}{\alpha}=\int_{\mathbb{R}_{+}}\left\\{H(\alpha)-H(-\alpha)\right\\}\frac{d\alpha}{\alpha},\end{split}$
(4.15)

and the symmetrization formulas

$\begin{split}F(a)-F(b)&=(b^{2}-a^{2})\cdot F(a)F(b),\\\
F(a)+F(b)&=(2+a^{2}+b^{2})\cdot F(a)F(b),\end{split}$ (4.16)
$\begin{split}2(a_{+}b_{+}-a_{-}b_{-})&=(a_{+}-a_{-})(b_{+}+b_{-})+(a_{+}+a_{-})(b_{+}-b_{-}),\\\
2(a_{+}b_{+}+a_{-}b_{-})&=(a_{+}+a_{-})(b_{+}+b_{-})+(a_{+}-a_{-})(b_{+}-b_{-}),\end{split}$
(4.17)

we can rewrite $W[g]=W[g,g;g]$, where

$\begin{split}W[g_{1},g_{2};g_{3}]&=\frac{1}{\pi}\int_{\mathbb{R}_{+}}\delta_{\alpha}[g_{1},g_{2}]\cdot
F(\fint_{\alpha}g_{3})\cdot
F(\fint_{-\alpha}g_{3})\cdot(1+g_{3}^{2})\frac{d\alpha}{\alpha}\\\
&=\int_{\mathbb{R}_{+}}\delta_{\alpha}[g_{1},g_{2}]\cdot
G(g_{3},\fint_{\alpha}g_{3},\fint_{-\alpha}g_{3})\frac{d\alpha}{\alpha}\end{split}$

for some function $G$ analytic in a neighborhood of $(0,0,0)$. In fact, to
properly control the linear part, it will be convenient to decompose further
$W=W^{lo}+W^{hi}$ where

$\begin{split}W^{hi}[g_{1},g_{2};g_{3}]&=\int_{\mathbb{R}_{+}}\delta_{\alpha}[g_{1},g_{2}]\cdot
G(g_{3},\fint_{\alpha}g_{3},\fint_{-\alpha}g_{3})\varphi(\alpha)\frac{d\alpha}{\alpha},\\\
W^{lo}[g_{1},g_{2};g_{3}]&=\int_{\mathbb{R}_{+}}\delta_{\alpha}[g_{1},g_{2}]\cdot
G(g_{3},\fint_{\alpha}g_{3},\fint_{-\alpha}g_{3})(1-\varphi(\alpha))\frac{d\alpha}{\alpha}.\end{split}$
(4.18)

####  4.2.2. Control on symmetrized operators

######  Lemma 4.3.

There holds that for $\ast\in\\{hi,lo\\}$,

$\begin{split}\|W^{\ast}[L_{s}]\|_{L^{\infty}}+\|\langle
y\rangle\partial_{y}W^{\ast}[L_{s}]\|_{L^{\infty}}&\lesssim s^{2},\end{split}$
(4.19)

and

$\begin{split}\|W^{\ast}_{1}[g]\|_{L^{\infty}}+\|W^{hi}_{1}[g]\|_{L^{2}}+\|\partial_{y}(W^{lo}_{1}[g])\|_{L^{2}}&\lesssim
s\|g\|_{H^{1}}.\end{split}$ (4.20)

Finally, we have the nonlinear estimates assuming (4.14)

$\begin{split}\|W^{\ast}_{\geq 2}[g]\|_{L^{2}\cap
L^{\infty}}&\lesssim\|g\|_{H^{1}}^{2},\\\ \|W^{\ast}_{\geq
2}[g_{1}]-W^{\ast}_{\geq
2}[g_{2}]\|_{L^{\infty}}&\lesssim(s+\|g_{1}\|_{H^{1}}+\|g_{2}\|_{H^{1}})\|g_{1}-g_{2}\|_{H^{1}}.\end{split}$
(4.21)

###### Proof of Lemma 4.3.

We start from the formula (4.18). For (4.19), it suffices to show that

$\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s}]|\frac{d\alpha}{\alpha}\|_{L^{\infty}}+\|\int_{\mathbb{R}_{+}}|\langle
y\rangle\partial_{y}\delta_{\alpha}[L_{s}]|\frac{d\alpha}{\alpha}\|_{L^{\infty}}\lesssim
s^{2},$ (4.22)

which follows from (4.2) and the bounds

$\|L_{s}\|_{L^{\infty}}+\|\langle y\rangle\partial_{y}L_{s}\|_{L^{\infty}}\leq
s.$

For (4.20), inspecting (2.4), and using (4.9) and (4.22), we see that it
suffices to show that

$\begin{split}\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s},g]|\frac{d\alpha}{\alpha}\|_{L^{\infty}}+\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s},g]|\varphi(\alpha)\frac{d\alpha}{\alpha}\|_{L^{2}}+\|\int_{\mathbb{R}_{+}}|\partial_{y}\delta_{\alpha}[L_{s},g]|(1-\varphi(\alpha))\frac{d\alpha}{\alpha}\|_{L^{2}}&\lesssim
s\|g\|_{H^{1}},\end{split}$ (4.23)

and

$\begin{split}\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s}]|\cdot|\fint_{\pm\alpha}g|\varphi(\alpha)\frac{d\alpha}{\alpha}\|_{L^{2}}+\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s}]|\cdot|\partial_{y}\fint_{\pm\alpha}g|(1-\varphi(\alpha))\frac{d\alpha}{\alpha}\|_{L^{2}}\lesssim
s\|g\|_{L^{2}}.\end{split}$

We start with the above bound. First, we see from (4.2) that
$s^{-1}|\delta_{\alpha}[L_{s}]|\lesssim s\min\\{1,\sqrt{|\alpha|}\\}$ and we
compute that

$\begin{split}\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s}]|\cdot|\fint_{\pm\alpha}g|\varphi(\alpha)\frac{d\alpha}{\alpha}\|_{L^{2}}&\lesssim
s\iint_{\\{|t|\leq|\alpha|\leq
2\\}}\|g(y+t)\|_{L^{2}}\varphi(\alpha)\frac{d\alpha}{\alpha^{\frac{3}{2}}}dt\lesssim
s\|g\|_{L^{2}},\\\ \end{split}$

and similarly,

$\begin{split}\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s}]|\cdot|\partial_{y}\fint_{\pm\alpha}g|(1-\varphi(\alpha))\frac{d\alpha}{\alpha}\|_{L^{2}}&\lesssim
s\int_{\\{\alpha\geq
1/2\\}}\left[\|g(y\mp\alpha)\|_{L^{2}_{y}}+\|g(y)\|_{L^{2}_{y}}\right]\frac{d\alpha}{\alpha^{2}}\lesssim
s\|g\|_{L^{2}}.\end{split}$

The first term in (4.23) can be controlled through (4.3), the second term can
be controlled through (4.4) and the last term through (4.5).

Finally, we consider (4.21). Inspecting (2.3), (2.5) and (2.6), we see that we
need to show

$\begin{split}\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[g_{1},g_{2}]|\frac{d\alpha}{\alpha}\|_{L^{\infty}\cap
L^{2}}&\lesssim\|g_{1}\|_{H^{1}}\|g_{2}\|_{H^{1}},\\\
\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s},g_{1}]\cdot\fint_{\pm\alpha,0}g_{2}|\frac{d\alpha}{\alpha}\|_{L^{\infty}\cap
L^{2}}&\lesssim s\|g_{1}\|_{H^{1}}\|g_{2}\|_{H^{1}},\\\
\|\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s}]\cdot\fint_{\pm\alpha,0}g_{1}\cdot\fint_{\pm\alpha,0}g_{2}|\frac{d\alpha}{\alpha}\|_{L^{\infty}\cap
L^{2}}&\lesssim s^{2}\|g_{1}\|_{H^{1}}\|g_{2}\|_{H^{1}}.\end{split}$ (4.24)

The $L^{\infty}$ bounds follow from (4.3) and (4.22) together with the simple
bound $\|g\|_{L^{\infty}}\lesssim\|g\|_{H^{1}}$. The $L^{2}$ bound follows
from (4.4) and (4.3) for the first two estimates while for the last, we see
that

$\begin{split}\int_{\mathbb{R}_{+}}|\delta_{\alpha}[L_{s}]\cdot\fint_{\pm\alpha,0}g_{1}\cdot\fint_{\pm\alpha,0}g_{2}|\frac{d\alpha}{\alpha}\lesssim
I_{1}+I_{2}+I_{3},\qquad
I_{j}:=\iiint_{S_{j}}|\delta_{\alpha}[L_{s}](y)|\cdot|g_{1}(y+t)g_{2}(y+u)|\frac{d\alpha}{\alpha^{3}}dtdu,\\\
S_{1}:=\\{|u|\leq|t|\leq\alpha\leq y\\},\qquad
S_{2}:=\\{|u|,y\leq|t|\leq\alpha\\},\qquad S_{3}:=\\{|u|\leq|t|\leq
y\leq\alpha\\},\end{split}$

and using (4.2), we can compute that

$\begin{split}\|I_{1}\|_{L^{2}}&\lesssim
s^{2}\iint_{\\{|u|\leq|t|\\}}\frac{\ln\langle t\rangle}{t\langle
t\rangle^{2}}\cdot\|g_{1}(y+t)g_{2}(y+u)\|_{L^{2}_{y}}dtdu\lesssim
s^{2}\|g_{1}\|_{L^{2}}\|g_{2}\|_{L^{\infty}},\\\ \end{split}$

while

$\begin{split}|I_{2}|&\lesssim s^{2}\iint\langle
t\rangle^{-2}|g_{1}(y+t)g_{2}(y+u)|dtdu\lesssim
s^{2}\|g_{2}\|_{L^{2}}\iint\langle t\rangle^{-\frac{3}{2}}|g_{1}(y+t)|dt,\\\
|I_{3}|&\lesssim s^{2}\langle y\rangle^{-2}\iint_{\\{|u|\leq|t|\leq
y\\}}|g_{1}(y+t)g_{2}(y+u)|dudt\lesssim s^{2}\langle
y\rangle^{-1}\|g_{1}\|_{L^{2}}\|g_{2}\|_{L^{2}},\end{split}$

and we see that $I_{2}$ and $I_{3}$ are in $L^{2}$. This finishes the proof.

∎

We now turn to the first semilinear term.

######  Lemma 4.4.

Assume (4.14) and consider $R$ defined in (3.3). There holds that

$\begin{split}\|R[L_{s}]\|_{L^{\frac{4}{3}}\cap L^{2}}&\lesssim
s^{3},\qquad\|R_{1}[g]\|_{L^{\frac{4}{3}}\cap L^{2}}\lesssim
s^{2}\|g\|_{H^{1}},\\\ \|R_{\geq 2}[g_{1}]-R_{\geq 2}[g_{2}]\|_{L^{1}\cap
L^{2}}&\lesssim\left[s^{2}+\|g_{1}\|_{H^{1}}^{2}+\|g_{2}\|_{H^{1}}^{2}\right]\|g_{2}-g_{1}\|_{H^{1}}.\end{split}$
(4.25)

###### Proof of Lemma 4.4.

From (4.16) and (4.17) we deduce that

$\begin{split}4(a_{+}b_{+}c_{+}-a_{-}b_{-}c_{-})&=(a_{+}-a_{-})(b_{+}+b_{-})(c_{+}+c_{-})+(a_{+}+a_{-})(b_{+}-b_{-})(c_{+}+c_{-})\\\
&\quad+(a_{+}+a_{-})(b_{+}+b_{-})(c_{+}-c_{-})+(a_{+}-a_{-})(b_{+}-b_{-})(c_{+}-c_{-}).\end{split}$
(4.26)

We can use (4.15), (4.26) and (4.16) to decompose

$\begin{split}R[h]&=-\frac{1}{\pi}\int_{\mathbb{R}}h^{\prime}(y-\alpha)\cdot\frac{(\fint_{\alpha}h)^{2}-h^{2}}{1+(\fint_{\alpha}h)^{2}}\frac{d\alpha}{\alpha}=\frac{1}{4\pi}(R^{1}-2R^{2}-R^{3}),\end{split}$

where $R^{j}=R^{j}[h;h]$ is given by

$\begin{split}R^{1}[h_{1};h_{2}]&=\int_{\mathbb{R}_{+}}\partial_{\alpha}\left\\{h_{1}(y+\alpha)+h_{1}(y-\alpha)\right\\}\cdot
J_{\alpha}[h_{2}]\cdot(F_{\alpha}+F_{-\alpha})\frac{d\alpha}{\alpha},\\\
R^{2}[h_{1};h_{2}]&=\int_{\mathbb{R}_{+}}\partial_{\alpha}\left\\{h_{1}(y+\alpha)-h_{1}(y-\alpha)\right\\}\cdot\delta_{\alpha}[h_{2}](y)\cdot(1+h^{2}_{2})\left(F_{\alpha}F_{-\alpha}\right)\frac{d\alpha}{\alpha},\\\
R^{3}[h_{1};h_{2}]&=\int_{\mathbb{R}_{+}}\partial_{\alpha}\left\\{h_{1}(y+\alpha)+h_{1}(y-\alpha)\right\\}\cdot\left[\delta_{\alpha}[h_{2}](y)\right]^{2}\cdot(F_{\alpha}F_{-\alpha})\frac{d\alpha}{\alpha},\\\
\end{split}$

with the notation

$\begin{split}F_{\alpha}:=F(\fint_{\alpha}h_{2}).\end{split}$

From now on, we will denote $F_{\alpha}=F(\fint_{\pm\alpha}h)$ without
distinction on $h$.

We start with the first estimate in (4.25). First, using (4.2) we observe that

$\begin{split}|R^{2}[L_{s};L_{s}]|+|R^{3}[L_{s};L_{s}]|&\lesssim
s\int_{\mathbb{R}_{+}}\langle
y-\alpha\rangle^{-2}|\delta_{\alpha}[L_{s}]|\frac{d\alpha}{\alpha}\lesssim
s^{3}\langle y\rangle^{-1}.\end{split}$ (4.27)

In addition, using that

$\begin{split}L_{s}^{\prime}(y-\alpha)-L_{s}^{\prime}(y+\alpha)&=\frac{2s\kappa}{\pi}\frac{4y\alpha}{(\kappa^{2}+(y-\alpha)^{2})(\kappa^{2}+(y+\alpha)^{2})},\end{split}$

and (4.10), we see that

$\begin{split}|R^{1}[L_{s};h](y)|&\lesssim
s\|h\|_{L^{\infty}}^{2}y\int\frac{d\alpha}{(\kappa^{2}+(y-\alpha)^{2})(\kappa^{2}+(y+\alpha)^{2})}\lesssim
s\|h\|_{L^{\infty}}^{2}\cdot y\langle y\rangle^{-2}.\end{split}$ (4.28)

We now turn to the other estimates in (4.25) and we start with
$\widetilde{R}^{j}[L_{s}+g]:=R^{j}[L_{s};L_{s}+g]$. Inspecting (2.4), we see
that the linear component follows from (4.27), (4.28) and from the bound

$\begin{split}s\int_{\mathbb{R}_{+}}\langle
y-\alpha\rangle^{-2}|\delta_{\alpha}[L_{s},g]|\frac{d\alpha}{\alpha}\lesssim
s^{2}\|g\|_{H^{1}}\cdot\langle y\rangle^{-3/2},\end{split}$

which follows from (4.3). Similarly, inspecting (2.3) and (2.5)-(2.6), we see
that the higher order terms can be controlled similarly since we can easily
estimate

$\begin{split}s\int_{\mathbb{R}_{+}}\langle
y-\alpha\rangle^{-2}|\delta_{\alpha}[g_{1},g_{2}]|\frac{d\alpha}{\alpha}\lesssim
s\|g_{1}\|_{H^{1}}\|g_{2}\|_{H^{1}}\cdot\langle y\rangle^{-3/2}\end{split}$

using (4.3) again.

We now consider $R^{j}[g_{1};L_{s}]$. Using Cauchy-Schwarz and (4.2), (4.10),
we see that

$\begin{split}R^{1,a}[h]&:=\int_{\mathbb{R}^{+}}\partial_{y}\left\\{h(y+\alpha)-h(y-\alpha)\right\\}\cdot
J_{\alpha}[L_{s}]\cdot(F_{\alpha}+F_{-\alpha})\cdot\varphi(4\langle
y\rangle^{-1}\alpha)\frac{d\alpha}{\alpha},\\\
R^{2,a}[h]&:=\int_{\mathbb{R}^{+}}\partial_{y}\left\\{h(y+\alpha)+h(y-\alpha)\right\\}\cdot\delta_{\alpha}[L_{s}]\cdot(1+L_{s}^{2})(F_{\alpha}F_{-\alpha})\cdot\varphi(4\langle
y\rangle^{-1}\alpha)\frac{d\alpha}{\alpha},\\\ \end{split}$

satisfy

$\begin{split}|R^{1,a}[h]|+|R^{2,a}[h]|&\lesssim
s^{2}\|\partial_{y}h\|_{L^{2}}\cdot\langle y\rangle^{-\frac{3}{2}}\ln\langle
y\rangle,\end{split}$ (4.29)

which gives an acceptable contribution. On the other hand, integrating by
parts and using (4.9), we see that

$\begin{split}R^{1,b}&=\int_{\mathbb{R}_{+}}\partial_{\alpha}\left\\{h(y+\alpha)+h(y-\alpha)\right\\}\cdot
J_{\alpha}[L_{s}]\cdot(F_{\alpha}+F_{-\alpha})\cdot(1-\varphi(4\langle
y\rangle^{-1}\alpha))\frac{d\alpha}{\alpha}\\\
&=-\int_{\mathbb{R}_{+}}\left\\{h(y+\alpha)+h(y-\alpha)\right\\}\cdot\alpha\partial_{\alpha}\left\\{J_{\alpha}[L_{s}]\cdot(F_{\alpha}+F_{-\alpha})\cdot\frac{1-\varphi(4\langle
y\rangle^{-1}\alpha)}{\alpha}\right\\}\frac{d\alpha}{\alpha}\\\ \end{split}$

satisfies

$\begin{split}|R^{1,b}(y)|&\lesssim s^{2}\|h\|_{L^{2}}\langle
y\rangle^{-\frac{3}{2}}\end{split}$

and similarly for $R^{2,b}$. $R^{3}$ can be treated similarly as $R^{2}$. To
finish the proof, it only remains to show

$\begin{split}\|R^{j}[g_{1};L_{s}\\!+\\!g_{1}]\\!-\\!R^{j}[g_{1};L_{s}]\\!-\\!(R^{j}[g_{2};L_{s}\\!+\\!g_{2}]\\!-\\!R^{j}[g_{2};L_{s}])\|_{L^{1}\cap
L^{2}}&\lesssim\left[s^{2}\\!+\\!\|g_{1}\|_{H^{1}}^{2}\\!+\\!\|g_{2}\|_{H^{1}}^{2}\right]\|g_{1}\\!-\\!g_{2}\|_{H^{1}}.\end{split}$

We start with the case $j=2$. The case $j=3$ is similar and will not be
detailed. We decompose

$\begin{split}R^{2}[g_{1};L_{s}+g_{2}]&=R^{2,lo}[g_{1};L_{s}+g_{2}]+R^{2,hi}[g_{1};L_{s}+g_{2}],\\\
R^{2,hi}[g_{1};h]&:=\int_{\mathbb{R}_{+}}\partial_{\alpha}\left\\{g_{1}(y+\alpha)-g_{1}(y-\alpha)\right\\}\cdot\delta[h,h](y)\cdot(1+h^{2})\left(F_{\alpha}F_{-\alpha}\right)\cdot\varphi(\alpha)\frac{d\alpha}{\alpha},\\\
R^{2,lo}[g_{1};h]&:=\int_{\mathbb{R}_{+}}\left\\{g_{1}(y+\alpha)-g_{1}(y-\alpha)\right\\}\cdot\alpha\partial_{\alpha}\left\\{\delta[h,h](y)\cdot(1+h^{2})\left(F_{\alpha}F_{-\alpha}\right)\cdot\frac{1-\varphi(\alpha)}{\alpha}\right\\}\frac{d\alpha}{\alpha}.\end{split}$

Inspecting (2.3) and (2.5)-(2.6), we see that to control $R^{2,hi}$, it
suffices to show that

$\begin{split}\|\int_{\mathbb{R}_{+}}|g_{1}^{\prime}(y\pm\alpha)|\cdot|\delta_{\alpha}[L_{s}](y)|\varphi(\alpha)\frac{d\alpha}{\alpha}\|_{L^{1}\cap
L^{2}}&\lesssim s^{2}\|g_{1}^{\prime}\|_{L^{2}},\\\
\|\int_{\mathbb{R}_{+}}|g_{1}^{\prime}(y\pm\alpha)|\cdot|\delta[g_{2},L_{s}](y)|\varphi(\alpha)\frac{d\alpha}{\alpha}\|_{L^{1}\cap
L^{2}}&\lesssim s\|g_{1}^{\prime}\|_{L^{2}}\|g_{2}\|_{H^{1}},\\\
\|\int_{\mathbb{R}_{+}}|g_{1}^{\prime}(y\pm\alpha)|\cdot|\delta[g_{2},g_{3}](y)|\varphi(\alpha)\frac{d\alpha}{\alpha}\|_{L^{1}\cap
L^{2}}&\lesssim\|g_{1}^{\prime}\|_{L^{2}}\|g_{2}\|_{H^{1}}\|g_{3}\|_{H^{1}}.\end{split}$

The first estimate follows by Cauchy-Schwarz as in (4.29); the last two follow
using (4.3) and (4.4). Independently,

$\begin{split}|R^{2,lo}_{\geq
1}[g_{1};L_{s}+g_{2}]|&\lesssim\sum_{\pm}\left[s+\|g_{2}\|_{L^{\infty}}\right]\int_{\mathbb{R}_{+}}|g_{1}(y\pm\alpha)\cdot\left[|\fint_{\pm\alpha,0}g_{2}(y)|+|\alpha\partial_{\alpha}\fint_{\pm\alpha}g_{2}(y)|\right]\cdot\frac{1-\varphi(\alpha)}{\alpha^{2}}d\alpha,\end{split}$

so that this term can easily be handled using (4.9).

It remains to consider $R^{1}[g_{1};L_{s}+g_{2}]$. The broad ideas are the
same as for $R^{2}$, but we need to be slightly more careful because
$J_{\alpha}$ has less good properties than $\delta_{\alpha}$ and we need to
take advantage of the difference of derivative in $g_{1}$ to compensate for
this.

We claim that it suffices to show that for any dyadic number $A>0$, there
exists a decomposition
$R^{1}[g_{1};L_{s}+g_{2}]=R^{1,hi}_{A}[g_{1};L_{s}+g_{2}]+R^{1,lo}_{A}[g_{1};L_{s}+g_{2}]$
such that

$\begin{split}\|R^{1,\ast}[g;L_{s}+h]\|_{L^{1}\cap L^{2}}&\lesssim
C(\ast,A,g)\left[s^{2}+\|h\|_{H^{1}}^{2}\right],\\\
\|R^{1,\ast}[g;L_{s}+h_{2}]-R^{1,\ast}[g;L_{s}+h_{1}]\|_{L^{1}\cap
L^{2}}&\lesssim
C(\ast,A,g)\|h_{2}-h_{1}\|_{H^{1}}\left[s+\|h_{1}\|_{H^{1}}+\|h_{2}\|_{H^{1}}\right],\\\
\end{split}$ (4.30)

where

$\begin{split}C(hi,A,g):=A^{-1}\langle
A\rangle^{-\frac{1}{2}}\|g^{\prime\prime}\|_{L^{2}},\qquad
C(lo,A,g):=A\|g\|_{L^{2}}.\end{split}$ (4.31)

Indeed, to obtain (4.25), we only need to replace $C(\ast,A,g)$ by
$\|g\|_{H^{1}}$. To do this, we decompose $g=\sum_{B}g_{B}$ using a
Littlewood-Paley projection such that

$\begin{split}\|g_{B}\|_{L^{2}}\lesssim\min\\{1,B^{-1}\\}\|g\|_{H^{1}},\qquad\|g_{B}^{\prime\prime}\|_{L^{2}}\lesssim\min\\{B^{2},B\\}\|g\|_{H^{1}},\end{split}$

and for each $g_{B}$, we choose $A:=B^{\delta}+B^{1-\delta}$ for
$0<\delta<1/5$ and we compute that

$\begin{split}\sum_{B}\left\\{C(lo,A,g_{B})+C(hi,A,g_{B})\right\\}\lesssim\sum_{B}\min\\{B^{\delta},B^{-\delta}\\}\|g\|_{H^{1}}\lesssim\|g\|_{H^{1}}.\end{split}$

It now remains to prove (4.30)-(4.31). We decompose, for $A>0$ dyadic,
$R^{1}[g_{1};L_{s}+g_{2}]=R^{1,hi}_{A}[g_{1};L_{s}+g_{2}]+R^{1,lo}_{A}[g_{1};L_{s}+g_{2}]$
where

$\begin{split}R^{1,hi}_{A}[g_{1},h]&:=\int_{\mathbb{R}_{+}}\left\\{g_{1}^{\prime}(y+\alpha)-g_{1}^{\prime}(y-\alpha)\right\\}\cdot
J_{\alpha}[h]\cdot(F_{\alpha}+F_{-\alpha})\cdot\varphi(A\alpha)\frac{d\alpha}{\alpha},\\\
R^{1,lo}_{A}[g_{1},h]&:=\int_{\mathbb{R}_{+}}\left\\{g_{1}(y+\alpha)+g_{1}(y-\alpha)\right\\}\cdot\alpha\partial_{\alpha}\left\\{J_{\alpha}[h]\cdot(F_{\alpha}+F_{-\alpha})\cdot\frac{1-\varphi(A\alpha)}{\alpha}\right\\}\frac{d\alpha}{\alpha}.\end{split}$
(4.32)

For $R^{1,hi}_{A}$, using (2.3) and (2.5)-(2.6), it suffices to show that

$\begin{split}\|\int_{\mathbb{R}_{+}}\int_{\\{|t|\leq\alpha\\}}|g_{1}^{\prime\prime}(y+t)|\cdot|J_{\alpha}[L_{s}]|\cdot|\fint_{\pm\alpha,0}g_{2}(y)|\cdot\varphi(A\alpha)\frac{d\alpha}{\alpha}\|_{L^{1}\cap
L^{2}}&\lesssim
s^{2}A^{-1}(1+A)^{-1}\|g_{1}^{\prime\prime}\|_{L^{2}}\|g_{2}\|_{H^{1}},\\\
\|\int_{\mathbb{R}_{+}}\int_{\\{|t|\leq\alpha\\}}|g_{1}^{\prime\prime}(y+t)|\cdot|J_{\alpha}[L_{s},g_{2}]|\cdot\varphi(A\alpha)\frac{d\alpha}{\alpha}\|_{L^{1}\cap
L^{2}}&\lesssim
sA^{-1}(1+A)^{-\frac{1}{2}}\|g_{1}^{\prime\prime}\|_{L^{2}}\|g_{2}\|_{H^{1}},\\\
\|\int_{\mathbb{R}_{+}}\int_{\\{|t|\leq\alpha\\}}|g_{1}^{\prime\prime}(y+t)|\cdot|J_{\alpha}[g_{2},g_{3}]|\cdot\varphi(A\alpha)\frac{d\alpha}{\alpha}\|_{L^{1}\cap
L^{2}}&\lesssim
A^{-1}(1+A)^{-\frac{1}{2}}\|g_{1}^{\prime\prime}\|_{L^{2}}\|g_{2}\|_{H^{1}}\|g_{3}\|_{H^{1}}.\end{split}$

The first estimate follows from (4.9) and the bound (4.10). The second and
third follow from (4.9) and (4.11). We now consider the contribution of
$R^{1,lo}_{A}$. Inspecting (4.32), it suffices to show that

$\begin{split}\int_{\mathbb{R}_{+}}\|g_{1}(y\pm\alpha)\fint_{\pm\alpha}g_{2}(y)\left[|J_{\alpha}[L_{s}]|+|\alpha\partial_{\alpha}(J_{\alpha}[L_{s}])|\right]\|_{L^{2}\cap
L^{1}}\cdot(1-\varphi(A\alpha))\frac{d\alpha}{\alpha^{2}}&\lesssim
s^{2}A\|g_{1}\|_{L^{2}}\|g_{2}\|_{H^{1}},\\\
\int_{\mathbb{R}_{+}}\|g_{1}(y\pm\alpha)\cdot\left(\alpha\partial_{\alpha}\fint_{\pm\alpha}g_{2}(y)\right)|J_{\alpha}[L_{s}]|\|_{L^{2}\cap
L^{1}}\cdot(1-\varphi(A\alpha))\frac{d\alpha}{\alpha^{2}}&\lesssim
s^{2}A\|g_{1}\|_{L^{2}}\|g_{2}\|_{H^{1}},\\\
\int_{\mathbb{R}_{+}}\|g_{1}(y\pm\alpha)\left[|J_{\alpha}[g_{2},g_{3}]|+|\alpha\partial_{\alpha}(J_{\alpha}[g_{2},g_{3}])|\right]\|_{L^{2}\cap
L^{1}}\cdot(1-\varphi(A\alpha))\frac{d\alpha}{\alpha^{2}}&\lesssim
A\|g_{1}\|_{L^{2}}\|g_{2}\|_{H^{1}}\|g_{3}\|_{L^{\infty}},\end{split}$

where in the last estimate we consider $g_{3}=L_{s}$ or $g_{3}\in H^{1}$.
These estimates all follow from (4.9) and direct integration.

∎

######  Lemma 4.5.

Assume (4.14) and recall $T$ and $\mathcal{T}$ defined in (3.3). There holds
that

$\begin{split}\|T[L_{s}]\|_{L^{\frac{4}{3}}\cap L^{2}}&\lesssim
s^{3},\qquad\|T_{1}[g]\|_{L^{\frac{4}{3}}\cap L^{2}}\lesssim
s^{2}\|g\|_{H^{1}},\end{split}$ (4.33)

and

$\begin{split}\|T_{\geq 2}[g_{2}]-T_{\geq 2}[g_{1}]\|_{L^{\frac{4}{3}}\cap
L^{2}}&\lesssim\left[s^{2}+\|g_{1}\|_{H^{1}}^{2}+\|g_{2}\|_{H^{1}}^{2}\right]\|g_{2}-g_{1}\|_{H^{1}}.\end{split}$
(4.34)

Besides the same bound hold if we replace $T[h]$ by
$\mathcal{T}[h]=(1+h^{2})T[h]$.

###### Proof of Lemma 4.5.

We have that

$\begin{split}T[h]=T[h,h;h]&=-\frac{4}{\pi}\int_{\mathbb{R}}(\Delta_{\alpha}h)\cdot(\Delta_{\alpha}h)\cdot\fint_{\alpha}h\cdot
F^{2}(\fint_{\alpha}h)d\alpha,\end{split}$

and using that

$\begin{split}|\Delta_{\alpha}L_{s}|\lesssim s(\langle\alpha\rangle+\langle
y\rangle)^{-1},\qquad|\Delta_{\alpha}g|\lesssim\min\\{\alpha^{-\frac{1}{2}},\alpha^{-1}\\}\|g\|_{H^{1}},\end{split}$

we see that

$\begin{split}\sqrt{1+y^{2}}|T[L_{s},L_{s};g](y)|&\lesssim
s^{2}\int\frac{\sqrt{1+y^{2}}}{1+\alpha^{2}+y^{2}}d\alpha\cdot\|g\|_{L^{\infty}}\lesssim
s^{2}\|g\|_{L^{\infty}},\\\ \sqrt{1+y^{2}}|T[L_{s},g;h](y)|&\lesssim
s\int\frac{\sqrt{1+y^{2}}}{\sqrt{1+\alpha^{2}+y^{2}}}\min\\{\alpha^{-\frac{1}{2}},\alpha^{-1}\\}d\alpha\cdot\|g\|_{H^{1}}\|h\|_{L^{\infty}}\lesssim
s\|g\|_{H^{1}}\|h\|_{L^{\infty}}\ln\langle y\rangle.\end{split}$

and we deduce (4.33). For (4.34) it suffices to also show that

$\begin{split}\|\int_{\mathbb{R}}|(\Delta_{\alpha}g_{1})\cdot(\Delta_{\alpha}g_{2})|d\alpha\|_{L^{\frac{4}{3}}\cap
L^{2}}&\lesssim\|g_{1}\|_{H^{1}}\|g_{2}\|_{H^{1}},\end{split}$

but this follows from

$\begin{split}|\Delta_{\alpha}g(y)|&\leq\frac{1}{|\alpha|}\int_{\\{|t|\leq|\alpha|\\}}|g^{\prime}(y+t)|dt,\qquad\|\Delta_{\alpha}g\|_{L^{2}_{y}}\lesssim|\alpha|^{-1}\|g\|_{L^{2}},\end{split}$

which gives

$\begin{split}\|(\Delta_{\alpha}g_{1})(\Delta_{\alpha}g_{2})\|_{L^{1}_{y}\cap
L^{2}_{y}}&\lesssim\min\\{\alpha^{-\frac{1}{2}},\alpha^{-\frac{3}{2}}\\}\|g_{1}\|_{H^{1}}\|g_{2}\|_{H^{1}},\\\
\end{split}$

and the proof is complete.

∎

###  4.3. Proof of the main estimates for the fixed-point formulation

####  4.3.1. Proof of Lemma 3.4

Starting from

$\begin{split}(\frac{2}{\pi}\arctan(z))^{2}-1=(1-\frac{2}{\pi}\arctan(\frac{1}{z}))^{2}-1=-\frac{4}{\pi}\arctan(\frac{1}{z})+(\frac{2}{\pi}\arctan(\frac{1}{z}))^{2},\end{split}$

we see that, for $z=(1+s^{2}/3)y\geq 1$,

$\begin{split}\pi_{s}&=-s^{3}\frac{4}{\pi}\arctan(z)\cdot\frac{2}{\pi}\arctan(\frac{1}{z})+s^{3}(\frac{2}{\pi}\arctan(\frac{1}{z}))^{2}\cdot\frac{2}{\pi}\arctan(z),\end{split}$

and in particular, $\pi_{s}$ is $C^{\infty}$ and

$\pi_{s}(y)=O_{y\to
0}(y),\qquad|\partial_{y}^{a}\pi_{s}(y)|=O_{x\to\infty}(\langle
y\rangle^{-1-a}).$

In particular, we observe that

$\begin{split}\|\pi_{s}\|_{H^{1}}\lesssim s^{3}.\end{split}$

In addition, using (4.19), using (4.25) and (4.33) and direct computations, we
see that

$\begin{split}\|W[L_{s}]\partial_{y}L_{s}\|_{L^{1}\cap
H^{1}}+\|R[L_{s}]\|_{L^{\frac{4}{3}}\cap
L^{2}}+\|\mathcal{T}[L_{s}]\|_{L^{\frac{4}{3}}\cap L^{2}}&\lesssim
s^{3},\end{split}$

and using Lemma 3.2, we see that

$\begin{split}\|\widehat{{\bf
T}}_{-1}(W[L_{s}]\partial_{y}L_{s})\|_{H^{1}}+\|\widehat{{\bf
T}}_{-1}R[L_{s}]\|_{H^{1}}+\|\widehat{{\bf
T}}_{-1}\mathcal{T}[L_{s}]\|_{H^{1}}&\lesssim s^{3}.\end{split}$

∎

####  4.3.2. Proof of Lemma 3.5

The proof follows by Neumann series. Using Lemma 3.2, we see that

$\|(L_{s}^{2}-s^{2})g\|_{H^{1}}+\|\widehat{{\bf
T}}_{-1}(|\nabla|(L_{s}^{2}-s^{2}/3)g)\|_{H^{1}}\lesssim s^{2}\|g\|_{H^{1}}.$

In addition, using Lemma 3.2, and Lemma 4.3, we see that

$\begin{split}\|\widehat{{\bf
T}}_{0}[W[L_{s}]g\|_{H^{1}}&\lesssim\|W[L_{s}]\|_{L^{\infty}}\|g\|_{H^{1}}+\|\partial_{y}W[L_{s}]\|_{L^{2}}\|g\|_{L^{\infty}},\\\
\|\widehat{{\bf
T}}_{-1}[g\partial_{y}W[L_{s}]]\|_{H^{1}}&\lesssim\|g\|_{L^{2}\cap
L^{\infty}}\|\partial_{y}W[L_{s}]\|_{L^{2}},\\\ \|\widehat{{\bf
T}}_{-1}[W_{1}[g]\partial_{y}L_{s}]\|_{H^{1}}&\lesssim\|W_{1}[g]\|_{L^{\infty}}\|\partial_{y}L_{s}\|_{L^{1}\cap
L^{2}}.\end{split}$

Finally, using Lemma 4.4 and Lemma 4.5, we see that

$\begin{split}\|\widehat{{\bf T}}_{-1}[R_{1}[g]]\|_{H^{1}}+\|\widehat{{\bf
T}}_{-1}[\mathcal{T}_{1}[g]]\|_{H^{1}}&\lesssim
s^{2}\|g\|_{H^{1}}.\end{split}$

This gives the estimate of $A+Id$.

Next, we compute that

$\begin{split}\partial_{s}A[g]&=-\left[12s/(3+s^{2})^{2}+\kappa\partial_{s}(L_{s}^{2}-s^{2})\right]g+\kappa^{2}\widehat{\bf
T}_{-1}[|\nabla|(4s/3+\partial_{s}(L_{s}^{2}-s^{2}))g]-\kappa\widehat{\bf
T}_{0}[g\partial_{s}W[L_{s}]]\\\ &\quad+\kappa\widehat{\bf
T}_{-1}[g\partial_{s}\partial_{y}W[L_{s}]-W_{1}[g]\partial_{y}\partial_{s}L_{s}-\partial_{y}L_{s}\partial_{s}W_{1}[g]+\partial_{s}R_{1}[g]]+\kappa\widehat{\bf
T}_{-1}[\partial_{s}\mathcal{T}_{1}[g]]\end{split}$

and using that

$\begin{split}\partial_{s}L_{s}=s^{-1}L_{s}+r,\qquad\|r\|_{H^{1}}\lesssim
s^{2},\end{split}$

direct calculations and adaptations of Lemma 4.3, Lemma 4.4 and Lemma 4.5 give
the bound on $\partial_{s}A$.

∎

####  4.3.3. Proof of Lemma 3.6

Using the fact that $H^{1}$ is an algebra, we see that

$\begin{split}\|L_{s}g^{2}\|_{H^{1}}+\|g^{3}\|_{H^{1}}+\|\widehat{{\bf
T}}_{-1}\left[|\nabla|(L_{s}g^{2}+g^{3}/3)\right]\|_{H^{1}}&\lesssim\|L_{s}\|_{L^{\infty}}\|g\|_{H^{1}}^{2}+\|g\|_{H^{1}}^{3},\end{split}$

and since these expressions are multilinear they extend to differences. In
addition, using Lemma 4.3, we see that

$\begin{split}\|(W_{\geq 2}[g_{2}]-W_{\geq
2}[g_{1}])\partial_{y}L_{s}\|_{L^{1}\cap L^{2}}&\lesssim\|W_{\geq
2}[g_{2}]-W_{\geq 2}[g_{1}]\|_{L^{\infty}}\|\partial_{y}L_{s}\|_{L^{1}\cap
L^{2}}\\\ &\lesssim
s\left[s^{2}+\|g_{1}\|_{H^{1}}^{2}+\|g_{2}\|_{H^{1}}^{2}\right]\|g_{2}-g_{1}\|_{H^{1}},\\\
\|W_{\geq 1}[g_{1}]\partial_{y}(g_{2}-g_{1})\|_{L^{1}\cap
L^{2}}&\lesssim\|W_{\geq 1}[g_{1}]\|_{L^{2}\cap
L^{\infty}}\|\partial_{y}(g_{2}-g_{1})\|_{L^{2}},\\\ \|(W_{\geq
1}[g_{2}]-W_{\geq 1}[g_{1}])\partial_{y}g_{2}\|_{L^{1}\cap
L^{2}}&\lesssim\|W_{\geq 1}[g_{2}]-W_{\geq 1}[g_{1}]\|_{L^{2}\cap
L^{\infty}}\|\partial_{y}g_{2}\|_{L^{2}}.\end{split}$

Similarly, using Lemma 4.4 and Lemma 4.5,

$\begin{split}\|R_{\geq 2}[g_{2}]-R_{\geq 2}[g_{1}]\|_{L^{1}\cap
L^{2}}+\|\mathcal{T}_{\geq 2}[g_{2}]-\mathcal{T}_{\geq 2}[g_{1}]\|_{L^{1}\cap
L^{2}}&\lesssim\left[s^{2}+\|g_{1}\|_{H^{1}}^{2}+\|g_{2}\|_{H^{1}}^{2}\right]\|g_{2}-g_{1}\|_{H^{1}},\end{split}$

and the proof is complete.

∎

##  5\. Numerical Results

In this section, we describe how to numerically compute the branch of
solutions $k_{s}$ starting from the zero solution. Their existence for small
$s$ was proved in Theorem 1.1. See Figures 1 and 2 below for different
depictions of the solutions.

The main advantage of working with the formulation of equation (1.3), as
opposed to a self-similar equation for the function $f$, is that all the
quantities involved are bounded. Nonetheless there are a few technicalities
which we outline below.

The first step consists in changing variables and transforming the infinite
domain into a finite one. We do so by setting $y=\tan(z)$ and
$\tilde{k}(z):=k(\tan(z))=k(y)$ so that the domain of definition is mapped
into $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$. This change of variables has
been used successfully for other problems in fluid mechanics (see for example
[40] and references therein).

Moreover, we exploit the symmetry to gain an extra cancellation at
$\frac{\pi}{2}$. After performing the change of variables and a lengthy
calculation, (1.3) is transformed into

$\displaystyle 0$
$\displaystyle=\sin(z)\cos(z)\tilde{k}^{\prime}(z)+\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{\tilde{k}^{\prime}(z)\cos(z)^{2}-\tilde{k}^{\prime}(y)\cos(y)^{2}}{\tan(z)-\tan(y)}\frac{\sec(y)^{2}}{1+(\tilde{\Delta}_{y}\tilde{k}(z))^{2}}dy$
$\displaystyle-\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\left(\frac{\tilde{k}(z)-\tilde{k}(y)}{\tan(z)-\tan(y)}\right)^{2}\frac{\tilde{\Delta}_{y}\tilde{k}(z)}{(1+(\tilde{\Delta}_{y}\tilde{k}(z))^{2})^{2}}\sec(y)^{2}dy$
$\displaystyle+\frac{1}{\pi}\int_{0}^{\frac{\pi}{2}}\frac{\tilde{k}^{\prime}(z)\cos(z)^{2}-\tilde{k}^{\prime}(y)\cos(y)^{2}}{\tan(z)+\tan(y)}\frac{1}{1+(\tilde{\Delta}_{-y}\tilde{k}(z))^{2}}\sec(y)^{2}dy$
$\displaystyle-\frac{2}{\pi}\int_{0}^{\frac{\pi}{2}}\left(\frac{\tilde{k}(z)+\tilde{k}(y)}{\tan(z)+\tan(y)}\right)^{2}\frac{\tilde{\Delta}_{-y}\tilde{k}(z)}{(1+(\tilde{\Delta}_{-y}\tilde{k}(z))^{2})^{2}}\sec(y)^{2}dy,$
(5.1)

where

$\displaystyle\tilde{\Delta}_{y}\tilde{k}(z)=\int_{z}^{y}\frac{\sec(w)^{2}\tilde{k}(w)}{\tan(y)-\tan(z)}dw.$
(5.2)

We performed continuation in $s$ in increments of $\Delta s=0.1$ and did $70$
iterations, using as initial guess $\tilde{k}=\Delta s\frac{2}{\pi}z$ (the
linear approximation) at the first iteration, and for the subsequent ones the
result of the previous iteration plus $\Delta s\frac{2}{\pi}z$ to ensure that
the updated boundary condition at $z=\frac{\pi}{2}$ is satisfied. In the range
we computed, we did not see any impediment towards advancing in $s$, other
than computation time, although the errors become bigger as $s$ grows and the
algorithm may take one or two more iterations to converge for $s\sim 5$ than
for $s\sim 0.1$.

To compute a solution for a fixed $s$, we used the Levenberg-Marquardt
algorithm [39, 41]. Our discretization variables consist on the values of
$\tilde{k}$ at gridpoints $z_{i}=\frac{\pi}{2(N-1)}i$, where we are using that
$\tilde{k}$ is odd to solve for positive $z$ only. Other strategies such as
non-uniform meshes (concentrating points towards 0) or $s$-dependent
compactifications would perhaps improve the performance for large $s$ since
$\tilde{k}^{\prime}(0)$ grows with $s$ (see Figure 1), but we did not explore
them here. We took $N=129$ and the discrete system we solved was equation
(5.1) evaluated at $z_{i}$, $i=1,\ldots,N-1$ plus the boundary conditions
$\tilde{k}(0)=0,\tilde{k}\left(\frac{\pi}{2}\right)=s$. In order to compute
the derivatives we calculated a spline of degree 4 interpolating through the
discrete grid and approximated the derivatives of $\tilde{k}$ by the
derivatives of the spline. To perform the integration, we integrated in (5.1)
in the variable $y$ using trapezoidal integration and a grid of $10N=1290$
points. We also tried finer grids and saw virtually no difference with respect
to the results. In order to get stable results, we took care of splitting the
domain into 3 regions $(z\sim\frac{\pi}{2},z\sim y$ and the rest$)$ and
computing separately the integrand in each of them, carefully computing the
limit. For example, note that despite being bounded at $z\sim\frac{\pi}{2}$,
there is a strong instability coming from the fact that the integrand is of
the form $\infty-\infty$ if not dealt with properly. The inner integrals in
(5.1) (i.e. the $\tilde{\Delta}$ integrals) were computed using an adaptive
Gauss-Kronrod quadrature of 15 points, also taking care of the limits at
$\frac{\pi}{2}$ and at $y$.

Figure 1.  Plot of the difference between the numerically computed solutions $\tilde{k}(z)$ and the linear approximation $\frac{2}{\pi}sz=\frac{2}{\pi}s\tan(y)$ for $s=1,2,4,7$. The branch continues beyond what is calculated. Figure 2.  Comparison of the numerically computed solutions $k(y)$, $y\in[0,10]$ for $s=0.1,0.2,0.4,1,2,4,7$. All functions are normalized to value 1 at infinity. The branch continues beyond what is calculated. The curve corresponding to $s=0.1$ is the lowest one and monotonically increase with $s$. |
---|---
Figure 3. Comparison of the numerically computed solutions
$\int_{0}^{y}k(s)ds$, for $s=0.1,0.2,0.4,1,2,4,7$. Panel (a): $y\in[-50,50]$,
panel (b): Close-up, $y\in[-0.5,0.5]$. The curves increase with $s$.

### Acknowledgments

EGJ and JGS were partially supported by the ERC Starting Grant ERC-StG-
CAPA-852741. EGJ was partially supported by the ERC Starting Grant project
H2020-EU.1.1.-639227. HQN was partially supported by NSF grant DMS-1907776. BP
was supported by NSF grant DMS-1700282 and the CY-Advanced Studies fellow
program. We thank Princeton University for computing facilities (Polar
Cluster). This project has received funding from the European Union’s Horizon
2020 research and innovation programme under the Marie Sklodowska-Curie grant
agreement CAMINFLOW No 101031111.

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