#  $C^{\gamma}$ well-posedness of some non-linear transport equations

J.C. Cantero

###### Abstract

Given $k:\mathbb{R}^{n}\setminus\\{0\\}\to\mathbb{R}^{n}$ a kernel of class
$C^{2}$ and homogeneous of degree $1-n$, we prove existence and uniqueness of
Hölder regular solutions for some non-linear transport equations with velocity
fields given by convolution of the density with $k$. The Aggregation, the 3D
quasi geostrophic, and the 2D Euler equations can be recovered for particular
choices of $k$.

AMS 2010 Mathematics Subject Classification: 35Q35 (primary) 35F20, 35B30
(secondary).

Keywords: non-linear transport equation, vortex patches, Euler equation,
quasi-geostrophic equation, compressible flow.

##  1 Introduction

Let $\rho(x,t)$ a scalar quantity usually known as the density and let
$v(x,t)$ a vector field called velocity both depending on the position
$x\in\mathbb{R}^{n}$ and on the time $t\in\mathbb{R}$ . The (homogeneous)
transport equation for the pair ($\rho$,$v$) is the partial differential
equation defined by

(1) $\begin{cases}\rho_{t}+v\cdot\nabla\rho=0,\\\
\rho(\cdot,0)=\rho_{0}.\end{cases}$

Given a velocity field $v$ and a point $\alpha\in\mathbb{R}^{n}$ we set,
whenever it is well defined, the flow map

$\begin{split}X(\alpha,\cdot):\mathbb{R}&\rightarrow\mathbb{R}^{n},\\\
t&\rightarrow X(\alpha,t)\end{split}$

as the solution of the ordinary differential equation

(2)
$\begin{cases}\frac{\mathrm{d}}{\mathrm{d}t}X(\alpha,t)=v(X(\alpha,t),t),\\\
X(\alpha,0)=\alpha.\end{cases}$

This map indicates the position at time $t$ of the particle that was initially
at $\alpha$ and that has moved following the velocity field at every moment.
It is also called the trajectory of the particle initially at $\alpha$. If
$\rho(\cdot,t)$ is smooth enough, then a straightforward computation allow us
to check that $\rho(X(\alpha,t),t)=\rho_{0}(\alpha).$ That means that the
density at time 0 and at position $\alpha$ takes the same value as the density
evaluated at $t$ and at the future position of $\alpha$ at time $t$. So, it
can be said that $\rho$ is transported with the flow defined by the velocity
field $v$. This is a good reason to call (1) a transport equation.

For a fixed time, the functions $\rho$ and $v$ in (1) can be related by some
functional $T$ so that $v(\cdot,t)=T(\rho(\cdot,t))$ and often $T$ can be
expressed as a convolution with a given kernel. In this situation, the
equation is not lineal for sure. The most important example of a transport
equation of this kind is the Euler equation in the plane. Let
$N(x)=\frac{1}{2\pi}\ln(\left|x\right|)$ be the fundamental solution of the
Laplacian and let

$K_{BS}(x_{1},x_{2})=\frac{1}{2\pi\left|x\right|^{2}}(-x_{2},x_{1})=\nabla^{\perp}N(x).$

We call $K_{BS}$ the Biot-Savart kernel. Let $\omega(\cdot,t)$ denote the
vorticity, which is the scalar curl of the velocity field $v(x,t)$, i.e.,

$\omega(x,t)=\partial_{1}v_{2}(x,t)-\partial_{2}v_{1}(x,t).$

Then the vorticity formulation of the Euler equation is

$\begin{cases}\omega_{t}+v\cdot\nabla\omega=0,\\\
v(\cdot,t)=K_{BS}\ast\omega(\cdot,t),\\\
\omega(\cdot,0)=\omega_{0},\end{cases}$

which is a non-linear transport equation for $(\omega,v)$.

Another example in $\mathbb{R}^{n}$ (see [BLL, Section 4.2] for more details
of its derivation) is the aggregation equation when the initial condition
$\rho_{0}$ is the characteristic function of some domain $D_{0}$,
$\chi_{D_{0}}$. We will call solutions for this type of initial data density
patches. Let $w_{n}$ the volume of the $n$-dimensional unit ball and set

(3) $N(x)=-\frac{1}{n(n-2)w_{n}}\frac{1}{\left|x\right|^{n-2}},\,\,\,\,\,n\geq
3,$

the fundamental solution of the Laplacian in $\mathbb{R}^{n}$. In this case,
given $K_{Ag}=-\nabla N$ we get

$\begin{cases}\rho_{t}+v\cdot\nabla\rho=0,\\\
v(\cdot,t)=K_{Ag}\ast\rho(\cdot,t),\\\
\rho(\cdot,0)=\rho_{0}=\chi_{D_{0}}.\end{cases}$

We would like to remark that the original aggregation equation is not a
transport equation. The equivalence with the equation above is just for the
initial condition equal to the characteristic function of a domain, as
mentioned above.

In the spirit of generalizing these example equations we will consider
throughout this paper a kernel
$k:\mathbb{R}^{n}\setminus\\{0\\}\to\mathbb{R}^{n}$, homogeneous of degree
$-(n-1)$ and of class $C^{2}(\mathbb{R}^{n}\setminus\\{0\\}).$ That is, for
such $k$, we consider the following general transport equation:

$\begin{cases}\rho_{t}+v\cdot\nabla\rho=0,\\\
v(\cdot,t)=k\ast\rho(\cdot,t),\\\
\rho(\cdot,0)=\rho_{0}=\chi_{D_{0}}.\end{cases}$

Our goal is to prove a well-posedness result for the transport equation and
for the kernel $k$ in some space of functions that will be defined in a
moment. We would like to anticipate that the divergence of $v$ is an important
quantity appearing in the computations and in the proofs we are going to
develop. For the Euler equation the divergence vanishes everywhere for any
time and for the aggregation equation the divergence at a given time $t$ is
equal to $-\rho(\cdot,t)$. Owing to the special simple form of the divergence
the proofs of well-posedness in the above two cases are relatively fluent.

The well-posedness will be proved in spaces of Hölder smooth functions. We
define them now.

######  Definition 1.

Given $0<\gamma<1$ and $f:\mathbb{R}^{n}\to\mathbb{R}$ let

$\left|\left|f\right|\right|_{L^{\infty}}=\sup_{x\in\mathbb{R}^{n}}\left|f(x)\right|\,\,\,\,\text{and}\,\,\,\,\left|f\right|_{\gamma}=\sup_{\stackrel{{\scriptstyle
x,y\in\mathbb{R}^{n}}}{{x\neq
y}}}\frac{\left|f(x)-f(y)\right|}{\left|x-y\right|^{\gamma}}.$

We define the norm

$\left|\left|f\right|\right|_{\gamma}:=\left|\left|f\right|\right|_{L^{\infty}}+\left|f\right|_{\gamma}.$

For $F:\mathbb{R}^{n}\to\mathbb{R}^{d}$, $x\to
F(x)=(f_{1}(x),\ldots,f_{d}(x))$, we define

$\left|\left|F\right|\right|_{\gamma}:=\sup_{i=1,\ldots,d}\left|\left|f_{i}\right|\right|_{\gamma}$

and then the space

$C^{\gamma}(\mathbb{R}^{n};\mathbb{R}^{d})=\left\\{f:\mathbb{R}^{n}\to\mathbb{R}^{d}:\left|\left|f\right|\right|_{\gamma}<\infty\right\\}.$

Finally, we define

$\left|F\right|_{1,\gamma}=\left|F(0)\right|+\left|\left|\nabla
F\right|\right|_{L^{\infty}}+\left|\nabla F\right|_{\gamma},$

where

$\left|\left|\nabla
F\right|\right|_{L^{\infty}}=\sup_{i=1,\ldots,d}\left(\sup_{j=1,\ldots,n}\left|\left|\frac{\partial}{\partial
x_{j}}f_{i}\right|\right|_{L^{\infty}}\right),$ $\left|\nabla
F\right|_{\gamma}=\sup_{i=1,\ldots,d}\left(\sup_{j=1,\ldots,n}\left|\frac{\partial}{\partial
x_{j}}f_{i}\right|_{\gamma}\right).$

We define the Hölder space $C^{1,\gamma}(\mathbb{R}^{n};\mathbb{R}^{d})$ as

$C^{1,\gamma}(\mathbb{R}^{n};\mathbb{R}^{d})=\left\\{f:\mathbb{R}^{n}\to\mathbb{R}^{d}:\left|\left|f\right|\right|_{1,\gamma}<\infty\right\\}.$

When it is clear enough, we will just write $C^{1,\gamma}$.

Additionally, we define $C^{\gamma}_{c}$ as the space of functions in
$C^{\gamma}$ which are also compactly supported.

We are ready to anticipate the main theorem of this paper.

######  Theorem 2.

Let $N$ the fundamental solution of the Laplacian in $\mathbb{R}^{n}$. Let
$k\in C^{2}(\mathbb{R}^{n}\setminus\\{0\\};\mathbb{R}^{n})$ a kernel
homogeneous of degree $1-n$. For $0<\gamma<1$, if $\rho_{0}\in
C^{\gamma}_{c}(\mathbb{R}^{n},\mathbb{R})$, then the transport equation

$\begin{cases}\rho_{t}+v\cdot\nabla\rho=0,\\\
v(\cdot,t)=k\ast\rho(\cdot,t),\\\ \rho(\cdot,0)=\rho_{0},\end{cases}$

has a unique weak solution $\rho(\cdot,t)\in
C^{\gamma}_{c}(\mathbb{R}^{n},\mathbb{R})$ for any time $t\in\mathbb{R}$.

The reason we have chosen this space is double. Firstly, the result was proved
for the Euler equation (see [MB, Chapter 4]) and for the Aggregation kernel
(see [CGK, Theorem 5.3]). Secondly, we wanted to be sure about having well-
posedness in the smooth case before moving to other situations (for instance,
density patches). In fact, if one wants to deal with the Yudovich problem,
that is, proving well-posedness for an initial data in $L^{1}\cap L^{\infty}$,
a strategy would be first to smoothen the initial data via convolution with a
mollifier, then apply the smooth Theorem presented here, check some
compactness properties, take limit of the solutions of the mollified equations
and verify that they indeed satisfy the original equation. So, the result
presented here can be seen as the first necessary step to prove existence of
weak solutions of the equation.

The fact that the divergence is as general as the situation envisaged allows
makes the known argument much more involved; some differences arise and
overcoming them requires often a delicate treatment. We will stress this fact
in the next sections whenever those differences appear.

###  1.1 Outline of the paper

The present paper is structured as follows. In Section 2 we give some
preliminary results on Hölder spaces and on Calderón-Zygmund Operators (CZO)
acting on them. In Section 3 we prove a local-in-time version of Theorem 2. In
Section 4 we prove that this local solution is actually global via appropiate
a priori estimates.

##  2 Preliminaries

First of all, we have the following elementary properties for elements of the
Hölder spaces.

######  Lemma 3.

Let $f,g$ be $C^{\gamma}$ functions, $0<\gamma<1$. Then

(4) $\displaystyle\left|fg\right|_{\gamma}$
$\displaystyle\leq\left|\left|f\right|\right|_{L^{\infty}}\left|g\right|_{\gamma}+\left|f\right|_{\gamma}\left|\left|g\right|\right|_{L^{\infty}},$
(5) $\displaystyle\left|\left|fg\right|\right|_{\gamma}$
$\displaystyle\leq\left|\left|f\right|\right|_{\gamma}\left|\left|g\right|\right|_{\gamma}.$

If moreover $X$ is a smooth invertible transformation in $\mathbb{R}^{n}$
satisfying

$\left|\det\nabla X(\alpha)\right|\geq c_{1}>0,$

then there exists $c>0$ such that

(6) $\displaystyle\left|\left|(\nabla X)^{-1}\right|\right|_{\gamma}$
$\displaystyle\leq c\left|\left|\nabla X\right|\right|_{\gamma}^{2n-1},$ (7)
$\displaystyle\left|X^{-1}\right|_{1,\gamma}$ $\displaystyle\leq
c\left|X\right|_{1,\gamma}^{2n-1}$ (8) $\displaystyle\left|f\circ
X\right|_{\gamma}$ $\displaystyle\leq\left|f\right|_{\gamma}\left|\left|\nabla
X\right|\right|_{L^{\infty}}^{\gamma}$ (9) $\displaystyle\left|\left|f\circ
X\right|\right|_{\gamma}$
$\displaystyle\leq\left|\left|f\right|\right|_{\gamma}(1+\left|X\right|_{1,\gamma}^{\gamma}),$
(10) $\displaystyle\left|\left|f\circ X^{-1}\right|\right|_{\gamma}$
$\displaystyle\leq\left|\left|f\right|\right|_{\gamma}(1+\left|X\right|_{1,\gamma}^{\gamma(2n-1)}).$

The proof of Lemma 3 can be found in [MB, p. 159] (see Lemmas 4.1, 4.2 and
4.3). Note that (5) implies that $C^{\gamma}$ is an algebra.

We also have the following bounds for Calderón-Zygmund operators acting on
Hölder spaces. They will be used repeatedly in the proofs developed in the
upcoming sections.

######  Lemma 4.

Let $k:\mathbb{R}^{n}\to\mathbb{R}$, $k\in
C^{2}(\mathbb{R}^{n}\setminus\\{0\\})$ a kernel homogeneous of degree $1-n$.
That is,

(11) $\displaystyle k(\lambda
x)=\frac{1}{\lambda^{n-1}}k(x),\,\,\,\,\,\forall\,\lambda>0.\,\,\forall x\neq
0,$

Let $P=\partial_{i}k$, $i=1,\ldots,n$. Set

$Tf(x)=\int_{\mathbb{R}^{n}}k(x-x^{\prime})f(x^{\prime})\,\mathrm{d}x^{\prime};\phantom{heyy}Sf(x)=\text{p.v.}\int_{\mathbb{R}^{n}}P(x-x^{\prime})f(x^{\prime})\,\mathrm{d}x^{\prime}.$

For $0<\gamma<1$ let $f\in C_{c}^{\gamma}(\mathbb{R}^{n};\mathbb{R})$. Set
$R^{n}:=m(\text{supp}(f))<\infty$, that is, the measure of the support of $f$.
Then, there exists a constant $c$, independent of $f$ and $R$, such that

(12) $\displaystyle\left|\left|Tf\right|\right|_{L^{\infty}}$
$\displaystyle\leq cR\left|\left|f\right|\right|_{L^{\infty}},$ (13)
$\displaystyle\left|\left|Sf\right|\right|_{L^{\infty}}$ $\displaystyle\leq
c\left\\{\left|f\right|_{\gamma}\varepsilon^{\gamma}+\max\left(1,\ln\frac{R}{\varepsilon}\right)\left|\left|f\right|\right|_{L^{\infty}}\right\\},\,\,\,\,\,\forall\varepsilon>0,$
(14) $\displaystyle\left|Sf\right|_{\gamma}$ $\displaystyle\leq
c\left|f\right|_{\gamma}.$

######  Remark 5.

The proof of Lemma 4 can be found in [MB, pp. 159-163] (see Lemmas 4.5 and
4.6). There more hypothesis on the kernels $k$ and $P$ are required but we
remark here that they are not needed. In particular, for $k$ as in the
previous lemma, by differentiating with respect to $x_{i}$ the equation (11)
it is clear that this derivative is homogeneous of degree $-n$. Also, we can
see that $\partial_{i}k$ has zero mean integral on the sphere. Let $0<a<b$. By
Stokes’s theorem we can write

(15) $\begin{split}\int_{a\leq\left|x\right|\leq
b}&\partial_{i}k(x)\,\mathrm{d}x=\\\ &=\int_{\partial
B(0,b)}k(x)n_{i}(x)\,\mathrm{d}\sigma(x)-\int_{\partial
B(0,a)}k(x)n_{i}(x)\,\mathrm{d}\sigma(x),\end{split}$

where $n_{i}(x)$ is the $i$-th component of the unitary normal vector to each
surface at the point $x$. By homogeneity of the kernel $k$ it is clear that
the two integrals in the second line of (15) are equal and then the difference
is 0. By doing a hyperspherical coordinates change of variables and again by
homogeneity of the kernel, the first line of (15) can be written as

$\int_{a\leq\left|x\right|\leq
b}\partial_{i}k(x)\,\mathrm{d}x=(\log(b)-\log(a))\int_{\partial
B(0,1)}\partial_{i}k(w)\,\mathrm{d}\sigma(w),$

and so we can conclude that

$\int_{\partial B(0,1)}\partial_{i}k(w)\,\mathrm{d}\sigma(w)=0$

as it is required in the mentioned proof done in [MB].

##  3 Local Theorem

As in the case of Euler equation, a good way to prove an existence and
uniqueness result is by dealing with an, in some sense, equivalent equation
rather than the one presented in Theorem 2. Recall that $\rho$ is transported
with the flow, i.e., $\rho(x,t)=\rho_{0}(X^{-1}(x,t))$ for $X^{-1}(\cdot,t)$
the inverse of the flow $X(\cdot,t)$. Therefore by (10) we have

$\left|\left|\rho(\cdot,t)\right|\right|_{\gamma}\leq\left|\left|\rho_{0}\right|\right|_{\gamma}\left(1+\left|X(\cdot,t)\right|_{1,\gamma}^{\gamma(2n-1)}\right).$

Thus, $\rho(\cdot,t)\in C^{\gamma}$ provided $X(\cdot,t)\in C^{1,\gamma}$.

Furthermore, eventually we will need to control the measure of the support of
$\rho(\cdot,t)$. In order to do it, we have the next lemma, which will be also
needed in the following section. Note that in the zero divergence case there
is no need to control the support of $\rho(\cdot,t)$ since its measure is
conserved with time and therefore it is equal to the measure of the support of
$\rho_{0}$.

######  Lemma 6.

Let $(\rho,v)$ be a solution of (1) and let $X$ be the flow map associated to
$v(\cdot,t)$ as in (2). Then

$m(\text{supp}(\rho(\cdot,t)))\leq
c(n)m(\text{supp}(\rho_{0}))\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}^{n}.$

###### Proof.

Given $A\subseteq\mathbb{R}^{n}$, let $\mathbb{1}_{A}$ be the function taking
value $1$ in $A$ and $0$ otherwise. Then

$m(\text{supp}(\rho(\cdot,t)))=\int_{\mathbb{R}^{n}}\mathbb{1}_{\text{supp}(\rho(\cdot,t))}(x)\,\mathrm{d}x.$

Taking the change of variables $x=X(\alpha,t)$ we get

$m(\text{supp}(\rho(\cdot,t)))=\int_{\mathbb{R}^{n}}\mathbb{1}_{\text{supp}(\rho(\cdot,t))}(X(\alpha,t))\det
DX(\alpha,t)\,\mathrm{d}\alpha.$

Since $\rho$ is transported with the flow, it is clear that
$X(\alpha,t)\in\text{supp}(\rho(\cdot,t))$ if and only if
$\alpha\in\text{supp}(\rho_{0})$. Thus,

$m(\text{supp}(\rho(\cdot,t)))=\int_{\mathbb{R}^{n}}\mathbb{1}_{\text{supp}(\rho_{0})}(\alpha)\det
DX(\alpha,t)\,\mathrm{d}\alpha.$

Taking absolute value on the previous equation and having into account that
$\left|\left|\det DX(\cdot,t)\right|\right|_{L^{\infty}}\leq
c(n)\left|\left|\nabla X(\cdot,t)\right|\right|_{L^{\infty}}^{n}$ we get

$m(\text{supp}(\rho(\cdot,t)))\leq
c(n)m(\text{supp}(\rho_{0}))\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}^{n}.$

∎

We can focus then on proving existence, uniqueness and regularity for $X$. We
know $X$ satisfies (2) and then, as $v(\cdot,t)=k\ast\rho(\cdot,t)$, we obtain

$\frac{\mathrm{d}X}{\mathrm{d}t}(\alpha,t)=v(X(\alpha,t),t)=\int_{\mathbb{R}^{n}}k(X(\alpha,t)-x^{\prime})\rho(x^{\prime},t)\,\mathrm{d}x^{\prime}.$

Applying a change of variables $x^{\prime}=X(\alpha^{\prime},t)$

$\begin{split}\frac{\mathrm{d}X}{\mathrm{d}t}(\alpha,t)&=\int_{\mathbb{R}^{n}}k(X(\alpha,t)-X(\alpha^{\prime},t))\rho(X(\alpha^{\prime},t))\,\det[DX(\alpha^{\prime},t)]\,\mathrm{d}\alpha^{\prime}=\\\
&=\int_{\mathbb{R}^{n}}k(X(\alpha,t)-X(\alpha^{\prime},t))\rho_{0}(\alpha^{\prime})\,\det[DX(\alpha^{\prime},t)]\,\mathrm{d}\alpha^{\prime},\end{split}$

where, in the last equality, we have used that $\rho$ is conserved along the
flow.

Consequently, we have an ordinary differential equation (ODE) for $X$. A
standard way to prove existence and uniqueness for an ODE is to apply Picard-
Lindelöf’s theorem. It can be stated as follows.

######  Theorem 7 (Picard-Lindelöf).

Let $O\subseteq B$ be an open subset of a Banach space $B$ and let $F:O\to B$
be a locally Lipschitz continuous mapping.

Then given $X_{0}\in O$, there exists a time $T>0$ such that the ordinary
differential equation

$\frac{\mathrm{d}X}{\mathrm{d}t}=F(X),\phantom{aaaaaa}X(\cdot,t=0)=X_{0}\in
O,$

has a unique (local) solution $X\in C^{1}\left[(-T,T);O\right]$.

So, in order to apply Theorem 7, we first need an equation of type
$\frac{\mathrm{d}X}{\mathrm{d}t}=F(X)$. As we have seen, we have it for

(16)
$F(X(\alpha,t)):=\int_{\mathbb{R}^{n}}k(X(\alpha,t)-X(\alpha^{\prime},t))\rho_{0}(\alpha^{\prime})\,\det[DX(\alpha^{\prime},t)]\,\mathrm{d}\alpha^{\prime}.$

Then we need a Banach space $B$ and an open subspace of $B$ such that the flow
maps $X(\cdot,t)$ belong to $O_{M}$. We also need a functional $F$ mapping
$O_{M}$ to $B$ being this map locally Lipschitz continuous and satisfying that
$F(X(\alpha,t))$ is equal to (16). Let
$B=C^{1,\gamma}(\mathbb{R}^{n};\mathbb{R}^{n})$ and

(17)
$O_{M}=B\cap\left\\{X:\mathbb{R}^{n}\to\mathbb{R}^{n}:\,\frac{1}{M}<\sup_{\alpha\neq\beta}\frac{\left|X(\alpha)-X(\beta)\right|}{\left|\alpha-\beta\right|}<M\right\\}.$

Then we have:

  * •

$O_{M}$ is non-empty: $Id\in O_{M}\,\,\forall M>1$.

  * •

It is an open set since it is the preimage of the open set $(\frac{1}{M},M)$
for some norm function (which is continuous).

  * •

If $X\in O_{M}$, then the image of $X$ is open because $X$ is locally a
diffeomorphism and it is also closed because it is complete ($X$ is a
bilipschitz function) . Then the image of $X$ is the whole space and so $X$ is
a homeomorphism.

After this, we have to check the hypothesis in Picard-Lindelöf’s theorem.
Since computations of derivatives of $F(X)$ will be needed, we first look how
distributional derivatives of our kernels are.

######  Lemma 8.

Given $k=(k_{1},\ldots,k_{n}):\mathbb{R}^{n}\setminus\\{0\\}$, $k\in
C^{2}(\mathbb{R}^{n}\setminus\\{0\\})$ homogeneous of degree $1-n$, we have,
distributionally, for $i,j\in\\{1,\ldots,n\\}$

$\partial_{i}k_{j}=\text{p.v. }\partial_{i}k_{j}+c_{ij}\delta_{0},$

where

(18) $c_{ij}=\int_{\partial B(0,1)}k_{j}(s)s_{i}\,\mathrm{d}\sigma(s).$

###### Proof.

Let $f$ be such that $k_{j}\ast f$ makes perfect sense and $\varphi$ a
$C^{\infty}$ and compactly supported test function. Then, computing the
directional derivative $\partial_{i}$,

$\begin{split}\langle\partial_{i}(k_{j}\ast f),\varphi\rangle&=-\langle
k_{j}\ast
f,\partial_{i}\varphi\rangle=-\int_{\mathbb{R}^{n}}\left\\{\int_{\mathbb{R}^{n}}k_{j}(x-y)f(y)\,\mathrm{d}y\right\\}\partial_{i}\varphi(x)\,\mathrm{d}x=\\\
&=-\int_{\mathbb{R}^{n}}\left\\{\int_{\mathbb{R}^{n}}k_{j}(x-y)\partial_{i}\varphi(x)\,\mathrm{d}x\right\\}f(y)\,\mathrm{d}y=\\\
&=-\int_{\mathbb{R}^{n}}\left\\{\lim_{\varepsilon\to
0}\int_{\mathbb{R}^{n}\setminus
B(y,\varepsilon)}k_{j}(x-y)\partial_{i}\varphi(x)\,\mathrm{d}x\right\\}f(y)\,\mathrm{d}y=\\\
&=-\int_{\mathbb{R}^{n}}\left\\{\lim_{\varepsilon\to
0}\int_{\mathbb{R}^{n}\setminus
B(y,\varepsilon)}\partial_{i}[k_{j}(x-y)\varphi(x)]\,\mathrm{d}x\right\\}f(y)\,\mathrm{d}y+\\\
&+\int_{\mathbb{R}^{n}}\left\\{\lim_{\varepsilon\to
0}\int_{\mathbb{R}^{n}\setminus
B(y,\varepsilon)}\partial_{i}k_{j}(x-y)\varphi(x)\,\mathrm{d}x\right\\}f(y)\,\mathrm{d}y=\\\
&=-\int_{\mathbb{R}^{n}}\left\\{\lim_{\varepsilon\to
0}\int_{\mathbb{R}^{n}\setminus
B(y,\varepsilon)}\partial_{i}[k_{j}(x-y)\varphi(x)]\,\mathrm{d}x\right\\}f(y)\,\mathrm{d}y+\\\
&+\langle\text{p.v. }\partial_{i}k_{j}\ast
f,\varphi\rangle=\int_{\mathbb{R}^{n}}g(y)f(y)\,\mathrm{d}y+\langle\text{p.v.
}\partial_{i}k_{j}\ast f,\varphi\rangle,\end{split}$

for $g(y):=-\lim_{\varepsilon\to 0}\int_{\mathbb{R}^{n}\setminus
B(y,\varepsilon)}\partial_{i}[k_{j}(x-y)\varphi(x)]\,\mathrm{d}x$. Applying
Stokes’ Theorem to the integral defining $g(y)$ we get

$\begin{split}g(y)&=-\lim_{\varepsilon\to 0}\int_{\mathbb{R}^{n}\setminus
B(y,\varepsilon)}\partial_{i}[k_{j}(x-y)\varphi(x)]\,\mathrm{d}x=\\\
&=\lim_{\varepsilon\to 0}\int_{\partial
B(y,\varepsilon)}k_{j}(x-y)\varphi(x)n_{i}(x)\,\mathrm{d}\sigma(x),\end{split}$

where $n_{i}(x)$ is the $i$-th component of $n(x)$, the exterior normal vector
to the surface $\partial B(y,\varepsilon)$ at the point $x$. Subtracting and
adding $k_{j}(x-y)\varphi(y)n_{i}(x)$ in the integrand we have

(19) $\begin{split}g(y)&=\lim_{\varepsilon\to 0}\int_{\partial
B(y,\varepsilon)}k_{j}(x-y)(\varphi(x)-\varphi(y))n_{i}(x)\,\mathrm{d}\sigma(x)+\\\
&+\varphi(y)\lim_{\varepsilon\to 0}\int_{\partial
B(y,\varepsilon)}k_{j}(x-y)n_{i}(x)\,\mathrm{d}\sigma(x),\end{split}$

The first integral vanishes when taking the limit by the continuity of
$\varphi$. Hence, we can write $g(y)=\varphi(y)h(y)$ and we need to compute
the value of the integral in $h(y)$. By a change of variable $x=y+\varepsilon
s$, $s\in\partial B(0,1)$ we get

$\begin{split}h(y)=\lim_{\varepsilon\to 0}\int_{\partial
B(0,1)}k_{j}(\varepsilon s)n_{i}(y+\varepsilon
s)\varepsilon^{n-1}\,\mathrm{d}\sigma(s)=\int_{\partial
B(0,1)}k_{j}(s)s_{i}\,\mathrm{d}\sigma(s)=:c_{ij},\end{split}$

by making use of the homogeneity of the kernel $k_{j}$ and the fact that
$n(x)=x$ for $x\in\partial B(0,1)$. Since $k$ is locally integrable out of the
origin, then $c_{ij}$ is a well defined quantity independent of $y$. Then
$g(y)=c_{ij}\varphi(y)$ and therefore,

$\langle\partial_{i}(k_{j}\ast f),\varphi\rangle=\langle\text{p.v.
}\partial_{i}k_{j}\ast f,\varphi\rangle+\langle c_{ij}\delta_{0}\ast
f,\varphi\rangle,$

which completes the proof of the lemma. ∎

We are now in position to show that $F:O_{M}\to B$.

######  Proposition 9.

Let $O_{M}$ as defined in (17). Then, the functional $F$ defined by

(20)
$F(X)(\alpha)=\int_{\mathbb{R}^{n}}k(X(\alpha)-X(\alpha^{\prime}))\rho_{0}(\alpha^{\prime})\,\det[DX(\alpha^{\prime})]\,\mathrm{d}\alpha^{\prime}$

maps $O_{M}$ to $C^{1,\gamma}(\mathbb{R}^{n};\mathbb{R}^{n})$.

###### Proof.

Let $X\in O_{M}$. In order to prove the proposition, we need to verify

(21)
$\left|\left|F(X)\right|\right|_{L^{\infty}}+\sup_{i\in\\{1,\ldots,n\\}}\left(\left|\left|\frac{\mathrm{d}}{\mathrm{d}\alpha_{i}}F(X)\right|\right|_{\gamma}\right)<\infty.$

If we consider the change of variables $x^{\prime}=X(\alpha^{\prime})$ in (20)
we get, for the $j$-th component

(22)
$\begin{split}F_{j}(X)(\alpha)&=\int_{\mathbb{R}^{n}}k_{j}(X(\alpha)-x^{\prime})\rho_{0}(X^{-1}(x^{\prime}))\,\mathrm{d}x^{\prime}=\\\
&=(k_{j}\ast(\rho_{0}\circ X^{-1}))(X(\alpha)).\end{split}$

Let $R=\textrm{m}(\textrm{supp }(\rho_{0}\circ X^{-1}))^{1/n}$. Then, as in
Lemma 6, we have, for $R_{0}=\textrm{m}(\textrm{supp }(\rho_{0}))^{1/n}$ that

$R\leq c_{n}R_{0}\left|\left|\nabla X\right|\right|_{L^{\infty}}.$

Since the kernel $k$ satisfies (11) then by (12) in Lemma 4 we have

$\left|\left|F(X)\right|\right|_{L^{\infty}}\leq cR\left|\left|\rho_{0}\circ
X^{-1}\right|\right|_{L^{\infty}}=cR\left|\left|\rho_{0}\right|\right|_{L^{\infty}}\leq
c_{n}R_{0}\left|\left|\nabla
X\right|\right|_{L^{\infty}}\left|\left|\rho_{0}\right|\right|_{L^{\infty}},$

which is bounded for $X\in C^{1,\gamma}$ and $\rho_{0}\in C^{\gamma}_{c}$.

We focus then on the norms of derivatives of $F(X)$. We write
$\partial_{i}=\frac{\partial}{\partial\alpha_{i}}$. We have, by definition of
the norm,
$\left|\left|\partial_{i}F(X)\right|\right|_{\gamma}=\sup_{j\in\\{1,\ldots,n\\}}\left|\left|\partial_{i}F_{j}(X)\right|\right|_{\gamma}$.
We work then with $\partial_{i}F_{j}(X)$ for $i,j\in\\{1,\ldots,n\\}$. An
application of the chain rule, combined with Lemma 18, yields

(23)
$\begin{split}\partial_{i}&F_{j}(X)(\alpha)=\nabla(k_{j}\ast(\rho_{0}\circ
X^{-1}))(X(\alpha))\cdot\partial_{i}X(\alpha)=\\\
&=\sum_{r=1}^{n}\partial_{r}(k_{j}\ast(\rho_{0}\circ
X^{-1}))(X(\alpha))\partial_{i}X_{r}(\alpha)=\\\
&=\sum_{r=1}^{n}\left(c_{rj}\rho_{0}(\alpha)+\text{p.v.}(\partial_{r}k_{j}\ast(\rho_{0}\circ
X^{-1}))(X(\alpha))\right)\partial_{i}X_{r}(\alpha)=\\\
&=\sum_{r=1}^{n}\left(c_{rj}\rho_{0}(\alpha)+S_{rj}(\alpha)\right)\partial_{i}X_{r}(\alpha),\end{split}$

where $S_{rj}(\alpha):=\text{p.v.}(\partial_{r}k_{j}\ast(\rho_{0}\circ
X^{-1}))(X(\alpha))$ and $\cdot$ stands for the usual scalar product.

Since $\rho_{0},\partial_{i}X_{r}\in C^{\gamma}$ and $C^{\gamma}$ is an
algebra, then it suffices to control the $C^{\gamma}$ norm of $S_{rj}$.
Clearly, hypothesis in Lemma 4 are satisfied if $P=\partial_{r}k_{j}$. Then we
set $\varepsilon=\left|\rho_{0}\circ X^{-1}\right|_{\gamma}^{1/\gamma}$and
apply bound (13) in Lemma 4 to get

(24) $\left|\left|S_{rj}\right|\right|_{L^{\infty}}\leq
c\left\\{1+\max(1,\frac{1}{\gamma}\ln(R\left|\rho_{0}\circ
X^{-1}\right|_{\gamma}))\left|\left|\rho_{0}\circ
X^{-1}\right|\right|_{L^{\infty}}\right\\}$

where $R=\textrm{m}(\textrm{supp }(\rho_{0}\circ X^{-1}))^{1/n}$. As
previously, $R$ is bounded by
$c_{n}R_{0}\left|\left|X\right|\right|_{L^{\infty}}$. Since both
$\left|\left|\rho_{0}\circ X^{-1}\right|\right|_{L^{\infty}}$ and
$\left|\rho_{0}\circ X^{-1}\right|_{\gamma}$ are bounded above by
$\left|\left|\rho_{0}\circ X^{-1}\right|\right|_{\gamma}$ and taking also into
account that

$\left|\left|\rho_{0}\circ
X^{-1}\right|\right|_{\gamma}\leq\left|\left|\rho_{0}\right|\right|_{\gamma}(1+\left|X\right|_{1,\gamma}^{\gamma(2n-1)})<\infty,$

then we have that the right hand side of (24) is finite. On the other hand, by
(14) we have

$\left|S_{rj}\right|_{\gamma}\leq c\left|\rho_{0}\circ
X^{-1}\right|_{\gamma}\left|\left|\nabla
X\right|\right|_{L^{\infty}}^{\gamma}\leq
c\left|\left|\rho_{0}\right|\right|_{\gamma}(1+\left|X\right|_{1,\gamma}^{\gamma(2n-1)})\left|X\right|_{1,\gamma}^{\gamma}.$

Then, as we argued before, the $C^{\gamma}$ norm of $\partial_{i}F_{j}(X)$ is
finite for any $i,j$ so the supremum in (21) is finite as well, completing the
proof of the proposition. ∎

Then we have that $F$ satisfies the first hypothesis in Picard-Lindelöf’s
theorem. It remains to check that $F$ is locally Lipschitz. We claim (and
prove later) that if the directional derivative $F^{\prime}(X)$ is bounded as
a linear operator between $O_{M}$ and $B$ then $F$ is locally Lipschitz. So,
first of all we have to compute this directional derivative. An auxiliary
lemma is useful for this computation and we need to give a previous definition
to write it.

######  Definition 10.

Given $A\in M_{n\times n}(\mathbb{R}^{n})$ we define $A_{i,j}^{c}\in
M_{(n-1)\times(n-1)}(\mathbb{R}^{n})$ as the submatrix of A obtained by
erasing the $i$-th row and the $j$-th column .

The following lemma is not needed whenever the velocity field is divergence
free, as in the Euler equation. In that case, $\det(DX(\cdot,t))\equiv 1$ and
consequently the functional in (20) does not contain the determinant inside
the integral. In the general case, as we will see later, an expression for the
sum of determinants will be necessary.

######  Lemma 11.

Given $X,Y:\mathbb{R}^{n}\to\mathbb{R}^{n}$ differentiable homeomorphisms.

$\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\det(DX+\varepsilon
DY)\big{|}_{\varepsilon=0}=\sum_{i,j=1}^{n}(-1)^{i+j}\partial_{j}Y_{i}\det(DX_{i,j}^{c}).$

###### Proof.

First we use a formula for the determinant of a sum of square matrices. The
proof can be found in [Ma, pp. 162-163]. Let $A,B\in M_{n\times
n}(\mathbb{R})$ and let $\alpha,\beta$ strictly increasing integer sequences
chosen from $\\{1,\ldots,n\\}$. Let $\left|\alpha\right|$ (resp.
$\left|\beta\right|$) the number of elements of $\alpha$ (resp. $\beta$). If
$\left|\alpha\right|=\left|\beta\right|$ then let $A[\alpha|\beta]\in
M_{\left|\alpha\right|\times\left|\alpha\right|}(\mathbb{R})$ the submatrix of
$A$ lying in rows $\alpha$ and columns $\beta$ and $B[\alpha|\beta]\in
M_{(n-\left|\alpha\right|)\times(n-\left|\alpha\right|)}(\mathbb{R})$ the
submatrix of $B$ lying in rows complementary to $\alpha$ and columns
complementary to $\beta$. Let $s(\alpha)$ (resp. $s(\beta)$) the sum of the
integers in $\alpha$ (resp. $\beta$). Then

(25)
$\det(A+B)=\sum_{r=0}^{n}\sum_{\stackrel{{\scriptstyle\alpha,\beta}}{{\left|\alpha\right|=\left|\beta\right|=r}}}(-1)^{s(\alpha)+s(\beta)}\det(A[\alpha|\beta])\det(B[\alpha|\beta]).$

Note that for a matrix $M\in M_{s\times s}(\mathbb{R})$ and a constant
$c\in\mathbb{R}$ we have $det(c\cdot M)=c^{s}\det(M)$. Then setting $A=DX$ and
$B=\varepsilon DY$ in (25) we get

$\det(DX+\varepsilon
DY)=\sum_{r=0}^{n}\varepsilon^{n-r}\sum_{\left|\alpha\right|=\left|\beta\right|=r}(-1)^{s(\alpha)+s(\beta)}\det(DX[\alpha|\beta])\det(DY[\alpha|\beta]).$

Differentiating with respect to $\varepsilon$ the previous equation and
setting $\varepsilon=0$ make some terms vanish and, in consequence,

$\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\det(DX+\varepsilon
DY)\big{|}_{\varepsilon=0}=\sum_{\left|\alpha\right|=\left|\beta\right|=n-1}(-1)^{s(\alpha)+s(\beta)}\det(DX[\alpha|\beta])\det(DY[\alpha|\beta]).$

Note that a strictly increasing sequence taking $n-1$ elements of
$\\{1,\ldots,n\\}$ is a sequence avoiding just one of them. So

$\alpha=(1,\ldots,i-1,i+1,\ldots,n),\,\,\,\,\beta=(1,\ldots,j-1,j+1,\ldots,n),$

for $i,j=1,\ldots,n$. Then $s(\alpha)+s(\beta)=n(n+1)-(i+j)$ and hence
$(-1)^{s(\alpha)+s(\beta)}=(-1)^{i+j}$. For these special sequences we can
simplify and write $DY[\alpha|\beta]=\partial_{j}Y_{i}$ and
$DX[\alpha|\beta]=DX_{i,j}^{c}$ as in Definition 10.

Summing up,

$\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\det(DX+\varepsilon
DY)\big{|}_{\varepsilon=0}=\sum_{i,j=1}^{n}(-1)^{i+j}\partial_{j}Y_{i}\det(DX_{i,j}^{c})$

and the lemma is proved.

∎

We already have the tools to compute the directional derivative of $F$.

######  Proposition 12.

Let $X\in O_{M},Y\in B$ . For $F=(F_{j})_{j=1}^{n}$ defined in (20) we have
$F_{j}^{\prime}(X)Y=$I+II where

$\begin{split}\text{I}:&=\int_{\mathbb{R}^{n}}\nabla
k_{j}(X(\alpha)-X(\alpha^{\prime}))\cdot(Y(\alpha)-Y(\alpha^{\prime}))\rho_{0}(\alpha^{\prime})\det(DX)(\alpha^{\prime})\,\mathrm{d}\alpha^{\prime},\\\
\text{II}:&=\sum_{r,s=1}^{n}(-1)^{r+s}\int_{\mathbb{R}^{n}}k_{j}(X(\alpha)-X(\alpha^{\prime}))\rho_{0}(\alpha^{\prime})\partial_{s}Y_{r}(\alpha^{\prime})\det(DX_{r,s}^{c})(\alpha^{\prime})\,\mathrm{d}\alpha^{\prime}.\end{split}$

###### Proof.

Let $j=1,\ldots,n$. Consider $X\in O_{M}$ and $Y\in B$. Firstly, we apply the
chain rule to see

$\begin{split}\frac{\mathrm{d}}{\mathrm{d}\varepsilon}&\left(k_{j}(X(\alpha)-X(\alpha^{\prime})+\varepsilon(Y(\alpha)-Y(\alpha^{\prime})))\right)_{\varepsilon=0}=\\\
&=\sum_{i=1}^{n}\partial_{i}k_{j}(X(\alpha)-X(\alpha^{\prime}))(Y_{i}(\alpha)-Y_{i}(\alpha^{\prime}))=\\\
&=\nabla
k_{j}(X(\alpha)-X(\alpha^{\prime}))\cdot(Y(\alpha)-Y(\alpha^{\prime})),\end{split}$

where $\cdot$ is the usual scalar product. Thus having into account the above
computation and applying Lemma 11 we get

$\begin{split}&(F_{j}^{\prime}(X)Y)(\alpha)=\frac{\mathrm{d}}{\mathrm{d}\varepsilon}(F_{j}(X+\varepsilon
Y)(\alpha))_{\varepsilon=0}=\\\ &=\int_{\mathbb{R}^{n}}\nabla
k_{j}(X(\alpha)-X(\alpha^{\prime}))\cdot(Y(\alpha)-Y(\alpha^{\prime}))\rho_{0}(\alpha^{\prime})\det(DX)(\alpha^{\prime})\,\mathrm{d}\alpha^{\prime}+\\\
&+\sum_{r,s=1}^{n}(-1)^{r+s}\int_{\mathbb{R}^{n}}k_{j}(X(\alpha)-X(\alpha^{\prime}))\rho_{0}(\alpha^{\prime})\partial_{s}Y_{r}(\alpha^{\prime})\det(DX_{r,s}^{c})(\alpha^{\prime})\,\mathrm{d}\alpha^{\prime},\end{split}$

as we wanted to prove. ∎

######  Remark 13.

Note that there is no need to write principal value in the term I of the
proposition because the singularity of $\nabla k_{j}$ when
$\alpha=\alpha^{\prime}$ is compensated with the term
$Y(\alpha)-Y(\alpha^{\prime})$.

The directional derivative computed in Proposition 12 has been decomposed as
the sum of two terms. The second one is very similar to the one treated in
Proposition 9, but the first one looks different. It can be written as an
integral with respect to a kernel which is not of convolution type and so its
derivatives may be tricky to handle. In the following lemma, which is somehow
technical, we compute exactly those derivatives.

######  Lemma 14.

Let $I$ defined in Proposition 12. Then $I=\sum_{i=1}^{n}I_{i}(\alpha)$ where

$I_{i}(\alpha):=\int_{\mathbb{R}^{n}}\partial_{i}k_{j}(X(\alpha)-X(\alpha^{\prime}))(Y_{i}(\alpha)-Y_{i}(\alpha^{\prime}))\rho_{0}(\alpha^{\prime})\det(DX)(\alpha^{\prime})\,\mathrm{d}\alpha^{\prime},$

and its distributional derivatives are
$\partial_{l}I_{i}(\alpha)=\nabla\tilde{I}_{i}(X(\alpha))\cdot\partial_{l}X(\alpha)$
where

$\begin{split}&\partial_{k}\tilde{I}_{i}(x)=\\\
&=\text{p.v.}\int_{\mathbb{R}^{n}}\partial_{k}\partial_{i}k_{j}(x-x^{\prime})(Y_{i}(X^{-1}(x))-Y_{i}(X^{-1}(x^{\prime})))(\rho_{0}\circ
X^{-1})(x^{\prime})\,\mathrm{d}x^{\prime}+\\\ &+\partial_{k}[Y_{i}\circ
X^{-1}](x)\text{p.v.}\int_{\mathbb{R}^{n}}\partial_{i}k_{j}(x-x^{\prime})(\rho_{0}\circ
X^{-1})(x^{\prime})\,\mathrm{d}x^{\prime}+\\\ &+c_{k}(\nabla[Y_{i}\circ
X^{-1}](x)\cdot\xi_{k})(\rho_{0}\circ X^{-1})(x),\end{split}$

where $\xi_{k}$ is a vector in $\mathbb{R}^{n}$ depending on $k$.

###### Proof.

Consider $\alpha=X^{-1}(x)$, then after the change of variables
$\alpha^{\prime}=X^{-1}(x^{\prime})$, we have

$\begin{split}\tilde{I}_{i}(x)&=I_{i}(X^{-1}(x))=\\\
&=\int_{\mathbb{R}^{n}}\partial_{i}k_{j}(x-x^{\prime})(Y_{i}(X^{-1}(x))-Y_{i}(X^{-1}(x^{\prime})))\rho_{0}(X^{-1}(x^{\prime}))\,\mathrm{d}x^{\prime}.\end{split}$

Following the scheme in [MB, p. 165] for the Euler equation, let
$R(x,x^{\prime})=\partial_{i}k_{j}(x-x^{\prime})(Y_{i}(X^{-1}(x))-Y_{i}(X^{-1}(x^{\prime})))$.
Firstly, we compute the partial distributional derivative with respect to
$x_{k}$ of $R(x+x^{\prime},x^{\prime})$. In order to do that, note previously
that, given $h>0$ and $a\in\mathbb{R}^{n},$ $\left|a\right|=1$, we have, by
Taylor expansion of $Y\circ X^{-1}$,

(26) $\begin{split}\lim_{h\to 0}R(x+&ah,x)h^{n-1}=\\\ &=\lim_{h\to
0}\partial_{i}k_{j}(ah)(Y_{i}(X^{-1}(x+ah))-Y_{i}(X^{-1}(x)))h^{n-1}=\\\
&=\lim_{h\to 0}\partial_{i}k_{j}(a)h^{-n}h^{n-1}(\nabla[Y_{i}\circ
X^{-1}](x)\cdot ah+o(h))=\\\ &=\partial_{i}k_{j}(a)\nabla[Y_{i}\circ
X^{-1}](x)\cdot a.\end{split}$

We then compute the distributional derivative of $R(x+x^{\prime},x^{\prime})$.
Let $\varphi\in C^{\infty}_{c}(\mathbb{R}^{n};\mathbb{R})$ a test function.
Then

$\begin{split}\langle&\partial_{x_{k}}R(\cdot+x^{\prime},x^{\prime}),\varphi\rangle=-\int_{\mathbb{R}^{n}}\partial_{x_{k}}\varphi(x)R(x+x^{\prime},x^{\prime})\,\mathrm{d}x=\\\
=&\lim_{\varepsilon\to 0}\int_{\mathbb{R}^{n}\setminus
B(0,\varepsilon)}\varphi(x)\partial_{x_{k}}R(x+x^{\prime},x^{\prime})\,\mathrm{d}x-\\\
-&\lim_{\varepsilon\to 0}\int_{\mathbb{R}^{n}\setminus
B(0,\varepsilon)}\partial_{x_{k}}\left[\varphi(x)R(x+x^{\prime},x^{\prime})\right]\,\mathrm{d}x.\\\
\end{split}$

Applying Stokes’s theorem, and since $\varphi$ has compact support we obtain

$\begin{split}\langle&\partial_{x_{k}}R(\cdot+x^{\prime},x^{\prime}),\varphi\rangle=p.v.\langle\partial_{x_{k}}R(\cdot+x^{\prime},x^{\prime}),\varphi\rangle+\\\
&+\lim_{\varepsilon\to 0}\int_{\partial
B(0,\varepsilon)}\varphi(x)R(x+x^{\prime},x^{\prime})n_{k}(x)\,\mathrm{d}\sigma(x),\end{split}$

where $n_{k}(x)$ is the $k$-th component of the unitary normal vector to
$\partial B(0,\varepsilon)$ at the point $x$. We apply the observation made in
(26) to conclude that

$\lim_{\varepsilon\to 0}\int_{\partial
B(0,\varepsilon)}\varphi(x)R(x+x^{\prime},x^{\prime})n_{k}(x)\,\mathrm{d}\sigma(x)=\varphi(0)\nabla[Y_{i}\circ
X^{-1}](x^{\prime})\cdot\xi_{k},$

where the $l$-th component of $\xi_{k}$ is

$(\xi_{k})_{l}=\int_{\partial
B(0,1)}\partial_{i}k_{j}(a)n_{k}(a)\,a_{l}\,\mathrm{d}\sigma(a).$

Therefore, distributionally

$\partial_{x_{k}}R(\cdot+x^{\prime},x^{\prime})=\text{p.v.}\partial_{x_{k}}R(\cdot+x^{\prime},x^{\prime})+(c_{k}\nabla[Y_{i}\circ
X^{-1}](x^{\prime})\cdot\xi_{k})\delta_{0}.$

Then, since
$\partial_{x_{k}}R(x,x^{\prime})=\partial_{x_{k}}[R(\cdot+x^{\prime},x^{\prime})](x-x^{\prime})$
we finally get for
$H(x)=\int_{\mathbb{R}^{n}}R(x,x^{\prime})f(x^{\prime})\,\mathrm{d}x^{\prime}$,

$\begin{split}\partial_{k}H(x)&=\text{p.v.}\int_{\mathbb{R}^{n}}\partial_{k}R(x,x^{\prime})f(x^{\prime})\,\mathrm{d}x^{\prime}+\\\
&+\int_{\mathbb{R}^{n}}\delta_{0}(x-x^{\prime})c_{k}\nabla[Y_{i}\circ
X^{-1}](x^{\prime})\cdot\xi_{k}f(x^{\prime})\,\mathrm{d}x^{\prime}=\\\
&=\text{p.v.}\int_{\mathbb{R}^{n}}\partial_{k}\partial_{i}k_{j}(x-x^{\prime})(Y_{i}(X^{-1}(x))-Y_{i}(X^{-1}(x^{\prime})))f(x^{\prime})\,\mathrm{d}x^{\prime}+\\\
&+\partial_{k}[Y_{i}\circ
X^{-1}](x)\text{p.v.}\int_{\mathbb{R}^{n}}\partial_{i}k_{j}(x-x^{\prime})f(x^{\prime})\,\mathrm{d}x^{\prime}+\\\
&+c_{k}\nabla[Y_{i}\circ X^{-1}](x)\cdot\xi_{k}f(x).\end{split}$

The proof is completed setting $f=\rho_{0}\circ X^{-1}$ in the previous
expression and applying the chain rule to
$I_{i}(\alpha)=\tilde{I}_{i}(X(\alpha))$.

∎

Remember that our goal is to bound the $C^{1,\gamma}$ norm of the map
$\alpha\to F^{\prime}(X)Y(\alpha)$ in such a way that we get

$\left|F^{\prime}(X)Y\right|_{1,\gamma}\leq c\left|Y\right|_{1,\gamma}$

for $c$ depending maybe on $n,\rho_{0}$ and $X$ in order to have that
$F^{\prime}(X)$ is bounded as a linear operator. Note that the first term in
$\partial_{k}\tilde{I}_{i}$ in Lemma 14 is an integral containing
$\partial_{k}\partial_{i}k_{j}$, which is a hypersingular kernel.
Nevertheless, the term $Y(\alpha)-Y(\alpha^{\prime})$ will, in some sense,
kill this excess of singularity. We quantify this effect in the following
Lemma.

######  Lemma 15.

Let $H:\mathbb{R}^{n}\to\mathbb{R}$, $H\in
C^{1}(\mathbb{R}^{n}\setminus\\{0\\})$, be a kernel homogeneous of degree
$-n-1$ such that $H_{1}^{i}(x)=x_{i}H(x)$, $i=1,\ldots,n$, define a CZO of
convolution type. Let $g\in C^{1,\gamma}(\mathbb{R}^{n};\mathbb{R})$ and $f\in
C^{\gamma}_{c}(\mathbb{R}^{n};\mathbb{R})$. Then for

$Tf(x)=\text{p.v.
}\int_{\mathbb{R}^{n}}H(x-x^{\prime})(g(x)-g(x^{\prime}))f(x^{\prime})\,\mathrm{d}x^{\prime}$

we have

$\left|\left|Tf\right|\right|_{\gamma}\leq
c\left|g\right|_{1,\gamma}\left|\left|f\right|\right|_{\gamma},$

for $c$ depending on $\text{m(supp(}f))$.

###### Proof.

Since $g\in C^{1,\gamma}$ we can write its Taylor series centered at
$x^{\prime}$ as

$g(x)=g(x^{\prime})+\sum_{i=1}^{n}\partial_{i}g(x^{\prime})(x_{i}-x_{i}^{\prime})+R(x,x^{\prime})$

with $\left|R(x,x^{\prime})\right|\leq
c\left|g\right|_{1,\gamma}\left|x-x^{\prime}\right|^{1+\gamma}$. Now, if we
add and subtract some term we obtain

$\begin{split}Tf(x)=&\int_{\mathbb{R}^{n}}H(x-x^{\prime})(g(x)-g(x^{\prime})-\nabla
g(x^{\prime})\cdot(x-x^{\prime}))f(x^{\prime})\,\mathrm{d}x^{\prime}+\\\
+\sum_{i=1}^{n}\text{p.v.
}&\int_{\mathbb{R}^{n}}(x_{i}-x_{i}^{\prime})H(x-x^{\prime})\partial_{i}g(x^{\prime})f(x^{\prime})\,\mathrm{d}x^{\prime}=:T_{1}f(x)+T_{2}f(x),\end{split}$

The kernel $H_{g}(x,x^{\prime}):=H(x-x^{\prime})(g(x)-g(x^{\prime})-\nabla
g(x^{\prime})\cdot(x-x^{\prime}))$ satisfies the bound
$\left|H_{g}(x,x^{\prime})\right|\leq\frac{c\left|g\right|_{1,\gamma}}{\left|x-x^{\prime}\right|^{n-\gamma}}$
and its gradient

$\begin{split}\nabla_{x}H_{g}(x,x^{\prime})&=\nabla
H(x-x^{\prime})(g(x)-g(x^{\prime})-\nabla g(x^{\prime})(x-x^{\prime}))+\\\
&+H(x-x^{\prime})(\nabla g(x)-\nabla g(x^{\prime}))\end{split}$

satisfies
$\left|\nabla_{x}H_{g}(x,x^{\prime})\right|\leq\frac{c\left|g\right|_{1,\gamma}}{\left|x-x^{\prime}\right|^{n+1-\gamma}}$.
To check that $T_{1}f$ belongs to $C^{\gamma}$ we use an usual argument. We
can see $\left|T_{1}f(x)\right|\leq
c\left|\left|f\right|\right|_{L^{\infty}}\left|g\right|_{1,\gamma}$ for every
$x\in\mathbb{R}^{n}$. Then $\left|\left|T_{1}f\right|\right|_{L^{\infty}}\leq
c\left|\left|f\right|\right|_{L^{\infty}}\left|g\right|_{1,\gamma}$. Now, let
$x_{1},x_{2}\in\mathbb{R}^{n}$ and $B:=B(x_{1},3\left|x_{1}-x_{2}\right|)$. We
decompose

$\begin{split}T_{1}f(x_{1})-T_{1}f(x_{2})&=\int_{\mathbb{R}^{n}\setminus
B}(H_{g}(x_{1},x^{\prime})-H_{g}(x_{2},x^{\prime}))f(x^{\prime})\,\mathrm{d}x^{\prime}+\\\
&+\int_{B}H_{g}(x_{1},x^{\prime})f(x^{\prime})\,\mathrm{d}x^{\prime}-\int_{B}H_{g}(x_{2},x^{\prime})f(x^{\prime})\,\mathrm{d}x^{\prime}.\end{split}$

Then, by the Mean Value Theorem and the bounds for $H_{g}$ and
$\nabla_{x}H_{g}$, we have

$\begin{split}&\left|T_{1}f(x_{1})-T_{1}f(x_{2})\right|\leq
c\left|g\right|_{1,\gamma}\Big{\\{}\left|x_{1}-x_{2}\right|\int_{\mathbb{R}^{n}\setminus
B}\frac{\left|f(x^{\prime})\right|}{\left|x_{1}-x^{\prime}\right|^{n+1-\gamma}}\,\mathrm{d}x^{\prime}+\\\
&+\int_{B}\frac{\left|f(x^{\prime})\right|}{\left|x_{1}-x^{\prime}\right|^{n-\gamma}}\,\mathrm{d}x^{\prime}+\int_{B}\frac{\left|f(x^{\prime})\right|}{\left|x_{2}-x^{\prime}\right|^{n-\gamma}}\,\mathrm{d}x^{\prime}\Big{\\}}\leq\\\
&\leq
c\left|g\right|_{1,\gamma}\left\\{\left|x_{1}-x_{2}\right|\left|\left|f\right|\right|_{L^{\infty}}\left|x_{1}-x_{2}\right|^{\gamma-1}+\left|\left|f\right|\right|_{L^{\infty}}\left|x_{1}-x_{2}\right|^{\gamma}\right\\}\leq\\\
&\leq
c\left|g\right|_{1,\gamma}\left|\left|f\right|\right|_{L^{\infty}}\left|x_{1}-x_{2}\right|^{\gamma}.\end{split}$

So $\left|T_{1}f\right|_{\gamma}\leq
c\left|\left|f\right|\right|_{L^{\infty}}\left|g\right|_{1,\gamma}$. To finish
we need to bound $T_{2}f$. If we set $H_{1}^{i}(x)=x_{i}H(x)$, which is a CZO
of convolution type by hypothesis, then since
$T_{2}f=\sum_{i=1}^{n}H_{1}^{i}\ast(f\partial_{i}g)$, by Lemma 4 we have

$\left|\left|T_{2}f\right|\right|_{\gamma}\leq c\left|\left|f\nabla
g\right|\right|_{\gamma}\leq
c\left|g\right|_{1,\gamma}\left|\left|f\right|\right|_{\gamma},$

finishing the proof of the lemma.

∎

The constant $c$ in Lemma 15 is finite whenever $f$ is compactly supported,
but we know this is true by Lemma 6. In a few words, Lemma 6 states that
$\rho(\cdot,t)$ is compactly supported when it is transported by a flow with
bounded gradient as it happens, in particular, if $X\in O_{M}$. We are ready
to check that $F^{\prime}(X)$ is bounded. As promised, taking into account
that boundedness, we can verify that the second hypothesis in Picard-Lindelöf
holds for $F$.

######  Proposition 16.

Let $O_{M}$ as defined in (17). Then, the functional $F:O_{M}\to
C^{1,\gamma}(\mathbb{R}^{n};\mathbb{R}^{n})$ defined in (20) is locally
Lipschitz.

###### Proof.

First of all, by the Fundamental Theorem of Calculus, given $X_{1},X_{2}\in
O_{M}$,

$\begin{split}\left|F(X_{1})-F(X_{2})\right|_{1,\gamma}&=\left|\int_{0}^{1}\frac{\mathrm{d}}{\mathrm{d}\varepsilon}F(X_{1}+\varepsilon(X_{2}-X_{1}))\,\mathrm{d}\varepsilon\right|_{1,\gamma}=\\\
&=\left|\left(\int_{0}^{1}F^{\prime}(X_{1}+\varepsilon(X_{2}-X_{1}))\cdot(X_{2}-X_{1})\,\mathrm{d}\varepsilon\right)\right|_{1,\gamma}\leq\\\
&\leq\left\\{\int_{0}^{1}\left|\left|F^{\prime}(X_{1}+\varepsilon(X_{2}-X_{1}))\right|\right|_{B\to
B}\,\mathrm{d}\varepsilon\right\\}\left|X_{2}-X_{1}\right|_{1,\gamma},\end{split}$

where $F^{\prime}(X)$ is the operator defined by $Y\to F^{\prime}(X)Y$. So,
provided $\left|\left|F^{\prime}(X)\right|\right|_{B\to B}<\infty$ for every
$X\in B$ then the integral in the previous expression is finite and therefore
$F$ is Lipschitz. Thus, it suffices to prove this boundedness in order to
prove the Proposition.

By Proposition 12 we have seen that every component of $F^{\prime}(X)Y$ can be
written as the sum of two terms $I$ and $II$. The arguments in Proposition 9
can be repeated for each element appearing in the sum in which $II$ is
decomposed, just by changing the role of $\rho_{0}$ to
$\rho_{0}\partial_{s}Y_{r}$. Then we can conclude, similarly, that

$\left|II\right|_{1,\gamma}\leq
c(n,\rho_{0})\left|Y\right|_{1,\gamma}\left|X\right|_{1,\gamma}^{n}.$

Hence, we just have to work with the first term, $I$, and bound its
$C^{1,\gamma}$ norm. Before considering derivatives, note that $I$ can be
compared to any derivative of $II$, so also in similar fashion to Proposition
9, we get

$\left|\left|I\right|\right|_{L^{\infty}}\leq
c(n,\rho_{0})\left|Y\right|_{1,\gamma}\left|X\right|_{1,\gamma}^{n}.$

We then need to consider derivatives of $I$. By Lemma 14 we write
$I=\sum_{i=1}^{n}I_{i}$ and also
$\partial_{l}I_{i}(\alpha)=\sum_{k=1}^{n}\partial_{k}\tilde{I}_{i}(X(\alpha))\partial_{l}X_{k}(\alpha)$.
Since $C^{\gamma}$ is an algebra and also, by (9), we have

$\begin{split}\left|\left|\partial_{i}I_{i}\right|\right|_{\gamma}&\leq\sum_{k=1}^{n}\left|\left|\partial_{k}\tilde{I}_{i}\circ
X\right|\right|_{\gamma}\left|\left|\partial_{l}X_{k}\right|\right|_{\gamma}\leq\\\
&\leq\sum_{k=1}^{n}\left|\left|\partial_{k}\tilde{I}_{i}\right|\right|_{\gamma}(1+\left|X\right|_{1,\gamma}^{\gamma})\left|X\right|_{1,\gamma}.\end{split}$

Now we focus on
$\left|\left|\partial_{k}\tilde{I}_{i}\right|\right|_{\gamma}$. We consider
the expression given by Lemma 14

$\begin{split}&\partial_{k}\tilde{I}_{i}(x)=\\\
&=\text{p.v.}\int_{\mathbb{R}^{n}}\partial_{k}\partial_{i}k_{j}(x-x^{\prime})(Y_{i}(X^{-1}(x))-Y_{i}(X^{-1}(x^{\prime})))(\rho_{0}\circ
X^{-1})(x^{\prime})\,\mathrm{d}x^{\prime}+\\\ &+\partial_{k}[Y_{i}\circ
X^{-1}](x)\text{p.v.}\int_{\mathbb{R}^{n}}\partial_{i}k_{j}(x-x^{\prime})(\rho_{0}\circ
X^{-1})(x^{\prime})\,\mathrm{d}x^{\prime}+\\\ &+c_{k}(\nabla[Y_{i}\circ
X^{-1}](x)\cdot\xi_{k})(\rho_{0}\circ X^{-1})(x)=A(x)+B(x)+C(x).\end{split}$

A straightforward repetition of the arguments done before let us verify that

$\left|\left|C\right|\right|_{\gamma}\leq
c(n,\rho_{0})\left|Y\right|_{1,\gamma}\left|X\right|_{1,\gamma}^{m},$

where $m$ is a finite constant depending on $\gamma$ and $n$. Also, as in
Proposition 9, we get

$\left|\left|B\right|\right|_{\gamma}\leq
c(n,\rho_{0})\left|Y\right|_{1,\gamma}\left|X\right|_{1,\gamma}^{m}.$

The term $A$ is more involved than the rest in the decomposition of
$\partial_{k}\tilde{I}_{i}$. Nevertheless, we claim that we can apply Lemma 15
for $H=\partial_{k}\partial_{i}k_{j}$, $g=Y_{i}\circ X^{-1}$ and
$f=\rho_{0}\circ X^{-1}$ to obtain also

$\left|\left|A\right|\right|_{\gamma}\leq
c(n,\rho_{0},R)\left|Y\right|_{1,\gamma}\left|X\right|_{1,\gamma}^{m},$

where $R=\text{m(supp(}\rho_{0}\circ X^{-1}))=\text{m(supp(}\rho))$. We
conclude that

$\left|I\right|_{1,\gamma}\leq
c(n,\rho_{0},R)\left|Y\right|_{1,\gamma}\left|X\right|_{1,\gamma}^{n},$

where $c(n,\rho,R)$ is finite by Lemma 6.

Summing up, for any $X\in B$ and any $Y\in O_{M}$ we have seen

$\left|F^{\prime}(X)Y\right|_{1,\gamma}\leq
c(n,\rho_{0},R)\left|X\right|_{1,\gamma}^{m}\left|Y\right|_{1,\gamma},$

so $F^{\prime}(X)$ is bounded as a linear operator from $O_{M}$ to $B$ and
then the Proposition is proved. ∎

######  Remark 17.

We have been able to apply Lemma 15 since the kernels
$x_{i}\partial_{j}\partial_{l}k$ are CZO. In general, if $k$ satisfies the
hypothesis in Lemma 4 then its second derivatives $\partial_{j}\partial_{l}k$
satisfies Lemma 15. It is clear that $\partial_{j}\partial_{l}k$ is
homogeneous of degree $-n-1$. Also, if $i\neq j$ (and similarly if $i\neq l$),

$\int_{\left|w\right|=1}w_{i}\partial_{j}\partial_{l}k(w)\,\mathrm{d}\sigma(w)=\int_{\left|w\right|=1}\partial_{j}[w_{i}\partial_{l}k(w)]\,\mathrm{d}\sigma(w)=0.$

The last integral vanishes for similar reasons as explained in Remark 5.
Otherwise, if $i=j=l$,

$\begin{split}\int_{\left|w\right|=1}&w_{i}\partial_{i}^{2}k(w)\,\mathrm{d}\sigma(w)=\\\
&=\int_{\left|w\right|=1}\partial_{i}[w_{i}\partial_{i}k(w)]\,\mathrm{d}\sigma(w)-\int_{\left|w\right|=1}\partial_{i}k(w)\,\mathrm{d}\sigma(w)=0\end{split}$

since both integrals in the right vanishes, similarly as before.

Finally, since all the hypothesis in Theorem 7 are verified, we prove the
existence result for the trajectory maps.

######  Theorem 18.

Let $\rho_{0}\in C^{\gamma}_{c}(\mathbb{R}^{n};\mathbb{R})$. Then there exists
$T^{*}>0$ such that the ordinary differential equation

$\begin{cases}\frac{\mathrm{d}}{\mathrm{d}t}X(\alpha,t)=F(X(\alpha,t)),\\\
X(\alpha,0)=\alpha,\end{cases}$

for

$F(X(\alpha,t))=\int_{\mathbb{R}^{n}}k(X(\alpha,t)-X(\alpha^{\prime},t))\rho_{0}(\alpha^{\prime})\,\det[DX(\alpha^{\prime},t)]\,\mathrm{d}\alpha^{\prime},$

has a unique solution $X(\cdot,t)\in
C^{1,\gamma}(\mathbb{R}^{n};\mathbb{R}^{n})$ for $t\in(-T^{*},T^{*})$.

###### Proof.

Let $B=C^{1,\gamma}(\mathbb{R}^{n};\mathbb{R}^{n})$ and let $O_{M}$ defined in
(17). Then, by Propositions 9 and 16 the functional $F$ satisfies the
hypothesis of Picard-Lindelöf’s theorem 7 and therefore we conclude that the
statement holds. ∎

Given the flow map $X(\cdot,t)$ we can define the solution to (1) in an unique
way: since the velocity field is smooth enough, any solution of the transport
equation (1) can be described through the trajectories. So we have the main
result of this section: well-posedness in the Hölder class for the transport
equation and for the class of kernels described above.

######  Theorem 19.

Let $\rho_{0}\in C^{\gamma}_{c}(\mathbb{R}^{n};\mathbb{R})$. Let $k\in
C^{2}(\mathbb{R}^{n}\setminus\\{0\\};\mathbb{R}^{n})$ be homogeneous of degree
$1-n$ . Then there exists $T^{*}>0$ such that the transport equation

(27) $\begin{cases}\rho_{t}+v\cdot\nabla\rho=0,\\\
v(\cdot,t)=k\ast\rho(\cdot,t),\\\ \rho(\cdot,0)=\rho_{0}.\end{cases}$

has a unique solution $\rho(\cdot,t)\in
C^{\gamma}_{c}(\mathbb{R}^{n};\mathbb{R})$, $v(\cdot,t)\in
C^{1+\gamma}(\mathbb{R}^{n};\mathbb{R}^{n})$ for $t\in(-T^{*},T^{*})$.

###### Proof.

By Theorem 18 up to time $T^{*}$ there exists a unique solution $X(\cdot,t)\in
C^{1+\gamma}$. Let $\tilde{\rho}\in C^{\gamma}_{c}$ and $\tilde{v}\in
C^{1+\gamma}$ satisfying (27). Then we can find a trajectory
$\tilde{X}(\cdot,t)$ associated to $\tilde{v}(\cdot,t)$ such that
$\tilde{\rho}$ is transported by $\tilde{X}$. By the uniqueness of trajectory
by Theorem 18 then $\tilde{X}=X$. Then
$\tilde{\rho}(\cdot,t)=\rho_{0}(\tilde{X}^{-1}(\cdot,t))=\rho_{0}(X^{-1}(\cdot,t))=\rho(\cdot,t)$
and hence, by convoluting the density with the kernel $k$ we can see that
$\tilde{v}=v$. ∎

##  4 Global Theorem

We want to show that the solution defined in Theorem 18 does exist for any
time, that is, we want to show that $T^{*}=\infty$. In order to do that, we
need to invoke a Continuation Theorem which gives us a necessary condition for
that to happen. The theorem is stated as in [MB, p. 148] and a proof for a
general version of it can be found in [LL, p. 161]. We would like to remark
that it is valid since we have been able to state the problem with a
functional $F$ which does not depend explicitly on time.

######  Theorem 20.

In the situation of Theorem 7 the unique solution $X\in C^{1}((-T,T);O)$
either exists globally in time or $T$ is finite and $X(t)$ leaves the open set
$O$ as $\left|t\right|$ approaches $T$.

In a nutshell, we need to check that at time $T^{*}$ the flow $X(\cdot,T^{*})$
still belongs to $O_{M}$. As we will see later, it is enough to verify that
the $C^{1,\gamma}$ norm of $X(\cdot,T^{*})$ is a priori bounded. The following
lemma is an auxiliary result needed in order to achieve bounds that allow us
to prove that boundedness.

######  Lemma 21.

Let $X(\cdot,t)$ defined in (2). Then for the inverse flow at time $t$, we
have $X^{-1}(\cdot,t)=\tilde{X}(0,t,\cdot)$, where $\tilde{X}(s,t,x)$ is the
solution of the integro-differential equation

$\tilde{X}(s,t,x)=x-\int_{s}^{t}v(\tilde{X}(r,t,x),r)\,\mathrm{d}r.$

###### Proof.

We define a generalized flow map
$\hat{X}:\mathbb{R}\times\mathbb{R}\times\mathbb{R}^{n}\to\mathbb{R}^{n}$
satisfying

(28) $\hat{X}(s,t,x)=x+\int_{s}^{t}v(\hat{X}(s,r,x),r)\,\mathrm{d}r.$

Then $\hat{X}(s,t,x)$ is the position at time $t$ of the particle that was at
the position $x$ at time $s$. It is clear that $X(x,t)=\hat{X}(0,t,x).$

First, we can check that $\hat{X}$ has the semigroup structure

(29) $\hat{X}(s,t,x)=\hat{X}(\tau,t,\hat{X}(s,\tau,x)).$

By definition of $\hat{X}$ in (28), the right hand side of (29) can be
expressed as

(30)
$\hat{X}(\tau,t,\hat{X}(s,\tau,x))=\hat{X}(s,\tau,x)+\int_{\tau}^{t}v(\hat{X}(\tau,u,\hat{X}(s,\tau,x)),u)\,\mathrm{d}u.$

We differentiate (30) with respect to $\tau$ to get

$\begin{split}\partial_{\tau}(\hat{X}(\tau,t,\hat{X}(s,\tau,x)))&=\partial_{\tau}\hat{X}(s,\tau,x)-v(\hat{X}(\tau,\tau,\hat{X}(s,\tau,x)),\tau)=\\\
&=v(\hat{X}(s,\tau,x),\tau)-v(\hat{X}(s,\tau,x),\tau)=0.\end{split}$

So $\hat{X}(\tau,t,\hat{X}(s,\tau,x))$ does not depend on $\tau$ and hence

$\hat{X}(\tau,t,\hat{X}(s,\tau,x))=\hat{X}(\tau,t,\hat{X}(s,\tau,x))|_{\tau=t}=\hat{X}(t,t,\hat{X}(s,t,x))=\hat{X}(s,t,x),$

which proves equation (29).

Secondly, we want to see that $\hat{X}$ satisfies a transport equation.
Differentiating (29) with respect to $s$ we obtain

(31)
$\begin{split}\partial_{s}\hat{X}(s,t,x)&=\nabla\hat{X}(\tau,t,\hat{X}(s,\tau,x))\partial_{s}\hat{X}(s,\tau,x)=\\\
&=-\nabla\hat{X}(\tau,t,\hat{X}(s,\tau,x))v(\hat{X}(s,s,x),s)=\\\
&=-\nabla\hat{X}(\tau,t,\hat{X}(s,\tau,x))v(x,s),\end{split}$

and putting $\tau=s$ in (31)

(32) $\partial_{s}\hat{X}(s,t,x)+v(x,s)\nabla\hat{X}(s,t,x)=0.$

Thus, $\hat{X}(\cdot,t,\cdot):\mathbb{R}\times\mathbb{R}^{n}\to\mathbb{R}^{n}$
satisfies a transport equation.

Then, we define a map
$\tilde{X}:\mathbb{R}\times\mathbb{R}\times\mathbb{R}^{n}$ via

(33) $\tilde{X}(s,t,x)=x-\int_{s}^{t}v(\tilde{X}(r,t,x),r)\,\mathrm{d}r.$

We want to check that for any $t,s\in\mathbb{R}$ and any $x\in\mathbb{R}^{n}$
the maps $\hat{X}$ and $\tilde{X}$ are inverse in the following sense

(34) $\hat{X}(s,t,\tilde{X}(s,t,x))=x.$

In order to prove (34) we differentiate its left hand side with respect to $s$

$\partial_{s}\hat{X}(s,t,\tilde{X}(s,t,x))+\nabla\hat{X}(s,t,\tilde{X}(s,t,x))\partial_{s}\tilde{X}(s,t,x).$

By (33) we can see that $\partial_{s}\tilde{X}(s,t,x)=v(\tilde{X}(s,t,x),s)$
and then the above expression can be written as

$\begin{split}\partial_{s}&\hat{X}(s,t,\tilde{X}(s,t,x))+v(\tilde{X}(s,t,x),s)\nabla\hat{X}(s,t,\tilde{X}(s,t,x))=\\\
&=\left[\partial_{s}\hat{X}(\cdot,t,\cdot)+v(\cdot,\cdot)\nabla\hat{X}(\cdot,t,\cdot)\right](x,\tilde{X}(s,t,x))=0\end{split}$

since $\hat{X}(\cdot,t,\cdot)$ satisfies the transport equation (32).

Thus, the left hand side of (34) does not depend on $s$ and hence

$\hat{X}(s,t,\tilde{X}(s,t,x))=\hat{X}(t,t,\tilde{X}(t,t,x))=x,$

as we wanted to prove.

Taking $s=0$ we get $\hat{X}(0,t,\tilde{X}(0,t,x))=X(\tilde{X}(0,t,x),t),$ so
$X(x,t)^{-1}=\tilde{X}(s,t,x)|_{s=0}$ and the Lemma is proved. ∎

As a consequence of the previous Lemma, we see that the flow map $X(\cdot,t)$
and its inverse $X^{-1}(\cdot,t)$ share the same regularity properties since
its integral expressions are similar. We will use this fact later on.

The next lemma is a well known and very classical fact in the theory of
ordinary differential differential equations. For its proof see, for instance,
[Pa, p. 13].

######  Lemma 22 (Gronwall Lemma).

Let $u$ and $f$ be continuous and nonnegative functions defined on $I=[a,b]$,
and let $n$ be a continuous, positive and non-decreasing function defined on
$I$. If

$u(t)\leq n(t)+\int_{a}^{t}f(s)u(s)\,\mathrm{d}s$

for $t\in I$, then

$u(t)\leq n(t)\exp\left(\int_{a}^{t}f(s)\,\mathrm{d}s\right).$

We are ready to give a first condition ensuring the a priori $C^{1,\gamma}$
boundedness of the flow map at any time. The following bounds are very similar
to those obtained for the Euler equation. The only difference that appears in
the proof is the fact that the measure of the support of $\rho(\cdot,t)$ is
not constant over time, because the divergence is not zero in general.

######  Proposition 23.

Let $X(\cdot,t)$ be the solution given in Theorem 18 and $c(n)$ a constant
depending on the dimension $n$. Then, for a certain function
$G:\mathbb{R}\to\mathbb{R}^{+}$ with $G(t)<\infty$ whenever
$\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s<\infty$, we have the
following inequalities

$\begin{split}&\left|X(0,t)\right|\leq
c(n)m(\text{supp}(\rho_{0}))^{1/n}\left|\left|\rho_{0}\right|\right|_{L^{\infty}}\int_{0}^{t}\left|\left|\nabla
X(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s,\\\ &\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}\leq\exp\left(\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s\right),\\\ &\left|\nabla
X(\cdot,t)\right|_{\gamma}\leq G(t)\exp\left(c\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s\right).\end{split}$

In particular, $\left|X(\cdot,t)\right|_{1,\gamma}$ is bounded provided
$\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s$ also is.

###### Proof.

By definition (2) we have, for any $\alpha\in\mathbb{R}^{n}$,

$X(\alpha,t)=\alpha+\int_{0}^{t}v(X(\alpha,s),s)\,\mathrm{d}s.$

Setting $\alpha=0$ and taking absolute value we get

$\left|X(0,t)\right|=\left|\int_{0}^{t}v(X(0,s),s)\,\mathrm{d}s\right|\leq\int_{0}^{t}\left|\left|v(X(\cdot,s),s)\right|\right|_{L^{\infty}}\,\mathrm{d}s.$

Since $X(\cdot,s)$ is an homeomorphism then
$\left|\left|v(X(\cdot,s),s)\right|\right|_{L^{\infty}}=\left|\left|v(\cdot,s)\right|\right|_{L^{\infty}}$
and therefore we simply have

$\left|X(0,t)\right|\leq\int_{0}^{t}\left|\left|v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s.$

Now let $R(s)=m(\text{supp}(\rho(\cdot,s)))^{1/n}$. Then as
$v(\cdot,s)=k\ast\rho(\cdot,s)$ by (12) we get the bound

$\left|\left|v(\cdot,s)\right|\right|_{L^{\infty}}\leq
cR(s)\left|\left|\rho(\cdot,s)\right|\right|_{L^{\infty}}=cR(s)\left|\left|\rho_{0}\right|\right|_{L^{\infty}},$

where the last equality stands provided $\rho$ is transported with the flow
and so, the $L^{\infty}$ norm is conserved in time.

By Lemma 6 we have a control for $R(s)$ and then

$\left|X(0,t)\right|\leq
c(n)m(\text{supp}(\rho_{0}))^{1/n}\left|\left|\rho_{0}\right|\right|_{L^{\infty}}\int_{0}^{t}\left|\left|\nabla
X(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s.$

In order to achieve bounds on derivatives of the flow map, we compute the
partial derivative with respect to $\alpha_{i}$ (denoted as $\partial_{i}$
from now on) of the $j$-th component of (2). By the chain rule, we get

(35)
$\begin{split}\frac{\mathrm{d}}{\mathrm{d}t}(\partial_{i}X_{j}(\alpha,t))&=\sum_{k=1}^{n}\frac{\partial
v_{j}(X(\alpha,t),t)}{\partial X_{k}(\alpha,t)}\partial_{i}X_{k}(\alpha,t)=\\\
&=\nabla v_{j}(X(\alpha,t))\cdot\partial_{i}X(\alpha,t),\end{split}$

where $\cdot$ stands for the scalar product between vectors in
$\mathbb{R}^{n}$.

Taking $L^{\infty}$ norm on (35) and considering supremum over $i,j$ we get

(36) $\begin{split}\frac{\mathrm{d}}{\mathrm{d}t}\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}&\leq\left|\left|\nabla
v(X(\cdot,t),t)\right|\right|_{L^{\infty}}\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}=\\\ &=\left|\left|\nabla
v(\cdot,t)\right|\right|_{L^{\infty}}\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}.\end{split}$

Therefore, by direct integration on (36) we have the desired bound

(37) $\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}\leq\exp\left(\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s\right).$

Finally, taking the $\left|\cdot\right|_{\gamma}$ seminorm and considering
supremum over $i,j$ on (35) we have

(38) $\begin{split}\frac{\mathrm{d}}{\mathrm{d}t}&\left|\nabla
X(\cdot,t)\right|_{\gamma}\leq\sup_{i,j}\left|\nabla
v_{j}(X(\cdot,t))\cdot\partial_{i}X(\cdot,t)\right|_{\gamma}\leq\\\ &\leq
c\left(\left|\left|\nabla v(\cdot,t)\right|\right|_{L^{\infty}}\left|\nabla
X(\cdot,t)\right|_{\gamma}+\left|\nabla
v(X(\cdot,t),t)\right|_{\gamma}\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}\right),\end{split}$

where we have used inequality (4) to bound the $\left|\cdot\right|_{\gamma}$
seminorm of a product.

By (8) we have

(39) $\left|\nabla v(X(\cdot,t),t)\right|_{\gamma}\leq\left|\nabla
v(\cdot,t)\right|_{\gamma}\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}^{\gamma}.$

Also, by (13),

(40) $\left|\nabla v(\cdot,t)\right|_{\gamma}\leq
c\left|\rho(\cdot,t)\right|_{\gamma}\leq
c\left|\rho_{0}\right|_{\gamma}\left|\left|\nabla
X^{-1}(\cdot,t)\right|\right|_{L^{\infty}}^{\gamma},$

where we have used that $\rho(\cdot,t)=\rho_{0}(X^{-1}(\cdot,t))$. Using the
equation for $X^{-1}(\cdot,t)$ described in Lemma 21 and similarly as done in
(36) we have

(41) $\left|\left|\nabla
X^{-1}(\cdot,t)\right|\right|_{L^{\infty}}\leq\exp\left(\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s\right).$

Combining inequalities (36), (39), (40), (41), we get a bound for (38)

(42) $\begin{split}\frac{\mathrm{d}}{\mathrm{d}t}\left|\nabla
X(\cdot,t)\right|_{\gamma}&\leq c\left(\left|\left|\nabla
v(\cdot,t)\right|\right|_{L^{\infty}}\left|\nabla
X(\cdot,t)\right|_{\gamma}+\right.\\\
&+\left.\left|\rho_{0}\right|_{\gamma}\exp\left((1+2\gamma)\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s\right)\right).\end{split}$

Setting

$\begin{split}g(t):=c\left|\rho_{0}\right|_{\gamma}\exp\left((1+2\gamma)\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s\right)\end{split}$

and $G(t):=\int_{0}^{t}g(s)\,\mathrm{d}s$, and then applying Lemma 22 to (42)
we get

$\left|\nabla X(\cdot,t)\right|_{\gamma}\leq
G(t)\exp\left(c\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s\right),$

which completes the proof of the proposition. ∎

Having proved the inequalities in Proposition 23 we can see that, in fact, the
$C^{1,\gamma}$ norm of the flow is finite for any time which was our first
goal.

######  Proposition 24.

Let $X(\cdot,t)$ be the solution given in Theorem 18. Then
$\left|X(\cdot,t)\right|_{1,\gamma}$ is finite for any time.

###### Proof.

By Proposition 23 $\left|X(\cdot,t)\right|_{1,\gamma}$ is finite provided
$\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s$ is. Then it suffices to
check that this integral is bounded for any time. Let
$i,j\in\left\\{1,\ldots,n\right\\}$. By Lemma 18 we have

(43)
$\partial_{i}v_{j}(\cdot,t)=c_{ij}\rho(\cdot,t)+p.v.\partial_{i}k_{j}\ast\rho(\cdot,t),$

where $c_{ij}$ are defined in (18). Let
$\varepsilon=\left|\rho(\cdot,t)\right|_{\gamma}^{-1/\gamma}$ and let
$R(t)=m(\text{supp}(\rho(\cdot,t)))^{1/n}$ and apply inequality (13) to the
equation (43) to obtain

(44) $\left|\left|\partial_{i}v_{j}(\cdot,t)\right|\right|_{L^{\infty}}\leq
c\\{1+\ln[R(t)\left|\rho(\cdot,t)\right|_{\gamma}^{1/\gamma}]\left|\left|\rho_{0}\right|\right|_{L^{\infty}}\\}.$

Since $\rho(\cdot,t)=\rho_{0}(X^{-1}(\cdot,t))$ then
$\left|\rho(\cdot,t)\right|_{\gamma}\leq\left|\rho_{0}\right|_{\gamma}\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}^{\gamma}$. Also, taking into account
Lemma 6 we get

$R(t)\left|\rho(\cdot,t)\right|_{\gamma}^{1/\gamma}\leq
c(n)m(\text{supp}(\rho_{0}))^{1/n}\left|\rho_{0}\right|_{\gamma}^{1/\gamma}\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}^{2}.$

By Proposition 23 we can bound $\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}$ and then

$R(t)\left|\rho(\cdot,t)\right|_{\gamma}^{1/\gamma}\leq
c(n)m(\text{supp}(\rho_{0}))^{1/n}\left|\rho_{0}\right|_{\gamma}^{1/\gamma}\exp\left(2\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s\right).$

Therefore inequality (44) can be written as

$\left|\left|\partial_{i}v_{j}(\cdot,t)\right|\right|_{L^{\infty}}\leq
c(n,\rho_{0})+c\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s.$

Taking supremum over $i,j\in\\{1,\ldots,n\\}$

$\left|\left|\nabla v(\cdot,t)\right|\right|_{L^{\infty}}\leq
c(n,\rho_{0})+c\int_{0}^{t}\left|\left|\nabla
v(\cdot,s)\right|\right|_{L^{\infty}}\,\mathrm{d}s.$

and applying Gronwall’s Lemma (22) we finally get

$\left|\left|\nabla v(\cdot,t)\right|\right|_{L^{\infty}}\leq
c(n,\rho_{0})\exp(ct),$

which is finite for any time. ∎

Finally, as we anticipated at the beginning of the section, using the a priori
bound for the flow and by the Continuation Theorem 20 we can prove that the
solution $X(\cdot,t)$ is global in time.

######  Theorem 25.

Let $\rho_{0}\in C^{\gamma}_{c}(\mathbb{R}^{n};\mathbb{R})$. Then the ordinary
differential equation

$\begin{cases}\frac{\mathrm{d}}{\mathrm{d}t}X(\alpha,t)=F(X(\alpha,t)),\\\
X(\alpha,0)=\alpha,\end{cases}$

for

$F(X(\alpha,t))=\int_{\mathbb{R}^{n}}k(X(\alpha,t)-X(\alpha^{\prime},t))\rho_{0}(\alpha^{\prime})\,\det[DX(\alpha^{\prime},t)]\,\mathrm{d}\alpha^{\prime}$

has a unique solution $X(\cdot,t)\in
C^{1,\gamma}(\mathbb{R}^{n};\mathbb{R}^{n})$ for any time $t\in\mathbb{R}$.

###### Proof.

We want to apply Theorem 20 in order to ensure the globalness of the solution
$X(\cdot,t)$. So, we need to check that, for any given $T$, the map
$X(\cdot,T)$ belongs to $O_{M}$ where

$O_{M}=B\cap\left\\{X:\mathbb{R}^{n}\to\mathbb{R}^{n}:\,\frac{1}{M}<\sup_{\alpha\neq\beta}\frac{\left|X(\alpha)-X(\beta)\right|}{\left|\alpha-\beta\right|}<M\right\\}.$

Let us first prove that we can avoid to check that the condition for $M$ is
satisfied at time $T$. By the Mean Value Theorem,

$\left|X(\alpha,t)-X(\beta,t)\right|\leq\left|\left|\nabla
X(\cdot,t)\right|\right|_{L^{\infty}}\left|\alpha-\beta\right|\leq\left|X(\cdot,t)\right|_{1,\gamma}\left|\alpha-\beta\right|,$

and also since we can express $\alpha=X^{-1}((X(\alpha,t),t))$ and
$\beta=X^{-1}((X(\beta,t),t))$,

$\left|\alpha-\beta\right|\leq\left|\left|\nabla
X^{-1}(\cdot,t)\right|\right|_{L^{\infty}}\left|X(\alpha,t)-X(\beta,t)\right|\leq\left|X(\cdot,t)\right|_{1,\gamma}^{2n-1}\left|X(\alpha)-X(\beta)\right|.$

Consider now $M^{\prime}$ such that

$\sup_{t\in[-T,T]}\max\\{\left|X(\cdot,t)\right|_{1,\gamma},\left|X(\cdot,t)\right|_{1,\gamma}^{2n-1}\\}\leq
M^{\prime}<\infty.$

Such an $M^{\prime}$ exists since $\left|X(\cdot,t)\right|_{1,\gamma}$ is
finite for every time by Proposition 24. For this choice of $M^{\prime}$ it is
sure that $X(\cdot,t)\in O_{M^{\prime}}$ for every time $t\in[-T,T]$.

Throughout the proofs of Proposition 9 and 16 we can check that those
statements are independent of $M$. Due to this independence we can modify
$O_{M}$ to $O_{M^{\prime}}$ without changing neither the solution nor the
maximal time of existence given by Picard-Lindelöf’s theorem.

Then as soon as $\left|X(\cdot,t)\right|_{1,\gamma}$ is finite at time $T$,
$X(\cdot,T)\in O_{M^{\prime}}$. But we know that this is true by Proposition
24. So $X(\cdot,T)\in O_{M^{\prime}}$ and this does not depend on the choice
of $T$ so we have existence and uniqueness of $X(\cdot,t)\in C^{1,\gamma}$ for
any time $t$ by Theorem 20. ∎

As a direct consequence of Theorems 19 and 25 we have finally have Theorem 2.

######  Acknowledgements.

The author acknowledge support by 2017-SGR-395 (AGAUR, Generalitat de
Catalunya), and MTM2016-75390 (MINECO, Spanish Government). The author also
acknowledge professors Joan Mateu, Joan Orobitg and Joan Verdera for their
priceless comments concerning the writing of this paper.

## References

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J.C. Cantero,
---
Departament de Matemàtiques, Universitat Autònoma de Barcelona.
E-mail<EMAIL_ADDRESS>


# Biexcitons are bound in CsPbBr3 Perovskite Nanocrystals

Yoonjae Park  Department of Chemistry, University of California, Berkeley,
California 94720, USA  Materials Science Division, Lawrence Berkeley National
Laboratory, Berkeley, California 94720, USA  David T. Limmer
<EMAIL_ADDRESS>Department of Chemistry, University of California,
Berkeley, California 94720, USA  Materials Science Division, Lawrence Berkeley
National Laboratory, Berkeley, California 94720, USA  Chemical Science
Division, Lawrence Berkeley National Laboratory, Berkeley, California 94720,
USA  Kavli Energy NanoScience Institute, Berkeley, California 94720, USA

###### Abstract

We study the energetics of quasi-particle excitations in CsPbBr3 perovskite
nanocrystals using path integral molecular dynamics simulations. Employing
detailed molecular models, we elucidate the interplay of anharmonic lattice
degrees of freedom, dielectric confinement, and electronic correlation on
exciton and biexciton binding energies over a range of nanocrystal sizes. We
find generally good agreement with some experimental observations on binding
energies, and additionally explain the observed size dependent Stokes shift.
The explicit model calculations are compared to simplified approximations to
rationalize the lattice contributions to binding. We find that polaron
formation significantly reduces exciton binding energies, whereas these
effects are negligible for biexciton interactions. While experimentally, the
binding energy of biexcitons is uncertain, based on our study we conclude that
biexcitons are bound in CsPbBr3.

Lead halide perovskite nanocrystals are currently the subject of significant
interest due to their exceptionally high photoluminescence quantum yields,
which make them ideal materials for light emission, lasing and photodetection
1; 2; 3; 4; 5. The electronic properties of the lead halides depend strongly
on the coupling between charges and their surrounding soft, polar lattices.6;
7; 8; 9; 10; 11 This coupling has been implicated in a number of novel
phenomena including photo-induced phase transitions, long radiative
recombination rates, and anomalous temperature dependent mobilities12; 13; 14;
15; 16. In nanocrystals, optical properties are largely determined by the
behavior of exciton complexes 17. Recently, Dana et al. 18 proposed a
potential anti-binding of biexcitons in perovskite nanocrystals, implicating
the potential role of the lattice in mediating this interaction. Here, we
study the energetics of quasiparticle excitations including excitons and
biexcitons with an explicit description of the lattice, over a range of
perovskite nanocrystal sizes. We find that while polaron formation weakens the
exciton binding energy, biexciton energetics are largely unaffected, leading
to an expectation that they are bound in nanocrystals and in bulk.

Excitons and biexcitons are both characterized by a binding energy, the energy
required to dissociate the pair of quasiparticles–free charges or excitons.
While there is reasonable consensus on the bulk exciton binding energy19; 20;
21; 22; 23; 24, its dependence on nanocrystal size is less certain. Further, a
number of biexciton binding energies have been reported experimentally for
lead halide perovskites, but their values span a large range, and even
disagree in sign3; 2; 18; 25; 26; 27; 28; 29; 30; 31; 18; 25. Experimental
measurements are hampered by the nanocrystal polydispersity, spectral drift,
and thermal broadening of spectral lines 32; 33; 34. These ambiguities could
be clarified theoretically, however there are few suitable approaches
available. Unlike traditional semiconductors where structural fluctuations can
be ignored or described within a harmonic approximation, the perovskite
lattice structure with its anharmonic tilting and rocking motions result in
significant renormalization of electronic properties.35; 36; 37 Given that
incorporating the effects from the lattice is important in understanding
excitonic properties, there have been some attempts to include them into a
theoretical description. For example, extensions to GW and Bethe-Salpeter
equations have been developed to include phonon effects on excitonic
properties perturbatively.38; 39; 40; 41; 42; 43; 44 At the same time,
quasiparticle excitations requires a balanced description of electron
correlation, making _ab initio_ models difficult to apply in nanocrystals
where the number of atoms are large. Tight binding and pseudo-potential models
have been developed to study excitonic structure, but these have not yet been
unified with approaches to describe electron-phonon effects.45; 46; 47; 48

Here, we use quasiparticle path integral molecular dynamics49 with an explicit
atomistic description of the lattice, which allows us to include all orders of
anharmonicity from the lattice, and treat electron correlation exactly. This
approach has been previously successful at describing the excitonic properties
of bulk systems,15; 37 and the lattice model we employ has been used
extensively to describe both vibrational and nonlinear properties of a variety
of lead-halide systems.48; 50; 51; 52 In this study, we consider a system of a
single biexciton interacting with CsPbBr3 cubic perovskite nanocrystals. The
model Hamiltonian consists of three pieces,
$\mathcal{H}=\mathcal{H}_{\mathrm{el}}+\mathcal{H}_{\mathrm{lat}}+\mathcal{H}_{\mathrm{int}}$.
Within an effective mass approximation, valid because of the highly dispersive
bands,53 the electronic Hamiltonoan $\mathcal{H}_{\mathrm{el}}$ is defined by
kinetic energies and Coulomb interactions between electrons and holes,

${\mathcal{H}}_{\mathrm{el}}=\sum_{i}\frac{\hat{\mathbf{p}}_{i}^{2}}{2m_{i}}+\sum_{i\neq
j}\frac{q_{i}q_{j}}{4\pi\varepsilon_{\infty}|\hat{\mathbf{x}}_{i}-\hat{\mathbf{x}}_{j}|}$
(1)

where the subscripts $i,j\in\\{e_{1},e_{2},h_{1},h_{2}\\}$ indicate two
electrons and two holes respectively, $\hat{\mathbf{p}}$ and
$\hat{\mathbf{x}}$ are momentum and position operators, $m_{i}$ is the band
mass set to 0.22 $m_{0}$ and 0.24 $m_{0}$ for electrons and holes in terms of
the bare electron mass $m_{0}$, $\varepsilon_{\infty}=4.3$ is the optical
dielectric constant in the unit of vacuum permittivity $\varepsilon_{0}$, and
$q$ is the charge of quantum particle which is $-e$ for electrons and $+e$ for
holes.54 We consider a singlet biexciton, so electrons and holes are
distinguishable particles.55

For the lattice, we model the CsPbBr3 nanocrystals explicitly using a
previously validated _ab initio_ derived forcefield.50; 48 Its Hamiltonian is
given by

${\mathcal{H}}_{\mathrm{lat}}=\sum_{i=1}^{N}\frac{\mathbf{p}_{i}^{2}}{2m_{i}}+U_{\mathrm{lat}}(\mathbf{x}_{\mathrm{lat}})$
(2)

where
$\mathbf{x}_{\mathrm{lat}}=\\{\mathbf{x}_{1},\mathbf{x}_{2},\dots,\mathbf{x}_{N}\\}$
are the positions of the $N$ atoms in the lattice, $\mathbf{p}_{i}$ and
$m_{i}$ are the momentum and mass for $i^{th}$ atom. The atomistic forcefield
50, $U_{\mathrm{lat}}$, is the sum of pairwise interactions with distance
$x_{ij}=|\mathbf{x}_{i}-\mathbf{x}_{j}|$ consisting of a Coulomb potential and
Lennard-Jones potential

$U_{\mathrm{lat}}=\sum_{i,j=1}^{N}\frac{q_{i}q_{j}}{4\pi\varepsilon_{0}x_{ij}}+4\varepsilon_{ij}\left[\left(\frac{\sigma_{ij}}{x_{ij}}\right)^{12}-\left(\frac{\sigma_{ij}}{x_{ij}}\right)^{6}\right]$
(3)

where the parameters $q_{i}$, $\varepsilon_{ij}$ and $\sigma_{ij}$ are
summarized in Table S1. The interaction between the quasiparticles and lattice
is written as the sum of pseudopotentials between each quantum particle $i$
and the lattice particle $j$,

$\mathcal{H}_{\mathrm{int}}=\sum_{i}\sum_{j=1}^{N}\frac{q_{i}q_{j}}{4\pi\varepsilon_{0}\sqrt{r_{\mathrm{cut}}^{2}+|\hat{\mathbf{x}}_{i}-\hat{\mathbf{x}}_{j}|^{2}}}$
(4)

where the cutoff distances $r_{\mathrm{cut}}$ are chosen using the atomic
radii of each atoms 15; 49; 56; 57.

The quasiparticle interactions are screened by the optical dielectric constant
of the nanocrystal, which changes discontinuously at the boundary between the
perovskite and surrounding solution. To account for the dielectric
discontinuity, we use a multipole expansion,58 resulting in an effective
potential for each charge,

$U_{\mathrm{wall}}=\sum_{i,k}\frac{q_{i}^{2}}{8\pi\varepsilon_{0}|\hat{\mathbf{x}}_{i}-\mathbf{x}_{\mathrm{wall},k}|}\cdot\frac{\varepsilon_{\infty}-\varepsilon_{0}}{\varepsilon_{\infty}+\varepsilon_{0}}$
(5)

where $i\in\\{e_{1},e_{2},h_{1},h_{2}\\}$, $k\in\\{\pm x,\pm y,\pm z\\}$, and
$\mathbf{x}_{\mathrm{wall},k}=\pm L/2$ is the position of the wall with $L$ as
the edge length of the nanocrystal. This external potential results in a
dielectric confinement. Additional details on the atomistic force field,
pseudopotentials, and wall potentials are described in the Supporting
information (SI).

For electrons and holes, we use imaginary time path integrals,59; 60 while the
heavy atoms of the lattice are treated classically. The resultant partition
function $\mathcal{Z}$ can be written as

$\mathcal{Z}=\int\mathcal{D}[\mathbf{x}_{eh}(\tau),\,\mathbf{x}_{\mathrm{lat}}]\,e^{-\int_{\tau=0}^{\beta\hbar}\mathcal{H}[\mathbf{x}_{eh}(\tau),\,\mathbf{x}_{\mathrm{lat}}]/\hbar}$
(6)

where
$\mathbf{x}_{eh}(\tau)=\\{\mathbf{x}_{e_{1}}(\tau),\mathbf{x}_{e_{2}}(\tau),\mathbf{x}_{h_{1}}(\tau),\mathbf{x}_{h_{2}}(\tau)\\}$
where $\mathbf{x}_{i}(\tau)$ is the position of quantum particle $i$ at
imaginary time $\tau$, $\hbar$ is the Planck’s constant, and
$\beta^{-1}=k_{\mathrm{B}}T$ with $k_{\mathrm{B}}$ and $T$ as the Boltzmann
constant and temperature. Discretizing the path action renders each quantum
particle isomorphic to a ring polymer, where neighboring timeslices are
connected by harmonic springs61; 55. We use molecular dynamics simulations
with a second order discretization of the path integral to compute expectation
values of this system, for which each quant particle is represented by 1000
timeslices. Representative simulation snapshots of both the biexciton and
exciton are shown in Fig. 1. Throughout we consider nano-crystalline cubes,
where $L$ is the edge length.

Figure 1: Simulation snapshot of biexciton (left) and exciton (right)
interacting with CsPbBr3 perovskite nanocrystal where $L=3.56$ nm represents
the edge length of the nanocrystal.

For comparison, we also consider two approximate models that incorporate
harmonic and static lattice effects. To incorporate effects from harmonic
phonons, we adopt a dynamic screening model described previously15; 37, which
is a model for charges that are coupled linearly with the polarization field
generated by a collection of harmonic modes. In this model, lattice variables
are analytically integrated out, resulting in an imaginary time influence
functional that can be studied straightforwardly numerically. We parameterize
the influence functional with a Frohlich coupling for the electrons and holes
as $\alpha_{e}=2.65$ and $\alpha_{h}=2.76$ and an optical phonon mode with
energy $\hbar\omega=16.8$ m _e_ V. 54 We also compare a static lattice, where
only the electronic Hamiltonian and confinement effects are considered. All
simulations are done in LAMMPS 62 and the details of the discretized
Hamiltonian can be found in the SI.

With an explicit lattice, the exciton binding energy is

$\Delta_{{X}}=\lim_{T\rightarrow 0}\langle E\rangle_{\mathrm{e}}+\langle
E\rangle_{\mathrm{h}}-\langle E\rangle_{\mathrm{ex}}-\langle
U_{\mathrm{lat}}\rangle_{\mathrm{ex}}\,,$

while the biexciton binding energy is

$\Delta_{{XX}}=\lim_{T\rightarrow 0}2\langle E\rangle_{\mathrm{ex}}-\langle
E\rangle_{\mathrm{biex}}-\langle U_{\mathrm{lat}}\rangle_{\mathrm{ex}}\,,$

where the subscripts $\mathrm{biex}$ and $\mathrm{ex}$ refers to a simulation
of a biexciton (two electrons and two holes) and exciton (electron and hole),
while $\mathrm{e}/\mathrm{h}$ indicates a simulation with only electron/hole
interacting with the surrounding lattice. Simulations are performed at 50K,
which is low enough to extract the ground state energy. The binding energies
is computed from the average energy 63,

$\langle
E\rangle=-\frac{\partial}{\partial\beta}\ln\mathcal{Z}[\mathbf{x}_{eh}(\tau),\mathbf{x}_{\mathrm{lat}}(\tau)]$
(7)

where $\mathbf{x}_{eh}=\\{\mathbf{x}_{e_{1}},\mathbf{x}_{h_{1}}\\}$ or
$\mathbf{x}_{eh}=\\{\mathbf{x}_{e_{1}},\mathbf{x}_{e_{2}},\mathbf{x}_{h_{1}},\mathbf{x}_{h_{2}}\\}$
for exciton and biexciton, respectively. The derivative above produces two
terms, an average kinetic energy and an average potential energy, where we use
a virial estimator64 to efficiently estimate the kinetic energy. For the
evaluation of binding energies from the simulations with only quasiparticles,
in the static or dynamic approximation, the same definitions are used without
the lattice relevant terms. Within the static approximation, the binding
energies are determined by only $\mathcal{H}_{e}$ and the confining potential.
In order to efficiently extract the binding energies with dynamic
approximation, we use a thermodynamic perturbative theory approach (see SI).

Figure 2: Energetics of exciton formation. (a) Exciton binding energy under
static ($\Delta_{X,S}$ black circles) dynamic ($\Delta_{X,D}$ blue triangles)
and explicit lattice (red stars) models. Comparison to Weinberg et al48 is
shown in green squares. (b) The difference in the exciton binding energies
from dynamic and static screenings, which is defined as
$\delta_{\mathrm{X}}=\Delta_{\mathrm{X,S}}-\Delta_{\mathrm{X,D}}$ compared to
Stokes shift measurements from Brennan et al65 and Protesescu et. al66. Solid
lines are fits to
$\Delta_{X}=\Delta_{X}^{\circ}(1+\ell_{1}/L+\ell_{2}^{2}/L^{2})$ and
$\delta_{X}=\delta_{X}^{\circ}(1+\ell^{\prime}/L)$.

Figure 2 (a) shows the exciton binding energy $\Delta_{\mathrm{X}}$ computed
from our molecular model, in addition to those approximated from with the
static or dynamics models for the nanocubes considered, ranging from $L=$ 2.4
nm to 6 nm. Given the reduced mass of the exciton, $\mu=0.11$ and optical
dielectric constant, the Bohr radius of the exciton is $R_{X}=2.07$nm. Thus
for the range of nanocrystals considered, we are in a moderate to strong
confinement regime. As a consequence, all binding energy estimates exhibit a
strong $L$ dependence, which we model in Fig. 2(a) as
$\Delta_{X}=\Delta_{X}^{\circ}(1+\ell_{1}/L+\ell_{2}^{2}/L^{2})$ where
$\Delta_{X}^{\circ}$ is the bulk binding energy and $\ell_{1}$ and $\ell_{2}$
are fit parameters.67; 68

Over the full range of nanocrystal sizes considered, the static approximation
yields a binding energy much larger than either the explicit model calculation
or the dynamic approximation. The value from the static approximation with the
largest nanocrystal gives a reasonable agreement with the expectation from a
Wannier-Mott model $\Delta_{X,\textrm{WM}}=R_{H}\mu/\varepsilon_{\infty}^{2}=$
84.5 m _e_ V, where $R_{H}$ is a Rydberg energy. The values at finite $L$
agree well with recent pseudo-potential based calculations of CsPbI3, up to a
shift of the bulk binding energy by 20 m _e_ V to account for the change in
halide.48

The explicit model calculations are well reproduced by the dynamic
approximation. The suppression of the binding energy due to lattice effects
could arise because of two distinct mechanisms. First the lattice could screen
the electron-hole interactions, weakening them. Second the lattice can lower
the self-energies of the free charges, bringing them closer to the exciton
self-energy, effectively reducing the binding energy. In agreement with
previous calculations on bulk MAPbI3,37 we find the latter effect is more
prominent. This effect is outside of first order perturbation, which over
estimates the decrease in the self-energy of the free charges which we find is
-30 m _e_ V for both charges, and ignores the decrease in self-energy of the
polaron-exciton, which is -12 m _e_ V. With the explicit perovskite lattice,
the extrapolated value to the large nanocrystal size limit is in good
agreement with bulk CsPbBr3 exciton binding energy of 40 meV, reflecting
slight anharmonic weakening of the optical phonon19; 20; 21; 22; 23; 24.

Shown in Fig. 2 (b) is the difference between the dynamic and static screening
models, $\delta_{X}=\Delta_{X,S}-\Delta_{X,D}$. This energy is due to polaron
formation. The strong nanocrystal size dependence reflects the increasing
confinement of the charges, which lower their self-energy by increasing their
localization, as the energy of a charge in a dielectric will decrease like
$1/L$, which fits $\delta_{X}$ well. We find the polaron formation energy
agrees well with size-dependent Stokes shift measurements made from the
difference between absorption and emission spectra.69 Results from two
different experiments on CsPbBr3 nanocrystals 66; 70, compare quantitatively
well with our computed $\delta_{X}$, increasing from 10 m _e_ V to over 450 m
_e_ V. Previously, a lattice origin of the Stokes shift had been disregarded
due to the relatively small polaron formation energy in bulk.70 However, we
find the polaron formation energy can increase substantially in small
nanocrystals.

With the excitonic energetics understood, we now turn to the binding of
biexcitons. The biexciton binding energy $\Delta_{XX}$ for the explicit model,
as well as the static and dynamic approximations, are shown in Fig. 3(a). All
three models agree quantitatively, indicating that contrary from expectations
from the exciton calculations, lattice effects are unimportant for biexciton
binding. We rationalize this by noting that the primary contribution from the
lattice for the exciton was the stabilization of the free charges due to the
polaron formation, which does not contribute to the biexciton energy. In Fig.
3 (b) is the electron-hole and electron-electron radial probability
distributions, $p(r)$, from the simulations with explicit lattice where
electron-electron repulsion makes the average distance slightly larger than
the average electron-hole distance. However this increase is small and the
their size is largely determined by confinement rather than electron
correlation. The similar masses for the electron and hole, produce a very weak
dipole in the exciton or quadrapole in the biexcton that is not large enough
to generate a polarization response from the lattice. Further, the similar
results in both binding energies from explicit lattice and dynamic models
imply that the anharmonicity coming from the explicit perovskite lattice does
not play a crucial role in determining the behavior of biexcitons, likely due
to the relatively small amplitude distortion. However, $\Delta_{XX}$ is
relatively large, 1/2-1/4 of $\Delta_{X}$, indicating the importance of
electron correlation.

As with $\Delta_{X}$, we find a strong system size dependence of
$\Delta_{XX}$, increasing strongly with decreasing $L$. We model this
dependence as
$\Delta_{XX}=\Delta_{XX}^{\circ}(1+\tilde{\ell}_{1}/L+\tilde{\ell_{2}}^{2}/L^{2})$,
with $\tilde{\ell}_{1}$ and $\tilde{\ell}_{2}$ employed as fitting parameters.
While experimentally values of the $\Delta_{XX}$ have been reported between
-100 and 100 m _e_ V,3; 2; 18; 25; 26; 27; 28; 29; 30; 31 recent size
dependent measurements on CsPbBr3 nanocubes from two different groups26; 28
are shown in Fig. 3(a) and agree well with our results.

Figure 3: Energetics and structure of biexcitons. (a) Biexciton binding
energies for our explicit model (red stars) and its dynamical (blue squares)
and static (black circles) approximations. Comparisons to Cho. et al 28 and
Amara et. al26 are shown in green squares. The solid line is a fit to
$\Delta_{XX}=\Delta_{XX}^{\circ}(1+\tilde{\ell_{1}}/L+\tilde{\ell_{2}}^{2}/L^{2})$.
(b) Electron-hole (blue) and electron-electron (red) distributions of
biexciton from the simulations with explicit lattice.

In summary, we employed path integral molecular dynamics simulations to study
quasiparticle energetics in perovskite nanocrystals. Using atomic models and
simple approximations, we are able to systematically explore the roles of the
lattice on effects on exciton and biexciton binding energies. We found that
the effects from the lattice are large in exciton binding energy. but
negligible in biexciton interaction, attributed to the weak coupling between
the lattice and the neutral exciton complex. For all nanocrystals, regardless
of the size and the type of screening employed, we found that biexcitons are
bound, strongly suggestive that reports of anti-binding are misinterpreted.

_Acknowledgements._ This work was supported by the U.S. Department of Energy,
Office of Science, Office of Basic Energy Sciences, Materials Sciences and
Engineering Division, under Contract No. DEAC02-05-CH11231 (Grant No. KC3103).
Y.P. acknowledges the Kwanjeong Educational Foundation. D.T.L. acknowledges
the Alfred P. Sloan Foundation.

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