# Fractional Burgers equation
with singular initial condition 22footnotetext: 2020 Mathematics Subject
Classification: 35A01, 35B40; 35K55; 35S10. 22footnotetext: Key words and
phrases: Fractional Burgers equation, Singular initial condition, Existence of
solutions, Estimates of solutions. 22footnotetext: The first author was
partially supported by Wrocław University of Science and Technology grant
8211104160 MPK: 9130730000 and by the grant 2015/18/E/ST1/00239 of National
Science Centre, Poland. The second author was partially supported by the grant
2015/18/E/ST1/00239 of National Science Centre, Poland.

T. Jakubowski111Wrocław University of Science and Technology, ul. Wybrzeże
Wyspiańskiego 27, Wrocław, Poland, email<EMAIL_ADDRESS>G.
Serafin222Wrocław University of Science and Technology, ul. Wybrzeże
Wyspiańskiego 27, Wrocław, Poland, email<EMAIL_ADDRESS>

###### Abstract

We consider the fractional Burgers equation
$\Delta^{\alpha/2}u+b\cdot\nabla(u|u|^{(\alpha-1)/\beta})$ on
$\mathbb{R}^{d}$, $d\geq 2$, with $\alpha\in(1,2)$ and $\beta>1$ and prove the
existence of a solution for a large class of initial conditions, which
contains functions that do not belong to any $L^{p}(\mathbb{R}^{d})$, $1\leq
p\leq\infty$. Next, we apply the general results to the initial condition
$u_{0}(x)=M|x|^{-\beta}$, $1<\beta<d$, and show the existence of a selfsimilar
solution and derive its properties such as smoothness, two-sided estimates,
asymptotics and gradient estimates.

##  1 Introduction

Let $d\geq 2$ and $\alpha\in(1,2)$. Consider the fractional Burgers equation

$\begin{cases}u_{t}=\Delta^{\alpha/2}u+b\cdot\nabla(u|u|^{q}),&\qquad t>0,\\\
u(0,\cdot)=u_{0}\end{cases}$ (1.1)

in $\mathbb{R}^{d}$, where $\beta>1$, $b\in\mathbb{R}^{d}$ and
$q=\frac{\alpha-1}{\beta}$ are fixed. For technical reasons, we assume
(without loss of generality) $b=(|b|,0,0,...,0)$. Here, $\Delta^{\alpha/2}$ is
the fractional Laplacian

$\displaystyle\Delta^{\alpha/2}f(x)=\lim_{\varepsilon\to
0^{+}}c_{d,\alpha}\int_{|y|>\varepsilon}\frac{f(x+y)-f(x)}{|y|^{d+\alpha}}dy,\hskip
25.0ptf\in C_{c}^{\infty}(\mathbb{R}^{d}),$

where $c_{d,\alpha}$ is some constant. It may be also defined by the Fourier
transform

$\displaystyle\widehat{\Delta^{\alpha/2}f}(\xi)=-|\xi|^{\alpha}\hat{f}(\xi).$

We put

$\displaystyle
p(t,x)=\frac{1}{(2\pi)^{d}}\int_{\mathbb{R}^{d}}e^{ix\cdot\xi}e^{-t|\xi|^{\alpha}},\qquad
t>0,\;x\in\mathbb{R}^{d}.$

and denote $p(t,x,y)=p(t,y-x)$, $t>0,\ x,y\in\mathbb{R}^{d}$. The function $p$
is the density of the semigroup $P_{t}=e^{t\Delta^{\alpha/2}}$ generated by
the fractional Laplacian,

$P_{t}f(x)=e^{t\Delta^{\alpha/2}}f(x)=\int_{\mathbb{R}^{d}}p(t,x,y)f(y)dy.$

By a solution to the Cauchy problem (1.1) we mean a mild solution, i.e. a
function $u\colon\mathbb{R}\times\mathbb{R}^{d}\to\mathbb{R}$ satisfying the
Duhamel formula

$\displaystyle
u(t,x)=P_{t}u_{0}(x)+\int_{0}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-s,x,z)u(s,z)|u(s,z)|^{q}\,dz\,ds.$
(1.2)

The fractional Burgers equation was intensely studied in recent years, see
e.g. [MR2659155, MR3789847, MR1637513, MR1849690, MR1881259, MR2407204, KMX,
AIK, KNS, WW, MR3382709]. It is a generalization of the classical Burgers
equation

$\displaystyle u_{t}=\nu u_{xx}-\tfrac{1}{2}(u^{2})_{x},$

which was introduced as a simplest model describing the turbulence phenomena
(see e.g. [Burgers]). The fractional counterpart (1.1) of this equation was
studied for the first time by Biler and Funaki in [MR1637513]. Later on, in
[MR1881259] Biler, Karch and Woyczyński considered some further
generalizations, where the fractional Laplacian was replaced by some Lévy
operator and nonlinearity $f(u)$ was given by a smooth function $f$. In
particular, they proved that for $u_{0}\in L^{1}(\mathbb{R}^{d})\cap
L^{\infty}(\mathbb{R}^{d})$ there exists a unique solution to (1.1) such that
(see [MR1881259, Theorem 3.1])

$\displaystyle u\in C\big{(}(0,\infty);W^{2,2}(\mathbb{R}^{d})\big{)}\cap
C^{1}\big{(}(0,\infty);L^{2}(\mathbb{R}^{d})\big{)},$ (1.3)

and

$\displaystyle\|u(t,\cdot)\|_{\gamma}\leq\|u_{0}\|_{\gamma},\ \ \ \ \ t>0,\ \
\gamma\in[1,\infty].$ (1.4)

We point out that in both papers the standing assumptions was $u_{0}\in
L^{1}(\mathbb{R}^{d})$. In [MR1849690] the authors studied the equation (1.1)
with the critical exponent $q=\frac{(\alpha-1)}{d}$, i.e.

$\displaystyle\begin{cases}u_{t}=\Delta^{\alpha/2}u+b\cdot\nabla
u|u|^{(\alpha-1)/d},&\qquad t>0,\\\ u(0,\cdot)=u_{0}.\end{cases}$ (1.5)

They showed that for $u_{0}=M\delta_{0}$, $M>0$, there is a unique selfsimilar
solution (called source solution) to (1.5) satisfying the scaling property
$U(t,x)=t^{-d/\alpha}U(1,t^{-1/\alpha}x)$ (the same as the density $p$). It
turns out that this self-similar solution determines the long time behavior of
solutions to a large class of Cauchy problems (1.5) with $u_{0}\in
L^{1}(\mathbb{R}^{d})$ and $\|u_{0}\|_{1}=M$ (see [MR1849690, Theorem 2.2])

$\displaystyle\lim_{t\to\infty}t^{\frac{d}{\alpha}(1-\frac{1}{\gamma})}\|u(t,\cdot)-U(t,\cdot))\|_{\gamma}=0,\qquad\gamma\in[1,\infty].$
(1.6)

If $u_{0}\in L^{1}(\mathbb{R}^{d})\cap L^{\infty}(\mathbb{R}^{d})$, the
existence of the solution to the problem (1.5) can by proved by using the
Banach fixed point theorem (see [MR1881259]). The case $u_{0}=M\delta_{0}$ is
in some sense similar to $u_{0}\in L^{1}(\mathbb{R}^{d})$ since such $u_{0}$
is still integrable and the solution may be constructed by using the
approximation by the solutions with initial conditions from
$L^{1}(\mathbb{R}^{d})$. Nevertheless, if $u_{0}$ does not belong to any
$L^{\gamma}(\mathbb{R}^{d})$, $\gamma\in[1,\infty]$, the construction by fixed
point theorems fails.

In this paper we assume that the initial condition
$u_{0}\colon\mathbb{R}^{d}\to\mathbb{R}$ satisfies

$\displaystyle\sup_{t>0,\,x\in\mathbb{R}^{d}}t^{(\beta-1)/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|p^{(d-1)}(t,\tilde{x},\tilde{y})dy\leq\mu,$
(A)

where $\mu\in(0,\infty)$ is some constant. Here, $p^{(d-1)}$ is the density of
$P_{t}$ in dimension $d-1$ and
$\tilde{x}={(x_{2},\ldots,x_{d})}\in\mathbb{R}^{d-1}$ (see (2.2) and Notation
in Section 2). The main feature of the class of functions satisfying (A) is
that it contains some functions not belonging to any
$L^{\gamma}(\mathbb{R}^{d})$, $\gamma\in[1,\infty]$. This allows us to drop
the common assumption $u_{0}\in L_{1}(\mathbb{R}^{d})$, or even $u_{0}\in
L_{1}(\mathbb{R}^{d})\cap L^{\infty}(\mathbb{R}^{d})$, and consider some
singular initial conditions. Our first result is

######  Theorem 1.1.

Let $\alpha\in(1,2)$, $\beta>1$ and $q=(\alpha-1)/\beta$. If $u_{0}$ satisfies
(A), then there exists a solution $u(t,x)$ to the problem (1.1). Furthermore,
there is a constant $C>0$ depending only on $d,\alpha,\beta,b,\mu$ such that

$\displaystyle|u(t,x)|\leq
C\,t^{-1/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|p^{(d-1)}(t,\tilde{x},\tilde{y})dy\leq
C\mu t^{-\beta/\alpha},\qquad t>0,\;x\in\mathbb{R}^{d}.$ (1.7)

In the proof of Theorem 1.1 we proceed similarly as in the case of source
solutions and approximate the initial condition by the monotone sequence of
functions from $L^{1}(\mathbb{R}^{d})\cap L^{\infty}(\mathbb{R}^{d})$.
However, the convergence of the corresponding sequence of solutions is a major
problem here. We resolve it by showing that a certain functional acting on
these solutions is uniformly bounded, which ensures the point-wise convergence
of the sequence almost everywhere (see Lemma 3.3). This approach, however,
requires the technical assumptions $\beta>1$ and $d\geq 2$.

The first bound in Theorem 1.1 has one weakness - it does not depend on
$x_{1}$, and therefore it is not optimal. Nevertheless, in many cases this
inconvenience could be removed. We present it on an example, which turns out
to be quite special. Namely, in the second part of the paper we focus on the
solution to the problem (1.1) with $u_{0}(x)=M|x|^{-\beta}$, $M>0$,
$1<\beta<d$, constructed in Theorem 1.1. Its first interesting property is
selfsimilarity, in which one can see an analogy to the source solution in the
case $\beta=d$ and $u_{0}=M\delta_{0}$. Next, we turn our attention to the
two-sided pointwise estimates, which not only improve (1.7) in this case, but
complement it with the lower bound of the same form. Such estimates were
obtained for the source solution of (1.5). In [MR2407204] Karch and Brandolese
showed that for $q=\frac{\alpha-1}{d}$ and $u_{0}=M\delta_{0}$ with $M>0$
sufficiently small, the solution $u$ to (1.5) admits the estimates
$0\leq{u(t,x)}\leq cp(t,x)$. In [JS1] we generalized this result to any $M>0$
and obtained two-sided estimates $c^{-1}p(t,x)\leq{u(t,x)}\leq cp(t,x)$.
Furthermore, in [JS2, Theorem 1.1] we showed that for nonnegative $u_{0}\in
L^{1}(\mathbb{R}^{d})\cap L^{\infty}(\mathbb{R}^{d})$ there is a constant
$C>0$ such that the solution to (1.1) with $q\geq\frac{\alpha-1}{d}$ satisfies

$\displaystyle C^{-1}\,P_{t}u_{0}(x)\leq u(t,x)\leq CP_{t}u_{0}(x),\ \ \ \ \ \
t>0,\ x\in\mathbb{R}^{d}.$ (1.8)

Note that the pointwise estimates of solutions are rather rare in the
literature, as deriving them is more challenging than e.g. the $L^{p}$ bounds.
Nevertheless, they give a better insight into the behaviour of the solution.
For instance the estimates (1.8) permitted the authors to improve the
asymptotics given in (1.6).

In Theorem 1.2 below we prove the existence of the selfsimilar solution to the
problem (1.1) for $u_{0}(x)=M|x|^{-\beta}$ with $M>0$, $1<\beta<d$ and provide
its properties. Note that the assumption $\beta<d$ is required in order to get
local integrability of $u_{0}$.

######  Theorem 1.2.

Let $\alpha\in(1,2)$, $1<\beta<d$, $q=(\alpha-1)/\beta$, $M>0$ and
$u_{0}(x)=M|x|^{-\beta}$. Then, there exists a function $U(x)$ such that

  1. 1.

$u(t,x)=t^{-\beta/\alpha}U(t^{-1/\alpha}x)$ is the solution to the problem
(1.1).

  2. 2.

There exists a constant $C_{1}>1$ such that

$\displaystyle C_{1}^{-1}\frac{1}{(1+|x|)^{\beta}}\leq U(x)\leq
C_{1}\frac{1}{(1+|x|)^{\beta}},\qquad x\in\mathbb{R}^{d}.$ (1.9)

  3. 3.

There is a constant $C_{2}$ such that for all $x\in\mathbb{R}^{d}$ we have

$\displaystyle\left|U(x)-P_{1}u_{0}(x)\right|\leq
C_{2}\frac{1}{(1+|x|)^{\beta+(\alpha-1)}},$ (1.10)
$\displaystyle\left|U(x)-P_{1}u_{0}(x)-\alpha\int_{0}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}\nabla_{x}p(1-r^{\alpha},x,rw)[P_{1}u_{0}(w)]^{1+q}dwdr\right|$
(1.11)
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\leq\frac{C_{2}}{(1+|x|)^{\beta+2(\alpha-1)}}.$

  4. 4.

$U\in\mathcal{C}^{1}\left(\mathbb{R}^{d}\right)$ and there is a constant
$C_{3}$ such that

$\displaystyle|\nabla U(x)|\leq C_{3}\frac{1}{(1+|x|)^{\beta}},\qquad
x\in\mathbb{R}^{d}.$ (1.12)

All the constants $C_{1},C_{2},C_{3}$ depend only on $d,\alpha,\beta,M,b$ and
might be calculated explicitly.

######  Remark 1.3.

By Lemma 2.3, the bounds in (1.9) may be equivalently written in the form

$\displaystyle u(t,x)\approx\frac{1}{(t^{1/\alpha}+|x|)^{\beta}}\approx
P_{t}u_{0}(x),\qquad t>0,\;x\in\mathbb{R}^{d},$

which resembles both: (1.8) by representation involving the semi-group, and
the estimates of the source solution in the critical case derived in [JS2] by
the exact form. Similarly, since
$\nabla_{x}u(t,x)=\lambda^{\beta+1}\nabla_{x}(\lambda^{\alpha}t,\lambda x)$,
by (1.12) we have

$|\nabla_{x}u(t,x)|\lesssim\frac{1}{t^{1/\alpha}(t^{1/\alpha}+|x|)^{\beta}},\qquad
t>0,\;x\in\mathbb{R}^{d}.$

The paper is organized as follows. In Section 2 we introduce the notation used
in the paper and give some preliminary results on the density $p(t,x,y)$ and
solutions to (1.1). Section 3 is devoted to some general estimates and proof
of Theorem 1.1. In Section 4 we prove Theorem 1.2.

##  2 Preliminaries

###  2.1 Notation

We will use the following notation. We write ”:=” to indicate definitions,
e.g. $a\wedge b:=\min\\{a,b\\}$ and $a\vee b:=\max\\{a,b\\}.$ All the
considered functions are tacitly assumed to be Borel measurable. We write
$f(x)\lesssim g(x)$ if $f,g\geq 0$ and there is a number $c>0$ not depending
on other parameters than $d$, $\alpha$, $\beta$, $M$ and $b$ such that
$f(x)\leq cg(x)$ for all arguments $x$. If $f\lesssim g$ and $g\lesssim f$, we
write $f(x)\approx g(x)$. If the comparability constant depends additionally
on some other parameters (e.g. on $\gamma$), we indicate the parameter over
the comparison sign (e.g. $f\stackrel{{\scriptstyle\gamma}}{{\approx}}g$).

By $\nabla=(\partial_{x_{1}},\ldots,\partial_{x_{d}})$ we denote the standard
gradient operator, while
$\nabla^{\mathbf{k}}=\partial^{k_{1}}_{x_{1}}\ldots\partial^{k_{d}}_{x_{d}}$,
where ${\mathbf{k}}=(k_{1},\ldots,k_{d})\in\mathbb{N}_{0}^{d}$. The norm of
the multi-index ${\mathbf{k}}$ is given by
$|{\mathbf{k}}|=k_{1}+\ldots+k_{d}$. By writing $\nabla_{x}$ and
$\nabla^{\mathbf{k}}_{x}$ we indicate variable with respect to which the
derivatives are taken.

For any $x=(x_{1},x_{2},\ldots,x_{d})\in\mathbb{R}^{d}$ we denote
$\tilde{x}:=(x_{2},x_{3},...,x_{d})\in\mathbb{R}^{d-1}$. For any function
$f\colon(0,\infty)\times\mathbb{R}^{d}\rightarrow\mathbb{R}$, we define its
rescaled version by

$\displaystyle f^{*}(t,x):=t^{\beta/\alpha}f(t,t^{1/\alpha}x).$ (2.1)

Similarly, we put

$\displaystyle
P_{t}^{*}f(x):=t^{\beta/\alpha}P_{t}f(t^{1/\alpha}x)=t^{\beta/\alpha}\int_{\mathbb{R}^{d}}p(t,t^{1/\alpha}x,y)f(y)dy.$

For any $a\in\mathbb{R}$ and $\gamma>1$ we define the french power
$a^{\left\langle\gamma\right\rangle}=a|a|^{\gamma-1}$. In particular,
$|u(s,z)|u(s,z)^{q}=u(s,z)^{\left\langle q+1\right\rangle}$.

###  2.2 Fractional Laplacian semigroup

Although in the paper we generally assume $\alpha\in(1,2)$ and $d\geq 2$, the
results of this section are valid for $\alpha\in(0,2)$ and $d\geq 1$. The
methodology we propose in the next sections requires consideration of the
density of the semigroup $P_{t}$ in various dimensions at the same time.
Therefore, for $m\in\\{1,2,3,\ldots\\}$ we put

$\displaystyle
p^{(m)}(t,x)=\frac{1}{(2\pi)^{m}}\int_{\mathbb{R}^{m}}e^{ix\cdot\xi}e^{-t|\xi|^{\alpha}},\qquad
t>0,\;x\in\mathbb{R}^{m},$ (2.2)

and denote $p^{(m)}(t,x,y)=p^{(m)}(t,y-x)$, $t>0,\,x,y\in\mathbb{R}^{m}$. In
the case $m=d$ we simply write $p^{(d)}(t,x,y)=p(t,x,y)=p(t,y-x)$. Below, we
collect some basic properties of the density $p$. The following scaling
property holds

$\displaystyle p(\lambda^{\alpha}t,\lambda x,\lambda
y)=\lambda^{-d}p(t,x,y),\qquad\lambda>0.$ (2.3)

Furthermore, the function $p^{(m)}$ admits the two-sided estimates

$\displaystyle
p^{{(m)}}(t,x,y)\approx\frac{t}{\left(t^{1/\alpha}+|x-y|\right)^{{m}+\alpha}},\qquad
t>0,\,x,y\in\mathbb{R}^{m},$ (2.4)

where the constants in the bounds depend on $m$ and $\alpha$. In particular,
this implies

$\displaystyle\|p(t,\cdot)\|_{\gamma}$
$\displaystyle=t^{-\frac{d}{\alpha}(1-1/\gamma)}\|p(1,\cdot)\|_{\gamma},\qquad
t>0,$ (2.5) $\displaystyle p(t,x,y)$ $\displaystyle\lesssim
t^{-1/\alpha}p^{(d-1)}(t,\tilde{x},\tilde{y}),\hskip
29.0ptt>0,\,x,y,\in\mathbb{R}^{d}.$ (2.6)

######  Lemma 2.1.

For $t>0$, $\tilde{x}\in\mathbb{R}^{d-1}$, $y\in\mathbb{R}^{d}$, we have

$\displaystyle\int_{\mathbb{R}}p(t,x,y)dx_{1}=p^{(d-1)}(t,\tilde{x},\tilde{y}).$
(2.7)

###### Proof.

By the subordination formula (see [MK]) there is a nonnegative function
$\eta_{t}(s)$ such that

$\displaystyle p^{(m)}(t,x,y)=\int_{0}^{\infty}(4\pi
s)^{-m/2}e^{-|x-y|^{2}/4s}\eta_{t}(s)ds,\qquad
m\in\mathbb{N},t>0,\,x,y\in\mathbb{R}^{m}.$ (2.8)

Hence,

$\displaystyle\int_{\mathbb{R}}p(t,x,y)dx_{1}$
$\displaystyle=\int_{0}^{\infty}\int_{\mathbb{R}}(4\pi
s)^{-d/2}e^{-|x-y|^{2}/4s}\eta_{t}(s)dx_{1}ds$
$\displaystyle=\int_{0}^{\infty}(4\pi
s)^{-(d-1)/2}e^{-|\tilde{x}-\tilde{y}|^{2}/4s}\eta_{t}(s)ds=p^{(d-1)}(t,\tilde{x},\tilde{y}).$

∎

The subordination formula (2.8) allows us also to derive the following
estimates of $\nabla^{\mathbf{k}}p$.

######  Lemma 2.2.

For any ${\mathbf{k}}\in\mathbb{N}_{0}^{d}$, $t>0$ and $x\in\mathbb{R}^{d}$,
we have

$\displaystyle|\nabla^{{\mathbf{k}}}_{x}p(t,x)|\stackrel{{\scriptstyle{\mathbf{k}}}}{{\lesssim}}\frac{p(t,x)}{(t^{1/\alpha}+|x|)^{|{\mathbf{k}}|}}.$
(2.9)

###### Proof.

Let ${\mathbf{k}}=(k_{1},\ldots,k_{d})\in\mathbb{N}_{0}^{d}$ and $g(s,x)=(4\pi
s)^{-d/2}e^{-|x|^{2}/(4s)}$ be the Gaussian kernel. Then,

$\displaystyle\nabla^{{\mathbf{k}}}g(s,x)=(4s)^{-|{\mathbf{k}}|/2}g(s,x)\prod_{i=1}^{d}H_{k_{i}}(x_{i}/2\sqrt{s}),$

where $H_{n}$ are the Hermite polynomials. By subordination formula (2.8) and
the Fubini theorem, we have

$\displaystyle\nabla^{{\mathbf{k}}}p(t,x)$
$\displaystyle=\int_{0}^{\infty}(4s)^{-|{\mathbf{k}}|/2}g(s,x)\prod_{i=1}^{d}H_{k_{i}}(x_{i}/2\sqrt{s})\eta_{t}(s)\,ds.$

Since every $H_{k_{i}}$ is a polynomial of degree $k_{i}$, we bound

$\left|\prod_{i=1}^{d}H_{k_{i}}(x_{i}/2\sqrt{s})\right|\stackrel{{\scriptstyle{\mathbf{k}}}}{{\lesssim}}\prod_{i=1}^{d}[1+|x_{i}/2\sqrt{s}|^{k_{1}}]\lesssim
1+|x/2\sqrt{s}|^{|{\mathbf{k}}|}.$

Consequently

$\displaystyle|\nabla^{{\mathbf{k}}}p(t,x)|$
$\displaystyle\stackrel{{\scriptstyle{\mathbf{k}}}}{{\lesssim}}\int_{0}^{\infty}(4s)^{-|{\mathbf{k}}|/2}g(s,x){\eta_{t}(s)}\,ds+|x|^{|{\mathbf{k}}|}\int_{0}^{\infty}(4s)^{-|{\mathbf{k}}|}g(s,x){\eta_{t}(s)}\,ds$
$\displaystyle=p^{(d+|{\mathbf{k}}|)}(t,x_{+})+|x|^{|{\mathbf{k}}|}p^{(d+2|{\mathbf{k}}|)}(t,x_{++}),$

where
${x_{+}}=(x_{1},\ldots,x_{d},0,\ldots,0)\in\mathbb{R}^{d+|{\mathbf{k}}|}$ and
${x_{++}}=(x_{1},\ldots,x_{d},0,\ldots,0)\in\mathbb{R}^{d+2|{\mathbf{k}}|}$.
Eventually, by (2.4), we obtain

$\displaystyle|\nabla^{{\mathbf{k}}}p(t,x)|\stackrel{{\scriptstyle{\mathbf{k}}}}{{\lesssim}}\frac{t}{(t^{1/\alpha}+|x|)^{d+\alpha+|{\mathbf{k}}|}}+\frac{|x|^{|{\mathbf{k}}|}t}{(t^{1/\alpha}+|x|)^{d+\alpha+2|{\mathbf{k}}|}}\approx\frac{1}{(t^{1/\alpha}+|x|)^{|{\mathbf{k}}|}}p(t,x),$

as required. ∎

The next lemma describes the behaviour of $P_{t}$ acting on the function
$(r^{1/\alpha}+|y|)^{-\gamma}$, which, as we will show, is comparable with a
solution to (1.1) for $u_{0}(x)=M|x|^{-\gamma}$.

######  Lemma 2.3.

Let $0<\gamma<d$. For $x\in\mathbb{R}^{d}$, $t>0$ and $r\geq 0$ we have

$\displaystyle\int_{\mathbb{R}^{d}}p(t,x,y)(r^{1/\alpha}+|y|)^{-\gamma}dy\stackrel{{\scriptstyle\gamma}}{{\approx}}\frac{1}{(r^{1/\alpha}+t^{1/\alpha}+|x|)^{\gamma}}.$
(2.10)

###### Proof.

By [MR3933622, Lemma 2.3],

$\displaystyle\int_{\mathbb{R}^{d}}p(t,x,y)|y|^{-\gamma}dy\stackrel{{\scriptstyle\gamma}}{{\approx}}\frac{1}{(t^{1/\alpha}+|x|)^{\gamma}},$
(2.11)

which proves (2.10) for $r=0$. Now, let $r>0$. By (2.11), we get

$\displaystyle\int_{\mathbb{R}^{d}}p(t,x,y)(r^{1/\alpha}+|y|)^{-\gamma}dy$
$\displaystyle\approx\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}p(t,x,y)p(r,y,z)|z|^{-\gamma}dzdy$
$\displaystyle=\int_{\mathbb{R}^{d}}p(t+r,x,z)|z|^{-\gamma}dz\approx\frac{1}{((r+t)^{1/\alpha}+|x|)^{\gamma}}.$

Since $(r+t)^{1/\alpha}\approx r^{1/\alpha}+t^{1/\alpha}$, we obtain (2.10). ∎

###  2.3 Basic properties of solutions to (1.1)

Let us observe that if $u$ and $v$ are the solutions to 1.1 with initials
conditions $u_{0}$ and $-u_{0}$, respectively, then $u(t,x)=-v(t,x)$ for all
$t>0$ and $x\in\mathbb{R}^{d}$.

As mentioned in Introduction, if $u_{0}\in L^{1}(\mathbb{R}^{d})\cap
L^{\infty}(\mathbb{R}^{d})$ then there exists a unique solution $u(t,x)$ to
the problem (1.1). Furthermore, [MR1881259, Corollary 3.1] implies the
monotonicity property of solutions. Namely, if $u$ and $v$ are the solutions
of (1.1) with initials conditions $u_{0},v_{0}\in L^{1}(\mathbb{R}^{d})\cap
L^{\infty}(\mathbb{R}^{d})$ and $u_{0}\geq v_{0}$, then

$\displaystyle u(t,x)\geq v(t,x),\qquad t>0,x\in\mathbb{R}^{d}.$ (2.12)

Let us now denote by $\overline{u}$ and $\underline{u}$ the solutions to 1.1
with initials conditions $|u_{0}|$ and $-|u_{0}|$, respectively. Since
$\overline{u}=-\underline{u}$, we get by (2.12)

$\displaystyle|u|\leq|\overline{u}|\vee|\underline{u}|=\overline{u}.$ (2.13)

For technical reasons we will often consider the rescaled version of
solutions. In order to represent $u^{*}$ (cf. (2.1)) in a recursive manner as
in (1.2), we rewrite the integral therein as follows.

$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{z}p(t-s,t^{1/\alpha}x,z)u(s,z)^{\left\langle
q+1\right\rangle}\,dz\,ds$
$\displaystyle=\int_{0}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{z}p(t-s,t^{1/\alpha}x,s^{1/\alpha}w)s^{d/\alpha}u(s,s^{1/\alpha}w)^{\left\langle
q+1\right\rangle}\,dw\,ds$
$\displaystyle=\int_{0}^{1}\int_{\mathbb{R}^{d}}b\cdot\nabla_{w}p(t-st,t^{1/\alpha}x,(st)^{1/\alpha}w)t(st)^{d/\alpha}u(st,(st)^{1/\alpha}w)^{\left\langle
q+1\right\rangle}\,dw\,ds$
$\displaystyle=\int_{0}^{1}\int_{\mathbb{R}^{d}}b\cdot\nabla_{w}p(1-s,x,s^{1/\alpha}w)t^{1-\frac{1}{\alpha}}s^{d/\alpha}u(st,(st)^{1/\alpha}w)^{\left\langle
q+1\right\rangle}\,dw\,ds$
$\displaystyle=\alpha\int_{0}^{1}\int_{\mathbb{R}^{d}}t^{1-\frac{1}{\alpha}}r^{d+\alpha-1}b\cdot\nabla_{w}p(1-r^{\alpha},x,rw)u(r^{\alpha}t,rt^{1/\alpha}w)^{\left\langle
q+1\right\rangle}\,dw\,dr$
$\displaystyle=\alpha\int_{0}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}b\cdot\nabla_{w}p(1-r^{\alpha},x,rw)u^{*}(r^{\alpha}t,w)^{\left\langle
q+1\right\rangle}\,dw\,dr.$

Thus, we have

$\displaystyle
u^{*}(t,x)=P_{t}^{*}u_{0}(x)+\alpha\int_{0}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}b\cdot\nabla_{x}p(1-r^{\alpha},x,rw)u^{*}(r^{\alpha}t,w)^{\left\langle
q+1\right\rangle}\,dw\,dr.$ (2.14)

##  3 General results

###  3.1 An integral conservation law

Up till the end of the paper we assume that $\alpha\in(1,2)$ and $d\geq 2$.
Let $F:\mathbb{R}^{d-1}\rightarrow\mathbb{R}$ be any function. For a function
$h:\mathbb{R}^{d}\rightarrow\mathbb{R}$ we denote

$h_{F}:=\int_{\mathbb{R}^{d}}h(x)F(\tilde{x})dx,$

whenever the integral is convergent.

######  Lemma 3.1.

Let $F:\mathbb{R}^{d-1}\rightarrow\mathbb{R}$. Then, for any $u$ satisfying
(1.2) with $|(P_{t}u_{0})_{F}|<\infty$ for every $t>0$ and such that

$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|\nabla_{x}p(t-s,x,z)||u(s,z)|^{q+1}|F(\tilde{x})|\,dx\,dz\,ds<\infty,$
(3.1)

we have

$\displaystyle u(t,\cdot)_{F}=(P_{t}u_{0})_{F},\qquad t>0.$

###### Proof.

Since $b=(|b|,0,\ldots,0)$, by (1.2) and Fubini theorem, we have

$\displaystyle u(t,\cdot)_{F}$
$\displaystyle=\int_{\mathbb{R}^{d}}\left(P_{t}u_{0}(x)+\int_{0}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-s,x,z)u(s,z)^{\left\langle
q+1\right\rangle}\,dz\,ds\right)F(\tilde{x})dx$
$\displaystyle=(P_{t}u_{0})_{F}+\int_{0}^{t}\int_{\mathbb{R}^{d}}\left(\int_{\mathbb{R}^{d}}|b|\partial_{x_{1}}p(t-s,x,z)F(\tilde{x})dx\right)u(s,z)^{\left\langle
q+1\right\rangle}\,dz\,ds$ $\displaystyle=(P_{t}u_{0})_{F},$

where we used (2.4) and the fact that $F(\tilde{x})$ does not depend on
$x_{1}$. ∎

In particular, we show that the assumptions of Lemma 3.1 are satisfied if
$u_{0}\in L^{1}(\mathbb{R}^{d})\cap L^{\infty}(\mathbb{R}^{d})$ and $F$ is
bounded.

######  Corollary 3.2.

Suppose $u$ is a solution to (1.1) with $u_{0}\in L^{1}(\mathbb{R}^{d})\cap
L^{\infty}(\mathbb{R}^{d})$. Let $F\colon\mathbb{R}^{d-1}\to\mathbb{R}$ be
bounded. Then,

$\displaystyle\int_{\mathbb{R}^{d}}F(\tilde{w})u(t,w)\,dw=\int_{\mathbb{R}^{d}}F(\tilde{w})P_{t}u_{0}(w)\,dw,\qquad
t>0.$

###### Proof.

Let $K=\sup_{\tilde{w}\in\mathbb{R}^{d-1}}|F(\tilde{w})|$. Since
$|P_{t}u_{0}|_{F}\leq K\|u_{0}\|_{1}<\infty$, by Lemma 3.1, we only need to
show that the condition (3.1) holds. By (1.4), (2.9) and (2.7) we have

$\displaystyle\int_{0}^{t}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}|\nabla
p(t-s,w,z)||u(s,z)|^{1+q}|F(\tilde{w})|dw\,dz\,ds$ $\displaystyle\lesssim
K\int_{0}^{t}\|u_{0}\|^{q}_{\infty}\int_{\mathbb{R}^{d}}\int_{\mathbb{R}^{d}}\frac{p(t-s,w,z)}{(t-s)^{1/\alpha}}|u(s,z)|dw\,dz\,ds$
$\displaystyle=K\|u_{0}\|^{q}_{\infty}\int_{0}^{t}\frac{1}{(t-s)^{1/\alpha}}\int_{\mathbb{R}^{d}}|u(s,z)|dz\,ds=K\|u_{0}\|^{q}_{\infty}\|u_{0}\|_{1}\frac{\alpha
t^{1-1/\alpha}}{\alpha-1}<\infty,$

which ends the proof. ∎

If in Corollary 3.2 we take
$F(\tilde{w})=p^{(d-1)}(t_{0},\tilde{x},\tilde{w})$ with $t_{0}>0$ and
$x\in\mathbb{R}^{d}$, by (2.4) we get

$\displaystyle\int_{\mathbb{R}^{d}}p^{(d-1)}(t_{0},\tilde{x},\tilde{w})u(t,w)\,dw=\int_{\mathbb{R}^{d}}p^{(d-1)}(t_{0},\tilde{x},\tilde{w})P_{t}u_{0}(w)\,dw,\qquad
t>0.$ (3.2)

The following lemma will be frequently exploit in the sequel. It also
partially explains the introduction of the assumption (A).

######  Lemma 3.3.

Suppose $u$ is a solution to (1.1) with $u_{0}\in L^{1}(\mathbb{R}^{d})\cap
L^{\infty}(\mathbb{R}^{d})$. Then, for $x\in\mathbb{R}^{d}$, $t>0$ and
$r\in(0,1)$ we have

$\displaystyle\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)\left|u^{*}(r^{\alpha}t,w)\right|\,dw\lesssim\frac{r^{\beta-d}}{(1-r)^{1/\alpha}}\,t^{(\beta-1)/\alpha}\int_{\mathbb{R}^{d}}p^{(d-1)}(t,t^{1/\alpha}\tilde{x},\tilde{y})|u_{0}(y)|dy.$
(3.3)

###### Proof.

In view of (2.13), it is enough to consider $u_{0}\geq 0$. In that case,
$u\geq 0$ and by (2.6), (2.3) and (3.2) we get

$\displaystyle\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)u^{*}(r^{\alpha}t,w)\,dw$
$\displaystyle\lesssim(1-r^{\alpha})^{-1/\alpha}\int_{\mathbb{R}^{d}}p^{(d-1)}(1-r^{\alpha},\tilde{x},r\tilde{w})\left({r^{\alpha}t}\right)^{\beta/\alpha}u(r^{\alpha}t,rt^{1/\alpha}w)\,dw$
$\displaystyle=(1-r^{\alpha})^{-1/\alpha}(r^{\alpha}t)^{(\beta-d)/\alpha}t^{d/\alpha}\int_{\mathbb{R}^{d}}p^{(d-1)}(t-tr^{\alpha},t^{1/\alpha}\tilde{x},\tilde{w})u(r^{\alpha}t,w)\,dw.$
$\displaystyle\lesssim(1-r)^{-1/\alpha}\int_{\mathbb{R}^{d}}p^{(d-1)}(1-r^{\alpha},\tilde{x},r\tilde{w})\left({r^{\alpha}t}\right)^{\beta/\alpha}P_{r^{\alpha}t}u_{0}(rt^{1/\alpha}w)\,dw$
$\displaystyle=(1-r)^{-1/\alpha}\int_{\mathbb{R}^{d}}p^{(d-1)}(1-r^{\alpha},\tilde{x},r\tilde{w})\int_{\mathbb{R}^{d}}\left({r^{\alpha}t}\right)^{\beta/\alpha}p(r^{\alpha}t,rt^{1/\alpha}w,y)u_{0}(y)dy\,dw$
$\displaystyle=(1-r)^{-1/\alpha}\int_{\mathbb{R}^{d-1}}p^{(d-1)}(1-r^{\alpha},\tilde{x},r\tilde{w})\int_{\mathbb{R}^{d}}\left({r^{\alpha}t}\right)^{(\beta-1)/\alpha}p^{(d-1)}(r^{\alpha}t,rt^{1/\alpha}\tilde{w},\tilde{y})u_{0}(y)dy\,d\tilde{w},$

where, to obtain the last equality, we used (2.7) and performed the
integration with respect to $w_{1}$. Finally, using the Tonelli theorem,
(2.3), and the semi-group properties of $p$, we arrive at

$\displaystyle\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)u^{*}(r^{\alpha}t,w)\,dw$
$\displaystyle\lesssim\frac{r^{\beta-2d+1}t^{(\beta-d)/\alpha}}{(1-r)^{1/\alpha}}\int_{\mathbb{R}^{d}}u_{0}(y)\int_{\mathbb{R}^{d-1}}p^{(d-1)}(r^{-\alpha}-1,r^{-1}\tilde{x},\tilde{w})p^{(d-1)}(1,\tilde{w},(rt^{1/\alpha})^{-1}\tilde{y})\,d\tilde{w}\,dy$
$\displaystyle=\frac{r^{\beta-2d+1}t^{(\beta-d)/\alpha}}{(1-r)^{1/\alpha}}\int_{\mathbb{R}^{d}}u_{0}(y)p^{(d-1)}(r^{-\alpha},r^{-1}\tilde{x},(rt^{1/\alpha})^{-1}\tilde{y})\,dy$
$\displaystyle=\frac{{r^{\beta-d}}t^{(\beta-1)/\alpha}}{(1-r)^{1/\alpha}}\int_{\mathbb{R}^{d}}u_{0}(y)p^{(d-1)}(t,t^{1/\alpha}\tilde{x},\tilde{y})\,dy,$

which ends the proof. ∎

###  3.2 Proof of Theorem 1.1

######  Theorem 3.4.

Suppose $u$ is a solution to the problem (1.1). If the assumption (A) is
satisfied and

$\sup_{t>0,\,x\in\mathbb{R}^{d}}|u^{*}(t,x)|<\infty,$

then there exists a constant $C>0$ depending only on $\alpha,d$ such that

$\sup_{t>0,\,x\in\mathbb{R}^{d}}|u^{*}(t,x)|\leq
1\vee\big{(}C(\mu+1)(|b|+1)\big{)}^{\frac{\beta}{(\alpha-1)(\beta-1)}}.$

###### Proof.

Denote $\kappa=\sup\limits_{t>0,\,x\in\mathbb{R}^{d}}|u^{*}(t,x)|$ and assume
$\kappa>1$. By (2.14), we have

$\displaystyle|u^{*}(t,x)|\leq\|P^{*}_{t}|u_{0}|\|_{\infty}$
$\displaystyle+C|b|\left(\int_{0}^{1-1/2\kappa^{\alpha}}+\int_{1-1/2\kappa^{\alpha}}^{1}\right)\int_{\mathbb{R}^{d}}\frac{r^{d-\beta}}{(1-r^{\alpha})^{1/\alpha}}p(1-r,x,rw)|u^{*}(r^{\alpha}t,w)|^{(\alpha-1+\beta)/\beta}\,dw\,dr$
$\displaystyle=:\|P^{*}_{t}|u_{0}|\|_{\infty}+C|b|\left(I_{1}+I_{2}\right),$

for some $C=C(\alpha,d)$. First, by the inequality (2.6) and the assumption
(A) we have

$\displaystyle
P^{*}_{t}|u_{0}|(x)=t^{\beta/\alpha}\int_{\mathbb{R}^{d}}p(t,t^{1/\alpha}x,y)|u_{0}(y)|dy\lesssim
t^{(\beta-1)/\alpha}\int_{\mathbb{R}^{d}}p^{(d-1)}(t,t^{1/\alpha}\tilde{x},\tilde{y})|u_{0}(y)|dy\leq\mu.$

Let us pass to estimating the integral $I_{2}$:

$\displaystyle I_{2}$
$\displaystyle\leq\kappa^{(\alpha-1+\beta)/\beta}\int_{1-1/2\kappa^{\alpha}}^{1}r^{d-\beta}(1-r)^{-1/\alpha}\int_{\mathbb{R}^{d}}p(1-r,x,rw)\,dw\,dr$
$\displaystyle=\kappa^{(\alpha-1+\beta)/\beta}\int_{1-1/2\kappa^{\alpha}}^{1}r^{-\beta}(1-r)^{-1/\alpha}dr.$

Then, since $\kappa>1$, we have $1-1/2\kappa^{\alpha}>1/2$, and consequently

$\displaystyle I_{2}$
$\displaystyle\leq\kappa^{(\alpha-1+\beta)/\beta}2^{\beta}\int_{1-1/2\kappa^{\alpha}}^{1}(1-r)^{-1/\alpha}dr$
$\displaystyle=\kappa^{(\alpha-1+\beta)/\beta}2^{\beta}\frac{\alpha}{\alpha-1}\left(\frac{1}{2\kappa^{\alpha}}\right)^{1-1/\alpha}=\kappa^{1-(\beta-1)\frac{(\alpha-1)}{\beta}}\,2^{\beta-1+1/\alpha}\frac{\alpha}{\alpha-1}.$

Concerning the integral $I_{1}$, we have

$\displaystyle
I_{1}\leq\kappa^{\frac{\alpha-1}{\beta}}\int_{0}^{1-1/2\kappa^{\alpha}}r^{d-\beta}(1-r)^{-1/\alpha}\int_{\mathbb{R}^{d}}p(1-r,x,rw)|u^{*}(r^{\alpha}t,w)|\,dw\,dr.$
(3.4)

Applying (3.3) to (3.4) and taking advantage of the assumption (A), we
eventually obtain

$\displaystyle
I_{1}\lesssim\mu\kappa^{\frac{\alpha-1}{\beta}}\int_{0}^{1-1/2\kappa^{\alpha}}(1-r)^{-2/\alpha}\,dr=\mu\,2^{2/\alpha-1}\kappa^{1-(\beta-1)\frac{(\alpha-1)}{\beta}},$

which leads to

$\kappa\lesssim\mu+|b|\left(1+\mu\right)\left(\kappa^{1-(\beta-1)\frac{(\alpha-1)}{\beta}}\right).$

Dividing both sides by $\kappa^{1-(\beta-1)\frac{(\alpha-1)}{\beta}}$ and
keeping in mind the assumptions $\alpha,\beta>1$ and $\kappa>1$, we obtain

$\kappa^{(\alpha-1)(\beta-1)/\beta}\lesssim\mu+|b|\left(1+\mu\right)\leq(1+|b|)(\mu+1),$

which implies the bound in the assertion of the theorem. ∎

Next we provide an improved upper bound, which depends on the space arguments
as well.

######  Proposition 3.5.

Under the assumptions of Theorem 3.4, we have

$\displaystyle|u(t,x)|\leq
C\,t^{-1/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|p^{(d-1)}(t,\tilde{x},\tilde{y})dy$
(3.5)

for some $C=C(\alpha,\beta,d,|b|,\mu)$.

###### Proof.

As in the proof of the previous theorem, we denote
$\kappa=\sup\limits_{t>0,\,x\in\mathbb{R}^{d}}|u^{*}(t,x)$|. Let
$0<\varepsilon<1/2$ and put

$\displaystyle H_{\varepsilon}(t,x)$
$\displaystyle=\int_{0}^{1-\varepsilon}(1-r^{\alpha})^{-1/\alpha}r^{d-\beta}\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)|u^{*}(r^{\alpha}t,w)|\,dw\,dr,$
$\displaystyle h_{\varepsilon}(t,x)$
$\displaystyle=\int_{1-\varepsilon}^{1}(1-r^{\alpha})^{-1/\alpha}r^{d-\beta}\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)|u^{*}(r^{\alpha}t,w)|\,dw\,dr.$

We also let $H(t,x)=H_{\varepsilon}(t,x)+h_{\varepsilon}(t,x)$. By Theorem 3.4
and the Duhamel formula (2.14), we have

$\displaystyle|u^{*}(t,x)|\leq P_{t}^{*}|u_{0}|(x)+c\kappa^{q}H(t,x).$ (3.6)

By (2.6),

$\displaystyle P^{*}_{t}|u_{0}|(x)$
$\displaystyle=t^{\beta/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|p(t,t^{1/\alpha}x,y)dy\lesssim
t^{(\beta-1)/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|p^{(d-1)}(t,t^{1/\alpha}\tilde{x},\tilde{y})dy.$

Next, applying (3.3) we get

$\displaystyle H_{\varepsilon}(t,x)$
$\displaystyle\lesssim\int_{0}^{1-\varepsilon}(1-r)^{-2/\alpha}t^{(\beta-1)/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|p^{(d-1)}(t,t^{1/\alpha}\tilde{x},\tilde{y})dy\,dr$
$\displaystyle=\frac{(2-\alpha)}{\alpha}\varepsilon^{1-2/\alpha}\,t^{(\beta-1)/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|p^{(d-1)}(t,t^{1/\alpha}\tilde{x},\tilde{y})\,dy.$

Hence, it is enough to show that for some $\varepsilon>0$,

$\displaystyle h_{\varepsilon}(t,x)\leq
C\left(P_{t}^{*}|u_{0}|(x)+H_{\varepsilon}(t,x)\right)$ (3.7)

holds with some $C=C(\alpha,\beta,d,|b|,\mu,\varepsilon)$. By virtue of (2.6)
and (3.6),

$\displaystyle h_{\varepsilon}(t,x)$
$\displaystyle\leq\int_{1-\varepsilon}^{1}(1-r^{\alpha})^{-1/\alpha}r^{d-\beta}\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)\Big{(}P_{r^{\alpha}t}^{*}|u_{0}|(w)+c\kappa^{q}H(r^{\alpha}t,w)\Big{)}\,dw\,dr$
$\displaystyle=:I_{1}+c\kappa^{q}I_{2}.$

Now, by the Tonelli theorem, (2.3) and the semigroup properties of $p$,

$\displaystyle I_{1}$
$\displaystyle=\int_{1-\varepsilon}^{1}(1-r^{\alpha})^{-1/\alpha}r^{d-\beta}\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)P_{r^{\alpha}t}^{*}|u_{0}(w)|\,dw\,dr$
$\displaystyle=\int_{1-\varepsilon}^{1}(1-r^{\alpha})^{-1/\alpha}r^{d}t^{\beta/\alpha}\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)\int_{\mathbb{R}^{d}}p(r^{\alpha}t,rt^{1/\alpha}w,y)|u_{0}(y)|\,dy\,dw\,dr$
$\displaystyle=\int_{1-\varepsilon}^{1}(1-r^{\alpha})^{-1/\alpha}r^{-d}t^{(\beta-d)/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|\int_{\mathbb{R}^{d}}p(r^{-\alpha}-1,r^{-1}x,w)p(1,w,(rt^{1/\alpha})^{-1}y)\,dw\,dy\,dr$
$\displaystyle=\int_{1-\varepsilon}^{1}(1-r^{\alpha})^{-1/\alpha}r^{-d}t^{(\beta-d)/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|p(r^{-\alpha},r^{-1}x,(rt^{1/\alpha})^{-1}y)\,dy\,dr$
$\displaystyle=\int_{1-\varepsilon}^{1}(1-r^{\alpha})^{-1/\alpha}t^{\beta/\alpha}\int_{\mathbb{R}^{d}}p(t,t^{1/\alpha}x,y)|u_{0}(y)|\,dy\,dr\leq\frac{\alpha\varepsilon^{(\alpha-1)/\alpha}}{\alpha-1}P_{t}^{*}|u_{0}|(x).$

Next,

$\displaystyle{I_{2}}=\int_{1-\varepsilon}^{1}(1-r^{\alpha})^{-1/\alpha}r^{d-\beta}\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)H(r^{\alpha}t,w)\,dw\,dr$
$\displaystyle=\int_{1-\varepsilon}^{1}(1-r^{\alpha})^{-1/\alpha}r^{d-\beta}\int_{\mathbb{R}^{d}}p(1-r^{\alpha},x,rw)$
$\displaystyle\ \ \
\times\int_{0}^{1}(1-s^{\alpha})^{-1/\alpha}s^{d-\beta}\int_{\mathbb{R}^{d}}p(1-s^{\alpha},w,sy)|u^{*}(r^{\alpha}s^{\alpha}t,w)|\,dy\,ds\,dw\,dr$
$\displaystyle=\int_{1-\varepsilon}^{1}\int_{0}^{1}(1-r^{\alpha})^{-1/\alpha}(1-s^{\alpha})^{-1/\alpha}(rs)^{d-\beta}\int_{\mathbb{R}^{d}}p(1-(rs)^{\alpha},x,rsy)|u^{*}((rs)^{\alpha}t,w)|\,dy\,ds\,dr.$

Substituting $s=v/r$ leads to

$\displaystyle{I_{2}}$
$\displaystyle=\int_{1-\varepsilon}^{1}\int_{0}^{r}\frac{(1-(v/r)^{\alpha})^{-1/\alpha}}{r(1-r^{\alpha})^{1/\alpha}}v^{d-\beta}\int_{\mathbb{R}^{d}}p(1-v^{\alpha},x,vy)|u^{*}(v^{\alpha}t,w)|\,dy\,dr\,dv.$

By changing the order of integration we get

$\displaystyle\int_{1-\varepsilon}^{1}\int_{0}^{r}\int_{\mathbb{R}^{d}}(\ldots)\,dy\,dv\,dr$
$\displaystyle=\int_{0}^{1-\varepsilon}\int_{1-\varepsilon}^{1}\int_{\mathbb{R}^{d}}(\ldots)\,dy\,dr\,dv$
$\displaystyle+\int_{1-\varepsilon}^{1}\int_{v}^{1}\int_{\mathbb{R}^{d}}(\ldots)\,dy\,dr\,dv=J_{1}+J_{2}.$

Since for $v\in(0,1-\varepsilon)$

$\displaystyle\int_{1-\varepsilon}^{1}\frac{(1-(v/r)^{\alpha})^{-1/\alpha}}{r(1-r^{\alpha})^{1/\alpha}}dr$
$\displaystyle\leq\int_{1-\varepsilon}^{1}\frac{dr}{(1-r^{\alpha})^{1/\alpha}(r^{\alpha}-(1-\varepsilon)^{\alpha})^{1/\alpha}}=c_{\varepsilon}\leq
c_{\varepsilon}(1-v^{\alpha})^{-1/\alpha},$

we obtain

$\displaystyle J_{1}\leq c_{\varepsilon}H_{\varepsilon}(t,x),$

for some constant $c_{\varepsilon}$ depending on $\varepsilon$ and $\alpha$.
In order to estimate $J_{2}$, let us recall [MR4105374, Lemma 4.3 ], which
says that for $v\in(1-\varepsilon,1)$,

$\displaystyle\int_{v}^{1}\frac{(1-(v/r)^{\alpha})^{-1/\alpha}}{r(1-r^{\alpha})^{1/\alpha}}dr$
$\displaystyle=\int_{v}^{1}(1-r^{\alpha})^{-1/\alpha}(r^{\alpha}-v^{\alpha})^{-1/\alpha}dr\leq
c\varepsilon^{(\alpha-1)/\alpha}(1-v^{\alpha})^{-1/\alpha},$

where $c$ does not depend on $\varepsilon$ and $v$. Hence, $J_{2}\leq
c\varepsilon^{(\alpha-1)/\alpha}h_{\varepsilon}(t,x)$ and consequently

$\displaystyle h_{\varepsilon}(t,x)\leq
c(\varepsilon)\big{(}P_{t}^{*}|u_{0}|(x)+H_{\varepsilon}(t,x)\big{)}+c\varepsilon^{(\alpha-1)/\alpha}h_{\varepsilon}(t,x),$

for some $c,c(\varepsilon)>0$, where $c$ does not depend on $\varepsilon$.
Finally, taking sufficiently small $\varepsilon$ we get

$h_{\varepsilon}(t,x)\leq\frac{c(\varepsilon)}{1-c\varepsilon^{(\alpha-1)/\alpha}}\big{(}P_{t}^{*}|u_{0}|(x)+H_{\varepsilon}(t,x)\big{)},$

which yields (3.7), what was to be shown.

∎

###### Proof of Theorem 1.1.

Let $u_{0}^{(k,n)}=u_{0}\mathbf{1}_{B(0,k)}\mathbf{1}_{\\{-n\leq u_{0}\leq
k\\}}$. Then, the functions $u_{0}^{(k,n)}\in L^{1}\cap L^{\infty}$ and, by
(1.3), there exists a double sequence of solutions $u^{(k,n)}(t,x)$ to the
problems

$\begin{cases}u_{t}=\Delta^{\alpha/2}u+b\cdot\nabla\left(u^{1+q}\right),\qquad
t>0,\;x\in\mathbb{R}^{d},\\\ u(0,x)=u_{0}^{(k,n)}(x).\end{cases}$

By (3.5) and (A), each of them satisfies

$\displaystyle|u^{({k,}n)}(t,x)|\leq
C\,t^{-1/\alpha}\int_{\mathbb{R}^{d}}|u_{0}(y)|p^{(d-1)}(t,\tilde{x},\tilde{y})dy\leq
C\mu t^{-\beta/\alpha},$ (3.8)

where $C>0$ does not depend on $k$ and $n$. Thus, due to the monotonicity
property (2.12), the sequence $(u^{(k,n)})_{k\geq 1}$ is non-decreasing. This
ensures existence of the limit
$u^{(\infty,n)}(t,x):=\lim_{k\to\infty}u^{(k,n)}(t,x)$. Similarly, by (2.12)
$(u^{(\infty,n)})_{n\geq 1}$ is decreasing, hence there exists the limit
$u(t,x):=\lim_{n\to\infty}u^{(\infty,n)}(t,x)$. We will show that this is the
solution we are looking for.

In view of (3.8), $u(t,x)$ satisfies (1.7). It remains to prove that $u$
satisfies (1.2) as well. Clearly $u(0,x)=u_{0}(x).$ Since all $u^{(k,n)}$
satisfy (1.2), we have

$\displaystyle
u(t,x)=\lim_{n\to\infty}{\lim_{k\to\infty}}\left(P_{t}u_{0}^{(k,n)}(x)+\int_{0}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-s,x,z){[u^{(k,n)}(s,z)]^{\left\langle
1+q\right\rangle}}dzds\right).$ (3.9)

By (2.6),

$\displaystyle|p(t,x,y)u_{0}^{(k,n)}(y)|\leq
t^{-1/\alpha}p^{(d-1)}(t,\tilde{x},\tilde{y})|u_{0}(y)|.$

Since the right-hand side is integrable thanks to the condition (A), the
dominated convergence theorem gives us
$\lim_{n\to\infty}\lim_{k\to\infty}P_{t}u_{0}^{(k,n)}(x)=P_{t}u_{0}(x)$, and
therefore we only need to show that we can pass with the limit under the
integral in (3.9). By Theorem 3.4,
$|b\cdot\nabla_{x}p(t-s,x,z)u^{(k,n)}(s,z)^{1+q}|\leq
C|b|(t-s)^{-1/\alpha}s^{-(\beta+\alpha-1)/\alpha}p(t-s,x,z)$ with $C>0$
independent of $k$ and $n$. Thus, for every $s\in(0,t)$,

$\displaystyle\lim_{n\to\infty}\lim_{k\to\infty}\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-s,x,z){[u^{(k,n)}(s,z)]^{\left\langle
1+q\right\rangle}}dz=\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-s,x,z){u(s,z)^{\left\langle
1+q\right\rangle}}dz.$

Like in the proof of Lemma 3.3 let $\overline{u}^{(k,n)}$ be the solution to
(1.1) with the initial condition $|u_{0}^{(k,n)}|$. Clearly,
$|u^{(k,n)}|\leq\overline{u}^{(k,n)}$. By Theorem 3.4,
$\overline{u}^{(k,n)}(s,z)\leq cs^{-\beta/\alpha}$ with $c$ not depending on
$k$ and $n$. Hence, by (2.9), (2.3), substituting $z={s^{1/\alpha}}w$ and then
using Lemma 3.3 and the assumption (A), for $s\in(0,t/2)$ we get

$\displaystyle\left|\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-s,x,z){[u^{(k,n)}(s,z)]^{\left\langle
1+q\right\rangle}}dz\right|$
$\displaystyle\lesssim|b|(t-s)^{-1/\alpha}s^{-(\beta+\alpha-1)/\alpha}\int_{\mathbb{R}^{d}}p(t-s,{x},{z})\big{(}\overline{u}^{(k,n)}\big{)}^{*}(s,s^{-1/\alpha}z)\,dz$
$\displaystyle=|b|(t-s)^{-1/\alpha}s^{(d+1-\beta-\alpha)/\alpha}t^{-d/\alpha}\int_{\mathbb{R}^{d}}p(1-(s/t),t^{-1/\alpha}{x},(s/t)^{1/\alpha}{w})\big{(}\overline{u}^{(k,n)}\big{)}^{*}(s,w)\,dw$
$\displaystyle\lesssim|b|t^{-1/\alpha}s^{(d+1-\beta-\alpha)/\alpha}t^{-d/\alpha}\frac{(s/t)^{(\beta-d)/\alpha}}{(1-(s/t)^{1/\alpha})^{1/\alpha}}t^{(\beta-1)/\alpha}\int_{\mathbb{R}^{d}}p^{(d-1)}(t,\tilde{x},\tilde{y})|u_{0}(y)|dy$
$\displaystyle\leq\frac{|b|t^{-(\beta+1)/\alpha}\mu}{(1-1/2^{1/\alpha})^{1/\alpha}}s^{-(\alpha-1)/\alpha}.$

On the other hand, for $s\in[t/2,t)$, by Theorem 3.4,

$\displaystyle\left|\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-s,x,z){[u^{(k,n)}(s,z)]^{\left\langle
1+q\right\rangle}}dz\right|$
$\displaystyle\lesssim|b|(t-s)^{-1/\alpha}s^{-(\beta+\alpha-1)/\alpha}\int_{\mathbb{R}^{d}}p(t-s,x,z)dz$
$\displaystyle\lesssim|b|t^{-(\beta+\alpha-1)/\alpha}(t-s)^{-1/\alpha}.$

Since $\int_{0}^{t}(s^{-(\alpha-1)/\alpha}\vee(t-s)^{-1/\alpha})ds<\infty$, we
apply the dominated convergence theorem once again and get

$\displaystyle\lim_{n\to\infty}\lim_{k\to\infty}\int_{0}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-s,x,z){[u^{(k,n)}(s,z)]^{\left\langle
1+q\right\rangle}}dzds$
$\displaystyle=\int_{0}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-s,x,z){u(s,z)^{\left\langle
1+q\right\rangle}}dzds,$

which ends the proof. ∎

##  4 The self-similar solution

This section is devoted to the proof of Theorem 1.2, where the initial
condition $u_{0}(x)=M|x|^{-\beta}$, $1<\beta<d$ and $M>0$ is considered.

###  4.1 Existence and selfsimilarity

Let us start with the following observation on how scaling of the initial
condition is transferred to scaling of the solution.

######  Lemma 4.1.

If $u$ is a solution to (1.1) with initial condition $u_{0}$, then
$v(t,x)=\lambda^{\beta/\alpha}u(\lambda t,\lambda^{1/\alpha}x)$ is a solution
with the initial condition
$v_{0}(x)=\lambda^{\beta/\alpha}u_{0}(\lambda^{1/\alpha}x).$

###### Proof.

By scaling property (2.3), we have

$\displaystyle\lambda^{\beta/\alpha}P_{\lambda t}u_{0}(\lambda^{1/\alpha}x)$
$\displaystyle=\lambda^{\beta/\alpha}\int_{\mathbb{R}^{d}}p(\lambda
t,\lambda^{1/\alpha}x,y)u_{0}(y)\,dy$
$\displaystyle=\int_{\mathbb{R}^{d}}\lambda^{d/\alpha}p(\lambda
t,\lambda^{1/\alpha}x,\lambda^{1/\alpha}w)\lambda^{\beta/\alpha}u_{0}(\lambda^{1/\alpha}w)\,dw=P_{t}v_{0}(x)\,.$

Similarly,

$\displaystyle\lambda^{\beta/\alpha}\int_{0}^{\lambda
t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(\lambda
t-s,\lambda^{1/\alpha}x,y)u^{q+1}(s,y)\,dy\,ds$
$\displaystyle=\int_{0}^{t}\int_{\mathbb{R}^{d}}\lambda^{d/\alpha+1}b\cdot\nabla_{x}p(\lambda
t-\lambda
r,\lambda^{1/\alpha}x,\lambda^{1/\alpha}w)\lambda^{(1-\alpha)/\alpha}\lambda^{\beta(q+1)/\alpha}u^{q+1}(\lambda
r,\lambda^{1/\alpha}w)\,dw\,dr$
$\displaystyle=\int_{0}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-r,x,w)v(r,w)\,dw\,dr.$

Hence,

$\displaystyle v(t,x)=\lambda^{\beta/\alpha}u(\lambda
t,\lambda^{1/\alpha}x)=P_{t}v_{0}(x)+\int_{0}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{x}p(t-r,x,w)v(r,w)\,dw\,dr,$

as required. ∎

Next we verify the condition (A) in the case $u_{0}(x)=M|x|^{-\beta}$.

######  Lemma 4.2.

The function $u_{0}(x)=M|x|^{-\beta}$, $M>0$, satisfies the condition (A).

###### Proof.

Since

$\displaystyle\int_{\mathbb{R}}|y|^{-\beta}dy_{1}\leq\int_{\mathbb{R}}[\tfrac{1}{2}(|\tilde{y}|+|y_{1}|)]^{-\beta}dy_{1}=\frac{2^{\beta+1}}{\beta-1}|\tilde{y}|^{1-\beta},$
(4.1)

Lemma 2.3 gives us

$\displaystyle\int_{\mathbb{R}^{d}}u_{0}(y)p^{(d-1)}(t,\tilde{x},\tilde{y})d\tilde{y}\lesssim\int_{\mathbb{R}^{d-1}}|\tilde{y}|^{1-\beta}p^{(d-1)}(t,\tilde{x},\tilde{y})d\tilde{y}\lesssim(t^{1/\alpha}+|\tilde{x}|)^{1-\beta},$
(4.2)

which yields (A). ∎

Therefore, by Theorem 1.1 there exists a solution $u(t,x)$ to the initial
problem (1.1) with $u_{0}(x)=M|x|^{-\beta}$, $M>0$. From Lemma 4.1 we know
that its rescaled version $\lambda^{\beta}u(\lambda^{\alpha}t,\lambda x)$,
$\lambda>0$, is also a solution to the same problem. Nevertheless, these may
be two different solutions. In Lemma 4.3 we show that the solution from
Theorem 1.1 is self-similar i.e. coincides with its rescaled versions.

Throughout the whole section we exploit an analogous notation to the one from
the proof of Theorem 1.1. Namely, $u^{(n)}$, $n\geq 1$, is defined as the
solution to the problem (1.1) with

$\displaystyle
u_{0}^{(n)}(x)=\left(n\wedge\frac{M}{|x|^{\beta}}\right)\mathbf{1}_{B(0,n)}(x),$
(4.3)

and $u(t,x):=\lim_{n\nearrow\infty}u^{(n)}(t,x)$, $x\in\mathbb{R}^{d}$. The
initial condition is positive in this case, so there is no need to consider
the double sequence $u_{0}^{(k,n)}$.

######  Lemma 4.3.

For any $\lambda>0$ we have

$\displaystyle u(t,x)=\lambda^{\beta}u(\lambda^{\alpha}t,\lambda x),\ \ \ \ \
\ t>0,x\in\mathbb{R}^{d}.$ (4.4)

###### Proof.

It suffices to consider $\lambda>1$. Indeed, if we substitute
$t=s/\lambda^{\alpha}$ and $x=y/\lambda$ in (4.4), we get the result for
$\lambda<1$. We note that

$\lambda^{\beta}u(\lambda^{\alpha}t,\lambda
x)=\lim_{n\to\infty}\lambda^{\beta}u^{(n)}(\lambda^{\alpha}t,\lambda x),$

where, by Lemma 4.1, $\lambda^{\beta}u^{(n)}(\lambda^{\alpha}t,\lambda x)$ is
the solution to the problem (1.1) with the rescaled initial condition
$\lambda^{\beta}u^{(n)}_{0}(\lambda
x)=\left(\lambda^{\beta}n\wedge\frac{M}{|x|^{\beta}}\right)\mathbf{1}_{B(0,n/\lambda)}$.
Observe that for $i_{n}=\lfloor n/\lambda\rfloor$ and
$j_{n}=\lceil\lambda^{\beta}n\rceil$,

$u^{(i_{n})}_{0}(x)\leq\lambda^{\beta}u^{(n)}_{0}(\lambda x)\leq
u^{(j_{n})}_{0}(x),\ \ \ \ \ x\in\mathbb{R}^{d}.$

Here $\lfloor\cdot\rfloor$ and $\lceil\cdot\rceil$ are the standard floor and
ceiling functions, respectively. Hence, by the monotonicity property (2.12),
we get

$u^{(i_{n})}(t,x)\leq\lambda^{\beta}u^{(n)}(\lambda^{\alpha}t,\lambda x)\leq
u^{(j_{n})}(t,x),\qquad t>0,\,x\in\mathbb{R}^{d},$

which yields
$\lim_{n\to\infty}\lambda^{\beta}u^{(n)}(\lambda^{\alpha}t,\lambda x)=u(t,x)$,
as required. ∎

In order to study the self-similar solution from Lemma 4.3, it is clearly
enough to study the case $t=1$. Namely, denoting $U(x):=u(1,x)$ we get

$\displaystyle u(t,x)=t^{-\beta/\alpha}U(t^{-1/\alpha}x).$ (4.5)

In particular, this implies $u^{*}(t,x)=U(x)$, and the Duhamel formula (2.14)
takes the form

$U(x)=M\int_{\mathbb{R}^{d}}\frac{p(1,x,y)}{|y|^{\beta}}dy+\alpha\int_{0}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}\nabla_{x}p(1-r^{\alpha},x,rw)U(w)^{q+1}\,dw\,dr.$
(4.6)

###  4.2 Pointwise estimates and asymptotics

From Proposition 3.5 and the estimate (4.2) we have the following upper bound

$\displaystyle U(x)\lesssim\frac{1}{(1+|\tilde{x}|)^{\beta-1}}.$ (4.7)

However, it is not optimal. In particular, the right-hand side does not depend
on $x_{1}$. In the sequel, we provide precise two-sided estimates. We start
with $L^{p}$ bounds.

######  Lemma 4.4.

For any $\gamma\in(d/\beta,\infty]$, $U\in L^{\gamma}(\mathbb{R}^{d})$.

###### Proof.

For $\gamma=\infty$ the result follows by (4.7). Furthermore, by monotonicity
of $u^{(n)}$, (4.7), Corollary 3.2 and Lemma 2.3, for
$\gamma\in\big{(}(d-1)/(\beta-1),\infty\big{)}$ we have

$\displaystyle\int_{\mathbb{R}^{d}}U^{\gamma}(x)dx$
$\displaystyle=\lim_{n\rightarrow\infty}\int_{\mathbb{R}^{d}}[u^{(n)}(1,x)]^{\gamma}dx$
$\displaystyle\lesssim\lim_{n\rightarrow\infty}\int_{\mathbb{R}^{d}}\frac{1}{(1+|\tilde{x}|)^{(\beta-1)(\gamma-1)}}u^{(n)}(1,x)\,dx$
$\displaystyle=\lim_{n\rightarrow\infty}\int_{\mathbb{R}^{d}}\frac{1}{(1+|\tilde{x}|)^{(\beta-1)(\gamma-1)}}P_{1}(u_{0}^{(n)})(x)\,dx$
$\displaystyle\lesssim\int_{\mathbb{R}^{d}}\frac{1}{(1+|\tilde{x}|)^{(\beta-1)(\gamma-1)}}\frac{1}{(1+|\tilde{x}|+|x_{1}|)^{\beta}}\,dx$
$\displaystyle=\frac{2}{\beta-1}\int_{\mathbb{R}^{d-1}}\frac{1}{(1+|\tilde{x}|)^{(\beta-1)\gamma}}\,d\tilde{x}<\infty.$
(4.8)

Now, let $\gamma\in(\frac{d}{\beta},\frac{d-1}{\beta-1})$. By the integral
Minkowski inequality, we have

$\displaystyle\|U\|_{\gamma}$ $\displaystyle\lesssim\|P_{1}u_{0}\|_{\gamma}$
$\displaystyle\ \ \
+\left(\int_{\mathbb{R}^{d}}\left(\int_{0}^{1}\int_{\mathbb{R}^{d}}\frac{r^{d-\beta}}{(1-r)^{1/\alpha}}p(1-r,rw)[U(w+\frac{x}{r})]^{1+q}\,dw\,dr\right)^{\gamma}dx\right)^{1/\gamma}$
$\displaystyle\leq
c_{\gamma}+\|U^{1+q}\|_{\gamma}\int_{0}^{1}\int_{\mathbb{R}^{d}}\frac{r^{d-\beta+d/\gamma}}{(1-r)^{1/\alpha}}p(1-r,rw)\,dw\,dr$
$\displaystyle=c_{\gamma}+\|U\|_{\gamma(1+q)}^{1+q}\int_{0}^{1}r^{-\beta+d/\gamma}(1-r)^{-1/\alpha}\,dr\leq
c_{0}\left(1+\|U\|_{\gamma(1+q)}^{1+q}\right).$ (4.9)

Thus, for every $m\in\mathbb{N}$ such that $\gamma(1+q)^{m}>(d-1)/(\beta-1)$,
by iterating (4.9) we get

$\displaystyle\|U\|_{\gamma}$ $\displaystyle\leq
c_{m}\left(1+\|U\|_{\gamma(1+q)^{m+1}}^{(1+q)^{m+1}}\right).$ (4.10)

We take the smallest $m\in\mathbb{N}$ such that that
$\gamma(1+q)^{m+1}>(d-1)/(\beta-1)$. Then, by (4.10) and (4.8), $U\in
L^{\gamma}(\mathbb{R}^{d})$.

∎

The next step is to show that $U$ vanishes at infinity.

######  Lemma 4.5.

We have

$\lim_{R\rightarrow\infty}\sup_{|x|>R}U(x)=0.$

###### Proof.

Estimates (2.4) and Lemma 2.3 applied to (4.6) give us

$\displaystyle U(x)$
$\displaystyle\lesssim\frac{M}{(1+|x|)^{\beta}}+\int_{0}^{1}\int_{\mathbb{R}^{d}}{r^{d-\beta}}\frac{p(1-r,x,rw)}{{(1-r)^{1/\alpha}}+|x-rw|}U(w)^{q+1}\,dw\,dr.$
(4.11)

Let us split the integral above into
$I_{1}+I_{2}:=\int_{0}^{\varepsilon}(\ldots)dr+\int_{\varepsilon}^{1}(\ldots)dr,$
for some $\varepsilon\in(0,1/2)$. First, note that

$\displaystyle\int_{\mathbb{R}}|y|^{-\beta}dy_{1}=\int_{\mathbb{R}}(y_{1}^{2}+|\tilde{y}|^{2})^{-\beta/2}dy=c|\tilde{y}|^{1-\beta}.$

Hence, by (3.3), (4.1) and Lemma 2.3 we have

$\displaystyle I_{1}$
$\displaystyle\lesssim\|U\|_{\infty}^{q}\int_{0}^{\varepsilon}(1-r)^{-2/\alpha}\int_{\mathbb{R}^{d}}p^{(d-1)}(1,\tilde{x},\tilde{y})|y|^{-\beta}dy$
$\displaystyle\lesssim\varepsilon\|U\|_{\infty}^{q}\int_{\mathbb{R}^{d-1}}p^{(d-1)}(1,\tilde{x},\tilde{y})|\tilde{y}|^{1-\beta}d\tilde{y}\lesssim\varepsilon\|U\|_{\infty}^{q}.$

Next, using Hölder inequality and the bound (2.5) we get

$\displaystyle I_{2}$
$\displaystyle\lesssim\int_{\varepsilon}^{1}{r^{d-\beta}}\int_{B(0,|x|/2)}\frac{p(1-r,x,rw)}{|x|}\|U\|_{\infty}^{q+1}\,dw\,dr$
$\displaystyle\ \ \
+\int_{\varepsilon}^{1}\frac{{r^{d-\beta}}}{(1-r)^{1/\alpha}}\int_{B(0,|x|/2)^{c}}p(1-r,x,rw)U(w)\|U\|_{\infty}^{q}\,dw\,dr$
$\displaystyle\leq\frac{\|U\|_{\infty}^{q+1}}{|x|}\int_{\varepsilon}^{1}{r^{-\beta}}dr$
$\displaystyle\ \ \
+\|U\|_{\infty}^{q}\int_{\varepsilon}^{1}\frac{{r^{d-\beta}}}{(1-r)^{1/\alpha}}\|p(1-r,x,r(\cdot))\|_{2d/(2d-\alpha+1)}\|\mathbf{1}_{\\{B(0,|x|/2)^{c}\\}}U\|_{2d/(\alpha-1)}\,dr$
$\displaystyle\lesssim\frac{\|U\|_{\infty}^{q+1}}{\varepsilon^{\beta}|x|}+\|U\|_{\infty}^{q}\|\mathbf{1}_{\\{B(0,|x|/2)^{c}\\}}U\|_{2d/(\alpha-1)}\int_{\varepsilon}^{1}\frac{{r^{(\alpha-1)/2-\beta}}}{(1-r)^{1-(\alpha-1)/2\alpha}}\,dr,$

which tends to zero as $|x|\rightarrow\infty$ by virtue of Lemma 4.4 applied
with $\gamma=2d/(\alpha-1)>d/\beta$. Thus, the integral in (4.11) is
arbitrarly small for large $|x|$, which ends the proof. ∎

We are now ready to derive the upper bound of $U$.

######  Proposition 4.6.

There is a constant $C=C(d,\alpha,b,M,\beta)$ such that

$\displaystyle U(x)$ $\displaystyle\leq C\frac{1}{(1+|x|)^{\beta}},\ \ \ \ \
x\in\mathbb{R}^{d}.$

###### Proof.

Recall that $u^{(n)}$ is a solution to the problem (1.1) with initial
condition given by (4.3) and $\left(u^{(n)}\right)^{*}(t,x)$ is defined by
(2.1). Note also that $\left(u^{(n)}\right)^{*}(t,x)\nearrow u^{*}(t,x)=U(x)$.
From (1.8) and Lemma 2.3 we have

$\displaystyle A_{n}:=\sup_{\begin{subarray}{c}t>0\\\
x\in\mathbb{R}^{d}\end{subarray}}{\left(u^{(n)}\right)^{*}(t,x)}(1+|x|)^{\beta}<\infty.$

It is enough to show that $A_{n}$’s are uniformly bounded. Duhamel formula
(2.14) combined with (2.9) give us

$\displaystyle\left(u^{(n)}\right)^{*}(t,x)\lesssim$
$\displaystyle\,C_{0}\left(\frac{1}{(1+|x|)^{\beta}}+I_{n}(t,x)\right),$

where $C_{0}=C_{0}(d,\alpha,b,M)>0$ is some constant and

$\displaystyle
I_{n}(t,x)=\int_{0}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}\frac{p(1-r^{\alpha},x,rw)}{(1-r^{\alpha})^{1/\alpha}+|x-rw|}\left[\left(u^{(n)}\right)^{*}(r^{\alpha}t,w)\right]^{(\alpha-1+\beta)/\beta}\,dw\,dr.$

Let $0<\varepsilon<1/2$. By Lemma 4.5, we may and do choose $R>0$ such that
$|U(z)|<\varepsilon^{\beta/(\alpha-1)}$ for $|z|\geq R$ and such that
${\|U\|^{(\alpha-1)/\beta}_{\infty}}/{|x|}<\varepsilon$ for $|x|>2R$.

Now let $|x|>2R$. Since $|x-rw|\geq|x|/2$ for $|w|<R$ and $0<r<1$, we get

$\displaystyle I_{n}(t,x)$
$\displaystyle\leq\int_{0}^{1}\int_{B(0,R)}r^{d-\beta}\frac{p(1-r^{\alpha},x,rw)}{|x|/2}\left(u^{(n)}\right)^{*}(r^{\alpha}t,w)\|U\|_{\infty}^{(\alpha-1)/\beta}dw\,dr$
$\displaystyle\ \ \
+\int_{0}^{1}\int_{B(0,R)^{c}}r^{d-\beta}\frac{p(1-r^{\alpha},x,rw)}{(1-r^{\alpha})^{1/\alpha}}\left(u^{(n)}\right)^{*}(r^{\alpha}t,w)\left(\sup_{|z|\geq
R}U(z)\right)^{(\alpha-1)/\beta}dw\,dr$ $\displaystyle\leq\varepsilon
A_{n}\left(2\|U\|_{\infty}^{(\alpha-1)/\beta}+1\right)\int_{0}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}(1-r^{\alpha})^{-1/\alpha}\frac{p(1-r^{\alpha},x,rw)}{(1+|w|)^{\beta}}\,dw\,dr.$

By (2.3) and Lemma 2.3,

$\displaystyle\int_{\mathbb{R}^{d}}\frac{p(1-r^{\alpha},x,rw)}{(1+|w|)^{\beta}}\,dw=\int_{\mathbb{R}^{d}}r^{-d}\frac{p(r^{-\alpha}-1,r^{-1}x,w)}{(1+|w|)^{\beta}}\,dw\approx\frac{r^{\beta-d}}{(1+|x|)^{\beta}}.$
(4.12)

Hence,

$\displaystyle I_{n}(t,x)$ $\displaystyle\lesssim\frac{\varepsilon
A_{n}}{(1+|x|)^{\beta}}\int_{0}^{1}(1-r^{\alpha})^{-1/\alpha}dr\approx\frac{\varepsilon
A_{n}}{(1+|x|)^{\beta}}.$

On the other hand, for $|x|\leq 2R$,

$\displaystyle\left(u^{(n)}\right)^{*}(t,x)\leq
U(x)\leq\|U\|_{\infty}\leq\frac{\|U\|_{\infty}(1+2R)^{\beta}}{(1+|x|)^{\beta}}.$

Finally, there is a constant $c=c(d,\alpha,b,M,\beta)$ such that for any $t>0$
and $x\in\mathbb{R}^{d}$ we have

$\left(u^{(n)}\right)^{*}(t,x)\leq\frac{c}{(1+|x|)^{\beta}}\left(1+\varepsilon
A_{n}+\|U\|_{\infty}(1+2R)^{\beta}\right),$

and consequently

$A_{n}\leq c(1+\varepsilon A_{n}+\|U\|_{\infty}(1+2R)^{\beta}).$

Taking $\varepsilon=1/(2c)$, we obtain

$A_{n}\leq{2(1+\|U\|_{\infty}(1+2R)^{\beta})},$

as required.

∎

Next we provide an asymptotic behaviour of $U$, which is driven by the
functions $h_{n}$ below. First, let us define the kernel

$k(x,w)=\alpha\int_{0}^{1}r^{d-\beta}\nabla_{x}p(1-r^{\alpha},x,rw)dr,$

and then, for a function $f:\mathbb{R}^{d}\rightarrow\mathbb{R}$, the operator

$\displaystyle(Kf)(x)$
$\displaystyle=P_{1}u_{0}(x)+\int_{\mathbb{R}^{d}}k(x,w)f(w)|f(w)|^{q}\,dw.$

In view of (4.6), we clearly have $KU=U$ . Furthermore, we put $h_{0}\equiv 0$
and

$\displaystyle h_{n}(x)={K}^{n}h_{0}(x),\qquad x\in\mathbb{R}^{d},n\geq 1.$

In particular,

$\displaystyle h_{1}(x)$ $\displaystyle=P_{1}u_{0}(x),$ $\displaystyle
h_{2}(x)$
$\displaystyle=P_{1}u_{0}(x)+\int_{\mathbb{R}^{d}}k(x,w)[P_{1}u_{0}(w)]^{q+1}dw.$

One can show inductively that for any $n\geq 0$ there is a constant $c_{n}>0$
such that

$\displaystyle|h_{n}(x)|\leq\frac{c_{n}}{(1+|x|)^{\beta}},\ \ \ \ \
x\in\mathbb{R}^{d}.$ (4.13)

Indeed, by (4.12) we get

$\displaystyle|h_{n+1}(x)|$ $\displaystyle=|Kh_{n}(x)|$ $\displaystyle\lesssim
P_{1}u_{0}(x)+\int_{0}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}|\nabla_{x}p(1-r^{\alpha},x,rw)||h_{n}(w)|^{q+1}\,dw\,dr$
$\displaystyle\lesssim\frac{1}{(1+|x|)^{\beta}}+c_{n}^{q+1}\int_{0}^{1}\frac{r^{d-\beta}}{(1-r)^{1/\alpha}}\int_{\mathbb{R}^{d}}\frac{p(1-r,x,rw)}{(1+|w|)^{\beta}}dw\,dr$
$\displaystyle\lesssim\frac{c_{n}^{q+1}+1}{(1+|x|)^{\beta}}.$

Note that the functions $h_{n}$ might be considered as Picard approximations
of $U$. The next result supplements Proposition 4.6 precisely describing
behaviour of $U(x)$ for large $|x|$. Although asymptotics of the first two
orders only are mentioned in Theorem 1.2, we provide below more general
result, as the arguments are similar.

######  Proposition 4.7.

For every $n\geq 0$ we have

$|U(x)-h_{n}(x)|\stackrel{{\scriptstyle
n}}{{\lesssim}}{(1+|x|)^{-[\beta+n(\alpha-1)]\wedge[d+\alpha+1]}}.$

###### Proof.

We will use induction. For $n=0$ the assertion follows from Theorem 4.6.
Consider some $n\geq 0$. Since $U$ and every $h_{n}$ are bounded, we may focus
only on large $|x|$.

Recall that for any $a\in\mathbb{R}$ and $\gamma>1$ we denote
$a^{\left\langle\gamma\right\rangle}=a|a|^{\gamma-1}$. Note that
$\frac{d}{dx}x^{\left\langle\gamma\right\rangle}=\gamma|x|^{\gamma-1}$. Hence,
for any $a,b\in\mathbb{R}$ and $\gamma>0$ we have

$\displaystyle\left|b^{\left\langle\gamma+1\right\rangle}-a^{\left\langle\gamma+1\right\rangle}\right|$
$\displaystyle=\left|\int_{a}^{b}\frac{d}{dx}x^{\left\langle\gamma+1\right\rangle}dx\right|=(\gamma+1)\left|\int_{a}^{b}|x|^{\gamma}dx\right|$
$\displaystyle\leq(\gamma+1)|b-a|(|a|+|b|)^{\gamma}.$ (4.14)

Thus, by Theorem 4.6, (4.13) and (4.14)

$\displaystyle|U(x)-h_{n+1}(x)|$
$\displaystyle\lesssim\int_{\mathbb{R}^{d}}|k(x,w)|\left|U(w)-h_{n}(w)\right|\left(|U(w)|+|h_{{n}}(w)|\right)^{q}\,dw$
$\displaystyle\stackrel{{\scriptstyle
n}}{{\lesssim}}C_{n}\int_{\mathbb{R}^{d}}|k(x,w)|\frac{1}{(1+|w|)^{[\beta+n(\alpha-1)]\wedge[d+\alpha+1]}}\frac{1}{(1+|w|)^{\beta
q}}dw\,dr$
$\displaystyle\lesssim\int_{0}^{1}r^{d-\beta}\int_{\mathbb{R}^{d}}\frac{p(1-r^{\alpha},x,rw)}{(1-r^{\alpha})^{1/\alpha}+|x-rw|}\;\frac{1}{(1+|w|)^{[\beta+(n+1)(\alpha-1)]\wedge[d+2\alpha}]}dw\,dr.$

Let $a_{n}=\beta+(n+1)(\alpha-1)$. In the case ${a_{n}}<d$, we bound $|x-rw|$
by zero and take advantage of (4.12). Assume therefore ${a_{n}}\geq d$. Since
$|x-rw|\geq|x|/2$ for $w\in B(0,|x|/2r)$ and $0<r<1$, we split the inner
integral above and estimate (applying (2.4)) it as follows

$\displaystyle\int_{\mathbb{R}^{d}}(\ldots)dw$
$\displaystyle\stackrel{{\scriptstyle
n}}{{\lesssim}}\int_{B(0,|x|/2r)}\frac{1-r^{\alpha}}{((1-r^{\alpha})^{1/\alpha}+|x|)^{d+\alpha+1}(1+|w|)^{{a_{n}\wedge[d+2\alpha]}}}dw$
$\displaystyle\ \ \ \
+\int_{B(0,|x|/2r)^{c}}\frac{p(1-r^{\alpha},x,rw)}{(1-r^{\alpha})^{1/\alpha}(1+|x|/r)^{{a_{n}\wedge[d+2\alpha]}}}dw$
$\displaystyle\stackrel{{\scriptstyle
n}}{{\lesssim}}\frac{1}{|x|^{d+\alpha+1}}\int_{B(0,|x|/2{r})}\frac{dw}{(1+|w|)^{{a_{n}\wedge[d+2\alpha]}}}+\frac{r^{-d+\left({a_{n}}\wedge[d+2\alpha]\right)}}{(1-r^{\alpha})^{1/\alpha}|x|^{{a_{n}\wedge[d+2\alpha]}}}$
$\displaystyle\stackrel{{\scriptstyle
n}}{{\lesssim}}\frac{1+\mathbf{1}_{{a_{n}}=d}\ln|x/2r|}{|x|^{d+\alpha+1}}+\frac{{1}}{(1-r^{\alpha})^{1/\alpha}|x|^{{a_{n}\wedge[d+\alpha+1]}}}$
$\displaystyle{\lesssim}\frac{1+{\ln(1/r)}}{(1-r^{\alpha})^{1/\alpha}|x|^{{a_{n}}\wedge[d+\alpha+1]}}.$

Thus, we arrive at

$\displaystyle|U(x)-h_{n+1}(x)|$ $\displaystyle\stackrel{{\scriptstyle
n}}{{\lesssim}}\frac{1}{|x|^{[\beta+(n+1)(\alpha-1)]\wedge[d+\alpha+1]}}\int_{0}^{1}\frac{r^{d-\beta}{(1+\ln(1/r))}}{(1-r^{\alpha})^{1/\alpha}}dr$
$\displaystyle\stackrel{{\scriptstyle
n}}{{\approx}}\frac{1}{|x|^{[\beta+(n+1)(\alpha-1)]\wedge[d+\alpha+1]}},$

which ends the proof. ∎

Proposition 4.7 directly imples the lower bound for $U(x)$.

######  Corollary 4.8.

We have $U(x)\gtrsim\dfrac{1}{(1+|x|)^{\beta}}$ for $x\in\mathbb{R}^{d}$.

###### Proof.

Proposition 4.7 with $n=1$ and Lemma 2.3 applied to Proposition 4.6 imply the
lower bound for large $|x|$. For any fixed $R>0$ and $|x|<R$ we employ (1.8)
and bound

$\displaystyle U(x)$ $\displaystyle\geq u^{(1)}(1,x)\gtrsim(M\wedge
1)(P_{1}\mathbf{1}_{B(0,1)})(x)\approx\int_{|w|<1}\frac{dw}{(1+|w-x|)^{d+\alpha}}$
$\displaystyle\gtrsim\frac{1}{(2+R)^{d+\alpha}}\geq\frac{1}{(2+R)^{d+\alpha}}\frac{1}{(1+|x|)^{\beta}},$

which proves the assertion. ∎

###  4.3 Gradient of $U$

Eventually, we turn our attention to the gradient of $U$. Note that the
estimates in (4.15), that we are going to prove, coincides with the ones of
$U$ from Proposition 4.6. Nevertheless, it is not true for the whole range of
arguments of the solution $u(t,x)$, as the scaling property of the gradient
$\nabla u(t,x)$ is slightly different, i.e. we have

$\displaystyle(\nabla_{x}u)(t,x)=\nabla_{x}\left(\lambda^{\beta}u(\lambda^{\alpha}t,\lambda
x)\right)=\lambda^{\beta+1}(\nabla_{x}u)(\lambda^{\alpha}t,\lambda x),$

and consequently, for $\lambda=t^{-1/\alpha}$, we have (cf. (4.5))

$\displaystyle(\nabla_{x}u)(t,x)=t^{-(\beta+1)/\alpha}(\nabla_{x}u)(1,x/t^{1/\alpha})=t^{-(\beta+1)/\alpha}(\nabla
U)(x/t^{1/\alpha}).$

######  Proposition 4.9.

We have $U\in\mathcal{C}^{1}\left(\mathbb{R}^{d}\right)$ and

$\displaystyle|\nabla U(x)|\lesssim\frac{1}{(1+|x|)^{\beta}}.$ (4.15)

###### Proof.

By [MR2407204, Theorem 3.5], since $|\nabla
u_{0}^{(n)}(x)|\leq\frac{M\beta}{|x|^{\beta+1}}\mathbf{1}_{\\{1/n<|x|<n\\}}$,
we have $|\nabla u^{(n)}(t,x)|\leq c(1+t)(1+|x|)^{-d-\alpha-1}$, where $c$ may
depend on $n$. Hence, for every $n$ we have

$A_{n}:=\sup_{\begin{subarray}{c}x\in\mathbb{R}^{d}\\\ 0<t\leq
1\end{subarray}}\left|t^{1/\alpha}(t^{1/\alpha}+|x|)^{\beta}\nabla
u^{(n)}(t,x)\right|<\infty.$

We will show that

$\displaystyle A:=\sup_{n\in\mathbb{N}}A_{n}<\infty.$ (4.16)

From (1.2) and by applying integration by parts, for any
$\varepsilon\in(0,1/2)$ we get

$\displaystyle\nabla u^{(n)}(t,x)$ $\displaystyle=\nabla
P_{t}u^{(n)}_{0}(x)+\int_{0}^{(1-\varepsilon)t}\int_{\mathbb{R}^{d}}\nabla\left(b\cdot\nabla
p(t-s,x,z)\right)[u^{(n)}(s,z)]^{q+1}\,dz\,ds$ $\displaystyle\ \ \
+(q+1)\int_{(1-\varepsilon)t}^{t}\int_{\mathbb{R}^{d}}b\cdot\nabla_{z}p(t-s,x,z)u^{(n)}(s,z)^{q}\nabla
u^{(n)}(s,z)\,dz\,ds.$ (4.17)

Let us note that the first integrand is not absolutely integrable on
$\big{(}(1-\varepsilon)t,t\big{)}\times\mathbb{R}^{d}$, which explains the
above decomposition. We estimate the term $\nabla P_{t}u^{(n)}_{0}(x)$ as
follows

$\displaystyle|\nabla_{x}P_{t}u^{(n)}_{0}(x)|$
$\displaystyle=\left|\nabla_{x}\int_{\mathbb{R}^{d}}p(t,x,y)u^{(n)}_{0}(y)dy\right|=\left|\int_{\mathbb{R}^{d}}\nabla_{x}p(t,x,y)u^{(n)}_{0}(y)dy\right|$
$\displaystyle\lesssim
t^{-1/\alpha}\int_{\mathbb{R}^{d}}p(t,x,y)|y|^{-\beta}dy\lesssim\frac{1}{t^{1/\alpha}\left(t^{1/\alpha}+|x|\right)^{\beta}},$

where we used (2.9) and Lemma 2.3. Next, applying this, (2.9), (4.5), and
Proposition 4.6 to (4.17), we obtain for $0<t\leq 1$

$\displaystyle|\nabla u^{(n)}(t,x)|$
$\displaystyle\lesssim\frac{1}{t^{1/\alpha}(t^{1/\alpha}+|x|)^{\beta}}$
$\displaystyle\ \ \
+\int_{0}^{(1-\varepsilon)t}\frac{1}{(t-s)^{2/\alpha}}\left(\frac{\|U\|_{\infty}}{s^{\beta/\alpha}}\right)^{q}P_{t-s}((s^{1/\alpha}+|\cdot|)^{-\beta})(x)ds$
$\displaystyle\ \ \
+A_{n}\int_{(1-\varepsilon)t}^{t}\frac{1}{(t-s)^{1/\alpha}}\left(\frac{\|U\|_{\infty}}{s^{\beta/\alpha}}\right)^{q}s^{-1/\alpha}P_{t-s}((s^{1/\alpha}+|\cdot|)^{-\beta})(x)\,ds.$

By Lemma 2.3, $P_{t-s}(1/(s^{1/\alpha}+|\cdot|)^{\beta})(x)\approx
1/(t^{1/\alpha}+|x|)^{\beta}$, and therefore

$\displaystyle|\nabla u^{(n)}(t,x)|$
$\displaystyle\lesssim\frac{1}{t^{1/\alpha}(t^{1/\alpha}+|x|)^{\beta}}$
$\displaystyle\ \ \
\times\left[1+\int_{0}^{1-\varepsilon}\frac{\|U\|_{\infty}^{q}}{(1-r)^{2/\alpha}r^{(\alpha-1)/\alpha}}dr+A_{n}\int_{1-\varepsilon}^{1}\frac{\|U\|_{\infty}^{q}}{(1-r)^{1/\alpha}r}dr\right]$
$\displaystyle\lesssim\frac{1}{t^{1/\alpha}(t^{1/\alpha}+|x|)^{\beta}}\left[1+\|U\|_{\infty}^{q}\left(\varepsilon^{-2/\alpha}+\varepsilon^{1-1/\alpha}A_{n}\right)\right].$

Consequently there is a constant $c$ not depending on $n$ and $\varepsilon$
such that

$\displaystyle A_{n}\leq
c\left(1+\|U\|_{\infty}^{q}\left(\varepsilon^{-2/\alpha}+\varepsilon^{1-1/\alpha}A_{n}\right)\right).$

Thus, taking $\varepsilon$ small enough, we get $A_{n}\leq
c\left(1+\varepsilon^{-2/\alpha}\|U\|_{\infty}^{q}\right)/\left(1-c\|U\|_{\infty}^{q}\varepsilon^{1-1/\alpha}\right)$,
which proves (4.16). In particular, this implies

$\displaystyle\left|U(x+z)-U(x)\right|$
$\displaystyle=\lim_{n\nearrow\infty}|u^{(n)}(1,x+z)-u^{(n)}(1,x)|$
$\displaystyle\leq|z|\sup_{0\leq v\leq 1}|\nabla
u^{(n)}(1,x+vz)|\lesssim\frac{A|z|}{(1+|x|)^{\beta}},$ (4.18)

whenever $|z|<1$. Let $e_{1},\ldots e_{n}$ be the standard basis in
$\mathbb{R}^{d}$. Now, rewriting (4.6) in a similar manner as it was done in
(4.17), for any $\varepsilon\in(0,1)$, $1\leq k\leq d$ and
$h\in\mathbb{R}-\\{0\\}$ we get

$\displaystyle\frac{U(x+he_{k})-U(x)}{|h|}=M\int_{\mathbb{R}^{d}}\frac{p(1,x+he_{k}-y)-p(1,x-y)}{|h||y|^{\beta}}dy$
(4.19)
$\displaystyle+\alpha\int_{0}^{1-\varepsilon}\int_{\mathbb{R}^{d}}r^{d-\beta}b\cdot\frac{\nabla_{w}p(1-r^{\alpha},x+he_{k},rw)-\nabla_{w}p(1-r^{\alpha},x,rw)}{|h|}[U(w)]^{q+1}\,dw\,dr$
$\displaystyle+\alpha\int_{1-\varepsilon}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}b\cdot\nabla_{w}p(1-r^{\alpha},0,rw)\frac{[U(w+(x+he_{k})/r)]^{q+1}-[U(w+x/r)]^{q+1}}{|h|}\,dw\,dr.$

By (2.9), we may pass with $|h|$ to zero under the first two integrals.
Furthermore, (4.18) together with (2.3) and (2.10) let us bound the third
integral for $|h|<1-\varepsilon$ by

$\displaystyle
C\|U\|^{q}\int_{1-\varepsilon}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}\frac{p(1-r^{\alpha},0,rw)}{(1-r)^{1/\alpha}}\frac{A}{(1+|w+x/r|)^{\beta}}\,dw\,dr$
$\displaystyle=AC\|U\|^{q}\int_{1-\varepsilon}^{1}\int_{\mathbb{R}^{d}}\frac{r^{-\beta}}{(1-r)^{1/\alpha}}P_{r^{-\alpha}-1}\left((1+|\cdot|)^{-\beta}\right)(xr^{-1})\,dw\,dr$
$\displaystyle\lesssim\frac{A}{(1+|x|)^{\beta}}\int_{1-\varepsilon}^{1}(1-r)^{-1/\alpha}\,dr\leq\frac{\varepsilon^{1-1/\alpha}}{1-1/\alpha}A.$

This gives us

$\displaystyle\limsup_{|h|\rightarrow
0}\frac{U(x+he_{k})-U(x)}{|h|}-\liminf_{|h|\rightarrow
0}\frac{U(x+he_{k})-U(x)}{|h|}\lesssim\varepsilon^{1-1/\alpha}A,$

for any $\varepsilon\in(0,1)$. Therefore the limit exists, and is bounded by
(4.15) due to (4.18). Furthermore, by (4.19) for any $\varepsilon\in(0,1)$ we
may express the gradient by

$\displaystyle\nabla U(x)$ $\displaystyle=\nabla
P_{1}u_{0}(x)+\alpha\int_{0}^{1-\varepsilon}\int_{\mathbb{R}^{d}}r^{d-\beta}\nabla\left(b\cdot\nabla_{w}p(1-r^{\alpha},x,rw)\right)U^{q+1}(w)\,dw\,dr$
$\displaystyle\ \ \
+\alpha\int_{1-\varepsilon}^{1}\int_{\mathbb{R}^{d}}r^{d-\beta}b\cdot\nabla_{w}p(1-r^{\alpha},x,rw)U^{q}(w)\nabla
U(w)\,dw\,dr,$

which ensures its continuity by the estimates of $U$ and $\nabla U$. ∎

Below we summarize the results of this section.

###### Proof of Theorem 1.2.

Let $u_{0}(x)=M|x|^{-\beta}$ with $M>0$ and $\beta\in(1,d)$. A solution
$u(t,x)$ to the problem (1.1) exists by Theorem 1.1 and Lemma 4.2. By Lemma
4.3 this solution is selfsimilar and there exists the function $U(x)$ such
that $u(t,x)=t^{-\beta/\alpha}U(t^{-1/\alpha}x)$. The estimates (1.9) follows
by Proposition 4.6 and Corollary 4.8. Proposition 4.7 yields (1.10) and
(1.11). Finally, the regularity of $U$ and (1.12) follows by Proposition 4.9.
∎

#### Acknowledgement

We thank Grzegorz Karch for introducing us into the subject and for helpful
discussions.

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