# Fractional Keller–Segel Equation: Global Well-posedness and Finite Time Blow-up

Laurent Lafleche, Samir Salem

► To cite this version:

Laurent Lafleche, Samir Salem. Fractional Keller–Segel Equation: Global Well-posedness and Finite Time Blow-up. Communications in Mathematical Sciences, 2019, 17 (8), pp.2055-2087. 10.4310/CMS.2019.v17.n8.a1. hal-01875506

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Distributed under a Creative Commons Attribution 4.0 International License
# FRACTIONAL KELLER-SEGEL EQUATION: GLOBAL WELL-POSEDNESS AND FINITE TIME BLOW-UP

LAURENT LAFLECHE AND SAMIR SALEM

ABSTRACT. This article studies the aggregation diffusion equation

$$ \partial_t \rho = \Delta^{\frac{\alpha}{2}} \rho + \lambda \operatorname{div}((K * \rho)\rho), $$

where $\Delta^{\frac{\alpha}{2}}$ denotes the fractional Laplacian and $K = \frac{x}{|x|^\alpha}$ is an attractive kernel. This equation is a generalization of the classical Keller-Segel equation, which arises in the modeling of the motion of cells. In the diffusion dominated case $a < \alpha$ we prove global well-posedness for an $L_k^1$ initial condition, and in the fair competition case $a = \alpha$ for an $L_k^1 \cap L \ln L$ initial condition. In the aggregation dominated case $a > \alpha$, we prove global or local well posedness for an $L^p$ initial condition, depending on some smallness condition on the $L^p$ norm of the initial condition. We also prove that finite time blow-up of even solutions occurs, under some initial mass concentration criteria.

## TABLE OF CONTENTS

<table><tr><td>1. Introduction</td><td>1</td></tr><tr><td>2. Main Results</td><td>3</td></tr><tr><td>3. Proof of Theorem 1 and Theorem 2</td><td>9</td></tr><tr><td>3.1. A Priori estimates</td><td>9</td></tr><tr><td>3.2. Tightness and coupling</td><td>19</td></tr><tr><td>4. Proof of Theorem 3</td><td>26</td></tr><tr><td>4.1. Fractional Laplacian of truncated polynomials</td><td>26</td></tr><tr><td>4.2. Blow-up for even solutions</td><td>27</td></tr><tr><td>Acknowledgements</td><td>29</td></tr><tr><td>References</td><td>29</td></tr></table>

## 1. INTRODUCTION

The models arising in the context of the chemotaxis of cells have been thoroughly studied in recent years. Among those, the (parabolic-elliptic) Keller-Segel equation models the competition between the aggregation and diffusion of cells (see [8] and references therein for a proper biological and mathematical introduction on the topic). In this paper we consider a variant of this classical model where the diffusion

<footnote>Date: September 17, 2018.</footnote>

<footnote n="1991">1991 Mathematics Subject Classification. 35R11, 35A01, 35A02, 35B44, 35B40.</footnote>

<footnote>Key words and phrases. fractional diffusion with drift, fractional Laplacian, aggregation diffusion, mean field equation.</footnote>
is modeled with a fractional laplacian. Such a choice is biologically motivated (see for instance [17, 9] and references therein). From a mathematical point of view, it is then interesting to study how such a diffusion competes with an aggregation field which singularity is up to the Newtonian one.

More precisely for some $(\alpha, a) \in \mathbb{R}_+^2$, we consider the fractional Keller-Segel equation

$$ (\text{FKS}) \qquad \partial_t \rho = \Delta^{\frac{\alpha}{2}} \rho + \lambda \operatorname{div}((K * \rho)\rho), $$

where $\lambda > 0$ is a parameter encoding the chemosensitivity, or the intensity of the aggregation. The interaction kernel is given by

$$ K(x) := \frac{x}{|x|^a}, $$

and $I := \Delta^{\frac{\alpha}{2}}$ denotes the fractional Laplacian defined by

$$ (1) \qquad I(u) = \Delta^{\frac{\alpha}{2}} u := c_{d,\alpha} \int_{\mathbb{R}^d} \frac{u(y) - u(x)}{|x - y|^{d+\alpha}} dy. $$

The constant $c_{d,\alpha}$ can be written $c_{d,\alpha} = -(2\pi)^{\alpha} \frac{\omega_{-\alpha}}{\omega_{d+\alpha}} > 0$ where $\omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$ is the size of the unit sphere in $\mathbb{R}^d$ when $d \in \mathbb{N}^*$.

Particular cases of equation (FKS) have been studied by numerous authors recently. The classical case corresponds to the choice $\alpha = a = d = 2$ and has been thoroughly studied in the past years. In [8], the authors show the global well-posedness when the initial mass $M_0$ is smaller than the critical one $M_c = \frac{4\pi}{\lambda}$. Above this mass, a finite time blow-up is shown to appear. In [14] is also established the well posedness for an $L^\infty$ initial condition. This assumption is sufficient to enjoy the Log-Lipschitz regularity of the nonlinear drift $K * \rho$, as in this case $K$ is the Newtonian kernel (see for instance [29]). It is possible to relax this assumption to $L \ln L$ initial data [16] or even measure initial data [1]. Large time behavior is also studied in [8, 10, 16]. In higher dimension, the variant case $\alpha = 2$, $a = d = 3$ is studied in [15], where a finite time blow-up is obtained under a concentration of initial mass condition.

The literature on the fractional case $\alpha < 2$, is also large and growing. In a significant part of it, the kernel $K$ is the Newtonian one ($a = d$). In the one dimensional case, [9] provides a well posedness result for an $L^p$ initial condition with $p > \frac{1}{\alpha}$ when $\alpha \in (0, 1)$ and $p > 1$ when $a \in (0, 1)$, as well as a finite time blow-up of even solutions under some concentration of initial mass criteria. In the case $d \ge 2$, [6] also provides some concentration of initial mass criteria leading to a blow-up of even solutions when $\alpha \in [1, 2)$ and non even when $\alpha \in (0, 2)$. Still in the Newtonian case, [25] provides similar results in the range $\alpha \in (0, 2)$. See also on the limiting case $\alpha = 0$, [2] for $a \in [0, 1)$, [3] for $a = 1$, and [26] for $a \in (1, 2)$. For $\alpha = 2$ and $a \in (0, 2)$, see [20] and [18], and for $a = 1$ and $\alpha \in (0, 1)$, see [23, 24] and [5]. For a wider than segment range of results see [30] of the second author, and [7, 5] which results are summarized in Figure 1 below.
<figure><img src="image_2.png" /><figcaption>FIGURE 1. Existing results of (FKS).</figcaption></figure>

## 2. MAIN RESULTS

We will work on weighted spaces defined by

$$ \mathcal{M}_k := \{\rho \in \mathcal{M}, \langle x \rangle^k \rho \in \mathcal{M}\} $$

$$ L_k^p := \{\rho \in L^p, \langle x \rangle^k \rho \in L^p\}, $$

where $\langle x \rangle = \sqrt{1 + |x|^2}$, $L^p = L^p(\mathbb{R}^d)$ and $\mathcal{M} = \mathcal{M}(\mathbb{R}^d)$ denote the space of bounded measures. We also define the space of functions with finite entropy by

$$ (2) \qquad L \ln L := \{\rho \in L^1, \rho \ln(\rho) \in L^1\}. $$

For $s \in (0, 1)$, we will denote by $C_{d,s}^S$ the best Sobolev's constant such that for any $f \in H^s$

$$ C_{d,s}^S \|f\|_L^{2 \frac{2d}{d-2s}} \le |f|_{H^s}^2, $$

and for $a \in (0, d)$ and $p, q > 1$ satisfying $2 - \frac{a}{d} = \frac{1}{p} + \frac{1}{q}$, we will denote by $C_{d,a,p}^{\text{HLS}}$ the best Hardy-Littlewood-Sobolev's constant such that for any $f \in L^p, g \in L^q$,

$$ (3) \qquad \left| \iint_{\mathbb{R}^{2d}} |x - y|^{-a} f(x) g(y) \, dx \, dy \right| \le C_{d,a,p}^{\text{HLS}} \|f\|_{L^p} \|g\|_{L^q}. $$
Finally for $s \in [0, d)$ and $r = \frac{2d}{d-s}$, we denote $C_{d,s}^{\text{GNS}}$ the best Gagliardo-Nirenberg-Sobolev's constant such that it holds

$$C_{d,s}^{\text{GNS}} \|f\|_{L^r}^2 \le \|f\|_{L^2} \|f\|_{H^s}.$$

For a given couple $(a, \alpha)$ we define the following exponents for the $L^p$ spaces which will characterize the integrability of the density

$$ (4) \qquad p_{a,\alpha} := \frac{d}{d + \alpha - a} $$

$$ (5) \qquad p_a := p_{a,0} = \frac{d}{d-a}. $$

Taking $K = \frac{x}{|x|^\alpha}$ let appear two main difficulties. The first one is the singularity at $x = 0$ and the second is the behavior when $x \to \infty$. We will therefore write

$$ K = K_0 + K_c = \chi K + (1 - \chi)K, $$

where $\chi \in C_c^\infty$ verifies $\mathbb{1}_{B_1} \le \chi \le \mathbb{1}_{B_2}$. Several parts of our analysis could be easily generalized to more general kernels with similar behavior.

**Definition 2.1.** For any $T > 0$, we say that $\rho$ is a weak solution to the (FKS) equation on $(0, T)$ with initial condition $\rho^{\text{in}} \in \mathcal{M}$ if it satisfies

$$ \rho \in C^0([0, T], \mathcal{M}_{(1-a)_+}) \qquad \text{if } a \in (0, 2] $$

$$ \rho \in C^0([0, T], \mathcal{M}) \cap L_{\text{loc}}^1((0, T), L^{p_{a,2}}) \qquad \text{if } a \in (2, d+2), $$

and for any $\varphi \in C_c^2$

$$ (6) \qquad \int_{\mathbb{R}^d} (\rho(t) - \rho^{\text{in}}) \varphi = \int_0^t \int_{\mathbb{R}^d} \rho(s) (I(\varphi) - K_c * (\rho(s) \cdot \nabla \varphi)) \\ \qquad + \iint_{\mathbb{R}^{2d}} K_0(x-y)(\nabla \varphi(x) - \nabla \varphi(y))\rho(s, dx)\rho(s, dy) \, ds. $$

We say that this solution is global if we can take $T = +\infty$.

The definition makes sense since it is easy to notice that

$$ K_c * (\rho \nabla \varphi) \in C^0 \cap L^\infty(\langle x \rangle^{a-1}) $$

$$ K_0(x-y)(\nabla\varphi(x) - \nabla\varphi(y)) \in C^0 \cap L^\infty(\mathbb{R}^{2d}) \qquad \text{if } a \in (0, 2). $$

Moreover, if $a \in (2, d+2)$, the last term in Definition 2.1 is bounded thanks to Hardy-Littlewood-Sobolev inequality. Remark that at least formally, this equation conserves the mass which we will denote by

$$ M_0 := \int_{\mathbb{R}^d} \rho^{\text{in}}. $$

First we obtain a global or local well-posedness result, depending on the regime, given in the

**Theorem 1.** Let $(\alpha, a) \in [0, 2) \times [0, d)$ be such that $a+\alpha > 1$ and $k \in [(1-a)_+, \alpha)$. 
• When $a < \alpha$ and $\rho^{\text{in}} \in L_k^1$, there exists a unique and global weak solution to the (FKS) equation.
• When $a = \alpha$, if $\rho^{\text{in}} \in L_k^1 \cap L \ln L$ satisfies

$$ (7) \qquad \lambda M_0 < C_{a,d} = \frac{4(2\pi)^a}{(d-a)} \left(\frac{\omega_{2d}}{\omega_d}\right) \frac{\omega_{d-a}}{\omega_{2d-a}} \max\left(\frac{\omega_{d-a}}{\omega_{d+a}}, \frac{\omega_{d-a/2}^2}{\omega_{d+a/2}^2}\right), $$

then there exists a unique and global weak solution to the (FKS) equation.

• When $a > \alpha$ and $\rho^{\text{in}} \in L_k^1 \cap L^p$ with $p \in (p_{a,\alpha}, p_a)$, there exists a time $T > 0$ such that there is a unique solution to the (FKS) equation on $(0,T)$. Moreover there is a constant $C_{\lambda,p}(M_0)$ such that if

$$ (8) \qquad \|\rho^{\text{in}}\|_{L^p} \le C_{\lambda,p}(M_0), $$

then the solution is global.

**Remark 2.1.** The constraint $a + \alpha > 1$ comes from the necessity to propagate moments, which is necessary for our notion of solution and gives us compactness. Remark that it is only due to the behaviour at infinity of the interaction kernel, which we denoted by $K_c$, and not to the singularity. Therefore, our Theorem would hold also for example for the following kernel

$$ K(x) = \frac{x}{|x|^a} \chi(x) + \frac{x}{|x|^\beta} (1 - \chi(x)), $$

for any $\beta > 1 - \alpha$ and which relaxes the condition $a+\alpha > 1$. As it can be seen in the proof, this condition can also be removed by assuming $\rho$ even. It is interesting also to notice that formula (20) could also provide an alternative definition of solution which does not need moments. However, it is not clear whether it is sufficient to provide compactness.

**Remark 2.2.** The explicit value of $C_{d,a,p}^{\text{HLS}}$ for $a \in (0, d)$ and $p = q$ in (3) and $C_{d,s}^{\text{S}}$ for $s \in (0, 1)$ are known, see for instance [27, 28]. Remarking that the HLS conjugate as defined in (3) of $p_{a/2}$ is itself, it holds

$$ C_{d,a, p_{a/2}}^{\text{HLS}} = \pi^{\frac{a}{2}} \frac{\Gamma\left(\frac{d-a}{2}\right)}{\Gamma\left(d-\frac{a}{2}\right)} \left(\frac{\Gamma\left(\frac{d}{2}\right)}{\Gamma(d)}\right)^{-1+\frac{a}{d}} = \frac{\omega_{2d-a}}{\omega_{d-a}} \left(\frac{\omega_{2d}}{\omega_d}\right)^{\frac{a-d}{d}} $$

$$ C_{d,s}^{\text{S}} = \frac{2^{2s} \pi^s \Gamma\left(\frac{d+2s}{2}\right)}{\Gamma\left(\frac{d-2s}{2}\right)} \left(\frac{\Gamma\left(\frac{d}{2}\right)}{\Gamma(d)}\right)^{\frac{2s}{d}} = (2\pi)^{2s} \frac{\omega_{d-2s}}{\omega_{d+2s}} \left(\frac{\omega_{2d}}{\omega_d}\right)^{\frac{2s}{d}}, $$

where we recall that $\omega_d = \frac{2\pi^{d/2}}{\Gamma(d/2)}$.

In the case $a \le \alpha$, this theorem enlarges the existing result by Biler et al. [7], where global existence is proved for $d = 2, 3$ in the case $\alpha \le \frac{d}{2}$, and is a novelty in higher dimension. Also it is provided with larger class of initial condition, and a uniqueness result. Note that the case $\alpha = a$ is only the object of some remark in [7, Remark 3.2]. As for the case $\alpha < \alpha < 2$, it seems it has not been treated yet to the best of the authors' knowledges. See also [6] and [25] for the case $a = 2$.

Let us briefly sketch the proof of this theorem in the case of an $L \ln L$ initial condition. Formally differentiating the Boltzmann's entropy along (FKS) (see for instance ([8, Section 2.2]) provides a control of the $L^1([0, T], L^p)$ for $p \in [1, p_\alpha]$ by fractional Sobolev's embedding, for any initial mass in the diffusion dominated case and for small initial mass in the fair competition case. Then a slight modification of standard coupling argument enables to obtain stability in this space when $p \in [1, p_a)$
and uniqueness when $p = p_a$. The other assumption on the initial condition are meant to control the $L^1([0, T], L^{p_a})$ norm of the solution in the different regimes.

When global existence holds, we also retrieve some additional properties as a quantitative rate of convergence to 0 in the *aggregation dominated* case and a gain of local integrability in the *diffusion dominated* case.

**Theorem 2.** Let $(\alpha, a) \in [0, 2) \times [0, d)$ and $\rho$ be a solution of the (FKS) equation as given by Theorem 1.

* When $a < \alpha$, the gain of integrability is given for any $p \in (1, p_a)$ by

$$ \|\rho(t)\|_{L^p} \le CM_0t^{-\frac{d}{\alpha q}} + C_{\lambda}(M_0). $$

* When $\alpha < a$ and for a given $p \in (p_{a,\alpha}, p_a)$, $\|\rho^{\text{in}}\|_{L^p} < C_{\lambda,p}(M_0)$ defined by (8), then there exists a constant $C = C_{\alpha,\alpha,p}(\rho^{\text{in}}) > 0$ such that

$$ \|\rho\|_{L^p} \le CM_0t^{-\frac{d}{\alpha q}}. $$

* When $a = \alpha$, the condition becomes

$$ \lambda M_0 < C_{a,d,p} = \frac{4C_{d,a/2}^S}{p(d-a)C_{d,a,r}^{\text{HLS}}}, \quad \frac{1}{r} := \frac{p}{p+1} + \frac{1}{p+1}, $$

which gives both a gain of integrability and an asymptotic behavior for any $p \in (1, p_a)$

$$ (9) \qquad \lambda M_0 \le C_{a,d,p} \implies \|\rho\|_{L^p} \le CM_0t^{-\frac{d}{\alpha q}}, $$

where $C$ depends only on $M_0$, $d$, $p$, $a$ and $\alpha$.

**Remark 2.3.** If $\rho$ is a weak solution to the (FKS) equation as given by definition 2.1 with $a = \alpha$ and $\lambda M_0 < C_{d,a,p}$ for a given $p > 1$, we are not able to assert the uniqueness unless we assume that $\rho^{\text{in}} \in L \ln L$.

Finally we obtain a finite time blow-up for even solutions to (FKS) under some concentration of mass condition stated in the

**Theorem 3.** Let $(\alpha, a) \in [0, 2) \times [0, d)$ be such that $a > \alpha$ and $\rho \in C^0(\mathbb{R}_+, L_k^1)$ be an even weak solution to the (FKS) equation with initial condition $\rho^{\text{in}} \in L_k^1$ verifying

$$ (10) \qquad \int_{\mathbb{R}^d} \rho^{\text{in}} \langle x \rangle^k \le C^* \lambda^{\frac{k}{2(a-k)}} M_0^{\frac{2a-k}{2(a-k)}}, $$

for a given universal constant $C^*$ depending only on $a$, $\alpha$, $k$ and $d$. Then the solution ceases to be in $L^1$ in finite time.

The proof of this theorem relies on the time differentiation of an adequate moment, which is adapted to the fractional diffusion and not Newtonian aggregation case, and which leads to a contradiction. We summarize our results in the following figure
<figure><img src="image_3.png" /><figcaption>FIGURE 2. Range of application of Theorems 1 and 3. We emphasize that for $d > 2$ the results extend to the segment $(\alpha, a) \in \{2\} \times (0, d)$.</figcaption></figure>

One of the strength of the result of Theorem 3, even if it deals only with even solutions, is that it applies to weakly singular interactions, i.e. $a < 2$. Indeed it seems that so far most of finite time blow-up results for aggregation fractional diffusion equation dealt with the case of a Newtonian interaction at the exception of [5, Theorem 2.2], which deals with interactions of the from $\frac{x}{|x|}$ near the origin. Considering a less singular kernel than the Newtonian erases some algebraic facilities and requires a thinner estimation of the competing terms. We emphasize that it also covers the purely aggregative case $\alpha = 0$, giving stronger results than [3, 26] for the case $a \ge 2$. For $a \le 2$, the blow-up was already proved in [2] using a Lagrangian point of view.

Finally, let us comment about the disjunction of the different global existence and finite time blow-up conditions. Condition (8) in Theorem 1 is heuristically in contradiction with the assumption of Theorem 3. First remark that if we require that $\rho^{\text{in}}$ is concentrated around zero, for instance with a condition of the type $\|\rho^{\text{in}}\|_{L_k^1} < CM_0$ for a given constant $C$ which does not depend on $\rho^{\text{in}}$, then the
condition of blow-up (10) is equivalent to

$$ \lambda M_0 \geq C', $$

where $C'$ is a positive constant that depends only on $a$, $\alpha$, $k$ and $d$. Moreover, in a more general setting, for $k > 0$, $q = p' \in (1, \infty)$ and $\rho \in L_k^1 \cap L^p$, the following inequality

$$ \int \rho \leq C \left( \int \rho(x)^k \right)^{\frac{d}{d+kq}} \|\rho\|_{L^p}^{\frac{kq}{d+kq}}, $$

holds with $C$ depending only on $d$, $k$ and $q$. With fixed $M_0$, this inequality is enough to exclude a priori (8) from (10), at least in the range of arbitrarily large (or small) $\|\rho^{\text{in}}\|_{L^p}$ or $\int \rho^{\text{in}}(x)^k$. When this is not the case, we expect that no other behavior appear in the remaining cases.

We bound ourselves to check that in the simple case $\alpha = a = 2 < d$, the global well-posedness condition (7) is coherent with the classical large mass blow-up criteria. Indeed take a solution to (FKS) in that case, it is possible to consider initial condition $\rho^{\text{in}} \in L_2^1$ and then classically

$$ \begin{aligned} \frac{d}{dt} \int \rho |x|^2 &= \int_{\mathbb{R}^d} \rho \Delta(|x|^2) - \lambda \iint_{\mathbb{R}^{2d}} K(x-y) \cdot (x-y) \rho(dx) \rho(dy) \\ &= 2dM_0 - \lambda M_0^2 \\ &= 2dM_0 \left(1 - \frac{\lambda M_0}{2d}\right), \end{aligned} $$

so that the condition $\lambda M_0 > 2d$ yields to final time blow-up. And since $\omega_{a+2} = \frac{2\pi}{a}\omega_a$, it holds

$$ \begin{aligned} C_{2,d} &= \frac{4(2\pi)^2}{(d-2)} \frac{\omega_{2d}}{\omega_d} \frac{\omega_{d-2}}{\omega_{2d-2}} \max\left(\frac{\omega_{d-2}}{\omega_{d+2}}, \frac{\omega_{d-1}^2}{\omega_{d+1}^2}\right) \\ &= \frac{4(2\pi)^2}{(d-2)} \frac{d-2}{2d-2} \max\left(\frac{d(d-2)}{(2\pi)^2}, \frac{(d-1)^2}{(2\pi)^2}\right) \\ &= 2(d-1) < 2d, \end{aligned} $$

so that the two conditions can not be realized simultaneously.
<figure><img src="image_4.png" /><figcaption>FIGURE 3. Lower bound of the threshold of condition (7) for $d = 2, 3, 4$ and $a \in (0, 2)$. For the case $a \le \frac{1}{2}$ see Remark 2.1.</figcaption></figure>

### 3. PROOF OF THEOREM 1 AND THEOREM 2

#### 3.1. A Priori estimates.
We begin this section with an a priori moment estimate given in the

**Proposition 3.1 (Propagation of weight).** Assume $1-a < \alpha$ and let $k \in [1-a, \alpha)$ and $\rho$ be a solution of the (FKS) equation with initial condition $\rho^{\text{in}} \in L_k^1$. Then

$$ \rho \in L_{\text{loc}}^{\infty}(\mathbb{R}_+, L_k^1). $$

**Proof.** Let $m = \langle x \rangle^k$ and $M_k = \|\rho\|_{L_k^1}$. When $k \ge 1$, the convexity of $m$ leads to

$$ \begin{align} (11) \qquad \frac{dM_k}{dt} &= \int_{\mathbb{R}^d} \rho I(m) - \lambda \iint_{\mathbb{R}^{2d}} h_m(x, y)\rho(dx)\rho(dy) \nonumber \\ &\le \int_{\mathbb{R}^d} \rho I(m), \end{align} $$

where $h_m(x, y) = \frac{(\nabla m(x) - \nabla m(y))(x-y)}{|x-y|^\alpha} \ge 0$. From [4, Remark 4.2] and [21, Proposition 2.2], we know that for any $k \in (0, \alpha)$,

$$ (12) \qquad I(m) \le C_{\alpha,k} m(x) \langle x \rangle^{-\alpha}. $$

Since $m(x)\langle x \rangle^{-\alpha} \le 1$, the following inequality holds

$$ \frac{dM_k}{dt} \le C_{\alpha,k} M_0. $$
When $k \in [1-a, \alpha \wedge 1)$, there exists $R > 0$ such that $m$ is strictly convex on $B_R$. Therefore, a part of the second term in (11) is still nonpositive

$$ - \iint_{|x-y| \le R} h_m(x, y) \rho(dx) \rho(dy) \le - \iint_{|x-y| \le R} |x-y|^{2-a} \rho(dx) \rho(dy). $$

The other part can be controlled as follows

$$ \begin{aligned} - \iint_{|x-y|>R} h_m(x,y)\rho(dx)\rho(dy) &\le k \iint_{|x-y|>R} \frac{(x \cdot y)(\langle x \rangle^{k-2} + \langle y \rangle^{k-2})}{|x-y|^a} \rho(dx)\rho(dy) \\ &\le 2k \iint_{|x-y|>R} \frac{(x \cdot y)\langle x \rangle^{k-2}}{|x-y|^a} \rho(dx)\rho(dy) \\ &\le 4k \iint_{|x-y|>R, |x|>|y|} \frac{|x||y|\langle x \rangle^{k-2}}{|x-y|^a} \rho(dx)\rho(dy) \\ &\le 4k(I_1 + I_2), \end{aligned} $$

where

$$ I_1 = \iint_{|x-y|>R, 2|x|<|y|} \frac{|y|\langle x \rangle^{k-1}}{|x-y|^a} \rho(dx)\rho(dy) $$

$$ I_2 = \iint_{|x-y|>R, |x|<|y|<2|x|} \frac{|y|\langle x \rangle^{k-1}}{|x-y|^a} \rho(dx)\rho(dy). $$

Since $|x-y| > ||y| - |x|| > |y||1 - |x|/|y|| > |y|/2$ when $|y| > 2|x|$, we get

$$ \begin{aligned} I_1 &\le 2^a \iint_{|x-y|>R, 2|x|<|y|} |y|^{1-a} \langle x \rangle^{k-1} \rho(dx)\rho(dy) \\ &\le 2^a M_{1-a} M_0. \end{aligned} $$

For $I_2$, we use the fact that $|y| < 2\langle x \rangle$ to obtain

$$ \begin{aligned} I_2 &\le \frac{2}{R^a} \iint_{|x-y|>R, |x|<|y|<2|x|} \langle x \rangle^k \rho(dx)\rho(dy) \\ &\le \frac{2}{R^a} M_k M_0. \end{aligned} $$

Combining these three inequalities with (11) and (12), we obtain

$$ \frac{dM_k}{dt} \le C_{\alpha,k} M_{k-\alpha} + \lambda M_0 \left( 2^{1-a} M_{1-a} + \frac{2}{R^a} M_k \right). $$

In particular, since $1 - a \le k$ and $k - \alpha < 0$, we get

$$ \frac{dM_k}{dt} M_k \le M_0 \left( C_{\alpha,k} + \lambda \left( 2^{1-a} + \frac{2}{R^a} \right) M_k \right). $$

By Gronwall's Lemma, this leads to

$$ M_k \le \left( M_k^{\text{in}} + \frac{C_{\alpha,k}}{\lambda C_{a,R}} \right) e^{\lambda C_{a,R} M_0 t}, $$

which proves the result.

<figure><img src="image_5.png" /></figure>

The second type of estimates are a priori bounds of integrability. Let us first briefly emphasize that the quantities we estimate will take the form

$$ \int_{\mathbb{R}^d} \Phi(u(x)) dx, $$
where $u \ge 0$ and $\Phi: \mathbb{R}_+ \to \mathbb{R}_+$ is a non-decreasing convex mapping such that $\Phi(0) = 0$ and $u \mapsto u\Phi''(u) \in L_{\text{loc}}^1$. Then we can define

$$ (13) \qquad \Psi(u) := \int_0^u v \Phi''(v) \, dv $$

$$ (14) \qquad \psi(u) := \frac{1}{2} \int_0^u \sqrt{\Phi''}. $$

For $p = q' > 1$ and $u \ge 0$, we recover Lebesgue norms and Boltzmann's entropy as follow

$$ \begin{aligned} \Phi_p(u) &:= \frac{1}{p-1} u^p & \implies \Psi_p(u) &= u^p \\ \psi_p(u) &= \frac{2}{\sqrt{p}} u^{p/2} \\ \Phi_1(u) &:= u \ln(u) & \implies \Psi(u) &= u \\ \psi_1(u) &= 2u^{1/2}. \end{aligned} $$

**Lemma 3.1 (General estimate).** Assume that $(\alpha, a) \in (0, 2] \times (0, d)$ (with $\alpha \neq 2$ if $d=2$) and let $\rho$ be a smooth solution to the (FKS) equation, $\Phi$ be a non-decreasing convex mapping, $\Psi$ and $\psi$ be defined respectively by (13) and (14) and $b \in (1, p_a)$. Then there holds

$$ (15) \qquad \frac{d}{dt} \left( \int_{\mathbb{R}^d} \Phi(\rho) \right) \le \lambda(d-a) C_{d,a,b}^{\text{HLS}} \|\rho\|_{L^s} \|\Psi(\rho)\|_{L^b} - |\psi(\rho)|_{H^{\frac{\alpha}{2}}}^2, $$

$$ (16) \qquad \le \lambda(d-a) C_{d,a,b}^{\text{HLS}} \|\rho\|_{L^s} \|\Psi(\rho)\|_{L^b} - C_{d,\alpha/2}^{\text{S}} \|\psi(\rho)\|_{L^b}^2, $$

where

$$ \frac{1}{s} = 2 - \frac{a}{d} - \frac{1}{b}, \quad \frac{2}{b} = 1 - \frac{\alpha}{d}. $$

**Proof.** We define the "Carré du Champs" and the $\Phi$-dissipation by

$$ (17) \qquad \Gamma(u, v) := \frac{c_{d,\alpha}}{2} \int_{\mathbb{R}^d} \frac{(u(y) - u(x))(v(y) - v(x))}{|x - y|^{d+\alpha}} dy dx $$

$$ (18) \qquad \mathfrak{D}_{\Phi}(u) := \Gamma(u, \Phi'(u)), $$

where $c_{d,\alpha}$ is defined in (1). With these definitions, we have

$$ \int_{\mathbb{R}^d} I(u)v = \int_{\mathbb{R}^d} u I(v) = - \int_{\mathbb{R}^d} \Gamma(u, v). $$

In particular, since $\Phi$ is convex,

$$ \int_{\mathbb{R}^d} I(u)\Phi'(u) = - \int_{\mathbb{R}^d} \mathfrak{D}_{\Phi}(u) \le 0. $$

We remark that

$$ \begin{aligned} |\psi(u) - \psi(v)|^2 &= \left| \int_u^v \sqrt{\Phi''} \right|^2 \\ &\le \left( \int_u^v ds \right) \left( \int_u^v \Phi'' \right) \\ &\le (u-v)(\Phi'(u) - \Phi'(v)), \end{aligned} $$

which by definition (17) leads to

$$ \Gamma(\psi(u), \psi(u)) \le \Gamma(u, \Phi'(u)). $$
Therefore

$$ (19) \qquad |\psi(u)|_{H^{\frac{\alpha}{2}}}^2 = \int_{\mathbb{R}^d} \Gamma(\psi(u), \psi(u)) \le \int_{\mathbb{R}^d} \Gamma(u, \Phi'(u)) = \int_{\mathbb{R}^d} \mathfrak{D}\Phi(u). $$

Let $\rho$ be a nonegative solution to the (FKS) equation. Then formally

$$ \begin{align} \frac{d}{dt} \left( \int_{\mathbb{R}^d} \Phi(\rho) \right) &= \int_{\mathbb{R}^d} \Phi'(\rho) I(\rho) - \lambda \Phi''(\rho) \nabla \rho \cdot (K * \rho) \rho \nonumber \\ &= - \int_{\mathbb{R}^d} \mathfrak{D}\Phi(\rho) - \int_{\mathbb{R}^d} \lambda \nabla(\Psi(\rho)) \cdot (K * \rho) \nonumber \\ &= - \int_{\mathbb{R}^d} \mathfrak{D}\Phi(\rho) + \lambda \int_{\mathbb{R}^d} \Psi(\rho) (\mathrm{div}(K) * \rho) \nonumber \\ &= - \int_{\mathbb{R}^d} \mathfrak{D}\Phi(\rho) + \lambda(d-a) \int_{\mathbb{R}^d} \left( \frac{1}{|x|^a} * \rho \right). \tag{20} \end{align} $$

We remark that by Hardy-Littlewood-Sobolev inequality, we have

$$ (d-a) \int_{\mathbb{R}^d} \left( \frac{1}{|x|^a} * \rho \right) \Psi(\rho) \le (d-a) C_{d,a,b}^{\mathrm{HLS}} \|\rho\|_{L^s} \|\Psi(\rho)\|_{L^b}, $$

and by (19) and Sobolev embeddings, we have

$$ - \int_{\mathbb{R}^d} \mathfrak{D}\Phi(\rho) \le -|\psi(\rho)|_{H^{\frac{\alpha}{2}}}^2 \le -C_{d,\alpha/2}^{\mathrm{S}} \|\psi(\rho)\|_{L^b}^2, $$

which ends the proof.

<figure><img src="image_6.png" /></figure>

**Proposition 3.2** (L $\ln L$ estimate). Let $a = \alpha$ and $\rho$ be a smooth function satisfying the (FKS) equation with initial condition $\rho^{\mathrm{in}} \in L \ln L$. Then it holds

$$ \int_{\mathbb{R}^d} \rho \ln(\rho) + 4C_{a,d}^{-1} (\lambda M_0 - C_{a,d}) \int_0^t |\sqrt{\rho}|_{H^{\frac{\alpha}{2}}}^2 \le \int_{\mathbb{R}^d} \rho^{\mathrm{in}} \ln(\rho^{\mathrm{in}}), \\ \text{with } \rho = \rho(t, \cdot) \text{ and} $$

$$ C_{a,d} = \frac{4(C_{d,a/2}^{\mathrm{GNS}})^2}{(d-a)C_{d,a,p_{a/2}}^{\mathrm{HLS}}}. $$

Moreover if $\lambda M_0 < C_{a,d}$ and for some $T, k > 0$, $\rho \in L^\infty((0,T), L_k^1)$, then

$$ (21) \qquad \rho \in L^1((0, T), L^{p_a}). $$

**Remark 3.1.** The explicit value for $C_{d,a/2}^{\mathrm{GNS}}$ does not seem to be known, however the following lower bound holds

$$ (22) \qquad C_{d,s}^{\mathrm{GNS}} \ge \max \left( C_{d,s/2}^{\mathrm{S}}, (C_{d,s}^{\mathrm{S}})^{1/2} \right). $$

Indeed, a first way to get the Gagliardo-Nirenberg-Sobolev inequality is to first use Sobolev's inequality and then interpolation between $H^s$ spaces

$$ C_{d,s/2}^{\mathrm{S}} \|f\|_{L^r}^2 \le |f|_{H^{\frac{\alpha}{2}}}^2 \le \|f\|_{L^2} |f|_{H^s}. $$

A second way is to first interpolate between Lebesgue spaces and then to use Sobolev's inequality

$$ (C_{d,s}^{\mathrm{S}})^{1/2} \|f\|_{L^r}^2 \le (C_{d,s}^{\mathrm{S}})^{1/2} \|f\|_{L^2} \|f\|_{L^{r_2}} \le \|f\|_{L^2} \|f\|_{H^s}, $$

where $r_2 := \frac{2d}{d-2s}$.
**Proof.** We use inequality (15) for $\Phi = \Phi_1$, $\psi_1(u) = 2u^{1/2}$ and $b = s = p_{a/2}$ to obtain

$$ \frac{d}{dt} \int_{\mathbb{R}^d} \rho \ln(\rho) \le \lambda(d-a) C_{d,a,b}^{\text{HLS}} \|\rho\|_{L^b}^2 - |\psi_1(\rho)|_{H^{\frac{a}{2}}}^2. $$

Then, by Gagliardo-Nirenberg-Sobolev's inequality, we have

$$ \begin{aligned} (C_{d,a/2}^{\text{GNS}})^2 \|\rho\|_{L^b}^2 &= (C_{d,a/2}^{\text{GNS}})^2 \|\rho^{1/2}\|_{L^{2b}}^4 \\ &\le \|\rho^{1/2}\|_{L^2}^2 |\rho^{1/2}|_{H^{\frac{a}{2}}}^2 = M_0 |\rho^{1/2}|_{H^{\frac{a}{2}}}^2. \end{aligned} $$

Hence, since $\psi_1(u) = 2u^{1/2}$, we have

$$ 4(C_{d,a/2}^{\text{GNS}})^2 \|\rho\|_{L^b}^2 \le M_0 |\psi_1(\rho)|_{H^{\frac{a}{2}}}^2. $$

This yields

$$ \frac{d}{dt} \int_{\mathbb{R}^d} \rho \ln(\rho) \le C_{a,d}^{-1} (\lambda M_0 - C_{a,d}) |\psi_1(\rho)|_{H^{\frac{a}{2}}}^2, $$

which proves the first assertion. Formula (21) comes from the fact for $k > 0$, defining $\lambda_k > 0$ such that $\int_{\mathbb{R}^d} e^{-\lambda_k \langle x \rangle^k} dx = 1$, with $h(u) = u \ln u - u + 1 \ge 0$ it holds

$$ \int_{\mathbb{R}^d} \rho \log \rho = \int_{\mathbb{R}^d} h \left( \rho e^{\lambda_k \langle x \rangle^k} \right) e^{-\lambda_k \langle x \rangle^k} + \int_{\mathbb{R}^d} \rho \ln(e^{-\lambda_k \langle x \rangle^k}) \ge -\lambda_k \int_{\mathbb{R}^d} \rho \langle x \rangle^k. $$

Combined with the following Sobolev's inequality

$$ 4C_{d,a/2}^{S} \|\rho\|_{L^{p_a}} = C_{d,a/2}^{S} \|\psi_1(\rho)\|_{L^{2p_a}}^2 \le |\psi_1(\rho)|_{H^{\frac{a}{2}}}^2, $$

it yields

$$ \begin{aligned} 0 \le & \left( \int_{\mathbb{R}^d} \rho \ln(\rho) + \lambda_k \langle x \rangle^k \rho \right) + 4C_{d,a/2}^S C_{a,d}^{-1} (C_{a,d} - \lambda M_0) \int_0^t \|\rho\|_{L^{p_a}} \\ & \le \int_{\mathbb{R}^d} \rho^{\text{in}} \ln(\rho^{\text{in}}) + \lambda_k \|\rho\|_{L^{\infty}(0,T;L_k^1)}, \end{aligned} $$

and the conclusion follows. □

**Proposition 3.3** ($L^p$ estimates). Let $(\alpha, a) \in [0, 2) \times [0, d)$. Then, when $a < \alpha$ and $p = q' \in (1, p_a)$, we get a gain of integrability from $L^1$ to $L^p$ and a global in time propagation of the $L^p$ norm

$$ (23) \qquad \|\rho(t)\|_{L^p} \le C M_0 \max \left( t^{-\frac{d}{\alpha q}}, M_0^{\frac{d}{q(\alpha-a)}} \right), $$

where $C > 0$ is a constant depending on $d, a, \alpha, p$ and $\lambda$. When $a > \alpha$, then for any $p \in (p_{a,\alpha}, p_a)$, there exists two constants $C = C_{a,\alpha,p} > 0$ and $C^{\text{in}} = C_{a,\alpha,p}(\|\rho^{\text{in}}\|_{L^p})$ such that

$$ (24) \qquad \|\rho^{\text{in}}\|_{L^p} < C M_0 (\lambda M_0)^{-\frac{d}{(\alpha-a)q}} \implies \|\rho\|_{L^p} \le C^{\text{in}} M_0 t^{-\frac{d}{\alpha q}} $$

$$ (25) \qquad \|\rho^{\text{in}}\|_{L^p} > C M_0 (\lambda M_0)^{-\frac{d}{(\alpha-a)q}} \implies \rho \in L^{\infty}((0,T), L^p) $$

$$ (26) \qquad \|\rho^{\text{in}}\|_{L^p} = C M_0 (\lambda M_0)^{-\frac{d}{(\alpha-a)q}} \implies \rho \in L^{\infty}(\mathbb{R}_+, L^p), $$

where $T < C_{a,\alpha,p}(\lambda, M_0) \|\rho^{\text{in}}\|_{L^p}^{-pb}$ with

$$ b = \frac{\alpha}{p(\alpha - a) + d(p - 1)}. $$
When $a = \alpha$, then there exists a constant

$$C_{a,d,p} = \frac{4C_{d,\alpha/2}^{S}}{(d-a)C_{d,a,r}^{\text{HLS}}}$$

such that for any $p \in (1, p_a)$,

$$(27) \qquad \lambda M_0 \le C_{a,d,p} \implies \|\rho\|_{L^p} \le M_0(C^{\text{in}}b)^{-\frac{1}{b}} t^{-\frac{d}{\alpha q}}$$

$$(28) \qquad \lambda M_0 \ge C_{a,d,p} \implies \rho \in L^{\infty}((0,T), L^p),$$

where $C^{\text{in}}$ is a nonnegative constant depending on the initial data and

$$T > \frac{1}{bC^{\text{in}}} \left( \frac{M_0}{\|\rho^{\text{in}}\|_{L^p}} \right)^{\frac{\alpha q}{d}}.$$

**Remark 3.2.** The critical mass is clearly not optimal since we could use optimal constants in the Gagliardo-Nirenberg type embeddings, as it is done in the $L \ln L$ estimate, instead of using Sobolev's embeddings and interpolation between Lebesgue spaces.

**Proof.** We will separate the proof into several steps.

**Step 1. Differential inequality for the $L^p$ norm.** We recall that

$$\frac{1}{r} = \frac{p}{p+1} \frac{1}{p} + \frac{1}{p+1} \frac{1}{p_a}.$$

Since $p < p_a$, it implies that $r \in (p, p_a)$ and in particular $r/p > 1$. Therefore, by taking $\Phi = \Phi_p$, $r = s$ and $b = r/p$ in inequality (16) and defining $\tilde{r} = \frac{p\tilde{b}}{2}$, we obtain

$$(29) \qquad \frac{d}{dt} \left( \int_{\mathbb{R}^d} \Phi_p(\rho) \right) \le \lambda C_{a,r} \|\rho\|_{L^r}^{p+1} - \frac{c_\alpha}{p} \|\rho\|_{L^{\tilde{r}}}^p,$$

where $C_{a,r} = (d-a)C_{d,a,r}^{\text{HLS}}$, $C_\alpha = 4C_{d,\alpha/2}^{S}$ and

$$(30) \qquad \frac{p+1}{r} = 2 - \frac{a}{d}$$

$$(31) \qquad \frac{p}{\tilde{r}} = 1 - \frac{\alpha}{d}.$$

We also remark that

$$\begin{align*}
r \le \tilde{r} &\Leftrightarrow \frac{1}{p} \left(1 - \frac{\alpha}{d}\right) \le \frac{1}{p+1} \left(2 - \frac{a}{d}\right) \\
&\Leftrightarrow \left(1 + \frac{1}{p}\right) \left(d - \alpha\right) \le \left(2d - a\right) \\
&\Leftrightarrow p \ge \frac{d - \alpha}{d + \alpha - a}.
\end{align*}$$

Since $p \ge p_{a,\alpha} \ge \frac{d-\alpha}{d+\alpha-a}$, we deduce that $r \le \tilde{r}$.

We will now use interpolation between Lebesgue spaces to express the left hand side of (29) in terms of $M_0$ and the $L^p$ norm only. Let $\varepsilon \in (0, 1)$ to be choosen later
and

$$ (32) \qquad b_0 := \frac{a - \alpha(1 - \varepsilon)}{\varepsilon d(p-1)} = \frac{\alpha}{d(p-1)} + \frac{a - \alpha}{\varepsilon d(p-1)} $$

$$ (33) \qquad \theta_1 := \frac{\varepsilon p}{p+1} (1 + b_0) $$

$$ (34) \qquad \theta_2 := \frac{(1 - \varepsilon)p}{p + 1} $$

$$ (35) \qquad \theta_0 := 1 - \theta_1 - \theta_2. $$

Since $p > 1$ and $\varepsilon \in (0, 1)$, we deduce that $\theta_2 \in [0, 1)$. Moreover, using the respective definitions (30) and (31) of $r$ and $\tilde{r}$, we have

$$ \begin{aligned} \frac{\theta_1}{p} + \frac{\theta_2}{\tilde{r}} + \theta_0 &= \frac{\varepsilon(1 + b_0)p}{p + 1} \left(\frac{1}{p} - 1\right) + \frac{(1 - \varepsilon)p}{p + 1} \left(\frac{1}{\tilde{r}} - 1\right) + 1 \\ &= \frac{1}{p + 1} \left(\varepsilon(1 - p)(1 + b_0) + (1 - \varepsilon) \left(1 - \frac{\alpha}{d} - p\right) + p + 1\right) \\ &= \frac{1}{p + 1} \left(\varepsilon(1 - p) - \frac{a - \alpha(1 - \varepsilon)}{d} + 2 - \frac{\alpha}{d} - \varepsilon(1 - p) + \varepsilon \frac{\alpha}{d}\right) \\ &= \frac{1}{p + 1} \left(2 - \frac{a}{d}\right) = \frac{1}{r}. \end{aligned} $$

Therefore, if we can choose $\varepsilon \in (0, 1)$ such that $(\theta_0, \theta_1) \in [0, 1]^2$, we obtain by interpolation

$$ \|\rho\|_{L^r}^{p+1} \le M_0^{\theta_0(p+1)} \|\rho\|_{L^p}^{p\varepsilon(1+b_0)} \|\rho\|_{L^{\tilde{r}}}^{p(1-\varepsilon)} = A^\varepsilon B^{1-\varepsilon}. $$

Then, by using the standard Young's inequality $a^\varepsilon b^{1-\varepsilon} \le \varepsilon a + (1-\varepsilon)b$, for any $\varepsilon_0 > 0$, we have

$$ A^{\varepsilon} B^{1-\varepsilon} = \left( \left( \frac{1-\varepsilon}{\varepsilon_0} \right)^{\frac{1-\varepsilon}{\varepsilon}} A \right)^{\varepsilon} \left( \frac{\varepsilon_0 B}{1-\varepsilon} \right)^{1-\varepsilon} \le C_{\varepsilon, \varepsilon_0} A + \varepsilon_0 B, $$

with $C_{\varepsilon, \varepsilon_0} = \varepsilon \left( \frac{1-\varepsilon}{\varepsilon_0} \right)^{\frac{1-\varepsilon}{\varepsilon}}$. Coming back to (29), it yields

$$ (36) \quad \frac{d}{dt} \left( \int_{\mathbb{R}^d} \Phi_p(\rho) \right) \le \left( \lambda C_{a,r} \right)^{1/\varepsilon} C_{\varepsilon, \varepsilon_0} M_0^{\theta_0(p+1)/\varepsilon} \|\rho\|_{L^p}^{p(1+b_0)} + \left( \varepsilon_0 - \frac{c_\alpha}{p} \right) \|\rho\|_{L^{\tilde{r}}}^p, $$

where we take $\varepsilon_0$ smaller than $C_\alpha/p$. Since $1 \le p \le \tilde{r}$, again by interpolation, we get

$$ \|\rho\|_{L^p}^{p(1+b_1)} \le M_0^{pb_1} \|\rho\|_{L^{\tilde{r}}}^p, $$

with

$$ b_1 = \frac{\alpha}{d(p-1)}. $$

Thus, inequality (36) becomes

$$ (37) \quad \frac{d}{dt} \|\rho\|_{L^p}^p \le C_1 M_0^{\theta_0(p+1)/\varepsilon} \|\rho\|_{L^p}^{p(1+b_0)} - C_2 M_0^{-pb_1} \|\rho\|_{L^p}^{p(1+b_1)}, $$

where $C_1 = (p-1)(\lambda C_{a,r})^{1/\varepsilon} C_{\varepsilon, \varepsilon_0}$ and $C_2 = (p-1) \left( \frac{c_\alpha}{p} - \varepsilon_0 \right)$.
**Step 2.** *Conditions on ε.* We still have to verify that we can choose ε so that $(\theta_0, \theta_1) \in [0, 1]^2$. By definition (33) of $\theta_1$, we get

$$ 
\begin{aligned}
\theta_1 \ge 0 &\Leftrightarrow b_0 \ge -1 \\
&\Leftrightarrow a - \alpha + \alpha\varepsilon > -\varepsilon d(p-1) \\
&\Leftrightarrow \varepsilon \ge \frac{\alpha - a}{\alpha + d(p-1)} = \varepsilon_m.
\end{aligned}
$$

Moreover, by definition (35) of $\theta_0$

$$ 
\begin{aligned}
\theta_0 \ge 0 &\Leftrightarrow \theta_1 + \theta_2 \le 1 \\
&\Leftrightarrow \frac{p}{p+1} (1 + \varepsilon b_0) \le 1 \\
&\Leftrightarrow \varepsilon b_0 \le \frac{1}{p} \\
&\Leftrightarrow \frac{a - \alpha(1 - \varepsilon)}{d(p-1)} \le \frac{1}{p} \\
&\Leftrightarrow \varepsilon \le 1 - \frac{1}{\alpha} \left( a - \frac{d}{q} \right) = \varepsilon_M.
\end{aligned}
$$

Since $p < p_a$, $\varepsilon_M < 1$. Let us check that it is nonnegative. We have

$$ \varepsilon_M \ge 0 \Leftrightarrow a - \frac{d}{q} \le \alpha \Leftrightarrow \frac{1}{q} \ge \frac{a - \alpha}{d}. $$

Since $q = p' \ge 1$, this is always verified when $a \le \alpha$. When $a > \alpha$, it is verified by hypothesis since we can also read previous formula as

$$ \varepsilon_M \ge 0 \Leftrightarrow p \ge \frac{d}{d + \alpha - a} = p_{a,\alpha}. $$

When $a < \alpha$, we also have to verify that $\varepsilon_m \le \varepsilon_M$. We have, indeed

$$ 
\begin{aligned}
\frac{\varepsilon_M}{\varepsilon_m} &= \frac{(p(\alpha - a) + d(p-1))(\alpha + d(p-1))}{p\alpha(\alpha - a)} \\
&= \frac{p\alpha(\alpha - a) + d(p-1)(\alpha - a) + \alpha(p-1) + d(p-1)^2}{p\alpha(\alpha - a)} \\
&= 1 + d(p-1) \frac{p(\alpha - a) + \alpha + d(p-1)}{p\alpha(\alpha - a)} > 1.
\end{aligned}
$$

Therefore, since $\theta_2 \ge 0$ and $\theta_0 + \theta_1 + \theta_2 = 1$, we proved that for any $\varepsilon \in [\max(\varepsilon_m, 0), \min(\varepsilon_M, 1)]$,

$$ (\theta_0, \theta_1, \theta_2) \in [0, 1]^3. $$

By looking at (37), we want to take $\varepsilon$ which minimizes $b_0$. Hence, we take

$$ \varepsilon = \varepsilon_m \text{ when } a < \alpha, $$

$$ \varepsilon = \varepsilon_M \text{ when } a > \alpha. $$

**Step 3.** *Case a < α.* In this case, we have $\varepsilon = \varepsilon_m$, which yields $b_0 = -1$. Moreover, since

$$ \theta_0(p+1) = p+1 - (1-\varepsilon)p = 1 + \varepsilon p, $$

by (37), we obtain

$$ \frac{d}{dt} \|\rho\|_{L^p}^p \le C_1 M_0^{p+1/\varepsilon} - C_2 M_0^{-pb_1} \|\rho\|_{L^p}^{p(1+b_1)}. $$
Then, either

$$ (38) \qquad C_2 M_0^{-pb_1} \|\rho\|_{L^p}^{p(1+b_1)} \le 2C_1 M_0^{p+1/\epsilon}, $$

or

$$ (39) \qquad \frac{d}{dt} \|\rho\|_{L^p}^p \le -\frac{1}{2} C_2 M_0^{-pb_1} \|\rho\|_{L^p}^{p(1+b_1)}. $$

Inequality (38) is equivalent to

$$ \|\rho\|_{L^p}^p \le \left( \frac{2C_1}{C_2} \right)^{\frac{1}{1+b_1}} M_0^{p+\frac{1}{\epsilon(1+b_1)}} =: C(M_0), $$

and by Gronwall's inequality, (39) leads to

$$ \|\rho\|_{L^p}^p \le \left( \frac{1}{2} C_2 M_0^{-pb_1} b_1 t \right)^{-1/b_1} = M_0^p \left( \frac{b_1}{2} C_2 t \right)^{-1/b_1} $$

**Step 4.** Case $a > \alpha$. In this case, we have

$$ \varepsilon = \varepsilon_M = \frac{p(\alpha - a) + d(p - 1)}{\alpha p} = \frac{1}{pb}, $$

which by definition (32) leads to

$$ \begin{aligned} b_0 &= \frac{1}{d(p-1)} \left( \alpha + \frac{a-\alpha}{\varepsilon} \right) \\ &= \frac{\alpha}{d(p-1)} \left( \frac{p(\alpha - a) + d(p-1) + p(a-\alpha)}{p(\alpha - a) + d(p-1)} \right) \\ &= \frac{\alpha}{p(\alpha - a) + d(p-1)} = b, \end{aligned} $$

and by inequality (37), to

$$ \frac{d}{dt} \|\rho\|_{L^p}^p \le C_1 M_0^{\theta_0(p+1)/\varepsilon} \|\rho\|_{L^p}^{p(1+b)} - C_2 M_0^{-pb_1} \|\rho\|_{L^p}^{p(1+b_1)}. $$

As remarked previously, $\varepsilon = \varepsilon_M \ge 0$. Therefore, since $a > \alpha$,

$$ (40) \qquad b = b_1 + \frac{a-\alpha}{\varepsilon d(p-1)} \ge b_1. $$

The estimate on the $L^p$ norm is then obtained by analyzing the corresponding ODE which is of the form

$$ y'(t) = Ay(t)^{1+b} - By(t)^{1+b_1}, $$

with $A$ and $B$ nonnegative. It has a fixed point at $y = 0$ and at

$$ y^{\#} = \left( \frac{B}{A} \right)^{\frac{1}{b-b_1}} \ge 0. $$

Therefore, when $y(0) \in [0, y^{\#})$, it yields $y(t) \in [0, y^{\#})$ for any $t > 0$, and since $y' \le 0$ in this interval, it implies the existence of a constant $C = C(y(0)^{\text{in}}) < 1$ such that

$$ Ay^{1+b} \le CB y^{1+b_1}. $$

It implies that

$$ y' \le -(1-C)By^{1+b_1}, $$

which, by Gronwall's inequality, leads to

$$ \forall t \in \mathbb{R}_+, y \le \frac{1}{(y(0)^{-b_1} + b_1(1-C)Bt)^{\frac{1}{b_1}}} \le \frac{M_0^p}{(b_1(1-C)C_2t)^{\frac{1}{b_1}}}. $$
When $y(0) > y^{\#}$, we can still write that

$$y' \le A y^{1+b}.$$

It implies that the solution is bounded in $[0, T]$ for some $T > 0$ and

$$\forall t \in [0, T], y(t) \le \frac{1}{(y(0)^{-b} - bAt)^{\frac{1}{b}}} \\ T < \frac{1}{bAy(0)^{b}}.$$

We deduce the corresponding results for the $L^p$ norm of $\rho$ by Gronwall's inequality. When $y = y^{\#}$, all we get that $y$ is constant and therefore that $\|\rho\|_{L^p}^p \le y^{\#}$ for any $t > 0$. We can compute more precisely

$$y^{\#} = \left( \frac{C_2 M_0^{-pb_1}}{C_1 M_0^{\theta_0(p+1)/\epsilon}} \right)^{\frac{1}{b-b_1}} = \left( \frac{C_2}{C_1} \right)^{\frac{1}{b-b_1}} \left( M_0^{-\theta_0(p+1)/\epsilon - pb_1} \right)^{\frac{1}{b-b_1}}.$$

Now by the definitions of $C_1$ and $C_2$ in step 1, by (40) and the definition (35) of $\theta_0$, we have

$$\begin{aligned} \theta_0(p+1) &= 1 - \epsilon pb = 0 \\ (b-b_1)\epsilon &= \frac{a-\alpha}{d(p-1)} \\ C_1 &= (p-1)(\lambda C_{a,r})^{1/\epsilon} C_{\epsilon, \epsilon_0} \\ C_2 &= (p-1) \left( \frac{C_\alpha}{p} - \epsilon_0 \right). \end{aligned}$$

This leads to

$$\begin{aligned} y^{\#} &= \left( \frac{C_{\alpha} - \epsilon_0 p}{(\lambda C_{a,r})^{1/\epsilon} C_{\epsilon, \epsilon_0} p} \right)^{\frac{1}{b-b_1}} \left( M_0^{p(b-b_1)-1/\epsilon} \right)^{\frac{1}{b-b_1}} \\ &= C_{a, \alpha, p}^p M_0^{p-\frac{d(p-1)}{a-\alpha}} \lambda^{\frac{d(p-1)}{a-\alpha}}. \end{aligned}$$

**Step 5.** Case $a = \alpha$. When $a = \alpha$, by definition (32), $b_0$ does not depend on $\epsilon$ and

$$\begin{aligned} b &= b_0 = b_1 = \frac{\alpha}{d(p-1)} \\ \theta_0(p+1) &= 1 - \epsilon pb. \end{aligned}$$

Moreover, we can take any $\epsilon \in (\epsilon_m, \epsilon_M] = (0, d/(\alpha q)]$. Thus, by inequality (37), we get

$$\begin{aligned} \frac{d}{dt} \|\rho\|_{L^p}^p &\le C_1 M_0^{(1-\epsilon pb)/\epsilon} \|\rho\|_{L^p}^{p(1+b)} - C_2 M_0^{-pb} \|\rho\|_{L^p}^{p(1+b)} \\ &\le \|\rho\|_{L^p}^{p(1+b)} M_0^{-pb} (C_1 M_0^{1/\epsilon} - C_2). \end{aligned}$$

The left hand side will be negative when

$$(41) \qquad M_0 \le \left( \frac{C_2}{C_1} \right)^{\epsilon} = \frac{(C_{\alpha}/p - \epsilon_0)^{\epsilon}}{\lambda C_{a,r}} \left( \frac{\epsilon_0}{1-\epsilon} \right)^{1-\epsilon} \epsilon^{-\epsilon} = u_{\epsilon}(\epsilon_0).$$
Taking $\varepsilon_0$ maximizing the right hand side, we get

$$ \varepsilon_0 = (1 - \varepsilon)C_{\alpha}/p $$ 
$$ u_{\varepsilon}(\varepsilon_0) = \frac{C_{\alpha}}{p\lambda C_{a,r}} = \frac{C_{a,d,p}}{\lambda}. $$

When this is the case, then $C^{\text{in}} := |C_1 M_0^{1/\varepsilon} - C_2| > 0$ and by Gronwall's inequality

$$ \forall t \in \mathbb{R}_+, \|\rho\|_{L^p}^p \le \frac{1}{(\|\rho^{\text{in}}\|_{L^p}^{-pb} + bM_0^{-pb}C^{\text{in}}t)^{\frac{1}{b}}} \le \frac{M_0^p}{(bC^{\text{in}}t)^{\frac{1}{b}}}, $$

which proves (27). When $M_0 > M_0^*$ we only get the existence of $T > 0$ such that

$$ \forall t \in [0, T], \|\rho\|_{L^p}^p \le \frac{1}{(\|\rho^{\text{in}}\|_{L^p}^{-pb} - bM_0^{-pb}C^{\text{in}}t)^{\frac{1}{b}}}. $$

Moreover, $T$ verifies

$$ T > \frac{1}{bC^{\text{in}}} \left( \frac{M_0}{\|\rho^{\text{in}}\|_{L^p}} \right)^{pb}, $$

which proves (28).

□

**Corollary 3.1.** When $a < \alpha$ and $\rho^{\text{in}} \in L^1$, then for any $p < p_\alpha$

$$ (42) \qquad \rho \in L_{\text{loc}}^{1}(\mathbb{R}_{+}, L^{p}), $$

which holds in particular if $p = p_a$. When $a > \alpha$ and $\rho^{\text{in}} \in L^p$ for a given $p \in (p_{a,\alpha}, p_a)$, then there exists $T > 0$ such that

$$ (43) \qquad \rho \in L^p([0, T], L^{\bar{r}}), $$

where $\bar{r} = pp_\alpha \ge p_a$. Moreover, if (24) is verified,

$$ \rho \in L_{\text{loc}}^{p}(\mathbb{R}_{+}, L^{\bar{r}}). $$

**Proof of Corollary 3.1.** Equation (42) comes from inequality (23) by remarking that $p < p_\alpha$ implies $d/(\alpha q) \le 1$ and integrating in time. Equation (43) is a consequence of (36), which by integrating in time leads to

$$ \|\rho\|_{L^p}^p(t) + C_2 \int_0^t \|\rho(s)\|_{L^{\bar{r}}}^p \mathrm{d}s \le \|\rho^{\text{in}}\|_{L^p}^p + C_1 M_0^{\theta_0(p+1)/\varepsilon} \int_0^t \|\rho(s)\|_{L^p}^{p(1+b_0)} \mathrm{d}s. $$

If $\rho \in L^\infty([0, T], L^p)$, then we deduce that

$$ C_2 \int_0^t \|\rho(s)\|_{L^{\bar{r}}}^p \mathrm{d}s \le \|\rho^{\text{in}}\|_{L^p}^p + C_1 M_0^{\theta_0(p+1)/\varepsilon} T \|\rho\|_{L^\infty([0, T], L^p)}^{p(1+b_0)}, $$

and we conclude by using (24) or (25).

□

3.2. **Tightness and coupling.** For the rest of the section we consider some given stochastic basis $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t \ge 0}, \mathbb{P})$. The expectation with respect to $\mathbb{P}$ will be denoted $\mathbb{E}$. We first provide a generalization of [13, Proposition 3.1] in the

**Lemma 3.2.** Let $a \le d, k \ge 1$ and $p \ge p_a$. There exists a constant $C$ depending only on $d, p, a$ such that for any $\rho_1, \rho_2 \in \mathcal{P}_k \cap L^p$ and $X, \bar{X}$ two i.i.d. random variables of law $\rho_1$ (respectively $Y, \bar{Y}$ two i.i.d. of law $\rho_2$), it holds when $p > p_a$

$$ (i) \quad \mathbb{E} [|X - Y|^{k-1} |K(X - \bar{X}) - K(Y - \bar{Y})|] \le CC_{\rho_1, \rho_2} \mathcal{E}_k, $$
and when $p = p_a$,

$$ (ii) \quad \mathbb{E} [|X - Y|^{k-1} |K(X - \bar{X}) - K(Y - \bar{Y})|] \le CC_{\rho_1, \rho_2} \mathcal{E}_k \left(1 + \frac{\ln_-(\mathcal{E}_k)}{k}\right), $$

$$ \text{where } C_{\rho_1, \rho_2} = 1 + \|\rho_1\|_{L^p} + \|\rho_2\|_{L^p} \text{ and } \mathcal{E}_k = \mathbb{E} [|X - Y|^k]. $$

**Remark 3.3.** The point (i) of this Lemma has been extensively used in the literature (See for instance [12, 11, 18, 30]). So has the point (ii) in the Newtonian case $a = d$ and thus $p_a = \infty$ (see for instance [29, 19, 13]). However to the best of the authors' knowledge, its generalization to a general Riesz interaction kernel $a \in (0, d)$ is a novelty. A similar technique can be found in [22]. Since this observation is crucial with respect to the competition between diffusion and aggregation, we detail the proof.

**Proof.** We start with the classical inequality (see [19, (3.9)], [18, Lemma 2.5], [11, (3.26)], [12, (3.5)]) which holds for any $(x, y) \in (\mathbb{R}^d)^2$

$$ |K(x) - K(y)| \le (|\nabla K(x)| + |\nabla K(y)|) |x - y|. $$

Then denote $\pi = \mathcal{L}(X, Y) = \mathcal{L}(\bar{X}, \bar{Y}) \in \mathcal{P}(\mathbb{R}^{2d})$.

**Step 1. Proof of (i).** We assume here that $p > p_a$. Then we have

$$ \begin{aligned} & \mathbb{E} [|X - Y|^{k-1} |K(X - \bar{X}) - K(Y - \bar{Y})|] \\ & \le \mathbb{E} [|X - Y|^{k-1} (|X - Y| + |\bar{X} - \bar{Y}|) (|\nabla K(X - \bar{X})| + |\nabla K(Y - \bar{Y})|)] \\ & := \mathcal{I}_1 + \mathcal{I}_2. \end{aligned} $$

We first estimate $\mathcal{I}_1$. Since $X$ and $\bar{X}$ are independent we get

$$ \begin{aligned} \mathcal{I}_1 &= \mathbb{E} [|X - Y|^k \mathbb{E} [(|\nabla K(X - \bar{X})| + |\nabla K(Y - \bar{Y})|) |X, Y|]] \\ &= \mathbb{E} \left[ |X - Y|^k \left( \iint_{\mathbb{R}^{2d}} (|\nabla K(X - x)| + |\nabla K(Y - y)|) \pi(dx, dy) \right) \right] \\ &= \mathbb{E} \left[ |X - Y|^k \left( \int_{\mathbb{R}^d} |\nabla K(X - x)| \rho_1(x) dx + \int_{\mathbb{R}^d} |\nabla K(Y - y)| \rho_2(y) dy \right) \right]. \end{aligned} $$

But since $|\nabla K| \le C_a |x|^{-a}$ with $C_a = \max(1 - a, a)$, we obtain

$$ \begin{aligned} C_a^{-1} \int_{\mathbb{R}^d} |\nabla K(X - x)| \rho_1(x) dx &\le \int_{\mathbb{R}^d} |X - x|^{-a} \rho_1(x) dx \\ &\le \int_{|X - x| \le r} |X - x|^{-a} \rho_1(x) dx + r^{-a} \|\rho_1\|_{L^1} \\ &\le \|\rho_1\|_{L^p} \left( \int_{|x| < r} |x|^{-aq} dx \right)^{1/q} + r^{-a} \|\rho_1\|_{L^1}, \end{aligned} $$

where $q = p'$ and $r > 0$. Since $p > p_a$, we get $aq < d$ so that $|x|^{-aq}$ is locally integrable and we obtain

$$ C_a^{-1} \int_{\mathbb{R}^d} |\nabla K(X - x)| \rho_1(x) dx \le C_K r^{d/q} \|\rho_1\|_{L^p} + r^{-a} \|\rho_1\|_{L^1}, $$

where $C_K = \left( \frac{\omega_d}{d - aq} \right)^{1/q}$.
**Step 2.** *Proof of* (ii). Note that for any $(x, y) \in (\mathbb{R}^d)^2$ and $r > 0$, it holds

$$|K(x) - K(y)| \le \begin{cases} |K(x)| + |K(y)| & \text{if } |x| \wedge |y| \le r \\ (|\nabla K(x)| + |\nabla K(y)|) |x - y| & \text{else.} \end{cases}$$

So that

$$\begin{aligned} \mathbb{E} [|X - Y|^{k-1} |K(X - \bar{X}) - K(Y - \bar{Y})|] &\le \\ & \mathbb{E} [|X - Y|^{k-1} (|K(X - \bar{X})| + |K(Y - \bar{Y})|) 1_{|X - \bar{X}| \wedge |Y - \bar{Y}| \le r}] \\ & + \mathbb{E} [|X - Y|^{k-1} (|X - Y| + |\bar{X} - \bar{Y}|) \\ & \qquad (|\nabla K(X - \bar{X})| + |\nabla K(Y - \bar{Y})|) 1_{|X - \bar{X}| \wedge |Y - \bar{Y}| > r}] \\ &:= \mathcal{I}_1 + \mathcal{I}_2. \end{aligned}$$

To estimate $\mathcal{I}_1$, we write

$$\begin{aligned} \mathcal{I}_1 &= I_1^1 + I_1^2 + I_1^3 \\ &:= \mathbb{E} \left[ |X - Y|^{k-1} (|K(X - \bar{X})| + |K(Y - \bar{Y})|) 1_{|X - \bar{X}| \vee |Y - \bar{Y}| \le r} \right] \\ &\quad + \mathbb{E} \left[ |X - Y|^{k-1} (|K(X - \bar{X})| + |K(Y - \bar{Y})|) 1_{|X - \bar{X}| > r \ge |Y - \bar{Y}|} \right] \\ &\quad + \mathbb{E} \left[ |X - Y|^{k-1} (|K(X - \bar{X})| + |K(Y - \bar{Y})|) 1_{|X - \bar{X}| \le r < |Y - \bar{Y}|} \right]. \end{aligned}$$

Then, for the estimate of $I_1^1$, we get by independence of $\bar{X}$ and $X$ (resp. $\bar{Y}$ and $Y$)

$$\begin{aligned} I_1^1 &= \mathbb{E} \left[ \mathbb{E} \left[ (|K(X - \bar{X})| + |K(Y - \bar{Y})|) 1_{|X - \bar{X}| \vee |Y - \bar{Y}| \le r} |X, Y| |X - Y|^{k-1} \right] \right] \\ &\le \mathbb{E} \left[ \left( \int_{|X-x| \le r} \frac{\rho_1(x)}{|X-x|^{a-1}} dx + \int_{|Y-y| \le r} \frac{\rho_2(y)}{|Y-y|^{a-1}} dy \right) |X - Y|^{k-1} \right] \\ &\le (\|\rho_1\|_{L^{p_a}} + \|\rho_2\|_{L^{p_a}}) \left( \int_{|z| \le r} |z|^{-(a-1)\frac{d}{a}} dz \right)^{\frac{d}{a}} \mathbb{E} [|X - Y|^{k-1}|. \end{aligned}$$

Since

$$\int_{|z| \le r} |z|^{-(a-1)\frac{d}{a}} dz = \omega_d \int_0^r u^{d-1-(a-1)\frac{d}{a}} ds = \frac{a\omega_d}{d} r^{\frac{d}{a}} =: (C'_{d,a} r)^{\frac{d}{a}},$$

we get

$$I_1^1 \le C'_{d,a} r (\|\rho_1\|_{L^{p_a}} + \|\rho_2\|_{L^{p_a}}) \mathbb{E} [|X - Y|^{k-1}|.$$

For $I_1^2$, we have

$$\begin{aligned} I_1^2 &\le \mathbb{E} \left[ \frac{2}{|Y - \bar{Y}|^{a-1}} 1_{|Y - \bar{Y}| \le r} |X - Y|^{k-1} \right] \\ &= 2 \mathbb{E} \left[ \left( \int_{|Y-y| \le r} \frac{\rho_2(y)}{|Y-y|^{a-1}} dy \right) |X - Y|^{k-1} \right] \\ &\le 2 C'_{d,a} r \|\rho_2\|_{L^{p_a}} \mathbb{E} [|X - Y|^{k-1}|. \end{aligned}$$
We then estimate $I_1^3$ similarly. Combining the above estimates, we obtain

$$
\begin{aligned}
\mathcal{I}_1 &\le 3C'_{d,a}r (\|\rho_1\|_{L^{p_a}} + \|\rho_2\|_{L^{p_a}}) \mathbb{E}[|X - Y|^{k-1}] \\
&\le 3C'_{d,a}r (\|\rho_1\|_{L^{p_a}} + \|\rho_2\|_{L^{p_a}}) \mathbb{E}[|X - Y|^k]^{(k-1)/k}.
\end{aligned}
$$

Next, we estimate $\mathcal{I}_2$ by writing

$$
\begin{aligned}
\mathcal{I}_2 &= C_a(I_2^1 + I_2^2) \\
&:= C_a \mathbb{E} \left[ |X - Y|^k \left( \frac{1}{|X - \bar{X}|^a} + \frac{1}{|Y - \bar{Y}|^a} \right) \mathbf{1}_{|X - \bar{X}| \wedge |Y - \bar{Y}| > r} \right] \\
&\quad + C_a \mathbb{E} \left[ |X - Y|^{k-1} |\bar{X} - \bar{Y}| \left( \frac{1}{|X - \bar{X}|^a} + \frac{1}{|Y - \bar{Y}|^a} \right) \mathbf{1}_{|X - \bar{X}| \wedge |Y - \bar{Y}| > r} \right].
\end{aligned}
$$

First we easily obtain since $\mathbf{1}_{a \wedge b \ge r} = \mathbf{1}_{a \ge r} \mathbf{1}_{b \ge r}$

$$
\begin{aligned}
I_2^1 &= \mathbb{E} \left[ |X - Y|^k \mathbb{E} \left[ \left( \frac{1}{|X - \bar{X}|^a} + \frac{1}{|Y - \bar{Y}|^a} \right) \mathbf{1}_{|X - \bar{X}| \wedge |Y - \bar{Y}| > r} | X, Y \right] \right] \\
&\le \mathbb{E} \left[ |X - Y|^k \left( \int_{|X-x| \ge r} \frac{\rho_1(x)}{|X-x|^a} dx + \int_{|Y-y| \ge r} \frac{\rho_2(y)}{|Y-y|^a} dy \right) \right].
\end{aligned}
$$

We then consider two cases: $r > 1$ and $0 < r \le 1$. For $r \le 1$, we get

$$
\begin{aligned}
\int_{|X-x| \ge r} \frac{\rho_1(x)}{|X-x|^a} dx &= \int_{|X-x| > 1} \frac{\rho_1(x)}{|X-x|^a} dx + \int_{|X-x| \in [r, 1]} \frac{\rho_1(x)}{|X-x|^a} dx \\
&\le \|\rho_1\|_{L^1} + \|\rho_1\|_{L^{p_a}} \left( \int_{|X-x| \in [r, 1]} \frac{1}{|X-x|^d} dx \right)^{\frac{a}{d}} \\
&\le \|\rho_1\|_{L^1} + \omega_d \|\rho_1\|_{L^{p_a}} \ln_-(r)^{\frac{a}{d}} \\
&\le C_d (\|\rho_1\|_{L^{p_a}} + \|\rho_1\|_{L^1}) (1 + \ln_-(r)).
\end{aligned}
$$

For the case $r > 1$, it is clear to obtain

$$
\int_{|X-x| \ge r} \frac{\rho_1(x)}{|X-x|^a} dx \le \|\rho_1\|_{L^1}.
$$

This yields

$$
I_2^1 \le C_d (\|\rho_1\|_{L^{p_a}} + \|\rho_2\|_{L^{p_a}} + 2) \mathbb{E}[|X - Y|^k] (1 + \ln_- r).
$$

On the other hand, by Hölder's inequality

$$
\begin{aligned}
I_2^2 \le & \mathbb{E} \left[ |\bar{X} - \bar{Y}|^k \left( \frac{1}{|\bar{X} - \bar{X}|^a} + \frac{1}{|\bar{Y} - \bar{Y}|^a} \right) \mathbf{1}_{|\bar{X} - \bar{X}| \wedge |\bar{Y} - \bar{Y}| > r} \right]^{1/k} \\
& \times \mathbb{E} \left[ |X - Y|^k \left( \frac{1}{|X - \bar{X}|^a} + \frac{1}{|\bar{Y} - \bar{Y}|^a} \right) \mathbf{1}_{|X - \bar{X}| \wedge |\bar{Y} - \bar{Y}| > r} \right]^{1-1/k}.
\end{aligned}
$$

The second term of the product is some power of the term $I_2^1$ which has already been dealt with, and so is the second term by symmetry of the roles of $(X, Y)$ and $(\bar{X}, \bar{Y})$. So that

$$
\mathcal{I}_2 \le C_{d,a} (\|\rho_1\|_{L^{p_a}} + \|\rho_2\|_{L^{p_a}} + 2) \mathbb{E}[|X - Y|^k] (1 + \ln_-(r)).
$$
Putting all these estimates together yields for any $r > 0$

$$ 
\begin{aligned}
& \mathbb{E}[|X - Y|^{k-1} | K(X - \bar{X}) - K(Y - \bar{Y})|] \\
& \le C'_{d,a} (\|\rho_1\|_{L^{p_a}} + \|\rho_2\|_{L^{p_a}}) r \mathbb{E} [|X - Y|^k|^{1-1/k} \\
& \qquad + C_{d,a} (\|\rho_1\|_{L^{p_a}} + \|\rho_2\|_{L^{p_a}} + 2) \mathbb{E} [|X - Y|^k] (1 + \ln_+ r).
\end{aligned}
$$

Choosing $r = \mathbb{E} [|X - Y|^k|^{1/k}$ yields the desired result. $\square$

<figure><img src="image_7.png" /></figure>

**Proof of Theorem 1.** Let $\rho^{\text{in}}$ be such as the assumptions of Theorem 1. For $\varepsilon > 0$ define

$$ K_{\varepsilon}(x) = \begin{cases} K(x) & \text{if } |x| \ge \varepsilon \\ \varepsilon^{-a}x & \text{else,} \end{cases} $$

and consider the following nonlinear PDE with smooth coefficient

$$ (44) \qquad \partial_t \rho_{\varepsilon} = I(\rho_{\varepsilon}) + \lambda \operatorname{div}((K_{\varepsilon} * \rho_{\varepsilon})\rho_{\varepsilon}) $$

with the initial condition $\rho_{\varepsilon}^{\text{in}} = \rho^{\text{in}}$. Since the kernel $K_{\varepsilon}$ is $(\varepsilon^{-a})$-Lipschitz, the difficulty for the well posedness of (44) does not come from the quadratic nonlinear term. Existence and uniqueness of solution for this nonlinear problem is straightforward in the case $a \in (1, 2)$. Indeed it is sufficient to apply a standard fix point in $C([0, T], \mathcal{P}_k)$ technique using Wasserstein metric, since in this case the solution a priori enjoys some $k \in (1, a)$ moment. In the case $a \in (0, 1]$, it is no more possible to use the completeness of $C([0, T], \mathcal{P}_k)$, $\kappa > 1$, and we have to proceed by compactness (see [30, Appendix B]).

Then due to Proposition 3.2 (if $\alpha = a$), Corollary 3.1 (if $a \neq \alpha$), and Proposition 3.1, $\rho_{\varepsilon} \in L^1([0, T], L^p) \cap L^{\infty}([0, T], L_k^1)$ for some $p \ge p_a$ and $T > 0$ depending or not on $\rho^{\text{in}}$, uniformly w.r.t. $\varepsilon > 0$.

**Step 1. Tightness.** Let $X_0$ be a random variable on $\mathbb{R}^d$ of law $M_0^{-1}\rho^{\text{in}}$ and $(Z_t^{\alpha})_{t \ge 0}$ be an $\alpha$-stable Lévy process independent of $X_0$. We denote by $(X_t^{\varepsilon})_{t \ge 0}$ (respectively $(X_t^{\varepsilon'})_{t \ge 0}$) the solution to the following SDE

$$ X_t^{\varepsilon} = X_0 - \lambda \int_0^t \int_{\mathbb{R}^d} K_{\varepsilon}(X_s^{\varepsilon} - x) \rho_{\varepsilon}(dx) ds + Z_t^{\alpha}. $$

Note that $(\mu_{\varepsilon}(t))_{t \ge 0} := (\mathcal{L}(X_t^{\varepsilon}))_{t \ge 0}$ solves the linear PDE

$$ \partial_t \mu_{\varepsilon} = I(\mu_{\varepsilon}) + \lambda \operatorname{div}((K_{\varepsilon} * \rho_{\varepsilon})\mu_{\varepsilon}), $$

with initial condition $\mu_{\varepsilon}^{\text{in}} = M_0^{-1}\rho^{\text{in}}$. Therefore $\mathcal{L}(X_t^{\varepsilon}) = M_0^{-1}\rho_{\varepsilon}(t)$ by uniqueness of solution to this linear PDE with smooth coefficient.

Assume first $0 < 1 - a < \alpha$. It is direct to obtain in this case for any $\beta > 1$

$$ 
\begin{aligned}
\mathcal{K}_{\varepsilon}^{\beta} &:= \iint_{\mathbb{R}^{2d}} |K_{\varepsilon}(x-y)|^{\beta} \rho_{\varepsilon}(dx) \rho_{\varepsilon}(dy) \\
&\le C_{a,\beta} \iint_{\mathbb{R}^{2d}} (|x-y| + \varepsilon)^{\beta(1-a)} \rho_{\varepsilon}(dx) \rho_{\varepsilon}(dy) \\
&\le C_{a,\beta} \int_{\mathbb{R}^d} (|x|^{(1-a)\beta} + \varepsilon^{(1-a)\beta}) \rho_{\varepsilon}(dx).
\end{aligned}
$$
Then choose $\beta = \frac{k}{1-a} > 1$ and use the symmetry between $x$ and $y$ to get

$$ \sup_{0<\varepsilon<1} \int_0^T \mathcal{K}_\varepsilon^\beta(t) \, dt \le \sup_{0<\varepsilon<1} \int_0^T \iint_{\mathbb{R}^{2d}} C_{a,\beta} \left( |x|^{(1-a)\beta} + \varepsilon^{(1-a)\beta} \right) \rho_\varepsilon(dx) \rho_\varepsilon(dy) \, dt \\ \le C_{a,\beta,T} \left( \sup_{\varepsilon>0} \|\rho_\varepsilon\|_{L^\infty((0,T),L_k^1)} + 1 \right) < \infty. $$

Assume now that $a > 1$. First note that Hardy-Littlewood-Sobolev inequality yields for any $\varepsilon > 0$ and $\beta > 1$ to be fixed later

$$ \mathcal{K}_\varepsilon^\beta \le \iint_{\mathbb{R}^{2d}} |x-y|^{-(a-1)\beta} \rho_\varepsilon(dx) \rho_\varepsilon(dy) \le C \|\rho_\varepsilon\|_{L^{\frac{d}{d+\beta(1-a)/2}}}^2. $$

By interpolation between Lebesgue spaces, if $\beta < \frac{2(p-1)d}{a-1}$, then

$$ \|\rho_\varepsilon\|_{L^{\frac{d}{d+\beta(1-a)/2}}} \le \|\rho_\varepsilon\|_{L^p}^\theta \|\rho_\varepsilon\|_{L^1}^{1-\theta}, $$

where $\theta = \beta \frac{(a-1)q}{2d}$ with $q = p'$. Therefore

$$ \sup_{\varepsilon>0} \int_0^T \mathcal{K}_\varepsilon^\beta(t) \, dt \le \sup_{\varepsilon>0} \int_0^T \|\rho_\varepsilon\|_{L^{\frac{d}{d+\beta(1-a)/2}}}^2 \, dt \\ \le \sup_{\varepsilon>0} \int_0^T \|\rho_\varepsilon\|_{L^p}^{\beta \frac{(a-1)q}{d}} \, dt < \infty, $$

provided that $\beta \in (1, \frac{d}{(a-1)q})$. Then in both cases, denote the stochastic process

$$ J_t^\varepsilon = -\lambda \int_0^t \int_{\mathbb{R}^d} K_\varepsilon(X_s^\varepsilon - x) \rho_\varepsilon(dx) \, ds, $$

and observe that for any $0 \le s < t \le T$, it holds by Holder's inequality

$$ \begin{aligned} |J_t^\varepsilon - J_s^\varepsilon| & \le \left| \int_s^t \int_{\mathbb{R}^d} K_\varepsilon(X_u^\varepsilon - x) \rho_\varepsilon(dx) \, du \right| \\ & \le \int_s^t \int_{\mathbb{R}^d} |K_\varepsilon(X_u^\varepsilon - x)| \rho_\varepsilon(dx) \, du \\ & \le |t-s|^{1/\beta'} \int_0^T \left( \int_{\mathbb{R}^d} |K_\varepsilon(X_u^\varepsilon - x)|^\beta \rho_\varepsilon(dx) \right)^{1/\beta} \, du, \end{aligned} $$

so that by the estimates carried out in the beginning of this step and Jensen's inequality

$$ \begin{aligned} \sup_{0<\varepsilon<1} \mathbb{E} \left[ \sup_{0 \le s < t \le T} \frac{|J_t^\varepsilon - J_s^\varepsilon|}{|t-s|^{1/\beta'}} \right] & \le \int_0^T \mathbb{E} \left[ \left( \int_{\mathbb{R}^d} |K_\varepsilon(X_u^\varepsilon - x)|^\beta \rho_\varepsilon(dx) \right)^{1/\beta} \right] \, du \\ & \le \int_0^T \left( \mathbb{E} \left[ \int_{\mathbb{R}^d} |K_\varepsilon(X_u^\varepsilon - x)|^\beta \rho_\varepsilon(dx) \right] \right)^{1/\beta} \, du \\ & \le T^{1/\beta'} \left( \int_0^T \mathcal{K}_\varepsilon^\beta(t) \, dt \right)^{1/\beta} < \infty. \end{aligned} $$
We then deduce that the family of law of the processes $(J_t^\epsilon)_{t \in [0,T]}$ is tight in $\mathcal{P}(C([0,T], \mathbb{R}^d))$. Indeed let us denote

$$ \mathcal{K}_R := \left\{ f \in C([0, T], \mathbb{R}^d), f(0) = 0, \sup_{0 \le s < t \le T} \frac{|f(t) - f(s)|}{|t - s|^{1/\beta'}} \le R \right\}, $$

which is compact due to Ascoli-Azerla's Theorem. By Markov's inequality we get for any $\epsilon > 0$

$$ \begin{aligned} \mathbb{P}((J_t^\epsilon)_{0 \le t \le T} \notin \mathcal{K}_R) &= \mathbb{P} \left( \sup_{0 \le s < t \le T} \frac{|J_t^\epsilon - J_s^\epsilon|}{|t - s|^{1/\beta'}} > R \right) \\ &\le R^{-1} \sup_{1 > \epsilon > 0} \mathbb{E} \left[ \sup_{0 \le s < t \le T} \frac{|J_t^\epsilon - J_s^\epsilon|}{|t - s|^{1/\beta'}} \right]. \end{aligned} $$

Hence the family of law of the processes $\mathcal{L}^\epsilon = \mathcal{L}((X_t^\epsilon = X_0 + J_t^\epsilon + Z_t^\alpha)_{0 \le t \le T}) \in \mathcal{P}(D([0,T], \mathbb{R}^d))$ is tight. Thus, we can find a sequence $\epsilon_n$ going to 0 such that $\mathcal{L}_{\epsilon_n}$ goes weakly to some $\pi \in \mathcal{P}(D([0,T], \mathbb{R}^d))$. For any $t \in [0,T]$, we define $\mathbf{e}_t : \gamma \in D([0,T], \mathbb{R}^d) \mapsto \gamma(t) \in \mathbb{R}^d$ and $\rho(t) := (\mathbf{e}_t)\#\pi \in \mathcal{P}$ the push-forward of $\rho$ by $\mathbf{e}_t$. Since for any $t \in [0,T]$, $(\mathbf{e}_t)\#\mathcal{L}^\epsilon = \rho_\epsilon(t)$, $\rho_{\epsilon_n}(t)$ goes weakly to $\rho(t)$ in $\mathcal{P}_k$, **Step 2. A priori properties of the limit point.** By lower semi continuity of $\|\cdot\|_{L^p}$ and $\|\cdot\|_{L_k^1}$ with respect to the weak convergence of measures and Fatou's Lemma, it holds $\rho \in L^1([0,T], L^p) \cap L^\infty([0,T], L_k^1)$. We now show that $\rho$ satisfies (6). Indeed for $\varphi \in C_c^2$ denote

$$ \begin{aligned} \mathcal{F}(\rho, t) &= \int_{\mathbb{R}^d} (\rho(t) - \rho^{\text{in}}) \varphi - \int_0^t \int_{\mathbb{R}^d} \rho(s) (I(\varphi) - K_c * (\rho(s) \cdot \nabla \varphi)) \, ds \\ &\quad - \int_0^t \iint_{\mathbb{R}^{2d}} K_0(x-y)(\nabla \varphi(x) - \nabla \varphi(y)) \rho(s, dx) \rho(s, dy) \, ds. \end{aligned} $$

Since $\rho_\epsilon$ solves (44), it holds for any $t > 0$

$$ \mathcal{F}_\epsilon(\rho_\epsilon, t) = 0, $$

where $\mathcal{F}_\epsilon$ is the same functional as $\mathcal{F}$ with $K$ replaced with $K_\epsilon$. So that for any $t \in [0, T]$

$$ |\mathcal{F}(\rho, t)| \le |\mathcal{F}(\rho, t) - \mathcal{F}_\eta(\rho, t)| + |\mathcal{F}_\eta(\rho, t) - \mathcal{F}_\eta(\rho_\epsilon, t)| + |\mathcal{F}_\eta(\rho_\epsilon, t) - \mathcal{F}_\epsilon(\rho_\epsilon, t)|. $$

But note that for $\eta > \epsilon \ge 0$

$$ |K_\epsilon(x) - K_\eta(x)| \le \mathbb{1}_{\epsilon \le |x| \le \eta} |x|^{1-a} \le \eta |x|^{-a}. $$

We deduce that for any $\varrho \in L^1([0,T]; L^{p_a})$, by (3), it holds

$$ \begin{aligned} |\mathcal{F}_\eta(\varrho, t) - \mathcal{F}_\epsilon(\varrho, t)| &\le \eta \int_0^t \iint |x - y|^{-a} \varrho_s(dx) \varrho_s(dy) \, ds \\ &\le \eta C_{d,a,p_a/2}^{\text{HLS}} \int_0^t \|\varrho\|_{L^{\frac{2d}{2d-a}}}^2 \, ds \le \eta C_{d,a,p_a/2}^{\text{HLS}} \int_0^t \|\varrho\|_{L^{p_a}} \, ds. \end{aligned} $$

So that

$$ \begin{aligned} |\mathcal{F}(\rho, t)| &\le \eta C_{d,a,p_a/2}^{\text{HLS}} \left( \int_0^t \|\rho\|_{L^{p_a}} \, ds + \sup_{0 < \epsilon < 1} \int_0^t \|\rho_\epsilon\|_{L^{p_a}} \, ds \right) \\ &\quad + |\mathcal{F}_\eta(\rho, t) - \mathcal{F}_\eta(\rho_\epsilon, t)|. \end{aligned} $$

Letting first $\epsilon$ go to 0 makes the second term in the r.h.s. vanishes, since for fixed $\eta > 0$, $\mathcal{F}_\eta$ is a smooth function on $L^1([0,T]; L^{p_a})$ and $\rho_\epsilon$ goes weakly to $\rho$ as $\epsilon$ goes
to 0, then letting $\eta$ go to 0 yields $\mathcal{F}(\rho, t) = 0$, and $\rho$ is a solution to (FKS) in the sense of Definition 2.1.

**Step 3. Uniqueness of the limiting point.** We now show that there exists at most one such solution. Let $\rho, \tilde{\rho} \in L^1([0, T], L^p) \cap L^\infty([0, T], L_k^1)$ for some $p \ge p_a$ and $T > 0$ be two solutions to the (FKS) equation with initial condition $\rho^{\text{in}}$. We argue by a coupling argument. Define

$$X_t := X_0 - \lambda \int_0^t \int_{\mathbb{R}^d} K(X_s - y)\rho(dy) \, ds + Z_t^\alpha$$
$$Y_t := X_0 - \lambda \int_0^t \int_{\mathbb{R}^d} K(Y_s - y)\tilde{\rho}(dy) \, ds + Z_t^\alpha.$$

Due to the $L^p$ regularity of $\rho$ and $\tilde{\rho}$ and Lemma 3.2, $K * \rho$ and $K * \tilde{\rho}$ are Lipschitz if $p > p_a$ and log-Lipschitz if $p = p_a$. But $\mu(t) := \mathcal{L}(X_t)$ solves the linear PDE

$$\partial_t \mu = I(\mu) + \lambda \operatorname{div}((K * \rho)\mu),$$

for the initial condition $\mu(0) = M_0^{-1}\rho^{\text{in}}$. By uniqueness of solution to this linear PDE with Lipschitz or log-Lipschitz coefficient, $\mathcal{L}(X_t) = M_0^{-1}\rho(t)$ (respectively $\mathcal{L}(Y_t) = M_0^{-1}\tilde{\rho}(t)$). Denoting $Z_s = X_s - Y_s$, and $\pi_s = \mathcal{L}(X_s, Y_s)$ yields

$$|Z_t|^2 = -2\lambda \int_0^t \iint_{\mathbb{R}^{2d}} Z_s \cdot (K(X_s - x) - K(Y_s - y))\pi_s(dx, dy) \, ds.$$

Introducing $\bar{X}_s$ i.i.d. from $X_s$ (resp. $\bar{Y}_s$ i.i.d. from $Y_s$) and taking the expectation yields

$$\mathbb{E}[|Z_t|^2] \le 2\lambda \int_0^t \mathbb{E}[|Z_s||K(X_s - \bar{X}_s) - K(Y_s - \bar{Y}_s)|] \, ds \\
\le \begin{cases}
C \int_0^t (\|\rho\|_{L^p} + \|\tilde{\rho}\|_{L^p} + 2) \mathbb{E}[|Z_s|^2] \, ds, & \text{if } p > p_a \\
C \int_0^t (\|\rho\|_{L^{p_a}} + \|\tilde{\rho}\|_{L^{p_a}} + 2) \mathbb{E}[|Z_s|^2] \left(1 + \frac{\ln^{-}(\mathbb{E}[|Z_s|^2])}{2}\right) \, ds & \text{else.}
\end{cases}$$

where we used Lemma 3.2. By Gronwall's inequality, we get

$$\forall t \in [0, T], \mathbb{E}[|Z_t|^2] = 0, \text{ i.e. } \forall t \in [0, T], \rho(t) = \tilde{\rho}(t),$$

which yields the desired results. $\square$

<figure><img src="image_8.png" /></figure>

## 4. PROOF OF THEOREM 3

### 4.1. Fractional Laplacian of truncated polynomials.

**Proposition 4.1.** Let $\varphi \in C_c^\infty$ be such that $\int_{\mathbb{R}^d} \varphi = 1$. Then for any $a > \alpha$

$$|I(|x|^a \varphi)| \le C \langle x \rangle^{-(d+\alpha)} \quad (45)$$

**Proof.** Let $\varphi_a := |x|^a \varphi$ and $R > 0$ be such that $\text{supp}(\varphi) \subset B_R$. For any $x \in B_R^c$, we get

$$I(\varphi_a)(x) = \int_{B_R} \frac{\varphi_a(y)}{|x-y|^{d+\alpha}} dy \in \left( \frac{m_{\varphi_a}}{(|x|+R)^{d+\alpha}}, \frac{m_{\varphi_a}}{(|x|-R)^{d+\alpha}} \right). \quad (46)$$

Now, assume $x \in B_R$ for a given $r > R$. Then we write the fractional Laplacian as

$$I(\varphi_a) = \int_{\mathbb{R}^d} h_{\alpha,a}(y) dy,$$
where

$$h_{\alpha,a}(y) = \frac{\varphi_a(y) - \varphi_a(x)}{|x - y|^{d+\alpha}} \qquad \text{when } \alpha \in (0, 1)$$

$$h_{\alpha,a}(y) = \frac{\varphi_a(y) - \varphi_a(x) - (y-x) \cdot \nabla \varphi_a(x)}{|x-y|^{d+\alpha}} \quad \text{when } \alpha \in [1, 2).$$

Then since $\varphi_a \in W^{a,\infty}$, we obtain that $h_{\alpha,a}(y) \le C|x-y|^{-d+a-\alpha}$, which, since $a > \alpha$, implies that $h_{\alpha,a} \in L_{\text{loc}}^1$. Moreover, when $|y| > r$, then

$$h_{\alpha,a}(y) \le \frac{C_{\varphi} r^{\alpha}}{(|y| - r)^{d+\alpha}} \in L^1(B_r^c).$$

Therefore, $h_{\alpha,a} \in L^1$ uniformly in $x \in B_r$. Hence $I(\varphi_a) \in L^\infty(B_r)$, which, combined with (46), leads to the expected result. □

**Proposition 4.2.** Let $\varphi \in C_c^\infty$ be such that $\int_{\mathbb{R}^d} \varphi = 1$ and $1_{B_r} \le \varphi \le 1_{B_{2r}}$. Then for any $k \in (0, \alpha)$

$$(47) \qquad |I(|x|^k \varphi^c)| \le C \langle x \rangle^{k-\alpha},$$

where $\varphi^c = 1 - \varphi$.

**Proof.** The proof is similar to [4, Remark 4.2] for $k > 1$ and [21, Proposition 2.2] for $k < 1$. □

## 4.2. Blow-up for even solutions.

**Proposition 4.3.** Assume $0 < k < \alpha < a$ and let $\rho \in L^\infty((0,T), L_{(1-a)_+}^1)$ be an even solution of the (FKS) equation with initial condition $\rho^{\text{in}}$ and verifying

$$(48) \qquad \left( \int_{\mathbb{R}^d} \rho^{\text{in}} \langle x \rangle^k \right)^{2\left(\frac{a}{k}-1\right)} < C^* \lambda M_0^{\frac{2a}{k}-1},$$

for a universal constant $C^*$ depending only on $d, a, \alpha$ and $k$. Then the solution ceases to exist in finite time.

**Proof.** Let $\varphi \in C_c^\infty(\mathbb{R})$ even and nonincreasing be such that $\int_{\mathbb{R}} \varphi = 1$ and $1_{B_r} \le \varphi \le 1_{B_{2r}}$ for a given $r \in (0, 1/2)$ and $\varphi^c = 1 - \varphi$. We define

$$m(x) := \varphi(|x|)|x|^a + \varphi^c(|x|)|x|^k.$$

By (45) and (47), we get

$$(49) \qquad I(m) \le C \langle x \rangle^{k-\alpha}.$$

Assuming the existence of $\rho \in L^\infty((0,T), L_k^1)$ to the (FKS) equation, we get

$$\begin{align*} \frac{d}{dt} \left( \int_{\mathbb{R}^d} \rho m \right) &= \int_{\mathbb{R}^d} \rho I(m) - \lambda \iint_{\mathbb{R}^{2d}} \frac{(\nabla m(x) - \nabla m(y)) \cdot (x-y)}{|x-y|^a} \rho(dx)\rho(dy) \\ &= \int_{\mathbb{R}^d} \rho I(m) - \lambda \iint_{\mathbb{R}^{2d}} \frac{g(x,y) - h(x,y)x \cdot y}{|x-y|^a} \rho(dx)\rho(dy), \end{align*}$$

where $m'(|x|) = \nabla m(x) \cdot \frac{x}{|x|}$ and

$$g(x,y) = m'(|x|)|x| + m'(|y|)|y|$$

$$h(x,y) = m'(|x|)|x|^{-1} + m'(|y|)|y|^{-1}.$$
Since $|x - y|^a \le 2^a(|x|^a + |y|^a)$, $h$ is symmetric and $\rho$ is even, we get

$$ (50) \qquad \iint_{\mathbb{R}^{2d}} \frac{g(x, y)}{|x - y|^a} (1 - h(x, y)x \cdot y) \rho(dx)\rho(dy) \ge \iint_{\mathbb{R}^{2d}} \frac{g(x, y) - h(x, y)x \cdot y}{2^a(|x|^a + |y|^a)} \rho(dx)\rho(dy) \\ \qquad \ge \iint_{\mathbb{R}^{2d}} \frac{g(x, y)}{2^a(|x|^a + |y|^a)} \rho(dx)\rho(dy). $$

We remark that if $(x, y) \in B_r^2$,

$$ \frac{g(x, y)}{2^a(|x|^a + |y|^a)} = \frac{a}{2^a}. $$

If $(x, y) \in (B_{2r}^c)^2$,

$$ \frac{g(x, y)}{2^a(|x|^a + |y|^a)} = \frac{k(|x|^k + |y|^k)}{2^a(|x|^a + |y|^a)} \ge \frac{k(2r)^{a-k}}{2^a(|x||y|)^{a-k}}. $$

If $(x, y) \in B_r \times B_{2r}^c$,

$$ \frac{g(x, y)}{2^a(|x|^a + |y|^a)} = \frac{a|x|^a + k|y|^k}{2^a(|x|^a + |y|^a)} \ge \frac{k|y|^k}{2^a(r + |y|^a)}. $$

Moreover, when $x \in B_{2r} \setminus B_r$,

$$ m'(|x|)|x| = \varphi'(|x|)(|x|^{a+1} - |x|^{k+1}) + a\varphi(|x|)|x|^a + k\varphi^c(|x|)|x|^k. $$

Remarking that we can take $\varphi$ decreasing and $r < 1/2$, which implies that $|x| \le 1$ and

$$ m'(|x|)|x| \ge a\varphi(|x|)|x|^a + k\varphi^c(|x|)|x|^k \ge k|x|^a, $$

it allows us to do the same kind of estimates for the remaining $(x, y) \in \mathbb{R}^{2d}$ and obtain

$$ (51) \qquad \frac{g(x, y)}{2^a(|x|^a + |y|^a)} \ge C \langle x \rangle^{k-a} \langle y \rangle^{k-a}. $$

Combining (49), (50) and (51), we obtain

$$ (52) \qquad \begin{aligned} \frac{d}{dt} \left( \int_{\mathbb{R}^d} \rho m \right) &\le C_1 \int_{\mathbb{R}^d} \rho \langle x \rangle^{k-\alpha} - C_2 \lambda \iint_{\mathbb{R}^{2d}} \langle x \rangle^{k-\alpha} \langle y \rangle^{k-\alpha} \rho(dx)\rho(dy) \\ &\le C_1 M_{k-\alpha} - C_2 \lambda M_{k-\alpha}^2, \end{aligned} $$

where $M_k = \int_{\mathbb{R}^d} \rho \langle x \rangle^k$. We define

$$ Y := M_0 + \int_{\mathbb{R}^d} \rho m = \int_{\mathbb{R}^d} \rho(1 + m). $$

Remarking that

$$ \frac{1}{2}(1 + m) \le \langle x \rangle^k \le 2^{k/2}(1 + m), $$

we obtain that $Y$ can always be compared to $M_k$ up to a constant depending on $k$. Therefore, Hölder's inequality yields

$$ M_0 \le M_{k-a}^{\frac{k}{a}} M_k^{1-\frac{k}{a}} \le C M_{k-a}^{\frac{k}{a}} Y^{1-\frac{k}{a}}. $$

Thus, using the fact that $M_{k-\alpha} < M_0$ because $k - \alpha < 0$ and the conservation of the mass $M_0$, we obtain

$$ \frac{dY}{dt} \le C_1 M_0 - C_2' \lambda M_0^{\frac{2\alpha}{k}} Y^{2(1-\frac{\alpha}{k})}. $$
By assumption (48) for the appropriate $C^*$,

$$ \varepsilon := C_2 \left( 1 - \frac{C_1 Y^{2(\frac{a}{k}-1)}(0)}{C_2 \lambda M_0^{\frac{2a}{k}-1}} \right) > 0. $$

Then for any $t \ge 0$, $\frac{dY}{dt} \le 0$ and

$$ Y^{2(\frac{a}{k}-1)}(t) \le Y^{2(\frac{a}{k}-1)}(0) = \frac{C_2 - \varepsilon}{C_1} \lambda M_0^{\frac{2a}{k}-1}, $$

and

$$ \frac{dY}{dt} \le -\varepsilon \lambda M_0^{\frac{2a}{k}} Y^{2(1-\frac{a}{k})}. $$

By Gronwall's inequality, we deduce

$$ Y(t) \le \left( Y(0)^{\frac{2a}{k}-1} - \varepsilon \lambda \left( \frac{2a}{k} - 1 \right) M_0^{\frac{2a}{k}} t \right)^{\frac{k}{2a-k}}. $$

Since $Y$ is positive and the above inequality goes to 0 in finite time, we deduce that the solution ceases to be well defined in $L^1$ in a finite time $T^*$ verifying

$$ T^* < \frac{k Y(0)^{\frac{2a}{k}-1}}{\varepsilon \lambda(2a-k) M_0^{\frac{2a}{k}}} = \frac{k Y(0)^{\frac{2a}{k}-1}}{2a-k \left( C_2 \lambda M_0^{\frac{2a}{k}} - C_1 Y^{2(\frac{a}{k}-1)}(0) \right) M_0}. \quad \square $$

## ACKNOWLEDGEMENTS

The second author was supported by the Fondation des Sciences Mathématiques de Paris and Paris Sciences & Lettres Université.

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LAURENT LAFLECHE

UNIVERSITÉ PARIS-DAUPHINE, PSL RESEARCH UNIVERSITY
CEREMADE, UMR CNRS 7534
PLACE DU MARÉCHAL DE LATTRE DE TASSIGNY
75775 PARIS CEDEX 16 FRANCE

E-MAIL: lafleche@ceremade.dauphine.fr

SAMIR SALEM

UNIVERSITÉ PARIS-DAUPHINE, PSL RESEARCH UNIVERSITY
CEREMADE, UMR CNRS 7534
PLACE DU MARÉCHAL DE LATTRE DE TASSIGNY
75775 PARIS CEDEX 16 FRANCE

E-MAIL: salem@ceremade.dauphine.fr