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# Mass concentration in a spatially inhomogeneous coagulation model with fast sedimentation

Iulia Cristian, Juan J. L. Velázquez

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## To cite this version:

Iulia Cristian, Juan J. L. Velázquez. Mass concentration in a spatially inhomogeneous coagulation model with fast sedimentation. 2025. hal-05298152

HAL Id: hal-05298152

https://hal.science/hal-05298152v1

Preprint submitted on 5 Oct 2025

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Distributed under a Creative Commons Attribution 4.0 International License
<table>
  <tbody>
    <tr>
      <td><b>3</b></td>
      <td><b>Region</b> y ≥ (<sup>v</sup>&frasl;<sub>3</sub>)<sup>α</sup></td>
      <td><b>26</b></td>
    </tr>
    <tr>
      <td></td>
      <td>3.1 Continuity argument</td>
      <td>26</td>
    </tr>
    <tr>
      <td></td>
      <td>3.2 Iteration argument</td>
      <td>28</td>
    </tr>
    <tr>
      <td></td>
      <td>3.3 Proof of Proposition 3.9</td>
      <td>33</td>
    </tr>
    <tr>
      <td></td>
      <td>3.4 Proof of Proposition 3.10</td>
      <td>42</td>
    </tr>
    <tr>
      <td><b>4</b></td>
      <td><b>Region</b> y ≤ (<sup>v</sup>&frasl;<sub>3</sub>)<sup>α</sup></td>
      <td><b>50</b></td>
    </tr>
    <tr>
      <td></td>
      <td>4.1 Moment estimates</td>
      <td>50</td>
    </tr>
    <tr>
      <td></td>
      <td>4.2 Properties of the characteristics of the equation satisfied by the supersolution</td>
      <td>54</td>
    </tr>
    <tr>
      <td></td>
      <td>4.3 Properties of the supersolution in the region y ≤ (<sup>v</sup>&frasl;<sub>3</sub>)<sup>α</sup></td>
      <td>57</td>
    </tr>
    <tr>
      <td></td>
      <td>4.4 Proof of existence of a supersolution in the region y ≤ (<sup>v</sup>&frasl;<sub>3</sub>)<sup>α</sup></td>
      <td>63</td>
    </tr>
    <tr>
      <td><b>5</b></td>
      <td><b>Proof of Theorem 2.7 - Existence</b></td>
      <td><b>66</b></td>
    </tr>
    <tr>
      <td><b>6</b></td>
      <td><b>Proof of Theorem 2.12 - Convergence to a Dirac measure</b></td>
      <td><b>73</b></td>
    </tr>
    <tr>
      <td><b>A</b></td>
      <td><b>Rescaled version of the results</b></td>
      <td><b>79</b></td>
    </tr>
    <tr>
      <td><b>B</b></td>
      <td><b>Construction of solutions with compact support</b></td>
      <td><b>80</b></td>
    </tr>
    <tr>
      <td></td>
      <td><b>References</b></td>
      <td><b>80</b></td>
    </tr>
  </tbody>
</table>

# 1 Introduction

## 1.1 Background

It is known that gelation (loss of mass) occurs for the standard one-dimensional coagulation model,

$$ \partial_{\tau} f(v, \tau) = \frac{1}{2} \int_{(0,v)} K(v-w, w) f(v-w, \tau) f(w, \tau) dw - \int_{(0,\infty)} K(v, w) f(v, \tau) f(w, \tau) dw, $$

when the coagulation kernel behaves like a power law of homogeneity $\gamma > 1$, see [13, 15, 18, 21]. In addition, for sum kernels of the form

$$ K(v, w) = v^{\gamma} + w^{\gamma}, \quad \gamma > 1, \qquad (1.1) $$

gelation happens instantaneously and one can prove that solutions belonging to $L^1$ do not exist, see [1, 2, 6, 12]. Nonetheless, there are some results concerning the existence of solutions for sum-type kernels as in (1.1) in the spatially inhomogeneous case,

$$ \begin{aligned} \partial_{\tau} f(x, v, \tau) + v^{\alpha} \partial_{x} f(x, v, \tau) &= \frac{1}{2} \int_{(0,v)} K(v-w, w) f(x, v-w, \tau) f(x, w, \tau) dw \\ &\quad - \int_{(0,\infty)} K(v, w) f(x, v, \tau) f(x, w, \tau) dw \quad (1.2) \\ &=: \mathbb{K}_1[f](x, v, \tau) - \mathbb{K}_2[f](x, v, \tau) =: \mathbb{K}[f](x, v, \tau), \quad x \in \mathbb{R}. \end{aligned} $$
More precisely, when $\gamma < \alpha + 1, \alpha \ge 0$, existence of solutions (which may not preserve mass) for a discrete version of (1.2) with a coagulation kernel as in (1.1) was proved in [19], while existence of mass-conserving solutions for short times for the model (1.2) when $\gamma < \alpha + 1, \alpha \in (0, 1)$ was proved in [9]. It is also worth mentioning that the relevance of the model in (1.2) with coagulation kernels of homogeneity $\gamma > 1$ stems from the fact that it was first introduced in order to describe the formation of rain or the behavior of air bubbles in water, see [17, 20, 27, 29]. In this case, the coagulation kernel has the form

$$K(v, w) = |v^{\alpha} - w^{\alpha}|(v^{\frac{1}{3}} + w^{\frac{1}{3}})^2. \qquad (1.3)$$

The main idea in [9] in order to prove existence of mass-conserving solutions for the model (1.2) is that the coagulation operator $\mathbb{K}[f]$ gives a negligible contribution for sufficiently small times. It is thus natural to ask what is the time range for which mass-conserving solutions exist if we work with a model of the form

$$\partial_{\tau} f_{\epsilon}(x, v, \tau) + v^{\alpha} \partial_{x} f_{\epsilon}(x, v, \tau) = \epsilon \mathbb{K}[f_{\epsilon}](x, v, \tau), \qquad (1.4)$$

for some sufficiently small $\epsilon \in (0, 1)$ and when $K$ is as in (1.1) or (1.3). We expect that this modification of our model will offer clues on how to prove global in time existence of mass-conserving solutions. The main goal of this paper is thus to prove that mass-conserving solutions of (1.4) exist for times of order $O(e^{\frac{1}{\epsilon}})$. While kernels of homogeneity $\gamma \le 1$ are well-studied in the mathematical literature, see for example [14, 16, 28], this is the first result of existence of mass-conserving solutions involving kernels of homogeneity $\gamma > 1$, with the exceptions being the product kernel [25, 31] and our previous result in [9].

Notice also that since we can rescale the time $\tau \to \epsilon\tau$, (1.4) is equivalent to the following form.

$$\partial_{\tau} f_{\epsilon}(x, v, \tau) + \frac{1}{\epsilon} v^{\alpha} \partial_{x} f_{\epsilon}(x, v, \tau) = \mathbb{K}[f_{\epsilon}](x, v, \tau). \qquad (1.5)$$

## 1.2 Connection with coagulation models with diagonal kernel

The first question is whether the proof in [9] can be reproduced in order to extend the existence result for the model (1.4). It was argued in [8, Section 6.4] that the steps to tackle local in time existence cannot be reproduced in order to prove global existence of solutions (or even existence for times of order one) for the model (1.4). As such, a new strategy needs to be adopted.

Different forms of solutions for the model (1.4) have then been considered in [8, Section 6.5] (such as polynomial decaying solutions, Dirac measure solutions, and Dirac measure solutions with sufficiently fast decay in the $v$ variable) and it has been noticed at a formal level that the solutions can be expected to portray a self-similar behavior of the form $f_{\epsilon}(x, v, \tau) \approx \frac{1}{\tau} S_{\epsilon}(\frac{x}{\tau}, v)$ as $\tau \to \infty$.

Motivated by the computations in [8], we make the following rescaling for solutions of (1.2).

$$f(x, v, \tau) = \frac{1}{\tau + 1} H(y, v, t), \quad y := \frac{x}{\tau + 1}, \quad t := \ln(\tau + 1), \qquad (1.6)$$
or, in other words, we have that

$$H(y, v, t) = e^t f(x, v, \tau), \quad x := e^t y, \quad \tau := e^t - 1. \qquad (1.7)$$

Then $H$ satisfies

$$\partial_t H(y, v, t) + \partial_y[(v^\alpha - y)H(y, v, t)] = \mathbb{K}[H](y, v, t). \qquad (1.8)$$

Since the rain kernel (1.3) vanishes on the diagonal $\{v = w\}$, it is straightforward to check that a steady state of the model (1.8) is given by $G(v)\delta(y - v^\alpha)$.

**Ansatz 1.1.** We thus make the following ansatz in the case when the coagulation kernel in (1.8) is $K_\epsilon(v, w) = \epsilon(v^\gamma + w^\gamma)$, for some $\epsilon < 1$ small. For large times, solutions of the model (1.8) can be expected to concentrate around $|y - v^\alpha| \ll 1$. Notice however that $G(v)\delta(y - v^\alpha)$ are not steady states of (1.8) in this case. Nonetheless, our assumptions are motivated by the fact that $K_\epsilon$ and the rain kernel (1.3) both behave like sum-type kernels with a small contribution on the diagonal $\{v = v'\}$.

We thus formally assume $H$ is of the form

$$H(y, v, t) = G(v, t)\delta(y - v^\alpha). \qquad (1.9)$$

We integrate the coagulation operator $\mathbb{K}$ in the $y$ variable. For the loss term $\mathbb{K}_2$ in (1.2), we have

$$\begin{aligned}
\int_{\mathbb{R}} \mathbb{K}_2[H](y, v, t)dy &= \int_{\mathbb{R}} \int_{(0,\infty)} K(v, w)G(v, t)\delta(y - v^\alpha)G(w, t)\delta(y - w^\alpha)dw dy \\
&= \int_{(0,\infty)} K(v, w)G(v, t)G(w, t)\delta(w^\alpha - v^\alpha)dw \\
&= \frac{v^{1-\alpha}}{\alpha} \int_{(0,\infty)} (v^\gamma + w^\gamma)G(v, t)G(w, t)\delta(w - v)dw \\
&= \frac{2}{\alpha} v^{\gamma+1-\alpha} (G(v, t))^2.
\end{aligned}$$

For the gain term $\mathbb{K}_1$ in (1.2), it holds that

$$\begin{aligned}
\int_{\mathbb{R}} \mathbb{K}_1[H](y, v, t)dy &= \frac{1}{2} \int_{\mathbb{R}} \int_{(0,v)} K(v-w, w)G(v-w, t)\delta(y-(v-w)^\alpha)G(w, t)\delta(y-w^\alpha)dw dy \\
&= \frac{1}{2} \int_{(0,v)} K(v-w, w)G(v-w, t)G(w, t)\delta((v-w)^\alpha - w^\alpha)dw \\
&= \frac{1}{4\alpha} \left(\frac{v}{2}\right)^{1-\alpha} \int_{(0,v)} ((v-w)^\gamma + w^\gamma)G(v-w, t)G(w, t)\delta(w-\frac{v}{2})dw \\
&= \frac{1}{2\alpha} \left(\frac{v}{2}\right)^{\gamma+1-\alpha} \left(G\left(\frac{v}{2}, t\right)\right)^2.
\end{aligned} \qquad (1.10)$$

Thus, $G$ satisfies the equation

$$\partial_t G(v, t) = \frac{2}{\alpha} \left[ \left( \frac{v}{2} \right)^{\gamma+1-\alpha} \left( G\left( \frac{v}{2}, t \right) \right)^2 - v^{\gamma+1-\alpha} (G(v, t))^2 \right]. \qquad (1.11)$$
Remark 1.2. Thus, G satisfies a coagulation equation in which the coagulation kernel is a diagonal kernel of the form $v^{\gamma+1-\alpha}\delta(v-w)$ multiplied by some constant. Notice that this means the kernel has homogeneity $\gamma - \alpha$ and thus the existence of solutions for this type of equation is well-known and has been proved in [23, 24, 26] when $\gamma - \alpha < 1$.

## 1.3 Self-consistent argument for behavior of solutions as $t \to \infty$

Equation (1.11) suggests (at least formally) that the function $G(v, t) := \int H(y, v, t)dy$, with $H$ as in (1.8), satisfies a one-dimensional coagulation model with a diagonal kernel of homogeneity $\gamma - \alpha < 1$. We denote by $[\cdot]$ the dimensional properties of a quantity. We analyze the scaling properties of mass-conserving solutions to (1.11). Mass-conservation $\int vG(v, t)dv \equiv \text{constant}$ implies that

$$[G][v]^2 = 1. \qquad (1.12)$$

From (1.11), we further deduce that

$$\frac{[G]}{[t]} = [v]^{\gamma-\alpha+1}[G]^2. \qquad (1.13)$$

From (1.12) and (1.13), we have that

$$[G] = \frac{1}{[v]^{\gamma-\alpha+1}[t]} \quad \text{and} \quad [v] = [t]^{\frac{1}{1-(\gamma-\alpha)}}.$$

With Ansatz 1.1, we have that $y$ should rescale as $v^\alpha$ and thus

$$[y] = [v]^\alpha = [t]^{\frac{\alpha}{1-(\gamma-\alpha)}}.$$

Since we also want mass-conservation $\iint vH(y, v, \tau)dvdy \equiv \text{constant}$, then the long-time behavior of solutions $H$ in (1.8) is described by $g$, with $g$ having the following form.

$$H(y, v, t) = \frac{1}{\tau^{\frac{2+\alpha}{1-(\gamma-\alpha)}}} g(Y, V, s), \quad \text{with} \quad Y := \frac{y}{t^{\frac{\alpha}{1-(\gamma-\alpha)}}}, \quad V := \frac{v}{t^{\frac{1}{1-(\gamma-\alpha)}}}, \quad s = \ln t.$$

Denoting by $\beta := \frac{1}{1-(\gamma-\alpha)}$, we obtain that $g$ satisfies

$$\frac{1}{\tau^{\beta(2+\alpha)+1}} [\partial_s g(Y, V, s) - \beta(2+\alpha)g - \beta V \partial_V g - \alpha \beta Y \partial_Y g] + \tau^{-\beta(2+\alpha)} \partial_Y [(V^\alpha - Y)g] = \frac{\tau^{\beta(\gamma+1)}}{\tau^{2\beta(2+\alpha)}} \mathbb{K}[g]$$

or alternatively since $\beta(\gamma - \alpha - 1) + 1 = 0$, it holds that

$$\partial_s g(Y, V, s) - \beta(2 + \alpha)g - \beta V \partial_V g - \alpha \beta Y \partial_Y g + e^s \partial_Y [(V^\alpha - Y)g] = \mathbb{K}[g](Y, V, s),$$

where we remember that we are interested in the behavior for large times.

Notice that the transport term $e^s \partial_Y [(V^\alpha - Y)g]$ can be expected to give a large contribution as $s \to \infty$. This justifies the study of the equation

$$\partial_t H_\epsilon(x, v, t) + \frac{1}{\epsilon} \partial_x [(v^\alpha - x)H_\epsilon(x, v, t)] = \mathbb{K}[H_\epsilon](x, v, t)$$

as a first natural step in order to study the long-time behavior of solutions to (1.2).
## 1.4 Connection with previous results for spatially inhomogeneous coagulation models

Under these conditions, the natural approach in order to extend the existence result of mass-conserving solutions is thus to analyze the following rescaled version of the model (1.4). Let $f_\epsilon$ be a solution of (1.4). Take $H_\epsilon$ to be such that

$$f_{\epsilon}(x, v, \tau) = \frac{1}{\tau + 1} H_{\epsilon}(y, v, t), \quad y := \frac{x}{\tau + 1}, \quad t := \epsilon \ln(\tau + 1), \qquad (1.14)$$

or, in other words, we have that

$$H_{\epsilon}(y, v, t) = e^{\frac{t}{\epsilon}} f_{\epsilon}(x, v, \tau), \quad x := e^{\frac{t}{\epsilon}} y, \quad \tau := e^{\frac{t}{\epsilon}} - 1. \qquad (1.15)$$

Then $H_\epsilon$ satisfies

$$\partial_t H_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y [(v^\alpha - y)H_\epsilon(y, v, t)] = \mathbb{K}[H_\epsilon](y, v, t). \qquad (1.16)$$

We analyze equation (1.16) in more detail. We first remember the existence result for mass-conserving solutions in [9].

**Theorem 1.3** (Local existence of mass-conserving solutions for small times). Let $K$ be as in (1.1) or (1.3) with $\alpha \in (0, 1)$, $\gamma \in [0, 1+\alpha)$. Let $T > 0$ be sufficiently small. Let $f_{\text{in}} \in C^1(\mathbb{R} \times \mathbb{R}_{>0})$ such that

$$f_{\text{in}}(x, v) \le \frac{A}{1 + |x|^m + v^p},$$

for some $A > 0$, $x \in \mathbb{R}$, $v \in (0, \infty)$, and where $m = \alpha p$, $m$ even. Then there exists a mass-conserving solution $f$ for the equation (1.4).

In our case, this translates to

<table><caption>**Table 1** Connection between the two equations</caption><thead><tr><th>$f_\epsilon$ satisfying (1.4)</th><th>$H_\epsilon$ satisfying (1.16)</th></tr></thead><tbody><tr><td>existence for $\tau \in [0, T]$, with $T$ small</td><td>existence for $t \in [0, \epsilon \ln(T+1)]$, with $T$ small</td></tr></tbody></table>

**Remark 1.4.** Notice that the existence from the previous paper translates for our equation (1.16) to a time of existence of order $\epsilon$. This means that as $\epsilon \to 0$ (so in the limit) the time of existence goes to zero. Thus, we prove in this paper existence of mass-conserving solutions for equation (1.16) for a time of order one which is independent of $\epsilon$. This will imply existence of mass-conserving solutions for times of order $O(e^{\frac{1}{\epsilon}})$ to the equation (1.4).

There is a simple heuristic argument that explains why mass-conserving solutions to equation (1.16) should exist. Assuming that the coagulation operator gives a negligible contribution, we look at the following PDE (for some suitable test function $\varphi$).

$$\partial_t \int H_\epsilon(y, v, t)\varphi(y, v, t)dydv - \frac{1}{\epsilon} \int (v^\alpha - y)H_\epsilon(y, v, t)\partial_y\varphi(y, v, t)dydv = 0, \\ H_\epsilon(y, v, 0) = H_{\text{in}}(y, v),$$
for which we know the limiting behavior (in some weak sense)

$$ \lim_{\epsilon \to 0} H_{\epsilon}(y, v, t) = H_{\text{in}}(v^{\alpha}, v). \qquad (1.17) $$

We can thus assume formally that solutions to (1.16) converge to a Dirac measure. More precisely, let us denote the limit of $H_{\epsilon}$ (in whichever sense and if it exists) by $H$ and that $H(y, v, t) = G(v, t)\delta(y - v^{\alpha})$. By the computations above, we obtain that $G$ can be expected to satisfy a coagulation equation in which the coagulation kernel is a kernel of homogeneity $\gamma - \alpha < 1$ for which existence is well-known.

It is also worthwhile to mention that despite the fact that one-dimensional coagulation equations with a diagonal collision kernel have already been the subject of study in the mathematical literature [3–5, 23, 24, 26], a motivation of how this type of kernels could appear has not yet been made. In addition, we remember that at the moment there are only few coagulation kernels for which the long-time behavior of solutions is known. These kernels are the constant kernel, the product kernel $K(v, w) = vw$, the sum kernel $K(v, w) = v + w$, the diagonal kernel, and perturbations of the constant kernel of the form $K(v, w) = 2 + \epsilon W(v, w)$, with $W(v, w)$ continuous, symmetric, bounded, and homogeneous of degree zero. The long-time behavior for general classes of kernels for coagulation equations remains an open problem. We refer to [25] for the first three kernels, to [23, 24, 26] for the diagonal kernel, and to [7, 30] for perturbations of the constant kernel. We thus obtain a first physical intuition on how coagulation models with a diagonal kernel could emerge: as the limiting behavior of coagulation models describing the sedimentation of particles (1.4) in the case of very fast sedimentation speed.

We conjecture that this limiting behavior can be proved rigorously. In the present paper, we rigorously prove that the sequence $\{H_{\epsilon}\}$ in (1.16) converges to $H$ as in (1.9) as $\epsilon \to 0$. The fact that $G$ in (1.9) satisfies (1.11) will be the study of a future work. Nonetheless, as noticed in Remark 1.4, there is no result in the present literature proving that there is a uniform time (in $\epsilon$) of existence of mass-conserving solutions for (1.16). The first step for proving that $H_{\epsilon}$ converges to $H$ as in (1.9), with $G$ satisfying (1.11), is thus to prove existence of mass-conserving solutions of (1.16) for times of order one. This result is also included in the current work.

## 1.5 Connection with coagulation models with fast fusion

It is also worth to analyze the similarities of the model (1.4) with coagulation models with fusion.

$$ \begin{aligned} & \partial_t P_{\epsilon}(a, v, t) + \frac{1}{\epsilon} \partial_a [(c_0 v^{\frac{2}{3}} - a) P_{\epsilon}(a, v, t)] \\ & = \int_{(0,a)} \int_{(0,v)} K(a - a', v - v', a', v') P_{\epsilon}(a - a', v - v', t) P_{\epsilon}(a', v', t) dv' da' \\ & \quad - \int_{(0,\infty)} \int_{(0,\infty)} K(a, v, a', v') P_{\epsilon}(a, v, t) P_{\epsilon}(a', v', t) dv' da'. \end{aligned} \qquad (1.18) $$

The additional variable $a$ represents in this case the total surface area of the cluster and the transport term $\partial_a[(c_0 v^{\frac{2}{3}} - a)P_{\epsilon}(a, v, t)]$ indicates an evolution of the particles towards a spherical shape. The term $c_0 v^{\frac{2}{3}} - a$ decreases the surface area of the clusters as long as it is larger than that
of a sphere $c_0v^{2/3}$. Physically, the transport term describes the merging of two colliding particles up to when they become a sphere and it is hence termed *fusion*.

It was proved in [11] that for the model (1.18) we have that the sequence $\{P_\epsilon\}_{\epsilon\in(0,1)}$ converges to $\bar{P} = P(v, t)\delta(a - c_0v^{2/3})$ as $\epsilon \to 0$, with $P$ satisfying the weak formulation of the standard one-dimensional coagulation equation

$$ \int_{(0,\infty)} P(v,t)\varphi(v)dv - \int_{(0,\infty)} P(v,0)\varphi(v)dv \qquad (1.19) \\ = \int_0^t \int_{(0,\infty)} \int_{(0,\infty)} K(c_0v^{2/3}, v, c_0v'^{2/3}, v')P(v,s)P(v',s)[\varphi(v+v') - \varphi(v) - \varphi(v')]dv'dvds, $$

for all appropriately chosen test functions $\varphi$. A clear similarity can be noticed between the model in (1.16) and the equation (1.18). However, the spatial inhomogeneity of the model (1.16) justifies the presence of the diagonal kernel in the limit (1.11), as opposed to the model in (1.19). More precisely, assuming that the limit of $H_\epsilon$, with $H_\epsilon$ as in (1.16), behaves indeed like a Dirac measure in the space variable, this implies that as we pass to the limit as $\epsilon \to 0$ in (1.16), only particles having the same size can interact due to the space inhomogeneity of the model.

Moreover, the same methods as in [11] cannot be used to prove the result of this paper. One issue stems from the fact that our results in relation to the model (1.18) are valid for Radon measure solutions. While it is possible to define Radon measure solutions for the model (1.18), we need $G$ to be continuous in (1.11) in order to rigorously pass to the limit as $\epsilon \to 0$ in (1.16). Nonetheless, comparing the two results offers a clear overview into the understanding of general existence theory for coagulation models. If we work with a coagulation kernel of form $K(v, v') = v^\gamma + v'^\gamma$, $\gamma > 1$, in (1.18), we obtain in the limit a coagulation kernel of the same homogeneity. This implies that, since solutions do not exist for the model (1.19) when $K(v, v') = v^\gamma + v'^\gamma$, in (1.18), we cannot expect solutions of (1.18) to exist in that case (unless we choose a stronger fusion speed). On the other hand, when we choose a coagulation kernel of the form $K(v, v') = v^\gamma + v'^\gamma$, $1 < \gamma < 1 + \alpha$, in (1.16), we obtain in the limit (1.11) a coagulation kernel of homogeneity $\gamma - \alpha < 1$ as noticed in Remark 1.2. We can thus expect solutions to exist and to behave like a product between a Dirac measure and a function in the volume (or a mollified version of this).

This clearly establishes the following point. The contribution given by a transport term does not suffice to assure existence of mass-conserving solutions of coagulation models (in the case where existence of solutions is not known in the standard case). Rather, both the presence of a transport term and a spatially inhomogeneous model are needed.

# 2 Local in time existence

## 2.1 Setting and main results

Assumption 2.1 (Assumptions on the initial data). Let $\epsilon \in (0, 1)$, $\alpha \in (0, 1)$. Assume that $H_{\epsilon,\text{in}} \in C^1(\mathbb{R} \times (0, \infty))$ and that there exists some constant $A > 0$ such that

$$ (1 + v^b)H_{\epsilon,\text{in}}(y, v) \le \frac{A}{1 + |y - v^\alpha|^m}, \qquad (2.1) $$
for all $y \in \mathbb{R}$, $v \in (0, \infty)$, and for sufficiently large $m, b > 2$.

**Assumption 2.2 (Assumptions on the coagulation kernel).** We work with a coagulation kernel

$$0 \le K(v, v') = K(v', v) \le K_0(v^\gamma + v'^\gamma), \quad 1 < \gamma < 1 + \alpha, \qquad (2.2)$$

such that

$$K(v - v', v') \le K(v, v'), \text{ for all } 0 \le v' \le \frac{v}{2}. \qquad (2.3)$$

**Remark 2.3.** Condition (2.3) is a rather standard assumption in the study of coagulation equations, see for example [22], and most of the kernels used in applications satisfy this condition. Notice that Assumption 2.2 implies that our results hold for sum-type kernels of homogeneity $\gamma > 1$ which are relevant since there are no solutions to the standard coagulation model for these type of kernels. Moreover, Assumption 2.2 is also valid for sum-type kernels which vanish on the diagonal, such as (1.3), which are used in the physical literature to describe droplet formation and the sedimentation of particles. We refer to [20] for more details on how these kernels are derived.

**Definition 2.4 (Mild solutions).** Let $\epsilon \in (0, 1)$, $\alpha \in (0, 1)$, $\gamma \in [0, 1 + \alpha)$, and $m \in \mathbb{N}$, sufficiently large. Let $T > 0$ and $K$ satisfy Assumption 2.2. We say that a non-negative function $H_\epsilon \in C([0, T]; L^\infty(\mathbb{R} \times (0, \infty)))$ is a mild solution of equation (1.16) if

$$H_{\epsilon}(y, v, t) - S_{\epsilon}(t)[H_{\epsilon}(y, v, 0)]D[H_{\epsilon}](y, v, 0, t) = \frac{1}{2} \int_{0}^{t} \int_{(0,v)} D[H_{\epsilon}](y, v, s, t)S_{\epsilon}(t-s) \left[ K(v-v', v')H_{\epsilon}(y, v-v', s)H_{\epsilon}(y, v', s) \right] ds, \qquad (2.4)$$

for all $t \in [0, T]$, $v \in (0, \infty)$ and $y \in \mathbb{R}$, where

$$D[H_{\epsilon}](y, v, s, t) := e^{-\int_{s}^{t} a[H_{\epsilon}](e^{\frac{t-s}{\epsilon}}(y-v^{\alpha})+v^{\alpha}, v, \tau)d\tau}, \qquad (2.5)$$

with

$$a[H_{\epsilon}](y, v, t) := \int_{(0,\infty)} K(v, v')H_{\epsilon}(y, v', t)dv', \qquad (2.6)$$

and

$$S_{\epsilon}(s)H_{\epsilon}(y, v) := e^{\frac{s}{\epsilon}} H_{\epsilon}(e^{\frac{s}{\epsilon}}(y - v^{\alpha}) + v^{\alpha}, v). \qquad (2.7)$$

**Definition 2.5.** Let $\epsilon \in (0, 1)$. We call $H_\epsilon$ a mass-conserving solution of equation (1.16) if $H_\epsilon$ is as in Definition 2.4 and satisfies in addition

$$\int_{\mathbb{R}} \int_{(0,\infty)} v H_{\epsilon}(y, v, t) dv dy = \int_{\mathbb{R}} \int_{(0,\infty)} v H_{\epsilon}(y, v, 0) dv dy \text{ for all } t \in [0, T].$$
**Some notation.** For two quantities x, y, we use the notation x ≈ y to say that there exists a constant M > 1 such that 1_M y ≤ x ≤ My. Additionally, we denote x ≫ y or y ≪ x to mean that My ≤ x, for a sufficiently large constant M > 1.

In Subsection 3.2, the dependence on the parameters plays an important role. We thus introduce the following notation which is valid throughout Subsection 3.2. We write C for a numerical constant that can change from line to line but which does not depend on the parameters α, γ, b, m, where α, γ, b, m are as in Assumption 2.1 and Assumption 2.2. In addition, we write C(a₁, ..., aₙ) for a constant that can change from line to line which depends on the parameters a₁, ..., aₙ.

In Subsection 2.2 and in Subsection 2.3, we offer some formal arguments to justify the proofs in this paper. For this purpose, for two quantities x and y, we introduce the notation x ~ y to indicate that x can be approximated at a formal level by y. This notation will be used in Subsection 2.2 and in Subsection 2.3.

Let A ≥ 1. Let j ∈ {1, 2}. We denote by

$M_j := M_j(\alpha, \gamma, b, m)$, a sufficiently large but fixed constant which depends on $\alpha, \gamma, b, m$; (2.8)

$$C_0 := C_0(A), \text{ a constant such that } C_0^{m-1} := M_1(\alpha, \gamma, b, m)A. \quad (2.9)$$

We remark that $M_2$ is chosen as in (4.43) and that $M_1$ depends on $M_2$ through (3.16). We then define the functions $\psi, T_1, T_2$, and $T_3$ as below.

$$\psi(z) = \frac{1}{1 + |z|^m}; \quad (2.10)$$

$$T_1(y, v, t) := \frac{2A}{1 + v^b} e^{\frac{t}{\epsilon}} \psi(e^{\frac{t}{\epsilon}}(y - v^\alpha)) \chi_1(y, v, t); \quad (2.11)$$

$$T_2(y, v, t) := \frac{2M_1 A^3 \epsilon}{(1 + v^b) |y - v^\alpha|} \chi_2(y, v, t); \quad (2.12)$$

$$T_3(y, v, t) := \frac{M_2 A}{1 + v^b} e^{\frac{t}{\epsilon}} \psi(e^{\frac{t}{\epsilon}}(y - v^\alpha)) \chi_3(y, v, t); \quad (2.13)$$

$$\chi_1(y, v, t) := \mathbb{1}_{\{y \ge v^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}}, \quad (2.14)$$

$$\chi_2(y, v, t) := \mathbb{1}_{\{(\frac{v}{3})^\alpha \le y \le v^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}}, \quad (2.15)$$

$$\chi_3(y, v, t) := \mathbb{1}_{\{y \le (\frac{v}{3})^\alpha\}}. \quad (2.16)$$

Moreover, we define

$$\bar{d} := 2 \left\lceil \frac{1}{2} \left( \frac{\alpha}{\alpha + 1 - \gamma} - 1 \right) \right\rceil, \quad (2.17)$$

where ⌈·⌋ denotes the ceiling function. We then define $\bar{b}(\gamma, \alpha)$ to be

$$\bar{b}(\gamma, \alpha) = \alpha \bar{d} + 2. \quad (2.18)$$

Remark 2.6. Notice that

$$T_2(y, v, t) := \frac{2C_0^{m-1} A^2 \epsilon}{(1 + v^b) |y - v^\alpha|} \chi_2(y, v, t).$$
The reason for introducing $C_0(A)$ is since the cut-offs $\chi_1, \chi_2$ in (2.14)-(2.15) (and regions included in the respective sets) will play an important role in the estimates. We refer to (3.20) and (3.23). Thus, we introduce $C_0$ in order to simplify the notation.

The main result of this paper is as follows.

**Theorem 2.7.** Let $H_{\epsilon,in}$ be as in Assumption 2.1 and $K$ as in Assumption 2.2. Let $b \ge \max\{\bar{b}(\gamma, \alpha), 2\gamma + 1\}$ and $m > \max\{\frac{2(\gamma+1)}{\alpha}, \frac{b}{\alpha} + 1\}$, where $\bar{b}(\gamma, \alpha)$ is as in (2.18). There exist two constants $M_1(\alpha, \gamma, b, m)$, $M_2(\alpha, \gamma, b, m)$, depending only on $\alpha, \gamma, b, m$, such that for any $A \ge 1$, there exist $\bar{\epsilon} \in (0, 1)$, sufficiently small, and $T > 0$, sufficiently small, which is independent of $\epsilon \in (0, 1)$, such that for all $\epsilon \in (0, \bar{\epsilon}]$ there exists a mass-conserving mild solution $H_{\epsilon} \in C([0, T]; L^{\infty}(\mathbb{R} \times (0, \infty)))$ to (1.16) as in Definition 2.5. Moreover, we have that $H_{\epsilon}$ satisfies the following estimates for $C_0$ as in (2.9).

$$ \text{If } e_{\epsilon}^{l} v^{\alpha} [1 - \frac{1}{3^{\alpha}}] \ge C_{0}^{-1} \epsilon^{-\frac{1}{m-1}}, \text{ then} \\ 0 \le (1 + v^{b}) H_{\epsilon}(y, v, t) \le T_{1}(y, v, t) + T_{2}(y, v, t) + T_{3}(y, v, t). \quad (2.19) $$

$$ \text{Otherwise, if } e_{\epsilon}^{l} v^{\alpha} [1 - \frac{1}{3^{\alpha}}] \le C_{0}^{-1} \epsilon^{-\frac{1}{m-1}}, \text{ then} \\ 0 \le (1 + v^{b}) H_{\epsilon}(y, v, t) \le 2A e_{\epsilon}^{l} \psi(e_{\epsilon}^{l} (y - v^{\alpha})). \quad (2.20) $$

The bounds in (2.19) can be explained as follows. Since our functions $H_{\epsilon}$, $\epsilon \in (0, 1)$, approach a Dirac measure supported on the curve $\{y = v^{\alpha}\}$ as $\epsilon \to 0$, we expect that most of the mass of the solution is contained in the region $|y - v^{\alpha}| \ll 1$. Outside this region, we expect the function to have small values and to approach 0 as $\epsilon \to 0$. In other words, since $H_{\epsilon}$ can be thought of as a mollifier approaching the region $\{y = v^{\alpha}\}$, most of the mass of $H_{\epsilon}$ will be contained in the region where $e_{\epsilon}^{l}|y - v^{\alpha}| \ll 1$. In other words, our choice of space is a rescaled version of the function

$$ \frac{1}{1 + |y - v^{\alpha}|^m} \mathbb{1}_{\{|y - v^{\alpha}| \le 1\}} + \frac{\epsilon}{|y - v^{\alpha}|} \mathbb{1}_{\{|y - v^{\alpha}| \ge 1\}}. $$

**Remark 2.8.** If we denote by

$$ T_{\min} := \sup_{t \ge 0} \left\{ \int_{\mathbb{R}} \int_{(0, \infty)} v H_{\epsilon}(y, v, t) dv dy = \int_{\mathbb{R}} \int_{(0, \infty)} v H_{\epsilon}(y, v, 0) dv dy \right\}, $$

then $T_{\min} > 0$ and $T_{\min}$ is independent of the choice of $\epsilon \in (0, \bar{\epsilon})$, where $\bar{\epsilon}$ is as in Theorem 2.7. In addition, the choice of $T_{\min}$ and $\bar{\epsilon}$ depends on the total mass of particles at initial time through the choice of $A$. For the exact choices of $T_{\min}$ and $\bar{\epsilon}$, we refer to the Proof of Proposition 3.6.

**Remark 2.9.** Notice that $\bar{d}$ in (2.17) is even and $\gamma \le \frac{\bar{d}\alpha}{d+1} + 1$. This will be needed in Proposition 4.7 and in Proposition 4.1.

**Remark 2.10.** As already mentioned in Remark 1.4, Theorem 2.7 yields existence of mass-conserving solutions for times of order $O(e^{\frac{1}{\epsilon}})$ to the equation (1.4). We refer to Appendix A for more details.
We summarize below the results in [9] and Theorem 2.7. We notice that the order of the time of existence can be improved if the upper bound of $H_\epsilon$ contains instead the error term $T_2$ as defined in (2.12). While $T_2(y, v, t)$ is larger than $\frac{Ae^{\frac{t}{\epsilon}}\chi_2(y,v,t)}{(1+v^b)(1+e^{\frac{tm}{\epsilon}}|y-v^\alpha|m)}$, it still vanishes as $\epsilon \to 0$ when $y$ and $v$ are supported away from the line $\{y = v^\alpha\}$.

<table><caption>**Table 2** Results in [9] in comparison to Theorem 2.7</caption><thead><tr><th>Results in [9]</th><th>Theorem 2.7</th></tr></thead><tbody><tr><td>existence for t &isin; [0, &epsilon; ln(T&#x0303; + 1)], with T&#x0303; small</td><td>existence for t &isin; [0, T], with T small</td></tr><tr><td>$H_{\epsilon,in}(y, v) &le; \frac{A}{1+|y|^m+v^p}$</td><td>$H_{\epsilon,in}(y, v) &le; \frac{A}{(1+v^b)(1+|y-v^\alpha|m)}$</td></tr><tr><td>$H_\epsilon(y, v, t) &le; \frac{CAe^{\frac{t}{\epsilon}}}{1+e^{\frac{t}{\epsilon}}(y-v^\alpha)+v^\alpha|m+v^p}$, for some C &gt; 0</td><td>$H_\epsilon(y, v, t) &le; \frac{CAe^{\frac{t}{\epsilon}}[\chi_1(y,v,t)+\chi_3(y,v,t)]}{(1+v^b)(1+e^{\frac{tm}{\epsilon}}|y-v^\alpha|m)} + T_2$</td></tr></tbody></table>

*Remark 2.11.* Notice that if in (2.1) we have that $(1 + v^b)H_{\epsilon,in}(y, v) \le \frac{B}{1+|y-v^\alpha|m}$, with $B < 1$, then $(1 + v^b)H_{\epsilon,in}(y, v) \le \frac{1}{1+|y-v^\alpha|m}$ and we can apply Theorem 2.7 with $A = 1$ in order to obtain existence of mass-conserving solutions in this case.

Actually, we can argue in a similar manner as in the case $A \ge 1$ in order to obtain that if $(1 + v^b)H_{\epsilon,in}(y, v) \le \frac{A}{1+|y-v^\alpha|m}$, with $A < 1$, then the following holds. For $C_0^{m-1} = M_1(\alpha, \gamma, b, m)$, if $e^{\frac{t}{\epsilon}}v^\alpha[1 - \frac{1}{3^\alpha}] \ge C_0^{-1}\epsilon^{-\frac{1}{m-1}}$, then

$$(1 + v^b)H_{\epsilon}(y, v, t) \le T_1(y, v, t) + T_3(y, v, t) + \frac{2C_0^{m-1}\epsilon}{(1 + v^b)|y - v^\alpha|} \chi_2(y, v, t), \quad (2.21)$$

where $T_1, T_3$ are as in (2.11) and (2.13). Otherwise, if $e^{\frac{t}{\epsilon}}v^\alpha[1 - \frac{1}{3^\alpha}] \le C_0^{-1}\epsilon^{-\frac{1}{m-1}}$, then

$$0 \le (1 + v^b)H_{\epsilon}(y, v, t) \le 2Ae^{\frac{t}{\epsilon}}\psi(e^{\frac{t}{\epsilon}}(y - v^\alpha)). \quad (2.22)$$

Notice that this estimate provides accurate information about the distribution of mass of the solution $H_\epsilon(y, v, t)$. The term in (2.21) which plays a role similar to $T_2$ does not contain any dependence on $A$. Therefore, if $A$ is very small, it yields a contribution that does not vary depending on $A$.

Nevertheless, as explained above, $T_2$ is to be understood as an error term since it vanishes as $\epsilon \to 0$ when $y$ and $v$ are supported away from the line $\{y = v^\alpha\}$. Thus, despite the fact that $T_2$ loses the dependence on $A$ when $A < 1$, it has a small contribution. In other words, we can consider at a formal level that we stay close to the mass at initial time also in the case $A < 1$.

Nonetheless, as stated in Remark 2.8, the time of existence varies depending on $A$ when $A \ge 1$. Because $T_2$ does not depend on $A$ when $A < 1$, the minimal time of existence will not vary depending on $A$ when $A < 1$. More precisely, for $T_{\min}$ as in Remark 2.8, we have that $T_{\min}$ when $A < 1$ is equal to $T_{\min}$ when $A = 1$.

As conjectured in the Introduction, we obtain that the solutions $H_\epsilon$ found in Theorem 2.7 converge to a Dirac measure.
**Theorem 2.12.** Let $H_{\epsilon, \text{in}}$ be as in Assumption 2.1 and $K$ as in Assumption 2.2. Let $T > 0$ and $H_{\epsilon} \in C([0, T]; L^{\infty}(\mathbb{R} \times (0, \infty)))$ be as in Theorem 2.7. Let $\sigma > 0$ and assume in addition that $b > \gamma + 3$. There exists a subsequence (which we do not relabel) and $H \in C([\sigma, T]; \mathcal{M}_+(\mathbb{R} \times (0, \infty)))$, where we denoted by $\mathcal{M}_+(\mathbb{R} \times (0, \infty))$ the space of non-negative Radon measures, such that

$$ \int_{\mathbb{R}} \int_{(0,\infty)} H_{\epsilon}(y, v, t) \varphi(y, v) dv dy \rightarrow \int_{\mathbb{R}} \int_{(0,\infty)} H(y, v, t) \varphi(y, v) dv dy, $$

as $\epsilon \to 0$, for every $t \in [\sigma, T]$, and every $\varphi \in C_0(\mathbb{R} \times (0, \infty))$. Moreover, for every $t \in [\sigma, T]$, $H$ has the form

$$ H(y, v, t) = \bar{H}(v, t)\delta(y - v^{\alpha}) \qquad (2.23) $$

in the sense of measures. More precisely, there exists $\bar{H} \in C([\sigma, T]; \mathcal{M}_+((0, \infty)))$ such that

$$ \int_{\mathbb{R}} \int_{(0,\infty)} H(y, v, t) \varphi(y, v) dv dy = \int_{\mathbb{R}} \int_{(0,\infty)} \bar{H}(v, t) \delta(y - v^{\alpha}) \varphi(y, v) dv dy, $$

for every $t \in [\sigma, T]$ and $\varphi \in C_0(\mathbb{R} \times (0, \infty))$.

However, it is not clear whether the limit $\bar{H}$ satisfies a one-dimensional coagulation equation with a diagonal kernel as in (1.11). From the bound in (2.19), we are able to prove that the sequence $\{H_{\epsilon}\}_{\epsilon \in (0,1)}$ is a uniformly bounded sequence in $L^1$, but it is not uniformly bounded in $L^{\infty}$. This, together with the spatial inhomogeneity of the model (1.16), will create difficulties when trying to prove that the marginals $\int_{\mathbb{R}} H_{\epsilon}(y, v, t) dy$ are uniformly equicontinuous in the $v$ variable. As such, we do not know whether $\bar{H}$ is continuous in $v$ in order to rigorously derive (1.11).

## 2.2 Justification of the main results

### Existence of solutions - Theorem 2.7

We first offer some insight into the proof of existence stated in Theorem 2.7 and the choice of $T_1, T_2, T_3$ in (2.11)-(2.13). We start by approximating the contribution of the gain term in the coagulation kernel assuming that our solutions behave like Dirac measures supported on the curve $\{y = v^{\alpha}\}$ with sufficiently fast decay for large values of $v$. The decay in $v$ is so that we obtain mass-conserving solutions. More precisely, we have that

$$ \begin{aligned} & v^{\gamma} \int_{0}^{v} \frac{1}{1 + w^{b}} \delta(y - w^{\alpha}) \frac{1}{1 + (v - w)^{b}} \delta(y - (v - w)^{\alpha}) dw \\ & \approx \frac{v^{\gamma+1-\alpha}}{(1 + v^{b})^{2}} \delta(y - (\frac{v}{2})^{\alpha}) \int_{0}^{v} \delta(w - \frac{v}{2}) dw = \frac{v^{\gamma+1-\alpha}}{(1 + v^{b})^{2}} \delta(y - (\frac{v}{2})^{\alpha}). \end{aligned} $$
Applying the semigroup action to the obtained term, we have that

$$
\begin{align*}
\int_0^t S_\epsilon(t-s) \left[ \frac{v^{\gamma+1-\alpha}}{(1+v^b)^2} \delta\left(y-\left(\frac{v}{2}\right)^\alpha\right) \right] ds &= \frac{v^{\gamma+1-\alpha}}{(1+v^b)^2} \int_0^t e^{\frac{t-s}{\epsilon}} \delta\left(e^{\frac{t-s}{\epsilon}}(y-v^\alpha) + v^\alpha - \left(\frac{v}{2}\right)^\alpha\right) ds \\
&= \frac{v^{\gamma+1-\alpha}}{(1+v^b)^2} \int_0^t e^{\frac{s}{\epsilon}} \delta\left(e^{\frac{s}{\epsilon}}(y-v^\alpha) + v^\alpha - \left(\frac{v}{2}\right)^\alpha\right) ds \\
&= \epsilon \frac{v^{\gamma+1-\alpha}}{(1+v^b)^2} \int_0^\frac{t}{\epsilon} e^s \delta\left(e^s(y-v^\alpha) + v^\alpha - \left(\frac{v}{2}\right)^\alpha\right) ds.
\end{align*}
$$

Making the change of variables $z = e^s(y - v^\alpha) + v^\alpha - (\frac{v}{2})^\alpha$, we further obtain that

$$
\begin{align*}
\int_0^t S_\epsilon(t-s) \left[ \frac{v^{\gamma+1-\alpha}}{(1+v^b)^2} \delta\left(y-\left(\frac{v}{2}\right)^\alpha\right) \right] ds &= \frac{\epsilon}{v^\alpha - y} \frac{v^{\gamma+1-\alpha}}{(1+v^b)^2} \int_{e^{\frac{t}{\epsilon}}(y-v^\alpha)+v^\alpha-\left(\frac{v}{2}\right)^\alpha}^{y-\left(\frac{v}{2}\right)^\alpha} \delta(z) dz \\
&= \frac{\epsilon}{v^\alpha - y} \frac{v^{\gamma+1-\alpha}}{(1+v^b)^2} \mathbb{1}_{\left\{\left(\frac{v}{2}\right)^\alpha \le y \le v^\alpha - e^{-\frac{t}{\epsilon}}(v^\alpha - \left(\frac{v}{2}\right)^\alpha)\right\}}
\end{align*}
$$

This explains the choice of the term $T_2$ in (2.12).

Choice of the estimate in (2.19)

As noticed in the previous computations the gain term of the coagulation operator shifts the
support from {$y = v^\alpha$} to {$y = (\frac{v}{2})^\alpha$} if we apply it on a Dirac mass $\delta(y - v^\alpha)$. For convenience,
we approximate the term $\frac{\epsilon}{v^\alpha - y} \frac{v^{\gamma+1-\alpha}}{(1+v^b)^2} \mathbb{1}_{\{(\frac{v}{2})^\alpha \le y \le v^\alpha - e^{-\frac{t}{\epsilon}}(v^\alpha - (\frac{v}{2})^\alpha)\}}$ by $\frac{\epsilon}{v^\alpha - y} \frac{v^{\gamma+1-\alpha}}{(1+v^b)^2} \mathbb{1}_{\{(\frac{v}{2})^\alpha \le y \le v^\alpha\}}$ in the fol-
lowing. However, the fact that $T_2$ is at a distance of order $v^\alpha e^{-\frac{t}{\epsilon}}$ from the curve {$y = v^\alpha$} will be
important in order to avoid unbounded terms. Because of the shift of the support, it is natural to
look for a decomposition of the form $H_\epsilon = h_0 + h_1 + \dots$, with $h_0(y, v, t) = g(v, t)\delta(y - v^\alpha)$ and
each $h_i$, $i \in \mathbb{N}$, supported in a region of the form $y \in [(\frac{v}{2})^\alpha, (\frac{v}{2i-1})^\alpha]$, with a decay in $v$ of larger
order, and $h_i \approx \mathcal{O}(\epsilon^i)$. It then holds that

$$
\epsilon \partial_t g(v, t) \delta(y - v^\alpha) + \partial_y((v^\alpha - y)h_1) = \epsilon v^\gamma \int_0^{\frac{v}{2}} g(v - w, t) \delta(y - (v - w)^\alpha) g(w, t) \delta(y - w^\alpha) dw \\
\approx \epsilon v^{\gamma+1-\alpha} \left(g(\frac{v}{2}, t)\right)^2 \delta(y - (\frac{v}{2})^\alpha).
$$

Thus, we have that

$$
\partial_y((v^\alpha - y)h_1) = \epsilon v^{\gamma+1-\alpha} \left(g\left(\frac{v}{2}, t\right)\right)^2 \delta\left(y - \left(\frac{v}{2}\right)^\alpha\right),
$$

which implies that

$$
(v^{\alpha} - y)h_1 = \epsilon v^{\gamma+1-\alpha} \left(g\left(\frac{v}{2}, t\right)\right)^2 \int_{-\infty}^{y} \delta\left(\xi - \left(\frac{v}{2}\right)^{\alpha}\right) d\xi
$$
or in other words

$$h_1 = \frac{\epsilon v^{\gamma+1-\alpha}}{(v^{\alpha} - y)} \left(g\left(\frac{v}{2}, t\right)\right)^2 \mathbb{1}_{\left\{\left(\frac{v}{2}\right)^{\alpha} \le y \le v^{\alpha}\right\}}.$$

Thus it holds that $h_0(y, v, t) = g(v, t)\delta(y - v^\alpha)$ and $h_1(y, v, t) = \frac{\epsilon v^{\gamma+1-\alpha}}{(v^{\alpha} - y)} \left(g\left(\frac{v}{2}, t\right)\right)^2 \mathbb{1}_{\left\{\left(\frac{v}{2}\right)^{\alpha} \le y \le v^{\alpha}\right\}}$. We further obtain that

$$\begin{aligned} & \int_0^v h_0(y, v-w, t) h_1(y, w, t) dw \\ &= \epsilon v^\gamma \int_0^v g(v-w, t) \delta(y - (v-w)^\alpha) w^{\gamma+1-\alpha} \frac{(g(\frac{w}{2}, t))^2}{w^\alpha - y} \mathbb{1}_{\left\{\left(\frac{w}{2}\right)^\alpha \le y \le w^\alpha\right\}} dw. \end{aligned}$$

Since if $y = (v - w)^\alpha$ and $(\frac{w}{2})^\alpha \le y \le w^\alpha$ implies that $\frac{2v}{3} \ge w \ge \frac{v}{2}$ and thus $\frac{v}{3} \le v - w \le \frac{v}{2}$, we have that

$$v^{\gamma} \int_{0}^{v} h_{0}(y, v - w, t) h_{1}(y, w, t) dw \approx \frac{\epsilon v^{2(y+1-\alpha)} g(\frac{v}{3}, t) (g(\frac{v}{4}, t))^{2}}{((v - y^{\frac{1}{\alpha}})^{\alpha} - y)} \mathbb{1}_{\{\left(\frac{v}{3}\right)^{\alpha} \le y \le \left(\frac{v}{2}\right)^{\alpha}\}}.$$

Thus, arguing as before, it follows that

$$\begin{aligned} \partial_y((v^\alpha - y)h_2) &= \frac{\epsilon^2 v^2 (y+1-\alpha) g(\frac{v}{3}, t) (g(\frac{v}{4}, t))^2}{((v - y^{\frac{1}{\alpha}})^\alpha - y)} \mathbb{1}_{\{\left(\frac{v}{3}\right)^\alpha \le y \le \left(\frac{v}{2}\right)^\alpha\}} \\ &\sim \frac{\epsilon^2 v^2 (y+1-\alpha) g(\frac{v}{3}, t) (g(\frac{v}{4}, t))^2}{(\frac{v}{2})^\alpha - y} \mathbb{1}_{\{\left(\frac{v}{3}\right)^\alpha \le y \le \left(\frac{v}{2}\right)^\alpha\}}, \end{aligned}$$

which by integrating in y implies that

$$h_2 = \frac{\epsilon^2 v^2 (y+1-\alpha) g(\frac{v}{3}, t) (g(\frac{v}{4}, t))^2}{(v^\alpha - y)} \int_{-\infty}^{y} \frac{1}{(\frac{v}{2})^\alpha - \xi} \mathbb{1}_{\{\left(\frac{v}{3}\right)^\alpha \le \xi \le \left(\frac{v}{2}\right)^\alpha\}} \quad (2.24)$$

The integral will give a logarithmic contribution which explodes at $y = (\frac{v}{2})^\alpha$. However, due to the cut-off in $h_1$ noticed above, namely $\mathbb{1}_{\{\left(\frac{v}{2}\right)^\alpha \le y \le v^{\alpha} - e^{-\frac{t}{\epsilon}}(v^{\alpha} - (\frac{v}{2})^\alpha)\}}$, the integral turns out to be convergent. Taking into account the effect of this cut-off, we are able to estimate the integral in (2.24) as $C(\frac{t}{\epsilon} + \log(v^\alpha))$ and thus

$$h_2 \approx \frac{t \epsilon v^{2(y+1-\alpha)} g(\frac{v}{3}, t) (g(\frac{v}{4}, t))^2}{(v^{\alpha} - y)} + \frac{\epsilon^2 v^2 (y+1-\alpha) g(\frac{v}{3}, t) (g(\frac{v}{4}, t))^2}{(v^{\alpha} - y)} \log(v^{\alpha}).$$

Then, by choosing a sufficiently good decay in $v$ for large values of $v$ for the function $g(v, t)$ and for times $t$ of order one, we should be able to approximate $h_2$ for $v \ge 1$ by

$$h_2 \sim \frac{\epsilon v^{2(y+1)} g(\frac{v}{3}, t) (g(\frac{v}{4}, t))^2}{(v^{\alpha} - y)}.$$
This is the expected choice for the estimate in (2.19) and explains the choice of the truncation in $T_1$ and $T_2$ in (2.11)-(2.12).

Another problem that will arise when passing to the rigorous computations is the control of the decay in $v$ in order to obtain mass-conserving solutions. We remark that if the function $g$ is chosen to have sufficiently fast decay for large $v$, then $h_0, h_1, h_2$ also have sufficiently fast decay in $v$. However, the terms of the series become difficult to study after multiple iterations. Moreover, our functions will behave like mollifiers of a Dirac measure supported on the curve $\{y = v^\alpha\}$ and because of this their support will become larger, creating the possibility of losing the decay in $v$ after their convolution. More precisely, for $\frac{y}{v^\alpha} \ll 1$, we may lose the decay in $v$. Nonetheless, we notice that for $\frac{y}{v^\alpha} \ll 1$, our equation can be approximated by

$$ \partial_t H_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)H_\epsilon(y, v, t)] - \mathbb{K}[H_\epsilon](y, v, t) \quad (2.25) $$

$$ \sim \partial_t H_\epsilon(y, v, t) + \frac{1}{\epsilon} v^\alpha \partial_y H_\epsilon(y, v, t) - \mathbb{K}[H_\epsilon](y, v, t). \quad (2.26) $$

For fixed $\epsilon$, existence of the model (2.26) has been studied in [9]. More precisely, in this case the coagulation operator can be approximated by a transport term in the $v$ variable and one can prove that $G_\epsilon$ is a supersolution of the model (2.26), where $G_\epsilon$ satisfies the following PDE.

$$ \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} v^\alpha \partial_y G_\epsilon(y, v, t) + \frac{L v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) = 0, \quad (2.27) $$

for some suitably chosen $d \in \mathbb{N}$ and $L > 0$. For completeness, we present the idea behind passing from (2.26) to (2.27) in Subsection 2.3. This idea was also used in [9, Section 2.1] and a rigorous proof can be found in Subsection 4.4 of this paper.

Using (2.27), the precise form of $G_\epsilon$ can then be studied since the associated characteristics to (2.27) are given by

$$ \left\{ \begin{array}{ll} \partial_t Y_\epsilon(y, v, t) = -\frac{1}{\epsilon} V_\epsilon^\alpha, & Y_\epsilon(y, v, 0) = y, \\ \\ \partial_t V_\epsilon(y, v, t) = -\frac{L V_\epsilon^\gamma}{1 + |Y_\epsilon|^d}, & V_\epsilon(y, v, 0) = v. \end{array} \right. \quad (2.28) $$

It immediately follows from (2.28) that we can use separation of variables to obtain more information about the form of the characteristics since

$$ \frac{dV_\epsilon}{dY_\epsilon} = \frac{\epsilon L V_\epsilon^{\gamma-\alpha}}{1 + |Y_\epsilon|^d} \Rightarrow V_\epsilon^{-\gamma+\alpha} dV_\epsilon = \frac{\epsilon L dY_\epsilon}{1 + |Y_\epsilon|^d}. $$

However, for a rigorous proof, a supersolution of (2.25) has the following form (up to minor modifications)

$$ \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon}(v^\alpha - y)\partial_y G_\epsilon(y, v, t) + \frac{L v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) = 0. \quad (2.29) $$

The associated characteristics for (2.29) satisfy

$$ \frac{dV_\epsilon}{dY_\epsilon} = \frac{\epsilon L V_\epsilon^\gamma}{(1 + |Y_\epsilon|^d)(V_\epsilon^\alpha - Y_\epsilon)} \quad (2.30) $$
and it is not possible to solve (2.30) using the separation of variables and implicitly we are not able to reproduce the methods in [9]. Nonetheless, we will prove that for $\frac{y}{v^\alpha} \ll 1$, namely for $y \le (\frac{y}{3})^\alpha$, the associated characteristics behave similarly to the ones in (2.28). In particular, the characteristics do not go to infinity in finite time. This explains the choice of $T_3$ in (2.13).

## Convergence to a Dirac measure - Theorem 2.12

We now present some justification for the proof of Theorem 2.12. The solution found in Theorem 2.7 satisfies the following upper bound,

$$H_{\epsilon}(y, v, t) \le T_1(y, v, t) + T_2(y, v, t) + T_3(y, v, t), \text{ with } T_1, T_2, T_3 \text{ as in (2.11) - (2.13).} \quad (2.31)$$

With this upper bound, it is straightforward to prove that $H_\epsilon$ vanishes in the region $\{|y - v^\alpha| \ge \delta\}$, for some $\delta > 0$. More precisely, we prove that if $t > 0$, $|y - v^\alpha| \ge \delta$, it holds that

$$H_{\epsilon}(y, v, t) \le C(\delta)[\epsilon + e^{-\frac{t(m-1)}{\epsilon}}] \to 0 \text{ as } \epsilon \to 0. \quad (2.32)$$

If in addition to (2.32), $H_\epsilon$ is also equicontinuous in time, then Theorem 2.12 follows. At a first glance, the transport term in (1.16) may bring a contribution that is too large because of the presence of $\frac{1}{\epsilon}$ and a more detailed analysis of the equation is then required in order to prove equicontinuity in time.

Assume we are at a time $t \ge \sigma > 0$. At a formal level, $H_\epsilon$ behaves like a mollifier of a Dirac measure at $\{y = v^\alpha\}$ with sufficiently fast decay in the $v$ variable. In other words, the part of $H_\epsilon$ supported in the same region as $T_1$ in (2.11) contains most of the mass. We test with $\varphi$ in (1.16) and integrate in $y$ and $v$. It follows that

$$\left| \int_{(0,\infty)} \int_{\mathbb{R}} \frac{e^{\frac{t}{\epsilon}} (y - v^\alpha)}{(1 + v^b)(1 + e^{\frac{t m}{\epsilon}} |y - v^\alpha|^m)} \partial_y \varphi(y, v) dy dv \right| \le C e^{-\frac{t}{\epsilon}} \int_{(0,\infty)} \frac{dv}{1 + v^b} \int_{\mathbb{R}} \frac{|\xi| d\xi}{1 + |\xi|^m} \\ \le C e^{-\frac{\sigma}{\epsilon}}.$$

In other words, the transport term in (1.16) behaves like

$$\frac{1}{\epsilon} \int_{s}^{t} \int_{\mathbb{R}} \int_{(0,\infty)} (v^{\alpha} - y) H_{\epsilon}(y, v, z) \partial_{y} \varphi(y, v) dv dy dz \sim \frac{C e^{-\frac{\sigma}{\epsilon}} |t - s|}{\epsilon} \le C |t - s|.$$

for all $t, s \ge \sigma > 0$. This offers a sufficiently good bound in order to obtain the desired equicontinuity in time. Moreover, the contribution due to the coagulation operator can be easily controlled using (2.31).

## 2.3 Approximation of the model by a transport term in the region $y \le (\frac{y}{3})^\alpha$

Finally, we present formally the approximation of (2.26) by (2.27) and refer to Subsection 4.4 for a rigorous proof. We can approximate (2.26) via

$$\partial_t H_\epsilon(y, v, t) + \frac{1}{\epsilon} v^\alpha \partial_y H_\epsilon(y, v, t) \\ \sim \int_{(0, \frac{v}{2})} [K(v - w, w)H_\epsilon(y, v - w, t) - K(v, w)H_\epsilon(y, v, t)]H_\epsilon(y, w, t) dw. \quad (2.33)$$
Since our strategy relies on finding a suitable supersolution, it suffices to find a lower bound for (2.33). We use (2.3), i.e., $K(v-w, w) \le K(v, w)$ when $w \in [0, \frac{v}{2}]$, and obtain that

$$ 
\begin{aligned}
& \partial_t H_\epsilon(y, v, t) + \frac{1}{\epsilon} v^\alpha \partial_y H_\epsilon(y, v, t) \\
& \quad - \int_{(0, \frac{v}{2})} [K(v-w, w)H_\epsilon(y, v-w, t) - K(v, w)H_\epsilon(y, v, t)]H_\epsilon(y, w, t)dw \\
& \ge \partial_t H_\epsilon(y, v, t) + \frac{1}{\epsilon} v^\alpha \partial_y H_\epsilon(y, v, t) - \int_{(0, \frac{v}{2})} K(v, w)[H_\epsilon(y, v-w, t) - H_\epsilon(y, v, t)]H_\epsilon(y, w, t)dw.
\end{aligned} 
\quad (2.34) $$

Since $K(v, w) \le K_0(w^\gamma + v^\gamma)$ and since $w \in [0, \frac{v}{2}]$, we further deduce that

$$ 
\begin{aligned}
& \partial_t H_\epsilon(y, v, t) + \frac{1}{\epsilon} v^\alpha \partial_y H_\epsilon(y, v, t) \\
& \sim \int_{(0, \frac{v}{2})} v^\gamma [H_\epsilon(y, v-w, t) - H_\epsilon(y, v, t)]H_\epsilon(y, w, t)dw \\
& = -v^\gamma \int_{(0, \frac{v}{2})} \int_{v-w}^{v} \partial_v H_\epsilon(y, z, t) dz H_\epsilon(y, w, t)dw.
\end{aligned} 
\quad (2.35) $$

Assuming that $\partial_v H_\epsilon(y, z, t)$ behaves similarly for $z \in [\frac{v}{2}, v]$, it follows that

$$ \partial_t H_\epsilon(y, v, t) + \frac{1}{\epsilon} v^\alpha \partial_y H_\epsilon(y, v, t) \sim -v^\gamma \partial_v H_\epsilon(y, v, t) \int_{(0, \frac{v}{2})} w H_\epsilon(y, w, t) dw. \quad (2.36) $$

We denote by $M_1(y, t) := \int_{(0,\infty)} w H_\epsilon(y, w, t) dw$ the first moment in $v$ of $H_\epsilon$. We consider only large values of $v$ so that we can safely assume that $\int_{(0, \frac{v}{2})} w H_\epsilon(y, w, t) dw$ contains most of the mass. Suppose now that $M_1(y, t)$ decays sufficiently fast for large values of $y$, that is assume that

$$ M_1(y, t) \le \frac{L}{1 + |y|^d}, \quad (2.37) $$

for some sufficiently large $d$ and some $L > 0$. Combining (2.37) with (2.36), we obtain that $H_\epsilon$ can be estimated formally from above by a solution of the equation

$$ \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} v^\alpha \partial_y G_\epsilon(y, v, t) + \frac{L v^\gamma \partial_v G_\epsilon(y, v, t)}{1 + |y|^d} = 0. \quad (2.38) $$

## 2.4 Mathematical toolbox

The proof of Theorem 2.7 relies on the fact that for small values of $\epsilon$, $H_\epsilon$ should behave like a mollified Dirac measure in the space variable of the form $\delta(y-v^\alpha)$. Following the computations in (1.10), the gain term in the coagulation operator will behave like a Dirac measure of the form $\delta(y - (\frac{v}{2})^\alpha)$ multiplied by some function depending on the $v$ variable. In other words, when we are at a positive distance from the line $\{y = (\frac{v}{2})^\alpha\}$, we should obtain sufficiently fast decay for our solutions in a straightforward manner. The terms $T_1$ and $T_3$ in (2.11), (2.13) represent the action of the semigroup generated by the sedimentation term on the initial condition in (2.1). The term $T_2$ in (2.12) is an additional term to control the problematic region approaching $\{y = (\frac{v}{2})^\alpha\}$.
We gather here some technical lemmas that will be useful later in the proof of Theorem 2.7 and that provide some insight into the choice of space for existence. Firstly, we need estimates for the gain term in the coagulation operator. As noticed in Subsection 2.2, we often use the fact that

$$ \int_0^v \delta(y - w^\alpha)\delta(y - (v - w)^\alpha)dw = \int_0^v \delta(y - w^\alpha)\delta(w^\alpha - (v - w)^\alpha)dw \quad (2.39) $$

$$ = \frac{1}{2\alpha} \left(\frac{v}{2}\right)^{1-\alpha} \delta\left(y - \left(\frac{v}{2}\right)^{\alpha}\right) \int_0^v \delta\left(w - \frac{v}{2}\right)dw, \quad (2.40) $$

or alternatively, in (2.39) we use the fact that for $\xi, B_1, B_2 \in \mathbb{R}$

$$ \delta(\xi - B_1)\delta(\xi - B_2) = \delta(B_1 - B_2)\delta(\xi - B_1) = \delta(B_1 - B_2)\delta(\xi - B_2). \quad (2.41) $$

Our solutions behave like a mollified version of a Dirac measure at $y = v^\alpha$. While we cannot expect (2.41) to be valid, we need a similar estimate to the one in (2.41) to hold for our solutions. We prove an estimate of the form

$$ H_{\epsilon}(\xi - B_1)H_{\epsilon}(\xi - B_2) \sim H_{\epsilon}(B_1 - B_2)[H_{\epsilon}(\xi - B_1) + H_{\epsilon}(\xi - B_2)]. $$

More precisely, we prove that

**Lemma 2.13.** Let $\xi, B_1, B_2 \in \mathbb{R}$. It holds that

$$ (1 + |\xi - B_1|)(1 + |\xi - B_2|) \ge \frac{(1 + \min\{|\xi - B_1|, |\xi - B_2|\})(1 + |B_1 - B_2|)}{2}. \quad (2.42) $$

Let $m \in \mathbb{N}$. As an immediate consequence of (2.42), we have that

$$ \frac{1}{(1 + |\xi - B_1|)^m} \frac{1}{(1 + |\xi - B_2|)^m} \le \frac{2^{2m-1}}{(1 + |B_1 - B_2|)^m} \left[ \frac{1}{(1 + |\xi - B_1|)^m} + \frac{1}{(1 + |\xi - B_2|)^m} \right], \quad (2.43) $$

for any $\xi, B_1, B_2 \in \mathbb{R}$.

This will in turn imply that for $\psi$ as in (2.10), it holds that

$$ e^{\frac{2t}{\epsilon}} \psi(e^{\frac{t}{\epsilon}}(\xi - B_1)) \psi(e^{\frac{t}{\epsilon}}(\xi - B_2)) \le C e^{\frac{t}{\epsilon}} \psi(e^{\frac{t}{\epsilon}}(B_1 - B_2)) \left[ e^{\frac{t}{\epsilon}} \psi(e^{\frac{t}{\epsilon}}(\xi - B_1)) + e^{\frac{t}{\epsilon}} \psi(e^{\frac{t}{\epsilon}}(\xi - B_2)) \right]. $$

**Proof of Lemma 2.13.** Let $\xi, B_1, B_2 \in \mathbb{R}$ and assume without loss of generality that $B_1 \le B_2$.

If $\xi \le B_1$, it holds that

$$ \begin{aligned} (1 + |\xi - B_1|)(1 + |\xi - B_2|) &\ge (1 + |\xi - B_1|)(1 + |B_1 - B_2|) \\ &= (1 + \min\{|\xi - B_1|, |\xi - B_2|\})(1 + |B_1 - B_2|). \end{aligned} $$

If $\xi \ge B_2$, it holds that

$$ \begin{aligned} (1 + |\xi - B_1|)(1 + |\xi - B_2|) &\ge (1 + |B_1 - B_2|)(1 + |\xi - B_2|) \\ &= (1 + \min\{|\xi - B_1|, |\xi - B_2|\})(1 + |B_1 - B_2|). \end{aligned} $$
If $\xi \in [B_1, B_2]$, it holds that

$$ (1 + |\xi - B_1|)(1 + |\xi - B_2|) \geq \frac{1}{2}(1 + \min\{|\xi - B_1|, |\xi - B_2|\})(1 + |B_1 - B_2|), $$

since

$$ |B_1 - B_2| \leq |\xi - B_1| + |\xi - B_2| \leq 2 \max\{|\xi - B_1|, |\xi - B_2|\}. $$

Thus, (2.42) holds. Moreover, we have that

$$ \begin{aligned} & [(1 + |\xi - B_1|) + (1 + |\xi - B_2|)](1 + \min\{|\xi - B_1|, |\xi - B_2|\}) \\ & \geq (1 + \max\{|\xi - B_1|, |\xi - B_2|\})(1 + \min\{|\xi - B_1|, |\xi - B_2|\}) \\ & = (1 + |\xi - B_1|)(1 + |\xi - B_2|) \end{aligned} $$

which implies that

$$ \frac{1}{1 + \min\{|\xi - B_1|, |\xi - B_2|\}} \leq \frac{1}{1 + |\xi - B_1|} + \frac{1}{1 + |\xi - B_2|}. $$

By (2.42), it follows that

$$ \begin{aligned} \frac{1}{(1 + |\xi - B_1|)^m} \frac{1}{(1 + |\xi - B_2|)^m} &\le \frac{2^m}{(1 + \min\{|\xi - B_1|, |\xi - B_2|\})^m (1 + |B_1 - B_2|)^m} \\ &\le \frac{2^m}{(1 + |B_1 - B_2|)^m} \left[ \frac{1}{1 + |\xi - B_1|} + \frac{1}{1 + |\xi - B_2|} \right]^m \\ &\le \frac{2^{2m-1}}{(1 + |B_1 - B_2|)^m} \left[ \frac{1}{(1 + |\xi - B_1|)^m} + \frac{1}{(1 + |\xi - B_2|)^m} \right] \end{aligned} $$

and therefore (2.43) holds. $\square$

We also use in (2.40) the fact that

$$ \delta((v-w)^\alpha - w^\alpha) = \frac{1}{2\alpha} \left(\frac{v}{2}\right)^{1-\alpha} \delta\left(w - \frac{v}{2}\right). $$

Similarly, the following inequality holds for the function $\psi$ in (2.10).

**Lemma 2.14 (Change of variables).** Let $\alpha \in (0, 1)$, $\epsilon \in (0, 1)$, $\beta > 1$, $s \ge 0$, $y \in \mathbb{R}$, and $v > 0$. Then there exists a constant $C(\alpha) > 0$, which depends on $\alpha$, such that

$$ \int_0^{\frac{v}{2}} \frac{e^{\frac{s}{\epsilon}} dw}{1 + e^{\frac{sn}{\epsilon}} |(v-w)^\alpha - w^\alpha|^{\beta}} \le C(\alpha) v^{1-\alpha}; \quad (2.44) $$

$$ \int_0^{\frac{v}{2}} \frac{e^{\frac{s}{\epsilon}} dw}{1 + e^{\frac{sn}{\epsilon}} |y - (v-w)^{\alpha}|^{\beta}} \le C(\alpha) v^{1-\alpha}. \quad (2.45) $$
*Proof.* We only prove (2.44) since (2.45) can be proved similarly. We make the change of variables $z = e^{\frac{\xi}{\epsilon}}[(v-w)^{\alpha} - w^{\alpha}]$. Then

$$ \frac{dw}{dz} = e^{-\frac{\xi}{\epsilon}}(-\alpha(v-w)^{\alpha-1} - \alpha w^{\alpha-1})^{-1}. $$

Since $\alpha \in (0, 1)$ and $w \in [0, \frac{\nu}{2}]$, it holds that

$$ \left| \frac{dw}{dz} \right| \le C e^{-\frac{\xi}{\epsilon} v^{1-\alpha}}. $$

Thus

$$ \int_{0}^{\frac{\nu}{2}} \frac{e^{\frac{\xi}{\epsilon}} dw}{1 + e^{\frac{\alpha n}{\epsilon}} (v-w)^{\alpha} - w^{\alpha} |\beta|} \le C v^{\alpha-1} \int_{0}^{e^{\frac{\xi}{\epsilon}} v^{\alpha}} \frac{dz}{1 + |z|^{\beta}} \le C v^{1-\alpha}. $$

□

Moreover, we need to derive some estimates for the action of the semigroup generated by the transport term on the function $\psi$ in (2.10). This will further justify the term $T_2$ in the estimate (2.19).

**Proposition 2.15.** Let $\epsilon \in (0, 1)$, $s \ge 0$, $y \in \mathbb{R}$, $v \in (0, \infty)$, and $S_\epsilon$ be as in (2.7). It holds that

$$ \int_0^t S_\epsilon(t-s)\psi(y-v^\alpha)ds \le \frac{C\epsilon \min\{1, e^{\frac{l}{\epsilon}}|y-v^\alpha|\}}{|y-v^\alpha|(1+|y-v^\alpha|^{m-1})}, \quad (2.46) $$

for $\psi$ as in (2.10).

*Proof.* We make the change of variables $z = t - s$ and thus (2.46) is equivalent to proving that

$$ \int_0^t S_\epsilon(z)\psi(y-v^\alpha)dz \le \frac{C\epsilon \min\{1, e^{\frac{l}{\epsilon}}|y-v^\alpha|\}}{|y-v^\alpha|(1+|y-v^\alpha|^{m-1})}. $$

We make the change of variables $s = e^{\frac{\xi}{\epsilon}}|y - v^\alpha|$ and we obtain that

$$ \int_0^t \frac{e^{\frac{\xi}{\epsilon}}}{1 + e^{\frac{m\epsilon}{\epsilon}}|y - v^\alpha|m} dz = \frac{\epsilon}{|y - v^\alpha|} \int_{|y - v^\alpha|}^{e^{\frac{l}{\epsilon}}|y - v^\alpha|} \frac{1}{1 + |s|^m} ds. $$

It holds that

$$ \int_{|y-v^\alpha|}^{e^{\frac{l}{\epsilon}}|y-v^\alpha|} \frac{1}{1+|s|^m} ds \le \int_{|y-v^\alpha|}^{\infty} \frac{1}{1+|s|^m} ds \le \frac{C}{1+|y-v^\alpha|^{m-1}}; \quad (2.47) $$

$$ \int_{|y-v^\alpha|}^{e^{\frac{l}{\epsilon}}|y-v^\alpha|} \frac{1}{1+|s|^m} ds \le \frac{1}{1+|y-v^\alpha|m} \int_{|y-v^\alpha|}^{e^{\frac{l}{\epsilon}}|y-v^\alpha|} ds \le \frac{Ce^{\frac{l}{\epsilon}}|y-v^\alpha|}{1+|y-v^\alpha|m} \le \frac{Ce^{\frac{l}{\epsilon}}|y-v^\alpha|}{1+|y-v^\alpha|^{m-1}}. \quad (2.48) $$

Combining the above estimates, we obtain the desired inequality. □
**Lemma 2.16** (Monotonicity properties of the semigroup). Let $y \in \mathbb{R}$ and $v \in (0, \infty)$. The following properties hold.

1. Let $S_\epsilon$ as in (2.7). Assume that $f(y, v) \ge g(y, v)$ for all $y \ge y_1$, where $y_1 \ge v^\alpha$. In addition, assume that $f(y, v) \le g(y, v)$ for all $y \le y_2$, where $y_2 \le v^\alpha$. It holds that

$$S_{\epsilon}(s)[f(y, v)] \ge S_{\epsilon}(s)[g(y, v)] \text{ if } y \ge y_1, \text{ for all } s \ge 0; \quad (2.49)$$

$$S_{\epsilon}(s)[f(y, v)] \le S_{\epsilon}(s)[g(y, v)] \text{ if } y \le y_2, \text{ for all } s \ge 0. \quad (2.50)$$

2. Let $y \in \mathbb{R}$, $v \in (0, \infty)$, and $w \in [0, \frac{v}{2}]$. It holds that either $|y - w^\alpha| \ge |y - (\frac{v}{2})^\alpha|$ or $|y - (v - w)^\alpha| \ge |y - (\frac{v}{2})^\alpha|$.

*Proof.* 1. We only prove that (2.50) holds. We have to prove that $e^{\frac{v}{2}}(y - v^\alpha) + v^\alpha \le y_2$ and (2.50) follows. This is true since $y_2 \le v^\alpha$ implies that $e^{\frac{v}{2}}(y - v^\alpha) \le y - v^\alpha \le y_2 - v^\alpha$, for all $y \le y_2 \le v^\alpha$ and $s \ge 0$.

2. Since $w \in [0, \frac{v}{2}]$, if $y \ge (\frac{v}{2})^\alpha$, then $y - w^\alpha \ge y - (\frac{v}{2})^\alpha \ge 0$. Otherwise, since $v - w \in [\frac{v}{2}, v]$, then $y - (v - w)^\alpha \le y - (\frac{v}{2})^\alpha \le 0$. This proves our claim. $\square$

## 2.5 Plan of the proof of Theorem 2.7

We define inductively a sequence of solutions that solve a linear version of our coagulation model. More precisely, let $H_{n+1}$ be defined in the following manner.

$$\partial_t H_{n+1}(y, v, t) + \frac{1}{\epsilon} \partial_y [(v^\alpha - y)H_{n+1}(y, v, t)] = \mathbb{K}[H_{n+1}, H_n](y, v, t) \\ := \int_0^{\frac{v}{2}} K(v - w, w)H_{n+1}(y, v - w, t)H_n(y, w, t)dw \\ - \int_0^{\infty} K(v, w)H_{n+1}(y, v, t)H_n(y, w, t)dw, \quad (2.51)$$

with $H_n(y, v, 0) \in C^1(\mathbb{R} \times (0, \infty))$, for all $n \in \mathbb{N}$ and such that

$$(1 + v^b)H_{n+1}(y, v, 0) = (1 + v^b)H_n(y, v, 0) \le \frac{A}{1 + |y - v^\alpha|^m}, \quad \text{for all } n \in \mathbb{N}. \quad (2.52)$$

At $n = 0$, we take $H_0$ to be

$$(1 + v^b)H_0(y, v, t) = \frac{Ae^{\frac{v}{\epsilon}}}{1 + e^{\frac{mv}{\epsilon}}|y - v^\alpha|^m} \quad (2.53)$$

or in other words

$$\partial_t H_0 + \partial_y[(v^\alpha - y)H_0] = 0; \quad (2.54) \\ (1 + v^b)H_0(y, v, 0) = \frac{A}{1 + |y - v^\alpha|^m}.$$

It holds that there exists $T \le 1$ sufficiently small and independent of $\epsilon$ such that there exists a solution $H_{n+1} \in C^1([0, T] \times \mathbb{R} \times (0, \infty))$ which solves (2.51) with $H_{n+1}(y, v, 0)$ as in (2.52).
**Proposition 2.17.** Let $N > 1$, $\epsilon > 0$, $n \in \mathbb{N}$. Then there exists $T \le 1$, sufficiently small but independent of $\epsilon$, $n$, or $N$, such that if $H_{\epsilon,in} \in C^1(\mathbb{R} \times (0, \infty))$ such that $\text{supp}H_{\epsilon,in} \subseteq \{[-N^\alpha, N^\alpha] \times [0, N]\}$ and if $H_n \in C^1([0, T] \times \mathbb{R} \times (0, \infty))$ is such that $\text{supp}H_n(t) \subseteq \{[-N^\alpha, N^\alpha] \times [0, N]\}$, for all $t \in [0, T]$, then there exists $H_{n+1} \in C^1([0, T] \times \mathbb{R} \times (0, \infty))$ that solves

$$ \partial_t H_{n+1}(y, v, t) + \frac{1}{\epsilon} \partial_y [(v^\alpha - y)H_{n+1}(y, v, t)] = K_N[H_{n+1}, H_n](y, v, t) \quad (2.55) $$

$$ \begin{aligned} &:= \int_0^{\frac{v}{2}} K_N(v-w, w)H_{n+1}(y, v-w, t)H_n(y, w, t)dw \\ &\quad - \int_0^{\infty} K_N(v, w)H_{n+1}(y, v, t)H_n(y, w, t)dw, \end{aligned} $$

for all $t \in [0, T]$, where

$$ K_N(v, w) := K(v, w)\chi_N(v + w), \quad \text{with} \quad (2.56) $$

$$ \chi_N \in C([0, \infty)), \quad \chi_N: [0, \infty) \to [0, 1] \quad \text{such that} \quad \chi_N(v) = 1 \quad \text{if } v \le \frac{N}{2} \quad \text{and} \quad \chi_N(v) = 0 \quad \text{if } v \ge N. $$

Moreover, it holds that

$$ \text{supp} H_{n+1}(t) \subseteq \{[-N^\alpha, N^\alpha] \times [0, N]\}, \text{ for all } t \in [0, T]. \quad (2.57) $$

We prove Proposition 2.17 in Appendix B.

**Remark 2.18.** In principle, since the functions in Proposition 2.17 depend on the iteration $n$ and on the truncation $N$, we have to keep in mind that we work with a sequence $H_{n,N}$. The compact support of functions will be used in Proposition 3.1. We will then obtain a uniform fast decaying bound for the sequence of solutions $H_{n,N}$, which is in particular independent of the truncation $N$ in (2.56) and the iteration in $n \in \mathbb{N}$. Then one can pass to the limit $N \to \infty$. Since this procedure is standard once one has the bounds on the solution, we omit the dependence on $N$ and work directly with $K$ and equation (2.51) in order to simplify the notation.

We now focus on proving Theorem 2.7.

**Plan of the proof of Theorem 2.7.** Let $\epsilon \in (0, 1)$, fixed. We will prove that there exists $T \le 1$ which is independent of $\epsilon$ such that for all $t \in [0, T]$, it holds that if

$$ H_n(y, v, t) \le T_1(y, v, t) + T_2(y, v, t) + T_3(y, v, t), \quad \text{for all } y \in \mathbb{R}, v \in (0, \infty), $$

where $T_1, T_2, T_3$ are as in (2.11)-(2.13), then

$$ H_{n+1}(y, v, t) \le T_1(y, v, t) + T_2(y, v, t) + T_3(y, v, t), \quad \text{for all } y \in \mathbb{R}, v \in (0, \infty). \quad (2.58) $$

• **Step 1. Region $y \le (\frac{v}{3})^\alpha$**.

We will prove that in this region $G_\epsilon$ is a supersolution of (2.51), where $G_\epsilon$ is defined as follows.

$$ \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y [(v^\alpha - y)G_\epsilon(y, v, t)] + \frac{L_\xi(v)v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) = 0, $$
for some suitably chosen $d \in \mathbb{N}$, some sufficiently large $L > 1$, and where $\xi$ is such that

$$ \xi \in C([0, \infty)), \quad \xi: [0, \infty) \to [0, 1] \text{ such that } \xi(v) = 1 \text{ if } v \ge 1 \text{ and } \xi(v) = 0 \text{ if } v \le \frac{1}{2}. \quad (2.59) $$

We then prove that when $y \le (\frac{v}{3})^\alpha$, it holds that $G_\epsilon$ satisfies the following upper bound

$$ G_{\epsilon}(y, v, t) \le T_3(y, v, t), \quad (2.60) $$

where $T_3$ is as in (2.13).

This is the content of Section 4.

• **Step 2. Region $y \ge (\frac{v}{3})^\alpha$**.

**Continuity argument:** We first prove by continuity that if at time $s \ge 0$, we have that $H_{n+1}(y, v, s) \le T_1(y, v, s) + T_2(y, v, s) + T_3(y, v, s)$, then for a sufficiently small $\delta$ (which depends on $\epsilon$) it holds that $H_{n+1}(y, v, z) \le 4[T_1(y, v, z) + T_2(y, v, z) + T_3(y, v, z)]$, for all $z \in [s, s+\delta]$. This is the content of Proposition 3.1.

**Iteration argument:** We then work with the mild formulation of the linear version of our coagulation model (2.51). More precisely, we have that

$$ H_{n+1}(y, v, t) = S_{\epsilon}(t)[H_{n+1}(y, v, 0)]D[H_n](y, v, 0, t) \quad (2.61) \\ + \int_0^t \int_{(0, \frac{v}{2})} D[H_n](y, v, s, t)S_{\epsilon}(t-s)[K(v-v', v')H_{n+1}(y, v-v', s)H_n(y, v', s)]ds, $$

with $D[H_n]$ defined as in (2.5). In other words, we have that

$$ H_{n+1}(y, v, t) \le S_{\epsilon}(t)[H_{n+1}(y, v, 0)] + 2K_0 v^{\gamma} \int_0^t \int_{(0, \frac{v}{2})} S_{\epsilon}(t-s)[H_{n+1}(y, v-v', s)H_n(y, v', s)]ds. $$

For $t \le s + \tilde{\delta}$, the following holds.

• By induction, we have that $H_n(y, v, t) \le T_1 + T_2 + T_3$.

• By the continuity argument, we have that $H_{n+1}(y, v, t) \le 4(T_1 + T_2 + T_3)$.

We then prove that as long as $t \le 1$ and $y \ge (\frac{v}{3})^\alpha$, it holds that

$$ C v^{\gamma} \int_{0}^{t} \int_{(0, \frac{v}{2})} S_{\epsilon}(t-s) [H_{n+1}(y, v-v', s) H_{n}(y, v', s)] ds \le \frac{T_1 + T_2}{2}. $$

Since if $y \ge (\frac{v}{3})^\alpha$ we also have that $S_\epsilon(t)[H_{n+1}(y, v, 0)] \le \frac{T_1+T_2}{2}$, we deduce that

$$ H_{n+1}(y, v, t) \le T_1 + T_2, \text{ for all } t \le \min\{1, s + \tilde{\delta}\}, y \ge (\frac{v}{3})^\alpha. \quad (2.62) $$

From Step 1, we have

$$ H_{n+1}(y, v, t) \le T_3, \text{ for all } t \le 1, y \le (\frac{v}{3})^\alpha. $$

We can thus conclude that $H_{n+1}(y, v, t) \le T_1 + T_2 + T_3$, for all $t \le \min\{1, s + \tilde{\delta}\}$ and then iterate the argument in order to obtain that the upper bound (2.62) holds true for all $t \le 1$.

This is the content of Section 3 of this paper.
• Step 3. Limit as $n \to \infty$

Let $t \ge 0$, $y \in \mathbb{R}$ and $v > 0$. For $n \in \mathbb{N}$, we denote by

$$R_n(y, v, t) := H_{n+1}(y, v, t) - H_n(y, v, t). \qquad (2.63)$$

From (2.51), we find that $R_n$ satisfies

$$\left\{ \begin{array}{l} \partial_t R_n(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)R_n(y, v, t)] = \mathbb{K}[R_n, H_n] + \mathbb{K}[H_n, R_{n-1}]; \\ R_n(y, v, 0) = 0, \end{array} \right. \qquad (2.64)$$

where $\mathbb{K}[f, g]$, for two functions $f, g$, is as in (2.51). Since $\mathbb{K}[R_n, H_n]$ is a linear operator in $R_n$ and $\mathbb{K}[H_n, R_{n-1}]$ does not depend on $H_n$, we use Duhamel's principle in order to deduce that in order to find estimates for $R_n$ it is sufficient to find estimates for the system

$$\left\{ \begin{array}{l} \partial_t R_n^s(x, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)R_n^s(y, v, t)] = \mathbb{K}[R_n^s, H_n], \quad \text{for } t > s; \\ R_n^s(y, v, s) = \mathbb{K}[H_n, R_{n-1}^s]. \end{array} \right. \qquad (2.65)$$

We then use induction in order to prove that

$$|R_n(t)| \le (Cte^{\frac{1}{\epsilon}})^n, \quad \text{for all } n \in \mathbb{N}. \qquad (2.66)$$

We can thus conclude that

$$\text{for all } t \approx \mathcal{O}(e^{-\frac{1}{\epsilon}}), R_n(t) \to 0 \text{ as } n \to \infty. \qquad (2.67)$$

This was also the strategy in [9]. Nonetheless, the difference is that functions in [9] were uniformly bounded in $\epsilon$ in time, while in our case $H_n(t) \le Ce^{\frac{1}{\epsilon}}$, which is the reason why the exponential term appears in (2.66). We then wish to extend (2.67) to hold for times of order $\mathcal{O}(1)$. By (2.66), we have that $R_n(\frac{e^{-\frac{1}{\epsilon}}}{2}) \le 2^{-n}$ and we can use the invariance under translations in time in order to iterate the argument. Let

$$\tilde{R}_n(t) = R_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right).$$

$\tilde{R}_n$ then satisfies

$$\left\{ \begin{array}{l} \partial_t \tilde{R}_n(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)\tilde{R}_n(y, v, t)] = \mathbb{K}[\tilde{R}_n, H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right)] + \mathbb{K}[H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right), \tilde{R}_{n-1}]; \\ \tilde{R}_n(y, v, 0) = R_n\left(\frac{e^{-\frac{1}{\epsilon}}}{2}\right). \end{array} \right. \qquad (2.68)$$

Since $R_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right) \le 2^{-n} \to 0$ as $n \to \infty$, then (2.68) is to be understood at a formal level as

$$\left\{ \begin{array}{l} \partial_t \tilde{R}_n(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)\tilde{R}_n(y, v, t)] = \mathbb{K}[\tilde{R}_n, H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right)] + \mathbb{K}[H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right), \tilde{R}_{n-1}]; \\ \tilde{R}_n(y, v, 0) = 0. \end{array} \right. \qquad (2.69)$$
We are able to use again Duhamel's principle and find suitable estimates that allow us to extend the argument to hold for longer times. We refer to Section 5 for a rigorous proof.

The choice of presentation, namely we present the proof of Step 2 before Step 1, is since the region $y \ge (\frac{v}{3})^\alpha$ contains the part of the solution where most of the mass will concentrate and is thus more difficult to prove. □

# 3 Region $y \ge (\frac{v}{3})^\alpha$

## 3.1 Continuity argument

We first prove the continuity argument.

**Proposition 3.1 (Continuity argument).** Assume that at time $s \ge 0$, we have that

$$H_{n+1}(y, v, s) \le T_1(y, v, s) + T_2(y, v, s) + T_3(y, v, s), \quad (3.1)$$

where $T_1, T_2, T_3$ are as in (2.11)-(2.13). Then there exists $\tilde{\delta}$, which may depend on $n, \epsilon$, but is independent of $s, y, v$, such that for all $z \in [s, \min\{s + \frac{\epsilon}{m}, 1, s + \tilde{\delta}\}]$, where $m$ is as in (2.1), it holds that

$$H_{n+1}(y, v, z) \le 4[T_1(y, v, z) + T_2(y, v, z) + T_3(y, v, z)]. \quad (3.2)$$

**Remark 3.2.** We will need in the proof of Proposition 3.3 that $\tilde{\delta}$ is independent of $y, v$. By Proposition 2.17, it holds that $H_{n+1}$ has compact support and this will imply that $\tilde{\delta}$ does not depend on $y$ or $v$. We refer to Remark 2.18. We remember we work with a sequence $H_{n,N}$, which is compactly supported and the support depends on $N$. We will then obtain a uniform fast decaying bound for the sequence of solutions $H_{n,N}$, which is in particular independent of the truncation $N$ and then one can pass to the limit as $N \to \infty$. The rest of the estimates in this paper do not depend on the truncation $N$ and thus do not depend on the compact support. As mentioned in Remark 2.18, we omit the dependence on $N$ in order to avoid complicated notation.

**Proof of Proposition 3.1.** We have that (3.1) holds. Let

$$\tilde{\epsilon} := \min\left\{\frac{A\psi(e^{\frac{1}{\tilde{\epsilon}}}(y - v^{\alpha}))}{1 + v^b}, \frac{A\epsilon}{|y - v^{\alpha}|(1 + v^b)}\right\}. \quad (3.3)$$

Since $H_{n+1}$ has compact support, it holds that there exists $\epsilon_N$, where $N$ is as in (2.56), which depends on $N$ but is otherwise independent on $y, v$, such that $\tilde{\epsilon} \ge \epsilon_N$. By continuity in time of $H_{n+1}$, we have that there exists $\tilde{\delta}$ such that for all $z \in [s, s + \tilde{\delta}]$ it holds that $|H_{n+1}(y, v, z) - H_{n+1}(y, v, s)| \le \epsilon_N$. It thus holds that

$$H_{n+1}(y, v, z) \le T_1(y, v, s) + T_2(y, v, s) + T_3(y, v, s) + \epsilon_N.$$

Take $z \in [s, \min\{1, s + \frac{\epsilon}{m}, s + \tilde{\delta}\}]$. We analyze each term separately.

$$T_1(y, v, s) = \frac{2A}{1 + v^b} e^{\frac{s}{\tilde{\epsilon}}} \psi(e^{\frac{s}{\tilde{\epsilon}}}(y - v^{\alpha})) \chi_1(y, v, s) \le \frac{2A}{1 + v^b} e^{\frac{z}{\tilde{\epsilon}}} \psi(e^{\frac{s}{\tilde{\epsilon}}}(y - v^{\alpha})) \chi_1(y, v, s),$$
with $\chi_1$ as in (2.14). Since $z \le s + \frac{\epsilon}{m}$, it holds that

$$1 + e^{\frac{z\epsilon}{\epsilon}} |y - v^{\alpha}|^m = 1 + e^{\frac{sm}{\epsilon}} |y - v^{\alpha}|^m e^{\frac{(z-s)m}{\epsilon}} \le 3(1 + e^{\frac{sm}{\epsilon}} |y - v^{\alpha}|^m) \quad (3.4)$$

and thus we further have that

$$T_1(y, v, s) \le \frac{6A}{1 + v^b} e^{\frac{z}{\epsilon}} \psi(e^{\frac{z}{\epsilon}}(y - v^{\alpha})) \chi_1(y, v, s).$$

We define the set

$$U(s) := \{y \le v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{z}{\epsilon}}\}.$$

Since $z \ge s$, we have $v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{z}{\epsilon}} \le v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{z}{\epsilon}}$. Thus

$$T_1(y, v, s) \le 3T_1(y, v, z) + \frac{6A}{1 + v^b} e^{\frac{z}{\epsilon}} \psi(e^{\frac{z}{\epsilon}}(y - v^{\alpha})) \mathbb{1}_{U(z)\setminus U(s)}, \text{ for } s \le z. \quad (3.5)$$

However, in the region where $\{y \le v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{z}{\epsilon}}\}$, it holds that $e^{\frac{z}{\epsilon}}|y - v^{\alpha}| \ge C_0^{-1} \epsilon^{-\frac{1}{m-1}}$. Thus

$$e^{\frac{z}{\epsilon}} \psi(e^{\frac{z}{\epsilon}}(y - v^{\alpha})) \mathbb{1}_{U(z)\setminus U(s)} \le \frac{1}{|y - v^{\alpha}|} \frac{1}{e^{\frac{z(m-1)}{\epsilon}} |y - v^{\alpha}|^{m-1}} \mathbb{1}_{U(z)\setminus U(s)} \le \frac{C_0^{m-1} \epsilon}{|y - v^{\alpha}|} \mathbb{1}_{U(z)\setminus U(s)}$$

from which we can deduce that

$$\frac{6A}{1 + v^b} e^{\frac{z}{\epsilon}} \psi(e^{\frac{z}{\epsilon}}(y - v^{\alpha})) \mathbb{1}_{U(z)\setminus U(s)} \le \frac{6AC_0^{m-1} \epsilon}{(1 + v^b)|y - v^{\alpha}|} \mathbb{1}_{U(z)\setminus U(s)} \le 3T_2(y, v, z). \quad (3.6)$$

From (3.5) and (3.6), we deduce that

$$T_1(y, v, s) \le 3[T_1(y, v, z) + T_2(y, v, z)], \quad s \le z.$$

Moreover, since $z \ge s$, we also have that

$$T_2(y, v, s) = \frac{2C_0^{m-1}A^2\epsilon}{(1 + v^b)|y - v^{\alpha}|} \chi_2(y, v, s) \le T_2(y, v, z), \quad s \le z,$$

where $\chi_2$ is as in (2.15). Using (3.4), we also have that $T_3(y, v, s) \le 3T_3(y, v, z)$. We now find an upper bound for $\epsilon_N$. Let $\tilde{\epsilon}$ be as in (3.3). We have that

$$\epsilon_N \le \tilde{\epsilon} \le \frac{1}{1 + v^b} \psi(e^{\frac{1}{\epsilon}}(y - v^{\alpha})) \mathbb{1}_{\{y \ge v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{z}{\epsilon}}\} \cup \{y \le (\frac{v}{3})^{\alpha}\}} + T_2(y, v, z)$$

and moreover since $z \le 1$ it follows that

$$\frac{A}{1 + v^b} \psi(e^{\frac{1}{\epsilon}}(y - v^{\alpha})) \le \frac{e^{\frac{z}{\epsilon}}}{1 + v^b} \psi(e^{\frac{z}{\epsilon}}(y - v^{\alpha})).$$

In other words,

$$\tilde{\epsilon} \le T_1(y, v, z) + T_2(y, v, z) + T_3(y, v, z).$$

Thus, we have that (3.2) holds for all $z \in [s, \min\{s + \epsilon, s + \tilde{\delta}, 1\}]$. This concludes our proof. $\square$
## 3.2 Iteration argument

We are now able to find an upper bound for $H_{n+1}$ in the region $y \ge (\frac{v}{3})^{\alpha}$.

**Proposition 3.3** (Region $y \ge (\frac{v}{3})^{\alpha}$). There exist $\epsilon_1 \in (0, 1)$ and $T(A) \in [0, 1]$, sufficiently small and depending on A in (2.11), such that for all $\epsilon \le \epsilon_1$ and for all $n \in \mathbb{N}$, the following statement holds. Let $H_n$ be as in (2.51). Let $T_1, T_2, T_3$ as in (2.11)-(2.13) with $b \ge \max\{\bar{b}(y, \alpha), \gamma + 1\}$ and $m > \max\{\frac{2(y+1)}{\alpha}, \frac{b}{\alpha} + 1\}$, where $\bar{b}(y, \alpha)$ is as in (2.18). Assume that $H_n \le \sum_{i=1}^3 T_i$. Then it holds that $H_{n+1}(y, v, t) \le T_1 + T_2$, for all $t \in [0, T(A)]$ and $y \ge (\frac{v}{3})^{\alpha}$.

*Proof.* Let $y \ge (\frac{v}{3})^{\alpha}$. Using the mild formulation of (2.51), it holds that

$$H_{n+1}(y, v, t) = S_{\epsilon}(t)[H_{n+1}(y, v, 0)]D[H_n](y, v, 0, t) + \int_{0}^{t} \int_{(0, \frac{v}{2})} D[H_n](y, v, s, t)S_{\epsilon}(t-s)[K(v-v', v')H_{n+1}(y, v-v', s)H_n(y, v', s)]ds,$$

with $D[H_n]$ defined as in (2.5). Since $D[H_n] \le 1$ and $K(v-w, w) \le 2K_0v^\gamma$ when $w \in [0, \frac{v}{2}]$, we further have that

$$H_{n+1}(y, v, t) \le S_{\epsilon}(t)[H_{n+1}(y, v, 0)] + 2K_0v^{\gamma} \int_{0}^{t} \int_{(0, \frac{v}{2})} S_{\epsilon}(t-s)[H_{n+1}(y, v-v', s)H_n(y, v', s)]ds.$$

We prove that if $y \ge (\frac{v}{3})^{\alpha}$, then

$$S_{\epsilon}(t)H_{\epsilon, \text{in}}(y, v) \le \frac{T_1 + T_2}{2} \qquad (3.7)$$

and

$$2K_0v^{\gamma} \int_{0}^{t} \int_{(0, \frac{v}{2})} S_{\epsilon}(t-s)[H_{n+1}(y, v-v', s)H_n(y, v', s)]ds \le \frac{T_1 + T_2}{2}. \qquad (3.8)$$

We then conclude that $H_{n+1} \le T_1 + T_2$ when $y \ge (\frac{v}{3})^{\alpha}$ as desired. We first prove (3.7). We thus bound $S_{\epsilon}(t)H_{\epsilon, \text{in}}(y, v)$ when $y \ge (\frac{v}{3})^{\alpha}$. From (2.1), it holds that

$$S_{\epsilon}(t)H_{\epsilon, \text{in}}(y, v) = e^{\frac{t}{\epsilon}}H_{\epsilon, \text{in}}(e^{\frac{t}{\epsilon}}(y - v^{\alpha}) + v^{\alpha}, v) \le \frac{Ae^{\frac{t}{\epsilon}}}{(1 + v^b)(1 + e^{\frac{mt}{\epsilon}}|y - v^{\alpha}|^m)} = \frac{Ae^{\frac{t}{\epsilon}}\psi(e^{\frac{t}{\epsilon}}(y - v^{\alpha}))}{1 + v^b}.$$

On the one hand,

$$\frac{Ae^{\frac{t}{\epsilon}}\psi(e^{\frac{t}{\epsilon}}(y - v^{\alpha}))}{1 + v^b}\chi_1(y, v, t) \le \frac{T_1}{2},$$

where $\chi_1$ is as in (2.14). On the other hand, if $y \le v^{\alpha} - C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{t}{\epsilon}}$ then it holds that $e^{\frac{t}{\epsilon}}|y - v^{\alpha}| \ge C_0^{-1}\epsilon^{-\frac{1}{m-1}}$ and thus

$$Ae^{\frac{t}{\epsilon}}\psi(e^{\frac{t}{\epsilon}}(y - v^{\alpha}))\chi_2(y, v, t) \le \frac{A}{e^{\frac{t(m-1)}{\epsilon}}|y - v^{\alpha}|^m}\chi_2(y, v, t) \le \frac{AC_0^{m-1}\epsilon}{|y - v^{\alpha}|}\chi_2(y, v, t) \le \frac{T_2}{2},$$
where $\chi_2$ is as in (2.15). We now focus on (3.8). By assumption it holds that $H_n \le T_1 + T_2 + T_3$, where $T_1, T_2, T_3$ are as in (2.11)-(2.13). We also have that $H_{\epsilon, \text{in}}(y, v) \le T_1(y, v, 0) + T_2(y, v, 0) + T_3(y, v, 0)$. By Proposition 3.1 and Remark 3.2, this implies that $H_{n+1}(y, v, z) \le 4[T_1(y, v, z) + T_2(y, v, z) + T_3(y, v, z)]$, for all $z \in [0, \min\{\epsilon_m, 1, \tilde{\delta}\}]$, with $\tilde{\delta}$ as in Proposition 3.1. Thus, for all $t \in [0, \min\{\epsilon_m, 1, \tilde{\delta}\}]$, it holds that

$$H_{n+1}(y, v, t) \le S_{\epsilon}(t)[H_{n+1}(y, v, 0)] + 2K_0 v^{\gamma} \int_0^t \int_{(0, \frac{v}{2})} S_{\epsilon}(t-s) [H_{n+1}(y, v-v', s)H_n(y, v', s)] ds \\ \le \frac{T_1 + T_2}{2} + 8K_0 v^{\gamma} \sum_{i,j=1}^3 \int_0^t \int_{(0, \frac{v}{2})} S_{\epsilon}(t-s) [T_i(y, v-v', s)T_j(y, v', s)] ds.$$

We will prove in Proposition 3.6 that for $m > \frac{2(\gamma+1)}{\alpha}$ and $b \ge \gamma+1$, there exist a time $T(A)$, which depends on $A$, and $\epsilon_1 \in (0, 1)$, which depends on $A$, such that

$$8K_0 v^{\gamma} \sum_{i,j=1}^{3} \int_{0}^{t} \int_{(0, \frac{v}{2})} S_{\epsilon}(t-s) [T_i(y, v-v', s)T_j(y, v', s)] ds \le \frac{T_1 + T_2}{2}, \quad (3.9)$$
for all $y \ge (\frac{v}{3})^{\alpha}, t \le \frac{1}{T(A)}, \epsilon \le \epsilon_1$

and thus

$$H_{n+1}(y, v, t) \le T_1 + T_2, \text{ for all } t \in [0, \min\{\epsilon, 1, \tilde{\delta}\}], \quad y \ge (\frac{v}{3})^{\alpha}.$$

From Section 4, we also have that if $b \ge \bar{b}(\gamma, \alpha)$, $m \ge \frac{b}{\alpha} + 1$ it holds that

$$H_{n+1}(y, v, t) \le T_3, \quad y \le (\frac{v}{3})^{\alpha}.$$

Thus $H_{n+1}(y, v, t) \le T_1 + T_2 + T_3$ at $t := \min\{\epsilon, 1, \tilde{\delta}\}$ and we can then use Proposition 3.1 and iterate the argument in order to extend the result for all times $t \le \frac{1}{T(A)}$, where $T(A)$ is as in (3.9). $\square$

**Remark 3.4.** Here and in all the following, we only treat the case when $(\frac{v}{3})^{\alpha} \le v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}$ since the case $(\frac{v}{3})^{\alpha} \ge v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}$ follows similarly and with fewer technicalities. On the one hand, if $(\frac{v}{3})^{\alpha} \ge v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}$, then $\{y \le v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\} \subseteq \{y \le (\frac{v}{3})^{\alpha}\}$ and thus we can bound the region $\{y \le v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}$ as in Section 4. On the other hand, we can bound the region $\{y \ge v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}$ using the estimates from Proposition 3.9 and then we replace $\mathbb{1}_{\{y \ge (\frac{v}{3})^{\alpha}\}}$ in the proof of Proposition 3.6 by $\mathbb{1}_{\{y \ge v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}}$. Notice we do not use Proposition 3.10 in this case.

**Remark 3.5.** Notice that in order to prove (3.9), since we are estimating the function on the set $\{y \ge (\frac{v}{3})^{\alpha}\}$, we actually only need to bound

$$S_{\epsilon}(t)H_{\epsilon, \text{in}}(y, v) + \mathbb{1}_{\{y \ge (\frac{v}{3})^{\alpha}\}} v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{v}{2}} T_{i}(y, v-w, s) T_{j}(y, w, s) dw ds.$$
We now focus on proving (3.9). More precisely, we prove that

**Proposition 3.6.** Let $\epsilon \in (0, 1)$ and $S_\epsilon$ as in (2.7). Let $m > \frac{2(\gamma+1)}{\alpha}$ and $b \ge \gamma + 1$. There exist a time $T(A) \in [0, 1]$, $\epsilon_1 \in (0, 1)$, and $C_0$, which are dependent on $A$ as in (2.1) and on the parameters $\alpha, \gamma, b, m$, such that if $t \le T(A)$, $\epsilon \le \epsilon_1$, then the following estimate holds for all $y \in \mathbb{R}$ and $v \in (0, \infty)$.

$$ \mathbb{1}_{\{y \ge \left(\frac{v}{3}\right)^{\alpha}\}} 8K_0 \sum_{i,j=1}^{3} v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{v}{3}} T_i(y, v-w, s) T_j(y, w, s) \mathrm{dw} \mathrm{ds} \le \frac{T_1(y, v, t) + T_2(y, v, t)}{2}. \quad (3.10) $$

*Remark 3.7.* We remember that $T_2$ as in (2.12) depends on $C_0$ and this is the reason why we require $C_0$ to depend on $A$ in the statement of Proposition 3.6 in order for (3.10) to hold. We refer to (2.9) for the exact choice of $C_0$ and to Proposition 3.10 and to (3.21) in order to see why this definition was required.

We first prove upper bounds for $\int_0^{\frac{v}{2}} T_i(y, v-w, s) T_j(y, w, s) \mathrm{dw}$ and then make use of Lemma 2.16. However, despite the fact that we only need to analyze the region $\{y \ge (\frac{v}{3})^\alpha\}$ as noticed in Remark 3.5, when proving the inequalities before applying the semigroup, we need to analyze a wider range of values of $y$. More precisely, we have the following.

**Lemma 3.8.** Let $\epsilon \in (0, 1)$, $t \in [0, 1]$, and $s \in [0, \frac{t}{2}]$. Let $S_\epsilon$ as in (2.7), $y \in \mathbb{R}$, and $v \in (0, \infty)$. There exists $\bar{\epsilon} \in (0, 1)$ sufficiently small such that for all $\epsilon \le \bar{\epsilon}$, there exists $X(\epsilon) \le 0$ such that $S_\epsilon(t-s)[\mathbb{1}_{\{y \ge X(\epsilon)\}}] = \mathbb{1}_{\{y \ge (\frac{v}{3})^\alpha\}}$.

*Proof.* It holds that

$$ S_{\epsilon}(t-s)[\mathbb{1}_{\{y \ge X(\epsilon)\}}] = \mathbb{1}_{\{y \ge (\frac{v}{3})^{\alpha}\}} \text{ if and only if } \mathbb{1}_{\{e^{\frac{t-s}{\epsilon}}(y-v^{\alpha})+v^{\alpha} \ge X(\epsilon)\}} = \mathbb{1}_{\{y \ge (\frac{v}{3})^{\alpha}\}}. $$

We then have that $e^{-\frac{t-s}{\epsilon}}(X(\epsilon) - v^\alpha) + v^\alpha = (\frac{v}{3})^\alpha$ if and only if $X(\epsilon) = e^{\frac{t-s}{\epsilon}}\left((\frac{v}{3})^\alpha - v^\alpha\right) + v^\alpha = v^\alpha(1 - e^{\frac{t-s}{\epsilon}}) + e^{\frac{t-s}{\epsilon}}(\frac{v}{3})^\alpha$. In other words, for sufficiently small $\epsilon$ we can have that $X(\epsilon) \le 0$. $\square$

We have the following estimates. We first prove estimates for the convolution of $T_1$, $T_3$ with themselves considering different regions of $y$.

**Proposition 3.9** (Estimates for $T_1$ and $T_3$). Let $\epsilon \in (0, 1)$, $t \in [0, 1]$, and $s \in [0, t]$. Let $S_\epsilon$ as in (2.7). Let $i, j \in \{1, 3\}$. Let $b \ge \gamma + 1 - \alpha$ and $m > \frac{2(\gamma+1)}{\alpha}$. Let $T_1$ and $T_3$ be as in (2.11) and (2.13). There exists a constant $C_1(\alpha, \gamma, b, m) > 0$, which depends on $\alpha, \gamma, b, m$, such that for all
$y \in \mathbb{R}$ and $v \in (0, \infty)$ the following estimates hold.

$$v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \left[ \int_{0}^{\frac{\gamma}{2}} T_{i}(y, v-w, s) T_{j}(y, w, s) \mathbb{1}_{\{y \ge v^{\alpha}\}} dw \right] ds \le \frac{C_{1}(\alpha, \gamma, b, m)}{1 + v^{b}} \frac{A^{2} t e^{\frac{t}{\epsilon}} \mathbb{1}_{\{y \ge v^{\alpha}\}}}{(1 + e^{\frac{t}{\epsilon}} |y - v^{\alpha}|)^{m}}; \quad (3.11)$$

$$v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \left[ \int_{0}^{\frac{\gamma}{2}} T_{i}(y, v-w, s) T_{j}(y, w, s) \mathbb{1}_{\{y \le v^{\alpha}\}} dw \right] ds \\ \le \frac{C_{1}(\alpha, \gamma, b, m) M_{2}^{2} A^{2} \epsilon \min\{1, e^{\frac{t}{\epsilon}}|y - v^{\alpha}|\}}{1 + v^{b} |y - v^{\alpha}|}. \quad (3.12)$$

We then focus on proving estimates for the remaining possible combinations of the convolution of $T_i$ with $T_j$.

**Proposition 3.10** (Estimates for the rest of combinations). Let $\epsilon \in (0, 1)$, $t \in [0, 1]$, and $s \in [0, t]$. Let $S_\epsilon$ as in (2.7). Let $i \in \{2, 3\}$. Let $b \ge \gamma+1$ and $m > \frac{\gamma+1}{\alpha}+2$. Let $T_1, \dots, T_3$ as in (2.11)-(2.13). There exists a constant $C_2(\alpha, \gamma, b, m) > 0$, which depends on $\alpha, \gamma, b, m$, such that for all $y \in \mathbb{R}$ and $v \in (0, \infty)$ the following estimates hold.

$$v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{\gamma}{2}} T_{i-1}(y, v-w, s) T_{i}(y, w, s) dw ds = 0; \quad (3.13)$$

$$v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{\gamma}{2}} T_{i}(y, v-w, s) T_{i-1}(y, w, s) dw ds \\ \le \frac{C_{2}(\alpha, \gamma, b, m) M_{2}}{1 + v^{b}} \frac{A^{3} C_{0}^{m-1} \epsilon [t + \epsilon + C_{0}^{-(m-1)}]}{v^{\alpha} - y} \mathbb{1}_{\{y \le v^{\alpha} - e^{-\frac{t}{\epsilon}} C_{0}^{-1} \epsilon^{-\frac{1}{m-1}}\}}; \quad (3.14)$$

$$v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{\gamma}{2}} T_{2}(y, v-w, s) T_{2}(y, w, s) dw ds \\ \le \frac{C_{2}(\alpha, \gamma, b, m) A^{4} C_{0}^{2(m-1)} \epsilon (t+\epsilon)^{2}}{1+v^{b}} \mathbb{1}_{\{y \le v^{\alpha} - e^{-\frac{t}{\epsilon}} C_{0}^{-1} \epsilon^{-\frac{1}{m-1}}\}}. \quad (3.15)$$

Assuming the estimates in Proposition 3.9 and Proposition 3.10 hold true, we are now able to prove Proposition 3.6.

*Proof of Proposition 3.6.* We wish to prove (3.10). We remember that $C_0^{m-1} = M_1 A$, see (2.9). More precisely, $C_0$ is of order $A^{\frac{1}{m-1}}$. We choose $M_1$, depending on $M_2$, such that

$$8K_0 \max\{C_1(\alpha, \gamma, b, m), C_2(\alpha, \gamma, b, m)\} M_2^2 \le \frac{M_1}{24}, \quad (3.16)$$

where $C_1(\alpha, \gamma, b, m)$ is as in Proposition 3.9 and $C_2(\alpha, \gamma, b, m)$ is as in Proposition 3.10.
From (3.11), we can choose a sufficiently small $T$, which depends on $A$, such that for all $t \le T$, it holds that

$$8K_0 \sum_{i,j \in \{1,3\}} v^\gamma \int_0^t S_\epsilon(t-s) \left[ \int_0^{\frac{\gamma}{2}} T_i(y, v-w, s) T_j(y, w, s) \mathbb{1}_{\{y \ge v^\alpha\}} dw \right] ds \\ \le 8K_0 C_1(\alpha, \gamma, b, m) A t T_1(y, v, t) \le \frac{T_1(y, v, t)}{12}.$$

We now estimate (3.12). We prove in this case that

$$\mathbb{1}_{\{y \ge (\frac{\gamma}{3})^\alpha\}} \frac{8K_0 C_1(\alpha, \gamma, b, m) M_2^2 A^2 \epsilon \min\{1, e_{\epsilon}^t |y - v^\alpha|\}}{|y - v^\alpha|} \le \frac{T_1(y, v, t)}{12} + \frac{T_2(y, v, t)}{12}. \quad (3.17)$$

• Let us assume first that $e_{\epsilon}^t |y - v^\alpha| \le 1$. Then

$$\frac{\epsilon}{|y - v^\alpha|(1 + v^b)} \min\{1, e_{\epsilon}^t |y - v^\alpha|\} = \frac{\epsilon e_{\epsilon}^t}{(1 + v^b)} \le \frac{2\epsilon e_{\epsilon}^t}{(1 + e_{\epsilon}^{\frac{m}{\epsilon}} |y - v^\alpha|^m)(1 + v^b)}. \quad (3.18)$$

Thus, if $e_{\epsilon}^t |y - v^\alpha| \le 1$, then for sufficiently small $\epsilon$, which depends on $A$, it holds that

$$\frac{8K_0 C_1(\alpha, \gamma, b, m) M_2^2 A^2 \epsilon \min\{1, e_{\epsilon}^t |y - v^\alpha|\}}{1 + v^b} \frac{|y - v^\alpha|}{|y - v^\alpha|} \le 8K_0 C_1(\alpha, \gamma, b, m) M_2^2 A \epsilon T_1(y, v, t) \le \frac{T_1(y, v, t)}{24}. \quad (3.19)$$

• If $C_0^{-1} \epsilon^{-\frac{1}{m-1}} \ge e_{\epsilon}^t |y - v^\alpha| \ge 1$, then

$$\begin{align*}
\frac{\epsilon}{|y - v^\alpha|} \min\{1, e_{\epsilon}^t |y - v^\alpha|\} \mathbb{1}_{\{1 \le e_{\epsilon}^t |y - v^\alpha| \le C_0^{-1} \epsilon^{-\frac{1}{m-1}}\}} &= \frac{\epsilon}{|y - v^\alpha|} \mathbb{1}_{\{1 \le e_{\epsilon}^t |y - v^\alpha| \le C_0^{-1} \epsilon^{-\frac{1}{m-1}}\}} \\
&= \frac{\epsilon e_{\epsilon}^t}{e_{\epsilon}^t |y - v^\alpha|} \mathbb{1}_{\{1 \le e_{\epsilon}^t |y - v^\alpha| \le C_0^{-1} \epsilon^{-\frac{1}{m-1}}\}} \le \frac{C_0^{-(m-1)} e_{\epsilon}^t}{(e_{\epsilon}^t |y - v^\alpha|)^m} \mathbb{1}_{\{1 \le e_{\epsilon}^t |y - v^\alpha| \le C_0^{-1} \epsilon^{-\frac{1}{m-1}}\}} \\
&\le \frac{2C_0^{-(m-1)} e_{\epsilon}^t}{1 + (e_{\epsilon}^t |y - v^\alpha|)^m} \mathbb{1}_{\{1 \le e_{\epsilon}^t |y - v^\alpha| \le C_0^{-1} \epsilon^{-\frac{1}{m-1}}\}}. \tag{3.20}
\end{align*}$$

It thus follows by (3.16) that

$$\frac{8K_0 C_1(\alpha, \gamma, b, m) M_2^2 A^2 \epsilon \min\{1, e_{\epsilon}^t |y - v^\alpha|\}}{1 + v^b} \frac{|y - v^\alpha|}{|y - v^\alpha|} \le 8K_0 C_1(\alpha, \gamma, b, m) M_2^2 A C_0^{-(m-1)} T_1(y, v, t) \le \frac{T_1(y, v, t)}{24}. \quad (3.21)$$

• If $e_{\epsilon}^t |y - v^\alpha| \ge C_0^{-1} \epsilon^{-\frac{1}{m-1}}$, then by the choice of $T_2$ in (2.12)
$$
\begin{equation}
\begin{aligned}
& \mathbb{1}_{\{v^\alpha \ge y \ge (\frac{y}{3})^\alpha\}} \frac{8K_0 C_1(\alpha, \gamma, b, m) M_2^2}{1 + v^b} \frac{A^2 \epsilon}{|y - v^\alpha|} \mathbb{1}_{\{e^{\frac{t}{\epsilon}} |y - v^\alpha| \ge C_0^{-1} \epsilon^{-\frac{1}{m-1}}\}} \\
& \le 8K_0 C_1(\alpha, \gamma, b, m) M_2^2 C_0^{-(m-1)} T_2(y, v, t) \le \frac{T_2(y, v, t)}{12}.
\end{aligned}
\tag{3.22}
\end{equation}
$$

Combining estimates (3.33)-(3.22) together with the fact that $S_{\epsilon}(t-s)(\mathbb{1}_{\{y \le v^{\alpha}\}}) = \mathbb{1}_{\{y \le v^{\alpha}\}}$
which follows by Item 1 of Lemma 2.16, we obtain (3.17).

We now proceed with (3.14). If $t, \epsilon$ are sufficiently small and for sufficiently large $C_0$, then

$$
\begin{align*}
& \mathbb{1}_{\{y \ge (\frac{y}{3})^{\alpha}\}} 8K_0 v^{\gamma} \sum_{i \in \{2,3\}} \int_0^t S_{\epsilon}(t-s) \int_0^{\frac{\gamma}{2}} T_i(y, v-w, s) T_{i-1}(y, w, s) dw ds \\
& \le \frac{8K_0 C_2(\alpha, \gamma, b, m) M_2}{1+v^b} \frac{A^3 C_0^{m-1} \epsilon [t+\epsilon + C_0^{-(m-1)}]}{v^{\alpha}-y} \chi_2(y, v, t) \le \frac{T_2(y, v, t)}{24}.
\end{align*}
$$

This is since by (3.16), it holds that

$$
\begin{equation}
\begin{aligned}
& \frac{8K_0 C_2(\alpha, \gamma, b, m) M_2 C_0^{-(m-1)}}{1 + v^b} \frac{A^2 C_0^{m-1} \epsilon}{v^\alpha - y} \chi_2(y, v, t) \\
& \qquad = 4K_0 C_2(\alpha, \gamma, b, m) M_2 C_0^{-(m-1)} A T_2(y, v, t) \\
& \qquad \le \frac{T_2(y, v, t)}{48}.
\end{aligned}
\tag{3.23}
\end{equation}
$$

Similarly, for (3.15), there exist $T$ and $\epsilon_1$, which depend on $A$, such that for all $t \le T$ and
$\epsilon \le \epsilon_1$, then

$$
\begin{align*}
& \mathbb{1}_{\{y \ge (\frac{y}{3})^{\alpha}\}} 8K_0 v^{\gamma} \sum_{i \in \{2,3\}} \int_0^t S_{\epsilon}(t-s) \int_0^{\frac{\gamma}{2}} T_2(y, v-w, s) T_2(y, w, s) dw ds \\
& \le \frac{8K_0 C_2(\alpha, \gamma, b, m) A^4 C_0^{2(m-1)} \epsilon (t+\epsilon)^2}{1+v^b (v^\alpha - y)} \chi_2(y, v, t) \\
& \le 8K_0 C_2(\alpha, \gamma, b, m) A^2 C_0^{m-1} (t+\epsilon)^2 T_2(y, v, t) \le \frac{T_2(y, v, t)}{24}.
\end{align*}
$$

This concludes our proof.

<figure><img src="image_4.png" /></figure>

The rest of this section is dedicated to proving Proposition 3.9 and Proposition 3.10.

**3.3 Proof of Proposition 3.9**

*Proof of Proposition 3.9.* Let $i, j \in \{1, 3\}$. We focus on finding an upper bound for

$$
v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{\gamma}{2}} T_{i}(y, v-w, s) T_{j}(y, w, s) dw ds. \quad (3.24)
$$
We notice that since $w \in [0, \frac{v}{2}]$, it holds that

$$ 
\begin{aligned}
& v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{v}{2}} T_{i}(y, v-w, s) T_{j}(y, w, s) dw ds \\
& \leq \int_{0}^{\frac{v}{2}} \frac{C(\alpha, \gamma, b, m) A^{2} v^{\gamma}}{(1+w^{b})(1+(v-w)^{b})} \psi(e^{\frac{s}{\epsilon}}(y-w^{\alpha})) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^{\alpha})) dw \\
& \leq \frac{C(\alpha, \gamma, b, m) A^{2} v^{\gamma}}{1+v^{b}} \int_{0}^{\frac{v}{2}} \frac{1}{1+w^{b}} \psi(e^{\frac{s}{\epsilon}}(y-w^{\alpha})) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^{\alpha})) dw.
\end{aligned}
$$

Since $w \in [0, \frac{v}{2}]$, it holds that $w^{\alpha} \leq (v-w)^{\alpha}$, for all $v > 0$, $w \in [0, \frac{v}{2}]$. By (2.43), we have that

$$ 
\begin{aligned}
& \int_{0}^{\frac{v}{2}} \frac{v^{\gamma}}{1+w^{b}} e^{\frac{2s}{\epsilon}} \psi(e^{\frac{s}{\epsilon}}(y-w^{\alpha})) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^{\alpha})) dw \\
&= \int_{0}^{\frac{v}{2}} \frac{v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|y-(v-w)^{\alpha}|)^{m}} \\
&\leq \int_{0}^{\frac{v}{2}} \frac{C(m)v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} \\
&\quad + \int_{0}^{\frac{v}{2}} \frac{C(m)v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-(v-w)^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} =: I_{1} + I_{2}. \quad (3.25)
\end{aligned}
$$

**Proof of (3.11).** We will prove that there exists a constant $C(\alpha, \gamma, b, m)$, which depends on $\alpha, \gamma, b, m$, such that

$$ 
\begin{aligned}
& v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \left[ \int_{0}^{\frac{v}{2}} T_{1}(y, v-w, s) T_{1}(y, w, s) dw \mathbb{1}_{\{y \ge v^{\alpha}\}} \right] ds \\
& \le 4A^{2} \int_{0}^{t} S_{\epsilon}(t-s) \left[ \int_{0}^{\frac{v}{2}} \frac{e^{\frac{2s}{\epsilon}} v^{\gamma}}{(1+w^{b})(1+(v-w)^{b})} \psi(e^{\frac{s}{\epsilon}}(y-w^{\alpha})) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^{\alpha})) dw \mathbb{1}_{\{y \ge v^{\alpha}\}} \right] ds \\
& \le \frac{C(\alpha, \gamma, b, m)}{1+v^{b}} \frac{A^{2} t e^{\frac{t}{\epsilon}}}{1+e^{\frac{m}{\epsilon}}|y-v^{\alpha}|^{m}} \mathbb{1}_{\{y \ge v^{\alpha}\}}.
\end{aligned}
$$

In order to do this, we will first prove some estimates for

$$ e^{\frac{2s}{\epsilon}} \int_{0}^{\frac{v}{2}} \frac{v^{\gamma}}{(1+w^{b})(1+(v-w)^{b})} \psi(e^{\frac{s}{\epsilon}}(y-w^{\alpha})) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^{\alpha})) dw \quad (3.26) $$

and then use Item 1 of Lemma 2.16 to obtain a bound for (3.24). Let $\delta \in (0, \frac{1}{2})$ be sufficiently small. We start by estimating $I_1$ in (3.25). We analyze separately the regions $w \in [0, \delta v]$ and $w \in [\delta v, \frac{v}{2}]$. Moreover, since we are in the case when $y \ge v^{\alpha}$, we have that $y - w^{\alpha} \ge y - (\frac{v}{2})^{\alpha} \ge$
$C(\alpha)v^\alpha$ and $y - w^\alpha \geq y - v^\alpha \geq 0$. We analyze first $w \in [\delta v, \frac{v}{2}]$. In this case

$$
\begin{align*}
& \mathbb{I}_{\{y \ge v^\alpha\}} \int_{\delta v}^{\frac{v}{2}} \frac{v^\gamma}{1+w^b} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^\alpha|)^m} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^m} \\
& \le \mathbb{I}_{\{y \ge v^\alpha\}} \frac{C(\delta, b)e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-v^\alpha|^m)^m} \int_{\delta v}^{\frac{v}{2}} \frac{v^\gamma}{1+v^b} \frac{e^{\frac{s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|^m)^m} \\
& \le \mathbb{I}_{\{y \ge v^\alpha\}} \frac{C(\delta, \alpha, b)e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-v^\alpha|^m)^m} \frac{v^{\gamma+1-\alpha}}{1+v^b} \le \mathbb{I}_{\{y \ge v^\alpha\}} \frac{C(\delta, \alpha, b)e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-v^\alpha|^m)^m}, \tag{3.27}
\end{align*}
$$

where in the last line we used Lemma 2.14 for $\beta = m$ and then the fact that $b \ge \gamma + 1 - \alpha$. Notice also that since $\gamma > 1$ and $\alpha \in (0, 1)$, we have that $\gamma + 1 - \alpha > 0$.

We now analyze the region where $w \le \delta v$. Then $(v-w)^\alpha - w^\alpha > (1-\delta)^\alpha v^\alpha - \delta^\alpha v^\alpha \ge C(\delta, \alpha)v^\alpha$
since $\delta < \frac{1}{2}$. Then, since $m > \frac{\gamma+1}{\alpha}$, it holds that

$$
\begin{align*}
& \mathbb{I}_{\{y \ge v^\alpha\}} \int_0^{\delta v} \frac{v^\gamma}{1+w^b} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^\alpha|^m)(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|^m)} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|^m)^m} \\
& \le \mathbb{I}_{\{y \ge v^\alpha\}} \frac{e^{\frac{s}{\epsilon}} v^\gamma}{(1+e^{\frac{s}{\epsilon}}|y-v^\alpha|^m)^m} \int_0^{\delta v} \frac{e^{\frac{s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|^m)^m} \\
& \le \mathbb{I}_{\{y \ge v^\alpha\}} \frac{C(\delta, \alpha, \gamma)e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-v^\alpha|^m)^m} \int_0^{\delta v} \frac{v^\gamma}{1+v^{\gamma+1-\alpha}} \frac{e^{\frac{s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|^m)^{m+1-\frac{\gamma+1}{\alpha}}} \\
& \le \mathbb{I}_{\{y \ge v^\alpha\}} \frac{C(\delta, \alpha, \gamma)e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-v^\alpha|^m)^m} \frac{v^{\gamma+1-\alpha}}{1+v^{\gamma+1-\alpha}}, \tag{3.28}
\end{align*}
$$

where in the last line we used Lemma 2.14 for $\beta = m - \frac{\gamma+1}{\alpha} + 1$.

We now estimate $I_2$ in (3.25). We start by analyzing the region $w \in [\delta v, \frac{v}{2}]$. Following the estimates for $I_1$, when $y \ge v^\alpha$ it holds that $y - (v-w)^\alpha \ge y - v^\alpha \ge 0$. By Lemma 2.14, we thus have as before that

$$
\mathbb{I}_{\{y \ge v^{\alpha}\}} \int_{\delta v}^{\frac{v}{2}} \frac{v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-(v-w)^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} \le \mathbb{I}_{\{y \ge v^{\alpha}\}} \frac{C(\delta, \alpha, b)e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-v^{\alpha}|)^{m}}
$$

In the region where $w \in [0, \delta v]$, we have that $(v-w)^\alpha - w^\alpha \ge C(\delta)v^\alpha$ and thus we obtain as
before that

$$
\begin{align*}
& \mathbb{I}_{\{y \ge v^{\alpha}\}} \int_{0}^{\delta v} \frac{v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-(v-w)^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} \\
& \le \mathbb{I}_{\{y \ge v^{\alpha}\}} \frac{C(\delta, \alpha, \gamma)e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-v^{\alpha}|)^{m}}.
\end{align*}
$$

Combining all the above estimates and since $b > 2$, we obtain that

$$
\begin{align*}
& \mathbb{I}_{\{y \ge v^{\alpha}\}} e^{\frac{2s}{\epsilon}} \int_{0}^{\frac{v}{2}} \frac{v^{\gamma}}{(1+w^{b})(1+(v-w)^{b})} \psi(e^{\frac{s}{\epsilon}}(y-w^{\alpha})) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^{\alpha})) dw \\
& \le \mathbb{I}_{\{y \ge v^{\alpha}\}} \frac{e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-v^{\alpha}|)^m} \frac{C(\delta, \alpha, \gamma, b, m)}{1+v^{b}}
\end{align*}
$$
and then using Item 1 of Lemma 2.16 it follows that when $y \ge v^\alpha$, it holds that

$$v^\gamma \int_0^t S_\epsilon(t-s) \left[ \int_0^{\frac{\gamma}{2}} T_1(y, v-w, s) T_1(y, w, s) dw \mathbb{1}_{\{y \ge v^\alpha\}} \right] ds \le \mathbb{1}_{\{y \ge v^\alpha\}} \frac{C(\delta, \alpha, \gamma, b, m) t e^{\frac{t}{\epsilon}}}{(1 + e^{\frac{t}{\epsilon}} |y - v^\alpha|^m)^m} \frac{A^2}{1 + v^b}.$$

**Proof of (3.12).**

* Case $y \le 0$: We will prove estimates for (3.26) in this case too.

We have that $0 \ge 2^{-\alpha}(y-v^\alpha) \ge y-(\frac{\gamma}{2})^\alpha$ and thus $|y-(\frac{\gamma}{2})^\alpha| \ge 2^{-\alpha}|y-v^\alpha|$. Moreover, for $w \in [0, \frac{\gamma}{2}]$, it holds that

$$y - (v - w)^\alpha \le y - (\frac{\gamma}{2})^\alpha \le -(\frac{\gamma}{2})^\alpha \le 0. \quad (3.29)$$

We thus have that

$$|y - (v - w)^\alpha| \ge |y - (\frac{\gamma}{2})^\alpha| \ge 2^{-\alpha}|y - v^\alpha| \quad \text{and} \quad |y - (v - w)^\alpha| \ge C(\alpha)v^{-\alpha}. \quad (3.30)$$

As in (3.25), by (2.43), it holds that

$$e^{\frac{2s}{\epsilon}} \int_0^{\frac{\gamma}{2}} \frac{v^\gamma}{1+w^b} \psi(e^{\frac{s}{\epsilon}}(y-w^\alpha)) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^\alpha)) dw \le I_1 + I_2,$$

with $I_1$ and $I_2$ as in (3.25). We start by estimating $I_2$ in (3.25). We first analyze the region where $w \in [\delta v, \frac{\gamma}{2}]$. By (3.30), we have that

$$\mathbb{1}_{\{y \le 0\}} \int_{\delta v}^{\frac{\gamma}{2}} \frac{v^\gamma}{1+w^b} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-(v-w)^\alpha|)^m} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^m} \\ \le \mathbb{1}_{\{y \le 0\}} \frac{C(\delta, \alpha, b)}{(1+e^{\frac{s}{\epsilon}}|y-v^\alpha|)^{m-1}} \int_{\delta v}^{\frac{\gamma}{2}} \frac{v^{\gamma-\alpha}}{1+v^b} \frac{e^{\frac{s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^m} \quad (3.31)$$

and we use Lemma 2.14 in order to conclude that

$$\mathbb{1}_{\{y \le 0\}} \int_{\delta v}^{\frac{\gamma}{2}} \frac{v^{\gamma}}{1+w^b} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-(v-w)^{\alpha}|)^m} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^m} \\ \le \mathbb{1}_{\{y \le 0\}} \frac{C(\delta, \alpha, b)}{(1+|y-v^{\alpha}|)^{m-1}} \frac{v^{\gamma+1-2\alpha}}{1+v^b} \quad (3.32)$$

in this case.

If $w \in [0, \delta v]$, by (3.30) and (3.28), it follows that

$$\mathbb{1}_{\{y \le 0\}} \int_{0}^{\delta v} \frac{v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-(v-w)^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} \le \mathbb{1}_{\{y \le 0\}} \frac{C(\delta, \alpha, \gamma)}{(1+|y-v^{\alpha}|)^{m-1}}.$$
We now estimate $I_1$ in (3.25). We analyze the region $w \in [\delta v, \frac{v}{2}]$. In this region, it holds that $y-w^\alpha \le y - \delta^\alpha v^\alpha \le -\delta^\alpha v^\alpha \le 0$ and $y-w^\alpha \le y - \delta^\alpha v^\alpha \le \delta^\alpha (y-v^\alpha) \le 0$. As such, we can proceed as in (3.31) and (3.32) in order to prove that

$$ \mathbb{1}_{\{y \le 0\}} \int_{\delta v}^{\frac{v}{2}} \frac{v^\gamma}{1+w^b} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^\alpha|)^m} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^m} \le \mathbb{1}_{\{y \le 0\}} \frac{C(\delta, \alpha, b)}{(1+|y-v^\alpha|)^{m-1}} $$

We now estimate $I_1$ in (3.25) in the region $w \in [0, \delta v]$. We have that

$$ \begin{aligned} & \mathbb{1}_{\{y \le 0\}} \int_{0}^{\delta v} \frac{v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} \\ & = \mathbb{1}_{\{y \in [-v^{\alpha}, 0]\}} \int_{0}^{\delta v} \frac{v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} \\ & + \mathbb{1}_{\{y \le -v^{\alpha}\}} \int_{0}^{\delta v} \frac{v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}}. \end{aligned} $$

For the region $y \le -v^\alpha$ we have that

$$ y - w^{\alpha} \le y = \frac{y}{2} + \frac{y}{2} \le \frac{y - v^{\alpha}}{2} \le 0 \text{ and } y - w^{\alpha} \le -v^{\alpha} \le 0. $$

Thus, when $y \le -v^\alpha$, we have that

$$ \begin{aligned} & \mathbb{1}_{\{y \le -v^{\alpha}\}} \int_{0}^{\delta v} \frac{v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} \\ & \le \mathbb{1}_{\{y \le -v^{\alpha}\}} \frac{C(m)v^{\gamma-\alpha}}{(1+e^{\frac{s}{\epsilon}}|y-v^{\alpha}|)^{m-1}} \int_{0}^{\delta v} \frac{e^{\frac{s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} \end{aligned} $$

and we can proceed as in (3.28) in order to show that

$$ \mathbb{1}_{\{y \le -v^{\alpha}\}} \int_{0}^{\delta v} \frac{v^{\gamma}}{1+w^{b}} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^{\alpha}|)^{m}} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^{\alpha}-w^{\alpha}|)^{m}} \le \frac{C(\delta, \alpha, \gamma, m)}{(1+|y-v^{\alpha}|)^{m-1}} $$

On the other hand, if $y \in [-v^\alpha, 0]$ and $w \in [0, \delta v]$, we have that $v^\alpha \le |y - v^\alpha| \le 2v^\alpha$ and that
$|(v-w)^\alpha - w^\alpha| \ge C(\delta, \alpha)v^\alpha$. Thus

$$
\begin{align*}
& \mathbb{1}_{\{y \in [-v^\alpha, 0]\}} \int_0^{\delta v} \frac{v^\gamma e^{\frac{2s}{\epsilon}}}{1+w^b (1+e^{\frac{s}{\epsilon}}|y-w^\alpha|)^m (1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^m} dw \\
& \le \mathbb{1}_{\{y \in [-v^\alpha, 0]\}} v^\gamma \int_0^{\delta v} \frac{e^{\frac{2s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^m} \\
& \le \mathbb{1}_{\{y \in [-v^\alpha, 0]\}} \frac{C(\delta, \alpha, m)v^\gamma}{1+|y-v^\alpha|^{\frac{m}{2}}} \int_0^{\delta v} \frac{e^{\frac{2s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^{\frac{m}{2}}} \\
& \le \mathbb{1}_{\{y \in [-v^\alpha, 0]\}} \frac{C(\delta, \alpha, m)v^{\gamma-\alpha}}{1+|y-v^\alpha|^{\frac{m}{2}}} \int_0^{\delta v} \frac{e^{\frac{s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^{\frac{m}{2}-1}} \\
& \le \mathbb{1}_{\{y \in [-v^\alpha, 0]\}} \frac{C(\delta, \alpha, \gamma, m)}{1+|y-v^\alpha|^{\frac{m}{2}}} \frac{v^{\gamma-\alpha}}{1+v^{\gamma+1-2\alpha}} \int_0^{\delta v} \frac{e^{\frac{s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^{\frac{m}{2}+1-\frac{\gamma+1}{\alpha}}} \\
& \le \mathbb{1}_{\{y \in [-v^\alpha, 0]\}} \frac{C(\delta, \alpha, \gamma, m)}{1+|y-v^\alpha|^{\frac{m}{2}}},
\end{align*}
$$

where in the last line we used Lemma 2.14 for $\beta = \frac{m}{2} + 1 - \frac{\gamma+1}{\alpha}$.

Combining all the above estimates and performing similar computations as the ones in Proposition 2.15 and since $m > \frac{2(\gamma+1)}{\alpha} > 2$, we obtain that

$$
v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \left[ \int_{0}^{\frac{v}{2}} T_{1}(y, w, s) T_{1}(y, v-w, s) dw \mathbb{1}_{\{y \le 0\}} \right] ds \le \frac{C(\alpha, \gamma, b, m) A^2 \epsilon \min\{1, e^{\frac{l}{\epsilon}}|y-v^{\alpha}|\}}{(1+v^b)|y-v^{\alpha}|} \quad (3.33)
$$

• **Case $y \in [0, v^\alpha]$**: We have in this case that $|y - v^\alpha| \le v^\alpha$.

• **Subcase $v \ge 1$**: We will prove that

$$
\begin{align}
& \mathbb{1}_{\{0 \le y \le 1^\alpha\}} \int_0^{\frac{v}{2}} \frac{e^{\frac{2s}{\epsilon}} v^\gamma}{(1+w^b)(1+(v-w)^b)} \psi(e^{\frac{s}{\epsilon}}(y-w^\alpha)) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^\alpha)) dw \nonumber \\
& \le \frac{C(\alpha, \gamma, b, m) e^{\frac{s}{\epsilon}}}{(1+v^b)(1+e^{\frac{s}{\epsilon}}|y-(\frac{v}{2})^\alpha|)^l}, \tag{3.34}
\end{align}
$$

for $l = \frac{m}{2}$. Notice that $l > 2$ since $m > \frac{2(\gamma+1)}{\alpha} > 4$. If (3.34) holds true, then we prove that

$$
\int_0^t S_\epsilon(t-s) \left[ \frac{e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-(\frac{v}{2})^\alpha|)^l} \right] ds \le \frac{C(\alpha) \epsilon \min\{1, e^{\frac{l}{\epsilon}}|y-v^\alpha|\}}{|y-v^\alpha|} \quad (3.35)
$$
We now prove (3.35). Assume first that $|y - v^\alpha| \ge e^{-\frac{t}{\epsilon}}$. It follows that

$$
\begin{align*}
\int_0^t S_\epsilon(t-s) \left[ \frac{e^{\frac{s}{\epsilon}}}{(1 + e^{\frac{sl}{\epsilon}}|y - (\frac{v}{2})^\alpha|)^l} \right] ds &= \int_0^t \frac{e^{\frac{l}{\epsilon}} ds}{1 + e^{\frac{sl}{\epsilon}}(y - v^\alpha)e^{\frac{t-s}{\epsilon}} + v^\alpha - (\frac{v}{2})^\alpha|l|} \\
&= \int_0^t \frac{e^{\frac{l}{\epsilon}} ds}{1 + e^{\frac{tl}{\epsilon}}(y - v^\alpha) + e^{-\frac{t-s}{\epsilon}}(v^\alpha - (\frac{v}{2})^\alpha)|l|} \\
&= \int_0^t \frac{e^{\frac{l}{\epsilon}} ds}{1 + e^{\frac{tl}{\epsilon}}(y - v^\alpha) + e^{-\frac{s}{\epsilon}}(v^\alpha - (\frac{v}{2})^\alpha)|l|},
\end{align*}
$$

where in the last line we made the change of variables $z = t - s$. Making now the change of variable $z = \frac{s}{\epsilon}$, we further obtain that

$$
\int_0^t S_\epsilon(t-s) \left[ \frac{e^{\frac{s}{\epsilon}}}{\left(1 + e^{\frac{sl}{\epsilon}}|y - \left(\frac{v}{2}\right)^{\alpha}\right)^l} \right] ds = \epsilon \int_0^t \frac{e^{\frac{l}{\epsilon}} ds}{1 + e^{\frac{tl}{\epsilon}}(y - v^{\alpha}) + e^{-s}(v^{\alpha} - \left(\frac{v}{2}\right)^{\alpha})|l|}.
$$

We further make the change of variables $z = y - v^\alpha + e^{-s}(v^\alpha - (\frac{v}{2})^\alpha)$. Notice that

$$
\begin{align*}
dz &= -e^{-s}(v^{\alpha} - (\frac{v}{2})^{\alpha})ds; \\
e^{-s} &= \frac{z + (v^{\alpha} - y)}{v^{\alpha} - (\frac{v}{2})^{\alpha}}; \\
\left|\frac{ds}{dz}\right| &= \frac{1}{z + (v^{\alpha} - y)}.
\end{align*}
$$

It then follows that

$$
\begin{align*}
\int_0^t S_\epsilon(t-s) \left[ \frac{e^{\frac{s}{\epsilon}}}{\left(1 + e^{\frac{sl}{\epsilon}}|y - \left(\frac{v}{2}\right)^{\alpha}\right)^l} \right] ds &= \epsilon \int_{y-v^\alpha+e^{-\frac{l}{\epsilon}}(v^\alpha-\left(\frac{v}{2}\right)^\alpha)}^{y-\left(\frac{v}{2}\right)^\alpha} \frac{e^{\frac{l}{\epsilon}} dz}{1+e^{\frac{tl}{\epsilon}}|z|^l} \left|\frac{ds}{dz}\right| \\
&= \epsilon \int_{y-v^\alpha+e^{-\frac{l}{\epsilon}}(v^\alpha-\left(\frac{v}{2}\right)^\alpha)}^{y-\left(\frac{v}{2}\right)^\alpha} \frac{e^{\frac{l}{\epsilon}} dz}{1+e^{\frac{tl}{\epsilon}}|z|^l} \frac{1}{z+(v^\alpha-y)}.
\end{align*}
$$

It follows that

$$
\begin{align*}
\int_0^t S_\epsilon(t-s) \left[ \frac{e^{\frac{s}{\epsilon}}}{\left(1 + e^{\frac{sl}{\epsilon}}|y - \left(\frac{v}{2}\right)^{\alpha}\right)^l} \right] ds &= \epsilon e^{\frac{l}{\epsilon}} \int_{e^{\frac{l}{\epsilon}}(y-v^\alpha) + (v^\alpha - \left(\frac{v}{2}\right)^\alpha)}^{e^{\frac{l}{\epsilon}}(y-\left(\frac{v}{2}\right)^\alpha)} \frac{dz}{1+|z|^l} \frac{1}{z+e^{\frac{l}{\epsilon}}(v^\alpha-y)} \\
&= \epsilon e^{\frac{l}{\epsilon}} \int_{(v^\alpha - \left(\frac{v}{2}\right)^\alpha)}^{e^{\frac{l}{\epsilon}}(v^\alpha - \left(\frac{v}{2}\right)^\alpha)} \frac{dz}{1+|z-e^{\frac{l}{\epsilon}}(v^\alpha-y)|^l z} \frac{1}{z}.
\end{align*}
$$

We thus have that

$$
\epsilon e^{\frac{l}{\epsilon}} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha})}^{e^{\frac{l}{\epsilon}}(v^{\alpha} - (\frac{v}{2})^{\alpha})} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^l z} = \epsilon e^{\frac{l}{\epsilon}} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha})}^{e^{\frac{l}{\epsilon}}(v^{\alpha} - (\frac{v}{2})^{\alpha})} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^l z} \\
+ \epsilon e^{\frac{l}{\epsilon}} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \ge \frac{\epsilon}{2}(v^{\alpha} - y)} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^l z}.
$$
For the first term, since $v \ge 1$, we have that

$$ \epsilon e^{\frac{l}{\epsilon}} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \le \frac{l}{\epsilon} \frac{1}{z}} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^l} \le \frac{C(\alpha) \epsilon e^{\frac{l}{\epsilon}}}{v^{\alpha}} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \le \frac{l}{\epsilon} \frac{1}{z}} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^l} \\ \le \frac{C(\alpha) \epsilon e^{\frac{l}{\epsilon}}}{e^{\frac{l}{\epsilon}}(v^{\alpha} - y)} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \le \frac{l}{\epsilon} \frac{1}{z}} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^{l-1}} \le \frac{C(\alpha) \epsilon}{(v^{\alpha} - y)}, $$

since $l > 2$, which is the correct estimate for the region $|y - v^{\alpha}| \ge e^{-\frac{l}{\epsilon}}$.

For the second term, we have that

$$ \epsilon e^{\frac{l}{\epsilon}} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \ge \frac{l}{\epsilon} \frac{1}{z}} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^l} \le \frac{2\epsilon}{v^{\alpha} - y} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \ge \frac{l}{\epsilon} \frac{1}{z}} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^l} \\ \le \frac{C\epsilon}{v^{\alpha} - y}, $$

which concludes our proof in the case when $v \ge 1$, $|y - v^{\alpha}| \ge e^{-\frac{l}{\epsilon}}$.

We now focus on the case $v \ge 1$, $|y - v^{\alpha}| \le e^{-\frac{l}{\epsilon}}$. Since $v \ge 1$, we have that

$$ \int_0^t S_\epsilon(t-s) \left[ \frac{e^{\frac{s}{\epsilon}}}{\left(1 + e^{\frac{s}{\epsilon}}|y - (\frac{v}{2})^\alpha|\right)^l} \right] ds \le \frac{C(\alpha) \epsilon e^{\frac{l}{\epsilon}}}{v^\alpha} \int_{(v^\alpha - (\frac{v}{2})^\alpha)}^{e^{\frac{l}{\epsilon}}(v^\alpha - (\frac{v}{2})^\alpha)} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^\alpha - y)|^l} \\ \le C(\alpha) \epsilon e^{\frac{l}{\epsilon}} \int_{(v^\alpha - (\frac{v}{2})^\alpha)}^{e^{\frac{l}{\epsilon}}(v^\alpha - (\frac{v}{2})^\alpha)} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^\alpha - y)|^l} \le C(\alpha) \epsilon e^{\frac{l}{\epsilon}}. $$

We now focus on proving (3.34) when $v \ge 1$. We assume without loss of generality that $|y - w^\alpha| \ge |y - (\frac{v}{2})^\alpha|$ since the region $|y - (v - w)^\alpha| \ge |y - (\frac{v}{2})^\alpha|$ can be treated similarly. As before, we analyze separately the regions $w \in [\delta v, \frac{v}{2}]$ and $w \in (0, \delta v)$, for some $\delta < \frac{1}{2}$. First, we notice that

$$ \int_{\delta v}^{\frac{v}{2}} \frac{e^{\frac{2s}{\epsilon}} v^{\gamma}}{(1 + w^b)(1 + (v - w)^b)} \psi(e^{\frac{s}{\epsilon}}(y - w^{\alpha})) \psi(e^{\frac{s}{\epsilon}}(y - (v - w)^{\alpha}) dw \\ \le \frac{C(\delta, b) e^{\frac{s}{\epsilon}}}{(1 + e^{\frac{s}{\epsilon}}|y - (\frac{v}{2})^{\alpha}|)^m} \int_{\delta v}^{\frac{v}{2}} \frac{v^{\gamma}}{(1 + v^b)^2} \frac{e^{\frac{s}{\epsilon}} dw}{(1 + e^{\frac{s}{\epsilon}}|y - (v - w)^{\alpha}|)^m} \\ \le \frac{e^{\frac{s}{\epsilon}}}{(1 + e^{\frac{s}{\epsilon}}|y - (\frac{v}{2})^{\alpha}|)^m} \frac{C(\delta, \alpha, b) v^{\gamma+1-\alpha}}{(1 + v^b)^2} \le \frac{C(\delta, \alpha, b) e^{\frac{s}{\epsilon}}}{(1 + v^b)(1 + e^{\frac{s}{\epsilon}}|y - (\frac{v}{2})^{\alpha}|)^m}, $$

where we used Lemma 2.14.

For the region where $w \in [0, \delta v]$, we will prove that

$$ \int_0^{\delta v} \frac{v^\gamma}{1 + w^b} \frac{e^{\frac{2s}{\epsilon}}}{(1 + e^{\frac{s}{\epsilon}}|y - w^\alpha|)^m} \frac{dw}{(1 + e^{\frac{s}{\epsilon}}|(v - w)^\alpha - w^\alpha|)^m} \\ + \int_0^{\delta v} \frac{v^\gamma}{1 + w^b} \frac{e^{\frac{2s}{\epsilon}}}{(1 + e^{\frac{s}{\epsilon}}|y - (v - w)^\alpha|)^m} \frac{dw}{(1 + e^{\frac{s}{\epsilon}}|(v - w)^\alpha - w^\alpha|)^m} \\ \le \frac{C(\delta, \alpha, \gamma, m) e^{\frac{s}{\epsilon}}}{(1 + e^{\frac{s}{\epsilon}}|y - (\frac{v}{2})^\alpha|)^l}, $$
for some $l = \frac{m}{2} > 2$ as before and then (3.34) follows by (3.25). We can assume without loss of generality that $|y - w^\alpha| \ge |y - (\frac{v}{2})^\alpha|$. Moreover, since when $w \in [0, \delta v]$ it holds that $(v - w)^\alpha - w^\alpha \ge C(\delta, \alpha)v^\alpha$, we have that

$$
\begin{aligned}
& \int_0^{\delta v} \frac{v^\gamma}{1+w^b} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-w^\alpha|)^m} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^m} \\
& \le \frac{C(\delta, \alpha, \gamma)e^{\frac{s}{\epsilon}}v^\gamma}{(1+e^{\frac{s}{\epsilon}}|y-(\frac{v}{2})^\alpha|)^m} \int_0^{\delta v} \frac{e^{\frac{s}{\epsilon}}dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^{m-\frac{\gamma+1}{\alpha}+1}} \frac{1}{1+v^{\gamma+1-\alpha}} \\
& \le \frac{C(\delta, \alpha, \gamma)e^{\frac{s}{\epsilon}}v^{\gamma+1-\alpha}}{(1+v^{\gamma+1-\alpha})(1+e^{\frac{s}{\epsilon}}|y-(\frac{v}{2})^\alpha|)^m} \le \frac{C(\delta, \alpha, \gamma)e^{\frac{s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-(\frac{v}{2})^\alpha|)^m},
\end{aligned}
$$

where in the last line we used Lemma 2.14.

We now estimate $I_2$. When $w \in [0, \delta v]$, then $(v-w)^\alpha - w^\alpha \ge C(\delta, \alpha)v^\alpha \ge C(\delta, \alpha)|y - (\frac{v}{2})^\alpha|$ and we have that

$$
\begin{aligned}
& \mathbb{I}_{\{0 \le y \le v^\alpha\}} \int_0^{\delta v} \frac{v^\gamma}{1+w^b} \frac{e^{\frac{2s}{\epsilon}}}{(1+e^{\frac{s}{\epsilon}}|y-(v-w)^\alpha|)^m} \frac{dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^m} \\
& \le \mathbb{I}_{\{0 \le y \le v^\alpha\}} C(\delta, \alpha, \gamma, m) e^{\frac{s}{\epsilon}} v^\gamma \int_0^{\delta v} \frac{e^{\frac{s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|(v-w)^\alpha - w^\alpha|)^{\frac{m}{2}-\frac{\gamma+1}{\alpha}+1}} \frac{1}{1+v^{\gamma+1-\alpha}} \frac{1}{1+e^{\frac{sm}{2\epsilon}} v^{\frac{am}{2}}} \\
& \le \mathbb{I}_{\{0 \le y \le v^\alpha\}} \frac{C(\delta, \alpha, \gamma, m) e^{\frac{s}{\epsilon}}}{1+e^{\frac{sm}{2\epsilon}} v^{\frac{am}{2}}} \le \mathbb{I}_{\{0 \le y \le v^\alpha\}} \frac{C(\delta, \alpha, \gamma, m) e^{\frac{s}{\epsilon}}}{1+e^{\frac{sm}{2\epsilon}} |y-(\frac{v}{2})^\alpha|^{\frac{m}{2}}},
\end{aligned}
$$

which concludes our proof.

• Subcase $v \le 1$: We will prove that

$$
\begin{aligned}
& \mathbb{I}_{\{0 \le y \le v^\alpha\}} \int_0^{\frac{v}{2}} \frac{e^{\frac{2s}{\epsilon}} v^\gamma}{(1+w^b)(1+(v-w)^b)} \psi(e^{\frac{s}{\epsilon}}(y-w^\alpha)) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^\alpha)) dw \\
& \le \frac{C(\alpha, b)v^\gamma e^{\frac{s}{\epsilon}}}{(1+v^b)(1+e^{\frac{s}{\epsilon}}|y-(\frac{v}{2})^\alpha|^m)}.
\end{aligned}
\qquad (3.36)
$$

If (3.36) holds true, then as in the case $v \ge 1$, we have that

$$
\int_0^t S_\epsilon(t-s) \left[ \frac{e^{\frac{s}{\epsilon}} v^\gamma}{(1 + e^{\frac{sm}{\epsilon}} |y - (\frac{v}{2})^\alpha|)^m} \right] ds = \epsilon e^{\frac{l}{\epsilon}} v^\gamma \int_{(v^\alpha - (\frac{v}{2})^\alpha)}^{e^{\frac{l}{\epsilon}} (v^\alpha - (\frac{v}{2})^\alpha)} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}} (v^\alpha - y)|^m} \frac{1}{z}.
$$

For the region $v \le 1$, $|y - v^\alpha| \le e^{-\frac{l}{\epsilon}}$, we have that

$$
\begin{aligned}
& v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \left[ \frac{e^{\frac{s}{\epsilon}}}{(1+e^{\frac{sm}{\epsilon}}|y-(\frac{v}{2})^{\alpha}|)^m} \right] ds \le C(\alpha) \epsilon e^{\frac{l}{\epsilon}} v^{\gamma-\alpha} \int_{(v^{\alpha}-(\frac{v}{2})^{\alpha})}^{e^{\frac{l}{\epsilon}}(v^{\alpha}-(\frac{v}{2})^{\alpha})} \frac{dz}{1+|z-e^{\frac{l}{\epsilon}}(v^{\alpha}-y)|^m} \\
& \le C(\alpha) \epsilon e^{\frac{l}{\epsilon}} v^{\gamma-\alpha} \le C(\alpha) \epsilon e^{\frac{l}{\epsilon}},
\end{aligned}
$$
since $\gamma > 1 > \alpha$ and $v \le 1$.

For the region $|y - v^\alpha| \ge e^{-\frac{l}{\epsilon}}$, it follows as before that

$$ \epsilon e^{\frac{l}{\epsilon}} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \le \frac{l}{\epsilon}} \frac{v^{\gamma} dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^m} \frac{1}{z} \le C(\alpha) \epsilon e^{\frac{l}{\epsilon}} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \le \frac{l}{\epsilon}} \frac{v^{\gamma-\alpha} dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^m} \\ \le \frac{C(\alpha) \epsilon e^{\frac{l}{\epsilon}} v^{\gamma-\alpha}}{e^{\frac{l}{\epsilon}}(v^{\alpha} - y)} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \le \frac{l}{\epsilon}} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^{m-1}} \le \frac{C(\alpha) \epsilon}{(v^{\alpha} - y)}, $$

since $v \le 1$. In the same manner, we have that

$$ v^{\gamma} \epsilon e^{\frac{l}{\epsilon}} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \ge \frac{l}{\epsilon}} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^m} \frac{1}{z} \le \frac{2\epsilon v^{\gamma}}{v^{\alpha} - y} \int_{(v^{\alpha} - (\frac{v}{2})^{\alpha}), |z| \ge \frac{l}{\epsilon}} \frac{dz}{1 + |z - e^{\frac{l}{\epsilon}}(v^{\alpha} - y)|^m} \\ \le \frac{C\epsilon}{v^{\alpha} - y}, $$

since $\gamma > 1$ and $v \le 1$. We now focus on proving (3.36) when $v \le 1$. We assume without loss of generality that $|y - w^\alpha| \ge |y - (\frac{v}{2})^\alpha|$. It holds that

$$ \int_0^{\frac{v}{2}} \frac{e^{\frac{2s}{\epsilon}} v^{\gamma}}{(1+w^b)(1+(v-w)^b)} \psi(e^{\frac{s}{\epsilon}}(y-w^{\alpha})) \psi(e^{\frac{s}{\epsilon}}(y-(v-w)^{\alpha})) dw \\ \le \frac{e^{\frac{s}{\epsilon}} v^{\gamma}}{(1+e^{\frac{s}{\epsilon}}|y-(\frac{v}{2})^{\alpha}|)^m} \int_0^{\frac{v}{2}} \frac{C(b)}{1+v^b} \frac{e^{\frac{s}{\epsilon}} dw}{(1+e^{\frac{s}{\epsilon}}|y-(v-w)^{\alpha}|)^m} \\ \le \frac{e^{\frac{s}{\epsilon}} v^{\gamma}}{(1+e^{\frac{s}{\epsilon}}|y-(\frac{v}{2})^{\alpha}|)^m} \frac{C(\alpha, b)v^{1-\alpha}}{1+v^b} \le \frac{C(\alpha, b)e^{\frac{s}{\epsilon}} v^{\gamma}}{(1+e^{\frac{s}{\epsilon}}|y-(\frac{v}{2})^{\alpha}|)^m} \frac{1}{1+v^b}, $$

where in the last line we used Lemma 2.14 and the fact that $\alpha < 1$.

□

## 3.4 Proof of Proposition 3.10

*Proof of Proposition 3.10. Proof of* (3.13). Let $i \in \{2, 3\}$. We want to bound from above the term

$$ v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{v}{2}} T_{i-1}(y, v-w, s) T_{i}(y, w, s) dw ds. \quad (3.37) $$

We start by estimating

$$ v^{\gamma} \int_{0}^{\frac{v}{2}} T_{1}(y, v-w, s) T_{2}(y, w, s) dw \\ \le v^{\gamma} \int_{0}^{\frac{v}{2}} \frac{4A^{3}C_{0}^{m-1}}{(1+w^{b})(1+(v-w)^{b})} \frac{e^{\frac{s}{\epsilon}} \mathbb{1}_{\{y \ge (v-w)^{\alpha} - C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}{1+e^{\frac{ms}{\epsilon}}|y-(v-w)^{\alpha}|^{m}} \frac{\epsilon \mathbb{1}_{\{y \le w^{\alpha} - C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}{|y-w^{\alpha}|} dw = 0. $$
Thus, this term does not play any role. Similarly

$$ 
\begin{aligned}
& v^{\gamma} \int_{0}^{\frac{v}{2}} T_2(y, v-w, s)T_3(y, w, s)dw \\
\leq & v^{\gamma} \int_{0}^{\frac{v}{2}} \frac{2M_2 A^3 C_0^{m-1}}{(1+w^b)(1+(v-w)^b)} \frac{\epsilon \mathbb{1}_{\{y \geq (\frac{v-w}{3})^\alpha\}}}{|y-(v-w)^\alpha|} \frac{e^{\frac{\epsilon s}{\epsilon}} \mathbb{1}_{\{y \leq (\frac{w}{3})^\alpha\}}}{1+e^{\frac{ms}{\epsilon}} |y-w^\alpha|^{m}} dw = 0.
\end{aligned}
$$

**Proof of (3.14).**

i) We continue by estimating from above the term

$$ v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{v}{2}} T_2(y, v-w, s) T_1(y, w, s) dw ds. \quad (3.38) $$

In the following, we bound first

$$ v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{\delta v}^{\frac{v}{2}} T_2(y, v-w, s) T_1(y, w, s) dw ds, \quad (3.39) $$

for some sufficiently small but fixed $\delta > 0$. We focus then on bounding

$$ v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\delta v} T_2(y, v-w, s) T_1(y, w, s) dw ds. $$

We start by estimating

$$ 
\begin{aligned}
& v^{\gamma} \int_{\delta v}^{\frac{v}{2}} T_2(y, v-w, s) T_1(y, w, s) dw \\
\leq & \frac{C(\delta, b) A^3 C_0^{m-1} v^{\gamma}}{(1+v^b)^2} \int_{\delta v}^{\frac{v}{2}} \frac{e^{\frac{\epsilon s}{\epsilon}} \mathbb{1}_{\{y \geq w^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{1+e^{\frac{ms}{\epsilon}} |y-w^{\alpha}|^m} \frac{\epsilon \mathbb{1}_{\{y \leq (v-w)^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}{|y-(v-w)^{\alpha}|} dw \\
\leq & \frac{C(\delta, b) A^3 C_0^{m-1} v^{\gamma}}{(1+v^b)^2} \int_{\delta v}^{\frac{v}{2}} \frac{e^{\frac{\epsilon s}{\epsilon}}}{1+e^{\frac{ms}{\epsilon}} |y-w^{\alpha}|^m |y-(v-w)^{\alpha}|} dw \\
& \times \mathbb{1}_{\{-C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}} \leq y \leq v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}
\end{aligned}
$$

**Remark 3.11.** Notice that the region $y \leq v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}$ implies that after the action of the semigroup we have that $e^{\frac{t-s}{\epsilon}}(y - v^{\alpha}) + v^{\alpha} \leq v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}$ or in other words after the action of the semigroup we are in the region where $e^{\frac{t}{\epsilon}}|y - v^{\alpha}| \geq C_0^{-1} \epsilon^{-\frac{1}{m-1}}$.

We can write $\mathbb{1}_{\{-C_0^{-1} e^{-\frac{s}{\epsilon}} \epsilon^{-\frac{1}{m-1}} \leq y \leq v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}$ as a sum of the form

$$ 
\begin{aligned}
\mathbb{1}_{\{-C_0^{-1} e^{-\frac{s}{\epsilon}} \epsilon^{-\frac{1}{m-1}} \leq y \leq v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}} = & \mathbb{1}_{\{-C_0^{-1} e^{-\frac{s}{\epsilon}} \epsilon^{-\frac{1}{m-1}} \leq y \leq (\frac{v}{2})^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}} \\
& + \mathbb{1}_{\{(\frac{v}{2})^{\alpha} + C_0^{-1} e^{-\frac{s}{\epsilon}} \epsilon^{-\frac{1}{m-1}} \leq y \leq v^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}} + \mathbb{1}_{\{(\frac{v}{2})^{\alpha} - C_0^{-1} e^{-\frac{s}{\epsilon}} \epsilon^{-\frac{1}{m-1}} \leq y \leq (\frac{v}{2})^{\alpha} + C_0^{-1} e^{-\frac{s}{\epsilon}} \epsilon^{-\frac{1}{m-1}}\}}
\end{aligned}
$$
We analyze first the region where $\{-C_0^{-1}e^{-\frac{s}{\epsilon}}\epsilon^{-\frac{1}{m-1}} \le y \le (\frac{v}{2})^{\alpha} - C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{s}{\epsilon}}\}$. In this region, since $|y - (v - w)^{\alpha}| \ge |y - (\frac{v}{2})^{\alpha}|$, then

$$ \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{ms}{\epsilon}} |y - w^{\alpha}|^m} \frac{\epsilon \mathbb{1}_{\{y \le (v-w)^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{|y - (v-w)^{\alpha}|} \le \frac{\epsilon}{|y - (\frac{v}{2})^{\alpha}|} \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{ms}{\epsilon}} |y - w^{\alpha}|^m} \mathbb{1}_{\{y \le (v-w)^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}} $$

Thus, it holds that

$$ \begin{aligned} I_1(y, v, s) &:= v^{\gamma} \int_{\delta v}^{\frac{v}{2}} T_2(y, v-w, s) T_1(y, w, s) dw \mathbb{1}_{\{-C_0^{-1}e^{-\frac{s}{\epsilon}}\epsilon^{-\frac{1}{m-1}} \le y \le (\frac{v}{2})^{\alpha} - C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{s}{\epsilon}}\}} \\ &\le \frac{C(\delta, b)A^3 C_0^{m-1} v^{\gamma}}{(1+v^b)^2} \int_{\delta v}^{\frac{v}{2}} \frac{\epsilon}{|y-(\frac{v}{2})^{\alpha}|} \frac{e^{\frac{s}{\epsilon}}}{1+e^{\frac{ms}{\epsilon}}|y-w^{\alpha}|^m} dw \\ &\qquad \times \mathbb{1}_{\{-C_0^{-1}e^{-\frac{s}{\epsilon}}\epsilon^{-\frac{1}{m-1}} \le y \le (\frac{v}{2})^{\alpha} - C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{s}{\epsilon}}\}} \\ &\le \frac{C(\delta, \alpha, b)\epsilon A^3 C_0^{m-1} v^{\gamma+1-\alpha}}{|y-(\frac{v}{2})^{\alpha}| (1+v^b)^2} \mathbb{1}_{\{-C_0^{-1}e^{-\frac{s}{\epsilon}}\epsilon^{-\frac{1}{m-1}} \le y \le (\frac{v}{2})^{\alpha} - C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{s}{\epsilon}}\}} \end{aligned} $$

where in the last line we used Lemma 2.14.

Then

$$ \begin{aligned} v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) [I_{1}(y, v, s)] ds &\le \frac{C(\delta, \alpha, b) A^{3} C_{0}^{m-1} v^{\gamma+1-\alpha} \epsilon}{(1+v^{b})^{2}} \int_{0}^{t} \frac{e^{\frac{l-s}{\epsilon}}}{|e^{\frac{l-s}{\epsilon}}(y-v^{\alpha}) + v^{\alpha} - (\frac{v}{2})^{\alpha}|} \\ &\qquad \times \mathbb{1}_{\{-C_{0}^{-1}e^{-\frac{s}{\epsilon}}\epsilon^{-\frac{1}{m-1}} \le e^{\frac{l-s}{\epsilon}}(y-v^{\alpha}) + v^{\alpha} \le (\frac{v}{2})^{\alpha} - C_{0}^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{s}{\epsilon}}\}} ds \\ &= \frac{C(\delta, \alpha, b) A^{3} C_{0}^{m-1} v^{\gamma+1-\alpha} \epsilon}{(1+v^{b})^{2}} \int_{0}^{t} \frac{e^{\frac{l}{\epsilon}}}{|e^{\frac{l}{\epsilon}}(y-v^{\alpha}) + e^{\frac{s}{\epsilon}}(v^{\alpha} - (\frac{v}{2})^{\alpha})|} \\ &\qquad \times \mathbb{1}_{\{-e^{\frac{s}{\epsilon}}v^{\alpha} - C_{0}^{-1}\epsilon^{-\frac{1}{m-1}} \le e^{\frac{l}{\epsilon}}(y-v^{\alpha}) \le e^{\frac{s}{\epsilon}}((\frac{v}{2})^{\alpha}-v^{\alpha}) - C_{0}^{-1}\epsilon^{-\frac{1}{m-1}}\}} ds \\ &= \frac{C(\delta, \alpha, b) A^{3} C_{0}^{m-1} v^{\gamma+1-\alpha} \epsilon}{(1+v^{b})^{2}} \int_{0}^{t} \frac{e^{\frac{l}{\epsilon}}}{|e^{\frac{l}{\epsilon}}(v^{\alpha}-y) - e^{\frac{s}{\epsilon}}(v^{\alpha} - (\frac{v}{2})^{\alpha})|} \\ &\qquad \times \mathbb{1}_{\{e^{\frac{s}{\epsilon}}v^{\alpha} + C_{0}^{-1}\epsilon^{-\frac{1}{m-1}} \ge e^{\frac{l}{\epsilon}}(v^{\alpha}-y) \ge e^{\frac{s}{\epsilon}}(v^{\alpha} - (\frac{v}{2})^{\alpha}) + C_{0}^{-1}\epsilon^{-\frac{1}{m-1}}\}} ds \\ &=: C(\delta, \alpha, b)L(y, v, t). \end{aligned} \quad (3.40) $$

We make the change of variables $z = e^{\frac{l}{\epsilon}}(v^{\alpha} - y) - e^{\frac{s}{\epsilon}}(v^{\alpha} - (\frac{v}{2})^{\alpha})$. We thus have that

$$ L(y, v, t) = \frac{A^3 C_0^{m-1} v^{\gamma+1-\alpha} \epsilon^2}{(1+v^b)^2} \int_{e^{\frac{l}{\epsilon}}((\frac{v}{2})^{\alpha}-y)}^{e^{\frac{l}{\epsilon}}(v^{\alpha}-y)-(\frac{v}{2})^{\alpha}} \frac{e^{\frac{l}{\epsilon}}}{e^{\frac{l}{\epsilon}}(v^{\alpha}-y)-z} \frac{1}{|z|} \\ \times \mathbb{1}_{\{e^{\frac{s}{\epsilon}}(\frac{v}{2})^{\alpha} + C_0^{-1}\epsilon^{-\frac{1}{m-1}} \ge z \ge C_0^{-1}\epsilon^{-\frac{1}{m-1}}\}} dz. \quad (3.41) $$
On the one hand, it holds that

$$
\begin{align*}
& \frac{\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2} \int \frac{e^{\frac{t}{\epsilon}(\nu^\alpha - y)} - (\nu^\alpha - (\frac{\nu}{2})^\alpha)}{e^{\frac{t}{\epsilon}}((\frac{\nu}{2})^\alpha - y), z \le \frac{e^{\frac{t}{\epsilon}}(\nu^\alpha - y)}{2}} \frac{e^{\frac{t}{\epsilon}}}{e^{\frac{t}{\epsilon}}(\nu^\alpha - y) - z} \frac{1}{|z|} \mathbb{1}_{\{e^{\frac{t}{\epsilon}}(\frac{\nu}{2})^\alpha + C_0^{-1}\epsilon^{-\frac{1}{m-1}} \ge z \ge C_0^{-1}\epsilon^{-\frac{1}{m-1}}\}} dz \\
& \le \frac{2\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)} \int \frac{e^{\frac{t}{\epsilon}(\nu^\alpha - y)} - (\nu^\alpha - (\frac{\nu}{2})^\alpha)}{e^{\frac{t}{\epsilon}}((\frac{\nu}{2})^\alpha - y), z \le \frac{e^{\frac{t}{\epsilon}}(\nu^\alpha - y)}{2}} \frac{1}{|z|} \mathbb{1}_{\{e^{\frac{t}{\epsilon}}(\frac{\nu}{2})^\alpha + C_0^{-1}\epsilon^{-\frac{1}{m-1}} \ge z \ge C_0^{-1}\epsilon^{-\frac{1}{m-1}}\}} dz \\
& \le \frac{2\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)} \int \frac{e^{\frac{t}{\epsilon}(\nu^\alpha - y)} - (\nu^\alpha - (\frac{\nu}{2})^\alpha)}{1} \frac{1}{z} dz \le \frac{C(\alpha)\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)} \left[ \frac{t}{\epsilon} + |\ln \nu^\alpha| \right], \quad (3.42)
\end{align*}
$$

where in (3.42) we used that we are in the region $e^{\frac{t}{\epsilon}}v^{\alpha} + C_0^{-1}\epsilon^{-\frac{1}{m-1}} \ge e^{\frac{t}{\epsilon}}(v^{\alpha} - y) \ge e^{\frac{t}{\epsilon}}(v^{\alpha} - (\frac{\nu}{2})^{\alpha}) + C_0^{-1}\epsilon^{-\frac{1}{m-1}}$ from (3.40) and that $C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{t}{\epsilon}} \le v^{\alpha}[1 - 3^{-\alpha}]$. We then have that

$$
\frac{\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)} \left[ \frac{t}{\epsilon} + |\ln \nu^\alpha| \right] \le \frac{\nu^{\gamma+1-\alpha} t \epsilon}{(1+\nu^b)^2(\nu^\alpha - y)} + \frac{\nu^{\gamma+1} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)}; \quad \nu \ge 1, \quad (3.43)
$$

$$
\frac{\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)} \left[ \frac{t}{\epsilon} + |\ln \nu^\alpha| \right] \le \frac{\nu^{\gamma+1-\alpha} t \epsilon}{(1+\nu^b)^2(\nu^\alpha - y)} + \frac{\nu^{\gamma-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)}; \quad \nu \le 1. \quad (3.44)
$$

We then use that $b \ge \gamma + 1$ in (3.43).

On the other hand, it holds that

$$
\begin{align*}
& \frac{\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2} \int \frac{e^{\frac{t}{\epsilon}(\nu^\alpha - y)} - (\nu^\alpha - (\frac{\nu}{2})^\alpha)}{e^{\frac{t}{\epsilon}}((\frac{\nu}{2})^\alpha - y), z \ge \frac{e^{\frac{t}{\epsilon}}(\nu^\alpha - y)}{2}} \frac{e^{\frac{t}{\epsilon}}}{e^{\frac{t}{\epsilon}}(\nu^\alpha - y) - z} \frac{1}{|z|} \mathbb{1}_{\{e^{\frac{t}{\epsilon}}(\frac{\nu}{2})^\alpha + C_0^{-1}\epsilon^{-\frac{1}{m-1}} \ge z \ge C_0^{-1}\epsilon^{-\frac{1}{m-1}}\}} dz \\
& \le \frac{2\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)} \int \frac{e^{\frac{t}{\epsilon}(\nu^\alpha - y)} - (\nu^\alpha - (\frac{\nu}{2})^\alpha)}{e^{\frac{t}{\epsilon}}((\frac{\nu}{2})^\alpha - y), z \ge \frac{e^{\frac{t}{\epsilon}}(\nu^\alpha - y)}{2}} \frac{1}{e^{\frac{t}{\epsilon}}(\nu^\alpha - y) - z} \mathbb{1}_{\{e^{\frac{t}{\epsilon}}(\frac{\nu}{2})^\alpha + C_0^{-1}\epsilon^{-\frac{1}{m-1}} \ge z \ge C_0^{-1}\epsilon^{-\frac{1}{m-1}}\}} dz \\
& \le \frac{2\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)} \int \frac{e^{\frac{t}{\epsilon}(\nu^\alpha - y)} - (\nu^\alpha - (\frac{\nu}{2})^\alpha)}{1} \frac{1}{e^{\frac{t}{\epsilon}}(\nu^\alpha - y) - z} dz \\
& \le \frac{2\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)} \int \frac{e^{\frac{t}{\epsilon}(\nu^\alpha - y)} - 1}{\nu^\alpha - (\frac{\nu}{2})^\alpha} \frac{1}{z} dz \le \frac{C(\alpha)\nu^{\gamma+1-\alpha} \epsilon^2}{(1+\nu^b)^2(\nu^\alpha - y)} \left[ \frac{t}{\epsilon} + |\ln \nu^\alpha| \right] \quad (3.45)
\end{align*}
$$

which concludes our proof.

We now estimate the region where $\{e^{\frac{t}{\epsilon}}|y - (\frac{\nu}{2})^{\alpha}| \le C_0^{-1}\epsilon^{-\frac{1}{m-1}}\}$. In this region, it holds that

$$
\frac{\epsilon^{\frac{m}{m-1}}}{2} \le \frac{C_0^{-m}}{1 + e^{\frac{sm}{\epsilon}} |y - (\frac{\nu}{2})^{\alpha}|^m}.
$$
Thus

$$
\begin{align*}
& \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{ms}{\epsilon}} |y - w^{\alpha}|^m} \frac{\epsilon \mathbb{1}_{\{y \le (v-w)^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{|y - (v-w)^\alpha|} \\
& \le \frac{2\epsilon \mathbb{1}_{\{y \le (v-w)^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{|y - (v-w)^\alpha|} \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{ms}{\epsilon}} |y - w^{\alpha}|^m} \frac{C_0^{-m} \epsilon^{-\frac{m}{m-1}}}{1 + e^{\frac{sm}{\epsilon}} |y - (\frac{v}{2})^\alpha|^m} \\
& \le \frac{2C_0^{-(m-1)} e^{\frac{s}{\epsilon}}}{1 + e^{\frac{ms}{\epsilon}} |y - w^{\alpha}|^m} \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{sm}{\epsilon}} |y - (\frac{v}{2})^\alpha|^m}.
\end{align*}
$$

Denote by

$$
I_2(y, v, s) := v^{\gamma} \int_{\delta v}^{\frac{v}{2}} T_2(y, v-w, s) T_1(y, w, s) dw \mathbb{1}_{\{e^{\frac{s}{\epsilon}} |y-(\frac{v}{2})^{\alpha}| \le C_0^{-1} \epsilon^{-\frac{1}{m-1}}\}}
$$

Thus, in this region it holds that

$$
\begin{align*}
I_2(y, v, s) &\le \frac{C(\delta, b) C_0^{m-1} A^3 C_0^{-(m-1)} v^\gamma}{(1 + v^b)^2} \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{sm}{\epsilon}} |y - (\frac{v}{2})^\alpha|^m} \int_{\delta v}^{\frac{v}{2}} \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{ms}{\epsilon}} |y - w^\alpha|^m} dw \\
&\le \frac{C(\delta, \alpha, b) C_0^{m-1} A^3 e^{\frac{s}{\epsilon}} C_0^{-(m-1)} v^{\gamma+1-\alpha}}{1 + e^{\frac{sm}{\epsilon}} |y - (\frac{v}{2})^\alpha|^m (1 + v^b)^2},
\end{align*}
$$

where in the last line we used Lemma 2.14.

Thus, in this region, we have that

$$
\int_0^t S_{\epsilon}(t-s)[I_2(y, v, s)]ds \\
\le \frac{C(\delta, \alpha, b) C_0^{m-1} A^3 C_0^{-(m-1)} v^{\gamma+1-\alpha}}{(1+v^b)^2} \int_0^t \frac{e^{\frac{l}{\epsilon}} \mathbb{1}_{\{|e^{\frac{l}{\epsilon}}(y-v^{\alpha})+e^{\frac{s}{\epsilon}}(v^{\alpha}-\frac{1}{2})^{\alpha}| \le C_0^{-1} \epsilon^{-\frac{1}{m-1}}\}}}{1+|e^{\frac{l}{\epsilon}}(y-v^{\alpha})+e^{\frac{s}{\epsilon}}(v^{\alpha}-\frac{1}{2})^{\alpha}|^m} ds.
$$

We then use (3.35), $m > 2$, in order to deduce that

$$
\int_0^t S_\epsilon(t-s)[I_2(y, v, s)]ds \le \frac{C(\delta, \alpha, b) C_0^{m-1} A^3 C_0^{-(m-1)} \epsilon}{(1+v^b)|y-v^\alpha|} \min\{1, e^{\frac{l}{\epsilon}}|y-v^\alpha|\}.
$$

By Remark 3.11 we further obtain that

$$
\int_0^t S_\epsilon(t-s)[I_2(y, v, s)]ds \le \frac{C(\delta, \alpha, b) C_0^{m-1} A^3 C_0^{-(m-1)} \epsilon}{(1+v^b)|y-v^\alpha|} \mathbb{1}_{\{y \le v^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{l}{\epsilon}}\}} \quad (3.46)
$$

Remark 3.12. Notice that since $C_0^{m-1}$ is of order $\mathcal{O}(A)$ by (2.9), this implies in particular that

$$
\begin{align*}
\int_0^t S_\epsilon(t-s)[I_2(y, v, s)]ds &\le \frac{C(\delta, \alpha, b)A^3 \epsilon}{(1+v^b)|y-v^\alpha|} \mathbb{1}_{\{y \le v^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{l}{\epsilon}}\}} \\
&= \frac{C(\delta, \alpha, \gamma, b, m)C_0^{m-1} A^2 \epsilon}{(1+v^b)|y-v^\alpha|} \mathbb{1}_{\{y \le v^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{l}{\epsilon}}\}}.
\end{align*}
$$
The reason we keep the form (3.46) is because this estimate is part of the more general estimate in (3.14).

We now estimate the region where $\{(\frac{v}{2})^{\alpha} + C_{0}^{-1}e^{-\frac{s}{\epsilon}}e^{-\frac{1}{m-1}} \le y \le v^{\alpha} + C_{0}^{-1}e^{-\frac{s}{\epsilon}}e^{-\frac{1}{m-1}}\}$. In this region, it holds that $|y - w^{\alpha}| \ge |y - (\frac{v}{2})^{\alpha}|$. Then

$$ \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{ms}{\epsilon}} |y - w^{\alpha}|^m} \frac{\epsilon \mathbb{1}_{\{y \le (v-w)^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{|y - (v-w)^{\alpha}|} \\ \le \frac{\epsilon}{|y - (v-w)^{\alpha}|} \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{ms}{\epsilon}} |y - (\frac{v}{2})^{\alpha}|^m} \mathbb{1}_{\{y \le (v-w)^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}} $$

We have that

$$ \int_0^{\frac{v}{2}} \frac{dw}{|y - (v - w)^{\alpha}|} \mathbb{1}_{\{y \le (v-w)^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}} \le C(\alpha) v^{1-\alpha} \int_{y-v^{\alpha}}^{y-(\frac{v}{2})^{\alpha}} \frac{dz}{|z|} \mathbb{1}_{\{z \le -C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}} \\ \le C(\alpha) v^{1-\alpha} \int_{(\frac{v}{2})^{\alpha}-y}^{v^{\alpha}-y} \frac{dz}{z} \mathbb{1}_{\{z \ge C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}} \\ \le C(\alpha) v^{1-\alpha} \left[ \frac{t}{\epsilon} + |\ln v^{\alpha}| \right] \quad (3.47) $$

since $0 \le y \le v^{\alpha}$. We can thus conclude that in this region, we have that

$$ v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \left[ \int_{\delta v}^{\frac{v}{2}} T_2(y, v-w, s) T_1(y, w, s) dw \mathbb{1}_{\{(\frac{v}{2})^{\alpha} + C_0^{-1} e^{-\frac{s}{\epsilon}} \epsilon^{-\frac{1}{m-1}} \le y \le v^{\alpha} + C_0^{-1} e^{-\frac{s}{\epsilon}} \epsilon^{-\frac{1}{m-1}}\}} \right] ds \\ \le \frac{C(\delta, \alpha, b) A^3 C_0^{m-1} [v^{\gamma+1-\alpha} t + v^{\gamma+1} \epsilon \mathbb{1}_{\{v \ge 1\}} + v^{\gamma-\alpha} \epsilon \mathbb{1}_{\{v \le 1\}}]}{(1 + v^b)^2} \\ \times \int_0^t \frac{e^{\frac{l}{\epsilon}}}{1 + |e^{\frac{l}{\epsilon}}(y-v^{\alpha}) + e^{\frac{s}{\epsilon}}(v^{\alpha} - (\frac{v}{2})^{\alpha})|^m} \mathbb{1}_{\{e^{\frac{l}{\epsilon}}(y-v^{\alpha}) + e^{\frac{s}{\epsilon}}(v^{\alpha} - (\frac{v}{2})^{\alpha}) \ge C_0^{-1} \epsilon^{-\frac{1}{m-1}}\}} ds. $$

We can now conclude as before using (3.35). This concludes this case. We now bound

$$ v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\delta v} T_{2}(y, v-w, s) T_{1}(y, w, s) dw ds. $$

Notice that we only have to make sure that we have sufficiently fast decay in the $v$ variable. We then follow the computations for the region $w \in [\delta v, \frac{v}{2}]$ in order to conclude. More precisely, it suffices to show that

$$ I_3(y, v, s) := v^{\gamma} \int_{0}^{\delta v} T_2(y, v-w, s) T_1(y, w, s) dw \\ \le \frac{C(\delta, \alpha, \gamma, b) A^3 C_0^{m-1} v^{\gamma}}{(1 + v^b)(1 + v^{\gamma+1})} \int_{0}^{\delta v} \frac{e^{\frac{s}{\epsilon}} \mathbb{1}_{\{y \ge w^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{1 + e^{\frac{(m-\gamma+1)}{\epsilon}s} |y - w^{\alpha}|^{m-\frac{\gamma+1}{\alpha}}} \mathbb{1}_{\{y \le (v-w)^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}} dw, $$
for a sufficiently small $\delta < \frac{1}{2}$. We have that

$$
\begin{aligned}
I_3(y, v, s) &\le \frac{C(b)A^3 C_0^{m-1} v^\gamma}{1 + v^b} \int_0^{\delta v} \frac{e^{\frac{s}{\epsilon}} \mathbb{1}_{\{y \ge w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{1 + w^b \left(1 + e^{\frac{ms}{\epsilon}}|y - w^\alpha|\right)} \frac{\epsilon \mathbb{1}_{\{\frac{v-w}{3}\}^\alpha \le y \le (v-w)^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}{|y - (v-w)^\alpha|} \mathrm{d}w \\
&\le \frac{C(b)A^3 C_0^{m-1} v^\gamma}{1 + v^b} \int_0^{\delta v} \frac{e^{\frac{s}{\epsilon}} \mathbb{1}_{\{y \ge w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{1 + e^{\frac{ms}{\epsilon}}|y - w^\alpha|\left(1 + \frac{(\frac{v-w}{3})^\alpha \le y \le (v-w)^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\right)\right)} \frac{\epsilon \mathbb{1}_{\{\frac{v-w}{3}\}^\alpha \le y \le (v-w)^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}{|y - (v-w)^\alpha|} \mathrm{d}w.
\end{aligned}
$$

We have that $(\frac{v-w}{3})^\alpha \le y$, which implies that for $\delta$ sufficiently small it holds that $y - w^\alpha \ge (\frac{v-w}{3})^\alpha - w^\alpha \ge v^\alpha [(\frac{1-\delta}{3})^\alpha - \delta^\alpha] \ge C(\delta, \alpha)v^\alpha$. We thus have that

$$
\begin{aligned}
I_3(y, v, s) &\le \frac{C(\delta, \alpha, \gamma, b) A^3 C_0^{m-1} v^\gamma}{(1 + v^b)(1 + v^{\gamma+1})} \int_0^{\delta v} \frac{e^{\frac{s}{\epsilon}} \mathbb{1}_{\{y \ge w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{1 + e^{\frac{(m-\frac{\gamma+1}{\alpha})s}{\epsilon}} |y - w^\alpha|^{m-\frac{\gamma+1}{\alpha}}} \\
&\qquad \times \frac{\epsilon \mathbb{1}_{\{(\frac{v-w}{3})^\alpha \le y \le (v-w)^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{|y - (v-w)^\alpha|} \mathrm{d}w
\end{aligned}
$$

and we can proceed as in the region $w \in [\delta v, \frac{v}{2}]$ since $m > \frac{\gamma+1}{\alpha} + 2$.

ii) We continue by bounding from above the term

$$ v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{v}{2}} T_{3}(y, v-w, s) T_{2}(y, w, s) dw ds. \quad (3.48) $$

We will show that

$$
\begin{aligned}
I_4(y, v, s) &:= v^{\gamma} \int_0^{\frac{v}{2}} T_3(y, v-w, s) T_2(y, w, s) dw \\
&\le \frac{C(\alpha, \gamma, b) M_2 A^3 C_0^{m-1} v^{\gamma}}{(1 + v^b)(1 + v^{\gamma+1})} \int_0^{\frac{v}{2}} \frac{e^{\frac{s}{\epsilon}} \mathbb{1}_{\{y \le (\frac{v-w}{3})^{\alpha}\}}}{1 + e^{\frac{(m-\frac{\gamma+1}{\alpha})s}{\epsilon}} |y - (v-w)^{\alpha}|^{m-\frac{\gamma+1}{\alpha}}} \\
&\qquad \times \frac{\epsilon \mathbb{1}_{\{(\frac{v}{3})^{\alpha} \le y \le w^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{|y - w^{\alpha}|} dw.
\end{aligned}
$$

Since $0 \le y \le (\frac{v-w}{3})^\alpha$, we have that $|y - (v-w)^\alpha| \ge C(\alpha)(v-w)^\alpha \ge C(\alpha)v^\alpha$. Thus

$$
\begin{aligned}
I_4(y, v, s) &\le \frac{2 M_2 A^3 C_0^{m-1} v^\gamma}{1 + v^b} \int_0^{\frac{v}{2}} \frac{e^{\frac{s}{\epsilon}} \mathbb{1}_{\{y \le (\frac{v-w}{3})^\alpha\}}}{1 + e^{\frac{ms}{\epsilon}} |y - (v-w)^\alpha|^{m-1}} \frac{\epsilon \mathbb{1}_{\{(\frac{w}{3})^\alpha \le y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}{|y - w^\alpha|} \mathrm{d}w \\
&\le \frac{C(\alpha, \gamma, b) M_2 A^3 C_0^{m-1} v^\gamma}{(1 + v^b)(1 + v^{\gamma+1})} \int_0^{\frac{v}{2}} \frac{e^{\frac{s}{\epsilon}} \mathbb{1}_{\{y \le (\frac{v-w}{3})^\alpha\}}}{1 + e^{\frac{(m-\frac{\gamma+1}{\alpha})s}{\epsilon}} |y - (v-w)^\alpha|^{m-\frac{\gamma+1}{\alpha}}} \\
&\qquad \times \frac{\epsilon \mathbb{1}_{\{(\frac{w}{3})^\alpha \le y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}{|y - w^\alpha|} \mathrm{d}w.
\end{aligned}
$$
We further have that

$$
I_4(y, v, s) \le \frac{C(\alpha, \gamma, b) M_2 A^3 C_0^{m-1} v^\gamma}{(1 + v^b)(1 + v^{\gamma+1})} \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{(m-\frac{\gamma+1}{\alpha})s}{\epsilon}} |y - (\frac{v}{2})^\alpha|^{m-\frac{\gamma+1}{\alpha}}} \\
\times \int_0^{\frac{v}{2}} \frac{\epsilon \mathbb{1}_{\{y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{|y - w^\alpha|} dw \mathbb{1}_{\{0 \le y \le (\frac{v}{2})^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}
$$

We bound $\int_0^{\frac{v}{2}} \frac{\mathbb{1}_{\{y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{|y - w^\alpha|} dw$ as in (3.47). Thus

$$
\int_0^{\frac{v}{2}} \frac{\mathbb{1}_{\{y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}}{|y - w^\alpha|} dw \le C(\alpha) v^{1-\alpha} \left[ \frac{t}{\epsilon} + |\ln v^\alpha| \right]. \quad (3.49)
$$

It follows that

$$
I_4(y, v, s) \le \frac{C(\alpha, \gamma, b) M_2 A^3 C_0^{m-1} \max\{v^{\gamma+1}, v^{\gamma-\alpha}\} (\epsilon + t)}{(1 + v^b)(1 + v^{\gamma+1})} \\
\times \frac{e^{\frac{s}{\epsilon}}}{1 + e^{\frac{(m-\frac{\gamma+1}{\alpha})s}{\epsilon}} |y - (\frac{v}{2})^{\alpha}|^{m-\frac{\gamma+1}{\alpha}} \{0 \le y \le (\frac{v}{2})^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}}\}}
$$

We then use (3.35) since $m > \frac{\gamma+1}{\alpha} + 2$ in order to deduce that

$$
\int_0^t S_\epsilon(t-s) [I_4(y, v, s)] ds \le \frac{C(\alpha, \gamma, b) M_2 A^3 C_0^{m-1} (\epsilon + t) \epsilon}{(1 + v^b) |y - v^\alpha|} \min\{1, e^{\frac{t}{\epsilon}} |y - v^\alpha|\}.
$$

**Proof of (3.15).** We now bound from above the term

$$
v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{0}^{\frac{v}{2}} T_{2}(y, v-w, s) T_{2}(y, w, s) dw ds. \quad (3.50)
$$

We notice that $(\frac{v-w}{3})^\alpha \le y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{s}{\epsilon}} \le w^\alpha$, which implies that $w \ge \frac{v}{4}$. In other words, it suffices to bound from above the following term

$$
v^{\gamma} \int_{0}^{t} S_{\epsilon}(t-s) \int_{\frac{v}{4}}^{\frac{v}{2}} T_{2}(y, v-w, s) T_{2}(y, w, s) dw ds. \quad (3.51)
$$
We start by estimating

$$
\begin{align*}
I_5(y, v, s) &:= v^\gamma \int_{\frac{v}{4}}^{\frac{v}{2}} T_2(y, v-w, s) T_2(y, w, s) dw \\
&\le 4A^4 C_0^{2(m-1)} v^\gamma \int_{\frac{v}{4}}^{\frac{v}{2}} \frac{1}{(1+w^b)(1+(v-w)^b)} \frac{\epsilon \mathbb{1}_{\{y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{\delta}{\epsilon}}\}}{|y-w^\alpha|} \\
&\quad \times \frac{\epsilon \mathbb{1}_{\{y \le (v-w)^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{\delta}{\epsilon}}\}}{|y-(v-w)^\alpha|} dw \\
&\le \frac{C(b) A^4 C_0^{2(m-1)} v^\gamma}{(1+v^b)^2} \int_{\frac{v}{4}}^{\frac{v}{2}} \frac{\epsilon \mathbb{1}_{\{y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{\delta}{\epsilon}}\}}{|y-w^\alpha|} \frac{\epsilon \mathbb{1}_{\{y \le (v-w)^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{\delta}{\epsilon}}\}}{|y-(v-w)^\alpha|} dw \\
&\le \frac{C(b) A^4 C_0^{2(m-1)} v^\gamma}{(1+v^b)^2} \frac{1}{|y-(\frac{v}{2})^\alpha|} \int_{\frac{v}{4}}^{\frac{v}{2}} \frac{\epsilon^2 \mathbb{1}_{\{y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{\delta}{\epsilon}}\}}{|y-w^\alpha|} dw \mathbb{1}_{\{y \le (\frac{v}{2})^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{\delta}{\epsilon}}\}}
\end{align*}
$$

Following the computations in (3.47), it holds that

$$
\int_0^{\frac{v}{2}} \frac{\mathbb{1}_{\{y \le w^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{\delta}{\epsilon}}\}}{|y - w^\alpha|} dw \le C(\alpha) v^{1-\alpha} \left[ \frac{t}{\epsilon} + |\ln v^\alpha| \right]. \quad (3.52)
$$

Thus, we have that

$$
\begin{align*}
I_5(y, v, s) &\le \frac{C(\alpha, b) A^4 C_0^{2(m-1)} \epsilon [v^{\gamma+1-\alpha}t + v^{\gamma+\frac{1}{2}} \epsilon \mathbb{1}_{\{v \ge 1\}} + v^{\gamma-\frac{\alpha}{2}} \epsilon \mathbb{1}_{\{v \le 1\}}]}{(1+v^b)^2} \frac{1}{|y-(\frac{v}{2})^\alpha|} \mathbb{1}_{\{y \le (\frac{v}{2})^\alpha - \epsilon^{-\frac{1}{m-1}} e^{-\frac{\delta}{\epsilon}}\}} \\
&\le \frac{C(\alpha, b) \max\{v^{\gamma+\frac{1}{2}}, v^{\gamma-\frac{\alpha}{2}}\} (t+\epsilon) A^4 C_0^{m-1} \epsilon}{(1+v^b)^2 |y-(\frac{v}{2})^\alpha|} \mathbb{1}_{\{y \le (\frac{v}{2})^\alpha - \epsilon^{-\frac{1}{m-1}} e^{-\frac{\delta}{\epsilon}}\}}
\end{align*}
$$

and then we can bound $v^\gamma \int_0^t S_\epsilon(t-s)[I_5(y, v, s)]ds$ in the same manner as $L(y, v, t)$, where $L(y, v, t)$ is as in (3.41). We then conclude that (3.15) holds using estimates similar to (3.43) and (3.45).

This concludes our proof.

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

4 Region $y \le (\frac{v}{3})^{\alpha}$

4.1 Moment estimates

In this section we prove that

**Proposition 4.1.** There exists a sufficiently small $T \le 1$ and a sufficiently small $\epsilon_1 \in (0, 1)$, such that for all $\epsilon \le \epsilon_1$ and for all $n \in \mathbb{N}$ the following statement holds. Let $b \ge \bar{b}(\gamma, \alpha)$, $m \ge \frac{b}{\alpha} + 1$ in (2.11)-(2.13), where $\bar{b}(\gamma, \alpha)$ is as in (2.18). Let $H_n$ be as in (2.51). If

$$
H_n(y, v, t) \le T_1(y, v, t) + T_2(y, v, t) + T_3(y, v, t),
$$

where $T_1, T_2, T_3$ are as in (2.11)-(2.13), then it holds that $H_{n+1}(y, v, t) \le T_3(y, v, t)$, for all $t \in [0, T]$ and $y \le (\frac{v}{3})^\alpha, v \in (0, \infty)$.
Remark 4.2. T will be chosen in (4.42) and depends on A in (2.1).

Let

$$ \xi \in C([0, \infty)), \quad \xi: [0, \infty) \to [0, 1] \text{ such that } \xi(v) = 1 \text{ if } v \ge 1 \text{ and } \xi(v) = 0 \text{ if } v \le \frac{1}{2}. \quad (4.1) $$

The idea behind the proof of Proposition 4.1 is to show that $G_{\epsilon}(y, v, t) = e^{\frac{1}{\epsilon}} \tilde{G}_{\epsilon}(y, v, t)$ (or a suitably modified version of this function) is a supersolution for (2.51), where $\tilde{G}_{\epsilon}$ solves

$$ \begin{gathered} \partial_t \tilde{G}_\epsilon(y, v, t) + \frac{1}{\epsilon}(v^\alpha - y)\partial_y \tilde{G}_\epsilon(y, v, t) + \frac{L v^\gamma \xi(v)}{1 + |y|^d} \partial_v \tilde{G}_\epsilon(y, v, t) = 0, \\ (1 + v^b) \tilde{G}_\epsilon(y, v, 0) = \frac{A}{1 + |y - v^\alpha|^m}, \end{gathered} $$

with A as in (2.1) and for some suitably chosen $L > 0$ and $d \in \mathbb{N}$. We will define $d$ in (4.7) and $L$ in (4.37) and notice that $L$ depends on A. We will then choose $\epsilon_1$ in Proposition 4.1 to depend on $L$. For the choice of $\epsilon_1$, we refer to Proposition 4.4 and Proposition 4.7. Notice that $G_{\epsilon}$ satisfies the following equation

$$ \partial_t G_{\epsilon}(y, v, t) + \frac{1}{\epsilon}(v^{\alpha} - y)\partial_y G_{\epsilon}(y, v, t) + \frac{L v^{\gamma} \xi(v)}{1 + |y|^d} \partial_v G_{\epsilon}(y, v, t) - \frac{1}{\epsilon} G_{\epsilon}(y, v, t) = 0. \quad (4.2) $$

We first prove some moment estimates.

**Proposition 4.3 (Moment estimates).** Let $T \le 1$, $k \ge 0$. Assume that $P: \mathbb{R} \times (0, \infty) \times [0, T] \to [0, \infty)$ is such that

$$ P(y, v, t) \le T_1(y, v, t) + T_2(y, v, t) + T_3(y, v, t), $$

for all $t \in [0, T]$, where $T_1, T_2, T_3$ are as in (2.11)-(2.13) with $b > k + 1$, $m \ge \frac{b}{\alpha} + 1$ in (2.11)-(2.12). It then holds that there exists a constant $K_1 > 0$, which depends on $\alpha, \gamma, b, m$, such that

$$ \int_{(0,\infty)} w^k P(y, w, t) dw \le \frac{K_1 A^3}{1 + |y|^{\frac{b-k-1}{\alpha}}}, \quad \text{for all } t \in [0, T], y \in \mathbb{R}, \quad (4.3) $$

where A is as in (2.1). In particular, for $k = 1$, it holds that if $b > 2$, then

$$ \int_{(0,\infty)} w P(y, w, t) dw \le \frac{K_1 A^3}{1 + |y|^{\frac{b-2}{\alpha}}}, \quad \text{for all } t \in [0, T], y \in \mathbb{R}. \quad (4.4) $$

*Proof.* We begin by noticing that

$$ (1 + v^b)[T_1(y, v, t) + T_2(y, v, t) + T_3(y, v, t)] \le C(\alpha, \gamma, b, m)A^3[e^{\frac{1}{\epsilon}}\psi(e^{\frac{1}{\epsilon}}(y - v^\alpha))\chi_1(y, v, t) + \frac{\epsilon}{|y - v^\alpha|}\chi_2(y, v, t) + e^{\frac{1}{\epsilon}}\psi(e^{\frac{1}{\epsilon}}(y - v^\alpha))\chi_3(y, v, t)], $$

where $\chi_1, \chi_2, \chi_3$ are as in (2.14)-(2.16). This is since by the definition of $C_0$ in (2.9), we have that $C_0$ is of order $A^{\frac{1}{m-1}}$.

Let $y \ge 0$. We distinguish between different regions.
* Assume first that {$y \ge w^{\alpha} - C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}$}. Then

$$ \int_{\{y \ge w^{\alpha} - C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}} w^k P(y, w, t) dw \le \int_{(0, \infty)} \frac{2Aw^k e^{\frac{t}{\epsilon}}}{(1 + w^b)(1 + e^{\frac{m}{\epsilon}} |y - w^{\alpha}|^m)} dw. $$

We make the change of variables $z = e^{\frac{t}{\epsilon}}(y - w^{\alpha})$ and we obtain that

$$ \int_{\{y \ge w^{\alpha} - C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}} w^k P(y, w, t) dw \le \int_{(-\infty, \infty)} \frac{2A(y - e^{-\frac{t}{\epsilon}}z)^{\frac{k+1-\alpha}{\alpha}}}{(1 + (y - e^{-\frac{t}{\epsilon}}z)^{\frac{b}{\alpha}})(1 + |z|^m)} dz. $$

On one side, if $|z| \le \frac{e^{\frac{t}{\epsilon}}y}{2}$, it holds that $\frac{y}{2} \le y - e^{-\frac{t}{\epsilon}}z \le \frac{3y}{2}$. Thus

$$ \int_{|z| \le \frac{e^{\frac{t}{\epsilon}}y}{2}} \frac{(y - e^{-\frac{t}{\epsilon}}z)^{\frac{k+1-\alpha}{\alpha}}}{(1 + (y - e^{-\frac{t}{\epsilon}}z)^{\frac{b}{\alpha}})(1 + |z|^m)} dz \le \frac{C(\alpha, b)|y|^{\frac{k+1-\alpha}{\alpha}}}{1 + |y|^{\frac{b}{\alpha}}} \int_{(-\infty, \infty)} \frac{C(\alpha, b)}{1 + |z|^m} \le \frac{C(\alpha, b)|y|^{\frac{k+1-\alpha}{\alpha}}}{1 + |y|^{\frac{b}{\alpha}}}. \quad (4.5) $$

On the other hand, $\frac{(y - e^{-\frac{t}{\epsilon}}z)^{\frac{k+1-\alpha}{\alpha}}}{(1 + (y - e^{-\frac{t}{\epsilon}}z)^{\frac{b}{\alpha}})} \le 1$. Thus

$$ \int_{|z| \ge \frac{e^{\frac{t}{\epsilon}}y}{2}} \frac{(y - e^{-\frac{t}{\epsilon}}z)^{\frac{k+1-\alpha}{\alpha}}}{(1 + (y - e^{-\frac{t}{\epsilon}}z)^{\frac{b}{\alpha}})(1 + |z|^m)} dz \le \int_{|z| \ge \frac{e^{\frac{t}{\epsilon}}y}{2}} \frac{dz}{1 + |z|^m} \le \frac{C}{1 + (e^{\frac{t}{\epsilon}}|y|)^{m-1}} \le \frac{C}{1 + |y|^{m-1}}. \quad (4.6) $$

Combining (4.5) with (4.6) and the fact that $m \ge \frac{b}{\alpha} + 1$, we obtain that (4.3) holds in the region $\{y \ge w^{\alpha} - C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}$.

* We now analyze the contribution of the region $(\frac{w}{3})^{\alpha} \le y \le w^{\alpha} - C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}$. Notice that in this region it holds that $y^{\frac{1}{\alpha}} \le (y + C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}})^{\frac{1}{\alpha}} \le w \le 3y^{\frac{1}{\alpha}}$. We remember that $C_0$ is of order $A^{\frac{1}{m-1}}$. Thus

$$ \int_{\{(\frac{w}{3})^{\alpha} \le y \le w^{\alpha} - C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}} w^k P(y, w, t) dw \le \int_{(y + C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}})^{\frac{1}{\alpha}}}^{3y^{\frac{1}{\alpha}}} \frac{C(\alpha, \gamma, b, m) A^3 w^k \epsilon}{(1 + w^b) |y - w^{\alpha}|} dw \\ \le \frac{C(\alpha, \gamma, b, m) A^3 \epsilon y^{\frac{k}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} \int_{(y + C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}})^{\frac{1}{\alpha}}}^{3y^{\frac{1}{\alpha}}} \frac{1}{|y - w^{\alpha}|} dw. $$

We make the change of variables $z = y - w^{\alpha}$ and obtain that

$$ \frac{\epsilon y^{\frac{k}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} \int_{(y + C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}})^{\frac{1}{\alpha}}}^{3y^{\frac{1}{\alpha}}} \frac{1}{|y - w^{\alpha}|} dw \le \frac{C(\alpha) \epsilon y^{\frac{k+1-\alpha}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} \int_{-(3^{\alpha}-1)y}^{-C_{0}^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}} \frac{dz}{|z|} \\ \le \frac{C(\alpha) \epsilon y^{\frac{k+1-\alpha}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} [\frac{t}{\epsilon} + |\ln y|] \le \frac{C(\alpha)}{1 + y^{\frac{b-k-1}{\alpha}}}, $$

where in the last inequality we used the fact that $t \le 1$.
• We now look at the contribution of the region $\{0 \le y \le (\frac{w}{3})^\alpha\}$. In this region, we have that $|y - w^\alpha| = w^\alpha - y$ and $[1 - \frac{1}{3^\alpha}]w^\alpha \le w^\alpha - y \le w^\alpha$. Then

$$ \begin{aligned} \int_{\{0 \le y \le (\frac{w}{3})^\alpha\}} w^k P(y, w, t) dw &\le \int_{\{0 \le y \le (\frac{w}{3})^\alpha\}} \frac{C(\alpha, \gamma, b, m) A e^{\frac{t}{\epsilon}}}{1 + e^{\frac{m}{\epsilon}}(w^\alpha - y)^m} \frac{w^k}{1 + w^b} dw \\ &\le \int_{\{0 \le y \le (\frac{w}{3})^\alpha\}} \frac{C(\alpha, \gamma, b, m) A e^{\frac{t}{\epsilon}}}{1 + e^{\frac{m}{\epsilon}} w^{\alpha m}} \frac{w^k}{1 + w^b} dw. \end{aligned} $$

We make the change of variable $z = e^{\frac{t}{\epsilon}} w^\alpha$ and obtain that

$$ \begin{aligned} \int_{\{0 \le y \le (\frac{w}{3})^\alpha\}} w^k P(y, w, t) dw &\le C(\alpha, \gamma, b, m) A \int_{(3^\alpha e^{\frac{t}{\epsilon}} y, \infty)} \frac{(ze^{-\frac{t}{\epsilon}})^{\frac{k+1-\alpha}{\alpha}}}{1 + z^m} dz \\ &\le C(\alpha, \gamma, b, m) A \int_{(3^\alpha e^{\frac{t}{\epsilon}} y, \infty)} \frac{z^{\frac{k+1-\alpha}{\alpha}}}{1 + z^m} dz \le \frac{C(\alpha, \gamma, b, m) A}{1 + y^{m-\frac{k+1-\alpha}{\alpha}}}, \end{aligned} $$

This concludes our proof in the case $y \ge 0$ since $m \ge \frac{b}{\alpha}$. If $y \le 0$, by making the change of variable $z = e^{\frac{t}{\epsilon}}(w^\alpha - y)$, we obtain that

$$ \begin{aligned} \int w^k P(y, w, t) dw &\le \int_{(0, \infty)} \frac{C(\alpha, \gamma, b, m) A w^k e^{\frac{t}{\epsilon}}}{(1 + w^b)(1 + e^{\frac{m}{\epsilon}}|y - w^\alpha|^m)} dw \\ &\le \int_{(e^{\frac{t}{\epsilon}}|y|, \infty)} \frac{C(\alpha, \gamma, b, m) A}{1 + |z|^m} dz \le \frac{C(\alpha, \gamma, b, m) A}{1 + (e^{\frac{t}{\epsilon}}|y|)^{m-1}}. \end{aligned} $$

This concludes our proof.

□

We continue by analyzing properties of the transport equation (4.2). Define $d$ via

$$ \begin{cases} d = \lfloor \frac{b-2}{\alpha} \rfloor - 1; & \text{if } \lfloor \frac{b-2}{\alpha} \rfloor \text{ odd;} \\ d = \lfloor \frac{b-2}{\alpha} \rfloor; & \text{if } \lfloor \frac{b-2}{\alpha} \rfloor \text{ even,} \end{cases} \quad (4.7) $$

where $\lfloor t \rfloor$ denotes the floor function. We use the method of characteristics in (4.2)

$$ \left\{ \begin{aligned} \partial_t Y_\epsilon(y, v, t) &= -\frac{1}{\epsilon}(V_\epsilon^\alpha - Y), & Y_\epsilon(y, v, 0) &= y \le (\frac{v}{3})^\alpha, \\ \partial_t V_\epsilon(y, v, t) &= -\frac{LV_\epsilon^\gamma \xi(V_\epsilon)}{1 + |Y_\epsilon|^d}, & V_\epsilon(y, v, 0) &= v. \end{aligned} \right. \quad (4.8) $$

Notice that for $b > 4$ then $d > 1$ and even. It immediately follows from (4.8) that

$$ \frac{dV_{\epsilon}}{dY_{\epsilon}} = \frac{\epsilon LV_{\epsilon}^{\gamma} \xi(V_{\epsilon})}{(1 + |Y_{\epsilon}|^d)(V_{\epsilon}^{\alpha} - Y_{\epsilon})}. \quad (4.9) $$

Notice that this is different from the analysis of the characteristics performed in [9] since in that case we could make use of the separation of variables in (4.9) since the transport term in the space variable contributed with $V_\epsilon^\alpha$ instead of $V_\epsilon^\alpha - Y_\epsilon$. We thus have to apply different methods than in [9] to prove the desired estimates in our case.
## 4.2 Properties of the characteristics of the equation satisfied by the supersolution

**Proposition 4.4.** Let $\alpha \in (0, 1)$, $\gamma \in (1, 1+\alpha)$, and $d > 1$ even. Let $L > 0$ and let $Y_\epsilon, V_\epsilon$ be as in (4.8). Then there exists a sufficiently small $\epsilon_1 \in (0, 1)$, which depends on $L$, such that for all $t \ge 0$ and $\epsilon \le \epsilon_1$, the following estimates hold for all $v > 0$ and $y \le (\frac{v}{3})^\alpha$.

$$ \frac{9}{10} v \leq V_{\epsilon}(y, v, t) \leq v; \qquad (4.10) $$

$$ Y_{\epsilon}(y, v, t) \leq \left(\frac{v}{3}\right)^{\alpha}; \qquad (4.11) $$

$$ e_{\epsilon}^{t}(y - v^{\alpha}) \leq Y_{\epsilon} - V_{\epsilon}^{\alpha}; \qquad (4.12) $$

$$ Y_{\epsilon} - V_{\epsilon}^{\alpha} \leq e_{\epsilon}^{t}(y - \left(\frac{9}{10}\right)^{\alpha} v^{\alpha}). \qquad (4.13) $$

*Remark 4.5.* We will later fix $L$ in (4.37) to depend on $A$ and other constants depending only on the parameters $\alpha, \gamma, b, m$. Thus $\epsilon_1$ will depend on $A$.

*Proof of Proposition 4.4.* In order to simplify the notation, we replace the notation of the pair $(Y_\epsilon, V_\epsilon)$ in (4.8) with $(Y, V)$, while keeping in mind the dependence on $\epsilon$. Moreover, we will assume in all the following that $v \ge 1$ and will work with the following system of ODE's.

$$ \left\{ \begin{aligned} \partial_t Y(y, v, t) &= -\frac{1}{\epsilon}(V^\alpha - Y), & Y(y, v, 0) &= y \le \left(\frac{v}{3}\right)^\alpha, \\ \partial_t V(y, v, t) &= -\frac{LV^\gamma}{1 + |Y|^d}, & V(y, v, 0) &= v. \end{aligned} \right. \qquad (4.14) $$

This is since there is nothing to prove in the case $v \le \frac{1}{2}$ since by the definition of $\xi$ in (4.1), it holds that $\partial_t V_\epsilon = 0$ when $v \le \frac{1}{2}$. Moreover, assume that (4.10)-(4.13) hold for $v \ge \frac{1}{2}$ and for (4.14). We notice that the following inequalities hold. We have that $\partial_t V_\epsilon(y, v, t) = -\frac{LV_\epsilon^\gamma \xi(V_\epsilon)}{1+|Y_\epsilon|^d} \ge -\frac{LV_\epsilon^\gamma}{1+|Y_\epsilon|^d}$ and that $\partial_t V_\epsilon(y, v, t) \le 0$. Thus, (4.10) follows. If (4.10) holds, then (4.11) follows from (4.8). (4.12)-(4.13) follow from (4.10)-(4.11). Thus, it suffices to analyze the system (4.14) with $v \ge \frac{1}{2}$. We assume in addition that $v \ge 1$ for simplicity of notation.

From (4.8), we have that

$$ \partial_t V(y, v, t) = - \frac{LV^\gamma}{1 + |Y|^d} \ge -LV^\gamma. \qquad (4.15) $$

Thus, there exists a time $t_1 := t_1(v)$ sufficiently small, which is independent of $\epsilon$ but dependent on $v$, such that

$$ \frac{9}{10} v \leq V(y, v, s) \leq v, $$

for all $0 \le s \le t_1$. This implies that for all $s \le t_1$ it holds that

$$ \partial_s Y(y, v, s) = \frac{1}{\epsilon}(Y - V^{\alpha}) \le \frac{1}{\epsilon}(Y - \left(\frac{9}{10}\right)^{\alpha} v^{\alpha}). \qquad (4.16) $$
We analyze the sign of the following ODE.

$$ \partial_z \bar{Y}(y, v, z) = \frac{1}{\epsilon} \left(\bar{Y} - \left(\frac{9}{10}\right)^{\alpha} v^{\alpha}\right), \quad \bar{Y}(y, v, 0) = y \le \left(\frac{v}{3}\right)^{\alpha}. \qquad (4.17) $$

Since $\bar{Y} = \left(\frac{9}{10}\right)^{\alpha} v^{\alpha}$ is the zero point of (4.17) and $\bar{Y}(y, v, 0) = y \le \left(\frac{v}{3}\right)^{\alpha} \le \left(\frac{9}{10}\right)^{\alpha} v^{\alpha}$, then we have that $\bar{Y}(y, v, z) \le \left(\frac{9}{10}\right)^{\alpha} v^{\alpha}$, for all $z \ge 0$ and furthermore $\bar{Y}(y, v, z) \le y \le \left(\frac{v}{3}\right)^{\alpha}$, for all $z \ge 0$.

By (4.16) and using a comparison principle, this implies that $Y(y, v, s) \le \left(\frac{v}{3}\right)^{\alpha}$, for all $s \le t_1$ and in particular $Y(y, v, t_1) \le \left(\frac{v}{3}\right)^{\alpha}$. We thus have that $V(s) \ge \frac{9}{10}v$ and $Y(s) \le \left(\frac{v}{3}\right)^{\alpha}$, for all $s \le t_1$. Then (4.16) becomes

$$ \partial_s Y \le \frac{v^\alpha}{\epsilon} \left[ \frac{1}{3^\alpha} - \left(\frac{9}{10}\right)^\alpha \right] \le -\frac{c v^\alpha}{\epsilon}, \qquad (4.18) $$

for some $c > 0$.

i) Assume first that $Y(y, v, 0) = y \in [0, (\frac{v}{3})^\alpha]$ and $Y(y, v, t_1) < 0$.

We have that for any $y \in \mathbb{R}$ and $v > 0$ there exists $\tau := \tau(y, v) \in [0, t_1)$ such that $Y(y, v, \tau(y, v)) = 0$. Thus, for all $s \in [0, t_1]$, it holds that

$$ |Y(y, v, s)| = |Y(y, v, s) - Y(y, v, \tau(y, v))| \ge \frac{c v^{\alpha}}{\epsilon} |s - \tau(y, v)|. $$

Let $s \in [0, t_1]$. Since $\frac{9}{10}v \le V \le v$, it holds that

$$ \partial_s V(y, v, s) = - \frac{LV^\gamma}{1 + |Y|^d} \ge - \frac{LV^\gamma}{1 + \frac{c v^{\alpha d}}{\epsilon^d} |s - \tau(y, v)|^d} \ge - \frac{CL v^\gamma}{1 + \frac{c v^{\alpha d}}{\epsilon^d} |s - \tau(y, v)|^d}. $$

In other words, we have that

$$ V(y, v, s) - v \ge -v^\gamma \int_0^s \frac{CL dz}{1 + \frac{c v^{\alpha d}}{\epsilon^d} |z - \tau(y, v)|^d}. \qquad (4.19) $$

By making the change of variables $\xi = \frac{v^\alpha}{\epsilon}(z - \tau(y, v))$ in (4.19), it follows that

$$ V(y, v, s) \ge v - \epsilon v^{\gamma - \alpha} \int_{-\infty}^{\infty} \frac{CL d\xi}{1 + c|\xi|^d} \ge v \left[1 - CL\epsilon v^{\gamma - \alpha - 1}\right]. \qquad (4.20) $$

Since $v \ge 1$ and $\gamma < \alpha + 1$, there exists $\epsilon_1 \in (0, 1)$ sufficiently small, depending on $L$, but independent of $v$ such that $\left[1 - CL\epsilon v^{\gamma - \alpha - 1}\right] \ge \frac{19}{20}$ and thus

$$ V(y, v, s) \ge \frac{19}{20} v, \qquad (4.21) $$

for all $s \in [0, t_1]$ and all $\epsilon \le \epsilon_1$.

From (4.21), (4.15), and by continuity, we deduce that there exists $t_2 := t_2(v) > t_1$ such that

$$ \frac{9}{10} v \le V(y, v, s) \le v, \text{ for all } s \in [t_1, t_2]. $$
ii) Assume $Y(y, v, 0) = y \in [0, (\frac{v}{3})^\alpha]$ and $Y(y, v, t_1) \ge 0$.

Since $Y(y, v, s) \ge Y(y, v, t_1) \ge 0$, for all $s \in [0, t_1]$, it holds that

$$|Y(y, v, s)| = Y(y, v, s) \ge Y(y, v, s) - Y(y, v, t_1) \ge |Y(y, v, s) - Y(y, v, t_1)| \ge \frac{cv^\alpha}{\epsilon}|s - t_1|.$$

We can then perform similar computations as in the case $Y(y, v, 0) = y \in [0, (\frac{v}{3})^\alpha]$ and $Y(y, v, t_1) < 0$ in order to deduce that $V(y, v, s) \ge \frac{19}{20}v$, for all $s \le t_1$ and all $\epsilon \le \epsilon_1$.

iii) Assume $Y(y, v, 0) = y < 0$.

Then for all $s \in [0, t_1]$, it holds by (4.16) that

$$Y(y, v, s) \le -\left(\frac{9}{10}\right)^\alpha \frac{v^\alpha s}{\epsilon} \quad (4.22)$$

and thus, as before, we can conclude that

$$\partial_s V(y, v, s) = -\frac{LV^\gamma}{1 + |Y|^d} \ge -\frac{LV^\gamma}{1 + \frac{cv^{\alpha d}}{\epsilon^d} s^d} \ge -\frac{CLv^\gamma}{1 + \frac{cv^{\alpha d}}{\epsilon^d} s^d}.$$

As before, we obtain that

$$V(y, v, s) \ge \frac{19}{20}v, \quad (4.23)$$

for all $s \in [0, t_1]$ and all $\epsilon \le \epsilon_1$.

We are now able to iterate the argument in order to obtain that (4.10) holds for all $t \ge 0$. Notice that at each iteration the times $t_i$, $i \in \mathbb{N}$, depend on $v$, but after the iteration we obtain that the estimate (4.10) holds for all times $t \ge 0$. Moreover, $\epsilon_1$ in (4.20) does not depend on $v$. The inequality (4.11) follows from (4.10) and (4.16).

By (4.10) and (4.8) it follows that

$$\partial_t Y - \frac{Y}{\epsilon} = -\frac{V^\alpha}{\epsilon} \ge -\frac{v^\alpha}{\epsilon}$$

and thus

$$\partial_t \left( e^{-\frac{t}{\epsilon}} Y \right) \ge \partial_t \left( e^{-\frac{t}{\epsilon}} v^\alpha \right).$$

Integrating in time and using the fact that $V \le v$ we obtain (4.12). (4.13) follows in a similar manner by noticing that

$$\partial_t Y - \frac{Y}{\epsilon} = -\frac{V^\alpha}{\epsilon} \le -\left(\frac{9}{10}\right)^\alpha \frac{v^\alpha}{\epsilon}.$$

This concludes our proof. □
Remark 4.6. From (4.13) we immediately deduce that for all $y \le (\frac{v}{3})^{\alpha}$, the following inequality holds.

$$Y_{\epsilon} - V_{\epsilon}^{\alpha} \le \left[ \left( \frac{9}{10} \right)^{\alpha} - \frac{1}{3^{\alpha}} \right] e^{\frac{t}{\epsilon}} (y - v^{\alpha}).$$

*Proof.* For $y \le 0$, it immediately follows that

$$Y_{\epsilon} - V_{\epsilon}^{\alpha} \le e^{\frac{t}{\epsilon}}(y - \left(\frac{9}{10}\right)^{\alpha}v^{\alpha}) \le \left(\frac{9}{10}\right)^{\alpha}e^{\frac{t}{\epsilon}}(y - v^{\alpha}) \le \left[\left(\frac{9}{10}\right)^{\alpha} - \frac{1}{3^{\alpha}}\right]e^{\frac{t}{\epsilon}}(y - v^{\alpha}).$$

If $y \in (0, (\frac{v}{3})^{\alpha})$, then

$$Y_{\epsilon} - V_{\epsilon}^{\alpha} \le e^{\frac{t}{\epsilon}}(y - \left(\frac{9}{10}\right)^{\alpha}v^{\alpha}) \le -\left[\left(\frac{9}{10}\right)^{\alpha} - \frac{1}{3^{\alpha}}\right]e^{\frac{t}{\epsilon}}v^{\alpha} \le \left[\left(\frac{9}{10}\right)^{\alpha} - \frac{1}{3^{\alpha}}\right]e^{\frac{t}{\epsilon}}(y - v^{\alpha}).$$

□

## 4.3 Properties of the supersolution in the region $y \le (\frac{v}{3})^{\alpha}$

**Proposition 4.7.** Fix $\alpha \in (0, 1)$, $\gamma \in (1, 1 + \alpha)$, $L > 0$ and $\epsilon \in (0, 1)$. Let $d > 1$ in (4.7) even and sufficiently large such that $\gamma \le \frac{d\alpha}{d+1} + 1$. Let $G_{\epsilon,L}$ solve equation (4.2). Then there exists a sufficiently small $\epsilon_1 \in (0, 1)$ such that for all $t \ge 0$ and $\epsilon \le \epsilon_1$, we have that for all $v > 0$ and $y \le (\frac{v}{3})^{\alpha}$,

$$\partial_v G_{\epsilon,L}(y, v, t) \le 0.$$

*Proof.* Let $Y_{\epsilon}$, $V_{\epsilon}$ be as in (4.8). We have that

$$G_{\epsilon,L}(y, v, t) = \frac{A e^{\frac{t}{\epsilon}}}{(1 + V_{\epsilon}^{b})(1 + |Y_{\epsilon} - V_{\epsilon}^{\alpha}|^{m})},$$

with $A$ as in (2.1). As before, in order to simplify the notation, we replace the notation of the pair $(Y_{\epsilon}, V_{\epsilon})$ in (4.8) with $(Y, V)$, while keeping in mind the dependence on $\epsilon$. Moreover, from Proposition 4.4, we know that $Y - V^{\alpha} \le 0$, for all $t \ge 0$. Thus

$$\partial_v G_{\epsilon,L}(y, v, t) = - \frac{m A e^{\frac{t}{\epsilon}} (V^{\alpha} - Y)^{m-1} [\alpha V^{\alpha-1} \partial_v V - \partial_v Y]}{(1 + V^{b})(1 + (V^{\alpha} - Y)^m)^2} - \frac{b A e^{\frac{t}{\epsilon}} V^{b-1} \partial_v V}{(1 + V^{b})^2 (1 + (V^{\alpha} - Y)^m)}.$$

Thus, in order to prove that $\partial_v G_{\epsilon,L}(y, v, t) \le 0$ it suffices to prove that $\alpha V^{\alpha-1} \partial_v V - \partial_v Y \ge 0$ and $\partial_v V \ge 0$. Fix $t > 0$. We know that $V \ge 0$ by Proposition 4.4. We will prove that

$$\partial_v V(y, v, t) \ge 0, \text{ for all } v > 0, y \le (\frac{v}{3})^{\alpha},$$

and that

$$\partial_v Y(y, v, t) \le 0, \text{ for all } v > 0, y \le (\frac{v}{3})^{\alpha},$$
thus concluding our proof. Since $\partial_v V(y, v, 0) = 1$, by continuity there exists $t_1 > 0$ such that $\partial_v V(y, v, s) \in [\frac{1}{2}, 1]$, for all $s \in [0, t_1]$. By (4.8), we have that

$$ \partial_s \partial_v Y(y, v, s) = \frac{\partial_v Y}{\epsilon} - \frac{\alpha V^{\alpha-1} \partial_v V}{\epsilon}, \quad \partial_v Y(y, v, 0) = 0. $$

In other words, using Proposition 4.4 and since $\alpha < 1$, it holds that for all $s \in [0, t_1]$

$$ \partial_s(e^{-\frac{s}{\epsilon}} \partial_v Y(y, v, s)) = -e^{-\frac{s}{\epsilon}} \frac{\alpha V^{\alpha-1} \partial_v V}{\epsilon} \leq -e^{-\frac{s}{\epsilon}} \frac{\alpha v^{\alpha-1}}{2\epsilon}. $$

Thus, by integrating in time, we obtain that

$$ e^{-\frac{s}{\epsilon}} \partial_v Y \leq \frac{\alpha}{2} e^{-\frac{s}{\epsilon}} v^{\alpha-1} - \frac{\alpha}{2} v^{\alpha-1}, $$

for all $s \in [0, t_1]$, and thus

$$ \partial_v Y(y, v, s) \leq \frac{\alpha}{2} v^{\alpha-1} (1 - e^{\frac{s}{\epsilon}}) \leq 0. $$

Using similar arguments, we obtain that

$$ \partial_v Y(y, v, s) \geq \frac{10^{1-\alpha} \alpha}{9^{1-\alpha} v^{\alpha-1}} (1 - e^{\frac{s}{\epsilon}}). \quad (4.24) $$

By (4.8), we also have that for all $s \in [0, t_1]$,

$$ \partial_s \partial_v V(y, v, s) = - \frac{\gamma L V^{\gamma-1} \partial_v V}{1 + |Y|^d} + \frac{d L V^{\gamma} |Y|^{d-2} \partial_v Y}{(1 + |Y|^d)^2}, \quad \partial_v V(y, v, 0) = 1. \quad (4.25) $$

We will prove that

$$ \int_0^s \frac{V^{\gamma-1}(z) dz}{1 + |Y|^d(z)} \leq C\epsilon \quad (4.26) $$

and that

$$ \int_0^s \frac{V^\gamma |Y|^{d-1} |\partial_v Y|}{(1 + |Y|^d)^2} dz \leq C\epsilon, \quad (4.27) $$

for all $s \in [0, t_1]$. We first prove that (4.26) holds. By (4.10) we have that

$$ \frac{V^{\gamma-1}}{1 + |Y|^d} \leq \frac{v^{\gamma-1}}{1 + |Y|^d}. $$

Moreover, we remember that

$$ \partial_s Y(y, v, s) = \frac{Y}{\epsilon} - \frac{V^\alpha}{\epsilon} \leq \frac{v^\alpha}{\epsilon} \left[ \frac{1}{3^\alpha} - \left( \frac{9}{10} \right)^\alpha \right] \leq -\frac{cV^\alpha}{\epsilon} $$

and then as before we have the following cases.
i) $Y(y, v, 0) = y \in [0, (\frac{v}{3})^\alpha]$ and $Y(y, v, t_1) < 0$.

Then there exists $\tau := \tau(y, v) \in [0, t_1)$ such that $Y(y, v, \tau(y, v)) = 0$ and, for all $s \in [0, t_1]$, it holds that

$$|Y(y, v, s)| \geq \frac{c v^{\alpha}}{\epsilon} |s - \tau(y, v)|. \quad (4.28)$$

Then

$$\int_0^s \frac{V^{\gamma-1}(z)dz}{1 + |Y|^d(z)} \leq \int_0^s \frac{C v^{\gamma-1} dz}{1 + \frac{v^{\alpha d}}{\epsilon^d} |z - \tau(y, v)|^d}.$$

We make the change of variable $\xi = \frac{v^\alpha}{\epsilon}(z - \tau)$ and use the fact that we can consider without loss of generality that $v \geq 1$ in order to obtain that

$$\int_0^s \frac{V^{\gamma-1}(z)dz}{1 + |Y|^d(z)} \leq \int_{-\infty}^{\infty} \frac{C \epsilon v^{\gamma-\alpha-1} d\xi}{1 + |\xi|^d} \leq C\epsilon.$$

ii) If $Y(y, v, 0) = y \in [0, (\frac{v}{3})^\alpha]$ and $Y(y, v, t_1) \geq 0$, we have that

$$|Y(y, v, s)| = Y(y, v, s) \geq Y(y, v, s) - Y(y, v, t_1) = |Y(y, v, s) - Y(y, v, t_1)| \geq \frac{c v^{\alpha}}{\epsilon} |s - t_1|$$

and the result follows as before.

iii) Otherwise $Y(y, v, 0) = y < 0$. Then for all $s \in [0, t_1]$, it holds by (4.16) that

$$Y(y, v, s) \leq -\left(\frac{9}{10}\right)^{\alpha} \frac{v^{\alpha} s}{\epsilon}$$

and proceed as in the previous case in order to show that (4.26) holds.

We now prove (4.27). We only analyze the case when $Y(y, v, 0) = y \in [0, (\frac{v}{3})^\alpha]$ and $Y(y, v, t_1) < 0$ as the other two cases can be treated in a similar, more straightforward manner. Thus, let us assume $Y(y, v, 0) = y \in [0, (\frac{v}{3})^\alpha]$ and $Y(y, v, t_1) < 0$. Then, for every $y$ and every $v > 0$, there exists $\tau := \tau(y, v) \in [0, t_1)$ such that $Y(y, v, \tau(y, v)) = 0$. We can write

$$\int_0^s \frac{V^\gamma |Y|^{d-1} |\partial_v Y|}{(1 + |Y|^d)^2} dz = \int_0^{\tau(y,v)} \frac{V^\gamma |Y|^{d-1} |\partial_v Y|}{(1 + |Y|^d)^2} dz + \int_{\tau(y,v)}^s \frac{V^\gamma |Y|^{d-1} |\partial_v Y|}{(1 + |Y|^d)^2} dz =: (I) + (II).$$

We bound (II). We have that

$$\partial_s Y = \frac{Y}{\epsilon} - \frac{V^\alpha}{\epsilon} \leq \frac{Y}{\epsilon} - \left(\frac{9}{10}\right)^\alpha \frac{v^\alpha}{\epsilon}$$

and thus

$$\partial_s (e^{-\frac{s}{\epsilon}} Y) \leq -e^{-\frac{s}{\epsilon}} \left(\frac{9}{10}\right)^{\alpha} \frac{v^{\alpha}}{\epsilon}.$$
We integrate from $\tau(y, v)$ to $s \in [\tau(y, v), t_1]$ and use the fact that $Y(y, v, \tau(y, v)) = 0$ in order to obtain that

$$e^{-\frac{s}{\epsilon}} Y(s) \leq e^{-\frac{s}{\epsilon}} \left(\frac{9}{10}\right)^{\alpha} v^{\alpha} - e^{-\frac{\tau}{\epsilon}} \left(\frac{9}{10}\right)^{\alpha} v^{\alpha}$$

or in other words

$$Y(y, v, s) \leq \left(\frac{9}{10}\right)^{\alpha} v^{\alpha} \left(1 - e^{-\frac{s-\tau}{\epsilon}}\right) \leq 0. \qquad (4.29)$$

Combining (4.29) with (4.24), we have that

$$\begin{aligned} \left| \frac{V^{\gamma}|Y|^{d-1} \partial_{v} Y}{(1 + |Y|^{d})^{2}} \right| & \leq \frac{C v^{\gamma} |\partial_{v} Y|}{1 + |Y|^{d+1}} \leq \frac{C v^{\gamma} |\partial_{v} Y|}{1 + C v^{\alpha(d+1)} (e^{\frac{s-\tau}{\epsilon}} - 1)^{d+1}} \\ & \leq \frac{C v^{\gamma+\alpha-1} (e^{\frac{s}{\epsilon}} - 1)}{1 + v^{\alpha(d+1)} (e^{\frac{s-\tau}{\epsilon}} - 1)^{d+1}} \\ & \leq \frac{C v^{\gamma-1} (e^{\frac{s}{\epsilon}} - 1)}{1 + v^{\alpha d} (e^{\frac{s-\tau}{\epsilon}} - 1)^{d+1}}. \end{aligned}$$

Thus

$$\int_{\tau}^{s} \left| \frac{V^{\gamma}|Y|^{d-2} Y \partial_{v} Y}{(1 + |Y|^{d})^{2}} \right| dz \leq \int_{\tau}^{s} \frac{C v^{\gamma-1} (e^{\frac{z}{\epsilon}} - 1)}{1 + v^{\alpha d} (e^{\frac{z-\tau}{\epsilon}} - 1)^{d+1}} dz.$$

We make the change of variables $\xi = v^{\frac{d\alpha}{d+1}}(e^{\frac{z-\tau}{\epsilon}} - 1)$ and thus

$$d\xi = \frac{v^{\frac{d\alpha}{d+1}}}{\epsilon} e^{\frac{z-\tau}{\epsilon}} dz.$$

We have that $\tau \leq C\epsilon$. This is since if we choose $s = 0$ in (4.28), then $|y| \geq \frac{c\epsilon^\alpha}{\epsilon}|\tau(y, v)|$ and we are in the case when $y \in [0, (\frac{v}{3})^\alpha]$. Using that $\tau \leq C\epsilon$ and $v \geq 1$, we further obtain that

$$\begin{aligned} \int_{\tau}^{s} \left| \frac{V^{\gamma}|Y|^{d-2} Y \partial_{v} Y}{(1 + |Y|^{d})^{2}} \right| dz & \leq \int_{0}^{\infty} \frac{C \epsilon v^{\gamma - \frac{ad}{d+1} - 1} e^{-\frac{z-\tau}{\epsilon}} (e^{\frac{z}{\epsilon}} - 1)}{1 + |\xi|^{d+1}} d\xi \leq \int_{0}^{\infty} \frac{C \epsilon v^{\gamma - \frac{ad}{d+1} - 1} e^{\frac{\tau}{\epsilon}} (1 - e^{-\frac{z}{\epsilon}})}{1 + |\xi|^{d+1}} d\xi \\ & \leq \int_{0}^{\infty} \frac{C \epsilon v^{\gamma - \frac{ad}{d+1} - 1} e^{\frac{\tau}{\epsilon}}}{1 + |\xi|^{d+1}} d\xi \leq \int_{0}^{\infty} \frac{C \epsilon v^{\gamma - \frac{ad}{d+1} - 1}}{1 + |\xi|^{d+1}} d\xi \leq C\epsilon \end{aligned} \qquad (4.30)$$

since we chose $d$ sufficiently large such that $\gamma \leq \frac{d\alpha}{d+1} + 1$.

We now bound (I). For $s \leq \tau$, we only have that (4.28) holds. Using in addition (4.24), we have that

$$\begin{aligned} \int_{0}^{\tau(y,v)} \frac{V^{\gamma}|Y|^{d-1} \partial_{v} Y}{(1 + |Y|^{d})^{2}} dz & \leq \int_{0}^{\tau} \frac{C v^{\gamma} |\partial_{v} Y|}{1 + |Y|^{d+1}} dz \leq \int_{0}^{\tau} \frac{C v^{\gamma+\alpha-1} (e^{\frac{z}{\epsilon}} - 1)}{1 + \frac{v^{\alpha(d+1)}}{\epsilon^{d+1}} |z - \tau|^{d+1}} dz \\ & \leq \int_{0}^{\tau} \frac{C v^{\gamma-1} (e^{\frac{z}{\epsilon}} - 1)}{1 + \frac{v^{d\alpha}}{\epsilon^{d+1}} |z - \tau|^{d+1}} dz. \end{aligned}$$
We now make the change of variables

$$ \xi = \frac{v^{\frac{ad}{d+1}}}{\epsilon}(\tau - z); $$

$$ d\xi = -\frac{v^{\frac{ad}{d+1}}}{\epsilon} dz; $$

$$ \frac{z}{\epsilon} = \frac{\tau - \epsilon v^{-\frac{ad}{d+1}} \xi}{\epsilon} \le \frac{\epsilon \xi + \tau}{\epsilon} = \xi + \frac{\tau}{\epsilon} \le \xi + 1. $$

Thus, using that $e^x - 1 \le Cx$, when $x \in [0, 1]$, we further obtain that

$$ (I) \le \int_{0}^{\infty} \frac{C \epsilon v^{\gamma - \frac{d\alpha}{d+1} - 1} \frac{z}{\epsilon}}{1 + |\xi|^{d+1}} d\xi \le \int_{0}^{\infty} \frac{C \epsilon v^{\gamma - \frac{d\alpha}{d+1} - 1} (\xi + 1)}{1 + |\xi|^{d+1}} d\xi \le C\epsilon, \quad (4.31) $$

since $v \ge 1$ and $\gamma \le \frac{d\alpha}{d+1} + 1$. Combining (4.31) with (4.30), we obtain that (4.27) holds. We denote by

$$ B(s) := \int_{0}^{s} \frac{L \gamma V^{\gamma-1}}{1 + |Y|^d} dz. $$

Notice that by (4.26), it holds that $0 \le B(s) \le CL$, for all $s \in [0, t_1]$. From (4.25), we obtain that

$$ \partial_s(e^{B(s)} \partial_v V(s)) \ge e^{B(s)} \frac{L d V^\gamma |Y|^{d-2} Y \partial_v Y}{(1 + |Y|^d)^2} \ge -e^{B(s)} \left| \frac{L d V^\gamma |Y|^{d-2} Y \partial_v Y}{(1 + |Y|^d)^2} \right| \ge -C(L) \left| \frac{V^\gamma |Y|^{d-2} Y \partial_v Y}{(1 + |Y|^d)^2} \right|, $$

for some constant $C(L) > 0$ which depends on $L$. Making use of (4.27), we further obtain that

$$ e^{B(s)} \partial_v V(s) - 1 \ge -C(L)\epsilon $$

and since by (4.26) it holds that $0 \le B(s) \le CL\epsilon$, for all $s \in [0, t_1]$, then there exists $\epsilon_1 \in (0, 1)$, which depends on $L$, such that for all $\epsilon \le \epsilon_1$, the following holds

$$ \partial_v V(s) \ge e^{-B(s)}(1 - C(L)\epsilon) \ge e^{-CL\epsilon}(1 - C(L)\epsilon) \ge \frac{3}{4}, $$

for all $s \in [0, t_1]$. We can thus iterate the argument in order to obtain that for all times $t \ge 0$, we have that $\partial_v V(y, v, t) \ge \frac{1}{2} \ge 0$ and $\partial_v Y(y, v, t) \le \frac{\alpha}{2} v^{\alpha-1} (1 - e^{\frac{t}{\epsilon}}) \le 0$. This concludes our proof. $\square$

**Remark 4.8.** As an immediate consequence, we have that

$$ \frac{10^{1-\alpha}\alpha}{9^{1-\alpha}} v^{\alpha-1} (1 - e^{\frac{t}{\epsilon}}) \le \partial_v Y(y, v, t) \le \frac{\alpha v^{\alpha-1}}{2} (1 - e^{\frac{t}{\epsilon}}); $$

$$ \frac{1}{2} \le \partial_v V(y, v, t) \le 1; $$

$$ \frac{\alpha v^{\alpha-1}}{2} e^{\frac{t}{\epsilon}} \le \alpha V^{\alpha-1} \partial_v V - \partial_v Y \le \frac{10^{1-\alpha}\alpha}{9^{1-\alpha}} v^{\alpha-1} e^{\frac{t}{\epsilon}}. $$
**Proposition 4.9.** Let $L > 0$. Let $G_{\epsilon,L}$ be as in (4.2) and $\epsilon_1 \in (0, 1)$ and $d > 1$ be as in Proposition 4.7. It holds that there exists $K_2 > 0$, which is independent of $\epsilon \le \epsilon_1$, $L > 0$, but which can depend on $\alpha, \gamma, b, m$, such that for all $t \ge 0$ and $\epsilon \le \epsilon_1$, we have that

$$ -\partial_v G_{\epsilon,L}(y, w, t) \le -K_2 \partial_v G_{\epsilon,L}(y, v, t), \text{ for all } w \in [\frac{v}{2}, v] \text{ and all } y \le (\frac{v}{3})^{\alpha}. $$

*Proof.* We remember that

$$ \partial_v G_{\epsilon,L}(y, v, t) = - \frac{m A e^{\frac{t}{\epsilon}} (V^{\alpha} - Y)^{m-1} [\alpha V^{\alpha-1} \partial_v V - \partial_v Y]}{(1 + V^b)(1 + (V^{\alpha} - Y)^m)^2} - \frac{b A e^{\frac{t}{\epsilon}} V^{b-1} \partial_v V}{(1 + V^b)^2 (1 + (V^{\alpha} - Y)^m)}, $$

with $A$ as in (2.1). By Proposition 4.4 and Remark 4.6, we have that

$$ 0 \le \left[ \left(\frac{9}{10}\right)^{\alpha} - \frac{1}{3^{\alpha}} \right] e^{\frac{t}{\epsilon}} (v^{\alpha} - y) \le V^{\alpha} - Y \le e^{\frac{t}{\epsilon}} (v^{\alpha} - y). \quad (4.32) $$

By Remark 4.8, it holds that

$$ \frac{\alpha v^{\alpha-1}}{2} e^{\frac{t}{\epsilon}} \le \alpha V^{\alpha-1} \partial_v V - \partial_v Y \le \frac{10^{1-\alpha}}{9^{1-\alpha}} v^{\alpha-1} e^{\frac{t}{\epsilon}}. \quad (4.33) $$

By (4.32) and (4.33), we have that

$$ \frac{m e^{\frac{t}{\epsilon}} (V^{\alpha} - Y)^{m-1} [\alpha V^{\alpha-1} \partial_v V - \partial_v Y]}{(1 + (V^{\alpha} - Y)^m)^2} \ge C(\alpha, m) \frac{e^{\frac{t(m+1)}{\epsilon}} (v^{\alpha} - y)^{m-1} v^{\alpha-1}}{(1 + e^{\frac{t m}{\epsilon}} (v^{\alpha} - y)^m)^2}. $$

It also holds that

$$ \frac{b e^{\frac{t}{\epsilon}} V^{b-1} \partial_v V}{(1 + V^b)^2 (1 + (V^{\alpha} - Y)^m)} \ge C(\alpha, b, m) \frac{e^{\frac{t}{\epsilon}} V^{b-1}}{(1 + V^b)^2 (1 + (v^{\alpha} - y)^m)}. $$

Thus

$$ -\partial_v G_{\epsilon,L}(y, v, t) \ge C(\alpha, b, m) A \left[ \frac{e^{\frac{t(m+1)}{\epsilon}} (v^{\alpha} - y)^{m-1} v^{\alpha-1}}{(1 + v^b)(1 + e^{\frac{t m}{\epsilon}} (v^{\alpha} - y)^m)^2} + \frac{e^{\frac{t}{\epsilon}} V^{b-1}}{(1 + v^b)^2 (1 + (v^{\alpha} - y)^m)} \right]. $$

We now look at the following system of ODE's.

$$ \left\{ \begin{array}{ll} \partial_t Y(y, w, t) = -\frac{1}{\epsilon}(V^{\alpha} - Y), & Y(y, w, 0) = y \le \left(\frac{v}{3}\right)^{\alpha}, \\ \\ \partial_t V(y, w, t) = -\frac{LV^{\gamma}}{1 + |Y|^d}, & V(y, w, 0) = w \in \left[\frac{v}{2}, v\right]. \end{array} \right. \quad (4.34) $$

We are able to reproduce the computations from Proposition 4.4 and Proposition 4.7 in order to obtain similar estimates for the solution $(\bar{Y}, \bar{V})$ of the system (4.34). This is since $y \le (\frac{v}{3})^{\alpha} \le (\frac{2}{3})^{\alpha} w^{\alpha}$. This implies that (4.18) becomes

$$ \partial_s \bar{Y} \le \frac{w^\alpha}{\epsilon} \left[ \frac{2^\alpha}{3^\alpha} - \left(\frac{9}{10}\right)^\alpha \right] \le -\frac{\tilde{c} w^\alpha}{\epsilon}. \quad (4.35) $$
The rest of the proof follows similarly, but some constants need to be modified. We obtain that

$$
\begin{align*}
\frac{9}{10}w &\le \bar{V}(y, w, t) \le w; \\
\bar{Y}(y, w, t) &\le \frac{2^\alpha}{3^\alpha}w^\alpha; \\
e^{\frac{l}{\epsilon}}(y-w^\alpha) &\le \bar{Y} - \bar{V}^\alpha; \\
\bar{Y} - \bar{V}^\alpha &\le \left[ \left(\frac{9}{10}\right)^\alpha - \frac{2^\alpha}{3^\alpha} \right] e^{\frac{l}{\epsilon}}(y-w^\alpha).
\end{align*}
$$

The bounds for $\partial_v \bar{Y}(y, w, t)$ and $\partial_v \bar{V}(y, w, t)$ do not change. We thus obtain that

$$
-\partial_v G_{\epsilon,L}(y, w, t) \le C(\alpha, b, m)A \left[ \frac{e^{\frac{l(m+1)}{\epsilon}} (w^{\alpha} - y)^{m-1} w^{\alpha-1}}{(1+w^b)(1+e^{\frac{lm}{\epsilon}}(w^{\alpha}-y)^m)^2} + \frac{e^{\frac{l}{\epsilon}} w^{b-1}}{(1+w^b)^2(1+e^{\frac{lm}{\epsilon}}(w^{\alpha}-y)^m)} \right].
$$

If $y \le 0$, we have that

$$
v^{\alpha} - y \geq w^{\alpha} - y \geq \left(\frac{v}{2}\right)^{\alpha} - y \geq \frac{v^{\alpha} - y}{2^{\alpha}}.
$$

If $y \in [0, (\frac{v}{3})^\alpha]$, then

$$
v^{\alpha} - y \geq w^{\alpha} - y \geq \left(\frac{v}{2}\right)^{\alpha} - y \geq \frac{3^{\alpha} - 2^{\alpha}}{6^{\alpha}} v^{\alpha} \geq \frac{3^{\alpha} - 2^{\alpha}}{6^{\alpha}} (v^{\alpha} - y).
$$

Concluding, we have that

$$
\frac{e^{\frac{l(m+1)}{\epsilon}} (w^{\alpha} - y)^{m-1} w^{\alpha-1}}{(1 + e^{\frac{lm}{\epsilon}} (w^{\alpha} - y)^m)^2} \le C(\alpha, b, m) \frac{e^{\frac{l(m+1)}{\epsilon}} (v^{\alpha} - y)^{m-1} v^{\alpha-1}}{(1 + e^{\frac{lm}{\epsilon}} (v^{\alpha} - y)^m)^2}
$$

and further that

$$
-\partial_v G_{\epsilon,L}(y, w, t) \leq -K_2 \partial_v G_{\epsilon,L}(y, v, t).
$$

□

4.4 Proof of existence of a supersolution in the region $y \le (\frac{v}{3})^\alpha$

We are now in a position to prove the main statement of this section.

*Proof of Proposition 4.1.* Assume $H_n(y, v, t) \le T_1(y, v, t) + T_2(y, v, t) + T_3(y, v, t)$. Let $C_y := 1 + \frac{1}{2y}$, $K_0$ as in (2.2), $K_1$ as in Proposition 4.3, $K_2 > 0$ be as in Proposition 4.9, and $A$ as in (2.1). Let

$$
G_{\epsilon} := e^{K_3(\alpha,b,m)Lt} G_{\epsilon,L}, \text{ where } G_{\epsilon,L} \text{ solves (4.2)} \quad (4.36)
$$

with

$$
L = C_{\gamma} K_{0} K_{1} K_{2} A^{3}. \tag{4.37}
$$
Moreover, $d$ is chosen as in (4.7) such that $d > 1$, even, and sufficiently large such that $\gamma \le \frac{d\alpha}{d+1} + 1$, and $K_3(\alpha, b, m)$ is a fixed constant depending on the parameters $\alpha, b, m$ which will be determined later. Notice that $d$ depends on $b$ and the minimum value of $b$ for which these conditions are satisfied determines the value of $\bar{b}(\gamma, \alpha)$ in (2.18) as noticed in Remark 2.9.

We prove Proposition 4.1 by first proving that $H_{n+1}(y, v, t) \le G_\epsilon(y, v, t)$, for all $t \in [0, T]$ and $y \le (\frac{v}{3})^\alpha$, $v \in (0, \infty)$, and then that $G_\epsilon(y, v, t) \le T_3(y, v, t)$ for all $t \in [0, T]$ and $y \le (\frac{v}{3})^\alpha$, $v \in (0, \infty)$.

Let $n \in \mathbb{N}$. We prove that $G_\epsilon$ is a supersolution for (2.51). In other words, we prove that

$$ \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)G_\epsilon(y, v, t)] - \int_0^{\frac{v}{2}} K(v-w, w)G_\epsilon(y, v-w, t)H_n(y, w, t)dw \\ + \int_0^\infty K(v, w)G_\epsilon(y, v, t)H_n(y, w, t)dw \ge 0. \quad (4.38) $$

Since $K(v-w, w) \le K(v, w)$, for all $w \le v$, it follows that

$$ \begin{aligned} & \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)G_\epsilon(y, v, t)] - \int_0^{\frac{v}{2}} K(v-w, w)G_\epsilon(y, v-w, t)H_n(y, w, t)dw \\ & \qquad + \int_0^\infty K(v, w)G_\epsilon(y, v, t)H_n(y, w, t)dw \\ & \ge \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)G_\epsilon(y, v, t)] - \int_0^{\frac{v}{2}} K(v-w, w)G_\epsilon(y, v-w, t)H_n(y, w, t)dw \\ & \qquad + \int_0^{\frac{v}{2}} K(v, w)G_\epsilon(y, v, t)H_n(y, w, t)dw \\ & \ge \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)G_\epsilon(y, v, t)] - \int_0^{\frac{v}{2}} K(v, w)G_\epsilon(y, v-w, t)H_n(y, w, t)dw \\ & \qquad + \int_0^{\frac{v}{2}} K(v, w)G_\epsilon(y, v, t)H_n(y, w, t)dw \\ & = \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)G_\epsilon(y, v, t)] - \int_0^{\frac{v}{2}} K(v, w)[G_\epsilon(y, v-w, t) - G_\epsilon(y, v, t)]H_n(y, w, t)dw \\ & = \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)G_\epsilon(y, v, t)] + \int_0^{\frac{v}{2}} K(v, w) \int_{v-w}^v \partial_v G_\epsilon(y, z, t) dz H_n(y, w, t) dw =: (*). \end{aligned} $$

From Proposition 4.4, Proposition 4.7, and Proposition 4.9, we have that $G_\epsilon$ has the following properties.

* $\partial_v G_\epsilon(y, z, t) \ge K_2 \partial_v G_\epsilon(y, v, t)$, for all $z \in [v-w, v]$, $w \in [0, \frac{v}{2}]$;

* $\partial_v G_\epsilon(y, v, t) \le 0$, for all $y \le (\frac{v}{3})^\alpha$.

Since also $K(v, w) \le C_\gamma K_0 v^\gamma$ by (2.2) when $w \in [0, \frac{v}{2}]$, we obtain that

$$ (*) \ge \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)G_\epsilon(y, v, t)] + C_\gamma K_0 K_2 v^\gamma \partial_v G_\epsilon(y, v, t) \int_0^{\frac{v}{2}} w H_n(y, w, t) dw. \quad (4.39) $$
Making use of Proposition 4.3, the non-positivity of $\partial_v G_\epsilon$ in (4.39), and since by assumption $H_n \le T_1 + T_2 + T_3$, we obtain that

$$ 
\begin{align*}
(*) &\ge \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y [(v^\alpha - y)G_\epsilon(y, v, t)] \\
&\quad + C_\gamma K_0 K_2 v^\gamma \partial_v G_\epsilon(y, v, t) \int_0^{\frac{v}{2}} w[T_1 + T_2 + T_3](y, w, t) dw \\
&\ge \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} \partial_y [(v^\alpha - y)G_\epsilon(y, v, t)] + \frac{C_\gamma K_0 K_1 K_2 A^3 v^\gamma}{1 + |y|^{\frac{b-2}{\alpha}}} \partial_v G_\epsilon(y, v, t) \\
&\ge \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} (v^\alpha - y) \partial_y G_\epsilon(y, v, t) + \frac{L v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) - \frac{1}{\epsilon} G_\epsilon(y, v, t),
\end{align*}
$$

where in the last line we used that $\frac{b-2}{\alpha} \ge d$. Notice that this is equation (4.2) when $v \ge 1$ and thus $G_\epsilon$ is a supersolution for (2.51). Thus

$$ 
\begin{align*}
(*) &\ge \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} (v^\alpha - y) \partial_y G_\epsilon(y, v, t) + \frac{L \xi_R(v) v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) - \frac{1}{\epsilon} G_\epsilon(y, v, t) \\
&\quad - \left| \frac{L[1 - \xi_R](v) v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) \right| \\
&\ge \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} (v^\alpha - y) \partial_y G_\epsilon(y, v, t) + \frac{L \xi_R(v) v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) - \frac{1}{\epsilon} G_\epsilon(y, v, t) \\
&\quad - \left| \frac{L \mathbb{1}_{[0,1]}(v) v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) \right|.
\end{align*}
$$

Assume that

$$ \left| \frac{L \mathbb{1}_{[0,1]}(v) v^{\gamma}}{1 + |y|^d} \partial_v G_{\epsilon}(y, v, t) \right| \le K_3(\alpha, b, m) L G_{\epsilon}(y, v, t). \quad (4.40) $$

Then

$$ 
\begin{align*}
(*) &\ge \partial_t G_\epsilon(y, v, t) + \frac{1}{\epsilon} (v^\alpha - y) \partial_y G_\epsilon(y, v, t) + \frac{L \xi_R(v) v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) - \frac{1}{\epsilon} G_\epsilon(y, v, t) \\
&\quad - K_3(\alpha, b, m) L G_\epsilon(y, v, t).
\end{align*}
$$

By (4.36), it holds that $\partial_t G_\epsilon(y, v, t) = e^{K_3(\alpha, b, m)Lt} \partial_t G_{\epsilon,L}(y, v, t) + K_3(\alpha, b, m)LG_\epsilon(y, v, t)$. Thus

$$ (*) \ge e^{K_3 Lt} \left[ \partial_t G_{\epsilon,L}(y, v, t) + \frac{1}{\epsilon} (v^\alpha - y) \partial_y G_{\epsilon,L}(y, v, t) + \frac{L \xi_R(v) v^\gamma}{1 + |y|^d} \partial_v G_{\epsilon,L}(y, v, t) - \frac{1}{\epsilon} G_{\epsilon,L}(y, v, t) \right] = 0. $$

This concludes our proof.

It remains to prove that (4.40) holds. We have proved in Proposition 4.9 that

$$ -\partial_v G_{\epsilon,L}(y, v, t) \le C(m, \alpha, b) A \left[ \frac{e^{\frac{t(m+1)}{\epsilon}} (v^\alpha - y)^{m-1} v^{\alpha-1}}{(1 + v^b)(1 + e^{\frac{tb}{\epsilon}} (v^\alpha - y)^m)^2} + \frac{e^{\frac{t}{\epsilon}} v^{b-1}}{(1 + v^b)^2 (1 + e^{\frac{tb}{\epsilon}} (v^\alpha - y)^m)} \right]. $$
Thus

$$ \left| \frac{L \mathbb{1}_{[0,1]}(v)v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) \right| \le C(\alpha, b, m) L v^\gamma \mathbb{1}_{[0,1]}(v) G_\epsilon(y, v, t) \left[ \frac{e^{\frac{im}{\epsilon}} (v^\alpha - y)^{m-1} v^{\alpha-1}}{1 + e^{\frac{im}{\epsilon}} (v^\alpha - y)^m} + \frac{v^{b-1}}{1 + v^b} \right]. $$

Since $y \le (\frac{v}{3})^\alpha$ and thus $v^\alpha - y \ge C(\alpha)v^\alpha$, we further have that

$$ \begin{aligned} \left| \frac{L \mathbb{1}_{[0,1]}(v)v^\gamma}{1 + |y|^d} \partial_v G_\epsilon(y, v, t) \right| &\le C(\alpha, b, m) L v^{\gamma-1} \mathbb{1}_{[0,1]}(v) G_\epsilon(y, v, t) \left[ \frac{e^{\frac{im}{\epsilon}} (v^\alpha - y)^m}{1 + e^{\frac{im}{\epsilon}} (v^\alpha - y)^m} + \frac{v^b}{1 + v^b} \right] \\ &\le C(\alpha, b, m) L v^{\gamma-1} \mathbb{1}_{[0,1]}(v) G_\epsilon(y, v, t) \le C(\alpha, b, m) L G_\epsilon(y, v, t), \quad (4.41) \end{aligned} $$

since $\gamma > 1$ and thus (4.40) holds if we choose $K_3$ as above. Thus, we have that $H_{n+1} \le G_\epsilon$ when $y \le (\frac{v}{3})^\alpha$.

We now prove that $G_\epsilon(y, v, t) \le T_3(y, v, t)$ for all $t \in [0, T]$ and $y \le (\frac{v}{3})^\alpha$, $v \in (0, \infty)$. Let $y \le (\frac{v}{3})^\alpha$. If

$$ t \le \frac{1}{K_3(\alpha, b, m)L}, \quad (4.42) $$

where $K_3(\alpha, b, m)$ is as in (4.41) and $L$ as in (4.37), then $e^{K_3(\alpha,b,m)Lt} \le 3$. From Proposition 4.4, we have that

$$ \begin{aligned} (1 + v^b) G_\epsilon(y, v, t) &\le \frac{C(\alpha, b, m) e^{K_3(\alpha, b, m)Lt} A e^{\frac{l}{\epsilon}}}{1 + e^{\frac{lm}{\epsilon}} |v^\alpha - y|^m} \\ &\le \frac{3C(\alpha, b, m) A e^{\frac{l}{\epsilon}}}{1 + e^{\frac{lm}{\epsilon}} |v^\alpha - y|^m} \\ &= \frac{M_2 A e^{\frac{l}{\epsilon}}}{1 + e^{\frac{lm}{\epsilon}} |v^\alpha - y|^m} = T_3(y, v, t) \end{aligned} \quad (4.43) $$

by the definition of $T_3$ is (2.13). We also remark that the constant $M_2$ in the above inequality depends only on the choice of parameters $\alpha, b, m$ since it appears due to the bounds found in Proposition 4.4 and does not depend on $n$. This concludes our proof. $\square$

# 5 Proof of Theorem 2.7 - Existence

This section is dedicated to the proof of Theorem 2.7. We remember the following.

Let $b \ge \max\{\bar{b}(\gamma, \alpha), \gamma + 1\}$ and $m > \max\{\frac{2(\gamma+1)}{\alpha}, \frac{b}{\alpha} + 1\}$, where $\bar{b}(\gamma, \alpha)$ is as in (2.18). By Proposition 3.3, there exists $\epsilon_1 \in (0, 1)$ and $t \in [0, 1]$ sufficiently small such that for all $\epsilon \le \epsilon_1$, such that $H_n(y, v, s) \le T_1(s) + T_2(s)$, for all $s \in [0, t]$, $y \ge (\frac{v}{3})^\alpha$, and $n \in \mathbb{N}$. By Proposition 4.1, there exist $\epsilon_1 \in (0, 1)$ and $T \le 1$ sufficiently small such that for all $\epsilon \le \epsilon_1$, $H_n(y, v, t) \le T_3(y, v, t)$, for all $t \in [0, T]$, $y \le (\frac{v}{3})^\alpha$, $v \in (0, \infty)$, and for all $n \in \mathbb{N}$. In other words, we have that

$$ H_n(y, v, t) \le T_1(y, v, t) + T_2(y, v, t) + T_3(y, v, t), \quad \text{for all } n \in \mathbb{N}, \quad (5.1) $$

for all $t \in [0, T]$, with $T \le 1$, sufficiently small, and all $y \in \mathbb{R}$, $v \in (0, \infty)$. We will prove that