**Claim:** Assume in addition that $b > 2\gamma + 1$. For all $n \in \mathbb{N}$

$$ \sup_{t \in [0, T], y \in \mathbb{R}, v \in (0, \infty)} |H_{n+1}(y, v, t) - H_n(y, v, t)| \le (CT)^n \to 0 \text{ as } n \to 0 \text{ for } T < \frac{1}{2C}. \quad (5.2) $$

A similar strategy can then be employed in order to prove that for all $n, \tilde{n} \in \mathbb{N}, \tilde{n} \ge n + 1$, it holds that

$$ \sup_{t \in [0, T], y \in \mathbb{R}, v \in (0, \infty)} |H_{\tilde{n}}(y, v, t) - H_n(y, v, t)| \le (CT)^n \to 0 \text{ as } n \to 0 \text{ for } T < \frac{1}{2C}. $$

Assuming that (5.2) is true, we have that for fixed $\epsilon$ there exists a limit $H_\epsilon(y, v, t) \in C([0, T]; L^\infty(\mathbb{R} \times (0, \infty))$ such that $H_n \to H_\epsilon$ as $n \to \infty$ and moreover

$$ H_{\epsilon}(y, v, t) \le T_{1}(y, v, t) + T_{2}(y, v, t) + T_{3}(y, v, t). $$

Thus, assuming (5.2) holds, what is left to prove is that $H_\epsilon$ solves (2.4). We have that $H_n$ satisfies

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

and wish to pass to the limit in the equation. Notice that

$$ H_n(e^{\frac{t-s}{\epsilon}}(y - v^{\alpha}) + v^{\alpha}, w, s) \le [T_1 + T_2 + T_3](e^{\frac{t-s}{\epsilon}}(y - v^{\alpha}) + v^{\alpha}, w, s), \quad (5.3) $$

where $T_1, T_2, T_3$ are as in (2.11)-(2.13). Denote by

$$ X_{\epsilon}(y, v, t, s) := e^{\frac{t-s}{\epsilon}}(y - v^{\alpha}) + v^{\alpha}. $$

We can then replace $y$ by $X_\epsilon$ in the proof of Proposition 4.3 in order to obtain the following.

**Proposition 5.1 (Moment estimates).** Let $b > \gamma + 1$ and $m \ge \frac{b}{\alpha} + 1$ in (2.11)-(2.13). Let $T \le 1$ and $H_n$ be as in (5.3). It then holds that

$$ \int_{(0,\infty)} w^k H_n(X_\epsilon(y, v, t, s), w, s) dw \le \frac{K_1 A^3}{1 + |X_\epsilon|^d}, \quad \text{for some } d \in \mathbb{N} \quad (5.4) $$

and all $k \in [0, \gamma + 1]$, $t \in [0, T]$, $s \in [0, t]$, $y \in \mathbb{R}$, $v \in (0, \infty)$.

**Proposition 5.2.** $H_\epsilon$ solves (2.4).

*Proof.* We prove first that

$$ D[H_n](y, v, s, t) \to D[H_\epsilon](y, v, s, t). $$

Using $|e^{-x} - e^{-y}| \le |x - y|$, for $x, y \ge 0$, we have that

$$ \left| e^{-\int_s^t a[H_n](e^{\frac{t-\tau}{\epsilon}}(y-v^\alpha)+v^\alpha, v, \tau)d\tau} - e^{-\int_s^t a[H_\epsilon](e^{\frac{t-\tau}{\epsilon}}(y-v^\alpha)+v^\alpha, v, \tau)d\tau}} \right| \\ \le \left| \int_s^t a[H_n](e^{\frac{t-\tau}{\epsilon}}(y-v^\alpha) + v^\alpha, v, \tau)d\tau - \int_s^t a[H_\epsilon](e^{\frac{t-\tau}{\epsilon}}(y-v^\alpha) + v^\alpha, v, \tau)d\tau \right|, $$
where $a[H_\epsilon]$ is as in (2.6). However

$$a[H_n](e^{\frac{t-\tau}{\epsilon}}(y - v^\alpha) + v^\alpha, v, \tau) = \int_{(0,\infty)} K(v, w)H_n(X_\epsilon(y, v, t, \tau), w, \tau)dw.$$

Let $M > 1$. Since $H_n \to H_\epsilon$ as $\epsilon \to 0$, we have that

$$\int_{(0,M)} K(v, w)H_n(X_\epsilon(y, v, t, \tau), w, \tau)dw \to \int_{(0,M)} K(v, w)H_\epsilon(X_\epsilon(y, v, t, \tau), w, \tau)dw$$

as $n \to \infty$, while by Proposition 5.1, it follows that

$$\int_{(M,\infty)} K(v, w)H_n(X_\epsilon(y, v, t, \tau), w, \tau)dw \le M^{-1} \int (v^\gamma w + w^{\gamma+1})H_n(X_\epsilon(y, v, t, \tau), w, \tau)dw \\ \le CM^{-1}(1 + v^\gamma) \to 0 \text{ as } M \to \infty.$$

The rest of the proof follows the same idea. For example, we have that

$$\begin{aligned} & \left| \int_0^t \int_{(0,v)} S_\epsilon(t-s) \left[ K(v-w,w)H_{n+1}(y,v-w,s)H_n(y,w,s) \right] ds \right. \\ & \qquad - \int_0^t \int_{(0,v)} S_\epsilon(t-s) \left[ K(v-w,w)H_\epsilon(y,v-w,s)H_\epsilon(y,w,s) \right] ds \\ & \le \int_0^t \int_{(0,v)} S_\epsilon(t-s) \left[ K(v-w,w)\left|H_{n+1} - H_\epsilon\right|(y,v-w,s)H_n(y,w,s) \right] ds \\ & \qquad \left. + \int_0^t \int_{(0,v)} S_\epsilon(t-s) \left[ K(v-w,w)H_\epsilon(y,v-w,s)\left|H_n - H_\epsilon\right|(y,w,s) \right] ds. \right\} $$

We can use the convergence of $H_n$ to $H_\epsilon$ together with the moment estimates from Proposition 5.1 in order to conclude as above. More precisely, by Proposition 5.1, we have

$$\begin{aligned} & \int_0^t \int_{(0,v)} S_\epsilon(t-s) \left[ K(v-w,w)\left|H_{n+1} - H_\epsilon\right|(y,v-w,s)H_n(y,w,s) \right] ds \\ &= \int_0^t \int_{(0,v)} e^{\frac{t-s}{\epsilon}} K(v-w,w)\left|H_{n+1} - H_\epsilon\right|(X_\epsilon(y,v,t,s),v-w,s)H_n(X_\epsilon(y,v,t,s),w,s)dwds \\ &\le C(\epsilon,v)\|H_{n+1} - H_\epsilon\|_{L^\infty} \int_0^t \int_{(0,v)} H_n(X_\epsilon(y,v,t,s),w,s)dwds \\ &\le C(\epsilon,v)\|H_{n+1} - H_\epsilon\|_{L^\infty} \to 0 \text{ as } n \to \infty. \end{aligned}$$

The rest of the steps are done in the same manner using in addition that

$$(1 + v^b)H_\epsilon(y, v, t) \le Ce^L.$$

This concludes our proof.

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

The rest of this subsection is dedicated to proving that (5.2) is true.
Proof of (5.2). **Step 1: (Set-up)** Let $t \ge 0$, $y \in \mathbb{R}$ and $v > 0$. For $n \in \mathbb{N}$, we denote by

$$R_n(y, v, t) := H_{n+1}(y, v, t) - H_n(y, v, t). \qquad (5.5)$$

For two functions $f, g$, we denote by

$$C_1[f, g] := \int_0^t \int_0^{\frac{\gamma}{2}} e^{\frac{t-s}{\epsilon}} K(v-w, w) f(e^{\frac{t-s}{\epsilon}}(y-v^\alpha) + v^\alpha, w, s) g(e^{\frac{t-s}{\epsilon}}(y-v^\alpha) + v^\alpha, v-w, s) dw ds, \qquad (5.6)$$

$$\mathcal{K}_1[f, g] := \int_0^{\frac{\gamma}{2}} K(v-w, w) f(y, w, t) g(y, v-w, t) dw, \qquad (5.7)$$

$$C_2[f, g] := \int_0^t \int_0^\infty e^{\frac{t-s}{\epsilon}} K(v, w) f(e^{\frac{t-s}{\epsilon}}(y-v^\alpha) + v^\alpha, w, s) g(e^{\frac{t-s}{\epsilon}}(y-v^\alpha) + v^\alpha, v, s) dw ds, \qquad (5.8)$$

$$\mathcal{K}_2[f, g] := \int_0^\infty K(v, w) f(y, w, t) g(y, v, t) dw, \qquad (5.9)$$

and

$$C[f, g] := C_1[f, g] - C_2[f, g], \qquad (5.10)$$

$$\mathcal{K}[f, g] := \mathcal{K}_1[f, g] - \mathcal{K}_2[f, g]. \qquad (5.11)$$

Using this notation, it holds that

$$R_n(y, v, t) = C[H_n, R_n] + C[R_{n-1}, H_n]. \qquad (5.12)$$

We notice that $C[H_n, R_n]$ is linear in $H_n$, while $C[R_{n-1}, H_n]$ does not depend on $H_n$. We can thus use Duhamel's principle in order to deduce that in order to find estimates for $R_n$ which solves

$$\left\{ \begin{array}{l} \partial_t R_n(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)R_n(y, v, t)] = \mathcal{K}[H_n, R_n] + \mathcal{K}[R_{n-1}, H_n]; \\ R_n(y, v, 0) = 0 \end{array} \right. \qquad (5.13)$$

it is sufficient to find estimates for the system

$$\left\{ \begin{array}{l} \partial_t R_n^s(x, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)R_n(y, v, t)] = \mathcal{K}[H_n, R_n], \text{ for } t > s; \\ R_n^s(y, v, s) = \mathcal{K}[R_{n-1}, H_n]. \end{array} \right. \qquad (5.14)$$

**Step 2: (Bound for inhomogeneous term)**

We wish to bound $C[R_{n-1}, H_n]$ by induction. We only bound from above $C_2[R_{n-1}, H_n]$ as $C_1[R_{n-1}, H_n]$ is bounded using similar computations.

**Induction basis:** By Proposition 5.1, (2.53)-(2.54), and since $(1 + v^b)|H_n(y, v, t)| \le Ce^{\frac{t}{\epsilon}}$, for all $n \in \mathbb{N}$, we have that

$$\begin{aligned} C_2[H_0, H_1] & \le (1 + v^\gamma) \int_0^t e^{\frac{t-s}{\epsilon}} H_1(X_\epsilon(y, v, t, s), v, s) \int_0^\infty (1 + w^\gamma)|H_0|(X_\epsilon(y, v, t, s), w, s) dw \\ & \le (1 + v^\gamma) \int_0^t e^{\frac{t-s}{\epsilon}} H_1(X_\epsilon(y, v, t, s), v, s) \frac{K_1 A^3}{1 + |X|^d} \le Cte^{\frac{t}{\epsilon}} \le Cte^{\frac{1}{\epsilon}}, \end{aligned}$$
where C does not depend on n or $\epsilon$.

**Induction step:** Assume by induction that

$$|R_{n-1}|(y, v, t) \leq (Cte^{\frac{1}{\epsilon}})^{n-1} \frac{1 + v^{\gamma}}{1 + v^{b}},$$

where C does not depend on n or on $\epsilon$. Then, since $(1+v^b)|H_n(y, v, t)| \leq Ce^{\frac{1}{\epsilon}}$ and since $b > 2\gamma+1$, it holds that

$$
\begin{aligned}
\mathcal{C}_2[R_{n-1}, H_n] &\leq (1 + v^\gamma) \int_0^t e^{\frac{t-s}{\epsilon}} H_n(X_\epsilon(y, v, t, s), v, s) \int_0^\infty (1 + w^\gamma)|R_{n-1}|(X_\epsilon(y, v, t, s), w, s) dw \\
&\leq (1 + v^\gamma) \int_0^t e^{\frac{t-s}{\epsilon}} H_n(X_\epsilon(y, v, t, s), v, s) \int_0^\infty \frac{1 + w^{2\gamma}}{1 + w^b} dw (Cse^{\frac{1}{\epsilon}})^{n-1} ds \\
&\leq \int_0^t e^{\frac{t-s}{\epsilon}} (1 + v^\gamma) H_n(X_\epsilon(y, v, t, s), v, s) (Cse^{\frac{1}{\epsilon}})^{n-1} ds \\
&\leq C^n t^n e^{\frac{n-1}{\epsilon}} e^{\frac{t}{\epsilon}} \frac{1 + v^\gamma}{1 + v^b} \leq (Cte^{\frac{1}{\epsilon}})^n \frac{1 + v^\gamma}{1 + v^b}. \quad (5.15)
\end{aligned}
$$

**Step 3: (Iteration in time in strong formulation)** By Step 2, we deduce that for every $n \in \mathbb{N}$, it holds that

$$R_n(y, v, t) \leq (Cte^{\frac{1}{\epsilon}})^n. \quad (5.16)$$

Thus, for $t < \frac{e^{-\frac{1}{\epsilon}}}{C}$, we have that

$$R_n(y, v, t) \to 0 \text{ as } n \to \infty. \quad (5.17)$$

A similar argument has been used in [9].

We now wish to extend (5.17) to hold for times of order $O(1)$. Let us assume for simplicity of notation that $C = 1$ in (5.16). The idea is to iterate the argument in Step 2. to hold for longer times. For simplicity, we first show the argument using the strong form of (2.61). In order to obtain a rigorous proof, the same idea needs to be applied directly to the mild formulation (2.61). This will be done in Step 4.

**Translation in time:** We first do a translation in time. For $t \in [0, \frac{e^{-\frac{1}{\epsilon}}}{2}]$, define

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

Notice that $\tilde{R}_n(t) := R_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right) = H_{n+1}\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right) - H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right)$ and thus

$$
\left\{
\begin{array}{l}
\partial_t \tilde{R}_n(y, v, t) + \frac{1}{\epsilon} \partial_y [(v^\alpha - y) \tilde{R}_n(y, v, t)] = \mathcal{K}[H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right), \tilde{R}_n] + \mathcal{K}[\tilde{R}_{n-1}, H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right)]; \\
\\
\tilde{R}_n(y, v, 0) = H_{n+1}\left(\frac{e^{-\frac{1}{\epsilon}}}{2}\right) - H_n\left(\frac{e^{-\frac{1}{\epsilon}}}{2}\right).
\end{array}
\right.
\quad (5.18)
$$
Initial condition becomes zero: Let now

$$P_n(y, v, t) := \tilde{R}_n(y, v, t) - e^{\frac{t}{\epsilon}} R_n \left( e^{\frac{t}{\epsilon}} (y - v^{\alpha}) + v^{\alpha}, v, \frac{e^{-\frac{1}{\epsilon}}}{2} \right);$$

$$\tilde{P}_n(y, v, t) := e^{\frac{t}{\epsilon}} R_n \left( e^{\frac{t}{\epsilon}} (y - v^{\alpha}) + v^{\alpha}, v, \frac{e^{-\frac{1}{\epsilon}}}{2} \right).$$

Then $P_n$ satisfies

$$\left\{
\begin{aligned}
\partial_t P_n(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)P_n(y, v, t)] &= \mathcal{K}\left[H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right), P_n\right] + \mathcal{K}\left[P_{n-1}, H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right)\right] \\
&\quad + \mathcal{K}\left[H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right), \tilde{P}_n\right] + \mathcal{K}\left[\tilde{P}_{n-1}, H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right)\right] \\
P(y, v, 0) &= 0.
\end{aligned}
\right.
\quad (5.19)$$

By Duhamel's principle, this becomes equivalent to solving

$$\left\{
\begin{aligned}
\partial_t P_n(y, v, t) + \frac{1}{\epsilon} \partial_y[(v^\alpha - y)P_n(y, v, t)] &= \mathcal{K}\left[H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right), P_n\right] \quad \text{for } t > s \\
P_n(y, v, s) &= \mathcal{K}\left[P_{n-1}, H_n\left(s + \frac{e^{-\frac{1}{\epsilon}}}{2}\right)\right] + \mathcal{K}\left[H_n\left(s + \frac{e^{-\frac{1}{\epsilon}}}{2}\right), \tilde{P}_n\right] \quad (5.20) \\
&\quad + \mathcal{K}\left[\tilde{P}_{n-1}, H_n\left(s + \frac{e^{-\frac{1}{\epsilon}}}{2}\right)\right] =: I_1 + I_2 + I_3.
\end{aligned}
\right.
$$

We are now able to reproduce the estimates we performed in Step 2 in order to obtain that there exists a constant $C > 0$ such that

$$P_n(t) \le (Cte^{\frac{1}{\epsilon}})^n + 2^{-n} \quad \text{for all } n \in \mathbb{N}. \quad (5.21)$$

Estimates for the iteration in time: The term $I_1$ we bound by induction as in Step 2, (5.15), using in addition that

$$\begin{aligned}
P_n(t) &= R_n\left(y, v, t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right) - e^{\frac{t}{\epsilon}} R_n\left(\frac{e^{-\frac{1}{\epsilon}}}{2}\right) \le R_n\left(y, v, t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right) \\
&\le H_{n+1}\left(y, v, t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right) + H_n\left(y, v, t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right).
\end{aligned}$$

$I_2$ and $I_3$ can then bounded in a similar manner. We thus only show $I_2$. More precisely, we have that

$$C_2 \left[ H_n \left( t + \frac{e^{-\frac{1}{\epsilon}}}{2} \right), \tilde{P}_n \right] \\
:= \int_0^t \int_0^\infty e^{\frac{t-s}{\epsilon}} K(v, w) H_n(X_\epsilon(y, v, t, s), w, s + \frac{e^{-\frac{1}{\epsilon}}}{2}) e^{\frac{s}{\epsilon}} R_n \left( X_\epsilon(y, v, t, 0), v, \frac{e^{-\frac{1}{\epsilon}}}{2} \right) dw ds.$$
By the upper bound in Theorem 2.7, we have that $(1 + v^b)|H_n(y, v, s)| \le Ce^{\frac{s}{\epsilon}}$, for all $t \le 1$. Moreover, by Step 2 it holds that $R_n(t) \le (te^{\frac{1}{\epsilon}})^n \frac{1+v^\gamma}{1+v^b}$. Thus, we have that $R_n(X_\epsilon(y, v, t, 0), v, \frac{e^{-\frac{1}{\epsilon}}}{2}) \le \frac{1}{2^n} \frac{1+v^\gamma}{1+v^b}$. This implies that as long as $2t \le 1$, it holds that

$$C_2[H_n(t + \frac{e^{-\frac{1}{\epsilon}}}{2}), \tilde{P}_n] \le \frac{1}{2^n} \int_0^t \int_0^\infty e^{\frac{t+s+\frac{e^{-\frac{1}{\epsilon}}}{2}}{\epsilon}} ds \le e^{\frac{1}{2}} \frac{te^{\frac{1}{\epsilon}}}{2^n} \le \frac{\sqrt{e}}{2^{n+1}}.$$

We bound $I_3$ similarly and conclude that (5.21) holds. Since

$$R_n(y, v, t + \frac{e^{-\frac{1}{\epsilon}}}{2}) = P_n(t) + e^{\frac{t}{\epsilon}} R_n(X_{\epsilon}(y, v, t, 0), v, \frac{e^{-\frac{1}{\epsilon}}}{2})$$

we then obtain that

$$R_n(y, v, t + \frac{e^{-\frac{1}{\epsilon}}}{2}) \le e^{\frac{t}{\epsilon}}(Ct)^n + 2^{-n+1} \to 0 \text{ as } n \to \infty, \text{ for all } t \le \frac{e^{-\frac{1}{\epsilon}}}{2}.$$

We can then iterate Step 3 to extend the result to hold for all times as long as $t \le \min\{2^{-1}, T\}$ with $T$ as in Theorem 2.7.

**Step 4: (Iteration in time in mild formulation)** As mentioned in Step 3, we now show the translation in time argument directly for the mild formulation (2.61).

*Mild formulation. Translation in time* Since $R_n(y, v, 0) = 0$, $R_n$ satisfies the following mild formulation

$$R_n(y, v, t) = C[H_n, R_n] + C[R_{n-1}, H_n].$$

Define now

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

Notice as before that $\tilde{R}_n(t) := R_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right) = H_{n+1}\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right) - H_n\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right)$. We show the argument for $C_2[H_n, R_n]$ in (5.9) as the other terms are obtained in the same manner.

$$\begin{align*}
C_2[H_n, R_n]\left(t + \frac{e^{-\frac{1}{\epsilon}}}{2}\right) \\
&= \int_0^{t+\frac{e^{-\frac{1}{\epsilon}}}{2}} \int_0^{\infty} e^{\frac{t+\frac{e^{-\frac{1}{\epsilon}}}{2}-s}{\epsilon}} K(v, w) H_n\left(e^{\frac{t+\frac{e^{-\frac{1}{\epsilon}}}{2}-s}{\epsilon}} (y-v^\alpha) + v^\alpha, v, s\right) R_n\left(e^{\frac{t+\frac{e^{-\frac{1}{\epsilon}}}{2}-s}{\epsilon}} (y-v^\alpha) + v^\alpha, v, s\right) dw ds \\
&= \int_0^{\frac{e^{-\frac{1}{\epsilon}}}{2}} \int_0^{\infty} e^{\frac{t+\frac{e^{-\frac{1}{\epsilon}}}{2}-s}{\epsilon}} K(v, w) H_n\left(e^{\frac{t+\frac{e^{-\frac{1}{\epsilon}}}{2}-s}{\epsilon}} (y-v^\alpha) + v^\alpha, v, s\right) R_n\left(e^{\frac{t+\frac{e^{-\frac{1}{\epsilon}}}{2}-s}{\epsilon}} (y-v^\alpha) + v^\alpha, v, s\right) dw ds \\
&\quad + \int_0^{t+\frac{e^{-\frac{1}{\epsilon}}}{2}} \int_0^{\infty} e^{\frac{t+\frac{e^{-\frac{1}{\epsilon}}}{2}-s}{\epsilon}} K(v, w) H_n\left(e^{\frac{t+\frac{e^{-\frac{1}{\epsilon}}}{2}-s}{\epsilon}} (y-v^\alpha) + v^\alpha, v, s\right) R_n\left(e^{\frac{t+\frac{e^{-\frac{1}{\epsilon}}}{2}-s}{\epsilon}} (y-v^\alpha) + v^\alpha, v, s\right) dw ds \\
&:= L_1 + L_2.
\end{align*}$$
For $L_2$ we make the following change of variable $\tilde{s} = s - \frac{e^{-\frac{1}{\epsilon}}}{2}$. It thus holds that

$$L_2 = \int_0^t \int_0^\infty e^{\frac{t-\bar{s}}{\epsilon}} K(v, w) H_n \left( e^{\frac{t-\bar{s}}{\epsilon}} (y - v^\alpha) + v^\alpha, w, \bar{s} + \frac{e^{-\frac{1}{\epsilon}}}{2} \right) R_n \left( e^{\frac{t-\bar{s}}{\epsilon}} (y - v^\alpha) + v^\alpha, v, \bar{s} + \frac{e^{-\frac{1}{\epsilon}}}{2} \right) dw d\bar{s}.$$

Thus, $\tilde{R}_n$ satisfies the following equation.

$$\tilde{R}_n(y, v, t) = e^{\frac{t}{\epsilon}} R_n \left( e^{\frac{t}{\epsilon}} (y - v^{\alpha}) + v^{\alpha}, v, \frac{e^{-\frac{1}{\epsilon}}}{2} \right) + \mathcal{C} \left[ H_n \left( \cdot + \frac{e^{-\frac{1}{\epsilon}}}{2} \right), \tilde{R}_n \right] + \mathcal{C} \left[ \tilde{R}_{n-1}, H_n \left( \cdot + \frac{e^{-\frac{1}{\epsilon}}}{2} \right) \right].$$

*Mild formulation. Initial condition becomes 0.* Lastly, let

$$P_n(y, v, t) := \tilde{R}_n(y, v, t) - e^{\frac{t}{\epsilon}} R_n \left( e^{\frac{t}{\epsilon}} (y - v^{\alpha}) + v^{\alpha}, v, \frac{e^{-\frac{1}{\epsilon}}}{2} \right)$$

$$\tilde{P}_n(y, v, t) := e^{\frac{t}{\epsilon}} R_n \left( e^{\frac{t}{\epsilon}} (y - v^{\alpha}) + v^{\alpha}, v, \frac{e^{-\frac{1}{\epsilon}}}{2} \right).$$

Then $P_n$ satisfies

$$\tilde{P}_n(y, v, t) = \mathcal{C} \left[ H_n \left( \cdot + \frac{e^{-\frac{1}{\epsilon}}}{2} \right), \tilde{R}_n \right] + \mathcal{C} \left[ \tilde{R}_{n-1}, H_n \left( \cdot + \frac{e^{-\frac{1}{\epsilon}}}{2} \right) \right].$$

In other words, $P_n$ satisfies

$$\left\{
\begin{array}{l}
P_n(y, v, t) = \mathcal{C} \left[ H_n \left( \cdot + \frac{e^{-\frac{1}{\epsilon}}}{2} \right), \tilde{R}_n \right] + \mathcal{C} \left[ \tilde{R}_{n-1}, H_n \left( \cdot + \frac{e^{-\frac{1}{\epsilon}}}{2} \right) \right] \\
\qquad + \mathcal{C} \left[ H_n \left( \cdot + \frac{e^{-\frac{1}{\epsilon}}}{2} \right), \tilde{P}_n \right] + \mathcal{C} \left[ \tilde{P}_{n-1}, H_n \left( \cdot + \frac{e^{-\frac{1}{\epsilon}}}{2} \right) \right] \\
P(y, v, 0) = 0.
\end{array}
\right.
\qquad (5.22)$$

We are now able to reproduce the estimates from Step 3. This concludes our proof. □

# 6 Proof of Theorem 2.12 - Convergence to a Dirac measure

**Proposition 6.1** (Moment estimates - general version). Let $b > \gamma + 3$ and $m > \gamma + 3 - \alpha$. Let $T \le 1$. There exist $r > 0$ and $\omega := \min\{m - \gamma - 1 + \alpha, b - \gamma - 1\} > 2$ such that

$$\int_{(0,\infty)} (w^{-r} + w^{\gamma}) H_{\epsilon}(y, w, t) dw \le \frac{C}{1 + |y|^{\omega}}, \quad \text{for all } t \in [0, T], y \in \mathbb{R}. \qquad (6.1)$$

and thus

$$\int_{\mathbb{R}} \int_{(0,\infty)} (1 + |y|) H_{\epsilon}(y, w, t) dw dy \le C, \quad \text{for all } t \in [0, T]. \qquad (6.2)$$
Proof. Since $\alpha < 1$, there exists $r \in (0, 1)$ such that $\alpha + 2r \le 1$.

(6.1) is a more general version of the proof of Proposition 4.3. We only have to prove that $\int_{(0,\infty)} w^{-r}H_{\epsilon}(y, w, t)dw \le \frac{C}{1+|y|^\alpha}$. Let $C_0$ be as in Proposition 3.3. 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

$$ \int_{\{y \ge w^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}} w^{-r} P(y, w, t) dw \le \int_{(0, \infty)} \frac{C w^{-r} e^{\frac{t}{\epsilon}}}{(1 + w^b)(1 + e^{\frac{t}{\epsilon}} |y - w^{\alpha}|^m)} dw. $$

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

$$ \int_{\{y \ge w^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}} w^{-r} P(y, w, t) dw \le \int_{(-\infty, e^{\frac{t}{\epsilon}} y)} \frac{C(y - e^{-\frac{t}{\epsilon}} z)^{\frac{1-\alpha-r}{\alpha}}}{(1 + (y - e^{-\frac{t}{\epsilon}} z)^{\frac{b}{\alpha}})(1 + |z|^m)} dz. $$

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

$$ \int_{|z| \le \frac{e^{\frac{t}{\epsilon}} y}{2}} \frac{C(y - e^{-\frac{t}{\epsilon}} z)^{\frac{1-\alpha-r}{\alpha}}}{(1 + (y - e^{-\frac{t}{\epsilon}} z)^{\frac{b}{\alpha}})(1 + |z|^m)} dz \le \frac{Cy^{\frac{1-\alpha-r}{\alpha}}}{1 + |y|^{\frac{b}{\alpha}}} \int_{(-\infty, \infty)} \frac{C}{1 + |z|^m} \le \frac{C}{1 + |y|^{\frac{b+r-1+\alpha}{\alpha}}}. \quad (6.3) $$

On the other hand, since $1 - \alpha - r \ge 0$, it holds that $\frac{(y - e^{-\frac{t}{\epsilon}} z)^{\frac{1-\alpha-r}{\alpha}}}{(1 + (y - e^{-\frac{t}{\epsilon}} z)^{\frac{b}{\alpha}})} \le C$. Thus

$$ \int_{|z| \ge \frac{e^{\frac{t}{\epsilon}} y}{2}} \frac{C(y - e^{-\frac{t}{\epsilon}} z)^{\frac{1-\alpha-r}{\alpha}}}{(1 + (y - e^{-\frac{t}{\epsilon}} z)^{\frac{b}{\alpha}})(1 + |z|^m)} dz \le \int_{|z| \ge \frac{e^{\frac{t}{\epsilon}} y}{2}} \frac{C}{1 + |z|^m} dz \le \frac{C}{1 + (e^{\frac{t}{\epsilon}} y)^{m-1}} \le \frac{C}{1 + y^{m-1}}. \quad (6.4) $$

Combining (6.3) and (6.4), we obtain that (6.1) holds in the region $\{y \ge w^{\alpha} - C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{t}{\epsilon}}\}$.

* We now analyze 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 $y^{\frac{1}{\alpha}} \le (y + C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{t}{\epsilon}})^{\frac{1}{\alpha}} \le w \le 3y^{\frac{1}{\alpha}}$. Thus

$$ \begin{aligned} \int_{\{(\frac{w}{3})^{\alpha} \le y \le w^{\alpha} - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}} w^{-r} P(y, w, t) dw &\le \int_{(y+C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{t}{\epsilon}})^{\frac{1}{\alpha}}}^{3y^{\frac{1}{\alpha}}} \frac{Cw^{-r}\epsilon}{(1+w^b)|y-w^{\alpha}|} dw \\ &\le \frac{C\epsilon y^{-\frac{r}{\alpha}}}{1+y^{\frac{b}{\alpha}}} \int_{(y+C_0^{-1}\epsilon^{-\frac{1}{m-1}}e^{-\frac{t}{\epsilon}})^{\frac{1}{\alpha}}}^{3y^{\frac{1}{\alpha}}} \frac{1}{|y-w^{\alpha}|} dw. \end{aligned} $$

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

$$ \begin{aligned} \frac{C \epsilon y^{-\frac{r}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} \int_{(y+C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}})^{\frac{1}{\alpha}}}^{3y^{\frac{1}{\alpha}}} \frac{1}{|y-w^{\alpha}|} dw &\le \frac{C \epsilon y^{\frac{1-\alpha-r}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} \int_{-(3^{\alpha}-1)y}^{-C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}} \frac{dz}{|z|} \\ &\le \frac{C \epsilon y^{\frac{1-\alpha-r}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} [\frac{t}{\epsilon} + |\ln y|]. \end{aligned} $$
If $y \ge 1$, then

$$ \frac{C \epsilon y^{\frac{1-\alpha-r}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} \left[ \frac{t}{\epsilon} + \ln y \right] \le \frac{C t y^{\frac{1-\alpha-r}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} + \frac{C \epsilon y^{\frac{1-r}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} y^{-1} \ln y \le \frac{C}{1 + y^{\frac{b+r-1}{\alpha}}}, $$

where in the last inequality we used the fact that $t \le 1$. If $y \le 1$, we use the fact that $1-2r-\alpha \ge 0$, together with the fact that $y^r \ln \frac{1}{y} \le C$, in order to conclude in the same manner that

$$ \frac{C \epsilon y^{\frac{1-\alpha-r}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} \left[ \frac{t}{\epsilon} - \ln y \right] \le \frac{C t y^{\frac{1-\alpha-r}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} + \frac{C \epsilon y^{\frac{1-\alpha-2r}{\alpha}}}{1 + y^{\frac{b}{\alpha}}} y^r \ln \frac{1}{y} \le \frac{C}{1 + y^{\frac{b+2r-1}{\alpha}}}. $$

The region $\{y \le (\frac{w}{3})^\alpha\}$ follows as in the proof of Proposition 4.3 using in addition the fact that $1 - \alpha - r \ge 0$ as in (6.4). This concludes our proof. $\square$

We notice that $H_\epsilon$ found in Theorem 2.7 satisfies the weak version of (2.4).

**Weak formulation:** From (2.4), it immediately follows that for every $t \in [0, T]$ and for every $\varphi \in C_c^1(\mathbb{R} \times (0, \infty))$, it holds that

$$ \partial_t \int_{\mathbb{R}} \int_{(0,\infty)} H_{\epsilon}(y, v, t)\varphi(y, v)dvdy = \frac{1}{\epsilon} \int_{\mathbb{R}} \int_{(0,\infty)} H_{\epsilon}(y, v, t)(v^{\alpha} - y)\partial_y\varphi(y, v)dvdy \\ + \frac{1}{2} \int_{\mathbb{R}} \int_{(0,\infty)} \int_{(0,\infty)} K(v, w)H_{\epsilon}(y, v, t)H_{\epsilon}(y, w, t)\chi_{\varphi}(y, v, w)dwdvdy, \quad (6.5) $$

where we denoted

$$ \chi_{\varphi}(y, v, w) := \varphi(y, v + w) - \varphi(y, v) - \varphi(y, w). $$

We are now in a position to prove Theorem 2.12.

**Proof of Theorem 2.12.** We first prove that for every fixed $t > 0$, $H_\epsilon$ converges to a Dirac measure.

**Convergence to a Dirac measure for fixed time:** If $|y - v^\alpha| \ge \delta$ and $t > 0$, then by (2.11)-(2.13), it holds that

$$ T_1(y, v, t) = \frac{2A}{1 + v^b} e^{\frac{t}{\epsilon}} \psi(e^{\frac{t}{\epsilon}}(y - v^\alpha)) \mathbb{1}_{\{|y \ge v^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}|} \le \frac{Ce^{\frac{t}{\epsilon}}}{1 + e^{\frac{t}{\epsilon}} |y - v^\alpha|} \le C(\delta)e^{-\frac{t(m-1)}{\epsilon}}; $$

$$ T_2(y, v, t) = \frac{2C_0^{m-1} \max\{1, A^2\}\epsilon}{(1 + v^b)|y - v^\alpha|} \mathbb{1}_{\{(\frac{v}{3})^\alpha \le y \le v^\alpha - C_0^{-1} \epsilon^{-\frac{1}{m-1}} e^{-\frac{t}{\epsilon}}\}} \le C(\delta)\epsilon; $$

$$ T_3(y, v, t) = \frac{M_2 A}{1 + v^b} e^{\frac{t}{\epsilon}} \psi(e^{\frac{t}{\epsilon}}(y - v^\alpha)) \mathbb{1}_{\{y \le (\frac{v}{3})^\alpha\}} \le C(\delta) e^{-\frac{t(m-1)}{\epsilon}}. $$

Thus, if $t > 0$, $\text{supp}\varphi \subseteq \{|y - v^\alpha| \ge \delta\}$, and $\varphi$ is compactly supported, it holds that

$$ \int_{\mathbb{R}} \int_{(0,\infty)} H_{\epsilon}(y, v, t)\varphi(y, v)dvdy \le C(\delta)[\epsilon + e^{-\frac{t(m-1)}{\epsilon}}] \to 0 \text{ as } \epsilon \to 0. \quad (6.6) $$
We continue by proving equicontinuity in time.

**Equicontinuity:** Let $\varphi \in C_0(\mathbb{R} \times (0, \infty))$ and $\sigma > 0$. We want to prove that for any $\delta \in (0, 1)$, there exists $\delta_1$ such that if $t, s \in [\sigma, T]$ with $|t-s| \le \delta_1$, then

$$ \left| \int_{\mathbb{R}} \int_{(0,\infty)} [H_{\epsilon}(y, v, t) - H_{\epsilon}(y, v, s)] \varphi(y, v) dv dy \right| \le \delta. \quad (6.7) $$

*Estimates for the transport term:* We first prove some estimates for the transport term in (6.5). Let $\varphi \in C_c^1(\mathbb{R} \times (0, \infty))$ and $t \ge \sigma$. It holds that

$$ \int_{\mathbb{R}} \int_{(0,\infty)} H_{\epsilon}(y, v, t) |y - v^{\alpha}| \partial_y \varphi(y, v) dv dy \le C \int_{\mathbb{R}} \int_{(0,\infty)} [T_1 + T_2 + T_3](y, v, t) |y - v^{\alpha}| dv dy. \quad (6.8) $$

We estimate first $T_1 + T_3$. We have that

$$ \begin{aligned} \int_{\mathbb{R}} \int_{(0,\infty)} [T_1 + T_3](y, v, t) |y - v^{\alpha}| dv dy &\le \int_{\mathbb{R}} \int_{(0,\infty)} \frac{C |y - v^{\alpha}|}{1 + v^b} e^{\frac{t}{\epsilon}} \psi(e^{\frac{t}{\epsilon}} (y - v^{\alpha})) dv dy \\ &\le \int_{\mathbb{R}} \int_{(0,\infty)} \frac{C e^{\frac{t}{\epsilon}} |y - v^{\alpha}|}{1 + v^b} \frac{1}{1 + e^{\frac{t m}{\epsilon}} |y - v^{\alpha}|^m} dv dy \\ &\le \int_{\mathbb{R}} \int_{(0,\infty)} \frac{C}{1 + v^b} \frac{1}{1 + e^{\frac{t(m-1)}{\epsilon}} |y - v^{\alpha}|^{m-1}} dv dy. \end{aligned} \quad (6.9) $$

We make the change of variables $\xi = e^{\frac{t}{\epsilon}}(y - v^{\alpha})$. (6.9) then becomes for all $t \ge \sigma$

$$ \begin{aligned} \int_{\mathbb{R}} \int_{(0,\infty)} [T_1 + T_3](y, v, t) |y - v^{\alpha}| dv dy &\le \int_{\mathbb{R}} \int_{(0,\infty)} \frac{C}{1 + v^b} \frac{e^{-\frac{t}{\epsilon}}}{1 + |\xi|^{m-1}} dv d\xi \\ &\le e^{-\frac{\sigma}{\epsilon}} \int_{\mathbb{R}} \int_{(0,\infty)} \frac{C}{1 + v^b} \frac{1}{1 + |\xi|^{m-1}} dv d\xi \le C e^{-\frac{\sigma}{\epsilon}}. \end{aligned} $$

We continue by estimating $T_2$. In this case, it follows that

$$ \begin{aligned} \int_{\mathbb{R}} \int_{(0,\infty)} T_2(y, v, t) |y - v^{\alpha}| dv dy &\le \int_{\mathbb{R}} \int_{(0,\infty)} \frac{C\epsilon|y - v^{\alpha}|}{(1 + v^b)|y - v^{\alpha}|} \mathbb{1}_{\{|\frac{t}{\epsilon}|^\alpha \le y \le v^{\alpha}\}} dv dy \\ &\le \int_{(0,\infty)} \frac{C\epsilon}{1 + v^{b-\alpha}} dv \le C\epsilon. \end{aligned} $$

Combining all the above estimates, we have that

$$ \int_{\mathbb{R}} \int_{(0,\infty)} H_{\epsilon}(y, v, t) |y - v^{\alpha}| \partial_y \varphi(y, v) dv dy \le C e^{-\frac{\sigma}{\epsilon}} + C\epsilon. \quad (6.10) $$

*Equicontinuity, test functions in $C_c^1(\mathbb{R} \times (0, \infty))$:* Let $M > 1$. We first prove (6.7) for a fixed function $\varphi \in C_c^1(\mathbb{R} \times (0, \infty))$ such that $\text{supp}\varphi \in \{[-M, M] \times [\frac{1}{M}, M]\}, \|\varphi\|_{L^\infty} \le 1$. Let $s, t \in [\sigma, T]$.
By (6.5), it holds that

$$
\begin{aligned}
& \left| \int_{\mathbb{R}} \int_{(0,\infty)} [H_{\epsilon}(y, v, t) - H_{\epsilon}(y, v, s)] \varphi(y, v) dv dy \right| \\
& \leq \frac{1}{\epsilon} \int_{s}^{t} \int_{\mathbb{R}} \int_{(0,\infty)} H_{\epsilon}(y, v, z) |v^{\alpha} - y| \|\partial_{y}\varphi(y, v)\| dv dy dz \\
& \qquad + \frac{1}{2} \int_{s}^{t} \int_{\mathbb{R}} \int_{(0,\infty)} \int_{(0,\infty)} K(v, w) H_{\epsilon}(y, v, z) H_{\epsilon}(y, w, z) |\chi_{\varphi}(y, v, w)| dw dv dy dz \\
& \leq \frac{1}{\epsilon} \int_{s}^{t} \int_{\mathbb{R}} \int_{(0,\infty)} H_{\epsilon}(y, v, z) |v^{\alpha} - y| \|\partial_{y}\varphi(y, v)\| dv dy dz \\
& \qquad + C \int_{s}^{t} \int_{\mathbb{R}} \int_{(0,\infty)} (1 + v^{\gamma}) H_{\epsilon}(y, v, z) dv \int_{(0,\infty)} (1 + w^{\gamma}) H_{\epsilon}(y, w, z) dw dy dz \\
& =: J_1 + J_2.
\end{aligned}
$$

By (6.10) and since $t, s \ge \sigma$, it follows that

$$ J_1 \le \frac{C}{\epsilon} [e^{-\frac{\sigma}{\epsilon}} + \epsilon] |t-s| \le C|t-s|, \quad (6.11) $$

for all $\epsilon \le \epsilon_\sigma$, where $\epsilon_\sigma$ only depends on $\sigma$. By (6.1) it follows that

$$ J_2 \le \int_s^t \int_{\mathbb{R}} \frac{C}{1 + |y|^{2\omega}} dy dz \le C|t-s|. \quad (6.12) $$

By (6.11) and (6.12), we obtain that

$$ \left| \int_{\mathbb{R}} \int_{(0,\infty)} [H_{\epsilon}(y, v, t) - H_{\epsilon}(y, v, s)] \varphi(y, v) dv dy \right| \le C|t-s|, \quad (6.13) $$

when $\varphi \in C_c^1(\mathbb{R} \times (0, \infty))$.

From $C_c^1(\mathbb{R} \times (0, \infty))$ to $C_c(\mathbb{R} \times (0, \infty))$: Let $\varphi \in C_c(\mathbb{R} \times (0, \infty))$ such that $\text{supp}\varphi \in [-M, M] \times [\frac{1}{M}, M]$. There exists $\varphi_n \in C_c^1(\mathbb{R} \times (0, \infty))$ such that $\|\varphi - \varphi_n\|_{\infty} \le \frac{1}{n}$. It then holds that

$$
\left| \int_{\mathbb{R}} \int_{(0,\infty)} [H_{\epsilon}(y, v, t) - H_{\epsilon}(y, v, s)] \varphi(y, v) dv dy \right| \le \left| \int_{\mathbb{R}} \int_{(0,\infty)} [H_{\epsilon}(y, v, t) - H_{\epsilon}(y, v, s)] \varphi_n(y, v) dv dy \right| \\
+ \int_{\mathbb{R}} \int_{(0,\infty)} [H_{\epsilon}(y, v, t) + H_{\epsilon}(y, v, s)] |\varphi(y, v) - \varphi_n(y, v)| dv dy.
$$

By (6.13), it follows that

$$
\begin{aligned}
& \left| \int_{\mathbb{R}} \int_{(0,\infty)} [H_{\epsilon}(y, v, t) - H_{\epsilon}(y, v, s)] \varphi(y, v) dv dy \right| \\
& \le C|t-s| + \frac{1}{n} \int_{\mathbb{R}} \int_{(0,\infty)} [H_{\epsilon}(y, v, t) + H_{\epsilon}(y, v, s)] dv dy \le C|t-s| + \frac{C}{n}. \quad (6.14)
\end{aligned}
$$

We can then conclude by choosing $n \in \mathbb{N}$ sufficiently large.
From $C_c(\mathbb{R} \times (0, \infty))$ to $C_0(\mathbb{R} \times (0, \infty))$: Let $\varphi \in C_0(\mathbb{R} \times (0, \infty))$. Let $M > 1$, then

$$ \begin{aligned} & \left| \int_{\mathbb{R}} \int_{(0,\infty)} [H_{\epsilon}(y, v, t) - H_{\epsilon}(y, v, s)] \varphi(y, v) dv dy \right| \\ & \le \left| \int_{[-M,M]} \int_{[\frac{1}{M},M]} [H_{\epsilon}(y, v, t) - H_{\epsilon}(y, v, s)] \varphi(y, v) dv dy \right| \\ & \quad + \int_{\mathbb{R}} \int_{\{v \ge M\} \cup \{v \le \frac{1}{M}\}} [H_{\epsilon}(y, v, t) + H_{\epsilon}(y, v, s)] |\varphi(y, v) - \varphi_n(y, v)| dv dy \\ & \quad + \int_{|y| \ge M} \int_{(M,\infty)} [H_{\epsilon}(y, v, t) + H_{\epsilon}(y, v, s)] |\varphi(y, v) - \varphi_n(y, v)| dv dy =: L_1 + L_2 + L_3. \end{aligned} $$

We have that the following inequalities hold by choosing $M$ sufficiently large.

$$ L_1 \le \frac{\delta}{3} \quad \text{by (6.14)}; $$

$$ \begin{aligned} L_2 \le M^{-1} & \int_{\mathbb{R}} \int_{\{v \ge M\}} v[H_{\epsilon}(y, v, t) + H_{\epsilon}(y, v, s)]dv dy \\ & + M^{-r} \int_{\mathbb{R}} \int_{\{v \le \frac{1}{M}\}} v^{-r}[H_{\epsilon}(y, v, t) + H_{\epsilon}(y, v, s)]dv dy \le \frac{\delta}{3} \quad \text{by (6.1)} \end{aligned} $$

$$ L_3 \le M^{-1} \int_{|y| \ge M} \int_{(M, \infty)} |y|[H_{\epsilon}(y, v, t) + H_{\epsilon}(y, v, s)]dv dy \le \frac{\delta}{3} \quad \text{by (6.2).} $$

This concludes the proof of equicontinuity. Similar computations for the passage from $\varphi \in C_c^1(\mathbb{R} \times (0, \infty))$ to $\varphi \in C_0(\mathbb{R} \times (0, \infty))$ have also been used in [10].

**Convergence to a Dirac measure:** By (6.2), we have that for $t \in [0, T]$, there exists a convergent subsequence of $H_\epsilon(t)$ in the weak-$^*$ topology as $\epsilon \to 0$. Let $\sigma > 0$. Together with (6.7), we deduce from Arzelà-Ascoli theorem that there exists a subsequence of $\{H_\epsilon\}$, which we do not relabel, and an $H \in C([\sigma, T]; \mathcal{M}_+(\mathbb{R} \times (0, \infty)))$, such that $H_\epsilon(t)$ converge to $H(t)$ in the weak-$^*$ topology as $\epsilon \to 0$, for every $t \in [\sigma, T]$.

By (6.6), we have that for every $t > 0$ and $\varphi$ such that $\text{supp}\varphi \subseteq \{|y - v^\alpha| \ge \delta\}$, $\varphi$ is compactly supported, it holds that

$$ \int_{\mathbb{R}} \int_{(0,\infty)} H_{\epsilon}(y, v, t) \varphi(y, v) dv dy \to 0 \quad \text{as} \quad \epsilon \to 0. \quad (6.15) $$

The extension to functions $\varphi \in C_0(\mathbb{R} \times (0, \infty))$ is as before. In other words, for any $\delta > 0$ and for every $\varphi \in C_0(\mathbb{R} \times (0, \infty))$ such that $\text{supp}\varphi \subseteq \{|y - v^\alpha| \ge \delta\}$, we have that

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

This concludes our proof. $\square$
A Rescaled version of the results

Existence of mass-conserving solutions has been proved in [9] for the rain model (1.4) for times of order one. Theorem 2.7 extends the existence of mass-conserving solutions in [9] to hold for times of order $O(e^{\frac{1}{\epsilon}})$ in the case of coagulation kernels with negligible contribution. More precisely, making use of (1.14), (1.15), Table 1, and Theorem 2.7, we obtain that

**Proposition A.1.** 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 $M_1(\alpha, \gamma, b, m)$, $M_2(\alpha, \gamma, b, m)$, 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 $f_{\epsilon} \in C([0, e^{\frac{T}{\epsilon}} - 1]; L^{\infty}(\mathbb{R} \times (0, \infty)))$ to (1.4). Moreover, we have that $f_{\epsilon}$ satisfies the following estimates for $C_0$ as in (2.9).

If $v^{\alpha}(\tau + 1)[1 - (\frac{1}{3})^{\alpha}] \ge C_0^{-1} \epsilon^{-\frac{1}{m-1}}$, then

$$0 \le (1 + v^b) f_{\epsilon}(y, v, \tau) \le \frac{2A}{1 + |x - (\tau + 1)v^{\alpha}|^m} \mathbb{1}_{A_{\epsilon,y,v,\tau}} + \frac{2M_1 A^3 \epsilon}{|x - (\tau + 1)v^{\alpha}|} \mathbb{1}_{(A_{\epsilon,y,v,\tau} \cup B_{\epsilon,y,v,\tau})^c} \quad (A.1) \\ + \frac{M_2 A}{1 + |x - (\tau + 1)v^{\alpha}|^m} \mathbb{1}_{B_{\epsilon,y,v,\tau}},$$

where

$$A_{\epsilon,y,v,\tau} := \{x - (\tau + 1)v^{\alpha} \ge -C_0^{-1} \epsilon^{-\frac{1}{m-1}}\};$$

$$B_{\epsilon,y,v,\tau} := \left\{ \frac{x}{\tau + 1} \le \left(\frac{v}{3}\right)^{\alpha} \right\}.$$

Otherwise, if $0 \le v^{\alpha}(\tau + 1)[1 - (\frac{1}{3})^{\alpha}] \le C_0^{-1} \epsilon^{-\frac{1}{m-1}}$, then

$$0 \le (1 + v^b) f_{\epsilon}(x, v, \tau) \le \frac{2A}{1 + |x - (\tau + 1)v^{\alpha}|^m}. \quad (A.2)$$

With the form of the bounds in (A.1), the reason behind the choice of (2.19) becomes clear. More precisely,

**Remark A.2.** Denote by $r(x, v, \tau) := \frac{1}{1+x-\tau v^{\alpha}|m+v^p}$ and by $p = \alpha m$. Then

$$\partial_v r(x, v, \tau) = - \frac{p v^{p-1} - p(x - v^\alpha \tau)^{m-1} v^{\alpha-1} \tau}{(1 + |x - \tau v^\alpha|^m + v^p)^2}.$$

In other words, $\partial_v r(x, v, \tau) = 0$ if and only if $x\tau^{\frac{1}{m-1}} = v^{\alpha}(1 + \tau^{\frac{m}{m-1}})$. Thus, for fixed $x$ and $\tau$, the point where the function $r$ changes its sign can be approximated for large times by $v^{\alpha} \approx \frac{x}{\tau}$ and for small times by $v^{\alpha} \approx x\tau^{\frac{1}{m-1}}$. The coagulation operator can be approximated at a formal level by a term of the form $v^{\gamma}\partial_v r(x, v, \tau) \int_0^{\infty} r(x, v', \tau)dv'$. The change of sign of $r$ was dealt with in the proof made in [9] by taking a cut-off at $v^{\alpha}_{\max} \approx x\tau^{\frac{1}{m-1}}$. While at a first glance there is no similarity between the methods to prove existence in [9] with the methods in the current paper, the bounds in (2.19) and (A.1), respectively, can be understood at an intuitive level as taking a cut-off at $v^{\alpha} \approx \frac{x}{\tau}$. Thus, our existence result can be understood as the natural continuation of the proof in [9] for large times.
## B Construction of solutions with compact support

We now prove that $H_{n+1}$ defined in (2.51) has compact support assuming that it has compact support at initial time. This is in order to find a time $\delta$ as in Proposition 3.1 that is independent of y and v.

*Proof of Proposition 2.17.* We first analyze the following equation.

$$ \partial_t f_{n+1}(y, v, t) + \frac{1}{\epsilon} v^\alpha \partial_x f(x, v, t) = \mathbb{K}[f_{n+1}, f_n](y, v, t), \quad (B.1) $$

where $\mathbb{K}_N[f_{n+1}, f_n]$ is as in (2.55). Existence of $f_n \in C^1([0, T] \times \mathbb{R} \times (0, \infty))$ for small times independent of $\epsilon$ follows via Banach's fixed point theorem.

We then define

$$ H_n(y, v, \tau) := e^\tau f_n(x, v, t), \quad x := e^\tau y, \quad t := e^\tau - 1, \text{ for all } n \in \mathbb{N}. $$

Thus, $H_n \in C^1([0, T] \times \mathbb{R} \times (0, \infty))$ and solves (2.55). Compact support follows directly from (2.55) and the choice of $K_N$ in (2.56). □

## Acknowledgements

The authors sincerely thank B. Niethammer for reviewing a draft of the manuscript and contributing to the improvement of its presentation. The authors would like to thank N. Fournier and B. Perthame for fruitful discussions about gelation theory and for explaining the arguments in [18].

I.C. was supported by the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation program Grant No. 637653, project BLOC "Mathematical Study of Boundary Layers in Oceanic Motion" and by the project BOURGEONS "Boundaries, Congestion and Vorticity in Fluids: A connection with environmental issues", grant ANR-23-CE40-0014-01 of the French National Research Agency (ANR).

J.J.L.V. gratefully acknowledges the financial support of the collaborative research centre The mathematics of emerging effects (CRC 1060, Project-ID 211504053) and of the Hausdorff Center for Mathematics (EXC 2047/1, Project-ID 390685813) funded through the Deutsche Forschungsgemeinschaft (DFG, German Research Foundation).

## Statements and Declarations

**Conflict of interest** The authors declare that they have no conflict of interest.

**Data availability** Data sharing not applicable to this article as no datasets were generated or analyzed during the current study.
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