# Analysis of a polarization model with exchange at the boundary

Thomas Lepoutre

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Thomas Lepoutre. Analysis of a polarization model with exchange at the boundary. 2025. hal-05380243

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Distributed under a Creative Commons Attribution 4.0 International License
# Analysis of a polarization model with exchange at the boundary

Thomas Lepoutre

Univ Lyon, Inria, Université Claude Bernard Lyon 1, CNRS
UMR5208, Institut Camille Jordan, F-69603 Villeurbanne, France

...

## Abstract

Here we analyze a drift-diffusion model on the half-line. The drift is linked to exchanges at 0. This corresponds to the modeling of filament generation following protein exchanges between the membrane and the cytoplasm of a cell. This model has a critical mass and the asymptotic behaviors can be divided into three categories. In subcritical mass regime, diffusive behavior dominates and convergence in self-similar variables can be obtained with parabolic scaling. In supercritical mass regime, drift compensates for diffusion and convergence towards equilibrium distributions occurs. In the critical case, convergence towards 0 for the drift occurs over a long period of time, the behavior exhibits a distinct relevant scaling. We quantify the convergences in the relevant variables in the three regimes.

**keywords:** Non-linear stability, asymptotic convergence, logarithmic Sobolev inequality, self-similar behaviour

## 1 Introduction

This work deals with the analysis of a model that was introduced in [12]. The initial purpose of the model was to describe the organization of the cytoskeleton. This approach has been exploited in the description of spontaneous polarization leading to a mathematical analysis of one dimensional models in [5, 7]. It describes in a rough way the exchange between the cytoplasm and a membrane in the framework of spontaneous polarization of yeast cells. This work is dedicated to a fine mathematical analysis of the model. For more recent works on polarization we refer for instance to [11]. Our work can be related to a number of nonlocal drift-diffusion models, describing cell motility [19, 20, 16] It can be linked to other works on drift diffusion on half line or on bounded domain with drift linked to the boundary [18, 15, 14] or to nonlocal quantities [2]. A closely related model (with a bistable behaviour) was studied in [6] as representing the
communication between two polarizing yeast cells. Drift diffusion models (but notably with a singular source) can be found in the mathematical modeling of neuronal activity, namely the integrate and fire model [1, 4]. Quite interestingly to some more recent work on field road diffusion model [21]. This model presents a critical mass phenomenon and, whereas the subcritical and supercritical behaviour follow similar lines of other works, the critical behaviour exhibits and intermediate scaling. Compared to the partial analysis derived in [5], our work provides

* Quantitative convergence in self similar variables for the subcritical behaviour

* Proof of vanishing mass on the membrane for the critical case

* Quantitative convergence in self similar variables for the critical case.

## 1.1 The model

We consider a density $n$ and a quantity $\mu$, linked by the following equation

$$ \partial_t n - \partial_{xx} n - \mu \partial_x n = 0, x > 0. \quad (1a) $$

$$ \partial_x n(t, 0) + \mu(t)n(t, 0) = \dot{\mu}, \quad (1b) $$

$$ \dot{\mu} = n(t, 0) - \mu(t), \quad (1c) $$

This will be supplemented with initial data satisfying

$$ n^0 \ge 0, \mu^0 > 0, \quad \int_0^\infty (n^0 \log n^0 + x^2 n^0) dx < +\infty. \quad (2) $$

**Remark 1.1.** Note that the condition $\mu^0 \ge 0$ can be relaxed to $\mu^0 \ge 0$ but since standard parabolic theory ensures that $\mu > 0$ for $t > 0$, we keep with this hypothesis that makes some estimates more quantitative.

**Remark 1.2.** With this hypothesis, it is easy to establish that the model has a unique nonnegative global solution that satisfies in particular

$$ \frac{d}{dt} \left( \mu + \int_0^\infty ndx \right) = 0. $$

So that the initial mass is preserved

$$ M = \int_0^\infty n^0 dx + \mu^0 = \int_0^\infty n(t, x) dx + \mu(t) \quad (3) $$

And in particular

$$ \forall t \ge 0, \quad \int_0^\infty n(t, x) dx = M - \mu(t) $$
## 1.2 Critical mass

A simple computation enlightens the existence of a critical mass 1 observed in [5]

$$ 
\begin{aligned}
\frac{d}{dt} \int (1+x)n(t,x)dx &= -\dot{\mu} - \int_0^\infty (\partial_x n + \mu(t)n(t,x))dx, \\
&= \mu(t) - n(t,0) + n(t,0) - \mu \int n, \\
\frac{d}{dt} \int (1+x)n(t,x)dx &= \mu(1-M+\mu)
\end{aligned}
\quad (4) $$

For $M > 1$ it has been established in [5], that the solution converges towards the steady state $\bar{\mu} = (M - 1)$, $\bar{n}(x) = \bar{\mu}e^{-\bar{\mu}x}$. For $M \le 1$ the reader can notice that the closer the mass gets to the critical value 1, the closer the limit $(M - 1)$ for $\mu$ gets to 0 (which will be a key point later), so that we can conjecture reasonably $\mu \to 0$ for $M \le 1$ and this is established in the next section. Assuming that this convergence holds true, one can investigate for self similar behavior since a vanishing drift could lead to a heat like behavior (as in [17]). It seems to be the case for $M < 1$ but we face a difficulty that is quite common: there exists no self similar solution to (1) and this is visible when we perform the (self similar) change of variable

$$ n(t,x) = \frac{1}{\sqrt{1+2t}} u \left( \underbrace{\log(\sqrt{1+2t})}_{\tau}, \frac{x}{\sqrt{1+2t}} \right) $$

And denoting $\tilde{\mu}(\tau) = \mu(\frac{e^{2\tau}-1}{2})$ (so that $\tilde{\mu}(\tau(t)) = \mu(t)$). Leading to

$$ \partial_{\tau} u - \partial_{yy} u - \partial_{y}(y u) - \tilde{\mu} e^{\tau} \partial_{y} u = 0, \quad (5a) $$

$$ \partial_y u(\tau, 0) + \tilde{\mu} e^{\tau} u(\tau, 0) = \frac{d}{d\tau} \tilde{\mu}, \quad (5b) $$

$$ \frac{d}{d\tau} \tilde{\mu} = e^{\tau} (u(\tau, 0) - \tilde{\mu} e^{\tau}) \quad (5c) $$

However, using educated guess $\tilde{\mu}e^{\tau}, u(\tau, 0) \to C$ we can derive the conjecture that the limit has the shape $\tilde{\mu} = 0, u = Ce^{-\frac{\mu^2}{2}-Cy}$. Then we can notice that the conservation of mass leads to the constraint

$$ M = m(C) = \int_{0}^{\infty} C e^{-\frac{\mu^2}{2} - Cy}. \quad (6) $$

Since it is straightforward to establish that $m$ is a monotone bijection between $]0, \infty[$ and $]0, 1[$, we can conjecture that the scaling does not capture well the behaviour for $M = 1$. This suggests an important difference between the subcritical and critical case. In order to give a first hint on the critical case, we introduce $J = \int (1+x)n$. If $J$ becomes large, one would expect the solution to flatten, $n(t, 0)$ and $\mu$ to vanish (diffusion behaviour dominates). We expect
a self similar behaviour $n \sim t^{-\alpha}\bar{n}(t^{-\alpha}x)$ (for the parabolic scaling $\alpha = 1/2$). If we plug such an ansatz, we expect $\mu \sim t^{-\alpha}$ and $J \sim t^{\alpha}$. The computations leads then formally, once injected into (4) to

$$ 
\begin{aligned}
t^{\alpha-1} &\propto (1-M)t^{-\alpha} \quad \text{for } M < 1, \\
t^{\alpha-1} &\propto t^{-2\alpha} \quad \text{for } M = 1,
\end{aligned}
$$

• for $M < 1$, we expect a diffusive scaling $\alpha = 1/2$,

• for $M = 1$ a specific scaling appears $\alpha = 1/3$.

## 1.3 Main results

Here we establish, using relative entropy the following convergence results that are here summarized in $L^1$.

**Theorem 1.3.** Assume condition (2) is fullfilled, and $M = \mu^0 + \int n^0$, then, using notation $v_{\lambda,\eta} = \lambda e^{-\lambda x - \eta x^2/2}$

1. if $M > 1$, $\mu(t) \to \bar{\mu} = M - 1$ as $t \to \infty$ and, denoting $\bar{n} = \bar{\mu}e^{-\bar{\mu}x}$, we have as well

$$ \|n - \bar{n}\|_1 + |\mu - \bar{\mu}| = O\left(\sqrt{\frac{\log(1 + 2t)}{1 + 2t}}\right) \quad (7) $$

2. for $M < 1$, we denote $\lambda$ the unique solution to equation

$$ \int_0^\infty \lambda e^{-\lambda x - \frac{x^2}{2}} dx = \int_0^\infty v_{\lambda,1} = M, $$

then we have

$$ \int_0^\infty \left| n(t, x) - \frac{1}{\sqrt{1 + 2t}} v_{\lambda,1} \left( \frac{x}{\sqrt{1 + 2t}} \right) \right| dx = O \left( \frac{\log(1 + 2t)}{\sqrt{1 + 2t}} \right) \quad (8) $$

3. in the critical case $M = 1$, we have

$$ \int_0^\infty \left| n(t, x) - \frac{1}{(1 + 3t)^{1/3}} v_{0,1} \left( \frac{x}{(1 + 3t)^{1/3}} \right) \right| dx = O \left( \frac{\log(1 + 3t)}{(1 + 3t)^{1/3}} \right) \quad (9) $$

**Remark 1.4.** The rate is not as good as the rate one expects in heat equation for the subcritical case. It remains a question wether is can be improve up to $O(t^{-1/2})$ in the subcritical case (a similar rate was obtained in [6]).

**Remark 1.5.** Our results might be compared to those of [21]. Even if there are no mass exchange in their model it contains quite converging ideas.
## 2 Quantitative estimates for the supercritical case

In [5], the authors have established the specific role of critical threshold 1 for the mass. Denote $H(x) = x \log x - x + 1$. For $M > 1$ and data satisfying (2), it has been established

**Theorem 2.1.** Assume (2) and $M > 1$, then

$$ \bar{\mu} = M - 1 > 0, \quad \bar{n} = \bar{\mu}e^{-\bar{\mu}x} $$

define the only stationary solution to (1) with mass $M$. Moreover, the relative entropy

$$ \int_{0}^{\infty} \bar{n}H\left(\frac{n}{\bar{n}}\right)dx+\bar{\mu}H\left(\frac{\mu}{\bar{\mu}}\right) $$

is decaying to 0.

### 2.1 Precising the asymptotics for the supercritical mass

This follows quite sharply the lines of [17]. We recall that in [5], the following Lyapunov functional was introduced (although written a bit differently)

$$ \mathcal{L}(t) = \underbrace{\int_{0}^{\infty} \bar{n}H\left(\frac{n}{\bar{n}}\right)dx}_{\mathcal{L}_1} + \bar{\mu}H\left(\frac{\mu}{\bar{\mu}}\right) + \frac{1}{2}(\mu - \bar{\mu})^2. $$
Proof. We differentiate the functional

$$
\begin{align*}
\mathcal{L}'(t) &= \int_0^\infty (\log \frac{n}{\bar{n}}) \partial_t n + \dot{\mu}(\log \frac{\mu}{\bar{\mu}} + \mu - \bar{\mu}), \\
&= \dot{\mu} \left( -\log \frac{n(t, 0)}{\bar{\mu}} + \log \frac{\mu}{\bar{\mu}} + \mu - \bar{\mu} \right) \\
&\quad - \int n(\partial_x \log n + \bar{\mu})(\partial_x \log n + \mu) \\
&= \dot{\mu} \left( -\log \frac{n}{\bar{\mu}} + \log \frac{\mu}{\bar{\mu}} + \mu - \bar{\mu} \right) \\
&\quad - \int n(\partial_x \log n + \mu)^2 \\
&\quad + (\mu - \bar{\mu}) \int n(\partial_x \log n + \mu) \\
&= \dot{\mu} \left( -\log \frac{n}{\bar{\mu}} + \log \frac{\mu}{\bar{\mu}} + \mu - \bar{\mu} \right) \\
&\quad - \int n(\partial_x \log n + \mu)^2 \\
&\quad + (\mu - \bar{\mu})(-n(t, 0) + \mu \int n) \\
&= -\log \frac{n(t, 0)}{\mu} (n(t, 0) - \mu) \\
&\quad - \int n(\partial_x \log n + \mu)^2 \\
&\quad + (\mu - \bar{\mu})(-\dot{\mu} - \mu + \mu(M - \mu)) \\
&= -\log \frac{n(t, 0)}{\mu} (n(t, 0) - \mu) \\
&\quad - \int n(\partial_x \log n + \mu)^2 - \mu(\mu - \bar{\mu})^2
\end{align*}
$$

The decay of this functional to 0 is sufficient to establish the convergence (see
[5]). To get a rate, we follow the trick of [17] and introduce a modified functional.
Let $c : \mathbb{R}_+ \mapsto \mathbb{R}_+$ a decreasing function of time (to be precised later). We denote

$$
\mathcal{L}_c(t) = \mathcal{L}(t) + c(t) \int_0^\infty x^2 n(t, x) dx.
$$
We have

$$
\begin{align*}
\frac{d}{dt} \mathcal{L}_c(t) &= \mathcal{L}'(t) + c'(t) \int x^2 n + c(t) \int x^2 \partial_t n \\
&= -\log \frac{n(t, 0)}{\mu} (n(t, 0) - \mu) - \mu(\mu - \bar{\mu})^2 + c'(t) \int x^2 n \\
&\quad - \int n (\partial_x \log n + \mu)^2 - \int 2cx (\partial_x n + \mu n) \\
&= -\log \frac{n(t, 0)}{\mu} (n(t, 0) - \mu) - \mu(\mu - \bar{\mu})^2 + (c'(t) + c^2(t)) \int x^2 n \\
&\quad - \int n (\partial_x \log n + \mu + cx)^2.
\end{align*}
$$

Next we apply Logarithmic Sobolev inequality (see lemma A.1 in the appendix)
so that (with notations $J_k = \int_0^\infty \frac{x^k}{k!} n$).

$$
\begin{align*}
\frac{1}{2c} \int n (\partial_x \log n + \mu + cx)^2 &\ge \int n \log \frac{n}{\mu e^{-\mu x - cx^2/2}} - \int n \log \frac{\int n}{\int \mu e^{-\mu x - cx^2/2}} \\
&= \int n \log \frac{n}{\bar{\mu} e^{-\bar{\mu} x}} + \int n \log e^{(\mu - \bar{\mu})x + cx^2/2} \\
&\quad - \int n \log \int n + \int n \log \int \mu e^{-\mu x - cx^2/2} \\
&= \mathcal{L}_1 - H(J_0) \\
&\quad + (\mu - \bar{\mu})J_1 + cJ_2 + J_0 \log \int \bar{\mu} e^{-\bar{\mu} x - cx^2/2}
\end{align*}
$$

Then we use Jensen inequality as in lemma A.2 to notice that

$$
\begin{align*}
\log \int \bar{\mu} e^{-\mu x - cx^2/2} &= \log \int e^{(\bar{\mu} - \mu)x - cx^2/2} \bar{\mu} e^{-\bar{\mu}x} \\
&\geq \int ((\bar{\mu} - \mu)x - cx^2/2) \bar{\mu} e^{-\bar{\mu}x} \\
&= \frac{\bar{\mu} - \mu}{\bar{\mu}} - \frac{c}{\bar{\mu}^2}
\end{align*}
$$

This leads to

$$
\frac{1}{2c} \int n (\partial_x \log n + \mu + cx)^2 \geq \mathcal{L}_1 - H(J_0) \\
+ (\mu - \bar{\mu}) \left(J_1 - \frac{J_0}{\bar{\mu}}\right) + c \left(J_2 - \frac{J_0}{\bar{\mu}^2}\right)
$$
Combining everything together, we end up with

$$
\begin{align*}
\frac{d}{dt}\mathcal{L}_c(t) &\leq -\mu(\mu - \bar{\mu})^2 + (c'(t) + c(t)^2) \int x^2n + 2c^2 \left(\frac{J_0}{\bar{\mu}^2} - J_2\right) \\
&\quad - 2c\mathcal{L}_1 + 2cH(J_0) - 2c(\mu - \bar{\mu}) \left(J_1 - \frac{J_0}{\bar{\mu}}\right) \\
&= -2c\mathcal{L}_c + (c' + 2c^2) \int x^2n + 2c^2 \frac{J_0}{\bar{\mu}^2} \\
&\quad - \mu(\mu - \bar{\mu})^2 + 2cH(J_0) - 2c(\mu - \bar{\mu}) \left(J_1 - \frac{1}{\bar{\mu}}\right) + c(\mu - \bar{\mu})^2 - 2c \frac{(\mu - \bar{\mu})^2}{\bar{\mu}} + 2c\bar{\mu}H\left(\frac{\mu}{\bar{\mu}}\right) \\
&\quad - 2c(\mu - \bar{\mu}) \left(J_1 - \frac{1}{\bar{\mu}}\right)
\end{align*}
$$

Using the ine quality $H(x) \le (x - 1)^2$ and noticing $J_0 - 1 = \bar{\mu} - \mu$, we get to

$$
\begin{align*}
\frac{d}{dt}\mathcal{L}_c(t) &\leq -2c\mathcal{L}_c + (c' + 2c^2) \int x^2n + 2c^2 \frac{J_0}{\bar{\mu}^2} + 3c(\mu - \bar{\mu})^2 \\
&\quad - \mu(\mu - \bar{\mu})^2 - 2c(\mu - \bar{\mu}) \left(J_1 - \frac{1}{\bar{\mu}}\right)
\end{align*}
$$

We focus here on the obtention of a rate. Choosing $c = (1 + 2t)^{-1}$, we obtain

$$
\frac{d}{dt}((1+2t)\mathcal{L}_c(t)) \leq (\mu - \bar{\mu})^2 \left(3 - \mu \cdot (1+2t) + \frac{\bar{\mu}}{2}(1+2t)\right) + \frac{1}{1+2t} \left(\frac{2J_0}{\bar{\mu}^2} + \frac{2}{\bar{\mu}}\left|J_1 - \frac{1}{\bar{\mu}}\right|\right)
$$

From the boudedness of the entropy, we know that $\mu \to \bar{\mu}$, $J_0 = 1 + o(1)$ and $J_1 - \frac{1}{\bar{\mu}} \to 0$. Therefore, the first term has a negative sign for $t$ large and we have

$$
(1 + 2t)\mathcal{L}_c(t) \le \mathcal{L}_c(0) + \frac{1}{\bar{\mu}^2} \log(1 + 2t) + o(\log(1 + 2t)).
$$

From this, we deduce,

$$
\mathcal{L}_1 + \frac{1}{2}(\mu - \bar{\mu})^2 = O\left(\frac{\log(1 + 2t)}{1 + 2t}\right),
$$

and using the Czisar-Kullback inequality as stated in lemma A.5, we obtain the
conclusion of the point (7) of theorem 1.3. $\square$

3 Concentration comparison principle and ap-
plications

3.1 Statement and first consequence

Integrating the equation (1a), we have easily, denoting $N(t,x) = \mu + \int_0^x ndx$,

$$
\begin{cases}
\partial_t N - \partial_{xx} N - \mu \partial_x N = 0, \\
N(t, 0) = \mu(t)
\end{cases}
\tag{10}
$$
This idea ressembles a lot to what is used for some analysis of radial solutions of Keller Segel model [8] for instance. We can note a version named average comparison. The main advantage is that it gives a tool that allows the comparison between some solutions (even with the same mass). As a very useful byproduct, it leads to a notions of sub and supersolutions

**Definition 3.1.** We call a supersolution, a couple of a nonnegative (smooth?) $\bar{\mu}$ and a nondecreasing function $\bar{N}$ that satisfy

$$ \partial_t \bar{N} - \partial_{xx} \bar{N} - \bar{\mu}(t) \partial_x \bar{N} \geq 0, \quad (11a) $$

$$ \bar{N}(t, 0) = \bar{\mu}(t) \quad (11b) $$

$$ \frac{d}{dt} \bar{\mu} \geq \partial_x \bar{N}(t, 0) - \bar{\mu} \quad (11c) $$

A subsolution is defined replacing all $\geq$ by $\leq$.

The fundamental result is the following

**Lemma 3.2.** Let $(\bar{N}, \bar{\mu})$, $(\underline{N}, \underline{\mu})$ be a supersolution and a subsolution satisfying $\bar{N}(0, x) \geq \underline{N}(0, x)$ and $\bar{\mu}(0) > \underline{\mu}(0)$. Assume they are defined on $[0, T]$, then the following holds

$$ \bar{N}(t, x) \geq \underline{N}(t, x), \quad \bar{\mu}(t) \geq \underline{\mu}(t), \quad t < T. $$

*Proof.* The proof is very similar to [17] Let us consider the difference $\delta N = \bar{N} - \underline{N}$, $\delta N = \bar{N} - \underline{N}$. From the hypothesis, we have initially $\delta N \geq 0$ and $\delta \mu > 0$, we have by construction

$$ \left\{ \begin{array}{l} \partial_t \delta N - \partial_{xx} \delta N - \bar{\mu} \partial_x \delta N = (\bar{\mu} - \underline{\mu}) \partial_x \underline{N} \geq 0, \\ \delta N(t, 0) = \delta \mu, \\ \frac{d}{dt} \delta \mu \geq \partial_x \delta N(t, 0) - \delta \mu. \end{array} \right. $$

As long as $\delta \mu > 0$, the persistence of the inequality is a straightforward consequence of the maximum principle. If $\delta \mu$ vanishes, then consider the first time $t^*$ for which it vanishes, then the Hopf's lemma (REF) ensures $\delta \partial_x N(t, 0) > 0$ and thereby $\frac{d}{dt} \delta \mu(t^*) > 0$ leading to a contradiction. $\square$

This particularly relevant if one these is corresponding to a solution of (1). It is also very convenient We give here a fundamental example

**Lemma 3.3.** Let $n^0$ satisfy (2) and let $\mu_1^0 < \mu_2^0$. Denote $n_i(t, x)$, $\mu_i(t)$ the solution to (1) with initial data $n_i^0 = n^0$, $\mu_i^0$. Then we have (denoting $N_i(t, x) = \mu_i(t) + \int_0^x n_i(t, y)dy$,

$$ \forall t, x \geq 0, N_1(t, x) \leq N_2(t, x) $$

In particular, we have

$$ \mu_1(t) \leq \mu_2(t). $$
And as a consequence, we give the result from binding lemma 3.3 and theorem 2.1.

**Theorem 3.4.** Assume (2) with $M \le 1$, then

$$ \mu(t) \to 0. $$

*Proof.* Consider $\varepsilon > 0$ and consider $\mu_1^0 = \mu^0 + 1 + \varepsilon - M$, $\mu_2^0 = \mu^0$. By lemma 3.3, we have

$$ \mu(t) \le \mu_1(t). $$

Moreover by construction, $n_1^0, \mu_1^0$ has mass $1 + \varepsilon$, and we have by theorem 2.1

$$ \mu_1(t) \to 1 + \varepsilon - 1 = \varepsilon, $$

leading to

$$ \limsup \mu \le \varepsilon. $$

Since we can choose $\varepsilon$ arbitrarily, we can conclude $\limsup \mu \le 0$ and the result follows from nonnegativity. $\square$

We end up with an easy but useful corollary for applications

**Lemma 3.5.** Let Let $(\bar{N}, \bar{\mu})$, $(N, \mu)$ be a supersolution and a subsolution, then for any $0 < \lambda \le 1$, $\lambda(\bar{N}, \bar{\mu}) \frac{1}{\lambda}(N, \mu)$ are still a supersolution and a subsolution

*Proof.* We simply need to check the PDE (the relations (11b, 11c) are satisfied by linearity). For instance, if $\bar{N}$ is a supersolution, we get for $0 < \lambda \le 1$,

$$ \partial_t(\lambda \bar{N}) - \partial_{xx}(\lambda \bar{N}) - (\lambda \bar{\mu}) \partial_x(\lambda \bar{N}) \ge \bar{\mu}(1 - \lambda) \partial_x(\lambda \bar{N}) \ge 0 $$

<figure><img src="image_3.png" /></figure>

## 3.2 Remarkable sub/supersolutions for the subcritical case

We first notice that the concentration comparison principle can be translated in self-similar variables. Sub and supersolutions $U(\tau, y)$ can be defined as before mutatis mutandis in self similar variable (or equivalently, checking that $U(\frac{1}{2} \log(1+2t), \frac{x}{\sqrt{1+2t}})$ is a sub or super solution in original variables. We begin with a subsolution based on Fokker Planck equation.

**Lemma 3.6 (Parametrized subsolutions).** Denote

$$ \gamma(y) = \sqrt{\frac{2}{\pi}} e^{-y^2/2} $$

the normalized gaussian on the half line and $\Gamma(y) = \int_0^y \gamma$. Consider $\eta > 0$, then, $U(\tau, y) = \eta(\gamma(0)e^{-\tau} + \Gamma)$ defines a subsolution to (5)
Proof. By construction, denoting $\tilde{\mu} = \eta\gamma(0)e^{-\tau}$,

$$ \partial_{\tau} U - \partial_{yy} U - (y + \tilde{\mu} e^{\tau}) \partial_{y} U = -\eta \sqrt{\frac{2}{\pi}} \gamma(y) - \eta \sqrt{\frac{2}{\pi}} e^{-\tau} \le 0, $$

$$ U(\tau, y) = \tilde{\mu}, $$

$$ \begin{aligned} \tilde{\mu}'(\tau) - e^{\tau} (u(\tau, 0) - \tilde{\mu} e^{\tau}) &= -\eta\gamma(0)e^{-\tau} - e^{\tau} (\eta\gamma(0) - \eta\gamma(0)) \\ &= -\eta\gamma(0)e^{-\tau} \le 0 \end{aligned} $$

□

This leads immediately to the

**Corollary 3.7.** Assume $\mu^0 > 0, u^0 \ge 0$, then the solution of (5) associated to $u^0, \mu^0$ satisfies

$$ \tilde{\mu} e^{\tau} \ge \frac{\mu^0}{\gamma(0) + 1}. \qquad (12) $$

Proof. Consider the subsolution $\underline{U}$ of the previous lemma with $\eta = \frac{\mu^0}{\gamma(0)+1}$. In this case, we have initially by construction

$$ \underline{U}(0, y) \le \underline{U}(0, \infty) = \eta(\gamma(0) + 1) \le \mu^0 \le \underline{U}(0, y). $$

And we apply the concentration comparison principle.

□

In the same spirit, we can also build supersolution. We denote

$$ g_{C,\lambda}(y) = C e^{-\frac{y^2}{2} - \lambda y}, $$

(in particular $m(\lambda) = \int_0^\infty g_{\lambda,\lambda}(y)$. We also denote $G_{C,\lambda}(y) = \int_0^y g_{C,\lambda}$. In particular, $G_{C,\lambda}$ increases from 0 to $\frac{Cm(\lambda)}{\lambda}$. We will also widely use the following property (omitting the indices $C, \lambda$ when not necessary)

$$ -\partial_{yy} g - \partial_y(yg) - \lambda \partial_y g = 0, $$

$$ -\partial_{yy} G - y \partial_y G - \lambda \partial_y G = 0. $$

In the following, we will consider the following

**Lemma 3.8.** Denote the function

$$ U(\tau, y) = G_{C,\lambda}(y + e^{-\tau}) \quad \text{and} \quad \tilde{\mu} = G_{C,\lambda}(e^{-\tau}). $$

Then for $\lambda \ge C + 1$, $U$ is a supersolution

Proof. Throughout the proof, we denote $G = G_{C,\lambda}$ and $g = g_{C,\lambda}$,

$$ \begin{aligned} \partial_{\tau} U - \partial_{yy} U - y \partial_{y} U - \tilde{\mu} e^{\tau} \partial_{y} U &= (\lambda - \tilde{\mu} e^{\tau} - e^{-\tau}) \partial_{y} U, \\ &= \left( \lambda - \frac{G(e^{-\tau})}{e^{-\tau}} - e^{-\tau} \right) \underbrace{\partial_{y} U}_{\ge 0} \\ &\ge (\lambda - C - 1) \underbrace{\partial_{y} U}_{\ge 0} = 0 \end{aligned} $$
Moreover, we notice that

$$g'(x) = -(x + \lambda)g(x).$$

By construction, we have

$$ 
\begin{aligned}
\frac{d}{d\tau} \tilde{\mu} &= -e^{-\tau} g(e^{-\tau}) \\
\frac{d}{d\tau} \tilde{\mu} - e^{\tau}(u(\tau, 0) - \tilde{\mu}e^{\tau}) &= -e^{-\tau}g(e^{-\tau}) - e^{\tau}(g(e^{-\tau})) - G(e^{-\tau})e^{\tau} \\
&= e^{2\tau}(G(e^{-\tau}) - e^{-\tau}g(e^{-\tau}) - e^{-3\tau}g(e^{-\tau})) \\
&= e^{2\tau}F(e^{-\tau}).
\end{aligned}
$$

Where $F(x) = G(x) - xg(x) - x^3g(x)$. Then, we notice that $F(0) = 0$ and

$$ 
\begin{aligned}
F'(x) &= g(x) - (3x^2 + 1)g(x) - (x^3 + x)g'(x), \\
&= g(x)(1 - 3x^2 - 1 + (x^3 + x)(x + \lambda)), \\
&= xg(x)(x^3 + \lambda x^2 - 2x + \lambda), \\
&= xg(x)\left(x^3 + \underbrace{(\lambda - 1)}_{\ge 0, \text{ if } \lambda \ge 1} 2x + \lambda(x - 1)^2\right)
\end{aligned}
$$

Therefore $F \ge 0$ if $\lambda \ge 1$ which is true if $\lambda \ge C + 1$. $\square$

The interest of this example can be seen in the

**Lemma 3.9.** Let $0 < M < 1$ and an initial data such that $u^0, \mu^0$ satisfies $\mu^0 + \int u^0 \le M$, then there exists $C$ large enough

$$U(0, y) = \mu^0 + \int_0^y u(0, z) dz \le U^{C, C+1}(0, y),$$

where $U^{C,C+1}$ the supersolution defined as above. In particular,

$$\tilde{\mu} \le G^{C,C+1}(e^{-\tau}) \le Ce^{-\tau}.$$

*Proof.* We simply notice firstly that for $C$ large enough

$$U^{C,C+1}(0,y) \ge U^{C,C+1}(0,0) = G^{C,C+1}(1) \ge M \ge U(0,y).$$

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

Together, up to starting at time $t = 1$ to ensure we satisfy the condition $\tilde{\mu}(0) > 0$, combination of these lemma imply in particular

$$0 < \liminf \tilde{\mu} e^{\tau} \le \limsup \tilde{\mu} e^{\tau} < +\infty. \qquad (13)$$
## 4 Convergence in the subcritical case

We recall the LlogL functional $h(x) = x \log x - x + 1$. For $M < 1$ we recall that there exists a unique $\lambda$ such that

$$ \int_{0}^{\infty} \lambda e^{-\lambda x - x^2/2} dx = M $$

We denote $\bar{u}(y) = \lambda e^{-\lambda y - y^2/2}$. We introduce a few more notation

$$ J_k(\tau) = \int_0^\infty \frac{y^k}{k!} u(\tau, y) dy, \quad \bar{J}_k = \int_0^\infty \frac{y^k}{k!} \bar{u}(y) dy $$

It will be useful to notice

$$ \bar{J}_1 = \lambda(1 - M) \qquad (14) $$

We also introduce the notation

$$ \mathcal{J} = J_1 + e^{-\tau} J_0, $$

Which has the nice closed equation on its evolution:

$$ \frac{d}{d\tau} \mathcal{J} = \mu e^{\tau} (1 - J_0) - \mathcal{J} \qquad (15) $$

Note that this, combined with the use of (13), ensures the following

**Lemma 4.1.** *The quantity $\mathcal{J}$ is bounded*

*Proof.* Simply a consequence of the fact that $\mu e^{\tau}$ is bounded (from (13)). Combined with the equation on $\mathcal{J}$ we can conclude. □

We define the following functional

$$ \begin{aligned} \mathcal{L}(\tau) &= \int_0^\infty \bar{u}h\left(\frac{u(\tau, y)}{\bar{u}(y)}\right) dy + \lambda e^{-\tau}h\left(\frac{\mu}{\lambda e^{-\tau}}\right) + \frac{1}{2}(\mu - \lambda e^{-\tau})^2 + \frac{1}{2(1-M)}(\mathcal{J} - \bar{J}_1)^2, \tag{16} \\ &= \mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3 + \mathcal{L}_4 \end{aligned} $$

Note that from computations above, we already know that

$$ \mathcal{L}_2 = O(e^{-\tau}), \quad \mathcal{L}_3(\tau) = O(e^{-2\tau}). \qquad (17) $$
We split the computations for sake of clarity.

$$
\begin{align*}
\mathcal{L}'_1 &= \int_0^\infty \log \frac{u}{\bar{u}} \partial_\tau u, \\
&= -\mu' \log \frac{u(\tau, 0)}{\bar{u}(0)} - \int_0^\infty u (\partial_y \log u + y + \lambda) (\partial_y \log u + y + \mu e^\tau) \\
&= -\mu' \log \frac{u(\tau, 0)}{\lambda} - \underbrace{\int_0^\infty u (\partial_y \log u + y + \mu e^\tau)^2}_{I(\mu e^\tau)} \\
&\quad + (\mu e^\tau - \lambda) \int_0^\infty u (\partial_y \log u + y + \mu e^\tau) \\
&= -\mu' \log \frac{u(\tau, 0)}{\lambda} - I(\mu e^\tau) \\
&\quad + (\mu e^\tau - \lambda) (-u(\tau, 0) + J_1 + \mu e^\tau J_0) \\
&= -\mu' \log \frac{u(\tau, 0)}{\lambda} - I(\mu e^\tau) - \mu' e^{-\tau} (\mu e^\tau - \lambda) \\
&\quad + (\mu e^\tau - \lambda) (-\mu e^\tau + J_1 + \mu e^\tau J_0) \\
&= -\mu' \left( \log \frac{u(\tau, 0)}{\lambda} + \mu - \lambda e^{-\tau} \right) - I(\mu e^\tau) \\
&\quad + (\mu e^\tau - \lambda) (-\mu e^\tau (1 - J_0) + J_1).
\end{align*}
$$

Then, we differentiate

$$
\begin{align*}
\mathcal{L}'_2 &= \mu' \log \frac{\mu}{\lambda e^{-\tau}} + \mu - \lambda e^{-\tau} \\
\mathcal{L}'_3 &= (\mu' + \lambda e^{-\tau})(\mu - \lambda e^{-\tau})
\end{align*}
$$

Leading to

$$
\begin{align}
\frac{d}{d\tau} (\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3) &= -\mu' \log \frac{u(\tau, 0)}{\mu e^{\tau}} - I(\mu e^{\tau}) \nonumber \\
&\quad + (\mu e^{\tau} - \lambda) (-\mu e^{\tau}(1 - J_0) + J_1 + e^{-\tau} + \lambda e^{-2\tau}) \tag{18}
\end{align}
$$

On the other hand

$$
\begin{align}
\mathcal{L}'_4 &= \frac{1}{1-M} (\mathcal{J} - \bar{J}_1) (\mu e^\tau (1 - J_0) - \mathcal{J}) \nonumber \\
&= -2\mathcal{L}_4 + \left(\frac{\mathcal{J}}{1-M} - \lambda\right) (\mu e^\tau (1 - J_0) - \lambda(1-M)) \nonumber \\
&= -2\mathcal{L}_4 + (\mu e^\tau - \lambda) (\mathcal{J} - \bar{J}_1) + \frac{\mu^2 e^\tau}{1-M} (\mathcal{J} - \bar{J}_1) \tag{19}
\end{align}
$$

We recall

$$
D_{\text{boundary}}(\tau) = \mu' \log \frac{u(\tau, 0)}{\mu e^{\tau}} = e^{\tau} (u(\tau, 0) - \mu e^{\tau}) \log \frac{u(\tau, 0)}{\mu e^{\tau}} \geq 0.
$$
We end up with

$$ \mathcal{L}' = -D_{\text{boundary}} - I(\mu e^{\tau}) - 2\mathcal{L}_4 \\ + (\mu e^{\tau} - \lambda)(-\mu e^{\tau}(1 - J_0) + J_1 + e^{-\tau} + \lambda e^{-2\tau} + \mathcal{J} - \bar{J}_1) \\ + \frac{\mu^2 e^{\tau}}{1 - M}(\mathcal{J} - \bar{J}_1). $$

We now compute using(18),(19)

$$ \mathcal{L}' + 2\mathcal{L} \le 2\mathcal{L}_2 + 2\mathcal{L}_3 \\ + 2\mathcal{L}_1 - I(\mu e^{\tau}) \qquad (20) $$

$$ + (\mu e^{\tau} - \lambda)(-\mu e^{\tau}(1 - J_0) + J_1 + e^{-\tau} + \lambda e^{-2\tau} + \mathcal{J} - \bar{J}_1) \quad (21) $$

$$ + \frac{\mu^2 e^{\tau}}{1 - M}(\mathcal{J} - \bar{J}_1) \qquad (22) $$

We treat the terms (20) and (21)separately. First we establish

**Lemma 4.2.** The following holds true

$$ 2\mathcal{L}_1 - I(\mu e^{\tau}) \le 2Mh \left(\frac{J_0}{M}\right) + 2(\mu e^{\tau} - \lambda) \left(\frac{J_0}{M}\bar{J}_1 - J_1\right) \quad (23) $$

*Proof.* Let us denote

$$ E(\eta) = \int_{0}^{\infty} e^{-\eta y - \frac{y^2}{2}} dy. $$

In particular, by construction $\lambda = \frac{M}{E(\lambda)}$. Consider

$$ \gamma_{\tau} = \frac{J_0}{E(\mu e^{\tau})} e^{-\mu e^{\tau} y - \frac{y^2}{2}} $$

the Gaussian normalized so that $\int \gamma_{\tau} = J_0$. We recall the computations from [17] based on logarithmic Sobolev inequality, that reads in our case

$$ 2 \left( \int_{0}^{\infty} u \log \frac{u}{\gamma_{\tau}} - u + \gamma_{\tau} \right) - I(\mu e^{\tau}) = 2 \left( \int_{0}^{\infty} u \log \frac{u}{\gamma_{\tau}} \right) - I(\mu e^{\tau}) \le 0. $$

and this leads to

$$ \begin{aligned} 2\mathcal{L}_1 - I(\mu e^{\tau}) &= 2 \int_0^\infty \left( u \log \frac{u}{\bar{u}} - u + \bar{u} - u \log \frac{u}{\gamma_\tau} \right) + \left( 2 \int_0^\infty u \log \frac{u}{\gamma_\tau} - I(\mu e^{\tau}) \right) \\ &\le 2(M - J_0) + 2 \int_0^\infty u \log \frac{\gamma_\tau}{\bar{u}} \\ &= 2M \left( 1 - \frac{J_0}{M} \right) + 2 \int_0^\infty u \left( (\lambda - \mu e^{\tau})y + \log \frac{J_0}{E(\mu e^{\tau})} \frac{E(\lambda)}{M} \right) \\ &= 2M \left( \frac{J_0}{M} \log \frac{J_0}{M} - \frac{J_0}{M} + 1 \right) + 2J_0 \log \frac{E(\lambda)}{E(\mu e^{\tau})} + 2J_1(\lambda - \mu e^{\tau}) \\ &= 2Mh \left( \frac{J_0}{M} \right) + 2J_0 \log \frac{E(\lambda)}{E(\mu e^{\tau})} + 2J_1(\lambda - \mu e^{\tau}) \end{aligned} $$
We recall that from Jensen inequality (– log is convex)

$$
\begin{align*}
\log \frac{E(\lambda)}{E(\mu e^{\tau})} &= -\log \left( \int_0^{\infty} e^{(\lambda-\mu e^{\tau})x} \frac{e^{-\lambda x - \frac{x^2}{2}} dx}{\int_0^{\infty} e^{-\lambda x - \frac{x^2}{2}} dx} \right) \\
&\leq (\mu e^{\tau} - \lambda) \left( \int_0^{\infty} x \frac{e^{-\lambda x - \frac{x^2}{2}} dx}{\int_0^{\infty} e^{-\lambda x - \frac{x^2}{2}} dx} \right) \\
&= (\mu e^{\tau} - \lambda) \frac{\bar{J}_1}{M}
\end{align*}
$$

This leads to

$$
2\mathcal{L}_1 - I(\mu e^{\tau}) \leq 2Mh \left(\frac{J_0}{M}\right) + (\mu e^{\tau} - \lambda) \left(2\frac{J_0}{M}\bar{J}_1 - 2J_1\right)
$$

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

Combining this with (21) and $2\mathcal{L}_3 = (\mu e^{\tau} - \lambda)(\mu e^{-\tau} - \lambda e^{-2\tau})$, we arrive at

$$
\begin{align*}
\mathcal{L}' + 2\mathcal{L} \le 2\mathcal{L}_2 + 2Mh \left(\frac{J_0}{M}\right) & \\
& + (\mu e^{\tau} - \lambda) \left(-\mu e^{\tau}(1-J_0) + J_1 + e^{-\tau} + \mu e^{-\tau} + \mathcal{J} - \bar{J}_1 - 2J_1 + 2\frac{J_0}{M}\bar{J}_1\right) \\
& + \frac{\mu^2 e^{\tau}}{1-M}(\mathcal{J} - \bar{J}_1)
\end{align*}
$$

We simplify

$$
\begin{align*}
A &= \left( -\mu e^{\tau}(1 - J_0) + J_1 + e^{-\tau} + \mu e^{-\tau} + \mathcal{J} - \bar{J}_1 - 2J_1 + 2\frac{J_0}{M}\bar{J}_1 \right) \\
&= \left( -\mu e^{\tau}(1 - M) - \mu^2 e^{\tau} + e^{-\tau} + \mu e^{-\tau} + e^{-\tau}J_0 + \bar{J}_1 - 2\frac{\mu}{M}\bar{J}_1 \right) \\
&= (1 - M)(\lambda - \mu e^{\tau}) + \left( -\mu^2 e^{\tau} + (1 + M)e^{-\tau} - 2\frac{\lambda\mu(1 - M)}{M} \right)
\end{align*}
$$

We regroup again the various terms and get

$$
\begin{align*}
\mathcal{L}' + 2\mathcal{L} \le {}& -D_{\text{boundary}} + 2\mathcal{L}_2 + 2Mh \left(\frac{J_0}{M}\right) \\
& + (\mu - \lambda e^{-\tau}) \left(1 + M - 2\frac{\mu e^{\tau}}{M}\lambda(1 - M) - \mu^2 e^{2\tau}\right) \\
& + \frac{\mu^2 e^{\tau}}{1 - M} [\mathcal{J} - \lambda(1 - M)] \\
& - (1 - M)(\lambda - \mu e^{\tau})^2
\end{align*}
$$

This leads us to a quantitative rate of convergence
**Lemma 4.3.** *The quantity*

$$ \frac{\mathcal{L}e^{2\tau}}{(1+\tau)^2} $$

is bounded

*Proof.* Now we notice that since we already established the boundedness of $\mu e^{\tau}$, we have, denoting $\delta(\tau) = |\mu e^{\tau} - \lambda|$ and $C$ a constant that can vary from line to line (but independent of $\tau$)

$$ \mathcal{L}_2 = \lambda e^{-\tau} h \left( \frac{\mu e^{\tau}}{\lambda} \right) \leq C e^{-\tau} \delta^2 $$

$$ 2Mh \left( \frac{J_0}{M} \right) = 2Mh \left( 1 - \frac{\mu}{M} \right) \leq C\mu^2 \leq Ce^{-2\tau} $$

$$ |(\mu - \lambda e^{-\tau})(1 + M - 2\frac{\mu e^{\tau}}{M}\lambda(1 - M) + \mu^2 e^{2\tau})| \leq Ce^{-\tau}\delta $$

$$ \frac{\mu^2 e^{\tau}}{1 - M} [\mathcal{J} - \lambda(1 - M)] \leq Ce^{-\tau} \sqrt{\mathcal{L}} $$

Alltogether this leads to, denoting $\mathcal{F} = \mathcal{L}e^{2\tau}$,

$$ \mathcal{F}'(\tau) \leq Ce^{\tau}\delta^2 + C + Ce^{\tau}\delta + C\sqrt{\mathcal{F}} - (1 - M)e^{2\tau}\delta^2 $$

$$ \leq C + C\sqrt{\mathcal{F}} - \frac{(1 - M)}{2}e^{2\tau}\delta^2 $$

In particular, we have

$$ \mathcal{F}'(\tau) \leq C\sqrt{1 + \mathcal{F}} $$

which leads easily to

$$ \sqrt{1 + \mathcal{F}} \leq C(1 + \tau) $$

leading to the result.

□

As a side result we get the sharp convergence of $\mu e^{\tau}$ to $\lambda$. Indeed, since $\mu e^{\tau}$ is bounded, it is easy to recall

$$ \mathcal{L}_2 \leq C(1 + \tau)^2 e^{-2\tau} $$

$$ h \left( \frac{\mu e^{\tau}}{\lambda} \right) \leq \frac{C}{\lambda} (1 + \tau)^2 e^{-\tau} $$

This leads immediately to

$$ (\mu e^{\tau} - \lambda)^2 \leq C(1 + \tau)^2 e^{-\tau} $$

This has a useful consequence for what follows

**Lemma 4.4.** *With these hypotheses we have as a by product $J_2 \to \frac{1-\lambda^2(1-M)}{2}$*
Proof. We apply lemma A.3 with $\eta = 1$ and for $\tau$ large enough, we can choose $\gamma = \sqrt{\mathcal{H}} < 1$ in (36). And this leads to

$$ \left| \int \frac{x^2}{2} (u - \bar{u}) \right| \le \sqrt{\mathcal{H}} \left( \int x^4 \lambda x e^{-\lambda x} dx + 1 \right) $$

and we end up noticing

$$ \begin{aligned} \int \frac{x^2}{2} \bar{u} &= \int \frac{x^2}{2} \lambda e^{-\lambda x - \frac{x^2}{2}} dx \\ &= \frac{M - \lambda^2(1 - M)}{2} \end{aligned} $$

□

# 5 Critical case

## 5.1 Arguments for the scaling

We start with some heuristics on the critical case. We consider data satisfying

$$ \int_{0}^{\infty} n^{0} \log n^{0} + x^{2} < +\infty $$

Based on our last remark, we notice that comparing the critical initial data with subcritical ones and supercritical (already stated in theorem 3.4) ones leads to the two results

$$ \mu(t) \to 0, \qquad (24) $$

$$ \mu(t)\sqrt{1+2t} \to +\infty \qquad (25) $$

Additionnally, we make some moments computations

$$ \frac{d}{dt}(J_0 + J_1) = \mu^2 \qquad (26) $$

$$ \frac{d}{dt} J_2 = 1 - \mu J_1 \qquad (27) $$

Moreover, by scaling, we know from lemma 3.5 that $(1 - \varepsilon)(\mu(t) + \int_0^x n) = (1 - \varepsilon)N$ defines a supersolution for any $\varepsilon$. This can be compared to solution with the same mass $n_\varepsilon$ that converges in selfsimilar variables to $\bar{u}_\varepsilon$ up to the second moment. Thereby

$$ \frac{1}{\sqrt{1+2t}}(1-\varepsilon) \int_{0}^{\infty} \frac{x^2}{2} n \le \frac{1}{\sqrt{1+2t}} \int_{0}^{\infty} \frac{x^2}{2} n_{\varepsilon} \to \int_{0}^{\infty} \frac{x^2}{2} \bar{u}_{\varepsilon}. $$

Thereby, we have, for any $\varepsilon \in ]0, 1[$

$$ \limsup \frac{J_2(t)}{1+2t} \le \frac{1}{1-\varepsilon} \int_0^\infty \frac{x^2}{2} \bar{u}_\varepsilon $$
Taking $\epsilon \to 0$, we deduce

$$ \limsup \frac{J_2(t)}{1+2t} = 0 \qquad (28) $$

From this we can give a shaper argument for the scaling. Indeed, we compute easily

$$ \begin{aligned} \frac{d}{dt} J_2 + \frac{1}{6}(J_0 + J_1)^3 &= 1 - \mu(J_0 + J_1) + \frac{1}{2}\mu^2(J_0 + J_1)^2 + \mu J_0 \\ &= \frac{1}{2}(\mu(J_0 + J_1) - 1)^2 + \frac{1}{2} + \mu(1 - \mu) \end{aligned} $$

In particular, knowing that $J_2 = o(t)$, we can deduce after integration

$$ \frac{1}{6}(J_0 + J_1)^3(t) + J_2(t) \ge \frac{1}{6}(J_0 + J_1)^3(0) + J_2(0) + \frac{1}{2}t $$

Leading to

$$ \liminf \frac{(J_1 + J_0)^3}{3t} \ge 1 $$

which leads our argument towards scaling by $t^{1/3}$.

## 5.2 Rate of convergence in selfsimilar variables

We introduce the change of variables

$$ \begin{aligned} \tau(t) &= \frac{1}{3}\log(1 + 3t), \quad \tau' = e^{-3\tau}, \\ n(t, x) &= e^{-\tau}u(\tau(t), e^{-\tau}x) \\ \tilde{\mu}(\tau(t)) &= \mu(t). \end{aligned} $$

The model can be rewritten as (omitting the $\tilde{\cdot}$ in the sequel)

$$ e^{-4\tau} (\partial_{\tau} u - \partial_{y}(yu)) - e^{-3\tau} (\partial_{yy} u + \mu e^{\tau} \partial_{y} u) = 0, $$

Leading to

$$ \left\{ \begin{array}{l} (\partial_{\tau} u - \partial_{y}(yu)) - e^{\tau} (\partial_{yy} u + \tilde{\mu} e^{\tau} \partial_{y} u) = 0, \\ e^{\tau} (\partial_{y} u + \tilde{\mu} e^{\tau} u)(0) = \frac{d}{d\tau} \tilde{\mu}, \\ \frac{d}{d\tau} \tilde{\mu} = e^{2\tau} (u(\tau, 0) - e^{\tau} \tilde{\mu}) \end{array} \right. \qquad (29) $$

**Remark 5.1.** An essential distinction with the subcritical case is the fact that we do not control $\mu e^{\tau}$ from above or below.

From these equations, we have as before, denoting $J = J_1 + e^{-\tau}J_0$,

$$ \frac{d}{d\tau} J = \mu^2 e^{2\tau} - J \qquad (30) $$

We can see that the convenient point of this rescaling is the fact that the candidate limit for $\mu e^{\tau}$ is 1. We introduce the following function

$$ v(\tau, y) = e^{-y - e^{-\tau}y^2/2}. $$
We introduce firstly

$$ \mathcal{L}_1 = \int u \log \frac{u}{v} - u + v, $$

$$ \mathcal{L}_2 = e^{-\tau} h(\mu e^{\tau}), $$

$$ \mathcal{L}_3 = \frac{1}{2}(\mu - e^{-\tau})^2 $$

$$ \mathcal{L}_4 = \frac{(\mathcal{J} - 1)^2}{4}. $$

We will often use the simple remark

$$ u(\tau, 0) = e^{-2\tau}\mu' + e^{\tau}\mu. $$

As before, we differentiate

$$ \begin{aligned} \mathcal{L}'_1 &= \int \partial_{\tau} u \log \frac{u}{v} + \partial_{\tau} v (1 - \frac{u}{v}) \\ &= -\mu' \log u(\tau, 0) - \int e^{\tau} u (\partial_y \log u + 1 + e^{-\tau} y) (\partial_y \log u + \mu e^{\tau} + e^{-\tau} y) \\ &\quad + \int e^{-\tau} \frac{y^2}{2} (v - u) \\ &= -\mu' \log u(\tau, 0) + \int e^{-\tau} \frac{y^2}{2} (v - u) \\ &\quad - \int e^{\tau} u (\partial_y \log u + \mu e^{\tau} + e^{-\tau} y)^2 + (\mu e^{\tau} - 1) \int u e^{\tau} (\partial_y \log u + \mu e^{\tau} + e^{-\tau} y) \\ &= -\mu' \log u(\tau, 0) + \int e^{-\tau} \frac{y^2}{2} (v - u) - \int u e^{\tau} (\partial_y \log u + \mu e^{\tau} + e^{-\tau} y)^2 \\ &\quad + (\mu e^{\tau} - 1)(-e^{\tau} u(\tau, 0) + \mu e^{2\tau} (1 - \mu) + J_1) \\ &= -\mu' \log u(\tau, 0) + \int e^{-\tau} \frac{y^2}{2} (v - u) - \int u e^{\tau} (\partial_y \log u + \mu e^{\tau} + e^{-\tau} y)^2 \\ &\quad + (\mu e^{\tau} - 1)(-e^{-\tau} \mu' - e^{2\tau} \mu + \mu e^{2\tau} (1 - \mu) + J_1) \\ &= -\mu' \log u(\tau, 0) + \int e^{-\tau} \frac{y^2}{2} (v - u) - \int u e^{\tau} (\partial_y \log u + \mu e^{\tau} + e^{-\tau} y)^2 \\ &\quad - \mu' (\mu - e^{-\tau}) + (\mu e^{\tau} - 1)(J_1 - \mu^2 e^{2\tau}) \\ &= -\mu' \log u(\tau, 0) + \int e^{-\tau} \frac{y^2}{2} (v - u) - \int u e^{\tau} (\partial_y \log u + \mu e^{\tau} + e^{-\tau} y)^2 \\ &\quad - \mu' (\mu - e^{-\tau}) + (\mu e^{\tau} - 1)(\mathcal{J} - \mu^2 e^{2\tau}) - e^{-\tau} J_0(\mu e^{\tau} - 1) \end{aligned} $$

As before, we may write

$$ \mathcal{L}'_2 = \mu' \log(\mu e^{\tau}) + \mu - e^{-\tau}, $$

$$ \mathcal{L}'_3 = \mu'(\mu - e^{-\tau}) + e^{-\tau}(\mu - e^{-\tau}), $$
And this leads to, denoting $I = \int u e^{\tau} (\partial_y \log u + \mu e^{\tau} + e^{-\tau} y)^2$ and

$$
\begin{align*}
\frac{d}{d\tau}(\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3) &= -\mu' \log \frac{u(\tau, 0)}{\mu e^{\tau}} + \int e^{-\tau} \frac{y^2}{2}(v-u) - I \\
&\quad (\mu e^{\tau} - 1)(J_1 - \mu^2 e^{2\tau} + e^{-\tau} + e^{-2\tau}) \\
&= -\mu' \log \frac{u(\tau, 0)}{\mu e^{\tau}} + \int e^{-\tau} \frac{y^2}{2}(v-u) - I \\
&\quad (\mu e^{\tau} - 1)(\mathcal{J} - \mu^2 e^{2\tau}) + (\mu - e^{-\tau})(1 + e^{-\tau} - J_0) \\
&\le \int e^{-\tau} \frac{y^2}{2}(v-u) - I \\
&\quad - (\mu e^{\tau} - 1)^2(\mu e^{\tau} + 1) + (\mu e^{\tau} - 1)(\mathcal{J} - 1) + (\mu - e^{-\tau})(e^{-\tau} + \mu)
\end{align*}
$$

Using logarithmic Sobolev inequality, we have (computations similar than above)

$$ -I \leq -2\mathcal{L}_1 + 2 \left( J_0 \log \frac{J_0}{\int v} - J_0 + \int v \right) + 2(1 - \mu e^{\tau}) \left( J_1 - \frac{\int yv}{\int v} \right) $$

This can be very useful since then, we can establish from the first lines of the derivatives

$$
\begin{align}
\frac{d}{d\tau}(\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3) &\le -\mu' \log \frac{u(\tau, 0)}{\mu e^{\tau}} + \int e^{-\tau} \frac{y^2}{2}(v-u) - \frac{I}{2} - \mathcal{L}_1 \nonumber \\
&\quad (\mu e^{\tau} - 1) \left( \frac{\int yv}{\int v} - \mu^2 e^{2\tau} + e^{-\tau} + e^{-2\tau} \right) \nonumber \\
&\quad + \left( J_0 \log \frac{J_0}{\int v} - J_0 + \int v \right) \tag{31}
\end{align}
$$

In order to exploit (31), we then need the following consequences of the fact that all moments of $v$ converge to the ones of the exponential distribution.

**Lemma 5.2.**

$$
\begin{align*}
\frac{d}{d\tau} \int v &= e^{-\tau} \int \frac{y^2}{2} v = e^{-\tau} + o(e^{-\tau}). \\
\int v &= 1 - e^{-\tau} + o(e^{-\tau}) \\
\frac{d}{d\tau} \int yv &= e^{-\tau} \int \frac{y^3}{2} v = 3e^{-\tau} + o(e^{-\tau}) \\
\int yv &= 1 - 3e^{-\tau} + o(e^{-\tau}) \\
\frac{\int yv}{\int v} &= 1 - 2e^{-\tau} + o(e^{-\tau})
\end{align*}
$$

Then we summarize some auxiliary results of comparison to $\mu e^{\tau} - 1$ in the following lemma.
**Lemma 5.3.** The following holds true: denoting $\delta(\tau) = \mu e^{\tau} - 1$

$$ \int e^{-\tau} \frac{y^2}{2} (v-u) \le 2\sqrt{3} e^{-\tau} \sqrt{\mathcal{L}_1} \\ 2 \left( J_0 \log \frac{J_0}{\int v} - J_0 + \int v \right) = O(e^{-2\tau}(\delta^2+1)), \\ \mathcal{L}_2 = O(e^{-\tau} \delta^2) $$

*Proof.* First line is a direct consequence of lemma A.3. Second line stems for the remark that for large time

$$ \begin{aligned} 2 \left( J_0 \log \frac{J_0}{\int v} - J_0 + \int v \right) &= 2(1 - e^{-\tau} + o(e^{-\tau}))h \left( \frac{1 - \mu}{1 - e^{-\tau} + o(e^{-\tau})} \right) \\ &= 2(1 - e^{-\tau} + o(e^{-\tau}))h (1 + (e^{-\tau} - \mu + o(e^{-\tau}))) \\ &\le C (e^{-\tau} - \mu + o(e^{-\tau}))^2 \\ &\le O(\delta^2 e^{-2\tau} + e^{-2\tau}) \end{aligned} $$

The third line simply stems from the nonoptimal bound $h(x) \le C(x-1)^2$. For the fourth line, we use the estimates of the previous lemma. $\square$

Based on this, we can establish the first crucial (though not quantitative) result

**Lemma 5.4.** The following holds true: for $\varepsilon \in ]0, 1[$,

$$ (\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3) e^{(1-\varepsilon)\tau} \to 0. $$

*Proof.* We simply start from (31) and insert the estimates

$$ \begin{aligned} \frac{d}{d\tau}(\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3) = & -\mu' \log \frac{u(\tau, 0)}{\mu e^{\tau}} + \int e^{-\tau} \frac{y^2}{2}(v-u) - \frac{I}{2} - \mathcal{L}_1 \\ & (\mu e^{\tau} - 1) \left( \frac{\int yv}{\int v} - \mu^2 e^{2\tau} + e^{-\tau} + e^{-2\tau} \right) \\ & + \left( J_0 \log \frac{J_0}{\int v} - J_0 + \int v \right) \end{aligned} $$

We recall

$$ \begin{aligned} (\mu e^{\tau} - 1) \left( \frac{\int yv}{\int v} - \mu^2 e^{2\tau} + e^{-\tau} + e^{-2\tau} \right) &= \delta(\tau) (1 - 2e^{-\tau} + o(e^{-\tau}) - \mu^2 e^{2\tau} + e^{-\tau} + e^{-2\tau}) \\ &= -\delta^2(\tau)(\mu e^{\tau} + 1) - \delta(\tau)(e^{-\tau} + o(e^{-\tau})) \end{aligned} $$

Since we have

$$ \mathcal{L}_2 + \mathcal{L}_3 = O(e^{-\tau} \delta^2) $$
We end up using corollary A.4

$$
\begin{aligned}
\frac{d}{d\tau}(\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3) &\le 2\sqrt{3}e^{-\tau}\sqrt{\mathcal{L}_1} - (\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3) \\
&\quad - \delta^2(\mu e^{\tau} + 1) + O(\delta^2 e^{-\tau} + e^{-2\tau} + |\delta|e^{-\tau}) \\
&\le 2\sqrt{3}e^{-\tau}\sqrt{\mathcal{L}_1} - (\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3) \\
&\quad - \delta^2/2 + O(e^{-2\tau})
\end{aligned}
$$

Leading immediately to

$$
\frac{d}{d\tau}(\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3) \le \frac{12}{\varepsilon}e^{-2\tau} + O(e^{-2\tau}) - (1-\varepsilon)(\mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3)
$$

And to the result.

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

A crucial consequence of this is the following

**Corollary 5.5.** *The following holds true*

$$
\mathcal{J} - 1 = O(e^{-\tau} + \sqrt{\mathcal{L}_1}) \to 0.
$$

*Proof.* Firstly, we recall

$$
\mathcal{J} - 1 = J_1 - 1 - e^{-\tau} J_0 = J_1 - 1 + O(e^{-\tau}).
$$

Moreover, using lemma A.3, we have for $\tau$ large enough, choosing $\mu = \sqrt{\mathcal{L}_1}(\tau) = < 1$

$$
\begin{aligned}
\left| \int x(u - v_{1,e^{-\tau}}) \right| &\le \int \frac{1 - \sqrt{\mathcal{L}_1} x e^{-\sqrt{\mathcal{L}_1} x} - e^{-\sqrt{\mathcal{L}_1} x}}{\sqrt{\mathcal{L}_1}} v_{1-\sqrt{\mathcal{L}_1},e^{-\tau}} + \frac{\mathcal{L}_1}{\sqrt{\mathcal{L}_1}} \\
&= \sqrt{\mathcal{L}_1} \left( \int \frac{x^2}{2} v_{1,0} + 1 + o(1) \right) \\
|J_1 - 1 + O(e^{-\tau})| &= \sqrt{\mathcal{L}_1} (2 + o(1))
\end{aligned}
$$

which leads to the result.

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

We introduce a few notations

$$
\mathcal{L}_4 = (\mathcal{J} - 1)^2 / 4, \quad \mathcal{L}'_4 = \frac{1}{2}(\mathcal{J} - 1)(\mu^2 e^{2\tau} - 1) - 2\mathcal{L}_4.
$$

So that if we differentiate $\mathcal{L} = \mathcal{L}_1 + \mathcal{L}_2 + \mathcal{L}_3 + \mathcal{L}_4$, we know that $\mathcal{L} \to 0$ for large $\tau$ and moreover

$$
\mathcal{L}' + 2\mathcal{L} \le R
$$

with
$$
\begin{align*}
R = \int e^{-\tau} \frac{y^2}{2} (v-u) \\
& + 2\mathcal{L}_1 - I \\
& - (\mu e^{\tau} - 1)^2 (\mu e^{\tau} + 1) + (\mu e^{\tau} - 1)(\mathcal{J} - 1) + (\mu - e^{-\tau})(e^{-\tau} + \mu) \\
& + 2\mathcal{L}_2 + 2\mathcal{L}_3 \\
& + \frac{1}{2}(\mathcal{J} - 1)(\mu^2 e^{2\tau} - 1).
\end{align*}
$$

It is worth regrouping some terms by growth.

$$
\begin{align*}
2\mathcal{L}_1 - I &\le O(e^{-2\tau}\delta^2 + e^{-2\tau}) - 2\delta(\mathcal{J}-1) + O(e^{-\tau}|\delta|). \\
\mathcal{L}_2 + \mathcal{L}_3 &= O(e^{-\tau}\delta^2) \\
(\mu - e^{-\tau})(e^{-\tau} + \mu) &= o(e^{-\tau}|\delta|).
\end{align*}
$$

We arrive at

$$
\begin{align*}
\mathcal{L}' + 2\mathcal{L} &\le C e^{-\tau} \sqrt{\mathcal{L}_1} - \delta^2 (\mu e^{\tau} + 1) \\
&\quad + O \left( e^{-\tau} \delta^2 + e^{-\tau} |\delta| + e^{-2\tau} \right) \\
&\quad (\mathcal{J} - 1) \underbrace{\left( -2\delta + \delta + \frac{\mu^2 e^{2\tau} - 1}{2} \right)}_{= \frac{\delta^2}{2}}
\end{align*}
$$

Using the fact $\mathcal{J} = o(1)$

$$
e^{-\tau}|\delta| \le \varepsilon\delta^2 + \frac{1}{\varepsilon}e^{-2\tau},
$$

we can bound the term in $O(e^{-\tau}|\delta|)$ by $\delta^2/2 + O(e^{-2\tau})$. So that we end up with

$$
\mathcal{L}' + 2\mathcal{L} \leq C e^{-\tau} \sqrt{\mathcal{L}} - \frac{\delta^2}{2} + C e^{-2\tau}
$$

This leads to (still with non explicit constant that might change from line to line)

$$
\mathcal{L}' + 2\mathcal{L} \leq C e^{-2\tau} (\sqrt{\mathcal{L}_1} e^{2\tau} + 1) - \frac{\delta^2}{2}
$$

And exactly as in the previous section this leads to

$$
\mathcal{L} \leq C(1 + \tau)^2 e^{-2\tau}. \tag{32}
$$

Using Czisar Kulback inequality, we can deduce

$$
\int_{0}^{\infty} |u - v_{1, e^{-\tau}}| = O((1 + \tau)e^{-\tau}).
$$
Since it is straightforward that

$$ \int_{0}^{\infty} |v_{1,e^{-\tau}} - v_{1,0}| \le e^{-\tau}, $$

we can conclude

$$ \int_{0}^{\infty} |u - v_{1,0}| = O((1 + \tau)e^{-\tau}). $$

which corresponds to (9)

## A Logarithmic Sobolev inequality and other technical lemmas

We recall here the LSI inequality (see [10, 3]) for Gaussian on the half-line.

**Lemma A.1.** Consider for $\lambda, \eta > 0$ the distribution

$$ v_{\lambda, \eta} = \lambda e^{-\lambda x - \eta x^2/2}, $$

then for sufficiently smooth $u$, the following inequality holds true

$$ \int_{0}^{\infty} u |\partial_{x} \log \frac{u}{v_{\lambda, \eta}}|^{2} \ge 2\eta \left( \int_{0}^{\infty} u \log \frac{u}{v_{\lambda, \eta}} - \int_{0}^{\infty} u \log \frac{\int_{0}^{\infty} u}{\int_{0}^{\infty} v_{\lambda, \eta}} \right) \quad (33) $$

The following lemma is used several times

**Lemma A.2.** The following inequality holds true for $f \in L^1$ and $\lambda, \mu \ge 0$

$$ (\lambda - \mu) \frac{\int_{0}^{\infty} x e^{-\lambda x} f(x) dx}{\int_{0}^{\infty} e^{-\lambda x} f(x) dx} \le - \log \frac{\int_{0}^{\infty} e^{-\lambda x} f(x) dx}{\int_{0}^{\infty} e^{-\mu x} f(x) dx} \le (\lambda - \mu) \frac{\int_{0}^{\infty} x e^{-\mu x} f(x) dx}{\int_{0}^{\infty} e^{-\mu x} f(x) dx} \quad (34) $$

*Proof.* We simply establish one side. By Jensen inequality, we have

$$ \begin{align*} - \log \frac{\int_{0}^{\infty} e^{-\lambda x} f(x) dx}{\int_{0}^{\infty} e^{-\mu x} f(x) dx} &= - \log \int_{0}^{\infty} \frac{e^{-\mu x} f(x) dx}{\int_{0}^{\infty} e^{-\mu x} f(x) dx} \\ &\le \int_{0}^{\infty} (-\log e^{(\mu-\lambda)x}) \frac{e^{-\mu x} f(x) dx}{\int_{0}^{\infty} e^{-\mu x} f(x) dx} \\ &= (\lambda - \mu) \frac{\int_{0}^{\infty} x e^{-\mu x} f(x) dx}{\int_{0}^{\infty} e^{-\mu x} f(x) dx} \end{align*} $$

□

Keeping our notations $v_{\lambda, \eta}$ we have in particular
**Lemma A.3.** Let $0 < \mu \le \lambda$ and $0 < \gamma \le \eta$, then the following hold true

$$ \left| \int_{0}^{\infty} x(u - v_{\lambda, \eta}) \right| \le \int \frac{e^{\mu x} - \mu x - 1}{\mu} v_{\lambda, \eta} + \frac{\mathcal{H}(u|v_{\lambda, \eta})}{\mu} \quad (35) $$

$$ \left| \int_{0}^{\infty} \frac{x^2}{2}(u - v_{\lambda, \eta}) \right| \le \int \frac{e^{\gamma \frac{x^2}{2}} - \gamma \frac{x^2}{2} - 1}{\gamma} v_{\lambda, \eta} + \frac{\mathcal{H}(u|v_{\lambda, \eta})}{\gamma} \quad (36) $$

Additionally, we have for any $f \ge 0$

$$ \int_{0}^{\infty} f(v_{\lambda, \eta} - u) \le \|f\|_{L^2(v_{\lambda, \eta})} \sqrt{2\mathcal{H}} \quad (37) $$

*Proof.* The two first inequalities come from a direct application of the former lemma to $x, -x, \frac{x^2}{2}, -\frac{x^2}{2}$. For the second point, we write

$$ \begin{aligned} \int_{0}^{\infty} f(v_{\lambda, \eta} - u) &\le \int_{0}^{\infty} f(v_{\lambda, \eta} - u)_{+} \\ &\le \int_{0}^{\infty} f\left(1 - \frac{u}{v_{\lambda, \eta}}\right)_{+} v_{\lambda, \eta} \\ &\le \|f\|_{L^2(v_{\lambda, \eta})} \left\|\left(1 - \frac{u}{v_{\lambda, \eta}}\right)_{+}\right\|_{L^2(v_{\lambda, \eta})} \end{aligned} $$

Then we use the elementary inequality

$$ \forall x \in [0, 1], x \log x - x + 1 \ge \frac{1}{2}(1 - x)^2, $$

leading to

$$ \forall x \ge 0, x \log x - x + 1 \ge \frac{1}{2}(1 - x)^2, $$

and to

$$ \|(1 - \frac{u}{v_{\lambda, \eta}})_{+}\|_{L^2(v_{\lambda, \eta})}^2 \le 2 \int (u \log \frac{u}{v} - u + v) = 2\mathcal{H}. \quad \square $$

We end up with the specific

**Corollary A.4.** The following holds true

$$ \int_{0}^{\infty} \frac{x^2}{2}(v_{1, \eta} - u) \le 2\sqrt{3}\sqrt{\mathcal{H}} \quad (38) $$

*Proof.* We simply notice that

$$ \int_{0}^{\infty} \frac{x^2}{2}(v_{1, \eta} - u) \le \left( \int_{0}^{\infty} \frac{x^4}{4} e^{-x - \eta x^2/2} dx \right)^{1/2} \sqrt{2\mathcal{H}} $$

and notice

$$ \left( \int_{0}^{\infty} \frac{x^4}{4} e^{-x - \eta x^2/2} dx \right)^{1/2} \le \left( \int_{0}^{\infty} \frac{x^4}{4} e^{-x} dx \right)^{1/2} = 6^{1/2}. \quad \square $$
We recall a version of Czsisar Kullback inequality [13] that does not require same mass for the functions (proof borrowed from [9])

**Lemma A.5.** Let $f, g \ge 0$ be $L^1(\mathbb{R}_+)$ functions such that

$$ \mathcal{H}(f|g) = \int_{0}^{\infty} f \log \frac{f}{g} - f + g < +\infty $$

then we have

$$ \left( \int_{0}^{\infty} |f - g| \right)^{2} \le 2\mathcal{H}(f|g) \frac{\int f + 2 \int g}{3} $$

## References

[1] José Antonio Carrillo, Benoît Perthame, Delphine Salort, and Didier Smets. Qualitative properties of solutions for the noisy integrate and fire model in computational neuroscience. *Nonlinearity*, 28(9):3365, August 2015.

[2] Ferdinando Auricchio, Giuseppe Toscani, and Mattia Zanella. Trends to equilibrium for a nonlocal Fokker-Planck equation. *Applied Mathematics Letters*, 145:108746, November 2023.

[3] Dominique Bakry, Ivan Gentil, and Michel Ledoux. Markov Semigroups. In Dominique Bakry, Ivan Gentil, and Michel Ledoux, editors, *Analysis and Geometry of Markov Diffusion Operators*, pages 3–75. Springer International Publishing, Cham, 2014.

[4] María J. Cáceres and Ricarda Schneider. Blow-up, steady states and long time behaviour of excitatory-inhibitory nonlinear neuron models. *KRM*, 10(3):587–612, Fri Sep 01 00:00:00 UTC 2017.

[5] V. Calvez, R. Hawkins, N. Meunier, and R. Voituriez. Analysis of a non-local model for spontaneous cell polarization. *SIAM Journal on Applied Mathematics*, 72(2):594–622, 2012.

[6] Vincent Calvez, Thomas Lepoutre, Nicolas Meunier, and Nicolas Muller. Non-linear analysis of a model for yeast cell communication. *ESAIM: Mathematical Modelling and Numerical Analysis*, 54(2):619–648, 2020.

[7] Vincent Calvez, Nicolas Meunier, and Raphael Voituriez. A one-dimensional Keller-Segel equation with a drift issued from the boundary. *C. R. Math. Acad. Sci. Paris*, 348(11-12):629–634, 2010.

[8] Jesus Ildefonso Diaz and Toshitaka Nagai. Symmetrization Techniques on Unbounded Domains: Application to a Chemotaxis System on RN.

[9] Gustavo L. Gilardoni. On Pinsker's and Vajda's Type Inequalities for Csiszár's f-Divergences. *IEEE Transactions on Information Theory*, 56(11):5377–5386, November 2010.
[10] Leonard Gross. Logarithmic Sobolev Inequalities. *American Journal of Mathematics*, 97(4):1061–1083, 1975.

[11] Stephan Hausberg and Matthias Röger. Well-posedness and fast-diffusion limit for a bulk-surface reaction-diffusion system. *Nonlinear Differ. Equ. Appl.*, 25(3):17, April 2018.

[12] R. J. Hawkins, O. Bénichou, M. Piel, and R. Voituriez. Rebuilding cytoskeleton roads: Active-transport-induced polarization of cells. *Phys. Rev. E*, 80(4):040903, October 2009.

[13] S. Kullback. On the convergence of discrimination information. *IEEE Trans. Information Theory*, IT-14:765–766, 1968.

[14] Ido Lavi, Nicolas Meunier, Raphael Voituriez, and Jaume Casademunt. Motility and morphodynamics of confined cells. *Phys. Rev. E*, 101(2):022404, February 2020.

[15] Sean Ledger and Andreas Søjmark. Uniqueness for contagious McKean-Vlasov systems in the weak feedback regime. *Bulletin of the London Mathematical Society*, 52(3):448–463, 2020.

[16] Thomas Lepoutre and Nicolas Meunier. Analysis of a model of cell crawling migration. *Comm. Math. Sci.*, 20(6):1589–1611, September 2022.

[17] Thomas Lepoutre, Nicolas Meunier, and Nicolas Muller. Cell polarisation model: The 1D case. *Journal de Mathématiques Pures et Appliquées*, 101(2):152–171, 2014.

[18] Nicolas Meunier and Philippe Souplet. Convergence, concentration and critical mass phenomena in a chemotaxis model with boundary. *ANNALI SCUOLA NORMALE SUPERIORE-CLASSE DI SCIENZE*, pages 31–31, 2022.

[19] Pierre Recho, Thibaut Putelat, and Lev Truskinovsky. Mechanics of motility initiation and motility arrest in crawling cells. *Journal of the Mechanics and Physics of Solids*, 84:469–505, November 2015.

[20] C. Alex Safsten, Volodmyr Rybalko, and Leonid Berlyand. Asymptotic stability of contraction-driven cell motion. *Phys. Rev. E*, 105(2):024403, February 2022.

[21] Samuel Tréton and Mingmin Zhang. A piston to counteract diffusion: The influence of an inward-shifting boundary on the heat equation in half-space, May 2025.