# On the dependence of the nonlinear Schrodinger flow upon the power of the nonlinearity

Rémi Carles, Quentin Chauleur, Guillaume Ferriere

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## To cite this version:

Rémi Carles, Quentin Chauleur, Guillaume Ferriere. On the dependence of the nonlinear Schrodinger flow upon the power of the nonlinearity. 2025. hal-05288185

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# ON THE DEPENDENCE OF THE NONLINEAR SCHRÖDINGER FLOW UPON THE POWER OF THE NONLINEARITY

RÉMI CARLES, QUENTIN CHAULEUR, AND GUILLAUME FERRIERE

ABSTRACT. We prove continuity properties for the flow map associated to the defocusing energy-subcritical power-like nonlinear Schrödinger equation, when the power varies. We show local in time continuity in the energy space for any power, and global in time continuity for sufficiently large powers. When the linear dispersive rate is counterbalanced by a time-dependent rescaling, we show a uniform in time continuity of the squared modulus of this rescaled function, in Kantorovich distance, for any power, including long range cases in terms of scattering. The most difficult result addresses the convergence of suitably renormalized solutions to the solution of the logarithmic Schrödinger equation, when the power goes to zero, uniformly in time, in Kantorovich distance. The proof relies on estimates for perturbed porous medium equations, involving the harmonic Fokker-Planck operator.

## 1. INTRODUCTION

1.1. **Setting.** We consider the Cauchy problem associated to the defocusing non-linear Schrödinger equation with power-like nonlinearity,

$$ (1.1) \qquad i\partial_t u + \frac{1}{2}\Delta u = |u|^{2\sigma}u \quad ; \quad u|_{t=0} = \phi, $$

for $x \in \mathbb{R}^d$, $d \ge 1$, in the energy-subcritical case, $0 < \sigma < \frac{2}{(d-2)_+}$. For clarity, we denote by $u_\sigma$ the above solution, as the main goal is to understand the dependence of $u_\sigma$ with respect to $\sigma$, locally in time and, especially, globally in time.

1.1.1. **Power nonlinearity.** Several papers have considered the continuity of the flow map with respect to the initial data, see e.g. [12, 10, 23, 24] for positive results, and [25, 15, 6, 8] for lack of continuity (according to the function space considered). In the present paper, we address the dependence of the flow map $\phi \mapsto u$ with respect to the nonlinearity, that is with respect to $\sigma$. The initial value $\phi$ may depend on $\sigma$ as well. We consider a defocusing nonlinearity in order to ensure that the solution is defined globally in time, and enjoys a dispersive behavior. In the focusing case, dynamical properties may be different.

Since we consider energy-subcritical nonlinearities, it is reasonable to expect that at least locally in time, the flow map is continuous in $H^1$. This is indeed what we prove in Theorem 1.3 by using Strichartz estimates. The question of the interplay with the large time behavior is less obvious.

<footnote>A CC-BY public copyright license has been applied by the authors to the present document and will be applied to all subsequent versions up to the Author Accepted Manuscript arising from this submission.</footnote>
When $\sigma > 1/d$, the solution to (1.1) is asymptotically linear, in the sense that there exists $u_+ \in L^2(\mathbb{R}^d)$ such that

$$ (1.2) \qquad \|u(t) - e^{i\frac{t}{2}\Delta}u_+\|_{L^2(\mathbb{R}^d)} \xrightarrow{t \to \infty} 0, $$

see e.g. [4] and references therein. We show that when $\sigma$ is sufficiently large ($\sigma > \sigma_0(d) > 1/d$), the flow map is continuous with respect to $\sigma$, uniformly in time (Theorem 1.4), which implies that the nonlinear scattering map is continuous with respect to $\sigma$ (Corollary 1.5).

It is well known that as soon as $\sigma \le 1/d$, (1.2) can hold only in the trivial case of the zero solution, $\phi = u = u_+ = 0$, as proven in [2]. In the critical case $\sigma = 1/d$, as established initially in [28, 18, 22], the large time behavior of $u$ is described at leading order by a nonlinear phase modification of the free dynamics. If $d = 1$ and $\phi$ is sufficiently small, smooth and localized, there exists $u_+ \in L^2(\mathbb{R})$ such that

$$ \|u(t) - e^{i\Phi_+(t)}e^{i\frac{t}{2}\Delta}u_+\|_{L^2(\mathbb{R}^d)} \xrightarrow{t \to \infty} 0, \quad \text{where } \Phi_+(t,x) = -\left|\hat{u}_+\left(\frac{x}{t}\right)\right|^2 \ln t, $$

and the Fourier transform (on $\mathbb{R}^d$) is normalized as

$$ \hat{f}(\xi) = \frac{1}{(2\pi)^{d/2}} \int_{\mathbb{R}^d} f(x)e^{-ix \cdot \xi} dx. $$

This implies for instance that the flow map cannot be continuous uniformly in time at $\sigma = 1/d$ (as long range effects are absent for $\sigma > 1/d$, see [4]). The general consensus is that even for $\sigma < 1/d$, modified scattering should be characterized at leading order by a (nonlinear) phase modification, even though no general proof of this fact seems to be available so far. Our approach supports this train of thought: instead of considering $u$, we first introduce a time dependent rescaling which counterbalances the linear dispersive behavior, and we consider the squared modulus of this new function, that is

$$ \langle t \rangle^d |u_\sigma(t, x \langle t \rangle)|^2, \quad \text{where} \quad \langle t \rangle = \sqrt{1 + t^2}. $$

Dividing this quantity by $\|\phi\|_{L^2}^2$ yields a time dependent family of probability densities $(\rho_\sigma(t))_{t \in \mathbb{R}}$. We show that this family is continuous with respect to $\sigma$, uniformly in time $t \in \mathbb{R}$, at any $\sigma \in (0, \frac{2}{(d-2)_+})$, in Wasserstein distance $W_1$ (Theorem 1.6), which implies uniform continuity in some negative order Sobolev spaces (Corollary 1.7). In particular, this property is insensitive to long range effects (when $\sigma \le 1/d$).

### 1.1.2. Logarithmic nonlinearity.
In the limit $\sigma \to 0$, at points where $u \ne 0$,

$$ |u|^{2\sigma} = \exp(\sigma \ln|u|^2) = 1 + \sigma \ln|u|^2 + \mathcal{O}(\sigma^2). $$

Up to a gauge transform ($u \to u e^{it}$), we may replace the nonlinearity in (1.1) with $\langle |u|^{2\sigma} - 1 \rangle u$. We may then proceed like in [17] where the stationary, focusing case is considered, and let $\sigma \to 0$ in

$$ (1.3) \qquad i \partial_t \mathbf{u}_\sigma + \frac{1}{2} \Delta \mathbf{u}_\sigma = \frac{1}{\sigma} (|\mathbf{u}_\sigma|^{2\sigma} - 1) \mathbf{u}_\sigma \quad ; \quad \mathbf{u}_{\sigma|t=0} = \phi_\sigma. $$

Formally, if $\phi_\sigma \to \phi$ as $\sigma \to 0$, $\mathbf{u}_\sigma$ is expected to converge to the solution of the logarithmic Schrödinger equation

$$ (1.4) \qquad i \partial_t \mathbf{u} + \frac{1}{2} \Delta \mathbf{u} = \mathbf{u} \ln(|\mathbf{u}|^2) \quad ; \quad \mathbf{u}|_{t=0} = \phi. $$
We denote by $\mathbf{u}_{\log}$ or $\mathbf{u}_0$ this solution. This equation was introduced in [3], and the mathematical study of the Cauchy problem (1.4) started in [11] (with a different sign in front of the nonlinearity though).

*Remark 1.1.* One may adopt an alternative point of view. If $u$ solves (1.1), then

$$ \tilde{u}_{\sigma}(t, x) = \sigma^{1/(2\sigma)} u(t, x) e^{it/\sigma} $$

solves the PDE in (1.3), but with initial value $\sigma^{1/(2\sigma)}\phi$. The fact that the factor $\sigma^{1/(2\sigma)}$ is unbounded as $\sigma \to 0$ is an alternative evidence that nonlinear effects must be enhanced in the nonlinear Schrödinger equation in order to get some convergence toward the logarithmic Schrödinger equation.

*Remark 1.2.* If the continuity of the flow map is considered at $\sigma > 0$ (instead of $\sigma = 0$), then it is essentially equivalent to consider (1.1) or (1.3), since $\sigma \mapsto 1/\sigma$ is continuous on $(0, \infty)$.

Several aspects indicate that the limit $\sigma \to 0$ in (1.3) is more involved than in the previous case. Indeed, it was proven in [7] that solutions to (1.4) obey an enhanced dispersive rate: roughly speaking, instead of a decay rate of order $t^{-d/2}$ in (1.1), solutions to (1.4) decay like $\tau_0(t)^{-d/2} \approx t^{-d/2} (\ln t)^{-d/4}$. Moreover, rescaling the solution to (1.4) to compensate this new dispersive rate, the large time behavior exhibits a universal profile,

$$ \tau_0(t)^d |\mathbf{u}_0(t, x\tau_0(t))|^2 \underset{t \to \infty}{\longrightarrow} \|\phi\|_{L^2}^2 \gamma^2 \text{ in Wasserstein distance } W_2, $$

where $\gamma(x) = e^{-|x|^2/2}$ is independent of $\phi$. Finally, the $H^1$-norm of any nonzero solution to (1.4) is unbounded in the large time limit. These three properties are in sharp contrast with the dynamical properties associated to (1.1).

## 1.2. Main results. Denote

$$ \Sigma = \left\{ f \in H^1(\mathbb{R}^d), \int_{\mathbb{R}^d} |x|^2 |f(x)|^2 dx < \infty \right\}, $$

equipped with the norm

$$ \|f\|_{\Sigma} = \|f\|_{L^2(\mathbb{R}^d)} + \|\nabla f\|_{L^2(\mathbb{R}^d)} + \|xf\|_{L^2(\mathbb{R}^d)}. $$

### 1.2.1. Power nonlinearity. Our first result addresses the continuity of the flow map on bounded time intervals:

**Theorem 1.3** (Local in time continuity). Let $0 < \sigma < \frac{2}{(d-2)_+}$ and $(\phi_\nu)|_{\nu-\sigma|\le\epsilon}$ bounded in $H^1(\mathbb{R}^d)$ for some $\epsilon > 0$ such that $0 < \sigma - \epsilon$ and $\sigma + \epsilon < \frac{2}{(d-2)_+}$. Denote by $u_\nu$ the corresponding solutions to the nonlinear Schrödinger equation (1.1), $u_\nu \in C(\mathbb{R}; H^1(\mathbb{R}^d))$. Let $T > 0$:

* There exists $C$ such that, as $\nu \to \sigma$,

$$ \sup_{t \in [-T, T]} \|u_\nu(t) - u_\sigma(t)\|_{L^2(\mathbb{R}^d)} \le C \|\phi_\nu - \phi_\sigma\|_{L^2(\mathbb{R}^d)} + C|\nu - \sigma|, $$

$$ \sup_{t \in [-T, T]} \|u_\nu(t) - u_\sigma(t)\|_{H^1(\mathbb{R}^d)} \le C \|\phi_\nu - \phi_\sigma\|_{H^1(\mathbb{R}^d)} + C|\nu - \sigma|^\theta, $$

for some $\theta > 0$ (given by (3.2)-(3.3)), which can be taken equal to one if $2\sigma > 1$.
• If in addition $(\phi_\nu)|_{\nu-\sigma|\le\epsilon}$ is bounded in $\Sigma$, then

$$ \sup_{t \in [-T, T]} \|u_\nu(t) - u_\sigma(t)\|_{\Sigma} \le C \|\phi_\nu - \phi_\sigma\|_{\Sigma} + C |\nu - \sigma|^\theta, $$

for the same $\theta$ as above.

In the case where the power $\sigma$ is sufficiently large, the previous continuity can be made uniform in time:

**Theorem 1.4** (Global continuity in the (very) short range case). Let $\sigma_0(d) < \sigma < \frac{2}{(d-2)_+}$ and $(\phi_\nu)|_{\nu-\sigma|\le\epsilon}$ bounded in $\Sigma$ with $\epsilon > 0$ such that $\sigma - \epsilon > \sigma_0(d)$ and $\sigma + \epsilon < \frac{2}{(d-2)_+}$, where

$$ \sigma_0(d) = \frac{2 - d + \sqrt{d^2 + 12d + 4}}{4d} \in \left(\frac{1}{d}, \frac{2}{d}\right). $$

Then we have the global in time estimate:

$$ \sup_{t \in \mathbb{R}} \left\| e^{-i \frac{t}{2} \Delta} (u_\nu(t) - u_\sigma(t)) \right\|_{\Sigma} \le C \|\phi_\nu - \phi_\sigma\|_{\Sigma} + C |\nu - \sigma|^\theta, $$

for the same $\theta$ as in Theorem 1.3.

We briefly recall the notion of nonlinear scattering operator. For a given asymptotic state $u_-$, consider the equation (1.1) where the initial value is replaced by a final value (at infinite time),

$$ e^{-i \frac{t}{2} \Delta} u(t) \Big|_{t \to -\infty} = u_{-}. $$

If this new problem has a solution defined up to $t = 0$, then the map $W_- : u_- \mapsto u_{|t=0}$ is called wave operator. Conversely, if the solution to the Cauchy problem (1.1) behaves asymptotically linearly,

$$ u(t) \underset{t \to +\infty}{\sim} e^{i \frac{t}{2} \Delta} u_{+}, $$

for some asymptotic state $u_+$, then $u_+ = W_+^{-1}u_{|t=0}$ and the scattering operator $S = W_+^{-1} \circ W_-$ is the map $u_- \mapsto u_+$. This map is well defined on $\Sigma$ for $\sigma > \sigma_0(d)$, as established initially in [19] (see also [13, 26]). In view of the present context, we emphasize its dependence upon $\sigma$ by using the notation $S_\sigma$.

**Corollary 1.5** (Continuity of the scattering operator). Let $\sigma_0(d) < \sigma < \frac{2}{(d-2)_+}$ and $(u_{-,\nu})|_{\nu-\sigma|\le\epsilon}$ be a family in $\Sigma$ for some $\epsilon > 0$. The scattering operator is continuous at $\sigma$ in the sense that if

$$ \|u_{-,\sigma} - u_{-,\nu}\|_{\Sigma} \xrightarrow{\nu \to \sigma} 0, $$

then

$$ \|S_\sigma u_{-,\sigma} - S_\nu u_{-,\nu}\|_{\Sigma} = \|u_{+,\sigma} - u_{+,\nu}\|_{\Sigma} \xrightarrow{\nu \to \sigma} 0, $$

For $\mu_1$ and $\mu_2$ probability measures on $\mathbb{R}^d$, the Wasserstein (or Kantorovich–Rubinstein) distance $W_1$ can be characterized by

$$ (1.5) \quad W_1(\mu_1, \mu_2) := \sup \left\{ \int_{\mathbb{R}^d} \psi \, d(\mu_1 - \mu_2) \mid \psi \in C(\mathbb{R}^d) \text{ and } \|\psi\|_{\text{Lip}} \le 1 \right\}, $$
where the Lipschitz semi-norm is defined by

$$ \|\psi\|_{\text{Lip}} = \sup_{x \neq y} \frac{|\psi(x) - \psi(y)|}{|x - y|}. $$

See for instance [31]. As announced in the previous subsection, we set

$$ (1.6) \qquad \varrho_{\sigma}(t, y) = \langle t \rangle^d |u_{\sigma}(t, y \langle t \rangle)|^2 \|\varrho_{\sigma}\|_{L^2}^{-2}. $$

For each $t \in \mathbb{R}$, $\varrho_{\sigma}(t, \cdot)$ is a probability density, and the change of space variable compensates the dispersion of the linear Schrödinger equation.

**Theorem 1.6** (Global continuity for the rescaled modulus). Let $0 < \sigma < \frac{2}{(d-2)_+}$ and $(\varrho_{\nu})_{|\nu-\sigma|\le\epsilon}$ bounded in $\Sigma$ for some $\epsilon > 0$ such that $0 < \sigma-\epsilon$ and $\sigma+\epsilon < \frac{2}{(d-2)_+}$. If

$$ \varphi_{\nu} \underset{\nu \to \sigma}{\longrightarrow} \varphi_{\sigma} \quad \text{in } L^2(\mathbb{R}^d), $$

then, with $\varrho_{\sigma}$ given by (1.6),

$$ \sup_{t \in \mathbb{R}} W_1(\varrho_{\nu}(t), \varrho_{\sigma}(t)) \underset{\nu \to \sigma}{\longrightarrow} 0. $$

**Corollary 1.7.** Under the assumptions of Theorem 1.6, for any $s > \frac{1+d}{2}$,

$$ \sup_{t \in \mathbb{R}} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{H^{-s}(\mathbb{R}^d)} \underset{\nu \to \sigma}{\longrightarrow} 0. $$

We infer,

* If $\sigma > 2/d$,

- When $d = 1$, for any $s \in [0, 1)$ and $p \in (1, \infty]$,

$$ \sup_{t \in \mathbb{R}} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{H^s(\mathbb{R})} + \sup_{t \in \mathbb{R}} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{L^p(\mathbb{R})} \underset{\nu \to \sigma}{\longrightarrow} 0. $$

- When $d = 2$, for any $p \in (1, \infty)$,

$$ \sup_{t \in \mathbb{R}} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{L^p(\mathbb{R}^2)} \underset{\nu \to \sigma}{\longrightarrow} 0. $$

- When $d \ge 3$, there exists $p(d) \in (1, 2)$ such that for all $p \in (1, p(d)]$,

$$ \sup_{t \in \mathbb{R}} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{L^p(\mathbb{R}^d)} \underset{\nu \to \sigma}{\longrightarrow} 0. $$

* If $\sigma \le 2/d$, for all $0 < \varepsilon \ll 1$, there exists $q(\varepsilon) \in (1, 1+\sigma)$ such that

$$ \sup_{t \in \mathbb{R}} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{W^{-\varepsilon,q}(\mathbb{R}^d)} \underset{\nu \to \sigma}{\longrightarrow} 0, \quad \forall q \in (1, q(\varepsilon)]. $$

The values of the above Lebesgue indices $p(d)$ and $q(\varepsilon)$ can actually be computed, as it should be clear from the proof in Section 5.2. Similarly, convergence holds in other Sobolev spaces, but we have chosen to make the statements as simple as possible in Corollary 1.7.

### 1.2.2. Logarithmic nonlinearity.
Like for the case of power nonlinearity, we first start with strong convergence locally in time, before considering global in time convergence, in Wasserstein distance. Recall that for the convergence toward the logarithmic Schrödinger equation, we consider (1.3) (as opposed to (1.1)).
**Theorem 1.8.** Let $\varepsilon > 0$ and $(\phi_\sigma)_{0\le\sigma\le\varepsilon}$ bounded in $\Sigma$. Let $\mathbf{u}_\sigma$ denote the solution to (1.3), and $\mathbf{u}_0$ the solution to (1.4), with initial data $\phi_\sigma$ and $\phi_0$ respectively. There exists $\varepsilon_0 > 0$ such that for all $\sigma \in (0, \varepsilon_0)$, the following holds. There exist $C_0, C_1 > 0$ such that for any $T > 0$,

$$ \sup_{t \in [-T, T]} \|\mathbf{u}_\sigma(t) - \mathbf{u}_0(t)\|_{L^2(\mathbb{R}^d)} \le (C_1\sigma + \|\phi_\sigma - \phi_0\|_{L^2(\mathbb{R}^d)}) e^{C_0T}, $$

and, if $\phi_\sigma \to \phi_0$ in $L^2(\mathbb{R}^d)$, for any $\theta \in (0, 1)$,

$$ \sup_{|t| \le \frac{\varrho}{C_0} \ln \frac{1}{\sigma}} \|\mathbf{u}_\sigma(t) - \mathbf{u}_0(t)\|_{L^2(\mathbb{R}^d)} \xrightarrow{\sigma \to 0} 0. $$

The last convergence stated in Theorem 1.8 can be understood as a convergence up to some Ehrenfest time, by analogy with the definition from the semiclassical propagation of coherent states in (linear) Schrödinger equations, see e.g. [29].

**Theorem 1.9.** For $\sigma > 0$, let $\tau_\sigma, \tau_0 \in C^2(\mathbb{R})$ solve the ordinary differential equations

$$ \ddot{\tau}_\sigma = \frac{1}{2\tau_\sigma^{d\sigma+1}}, \quad \dot{\tau}_0 = \frac{1}{2\tau_0}, \quad \tau_\sigma(0) = \tau_0(0) = 1, \quad \dot{\tau}_\sigma(0) = \dot{\tau}_0(0) = 0. $$

Their large time behavior is given by

$$ \tau_\sigma(t) \underset{t \to \infty}{\sim} \frac{t}{\sqrt{\sigma}}, \quad \tau_0(t) \underset{t \to \infty}{\sim} t\sqrt{\ln t}, $$

and $\tau_\sigma$ converge to $\tau_0$: there exists $C > 0$ such that

$$ |\tau_0(t) - \tau_\sigma(t)| \le C\sigma t (\ln(t+2))^{3/2}, \quad \text{hence } \tau_0(t) - \tau_\sigma(t) = o(\tau_0(t)) \text{ for } \sigma \ln t \to 0. $$

Let $\varepsilon > 0$, $(\phi_\sigma)_{0\le\sigma\le\varepsilon}$ bounded in $\Sigma$, with $\phi_\sigma \to \phi_0$ in $L^2(\mathbb{R}^d)$ as $\sigma \to 0$. Define

$$ \varrho_0(t, y) = \tau_0(t)^d |\mathbf{u}_0(t, y\tau_0(t))|^2 \|\phi_0\|_{L^2}^{-2}, \quad \varrho_\sigma(t, y) = \tau_\sigma(t)^d |\mathbf{u}_\sigma(t, y\tau_\sigma(t))|^2 \|\phi_\sigma\|_{L^2}^{-2}. $$

We have the uniform in time convergence in Wasserstein distance:

$$ (1.7) \qquad \sup_{t \ge 0} W_1(\varrho_\sigma, \varrho_0) \xrightarrow{\sigma \to 0} 0. $$

Moreover, if $\|\phi_\sigma - \phi_0\|_{L^2(\mathbb{R}^d)} = O(\sigma)$, we have the convergence rate

$$ (1.8) \qquad \sup_{t \ge 0} W_1(\varrho_\sigma, \varrho_0) \lesssim \frac{1}{\sqrt{\ln \ln \frac{1}{\sigma}}}. $$

We note that the convergence of $\varrho_\sigma$ toward $\varrho_0$ holds uniformly in time, and so supersedes the range on which $\tau_\sigma$ converges to $\tau_0$. Indeed, the asymptotic behaviors of $\tau_0$ and $\tau_\sigma$ become different for $\sigma \approx \frac{1}{\ln t}$ (and then the control of $\tau_0 - \tau_\sigma$ becomes of the same order as $\tau_0$, indicating some sharpness in this error estimate).

**1.3. Content of the article.** In Section 2, we introduce standard tools and results related to (1.1). Based on those, Theorem 1.3 is proven in Section 3. The case of short range nonlinearity $\sigma > \sigma_0(d)$ is addressed in Section 4, containing the proof of Theorem 1.4 and Corollary 1.5. The uniform in time convergence of rescaled densities, stated in Theorem 1.6, is given in Section 5. The rest of the paper deals with the convergence to the logarithmic Schrödinger equation, from (1.3) to (1.4). In Section 6, we focus our attention on the ordinary differential equations for $\tau_0$ and $\tau_\sigma$, proving the first conclusions of Theorem 1.9. In Section 7, we prove a priori estimates for the solution $\mathbf{u}_\sigma$, with emphasis on large time. The convergence
toward the logarithmic Schrödinger equation up to Ehrenfest time, Theorem 1.8, is proven in Section 8, where we also establish convergence in Wasserstein distance $W_1$ on such time intervals, serving as a preparatory step for the final Section 9, where the proof of Theorem 1.9 is completed.

## 2. PRELIMINARY ESTIMATES

### 2.1. Some technical tools.
We recall Gagliardo-Nirenberg inequalities, and standard inequalities which may be viewed as a dual version of Gagliardo-Nirenberg inequalities: for all $v \in S(\mathbb{R}^d)$,

$$ (2.1) \qquad \|v\|_{L^p} \lesssim \|v\|_{L^2}^{1-\delta(p)} \|\nabla v\|_{L^2}^{\delta(p)}, \quad \text{for } 2 \le p < \frac{2d}{(d-2)_+}, $$

$$ (2.2) \qquad \|v\|_{L^p} \lesssim \|v\|_{L^2}^{1-\delta(p')} \|yv\|_{L^2}^{\delta(p')}, \quad \text{for } \max\left(1, \frac{2d}{d+2}\right) < p \le 2, $$

where

$$ \delta(p) := d \left( \frac{1}{2} - \frac{1}{p} \right), \quad \frac{1}{p} + \frac{1}{p'} = 1. $$

We now recall classical Strichartz estimates:

**Definition 2.1.** A pair $(q, r)$ is admissible if $2 \le r \le \infty$ if $d = 1, 2 \le r < \infty$ if $d = 2, 2 \le r \le \frac{2d}{d-2}$ if $d \ge 3$, and

$$ \frac{2}{q} = \delta(r) := d \left( \frac{1}{2} - \frac{1}{r} \right). $$

**Lemma 2.2 (Strichartz estimates).** Denote $U_0(t) = e^{i\frac{t}{2}\Delta}$.

(1) Homogeneous Strichartz estimate. For any admissible pair $(q, r)$, there exists $C_q$ such that

$$ \|U_0(t)\varphi\|_{L^q(\mathbb{R};L^r)} \le C_q \|\varphi\|_{L^2}, \quad \forall \varphi \in L^2(\mathbb{R}^d). $$

(2) Inhomogeneous Strichartz estimate. For a time interval $I$, denote

$$ D_I(F)(t, x) = \int_{I \cap \{\tau \le t\}} U_0(t - \tau)F(\tau, x)d\tau. $$

For all admissible pairs $(q_1, r_1)$ and $(q_2, r_2)$, and any interval $I$, there exists $C = C_{r_1, r_2}$ independent of $I$ such that

$$ (2.3) \qquad \|D_I(F)\|_{L^{q_1}(I;L^{r_1})} \le C \|F\|_{L^{q'_2}(I;L^{r'_2})}, $$

for all $F \in L^{q'_2}(I; L^{r'_2})$.

For a time interval $I$, we introduce the norms

$$ \|u\|_{X(I)} = \sup_{(q,r) \text{ admissible}} \|u\|_{L^q(I;L^r)}, $$

$$ \|u\|_{Y(I)} = \max_{A \in \{\mathrm{Id}, \nabla\}} \sup_{(q,r) \text{ admissible}} \|Au\|_{L^q(I;L^r)}, $$

$$ \|u\|_{Z(I)} = \max_{A \in \{\mathrm{Id}, \nabla, x+it\nabla\}} \sup_{(q,r) \text{ admissible}} \|Au\|_{L^q(I;L^r)}. $$
2.2. **A priori estimates.** Suppose that $\phi_\sigma \in H^1(\mathbb{R}^d)$. Both for the power case (1.1) and the logarithmic case (1.4), the mass is conserved:

$$M_{\sigma} = \|u_{\sigma}(t)\|_{L^2(\mathbb{R}^d)}^2 = \|\mathbf{u}_{\sigma}(t)\|_{L^2(\mathbb{R}^d)}^2 = \|\phi_{\sigma}\|_{L^2(\mathbb{R}^d)}^2, \quad \forall t \in \mathbb{R}, \forall \sigma \ge 0.$$

Both equations are Hamiltonian, and the conserved energies read

$$\begin{aligned}
E_{\sigma} &= \frac{1}{2} \|\nabla u_{\sigma}(t)\|_{L^2(\mathbb{R}^d)}^2 + \frac{1}{\sigma+1} \|u_{\sigma}(t)\|_{L^{2^{\sigma+2}}(\mathbb{R}^d)}^{2^{\sigma+2}}, \\
E_0 &= \frac{1}{2} \|\nabla \mathbf{u}_0(t)\|_{L^2(\mathbb{R}^d)}^2 + \int_{\mathbb{R}^d} |\mathbf{u}_0(t, x)|^2 \ln |\mathbf{u}_0(t, x)|^2 dx.
\end{aligned}$$

In the case of (1.3), the second factor in $E_\sigma$ must be divided by $\sigma$ (recalling that the $L^2$-norm is conserved); we study this case in more details in Section 7. Suppose in addition that $\phi \in \Sigma$ (we leave out the possible dependence upon $\sigma$ at this stage). For the power case (1.1), the pseudoconformal evolution law, discovered in [19], reads:

$$(2.4) \quad \frac{d}{dt} \left( \frac{1}{2} \|(x + it\nabla)u\|_{L^2}^2 + \frac{t^2}{\sigma+1} \|u\|_{L^{2^{\sigma+2}}(\mathbb{R}^d)}^{2^{\sigma+2}} \right) = \frac{t}{\sigma+1} (2 - d\sigma) \|u\|_{L^{2^{\sigma+2}}(\mathbb{R}^d)}^{2^{\sigma+2}}.$$

For future references, we recall some important properties associated with the operator involved in the pseudoconformal evolution law:

**Proposition 2.3.** The operator

$$J(t) = x + it\nabla$$

satisfies the following properties:

* $J(t) = U_0(t)xU_0(-t)$, where $U_0$ denotes the Schrödinger group $U_0(t) = e^{i\frac{\Delta}{2}}$. Therefore, $J$ commutes with the linear part of (1.1),

$$\left[ J(t), i\partial_t + \frac{1}{2}\Delta \right] = 0.$$

* It can be factorized as

$$J(t) = it e^{i \frac{|x|^2}{2t}} \nabla \left( e^{-i \frac{|x|^2}{2t}} \cdot \right).$$

As a consequence, $J$ yields weighted Gagliardo-Nirenberg inequalities. For $2 \le r < \frac{2d}{(d-2)_+}$ ($2 \le r \le \infty$ if $d=1$), there exists $C(d,r)$ depending only on $d$ and $r$ such that

$$\|f\|_{L^r} \le \frac{C(d,r)}{|t|^{\delta(r)}} \|f\|_{L^2}^{1-\delta(r)} \|J(t)f\|_{L^2}^{\delta(r)}, \quad \delta(r) := d \left( \frac{1}{2} - \frac{1}{r} \right).$$

Also, if $F(z) = G(|z|^2)z$ is $C^1$, then $J(t)$ acts like a derivative on $F(w)$:

$$J(t) (F(w)) = \partial_z F(w) J(t) w - \partial_{\bar{z}} F(w) \overline{J(t) w}.$$

If we assume $\sigma \ge 2/d$, the pseudoconformal evolution law (2.4) yields the a priori bound

$$\|(x + it\nabla)u_{\sigma}(t)\|_{L^2} \le C (\|\phi\|_{\Sigma}),$$

hence, by Proposition 2.3,

$$\|u_{\sigma}(t)\|_{L^r} \le \frac{C}{\langle t \rangle^{\delta(r)}}, \quad 2 \le r < \frac{2}{(d-2)_+}, \quad \delta(r) = d \left( \frac{1}{2} - \frac{1}{r} \right),$$
for some constant C depending only on d and $||\phi||_{\Sigma}$. In particular,

$$ (2.5) \qquad \|u_{\sigma}(t)\|_{L^{2\sigma+2}} \lesssim \frac{1}{\langle t \rangle^{\frac{d\sigma}{2\sigma+2}}}. $$

Integrating (2.4) in time, and discarding the (positive) term $||J(t)u_{\sigma}||_{L^2}^2$, Grönwall lemma shows that for

$$ \sigma > \frac{2 - d + \sqrt{d^2 + 12d + 4}}{4d} = \sigma_0(d), $$

(2.5) stills holds. The exponent $\sigma_0(d)$ is standard in the scattering theory in $\Sigma$ for (1.1) (see e.g [9]). We note that $\sigma \ge 2/d > \sigma_0(d)$.

In the case $\sigma < 2/d$, (2.4) and Grönwall lemma yield

$$ \|J(t)u\|_{L^2}^2 + t^2 \|u(t)\|_{L^{2\sigma+2}}^{2\sigma+2} \lesssim \langle t \rangle^{2-d\sigma}. $$

2.3. **Existence results.** The following result is classical (see e.g. [9]):

**Proposition 2.4.** Let $d \ge 1$, $0 < \sigma < \frac{2}{d-2}_+$ and $\phi \in H^1(\mathbb{R}^d)$. Then (1.1) has a unique solution

$$ u_{\sigma} \in C(\mathbb{R}; H^1(\mathbb{R}^d)) \cap C^1(\mathbb{R}; H^{-1}(\mathbb{R}^d)). $$

For all bounded interval $I$, there exists $C = C(I, ||\phi||_{H^1})$ such that

$$ \|u_{\sigma}\|_{Y(I)} \le C, $$

and the mass $M$ and the energy $E_{\sigma}$ are independent of $t \in \mathbb{R}$.

If in addition $\phi \in \Sigma$, then $u_{\sigma} \in C(\mathbb{R}; \Sigma)$ and for all bounded interval $I$, there exists $C = C(I, ||\phi||_{\Sigma})$ such that

$$ \|u_{\sigma}\|_{Z(I)} \le C, $$

and (2.4) holds.

The next result follows from [7] (case $\Sigma$) and [21] (case $H^1$) for (1.4):

**Proposition 2.5.** Let $d \ge 1$. If $\phi \in H^1(\mathbb{R}^d)$, then (1.4) has a unique solution $u_0 \in C(\mathbb{R}; H^1(\mathbb{R}^d))$. If in addition $\phi \in \Sigma$, then $u_0 \in C(\mathbb{R}; \Sigma)$, and the mass $M_0$ and the energy $E_0$ are independent of time.

For future reference, we gather a priori estimates stemming from (2.4) and the conservation of the energy in the following lemma.

**Lemma 2.6.** Let $\phi \in \Sigma$ and $u \in C(\mathbb{R}; \Sigma)$ solve (1.1), with $0 < \sigma < \frac{2}{d-2}_+$.

* If $\sigma \ge 2/d$, then for all $t \in \mathbb{R}$,

$$ \|J(t)u\|_{L^2(\mathbb{R}^d)} \le \|x\phi\|_{L^2(\mathbb{R}^d)}, $$

$$ \|u(t)\|_{L^r(\mathbb{R}^d)} \le \frac{C}{|t|^{\delta(r)}} \|x\phi\|_{L^2(\mathbb{R}^d)}, \quad 2 \le r < \frac{2d}{(d-2)_+}, $$

where $\delta(r)$ is defined in Proposition 2.3.

* If $\sigma < 2/d$, then for all $t \in \mathbb{R}$,

$$ \|J(t)u\|_{L^2(\mathbb{R}^d)} \lesssim \langle t \rangle^{1-\frac{d\sigma}{2}}, $$

$$ \|u(t)\|_{L^r(\mathbb{R}^d)} \lesssim \langle t \rangle^{-\frac{d\sigma}{2}\delta(r)}, \quad 2 \le r < \frac{2d}{(d-2)_+}. $$
### 3. BOUNDED TIME INTERVALS

In this section, we prove Theorem 1.3. Since the equation (1.1) is reversible, we consider positive time only.

**3.1. L²-estimate.** We fix $\sigma > 0$, and consider $\nu \in (0, \frac{2}{(d-2)_+})$ in a neighborhood of $\sigma$. The difference $u_\nu - u_\sigma$ solves

$$ (3.1) \quad i \partial_t (u_\nu - u_\sigma) + \frac{1}{2} \Delta (u_\nu - u_\sigma) = |u_\nu|^{2\nu} u_\nu - |u_\sigma|^{2\nu} u_\sigma + |u_\sigma|^{2\nu} u_\sigma - |u_\sigma|^{2\sigma} u_\sigma. $$

The pair

$$ (q_\sigma, r_\sigma) := \left( \frac{4\sigma + 4}{d\sigma}, 2\sigma + 2 \right) $$

is admissible, and (with the same implicit time interval in each term)

$$ \| |u_\sigma|^{2\nu} u_\sigma - |u_\sigma|^{2\sigma} u_\sigma \|_{L^{q'_\sigma} L^{r'_\sigma}} \le \| |u_\sigma|^{2\nu} - |u_\sigma|^{2\sigma} \|_{L^\theta L^{\frac{\nu\sigma}{2\sigma}}} \| u_\sigma \|_{L^{q_\sigma} L^{r_\sigma}}, $$

where

$$ \frac{1}{r'_\sigma} = \frac{2\sigma + 1}{r_\sigma} \quad ; \quad \frac{1}{q'_\sigma} = \frac{1}{q_\sigma} + \frac{1}{\theta} \iff \theta = \frac{2\sigma + 2}{2 - (d-2)\sigma}. $$

We have $1 < \theta < \infty$ since we consider energy-subcritical nonlinearities. Write $\nu = \sigma(1+h)$, with $|h| \ll 1$. Taylor formula yields for $y \ge 0$ (a placeholder for $|u_\sigma|^{2\sigma})$,

$$ y - y^{1+h} = -hy \ln y \int_0^1 y^{sh} ds, $$

and

$$ \int_0^1 y^{sh} ds \le 1 + y^{-|h|} 1_{0<y\le 1} + y^{|h|} 1_{y>1}, $$

hence

$$ |y - y^{1+h}| \lesssim |h|y \ln y \left(1 + y^{-|h|} + y^{|h|}\right) \lesssim |h| (y^{1-\eta} + y^{1+\eta}), $$

as soon as $|h| < \eta$. Denoting $\nu_\pm = \sigma(1 \pm \eta)$, we have, for $0 < \eta \ll 1$,

$$ |u_\sigma|^{2\nu_\pm} \in L^\theta L^{\frac{\nu\sigma}{2\sigma}}, $$

since for $0 < \eta \ll 1$, $r_\sigma^{\nu_\pm/\sigma} = (1 \pm \eta)r_\sigma \in (2, 2^*)$, where

$$ 2^* = \frac{2d}{(d-2)_+}. $$

Indeed, $r_\sigma \in (2, 2^*)$, and since this condition is open, it remains true for nearby indices. Therefore,

$$ \| |u_\sigma|^{2\nu} u_\sigma - |u_\sigma|^{2\sigma} u_\sigma \|_{L^{q'_\sigma} L^{r'_\sigma}} \lesssim |\sigma - \nu| \| u_\sigma \|_{L^{\infty H^1}}^{2\sigma+1}. $$

For the term $|u_\nu|^{2\nu} u_\nu - |u_\sigma|^{2\nu} u_\sigma$, we use the pair $(q_\nu, r_\nu)$, and write

$$ \| |u_\nu|^{2\nu} u_\nu - |u_\sigma|^{2\nu} u_\sigma \|_{L^{q'_\nu} L^{r'_\nu}} \lesssim (\|u_\nu\|_{L^k L^{r_\nu}}^{2\nu} + \|u_\sigma\|_{L^k L^{r_\nu}}^{2\nu}) \|u_\nu - u_\sigma\|_{L^{q_\nu} L^{r_\nu}}, $$

where

$$ \frac{1}{q'_\nu} = \frac{1}{q_\nu} + \frac{2\nu}{k} \iff k = \frac{2\nu(2\nu + 2)}{2 - (d-2)\nu}. $$

For a bounded time interval, the conservation of the energy implies

$$ \|u_\sigma\|_{L^\infty H^1} + \|u_\nu\|_{L^\infty H^1} \le C (\|\phi\|_{H^1}), $$
and so, thanks to Sobolev embedding,

$$ \|u_{\nu}\|_{L^k(I; L^{r_{\nu}})}^{2\nu} + \|u_{\sigma}\|_{L^k(I; L^{r_{\sigma}})}^{2\nu} \le |I|^{1/k} C (\|\phi\|_{H^1}). $$

Let $I_n = [t_n, t_{n+1}]$:

$$ L^{q_{\nu}}(I_n; L^{r_{\nu}}) \cap L^{q_{\sigma}}(I_n; L^{r_{\sigma}}) \cap L^{\infty}(I_n; L^2) \subset X(I_n). $$

Strichartz estimates yield

$$ \|u_{\nu} - u_{\sigma}\|_{X(I_n)} \lesssim \|u_{\nu}(t_n) - u_{\sigma}(t_n)\|_{L^2} + |I_n|^{1/k} \|u_{\nu} - u_{\sigma}\|_{X(I_n)} \\ + \|u_{\sigma}\|_{L^{\theta}(I_n; L^{\frac{r_{\sigma}}{2\sigma}})}^{2\nu} \|u_{\sigma}\|_{L^{q_{\sigma}} L^{r_{\sigma}}}. $$

We decompose $[0, T]$ into finitely many $I_n$ so we infer, up to doubling the implicit constant,

$$ \|u_{\nu} - u_{\sigma}\|_{X(I_n)} \lesssim \|u_{\nu}(t_n) - u_{\sigma}(t_n)\|_{L^2} + \| |u_{\sigma}|^{2\nu} - |u_{\sigma}|^{2\sigma} \|_{L^{\theta}(I_n; L^{\frac{r_{\sigma}}{2\sigma}})} \|u_{\sigma}\|_{L^{q_{\sigma}} L^{r_{\sigma}}}. $$

We conclude that, as $\nu \to \sigma$,

$$ \|u_{\nu} - u_{\sigma}\|_{X([0,T])} = \mathcal{O}_T (\|\phi_{\nu} - \phi_{\sigma}\|_{L^2} + |\nu - \sigma|). $$

3.2. **Estimate in $\Sigma$**. The convergence in $H^1$ is interesting, as well as in $\mathcal{F}(H^1)$: the latter essentially needs the former, as either we apply the multiplication operator $x$ to (3.1), and a source term $\nabla(u_{\nu} - u_{\sigma})$ appears due to the commutator $[x, \Delta]$, or we apply $x + it\nabla$, which commutes with the left hand side of (3.1), but acts on the nonlinearity like the gradient.

Let $A \in \{i\nabla, x + it\nabla\}$, and apply $A$ to (3.1):

$$ i \partial_t A (u_{\nu} - u_{\sigma}) + \frac{1}{2} \Delta A (u_{\nu} - u_{\sigma}) = L_A(\nu, \sigma) + S_A(\nu, \sigma), $$

where the term which is always locally Lipschitz when $A = \text{Id}$ is

$$ L_A(\nu, \sigma) = A (|u_{\nu}|^{2\nu} u_{\nu} - |u_{\sigma}|^{2\nu} u_{\sigma}), $$

and the source term is

$$ S_A(\nu, \sigma) = A (|u_{\sigma}|^{2\nu} u_{\sigma} - |u_{\sigma}|^{2\sigma} u_{\sigma}). $$

We have

$$ A (|u|^{2\nu} u) = (\nu + 1) |u|^{2\nu} Au - \nu u^{\nu+1} \bar{u}^{\nu-1} \overline{Au}, $$

and we write

$$ L_A(\nu, \sigma) = (\nu + 1)|u_{\nu}|^{2\nu} Au_{\nu} - \nu u_{\nu}^{\nu+1} \bar{u}_{\nu}^{\nu-1} \overline{Au_{\nu}} - (\nu + 1)|u_{\sigma}|^{2\nu} Au_{\sigma} \\ - \nu u_{\sigma}^{\nu+1} \bar{u}_{\sigma}^{\nu-1} \overline{Au_{\sigma}} \pm ((\nu + 1)|u_{\nu}|^{2\nu} Au_{\sigma} - \nu u_{\nu}^{\nu+1} \bar{u}_{\nu}^{\nu-1} \overline{Au_{\sigma}}), $$

so we have the pointwise estimate

$$ |L_A(\nu, \sigma)| \lesssim |u_{\nu}|^{2\nu} |A(u_{\nu} - u_{\sigma})| + |||u_{\nu}|^{2\nu} - |u_{\sigma}|^{2\nu}|| |Au_{\sigma}|. $$

Resuming the Lebesgue indices from the previous subsection,

$$ \|L_A(\nu, \sigma)\|_{L^{q_{\nu}} L^{r_{\nu}}} \lesssim \|u_{\nu}\|_{L^k L^{r_{\nu}}}^{2\nu} \|A(u_{\nu} - u_{\sigma})\|_{L^{q_{\nu}} L^{r_{\nu}}} \\ + \| |u_{\nu}|^{2\nu} - |u_{\sigma}|^{2\nu} \|_{L^{\frac{k}{2\nu}} L^{\frac{r_{\nu}}{2\nu}}} \|Au_{\sigma}\|_{L^{q_{\nu}} L^{r_{\nu}}}, $$

where

$$ \frac{1}{q'_{\nu}} = \frac{1}{q_{\nu}} + \frac{2\nu}{k} \iff k = \frac{2\nu(2\nu + 2)}{2 - (d-2)\nu}. $$
The second term in the above inequality is actually considered as a source term in view of the convergence established in the previous subsection. Like in e.g. [10], we distinguish two cases:

• If $2\nu \ge 1$, the application $z \mapsto |z|^{2\nu}$ is (locally) Lipschitz continuous,
$$|||u_\nu|^{2\nu} - |u_\sigma|^{2\nu}| \lesssim (|u_\nu|^{2\nu-1} + |u_\sigma|^{2\nu-1}) |u_\nu - u_\sigma|,$$

hence, by Hölder inequality,

$$|||u_\nu|^{2\nu} - |u_\sigma|^{2\nu}| \| u_\nu - u_\sigma \|_{L^k L^{r_\nu}} \| u_\nu + u_\sigma \|_{L^k L^{r_\nu}}.$$

• If $0 < 2\nu < 1$, the application $z \mapsto |z|^{2\nu}$ is Hölder continuous,
$$|||u_\nu|^{2\nu} - |u_\sigma|^{2\nu}| \lesssim |u_\nu - u_\sigma|^{2\nu},$$

and

$$|||u_\nu|^{2\nu} - |u_\sigma|^{2\nu}| \| u_\nu - u_\sigma \|_{L^k L^{r_\nu}} \lesssim \| u_\nu - u_\sigma \|_{L^k L^{r_\nu}}^{2\nu}.$$

In view of Proposition 2.4, we infer

$$|||u_\nu|^{2\nu} - |u_\sigma|^{2\nu}| \| u_\nu - u_\sigma \|_{L^k L^{r_\nu}}^{\min(1,2\nu)}.$$

In view of Sobolev embedding, when $\nu < \frac{2d}{(d-2)_+}$,

$$H^{s_\nu}(\mathbb{R}^d) \hookrightarrow L^{2\nu+2}(\mathbb{R}^d), \quad s_\nu = \frac{d\nu}{2\nu+2} < 1,$$

for $\nu \in ((1-\eta)\sigma, (1+\eta)\sigma)$ and $\eta > 0$ sufficiently small. Therefore, in view of the previous subsection,

$$\begin{aligned} \|u_\nu - u_\sigma\|_{L^\infty L^{r_\nu}} &\lesssim \|u_\nu - u_\sigma\|_{L^\infty H^{s_\nu}} = \|u_\nu - u_\sigma\|_{L^\infty L^2} + \|u_\nu - u_\sigma\|_{L^\infty \dot{H}^{s_\nu}} \\ &\lesssim |\nu - \sigma| + \|u_\nu - u_\sigma\|_{L^\infty \dot{H}^{s_\nu}}, \end{aligned}$$

where $\dot{H}^s$ denotes the homogeneous Sobolev space. By interpolation,

$$\|f\|_{\dot{H}^{s_\nu}} \le \|f\|_{L^2}^{1-s_\nu} \|f\|_{\dot{H}^1}^{s_\nu},$$

and from the previous subsection,

$$\|u_\nu - u_\sigma\|_{L^\infty L^{r_\nu}} \lesssim |\nu - \sigma| + |\nu - \sigma|^{1-s_\nu} \|u_\nu - u_\sigma\|_{L^\infty \dot{H}^1}^{s_\nu}.$$

Therefore, setting $\gamma := \min(1, 2\nu)$, for any $I \subset [0, T]$,

$$|||u_\nu|^{2\nu} - |u_\sigma|^{2\nu}| \| u_\nu - u_\sigma \|_{L^k (I; L^{r_\nu})} \lesssim |\nu - \sigma|^\gamma + |I|^{\gamma/k} |\nu - \sigma|^{\gamma(1-s_\nu)} \| u_\nu - u_\sigma \|_{L^\infty \dot{H}^1}^{\gamma s_\nu}.$$

We also have

$$\|u_\nu\|_{L^k(I; L^{r_\nu})}^{2\nu} \|A(u_\nu - u_\sigma)\|_{L^{q_\nu}(I; L^{r_\nu})} \le |I|^{1/k} \|u_\nu\|_{L^\infty H^1}^{2\nu} \|A(u_\nu - u_\sigma)\|_{X(I)},$$

and so

$$\|L_A(\nu, \sigma)\|_{L^{q'_\nu}(I; L^{r'_\nu})} \lesssim |\nu - \sigma|^\gamma + |I|^{1/k} \|A(u_\nu - u_\sigma)\|_{X(I)} \\ + |I|^{\gamma/k} |\nu - \sigma|^{\gamma(1-s_\nu)} \|u_\nu - u_\sigma\|_{L^\infty \dot{H}^1}^{\gamma s_\nu}.$$

On the other hand, we write

$$S_A(\nu, \sigma) = (\nu + 1)|u_\sigma|^{2\nu} A u_\sigma - \nu u_\sigma^{\nu+1} \bar{u}_\sigma^{\nu-1} \overline{A u_\sigma} - (\sigma + 1)|u_\sigma|^{2\sigma} A u_\sigma \\ + \sigma u_\sigma^{\sigma+1} \bar{u}_\sigma^{\sigma-1} \overline{A u_\sigma} \pm ((\nu + 1)|u_\sigma|^{2\sigma} A u_\sigma - \nu u_\sigma^{\sigma+1} \bar{u}_\sigma^{\sigma-1} \overline{A u_\sigma}),$$

hence

$$|S_A(\nu, \sigma)| \lesssim |||u_\sigma|^{2\nu} - |u_\sigma|^{2\nu}| |A u_\sigma| + |\nu - \sigma||u_\sigma|^{2\sigma} |A u_\sigma|.$$
Resuming the notation $I_n = [t_n, t_{n+1}]$, Strichartz estimates yield

$$
\begin{align*}
\|A(u_\nu - u_\sigma)\|_{X(I_n)} &\lesssim \|A(u_\nu - u_\sigma)(t_n)\|_{L^2} + \|L_A(\nu, \sigma)\|_{L^{q'_\nu}(I_n; L^{r'_\nu})} \\
&\qquad + \|S_A(\nu, \sigma)\|_{L^{q'_\sigma}(I_n; L^{r'_\sigma})} \\
&\lesssim \|A(u_\nu - u_\sigma)(t_n)\|_{L^2} + |\nu - \sigma|^\gamma + |I_n|^{1/k} \|A(u_\nu - u_\sigma)\|_{X(I_n)} \\
&\qquad + |I_n|^{\gamma/k} |\nu - \sigma|^{\gamma(1-s_\nu)} \|u_\nu - u_\sigma\|_{L^\infty \dot{H}^1}^{\gamma s_\nu} \\
&\qquad + \|u_\sigma|^{2\nu} - |u_\sigma|^{2\sigma}\|_{L^\theta(I_n; L^{\frac{r\sigma}{2\sigma}})} \|Au_\sigma\|_{L^{q_\sigma}(I_n; L^{r_\sigma})} \\
&\qquad + |\nu - \sigma| \|u_\sigma\|_{L^{2\sigma\theta}(I_n; L^{r_\sigma})}^{2\sigma} \|Au_\sigma\|_{L^{q_\sigma}(I_n; L^{r_\sigma})} \\
&\lesssim \|A(u_\nu - u_\sigma)(t_n)\|_{L^2} + |\nu - \sigma|^\gamma + |I_n|^{1/k} \|A(u_\nu - u_\sigma)\|_{X(I_n)} \\
&\qquad + |I_n|^{\gamma/k} |\nu - \sigma|^{\gamma(1-s_\nu)} \|u_\nu - u_\sigma\|_{L^\infty \dot{H}^1}^{\gamma s_\nu} \\
&\qquad + \|u_\sigma|^{2\nu} - |u_\sigma|^{2\sigma}\|_{L^\theta(I_n; L^{\frac{r\sigma}{2\sigma}})} \|Au_\sigma\|_{X([0,T])} \\
&\qquad + |\nu - \sigma| \|u_\sigma\|_{L^\infty \dot{H}^1}^{2\sigma} \|Au_\sigma\|_{X([0,T])}.
\end{align*}
$$

We note that $\gamma s_\nu < 1$, and Young inequality implies

$$
|\nu - \sigma|^{\gamma(1-s_{\nu})} \|u_{\nu} - u_{\sigma}\|_{L^{\infty}\dot{H}^{1}}^{\gamma s_{\nu}} \lesssim |\nu - \sigma|^{\gamma(1-s_{\nu})/(1-\gamma s_{\nu})} + \|u_{\nu} - u_{\sigma}\|_{L^{\infty}\dot{H}^{1}}.
$$

Since $\gamma \le 1$,

$$
\frac{1 - s_{\nu}}{1 - \gamma s_{\nu}} \le 1,
$$

for $|\nu - \sigma| \le 1$,

$$
|\nu - \sigma| \le |\nu - \sigma|^\gamma \le |\nu - \sigma|^{\gamma(1-s_\nu)/(1-\gamma s_\nu)}.
$$

The term $|||u_\sigma|^{2\nu} - |u_\sigma|^{2\sigma}||_{L^\theta(I_n; L^{\frac{r\sigma}{2\sigma}})}$ was estimated in the previous subsection,

$$
\| |u_\sigma|^{2\nu} - |u_\sigma|^{2\sigma} \|_{L^\theta(I_n; L^{\frac{r\sigma}{2\sigma}})} \lesssim |\nu - \sigma|,
$$

and we have

$$
\begin{align*}
\|A(u_\nu - u_\sigma)\|_{X(I_n)} &\lesssim \|A(u - u_\sigma)(t_n)\|_{L^2} + |\nu - \sigma|^{\gamma(1-s_\nu)/(1-\gamma s_\nu)} \\
&\qquad + |I_n|^{1/k} \|A(u_\nu - u_\sigma)\|_{X(I_n)} + |I_n|^{\gamma/k} \|u_\nu - u_\sigma\|_{L^\infty(I_n; \dot{H}^1)} \\
&\lesssim \|A(u - u_\sigma)(t_n)\|_{L^2} + |\nu - \sigma|^{\gamma(1-s_\nu)/(1-\gamma s_\nu)} \\
&\qquad + |I_n|^{1/k} \|A(u_\nu - u_\sigma)\|_{X(I_n)} + |I_n|^{\gamma/k} \|\nabla(u_\nu - u_\sigma)\|_{X(I_n)}.
\end{align*}
$$

We first let $A = \nabla$, and decompose $[0, T]$ into finitely many $I_n$, to get

$$
\| \nabla (u_\nu - u_\sigma) \|_{X([0,T])} = \mathcal{O}_T \left( \| \nabla (\phi_\nu - \phi_\sigma) \|_{L^2} + |\nu - \sigma|^{\gamma(1-s_\nu)/(1-\gamma s_\nu)} \right).
$$

Considering then $A = x + it\nabla = J(t)$, we infer

$$
\| J(u_\nu - u_\sigma) \|_{X([0,T])} = \mathcal{O}_T \left( \| J(\phi_\nu - \phi_\sigma) \|_{L^2} + |\nu - \sigma|^{\gamma(1-s_\nu)/(1-\gamma s_\nu)} \right).
$$

In view of the expressions of $s_\nu$ and $\gamma$,

$$
\gamma \frac{1 - s_{\nu}}{1 - \gamma s_{\nu}} = \begin{cases} 1 & \text{if } 2\nu \ge 1, \\ \frac{2 - (d-2)\nu}{\nu + 1 - d\nu^2}\nu & \text{if } 2\nu < 1, \end{cases}
$$

and
$$ (3.2) \qquad \gamma \frac{1 - s_{\nu}}{1 - \gamma s_{\nu}} = \min \left( 1, \frac{2 - (d-2)\nu}{\nu + 1 - d\nu^2} \right). $$

The final rate is

$$ (3.3) \qquad \theta = \inf_{|\nu - \sigma| \le \eta} \gamma \frac{1 - s_{\nu}}{1 - \gamma s_{\nu}}. $$

## 4. GLOBAL DYNAMICS IN THE (VERY) SHORT RANGE CASE

### 4.1. From local in time to global in time.
To make the estimate from Section 3 global in time, we essentially proceed the same way, except that to estimate the “nonlinear potential” terms, $||u_{\nu}||_{L^k(I;L^{r_{\nu}})}$ and $||u_{\sigma}||_{L^k(I;L^{r_{\sigma}})}$, we want to let $I = [T, \infty)$ for $T$ sufficiently large, by using an explicit time decay, provided in Proposition 2.3,

$$ (4.1) \qquad \|u_{\nu}(t)\|_{L^{r_{\nu}}} \lesssim \frac{1}{|t|^{\delta}} \|u_{\nu}(t)\|_{L^2}^{1-\delta} \|J(t)u_{\nu}\|_{L^2}^{\delta}, \quad \delta = \frac{d\nu}{2\nu + 2}. $$

Resuming the same notation as in the previous section, we note the equivalence

$$ k \times \delta = \frac{2d\nu^2}{2 - (d-2)\nu} > 1 \iff \nu > \frac{2 - d + \sqrt{d^2 + 12d + 4}}{4d} = \sigma_0(d). $$

We note that $\sigma > \sigma_0(d)$, so the condition is fulfilled when $\nu = \sigma$. As the condition is open, for $\eta > 0$ sufficiently small, the above inequality is satisfied for $|\nu - \sigma| \le \eta$.

For the $L^2$-estimate, the previous computations yield, for $I = (t_i, t_f) \subset \mathbb{R}$,

$$ \begin{aligned} \|u_{\nu} - u_{\sigma}\|_{X(I)} & \lesssim \|u_{\nu}(t_i) - u_{\sigma}(t_i)\|_{L^2} \\ & \qquad + \| |u_{\sigma}|^{2\nu} - |u_{\sigma}|^{2\sigma} \|_{L^{\theta}(I; L^{\frac{r_{\sigma}}{2\sigma}})} \|u_{\sigma}\|_{L^{q_{\sigma}}(I; L^{r_{\sigma}})} \\ & \qquad + \left( \|u_{\nu}\|_{L^k(I; L^{r_{\nu}})}^{2\nu} + \|u_{\sigma}\|_{L^k(I; L^{r_{\sigma}})}^{2\nu} \right) \|u_{\nu} - u_{\sigma}\|_{L^{q_{\nu}}(I; L^{r_{\nu}})} \\ & \lesssim \|u_{\nu}(t_i) - u_{\sigma}(t_i)\|_{L^2} \\ & \qquad + |\nu - \sigma| \left( \| |u_{\sigma}|^{2\sigma(1-\eta)} \|_{L^{\theta}(I; L^{\frac{r_{\sigma}}{2\sigma}})} + \| |u_{\sigma}|^{2\sigma(1+\eta)} \|_{L^{\theta}(I; L^{\frac{r_{\sigma}}{2\sigma}})} \right) \|u_{\sigma}\|_{L^{q_{\sigma}}(I; L^{r_{\sigma}})} \\ & \qquad + \left( \|u_{\nu}\|_{L^k(I; L^{r_{\nu}})}^{2\nu} + \|u_{\sigma}\|_{L^k(I; L^{r_{\sigma}})}^{2\nu} \right) \|u_{\nu} - u_{\sigma}\|_{L^{q_{\nu}}L^{r_{\nu}}} \\ & \lesssim \|u_{\nu}(t_i) - u_{\sigma}(t_i)\|_{L^2} \\ & \qquad + |\nu - \sigma| \left( \| |u_{\sigma}|^{1-\eta} \|_{L^{2\sigma\theta}(I; L^{r_{\sigma}})} + \| |u_{\sigma}|^{1+\eta} \|_{L^{2\sigma\theta}(I; L^{r_{\sigma}})} \right) \|u_{\sigma}\|_{L^{q_{\sigma}}(I; L^{r_{\sigma}})} \\ & \qquad + \left( \|u_{\nu}\|_{L^k(I; L^{r_{\nu}})}^{2\nu} + \|u_{\sigma}\|_{L^k(I; L^{r_{\sigma}})}^{2\nu} \right) \|u_{\nu} - u_{\sigma}\|_{L^{q_{\nu}}L^{r_{\nu}}}. \end{aligned} $$

In view of the above preliminary, for $\eta \in (0, 1)$ sufficiently small,

$$ |u_{\sigma}|^{1-\eta}, |u_{\sigma}|^{1+\eta} \in L^{2\sigma\theta}(\mathbb{R}; L^{r_{\sigma}}), \quad u_{\nu}, u_{\sigma} \in L^k(\mathbb{R}; L^{r_{\sigma}}), \quad u_{\sigma} \in L^{q_{\sigma}}(\mathbb{R}; L^{r_{\sigma}}), $$

so we can decompose $\mathbb{R}$ into finitely many intervals and repeat the previous argument. The approach also makes it possible to readily adapt the proof of the convergence in $\Sigma$, as the “nonlinear potential” terms that need to be absorbed by the left hand side are the same.

Theorem 1.4 follows, by noticing the standard identity

$$ \|xe^{-i\frac{t}{2}\Delta}w\|_{L^2} = \|J(t)w\|_{L^2}, $$
due to the identity $J(t) = U_0(t)xU_0(-t)$ and the fact that $U_0(t)$ is unitary on $L^2$.

4.2. Continuity of the scattering operator. Corollary 1.5 is a rather direct consequence of the previous estimates. We first note that since $u_{+,\sigma} = \lim_{t \to +\infty} e^{-i\frac{t}{2}\Delta}u(t)$ as $t \to +\infty$, Theorem 1.4 implies that if $\|\phi_\sigma - \phi_\nu\|_\Sigma \to 0$ as $\nu \to \sigma$, then $\|u_{+,\sigma} - u_{+,\nu}\|_\Sigma \to 0$. So all we have to explain is why if $\|u_{-,\sigma} - u_{-,\nu}\|_\Sigma \to 0$, then $\|u_{\sigma|t=0} - u_{\nu|t=0}\|_\Sigma \to 0$. We recall that wave operators are constructed via a fixed point argument on Duhamel's formula, like for the initial value problem,

$$u_{\sigma}(t) = e^{i\frac{t}{2}\Delta}u_{-,\sigma} - i \int_{-\infty}^{t} e^{i\frac{t-s}{2}\Delta} (|u_{\sigma}|^{2\sigma}u_{\sigma}) (s)ds.$$

The convergence of the integral near $t = -\infty$ is granted by the same estimate as before, (4.1). Therefore, no real modification is needed in order to prove the continuity of $S_\sigma$ in $\Sigma$ at $\sigma > \sigma_0(d)$.

5. GLOBAL DYNAMICS FOR GENERAL POWERS

5.1. Convergence in Wasserstein distance: proof of Theorem 1.6. Again, as the equation (1.1) is reversible, we focus on positive time only. The general principle is the following. It follows from Theorem 1.3 that for any given $T > 0$,

$$\limsup_{\nu \to \sigma} \sup_{t \in [0,T]} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{L^1} \lesssim \limsup_{\nu \to \sigma} \|\phi_{\nu} - \phi_{\sigma}\|_{L^2}.$$

Details are given below. For $T$ sufficiently large, the variations of $\varrho_\nu$ are negligible when $t \ge T$. If we worked in a normed space $X$, we would write, for $t > T$,

$$\|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{X} = \left\| \varrho_{\nu}(T) - \varrho_{\sigma}(T) + \int_{T}^{t} \partial_{t} (\varrho_{\nu}(s) - \varrho_{\sigma}(s)) ds \right\|_{X} \\ \le \|\varrho_{\nu}(T) - \varrho_{\sigma}(T)\|_{X} + \int_{T}^{t} \|\partial_{t}\varrho_{\nu}(s)\|_{X} ds + \int_{T}^{t} \|\partial_{t}\varrho_{\sigma}(s)\|_{X} ds.$$

If $X$ is such that we can invoke Theorem 1.3,

$$\limsup_{\nu \to \sigma} \sup_{t \in \mathbb{R}} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{X} \le \sup_{\nu} \int_{T}^{\infty} \|\partial_{t}\varrho_{\nu}(s)\|_{X} ds + \int_{T}^{\infty} \|\partial_{t}\varrho_{\sigma}(s)\|_{X} ds,$$

and it suffices to show that the right hand side is an $o(1)$ as $T \to \infty$. This in turn follows from the continuity equation

$$\partial_t \varrho_\sigma + \frac{1}{\langle t \rangle^2} \operatorname{div} j_\sigma = 0, \quad \text{where } j_\sigma(t, y) = \frac{1}{\|\phi_\sigma\|_{L^2}^2} \operatorname{Im} (\bar{v}_\sigma \nabla v_\sigma),$$

where $v_\sigma = v_\sigma(t, y)$ is defined by the transform

$$u_{\sigma}(t, x) = \frac{1}{\langle t \rangle^{d/2}} v_{\sigma} \left( t, \frac{x}{\langle t \rangle} \right) \exp \left( i \frac{t}{1+t^2} \frac{|x|^2}{2} \right),$$

and solves

$$i \partial_t v_\sigma + \frac{1}{2 \langle t \rangle^2} \Delta v_\sigma = \frac{|y|^2}{2 \langle t \rangle^2} v_\sigma + \frac{1}{\langle t \rangle^{d\sigma}} |v_\sigma|^{2\sigma} v_\sigma.$$
We obviously have $\|v_\sigma(t)\|_{L^2} = \|\phi_\sigma\|_{L^2}$. In view of Lemma 2.6, we also have the a priori estimates

$$ (5.1) \quad \|y v_\sigma(t)\|_{L^2} = \frac{1}{\langle t \rangle} \|x u_\sigma(t)\|_{L^2} \le \frac{1}{\langle t \rangle} \|J(t) u_\sigma(t)\|_{L^2} + \|\nabla u_\sigma(t)\|_{L^2} \le C (\|\phi_\sigma\|_{\Sigma}), $$

$$ (5.2) \quad \|\nabla v_\sigma(t)\|_{L^2} = \|(x + i \langle t \rangle \nabla) u_\sigma\|_{L^2} \le \langle t \rangle^{\max(0,1-d\sigma/2)} C (\|\phi_\sigma\|_{\Sigma}). $$

First, we show that Theorem 1.3 implies that for any $T > 0$,

$$ \limsup_{\nu \to \sigma} \sup_{t \in [0, T]} W_1(\varrho_\nu(t), \varrho_\sigma(t)) = 0. $$

In view of (5.1), for $\eta > 0$ sufficiently small, $0 < (1 - \eta)\sigma < (1 + \eta)\sigma < \frac{2}{(d-2)_+}$, and there exists $M_2$ such that

$$ \sup_{|\nu-\sigma| \le \eta} \sup_{t \in \mathbb{R}} \int_{\mathbb{R}^d} |y|^2 \varrho_\nu(t, y) dy \le M_2. $$

In the integral from (1.5) for the characterization of the Wasserstein distance $W_1$, since $\mu_1$ and $\mu_2$ have the same mass, we may replace $\psi$ with $\psi - \psi(0)$, and thus suppose that $\psi(0) = 0$. We have in particular

$$ |\psi(x)| \le |x|, \quad \forall x \in \mathbb{R}^d. $$

We thus write, for $t \ge 0$,

$$ \begin{align*} W_1(\varrho_\nu(t), \varrho_\sigma(t)) &= \sup_{\|\psi\|_{\text{Lip}} \le 1} \int_{\mathbb{R}^d} \psi(y) (\varrho_\nu(t, y) - \varrho_\sigma(t, y)) dy \\ &\le \int_{\mathbb{R}^d} |y| |\varrho_\nu(t, y) - \varrho_\sigma(t, y)| dy \\ &\le R \int_{|y|<R} |\varrho_\nu(t, y) - \varrho_\sigma(t, y)| dy \\ &\quad + \frac{1}{R} \int_{|y| \ge R} |y|^2 |\varrho_\nu(t, y) - \varrho_\sigma(t, y)| dy \\ &\le R \|\varrho_\nu(t) - \varrho_\sigma(t)\|_{L^1} + 2 \frac{M_2}{R}. \end{align*} $$

Optimizing in $R$ by equating the last two terms, we infer

$$ W_1(\varrho_\nu(t), \varrho_\sigma(t)) \le 2 \sqrt{2M_2 \|\varrho_\nu(t) - \varrho_\sigma(t)\|_{L^1}}. $$

Now we show that Theorem 1.3 and the assumptions that $\phi_\nu$ converges to $\phi_\sigma$ in $L^2$ imply that for any $T > 0$,

$$ \limsup_{\nu \to \sigma} \sup_{t \in [0, T]} \|\varrho_\nu(t) - \varrho_\sigma(t)\|_{L^1} = 0. $$
Indeed,

$$
\begin{align*}
\|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{L^1} &= \left\| \langle t \rangle^d \frac{|u_{\nu}(t, y \langle t \rangle)|^2}{\|\phi_{\nu}\|_{L^2}^2} - \langle t \rangle^d \frac{|u_{\sigma}(t, y \langle t \rangle)|^2}{\|\phi_{\sigma}\|_{L^2}^2} \right\|_{L^1} \\
&= \left\| \frac{|u_{\nu}(t)|^2}{\|\phi_{\nu}\|_{L^2}^2} - \frac{|u_{\sigma}(t)|^2}{\|\phi_{\sigma}\|_{L^2}^2} \pm \frac{|u_{\nu}(t)|^2}{\|\phi_{\sigma}\|_{L^2}^2} \right\|_{L^1} \\
&\le \frac{1}{\|\phi_{\sigma}\|_{L^2}^2} \left\| |u_{\nu}(t)|^2 - |u_{\sigma}(t)|^2 \right\|_{L^1} + \|\phi_{\nu}\|_{L^2} \left| \frac{1}{\|\phi_{\nu}\|_{L^2}^2} - \frac{1}{\|\phi_{\sigma}\|_{L^2}^2} \right| \\
&\le \frac{1}{\|\phi_{\sigma}\|_{L^2}^2} (\|\phi_{\nu}\|_{L^2} + \|\phi_{\sigma}\|_{L^2}) \|u_{\nu}(t) - u_{\sigma}(t)\|_{L^2} + o_{\nu \to \sigma}(1),
\end{align*}
$$

hence the claimed convergence.

Now for $t > T$, we write

$$
\begin{align*}
W_1 (\varrho_\nu(t), \varrho_\nu(T)) &= \sup_{\|\psi\|_{\text{Lip}} \le 1} \int_{\mathbb{R}^d} \psi(y) (\varrho_\nu(t, y) - \varrho_\nu(T, y)) \, dy \\
&= \sup_{\|\psi\|_{\text{Lip}} \le 1} \int_T^t \int_{\mathbb{R}^d} \psi(y) \partial_t \varrho_\nu(s, y) \, ds dy \\
&= \sup_{\|\psi\|_{\text{Lip}} \le 1} \int_T^t \int_{\mathbb{R}^d} \psi(y) \frac{1}{\langle s \rangle^2} \, \text{div} \, j_\nu(s, y) \, ds dy \\
&= \sup_{\|\psi\|_{\text{Lip}} \le 1} \int_T^t \int_{\mathbb{R}^d} \nabla \psi(y) \frac{1}{\langle s \rangle^2} j_\nu(s, y) \, ds dy \\
&\le \int_T^\infty \frac{1}{\langle s \rangle^2} \|j_\nu(s)\|_{L^1} \, ds \lesssim \int_T^\infty \frac{1}{\langle s \rangle^2} \times \langle s \rangle^{1-\min(1, \frac{d\nu}{2})} \, ds,
\end{align*}
$$

where we have used Cauchy-Schwarz inequality and (5.2). We infer

$$
\sup_{t \ge T} W_1 (\varrho_\nu(t), \varrho_\nu(T)) = o(1) \quad \text{as } T \to \infty,
$$

uniformly in $\nu \in [(1 - \eta)\sigma, (1 + \eta)\sigma]$, hence Theorem 1.6.

5.2. **Interpolation.** We now prove Corollary 1.7. We know from [20, Lemma 2.1]
that for any $s > \frac{d+1}{2}$, there exists $C = C(s, d)$ such that any probability measures
$f$ and $g$,

$$
\|f - g\|_{H^{-s}(\mathbb{R}^d)} \leq C W_1(f, g)^{1/2}.
$$

Therefore, Theorem 1.6 implies that

$$
(5.3) \quad \forall s > \frac{d+1}{2}, \quad \sup_{t \in \mathbb{R}} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{H^{-s}(\mathbb{R}^d)} \xrightarrow{\nu \to \sigma} 0.
$$

According to the value of $\sigma$ relative to $2/d$, we have an extra boundedness property
for $(\varrho_\nu)_\nu$ in some Sobolev space $W^{k,p}$ (with $k=0$ or $1$), and by interpolation, we
infer convergence to zero in intermediate Sobolev spaces.

Case $\sigma > 2/d$. If $\sigma > 2/d$, we know in addition that for $\nu$ in a neighborhood of $\sigma$, $\nu \ge 2/d$, hence $v_\nu \in L^\infty(\mathbb{R}; H^1)$.

• In the case $d = 1$, $H^1$ is an algebra, so $\varrho_\nu \in L^\infty(\mathbb{R}; H^1)$. Combined with the
convergence (5.3), this property implies that

$$
\forall s < 1, \quad \sup_{t \in \mathbb{R}} \|\varrho_{\nu}(t) - \varrho_{\sigma}(t)\|_{H^{s}(\mathbb{R}^{d})} \xrightarrow{\nu \to \sigma} 0.
$$
In particular, the convergence holds in $L^\infty(\mathbb{R}; L^2(\mathbb{R}))$, and since $\varrho_\nu \in L^\infty(\mathbb{R}; L^1)$,

$$ \forall p \in (1, \infty], \quad \sup_{t \in \mathbb{R}} \|\varrho_\nu(t) - \varrho_\sigma(t)\|_{L^p(\mathbb{R}^d)} \xrightarrow{\nu \to \sigma} 0. $$

• In the case $d = 2$, $H^1(\mathbb{R}^2)$ barely fails to be an algebra, and we check that $\varrho_\nu \in L^\infty(\mathbb{R}; W^{1,p} \cap L^q)$ for all $p \in [1, 2)$ and all $q \in [1, \infty)$. Indeed, $v_\nu \in L^\infty(\mathbb{R}; L^p)$ for all $p \in [2, \infty)$, and $\nabla v_\nu \in L^\infty(\mathbb{R}; L^2)$, hence the announced property by Hölder inequality. Therefore, $\varrho_\nu$ converge toward $\varrho_\sigma$ in $L^\infty(\mathbb{R}; X)$ for any space $X$ obtained by (real) interpolation between $H^{-s}$ and $W^{1,p}$, and different from $W^{1,p}$. We find in [30], Theorem 1 in Section 2.4.2 (pp. 184-185) that this space is $W^{s_\theta, p_\theta}$, where

$$ s_\theta = -s\theta + (1 - \theta), \quad \frac{1}{p_\theta} = \frac{\theta}{2} + \frac{1 - \theta}{p}, \quad \theta \in [0, 1). $$

We do not pursue the computation in full detail, and simply note that for fixed $s$, we can find $\theta_0 \in (0, 1)$ such that $s_{\theta_0} = 0$. By interpolating with the boundedness in $L^\infty L^q$ for all $q \in [1, \infty)$, we obtain the convergence in $L^\infty L^q$ for all $q \in (1, \infty)$.

• In the case $d \ge 3$, $H^1(\mathbb{R}^d) \hookrightarrow L^{\frac{2d}{d-2}}(\mathbb{R}^d)$, and Hölder inequality implies that $\varrho_\nu \in L^\infty(\mathbb{R}; W^{1, \frac{d}{d-1}})$, since

$$ \frac{1}{2} + \frac{d-2}{2d} = \frac{d-1}{d}. $$

Like in the case $d = 2$, we simply note that there exists $\theta \in (0, 1)$ such that

$$ [H^{-s}, W^{1, \frac{d}{d-1}}]_{\theta} = L^{p(d)}, $$

for some $p(d) \in (1, 2)$ whose exact value we do not compute. Combining this with the boundedness in $L^\infty(\mathbb{R}; L^1)$, we obtain the convergence in $L^p$ for all $p \in (1, p(d)]$.

Case $\sigma < 2/d$. In this case, we only know that $v_\nu \in L^\infty(\mathbb{R}; L^{2\sigma+2} \cap L^2)$, hence $\varrho_\nu \in L^\infty(\mathbb{R}; L^{\sigma+1} \cap L^1)$, that is $\varrho_\nu \in L^\infty(\mathbb{R}; W^{0,p})$ for all $p \in [1, 1 + \sigma]$. The interpolation between $H^{-s}$ and $W^{0,p}$ is now given by

$$ [H^{-s}, W^{0,p}]_{\theta} = W^{s_{\theta}, p_{\theta}}, \quad s_{\theta} = -\theta s, \quad \frac{1}{p_{\theta}} = \frac{\theta}{2} + \frac{1 - \theta}{p}, \quad \theta \in [0, 1]. $$

The result stated in Corollary 1.7 follows, by considering the limit $\theta \to 0$.

In the case $\sigma = 2/d$, considering the case $\nu > \sigma$ makes it possible to use the results from the case $\sigma > 2/d$, while if $\nu < 2/d$, we have to use the results from the case $\sigma < 2/d$. As the latter are weaker than the former, the case $\sigma = 2/d$ is included in the statement corresponding to $\sigma < 2/d$ in Corollary 1.7.

## 6. ANALYSIS OF THE DISPERSION EQUATIONS

The proof of the convergence of the solution $u_\sigma$ of the power nonlinear Schrödinger equation (1.3) to the solution $u_0$ of the logarithmic Schrödinger equation (1.4) involves ordinary differential equations, which were used to describe the dispersive rate for $u_0$ in [7], and were generalized in [5] in order to consider the power case. In this section, we recall some properties of the solutions to these ordinary differential equations, and show convergence results at this level when $\sigma \to 0$.
6.1. Power nonlinearity. Like in [5], for $\alpha > 0$, consider

$$ (6.1) \qquad \ddot{r}_{\alpha} = \frac{\alpha}{2r_{\alpha}^{\alpha+1}}; \quad r_{\alpha}(0) = 1, \dot{r}_{\alpha}(0) = 0. $$

Multiply by $\dot{r}_{\alpha}$ and integrate between 0 and t:

$$ (6.2) \qquad (\dot{r}_{\alpha})^2 = 1 - \frac{1}{r_{\alpha}^{\alpha}}. $$

So long as $r_{\alpha} \ge 0$, $\ddot{r}_{\alpha} \ge 0$: $\dot{r}_{\alpha}$ is nondecreasing, hence nonnegative, so $r_{\alpha}$ is nondecreasing (and remains positive). If $r_{\alpha}$ is bounded, then $\ddot{r}_{\alpha} \ge 1$, hence a contradiction by integrating. Therefore, since $r_{\alpha}$ is nondecreasing,

$$ r_{\alpha}(t) \underset{t \to +\infty}{\longrightarrow} +\infty. $$

Therefore, (6.2) implies

$$ \dot{r}_{\alpha}(t) \underset{t \to +\infty}{\longrightarrow} 1, $$

and so

$$ r_{\alpha}(t) \underset{t \to +\infty}{\sim} t. $$

We go one step further compared to [5], and examine the next term in the asymptotic expansion of $r_{\alpha}$ as $t \to +\infty$. Write

$$ r_{\alpha}(t) = t + w_{\alpha}(t), $$

so (6.2) yields

$$ (1 + \dot{w}_{\alpha})^2 = 1 - \frac{1}{(t + w_{\alpha})^{\alpha}}. $$

As $w_{\alpha}(t) = o(t)$,

$$ (t + w_{\alpha})^{-\alpha} = t^{-\alpha} \left(1 + \frac{w_{\alpha}}{t}\right)^{-\alpha} = t^{-\alpha} \left(1 - \alpha \frac{w_{\alpha}}{t} + \mathcal{O}\left(\left(\frac{w_{\alpha}}{t}\right)^2\right)\right). $$

At leading order, we have

$$ 2\dot{w}_{\alpha} = -\frac{1}{t^{\alpha}}, \quad \text{hence } w_{\alpha}(t) = \frac{1}{2(1-\alpha)}t^{1-\alpha}. $$

We see that for $\alpha \approx \frac{1}{\ln t}$, $w_{\alpha}$ becomes comparable with the leading order term, since $t^{-\alpha}$ becomes $\mathcal{O}(1)$.

Remark 6.1. We note in passing that the function $r_{\text{lin}}(t) = \langle t \rangle$ considered in Theorem 1.6 (and Section 5) is equal to $r_2$, since it solves

$$ \ddot{r}_{\text{lin}} = \frac{1}{r_{\text{lin}}^3}, \quad r_{\text{lin}}(0) = 1, \dot{r}_{\text{lin}}(0) = 0. $$

This function appears for instance when considering the evolution of Gaussian initial data under the linear Schrödinger flow $e^{i\frac{t}{2}\Delta}$ (see e.g. [7, Section 3]).
6.2. **Logarithmic case and transition.** The equation considered in [7] reads

$$ (6.3) \qquad \ddot{\tau}_0 = \frac{1}{2\tau_0} \quad ; \quad \tau_0(0) = 1, \dot{\tau}_0(0) = 0, $$

and we recall the main features of the solution (see [7] for a complete justification). Multiply by $\dot{\tau}_0$ and integrate between 0 and $t$:

$$ (6.4) \qquad (\dot{\tau}_0)^2 = \ln \tau_0, $$

which readily implies $\tau_0 \ge 1$. Like above, $\tau_0$ is convex, and

$$ \tau_0(t) \underset{t \to +\infty}{\longrightarrow} +\infty. $$

This is the beginning of the analysis presented in [7], which also yields:

$$ (6.5) \qquad \tau_0(t) \underset{t \to +\infty}{\sim} t\sqrt{\ln t}, \quad \dot{\tau}_0(t) \underset{t \to +\infty}{\sim} \sqrt{\ln t}. $$

We set

$$ \tau_{\sigma}(t) = r_{d\sigma} \left( \frac{t}{\sqrt{\sigma}} \right). $$

It solves

$$ (6.6) \qquad \ddot{\tau}_{\sigma} = \frac{1}{2\tau_{\sigma}^{d\sigma+1}} \quad ; \quad \tau_{\sigma}(0) = 1, \dot{\tau}_{\sigma}(0) = 0. $$

We have

$$ \tau_{\sigma}(t) \underset{t \to +\infty}{\sim} \frac{t}{\sqrt{\sigma}}. $$

Taking the boundary layer $\sigma \approx \frac{1}{\ln t}$ identified above, we have

$$ \tau_{\frac{1}{\ln t}}(t) \approx t\sqrt{\ln t}, $$

which is the same order of magnitude as $\tau_0$.

The choice $\alpha = d\sigma$ is motivated by the forthcoming analysis of the pseudo-energy $\mathcal{E}_\sigma$ defined in (7.8) in the proof of Lemma 7.1 (see also [5]).

6.3. **Convergence.** We want to prove that $\tau_\sigma \to \tau_0$ as $\sigma \to 0$. The convergence cannot hold uniformly in time, since the large time behavior of $\tau_0$ is different from that of $\tau_\sigma$ for $\sigma > 0$, and the above approach suggests that a transition occurs for $\sigma$ of order $1/\ln t$. Subtracting the analogue of (6.2) for $\tau_\sigma$, from (6.4), we find

$$ \begin{aligned} (\dot{\tau}_0)^2 - (\dot{\tau}_\sigma)^2 &= \ln \tau_0 - \frac{1}{d\sigma} \left(1 - \frac{1}{\tau_\sigma^{d\sigma}}\right) \\ &= \frac{1}{d\sigma} \left(\frac{1}{\tau_\sigma^{d\sigma}} - \frac{1}{\tau_0^{d\sigma}}\right) + \underbrace{\frac{1}{d\sigma} \left(\frac{1}{\tau_0^{d\sigma}} - 1 + d\sigma \ln \tau_0\right)}_{=:S_\sigma}. \end{aligned} $$

As $\tau_0(t) \ge 1$ for all $t \ge 0$, and since $e^{-z} - 1 + z \ge 0$ for all $z \ge 0$, the source term $S_\sigma$ is nonnegative for all $t \ge 0$. This implies $\tau_0(t) \ge \tau_\sigma(t)$ and $\dot{\tau}_0(t) \ge \dot{\tau}_\sigma(t)$ for all $t \ge 0$. Let

$$ w = \tau_0 - \tau_\sigma, \quad y = w^2 + (\dot{w})^2. $$

We have

$$ \ddot{w} = \frac{1}{2\tau_0} - \frac{1}{2\tau_\sigma^{1+d\sigma}} = \frac{1}{2\tau_0^{1+d\sigma}} (\tau_0^{d\sigma} - 1) + \frac{\tau_\sigma^{1+d\sigma} - \tau_0^{1+d\sigma}}{2\tau_\sigma^{1+d\sigma} \tau_0^{1+d\sigma}}. $$
Writing

$$ \tau_{\sigma}^{1+d\sigma} = \tau_{0}^{1+d\sigma} \left(1 - \frac{w}{\tau_{0}}\right)^{1+d\sigma}, $$

Taylor's formula yields

$$ \tau_{\sigma}^{1+d\sigma} - \tau_{0}^{1+d\sigma} = -(1+d\sigma)\tau_{0}^{1+d\sigma} \frac{w}{\tau_{0}} \underbrace{\int_{0}^{1} \left(1-\theta \frac{w}{\tau_{0}}\right)^{d\sigma} d\theta}_{\le 1, \text{ since } w \ge 0}. $$

On the other hand, for any $T > 0$ (independent of $\sigma$), we have, uniformly in $t \in [0, T]$,

$$ 0 \le \frac{1}{2\tau_0^{1+d\sigma}} (\tau_0^{d\sigma} - 1) \le C(T) \frac{\sigma}{\tau_0^{1+d\sigma}} \le C(T)\sigma. $$

Therefore,

$$ \dot{y} = 2w\dot{w} + 2\dot{w}\ddot{w} \le 2w\dot{w} + C(T)\sigma|\dot{w}| + \tau_0^{d\sigma}|w\dot{w}| $$

Using again the fact that $\tau_0$ is bounded on $[0, T]$, Young inequality yields

$$ \dot{y} \le y + C(T)^2 \sigma^2 + (\dot{w})^2 + C(T)y \le K(T)y + C(T)^2 \sigma^2, $$

and we conclude by Grönwall lemma:

**Lemma 6.2.** Let $\sigma > 0$. For all $t \ge 0$,

$$ (6.7) \qquad \tau_0(t) \ge \tau_\sigma(t), \quad \dot{\tau}_0(t) \ge \dot{\tau}_\sigma(t). $$

Let $T > 0$. There exists $C = C(T) > 0$ such that for all $\sigma \in (0, 1)$,

$$ \sup_{t \in [0,T]} |\tau_0(t) - \tau_\sigma(t)| + \sup_{t \in [0,T]} |\dot{\tau}_0(t) - \dot{\tau}_\sigma(t)| \le C\sigma. $$

Using this first result, we now address the convergence on larger time intervals:

**Proposition 6.3.** There exists $C > 0$ such that for all $\sigma \in (0, 1]$,

$$ |\tau_0(t) - \tau_\sigma(t)| \le C\sigma t (\ln t)^{3/2}, \quad |\dot{\tau}_0(t) - \dot{\tau}_\sigma(t)| \le C\sigma (\ln t)^{3/2}, \quad \forall t \ge 2. $$

Recall that $w = \tau_0 - \tau_\sigma \ge 0$. As we know that

$$ \dot{\tau}_{\sigma}^{2} = \frac{1}{d\sigma} \left(1 - \frac{1}{\tau_{\sigma}^{d\sigma}}\right) \quad \text{and} \quad \dot{\tau}_{0}^{2} = \ln \tau_{0}, $$

we can write

$$ \dot{w} = \sqrt{\ln \tau_0} - \frac{1}{\sqrt{d\sigma}} \sqrt{1 - \frac{1}{\tau_\sigma^{d\sigma}}} = \underbrace{f(0, \tau_0) - f(0, \tau_\sigma)}_{=A(\tau_0, \tau_\sigma)} + \underbrace{f(0, \tau_\sigma) - f(\sigma, \tau_\sigma)}_{=B}, $$

where $f(\cdot, z)$ is continuous at every $z > 0$,

$$ f(\sigma, z) := \frac{1}{\sqrt{d\sigma}} \sqrt{1 - \frac{1}{z^{d\sigma}}} \underset{\sigma \to 0}{\longrightarrow} \sqrt{\ln z} =: f(0, z). $$

To estimate $A$, we write that

$$ \sqrt{\ln \tau_0} - \sqrt{\ln \tau_\sigma} = \sqrt{\ln \tau_0} \left( 1 - \sqrt{\frac{\ln \tau_\sigma}{\ln \tau_0}} \right) = \sqrt{\ln \tau_0} \left( 1 - \sqrt{1 - \frac{\ln(\tau_0/\tau_\sigma)}{\ln \tau_0}} \right). $$
Denote $X = \frac{\ln(\tau_0/\tau_\sigma)}{\ln \tau_0}$ and remark that as $1 \le \tau_\sigma \le \tau_0$, we have $0 \le X \le 1$. Then from the estimate $\frac{1}{2}X \le 1 - \sqrt{1-X} \le X$ for all $X \in [0, 1]$, we get

$$0 \le A(\tau_0, \tau_\sigma) \le \frac{\ln\left(\frac{\tau_0}{\tau_\sigma}\right)}{\sqrt{\ln \tau_0}} = \frac{\ln\left(1 + \frac{\tau_0 - \tau_\sigma}{\tau_\sigma}\right)}{\sqrt{\ln \tau_0}}.$$

We then denote $Y = \frac{\tau_0 - \tau_\sigma}{\tau_\sigma}$, so that using the inequality

$$\ln(1+Y) \le \frac{Y}{1+Y} (1+\ln(1+Y/2)),$$

which is easily proved by studying the map $Y \mapsto (1+Y)\ln(1+Y) - Y(1+\ln(1+Y/2))$ (its second order derivative is negative for $Y > 0$), we infer

$$A(\tau_0, \tau_\sigma) \le \frac{\tau_0 - \tau_\sigma}{\tau_0 \sqrt{\ln \tau_0}} \left( 1 + \ln \left( \frac{\tau_0 + \tau_\sigma}{2\tau_\sigma} \right) \right).$$

We now need to bound $\tau_\sigma$ from below. By definition, $\tau_\sigma(0) = 1$ and $\dot{\tau}_\sigma(0) = 0$, so using the previous estimate on compact time intervals, we know that for all $\epsilon > 0$, there exists $T_0 > 0$ independent of $\sigma > 0$ such that for $t \ge T_0$,

$$\dot{\tau}_\sigma(t)^2 = \frac{1}{d\sigma} \left(1 - \frac{1}{\tau_\sigma(t)^{d\sigma}}\right) \ge \frac{1}{d\sigma} \left(1 - \frac{1}{(1+\epsilon)^{d\sigma}}\right) \ge C_0,$$

with $C_0 > 0$ independent of $\sigma \in (0, 1]$. Taking the square root and integrating over time, we then get that there exists $c_0 > 0$ independent of $\sigma$ such that

$$\tau_\sigma(t) \ge c_0 t$$

for all $t \ge T_0$. We then infer from the behavior of $\tau_0$ that

$$\ln\left(\frac{1}{2} + \frac{\tau_0}{\tau_\sigma}\right) \le \ln\left(\frac{1}{2} + c_0^{-1}\sqrt{\ln t}\right) \le C \ln \ln t,$$

for a constant $C > 0$ independent of $\sigma$. This entails that there exists a constant $C_A > 0$ uniform in $\sigma \in (0, 1]$ such that for all $t \ge T_0$,

$$A(\tau_0, \tau_\sigma) \le C_A \frac{\tau_0 - \tau_\sigma}{\tau_0} \frac{\ln \ln t}{\sqrt{\ln \tau_0}}.$$

We now turn to the analysis of $B$: for $z > 0$ fixed, Taylor expansion in $\sigma$ yields

$$\frac{1}{z^{d\sigma}} = 1 - d\sigma \ln z + \int_{0}^{d\sigma} (d\sigma - \beta) \frac{(\ln z)^2}{z^\beta} d\beta,$$

and we write

$$f(\sigma, z) = \sqrt{\ln z - \frac{1}{d\sigma} \int_{0}^{d\sigma} (d\sigma - \beta) \frac{(\ln z)^2}{z^\beta} d\beta},$$

so

$$f(0, z) - f(\sigma, z) = \sqrt{\ln z} \left( 1 - \sqrt{1 - \underbrace{\frac{\ln z}{d\sigma} \int_{0}^{d\sigma} \frac{d\sigma - \beta}{z^\beta} d\beta}_{=: \zeta_\sigma(z)}} \right).$$

Note that we have $0 \le \zeta_\sigma(z) \le 1$ for all $z \ge 1$. Thanks to the estimate

$$1 - \sqrt{1 - X} \le X, \quad 0 \le X \le 1,$$
we infer that

$$f(0, \tau_{\sigma}) - f(\sigma, \tau_{\sigma}) \le \frac{(\ln \tau_{\sigma})^{3/2}}{d\sigma} \int_{0}^{d\sigma} \frac{d\sigma - \beta}{\tau_{\sigma}^{\beta}} d\beta \le \frac{d\sigma}{2} (\ln \tau_{0})^{3/2},$$

using that $\tau_{\sigma} \le \tau_0$ and $\tau_{\sigma} \ge 1$. To summarize, we have proved that

$$(6.8) \qquad 0 \le \dot{w} \le \underbrace{\frac{\ln \ln t}{2\tau_0 \sqrt{\ln \tau_0}}}_{=: \psi(t)} w + \underbrace{\frac{d\sigma}{2}(\ln \tau_0)^{3/2}}_{=: K(t)}, \quad \forall t \ge T_0.$$

Grönwall lemma yields

$$w(t) \le \exp \left( \int_{T_0}^{t} \psi(s) ds \right) w(T_0) + \int_{T_0}^{t} K(t) \exp \left( \int_{s}^{t} \psi(\gamma) d\gamma \right) ds.$$

In view of [7, Lemma 1.6],

$$\tau_0(s) = s\sqrt{\ln s} \left( 1 + \mathcal{O}\left( \frac{\ln \ln s}{\ln s} \right) \right),$$

and so

$$\ln \tau_0(s) = \ln s + \ln \sqrt{\ln s} + \mathcal{O}\left(\frac{\ln \ln s}{\ln s}\right).$$

We then compute

$$\begin{align*}
\psi(s) &= \frac{\ln \ln s}{2\tau_0(s)\sqrt{\ln \tau_0(s)}} \\
&= \frac{\ln \ln s}{2s\sqrt{\ln s}\left(1 + \mathcal{O}\left(\frac{\ln \ln s}{\ln s}\right)\right)\sqrt{\ln s + \ln \sqrt{\ln s} + \mathcal{O}\left(\frac{\ln \ln s}{\ln s}\right)}} \\
&= \frac{\ln \ln s}{2s\ln s\left(1 + \mathcal{O}\left(\frac{\ln \ln s}{\ln s}\right)\right)\sqrt{1 + \frac{\ln \ln s}{2\ln s} + \left(1 + \mathcal{O}\left(\frac{\ln \ln s}{(\ln s)^2}\right)\right)}} \\
&= \frac{\ln \ln s}{2s\ln s\left(1 + \mathcal{O}\left(\frac{\ln \ln s}{\ln s}\right)\right)} = \frac{1}{2s\ln s}\left(1 + \mathcal{O}\left(\frac{\ln \ln s}{\ln s}\right)\right),
\end{align*}$$

hence, since $s \mapsto \frac{(\ln \ln s)^2}{s(\ln s)^2}$ is integrable as $s \to \infty$ (it is controlled by $\frac{1}{s(\ln s)^{3/2}}$ for instance),

$$\int_{T_0}^{t} \psi(\gamma) d\gamma = \frac{1}{4} (\ln \ln t)^2 + \mathcal{O}(1) \quad \text{as } t \to \infty.$$

We can then estimate the first term in the Grönwall inequality, using Lemma 6.2 on $[0, T_0]$, so that

$$\exp \left( \int_{T_0}^{t} \psi(s) ds \right) w(T_0) \le C(T_0) \sigma e^{\frac{1}{4}(\ln \ln t)^2},$$

for a constant $C(T_0) > 0$ independent of $\sigma$ and $t$. For the second term, we once again use the asymptotic behavior of $\tau_0$ together with the previous bound on $\psi$ so
that

$$
\begin{align*}
\int_{T_0}^{t} K(t) \exp \left( \int_{s}^{t} \psi(\gamma) d\gamma \right) ds &\lesssim \sigma e^{\frac{1}{4}(\ln \ln t)^2} \int_{T_0}^{t} (\ln s)^{\frac{3}{2}} e^{-\frac{1}{4}(\ln \ln s)^2} ds \\
&\lesssim \sigma e^{\frac{1}{4}(\ln \ln t)^2} (\ln t)^{\frac{3}{2}} \int_{T_0}^{t} e^{-\frac{1}{4}(\ln \ln s)^2} ds.
\end{align*}
$$

The last integral diverges when $t \to +\infty$, as can be seen by integrating by parts:

$$
\int_{T_0}^{t} e^{-\frac{1}{4}(\ln \ln s)^2} ds = \left[se^{-\frac{1}{4}(\ln \ln s)^2}\right]_{T_0}^{t} + \frac{1}{2} \int_{T_0}^{t} \frac{\ln \ln s}{\ln s} e^{-\frac{1}{4}(\ln \ln s)^2} ds.
$$

Since $\frac{\ln \ln s}{\ln s} \to 0$ as $s \to \infty$, the boundary term (at $s = t$) provides the leading order behavior in the asymptotic expansion as $t \to \infty$, so finally we have

$$
\int_{T_0}^{t} K(t) \exp \left( \int_{s}^{t} \psi(\gamma) d\gamma \right) ds \lesssim \sigma t (\ln t)^{3/2}.
$$

Going back to the Grönwall inequality, we finally get by comparing behaviors that

$$
w(t) \lesssim \sigma t (\ln t)^{3/2}.
$$

In the regime $\sigma = \frac{1}{\ln t}$, we recover the time growth of $w$, induced by $\tau_0$. Going back to (6.8), we infer the announced estimate for $\dot{w} = \dot{\tau}_0 - \tau_0$.

7. UNIFORM ESTIMATES

In this section, we prove large time a priori estimates for the solutions to (1.3)
and (1.4). To do so, we modulate the corresponding solutions **u**_σ and **u**_0 by using
the dispersion rates τ_σ and τ_0 considered in Section 6.

7.1. Dispersion rescaling: logarithmic case. Let $\tau_0$ denote the solution to (6.3). As established in [7] (with a different normalization for $\tau_0$), the function $v_0$, related to the solution $\mathbf{u}_0$ to (1.4) via

$$
(7.1) \qquad \mathbf{u}_0(t, x) = \frac{1}{\tau_0(t)^{d/2}} v_0 \left( t, \frac{x}{\tau_0(t)} \right) e^{i \frac{\dot{\tau}_0(t)}{\tau_0(t)} \frac{|x|^2}{2}},
$$

which is nothing but (7.5) with indices 0, solves:

$$
i \partial_t v_0 + \frac{1}{2\tau_0(t)^2} \Delta v_0 = \frac{|y|^2}{4} v_0 + v_0 \ln(|v_0|^2) - dv_0 \ln \tau_0.
$$

The last term is removed by an explicit time-dependent gauge transform. The
pseudo-energy

$$
\mathcal{E}_0(t) = \frac{1}{2\tau_0(t)^2} \|\nabla v_0(t)\|_{L^2}^2 + \frac{1}{4} \|yv_0(t)\|_{L^2}^2 + \int_{\mathbb{R}^d} |v_0(t, y)|^2 \ln(|v_0(t, y)|^2) dy
$$

satisfies

$$
\dot{\mathcal{E}}_0 = -2 \frac{\dot{\tau}_0}{\tau_0} \times \frac{1}{2\tau_0(t)^2} \|\nabla_y v_0(t)\|_{L^2}^2.
$$

We infer the a priori estimates (see [7] for details)

$$
(7.2) \quad \sup_{t \ge 0} \left( \int_{\mathbb{R}^d} \left( 1 + |y|^2 + |\ln|v_0(t, y)||^2 \right) |v_0(t, y)|^2 dy + \frac{1}{\tau_0(t)^2} \|\nabla v_0(t)\|_{L^2}^2 \right) < \infty
$$

and

$$
(7.3) \qquad \int_0^\infty \frac{\dot{\tau}_0(t)}{\tau_0^3(t)} \|\nabla v_0(t)\|_{L^2(\mathbb{R}^d)}^2 dt < \infty.
$$
Back to $\mathbf{u}_0$, (7.1), (6.5) and (7.2) yield the bound

$$ (7.4) \qquad \|\nabla \mathbf{u}_0(t)\|_{L^2(\mathbb{R}^d)} \lesssim \langle \ln t \rangle^{1/2}. $$

If is shown in [7] that this bound is sharp, a property that we do not use here.

## 7.2. Toward the logarithmic nonlinearity.
If we consider (1.3), the change of unknown

$$ (7.5) \qquad \mathbf{u}_{\sigma}(t, x) = \frac{1}{\tau_{\sigma}(t)^{d/2}} v_{\sigma}\left(t, \frac{x}{\tau_{\sigma}(t)}\right) e^{i \frac{\dot{\tau}_{\sigma}(t)}{\tau_{\sigma}(t)} \frac{|x|^2}{2}}, $$

leads to

$$ i \partial_t v_{\sigma} + \frac{1}{2\tau_{\sigma}^2} \Delta v_{\sigma} = \frac{|y|^2}{4\tau_{\sigma}^{d\sigma}} v_{\sigma} + \frac{1}{\sigma\tau_{\sigma}^{d\sigma}} |v_{\sigma}|^{2\sigma} v_{\sigma} - \frac{1}{\sigma} v_{\sigma}. $$

Up to another gauge transform, we get

$$ (7.6) \qquad i \partial_t v_{\sigma} + \frac{1}{2\tau_{\sigma}^2} \Delta v_{\sigma} = \frac{|y|^2}{4\tau_{\sigma}^{d\sigma}} v_{\sigma} + \frac{1}{\sigma\tau_{\sigma}^{d\sigma}} (|v_{\sigma}|^{2\sigma} - 1) v_{\sigma}. $$

This means that we have replaced the initial $v_{\sigma}$ with

$$ (7.7) \qquad \tilde{v}_{\sigma}(t, y) = v_{\sigma}(t, y) \exp \left( -i \frac{t}{\sigma} + i \int_{0}^{t} \frac{ds}{\sigma \tau_{\sigma}(s)^{d\sigma}} \right). $$

We have dropped the tildas to lighten notations. We note that if we forget that $\tau_{\sigma}$ solves (6.6), then the coefficient of the harmonic potential is given more generally by $\tau_{\sigma}\ddot{\tau}_{\sigma}/2$. The choice $\alpha = d\sigma$ in Section 6 is made to ensure that the power of $\tau_{\sigma}$ in front of the two terms on the right hand side of (7.6) is the same, a property which is crucial in the proof of the following lemma:

**Lemma 7.1.** Let $\varepsilon \in (0, 1/d)$, and $(\phi_{\sigma})_{0 \le \sigma \le \varepsilon}$ bounded in $\Sigma$. For $\mathbf{u}_{\sigma}$ solution to (1.3), define $v_{\sigma}$ by

$$ \mathbf{u}_{\sigma}(t, x) = \frac{1}{\tau_{\sigma}(t)^{d/2}} v_{\sigma}\left(t, \frac{x}{\tau_{\sigma}(t)}\right) \exp\left(i \frac{\dot{\tau}_{\sigma}(t)}{\tau_{\sigma}(t)} \frac{|x|^2}{2}\right), $$

where $\tau_{\sigma}$ solves (6.6). There exists $C$ independent of $t \ge 0$ and $\sigma \in (0, 1/d)$ such that

$$ \frac{1}{\tau_{\sigma}(t)^{2-d\sigma}} \|\nabla v_{\sigma}(t)\|_{L^2}^2 + \|y v_{\sigma}(t)\|_{L^2}^2 + \int_{\mathbb{R}^d} \left| \frac{|v_{\sigma}(t, y)|^{2\sigma} - 1}{\sigma} \right| |v_{\sigma}(t, y)|^2 dy \le C, $$

and

$$ \int_{0}^{\infty} \frac{\dot{\tau}_{\sigma}(t)}{\tau_{\sigma}(t)^{3-d\sigma}} \|\nabla v_{\sigma}(t)\|_{L^2}^2 dt \le C. $$

*Proof.* Introduce the pseudo-energy

$$ (7.8) \qquad \begin{aligned} \mathcal{E}_{\sigma}(t) = & \frac{1}{2\tau_{\sigma}(t)^2} \|\nabla v_{\sigma}(t)\|_{L^2}^2 + \frac{1}{4\tau_{\sigma}(t)^{d\sigma}} \|y v_{\sigma}(t)\|_{L^2}^2 \\ & + \frac{1}{(\sigma+1)\tau_{\sigma}(t)^{d\sigma}} \int_{\mathbb{R}^d} \left( \frac{|v_{\sigma}(t, y)|^{2\sigma} - 1}{\sigma} \right) |v_{\sigma}(t, y)|^2 dy. \end{aligned} $$
Since the purely time dependent phase function in (7.7) does not affect $\mathcal{E}_\sigma$, we may consider that $v_\sigma$ solves (7.6). We compute

$$ \dot{\mathcal{E}}_{\sigma}(t) = -\frac{\dot{\tau}_{\sigma}(t)}{\tau_{\sigma}(t)} \left( \frac{1}{\tau_{\sigma}(t)^2} \|\nabla v_{\sigma}(t)\|_{L^2}^2 + \frac{d\sigma}{4\tau_{\sigma}(t)^{d\sigma}} \|yv_{\sigma}(t)\|_{L^2}^2 \right. \\ \left. + \frac{d\sigma}{(\sigma+1)\tau_{\sigma}(t)^{d\sigma}} \int_{\mathbb{R}^d} \left( \frac{|v_{\sigma}(t, y)|^{2\sigma} - 1}{\sigma} \right) |v_{\sigma}(t, y)|^2 dy \right). $$

We note that, like $\mathcal{E}_0$, $\mathcal{E}_\sigma$ is not the sum of positive terms, and decompose it as $\mathcal{E}_\sigma = \mathcal{E}_\sigma^+ - \mathcal{E}_\sigma^-$, where

$$ \begin{aligned} \mathcal{E}_{\sigma}^{+}(t) &= \frac{1}{2\tau_{\sigma}(t)^2} \|\nabla v_{\sigma}(t)\|_{L^2}^2 + \frac{1}{4\tau_{\sigma}(t)^{d\sigma}} \|yv_{\sigma}(t)\|_{L^2}^2 \\ &\quad + \frac{1}{(\sigma+1)\tau_{\sigma}(t)^{d\sigma}} \int_{|v_{\sigma}| \ge 1} \left( \frac{|v_{\sigma}(t, y)|^{2\sigma} - 1}{\sigma} \right) |v_{\sigma}(t, y)|^2 dy, \\ \mathcal{E}_{\sigma}^{-}(t) &= \frac{1}{(\sigma+1)\tau_{\sigma}(t)^{d\sigma}} \int_{|v_{\sigma}| < 1} \left( \frac{1 - |v_{\sigma}(t, y)|^{2\sigma}}{\sigma} \right) |v_{\sigma}(t, y)|^2 dy. \end{aligned} $$

Now $\mathcal{E}_\sigma^+$ is the sum of three nonnegative terms, and $\mathcal{E}_\sigma^-$ is nonnegative. Taylor formula for the function $f(\sigma) = y^\sigma$ yields

$$ \frac{1 - y^{\sigma}}{\sigma} = \ln \frac{1}{y} \int_{0}^{1} y^{\theta\sigma} d\theta. $$

We infer

$$ \begin{aligned} \mathcal{E}_{\sigma}^{-}(t) &\le \frac{1}{(\sigma+1)\tau_{\sigma}(t)^{d\sigma}} \int_{|v_{\sigma}|<1} |v_{\sigma}(t, y)|^2 \ln \frac{1}{|v_{\sigma}(t, y)|^2} dy \\ &\lesssim \frac{C(\eta)}{(\sigma+1)\tau_{\sigma}(t)^{d\sigma}} \int_{\mathbb{R}^d} |v_{\sigma}(t, y)|^{2-2\eta} dy, \end{aligned} $$

where $\eta > 0$ is arbitrarily small. In view of the conservation of the mass and (2.2), this implies

$$ (7.9) \qquad \mathcal{E}_{\sigma}^{-}(t) \lesssim \frac{1}{\tau_{\sigma}(t)^{d\sigma}} \|yv_{\sigma}(t)\|_{L^2}^{\frac{d\eta}{2-2\eta}} \lesssim \frac{1}{\tau_{\sigma}(t)^{d\sigma}} (\tau_{\sigma}^{d\sigma} \mathcal{E}_{\sigma}^{+})^{\frac{d\eta}{1-\eta}}. $$

This implies in particular that $\mathcal{E}_\sigma(0)$ is bounded uniformly in $\sigma \in (0, 1/d)$. In view of the derivative of $\mathcal{E}_\sigma$, we also have

$$ \frac{d}{dt} (\tau_{\sigma}^{d\sigma} \mathcal{E}_{\sigma}) = -\dot{\tau}_{\sigma} \tau^{d\sigma-1} \left(1 - \frac{d\sigma}{2}\right) \frac{1}{\tau_{\sigma}^2} \|\nabla v_{\sigma}\|_{L^2}^2 \le -\dot{\tau}_{\sigma} \tau_{\sigma}^{d\sigma-1} \frac{1}{2\tau_{\sigma}^2} \|\nabla v_{\sigma}\|_{L^2}^2, $$

where we have used the facts that $\tau_\sigma, \dot{\tau}_\sigma \ge 0$ and $\sigma < 1/d$. We infer the uniform bound

$$ \mathcal{E}_{\sigma}(t) \le \frac{\mathcal{E}_{\sigma}(0)}{\tau_{\sigma}(t)^{d\sigma}} \le \frac{C}{\tau_{\sigma}(t)^{d\sigma}}. $$

Invoking (7.9), we have

$$ \tau_{\sigma}^{d\sigma} \mathcal{E}_{\sigma}^{+} \le C + \tau_{\sigma}^{d\sigma} \mathcal{E}_{\sigma}^{-} \lesssim 1 + (\tau_{\sigma}^{d\sigma} \mathcal{E}_{\sigma}^{+})^{\frac{d\eta}{1-\eta}}. $$

Taking $\sigma > 0$ such that $\frac{d\eta}{1-\eta} < 1$ shows that $\tau_\sigma^{d\sigma} \mathcal{E}_\sigma^+$ is bounded uniformly in $t \ge 0$ and $\sigma \in (0, 1/d)$. Again from (7.9), this implies that so is $\tau_\sigma^{d\sigma} \mathcal{E}_\sigma^-$, hence
$\tau_{\sigma}^{d\sigma} (\mathcal{E}_{\sigma}^{+} + \mathcal{E}_{\sigma}^{-}) \le C$, which is the first claim of the lemma. We infer that $\tau_{\sigma}^{d\sigma} \mathcal{E}_{\sigma}$ is uniformly bounded from below, so its (nonpositive) derivative is integrable,

$$ \int_{0}^{\infty} \frac{\dot{\tau}_{\sigma}(t)}{\tau_{\sigma}(t)^{3-d\sigma}} \|\nabla v_{\sigma}(t)\|_{L^2}^2 dt \le C, $$

which completes the proof. □

## 8. FROM POWER TO LOGARITHMIC NONLINEARITY: EHRENFEST TIME

Since the rest of the paper is dedicated to the limit $\sigma \to 0$, we shall assume from now on that $0 < \sigma < 1/(2d)$, and $\varepsilon \le 1/(2d)$ in the assumption of Theorem 1.8.

**8.1. Proof of Theorem 1.8.** In view of Lemma 7.1, since $\tau_{\sigma}(t), \dot{\tau}_{\sigma}(t) \ge 0$ for all $t \ge 0$, we have

$$ \begin{array}{l} (8.1) \\ \|\nabla \mathbf{u}_{\sigma}(t)\|_{L^2} \le \frac{1}{\tau_{\sigma}(t)} \|\nabla v_{\sigma}(t)\|_{L^2} + \dot{\tau}_{\sigma}(t) \|y v_{\sigma}(t)\|_{L^2} \le C (\tau_{\sigma}(t)^{d\sigma/2} + \dot{\tau}_{\sigma}(t)), \\ \|x \mathbf{u}_{\sigma}(t)\|_{L^2} = \tau_{\sigma}(t) \|y v_{\sigma}(t)\|_{L^2} \le C \tau_{\sigma}(t), \end{array} $$

where $C$ is independent of $\sigma \in (0, \varepsilon)$ and $t \ge 0$. The difference $w = \mathbf{u}_{\sigma} - \mathbf{u}_{0}$ solves

$$ i \partial_t w + \frac{1}{2} \Delta w = \frac{1}{\sigma} (|\mathbf{u}_{\sigma}|^{2\sigma} - 1) \mathbf{u}_{\sigma} - \mathbf{u}_{\sigma} \ln(|\mathbf{u}_{\sigma}|^2) + \mathbf{u}_{\sigma} \ln(|\mathbf{u}_{\sigma}|^2) - \mathbf{u}_{0} \ln(|\mathbf{u}_{0}|^2), $$

with initial value $w|_{t=0} = \phi_{\sigma} - \phi_{0}$. The source term is

$$ (8.2) \qquad S_{\sigma} = \frac{1}{\sigma} (|\mathbf{u}_{\sigma}|^{2\sigma} - 1) \mathbf{u}_{\sigma} - \mathbf{u}_{\sigma} \ln(|\mathbf{u}_{\sigma}|^2). $$

Taylor formula applied to $\sigma \mapsto (|z|^{2\sigma} - 1)z = (e^{\sigma \ln|z|} - 1)z$ yields

$$ (8.3) \qquad S_{\sigma} = \sigma \mathbf{u}_{\sigma} (\ln(|\mathbf{u}_{\sigma}|^2))^2 \int_{0}^{1} (1 - \theta) |\mathbf{u}_{\sigma}|^{2\theta} d\theta, $$

hence the pointwise bound

$$ |S_{\sigma}| \lesssim \sigma (\ln(|\mathbf{u}_{\sigma}|^2))^2 (|\mathbf{u}_{\sigma}| + |\mathbf{u}_{\sigma}|^{2\sigma+1}). $$

Let $\eta > 0$: for $0 < \sigma \le \eta/4$, we have the uniform pointwise estimate

$$ |S_{\sigma}| \lesssim \sigma (|\mathbf{u}_{\sigma}|^{1-\eta} + |\mathbf{u}_{\sigma}|^{1+\eta}), $$

where the implicit constant depends on $\eta > 0$. Gagliardo-Nirenberg inequality (2.1) yields, in view of (8.1),

$$ \|\|\mathbf{u}_{\sigma}\|_{L^2}^{1+\eta}\| = \|\mathbf{u}_{\sigma}\|_{L^{2+2\eta}}^{1+\eta} \lesssim \|\mathbf{u}_{\sigma}\|_{L^2}^{1+\eta-d\eta/2} \|\nabla \mathbf{u}_{\sigma}\|_{L^2}^{d\eta/2} \lesssim (\tau_{\sigma}(t)^{d\sigma/2} + \dot{\tau}_{\sigma}(t))^{d\eta/2}. $$

Similarly, (2.2) yields, in view of (8.1),

$$ \|\|\mathbf{u}_{\sigma}\|_{L^2}^{1-\eta}\| = \|\mathbf{u}_{\sigma}\|_{L^{2-2\eta}}^{1-\eta} \lesssim \|\mathbf{u}_{\sigma}\|_{L^2}^{1-\eta-d\eta/2} \|x \mathbf{u}_{\sigma}\|_{L^2}^{d\eta/2} \lesssim \tau_{\sigma}(t)^{d\eta/2}. $$

We infer

$$ (8.4) \qquad \|S_{\sigma}(t)\|_{L^2} \lesssim \sigma (\tau_{\sigma}(t)^{d\sigma/2} + \tau_{\sigma}(t) + \dot{\tau}_{\sigma}(t))^{d\eta/2}. $$

To complete the energy estimate, recall an identity discovered in [11]:

**Lemma 8.1** (From Lemma 1.1.1 in [11]). There holds

$$ \left| \operatorname{Im} \left( \left( z_2 \log |z_2|^2 - z_1 \log |z_1|^2 \right) \left( \overline{z_2} - \overline{z_1} \right) \right) \right| \le 2 |z_2 - z_1|^2, \quad \forall z_1, z_2 \in \mathbb{C}. $$
The energy estimate then yields, together with Cauchy-Schwarz inequality,

$$ \frac{d}{dt} \|w\|_{L^2}^2 \le 4 \|w\|_{L^2}^2 + 2 \|w\|_{L^2} \|S_\sigma\|_{L^2} \le 5 \|w\|_{L^2}^2 + \|S_\sigma\|_{L^2}^2, $$

where we have used Young inequality. Grönwall lemma and (8.4) imply:

$$ \|w(t)\|_{L^2}^2 \lesssim \|w(0)\|_{L^2}^2 e^{5t} + \sigma^2 \int_0^t e^{5(t-s)} \left( \tau_\sigma(s)^{d\sigma/2} + \tau_\sigma(s) + \dot{\tau}_\sigma(s) \right)^{d\eta} ds. $$

In view of (6.5) and (6.7), we infer the bound, uniform in $\sigma \in (0, \varepsilon)$,

$$ \|w(t)\|_{L^2}^2 \lesssim \|w(0)\|_{L^2}^2 e^{5t} + \sigma^2 \int_0^t e^{5(t-s)} \langle s\sqrt{\ln s} \rangle^{d\eta} ds. $$

This yields Theorem 1.8.

## 8.2. Convergence in Wasserstein distance.
Recall that the probability densities $\varrho_0$ and $\varrho_\sigma$ are given by

$$ \varrho_0(t, y) = \tau_0(t)^d |\mathbf{u}_0(t, y\tau_0(t))|^2 \|\phi_0\|_{L^2}^{-2}, \quad \varrho_\sigma(t, y) = \tau_\sigma(t)^d |\mathbf{u}_\sigma(t, y\tau_\sigma(t))|^2 \|\phi_\sigma\|_{L^2}^{-2}, $$

and, to lighten notations, introduce $\rho_0$ and $\rho_\sigma$ are given by

$$ \rho_0(t, y) = \tau_0(t)^d |\mathbf{u}_0(t, y\tau_0(t))|^2, \quad \rho_\sigma(t, y) = \tau_\sigma(t)^d |\mathbf{u}_\sigma(t, y\tau_\sigma(t))|^2. $$

We have established in Section 7 the uniform bounds for the second momenta of these densities: there exists $C (\|\phi\|_{\Sigma})$ independent of $t \in \mathbb{R}$ and $\sigma < 1/(2d)$ such that

$$ (8.5) \quad \int_{\mathbb{R}^d} (1 + |y|^2) (\rho_0(t, y) + \rho_\sigma(t, y)) dy \le C (\|\phi\|_{\Sigma}), \quad \forall t \ge 0, \forall \sigma \in (0, \frac{1}{2d}). $$

Using the triangle inequality and changing variables in the integrals,

$$ \begin{aligned} \int_{\mathbb{R}^d} |\rho_0(t, y) - \rho_\sigma(t, y)| dy &\le \int_{\mathbb{R}^d} \left| \tau_0(t)^d |\mathbf{u}_0(t, y\tau_0(t))|^2 - \tau_\sigma(t)^d |\mathbf{u}_0(t, y\tau_\sigma(t))|^2 \right| dy \\ &\quad + \int_{\mathbb{R}^d} \left| |\mathbf{u}_0(t, x)|^2 - |\mathbf{u}_\sigma(t, x)|^2 \right| dx \\ &\lesssim \int_{\mathbb{R}^d} \left| |\mathbf{u}_0(t, y)|^2 - \left(\frac{\tau_\sigma(t)}{\tau_0(t)}\right)^d \left|\mathbf{u}_0\left(t, y\frac{\tau_\sigma(t)}{\tau_0(t)}\right)\right|^2 \right| dy \\ &\quad + \|\mathbf{u}_0(t) - \mathbf{u}_\sigma(t)\|_{L^2}^2 \\ &\lesssim \left| 1 - \left(\frac{\tau_\sigma(t)}{\tau_0(t)}\right)^d \right| \|\mathbf{u}_0(t)\|_{L^2}^2 \\ &\quad + \int_{\mathbb{R}^d} \left| |\mathbf{u}_0(t, y)|^2 - \left|\mathbf{u}_0\left(t, y\frac{\tau_\sigma(t)}{\tau_0(t)}\right)\right|^2 \right| dy + C_1 \sigma e^{C_0 t}, \end{aligned} $$

where we have used (6.7) and Theorem 1.8 for the last inequality. Proposition 6.3 implies, in view of the behavior of $\tau_0$,

$$ \left| 1 - \frac{\tau_{\sigma}(t)}{\tau_{0}(t)} \right| \lesssim \sigma \ln(t+4), \quad \forall t \ge 0, $$
where we add 4 to simplify notations for small $t$. In view of (6.7), the first term of the previous right hand side is controlled by $\sigma \ln(t+4)$ for $t \ge 0$. For the second term, in view of Cauchy-Schwarz inequality,

$$ \int_{\mathbb{R}^d} \left| \mathbf{u}_0(t, y) \right|^2 - \left| \mathbf{u}_0 \left( t, y \frac{\tau_\sigma(t)}{\tau_0(t)} \right) \right|^2 \left| dy \le \left( 1 + \left( \frac{\tau_0(t)}{\tau_\sigma(t)} \right)^{d/2} \right) \|\mathbf{u}_0(t)\|_{L^2} \times \left\| \mathbf{u}_0(t, y) - \mathbf{u}_0 \left( t, y \frac{\tau_\sigma(t)}{\tau_0(t)} \right) \right\|_{L^2} \right. $$

Taylor formula yields

$$ \mathbf{u}_0(t, y) - \mathbf{u}_0 \left( t, y \frac{\tau_\sigma(t)}{\tau_0(t)} \right) = \left( 1 - \frac{\tau_\sigma(t)}{\tau_0(t)} \right) \int_0^1 \nabla \mathbf{u}_0 \left( t, y \left( 1 + \kappa \left( \frac{\tau_\sigma(t)}{\tau_0(t)} - 1 \right) \right) \right) d\kappa. $$

In view of (6.7),

$$ 1 + \kappa \left( \frac{\tau_\sigma(t)}{\tau_0(t)} - 1 \right) \ge \frac{\tau_\sigma(t)}{\tau_0(t)}, \quad \forall \kappa \in [0, 1], \forall t \ge 0. $$

Using (7.4), we infer, for $0 \le t \le \ln \frac{1}{\sigma}$ (which implies $\frac{\tau_0(t)}{\tau_\sigma(t)} \le 1$),

$$ \int_{\mathbb{R}^d} \left| \mathbf{u}_0(t, y) \right|^2 - \left| \mathbf{u}_0 \left( t, y \frac{\tau_\sigma(t)}{\tau_0(t)} \right) \right|^2 \left| dy \le \sigma \ln(t+4) \| \nabla \mathbf{u}_0(t) \|_{L^2} \le \sigma (\ln(t+4))^{3/2} \right. $$

This proves strong convergence in $L^1$ for $0 \le t \le \ln \frac{1}{\sigma}$. Along with the uniform boundedness of second momenta (8.5), we have in particular, for all $\theta \in (0, 1)$,

$$ \sup_{|t| \le \frac{\theta}{C_0} \ln \frac{1}{\sigma}} \int_{\mathbb{R}^d} |y| |\rho_0(t, y) - \rho_\sigma(t, y)| dy \xrightarrow{\sigma \to 0} 0. $$

As the same remains true for $\varrho_0$ and $\varrho_\sigma$, this implies that for all $\theta \in (0, 1)$,

$$ \sup_{|t| \le \frac{\theta}{C_0} \ln \frac{1}{\sigma}} W_1(\varrho_0(t), \varrho_\sigma(t)) \xrightarrow{\sigma \to 0} 0. $$

## 9. FROM POWER TO LOGARITHMIC NONLINEARITY: GLOBAL DYNAMICS

We keep the notations from the previous section. Recall that changing the unknown function like in (7.5), up to a purely time dependent phase function (7.7), the new wave function solves (7.6). From Madelung transform $\rho_\sigma = |v_\sigma|^2$ and $J_\sigma = \text{Im}(\bar{v}_\sigma \nabla v_\sigma)$, we infer the system

$$ \left\{ \begin{aligned} \partial_t \rho_\sigma + \frac{1}{\tau_\sigma^2} \operatorname{div} J_\sigma &= 0, \\ \partial_t J_\sigma + \frac{1}{(\sigma+1)\tau_\sigma^{d\sigma}} \nabla \rho_\sigma^{\sigma+1} + \frac{2}{\tau_\sigma^{d\sigma}} y \rho_\sigma &= \frac{1}{4\tau_\sigma^2} \Delta \nabla \rho_\sigma - \frac{1}{\tau_\sigma^2} \operatorname{div} \operatorname{Re}(\nabla v_\sigma \otimes \nabla \bar{v}_\sigma). \end{aligned} \right. $$

Combining these equations and writing $\nu_\sigma = \text{Re}(\nabla v_\sigma \otimes \nabla \bar{v}_\sigma)$ leads to the following non-autonomous fractional Fokker-Planck equation with source terms

$$ (9.1) \quad \partial_t (\tau_\sigma^2 \partial_t \rho_\sigma) = \frac{1}{(\sigma+1)\tau_\sigma^{d\sigma}} \Delta \rho_\sigma^{\sigma+1} + \frac{2}{\tau_\sigma^{d\sigma}} \operatorname{div}(y \rho_\sigma) - \frac{1}{4\tau_\sigma^2} \Delta^2 \rho_\sigma + \frac{1}{\tau_\sigma^2} \operatorname{div} \operatorname{div} \nu_\sigma. $$

Note that we have

$$ \partial_t (\tau_\sigma^2 \partial_t \rho_\sigma) = \tau_\sigma^2 \partial_t^2 \rho_\sigma + 2 \dot{\tau}_\sigma \tau_\sigma \partial_t \rho_\sigma. $$
Here we recall that, as $t \to \infty$,

$$ \tau_{\sigma}(t) \sim \frac{t}{\sqrt{\sigma}} \quad \text{and} \quad \dot{\tau}_{\sigma}(t) \to \frac{1}{\sqrt{\sigma}}, $$

so $\tau_{\sigma}^2 \sim (\dot{\tau}_{\sigma} \tau_{\sigma})^2$ and both terms are of the same order in time, but not in $\sigma$.

**9.1. Change of time variable.** We now make the change of variables $s = s_{\sigma}(t)$ such that

$$ \partial_s = (1 - d\sigma)\dot{\tau}_{\sigma}(t)\tau_{\sigma}(t)^{d\sigma+1}\partial_t, $$

hence

$$ \begin{aligned} s_{\sigma}(t) &= \int \frac{1}{(1 - d\sigma)\dot{\tau}_{\sigma}(\gamma)\tau_{\sigma}(\gamma)^{d\sigma+1}} d\gamma = \int \frac{\ddot{\tau}_{\sigma}(\gamma)}{2(1 - d\sigma)\dot{\tau}_{\sigma}(\gamma)} d\gamma \\ &= \frac{1}{2(1 - d\sigma)} \ln \dot{\tau}_{\sigma}(t) + C_0, \end{aligned} $$

so taking $C_0 = 0$ and using that

$$ \dot{\tau}_{\sigma}(t) = \sqrt{\frac{1}{d\sigma}\left(1 - \frac{1}{\tau_{\sigma}(t)^{d\sigma}}\right)}, $$

we infer

$$ s_{\sigma}(t) = \frac{1}{4(1 - d\sigma)} \ln \left( \frac{1}{d\sigma} \left( 1 - \frac{1}{\tau_{\sigma}(t)^{d\sigma}} \right) \right). $$

In particular we see that as $t \to +\infty$,

$$ s_{\sigma}(t) \to -\frac{1}{4(1 - d\sigma)} \ln(d\sigma) =: s_{\sigma}^{\max}, $$

so this new change of variable compactifies time. On the other hand, defining $\tilde{\tau}_{\sigma}(s_{\sigma}(t)) = \tau_{\sigma}(t)$, we have

$$ (9.2) \qquad \tilde{\tau}_{\sigma}(s_{\sigma}) = \left(1 - d\sigma e^{4s_{\sigma}(1-d\sigma)}\right)^{-\frac{1}{d\sigma}}, $$

and

$$ \dot{\tilde{\tau}}_{\sigma}(s_{\sigma}) = \frac{d}{dt} \tilde{\tau}_{\sigma}(s_{\sigma}(t)) = \dot{\tau}_{\sigma}(t) = e^{2s_{\sigma}(1-d\sigma)}. $$

From this change of variable we infer a new integrable bound from Lemma 7.1 on the quantity

$$ V_{\sigma}(t) := \frac{1}{\tau_{\sigma}(t)^{2-d\sigma}} \|\nabla v_{\sigma}(t)\|_{L^2}^{2}, $$

writing that

$$ \begin{aligned} \int_{0}^{\infty} \frac{\dot{\tau}_{\sigma}(t)}{\tau_{\sigma}(t)} V_{\sigma}(t) dt &= (1 - d\sigma) \int_{-\infty}^{s_{\sigma}^{\max}} \dot{\tilde{\tau}}_{\sigma}(s)^2 \tilde{\tau}_{\sigma}(s)^{d\sigma} \tilde{V}_{\sigma}(s) ds \\ &= (1 - d\sigma) \int_{-\infty}^{s_{\sigma}^{\max}} e^{4s(1-d\sigma)} (1 - d\sigma e^{4s(1-d\sigma)})^{-1} \tilde{V}_{\sigma}(s) ds, \end{aligned} $$

with the notation $\tilde{V}_{\sigma}(s_{\sigma}) = V_{\sigma}(t(s_{\sigma}))$, which implies the uniform bound

$$ (9.3) \qquad \int_{-\infty}^{s_{\sigma}^{\max}} e^{4s(1-d\sigma)} \tilde{V}_{\sigma}(s) ds \le \frac{C}{1-d\sigma}. $$
Now making this change of time variable in equation (9.1) we then get the following equation on $\tilde{\rho}_{\sigma}(s_{\sigma}(t)) = \rho_{\sigma}(t)$:

$$ (9.4) \qquad \partial_{s_{\sigma}} \tilde{\rho}_{\sigma} = \frac{1}{\sigma+1} \Delta \tilde{\rho}_{\sigma}^{\sigma+1} + 2 \operatorname{div}(y \tilde{\rho}_{\sigma}) + \mathcal{R}, $$

where

$$ \begin{aligned} \mathcal{R} = & \frac{2}{1-d\sigma} e^{-4s_{\sigma}(1-d\sigma)} (1-d\sigma e^{4s_{\sigma}(1-d\sigma)}) \partial_{s_{\sigma}} \tilde{\rho}_{\sigma} \\ & - \frac{e^{-4s_{\sigma}(1-d\sigma)}}{(1-d\sigma)^2} (1-d\sigma e^{4s_{\sigma}(1-d\sigma)}) \partial_{s_{\sigma}}^2 \tilde{\rho}_{\sigma} \\ & - \frac{1}{4} (1-d\sigma e^{4s_{\sigma}(1-d\sigma)})^{\frac{2-d\sigma}{d\sigma}} \Delta^2 \tilde{\rho}_{\sigma} \\ & + (1-d\sigma e^{4s_{\sigma}(1-d\sigma)})^{\frac{2-d\sigma}{d\sigma}} \operatorname{div} \operatorname{div} \tilde{\nu}_{\sigma} \\ = & R_1 + R_2 + R_3 + R_4, \end{aligned} $$

with the notation $\tilde{\nu}_{\sigma}(s_{\sigma}(t)) = \nu_{\sigma}(t)$. Note that at leading order in time, discarding the term $\mathcal{R}$, we formally get the porous medium equation with drift

$$ \partial_t f = \frac{1}{\sigma+1} \Delta f^{\sigma+1} + 2 \operatorname{div}(y f), $$

whose solutions are known to converge exponentially in time towards a Barenblatt profile in Wasserstein distance [27]. Of course as our time range in $s_{\sigma}$ is compact, we do not expect such property for our system. However it is known that such Barenblatt functions converge towards a limit Gaussian profile as $\sigma \to 0$, see for instance [14, Theorem 2.10]. This motivates our upcoming convergence result of $\tilde{\rho}_{\sigma}(s_{\sigma})$ towards a limit Gaussian profile in Wasserstein distance as both $t \to +\infty$ and $\sigma \to 0$. Our strategy of proof will be to look at the nonlinear porous medium equation with drift as a perturbation of a linear harmonic Fokker-Planck operator in the limit $\sigma \to 0$.

## 9.2. Harmonic Fokker-Planck operator.
We recall that the Wasserstein distance between two probability measures $\nu_1$ and $\nu_2$ is given by

$$ W_p(\nu_1, \nu_2) = \inf \left\{ \left( \int_{\mathbb{R}^d \times \mathbb{R}^d} |x-y|^p d\mu(x,y) \right)^{\frac{1}{p}} \middle| (\pi_j)_{\#} \mu = \nu_j \right\} $$

for any $1 \le p < \infty$, where $\mu$ varies among all probability measures on $\mathbb{R}^d \times \mathbb{R}^d$, and where $\pi_j: \mathbb{R}^d \times \mathbb{R}^d \to \mathbb{R}^d$ denotes the canonical projection onto the j-th factor (see for instance [31]). In particular, for all $1 \le p \le q < \infty$, we directly get that

$$ W_p(\nu_1, \nu_2) \le W_q(\nu_1, \nu_2). $$

We now introduce the well-known Fokker-Planck operator

$$ L := \Delta + 2 \operatorname{div}(y \cdot), $$

and recall some of its useful properties, starting from the convergence of its associated linear flow towards the universal Gaussian profile

$$ \Gamma(x) := \frac{1}{\pi^{d/2}} e^{-|x|^2}. $$
**Lemma 9.1** (see e.g. Theorem 11.2.1 in [1] or Theorem 24.7 in [32]). Let $\phi$ be a probability density such that

$$ \int_{\mathbb{R}^d} |x|^2 \phi(x) dx < \infty \quad \text{and} \quad \int_{\mathbb{R}^d} \phi(x) |\ln \phi(x)| dx < \infty. $$

Then for all $s \ge 0$, we have

$$ W_1(e^{sL}\phi, \Gamma) \le W_2(e^{sL}\phi, \Gamma) \le e^{-2s}W_2(\phi, \Gamma). $$

**Lemma 9.2.** Let $\phi \in L^1(\mathbb{R}^d)$, then

$$ \|x\|^2 e^{sL} \phi \|_{L^1} \le C (\|\phi\|_{L^1} + \|x\|^2 \phi \|_{L^1}). $$

*Proof.* The kernel of the harmonic Fokker-Planck operator writes

$$ K(t, x, y) := \pi^{-\frac{d}{2}}(1 - e^{-4s})^{-\frac{d}{2}} \Gamma\left((x - e^{-2s})(1 - e^{-4s})^{-\frac{1}{2}}\right), $$

where $\Gamma(z) = \exp(-|z|^2)$, see for instance [16, Lemma 4.7], so that

$$ (e^{sL}\phi)(x) = \int_{\mathbb{R}^d} K(s, x, y)\phi(y)dy $$

for all $x \in \mathbb{R}^d$. We then compute

$$ \begin{aligned} \int_{\mathbb{R}^d} |x|^2 |(e^{sL}\phi)(x)| dx & \lesssim \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} |x - e^{-2s}y|^2 K(s, x, y) |\phi(y)| dy dx \\ & \qquad + e^{-4s} \int_{\mathbb{R}^d} \int_{\mathbb{R}^d} K(s, x, y) |y|^2 |\phi(y)| dy dx \\ & \lesssim (1 - e^{-4s}) \int_{\mathbb{R}^d} |\phi(y)| \int_{\mathbb{R}^d} \frac{|z|^2}{\pi^{d/2}} \Gamma(z) dz dy \\ & \qquad + e^{-4s} \int_{\mathbb{R}^d} |e^{sL}(|x|^2 \phi(x))| dx, \end{aligned} $$

where we have performed the change of variable $z = (x - e^{-2s}y)(1 - e^{-4s})^{-\frac{1}{2}}$ in the first integral on the right hand side. The result follows, as

$$ \|x\|^2 e^{sL} \phi \|_{L^1} \le \|\phi\|_{L^1} \|x\|^2 \Gamma \|_{L^1} + \|e^{sL}(|x|^2 \phi)\|_{L^1}. \quad \square $$

We also provide the following property, which allows to trade differentiation in space with exponential decay in time.

**Lemma 9.3** (Lemma 4.8 in [16]). Let $\phi \in L^1(\mathbb{R}^d)$, $n \in \mathbb{N}$ and $K \in \{1, \dots, d\}^n$. Then $f(s) = e^{sL}(\partial_K \phi)$ is a $W^{\infty, 1}$ function for all $s > 0$, and for all $m \in \mathbb{N}$ we have

$$ f(s) = e^{-2ns} \partial_K (e^{sL} \phi) \quad \text{and} \quad \|f(s)\|_{W^{m-n,1}} \le \frac{e^{-2ns}}{(1-e^{-4s})^{\frac{m}{2}}} \|\Gamma\|_{W^{m,1}} \|\phi\|_{L^1}. $$

**9.3. Duhamel formula and perturbation of the porous medium equation.**
We can express Equation (9.4) as a perturbation of the Fokker-Planck evolution equation with source term

$$ \partial_{s\sigma} \tilde{\rho}_{\sigma} = \frac{1}{1+\sigma} L \tilde{\rho}_{\sigma} + \left(1 - \frac{1}{1+\sigma}\right) \underbrace{\text{div}(2y \tilde{\rho}_{\sigma})}_{=: R_0^{(1)}} + \frac{1}{1+\sigma} \underbrace{\Delta (\tilde{\rho}_{\sigma}^{1+\sigma} - \tilde{\rho}_{\sigma})}_{=: R_0^{(2)}} + \mathcal{R}. $$
Let $s_{\sigma} \in [0, s_{\sigma}^{\max}]$. Denoting $R_0 := R_0^{(1)} + \sigma^{-1}R_0^{(2)}$, this equation writes in Duhamel form as

$$ (9.5) \quad \tilde{\rho}_{\sigma}(s_{\sigma}) = e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0) + \frac{\sigma}{1+\sigma} \int_{0}^{s_{\sigma}} e^{(s_{\sigma}-r)\frac{L}{1+\sigma}} R_{0} dr + \int_{0}^{s_{\sigma}} e^{(s_{\sigma}-r)\frac{L}{1+\sigma}} \mathcal{R} dr. $$

Note that even if $s_{\sigma} = 0$ does not correspond to $t = 0$ back to the original time variable, it still leads to an initial time which is bounded in $\sigma$, hence it is harmless in the following long time analysis of our equation.

Our strategy of proof is the following: even if $s_{\sigma}$ is limited to a compact set $[0, s_{\sigma}^{\max}]$, we remark that as $\sigma \to 0$, we have $s_{\sigma}^{\max} \to +\infty$ hence $e^{s_{\sigma}L_{\sigma}}\tilde{\rho}_{\sigma}(0)$ should be arbitrary near the limiting Gaussian profile at large $s_{\sigma}$ when $\sigma \to 0$. Our aim is then to prove that remainder terms are indeed negligible, thanks to vanishing time coefficients for $\mathcal{R}$, or in the limit $\sigma \to 0$ for $R_0$.

The rest of this section consists essentially in showing the following result:

**Theorem 9.4.** Under the assumptions of Theorem 1.9, there exists $C > 0$ independent of $\sigma$ such that for all $0 \le s_{\sigma} \le s_{\sigma}^{\max}$,

$$ \sup \left\{ \int_{\mathbb{R}^d} \varphi \left( \tilde{\rho}_{\sigma}(s_{\sigma}) - e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0) \right) \middle| \varphi \in \mathcal{C}(\mathbb{R}^d) \text{ and } \|\varphi\|_{\text{Lip}} \le 1 \right\} \\ \le C(\sigma + e^{-2s_{\sigma}(1-d\sigma)}). $$

Even though the supremum is reminiscent of the definition of the distance $W_1$, we do not use this notation in the above statement, because the densities $\tilde{\rho}_{\sigma}$ (or, equivalently, $\rho_{\sigma}$) are not normalized to be probability densities, contrary to $\rho_{\sigma}$.

**9.4. Duality and regularization.** In order to prove Theorem 9.4, we will rely on the dual representation of the Wasserstein distance $W_1$ recalled in the introduction, (1.5). Now the test functions $\varphi$ must be regularized in order to absorb space derivatives by integration by parts, and also need to be truncated as they may grow linearly at infinity. The aim of this paragraph is to provide precise bounds on such regularization and cut-off processes.

Let $\varphi$ be any 1-Lipschitz function. Let $\zeta \in S(\mathbb{R}^d)$ such that $\zeta \ge 0$ and $\int_{\mathbb{R}^d} \zeta = 1$, and consider the mollifier $\zeta_{\epsilon}(x) = \epsilon^{-d}\zeta(x/\epsilon)$. We can then define the regularized function

$$ \tilde{\varphi}_{\epsilon} := \varphi * \zeta_{\epsilon}, $$

which is still 1-Lipschitz and satisfies

$$ \|\tilde{\varphi}_{\epsilon} - \varphi\|_{L^{\infty}} \le \epsilon \|\cdot\|_{L^1}. $$

We now introduce the cut-off function $\chi \in C_c^{\infty}(\mathbb{R}^d)$ such that $\chi \equiv 1$ on $B(0, 1)$ and $0 \le \chi \le 1$. We then define $\chi_{\epsilon}(x) = \chi(\epsilon x)$ and

$$ \varphi_{\epsilon} := (\tilde{\varphi}_{\epsilon} - \tilde{\varphi}_{\epsilon}(0))\chi_{\epsilon} \in C_c(\mathbb{R}^d). $$

We similarly get that $\varphi_{\epsilon}$ is Lipschitz continuous, uniformly in $\epsilon$, and that

$$ \left\| \frac{\tilde{\varphi}_{\epsilon} - \tilde{\varphi}_{\epsilon}(0) - \varphi_{\epsilon}}{|\cdot|^2} \right\|_{L^{\infty}} \le \epsilon. $$

We now state the main property which allows to trade space derivatives into negative powers of $\epsilon$.
**Lemma 9.5** (Equation 4.21 in [16]). Let $n \in \mathbb{N}$. There exists $C_n > 0$ such that for all $\epsilon > 0$,

$$ \|\varphi_{\epsilon}\|_{W^{1+n,\infty}} \le C_n \epsilon^{-n}. $$

9.5. **Proof of Theorem 9.4.** Let $\varphi$ be a 1-Lipschitz function. Following the previous discussion, we first introduce the decomposition

$$ (9.6) \qquad \int_{\mathbb{R}^d} \varphi(\tilde{\rho}_{\sigma}(s_{\sigma}) - e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)) = \int_{\mathbb{R}^d} (\varphi - \tilde{\varphi}_{\epsilon})(\tilde{\rho}_{\sigma}(s_{\sigma}) - e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)) \\ \qquad \qquad + \int_{\mathbb{R}^d} (\tilde{\varphi}_{\epsilon} - \varphi_{\epsilon})(\tilde{\rho}_{\sigma}(s_{\sigma}) - e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)) \\ \qquad \qquad + \int_{\mathbb{R}^d} \varphi_{\epsilon}(\tilde{\rho}_{\sigma}(s_{\sigma}) - e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)). $$

To bound the first term, we make use of the previous estimates from Section 9.4, so that

$$ \left| \int_{\mathbb{R}^d} (\varphi - \tilde{\varphi}_{\epsilon})(\tilde{\rho}_{\sigma}(s_{\sigma}) - e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)) \right| \le \|\tilde{\varphi}_{\epsilon} - \varphi\|_{L^{\infty}} (\|\tilde{\rho}_{\sigma}\|_{L^1} + \|e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)\|_{L^1}) \lesssim \epsilon. $$

Similarly we estimate the second term, recalling that $\int_{\mathbb{R}^d} \tilde{\rho}_{\sigma}(s_{\sigma}) = \int_{\mathbb{R}^d} e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)$ by the conservation of mass, and writing

$$ \left| \int_{\mathbb{R}^d} (\tilde{\varphi}_{\epsilon} - \varphi_{\epsilon})(\tilde{\rho}_{\sigma}(s_{\sigma}) - e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)) \right| \le \left| \int_{\mathbb{R}^d} (\tilde{\varphi}_{\epsilon} - \tilde{\varphi}_{\epsilon}(0) - \varphi_{\epsilon})(\tilde{\rho}_{\sigma}(s_{\sigma}) - e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)) \right| + \left| \tilde{\varphi}_{\epsilon}(0) \int_{\mathbb{R}^d} (\tilde{\rho}_{\sigma}(s_{\sigma}) - e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)) \right| \le \left\| \frac{\tilde{\varphi}_{\epsilon} - \tilde{\varphi}_{\epsilon}(0) - \varphi_{\epsilon}}{1 + |y|^2} \right\|_{L^{\infty}} \left( \|y|^2 \tilde{\rho}_{\sigma}\|_{L^1} + \|y|^2 e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0) \|_{L^1} \right) \lesssim \epsilon. $$

In the previous estimate, we used that, from Lemma 7.1, $\|y|^2 \tilde{\rho}_{\sigma}\|_{L^1}$ is uniformly bounded with respect to $\sigma$ and $s_{\sigma}$, as well as Lemma 9.2 in order to estimate $\|y|^2 e^{s_{\sigma} \frac{L}{1+\sigma}} \tilde{\rho}_{\sigma}(0)\|_{L^1}$. We now turn our attention to the second term of (9.6), using the Duhamel formulation from equation (9.5) to express $\tilde{\rho}_{\sigma}(s_{\sigma})$, and we control each error term.

The first error term corresponds to

$$ \begin{aligned} E_0 &= E_0^{(1)} + E_0^{(2)} \\ &= \frac{\sigma}{1+\sigma} \int_{\mathbb{R}^d} \varphi_{\epsilon} \int_0^{s_{\sigma}} e^{\frac{s_{\sigma}-r}{1+\sigma} L} R_0^{(1)} dr + \frac{1}{1+\sigma} \int_{\mathbb{R}^d} \varphi_{\epsilon} \int_0^{s_{\sigma}} e^{\frac{s_{\sigma}-r}{1+\sigma} L} R_0^{(2)} dr. \end{aligned} $$

We easily estimate

$$ \begin{aligned} E_0^{(1)} &= \frac{\sigma}{1+\sigma} \int_{\mathbb{R}^d} \varphi_{\epsilon} \int_0^{s_{\sigma}} e^{\frac{s_{\sigma}-r}{1+\sigma} L} \operatorname{div}(2y \tilde{\rho}_{\sigma}) dr \\ &= \frac{2\sigma}{1+\sigma} \int_0^{s_{\sigma}} e^{-2\frac{s_{\sigma}-r}{1+\sigma} L} \int_{\mathbb{R}^d} \nabla \varphi_{\epsilon} \cdot e^{\frac{s_{\sigma}-r}{1+\sigma} L} (y \tilde{\rho}_{\sigma}) dr, \end{aligned} $$
where we have used Lemma 9.3 and an integration by parts. From Lemma 7.1, we
then infer the uniform bound

$$
|E_0^{(1)}| \lesssim \sigma \|\nabla \varphi_\epsilon\|_{L^\infty} \int_0^{s_\sigma} e^{-2\frac{s_\sigma-r}{1+\sigma}} \|y \tilde{\rho}_\sigma(r)\|_{L^1} dr \lesssim \sigma.
$$

We similarly write, using Lemma 9.3 again,

$$
\begin{align*}
E_0^{(2)} &= \frac{1}{1+\sigma} \int_{\mathbb{R}^d} \varphi_{\epsilon} \int_0^{s_{\sigma}} e^{\frac{s_{\sigma}-r}{1+\sigma} L} R_0^{(2)} dr \\
&= \frac{1}{1+\sigma} \int_{\mathbb{R}^d} \varphi_{\epsilon} \int_0^{s_{\sigma}} e^{\frac{s_{\sigma}-r}{1+\sigma} L} \Delta (\tilde{\rho}_{\sigma}^{1+\sigma} - \tilde{\rho}_{\sigma}) dr \\
&= \frac{1}{1+\sigma} \int_{\mathbb{R}^d} \varphi_{\epsilon} \int_0^{s_{\sigma}} e^{-\frac{4}{1+\sigma}(s_{\sigma}-r)} \Delta e^{\frac{s_{\sigma}-r}{1+\sigma} L} (\tilde{\rho}_{\sigma}^{1+\sigma} - \tilde{\rho}_{\sigma}) dr \\
&= \frac{1}{1+\sigma} \int_0^{s_{\sigma}} e^{-\frac{4}{1+\sigma}(s_{\sigma}-r)} \int_{\mathbb{R}^d} \nabla \varphi_{\epsilon} \cdot \nabla e^{\frac{s_{\sigma}-r}{1+\sigma} L} (\tilde{\rho}_{\sigma}^{1+\sigma} - \tilde{\rho}_{\sigma}) dr.
\end{align*}
$$

In view of Lemma 7.1, there exists $C > 0$ independent of $\sigma \in (0, \epsilon)$ and $t \ge 0$ such that

$$
\sup_{0 \le s_{\sigma} \le s_{\sigma}^{\max}} \| \tilde{\rho}_{\sigma}(s_{\sigma})^{1+\sigma} - \tilde{\rho}_{\sigma}(s_{\sigma}) \|_{L^1} \le C\sigma.
$$

From Hölder inequality, Lemma 9.3 with $m = 1$, we infer

$$
|E_0^{(2)}| \le \frac{1}{1+\sigma} \int_0^{s_\sigma} \frac{e^{-\frac{4}{1+\sigma}(s_\sigma-r)}}{\left(1-e^{-\frac{4}{1+\sigma}(s_\sigma-r)}\right)^{1/2}} \|\nabla\varphi_\epsilon\|_{L^\infty} \|\tilde{\rho}_\sigma^{1+\sigma} - \tilde{\rho}_\sigma\|_{L^1} dr \le C\sigma,
$$

thanks to Lemma 9.5.

The second error term, associated to $R_1$ defined in equation (9.4), writes

$$
\begin{align*}
E_1 &= \frac{2}{1-d\sigma} \int_{\mathbb{R}^d} \varphi_{\epsilon} \int_0^{s_{\sigma}} e^{\frac{s_{\sigma}-r}{1+\sigma} L} e^{-4r(1-d\sigma)} (1-d\sigma e^{4r(1-d\sigma)}) \partial_{s_{\sigma}} \tilde{\rho}_{\sigma}(r) dr \\
&= \frac{2}{1-d\sigma} \int_0^{s_{\sigma}} e^{-4r(1-d\sigma)} (1-d\sigma e^{4r(1-d\sigma)}) \int_{\mathbb{R}^d} \varphi_{\epsilon} e^{\frac{s_{\sigma}-r}{1+\sigma} L} (\partial_{s_{\sigma}} \tilde{\rho}_{\sigma}(r)) dr.
\end{align*}
$$

To get further estimates on $\partial_{s_\sigma} \tilde{\rho}_\sigma$, we recall that in the time variable $t$ we have

$$
\partial_t \rho_\sigma(t) + \frac{1}{\tau_\sigma(t)^2} \mathrm{div} \, J_\sigma(t) = \partial_t \rho_\sigma(t) + \frac{1}{\tau_\sigma(t)^{1+\frac{d\sigma}{2}}} \mathrm{div} \, \mu_\sigma(t) = 0,
$$

with

$$
\mu_{\sigma}(t) := \frac{1}{\tau_{\sigma}(t)^{1-\frac{d\sigma}{2}}} \operatorname{Im}(\bar{v}_{\sigma}(t) \nabla v_{\sigma}(t)),
$$

which satisfies a uniform $L^1$-estimate from Cauchy-Schwarz inequality, as

$$
(9.7) \quad \|\mu_\sigma(t)\|_{L^1} \lesssim \frac{1}{\tau_\sigma(t)^{1-\frac{d\sigma}{2}}} \|\nabla v_\sigma(t)\|_{L^2} \|v_\sigma(t)\|_{L^2} = \|\phi\|_{L^2} \sqrt{V_\sigma(t)}.
$$

Moreover, we recall that $V_\sigma$ satisfies a uniform (both in $\sigma$ and $t$) bound and a
uniform integrability property, from Lemma 7.1. Hence we end up, in the $s_\sigma$ time
variable, with

$$
\begin{align*}
\partial_{s_{\sigma}} \tilde{\rho}_{\sigma} &= (1 - d\sigma) \dot{\tau}_{\sigma}(t(s_{\sigma})) \tau_{\sigma}(t(s_{\sigma}))^{1+d\sigma} \partial_t \rho_{\sigma}(t(s_{\sigma})) \\
&= -(1 - d\sigma) \dot{\tau}_{\sigma}(t(s_{\sigma})) \tau_{\sigma}(t(s_{\sigma}))^{\frac{d\sigma}{2}} \operatorname{div} \mu_{\sigma}(t(s_{\sigma})) \\
&= -(1 - d\sigma) e^{2s_{\sigma}(1-d\sigma)} (1 - d\sigma e^{4s_{\sigma}(1-d\sigma)})^{-1/2} \operatorname{div} \mu_{\sigma}(t(s_{\sigma})),
\end{align*}
$$

so using properties of the Fokker-Planck flow from Lemma 9.3, and integrating by parts, we get

$$
\begin{align*}
E_1 &= -2 \int_0^{s_\sigma} e^{-2r(1-d\sigma)} (1-d\sigma e^{4r(1-d\sigma)})^{1/2} \int_{\mathbb{R}^d} \varphi_\epsilon e^{\frac{s_\sigma-r}{1+\sigma}L} \operatorname{div} \mu_\sigma(r) \, dr \\
&= 2 \int_0^{s_\sigma} e^{-2r(1-d\sigma)} e^{-\frac{2}{1+\sigma}(s_\sigma-r)} (1-d\sigma e^{4r(1-d\sigma)})^{1/2} \int_{\mathbb{R}^d} \nabla \varphi_\epsilon \cdot \left( e^{\frac{s_\sigma-r}{1+\sigma}L} \mu_\sigma(r) \right) \, dr.
\end{align*}
$$

By taking the absolute value and using (9.7) and Lemma 9.5, we infer

$$
\begin{align*}
|E_1| &\le 2 \|\nabla \varphi_\epsilon\|_{L^\infty} \|\phi\|_{L^2} \int_0^{s_\sigma} e^{-2r(1-d\sigma)} e^{-\frac{2}{1+\sigma}(s_\sigma-r)} \left(1 - \frac{d\sigma}{4} e^{4s(1-d\sigma)}\right)^{1/2} \sqrt{\tilde{V}_\sigma(r)} dr \\
&\lesssim \left( \int_0^{s_\sigma} e^{-8r(1-d\sigma)} e^{-\frac{4}{1+\sigma}(s_\sigma-r)} dr \right)^{1/2} \left( \int_0^{s_\sigma} e^{4r(1-d\sigma)} \tilde{V}_\sigma(r) dr \right)^{1/2} \\
&\lesssim \frac{\mathcal{E}(0)}{1-d\sigma} \left( \int_0^{s_\sigma} e^{-8r(1-d\sigma)} e^{-\frac{4}{1+\sigma}(s_\sigma-r)} dr \right)^{1/2},
\end{align*}
$$

from Cauchy-Schwarz inequality and using the uniform bound from equation (9.3).
Since we assume $\sigma < 1/(2d)$, $2 - 2d\sigma - \frac{1}{1+\sigma} \ge \frac{1}{2d+1} > 0$, we then get the decreasing
in time estimate

$$
(9.8) \quad |E_1| \lesssim \left( \int_0^{s_\sigma} e^{-4r(2-2d\sigma-\frac{1}{1+\sigma})} e^{-\frac{4}{1+\sigma}s_\sigma} dr \right)^{1/2} \lesssim e^{-\frac{2}{1+\sigma}s_\sigma}.
$$

We now turn to the fourth error term. We directly write

$$
\begin{align*}
E_3 &= -\frac{1}{4} \int_{\mathbb{R}^d} \varphi_\epsilon \int_0^{s_\sigma} e^{-\frac{s_\sigma-r}{1+\sigma}L} \left(1 - d\sigma e^{4r(1-d\sigma)}\right)^{\frac{2-d\sigma}{d\sigma}} \Delta^2 \tilde{\rho}_\sigma(r) dr \\
&= -\frac{1}{4} \int_0^{s_\sigma} e^{-\frac{s}{1+\sigma}(s_\sigma-r)} \left(1 - d\sigma e^{4r(1-d\sigma)}\right)^{\frac{2-d\sigma}{d\sigma}} \int_{\mathbb{R}^d} \Delta^2 \varphi_\epsilon e^{-\frac{s_\sigma-r}{1+\sigma}L} \tilde{\rho}_\sigma(r) dr,
\end{align*}
$$

so by Hölder inequality, and from Lemma 9.5, we infer

$$
\begin{align*}
|E_3| &\le \frac{1}{4} \|\Delta^2 \varphi_\epsilon\|_{L^\infty} \|\tilde{\rho}_\sigma(0)\|_{L^1} \int_0^{s_\sigma} e^{-\frac{s}{1+\sigma}(s_\sigma-r)} \exp\left(\frac{2-d\sigma}{d\sigma} \ln\left(1-d\sigma e^{4r(1-d\sigma)}\right)\right) dr \\
&\lesssim \frac{1}{\epsilon^3} e^{-\frac{s}{1+\sigma}s_\sigma} \int_0^\infty e^{\frac{8r}{1+\sigma}} e^{-\frac{3}{2}e^{2r}} dr \\
&\lesssim \frac{1}{\epsilon^3} e^{-\frac{s}{1+\sigma}s_\sigma}.
\end{align*}
$$

We similarly estimate the fifth error term. We first write

$$
\begin{align*}
E_4 &= \int_{\mathbb{R}^d} \varphi_\epsilon \int_0^{s_\sigma} \left(1 - d\sigma e^{4r(1-d\sigma)}\right)^{\frac{2-d\sigma}{d\sigma}} e^{-\frac{s_\sigma-r}{1+\sigma}L} \operatorname{div} \operatorname{div} \tilde{\nu}_\sigma(r) dr \\
&= \int_0^{s_\sigma} \left(1 - d\sigma e^{4r(1-d\sigma)}\right)^{\frac{2-d\sigma}{d\sigma}} e^{-\frac{4}{1+\sigma}(s_\sigma-r)} \int_{\mathbb{R}^d} \nabla^2 \varphi_\epsilon \cdot e^{-\frac{s_\sigma-r}{1+\sigma}L} \tilde{\nu}_\sigma(r) dr,
\end{align*}
$$
where “.” denotes here the Frobenius scalar product between matrices. In view of the inequality $||\tilde{\nu}_{\sigma}||_{L^1} \le ||\nabla v_{\sigma}||_{L^2}^2$, applying (9.3), we have

$$
\begin{align*}
|E_4| &\le ||\nabla^2 \varphi_{\epsilon}||_{L^{\infty}} \int_0^{s_{\sigma}} e^{-\frac{4}{1+\sigma}(s_{\sigma}-r)} \tilde{V}_{\sigma}(r) dr \\
&\lesssim \frac{1}{\epsilon} e^{-4s_{\sigma}(1-d\sigma)} \int_0^{s_{\sigma}} \underbrace{e^{-4(\frac{1}{1+\sigma}-1+d\sigma)(s_{\sigma}-r)}}_{\lesssim 1} e^{4r(1-d\sigma)} \tilde{V}_{\sigma}(r) dr \\
&\lesssim \frac{1}{\epsilon} e^{-4s_{\sigma}(1-d\sigma)} \frac{C}{1-d\sigma}.
\end{align*}
$$

We finally estimate the third error term

$$
E_2 = \int_{\mathbb{R}^d} \varphi_{\epsilon} \int_0^{s_{\sigma}} e^{-\frac{s_{\sigma}-r}{1+\sigma} L} R_2(r) dr,
$$

which is the most delicate to treat due to the presence of the double time derivative:

$$
R_2(s_\sigma) = - \frac{e^{-4s_\sigma(1-d\sigma)}}{(1-d\sigma)^2} \left(1 - d\sigma e^{4s_\sigma(1-d\sigma)}\right) \partial_{s_\sigma}^2 \tilde{\rho}_\sigma.
$$

Writing

$$
\begin{align*}
R_2(s_\sigma) &= -\partial_s \left( \frac{e^{-4s_\sigma(1-d\sigma)}}{(1-d\sigma)^2} \left(1 - d\sigma e^{4s_\sigma(1-d\sigma)}\right) \partial_s \tilde{\rho}_\sigma \right) - \frac{4}{1-d\sigma} e^{-4s_\sigma(1-d\sigma)} \partial_s \tilde{\rho}_\sigma(s_\sigma) \\
&=: R_2^{(1)} + R_2^{(2)},
\end{align*}
$$

we treat each term separately. We begin with

$$
E_2^{(1)} = \frac{1}{(1-d\sigma)^2} \int_{\mathbb{R}^d} \varphi_{\epsilon} \int_{0}^{s_{\sigma}} e^{-\frac{s_{\sigma}-r}{1+\sigma} L} \partial_s \left( e^{-4r(1-d\sigma)} \left(1 - d\sigma e^{4r(1-d\sigma)}\right) \partial_s \tilde{\rho}_{\sigma}(r) \right) dr.
$$

We will use a result which is a direct consequence of [16, Corollary 4.10].

**Lemma 9.6.** Let $T > 0$ and $f \in L^\infty((0,T); W^{1,1}(\mathbb{R}^d)) \cap C((0,T); L^1(\mathbb{R}^d))$. For all $s \in (0,T)$, we have

$$
\int_0^s e^{(s-r)L} \partial_s f(r) dr = \int_0^s L e^{(s-r)L} f(r) dr + f(s) - e^{sL} f(0).
$$

Such a result allows to rewrite $E_2^{(1)}$ as a sum of four error terms, as $L$ is the sum of two operators,

$$
\begin{align*}
E_2^{(1)} &= -\frac{1}{(1-d\sigma)^2} \int_0^{s_\sigma} e^{-4r(1-d\sigma)} (1-d\sigma e^{4r(1-d\sigma)}) \int_{\mathbb{R}^d} \varphi_\epsilon e^{-\frac{s_\sigma-r}{1+s_\sigma}L} \Delta \partial_s \tilde{\rho}_\sigma(r) dr \\
&\quad -\frac{1}{(1-d\sigma)^2} \int_0^{s_\sigma} e^{-4r(1-d\sigma)} (1-d\sigma e^{4r(1-d\sigma)}) \int_{\mathbb{R}^d} \varphi_\epsilon e^{-\frac{s_\sigma-r}{1+s_\sigma}L} \operatorname{div}(2y \partial_s \tilde{\rho}_\sigma(r)) dr \\
&\quad -\frac{1}{(1-d\sigma)^2} e^{-4s_\sigma(1-d\sigma)} (1-d\sigma e^{4s_\sigma(1-d\sigma)}) \int_{\mathbb{R}^d} \varphi_\epsilon \partial_s \tilde{\rho}_\sigma(s_\sigma) dr \\
&\quad -\frac{1}{(1-d\sigma)^2} (1-d\sigma) \int_{\mathbb{R}^d} \varphi_\epsilon e^{\frac{s_\sigma}{1+s_\sigma}L} \partial_s \tilde{\rho}_\sigma(0) dr \\
&=: E_2^{(1,1)} + E_2^{(1,2)} + E_2^{(1,3)} + E_2^{(1,4)}.
\end{align*}
$$

To estimate $E_2^{(1,1)}$ we recall that

$$
(9.9) \qquad \partial_s \tilde{\rho}_\sigma(r) = -(1-d\sigma)e^{2r(1-d\sigma)} (1-d\sigma e^{4r(1-d\sigma)})^{-1/2} \operatorname{div} \mu_\sigma,
$$
which implies

$$E_2^{(1,1)} = \frac{e^{-\frac{6}{1+\sigma}}}{1-d\sigma} \int_0^{s_{\sigma}} e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} \sqrt{1-d\sigma e^{4r(1-d\sigma)}} \int_{\mathbb{R}^d} \nabla\varphi_{\epsilon} \cdot \Delta e^{-\frac{s_{\sigma}-r}{1+\sigma}L} \mu_{\sigma}(r) dr,$$

from Lemma 9.3. Since the square root term in $E_2^{(1,1)}$ is controlled by one,

$$(9.10) \quad |E_2^{(1,1)}| \le \frac{e^{-\frac{6s_{\sigma}}{1+\sigma}}}{(1-d\sigma)^2} \int_0^{s_{\sigma}} e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} \left| \int_{\mathbb{R}^d} \nabla\varphi_{\epsilon} \cdot \Delta \left( e^{-\frac{s_{\sigma}-r}{1+\sigma}L} \mu_{\sigma}(r) \right) \right| dr.$$

We then remark that, for any $r \in (0, s_{\sigma})$, Lemma 9.3 with $m = 1$ leads to

$$\begin{align} \left| \int_{\mathbb{R}^d} \nabla \varphi_{\epsilon} \cdot \Delta \left( e^{-\frac{s_{\sigma}-r}{1+\sigma}L} \mu_{\sigma}(r) \right) \right| &\le \|\nabla \varphi_{\epsilon}\|_{L^{\infty}} \left\| \Delta \left( e^{-\frac{s_{\sigma}-r}{1+\sigma}L} \mu_{\sigma}(r) \right) \right\|_{L^1} \nonumber \\ &\lesssim \frac{1}{1 - e^{-4 \frac{s_{\sigma}-r}{1+\sigma}}} \|\mu_{\sigma}(r)\|_{L^1} \nonumber \\ &\lesssim \frac{1}{1 - e^{-4 \frac{s_{\sigma}-r}{1+\sigma}}} \|\phi\|_{L^2} \sqrt{\tilde{V}_{\sigma}(r)}, \tag{9.11} \end{align}$$

thanks to (9.7). This last estimate suggests adding and subtracting the factor $e^{-4 \frac{s_{\sigma}-r}{1+\sigma}}$ inside the estimating term for $E_2^{(1,1)}$. In fact, this leads to

$$(9.12) \quad \int_0^{s_{\sigma}} e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} (1 - e^{-4\frac{s_{\sigma}-r}{1+\sigma}}) \left| \int_{\mathbb{R}^d} \nabla \varphi_{\epsilon} \cdot \Delta \left( e^{-\frac{s_{\sigma}-r}{1+\sigma}L} \mu_{\sigma}(r) \right) \right| dr \\ \lesssim \int_0^{s_{\sigma}} e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} \sqrt{\tilde{V}_{\sigma}(r)} dr \\ \lesssim \left( \int_0^{s_{\sigma}} e^{2r[\frac{6}{1+\sigma}-4(1-d\sigma)]} dr \right)^{1/2} \left( \int_0^{s_{\sigma}} e^{4r(1-d\sigma)} \tilde{V}_{\sigma}(r) dr \right)^{1/2} \\ \lesssim e^{6\frac{s_{\sigma}}{1+\sigma}} \frac{e^{-4s_{\sigma}(1-d\sigma)}}{2\sqrt{\frac{3}{1+\sigma} - 2(1-d\sigma)}} \sqrt{\mathcal{E}(0)}.$$
On the other hand, take some $S \in (0, s_\sigma)$ to be optimized later. Then we also have

$$
\begin{align*}
(9.13) \quad & \int_0^S e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} e^{-4\frac{s\sigma-r}{1+\sigma}} \left| \int_{\mathbb{R}^d} \nabla\varphi_\epsilon \cdot \Delta \left( e^{-\frac{s\sigma-r}{1+\sigma}L} \mu_\sigma(r) \right) \right| dr \\
& \lesssim \int_0^S e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} \frac{e^{-4\frac{s\sigma-r}{1+\sigma}}}{1-e^{-4\frac{s\sigma-r}{1+\sigma}}} \sqrt{V_\sigma(r)} dr \\
& \lesssim e^{-2\frac{s\sigma}{1+\sigma}} e^{4S(\frac{2}{1+\sigma}-1+d\sigma)} \left( \int_0^S \frac{e^{-4\frac{s\sigma-r}{1+\sigma}}}{\left(1-e^{-4\frac{s\sigma-r}{1+\sigma}}\right)^2} dr \right)^{1/2} \left( \int_0^{s_\sigma} e^{4r(1-d\sigma)} \tilde{V}_\sigma(r) dr \right)^{1/2} \\
& \lesssim e^{-2\frac{s\sigma}{1+\sigma}} e^{4s_\sigma(\frac{2}{1+\sigma}-1+d\sigma)} \left( \int_0^S \frac{e^{-4\frac{s\sigma-r}{1+\sigma}}}{\left(1-e^{-4\frac{s\sigma-r}{1+\sigma}}\right)^2} dr \right)^{1/2} \sqrt{\frac{\mathcal{E}(0)}{1-d\sigma}} \\
& \lesssim e^{-2\frac{s\sigma}{1+\sigma}} e^{4s_\sigma(\frac{2}{1+\sigma}-1+d\sigma)} (1+\sigma) \left( \frac{1}{1-e^{-4\frac{s\sigma-S}{1+\sigma}}} - \frac{1}{1-e^{-4\frac{s\sigma}{1+\sigma}}} \right)^{1/2} \\
& \lesssim e^{-2\frac{s\sigma}{1+\sigma}} e^{4s_\sigma(\frac{2}{1+\sigma}-1+d\sigma)} \frac{1}{\left(1-e^{-4\frac{s\sigma-S}{1+\sigma}}\right)^{1/2}} \\
& \lesssim e^{4s_\sigma(\frac{2}{1+\sigma}-1+d\sigma)} \frac{1}{\left(e^{\frac{4s\sigma}{1+\sigma}} - e^{\frac{4S}{1+\sigma}}\right)^{1/2}}.
\end{align*}
$$

In order to handle the remaining integral on $[S, s_\sigma]$, similarly to estimate (9.11) we use Lemma 9.3 this time with $m = 2$, which implies that, for all $r \in (S, s_\sigma)$,

$$
\left| \int_{\mathbb{R}^d} \nabla \varphi_\epsilon \cdot \Delta e^{-\frac{s\sigma-r}{1+\sigma}L} \mu_\sigma(r) \right| \lesssim \|\Delta \varphi_\epsilon\|_{L^\infty} \|\nabla e^{-\frac{s\sigma-r}{1+\sigma}L} \mu_\sigma(r)\|_{L^1} \\
\lesssim \frac{1}{\epsilon} \frac{1}{\left(1 - e^{-4\frac{s\sigma-r}{1+\sigma}}\right)^{1/2}} \|\phi\|_{L^2} \sqrt{\tilde{V}_\sigma(r)}.
$$

Therefore, from the uniform bound on $V_\sigma$ and by direct integration, we also have

$$
\begin{align*}
\int_S^{s_\sigma} e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} e^{-4\frac{s\sigma-r}{1+\sigma}} \left| \int_{\mathbb{R}^d} \Delta\varphi_\epsilon \cdot \nabla e^{-\frac{s\sigma-r}{1+\sigma}L} \mu_\sigma(r) \right| dr & \\
& \lesssim \frac{1}{\epsilon} \int_S^{s_\sigma} e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} \frac{e^{-4\frac{s\sigma-r}{1+\sigma}}}{\left(1-e^{-4\frac{s\sigma-r}{1+\sigma}}\right)^{1/2}} \sqrt{\tilde{V}_\sigma(r)} dr \\
(9.14) \qquad & \lesssim \frac{1}{\epsilon} e^{s_\sigma[\frac{6}{1+\sigma}-2(1-d\sigma)]} \int_S^{s_\sigma} \frac{e^{-4\frac{s\sigma-r}{1+\sigma}}}{\left(1-e^{-4\frac{s\sigma-r}{1+\sigma}}\right)^{1/2}} dr \\
& \lesssim \frac{1}{\epsilon} \frac{\sigma+1}{2} e^{s_\sigma[\frac{6}{1+\sigma}-2(1-d\sigma)]} \left(1-e^{-4\frac{s\sigma-S}{1+\sigma}}\right)^{1/2} \\
& \lesssim \frac{1}{\epsilon} e^{s_\sigma[\frac{6}{1+\sigma}-4(1-d\sigma)]} \left(e^{\frac{4s\sigma}{1+\sigma}} - e^{\frac{4S}{1+\sigma}}\right)^{1/2}.
\end{align*}
$$
Taking S such that $e^{\frac{4s\sigma}{1+\sigma}} - e^{\frac{4S}{1+\sigma}} = 1$, estimate (9.13) multiplied by the factor $e^{-\frac{6s\sigma}{1+\sigma}}$ leads to the bound

$$ 
\begin{aligned}
& e^{-\frac{6s\sigma}{1+\sigma}} \int_0^S e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} e^{-\frac{4s\sigma-r}{1+\sigma}} \left| \int_{\mathbb{R}^d} \nabla\varphi_\epsilon \cdot \Delta e^{-\frac{s\sigma-r}{1+\sigma}L} \mu_\sigma(r) \right| dr \\
& \lesssim e^{-4s\sigma(1-d\sigma)+2\frac{s\sigma}{1+\sigma}} = e^{-2s\sigma(1-d\sigma)} e^{2\sigma s\sigma \frac{d\sigma+d-1}{1+\sigma}} \\
& \lesssim e^{-2s\sigma(1-d\sigma)} e^{2\sigma s\sigma^{\max} \frac{d\sigma+d-1}{1+\sigma}} = e^{-2s\sigma(1-d\sigma)} (d\sigma)^{-\sigma} e^{-\frac{d\sigma+d-1}{4(1+\sigma)(1-d\sigma)}} \\
& \lesssim e^{-2s\sigma(1-d\sigma)},
\end{aligned}
$$

while (9.14) leads to

$$ e^{-\frac{6s\sigma}{1+\sigma}} \int_S e^{r[\frac{6}{1+\sigma}-2(1-d\sigma)]} e^{-\frac{4s\sigma-r}{1+\sigma}} \left| \int_{\mathbb{R}^d} \nabla\varphi_\epsilon \cdot \Delta e^{-\frac{s\sigma-r}{1+\sigma}L} \mu_\sigma(r) \right| dr \lesssim \frac{1}{\epsilon} e^{-4s\sigma(1-d\sigma)}. $$

Summing these last two estimates with (9.12) multiplied by $e^{-\frac{6s\sigma}{1+\sigma}}$ leads, in view of (9.10), to the bound

$$ |E_2^{(1,1)}| \lesssim e^{-2s\sigma(1-d\sigma)} + \frac{1}{\epsilon} e^{-4s\sigma(1-d\sigma)}. $$

We now turn to $E_2^{(1,2)}$: invoking (9.9), integrating by parts and reordering terms, we find

$$ 
\begin{aligned}
E_2^{(1,2)} &= \frac{2}{1-d\sigma} \int_0^{s\sigma} e^{-2r(1-d\sigma)} (1-d\sigma e^{4r(1-d\sigma)})^{1/2} e^{-2\frac{s\sigma-r}{1+\sigma}} \\
&\qquad \times \int_{\mathbb{R}^d} \nabla\varphi_\epsilon \cdot e^{-\frac{s\sigma-r}{1+\sigma}L} (y \operatorname{div} \mu_\sigma(r)) dr \\
&= -\frac{2}{1-d\sigma} \int_0^{s\sigma} e^{-2r(1-d\sigma)} (1-d\sigma e^{4r(1-d\sigma)})^{1/2} e^{-4\frac{s\sigma-r}{1+\sigma}} \\
&\qquad \times \int_{\mathbb{R}^d} \nabla\varphi_\epsilon \cdot \operatorname{div} e^{-\frac{s\sigma-r}{1+\sigma}L} (\mu_\sigma(r) \otimes y) dr \\
&\quad - \frac{2d}{1-d\sigma} \int_0^{s\sigma} e^{-2r(1-d\sigma)} (1-d\sigma e^{4r(1-d\sigma)})^{1/2} e^{-2\frac{s\sigma-r}{1+\sigma}} \\
&\qquad \times \int_{\mathbb{R}^d} \nabla\varphi_\epsilon \cdot e^{-\frac{s\sigma-r}{1+\sigma}L} \mu_\sigma(r) dr \\
&=: E_2^{(1,2,1)} + E_2^{(1,2,2)},
\end{aligned}
$$

which allows to define two new error terms $E_2^{(1,2,1)}$ and $E_2^{(1,2,2)}$ that we need to estimate. The second one is straightforward from the same techniques as above, using (9.7) and (9.3), so that

$$ |E_2^{(1,2,2)}| \lesssim e^{-\frac{2s\sigma}{1+\sigma}} \left( \int_0^\infty e^{-4r(2-2d\sigma-\frac{1}{1+\sigma})} dr \right)^{1/2}, $$

where the last integral is convergent since, as we have seen before, the assumption $\sigma < 1/(2d)$ implies $2 - 2d\sigma - \frac{1}{1+\sigma} > \frac{1}{2d+1}$. For the first one we rather use the identity

$$ y \cdot \mu_\sigma = \operatorname{Im} \left( \frac{1}{\tau_\sigma} \overline{v_\sigma} y \cdot \nabla v_\sigma \right), $$
and that $\|yv_{\sigma}\|_{L^2} \le C$ from Lemma 7.1, along with Lemma 9.3 with $m = 1$, so that

$$
\begin{align*}
|E_2^{(1,2,1)}| & \lesssim \int_0^{s_\sigma} e^{-2r(1-d\sigma)} \underbrace{(1-d\sigma e^{4r(1-d\sigma)})} \limits_{\le 1}^{1/2} \frac{e^{-4\frac{s_\sigma-r}{1+\sigma}}}{(1-e^{-4\frac{s_\sigma-r}{1+\sigma}})^{1/2}} \|y\mu_\sigma(r)\|_{L^1} dr \\
& \lesssim \int_0^{s_\sigma} e^{-2r(1-d\sigma)} \frac{e^{-4\frac{s_\sigma-r}{1+\sigma}}}{(1-e^{-4\frac{s_\sigma-r}{1+\sigma}})^{1/2}} \|yv_\sigma\|_{L^2} \sqrt{V_\sigma(r)} dr \\
& \lesssim \int_0^{s_\sigma} e^{-2r(1-d\sigma)} \frac{e^{-4\frac{s_\sigma-r}{1+\sigma}}}{(1-e^{-4\frac{s_\sigma-r}{1+\sigma}})^{1/2}} dr \\
& \lesssim \int_0^{s_\sigma} e^{-2(s_\sigma-q)(1-d\sigma)} \frac{e^{-\frac{4q}{1+\sigma}}}{(1-e^{-\frac{4q}{1+\sigma}})^{1/2}} dq \\
& \lesssim e^{-2s_\sigma(1-d\sigma)} \int_0^{s_\sigma} \frac{e^{-2q(\frac{2}{1+\sigma}-(1-d\sigma))}}{(1-e^{-\frac{4q}{1+\sigma}})^{1/2}} dq \\
& \lesssim e^{-2s_\sigma(1-d\sigma)}.
\end{align*}
$$

We can now go back to $E_2^{(1,3)}$ and $E_2^{(1,4)}$, which are straightforward in view of the above analysis: as

$$E_2^{(1,3)} = \frac{1}{1-d\sigma} e^{-2s_{\sigma}(1-d\sigma)} (1-d\sigma e^{4s_{\sigma}(1-d\sigma)})^{1/2} \int_{\mathbb{R}^d} \varphi_{\epsilon} \operatorname{div} \mu_{\sigma} dr,$$

we directly get

$$|E_2^{(1,3)}| \lesssim e^{-2s_{\sigma}(1-d\sigma)},$$

and similarly, writing

$$E_2^{(1,4)} = -\frac{1}{1-d\sigma} e^{-2\frac{s_\sigma}{1+\sigma}} (1-d\sigma)^{1/2} \int_{\mathbb{R}^d} \nabla \varphi_\epsilon \cdot e^{\frac{s_\sigma}{1+\sigma} L} \mu_\sigma(0),$$

from Lemma 9.3, we infer

$$|E_2^{(1,4)}| \lesssim e^{-\frac{2s_\sigma}{1+\sigma}}.$$

Gathering the bounds on $E_0^{(1)}$, $E_0^{(2)}$, $E_1$, $E_3$, $E_4$, $E_2^{(1,1)}$, $E_2^{(1,2,1)}$, $E_2^{(1,2,2)}$, $E_2^{(1,3)}$ and $E_2^{(1,4)}$, we get the final estimate

$$\int_{\mathbb{R}^d} \varphi_\epsilon(\tilde{\rho}_\sigma(s_\sigma) - e^{s_\sigma \frac{L}{1+\sigma}} \tilde{\rho}_\sigma(0)) \\
\lesssim \epsilon + \sigma + 3e^{-\frac{2s_\sigma}{1+\sigma}} + \frac{1}{\epsilon^3}e^{-\frac{8s_\sigma}{1+\sigma}} + \frac{2}{\epsilon}e^{-4s_\sigma(1-d\sigma)} + 4e^{-2s_\sigma(1-d\sigma)}.$$

We optimize this estimate by setting $\epsilon = e^{-2s_\sigma(1-d\sigma)}$, which yields

$$\int_{\mathbb{R}^d} \varphi_\epsilon(\tilde{\rho}_\sigma(s_\sigma) - e^{s_\sigma \frac{L}{1+\sigma}} \tilde{\rho}_\sigma(0)) \lesssim \sigma + e^{-2s_\sigma(1-d\sigma)},$$

which ends the proof of Theorem 9.4.

## 9.6. Conclusion: proof of Theorem 1.9.
9.6.1. Convergence in Wasserstein distance (1.7). Recall that $\Gamma(x) = \frac{e^{-|x|^2}}{(2\pi)^{d/2}}$, so

$$ \int_{\mathbb{R}^d} \tilde{\rho}_{\sigma}(s_{\sigma}) = \frac{1}{\|\phi_{\sigma}\|_{L^2}^2} \int_{\mathbb{R}^d} \tilde{\rho}_{\sigma}(s_{\sigma}) = 1 = \int_{\mathbb{R}^d} \Gamma, $$

with the obvious notation $\tilde{\rho}_{\sigma}(s_{\sigma}) = \rho_{\sigma}(s_{\sigma}(t))$. Using Theorem 9.4 (as the same remains valid for $\tilde{\rho}_{\sigma}$) alongside Lemma 9.1, we can then write that

$$ \begin{aligned} W_1(\tilde{\rho}_{\sigma}(s_{\sigma}), \Gamma) &\le W_1(\tilde{\rho}_{\sigma}(s_{\sigma}), e^{\frac{2s_{\sigma}}{1+\sigma}L} \tilde{\rho}_{\sigma}(0)) + W_1(e^{\frac{2s_{\sigma}}{1+\sigma}L} \tilde{\rho}_{\sigma}(0), \Gamma) \\ &\lesssim \sigma + e^{-2s_{\sigma}(1-d\sigma)} + e^{-\frac{2s_{\sigma}}{1+\sigma}} W_2(\tilde{\rho}_{\sigma}(0), \Gamma). \end{aligned} $$

Going back to the time variable $t$, we then get that for all $t \ge 2$,

$$ W_1(\rho_{\sigma}(t), \Gamma) \lesssim \sigma + \frac{1}{\dot{\tau}_{\sigma}(t)}. $$

On the other hand, we recall that from [16, Theorem 1.5], there exists $C > 0$ such that for all $t \ge 2$,

$$ W_1(\rho_0(t), \Gamma) \le \frac{C}{\sqrt{\ln t}}. $$

We finally need a last property on the dispersive rate to conclude:

**Lemma 9.7.** For all $\epsilon > 0$, there exists $T > 0$ and $\sigma_0 > 0$ such that for all $0 < \sigma \le \sigma_0$ and for all $t \ge T$, $\dot{\tau}_{\sigma}(t)^{-1} \le \epsilon$.

*Proof.* We know from Proposition 6.3 that there exists a constant $C > 0$ such that

$$ \dot{\tau}_0(t) - C\sigma(\ln t)^{3/2} \le \dot{\tau}_{\sigma}(t) $$

for all $t \ge 2$. As $\dot{\tau}_0(t) \sim \sqrt{\ln t}$ as $t \to +\infty$, there exists $T = T(\epsilon) > 0$ large enough such that

$$ \dot{\tau}_0(T) \ge 2\epsilon^{-1}. $$

On the other hand, for $\sigma_0 = \sigma_0(T, \epsilon) > 0$ small enough we have

$$ C\sigma(\ln T)^{3/2} \le \epsilon^{-1}, \quad \forall \sigma \le \sigma_0. $$

We then deduce that $\dot{\tau}_{\sigma}(T) \le \epsilon^{-1}$, and as $\dot{\tau}$ is a non-decreasing function since $\ddot{\tau}_{\sigma} \ge 0$, we get the result. $\square$

We now fix $\epsilon > 0$. We take $T > 0$ and $\sigma_0 > 0$ given by Lemma 9.7 so that

$$ W_1(\rho_{\sigma}(t), \rho_0(t)) \lesssim \sigma + \frac{1}{\dot{\tau}_{\sigma}(t)} + \frac{1}{\sqrt{\ln t}} \le \epsilon, $$

for all $t \ge T$ and $\sigma \le \sigma_0$. On the other hand, using Theorem 1.8, there exists $\sigma_1 > 0$ such that

$$ W_1(\rho_{\sigma}(t), \rho_0(t)) \lesssim \sqrt{C_1\sigma + \|\phi_{\sigma} - \phi_0\|_{L^2}^2} e^{\frac{C_0T}{2}} \le \epsilon, $$

for all $0 \le t \le T$ taking $\sigma \le \min(\sigma_0, \sigma_1)$ small enough. This ends the proof of Theorem 1.9, up to the final claim concerning the convergence rate.
9.6.2. Towards the convergence rate (1.8). In this section we perform a refinement of the previous analysis in order to obtain a convergence rate for the quantity of interest $\sup_{t \ge 0} W_1(\varrho_\sigma(t), W_1(\varrho_0(t)))$, assuming that $\|\phi_\sigma - \phi_0\|_{L^2} = \mathcal{O}(\sigma)$. First we give a more precise asymptotic expansion for $\dot{\tau}_\sigma^{-1}$.

**Lemma 9.8.** There exists a constant $C > 0$ such that for all $t \ge 2$, we have

$$ \left| \frac{1}{\dot{\tau}_{\sigma}(t)} - \sqrt{d\sigma} \right| \le C \frac{1}{\sqrt{\ln t}}. $$

*Proof.* We first recall that $\tau_0(t) \ge \tau_\sigma(t)$ for all $t \ge 0$ hence

$$ \dot{\tau}_{\sigma}^{2} = \frac{1}{d\sigma} \left( 1 - \frac{1}{\tau_{\sigma}^{d\sigma}} \right) \le \frac{1}{d\sigma} \left( 1 - \frac{1}{\tau_{0}^{d\sigma}} \right) \le \ln \tau_{0} = \dot{\tau}_{0}^{2}. $$

Denote $X_\sigma(t) = \dot{\tau}_\sigma(t)^{-1} - \sqrt{d\sigma} \ge 0$, so we have

$$ \begin{aligned} \dot{X}_{\sigma} &= -\frac{\ddot{\tau}_{\sigma}}{\dot{\tau}_{\sigma}^{2}} = -\frac{1}{2\dot{\tau}_{\sigma}^{2}\tau_{\sigma}^{1+d\sigma}} = -\frac{1}{2\dot{\tau}_{\sigma}^{2}\tau_{\sigma}}(1-d\sigma\dot{\tau}_{\sigma}^{2}) = -\frac{\dot{\tau}_{\sigma}^{-1} + \sqrt{d\sigma}}{2\tau_{\sigma}}X_{\sigma} \\ &\le -\frac{\dot{\tau}_{0}^{-1} + \sqrt{d\sigma}}{2\tau_{0}}X_{\sigma} = -\ddot{\tau}_{0}(\dot{\tau}_{0}^{-1} + \sqrt{d\sigma})X_{\sigma}, \end{aligned} $$

as $\dot{\tau}_0 \ge \dot{\tau}_\sigma$. We then compute, for any $t_0 > 1$,

$$ \begin{aligned} X_{\sigma}(t) &\le X_{\sigma}(t_0) \exp \left( - \int_{t_0}^{t} \ddot{\tau}_0(s) (\dot{\tau}_0(s))^{-1} + \sqrt{d\sigma} ds \right) \\ &= X_{\sigma}(t_0) \exp \left( - \left[ \ln(\dot{\tau}_0(s)) + \sqrt{d\sigma} \dot{\tau}_0(s) \right]_{t_0}^{t} \right) \\ &\le X_{\sigma}(t_0) \dot{\tau}_0(t_0) e^{\sqrt{d\sigma} \dot{\tau}_0(t_0)} \frac{1}{\dot{\tau}_0(t)} e^{-\sqrt{d\sigma} \dot{\tau}_0(t)}, \end{aligned} $$

which provides the announced bound recalling that $\dot{\tau}_0(t) \sim \sqrt{\ln t}$. $\square$

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

The convergence rate from Lemma 9.8 then ensures that there exists $C_1 > 0$ such that

$$ (9.15) \qquad W_1(\varrho_\sigma(t), \varrho_0(t)) \le C_1 \left( \frac{1}{\sqrt{\ln t}} + \sqrt{\sigma} \right), $$

for all $t \ge 2$ from the previous analysis. On the other hand, for any 1-Lipschitz function $\varphi$,

$$ \begin{aligned} & \int_{\mathbb{R}^d} \varphi(y)(\varrho_\sigma(t, y) - \varrho_0(t, y))dy \\ &= \int_{\mathbb{R}^d} \varphi(y) (\|\phi_\sigma\|_{L^2}^{-2} \tau_\sigma(t)^d |\mathbf{u}_\sigma(t, y\tau_\sigma(t))|^2 - \|\phi_0\|_{L^2}^{-2} \tau_0(t)^d |\mathbf{u}_0(t, y\tau_0(t))|^2) dy \\ &= (\|\phi_\sigma\|_{L^2}^{-2} - \|\phi_0\|_{L^2}^{-2}) \int_{\mathbb{R}^d} \varphi(y) \tau_\sigma(t)^d |\mathbf{u}_\sigma(t, y\tau_\sigma(t))|^2 dy \\ &\quad + \|\phi_0\|_{L^2}^{-2} \int_{\mathbb{R}^d} \varphi(y) (\tau_\sigma(t)^d |\mathbf{u}_\sigma(t, y\tau_\sigma(t))|^2 - \tau_0(t)^d |\mathbf{u}_0(t, y\tau_0(t))|^2) dy. \end{aligned} $$

Since $\varphi$ is 1-Lipschitz, $\varphi(y) \le |\varphi(0)| + |y|$. It is clear that the first term on the right hand side is $\mathcal{O}(\sigma)$ from the assumption $\|\phi_\sigma - \phi_0\|_{L^2} = \mathcal{O}(\sigma)$, and using
the boundedness of the second momenta (8.5). Resuming the computations from Section 8.2,

$$ \int_{\mathbb{R}^d} |\tau_\sigma(t)^d |\mathbf{u}_\sigma(t, y\tau_\sigma(t))|^2 - \tau_0(t)^d |\mathbf{u}_0(t, y\tau_0(t))|^2 dy \\ \lesssim \left( 1 + \left( \frac{\tau_0(t)}{\tau_\sigma(t)} \right)^{d/2} \right) \left( \sigma (\ln(t+4))^{3/2} + \sigma e^{C_0 t} \right). $$

We have a similar estimate for

$$ \int_{\mathbb{R}^d} |y| |\tau_\sigma(t)^d |\mathbf{u}_\sigma(t, y\tau_\sigma(t))|^2 - \tau_0(t)^d |\mathbf{u}_0(t, y\tau_0(t))|^2 dy, $$

by distributing the weight $|y|$ as follows when invoking Cauchy-Schwarz inequality,

$$ \int_{\mathbb{R}^d} |y| \times ||f(y)|^2 - |g(y)|^2 dy \le (\|yf\|_{L^2} + \|yg\|_{L^2}) \|f - g\|_{L^2}, $$

and using the boundedness of momenta in $L^2$, (8.5). Up to changing the constants, we infer that there exist $C_0$ and $C_2 > 0$ such that for all $t \ge 0$,

$$ (9.16) \qquad W_1(\varrho_\sigma(t), \varrho_0(t)) \le C_2 \sigma e^{C_0 t}. $$

We interpolate between estimates (9.15) and (9.16). Denote by $t_\sigma \ge 2$ the unique time such that we have the equality

$$ C_1 \left( \frac{1}{\sqrt{\ln t_\sigma}} + \sqrt{\sigma} \right) = C_2 \sigma e^{C_0 t_\sigma}. $$

Writing $Y = \sqrt{\sigma}$, we consider the quadratic equation

$$ C_2 e^{C_0 t_\sigma} Y^2 - C_1 Y - \frac{C_1}{\sqrt{\ln t_\sigma}} = 0, $$

whose only positive root is

$$ Y_+ = \frac{C}{2} e^{-C_0 t_\sigma} + \frac{1}{2} \left( C^2 e^{-2C_0 t_\sigma} + 4C \frac{e^{-C_0 t_\sigma}}{\sqrt{\ln t_\sigma}} \right)^{1/2}, \quad \text{where } C = C_1/C_2. $$

With rough upper and lower estimates, we infer that $Y_+^2 = \sigma$ satisfies

$$ c_1 \frac{e^{-C_0 t_\sigma}}{\sqrt{\ln t_\sigma}} \le \sigma \le c_2 \frac{e^{-C_0 t_\sigma}}{\sqrt{\ln t_\sigma}}, $$

for some constants $c_1, c_2 > 0$. The lower bound provides that $\ln \ln(\sigma^{-1}) \le c_3 \ln t_\sigma$ for a constant $c_3 > 0$, which combined with the upper bound allows to write that

$$ e^{C_0 t_\sigma} \le \frac{c_2 \sqrt{c_3}}{\sigma \sqrt{\ln \ln \frac{1}{\sigma}}}. $$

Plugging such estimate into (9.16) with $t = t_\sigma$ gives the result.

**Acknowledgments.** The authors are grateful to François Bolley for helpful discussions on Wasserstein distance. Q.C. and G.F. acknowledge the support of the CDP C2EMPI, together with the French State under the France-2030 programme, the University of Lille, the Initiative of Excellence of the University of Lille, the European Metropolis of Lille for their funding and support of the R-CDP-24-004-C2EMPI project.
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CNRS, IRMAR - UMR 6625, F-35000 RENNES, FRANCE

Email address: Remi.Carles@math.cnrs.fr

UNIV. LILLE, CNRS, INRIA, UMR 8524 - LABORATOIRE PAUL PAINLEVÉ, F-59000 LILLE, FRANCE

Email address: quentin.chauleur@inria.fr

UNIV. LILLE, CNRS, INRIA, UMR 8524 - LABORATOIRE PAUL PAINLEVÉ, F-59000 LILLE, FRANCE

Email address: guillaume.ferriere@inria.fr