# Wellposedness of a nonlinear advective-diffusive model governing the adipocyte size distribution

Aleksandra Tomaszek

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Aleksandra Tomaszek. Wellposedness of a nonlinear advective-diffusive model governing the adipocyte size distribution. 2025. hal-05351398

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# Wellposedness of a nonlinear advective-diffusive model governing the adipocyte size distribution*

Aleksandra Tomaszek

Sorbonne Université, Université Paris Cité, CNRS,
Laboratoire Jacques-Louis Lions, LJLL, F-75005 Paris, France

aleksandra.tomaszek@sorbonne-universite.fr

## Abstract

In this paper, we investigate the wellposedness of a continuously structured population dynamics model, formalized as an advection-diffusion partial differential equation governing the size distribution of adipocytes in a living being. The model specificity lies in the non-local nonlinearities on both the convective term and the boundary condition. It is motivated by the way the adipose tissue manages the storage of energy resources. The first-order moment of the unknown function, which holds the nonlinearity, represents the energy expenditure and is compared at each time with a given time-dependent energy intake. Their difference determines which of the two evolutionary regimes (lipogenesis vs. lipolysis) runs in the population. It also prescribes the corresponding (algebraic) growth rate. The proof is based on a linearization method. It uses the Aubin-Lions lemma to be applied to the underlying sequence of the solutions to the linearized version of the problem, in addition to estimates of some norms of the solution at infinity.

## 1 Introduction

The adipocytes, main component of the adipose tissue, serve as an energy reserve for the body thanks to their remarkable capacity to store and release lipids by adapting their size depending on the current energy balance. Indeed, the causal relationship between the energy equilibrium, expressed as the difference between the energy intake and expenditure, and the weight change is well established [21, 9]. When the energy intake equals the energy expenditure, the organism is in a state of homeostasis and its weight should not change [23].

Under the energy surplus conditions, adipocytes store excess calories through a process called lipogenesis [6, 14], leading to the accumulation of body fat [18, 26, 8]. Adipose tissue stores energy with two mechanisms, increasing the content of cells and recruiting new cells through the differentiation of pre-adipocyte cells [10]. During periods of insufficient energy intake, the adipocytes release the required energy into the blood circulation. This process is called lipolysis [28, 14], and leads to the shrinkage of the adipose tissue and a subsequent weight loss [9].

In our nonlinear advective-diffusive model, the unknown function $a$ is the number of adipocytes, and the trait $x \ge 0$ denotes the lipid content of an adipose cell. We here assume that the total amount of lipids absorbed by the adipose cells can be expressed as

$$ \int_{0}^{+\infty} xa(t, x) dx. $$

It represents the energy expenditure. This is due to the fact that the resting energy expenditure (expenditure when an individual lies quietly) accounts for approximately 60% to 70% of an individual's

<footnote>*This work was partially financially supported by the ANR MATIDY grant (ANR-20-CE45-0003) headed by Chloé Audebert.</footnote>
total energy expenditure [13, 25, 5]. The energy intake of an individual, represented by the time-dependent function $S$, is equivalent to the caloric content of the consumed food, converted into a corresponding amount of lipids. If $S - \int_{\mathbb{R}_+} xa \, dx > 0$, each cell increases its lipid content and the corresponding growth rate is positive. This rate can be seen as the product of two factors: the surplus $S - \int_{\mathbb{R}_+} xa \, dx$ and a trait-dependent lipogenesis term given by function $v_G$ [17]. In the same time, we assume that the recruitment of new cells with no lipid inside happens at a rate depending on energy balance through the function $f$. As a result, $\int_{\mathbb{R}_+} xa \, dx$ increases. On the other hand, when $S - \int_{\mathbb{R}_+} xa \, dx < 0$, the growth rate becomes negative. The lipid content of all cells decreases with an intensity given by function $v_L$, see [12], and proportional to $S - \int_{\mathbb{R}_+} xa \, dx$. Thus the system strives to reach the equilibrium $S = \int_{\mathbb{R}_+} xa \, dx$, and whenever it is reached, the dynamics of the model become purely diffusive. Besides, in the human body, the adipocyte average age is close to ten years [22], and here we focus on the evolution over a few months. Thus, we do not account for cell death in our model, and choose to work in a finite-time horizon.

Let us now write the continuously structured population dynamics model which we study here. It is formalized as an advection-diffusion partial differential equation, with non-local nonlinearities in the convection term and in the boundary condition. More precisely, let $T > 0$ and consider the following problem

$$ \partial_t a(t,x) + \left[ S(t) - \int_{\mathbb{R}_+} ya(t,y)dy \right] \partial_x \left[ v_a(t,x)a(t,x) \right] = \varepsilon \partial_{xx}^2 a(t,x), \quad t \in (0,T), x > 0, \quad (1) $$

$$ v_a(t,x) = 1_{\{S(t) \ge \int_{\mathbb{R}_+} ya(t,y)dy\}}(t)v_G(x) + 1_{\{S(t) < \int_{\mathbb{R}_+} ya(t,y)dy\}}(t)v_L(x), \quad t \in (0,T), x > 0, \quad (2) $$

$$ \left[ S(t) - \int_{\mathbb{R}_+} ya(t,y)dy \right] v_a(t,0) a(t,0) = \varepsilon \partial_x a(t,0) + f \left( S(t) - \int_{\mathbb{R}_+} ya(t,y)dy \right), \quad t \in (0,T), \quad (3) $$

where the (small) diffusion coefficient $\varepsilon > 0$ is given. The problem is supplemented with the initial condition

$$ a(0,x) = a^0(x), \quad x > 0. \quad (4) $$

Here, $a(t,x)$ represents the number of cells possessing trait $x$ at time $t$. The advective flux in (1) features a switching mechanism based on the comparison between the first-order moment $\int_{\mathbb{R}_+} ya \, dy$ of $a$ and the external function $S$. The function $v_a$, being a sum of functions with separated variables, is built in such a way that the sign of $S - \int_{\mathbb{R}_+} ya \, dy$ with respect to time selects between two evolution regimes, driven by the trait-dependent functions $v_G$ and $v_L$, while its value of $S(t) - \int_{\mathbb{R}_+} ya(t,y)dy$ is involved in the growth rate expressions. The recruitment of new cells is governed by this first-order moment term as well.

The dependence of the advective flux with respect to a moment of the unknown function rises in various application fields, such as epidemiology [16], where an existence result is established through semigroup theory, or in a more general context [11], where the non-local property is obtained by convolution approximations. The moment dependence happens in more situations for the diffusive part, see for instance [24] and the references therein. We must emphasize that close models to the one studied in this paper are of Lifshitz-Slyozov type, see for instance [7]. In fact, in the adipocyte context, in [15], we can find a second-order diffusive Lifshitz-Slyozov model, of advection-diffusion type, to describe the adipose tissue dynamics focusing on stationary solution and numerical simulation. This model, derived from a Becker-Döring model, presents an advective flux and a diffusive term which both depend on a moment of the unknown function. One of the main differences between this last model and ours lies in the presence of a recruitment term in the boundary condition driven by $S - \int_{\mathbb{R}_+} ya \, dy$.

We prove here that, under some suitable assumptions, the problem (1)-(4) is well-posed in the class of weak solutions. To this end, we iteratively build a sequence of approximations in the spirit of Picard's method, where each element solves a linear problem. Then, using uniform estimates and, consequently, a compactness argument, we can extract a convergent subsequence and identify its limit.

The paper is structured as follows. In the next section, the precise form of our wellposedness result is stated in Theorem 1, along with all the required assumptions. Section 3 provides the part of the
proof of Theorem 1 concerning the existence of the solutions. The proof is given in a self-contained manner, assuming a preliminary result about a linearized version of our problem. Section 4, in turn, is devoted to the rigorous proof of that linear preliminary result. The existence and uniqueness of the solution to the linear problem relies on a well-known result from the theory of parabolic partial differential equations. The second part of the proof aims to obtain useful properties of the solution to the linear problem, including nonnegativity, control of tails, and norm estimates. These are established through a careful regularization procedure. Eventually, Section 5 completes the proof of Theorem 1 by establishing the uniqueness.

## 2 Mathematical framework and main results

In this section, we first recall the model we investigate and add all the assumptions required for our study. We then provide our main result and present the sketch of the proof.

Our unknown function $a := a(t, x)$ represents the number of cells with trait $x > 0$ at time $t \in (0, T)$. It must satisfy the following advection-diffusion equation on $(0, T) \times \mathbb{R}_+^*$

$$ \partial_t a + \left[ S - \int_{\mathbb{R}_+} ya \, dy \right] \partial_x (v_a a) = \varepsilon \partial_{xx}^2 a, \qquad (5) $$

where the constant diffusion coefficient $\varepsilon > 0$ that drives the diffusive part is assumed to remain small. The energy intake $S$ is a given nonnegative time-dependent function lying in $L^\infty(0, T)$ and the function $v_a$ mainly depends on the sign of $S - \int_{\mathbb{R}_+} ya \, dy$. More precisely, for almost every $t \in (0, T)$ and $x \ge 0$, we have

$$ v_a(t, x) = \begin{cases} v_G(x) & \text{if } S(t) \ge \int_{\mathbb{R}_+} ya(t, y) \, dy, \\ v_L(x) & \text{if } S(t) < \int_{\mathbb{R}_+} ya(t, y) \, dy, \end{cases} $$

the corresponding rates $v_G$ and $v_L$ being nonnegative $W^{1,\infty}(\mathbb{R}_+^*)$ functions respectively satisfying $v_G(0) > 0$ and $v_L(0) = 0$, and both with compact supports.

Equation (5) is then supplemented with the boundary condition

$$ \left[ S(t) - \int_{\mathbb{R}_+} ya(t, y) \, dy \right] v_a(t, 0)a(t, 0) = \varepsilon \partial_x a(t, 0) + f \left( S(t) - \int_{\mathbb{R}_+} ya(t, y) \, dy \right), \qquad (6) $$

where $f$ is a nonnegative Lipschitz-continuous function on $\mathbb{R}$ with Lipschitz constant denoted by $L_f > 0$, and which is chosen to be zero on $(-\infty, 0]$. The unknown $a$ also satisfies an initial condition for almost every $x > 0$,

$$ a(0, x) = a^0(x), \qquad (7) $$

where $a^0 \in L^2(\mathbb{R}_+^*)$ is assumed to be nonnegative and compactly supported.

For the sake of simplicity, we choose $x_R > 0$ such that the compact supports of $a^0$, $v_G$ and $v_L$ are all subsets of $[0, x_R]$.

Our hypotheses ensure that the problem remains consistent with the biological mechanisms of adipose tissues. Indeed, the boundedness of the function $S$ reflects the fact that the energy intake of a living organism is limited. Furthermore, as functions $v_G$ and $v_L$ describe the rate of physiological processes, it is natural to assume that they vanish outside a relevant trait range and that they remain bounded. Their nonnegativity implies that the physiological traits of all individuals will change in the same direction, consistently with the energy balance. Condition $v_L(0) = 0$ guarantees a reflecting boundary at $x = 0$ under insufficient energy conditions, confining population to nonnegative trait values. When an energy surplus arises, condition $v_G(0) > 0$ enforces a positive velocity at the boundary $x = 0$, thereby ensuring an increase in the smallest traits. As for the function $f$, governing the recruitment of new cells, it is assumed to be active only in response to an energy surplus. The assumption of Lipschitz continuity reflects the fact that such responses in living organisms are often gradual. Finally, as $a^0$ represents a population distribution with respect to the physiological trait, it is assumed to be nonnegative and supported on a bounded range of traits.
In this work, we investigate the existence of weak solutions to (1)–(4) generalized as follows. Let us suppose, for the time being, that the problem (1)–(4) admits a smooth solution $a$. For $t \in (0, T)$ we set

$$ \Gamma(t) = S(t) - \int_{\mathbb{R}_+} ya(t, y)dy. $$

Further, let $\psi \in C_c^\infty(\mathbb{R})$. One can observe, that for any $t \in (0, T)$, multiplying (1) by $\psi$, and integrating with respect to the trait variable on the positive half-line, results in

$$ \begin{aligned} \int_{\mathbb{R}_+} \partial_t a(t, x) \psi(x) dx &= - \int_{\mathbb{R}_+} \partial_x [\Gamma(t) v_a(t, x) a(t, x) - \varepsilon \partial_x a(t, x)] \psi(x) dx \\ &= - [\Gamma(t) v_a(t, x) a(t, x) - \varepsilon \partial_x a(t, x)] \psi(x) \Big|_{0}^{\infty} + \int_{\mathbb{R}_+} [\Gamma(t) v_a(t, x) a(t, x) - \varepsilon \partial_x a(t, x)] \partial_x \psi(x) dx \\ &= f(\Gamma(t)) \psi(0) + \int_{\mathbb{R}_+} [\Gamma(t) v_a(t, x) a(t, x) - \varepsilon \partial_x a(t, x)] \partial_x \psi(x) dx, \quad (8) \end{aligned} $$

where the boundary term is evaluated using condition (3) and the fact that $\psi$ has compact support. By the density of $\{\psi|_{\mathbb{R}_+^*} : \psi \in C_c^\infty(\mathbb{R})\}$ in $H^1(\mathbb{R}_+^*)$, see [4, p. 211], we find that the same identity holds when the smooth function $\psi$ is substituted with $w \in H^1(\mathbb{R}_+^*)$. In fact, functions with regularity $H^1$ are in particular Hölder continuous. It follows that they can be continuously extended to the boundary of domain, here $x=0$, and also that they disappear at infinity. One can further notice, that identity (8) is actually well-defined only under the assumption $a(t, \cdot) \in H^1(\mathbb{R}_+^*)$. Consequently, we search for a function $a : (0, T) \to H^1(\mathbb{R}_+^*)$ satisfying for almost every $t \in (0, T)$ and for all $w \in H^1(\mathbb{R}_+^*)$

$$ \frac{d}{dt} \int_{\mathbb{R}_+} a(t, x) w(x) dx + \varepsilon \int_{\mathbb{R}_+} \partial_x a(t, x) \partial_x w(x) dx - \int_{\mathbb{R}_+} \Gamma(t) v_a(t, x) a(t, x) \partial_x w(x) dx = f(\Gamma(t)) w(0), $$

with the initial condition

$$ a(0, x) = a^0(x), \quad \text{for a.e. } x > 0. $$

More precisely, we are interested in the following problem: find a smooth enough such that, for any $w \in H^1(\mathbb{R}_+^*)$, and for almost every $t \in (0, T)$,

$$ \begin{aligned} \frac{d}{dt} \int_{\mathbb{R}_+} a(t, x) w(x) dx + \varepsilon \int_{\mathbb{R}_+} \partial_x a(t, x) \partial_x w(x) dx \\ = \left( S(t) - \int_{\mathbb{R}_+} ya(t, y) dy \right) \int_{\mathbb{R}_+} v_a(t, x) a(t, x) \partial_x w(x) dx + f \left( S(t) - \int_{\mathbb{R}_+} ya(t, y) dy \right) w(0), \quad (9) \end{aligned} $$

with the initial condition (7). Our main result is then the following.

**Theorem 1.** Let $\varepsilon_0 > 0$. Under the assumptions provided for the data $a^0$, $v_G$, $v_L$, $S$ and $f$, for any $\varepsilon \in (0, \varepsilon_0)$, the problem (7)–(9) admits a unique nonnegative solution $a$ satisfying

$$ a \in L^2(0, T; H^1(\mathbb{R}_+^*)), \quad \partial_t a \in L^2(0, T; H^1(\mathbb{R}_+^*)'). \quad (10) $$

Moreover, there exists a constant $C \ge 0$ only depending on the data, such that

$$ 0 \le a(t, x) \le C, \quad \text{a.e. } t, x, \qquad \|a(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 \le C, \quad \text{a.e. } t, \quad (11) $$

$$ \| \partial_x a \|_{L^2(0, T; L^2(\mathbb{R}_+^*))} \le \frac{C}{\varepsilon}, \qquad \| \partial_t a \|_{L^2(0, T; H(\mathbb{R}_+^*))'} \le (1 + \varepsilon_0)C, \quad (12) $$

$$ \|a\|_{L^\infty(0, T; L^1(\mathbb{R}_+^*))} \le C, \qquad (13) $$

$$ \int_r^{+\infty} a(t, y) dy \le \frac{\varepsilon_0 C}{\sqrt{r}}, \quad \int_r^{+\infty} ya(t, y) dy \le \frac{\varepsilon_0 C}{r}, \quad \text{a.e. } t, \forall r \ge 2x_R. \quad (14) $$
The proof of Theorem 1 is based on the construction of an approximating sequence, with each element being a solution to the linear analogue of the problem (7)–(9). Replacing the non-local term

$$ \Gamma(t) := S(t) - \int_{\mathbb{R}_+} ya(t, y)dy $$

with a given function

$$ \tilde{\Gamma} \in L^{\infty}(0, T) $$

yields a linear form of the weak formulation (9), that is

$$ \begin{aligned} & \frac{d}{dt} \int_{\mathbb{R}_+} a(t, x)w(x)dx + \varepsilon \int_{\mathbb{R}_+} \partial_x a(t, x)\partial_x w(x)dx \\ & = \int_{\mathbb{R}_+} (\tilde{\Gamma}_+(t)v_G(x) - \tilde{\Gamma}_-(t)v_L(x))a(t, x)\partial_x w(x)dx + f(\tilde{\Gamma}(t))w(0), \end{aligned} \quad (15) $$

with the same initial condition (7). The uniform estimates on the solution to the above linear problem, required to extract a convergent subsequence, are the topic of the proposition below.

**Proposition 1.** Let $\varepsilon_0 > 0$. Under the assumptions provided for $a^0$, $v_G$, $v_L$, $\tilde{\Gamma}$ and $f$, and for any $\varepsilon \in (0, \varepsilon_0)$ the problem (15) with initial condition (7) admits a unique nonnegative solution

$$ a \in L^2(0, T; H^1(\mathbb{R}_+^*)) \cap C([0, T]; L^2(\mathbb{R}_+^*)), \quad \partial_t a \in L^2(0, T; H^1(\mathbb{R}_+^*)'). \quad (16) $$

Furthermore, there exist constants $\tilde{C}_1 \ge 0$ depending on $a^0$, $v_G$, $v_L$, $\tilde{\Gamma}$, $f$ and $\tilde{C}_2 \ge 0$ only depending on $a^0$, $\|\tilde{\Gamma}_+\|_{L^\infty(0,T)}$ and $f$ such that, for almost every $t \in (0,T)$, the following estimates hold:

$$ 0 \le a(t, x) \le \tilde{C}_1, \quad \text{a.e. } t \in (0, T), x > 0, \qquad \|a(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 \le \tilde{C}_1, \quad \text{a.e. } t \in (0, T), \quad (17) $$

$$ \|\partial_x a\|_{L^2(0,T;L^2(\mathbb{R}_+^*))}^2 \le \frac{\tilde{C}_1}{\varepsilon}, \qquad \|\partial_t a\|_{L^2(0,T;H^1(\mathbb{R}_+^*))'} \le (1+\varepsilon_0)\tilde{C}_1, \quad (18) $$

$$ \|a\|_{L^\infty((0,T;L^1(\mathbb{R}_+^*))} \le \tilde{C}_2 \quad (19) $$

$$ \int_r^{+\infty} a(t, y)dy \le \frac{\varepsilon_0 \tilde{C}_1}{\sqrt{r}}, \quad \int_r^{+\infty} ya(t, y)dy \le \frac{\varepsilon_0 \tilde{C}_2}{r}, \quad \text{a.e. } t \in (0, T), \forall r \ge 2x_R. \quad (20) $$

Let us briefly explain how the proof of Theorem 1 is handled. We start by building an approximating sequence $(a_n)_{n \ge 1}$ upon a Picard-type iterative method. Given an element $a_n$, the next iterate $a_{n+1}$ is defined as the solution to a problem in which the nonlocal term is replaced by a function $\Gamma_n$ only depending on $a_n$ and the data. Since such an equation becomes linear, Proposition 1 allows to derive by induction the sequence $(a_n)_{n \ge 1}$, and a collection of estimates which are uniform with respect to $n$. Firstly, thanks to those estimates, we can apply the Aubin-Lions lemma [1] to extract a subsequence which strongly converges in $L^2(0, T; L^2_{\text{loc}}(\mathbb{R}_+^*))$. Furthermore, the uniform control of the tails, as in (20), enables this convergence to be extended to $L^2(0, T; L^2(\mathbb{R}_+^*))$. Let us denote the corresponding limit by $a$. We still have to deal with the convergence of $(\Gamma_n)_{n \ge 1}$. Given the almost-everywhere convergence $(xa_n)$ towards $xa$, the available estimates, together with the Vitali convergence theorem, allow to deduce the convergence of $(\Gamma_n)_{n \ge 1}$ to $\Gamma: t \mapsto S(t) - \int_{\mathbb{R}_+} ya(t, y)dy$. Finally, the Banach-Alaouglu theorem ensures the existence of weak limits of $(\partial_x a_n)_{n \ge 1}$ and $(\partial_t a_n)_{n \ge 1}$, again up to a subsequence. This leads to the conclusion that, in fact, $a \in L^2(0, T; H^1(\mathbb{R}_+^*))$ and $\partial_t a \in L^2(0, T; H^1(\mathbb{R}_+^*)')$. All these facts allow to conclude that $a$ satisfies (9) with initial condition (7).

## 3 Existence proof for Theorem 1

We now proceed with the existence part of the proof of Theorem 1. The argument is developed in a self-contained manner, except for the use of Proposition 1. Let us recall the assumptions satisfied by
the data of Problem (7)-(9), which are

$$S \in L^{\infty}(0, T), \qquad (21)$$

$$a^0 \in L^2(\mathbb{R}_+^*), \quad a^0 \ge 0, \quad \text{supp } a^0 \subset [0, x_R], \qquad (22)$$

$$v_G \in W^{1,\infty}(\mathbb{R}_+^*), \quad v_G(x) \ge 0 \text{ a.e. } x > 0, \quad v_G(0) > 0, \quad \text{supp } v_G \subset [0, x_R], \qquad (23)$$

$$v_L \in W^{1,\infty}(\mathbb{R}_+^*), \quad v_L(x) \ge 0 \text{ a.e. } x > 0, \quad v_L(0) = 0, \quad \text{supp } v_L \subset [0, x_R], \qquad (24)$$

$$f \in \text{Lip}(\mathbb{R}) \text{ with } L_f \text{ as Lipschitz constant, } f(z) \ge 0 \text{ if } z > 0, f(z) = 0 \text{ otherwise,} \qquad (25)$$

where $x_R > 0$ is given.

As already emphasized, the core idea consists in building an approximating sequence through a Picard iteration scheme. For any $n \ge 1$, from $a_{n-1}$, we define $a_n$ as the function satisfying, for almost every $t \in (0, T)$ and all $w \in H^1(\mathbb{R}_+^*)$,

$$\frac{d}{dt} \int_{\mathbb{R}_+} a_n(t,x)w(x)dx + \varepsilon \int_{\mathbb{R}_+} \partial_x a_n(t,x)\partial_x w(x)dx \\ - \Gamma_{n-1}(t) \int_{\mathbb{R}_+} v_{a_{n-1}}(t,x) a_n(t,x)\partial_x w(x)dx = f(\Gamma_{n-1}(t))w(0), \qquad (26)$$

and the initial condition, for almost every $x > 0$,

$$a_n(0, x) = a^0(x), \qquad (27)$$

where $\Gamma_{n-1}(t) = S(t) - \int_{\mathbb{R}_+} ya_{n-1}(t, y)dy$, with the convention $a_0(t, x) = a^0(x)$ for almost every $t \in (0, T)$ and $x > 0$. The following lemma explains why $(a_n)_{n \ge 1}$ is well defined and which estimates it satisfies.

**Lemma 1.** Let $\varepsilon_0 > 0$. Under the assumptions (21)-(25) on the data, for any $\varepsilon \in (0, \varepsilon_0)$ the problem (26)-(27) defines a sequence $(a_n)_{n \ge 1}$ of nonnegative functions which satisfy the following uniform estimates with respect to $n \ge 1$

$$0 \le a_n(t, x) \le M_1, \quad \text{a.e. } t \in (0, T), x > 0, \qquad \|a_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} \le M_1, \quad \text{a.e. } t \in (0, T), \qquad (28)$$

$$\|\partial_x a_n\|_{L^2(0,T;L^2(\mathbb{R}_+^*))} \le \frac{M_1}{\varepsilon}, \qquad \|\partial_t a_n\|_{L^2(0,T;H^1(\mathbb{R}_+^*))} \le (1+\varepsilon_0)M_1, \qquad (29)$$

$$\|a\|_{L^{\infty}(0,T;L^1(\mathbb{R}_+^*))} \le M_2, \qquad \int_r^{+\infty} ya_n(t,y)dy \le \frac{\varepsilon_0 M_2}{r}, \quad \forall r \ge 2x_R, \qquad (30)$$

where the constant $M_1 \ge 0$ depends on $a^0$, $v_G$, $v_L$, $S$ and $f$, and the constant $M_2 \ge 0$ only on $a^0$, $S$ and $f$.

*Proof.* In order to apply Proposition 1, we just need to prove that $(\Gamma_n)_{n \in \mathbb{N}}$ is $L^\infty(0, T)$. In fact, we shall prove the stronger result that $(\Gamma_n)_{n \in \mathbb{N}}$ is uniformly bounded in $L^\infty(0, T)$ with respect to $n \in \mathbb{N}$. First, we note that, for almost every $t \in (0, T)$,

$$|\Gamma_0(t)| \le |S(t)| + \int_{\mathbb{R}_+} ya^0(y)dy \le \|S\|_{L^\infty(0,T)} + x_R^{3/2} \|a^0\|_{L^2(\mathbb{R}_+^*)},$$

thanks to Assumption (22). Consequently, using Proposition 1, we obtain a unique nonnegative solution $a_1$ satisfying all the properties listed in Proposition 1. In particular, there exists a constant $\tilde{C}_2^0 \ge 0$ only depending on $a^0$, $\|(\tilde{\Gamma}_0)_+\|_{L^\infty(0,T)}$ and $f$, such that, for almost every $t \in (0, T)$ and any $r \ge 2x_R$,

$$0 \le \int_{\mathbb{R}_+} ya_1(t, y)dy \le r \|a_1\|_{L^\infty(0,T)} + \int_r^\infty ya_1(t, y)dy \le \tilde{C}_2^0 \left(r + \frac{\varepsilon_0}{r}\right).$$

In particular, by taking $r = 2x_R$, for almost every $t \in (0, T)$, we get

$$|\Gamma_1(t)| \le \|S\|_{L^\infty(0,T)} + \tilde{C}_2^0 \left(2x_R + \frac{\varepsilon_0}{2x_R}\right).$$
We claim that the estimate

$$ \|\Gamma_n\|_{L^\infty(0,T)} \le \|S\|_{L^\infty(0,T)} + \tilde{C}_2^0 \left( 2x_R + \frac{\varepsilon_0}{2x_R} \right) \quad (31) $$

actually holds for any $n \ge 1$. With the base case $n = 1$, addressed above, we turn to the inductive step. Assume that (31) holds for, some $n \ge 1$. Then Proposition 1 can be applied yielding existence of unique, nonnegative solution $a_{n+1}$ along with the constant $\tilde{C}_2^n \ge 0$ depending only on $a^0$, $\|(\tilde{\Gamma}_n)_+\|_{L^\infty(0,T)}$ and $f$, such that, for almost every $t \in (0,T)$

$$ 0 \le \int_{\mathbb{R}_+} ya_{n+1}(t, y)dy \le \tilde{C}_2^n \left( 2x_R + \frac{\varepsilon_0}{2x_R} \right). $$

However, since both $a_0$ and $a_n$ are nonnegative, for almost every $t \in (0,T)$, it holds that $(\Gamma_0)_+(t) \le S(t)$ and $(\Gamma_n)_+(t) \le S(t)$. This implies that $\tilde{C}_2^n$ can be chosen equal to $\tilde{C}_2^0$. $\square$

At that point, $(a_n)_{n \ge 1}$ appears as a bounded sequence in $L^2(0,T; H^1(\mathbb{R}_+^*))$ which also satisfies the fact that $(\partial_t a_n)_{n \ge 1}$ is bounded in $L^2(0,T; H^1(\mathbb{R}_+^*))'$.

In order to apply the Aubin-Lions lemma and extract a convergent subsequence, it is however mandatory to work with functions defined on compact subsets of the space. Let $K > 0$, and, for any $n \ge 1$, define the restriction of $a_n$ to $(0,T) \times (0,K)$,

$$ b_n = a_n|_{(0,T) \times (0,K)}. \quad (32) $$

We clearly have, for any $n \ge 1$,

$$ \|b_n\|_{L^2(0,T;H^1(0,K))} \le \|a_n\|_{L^2(0,T;H^1(\mathbb{R}_+^*))} \le \left(1 + \frac{1}{\varepsilon}\right) M_1. $$

The identification and uniform boundedness of $(\partial_t b_n)_{n \ge 1}$ is established in the following lemma.

**Lemma 2.** Let $(b_n)_{n \ge 1}$ be as in (32). The sequence $(\partial_t b_n)_{n \ge 1}$ is bounded in $L^2(0,T; H^{-1}(0,K))$.

*Proof.* Let $P: H_0^1(0,K) \to H^1(\mathbb{R}_+^*)$ be the trivial extension operator given by

$$ Pu(x) = \begin{cases} u(x) & \text{if } x \in (0, K), \\ 0 & \text{if } x \ge K. \end{cases} $$

Further, let $P^*: H^1(\mathbb{R}_+^*)' \to H^{-1}(0,K)$ be its adjoint operator, defined by

$$ \langle z, Pu \rangle_{H^1(\mathbb{R}_+^*)', H^1(\mathbb{R}_+^*)} = \langle P^*z, u \rangle_{H^{-1}(0,K), H_0^1(0,K)}, \quad z \in H^1(\mathbb{R}_+^*)', u \in H_0^1(0,K). $$

In particular,

$$ \|P^*\|_{\mathcal{L}(t; H^1(\mathbb{R}_+^*)', H^{-1}(0,K))} = \|P\|_{\mathcal{L}(t; H_0^1(0,K), H^1(\mathbb{R}_+^*))} = 1. $$

We claim that, for any $n \ge 1$, it holds that $\partial_t b_n = P^* \partial_t a_n \in H^{-1}(0,K)$. Indeed, for all $\phi \in C_c^1(0,T)$, the space of continuously differentiable functions whose support is a compact subset of $(0,T)$, and all $w \in H_0^1(0,K)$, one may compute

$$ \begin{aligned} \int_0^T \phi' \langle b_n, w \rangle_{H^1(0,K), H^1(0,K)} dt &= \int_0^T \phi' \langle a_n, Pw \rangle_{H^1(\mathbb{R}_+^*), H^1(\mathbb{R}_+^*)} dt \\ &= - \int_0^T \phi \langle \partial_t a_n, Pw \rangle_{H^1(\mathbb{R}_+^*), H^1(\mathbb{R}_+^*)} dt = - \int_0^T \phi \langle P^* \partial_t a_n, w \rangle_{H^{-1}(0,K), H_0^1(0,K)} dt. \end{aligned} $$

Thus, it follows that, for any $n \ge 1$,

$$ \|\partial_t b_n\|_{L^2(0,T; H^{-1}(0,K))} \le \|P^*\|_{\mathcal{L}(t; H^1(\mathbb{R}_+^*)', H^{-1}(0,K))} \|\partial_t a_n\|_{L^2(0,T; H^1(\mathbb{R}_+^*)')} \le (1 + \varepsilon_0)M_1, $$

which concludes the proof of Lemma 2. $\square$
Since $H^1(0, K) \hookrightarrow L^2(0, K) \hookrightarrow H^{-1}(0, K)$ with the compact embedding of $H^1(0, K)$ in $L^2(0, K)$ and the continuous one of $L^2(0, K)$ in $H^{-1}(0, K)$, the Aubin-Lions lemma can be applied in the form which can be found in [3, p. 102], to state that $(b_n)_{n \ge 1}$ admits a strongly convergent subsequence in $L^2(0, T; L^2(0, K))$. To simplify the notation, slightly abusing it, this subsequence will be still denoted by $(b_n)_{n \ge 1}$.

We now want to prove that $(a_n)_{n \ge 1}$ is itself convergent in $L^2(0, T; L^2(\mathbb{R}_+^*))$. It is enough to check that it is a Cauchy sequence. Let $\delta > 0$ and $n, m \ge 1$. We then want to estimate

$$ \begin{aligned} \|a_n - a_m\|_{L^2(0,T;L^2(\mathbb{R}_+^*))}^2 &= \int_0^T \int_{\mathbb{R}_+} |a_n(t,x) - a_m(t,x)|^2 dx dt \\ &\le \int_0^T \int_0^K |b_n(t,x) - b_m(t,x)|^2 dx dt + \int_0^T \int_K^{+\infty} |a_n(t,x) - a_m(t,x)|^2 dx dt. \end{aligned} $$

The first term is readily controlled thanks to the Cauchy property of the sequence $(b_n)_{n \ge 1}$ in $L^2(0, T; L^2(0, K))$. Indeed, there exists $N_\delta \in \mathbb{N}$, such that, for any $n, m \ge N_\delta$,

$$ \|b_n - b_m\|_{L^2(0,T;L^2(0,K))}^2 < \frac{\delta}{2}. $$

The second term can be upper-bounded using tail estimates (30) together with the pointwise bound (28). More precisely, we have

$$ \begin{aligned} \int_0^T \int_K^{+\infty} |a_n(t,x) - a_m(t,x)| dx dt &= \int_0^T \int_K^{+\infty} \frac{|a_n(t,x) - a_m(t,x)|}{x} |a_n(t,x) - a_m(t,x)| dx dt \\ &\le \frac{1}{K} \left( \|a_n\|_{L^\infty(0,T;L^\infty(\mathbb{R}_+^*))} + \|a_m\|_{L^\infty(0,T;L^\infty(\mathbb{R}_+^*))} \right) \int_0^T \int_K^{+\infty} x (a_n(t,x) + a_m(t,x)) dx dt \\ &\le \frac{2M_1}{K} \times \frac{2\epsilon_0 M_1 T}{K}, \end{aligned} $$

for any $K \ge 2x_R$. Since one can always choose

$$ K \ge \max \left( 2x_R, \sqrt{\frac{1}{8\delta\epsilon_0 T M_1^2}} \right), $$

we conclude that $(a_n)_{n \ge 1}$ is indeed Cauchy, and therefore a convergent sequence in the Hilbert space $L^2(0, T; L^2(\mathbb{R}_+^*))$. Denote by $a$ its limit. In particular, bounds (28) and (30) are preserved at the limit when $n$ goes to infinity, so that $a$ satisfies the same bounds.

The next step is to prove that $\Gamma: t \mapsto S(t) - \int_{\mathbb{R}_+} ya(t, y)dy$, which, at this stage, may not even be well-defined, is a limit of $(\Gamma_n)_{n \ge 1}$ in $L^2(0, T)$. To this end, we aim to apply the Vitali convergence theorem, and begin by checking its assumptions.

First, the convergence of $(a_n)_{n \ge 1}$ towards $a$ in $L^2(0, T; L^2(\mathbb{R}_+^*))$ entails that, for almost every $t \in (0, T)$, when $n$ goes to $+\infty$,

$$ (ya_n(t, y)) \to ya(t, y), \quad \text{a.e. } y > 0. $$

Remembering property (30), it is clear that, for any $\delta > 0$, the choice $r \ge \max(2x_R, \frac{\epsilon_0 M_2}{\delta})$ allows to obtain, for almost every $t$, that

$$ \int_r^{+\infty} ya_n(t, y) dy < \delta, \quad \forall n \ge 1. $$

Finally, we claim that the sequence $(y \mapsto ya_n(t, y))_{n \ge 1}$ has uniformly absolutely continuous integrals. Let $\delta > 0$. We aim to prove that, for $A \subset \mathbb{R}_+^*$ of measure small enough, the following quantity

$$ \int_A ya_n(t, y) dy = \int_{A \cap \{y < r\}} ya_n(t, y) dy + \int_{A \cap \{y > r\}} ya_n(t, y) dy \quad (33) $$
is upper-bounded by $\delta$.

As previously, the second term of the right-hand side of (33) can be estimated using property (30). Indeed, if we choose this time $r > \max(2x_R, \frac{2\epsilon_0 M_2}{\delta})$, we obtain

$$ \int_{A \cap \{y > r\}} ya_n(t, y) \, dy \le \int_r^{+\infty} ya_n(t, y) \, dy < \frac{\delta}{2}. $$

And the first term is estimated as follows

$$ \int_{A \cap \{x < r\}} ya_n(t, y) \, dy \le r \int_A |a_n(t, y)| \, dy \le r |A|^{\frac{1}{2}} \|a_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} \le r |A|^{\frac{1}{2}} M_1 $$

Thus, it is enough to take $A \subset \mathbb{R}_+^*$, such that $|A|^{\frac{1}{2}} < \frac{\delta}{2rM_1}$ to obtain the required bound. Hence, the Vitali convergence theorem can be applied, to ensure that

$$ \lim_{n \to +\infty} \Gamma_n(t) = S(t) - \lim_{n \to +\infty} \int_{\mathbb{R}_+} ya_n(t, y) \, dy = S(t) - \int_{\mathbb{R}_+} ya(t, y) \, dy. $$

Since $t$ is an arbitrary point within a set of full measure, we conclude that $(\Gamma_n)_{n \ge 1}$ converges to $\Gamma: t \mapsto S(t) - \int_{\mathbb{R}_+} ya(t, y) \, dy$ almost everywhere. As this is also a uniformly bounded sequence with respect to $n \ge 1$, the assumptions of the Lebesgue dominated convergence theorem are satisfied. Hence, it follows that in fact $(\Gamma_n)_{n \ge 1}$ converges to $\Gamma$ in $L^2(0, T)$.

On the other hand, since the sequence $(\partial_t a_n)_{n \ge 1}$ is uniformly bounded with respect to $n \ge 1$ in $L^2(0, T; H^1(\mathbb{R}_+^*)')$, the Banach-Alaoglu theorem, see [4, p. 66], guarantees the existence of a weakly-* convergent subsequence in $L^2(0, T; H^1(\mathbb{R}_+^*)')$. That is, slightly abusing the notation and still denoting the sequence by $(\partial_t a_n)_{n \ge 1}$, we have $\partial_t a_n \rightharpoonup^* G$ in $L^2(0, T; L^2(\mathbb{R}_+^*))$. In particular, for any $w \in C_c^1(0, T; H^1(\mathbb{R}_+^*))$, one may compute

$$ \begin{aligned} \int_0^T \langle G(t), w(t) \rangle_{H^1(\mathbb{R}_+^*)', H^1(\mathbb{R}_+^*)} dt &= \lim_{n \to +\infty} \int_0^T \langle \partial_t a_n(t), w(t) \rangle_{H^1(\mathbb{R}_+^*)', H^1(\mathbb{R}_+^*)} dt \\ &= - \lim_{n \to +\infty} \int_0^T \langle a_n(t), \partial_t w(t) \rangle_{H^1(\mathbb{R}_+^*)', H^1(\mathbb{R}_+^*)} dt = - \int_0^T \langle a(t), \partial_t w(t) \rangle_{H^1(\mathbb{R}_+^*)', H^1(\mathbb{R}_+^*)} dt. \end{aligned} $$

It follows that the distributional derivative $\partial_t a$ lies in $L^2(0, T; H^1(\mathbb{R}_+^*)')$, with $\partial_t a(t, \cdot) = G(t, \cdot)$ for almost every $t \in (0, T)$. Since passing to the limit preserves the inequality for the norm, it holds that $\|\partial_t a\|_{L^2(0, T; H^1(\mathbb{R}_+^*)')} \le M_1$.

Similarly, by a standard consequence of the Banach-Alaoglu theorem, see [4, p. 69], the uniform boundness of $(\partial_x a_n)_{n \ge 1}$ in $L^2(0, T; L^2(\mathbb{R}_+^*))$ implies the existence of a weakly convergent subsequence. As a result, there exists $F \in L^2(0, T; L^2(\mathbb{R}_+^*))$ such that $\partial_x a_n \rightharpoonup F$ in $L^2(0, T; L^2(\mathbb{R}_+^*))$, again up to a subsequence. In particular, for any $w \in L^2(0, T; C_c^1(\mathbb{R}_+^*))$, we may compute

$$ \begin{aligned} \int_0^T \langle F(t), w(t) \rangle_{L^2(\mathbb{R}_+^*), L^2(\mathbb{R}_+^*)} dt &= \lim_{n \to +\infty} \int_0^T \langle \partial_x a_n(t), w(t) \rangle_{L^2(\mathbb{R}_+^*), L^2(\mathbb{R}_+^*)} dt \\ &= - \lim_{n \to +\infty} \int_0^T \langle a_n(t), \partial_x w(t) \rangle_{L^2(\mathbb{R}_+^*), L^2(\mathbb{R}_+^*)} dt = - \int_0^T \langle a(t), \partial_x w(t) \rangle_{L^2(\mathbb{R}_+^*), L^2(\mathbb{R}_+^*)} dt. \end{aligned} $$

It follows that, in fact, $a \in L^2(0, T; H^1(\mathbb{R}_+^*))$, with $\partial_x a(t, x) = F(t, x)$ for almost every $t \in (0, T)$ and $x > 0$. As before, the norm estimate is preserved in the limit and hence (12) holds.

With the above results in place, we can now take the limit with respect to $n$ in (26) and observe, that this yields the relation (9). It remains to verify that the initial condition is satisfied. Because of the continuous embedding [19, p. 202]

$$ \{f \in L^2(0, T; H^1(\mathbb{R}_+^*)) : \partial_t f \in L^2(0, T; H^1(\mathbb{R}_+^*)')\} \hookrightarrow C([0, T]; H^1(\mathbb{R}_+^*)'), $$

it follows that the convergence of $(a_n)_{n \ge 1}$ to $a$ holds in $C([0, T]; H^1(\mathbb{R}_+^*)')$. Since $L^2(\mathbb{R}_+^*)' \subset H^1(\mathbb{R}_+^*)'$ and $a^0 \in L^2(\mathbb{R}_+^*)$, for every $w \in L^2(\mathbb{R}_+^*)$. In particular,

$$ a(0, \cdot) = \lim_{n \to \infty} a_n(0, \cdot) = a^0 \quad \text{in } H^1(\mathbb{R}_+^*)'. $$
As $a^0$ is in fact an element of $L^2(\mathbb{R}_+^*)$, we may identify $a(0, \cdot)$ with $a^0$ in $L^2$, by viewing it as the preimage under the canonical embedding $L^2(\mathbb{R}_+^*) \hookrightarrow H^1(\mathbb{R}_+^*)'$. Hence, $a(0, x) = a^0(x)$ for almost every $x > 0$. In the end, we conclude that $a$ indeed solves (7)-(9).

The uniqueness of the nonnegative solution can only be proven in an upcoming section. Indeed, to deal with it, we need further results about the linear problem, which are now discussed in the following section.

# 4 Proof of Proposition 1

We now turn to the analysis of the linear problem (15) with initial condition (7). The assumptions considered in this section are (22)–(25), together with the additional hypothesis

$$ \tilde{\Gamma} \in L^{\infty}(0, T). \qquad (34) $$

## 4.1 Wellposedness of the linear problem

The well-posedness of the problem comes from the application of a classical result, here formulated as in [27, pp. 423–425].

**Theorem 2.** Let $V$ and $H$ be real, separable Hilbert spaces, such that $V$ is infinite dimensional and $V \hookrightarrow H \simeq H' \hookrightarrow V'$ and the embedding $V \hookrightarrow H$ is continuous and dense. Consider the initial value problem

$$ \frac{d}{dt}\langle u(t), w \rangle_H + b(t; u(t), w) = \langle g(t), w \rangle_V, \quad \text{for all } w \in V \text{ and a.e. } t \in (0, T), \qquad (35) $$

$$ u(0) = u_0. \qquad (36) $$

Assume the following:

(i) $b(t; \cdot, \cdot) : V \times V \to \mathbb{R}$ is a bilinear form, that is measurable in $t$ and satisfies a Gårding-type inequality, i.e. there exist constants $d_1, d_2 > 0, d_3 \ge 0$ such that

$$ |b(t; u, v)| \le d_1 \|u\|_V \|v\|_V, \qquad (37) $$

$$ b(t; u, u) \ge d_2 \|u\|_V^2 - d_3 \|u\|_H^2, \qquad (38) $$

for all $u, v \in V$ and almost every $t \in (0, T)$,

(ii) $g \in L^2(0, T; V')$,

(iii) $u_0 \in H$.

Then there exists a unique solution $u \in L^2(0, T; V) \cap C([0, T]; H)$, $\partial_t u \in L^2(0, T; V')$ to the problem (35)–(36). Furthermore the solution depends continuously on the data. Specifically, there exists a constant $D > 0$, such that

$$ \|u\|_{L^2(0,T;V)} + \|\partial_t u\|_{L^2(0,T;V')} \le D (\|u_0\|_H + \|g\|_{L^2(0,T;V')}), $$

Given that both $L^2(\mathbb{R}_+^*)$ and $H^1(\mathbb{R}_+^*)$ are infinite-dimensional separable Hilbert spaces, we identify the spaces appearing in the statement of Theorem 2 as $V = H^1(\mathbb{R}_+^*)$ and $H = L^2(\mathbb{R}_+^*)$. From the embeddings $C_c^\infty(\mathbb{R}_+^*) \hookrightarrow H^1(\mathbb{R}_+^*) \hookrightarrow L^2(\mathbb{R}_+^*)$ and the fact that $C_c^\infty(\mathbb{R}_+^*)$ is dense in $L^2(\mathbb{R}_+^*)$ [4, pp. 109–110], follows the density of the embedding $H^1(\mathbb{R}_+^*) \hookrightarrow L^2(\mathbb{R}_+^*)$. From the stated density subsequently arises the embedding $L^2(\mathbb{R}_+^*)' \hookrightarrow H^1(\mathbb{R}_+^*)'$, see [20, p. 97].

Further, for almost every $t \in (0, T)$, we define a family of bilinear maps

$$ b(t; \cdot, \cdot) : H^1(\mathbb{R}_+^*) \times H^1(\mathbb{R}_+^*) \to \mathbb{R}, $$

$$ (u, w) \mapsto \varepsilon \int_{\mathbb{R}_+} \partial_x u(x) \partial_x w(x) \, dx - \int_{\mathbb{R}_+} (\tilde{\Gamma}_+(t) v_G(x) - \tilde{\Gamma}_-(t) v_L(x)) u(x) \partial_x w(x) \, dx $$
and a linear functional

$$g(t): H^1(\mathbb{R}_+^*) \to \mathbb{R}, \quad u \to f(\tilde{\Gamma}(t))u(0).$$

We observe that, with these choices, (15) can be rewritten as

$$\frac{d}{dt}\langle u(t), w\rangle_{L^2(\mathbb{R}_+^*)} + b(t; u(t), w) = \langle g(t), w\rangle_{H^1(\mathbb{R}_+^*)'} \cdot H^1(\mathbb{R}_+^*), \quad \text{a.e. } t \in (0, T).$$

We proceed to check the assumptions of Theorem 2. The measurability with respect to $t$ of the bilinear form $b(t)$ is derived from the measurability of $\tilde{\Gamma}$. Further, let $u, w \in H^1(\mathbb{R}_+^*)$ and set $M_v = \|v_G\|_{W^{1,\infty}(\mathbb{R}_+^*)} + \|v_L\|_{W^{1,\infty}(\mathbb{R}_+^*)}$. We straightforwardly have, for almost every $t \in (0, T)$,

$$\begin{aligned} |b(t; u, w)| &\le \varepsilon \int_{\mathbb{R}_+} |\partial_x u(x) \partial_x w(x)| dx + \|\tilde{\Gamma}\|_{L^\infty(0,T)} M_v \int_{\mathbb{R}_+} |u(x) \partial_x w(x)| dx \\ &\le \varepsilon \|\partial_x u\|_{L^2(\mathbb{R}_+^*)} \|\partial_x w\|_{L^2(\mathbb{R}_+^*)} + M_v \|\tilde{\Gamma}\|_{L^\infty(0,T)} \|u\|_{L^2(\mathbb{R}_+^*)} \|\partial_x w\|_{L^2(\mathbb{R}_+^*)} \\ &\le (\varepsilon + M_v \|\tilde{\Gamma}\|_{L^\infty(0,T)}) \|u\|_{H^1(\mathbb{R}_+^*)} \|w\|_{H^1(\mathbb{R}_+^*)}. \end{aligned}$$

Hence (37) is satisfied. Further, we proceed to derive a lower bound. For almost every $t \in (0, T)$, we can estimate

$$b(t; u, u) \ge \varepsilon \|\partial_x u\|_{L^2(\mathbb{R}_+^*)}^2 - M_v \|\tilde{\Gamma}\|_{L^\infty(0,T)} \|u\|_{L^2(\mathbb{R}_+^*)} \|\partial_x u\|_{L^2(\mathbb{R}_+^*)}.$$

Completing the square then leads to

$$\begin{aligned} & \varepsilon \|\partial_x u\|_{L^2(\mathbb{R}_+^*)}^2 - \|\tilde{\Gamma}\|_{L^\infty(0,T)} M_v \|u\|_{L^2(\mathbb{R}_+^*)} \|\partial_x u\|_{L^2(\mathbb{R}_+^*)} \\ & = \frac{\varepsilon}{2} \|\partial_x u\|_{L^2(\mathbb{R}_+^*)}^2 + \left( \sqrt{\frac{\varepsilon}{2}} \|\partial_x u\|_{L^2(\mathbb{R}_+^*)} - \frac{\|\tilde{\Gamma}\|_{L^\infty(0,T)} M_v}{\sqrt{2\varepsilon}} \|u\|_{L^2(\mathbb{R}_+^*)} \right)^2 - \frac{M_v^2 \|\tilde{\Gamma}\|_{L^\infty(0,T)}^2}{2\varepsilon} \|u\|_{L^2(\mathbb{R}_+^*)}^2, \end{aligned}$$

from which we ultimately conclude that

$$\begin{aligned} b(t; u, u) &\ge \frac{\varepsilon}{2} \|\partial_x u\|_{L^2(\mathbb{R}_+^*)}^2 - \frac{M_v^2 \|\tilde{\Gamma}\|_{L^\infty(0,T)}^2}{2\varepsilon} \|u\|_{L^2(\mathbb{R}_+^*)}^2 \\ &= \frac{\varepsilon}{2} \|u\|_{H^1(\mathbb{R}_+^*)}^2 - \left( \frac{M_v^2 \|\tilde{\Gamma}\|_{L^\infty(0,T)}^2}{2\varepsilon} - \frac{\varepsilon}{2} \right) \|u\|_{L^2(\mathbb{R}_+^*)}^2. \end{aligned}$$

Thus (38) is satisfied for

$$d_2 = \frac{\varepsilon}{2}, \quad d_3 = \max \left( \frac{M_v^2 \|\tilde{\Gamma}\|_{L^\infty(0,T)}^2}{2\varepsilon} - \frac{\varepsilon}{2}, 0 \right).$$

We now turn to the functional $g$. For each $u \in H^1(\mathbb{R}_+^*)$, we denote by $\tilde{u}$ an extension of $u$ to $\mathbb{R}$, that is, a function $\tilde{u} \in H^1(\mathbb{R})$ such that $\tilde{u}|_{\mathbb{R}_+^*} = u$ and

$$\|\tilde{u}\|_{H^1(\mathbb{R})} \le \tilde{c}_1 \|u\|_{H^1(\mathbb{R}_+^*)},$$

where the constant $\tilde{c}_1$ does not depend on $u$, see [4, p. 209]. We then know that $\tilde{u} \in C^{0, \frac{1}{2}}(\mathbb{R})$, and moreover,

$$\|\tilde{u}\|_{L^\infty(\mathbb{R})} \le \tilde{c}_2 \|\tilde{u}\|_{H^1(\mathbb{R})},$$

where the constant $\tilde{c}_2$ does not depend on $\tilde{u}$, as shown again in [4, p. 282]. It implies that $u$ itself also belongs to $L^\infty$, and that

$$\|u\|_{L^\infty(\mathbb{R}_+^*)} = \left\| \tilde{u}|_{\mathbb{R}_+^*} \right\|_{L^\infty(\mathbb{R}_+^*)} \le \|\tilde{u}\|_{L^\infty(\mathbb{R})} \le \tilde{c}_2 \|\tilde{u}\|_{H^1(\mathbb{R})} \le \tilde{c}_2 \tilde{c}_1 \|u\|_{H^1(\mathbb{R}_+^*)}.$$
Similarly, $u \in C^{0, \frac{1}{2}}(\mathbb{R}_+^*)$ and thus, as a uniformly continuous function, can be prolonged continuously to $\mathbb{R}_+$. Consequently, we can write

$$|u(0)| = \left| \lim_{x \to 0^+} u(x) \right| = \lim_{x \to 0^+} |u(x)| \le \tilde{c}_2 \tilde{c}_1 \|u\|_{H^1(\mathbb{R}_+^*)}. \quad (39)$$

It follows that, for almost every $t \in (0, T)$, the linear operator $g(t)$ is bounded, since

$$\|g\|_{H^1(\mathbb{R}_+^*)}' = \sup_{\|u\|_{H^1(\mathbb{R}_+^*)} \le 1} \left|f(\tilde{\Gamma}(t))u(0)\right| \le \sup_{\|u\|_{H^1(\mathbb{R}_+^*)} \le 1} \tilde{c}_2 \tilde{c}_1 L_f |\tilde{\Gamma}(t)| \|u\|_{H^1(\mathbb{R}_+^*)} \le \tilde{c}_2 \tilde{c}_1 L_f \|\tilde{\Gamma}\|_{L^\infty(0,T)}.$$

Thus we can write

$$\|g\|_{L^2(0,T;H^1(\mathbb{R}_+^*))'} \le \sqrt{T} \tilde{c}_2 \tilde{c}_1 L_f \|\tilde{\Gamma}\|_{L^\infty(0,T)}$$

and hypothesis (2) is satisfied. All of the assumptions of Theorem 2 are thus fulfilled and we can conclude the existence and uniqueness of solution $a \in L^2(0, T; H^1(\mathbb{R}_+^*)) \cap C([0, T]; L^2(\mathbb{R}_+^*))$, with $\partial_t a \in L^2(0, T; H^1(\mathbb{R}_+^*))'$ to problem (15) with initial condition (7).

## 4.2 Estimates on the solution to the linear problem

Various properties of solution $a$, including nonnegativity and norm estimates, are based on the following proposition. Some proofs are close to the similar ones which can be found in [2].

**Proposition 2.** Let $\varepsilon_0 > 0$ and $\varepsilon \in (0, \varepsilon_0)$ fixed. Consider $g := g(t, x)$ compactly supported in $x$ such that

$$g \in L^2(0, T; L^2(\mathbb{R}_+^*)) , \quad \text{supp } g(t, \cdot) \subset [0, x_R], \quad \text{a.e. } t \in (0, T), \quad (40)$$

and $h := h(t)$ such that

$$h \in L^1(0, T). \quad (41)$$

Introduce $\bar{v} := \bar{v}(t, x)$ such that

$$\bar{v} \in L^{\infty}(0, T; W^{1, \infty}(\mathbb{R}_+^*)), \quad \bar{v}(t, 0) \ge 0 \quad \text{a.e. } t \in (0, T), \quad (42)$$

Then define the time-dependent first-order differentiable operator

$$\mathcal{L}(t; \sigma) = \varepsilon \partial_x \sigma - \bar{v}\sigma, \quad (43)$$

for any $\sigma := \sigma(t, x)$ such that (43) makes sense. Eventually, let $u \in L^2(0, T; H^1(\mathbb{R}_+^*))$ with $\partial_t u \in L^2(0, T; H^1(\mathbb{R}_+^*))'$, satisfying, for all $w \in H^1(\mathbb{R}_+^*)$ and almost every $t \in (0, T)$,

$$\frac{d}{dt} \int_{\mathbb{R}_+} u(t, x)w(x)dx + \int_{\mathbb{R}_+} \mathcal{L}(t; u)(t, x)\partial_x w(x)dx = h(t)w(0) + \int_{\mathbb{R}_+} g(t, x)w(x)dx, \quad (44)$$

supplemented with the initial condition

$$u(0, x) = u^0(x), \quad \text{a.e. } x \ge 0,$$

where

$$u^0 \in L^2(\mathbb{R}_+^*), \quad \text{supp}(u^0) \subseteq [0, x_R]. \quad (45)$$

Then $u$ satisfies the following estimates for almost every $t \in (0, T)$

$$\|u(t, \cdot)\|_{L^1(\mathbb{R}_+^*)} \le \|u^0\|_{L^1(\mathbb{R}_+^*)} + \int_0^t (|h(s)| + \|g(s, \cdot)\|_{L^1(\mathbb{R}_+^*)}) \, ds, \quad (46)$$

$$\int_r^{+\infty} |u(t, y)| dy \le \frac{\varepsilon c}{\sqrt{r}} \int_0^t \|\partial_x u(s, \cdot)\|_{L^2} ds, \quad \forall r \ge 2x_R, \quad (47)$$

$$\int_r^{+\infty} y|u(t, y)|dy \le \frac{\varepsilon c}{r} \int_0^t \|u(s, \cdot)\|_{L^1(\mathbb{R}_+^*)} ds, \quad \forall r \ge 2x_R, \quad (48)$$
where $c > 0$ is an absolute constant.
Moreover, if

$$ (\bar{v}(t, 0) = 0 \Rightarrow h(t) = 0) \quad \text{a.e. } t \in (0, T), \qquad (49) $$

then, upon setting for almost every $t \in (0, T)$

$$ \rho(t) = \begin{cases} \frac{h(t)^2}{\bar{v}(t, 0)} & \text{if } \bar{v}(t, 0) > 0, \\ 0 & \text{if } \bar{v}(t, 0) = 0, \end{cases} $$

u satisfies the following estimates for almost every $t \in (0, T)$

$$ \|u(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 \le B \left[ \|u^0\|_{L^2(\mathbb{R}_+^*)}^2 + \int_0^t \left( \rho(s) + \|g(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} \right) ds \right], \qquad (50) $$

$$ \int_0^t \|\partial_x u(s, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 ds \le \frac{1}{\varepsilon} \left( \frac{1}{2} \|u^0\|_{L^2(\mathbb{R}_+^*)}^2 + \int_0^t \rho(s) ds + B \int_0^t \|u(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} ds \right), \qquad (51) $$

where $B \ge 0$ is a constant which only depends on $\|\bar{v}\|_{L^\infty(0,T;W^{1,\infty}(\mathbb{R}_+^*))}$ and $\|g\|_{L^2(0,T;L^2(\mathbb{R}_+^*))}$. Finally, if $u^0$, $h$ and $g$ are nonnegative, then $u$ is also nonnegative.

The proof of Proposition 2 is provided in Subsection 4.3. Let us apply Proposition 2 to our situation. Equation (15) can be seen as a particular case of (44) by setting

$$ \bar{v}(t, x) = \tilde{\Gamma}_+(t)v_G(x) - \tilde{\Gamma}_-(t)v_L(x), \text{ a.e. } t, x, \quad h = f \circ \tilde{\Gamma}, \quad g = 0. $$

We begin by verifying that $h = f \circ \tilde{\Gamma}$ belongs to $L^1(0, T)$. In fact, $h \in L^\infty(0, T)$, since, thanks to (25), we have, for almost every $t \in (0, T)$,

$$ |f \circ \tilde{\Gamma}(t)| \le f \circ \tilde{\Gamma}_+(t) \le \underbrace{f(0)+L_f|\tilde{\Gamma}_+(t)|}_{=0} \le L_f \|\tilde{\Gamma}\|_{L^\infty(0,T)}. \qquad (52) $$

Further, Assumptions (23)–(24) on $v_G$ and $v_L$, together with Assumption (34) on $\tilde{\Gamma}$, ensure that $\bar{v} \in L^\infty(0, T; W^{1,\infty}(\mathbb{R}_+^*))$. Moreover, for any $t \in (0, T)$, we have

$$ \bar{v}(t, 0) = \tilde{\Gamma}_+(t)v_G(0) \ge 0, $$

so that (42) is satisfied. Assume now that $\bar{v}(t, 0) = 0$ for some $t \in (0, T)$. Then the condition $v_G(0) > 0$ implies $\tilde{\Gamma}(t) \le 0$. Thus, $h(t) = f(\tilde{\Gamma}(t)) = 0$ by Assumption (25), and (49) is satisfied. Last, the hypothesis on the initial datum are automatically fulfilled under Assumption (22). Therefore, Proposition 2 can be applied to directly infer the collection of estimates (50)–(48). First, for almost every $t \in (0, T)$, we obtain

$$ \|a(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 \le B \left( \|a^0\|_{L^2(\mathbb{R}_+^*)}^2 + \int_0^t \frac{f(\tilde{\Gamma}_+(s))^2}{2\tilde{\Gamma}_+(s)v_G(0)} ds \right) \le B \left( \|a^0\|_{L^2(\mathbb{R}_+^*)}^2 + \frac{L_f}{2v_G(0)} \int_0^t \tilde{\Gamma}_+(s) ds \right). $$

Similarly, for almost every $t \in (0, T)$, it holds that

$$ \int_0^t \|\partial_x a(s, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 ds \le \frac{1}{\varepsilon} \left( \frac{1}{2} \|a^0\|_{L^2(\mathbb{R}_+^*)}^2 + \frac{L_f}{2v_G(0)} \int_0^t \tilde{\Gamma}_+(s) ds + B \int_0^t \|a(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} ds \right). $$

We are now in a position to bound the norm of $\partial_t a$ in $L^2(0, T; H^1(\mathbb{R}_+^*)')$. For this purpose, let $w \in H^1(\mathbb{R}_+^*)'$. Equation (15) gives, for almost every $t \in (0, T)$,

$$ \langle \partial_t a(t, \cdot), w \rangle_{H^1(\mathbb{R}_+^*)', H^1(\mathbb{R}_+^*)} = \int_{\mathbb{R}_+} \left[ (\tilde{\Gamma}_+(t)v_G(x) - \tilde{\Gamma}_-(t)v_L(x)) a(t, x) - \varepsilon \partial_x a(t, x) \right] \partial_x w(x) dx + f(\tilde{\Gamma}(t)) w(0). \qquad (53) $$
Setting $M_v = \|v_G\|_{L^\infty(\mathbb{R}_+^*)} + \|v_L\|_{L^\infty(\mathbb{R}_+^*)}$, we bound the previous integral term as

$$ \begin{aligned} & \int_{\mathbb{R}_+} \left[ (\tilde{\Gamma}_+(t)v_G(x) - \tilde{\Gamma}_-(t)v_L(x)) a(t,x) - \varepsilon \partial_x a(t,x) \right] \partial_x w(x) dx \\ & \le \left( \|\tilde{\Gamma}\|_{L^\infty(0,T)} M_v \|a(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} + \varepsilon \|\partial_x a(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} \right) \|\partial_x w\|_{L^2(\mathbb{R}_+^*)} \\ & \le (M_v \|\tilde{\Gamma}\|_{L^\infty(0,T)} + \varepsilon) \|a(t, \cdot)\|_{H^1(\mathbb{R}_+^*)} \|w\|_{H^1(\mathbb{R}_+^*)}. \end{aligned} \quad (54) $$

To estimate the second term, we recall that every $w \in H^1(\mathbb{R}_+^*)$ is actually a uniformly continuous function and can be extended to $x = 0$ in such a way that

$$ |w(0)| \le \tilde{c} \|w\|_{H^1(\mathbb{R}_+^*)}, $$

where $\tilde{c} > 0$ is an absolute constant. Thus, using (52), we get

$$ f(\tilde{\Gamma}(t)) w(0) \le \tilde{c} L_f |\tilde{\Gamma}_+(t)| \|w\|_{H^1(\mathbb{R}_+^*)} \le \tilde{c} L_f \|\tilde{\Gamma}\|_{L^\infty(0,T)} \|w\|_{H^1(\mathbb{R}_+^*)}. \quad (55) $$

Thus, for almost every $t \in (0, T)$, from (53)–(55), we get

$$ \|\partial_t a(t, \cdot)\|_{H^1(\mathbb{R}_+^*)}' \le (M_v \|\tilde{\Gamma}\|_{L^\infty(0,T)} + \varepsilon) \|a(t, \cdot)\|_{H^1(\mathbb{R}_+^*)} + C_0 L_f \|\tilde{\Gamma}\|_{L^\infty(0,T)} $$

and subsequently, by the triangle inequality,

$$ \|\partial_t a\|_{L^2(0,T;H^1(\mathbb{R}_+^*))}' \le (M_v \|\tilde{\Gamma}\|_{L^\infty(0,T)} + \varepsilon) \|a\|_{L^2(0,T;H^1(\mathbb{R}_+^*))} + C_0 L_f \sqrt{T} \|\tilde{\Gamma}\|_{L^\infty(0,T)}. $$

Further, Proposition 2 yields for almost every $t \in (0, T)$

$$ \|a(t, \cdot)\|_{L^1(\mathbb{R}_+^*)} \le \|a^0\|_{L^1(\mathbb{R}_+^*)} + \int_0^t f(\tilde{\Gamma}(s)) ds \le \|a^0\|_{L^1(\mathbb{R}_+^*)} + L_f \int_0^t \tilde{\Gamma}_+(s) ds. $$

As for (20), those bounds are a straightforward consequence of the aforementioned ones. Finally, the nonnegativity of $a$ follows from the nonnegativity of the provided data. It remains to establish the pointwise upper-bound for $a$. For $t \in (0, T)$, we define the almost-everywhere differentiable function

$$ M(t) = \left( \|a^0\|_{L^\infty(\mathbb{R}_+^*)} + \frac{L_f}{v_G(0)} \right) \exp \left( M_v \int_0^t |\tilde{\Gamma}(s)| ds \right), $$

and claim that, for almost every $t \in (0, T)$ and $x > 0$, we have $a(t, x) \le M(t)$. We observe that it suffices to apply Proposition 2 once more, this time to the function

$$ \bar{a}(t, x) := M(t) - a(t, x). $$

A direct computation shows that, for all $w \in H^1(\mathbb{R}_+^*)$ and almost every $t \in (0, T)$, it holds that

$$ \begin{aligned} & \frac{d}{dt} \int_{\mathbb{R}_+} M(t) w(x) dx - M(t) \int_{\mathbb{R}_+} (\tilde{\Gamma}_+(t) v_G(x) - \tilde{\Gamma}_-(t) v_L(x)) \partial_x w(x) dx \\ & = M'(t) \int_{\mathbb{R}_+} w(x) dx + M(t) \tilde{\Gamma}_+(t) v_G(0) w(0) + M(t) \int_{\mathbb{R}_+} \partial_x v(\tilde{\Gamma}(t), x) w(x) dx. \end{aligned} $$

Subtracting (15) from the above equality, we are led to

$$ \begin{aligned} & \frac{d}{dt} \int_{\mathbb{R}_+} \bar{a}(t, x) w(x) dx + \varepsilon \int_{\mathbb{R}_+} \partial_x \bar{a}(t, x) \partial_x w(x) dx - \int_{\mathbb{R}_+} (\tilde{\Gamma}_+(t) v_G(x) - \tilde{\Gamma}_-(t) v_L(x)) \bar{a}(t, x) \partial_x w(x) dx \\ & = \int_{\mathbb{R}_+} [M'(t) + M(t) (\tilde{\Gamma}_+(t) v'_G(x) - \tilde{\Gamma}_-(t) v'_L(x))] w(x) dx + [M(t) \tilde{\Gamma}_+(t) v_G(0) - f(\tilde{\Gamma}(t))] w(0), \end{aligned} $$
again for all $w \in H^1(\mathbb{R}_+^*)$ and almost every $t \in (0, T)$. Thus, we apply Proposition 2 with

$$ \begin{aligned} \bar{v}(t,x) &= \tilde{\Gamma}_+(t)v_G(x) - \tilde{\Gamma}_-(t)v_L(x), \quad h(t) = M(t)\tilde{\Gamma}_+(t)v_G(0) - f(\tilde{\Gamma}_+(t)), \\ g(t,x) &= M'(t) + M(t)(\tilde{\Gamma}_+(t)v'_G(x) - \tilde{\Gamma}_-(t)v'_L(x)). \end{aligned} $$

The hypotheses regarding $\bar{v}$ were already verified. These concerning $g$ in (40) and $h$ in (41) are straightforward to check. In order to ensure nonnegativity, we still need functions $\bar{a}(0, \cdot)$, $h$ and $g$ to be nonnegative themselves. As for the initial datum, we easily derive

$$ \bar{a}(0, x) = M(0) - a^0(x) = \frac{L_f}{v_G(0)} + \|a^0\|_{L^\infty(\mathbb{R}_+^*)} - a^0(x) \geq 0, \quad \text{a.e. } x > 0. $$

Further, remembering that $f$ is Lipschitz continuous and such that $f(\gamma) = 0$ if $\gamma \le 0$, we can bound $h$ from below as follows, for almost every $t \in (0, T)$,

$$ \begin{aligned} h(t) &= \left( \|a^0\|_{L^\infty(\mathbb{R}_+^*)} + \frac{L_f}{v_G(0)} \right) \exp \left( M_v \int_0^t |\tilde{\Gamma}(s)| ds \right) \tilde{\Gamma}_+(t) v_G(0) - f(\tilde{\Gamma}_+(t)) \\ &\ge \left( \|a^0\|_{L^\infty(\mathbb{R}_+^*)} + \frac{L_f}{v_G(0)} \right) \tilde{\Gamma}_+(t) v_G(0) - L_f \tilde{\Gamma}_+(t) \ge 0. \end{aligned} $$

At last, for almost every $t \in (0, T)$ and $x > 0$, it immediately follows that

$$ g(t, x) = M'(t) + M(t) \partial_x v(\tilde{\Gamma}(t), x) = M(t) \left( M_v |\tilde{\Gamma}(t)| + \tilde{\Gamma}_+(t) v'_G(x) - \tilde{\Gamma}_-(t) v'_L(x) \right) \ge 0. $$

We can therefore apply Proposition 2 and, observing that $M$ is non-decreasing function, conclude that

$$ a(t, x) \le M(T), \quad \text{a.e. } t \in (0, T), x > 0. $$

## 4.3 Proof of Proposition 2

The proof of Proposition 2 heavily relies on multiple applications of the differential form of Gronwall's lemma. We start by introducing a sequence of smooth approximations for $u$. In light of [27, p. 407], there exists a sequence $(u_n)_{n \ge 1} \subset C^1(0, T; H^1(\mathbb{R}_+^*))$, satisfying, without loss of generality, for any $n \ge 1$ and for almost every $t \in (0, T)$,

$$ \|u(t, \cdot) - u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)} \le \frac{1}{n}, \quad \|\partial_t u(t, \cdot) - \partial_t u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)'} \le \frac{1}{n}. \quad (56) $$

The following lemma states a fundamental variational inequality satisfied by any $u_n$, $n \ge 1$.

**Lemma 3.** Let $n \ge 1$. Let $u_n \in C^1(0, T; H^1(\mathbb{R}_+^*))$ such that (56) is satisfied. Then, for almost every $t \in (0, T)$ and any $w \in H^1(\mathbb{R}_+^*)$, the following estimate holds

$$ \begin{aligned} & \int_{\mathbb{R}_+} \partial_t u_n(t, x) w(x) dx + \int_{\mathbb{R}_+} \mathcal{L}(t; u_n)(t, x) \partial_x w(x) dx \\ & \le h(t) w(0) + \int_{\mathbb{R}_+} g(t, x) w(x) dx + \frac{1}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^\infty(0, T; W^{1, \infty}(\mathbb{R}_+^*))} \right) \|w\|_{H^1(\mathbb{R}_+^*)}. \end{aligned} \quad (57) $$

*Proof.* Let $w \in H^1(\mathbb{R}_+^*)$. From (44), and thanks to the linearity of $\mathcal{L}$, we can write, for almost every $t \in (0, T)$ and any $n \ge 1$,

$$ \begin{aligned} & \frac{d}{dt} \int_{\mathbb{R}_+} u_n(t, x) w(x) dx + \int_{\mathbb{R}_+} \mathcal{L}(t; u_n)(t, x) \partial_x w(x) dx - h(t) w(0) - \int_{\mathbb{R}_+} g(t, x) w(x) dx \\ & = \frac{d}{dt} \int_{\mathbb{R}_+} (u_n - u)(t, x) w(x) dx - \int_{\mathbb{R}_+} \mathcal{L}(t; u_n - u)(t, x) \partial_x w(x) dx. \end{aligned} $$
First, we have

$$ \frac{d}{dt} \int_{\mathbb{R}_+} (u(t,x) - u_n(t,x))w(x)dx = \langle \partial_t(u_n - u)(t, \cdot), w \rangle_{H^1(\mathbb{R}_+^*)', H^1(\mathbb{R}_+^*)} \le \frac{1}{n} \|w\|_{H^1(\mathbb{R}_+^*)}. $$

Further, using the definition (43) of $\mathcal{L}$ and the Cauchy-Schwarz inequality, we get

$$ \int_{\mathbb{R}_+} \mathcal{L}(t; u-u_n)(t,x) \partial_x w(x) dx = \varepsilon \int_{\mathbb{R}_+} \partial_x (u-u_n)(t,x) \partial_x w(x) dx + \int_{\mathbb{R}_+} \bar{v}(t,x) (u-u_n)(t,x) \partial_x w(x) dx \\ \le \left( \varepsilon \|\partial_x(u_n-u)(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} + \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \|(u_n-u)(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} \right) \|\partial_x w\|_{L^2(\mathbb{R}_+^*)}, $$

which show thanks to (56), that

$$ \int_{\mathbb{R}_+} \mathcal{L}(t; u - u_n)(t, x) \partial_x w(x) dx \le \frac{1}{n} \left( \varepsilon_0 + \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \right) \|w\|_{H^1(\mathbb{R}_+^*)} $$

Thus, (57) immediately follows, which ends the proof.

<figure><img src="image_2.png" /></figure>

### 4.3.1 Nonnegativity

Let us focus on the nonnegativity of $u$ when assuming that $u^0$, $g$ and $h$ are nonnegative. We intend to prove that $\|u_-(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} = 0$ for almost every $t \in (0, T)$. Noticing that it is indeed the case for $t = 0$, it is enough to show that this expression is non-increasing over time. In order to proceed rigorously, we need to introduce the auxiliary functions $\chi_\delta$, $\delta > 0$,

$$ \chi_\delta(z) = \begin{cases} 0 & \text{if } z \ge 0, \\ \sqrt{z^2 + \delta^2} - \delta & \text{if } z < 0. \end{cases} \qquad (58) $$

Functions $\chi_\delta$ form a decreasing sequence of $\mathcal{C}^\infty$ functions that converges pointwise to the negative part function, $z \mapsto z_-$, when $\delta$ vanishes. In particular,

$$ \chi'_{\delta}(z) = \begin{cases} 0 & \text{if } z \ge 0, \\ \frac{z}{\sqrt{z^2 + \delta^2}} & \text{if } z < 0. \end{cases} $$

Using the chain rule, we compute, for every $t \in (0, T)$,

$$ \frac{1}{2} \frac{d}{dt} \int_{\mathbb{R}_+} \chi_\delta(u_n(t,x))^2 dx = \int_{\mathbb{R}_+} \partial_t u_n(t,x) \chi_\delta \circ u_n(t,x) \chi'_\delta \circ u_n(t,x) dx. \qquad (59) $$

As explained in [4, pp. 215-216], from the fact that $\chi_\delta(0) = \chi'_\delta(0) = 0$, it follows that both $\chi_\delta \circ u_n(t, \cdot)$ and $\chi'_\delta \circ u_n(t, \cdot)$ belong to $H^1(\mathbb{R}_+^*)$, and subsequently their product $\chi_\delta \circ u_n(t, \cdot) \chi'_\delta \circ u_n(t, \cdot)$ too. Thus we fix $t \in (0, T)$, which is, without loss of generality, a point where (57) holds. Lemma 3 applied with $w = \chi_\delta \circ u_n(t, \cdot) \chi'_\delta \circ u_n(t, \cdot)$, gives then

$$ \int_{\mathbb{R}_+} \partial_t u_n(t, x) \chi_\delta(u_n)(t, x) \chi'_\delta(u_n)(t, x) dx \le - \int_{\mathbb{R}_+} \mathcal{L}(t; u_n)(t, x) \partial_x (\chi_\delta(u_n)(t, x) \chi'_\delta(u_n)(t, x)) dx \\ + h(t) \chi_\delta(u_n)(t, 0) \chi'_\delta(u_n)(t, 0) + \int_{\mathbb{R}_+} g(t, x) \chi_\delta(u_n)(t, x) \chi'_\delta(u_n)(t, x) dx \\ + \frac{1}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \right) \|\chi_\delta \circ u_n(t, \cdot) \chi'_\delta \circ u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)}. \quad (60) $$

Besides, we notice that $\chi_\delta \chi'_\delta$, which is given by

$$ \chi_\delta(z) \chi'_\delta(z) = \begin{cases} 0 & \text{if } z \ge 0, \\ z \left(1 - \frac{\delta}{\sqrt{z^2 + \delta^2}}\right) & \text{if } z < 0, \end{cases} $$
is a non-positive function. Hence, it follows from our assumptions that,

$$h(t)\chi_{\delta}\chi'_{\delta}(u_n)(t, 0) \le 0, \qquad \int_{\mathbb{R}_+} g(t, x)\chi_{\delta}\chi'_{\delta}(u_n)(t, x)dx \le 0. \tag{61}$$

Further, since

$$(\chi_{\delta}\chi'_{\delta})'(z) = \begin{cases} 0 & \text{if } z \ge 0, \\ 1 - \frac{\delta^3}{(z^2+\delta^2)^{3/2}} & \text{if } z < 0, \end{cases}$$

we find that

$$\partial_x (\chi_{\delta}\chi'_{\delta}(u_n)) = \left(1 - \frac{\delta^3}{(u_n^2 + \delta^2)^{3/2}}\right) \partial_x u_n.$$

Setting $D_t = \{x \in \mathbb{R}_+: u_n(t, x) < 0\}$, which is an open subset of $\mathbb{R}_+$, we thus have, by definition of $\mathcal{L}$,

$$\begin{aligned} & -\int_{\mathbb{R}_+} \partial_x (\chi_{\delta}(u_n)\chi'_{\delta}(u_n)) \mathcal{L}(t; u_n)dx \\ &= -\varepsilon \int_{D_t} \left(1 - \frac{\delta^3}{(u_n^2 + \delta^2)^{3/2}}\right) (\partial_x u_n)^2 dx + \int_{D_t} \bar{v} \left(1 - \frac{\delta^3}{(u_n^2 + \delta^2)^{3/2}}\right) \partial_x u_n u_n dx := -\varepsilon I_1 + I_2. \end{aligned} \tag{62}$$

The first integral $I_1$ is straightforwardly nonnegative. Regarding the second one, $I_2$, we note

$$\partial_x u_n u_n = \frac{1}{2} \partial_x (u_n^2), \quad -\frac{\partial_x u_n u_n}{(u_n^2 + \delta^2)^{\frac{3}{2}}} = \partial_x \left( (u_n^2 + \delta^2)^{-\frac{1}{2}} \right),$$

so that

$$I_2 = \int_{D_t} \bar{v} \partial_x \left( \frac{u_n^2}{2} + \frac{\delta^3}{\sqrt{u_n^2 + \delta^2}} \right) dx.$$

As an open set of $\mathbb{R}_+$, $D_t$ is a countable (at most) union of disjoint open intervals of $\mathbb{R}_+$, more precisely, there exist two real sequences $(a_k)_{k \in \mathbb{N}}$, $(b_k)_{k \in \mathbb{N}}$ such that $a_0 \ge 0$ and

$$a_k < b_k \le a_{k+1} < b_{k+1}, \quad k \in \mathbb{N}.$$

Hence, we have

$$D_t = \bigcup_{k \in \mathbb{N}} (a_k, b_k) \text{ or } D_t = [0, b_0) \cup \bigcup_{k \in \mathbb{N}^*} (a_k, b_k).$$

In any case, $I_2$ writes

$$\begin{aligned} I_2 &= \sum_{k=0}^{+\infty} \left( \bar{v}(t, b_k) \left[ \frac{u_n(t, b_k)^2}{2} + \frac{\delta^3}{\sqrt{u_n(t, b_k)^2 + \delta^2}} \right] - \bar{v}(t, a_k) \left[ \frac{u_n(t, a_k)^2}{2} + \frac{\delta^3}{\sqrt{u_n(t, a_k)^2 + \delta^2}} \right] \right) \\ &\quad - \int_{D_t} \partial_x \bar{v} \left( \frac{u_n^2}{2} + \frac{\delta^3}{\sqrt{u_n^2 + \delta^2}} \right) dx \\ &:= I_{2,1} - I_{2,2}. \end{aligned}$$

It is clear, by continuity of $u_n(t, \cdot)$, that

$$u_n(t, b_k) = 0, \quad \forall k \in \mathbb{N}, \qquad u_n(t, a_k) = 0, \quad \forall k \ge 1,$$
and if $a_0 > 0$, we also have $u_n(t, a_0) = 0$. In that case, since $\bar{v}(t, \cdot)$ is Lipschitz continuous and compactly supported on $[0, x_R]$, $I_{2,1}$ satisfies

$$I_{2,1} = \delta^2 \sum_{\substack{k=0 \\ b_k \le x_R}}^{+\infty} (\bar{v}(t, b_k) - \bar{v}(t, a_k)) \le \delta^2 x_R \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \quad (63)$$

When $a_0 = 0$, the corresponding term satisfies

$$-\bar{v}(t, 0) \left[ \frac{u_n(t, 0)^2}{2} + \frac{\delta^3}{\sqrt{u_n(t, 0)^2 + \delta^2}} \right] \le 0,$$

which allows to recover the same upper bound of $I_{2,1}$ as (63).

Eventually, $I_{2,2}$ can be written as

$$I_{2,2} = \frac{1}{2} \int_{\mathbb{R}_+} \partial_x \bar{v} \chi_\delta(u_n)^2 dx + \delta I_{2,2,1}.$$

Since

$$\frac{1}{2} \int_{\mathbb{R}_+} \partial_x \bar{v} \chi_\delta(u_n)^2 dx = \int_{D_t} \partial_x \bar{v} \left( \frac{u_n^2}{2} - \delta \sqrt{u_n^2 + \delta^2} + \delta^2 \right) dx,$$

we get

$$I_{2,2,1} = \int_{D_t} \partial_x \bar{v} \left( \sqrt{u_n^2 + \delta^2} - \delta + \frac{\delta^2}{\sqrt{u_n^2 + \delta^2}} \right) dx \le \int_{D_t} |\partial_x \bar{v}| (|u_n| + 3\delta) dx.$$

Using the Cauchy-Schwarz inequality, we obtain

$$\int_{D_t} |\partial_x \bar{v}| |u_n| dx \le \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \int_0^{x_R} |u_n| dx \le \sqrt{x_R} \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \|u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}.$$

It follows directly that

$$I_{2,2,1} \le \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty})} \left( \sqrt{x_R} \|u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} + 3\delta x_R \right).$$

Consequently, combining the estimate for the boundary and volume integrals, we get

$$I_2 = I_{2,1} - I_{2,2} \\ \le -\frac{1}{2} \int_{\mathbb{R}_+} \partial_x \bar{v} \chi_\delta(u_n)^2 dx + \delta \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \left( \delta \sqrt{x_R} \|u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} + 4\delta x_R \right). \quad (64)$$

Using Lemma 3 and (60)-(64) in (59) gives

$$\frac{1}{2} \frac{d}{dt} \int_{\mathbb{R}_+} \chi_\delta(u_n(t, x))^2 dx \le \frac{1}{2} \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \int_{\mathbb{R}_+} \chi_\delta(u_n(t, x))^2 dx \\ + \delta \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \left( \delta \sqrt{x_R} \int_0^t \|u_n(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} ds + 4t \delta x_R \right) \\ + \frac{1}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \right) \|\chi_\delta \circ u_n(t, \cdot)\chi'_\delta \circ u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)}.$$

Applying the Grönwall inequality leads to

$$\int_{\mathbb{R}_+} \chi_\delta(u_n(t, x))^2 dx \le \exp \left( \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} t \right) \int_{\mathbb{R}_+} \chi_\delta(u_n(0, x))^2 dx \\ + \delta \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \left( \delta \sqrt{x_R} \int_0^t \|u_n(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} ds + 4t \delta x_R \right) \exp \left( \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} t \right) \\ + \frac{2}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \right) \left( \int_0^t \|\chi_\delta \chi'_\delta \circ u_n(s, \cdot)\|_{H^1(\mathbb{R}_+^*)} ds \right) \exp \left( \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} t \right).$$
By the monotone convergence theorem, as $\delta$ vanishes, we find

$$ \lim_{\delta \to 0} \int_{\mathbb{R}_+} \chi_{\delta}(u_n(t, x))^2 dx = \int_{\mathbb{R}_+} \lim_{\delta \to 0} \chi_{\delta}(u_n(t, x))^2 dx = \int_{\mathbb{R}_+} (u_n)_-(t, x)^2 dx. $$

Thus, noticing

$$ \|u_n\|_{L^1(0,T;L^2(\mathbb{R}_+^*))} \le \sqrt{T} \|u_n\|_{L^2(0,T;L^2(\mathbb{R}_+^*))} \le \sqrt{T} \left( \|u\|_{L^2(0,T;H^1(\mathbb{R}_+^*))} + 1 \right), $$

we pass to the limit with respect to $\delta$ on both sides of the above inequality, obtaining

$$ \int_{\mathbb{R}_+} (u_n)_-(t, x)^2 dx \le \exp \left( 2 \|\bar{v}\|_{L^\infty(0,T;W^{1,\infty}(\mathbb{R}_+^*))} t \right) \int_{\mathbb{R}_+} (u_n)_-(0, x)^2 dx \\ + \frac{C}{n} \exp(Ct) \int_0^t \|(u_n)_-(s, \cdot)\|_{H^1(\mathbb{R}_+^*)} ds, $$

where $C > 0$ is a constant only depending on $\bar{v}$. Finally, observing as before, that

$$ \|u_n\|_{L^1(0,T;H^1(\mathbb{R}_+^*))} \le \sqrt{T} \left( \|u\|_{L^2(0,T;H^1(\mathbb{R}_+^*))} + 1 \right), $$

we let $n$ go to $+\infty$ to find that

$$ \int_{\mathbb{R}_+} u_-(t, x)^2 dx \le \exp \left( 2 \|\bar{v}\|_{L^\infty(0,T;W^{1,\infty}(\mathbb{R}_+^*))} t \right) \int_{\mathbb{R}_+} u_-(0, x)^2 dx. $$

Since we have assumed the nonnegativity of the initial datum, the right-hand side of the equation above vanishes. Consequently, we conclude that $u(t, x) \ge 0$ for almost every $t \in (0, T)$, $x \in \mathbb{R}_+$.

### 4.3.2 $L_x^1$ norm

We now seek the bound (46) for the $L^1(\mathbb{R}_+^*)$ norm of $u(t, \cdot)$, for almost every $t \in (0, T)$. We make use of the approximating sequence satisfying (56). Further, for $\delta > 0$ we introduce the smooth auxiliary functions

$$ \varphi_{\delta}(z) = \sqrt{z^2 + \delta^2} - \delta, \quad z \in \mathbb{R}. \qquad (65) $$

By the monotone convergence theorem, $(\varphi_{\delta})_{\delta>0}$ converges pointwise to the absolute value function as $\delta$ goes to 0. The objective is to bound

$$ \frac{d}{dt} \int_{\mathbb{R}_+} \varphi_{\delta}(u_n(t, x)) dx = \int_{\mathbb{R}_+} \partial_t u_n(t, x) \varphi'_{\delta} \circ u_n(t, x) dx. $$

It can be observed that

$$ \varphi_{\delta}(z) = \chi_{\delta}(z) + \chi_{\delta}(-z), \quad z \in \mathbb{R}, $$

with $\chi_{\delta}$ defined in (58). Since for every $t \in (0, T)$, $\chi'_{\delta} \circ u_n(t, \cdot) \in H^1(\mathbb{R}_+^*)$, it follows that $\varphi'_{\delta} \circ u_n(t, \cdot) \in H^1(\mathbb{R}_+^*)$ as well. Consequently, having fixed $t \in (0, T)$ from the full-measure set where (57) applies, we may use it with $w = \varphi'_{\delta} \circ u_n(t, \cdot)$ to find that

$$ \int_{\mathbb{R}_+} \partial_t u_n(t, x) \varphi'_{\delta}(u_n)(t, x) dx \le - \int_{\mathbb{R}_+} \mathcal{L}(t; u_n)(t, x) \partial_x \varphi'_{\delta}(u_n)(t, x) dx + h(t) \varphi'_{\delta}(u_n)(t, 0) \\ + \int_{\mathbb{R}_+} g(t, x) \varphi'_{\delta}(u_n)(t, x) dx + \frac{1}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^\infty(0,T;W^{1,\infty}(\mathbb{R}_+^*))} \right) \|\varphi'_{\delta} \circ u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)}. \quad (66) $$

First, since the derivative of $\varphi_{\delta}$ is given by

$$ \varphi'_{\delta}(z) = \frac{z}{\sqrt{z^2 + \delta^2}}, \quad z \in \mathbb{R}, $$
it is straightforward to get

$$h(t)\varphi'_{\delta}(u_n)(t, 0) \le |h(t)|, \quad \int_{\mathbb{R}_+} g\varphi'_{\delta}(u_n) \, dx \le \|g(t, \cdot)\|_{L^1(\mathbb{R}_+^*)}. \quad (67)$$

Further, we estimate

$$\int_{\mathbb{R}_+} \mathcal{L}(t; u_n) \partial_x \varphi'_{\delta}(u_n) \, dx = \varepsilon \int_{\mathbb{R}_+} (\partial_x u_n)^2 \varphi''_{\delta}(u_n) \, dx - \int_{\mathbb{R}_+} \bar{v} u_n \partial_x u_n \varphi''_{\delta}(u_n) \, dx := \varepsilon J_1 - J_2. \quad (68)$$

Noticing that

$$\varphi''_{\delta}(z) = \frac{\delta^2}{(z^2 + \delta^2)^{3/2}} \ge 0, \quad z \in \mathbb{R},$$

we find that $J_1$ is nonnegative. Turning to the second integral, $J_2$, we compute

$$J_2 = \int_{\mathbb{R}_+} \bar{v} u_n \partial_x u_n \varphi''_{\delta}(u_n) \, dx = \delta^2 \int_{\mathbb{R}_+} \bar{v} \frac{u_n \partial_x u_n}{(u_n^2 + \delta^2)^{3/2}} \, dx.$$

Given that $\frac{u_n \partial_x u_n}{(u_n^2 + \delta^2)^{3/2}} = \partial_x \left( -(u_n^2 + \delta^2)^{-1/2} \right)$, an integration by parts, owing to the compact support of $\bar{v}(t, \cdot)$, leads to

$$J_2 = \delta \left( \bar{v}(t, 0) \frac{\delta}{\sqrt{u_n(t, 0)^2 + \delta^2}} + \int_{\mathbb{R}_+} \partial_x \bar{v} \frac{\delta}{\sqrt{u_n^2 + \delta^2}} \, dx \right) \le \delta \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} (1+x_R). \quad (69)$$

Substituting (67)–(69) in (66) yields

$$\frac{d}{dt} \int_{\mathbb{R}_+} \varphi_{\delta}(u_n(t, x)) \, dx \le \delta \|\bar{v}\|_{L^{\infty}(0,T; W^{1,\infty}(\mathbb{R}_+^*))} (1+x_R) + |h(t)| + \|g(t, \cdot)\|_{L^1(\mathbb{R}_+^*)} \, dx \\ + \frac{1}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^{\infty}(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \right) \|\varphi'_{\delta} \circ u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)}.$$

By integrating the above inequality from 0 to $t \in (0, T)$, we obtain

$$\int_{\mathbb{R}_+} \varphi_{\delta}(u_n(t, x)) \, dx - \int_{\mathbb{R}_+} \varphi_{\delta}(u_n(0, x)) \, dxv \le \delta Ct + \int_0^t \left( |h(s)| + \|g(s, \cdot)\|_{L^1(\mathbb{R}_+^*)} \right) \, ds \\ + \frac{C}{n} \int_0^t \|\varphi'_{\delta} \circ u_n(s, \cdot)\|_{H^1(\mathbb{R}_+^*)} \, ds,$$

where the constant $C > 0$ only depends on $\bar{v}$. Since

$$\|\varphi'_{\delta} \circ u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)} \le \|u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)} \le \|u(t, \cdot)\|_{H^1(\mathbb{R}_+^*)} + 1,$$

we pass to the limit when $n$ goes to $+\infty$, which yields, for almost every $t \in (0, T)$,

$$\int_{\mathbb{R}_+} \varphi_{\delta}(u(t, x)) \, dx \le \int_{\mathbb{R}_+} \varphi_{\delta}(u(0, x)) \, dx + \delta Ct + \int_0^t \left( |h(s)| + \|g(s, \cdot)\|_{L^1(\mathbb{R}_+^*)} \right) \, ds.$$

Finally, as $\delta$ vanishes we obtain

$$\int_{\mathbb{R}_+} |u(t, x)| \, dx \le \int_{\mathbb{R}_+} |u(0, x)| \, dx + \int_0^t \left( |h(s)| + \|g(s, \cdot)\|_{L^1(\mathbb{R}_+^*)} \right) \, ds$$

and (46) is proven.
### 4.3.3 Estimate of the tail integral

Our goal here is to derive an estimate (47) for the tail contribution $\int_{r}^{\infty} |u(t, x)|dx$, valid for almost every $t \in (0, T)$. For $r$ and $R$ satisfying $2x_R < r < R$, let $\vartheta_{r,R} \in C_c^{\infty}(\mathbb{R}_+)$, such that

$$ 
\begin{array}{l}
0 \le \vartheta_{r,R} \le 1, \quad \text{supp } \vartheta_{r,R} = \left[ \frac{r}{2}, 2R \right], \\
|\vartheta'_{r,R}(x)| \le \frac{c_1}{r} \quad \text{for } x \in \left[ \frac{r}{2}, r \right], \quad \vartheta_{r,R}(x) = 1 \quad \text{for } x \in [r, R], \quad |\vartheta'_{r,R}(x)| \le \frac{c_2}{R} \quad \text{for } x \in [R, 2R],
\end{array}
$$

where $c_1, c_2 > 0$ are absolute constants. Further, we work with $\varphi_{\delta} \circ u_n$, where the sequence $(u_n)_{n \ge 1}$ satisfies (56) and $(\varphi_{\delta})_{\delta > 0}$ is defined in (65). We intend to estimate

$$ \frac{d}{dt} \int_{\mathbb{R}_+} \varphi_{\delta}(u_n)(t, x) \vartheta_{r,R}(x) dx = \int_{\mathbb{R}_+} \partial_t u_n(t, x) \varphi'_{\delta}(u_n)(t, x) \vartheta_{r,R}(x) dx, \quad (70) $$

and then integrate this estimate and pass to the limit. Since, for every $t \in (0, T)$, as explained in the subsection 4.3.2, $\varphi_{\delta} \circ u_n(t, \cdot) \in H^1(\mathbb{R}_+^*)$, the product $\varphi'_{\delta}(u_n)(t, \cdot)\vartheta_{r,R}$ belongs to $H^1(\mathbb{R}_+^*)$ too. Next, as (57) holds with $w\varphi'_{\delta}(u_n)(t, \cdot)\vartheta_{r,R}$ for almost every $t \in (0, T)$, we fix one such $t$ and apply Lemma 3, obtaining

$$ 
\begin{aligned}
\int_{\mathbb{R}_+} \partial_t u_n(t, x) \varphi'_{\delta}(u_n)(t, x) \vartheta_{r,R}(x) dx &\le - \int_{\mathbb{R}_+} \mathcal{L}(t; u_n)(t, x) \partial_x(\varphi'_{\delta}(u_n)(t, x) \vartheta_{r,R}(x)) dx \\
&\quad + h(t) \varphi'_{\delta}(u_n)(t, 0) \vartheta_{r,R}(0) + \int_{\mathbb{R}_+} g(t, x) \varphi'_{\delta}(u_n)(t, x) \vartheta_{r,R}(x) dx \\
&\quad + \frac{1}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^{\infty}(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \right) \|\varphi'_{\delta} \circ u_n(t, \cdot) \vartheta_{r,R}\|_{H^1(\mathbb{R}_+^*)}.
\end{aligned}
\quad (71) $$

As $0 \notin \text{supp } \vartheta_{r,R}$ and $\text{supp } g(t, \cdot) \cap \text{supp } \vartheta_{r,R} = \emptyset$, the second and third term on the right-hand side of (71) vanish. Similarly, by definition of operator $\mathcal{L}$, since $\text{supp } \bar{v}(t, \cdot) \cap \text{supp } \vartheta_{r,R} = \emptyset$,

$$ 
\begin{aligned}
\int_{\mathbb{R}_+} \mathcal{L}(t; u_n) \partial_x (\varphi'_{\delta}(u_n) \vartheta_{r,R}) dx &= \varepsilon \int_{\frac{r}{2}}^{2R} (\partial_x u_n)^2 \varphi''_{\delta}(u_n) \vartheta_{r,R} dx \varepsilon \int_{\frac{r}{2}}^{2R} \partial_x u_n \varphi'_{\delta}(u_n) \vartheta'_{r,R} dx \\
&:= \varepsilon J_3 + \varepsilon J_4.
\end{aligned}
\quad (72) $$

The first integral, $J_3$, is nonnegative because both $\varphi''_{\delta}$ and $\vartheta_{r,R}$ are nonnegative. Regarding $J_4$, we recall that $|\varphi'(z)| < 1$ for all $z \in \mathbb{R}$. Further, since $\vartheta_{r,R}$ is constant on $[r, R]$, and in addition, its derivative is controlled over the whole domain, we have

$$ 
\begin{aligned}
J_4 = \int_{\frac{r}{2}}^{r} \partial_x u_n \varphi'_{\delta}(u_n) \vartheta'_{r,R} dx &+ \int_{R}^{2R} \partial_x u_n \varphi'_{\delta}(u_n) \vartheta'_{r,R}(x) dx \\
&\le \frac{c_1}{r} \int_{\frac{r}{2}}^{r} |\partial_x u_n| dx + \frac{c_2}{R} \int_{R}^{2R} |\partial_x u_n| dx \\
&\le \frac{c_1}{\sqrt{2r}} \|\partial_x u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} + \frac{c_2}{\sqrt{R}} \|\partial_x u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}.
\end{aligned}
\quad (73) $$

Thus, plugging (71)–(73) in (70) leads to

$$ \frac{d}{dt} \int_{\mathbb{R}_+} \varphi_{\delta}(u_n)(t, x) \vartheta_{r,R}(x) dx \le \varepsilon \left( \frac{c_1}{\sqrt{2r}} + \frac{c_2}{\sqrt{R}} \right) \|\partial_x u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} + \frac{C}{n} \|\varphi'_{\delta} \circ u_n(t, \cdot) \vartheta_{r,R}\|_{H^1(\mathbb{R}_+^*)} $$

so that, integrating this differential inequality from 0 to $t \in (0, T)$, we derive

$$ 
\begin{aligned}
\int_{\mathbb{R}_+} \varphi_{\delta}(u_n)(t, x) \vartheta_{r,R}(x) dx &\le \int_{\mathbb{R}_+} \varphi_{\delta}(u_n)(0, x) \vartheta_{r,R}(x) dx + \varepsilon \left( \frac{c_1}{\sqrt{2r}} + \frac{c_2}{\sqrt{R}} \right) \int_0^t \|\partial_x u_n(s, \cdot)\|_{L^2} ds \\
&\quad + \frac{C}{n} \int_0^t \|\varphi'_{\delta} \circ u_n(s, \cdot) \vartheta_{r,R}\|_{H^1(\mathbb{R}_+^*)} ds,
\end{aligned}
$$
where the constant $C > 0$ is absolute. Since, for almost every $s \in (0, T)$,

$$ \|\varphi'_{\delta} \circ u_n(s, \cdot) \vartheta_{r,R}\|_{H^1(\mathbb{R}_+^*)} \le \|\varphi'_{\delta} \circ u_n(s, \cdot)\|_{H^1(\mathbb{R}_+^*)} \le \|u(s, \cdot)\|_{H^1(\mathbb{R}_+^*)} + 1, $$

we pass to the limit when $n$ tends to $+\infty$, obtaining

$$ \int_{\mathbb{R}_+} \varphi_{\delta}(u)(t, x) \vartheta_{r,R}(x) dx \le \int_{\mathbb{R}_+} \varphi_{\delta}(u)(0, x) \vartheta_{r,R}(x) dx + \varepsilon \left( \frac{c_1}{\sqrt{2r}} + \frac{c_2}{\sqrt{R}} \right) \int_0^t \| \partial_x u(s, \cdot) \|_{L^2} ds. $$

Further, letting $R$ go to $+\infty$, we derive

$$ \int_{r}^{+\infty} \varphi_{\delta}(u)(t, x) dx = \lim_{R \to +\infty} \int_{\mathbb{R}_+} \varphi_{\delta}(u)(t, x) \vartheta_{r,R}(x) dx \le \lim_{R \to +\infty} \int_{\mathbb{R}_+} \varphi_{\delta}(u)(t, x) \vartheta_{r,R}(x) dx \\ \le \varepsilon \frac{c_1}{\sqrt{2r}} \int_{0}^{t} \| \partial_x u(s, \cdot) \|_{L^2} ds. $$

Finally, as $\delta$ vanishes we obtain, for almost every $t \in (0, T)$,

$$ \int_{r}^{+\infty} |u(t, x)| dx \le \varepsilon \frac{c_1}{\sqrt{2r}} \int_{0}^{t} \| \partial_x u(s, \cdot) \|_{L^2} dx $$

and conclude that (47) holds.

### 4.3.4 Estimate of $\int_{r}^{+\infty} xu \, dx$

We now turn to the estimate (48) regarding the control of $\int_{r}^{\infty} x|u(t, x)|dx$ for almost every $t \in (0, T)$. Let $r$ and $R$ satisfying $2x_R < r < R$. Further, let $\omega_{r,R} \in C_c^{\infty}(\mathbb{R}_+)$ such that

$$ \omega_{r,R} \ge 0, \quad \supp \omega_{r,R} \subseteq [\frac{r}{2}, 2R], \\ |\omega''_{r,R}(x)| \le \frac{c_3}{r} \quad \text{for } x \in [\frac{r}{2}, r], \quad \omega_{r,R}(x) = x \quad \text{for } x \in [r, R], \quad |\omega''_{r,R}(x)| \le \frac{c_4}{R} \quad \text{for } x \in [R, 2R]. $$

where $c_3, c_4 > 0$ are absolute constants. Similarly as in previous sections, we aim to bound

$$ \frac{d}{dt} \int_{\mathbb{R}_+} \varphi_{\delta}(u_n)(t, x) \omega_{r,R}(x) dx = \int_{\mathbb{R}_+} \partial_t u_n(t, x) \varphi'_{\delta}(u_n)(t, x) \omega_{r,R}(x) dx, \quad (74) $$

with $u_n, n \ge 1$, as in (56) and $\varphi_{\delta}, \delta > 0$, defined in (65). For every $t \in (0, T)$, we have $\varphi_{\delta} \circ u_n(t, \cdot) \in H^1(\mathbb{R}_+^*)$, and thus $\varphi'_{\delta}(u_n)(t, \cdot)\omega_{r,R}$ belongs to $H^1(\mathbb{R}_+^*)$ as well. Hence, applying Lemma 3 for $w = \varphi'_{\delta}(u_n)(t, \cdot)\omega_{r,R}$ at a fixed $t \in (0, T)$ where (57) is valid leads to

$$ \int_{\mathbb{R}_+} \partial_t u_n(t, x) \varphi'_{\delta}(u_n)(t, x) \omega_{r,R}(x) dx \le - \int_{\mathbb{R}_+} \mathcal{L}(t; u_n)(t, x) \partial_x (\varphi'_{\delta}(u_n)(t, x) \omega_{r,R}(x)) dx \\ + h(t) \varphi'_{\delta}(u_n)(t, 0) \omega_{r,R}(0) + \int_{\mathbb{R}_+} g(t, x) \varphi'_{\delta}(u_n)(t, x) \omega_{r,R}(x) dx \\ + \frac{1}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^{\infty}(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \right) \|\varphi'_{\delta} \circ u_n(t, \cdot) \omega_{r,R}\|_{H^1(\mathbb{R}_+^*)}. \quad (75) $$

Since $\supp \omega_{r,R} \subseteq [x_R, 2R]$, we straightforwardly have

$$ h(t)\omega_{r,R}(0)\varphi'_{\delta}(u_n)(t, 0) = 0, \quad \int_{\mathbb{R}_+} g\varphi'_{\delta}(u_n)\omega_{r,R}dx = 0. \quad (76) $$

For the same reason, by definition (43) of $\mathcal{L}$, we write

$$ \int_{\mathbb{R}_+} \mathcal{L}(t; u_n) \partial_x (\varphi'_{\delta}(u_n) \omega_{r,R}) dx = \varepsilon \int_{\mathbb{R}_+} (\partial_x u_n)^2 \varphi''_{\delta}(u_n) \omega_{r,R} dx + \varepsilon \int_{\mathbb{R}_+} \partial_x u_n \varphi'_{\delta}(u_n) \omega'_{r,R} dx \\ := \varepsilon J_5 + \varepsilon J_6. \quad (77) $$
Since $\varphi_{\delta}''$ and $\omega_{r,R}$ are nonnegative, $J_5$ is nonnegative too. Further, we first rewrite $J_6$, then integrate by parts, to obtain

$$J_6 = \int_{\mathbb{R}_+} \partial_x (\varphi_{\delta}(u_n)) \omega'_{r,R} dx = - \int_{\mathbb{R}_+} \varphi_{\delta}(u_n) \omega''_{r,R} dx.$$

Noticing that $|\varphi_{\delta}(z)| = \sqrt{z^2 + \delta^2} - \delta \le |z|$ for every $z \in \mathbb{R}$ and $\delta > 0$, we estimate

$$J_6 \le \int_{\mathbb{R}_+} |u_n||\omega''_{r,R}|dx = \int_{\frac{r}{2}}^{r} |u_n||\omega''_{r,R}|dx + \int_{R}^{2R} |u_n||\omega''_{r,R}|dx \le \frac{c_3}{r} \int_{\frac{r}{2}}^{r} |u_n|dx + \frac{c_4}{R} \int_{R}^{2R} |u_n|dx. \quad (78)$$

Combining (74) with (75)–(78) gives

$$\frac{d}{dt} \int_{\mathbb{R}_+} \varphi_{\delta}(u_n)(t, x) \omega_{r,R}(x) dx \le \varepsilon \left( \frac{c_3}{r} + \frac{c_4}{\sqrt{R}} \right) \|u_n(t, \cdot)\|_{L^1(\mathbb{R}_+^*)} \\ + \frac{1}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^{\infty}(0,T;W^{1,\infty}(\mathbb{R}_+^*))} \right) \|\varphi'_{\delta} \circ u_n(t, \cdot) \omega_{r,R}\|_{H^1(\mathbb{R}_+^*)}.$$

Accordingly, the integration of the previous estimate with respect to time, from 0 to $t \in (0, T)$, yields

$$\int_{\mathbb{R}_+} \varphi_{\delta}(u_n)(t, x) \omega_{r,R}(x) dx \le \int_{\mathbb{R}_+} \varphi_{\delta}(u_n)(0, x) \omega_{r,R}(x) dx + \varepsilon \left( \frac{c_3}{r} + \frac{c_4}{\sqrt{R}} \right) \int_0^t \|u_n(s, \cdot)\|_{L^1(\mathbb{R}_+^*)} ds \\ + \frac{C}{n} \int_0^t \|\varphi'_{\delta} \circ u_n(s, \cdot) \omega_{r,R}\|_{H^1(\mathbb{R}_+^*)} ds,$$

where the constant $C > 0$ only depends on $\bar{v}$. Since $H^1$ is an algebra, we have

$$\|\varphi'_{\delta} \circ u_n(s, \cdot) \omega_{r,R}\|_{H^1(\mathbb{R}_+^*)} \le \|u(s, \cdot) \omega_{r,R}\|_{H^1(\mathbb{R}_+^*)} + c \|\omega_{r,R}\|_{H^1(\mathbb{R}_+^*)},$$

with an absolute constant $c > 0$. Hence, in the limits as $n$ goes to $+\infty$ and $\delta$ vanishes, we find

$$\int_{\mathbb{R}_+} |u(t, x))|\omega_{r,R}(x)dx = \lim_{\delta \to 0} \lim_{n \to \infty} \int_{\mathbb{R}_+} \varphi_{\delta}(u_n(t, x))\omega_{r,R}(x)dx \\ \le \int_{\mathbb{R}_+} |u^0(x)|\omega_{r,R}(x)dx + \varepsilon \left( \frac{c_3}{r} + \frac{c_4}{\sqrt{R}} \right) \int_0^t \|u(s, \cdot)\|_{L^1(\mathbb{R}_+^*)} ds.$$

To conclude we first observe that $\supp u^0 \cap \supp \omega_{r,R} = \emptyset$, and then let $R$ go to $+\infty$, finding

$$\int_{r}^{+\infty} x|u(t, x)|dx = \lim_{R \to +\infty} \int_{r}^{+\infty} |u(t, x)|\omega_{r,R}(x)dx \le \lim_{R \to +\infty} \int_{\mathbb{R}_+} |u(t, x)|\omega_{r,R}(x)dx \\ \le \frac{\varepsilon c_3}{r} \int_{0}^{t} \|u(s, \cdot)\|_{L^1(\mathbb{R}_+^*)} ds.$$

Thus (48) is established.

### 4.3.5 $L_t^2(H_x^1)$ norm

We now address the estimates (50)–(51). We first establish a pointwise-in-time estimate for the $L^2$ norm of $u(t, \cdot)$, and then, we use this result to bound $\int_0^t \|\partial_x u(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} ds$ for almost every $t \in (0, T)$. As we intend to apply the Grönwall lemma, we want to upper-bound

$$\frac{1}{2} \frac{d}{dt} \int_{\mathbb{R}_+} u_n(t, x)^2 dx = \int_{\mathbb{R}_+} \partial_t u_n(t, x) u_n(t, x) dx. \quad (79)$$
Without loss of generality, we fix $t \in (0, T)$ to be a point where (57) holds. Then, applying Lemma 3 with $w = u_n(t, \cdot) \in H^1(\mathbb{R}_+^*)$ yields

$$ \int_{\mathbb{R}_+} \partial_t u_n(t, x) u_n(t, x) dx \leq - \int_{\mathbb{R}_+} \mathcal{L}(t; u_n)(t, x) \partial_x u_n(t, x) dx + h(t) u_n(t, 0) + \int_{\mathbb{R}_+} g(t, x) u_n(t, x) dx \\ + \frac{1}{n} \left( 1 + \varepsilon_0 + \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \right) \|u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)}^2. \quad (80) $$

Firstly, by definition of $\mathcal{L}$,

$$ - \int_{\mathbb{R}_+} \mathcal{L}(t; u_n) \partial_x u_n dx \leq -\varepsilon \int_{\mathbb{R}_+} (\partial_x u_n)^2 dx + \int_{\mathbb{R}_+} \bar{v} u_n \partial_x (u_n) dx. \quad (81) $$

Integrating by parts the second integral yields

$$ \int_{\mathbb{R}_+} \bar{v} u_n \partial_x (u_n) dx = \frac{1}{2} \int_{\mathbb{R}_+} \bar{v} \partial_x (u_n^2) dx = - \frac{1}{2} \bar{v}(t, 0) u_n(t, 0)^2 - \frac{1}{2} \int_{\mathbb{R}_+} \partial_x \bar{v} u_n^2 dx \\ \leq - \frac{1}{2} \bar{v}(t, 0) u_n(t, 0)^2 + \frac{1}{2} \|\bar{v}\|_{L^\infty(0,T; W^{1,\infty}(\mathbb{R}_+^*))} \|u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2. \quad (82) $$

Thus, combining (80)-(82) with (79) yields

$$ \frac{1}{2} \frac{d}{dt} \int_{\mathbb{R}_+} u_n(t, x)^2 dx + \varepsilon \int_{\mathbb{R}_+} (\partial_x u_n)^2 dx \leq u_n(t, 0) \left( h(t) - \frac{\bar{v}(t, 0)}{2} u_n(t, 0) \right) + \frac{C}{2} \|u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 \\ + \|g(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} \|u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} + \frac{C}{n} \|u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)}, \quad (83) $$

for a constant $C > 0$ only depending on $\bar{v}$. If $\bar{v}(t, 0) = 0$, then, thanks to (49), the term corresponding to $x = 0$ in (83) is zero. Otherwise, if $\bar{v}(t, 0) > 0$, we can write

$$ u_n(t, 0) \left( h(t) - \frac{\bar{v}(t, 0)}{2} u_n(t, 0) \right) \leq \frac{h(t)^2}{2\bar{v}(t, 0)}, \quad (84) $$

which follows from identifying the point of maximum of the quadratic function $q(z) = -\frac{\bar{v}(t, 0)}{2}z^2 + h(t)z$, $\bar{v}(t, 0) > 0$. In what remains in this subsection, we assume that $\bar{v}(t, 0) > 0$. The case when $\bar{v}(t, 0) = 0$ can be dealt with by simply dropping the term depending on $\frac{h(t)^2}{2\bar{v}(t, 0)}$.

Plugging (84) in (83), we find

$$ \frac{1}{2} \frac{d}{dt} \int_{\mathbb{R}_+} u_n(t, x)^2 dx + \varepsilon \int_{\mathbb{R}_+} (\partial_x u_n)^2 dx \leq \frac{C}{2} \|u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 \\ + \|g(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} \|u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} + \frac{h(t)^2}{2\bar{v}(t, 0)} + \frac{C}{n} \|u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)}, \quad (85) $$

and, in particular,

$$ \frac{1}{2} \frac{d}{dt} \int_{\mathbb{R}_+} u_n(t, x)^2 dx \leq \frac{1}{2} \left( C + \|g(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} \right) \|u_n(t, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 \\ + \frac{1}{2} \|g(t, \cdot)\|_{L^2(\mathbb{R}_+^*)} + \frac{h(t)^2}{2\bar{v}(t, 0)} + \frac{C}{n} \|u_n(t, \cdot)\|_{H^1(\mathbb{R}_+^*)}. $$

The Grönwall lemma yields, in this case,

$$ \int_{\mathbb{R}_+} u_n(t, x)^2 dx \leq \exp \left( Ct + \int_0^t \|g(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} ds \right) \int_{\mathbb{R}_+} u_n(0, x)^2 dx \\ + \exp \left( Ct + \int_0^t \|g(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} ds \right) \int_0^t \left( \frac{h(s)^2}{\bar{v}(s, 0)} + \|g(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} + \frac{2C}{n} \|u_n(s, \cdot)\|_{H^1(\mathbb{R}_+^*)} \right) ds. $$
Letting $n$ go to $+\infty$, we obtain

$$ \int_{\mathbb{R}_+} u(t,x)^2 dx \le B \left[ \int_{\mathbb{R}_+} u^0(x)^2 dx + \int_0^t \left( \frac{h(s)^2}{\bar{v}(s,0)} + \|g(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} \right) ds \right], $$

for a constant $B > 0$ only depending on $\bar{v}$ and $g$, and thus (50) is retrieved.

To obtain the $H^1$ norm, we now go back to (85). Integrating it over the time interval $(0, t)$, for almost every $t \in (0, T)$, we obtain

$$ \begin{aligned} & \frac{1}{2} \int_{\mathbb{R}_+} u_n(t,x)^2 dx - \frac{1}{2} \int_{\mathbb{R}_+} u_n(0,x)^2 dx + \varepsilon \int_0^t \int_{\mathbb{R}_+} (\partial_x u_n(s,x))^2 dx ds \\ & \le \frac{1}{2} \left( C + \|g\|_{L^2(0,T;L^2(\mathbb{R}_+^*))} \right) \int_0^t \|u_n(s, \cdot)\|_{L^2(\mathbb{R}_+^*)} ds + \frac{1}{2} \int_0^t \left[ \frac{h(s)^2}{\bar{v}(s,0)} + \frac{2C}{n} \|u_n(s, \cdot)\|_{H^1(\mathbb{R}_+^*)} \right] ds. \end{aligned} $$

That implies

$$ \begin{aligned} \varepsilon \int_0^t \|\partial_x u_n(s, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 ds &\le \frac{1}{2} \|u_n(0, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 + B \int_0^t \|u_n(s, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 ds \\ &\quad + \frac{1}{2} \int_0^t \frac{h(s)^2}{\bar{v}(s,0)} ds + \frac{C}{n} \int_0^t \|u_n(s, \cdot)\|_{H^1(\mathbb{R}_+^*)} ds. \end{aligned} $$

In the limit when $n$ goes to $+\infty$, we eventually get

$$ \varepsilon \int_0^t \|\partial_x u(s, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 ds \le \frac{1}{2} \|u^0\|_{L^2(\mathbb{R}_+^*)}^2 + B \int_0^t \|u(s, \cdot)\|_{L^2(\mathbb{R}_+^*)}^2 ds + \frac{1}{2} \int_0^t \frac{h(s)^2}{\bar{v}(s,0)} ds, $$

which establishes (51).

# 5 Uniqueness proof for Theorem 1

We now address the uniqueness of the solution to problem (7)–(9). Assume that two nonnegative solutions exist, $a_1$ and $a_2$ in $L^2(0, T; H^1(\mathbb{R}_+^*))$ with $\partial_t a_1$ and $\partial_t a_2$ in $L^2(0, T; H^1(\mathbb{R}_+^*)')$, and that they satisfy properties (11)–(14). We denote, for almost every $t \in (0, T)$,

$$ \Gamma_k(t) := S(t) - \int_{\mathbb{R}_+} ya_k(t, x) dy \in L^\infty(0, T), \quad k \in \{1, 2\}. $$

Function $Z = a_1 - a_2$ satisfies, for almost every $t \in (0, T)$,

$$ \begin{aligned} & \frac{d}{dt} \int_{\mathbb{R}_+} Z(t,x)w(x)dx + \varepsilon \int_{\mathbb{R}_+} \partial_x Z(t,x)\partial_x w(x)dx - \int_{\mathbb{R}_+} v(\Gamma_1(t),x)Z(t,x)\partial_x w(x)dx \\ & = - \int_{\mathbb{R}_+} \partial_x [(v(\Gamma_1(t),x) - v(\Gamma_2(t),x))a_2(t,x)]w(x)dx \\ & \quad - (v(\Gamma_1(t),0) - v(\Gamma_2(t),0))a_2(t,0)w(0) + (f(\Gamma_1(t)) - f(\Gamma_2(t)))w(0), \end{aligned} \quad (86) $$

where, $v(\gamma, x) = \gamma_+ v_G(x) - \gamma_- v_L(x)$ for any $x \in \mathbb{R}_+$. Equation (86) is supplemented with the initial condition

$$ Z(0, x) = 0, \quad \text{a.e. } x > 0 \quad (87) $$

We intend to prove that

$$ \int_{\mathbb{R}_+} y|Z(t, y)|dy = 0 $$
by combining (48) and (46) from Proposition 2. From this, the uniqueness of the solution easily follows. One observes that (44) coincides with (86) when choosing

$$ \bar{v}(t, x) = v(\Gamma_1(t), x), \qquad (88) $$

$$ h(t) = -(v(\Gamma_1(t), 0) - v(\Gamma_2(t), 0)) a_2(t, 0) + f(\Gamma_1(t)) - f(\Gamma_2(t)), \qquad (89) $$

$$ g(t, x) = \partial_x [(v(\Gamma_1(t), x) - v(\Gamma_2(t), x)) a_2(t, x)]. \qquad (90) $$

Turning to the assumptions, these regarding $\bar{v}$ in (42) are straightforward to check. Before tackling the remaining hypotheses, we prove a supporting lemma.

**Lemma 4.** Let $\gamma, \gamma' \in \mathbb{R}$ and $\alpha, \beta \ge 0$. Then we have

$$ |\gamma_+\alpha - \gamma_-\beta - \gamma'_+\alpha + \gamma'_-\beta| \le |\gamma - \gamma'| \max(\alpha, \beta). \qquad (91) $$

*Proof.* Inequality (91) is straightforward if $\gamma\gamma'w \ge 0$. Assume now that $\gamma\gamma' < 0$ and, without loss of generality, that $\gamma < 0 < \gamma'$. The right hand side of (91) satisfies

$$ |-\gamma_{-}\beta - \gamma'_{+}\alpha| = -\gamma\beta + \gamma'\alpha \le -\gamma\max(\alpha, \beta) + \gamma'\max(\alpha, \beta) = |\gamma - \gamma'|\max(\alpha, \beta). $$

That establishes the proof of Lemma 4.

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

We can apply Lemma 4 with $\gamma = \Gamma_1(t)$, $\gamma' = \Gamma_2(t)$, $\alpha \in \{v_G(x), v'_G(x)\}$ and $\beta \in \{v_L(x), v'_L(x)\}$. Setting $M_v = \max(\|v_G\|_{W^{1,\infty}(\mathbb{R}_+^*)}, \|v_L\|_{W^{1,\infty}(\mathbb{R}_+^*)})$, we deduce that, for almost every $t \in (0, T)$ and every $x \in \mathbb{R}_+$,

$$ \begin{aligned} |v(\Gamma_1(t), x) - v(\Gamma_2(t), x)| &= |(\Gamma_1)_+(t)v_G(x) - (\Gamma_1)_-(t)v_L(x) - (\Gamma_2)_+(t)v_G(x) + (\Gamma_2)_-(t)v_L(x)| \\ &\le |\Gamma_1(t) - \Gamma_2(t)| \max_{x \in \mathbb{R}_+} (v_G(x), v_L(x)) \le M_v |\Gamma_1(t) - \Gamma_2(t)|, \end{aligned} $$

and, in the same way, for almost every $t, x$,

$$ |\partial_x v(\Gamma_1(t), x) - \partial_x v(\Gamma_2(t), x)| \le M_v |\Gamma_1(t) - \Gamma_2(t)|. $$

Let us now prove that the function $h$ defined by (89) belongs to $L^1(0, T)$. Firstly, for almost every $t \in (0, T)$, we clearly have

$$ \begin{aligned} |h(t)| &\le M_v |\Gamma_1(t) - \Gamma_2(t)| |a_2(t, 0)| + L_f |\Gamma_1(t) - \Gamma_2(t)| \\ &\le |\Gamma_1(t) - \Gamma_2(t)| \left( M_v \|a_2\|_{L^\infty(0,T;L^\infty(\mathbb{R}_+^*))} + L_f \right) \le |\Gamma_1(t) - \Gamma_2(t)| (M_v C + L_f), \quad (92) \end{aligned} $$

where the constant $C > 0$ depends only on the data, which shows that $h \in L^1(0, T)$.

We now turn to proving that $g$ is compactly supported in $x$ and belongs to $L^2(0, T; L^2(\mathbb{R}_+^*))$. Since $\text{supp } v(\gamma, \cdot) \subset [0, x_R]$ for almost every $\gamma \in \mathbb{R}$, it is also true that

$$ \text{supp} (\partial_x [(v(\Gamma_1(t), \cdot) - v(\Gamma_2(t), \cdot)) a_2(t, \cdot)]) \subset [0, x_R], \quad \text{a.e. } t \in (0, T). $$

Further, since $a_2 \in L^2(0, T; H^1(\mathbb{R}_+^*))$ and

$$ (t, x) \mapsto v(\Gamma_k(t), x) \in L^\infty(0, T; W^{1,\infty}(\mathbb{R}_+^*)), \quad k \in \{1, 2\}, $$

we have

$$ (t, x) \mapsto v(\Gamma_k(t), x) a_2(t, x) \in L^2(0, T; H^1(\mathbb{R}_+^*)), \quad k \in \{1, 2\}. $$

It follows that

$$ (t, x) \mapsto ((v(\Gamma_1(t), x) - v(\Gamma_2(t), x))a_2(t, x) \in L^2(0, T; H^1(\mathbb{R}_+^*)), $$

and therefore its weak derivative

$$ (t, x) \mapsto \partial_x [((v(\Gamma_1(t), x) - v(\Gamma_2(t), x))a_2(t, x))] $$
belongs to $L^2(0, T; L^2(\mathbb{R}_+^*))$.

In what follows, the $L^1(0, t; L^1(\mathbb{R}_+^*))$ norm of $g$ is useful as well. For any $t \in (0, T]$, it can be upper-bounded as

$$ \begin{aligned} \|g\|_{L^1(0,t;L^1(\mathbb{R}_+^*))} &= \int_0^t \int_{\mathbb{R}_+} |(\partial_x v(\Gamma_1(s), x) - \partial_x v(\Gamma_2(s), x))a_2(s, x)| dx ds \\ &\quad + \int_0^t \int_{\mathbb{R}_+} |(v(\Gamma_1(s), x) - v(\Gamma_2(s), x)) \partial_x a_2(s, x)| dx ds \\ &\le M_v \int_0^t |\Gamma_1(s) - \Gamma_2(s)| \int_0^{x_R} (a_2(s, x) + |\partial_x a_2(s, x)|) dx ds \\ &\le M_v \sqrt{x_R} \int_0^t |\Gamma_1(s) - \Gamma_2(s)| \|a_2(s, \cdot)\|_{H^1(\mathbb{R}_+)} ds \\ &\le M_v \sqrt{x_R} \|\Gamma_1 - \Gamma_2\|_{L^2(0,t)} \|a_2\|_{L^2(0,t;H^1(\mathbb{R}_+^*))}. \end{aligned} \quad (93) $$

Consequently, Assumptions (40)–(41) are satisfied in the setting of problem (86) supplied with initial condition (87). Therefore, combining (48) with (46), we obtain, for almost every $t \in (0, T)$ and all $r \ge 2x_R$,

$$ \begin{aligned} \int_{\mathbb{R}_+} x|Z(t,x)|dx &\le \int_0^r x|Z(t,x)|dx + \int_r^\infty x|Z(t,x)|dx \le r\|Z(t, \cdot)\|_{L^1(\mathbb{R}_+^*)} + \frac{\varepsilon c}{r} \int_0^t \|Z(s, \cdot)\|_{L^1(\mathbb{R}_+^*)} ds \\ &\le r \int_0^t (|h(s)| + \|g(s, \cdot)\|_{L^1(\mathbb{R}_+^*)}) ds + \frac{\varepsilon c}{r} \int_0^t \int_0^s (|h(s')| + \|g(s', \cdot)\|_{L^1(\mathbb{R}_+^*)}) ds' ds \\ &\le \left(r + \frac{\varepsilon c t}{r}\right) \int_0^t (|h(s)| + \|g(s, \cdot)\|_{L^1(\mathbb{R}_+^*)}) ds, \end{aligned} \quad (94) $$

where we have used estimate

$$ \begin{aligned} \int_0^t \int_0^s (|h(s')| + \|g(s', \cdot)\|_{L^1(\mathbb{R}_+^*)}) ds' ds &= \int_0^t \int_{s'}^t (|h(s')| + \|g(s', \cdot)\|_{L^1(\mathbb{R}_+^*)}) ds ds' \\ &= \int_0^t (t - s') (|h(s')| + \|g(s', \cdot)\|_{L^1(\mathbb{R}_+^*)}) ds' \le t \int_0^t (|h(s')| + \|g(s', \cdot)\|_{L^1(\mathbb{R}_+^*)}) ds'. \end{aligned} $$

Further, (92) and (93) directly imply that

$$ \begin{aligned} \int_0^t (|h(s)| + \|g(s, \cdot)\|_{L^1(\mathbb{R}_+^*)}) ds &\le (M_v C + L_f) \int_0^t |\Gamma_1(s) - \Gamma_2(s)|ds + M_v C \sqrt{x_R} \|\Gamma_1 - \Gamma_2\|_{L^2(0,T)} \\ &\le [(M_v C + L_f) \sqrt{T} + M_v C \sqrt{x_R}] \|\Gamma_1 - \Gamma_2\|_{L^2(0,T)}. \end{aligned} \quad (95) $$

Consequently, substituting (95) into (94) yields

$$ \int_{\mathbb{R}_+} x|Z(t,x)|dx \le C_1 \left(r + \frac{\varepsilon c t}{r}\right) \|\Gamma_1 - \Gamma_2\|_{L^2(0,t)}, $$

with the constant $C_1$ only depending on the data and with fixed $r > 2x_R$. We set $C_2 = C_1 \left(r + \frac{\varepsilon c T}{r}\right)$ for some $r > 2x_R$. Squaring both sides of above inequality yields, for almost every $t \in (0, T)$,

$$ \left( \int_{\mathbb{R}_+} x|Z(t,x)|dx \right)^2 \le C_2 \int_0^t (\Gamma_1(s) - \Gamma_2(s))^2 ds \le C_2 \int_0^t \left( \int_{\mathbb{R}_+} x|Z(s,x)|dx \right)^2 ds. \quad (96) $$

Hence, by the Grönwall inequality $\int_{\mathbb{R}_+} y|Z(t, y)|dy \le 0$ for almost every $t \in (0, T)$. It follows that $a_1 = a_2$ almost everywhere and the uniqueness of solutions is established.
# Acknowledgements

The author would like to express her gratitude to Magali Ribot for the fruitful discussions.

# References

[1] J.-P. Aubin. Un théorème de compacté. *C. R. Acad. Sci., Paris*, 256:5042-5044, 1963.

[2] L. Boudin. A solution with bounded expansion rate to the model of viscous pressureless gases. *SIAM J. Math. Anal.*, 32(1):172-193, 2000.

[3] F. Boyer and P. Fabrie. *Mathematical tools for the study of the incompressible Navier-Stokes equations and related models*, volume 183 of *Appl. Math. Sci*. New York, NY: Springer, 2013.

[4] H. Brezis. *Functional analysis, Sobolev spaces and partial differential equations*. Universitext. New York, NY: Springer, 2011.

[5] R. Fernández-Verdejo, G. Sanchez-Delgado, and E. Ravussin. Energy expenditure in humans: Principles, methods, and changes throughout the life course. *Annual Review of Nutrition*, 44(Vol-ume 44, 2024):51-76, 2024.

[6] J. Gliemann, J. Vinten, and A. Vølund. Lipogenesis and insulin sensitivity of single fat cells*: With an appendix. *J. Physiol. (London)*, 236(3):499-516, 1974.

[7] T. Goudon, F. Lagoutière, and L. M. Tine. The Lifschitz-Slyozov equation with space-diffusion of monomers. *Kinet. Relat. Models*, 5(2):325-355, 2012.

[8] F. Haczeyni, K. S. Bell-Anderson, and G. C. Farrell. Causes and mechanisms of adipocyte en-largement and adipose expansion. *Obesity Reviews: An Official Journal of the International Association for the Study of Obesity*, 19(3):406-420, 2018.

[9] J. O. Hill, H. R. Wyatt, and J. C. Peters. Energy Balance and Obesity. *Circulation*, 126(1):126-132, 2012. Publisher: American Heart Association.

[10] A. Horwitz and R. Birk. Adipose Tissue Hyperplasia and Hypertrophy in Common and Syndromic Obesity-The Case of BBS Obesity. *Nutrients*, 15(15):3445, 2023.

[11] L. I. Ignat and J. D. Rossi. A nonlocal convection-diffusion equation. *J. Funct. Anal.*, 251(2):399-437, 2007.

[12] J. Laurencikiene, T. Skurk, A. Kulytė, P. Hedén, G. Åström, E. Sjölin, M. Rydén, H. Hauner, and P. Arner. Regulation of Lipolysis in Small and Large Fat Cells of the Same Subject. *The Journal of Clinical Endocrinology & Metabolism*, 96(12):E2045-E2049, 2011.

[13] R. L. Leibel, M. Rosenbaum, and J. Hirsch. Changes in energy expenditure resulting from altered body weight. *New England Journal of Medicine*, 332(10):621-628, 1995.

[14] Q. Li and K. L. Spalding. The regulation of adipocyte growth in white adipose tissue. *Frontiers in Cell and Developmental Biology*, 10, 2022.

[15] L. Meyer, M. Ribot, and R. Yvinec. A lifshitz-slyozov type model for adipocyte size dynamics: limit from becker-döring system and numerical simulation. *J. Math. Biol.*, 88(16), 2024.

[16] S. Nag and S. K. Tumuluri. Analysis of a nonlinear age-structured SIS epidemic model with spatial diffusion. *Comput. Appl. Math.*, 44(5):20, 2025. Id/No 250.

[17] T. I. Rajjo, D. A. Harteneck, and M. D. Jensen. Direct Free Fatty Acid Storage in Different Sized Adipocytes from the Same Depot. *Obesity (Silver Spring, Md.)*, 22(5):1275, 2014.
[18] I. Romieu, L. Dossus, S. Barquera, H. M. Blottière, P. W. Franks, M. Gunter, N. Hwalla, S. D. Hursting, M. Leitzmann, B. Margetts, C. Nishida, N. Potischman, J. Seidell, M. Stepien, Y. Wang, K. Westerterp, P. Winichagoon, M. Wiseman, W. C. Willett, and On behalf of the IARC working group on Energy Balance and Obesity. Energy balance and obesity: what are the main drivers? *Cancer Causes & Control*, 28(3):247-258, 2017.

[19] T. Roubíček. *Nonlinear partial differential equations with applications*, volume 153 of *ISNM, Int. Ser. Numer. Math*. Basel: Birkhäuser, 2nd ed. edition, 2013.

[20] W. Rudin. *Functional analysis*. New York, NY: McGraw-Hill, 2nd ed. edition, 1991.

[21] Y. Schutz. Dietary fat, lipogenesis and energy balance. *Physiology & Behavior*, 83(4):557-564, 2004.

[22] K. L. Spalding, E. Arner, P. O. Westermark, S. Bernard, B. A. Buchholz, O. Bergmann, L. Blomqvist, J. Hoffstedt, E. Näslund, T. Britton, H. Concha, M. Hassan, M. Rydén, J. Frisén, and P. Arner. Dynamics of fat cell turnover in humans. *Nature*, 453(7196):783-787, 2008.

[23] B. M. Spiegelman and J. S. Flier. Obesity and the Regulation of Energy Balance. *Cell*, 104(4):531-543, 2001. Publisher: Elsevier.

[24] C. Walker. Well-posedness and stability analysis of an epidemic model with infection age and spatial diffusion. *J. Math. Biol.*, 87(3):46, 2023. Id/No 52.

[25] K. R. Westerterp. Control of energy expenditure in humans. *European Journal of Clinical Nutrition*, 71(3):340-344, 2017. Publisher: Nature Publishing Group.

[26] S. Yoo. Dynamic Energy Balance and Obesity Prevention. *Journal of Obesity & Metabolic Syndrome*, 27(4):203-212, 2018. Publisher: Korean Society for the Study of Obesity.

[27] E. Zeidler. *Nonlinear Functional Analysis and Its Applications: Linear Monotone Operators*, volume II/A. Springer New York, 1990.

[28] O. Zinder and B. Shapiro. Effect of cell size on epinephrine- and acth-induced fatty acid release from isolated fat cells. *J. Lipid Res.*, 12(1):91-95, 1971.