if we denote by $S_L$ the semigroup generated by the transport equation (8.58) with $G = \mathfrak{g} = 0$, by $S_B$ the semigroup when additionally $\mathcal{H} = 0$, and $Af = \mathcal{H}[f]$.

**Proof of Proposition 8.23.** We split the proof into five steps.

**Step 1.** A priori estimates. For a positive solution, we formally compute

$$ (8.64) \qquad \frac{1}{p} \frac{d}{dt} \int g^p m^p = \frac{1}{p} \int_{\Sigma_{-}} (\mathcal{R}[g, \gamma_+ g] + \mathfrak{g})^p m_{\Sigma}^p d\sigma_y - \frac{1}{p} \int_{\Sigma_{+}} (\gamma_+ g)^p m_{\Sigma}^p d\sigma_y \\ \qquad \qquad \qquad \qquad + \int_{\mathcal{O}} \{g^{p-1}(\mathcal{H}[g] + G) - g^p \varpi\} m^p. $$

Using the Young inequality and (8.61), we have

$$ \begin{aligned} \int_{\mathcal{O}} g^{p-1} \mathcal{H}[g] m^p &\le \frac{1}{p'} \int_{\mathcal{O}} g^p \langle \varpi_+ \rangle m^p + \frac{1}{p} \int_{\mathcal{O}} \mathcal{H}[g]^p \langle \varpi_+ \rangle^{-p/p'} m^p \\ &\le \left( \frac{1}{p'} + \frac{\alpha_{\mathcal{H}}}{p} \right) \int_{\mathcal{O}} g^p \langle \varpi_+ \rangle m^p + \frac{M_{\mathcal{H}}}{p} \int_{\mathcal{O}} g^p m^p. \end{aligned} $$

• When $\mathfrak{g} = 0$, using also (8.62) and once more the Young inequality, we then have

$$ \frac{1}{p} \frac{d}{dt} \int g^p m^p \le \frac{1}{p} (\beta_{\mathcal{H}} - 1) \int_{\Sigma_+} (\gamma_+ g)^p m^p a \cdot n + \left( \frac{M_{\mathcal{H}}}{p} + \frac{M_{\mathcal{H}}}{p} + \|\langle \varpi_- \rangle\|_{L^\infty} \right) \int_{\mathcal{O}} g^p m^p \\ + \left( \frac{\alpha_{\mathcal{H}}}{p} + \frac{1}{p'} + \frac{\alpha_{\mathcal{H}}}{p} + \frac{\varepsilon}{p'} - 1 \right) \int_{\mathcal{O}} g^p \langle \varpi_+ \rangle m^p + \frac{\varepsilon^{-p/p'}}{p} \int_{\mathcal{O}} G^p m^p \langle \varpi_+ \rangle^{-p/p'} $$

for any $\varepsilon > 0$. Making the choice $\varepsilon := \vartheta_{\mathcal{O}}p'/p$, we deduce

$$ \frac{d}{dt} \|g\|_{L_m^p} + \vartheta_{\mathcal{O}} \|g\|_{L_m^p}^p + \vartheta_{\Sigma} \|\gamma_+ g\|_{L_m^p}^p \le p\kappa \|g\|_{L_m^p}^p + C_{\mathcal{O}} \|G\|_{L_m^p}^p, $$

with

$$ \kappa := \frac{M_{\mathcal{H}}}{p} + \frac{M_{\mathcal{H}}}{p} + \|\langle \varpi_- \rangle\|_{L^\infty}, \quad C_{\mathcal{O}} := (\vartheta_{\mathcal{O}}p'/p)^{-p'/p}. $$

Using the Gronwall lemma, we then obtain

$$ (8.65) \qquad \|g(t)\|_{L_m^p}^p + \int_0^t e^{p\kappa(t-s)} (\vartheta_{\mathcal{O}} \|g_s\|_{L_m^p}^p + \vartheta_{\Sigma} \|\gamma_+ g_s\|_{L_m^p}^p) ds \\ \qquad \le e^{p\kappa t} \|\mathfrak{g}_0\|_{L_m^p}^p + C_{\mathcal{O}} \int_0^t e^{p\kappa(t-s)} \|G_s\|_{L_m^p}^p ds, \quad \forall t \ge 0. $$

• When $\mathfrak{g} \ne 0$ and thus $\vartheta_{\Sigma} > 0$, we control the ingoing boundary term by

$$ \int_{\Sigma_{-}} (\mathcal{R}[g, \gamma_+ g] + \mathfrak{g})^p m_{\Sigma}^p \le (1 + \varepsilon_1) \int_{\Sigma_{-}} \mathcal{R}[g, \gamma_+ g]^p m_{\Sigma}^p + C_{\varepsilon_1} \int_{\Sigma_{-}} \mathfrak{g}^p m_{\Sigma}^p, \quad \forall \varepsilon_1 > 0, $$

and a very similar computation as above leads to the a priori estimate

$$ (8.66) \qquad \|g(t)\|_{L_m^p}^p + \int_0^t e^{p\kappa(t-s)} (\vartheta'_{\mathcal{O}} \|g_s\|_{L_m^p}^p + \vartheta'_{\Sigma} \|\gamma_+ g_s\|_{L_m^p}^p) ds \\ \qquad \le e^{p\kappa t} \|\mathfrak{g}_0\|_{L_m^p}^p + \int_0^t e^{p\kappa(t-s)} (C_{\mathcal{O}} \|G_s\|_{L_m^p}^p + C_{\Sigma} \|\mathfrak{g}_s\|_{L_m^p}^p) ds, $$

for any $t \ge 0$, with

$$ \vartheta'_{\Sigma} := 1 - \beta_{\mathcal{H}}(1 + \varepsilon_1), \quad \vartheta'_{\mathcal{O}} := 1 - \alpha_{\mathcal{H}}(1 + \varepsilon_1) - \alpha_{\mathcal{H}} - \varepsilon \frac{p}{p'}, $$

$$ \kappa := \frac{M_{\mathcal{H}}}{p}(1 + \varepsilon_1) + \frac{M_{\mathcal{H}}}{p} + \|\langle \varpi_- \rangle\|_{L^\infty}, \quad C_{\mathcal{O}} := \varepsilon^{-p'/p}, \quad C_{\Sigma} := C_{\varepsilon_1}, $$

and where we have chosen $\varepsilon, \varepsilon_1 > 0$ small enough in such a way that $\vartheta'_{\Sigma} > 0$ and $\vartheta'_{\mathcal{O}} > 0$.
**Uniqueness.** We present now a uniqueness result.

**Lemma 8.14 (Uniqueness).** We assume that a and b satisfy (8.35) for some exponent $p \in [1, \infty)$ and some weight function $m: \bar{\mathcal{O}} \to [1, \infty)$ as well as for $p = 1$ and $m \equiv 1$. We additionally assume diva $\in L_{\text{loc}}^{\infty}(\bar{\mathcal{O}})$ (what is automatically true under assumption (8.33)). With obvious notations, for any $\lambda > \max(\lambda_p^*(m), \lambda_1^*(1))$, and any solution $g \in L_{m\bar{\mathcal{O}}}^p(\bar{\mathcal{O}})$ to the transport equation

$$ (8.54) \qquad \lambda g + a \cdot \nabla g + bg = 0 \quad \text{in } \mathcal{D}'(\bar{\mathcal{O}}), \quad \gamma_{-}g = 0 \quad \text{on } \Sigma_{-}, $$

we have $g \equiv 0$.

**Proof of Lemma 8.14.** We main follow the proof of [138, Cor. II.1]. We fix $\beta \in W^{1,\infty}(\mathbb{R})$, $\beta(0) = 0$, in such a way that $\beta(g) \in L_{m\bar{\mathcal{O}}}^p \cap L^\infty$ is a solution to

$$ (\lambda + b)g\beta'(g) + a \cdot \nabla\beta(g) = 0 \quad \text{in } \mathcal{D}'(\bar{\mathcal{O}}), \quad \gamma_{-}\beta(g) = 0 \quad \text{on } \Sigma_{-}. $$

For any $\psi \in C_c(\bar{\mathcal{O}})$ and any $\lambda > \lambda_{m,p}^*$, we solve in $L_{m-1(\bar{\omega}_+)}^{p'} \cap L^\infty$ the dual problem

$$ (8.55) \qquad \lambda\varphi - a \cdot \nabla\varphi + (b - \text{div}\,\varphi)\varphi = \psi \quad \text{in } \mathcal{D}'(\bar{\mathcal{O}}), \quad \gamma_+\varphi = 0 \quad \text{on } \Sigma_+, $$

thanks to Lemma 8.12 and Lemma 8.13, where we observe that, because of (8.37), the necessary condition on $\lambda$ in these results is precisely the one made here. For $\chi \in C_c^1(\mathbb{R}^D)$, $\mathbf{1}_{B_1} \le \chi \le \mathbf{1}_{B_2}$, and $R > 0$, we define $\chi_R(x) := \chi(x/R)$. Using the Green formula (8.21), we have

$$ 0 = \int_{\bar{\mathcal{O}}} ((\lambda + b)\varphi - \psi)\beta(g)\chi_R - \int_{\bar{\mathcal{O}}} (\lambda + b)\varphi g\beta'(g)\chi_R + \int_{\bar{\mathcal{O}}} \varphi\beta(g) \frac{a}{R} \cdot (\nabla\chi)_R. $$

Because on the one hand $\varphi\beta(g) \in L_{(\bar{\omega}_+)}^1 \cap L^\infty$ and on the other hand $a/R \cdot (\nabla\chi)_R \to 0$ a.e. and is bounded in $L_{(\bar{\omega}_+)}^\infty + L^1$ we deduce that the last term vanishes when $R \to \infty$. Using also that $b\varphi g \in L^1$ thanks to (8.35), we may pass to the limit $R \to \infty$ in the above equation and we get

$$ 0 = \int_{\bar{\mathcal{O}}} ((\lambda + b)\varphi - \psi)\beta(g) - \int_{\bar{\mathcal{O}}} (\lambda + b)\varphi g\beta'(g). $$

We take $\beta := \beta_\delta$ and we observe that $\langle b \rangle |\varpi||\beta_\delta(g) - g\beta'_\delta(g)| \le \langle b \rangle |\varphi g| \in L^1(\bar{\mathcal{O}})$. We may then pass to the limit $\delta \to 0$ in the last equation, and we get

$$ 0 = - \int_{\bar{\mathcal{O}}} \psi g, \quad \forall \psi \in C_c(\bar{\mathcal{O}}), $$

from which we conclude that $g \equiv 0$. $\square$

We come to the time dependent transport equation by formulating a general continuity result.

**Proposition 8.15.** Assume that $a \in W_{\text{loc}}^{1,1}(\bar{\mathcal{O}})$, $b \in L_{\text{loc}}^1(\bar{\mathcal{O}})$, $G \in L_{\text{loc}}^1([0, T] \times \bar{\mathcal{O}})$. Any renormalized solution $g \in L_{\text{loc}}^1([0, T] \times \bar{\mathcal{O}})$ to the first equation in (8.31), meaning

$$ \frac{\partial}{\partial t}\beta(g) + a \cdot \nabla\beta(g) + \beta'(g)bg = \beta'(g)G \quad \text{in } \mathcal{D}'((0, T) \times \bar{\mathcal{O}}), $$

for any renormalizing function $\beta \in C_*^1(\mathbb{R})$, satisfies $g \in C([0, T]; L^0(\bar{\mathcal{O}}))$, meaning that $\beta(g) \in C([0, T]; L_{\text{loc}}^1(\bar{\mathcal{O}}))$ for any $\beta \in C_b(\mathbb{R})$.

**Proof of Proposition 8.15.** The proof is a variant of the proof of [138, Thm. II.3] and we just allude it. Because $\beta(g) \in L^\infty((0, T) \times \bar{\mathcal{O}})$ is a solution to the transport equation with source term $\beta'(g)G - \beta'(g)bg \in L_{\text{loc}}^1([0, T] \times \bar{\mathcal{O}})$, we have $\beta(g) \in C([0, T]; \mathcal{D}'(\bar{\mathcal{O}}))$ for any $\beta \in C_*^1(\mathbb{R})$. Fixing $\beta_0 \in C_*^1(\mathbb{R})$ strictly increasing, we deduce that $\beta_0(g), \beta_0(g)^2 \in C([0, T]; \mathcal{D}'(\bar{\mathcal{O}}))$, so that $\beta_0(g) \in C([0, T]; L_{\text{loc}}^2(\bar{\mathcal{O}}))$, and the conclusion. $\square$

We consider now the time dependent transport equation (8.31).

**Proposition 8.16 (Renormalized solutions).** We assume (8.35) for some $p \in [1, \infty)$ and some weight function $m$. For any $g_0 \in L_m^p(\bar{\mathcal{O}})$, $G \in L_{m\bar{\mathcal{O}}}^p((0, T) \times \bar{\mathcal{O}})$, $g \in L_{m\Sigma}^p((0, T) \times \Sigma)$, there exists a unique $g \in C([0, T]; L_{\text{loc}}^1(\bar{\mathcal{O}}))$ satisfying the estimate (8.45) or (8.46) and being a solution to the transport equation (8.31) in the renormalized sense, namely

$$ (8.56) \qquad \begin{cases} \frac{\partial \beta(g)}{\partial t} + a \cdot \nabla \beta(g) + \beta'(g)bg = \beta'(g)G & \text{on } (0, T) \times \bar{\mathcal{O}}, \\ \gamma_- \beta(g) = \beta(g) & \text{on } (0, T) \times \Sigma_-, \quad \beta(g)(0, \cdot) = \beta(g_0) & \text{on } \bar{\mathcal{O}}, \end{cases} $$
• When $\vartheta_{\Sigma} = 0$ and thus $\mathfrak{g} = 0$, we further multiply the equation by $m^p \tau^{\pm}$, where $\tau^{\pm}$ is defined in (8.57), and integrating, we deduce

$$ \int_0^T \int_{\Sigma_\mp} \tau^{\pm} a \cdot n(\gamma g)^p m^p d\sigma dt = \left[ \int_{\mathcal{O}} g^p m^p \tau^{\pm} \right]_T^0 + \int_0^T \int_{\mathcal{O}} pg^{p-1}(\mathcal{H}[g] + G)m^p \tau^{\pm} + \int_0^T \int_{\mathcal{O}} g^p \left( \frac{\mathrm{div}(am^p)}{m^p} + 1 - \lambda_0 \tau^{\pm} - pKm^p \tau^{\pm} \right). $$

Together with (8.65) and $\tau^{\pm} \in L^{\infty}(\mathcal{O})$, we obtain

$$ (8.67) \qquad \int_0^T \int_{\Sigma_+} (\gamma_+ g)^p \tau^- m_\Sigma^p \le C_T \left( \|g_0\|_{L_m^p}^p + \|G\|_{L^{p}(0,T;L_m^p)}^p \right) $$

and

$$ (8.68) \qquad \int_0^T \int_{\Sigma_-} [\mathcal{R}_\Sigma(\gamma_+ g)]^p \tau^+ m_\Sigma^p \le C_T \left( \|g_0\|_{L_m^p}^p + \|G\|_{L^{p}(0,T;L_m^p)}^p \right), $$

for some constant $C_T \in (0, \infty)$. In particular, when $p=1$ and $\mathcal{R}_\Sigma$ is diffusive, we have

$$ c_{\Sigma} \int_{0}^{T} \int_{\Sigma_{+}} (\gamma_{+} g) m_{\Sigma} \leq \int_{0}^{T} \int_{\Sigma_{+}} (\gamma_{+} g) \mathcal{R}_{\Sigma}^{*}(\tau^{+} m_{\Sigma}) = \int_{0}^{T} \int_{\Sigma_{-}} \mathcal{R}_{\Sigma}(\gamma_{+} g) \tau^{+} m_{\Sigma}, $$

and together with (8.68), we deduce the additional estimate

$$ (8.69) \qquad c_{\Sigma} \int_{0}^{T} \int_{\Sigma_{+}} (\gamma_{+} g) m_{\Sigma} \le C_{T} \left( \|g_{0}\|_{L_{m}^{1}} + \|G\|_{L^{1}(0,T;L_{m}^{1})} \right). $$

**Step 2. Existence.** As a consequence of these a priori estimates, we may classically build a solution through an iterative scheme. For the sake of brevity, we only consider the (more interesting and more difficult) case $b_{\Sigma} = 1$ (so that $\vartheta_{\Sigma} = 0$ and $\mathfrak{g} = 0$) and $G = 0$. For a given $0 \le g_0 \in L_m^p(\mathcal{O})$, we define a sequence of solution $(h_n)$ starting from $h_0 \equiv 0$ thanks to the recursive definition

$$ \left\{ \begin{array}{ll} \displaystyle \frac{\partial h_{n+1}}{\partial t} + a \cdot \nabla h_{n+1} + Kh_{n+1} = \mathcal{K}[h_n] & \text{on } (0,T) \times \mathcal{O}, \\ \gamma_{-}h_{n+1} = \mathcal{R}[h_n, \gamma_{+}h_n] & \text{on } (0,T) \times \Sigma_{-}, \quad h_{n+1}(0, \cdot) = g_0 & \text{on } \mathcal{O}. \end{array} \right. $$

From Proposition 8.16, there exists a unique renormalized solution $h_{n+1} \in C([0, T]; L_m^p(\mathcal{O}))$ to the above equation satisfying the estimate (8.45) with $g := h_{n+1}$, $G := \mathcal{K}[h_n]$ and $\mathfrak{g} := \mathcal{R}[h_n, \gamma_+ h_n] \in L_{m\Sigma}^p$. We observe that $0 \le h_n \le h_{n+1}$ thanks to the weak maximum principle (see Remark 8.17) and that $h_n$ satisfies the estimates (8.65) and (8.68) where $g$ is replaced by $h_n$. Thanks to the monotonous convergence theorem of Beppo Levi, there exists $g$ satisfying estimates (8.65) and $h_n \to g$ in $L_{m\mathcal{O}}^p((0,T) \times \mathcal{O})$. We may pass to the limit in the equation satisfied by $(h_n)$ and we deduce that $g$ is a renormalized solution to

$$ \frac{\partial g}{\partial t} + a \cdot \nabla g + bg = \mathcal{K}[g] \quad \text{on } (0,T) \times \mathcal{O}. $$

From Theorem 8.8 and Remark 8.9-(5), the function $g$ admits a trace $\gamma g$ and thanks to Proposition 8.10, we have $\gamma h_n \to \gamma g$ a.e. on $\Sigma \setminus \Sigma_0$. Because of (8.4) and the Beppo Levi theorem again we deduce that $\mathcal{R}[h_n, \gamma_+ h_n] \to \mathcal{R}[g, \gamma_+ g]$ a.e. on $\Sigma_-$, Together with the fact that $\gamma_- h_n \to \gamma_- g$ a.e. on $\Sigma_-$, we have established that the boundary condition in (8.58) holds true. It is worth emphasizing here that $\gamma_+ g \in L^1(\Sigma_+; dr_{\Sigma}(y, \cdot))$ for a.e. $y \in \Sigma_-$ because of (8.68). For $g_0 \in L_m^p(\mathcal{O})$, we separate the positive and the negative parts $g_0 = g_{0+} - g_{0-}$ and we obtain two renormalized solutions $g^{\pm} \in L^{\infty}(0, T; L_m^p)$ associated to $g_{0\pm}$ respectively. By linearity, the function $g := g^{+} - g^{-} \in L^{\infty}(0, T; L_m^p)$ is a renormalized solution to the transport equation and the boundary condition is

$$ \begin{aligned} \gamma_{-}g &= \gamma_{-}g^{+} - \gamma_{-}g^{-} = \mathcal{R}_{\mathcal{O}}[g^{+}] - \mathcal{R}_{\mathcal{O}}[g^{-}] + \mathcal{R}_{\Sigma}[\gamma_{+}g^{+}] - \mathcal{R}_{\Sigma}[\gamma_{+}g^{-}] \\ &= \mathcal{R}_{\mathcal{O}}[g] + \mathcal{R}_{\Sigma}[\gamma_{+}g], \end{aligned} $$

where the last term is indeed well defined a.e. from the fact that $\gamma_+ g^{\pm} \in L^1(\Sigma_+; dr_{\Sigma}(y, \cdot))$ for a.e. $y \in \Sigma_-$ and thus $\gamma_+ g = \gamma_+ g^{+} - \gamma_+ g^{-}$ belongs to the same spaces. From Proposition 8.15, we already
for any $\beta \in C_{\text{pw},*}^1$. Furthermore, $g \in C([0, T]; L_m^p)$ when $p \in [1, \infty)$ and $g \ge 0$ if $g_0, G, \mathfrak{g} \ge 0$.

**Remark 8.17.** (1) The above result extends some previous results due to Bardos in [37, Chap. III], Boyer in [70, Thm. 4.1] and Crippa et al in [119, Thm. 1.1] and [118, Thm. 1.1], where the cases $p=2$ or $p=\infty$ are considered with always the additional assumption $a \in L^\infty$ (in the last paper however the present $W^{1,1}$ bound on a is relaxed into a BV condition) by adapting the DiPerna-Lions theory developed in [138, Sec. II].

(2) We immediately deduce from the above result and Lemma 8.7-(2) a weak maximum principle: $g_1 \le g_2$ if $g_i$ is renormalized solution to the transport equation (8.31) associated to the data $g_{0i}$, $G_i$, $\mathfrak{g}_i$ such that $g_{01} \le g_{02}$, $G_{01} \le G_{02}$, $\mathfrak{g}_{01} \le \mathfrak{g}_{02}$.

**Proof of Proposition 8.16.** We proceed similarly as during the proof of Lemma 8.12.

**Step 1. Characteristics.** We assume first $a \in C^1(\mathbb{R}^D)$, $g_0 \in C_c(\mathcal{O})$, $b \in C_b(\bar{\mathcal{O}})$, $\mathfrak{g} \in C_c((0, T) \times \Sigma_0)$, $G \in C_c^1((0, T) \times \mathcal{O})$. We use the characteristics representation (8.47). We may verify that $\tilde{g}$ both satisfies the transport equation in the renormalized sense and the boundary conditions in (8.56) and we may justify the computations leading to the a priori estimates (8.45) and (8.46).

**Step 2. Existence.** In the general case, we define some regularized sequence $(a_\varepsilon)$, $(g_{0,\varepsilon})$, $(b_\varepsilon)$, $(\mathfrak{g}_\varepsilon)$, $(G_\varepsilon)$ and thanks to the first step we deduce the existence of an associated function $g_\varepsilon \in C([0, T]; L_m^p)$ satisfies both the equation (8.56) in the renormalized sense and the a priori estimates (8.45) or (8.46). When $p > 1$, the sequence $(g_\varepsilon)$ is bounded in $L^\infty(0, T; L_m^p)$ and (up to the extraction of a subsequence) we may pass to the limit $\varepsilon \to 0$ using Proposition 8.10-(2) and Remark 8.11. We have established the existence of a renormalized solution to the transport equation which satisfies the estimate (8.45) or (8.46). When $p = 1$, we may for instance proceed in the following way by first assuming $0 \le g_0 \in L_m^1$, $0 \le G \in L_m^1((0, T) \times \mathcal{O})$, $0 \le \mathfrak{g} \in L_{m\Sigma}^1((0, T) \times \Sigma)$. We may thus consider some nonnegative approximating sequences $(g_{0,\varepsilon})$ in $L_m^p \cap L_m^1$, $G_\varepsilon \in L_{\tilde{m}\Sigma}^p \cap L_m^1$, $\mathfrak{g}_\varepsilon \in L_{m\Sigma}^p \cap L_{m\Sigma}^1$ such that $g_{0,\varepsilon} \nearrow g_0$, $G_\varepsilon \nearrow G$ and $\mathfrak{g}_\varepsilon \nearrow \mathfrak{g}$. The same construction as above implies the existence of $0 \le g_\varepsilon \in L^\infty(0, T; L_m^1 \cap L_m^p)$ renormalized solution to the transport equation associated to these data and such that $(g_\varepsilon)$ is increasing and uniformly bounded in $L^\infty(0, T; L_m^1)$ thanks to the a priori $L_m^1$ estimate (8.45). There thus exists $0 \le g \in L^\infty(0, T; L_m^1)$ such that $g_\varepsilon \nearrow g$, and we get that $g$ is a renormalized solution to the transport equation by using again Proposition 8.10-(2) and Remark 8.11. We remove the nonnegative condition on $g_0$, $G$ and $\mathfrak{g}$ by introducing the positive and negative parts of each function, using the preceding step in order to prove the existence of two solutions $0 \le g_\pm \in L^\infty(0, T; L_m^1)$ associated respectively to $(g_{0+}, G_+, \mathfrak{g}_+)$ and $(g_{0-}, G_-, \mathfrak{g}_-)$, and finally defining $g := g_+ - g_-$ which is a renormalized solution to the transport equation thanks to Lemma 8.7 and Remark 8.9-(5).

**Step 3. Continuity.** From Proposition 8.15, we already know that $g \in C([0, T]; L^0(\mathcal{O}))$. Together with the a priori estimate (8.45) or (8.46), we also have $g \in C([0, T]; L_{\text{loc}}^1(\mathcal{O}))$ when $p > 1$. When $p \in [1, \infty)$, we may improve the above continuity properties by arguing in the following way. We define $\tilde{g} := gm$ and we observe that it is a solution to the transport equation

$$ \partial_t \tilde{g} + a \cdot \nabla \tilde{g} + \tilde{b} \tilde{g} = \tilde{G}, \quad \gamma \tilde{g} = \tilde{\mathfrak{g}}, \quad \tilde{g}(0) = \tilde{g}_0, $$

with $\tilde{b} := b - a \cdot \nabla m/m$, $\tilde{G} := Gm$, $\tilde{\mathfrak{g}} := gm$ and $\tilde{g}_0 := g_0m$. We write the associated renormalized equation (8.44) for the renormalizing function $\beta_M(s) := (|s| \wedge M)^p$, $M > 0$, and the test function $\varphi := \chi_R$, with $\chi \in C_c^1(\mathbb{R}^D)$, $\mathbf{1}_{B_1} \le \chi \le \mathbf{1}_{B_2}$ and $\chi_R(y) := \chi(y/R)$. Observing in particular that

$$ \int_0^t \int_{\mathcal{O}} \beta_M(\tilde{g}) a \cdot \nabla \chi_R \to 0 \text{ as } R \to \infty, $$

because of (8.45) and (8.35) by arguing as in the proof of Lemma 8.14, we may pass to the limit in the associated renormalized equation as $R \to \infty$, and we obtain

$$ \begin{aligned} \left[ \int_{\mathcal{O}} \beta_M(\tilde{g}) dy \right]_0^t = & \int_0^t \int_{\mathcal{O}} \beta'_M(\tilde{g}) \tilde{G} dy ds - \int_0^t \int_{\Sigma} \beta_M(\gamma \tilde{g}) a \cdot n d\sigma_y ds \\ & + \int_0^t \int_{\mathcal{O}} \beta_M(\tilde{g}) \mathbf{1}_{\tilde{g}>M} \text{div} a dy ds - \int_0^t \int_{\mathcal{O}} p \beta_M(\tilde{g}) \mathbf{1}_{\tilde{g}\le M} \varpi dy ds. \end{aligned} $$
know that $g \in C([0, T]; L^0(\mathcal{O}))$ and thus using an interpolation argument $g \in C([0, T]; L_{m_1}^{p_1}(\mathcal{O}))$ for any $p_1 \in [1, p)$ and any weight function $m_1$ such that $m_1/m \in L^{pp_1/(p-p_1)}$ when $p > 1$.

**Step 3.** When $\beta_{\mathcal{R}} < 1$ and $p \in [1, \infty)$, we have (8.66), and we may just repeat the proof of Proposition 8.16 in order to get $g \in C([0, T]; L_m^p(\mathcal{O}))$ and the uniqueness of the solution.

**Step 4.** We assume $\beta_{\mathcal{R}} = 1$ and the structural assumption (2). From the estimate (8.64) on a solution $g$ and the renormalized formulation of the equation, we deduce that

$$ \frac{1}{p_0} \frac{d}{dt} \int |g|^{p_0} m^{p_0} = \frac{1}{p_0} \int_{\Sigma_{-}} |\mathcal{R}[g, \gamma_+ g]|^{p_0} m_{0\Sigma}^{p_0} - \frac{1}{p_0} \int_{\Sigma_{+}} |\gamma_+ g|^{p_0} m_{0\Sigma}^{p_0} \\ + \int_{\mathcal{O}} \{g|g|^{p_0-2}(\mathcal{H}[g] + G)m^{p_0} - |g|^{p_0} m_{0\mathcal{O}}^{p_0}\}, $$

with a RHS term in $L^1(0, T)$. As above, we thus deduce $g \in C([0, T]; L_{m_0}^{p_0}(\mathcal{O}))$ and next the uniqueness of the solution.

**Step 5.** We assume $\beta_{\mathcal{R}} = 1$ and the structural assumption (3). In that case, we have $p = 1$, $\gamma_+ g \in L_{m_\Sigma}^1((0, T) \times \Sigma_+)$ from (8.69) and then $\gamma_- g \in L_{m_\Sigma}^1((0, T) \times \Sigma_-)$ from (8.62). We may thus justify the same computation as in Step 4 with $p_0 = 1$, and we deduce $g \in C([0, T]; L_m^1(\mathcal{O}))$ and next the uniqueness of the solution. $\square$

As an immediate consequence of the above analysis, we may associate to the transport equation (8.58) a semigroup.

**Corollary 8.25.** Under the assumptions of Proposition 8.23, there exists a positive semigroup $S$ on $L_m^p$ such that for any $\mathfrak{g}_0 \in L_m^p(\mathcal{O})$, the function $t \mapsto g(t) := S(t)\mathfrak{g}_0 \in C(\mathbb{R}_+; L_{m_0}^{p_0}(\mathcal{O})) \cap L_{\text{loc}}^{\infty}(\mathbb{R}_+; L_m^p(\mathcal{O}))$ is the unique renormalized solution to the transport equation (8.58) associated to the initial datum $\mathfrak{g}_0$ (and with $G = \mathfrak{g} = 0$). Furthermore the growth bound satisfies $\omega(S) \le \kappa$.

We end this section by formulating the counterpart of the above result for the associated stationary problem

$$ (8.70) \qquad \begin{cases} \lambda g + a \cdot \nabla g + bg = \mathcal{H}[g] + G & \text{on } \mathcal{O}, \\ \gamma_- g = \mathcal{R}[g, \gamma_+ g] + \mathfrak{g} & \text{on } \Sigma_- \end{cases} $$

**Proposition 8.26.** We make exactly the same assumptions as in Proposition 8.23 on a, b, $\mathcal{H}$ and $\mathcal{R}$ for some weight function $m: \mathcal{O} \to [1, \infty)$ and some exponent $p \in [1, \infty)$ as well as either $\beta_{\mathcal{R}} < 1$ holds or $\beta_{\mathcal{R}} = 1$ holds with $\mathfrak{g} = 0$ and one of the additional structure assumptions (1) or (2). There exists $\lambda^{**} \in \mathbb{R}$ such that for any $\lambda > \lambda^{**}$, $G \in L_m^p(\mathcal{O})$ and $\mathfrak{g} \in L_{m_\Sigma}^p(\Sigma_+)$, there exists a unique solution $g \in L_{m_\Sigma}^p(\mathcal{O})$ satisfying the transport equation (8.70) in the renormalized sense and some additional a priori estimates listed during the proof.

*Proof of Proposition 8.26.* We just explain the main steps. We first establish an a priori estimate. We observe that any solution $g$ to the stationary problem (8.70) (at least formally) satisfies

$$ (8.71) \quad \int_{\mathcal{O}} |g|^p m^p (\lambda + \varpi) + \frac{1}{p} \int_{\Sigma_+} |\gamma_+ g|^p m_\Sigma^p = \int_{\mathcal{O}} (\mathcal{H}[g] + G) g |g|^{p-2} m^p + \frac{1}{p} \int_{\Sigma_-} |\mathcal{R}[g, \gamma_+ g]|^p m_\Sigma^p. $$

We then only consider the case $\mathfrak{g} = 0$. Repeating the same computations as in Step 1 of the proof of Proposition 8.23 and with the same notations, we get

$$ (8.72) \qquad p(\lambda - \kappa) \|g\|_{L_m^p}^p + \vartheta_{\mathcal{O}} \|g\|_{L_{m_\Sigma}^p}^p + \vartheta_{\Sigma} \|\gamma_+ g\|_{L_{m_\Sigma}^p}^p \le C_{\mathcal{O}} \|G\|_{L_m^p}^p. $$

For $\lambda > \lambda^{**} := \max(\kappa, \lambda_p^*)$ and $G \ge 0$, we next consider the sequence $(h_k)$ in $L_{m_\Sigma}^p$ defined iteratively as the solution given by Lemma 8.13 to

$$ \begin{cases} \lambda h_k + a \cdot \nabla h_k + bh_k = \mathcal{H}[h_{k-1}] + G & \text{on } \mathcal{O}, \\ \gamma_- h_k = \mathcal{R}[h_{k-1}, \gamma_+ h_{k-1}] & \text{on } \Sigma_-, \end{cases} $$

for $k \ge 1$ and starting from $h_0 \equiv 0$. We observe that $(h_k)$ is increasing and satisfies the estimate (8.72) where $g$ is replaced by $h_k$. We may pass to the limit in the above equation and estimate and we obtain a renormalized solution $g \in L_{m_\Sigma}^p$ to the transport equation (8.70) and satisfying the estimate (8.72). By linearity, the same holds without sign condition on $G$. Finally, considering the three different cases as in Steps 3, 4 and 5 in Proposition 8.23, we similarly show that $g \in L_{m_0\mathcal{O}}^{p_0}$
Using again (8.45) and (8.35), we may next pass to the limit as $M \to \infty$ in the above equation, and we get

$$ \frac{d}{dt} \int_{\mathcal{O}} |\tilde{g}|^p = -p \int_{\mathcal{O}} |\tilde{g}|^p \varpi + \int_{\mathcal{O}} p \tilde{G} \tilde{g} |\tilde{g}|^{p-2} + \int_{\Sigma} |\gamma \tilde{g}|^p a \cdot n \in L^1(0, T). $$

We deduce that $t \mapsto \|g(t)\|_{L_m^p} = \|\tilde{g}(t)\|_{L^p}$ is continuous. Consider then $t \in [0, T]$ and $t_k \to t$, so that in particular $\|g_{t_k}\|_{L_m^p} \to \|g_t\|_{L_m^p}$ as $k \to \infty$. On the other hand, we have yet established that $\|\beta_0(g_{t_k}) - \beta_0(g_t)\|_{L^1(\mathcal{O} \cap B_R)} \to 0$ as $k \to \infty$ for any $R > 0$. There exists thus a subsequence $(g_{t_{k'}})$ such that $g_{t_{k'}} \to g_t$ a.e. on $\mathcal{O}$. Thanks to Brézis-Lieb theorem [71], we deduce that $g_{t_{k'}} \to g_t$ in $L_m^p$ and it is the whole sequence which converges by uniqueness of the limit. We have thus established $g \in C([0, T]; L_m^p)$ when $p \in [1, \infty)$.

**Step 4. Uniqueness.** Because of Lemma 8.7 and Remark 8.9-(5), we just have to prove that $g \equiv 0$ if $g$ is a renormalized solution associated to vanishing data $g_0 = 0$, $G = 0$ and $g = 0$. When $p \in [1, \infty)$, the previous step implies that

$$ \frac{d}{dt} \int_{\mathcal{O}} |g|^p m^p = \int_{\mathcal{O}} |g|^p m^p \varpi \in L^1(0, T), \quad \int_{\mathcal{O}} |g(0)|^p m^p = 0, $$

and together with the Gronwall lemma, we deduce that $g = 0$. The case $p = \infty$ may be tackled thanks to a duality argument exactly as in the proof of Lemma 8.14. $\square$

**Corollary 8.18.** The semigroup $S_b$ defined by (8.49) extends to a positive semigroup of contractions in $L_m^p$.

*Proof of Corollary 8.18.* We just apply Proposition 8.16 with $G = g = 0$. When $p \in [1, \infty)$, we define in that way a mapping $L_m^p \to C(\mathbb{R}_+; L_m^p)$, $g_0 \mapsto g$, where $g$ denotes the unique renormalized solution. Defining then $S(t)g_0 := g(t)$ we have built a strongly continuous semigroup in $L_m^p$. The case $p = \infty$ is identical, except the fact that the semigroup is only weak $\ast\sigma(L_m^\infty, L_{m-1}^1)$ continuous. The positivity has been established in Proposition 8.16 and the contraction property comes from the estimates (8.45) and (8.46). $\square$

**Remark 8.19.** It is worth emphasizing that in Bardos [37] the semigroup is defined by its representation formula for smooth data and by Hille-Yosida theory for $L^2$ data. Here we proceed in another way, by rather following [138, 270, 260].

**8.3. Optimal weighted trace theorem and transport equation with reflection at the boundary.** We define the functions $\tau^\pm$ as the solutions to

$$ (8.57) \qquad \lambda_0 \tau^\pm \mp a \cdot \nabla \tau^\pm = 1 \quad \text{in } \mathcal{D}'(\mathcal{O}), \quad \gamma_\pm \tau^\pm = 0 \quad \text{on } \Sigma_\pm, \\ \text{with } \lambda_0 := 1 + \|\text{div}a\|_{L^\infty}. $$

**Lemma 8.20.** Each of the two equations (8.57) has a unique solution $\tau^\pm \in L^\infty(\mathcal{O})$ and
$$ 0 < \tau^\pm \le 1 \quad \text{a.e. in } \mathcal{O}, \quad 0 < \gamma_\pm \tau^\pm \le 1 \quad \text{a.e. on } \Sigma_\mp. $$

*Proof of Lemma 8.20.* We follow a similar proof as in [70, Prop. 5.1] (see also [270, Sec. 5]). We only deals with $\tau^-$ since the case of $\tau^+$ can be handled in the same way. The existence of $\tau^- \in L^\infty$, its non negativity and the upperbound are consequences of Lemma 8.12 while the uniqueness is ensured by Lemma 8.14. In order to prove the strict positivity we argue as follows. We first fix $A \in \mathcal{O}$, $|A| \in (0, \infty)$ and we solve

$$ \lambda_0 \varphi - \text{div}(a\varphi) = \mathbf{1}_A \quad \text{in } \mathcal{D}'(\mathcal{O}), \quad \gamma_+ \varphi = 0 \quad \text{on } \Sigma_+, $$

for which there exists a unique solution $\varphi \in L^1(\mathcal{O})$ thanks to Lemma 8.12 and Lemma 8.14, which furthermore satisfies $\varphi \ge 0$ and $\varphi \ne 0$. We observe that $\tau^- \varphi \in L^1(\mathcal{O})$ satisfies

$$ \text{div}(a\tau^-\varphi) = \varphi - \tau \mathbf{1}_A \quad \text{in } \mathcal{O}, \quad \gamma(\tau^-\varphi) = 0 \quad \text{on } \Sigma \setminus \Sigma_0. $$

Thanks to the Green formula, first written for $\beta_\delta(\tau^-\varphi)$ and next passing to the limit $\delta \to 0$, we deduce

$$ 0 = \int_{\Sigma} \gamma(\tau^{-} \varphi) a \cdot nd\sigma_y = \int_{\mathcal{O}} \text{div}(a\tau^{-} \varphi) dy = \int_{\mathcal{O}} \varphi - \int_{A} \tau_{-} dy, $$
and $\gamma_+g \in L_{m_{0\Sigma}}^{p_0}$ for suitable exponent $p_0 \in [1, p]$ and weight function $m_0$. For two such solutions $g_i$ to (8.70), the function $g := g_2 - g_1$ is also a renormalized solution to (8.70) for which we may justify the identity (8.71) with $p = p_0$, $m = m_0$, $G = 0$. We thus deduce that (8.72) holds with $p = p_0$, $m = m_0$, $G = 0$, and we conclude that $g = 0$, what ends the proof of the uniqueness. $\square$

### 8.4. Characteristics.

In this section we come back to the characteristics method for the evolution and the stationary transport equation. Our aim is in particular to discuss the representation formula (8.47). We consider a vector field $a : \mathcal{O} \to \mathbb{R}^D$ which extends to $\mathbb{R}^D$ and, denoting by the same letter $a$ its extension, we at least assume

$$ (8.73) \qquad a \in W_{\text{loc}}^{1,1}(\mathbb{R}^D), \quad \text{div} a \in L^\infty(\mathbb{R}^D), \quad a/\langle y \rangle \in L^1(\mathbb{R}^D) + L^\infty(\mathbb{R}^D). $$

After DiPerna and Lions [139, 255] (see also [204, Def. 1] or [203, Def. 1]), we introduce the following notion of flow.

**Definition 8.27.** We name almost everywhere flow associated to (8.6) a measurable function $Y : \mathbb{R} \times \mathbb{R}^D \to \mathbb{R}^D$, $(t, y) \mapsto Y_t(y)$, such that

(i) for a.e. $y \in \mathbb{R}^D$, the map $t \mapsto Y_t(y)$ is continuous and

$$ \dot{Y}_t(y) = a(Y_t(y)) \text{ in } \mathcal{D}'(\mathbb{R}), \quad Y_0(y) = y; $$

(ii) for a.e. $y \in \mathbb{R}^D$ and for any $s, t \in \mathbb{R}$, there holds $Y_{s+t}(y) = Y_s(Y_t(y))$;

(iii) there exists $C \ge 0$ such that

$$ (8.74) \qquad \forall t \in \mathbb{R}, \quad e^{-CT}\lambda \le Y(t, \cdot)_{\#}\lambda \le e^{CT}\lambda, $$

where $(Y(t, \cdot)_{\#}\lambda)(A) = \lambda(Y(-t, A))$, $A \subset \mathbb{R}^D$, is the pushforward of the Lebesgue measure $\lambda$.

From [139, Thm. III.1], [9, Thm. 31 & Remark 32] and [204, Sec. 3] (see also Theorem 8.33 below), we know the existence and uniqueness of such an a.e. flow for a satisfying (8.73). In the incompressible case ($\text{div} a = 0$), this one furthermore satisfies:

$$ (8.75) \qquad \int_{\mathbb{R}^D} \varphi(Y_t(y)) e^{\int_0^t (\text{div} a)(Y_\tau(y)) d\tau} dy = \int_{\mathbb{R}^D} \varphi(y) dy, $$

for any $t \in \mathbb{R}$ and any $\varphi \in L_c^\infty(\mathbb{R}^D)$, the space of $L^\infty$ functions with compact support. In the compressible case ($\text{div} a \ne 0$) and when $a$ only satisfies (8.73), it seems not clear that [139, Thm. III.2] or [9, Thm. 31 & Remark 32] provides an a.e. flow such that (8.75) holds. In that general case, the volume identity (8.75) must be replaced by the volume two sides estimate (or nearly-incompressible condition):

$$ e^{-t \|\text{div} a\|_{\infty}} \int_{\mathbb{R}^D} \varphi(y) dy \le \int_{\mathbb{R}^D} \varphi(Y_t(y)) dy \le e^{t \|\text{div} a\|_{\infty}} \int_{\mathbb{R}^D} \varphi(y) dy, $$

for any $0 \le \varphi \in L_c^\infty(\mathbb{R}^D)$ and $t \in \mathbb{R}$, what is nothing but (8.74) with $C := \|\text{div} a\|_{\infty}$. It is however quite straightforward to prove from [139, 9] that the a.e. flows $Y$ satisfies (8.75) when $a$ additionally satisfies $\text{div} a \in C(\mathbb{R}^D)$. One possible way to construct the a.e. flow $Y$ is to define $Y = Y_t(y)$ as the unique renormalized solution in $C(\mathbb{R}; L)$ to the transport equation

$$ \partial_t Y - a \cdot \nabla_y Y = 0 \text{ on } \mathbb{R} \times \mathbb{R}^D, \quad Y_0(y) = y \text{ on } \mathbb{R}^D, $$

or more explicitly

$$ \frac{\partial}{\partial t} \beta(Y) - a(Y) \cdot \nabla \beta(Y) = 0 \text{ on } \mathbb{R} \times \mathbb{R}^D, \quad \beta(Y_0) = \beta(y) \text{ on } \mathbb{R}^D, $$

in the distributional sense for any $\beta \in C^1(\mathbb{R}^D, \mathbb{R})$ such that $\beta$ and $|\nabla \beta(z)|(1+|z|)$ are uniformly bounded on $\mathbb{R}^D$. In particular, for any $g_0 \in C^1(\mathbb{R}^D)$ and next for any $g_0 \in L^0(\mathbb{R}^D)$ the function $g^\#(t, y) := g_0(Y_{-t}(y))$ is the unique renormalized solution to the transport equation

$$ \partial_t g^\# + a \cdot \nabla g^\# = 0 \text{ on } \mathbb{R} \times \mathbb{R}^D, \quad g^\#(0, \cdot) = g_0 \text{ on } \mathbb{R}^D. $$

We introduce some notations. We denote $y \in \mathcal{Y}$ if (i) and (ii) hold. In particular, $\mathcal{Y}$ is a measurable subset of $\mathbb{R}^D$ and $|\mathbb{R}^D \setminus \mathcal{Y}| = 0$. Because of (i), for $y \in \mathcal{V} := \mathcal{O} \cap \mathcal{Y}$, we may define the backward exit time

$$ t_b(y) := \sup\{\tau > 0 \mid Y_{-t}(y) \in \mathcal{O}, \forall s \in [0, \tau]\} \in (0, +\infty], $$
so that the last integral does not vanish. This being true for any $A \subset \mathcal{O}$, we get $\tau^{-} > 0$ a.e. on $\mathcal{O}$.
For $A \subset \Sigma_{+}$ such that

$$0 < \int_{A} (a \cdot n)_{+} d\sigma_{y} < \infty,$$

we solve

$$\lambda_{0}\varphi - \operatorname{div}(a\varphi) = 0 \quad \text{in } \mathcal{D}'(\mathcal{O}), \quad \gamma_{+}\varphi = \mathbf{1}_{A} \quad \text{on } \Sigma_{+},$$

thanks to Lemma 8.12 and Lemma 8.14, and we get a unique solution $0 \le \varphi \in L^{1}(\mathcal{O})$ such that $\varphi \neq 0$. The Green formula again implies

$$\int_{A} \gamma \tau^{-} (a \cdot n)_{+} d\sigma_{y} = \int_{\mathcal{O}} \operatorname{div}(a \tau^{-} \varphi) dy = \int_{\mathcal{O}} \varphi,$$

so that the first integral does not vanish. This being true for any $A \subset \Sigma_{+}$, we conclude that $\gamma_{+} \tau^{-} > 0$ a.e. on $\Sigma_{+}$. $\square$

**Lemma 8.21 (Optimal weight).** We assume that a satisfies (8.33) as well as $a \in W_{\text{loc}}^{1,p'}(\bar{\mathcal{O}})$ for some $1 \le p < \infty$. For any $g \in L^{p}(\mathcal{O})$ satisfying (8.9) in the distributional sense with $G \in L^{p}(\mathcal{O})$, the associated trace function $\gamma g$ defined in Theorem 8.8 satisfies

$$\gamma g \in L^p(\Sigma, |n \cdot a| \tau d\sigma).$$

*Proof of Lemma 8.21.* One fixes $\beta_M(z) = (|z| \wedge M)^p$. From the DiPerna-Lions renormalizing theory, we have

$$a \cdot \nabla (\beta_M(g) \tau^+) = \beta'_M(g) G \tau^+ + \beta_M(g) (\tau^+ - 1) \quad \text{in } \mathcal{D}'(\mathcal{O}).$$

Because $\beta_M(g_\varepsilon) \tau^+ \in L^1(\mathcal{O}) \cap L^\infty(\mathcal{O})$ and $a/\langle y \rangle \in L^1 + L^\infty$, we may use the Green formula (8.21) with $\phi \equiv 1$, and we get

$$\int_{\Sigma_{-}} \beta_{M}(\gamma g) \tau^{+} |n \cdot a| d\sigma = \int_{\mathcal{O}} \{(\text{div}a)\beta_{M}(g) \tau^{+} - \beta'_{M}(g) G \tau^{+} + \beta_{M}(g)(\tau^{+} - 1)\} \\ \lesssim \|g\|_{L^p}^{p-1} \{\|g\|_{L^p} + \|G\|_{L^p}\}.$$

Passing to the limit $M \to \infty$, we obtain $\gamma_- g \in L^p(\Sigma_-, |n \cdot a| \tau d\sigma)$. In a very same way, we prove $\gamma_+ g \in L^p(\Sigma_+, |n \cdot a| \tau d\sigma)$. $\square$

We give now a second version of an existence result in a $L^p$ framework with optimal assumption on the boundary condition in the sense that it is reverse with respect to Lemma 8.21. That also a posteriori justifies that Lemma 8.21 provides the optimal trace result in term of weight function on the boundary.

**Lemma 8.22 (Existence in $L^p$ - optimal assumption).** We make the same assumption on $a$, $b$ and $p$ as in Lemma 8.13. For any $\lambda > \lambda_{a,b,p} + 1/p$ and any given functions $G \in L^p(\mathcal{O})$ and $g \in L^p(\Sigma_-, \tau|a \cdot n|d\sigma)$, there exists $g \in L^p(\mathcal{O})$ solution to (8.29).

*Proof of Lemma 8.22.* We only sketch the proof in the case of equation (8.29), arguing along the lines of Lemma 8.12. We start with an a priori estimate. Observing that

$$\operatorname{div}(a\tau^{+}|g|^{p}) = (\operatorname{div}a)\tau^{+}g^{p} + (\tau^{+} - 1)g^{p} + p\tau^{+}(Gg|g|^{p-2} - b|g|^{p} - \lambda|g|^{p}),$$

we have

$$\int_{\mathcal{O}} |g|^p \{1 + p\tau^+(\lambda + b - \frac{1}{p}\operatorname{div}a - \frac{1}{p}\} = \int_{\Sigma_{-}} |\gamma_- g|^p \tau |a \cdot n| d\sigma + \int_{\mathcal{O}} Gg|g|^{p-2}\tau^+.$$

Using the condition on $\lambda$, the property $0 \le \tau^+ \le 1$ and the Young inequality, we deduce

$$\frac{1}{p} \int_{\mathcal{O}} |g|^p \le \int_{\Sigma_{-}} |\gamma_{-} g|^p \tau |a \cdot n| d\sigma + \frac{1}{p} \int_{\mathcal{O}} |G|^p.$$

We conclude in a similar way as in the proof of Lemma 8.13. $\square$
the subset $\mathcal{V}_b := \{y \in \mathcal{V}; t_b(y) < +\infty\}$ and the associated entering position

$$y_b(y) := Y_{-t_b(y)}(y) \text{ when } y \in \mathcal{V}_b.$$

We observe that the function $t_b: \mathcal{V} \to (0, +\infty]$ is measurable, $\mathcal{V}_b$ is a mesurable subset of $\mathcal{O}$ and

$$ (8.76) \qquad t_b(Y_s(y)) = t_b(y) + s, \quad y_b(Y_s(y)) = y_b(y), \quad \forall y \in \mathcal{V}_b, \forall s \in [0, t_b(y)). $$

$$ (8.77) \qquad \mathrm{meas}(\{y \in \mathcal{O}; t_b(y) = t\}) = 0, \quad \forall t > 0. $$

The properties (8.76) are straightforward while (8.77) is a consequence of the fact that $\{y \in \mathcal{V}; t_b(y) = t\} \subset Y_t(\Sigma)$ and of the nearly-incompressible condition (8.74).

We now introduce the following first regularity assumption on $a$ at the boundary

$$ (8.78) \qquad \forall y_0 \in \Sigma, y \mapsto a(y) \cdot n(y_0) \text{ is continuous on } \bar{\mathcal{O}}. $$

Let us present two examples.

- It may happen that $\mathcal{V}_b = \emptyset$. For instance, choosing $\mathcal{O} := \{y \in \mathbb{R}^2; |y| < 1\}$ the unit disk of the plane and $a(y) := |y|y^\perp \in C^{0,1}(\mathbb{R}^2; \mathbb{R}^2)$, $y^\perp := (y_2, -y_1)$, we have $\mathrm{div} a = 0$ and $a(y) \cdot n(y) = y^\perp \cdot y = 0$ for any $y \in \mathbb{R}^2$, so that the flows do not encounter the boundary set $\Sigma = \{y \in \mathbb{R}^2; |y| = 1\}$. In that situation $\mathcal{V} = \mathcal{O}$ and $\mathcal{V}_b = \emptyset$.

- In the kinetic case, namely $y = (x, v) \in \mathcal{O} := \Omega \times \mathbb{R}^d$, $\Omega \subset \mathbb{R}^d$ an open set with smooth boundary with unit normal outward $\nu_x$, so that $n(y) = (\nu_x, 0)$, and $a(y) = (v, F(x, v))$, we have $\mathrm{div} a = \mathrm{div}_v F$ and $\mathrm{div}_v F = 0$ when $F = E(x) + v \wedge B(x)$, and we have $a(y) \cdot n(y_0) = v \cdot \nu_{x_0}$ which is a smooth function on $\bar{\mathcal{O}} \times \Sigma$.

**Lemma 8.28.** *Under the condition (8.78), the mapping $y_b: \mathcal{V}_b \to \Sigma_- \cup \Sigma_0$ is measurable.*

*Proof of Lemma 8.28.* From the very definitions and composition rules, we have $y_b: \mathcal{V}_b \to \Sigma \cap \mathcal{Y}$ is mesurable. Take $y \in \mathcal{V}_b$, denote $y_0 := y_b(y)$ and consider a sequence $t_k \searrow 0$ so that $Y_{t_k}(y_0) \to y_0$ and $Y_{t_k}(y_0) \in \mathcal{O}$ for any $k \ge 1$. From (8.78), we deduce

$$ 0 \ge \limsup_{k \to \infty} \frac{Y_{t_k}(y_0) - y_0}{t_k} \cdot n(y_0) = \lim_{k \to 0} \frac{1}{t_k} \int_0^{t_k} a(Y_s(y_0)) \cdot n(y_0) ds = a(y_0) \cdot n(y_0), $$

which means that $y_0 \in \Sigma_- \cup \Sigma_0$. $\square$

For further references, we introduce the following second additional mild regularity assumption on $a$ in the domain

$$ (8.79) \qquad a \in L_{\mathrm{loc}}^{\infty}(\bar{\mathcal{O}}), \quad (a(y) \cdot y)_+ \lesssim \langle y \rangle^2. $$

This one may in fact replace the last boundedness condition on $a$ in (8.73). We establish a technical result which will be useful in the sequel.

**Lemma 8.29.** *Under assumptions (8.78) and (8.79), the following hold:*

(1) For any $R_0, T > 0$, there exists $R_T, L_T > 0$ such that for any $y \in B_{R_0}$ there hold

$$ (8.80) \qquad \sup_{[-T,T]} |Y_t(y)| \le R_T \quad \text{and} \quad |Y_{t_2}(y) - Y_{t_1}(y)| \le L_T |t_2 - t_1|, \quad \forall t_1, t_2 \in [-T,T]. $$

(2) For a sequence $(y_\epsilon)$ of $\mathcal{V}$ such that $y_\epsilon \to y_0 \in \Sigma_- \cap \mathcal{Y}$, we have $t_b(y_\epsilon) \to 0$ and $y_b(y_\epsilon) \to y_0$.

*Proof of Lemma 8.29.* Proof of (1). Take $y \in \mathcal{V} \cap B_{R_0}$. On the one hand, from (8.79), we have

$$ \langle Y_t(y) \rangle^2 - \langle y \rangle^2 = 2 \int_0^t a(Y_s(y)) \cdot Y_s(y) ds \le C \int_0^t \langle Y_s(y) \rangle^2 ds, $$

and we conclude to $Y_t(y) \in B_{R_T}$ thanks to the Gronwall lemma. As a consequence, we have

$$ |Y_{t_2}(y) - Y_{t_1}(y)| \le |t_2 - t_1| \|a\|_{L^\infty(B_{R_T})}, \quad \forall t_i \in [-T, T]. $$

*Proof of (2).* Assume by contradiction that $\limsup t_b(y_\epsilon) \ge \tau > 0$ and set $T := \tau + 1$. By assumption, there exists $R_0 > 0$ such that $y_\epsilon \in B_{R_0}$ and thus by step 1, (8.80) holds uniformly in $\epsilon \in (0, 1]$. Thanks to the Ascoli Theorem and the contradiction hypothesis, there exists $\epsilon_k \to 0$ such that $t_b(y_{\epsilon_k}) \ge \tau$ and a there exists $Y \in C([-T, T])$ such that $Y_\bullet(y_{\epsilon_k}) \to Y_\bullet$ in $C([-T, T])$. Next, passing to the limit in the conditions

$$ Y_{-t}(y_{\epsilon_k}) \in \mathcal{O} \quad \text{and} \quad Y_{-t}(y_{\epsilon_k}) - y_{\epsilon_k} = - \int_0^t a(Y_{-s}(y_{\epsilon_k})) ds, $$
We consider now the time dependent transport equation with positive abstract kernels

$$ (8.58) \qquad \begin{cases} \frac{\partial g}{\partial t} + a \cdot \nabla g + bg = \mathcal{H}[g] + G \quad \text{on } (0, T) \times \mathcal{O}, \\ \gamma_-g = \mathcal{R}[g, \gamma_+g] + \mathfrak{g} \quad \text{on } (0, T) \times \Sigma_-, \quad g(0, \cdot) = \mathfrak{g}_0 \quad \text{on } \mathcal{O}, \end{cases} $$

with notations introduced at the beginning of the Section. We will work in a weighted Lebesgue space $L_m^p$ with the same conditions on $p, m, a$ and $b$ as introduced at the beginning of Section 8.2. On the other hand, we assume

$$ (8.59) \qquad \mathcal{H} : L_{m\mathcal{O}}^p(\mathcal{O}) \to L_{\tilde{m}\mathcal{O}}^p(\mathcal{O}) \text{ linear and positive,} $$

$$ (8.60) \qquad \mathcal{R} : L_{m\mathcal{O}}^p(\mathcal{O}) \times L_{m\Sigma}^p(\Sigma_+) \to L_{m\Sigma}^p(\Sigma_-) \text{ linear and positive in each variable,} $$

where we recall that the weight functions $m_\mathcal{O}, \tilde{m}_\mathcal{O}$ and $m_\Sigma$ have been defined in (8.38). More precisely, recalling that $\mathcal{R} = \mathcal{R}_\mathcal{O} + \mathcal{R}_\Sigma$ with $\mathcal{R}_\mathcal{O}$ and $\mathcal{R}_\Sigma$ defined by (8.4), we assume

$$ (8.61) \qquad \|\mathcal{H}[g]\|_{L_{\tilde{m}\mathcal{O}}^p}^p \le \alpha_\mathcal{X} \|g\|_{L_{m\mathcal{O}}^p}^p + M_\mathcal{X} \|g\|_{L_m^p}^p, $$

$$ (8.62) \qquad \|\mathcal{R}_\mathcal{O}[g]\|_{L_{m\Sigma}^p}^p \le \alpha_\mathcal{X} \|g\|_{L_{m\mathcal{O}}^p}^p + M_\mathcal{X} \|g\|_{L_m^p}^p, \quad \|\mathcal{R}_\Sigma[h]\|_{L_{m\Sigma}^p}^p \le \beta_\mathcal{X} \|h\|_{L_{m\Sigma}^p}^p, $$

with $\alpha_\mathcal{X}, \alpha_\mathcal{R}, \beta_\mathcal{R} \in [0, 1], M_\mathcal{X}, M_\mathcal{R} \ge 0$ and

$$ (8.63) \qquad \vartheta_\mathcal{O} := (1 - \alpha_\mathcal{X} - \alpha_\mathcal{R})/2 > 0, \quad \vartheta_\Sigma := 1 - \beta_\mathcal{R} \ge 0. $$

Let us emphasize that when $p = 1$, the assumption (8.61) is equivalent to the Lyapunov type condition

$$ \mathcal{H}^*[m] \le \alpha_\mathcal{X} \varpi_+ m + M_\mathcal{X} m. $$

**Proposition 8.23.** We assume that $a, b, \mathcal{H}$ and $\mathcal{R}$ satisfy the conditions (8.35), (8.59), (8.60), (8.61), (8.62), and (8.63) for some weight function $m: \bar{\mathcal{O}} \to [1, \infty)$ and some exponent $p \in [1, \infty)$. We consider some data $g_0 \in L_m^{p_0}(\mathcal{O})$, $G \in L_m^p((0, T) \times \mathcal{O})$ and $\mathfrak{g} \in L_{m\Sigma}^p((0, T) \times \Sigma_+)$ with either
(1) $\beta_\mathcal{R} \in [0, 1]$;

or $\beta_\mathcal{R} = 1$. In the latter case, we assume that $\mathfrak{g} = 0$ and we make one of the following additional structural assumptions

(2) there exist an exponent $p_0 \in [1, p]$ and a weight function $m_0$ such that $\mathcal{H}$ and $\mathcal{R}$ satisfy (8.59) in $L_{m_0}^{p_0}$ and (8.60) in $L_{m_0\mathcal{O}}^{p_0}(\mathcal{O}) \times L_{m_0\Sigma}^{p_0}(\Sigma_+)$, with obvious definitions for the weight functions $m_{0\mathcal{O}}$ and $m_{0\Sigma}$, and with $L_m^p \subset L_{m_0}^{p_0}$, $L_{m\mathcal{O}}^p \subset L_{m_0\mathcal{O}}^{p_0}$, $L_{\tilde{m}\Sigma}^p \subset L_{m_0\Sigma}^{p_0}$, where $\tilde{m}_\Sigma := m(\tau^+ a \cdot n)^{2/p}$;

(3) $p = 1$ and $\mathcal{R}_\Sigma$ is diffusive, namely $\mathcal{R}_\Sigma^*[\tau^+ m_\Sigma] \ge c_\Sigma m_\Sigma$ a.e. on $\Sigma_+$ with $c_\Sigma > 0$. In the above three cases, there exists a unique solution $g \in L^\infty(0, T; L_m^p(\mathcal{O})) \cap C([0, T]; L_{m_0}^{p_0}(\mathcal{O}))$ satisfying the transport equation (8.58) in the renormalized sense as well as $g \in L^{p_0}(0, T; L_{m_0\mathcal{O}}^{p_0}(\mathcal{O}))$ and $\gamma g \in L^{p_0}(0, T; L_{m_0\mathcal{O}}^{p_0}(\Sigma))$, with $p_0 = p$ and $m_0 = m$ in the first and the third cases.

**Remark 8.24.** (1) The above result extends some previous results initiated by Bardos in [37, Chap. III] and Beals et al in [43, Thm. 1&7], where however only the kinetic case were considered. We refer to Section 10 for a discussion about that important model.

(2) When $\beta_\mathcal{R} = 1$, the existence part of the above result still holds (without any additional structural assumption).

(3) Similarly as observed in Remark 8.17, a weak maximum principle holds: $g_1 \le g_2$ if $g_1$ and $g_2$ are the renormalized solutions to two transport equations (8.58) such that (with obvious notations) $b_1 \ge b_2$, $\mathcal{H}_1 \le \mathcal{H}_2$, $\mathcal{R}_1 \le \mathcal{R}_2$, $g_{01} \le g_{02}$, $G_{01} \le G_{02}$ and $\mathfrak{g}_{01} \le \mathfrak{g}_{02}$. That is an immediate consequence of the way we build the solutions $g_i$ thanks to the iterative scheme we present in Step 2 of the proof of Proposition 8.23.

(4) Another immediate consequence of the iterative way of building the solution, together with the fact that the characteristics representation (8.47) in the very first step of the construction, is the validity of the Duhamel formula

$$ S_\mathcal{L} = S_\mathcal{B} + S_\mathcal{B}A * S_\mathcal{L} $$

if we denote by $S_\mathcal{L}$ the semigroup generated by the transport equation (8.58) with $G = \mathfrak{g} = 0$, by $S_\mathcal{B}$ the semigroup when additionally $\mathcal{H} = 0$, and $A\mathfrak{f} = \mathcal{H}[f]$.
for any $k \ge 1$ and any $t \in [0, \tau]$, we get

$$Y_{-t} \in \bar{\mathcal{O}} \quad \text{and} \quad Y_{-t} - y_0 = - \int_0^t a(Y_{-s}) ds.$$

We deduce

$$0 \ge \limsup_{t \to 0} \frac{Y_{-t} - y_0}{t} \cdot n(y_0) = \lim_{t \to 0} -\frac{1}{t} \int_0^t a(Y_{-s}) \cdot n(y_0) ds = -a(y_0) \cdot n(y_0),$$

which is in contradiction with the hypothesis $y_0 \in \Sigma_-$. Now, we may estimate

$$|y_b(y_\epsilon) - y_0| \le |y_\epsilon - y_0| + \int_0^{t_b(y_\epsilon)} |a(Y_s(y_\epsilon))| ds \to 0,$$

as $\epsilon \to 0$, as a consequence of the convergence $t_b(y_\epsilon) \to 0$, the first estimate in (8.80) and the first condition in (8.79). $\square$

We reformulate some “space continuity” of solution to the transport equation results picked up in [70, Sec. 7]. Defining

$$\mathcal{O}_\alpha := \{y \in \mathcal{O}; \delta(y) > \alpha\}, \quad \Sigma_\alpha := \{y \in \mathcal{O}; \delta(y) = \alpha\} = \partial \mathcal{O}_\alpha,$$

we know from [70, Sec. 2], that there exists $\alpha_\mathcal{O} > 0$ such that for any $\alpha \in (0, \alpha_\mathcal{O})$, the mapping

$$\theta_\alpha : \Sigma \to \Sigma_\alpha, \quad \theta_\alpha(z) := z - \alpha n(z)$$

is an isomorphim with associated jacobian function $J_\alpha$ and

$$(8.81) \qquad \int_{\Sigma_\alpha} h(z') d\sigma_\alpha(z') = \int_{\Sigma} h \circ \theta_\alpha(z) J_\alpha(z) d\sigma_z,$$

$$(8.82) \qquad \int_{\mathcal{O} \setminus \mathcal{O}_\alpha} g(y) dy = \int_0^\alpha \int_{\Sigma} g \circ \theta_{\alpha'}(z) J_{\alpha'}(z) d\sigma_z d\alpha',$$

for any $h \in L^1(\Sigma_\alpha)$ and $g \in L^1(\mathcal{O} \setminus \mathcal{O}_\alpha)$, where $d\sigma_\alpha$ denotes the Lebesgue measure on $\Sigma_\alpha$ and where the jacobian function $J_\alpha$ satisfies $1/2 \le J_\alpha \le 3/2$ as well as $J_\alpha \to I$ as $\alpha \to 0$.

**Lemma 8.30.** For any $g \in L^\infty(\mathcal{O})$ satisfying $a \cdot \nabla g \in L^1(\mathcal{O})$, we have

$$(8.83) \qquad g = \gamma_\alpha g \quad \text{a.e. on } \Sigma_\alpha \setminus \{a \cdot n = 0\} \quad \text{for a.e. } \alpha \in [0, \alpha_\mathcal{O}],$$

where we denote by $\gamma_\alpha g$ the trace of $g$ on $\Sigma_\alpha$, and

$$(8.84) \qquad \gamma_\alpha g \circ \theta_\alpha \to \gamma g \quad \text{as } \alpha \to 0, \quad \text{a.e. on } \Sigma \setminus \Sigma_0.$$

**Proof of Lemma 8.30.** For $\varphi \in C_c(\bar{\mathcal{O}}) \cap W^{1,1}(\mathcal{O})$ and $\beta \in C^1(\mathbb{R})$, the renormalized Green formula

(8.24) writes

$$\int_{\Sigma} \varphi \beta(\gamma g) a \cdot n \, d\sigma = \int_{\Sigma_\alpha} \varphi \beta(\gamma_\alpha g) a \cdot n \, d\sigma_\alpha + \int_{\mathcal{O} \setminus \mathcal{O}_\alpha} [\text{div}(a\varphi)\beta(g) + \varphi\beta'(g)a \cdot \nabla g] \, dy,$$

and thus

$$(8.85) \qquad \int_{\Sigma} \varphi \beta(\gamma g) a \cdot n \, d\sigma = \lim_{\alpha \to 0} \int_{\Sigma_\alpha} \varphi \beta(\gamma_\alpha g) a \cdot n \, d\sigma_\alpha.$$

Denoting $\psi := (1 - (\delta(x) - \alpha)/s)_+$, $\alpha + s \in (\alpha, \alpha_\mathcal{O})$, observing that $\psi|_{\Sigma_\alpha} = 1$, recalling that $n = -\nabla\delta$ and using $\varphi\psi$ as a test function, we similarly have

$$\int_{\Sigma_\alpha} \varphi \beta(\gamma_\alpha g) a \cdot n \, d\sigma_\alpha = \frac{1}{h} \int_{\mathcal{O}_\alpha \setminus \mathcal{O}_{\alpha+s}} \varphi \beta(g) a \cdot n \, dy + \int_{\mathcal{O}_\alpha \setminus \mathcal{O}_{\alpha+s}} [\text{div}(a\varphi)\beta(g) + \varphi\beta'(g)a \cdot \nabla g] \psi \, dy,$$

so that

$$\int_{\Sigma_\alpha} \varphi \beta(\gamma_\alpha g) a \cdot n \, d\sigma_\alpha = \lim_{s \to 0} \frac{1}{s} \int_{\mathcal{O}_\alpha \setminus \mathcal{O}_{\alpha+s}} \varphi \beta(g) a \cdot n \, dy.$$

We immediately deduce

$$\int_{\Sigma_\alpha} \varphi \beta(\gamma_\alpha g) a \cdot n \, d\sigma_\alpha = \frac{d}{d\alpha} \int_{\mathcal{O}_\alpha} \varphi \beta(g) a \cdot n \, dy$$
Proof of Proposition 8.23. We split the proof into five steps.

Step 1. A priori estimates. For a positive solution, we formally compute

$$
(8.64) \quad \frac{1}{p} \frac{d}{dt} \int g^p m^p = \frac{1}{p} \int_{\Sigma_{-}} (\mathcal{R}[g, \gamma_+ g] + \mathfrak{g})^p m_{\Sigma}^p d\sigma_y - \frac{1}{p} \int_{\Sigma_{+}} (\gamma_+ g)^p m_{\Sigma}^p d\sigma_y \\
\qquad + \int_{\mathcal{O}} \{g^{p-1}(\mathcal{H}[g] + G) - g^p \varpi\} m^p.
$$

Using the Young inequality and (8.61), we have

$$
\begin{aligned}
\int_{\mathcal{O}} g^{p-1} \mathcal{H}[g] m^p &\le \frac{1}{p'} \int_{\mathcal{O}} g^p \langle \bar{\varpi}_+ \rangle m^p + \frac{1}{p} \int_{\mathcal{O}} \mathcal{H}[g]^p \langle \bar{\varpi}_+ \rangle^{-p/p'} m^p \\
&\le \left( \frac{1}{p'} + \frac{\alpha_{\mathcal{H}}}{p} \right) \int_{\mathcal{O}} g^p \langle \bar{\varpi}_+ \rangle m^p + \frac{M_{\mathcal{H}}}{p} \int_{\mathcal{O}} g^p m^p.
\end{aligned}
$$

• When $\mathfrak{g} = 0$, using also (8.62) and once more the Young inequality, we then have

$$
\frac{1}{p} \frac{d}{dt} \int g^p m^p \le \frac{1}{p} (\beta_{\mathfrak{g}} - 1) \int_{\Sigma_+} (\gamma_+ g)^p m^p a \cdot n + \left( \frac{M_{\mathfrak{g}}}{p} + \frac{M_{\mathfrak{g}}}{p} + \|\langle \bar{\varpi}_- \rangle\|_{L^\infty} \right) \int_{\mathcal{O}} g^p m^p \\
+ \left( \frac{\alpha_{\mathfrak{g}}}{p} + \frac{1}{p'} + \frac{\alpha_{\mathfrak{g}}}{p} + \frac{\varepsilon}{p'} - 1 \right) \int_{\mathcal{O}} g^p \langle \bar{\varpi}_+ \rangle m^p + \frac{\varepsilon^{-p/p'}}{p} \int_{\mathcal{O}} G^p m^p \langle \bar{\varpi}_+ \rangle^{-p/p'}
$$

for any $\varepsilon > 0$. Making the choice $\varepsilon := \vartheta_{\mathcal{O}}p'/p$, we deduce

$$
\frac{d}{dt} \|g\|_{L_m^p} + \vartheta_{\mathcal{O}} \|g\|_{L_{m\mathcal{O}}^p}^p + \vartheta_{\Sigma} \|\gamma_+ g\|_{L_{m\Sigma}^p}^p \le p\kappa \|g\|_{L_m^p}^p + C_{\mathcal{O}} \|G\|_{L_m^p}^p,
$$

with

$$
\kappa := \frac{M_{\mathfrak{g}}}{p} + \frac{M_{\mathfrak{g}}}{p} + \|\langle \bar{\varpi}_- \rangle\|_{L^\infty}, \quad C_{\mathcal{O}} := (\vartheta_{\mathcal{O}}p'/p)^{-p'/p}.
$$

Using the Gronwall lemma, we then obtain

$$
(8.65) \quad \begin{aligned}
\|g(t)\|_{L_m^p}^p + \int_0^t e^{p\kappa(t-s)} (\vartheta_{\mathcal{O}} \|g_s\|_{L_m^p}^p + \vartheta_{\Sigma} \|\gamma_+ g_s\|_{L_{m\Sigma}^p}^p) \, ds \\
\le e^{p\kappa t} \|\mathfrak{g}_0\|_{L_m^p}^p + C_{\mathcal{O}} \int_0^t e^{p\kappa(t-s)} \|G_s\|_{L_{m\mathcal{O}}^p}^p \, ds, \quad \forall t \ge 0.
\end{aligned}
$$

• When $\mathfrak{g} \neq 0$ and thus $\vartheta_{\Sigma} > 0$, we control the ingoing boundary term by

$$
\int_{\Sigma_{-}} (\mathcal{R}[g, \gamma_{+}g] + \mathfrak{g})^{p} m_{\Sigma}^{p} \leq (1 + \varepsilon_{1}) \int_{\Sigma_{-}} \mathcal{R}[g, \gamma_{+}g]^{p} m_{\Sigma}^{p} + C_{\varepsilon_{1}} \int_{\Sigma_{-}} \mathfrak{g}^{p} m_{\Sigma}^{p}, \quad \forall \varepsilon_{1} > 0,
$$

and a very similar computation as above leads to the a priori estimate

$$
(8.66) \quad \begin{aligned}
\|g(t)\|_{L_m^p}^p + \int_0^t e^{p\kappa(t-s)} (\vartheta_{\mathcal{O}} \|g_s\|_{L_m^p}^p + \vartheta_{\Sigma}' \|\gamma_+ g_s\|_{L_{m\Sigma}^p}^p) \, ds \\
\le e^{p\kappa t} \|\mathfrak{g}_0\|_{L_m^p}^p + \int_0^t e^{p\kappa(t-s)} (C_{\mathcal{O}} \|G_s\|_{L_{m\mathcal{O}}^p}^p + C_{\Sigma} \|\mathfrak{g}_s\|_{L_{m\Sigma}^p}^p) \, ds,
\end{aligned}
$$

for any $t \ge 0$, with

$$
\begin{align*}
\vartheta'_{\Sigma} &:= 1 - \beta_{\mathcal{H}}(1 + \varepsilon_1), & \vartheta'_{\mathcal{O}} &:= 1 - \alpha_{\mathcal{H}}(1 + \varepsilon_1) - \alpha_{\mathcal{H}} - \varepsilon \frac{p}{p'}, \\
\kappa &:= \frac{M_{\mathcal{H}}}{p}(1 + \varepsilon_1) + \frac{M_{\mathcal{H}}}{p} + \|\langle \bar{\varpi}_- \rangle\|_{L^\infty}, & C_{\mathcal{O}} &:= \varepsilon^{-p'/p}, & C_{\Sigma} &:= C_{\varepsilon_1},
\end{align*}
$$

and where we have chosen $\varepsilon, \varepsilon_1 > 0$ small enough in such a way that $\vartheta'_{\Sigma} > 0$ and $\vartheta'_{\mathcal{O}} > 0$.

• When $\vartheta_{\Sigma} = 0$ and thus $\mathfrak{g} = 0$, we further multiply the equation by $m^p \tau^{\pm}$, where $\tau^{\pm}$ is defined in (8.57), and integrating, we deduce

$$
\int_0^T \int_{\Sigma_{\mp}} \tau^{\pm} a \cdot n(\gamma g)^p m^p \, d\sigma dt = \left[ \int_{\mathcal{O}} g^p m^p \tau^{\pm} \right]_T^0 \\
+ \int_0^T \int_{\mathcal{O}} pg^{p-1} (\mathcal{H}[g] + G) m^p \tau^{\pm} + \int_0^T \int_{\mathcal{O}} g^p \left( \frac{\operatorname{div}(am^p)}{m^p} + 1 - \lambda_0 \tau^{\pm} - pK m^p \tau^{\pm} \right) \, d\sigma dt.
$$
and next

$$ \int_0^\alpha \int_{\Sigma_{\alpha'}} \varphi\beta(\gamma_{\alpha'}g) a \cdot n \, d\sigma_{\alpha'} d\alpha' = \int_{\mathcal{O} \setminus \mathcal{O}_\alpha} \varphi\beta(g) a \cdot n \, dy. $$

Together with (8.82) and (8.81), we have established

$$ \int_0^\alpha \int_\Sigma [\varphi\beta(\gamma_{\alpha'}g) a \cdot n] \circ \theta_{\alpha'}(z) J_{\alpha'}(z) d\sigma_z d\alpha' = \int_0^\alpha \int_\Sigma [\varphi\beta(g) a \cdot n] \circ \theta_{\alpha'}(z) J_{\alpha'}(z) d\sigma d\alpha', $$

so that

$$ \beta(\gamma_{\alpha'}g) a \cdot n = \beta(g) a \cdot n \quad \text{a.e. on } \Sigma_{\alpha'} \text{ for a.e. } \alpha' \in (0, \alpha), $$

from what (8.83) follows. On the other hand, using (8.81) and (8.85) together, we have

$$ \int_{\Sigma} \varphi\beta(\gamma g) a \cdot n \, d\sigma = \lim_{\alpha \to 0} \int_{\Sigma} [\varphi\beta(\gamma_{\alpha}g) a \cdot n] \circ \theta_{\alpha} J_{\alpha} \, d\sigma_{\alpha}, $$

which implies

$$ (8.86) \qquad [\beta(\gamma_{\alpha}g) a \cdot n] \circ \theta_{\alpha} J_{\alpha} \to \beta(\gamma g) a \cdot n \quad * \sigma(L_{\text{loc}}^{\infty}(\Sigma), L_{\text{c}}^{1}(\Sigma)). $$

Repeating the same argument with $g := a \cdot n$ and using that $J_{\alpha} \to I$ uniformly, we get

$$ [\beta(a \cdot n) a \cdot n] \circ \theta_{\alpha} \to \beta(a \cdot n) a \cdot n \quad * \sigma(L_{\text{loc}}^{\infty}(\Sigma), L_{\text{c}}^{1}(\Sigma)), $$

for any $\beta \in C^1(\mathbb{R})$. Choosing $\beta(s) = 1$ and $\beta(s) = s$, we classically deduce that $[a \cdot n] \circ \theta_{\alpha} \to a \cdot n$ a.e. on $\Sigma$. We finally conclude to (8.84) by gathering that last information with (8.86) written for $\beta(s) = s$ and $\beta(s) = s^2$. $\square$

**Remark 8.31.** During the proof, we have in fact established that

$$ [0, \alpha_{\mathcal{O}}] \to L^1(\Sigma); \quad \alpha \mapsto [\gamma_{\alpha} g a \cdot n] \circ \theta_{\alpha} $$

is continuous.

**Lemma 8.32.** We make the additional assumptions (8.78) and (8.79). If $Y$ is an almost everywhere flow associated to (8.6), then the function $t_b \in L(\mathcal{O})$ is a renormalized solution to the equation

$$ (8.87) \qquad a \cdot \nabla t_b = 1 \quad \text{in } \mathcal{O}, \quad \gamma_{-} t_b = 0 \quad \text{on } \Sigma_{-} \cap \mathcal{Y}. $$

*Proof of Lemma 8.32. Step 1.* We fix $\beta \in C_s^1(\mathbb{R})$ and we recall that $\beta(t_b(Y_s(y))) = \beta(t_b(y) + s)$ for any $s \in \mathbb{R}$ for a.e. $y \in \mathcal{O}$. For any $\varphi \in C_c^1(\mathcal{O})$, we may compute

$$ \begin{align*} \int_{\mathbb{R}^d} \beta'(t_b(y) + s)\varphi(y) \, dy &= \frac{d}{ds} \int_{\mathbb{R}^d} \beta(t_b(y) + s)\varphi(y) \, dy \\ &= \frac{d}{ds} \int_{\mathbb{R}^d} \beta(t_b(Y_s(y)))\varphi(y) \, dy \\ &= \frac{d}{ds} \int_{\mathbb{R}^d} \beta(t_b(y))\varphi(Y_{-s}(y))e^{-\int_{-s}^0 (\text{div}\alpha)(Y_\tau(y))d\tau} \, dy \\ &= \int_{\mathbb{R}^d} \beta(t_b(y))[-a \cdot \nabla\varphi - (\text{div}\alpha)\varphi](Y_{-s}(y))e^{-\int_{-s}^0 (\text{div}\alpha)(Y_\tau(y))d\tau} \, dy. \end{align*} $$

Taking $s = 0$, we conclude to

$$ \int_{\mathbb{R}^d} \beta'(t_b)\varphi \, dy = \int_{\mathbb{R}^d} \beta(t_b)[-a \cdot \nabla\varphi - (\text{div}\alpha)\varphi] \, dy, $$

which is nothing but the distributional formulation of the equation

$$ a \cdot \nabla \beta(t_b) = \beta'(t_b). $$

That last family of equations is the renormalized formulation of equation (8.87) in the domain.

*Step 2.* Using lemma 8.30 with $g := \beta(t_b)$, we have

$$ \gamma_{\alpha} \beta(t_b) \circ \theta_{\alpha} \to \beta(\gamma t_b) \quad \text{a.e. on } \Sigma. $$

Using Lemma 8.29, we also have

$$ \gamma_{\alpha} \beta(t_b) \circ \theta_{\alpha} \to 0 \quad \text{a.e. on } \Sigma_{-} \cap \mathcal{Y}. $$

Both together, we find $\gamma_{-}t_b = 0$ on $\Sigma_{-} \cap \mathcal{Y}$. $\square$
Together with (8.65) and $\tau^{\pm} \in L^{\infty}(\mathcal{O})$, we obtain

$$ (8.67) \qquad \int_0^T \int_{\Sigma_+} (\gamma_+ g)^p \tau^- m_\Sigma^p \le C_T \left( \|g_0\|_{L_m^p}^p + \|G\|_{L^{p_0}(0,T; L_m^p)}^p \right) $$

and

$$ (8.68) \qquad \int_0^T \int_{\Sigma_-} [\mathcal{R}_\Sigma(\gamma_+g)]^p \tau^+ m_\Sigma^p \le C_T \left( \|g_0\|_{L_m^p}^p + \|G\|_{L^{p_0}(0,T; L_m^p)}^p \right), $$

for some constant $C_T \in (0, \infty)$. In particular, when $p=1$ and $\mathcal{R}_\Sigma$ is diffusive, we have

$$ c_\Sigma \int_0^T \int_{\Sigma_+} (\gamma_+ g) m_\Sigma \le \int_0^T \int_{\Sigma_+} (\gamma_+ g) \mathcal{R}_\Sigma^*(\tau^+ m_\Sigma) = \int_0^T \int_{\Sigma_-} \mathcal{R}_\Sigma(\gamma_+ g) \tau^+ m_\Sigma, $$

and together with (8.68), we deduce the additional estimate

$$ (8.69) \qquad c_\Sigma \int_0^T \int_{\Sigma_+} (\gamma_+ g) m_\Sigma \le C_T \left( \|g_0\|_{L_m^1} + \|G\|_{L^{1}(0,T; L_m^1)} \right). $$

**Step 2. Existence.** As a consequence of these a priori estimates, we may classically build a solution through an iterative scheme. For the sake of brevity, we only consider the (more interesting and more difficult) case $b_\Sigma = 1$ (so that $\vartheta_\Sigma = 0$ and $g = 0$) and $G = 0$. For a given $0 \le g_0 \in L_m^p(\mathcal{O})$, we define a sequence of solution $(h_n)$ starting from $h_0 \equiv 0$ thanks to the recursive definition

$$ \left\{ \begin{array}{l} \displaystyle \frac{\partial h_{n+1}}{\partial t} + a \cdot \nabla h_{n+1} + Kh_{n+1} = \mathcal{K}[h_n] \quad \text{on } (0,T) \times \mathcal{O}, \\ \displaystyle \gamma_- h_{n+1} = \mathcal{R}[h_n, \gamma_+ h_n] \quad \text{on } (0,T) \times \Sigma_-, \quad h_{n+1}(0, \cdot) = \mathfrak{g}_0 \quad \text{on } \mathcal{O}. \end{array} \right. $$

From Proposition 8.16, there exists a unique renormalized solution $h_{n+1} \in C([0, T]; L_m^p(\mathcal{O}))$ to the above equation satisfying the estimate (8.45) with $g := h_{n+1}$, $G := \mathcal{K}[h_n]$ and $g := \mathcal{R}[h_n, \gamma_+ h_n] \in L_{m\Sigma}^p$. We observe that $0 \le h_n \le h_{n+1}$ thanks to the weak maximum principle (see Remark 8.17) and that $h_n$ satisfies the estimates (8.65) and (8.68) where $g$ is replaced by $h_n$. Thanks to the monotonous convergence theorem of Beppo Levi, there exists $g$ satisfying estimates (8.65) and $h_n \to g$ in $L_{m\Sigma}^p((0, T) \times \mathcal{O})$. We may pass to the limit in the equation satisfied by $(h_n)$ and we deduce that $g$ is a renormalized solution to

$$ \frac{\partial g}{\partial t} + a \cdot \nabla g + bg = \mathcal{K}[g] \quad \text{on } (0,T) \times \mathcal{O}. $$

From Theorem 8.8 and Remark 8.9-(5), the function $g$ admits a trace $\gamma g$ and thanks to Proposition 8.10, we have $\gamma h_n \to \gamma g$ a.e. on $\Sigma \setminus \Sigma_0$. Because of (8.4) and the Beppo Levi theorem again we deduce that $\mathcal{R}[h_n, \gamma_+ h_n] \to \mathcal{R}[g, \gamma_+ g]$ a.e. on $\Sigma_-$, Together with the fact that $\gamma_- h_n \to \gamma_- g$ a.e. on $\Sigma_-$, we have established that the boundary condition in (8.58) holds true. It is worth emphasizing here that $\gamma_+ g \in L^1(\Sigma_+; dr_\Sigma(y, \cdot))$ for a.e. $y \in \Sigma_-$ because of (8.68). For $g_0 \in L_m^p(\mathcal{O})$, we separate the positive and the negative parts $g_0 = g_{0+} - g_{0-}$ and we obtain two renormalized solutions $g^{\pm} \in L^{\infty}(0, T; L_m^p)$ associated to $g_{0\pm}$ respectively. By linearity, the function $g := g^+ - g^-$ in $L^{\infty}(0, T; L_m^p)$ is a renormalized solution to the transport equation and the boundary condition is

$$ \begin{aligned} \gamma_- g &= \gamma_- g^+ - \gamma_- g^- = \mathcal{R}_\mathcal{O}[g^+] - \mathcal{R}_\mathcal{O}[g^-] + \mathcal{R}_\Sigma[\gamma_+ g^+] - \mathcal{R}_\Sigma[\gamma_+ g^-] \\ &= \mathcal{R}_\mathcal{O}[g] + \mathcal{R}_\Sigma[\gamma_+ g], \end{aligned} $$

where the last term is indeed well defined a.e. from the fact that $\gamma_+ g^{\pm} \in L^1(\Sigma_+; dr_\Sigma(y, \cdot))$ for a.e. $y \in \Sigma_-$ and thus $\gamma_+ g = \gamma_+ g^+ - \gamma_+ g^-$ belongs to the same spaces. From Proposition 8.15, we already know that $g \in C([0, T]; L^0(\mathcal{O}))$ and thus using an interpolation argument $g \in C([0, T]; L_{m_1}^{p_1}(\mathcal{O}))$ for any $p_1 \in [1, p)$ and any weight function $m_1$ such that $m_1/m \in L^{pp_1/(p-p_1)}$ when $p > 1$.

**Step 3.** When $\beta_{\mathcal{R}} < 1$ and $p \in [1, \infty)$, we have (8.66), and we may just repeat the proof of Proposition 8.16 in order to get $g \in C([0, T]; L_m^p(\mathcal{O}))$ and the uniqueness of the solution.
We establish the main result of this section.

**Theorem 8.33 (characteristics method).** Assume that $Y$ is an almost everywhere flow associated to (8.6) with a satisfying (8.73). For any $g_0 \in L^\infty(\mathcal{O})$, $g \in L^\infty((0,T) \times \Sigma_–)$, $T > 0$, $b \in L^\infty(\mathcal{O})$, the function

$$ (8.88) \qquad \bar{g}(t, y) := g_0(Y_{-t}(y))e^{-\int_0^t b(Y_{\tau-t}(y)) d\tau} \mathbf{1}_{t<t_b(y)} \\ \qquad \qquad + g(t-t_b(y), y_b(y))e^{-\int_0^{t_b(y)} b(Y_{\tau-t_b(y)}(y)) d\tau} \mathbf{1}_{t>t_b(y)} $$

is the unique solution in $C([0,T]; L_\text{loc}^1(\mathcal{O})) \cap L^\infty((0,T) \times \mathcal{O})$ to the evolution transport equation

$$ (8.89) \qquad \begin{cases} \frac{\partial g}{\partial t} + a \cdot \nabla g + bg = 0 & \text{on } (0,T) \times \mathcal{O}, \\ \gamma_- g = g & \text{on } (0,T) \times \Sigma_-, \quad g(0, \cdot) = g_0 & \text{on } \mathcal{O}. \end{cases} $$

*First proof of Theorem 8.33.* We additionally assume that (8.75), (8.78) and (8.79) hold, that $g \in C((0,T) \times \Sigma_–)$, $\text{supp } g \subset (0,T) \times (\Sigma_– \cap \mathcal{Y})$ and $b \in C(\bar{\mathcal{O}})$, and we mostly repeat the proof of [255, Prop. 1]. From the above definition, for a.e. $y \in \mathcal{O}$ and any $t \in (0, \infty)$, we have

$$ (8.90) \qquad \bar{g}(t+s, Y_s(y))e^{\int_0^s b(Y_\tau(y))d\tau} = \bar{g}(t, y), \quad \forall s \ge -t. $$

Let us then fix $\varphi \in \mathcal{D}((0,T) \times \mathcal{O})$ and let us extend $\bar{g}$ and $\varphi$ by 0 outside of $\mathcal{O}$. We compute

$$ \begin{aligned} 0 &= \frac{d}{ds} \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y) \varphi(t, y) dydt \\ &= \frac{d}{ds} \int_0^T \int_{\mathbb{R}^d} \bar{g}(t+s, Y_s(y)) e^{\int_0^s b(Y_\tau(y)) d\tau} \varphi(t, y) dydt \\ &= \frac{d}{ds} \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y) \varphi(t-s, Y_{-s}(y)) e^{\int_0^s (b-\text{diva})(Y_\tau(y)) ds} dydt \\ &= \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y) \frac{d}{ds} [\varphi(t-s, Y_{-s}(y)) e^{\int_0^s (b-\text{diva})(Y_\tau(y)) ds}] dydt \\ &= \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y)[-\partial_t \varphi - a \cdot \nabla \varphi + (b-\text{diva})\varphi](t-s, Y_{-s}(y)) dydt, \end{aligned} $$

where we have used the relation (8.90) in the second line and the change of variables property (8.75) in the third line. Taking $s = 0$, we get

$$ 0 = \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y)[-\partial_t \varphi - a \cdot \nabla \varphi + (b - \text{diva})\varphi](t, y) dydt, $$

which exactly means that $\bar{g}$ is a solution to equation (8.89) in the distributional sense. Now, because $t_b(y) > 0$ for any $y \in \mathcal{Y}$, we have $\bar{g}(0, y) = g_0(y)$. Take $t > 0$, and for $\alpha \in (0, \alpha_\mathcal{O})$, let us denote $A_\alpha := \{y \in \Sigma_- \cap \mathcal{Y}; t_b \circ \theta_\alpha(y) < t\}$, so that

$$ \bar{g}(t, \theta_\alpha(y)) = g(t - t_{b,\alpha}(y), y_{b,\alpha}(y))e^{-\int_0^{t_{b,\alpha}(y)} b(Y_{\tau-t_{b,\alpha}(y)}(\theta_\alpha(y)) d\tau)} \quad \text{on } A_\alpha, $$

where we use the shorthands $t_{b,\alpha} := t_b \circ \theta_\alpha$ and $y_{b,\alpha} := y_b \circ \theta_\alpha$. From $\theta_\alpha'(y) \to y$ as $\alpha' \to 0$ when $y \in \Sigma$ and Lemma 8.29, we deduce

$$ \bar{g}(t, \theta_{\alpha'}(\cdot)) \to g(t, \cdot) \quad \text{as } \alpha' \to 0, $$

on $A_\alpha$ for any fixed $\alpha > 0$. Because $\cup_{\alpha>0} A_\alpha = \Sigma_- \cap \mathcal{Y}$, the same convergence holds on $\Sigma_- \cap \mathcal{Y}$. On the other hand, we have $\gamma_\alpha \bar{g} \circ \theta_\alpha \to \gamma \bar{g}$ as $\alpha \to 0$ a.e. on $\Sigma \setminus \Sigma_0$, from Lemma 8.30. We deduce that $\gamma \bar{g} = g$ on $\Sigma_- \cap \mathcal{Y}$.

*Second proof of Theorem 8.33.* We do not make any additional assumption and we mainly repeat the proof presented in [204, Sec. 3]. Consider the unique solution $g \in C([0,T]; L_\text{loc}^1(\bar{\mathcal{O}})) \cap L^\infty((0,T) \times \mathcal{O})$ to the transport equation (8.89). Regularizing by convolution

$$ g_\varepsilon := g *_{t,x,\varepsilon} \rho_\varepsilon, $$
Step 4. We assume $\beta_{\mathcal{R}} = 1$ and the structural assumption (2). From the estimate (8.64) on a solution $g$ and the renormalized formulation of the equation, we deduce that

$$ \frac{1}{p_0} \frac{d}{dt} \int |g|^{p_0} m^{p_0} = \frac{1}{p_0} \int_{\Sigma_{-}} |\mathcal{R}[g, \gamma_+ g]|^{p_0} m_{0\Sigma}^{p_0} - \frac{1}{p_0} \int_{\Sigma_{+}} |\gamma_+ g|^{p_0} m_{0\Sigma}^{p_0} \\ + \int_{\mathcal{O}} \{g|g|^{p_0-2}(\mathcal{H}[g] + G)m^{p_0} - |g|^{p_0} m_{0\mathcal{O}}^{p_0}\}, $$

with a RHS term in $L^1(0,T)$. As above, we thus deduce $g \in C([0,T]; L_{m_0}^{p_0}(\mathcal{O}))$ and next the uniqueness of the solution.

Step 5. We assume $\beta_{\mathcal{R}} = 1$ and the structural assumption (3). In that case, we have $p = 1$, $\gamma_+g \in L_{m_\Sigma}^1((0,T) \times \Sigma_+)$ from (8.69) and then $\gamma_-g \in L_{m_\Sigma}^1((0,T) \times \Sigma_-)$ from (8.62). We may thus justify the same computation as in Step 4 with $p_0 = 1$, and we deduce $g \in C([0,T]; L_m^1(\mathcal{O}))$ and next the uniqueness of the solution. $\square$

As an immediate consequence of the above analysis, we may associate to the transport equation (8.58) a semigroup.

**Corollary 8.25.** Under the assumptions of Proposition 8.23, there exists a positive semigroup $S$ on $L_m^p$ such that for any $\mathfrak{g}_0 \in L_m^p(\mathcal{O})$, the function $t \mapsto g(t) := S(t)\mathfrak{g}_0 \in C(\mathbb{R}_+; L_{m_0}^{p_0}(\mathcal{O})) \cap L_{\text{loc}}^{\infty}(\mathbb{R}_+; L_m^p(\mathcal{O}))$ is the unique renormalized solution to the transport equation (8.58) associated to the initial datum $\mathfrak{g}_0$ (and with $G = \mathfrak{g} = 0$). Furthermore the growth bound satisfies $\omega(S) \le \kappa$.

We end this section by formulating the counterpart of the above result for the associated stationary problem

$$ (8.70) \qquad \begin{cases} \lambda g + a \cdot \nabla g + bg = \mathcal{K}[g] + G \quad \text{on } \mathcal{O}, \\ \gamma_- g = \mathcal{R}[g, \gamma_+ g] + \mathfrak{g} \quad \text{on } \Sigma_- \end{cases} $$

**Proposition 8.26.** We make exactly the same assumptions as in Proposition 8.23 on a, b, $\mathcal{K}$ and $\mathcal{R}$ for some weight function $m: \mathcal{O} \to [1, \infty)$ and some exponent $p \in [1, \infty)$ as well as either $\beta_{\mathcal{R}} < 1$ holds or $\beta_{\mathcal{R}} = 1$ holds with $\mathfrak{g} = 0$ and one of the additional structure assumptions (1) or (2). There exists $\lambda^{**} \in \mathbb{R}$ such that for any $\lambda > \lambda^{**}$, $G \in L_m^p(\mathcal{O})$ and $\mathfrak{g} \in L_{m_\Sigma}^p(\Sigma_+)$, there exists a unique solution $g \in L_{m_\Sigma}^p(\mathcal{O})$ satisfying the transport equation (8.70) in the renormalized sense and some additional a priori estimates listed during the proof.

*Proof of Proposition 8.26.* We just explain the main steps. We first establish an a priori estimate. We observe that any solution $g$ to the stationary problem (8.70) (at least formally) satisfies

$$ (8.71) \quad \int_{\mathcal{O}} |g|^p m^p (\lambda + \varpi) + \frac{1}{p} \int_{\Sigma_+} |\gamma_+ g|^p m_{\Sigma}^p = \int_{\mathcal{O}} (\mathcal{K}[g] + G) g |g|^{p-2} m^p + \frac{1}{p} \int_{\Sigma_-} |\mathcal{R}[g, \gamma_+ g] + \mathfrak{g}|^p m_{\Sigma}^p. $$

We then only consider the case $\mathfrak{g} = 0$. Repeating the same computations as in Step 1 of the proof of Proposition 8.23 and with the same notations, we get

$$ (8.72) \qquad p(\lambda - \kappa) \|g\|_{L_m^p}^p + \vartheta_{\mathcal{O}} \|g\|_{L_{m_\Sigma}^p}^p + \vartheta_{\Sigma} \|\gamma_+ g\|_{L_{m_\Sigma}^p}^p \le C_{\mathcal{O}} \|G\|_{L_m^p}^p. $$

For $\lambda > \lambda^{**} := \max(\kappa, \lambda_p^*)$ and $G \ge 0$, we next consider the sequence $(h_k)$ in $L_{m_\Sigma}^p$ defined iteratively as the solution given by Lemma 8.13 to

$$ \begin{cases} \lambda h_k + a \cdot \nabla h_k + bh_k = \mathcal{K}[h_{k-1}] + G \quad \text{on } \mathcal{O}, \\ \gamma_- h_k = \mathcal{R}[h_{k-1}, \gamma_+ h_{k-1}] \quad \text{on } \Sigma_-, \end{cases} $$

for $k \ge 1$ and starting from $h_0 \equiv 0$. We observe that $(h_k)$ is increasing and satisfies the estimate (8.72) where $g$ is replaced by $h_k$. We may pass to the limit in the above equation and estimate and we obtain a renormalized solution $g \in L_{m_\Sigma}^p$ to the transport equation (8.70) and satisfying the estimate (8.72). By linearity, the same holds without sign condition on $G$. Finally, considering the three different cases as in Steps 3, 4 and 5 in Proposition 8.23, we similarly show that $g \in L_{m_0}^{p_0}$ and $\gamma_+ g \in L_{m_0}^{p_0}$ for suitable exponent $p_0 \in [1, p]$ and weight function $m_0$. For two such solutions $g_i$ to (8.70), the function $g := g_2 - g_1$ is also a renormalized solution to (8.70) for which we may justify the identity (8.71) with $p = p_0$, $m = m_0$, $G = 0$. We thus deduce that (8.72) holds with $p = p_0$, $m = m_0$, $G = 0$, and we conclude that $g = 0$, what ends the proof of the uniqueness. $\square$
for a time and space dependent mollifier sequence $\rho_\epsilon$ similarly as in (8.20), we have $g_\epsilon \in C^1([0, T] \times \bar{O})$ and

$$ \frac{\partial g_\epsilon}{\partial t} + a \cdot \nabla g_\epsilon + bg_\epsilon = R_\epsilon \quad \text{on } (0, T) \times \mathcal{O}, $$

with the usual commutator $R_\epsilon := [a \cdot \nabla g + bg, \rho_\epsilon]$. We know from a classical time and space variant of Lemma 8.5 (see also [139, Lem. II.1], [271, Lem. 1], [70, Lem. 3.1] and the proof in [204, Sec. 3]) that $R_\epsilon \to 0$ in $L^1$. Because $g_\epsilon$ is smooth and $Y$ is an almost everywhere flow, we may set $H^\#(s, y) := H(t + s, Y_s(y))$, $\mathfrak{B}(s, y) := \int_0^s b(Y_\tau(y))d\tau$ and compute

$$ \frac{d}{ds}[(g_\epsilon^\# e^{\mathfrak{B}})(s, y)] = (R_\epsilon^\# e^{\mathfrak{B}})(s, y), $$

from what we get

$$ \begin{aligned} \tilde{g}_\epsilon(t, y) &:= g_\epsilon(t, y) - g_\epsilon(0, Y_{-t}(y))e^{\mathfrak{B}(-t, y)} \mathbf{1}_{t<t_b(y)} - g_\epsilon(t - t_b(y), y_b(y))e^{\mathfrak{B}(-t_{b(y)}, y)} \mathbf{1}_{t>t_b(y)} \\ &= \int_0^t (R_\epsilon^\# e^{\mathfrak{B}})(s, y)ds \mathbf{1}_{t<t_b(y)} + \int_{t-t_b(y)}^t (R_\epsilon^\# e^{\mathfrak{B}})(s, y)ds \mathbf{1}_{t>t_b(y)}, \end{aligned} $$

for a.e. $y \in \mathcal{O}$ and any $t > 0$ and where we use (8.77) for getting rid of the set $\{y \in \mathcal{O}; t_b(y) = t\}$. For $T, \varrho > 0$ and setting $\mathcal{U}_{T,\varrho} := (0, T) \times (\mathcal{O} \cap B_\varrho)$, we deduce

$$ \begin{aligned} \int_{\mathcal{U}_{T,\varrho}} |\tilde{g}_\epsilon(t, y)| dydt &\le T \int_{\mathcal{U}_{T,\varrho}} |(R_\epsilon^\# e^{\mathfrak{B}})(s, y)| dyds \\ &\le T e^{(C+\|b\|_{L^\infty})T} \int_{\mathcal{U}_{T,\varrho}} |R_\epsilon(s, y)| dyds, \end{aligned} $$

from the near-incompressibility condition (8.74) of the flow. From Proposition 8.10 and Remark 8.11, we know that

$$ g_\epsilon(t, \cdot) \to g(t, \cdot) \text{ in } L_{\text{loc}}^1(\bar{O}) \text{ as } \epsilon \to 0, \text{ for any } t \in [0, T]; $$

$$ g_{\epsilon|\Sigma_{-}} \to \gamma_{-}g = \mathfrak{g} \text{ in } L_{\text{loc}}^{1}(\Sigma_{-}) \text{ as } \epsilon \to 0. $$

Passing to the limit, we get

$$ \int_{\mathcal{U}_{T,\varrho}} |g(t, y) - g_0(Y_{-t}(y))e^{\mathfrak{B}(-t, y)} \mathbf{1}_{t<t_b(y)} - g(t - t_b(y), y_b(y))e^{\mathfrak{B}(-t_{b(y)}, y)} \mathbf{1}_{t>t_b(y)}| dtdy = 0, $$

for any $T, \varrho > 0$, what is nothing but (8.88). $\square$

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

**Corollary 8.34 (Representation formula).** Under the assumptions of Lemma 8.14 in a smooth domain $\mathcal{O} \neq \mathbb{R}^D$, for any $\lambda > \lambda_{a,b,p}$, $G \in L^p(\mathcal{O})$, the unique solution $g \in L^p(\mathcal{O})$ to the stationary transport equation (8.29) (with $\mathfrak{g} = 0$) satisfies

$$ (8.91) \qquad g(y) = \int_0^\infty e^{-\lambda t} (S_b(t)G)(y)dt, \quad \text{for a.e } y \in \mathcal{O}, $$

with $S_b$ is defined by (8.49) in which formula $Y$ and $t_b$ stand for the characteristics and backward exit time defined just as above.

**Proof of Corollary 8.34.** That is nothing but (2.13). $\square$

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

Adapting the second proof of Theorem 8.33, we obtain a more accurate characterization of the backward exit time $t_b$ with more general assumptions on the vector field.

**Lemma 8.35.** Assume that $Y$ is an almost everywhere flow associated to (8.6) with a satisfying (8.73). The backward exit time $t_b$ is the unique renormalized solution in $L(\mathcal{O})$ to the backward exit time problem

$$ (8.92) \qquad a \cdot \nabla \tau = 1 \text{ in } \mathcal{O}, \quad \gamma_{-} \tau = 0 \text{ on } \Sigma_{-}. $$

We also have

$$ (8.93) \qquad t_b = \int_0^\infty S_0(t) \mathbf{1} dt, $$

where $\mathbf{1}$ stands for the unit function in $\mathcal{O}$ and $S_0$ is defined by (8.49) with $b = 0$ and in which formula $Y$ and $t_b$ stand for the characteristics and backward exit time defined just as above.
## 8.4. Characteristics.

In this section we come back to the characteristics method for the evolution and the stationary transport equation. Our aim is in particular to discuss the representation formula (8.47).

We consider a vector field $a: \mathcal{O} \to \mathbb{R}^D$ which extends to $\mathbb{R}^D$ and, denoting its extension by the same letter $a$, we at least assume

$$ (8.73) \qquad a \in W_{\text{loc}}^{1,1}(\mathbb{R}^D), \quad \text{div} a \in L^\infty(\mathbb{R}^D), \quad a/\langle y \rangle \in L^1(\mathbb{R}^D) + L^\infty(\mathbb{R}^D). $$

After DiPerna and Lions [138, 254] (see also [203, Def. 1] or [202, Def. 1]), we introduce the following notion of flow.

**Definition 8.27.** We name almost everywhere flow associated to (8.6) a measurable function $Y: \mathbb{R} \times \mathbb{R}^D \to \mathbb{R}^D$, $(t, y) \mapsto Y_t(y)$, such that

(i) for a.e. $y \in \mathbb{R}^D$, the map $t \mapsto Y_t(y)$ is continuous and

$$ \dot{Y}_t(y) = a(Y_t(y)) \text{ in } \mathcal{D}'(\mathbb{R}), \quad Y_0(y) = y; $$

(ii) for a.e. $y \in \mathbb{R}^D$ and for any $s, t \in \mathbb{R}$, there holds $Y_{s+t}(y) = Y_s(Y_t(y))$;

(iii) there exists $C \ge 0$ such that

$$ (8.74) \qquad \forall t \in \mathbb{R}, \quad e^{-CT}\lambda \le Y(t, \cdot)_{\#}\lambda \le e^{CT}\lambda, $$

where $(Y(t, \cdot)_{\#}\lambda)(A) = \lambda(Y(-t, A))$, $A \subset \mathbb{R}^D$, is the pushforward of the Lebesgue measure $\lambda$.

From [138, Thm. III.1], [9, Thm. 31 & Remark 32] and [203, Sec. 3] (see also Theorem 8.33 below), we know the existence and uniqueness of such an a.e. flow for $a$ satisfying (8.73). In the incompressible case ($\text{div} a = 0$), this one furthermore satisfies:

$$ (8.75) \qquad \int_{\mathbb{R}^D} \varphi(Y_t(y)) e^{\int_0^t (\text{div} a)(Y_t(y)) d\tau} dy = \int_{\mathbb{R}^D} \varphi(y) dy, $$

for any $t \in \mathbb{R}$ and any $\varphi \in L_c^\infty(\mathbb{R}^D)$, the space of $L^\infty$ functions with compact support. In the compressible case ($\text{div} a \ne 0$) and when $a$ only satisfies (8.73), it seems not clear that [138, Thm. III.2] or [9, Thm. 31 & Remark 32] provides an a.e. flow such that (8.75) holds. In that general case, the volume identity (8.75) must be replaced by the volume two sides estimate (or nearly-incompressible condition):

$$ e^{-t \|\text{div} a\|_{\infty}} \int_{\mathbb{R}^D} \varphi(y) dy \le \int_{\mathbb{R}^D} \varphi(Y_t(y)) dy \le e^{t \|\text{div} a\|_{\infty}} \int_{\mathbb{R}^D} \varphi(y) dy, $$

for any $0 \le \varphi \in L_c^\infty(\mathbb{R}^D)$ and $t \in \mathbb{R}$, what is nothing but (8.74) with $C := \|\text{div} a\|_{\infty}$. It is however quite straightforward to prove from [138, 9] that the a.e. flows $Y$ satisfies (8.75) when $a$ additionally satisfies $\text{div} a \in C(\mathbb{R}^D)$. One possible way to construct the a.e. flow $Y$ is to define $Y = Y_t(y)$ as the unique renormalized solution in $C(\mathbb{R}; L)$ to the transport equation

$$ \partial_t Y - a \cdot \nabla_y Y = 0 \quad \text{on} \quad \mathbb{R} \times \mathbb{R}^D, \quad Y_0(y) = y \quad \text{on} \quad \mathbb{R}^D, $$

or more explicitly

$$ \frac{\partial}{\partial t} \beta(Y) - a(Y) \cdot \nabla \beta(Y) = 0 \quad \text{on} \quad \mathbb{R} \times \mathbb{R}^D, \quad \beta(Y_0) = \beta(y) \quad \text{on} \quad \mathbb{R}^D, $$

in the distributional sense for any $\beta \in C^1(\mathbb{R}^D, \mathbb{R})$ such that $\beta$ and $|\nabla \beta(z)|(1+|z|)$ are uniformly bounded on $\mathbb{R}^D$. In particular, for any $g_0 \in C^1(\mathbb{R}^D)$ and next for any $g_0 \in L^0(\mathbb{R}^D)$ the function $g^\#(t, y) := g_0(Y_{-t}(y))$ is the unique renormalized solution to the transport equation

$$ \partial_t g^\# + a \cdot \nabla g^\# = 0 \quad \text{on} \quad \mathbb{R} \times \mathbb{R}^D, \quad g^\#(0, \cdot) = g_0 \quad \text{on} \quad \mathbb{R}^D. $$

We introduce some notations. We denote $y \in \mathcal{Y}$ if (i) and (ii) hold. In particular, $\mathcal{Y}$ is a measurable subset of $\mathbb{R}^D$ and $|\mathbb{R}^D \setminus \mathcal{Y}| = 0$. Because of (i), for $y \in \mathcal{V} := \mathcal{O} \cap \mathcal{Y}$, we may define the backward exit time

$$ t_b(y) := \sup\{\tau > 0 \mid Y_{-t}(y) \in \mathcal{O}, \forall s \in [0, \tau]\} \in (0, +\infty], $$

the subset $\mathcal{V}_b := \{y \in \mathcal{V}; t_b(y) < +\infty\}$ and the associated entering position

$$ y_b(y) := Y_{-t_b(y)}(y) \quad \text{when} \quad y \in \mathcal{V}_b. $$
**Proof of Lemma 8.35. Step 1. Existence.** From Lemma 8.12, for any $\lambda > \lambda_{\infty}^* := 0$, there exists a (unique) solution $\tau_{\lambda} \in L^{\infty}(\mathcal{O})$ to the truncated backward exit time problem

$$ \lambda\tau_{\lambda} + a \cdot \nabla\tau_{\lambda} = 1 \text{ in } \mathcal{O}, \quad \gamma_{-}\tau_{\lambda} = 0 \text{ on } \Sigma_{-}. $$

From the weak maximum principle, we have $\tau_{\lambda} \ge 0$. As a consequence, for $0 < \lambda < \lambda'$, we have

$$ \lambda(\tau_{\lambda} - \tau_{\lambda'} + a \cdot \nabla(\tau_{\lambda} - \tau_{\lambda'} = (\lambda' - \lambda)\tau_{\lambda'} \ge 0 \text{ in } \mathcal{O}, \quad \gamma_{-}(\tau_{\lambda} - \tau_{\lambda'}) = 0 \text{ on } \Sigma_{-}. $$

From the weak maximum principle again, we deduce that $\tau_{\lambda} - \tau_{\lambda'} \ge 0$, and $(\tau_{\lambda_n})$ is an increasing sequence when $\lambda_n \searrow 0$. We set $\tau := \lim_{n \to \infty} \tau_{\lambda_n}$, so that $\tau \in L(\mathcal{O})$ is a nonnegative renormalized solution to the backward exit time problem.

**Step 2. Characterization.** By definition, for any $\beta \in C_*^1(\mathbb{R})$, $\beta(0) = 0$, the function $\tau$ satisfies

$$ a \cdot \nabla\beta(\tau) = \beta'(\tau) \text{ in } \mathcal{O}, \quad \gamma_{-}\beta(\tau) = 0 \text{ on } \Sigma_{-}. $$

With the notations of (8.20), we define $b_{\varepsilon} := \beta(\tau) *_{\varepsilon} \rho_{\varepsilon}$, $B_{\varepsilon} := \beta'(\tau) *_{\varepsilon} \rho_{\varepsilon}$, and thanks to Lemma 8.5, we have thus

$$ a \cdot \nabla b_{\varepsilon} = B_{\varepsilon} + r_{\varepsilon} \text{ in } \mathcal{D}'(\mathcal{O}), $$

with

$$ b_{\varepsilon} \to \beta(\tau), \quad B_{\varepsilon} \to \beta'(\tau), \quad r_{\varepsilon} \to 0, $$

respectively in $L_{\text{loc}}^p(\bar{\mathcal{O}})$, $L_{\text{loc}}^p(\bar{\mathcal{O}})$ and $L_{\text{loc}}^1(\bar{\mathcal{O}})$ for any $p \in [1, \infty)$. Because then

$$ \frac{d}{ds} b_{\varepsilon} \circ Y_s = (a \cdot \nabla b_{\varepsilon})(Y_s) = (B_{\varepsilon} + r_{\varepsilon})(Y_s), $$

and defining

$$ \tilde{b}_{t, \varepsilon}(y) := \left\{ \begin{aligned} & b_{\varepsilon}(Y_{-t}(y)) + \int_{-t}^{0} B_{\varepsilon}(Y_s(y)) ds \mathbf{1}_{t<t_{\mathbf{b}}(y)} \\ & + \left\{ (\gamma_{-}b_{\varepsilon})(y_{\mathbf{b}}(y)) + \int_{-t_{\mathbf{b}}(y)}^{0} B_{\varepsilon}(Y_s(y)) ds \right\} \mathbf{1}_{t>t_{\mathbf{b}}(y)}, \end{aligned} \right. $$

we have

$$ b_{\varepsilon}(y) - \tilde{b}_{t, \varepsilon}(y) = R_{t, \varepsilon} := \int_{-t}^{0} r_{\varepsilon}(Y_s(y)) ds \mathbf{1}_{t<t_{\mathbf{b}}(y)} + \int_{-t_{\mathbf{b}}(y)}^{0} r_{\varepsilon}(Y_s(y)) ds \mathbf{1}_{t>t_{\mathbf{b}}(y)}. $$

Arguing similarly as in the second proof of Theorem 8.33, we have $\tilde{b}_{t, \varepsilon} \to \tilde{b}_t$ and $R_{t, \varepsilon} \to 0$ in $L^1(\mathcal{U}_{\rho,T})$ as $\varepsilon \to 0$, with

$$ \tilde{b}_t(y) := \left\{ \beta(\tau(Y_{-t}(y))) + \int_{-t}^{0} \beta'(\tau(Y_s(y))) ds \right\} \mathbf{1}_{t<t_b(y)} + \int_{-t_b(y)}^{0} \beta'(\tau(Y_s(y))) ds \mathbf{1}_{t>t_b(y)}. $$

We deduce that

$$ \beta(\tau(y)) = \tilde{b}_t(y), \quad \text{for a.e. } t > 0, y \in \mathcal{O}. $$

Choosing a sequence $(\beta_n)$ of renormalizing function in $C_*^1(\mathbb{R})$ such that $0 \le \beta_n(s) \nearrow s$ and $0 \le \beta'_n(s) \nearrow 1$ locally uniformly, writing the above equation for $\beta = \beta_n$ and passing to the limit $n \to \infty$, we obtain

$$ \tau(y) = \left\{ \tau(Y_{-t}(y)) + t \right\} \mathbf{1}_{t<t_b(y)} + t_b(y) \mathbf{1}_{t>t_b(y)}, $$

and in particular $\tau = t_b$ a.e. on $\mathcal{O}$. That implies that $t_b$ is the unique renormalized solution to the backward exit time problem (8.92). From Corollary 8.34, for any $\lambda > 0$, we have

$$ \tau_{\lambda} = \int_{0}^{\infty} e^{-\lambda t} S_{0}(t) \mathbf{1} dt \quad \text{a.e on } \mathcal{O}, $$

and we deduce (8.93) by passing to the limit $\lambda \searrow 0$ in that identity. $\square$

<figure><img src="image_8.png" /></figure>
We observe that the function $t_b: \mathcal{V} \to (0, +\infty]$ is measurable, $\mathcal{V}_b$ is a measurable subset of $\mathcal{O}$ and

$$ (8.76) \qquad t_b(Y_s(y)) = t_b(y) + s, \quad y_b(Y_s(y)) = y_b(y), \quad \forall y \in \mathcal{V}_b, \forall s \in [0, t_b(y)). $$

$$ (8.77) \qquad \mathrm{meas}(\{y \in \mathcal{O}; t_b(y) = t\}) = 0, \quad \forall t > 0. $$

The properties (8.76) are straightforward while (8.77) is a consequence of the fact that $\{y \in \mathcal{V}; t_b(y) = t\} \subset Y_t(\Sigma)$ and of the nearly-incompressible condition (8.74).

We now introduce the following first regularity assumption on $a$ at the boundary

$$ (8.78) \qquad \forall y_0 \in \Sigma, y \mapsto a(y) \cdot n(y_0) \text{ is continuous on } \bar{\mathcal{O}}. $$

Let us present two examples.

- It may happen that $\mathcal{V}_b = \emptyset$. For instance, choosing $\mathcal{O} := \{y \in \mathbb{R}^2; |y| < 1\}$ the unit disk of the plane and $a(y) := |y|y^\perp \in C^{0,1}(\mathbb{R}^2; \mathbb{R}^2)$, $y^\perp := (y_2, -y_1)$, we have $\mathrm{div} a = 0$ and $a(y) \cdot n(y) = y^\perp \cdot y = 0$ for any $y \in \mathbb{R}^2$, so that the flows do not encounter the boundary set $\Sigma = \{y \in \mathbb{R}^2; |y| = 1\}$. In that situation $\mathcal{V} = \mathcal{O}$ and $\mathcal{V}_b = \emptyset$.

- In the kinetic case, namely $y = (x, v) \in \mathcal{O} := \Omega \times \mathbb{R}^d$, $\Omega \subset \mathbb{R}^d$ an open set with smooth boundary with unit normal outward $\nu_x$, so that $n(y) = (\nu_x, 0)$, and $a(y) = (v, F(x, v))$, we have $\mathrm{div} a = \mathrm{div}_v F$ and $\mathrm{div}_v F = 0$ when $F = E(x) + v \wedge B(x)$, and we have $a(y) \cdot n(y_0) = v \cdot \nu_{x_0}$ which is a smooth function on $\bar{\mathcal{O}} \times \Sigma$.

**Lemma 8.28.** Under the condition (8.78), the mapping $y_b: \mathcal{V}_b \to \Sigma_- \cup \Sigma_0$ is measurable.

*Proof of Lemma 8.28.* From the very definitions and composition rules, we have $y_b: \mathcal{V}_b \to \Sigma \cap \mathcal{Y}$ is measurable. Take $y \in \mathcal{V}_b$, denote $y_0 := y_b(y)$ and consider a sequence $t_k \searrow 0$ so that $Y_{t_k}(y_0) \to y_0$ and $Y_{t_k}(y_0) \in \mathcal{O}$ for any $k \ge 1$. From (8.78), we deduce

$$ 0 \ge \limsup_{k \to \infty} \frac{Y_{t_k}(y_0) - y_0}{t_k} \cdot n(y_0) = \lim_{k \to 0} \frac{1}{t_k} \int_0^{t_k} a(Y_s(y_0)) \cdot n(y_0) ds = a(y_0) \cdot n(y_0), $$

which means that $y_0 \in \Sigma_- \cup \Sigma_0$. $\square$

For further references, we introduce the following second additional mild regularity assumption on $a$ in the domain

$$ (8.79) \qquad a \in L_{\mathrm{loc}}^{\infty}(\bar{\mathcal{O}}), \quad (a(y) \cdot y)_+ \lesssim \langle y \rangle^2. $$

This one may in fact replace the last boundedness condition on $a$ in (8.73). We establish a technical result which will be useful in the sequel.

**Lemma 8.29.** Under assumptions (8.78) and (8.79), the following hold:

(1) For any $R_0, T > 0$, there exists $R_T, L_T > 0$ such that for any $y \in B_{R_0}$ there hold

$$ (8.80) \qquad \sup_{[-T,T]} |Y_t(y)| \le R_T \quad \text{and} \quad |Y_{t_2}(y) - Y_{t_1}(y)| \le L_T |t_2 - t_1|, \quad \forall t_1, t_2 \in [-T,T]. $$

(2) For a sequence $(y_\epsilon)$ of $\mathcal{V}$ such that $y_\epsilon \to y_0 \in \Sigma_- \cap \mathcal{Y}$, we have $t_b(y_\epsilon) \to 0$ and $y_b(y_\epsilon) \to y_0$.

*Proof of Lemma 8.29.* Proof of (1). Take $y \in \mathcal{V} \cap B_{R_0}$. On the one hand, from (8.79), we have

$$ \langle Y_t(y) \rangle^2 - \langle y \rangle^2 = 2 \int_0^t a(Y_s(y)) \cdot Y_s(y) ds \le C \int_0^t \langle Y_s(y) \rangle^2 ds, $$

and we conclude to $Y_t(y) \in B_{R_T}$ thanks to the Gronwall lemma. As a consequence, we have

$$ |Y_{t_2}(y) - Y_{t_1}(y)| \le |t_2 - t_1| \|a\|_{L^\infty(B_{R_T})}, \quad \forall t_i \in [-T, T]. $$

*Proof of (2).* Assume by contradiction that $\limsup t_b(y_\epsilon) \ge \tau > 0$ and set $T := \tau + 1$. By assumption, there exists $R_0 > 0$ such that $y_\epsilon \in B_{R_0}$ and thus by step 1, (8.80) holds uniformly in $\epsilon \in (0, 1]$. Thanks to the Ascoli Theorem and the contradiction hypothesis, there exists $\epsilon_k \to 0$ such that $t_b(y_{\epsilon_k}) \ge \tau$ and a there exists $Y \in C([-T, T])$ such that $Y_\bullet(y_{\epsilon_k}) \to Y_\bullet$ in $C([-T, T])$. Next, passing to the limit in the conditions

$$ Y_{-t}(y_{\epsilon_k}) \in \mathcal{O} \quad \text{and} \quad Y_{-t}(y_{\epsilon_k}) - y_{\epsilon_k} = - \int_0^t a(Y_{-s}(y_{\epsilon_k})) ds, $$
### 8.5. On the Krein-Rutman theorem for the transport equation with kernel terms.

In this section we carry on our analysis of the transport equation with kernel term (8.1)-(8.3) for which we establish a Krein-Rutman result under strong positivity assumption on the kernel acting on the domain. As in section 8.3, we assume that $a, b, \mathcal{K}$ and $\mathcal{R}$ satisfy the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) for some weight function $m: \bar{\mathcal{O}} \to [1, \infty)$ and some exponent $p \in [1, \infty)$.

On the kernel $\mathcal{K}$, we make the additional strong positivity hypothesis: for any $x \in \mathcal{O}$, there exist $r_x, \eta_x > 0$ such that

$$ (8.94) \qquad \forall f \ge 0, \forall y \in B(x, r_x), \quad \mathcal{K}[f](y) \ge \eta_x \int_{B(x, r_x)} f_* dy_* $$

and

$$ (8.95) \qquad \exists x_0, \ a, b \in L^\infty(B(x_0, r_0)), \ r_0 := r_{x_0}, $$

as well as one of the two following regularity assumptions

$$ (8.96) \qquad \mathcal{K} \in \mathcal{K}(L_m^p(\mathcal{O})) \quad \text{or} \quad \mathcal{K} : L_m^p(\mathcal{O}) \to L^{p_1}(\mathcal{O}) \cap L_{m_1}^p(\mathcal{O}), $$

with $p_1 > p$ and $m_1/m \to \infty$ when $y \to \infty$.

We thus consider the operator

$$ (8.97) \qquad \mathcal{L}f = -a \cdot \nabla f - bf + \mathcal{K}[f] = -\operatorname{div}(af) - Kf + \mathcal{K}[f] $$

with $K := b - \operatorname{div} a \ge 0$, which is complemented with the boundary condition

$$ (8.98) \qquad \gamma_-f = \mathcal{R}_\mathcal{O}[f] + \mathcal{R}_\Sigma[\gamma_+f] \quad \text{on } \Sigma_-. $$

More precisely, we define $\mathcal{L}$ in the Banach space $L_m^p(\mathcal{O})$ with domain

$$ D(\mathcal{L}) \subset \{f \in L^1(\mathcal{O}); a \cdot \nabla f \in L_{\text{loc}}^1(\bar{\mathcal{O}}), \gamma_-f = \mathcal{R}[f, \gamma_+f]\}. $$

Notice that because of Section 8.1 the trace function is well defined.

**Example.** The nonlocal operator with a drift

$$ (8.99) \qquad \partial_t f = -a \partial_x f - b + \mathcal{K}[f] \quad \text{in } \mathcal{O}, \quad \gamma_-f = 0 \quad \text{on } \Sigma_-, $$

with $\mathcal{O} \subset \mathbb{R}$ a bounded interval, $a \in W_{\text{loc}}^{1,1}(\mathcal{O})$, $a' \in L^\infty(\mathcal{O})$, $b \in L^\infty(\mathcal{O})$, and thus the boundary kernel is $\mathcal{R} \equiv 0$. Motivated by some non-local reaction-diffusion models, this problem was recently investigated in [109, 118, 249]. It is also used in the study of selection-mutation models in changing environment, see the even newer works [162, 207].

We start by checking that with the above assumptions, the conditions (H1)-(H5) presented in the abstract part hold true.

**Condition (H1).** From Proposition 8.26, we know that for any $\lambda > \lambda^{**}$ the stationary problem

$$ (\lambda - \mathcal{L})g = G \quad \text{in } \mathcal{O}, \quad \gamma_-g = \mathcal{R}[g, \gamma_+g] \quad \text{on } \Sigma_-, $$

has a unique solution. More precisely, the associated inverse operator denoted by $\mathcal{R}_\mathcal{L}$ (without reference to the boundary operator $\mathcal{R}$) satisfies $\mathcal{R}_\mathcal{L}: L_m^p \to L_m^p$ and $\mathcal{R}_\mathcal{L}G \ge 0$ if $G \ge 0$.

**Condition (H2).** We first consider the case when $\mathcal{R}_\mathcal{O} \equiv 0$ and we denote by $\mathcal{L}_0$ the associated generator. We fix $f_0 \in C_c^2(\mathcal{O})$, such that $\int_{\mathcal{B}_0} f_0 dy = 1$, $f_0 > 0$ on $\mathcal{B}_0$, $\text{supp} f_0 = \bar{\mathcal{B}}_0$, as well as

$$ \|f_0^{-1}\|_{L^\infty(\mathcal{B}_\varepsilon)} = \varepsilon^{-2}, \quad \|\nabla f_0\|_{L^\infty(\mathcal{B}_0 \setminus \mathcal{B}_\varepsilon)} = \varepsilon, \quad \forall \varepsilon \in (0, 1/2), $$

where we denote $\mathcal{B}_\varepsilon := B(x_0, (1-\varepsilon)r_0)$. We also define $C_0 := \|f_0\|_{L^\infty(\mathcal{B}_0)}$ and $C_1 := \|\nabla f_0\|_{L^\infty(\mathcal{B}_0)}$, both may be bounded by a constant which only depends on $r_0$ and $d$. Because of (8.94), we have

$$ \mathcal{K}[f_0](y) \ge \eta_0 \mathbf{1}_{\mathcal{B}_0} \ge \frac{\eta_0}{C_0} f_0. $$

We observe that $f_0 \in D(\mathcal{L}_0)$ and we compute

$$ (8.100) \qquad \mathcal{L}_0 f_0 \ge -\|a\|_{L^\infty(\mathcal{B}_0)} C_1 \mathbf{1}_{\mathcal{B}_0} - \|b\|_{L^\infty(\mathcal{B}_0)} C_0 \mathbf{1}_{\mathcal{B}_0} + \eta_0 \mathbf{1}_{\mathcal{B}_0} \ge \kappa_0 f_0, $$

if $\kappa_0 := \eta_0/C_0 - \|a\|_{L^\infty(\mathcal{B}_0)} C_1/C_0 - \|b\|_{L^\infty(\mathcal{B}_0)} \ge 0$. More generally, we have

$$ (8.101) \qquad \mathcal{L}_0 f_0 \ge -\|a\|_{L^\infty(\mathcal{B}_0)} C_1 \mathbf{1}_{\mathcal{B}_\varepsilon} - \|a\|_{L^\infty} \varepsilon \mathbf{1}_{\mathcal{B}_0 \setminus \mathcal{B}_\varepsilon} - \|b\|_{L^\infty(\mathcal{B}_0)} f_0 + \eta_0 \mathbf{1}_{\mathcal{B}_0} \ge \kappa_0 f_0, $$
for any $k \ge 1$ and any $t \in [0, \tau]$, we get

$$Y_{-t} \in \bar{\mathcal{O}} \quad \text{and} \quad Y_{-t} - y_0 = - \int_0^t a(Y_{-s}) ds.$$

We deduce

$$0 \ge \limsup_{t \to 0} \frac{Y_{-t} - y_0}{t} \cdot n(y_0) = \lim_{t \to 0} -\frac{1}{t} \int_0^t a(Y_{-s}) \cdot n(y_0) ds = -a(y_0) \cdot n(y_0),$$

which is in contradiction with the hypothesis $y_0 \in \Sigma_-$. Now, we may estimate

$$|y_{\mathbf{b}}(y_{\varepsilon}) - y_0| \le |y_{\varepsilon} - y_0| + \int_0^{t_{\mathbf{b}}(y_{\varepsilon})} |a(Y_s(y_{\varepsilon}))| ds \to 0,$$

as $\varepsilon \to 0$, as a consequence of the convergence $t_{\mathbf{b}}(y_{\varepsilon}) \to 0$, the first estimate in (8.80) and the first condition in (8.79). $\square$

We reformulate some “space continuity” of solutions to the transport equation results picked up in [70, Sec. 7]. Defining

$$\mathcal{O}_{\alpha} := \{y \in \mathcal{O}; \delta(y) > \alpha\}, \quad \Sigma_{\alpha} := \{y \in \mathcal{O}; \delta(y) = \alpha\} = \partial \mathcal{O}_{\alpha},$$

we know from [70, Sec. 2], that there exists $\alpha_{\mathcal{O}} > 0$ such that for any $\alpha \in (0, \alpha_{\mathcal{O}})$, the mapping

$$\theta_{\alpha} : \Sigma \to \Sigma_{\alpha}, \quad \theta_{\alpha}(z) := z - \alpha n(z)$$

is an isomorphism with associated jacobian function $J_{\alpha}$ and

$$(8.81) \qquad \int_{\Sigma_{\alpha}} h(z') d\sigma_{\alpha}(z') = \int_{\Sigma} h \circ \theta_{\alpha}(z) J_{\alpha}(z) d\sigma_z,$$

$$(8.82) \qquad \int_{\mathcal{O} \setminus \mathcal{O}_{\alpha}} g(y) dy = \int_{0}^{\alpha} \int_{\Sigma} g \circ \theta_{\alpha'}(z) J_{\alpha'}(z) d\sigma_z d\alpha',$$

for any $h \in L^1(\Sigma_{\alpha})$ and $g \in L^1(\mathcal{O} \setminus \mathcal{O}_{\alpha})$, where $d\sigma_{\alpha}$ denotes the Lebesgue measure on $\Sigma_{\alpha}$ and where the jacobian function $J_{\alpha}$ satisfies $1/2 \le J_{\alpha} \le 3/2$ as well as $J_{\alpha} \to I$ as $\alpha \to 0$.

**Lemma 8.30.** For any $g \in L^{\infty}(\mathcal{O})$ satisfying $a \cdot \nabla g \in L^1(\mathcal{O})$, we have

$$(8.83) \qquad g = \gamma_{\alpha} g \quad \text{a.e. on } \Sigma_{\alpha} \setminus \{a \cdot n = 0\} \quad \text{for a.e. } \alpha \in [0, \alpha_{\mathcal{O}}],$$

where we denote by $\gamma_{\alpha}g$ the trace of $g$ on $\Sigma_{\alpha}$, and

$$(8.84) \qquad \gamma_{\alpha} g \circ \theta_{\alpha} \to \gamma g \quad \text{as } \alpha \to 0, \quad \text{a.e. on } \Sigma \setminus \Sigma_0.$$

**Proof of Lemma 8.30.** For $\varphi \in C_c(\bar{\mathcal{O}}) \cap W^{1,1}(\mathcal{O})$ and $\beta \in C^1(\mathbb{R})$, the renormalized Green formula (8.17) writes

$$\int_{\Sigma} \varphi \beta(\gamma g) a \cdot n \, d\sigma = \int_{\Sigma_{\alpha}} \varphi \beta(\gamma_{\alpha} g) a \cdot n \, d\sigma_{\alpha} + \int_{\mathcal{O} \setminus \mathcal{O}_{\alpha}} [\text{div}(a\varphi)\beta(g) + \varphi\beta'(g)a \cdot \nabla g] \, dy,$$

and thus

$$(8.85) \qquad \int_{\Sigma} \varphi \beta(\gamma g) a \cdot n \, d\sigma = \lim_{\alpha \to 0} \int_{\Sigma_{\alpha}} \varphi \beta(\gamma_{\alpha} g) a \cdot n \, d\sigma_{\alpha}.$$

Denoting $\psi := (1 - (\delta(x) - \alpha)/s)_+$, $\alpha + s \in (\alpha, \alpha_{\mathcal{O}})$, observing that $\psi|_{\Sigma_{\alpha}} = 1$, recalling that $n = -\nabla\delta$ and using $\varphi\psi$ as a test function in the renormalized Green formula (8.24), we similarly have

$$\int_{\Sigma_{\alpha}} \varphi \beta(\gamma_{\alpha} g) a \cdot n \, d\sigma_{\alpha} = \frac{1}{s} \int_{\mathcal{O}_{\alpha} \setminus \mathcal{O}_{\alpha+s}} \varphi \beta(g) a \cdot n \, dy + \int_{\mathcal{O}_{\alpha} \setminus \mathcal{O}_{\alpha+s}} [\text{div}(a\varphi)\beta(g) + \varphi\beta'(g)a \cdot \nabla g]\psi \, dy,$$

so that

$$\int_{\Sigma_{\alpha}} \varphi \beta(\gamma_{\alpha} g) a \cdot n \, d\sigma_{\alpha} = \lim_{s \to 0} \frac{1}{s} \int_{\mathcal{O}_{\alpha} \setminus \mathcal{O}_{\alpha+s}} \varphi \beta(g) a \cdot n \, dy.$$

We immediately deduce

$$\int_{\Sigma_{\alpha}} \varphi \beta(\gamma_{\alpha} g) a \cdot n \, d\sigma_{\alpha} = - \frac{d}{d\alpha} \int_{\mathcal{O}_{\alpha}} \varphi \beta(g) a \cdot n \, dy$$
with $\kappa_0 := -\|a\|_{L^\infty(B_0)} C_1 \varepsilon^{-2} - \|b\|_{L^\infty(B_0)} \in \mathbb{R}$ when $\|a\|_{L^\infty} \varepsilon \le \eta_0$. Depending on how $\eta_0 > 0$ is large, we obtain in that way two constructive lower bounds of $\mathcal{I}$ thanks to Lemma 2.4-(ii) and we have thus established that $\mathcal{L}_0$ satisfies (H2). Because $f_0 \in D(\mathcal{L}_0)$, we have $S_{\mathcal{L}_0}(t)f_0 \ge e^{\kappa_0 t} f_0$ for any $t \ge 0$, from Remark 2.5-(2). On the other hand, we observe that $S_{\mathcal{L}}(t) \ge S_{\mathcal{L}_0}(t)$ for any $t \ge 0$, from the weak maximum principle mentioned in Remark 8.24-(3). These two last observations together imply $S_{\mathcal{L}}(t)f_0 \ge e^{\kappa_0 t} f_0$, for any $t \ge 0$. We deduce from Lemma 2.4-(iv) that (H2) holds.

**Condition (H3).** We introduce the semigroup $S_B$ associated to the transport equation

$$ \frac{\partial g}{\partial t} + a \cdot \nabla g + bg = 0, \quad \gamma_-g = \mathcal{R}[\gamma_+g], $$

which is well defined thanks to Corollary 8.25 and satisfies $\|S_B(t)g_0\|_{L^p_m} \le e^{\kappa_B t} \|g_0\|_{L^p_m}$ for any $t \ge 0$ and $g_0 \in L^p_m$ with $\kappa_B := \|\langle \bar{\varpi}_- \rangle\|_{L^\infty} + M_{\mathcal{R}}/p$ because of the a priori estimate (8.65) particularized to the present case (in particular where we can take $\varepsilon_1 = 0$ because the influx function is $g = 0$ here). We formulate the first hypothesis

$$ (8.102) \qquad \eta_0 > \|\langle \bar{\varpi}_- \rangle\|_{L^\infty} C_0 + M_{\mathcal{R}} C_0 / p + \|a\|_{L^\infty(B_0)} C_1 - \|b\|_{L^\infty(B_0)} C_0, $$

with the same definitions as above for $B_0$, $C_0$ and $C_1$, so that $\kappa_0 > \kappa_B$ because of (8.100). In a second case, we assume

$$ (8.103) \qquad \mathcal{R} \equiv 0, \quad \mathcal{O} \text{ is bounded and there exists } T_{\mathcal{O}} \text{ such that } t_{\mathbf{b}}(y) \le T_{\mathcal{O}} \text{ for a.e. } y \in \mathcal{O}. $$

In that case, the semigroup $S_B$ is explicitly given by

$$ (S_B(t)f_0)(y) = \begin{cases} f_0(Y_{-t}(y)) \exp(-\int_0^t K(Y_{\tau-t}(y))d\tau), & \text{if } t \in (0, t_{\mathbf{b}}(y)) \\ 0 & \text{otherwise,} \end{cases} $$

and in particular $S_B(t)f = 0$ for any $f$ and any $t > T_{\mathcal{O}}$. We immediately deduce $\kappa_B = -\infty$ and thus $\kappa_0 > \kappa_B$ because we have established that $\kappa_0 \in \mathbb{R}$. We next define $\mathcal{A}f := \mathcal{K}[f]$. Using Lemma 2.8 and Remark 2.9-(2) or Lemma 2.13 and Remark 2.14-(1) depending on the assumption (8.96) made on $\mathcal{K}$, we deduce that the condition (H3) holds in both cases discussed above. Under the first condition in (8.96), we conclude to the existence of eigenvalue triplet $(\lambda_1, f_1, \phi_1) \in \mathbb{R} \times L^p_m \times L^{p'_1}$. Under the second condition in (8.96), we may also get the same conclusion by using [337, Cor. 1 of Thm. II.9.9] when $p = 1$ or by observing that the dual problem is similar to the primal problem when $p > 1$ and thus we may apply the same arguments for the dual problem as those explained above for the primal problem.

**Condition (H4).** Let us consider $\lambda > \lambda^{**}$ and $0 \le f \in L^p_{m_\mathcal{O}}(\mathcal{O})$ a (renormalized) solution to

$$ \lambda f + a \cdot \nabla f + bf - \mathcal{K}[f] = F \quad \text{in } \mathcal{O}, \quad \gamma_-f = \mathcal{R}[f, \gamma_+f] \quad \text{on } \Sigma_-, $$

with $0 \le F \in L^p_m(\mathcal{O})$. If $f \ne 0$, there exists $x_1 \in \mathcal{O}$ such that $\int_{B(x_1, r_1)} f(z) dz > 0$. From (8.94), we deduce

$$ \mathcal{K}[f](y) \ge \int_{B(x_1, r_1)} \kappa(y, z)f(z) dz > 0, \quad \forall y \in B(x_1, r_1). $$

Now, we argue similarly as during the proof of Lemma 8.14 and in particular we use the same notations. For $A \subset B(x_1, r_1)$, we define the solution $0 \le \varphi \in L^{p'_1} \cap L^\infty$ to the equation

$$ \lambda\varphi - \operatorname{div}(a\varphi) + b\varphi = \mathbf{1}_A \quad \text{in } \mathcal{O}, \quad \gamma_+\varphi = 0 \quad \text{on } \Sigma_+, $$

thanks to Lemma 8.12 and Lemma 8.13, and we observe that $\varphi \ne 0$ on $B(x_1, r_1)$ if $|A| > 0$. For the renormalizing function $\beta_\delta$ and a truncation function $\chi_R$, we compute

$$ \begin{aligned} 0 &\ge \int_{\Sigma} a \cdot n \beta_{\delta}(\gamma f) \gamma \varphi \chi_R \\ &= \int_{\mathcal{O}} [\beta'_{\delta}(f)(F + \mathcal{K}[f])\varphi - \beta_{\delta}(f)\mathbf{1}_A]\chi_R \\ &\quad + \int_{\mathcal{O}} (\beta_{\delta}(f) - f\beta'_{\delta}(f))(\lambda + b)\varphi \chi_R + \int_{\mathcal{O}} \varphi \beta(f) \frac{a}{R} \cdot (\nabla \chi)_R. \end{aligned} $$
and next

$$ \int_0^\alpha \int_{\Sigma_{\alpha'}} \varphi\beta(\gamma_{\alpha'}g) a \cdot n \, d\sigma_{\alpha'} d\alpha' = \int_{\mathcal{O}\setminus\mathcal{O}_\alpha} \varphi\beta(g) a \cdot n \, dy. $$

Together with (8.82) and (8.81), we have established

$$ \int_0^\alpha \int_{\Sigma} [\varphi\beta(\gamma_{\alpha'}g) a \cdot n] \circ \theta_{\alpha'}(z) J_{\alpha'}(z) d\sigma_z d\alpha' = \int_0^\alpha \int_{\Sigma} [\varphi\beta(g) a \cdot n] \circ \theta_{\alpha'}(z) J_{\alpha'}(z) d\sigma d\alpha', $$

so that

$$ \beta(\gamma_{\alpha'}g) a \cdot n = \beta(g) a \cdot n \quad \text{a.e. on } \Sigma_{\alpha'} \text{ for a.e. } \alpha' \in (0, \alpha), $$

from what (8.83) follows. On the other hand, using (8.81) and (8.85) together, we have

$$ \int_{\Sigma} \varphi\beta(\gamma g) a \cdot n \, d\sigma = \lim_{\alpha \to 0} \int_{\Sigma} [\varphi\beta(\gamma_{\alpha}g) a \cdot n] \circ \theta_{\alpha} J_{\alpha} \, d\sigma_{\alpha}, $$

which implies

$$ (8.86) \qquad [\beta(\gamma_{\alpha}g) a \cdot n] \circ \theta_{\alpha} J_{\alpha} \to \beta(\gamma g) a \cdot n \quad * \sigma(L_{\text{loc}}^{\infty}(\Sigma), L_c^1(\Sigma)). $$

Repeating the same argument with $g := a \cdot n$ and using that $J_{\alpha} \to I$ uniformly, we get

$$ [\beta(a \cdot n) a \cdot n] \circ \theta_{\alpha} \to \beta(a \cdot n) a \cdot n \quad * \sigma(L_{\text{loc}}^{\infty}(\Sigma), L_c^1(\Sigma)), $$

for any $\beta \in C^1(\mathbb{R})$. Choosing $\beta(s) = 1$ and $\beta(s) = s$, we classically deduce that $[a \cdot n] \circ \theta_{\alpha} \to a \cdot n$ a.e. on $\Sigma$. We finally conclude to (8.84) by gathering that last information with (8.86) written for $\beta(s) = s$ and $\beta(s) = s^2$. $\square$

**Remark 8.31.** During the proof, we have in fact established that

$$ [0, \alpha_{\mathcal{O}}] \to L^1(\Sigma); \quad \alpha \mapsto [\gamma_{\alpha}g a \cdot n] \circ \theta_{\alpha} $$

is continuous.

**Lemma 8.32.** We make the additional assumptions (8.78) and (8.79). If $Y$ is an almost everywhere flow associated to (8.6), then the function $t_b \in L(\mathcal{O})$ is a renormalized solution to the equation

$$ (8.87) \qquad a \cdot \nabla t_b = 1 \quad \text{in } \mathcal{O}, \quad \gamma_{-}t_b = 0 \quad \text{on } \Sigma_{-} \cap \mathcal{Y}. $$

*Proof of Lemma 8.32. Step 1.* We fix $\beta \in C_*^1(\mathbb{R})$ and we recall that $\beta(t_b(Y_s(y))) = \beta(t_b(y) + s)$ for any $s \in \mathbb{R}$ for a.e. $y \in \mathcal{O}$. For any $\varphi \in C_c^1(\mathcal{O})$, we may compute

$$ \begin{align*} \int_{\mathbb{R}^d} \beta'(t_b(y) + s)\varphi(y) \, dy &= \frac{d}{ds} \int_{\mathbb{R}^d} \beta(t_b(y) + s)\varphi(y) \, dy \\ &= \frac{d}{ds} \int_{\mathbb{R}^d} \beta(t_b(Y_s(y)))\varphi(y) \, dy \\ &= \frac{d}{ds} \int_{\mathbb{R}^d} \beta(t_b(y))\varphi(Y_{-s}(y))e^{-\int_{-s}^0 (\text{diva})(Y_\tau(y))d\tau} \, dy \\ &= \int_{\mathbb{R}^d} \beta(t_b(y))[-a \cdot \nabla\varphi - (\text{diva})\varphi](Y_{-s}(y))e^{-\int_{-s}^0 (\text{diva})(Y_\tau(y))d\tau} \, dy, \end{align*} $$

where we have used the change of variables property (8.75) in the third line. Taking $s = 0$, we conclude to

$$ \int_{\mathbb{R}^d} \beta'(t_b)\varphi \, dy = \int_{\mathbb{R}^d} \beta(t_b)[-a \cdot \nabla\varphi - (\text{diva})\varphi] \, dy, $$

which is nothing but the distributional formulation of the equation

$$ a \cdot \nabla \beta(t_b) = \beta'(t_b). $$

That last family of equations is the renormalized formulation of equation (8.87) in the domain.

*Step 2.* Using lemma 8.30 with $g := \beta(t_b)$, we have

$$ \gamma_{\alpha}\beta(t_b) \circ \theta_{\alpha} \to \beta(\gamma t_b) \quad \text{a.e. on } \Sigma. $$

Using Lemma 8.29, we also have

$$ \gamma_{\alpha}\beta(t_b) \circ \theta_{\alpha} \to 0 \quad \text{a.e. on } \Sigma_{-} \cap \mathcal{Y}. $$
Passing first to the limit $R \to \infty$ and next to the limit $\delta \to 0$, we deduce

$$0 \geq \int_{\mathcal{O}} [(F + \mathcal{H}[f])\varphi - f \mathbf{1}_A],$$

so that in particular

$$\int_A f dy \geq \int_{B(x_1, r_1)} \varphi \mathcal{H}[f] > 0.$$

This being true for any $A \subset B(x_1, r_1)$, we deduce $f > 0$ a.e. on $B(x_1, r_1)$. By a classical continuity argument, we conclude that $f > 0$ a.e. on $\mathcal{O}$. We have thus established (H4) for $\lambda > \lambda^{**}$ from what we immediately and classically deduce the general case $\lambda \in \mathbb{R}$.

**Condition (H5).** Assume that $(\lambda, f) \in \mathbb{C} \times D(\mathcal{L})$ satisfies

$$\mathcal{L}|f| = (\Re e\lambda)|f| \quad \text{in } \mathcal{O}, \quad \mathcal{R}[|f|, \gamma_+|f|] = \gamma_-|f| \quad \text{on } \Sigma_-,$$

and

$$\mathcal{L}|f| = \Re e(\operatorname{sign} f)\mathcal{L}f \quad \text{in } \mathcal{O}, \quad \mathcal{R}[|f|, \gamma_+|f|] = \Re e(\operatorname{sign} \gamma_- f)\mathcal{R}[f, \gamma_+ f] \quad \text{on } \Sigma_-.$$

From (H4) and the first identity, we know that $|f| > 0$ a.e. on $\mathcal{O}$. Using the second identity, we get

$$\mathcal{H}[|f|] = \Re e(\operatorname{sign} f)\mathcal{H}[f].$$

Writing $f = e^{i\alpha}|f|$, we deduce

$$\int_{\mathcal{O}} k|f_*|(1 - \cos(\alpha - \alpha_*))dy_* = 0 \quad \text{a.e. on } \mathcal{O}.$$

Using (8.94), we deduce

$$\int_{B(y, r_\infty)} |f_*|(1 - \cos(\alpha - \alpha_*))dy_* = 0,$$

and thus $\alpha = \alpha_*$ a.e. on $\mathcal{O} \times \mathcal{O}$. That means $f = u|f|$, for a constant $u = S^1$, that completes the proof of the fact that $\mathcal{L}$ satisfies the reverse Kato's inequality condition (H5).

We summarize our analysis in the following result which is a straightforward consequence of the above checked conditions together with Theorem 2.21, Theorem 4.13, Theorem 5.16 and Theorem 5.23. We state the available result in that situation.

**Theorem 8.36.** We assume that $a, b, \mathcal{H}$ and $\mathcal{R}$ satisfy the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) for some weight function $m: \mathcal{O} \to [1, \infty)$ and some exponent $p \in [1, \infty)$. Consider the semigroup $S_{\mathcal{L}}$ associated to the transport equation (8.1)-(8.3) through Corollary 8.25. We assume further that $\mathcal{H}$ satisfies the strong positivity conditions (8.94) together with (8.95) and the first compactness property formulated in (8.96). We finally assume that (8.102) holds or (8.103) holds. In both cases, the conclusion (C3) holds as well as the ergodicity (E2) in $L_{\phi_1}^1$.

We are not aware of any similar result for such a general transport equation, see however the next sections where more specific transport like equations are discussed. We do not try to improve the convergence result in the general case, but rather we aim to make one step further in the following particular situation where Doblin approach may be used.

**Doblin condition.** We suppose here that $\mathcal{O}$ is bounded, $K \in L^\infty(\mathcal{O})$, $R_{\Sigma}^*\mathbf{1} = R_{\Sigma}\mathbf{1} = \mathbf{1}$, and $k(y, y_*) \geq k_0 > 0$. We aim at establishing the Doblin condition

$$S_{\mathcal{L}}(T)f_0 \geq \kappa \langle f_0, \mathbf{1} \rangle,$$

which is (6.2) with $\psi_0 = \mathbf{1}$ and $g_0 = \kappa\mathbf{1}$. From (8.64) we have

$$\frac{d}{dt} \int_{\mathcal{O}} f dy = \int_{\mathcal{O}} f K dy \geq -\|K\|_{\infty} \int_{\mathcal{O}} f dy$$

and so

$$\int_{\mathcal{O}} f(t, y) dy \geq e^{-\|K\|_{\infty} t} \int_{\mathcal{O}} f_0(y) dy.$$

Now we define, for $\varphi_0 \in C_c^1(\mathcal{O})$, $\varphi_0 \geq 0$, $\int \varphi_0 = 1$, the solution $\varphi$ to the equation

$$\left\{ \begin{array}{l} \partial_t \varphi + \operatorname{div}(a\varphi) = 0, \\ \gamma_+ \varphi = R_{\Sigma}^*[\gamma_- \varphi]. \end{array} \right.$$
Both together, we find $\gamma_{-}t_{\mathbf{b}} = 0$ on $\Sigma_{-} \cap \mathcal{Y}$.

<figure><img src="image_9.png" /></figure>

We establish the main result of this section.

**Theorem 8.33 (characteristics method).** Assume that $Y$ is an almost everywhere flow associated to (8.6) with a satisfying (8.73). For any $g_0 \in L^\infty(\mathcal{O})$, $\mathfrak{g} \in L^\infty((0,T) \times \Sigma_-)$, $T > 0$, $b \in L^\infty(\mathcal{O})$, the function

$$ (8.88) \qquad \bar{g}(t, y) := g_0(Y_{-t}(y))e^{-\int_0^t b(Y_{\tau-t}(y)) d\tau} \mathbf{1}_{t<t_{\mathbf{b}}(y)} \\ \qquad \qquad + \mathfrak{g}(t-t_{\mathbf{b}}(y), y_{\mathbf{b}}(y))e^{-\int_0^t b^{(y)}(Y_{\tau-t_{\mathbf{b}}(y)}(y)) d\tau} \mathbf{1}_{t>t_{\mathbf{b}}(y)} $$

is the unique solution in $C([0,T]; L_{\text{loc}}^1(\mathcal{O})) \cap L^\infty((0,T) \times \mathcal{O})$ to the evolution transport equation

$$ (8.89) \qquad \begin{cases} \frac{\partial g}{\partial t} + a \cdot \nabla g + bg = 0 & \text{on } (0,T) \times \mathcal{O}, \\ \gamma_-g = \mathfrak{g} & \text{on } (0,T) \times \Sigma_-, \\ g(0, \cdot) = g_0 & \text{on } \mathcal{O}. \end{cases} $$

*First proof of Theorem 8.33.* We additionally assume that (8.75), (8.78) and (8.79) hold, that $\mathfrak{g} \in C((0,T) \times \Sigma_-)$, $\text{supp} \mathfrak{g} \subset (0,T) \times (\Sigma_- \cap \mathcal{Y})$ and $b \in C(\mathcal{O})$, and we mostly repeat the proof of [254, Prop. 1]. From the above definition, for a.e. $y \in \mathcal{O}$ and any $t \in (0, \infty)$, we have

$$ (8.90) \qquad \bar{g}(t+s, Y_s(y))e^{\int_0^s b(Y_\tau(y))d\tau} = \bar{g}(t, y), \quad \forall s \ge -t. $$

Let us then fix $\varphi \in \mathcal{D}((0,T) \times \mathcal{O})$ and let us extend $\bar{g}$ and $\varphi$ by 0 outside of $\mathcal{O}$. We compute

$$ \begin{aligned} 0 &= \frac{d}{ds} \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y) \varphi(t, y) \, dydt \\ &= \frac{d}{ds} \int_0^T \int_{\mathbb{R}^d} \bar{g}(t + s, Y_s(y)) e^{\int_0^s b(Y_\tau(y)) d\tau} \varphi(t, y) \, dydt \\ &= \frac{d}{ds} \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y) \varphi(t - s, Y_{-s}(y)) e^{\int_0^s (b-\text{diva})(Y_\tau(y)) ds} \, dydt \\ &= \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y) \frac{d}{ds} [\varphi(t - s, Y_{-s}(y)) e^{\int_0^s (b-\text{diva})(Y_\tau(y)) ds}] \, dydt \\ &= \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y) [-\partial_t \varphi - a \cdot \nabla \varphi + (b - \text{diva})\varphi] (t - s, Y_{-s}(y)) \, dydt, \end{aligned} $$

where we have used the relation (8.90) in the second line and the change of variables property (8.75) in the third line. Taking $s = 0$, we get

$$ 0 = \int_0^T \int_{\mathbb{R}^d} \bar{g}(t, y) [-\partial_t \varphi - a \cdot \nabla \varphi + (b - \text{diva})\varphi](t, y) \, dydt, $$

which exactly means that $\bar{g}$ is a solution to equation (8.89) in the distributional sense. Now, because $t_{\mathbf{b}}(y) > 0$ for any $y \in \mathcal{Y}$, we have $\bar{g}(0, y) = g_0(y)$. Take $t > 0$, and for $\alpha \in (0, \alpha_\mathcal{O})$, let us denote $A_\alpha := \{y \in \Sigma_- \cap \mathcal{Y}; t_{\mathbf{b}} \circ \theta_\alpha(y) < t\}$, so that

$$ \bar{g}(t, \theta_{\alpha}(y)) = \mathfrak{g}(t - t_{\mathbf{b},\alpha}(y), y_{\mathbf{b},\alpha}(y))e^{-\int_0^t b^{(y)}(Y_{\tau-t_{\mathbf{b},\alpha}(y)}(\theta_{\alpha}(y)) d\tau} \quad \text{on } A_{\alpha}, $$

where we use the shorthands $t_{\mathbf{b},\alpha} := t_{\mathbf{b}} \circ \theta_\alpha$ and $y_{\mathbf{b},\alpha} := y_{\mathbf{b}} \circ \theta_\alpha$. From $\theta_{\alpha'}(y) \to y$ as $\alpha' \to 0$ when $y \in \Sigma$ and Lemma 8.29, we deduce

$$ \bar{g}(t, \theta_{\alpha'}(\cdot)) \to \mathfrak{g}(t, \cdot) \quad \text{as } \alpha' \to 0, $$

on $A_\alpha$ for any fixed $\alpha > 0$. Because $\cup_{\alpha>0} A_\alpha = \Sigma_- \cap \mathcal{Y}$, the same convergence holds on $\Sigma_- \cap \mathcal{Y}$. On the other hand, we have $\gamma_\alpha \bar{g} \circ \theta_\alpha \to \gamma \bar{g}$ as $\alpha \to 0$ a.e. on $\Sigma \setminus \Sigma_0$, from Lemma 8.30. We deduce that $\gamma \bar{g} = \mathfrak{g}$ on $\Sigma_- \cap \mathcal{Y}$.

*Second proof of Theorem 8.33.* We do not make any additional assumption and we mainly repeat the proof presented in [203, Sec. 3]. Consider the unique solution $g \in C([0, T]; L_{\text{loc}}^1(\mathcal{O})) \cap L^\infty((0, T) \times \mathcal{O})$ to the transport equation (8.89). Regularizing by convolution

$$ g_\varepsilon := g *_{t,x,\varepsilon} \rho_\varepsilon, $$
We have

$$ \frac{d}{dt} \int_{\mathcal{O}} \varphi = 0, \quad \text{and so} \quad \int_{\mathcal{O}} \varphi(t, y) dy = \int_{\mathcal{O}} \varphi_0(y) dy = 1, $$

and

$$ \frac{d}{dt} \int_{\mathcal{O}} f \varphi = \int_{\mathcal{O}} \mathcal{H}[f] \varphi - \int_{\mathcal{O}} K f \varphi \geq k_0 \int_{\mathcal{O}} f - \|K\|_{\infty} \int_{\mathcal{O}} f \varphi. $$

We deduce from Grönwall's inequality that, for any fixed $T > 0$,

$$ \begin{aligned} \int_{\mathcal{O}} f(T, y) \varphi_0(y) dy &\ge e^{-\|K\|_{\infty} t} \int_{\mathcal{O}} f_0(y) \varphi_0(y) dy + k_0 \int_0^T e^{-(T-t)\|K\|_{\infty}} \int_{\mathcal{O}} f(t, y) dy dt \\ &\ge k_0 T e^{-\|K\|_{\infty}} \int_{\mathcal{O}} f_0(y) dy =: \kappa \langle f_0, \mathbf{1} \rangle. \end{aligned} $$

This is nothing but the Doblin condition (6.2) since $\varphi_0$ is any non-negative function in $C_c^1(\mathcal{O})$ with $\int \varphi_0 = 1$.

In order to verify (6.3) in a quantitative way, we suppose that the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) are verified with the weight function $m = \mathbf{1}$ and the exponent $p = 1$. Note that in this case we have $\varpi = K \ge 0$. The first condition in (8.35) then imposes that $K \in L^{\infty}(\mathcal{O})$, and (8.102) reads

$$ \eta_0 > \|\langle K \rangle\|_{L^{\infty}} C_0 + M_{\mathcal{H}} C_0 + \|a\|_{L^{\infty}(\mathcal{B}_0)} C_1 - \|b\|_{L^{\infty}(\mathcal{B}_0)} C_0. $$

We also assume that

$$ (8.104) \qquad \mathcal{R}_{\Sigma}^{*} \mathbf{1} = \mathcal{R}_{\Sigma} \mathbf{1} = \mathbf{1}, $$

and

$$ (8.105) \qquad \forall y, y_* \in \mathcal{O}, \quad k_0 \le k(y, y_*) \le k_1 $$

for some $k_1 > k_0 > 0$,

**Theorem 8.37.** We assume that $\mathcal{O}$ is bounded and that the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) are satisfied by a, b, $\mathcal{H}$ and $\mathcal{R}$ for the weight function $m = \mathbf{1}$ and the exponent $p = 1$. We assume further that $\mathcal{H}$ satisfies the strong positivity conditions (8.94) together with (8.95) and the first compactness property formulated in (8.96). We finally assume that (8.102), (8.104) and (8.105) are satisfied. Then the exponential convergence in $(\mathbf{E}3_1)$ holds in $L^1$ with constructive constants $C$ and $\omega$.

*Proof of Theorem 8.37.* We work in $X = L^1(\mathcal{O})$ and we normalize $\phi_1$ by $\|\phi_1\|_{L^\infty} = 1$. We have proved above that (6.2) holds true with $\psi_0 = \mathbf{1}$ and $g_0 = \kappa \mathbf{1}$ for some explicit $\kappa > 0$, recalling that the assumption that $K \in L^\infty$ is nothing but the first condition in (8.35) when $m = \mathbf{1}$ and $p = 1$ since $b = K + \text{diva}$. Due to the normalization $\|\phi_1\|_{L^\infty} = 1$, the condition (6.4) holds with $R_0 = 1$. It only remains to check the validity of (6.3) in order to be able to apply Theorem 6.2. Since we assume that the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) are satisfied for the weight function $m = \mathbf{1}$ and the exponent $p = 1$, we have that $\mathcal{R}_B(\lambda_1) : L^1 \to L^1$ with $\|\mathcal{R}_B(\lambda_1)\|_{\mathcal{B}(L^1)} \le \frac{1}{\kappa_0 - \kappa_B}$. This yields by duality that $\mathcal{R}_B^*(\lambda_1) : L^\infty \to L^\infty$ with $\|\mathcal{R}_B^*(\lambda_1)\|_{\mathcal{B}(L^\infty)} \le \frac{1}{\kappa_0 - \kappa_B}$. Since $k$ is bounded by the constant $k_1$, we have on the other hand that $\mathcal{A}^* = \mathcal{H}^* : L^1 \to L^\infty$ with $\|\mathcal{A}^*\|_{\mathcal{B}(L^1, L^\infty)} \le k_1$. We thus get

$$ 1 = \|\phi_1\|_{L^\infty} \le \frac{k_1}{\kappa_0 - \kappa_B} \|\phi_1\|_{L^1}, $$

which yields (6.3) with $r_0 = \kappa(\kappa_0 - \kappa_B)/k_1$, and the proof is complete. $\square$

<figure><img src="image_9.png" /></figure>
for a time and space dependent mollifier sequence $\rho_\varepsilon$ similarly as in (8.20), we have $g_\varepsilon \in C^1([0, T] \times \bar{O})$ and

$$ \frac{\partial g_\varepsilon}{\partial t} + a \cdot \nabla g_\varepsilon + bg_\varepsilon = R_\varepsilon \quad \text{on } (0, T) \times \mathcal{O}, $$

with the usual commutator $R_\varepsilon := [a \cdot \nabla g + bg, \rho_\varepsilon]$. We know from a classical time and space variant of Lemma 8.5 (see also [138, Lem. II.1], [270, Lem. 1], [70, Lem. 3.1] and the proof in [203, Sec. 3]) that $R_\varepsilon \to 0$ in $L^1$. Because $g_\varepsilon$ is smooth and $Y$ is an almost everywhere flow, we may set $H^1(s, y) := H(t+s, Y_s(y))$, $\mathfrak{B}(s, y) := \int_0^s b(Y_\tau(y))d\tau$ and compute

$$ \frac{d}{ds}[(g_{\varepsilon}^{\sharp} e^{\mathfrak{B}})(s, y)] = (R_{\varepsilon}^{\sharp} e^{\mathfrak{B}})(s, y), $$

from what we get

$$ \begin{aligned} \tilde{g}_{\varepsilon}(t, y) &:= g_{\varepsilon}(t, y) - g_{\varepsilon}(0, Y_{-t}(y))e^{\mathfrak{B}(-t, y)}\mathbf{1}_{t<t_{\mathbf{b}}(y)} - g_{\varepsilon}(t - t_{\mathbf{b}}(y), y_{\mathbf{b}}(y))e^{\mathfrak{B}(-t_{\mathbf{b}(y)}, y)}\mathbf{1}_{t>t_{\mathbf{b}}(y)} \\ &= \int_{0}^{t} (R_{\varepsilon}^{\sharp} e^{\mathfrak{B}})(s, y)ds\mathbf{1}_{t<t_{\mathbf{b}}(y)} + \int_{t-t_{\mathbf{b}}(y)}^{t} (R_{\varepsilon}^{\sharp} e^{\mathfrak{B}})(s, y)ds\mathbf{1}_{t>t_{\mathbf{b}}(y)}, \end{aligned} $$

for a.e. $y \in \mathcal{O}$ and any $t > 0$ and where we have used (8.77) for getting rid of the set $\{y \in \mathcal{O}; t_{\mathbf{b}}(y) = t\}$. For $T, \varrho > 0$ and setting $\mathcal{U}_{T, \varrho} := (0, T) \times (\mathcal{O} \cap B_{\varrho})$, we deduce

$$ \begin{aligned} \int_{\mathcal{U}_{T, \varrho}} |\tilde{g}_{\varepsilon}(t, y)| dydt &\le T \int_{\mathcal{U}_{T, \varrho}} |(R_{\varepsilon}^{\sharp} e^{\mathfrak{B}})(s, y)| dyds \\ &\le T e^{(C+\|b\|_{L^{\infty}})T} \int_{\mathcal{U}_{T, \varrho}} |R_{\varepsilon}(s, y)| dyds, \end{aligned} $$

from the near-incompressibility condition (8.74) of the flow. From Proposition 8.10 and Remark 8.11, we know that

$$ g_{\varepsilon}(t, \cdot) \to g(t, \cdot) \text{ in } L_{\text{loc}}^{1}(\bar{\mathcal{O}}) \text{ as } \varepsilon \to 0, \text{ for any } t \in [0, T]; $$

$$ g_{\varepsilon|\Sigma_{-}} \to \gamma_{-}g = \mathfrak{g} \text{ in } L_{\text{loc}}^{1}(\Sigma_{-}) \text{ as } \varepsilon \to 0. $$

Passing to the limit, we get

$$ \int_{\mathcal{U}_{T, \varrho}} |g(t, y) - g_0(Y_{-t}(y))e^{\mathfrak{B}(-t, y)}\mathbf{1}_{t<t_{\mathbf{b}}(y)} - \mathfrak{g}(t - t_{\mathbf{b}}(y), y_{\mathbf{b}}(y))e^{\mathfrak{B}(-t_{\mathbf{b}(y)}, y)}\mathbf{1}_{t>t_{\mathbf{b}}(y)}|dtdy = 0, $$

for any $T, \varrho > 0$, what is nothing but (8.88). $\square$

**Corollary 8.34 (Representation formula).** Under the assumptions of Lemma 8.14 in a smooth domain $\mathcal{O} \neq \mathbb{R}^D$, for any $\lambda > \lambda_{a,b,p}$, $G \in L^p(\mathcal{O})$, the unique solution $g \in L^p(\mathcal{O})$ to the stationary transport equation (8.29) (with $\mathfrak{g} = 0$) satisfies

$$ (8.91) \qquad g(y) = \int_{0}^{\infty} e^{-\lambda t} (S_{\mathbf{b}}(t)G)(y)dt, \quad \text{for a.e } y \in \mathcal{O}, $$

with $S_{\mathbf{b}}$ is defined by (8.49) in which formula $Y$ and $t_{\mathbf{b}}$ stand for the characteristics and backward exit time defined just as above.

*Proof of Corollary 8.34.* That is nothing but (2.13). $\square$

Adapting the second proof of Theorem 8.33, we obtain a more accurate characterization of the backward exit time $t_{\mathbf{b}}$ with more general assumptions on the vector field.

**Lemma 8.35.** Assume that $Y$ is an almost everywhere flow associated to (8.6) with a satisfying (8.73). The backward exit time $t_{\mathbf{b}}$ is the unique renormalized solution in $L(\mathcal{O})$ to the backward exit time problem

$$ (8.92) \qquad a \cdot \nabla \tau = 1 \text{ in } \mathcal{O}, \quad \gamma_{-} \tau = 0 \text{ on } \Sigma_{-}. $$

We also have

$$ (8.93) \qquad t_{\mathbf{b}} = \int_{0}^{\infty} S_{0}(t) \mathbf{1} dt, $$

where $\mathbf{1}$ stands for the unit function in $\mathcal{O}$ and $S_0$ is defined by (8.49) with $b = 0$ and in which formula $Y$ and $t_{\mathbf{b}}$ stand for the characteristics and backward exit time defined just as above.
8.6. **A word about the renewal equation.** We look at the case $\mathcal{O} = (0, +\infty)$ and $a(y) = 1$, which corresponds to the equation

$$ (8.106) \qquad \partial_t f + \partial_y f + Kf = 0 $$

with the boundary condition

$$ (8.107) \qquad (\gamma_{-}f)(t, 0) = \int_{0}^{\infty} r_{\mathcal{O}}(y_{*})f(t, y_{*})dy_{*}. $$

This renewal age structured model is standard in structured population dynamics, and the Krein-Rutman theorem is well-known for it, see for instance [34, 158, 187, 196, 338, 358]. The existence and uniqueness of $(\lambda_1, f_1, \phi_1)$ can even be obtained by explicit computations. However, it is not covered by the cases considered in Section 8.5 because $\mathcal{K} = 0$ here.

The singularity of this transport equation lies in the fact that (H2) is only guaranteed by the boundary condition. To fall into our splitting framework, we may replace the boundary condition by a singular source term $Af = (\mathcal{R}_{\mathcal{O}}f)(0)\delta_0$, where $\delta_0$ is the Dirac mass at the origin, and write $\mathcal{L} = \mathcal{A} + \mathcal{B}$ with $\mathcal{B}$ the generator of the free transport equation with zero flux boundary condition. This forces working in a space of measures, as in [277, 280]. We briefly present an alternative approach, which is more in the spirit of [34, 168] and which consists in working in the Lebesgue space $L^1$, first to solve the dual problem in $L^\infty = (L^1)'$ and next to use for instance Doblin's contraction to solve the primal problem.

We assume here that

$$ (8.108) \qquad 0 \le K, r_{\mathcal{O}} \in L_{\text{loc}}^{\infty}(0, \infty), \quad (r_{\mathcal{O}} - \alpha K)_{+} \in L^{\infty}, $$

$$ (8.109) \qquad \lim_{y \to \infty} K(y) = +\infty, \quad \liminf_{y \to +\infty} r_{\mathcal{O}}(y) > 0, $$

for some $\alpha \in (0, 1)$, and we verify the usual conditions for the direct or the dual problem.

**Condition (H1).** Under assumption (8.108), the age structured equation (8.106)-(8.107) is well-posed in $L^1$ thanks to Proposition 8.23 and we may associate to it a positive semigroup $S_{\mathcal{L}}$ in $L^1$ with growth bound $\omega(S_{\mathcal{L}}) \le \kappa_1 := \|(r_{\mathcal{O}} - K)_+\|_{L^\infty}$ thanks to Corollary 8.25. We deduce that (H1) holds for the primal problem and thus also for the dual problem thanks to Lemma 2.2 and Lemma 2.3.

**Condition (H2).** The generator of the dual problem is

$$ \mathcal{L}^*\phi = \partial_y\phi - K(y)\phi + \phi(0)r_{\mathcal{O}}(y) $$

with domain $D(\mathcal{L}^*) \subset W_{\text{loc}}^{1,\infty}(\mathcal{O})$. From the second hypothesis in (8.109), there exist $y_0, \eta_0 \in (0, \infty)$ such that $r_{\mathcal{O}}(y) \ge \eta_0$ for any $y \ge y_0$. We then define

$$ \phi_0(y) = \mathbf{1}_{[0, y_0)} + \eta_0(y_1 - y)\mathbf{1}_{[y_0, y_1)}, \quad y_1 := y_0 + 1/\eta_0, $$

and we compute

$$ \mathcal{L}^*\phi_0 = r_{\mathcal{O}}(y) - K \ge -\|K - r_{\mathcal{O}}\|_{L^\infty(0, y_0)} \quad \text{on } (0, y_0), $$

$$ \mathcal{L}^*\phi_0 = r_{\mathcal{O}}(y) - \eta_0 - K\phi_0 \ge -\|K\|_{L^\infty(y_0, y_1)}\phi_0 \quad \text{on } (y_0, y_1), $$

$$ \mathcal{L}^*\phi_0 = 0 \quad \text{on } (y_1, \infty), $$

so that in the three case $\mathcal{L}^*\phi_0 \ge \kappa_0\phi_0$ with $\kappa_0 := -\max(\|K - r_{\mathcal{O}}\|_{L^\infty(0, y_0)}, \|K\|_{L^\infty(y_0, y_1)})$. Using Lemma 2.4-(i), we have thus established that $\mathcal{L}$ satisfies (H2) with constructive constant $\kappa_0$.

**Condition (H3) on the dual problem.** We define the splitting $\mathcal{L}^* = \mathcal{A}^* + \mathcal{B}^*$ with $\mathcal{A}^*\phi := (\mathcal{R}_{\mathcal{O}}^*\phi)(y) = \phi(0)r_{\mathcal{O}}(y)$. From the first hypothesis in (8.109), for any $\kappa_* \le 0$ there exists $y_* \in [0, \infty)$ such that $K(y) \ge -\kappa_*$ for any $y \ge y_*$. Defining $m_* := e^{\kappa_*y}\mathbf{1}_{[0, y_*)} + e^{\kappa_*y_*}\mathbf{1}_{[y_*, \infty)}$, we compute

$$ \mathcal{B}^*m_* = \kappa_*e^{\kappa_*y}\mathbf{1}_{[0, y_*)} - Km_* \le \kappa_*m_* $$

Together with Proposition 8.23 and Corollary 8.25, we deduce that the operator $\mathcal{B} - \kappa_*$, with domain $D(\mathcal{B}) := \{f \in L^1(\mathcal{O}); \partial_yf + Kf \in L^1(\mathcal{O}), f(0) = 0\}$, generates a contraction semigroup in $L^1_{m_*}(\mathcal{O})$, and thus a bounded semigroup in $L^1(\mathcal{O})$ because $m_*, m_*^{-1} \in L^\infty(\mathcal{O})$. In other words, we have established that $\omega(S_{\mathcal{B}}) = -\infty$. Now, we see that $\mathcal{R}_{\mathcal{B}*}(\lambda) : L^\infty \to D(\mathcal{B}^*) \subset W_{\text{loc}}^{1,\infty}([0, \infty))$ is bounded for any $\lambda \in \mathbb{R}$ and thus $\mathcal{A}^*\mathcal{R}_{\mathcal{B}*}(\lambda) : L^\infty \to L^\infty$ is compact for any $\lambda \in \mathbb{R}$. We deduce from Lemma 2.8 and Remark 2.10, that $\mathcal{L}^*$ satisfies (H3).
*Proof of Lemma 8.35. Step 1. Existence.* From Lemma 8.12, for any $\lambda > \lambda_{\infty}^* := 0$, there exists a (unique) solution $\tau_{\lambda} \in L^{\infty}(\mathcal{O})$ to the truncated backward exit time problem

$$ \lambda\tau_{\lambda} + a \cdot \nabla\tau_{\lambda} = 1 \text{ in } \mathcal{O}, \quad \gamma_{-}\tau_{\lambda} = 0 \text{ on } \Sigma_{-}. $$

From the weak maximum principle, we have $\tau_{\lambda} \ge 0$. As a consequence, for $0 < \lambda < \lambda'$, we have

$$ \lambda(\tau_{\lambda} - \tau_{\lambda'}) + a \cdot \nabla(\tau_{\lambda} - \tau_{\lambda'}) = (\lambda' - \lambda)\tau_{\lambda'} \ge 0 \text{ in } \mathcal{O}, \quad \gamma_{-}(\tau_{\lambda} - \tau_{\lambda'}) = 0 \text{ on } \Sigma_{-}. $$

From the weak maximum principle again, we deduce that $\tau_{\lambda} - \tau_{\lambda'} \ge 0$, and $(\tau_{\lambda_n})$ is an increasing sequence when $\lambda_n \searrow 0$. We set $\tau := \lim_{n \to \infty} \tau_{\lambda_n}$, so that $\tau \in L(\mathcal{O})$ is a nonnegative renormalized solution to the backward exit time problem.

**Step 2. Characterization.** By definition, for any $\beta \in C_*^1(\mathbb{R})$, $\beta(0) = 0$, the function $\tau$ satisfies

$$ a \cdot \nabla \beta(\tau) = \beta'(\tau) \text{ in } \mathcal{O}, \quad \gamma_{-} \beta(\tau) = 0 \text{ on } \Sigma_{-}. $$

With the notations of (8.20), we define $b_{\varepsilon} := \beta(\tau) *_{\varepsilon} \rho_{\varepsilon}$, $B_{\varepsilon} := \beta'(\tau) *_{\varepsilon} \rho_{\varepsilon}$, and thanks to Lemma 8.5, we have thus

$$ a \cdot \nabla b_{\varepsilon} = B_{\varepsilon} + r_{\varepsilon} \text{ in } \mathcal{D}'(\mathcal{O}), $$

with

$$ b_{\varepsilon} \to \beta(\tau), \quad B_{\varepsilon} \to \beta'(\tau), \quad r_{\varepsilon} \to 0, $$

respectively in $L_{\text{loc}}^p(\bar{\mathcal{O}})$, $L_{\text{loc}}^p(\bar{\mathcal{O}})$ and $L_{\text{loc}}^1(\bar{\mathcal{O}})$ for any $p \in [1, \infty)$. Because then

$$ \frac{d}{ds} b_{\varepsilon} \circ Y_s = (a \cdot \nabla b_{\varepsilon})(Y_s) = (B_{\varepsilon} + r_{\varepsilon})(Y_s), $$

and defining

$$ \tilde{b}_{t,\varepsilon}(y) := \left\{ \begin{aligned} & b_{\varepsilon}(Y_{-t}(y)) + \int_{-t}^{0} B_{\varepsilon}(Y_s(y)) ds \mathbf{1}_{t<t_{\mathbf{b}}(y)} \\ & + \left\{ (\gamma_{-}b_{\varepsilon})(y_{\mathbf{b}}(y)) \right\} + \int_{-t_{\mathbf{b}}(y)}^{0} B_{\varepsilon}(Y_s(y)) ds \mathbf{1}_{t>t_{\mathbf{b}}(y)}, \end{aligned} \right. $$

we have, as in the second proof of Theorem 8.33,

$$ b_{\varepsilon}(y) - \tilde{b}_{t,\varepsilon}(y) = R_{t,\varepsilon} := \int_{-t}^{0} r_{\varepsilon}(Y_s(y)) ds \mathbf{1}_{t<t_{\mathbf{b}}(y)} + \int_{-t_{\mathbf{b}}(y)}^{0} r_{\varepsilon}(Y_s(y)) ds \mathbf{1}_{t>t_{\mathbf{b}}(y)}. $$

Arguing similarly as in the second proof of Theorem 8.33, we have $\tilde{b}_{t,\varepsilon} \to \tilde{b}_t$ and $R_{t,\varepsilon} \to 0$ in $L^1(\mathcal{U}_{p,T})$ as $\varepsilon \to 0$, with

$$ \tilde{b}_t(y) := \left\{ \beta(\tau(Y_{-t}(y))) + \int_{-t}^{0} \beta'(\tau(Y_s(y))) ds \right\} \mathbf{1}_{t<t_b(y)} + \int_{-t_b(y)}^{0} \beta'(\tau(Y_s(y))) ds \mathbf{1}_{t>t_b(y)}. $$

We deduce that

$$ \beta(\tau(y)) = \tilde{b}_t(y), \quad \text{for a.e. } t > 0, y \in \mathcal{O}. $$

Choosing a sequence $(\beta_n)$ of renormalizing function in $C_*^1(\mathbb{R})$ such that $0 \le \beta_n(s) \nearrow s$ and $0 \le \beta'_n(s) \nearrow 1$ locally uniformly, writing the above equation for $\beta = \beta_n$ and passing to the limit $n \to \infty$, we obtain

$$ \tau(y) = \left\{ \tau(Y_{-t}(y)) + t \right\} \mathbf{1}_{t<t_b(y)} + t_b(y) \mathbf{1}_{t>t_b(y)}, \quad \forall t > 0, $$

and in particular $\tau = t_b$ a.e. on $\mathcal{O}$, by taking $t \to \infty$. That implies that $t_b$ is the unique renormalized solution to the backward exit time problem (8.92). From Corollary 8.34, for any $\lambda > 0$, we have

$$ \tau_{\lambda} = \int_{0}^{\infty} e^{-\lambda t} S_{0}(t) \mathbf{1} dt \quad \text{a.e on } \mathcal{O}, $$

and we deduce (8.93) by passing to the limit $\lambda \searrow 0$ in that identity. $\square$
Using Lemma 2.8-(1), we conclude to the existence of $(\lambda_1, \phi_1)$ solution to the dual eigenvalue problem. Now we turn to the existence, uniqueness, and exponential stability of $f_1 \in L^1$, by verifying that Doblin's condition (6.2) is satisfied.

**Doblin condition.** Denoting $S_t := S_L(t)$, we have from the characteristics method

$$S_t f(y) = f(y - t) e^{-\int_0^t K(y-s)ds} \mathbf{1}_{t<y} + N(t - y) e^{-\int_0^y K(s)ds} \mathbf{1}_{t>y}$$

with $N(t) = \int_0^\infty r_O(y_*)S_t f(y_*)dy_*$. Iterating this formula and using the positivity of $S_t$ we get that for any $f \ge 0$

$$S_t f(y) \ge \left( \int_0^{t-y} r_O(y_*) N(t - y - y_*) e^{-\int_0^{y_*} K(s)ds} dy_* \right) e^{-\int_0^y K(s)ds} \mathbf{1}_{0<y<t} \\ \ge \left( \int_0^{t-y} r_O(t - y - \tau) N(\tau) e^{-\int_0^{t-y-\tau} K(s)ds} d\tau \right) e^{-\int_0^y K(s)ds} \mathbf{1}_{0<y<t}.$$

Choosing $t_0 > 2y_0$ so that $r_O(y) \ge \eta_0 > 0$ for all $y \ge t_0/2$, we obtain

$$S_{t_0} f(y) \ge \eta_0 e^{-\int_0^{t_0} K(s)ds} \left( \int_0^{t_0/4} N(\tau) d\tau \right) e^{-\int_0^y K(s)ds} \mathbf{1}_{0<y<t_0/4}.$$

From the expression of $N(t)$ we get by duality, using that $r_O \ge \eta_0 \mathbf{1}_{(y_0, \infty)}$, that

$$(8.110) \quad S_{t_0} f(y) \ge \eta_0^2 e^{-2 \int_0^{t_0} K(s)ds} \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_\tau^* \mathbf{1}_{(y_0, \infty)}(y_*) d\tau \right) dy_* \right) \mathbf{1}_{0<y<t_0/4}.$$

Applying $S_{t_1}$ to this inequality we deduce that for any $t_1 > 0$

$$S_{t_0+t_1} f(y) \ge \eta_0^2 e^{-2 \int_0^{t_0} K(s)ds} \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_\tau^* \mathbf{1}_{(y_0, \infty)}(y_*) d\tau \right) dy_* \right) S_{t_1} \mathbf{1}_{0<y<t_0/4} \\ \ge \eta_0^2 e^{-2 \int_0^{t_0} K} e^{-\int_0^{t_1} K} \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_\tau^* \mathbf{1}_{(y_0, \infty)}(y_*) d\tau \right) dy_* \right) \mathbf{1}_{t_1<y<t_0/4+t_1}.$$

On the other hand, replacing $f$ by $S_{t_1}f$ in (8.110) we obtain

$$(8.111) \quad S_{t_0+t_1} f(y) \ge \eta_0^2 e^{-2 \int_0^{t_0} K(s)ds} \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_{\tau+t_1}^* \mathbf{1}_{(y_0, \infty)}(y_*) d\tau \right) dy_* \right) \mathbf{1}_{0<y<t_0/4}.$$

The fact that $S_t^* \phi(y) \ge \phi(t+y) e^{-\int_0^t K(y+s)ds}$ ensures that for $t_1 > y_0$

$$S_{t_1}^* \mathbf{1}_{(y_0, \infty)} \ge e^{-\sup_{y \in [0, y_0]} \int_0^{t_1} K(y+s)ds} \mathbf{1}_{[0, y_0]}.$$

All together, we have proved that for any $t_0 > 4t_1 > 4y_0$ we have

$$S_{t_0+t_1} f(y) \ge c_0 \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_\tau^* \mathbf{1}(y_*) d\tau \right) dy_* \right) \mathbf{1}_{t_1<y<t_0/4}$$

for some explicit constant $c_0$ and all $f \ge 0$. This is Doblin's condition (6.2) with $T = t_0 + t_1$, and the functions $\psi_0 = \int_0^{t_0/4} S_\tau^* \mathbf{1} d\tau$ and $g_0 = c_0 \mathbf{1}_{(t_1, t_0/4)}$. We are now in position to prove the following result.

**Theorem 8.38.** Under the assumptions (8.108) and (8.109), the renewal equation (8.106)-(8.107) enjoys the conclusions (C3) and (E31) with quantitative rate in $L^1$.

Despite the numerous results about the renewal age-structured model, we are not aware of any previous result with a constructive rate of convergence under such general assumptions.

*Proof of Theorem 8.38.* The conditions (H1), (H2) and (H3) for $L^*$ ensure the existence of $\lambda_1 \ge \kappa_0$ and $\phi_1 \in L^\infty$, $\phi_1 > 0$, that we normalize by $\|\phi_1\|_{L^\infty} = 1$. If we can prove that the conditions (6.2), (6.3) and (6.4) are verified, then the conclusions (C1) and (E31) follow by applying Theorem 6.2. Indeed, the contraction argument in the proof of Theorem 6.2 does not require the existence of $f_1$ and it can even be used for deriving the existence and uniqueness of $f_1$, see Remark 6.4. We have already proved (6.2) with the functions $\psi_0 = \int_0^{t_0/4} S_\tau^* \mathbf{1} d\tau$ and $g_0 = c_0 \mathbf{1}_{(t_1, t_0/4)}$. For proving (6.3), we start by recalling that $\phi_1 = R_B^*(\lambda_1)A^*\phi_1 \in W_{loc}^{1,\infty}$ due to the informations
8.5. On the Krein-Rutman theorem for the transport equation with kernel terms.

In this section we carry on our analysis of the transport equation with kernel term (8.1)-(8.3) for which we establish a Krein-Rutman result under strong positivity assumption on the kernel acting on the domain. As in section 8.3, we assume that $a, b, \mathcal{K}$ and $\mathcal{R}$ satisfy the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) for some weight function $m: \bar{O} \to [1, \infty)$ and some exponent $p \in [1, \infty)$.

On the kernel $\mathcal{K}$, we make the additional strong positivity hypothesis: for any $x \in \mathcal{O}$, there exist $r_x, \eta_x > 0$ such that

$$ (8.94) \qquad \forall f \ge 0, \forall y \in B(x, r_x), \quad \mathcal{K}[f](y) \ge \eta_x \int_{B(x, r_x)} f_* dy_* $$

and

$$ (8.95) \qquad \exists x_0, \ a, b \in L^\infty(B(x_0, r_0)), \ r_0 := r_{x_0}, $$

as well as one of the two following regularity assumptions

$$ (8.96) \qquad \mathcal{K} \in \mathcal{K}(L_m^p(\mathcal{O})) \quad \text{or} \quad \mathcal{K} : L_m^p(\mathcal{O}) \to L^{p_1}(\mathcal{O}) \cap L_{m_1}^p(\mathcal{O}), $$

with $p_1 > p$ and $m_1/m \to \infty$ when $y \to \infty$.

We thus consider the operator

$$ (8.97) \qquad \mathcal{L}f = -a \cdot \nabla f - bf + \mathcal{K}[f] = -\operatorname{div}(af) - Kf + \mathcal{K}[f] $$

with $K := b - \operatorname{div} a \ge 0$, which is complemented with the boundary condition

$$ (8.98) \qquad \gamma_-f = \mathcal{R}_\mathcal{O}[f] + \mathcal{R}_\Sigma[\gamma_+f] \quad \text{on } \Sigma_-. $$

More precisely, we define $\mathcal{L}$ in the Banach space $L_m^p(\mathcal{O})$ with domain

$$ D(\mathcal{L}) \subset \{f \in L^1(\mathcal{O}); a \cdot \nabla f \in L_{\text{loc}}^1(\bar{O}), \gamma_-f = \mathcal{R}[f, \gamma_+f]\}. $$

Notice that because of Section 8.1 the trace function is well defined.

**Example.** The nonlocal operator with a drift

$$ (8.99) \qquad \partial_t f = -a \partial_x f - b + \mathcal{K}[f] \quad \text{in } \mathcal{O}, \quad \gamma_-f = 0 \quad \text{on } \Sigma_-, $$

with $\mathcal{O} \subset \mathbb{R}$ a bounded interval, $a \in W_{\text{loc}}^{1,1}(\mathcal{O})$, $a' \in L^\infty(\mathcal{O})$, $b \in L^\infty(\mathcal{O})$, and thus the boundary kernel is $\mathcal{R} \equiv 0$. Motivated by some non-local reaction-diffusion models, this problem was recently investigated in [108, 117, 248]. It is also used in the study of selection-mutation models in changing environment, see the even newer works [161, 206].

We start by checking that with the above assumptions, the conditions (H1)-(H5) presented in the abstract part hold true.

**Condition (H1).** From Proposition 8.26, we know that for any $\lambda > \lambda^{**}$ the stationary problem

$$ (\lambda - \mathcal{L})g = G \text{ in } \mathcal{O}, \quad \gamma_-g = \mathcal{R}[g, \gamma_+g] \text{ on } \Sigma_-, $$

has a unique solution. More precisely, the associated inverse operator denoted by $\mathcal{R}_\mathcal{L}$ (without reference to the boundary operator $\mathcal{R}$) satisfies $\mathcal{R}_\mathcal{L}: L_m^p \to L_m^p$ and $\mathcal{R}_\mathcal{L}G \ge 0$ if $G \ge 0$.

**Condition (H2).** We first consider the case when $\mathcal{R}_\mathcal{O} \equiv 0$ and we denote by $\mathcal{L}_0$ the associated generator. We fix $f_0 \in C_c^2(\mathcal{O})$, such that $\int_{\mathcal{B}_0} f_0 dy = 1$, $f_0 > 0$ on $\mathcal{B}_0$, $\text{supp} f_0 = \mathcal{B}_0$, as well as

$$ \|f_0^{-1}\|_{L^\infty(\mathcal{B}_\varepsilon)} = \varepsilon^{-2}, \quad \|\nabla f_0\|_{L^\infty(\mathcal{B}_0 \setminus \mathcal{B}_\varepsilon)} = \varepsilon, \quad \forall \varepsilon \in (0, 1/2), $$

where we denote $\mathcal{B}_\varepsilon := B(x_0, (1-\varepsilon)r_0)$. We also define $C_0 := \|f_0\|_{L^\infty(\mathcal{B}_0)}$ and $C_1 := \|\nabla f_0\|_{L^\infty(\mathcal{B}_0)}$, both may be bounded by a constant which only depends on $r_0$ and $d$. Because of (8.94), we have

$$ \mathcal{K}[f_0](y) \ge \eta_0 \mathbf{1}_{\mathcal{B}_0} \ge \frac{\eta_0}{C_0} f_0. $$

We observe that $f_0 \in D(\mathcal{L}_0)$ and we compute

$$ (8.100) \qquad \mathcal{L}_0 f_0 \ge -\|a\|_{L^\infty(\mathcal{B}_0)} C_1 \mathbf{1}_{\mathcal{B}_0} - \|b\|_{L^\infty(\mathcal{B}_0)} C_0 \mathbf{1}_{\mathcal{B}_0} + \eta_0 \mathbf{1}_{\mathcal{B}_0} \ge \kappa_0 f_0, $$

if $\kappa_0 := \eta_0/C_0 - \|a\|_{L^\infty(\mathcal{B}_0)} C_1/C_0 - \|b\|_{L^\infty(\mathcal{B}_0)} \ge 0$. More generally, we have

$$ (8.101) \qquad \mathcal{L}_0 f_0 \ge -\|a\|_{L^\infty(\mathcal{B}_0)} C_1 \mathbf{1}_{\mathcal{B}_0} - \|a\|_{L^\infty} \varepsilon \mathbf{1}_{\mathcal{B}_0 \setminus \mathcal{B}_\varepsilon} - \|b\|_{L^\infty(\mathcal{B}_0)} f_0 + \eta_0 \mathbf{1}_{\mathcal{B}_0} \ge \kappa_0 f_0, $$
derived on $\mathcal{R}_B^*$ in (H3). Consequently, there exists $y_1 > 0$ such that $\phi_1(y_1) > 1/2$, and we deduce from $\phi'_1 \le (\lambda_1 + K)\phi_1$ that

$$ \phi_1(y) \ge \frac{1}{2} e^{-\int_{y_1}^{y_2} (\lambda_1 + K)} $$

for all $y \in (0, y_1)$. Choosing in the proof of the Doblin condition $t_0$ such that $y_1 < t_0/4$, we obtain that

$$ \langle \phi_1, g_0 \rangle \ge \frac{c_0}{2} \int_{y_1/2}^{y_1} e^{-\int_{y_1}^{y_2} (\lambda_1 + K)} dy, $$

which gives (6.3). For (6.4), we use that

$$ \phi_1 = e^{-\lambda_1 \tau} S_{\tau}^{*} \phi_1 \le e^{-\lambda_1 \tau} S_{\tau}^{*} \mathbf{1} $$

for any $\tau > 0$ to deduce that

$$ \phi_1 = \frac{4}{t_0} \int_{0}^{t_0} e^{-\lambda_1 \tau} S_{\tau}^{*} \phi_1 d\tau \le \frac{4e^{\lambda_1 |t_0|}}{t_0} \psi_0. $$

Finally, we check that the condition (H5') is verified, so that (C3) is valid by virtue of Theorem 5.18 and Remark 5.19. The condition (H5') is actually a direct consequence of the fact that (8.111) is verified for any $t_0 > 2y_0$ and $t_1 > 0$ together with the estimate

$$ S_{\tau+t_1}^{*} \mathbf{1}_{(y_0, \infty)}(y) \ge e^{-\int_{0}^{t_1} K(y+s)ds} > 0 $$

for any $t_1 > y_0$ and $\tau > 0$, and all $y > 0$. $\square$

## 9. THE GROWTH-FRAGMENTATION EQUATION

In this section, we are interested in the growth-fragmentation equation with equal mitosis kernel

$$ (9.1) \qquad \partial_t f(t, x) + \partial_x (a(x)f(t, x)) + K(x)f(t, x) = 4K(2x)f(t, 2x) $$

and to its variant with an additional “growth speed” variable

$$ (9.2) \qquad \partial_t f(t, x, v) + v \partial_x (a(x)f(t, x, v)) + K(x)f(t, x, v) = 4 \int_1^2 K(2x) \wp(v, v_*) f(t, 2x, v_*) dv_*, $$

with $x > 0$ and $v \in [1, 2]$. For both equations, we assume that the total fragmentation rate $K$ is a continuous function defined on $\mathbb{R}_+$ such that

$$ (9.3) \qquad \exists x_0 > 0, \quad K = 0 \text{ on } (0, 2x_0] \quad \text{and} \quad K > 0 \text{ on } (2x_0, \infty). $$

This condition ensures that no particle of size less than $x_0$ can be produced by division, and we thus consider the equations posed on the size space $(x_0, \infty)$ with zero flux boundary condition $f(t, x_0) = 0$ or $f(t, x_0, v) = 0$. The growth rate $a$ is supposed to be positive and globally Lipschitz on $[x_0, \infty)$, and we assume that

$$ (9.4) \qquad \lim_{x \to \infty} \frac{xK(x)}{a(x)} = +\infty. $$

For quantifying the positivity of the first eigenvalue, we also make the technical assumption that

$$ (9.5) \qquad \exists k > 0, \quad \lim_{x \to \infty} e^{x^k} K(x) = +\infty. $$
with $\kappa_0 := -\|a\|_{L^\infty(\mathcal{B}_0)} C_1 \epsilon^{-2} - \|b\|_{L^\infty(\mathcal{B}_0)} \in \mathbb{R}$ when $\|a\|_{L^\infty} \epsilon \le \eta_0$. Depending on how $\eta_0 > 0$ is large, we obtain in that way two constructive lower bounds of $\mathcal{I}$ thanks to Lemma 2.4-(ii) and we have thus established that $\mathcal{L}_0$ satisfies (H2). Because $f_0 \in D(\mathcal{L}_0)$, we have $S_{\mathcal{L}_0}(t)f_0 \ge e^{\kappa_0 t} f_0$ for any $t \ge 0$, from Remark 2.5-(2). On the other hand, we observe that $S_{\mathcal{L}}(t) \ge S_{\mathcal{L}_0}(t)$ for any $t \ge 0$, from the weak maximum principle mentioned in Remark 8.24-(3). These two last observations together imply $S_{\mathcal{L}}(t)f_0 \ge e^{\kappa_0 t} f_0$, for any $t \ge 0$. We deduce from Lemma 2.4-(iv) that (H2) holds.

**Condition (H3).** We introduce the semigroup $S_B$ associated to the transport equation

$$ \frac{\partial g}{\partial t} + a \cdot \nabla g + bg = 0, \quad \gamma_{-}g = \mathcal{R}[\gamma_{+}g], $$

which is well defined thanks to Corollary 8.25 and satisfies $\|S_B(t)g_0\|_{L_m^p} \le e^{\kappa_B t} \|g_0\|_{L_m^p}$ for any $t \ge 0$ and $g_0 \in L_m^p$ with $\kappa_B := \|\langle \bar{\omega}_- \rangle\|_{L^\infty} + M_{\mathcal{B}}/p$ because of the a priori estimate (8.65) particularized to the present case (in particular where we can take $\epsilon_1 = 0$ because the influx function is $g = 0$ here). We formulate the first hypothesis

$$ (8.102) \qquad \eta_0 > \|\langle \bar{\omega}_- \rangle\|_{L^\infty} C_0 + M_{\mathcal{B}} C_0 / p + \|a\|_{L^\infty(\mathcal{B}_0)} C_1 - \|b\|_{L^\infty(\mathcal{B}_0)} C_0, $$

with the same definitions as above for $\mathcal{B}_0$, $C_0$ and $C_1$, so that $\kappa_0 > \kappa_B$ because of (8.100). In a second case, we assume

$$ (8.103) \qquad \mathcal{R} \equiv 0, \mathcal{O} \text{ is bounded and there exists } T_{\mathcal{O}} \text{ such that } t_{\mathbf{b}}(y) \le T_{\mathcal{O}} \text{ for a.e. } y \in \mathcal{O}. $$

In that case, the semigroup $S_B$ is explicitly given by

$$ (S_B(t)f_0)(y) = \begin{cases} f_0(Y_{-t}(y)) \exp(-\int_0^t K(Y_{\tau-t}(y))d\tau), & \text{if } t \in (0, t_{\mathbf{b}}(y)) \\ 0 & \text{otherwise,} \end{cases} $$

and in particular $S_B(t)f = 0$ for any $f$ and any $t > T_{\mathcal{O}}$. We immediately deduce $\kappa_B = -\infty$ and thus $\kappa_0 > \kappa_B$ because we have established that $\kappa_0 \in \mathbb{R}$. We next define $\mathcal{A}f := \mathcal{K}[f]$. Using Lemma 2.8 and Remark 2.9-(2) or Lemma 2.13 and Remark 2.14-(1) depending on the assumption (8.96) made on $\mathcal{K}$, we deduce that the condition (H3) holds in both cases discussed above. Under the first condition in (8.96), we conclude to the existence of eigenvalue triplet $(\lambda_1, f_1, \phi_1) \in \mathbb{R} \times L_m^p \times L_{m-1}^{p'}$. Under the second condition in (8.96), we may also get the same conclusion by using [336, Cor. 1 of Thm. II.9.9] when $p = 1$ or by observing that the dual problem is similar to the primal problem when $p > 1$ and thus we may apply the same arguments for the dual problem as those explained above for the primal problem.

**Condition (H4).** Let us consider $\lambda > \lambda^{**}$ and $0 \le f \in L_{m\mathcal{O}}^p(\mathcal{O})$ a (renormalized) solution to

$$ \lambda f + a \cdot \nabla f + b f - \mathcal{K}[f] = F \quad \text{in } \mathcal{O}, \quad \gamma_{-}f = \mathcal{R}[f, \gamma_{+}f] \quad \text{on } \Sigma_{-}, $$

with $0 \le F \in L_m^p(\mathcal{O})$. If $f \ne 0$, there exists $x_1 \in \mathcal{O}$ such that $\int_{B(x_1, r_1)} f(z) dz > 0$. From (8.94), we deduce

$$ \mathcal{K}[f](y) \ge \int_{B(x_1, r_1)} \kappa(y, z) f(z) dz > 0, \quad \forall y \in B(x_1, r_1). $$

Now, we argue similarly as during the proof of Lemma 8.14 and in particular we use the same notations. For $A \subset B(x_1, r_1)$, we define the solution $0 \le \varphi \in L_{m-1}^{p'} \cap L^\infty$ to the equation

$$ \lambda\varphi - \operatorname{div}(a\varphi) + b\varphi = \mathbf{1}_A \quad \text{in } \mathcal{O}, \quad \gamma_+\varphi = 0 \quad \text{on } \Sigma_+, $$

thanks to Lemma 8.12 and Lemma 8.13, and we observe that $\varphi \ne 0$ on $B(x_1, r_1)$ if $|A| > 0$. For the renormalizing function $\beta_\delta$ and a truncation function $\chi_R$, we compute

$$ \begin{aligned} 0 &\ge \int_{\Sigma} a \cdot n \beta_{\delta}(\gamma f) \gamma \varphi \chi_R \\ &= \int_{\mathcal{O}} [\beta'_{\delta}(f)(F + \mathcal{K}[f])\varphi - \beta_{\delta}(f)\mathbf{1}_A]\chi_R \\ &\quad + \int_{\mathcal{O}} (\beta_{\delta}(f) - f\beta'_{\delta}(f))(\lambda + b)\varphi \chi_R + \int_{\mathcal{O}} \varphi \beta(f) \frac{a}{R} \cdot (\nabla \chi)_R. \end{aligned} $$
**9.1. The mitosis equation with mixing growth rate.** We are interested here in the growth-fragmentation equation (9.1) in the case where

$$ (9.6) \qquad \exists x_1 > x_0, \ a(2x_1) \neq 2a(x_1). $$

As we will see below, this condition ensures some mixing property for the trajectories that guarantees the triviality of the boundary point spectrum.

We work in the space $X = L_m^1$ with a weight $m$ that can be

$$ (9.7) \qquad \text{either } m(x) = x^r, \ r > 1, \quad \text{or } m(x) = \exp\left(\eta \int_{x_0}^x \frac{K}{a}\right), \ 0 < \eta < 1. $$

Note that due to assumption (9.4), the weight $\exp(\eta \int_{x_0}^x K/a)$ is always stronger than $x^r$.

**Theorem 9.1.** Suppose that (9.3), (9.4), (9.5) and (9.6) are satisfied. The first eigentriplet problem admits a unique solution $(\lambda_1, f_1, \phi_1) \in \mathbb{R} \times X_+ \times X'_+$ with the normalization $\|\phi_1\| = \langle \phi_1, f_1 \rangle = 1$, and this triplet additionally satisfies $\lambda_1 > 0$, $f_1 > 0$ and $\phi_1 > 0$. Besides, there are some constructive constants $C \ge 1$, $\omega > 0$ such that

$$ \|e^{-\lambda_1 t} S_{\mathcal{L}}(t) f - \langle \phi_1, f \rangle f_1\|_{X} \le C e^{-\omega t} \|f - \langle \phi_1, f \rangle f_1\|_{X} \quad \text{for any } f \in X \text{ and } t \ge 0. $$

This result is contained in the recent paper [354]. The novelty here is that all the constants are obtained constructively, which is not clear in [354]. We also provide what seems to us to be a more direct and comprehensive proof. We also refer to [35, 52, 80, 278] where the same result is obtained under stronger assumptions.

Before starting the proof of Theorem 9.1, let us briefly justify the relevance of the chosen weight functions $m$ in (9.7). The dual operator associated to equation (9.1) is given by

$$ \mathcal{L}^* \phi(x) = a(x) \phi'(x) - K(x) \phi(x) + 2K(x) \phi(x/2). $$

For $m(x) = x^r, r > 1$, we can compute

$$ (9.8) \qquad \mathcal{L}^* m(x) = \left[ r \frac{a(x)}{x} - (1 - 2^{1-r})K(x) \right] m(x), $$

and for $m(x) = \exp(\eta \int_{x_0}^x K/a)$, $0 < \eta < 1$,

$$ (9.9) \qquad \mathcal{L}^* m(x) = \left[ 2 \exp\left(-\eta \int_{x/2}^x \frac{K}{a}\right) - (1 - \eta) \right] K(x) m(x). $$

Assumption (9.3) then ensures that $\mathcal{L}^* m \sim -\xi K m$ as $x \to +\infty$, with $\xi = 1 - 2^{1-r} > 0$ in the first case and $\xi = 1 - \eta > 0$ in the second case. In both cases, we deduce that

$$ (9.10) \qquad \mathcal{L}^* m \le \kappa m + M \mathbf{1}_{(x_0, R)} m $$

for any $\kappa \ge 0$, by choosing $M > 0$ and $R > x_0$ large enough, and this type of Lyapunov inequality is pivotal in our analysis.

**Condition (H1).** Equation (9.1) is a particular case of equation (8.58) with $G = g = \mathbb{R} = 0$, $b = K + \text{diva}$ and $\mathcal{H}[g](x) = 4K(2x)g(2x)$. We may then use Proposition 8.23-(1) to infer the well-posedness of equation (9.1) in $X = L_m^1(x_0, \infty)$, provided that the conditions (8.35) and (8.61) are met, with $0 < \alpha_{\mathcal{H}} < 1$, which is nothing but (8.63) when $\mathbb{R} \equiv 0$. To do so, we define the function

$$ \varpi := K - a \frac{m'}{m}, $$

which corresponds to $\varpi_1$ in (8.34). When $m(x) = x^r$ with $r > 1$, we have

$$ \varpi(x) = K(x) - r \frac{a(x)}{x}, $$

and for $m(x) = \exp(\eta \int_{x_0}^x K/a)$ with $0 < \eta < 1$, we have

$$ \varpi(x) = (1 - \eta)K(x). $$
Passing first to the limit $R \to \infty$ and next to the limit $\delta \to 0$, we deduce

$$0 \geq \int_{\mathcal{O}} [(F + \mathcal{H}[f])\varphi - f \mathbf{1}_{A}],$$

so that in particular

$$\int_{A} f dy \geq \int_{B(x_1, r_1)} \varphi \mathcal{H}[f] > 0.$$

This being true for any $A \subset B(x_1, r_1)$, we deduce $f > 0$ a.e. on $B(x_1, r_1)$. By a classical continuity argument, we conclude that $f > 0$ a.e. on $\mathcal{O}$. We have thus established (H4) for $\lambda > \lambda^{**}$ from what we immediately and classically deduce the general case $\lambda \in \mathbb{R}$.

**Condition (H5).** Assume that $(\lambda, f) \in \mathbb{C} \times D(\mathcal{L})$ satisfies

$$\mathcal{L}|f| = (\Re e \lambda)|f| \quad \text{in } \mathcal{O}, \quad \mathcal{R}[|f|, \gamma_+|f|] = \gamma_-|f| \quad \text{on } \Sigma_-,$$

and

$$\mathcal{L}|f| = \Re e(\operatorname{sign} f)\mathcal{L}f \quad \text{in } \mathcal{O}, \quad \mathcal{R}[|f|, \gamma_+|f|] = \Re e(\operatorname{sign} \gamma_- f)\mathcal{R}[f, \gamma_+ f] \quad \text{on } \Sigma_-.$$

From (H4) and the first identity, we know that $|f| > 0$ a.e. on $\mathcal{O}$. Using the second identity, we get

$$\mathcal{H}[|f|] = \Re e(\operatorname{sign} f)\mathcal{H}[f].$$

Writing $f = e^{i\alpha}|f|$, we deduce

$$\int_{\mathcal{O}} k|f_*|(1 - \cos(\alpha - \alpha_*))dy_* = 0 \quad \text{a.e. on } \mathcal{O}.$$

Using (8.94), we deduce

$$\int_{B(y, r_{\mathcal{O}})} |f_*|(1 - \cos(\alpha - \alpha_*))dy_* = 0,$$

and thus $\alpha = \alpha_*$ a.e. on $\mathcal{O} \times \mathcal{O}$. That means $f = u|f|$, for a constant $u = \mathbb{S}^1$, that completes the proof of the fact that $\mathcal{L}$ satisfies the reverse Kato's inequality condition (H5).

We summarize our analysis in the following result which is a straightforward consequence of the above checked conditions together with Theorem 2.21, Theorem 4.13, Theorem 5.16 and Theorem 5.23. We state the available result in that situation.

**Theorem 8.36.** We assume that $a, b, \mathcal{H}$ and $\mathcal{R}$ satisfy the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) for some weight function $m: \mathcal{O} \to [1, \infty)$ and some exponent $p \in [1, \infty)$. Consider the semigroup $S_{\mathcal{L}}$ associated to the transport equation (8.1)-(8.3) through Corollary 8.25. We assume further that $\mathcal{H}$ satisfies the strong positivity conditions (8.94) together with (8.95) and the first compactness property formulated in (8.96). We finally assume that (8.102) holds or (8.103) holds. In both cases, the conclusions (S1), (S2) and (S32) hold in $L_p^m$ as well as the ergodicity (E2) in $L_{\phi_1}^1$.

We are not aware of any similar result for such a general transport equation, see however the next sections where more specific transport like equations are discussed. We do not try to improve the convergence result in the general case, but rather we aim to make one step further in the following particular situation where Doblin approach may be used.

**Doblin condition.** We suppose here that $\mathcal{O}$ is bounded, $K \in L^{\infty}(\mathcal{O})$, $R_{\Sigma}^*\mathbf{1} = R_{\Sigma}\mathbf{1} = \mathbf{1}$, and $k(y, y_*) \geq k_0 > 0$. We aim at establishing the Doblin condition

$$S_{\mathcal{L}}(T)f_0 \geq \kappa \langle f_0, \mathbf{1} \rangle,$$

which is (6.2) with $\psi_0 = \mathbf{1}$ and $g_0 = \kappa \mathbf{1}$. From (8.64) we have

$$\frac{d}{dt} \int_{\mathcal{O}} f dy = \int_{\mathcal{O}} f K dy \geq - \|K\|_{\infty} \int_{\mathcal{O}} f dy$$

and so

$$\int_{\mathcal{O}} f(t, y) dy \geq e^{-\|K\|_{\infty}t} \int_{\mathcal{O}} f_0(y) dy.$$
In both cases, the fact that $a \in \text{Lip}$ ensures that $\varpi_q := \varpi + (1 - 1/q)a'$ enjoys $(\varpi_q)_- \in L^\infty$ for any $q \in [1, \infty)$. On the other hand, (9.4) guarantees that $K \lesssim \langle \varpi_+ \rangle$ and $a/x \lesssim \langle \varpi_+ \rangle$, and finally (8.35) is verified. The condition (8.61) is equivalent to the Lyapunov type condition

$$ (9.11) \qquad \mathcal{H}^*[m] \le (\alpha_{\mathcal{H}} \varpi_+ + M_{\mathcal{H}})m, $$

where $\mathcal{H}^*[m](x) = 2K(x)m(x/2)$. For $m(x) = x^r$ with $r > 1$, we compute

$$ \mathcal{H}^*[m]/m = 2^{1-r}K, $$

and for $m(x) = \exp(\eta \int_{x_0}^x K/a)$ with $0 < \eta < 1$,

$$ \frac{\mathcal{H}^*[m](x)}{m(x)} = 2 \exp \left( - \eta \int_{x/2}^x \frac{K}{a} \right) K(x). $$

Using (9.4), we obtain that (9.11) is satisfied, for any $\alpha_{\mathcal{H}} \in (2^{1-r}, 1)$ in the first case, and for any $\alpha_{\mathcal{H}} \in (0, 1)$ in the second case, by choosing $M_{\mathcal{H}}$ large enough.

We can then apply Proposition 8.23-(1) for associating to equation (9.1) a strongly continuous semigroup $S$ in $X = L_m^1(x_0, \infty)$, and (H1) then follows from Lemma 2.2-(i). Moreover, we readily have that $\kappa_1 \le \kappa + M$ for any couple $(\kappa, M)$ such that (9.10) is verified.

**Condition (H2).** We aim at verifying (H2) for some $\kappa_0 > 0$. Recalling assumption (9.5), we pick up $\ell > k$ and we consider the function

$$ \phi_0(x) = xe^{-x^\ell/n} $$

with $n$ large enough to be chosen later. We compute

$$ \frac{\mathcal{L}^*\phi_0(x)}{\phi_0(x)} = \frac{a(x)}{x} \left(1 - \frac{\ell}{n}x^\ell\right) + K(x) \left(e^{\frac{1-2^{-\ell}}{n}}x^\ell - 1\right). $$

Choosing $R > x_0$ such that $xK(x)/a(x) \ge \frac{2\ell}{1-2^{-\ell}}$ and $K(x) \ge e^{-x^k}$ for all $x \ge R$, we get that

$$ \frac{\mathcal{L}^* \phi_0(x)}{\phi_0(x)} \ge \frac{a(x)}{x} + K(x) \left( e^{\frac{1-2^{-\ell}}{n}} x^{\ell} - 1 - \frac{1-2^{-\ell}}{2n} x^{\ell} \right) \\ \ge e^{-x^k} \left( e^{\frac{1-2^{-\ell}}{n}} x^{\ell} - e^{\frac{1-2^{-\ell}}{2n}} x^{\ell} \right) \ge e^{\frac{1-2^{-\ell}}{2n}} x^{\ell} - x^k \left( e^{\frac{1-2^{-\ell}}{2n}} R^{\ell} - 1 \right) $$

on $[R, \infty)$. Choosing then $n \ge \frac{\ell}{2}R^\ell$, we have

$$ \frac{\mathcal{L}^* \phi_0(x)}{\phi_0(x)} \ge \frac{a(x)}{2x} $$

on $(x_0, R)$. Gathering the two above estimates, we deduce the existence of an explicit $\kappa_0 > 0$ such that $\mathcal{L}^*\phi_0 \ge \kappa_0\phi_0$. We conclude by invoking Lemma 2.4-(i).

**Condition (H3).** We consider the weight function $m(x) = x^r$ for some $r > 1$ or $m(x) = \exp(\eta \int_{x_0}^x K/a)$ with $0 < \eta < 1$ and we define the stronger weight function $m_1(x) = \exp(\eta_1 \int_{x_0}^x K/a)$ for some $\eta_1 \in (\eta, 1)$. We fix $\kappa_B \in [0, \kappa_0)$, $M > 0$, and $R > x_0$ such that (9.10) is verified by $m_1$ with $\kappa = \kappa_B$. Using the splitting $\mathcal{L} = \mathcal{A} + \mathcal{B}$ with $\mathcal{A}f = M1_{(x_0, R)}f$, the inequality (9.10) for $m_1$ reads $\mathcal{B}^*m_1 \le \kappa_B m_1$ and this ensures (see the proof of Corollary 2.20) that $\kappa - \mathcal{B}$ is invertible in $L_{m_1}^1$ for any $\kappa > \kappa_B$, with positive inverse, and

$$ \|(\kappa - \mathcal{B})^{-1}\|_{\mathcal{B}(L_{m_1}^1)} \le \frac{1}{\kappa - \kappa_B}. $$

The operator $\mathcal{A}$ maps $L_m^1$ into $L_{m_1}^1$ with

$$ \|\mathcal{A}\|_{\mathcal{B}(L_m^1, L_{m_1}^1)} \le M \frac{m_1(R)}{m(x_0)}. $$

Besides, due to the derivative part $\partial_x(a \cdot)$ in the operator $\mathcal{B}$, we also have that $\mathcal{R}_B(\kappa)$ maps $L_{m_1}^1$ into $W_{\text{loc}}^{1,1}$. Finally, we have $\mathcal{R}_B(\kappa)\mathcal{A} : L_m^1 \to L_{m_1}^1 \cap W_{\text{loc}}^{1,1}$, and thus $\mathcal{R}_B(\kappa)\mathcal{A} \in \mathcal{H}(L_m^1)$, for any $\kappa \ge \kappa_0 > \kappa_B$. We deduce from Lemma 2.8-(2) that the condition (H3) holds for both the primal and the dual problems.
Now we define, for $\varphi_0 \in C_c^1(\mathcal{O})$, $\varphi_0 \ge 0$, $\int \varphi_0 = 1$, the solution $\varphi$ to the equation

$$ egin{cases} 
\partial_t \varphi + \operatorname{div}(a\varphi) = 0, \\
\gamma_+ \varphi = \mathcal{R}_{\Sigma}^*[\gamma_- \varphi].
\end{cases} $$

We have

$$ \frac{d}{dt} \int_{\mathcal{O}} \varphi = 0, \quad \text{and so} \quad \int_{\mathcal{O}} \varphi(t, y) dy = \int_{\mathcal{O}} \varphi_0(y) dy = 1, $$

and

$$ \frac{d}{dt} \int_{\mathcal{O}} f\varphi = \int_{\mathcal{O}} \mathcal{H}[f]\varphi - \int_{\mathcal{O}} Kf\varphi \ge k_0 \int_{\mathcal{O}} f - \|K\|_{\infty} \int_{\mathcal{O}} f\varphi. $$

We deduce from Grönwall's inequality that, for any fixed $T > 0$,

$$ \begin{aligned} \int_{\mathcal{O}} f(T, y) \varphi_0(y) dy &\ge e^{-\|K\|_{\infty} t} \int_{\mathcal{O}} f_0(y) \varphi_0(y) dy + k_0 \int_0^T e^{-(T-t)\|K\|_{\infty}} \int_{\mathcal{O}} f(t, y) dy dt \\ &\ge k_0 T e^{-\|K\|_{\infty}} \int_{\mathcal{O}} f_0(y) dy =: \kappa \langle f_0, \mathbf{1} \rangle. \end{aligned} $$

This is nothing but the Doblin condition (6.2) since $\varphi_0$ is any non-negative function in $C_c^1(\mathcal{O})$ with $\int \varphi_0 = 1$.

In order to verify (6.3) in a quantitative way, we suppose that the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) are verified with the weight function $m = \mathbf{1}$ and the exponent $p = 1$. Note that in this case we have $\varpi = K \ge 0$. The first condition in (8.35) then imposes that $K \in L^{\infty}(\mathcal{O})$, and (8.102) reads

$$ \eta_0 > \|\langle K \rangle\|_{L^{\infty}} C_0 + M_{\mathcal{H}} C_0 + \|a\|_{L^{\infty}(\mathcal{B}_0)} C_1 - \|b\|_{L^{\infty}(\mathcal{B}_0)} C_0. $$

We also assume that

$$ (8.104) \qquad \mathcal{R}_{\Sigma}^* \mathbf{1} = \mathcal{R}_{\Sigma} \mathbf{1} = \mathbf{1}, $$

and

$$ (8.105) \qquad \forall y, y_* \in \mathcal{O}, \quad k_0 \le k(y, y_*) \le k_1 $$

for some $k_1 > k_0 > 0$,

**Theorem 8.37.** We assume that $\mathcal{O}$ is bounded and that the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) are satisfied by $a$, $b$, $\mathcal{H}$ and $\mathcal{R}$ for the weight function $m = \mathbf{1}$ and the exponent $p = 1$. We assume further that $\mathcal{H}$ satisfies the strong positivity conditions (8.94) together with (8.95) and the first compactness property formulated in (8.96). We finally assume that (8.102), (8.104) and (8.105) are satisfied. Then the exponential convergence in $(\mathbf{E}3_1)$ holds in $L^1$ with constructive constants $C$ and $\omega$.

*Proof of Theorem 8.37.* We work in $X = L^1(\mathcal{O})$ and we normalize $\phi_1$ by $\|\phi_1\|_{L^\infty} = 1$. We have proved above that (6.2) holds true with $\psi_0 = \mathbf{1}$ and $g_0 = \kappa \mathbf{1}$ for some explicit $\kappa > 0$, recalling that the assumption that $K \in L^\infty$ is nothing but the first condition in (8.35) when $m = \mathbf{1}$ and $p = 1$ since $b = K + \text{diva}$. Due to the normalization $\|\phi_1\|_{L^\infty} = 1$, the condition (6.4) holds with $R_0 = 1$. It only remains to check the validity of (6.3) in order to be able to apply Theorem 6.2. Since we assume that the conditions (8.35), (8.59), (8.60), (8.61), (8.62) and (8.63) are satisfied for the weight function $m = \mathbf{1}$ and the exponent $p = 1$, we have that $\mathcal{R}_B(\lambda_1) : L^1 \to L^1$ with $\|\mathcal{R}_B(\lambda_1)\|_{\mathcal{B}(L^1)} \le \frac{1}{\kappa_0 - \kappa_B}$. This yields by duality that $\mathcal{R}_B^*(\lambda_1) : L^\infty \to L^\infty$ with $\|\mathcal{R}_B^*(\lambda_1)\|_{\mathcal{B}(L^\infty)} \le \frac{1}{\kappa_0 - \kappa_B}$. Since $k$ is bounded by the constant $k_1$, we have on the other hand that $A^* = \mathcal{H}^* : L^1 \to L^\infty$ with $\|A^*\|_{\mathcal{B}(L^1, L^\infty)} \le k_1$. We thus get

$$ 1 = \|\phi_1\|_{L^\infty} \le \frac{k_1}{\kappa_0 - \kappa_B} \|\phi_1\|_{L^1}, $$

which yields (6.3) with $r_0 = \kappa(\kappa_0 - \kappa_B)/k_1$, and the proof is complete. $\square$

<figure><img src="image_10.png" /></figure>
**Proof of the existence part of Theorem 9.1.** We deduce from Theorem 2.21 that the conclusion (C1) about the existence of a solution $(\lambda_1, f_1, \phi_1) \in \mathbb{R} \times X_+ \times X'_+$ to the first eigentriplet problem holds true. □

Moreover, we have $\lambda_1 \ge \kappa_0 > 0$ and $f_1 \in W_{\text{loc}}^{1,1} \cap L_m^1$ with $m(x) = \exp(\eta \int_{x_0}^x K/a)$ for any $\eta \in (0, 1)$. For deriving similar additional estimates on $\phi_1$, we can check directly that the condition (H3) holds for the dual operator $\mathcal{L}^*$.

**Condition (H3) for $\mathcal{L}^*$:** We consider the weight function $m(x) = x^r$ for some $r > 1$ or $m(x) = \exp(\eta \int_{x_0}^x K/a)$ with $0 < \eta < 1$ and we define the weaker weight function $m_0(x) = x^{r_0}$ for some $r_0 \in (1, r)$. We fix $\kappa_B \in [0, \kappa_0)$, $M > 0$, and $R > x_0$ such that (9.10) is verified by $m_0$. Using again the splitting $\mathcal{L} = \mathcal{A} + \mathcal{B}$ with $\mathcal{A}f = M\mathbf{1}_{(x_0, R)}f$, (9.10) means that $\mathcal{B}^*m_0 \le \kappa_B m_0$ and this ensures that for any $\kappa > \kappa_B$ the operator $\kappa - \mathcal{B}^*$ is invertible in $L_{m_0}^\infty$, with positive inverse, and

$$ \|(\kappa - \mathcal{B}^*)^{-1}\|_{\mathcal{B}(L_{m_0}^\infty)} \le \frac{1}{\kappa - \kappa_B}. $$

Because of the derivative part of $\mathcal{B}^*$, we also have that $\mathcal{R}_{\mathcal{B}^*}(\kappa) : L_{m_0}^\infty \to W_{\text{loc}}^{1,\infty}$. Besides, the operator $\mathcal{A}^* = \mathcal{A}$ maps $L_{m_0}^\infty$ into $L_{m_0}^\infty$ with

$$ \|\mathcal{A}^*\|_{\mathcal{B}(L_{m_0}^\infty, L_{m_0}^\infty)} \le M \frac{m(R)}{m_0(x_0)}. $$

Finally we have that $\mathcal{R}_{\mathcal{B}^*}(\kappa)\mathcal{A} : L_{m_0}^\infty \to L_{m_0}^\infty \cap W_{\text{loc}}^{1,\infty}$. Consequently $\phi_1 \in L_{m_0}^\infty \cap W_{\text{loc}}^{1,\infty}$ and

$$ (9.12) \qquad \|\phi_1\|_{L_{m_0}^\infty} = \|(\lambda_1 - \mathcal{B}^*)^{-1}\mathcal{A}^*\phi_1\|_{L_{m_0}^\infty} \le \frac{m(R)}{m_0(x_0)} \frac{M}{\kappa_0 - \kappa_B} \|\phi_1\|_{L_{m_0}^\infty}. $$

We also easily deduce quantitative estimates of $\phi_1$ in $W_{\text{loc}}^{1,\infty}$ from the identity

$$ \phi_1'(x) = \frac{1}{a(x)} \left[ \lambda_1 \phi_1(x) + K(x) \phi_1(x) - 2K(x) \phi_1(x/2) \right]. $$

**Condition (H4).** The operator $\mathcal{L}$ satisfies the strong maximum principle. Let $\lambda \in \mathbb{R}$ and $f \in X_+ \cap D(\mathcal{L}) \setminus \{0\}$ such that $(\lambda - \mathcal{L})f \ge 0$, i.e.

$$ \lambda f(x) + (af)'(x) + K(x)f(x) \ge 4K(2x)f(2x) \quad \forall x > x_0. $$

Denoting by $\Lambda_\lambda$ a function such that $\Lambda'_\lambda(x) = \frac{\lambda+K(x)}{a(x)}$, we get that

$$ (9.13) \qquad a(x)f(x) \ge 4 \int_{x_0}^{x} e^{\Lambda_{\lambda}(y)-\Lambda_{\lambda}(x)} K(2y)f(2y) dy. $$

Since $K(2y) > 0$ for all $y > x_0$, $f \in X_+ \setminus \{0\}$, and $a(x) > 0$ for all $x > x_0$, we deduce from (9.13) that the set $\{x > x_0, f(x) > 0\}$ is an interval of the form $(x, +\infty)$. Using again (9.13) we remark that we must have $x = \max(x_0, x/2)$, which enforces $x = x_0$ and finally $f > 0$.

**Proof of the uniqueness and positivity part of Theorem 9.1.** We deduce from Theorem 4.13 the validity of the conclusion (C2) about existence, uniqueness and positivity of a solution $(\lambda_1, f_1, \phi_1)$ to the first eigentriplet problem. □

For deriving the exponential stability, we start by verifying a quantified irreducibility and periodicity condition on $S$, given in the next lemma, which then allows us to prove that the Harris condition (6.8) is met.

**Lemma 9.2.** Assume that (9.6) is satisfied. Then for all $\varepsilon > 0$, $R_1 > x_0$, and $R_2 > x_0 + \varepsilon$, there exists $T_1 > 0$ such that for any $T > T_1$, there exists $c_T > 0$ such that

$$ S_T^* \phi \ge c_T \mathbf{1}_{(x_0, R_1)} \int_{x_0+\varepsilon}^{R_2} \phi \, dx, \quad \forall \phi \ge 0. $$
8.6. **A word about the renewal equation.** We look at the case $\mathcal{O} = (0, +\infty)$ and $a(y) = 1$, which corresponds to the equation

$$ (8.106) \qquad \partial_t f + \partial_y f + Kf = 0 $$

with the boundary condition

$$ (8.107) \qquad (\gamma_{-}f)(t, 0) = \int_{0}^{\infty} r_{\mathcal{O}}(y_{*})f(t, y_{*})dy_{*}. $$

This renewal age structured model is standard in structured population dynamics, and the Krein-Rutman theorem is well-known for it, see for instance [34, 157, 186, 195, 337, 357]. The existence and uniqueness of $(\lambda_1, f_1, \phi_1)$ can even be obtained by explicit computations. However, it is not covered by the cases considered in Section 8.5 because $\mathcal{K} = 0$ here.

The singularity of this transport equation lies in the fact that (H2) is only guaranteed by the boundary condition. To fall into our splitting framework, we may replace the boundary condition by a singular source term $Af = \mathcal{R}_{\mathcal{O}}[f](0)\delta_0$, where $\delta_0$ is the Dirac mass at the origin, and write $\mathcal{L} = \mathcal{A} + \mathcal{B}$ with $\mathcal{B}$ the generator of the free transport equation with zero flux boundary condition. This forces working in a space of measures, as in [276, 279]. We briefly present an alternative approach, which is more in the spirit of [34, 167] and which consists in working in the Lebesgue space $L^1$, first to solve the dual problem in $L^\infty = (L^1)'$ and next to use for instance Doblín's contraction to solve the primal problem.

We assume here that

$$ (8.108) \qquad 0 \le K, r_{\mathcal{O}} \in L_{\text{loc}}^{\infty}(0, \infty), \quad (r_{\mathcal{O}} - \alpha K)_{+} \in L^{\infty}, $$

$$ (8.109) \qquad \lim_{y \to \infty} K(y) = +\infty, \quad \liminf_{y \to +\infty} r_{\mathcal{O}}(y) > 0, $$

for some $\alpha \in (0, 1)$, and we verify the usual conditions for the direct or the dual problem.

**Condition (H1).** Under assumption (8.108), the age structured equation (8.106)-(8.107) is well-posed in $L^1$ thanks to Proposition 8.23 and we may associate to it a positive semigroup $S_{\mathcal{L}}$ in $L^1$ with growth bound $\omega(S_{\mathcal{L}}) \le \kappa_1 := \|(r_{\mathcal{O}} - K)_{+}\|_{L^{\infty}}$ thanks to Corollary 8.25. We deduce that (H1) holds for the primal problem and thus also for the dual problem thanks to Lemma 2.2 and Lemma 2.3.

**Condition (H2).** The generator of the dual problem is

$$ \mathcal{L}^*\phi = \partial_y\phi - K(y)\phi + \phi(0)r_{\mathcal{O}}(y) $$

with domain $D(\mathcal{L}^*) \subset W_{\text{loc}}^{1,\infty}(\mathcal{O})$. From the second hypothesis in (8.109), there exist $y_0, \eta_0 \in (0, \infty)$ such that $r_{\mathcal{O}}(y) \ge \eta_0$ for any $y \ge y_0$. We then define

$$ \phi_0(y) = \mathbf{1}_{[0, y_0)} + \eta_0(y_1 - y)\mathbf{1}_{[y_0, y_1)}, \quad y_1 := y_0 + 1/\eta_0, $$

and we compute

$$ \mathcal{L}^*\phi_0 = r_{\mathcal{O}}(y) - K \ge -\|K - r_{\mathcal{O}}\|_{L^{\infty}(0, y_0)} \quad \text{on } (0, y_0), $$

$$ \mathcal{L}^*\phi_0 = r_{\mathcal{O}}(y) - \eta_0 - K\phi_0 \ge -\|K\|_{L^{\infty}(y_0, y_1)}\phi_0 \quad \text{on } (y_0, y_1), $$

$$ \mathcal{L}^*\phi_0 = 0 \quad \text{on } (y_1, \infty), $$

so that in the three case $\mathcal{L}^*\phi_0 \ge \kappa_0\phi_0$ with $\kappa_0 := -\max(\|K - r_{\mathcal{O}}\|_{L^{\infty}(0, y_0)}, \|K\|_{L^{\infty}(y_0, y_1)})$. Using Lemma 2.4-(i), we have thus established that $\mathcal{L}$ satisfies (H2) with constructive constant $\kappa_0$.

**Condition (H3) on the dual problem.** We define the splitting $\mathcal{L}^* = \mathcal{A}^* + \mathcal{B}^*$ with $\mathcal{A}^*\phi := (\mathcal{R}_{\mathcal{O}}^*\phi)(y) = \phi(0)r_{\mathcal{O}}(y)$. From the first hypothesis in (8.109), for any $\kappa_* \le 0$ there exists $y_* \in [0, \infty)$ such that $K(y) \ge -\kappa_*$ for any $y \ge y_*$. Defining $m_* := e^{\kappa_*y}\mathbf{1}_{[0, y_*)} + e^{\kappa_*y_*}\mathbf{1}_{[y_*, \infty)}$, we compute

$$ \mathcal{B}^*m_* = \kappa_*e^{\kappa_*y}\mathbf{1}_{[0, y_*)} - Km_* \le \kappa_*m_* $$

Together with Proposition 8.23 and Corollary 8.25, we deduce that the operator $\mathcal{B} - \kappa_*$, with domain $D(\mathcal{B}) := \{f \in L^1(\mathcal{O}) ; \partial_yf + Kf \in L^1(\mathcal{O}) , f(0) = 0\}$, generates a contraction semigroup in $L^1_{m_*}(\mathcal{O})$, and thus a bounded semigroup in $L^1(\mathcal{O})$ because $m_*, m_*^{-1} \in L^\infty(\mathcal{O})$. In other words, we have established that $\omega(S_{\mathcal{B}}) = -\infty$. Now, we see that $\mathcal{R}_{\mathcal{B}*}(\lambda) : L^\infty \to D(\mathcal{B}^*) \subset W_{\text{loc}}^{1,\infty}([0, \infty))$ is bounded for any $\lambda \in \mathbb{R}$ and thus $\mathcal{A}^*\mathcal{R}_{\mathcal{B}*}(\lambda) : L^\infty \to L^\infty$ is compact for any $\lambda \in \mathbb{R}$. We deduce from Lemma 2.8 and Remark 2.10, that $\mathcal{L}^*$ satisfies (H3).
Proof of Lemma 9.2. Throughout the proof we denote by $c_t$ any positive constant that depend only on $t$. It is proved in [354, Prop. 5] the existence of $(x_2, x_3) \subset (x_0, \infty)$ such that for all $R_1 > x_0$ there exists $T_0 > 0$ such that for any $T > T_0$ and any $\phi \ge 0$

$$ (9.14) \qquad S_T^* \phi \ge c_T \mathbf{1}_{(x_0, R_1)} \int_{x_2}^{x_3} \phi(x) dx. $$

We may now extend the integral to $[x_0 + \varepsilon, R_2]$. The Duhamel formula

$$ S_{\mathcal{L}}^* = S_{B_0}^* + S_{B_0}^* A_0 * S_{\mathcal{L}}^* $$

for the splitting $\mathcal{L}^* = A_0^* + B_0^*$ with $A_0^*\phi = \mathcal{H}^*[\phi]$ and $B_0^*\phi = b\phi' - K\phi$, also reads

$$ (9.15) \quad S_t^* \phi(x) = \phi(X_t(x)) e^{-\int_0^t K(X_s(x))ds} + 2 \int_0^t K(X_{t-s}(x)) S_s^* \phi(X_{t-s}(x)/2) e^{-\int_0^{t-s} K(X_{s'}(x))ds'} ds, $$

where $X_t(x)$ is the solution to the characteristic equation

$$ (9.16) \qquad \dot{X}_t(x) = a(X_t(x)) \quad \text{with} \quad X_0(x) = x. $$

Applying (9.14) to $S_t^*\phi$, that we bound from below by the first in Duhamel's formula (9.15), we obtain

$$ S_{T+t}^* \phi \ge c_T \mathbf{1}_{(x_0, R_1)} \int_{x_2}^{x_3} \phi(X_t(x)) e^{-\int_0^t K(X_s(x))ds} dx \ge c_T c_t \mathbf{1}_{(x_0, R_1)} \int_{X_t(x_2)}^{X_t(x_3)} \phi(y) dy. $$

Choosing $t_0$ such that $X_{t_0}(x_2) = x_3$, we get that for all $T > T_0 + t_0$

$$ S_T^* \phi \ge c_T \mathbf{1}_{(x_0, R_1)} \int_{x_2}^{X_{t_0}(x_3)} \phi(x) dx. $$

Iterating this argument and using the strict positivity of $a$ we get for any $R_2 > x_2$ the existence of a time $t_1$ such that for all $T > T_0 + t_1$

$$ (9.17) \qquad S_T^* \phi \ge c_T \mathbf{1}_{(x_0, R_1)} \int_{x_2}^{R_2} \phi(x) dx. $$

For decreasing the lower bound of the integral from $x_2$ to $x_0 + \varepsilon$, we iterate once Duhamel's formula (9.15) to get

$$ S_t^* \phi(x) \ge 2 \int_0^t K(X_{t-s}(x)) \phi(X_s(X_{t-s}(x)/2)) e^{-\int_0^{t-s} K(X_{s'}(x))ds' - \int_0^s K(X_{s'}(x))ds'} ds $$

and then, using (9.17),

$$ S_{T+t}^* \phi \ge c_t c_T \mathbf{1}_{(x_0, R_1)} \int_0^t \int_{x_2}^{R_2} K(X_{t-s}(x)) \phi(X_s(X_{t-s}(x)/2)) dx ds. $$

We can assume that $x_2 > 2x_0$ and $R_2 > 2x_2$. The fact that $x_2 > 2x_0$ ensures, due to assumption (9.3), that $K$ is bounded from below by a positive constant on $[x_2, X_t(R_2)]$. We thus deduce, by using of a change of variables, that for any $t > 0$

$$ S_{T+t}^* \phi \ge c_t c_T \mathbf{1}_{(x_0, R_1)} \int_{X_t(X_t(x_2)/2)}^{R_2/2} \phi(y) dy. $$

Since $X_t(x) \to x$ when $t \to 0$, we deduce for all $\zeta > 0$ the existence of $t > 0$ such that

$$ S_{T+t}^* \dot{\phi} \ge c_t c_T \mathbf{1}_{(x_0, R_1)} \int_{x_2/2+\zeta}^{R_2/2} \phi(y) dy. $$

Since $R_2 > 2x_2$, we deduce by combining the above inequality with (9.17) that for all $T > T_0 + t_1 + t$

$$ S_T^* \phi \ge c_T \mathbf{1}_{(x_0, R_1)} \int_{x_2/2+\zeta}^{R_2} \phi(x) dx. $$

Let us take $\zeta = x_0$. Since the sequence $(u_n)$ defined by $u_0 = x_2$ and $u_{n+1} = u_n/2 + x_0$ converges to $2x_0$, we obtain by an iteration argument the existence of a time $t_2$ such that for all $T > T_0 + t_2$

$$ S_T^* \phi \ge c_T \mathbf{1}_{(x_0, R_1)} \int_{2x_0+\varepsilon}^{R_2} \phi(x) dx. $$
Using Lemma 2.8-(1), we conclude to the existence of $(\lambda_1, \phi_1)$ solution to the dual eigenvalue problem. Now, we turn to the existence, uniqueness, and exponential stability of $f_1 \in L^1$, by verifying that Doblín's condition (6.2) is satisfied.

**Doblín condition.** Denoting $S_t := S_L(t)$, we have from the characteristics method

$$S_t f(y) = f(y-t)e^{-\int_0^t K(y-s)ds} \mathbf{1}_{t<y} + N(t-y)e^{-\int_0^y K(s)ds} \mathbf{1}_{t>y}$$

with $N(t) = \int_0^\infty r_O(y_*)S_t f(y_*)dy_*$. Iterating this formula and using the positivity of $S_t$ we get that for any $f \ge 0$

$$\begin{aligned} S_t f(y) &\ge \left( \int_0^{t-y} r_O(y_*) N(t-y-y_*) e^{-\int_0^{y_*} K(s)ds} dy_* \right) e^{-\int_0^y K(s)ds} \mathbf{1}_{0<y<t} \\ &\ge \left( \int_0^{t-y} r_O(t-y-\tau) N(\tau) e^{-\int_0^{t-y-\tau} K(s)ds} d\tau \right) e^{-\int_0^y K(s)ds} \mathbf{1}_{0<y<t}. \end{aligned}$$

Choosing $t_0 > 2y_0$ so that $r_O(y) \ge \eta_0 > 0$ for all $y \ge t_0/2$, we obtain

$$S_{t_0} f(y) \ge \eta_0 e^{-\int_0^{t_0} K(s)ds} \left( \int_0^{t_0/4} N(\tau)d\tau \right) e^{-\int_0^y K(s)ds} \mathbf{1}_{0<y<t_0/4}.$$

From the expression of $N(t)$, we get by duality, using that $r_O \ge \eta_0 \mathbf{1}_{(y_0, \infty)}$, that

$$(8.110) \quad S_{t_0} f(y) \ge \eta_0^2 e^{-2 \int_0^{t_0} K(s)ds} \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_\tau^* \mathbf{1}_{(y_0, \infty)}(y_*) d\tau \right) dy_* \right) \mathbf{1}_{0<y<t_0/4}.$$

Applying $S_{t_1}$ to this inequality we deduce that for any $t_1 > 0$

$$\begin{aligned} S_{t_0+t_1} f(y) &\ge \eta_0^2 e^{-2 \int_0^{t_0} K(s)ds} \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_\tau^* \mathbf{1}_{(y_0, \infty)}(y_*) d\tau \right) dy_* \right) S_{t_1} \mathbf{1}_{0<y<t_0/4} \\ &\ge \eta_0^2 e^{-2 \int_0^{t_0} K} e^{-\int_0^{t_1} K} \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_\tau^* \mathbf{1}_{(y_0, \infty)}(y_*) d\tau \right) dy_* \right) \mathbf{1}_{t_1<y<t_0/4+t_1}. \end{aligned}$$

On the other hand, replacing $f$ by $S_{t_1}f$ in (8.110) we obtain

$$(8.111) \quad S_{t_0+t_1} f(y) \ge \eta_0^2 e^{-2 \int_0^{t_0} K(s)ds} \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_{\tau+t_1}^* \mathbf{1}_{(y_0, \infty)}(y_*) d\tau \right) dy_* \right) \mathbf{1}_{0<y<t_0/4}.$$

The fact that $S_t^* \phi(y) \ge \phi(t+y)e^{-\int_0^t K(y+s)ds}$ ensures that for $t_1 > y_0$

$$S_{t_1}^* \mathbf{1}_{(y_0, \infty)} \ge e^{-\sup_{y \in [0, y_0]} \int_0^{t_1} K(y+s)ds} \mathbf{1}_{[0, y_0]}.$$

All together, we have proved that for any $t_0 > 4t_1 > 4y_0$ we have

$$S_{t_0+t_1} f(y) \ge c_0 \left( \int_0^\infty f(y_*) \left( \int_0^{t_0/4} S_\tau^* \mathbf{1}(y_*) d\tau \right) dy_* \right) \mathbf{1}_{t_1<y<t_0/4}$$

for some explicit constant $c_0$ and all $f \ge 0$. This is Doblín's condition (6.2) with $T = t_0 + t_1$, and the functions $\psi_0 = \int_0^{t_0/4} S_\tau^* \mathbf{1} d\tau$ and $g_0 = c_0 \mathbf{1}_{(t_1, t_0/4)}$. We are now in position to prove the following result.

**Theorem 8.38.** Under the assumptions (8.108) and (8.109), the renewal equation (8.106)-(8.107) enjoys the conclusions (S1), (S2) and (S32) in $L^1$ as well as (E31) with quantitative rate in $L_{\psi_0}^1$.

Despite the numerous results about the renewal age-structured model, we are not aware of any previous result with a constructive rate of convergence under such general assumptions.

*Proof of Theorem 8.38.* The conditions (H1), (H2) and (H3) for $\mathcal{L}^*$ ensure the existence of $\lambda_1 \ge \kappa_0$ and $\phi_1 \in L^\infty$, $\phi_1 > 0$, that we normalize by $\|\phi_1\|_{L^\infty} = 1$. If we can prove that the conditions (6.2), (6.3) and (6.4) are verified, then the conclusions (S1), (S2) and (E31) follow by applying Theorem 6.2. We have already proved (6.2) with the functions $\psi_0 = \int_0^{t_0/4} S_\tau^* \mathbf{1} d\tau$ and $g_0 = c_0 \mathbf{1}_{(t_1, t_0/4)}$. For proving (6.3), we start by recalling that $\phi_1 = R_B^*(\lambda_1)A^*\phi_1 \in W_{loc}^{1,\infty}$ due to
Using a last time the argument with $\zeta = \varepsilon/2$ yields the desired result. $\square$

We now prove another positivity result which allows making the time $T$ independent of $R_1$ in Lemma 9.2.

**Lemma 9.3.** Let $R_1 > 2x_0$. Then there exists $t_0 > 0$ such that for any $R > R_1$ we have

$$S_{t_0}^* \mathbf{1}_{(x_0, R_1)} \geq c_R \mathbf{1}_{(x_0, R)}$$

for some $c_R > 0$.

*Proof of Lemma 9.3.* Since $a$ is Lipschitz continuous, we can find $t_0 > 0$ small enough so that $X_{t_0}(x) \leq \alpha x$ for all $x > x_0$, with $\alpha > 1$ to be determined later. Then for any $t \in (0, t_0]$ and any $x \in (x_0, \frac{R_1}{\alpha})$, we have by using the first term in (9.15)

$$S_t^* \mathbf{1}_{(x_0, R_1)}(x) \geq c_{t_0} \mathbf{1}_{(x_0, R_1)}(X_t(x)) = c_{t_0} > 0.$$

Iterating once (9.15) and keeping only the second term, we get that for any $t \in (0, t_0]$ and any $x \in (2x_0, \frac{2R_1}{\alpha^2})$

$$S_t^* \mathbf{1}_{(x_0, R_1)}(x) \geq c_{t_0} \int_0^t \mathbf{1}_{(x_0, R_1)}(X_s(X_{t-s}(x)/2)) ds = c_{t_0} t.$$

Choosing $\alpha > 1$ such that $\frac{R_1}{\alpha} > 2x_0$ and $\frac{2R_1}{\alpha^2} > R_1$, we deduce that for any $t \in (0, t_0]$ there exists $c_t > 0$ such that

$$S_t^* \mathbf{1}_{(x_0, R_1)} \geq c_t \mathbf{1}_{(x_0, 2\alpha^{-2}R_1)}.$$

Dividing $[0, t_0]$ into $n$ sub-intervals $[\frac{k}{n}t_0, \frac{k+1}{n}t_0]$, $0 \leq k \leq n-1$, and iterating the above inequality with $t = t_0/n$, we deduce for all integer $n \geq 1$ the existence of $c_n > 0$ such that

$$S_{t_0}^* \mathbf{1}_{(x_0, R_1)} \geq c_n \mathbf{1}_{(x_0, (2\alpha^{-2})^n R_1)}$$

and the proof is complete since $2\alpha^{-2} > 1$. $\square$

With Lemmas 9.2 and 9.3, we are now in position to prove the convergence result in Theorem 9.1.

*Proof of the exponential stability part of Theorem 9.1.* We apply Theorem 6.3. We start by proving that (6.9) is verified, in a quantitative way, for the function $g_0 = \mathbf{1}_{(x_0+\varepsilon, R_2)}$ with a suitable choice of $R_2$ and $\varepsilon$. Choosing $r_0 \in (1, r)$ if $m(x) = x^r$ or any $r_0 > 1$ is $m(x) = \exp(\eta \int_{x_0}^x K/a)$ and defining $m_0(x) = x^{r_0}$ we have from (9.12), because of the normalization $\|\phi_1\|_{L_{m-1}^\infty} = 1$,

$$\|\phi_1\|_{L_{m_0-1}^\infty} \leq C_0$$

for some explicit constant $C_0 > 0$. Defining

$$R_2 := \inf\{R > 0; m_0(x)/m(x) \leq 1/2C_0, \forall x > R\},$$

we have

$$1 = \|\phi_1\|_{L_{m-1}^\infty} = \sup_{(x_0, \infty)} \frac{\phi_1}{m} = \sup_{(x_0, R_2)} \frac{\phi_1}{m},$$

because

$$\sup_{(R_2, \infty)} \frac{\phi_1}{m} \leq \sup_{(R_2, \infty)} \frac{\phi_1}{m_0} \frac{m_0}{m} \leq C_0 \frac{1}{2C_0} < 1.$$

Together with the fact that $\phi_1' \in L_{\text{loc}}^\infty$, with a quantitative estimate on $\|\phi_1'\|_{L^\infty(x_0, R_2)}$, we see that $\phi_1$ has some quantifiable mass on $(x_0 + \varepsilon, R_2)$ for $\varepsilon > 0$ small enough, which exactly means that $\langle g_0, \phi_1 \rangle$ is quantified from below.

Now we prove that the Harris condition (6.8) is verified. Choosing $R_1 > 2x_0$ and combining Lemma 9.2 and Lemma 9.3, we have for any $\varepsilon > 0$ and $R_2 > x_0 + \varepsilon$ the existence of $T > 0$ such that for any $R > R_1$

$$(9.18) \qquad S_T^* \phi \geq c_R \mathbf{1}_{(x_0, R)} \int_{x_0+\varepsilon}^{R_2} \phi \, dx, \quad \forall \phi \geq 0.$$

Defining $g_0 = \mathbf{1}_{(x_0+\varepsilon, R_2)}$, we deduce by duality that for all $f \geq 0$,

$$S_T f \geq c_R \langle f, \mathbf{1}_{(x_0, R)} \rangle g_0.$$
the informations derived on $\mathcal{R}_B^*$ in (H3). Consequently, there exists $y_1 > 0$ such that $\phi_1(y_1) > 1/2$, and we deduce from $\phi'_1 \le (\lambda_1 + K)\phi_1$ that

$$ \phi_1(y) \ge \frac{1}{2} e^{-\int_y^{y_1} (\lambda_1 + K)} $$

for all $y \in (0, y_1)$. Choosing in the proof of the Doblin condition $t_0$ such that $y_1 < t_0/4$, we obtain that

$$ \langle \phi_1, g_0 \rangle \ge \frac{c_0}{2} \int_{y_1/2}^{y_1} e^{-\int_y^{y_1} (\lambda_1 + K)} dy, $$

which gives (6.3). For (6.4), we use that

$$ \phi_1 = e^{-\lambda_1 \tau} S_{\tau}^{*} \phi_1 \le e^{-\lambda_1 \tau} S_{\tau}^{*} \mathbf{1} $$

for any $\tau > 0$ to deduce that

$$ \phi_1 = \frac{4}{t_0} \int_{0}^{t_0} e^{-\lambda_1 \tau} S_{\tau}^{*} \phi_1 d\tau \le \frac{4e^{\lambda_1 |t_0|}}{t_0} \psi_0. $$

Finally, we check that the condition (H5') is verified, so that (S32) is valid by virtue of Theorem 5.18 and Remark 5.19. The condition (H5') is actually a direct consequence of the fact that (8.111) is verified for any $t_0 > 2y_0$ and $t_1 > 0$ together with the estimate

$$ S_{\tau+t_1}^* \mathbf{1}_{(y_0, \infty)}(y) \ge e^{-\int_0^{t_1} K(y+s)ds} > 0 $$

for any $t_1 > y_0$ and $\tau > 0$, and all $y > 0$. $\square$

## 9. THE GROWTH-FRAGMENTATION EQUATION

In this section, we are interested in the growth-fragmentation equation with equal mitosis kernel

$$ (9.1) \qquad \partial_t f(t, x) + \partial_x(a(x)f(t, x)) + K(x)f(t, x) = 4K(2x)f(t, 2x) $$

and to its variant with an additional “growth speed” variable

$$ (9.2) \qquad \partial_t f(t, x, v) + v \partial_x (a(x) f(t, x, v)) + K(x) f(t, x, v) = 4 \int_1^2 K(2x) \wp(v, v_*) f(t, 2x, v_*) dv_*, $$

with $x > 0$ and $v \in [1, 2]$. For both equations, we assume that the total fragmentation rate $K$ is a continuous function defined on $\mathbb{R}_+$ such that

$$ (9.3) \qquad \exists x_0 > 0, \quad K = 0 \text{ on } (0, 2x_0] \quad \text{and} \quad K > 0 \text{ on } (2x_0, \infty). $$

This condition ensures that no particle of size less than $x_0$ can be produced by division, and we thus consider the equations posed on the size space $(x_0, \infty)$ with zero flux boundary condition $f(t, x_0) = 0$ or $f(t, x_0, v) = 0$. The growth rate $a$ is supposed to be positive and globally Lipschitz on $[x_0, \infty)$, and we assume that

$$ (9.4) \qquad \lim_{x \to \infty} \frac{xK(x)}{a(x)} = +\infty. $$

For quantifying the positivity of the first eigenvalue, we also make the technical assumption that

$$ (9.5) \qquad \exists k > 0, \quad \lim_{x \to \infty} e^{x^k} K(x) = +\infty. $$
Let us now consider $A > 0$ and $f \in X_+$ such that $\|f\| \le A[f]_{\phi_1}$. Since $m_0(x)/m(x) \to 0$ as $x \to +\infty$ and $\|\phi_1/m_0\|_{\infty} \le C_0$, we have

$$ 
\begin{aligned}
[f]_{\phi_1} &= \int_{x_0}^{R} f \frac{\phi_1}{m} m \, dm + \int_{R}^{\infty} f m \frac{\phi_1}{m_0} \frac{m_0}{m} \, dm \\
&\le \langle f, \mathbf{1}_{(x_0, R)} \rangle \sup_{(x_0, R)} m + \|f\| C_0 \sup_{(R, \infty)} \frac{m_0}{m} \\
&\le \langle f, \mathbf{1}_{(x_0, R)} \rangle m(R) + \frac{1}{2} [f]_{\phi_1},
\end{aligned}
$$

by choosing $R$ large enough. We deduce that $S_T f \ge \frac{c_R}{2m(R)} [f]_{\phi_1} g_0$, which is equivalent to (6.8) where we recall the definition $\tilde{S}_t := S_t e^{-\lambda_1 t}$.

Finally, we prove the Lyapunov condition (6.7). On the one hand, we get from (9.10) that

$$ \frac{d}{dt} \tilde{S}_t^* m = \tilde{S}_t^* (\mathcal{L}^* - \lambda_1) m \le (\kappa_B - \lambda_1) \tilde{S}_t^* m + M \tilde{S}_t^* (\mathbf{1}_{(x_0, R)} m). $$

On the other hand, arguing as in (6.6), we infer from (9.18) that

$$ \phi_1 = e^{-\lambda_1 T} S_T^* \phi_1 \ge c_R e^{-\lambda_1 T} \langle g_0, \phi_1 \rangle \mathbf{1}_{(x_0, R)}. $$

Combining both we deduce that

$$ \frac{d}{dt} \tilde{S}_t^* m \le (\kappa_B - \lambda_1) \tilde{S}_t^* m + \tilde{M} \phi_1 $$

with $\tilde{M} = \frac{m(R)}{c_R} \frac{e^{\lambda_1 T}}{\langle g_0, \phi_1 \rangle}$, and Grönwall's inequality then yields

$$ \tilde{S}_t^* m \le e^{(\kappa_B - \lambda_1)t} m + \tilde{M}te^{(\kappa_B - \lambda_1)t} \phi_1. $$

This guarantees that (6.7) is verified with $\gamma_L = e^{(\kappa_B - \lambda_1)T} \in (0, 1)$ and $K = \tilde{M}T$.

We have proved that the conditions (6.13), (6.7) and (6.9) are verified. The conclusion of the proof then follows from Theorem 6.3. $\square$

## 9.2. The mitosis equation with non-mixing growth rate.
In this section, we investigate the case when the mixing condition (9.6) is not verified. In other words, we place ourselves under the singular condition that

$$ (9.19) \qquad \forall x > x_0, \quad a(2x) = 2a(x). $$

In this case, we still have the existence of a unique eigen-triplet $(\lambda_1, f_1, \phi_1)$ but the boundary point spectrum is not reduced to $\lambda_1$. As a consequence, the long time behavior of the semigroup does not stabilizes along $f_1$ but it exhibits periodic oscillations.

**Theorem 9.4.** Suppose that (9.3), (9.4), (9.5) and (9.19) are satisfied. The first eigentriplet problem admits a unique solution $(\lambda_1, f_1, \phi_1) \in \mathbb{R} \times X_+ \times X'_+$ with the normalization $\|\phi_1\| = \langle \phi_1, f_1 \rangle = 1$, and this triplet additionally satisfies $\lambda_1 > 0$, $f_1 > 0$ and $\phi_1 > 0$.

Besides, $\Sigma_P^+(\mathcal{L}) = \{\lambda_1 + ik\alpha, k \in \mathbb{Z}\}$ for some quantifiable $\alpha > 0$, there exists a family $(g_k, \psi_k)_{k \in \mathbb{Z}}$ of corresponding primal and dual eigenvectors that verifies $\langle \psi_k, g_\ell \rangle = \delta_{k\ell}$, and for all $f \in L_{\phi_1}^1$, we have the convergence

$$ \|e^{-\lambda_1 t} S_{\mathcal{L}}(t)(f - \Pi f)\|_{L_{\phi_1}^1} \to 0 \quad \text{as } t \to +\infty, $$

where $\Pi f = \lim_{n \to \infty} \frac{1}{n} \sum_{\ell=0}^{n} \sum_{k=-\ell}^{\ell} \langle \psi_k, f \rangle g_k$.

This new result complements the scarce literature on the long time behavior of equation (9.1) in the singular case (9.19) which, to the best of our knowledge, is limited to the references [50, 169, 188]. We will actually prove that the convergence in Theorem 9.4 also holds in other spaces, and this will be the occasion to compare our method and results to the three above mentioned papers.

The proof of the conclusion (C2) in Section 9.1 does not use the mixing assumption (9.6). It thus also proves the existence, uniqueness and strict positivity of eigentriplet $(\lambda_1, f_1, \phi_1)$ under the assumptions of Theorem 9.4, as well as the fact that equation (9.1) is associated with a semigroup
**9.1. The mitosis equation with mixing growth rate.** We are interested here in the growth-fragmentation equation (9.1) in the case where

$$ (9.6) \qquad \exists x_1 > x_0, \ a(2x_1) \neq 2a(x_1). $$

As we will see below, this condition ensures some mixing property for the trajectories that guarantees the triviality of the boundary point spectrum.

We work in the space $X = L_m^1$ with a weight $m$ that can be

$$ (9.7) \qquad \text{either } m(x) = x^r, \ r > 1, \quad \text{or } m(x) = \exp \left( \eta \int_{x_0}^x \frac{K}{a} \right), \ 0 < \eta < 1. $$

Note that due to assumption (9.4), the weight $\exp(\eta \int_{x_0}^x K/a)$ is always stronger than $x^r$.

**Theorem 9.1.** Suppose that (9.3), (9.4), (9.5) and (9.6) are satisfied. The first eigentriplet problem admits a unique solution $(\lambda_1, f_1, \phi_1) \in \mathbb{R} \times X_+ \times X'_+$ with the normalization $\|\phi_1\| = \langle \phi_1, f_1 \rangle = 1$, and this triplet additionally satisfies $\lambda_1 > 0$, $f_1 > 0$ and $\phi_1 > 0$. Besides, there are some constructive constants $C \ge 1$, $\omega > 0$ such that

$$ \|e^{-\lambda_1 t} S_{\mathcal{L}}(t) f - \langle \phi_1, f \rangle f_1\|_{X} \le C e^{-\omega t} \|f - \langle \phi_1, f \rangle f_1\|_{X} $$

for any $f \in X$ and $t \ge 0$. In other words, the conclusions (S1), (S2), (S3) and (E3) hold with constructive constants in $L_m^1$.

This result is contained in the recent paper [353]. The novelty here is that all the constants are obtained constructively, which is not clear in [353]. We also provide what seems to us to be a more direct and comprehensive proof. We also refer to [35, 52, 79, 277] where the same result is obtained under stronger assumptions.

Before starting the proof of Theorem 9.1, let us briefly justify the relevance of the chosen weight functions $m$ in (9.7). The dual operator associated to equation (9.1) is given by

$$ \mathcal{L}^* \phi(x) = a(x) \phi'(x) - K(x) \phi(x) + 2K(x) \phi(x/2). $$

For $m(x) = x^r, r > 1$, we can compute

$$ (9.8) \qquad \mathcal{L}^* m(x) = \left[ r \frac{a(x)}{x} - (1 - 2^{1-r}) K(x) \right] m(x), $$

and for $m(x) = \exp(\eta \int_{x_0}^x K/a)$, $0 < \eta < 1$,

$$ (9.9) \qquad \mathcal{L}^* m(x) = \left[ 2 \exp \left( -\eta \int_{x/2}^x \frac{K}{a} \right) - (1 - \eta) \right] K(x) m(x). $$

Assumption (9.3) then ensures that $\mathcal{L}^* m \sim -\xi K m$ as $x \to +\infty$, with $\xi = 1 - 2^{1-r} > 0$ in the first case and $\xi = 1 - \eta > 0$ in the second case. In both cases, we deduce that

$$ (9.10) \qquad \mathcal{L}^* m \le \kappa m + M \mathbf{1}_{(x_0, R)} m $$

for any $\kappa \ge 0$, by choosing $M > 0$ and $R > x_0$ large enough, and this type of Lyapunov inequality is pivotal in our analysis.

**Condition (H1).** Equation (9.1) is a particular case of equation (8.58) with $G = \mathfrak{g} = \mathcal{R} = 0$, $b = K + \text{diva}$ and $\mathcal{H}[g](x) = 4K(2x)g(2x)$. We may then use Proposition 8.23-(1) to infer the well-posedness of equation (9.1) in $X = L_m^1(x_0, \infty)$, provided that the conditions (8.35) and (8.61) are met, with $0 < \alpha_{\mathcal{H}} < 1$, which is nothing but (8.63) when $\mathcal{R} \equiv 0$. To do so, we define the function

$$ \varpi := K - a \frac{m'}{m}, $$

which corresponds to $\varpi_1$ in (8.34). When $m(x) = x^r$ with $r > 1$, we have

$$ \varpi(x) = K(x) - r \frac{a(x)}{x}, $$

and for $m(x) = \exp(\eta \int_{x_0}^x K/a)$ with $0 < \eta < 1$, we have

$$ \varpi(x) = (1 - \eta) K(x). $$
$S$ in $X$. For proving the long time convergence result, we start by verifying that this semigroup extends to other relevant Banach spaces.

**Well-posedness in entropic $L^p$ and $M^1$ spaces.** The dual eigenfunction $\phi_1$ satisfies by definition $\mathcal{L}^*\phi_1 = \lambda_1\phi_1$ and the rescaled semigroup $\tilde{S}_t = S_t e^{-\lambda_1 t}$ is thus a contraction for the norm of $L_{\phi_1}^1$. In particular $S_t$ is a bounded operator for this norm and, since $L_m^1$ is dense in $L_{\phi_1}^1$, we can uniquely extend the semigroup $S$ into a strongly continuous semigroup in $L_{\phi_1}^1$. Similarly, due to the weak-* density of $L_{\phi_1}^1$ into $M_{\phi_1}^1$, this semigroup extends uniquely into a weakly-* continuous semigroup in $M_{\phi_1}^1$. We still denote by $S$ these extensions.

The General Relative Entropy principle, see [269, 49], ensures that the weighted $L^p$ sub-spaces of $L_{\phi_1}^1$ defined by

$$X_p := L_{f_1^{1-p}\phi_1}^p (x_0, \infty) \quad \text{for } p \in [1, \infty) \quad \text{and} \quad X_\infty := L_{f_1^{-1}}^\infty (x_0, \infty)$$

are invariant under the semigroup $S$ and the restriction to these spaces is a contraction. Besides, Jensen's inequality yields that it is a decreasing sequence for the inclusion

$$p > q \implies X_p \supset X_q.$$

Since $X_\infty \subset X_p$ is dense, we can infer the strong continuity of $S$ in $X_p$ from the strong continuity in $X_1$ by writing for any $f \in X_\infty$

$$\|\tilde{S}_t f - f\|_{X_p}^p \le \|\tilde{S}_t f - f\|_{X_\infty}^{p-1} \|\tilde{S}_t f - f\|_{X_1} \le 2^{p-1} \|f\|_{X_\infty}^{p-1} \|\tilde{S}_t f - f\|_{X_1} \to 0,$$

as $t \to 0$.

**Long-time convergence in $M_{\phi_1}^1$.** We start by giving some useful properties of the dual semigroup $S^*$ in $X' = L_{m-1}^\infty$. Splitting $\mathcal{L}^*$ as $\mathcal{L}^* = \mathcal{A}_0^* + \mathcal{B}_0^*$ with $\mathcal{A}_0^*\phi = \mathcal{K}^*[\phi]$, so that $\mathcal{B}_0^*\phi = a\phi' - K\phi$, the Duhamel formula

$$S_{\mathcal{L}}^* = S_{\mathcal{B}_0}^* + S_{\mathcal{B}_0}^* \mathcal{A}_0 * S_{\mathcal{L}}^*$$

ensures that $\bar{\varphi}(t, x) := S_t^*\phi(x)$ is a fixed point of the operator $\Gamma$ defined by

$$(9.20) \quad \Gamma\varphi(t, x) := S_{\mathcal{B}_0}^*(t)\phi(x) + [S_{\mathcal{B}_0}^* \mathcal{A}_0 * \varphi(\cdot, x)](t) \\ = \phi(X_t(x))e^{-\int_0^t K(X_s(x))ds} + 2\int_0^t K(X_{t-s}(x))\varphi(s, X_{t-s}(x)/2)e^{-\int_0^{t-s} K(X_{s'}(x))ds'}ds,$$

where we recall that $X_t(x)$ is the solution to the characteristic equation (9.16). It turns out that $\Gamma$ has a unique fixed point in $L_{\text{loc}}^\infty([0, \infty) \times (x_0, \infty))$, and that this fixed point also lies in any closed subset of $L_{\text{loc}}^\infty([0, \infty) \times (x_0, \infty))$ which is left invariant by $\Gamma$. This property is proved in [169] or in [35, Sec. 6.3], by building $\bar{\varphi}$ thanks to the Banach-Picard fixed point theorem. It has very useful consequences, as for instance the fact that if $\phi \in C(x_0, \infty)$, then $\bar{\varphi} \in C([0, \infty) \times (x_0, \infty))$. In particular, this implies that $C(x_0, \infty) \cap L_{m-1}^\infty$ is invariant under the semigroup $S^*$. Since $C(x_0, \infty) \cap L_{m-1}^\infty$ is a dense subspace of $C_{0,\phi_1}(x_0, \infty)$, this ensures that $C_{0,\phi_1}$ is invariant under $S^*$ and that the duality relation

$$\langle S_t f, \phi \rangle = \langle f, S_t^* \phi \rangle$$

is valid for any $f \in M_{\phi_1}^1$ and $\phi \in C_{0,\phi_1}$. The proof of the next result crucially relies on another application of the fact that the fixed point of $\Gamma$ belongs to any closed invariant subset.

**Proposition 9.5.** Suppose that (9.3), (9.4), (9.5) and (9.19) are satisfied. Then $\Sigma_P^+(\mathcal{L}) = \{\lambda_1 + ik\alpha, k \in \mathbb{Z}\}$ for some $\alpha > 0$, there exists a family $(g_k, \psi_k)_{k \in \mathbb{Z}}$ of corresponding primal and dual eigenvectors that verifies $\langle \psi_k, g_\ell \rangle = \delta_{k\ell}$, and for all $f \in M_{\phi_1}^1$ we have the convergence

$$(9.21) \quad \tilde{S}_t f - \tilde{S}_t \Pi f \to 0, \quad \text{as } t \to \infty, \quad \Pi f := \lim_{n \to \infty} \frac{1}{n} \sum_{\ell=0}^{n} \sum_{k=-\ell}^{\ell} \langle \psi_k, f \rangle g_k,$$

both convergences having to be understood in the sense of the weak-* topology.

Note that we did not specify the space in which we define the boundary point spectrum $\Sigma_P^+(\mathcal{L})$ in Proposition 9.5. It is because this set is the same in all the Banach lattices we consider. Indeed, any $g \in M_{\phi_1}^1$ such that $\mathcal{L}g = \lambda g$ for some $\lambda \in \mathbb{C}$ with $\Re e(\lambda) = \lambda_1$ satisfies $|g| \in \text{Span}(f_1)$, so that $g \in X = L_m^1$ for any weight $m$ as in (9.7).