below). In order to apply Proposition B.3, we need that $\overline{\chi_{\infty}S^V}$ belongs to some space $K_{\beta}^{\infty}$ (since $\Omega_1 \setminus B(0, R_c) = \Pi \setminus B(0, R_c)$, $\chi_{\infty}S^V$ can be seen as a function defined on $\Pi$, which allows us to consider $\overline{\chi_{\infty}S^V}$). We will use again the notations $\langle \cdot, \cdot \rangle$ and $T_{\ge d}$ of Definitions 2.3 and 2.21.

**Proposition B.4:** Let $d \in \mathbb{R}_-$,

$$ S^A := T_{\ge d} \left[ \left\langle -R_{\partial_Y^2} \circ \partial_{X|\Lambda}^2, \frac{\mu_0}{\mu_1} R_N \circ \partial_{Y|\Gamma, Y=0^+} \right\rangle \left( \frac{1}{\mu_0} R_{\Delta}(F_{\Omega}^{\infty}) + \frac{1}{\mu_1} R_{\partial_Y^2}(F_{\Lambda}^{\infty}) + \sum_{d \in \frac{\pi}{\bar{\Omega}}\mathbb{N}^*} \sigma_d(S) \phi_d \right) \right] $$

in $\mathcal{A}(\Pi)$ and $S^V := S - \chi_{\infty}S^A$. There is $d \in \mathbb{R}_-$ s.t. $S^V \in V$ and: $\forall m \in \mathbb{N}$, $\overline{\chi_{\infty}S^V} \in K_{1/2}^{\infty}(\mathbb{R}, \mathcal{H}^m)$.

**Proof:** We consider any $d \in \mathbb{R}_-$ and we will fix it later. The proof has three steps: showing that $S^V \in V$, showing that $S^V$ is regular w.r.t $t$, and deducing that it is regular w.r.t. $\theta$. For any $r_1, r_2 \in \mathbb{R}$ s.t. $R_c < r_1 < r_2$, we will denote $\chi \in C^{\infty}(\mathbb{R}^2)$ a radial function equal to 0 on $B(0, r_1)$ and 1 on $\mathbb{R}^2 \setminus B(0, r_2)$ ($r_1, r_2$ are implicit in this notation).

**Step 1:** Let $\varphi : \Pi \to \mathbb{C}$ be equal to $T_{\le d}(F_D^{\infty})$ on $D$ for any $D \in \{\Omega, \Lambda\}$, and $F_v = F - \chi_{\infty}\varphi$. If $d < -2$ then $(1+r)F_v \in L^2(\Omega_1)$. So the same construction than in Theorem 3.11 shows that for $d$ small enough, the problem

$$ \left\{ \begin{array}{ll} \text{div}(\mu \nabla S^V) = f_1 := F_v + \chi_{\infty} \varphi - \text{div}(\mu \nabla (\chi_{\infty} \cdot S^A)) & \text{in } \Omega_1 \setminus (\Gamma \cap \{X > R_c\}) \\ [\mu \partial_Y S^V]_{\Gamma} = g_1 := -[\mu \partial_Y (\chi_{\infty} S^A)]_{\Gamma} & \text{on } \Gamma \cap \{X > R_c\} \\ S^V = 0 & \text{on } \partial \Omega_1 \end{array} \right. \quad (B.8) $$

has a unique solution in $V$ and that $S = S^V + \chi_{\infty}S^A$. In addition, it also gives that $(\mu\Delta S^A - \varphi)|_D \in \sum_{d'<d} \mathcal{A}_{d'}(D)$ for any $D \in \{\Omega, \Lambda\}$, and $[\mu\partial_Y S^A]_{\Gamma} \in \sum_{d'<d} \mathcal{A}_{d'}(\Gamma)$. And we have $F_v = o_{\partial}(\mathbf{r}^d)$ by hypothesis, so for $d$ small enough and up to increasing $R_c$ (without loss of generality): $\forall m \in \mathbb{N}$,

$$ (\overline{\chi f_1})_{|\bar{\Omega}} \in K_{-2}^{\infty}(\mathbb{R}, H^m(0, \Theta)) \quad \text{and} \quad (\overline{\chi f_1})_{|\bar{\Lambda}} \in K_{-3/2}^{\infty}(\mathbb{R}, H^m(-1, 0)) \quad \text{and} \quad \overline{\chi g_1} \in K_{-1/2}^{\infty}(\bar{\Gamma}) \quad (B.9) $$

**Step 2:** For any $\beta \in \mathbb{R}$, we denote $K_{\beta}^{0}(\bar{\Omega}) := K_{\beta}^{0}(\mathbb{R}, L^2(0, \Theta)) = \{(t, \theta) \mapsto e^{\beta t}u(t, \theta) \mid u \in L^2(\bar{\Omega})\}$ and $K_{\beta}^{0}(\bar{\Lambda}) := K_{\beta}^{0}(\mathbb{R}, L^2(-1, 0))$. And we define

$$ \bar{V} := \{w \in H_{\text{loc}}^{1}(\bar{\Pi}) \mid \nabla w_{|\bar{\Omega}} \in L^2(\bar{\Omega}), \partial_t w_{|\bar{\Lambda}} \in K_{1/2}^{0}(\bar{\Lambda}), \partial_{\theta} w_{|\bar{\Lambda}} \in K_{-1/2}^{0}(\bar{\Lambda}) \text{ and } w_{|\bar{\Sigma}_{\Omega} \cup \bar{\Sigma}_{\Lambda}} = 0\}. $$

Note that if $\Omega_1$ were equal to $\Pi$, $\bar{V}$ would simply be $\{\bar{u} \mid u \in V\}$. So it is the natural variational space for Poisson's problem transferred into $\bar{\Pi}$. Looking at $\chi S^V$ as a function defined on $\Pi$, we will show by induction on $n$ that:

$$ \forall n \in \mathbb{N}, \forall R_c < r_1 < r_2, \quad \partial_t^n \overline{\chi S^V} \in \bar{V}. $$

Since $S^V \in V$, the initial case is trivial, so only the inductive step remains to prove. Let us assume the property at rank $n$ and show it at rank $n+1$. We will use the method of finite differences. Letting $\mu := \mu_0$ on $\Omega$ and $\mu := \mu_1$ on $\Lambda$, (B.8) implies:

$$ \left\{ \begin{array}{ll} \mu\Delta(\chi S^V) = f_2 := \chi f_1 - 2\mu\nabla\chi \cdot \nabla S^V - \mu\Delta\chi \cdot S^V & \text{in } D, \forall D \in \{\Omega, \Lambda\} \\ [\mu\partial_Y(\chi S^V)]_{\Gamma} = g_2 := \chi g_1 & \text{on } \Gamma \\ \chi S^V = 0 & \text{on } \Sigma_{\Omega} \cup \Sigma_{\Lambda} \end{array} \right. $$

By induction hypothesis applied to $(r'_1, r'_2) := (\frac{R_c+r_1}{2}, r_1)$, $\partial_t^k \nabla S^V$ and $\partial_t^k S^V$ are $L^2$ on $\{r_1 < r < r_2\}$ for any $k \in [0, n]$. So (B.9) implies $\bar{f}_{2|\bar{\Omega}} \in K_{-2}^n(\mathbb{R}, L^2(0, \Theta))$ and $\bar{f}_{2|\bar{\Lambda}} \in K_{-3/2}^n(\mathbb{R}, L^2(-1, 0))$. Similarly $\bar{g}_2 \in K_{-1/2}^{\infty}(\bar{\Gamma})$. Next, we apply the change variables $(x, y) \rightsquigarrow (t, \theta)$ using (B.2), and then $\partial_t^n$. We get that $s := \partial_t^n(\overline{\chi S^V})$ satisfies:

$$ \left\{ \begin{array}{ll} -e^{-2t}\mu_0\Delta s = f_\Omega := e^{-2t}\partial_t^n(e^{2t}\bar{f}_{2|\Omega}) & \text{in } \bar{\Omega} \\ -\mu_1(e^{-2t}(\partial_t^2 - \partial_t) + \partial_\theta^2)s = f_\Lambda := \partial_t^n \bar{f}_{2|\Lambda} + \sum_{k=0}^{n-1}(-2)^{n-k}e^{-2t}\mu_1(\partial_t^2 - \partial_t)\partial_t^k(\overline{\chi S^V}) & \text{in } \bar{\Lambda} \\ s|_{\theta=0^+} - s|_{\theta=0^-} = 0 & \text{on } \bar{\Gamma} \\ e^{-t}\mu_0\partial_\theta s|_{\theta=0^+} - \mu_1\partial_\theta s|_{\theta=0^-} = g := \partial_t^n \bar{g}_2 & \text{on } \bar{\Gamma} \\ s = 0 & \text{on } \overline{\Sigma_{\Omega} \cup \Sigma_{\Lambda}} \end{array} \right. \quad (B.10) $$
hypothesis, so for $d$ small enough and up to increasing $R_c$ (without loss of generality): $\forall m \in \mathbb{N}$,

$$ (\overline{\chi f_1})_{|\bar{\Omega}} \in K_{-2}^{\infty}(\mathbb{R}, H^m(0, \Theta)) \quad \text{and} \quad (\overline{\chi f_1})_{|\bar{\Lambda}} \in K_{-3/2}^{\infty}(\mathbb{R}, H^m(-1, 0)) \quad \text{and} \quad \overline{\chi g_1} \in K_{-1/2}^{\infty}(\bar{\Gamma}) \quad (B.9) $$

**Step 2:** For any $\beta \in \mathbb{R}$, we denote $K_{\beta}^{0}(\bar{\Omega}) := K_{\beta}^{0}(\mathbb{R}, L^2(0, \Theta)) = \{(t, \theta) \mapsto e^{\beta t}u(t, \theta) \mid u \in L^2(\bar{\Omega})\}$ and $K_{\beta}^{0}(\bar{\Lambda}) := K_{\beta}^{0}(\mathbb{R}, L^2(-1, 0))$. And we define

$$ \bar{V} := \{w \in H_{\text{loc}}^{1}(\bar{\Pi}) \mid \nabla w_{|\bar{\Omega}} \in L^2(\bar{\Omega}), \partial_t w_{|\bar{\Lambda}} \in K_{1/2}^{0}(\bar{\Lambda}), \partial_{\theta} w_{|\bar{\Lambda}} \in K_{-1/2}^{0}(\bar{\Lambda}) \text{ and } w_{|\overline{\Sigma_{\Omega} \cup \Sigma_{\Lambda}}} = 0\}. $$

Note that if $\Omega_1$ were equal to $\Pi$, $\bar{V}$ would simply be $\{\bar{u} \mid u \in V\}$. So it is the natural variational space for Poisson's problem transferred into $\bar{\Pi}$. Looking at $\chi S^V$ as a function defined on $\Pi$, we will show by induction on $n$ that:

$$ \forall n \in \mathbb{N}, \forall R_c < r_1 < r_2, \quad \partial_t^n \overline{\chi S^V} \in \bar{V}. $$

Since $S^V \in V$, the initial case is trivial, so only the inductive step remains to prove. Let us assume the property at rank $n$ and show it at rank $n+1$. We will use the method of finite differences. Letting $\mu := \mu_0$ on $\Omega$ and $\mu := \mu_1$ on $\Lambda$, (B.8) implies:

$$ \left\{ \begin{array}{ll} \mu\Delta(\chi S^V) = f_2 := \chi f_1 - 2\mu\nabla\chi \cdot \nabla S^V - \mu\Delta\chi \cdot S^V & \text{in } D, \forall D \in \{\Omega, \Lambda\} \\ [\mu\partial_Y(\chi S^V)]_\Gamma = g_2 := \chi g_1 & \text{on } \Gamma \\ \chi S^V = 0 & \text{on } \Sigma_\Omega \cup \Sigma_\Lambda \end{array} \right. $$

By induction hypothesis applied to $(r'_1, r'_2) := (\frac{R_c+r_1}{2}, r_1)$, $\partial_t^k \nabla S^V$ and $\partial_t^k S^V$ are $L^2$ on $\{r_1 < r < r_2\}$ for any $k \in [0, n]$. So (B.9) implies $\bar{f}_2|_{\bar{\Omega}} \in K_{-2}^n(\mathbb{R}, L^2(0, \Theta))$ and $\bar{f}_2|_{\bar{\Lambda}} \in K_{-3/2}^n(\mathbb{R}, L^2(-1, 0))$. Similarly $\bar{g}_2 \in K_{-1/2}^{\infty}(\bar{\Gamma})$. Next, we apply the change variables $(x, y) \rightsquigarrow (t, \theta)$ using (B.2), and then $\partial_t^n$. We get that $s := \partial_t^n(\chi S^V)$ satisfies:

$$ \left\{ \begin{array}{ll} -e^{-2t}\mu_0\Delta s = f_\Omega := e^{-2t}\partial_t^n(e^{2t}\bar{f}_2|_{\bar{\Omega}}) & \text{in } \bar{\Omega} \\ \\ -\mu_1(e^{-2t}(\partial_t^2 - \partial_t) + \partial_\theta^2)s = f_\Lambda := \partial_t^n \bar{f}_2|_{\bar{\Lambda}} + \sum_{k=0}^{n-1} (-2)^{n-k}e^{-2t}\mu_1(\partial_t^2 - \partial_t)\partial_t^k(\overline{\chi S^V}) & \text{in } \bar{\Lambda} \\ \\ s|_{\theta=0^+} - s|_{\theta=0^-} = 0 & \text{on } \bar{\Gamma} \\ \\ e^{-t}\mu_0\partial_\theta s|_{\theta=0^+} - \mu_1\partial_\theta s|_{\theta=0^-} = g := \partial_t^n \bar{g}_2 & \text{on } \bar{\Gamma} \\ \\ s = 0 & \text{on } \overline{\Sigma_\Omega \cup \Sigma_\Lambda} \end{array} \right. \quad (B.10) $$

with $f_\Omega \in K_{-2}^0(\Omega)$, $f_\Lambda \in K_{-3/2}^0(\bar{\Lambda})$ and $g \in K_{-1/2}^{\infty}(\bar{\Gamma})$ by induction hypothesis. The variational formula- tion of (B.10) is: $\forall \varphi \in \bar{V}$,

$$ \int_{\bar{\Omega}} \mu_0 \nabla s \cdot \nabla \varphi + \int_{\bar{\Lambda}} \mu_1 (e^{-t} \partial_t s \partial_t \varphi + e^t \partial_\theta s \partial_\theta \varphi) = \int_{\bar{\Omega}} e^{2t} f_\Omega \varphi + \int_{\bar{\Lambda}} e^t f_\Lambda \varphi + \int_{\bar{\Gamma}} e^t g \varphi. $$

Let us denote $D_\eta\varphi(t, \theta) := \frac{\varphi(t+\eta, \theta)-\varphi(t, \theta)}{\eta}$ for any $\eta \in \mathbb{R}^*$ and any function $\varphi$. Taking $\varphi := D_{-\eta}D_\eta\bar{s}$ and discretely integrating by parts $D_{-\eta}$, we have:

$$ \int_{\bar{\Omega}} \mu_0 |D_\eta \nabla s|^2 + \int_{\bar{\Lambda}} \mu_1 (D_\eta(e^{-t} \partial_t s) \cdot D_\eta \partial_t \bar{s} + e^t |D_\eta \partial_\theta s|^2) = \int_{\bar{\Omega}} e^{2t} f_\Omega \cdot D_{-\eta} D_\eta \bar{s} + \int_{\bar{\Lambda}} e^t f_\Lambda \cdot D_{-\eta} D_\eta \bar{s} + \int_{\bar{\Gamma}} D_{-\eta} D_\eta (e^t g) \cdot \bar{s} $$

But for any functions $\varphi, \psi$, we have $D_\eta(\varphi\psi) = \varphi \cdot D_\eta\psi + D_\eta\varphi \cdot \psi$. Therefore:

$$ \bullet \quad D_\eta(e^{-t} \partial_t s) \cdot D_\eta \partial_t \bar{s} = e^{-t} |D_\eta \partial_t s|^2 + \frac{e^{-\eta-1}}{\eta} e^{-t} \partial_t s \cdot D_\eta \partial_t \bar{s} $$

$$ \bullet \quad D_{-\eta} D_{\eta}(e^t g) = e^t D_{-\eta} D_{\eta} g + 2 \frac{e^{\eta-1}}{\eta} e^t D_{\eta} g + \left(\frac{e^{\eta-1}}{\eta}\right)^2 e^t g $$

Let us assume that $\eta$ is small enough so that $|\frac{e^{\eta-1}}{\eta}| < 2$ and $|\frac{e^{-\eta-1}}{\eta}| < 2$. Then:

$$ \|D_{\eta} \nabla s\|_{L^2(\bar{\Omega})}^2 + \|D_{\eta} \partial_t s\|_{K_{1/2}^0(\bar{\Lambda})}^2 + \|D_{\eta} \partial_{\theta} s\|_{K_{-1/2}^0(\bar{\Lambda})}^2 + \int_{\bar{\Lambda}} \mu_1 \frac{e^{-\eta-1}}{\eta} e^{-t} \partial_t s \cdot D_{\eta} \partial_t \bar{s} \\ \lesssim \|f_{\Omega}\|_{K_{-2}^0(\bar{\Omega})} \|D_{-\eta} D_{\eta} s\|_{L^2(\bar{\Omega})} + \|f_{\Lambda}\|_{K_{-3/2}^0(\bar{\Lambda})} \|D_{-\eta} D_{\eta} s\|_{K_{1/2}^0(\bar{\Lambda})} + \|g\|_{K_{-1/2}^2(\bar{\Gamma})} \|s\|_{K_{-1/2}^0(\bar{\Gamma})} $$
with $f_{\Omega} \in K_{-2}^{0}(\Omega)$, $f_{\Lambda} \in K_{-3/2}^{0}(\bar{\Lambda})$ and $g \in K_{-1/2}^{\infty}(\bar{\Gamma})$ by induction hypothesis. The variational formula-
tion of (B.10) is: $\forall \varphi \in \bar{V}$,

$$
\int_{\Omega} \mu_0 \nabla s \cdot \nabla \varphi + \int_{\bar{\Lambda}} \mu_1 (e^{-t} \partial_t s \partial_t \varphi + e^t \partial_t s \partial_t \varphi) = \int_{\bar{\Omega}} e^{2t} f_{\Omega} \varphi + \int_{\bar{\Lambda}} e^t f_{\Lambda} \varphi + \int_{\bar{\Gamma}} e^t g \varphi.
$$

Let us denote $D_{\eta}\varphi(t, \theta) := \frac{\varphi(t+\eta, \theta)-\varphi(t, \theta)}{\eta}$ for any $\eta \in \mathbb{R}^*$ and any function $\varphi$. Taking $\varphi := D_{-\eta}D_{\eta}\bar{s}$ and discretely integrating by parts $D_{-\eta}$, we have:

$$
\begin{align*}
\int_{\bar{\Omega}} \mu_0 |D_{\eta} \nabla s|^2 + \int_{\bar{\Lambda}} \mu_1 (D_{\eta}(e^{-t} \partial_t s) \cdot D_{\eta} \partial_t \bar{s} + e^t |D_{\eta} \partial_t s|^2) \\
&= \int_{\bar{\Omega}} e^{2t} f_{\Omega} \cdot D_{-\eta} D_{\eta} \bar{s} + \int_{\bar{\Lambda}} e^t f_{\Lambda} \cdot D_{-\eta} D_{\eta} \bar{s} + \int_{\bar{\Gamma}} D_{-\eta} D_{\eta} (e^t g) \cdot \bar{s}
\end{align*}
$$

But for any functions $\varphi, \psi$, we have $D_{\eta}(\varphi\psi) = \varphi \cdot D_{\eta}\psi + D_{\eta}\varphi \cdot \psi$. Therefore:

$$
\bullet \quad D_{\eta}(e^{-t} \partial_t s) \cdot D_{\eta} \partial_t \bar{s} = e^{-t} |D_{\eta} \partial_t s|^2 + \frac{e^{-\eta}-1}{\eta} e^{-t} \partial_t s \cdot D_{\eta} \partial_t \bar{s}
$$

$$
\bullet \quad D_{-\eta} D_{\eta}(e^t g) = e^t D_{-\eta} D_{\eta} g + 2 \frac{e^{\eta}-1}{\eta} e^t D_{\eta} g + \left( \frac{e^{\eta}-1}{\eta} \right)^2 e^t g
$$

Let us assume that $\eta$ is small enough so that $|\frac{e^{\eta}-1}{\eta}| < 2$ and $|\frac{e^{-\eta}-1}{\eta}| < 2$. Then:

$$
\begin{align*}
\| D_{\eta} \nabla s \|_{L^2(\bar{\Omega})}^2 + \| D_{\eta} \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})}^2 + \| D_{\eta} \partial_t s \|_{K_{-1/2}^0(\bar{\Lambda})}^2 &+ \int_{\bar{\Lambda}} \mu_1 \frac{e^{-\eta}-1}{\eta} e^{-t} \partial_t s \cdot D_{\eta} \partial_t \bar{s} \\
&\lesssim \| f_{\Omega} \|_{K_{-2}^0(\bar{\Omega})} \| D_{-\eta} D_{\eta} s \|_{L^2(\bar{\Omega})} + \| f_{\Lambda} \|_{K_{-3/2}^0(\bar{\Lambda})} \| D_{-\eta} D_{\eta} s \|_{K_{1/2}^0(\bar{\Lambda})} + \| g \|_{K_{-1/2}^2(\bar{\Gamma})} \| s \|_{K_{-1/2}^0(\bar{\Gamma})} \\
&+ \| f_{\Lambda} \|_{K_{-3/2}^0(\bar{\Lambda})} \| D_{\eta} \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})} + \| g \|_{K_{-1/2}^2(\bar{\Gamma})} \| s \|_{K_{-1/2}^0(\bar{\Gamma})} + \| \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})} \| D_{\eta} \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})}
\end{align*}
$$

Then, moving the integral to the right-hand side and majorizing some $\|D_{\eta} \cdot \|$ by $\|\partial_t \cdot \|$ or $\|\nabla \cdot \|$, we get:

$$
\| D_{\eta} \nabla s \|_{L^2(\bar{\Omega})}^2 + \| D_{\eta} \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})}^2 + \| D_{\eta} \partial_t s \|_{K_{-1/2}^0(\bar{\Lambda})}^2 \lesssim \| f_{\Omega} \|_{K_{-2}^0(\bar{\Omega})} \| D_{\eta} \nabla s \|_{L^2(\bar{\Omega})} \\
+ \| f_{\Lambda} \|_{K_{-3/2}^0(\bar{\Lambda})} \| D_{\eta} \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})} + \| g \|_{K_{-1/2}^2(\bar{\Gamma})} \| s \|_{K_{-1/2}^0(\bar{\Gamma})} + \| \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})} \| D_{\eta} \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})}
$$

But, since $s|_{\bar{\Sigma}_{\Lambda}^{-}} = 0$, a Poincaré-type inequality gives $\|s\|_{K_{-1/2}^{0}(\bar{\Gamma})}^{2} \lesssim \|\partial_{\theta}s\|_{K_{-1/2}^{0}(\bar{\Lambda})}^{2}$. Finally, using Young's inequality $ab \lesssim \frac{1}{\delta}a + \delta b$ on norm products with $\delta$ small enough and moving the $\|D_{\eta}\partial_t s\|_{K_{1/2}^{0}(\bar{\Lambda})}^{2}$ and $\|D_{\eta}\partial_{\theta}s\|_{K_{-1/2}^{0}(\bar{\Lambda})}^{2}$ from the right-hand side to the left-hand one, we get:

$$
\begin{align*}
\| \nabla \partial_t s \|_{L^2(\bar{\Omega})}^2 + \| \partial_t^2 s \|_{K_{1/2}^0(\bar{\Lambda})}^2 + \| \partial_{\theta} \partial_t s \|_{K_{-1/2}^0(\bar{\Lambda})}^2 \\
&\le \limsup_{\eta \to 0} \| D_{\eta} \nabla s \|_{L^2(\bar{\Omega})}^2 + \| D_{\eta} \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})}^2 + \| D_{\eta} \partial_{\theta} s \|_{K_{-1/2}^0(\bar{\Lambda})}^2 \\
&\lesssim \| f_{\Omega} \|_{K_{-2}^0(\bar{\Omega})}^2 + \| f_{\Lambda} \|_{K_{-3/2}^0(\bar{\Lambda})}^2 + \| g \|_{K_{-1/2}^1(\bar{\Gamma})}^2 + \| \partial_{\theta} s \|_{K_{-1/2}^0(\bar{\Lambda})}^2 + \| \partial_t s \|_{K_{1/2}^0(\bar{\Lambda})}^2 \\
&< \infty
\end{align*}
$$

By definition of $\bar{V}$, it implies that $\partial_t s \in \bar{V}$ and completes the induction.

Step 3: Let $K_{1/2}^0(\bar{\Pi}) := K_{1/2}^0(\mathbb{R}, L^2(-1, \Theta))$. For now we have proven that, for any $(n, i) \in (\mathbb{N} \times \{0, 1\}) \setminus ((0, 0))$ and $R_c < r_1 < r_2$, $\partial_t^n \partial_\theta^i \overline{\chi S^V} \in K_{1/2}^0(\bar{\Pi})$. It generalizes to the case $(n, i) = (0, 0)$ thanks to a Poincaré inequality: $\|\overline{\chi S^V}\|_{K_{1/2}^0(\bar{\Pi})} \lesssim \|\partial_\theta \overline{\chi S^V}\|_{K_{1/2}^0(\bar{\Pi})} \lesssim \|\partial_\theta \overline{\chi S^V}\|_{L^2(\bar{\Omega})} + \|\partial_\theta \overline{\chi S^V}\|_{K_{-1/2}^0(\bar{\Lambda})} < \infty$.

To treat higher-order $\theta$-derivatives, we start from the equality $\mu\Delta(\chi S^V) = \chi f_1 - 2\mu\nabla\chi \cdot \nabla S^V - \mu\Delta\chi \cdot S^V$
in $\Omega \cup \Lambda$ proven in to step 2. Applying the change of variables $(x, y) \rightsquigarrow (t, \theta)$ gives by (B.2):

$$
\left\{
\begin{array}{ll}
-e^{2t} \mu_0 \Delta \overline{\chi S^V} = \overline{\chi} f_1 - 2 \mu_0 \overline{\nabla \chi} \cdot \overline{\nabla S^V} - \mu_0 \overline{\Delta \chi} \cdot \overline{S^V} & \text{in } \bar{\Omega} \\
-\mu_1 (e^{-2t} (\partial_t^2 - \partial_t) + \partial_\theta^2) (\overline{\chi S^V}) = \overline{\chi} f_1 - 2 \mu_1 \overline{\nabla \chi} \cdot \overline{\nabla S^V} - \mu_1 \overline{\Delta \chi} \cdot \overline{S^V} & \text{in } \bar{\Lambda}
\end{array}
\right.
\quad (B.11)
$$

Moreover, (B.9) implies that, for any $m \in \mathbb{N}$, $\overline{\chi} f_1|_{\bar{\Omega}} \in K_{-3/2}^{\infty}(\mathbb{R}, H^m(0, \Theta))$ and $\overline{\chi} f_1|_{\bar{\Lambda}} \in K_{1/2}^{\infty}(\mathbb{R}, H^m(-1, 0))$.
So deriving (B.11) w.r.t. $t$ and $\theta$ enough times gives by induction: $\forall (n, i) \in \mathbb{N}^2, \forall R_c < r_1 < r_2, \forall D \in$
$\{\bar{\Omega}, \bar{\Lambda}\}, \partial_t^n \partial_\theta^i \overline{\chi S^V}|_D \in K_{1/2}^0(D)$. Finally, applying this to $\chi := \chi_\infty$ completes the proof. $\square$
Then, moving the integral to the right-hand side and majorizing some $||D_\eta \cdot||$ by $||\partial_t \cdot||$ or $||\nabla \cdot||$, we get:

$$ \begin{aligned} & ||D_\eta \nabla s||_{L^2(\bar{\Omega})}^2 + ||D_\eta \partial_t s||_{K_{1/2}^0(\bar{\Lambda})}^2 + ||D_\eta \partial_\theta s||_{K_{-1/2}^0(\bar{\Lambda})}^2 \\ & \lesssim ||f_\Omega||_{K_{-2}^0(\bar{\Omega})} ||D_\eta \nabla s||_{L^2(\bar{\Omega})} \\ & + ||f_\Lambda||_{K_{-3/2}^0(\bar{\Lambda})} ||D_\eta \partial_t s||_{K_{1/2}^0(\bar{\Lambda})} + ||g||_{K_{-1/2}^2(\bar{\Gamma})} ||s||_{K_{-1/2}^0(\bar{\Gamma})} + ||\partial_t s||_{K_{1/2}^0(\bar{\Lambda})} ||D_\eta \partial_\theta s||_{K_{1/2}^0(\bar{\Lambda})} \end{aligned} $$

But, since $s|_{\bar{\Sigma}_\Lambda} = 0$, a Poincaré-type inequality gives $||s||_{K_{-1/2}^0(\bar{\Gamma})}^2 \lesssim ||\partial_\theta s||_{K_{-1/2}^0(\bar{\Lambda})}^2$. Finally, using Young's inequality $ab \lesssim \frac{1}{\delta}a + \delta b$ on norm products with $\delta$ small enough and moving the $||D_\eta \partial_t s||_{K_{1/2}^0(\bar{\Lambda})}^2$ and $||D_\eta \partial_\theta s||_{K_{-1/2}^0(\bar{\Lambda})}^2$ from the right-hand side to the left-hand one, we get:

$$ \begin{aligned} & ||\nabla \partial_t s||_{L^2(\bar{\Omega})}^2 + ||\partial_t^2 s||_{K_{1/2}^0(\bar{\Lambda})}^2 + ||\partial_\theta \partial_t s||_{K_{-1/2}^0(\bar{\Lambda})}^2 \\ & \le \limsup_{\eta \to 0} ||D_\eta \nabla s||_{L^2(\bar{\Omega})}^2 + ||D_\eta \partial_t s||_{K_{1/2}^0(\bar{\Lambda})}^2 + ||D_\eta \partial_\theta s||_{K_{-1/2}^0(\bar{\Lambda})}^2 \\ & \lesssim ||f_\Omega||_{K_{-2}^0(\bar{\Omega})}^2 + ||f_\Lambda||_{K_{-3/2}^0(\bar{\Lambda})}^2 + ||g||_{K_{-1/2}^1(\bar{\Gamma})}^2 + ||\partial_\theta s||_{K_{-1/2}^0(\bar{\Lambda})}^2 + ||\partial_t s||_{K_{1/2}^0(\bar{\Lambda})}^2 \\ & < \infty \end{aligned} $$

By definition of $\bar{V}$, it implies that $\partial_t s \in \bar{V}$ and completes the induction.

**Step 3:** Let $K_{1/2}^0(\bar{\Pi}) := K_{1/2}^0(\mathbb{R}, L^2(-1, \Theta))$. For now we have proven that, for any $(n, i) \in (\mathbb{N} \times \{0, 1\}) \setminus ((0, 0))$ and $R_c < r_1 < r_2$, $\partial_t^n \partial_\theta^i \chi \overline{S^V} \in K_{1/2}^0(\bar{\Pi})$. It generalizes to the case $(n, i) = (0, 0)$ thanks to a Poincaré inequality: $||\chi \overline{S^V}||_{K_{1/2}^0(\bar{\Pi})} \lesssim ||\partial_\theta \chi \overline{S^V}||_{K_{1/2}^0(\bar{\Pi})} \lesssim ||\partial_\theta \chi \overline{S^V}||_{L^2(\bar{\Omega})} + ||\partial_\theta \chi \overline{S^V}||_{K_{-1/2}^0(\bar{\Lambda})} < \infty$.

To treat higher-order $\theta$-derivatives, we start from the equality $\mu\Delta(\chi S^V) = \chi f_1 - 2\mu\nabla\chi \cdot \nabla S^V - \mu\Delta\chi \cdot S^V$ in $\Omega \cup \Lambda$ proven in to step 2. Applying the change of variables $(x, y) \rightsquigarrow (t, \theta)$ gives by (B.2):

$$ \left\{ \begin{array}{ll} -e^{-2t}\mu_0 \Delta \overline{\chi S^V} = \overline{\chi f_1} - 2\mu_0 \overline{\nabla \chi \cdot \nabla S^V} - \mu_0 \overline{\Delta \chi \cdot S^V} & \text{in } \bar{\Omega} \\ -\mu_1 (e^{-2t}(\partial_t^2 - \partial_t) + \partial_\theta^2)(\overline{\chi S^V}) = \overline{\chi f_1} - 2\mu_1 \overline{\nabla \chi \cdot \nabla S^V} - \mu_1 \overline{\Delta \chi \cdot S^V} & \text{in } \bar{\Lambda} \end{array} \right. \quad (B.11) $$

Moreover, (B.9) implies that, for any $m \in \mathbb{N}$, $\overline{\chi f_1|_{\bar{\Omega}}} \in K_{-3/2}^\infty(\mathbb{R}, H^m(0, \Theta))$ and $\overline{\chi f_1|_{\bar{\Lambda}}} \in K_{1/2}^\infty(\mathbb{R}, H^m(-1, 0))$. So deriving (B.11) w.r.t. $t$ and $\theta$ enough times gives by induction: $\forall (n, i) \in \mathbb{N}^2, \forall R_c < r_1 < r_2, \forall D \in \{\bar{\Omega}, \bar{\Lambda}\}, \partial_t^n \partial_\theta^i \chi \overline{S^V}|_D \in K_{1/2}^0(D)$. Finally, applying this to $\chi := \chi_\infty$ completes the proof. $\square$

**Proof of Theorem 2.23:** For any $q \in \frac{\pi}{\Theta}\mathbb{N}^*$, let $\sigma_q(S^\infty) := \sigma_q(S)$. We denote, for any $d \in \mathbb{R}$, $T_{>d} := \text{id} - T_{\le d}$ and

$$ S_d^\infty := \left\langle -R_{\partial_Y^2} \circ \partial_{x|\Lambda}^2, \frac{\mu_0}{\mu_1} R_{\mathbb{N}} \circ \partial_{y|\Gamma, y=0^+} \right\rangle \left( \frac{1}{\mu_0} R_\Delta(F_\Omega^\infty) + \frac{1}{\mu_1} R_{\partial_Y^2}(F_\Lambda^\infty) + \sum_{q \in \frac{\pi}{\Theta}\mathbb{Z}^* \cap (d, \infty)} \sigma_q(S^\infty) \phi_q \right) $$

and, for any $n \in \mathbb{N}$, $d_n := \frac{1}{2} - n$ and $S_{r,n} := \chi_\infty S - \chi_\infty T_{>d_n}(S_{d_n}^\infty)$. Let $m \in \mathbb{N}$. We will show by induction on $n$ that there are coefficients $(\sigma_q(S^\infty))_{q \in -\frac{\pi}{\Theta}\mathbb{N}^*}$ s.t.:

$$ \forall n \in \mathbb{N}, \forall \beta > d_n, \quad \overline{S_{r,n}} \in K_\beta^\infty(\mathbb{R}, \mathcal{H}^m). $$

For now we only know $\sigma_q(S^\infty)$ for $q \ge \frac{\pi}{\Theta} > d_0$, and $S_{r,n}$ involves $\sigma_q(S^\infty)$ only when $q > d_n$, so the $n$-th inductive step involves fixing $\sigma_q(S^\infty)$ for all $q \in \frac{\pi}{\Theta}\mathbb{Z}^* \cap (d_n, d_{n-1}]$.

**Initial case:** By Proposition B.4, there is $d \in \mathbb{R}_-$ and $S^V$ s.t. $S = S^V + \chi_\infty T_{\ge d}(S_{d_0}^\infty)$ and $\overline{\chi_\infty S^V} \in K_{1/2}^\infty(\mathbb{R}, \mathcal{H}^m)$. Hence

$$ \begin{aligned} S_{r,0} &= \chi_\infty S - \chi_\infty T_{>d_0}(S_{d_0}^\infty) \\ &= \chi_\infty (S^V + \chi_\infty T_{\ge d}(S_{d_0}^\infty)) - \chi_\infty T_{>d_0}(S_{d_0}^\infty) \\ &= \chi_\infty S^V + \chi_\infty \cdot (\chi_\infty - 1) \cdot T_{\ge d}(S_{d_0}^\infty) + \chi_\infty \cdot (T_{\ge d}(S_{d_0}^\infty) - T_{>0}(S_{d_0}^\infty)) \end{aligned} $$

So $\overline{S_{r,0}} \in K_{1/2}^\infty(\mathbb{R}, \mathcal{H}^m)$. But $\overline{S_{r,0}}$ is null in a vicinity of $-\infty$, so: $\forall \beta > d_0 = \frac{1}{2}$, $\overline{S_{r,0}} \in K_\beta^\infty(\mathbb{R}, \mathcal{H}^m)$.
**Proof of Theorem 2.23:** For any $q \in \frac{\pi}{\Theta}\mathbb{N}^*$, let $\sigma_q(S^\infty) := \sigma_q(S)$. We denote, for any $d \in \mathbb{R}$, $T_{>d} := \text{id} - T_{\le d}$ and

$$S_d^\infty := \left\langle -R_{\partial_Y^2} \circ \partial_{x|\Lambda}^2, \frac{\mu_0}{\mu_1} R_N \circ \partial_{y|\Gamma, y=0^+} \right\rangle \left( \frac{1}{\mu_0} R_\Delta(F_\Omega^\infty) + \frac{1}{\mu_1} R_{\partial_Y^2}(F_\Lambda^\infty) + \sum_{q \in \frac{\pi}{\Theta}\mathbb{Z}^* \cap (d, \infty)} \sigma_q(S^\infty) \phi_q \right)$$

and, for any $n \in \mathbb{N}$, $d_n := \frac{1}{2} - n$ and $S_{r,n} := \chi_\infty S - \chi_\infty T_{>d_n}(S_{d_n}^\infty)$. Let $m \in \mathbb{N}$. We will show by induction on $n$ that there are coefficients $(\sigma_q(S^\infty))_{q \in -\frac{\pi}{\Theta}\mathbb{N}^*}$ s.t.:

$$\forall n \in \mathbb{N}, \forall \beta > d_n, \quad \overline{S_{r,n}} \in K_\beta^\infty(\mathbb{R}, \mathcal{H}^m).$$

For now we only know $\sigma_q(S^\infty)$ for $q \ge \frac{\pi}{\Theta} > d_0$, and $S_{r,n}$ involves $\sigma_q(S^\infty)$ only when $q > d_n$, so the $n$-th inductive step involves fixing $\sigma_q(S^\infty)$ for all $q \in \frac{\pi}{\Theta}\mathbb{Z}^* \cap (d_n, d_{n-1}]$.

**Initial case:** By Proposition B.4, there is $d \in \mathbb{R}_-$ and $S^V$ s.t. $S = S^V + \chi_\infty T_{\ge d}(S_{d_0}^\infty)$ and $\overline{\chi_\infty S^V} \in K_{1/2}^\infty(\mathbb{R}, \mathcal{H}^m)$. Hence

$$\begin{aligned}
S_{r,0} &= \chi_\infty S - \chi_\infty T_{>d_0}(S_{d_0}^\infty) \\
&= \chi_\infty (S^V + \chi_\infty T_{\ge d}(S_{d_0}^\infty)) - \chi_\infty T_{>d_0}(S_{d_0}^\infty) \\
&= \chi_\infty S^V + \chi_\infty \cdot (\chi_\infty - 1) \cdot T_{\ge d}(S_{d_0}^\infty) + \chi_\infty \cdot (T_{\ge d}(S_{d_0}^\infty) - T_{>0}(S_{d_0}^\infty))
\end{aligned}$$

So $\overline{S_{r,0}} \in K_{1/2}^\infty(\mathbb{R}, \mathcal{H}^m)$. But $\overline{S_{r,0}}$ is null in a vicinity of $-\infty$, so: $\forall \beta > d_0 = \frac{1}{2}$, $\overline{S_{r,0}} \in K_\beta^\infty(\mathbb{R}, \mathcal{H}^m)$.

**Inductive step:** We assume the property at rank $n$ and will show it at rank $n+1$. Let $\tilde{S}_{r,n+1} := \chi_\infty S - \chi_\infty T_{>d_{n+1}}(S_{d_n}^\infty)$ and $\mathfrak{D} := \frac{\pi}{\Theta}\mathbb{Z}^* \cap (d_{n+1}, d_n]$. $\tilde{S}_{r,n+1}$ is a variant of $S_{r,n+1}$ that does not involve $\sigma_q(S^\infty)$ for $q \in \mathfrak{D}$. We will apply Proposition B.3 to $d := d_n$ and $s := \tilde{S}_{r,n+1}$. To do so, we must check that:

$$\left\{
\begin{array}{l}
\mu_0 \Delta \overline{\tilde{S}_{r,n+1|\bar{\Omega}}} \in \bigcap_{\beta > d_{n+1}} K_\beta^\infty(\mathbb{R}, H^{m-2}(0, \Theta)) \\
\\
\mu_1 (e^{-2t}(\partial_t^2 - \partial_t) + \partial_\theta^2) \overline{\tilde{S}_{r,n+1|\bar{\Lambda}}} \in \bigcap_{\beta > d_{n+1}} K_\beta^\infty(\mathbb{R}, H^{m-2}(-1, 0)) \\
\\
(e^{-t}\mu_0 \partial_{\theta|\theta=0^+} - \mu_1 \partial_{\theta|\theta=0^-}) \overline{\tilde{S}_{r,n+1}} \in \bigcap_{\beta > d_{n+1}} K_\beta^\infty(\bar{\Gamma})
\end{array}
\right.$$

Let us show only the first line, the others being similar.

$$\left\{
\begin{array}{ll}
\text{Since } \overline{\Delta \tilde{S}_{r,n+1}} = e^{-2t} \Delta \overline{\tilde{S}_{r,n+1}}, \text{ it suffices to have that: } \forall \beta > d_{n+1} - 2, \overline{\Delta \tilde{S}_{r,n+1}} \in K_\beta^\infty(\mathbb{R}, H^{m-2}(0, \Theta)). \text{ And this is true because we have in } \Omega \text{ in a vicinity of } r \to \infty:} & \\
\\
\mu_0 \Delta \tilde{S}_{r,n+1} = F - \mu_0 \Delta [T_{>d_{n+1}}(S_{d_n}^\infty)] & \text{by definition of } \tilde{S}_{r,n+1}, \text{ and } \mu_0 \Delta S = F \text{ in } \Omega \\
\\
= T_{\ge d_{n+1}-2} (F_\Omega^\infty) + o_\partial(r^{d_{n+1}-2}) & \text{by hypothesis on } F \\
\\
- \mu_0 T_{>d_{n+1}-2} (\Delta S_{d_n}^\infty) & \text{by deg } \Delta = -2 \text{ (Lemma 2.16)} \\
\\
= (T_{\le d_{n+1}-2} - T_{<d_{n+1}-2}) (F_\Omega^\infty) + o_\partial(r^{d_{n+1}-2}) & \text{by } \mu_0 \Delta S_{d_n}^\infty = F_\Omega^\infty \text{ (Lemma 2.19)}
\end{array}
\right.$$

Therefore, Proposition B.3 states that there are coefficients $(c_q)_{q \in \mathfrak{D}}$, $s_r$ and $d_{\min} := \min(\mathfrak{D})$ s.t.: $\forall \beta \in (d_{n+1}, d_{\min})$, $s_r \in K_\beta^\infty(\mathbb{R}, \mathcal{H}^m)$ and $\overline{\tilde{S}_{r,n+1}} = \sum_{q \in \mathfrak{D}} c_q \overline{\phi_q} + s_r$. Let $\sigma_q(S^\infty) := -c_q$ for any $q \in \frac{\pi}{\Theta}\mathbb{Z}^* \cap (d_{n+1}, d_n]$. Then:

$$\overline{\tilde{S}_{r,n+1}} = \overline{\tilde{S}_{r,n+1}} + \sum_{q \in \mathfrak{D}} \sigma_q(S^\infty) \overline{\chi_\infty \phi_q} = s_r + \sum_{q \in \mathfrak{D}} c_q (\overline{1 - \chi_\infty}) \overline{\phi_q} \in \bigcap_{\beta \in (d_{n+1}, d_{\min})} K_\beta^\infty(\mathbb{R}, \mathcal{H}^m)$$

Moreover, $\overline{\tilde{S}_{r,n+1}}$ is null in a vicinity of $-\infty$, so: $\forall \beta > d_{n+1}$, $\overline{\tilde{S}_{r,n+1}} \in K_\beta^\infty(\mathbb{R}, \mathcal{H}^m)$. This concludes the induction.

To complete the proof of the theorem, we must show that: $\forall d \in \mathbb{R}$, $S = T_{\ge d}(S^\infty) + o_\partial(\mathbf{r}^d)$ when $\mathbf{r} \to \infty$. We will do it in $\Omega$, but it works the same in $\Lambda$. Let $d \in \mathbb{R}$ and $n \in \mathbb{N}$ s.t. $d_n < d$. We have in $\Omega$:

$$s := \chi_\infty S - \chi_\infty T_{\ge d}(S^\infty) = S_{r,n} + \chi_\infty \cdot T_{<d} \circ T_{>d_n}(S^\infty) \quad \text{with} \quad T_{<d} \circ T_{>d_n}(S^\infty) \in \sum_{d' < d} A_{d'}(\Omega).$$
**Inductive step:** We assume the property at rank $n$ and will show it at rank $n+1$. Let $\tilde{S}_{r,n+1} := \chi_\infty S - \chi_\infty T_{>d_{n+1}}(S_{d_n}^\infty)$ and $\mathfrak{D} := \frac{\pi}{\Theta} \mathbb{Z}^* \cap (d_{n+1}, d_n]$. $\tilde{S}_{r,n+1}$ is a variant of $S_{r,n+1}$ that does not involve $\sigma_q(S^\infty)$ for $q \in \mathfrak{D}$. We will apply Proposition B.3 to $d := d_n$ and $s := \overline{\tilde{S}_{r,n+1}}$. To do so, we must check that:

$$ egin{cases} 
\mu_0 \Delta \overline{\tilde{S}_{r,n+1|\bar{\Omega}}} \in \bigcap_{\beta > d_{n+1}} K_\beta^\infty(\mathbb{R}, H^{m-2}(0, \Theta)) \\ 
\mu_1 (e^{-2t}(\partial_t^2 - \partial_t) + \partial_\theta^2) \overline{\tilde{S}_{r,n+1|\bar{\Lambda}}} \in \bigcap_{\beta > d_{n+1}} K_\beta^\infty(\mathbb{R}, H^{m-2}(-1, 0)) \\ 
(e^{-t} \mu_0 \partial_{\theta|\theta=0^+} - \mu_1 \partial_{\theta|\theta=0^-}) \overline{\tilde{S}_{r,n+1}} \in \bigcap_{\beta > d_{n+1}} K_\beta^\infty(\bar{\Gamma}) 
\end{cases} $$

Let us show only the first line, the others being similar.

$$ 
\left. 
\begin{array}{l}
\text{Since } \overline{\Delta \tilde{S}_{r,n+1}} = e^{-2t} \Delta \overline{\tilde{S}_{r,n+1}}, \text{ it suffices to have that: } \forall \beta > d_{n+1} - 2, \overline{\Delta \tilde{S}_{r,n+1}} \in K_\beta^\infty(\mathbb{R}, H^{m-2}(0, \Theta)). \text{ And this is true because we have in } \Omega \text{ in a vicinity of } r \to \infty:} \\
\\
\mu_0 \Delta \overline{\tilde{S}_{r,n+1}} = F - \mu_0 \Delta [T_{>d_{n+1}}(S_{d_n}^\infty)] \quad \text{by definition of } \tilde{S}_{r,n+1}, \text{ and } \mu_0 \Delta S = F \text{ in } \Omega \\
\qquad = T_{\ge d_{n+1}-2}(F_\Omega^\infty) + o_\partial(r^{d_{n+1}-2}) \quad \text{by hypothesis on } F \\
\qquad \quad - \mu_0 T_{>d_{n+1}-2}(\Delta S_{d_n}^\infty) \quad \text{by deg } \Delta = -2 \text{ (Lemma 2.16)} \\
\qquad = (T_{\le d_{n+1}-2} - T_{<d_{n+1}-2})(F_\Omega^\infty) + o_\partial(r^{d_{n+1}-2}) \quad \text{by } \mu_0 \Delta S_{d_n}^\infty = F_\Omega^\infty \text{ (Lemma 2.19)}
\end{array}
\right\} $$

Therefore, Proposition B.3 states that there are coefficients $(c_q)_{q \in \mathfrak{D}}$, $s_r$ and $d_{\min} := \min(\mathfrak{D})$ s.t.: $\forall \beta \in (d_{n+1}, d_{\min})$, $s_r \in K_\beta^\infty(\mathbb{R}, H^m)$ and $\overline{\tilde{S}_{r,n+1}} = \sum_{q \in \mathfrak{D}} c_q \bar{\phi}_q + s_r$. Let $\sigma_q(S^\infty) := -c_q$ for any $q \in \frac{\pi}{\Theta} \mathbb{Z}^* \cap (d_{n+1}, d_n]$. Then:

$$ \overline{\tilde{S}_{r,n+1}} = \overline{\tilde{S}_{r,n+1}} + \sum_{q \in \mathfrak{D}} \sigma_q(S^\infty) \overline{\chi_\infty \phi_q} = s_r + \sum_{q \in \mathfrak{D}} c_q (\overline{1 - \chi_\infty}) \overline{\phi_q} \in \bigcap_{\beta \in (d_{n+1}, d_{\min})} K_\beta^\infty(\mathbb{R}, H^m) $$

Moreover, $\overline{\tilde{S}_{r,n+1}}$ is null in a vicinity of $-\infty$, so: $\forall \beta > d_{n+1}$, $\overline{\tilde{S}_{r,n+1}} \in K_\beta^\infty(\mathbb{R}, H^m)$. This concludes the induction.

To complete the proof of the theorem, we must show that: $\forall d \in \mathbb{R}$, $S = T_{\ge d}(S^\infty) + o_\partial(\mathbf{r}^d)$ when $\mathbf{r} \to \infty$. We will do it in $\Omega$, but it works the same in $\Lambda$. Let $d \in \mathbb{R}$ and $n \in \mathbb{N}$ s.t. $d_n < d$. We have in $\Omega$:

$$ s := \chi_\infty S - \chi_\infty T_{\ge d}(S^\infty) = S_{r,n} + \chi_\infty \cdot T_{<d} \circ T_{>d_n}(S^\infty) \quad \text{with} \quad T_{<d} \circ T_{>d_n}(S^\infty) \in \sum_{d'<d} A_{d'}(\Omega). $$

So there is $d' < d$ s.t.: $\forall m \in \mathbb{N}$, $\bar{s} \in K_{d'}^\infty(\mathbb{R}, H^m(0, \Theta))$. Thus for any $(i, j) \in \mathbb{N}^2$:

$$ \overline{r^{-d'}(r \partial_r)^i \partial_\theta^j s} = e^{-d't} \partial_t^i \partial_\theta^j \bar{s} \in H^2(\mathbb{R}, H^2(0, \Theta)) \subset L^\infty(\bar{\Omega}). $$

So $r^{-d'+i} \partial_r^i \partial_\theta^j s$ is also bounded. By definition of $o_\partial$ (given in 2.20), this concludes the proof. $\square$

## B.2 Proof of Theorem 2.22: asymptotic behavior for far-and-layer fields-like problems

This proof is very similar to Section B.1. The main difference is that we look at the asymptotic expansion when $\mathbf{r} \to 0$ (i.e. $t \to -\infty$) instead of $\mathbf{r} \to \infty$ (i.e. $t \to +\infty$). So this time we use a truncation function $\chi_f \in C^\infty(\Pi)$ that is equal to 1 in a vicinity of $\mathbf{r} = 0$ and to 0 in a vicinity of infinity. In addition, we can assume that $\chi_f f = 0$. Moreover the Laplace transform of $\overline{\chi_f \mathbf{u}}$ is first defined in a left half-plane of the complex plan (instead of a right one), and then extended to the right.

Applying the change of variables $(x, y) \rightsquigarrow (t, \theta)$ on the equations satisfied by $\chi_f \mathbf{u}$ (that one can easily deduce from (2.23)) and then the Laplace transform yields that $s := \overline{\chi_f \mathbf{u}}$ satisfies:

$$ 
\left\{ 
\begin{array}{ll}
\mu_0(\partial_\theta^2 + \lambda^2)\hat{s}(\lambda, \theta) = \tilde{f} - \omega^2 \rho_0 \hat{s}(\lambda - 2, \theta) & \text{if } \theta \in (0, \Theta) \\
\partial_\theta^2 \hat{s}(\lambda, \theta) = \tilde{g} & \text{if } \theta \in (-1, 0) \\
\hat{s}(\lambda, 0^+) - \hat{s}(\lambda, 0^-) = 0 & \\
\partial_\theta \hat{s}(\lambda, 0^-) = \tilde{h} & \\
\hat{s}(\lambda, \Theta) = \hat{s}(\lambda, -1) = 0 & 
\end{array}
\right. 
\quad (B.12) $$
So there is $d' < d$ s.t.: $\forall m \in \mathbb{N}$, $\bar{s} \in K_{d'}^{\infty}(\mathbb{R}, H^m(0, \Theta))$. Thus for any $(i, j) \in \mathbb{N}^2$:

$$ \overline{r^{-d'}(r\partial_r)^i \partial_{\theta}^j s} = e^{-d't}\partial_t^i \partial_{\theta}^{j\bar{s}} \in H^2(\mathbb{R}, H^2(0, \Theta)) \subset L^{\infty}(\bar{\Omega}). $$

So $r^{-d'+i}\partial_r^i \partial_{\theta}^j s$ is also bounded. By definition of $o_{\partial}$ (given in 2.20), this concludes the proof. □

## B.2 Proof of Theorem 2.22: asymptotic behavior for far-and-layer fields-like problems

This proof is very similar to Section B.1. The main difference is that we look at the asymptotic expansion when $\mathbf{r} \to 0$ (i.e. $t \to -\infty$) instead of $\mathbf{r} \to \infty$ (i.e. $t \to +\infty$). So this time we use a truncation function $\chi_f \in C^{\infty}(\Pi)$ that is equal to 1 in a vicinity of $\mathbf{r} = 0$ and to 0 in a vicinity of infinity. In addition, we can assume that $\chi_f f = 0$. Moreover the Laplace transform of $\overline{\chi_f \mathbf{u}}$ is first defined in a left half-plane of the complex plan (instead of a right one), and then extended to the right.

Applying the change of variables $(x, y) \rightsquigarrow (t, \theta)$ on the equations satisfied by $\chi_f \mathbf{u}$ (that one can easily deduce from (2.23)) and then the Laplace transform yields that $s := \overline{\chi_f \mathbf{u}}$ satisfies:

$$ \left\{ \begin{array}{ll} \mu_0(\partial_{\theta}^2 + \lambda^2)\hat{s}(\lambda, \theta) = \tilde{f} - \omega^2 \rho_0 \hat{s}(\lambda - 2, \theta) & \text{if } \theta \in (0, \Theta) \\ \partial_{\theta}^2 \hat{s}(\lambda, \theta) = \tilde{g} & \text{if } \theta \in (-1, 0) \\ \hat{s}(\lambda, 0^+) - \hat{s}(\lambda, 0^-) = 0 & \\ \partial_{\theta} \hat{s}(\lambda, 0^-) = \tilde{h} & \\ \hat{s}(\lambda, \Theta) = \hat{s}(\lambda, -1) = 0 & \end{array} \right. \qquad (B.12) $$

for some functions $\tilde{f}, \tilde{g}, \tilde{h}$ depending on $g, h, \chi_f$. This system has the form of (B.4), so Lemma B.2 gives the tools to solve it and to extend $\hat{s}$ to the right (by steps of 2 here). This is stated in Proposition B.5, whose proof is very similar to Proposition B.3.

**Proposition B.5:** Let $d \in \mathbb{R}$, $m \in \mathbb{N} \setminus \{0, 1\}$, $s \in \bigcap_{\beta<d} K_{\beta}^{\infty}(\mathbb{R}, \mathcal{H}^m)$, $\tilde{f} \in \bigcap_{\beta<d+2} K_{\beta}^{\infty}(\mathbb{R}, H^{m-2}(0, \Theta))$, $\tilde{g} \in \bigcap_{\beta<d+2} K_{\beta}^{\infty}(\mathbb{R}, H^{m-2}(-1, 0))$, $\tilde{h} \in \bigcap_{\beta<d+2} K_{\beta}^{\infty}(\bar{\Gamma})$ and $d_{\max} = \max(\frac{\pi}{\bar{\Theta}}\mathbb{Z}^* \cap [d, d+2))$. We assume:

$$ \left\{ \begin{array}{ll} \mu_0 \Delta s + e^{2t} \omega^2 \rho_0 s = \tilde{f} & \text{in } \bar{\Omega} \\ \partial_{\theta}^2 s = \tilde{g} & \text{in } \bar{\Lambda} \\ s|_{\theta=0^+} - s|_{\theta=0^-} = 0 & \text{on } \bar{\Gamma} \\ \mu_1 \partial_{\theta} s|_{\theta=0^-} = \tilde{h} & \text{on } \bar{\Gamma} \\ s = 0 & \text{on } \bar{\Sigma}_{\Omega} \cup \bar{\Sigma}_{\Lambda} \end{array} \right. $$

Then there is $s_r \in \bigcap_{d_{\max}<\beta<d+2} K_{\beta}^{\infty}(\mathbb{R}, \mathcal{H}^m)$ and $(c_q) \in \mathbb{C}^{\frac{\pi}{\bar{\Theta}}\mathbb{Z}^* \cap [d, d+2)}$ s.t. $s = \sum_{q \in \frac{\pi}{\bar{\Theta}}\mathbb{Z}^* \cap [d, d+2)} c_q \bar{\phi}_q + s_r$.

**Assumption B.6:** Replacing $\mathbf{u}$ by $\chi_{\infty}\mathbf{u}$, we assume without loss of generality that $\mathbf{u}$ has a compact support.

For any $q \in -\frac{\pi}{\bar{\Theta}}\mathbb{N}^*$, let $\sigma_q(\mathbf{u}) := \sigma_q(\mathbf{u}|_{\Omega})$, where $\sigma_q(\mathbf{u}|_{\Omega})$ is set in Definition 3.5. This quantity vanishes when $q$ is small enough. Proposition B.7 is the analogue of Proposition B.4.

**Proposition B.7:** Let $d \in \mathbb{R}_+$,

$$ \mathbf{u}^A := T_{\le d} \left[ \langle -k_0^2 R_{\Delta} \rangle \left( \frac{1}{\mu_1} R_{\partial_Y^2}(g^0) + \frac{1}{\mu_1} R_N(h^0) + \sum_{q \in -\frac{\pi}{\bar{\Theta}}\mathbb{N}^*} \sigma_q(\mathbf{u}) \phi_q \right) \right] $$

in $\mathcal{A}(\Pi)$ and $\mathbf{u}^v := \mathbf{u} - \chi_f \mathbf{u}^A$. There is $d \in \mathbb{R}_+$ s.t. $\mathbf{u}|_{\Omega}^v \in H^1(\Omega)$ and: $\forall m \in \mathbb{N}$, $\overline{\chi_f \mathbf{u}^v} \in K_{-1}^{\infty}(\mathbb{R}, \mathcal{H}^m)$.
for some functions $\tilde{f}, \tilde{g}, \tilde{h}$ depending on $g, h, \chi_f$. This system has the form of (B.4), so Lemma B.2 gives the tools to solve it and to extend $\hat{s}$ to the right (by steps of 2 here). This is stated in Proposition B.5, whose proof is very similar to Proposition B.3.

**Proposition B.5:** Let $d \in \mathbb{R}$, $m \in \mathbb{N} \setminus \{0, 1\}$, $s \in \bigcap_{\beta<d} K_{\beta}^{\infty}(\mathbb{R}, \mathcal{H}^m)$, $\tilde{f} \in \bigcap_{\beta<d+2} K_{\beta}^{\infty}(\mathbb{R}, H^{m-2}(0, \Theta))$, $\tilde{g} \in \bigcap_{\beta<d+2} K_{\beta}^{\infty}(\mathbb{R}, H^{m-2}(-1, 0))$, $\tilde{h} \in \bigcap_{\beta<d+2} K_{\beta}^{\infty}(\bar{\Gamma})$ and $d_{\max} = \max(\bar{\Theta}\mathbb{Z}^* \cap [d, d+2))$. We assume:

$$ 
\left\{ 
\begin{array}{lll}
\mu_0 \Delta s + e^{2t} \omega^2 \rho_0 s = \tilde{f} & \text{in } \bar{\Omega} \\
\partial_{\theta}^2 s = \tilde{g} & \text{in } \bar{\Lambda} \\
s|_{\theta=0^+} - s|_{\theta=0^-} = 0 & \text{on } \bar{\Gamma} \\
\mu_1 \partial_{\theta} s|_{\theta=0^-} = \tilde{h} & \text{on } \bar{\Gamma} \\
s = 0 & \text{on } \bar{\Sigma}_{\Omega} \cup \bar{\Sigma}_{\Lambda}
\end{array}
\right. 
$$

Then there is $s_r \in \bigcap_{d_{\max}<\beta<d+2} K_{\beta}^{\infty}(\mathbb{R}, \mathcal{H}^m)$ and $(c_q) \in \mathbb{C}^{\bar{\Theta}\mathbb{Z}^* \cap [d, d+2)}$ s.t. $s = \sum_{q \in \bar{\Theta}\mathbb{Z}^* \cap [d, d+2]} c_q \bar{\phi}_q + s_r$.

**Assumption B.6:** Replacing $\mathbf{u}$ by $\chi_{\infty}\mathbf{u}$, we assume without loss of generality that $\mathbf{u}$ has a compact support.

For any $q \in -\bar{\Theta}\mathbb{N}^*$, let $\sigma_q(\mathbf{u}) := \sigma_q(\mathbf{u}|_{\Omega})$, where $\sigma_q(\mathbf{u}|_{\Omega})$ is set in Definition 3.5. This quantity vanishes when $q$ is small enough. Proposition B.7 is the analogue of Proposition B.4.

**Proposition B.7:** Let $d \in \mathbb{R}_+$,

$$ \mathbf{u}^A := T_{\le d} \left[ \langle -k_0^2 R_\Delta \rangle \left( \frac{1}{\mu_1} R_{\partial_Y^2}(g^0) + \frac{1}{\mu_1} R_N(h^0) + \sum_{q \in -\bar{\Theta}\mathbb{N}^*} \sigma_q(\mathbf{u}) \phi_q \right) \right] $$

in $\mathcal{A}(\Pi)$ and $\mathbf{u}^v := \mathbf{u} - \chi_f \mathbf{u}^A$. There is $d \in \mathbb{R}_+$ s.t. $\mathbf{u}_{|\Omega}^v \in H^1(\Omega)$ and: $\forall m \in \mathbb{N}$, $\overline{\chi_f \mathbf{u}^v} \in K_{-1}^{\infty}(\mathbb{R}, \mathcal{H}^m)$.

**Proof:** We consider any $d \in \mathbb{R}_+$ and we will fix it later. The proof has five steps: writing the equations satisfied by $\mathbf{u}^v$, and then showing that $\mathbf{u}^v$ is regular in the layer, that it is $H^1$ in $\Omega$, that it is regular in $\Omega$ w.r.t $t$, and that it is regular in $\Omega$ w.r.t. $\theta$. For any $r_1, r_2 \in \mathbb{R}$ s.t. $r_1 < r_2 < r_f$, we denote $\chi \in \mathcal{C}^{\infty}(\mathbb{R}^2)$ a radial function equal to 1 on $B(0, r_1)$ and 0 on $\mathbb{R}^2 \setminus B(0, r_2)$ ($r_1, r_2$ are implicit in this notation).

**Step 1:** Similarly to the proofs of theorems 3.6 and 3.11 and Proposition B.4, one can show that:

$$ 
\left\{ 
\begin{array}{lll}
(\mu_0 \Delta + \omega^2 \rho_0) \mathbf{u}^v = f_1 := f - (\mu_0 \Delta + \omega^2 \rho_0)(\chi_f \mathbf{u}^A) & \text{in } \Omega \\
\mu_1 \partial_Y^2 \mathbf{u}^v = g_1 := g - \mu_1 \partial_Y^2 (\chi_f \mathbf{u}^A) & \text{in } \Lambda \\
\mathbf{u}_{|Y=0^+}^v - \mathbf{u}_{|Y=0^-}^v = 0 & \text{on } \Gamma \\
\mu_1 \partial_Y \mathbf{u}_{|Y=0^-}^v = h_1 := h - \mu_1 \partial_Y (\chi_f \mathbf{u}^A) & \text{on } \Gamma \\
\mathbf{u}^v = 0 & \text{on } \Sigma_{\Omega} \cup \Sigma_{\Lambda}
\end{array}
\right. 
\quad \text{with} 
\quad 
\left\{ 
\begin{array}{lll}
(\mu_0 \Delta + \omega^2 \rho_0) \mathbf{u}^A \in \sum_{d'>d} \mathcal{A}_{d'}(\Omega) \\
\partial_Y^2 \mathbf{u}_{|\Lambda}^A - T_{\le d}(g^0) \in \sum_{d'>d} \mathcal{A}_{d'}(\Lambda) \\
\partial_Y \mathbf{u}_{|Y=0^-}^A - T_{\le d}(h^0) \in \sum_{d'>d} \mathcal{A}_{d'}(\Gamma)
\end{array}
\right. 
$$

Moreover, given that $\chi_f = 0$ for any $r_1 < r_2 < r_f$, one can check that for $d$ big enough and up to decreasing $r_f$ (without loss of generality), we have $\overline{\chi f_1} \in K_0^{\infty}(\mathbb{R}, H^m(0, \Theta))$, $\overline{\chi g_1} \in K_0^{\infty}(\mathbb{R}, H^m(-1, 0))$ and $\overline{\chi h_1} \in K_0^{\infty}(\bar{\Gamma})$ for any $m \in \mathbb{N}$.

**Step 2:** Let us show that: $\forall r_1 < r_2 < r_f, \forall m \in \mathbb{N}$, $\overline{\chi \mathbf{u}}_{|\Lambda}^v \in K_0^{\infty}(\mathbb{R}, H^m(-1, 0))$. By assumption, for any $x > 0$, $\mathbf{u}^v$ in $H^1$ in a vicinity of $\{x\} \times [-1, 0] \subset \Lambda$, so $\mathbf{u}_{|\{x\} \times [-1, 0]}^v$ is well-defined in $L^2(\{x\} \times [-1, 0])$. Moreover, $g_1$ and $h_1$ are $o_{\partial}(x^d)$, so they are differentiable on $(0, r_f) \times [-1, 0]$ (up to decreasing $r_f$). Hence, $Y \in (-1, 0) \mapsto \mathbf{u}^v(x, Y)$ is $\mathcal{C}^2$ and the ODE it satisfies (see step 1) is explicitly solvable:

$$ \forall (x, Y) \in (0, r_f) \times [-1, 0], \quad \mathbf{u}^v(x, Y) = \int_{-1}^{0} ((Y - Y')^+ - Y - 1) g_1(x, Y') \, dY' + h_1(x) \cdot (Y + 1). $$
**Proof:** We consider any $d \in \mathbb{R}_+$ and we will fix it later. The proof has five steps: writing the equations satisfied by $\mathbf{u}^v$, and then showing that $\mathbf{u}^v$ is regular in the layer, that it is $H^1$ in $\Omega$, that it is regular in $\Omega$ w.r.t $t$, and that it is regular in $\Omega$ w.r.t. $\theta$. For any $r_1, r_2 \in \mathbb{R}$ s.t. $r_1 < r_2 < r_f$, we denote $\chi \in C^\infty(\mathbb{R}^2)$ a radial function equal to 1 on $B(0, r_1)$ and 0 on $\mathbb{R}^2 \setminus B(0, r_2)$ ($r_1, r_2$ are implicit in this notation).

**Step 1:** Similarly to the proofs of theorems 3.6 and 3.11 and Proposition B.4, one can show that:

$$ egin{cases} (\mu_0\Delta + \omega^2\rho_0)\mathbf{u}^v = f_1 := f - (\mu_0\Delta + \omega^2\rho_0)(\chi_f\mathbf{u}^A) & \text{in } \Omega \\ \mu_1\partial_Y^2\mathbf{u}^v = g_1 := g - \mu_1\partial_Y^2(\chi_f\mathbf{u}^A) & \text{in } \Lambda \\ \mathbf{u}^v|_{y=0^+} - \mathbf{u}^v|_{y=0^-} = 0 & \text{on } \Gamma \\ \mu_1\partial_Y\mathbf{u}^v|_{Y=0^-} = h_1 := h - \mu_1\partial_Y(\chi_f\mathbf{u}^A) & \text{on } \Gamma \\ \mathbf{u}^v = 0 & \text{on } \Sigma_\Omega \cup \Sigma_\Lambda \end{cases} \quad \text{with} \quad 
\begin{cases} (\mu_0\Delta + \omega^2\rho_0)\mathbf{u}^A \in \sum_{d'>d} \mathcal{A}_{d'}(\Omega) \\ \partial_Y^2\mathbf{u}^A|_{\Lambda} - T_{\le d}(g^0) \in \sum_{d'>d} \mathcal{A}_{d'}(\Lambda) \\ \partial_Y\mathbf{u}^A|_{Y=0^-} - T_{\le d}(h^0) \in \sum_{d'>d} \mathcal{A}_{d'}(\Gamma) \end{cases} $$

Moreover, given that $\chi_f = 0$ for any $r_1 < r_2 < r_f$, one can check that for $d$ big enough and up to decreasing $r_f$ (without loss of generality), we have $\overline{\chi f_1} \in K_0^\infty(\mathbb{R}, H^m(0, \Theta))$, $\overline{\chi g_1} \in K_0^\infty(\mathbb{R}, H^m(-1, 0))$ and $\overline{\chi h_1} \in K_0^\infty(\bar{\Gamma})$ for any $m \in \mathbb{N}$.

**Step 2:** Let us show that: $\forall r_1 < r_2 < r_f, \forall m \in \mathbb{N}, \overline{\chi \mathbf{u}}|_{\Lambda}^v \in K_0^\infty(\mathbb{R}, H^m(-1, 0))$. By assumption, for any $x > 0$, $\mathbf{u}^v$ in $H^1$ in a vicinity of $\{x\} \times [-1, 0] \subset \Lambda$, so $\mathbf{u}^v|_{\{x\} \times [-1, 0]}$ is well-defined in $L^2(\{x\} \times [-1, 0])$. Moreover, $g_1$ and $h_1$ are $o_\theta(x^d)$, so they are differentiable on $(0, r_f) \times [-1, 0]$ (up to decreasing $r_f$). Hence, $Y \in (-1, 0) \mapsto \mathbf{u}^v(x, Y)$ is $C^2$ and the ODE it satisfies (see step 1) is explicitly solvable:

$$ \forall (x, Y) \in (0, r_f) \times [-1, 0], \quad \mathbf{u}^v(x, Y) = \int_{-1}^{0} ((Y - Y')^{+} - Y - 1) g_1(x, Y') \, dY' + h_1(x) \cdot (Y + 1). $$

Since $\overline{\chi g_1} \in K_0^\infty(\mathbb{R}, H^m(-1, 0))$ for any $m$ and $\overline{\chi h_1} \in K_0^\infty(\bar{\Gamma})$, we deduce $\overline{\chi \mathbf{u}}|_{\Lambda}^v \in K_0^\infty(\mathbb{R}, H^m(-1, 0))$.

**Step 3:** Let us show that $\mathbf{u}^v|_{\Omega} \in H^1(\Omega)$. First, step 2 implies $\overline{\chi \mathbf{u}}|_{\Gamma}^v \in K_0^\infty(\bar{\Gamma})$ so $\chi \mathbf{u}^v|_{\Gamma} \in H^{1/2}(\bar{\Gamma})$. Since $\mathbf{u} \in H_{\text{loc}}^1$ and $\mathbf{u}$ has a compact support by assumption B.6, we get $\mathbf{u}^v|_{\Gamma} \in H^{1/2}(\bar{\Gamma})$. So the following system has a solution in $H^1(\Omega)$ by Lemma 3.3:

$$ egin{cases} \mu_0\Delta u^v + \omega^2\rho_0 u^v = f_1 & \text{in } \Omega \\ u^v|_{y=0^+} = \mathbf{u}^v|_{y=0^-} & \text{on } \Gamma \\ u^v = 0 & \text{on } \Sigma_\Omega \end{cases} $$

But $\mathbf{u}^v$ also satisfies these equations, so $u^v + \chi_f \mathbf{u}^A|_{\Omega}$ satisfies the same problem as $\mathbf{u}|_{\Omega}$, and they both belong to $H^1(\Omega) + \chi_0 A(\Omega)$. Then by uniqueness in Theorem 3.6, $u^v + \chi_f \mathbf{u}^A|_{\Omega} = \mathbf{u}|_{\Omega}$, i.e. $u^v = \mathbf{u}^v|_{\Omega}$. Thus, $\mathbf{u}^v|_{\Omega} \in H^1(\Omega)$.

**Step 4:** Let $v: \Omega \to \mathbb{C}$ defined by $v := \mathbf{u}^v|_{\Omega} - v_{\text{lift}}$ with $v_{\text{lift}}(r, \theta) := (1 - \frac{\theta}{\Theta})\mathbf{u}^v|_{\Gamma}(r)$ in polar coordinates. Let $K_\beta^0(\bar{\Omega}) := K_\beta^0(\mathbb{R}, L^2(0, \Theta))$ for any $\beta \in \mathbb{R}$. We will show by induction on $n$ that:

$$ \forall n \in \mathbb{N}^*, \forall r_1 < r_2 < r_f, \quad \partial_t^n \overline{\chi v} \in K_{-1}^0(\bar{\Omega}) \quad \text{and} \quad \nabla \partial_t^n \overline{\chi v} \in L^2(\bar{\Omega}). $$

For the initial case, we have on the one hand $\chi \mathbf{u}^v|_{\Omega} \in H^1(\Omega)$ so $\overline{\chi \mathbf{u}}^v|_{\Omega} \in K_{-1}^0(\bar{\Omega})$ and $\nabla \overline{\chi \mathbf{u}}^v|_{\Omega} \in L^2(\bar{\Omega})$, and on the other $\overline{\chi \mathbf{u}}^v|_{\Gamma} \in K_0^1(\bar{\Gamma})$ so $\overline{\chi v}_{\text{lift}} \in K_{-1}^0(\bar{\Omega})$ and $\nabla \overline{\chi v}_{\text{lift}} \in L^2(\bar{\Omega})$. Thus the initial case is proven and only the inductive step remains to prove. We assume the property at rank $n$ and we will show it at rank $n+1$, using the method of finite differences. Step 1 implies that:

$$ egin{cases} (\mu_0\Delta + \omega^2\rho_0)(\chi v) = f_2 := \chi f_1 - \chi \mu_0\Delta v_{\text{lift}} - 2\mu_0 \nabla\chi \cdot \nabla\mathbf{u}^v - \mu_0\Delta\chi \cdot \mathbf{u}^v & \text{in } \Omega \\ \chi v = 0 & \text{on } \Sigma_\Omega \cup \Gamma \end{cases} $$

In addition, we have $\overline{f}_2 \in K_{-2}^n(\mathbb{R}, L^2(0, \Theta))$ since:

* by step 1, $\overline{\chi f_1} \in K_0^\infty(\mathbb{R}, L^2(0, \Theta))$, so $\overline{\chi f_1} \in K_{-2}^\infty(\mathbb{R}, L^2(0, \Theta))$,

* by step 2, $\overline{\chi \mathbf{u}}^v|_{\Gamma} \in K_0^\infty(\bar{\Gamma})$, so $\overline{\chi v}_{\text{lift}} \in K_0^\infty(\mathbb{R}, H^2(0, \Theta))$, and applying it at $(r'_1, r'_2) := (r_2, \frac{r_2+r_f}{2})$ gives $\overline{\chi \Delta v}_{\text{lift}} \in K_{-2}^\infty(\mathbb{R}, L^2(0, \Theta))$,
Since $\overline{\chi g_1} \in K_0^\infty(\mathbb{R}, H^m(-1, 0))$ for any $m$ and $\overline{\chi h_1} \in K_0^\infty(\bar{\Gamma})$, we deduce $\overline{\chi \mathbf{u}}_{|\bar{\Omega}}^v \in K_0^\infty(\mathbb{R}, H^m(-1, 0))$.

**Step 3:** Let us show that $\mathbf{u}_{|\bar{\Omega}}^v \in H^1(\bar{\Omega})$. First, step 2 implies $\overline{\chi \mathbf{u}}_{|\bar{\Gamma}}^v \in K_0^\infty(\bar{\Gamma})$ so $\chi \mathbf{u}_{|\bar{\Gamma}}^v \in H^{1/2}(\bar{\Gamma})$. Since $\mathbf{u} \in H_{\text{loc}}^1$ and $\mathbf{u}$ has a compact support by assumption **B.6**, we get $\mathbf{u}_{|\bar{\Gamma}}^v \in H^{1/2}(\bar{\Gamma})$. So the following system has a solution in $H^1(\bar{\Omega})$ by Lemma **3.3**:

$$ 
\left\{ 
\begin{array}{ll}
\mu_0 \Delta u^v + \omega^2 \rho_0 u^v = f_1 & \text{in } \Omega \\
u^v_{|y=0^+} = \mathbf{u}^v_{|y=0^-} & \text{on } \Gamma \\
u^v = 0 & \text{on } \Sigma_{\Omega}
\end{array}
\right. 
$$

But $\mathbf{u}^v$ also satisfies these equations, so $u^v + \chi_f \mathbf{u}_{|\bar{\Omega}}^A$ satisfies the same problem as $\mathbf{u}_{|\bar{\Omega}}$, and they both belong to $H^1(\bar{\Omega}) + \chi_0 A(\bar{\Omega})$. Then by uniqueness in Theorem **3.6**, $u^v + \chi_f \mathbf{u}_{|\bar{\Omega}}^A = \mathbf{u}_{|\bar{\Omega}}$, i.e. $u^v = \mathbf{u}_{|\bar{\Omega}}^v$. Thus, $\mathbf{u}_{|\bar{\Omega}}^v \in H^1(\bar{\Omega})$.

**Step 4:** Let $v : \bar{\Omega} \to \mathbb{C}$ defined by $v := \mathbf{u}_{|\bar{\Omega}}^v - v_{\text{lift}}$ with $v_{\text{lift}}(r, \theta) := (1 - \frac{\theta}{\Theta})\mathbf{u}_{|\bar{\Gamma}}^v(r)$ in polar coordinates. Let $K_\beta^0(\bar{\Omega}) := K_\beta^0(\mathbb{R}, L^2(0, \Theta))$ for any $\beta \in \mathbb{R}$. We will show by induction on $n$ that:

$$ \forall n \in \mathbb{N}^*, \forall r_1 < r_2 < r_f, \quad \partial_t^n \bar{\chi} v \in K_{-1}^0(\bar{\Omega}) \quad \text{and} \quad \nabla \partial_t^n \bar{\chi} v \in L^2(\bar{\Omega}). $$

For the initial case, we have on the one hand $\chi \mathbf{u}_{|\bar{\Omega}}^v \in H^1(\bar{\Omega})$ so $\overline{\chi \mathbf{u}}_{|\bar{\Omega}}^v \in K_{-1}^0(\bar{\Omega})$ and $\nabla \overline{\chi \mathbf{u}}_{|\bar{\Omega}}^v \in L^2(\bar{\Omega})$, and on the other $\overline{\chi \mathbf{u}}_{|\bar{\Gamma}}^v \in K_0^1(\bar{\Gamma})$ so $\overline{\chi v}_{\text{lift}} \in K_{-1}^0(\bar{\Omega})$ and $\nabla \overline{\chi v}_{\text{lift}} \in L^2(\bar{\Omega})$. Thus the initial case is proven and only the inductive step remains to prove. We assume the property at rank $n$ and we will show it at rank $n+1$, using the method of finite differences. Step 1 implies that:

$$ 
\left\{ 
\begin{array}{ll}
(\mu_0 \Delta + \omega^2 \rho_0)(\chi v) = f_2 & := \chi f_1 - \chi \mu_0 \Delta v_{\text{lift}} - 2\mu_0 \nabla \chi \cdot \nabla \mathbf{u}^v - \mu_0 \Delta \chi \cdot \mathbf{u}^v & \text{in } \Omega \\
\chi v = 0 & & \text{on } \Sigma_{\bar{\Omega}} \cup \Gamma
\end{array}
\right. 
$$

In addition, we have $\bar{f}_2 \in K_{-2}^n(\mathbb{R}, L^2(0, \Theta))$ since:

* by step 1, $\overline{\chi f_1} \in K_0^\infty(\mathbb{R}, L^2(0, \Theta))$, so $\overline{\chi f_1} \in K_{-2}^\infty(\mathbb{R}, L^2(0, \Theta))$,

* by step 2, $\overline{\chi \mathbf{u}}_{|\bar{\Gamma}}^v \in K_0^\infty(\bar{\Gamma})$, so $\overline{\chi v}_{\text{lift}} \in K_0^\infty(\mathbb{R}, H^2(0, \Theta))$, and applying it at $(r'_1, r'_2) := (r_2, \frac{r_2+r_f}{2})$ gives $\overline{\chi \Delta v}_{\text{lift}} \in K_{-2}^\infty(\mathbb{R}, L^2(0, \Theta))$,

* and by induction hypothesis applied to $(r'_1, r'_2) := (r_2, \frac{r_2+r_f}{2})$, $\partial_t^k \nabla v$ and $\partial_t^k v$ are $L^2$ on $\{r_1 < r < r_2\}$ for any $k \in [0, n]$.

Let $s := \partial_t^n(\overline{\chi v})$. By changing variables $(x, y) \rightsquigarrow (t, \theta)$ in the previous system and applying $\partial_t^n$, we get:

$$ 
\left\{ 
\begin{array}{ll}
e^{-2t} \mu_0 \Delta s + \omega^2 \rho_0 s = f_3 := e^{-2t} \partial_t^n (e^{2t} \bar{f}_2) - \sum_{k=0}^{n-1} \omega^2 \rho_0 2^{n-k} \partial_t^k (\overline{\chi v}) & \text{in } \bar{\Omega} \\
s = 0 & \text{on } \Sigma_{\bar{\Omega}} \cup \bar{\Gamma}
\end{array}
\right. 
$$

with $f_3 \in K_{-2}^0(\bar{\Omega})$ by induction hypothesis. And the variational formulation of this is:

$$ \forall \varphi \in K_{-1}^0(\bar{\Omega}), \nabla \varphi \in L^2(\bar{\Omega}) \text{ and } \varphi|_{\Sigma_{\bar{\Omega}} \cup \bar{\Gamma}} = 0 \Rightarrow \int_{\bar{\Omega}} (\mu_0 \nabla s \cdot \nabla \varphi + \omega^2 \rho_0 e^{2t} s \varphi) = \int_{\bar{\Omega}} e^{2t} f_3 \varphi. $$

Let us denote $D_\eta \varphi(t, \theta) := \frac{\varphi(t+\eta, \theta)-\varphi(t, \theta)}{\eta}$ for any $\eta \in \mathbb{R}^*$ and any function $\varphi$. Taking $\varphi := D_{-\eta} D_\eta \bar{s}$ and discretely integrating by parts $D_{-\eta}$ gives:

$$ \int_{\bar{\Omega}} (\mu_0 |D_\eta \nabla s|^2 - \omega^2 \rho_0 e^{2t} |s|^2) = \int_{\bar{\Omega}} e^{2t} f_3 \cdot D_{-\eta} D_\eta \bar{s} $$

Then by coercivity (since $\text{Im}(\omega) \neq 0$), we have for any $\delta > 0$:

$$ \|D_{\eta} \nabla s\|_{L^2(\bar{\Omega})}^2 + \|D_{\eta} s\|_{K_{-1}^0(\bar{\Omega})}^2 \lesssim \|f_3\|_{K_{-2}^0(\bar{\Omega})} \|D_{-\eta} D_{\eta} s\|_{K_{-1}^0(\bar{\Omega})} \lesssim \frac{1}{\delta} \|f_3\|_{K_{-2}^0(\bar{\Omega})}^2 + \delta \|D_{\eta} \partial_t s\|_{L^2(\bar{\Omega})}^2 $$

Taking $\delta$ small enough and moving $\|D_{\eta} \partial_t s\|_{L^2(\bar{\Omega})}^2$ from the right-hand side to the left-hand one, we get:

$$ \|\partial_t \nabla s\|_{L^2(\bar{\Omega})}^2 + \|\partial_t s\|_{K_{-1}^0(\bar{\Omega})}^2 \lesssim \limsup_{\eta \to 0} \|D_{\eta} \nabla s\|_{L^2(\bar{\Omega})}^2 + \|D_{\eta} s\|_{K_{-1}^0(\bar{\Omega})}^2 \lesssim \|f_3\|_{K_{-2}^0(\bar{\Omega})}^2 < \infty. $$
• and by induction hypothesis applied to $(r'_1, r'_2) := (r_2, \frac{r_2+r_f}{2})$, $\partial_t^k \nabla v$ and $\partial_t^k v$ are $L^2$ on $\{r_1 < r < r_2\}$ for any $k \in [0, n]$.

Let $s := \partial_t^n(\bar{\chi}v)$. By changing variables $(x, y) \rightsquigarrow (t, \theta)$ in the previous system and applying $\partial_t^n$, we get:

$$ \left\{ \begin{array}{ll} e^{-2t}\mu_0\Delta s + \omega^2\rho_0 s = f_3 := e^{-2t}\partial_t^n(e^{2t}\bar{f}_2) - \sum_{k=0}^{n-1} \omega^2\rho_0 2^{n-k} \partial_t^k(\bar{\chi}v) & \text{in } \bar{\Omega} \\ s = 0 & \text{on } \bar{\Sigma}_\Omega \cup \bar{\Gamma} \end{array} \right. $$

with $f_3 \in K_{-2}^0(\bar{\Omega})$ by induction hypothesis. And the variational formulation of this is:

$$ \forall \varphi \in K_{-1}^0(\bar{\Omega}), \nabla \varphi \in L^2(\bar{\Omega}) \text{ and } \varphi|_{\bar{\Sigma}_\Omega \cup \bar{\Gamma}} = 0 \Rightarrow \int_{\bar{\Omega}} (\mu_0 \nabla s \cdot \nabla \varphi + \omega^2 \rho_0 e^{2t} s \varphi) = \int_{\bar{\Omega}} e^{2t} f_3 \varphi. $$

Let us denote $D_\eta \varphi(t, \theta) := \frac{\varphi(t+\eta, \theta)-\varphi(t, \theta)}{\eta}$ for any $\eta \in \mathbb{R}^*$ and any function $\varphi$. Taking $\varphi := D_{-\eta}D_\eta \bar{s}$ and discretely integrating by parts $D_{-\eta}$ gives:

$$ \int_{\bar{\Omega}} (\mu_0 |D_{\eta} \nabla s|^2 - \omega^2 \rho_0 e^{2t} |s|^2) = \int_{\bar{\Omega}} e^{2t} f_3 \cdot D_{-\eta} D_{\eta} \bar{s} $$

Then by coercivity (since $\mathrm{Im}(\omega) \neq 0$), we have for any $\delta > 0$:

$$ \|D_{\eta} \nabla s\|_{L^2(\bar{\Omega})}^2 + \|D_{\eta} s\|_{K_{-1}^0(\bar{\Omega})}^2 \lesssim \|f_3\|_{K_{-2}^0(\bar{\Omega})} \|D_{-\eta} D_{\eta} s\|_{K_{-1}^0(\bar{\Omega})} \lesssim \frac{1}{\delta} \|f_3\|_{K_{-2}^0(\bar{\Omega})}^2 + \delta \|D_{\eta} \partial_t s\|_{L^2(\bar{\Omega})}^2 $$

Taking $\delta$ small enough and moving $\|D_{\eta} \partial_t s\|_{L^2(\bar{\Omega})}^2$ from the right-hand side to the left-hand one, we get:

$$ \|\partial_t \nabla s\|_{L^2(\bar{\Omega})}^2 + \|\partial_t s\|_{K_{-1}^0(\bar{\Omega})}^2 \le \limsup_{\eta \to 0} \|D_{\eta} \nabla s\|_{L^2(\bar{\Omega})}^2 + \|D_{\eta} s\|_{K_{-1}^0(\bar{\Omega})}^2 \lesssim \|f_3\|_{K_{-2}^0(\bar{\Omega})}^2 < \infty. $$

This completes the induction.

**Step 5:** We have shown that for any $(n, i) \in \mathbb{N} \times \{0, 1\}$ and $r_1 < r_2 < r_f$, $\partial_t^n \partial_\theta^i \overline{\chi u}_\Omega^v \in K_{-1}^0(\bar{\Omega})$. Since $\overline{\chi v}_{\text{lift}} \in K_{-1}^\infty(\mathbb{R}, H^1(0, \Theta))$, we also have $\partial_t^n \partial_\theta^i \overline{\chi u}_\Omega^v \in K_{-1}^0(\bar{\Omega})$ for these $(n, i)$.

To treat higher-order $\theta$-derivatives, we begin with the equality $(\mu_0\Delta + \omega^2\rho_0)(\chi u^v) = \chi f_1 - 2\mu_0 \nabla\chi \cdot \nabla u^v - \mu_0\Delta\chi \cdot u^v$ in $\bar{\Omega}$, which follows from step 1. Applying the change of variables $(x, y) \rightsquigarrow (t, \theta)$, we get:

$$ (e^{-2t}\mu_0\Delta + \omega^2\rho_0)(\chi u^v) = \overline{\chi f_1} - 2\mu_0 \overline{\nabla\chi \cdot \nabla u^v} - \mu_0 \overline{\Delta\chi \cdot u^v}. \quad (B.13) $$

Now $\overline{\chi f_1} \in K_0^\infty(\mathbb{R}, H^m(0, \Theta))$ for any $m \in \mathbb{N}$ by step 1. So deriving (B.13) w.r.t. $t$ and $\theta$ enough times gives by induction: $\forall (n, i) \in \mathbb{N}^2, \forall r_1 < r_2 < r_f, \partial_t^n \partial_\theta^i \overline{\chi u}_\Omega^v \in K_{-1}^0(\bar{\Omega})$. Combining it with step 2, we get $\overline{\chi u}^v \in K_{-1}^\infty(\mathbb{R}, H^m)$. Finally, we apply it to $\chi := \chi_\infty$ to conclude. $\square$

Finally, Propositions B.5 and B.7 give all the ingredients to prove Theorem 2.22. The proof is very similar to the one of Theorem 2.23 on page 44, so we do not go into details again.

## References

[1] Yves Achdou, O. Pironneau, and F. Valentin. “Effective boundary conditions for laminar flows over periodic rough boundaries”. In: J. Comput. Phys. 147.1 (1998), pp. 187–218. ISSN: 0021-9991. DOI: 10.1006/jcph.1998.6088. URL: https://doi.org/10.1006/jcph.1998.6088.

[2] H. Ammari and S. He. “Effective impedance boundary conditions for an inhomogeneous thin layer on a curved metallic surface”. In: IEEE Transactions on Antennas and Propagation 46.5 (1998), pp. 710–715. DOI: 10.1109/8.668915.

[3] H. Ammari and S. He. “Gerneralized Effective Impedance Boundary Conditions for an Inhomogeneous Thin Layer in Electromagnetic Scattering”. In: Journal of Electromagnetic Waves and Applications 11.9 (1997), pp. 1197–1212. DOI: 10.1163/156939397X01106. eprint: https://doi.org/10.1163/156939397X01106. URL: https://doi.org/10.1163/156939397X01106.
This completes the induction.

Step 5: We have shown that for any $(n, i) \in \mathbb{N} \times \{0, 1\}$ and $r_1 < r_2 < r_f$, $\partial_t^n \partial_\theta^i \overline{\chi v} \in K_{-1}^0(\bar{\Omega})$. Since $\overline{\chi v}_{\text{lift}} \in K_{-1}^\infty(\mathbb{R}, H^1(0, \Theta))$, we also have $\partial_t^n \partial_\theta^i \overline{\chi u}_\Omega^v \in K_{-1}^0(\bar{\Omega})$ for these $(n, i)$.

To treat higher-order $\theta$-derivatives, we begin with the equality $(\mu_0\Delta + \omega^2\rho_0)(\chi u^v) = \chi f_1 - 2\mu_0 \nabla\chi \cdot \nabla u^v - \mu_0\Delta\chi \cdot u^v$ in $\Omega$, which follows from step 1. Applying the change of variables $(x, y) \rightsquigarrow (t, \theta)$, we get:

$$(e^{-2t}\mu_0\Delta + \omega^2\rho_0)(\chi u^v) = \overline{\chi f_1} - 2\mu_0\overline{\nabla\chi \cdot \nabla u^v} - \mu_0\overline{\Delta\chi \cdot u^v}. \quad (B.13)$$

Now $\overline{\chi f_1} \in K_0^\infty(\mathbb{R}, H^m(0, \Theta))$ for any $m \in \mathbb{N}$ by step 1. So deriving (B.13) w.r.t. $t$ and $\theta$ enough times gives by induction: $\forall (n, i) \in \mathbb{N}^2, \forall r_1 < r_2 < r_f, \partial_t^n \partial_\theta^i \overline{\chi u}_\Omega^v \in K_{-1}^0(\bar{\Omega})$. Combining it with step 2, we get $\overline{\chi u}^v \in K_{-1}^\infty(\mathbb{R}, H^m)$. Finally, we apply it to $\chi := \chi_\infty$ to conclude. $\square$

Finally, Propositions B.5 and B.7 give all the ingredients to prove Theorem 2.22. The proof is very similar to the one of Theorem 2.23 on page 44, so we do not go into details again.

## References

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