We follow the proof in [1].

Let \( \alpha_+, \alpha_-, \beta, \delta_0, r_0, \tau_0 \) and \(C_{\mathrm{Carl}}\) denote the constants defined in (0x68ca8d6d) . Furthermore, \(\delta< \delta_0\).

Definition of \(R\). We adjust \(\alpha_+\) such that ¹

\[ \alpha_+> \alpha_-, \]

and we define

\[ r = \min\!\left\{ \tfrac{\alpha_-}{18 \beta}, \tfrac{2 \delta}{19 \alpha_-}, r_0 \right\}, \quad R = \min \{\alpha_- r, 1\}. \]

Thus,

\[ R\le \frac{\alpha_-^2}{18\beta}. \]

Choice of a cut-off function \(\vartheta\). Given \( 0 < R_1 < R_2 \leq R \), we define \( \vartheta_1 \in C^\infty(\mathbb{R}) \) with \( 0 \leq \vartheta_1(t) \leq 1 \), and

\[ \vartheta_1(t) = \begin{cases} 1, & t > -2R_2, \\ 0, & t \leq -3R_2. \end{cases} \]

Also, let \( \vartheta_2 \in C^\infty(\mathbb{R}) \) with \( 0 \leq \vartheta_2(y) \leq 1\), and

\[ \vartheta_2(y) = \begin{cases} 1, & y < \tfrac{R_1}{4 a}, \\ 0, & y \geq \tfrac{R_1}{2 a}. \end{cases} \]

We set

\[ \vartheta(x,y) = \vartheta_1(z(x,y)) \cdot \vartheta_2(y). \]

---

Use of a Carleman estimate. Applying (0x68ca8d6d) to \(\vartheta u\) and using the boundary condition

\[ u_+(x)-u_-(x) = 0, \quad \forall x\in \mathbb{R}^{d-1}\times \{0\}, \]

yields

\begin{equation}\label{eq:carleman} \begin{aligned} &\sum_{\pm }\Bigg( \tau^{-1}\int_{\mathbb{R}^d_\pm } |D^2 (\vartheta u_\pm) |^2 e^{2\tau \phi_\delta} + \tau\int_{\mathbb{R}^d_\pm } |\nabla (\vartheta u_\pm) |^2 e^{2\tau \phi_\delta} + \tau^3 \int_{\mathbb{R}^d_\pm } |(\vartheta u_\pm) |^2 e^{2\tau \phi_\delta} \Bigg) \\ &\quad\leq C_{\mathrm{carl}} (I_1+I_2) \end{aligned} \end{equation}

where

\begin{align*} I_1&=\sum_{\pm } \int_{\mathbb{R}^d_\pm } | \operatorname{div}(A_\pm \nabla (\vartheta u_\pm) )|^2 e^{2\tau \phi_\delta},\\ I_2&=\frac{\tau}{\delta} \int_{\mathbb{R}^{d-1}} |A_+(x,0)\nabla(\vartheta u_+)(x,0)\cdot \nu - A_-(x,0)\nabla (\vartheta u_-)(x,0)\cdot \nu |^2 e^{2\tau \phi_\delta(x,0)} dx + \\ &\quad + \Bigl[ e^{\tau \phi_\delta(\cdot,0)} \bigl(A_+(\cdot,0)\nabla(\vartheta u_+)(\cdot,0)\cdot \nu - A_-(\cdot,0)\nabla (\vartheta u_-)(\cdot,0)\cdot \nu \bigr)\Bigr]^2_{H^{1/2}(\mathbb{R}^{d-1})}, \end{align*}

\(\tau\ge \tau_0\), and

\[ \phi_\delta(x,y) = \begin{cases} \tfrac{\alpha_+y}{\delta} + \tfrac{\beta y^2}{2 \delta^2} - \tfrac{|x|^2}{2\delta}, & y \geq 0, \\ \tfrac{\alpha_- y}{\delta} + \tfrac{\beta y^2}{2 \delta^2} - \tfrac{|x|^2}{2\delta}, & y < 0. \end{cases} \]

Estimate of \(I_1 + I_2\). According to (0x68f5eeb1) ,

\begin{equation}\label{eq:i1_est} \begin{aligned} I_1 &\le C_1 \biggl(R_1^{-4} e^{-\tfrac{5\tau R_2}{2}} \int_{\Bigl\{-4R_2 \le z,\, y \le \tfrac{R_1}{a}\Bigr\}} |u|^2 \, dx \, dy + \\ &+ R_1^{-4} e^{2\tau R_1} \int_{\{-4R_2 \le z,\, \tfrac{R_1}{8a} \le y \le \tfrac{R_1}{2a}\}} |u|^2 \, dx \, dy\biggr), \end{aligned} \end{equation}

where

\[ C_1=\frac{C'M^4 \beta^2(1+\alpha_+^2)^2 }{\lambda^2 \delta^4} \]

and \(C'>0\) denotes a constant, depending on \(d\).

Furthermore, according to (0x68f786d5) ,

\begin{equation}\label{eq:i2_est} I_2< C_2 R_1^{-7} \tau^2 e^{- \tfrac{5}{2} \tau R_2} \int_{\Bigl\{-4R_2 \le z,\, y \le \tfrac{R_1}{a}\Bigr\}} |u|^2 \, dx \, dy, \end{equation}

where

\[ C_2 = \frac{C'M^4(1+\alpha_+)^3}{\lambda^2\delta^7}. \]

(see (0x68f786d5) ).

Together, we obtain

\[ I_1+I_2 \le C_3 R_1^{-7} \biggl( e^{2\tau R_1} \int_{\Bigl\{-4R_2 \le z,\, \tfrac{R_1}{8a} \le y \le \tfrac{R_1}{2a}\Bigr\}} |u|^2 \, dx \, dy + \tau^2 e^{-\tfrac{5}{2}\tau R_2} \int_{\Bigl\{-4R_2 \le z,\, y \le \tfrac{R_1}{a}\Bigr\}} |u|^2 \, dx \, dy \biggr), \]

where

\[ C_3 = \frac{C' M^4 \beta^2 (1+\alpha_+)^3}{\lambda^2 \delta^7}. \]

---

Putting everything together. Let

\begin{equation*} \begin{aligned} U_1&=\{-4R_2\le z,\ \tfrac{R_1}{8a}\le y\le \tfrac{R_1}{a}\},\\ U_2&=\{-R_2\le z\le \tfrac{R_1}{2a},\ y<\tfrac{R_1}{8a}\},\\ U_3&=\{-4R_2\le z,\ y<\tfrac{R_1}{a}\}. \end{aligned} \end{equation*}

On \( U_2 \),

\[ \phi_{\delta, \pm}(x, y) \ge -R_2, \]

since for \( (x, y) \in U_2 \cap \mathbb{R}^n_+ \),

\[ \phi_{\delta, +}(x, y) = z(x, y) + \frac{\alpha_+ - \alpha_-}{\delta}y + \frac{\beta}{\delta^2}y^2 \ge -R_2, \]

and for \( (x, y) \in U_2 \cap \mathbb{R}^n_- \),

\[ \phi_{\delta, -}(x, y) = z(x, y) + \frac{\beta}{\delta^2}y^2 \ge -R_2. \]

Then, by the previous estimates, we have

\begin{equation}\label{eq:u2_est1} \begin{aligned} \tau^{3}e^{-2\tau R_2}\int_{U_2}|u|^{2} &\le \tau^{3}\sum_{\pm}\int_{\Bigl\{-3R_2\le z,\ y\le \tfrac{R_1}{2a} \Bigr\}\cap\mathbb{R}^d_{+}} \bigl|\vartheta u_{\pm}\bigr|^{2}\,e^{2\tau\phi_{\pm}(x,y)}\,dx\,dy \\ &\lesssim C_{\mathrm{carl}}(I_1+I_2)\\ &\lesssim C_3C_{\mathrm{carl}}\,R_1^{-7}\!\left(e^{2\tau R_1}E+\tau^{2}e^{-\frac{5}{2}\tau R_2}F\right), \end{aligned} \end{equation}

where

\[ E=\int_{U_1}|u|^{2},\qquad F=\int_{U_3}|u|^{2}. \]

Dividing by \(\tau^{3}e^{-2\tau R_2}\) on both sides in \eqref{eq:u2_est1} and using \(\tau\ge1\), we obtain

\begin{equation}\label{eq:u2_est2} \int_{U_2}|u|^{2} \le C_4 R_1^{-7}\Bigl(e^{2\tau(R_1+R_2)}E+e^{-\frac12 \tau R_2}F\Bigr), \end{equation}

where \(C_4=C_{\mathrm{carl}}C_3\).

If \(E=0\), letting \(\tau\to\infty\) in \eqref{eq:u2_est2} yields \(\displaystyle \int_{U_2}|u|^{2}=0\).

Assume \(E\neq0\). If

\[ e^{2\tau_0(R_1+R_2)}E< e^{-\tfrac{1}{2} \tau_0R_2}F, \]

choose \(\tau'\ge \tau_0\) such that

\[ e^{2\tau'(R_1+R_2)}E=e^{- \tfrac{1}{2} \tau' R_2}F \quad\text{or}\quad E=e^{-\tau'\left(2R_1+ \tfrac{5}{2} R_2\right)}F . \]

Then, with

\[ \alpha = \frac{2R_1 + 2R_2}{2R_1 + \frac{5}{2}R_2}, \]

it follows that

\begin{align*} \int_{U_2} |u|^2 &\le 2C_4 R_1^{-7} e^{2\tau'(R_1 + R_2)} E \\[4pt] &= 2C_4 R_1^{-7} e^{2\tau'(R_1 + R_2)} E^{(1 - \alpha)} e^{-\tau'\alpha(2R_1 + \frac{5}{2}R_2)} F^{\alpha} \\[4pt] &= 2C_4 R_1^{-7} E^{\frac{\frac{1}{2}R_2}{2R_1 + \frac{5}{2}R_2}} F^{\frac{2R_1 + 2R_2}{2R_1 + \frac{5}{2} R_2}}. \end{align*}

On the other hand, if

\[ e^{2\tau_0(R_1 + R_2)}E > e^{-\frac{1}{2}\tau_0R_2}F, \quad \text{or equivalently} \quad F < e^{\tau_0(2R_1 + \frac{5}{2}R_2)}E, \]

then we estimate without involving \eqref{eq:u2_est2}.

Since \( U_2 \subset U_3 \),

\[ \int_{U_2} |u|^2 \le F^{\frac{\frac{1}{2}R_2}{2R_1 + \frac{5}{2}R_2}} F^{\frac{2R_1 + 2R_2}{2R_1 + \frac{5}{2}R_2}} \le e^{\frac{1}{2}\tau_0R_2} E^{\frac{\frac{1}{2}R_2}{2R_1 + \frac{5}{2}R_2}} F^{\frac{2R_1 + 2R_2}{2R_1 + \frac{5}{2}R_2}}. \]

Putting everything together, we deduce

\[ \int_{U_2} \lvert u\rvert^2\, dx \le (e^{\frac{1}{2}\tau_0 R_2} + 2C_4R_1^{-7}) \Biggl(\int_{U_1} \lvert u\rvert^2 \,dx\Biggr)^{\frac{\frac{1}{2}R_2}{2R_1 + \frac{5}{2}R_2}} \Biggl(\int_{U_3} \lvert u\rvert^2 \,dx\Biggr)^{\frac{2R_1 + 2R_2}{2R_1 + \frac{5}{2}R_2}}. \]