0x68f5eeb1

We write \(\lesssim\) if \(\le\) holds up to a constant depending on \(d\).

\begin{align*} |\div (a_\pm \nabla(\vartheta u_\pm))|^2 &= |a_\pm (\Delta \vartheta\, u_\pm + 2 \nabla \vartheta \cdot \nabla u_\pm + \vartheta\, \Delta u_\pm)|^2 \\[4pt] &\lesssim |a_\pm \Delta \vartheta\, u_\pm|^2 + |a_\pm \nabla \vartheta \cdot \nabla u_\pm|^2 \\[4pt] &\lesssim M^2 \left( |\Delta \vartheta|^2 |u_\pm|^2 + |\nabla \vartheta|^2 |\nabla u_\pm|^2 \right), \end{align*}

where we used \(a_\pm \Delta u_{\pm }=0\).

Then, by the bounds on \(\Delta \vartheta\) and \(\nabla \vartheta\) (see (0x68ed5419) ), we obtain

\begin{align*} I_1&=\sum_{\pm} \int_{\mathbb{R}^d_+} |\div (a_\pm \nabla(\vartheta u_\pm))|^2 e^{2\tau \phi_{\delta,\pm}(x,y)}\, dx\, dy \\ &\leq \sum_{\pm} \int_{\{-3R_2\le z\le -2R_2,\, y<\tfrac{R_1}{2a}\}\cap \mathbb{R}^d_\pm } \tfrac{M^2\beta^2(1+\alpha_-^2)^2}{\delta^4} R_2^{-4} |u_\pm|^2 e^{2\tau \phi_{\delta,\pm}} + \tfrac{M^2(1+\alpha_+^2)}{\delta^2} R_2^{-2} |\nabla u_\pm|^2 e^{2\tau \phi_{\delta,\pm}}\, dx\,dy \\[4pt] &+ \int_{\{-3R_2\le z,\, \tfrac{R_1}{4a} < y<\tfrac{R_1}{2a}\}} \tfrac{M^2(1+\alpha_+^2)^2}{\delta^4} R_1^{-4} |u_+|^2 e^{2\tau \phi_{\delta,+}}+ \tfrac{M^2(1+\alpha_+^2)}{\delta^2} R_1^{-2} |\nabla u_+|^2 e^{2\tau \phi_{\delta,+}}\, dx\,dy. \end{align*}

By applying the bounds of \(\phi_{\delta,\pm } \) (see (0x68f5f313) ), it follows that

\begin{equation}\label{eq:1} \begin{aligned} I_1&\lesssim \sum_{\pm} e^{-\frac{5\tau R_2}{2}} \int_{\{-3R_2\le z\le -2R_2,\, y<\tfrac{R_1}{2a}\}\cap \mathbb{R}^d_\pm } \tfrac{M^2\beta^2(1+\alpha_-^2)^2}{\delta^4} R_2^{-4} |u_\pm|^2 + \tfrac{M^2(1+\alpha_+^2)}{\delta^2} R_2^{-2} |\nabla u_\pm|^2 \, dx\,dy \\[4pt] &+ e^{2\tau R_1} \int_{\{-3R_2\le z,\, \tfrac{R_1}{4a} < y<\tfrac{R_1}{2a}\}} \tfrac{M^2(1+\alpha_+^2)^2}{\delta^4} R_1^{-4} |u_+|^2 + \tfrac{M^2(1+\alpha_+^2)}{\delta^2} R_1^{-2} |\nabla u_+|^2 \, dx\,dy. \end{aligned} \end{equation}

Caccioppoli-type estimate. We apply a Caccioppoli-type inequality on

\[ J_1 := \sum_{\pm} \int_{\{-3R_2\le z\le -2R_2,\, y<\tfrac{R_1}{2a}\}\cap \mathbb{R}^d_\pm } |\nabla u_\pm|^2 \, dx\,dy, \quad J_2:=\int_{\{-3R_2\le z,\, \tfrac{R_1}{4a} < y<\tfrac{R_1}{2a}\}} |\nabla u_+|^2\, dx\,dy. \]

For \( J_1 \), we choose

\[ \eta(x,y) = \eta_1(z(x,y)) \eta_2(y) \]

with \( \eta_1, \eta_2 \in C_c^\infty(\mathbb{R}) \), \( 0 \le \eta_1(t) \le 1 \), \( 0 \le \eta_2(y) \le 1 \),

\[ \eta_1(t) = \begin{cases} 1, & t \ge -3R_2, \\ 0, & t \le -4R_2, \end{cases} \qquad \eta_2(y) = \begin{cases} 1, & y \le \tfrac{R_1}{4a},\\[4pt] 0, & y \ge \tfrac{R_1}{a}. \end{cases} \]

Then, like in (0x68ed5419) , we deduce

\begin{align*} |\nabla \eta(x,y)|^2 &\lesssim \bigl|\eta_1'(z(x,y)) \nabla z(x,y)\bigr|^2 + |\eta_2'(y)|^2 \\[4pt] &\lesssim \frac{1+\alpha_-^2}{\delta^2 R_2^2} + \frac{\alpha_+^2}{\delta^2 R_1^2} \\[4pt] &\lesssim \frac{1+\alpha_+^2}{\delta^2 R_1^2}, \end{align*}

where we employed \( R_1 < R_2 \), and \(\alpha_-< \alpha_+\).

Applying (0x68f6035b) yields

\begin{equation}\label{eq:2} J_1 \lesssim \frac{M^2}{\lambda^2}\frac{1+\alpha_+^2}{\delta^2 R_1^2} \int_{\Bigl\{-4R_2 \le z,\, y \le \tfrac{R_1}{a}\Bigr\}} |u|^2 \, dx \, dy. \end{equation}

For \( J_2 \), choose \( \tilde{\eta}_2 \in C_c^\infty(\mathbb{R}) \), with \( 0 \le \tilde{\eta}_2(y) \le 1 \) and

\[ \tilde{\eta}_2(y) = \begin{cases} 1, & \dfrac{R_1}{4a} \le y \le \dfrac{R_1}{2a},\\[4pt] 0, & y \le \dfrac{R_1}{8a} \text{ or } y \ge \dfrac{R_1}{a}. \end{cases} \]

We set

\[ \tilde{\eta}(x,y) = \eta_1(z(x,y)) \tilde{\eta}_2(y). \]

Then, as before,

\[ |\nabla \tilde{\eta}(x,y)|^2 \lesssim \frac{1+\alpha_+^2}{\delta^2 R_1^2}, \]

and by Caccioppoli’s inequality¹

\begin{equation}\label{eq:3} J_2 \lesssim \frac{M^2}{\lambda^2} \frac{1+\alpha_+^2}{\delta^2 R_1^2} \int_{\{-4R_2 \le z,\, \tfrac{R_1}{8a} \le y \le \tfrac{R_1}{2a}\}} |u|^2 \, dx \, dy. \end{equation}

Combining \eqref{eq:1} with \eqref{eq:2} and \eqref{eq:3}

\begin{align*} I_1 &\le \tfrac{M^4 \beta^2 (1+\alpha_+)^2}{\lambda^2\delta^4} 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 + \\ &+ \tfrac{M^4 (1+\alpha_+)^2}{\lambda^2\delta^4} 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, \end{align*}

where we used \(M>1\).

This yields \eqref{eq:i1_est}.