0x68f786d5

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

Let

\begin{align*} J_1&=\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, \\ J_2&= \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*}

Estimate of \(J_1\). Let

\[ a(x)=A_+(x,0)-A_-(x,0). \]

Then

\[ \lVert a(x)\rVert_2^2 \lesssim M^2, \quad \lVert a(x)-a(y)\rVert_2^2\lesssim M^2, \]

for \(x,y\in \mathbb{R}^{d-1}\).

Using boundary conditions, the sub-multiplicativity of the Frobenius norm and the Cauchy–Schwarz inequality , we obtain

\begin{align*} &\quad | A_+(x,0) \nabla (\vartheta u_+)(x,0) \cdot \nu - A_-(x,0) \nabla (\vartheta u_-)(x,0) \cdot \nu |^2\\ &= | \bigl(a(x) \nabla \vartheta (x,0)\bigr) \cdot \nu |^2 \, |u_+(x,0)|^2\\ &\leq \| a(x) \|^2_2 \, \vert \nabla \vartheta(x,0)\vert^2 \, |u_+(x,0)|^2\\ &\lesssim M^2 \lvert \nabla \vartheta(x,0)\rvert^2 \lvert u_+(x,0)\rvert^2. \end{align*}

By (0x68cb9821) ,

\[ \supp \nabla \vartheta \cap (\mathbb{R}^{d-1}\times \{0\}) \subseteq D:=\Bigl\{(x,0)\in \mathbb{R}^{d-1}\times \{0\}\mid \sqrt{4\delta R_2}< \lvert x\rvert < \sqrt{6\delta R_2}\Bigr\}, \]

and

\begin{equation*}\label{eq:e_est} | e^{2\tau \phi_\delta(x,0)} | \leq e^{-4\tau R_2}, \end{equation*}

for \(x\in D\), it follows that

\begin{equation}\label{eq:j1_est} J_1\lesssim \tfrac{M^2 (1+\alpha_-^2)}{\delta^3 R_2^2} \,\tau e^{-4\tau R_2} \,\int_{D} |u_+(x,0)|^2 dx. \end{equation}

Estimate of \(J_2\). Again, by applying the boundary conditions of \(u\), we deduce

\[ J_2= [f]^2_{H^{1/2}(\mathbb{R}^{d-1})}, \]

where

\[ f(x)= e^{\tau \phi_\delta(x,0)}\bigl(a(x) \nabla \vartheta (x,0)\bigr) \cdot \nu \, u_+(x ,0). \]

Let \(\varepsilon \gtrsim \sqrt{\delta R_2}\), such that the \(\epsilon\)-neighbourhood \(D_\varepsilon\) satisfies

\[ D_{\varepsilon} \subseteq \Bigl\{ (x,0) \in \mathbb{R}^{d-1}\times \{0\} \;|\; |x| \in \bigl[\sqrt{ \tfrac{5}{4}\delta R_2}, \sqrt{7 \delta R_2}\bigr] \Bigr\} =: \hat{D}. \]

The inradius \(r\) of \(\hat{D}\) can be estimated by

\[ r\gtrsim \sqrt{\delta R_2}, \]

and \(\diam \hat{D} \lesssim \varepsilon\). Then, by (0x68cbf31f) , it follows that

\begin{equation}\label{eq:B3_est_1} J_2 \lesssim [f]_{H^{1/2}(\hat{D})}^2 + \tfrac{1}{\sqrt{\delta R_2}} \int_D |f|^2. \end{equation}

Using the mean value inequality and

\begin{equation*} | e^{\tau \phi_\delta(x,0)} |^2 \leq e^{-\tfrac{5}{2}\tau R_2}, \end{equation*}

we obtain for \(x,y\in \hat{D}\)

\begin{align*} |e^{\tau \phi_\delta(x,0)} - e^{\tau \phi_\delta(y,0)}|^2 & \leq \sup_{\xi \in \hat{D}} \left| \frac{\tau \xi}{\delta} e^{\tau \phi_\delta(x,0)} \right| 2|x-y|^2 \\ &\lesssim \frac{\delta R_2}{\delta^2}\, \tau^2 e^{-\tfrac{5}{2}\tau R_2} |x-y|^2. \end{align*}

Let

\[ g(x) = (a(x) \nabla \vartheta(x,0))\cdot \nu \,u_+(x,0). \]

Then \(f(x)=e^{\tau \phi_\delta(x,0)} g(x)\), and due to (0x68cbf438) and the estimate, for \(x\in D\),

\[ \lvert g(x)\rvert^2 \lesssim \frac{M^2(1+\alpha_-^2)}{\delta^2 R_2^2} \lvert u_+(x,0)\rvert^2, \]

we deduce

\begin{equation}\label{eq:f_est} \begin{aligned} \,[f]_{H^{1/2}(\hat{D})}^2 &\lesssim e^{-\tfrac{5}{2}\tau R_2} [g]_{H^{1/2}(\hat{D})}^2 +\sqrt{\delta R_2} \tfrac{\delta R_2}{\delta^2}\, \tau^2 e^{-\tfrac{5}{2}\tau R_2} \int_D |g|^2 \\ &\lesssim e^{-\tfrac{5}{2}\tau R_2} [g]_{H^{1/2}(\hat{D})}^2 + \tfrac{M^2(1+\alpha_-^2)}{\delta^{5/2}\sqrt{R_2}}\, \tau^2 e^{-\tfrac{5}{2}\tau R_2} \int_D |u_+|^2 . \end{aligned} \end{equation}

It follows from (0x68ed5419) that

\[ \lvert a(x) \nabla \vartheta(x,0) \cdot \nu\rvert^2\lesssim \tfrac{M^2 (1+\alpha_-^2)}{\delta^2 R_2^2}, \]

and

\begin{align*} &\quad\lvert (a(x)\nabla \vartheta(x,0))\cdot \nu - (a(y)\nabla \vartheta(y,0))\cdot \nu\rvert^2\\[4pt] &\le \lVert a(x)\rVert_2^2 \lvert \nabla \vartheta(x,0)-\nabla \vartheta(y,0)\rvert^2 + \lvert \nabla \vartheta(y,0)\rvert^2 \lVert a(x)-a(y)\rVert_2^2\\[4pt] &\lesssim \tfrac{M^2(1+\alpha_-^2)}{\delta^3 R_2^{3}} \lvert x-y\rvert^2, \end{align*}

for \(x,y\in \hat{D}\). This and the bounds \(R_2< 1\), \(\delta< 1\) yield

\begin{equation}\label{eq:g_est} [g]^2_{H^{1/2}(\hat{D})} \lesssim \tfrac{M^2(1+\alpha_-^2)}{\delta^2 R_2^2} [u_+]^2_{H^{1/2}(\hat{D})} + \sqrt{\delta R_2} \,\tfrac{M^2(1+\alpha_-^2)}{\delta^3 R_2^3} \int_{D} \lvert u_+\rvert^2 . \end{equation}

By combining \eqref{eq:B3_est_1}, \eqref{eq:f_est}, \eqref{eq:g_est} and the estimate

\[ \tfrac{1}{\sqrt{\delta R_2}} \int_{D} \lvert f\rvert \lesssim \tfrac{M^2 (1+\alpha_-^2)}{\delta^{5/2} R_2^{5/2}} e^{-\tfrac{5}{2}\tau R_2} \int_{D} \lvert u_+\rvert^2, \]

we obtain

\[ J_2 \lesssim C_{1} \int_{D} \lvert u_+\rvert^2 + C_{2} [u_+]^2_{H^{1/2}(\hat{D})} \]

with

\begin{align*} C_{1} &= \tfrac{M^2 (1+\alpha_-^2)}{\delta^{5/2} R_2^{5/2}} e^{-\tfrac{5}{2}\tau R_2} + \tfrac{M^2 (1+\alpha_-^2) }{\delta^{5/2}\sqrt{R_2}}\, \tau^2 e^{- \tfrac{5}{2} \tau R_2} + \tfrac{M^2(1+\alpha_-^2)}{\delta^{5/2} R_2^{5/2}} e^{- \tfrac{5}{2} \tau R_2}\\[4pt] &\lesssim \tfrac{M^2(1+\alpha_-^2)}{\delta^{5/2}R_2^{5/2}} \tau^2 e^{- \tfrac{5}{2} \tau R_2}, \end{align*}

and

\[ C_{2} = \tfrac{M^2 (1+\alpha_-^2)}{\delta^2 R_2^2} e^{- \tfrac{5}{2} \tau R_2}. \]

Reformulating gives

\begin{equation}\label{eq:j2_est} J_2 \lesssim \tfrac{M^2(1+\alpha_-^2)}{\delta^{5/2}R_2^{5/2}}\, \tau^2 e^{- \tfrac{5}{2} \tau R_2} \biggl( \int_{D} \lvert u_+\rvert^2 + [u_+]^2_{H^{1/2}(\hat{D})}\biggr). \end{equation}

Combining the estimates of \(J_1\) and \(J_2\). Together \eqref{eq:j1_est} and \(\eqref{eq:j2_est}\) imply

\begin{equation}\label{eq:i2_est1} \begin{aligned} I_2&\lesssim \tfrac{M^2 (1+\alpha_-^2)}{\delta^3 R_2^2} \,\tau e^{-4\tau R_2} \,\int_{D} |u_+(x,0)|^2 dx + \tfrac{M^2(1+\alpha_-^2)}{\delta^{5/2}R_2^{5/2}}\, \tau^2 e^{- \tfrac{5}{2} \tau R_2} \biggl( \int_{D} \lvert u_+\rvert^2 + [u_+]^2_{H^{1/2}(\hat{D})}\biggr)\\[4pt] &\lesssim \tfrac{M^2(1+\alpha_-^2)}{\delta^3 R_2^3} \tau^2 e^{- \tfrac{5}{2} \tau R_2} \biggl( \int_{D} \lvert u_+\rvert^2 + [u_+]^2_{H^{1/2}(\hat{D})}\biggr)\\ &\lesssim \tfrac{M^2(1+\alpha_-^2)}{\delta^3 R_2^3} \tau^2 e^{- \tfrac{5}{2} \tau R_2} \lVert u_+\rVert^2_{H^{1/2}(\{\lvert x\rvert\le \sqrt{7\delta R_2}\})}, \end{aligned} \end{equation}

where we used \(\tau\ge 1\), \(R_2< 1\), and \(\delta< 1\).

Dimension Increasing. We estimate \(\lVert u_+\rVert^2_{H^{1/2}(\{\lvert x\rvert\le \sqrt{7\delta R_2}\})}\) by applying a trace theorem (0x68f605f9) and a Caccioppoli type inequality (0x68f6035b) . Note, that

\[ \{x\in \mathbb{R}^{d-1} \;\mid\; \lvert x\rvert\le \sqrt{7\delta R_2}\} = \{x\in \mathbb{R}^{d-1} \;\mid\; z(x,0)\ge -\tfrac{7}{2} R_2\}. \]

First, we choose

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

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

\[ \eta_1(t) = \begin{cases} 1, & t \ge -\tfrac{7}{2} R_2, \\ 0, & t \le -\tfrac{15}{4} R_2, \end{cases} \qquad \eta_2(y) = \begin{cases} 1, & y \le \tfrac{R_1}{4a},\\[4pt] 0, & y \ge \tfrac{R_1}{2a}. \end{cases} \]

Then, like in (0x68f5eeb1) , it follows that

\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 (0x68f605f9) yields

\begin{equation}\label{eq:trace} \lVert u_+\rVert^2_{H^{1/2}(\{\lvert x\rvert\le \sqrt{7\delta R_2}\})} \lesssim \tfrac{1+\alpha_+^2}{\delta^2 R_1^2} \lVert u_+\rVert^2_{H^1(\{-\frac{15}{4} \le z,\, 0\le y\le \frac{R_1}{2a}\})}. \end{equation}

Next, let

\[ \tilde{\eta}(x,y) = \tilde{\eta}_1(z(x,y)) \tilde{\eta}_2(y) \]

with \( \tilde{\eta}_1, \tilde{\eta}_2 \in C_c^\infty(\mathbb{R}) \), \( 0 \le \tilde{\eta}_1(t) \le 1 \), \( 0 \le \tilde{\eta}_2(y) \le 1 \),

\[ \tilde{\eta}_1(t) = \begin{cases} 1, & t \ge -\tfrac{15}{4} R_2, \\ 0, & t \le -4 R_2, \end{cases} \qquad \tilde{\eta}_2(y) = \begin{cases} 1, & y \le \tfrac{R_1}{2a},\\[4pt] 0, & y \ge \tfrac{R_1}{a}. \end{cases} \]

As before,

\begin{equation*} |\nabla \tilde{\eta}(x,y)|^2 \lesssim \frac{1+\alpha_+^2}{\delta^2 R_1^2}. \end{equation*}

By (0x68f6035b) , we deduce

\begin{equation}\label{eq:caccioppoli} \sum_{\pm} \int_{\{-\tfrac{15}{4}R_2\le z,\, y<\tfrac{R_1}{2a}\}\cap \mathbb{R}^d_\pm } |\nabla u_\pm|^2 \, dx\,dy, \lesssim \tfrac{M^2(1+\alpha_+^2)}{\lambda^2\delta^2R_1^2} \int_{\Bigl\{-4R_2 \le z,\, y \le \tfrac{R_1}{a}\Bigr\}} |u|^2 \, dx \, dy. \end{equation}

By combining \eqref{eq:i2_est1} with \eqref{eq:trace}, \eqref{eq:caccioppoli}, we finally obtain

\[ I_2\lesssim \tfrac{M^4(1+\alpha_+)^3}{\lambda^2\delta^7 R_1^7} \tau^2 e^{- \tfrac{5}{2} \tau R_2} \int_{\Bigl\{-4R_2 \le z,\, y \le \tfrac{R_1}{a}\Bigr\}} |u|^2 \, dx \, dy. \]