Proof. Using polar coordinates transformation and Proposition 2.1, we have

\begin{equation*} \begin{aligned} \int _{\Gamma _h^{\pm }}e^{\rho \cdot {\mathbf{x}}}\mathrm{d} \sigma &=\frac{\Gamma (1)}{\tau }\frac{1}{({\mathbf{d}}+\mathrm{i}{\mathbf{d}}^{\perp })\cdot \hat{{\mathbf{x}}_{{\mathbf{1}}}}}-I_{R_1}+\frac{\Gamma (1)}{\tau }\frac{1}{({\mathbf{d}}+\mathrm{i}{\mathbf{d}}^{\perp })\cdot \hat{{\mathbf{x}}_{{\mathbf{2}}}}}-I_{R_2}, \end{aligned} \end{equation*}

where $\hat{{\mathbf{x}}_{{\mathbf{1}}}}$ and $\hat{{\mathbf{x}}_{{\mathbf{2}}}}$ are unit vector of ${\mathbf{x}}$ on $\Gamma ^-$ and $\Gamma ^+$ , and

\begin{equation*} I_{R_1}=\int _{\Gamma ^-\setminus \Gamma ^-_h}e^{-\tau ({\mathbf{d}}+\mathrm{i}{\mathbf{d}}^\perp )}\mathrm{d}\sigma,\quad I_{R_2}=\int _{\Gamma ^+\setminus \Gamma ^+_h}e^{-\tau ({\mathbf{d}}+\mathrm{i}{\mathbf{d}}^\perp )}\mathrm{d}\sigma . \end{equation*}

Hence, with the help of Proposition 2.1, for sufficiently large $\tau$ , we have the following integral inequality

(3.12) \begin{equation} \begin{aligned} \left \vert \int _{\Gamma _h^{\pm }}e^{\rho \cdot {\mathbf{x}}}\mathrm{d} \sigma \right \vert &\geq \frac{1}{\tau }\left \vert \frac{1}{({\mathbf{d}}+\mathrm{i}{\mathbf{d}}^{\perp })\cdot \hat{{\mathbf{x}}_{{\mathbf{1}}}}} + \frac{1}{({\mathbf{d}}+\mathrm{i}{\mathbf{d}}^{\perp })\cdot \hat{{\mathbf{x}}_{{\mathbf{2}}}}} \right \vert -\vert I_{R_1}\vert -\vert I_{R_2} \vert \\[5pt] &\geq \frac{C_h}{\tau }-O\left (\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ) \end{aligned} \end{equation}

by using (3.10).

Proof. Since $q \in H^2(\overline{ S}_h)$ , let $\widetilde q$ be the Sobolev extension of $q$ such that $\widetilde q \in H^2$ , then by this we have $\tilde q\in H^{1,1+\varepsilon _0},\varepsilon _0\in (0,1)$ . Then by Lemma 2.1, one readily has (3.16).

Proof. Taking $\textbf{y}=\tau {\mathbf{x}}$ , then using Hölder inequality and Lemma 3.2, one has

(3.18) \begin{align} \int _{\Gamma _h^\pm }\vert e^{\rho \cdot {\mathbf{x}}}\vert &\vert \psi ({\mathbf{x}})\vert \mathrm{d}\sigma \lesssim \frac{1}{\tau }\|e^{-{\mathbf{d}}\cdot \textbf{y}}\|_{L^{\frac{8}{7}}(\Gamma _{\tau h}^\pm )}\left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{L^8(\Gamma _{\tau h}^\pm )} \\[5pt] & \lesssim \frac{1}{\tau }\|e^{-{\mathbf{d}}\cdot \textbf{y}}\|_{L^{\frac{8}{7}}(\Gamma ^\pm )}\left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{H^{1,8}(S_{\tau h})} \lesssim \frac{1}{\tau }\|e^{-{\mathbf{d}}\cdot \textbf{y}}\|_{L^{\frac{8}{7}}(\Gamma ^\pm )}\left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{H^{1,8}(\mathcal{K})},\nonumber \end{align}

for sufficiently large $\tau$ . We have $\frac{1}{\tau }\lt 1$ , and it holds that

(3.19) \begin{align} \left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{H^{1,8}(\mathcal{K})} &\leq \tau ^{\frac{1}{4}}\left \|\psi ({\mathbf{x}})\right \|_{H^{1,8}(\mathcal{K})}=\mathcal{O}(\tau ^{-\frac{5}{12}}),\quad \mbox{as $\tau \rightarrow \infty $. } \end{align}

Furthermore,

(3.20) \begin{equation} \|e^{-{\mathbf{d}}\cdot \textbf{y}}\|_{L^{\frac{8}{7}}(\Gamma ^\pm )} \leq \left (\int _{\Gamma ^\pm }e^{-\frac{8}{7}\zeta \vert \textbf{y}\vert }\mathrm{d}\sigma \right )^{\frac{7}{8}} =2\left ( \frac{7}{8}\frac{1}{\zeta }\right )^{\frac{7}{8}}, \end{equation}

where $\zeta$ is defined in (3.10). Hence, $\|e^{-{\mathbf{d}}\cdot \textbf{y}}\|_{L^{\frac{8}{7}}(\Gamma ^\pm )}$ is a positive constant which only depends on $\zeta$ . Combining (3.19) and (3.20) with (3.18), we can prove Lemma 3.3.

Proof. Using polar coordinates transformation, (2.6a) and (3.10), we can obtain that

(3.22) \begin{equation} \|e^{\rho \cdot {\mathbf{x}}}\|_{ L^{t}(\Lambda _{h})} \lesssim e^{-\zeta h\tau }. \end{equation}

where $\rho$ is defined in (2.3) and $t$ is a positive constant.

According to (3.16) and Lemma 3.2, for sufficient large $\tau$ , one can show that

(3.23) \begin{align} \|\psi ({\mathbf{x}})\|_{L^{4}(\Lambda _{h})} \leq C \| \psi ({\mathbf{x}})\|_{H^{1,8}(S_h)}=O(\tau ^{-\frac{2}{3}}), \end{align}

where $C$ is a positive constant, which is not a function of $\tau$ .

By virtue of (3.23) and Hölder inequality, it can be directly verified that

(3.24) \begin{equation} \begin{aligned} \|u_{0}\|_{ L^{2}(\Lambda _h)} &\leq \|e^{\rho \cdot {\mathbf{x}}}\|_{ L^{2}(\Lambda _{h})}+\|e^{\rho \cdot {\mathbf{x}}}\|_{ L^{4}(\Lambda _{h})}\|\psi ({\mathbf{x}})\|_{ L^{4}(\Lambda _{h})}\\[5pt] &\lesssim \left (1+\tau ^{-\frac{2}{3}}\right )e^{-\zeta h\tau },\quad \mbox{as $\tau \rightarrow \infty $. } \end{aligned} \end{equation}

Similarly, using Cauchy-Schwarz inequality, (3.21a) and Proposition 2.1, we have

(3.25) \begin{equation} \begin{aligned} \|\nabla u_{0}({\mathbf{x}})\|_{L^{2}(\Lambda _h)} &\leq \sqrt{2}\tau \|u_{0}\|_{ L^{2}(\Lambda _h)}+\|e^{\rho \cdot {\mathbf{x}}}\|_{ L^{4}(\Lambda _h)} \|\nabla \psi ({\mathbf{x}})\|_{ L^{4}(\Lambda _h)}\\[5pt] &\lesssim (1+\tau )(1+\tau ^{-\frac{2}{3}})e^{-\zeta h\tau },\quad \mbox{as $\tau \rightarrow \infty $. } \end{aligned} \end{equation}

Moreover, by using Cauchy-Schwarz inequality, we know that

(3.26) \begin{equation} \begin{aligned} \int _{S_{h}}\vert {\mathbf{x}}\vert ^{\alpha }\vert u_{0} \vert \mathrm{d}{\mathbf{x}} &\leq \int _{S_{h}}\vert {\mathbf{x}}\vert ^{\alpha }\vert e^{\rho \cdot {\mathbf{x}}} \vert \mathrm{d}{\mathbf{x}}+\int _{S_{h}}\vert {\mathbf{x}}\vert ^{\alpha }\vert e^{\rho \cdot {\mathbf{x}}}\vert \vert \psi ({\mathbf{x}})\vert \mathrm{d}{\mathbf{x}}.\\[5pt] \end{aligned} \end{equation}

Using polar coordinates transformation and Proposition 2.1, we can deduce that

(3.27) \begin{equation} \int _{S_{h}}\vert {\mathbf{x}}\vert ^{\alpha }\vert e^{\rho \cdot {\mathbf{x}}} \vert \mathrm{d}{\mathbf{x}} \lesssim \left (\frac{1}{\tau ^{\alpha +2}}+\frac{1}{\tau }\right )e^{-\frac{1}{2}\zeta h\tau },\quad \mbox{as $\tau \rightarrow \infty $. } \end{equation}

Next, by letting $\textbf{y}=\tau {\mathbf{x}}$ and Hölder inequality, it can be calculated that

(3.28) \begin{equation} \begin{aligned} \int _{S_{h}}\vert {\mathbf{x}}\vert ^{\alpha }\vert e^{\rho \cdot {\mathbf{x}}}\vert \vert \psi ({\mathbf{x}})\vert \mathrm{d}{\mathbf{x}} &\leq \frac{1}{\tau ^{\alpha +2}}\int _{\mathcal{K}}\vert \textbf{y}\vert ^{\alpha }\vert e^{-{\mathbf{d}}\cdot \textbf{y}}\vert \left \vert \psi \left (\frac{\textbf{y}}{\tau }\right )\right \vert \mathrm{d}{\mathbf{x}}\\[5pt] &\leq \frac{1}{\tau ^{\alpha +2}}\|\vert \textbf{y}\vert ^{\alpha }\vert e^{-{\mathbf{d}}\cdot \textbf{y}}\vert \|_{L^{\frac{8}{7}}(\mathcal{K})}\left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{L^{8}(\mathcal{K})}. \end{aligned} \end{equation}

With the help of variable substitution and (3.16), we can calculate that

(3.29) \begin{equation} \left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{L^8(\mathcal{K})}=\tau ^{\frac{1}{4}}\|\psi \!\left ({\mathbf{x}}\right )\|_{L^8(\mathcal{K})}=\mathcal{O}(\tau ^{-\frac{5}{12}}),\quad \mbox{as $\tau \rightarrow \infty $. } \end{equation}

Similar to (3.20), by using polar coordinates transformation, we have $ \|{\vert \textbf{y}\vert }^\alpha \vert e^{-{\mathbf{d}}\cdot \textbf{y}}\vert \|_{L^{\frac{8}{7}}(\mathcal{K})}$ is a positive constant and not a function of $\tau$ . Therefore, combining (3.29), (3.28) and (3.27) with (3.26), we have (3.21c).

The proof is complete.

Proof of Theorem 3.1 . By Green’s formula (2.7) and (3.15), the following integral equality holds

(3.30) \begin{equation} \begin{aligned} \int _{\Lambda _{h}}(w-v)\partial _{\nu }u_{0}-u_{0}\partial _{\nu }(w-v)\mathrm{d}\sigma -\int _{\Gamma _h^{\pm }}\eta u_0v\mathrm{d}\sigma &=k^{2}\int _{S_{h}}(q-1)vu_{0}\mathrm{d}{\mathbf{x}}. \end{aligned} \end{equation}

Denote

\begin{equation*} f_{j}=(q-1)v_{j}. \end{equation*}

Since $q\in H^2(\overline S_h)$ , by Sobolev embedding property, one has $q\in C^{\alpha }(\overline S_h)$ where $\alpha \in (0,1]$ . Clearly, $v_{j}\in C^{\alpha }(\overline S_h)$ , hence $f_{j}\in C^{\alpha }(\overline S_h)$ . According $v_j\in C^{\alpha },\ \eta \in C^{\alpha }$ , we have the expansion

(3.31) \begin{equation} \begin{aligned} f_{j}&=f_{j}({\mathbf{0}})+\delta f_{j},\ \vert \delta f_{j}\vert \leq \| f_{j}\|_{C^{\alpha }({\overline S}_h)}\vert {\mathbf{x}} \vert ^{\alpha },\\[5pt] v_j&=v_j({\mathbf{0}})+\delta v_j,\ \vert \delta v_j\vert \leq \|v_j\|_{C^{\alpha }(\overline S_h)}\vert {\mathbf{x}}\vert ^{\alpha },\\[5pt] \eta &=\eta ({\mathbf{0}})+\delta \eta,\ \vert \delta \eta \vert \leq \|\eta \|_{C^{\alpha }(\overline{ \Gamma _h^\pm } )}\vert {\mathbf{x}}\vert ^{\alpha }. \end{aligned} \end{equation}

By virtue of (3.31) and (2.2), it yields that

(3.32) \begin{equation} \begin{aligned} &k^2\int _{S_{h}}(q-1)vu_0\mathrm{d}{\mathbf{x}} =-\sum _{m=1}^3 I_{m},\quad \int _{\Gamma _h^{\pm }}\eta u_0v\mathrm{d}\sigma =I-\sum _{m=4}^9I_m, \end{aligned} \end{equation}

where

\begin{align*} I_{1}&=-k^{2}\int _{S_{h}}(q-1)(v-v_{j})u_{0}\mathrm{d}{\mathbf{x}},\quad I_{2}=-\int _{S_{h}}\delta f_{j}u_{0}\mathrm{d}{\mathbf{x}},\quad \\[5pt] I_{3}&=-f_{j}({\mathbf{0}})\int _{S_{h}}u_0\mathrm{d}{\mathbf{x}},\quad I_4=-\eta ({\mathbf{0}})\int _{\Gamma _h^{\pm }}(v-v_j)u_0\mathrm{d}\sigma,\\[5pt] I_5&=-\int _{\Gamma _h^{\pm }}\delta \eta (v-v_j)u_0\mathrm{d}\sigma,\quad I_6=-\eta ({\mathbf{0}})v_j({\mathbf{0}})\int _{\Gamma _h^{\pm }}e^{\rho \cdot {\mathbf{x}}}\psi ({\mathbf{x}})\mathrm{d}\sigma,\\[5pt] I_{7}&=-\eta ({\mathbf{0}})\int _{\Gamma _h^{\pm }}\delta v_ju_0\mathrm{d}\sigma,\quad I_8=-v_j({\mathbf{0}}) \int _{\Gamma _h^{\pm }}\delta \eta u_0\mathrm{d}\sigma,\\[5pt] I_9&=-\int _{\Gamma _h^{\pm }}\delta \eta \delta v_j u_0\mathrm{d}\sigma,\quad I=\eta ({\mathbf{0}})v_j({\mathbf{0}})\int _{\Gamma _h^{\pm }}e^{\rho \cdot {\mathbf{x}}}\mathrm{d}\sigma . \end{align*}

Substituting (3.32) into (3.30), we have the following integral identity

\begin{align*} I=\sum _{m=1}^9I_m+J_1+J_2, \end{align*}

where

(3.33) \begin{equation} \begin{aligned} J_{1}&=\int _{\Lambda _{h}}(w-v)\partial _{\nu }u_{0}\mathrm{d}\sigma,\quad J_{2}=-\int _{\Lambda _{h}} u_{0}\partial _{\nu }(w-v)\mathrm{d}\sigma . \end{aligned} \end{equation}

Therefore, it yields that

(3.34) \begin{equation} \vert I\vert \leq \sum _{m=1}^9\vert I_m\vert +\vert J_1\vert +\vert J_2\vert . \end{equation}

In the following, we give detailed asymptotic estimates of $I_m,m=1,\cdots,9$ and $J_j,\ j=1,2$ as $\tau \to \infty$ , separately. With the help of Proposition 3.2, Hölder inequality and (3.16), it arrives at

(3.35) \begin{equation} \begin{aligned} \vert I_{1} \vert &\lesssim \|v-v_j\|_{L^2(S_h)}\left (\|e^{\rho \cdot {\mathbf{x}}}\|_{L^2(S_h)}+\| e^{\rho \cdot {\mathbf{x}}}\psi ({\mathbf{x}})\|_{L^2(S_h)}\right )\\[5pt] &\lesssim \|v-v_j\|_{L^2(S_h)}\left (\|e^{\rho \cdot {\mathbf{x}}}\|_{L^2(S_h)}+\| e^{\rho \cdot {\mathbf{x}}}\|_{L^4(S_h)}\|\psi ({\mathbf{x}})\|_{L^4(S_h)}\right )\\[5pt] &\lesssim j^{-\beta }\left [\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{3}} \right ] \end{aligned} \end{equation}

as $\tau \rightarrow \infty$ .

By virtue of (3.31), it yields that

(3.36) \begin{equation} \vert I_{2} \vert \leq \|f_{j} \|_{C^{\alpha }( \overline S_h)}\int _{S_{h}}\vert {\mathbf{x}}\vert ^{\alpha }\vert u_{0} \vert \mathrm{d}{\mathbf{x}}, \end{equation}

where

(3.37) \begin{equation} \begin{aligned} \|f_{j}\|_{C^{\alpha } (S_h)}& \leq k^2\left (\|q\|_{C^{\alpha }(\overline S_h)}\sup _{S_h}\vert v_{j} \vert +\|v_{j}\|_{C^{\alpha }( \overline S_h)}\sup _{S_h}\vert q-1\vert \right ). \end{aligned} \end{equation}

Using the property of compact embedding of Hölder spaces, we can derive that

(3.38) \begin{equation} \| v_{j}\|_{C^{\alpha }}\leq{\mathrm{diam} }\left (S_{h}\right )^{1-\alpha }\|v_{j}\|_{C^{1}(S_h)}, \end{equation}

where diam $\left (S_{h}\right )$ is the diameter of $S_{h}$ . By direct computations, we obtain

(3.39) \begin{equation} \|v_{j}\|_{C^{1}}\leq \sqrt{2\pi }(1+k)\|g\|_{L^{2}(\mathbb S^{1})}. \end{equation}

Furthermore, by Cauchy-Schwarz inequality, we also can deduce that

(3.40) \begin{equation} \vert v_{j}\vert \leq \sqrt{2\pi }\|g\|_{L^{2}(\mathbb S^{1})}. \end{equation}

Due to (3.5), by using the fact that $q\in C^{\alpha }(\overline S_h)$ , substituting (3.38), (3.39) and (3.40) into (3.37), we have

(3.41) \begin{equation} \|f_{j}\|_{C^{\alpha }(\overline S_h)}\lesssim j^{\gamma },\quad \|v_j\|_{C^\alpha (\overline S_h)}\lesssim j^{\gamma }, \end{equation}

where $\gamma$ is a given positive constant defined in (3.5). Substituting (3.21c) and (3.41) into (3.36), we can deduce that

(3.42) \begin{equation} \vert I_{2} \vert \lesssim j^{\gamma }\left [\tau ^{-(\alpha +\frac{29}{12})}+\left (\frac{1}{\tau ^{\alpha +2}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\right ] \end{equation}

as $\tau \rightarrow \infty$ .

Using Cauchy-Schwarz inequality, it can be easily calculated that

(3.43) \begin{equation} \vert I_{3} \vert \lesssim \int _{S_h}\vert e^{\rho \cdot {\mathbf{x}}}\vert \mathrm{d}{\mathbf{x}}+\int _{S_h}\vert e^{\rho \cdot {\mathbf{x}}}\psi ({\mathbf{x}})\vert \mathrm{d}{\mathbf{x}}\lesssim \int _{S_h}\vert e^{\rho \cdot {\mathbf{x}}}\vert \mathrm{d}{\mathbf{x}}+\int _{\mathcal{K}}\vert e^{\rho \cdot {\mathbf{x}}}\psi ({\mathbf{x}})\vert \mathrm{d}{\mathbf{x}}. \end{equation}

By integral substitution and using (3.29), we obtain that

(3.44) \begin{equation} \begin{aligned} \int _{\mathcal{K}}\vert e^{\rho \cdot {\mathbf{x}}}\psi ({\mathbf{x}})\vert \mathrm{d}{\mathbf{x}}&=\frac{1}{\tau ^2}\int _{\mathcal{K}}e^{-{\mathbf{d}}\cdot \textbf{y}}\left \vert \psi \left (\frac{\textbf{y}}{\tau }\right )\right \vert \mathrm{d}\textbf{y} \leq \frac{1}{\tau ^2} \|e^{-{\mathbf{d}}\cdot \textbf{y}}\|_{L^{\frac{8}{7}}(\mathcal{K})}\|\psi \left (\frac{\textbf{y}}{\tau }\right )\|_{L^{8}(\mathcal{K})}\\[5pt] &\lesssim \tau ^{-\frac{29}{12}},\quad \mbox{as $\tau \rightarrow \infty $. } \end{aligned} \end{equation}

With the help of Proposition 3.1, substituting (3.44) into (3.43), we can derive that

(3.45) \begin{equation} \vert I_{3} \vert \lesssim \tau ^{-\frac{29}{12}}+\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ), \mbox{as $\tau \rightarrow \infty $. } \end{equation}

Using Cauchy-Schwarz inequality, the trace theorem and Hölder inequality, we have

(3.46) \begin{equation} \begin{aligned} \vert I_4\vert &\lesssim \|v-v_j\|_{L^2(\Gamma _h^{\pm })}(\|e^{\rho \cdot {\mathbf{x}}}\|_{L^2(\Gamma _h^{\pm })}+\|e^{\rho \cdot {\mathbf{x}}}\|_{L^4(\Gamma _h^{\pm })}\|\psi ({\mathbf{x}})\|_{L^4(\Gamma _h^{\pm })})\\[5pt] &\lesssim j^{-\beta }\left [\left (\frac{1}{\tau }+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau }+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{3}}\right ], \end{aligned} \end{equation}

as $\tau \rightarrow \infty$ . Similarly, by virtue of Cauchy-Schwarz inequality, the trace theorem and Hölder inequality, it can be calculated that

\begin{equation*} \begin{aligned} \vert I_5\vert &\lesssim \|v-v_j\|_{H^1(S_h)}\left (\|e^{\rho \cdot {\mathbf{x}}}\vert \textbf x\vert ^{\alpha }\|_{L^2(\Gamma _h^{\pm })}+\|e^{\rho \cdot {\mathbf{x}}}\vert \textbf x\vert ^{\alpha }\|_{L^4(\Gamma _h^{\pm })}\|\psi ({\mathbf{x}})\|_{L^4(\Gamma _h^{\pm })}\right )\\[5pt] &\lesssim j^{-\beta }\left [ \left (\frac{1}{\tau ^{(2\alpha +1)}}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^{(4\alpha +1)}}+\frac{1}{\tau }e^{-2\zeta h\tau } \right )^{\frac{1}{4}}\tau ^{-\frac{2}{3}}\right ], \end{aligned} \end{equation*}

as $\tau \rightarrow \infty$ . By using Lemma 3.3, one can show that

(3.47) \begin{equation} \begin{aligned} I_6&\lesssim \int _{\Gamma _h^\pm }\vert e^{\rho \cdot {\mathbf{x}}}\psi ({\mathbf{x}})\vert \mathrm{d}{\mathbf{x}}\lesssim \tau ^{-\frac{17}{12}}, \quad \mbox{as $\tau \rightarrow \infty $. } \end{aligned} \end{equation}

Using (3.31), (3.41) and Proposition 2.1, we have the following inequality

(3.48) \begin{align} \vert I_7\vert &\lesssim j^{\gamma }\left [\int _{\Gamma _{h}^{\pm }}\vert {\mathbf{x}}\vert ^{\alpha }\vert e^{\rho \cdot {\mathbf{x}}}\vert \mathrm{d}\sigma +\|\vert {\mathbf{x}}\|^{\alpha }\vert e^{\rho \cdot {\mathbf{x}}}\vert \|_{L^2(\Gamma _h^{\pm })}\|\psi ({\mathbf{x}})\|_{L^2(\Gamma _h^{\pm })} \right ] \nonumber \\[5pt] &\lesssim j^{\gamma }\left [\left (\frac{1}{\tau ^{\alpha +1}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )+\left (\frac{1}{\tau ^{2\alpha +1}}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}\tau ^{-\frac{2}{3}}\right ], \end{align}

as $\tau \rightarrow \infty$ . According to (3.48), we can derive that

(3.49a) \begin{align} & \vert I_8\vert \lesssim \left (\frac{1}{\tau ^{\alpha +1}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )+\left (\frac{1}{\tau ^{2\alpha +1}}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}\tau ^{-\frac{2}{3}}, \end{align}

(3.49b) \begin{align} & \vert I_9\vert \lesssim j^{\gamma }\left [\left (\frac{1}{\tau ^{2\alpha +1}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )+\left (\frac{1}{\tau ^{4\alpha +1}}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}\tau ^{-\frac{2}{3}}\right ], \end{align}

as $\tau \rightarrow \infty$ . By the Cauchy-Schwarz inequality and the trace theorem, we deduce that

(3.50) \begin{align} \vert J_1\vert & \leq C\|u_{0}\|_{H^{1}(\Lambda _{h})} \|w-v\|_{H^{1}(\Lambda _{h})} \\[5pt] &\lesssim \|u_{0}\|_{H^{1}(\Lambda _{h})} \nonumber \end{align}

as $\tau \rightarrow \infty$ , where $C$ is a positive constant arising from the trace theorem. Hence, by virtue of (3.21a) and (3.21b), from (3.50), it is readily known that

(3.51) \begin{equation} \begin{aligned} \vert J_1 \vert \lesssim (1+\tau )(1+\tau ^{-\frac{2}{3}})e^{-\zeta h\tau } \end{aligned} \end{equation}

as $\tau \rightarrow \infty$ , where $\zeta$ is a positive constant given in (3.10).

Similarly, using Cauchy-Schwarz inequality, the trace theorem and (3.21b), we can obtain that

(3.52) \begin{align} \vert J_2 \vert & \leq \|\partial _{\nu } u_0\|_{ L^{2}(\Lambda _h)} \|w-v\|_{ L^{2}(\Lambda _h)} \leq C \|\partial _{\nu } u_0\|_{ L^{2}(\Lambda _h)} \|w-v\|_{ H^{1}(S_{h})} \nonumber\\[5pt] &\lesssim \|\nabla u_{0}\|_{ L^{2}(\Lambda _h)} \lesssim (1+\tau )(1+\tau ^{-\frac{2}{3}})e^{-\zeta h\tau }. \end{align}

Substituting (3.35), (3.42), (3.45), (3.51) and (3.52) into (3.34), by virtue of (3.11), we derive that

(3.53) \begin{align} \left (\frac{C_{S_h}}{\tau }-\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )&\vert \eta ({\mathbf{0}})v_j({\mathbf{0}})\vert \lesssim j^{-\beta }\left [\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{3}} \right ] \nonumber \\[5pt] & + j^{\gamma }\left [\tau ^{-(\alpha +\frac{29}{12})}+\left (\frac{1}{\tau ^{\alpha +2}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\right ] \nonumber \\[5pt] &+j^{-\beta }\left [\left (\frac{1}{\tau }+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau }+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{3}}\right ] \\[5pt] &+j^{-\beta }\left [ \left (\frac{1}{\tau ^{(2\alpha +1)}}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^{(4\alpha +1)}}+\frac{1}{\tau }e^{-2\zeta h\tau } \right )^{\frac{1}{4}}\tau ^{-\frac{2}{3}}\right ] \nonumber \\[5pt] &+(j^{\gamma }+1)\left [\left (\frac{1}{\tau ^{\alpha +1}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )+\left (\frac{1}{\tau ^{2\alpha +1}}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}\tau ^{-\frac{2}{3}}\right ] \nonumber \\[5pt] &+j^{\gamma }\left [\left (\frac{1}{\tau ^{2\alpha +1}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )+\left (\frac{1}{\tau ^{4\alpha +1}}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}\tau ^{-\frac{2}{3}}\right ] \nonumber \\[5pt] &+(1+\tau )(1+\tau ^{-\frac{2}{3}})e^{-\zeta h\tau }+ \tau ^{-\frac{29}{12}}+\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )+\tau ^{-\frac{17}{12}}\nonumber \end{align}

as $\tau \rightarrow \infty$ , where $C_{S_h}$ is a positive constant given in (3.11). Multiplying $\tau$ on both sides of (3.53) and letting $\tau =j^s$ , where $s\gt 0$ , it can be derived that

(3.54) \begin{equation} \begin{aligned} \left (C_h-e^{-\frac{1}{2}\zeta hj^s}\right )\vert \eta ({\mathbf{0}})v_j({\mathbf{0}})\vert &\lesssim j^{-\beta +s}+j^{\gamma -(\alpha +1)s}+j^{-\beta +\frac{1}{2}s}+j^{-\beta +(-\alpha +\frac{1}{2})s}\\[5pt] &+j^{\gamma -\alpha s}+j^{-\frac{13}{24}s}+j^{-\frac{17}{12}s}, \end{aligned} \end{equation}

as $\tau \rightarrow \infty$ . Under the assumption (3.4), we can choose $s \in (\gamma/ \alpha, \beta )$ . Hence in (3.54), let $j\rightarrow \infty$ it is readily to know that

\begin{equation*} \lim _{j \rightarrow \infty } \left \vert \eta (\textbf {0})v_j(\textbf {0})\right \vert =0.\end{equation*}

Since $\eta ({\mathbf{0}})\not = 0$ , one has $\lim _{j \rightarrow \infty }\vert v_j({\mathbf{0}})\vert =0$ . Using (3.5) and the integral mean value theorem, we can obtain (3.6).

The proof is complete.

Proof of Theorem 3.2 . Due to $q\in H^2(\overline S_h)$ , using the Sobolev embedding property, we know that $q\in C^\alpha (\overline{S_h})$ with $\alpha \in (0,1]$ . Under the assumption $v\in C^\alpha (\overline S_h)$ ( $\alpha \in (0,1]$ ), it readily has $f({\mathbf{x}})\;:\!=\;(q({\mathbf{x}})-1)v({\mathbf{x}}) \in C^{\alpha } (\overline{S_h})$ . Hence, we have the expansion of $f({\mathbf{x}}),\ \eta$ and $v({\mathbf{x}})$ near the origin as follows

(3.55) \begin{equation} \begin{aligned} f({\mathbf{x}})&= f({\mathbf{0}})+\delta f,\quad \vert \delta f\vert \leq \|f\|_{C^\alpha }\vert {\mathbf{x}} \vert ^\alpha \\[5pt] \eta &= \eta ({\mathbf{0}})+\delta \eta,\quad \vert \delta \eta \vert \leq \|\eta \|_{C^\alpha }\vert {\mathbf{x}}\vert ^\alpha \\[5pt] v({\mathbf{x}})&= v({\mathbf{0}})+\delta v,\quad \vert \delta v\vert \leq \|v\|_{C^\alpha }\vert {\mathbf{x}}\vert ^\alpha \end{aligned} \end{equation}

Plugging (3.55) into the integral identity (3.30), it yields that

(3.56) \begin{align} \eta ({\mathbf{0}})v({\mathbf{0}})\int _{\Gamma _h^{\pm }}e^{\rho \cdot {\mathbf{x}}}\mathrm{d}\sigma &=f({\mathbf{0}})\int _{S_h}u_0\mathrm{d}{\mathbf{x}}+\int _{S_h}\delta fu_0\mathrm{d}{\mathbf{x}} +\eta ({\mathbf{0}})v({\mathbf{0}})\int _{\Gamma _h^{\pm }}\psi ({\mathbf{x}})e^{\eta \cdot {\mathbf{x}}}\mathrm{d}\sigma \nonumber \\[5pt] &+\eta ({\mathbf{0}})\int _{\Gamma _h^{\pm }}\delta v u_0\mathrm{d}\sigma +v({\mathbf{0}})\int _{\Gamma _h^{\pm }}\delta \eta u_0\mathrm{d}\sigma +\int _{\Gamma _h^\pm }\delta v\delta \eta u_0\mathrm{d}\sigma \nonumber \\[5pt] &-\int _{\Lambda _h}(w-v)\partial _\nu u_0\mathrm{d}\sigma +\int _{\Lambda _h}u_0\partial _\nu (w-v)\mathrm{d}\sigma . \end{align}

By adopting similar asymptotic analysis for each integral in (3.56) with respect to the parameter $\tau$ as in the proof of Theorem 3.1, and letting $\tau \rightarrow \infty$ , we can prove Theorem 3.2.

Proof. Using polar coordinates transformation and (2.6b) in Proposition 2.1, we have

\begin{equation*} \begin{aligned} \int _{S_h} e^{\rho \cdot {\mathbf{x}}}\mathrm{d} {\mathbf{x}} &=\int _{\theta _{m}}^{\theta _M}\left [\frac{\Gamma (2)}{(\tau ({\mathbf{d}}+\mathrm{i} {\mathbf{d}}^\perp )\cdot \hat{\textbf x} )^2}-I_\textsf{R}\right ]\mathrm{d}\theta \\[5pt] &=\frac{\Gamma (2)}{\tau ^2}\int _{\theta _{m}}^{\theta _M}\frac{1}{\left ({\mathbf{d}}\cdot \hat{\textbf x}+\mathrm{i} {\mathbf{d}}^\perp \cdot \hat{\textbf x} \right )^2}\mathrm{d} \theta -\int _{\theta _{m}}^{\theta _M}I_{\textsf{R}}\mathrm{d}\theta, \end{aligned} \end{equation*}

where $I_\textsf{R}= \int _{h}^{\infty }e^{-\tau ({\mathbf{d}} +\mathrm{i}{\mathbf{d}})\cdot \hat{{\mathbf{x}}}r}r\mathrm{d} r$ . Hence, it can be directly calculated that

\begin{equation*} \int _{\theta _{m}}^{\theta _M}\frac{1}{\left ({\mathbf{d}}\cdot \hat{\textbf x}+\mathrm{i} {\mathbf{d}}^\perp \cdot \hat{\textbf x} \right )^2}\mathrm{d} \theta \geq \frac{\theta _M-\theta _m }{2} \end{equation*}

by using the integral mean value theorem.

With the help of Proposition 2.1, for sufficiently large $\tau$ , we have the following integral inequality

(3.58) \begin{equation} \begin{aligned} \left \vert \int _{S_h} e^{\rho \cdot {\mathbf{x}}}\mathrm{d} {\mathbf{x}} \right \vert &\geq \frac{\Gamma (2)(\theta _{M}-\theta _{m})}{\tau ^2} \frac{1}{\left \vert {\mathbf{d}}\cdot {\mathbf{x}}( \theta _{\xi })+ \mathrm{i} {\mathbf{d}}^{\perp} \cdot \hat{{\mathbf{x}}} (\theta _{\xi }) \right \vert ^2}- \left \vert \int _{\theta _{m}}^{\theta _M} I_{\textsf{R}} \mathrm{d}{\theta} \right \vert \\[5pt] &\geq \frac{\Gamma (2)}{(\theta _{M}-\theta _{m})}\frac{1}{\left ({\mathbf{d}}\cdot \hat{{\mathbf{x}}}(\theta _{\xi }) \right )^2+ \left ({\mathbf{d}}^{\perp} \cdot \hat{{\mathbf{x}}}(\theta _{\xi }) \right )^2}-\int _{\theta _{m}}^{\theta _M} \vert I_{\textsf{R}} \vert \mathrm{d}\theta \\[5pt] &\geq \frac{\widetilde{C_{S_h }}}{\tau ^2}-\frac{1}{\tau }e^{-\frac{1}{2}\zeta h \tau }, \end{aligned} \end{equation}

by using (3.10).

Proof of Corollary 3.3 (a) . Similar to the proof of Theorem 3.1, we have the following integral identity according to (3.30) by noting $\eta \equiv 0$ on $\Gamma _h^\pm$ ,

(3.59) \begin{equation} k^2f_j(0)\int _{S_h}e^{\rho \cdot {\mathbf{x}}}\mathrm{d} {\mathbf{x}}=I_1+I_2+I_3+J_1+J_2, \end{equation}

where

\begin{align*} I_1&=-k^2\int _{S_h}(q-1)(v-v_j)u_0\mathrm{d} {\mathbf{x}},\quad I_2=-k^2\int _{S_h}\delta f_ju_0\mathrm{d}{\mathbf{x}}, \\[5pt] I_3&=-k^2f_j({\mathbf{0}})\int _{S_h}e^{\rho \cdot {\mathbf{x}}}\psi ({\mathbf{x}})\mathrm{d}{\mathbf{x}}, \end{align*}

and $J_1$ , $J_2$ are defined in (3.33), respectively.

By the Sobolev embedding theorem and $q\in H^2(\overline S_{h})$ , we have $q\in C^\alpha (\overline S_{h} )$ , where $\alpha =1$ . Combining (3.59) with (3.35), (3.42) and (3.44), we can deduce that

(3.60) \begin{equation} \begin{aligned} k^2\left [ \frac{\widetilde{C_{S_h }}}{\tau ^2}-\frac{1}{\tau }e^{-\frac{1}{2}\zeta h \tau }\right ] &\vert f_j({\mathbf{0}})\vert \lesssim j^{-\beta }\left [\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{3}} \right ]\\[5pt] &+j^{\gamma }\left [\tau ^{-(\alpha +\frac{29}{12})}+\left (\frac{1}{\tau ^{\alpha +2}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\right ]\\[5pt] &+\tau ^{-\frac{29}{12}} +(1+\tau )(1+\tau ^{-\frac{2}{3}})e^{-\zeta h\tau } \end{aligned} \end{equation}

as $\tau \to \infty$ . Multiplying $\tau ^2$ on the both sides of (3.60), using the assumption (3.5), by letting $\tau =j^s$ , it is easy to see that

(3.61) \begin{equation} k^2\widetilde{C_{S_h }}\vert f_j({\mathbf{0}})\vert \lesssim j^{-\beta +s}+j^{\gamma - \alpha s}. \end{equation}

And under the assumptions $\gamma/\alpha \lt \beta$ , we choose $s\in (\gamma/\alpha,\beta )$ . Letting $j\to \infty$ in (3.61), we obtain that

\begin{equation*}\vert f_j(\textbf {0})\vert =0.\end{equation*}

Since $q({\mathbf{0}})\not =1$ , we finish the proof of this corollary.

Proof. Using polar coordinates transformation and the mean value theorem for integrals, we have

(4.12) \begin{equation} \begin{aligned} \int _{\Gamma _h}e^{\rho \cdot {\mathbf{x}}}\mathrm{d}\sigma &=\sin \theta _0\frac{2\pi \Gamma (2)}{\tau ^2}\frac{1}{(({\mathbf{d}}+\mathrm{i}{\mathbf{d}}^{\perp })\cdot \hat{ {\mathbf{x}}}(\theta _0,\varphi _\xi ))^2}-\sin \theta _0\int _{0}^{2\pi }I_{R}\mathrm{d}\varphi, \end{aligned} \end{equation}

where $I_{R}= \int _{h}^{\infty }e^{-\tau ({\mathbf{d}} +\mathrm{i}{\mathbf{d}})\cdot \hat{{\mathbf{x}}}r}r\mathrm{d} r$ . Furthermore, for sufficiently large $\tau$ , it is ready to know that

\begin{equation*}\frac {1}{\tau ^2}-\frac {1}{\tau }e^{-\frac {1}{2}\zeta h \tau }\gt 0.\end{equation*}

Hence, by virtue of (4.10) and Proposition 2.1, we have the following integral inequality

\begin{equation*} \begin{aligned} \left \vert \int _{\Gamma _h}e^{\rho \cdot {\mathbf{x}}}\mathrm{d}\sigma \right \vert &\geq \sin \theta _0 \frac{2\pi \Gamma (2)}{\tau ^2}\frac{1}{({\mathbf{d}}\cdot \hat{\textbf x}(\theta _{0},\varphi _{\xi }))^2+({\mathbf{d}}^\perp \cdot \hat{\textbf x}(\theta _{0},\varphi _\xi ))^2}-\sin \theta _0\int _{0}^{2\pi }|I_R|\mathrm{d}\varphi \\[5pt] &\geq \frac{C_{\mathcal{C}^h}}{\tau ^2}-\mathcal{O}(\frac{1}{\tau }e^{-\frac{1}{2}\zeta h \tau }), \end{aligned} \end{equation*}

which completes the proof of this proposition.

Proof. Using (4.15) and Lemma 3.2 about the trace theorem, it yields that

\begin{equation*} \|\psi ({\mathbf{x}})\|_{L^{4}(\Lambda _h)} \leq C \|\psi ({\mathbf{x}})\|_{H^{\frac{7}{8},8}(\Lambda _h)}\leq C\|\psi ({\mathbf{x}})\|_{H^{1,8}(\mathcal{C}^h)}= O(\tau ^{-\frac{2}{5}}), \end{equation*}

as $\tau \rightarrow \infty$ . By using polar coordinates transformation, (2.6a) and (4.10), one can derive that

(4.17) \begin{equation} \|e^{\rho \cdot {\mathbf{x}}}\|_{L^t(\Lambda _h)}\lesssim e^{-\zeta h\tau }, \end{equation}

where $\rho$ is defined in (2.3) and $t$ is a positive constant.

Due to polar coordinates transformation, (4.17), (4.15) and Hölder inequality, it can be calculated that

(4.18) \begin{equation} \begin{aligned} \|u_{0}\|_{ L^{2}(\Lambda _{h})} &\leq \|e^{\rho \cdot {\mathbf{x}}}\|_{ L^{2}(\Lambda _{h})}+\|e^{\rho \cdot {\mathbf{x}}}\|_{ L^{4}(\Lambda _{h})}\|\psi ({\mathbf{x}})\|_{ L^{4}(\Lambda _{h})}\\[5pt] &\lesssim \left (1+\tau ^{-\frac{2}{5}}\right )e^{-\zeta h\tau }, \quad \mbox{as } \tau \rightarrow \infty . \end{aligned} \end{equation}

By virtue of Cauchy-Schwarz inequality, (4.16a) and Proposition 2.1, we can deduce that

(4.19) \begin{equation} \begin{aligned} \|\nabla u_{0}\|_{L^{2}(\Lambda _{h})} &\leq \sqrt{2}\tau \|u_{0}\|_{ L^{2}(\Lambda _{h})}+\|e^{\rho \cdot {\mathbf{x}}}\|_{ L^{4}(\Lambda _{h})} \|\nabla \psi ({\mathbf{x}})\|_{ L^{4}(\Lambda _{h})}\\[5pt] &\lesssim (1+\tau )(1+\tau ^{-\frac{2}{5}})e^{-\zeta h\tau },\quad \mbox{as $\tau \rightarrow \infty $. } \end{aligned} \end{equation}

It is clear that we can get the following integral inequality,

(4.20) \begin{equation} \begin{aligned} \int _{\mathcal{C}^h}\vert {\mathbf{x}}\vert ^{\alpha }\vert u_{0} \vert \mathrm{d}{\mathbf{x}} &\leq \int _{\mathcal{C}^h}\vert {\mathbf{x}}\vert ^{\alpha }\vert e^{\rho \cdot {\mathbf{x}}} \vert \mathrm{d}{\mathbf{x}}+\int _{\mathcal{C}^h}(\vert {\mathbf{x}}\vert ^{\alpha }\vert e^{\rho \cdot {\mathbf{x}}}\vert )(\vert \psi ({\mathbf{x}})\vert )\mathrm{d}{\mathbf{x}}. \end{aligned} \end{equation}

By virtue of polar coordinates transformation and Proposition 2.1, it reveals that

(4.21) \begin{equation} \begin{aligned} \int _{\mathcal{C}^h}\vert {\mathbf{x}}\vert ^{\alpha }e^{-\zeta \tau \vert {\mathbf{x}}\vert }\mathrm{d}{\mathbf{x}} \lesssim \frac{1}{\tau ^{\alpha +3}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }, \end{aligned} \end{equation}

as $\tau \rightarrow \infty$ . Next, letting $\textbf{y}=\tau {\mathbf{x}}$ , using Cauchy-Schwarz inequality and H $\ddot{\mathrm{o}}$ lder inequality, it arrives that

(4.22) \begin{align} \int _{\mathcal{C}^h}(\vert {\mathbf{x}}\vert ^{\alpha }\vert e^{\rho \cdot {\mathbf{x}}}\vert )(\vert \psi ({\mathbf{x}})\vert )\mathrm{d}{\mathbf{x}} &\leq \frac{1}{\tau ^{\alpha +3}}\int _{\mathcal{C}}\vert \textbf{y}\vert ^\alpha \vert e^{-{\mathbf{d}}\cdot \textbf{y}}\vert \left \vert \psi \left (\frac{\textbf{y}}{\tau }\right )\right \vert \mathrm{d}\textbf{y}\nonumber \\[5pt] &\leq \frac{1}{\tau ^{\alpha +3}}\|\vert \textbf{y}\vert ^\alpha \vert e^{-{\mathbf{d}}\cdot \textbf{y}}\vert \|_{L^{\frac{8}{7}}(\mathcal{C})}\left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{L^{8}(\mathcal{C})}, \end{align}

as $\tau \rightarrow \infty$ , using variable substitution and (4.15), it arrives that

(4.23) \begin{equation} \left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{L^8(\mathcal{C})} =\tau ^{\frac{3}{8}}\|\psi ({\mathbf{x}})\|_{L^8(\mathcal{C})}\leq \tau ^{\frac{3}{8}}\|\psi ({\mathbf{x}})\|_{H^{1,8}(\mathcal{C})} =\mathcal{O}(\tau ^{-\frac{1}{40}}), \end{equation}

as $\tau \rightarrow \infty$ . Furthermore, one has

(4.24) \begin{equation} \|\vert \textbf{y}\vert ^\alpha \vert e^{\rho \cdot {\mathbf{x}}}\vert \|_{L^{\frac{8}{7}}(\mathcal{C})} =\left ( \int _{\mathcal{C}}\vert \textbf{y}\vert ^{\frac{8}{7}\alpha }e^{-\frac{8}{7}{\mathbf{d}}\cdot \textbf{y}}\mathrm{d}\textbf{y}\right )^{\frac{7}{8}} \leq \left ( \int _{\mathcal{C}}\vert \textbf{y}\vert ^{\frac{8}{7}\alpha }e^{-\frac{8}{7}\zeta \vert \textbf{y}\vert }\mathrm{d}\textbf{y} \right )^{\frac{7}{8}}\leq \frac{C}{\zeta ^{3+\frac{8}{7}\alpha }}, \end{equation}

where $C=2\pi \theta _0 \Gamma (3+\frac{8}{7}\alpha )(\frac{7}{8})^{3+\frac{8}{7}\alpha }$ . Hence, $\|\vert \textbf{y}\vert ^\alpha \vert e^{\rho \cdot {\mathbf{x}}}\vert \|_{L^{\frac{8}{7}}(\mathcal{C})}$ is a positive constant which only depends on $\theta _0$ , $\zeta$ and $\alpha$ . Combining (4.23), (4.22) and (4.21) with (4.20), one has (4.16c).

Furthermore, we have

(4.25) \begin{equation} \int _{\Gamma _h}\vert {\mathbf{x}}\vert ^\alpha \vert u_0\vert \mathrm{d} \sigma \leq \int _{\Gamma _h}\vert {\mathbf{x}}\vert \vert e^{\rho \cdot {\mathbf{x}}}\vert \mathrm{d}\sigma +\int _{\Gamma _h}\vert {\mathbf{x}}\vert ^\alpha \vert e^{\rho \cdot {\mathbf{x}}}\vert \vert \psi ({\mathbf{x}})\vert \mathrm{d}\sigma, \end{equation}

and we can easily get (4.26) by using polar coordinates transformation and Proposition 2.1,

(4.26) \begin{equation} \int _{\Gamma _h}\vert {\mathbf{x}}\vert \vert e^{\rho \cdot {\mathbf{x}}}\vert \mathrm{d}\sigma \lesssim \frac{1}{\tau ^{\alpha +2}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }, \end{equation}

as $\tau \rightarrow \infty$ . Then letting $\textbf{y}=\tau {\mathbf{x}}$ and utilizing Hölder inequality, it can be obtained that

(4.27) \begin{align} \int _{\Gamma _h}\vert {\mathbf{x}}\vert ^\alpha \vert e^{\rho \cdot {\mathbf{x}}}\vert \vert \psi ({\mathbf{x}})\vert \mathrm{d}\sigma &\leq \frac{1}{\tau ^{\alpha +2}}\int _{\partial \mathcal{C}}\vert \textbf{y} \vert ^\alpha \vert e^{-{\mathbf{d}} \cdot \textbf{y}}\vert \left \vert \psi \left (\frac{\textbf{y}}{\tau }\right )\right \vert \mathrm{d} \sigma \nonumber \\[5pt] &\leq \frac{1}{\tau ^{\alpha +2}} \|\vert \textbf{y} \vert ^\alpha \vert e^{-{\mathbf{d}}\cdot \textbf{y}}\vert \|_{L^{\frac{8}{7}}(\partial \mathcal{C})}\left \| \psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{L^8(\partial \mathcal{C})}. \end{align}

Similar to (4.24), we know that $\|\vert \textbf{y} \vert ^\alpha \vert e^{-{\mathbf{d}}\cdot \textbf{y}}\vert \|_{L^{\frac{8}{7}}(\partial \mathcal{C})}$ is a positive constant. By virtue of variable substitution, trace theorem and (4.15), it arrives that

(4.28) \begin{equation} \left \| \psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{L^8(\partial \mathcal{C})} \lesssim \tau ^{\frac{1}{4}}\|\psi ({\mathbf{x}})\|_{H^{\frac{7}{8},8}(\partial \mathcal{C})}\lesssim \tau ^{\frac{1}{4}}\|\psi ({\mathbf{x}})\|_{H^{1,8}( \mathcal{C})}\lesssim \tau ^{-\frac{3}{20}}, \end{equation}

as $\tau \rightarrow \infty$ . Combining (4.26), (4.27) and (4.28) with (4.25), one has (4.16d).

Proof of Theorem 4.1 . The proof of this theorem is similar to the counterpart of Theorem 3.1. Recall that $(v,w)$ is a pair of transmission eigenfunctions to (1.1). Using Green formula (2.7) and boundary conditions in (4.14), the following integral identity holds

(4.29) \begin{equation} \int _{\mathcal{C}^h}k^2(q-1)vu_0\mathrm{d}{\mathbf{x}}=\int _{\Lambda _h}(w-v)\partial _\nu u_0-u_0\partial (w-v)\mathrm{d}\sigma -\int _{\Gamma _h}\eta u_0v\mathrm{d}\sigma \end{equation}

where $\mathcal{C}^h$ , $\Lambda _h$ and $\Gamma _h$ are defined by (4.2). Let

\begin{equation*} f_{j}=(q-1)v_j. \end{equation*}

Due to $q\in H^2{(\overline{\mathcal{C}^h})}$ , we know that $q\in C^{1/2}(\overline{ \mathcal{C}^h})$ by using the property of embedding of Sobolev space. Recall that $\eta \in C^{\alpha _1}(\overline{ \Gamma _h})$ . Let $\alpha =\{ \alpha _1,1/2 \}$ . Furthermore, since the Herglotz wave function $v_j\in C^{\alpha }(\overline{ \mathcal{C}^h})$ , it yields that $f_j\in C^\alpha (\overline{\mathcal{C}}^h)$ . Hence, one has the expansion

(4.30) \begin{equation} \begin{aligned} f_{j}&=f_j({\mathbf{0}})+\delta f_j,\ \vert \delta f_{j}\vert \leq \| f_{j}\|_{C^{\alpha }(\overline{\mathcal{C}^h })}\vert {\mathbf{x}} \vert ^{\alpha },\\[5pt] v_{j}&=v_j({\mathbf{0}})+\delta v_j,\ \vert \delta v_j\vert \leq \| v_j\|_{C^\alpha (\overline{\mathcal{C}^h })}\vert {\mathbf{x}}\vert ^\alpha,\\[5pt] \eta &=\eta ({\mathbf{0}})+\delta \eta,\ \vert \delta \eta \vert \leq \|\eta \|_{C^\alpha (\overline{\Gamma _h})}\vert {\mathbf{x}}\vert ^\alpha . \end{aligned} \end{equation}

By virtue of (4.30), we have the following integral identities

(4.31) \begin{equation} k^2\int _{\mathcal{C}_h}(q-1)vu_0\mathrm{d}{\mathbf{x}}=-\sum _{m=1}^3I_m,\quad \int _{\Gamma _h}\eta u_0v\mathrm{d}\sigma =I-\sum _{m=4}^9I_m, \end{equation}

where

(4.32) \begin{equation} \begin{aligned} I_1&=-k^2\int _{\mathcal{C}^h}(q-1)(v-v_j)u_0\mathrm{d}{\mathbf{x}},\quad I_2=-\int _{\mathcal{C}^h}\delta f_j u_0\mathrm{d}{\mathbf{x}},\\[5pt] I_3&=-f_j({\mathbf{0}})\int _{\mathcal{C}^h}u_0\mathrm{d}{\mathbf{x}},\quad I_4=-\eta ({\mathbf{0}})\int _{\Gamma _h}(v-v_j)u_0\mathrm{d} \sigma \\[5pt] I_5&=-\int _{\Gamma _h}\delta \eta (v-v_j)u_0\mathrm{d}\sigma,\quad I_6=-\eta ({\mathbf{0}})v_j({\mathbf{0}})\int _{\Gamma _h}e^{\rho \cdot {\mathbf{x}}}\psi ({\mathbf{x}})\mathrm{d}\sigma, \\[5pt] I_7&=\eta ({\mathbf{0}})\int _{\Gamma _h}\delta v_ju_0\mathrm{d}\sigma,\quad I_8=-v_j({\mathbf{0}})\int _{\Gamma _h}\delta \eta u_0\mathrm{d}\sigma,\\[5pt] I_9&= \int _{\Gamma _h}\delta \eta \delta v_ju_0\mathrm{d}\sigma,\quad I=\eta ({\mathbf{0}})v_j({\mathbf{0}})\int _{\Gamma _h}e^{\rho \cdot {\mathbf{x}}}\mathrm{d}\sigma . \end{aligned} \end{equation}

Substituting (4.31) into (4.29), it yields that

\begin{equation*} I=\sum _{m=1}^9I_m+J_1+J_2, \end{equation*}

where

(4.33) \begin{equation} J_1=\int _{\Lambda _{h}}u_{0}\partial _{\nu }(w-v)\mathrm{d}\sigma,\quad J_2=-\int _{\Lambda _{h}}(w-v)\partial _{\nu }u_{0}\mathrm{d}\sigma . \end{equation}

Hence, it readily yields that

(4.34) \begin{equation} \vert I\vert \leq \sum _{m=1}^9\vert I_m\vert +\vert J_1\vert +\vert J_2\vert . \end{equation}

In the sequel, we derive the asymptotic estimates of $I_j$ $(j=1,\ldots,9)$ and $J_j,\ j=1,2$ with respect to the parameter $\tau$ in the CGO solution $u_0$ when $\tau \rightarrow \infty$ , separately. Using Hölder inequality, Proposition 4.2 and (4.15), it is clear that

(4.35) \begin{equation} \begin{aligned} \vert I_1\vert &\leq \|v-v_j\|_{L^2(\mathcal{C}^h)}\|e^{\rho \cdot {\mathbf{x}}}\|_{L^2(\mathcal{C}^h)}+\|v-v_j\|_{L^2(\mathcal{C}^h)}\|e^{\rho \cdot {\mathbf{x}}}\psi ({\mathbf{x}})\|_{L^2(\mathcal{C}^h)}\\[5pt] &\lesssim j^{-\beta }\left [\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{5}} \right ], \end{aligned} \end{equation}

as $\tau \rightarrow \infty$ .

With the help of (4.30), we have

(4.36) \begin{equation} \vert I_{2} \vert \leq k^{2}\|f_{j} \|_{C^{\alpha }}\int _{\mathcal{C}^h}\vert {\mathbf{x}}\vert ^{\alpha }\vert u_{0} \vert \mathrm{d}{\mathbf{x}}, \end{equation}

and

(4.37) \begin{equation} \begin{aligned} \|f_{j}\|_{C^{\alpha }(\mathcal{C}^h )}& \leq \|q\|_{C^{\alpha }(\mathcal{C}^h )}\sup _{\mathcal{C}^h }\vert v_{j}\vert +\|v_{j}\|_{C^{\alpha }(\mathcal{C}^h )}\sup _{\mathcal{C}^h }\vert q-1\vert . \end{aligned} \end{equation}

Moreover, due to the property of compact embedding of Hölder spaces, one has

(4.38) \begin{equation} \| v_{j}\|_{C^{\alpha }(\mathcal{C}^h )}\leq \mathrm{ diam}\left (\mathcal{C}^h\right )^{1-\alpha }\|v_{j}\|_{C^{1}(\mathcal{C}^h )}, \end{equation}

where diam( $\mathcal{C}^h$ ) is the diameter of $\mathcal{C}^h$ . It can be directly shown that

(4.39) \begin{equation} \|v_{j}\|_{C^{1}(\mathcal{C}^h )}\leq 4\sqrt{\pi }(1+k)\|g\|_{L^{2}(\mathbb S^{2})}. \end{equation}

On the other hand, we can obtain the following estimate by using the Cauchy-Schwarz inequality,

(4.40) \begin{equation} \vert v_{j}\vert \leq 4\sqrt{\pi }\|g\|_{L^{2}(\mathbb S^{2})}. \end{equation}

Using (4.7) and $q\in C^{\alpha }(\overline{\mathcal{C}^h})$ , plugging (4.7), (4.38), (4.39) and (4.40) into (4.37), one can arrive at

(4.41) \begin{equation} \|f_{j}\|_{C^{\alpha }(\mathcal{C}^h)}\lesssim j^{\gamma }, \end{equation}

where $\gamma$ is a given positive constant defined in (4.7). Substituting (4.16c) and (4.41) into (4.36), we obtain

(4.42) \begin{equation} \vert I_{2} \vert \lesssim j^{\gamma }\left [\tau ^{-(\alpha +\frac{121}{40})}+\left (\frac{1}{\tau ^{\alpha +3}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\right ] \end{equation}

as $\tau \to \infty$ .

With the help of Cauchy-Schwarz inequality and (4.7), it yields that

(4.43) \begin{equation} \vert I_{3} \vert \leq \int _{\mathcal{C}^h} \vert e^{\rho \cdot {\mathbf{x}}}\vert \mathrm{d}{\mathbf{x}}+\int _{\mathcal{C}}\vert e^{\rho \cdot {\mathbf{x}}}\vert \vert \psi ({\mathbf{x}})\vert \mathrm{d}{\mathbf{x}}, \end{equation}

Similar to (4.24), we have that ${\|e^{-{\mathbf{d}}\cdot \textbf{y}}\|}_{L^{\frac{8}{7}}(\mathcal{C})}$ is a positive constant depending only on $\zeta$ and $\theta _0$ . Letting $\textbf{y}=\tau {\mathbf{x}}$ and using (4.23), it can be calculated that

(4.44) \begin{equation} \begin{aligned} \int _{\mathcal{C}}\vert e^{\rho \cdot {\mathbf{x}}}\vert \vert \psi ({\mathbf{x}})\vert \mathrm{d}{\mathbf{x}} \leq \frac{1}{\tau ^3}\|e^{-{\mathbf{d}}\cdot \textbf{y}}\|_{L^{\frac{8}{7}}(\mathcal{C})}\left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{L^8(\mathcal{C})} \lesssim \tau ^{-\frac{121}{40}}, \end{aligned} \end{equation}

as $\tau \rightarrow \infty$ . Therefore, with the help of Proposition 4.2, and plugging (4.44) into (4.43), one has

(4.45) \begin{equation} \vert I_{3} \vert \lesssim \tau ^{-\frac{121}{40}}+\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ),\quad \mbox{ as } \tau \rightarrow \infty . \end{equation}

By virtue of Cauchy-Schwarz inequality and Lemma 3.2, we can obtain that

(4.46) \begin{equation} \begin{aligned} \vert I_4 \vert &\lesssim \|v-v_j\|_{L^{2}(\Gamma _h)}\|e^{\rho \cdot {\mathbf{x}}}\|_{L^2(\Gamma _h)}+\|v-v_j\|_{L^2(\Gamma _h)}\|e^{\rho \cdot {\mathbf{x}}}\|_{L^4(\Gamma _h)}\|\psi ({\mathbf{x}} )\|_{L^4(\Gamma _h)}\\[5pt] &\lesssim j^{-\beta }\left [\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{5}}\right ] \end{aligned} \end{equation}

as $\tau \to \infty$ .

With the help of Cauchy-Schwarz inequality, Lemma 3.2 and Hölder inequality, one has

(4.47) \begin{equation} \begin{aligned} \vert I_5\vert &\lesssim \|v-v_j\|_{L^2(\Gamma _h)}(\| \vert e^{\rho \cdot {\mathbf{x}}}\vert \vert {\mathbf{x}}\vert ^\alpha \|_{L^2(\Gamma _h)}+\|\vert e^{\rho \cdot {\mathbf{x}}}\vert \vert {\mathbf{x}}\vert ^\alpha \|_{L^4(\Gamma _h)}\|\psi ({\mathbf{x}})\|_{L^{4}(\Gamma _h)})\\[5pt] &\lesssim j^{-\beta }\left [\left (\frac{1}{\tau ^{(2\alpha +2)}}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^{(4\alpha +2)}}+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{5}}\right ], \end{aligned} \end{equation}

as $\tau \rightarrow \infty$ .

Similar to (3.19), it can be directly obtained that

(4.48) \begin{equation} \left \| \psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{H^{1,8}(\mathcal{C})}\leq \tau ^{\frac{3}{8}}\|\psi ({\mathbf{x}})\|_{H^{1,8}(\mathcal{C})}=\mathcal{O}(\tau ^{-\frac{1}{40}}), \end{equation}

as $\tau \rightarrow \infty$ . Therefore, following the proof of Lemma 3.3 and using Hölder inequality and Lemma 3.2, we have

(4.49) \begin{equation} \begin{aligned} \vert I_6\vert & \lesssim \frac{1}{\tau ^2}\|e^{-{\mathbf{d}}\cdot \textbf{y}}\|_{L^{\frac{8}{7}}(\Gamma )}\left \|\psi \left (\frac{\textbf{y}}{\tau }\right )\right \|_{L^8(\Gamma )}\\[5pt] &\lesssim \frac{1}{\tau ^2} \|e^{-{\mathbf{d}}\cdot \textbf{y}}\|_{L^{\frac{8}{7}}(\Gamma )} \tau ^{\frac{3}{8}}\|\psi ({\mathbf{x}})\|_{H^{1,8}(\mathcal{C})} \lesssim \tau ^{-\frac{81}{40}}, \quad \mbox{as $\tau \rightarrow \infty $. } \end{aligned} \end{equation}

Moreover, we have the following estimates for $I_7,\ I_8$ and $I_9$ by virtue of (4.16d) directly,

(4.50) \begin{align} \vert I_7\vert &\lesssim \|v_j\|_{C^\alpha (\Gamma _h)}\int _{\Gamma _h}\vert {\mathbf{x}}\vert ^\alpha \vert u_0\vert \mathrm{d}\sigma \lesssim j^\gamma \left [ \tau ^{-(\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ) \right ], \nonumber \\[5pt] \vert I_8\vert & \lesssim \tau ^{-(\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ),\nonumber \\[5pt] \vert I_9\vert &\lesssim j^\gamma \left [ \tau ^{-(2\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{2\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\right ],\quad \mbox{as $\tau \rightarrow \infty $. } \end{align}

Using Cauchy-Schwarz inequality and Lemma 3.2, we obtain that

(4.51) \begin{equation} \begin{aligned} \vert J_1 \vert &\leq \|u_{0}\|_{H^{\frac{1}{2}}(\Lambda _{h})} \|\partial _{\nu }(w-v)\|_{H^{-\frac{1}{2}}(\Lambda _{h})} \leq C\|u_{0}\|_{H^{1}(\Lambda _{h})} \|\partial _{\nu }(w-v)\|_{H^{1}(\Lambda _{h})}\\[5pt] &\lesssim \|u_{0}\|_{H^{1}(\Lambda _{h})} \end{aligned} \end{equation}

as $\tau \to \infty$ , where $C$ is a positive constant arising from the trace theorem. By virtue of (4.16a) and (4.16b), it can be calculated that

(4.52) \begin{equation} \begin{aligned} \vert J_1 \vert \lesssim (1+\tau )(1+\tau ^{-\frac{2}{5}})e^{-\zeta h\tau } \end{aligned} \end{equation}

as $\tau \to \infty$ , where $\zeta$ is a positive constant given in (4.10). Finally, using Cauchy-Schwarz inequality, the trace theorem and (4.16b), we can obtain that

(4.53) \begin{equation} \begin{aligned} \vert J_2 \vert &\leq \|\partial _{\nu } u_0\|_{ L^{2}(\Lambda _{h})} \|w-v\|_{ L^{2}(\Lambda _{h})} \leq C\|\partial _{\nu } u_0\|_{ L^{2}(\Lambda _{h})} \|w-v\|_{ H^{1}(\mathcal{C}^h)}\\[5pt] &\lesssim (1+\tau )(1+\tau ^{-\frac{2}{5}})e^{-\zeta h\tau }, \end{aligned} \end{equation}

as $\tau \rightarrow \infty$ .

Substituting (4.35), (4.42), (4.45) $-$ (4.50), (4.52) and (4.53) into (4.34), we have

(4.54) \begin{align} \vert \eta ({\mathbf{0}})v_j({\mathbf{0}})\vert &\left (\frac{C_{\mathcal{C}^h}}{\tau ^2}-\frac{1}{\tau }e^{-\frac{1}{2}\zeta h \tau }\right ) \lesssim j^{-\beta }\left [\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-4\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{5}} \right ]\nonumber \\[5pt] & +j^{\gamma }\left [\tau ^{-(\alpha +\frac{121}{40})}+\left (\frac{1}{\tau ^{\alpha +3}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\right ]\nonumber \\[5pt] & +j^{-\beta }\left [\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^2}+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{5}}\right ]\nonumber \\[5pt] & +j^{-\beta }\left [\left (\frac{1}{\tau ^{(2\alpha +2)}}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^{(4\alpha +2)}}+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{5}}\right ]\nonumber \\[5pt] & +(j^\gamma +1)\left [ \tau ^{-(\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{\alpha +2}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ) \right ]\nonumber \\[5pt] & +j^\gamma \left [\tau ^{-(2\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{2\alpha +2}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\right ]\nonumber \\[5pt] & +\tau ^{-\frac{81}{40}}+\tau ^{-\frac{121}{40}}+\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ) +(1+\tau )(1+\tau ^{-\frac{2}{5}})e^{-\zeta h\tau } \end{align}

as $\tau \to \infty$ , where $C_{\mathcal{C}^h}$ is a positive constant given in (4.11). Moreover, for sufficiently large $\tau$ , we know that

\begin{equation*}\frac {C_{\mathcal {C}^h}}{\tau ^2}-\frac {1}{\tau }e^{-\frac {1}{2}\zeta h \tau }\gt 0.\end{equation*}

Hence, multiplying $\tau ^2$ on both sides of (4.54) and taking $\tau =j^s$ and $s\gt 0$ , we derive that

(4.55) \begin{equation} \begin{aligned} \left (C_{\mathcal{C}^h}-j^s e^{-\frac{1}{2}\zeta hj^s}\right )\vert \eta ({\mathbf{0}})v_j({\mathbf{0}})\vert &\lesssim j^{-\beta +\frac{17}{20}s}+j^{\gamma -(\alpha +1)s}+j^{-\beta +\frac{11}{10}s}\\[5pt] &+j^{-\beta +(-\alpha +\frac{11}{10})s} +j^{\gamma -\alpha s}+j^{\gamma -2\alpha s} \end{aligned} \end{equation}

as $\tau \to \infty$ . Recalling that $\gamma/\alpha \lt \frac{10}{11}\beta$ , we can choose $s\in (\gamma/\alpha,\frac{10}{11}\beta )$ . Hence in (4.55), by letting $j\to \infty$ , we prove that

\begin{equation*}\lim _{j\to \infty }\vert \eta (\textbf {0})v_j(\textbf {0})\vert =0.\end{equation*}

Since $\eta ({\mathbf{0}})\not =0$ , we have $\lim _{j\to \infty }\vert v_j({\mathbf{0}}) \vert =0$ . Using (4.6) and integral mean value theorem, we can obtain (4.8).

The proof is complete.

Proof. By direct calculations, we have

(4.58) \begin{equation} \sum _{j=1}^3 \frac{1}{z_j^2}=\frac{S(\theta _\xi )}{z_1^2},\quad S(\theta _\xi )=1+\left (\frac{z_1}{\bar z_1}\right )^2 +c_1 z_4, \end{equation}

where

\begin{align*} c_1=&\frac{|z_1|^4}{||z_1|^2-\sin \theta _\xi \cos \theta _\xi +\mathrm{i} \cos \theta _\xi (\!\cos \theta _\xi +\sin \theta _\xi )|^4},\\[5pt] z_4=&\left (|z_1|^2-1/2\sin 2\theta _\xi -\mathrm{i} \cos \theta _\xi (\!\cos \theta _\xi +\sin \theta _\xi ) \right )^2. \end{align*}

By noting $\theta _\xi \in (0,\pi/2)$ , it yields that $c_1\geq 0.05\sin \theta _\xi$ and $\Re (z_4) \geq 2 \sin ^2 \theta _\xi$ . Hence according to (4.58), we obtain (4.57).

Proof. Since $\mathcal{K}$ is a cuboid, by the geometrical setup and notations in this subsection, we have $\Gamma _h=\Gamma _{h1}\cup \Gamma _{h2}\cup \Gamma _{h3}$ , where $\Gamma _{h1}\;:\!=\;\partial \mathcal{K}_1\cap B_h,\ \Gamma _{h2}\;:\!=\;\partial \mathcal{K}_2\cap B _h,\ \Gamma _{h3}\;:\!=\;\partial \mathcal{K}_3\cap B_h$ .

According to Proposition 2.1, it can be derived that

\begin{equation*} \begin{aligned} \int _{\Gamma _h}e^{\rho \cdot {\mathbf{x}}} \mathrm{d} \sigma &=\int _{\Gamma _{h1}}e^{\rho \cdot {\mathbf{x}}} \mathrm{d} \sigma +\int _{\Gamma _{h2}}e^{\rho \cdot {\mathbf{x}}} \mathrm{d} \sigma +\int _{\Gamma _{h3}}e^{\rho \cdot {\mathbf{x}}} \mathrm{d} \sigma \\[5pt] &= \frac{1}{2\pi \tau ^2} \sum _{j=1}^3 \frac{1}{(\rho _1\cdot \hat{{\mathbf{x}}}_j(\theta _\xi ))^2} -\mathcal{O}(\frac{1}{\tau }e^{-\frac{1}{2}\zeta h \tau }), \end{aligned} \end{equation*}

where $\theta _\xi \in (0,\pi/2)$ is fixed. By virtue of Proposition 4.3, we complete the proof.

The proof of Theorem 4.3 . Using the fact that $f=(q-1)v\in C^\alpha (\overline{\mathcal{K}^h}),\ v\in C^\alpha (\overline{\mathcal{K}^h}),\eta \in C^\alpha (\overline{\Gamma _h})$ , we have the following expansion

(4.60) \begin{equation} \begin{aligned} f&=f({\mathbf{0}})+\delta f,\quad \vert \delta f \vert \leq \|f\|_{C^\alpha }\vert {\mathbf{x}} \vert ^\alpha,\\[5pt] v&=v({\mathbf{0}})+\delta v,\quad \vert \delta v \vert \leq \|v\|_{C^\alpha }\vert {\mathbf{x}} \vert ^\alpha,\\[5pt] \eta &=\eta ({\mathbf{0}})+\delta \eta, \quad \vert \delta \eta \vert \leq \|\eta \|_{C^\alpha }\vert {\mathbf{x}}\vert ^\alpha . \end{aligned} \end{equation}

Combining the integral identity (4.29) with (4.60), it arrives that

(4.61) \begin{equation} \eta ({\mathbf{0}})v({\mathbf{0}})\int _{\Gamma _h}e^{\rho \cdot {\mathbf{x}}}\mathrm{d} \sigma =\sum _{i=1}^6 I_i+J_1+J_2, \end{equation}

where

(4.62) \begin{equation} \begin{aligned} I_1&=f({\mathbf{0}})\int _{\mathcal{K}^h}u_0\mathrm{d}{\mathbf{x}},\quad I_2= \int _{\mathcal{K}^h}\delta f u_0\mathrm{d}{\mathbf{x}},\quad I_3=\eta ({\mathbf{0}})v({\mathbf{0}})\int _{\Gamma _h}\psi ({\mathbf{x}})e^{\rho \cdot {\mathbf{x}}}\mathrm{d}\sigma,\\[5pt] I_4&=\eta ({\mathbf{0}})\int _{\Gamma _h}\delta v u_0\mathrm{d}\sigma,\quad I_5=v({\mathbf{0}})\int _{\Gamma _h}\delta \eta u_0\mathrm{d}\sigma,\quad I_6=\int _{\Gamma _h}\delta \eta \delta v u_0\mathrm{d}\sigma,\\[5pt] J_1&=-\int _{\Lambda _h}(w-v)\partial _{\nu }u_0\mathrm{d}\sigma,\quad J_2=\int _{\Lambda _h}u_0\partial _{\nu }(w-v)\mathrm{d}\sigma . \end{aligned} \end{equation}

There must exist a convex conic cone $\mathcal{C}$ contains the cuboid cone $\mathcal{K}$ , namely $\mathcal{K} \subset \mathcal{C}$ . Hence, by virtue of (4.45) and (4.16d), we have

(4.63) \begin{equation} \vert I_1\vert \leq \vert f({\mathbf{0}})\vert \int _{\mathcal{K}^h}\vert u_0\vert \mathrm{d}{\mathbf{x}} \leq \vert f({\mathbf{0}})\vert \int _{\mathcal{C}^h}\vert u_0\vert \mathrm{d}{\mathbf{x}} \lesssim \tau ^{-\frac{121}{40}}+\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ), \end{equation}

and

(4.64) \begin{equation} \vert I_2\vert \leq \int _{\mathcal{K}^h} \vert \delta f u_0\vert \mathrm{d}{\mathbf{x}} \leq \int _{\mathcal{C}^h} \vert \delta f u_0\vert \mathrm{d}{\mathbf{x}} \lesssim \tau ^{-(\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ), \end{equation}

as $\tau \rightarrow \infty$ .

In view of (4.49), we have

(4.65) \begin{equation} \vert I_3\vert \lesssim \tau ^{-\frac{81}{40}}. \end{equation}

In addition, by using (4.16d) in Lemma 4.2, we have the following inequalities:

(4.66) \begin{align} \vert I_4\vert &\lesssim \tau ^{-(\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ), \end{align}

(4.67) \begin{align} \vert I_5 \vert &\lesssim \tau ^{-(\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ), \end{align}

(4.68) \begin{align} \vert I_6\vert &\lesssim \tau ^{-(2\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{2\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ), \end{align}

as $\tau \rightarrow \infty$ . Moreover, by using (4.52) and (4.53), we have

(4.69) \begin{equation} \vert J_1 \vert \lesssim (1+\tau )(1+\tau ^{-\frac{2}{5}})e^{-\zeta h\tau } \end{equation}

and

(4.70) \begin{equation} \vert J_2 \vert \lesssim (1+\tau )(1+\tau ^{-\frac{2}{5}})e^{-\zeta h\tau }. \end{equation}

as $\tau \rightarrow \infty$ . Let $\rho$ be defined in (2.3) with ${\mathbf{d}}=(1,1,1)^\top$ and ${\mathbf{d}}^\perp =(1,-1,0)^\top$ . By Proposition 4.4, one has (4.59). Plugging (4.63)-(4.70) and (4.59) into (4.61), it arrives that

(4.71) \begin{equation} \begin{aligned} \vert \eta ({\mathbf{0}}) v({\mathbf{0}})\vert &\left (\frac{C^{'}_{\mathcal{K}^h}}{\tau ^2}-\frac{1}{\tau }e^{-\frac{1}{2}\zeta h \tau }\right ) \lesssim \tau ^{-\frac{121}{40}}+\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ) +\tau ^{-(\alpha +\frac{43}{20})}\\[5pt] &+\left (\frac{1}{\tau ^{\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ) +\tau ^{-\frac{81}{40}} +\tau ^{-(\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\\[5pt] &+\tau ^{-(2\alpha +\frac{43}{20})}+\left (\frac{1}{\tau ^{2\alpha +2}} +\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\\[5pt] &+(1+\tau )(1+\tau ^{-\frac{2}{5}})e^{-\zeta h\tau }, \end{aligned} \end{equation}

where the positive constant $C^{'}_{\mathcal{K}^h}$ not depending on $\tau$ is defined in (4.59). Multiplying $\tau ^2$ on both sides of (4.71) and letting $\tau \to \infty$ , one has

\begin{equation*}\vert \eta (\textbf {0})v(\textbf {0})\vert =0.\end{equation*}

Due to $\eta ({\mathbf{0}})\not =0$ , we complete the proof of Theorem 4.3.

Proof of Corollary 4.4 (a). The following integral identity can be obtained according to (4.29):

(4.73) \begin{equation} k^2f_j({\mathbf{0}})\int _{\mathcal{C}^h}e^{\rho \cdot {\mathbf{x}}}=I_1+I_2+I_3+J_1+J_2, \end{equation}

where $I_m,\ m=1,2,3$ , $J_1$ and $J_2$ defined in (4.32).

With the help of (4.35), (4.42), (4.45), (4.52) and Proposition 4.5, we have the following integral inequality

(4.74) \begin{equation} \begin{aligned} k^2\left [\frac{\widetilde{C_{\mathcal{C}^h}}}{\tau ^3}-\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right ]&\vert f_j({\mathbf{0}})\vert \lesssim j^{-\beta }\left [\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-\zeta h\tau }\right )^{\frac{1}{2}}+\left (\frac{1}{\tau ^3}+\frac{1}{\tau }e^{-2\zeta h\tau }\right )^{\frac{1}{4}}\tau ^{-\frac{2}{5}} \right ]\\[5pt] &+j^{\gamma }\left [\tau ^{-(\alpha +\frac{121}{40})}+\left (\frac{1}{\tau ^{\alpha +3}}+\frac{1}{\tau }e^{-\frac{1}{2}\zeta h\tau }\right )\right ]\\[5pt] &+\tau ^{-\frac{121}{40}}+(1+\tau )(1+\tau ^{-\frac{2}{5}})e^{-\zeta h\tau } \end{aligned} \end{equation}

as $\tau \to \infty$ . For sufficiently large $\tau$ , we know that

\begin{equation*} \frac {\widetilde {C_{\mathcal {C}^h}}}{\tau ^3}-\frac {1}{\tau }e^{-\frac {1}{2}\zeta h\tau }\gt 0.\end{equation*}

Then, multiplying $\tau ^3$ in the both sides of (4.74) and letting $\tau \to \infty$ and $\tau =j^s$ , one has

(4.75) \begin{equation} k^2\widetilde{C_{\mathcal{C}^h}}\vert f_j({\mathbf{0}})\vert \lesssim j^{-\beta +\frac{37}{20}s}+j^{\gamma -\alpha s}. \end{equation}

Due to the assumption that $\gamma \lt \frac{20}{37}\alpha \beta$ , we choose $s\in (\gamma/\alpha, \frac{20}{37}\beta )$ . By letting $j\to \infty$ , we have

\begin{equation*}\vert f_j(\textbf {0})\vert =0.\end{equation*}

Since $q({\mathbf{0}})\neq 1$ , the proof of this corollary is complete.