and we introduce the operator $M_{r_\ast}\equiv(M_{r_\ast,1},M_{r_\ast,2})$ from $C^{0,\alpha}(\partial \Omega^i_1)\times C^{0,\alpha}(\partial \Omega^i_2)$ to itself by setting $$\begin{split}
M_{r_\ast,1}&[\theta^{i}_1,\theta^{i}_2](t)\equiv \frac{1}{2}\theta^{i}_1(t)+\int_{\partial \Omega^i_1}DS_n(t-s)\nu_{\Omega^i_1}(t)\theta^{i}_1(s)\, d\sigma_s\\ 
&  +r_\ast^{n-1}\int_{\partial \Omega^i_2}DS_n\big((p^1-p^2)+r_\ast (t-s)\big)\nu_{\Omega^i_1}(t)\theta^{i}_2(s)\, d\sigma_s \qquad \forall t \in \partial \Omega^i_1\, ,
\end{split}$$ $$\begin{split}
M_{r_\ast,2}&[\theta^{i}_1,\theta^{i}_2](t)\equiv \frac{1}{2}\theta^{i}_2(t)+\int_{\partial \Omega^i_2}DS_n(t-s)\nu_{\Omega^i_2}(t)\theta^{i}_2(s)\, d\sigma_s\\ 
&  +r_\ast^{n-1}\int_{\partial \Omega^i_1}DS_n\big((p^2-p^1)+r_\ast(t-s)\big)\nu_{\Omega^i_2}(t)\theta^{i}_1(s)\, d\sigma_s \qquad \forall t \in \partial \Omega^i_2\, ,
\end{split}$$ for all $(\theta^i_1,\theta^i_2) \in C^{0,\alpha}(\partial \Omega^i_1)\times C^{0,\alpha}(\partial \Omega^i_2)$. We need to show that there exists a unique pair $(\theta^i_1,\theta^i_2)$ such that $$\begin{split}
M_{r_\ast,1}&[\theta^{i}_1,\theta^{i}_2](t) =\int_{\partial \Omega^o}DS_n(y)\nu_{\Omega^i_1}(t)\tilde{\theta}^{o}(y)\, d\sigma_y +f_1(t)\qquad \forall t \in \partial \Omega^i_1\, ,
\end{split}$$ $$\begin{split}
M_{r_\ast,2}&[\theta^{i}_1,\theta^{i}_2](t) =\int_{\partial \Omega^o}DS_n(y)\nu_{\Omega^i_2}(t)\tilde{\theta}^{o}(y)\, d\sigma_y +f_2(t)\qquad \forall t \in \partial \Omega^i_2\, .
\end{split}$$ In order to do so, it clearly suffices to show that the operator $M_{r_\ast}$ is invertible. If $r_\ast=0$, the invertibility follows immediately by Lemma \[lem:smp\] (iii). If $r_\ast > 0$, we note that $$\begin{split}
M_{r_\ast,1}&[\theta^{i}_1,\theta^{i}_2]\big((x-p^1)/r_{\ast}\big)= \frac{1}{2}\theta^{i}_1\big((x-p^1)/r_{\ast}\big)\\\ +&\int_{\partial (p^1+r_\ast \Omega^i_1)}DS_n(x-y)\nu_{p^1+r_\ast \Omega^i_1}(x)\theta^{i}_1\big((y-p^1)/r_{\ast}\big)\, d\sigma_y\\ 
 +&\int_{\partial (p^2+r_\ast\Omega^i_2)}DS_n(x-y)\nu_{p^1+r_\ast \Omega^i_1}(x)\theta^{i}_2\big((y-p^2)/r_{\ast}\big)\, d\sigma_y \\ 
 & \qquad \qquad \qquad \qquad \qquad  \qquad \qquad \qquad \forall x \in \partial (p^1+r_\ast \Omega^i_1)\, ,
\end{split}$$ $$\begin{split}
M_{r_\ast,2}&[\theta^{i}_1,\theta^{i}_2]\big((x-p^2)/r_{\ast}\big)= \frac{1}{2}\theta^{i}_2\big((x-p^2)/r_{\ast}\big)\\\ +&\int_{\partial (p^2+r_\ast \Omega^i_2)}DS_n(x-y)\nu_{p^2+r_\ast \Omega^i_2}(x)\theta^{i}_2\big((y-p^2)/r_{\ast}\big)\, d\sigma_y\\ 
 +&\int_{\partial (p^1+r_\ast\Omega^i_1)}DS_n(x-y)\nu_{p^2+r_\ast \Omega^i_2}(x)\theta^{i}_1\big((y-p^1)/r_{\ast}\big)\, d\sigma_y \\ 
 & \qquad \qquad \qquad \qquad \qquad  \qquad \qquad \qquad \forall x \in \partial (p^2+r_\ast \Omega^i_2)\, .
\end{split}$$ As a consequence, the invertibility of $M_{r_\ast}$ follows by Lemma 