## Appendix  
Let \(\psi(\tau,\mathbf{q})\) be the velocity potential at point \(\mathbf{q}\) at time \(\tau\). Applying [PERSON]'s theorem to \(\psi(\tau,\mathbf{q})\) and \(G(\mathbf{p},t;\mathbf{q},\tau;\varepsilon)\), one obtains  
\[\int_{0}^{t}\mathrm{d}\tau\int_{S_{\mathrm{F}}+S_{\mathrm{B}}+S_{\infty}}\! \left(\psi(\tau,\mathbf{q})\frac{\partial G}{\partial n_{q}}-G\frac{\partial\psi( \tau,\mathbf{q})}{\partial n_{q}}\right)\mathrm{d}s_{q}=2\pi\beta\psi(t,\mathbf{q}) \tag{10}\]  
where \(\beta\) is an unknown constant decided by the instantaneous term in \(G\). The free-surface term \(\widetilde{G}\) (see Equation (3)) is harmonic in the fluid domain, so applying [PERSON]'s theorem to \(\psi(\tau,\mathbf{q})\) and \(\widetilde{G}\) yields  
\[\int_{0}^{t}\mathrm{d}\tau\int_{S_{\mathrm{F}}+S_{\mathrm{B}}+S_{\infty}}\! \left(\psi(\tau,\mathbf{q})\frac{\partial\widetilde{G}}{\partial n_{q}}- \widetilde{G}\frac{\partial\psi(\tau,\mathbf{q})}{\partial n_{q}}\right)\mathrm{ d}s_{q}=0 \tag{11}\]  
Using the boundary conditions for \(\widetilde{G}\) on \(S_{\infty}\), it can be deduced that the integral (11) on \(S_{\infty}\) equals to \(0\). Using the initial and free-surface conditions for \(\psi(\tau,\mathbf{q})\) and \(\widetilde{G}(\mathbf{p},t;\mathbf{q},\tau;\varepsilon)\), the integral (11) on \(S_{\mathrm{F}}\) can be transformed to  
\[\begin{array}{l}\int_{0}^{t}\mathrm{d}\tau\int_{S_{\mathrm{F}}}\!\left(\psi \frac{\partial\widetilde{G}}{\partial n_{q}}-\widetilde{G}\frac{\partial\psi} {\partial n_{q}}\right)\mathrm{d}s_{q}\\ =-\frac{1}{8}\int_{0}^{t}\mathrm{d}\tau\int_{S_{\mathrm{F}}}\!\left(\psi\frac {\partial\widetilde{G}}{\partial\tau^{2}}-2\varepsilon\omega_{0}\frac{ \partial\widetilde{G}}{\partial\tau}\right)-\widetilde{G}\!\left(\frac{ \partial^{2}\psi}{\partial\tau^{2}}+2\varepsilon\omega_{0}\frac{\partial \psi}{\partial\tau}\right)\right)\mathrm{d}s_{q}\\ =-\frac{1}{8}\int_{0}^{t}\mathrm{d}\tau\int_{S_{\mathrm{F}}}\frac{\partial \tau}{\partial\tau}\left(\psi\frac{\partial\widetilde{G}}{\partial\tau}- \widetilde{G}\frac{\partial\psi}{\partial\tau}-2\varepsilon\omega_{0}\psi \widetilde{G}\right)\mathrm{d}s_{q}\\ =-\frac{1}{8}\!\left(\int_{S_{\mathrm{F}}}\!\left(\psi\frac{\partial \widetilde{G}}{\partial\tau}-\widetilde{G}\frac{\partial\psi}{\partial\taui.e.,  
\[\int_{S_{\rm B}}\biggl{(}\psi(t,\mathbf{q})\frac{\partial\overline{G}}{\partial n_{q}} -\overline{G}\frac{\partial\psi(t,\mathbf{q})}{\partial n_{q}}\biggr{)}\mathrm{d}s _{q}-2\pi\beta\psi(t,\mathbf{q})=\int_{S_{\rm F}+S_{\infty}}\biggl{(}\overline{G} \frac{\partial\psi(t,\mathbf{q})}{\partial n_{q}}-\psi(t,\mathbf{q})\frac{\partial \overline{G}}{\partial n_{q}}\biggr{)}\mathrm{d}s_{q} \tag{10}\]  
Subtracting Equation (10) from Equation (10) yields  
\[\int_{S_{\rm B}}\biggl{(}\psi\frac{\partial\overline{G}}{\partial n _{q}}-\overline{G}\frac{\partial\psi}{\partial n_{q}}\biggr{)}\mathrm{d}s_{q}- \int_{0}^{t}\mathrm{d}\tau\int_{S_{\rm B}}\biggl{(}\psi\frac{\partial\overline {G}}{\partial n_{q}}-\widetilde{G}\frac{\partial\psi}{\partial n_{q}}\biggr{)} \mathrm{d}s_{q}-2\pi\beta\psi \tag{11}\] \[+\frac{1}{8}\int_{0}^{t}\biggl{(}\psi\frac{\partial\overline{G}} {\partial\tau}-\widehat{G}\frac{\partial\psi}{\partial\tau}-2\epsilon\omega_{0 }\psi\widehat{G}\biggr{)}\hat{\xi}(\tau)\Bigg{|} \begin{array}{c}B\\ A\end{array} \mathrm{d}\tau\] \[=\int_{S_{\rm F}}\overline{G}\frac{\partial\psi}{\partial q} \mathrm{d}s_{q}-\int_{S_{\rm F}}\psi\Biggl{(}\frac{\partial\overline{G}}{ \partial n_{q}}+\frac{1}{8}\frac{\partial\overline{G}}{\partial\tau}\Bigg{|} \begin{array}{c}\tau=t\end{array} \Biggr{)}\mathrm{d}s_{q}-\int_{S_{\infty} }\biggl{(}\psi\frac{\partial\overline{G}}{\partial n_{q}}-\overline{G}\frac{ \partial\psi}{\partial n_{q}}\biggr{)}\mathrm{d}s_{q}\]  
If the right hand side of (11) equals to 0, i.e., \(\overline{G}=\frac{\partial\overline{G}}{\partial n_{q}}+\frac{1}{8}\frac{ \partial\overline{G}}{\partial\tau}\Bigg{|} \begin{array}{c}\tau=t\end{array}