## Appendix A  
Substitute Equation (9) into the first equation of Equation (17), one has  
\[L[w_{0}]=((A_{4}+A_{5}\hat{v})e^{i\omega t}+(A_{4}-A_{5}\hat{v})e^{-i\omega t }+A_{6})\delta(x-vt) \tag{18}\]Let \(z=x-v\,t\), and assume that Equation (A1) has the following formal solution:  
\[w_{0}=u_{1}(z)(e^{i\omega t}+e^{-i\omega t})+u_{2}(z)\] (A2)  
By substituting Equation (A2) into Equation (A1) simplifying, and separating the real part, we obtain  
\[ \begin{split}& EI\frac{d^{4}u_{1}}{dz^{4}}+mv^{2}\frac{d^{2}u_{1}}{dz^ {2}}-cv\frac{du_{1}}{dz}+(k-m\omega^{2})u_{1}=A_{4}\delta(z)\\ & EI\frac{d^{4}u_{2}}{dz^{4}}+mv^{2}\frac{d^{2}u_{2}}{dz^{2}}-cv \frac{du_{2}}{dz}+ku_{2}=A_{6}\delta(z)\end{split} \] (A3)  
By applying the following Fourier transform  
\[ \begin{split}\hat{u}_{j}(\xi)&=\int_{-\infty}^{ \infty}u_{j}(z)e^{-i\xi z}dz,\\ \hat{u}_{j}(z)&=\frac{1}{2\pi}\int_{-\infty}^{\infty }\hat{u}_{j}(\xi)e^{i\xi z}d\xi,\quad j=1,2,\end{split} \] (A4)  
to both sides of Equation (A3) and simplifying, we have  
\[ \begin{split}\hat{u}_{1}&=\frac{A_{4}}{EI\xi^{4}-mv^ {2}\xi^{2}-cvi\xi+(k-m\omega^{2})}\\ \hat{u}_{2}&=\frac{A_{6}}{EI\xi^{4}-mv^{2}\xi^{2}- cvi\xi+k}\end{split} \] (A5)  
Perform an inverse Fourier transform on both sides of above equation to obtain \(u_{1,2}\)  
\[ \begin{split} u_{1}&=\frac{1}{2\pi}\int_{-\infty}^{ +\infty}\frac{A_{4}}{EI\xi^{4}-mv^{2}\xi^{2}-cvi\xi+(k-m\omega^{2})}e^{i\xi z }d\xi\\ u_{2}&=\frac{1}{2\pi}\int_{-\infty}^{+\infty} \frac{A_{6}}{EI\xi^{4}-mv^{2}\xi^{2}-cvi\xi+k}e^{i\xi z}d\xi\end{split} \] (A6)  
Substitute Equation (A2) into the second equation in Equation (17), we have  
\[ \begin{split} L[w_{1}]&=-k_{N}[(6u_{1}^{2}u_{2}+u_ {2}^{3})+(3u_{1}^{3}+3u_{1}u_{2}^{2})(e^{i\omega t}+e^{-i\omega t})+3u_{1}^{2 }u_{2}(e^{i2\omega t}+e^{-i2\omega t})\\ &\quad+u_{1}^{3}(e^{i3\omega t}+e^{-i3\omega t})]\end{split} \] (A7)  
Based on the characteristics of Equation (A7), it can be assumed that the equation has the following form of solution  
\[w_{1}(x,t)=-k_{N}[u_{3}(z)+u_{4}(z)(e^{i\omega t}+e^{-i\omega t})+u_{5}(z)(e^{ i2\omega t}+e^{-i2\omega t})+u_{6}(z)(e^{i3\omega t}+e^{-i2\omega t})]\] (A8)  
Utilizing the superposition principle and Duhamel's integral, \(u_{3,4,5,6}\) can be determined as  
\[\begin{split} u_{3}(z)&=\int_{-\infty}^{+\infty}[6u_ {1}^{2}(z-\tau)u_{2}(z-\tau)+u_{2}^{3}(z-\tau)]\lambda_{0}(\tau)d\tau,\\ u_{4}(z)&=\int_{-\infty}^{+\infty}[3u_{1}^{3}(z- \tau)+3u_{1}(z-\tau)u_{2}^{2}(z-\tau)]\lambda_{1}(\tau)d\tau,\\ u_{5}(z)&=\int_{-\infty}^{+\infty}3u_{1}^{2 By employing a method akin to the one used for Equations (A3)-(A6), the expressions of \(\lambda_{j}\) can be ascertained. Substitute Equation (A2) into the third equation in Equation (17), we obtain  
\[ \begin{split} L[w_{2}]=& 3k_{N}^{2}[(2u_{1}^{2}u_{3}+2u_{1}^{2 }u_{5}+4u_{1}u_{2}u_{4}+u_{2}^{2}u_{3})\\ &+(u_{1}^{2}u_{6}+3u_{1}^{2}u_{4}+2u_{1}u_{2}u_{3}+2u_{1}u_{2}u_{ 5}+u_{2}^{2}u_{4})(e^{i\omega t}+e^{-i\omega t})\\ &+(u_{1}^{2}u_{3}+2u_{1}^{2}u_{5}+2u_{1}u_{2}u_{4}+2u_{1}u_{2}u_{ 6}+u_{2}^{2}u_{5})(e^{i2\omega t}+e^{-i2\omega t})\\ &+(u_{1}^{2}u_{4}+2u_{1}^{2}u_{6}+2u_{1}u_{2}u_{5}+u_{2}^{2}u_{ 6})(e^{i3\omega t}+e^{-i3\omega t})\\ &+(u_{1}^{2}u_{5}+2u_{1}u_{2}u_{6})(e^{i4\omega t}+e^{-i4\omega t })+u_{1}^{2}u_{6}(e^{i5\omega t}+e^{-i5\omega t})],\end{split}  \tag{17}\]  
Similarly, the solution to the equation can be expressed as  
\[ \begin{split} w_{2}(x,t)=& 3k_{N}{}^{2}[u_{7}+u_{8}(e^{i \omega t}+e^{-i\omega t})+u_{9}(e^{i2\omega t}+e^{-i2\omega t})\\ &+u_{10}(e^{i3\omega t}+e^{-i3\omega t})+u_{11}(e^{i4\omega t}+e^ {-i4\omega t})+u_{12}(e^{i5\omega t}+e^{-i5\omega t})],\end{split}