# A Study on a Mechanism of Lateral Pedestrian-Footbridge Interaction
## 3 Study on Dynamic Response of Pedestrian-Footbridge Interactions  
### Dynamic Coupling Equation Solution of the Lateral Interaction Pedestrian-Footbridge  
The multiscale method is used to solve Equations (1)-(3). Assume that the nonlinear systems (1)-(3) are weak nonlinear systems. Then, a small parameter is added to the damping term, external excitation term, and nonlinear term in the system. Thus, Equations (1)-(3) can be written as:  
\[\ddot{u}_{yp}(t)+\omega_{0}^{2}u_{yp}(t)-\varepsilon\eta\dot{u}_{yp}(t)+ \varepsilon\rho\dot{u}^{3}_{yp}(t)+\varepsilon\dot{\varepsilon}\dot{u}_{yp}(t )u_{yp}^{2}(t)+\varepsilon\ddot{u}_{ys}(t)=0 \tag{4}\]  
\[ \begin{array}{l}\ddot{u}_{ys}(t)+2\varepsilon\zeta_{\rm s}\omega_{\rm s}i_{ uys}(t)+\omega_{\rm s}^{2}u_{ys}(t)=\\ \frac{me\omega_{0}^{2}u_{yp}(t)-me\tau\dot{u}_{yp}(t)+me\rho\dot{u}^{3}_{yp}(t )+me\xi\dot{u}_{yp}(t)u_{yp}^{2}(t)}{m}\end{array} . \tag{5}\]  
It is assumed that the first-order approximate solutions of Equations (4) and (5) are:  
\[u_{yp}(t)=u_{0}(T_{0},T_{1})+\varepsilon u_{1}(T_{0},T_{1})+\cdots \tag{6}\]  
\[u_{ys}(t)=U_{0}(T_{0},T_{1})+\varepsilon U_{1}(T_{0},T_{1})+\cdots \tag{7}\]  
of which \(T_{0}=\tau\) and \(T_{1}=\varepsilon\tau\). Equations (6) and (7) are substituted into Equations (4) and (5) by adopting the multiscale method, aiming to achieve a situation where the coefficients of the small parameters \(\varepsilon^{0}\) and \(\varepsilon^{1}\) on both sides are equal. Then, the following partial differential equations can be defined as follows:  
\[\varepsilon^{0}\text{: }\left\{ \begin{array}{l}D_{0}^{2}u_{0}+\omega_{p}^{2}u_{0 }=0\\ D_{0}^{2}U_{0}+\omega_{\rm s}^{2}U_{0}=0\end{array} \right. \tag{8}\]  
Figure 1: Lateral coupling of an individual pedestrian and a footbridge. Equations (17) and (18) are substituted into Equations (15) and (16); thus, Equations (15) and (16) become:  
\[ \begin{array}{l}-j\omega_{p}D_{1}a_{p}(T_{1})e^{i\theta_{p}}+\omega_{p}a_{p}(T _{1})D_{1}\theta_{p}(T_{1})e^{i\theta_{p}}+\frac{1}{2}j\eta a_{p}(T_{1})\omega_ {p}e^{i\theta_{p}}\\ -\frac{3}{8}j\rho a_{p}^{3}(T_{1})\alpha_{p}^{3p}e^{i\theta_{p}}-\frac{1}{8}j \xi a_{p}^{3}(T_{1})\omega_{p}e^{i\theta_{p}}+\frac{a_{s}(T_{1})}{2}\alpha_{s} ^{2}e^{i\theta_{p}}\cos(\sigma T_{1}\\ -\theta_{p})+\frac{a_{s}(T_{1})}{2}j\omega_{s}^{2}e^{i\theta_{p}}\sin(\sigma T _{1}-\theta_{p})=0\\ \\ -2j\omega_{s}D_{1}a_{s}-\zeta_{s}a_{s}j\omega_{s}^{2}+\frac{m}{2M}\omega_{p}^ {2}a_{p}\cos(\theta_{p}-\sigma T_{1})+j\frac{m}{2M}\omega_{p}^{2}a_{p}\sin( \theta_{p}\\ -\sigma T_{1})-\eta\frac{m}{2M}a_{p}\omega_{p}\cos(\theta_{p}-\sigma T_{1})+ \eta\frac{m}{2M}a_{p}\omega_{p}\sin(\theta_{p}-\sigma T_{1})+\\ \frac{3m}{8M}\omega_{p}^{3}j\omega_{p}^{3}\cos(\theta_{p}-\sigma T_{1})-\frac {3m}{8M}\rho a_{p}^{3}\omega_{p}^{3}\sin(\theta_{p}-\sigma T_{1})+\\ \frac{m}{8M}\xi a_{p}^{3}\omega_{p}\cos(\theta_{p}-\sigma T_{1})-\frac{m}{8M} \xi a_{p}^{3}\omega_{p}\sin(\theta_{p}-\sigma T_{1})=0\\ \end{array} . \tag{20}\]  
If the coefficients of the real part and imaginary part of Equations (19) and (20) are zero, respectively, then in this case, one obtains:  
\[-\omega_{p}D_{1}a_{p}+\frac{1}{2}\eta a_{p}\omega_{p}-\frac{3}{8}\rho a_{p}^ {3}a_{p}^{3}-\frac{1}{8}\xi\omega_{p}a_{p}^{3}+\frac{a_{s}}{2}\omega_{s}^{2} \sin(\sigma T_{1}-\theta_{0})=0 \tag{21}\]  
\[a_{p}\omega_{p}D_{1}\theta_{p}+\frac{a_{s}}{2}\omega_{s}^{2}\cos(\sigma T_{1} -\theta_{0})=0 \tag{22}\]  
\[ \begin{array}{l}-2\omega_{s}D_{1}a_{s}-\zeta_{s}a_{s}\omega_{s}^{2}+\frac{m} {2M}\omega_{p}^{2}a_{p}\sin(\theta_{p}-\sigma T_{1})-\eta\frac{m}{2M}a_{p} \omega_{p}\cos(\theta_{p}-\sigma T_{1})\\ +\frac{3m}{8M}\rho a_{p}^{3}\omega_{p}^{3}\cos(\theta_{p}-\sigma T_{1})+\frac {m}{8M}\alpha_{p}^{3}\omega_{p}\cos(\theta_{p}-\sigma T_{1})=0\\ \\ \frac{m}{2M}\omega_{p}^{2}a_{p}\cos(\theta_{p}-\sigma T_{1})+\eta\frac{m}{2M}a_ {p}\omega_{p}\sin(\theta_{p}-\sigma T_{1})-\frac{3m}{8M}\rho a_{p}^{3}\omega _{p}^{3}\sin(\theta_{p}\\ -\sigma T_{1})-\frac{m}{8M}\xi a_{p}^{3}\omega_{p}\sin(\theta_{p}-\sigma T_{1} )=0\\ \end{array} .