## 2 The Basic Equations and Biparametric Perturbation Solution
### The Biparametric Perturbation Solution  
The basic equation, boundary conditions, and initial conditions are given above. Now, let us solve the basic equation. In order to solve Equation (12), the variable separation is required firstly, suppose the following:  
\[w(x,t)=v(x)Y(t). \tag{13}\]Substituting Equation (13) into Equation (12), it can be obtained that  
\[E^{\prime}I_{y}Y(t)\frac{\partial^{4}v(x)}{\partial x^{4}}+\overline{m}v(x)\frac {\partial^{2}Y(t)}{\partial l^{2}}+cv(x)\frac{\partial Y(t)}{\partial t}=0. \tag{14}\]  
By simplifying Equation (14), it can be transformed into the following:  
\[\frac{v^{(4)}(x)}{v(x)}+\frac{\overline{m}Y^{\prime\prime}(t)+cY^{\prime}(t)}{ E^{\prime}I_{y}Y(t)}=0. \tag{15}\]  
Here, we let  
\[\frac{v^{(4)}(x)}{v(x)}=-\frac{\overline{m}Y^{\prime\prime}(t)+cY^{\prime}(t)} {E^{\prime}I_{y}Y(t)}=D^{4}, \tag{16}\]  
in which \(D\) is the unknown constants. Two differential equations can be obtained, and they are as follows:  
\[v^{(4)}(x)-D^{4}v(x)=0 \tag{17}\]  
and  
\[Y^{\prime\prime}(t)+\frac{c}{\overline{m}}Y^{\prime}(t)+\omega^{2}Y(t)=0, \tag{18}\]  
in which \(\omega^{2}=D^{4}E^{\prime}I_{y}/\overline{m}\). It is assumed that the solution of Equation (17) is in the form of  
\[v(x)=K\epsilon^{x}. \tag{19}\]  
From Equations (17) and (19), we can obtain the following:  
\[v(x)=K_{1}\epsilon^{iDx}+K_{2}\epsilon^{-iDx}+K_{3}\epsilon^{Dx}+K_{4}\epsilon ^{-Dx}. \tag{20}\]  
Converting Equation (20) into the form of the trigonometric function and hyperbolic function, one has the following:  
\[v(x)=A_{1}\sin(Dx)+A_{2}\cos(Dx)+A_{3}\sinh(Dx)+A_{4}\cosh(Dx). \tag{21}\]  
Substituting Equation (21) into Equations (2) and (3), respectively, we can obtain the following  
\[A_{2}=-A_{4}, \tag{22}\]  
\[E^{\prime}I_{y}Y(t)\frac{\partial^{4}v(x)}{\partial x^{4}}+\overline{m}v(x)\frac {\partial^{2}Y(t)}{\partial l^{2}}+cv(x)\frac{\partial Y(t)}{\partial t}=0. \tag{14}\]0  
\[E^{\prime}I_{y}Y(t)\frac{\partial^{4}v(x)}{\partial x^{4}}+\overline{m}v(x)\frac {\partial^{2}Y(t)}{\partial l^{2}}+cv(x)\frac{\partial Y(t)}{\partial t}=0. \tag{14}\]1  
and  
\[E^{\prime}I_{y}Y(t)\frac{\partial^{4}v(x)}{\partial x^{4}}+\overline{m}v(x)\frac {\partial^{2}Y(t)}{\partial l^{2}}+cv(x)\frac{\partial Y(t)}{\partial t}=0. \tag{14}\]2  
Thus, the solution of Equation (17) is  
\[E^{\prime}I_{y}Y(t)\frac{\partial^{4}v(x)}{\partial x^{4}}+\overline{m}v(x)\frac {\partial^{2}Y(t)}{\partial l^{2}}+cv(x)\frac{\partial Y(t)}{\partial t}=0. \tag{14}\]3  
in which \(D=\sqrt[4]{\omega^{2}\overline{m}/I_{y}E^{\prime}}\). Substituting Equation (11) into Equation (18), we can obtain the following:  
\[E^{\prime}I_{y}Y(t)\frac{\partial^{4}v(x)}{\partial x^{4}}+\overline{m}v(x)\frac {\partial^{2}Y(t)}{\partial l^{2}}+cv(x)\frac{\partial Y(t)}{\partial t}=0. \tag{14}\]4Here, we used the biparametric perturbation method to solve Equation (27), and selected \(\left(d_{31}\right)^{2}\) and \(c\) as the perturbation parameters, thus the \(Y(t)\) can be expressed as follows  
\[E^{\prime}I_{y}Y(t)\frac{\partial^{4}v(x)}{\partial x^{4}}+\overline{m}v(x)\frac {\partial^{2}Y(t)}{\partial l^{2}}+cv(x)\frac{\partial Y(t)}{\partial t}=0. \tag{14}\]5  
Substituting Equation (28) into Equation (27), and comparing the coefficients of \(\left[\left(d_{31}\right)^{2}\right]^{0}\) and \(c^{0}\), we may obtain the zero-order perturbation equation  
\[E^{\prime}I_{y}Y(t)\frac{\partial^{4}v(x)}{\partial x^{4}}+\overline{m}v(x)\frac {\partial^{2}Y(t)}{\partial l^{2}}+cv(x)\frac{\partial Y(t)}{\partial t}=0. \tag{14}\]6  
The corresponding initial conditions are as follows  
\[\left\{ \begin{array}{l}v(x)Y_{0}^{0}(t)=w_{0}\\ v(x)Y_{0}^{0\prime}(t)=0\end{array} \right.,\text{ at }x=l,t=0.