## 2 Basic Equations  
Consider a rectangular thin plate with length a, width b and uniform thickness h, and the coordinate of plates under title boundary conditions is depicted in Figure 1. Through the derivation with the consideration of the classic Kirchhoff assumptions, the free vibration governing equation can be expressed as:  
\[\frac{\partial^{4}w(x,y,t)}{\partial x^{4}}+2\frac{\partial^{4}w(x,y,t)}{ \partial x^{2}\partial y^{2}}+\frac{\partial^{4}w(x,y,t)}{\partial y^{4}}+ \frac{\rho h}{D}\frac{\partial^{2}w(x,y,t)}{\partial t^{2}}=0 \tag{1}\]  
where \(\mu\) is Poisson's ratio, \(D=Eh^{3}/\left[12\left(1-\mu^{2}\right)\right]\) is flexural rigidity, \(E\) is elasticity modulus and \(\rho\) is the mass density of the plate. Resultant internal forces including bending moments, effective shear forces and twisting moments can be described by the deflection \(w\) (\(x\), \(y\)) as follow:  
\[ \begin{array}{l}M_{x}=-D\Big{(}\frac{\partial^{2}w}{\partial x^{2}}+\mu \frac{\partial^{2}w}{\partial y^{2}}\Big{)}\,\ V_{x}=-D\Big{[}\frac{\partial^{3}w}{ \partial x^{3}}+(2-\mu)\frac{\partial^{2}w}{\partial x\partial y^{2}}\Big{]} \\ M_{y}=-D\Big{(}\frac{\partial^{2}w}{\partial y^{2}}+\mu\frac{\partial^{2}w}{ \partial x^{2}}\Big{)}\,\ V_{y}=-D\Big{[}\frac{\partial^{3}w}{\partial y^{2}}+(2-\mu) \frac{\partial^{3}w}{\partial x^{2}\partial y}\Big{]}\\ M_{\rm xy}=-D(1-\mu)\frac{\partial^{2}w}{\partial x\partial y}\end{array}  \tag{2}\]  
On basis of the classical vibration theory, the deflection for a free vibration plate could be preselected as \(w(x,y,t)=\sin(\omega t)w(x,y)\). Then one will obtain  
\[\frac{\partial^{4}w(x,y)}{\partial x^{4}}+2\frac{\partial^{4}w(x,y)}{\partial x ^{2}\partial y^{2}}+\frac{\partial^{4}w(x,y)}{\partial y^{4}}-\frac{\rho h \omega^{2}}{D}w(x,y)=0 \tag{3}\]  
Expressions for the RRTF plate, in which rotationally-restrained at edges \(x=0\) and \(y=0\), and its opposite corner free, are described as:  
\[w|_{x=0}=\left.w\right|_{y=0}=0,\ V_{x}|_{x=a}=\left.V_{y}\right|_{y=b}=0,\ M_{xy}|_{x=a,y=b}=\left.-D(1-\mu)\frac{ \partial^{2}w}{\partial x\partial y}\right|_{x=a,y=b}=0 \tag{4}\]  
\[ \begin{array}{l}M_{x}|_{x=0}=-D\left.\frac{\partial^{2}w}{\partial x^{2}} \right|_{x=0}=-R_{x0}\left.\frac{\partial w}{\partial x}\right|_{x=0},\ M_{x}|_{x=a}= \left.-D\left(\frac{\partial^{2}w}{\partial x^{2}}+\mu\frac{\partial^{2}w}{ \partial y^{2}}\right)\right|_{x=a}=0\\ M_{y}\left|_{y=0}=-D\left.\frac{\partial^{2}w}{\partial y^{2}}\right|_{y=0}=-R _{y0}\left.\frac{\partial w}{\partial y}\right|_{y=0},\ M_{y}|_{y=b}=-D\left( \frac{\partial^{2}w}{\partial y^{2}}+\mu\frac{\partial^{2}w}{\partial x^{2}} \right)\right|_{y=b}=0\end{array}  \tag{5}\]  
Figure 1: Rectangular plate with one corner free and its opposite two adjacent edges rotationally restrained. where \(R_{x0},R_{y0}\) are the rotational springs that are reflected through setting the rotating fixed factor \(r\) from reference [36] as follows:  
\[\frac{a\cdot R_{x0}}{D_{x}}=\frac{3r_{x0}}{1-r_{x0}}\,\ \frac{b\cdot R_{y0}}{D_{y}}= \frac{3r_{y0}}{1-r_{y0}} \tag{6}\]  
As to problems of rectangular thin plates, Fourier series is usually adopted as the trial function for the deflection \(w(x,y)\) due to its simple form and orthogonality properties. To deal with title problem, \(w(x,y)\) is assumed to have the following form:  
\[w(x,y)=\sum\limits_{m=1,3}^{\infty}\sum\limits_{n=1,3}^{\infty}w_{mn}\sin \frac{\delta_{m}x}{2}\sin\frac{\lambda_{n}y}{2}\ \ m,n=1,3,5,\cdots \tag{7}\]  
where \(\delta_{m}=\frac{m\pi}{a}\), \(\lambda_{n}=\frac{n\pi}{b}\); \(w_{mn}=\frac{4}{ab}f_{0}^{a}\int_{0}^{b}\!\!w(x,y)\sin\frac{\delta_{m}x}{2} \sin\frac{\lambda_{n}y}{2}\mathrm{d}x\mathrm{d}y\) is the unknown Fourier coefficient for the deflection. The term-by-term differentiation of the trial function is achieved by utilizing the [PERSON]'s transformation [42], which is the main contribution of the current study. Now utilizing the [PERSON]'s transformation, the detail expressions for the first four order derivatives of the deflection are given as below:  
\[