# Influence of Non-Uniform Rail Loads on the Rotation of Railway Sleepers
## 2 Movement Equations and Their Solutions
### Solution for Oscillating Forces  
Thus, a set of equations necessary to determine the response of the system with load by time-invariable forces has been obtained. In the case of oscillating forces with the provided wheel frequency \(\omega\) in a mobile coordinate system, Equation (2) describing the system vibrations with a symmetric load of both rails has the following form:  
\[ \begin{split}& EI\frac{\partial^{2}y_{\alpha}}{\partial x^{4}}+N_{t} \frac{\partial^{2}y_{\alpha}}{\partial x^{2}}+m_{r}(\frac{\partial^{2}y_{ \alpha}}{\partial x^{2}}-2v\frac{\partial^{2}y_{\alpha}}{\partial x^{2}}+v^{2 }\frac{\partial^{2}y_{\alpha}}{\partial x^{2}})+c_{r}(\frac{\partial y_{ \alpha}}{\partial x}-v\frac{\partial y_{\alpha}}{\partial x}-\frac{\partial y_{ \alpha}}{\partial x})+v\frac{\partial y_{\alpha}}{\partial x})+\\ &+k_{r}(y_{\alpha}-y_{\beta\alpha})=q_{\alpha\alpha}^{\alpha}( \xi,t)\\ &\qquad\qquad\qquad m_{p}(\frac{\partial^{2}y_{\beta\alpha}}{ \partial x^{2}}-2v\frac{\partial^{2}y_{\alpha}}{\partial x^{2}}+v^{2}\frac{ \partial^{2}y_{\alpha}}{\partial x^{2}})+c_{p}(\frac{\partial y_{\alpha}}{ \partial x}-v\frac{\partial y_{\alpha}}{\partial x})+k_{p}y_{\alpha}=\\ &\qquad\qquad+c_{r}(\frac{\partial y_{\alpha}}{\partial t}-v \frac{\partial y_{\alpha}}{\partial x}-\frac{\partial y_{\alpha}}{\partial t} +v\frac{\partial y_{\alpha}}{\partial x})+k_{r}(y_{\alpha}-y_{\alpha})\end{split}  \tag{16}\]  
The load in the first equation of system (16) can be formulated as follows:  
\[q_{\alpha v}^{\alpha}(\xi,t)=q_{\alpha v}(\xi)\cdot\cos\omega t+q_{\alpha s}( \xi)\cdot\sin\omega t \tag{17}\]  
where qavc and qavs are the cosine and sine components of the symmetric load. The following form [16] is used to obtain the stationary solution of system (16):  
\[ \begin{split}& y_{\alpha}(\xi,t)=Y_{\alpha x}(\xi)\cdot\cos \omega t+Y_{\alpha s}(\xi)\cdot\sin\omega t\\ & y_{\alpha}(\xi,t)=Y_{\alpha x}(\xi)\cdot\cos\omega t+Y_{\alpha s }(\xi)\cdot\sin\omega t\end{split}  \tag{18}\]  
When differentiating Equation (18), taking into consideration relationships (17), system (16) has the form of a system of four ordinary equations taking into consideration the cosine and sine components of the solutions of the following system:  
\[ \begin{split}& L_{1}(Y_{\alpha x},Y_{\alpha s},Y_{\alpha x},Y_{ \alpha s},P)\cdot\cos\omega t=q_{\alpha v}\cdot\cos\omega t\\ & L_{2}(Y_{\alpha x},Y_{\alpha s},Y_{\alpha x},Y_{\alpha s},P) \cdot\sin\omega t=q_{\alpha s}\cdot\sin\omega t\\ & L_{3}(Y_{\alpha x},Y_{\alpha s},Y_{\alpha x},Y_{\alpha s},P) \cdot\cos\omega t=0\cdot\cos\omega t\\ & L_{4}(Y_{\alpha x},Y_{\alpha s},Y_{\alpha x},Y_{\alpha s},P) \cdot\sin\omega t=0\cdot\sin\omega t\end{split}  \tag{19}\]  
where L1-L4--the linear operators Yrac, Yras, Ypac, and Ypas of the functions and their derivatives; P--the vector of the model parameters (EI, m, N, k, r, etc., see Figure 2). For example, operator L1 has the following form:  
\[ \begin{split}& L_{1}=EI\frac{d^{4}Y_{\alpha x}}{d_{0}^{4}}-m_{r} \omega^{2}Y_{\alpha x}-2m_{r}\omega\frac{dY_{\alpha s}}{d_{0}^{6}}+(N_{t}+m_{r} v^{2})\frac{d^{2}Y_{\alpha x}}{d_{0}^{6}}+c_{r}\omega Y_{\alpha s}-c_{r}v\frac{dY_{ \alpha x}}{d_{0}^{6}}-\\ &-c_{r}\omega Y_{\alpha s}+c_{r}v\frac{dY_{\alpha x}}{d_{0}^{6}}+k _{r}Y_{\alpha x}-k_{r}Y_{\alpha x}\end{split}