## 2 Theoretical Method and Physical Model
### Solution of Oil Film Reynold Equation
#### 2.1.3 Solving Method  
The internal surface of a cylindrical bearing is flattened into a two-dimensional rectangle and the rectangle plane is partitioned. Then, several meshes are divided along with the axial and circumferential directions. The pressure value on each grid node is constituted by the difference quotient of each order, which approximately replaces the partial derivative of each order in the Reynolds equation. The continuous partial differential equation is discretized into \(n\) algebraic equations. Then, the pressure value at each node is solved iteratively. The pressure value obtained by the solution can approximately represent the pressure distribution of the oil film in the oil cavity. Figure 2 depicts the obtained discrete grids of the oil film. The grid nodes are numbered sequentially in rows and columns. The number of rows along the axis is coded as \(j\), and the number of columns along the circumferential direction is coded as \(i\). It is uniformly divided into n grids along the axial direction, \(j=1,2,\ldots,n+1\), where \(j=1\) and \(j=n+1\) are the same points. It is evenly divided into \(m\) grids along the circumferential direction, \(i=1,2,\ldots,m+1\), where \(i=1\) and \(i=m+1\) are the same nodes. Then, the length of each grid in the axial direction is \(\Delta\lambda=\lambda_{j}-\lambda_{j-1}=2/n\) and the length of each grid in the circumferential direction is \(\Delta\theta=\theta_{i}-\theta_{i-1}=2\pi R/m\). The coordinate number of each node is \((i,j)\), and the pressure value on each node is represented by \(P_{i,j}\). The partial derivative of the pressure on each node can be approximately equal to the pressure value of its adjacent nodes through the central difference, which can be deduced as  
\[(\frac{\partial P}{\partial\theta})_{i,j}=\frac{P_{i+1,j}-P_{i-1,j}}{2\Delta \theta},(\frac{\partial P}{\partial\lambda})_{i,j}=\frac{P_{i,j}-P_{i,j+1}}{2 \Delta\lambda} \tag{8}\]  
Equation (8) is the classical five-point difference format. In order to improve the calculation accuracy, it is required to use the half-step difference method, as shown in Figure 3.  
Figure 3: Half-step difference method. Figure 2: Discrete grid of the oil film. Equation (4) can be solved under the rectangular coordinate [28]  
\[\frac{\partial}{\partial\theta}(H^{3}\frac{\partial P}{\partial\theta})+\left( \frac{2R}{\mathcal{I}}\right)^{2}\frac{\partial}{\partial\lambda}(H^{3}\frac{ \partial P}{\partial\lambda})=6\frac{\partial H}{\partial\lambda} \tag{9}\]  
The first derivative is obtained as  
\[\left\{ \begin{matrix}(\frac{\partial P}{\partial\theta})_{ij}=\frac{P_{i+ \frac{1}{2},j}-P_{i-\frac{1}{2},j}}{\Delta\theta},(\frac{\partial P}{\partial \lambda})_{ij}=\frac{P_{ij+\frac{1}{2}}-P_{ij-\frac{1}{2}}}{\Delta\lambda}\\ (H^{3}\frac{\partial P}{\partial\theta})_{i+\frac{1}{2},j}=H^{3}_{i+\frac{1}{2 },j}\frac{P_{i+1}-P_{i-\frac{1}{2},j}}{\Delta\theta},(H^{3}\frac{\partial P}{ \partial\theta})_{i-\frac{1}{2},j}=H^{3}_{i+\frac{1}{2},j}\frac{P_{i,j}-P_{i -\frac{1}{2},j}}{\Delta\theta}\\ (H^{3}\frac{\partial P}{\partial\theta})_{ij+\frac{1}{2}}=H^{3}_{i,j+\frac{1} {2}}\frac{P_{i,j+1}-P_{i,j}}{\Delta\lambda},(H^{3}\frac{\partial P}{\partial \theta})_{ij-\frac{1}{2}}=H^{3}_{i,j-\frac{1}{2}}\frac{P_{i,j}-P_{i,j-1}}{ \Delta\theta}\\ \end{matrix} \right. \tag{10}\]  
The second derivative gives the following expression as  
\[\left\{ \begin{matrix}[\frac{\partial}{\partial\theta}(H^{3}\frac{\partial P}{ \partial\theta})]_{ij}=\frac{H^{3}_{i+\frac{1}{2},j}P_{i+\frac{1}{2},j}H^{3}_{ i-\frac{1}{2},j}P_{i-\frac{1}{2},j}P_{i-\frac{1}{2},j}P_{i,j}}{(\Delta\theta)^{2}} \\ [\frac{\partial}{\partial\lambda}(H^{3}\frac{\partial P}{\partial\lambda})]_{ ij}=\frac{H^{3}_{i+\frac{1}{2},j}P_{i+\frac{1}{2},j}P_{i-\frac{1}{2},j}P_{i- \frac{1}{2},j+\frac{1}{2},j}P_{i,j}}{(\Delta\lambda)^{2}}\\ \end{matrix} \right.