## 2 Mathematical Modeling
### Lubrication Model
#### 2.3.2 Pressure Distribution in the Liquid Film
\begin{array}{l}\int_{0}^{h}\rho\bigg{(}v\frac{\partial v}{\partial \tau}+w\frac{\partial v}{\partial z}\bigg{)}dn=-\text{h}\frac{\partial p}{ \partial\tau}+\int_{0}^{h}\mu\bigg{(}\frac{\partial^{2}v}{\partial n^{2}} \bigg{)}dn\\ \int_{0}^{h}\rho\bigg{(}v\frac{\partial w}{\partial\tau}+w\frac{\partial w}{ \partial z}\bigg{)}dn=-\text{h}\frac{\partial p}{\partial z}+\int_{0}^{h}\mu \bigg{(}\frac{\partial^{2}w}{\partial n^{2}}\bigg{)}dn\end{array} \right. \tag{10}\]  
The integral continuous equation is given by:  
\[\int_{0}^{h}\bigg{(}\frac{\partial v}{\partial\tau}+\frac{\partial w}{\partial z }\bigg{)}dn=0 \tag{11}\]  
Equations (10) and (11) are the governing equations of the liquid flow in the meshing pair clearance. However, solving the governing equations is still very difficult because of the inclusion of the nonlinear inertia item and the complex gap shape \(h\). To simplify this problem, the tooth flank is divided to many infinitesimals. The infinitesimal width is \(dl\), as shown in Figure 9. In infinitesimal \(dl\), \(\partial\)/\(\partial\)z is far smaller than \(\partial\)/\(\partial\)\(\tau\). Therefore, the tooth flank and the screw groove flank in the width \(dl\) can be viewed as an infinite-width thrust bearing. Accordingly, the whole flank is composed of many infinite-width thrust bearings in a row. Figure 9: The infinitesimal of the tooth flank. By introducing the mass mean velocity \(v_{m}\), [PERSON] and [PERSON] linearized the inertial item of the equations which is similar to the governing equations and deduced the pressure distribution [39]:  
\[\frac{\mathrm{d}p}{d\tau}=\frac{12\mu}{h^{2}}\Big{(}\frac{v_{\mathrm{r}}}{2}-v_{ m}\Big{)}+\frac{\rho}{h}\big{(}1.2v_{m}^{2}-0.133v_{\mathrm{r}}{}^{2}\big{)} \frac{\mathrm{d}h}{d\tau} \tag{12}\]  
where  
\[v_{m}=\frac{1}{\rho h}\int_{0}^{h}vdh \tag{13}\]  
The boundary conditions are given as follows:  
\[p(\tau_{0})=P_{0},p(\tau_{1})=P\big{(}\phi_{gr}\big{)} \tag{14}\]  
The analytic solution of the water film pressure distribution can be deduced by integrating Equation (12):  
\[ \begin{array}{ll}p(\tau)=&P_{0}+\int_{\tau_{0}}^{\tau}\frac{6\mu v_{\mathrm{ r}}}{h^{2}}d\tau-\frac{12\mu Q}{\rho}\int_{\tau_{0}}^{\tau}\frac{1}{h^{3}}d\tau-0.6 \frac{Q^{2}}{\rho}\Big{(}\frac{1}{h(\tau_{0})^{2}}-\frac{1}{h(\tau_{0})^{2}} \Big{)}-\\ &0.133\rho v_{\mathrm{r}}{}^{2}[\ln h(\tau)-\ln h(\tau_{0})]\end{array}  \tag{15}\]  
where  
\[Q=\frac{\frac{12\mu}{\rho}\int_{\tau_{0}}^{\tau_{1}}\frac{1}{h^{3}}d\tau- \sqrt{\Big{(}\frac{12\mu}{\rho}\int_{\tau_{0}}^{\tau_{1}}\frac{1}{h^{3}}d\tau \Big{)}^{2}+\frac{2}{\rho}\Big{(}\frac{1}{h(\tau_{1})^{2}}-\frac{1}{h(\tau_{0 })^{2}}\Big{)}T}}{-\frac{12}{\rho}\Big{(}\frac{1}{h(\tau_{1})^{2}}-\frac{1}{h( \tau_{0})^{2}}\Big{)}} \tag{16}\]  
where  
\[T=-P(\phi_{\mathrm{gr}})+P_{0}+\int_{\tau_{0}}^{\tau_{1}}\frac{6\mu v_{ \mathrm{r}}}{h^{2}}d\tau-0.133\rho v_{\mathrm{r}}{}^{2}[\ln h(\tau_{1})-\ln h (\tau_{0})] \tag{17}\]  
In conclusion, in the infinitesimal \(dl\), the water film pressure distributes in one dimension along \(\tau\) direction. Since the sliding velocity and the clearance shape in every infinitesimal are different, it is necessary to calculate pressure distribution in each infinitesimal to obtain the quasi-two-dimensional pressure information of the whole tooth flank.