## 3 General Mathematical Model of the Static State of a Conventional Bearing
### Boundary Value Problems for a Dimensionless Function of Pressure in the Bearing Gap  
When the axes of the shaft 1 and sleeve 3 are parallel, the pressure function \(P(Z,\varphi)\) in the bearing gap satisfies the stationary Reynolds equation [26]  
\[\frac{\partial}{\partial\varphi}\left(H^{3}P\frac{\partial P}{\partial\varphi }\right)+\frac{\partial}{\partial Z}\left(H^{3}P\frac{\partial P}{\partial Z }\right)=0, \tag{1}\]  
where \(Z\), \(\varphi\) are longitudinal and circumferential coordinates. \[H(\varphi)=1-\varepsilon\cos\varphi \tag{2}\]  
is a function of the thickness of the bearing gap; \(\varepsilon\) is an eccentricity. To formulate boundary conditions for Equation (1), we introduced local coordinate systems. We will consider the symmetrical right half of the bearing. In the inter-row zone \(0\leq Z\leq L_{2}\), in the end zone \(0\leq Z\leq L_{1}\), and for both of these zones \(0\leq\varphi\leq 2\pi\), where \(L_{2}=L-L_{1}\) is half the length of the inter-row zone, \(L_{1}\) is the length of the end zones, and \(L\) is half the bearing length. Let us take \(\Psi=P^{2}\). Then, the corresponding boundary value problem for Equation (1) takes the form  
\[\left\{ \begin{array}{l}\frac{\partial}{\partial\varphi}\left(H^{3}\frac{ \partial\Psi}{\partial\varphi}\right)+\frac{\partial}{\partial Z}\left(H^{3} \frac{\partial\Psi}{\partial Z}\right)=0,\\ \frac{\partial\Psi}{\partial Z}(0,\varphi)=0,\,\Psi(L_{1},\varphi)=\Psi_{k}( \varphi),\\ \Psi(0,\varphi)=\Psi_{k}(\varphi),\,\Psi(L_{2},\varphi)=1,\\ \Psi(Z,\varphi)=\Psi(Z,\varphi+2\pi),\\ \frac{\partial\Psi}{\partial\varphi}(Z,\varphi)=\frac{\partial\Psi}{\partial \varphi}(Z,\varphi+2\pi),\end{array} \right. \tag{3}\]  
where \(\Psi_{k}(\varphi)=P_{k}^{2}(\varphi)\) is the squared function of the pressure on the boost line at the air outlet from the damping annular diaphragms. The response to small loads \(F\) will be small changes in the bearing gap, eccentricity \(\Delta\varepsilon\), and the square of the pressures at the feeders' outlet  
\[\Psi_{k}(\varphi)=\Psi_{k0}+\Delta\Psi_{k}\cos\varphi, \tag{4}\]  
\[\Psi_{p}(\varphi)=\Psi_{p0}+\Delta\Psi_{p}\cos\varphi, \tag{5}\]  
where \(\Psi_{k0}=P_{k0}^{2},\Psi_{p0}=P_{p0}^{2}\) are the values of the function \(\Psi\) at the exit from the annular and simple diaphragms with the coaxial arrangement of the shaft and sleeve, \(\Delta\Psi_{k},\Delta\Psi_{p}\) are small deviations. By analogy, let us take  
\[\Psi(Z,\varphi)=\Psi_{0}(Z)+\Delta\Psi(Z)\cos\varphi. \tag{6}\]  
Substituting (6) into (1) and performing the separation of functions, taking into account (4), we obtain the boundary value problem for the function \(\Psi_{0}(Z)\)  
\[\left\{ \begin{array}{l}\frac{d^{2}\Psi_{0}}{dZ^{2}}=0,\\ \frac{d\Psi_{0}}{dZ}(0)=0,\Psi_{0}(L_{2})=\Psi_{k0},\\ \Psi_{0}(0)=\Psi_{k0},\Psi_{0}(L_{1})=1\end{array} \right. \tag{7}\]  
and the boundary value problem for the function of small deviations \(\Delta\Psi(Z)\)  
\[\left\{ \begin{array}{l}\frac{d^{2}\Delta\Psi}{dZ^{2}}-\Delta\Psi=0,\\ \Delta\Psi(0)=\Delta\Psi_{k},\Delta\Psi(L_{1})=0,\quad 0\leq Z\leq L_{1},\\ \frac{d\Delta\Psi}{dZ}(0)=0,\Delta\Psi(L_{2})=\Delta\Psi_{k},\quad 0\leq Z \leq L_{2}.\end{array} \right. \tag{8}\]  
The solution to Problem (7) is the function  
\[\Psi_{0}(Z)=\left\{ \begin{array}{l}\left(P_{k0}^{2}-1\right)\frac{L_{1}-Z}{ L_{1}}+1,\quad 0\leq Z\leq L_{1},\\ P_{k0}^{2},\quad 0\leq Z\leq L_{2}.\end{array} \right. \tag{9}\]  
The solution to Problem (8) is the function  
\[\Delta\Psi(Z)=\left\{ \begin{array}{l}\Delta\Psi_{k}\frac{\mathrm{sh}(L_{1}-Z )}{\mathrm{sh}L_{1}},\quad Z\leq L_{1},\\ \Delta\Psi_{k}\frac{\mathrm{ch}Z}{\mathrm{ch}L_{2}},\quad Z\leq L_{2}.\end{array} \right. \tag{10}\]  
The dimensional bearing capacity of the bearing is calculated by the following formula [26]:  
\[w=2\int\limits_{0}^{2\pi}\int\limits_{0}^{L}\left(p-p_{a}\right)\cos\varphi dzd\varphi. \tag{11}\]After reducing (11) to a dimensionless form, we obtain  
\[