## 3 Linear Mathematical Model of the Bearing Stationary State
### Linearization of the Stationary Model by the Small Parameter Method  
The response to small loads \(F\) will be the small changes in the clearance (2) and pressure at the outlet of the feeders:  
\[P_{k}(\varphi)=P_{R0}+U_{k}\varepsilon\text{cos}\varphi, \tag{20}\]  
\[P_{p}(\varphi)=P_{p0}+U_{p}\varepsilon\text{cos}\varphi, \tag{21}\]  
where eccentricity \(\varepsilon\) is a small parameter.  
Representing, by analogy, the pressure distributed in the lubricant gap in the form  
\[P(Z,\varphi)=P_{0}(Z)+U(Z)\varepsilon\text{cos}\varphi, \tag{22}\]  
substituting (25) into the boundary value of Problems (1)-(5) and performing its linearization, we obtain the boundary value problem  
\[\left\{\begin{array}{c}\frac{d^{2}(P_{0}U)}{dZ^{2}}-P_{0}U=0,\\ \frac{dU}{dZ}(0)=0,U(L_{1})=U_{k},\ 0\leq Z\leq L_{1},\\ U(0)=U_{k},U(L_{2})=0,\ 0\leq Z\leq L_{2}.\end{array}\right. \tag{23}\]  
The solution to Problem (23) is the function  
\[U(Z)=\left\{\begin{array}{c}U_{k}\frac{\text{sh}Z}{\kappa L_{1}},\ Z\leq L_{ 1},\\ \frac{P_{0}|U_{k}|\text{sh}(L_{1}-Z)}{P_{0}(Z)\text{sh}L_{2}},\ Z\leq L_{2}. \end{array}\right. \tag{24}\]  
The bearing capacity is calculated by the formula from [25]  
\[W=\frac{1}{\pi}\int_{0}^{2\pi}\int_{0}^{L}(P-1)\text{cos}\varphi dZd\varphi= \frac{1}{\pi}\int_{0}^{2\pi}\text{cos}^{2}\varphi d\varphi\int_{0}^{L}UdZ= \varepsilon\int_{0}^{L}U(Z)dZ. \tag{25}\]  
Substituting (24) into (25), we find  
\[W=A_{wk}U_{k}\varepsilon, \tag{26}\]  
where \(A_{wk}=\text{th}L_{1}+\frac{P_{0}}{\text{sh}L_{1}}\int_{0}^{L_{2}}\frac{ \text{sh}(L_{2}-Z)}{P_{0}(Z)}dZ\).  
In the inter-throttle region, the pressure is determined by Formula (21), so the reaction of the pressure forces is  
\[W_{p}=\frac{R_{1}}{\pi}\int_{0}^{2\pi}\int_{0}^{L_{2}}(P_{p}-1)\text{cos} \varphi dZd\varphi=A_{wp}U_{p}\varepsilon, \tag{27}\]  
where  
\[A_{wp}=L_{3}R_{1}.\]  
The flow rate at the inlet to the carrier gap is determined by the formula  
\[Q_{q}=H^{3}\frac{dP^{2}}{dZ}_{Z=L_{1}-}-H^{3}\frac{dP^{2}}{dZ}_{Z=L_{1}+} \tag{28}\]Performing linearization (28) and substituting (20) and (24) into the resulting expression, we found  
\[Q_{q}=(A_{0}U_{k}-A_{1})\varepsilon, \tag{29}\]  
where  
\[A_{0}=2P_{k0}(\text{th}L_{1}+\text{cth}L_{2}),A_{1}=\frac{3}{L_{2}}\Big{(}P_{k0 }^{2}-1\Big{)}.\]  
For the slotted feeders, the flow rates determined by Expressions (9) after linearization take the form  
\[\left\{\begin{array}{c}Q_{k}=\big{(}A_{2}U_{k}+A_{3}U_{p}-A_{4}\big{)} \varepsilon,\\ Q_{p}=A_{5}U_{p}\varepsilon,\end{array}\right. \tag{30}\]  
where  
\[A_{2}=-2A_{k}P_{k0},A_{3}=2A_{k}P_{p0},A_{4}=0,A_{5}=-2A_{p}P_{p0}.\]  
For the simple and annular diaphragms, taking into account (10), we obtain  
\[A_{2}=A_{k}\frac{\partial\Pi(P_{p0},P_{k0})}{\partial P_{k0}},A_{3}=A_{k}\frac {\partial\Pi(P_{p0},P_{k0})}{\partial P_{p0}},A_{4}=A_{k}\Pi(P_{p0},P_{k0}),A_ {5}=A_{p}\frac{\partial\Pi(P_{s},P_{p0})}{\partial P_{p0}}.\]  
Substituting (29) and (30) into the last two Equations (7), we get a system of linear equations for the unknown coefficients \(U_{k}\) and \(U_{p}\):  
\[\left[\begin{array}{cc}(A_{2}-A_{0})&A_{3}\\ A_{2}&(A_{3}-A_{5})\end{array}\right]\left[\begin{array}{c}U_{k}\\ U_{p}\end{array}\right]=\left[\begin{array}{c}A_{4}-A_{1}\\ A_{4}\end{array}\right]. \tag{31}\]  
Solving (31), we find  
\[U_{k}=\frac{(A_{4}-A_{1})(A_{3}-A_{5})-A_{3}A_{4}}{(A_{2}-A_{0})(A_{3}-A_{5}) -A_{2}A_{3}}J_{p}=\frac{A_{4}(A_{2}-A_{0})-A_{2}(A_{4}-A_{1})}{(A_{2}-A_{0})( A_{3}-A_{5})-A_{2}A_{3}}. \tag{32}\]  
The bearing compliance is  
\[K=\frac{\varepsilon_{s}}{F}=\frac{1+K_{e}\big{(}W-W_{p}\big{)}}{W}. \tag{33}\]  
From (33), it follows that the bearing has zero compliance at  
\[K_{e}=\frac{1}{W_{p}-W}. \tag{34}\]