## 5 Mathematical Model of the Unsteady State of the Bearing
### Method for Determining the Transfer Function of a Dynamic System  
The pressure function \(P(Z,\varphi,\tau)\) in the carrier gap satisfies the dimensionless nonstationary Reynolds equation [25]  
\[\frac{\partial}{\partial\varphi}\left(H^{3}P\frac{\partial P}{\partial\varphi} \right)+\frac{\partial}{\partial Z}\left(H^{3}P\frac{\partial P}{\partial Z} \right)=\sigma\frac{\partial(PH)}{\partial\tau}, \tag{75}\]  
where \(Z\) and \(\varphi\) are dimensionless longitudinal and circumferential coordinates, \(\tau\) is the dimensionless time, \(\sigma=\frac{12\mu_{0}^{2}}{h_{0}^{2}p_{a}t_{0}}\) is the so-called squeezing number [27] and \(\sigma=\frac{12\mu_{0}^{2}}{h_{0}^{2}p_{a}t_{0}}\) is the scale of the current time. Figure 10: Comparative nonlinear (solid curves) and linear (dashed straight lines) dependences of the total eccentricity \(\varepsilon_{s}\) of the throttle SECT on the external load F for different values of the elasticity coefficient \(K_{e}\). At low loads \(\Delta F(\tau)\), the changes in the gap and pressures on the injection line will be small:  
\[H(\varphi,\tau)=1-\Delta H=1-\Delta\varepsilon(\tau)\text{cos}\varphi, \tag{76}\]  
\[P_{k}(\varphi,\tau)=P_{k0}+\Delta P_{k}(\tau)\text{cos}\varphi, \tag{77}\]  
\[P_{p}(\varphi,\tau)=P_{p0}+\Delta P_{p}(\tau)\text{cos}\varphi. \tag{78}\]  
By an analogy with (77) and (78), we represent the pressure distributed in the lubricant gap in the form  
\[P(Z,\varphi,\tau)=P_{0}(Z)+\Delta P(Z,\tau)\text{cos}\varphi. \tag{79}\]  
Substituting (76) and (79) into (75), performing linearization and the subsequent Laplace transform [35], we obtain the problem  
\[\left\{ \begin{array}{c}\frac{d^{2}\overline{\Delta\overline{\tau}}}{dZ^{2}} -\overline{\Delta\overline{\Psi}}=\text{cos}\left(\frac{\overline{\Delta \overline{\tau}}}{P_{0}}-P_{0}\overline{\Delta\varepsilon}\right),\\ \frac{d\overline{\Delta\overline{\tau}}}{dZ}(0,s)=0,\,\overline{\Delta \overline{\Psi}}(L_{1},s)=\overline{\Delta\overline{\Psi}}_{k},\\ \overline{\Delta\overline{\Psi}}(L,s)=0,\\ \overline{\Delta\overline{\Psi}}=P_{0}\overline{\Delta\overline{\Psi}},\, \overline{\Delta\overline{\Psi}}_{k}=P_{k0}\overline{\Delta\overline{P}}_{k}, \end{array} \right. \tag{80}\]  
where \(s\) is the Laplace transform variable, and \(\overline{\Delta\varepsilon}(s),\overline{\Delta\overline{P}}_{k}(s), \overline{\Delta\overline{P}}(s)\) are the Laplace transformants of the deviations of pressures \(P\) and \(P_{k}\) and eccentricity \(\varepsilon\) from their equilibrium state. Problem (80) has no analytical solution; therefore, we apply the superposition method to it and represent the required function as a linear combination:  
\[\overline{\Delta\overline{\Psi}}=U_{k}(Z,s)\overline{\Delta\overline{\Psi}}_{ k}+U_{\varepsilon}(Z,s)\overline{\Delta\varepsilon}. \tag{81}\]  
Substituting (81) into (80) and separating the transformants, we obtain two boundary value problems for the functions U, which can be represented in the following general form:  
\[\left\{ \begin{array}{c}U^{\prime\prime}-G_{1}U+G_{2}=0,\\ U^{\prime}(0)=0,U(L_{1})=(1-\alpha),U(L)=0,\end{array} \right. \tag{82}\]  
where \(\alpha=0\) corresponds to \(U_{k}\), \(\alpha=1-U_{\varepsilon}\), \(G_{1}(Z)=1+\frac{\text{cos}}{P_{0}(Z)}\), \(G_{2}(Z)=\alpha\text{s}P_{0}(Z)\), and \(P_{0}(Z)\) is the stationary pressure distribution function (14). The solution to Problem (82) was obtained by the finite-difference sweep method [29]. To do this, we divided the integration segment \([0,L]\) into an even quantity \(n\) of equal parts and replaced the differential equations with their algebraic analogs:  
\[\frac{U_{i+1}-2U_{i}+U_{i-1}}{v^{2}}-G_{1,i}U_{i}+G_{2,i}=0, \tag{83}\]  
where \(\upnu=L/n\) is the grid step, \(i=1\), \(2\), \(\ldots\), \(n-1\) is the number of its node and \(G_{1,i}=1+\frac{\text{cos}}{P_{0,i}},G_{2,i}=\alpha\text{s}P_{0,i},P_{0,i}=P_{ 0}(Z_{i}),Z_{i}=i\
u\).