## 4 Nonlinear Mathematical Model of the Bearing Stationary State
### Bearing Model with Slotted SECT
Using the last equation of system (7) and taking into account that \(P_{k,j}^{2}=\Psi_{k}^{j}\), we write an analog of the last Equation (7) in the finite-difference form  
\[\Psi_{k+2}^{j}-4\Psi_{k+1}^{j}+b_{j}\Psi_{k}^{j}-4\Psi_{k-1}^{j}+\Psi_{k-2}^{ j}=c_{j}, \tag{49}\]  
where \(b_{j}=6+a_{j},c_{j}=a_{j}P_{s}^{2},a_{j}=\frac{2\gamma_{j}A_{kp}}{H_{h}^{3}}\). The vector form (49) is the equation  
\[\overline{\Psi}_{k+2}-4\overline{\Psi}_{k+1}+A_{2}\overline{\Psi}_{k}-4 \overline{\Psi}_{k-1}+\overline{\Psi}_{k+1}=\overline{C}_{1}, \tag{50}\]where  
\[\overline{C}_{1}=\left[ \begin{array}{c}c_{0}\\ c_{1}\\ \ldots\\ c_{m-1}\\ c_{m}\end{array} \right],\ A_{2}=\left[ \begin{array}{cccccccc}b_{0}&0&0&\ldots &0&0&0\\ 0&b_{1}&0&\ldots&0&0&0\\ \ldots&\ldots&\ldots&b_{j}&\ldots&\ldots&\ldots\\ 0&0&0&\ldots&0&b_{m-1}&0\\ 0&0&0&\ldots&0&0&b_{m}\end{array} \right].\]  
Let us write Equations (42) for \(i=k-1\) and \(i=k+1\) in the form  
\[\left\{ \begin{array}{l}\overline{\Psi}_{k+2}=-A_{1}\overline{\Psi}_{k+1}- \overline{\Psi}_{k},\\ \overline{\Psi}_{k-2}=-A_{1}\overline{\Psi}_{k-1}-\overline{\Psi}_{k}.\end{array} \right. \tag{51}\]  
Substituting (51) into (50), we obtain the vector equation  
\[-(4E+A_{1})\overline{\Psi}_{k+1}+A_{2}\overline{\Psi}_{k}-(4E+A_{1}) \overline{\Psi}_{k-1}=\overline{C_{1}}. \tag{52}\]  
Let us reduce (52) to the form (42)  
\[\overline{\Psi}_{k+1}+A_{3}\overline{\Psi}_{k}+\overline{\Psi}_{k-1}= \overline{C_{2}}, \tag{53}\]  
where \(A_{3}=-(4E+A_{1})^{-1}A_{2},\overline{C_{2}}=-(4E+A_{1})^{-1}\overline{C_{1}}\), \(E\) is the identity matrix. We write the vector Equations (42) and (53) in the general form  
\[\overline{\Psi}_{i+1}+A_{4}\overline{\Psi}_{i}+\overline{\Psi}_{i-1}= \overline{C_{3}}, \tag{54}\]  
where  
\[A_{4}=\left\{ \begin{array}{l}A_{1},\ i=1,2,\ldots,k-1,k+1,\ldots,n-1\\ A_{3},\ i=k\end{array} \right.\ \overline{C_{3}}=\left\{ \begin{array}{l}0,\ i=1,2, \ldots,k-1,k+1,\ldots,n-1\\ \overline{C_{2}},\ i=k\end{array} \right.\]  
The solution of the finite-difference problems (37)-(40) is found by the matrix sweep method [30]. For this, we will use the recurrence formula  
\[\overline{\Psi}_{i-1}=X_{i}\overline{\Psi}_{i}+\overline{y}_{i}, \tag{55}\]  
where \(X_{i},\overline{y}_{i}\) are matrices of size \((m+1)\times(m+1)\) and vectors of size \(m+1\), \(i=1\), \(2\), \(\ldots\), \(n\). Substituting (55) into (54), we obtain  
\[\overline{\Psi}_{i+1}+(A_{4}+X_{i})\overline{\Psi}_{i}+\overline{y}_{i}= \overline{C}_{3}. \tag{56}\]  
Let us reduce (56) to form (55):  
\[\overline{\Psi}_{i}=-(A_{4}+X_{i})^{-1}\overline{\Psi}_{i+1}-(A_{1}+X_{i})^{-1 }\big{(}\overline{y}_{i}-\overline{C}_{3}\big{)}. \tag{57}\]  
Comparing (57) with (55), we find recurrent formulas for calculating the sweep matrices and vectors  
\[X_{i+1}=-(A_{4}+X_{i})^{-1},\overline{y}_{i+1}=X_{i+1}\big{(}\overline{y}_{i}- \overline{C}_{3}\big{)},\ i=1,2,\ldots,n-1. \tag{58}\]  
We find the initial matrix and vector \(X_{1},\overline{y}_{1}\). To do this, write (42) for \(i=1\)  
\[\overline{\Psi}_{2}+A_{1}\overline{\Psi}_{1}+\overline{\Psi}_{0}=0. \tag{59}\]  
Adding Equations (45) and (59), we obtain the equation  
\[(A_{1}+4E)\overline{\Psi}_{1}-2\overline{\Psi}_{0}=0. \tag{60}\]Let us reduce (60) to the form (55):  
\[\overline{\Psi}_{0}=\left(\frac{1}{2}A_{1}+2E\right)\overline{\Psi}_{1}. \tag{61}\]  
Comparing (61) with (55), we find the initial sweep matrix and vector:  
\[X_{1}=\frac{1}{2}A_{1}+2E,\overline{y}_{1}=0. \tag{62}\]  
Using (62), sweep matrices and vectors were found by a forward sweep with the formulas (55) for \(i=1,\,2,\,\ldots,\,n-1\). The reverse sweep was performed using the vector (39) and formulas (55) for \(i=n,\,n-1,\,\ldots,\,1\). As a result, the field of the pressure square \(P^{2}\) in the carrier gap and the vector of the pressure square \(P_{p}{}^{2}\) in the throttle chambers were obtained.