## 4 A Close Form Solution to the Flow and Dynamic Force Coefficients in a PGB
Once found, integration of the pressure field (\(p\)) acting on the rotor surface gives the bearing reaction force (\(\mathbf{F}\)) with components  
\[-\left[ \begin{array}{c}F_{X}\\ F_{Y}\end{array} \right]=\oint\!\left(P_{0}-P_{a}+(\delta e_{X}\,P_{X}+\,\, \delta e_{Y}P_{Y})\,e^{i\,\omega\,t}\right)\left[ \begin{array}{c}\cos\theta \\ \sin\theta\end{array} \right]Rd\theta dz. \tag{20}\]  
As the bearing is centered, the zeroth order pressure field does not produce a static force, i.e., \(\mathbf{F}=0\). The first order pressure fields produce the matrix of complex dynamic stiffnesses \(\mathbf{H}\), where \(H_{ij}=\left(K_{ij}+i\,\,\omega\,\,C_{ij}\right)_{ij}\!\!=\!\!\mathbf{x}\). \[\left[ \begin{array}{c}H_{XX}\\ H_{YX}\end{array} \right]=\left[ \begin{array}{c}H_{YY}\\ -H_{XY}\end{array} \right]=\oint\!-p_{X}\!\left[ \begin{array}{c}\cos\theta \\ \sin\theta\end{array} \right]R\,d\theta dz=-\oint\!\left(\frac{\psi_{X}}{2\,p_{0} }\right)\!\left[ \begin{array}{c}\cos\theta\\ \sin\theta\end{array} \right]R\,d\theta dz. \tag{21}\]  
Then substituting Equation (15) above leads to  
\[H_{XX}=H_{YY}=-\pi R^{2}\int\limits_{0}^{L/D}\frac{\psi_{X_{c}\left(\overline{z }\right)}}{p_{0\left(\overline{z}\right)}}\,\,d\overline{z};\,\,H_{XY}=-H_{YX} =-\pi R^{2}\int\limits_{0}^{L/D}\frac{\psi_{Y_{c}\left(\overline{z}\right)}} {p_{0\left(\overline{z}\right)}}\,\,d\overline{z}. \tag{22}\]  
The exact solution for the first-order fields is as follows: Let \(\psi_{X}=\left[ \begin{array}{cc}\psi_{Xc}&\psi_{X_{s}}\end{array} \right]^{T}\), then Equation (16) has the general solution  
\[\psi_{X}=\psi_{XP1}\cosh(\gamma\overline{z})+\psi_{XP2}+\,\psi_{XH} \tag{23}\]  
where the coefficients of the particular solution are  
\[\psi_{XP1}=-\Big{[}\mathbf{A}-\mathbf{I}\gamma^{2}\Big{]}^{-1}\mathbf{C}_{1}=- \frac{1}{\Delta}\left[ \begin{array}{cc}1+i\overline{\Lambda}_{\omega}&- \overline{\Lambda}_{\Omega}\\ \overline{\Lambda}_{\Omega}&1+i\overline{\Lambda}_{\omega}\end{array} \right] \left[ \begin{array}{c}i\,2\overline{\Lambda}_{\omega}-3\,\gamma^{2}\\ -2\overline{\Lambda}_{\Omega}\end{array} \right]\frac{1}{\cosh\left(\gamma\, \frac{1}{D}\right)}\frac{\psi_{as}}{c} \tag{24}\]  
\[\psi_{XP2}=-\mathbf{A}^{-1}\mathbf{C}_{2}=-\frac{1}{\Delta_{\gamma}}\left[ \begin{array}{cc}(1+\gamma^{2})+i\overline{\Lambda}_{\omega}&-\overline{ \Lambda}_{\Omega}\\ \overline{\Lambda}_{\Omega}&(1+\gamma^{2})+i\overline{\Lambda}_{\omega}\end{array}  \right]\left[ \begin{array}{c}i\,2\overline{\Lambda}_{\omega}\\ -2\overline{\Lambda}_{\Omega}\end{array} \right]\frac{p_{S}^{2}}{c} \tag{25}\]with \(\Delta=\left[1+i\overline{\Lambda}_{\omega}\right]^{2}+\overline{\Lambda}_{\Omega}^{2}\) and \(\Delta_{\gamma}=\left[\left(1+\gamma^{2}\right)+i\overline{\Lambda}_{\omega} \right]^{2}+\overline{\Lambda}_{\Omega}^{2}\) as determinants of the system of equations. \(\psi_{XH}=\widehat{\psi}_{H}e^{\xi\overline{z}}\), the homogenous solution of Equation (16), satisfies  
\[\Big{[}\mathbf{I}\,s^{2}-\mathbf{A}\Big{]}\widehat{\psi}_{H}=\mathbf{0} \tag{26}\]  
whose roots are \(x_{1,2}=\ \left(s_{1,2}\right)^{2}=\left(1+\gamma^{2}\right)+i\big{(}\overline{ \Lambda}_{\omega}\pm\overline{\Lambda}_{\Omega}\big{)}\), and with corresponding eigenvectors \(\widehat{\psi}_{H_{1}}=\left[ \begin{array}{cc}1&i\end{array} \right]^{T}\), \(\widehat{\psi}_{H_{2}}=\left[ \begin{array}{cc}1&-i\end{array} \right]^{T}\). Thus,  
\[\psi_{XH}=\widehat{\psi}_{H}e^{\xi\overline{z}}=d_{1c}\ \widehat{\psi}_{H_{1c}}\cosh( \sqrt{x_{1}}\overline{z})+\ d_{2c}\ \widehat{\psi}_{H_{2c}}\cosh(\sqrt{x_{2}} \overline{z}). \tag{27}\]  
Satisfying the end condition \(\psi_{X(\overline{z}=\frac{1}{D})}=\mathbf{0}\) leads to  
\[\left[ \begin{array}{c}d_{1c}\\ d_{2c}\end{array} \right]=-\left[ \begin{array}{cc}\cosh(\sqrt{x_{1}}\frac{L}{ D})&\cosh(\sqrt{x_{2}}\frac{L}{D})\\ i\ \cosh(\sqrt{x_{1}}\frac{L}{D})&-i\cosh(\sqrt{x_{2}}\frac{L}{D})\end{array}  \right]^{-1}\bigg{[}\psi_{XP_{1}}\cosh(\gamma\frac{L}{D})+\psi_{XP2}\bigg{]}. \tag{28}\]  
Finally, the exact solution is  
\[\psi_{X}=\left[