= Solution
Put the fixed outer centre at the origin, and let the moving inner centre be $\mathbf d=(x,z)$. This reverses the sign of the vertical offset used in the previous parts: now $h=\Delta-\mathbf d\cdot\mathbf n$. At concentricity rotation produces no <force> and translation has isotropic resistance $R_0=12\pi\mu a^3/\Delta^3$. Balancing the external load gives
$$
\boxed{\dot x(0)=0,\qquad \dot z(0)=-\frac{F\Delta^3}{12\pi\mu a^3}.}
$$
There is no inertial acceleration in this <Stokes flow> model; <velocity> is determined instantaneously by <force> balance.
For the subsequent motion, the <full-film journal bearing> assumption matters: <pressure> is single-valued around the whole annulus, with no specified <cavitation> boundary. Let $r=|\mathbf d|$, $e=r/\Delta$, $\phi=\arg(x+iz)$, and $\mathbf e_r,\mathbf e_\phi$ be its radial and tangential unit <vectors>. The <squeeze resistance of an eccentric journal bearing> has radial and tangential coefficients
$$
R_r=\frac{R_0}{(1-e^2)^{3/2}},\qquad
R_t=\frac{R_0}{\sqrt{1-e^2}(1+e^2/2)}.
$$
For completeness, rotate coordinates until $\mathbf d$ points along $x$, so $h=\Delta(1-e\cos\theta)$. Radial translation gives squeezing flux $q=av_r\sin\theta$ and resistance proportional to $\int\sin^2\theta/(1-e\cos\theta)^3=\pi/(1-e^2)^{3/2}$. Tangential translation gives $q=av_t[-\cos\theta+3e/(2+e^2)]$; the constant enforces periodic <pressure>. Its resistance uses
$$
\int\frac{\cos^2\theta}{(1-e\cos\theta)^3}d\theta
-\frac{\left[\int\cos\theta/(1-e\cos\theta)^3d\theta\right]^2}{I_3}
=\frac{\pi}{\sqrt{1-e^2}(1+e^2/2)}.
$$
These expressions supply both translation directions; retaining only the vertical squeezing formula would miss the moving axle's horizontal dynamics.
The rotational fluid <force> from the earlier calculation points along $\mathbf e_\phi$, with magnitude $R_t\Omega r/2$. Superposing rotation and translation by <Linearity of Stokes flow>, and resolving the load, gives
$$
\boxed{R_r\dot r=-F\sin\phi,\qquad
R_t r\left(\dot\phi-\frac\Omega2\right)=-F\cos\phi.}
$$
For positive $\Omega$ the axle initially moves down, then to the right because the rotation-generated <force> is perpendicular to the downward eccentricity. The equilibrium is displaced horizontally to the right, at the same height as the outer axis. Its eccentricity $0<e_*<1$ is the unique solution
$$
\boxed{x_* =\Delta e_*,\quad z_*=0,\qquad
F=\frac{6\pi\mu\Omega a^3}{\Delta^2}\frac{e_*}{\sqrt{1-e_*^2}(1+e_*^2/2)}.}
$$
The rotational <force> then points upwards and balances the load. For negative $\Omega$ the displacement is to the left, using $|\Omega|$ in the magnitude equation. Uniqueness follows because the logarithmic derivative of $e/[\sqrt{1-e^2}(1+e^2/2)]$ has positive numerator $1-e^2/2+e^4$ after a common positive denominator. If $\Omega=0$ and $F>0$, no separated static equilibrium exists in this ideal model. If $F=0$, concentricity remains a solution.
One should not assert that this full-film model necessarily settles to the equilibrium. It predicts <full-film journal-bearing whirl>. For $F,\Omega>0$, eliminate time from the two boxed evolution equations and put $K=6\pi\mu\Omega a^3/\Delta^2$. The orbit starting at concentricity obeys
$$
\frac d{de}x+\frac{3e}{2(1-e^2)}x=\frac{\Omega\Delta^2R_0}{2F}\frac e{(1-e^2)^{3/2}},
$$
so, with its zero initial horizontal displacement,
$$
\boxed{\frac x\Delta=\frac{2K}{5F}\left[(1-e^2)^{-1/2}-(1-e^2)^{3/4}\right],\qquad
z=\pm\Delta\sqrt{e^2-(x/\Delta)^2}.}
$$
The orbit begins on the lower branch, turns at $z=0$ before contact, and returns along the upper branch, forming a closed orbit through concentricity. The ratio of the right-hand side to $e$ increases from zero to infinity, giving a unique turning eccentricity below one. Linearizing at equilibrium likewise gives $\delta\dot r=-(F/R_{r,*})\delta\phi$ and $\delta\dot\phi=c_*\delta r$ with $c_*>0$: the <eigenvalues> are imaginary, not damped. Thus the equilibrium position exists, but the initially concentric axle circulates around it in the stated ideal full-film approximation. <Cavitation> or additional mechanical constraints would change the model; none were specified.
The normalized centre trajectory below illustrates the full-film prediction for $K/F=1$. The dashed circle is the contact locus of the axle centre, not either cylinder wall.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2017/iii/paper-329-bearing-whirl.png]
{title=Journal-bearing geometry and whirl}
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