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For a vector-valued Compound Poisson process, its coordinate processes are independent exactly when its Lévy measure is supported on the union of the coordinate axes. A jump can then change only one coordinate. Splitting the marked Poisson process by the axis of its mark gives independent coordinate jump processes. Conversely, independent finite-activity coordinates have no simultaneous nonzero jumps almost surely, excluding Lévy mass away from the axes.

For a positive rate and two-dimensional marks $(g_1(U),g_2(U))$, with $U$ uniform on $[0,1]$, the condition is $g_1(U)g_2(U)=0$ almost surely. If both functions are continuous, it is equivalent toA nonzero value of the continuous product would persist on an interval of positive uniform probability. Without continuity only the almost-everywhere condition is necessary.

For example, marks $((1/2-U{)}^{+},(U-1/2{)}^{+})$ give independent coordinates by thinning into the two half-intervals. Marks $(U,1-U)$ produce common jumps in both coordinates almost surely, so the coordinates are dependent even though a single scalar uniform generates the marks. The positive-rate qualification is needed: a zero-rate process has independent constant coordinates irrespective of its unused mark functions.

Here $m{h}^2=b$, so the Binet equation reduces toThe initial conditions becomeMultiply the differential equation by $2u^{\prime }$ and use the chain rule:The initial conditions give $C=0$. Since $e^{-u}>0$, $u^{\prime }$ cannot change sign at a finite value of $u$. Thus $u^{\prime }=e^{-u/2}$ on the physical branch. Separation of variables givesThereforeThe natural logarithm must be positive because $r>0$. For the forward trajectory, $\theta \ge 0$ and $h=\sqrt{b/m}>0$, so $\dot{\theta }=h/r^2>0$: the particle travels counterclockwise while $r$ strictly decreases.

For $a=1$, it starts at $(x,y)=(1,0)$, with tangent vector proportional to $(-e^{-1/2},1)$. Its successive crossings of the positive $x$ coordinate axis have radiiIt makes infinitely many inward turns, approaching the origin as $\theta \to \infty$. There is no finite-time arrival at the origin: $dt/d\theta =r^2/h$ behaves as $1/(4h{\log }^2\theta )$, whose improper integral diverges. If extended backwards, it comes from infinity as $\theta \downarrow 2(1-\sqrt{e})$.