Solving the continuous mice problem

Solution 1:

I think one should use a time-dependant curve, to avoid division by zero.

Now let there be $n > 1$ entities chasing after one another in a circle like fashion. At the beginning they shall be positioned at the vertices of a regular polygon with $n$ edges.

This set up is symmetric to a rotation of $2π/n$. So the $n$ solution paths will also be symmetric in this manner. Let the centre of the rotation be the centre of an Cartesian coordinate system. Choose a single entity and name its path $φ: ℝ \ni t ↦ φ(t) \in ℝ^2$ with polar coordinates $α: ℝ → ℝ$ and $r: ℝ → ℝ$ such that $φ = r(\cos α, \sin α)$.

The centre, the position of the entity, $φ(t)$, and the position of the next entity, $φ(t)$ rotated by $2π/n$ (say $R_nφ(t)$), form an isosceles triangle. The entity is moving in the direction from $φ(t)$ to $R_nφ(t)$. So the angle between the vectors $φ(t)$ and $φ'(t)$ is $$2π/n + \frac{π - 2π/n}{2} = π\frac{2/n + 1}{2} = π(\frac1n + \frac12)$$ This means $$⟨φ', φ⟩ = ||φ'||\cdot||φ||\cos π(\frac1n+\frac12) ≕ ||φ'||\cdot||φ||\cdot C$$ where $-1≤C<0$. Now $$2⟨φ', φ⟩ = ⟨φ, φ⟩' = (||φ||^2)' = 2||φ||\cdot||φ||'$$ so that $$ ||φ||' = C||φ'||$$ This velocity is assumed to be constant $||φ'|| ≕ v > 0$. Together with the above we get $r' = Cv$. So $r = r_0 + Cvt$. Using the polar coordinates we calculate $$φ' = r'(\cos α, \sin α) + rα'(-\sin α, \cos α)$$ The two summands of $φ'$ are orthogonal so $$v = ||φ'|| = -r' + rα' = -Cv + rα'$$ This in turn gives us $$α' = \frac{v(1+C)}{r} = \frac{v(1+C)}{r_0 + Cvt}$$ if $C > -1$, i.e. $n > 2$ $$α = \frac{v(1+C)}{Cv}\ln r = (\frac1C+1)\ln r\\r = \exp\left(\frac{α}{\frac1C+1}\right)$$ when we start at $α_0 = 0$. For $n=2$ we have $C=-1, α' = 0$ and $α$ is constant.