Solving the continuous mice problem The classic Mice Problem considers 4 mice standing on the four vertices of a square, and at some point every mouse starts running after its clockwise neighbour in a clockwise manner.
It is not too difficult to prove that if the mice change their direction every fixed amount of time/distance traveled, then their paths are going to be line segments that approximate a logarithmic spiral.
However it seems to me that the spiral's properties depends heavily on the "sampling rate," and I'd like to find a solution for the general, continuous case (so their paths aren't straight at any point, because their targets keep moving). I tried to write the proper differential equations, but they seem too difficult for me to solve; has this problem been addressed before?
 A: 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.
A: You can solve the corresponding differential equation without even considering polar coordinates.
Let's assume that the mice are at the four corners $(1,1)^T$, $(1,-1)^T$, $(-1,-1)^T$ and $(-1,1)^T$.
Denote by $x_i(t)$ for $\,i=1,2,3,4\,$ the trajectories of the mice corresponding to each starting corner.
Then:
$$x_i'(t)=\frac{x_{i+1}(t)-x_i(t)}{||x_{i+1}(t)-x_i(t)||}\quad\mbox{with}\quad x_5=x_1.$$
Due to symmetry, $x_{i+1}(t)$ is $x_i(t)$ up to a 90-degree rotation, i.e.
$$x_{i+1}(t)=\begin{pmatrix}0 & 1\\-1 & 0\end{pmatrix}x_i(t).$$
Hence, we can focus on $\,i=1$ (the other curves follow from 90-degree rotations).
Let $\,x(t)=x_1(t)$, then
$$x'(t)=\frac{M\,x(t)}{||M\,x(t)||}\quad\mbox{with}\quad M=\begin{pmatrix}-1 & 1\\-1 & -1\end{pmatrix}.$$
Let's solve the following problem first:
$\,x_\ell'(t)=M\,x_\ell(t)\,$ with $\,x_\ell(0)=(1,1)^T.$
This is a linear ODE with the solution:
$$x_\ell(t)=\mathrm{e}^{-t}\begin{pmatrix}\cos(t)+\sin(t)\\\cos(t)-\sin(t)\end{pmatrix}.$$
The solution of the original problem follows now from the ansatz: $\,x(t)=x_\ell(\phi(t))$.
Note $\,||M\,x_h(\phi(t))||=2\,\mathrm{e}^{-\phi(t)}\,$ because $\,\sin^2(t)+\cos^2(t)=1$.
Hence, we need to solve $\,\phi'(t)=\frac{1}{2}\,\mathrm{e}^{\phi(t)}\,$ with $\,\phi(0)=0$.
This is a separable ODE with solution $\,\phi(t)=-\log(1-\frac{t}{2})$.
Overall, we get the solution:
$$x_1(t)=(1-\frac{t}{2})\begin{pmatrix}\cos(\log(1-\frac{t}{2}))-\sin(\log(1-\frac{t}{2}))\\\cos(\log(1-\frac{t}{2}))+\sin(\log(1-\frac{t}{2}))\end{pmatrix}.$$
