Does the image of a moving square under the map $(r,\theta) \mapsto (r,\theta +\log r)$ converges? Let $\phi:\mathbb{R}^2 \setminus\{0\} \to \mathbb{R}^2 \setminus\{0\}$ be defined in polar coordinates by $$ \phi:(r,\theta) \mapsto (r,\theta +\log r).$$
Let $S_n=[n-1,n+1]^2$ be a square of edge length $2$ centered around $(n,n)$.
Does $\phi(S_n)$ converges to a parallelogram when $n \to \infty$ w.r.t the Hausdorff distance (up to a translation and rotation which depend on $n$)? If so, what are the edge-lengths/angles of that parallelogram? If not, does it converge to something else? Can we characterize the limit somehow?
Note that $\phi$ is area-preserving. It is the one-time flow of the divergence-free vector field $\log(r)\, \partial_{\theta}$.
Heuristically, when $n$ is very large, $r$ and $\theta$ are approximately constant along the square $S_n$. I am not sure how to use this formally though.
Note that the corner $C=(x_n,y_n)=(n+1,n+1)"="(r_n,\theta_n)=(\sqrt 2(n+1),\pi/4)$ of $S_n$ is mapped into
$$
\tilde C=(\tilde r_n,\tilde  \theta_n)=\big(\sqrt 2(n+1),\pi/4+\log (\sqrt 2(n+1))\big).
$$
and the corner $B=(x_n,y_n)=(n+1,n-1)"="(r_n,\theta_n)=(\sqrt 2\sqrt{n^2+1},\arctan{\frac{n-1}{n+1}})$ is mapped into
$$
\tilde B=(\tilde r_n,\tilde  \theta_n)=\big(\sqrt 2\sqrt{n^2+1},\arctan{\frac{n-1}{n+1}}+\log (\sqrt 2\sqrt{n^2+1})\big).
$$
Thus, I guess we first need to decide whether the images of the four corners converge (up to translations) to the vertices of a parallelogram.
The  difference of angles of $\tilde B,\tilde C$ diverges, since
$$
\log\big( \frac{n+1}{\sqrt{n^2+1}}\big)\to -\infty
$$
when $n \to \infty$. I am not sure  if this has any relevance though.
Below are pictures of $\operatorname{Image}(\phi)=\phi(S_n)$ for different values of $n$:
$n=2$:

$n=80$:

$n=800$:

 A: Nice question. The easiest is perhaps to linearize the problem in local coordinates around the center of $S_n$. So let $r_n=\sqrt{2}n$ be the radius to the center of $S_n$ and let us consider the image after making a backward rotation by an angle $-\log r_n$, thus fixing the center of $S_n$. Introducing local coordinates  $(x,y)=(n,n)+(x',y')$ with $x',y'$ of order 1 and writing $r=r_n+\delta r_n$ for the corresponding radii we have to first order: $\delta r_n = \frac{1}{\sqrt{2}}(x'+ y') + O(\frac1n)$ and then
up to error terms of $O(\frac1n)$:
$$ V_n = \log \frac{r}{r_n} \frac{\partial}{\partial \theta} =\frac{\delta r_n}{r_n} \times \left( -r\sin\theta \frac{\partial}{\partial x} + r \cos \theta \frac{\partial}{\partial y}\right)  = \frac{x'+y'}{2} \left( - \frac{\partial}{\partial x'} +\frac{\partial}{\partial y'}\right) +O(\frac1n)$$
The limiting vector field $V_\infty$ gives rise to the linear shear flow, $V_\infty^t :(x',y')\mapsto (x',y')+\frac{t}{2} (x'+y')(-1,1)$ which at time one maps the square $S_n$ with corners $(-1,-1), (1,-1), (1,1),(-1,1)$ to a parallelogram with corners $(0,-2), (1,-1), (0,2), (-1,1)$.
Regarding rigorous bounds we use Gronwall's inequality: Consider a bounded convex neighborhood $B$ of the origin, say a ball of radius 3. For $n$ large we have that $V_n$ and $V_\infty$ are both uniformly $L$-Lipshitz in $B$ (the value depends on the chosen norm). We also have
$\|V_n - V_\infty\| \leq \frac{C}{n}$ as $n\to +\infty$. As long as the time $t$ flows of $p=(x',y')$ stays in such a neighborhood it follows that
$$ \| \frac{d}{dt} (V^t_n(p)-V^t_\infty(p)\| \leq \frac{C}{n} + L \|V^t_n(p)-V^t_\infty(p)\|. $$
By Gronwall's Lemma $\|V^t_n(p)-V^t_\infty(p)\|\leq \frac{C}{n} \frac{e^{Lt}-1}{L}$.
This implies that after rotation, $R_{-\log r_n} \circ\phi(S_n)=V_n^1(S_n)$ is at Hausdorff distance less than $\frac{C}{n} \frac{e^L-1}{L}$ from the parallelogram (for $n$ large enough).
