$\newcommand{\tr}{\operatorname{tr}}$
$\renewcommand{\div}{\operatorname{div}}$


Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Given a vector field $X$ on $D$, let $\psi\_t$ be its flow.


Does there exist a (non-homothetic) divergence-free vector field $X \in \Gamma(D)$ whose associated flow $\psi\_t \in \operatorname{diff}(D)$ has singular values which depend only on $t$ and not on the point in $D$? Equivalently, I want $|(d\psi\_t)\_p|^2$ to be independent of $p$.


By requiring that $X$ is not homothetic, I am excluding the possibility of $L\_x g=\lambda g$ for some constant $\lambda$.




---


Write $f(t,p)=|(d\psi\_t)\_p|^2=\tr\_g\big((\psi\_t^\*g)\_p\big)$.
A necessary and sufficient condition on $X$ is that $$\frac{\partial }{\partial t}f(t,p)=\tr\_g(\psi\_t^\*L\_Xg)$$ would be independent of $p$, for all $t$. Can we write this as an explicit PDE on $X$?


Some partial information [is obtained](https://math.stackexchange.com/questions/3804878/how-can-we-differentiate-twice-the-trace-of-pullback-metric-via-a-flow) from differentiating twice at $t=0$:


$$
\frac{\partial^2 }{\partial t^2}\left|\_{t=0}\right.\tr\_g\big((\psi\_t^\*g)\_p\big)=\operatorname{tr}\_g(\mathcal{L}\_X(\mathcal{L}\_Xg))=2X(\operatorname{div}X)+2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X)=
$$
$$
2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X),
$$
so $\|\nabla X\|^2+\operatorname{tr}(\nabla X\circ\nabla X)$ must be constant.




---


$X(r,\theta)=\log r \frac{\partial}{\partial \theta}$ is an example on $D\setminus\{0\}$:


Its flow is $\psi\_t: (r,\theta)\mapsto (r,\theta+t\log r)$;
$[d\psi\_t]\_{\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}}=\begin{pmatrix} 1 & 0 \\\ t & 1\end{pmatrix}$
is independent of $p$.



A vector field whose flow has constant singular values
The vector fields $X$ that satisfy this condition are highly constrained, as there exist only a finite-dimensional family of $C^5$ solutions in a neighborhood of any given point in the plane. Here is how one can see this:


Using the standard coordinates, we have $g = dx^2 + dy^2$. Suppose that $X$ is defined on a simply-connected domain $D$ in the plane. Then, since $X$ is divergence-free, it can be written in the form
$$
X = u\_y\,\frac{\partial}{\partial x} - u\_x\,\frac{\partial}{\partial y}
$$
for some function $u$ on $D$. Define $\mathcal{L}\_X^2g = \mathcal{L}\_X(\mathcal{L}\_X g)$ and, inductively $\mathcal{L}\_X^{k+1}g = \mathcal{L}\_X(\mathcal{L}^k\_X g)$. As has been observed, we have $\mathrm{tr}\_g(\mathcal{L}\_X g) = 0$, since $X$ is divergence free. A direct computation gives
$$
\mathrm{tr}\_g(\mathcal{L}^2\_X g) = 2\left((u\_{xx}-u\_{yy})^2 + (2u\_{xy})^2\right) = 8c^2
$$
for some constant $c\ge0$. If $c=0$, then either $X = a\_0\,(y-y\_0)\partial\_x - a\_0\,(x-x\_0)\partial\_y$ for some constants $a\_0,a\_1,a\_2$, and the flow of $X$ is rotation around $(x\_0,y\_0)$ or $X = a\_1\partial\_x + a\_2\partial\_y$ and hence the flow of $X$ is simply translation.


Thus, from now on assume that $c>0$. Then the Hessian matrix of $u$ is of the form
$$
\begin{pmatrix} u\_{xx} & u\_{xy}\\u\_{xy} & u\_{yy}\end{pmatrix}
= \begin{pmatrix} h + c\,\cos v & c\,\sin v\\c\,\sin v & h-c\,\cos v\end{pmatrix}
$$
for some functions $h$ (unique) and $v$ (unique up to an additive constant that is an integral multiple of $2\pi$). The identities $(u\_{xx})\_y = (u\_{xy})\_x$ and $(u\_{xy})\_y = (u\_{yy})\_x$ imply that
$$
\mathrm{d}h = c\bigl(v\_y\,\cos v-v\_x\,\sin v\bigr)\,dx 
 + c\bigl(v\_x\,\cos v+v\_y\,\sin v\bigr)\,dy,
$$
and, since $c\not=0$, the identity $\mathrm{d}(\mathrm{d}h) = 0$ gives
$$
(v\_{xx}-v\_{yy}+2v\_xv\_y)\,\cos v + (2v\_{xy}-{v\_x}^2+{v\_y}^2)\,\sin v = 0.
$$
Thus, the Hessian of $v$ can be parametrized as
$$
\begin{pmatrix} v\_{xx} & v\_{xy}\\v\_{xy} & v\_{yy}\end{pmatrix}
= \begin{pmatrix} 
q+p\,\sin v - v\_xv\_y & \tfrac12({v\_x}^2-{v\_y}^2)-p\,\cos v\\
\tfrac12({v\_x}^2-{v\_y}^2)-p\,\cos v & q-p\,\sin v + v\_xv\_y\end{pmatrix}
$$
for some functions $p$ and $q$. Now, a straightforward calculation yields that
$$
\mathrm{tr}\_g(\mathcal{L}^3\_X g) = 0
\quad\text{while}\quad
\mathrm{tr}\_g(\mathcal{L}^4\_X g) = 32\,c^4 - 8c^2\,(2h-u\_xv\_y+u\_yv\_x)^2.
$$
It follows that we must have $h = b + \tfrac12(u\_xv\_y-u\_yv\_x)$ for some constant $b$. Differentiating this equation and using the various formulae above allows one to solve for $p$ and $q$ in the form.
$$
p = \frac{P(\cos v, \sin v, v\_x,v\_y,u\_x,u\_y)}{\bigl(({u\_x}^2-{u\_y}^2)\cos v +2u\_xu\_y\sin v\bigr)}
$$
and
$$
q = \frac{Q(\cos v, \sin v, v\_x,v\_y,u\_x,u\_y)}{\bigl(({u\_x}^2-{u\_y}^2)\cos v +2u\_xu\_y\sin v\bigr)},
$$
where $P$ and $Q$ are certain polynomials in their arguments, which I won't write out here. Moreover, using these formulae, we find that
$$
\begin{align}
\mathcal{L}\_X g 
&= 2c\bigl(\sin v\,dx^2-2\,\cos v\,dx\,dy-\sin v\,dy^2\bigr)\\
\mathcal{L}^2\_X g 
&= 4c\bigl((c{+}b\,\cos v)\,dx^2+2b\sin v\,dx\,dy + (c{-}b\,\cos v)\,dx^2\bigr)
\end{align}
$$
and
$$
\mathcal{L}^3\_X g = 4(c^2{-}b^2)\,\mathcal{L}\_X g,
$$
from which it follows by induction that $\mathcal{L}^{k+2}\_X g = 4(c^2{-}b^2)\,\mathcal{L}^k\_X g$ for all $k\ge 1$, so that
$$
\mathrm{tr}\_g(\mathcal{L}^{2k+1}\_X g) = 0
\quad\text{and}\quad
\mathrm{tr}\_g(\mathcal{L}^{2k+2}\_X g) = 8c^2 \bigl(4(c^2{-}b^2)\bigr)^k
$$
for all $k\ge0$.


Moreover, because the equations express all of the second derivatives of $u$ and $v$ algebraically in terms of $u$, $v$ and their first derivatives, it follows that any solution $(u,v)$ that is $C^2$ is real-analytic. Thus, any $C^2$ solution $(u,v)$ of the equations above defines a vector field $X$ whose flow has singular values that depend only on $t$.


Now, it is easy to show that the equation $({u\_x}^2-{u\_y}^2)\cos v +2u\_xu\_y\sin v = 0$ cannot hold on any open domain. Moreover, given constants $(r\_1,r\_2,s,s\_1,s\_2)$ that satisfy $({r\_1}^2-{r\_2}^2)\cos s +2r\_1r\_2\sin s \not= 0$, and fixing constants $b$ and $c\not=0$, there is at most one solution on a neighborhood of $(x,y) = (0,0)$ of the above equations such that
$$
\bigl(u\_x(0,0),u\_y(0,0),v(0,0),v\_x(0,0),v\_y(0,0)\bigr) = (r\_1,r\_2,s,s\_1,s\_2).
$$


In fact, one finds that, in the case $b=0$, there exists such a local solution, as the PDE system is involutive (i.e., Frobenius). Unfortunately, there is good evidence that the 'generic' solution with $b=0$ cannot be written in terms of elementary functions. Sometimes, though, one can express a solution implicitly: As an example of a particular solution with $b=0$, one can show that when $x+i\,y$ is parametrized by a holomorphic parameter $w$ such that
$$
dx + i\,dy = \mathrm{e}^{w^2}\,dw,
$$
then, setting
$$
u\_x + i\,u\_y = \overline{w}\,\mathrm{e}^{w^2},
$$
one finds that, for all $t$, the time $t$ flow of the unimodular vector field $X = u\_y\,\partial\_y - u\_x\,\partial\_x$ has constant singular values. (This is most easily checked in the $w$ coordinate, where $X = 2i\left(w\,\partial\_{\bar w} - \bar w\,\partial\_w\right)$ and $g = \mathrm{e}^{w^2+\bar w^2}\,dw\circ d\bar w$.)


When $b\not=0$, it turns out that there are more restrictions on the initial conditions. In fact, an analysis shows that, when $b\not=0$, either $v$ is constant, or else the equation $(u\_xv\_x+u\_yv\_y)=4(c^2{-}b^2)$ must hold.


When $v$ is constant, the vector field $X$ is an affine vector field
whose flow is a $1$-parameter family of (unimodular) affine transformations of the plane, which is clearly a solution.


When $v$ is not constant and $b\not=0$, it turns out that one can integrate the equations explicitly. One finds that, up to a rigid motion in the plane, these solutions can be written in polar coordinates $(r,\theta)$ (where $g = dr^2 + r^2\,d\theta^2$) in the form
$$
X = a\_1\left(r\,\frac{\partial}{\partial r}-2\theta\,\frac{\partial}{\partial\theta}\right) + (a\_2 - 2b\log r)\,\frac{\partial}{\partial\theta}
$$
where $a\_1, a\_2, b$ are constants and $c^2=a\_1^2+b^2$. Note that the case $a\_1=a\_2=0$ is, up to a constant multiple, the example provide by the OP.


