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

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

Question: Does there exist a divergence-free vector field $X$ on $D$ such that:

1. $|(d\psi_t)_p|^2=\tr_g ((d\psi_t)_p^T(d\psi_t)_p)$ is independent of $p$, or equivalently the singular values of $d\psi_t$ are constant (independent of $p$).

2. $X$ is not homothetic-by homothetic I mean $L_x g=\lambda g$ for some constant $\lambda$. Since $X$ is assumed to be divergence-free, it is homothetic if and only if it is Killing.

If conditions $(1),(2)$ hold then $|(d\psi_t)_p|^2$ is not constant in $t$. (if it were constant in $t$ as well as in $p$, then since $\psi_t$ is volume-preserving $X$ would be Killing).

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 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, we must have $\|\nabla X\|^2+\operatorname{tr}(\nabla X\circ\nabla X)$ constant.

Is there a systematic way to obtain expressions for higher-order derivatives?

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

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

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

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 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 $p$.

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

Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Given a vector field $X$ on $D$, Let $\psi_t:D \to D$ be the flow of $X$.

Question: Does there exist a divergence-free vector field $X \in \Gamma(TD)$ such that:

1. $\|(d\psi_t)_p\|^2=\tr_g ((d\psi_t)_p^T(d\psi_t)_p)$ is independent of $p$.

The second condition is equivalent to the fact that $d\psi_t$ would have constant singular values. So, this question is a continuous version of this one.

2. $X$ is not homothetic- where by homothetic I refer to $L_x g=\lambda g$ for some constant $\lambda$. (In that case $\psi_t^*g=e^{\lambda t}g$. $\lambda=0$ corresponds to Killing fields). Since $X$ is divergence-free, it is homothetic if and only if it is Killing.

If all the conditions above hold, then $\|(d\psi_t)_p\|^2$ is not constant in $t$. (Indeed, if it were constant in $t$ as well as in $p$, then since $\psi_t$ is volume-preserving, we $d\psi_t$ would have constant singular values $1$, so $X$ would be Killing).

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 $t$. Is there a way to write this as an explicit PDE on $X$?

A partial information can be obtained from differentiating twice at $t=0$:

$$ \frac{\partial^2 }{\partial t^2}\left|_{t=0}\right.\tr_g\big((\psi_t^*g)_p\big)=\operatorname{tr}_g(\mathcal{L}_X(\mathcal{L}_Xg))=2X(\operatorname{div}X)+2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X)= $$ $$ 2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X) $$ since $X$ is divergence-free. So, we must have $\|\nabla X\|^2+\operatorname{tr}(\nabla X\circ\nabla X)$ constant.

Perhaps there is a systematic way to obtain expressions for the higher-order derivatives?

Here is an example for such an $X$ on $D\setminus{0}$:

$X=\log r \frac{\partial}{\partial \theta}$. Its flow, described in polar coordinates is $$ \psi_t: (r,\theta)\mapsto (r,\theta+t\log r). $$ Indeed w.r.t the orthonormal frame $( \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta})$, $d\psi_t$ is given by $$ [d\psi_t]_{\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}}=\begin{pmatrix} 1 & 0 \\\ t & 1\end{pmatrix}. $$ which is independent of $p$.

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

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

Question: Does there exist a divergence-free vector field $X$ on $D$ such that:

1. $|(d\psi_t)_p|^2=\tr_g ((d\psi_t)_p^T(d\psi_t)_p)$ is independent of $p$, or equivalently the singular values of $d\psi_t$ are constant (independent of $p$).

2. $X$ is not homothetic-by homothetic I mean $L_x g=\lambda g$ for some constant $\lambda$. Since $X$ is assumed to be divergence-free, it is homothetic if and only if it is Killing.

If conditions $(1),(2)$ hold then $|(d\psi_t)_p|^2$ is not constant in $t$. (if it were constant in $t$ as well as in $p$, then since $\psi_t$ is volume-preserving $X$ would be Killing).

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 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, we must have $\|\nabla X\|^2+\operatorname{tr}(\nabla X\circ\nabla X)$ constant.

Is there a systematic way to obtain expressions for higher-order derivatives?

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

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

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

Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Given a vector field $X$ on $D$, Let $\psi_t:D \to D$ be the flow of $X$.

Question: Does there exist a divergence-free vector field $X \in \Gamma(TD)$ such that:

1. $$\|(d\psi_t)_p\|^2=\langle (d\psi_t)_p,(d\psi_t)_p \rangle=\tr_g ((d\psi_t)_p^T(d\psi_t)_p)$$ is independent of $p$.

2. $X$ is not homothetic- where by homothetic I refer to $L_x g=\lambda g$ for some constant $\lambda$. (In that case $\psi_t^*g=e^{\lambda t}g$. $\lambda=0$ corresponds to Killing fields). In fact, since $X$ is divergence-free, it is homothetic if and only if it is Killing.

Note that if all the conditions above hold, then $\|(d\psi_t)_p\|^2$ is not constant in $t$.

Since $f(t,p):=\|(d\psi_t)_p\|^2=\tr_g\big((\psi_t^*g)_p\big)$ 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 $t$. Is there a way to write this as an explicit PDE on $X$?

Here is an example for such an $X$ on $D\setminus{0}$:

$X=\log r \frac{\partial}{\partial \theta}$. Its flow, described in polar coordinates is $$ \psi_t: (r,\theta)\mapsto (r,\theta+t\log r). $$ Indeed w.r.t the orthonormal frame $( \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta})$, $d\psi_t$ is given by $$ [d\psi_t]_{\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}}=\begin{pmatrix} 1 & 0 \\\ t & 1\end{pmatrix}. $$ which is independent of $p$.

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

Let $D \subseteq \mathbb{R}^2$ be the closed unit disk. Given a vector field $X$ on $D$, Let $\psi_t:D \to D$ be the flow of $X$.

Question: Does there exist a divergence-free vector field $X \in \Gamma(TD)$ such that:

1. $\|(d\psi_t)_p\|^2=\tr_g ((d\psi_t)_p^T(d\psi_t)_p)$ is independent of $p$.

The second condition is equivalent to the fact that $d\psi_t$ would have constant singular values. So, this question is a continuous version of this one.

2. $X$ is not homothetic- where by homothetic I refer to $L_x g=\lambda g$ for some constant $\lambda$. (In that case $\psi_t^*g=e^{\lambda t}g$. $\lambda=0$ corresponds to Killing fields). Since $X$ is divergence-free, it is homothetic if and only if it is Killing.

If all the conditions above hold, then $\|(d\psi_t)_p\|^2$ is not constant in $t$. (Indeed, if it were constant in $t$ as well as in $p$, then since $\psi_t$ is volume-preserving, we $d\psi_t$ would have constant singular values $1$, so $X$ would be Killing).

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 $t$. Is there a way to write this as an explicit PDE on $X$?

A partial information can be obtained from differentiating twice at $t=0$:

$$ \frac{\partial^2 }{\partial t^2}\left|_{t=0}\right.\tr_g\big((\psi_t^*g)_p\big)=\operatorname{tr}_g(\mathcal{L}_X(\mathcal{L}_Xg))=2X(\operatorname{div}X)+2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X)= $$ $$ 2\|\nabla X\|^2+2\operatorname{tr}(\nabla X\circ\nabla X) $$ since $X$ is divergence-free. So, we must have $\|\nabla X\|^2+\operatorname{tr}(\nabla X\circ\nabla X)$ constant.

Perhaps there is a systematic way to obtain expressions for the higher-order derivatives?

Here is an example for such an $X$ on $D\setminus{0}$:

$X=\log r \frac{\partial}{\partial \theta}$. Its flow, described in polar coordinates is $$ \psi_t: (r,\theta)\mapsto (r,\theta+t\log r). $$ Indeed w.r.t the orthonormal frame $( \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta})$, $d\psi_t$ is given by $$ [d\psi_t]_{\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}}=\begin{pmatrix} 1 & 0 \\\ t & 1\end{pmatrix}. $$ which is independent of $p$.