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

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 $p$.


1 Answer 1


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.

  • $\begingroup$ @AsafShachar: I don't know about specifying terms on the boundary, other than that you can't do with it $b\not=0$. Unfortunately, there is good evidence that the general solution with $b=0$ (and there is a $6$-parameter family of them in a neighborhood of each point in the plane) cannot be written down in elementary terms. However, these solutions are all real-analytic, and one can effectively compute the Taylor series to arbitrary order, so even without integrating the equations, one can tell whether an integral curve of the vector field is a circle, so there is hope to answer the question. $\endgroup$ Commented Apr 10, 2021 at 11:11
  • $\begingroup$ @AsafShachar: Yes. The point is, though, that constancy of the trace for $k=2$ and $k=4$ is sufficient, already, to guarantee that a $C^5$ solution is, in fact, real-analytic, so you don't need to assume it. $\endgroup$ Commented Apr 10, 2021 at 11:15
  • $\begingroup$ @AsafShachar: One more comment: Since $X$ is divergence free, if it were 'inward pointing' on the boundary of a compact domain $D$ where it is differentiable, it would have to be tangent to $\partial D$. If $D$ is the interior of a circle, this would force the boundary to be a union of flow lines of $X$ and hence those flow-lines would have constant curvature. I believe that the only local solutions that have flow lines of constant nonzero curvature have $b\not=0$ and have the form that I gave above with $a_1=0$, and these are not smooth at the origin. $\endgroup$ Commented Apr 11, 2021 at 19:46
  • $\begingroup$ Thank you for your last comment, and thanks again for this thorough and extremely beautiful analysis. I have now succeeded (after two days effort) verifying most of your computations. I am very interested in fully understanding everything you wrote, and I still have some questions about some parts: (1) Can you please elaborate on why the equation $ ({u_x}^2-{u_y}^2)\cos v +2u_xu_y\sin v = 0$ cannot hold on an open domain? (2) Can you elaborate on the analysis of the case $b\not=0$? How do you deduce that either $v$ is constant, or else the equation $(u_xv_x+u_yv_y)=4(c^2{-}b^2)$ must hold? $\endgroup$ Commented Apr 14, 2021 at 6:26
  • $\begingroup$ (3) Can you describe the explicit integration process which led you to the last formula in polar coordinates? I will appreciate any further elaboration you feel convenient to provide. I understand that I might be asking too much here for an MO setting, but I really do want to understand more, and I find your analysis intriguing. If you feel that the answer is already too long, and do not want to answer further in the comments (I deleted my previous comments to reduce the clutter), I am of course ready to continue the discussion outside of MO. Thanks again for the effort and patience. $\endgroup$ Commented Apr 14, 2021 at 6:26

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