Question. Let $\Omega \subset \mathbf{R}^2$ be a convex polygonal domain, equipped with a Riemannian metric $g$. Under which conditions on $g$ is there a vector field $X$ in $\Omega$ with $\mathrm{div}_g X = 0$, $g(X,X) = 1$, and tangent to the boundary?

  • Here an open domain $\Omega \subset \mathbf{R}^2$ is called polygonal if its boundary is a union of line segments. This is allowed to be unbounded: the special case I would be most interested in is when $\Omega = \{ (x_1,x_2) \in \mathbf{R}^2 \mid x_1,x_2 > 0 \}$.

  • If that helps, the metric may be assumed to be of the form $g = e^{2\varphi} (\mathrm{d} x_1^2 + \mathrm{d} x_2^2)$ for some function $\varphi: \Omega \to \mathbf{R}$. However $\varphi$ may go to $-\infty$ approaching the boundary - in fact this is the special case I am most interested in.

  • In the first place I would be looking for vector fields that are defined at all points in $\Omega$, and satisfy the hypotheses pointwise. For this $X$ would be $C^1$ in $\Omega$, and continuous up to the smooth portions of the boundary.

  • The tangency of $X$ is meant in the Euclidean sense: $X \cdot \nu = X^1 \nu^1 + X^2 \nu^2 = 0$ along $\partial \Omega$. This need only be satisfied at the smooth portions of the boundary. (The vector field cannot be defined continuously at the corners.)

Expressed in coordinates it looks like an overdetermined first-order PDE: \begin{equation} \begin{cases} \partial_i(\sqrt{\det g} X^i) = 0 \text{ in $\Omega$} \\ g_{ij} X^i X^j = 1 \text{ in $\Omega$} \\ \delta_{ij} X^i \nu^j = 0 \text{ on $\partial \Omega$}. \end{cases} \end{equation}

Example. Let $\Omega = \{ x_1 > 0, x_2 > 0 \}$ and let the metric be so that $e^{2\varphi} = 0$ on $\{ x_1 = 0 \} \cup \{ x_2 = 0 \}$. I thought that perhaps integrating the curvature $K_g = -e^{-2\varphi} \Delta \varphi$ over regions bounded by leaves of a geodesic foliation, combined with the Gauss--Bonnet theorem, would yield some constraints on the existence of the vector field, but that did not ultimately go anywhere.

  • $\begingroup$ Do you allow an interior singularity ? I mean, if $\Omega$ is the unit disk with the standard metric, a natural solution is $\vec e_\theta$ in polar coordinates, which is singular at the origin. $\endgroup$ Apr 24, 2021 at 14:24
  • $\begingroup$ Do you have a typo in your last paragraph? The domain $\Omega$ that you list is the right half-plane, and $X = \partial_2$ is tangent to the boundary line $x_1=0$, not normal to it. $\endgroup$ Apr 25, 2021 at 12:08
  • $\begingroup$ @LeoMoos: Three questions: 1) Why specify 'conformally flat' in the question, as all metrics in dimension $2$ are conformally flat? Did you mean to impose the stronger condition that $g$ be a multiple of the Euclidean metric? 2) If $g$ isn't assumed to be a multiple of the Euclidean metric, then why not just let $\Omega$ be an abstract surface with a boundary that consists of smooth arcs that meet at 'corners' and let $g$ be a Riemannian metric on $\Omega$ minus the 'corner points'. 3) If $g$ doesn't extend to the smooth parts of the boundary, then what do your boundary conditions on $X$ mean? $\endgroup$ Apr 26, 2021 at 10:52
  • $\begingroup$ @DenisSerre To be honest I was hoping for 'classical solutions', without singular points. However if you were aware of general existence results of weak solutions that would of course also be interesting - sorry for only replying to your comment now. $\endgroup$
    – Leo Moos
    Apr 27, 2021 at 17:22
  • $\begingroup$ @RobertBryant I apologise for taking so long to get back to you. I'll try to address your questions in order. 1) I misspoke - you're right, I'd be happy to restrict to the case where $g$ is a multiple of the Euclidean metric. 2) See the first point. 3) I've tried to fix this, but I couldn't quite figure out the right way. The question arose when studying metrics $g$ which do degenerate near the boundary (in the sense that the factor $e^{2 \varphi}$ goes to zero), and which additionally admit foliations by geodesics which intersect the boundary orthogonally. [...] $\endgroup$
    – Leo Moos
    Apr 27, 2021 at 17:28

1 Answer 1


Now that the question has been changed so extensively, the remarks that I made for the old version are no longer of any interest. Here is what I understand the problem to look like now:

First, $\Omega\subset\mathbb{R}^2$ is a convex open set in the plane whose boundary is polygonal, i.e., a union of line segements or rays that meet at a discrete set of 'corners'. In what I am going to discuss, these hypotheses seem a bit restrictive. Most or all of what I have to say would apply if $\Omega$ were simply-connected (and hence contractible) and its boundary $\partial\Omega$ minus a discrete set $C\subset\partial\Omega$ were a union of embedded (smooth) arcs.

Second, we are given a (smooth) function $\lambda>0$ on $\Omega$ that limits to $0$ at every smooth point of $\partial\Omega$. (We don't further specify the behavior of $\lambda$ near the corner points $C\subset\partial\Omega$.) We use $\lambda$ to define a metric $g = \lambda\,(\mathrm{d}{x_1}^2+\mathrm{d}{x_2}^2)$.

Then, we ask the following question: When does there exist a (smooth) foliation $\mathcal{F}$ by curves on an open set in $\mathbb{R}^2$ containing $\Omega\cup (\partial\Omega\setminus C)$ with the following two properties: First, in the interior of $\Omega$, the leaves of $\mathcal{F}$ are $g$-geodesics. Second, for each point $p\in \partial\Omega\setminus C$, the boundary of $\Omega$ and the leaf of $\mathcal{F}$ through $p$ meet orthogonally (at $p$).

Here are a few examples:

  1. Let $\Omega$ be the first quadrant, i.e., defined by $x_1>0$ and $x_2>0$. Then $\partial\Omega$ is the union of two rays, the positive $x_i$-axes, which meet at right angles at the origin (which is the unique corner point). Let $\lambda = {x_1}^2{x_2}^2({x_1}^2{+}{x_2}^2)$. Then a solution to the problem is defined by letting $\mathcal{F}$ be the foliation given by the level sets of the function $F = {x_1}^2-{x_2}^2$.

  2. Let $\Omega$ be defined by $x_1>0$ and $x_2>{x_1}^2$. Then $\partial\Omega$ is the union of a ray (the positive $x_2$-axis) and the right half of the parabola $x_2-{x_1}^2=0$. These meet at right angles at the origin (which is the unique corner point). Let $\lambda = {x_1}^2(x_2{-}{x_1}^2)^2({x_1}^2{+}(x_2{-}3{x_1}^2)^2)$. Then a solution to the problem is defined by letting $\mathcal{F}$ be the foliation given by the level sets of the function $F = \mathrm{e}^{6x_1}(1-6x_1+18x_2)$.

  3. Again, let $\Omega$ be the first quadrant, i.e., defined by $x_1>0$ and $x_2>0$. Then $\partial\Omega$ is the union of two rays, the positive $x_i$-axes, which meet at right angles at the origin (which is the unique corner point). Let $\lambda = {x_1}^2{x_2}^2/({x_1}^2{+}{x_2}^2)^{3}$. Then a solution to the problem is defined by letting $\mathcal{F}$ be the foliation given by the level sets of the function $F = {x_1}^2+{x_2}^2$. (Some interesting features of this example will be explained below, but note, already, that $\lambda$ does not limit to $0$ as one approaches the corner, although it does limit to $0$ as one approaches any smooth boundary point.)

To understand how these examples were constructed and to get an idea of some of the restrictions that the existence of a foliation $\mathcal{F}$ satisfying the two conditions places on the function $\lambda$, suppose that we have a given pair $(\Omega,\lambda)$ satisfying the above conditions and that there exists a foliation $\mathcal{F}$ with the desired properties.

Since $\Omega$ is contractible, we can write $g$ on $\Omega$ in the form $g=\eta_1^2+\eta_2^2$, where $\eta_1$ and $\eta_2$ are $1$-forms and where the leaves of $\mathcal{F}$ are the null curves of $\eta_2$. Then the condition that these leaves be $g$-geodesics is equivalent to the condition that $\mathrm{d}\eta_1 = 0$. Since $\Omega$ is simply-connected, there will exist a function $S:\Omega\to\mathbb{R}$, unique up to an additive constant, such that $\mathrm{d}S = \eta_1$.

Write $\mathrm{d}S = S_1\,\mathrm{d}x_1 + S_2\,\mathrm{d}x_2$. Since $g = \lambda\,({\mathrm{d}x_1}^2+{\mathrm{d}x_2}^2)=\eta_1^2+\eta_2^2$, it follows that $\eta_2 = \pm (S_2\,\mathrm{d}x_1 - S_1\,\mathrm{d}x_2)$, so we have $$ g = \bigl(S_1^2+S_2^2\bigr)\,({\mathrm{d}x_1}^2+{\mathrm{d}x_2}^2) = \lambda\,({\mathrm{d}x_1}^2+{\mathrm{d}x_2}^2). $$ In other words, $\lambda = \bigl(S_1^2+S_2^2\bigr)$. From this, it is not difficult to see that $S$ extends continuously to the smooth part of $\partial\Omega$ and is constant on each connected component of the smooth part of the boundary, while it has no critical points within $\Omega$ itself.

Because of the hypothesis that the foliation $\mathcal{F}$ extends smoothly to an open set $U$ containing $\Omega$ and the smooth part of its boundary, it follows that there is a smooth function $\theta:U\to S^1$ such that the leaves of $\mathcal{F}$ in $U$ are the null curves of the smooth $1$-form $\sin\theta\,\mathrm{d}x_1-\cos\theta\,\mathrm{d}x_2$ and so that $\eta_2 = \sqrt{\lambda}\bigl(\sin\theta\,\mathrm{d}x_1-\cos\theta\,\mathrm{d}x_2\bigr)$ in $\Omega$, and this extends at least continuously to the smooth part of the boundary of $\Omega$ by extending $\lambda$ to be $0$ on the smooth part.

Conversely, the function $S:\Omega\to\mathbb{R}$ gives enough information to recover both the metric $g$ and the foliation $\mathcal{F}$. If, as the OP has asked, one has $\lambda$ already specified, then $S$ must satisfy the first-order scalar PDE $|\nabla S|^2 = \lambda$ on $\Omega$ plus the smooth part of the boundary.

Sometimes, such an $S$ exists (as the examples above show, although they were found by starting with a candidate $S$), and sometimes it does not. I suspect that the OP would like an effectively computable necessary and sufficient condition for a given $\lambda$ to possess an admissable solution $S$, but that is likely to be hard to determine, because it really depends on delicate information about the geodesic flow in $\Omega$ endowed with the metric $g$.

For an example of an definitive condition, suppose that there is a simple closed $g$-geodesic $\gamma$ in $\Omega$. Then $\gamma$ cannot be a leaf of any geodesic foliation $\mathcal{F}$ of $\Omega$, since the function $S$ that corresponds to $\mathcal{F}$ could not then have a critical point on $\gamma$, which it must. Thus, $\gamma$ would have to be transverse to all of the leaves of $\mathcal{F}$, and we would then have a foliation of the disk $D\subset\Omega$ bounded by $\gamma$ that is transverse to the boundary of $D$, an obvious impossibility (by degree theory). (Probably, this argument works even if $\gamma$ is closed but not simple, but I leave that for the interested.)

It is certianly possible to chose an admissable $\lambda$ for which the metric $g$ would have a simple closed geodesic, thus showing that there are $\lambda$ for which there is no $g$-geodesic foliation, and hence no function $S$ with the desired properties. However, for a given $\lambda$, it is not clear how to tell whether there exists a simple closed $g$-geodesic in $\Omega$. There are checkable conditions that guarantee one and checkable conditions that forbid one, but the gap between these is large.

I am dubious that an effectively checkable necessary and sufficient condition for the solvability of the problem of given $\lambda$ exists.

  • $\begingroup$ If I understand your answer correctly, you're explaining how to construct a metric on a given domain, which admits a vector field with the required properties. That wasn't quite the aim of my question: both $\Omega$ and $g$ are given, and I am asking which metrics allow divergence-free unit vector fields. That there is always at least one metric is interesting, but weaker than what was intended. $\endgroup$
    – Leo Moos
    Apr 25, 2021 at 7:05
  • $\begingroup$ @LeoMoos: I have now replaced my original answer with one that may be of more interest to you. $\endgroup$ Apr 28, 2021 at 20:36
  • $\begingroup$ Thanks again for your detailed answer. If I were to think about this some more, would you have some suggestions of papers or books where related problems are discussed? $\endgroup$
    – Leo Moos
    May 6, 2021 at 17:00
  • $\begingroup$ @LeoMoos: Unfortunately, I don't know anything about what literature there might be on questions such as this. Metrics that degenerate at the boundary of a manifold with corners don't seem to be much discussed, and I don't know where they arise naturally in physical problems. Even the formulation that I gave for dealing with the foliation at the boundary seems a bit artificial to me, but I don't see any way around something like this in order to make sense of the leaves being perpendicular to the boundary if the metric doesn't extend to the boundary. $\endgroup$ May 7, 2021 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.