Existence of divergence-free unit vector field in conformally rescaled euclidean metric

Question

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.

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.