Let $f: U \subset \mathbb{R}^2 \rightarrow V \subset \mathbb{R}$. I am studying a parametrization $g$ of the integral curves of the gradient field $\nabla f$ with $r$ selecting an integral curve and $t$ running along the curve. As a consequence, for a given $t$, the corresponding level curve of $f$ is parameterized by $r$.

Notation and basic setup: $g(t, r): V \times I \rightarrow U \subset \mathbb{R}^2$, $g_t = \frac{\partial g}{\partial t}$, $g_r = \frac{\partial g}{\partial r}$, $g_r \perp g_t$, $\frac{g_t}{|g_t|} = \frac{\nabla f}{|\nabla f|}$, $|\nabla f| = \frac{1}{|g_t|}$.

Let $g_{tt}$, $g_{rt}$, $g_{tr}$, $g_{rr}$ denote the component-wise second derivatives.

My question actually concerns $g$. $f$ is just provided to give some more context. Since $f$ induces no parameterization velocity for its levelsets, I assume $|g_r|$ is some positive velocity in each point yielding a consistent parameterization.

By some variational calculus on my specific $g$ I derived (among others) the following properties:

(1) $\qquad g_t^T g_{tr} = - g_r^T g_{tt}$

(2) $\qquad g_r^T g_{rt} = - g_t^T g_{rr}$

I feel like these identities are not specific for my function but hold for many maps (every sufficiently differentiable?) in general. These terms have the following interpretation in terms of velocity change and curvature, i.e. relate these quantities:

$\frac{g_t^T g_{tr}}{|g_r|} = \frac{\partial}{\partial r} |g_t|$ (acceleration of $|g_t|$ perpendicular to the integral curve selected by $r$)

$\frac{g_r^T g_{tt}}{|g_r| \, |g_t|^2} = k_t(r)$ (signed curvature of the integral curve of $\nabla f$ selected by $r$)

$\frac{g_r^T g_{rt}}{|g_t|} = \frac{\partial}{\partial t} |g_r|$ (acceleration of $|g_r|$ perpendicular to the level curve selected by $t$)

$\frac{g_t^T g_{rr}}{|g_r|^2 \, |g_t|} = k_r(t)$ (signed curvature of the level set selected by $t$)

I think the identity (1) describes - so to say - the odometry of a two-wheeled robot with infinitesimal axis length/distance between the wheels. Each wheel drives along an integral curve and if there is a non-zero velocity change between the curves, i.e. $\frac{\partial}{\partial r} |g_t|$ the robot's wheels have unequal velocity and the robot drives a curve, therefore non-zero $k_t$. Identity (2) is the same rule in perpendicular view. (Disclaimer: The problem I study is not from odometry. This is really just a picture to illustrate why I suspect these identities might hold more generally.)

My questions:

1) Are (1) and (2) known general identities?

2) How are they called / is there a reference to work concerning them?

3) For what type of maps do they hold? (E.g. is it a property of maps constructed from a gradient like above or of conformal maps in general)

4) How can they be derived from the general properties of these maps?

5) Is there a generalization of this law to $\mathbb{R}^n$ and manifolds w.r.t. geodesic curvature?

My thoughts so far:

By construction, $g$ is a conformal map. I know, such a parameterization does not always exist globally. E.g. if $U$ is simply connected it exists by Riemann's mapping theorem. However, I think (1) and (2) are local properties and would hold for a more general case. Let's assume $U$ is of some form that permits the explained parameterization.

Thus, each component of $g$ is a harmonic map, therefore $Tr(Hess(g_x)) = 0$ and $Tr(Hess(g_y)) = 0$ for $g_x, g_y$ denoting the components of $g$. Due to its construction from a gradient I can derive that $Hess(g_x)$ is symmetric. I think that $Hess(g_y)$ is also symmetric, but that's less easy to see.

However, I was so far not able to derive the identities (1) and (2) just from these observations. I also looked into the Wikipedia articles about curvature and principal curvature but did not really find it. Also https://en.wikipedia.org/wiki/Centripetal_force#Alternative_approach looked promising, but didn't do the trick so far. Probably this is really something elementary, e.g. a continuation law of curvature. I'd just like to avoid reinventing the wheel regarding these identities.

Finally, there is a third identity which I believe is specific for my function. However if someone recognizes it as a known law, I'd be glad for hints:

(3) $\qquad \frac{g_t^T g_{tt}}{|g_t|^2} = \frac{g_r^T g_{rt}}{|g_r|^2}$

Because of identity (2) this can equivalently be stated as

(3) $\qquad \frac{g_t^T g_{tt}}{|g_t|^2} = - \frac{g_t^T g_{rr}}{|g_r|^2}$

  • $\begingroup$ This may be related to math.stackexchange.com/questions/707784/… but is not fully the same. While the linked question is about expressing curvature of the level set by divergence of the gradient, my question is about expressing it by velocity change between "neighboring" level sets. $\endgroup$ – stewori Jan 13 at 22:47
  • $\begingroup$ Another related question is math.stackexchange.com/questions/2386433/…. I think my question may be a variant of that question for infinitesimal $\eta$, but I'm not sure. $\endgroup$ – stewori Jan 13 at 22:48
  • $\begingroup$ Okay, I found that it is rather straight forward to show this law for polar coordinates, i.e. for $g(t, r) = r \begin{pmatrix} sin(t) \\ cos(t) \end{pmatrix}$. I think this implies the general result as curvature boils down to tangent circles. However I am not sure what formal argument is required to assert this implication. So far this only scoped the 2D case anyway. $\endgroup$ – stewori Jan 14 at 0:22
  • $\begingroup$ I think that identity (3) indicates that $f$ is a harmonic function. With some calculation I think I can show that it implies $\bigtriangleup f = 0$. Maybe there is a more high-level way to see this. $\endgroup$ – stewori Jan 16 at 1:38

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