0

I need a way to parametrise all continuous paths from the positive to the negative x-axis, which go through the upper half plane (in $2$ dimensions). I do not care about the speed of the parametrization, just as long as I can describe the set of points the curve passes through.

I have two parametrisations in mind and 2 questions about them.

$1.$ ($f(t)\cos(t),g(t)\sin(t))$, where $f,g$ are continuous with $|f|,|g|>0$ and $t\in [0,\pi]$.

$2.$ $r(t)e^{it}$, where $r$ is continuous with $|r|>0$ and $t\in [0,\pi]$.

First question are these parametrizations equivalent? The second one seems easier to work with and is a special case of the first one where $f=g$.

Is there a known way of doing this, based on some known result?

pureundergrad
  • 971
  • Am I correct in believing that you only care about paths that take on each $x$-value at most once? Otherwise, both these approaches miss out on almost all paths. Think, for example, of a random walk in the open upper half-plane, but starting at, say $(1,0)$ and ending at $(-1, 0)$. It's likely to cross the $y$-axis a great many times, but in both your first and second parameterizations, there's only one point with $t = \pi/2$. – John Hughes Jan 25 '19 at 13:56
  • Good point, I would need to think about that, but for now lets rule that possibility out – pureundergrad Jan 25 '19 at 14:02
  • I was mistaken about the first parameterization --- it DOES allow some kinds of backtracking, although not across the $y$-axis. More details in my answer below. – John Hughes Jan 25 '19 at 14:11

1 Answers1

1

The parameterizations are not equivalent. Consider $$ g(t) = 1 $$ and $$ f(t) = 1 + \sin(100 t) $$ For $t$ near $0$, this looks like a rapid left-right zigzag rising from the point $(1, 0)$. Because of this, the ray from the origin in the direction $(1,1)$ (shown in magenta in the image below) intersects it multiple timesenter image description here.

Hence in the second parameterization, you'd need multiple values for $r(\pi/4)$, which is impossible if $r$ is a function.

On the other hand, if you're willing to work with "at most one $y$-value for each $x$-value," then you can parameterize all such paths with the following:

\begin{align} x(t) = (1-t)a + t b & \text{where $a > 0, b < 0$}\\ y(t) = h(t) * \sin(\pi t) \end{align} where $h: [0, 1] \to \Bbb R^{+}$ is continuous.

This parameterization has the charm that for any continuous curve from the point $a$ (on the positive $x$-axis) to $b$ (on the negative $x$-axis) that meets each $x$-coordinate at most once, there is a unique function $h$, and vice versa.

[I'm assuming that by "upper half plane" you mean the open upper half plane, i.e., points with $y > 0$. If you want to allow the closed upper half plane ($y \ge 0$), then the target for $h$ should be the nonnegative reals rather than the positive reals. ]

John Hughes
  • 93,729
  • Excellent! Is there anywhere I can read more about that last parametrization? Does it have a name? – pureundergrad Jan 25 '19 at 14:28
  • Well...the $x$-part is just "linear interpolation from $a$ to $b$." The $y$-part was inspired by your sine-and-cosine parameterization. So...to be honest, it was just invented. But roughly speaking, it's almost exactly the graph of $h$, except for the sine factor that makes it touch the $x$-axis at the ends. There's nothing really clever or special about it, and it surely doesn't have a name. – John Hughes Jan 25 '19 at 14:39