# Location of the endpoints of two parametric curves

I have two curves, $C_1$ and $C_2$ parametrized by $\theta$, the angle of the outward normal with the X-axis.

$C_1$ is given by the following equations (say $r = 0.2$): \begin{align*} \frac{dx}{d\theta} &= - \frac{\sin \theta}{\kappa(\theta)} \\ \frac{dy}{d\theta} &= \frac{\cos \theta}{\kappa(\theta)}\\ \kappa(\theta) &= \max \left\{\frac{(1-x)\cos \theta + (1-y)\sin \theta}{\frac{r}{\cos^2\theta} + \frac{r}{\mu^2 \sin^2\theta}}, \frac{-(1+x)\cos \theta + (2-y) \sin \theta}{\frac{r}{\mu^2 \sin^2\theta}} \right\} \end{align*}

$C_2$ is given by the following equations (say $r = 0.2$): \begin{align*} \frac{dx}{d\theta} &= - \frac{\sin \theta}{\kappa(\theta)} \\ \frac{dy}{d\theta} &= \frac{\cos \theta}{\kappa(\theta)}\\ \kappa(\theta) &= \max \left\{\frac{(1-x)\cos \theta + (1-y)\sin \theta}{\frac{r}{\mu^2 \cos^2\theta} + \frac{r}{ \sin^2\theta}}, \frac{-(1+x)\cos \theta + (2-y) \sin \theta}{\frac{r}{ \sin^2\theta}} \right\} \end{align*}

That is, the only difference between the curvatures of $C_1$ and $C_2$ is that $\mu$ goes on $\sin$ in $C_1$ while on $\cos$ in $C_2$. Suppose $\mu > 1$.

Both the curves start from some point $(w,w)$ such that $w \in [0.5,1]$ with the initial angle $\theta_0^{C_1}$ for $C_1$ = $\pi/4 + \Delta$. For $C_2$, the initial angle $\theta_0^{C_2} = \pi/4 - \Delta$ for some $\Delta \in [0,\pi/12]$.

Domain of $\theta$ for curve $C_i$ is $[\theta_0^{C_i},\pi/2]$ and I am interested in $(x^{C_i}(\pi/2), y^{C_i}(\pi/2))$ for both the curves.

Suppose the initial point is chosen so that the curvature remains strictly positive. I want to claim the following: $$\text{If } x^{C_1}(\pi/2) = x^{C_2} (\pi/2) \Rightarrow y^{C_1}(\pi/2) > y^{C_2}(\pi/2).$$

Alternatively, if I can prove the following even that would do the job:

$$\text{If } y^{C_1}(\pi/2) = y^{C_2} (\pi/2) \Rightarrow x^{C_1}(\pi/2) > x^{C_2}(\pi/2).$$

Numerically this seems to hold for both large and small $\mu$. But I don't have any idea how to approach a proof. I don't have much of a clue about how to bound the distance between the two curves.

• What is the motivation for this question? Is it some scattering problem? – Konstantinos Kanakoglou Dec 27 '17 at 0:07
• The motivation comes from repeated games. I don't know what a scattering problem is besides a casual reading of the Wikipedia article. But, I won't be shocked if there's some connection. – avk255 Dec 27 '17 at 7:12