This question was inspired by a question at a trivia event: what is the length of Singapore's coastline? I knew that Singapore spanned roughly 50km across and 30km from North to South, and am curious whether this information is enough to derive a lower bound for the length of the coastline.
To be more precise: let $\Gamma$ be the set of piece-wise smooth simple closed curves in $\mathbb{R}^2$ parametrised over $[0,1]$. Let $p_1$ and $p_2$ be the projections onto the $x$ and $y$ co-ordinates respectively. Let the width of a curve $w: \Gamma \to \mathbb{R}$ be defined by $$ w(\gamma) = \sup_{x, y \in [0,1]} |(p_1 \circ \gamma)(x) - (p_1 \circ \gamma)(y)| $$ and let the height $h(\gamma)$ be defined similarly. Define the length $L(\gamma)$ of a curve in the usual way: $$L(\gamma) = \int_0^1 |\gamma'(x)|dx.$$
Here are my questions:
- Given $a, b > 0$, is there a simple expression (in terms of $a$ and $b$) or procedure to compute $$\inf \{L(\gamma): \gamma \in \Gamma, h(\gamma) \geq a, w(\gamma) \geq b\}?$$
- Does there exist a minimiser, and if so, is it unique (upto translation)?
- Do the solutions to (1) and (2) change when the constraints $h(\gamma) \geq a, w(\gamma) \geq b$ are changed to equality?
- How does the solution of this problem change under the regularity constraint on $\Gamma$, i.e., when we change "piecewise smooth" to "smooth" or even "rectifiable"?
Unfortunately my background in such variational problems is minimal, so I'd find detailed answers most helpful. Any pointers to relevant reading material would also be appreciated.
A scaling argument shows that if $h(\gamma)>a$ or $w(\gamma) > b$ then we can find a $\gamma'$ of smaller length. Hence we can presume that $h(\gamma)=a$ and $w(\gamma) = b$. This should answer 3.
We can find points $v_1 \in \operatorname{argmin}_{t \in [0,1]} p_1 \circ \gamma (t)$, $v_2 \in \operatorname{argmin}_{t \in [0,1]} p_2 \circ \gamma (t)$, $v_3 \in \operatorname{argmax}_{t \in [0,1]} p_1 \circ \gamma (t)$, $v_4 \in \operatorname{argmax}_{t \in [0,1]} p_2 \circ \gamma (t)$.
Note that $\gamma$ must pass through all 4 points. Furthermore, by drawing a little picture, it is easy to see that we can presume that $\gamma$ passes through the points in order (that is $p_1$ before $p_2$, etc.). In particular, $L(\gamma) \ge \sum_k \|p_k-p_{k-1}\|$ (where I take $p_4=p_0$ for my notational convenience).
We will demonstrate that $\sum_k \|p_k-p_{k-1}\| \ge 2\sqrt{a^2+b^2}$, and by taking $p_1=p_2 = (0,0), p_3=p_4=(w,h)$ this is attained, hence we have a result for 1. To see this, consider the following diagram:
Note that we can 'unroll' the polygonal path to see that can be expressed as a path between $p_1$ and $p_1'$. It follows that a straight line is the shortest path, and that the distance is $2 \sqrt{w^2+h^2}$. This answers 1.
Note that the straight line can be moved vertically up & down and the distance remains the same. This addresses 2.
By using a smooth transition function (see https://en.wikipedia.org/wiki/Non-analytic_smooth_function#Smooth_transition_functions for example), the polygonal path can be made smooth. This should address 4.