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I wonder if someone can explain to me how one goes about to calculating the first variation of functions defined on curves. For example, if $C$ is a curve in $\mathbb{R}^3$, and the functional $$F(C) = \frac{1}{\min_{x,y,z \in C}r(x,y,z)}$$ where you can assume whatever is necessary for $r$. How to calculate or go about calculating the first variation?

So we want to take the limit, if $\gamma$ is another curve (but how to define it exactly?), $$\lim_{\epsilon \to 0}\frac{F(C+\epsilon \gamma) - F(C)}{\epsilon}$$ but I don't know how to simplify this expression when we write it out..


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The variation is not nice, and may not be very helpful. The best I can make of it is this:

For a curve $C$ such that $r \circ C$ is minimized by a unique $t$, the first variation is $-(\nabla r(C(t)) \cdot \gamma(t)) r(C(t))^2$. For curves on which $r$ does not have a unique minimum, the variation does not exist, except for very special $\gamma$, because the upper and lower limits differ.

Assuming the objective is to find curves, with given endpoints, that maximize $\min(r\circ C)$, the only conclusion that can be drawn is that if $r$ has a unique minimum on sucb a curve, other than at an endpoint, then it must be at a stationary point of $r$.

That this conclusion is not very specific should not come as a big surprise, as generally $\min(r\circ C)$ is maximized by a huge family of curves. If we call the endpoints $a$ and $b$, then basically the problem is to find the largest $m$ such that $a$ and $b$ are in the same path-component of $r^{-1}([m, \infty)).$ For such $m$, any curve in $r^{-1}([m, \infty))$ between $a$ and $b$ minimizes $F$.

To make this easier to visualize, restrict to two dimensions and take $r(x, y) = xy$. Let $a = (1,1)$ and $b = (-1,-1)$. For $m > 0$, the set where $xy \ge m$ is disconnected and $a$ and $b$ lie in different components. The set where $xy \ge 0$ consists of the first and third quadrant, including axes. Any curve within this set that connects $a$ and $b$ minimizes $F$ and the only thing they have in common is that they pass through $(0, 0)$, the only stationary point of $r$.

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