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I have a number of basic questions.

1) Consider a curve $\gamma:I \to \mathbb{R}^n$. I understand that if the arclength is $s$, the change of variables requires $ds = |\gamma_x|dx$. But what does $\partial_s = \frac{\partial_x}{|\gamma_x|}$ mean? I am told that this is the arclength derivative. What is its use?

2) Also, if $\nabla_s f$ of a vector $f$ is defined as $\nabla_s f = f_s - (f_s\cdot T)T$ where $T$ is the tangent, then this is the normal component of $f_s.$ Presumably this means taking the element-wise derivative. Does this have some meaning other than being the normal component of $f_s$?

3) Finally, I know that for normal fields $f$ and $g$ $$\partial_s (f \cdot g) = \nabla_s f \cdot g + f \cdot \nabla_s g.$$

How can I get an integration by parts formula for this? Is it something like $$\int v\nabla_s u\;ds = -\int u \nabla_s v\;ds$$ for all (not necessarily normal $u$ and $v$)?

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Where did you see the notation $\partial_s = \partial_x / |\gamma_x|$? That is not generally how I would think of writing the expression you are describing. – Willie Wong May 31 '12 at 15:58
@WillieWong I saw it in the paper "Evolution of elastic curves in R^n: Existence and Computation". See… – Court May 31 '12 at 17:21
Ah, I misunderstood what you meant to write. – Willie Wong Jun 1 '12 at 6:27
up vote 1 down vote accepted
  1. For clarification, let $x\mapsto \gamma(x)$ be the curve mapping. Consider the change of variables $s\mapsto x(s)$ and the function $\tilde{\gamma}(s) = \gamma(s)$. The image set is the same: they define the same curve. Then the change of variable formulae give $$ \frac{d}{ds}\tilde{\gamma} = \frac{d}{dx}\gamma \cdot \frac{dx}{ds} $$ So if the function $x(s)$ is chosen such that $$ \frac{dx}{ds} = \frac{1}{\left| \frac{d\gamma}{dx}\right|} $$ you have that $$ \left| \frac{d\gamma}{ds}\right| = 1$$ and hence $s$ is an arclength parametrisation.

    Now let $f$ be a function on $\mathbb{R}^n$. Consider the compositions $f\circ \gamma$ which is now a function $x\mapsto f(\gamma(x))$ and $f\circ\tilde\gamma$ which is now a function $s\mapsto f(\tilde\gamma(s))$. Taking the derivatives we have $$ \frac{d}{ds} f\circ\tilde\gamma = \nabla f \cdot \frac{d\tilde{\gamma}}{ds} = \nabla f \cdot \frac{d\gamma}{dx} \frac{dx}{ds} = \frac{d}{dx} (f\circ\gamma) \frac{dx}{ds} $$ That is, the parameter derivative (derivative along the parameter for the curve) can be evaluated as the dot product between the gradient of the function $f$ on $\mathbb{R}^n$ with the velocity vector of the curve.

    With the arclength parametrisation, you chose a convenient normalisation of the velocity (you set it to have norm 1). This means that the evaluated quantity $\frac{d}{ds} f\circ\tilde\gamma$ is a directional derivative.

    Another way to think about it is this: if you take the curve and straighten it out without changing its length, a function $f$ along the curve now becomes a function defined on the real line. The arclength derivative coincides with the usual derivative on the real line in this case. The main point of the arclength derivative is to "factor out" an arbitrary scaling factor that arises from the arbitrariness of the choice of parametrisation.

  2. The normal component of the change of a vector along a curve is very useful in many computations, especially so when you need to deal with the curvature of the curve.

  3. No, you cannot apply that to "not necessarily normal $u$ and $v$", since the algebraic relation you wrote down for the Leibniz rule is only valid for normal fields. What you do is use the fundamental theorem of calculus, and integrate both sides in $s$, which gives that if $f$ and $g$ are normal fields along $\gamma:s\mapsto \gamma(s)$ which is parametrised via arclength, you have $$ (f\cdot g)(\gamma(s_1)) - (f\cdot g)(\gamma(s_0)) = \int_{s_0}^{s_1} (\nabla_s f\cdot g + \nabla_s g\cdot f)\circ\gamma ~\mathrm{d}s $$ That is, it is essentially identical to the integration by parts formula from single variable calculus. Of course, in the case where $f\cdot g$ evaluates to the same number at $\gamma(s_1)$ and $\gamma(s_0)$ (as would be the case where $\gamma$ is a closed curve and $\gamma(s_1) = \gamma(s_0)$) the left hand side will vanish.

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Thanks, great answer. – Court Jun 3 '12 at 10:29

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