2
$\begingroup$

To be more precise I am interested in questions similar to the one below (I asked the question below on math.stackexchange last week but got not answer.)

I have a $C^1$ function $f:[0,1]^2 \to \mathbb{R}$. The arclength of the graph of $f_y:x\mapsto f(x,y)$ is $$L(y)=\int_0^1 \sqrt{1+\left(\frac{\partial f}{\partial x}(x,y)\right)^2} dx$$ Deriving under the integral does not seem to be possible since it would involve a crossed derivative $\partial^2 f/\partial x\partial y$ whose existence is not assumed. Hence I do not expect L to always be a $C^1$ function of y. But so far I have not been able to find a counterexample. This question must have been already looked at. Any idea?

$\endgroup$
5
  • 2
    $\begingroup$ The way I see it, it doesn't even need to be continuous. An extreme example: take a cylinder and cut it by planes parallel to the axis: at some point, the length changes from $\infty$ to $0$. $\endgroup$ Oct 15, 2014 at 16:46
  • 2
    $\begingroup$ I think the difficult-looking appearance of your question has to do with your choice of coordinates. If instead of comparing the Riemann metric on your curves to the domain of $f$, i.e. $[0,1]$ you compare them directly, you get a much better sense for how the length varies. If your two curves are uniformly $C^1$-close, orthogonal projection from one to the other (using the normal line to the tangent space) gives a natural map between the two and allows you to compare their lengths with less bias. I think if you write out the details this answers your question. $\endgroup$ Oct 15, 2014 at 18:06
  • $\begingroup$ OK Alex, the title is bad, I will change it if I can. $\endgroup$ Oct 15, 2014 at 20:41
  • $\begingroup$ Alex's argument works just as well for any compact surface containing a line segment. Take the surface of a cube, and then slightly smooth out the corners and the edges. Length of intersection is not defined along planes that hit the faces. Or take a cigar: cap off a finite length cylinder with two spherical caps. Then your length of intersection jumps up from zero to positive suddenly. $\endgroup$
    – Ben McKay
    Oct 17, 2014 at 13:55
  • $\begingroup$ You are right, Ben. I am so sorry to be unable formulate a title right. What I meant is really the question that is asked, not its title. $\endgroup$ Oct 17, 2014 at 17:42

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.