3
$\begingroup$

In a recent question, I asked for examples of families of distinct smooth curves with fixed area and perimeter (which for this question I will dub as doubly-isometric). That wording allows $C^1$ examples and there are several piecewise-smooth examples of such in that question. However, my real hope was to see $C^\infty$ examples; as of yet (7/30/2014) there have been no examples of doubly-isometric curves which are $C^{n>1}$ smooth.

This isn't surprising: It's hard enough finding a doubly-isometric family of piecewise-functions; requiring that the curvatures be smooth as well is thus quite difficult, to say nothing of an analytic example. Because of this, my current feeling is that a $C^\infty$ example cannot be 'engineered' but must rather be derived from the fixed area/perimeter condition.

One strategy to this end is to require that the area and perimeter functionals $$A[r(\phi)]=\oint\frac{1}{2}r(\phi)^2\,d\phi,\hspace{1cm}L[(\phi)]=\oint \sqrt{r(\phi)^2+r'(\phi)^2}\,d\phi,$$ (specified in polar coordinates) be held as constrained while the curve $r=r(\phi)$ is perturbed. This is strongly reminiscent of Dido's problem: Among all curves of a given perimeter, the circle is the one with the largest area. This is a problem in the calculus of variations, whereby one extremalizes $A[r]$ while fixing $L[r]$; in this manner one derives the appropriate Euler-Lagrange equation and solves to find the correct $r(\phi)$.

Thus I hoped that the calculus of variations could be applied to this problem as well. The stumbling block is that area and perimeter are both held fixed, and there is no obvious functional to extremalize. Because of this it's not obvious to me how (or even if) a variational approach is appropriate. My question is therefore: Is there a viable variational strategy for discovering doubly-isometric curves? If so, how is this to be implemented?

$\endgroup$
  • $\begingroup$ Note that, fully carried out, a positive result for this question will also answer the linked question. $\endgroup$ – Semiclassical Jul 30 '14 at 19:48
  • $\begingroup$ Note, you will need some other constraint to make sure your curves are not all the same shape. For example, what's to stop your family of curves all happening to be circles of the same radius? So you probably want to constrain them to all pass through a couple of fixed points. $\endgroup$ – DavidButlerUofA Jul 30 '14 at 20:29
  • 1
    $\begingroup$ Well, I did say that they should be distinct curves. But I see your point: that needs to be implemented somehow. @DavidButlerUofA $\endgroup$ – Semiclassical Jul 30 '14 at 20:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.