An isoperimetric type maximization problem with a barrier. - MathOverflow most recent 30 from http://mathoverflow.net2013-05-22T17:42:08Zhttp://mathoverflow.net/feeds/question/99036http://www.creativecommons.org/licenses/by-nc/2.5/rdfhttp://mathoverflow.net/questions/99036/an-isoperimetric-type-maximization-problem-with-a-barrierAn isoperimetric type maximization problem with a barrier.Dorian2012-06-07T13:55:31Z2012-06-07T20:10:18Z
<p>I'm trying to minmize a particular functional which depends on a curve with fixed endpoints which lies below a fixed line in $\mathbb{R}^2$. Here are the details:</p>
<p>Let $(r(\theta), \theta)$ be a smooth segment of a curve in polar coordinates satisfying the following for $0 < \theta_1, \theta_2 < \pi$. </p>
<p>1) $r(\theta_i) \sin (\theta_i) = \lambda$ for $i=1,2$ where $\lambda$ is a fixed constant.</p>
<p>2) $r(\theta) \sin(\theta) \leq \lambda$ for all $\theta \in [\theta_1, \theta_2]$. </p>
<p>Thus $(r(\theta),\theta)$ is a curve which touches the line $y=\lambda = $ constant at its two endpoints and lies beneath this line for all other value of $\theta$. Let $\mathcal{A}$ denote the class of curves $(r(\theta),\theta)$ satisfying conditions $1)$ and $2)$. Then consider the maximization problem:</p>
<p><code>$\max_{ (r,\theta) \in \mathcal{A} } \int_{\theta_1}^{\theta_2} \frac{1}{\sin^2(\theta) \sqrt{1+(\dot r/r)^2} } d\theta$</code>.</p>
<p>Geometrically speaking, the above can be written as <code>$\int_{\theta_1}^{\theta_2} \frac{d\theta}{dS} \frac{r^2}{\sin^2(\theta)} d\theta$</code>, which makes it appear as a sort of inverse perimeter problem.
Clearly the curve would like to follow the trajectory of a circle, but due to the constraint it cannot do this everywhere. My suspicion is that the solution should be attained by $r(\theta) = \frac{\lambda}{\sin \theta}$ but it's not clear that it would not be favorable for the curve to descend below this line if it can follow the path of a circle for some rangle of angle. </p>
<p>If this type of problem is known then references would also be appreciated. </p>