What form should the minimal-length curve to $\int \sqrt{dr^2 +r^2d\theta^2\over{1-r^2}}$ take? I think I can use the Euler-Lagrange equations. So write the integral as $\int\sqrt{({dr\over d\theta})^2 +r^2\over{1-r^2}}d\theta$, then since the integrand is not explicitly dependent on $\theta$, we can say that I-r'{\partial I\over \partial r'}=c for some (complex?) constant $c$. But I don't seem to be getting anything useful. The equation reduced to r^2=c\sqrt{(r'^2+r^2)(1-r^2)} by my calculation. But I think I am expecting straight lines through the origin...
Lines through the origin and Euler-Lagrange
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0@WillieWong: You are right, I am indeed looking for geodesics on the disc, but the metric is one $1\over (1-r^2)$ factor different – 2012-02-26
1 Answers
If you expect straight lines through the origin, it makes no sense to consider $r$ as a function of $\theta$, since your entire line is at a single $\theta$ value. (Note also that using the Beltrami identity instead of the Euler-Lagrange equation as you did introduces a spurious solution r'=0 that leads to circles that don't solve the Euler-Lagrange equation.)
You can consider $\theta$ as a function of $r$ instead. Then you have
\int\sqrt{\frac{1+r^2\theta'^2}{1-r^2}}\mathrm dr\;,
and since $\theta$ doesn't occur in the integrand, the Euler-Lagrange equation becomes
\frac{\partial\mathcal L}{\partial\theta'}=C\;,\\ r^2\theta'\sqrt{\frac{1-r^2}{1+r^2\theta'^2}}=C\;.
This equation is indeed solved by $\theta=\text{const.}$, so lines through the origin are solutions, but it also has further solutions:
r^4\theta'^2(1-r^2)=C(1+r^2\theta'^2)\;,\\ \theta'=\sqrt{\frac C{r^4-r^6-Cr^2}}\;.
Wolfram|Alpha integrates this to a rather complicated function of $r$.
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0@joriki, Any chance for assistance here: http://math.stackexchange.com/questions/1066495/minimization-of-variational-total-variation-tv-deblurring – 2014-12-14