I have been studying Euler-Lagrange in Variation Calculus. I am comfortable with the formulation when the function under the integral is of the form f = f(x, y). But I am unsure as to how this change for a function given in polar coordinates f = f(r, theta)
say
$\begin{equation*} I\ ( u) \ =\ \int ^{1}_{0}\int ^{2\pi }_{0} F\left( r,\ \theta ,\ u,\ \frac{\partial u}{\partial r} ,\ \frac{\partial u}{\partial \theta }\right) \ rd\theta dr \end{equation*}$
so how would you calculate the variation, $\delta I $
Will it be
$ \delta I (u) = \int ^1_0 \int ^{2\pi}_0 (\frac {\partial F} {\partial u} + \frac {\partial F} {\partial u ^`_r} {\delta u ^`_r} + \frac {\partial F} {\partial u ^`_\theta} {\delta u ^`_\theta} ) rd\theta dr$
I am confused about this, this seems incorrect, what should be done with the $r$ at the end? Also how do I apply Gauss divergence theorem at the end to seperate the boundary conditions from the rest of the integral.
You can just define $G=F \cdot r$, use the standard methods (treating $r$ and $\theta$ the same as cartesian coordinates), and transform back to $F$ at the end.