# Euler-Lagrange equation in polar or cylindrical coordinates

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.
• Thank you so much. Doing as you say I would get the Euler-Lagrange equation and the following part $\begin{equation*} \int ^{1}_{0}\int ^{2\pi }_{0}\frac{\partial }{\partial r}\left(\frac{\partial G}{\partial u^{'}_{r}} \delta u\right) \ +\ \frac{\partial }{\partial \theta }\left(\frac{\partial G}{\partial u^{'}_{\theta }} \delta u\right) \ d\theta dr \end{equation*}$ Can you also help me with how I can apply the Gauss divergence theorem on this to get the natural boundary conditions? – user1887239 Dec 30 '18 at 12:02
• With cartesian I'd use Gauss divergence theorem at this point to get the natural BCs which are of the form $\begin{equation*} I\ ( u) \ =\ \int ^{1}_{0}\int ^{1}_{0}\frac{\partial }{\partial x}\left(\frac{\partial F}{\partial u^{'}_{x}} \delta u\right) \ +\ \frac{\partial }{\partial y}\left(\frac{\partial G}{\partial u^{'}_{y}} \delta u\right) \ dxdy \end{equation*}$ which then equals ... (second comment) – user1887239 Dec 30 '18 at 13:17
• $\begin{equation*} \oint _{s}\left(\left(\frac{\partial F}{\partial u^{'}_{x}}\right) \ \overrightarrow{n_{x}} +\ \left(\frac{\partial G}{\partial u^{'}_{y}}\right)\overrightarrow{n_{y}} \ \right) \delta u\ ds \end{equation*}$ (n being the unit vector) I don't know how to apply the same thing here – user1887239 Dec 30 '18 at 13:17
• But you now have an integral equation over a rectangle, its exactly the same. Just substitute $r=x$, $\theta=y$ and see how far you can get forgetting about polar coordinates. – Eddy Dec 30 '18 at 13:20