I think the question boils down to "Am I allowed to write the following" :
$$ \frac{d}{dx}\frac{\partial u(x,y)}{\partial x} = \frac{\partial^2 u}{\partial x^2} $$
I would say so because in the chain rule the partial derivative of y with respect to x should be equal to zero but nevertheless I feel the need to ask the question because in the context described below, I feel a little bit uneasy, especially when derivative with respect to derivatives appear
To put things into perspective, I aim to compute the Euler-Lagrange equation for the following functional $J[U]$ in polar coordinate :
$$J[U] = \int_{\Omega}\frac{1}{2} \left(\nabla \cdot U\right)^2d\Omega$$
where $U = \left[ \begin{matrix} u(r,\theta) \\ v(r,\theta) \end{matrix} \right] $
The Euler-Lagrange equations for this type of functional can be written as a system of two equations :
$$ \frac{\partial F}{\partial u} - \frac{d}{dr}\frac{\partial F}{\partial u_r} - \frac{d}{d \theta}\frac{\partial F}{\partial u_\theta} = 0\\ \frac{\partial F}{\partial v} - \frac{d}{dr}\frac{\partial F}{\partial v_r} - \frac{d}{d\theta}\frac{\partial F}{\partial v_\theta} = 0\\ $$
Note that in this context $u_x$ is a short notation for $\frac{\partial u}{\partial x}$
Assuming that wikipedia gives the proper formula for the del operator in polar coordinates, I assume $\nabla \cdot U$ to be equal to :
$$ \nabla \cdot U = \frac{1}{r} \frac{\partial (r u)}{\partial r} + \frac{1}{r}\frac{\partial v}{\partial \theta} $$
Hence we have :
$$ F = \frac{1}{2}\left(\frac{1}{r} \frac{\partial (r u)}{\partial r} + \frac{1}{r}\frac{\partial v}{\partial \theta} \right)^2 \\ F = \frac{1}{2}\left(\frac{1}{r} \left[u + r\frac{\partial u}{\partial r}\right]+ \frac{1}{r}\frac{\partial v}{\partial \theta}\right)^2\\ F = \frac{1}{2}\left(\frac{u}{r} + \frac{\partial u}{\partial r} +\frac{1}{r}\frac{\partial v}{\partial\theta} \right)^2 $$ I'm aiming to compute the different terms of the EL system and would like to be certain that my calculations are correct. In particular, I'm not completely at ease with partial differentiation, especially in this context.
Let's take the term $\frac{d}{dr}\frac{\partial F}{\partial u_r}$ for example. I would compute it as follows :
$$ \frac{d}{dr}\frac{\partial F}{\partial u_r} = \frac{d}{dr} \left( \frac{1}{2} \cdot 2 \cdot \frac{\partial}{\partial u_r} \left(\frac{u}{r} + \frac{\partial u}{\partial r} +\frac{1}{r}\frac{\partial v}{\partial\theta} \right) \cdot \left(\frac{u}{r} + \frac{\partial u}{\partial r} +\frac{1}{r}\frac{\partial v}{\partial\theta} \right)\right)\\ \frac{d}{dr}\frac{\partial F}{\partial u_r} = \frac{d}{dr} \left(\frac{u}{r} + \frac{\partial u}{\partial r} +\frac{1}{r}\frac{\partial v}{\partial\theta} \right)\\ \frac{d}{dr}\frac{\partial F}{\partial u_r} = \frac{d}{dr}\frac{u}{r} + \frac{d}{dr}\frac{\partial u}{\partial r} + \frac{d}{dr}\left(\frac{1}{r}\frac{\partial v}{\partial \theta}\right) $$
This last computation, thouth, is where I'm the least at ease. I would be tempted to write the following :
$$ \frac{d}{dr}\frac{\partial F}{\partial u_r} = \frac{1}{r^2}\left(r\frac{d u}{d r} - u\right) + \frac{\partial^2 u}{\partial r^2} + \frac{1}{r^2}\left(r\frac{\partial^2v}{\partial r \partial \theta} - \frac{\partial v}{\partial \theta}\right) $$
Is this correct ?