Functional derivative - understanding some basics

I have the following functional $$L[u] = \int_0^l dx [-\frac{\lambda}{2}u^2 + \frac{1}{4}u^4] = \int dx J[u]$$ Now, I need to calculate $$\frac{\delta L}{\delta u}$$ As I understand, since I can take an arbitrary $u$, I can use the fundamental lemma of the calculus of variations that states that $$\frac{\delta L}{\delta u} = \frac{\partial J}{\partial u} - \frac{d}{dx}\frac{\partial J}{\partial u'}$$ My problem is with the second part of the RHS. Can I change the order of differentiation, notice that $J$ does does not explicitly depend on $x$ and therefore discard it?

No, you can't change the order of differentiation in the term $$\frac{d}{dx} \frac{\partial J}{\partial u'}.$$ The idea here is that $u$ and $u'$ are functions of $x$, and the notation (writing $\frac{d}{dx}$ as a total derivative instead of a partial derivative) is supposed to suggest that the $\tfrac{d}{dx}$ sees that $x$-dependence. For instance, if $J[u, u'] = (u')^2 - 3u$, then $$\frac{\partial J}{\partial u'} = 2u'$$ and $$\frac{d}{dx} \frac{\partial J}{\partial u'} = 2u''.$$ Notice that, on the other hand, $$\frac{\partial}{\partial u'} \frac{d J}{dx} = \frac{\partial}{\partial u'} (2u'u'' - 3u'),$$ which, frankly, doesn't even really make sense since what's implied by the notion of "partial derivative" in $\tfrac{\partial}{\partial u'}$, in this context, assuming $J$ is written as a function of $u, u'$, is that $u$ is being held constant (it doesn't say anything about $u''$; that's not even on $\tfrac{\partial}{\partial u'}$'s radar). However, if it did make any sense, it'd probably be equal to $2u'' - 3$, which is not what we want.
You can, however, note that in your case $J$ doesn't depend explicitly on $u'$, so $\tfrac{\partial J}{\partial u'} = 0$.
• Great, thanks for the explanation. Just another verification - I can use the Euler-Lagrange equation because I can choose an arbitrary $u$, right?