Differentiation with respect to non-independent variable

Question

If a function is defined as $F(x(t), x'(t), t) = \int(E+V)dt$ would $dF/dE$ equal zero? We know that $E$ and $V$ are made up of $x, x', t$ but $F$ is a function of $x, x', t$ not $E$. So the confusion is that it seems like the function $F$ contains $E$, but the way the function is defined does not explicitly say so.

Answer 1

The answer is no: $\mathrm{d}F/\mathrm{d}E$ is never equal to zero. However, explaining the concept need some clarifications, so let's start from the very concept of functional dependence.

First of all, the writing $$ F\big(x(t), x'(t), t\big) = \int(E+V)\,\mathrm{d}t $$ is too imprecise to help in the understanding of what's going on. It is not safe to use the same $t$ symbol for the integration variable and the independent (time?) variable on the left side of the equality, thus in the following I'll use $s$ for the former and reserve $t$ for the latter.
Now let's try to analyze what can be thought the most general form of a integral function-functional as given in the original post: $$ F\big(x(t), x'(t), t\big) = \int\limits_{[a(t),b(t)]}\Big[E\big(t, x(t), x'(t), s, x(s), x'(s)\big)+ V\big(t, x(t), x'(t), s, x(s), x'(s)\big)\Big]\,\mathrm{d}s\label{1}\tag{1} $$ where $[a(t),b(t)]\subseteq\Bbb R$, meaning with this that $a=-\infty$ and $b=+\infty$ are acceptable choices. Then, calculating the functional derivative of \eqref{1} (as described, for example, in this long Q&A) we have $$ \begin{split} \frac{\delta F}{\delta E}& =\frac{\delta F}{\delta E}[E_o]\big(x(t), x'(t), t\big)\\ &=\frac{\mathrm{d}}{\mathrm{d} \epsilon}\int\limits_{[a(t),b(t)]}\Big[E\big(t, x(t), x'(t), s, x(s), x'(s)\big)+ \epsilon E_o\big(t, x(t), x'(t), s, x(s), x'(s)\big)\\ & \qquad \qquad\qquad \qquad\qquad+V\big(t, x(t), x'(t), s, x(s), x'(s)\big)\Big]\,\mathrm{d}s\Bigg|_{\epsilon=0}\\ &= \int\limits_{[a(t),b(t)]} E_o\big(t, x(t), x'(t), s, x(s), x'(s)\big)\,\mathrm{d}s \end{split}\label{2}\tag{2} $$ where $E_o=E_o\big(t, x(t), x'(t), s, x(s), x'(s)\big)$ is an admissible variation of the quantity (energy?) $E$.
The meaning of \eqref{2} is simply that the functional derivative of the functional $F$ is the integration on the given interval $[a(t), b(t)]$, i.e. $$ \frac{\delta F}{\delta E}[\cdot] = \int\limits_{[a(t),b(t)]}[\cdot]\,\mathrm{d}s $$

Notes

• In my opinion, \eqref{1} is the "most general" form of the functionals described in the OP since its value depend both directly on the pointwise values of $t, x(t)$ and $x'(t)$ than on their "global" shape as a whole.
• The calculations show that, if the structure of the functional is the one given in the OP, its functional derivative cannot be zero: in this case, it is the operator of integration over a given (fixed or variable, finite or infinite) interval. And the Euler-Lagrange equations loose their significance in this setting.
• Note also that, assuming even a slightly simpler structure i.e. choosing $$ E=E\big(s, x(s), x'(s)\big) $$ i.e. allowing $E$ to be independent on the actual values $t, x(t), x'(t)$, does not change the structure of the functional derivative, which is again the integration on the interval $[a(t), b(t)]$.