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Refering to the two sources given as hyperlinks in this question, there seems to me to be a discrepancy (or contradiction) in the definition of the functional derivative and its relation to the first variation which I cannot resolve.

Let $F[y]:M \rightarrow \mathbb R$ be a sufficiently regular functional with some set M of sufficiently regular functions y(x) defined on the intervall $[a,b]$. Let furthermore $\eta(x)$ be a test function and $\epsilon \in \mathbb R$ a variable. Let's assume that both y and $\eta$ are fixed.

The variation of y can be represented as $\delta y(x) = \epsilon \, \eta(x)$. According to the first source, which is an appendix to the book by Engel/Deizler, the increment $\triangle F:= F[y+\delta y]-F[y]= F[y+\epsilon \, \eta]-F[y]$ of the functional can be represented as the Taylor expansion

$ F[y+\epsilon \, \eta]-F[y]= \left[ \frac{dF[y+\epsilon \, \eta]}{d\epsilon} \right]_{\epsilon=0} \epsilon + \frac{1}{2} \left[ \frac{d^2F[y+\epsilon \, \eta]}{d\epsilon^2} \right]_{\epsilon=0} \epsilon^2 + \cdots. \qquad \qquad (1)$

Here my understanding is that the first term on the rhs represents the first variation $\delta F[y]$, etc., so that we have

$ F[y+\epsilon \, \eta]-F[y]= \delta F + \delta^2 F + \cdots. \qquad \qquad (2) $

Now in the second source, which is Wikipedia, the functional derivative $\frac{\delta F}{\delta y(x)}$ is defined by setting (in my notation)

$ \left[ \frac{dF[y+\epsilon \, \eta]}{d\epsilon} \right]_{\epsilon=0} = \int \frac{\delta F}{\delta y(x)} \eta \, dx = \delta F[y]. \qquad \qquad (3)$

The expression in the middle is explicitly identified with the differential or first variation $\delta F[y]$. It is explained that the functional derivative can be understood as something like a partial derivative in the case of infinitely many variables, and the integral, i.e. the first variation, as something like the total differential in the case of infinitely many variables.

The discrepancy is that in the Taylor expansion the first variation is the expression on the rhs of

$ \delta F[y] = \left[ \frac{dF[y+\epsilon \, \eta]}{d\epsilon} \right]_{\epsilon=0} \epsilon, \qquad \qquad (4) $

note the additional $\epsilon$.

Related to this contradiction I have a general question of understanding. If the functional derivative is the analogue to partial derivatives, and the first variation the analogue to a total differential, what, in calculus of variations, is the analogue to a total derivative? We speak of F[y] as being differentiable, but there is no total derivative.

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  • $\begingroup$ were you able to reconcile the definitions? $\endgroup$ Jul 25, 2023 at 20:08

1 Answer 1

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It might be, that there is a mistake in the Engl. Wiki in that the right equation of (3) is not correct. In the Italian Wiki the formula for the first variation in terms of the functional derivative is instead

$\delta F[y] = \int \frac{\delta F}{\delta y(x)} \delta y(x) \, dx $.

This would resolve the contradiction, because with $\delta y(x) = \epsilon \eta(x)$ and $\epsilon$ being constant over [a,b], the right equation of (3) is in line with (4).

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  • $\begingroup$ Maybe this Q&A will help you in finding some inexact statements that are found on the Wikipedia. $\endgroup$ Mar 9, 2021 at 21:12
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    $\begingroup$ @DanieleTampieri, my interest is not in finding inexact statements on Wikipedia, but in finding the truth. $\endgroup$ Mar 10, 2021 at 8:00

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