I have seen wikipedia use the product rule and maybe the chain rule for the variation of the Langragin as follows:

\begin{align} \dfrac{\delta [f(g(x,\dot{x}))h(x,\dot{x})] } {\delta x} = \left( \dfrac{\delta [f(g)] } {\delta g} \dfrac{\delta [g(x,\dot{x})] } {\delta x} \right) h(x,\dot{x}) + f(g(x,\dot{x})) \dfrac{\delta [h(x,\dot{x})] } {\delta x} \end{align} where the variation of the Lagrangian is defined \begin{align} \dfrac{\delta \mathcal{L} } {\delta x} = \dfrac{\partial \mathcal{L} } {\partial x} - \dfrac{d}{d \tau} \dfrac{\partial \mathcal{L} } {\partial \dot{x}} \end{align} and $\mathcal{L}=f(g(x,\dot{x}))h(x,\dot{x})$.

Does the variation of the Lagrangian satisfy the product rule and chain rule of the derivative?

  • 3
    $\begingroup$ Typically it is the action, $S=\int\,dt\,\mathcal L$, that is varied and you have to use integration by parts to prove your second formula. $\endgroup$
    – Kyle Kanos
    Sep 14, 2014 at 14:25

3 Answers 3

  1. OP considers the 'same-time' functional derivative (FD) $$\tag{1} \frac{\delta f(t)}{\delta x(t)}~:=~\frac{\partial f(t)}{\partial x(t)} - \frac{d}{dt} \frac{\partial f(t)}{\partial \dot{x}(t)} +\ldots. $$ Here $f(t)$ is shorthand for the function $f(x(t), \dot{x}(t), \ldots;t)$. Although the 'same-time' FD (1) can be notationally useful, it has various fallacies, cf. my Phys.SE answer here.

  2. The Leibniz rule $$\tag{2} \frac{\delta (f(t)g(t))}{\delta x(t)} ~=~\frac{\delta f(t)}{\delta x(t)} g(t) +f(t)\frac{\delta g(t)}{\delta x(t)}\qquad(\leftarrow \text{Wrong!}) $$ for the 'same-time' FD (1) does not hold. Counterexample: Take $f(t)=g(t)=\dot{x}(t)$.

  3. The chain rule $$\tag{3} \frac{\delta f(t)}{\delta x(t)} ~=~\frac{\delta f(t)}{\delta y(t)}\frac{\delta y(t)}{\delta x(t)}\qquad\qquad(\leftarrow \text{Wrong!}) $$ for the 'same-time' FD (1) does not hold. Counterexample: Take $f(t)=y(t)^2$ and $y(t)=\dot{x}(t)$.

  4. However, the usual FD $\frac{\delta F}{\delta x(t)}$ (where $F[x]$ is a functional) does satisfy a Leibniz rule $$\tag{4} \frac{\delta (FG)}{\delta x(t)} ~=~\frac{\delta F}{\delta x(t)} G +F\frac{\delta G}{\delta x(t)}, $$ and a chain rule $$\tag{5} \frac{\delta F}{\delta x(t)}~=~ \int dt^{\prime} ~\frac{\delta F}{\delta y(t^{\prime})}\frac{\delta y(t^{\prime})}{\delta x(t)}.$$

  • $\begingroup$ Good points, +1. I should've mentioned that this (unorthodox) definition does not work. $\endgroup$
    – Danu
    Sep 14, 2014 at 14:41

In general functional derivatives obey chain and product rules. If the concept troubles you you can always think of a function as a vector with an infinity of coordinates. Then a functional derivative is just a partial derivative.

If $F[h]$ is a functional of the function $h(x)$. You can think of this as

$$ h \to \vec{h} = \left(h(x_1), h(x_2), ..., h(x_n)\right) \equiv \left(h_1, h_2, ..., h_n\right)\, ,$$


$$ F[h] \to F\left(h_1, h_2, ..., h_n\right) \, .$$

The set of $x_i$ is infinitely large and covers all values of $x$. Then the analogy for functional derivatives is

$$ \frac{\delta F}{\delta h(x)} \to \frac{\partial F}{\partial h_i} \, .$$

$i$ is chosen such that $x_i = x$.

This analogy works well but beware of dimensions! The definition of a functional derivative is ($\delta(x-x')$ is the delta distribution),

$$ \frac{\delta F}{\delta h(x)} \equiv \lim_{\epsilon \to 0} \frac{F[h(x)+\epsilon \delta(x-x')]-F[h]}{\epsilon} \, .$$

This does not have the same dimension as what you would expect from the analogy,

$$ \frac{\partial F}{\partial h_i} \equiv \lim_{\delta h \to 0}\frac{F(h_1,...,h_i+\delta h,...,h_n)}{\delta h} \, ,$$

because the delta function as well as $\epsilon$ carry a dimension.

Note that what you call definition for the functional derivative

$$ \frac{\delta F}{\delta h(x)} = \frac{\partial F}{\partial h} - \frac{\text{d}}{\text{d}t} \frac{\partial F}{\partial \dot{h}}\, ,$$

only applies to the Lagrangian and is a property of classical mechanics.


Yes. Here, we are dealing with functional derivatives and these satisfy the chain rule and the product rule, which is really an important reason why it can be called a derivative to begin with.

Important note: The definition that you give for the functional derivative is not the standard one, and does not satisfy its usual properties (as shown by Qmechanic).

  • $\begingroup$ I would love to see that prove. I am very curious. $\endgroup$ Sep 14, 2014 at 14:28
  • $\begingroup$ @linuxfreebird I don't think it should be all too hard starting from the definition given in the wiki article I linked. Alternatively, you can try to obtain a copy of the book by Parr & Young that wikipedia cites. $\endgroup$
    – Danu
    Sep 14, 2014 at 14:40

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