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

Question

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?

Answer 1

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 $$\tag{5} \frac{\delta F}{\delta x(t)}~=~ \int dt^{\prime} ~\frac{\delta F}{\delta y(t^{\prime})}\frac{\delta y(t^{\prime})}{\delta x(t)}.$$

Answer 2

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

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

and

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

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.

Answer 3

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).