I do not understand, if the functional derivative is
- a function
- a generalized function (distribution)
- a functional itself
- something different (see Euler-Lagrange)
To clarify my question, I have seen multiple instances of functional derivative definitions
Functionals
When the Functional gets Taylor expanded (here using a "good" $\eta(x)$) we get
$$F[y(x)+\epsilon \eta(x)] = F[y(x)] + \frac{dF[y(x) + \epsilon \eta(x)]}{d\epsilon}\Big|_{\epsilon=0}\cdot \epsilon + ...$$
as I understood, the term on the RHS is the functional derivative. But since the LHS is a functional and the RHS is a functional + a real number ($\epsilon$) times the functional derivative, I conclude that the functional derivative must also be a functional.
Functions/Distributions
The english wikipedia page [2] states, that the functional derivative is defined as
$$\int{\frac{\delta F}{\delta \rho} (x)\phi(x)dx}=\frac{dF[\rho(x) + \epsilon \phi(x)]}{d\epsilon}\Big|_{\epsilon=0}$$
notice that the RHS is equivalent to the functional derivative defined above. However, it is $$\frac{\delta F}{\delta \rho} (x)$$ that is defined to be the functional derivative, and not the RHS (as I concluded above). Therefore I may as well assume that the functional derivative is a function/distribution.
Something else
The solution to the Euler-Lagrange Equation (one dimensional for simplicity) given an Energy Functional $J[y] = \int_{a}^{b}{L(x,y,y')}$ is
$$\frac{\delta J}{\delta y} = \frac{dL}{dy} - \frac{d}{dx}(\frac{dL}{dy'}) = 0$$
here, $\frac{\delta J}{\delta y}$ is supposedly the fractional derivative of the integral, which has to be stationary. RHS tells me that the functiona derivative is a differential equation - which has a function as a solution - but I am now completely unsure what the functional derivative in itself actualy is.
I have seen multiple viewpoints, each and every one cluttering my intuition even more. For instance the wikipedia article claims that $\frac{\delta F}{\delta \rho} (x)$ has to be seen as a "gradient" (which is a vector in multivariate calculus), while $\int{\frac{\delta F}{\delta \rho} (x)\phi(x)dx}$ has to be thought of like a directional derivative (which is the inner product of the gradient and the direction vector). But since there are no bounds on the integral the "directional derivative" is also a function, or am I mistaken?
[1] page 4
[2]
$\endgroup$ 11 Answer
$\begingroup$The expression$\delta F[\rho,\phi] := \frac{dF[\rho(x) + \epsilon \phi(x)]}{d\epsilon}\Big|_{\epsilon=0},$when defined, is a functional of $\rho$ and $\phi.$ The dependency on $\rho$ is usually non-linear, while the dependency on $\phi$ is usually linear.
If the expression is restricted to $\phi \in C_c^\infty(\mathbb R^n)$ and the dependency on $\phi$ is linear, then the mapping $\phi \mapsto \delta F[\rho,\phi]$ is usually a distribution. Often this distribution can be identified with a function.
Thus, $\delta F[\rho,\phi]$ is a functional, usually a distribution, and often a function.
Often we have $F[\rho] = \int L(x, \rho(x), \rho'(x)) \, dx$ for some Lagrangian $L.$ Then, if $\phi$ vanishes on the boundary of the domain, $$ \delta F[\rho,\phi] = \int \left( \frac{\partial L}{\partial \rho} \phi(x) + \frac{\partial L}{\partial \rho'} \phi'(x) \right) dx = \int \left( \frac{\partial L}{\partial \rho} - \frac{d}{dx}\frac{\partial L}{\partial \rho'} \right) \phi(x) \, dx. $$In this case, $\delta F[\rho,\phi]$ is given by an integral of a function (the parenthesis) times $\phi.$ Thus this falls into the case "Often this distribution can be identified with a function".
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