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


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.


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] http://lab.sentef.org/wp-content/uploads/2016/11/Tutorial_02.pdf page 4

[2] https://en.wikipedia.org/wiki/Functional_derivative

  • $\begingroup$ in this short article its very well explained $\endgroup$ – Masacroso Apr 4 at 12:20

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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  • $\begingroup$ "Thus, δF[ρ,ϕ] is a functional, usually a distribution, and often a function." - this confuses me Did you mean the mapping from $\phi$ or the $\delta F$ itself (in the case of distribution/function? $\endgroup$ – Murad Babayev Apr 8 at 22:10
  • $\begingroup$ @MuradBabayev. The mapping $\phi \mapsto \delta F[\rho,\phi]$ is usually a distribution, and this is often given by an integration against a function, $\phi \mapsto \int f_\rho(x) \, \phi(x) \, dx.$ $\endgroup$ – md2perpe Apr 9 at 5:44

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