# What is the functional derivative?

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  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?

• in this short article its very well explained – 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".
• "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? – Murad Babayev Apr 8 at 22:10
• @MuradBabayev. The mapping $\phi \mapsto \delta F[\rho,\phi]$ is usually a distribution, and this is often given by an integration against a function, $\phi \mapsto \int f_\rho(x) \, \phi(x) \, dx.$ – md2perpe Apr 9 at 5:44