# 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 [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?

• in this short article its very well explained Commented Apr 4, 2020 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? Commented Apr 8, 2020 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, $\phi \mapsto \int f_\rho(x) \, \phi(x) \, dx.$ Commented Apr 9, 2020 at 5:44
• Isn't $\phi$ just an arbitrary function from the same space as $\rho$? Why are we assuming that it vanishes on the boundary? Do we need to consider $\rho$ also vanishing on the boundary for this to be the case? Commented Nov 13, 2022 at 2:17
• @lightxbulb. It's a special case that is often used. Commented Nov 19, 2022 at 14:08
• @lightxbulb. That's certainly related. If the function vanishes on the boundary (sometimes also vanishing derivatives are required) then boundary terms from the integration disappear. Unfortunately I don't have any references other than Wikipedia and any book about calculus of variations. Commented Nov 20, 2022 at 13:57

Functionals of Smooth Fields

Suppose you have a smooth multivariable scalar field $${f}\in\mathcal{C}^{\infty}(\mathbb{R}^{n})$$. Then a functional $${W}\in\mathrm{Funct}(\mathcal{C}^{\infty}(\mathbb{R}^{n}))$$ is a mapping from $$\mathcal{C}^{\infty}(\mathbb{R}^{n})$$ to $$\mathbb{R}$$ that takes the form \begin{align*} {W}[f]=\int\mathcal{L}(x)[f(x)]{\,}\mathrm{d}^{n}{x}{\,}{,} \end{align*} where $$\mathcal{L}(x)[f(x)]$$ is the Lagrangian, that depends both on the coordinate variables $${x}$$, as well as functionally depends on the field in coordinates $${f(x)}$$ and its consecutive derivatives $$\big\lbrace\frac{\partial{f(x)}}{\partial{x}^{\alpha}},\frac{\partial^{2}{f(x)}}{\partial{x}^{\alpha_{1}}\partial{x}^{\alpha_{2}}},\frac{\partial^{3}{f(x)}}{\partial{x}^{\alpha_{1}}\partial{x}^{\alpha_{2}}\partial{x}^{\alpha_{3}}},\cdots\big\rbrace$$. Now, rather then looking at the variational derivative as a limit, try thinking of $$\frac{\delta}{\delta{f(x)}}\in\mathrm{Der}(\mathrm{Funct}(\mathcal{C}^{\infty}(\mathbb{R}^{n})))$$ as the derivative operator on $$\mathrm{Funct}(\mathcal{C}^{\infty}(\mathbb{R}^{n}))$$, similar to how $$\frac{\partial}{\partial{x}^{\alpha}}\in\mathrm{Der}(\mathcal{C}^{\infty}(\mathbb{R}^{n}))$$ is the derivative operator on $$\mathcal{C}^{\infty}(\mathbb{R}^{n})$$, that satisfies the algebra \begin{align*} (\mathrm{I}){\,\,}&\frac{\delta({a}\mathcal{L}_{\alpha}(x)[f(x)]+{b}\mathcal{L}_{\beta}(x)[f(x)])}{\delta{f(y)}}={a}\frac{\delta\mathcal{L}_{\alpha}(x)[f(x)]}{\delta{f(y)}}+{b}\frac{\delta\mathcal{L}_{\beta}(x)[f(x)]}{\delta{f(y)}}{\quad}{a}{,}{b}\in\mathbb{R}{\,}{,}\\ (\mathrm{II}){\,\,}&\frac{\delta(\mathcal{L}_{\alpha}(x)[f(x)]{\,}\mathcal{L}_{\beta}(x)[f(x)])}{\delta{f(y)}}=\mathcal{L}_{\alpha}(x)[f(x)]\frac{\delta\mathcal{L}_{\beta}(x)[f(x)]}{\delta{f(y)}}+{\,}\mathcal{L}_{\beta}(x)[f(x)]\frac{\delta\mathcal{L}_{\alpha}(x)[f(x)]}{\delta{f(y)}}{\,}{,}\\ (\mathrm{III}){\,\,}&\frac{\delta\mathcal{L}(x)[g(f(x))]}{\delta{f(y)}}=\frac{\delta\mathcal{L}(x)[g(f(x))]}{\delta{g(f(y))}}\frac{\mathrm{d}{g(z)}}{\mathrm{d}{z}}\bigg\vert_{{z}={f(y)}}{\,}{.} \end{align*} The functional derivative can now be written as \begin{align*} \frac{\delta{W}[f]}{\delta{f(y)}}&=\int\frac{\delta\mathcal{L}(x)[f(x)]}{\delta{f(y)}}{\,}\mathrm{d}^{n}{x}=\int\sum_{{m}={0}}^{\infty}\frac{\partial\mathcal{L}(x)[f(x)]}{\partial\partial^{m}{f(x)}/\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\frac{\delta}{\delta{f(y)}}\bigg[\frac{\partial^{m}{f(x)}}{\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\bigg]{\,}\mathrm{d}^{n}{x}\\ &=\int\sum_{{m}={0}}^{\infty}\bigg(\frac{\partial^{m}}{\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\bigg[\frac{\partial\mathcal{L}(x)[f(x)]}{\partial\partial^{m}{f(x)}/\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\frac{\delta{f(x)}}{\delta{f(y)}}\bigg]\\ &+(-{1})^{m}\frac{\partial^{m}}{\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\bigg[\frac{\partial\mathcal{L}(x)[f(x)]}{\partial\partial^{m}{f(x)}/\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\bigg]\frac{\delta{f(x)}}{\delta{f(y)}}\bigg){\,}\mathrm{d}^{n}{x}{\,}{,} \end{align*} where $$\frac{\delta{f(x)}}{\delta{f(y)}}=\Delta({x}-{y})$$ is the Dirac delta function that is simply defined as \begin{align*} \int{f(x)}\Delta({x}-{y}){\,}\mathrm{d}^{n}{x}={f(y)}. \end{align*} Furthermore, since $$\frac{\partial^{m}}{\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\big[\frac{\partial\mathcal{L}(x)[f(x)]}{\partial\partial^{m}{f(x)}/\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\Delta({x}-{y})\big]$$ is a total derivative, it vanishes when being integrated over a set without boundary. And with all that combined, the functional derivative finally becomes \begin{align*} \frac{\delta{W}[f]}{\delta{f(y)}}&=\int\sum_{{m}={0}}^{\infty}(-{1})^{m}\frac{\partial^{m}}{\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\bigg[\frac{\partial\mathcal{L}(x)[f(x)]}{\partial\partial^{m}{f(x)}/\partial{x}^{\alpha_{1}}\cdots\partial{x}^{\alpha_{m}}}\bigg]\Delta({x}-{y}){\,}\mathrm{d}^{n}{x}\\ &=\sum_{{m}={0}}^{\infty}(-{1})^{m}\frac{\partial^{m}}{\partial{y}^{\alpha_{1}}\cdots\partial{y}^{\alpha_{m}}}\bigg[\frac{\partial\mathcal{L}(y)[f(y)]}{\partial\partial^{m}{f(y)}/\partial{y}^{\alpha_{1}}\cdots\partial{y}^{\alpha_{m}}}\bigg]{\,}{.} \end{align*}

Euler Lagrange Equation

The Euler Lagrange equation of an action is given by simply setting the variational derivative of the functional equal to zero, being \begin{align*} {0}\equiv\frac{\delta{W}[f]}{\delta{f(y)}}=\sum_{{m}={0}}^{\infty}(-{1})^{m}\frac{\partial^{m}}{\partial{y}^{\alpha_{1}}\cdots\partial{y}^{\alpha_{m}}}\bigg[\frac{\partial\mathcal{L}(y)[f(y)]}{\partial\partial^{m}{f(y)}/\partial{y}^{\alpha_{1}}\cdots\partial{y}^{\alpha_{m}}}\bigg]{\,}{.} \end{align*}

Application in classical mechanics

In classical mechanics, the action functional of a curve has the form \begin{align*} {W}[x]=\int\mathcal{L}(t)[x(t)]{\,}\mathrm{d}{t}=\int\bigg(\frac{m}{2}\bigg\Vert\frac{\mathrm{d}{x(t)}}{\mathrm{d}{t}}\bigg\Vert^{2}-{m}\Phi(x(t))\bigg){\,}\mathrm{d}{t}{\,}{.} \end{align*} With the standard Euler Lagrange equation \begin{align*} {0}\equiv\frac{\delta{W}[x]}{\delta{{x}^{\alpha}(t)}}=\frac{\partial\mathcal{L}(t)[x(t)]}{\partial{{x}^{\alpha}(t)}}-\frac{\mathrm{d}}{\mathrm{d}{t}}\bigg[\frac{\partial\mathcal{L}(t)[x(t)]}{\partial\mathrm{d}{{x}^{\alpha}(t)}/\mathrm{d}{t}}\bigg]=-{m}\frac{\partial\Phi(x(t))}{\partial{{x}^{\alpha}(t)}}-{m}\frac{\mathrm{d}^{2}{{x}^{\alpha}(t)}}{\mathrm{d}{t}^{2}}{\,}{.} \end{align*}

• Welcome to the site. We appreciate this detailed answer Commented Dec 21, 2022 at 17:18
• Aren't you missing a summation $\sum_{i=1}^{n}$ over the n variables? Commented Sep 13, 2023 at 9:53