Functional derivative where the functional is not an integral

Question

Let $\phi(f)=F(f(x_1),f'(x_2),...,f^{(n)}(x_{n+1}))$ be a functional defined on $C^n([a,b])$: here $F$ is a given (say) smooth function and $x_1,...,x_{n+1}$ are given points in $[a,b]$. What is the functional derivative of $\phi$? I'm asking this question since usual texts about functional derivatives deals with functionals given as an integral.

Answer 1

The best way to calculate the functional derivative of a given functional is to use its basic definition, so let's recall it in the form introduced (under some restrictive conditions) by Vito Volterra (for the references see this Q&A, where the problem of definition of functional derivatives is extensively examined) $$ \delta\phi(f,\eta)=\lim_{\varepsilon \to 0}\frac{\phi[f+\varepsilon \eta]-\phi[f]}{\varepsilon} = \bigg{[}\frac{\mathrm{d}}{\mathrm{d}\varepsilon}\phi[f+\varepsilon \eta]\bigg{]}_{\varepsilon = 0} \label{1}\tag{1} $$ where $\eta$ belongs to the same function space as $f$.
By applying definition \eqref{1} to the functional $\phi:C^n([a,b])\to\Bbb R$ defined as $$ f\mapsto F\big(f(x_1),f^\prime(x_2),\ldots,f^{(n)}(x_{n+1})\big). $$ after calling the general arguments of the (sufficiently) smooth function $F:\Bbb R^{n+1}\to \Bbb R$ as $(y_0,\ldots,y_n)$ in order to avoid any confusion that may arise, we get $$ \begin{split} \delta\phi(f,\eta) & = \lim_{\varepsilon \to 0}\frac{\phi[f+\varepsilon \eta]-\phi[f]}{\varepsilon}\\ & = \lim_{\varepsilon \to 0}\frac{F\big(f(x_1)+\varepsilon \eta(x_1),\ldots,f^{(n)}(x_{n+1})+\varepsilon \eta^{(n)}(x_{n+1})\big)-F\big(f(x_1),,\ldots,f^{(n)}(x_{n+1})\big)}{\varepsilon}\\ & = \sum_{i=1}^{n+1}\frac{\partial}{\partial y_i}F\big(f(x_1),f^\prime(x_2),\ldots,f^{(n)}(x_{n+1})\big)\eta^{(i-1)}(x_i) \end{split}\label{2}\tag{2} $$

Notes

• In order for $\phi$ to have a functional derivative, it is necessary and sufficient that at least $F\in C^1(\Bbb R^{n+1})$
• Like the original functional $\phi$, also $\delta\phi$ doesn't have an integral representation.
• Edit: as correctly noted by truebaran, the functional derivative of $\phi$ is linear in this case, therefore it is at least a Gâteaux derivative. If the functional derivative is both linear and continuous, then it is a Frechet derivative.
• As said earlier, in this Q&A several issues related to currently found presentations of functional derivatives are analyzed.