Given a functional $S=S\{Y[X(r)]\}$, is the following "chain rule" valid?

$$\frac{\delta S\{Y[X]\}}{\delta X(r)}=\frac{\partial Y(r)}{\partial X(r)}\frac{\delta S[Y]}{\delta Y(r)}$$


2 Answers 2


The chain rule for functional differentiation is just the continuum generalisation of the usual chain rule for differentiation of a function of many variables $f(y_1,y_2,\ldots,y_N) = f(\mathbf{y})$, which reads $$ \frac{\partial f(\mathbf{y})}{\partial x_i(\mathbf{y})} = \sum\limits_{j=1}^N\frac{\partial y_j}{\partial x_i}\frac{\partial f}{\partial y_j}. $$ The continuum limit amounts to sending the number of variables $N\to\infty$, and defining a new continuous index $r$ such that $j\to rN$. Then you just change your notation to agree with the continuous nature of the new index $r$, e.g. $x_j \to X(r)$, $f(\mathbf{x}) = f(\{x_j\}) \to F[X(r)]$ etc. In our example, this means substituting the above sum over the discrete index $j$ by an integral over a continuous index, finding the chain rule for functional differentiation: $$\frac{\delta F[Y]}{\delta X(r)} = \int \mathrm{d}s\,\frac{\delta Y(s)}{\delta X(r)}\frac{\delta F[Y]}{\delta Y(s)}.$$ You can get any functional calculus identity you want along the same lines. Just think about what happens in ordinary multivariate calculus with a finite number of variables. Then take the number of variables to infinity.

  • $\begingroup$ Sorry to accept it so late. Thank you for your effort! $\endgroup$
    – Machine
    Dec 2, 2012 at 7:16
  • $\begingroup$ It is very difficult to find approachable resources on functional calculus beyond the basics, so this answer was incredibly helpful! There is one thing I still can't wrap my head around, though: what is the domain of integration of this integral? Is it from minus infinity to infinity? $\endgroup$
    – Godzilla
    Jun 8, 2020 at 17:03
  • $\begingroup$ it would be the domain of the functional. Ex: if the functional was $\int_{0}^{1} (f+f')$ then this domain of integration would be from $0$ to $1$. Note most functionals, that is functions which take functions as inputs and produce as output complex numbers, Are representable as an integral of a (function of functions) over some complex domain. In the case above the functional was integral 0 to 1 (f + f'). the Function of functions was just f -> f+f', and the domain was the interval from 0 to 1. $\endgroup$ Jan 3, 2021 at 4:39
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    $\begingroup$ what does $\frac{\delta Y(s)}{\delta X(r)}$ mean? $\endgroup$
    – wlad
    Jan 6, 2021 at 14:46
  • $\begingroup$ An example would help: what is $\frac{\delta F[Y[f](r)]}{\delta f(r)}$ where $Y[f](r) = \int_0^r X(t)\, dt$? $\endgroup$
    – wlad
    Jan 6, 2021 at 15:45

Yes. There is a chain rule for functional derivatives. As shown below, it takes a similar form to the chain rule for functions and in order to obtain it one has to be careful with the definitions. In the following the term functional is used as in the calculus of variations.

Consider the scalar function $f: (y_1,...,y_n) \mapsto g(x_1(\mathbf{y}),...,x_n(\mathbf{y}))$, which is the composition of $g$ with the coordinate functions $x_i(\mathbf{y})$.

The partial derivative of $f$ with respect to $y_j$ at the point $\mathbf{y}$ is given by the chain rule

$$ \frac{\partial f (\mathbf{y}) }{\partial y_j}= \sum_{i = 1}^N \frac{\partial g(\mathbf{x}(\mathbf{y}))}{\partial x_i} \frac{\partial x_i(\mathbf{y})}{\partial y_j}. $$

Now, the continuum generalization of the coordinate functions $x_i$ is the one-parameter family of functionals

$$ \lambda(s) = \Lambda_s[\psi], \quad s \in \mathbb{R}, $$

where $\Lambda_s[\psi]$ means that $\Lambda$ is a functional of $\psi$ and a function of $s$.

A functional $G$ can only act on $\Lambda$ considered as a function of the parameter $s$. Therefore, $G[\Lambda_s[\psi]]$ is meant to be understood as $G[\lambda]$.

Since a variation of $\psi$ will change the value of $\lambda$, ultimately it results in a variation of $G$. Therefore, $G[\Lambda_s[\psi]]=G[\lambda]$ can be considered as a functional $F[\psi]$. The partial functional derivative of $F$ with respect to the local change of $\psi$ at $t$ is given by [1] Appendix A:

$$ \frac{\delta F [\psi]}{\delta \psi (t)} = \int \mathrm{d} s \frac{\delta G [\lambda]}{\delta \lambda (s)} \frac{\delta \Lambda_s[\psi]}{\delta \psi (t)} . $$

This is the generalization of the chain rule for functionals.

Note: Some physics textbooks use the term "functional of a functional" for $G[\Lambda]$, but two mappings from a function to a scalar cannot be composed. As shown above, being careful with the definition of the mappings is essential to understanding the derivation of the chain rule for functional derivatives.

[1] E. Engel, R.M. Dreizler, Density Functional Theory: An Advanced Course, Springer Science & Business Media, 2011.

  • $\begingroup$ ... No, even by your needlessly restrictive definition, you can still have a functional of a functional. A functional by nature operates on the structure of its inputs, not just its final result (which is usually achieved by using an integral over the domain of the function rather than simply evaluating the function at a value.) A functional of a functional just needs to be able to operate on something that itself operates on functions, say by using a functional integral and integrating over all possible function inputs. $\endgroup$ Oct 13 at 4:00
  • $\begingroup$ There are different definitions of functional (see Wikipedia and How to interpret the functional of a functional?). I am using the definition of functional as an element of the dual space of a vector space, whereas you define functional as a higher-order function. In any case, functionals are mappings and mappings can only be composed when their domains and codomains are compatible. Integrals are an implementation detail. $\endgroup$
    – Marduk
    Oct 15 at 16:39
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    $\begingroup$ @linkhyrule5 I updated the answer avoiding statements that can be misleading. Thank you for your feedback. $\endgroup$
    – Marduk
    Oct 16 at 19:27
  • $\begingroup$ Fair enough, that works. $\endgroup$ Oct 17 at 3:45

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