Is there a chain rule for functional derivatives?

Question

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

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

Answer 1

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.

Answer 2

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.