# Is there a chain rule for functional derivatives?

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

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

## migrated from physics.stackexchange.comNov 12 '12 at 16:16

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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.

• Sorry to accept it so late. Thank you for your effort! – Machine Dec 2 '12 at 7:16

Yes. There is a chain rule for functional derivatives. But it is not a direct generalization of the chain rule for functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to a field) cannot.

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

$$\lambda(s) = \Lambda[f;s], \quad s \in \mathbb{R},$$

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

A functional $$G$$ can only act on $$\Lambda$$ considered as a function of the parameter $$s$$. Therefore, $$G[\Lambda[f;s]]$$ is meant to be understood as $$G[\lambda]$$. Contrary to the claims of many physics textbooks, there is no such thing as the functional of a functional.

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

$$\frac{\delta F [f]}{\delta f (x)} = \int \mathrm{d} y \frac{\delta G [\lambda]}{\delta \lambda (y)} \frac{\delta \Lambda [f ; y]}{\delta f (x)} .$$

This is the generalization of the chain rule for functionals.

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