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

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

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migrated from Nov 12 '12 at 16:16

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up vote 13 down vote accepted

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.

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

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