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In multivariable calculus we learn that partial derivatives are commutative: i.e. $\partial _{xy}F=\partial_{yx}F$

When dealing with multiple functional derivatives this doesn't seem to be the case.

Consider the simple functional equation: $\mathcal{I}[f]=\int f(x)dx$

It would seem that if we compute the functional derivative: $\large\frac{\delta^2\mathcal{I}[f^3]}{\delta f(x_0)\delta f(x_1)}$ the same rules of commutivity that apply in multivariable calculus are not present. Depending on the order we take the two derivatives, we would either get $6[f(x_0)]$ or $ 6[f(x_1)]$.

My question is obvious: which is the proper answer? It would seem to me that the derivative with respect to $f(x_1)$ would be computed last and thus the ladder of the two would be correct.

Seenathin
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  • I'm not clear on the notation, but failure of "shape" derivatives to commute is more a consequence of dependence (lack of independence) between the quantities being varied. Partial derivatives commute only when the variable not being differentiated with respect to is held constant. So I'd be curious if this condition of independence is met by your second order "functional derivative". – hardmath Dec 19 '14 at 00:15
  • The definition of the functional derivative that my book has is $\frac{\delta H[f(y)]}{\delta f(x)}=\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon}\huge( \normalsize H[f(y)+\epsilon \delta(y-x)]-H[f(y)] \huge)$ from this you can see how I got the two different answers that I posted. Is the term "functional derivative" that my book is using an incorrect term? – Seenathin Dec 19 '14 at 05:37

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