Derivative of a functional formula

Question

I would like the derive the following equation (eq. (25) in [1]):

${d F[\psi(\theta)] \over d \theta} = \int {\delta F[\psi] \over \delta \psi(x; \theta)} {\partial \psi(x; \theta) \over \partial \theta} d x$

For example $\psi(x; \theta) = \varphi(x) \cos(\theta) + \phi(x) \sin(\theta)$. The functional $F[\psi]$is for example $F[\psi] = \int \left({d \psi(x)\over d x}\right)^2 + \psi(x) V(x) d x$

Besides the fact that it works, and that it is used in [1], I found at least the following, hand-waving, explanation:

${d \over d \theta} F(\psi_1, \psi_2, \psi_3, ...) = \sum_i {\partial F \over \partial \psi_i} {d \psi_i \over d \theta}$

Which is a discrete form of the above formula. But I would like to derive it using the definition of the functional derivative, or some other more rigorous way.

[1] Jiang, H., & Yang, W. (2004). Conjugate-gradient optimization method for orbital-free density functional calculations. The Journal of Chemical Physics, 121(5), 2030–2036. doi:10.1063/1.1768163

Answer 1

I will sketch just the idea omitting several (actually important) mathematical details.

If $F=F[\psi]$ is a functional, we define, if exists, its functional derivative $\frac{\delta F}{\delta \psi}$ as the function (a distribution, more generally) such that

$$\left.\frac{dF[\psi + h \phi]}{dh}\right|_{h=0}= \int \frac{\delta F}{\delta \psi}(x) \phi(x) dx$$

for every test function $\phi$. Now suppose that $\psi$ parametrically depends on the parameter $\theta$. We have

$$\frac{dF[\psi_\theta]}{d\theta} =\lim_{h\to 0} \frac{1}{h}\left(F[\psi_{\theta + h}] - F[\psi_\theta]\right)$$

$$=\lim_{h\to 0} \frac{1}{h}\left(F[\psi_\theta + h \partial_\theta \psi + O_\theta(h^2)] - F[\psi_\theta]\right)$$

$$=  \lim_{h\to 0} \frac{1}{h}\left(F[\psi_\theta + h \partial_\theta \psi]  - F[\psi_\theta]\right) \:.$$

$$= \left.\frac{dF[\psi_\theta + h \partial_\theta \psi_\theta]}{dh}\right|_{h=0}\:.$$

Applying the definition of functional derivative given above, we have that

$$\frac{dF[\psi_\theta]}{d\theta} = \left.\frac{dF[\psi_\theta + h \partial_\theta \psi_\theta]}{dh}\right|_{h=0}= \int \frac{\delta F}{\delta \psi_\theta}(x) \partial_\theta\psi_\theta (x) dx\:.$$

There are several open mathematical issues in the outlined procedure (for instance dropping $O_\theta(h^2)$ in the formula above is not so easy as it could seem at first glance). Nevertheless, everything goes right (it can be proved by direct inspection) when dealing with functionals $F[\psi]$ of integral form  like the one you consider and assuming to work with a domain of suitably smooth and rapidly vanishing at infinity functions.

Comments

I think this is exactly the derivation that I was looking for! Thanks a lot. I will go over it slowly later today and if all works out, accept this as the correct answer.

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I went over this and this derivation is correct. You can actually do it even simpler like this:

${d F[\psi(\theta)] \over d \theta} =\left.{d F[\psi(\theta+\epsilon)] \over d \epsilon}\right|_{\epsilon=0} = \left.{d F[\psi(\theta)+\epsilon {d \psi(\theta) \over d \theta} + O(\epsilon^2)] \over d \epsilon}\right|_{\epsilon=0} = \left.{d F[\psi(\theta)+\epsilon {d \psi(\theta) \over d \theta}] \over d \epsilon}\right|_{\epsilon=0} = \int {\delta F[\psi] \over \delta \psi} {d \psi(\theta) \over d \theta} d x$

where we used the definition of a variation with $\delta \psi = {d \psi(\theta) \over d \theta}$:

$\delta F = \left. {d \over d\epsilon} F[\psi+\epsilon \delta\psi] \right|_{\epsilon=0} = \int {\delta F \over \delta \psi} \delta\psi d x$

I guess one cannot accept answers here at physicsoverflow. Also I was not allowed to upvote your answer, sorry about that.

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Hi @certik,

yes, we have disabled the feature of accepting answers known from SE, because in previous discussions most people thought that it is not really needed.

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