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  Derivative of a functional formula

+ 5 like - 0 dislike

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

asked May 14, 2014 in Mathematics by certik (25 points) [ no revision ]

1 Answer

+ 5 like - 0 dislike

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.

answered May 14, 2014 by Valter Moretti (2,085 points) [ revision history ]
edited May 14, 2014 by Valter Moretti

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.

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.

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.

To vote on answers, 50 rep points are needed as is for example explained in the chapter (11 Permissions) of the FAQ.

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