Talk:Functional derivative

‹See TfM›

> Another definition is in terms of a limit and the Dirac delta function, δ:

> ${\frac {\delta F[\phi (x)]}{\delta \phi (y)}}=\lim _{\varepsilon \to 0}{\frac {F[\phi (x)+\varepsilon \delta (x-y)]-F[\phi (y)]}{\varepsilon }}.$

But in general, F is a functional which is only defined over continuous/smooth functions. φ(x)+εδ(x-y) does not even count as a function. It's a distribution. Phys 22:03, 3 Dec 2004 (UTC)

Yes, I think that it should be noted that some physicians write the definition as above, but that it isn't a mathematically correct definition. Also, the correct definition (the first one) misses $f$ on the left side. I'll do these corrections. --Md2perpe 22:53, 30 July 2006 (UTC)

Could it be that due to the facts that

• physics problems most often find their mathematical statement with respect to a Hilbert space
• where on its dual space the Riesz representation theorem applies and
• due to the scalar product's symmetry
• the delta distribution is very often mistaken as the Dirac delta function

physicists (although this is mathematically not correct) calculate the gradient in the direction of the delta "function", yielding the physicist's version? If so, this could be pointed out more clearly. Greetings--Alexnullnullsieben (talk) 21:22, 5 November 2011 (UTC)

"where the arrow on the right handside, inside the functional F, indicates a function definition"--what arrow on the RHS? Steve Avery (talk) 02:18, 25 November 2007 (UTC)

Is this too close to Wolfram's description?

From wikipedia:
"the functional derivative is a generalization of the usual derivative that arises in the calculus of variations. In a functional derivative, instead of differentiating a function with respect to a variable, one differentiates a functional with respect to a function."

From MathWorld (http://mathworld.wolfram.com/FunctionalDerivative.html):
"The functional derivative is a generalization of the usual derivative that arises in the calculus of variations. In a functional derivative, instead of differentiating a function with respect to a variable, one differentiates a functional with respect to a function."

I don't know how different things need to be in order to avoid looking like someone was just copying their site. http://mathworld.wolfram.com/about/faq.html#copyright --anon

Thanks. I removed that text, just in case. Oleg Alexandrov (talk) 23:35, 23 October 2005 (UTC)

This text was added by User:Kumkee on 13 feb 2004. It was his first (non-anonymous) edit ever; he's only made some half-dozen non-user-space edits since. linas 23:39, 24 October 2005 (UTC)

Partial Integration?

In the first example in the Examples section, there seems to be a sign error when the "Partial integration of second term" happens. I also think that the text within the math is not good style. Since this is my first time editing a math-related article, I'm not too confident about either of these points. I've changed:

${\begin{matrix}\left\langle \delta F[\rho ],\phi \right\rangle &=&{\frac {d}{d\epsilon }}\left.\int f(\mathbf {r} ,\rho +\epsilon \phi ,\nabla \rho +\epsilon \nabla \phi )d^{3}r\right|_{\epsilon =0}\\&=&\int \left({\frac {\partial f}{\partial \rho }}\phi +{\frac {\partial f}{\partial \nabla \rho }}\cdot \nabla \phi \right)d^{3}r\\&=&[{\mbox{partial integration of second term}}]\\&=&\int \left({\frac {\partial f}{\partial \rho }}\phi +\nabla \cdot {\frac {\partial f}{\partial \nabla \rho }}\phi \right)d^{3}r\\&=&\left\langle {\frac {\partial f}{\partial \rho }}+\nabla \cdot {\frac {\partial f}{\partial \nabla \rho }},\phi \right\rangle \end{matrix}}$

to

${\begin{matrix}\left\langle \delta F[\rho ],\phi \right\rangle &=&{\frac {d}{d\epsilon }}\left.\int f(\mathbf {r} ,\rho +\epsilon \phi ,\nabla \rho +\epsilon \nabla \phi )\,d^{3}r\right|_{\epsilon =0}\\&=&\int \left({\frac {\partial f}{\partial \rho }}\phi +{\frac {\partial f}{\partial \nabla \rho }}\cdot \nabla \phi \right)d^{3}r\\&=&\int \left({\frac {\partial f}{\partial \rho }}\phi +\nabla \cdot \left[{\frac {\partial f}{\partial \nabla \rho }}\phi \right]-\left[\nabla \cdot {\frac {\partial f}{\partial \nabla \rho }}\right]\phi \right)d^{3}r\\&=&\int \left({\frac {\partial f}{\partial \rho }}\phi -\left[\nabla \cdot {\frac {\partial f}{\partial \nabla \rho }}\right]\phi \right)d^{3}r\\&=&\left\langle {\frac {\partial f}{\partial \rho }}-\nabla \cdot {\frac {\partial f}{\partial \nabla \rho }}\,,\phi \right\rangle \end{matrix}}$,

Eubene 22:13, 18 September 2006 (UTC)

You're correct; I had made an error. Good change! Md2perpe 20:58, 26 September 2006 (UTC)

Weizsäcker kinetic energy functional

Hi, when I follow the steps to the Weizsäcker derivative the first, eg. (1/8), term has a negative sign, as it comes from deriving 1/rho. Then, in the last equality, it is positive, and in accordance with all texts I know. Is it there a typo somewhere? Also, is it nabla^2 (eg. the sum of the second derivatives) or nabla.nabla ?

I've read the paper: T Gál, Á Nagy. A method to get an analytical expression for the non-interacting kinetic energy density functional. J. Mol. Struct. (Theochem) 1280, 167 - 171 (2000).

and I think I got confused. —Preceding unsigned comment added by 83.53.150.6 (talk) 04:00, 21 August 2008 (UTC)

Merge proposal

The following discussion is closed. Please do not modify it. Subsequent comments should be made in a new section. A summary of the conclusions reached follows.

The result was do not merge into Functional derivative. -- DarkCrow^Caw 19:14, 8 May 2012 (UTC)

---

I have proposed to merge the entry First variation into the Functional derivative entry, since they are the same concept: any standard text on the calculus of variation reports the definition of the first functional derivative, calling it "first variation". Also, the concept of functional derivative has its origins in the systematization of the calculus of variation in framework of functional analysis as done by Vito Volterra, i.e. as a concept in the theory of functionals, linear and non-linear. Daniele.tampieri (talk) 15:20, 4 July 2009 (UTC)

I don't actually disagree with you, but some authors do make a distinction between the two terms. In Gelfand and Formin's "Caluculus of Variations", The functional derivative (if I remember correctly) is essentially the left hand side of the Euler-Lagrange equations for the functional.--129.69.21.117 (talk) 09:48, 30 January 2012 (UTC)

As a practical matter, I think this should remain a separate topic. There are plenty of practical cases where the calculational machinery of the functional derivative can be applied without need for recourse to the First variation. — Preceding unsigned comment added by 68.232.116.79 (talk) 02:14, 11 February 2012 (UTC)

The discussion above is closed. Please do not modify it. Subsequent comments should be made on the appropriate discussion page. No further edits should be made to this discussion.

A different way to look at this

I think all the definitions that use delta functions are based on historical, but generally invalid, methods. In particular, since the delta function is not generally in the space of functions being varied, the limit in question doesn't even make sense.

In reality, two different things are happening. First, we're looking for a first approximation to the functional in question near a given function:

$\delta F[\phi ][\eta ]=F[\phi +\eta ]-F[\phi ]+O(\eta ^{2})$

Here, η is any "small" test function.

On the other hand, physicists like to define objects via integral kernels. Hence, we look for a function (or, more generally, a distribution) ${\frac {\delta F}{\delta \phi [x]}}$ such that:

$\delta F[\phi ][\eta ]=\int {\frac {\delta F}{\delta \phi [x]}}\eta (x)dx$

This gives the right results in all cases that I've ever checked, and is much less confusing. If it were up to me, I would write ${\frac {\delta F}{\delta \phi [x]}}$ as ${\frac {\delta F}{\delta \phi }}(x)$, which I consider much clearer.

Note that, generally, when a second variable is present, it's part of the definition of the functional and should be clearly shown as such, e.g., by making it a subscript on the functional, etc. For example, in the case given in the article, $\rho (r)=\int \rho (r')\delta (r-r')dr'$, what we really have is a functional $F_{r}$, where $F_{r}[\rho ]=\int \rho (r')\delta (r-r')dr'=\rho (r)$ for a fixed r - this is just the Dirac delta distribution, centered at r.

Then:

$\delta F_{r}[\rho ][\eta ]=F_{r}[\rho +\eta ]-F_{r}[\rho ]+O(\eta ^{2})=\int \eta (r')\delta (r-r')dr'=\int \eta (r'){\frac {\delta F_{r}}{\delta \rho [r']}}dr'$, so ${\frac {\delta F_{r}}{\delta \rho [r']}}=\delta (r-r')$, as expected.

As a matter of philosophy, one has to ask when Wikipedia should push a point of view and when it should just reflect what exists. I don't know a good answer to this. But I'd love people's thoughts on whether we couldn't incorporate this way of looking at functional derivatives into this article.

Rwilsker (talk) 19:36, 18 July 2010 (UTC)

Properties

Hi!

There are some examples - which is good. What is completely missing though is a section with "Properties" of the functional derivative.

I am thinking about the "product rule", the "chain rule" and things like that. (These rules, by the way, are quite easy to derive, especially if one "cheats" and use the physics treatment.)

I suggest a new section labelled "Properties" where properties of the functional derivative are listed. YohanN7 (talk) 17:36, 20 August 2012 (UTC)

Computational step

I added a computational step to the definition that would have saved me a lot of frustration and time had someone put it there for me. I didn't know this stuff before coming to this page, and I had to do a lot of digging to unravel what you're doing. Do we want people to learn, or are we going for brevity? I understand if you want to delete it (actually I don't), but you threw in the epsilon out of nowhere, and didn't clarify it further down the page. I know the traditional job of an epsilon, but coming at it blind, especially when you aren't familiar with functional derivative notation, most people who don't know it already won't automatically deduce that step, and so right off the bat you crush outsiders who are fluent in the language but not the mechanics.

173.25.54.191 (talk) 22:07, 16 February 2013 (UTC)

Re "but you threw in the epsilon out of nowhere, and didn't clarify it further down the page." — Thanks for pointing that out, although that was done according to the source. I just now added clarification of ε after the equations and deleted the right hand side that you added. Regarding ε, not clear why you didn't "clarify it further down the page" as you thought was needed, instead of making an edit that is just creating a right hand side of an equation by adding factors 1/ε and ε to the left hand side, which didn't seem useful. --Bob K31416 (talk) 15:42, 24 February 2013 (UTC)