Talk:Calculus of variations

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Needs more prose to accompany equations Tompw 17:54, 7 October 2006 (UTC)

1 Undefined symbol, references
2 Vote for new external link
3 Opening Paragraphs
4 notation bug in section on Fermat's principle in three dimensions
5 Merge article with Variational Principle page
6 action principle
7 Comparison with functional analysis
8 Connection with the wave equation
9 Linguistic ambiguity
10 Clarity of Euler Lagrange Section

[edit] Undefined symbol, references

The symbol L appears halfway through the discussion and is not defined. Is it the same as A, or something different?

Can someone give some books on Calculus of Variations?

"Calculus of Variations" by I M Gelfund and S V Fomin is a good book on this subject. Wilmot1 12:55, 18 February 2007 (UTC)

L is a functional, and A is the integral of L. Initially, the article defines A as the integral of a specific functional, but this is just an example. It switches to L to indicate that the same procedure works for other functionals.

The letters L and A were probably used because they correspond to the Lagrangian and the Action in Lagrangian Mechanics.

A and L are not the letter normally used in the literature S and F are more usual. However some authors use other letter. Wilmot1 12:55, 18 February 2007 (UTC)

[edit] Vote for new external link

Here is my site with calculus of variations example problems. Someone please put this link in the external links section if you think it's helpful and relevant. Tbsmith

http://www.exampleproblems.com/wiki/index.php/Calculus_of_Variations

If you are looking to extend the subject beyond its applications and look closely at the mathematical formalism try

Introduction to the Calculus of Variations by Bernard Dacorogna Imperial College Press 2004 ISBN: 186094499X

I found your examples clearly presented and useful.Dogchaser 09:02, 20 February 2007 (UTC)

[edit] Opening Paragraphs

The second paragraph says "The preceding examples have all involved unknown functions of a single variable, which may be identified with a time variable." I don't believe this is true. For example, in mechanics and optimal control, the functions can (and generally do) depend on space and time. (Cj67 17:44, 26 June 2006 (UTC))

While we're at it, "the well-known fact that electricity takes the path of least resistance" isn't true at all (see voltage division, for instance). I'm removing it. Vonspringer 04:43, 15 August 2007 (UTC)

[edit] notation bug in section on Fermat's principle in three dimensions

Is the x in $P = \frac{n(x) \dot X}{\sqrt{\dot X \cdot \dot X} }.\,$ meant to be X? -- njh 10:02, 11 July 2006 (UTC)

Shouldn't $\int_{x_1}^{x_2} \frac{ \frac{df_0}{dx} \frac{df_1}{dx} } {\sqrt{1 + \left(\frac{df_0}{dx}\right)^2}} =0, \,$ be $\int_{x_1}^{x_2} \frac{ \frac{df_0}{dx} \frac{df_1}{dx} } {\sqrt{1 + \left(\frac{df_0}{dx}\right)^2}}dx =0, \,$

[edit] Merge article with Variational Principle page

It seems like there is redundant content between this page and the variational principle page. I think they should be merged under the title of "calculus of variations", and have the article "Variational Principle" redirected to "Calculus of Variations." On a different note, does anyone know enough to write about about computational methods in the calculus of variations? I don't, but I think it's important. --69.180.18.247 14:48, 4 September 2006 (UTC)

[edit] action principle

The text says that Hamilton defines integral of T - V as the action. Pardon my nitpicking, but that seems not to be true. Charles Fox: Introduction to the Calculus of Variations (1963 printing, reprinted by Dover) says that the action is the integral of T.

Also if you look at the Feynman Lectures on Physics Volume II, chapter on The Principle of Least Action, he remarks that he (Feynman) calls the integral of T - V the action, but actually pedants call it Hamilton's first principle function. Historically something less convenient was first named the action. But Feynman hates to give a lecture on the principle of least Hamilton's-first-principle-function. Also, more and more people are calling integral T - V the action, and if you join them, soon EVERYBODY will be calling the more useful thing the action.

Point is, not Hamilton's definition, but common mid-20th century usage.

[edit] Comparison with functional analysis

I saw calculus of variations first, with its cryptic " $δ f$ " notation, then I was introduced to functional analysis, which seems to solve the same problems. Is the $δ f$ notation outdated? Are the two fields different? Was "calculus of variations" a precursor to modern functional analysis? —Ben FrantzDale 03:07, 3 May 2007 (UTC)

[edit] Connection with the wave equation

the pde for an inhomogeneous medium is

rho u_tt=-div(A grad u),

where rho and A are material parameters. Only if A is constant can this pde be written in the form given in the article. Cj67 18:54, 14 June 2007 (UTC)

[edit] Linguistic ambiguity

Does "any function with at least one derivative that vanishes at the endpoints" refer to a vanishing derivative or a vanishin function ("vanish" = "apporach zero"?)? 82.181.95.21 10:00, 16 June 2007 (UTC)

the function should be zero at the endpoints (technically, its extension should be zero, which is equivalent to the limit being zero as you approach the endpoints).Cj67 14:26, 21 June 2007 (UTC)

I find the wording "vanish" very ambiguous also. It took be a while of googling to understand that it means "being or approaching zero". JimQ (talk) 12:45, 8 May 2008 (UTC)

[edit] Clarity of Euler Lagrange Section

In the section titled "Euler Lagrange", I feel the following step is unjustified:

for any number ε close to 0. Therefore, the derivative of A[f0 + εf1] with respect to ε (the first variation of A) must vanish at ε=0. Thus
$\int_{x_1}^{x_2} \frac{ f_0'(x) f_1'(x) } {\sqrt{1 + [ f_0'(x) ]^2}}dx =0, \,$
for any choice of the function f1. We may interpret this condition as the vanishing of all directional derivatives of A[f0] in the space of differentiable functions. If we assume that f0 has two continuous derivatives (or if we consider weak derivatives), then it follows from integration by parts that,
$\int_{x_1}^{x_2} f_1(x) \frac{d}{dx}\left[ \frac{ f_0'(x) } {\sqrt{1 + [ f_0'(x) ]^2}} \right] \, dx =0,$

How does this follow from integration by parts?

I feel this step requires further justification. —Preceding unsigned comment added by Danielkwalsh (talk • contribs) 01:51, 19 September 2007 (UTC)

The intermediate step:
$\left[\frac{ f_1(x) f_0'(x) } {\sqrt{1 + [ f_0'(x) ]^2}}\right]_{x_1}^{x_2} - \int_{x_1}^{x_2} f_1(x) \frac{d}{dx}\left[ \frac{ f_0'(x) } {\sqrt{1 + [ f_0'(x) ]^2}}\right] \, dx =0$
Since f1 is a differentiable function that vanishes at the endpoints, the first term in the left hand side above vanishes. In a related manner, the qualification that is given, i.e. "for any choice of f1 with two derivatives that vanishes at the endpoints of the interval" is somewhat ambiguous, since it could be mistakenly taken as meaning that the derivatives vanish, rather than the function itself. A better wording would be: "for any choice of a twice-differentiable f1 that vanishes at the endpoints of the interval". —Preceding unsigned comment added by 128.83.162.226 (talk) 16:32, 20 September 2007 (UTC)

Thanks. I did some research and came to the same conclusion. I went ahead and added the intermediate step, and changed the ambiguous language. The vanishing term that is evaluated from $x 1$ to $x 2$ is an important step; it should be described. Danielkwalsh 20:13, 21 September 2007 (UTC)