Talk:Euler–Lagrange equation

message: 1. I think that section entitled "Statement" should be renamed
"Definition" because it is defining the Euler-Lagrange equation.
"Statement" seems kind of meaningless to me in this context.
2. I would restrict consideration in this section to Euclidean
spaces and ordinary time.  Tangent spaces and tangent bundles 
etc, should be presented later for those interested in these 
important but less commonly applied concepts.  The section
on generalization to manifolds is the obvious place for their
discussion.
3. There should be some reference to the Wiki article
"Action(physics)" and to the historic application of the
Euler-Lagrange equation for finding extrema of the action.
4. Lx and Lv should be Lq and Lq'.
x and v sound in conflict.  x is used in place of t, which
historically was the time variable.  v is ok in itself, but
in context I think it would be better to have it as q'.
Dot notation is not comkpatible with the section on Generalizations.
190.160.150.252 (talk) 02:37, 5 July 2018 (UTC)
thread_title: General Comments

message: 1.  The section entitled "Examples" has only one example,
suggesting the name "Example" would be more appropriate.
2.  The definition of ds as an integral is confusing.
Typically ds is a differential and appears to the right of the integral.
The result of the integration is the "Action".
3.  L and F are both used as names of the same thing.  
L should replace F.  
4.  L should be referred to as "the Lagrangian" as is customary,
not as the "integrand function".
5.  I think a less trivial example should be used -- not that I 
would want to go to war over it.  If you add an example,
retain the name "Examples" for this section.
190.160.150.252 (talk) 02:37, 5 July 2018 (UTC)
thread_title: Obvious oversights

message: The site referenced at the end
http://www.exampleproblems.com/
is not functional. It would be better to remove it.  —Preceding unsigned comment added by 87.120.6.146 (talk) 08:37, 8 February 2008 (UTC)
thread_title: Link does not work

message: Is anybody able to give me some hints about the connection between Euler-Lagrange Methods and Lagrange multipliers? At first glance they seem to be closely related. Maybe somebody can clarify this and maybe add a short note to the articles. Cyc 12:37, 22 September 2006 (UTC)
thread_title: Euler-Lagrange Methods & [[Lagrange multipliers]]

message: My concern is with this section:
\partial_\mu \left( \frac{\partial \mathcal{L}}{\partial ( \partial_\mu \psi )} \right) - ... 
    where ...
    \partial\, is a vector of derivatives:
        \partial_\mu = \left(\frac{1}{c} \frac{\partial}{\partial t}, \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right). \,
Seems like it could be interpreted as a set of N equations (N being the dimension of the "vector"), rather that the sum over mu (do we always assume summation over repeated indices ?  Apologies if this is nit-picking.
Many thanks
—Preceding unsigned comment added by JM516 (talk • contribs) 23:03, 4 October 2007 (UTC)
replies: text: In quantum mechanics, the sum is implicit due to Einstein summation notation which is taken as read. — ras52 (talk) 14:41, 5 December 2007 (UTC)
thread_title:  Does summation over repeated indices need to be made explicit ? 

message: I think for any given general points A and B , the shortest length should collapse to a point or a set of points, hence the result should be of the form (x-x1)^2 + (y-y1)^2=0 , or a product of the same (the locus of the combined individual loci of 2 points)
i.e the locus of points, The function need not be an explicit function, it can be implicit and it is still real valued.
Note that if one drops the assumption made in considering a small "functional" perturbation this is indeed the result if one tries to solve the fact that:
 d/dy (intergal (1+f`(x)^2)^1/2 . dx) =0
so (y-y1)=i.(x-x1) , f`(x)=i is a solution
The fact that you assume dy/dx is the distance to arrive at the formal sqrt(1+f`(x)^2) as the formula for curve length itself implies you are trying to assume the minimal distance is a line (dx^2+dy^2=ds^2). The logic seems circular.
Hence I am not sure why one puts in that perturbation. -Alok 08:35, 24 May 2011 (UTC)
thread_title:  Locus of shortest curve  

message: This article frequently states that a function must be differentiable.
Last I heard, a function is NOT "differentiable"; it is "differentiable with respect to <a specific variable (or function)>"  There are many examples here where the authors seem to think that the "with respect to X" can be neglected.  It can not, unless the wrt can only be understood one way.  L(q(t)) is "differentiable wrt WHAT? x? t? q(t)? something else?  This needs to be fixed, imho.  The instances of this sloppiness are too numerous to list, it is pervasive and detracts significantly from the utility of the article (imho)Abitslow (talk) 06:25, 20 December 2013 (UTC)
replies: text: The differentiability of a properly defined function is a well-defined notion because a function is only properly defined when at least its domain and range space are given. Functions of more than one variable are usually called differentiable if all the partial derivatives are continuous and, hence, for well-defined functions there is no ambiguity in the notion of differentiability. I agree with Abitslow, that this need not be a sufficiently precise classification for all purposes. However, in the given example, L(q(t)) defines the value of the composition of the function L with the function q. This composed function is unambiguously defined when the functions L and q are (and in such a way that the domain of L is in the range space of q, of course). The differentiability of "L after q" can be inferred from the differentiability of q and L. In this example, there really is no ambiguity at all because "L after q" is a function of just one variable. Bas Michielsen (talk) 12:47, 20 December 2013 (UTC)
thread_title:  differentiable 

message: Firstly in "Several functions of several variables" partial derivatives were replaced with ordinary derivatives. But that makes not much sense, since the functions are functions of several variables. Total derivatives don't come into question. Then the partial derivative of the functional would be a function of the single variable x_i and x_i would be a function of several variables. But that's not the case here.
Secondly in "Single function of two variables with higher derivatives" and the following section summation is taken over \mu_1,...,\mu_i. Then due to the theorem of Schwarz there would be
:2\frac{\partial^2}{\partial x_1\partial x_2}.
So I think summation should be taken over all combinations with repetition, so that
\mu_1\le...\le\mu_i. If this is the case, the sum should be noted with
:\sum_{i=1}^n (-1)^i \sum_{\mu_1\le...\le\mu_i}\ldots.
— Preceding unsigned comment added by 188.103.118.233 (contribs)  
:I disagree with the partial derivatives, and the current wording of this page confuses me every time I use it as reference (that is, about three times a year). Taking the full derivative is important in most cases. Consider this Lagrangian for one function of several variables:
::\mathcal{L} = \cfrac 1 2 \sum{f_{x_i}^2} \Rightarrow \cfrac{\partial \mathcal{L}}{\partial f_{x_i}} = f_{x_i}.
:That is, derivative of the Lagrangian wrt. image gradient is the gradient itself. Then, the partial derivative:
::\cfrac{\partial}{\partial x_i} \cfrac{\partial \mathcal{L}}{\partial f_{x_i}} = \cfrac{\partial}{\partial x_i} f_{x_i} = 0
:is zero simply because f_{x_i} and x_i are different variables. In contrast, taking the full derivative leads to the second derivative of f:
::\cfrac{\mathrm{d}}{\mathrm{d}x} \cfrac{\partial \mathcal{L}}{\partial f_{x_i}} = \cfrac{\mathrm{d}}{\mathrm{d}x} f_{x_i} = f_{x_i, x_i} also known as \cfrac{\partial^2 f}{{\partial x_i}^2}.
:In my example, this ultimately leads to the equation \cfrac{\partial \mathcal{L}}{\partial f} = \Delta f, which is the correct result (the heat equation).
:I am convinced that all the sections concerning several variables need to be fixed, and I will do so in three months.
:— Emu cz (talk) 10:26, 8 April 2016 (UTC)
replies: text:  The original version that I added several years ago was from the book by Courant and Hilbert, two of the greatest mathematicians the world has known.  My suggestion would be to track that source down and compare the current equations with those in that text.  The description in the Courant-Hilbert book is quite detailed and I would trust their results over almost any other source (modulo typesetting errors).  Bbanerje (talk) 22:04, 9 April 2016 (UTC)
replies: text:  The current notation conforms to the one by Hilbert and Courant (MMP vol. I, p. 192, Eq. 25). It feels awkward to me, but arguing with these two gentlemen is beyond the scope of this talk page ;) Anyway, I will not edit this article unless I find a super-reliable proof. — Emu cz (talk) 15:10, 11 April 2016 (UTC)
thread_title:  Notation Errors 

message: The subsection Classical mechanics (see this version at the time of this post) has a very long-winded way of explaining how to use the EL equations to solve problems in mechanics. This is the subject of the Lagrangian mechanics article, which gives the same example more compactly and in both Cartesian and spherical coordinates. The ordering of derivatives is clear also, without textbooky over-detailed explanations "evaluate this" then "treat this as" etc. I am going to deleted this subsection. 
This article is more mathematical and general, giving various extensions (e.g. higher derivatives, more variables). It would cut pointless duplication if mathematical examples are contained in this article, physics examples elsewhere. M∧Ŝc2ħεИτlk 17:29, 11 January 2016 (UTC)
thread_title:  Make the scope of this article mathematics 

message: There's no reference on the Manifold generalization part.
Everton.constantino (talk) 04:53, 16 January 2016 (UTC)
thread_title:  Manifold generalization 

message: Was Euler truly "Swiss-Russian"? Yes, so he spent much of his adult life in St. Petersburg, but I'm not sure (and I've never seen anything written down anywhere) that he actually took on dual nationality. (Admittedly the rules on what nationality you are may have changed since those days, and it may have been that all you needed to do to be declared "Russian" was to live there for a period of time -- I haven't a clue). But, as I say, I have never anywhere else seen a suggestion that he actually was "Swiss-Russian" -- only that he was Swiss. I'm asking the question because I don't actually know one way or the other, I'm asking in case anyone does know. --Matt Westwood 06:51, 6 May 2016 (UTC)
replies: text: I don't know much about him (aside from this long List of things named after Leonhard Euler). This is probably a question for talk:Leonhard Euler anyway. M∧Ŝc2ħεИτlk 08:56, 6 May 2016 (UTC)
thread_title:  Nationality of Euler 

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thread_title:  External links modified 

message: Hi all,
The article says function q is differentiable, but the domain of q is not open, hence q is not differentiable. Could anyone fix that please.
Thanks, Konstantin Pavlovskii (talk) 12:27, 12 May 2018 (UTC)
thread_title:  Differentiability of q(t) 

message: I am not complaining, but others seem to say "the proof isn't right".  the proof is somewhat long (it varies depending on who's is using what method and the "realm of influence" proven).
for me the most rapid and SIMPLE proof is as by farlow's PDE book and my solving (farlow asks the student to complete steps):
'''
    let Gamma[x] represent minimized y[x] of J[y[x]] == Integral[b,a] F dx
    (y is any y substituted, y is not a specific function)
    y[a]==A, y[b]==B (Gamma[x] found must comply)
    where F == F[x,y[x],y'[x]]
    let Xi be a value which is small and approaches zero
    let Eta[x] be a smooth curve (any function)
For minimization we must have:
    J[Gamma[x]] <= J[Gamma[x] + Xi Eta[x]]
The above is a calculus of variations premise.
    let Psi[Xi] == J[Gamma[x] + Xi Eta[x]]
You can visualize a graph of J[g+xe] now as a parabola, Psi' it's slope, which has a minimal value where Psi'[Xi]==0, when Psi'[Xi] is not 0 the we have the J[g+xe] that isn't minimized.  Furthermore note Psi is critical when Xi==0 (critical points are part of calculus derivative function minimization).  This gamma, if we find it, is the smallest of all y[x] in interval [a,b].
    (Psi'[Xi])[0] == 0 == [Eta][x] D[J[Gamma[x]],Xi]
    (the right we discard, so ignore; it is D[J[g+xe],xi] then xi->0)
This is the step that is more rapid than texts I've seen.  Expressing F in it's full form is key.  (this is just F with our variation substituted in)
    Integral[b,a]  F[x, Gamma[x] + Xi Eta[x], Gamma'[x] + Xi Eta'[x]]  dx
    Integral[b,a]  (@/Xi) F[x, Gamma[x] + Xi Eta[x], Gamma'[x] + Xi Eta'[x]]  dx  Xi->0
Just work in parallel: do to F what we did with Psi (to set them equal, the integral must undergo the same steps).  Take one derivative of F with respect to Xi (perform @/Xi).  Then substitute 0 for Xi after that.  The result is the same as seen at the end of proofs.  DONE, our F now matches the theorem's very simply.  (use of chain rule for partial derivative of a function with arguments is used, simple but is a pre-requisite).  Mathematica can do the step easily.  Many say the next step is "to use integration by parts to move to a simpler form".  I would say i.p. would be used in solving the integral - which is a solution we do not want.  We wish only an expression equating (F,Gamma,x).
    Integral[b,a] Eta'[x] F^{0,0,1} + Eta[x] F^{0,1,0} dx == 0 == @J[Gamma[x] + Xi Eta[x]]
    Eta'[x] F^{0,0,1} + Eta[x] F^{0,1,0}
I next would claim by ODE, "N[x] F dx + M[x] F dy == 0", we can move around coefficients by simple algebra principles and that n[x] qualifies.
    ((d/dx) Eta[x]) F^{0,0,1} + Eta[x] F^{0,1,0} ==
     Eta[x] (d/dx) F^{0,0,1} + Eta[x] F^{0,1,0} ==
    ( (d/dx) F^{0,0,1} + F^{0,1,0} ) Eta[x]
We assumed Eta[x] is a smooth differentiable curve, it is not zero.  For the integral to be zero:
    (d/dx) F^{0,0,1} == F^{0,1,0}
    -(d/dx) F^{0,0,1} + F^{0,1,0} == 0
In solving J[y] using the above it will become clear Gamma[x]==y[x], that is:  while solving problem use J[y] and F[x,y,y'] not J[Gamma[x] + Xi Eta[x]].
We could now write out Psi'[x] == 0 == Integral[ result ] but that is discarded too; we only sought an expression relating (F,Gamma[x],x).  (most) proofs use i.p. to do the last step (not solved as ODE) and it is "tricky" to find the (F,Gamma,x) form that way but well documented.
— Preceding unsigned comment added by 2601:143:480:A4C0:88F:739C:2A43:F405 (talk) 23:14, 26 January 2020 (UTC)
thread_title:  the proof is simple but not shown in article 

message: Proof box does not appear to be expandable.  — Preceding unsigned comment added by 173.206.33.141 (talk) 22:31, 29 August 2020 (UTC)
thread_title:  Proof box does not appear to be expandable 

message: Since there is a tangent space on X defined I assume that X must be a differential manifold. If so I think that should be mentioned.
Jyyb (talk) 17:31, 18 March 2021 (UTC)
thread_title:  Characteristic of the set X in section "Statement" 

message: The symbol \mathcal{L} is introduced in the section Euler–Lagrange_equation#Generalizations without definition -though one can deduce it is a differentiable enough function of several variables-, and then used also in Euler–Lagrange equation#Generalization to manifolds with a different definition which is here clearly, it is here the Lie derivative operator on differential forms. It would be good to define this symbol at its first use, and warn in its second appearance that it is redefined, or overloaded. Plm203 (talk) 10:33, 12 November 2023 (UTC)
thread_title:  The notation <math>\mathcal{L}</math> needs clarification