
#1
May1913, 12:13 AM

P: 273

I'm reading Classical Mechanics (Taylor), and the 6th chapter is a basic introduction to calculus of variations. I'm super confused
I've tried to go to other sources for an explanation, but they just make it even worse! So, let me see if I can get some help here. [tex]\int^{x_{2}}_{x_{1}} f(y(x), y'(x), x)dx[/tex]  the integration of a function of three variables. y(x) is an as yet unknown curve. I understand that although f(y, y', x) is a function of three variables, it is only dependent on one variable, x. (where do these come from, exactly?) Taylor then defines Y(x) = y(x) + η(x) is the WRONG path, where y(x) is the correct one. η is the variation of Y(x) from y(x).  why do we need to introduce the INCORRECT path? Next, he introduces α into Y(x) = y(x) + αη(x). If we set α = 0, we will have Y(x) = y(x)  why do we need α? Our integral now becomes: [tex]\int^{x_{2}}_{x_{1}} f(y(x) + αη(x), y'(x) + αη'(x), x)dx[/tex]  we're assuming that α is equal to 0? I'm not sure I 100% understand this step. We need to check that [itex]\frac{dS}{d\alpha} = 0[/itex] is this to check that α is a constant? or used as a way of making sure α = 0? Take partial derivative: [tex]\frac{\partial f ((y(x) + αη(x), y'(x) + αη '(x), x)}{\partial \alpha}= \eta \frac{\partial f}{\partial \alpha}+ \eta ' \frac{\partial f}{\partial y'} [/tex]  because of the chain rule [tex]\frac{dS}{dα}=\int^{x_{2}}_{x_{1}}\frac{\partial f}{\partial α}dx = 0[/tex]  Next he works some voodoo magic by using integration by parts on the integral. I haven't worked this step out myself, but I assume it's straight forward. So, in the end, we get: [tex]\frac{\partial f}{\partial y}\frac{d}{dx}\frac{\partial f}{\partial y'}= 0[/tex] So, I'm pretty lost. I think it would help if I understood the idea of what we're really doing here. This is essential arc length along the shortest curve, but all the additional variables and what not are confusing the hell out of me. Sorry for the long post, but any help is much appreciated. I'm so desperate, I'm offering a reward of one (1) virtual cookie to the first helpful post. NB  I put this in the Classical Physics section, because I'm more concerned with how this is used in mechanics right now. Although I'm interested, in what course is Calculus of Variation taught in at a rigorous level? 



#2
May1913, 12:24 AM

P: 2,491

This writeup on the Brachistochrone problem may help:
http://www.hep.caltech.edu/~fcp/math...alCalculus.pdf 



#3
May1913, 12:30 AM

C. Spirit
Sci Advisor
Thanks
P: 4,940

What part of Taylor is this exactly? Is it related to a problem? Or is it the derivation of the EulerLagrange equations? If it is the latter then I must say it is one of the most convoluted ways of deriving the equations I've ever seen; the standard method of taking a first variation is a lot more intuitive and straightforward.




#4
May1913, 12:37 AM

P: 273

Calculus of Variation  Classical MechanicsStarts on page 218 Here's your virtual cookie, by the way 



#5
May1913, 12:56 AM

C. Spirit
Sci Advisor
Thanks
P: 4,940

See here instead: http://www.colorado.edu/engineering/...20Equation.pdf




#6
May1913, 01:02 AM

P: 273

I'm looking into the link first posted. I'm befuddled by line 3 on page 4. But other than that, It it explained it much better than in Taylor's book.
Although I'm still confused by why the function is f(y,y', x). y = this is the curve, of which the constraints are on. y' is the rate of change of this curve, and x is what all of this is dependent on. [tex]\int^{x_{2}}_{x_{1}} f(y(x), y'(x), x)dx[/tex] this gives us the length of the shortest path? What exactly is f(y,y', x)? 



#7
May1913, 01:04 AM

P: 2,491





#8
May1913, 01:06 AM

P: 2,491





#9
May1913, 01:07 AM

C. Spirit
Sci Advisor
Thanks
P: 4,940





#10
May1913, 01:09 AM

P: 2,491





#11
May1913, 01:12 AM

C. Spirit
Sci Advisor
Thanks
P: 4,940





#12
May1913, 01:12 AM

C. Spirit
Sci Advisor
Thanks
P: 4,940





#13
May1913, 01:22 AM

P: 2,491





#14
May1913, 01:24 AM

P: 273

Many thanks for those links, they really cleared things up for me!
As for cookies, have some chemistry cookies: I need to go and work some problems out, if I have any other questions, I'll be sure to post. 



#16
May1913, 01:34 AM

P: 273

Alright alright. Have some Einstein chocolate Never happy! So, what kinda of class is this taught in, anyway? In a rigorous mathy style 


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