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Euler-Lagrange derivation (one query)

This may seem like an obvious question to most of you but for some reason I cannot seem to get my head around this; if you read here for instance, http://en.wikipedia.org/wiki/Calculus_of_variations, and scroll down to their derivation of this equation, then why is it necessary for η(x)\eta(x) to vanish at the end points x1x_1 and x2x_2?

Because I thought that even if it didn't, since JJ attains its minimum at ff then surely any small change in ff i.e. in JJ will always be greater than its minimum value regardless of whether η\eta vanishes at the end points. Though I do like this particular derivation, I simply don't understand that part of it.

Thanks very much in advance!
(edited 9 years ago)
Original post by Gamma
This may seem like an obvious question to most of you but for some reason I cannot seem to get my head around this; if you read here for instance, http://en.wikipedia.org/wiki/Calculus_of_variations, and scroll down to their derivation of this equation, then why is it necessary for η(x)\eta(x) to vanish at the end points x1x_1 and x2x_2?

Because I thought that even if it didn't, since JJ attains its minimum at ff then surely any small change in ff i.e. in JJ will always be greater than its minimum value regardless of whether η\eta vanishes at the end points. Though I do like this particular derivation, I simply don't understand that part of it.

Thanks very much in advance!


Just to verify, is it the beginning or the end of the derivation you're confused about?
Reply 2
Avatar for Gamma
Gamma
OP
Original post by Slowbro93
Just to verify, is it the beginning or the end of the derivation you're confused about?


Sorry - it's the beginning that confuses me. The part where it mentions that η\eta vanishes at the end points. Then, if y=f+ϵηy = f + \epsilon \eta surely, y(x1)=y(x2)=fy(x_1) = y(x_2) = f - is this saying that the functional JJ attains a minimum at both those points? Seems not which is why I am confused. I'm missing something I know, but it's probably my understanding that's gone awry..
Reply 3
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BabyMaths
Original post by Gamma
is this saying that the functional JJ attains a minimum at both those points?


No. Functionals don't attain minima or maxima at points, rather they do so on paths.
Original post by Gamma
Sorry - it's the beginning that confuses me. The part where it mentions that η\eta vanishes at the end points. Then, if y=f+ϵηy = f + \epsilon \eta surely, y(x1)=y(x2)=fy(x_1) = y(x_2) = f - is this saying that the functional JJ attains a minimum at both those points? Seems not which is why I am confused. I'm missing something I know, but it's probably my understanding that's gone awry..

As BabyMaths says, y is a path, not a point. The requirement that eta vanish at the endpoints is equivalent to insisting that our small variation from our "trial path" doesn't alter the ends of the path - think "I want to find the shortest distance between these two nails hammered into the wall. I stretch string between them, and start wiggling the string around to see when the shortest length of string is used. Oh, but I need to make sure the string stays attached to the pegs, otherwise my shortest path would be just a single point of no string at all, sitting somewhere in between the two pegs."
Reply 5
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Gamma
OP
Original post by Smaug123
As BabyMaths says, y is a path, not a point. The requirement that eta vanish at the endpoints is equivalent to insisting that our small variation from our "trial path" doesn't alter the ends of the path - think "I want to find the shortest distance between these two nails hammered into the wall. I stretch string between them, and start wiggling the string around to see when the shortest length of string is used. Oh, but I need to make sure the string stays attached to the pegs, otherwise my shortest path would be just a single point of no string at all, sitting somewhere in between the two pegs."


Thanks a lot for your help.

Apologies for the late-ish reply!

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