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Topic: calculus of variations (Euler-Lagrange) problem
Replies: 2   Last Post: Jun 27, 2011 8:38 PM

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Rob Johnson

Posts: 1,771
Registered: 12/6/04
Re: calculus of variations (Euler-Lagrange) problem
Posted: Jun 27, 2011 8:38 PM
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In article <20110625.070534@whim.org>,
Rob Johnson <rob@trash.whim.org> wrote:
>In article <f22efe88-a31c-454f-84e4-33cbb8607850@ct4g2000vbb.googlegroups.com>,
>hanrahan398@yahoo.co.uk wrote:

>>Writing E for epsilon, consider variations of the curve x(t) of the
>>form
>>x^E(t), defined as (x(t) + E*z(t) ) / || x(t) + E*z(t) ||,
>>which lie on the surface of the unit sphere S^2,
>>and require that d/dE (I [x^E]) = 0 at E = 0 for every smooth z(t),
>>where I[x] is defined as int (||x'||^2) dt for x(t) in S^2.
>>
>>From this consideration, how can we deduce the Euler-Lagrange equation
>>x'' + (||x'||^2)x = 0 for the problem of minimising I[x]?

>
>I worked this problem in a different way. I did note that it was
>necessary to make use of the fact that ||x||^2 = 1, i.e. <x,x'> = 0.
>
>The equation I got for the solution was x'' = kx. From this and the
>fact that ||x||^2 = 1, it can be deduced that k = -||x'||^2. This
>then implies that x'' + x ||x'||^2 = 0.


I had some spare time, so I worked the problem again using
Euler-Lagrange. L(u,v,w) in

|\
| L(t,x,x') dt
\|

|\ 2
= | ||x'|| dt
\|

is L(u,v,w) = ||w||^2. Note that L_1 = L_2 = 0. Also,

L (u,v,w) = 2w
3

Under the constraint that C(t,x,x') = ||x||^2 - 1 = 0, where C_1 = 0,
C_3 = 0 and

C (u,v,w) = 2v
2

the stationary E-L equation is

d d
0 = L (t,x,x') - -- L (t,x,x') + k (C (t,x,x') - -- C (t,x,x'))
2 dt 3 2 dt 3

= -2x'' + 2kx

For some k. This is the same equation I got before.

Rob Johnson <rob@trash.whim.org>
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