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Euler-Lagrange equation
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Extended explanation
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PROOF
Let us find the extrema of the functional

Such a functional could be arc length, for example. For the variation of v,
let Δy be an arbitrary differentiable function such that Δy(x1)=Δy(x2)=0.
Now, to find the extrema, the variation must be zero. i.e.
or

Using the chain rule of multiple variables, this simplifies to
We then split d(y+aΔy) and d(y'+aΔy') into dy+Δyda and dy'+Δy'da respectively. Remember that y and y' is independent of a, and da/da=1. We therefore get (using different notation: )

Using integration by parts on the right side with "u"=Fy' and "dv"=Δy'dx:

However, Δy(x1)=Δy(x2)=0. Thus the middle term is zero, so:

Applying the Fundamental Lemma of Calculus of Variations, we find

Or, more compactly,

where Dx is the differential operator with respect to x.
This is a second order differential equation which, when solved, gives the desired extrema of the functional. |
Commentary
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Pinu7 @ 12:02 AM Jun28-09
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Pinu7 @ 05:25 PM Jun27-09
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Pinu7 @ 04:24 PM Jun23-09
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