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Calculus of Variations: Nature of the Functional

  1. Apr 27, 2014 #1
    Let [itex] \normalsize S[y] = \int ^{a}_{b} f[y, \dot{y}, x] dx [/itex] be the functional i want to minimize. Why does [itex] \normalsize f [/itex] (inside the integral) take this specific form?

    Would i not be able to minimize the integral, [itex] \normalsize S[/itex] , if [itex] f [/itex] had any other form instead of [itex] f = f[x, y, \dot{y}] [/itex]?
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  3. Apr 28, 2014 #2


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    Do you understand what "f[x,y,y˙]" means? f is a function that can depend upon x, y, or the derivative of y but the "dependence" on any one can be 0- that is, this includes f(x), with f depending on x only, f(y) with f depending on y only, or f(y') with f depending on the derivative of y only. What more generality do you want?
  4. Apr 28, 2014 #3


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    Mathematically one can consider a functional of the form [tex]
    S[y] = \int_a^b f(x,y,y', \dots, y^{(n)})\,dx
    [/tex] for any [itex]n \geq 1[/itex], where the optimal solution satisfies [tex]
    \sum_{k=0}^n (-1)^k \frac{d^k}{dx^k}\left( \frac{\partial f}{\partial y^{(k)}}\right) = 0,
    [/tex] which is in principle a [itex]2n[/itex]-order ODE subject to boundary conditions on [itex]y[/itex], [itex]y'[/itex], ..., [itex]y^{(n-1)}[/itex] at both [itex]x = a[/itex] and [itex]x = b[/itex]. However in physical applications one generally has
    \frac{\partial f}{\partial y^{(k)}} = 0
    [/tex] for [itex]k \geq 2[/itex] so there is no point in going beyond [itex]n = 1[/itex]. Also the method of deriving the above ODE does not involve any ideas which are not required for the derivation of the Euler-Lagrange equation for the case [itex]n = 1[/itex]; it just requires more integrations by parts.
  5. Apr 28, 2014 #4
    Yes, i understand what f(x, y, y') means here. I was thinking about generalizations of the form that pasmith mentioned. Most of the texts are physically motivated, i guess. Probably that's why i didn't find the general form. Thanks, all! :)
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