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Calculus of variations question

  1. Jul 9, 2008 #1
    In my Classical Dynamics text (Thornton & Marion), there's one step in the derivation of Euler's equation that I don't understand. I think if I understood it, I'd be able to derive the equation on my own. I wrote out the entire derivation up to the point I don't get, just so you guys would understand the notation. You can skip to the section in the dotted lines for the specific step (it's about partial derivatives).

    So the derivation starts with a quantity, J, that you want to minimize (or maximize). J is a function of the dependent variables y(x), y'(x), and the independent variable x.

    [tex]\ J=\int{f(y(x),y'(x);x)dx}[/tex]

    y(x) is varied until a minimum of J is found. This is accomplished by adding to y(x) a "neighboring function" such that y can be given the parametric representation:
    [tex]y(\alpha, x) = y(0,x) + \alpha\eta(x)[/tex]
    where α is a parameter and η(x) is some function of x that vanishes at the endpoints of integration.

    Now, J can be written as a functional of α:
    [tex]\ J(\alpha)=\int{f(y(\alpha,x),y'(\alpha,x);x)dx}[/tex]

    The extreme value can be found, as in basic calculus, by taking the derivative of J with respect to α and setting it equal to zero:
    [tex]\frac{\delta J}{\delta\alpha}=\frac{\delta}{\delta\alpha}\int{f(y,y';x)dx}[/tex]

    And here's the part I don't understand. It moves the differential operator inside the integral and takes the derivative of the integrand.
    [tex]\frac{\delta J}{\delta\alpha}=\int{\left(\frac{\delta f}{\delta y}\frac{\delta y}{\delta\alpha} + \frac{\delta f}{\delta y'}\frac{\delta y'}{\delta\alpha}\right)dx}[/tex]

    [tex]f = f(y(\alpha,x), y'(\alpha,x); x) [/tex]

    [tex]\frac{\delta f}{\delta\alpha}=\frac{\delta f}{\delta y}\frac{\delta y}{\delta\alpha}+\frac{\delta f}{\delta y'}\frac{\delta y'}{\delta\alpha}[/tex]

    In case you're interested, here's the rest of the derivation:
    [tex]y(\alpha,x) = y(x) + \alpha\eta(x) [/tex]
    [tex]\frac{\delta y}{\delta\alpha}= \eta(x) [/tex]

    [tex]y'(\alpha,x) = \frac{dy}{dx} = y'(x) + \alpha\frac{d\eta}{dx} [/tex]
    [tex]\frac{\delta y'}{\delta\alpha} = \frac{d\eta}{dx} [/tex]

    And so...
    [tex]\frac{\delta J}{\delta\alpha}=\int{\left(\frac{\delta f}{\delta y}\frac{\delta y}{\delta\alpha} + \frac{\delta f}{\delta y'}\frac{\delta y'}{\delta\alpha}\right)dx} = \int{\left(\frac{\delta f}{\delta y}\eta(x) + \frac{\delta f}{\delta y'}\frac{d\eta}{dx}\right)dx}[/tex]

    Integrating the second integrand term by parts gives:
    [tex]\int{\frac{\delta f}{\delta y'}\frac{d\eta}{dx}dx} = \frac{\delta}{\delta y'}\eta(x) - \int{\frac{d}{dx}\left(\frac{\delta f}{\delta y'}\right)\eta(x)dx} [/tex]

    But η=0 at the limits of integration, and so...
    [tex]\frac{\delta J}{\delta\alpha} = \int{\left(\frac{\delta f}{\delta y}\eta(x) - \frac{d}{dx}\left(\frac{\delta f}{\delta y'}\right)\eta(x)\right)dx} [/tex]
    [tex]=\int{\left(\frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'}\right)\eta(x)dx} = 0 [/tex]

    Because η(x) is an arbitrary function (i.e. it can be anything, as long as it's differentiable and vanishes at the endpoints), the quantity in parentheses be equal to 0:
    [tex] \frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'} = 0 [/tex]

    This is Euler's equation.

    Last edited: Jul 9, 2008
  2. jcsd
  3. Jul 9, 2008 #2
    umm thats just how you do partial differentiation

    suppose there exists [itex] f(x(t),y(t),z(t)[/itex]


    [tex] \frac{\partial f(x(t),y(t),z(t))}{\partial t} = \frac{\partial f}{\partial x} \frac{d x}{dt } +\frac{\partial f}{\partial y} \frac{d y}{dt }+ \frac{\partial f}{\partial z} \frac{d z}{dt }[/tex]

    or maybe you're confused about how you're allowed to move the differentiation into the integral? in which case you should read this

  4. Jul 9, 2008 #3
    Oh man...I knew it would be something like that. My thermodynamics professor would be ashamed. But I was also confused by the fact that they moved the differential operator inside the integral. In their words: "Because the limits of integration are fixed, the differential operation affects only the integrand". I'll read that article you posted.

    Thanks for the help
  5. Jul 10, 2008 #4


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    You shouldn't be ashamed.
    Even when the limits ARE fixed, "moving the differential operator inside the integral" is just a hand wavy way of saying that we "switch the order of the two limiting processes, from integration, then differentiation, TO differentiation, then integration".

    That this is allowable is not at all that obvious.

    However, theorems exist that prove that under fairly lenient conditions, this switch of limiting processes is, indeed, allowable.

    And physicists assume that these conditions DO prevail.

    But I agree, your professor SHOULD be ashamed, if he was unable to explain it.
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