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Derivations of Euler equations

  1. Jun 4, 2010 #1
    Every text book i find breezes over the following point:

    [tex]\delta[/tex][tex]\partial[/tex] (x) =[tex]\partial[/tex] [tex]\delta[/tex] (x)

    where delta is just the variation. Someone asked me why thats true and i guessed the only thing i could say was that delta is an operation not a variable so this is more like an algebraric statment of the commutivity of delta and taking a derivative. Is this right or is there a more clear explanation.
  2. jcsd
  3. Jun 5, 2010 #2


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    I always view it as follows.
    What we do, is changing some function f(x) by an infinitesimal amount, i.e. replace
    [tex]f(x) \to f(x) + \delta f(x) \qquad\qquad(*)[/tex]
    (and ignore higher order variations).

    So in this notation, where delta indicates the infinitesimal change, it should also be true that
    [tex]\frac{\partial f(x)}{\partial x} \to \frac{\partial f(x)}{\partial x} + \delta \left(\frac{\partial f(x)}{\partial x} \right) [/tex]
    (which is just statement (*) again for another function g(x) = df(x)/dx).

    Now if you differentiate (*) you get an equation from which it follows immediately that
    [tex]\delta \left(\frac{\partial f(x)}{\partial x} \right) = \frac{\partial(\delta f(x))}{\partial x}[/tex]
  4. Jun 5, 2010 #3
    Is there a way to argue that (using the chain rule)
    [tex]\partial[/tex]([tex]\delta[/tex]f(x) ) = f(x)[tex]\partial[/tex][tex]\delta[/tex] + [tex]\delta[/tex][tex]\partial[/tex]f(x)
    and then argue that[tex]\partial[/tex][tex]\delta[/tex] is a second order differential that can be tossed out as approx 0 giving the result

    [tex]\partial[/tex]([tex]\delta[/tex]f(x) ) = [STRIKE]f(x)[tex]\partial[/tex][tex]\delta[/tex][/STRIKE] [tex]\rightarrow[/tex] 0 + [tex]\delta[/tex][tex]\partial[/tex]f(x)
  5. Jun 7, 2010 #4


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    Not if you want to be rigorous.
    Because [itex]\delta[/itex] in itself doesn't mean anything, just like [itex]\partial[/itex] doesn't mean anything.

    The object you are looking at is [itex]\delta f(x)[/itex].
    If you want to be precise, you can use a first order Taylor expansion
    [tex]f(x) \to F(x) = f(a) + f'(a) (x - a) + \mathcal{O}((x - a)^2)[/tex]
    where you call [tex]\delta = f'(a) (x - a)[/tex] and neglect the quadratic terms.
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