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Let $f(u)=(du/dx)^2$, where $u=u(x)$. Let's write f(u)=(u')^2 in the following.

Let's say we want to calculate $\partial/\partial\epsilon[f(u+\epsilon\delta u)]_{\epsilon=0},$ where $\delta u = \delta u(x)$. As I see it, two approaches are:

1) \frac{\partial}{\partial\epsilon}[f(u+\epsilon\delta u)]|_{\epsilon=0} = \frac{\partial}{\partial\epsilon}[((u+\epsilon\delta u)')^2]|_{\epsilon=0} = \frac{\partial}{\partial\epsilon}[(u'+\epsilon\delta u')^2]|_{\epsilon=0} = 2(u'+\epsilon\delta u')\frac{\partial}{\partial\epsilon}(u'+\epsilon\delta u')|_{\epsilon=0} = 2u'\delta u'

2)

$\frac{\partial}{\partial\epsilon}[f(u+\epsilon\delta u)]|_{\epsilon=0} = \frac{\partial f(u+\epsilon\delta u)}{\partial(u+\epsilon\delta u)}\frac{\partial(u+\epsilon\delta u)}{\partial\epsilon} = \frac{\partial f(u)}{\partial u}\delta u = 0*\delta u?$

I obviously end up wrong with 2), but where is the error? I suppose it has something to do with $f(u)$ being a functional, and with the use of the chain rule to change from a derivative wrt. a variable ($\epsilon$ in this case) to an expression also involving derivatives wrt. functions ($u$ in this case), but can someone please clarify?

Also, if the problem lies in is applying the chain rule in this way, why does the following (found in 1) ), where I'm essentially doing the same thing, work? \frac{\partial}{\partial\epsilon}[(u'+\epsilon\delta u')^2] = 2(u'+\epsilon\delta u')\frac{\partial}{\partial\epsilon}(u'+\epsilon\delta u') = 2u'\delta u'. Why don't I have to write \frac{\partial}{\partial\epsilon}[(u'+\epsilon\delta u')^2] = \frac{\partial}{\partial\epsilon}[(u')^2+2u'\epsilon\delta u'+\epsilon^2(\delta u')^2] = 2u'\delta u'? Does it work just by accident?

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    I'm faced with $\partial u'/\partial u$ in any case though, which I can't see how it can be anything other than zero (don't take that to mean that I'm sure, though!). And in any case, I need a $\delta u'$ at the end of 2), not a $\delta u$.2012-03-06

1 Answers 1

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  1. Where 's the error ?
    Let $ g(\epsilon) = u+ \epsilon \delta u $

in 2) Before using chain rule for $ f(g( \epsilon )) $ at $ \epsilon =0 $ you have to show that f is differentiable at g(0) which is u. Ironically from the way you define f, i don't think this is possible. Also what's with using $ \delta u(x) $ you could've just used v(x) instead to make the question more readable.

2.Does it work just by accident?
No since $ (a+b)^2 = a^2+b^2+2ab $
and
$ \frac{d}{dx} (a+b)^2 = \frac{d}{dx} (a^2+b^2+2ab) =\frac{d}{dx} (a^2) +\frac{d}{dx} (b^2) + \frac{d}{dx} (2ab) $

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    also, 2 follow from 1. it was correct whenever you have the explicit formula that is differentiable at g(ϵ) (in this case, f(g(ϵ)) = (u′+ϵδu′)^2 and g(ϵ) =u′+ϵδu′ ).2012-03-12