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


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

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?

  • $\begingroup$ why do you think ∂f(u)/∂u = 0 ? u and u' are not independent, so shouldn't you use chain rule one more time ? $\endgroup$ – Geralt of Rivia Mar 6 '12 at 18:10
  • $\begingroup$ 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 $\delta u$. $\endgroup$ – andreasdr Mar 6 '12 at 18:37
  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 $
$ \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) $

  • $\begingroup$ 1. That must mean $u'$ isn't differentiable wrt. $u$, right? Is "the derivative of $u'$ wrt. $u$ really a well-posed question? 2. You misunderstand me, by "doing the same thing" I mean using the chain rule on a derivative wrt. a variable ($\epsilon$) to an expression involving derivatives wrt. both variables ($\epsilon$) and functions ($u'+\epsilon\delta u'$). But nevermind, this was a minor point. $\endgroup$ – andreasdr Mar 8 '12 at 16:05
  • $\begingroup$ To ask whether u' is differentiable wrt u without knowing anything about u is an ambiguous question. Your problem originated from the fact that f(u) is not well defined through only u'. Consider a concrete case (u=x^2 for example), you will see that you have to find the explicit function f(u) first to apply chain rule. You will have similar error when you try to use chain rule on function that is not defined explicitly wrt the variable you are differentiating. Also, the correct answer for your question above is df/du*(delta u). $\endgroup$ – Geralt of Rivia Mar 12 '12 at 13:15
  • $\begingroup$ 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′ ). $\endgroup$ – Geralt of Rivia Mar 12 '12 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.