Here is a simple (probably trivial) step in the derivation of the Euler-Lagrange equation. If we denote $Y(x) = y(x) + \epsilon \eta(x) $, I want to know why is

$\dfrac{\partial f(Y,x)}{\partial Y} \Big\vert_{\epsilon = 0} = \dfrac{\partial f(y,x)}{\partial y} $

Could someone justify the steps involved in justifying this ? I am sure am missing something elementary.

EDIT: Could someone comment on correctness of the 'proof ' ? $\dfrac{\partial f(Y,x)}{\partial Y} \Big\vert_{\epsilon = 0} = \lim_{H\rightarrow 0} \dfrac{f(y + \epsilon \eta + H,x) -f(y + \epsilon \eta,x)}{H} \Big\vert_{\epsilon = 0} = \lim_{H\rightarrow 0} \dfrac{f(y + H,x) -f(y,x)}{H} = \dfrac{\partial f(y,x)}{\partial y} $

EDIT 2: The answers provided still leave me confused. Here is another attempt at a "proof". $\dfrac{\partial f(Y,x)}{\partial Y} \Big\vert_{\epsilon = 0} = \lim_{\epsilon\rightarrow 0} \dfrac{\partial f(Y,x)}{\partial Y} = \lim_{\epsilon \rightarrow 0} \left(\lim_{H\rightarrow 0} \dfrac{f(y + \epsilon \eta + H,x) -f(y + \epsilon \eta,x)}{H} \right)$

Now if I could interchange the limits, then it would make sense that I get $\lim_{H\rightarrow 0} \dfrac{f(y + H,x) -f(y,x)}{H} = \dfrac{\partial f(y,x)}{\partial y} $.

So is the interchange of limits allowed. What hypothesis must be satisfied for that to happen ?

  • $\begingroup$ Often, $f$ is a linear functional on the space of trajectories, so it will not see any dependence on $x$, only $y$ and $\eta$. So if you use chain rule, it should be clear. Remember that $y$ and $\eta$ are independent of each other when doing the chain rule otherwise you will end up with terms that are pretty awkward to deal with. $\endgroup$ – Cameron Williams Apr 25 '13 at 1:59
  • $\begingroup$ The edit seems correct, as when $\varepsilon =0$, $Y(x)=y(x)$ for all $x$, therefore when operated on $Y(x)$ will give the same result as $y(x)$. Also note that as $Y(x)=y(x)$, the operator $\frac{\partial}{\partial Y}=\frac{\partial}{\partial y}$. $\endgroup$ – Meow Apr 25 '13 at 15:51

The formula for $\frac{\partial f(Y,x)}{\partial Y}|_{\epsilon=0}$ is not correct: you should write

$$\frac{\partial f(Y,x)}{\partial Y}|_{\epsilon=0}:= \lim_{\epsilon\rightarrow 0}\frac{ f(y+\epsilon\eta,x)-f(y,x)}{\epsilon}= \lim_{\epsilon\rightarrow 0}\frac{ \langle \nabla f(y,x),h_\epsilon\rangle+O(\|h\|^2)}{\epsilon}, $$

where we suppose that $f$ is differentiable at $(y(x)+\epsilon\eta(x),x)$ for all $x$, denoting by $h_\epsilon$ the increment

$$h_\epsilon=( \epsilon\eta(x), 0)$$

and by $\nabla f(y,x)=\left(\frac{\partial f}{\partial y},\frac{\partial f}{\partial x}\right)$ the gradient of $f$ at $(y(x),x)$.


$$\frac{\partial f(Y,x)}{\partial Y}|_{\epsilon=0}= \lim_{\epsilon\rightarrow 0}\frac{\frac{\partial f}{\partial y}\epsilon\eta(x) +O(\epsilon^2)}{\epsilon}=\frac{\partial f}{\partial y}(y(x),x)\eta(x), $$ as claimed (for all small variations $\eta(x)$). What we are missing here are the boundary conditions on the increment $\eta$ of $y$ in order to talk about a "variational problem" and the explicit form of the functional $f$.

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  • $\begingroup$ Any chance for assistance here: math.stackexchange.com/questions/1066495/… $\endgroup$ – Royi Dec 14 '14 at 7:10
  • $\begingroup$ @Avitus:I dont understand how you got your starting definition. To define $\dfrac{\partial f(Y,x)}{\partial Y}$, surely I must start with $\dfrac{f(Y+H,x) - f(Y,x)}{H}$ as $H$ is small ? If I do that and then rewrite $Y= y+ \epsilon \eta(x)$, I am back at my original answer. What is the error in my definition ? $\endgroup$ – me10240 Feb 11 '15 at 19:08
  • $\begingroup$ I think that the misunderstanding is generated by the symbol $\partial$ used on the l.h.s. of the equation you want to prove. It is, in my opinion, not a partial derivative of a function $f$ of many variables, but a variational derivative of a functional $f$ around the "trajectory" $y(x)$. In my answer I start with the proof of it; the rest is just the use of differentiability. $\endgroup$ – Avitus Feb 12 '15 at 19:04

This Confusion with Euler-Lagrange Derivation was basically what convinced me that it was all a muddle with symbols and notation. Just for the benefit of future readers, I write it down in my own words.

So, the notation $ \dfrac{d f}{dx}\Big|_{x = x^*}$ basically means derivative of the function $f(\cdot)$ with respect to its argument, and then evaluation of the result when the argument is set to $x^*$.

Exactly the same reasoning is involved in the question, for we take the partial derivative with respect to $Y$ and the result is evaluated at $\epsilon =0$.

But evaluation of the result at $\epsilon =0$ is equivalent to $ \dfrac{\partial f(Y, x)}{\partial Y}\Big|_{Y = y}$ which looks silly, but it is blindingly obvious to me now that

$ \dfrac{\partial f(Y, x)}{\partial Y}\Big|_{\epsilon = 0} = \dfrac{\partial f(Y, x)}{\partial Y}\Big|_{Y = y} = \dfrac{\partial f(y, x)}{\partial y}$

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