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There is a step in what seems to be the standard derivation of the Euler-Lagrange equation that I do not see how can be justified (it is the first derivation in the Wikipedia article on the equation).

We have that $g(\epsilon) = f(t) + \epsilon \lambda(t)$

We are computing the total derivative $\frac{\partial F}{\partial\epsilon}(t,g(\epsilon),\dot{g}(\epsilon))$ and get that this is equal to $\frac{\partial F}{\partial g}\lambda(t) + \frac{\partial F}{\partial \dot{g}}\dot{\lambda}(t)$. The thing that does not make sense to me is that this is supposed to equal $\frac{\partial F}{\partial f}\lambda(t) + \frac{\partial F}{\partial \dot{f}}\dot{\lambda}(t)$ at $\epsilon = 0$.

To make some sense of it, I tried using the definition of the derivative on $\frac{\partial F}{\partial g}\lambda(t)$ to see if I can make it equal $\frac{\partial F}{\partial f}\lambda(t)$. However, what I do get is: $$\lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+(\epsilon+\Delta \epsilon) \lambda(t),\dot{g}) - F(t,f(t)+ \epsilon \lambda(t),\dot{g})}{\Delta \epsilon }$$ $$=\lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+(\epsilon+\Delta \epsilon) \lambda(t),\dot{g}) - F(t,f(t)+ \epsilon \lambda(t),\dot{g})}{\Delta g } \frac{\Delta g}{\Delta\epsilon}$$ Here we have that $\Delta g = \Delta \epsilon \lambda(t)$. This gives me: $$=\lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+(\epsilon+\Delta \epsilon) \lambda(t),\dot{g}) - F(t,f(t)+ \epsilon \lambda(t),\dot{g})}{\Delta g } \lambda(t)$$

Now I am just stuck. I cannot convince myself that $\Delta g = \Delta f$. I mean, the derivation is with respect to $\epsilon$, not t. So, why does this lead to the Euler-Lagrange equation?


Addendum

I think I may have to clarify my problem: $\frac{\partial F}{\partial f} \lambda$ looks to me to be equal to: $$\lim_{\Delta f \to 0}\frac{F(t,f(t)+\Delta f(t),\dot{g}) - F(t,f(t),\dot{g})}{\Delta f(t) } \lambda(t)$$

However, my computations above imply that the solution is $$\lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+(\epsilon+\Delta \epsilon) \lambda(t),\dot{g}) - F(t,f(t)+ \epsilon \lambda(t),\dot{g})}{\Delta g } \lambda(t) = \lim_{\Delta \epsilon \to 0}\frac{F(t,f(t)+\Delta \epsilon \lambda(t),\dot{g}) - F(t,f(t),\dot{g})}{\Delta \epsilon \lambda(t) } \lambda(t) $$

Clearly, my reasoning and computations are wrong. However, I don't see how I can get the solution above by computing $\frac{dF}{d\epsilon}$.

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    $\begingroup$ The answer here might help you a bit. $\endgroup$ – user99914 Aug 3 '17 at 6:52
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It's just the chain rule. You are given a function $F(\cdot,\cdot,\cdot):{\mathbb R}^3\to{\mathbb R}$ of three variables, say $t$, $p$, $q$, and then are studying the expression $$F\bigl(t,f(t),\dot f(t)\bigr)$$ for various known or unknown "test functions" $t\mapsto f(t)$. In particular you form the expression $$\psi(\epsilon):=F\bigl(t,f(t)+\epsilon g(t),\dot f(t)+\epsilon\dot g(t)\bigr)\ .$$ Then for any fixed $t$ you have $$\psi'(\epsilon)=F_{.2}\bigl(t,f(t)+\epsilon g(t),\dot f(t)+\epsilon\dot g(t)\bigr) \>g(t)+F_{.3}\bigl(t,f(t)+\epsilon g(t),\dot f(t)+\epsilon\dot g(t)\bigr)\>\dot g(t)\ .\tag{1}$$ Note that you do not "differentiate $F$ with respect to $f$" here. It is an abuse of notation to write ${\partial F\over\partial f}$, but we have learnt to live with it. $F$ has known partial derivatives $F_{.i}$ at each point $(t,p,q)\in{\mathbb R}^3$ before any "test function" $f$ is inserted.

Put $\epsilon:=0$ in $(1)$ and obtain $$F_{.2}\bigl(t,f(t),\dot f(t)\bigr) \>g(t)+F_{.3}\bigl(t,f(t),\dot f(t)\bigr)\>\dot g(t)\ .$$

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  • $\begingroup$ Well, that does not address the issue I am having. I understand it's the chain rule I am working with. The thing is, like you said, t is fixed, and f depends on t and not $\epsilon$. So, I don't understand how dF/df and dF/df' can be anything other than 0 in the equation dF/d$\epsilon$ (at least, the way it is derived in the proof I am referencing). I may have to edit my original question. $\endgroup$ – Avatrin Aug 3 '17 at 8:46
  • $\begingroup$ I see you have edited your answer. Isn't dF/df = $\lim{x \to a} \frac{F(f(x)) - F(f(a))}{f(x)-f(a)}$? Isn't that, in terms of calculation, differentiating F with respect to f? I hope this is not too much to ask, but can you show me why the step is true using the definition of the derivative? Or, do you know a website or book where I can find this computation having been done? Preferably with explanations.... $\endgroup$ – Avatrin Aug 3 '17 at 12:55

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