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For the integral $$I =\int_a^b F(y', y, x) \, \mathrm dx$$ I’ve seen the requiremts expressed for the Euler Lagrange equation expressed in 2 different ways, but I do not see how they are equivalent.

**First way:

For a perturbation given to $y$ given as

$$y(x)\rightarrow y(x)+\alpha\eta(x)$$

we require $$\left. \frac{\mathrm dI}{\mathrm d\alpha} \right|_{\alpha=0}=0 $$

Second Way:

$$\delta I = F[y+\alpha\eta]-F[y]=0$$

I sort of understand intuitively what each of these mean, but I can’t understand why they are essentially saying the same thing.

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  • $\begingroup$ The boxed section "Derivation of the one-dimensional Euler–Lagrange equation" at Euler-Lagrange Equation: Statement: 26 March 2025 17:59 UTC currently explains this in one direction. It's a bit terse. Do you follow that explanation or are there steps that would benefit from expansion? $\endgroup$ Commented Mar 26 at 18:01
  • $\begingroup$ I had a look at that derivation, but it seems to use only the first method which I mentioned $\endgroup$ Commented Mar 26 at 18:26
  • $\begingroup$ How do you understand the second way? To me, the rigorous definition of the second is precisely the first way. $\endgroup$ Commented Mar 27 at 1:59
  • $\begingroup$ I understand it visually, simply as the difference between the functions $\endgroup$ Commented Mar 27 at 20:55

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