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I'm having trouble understanding how Gelfand & Fomin go from

$$ \begin{align} \Delta J[h] &= J[y+h] - J[y] \\ &= \int_a^b \left( F_y h + F_y' h' \right) \, dx + \frac{1}{2}\int_a^b \left( \bar{F}_{yy}h^2 + 2\bar{F}_{yy'}hh' + \bar{F}_{y'y'}h'^2 \right) \, dx \tag{1}\label{eq1} \end{align} $$

to

$$ \Delta J[h] = \int_a^b \left( F_y h + F_y' h' \right) \, dx + \frac{1}{2}\int_a^b \left( F_{yy}h^2 + 2F_{yy'}hh' + F_{y'y'}h'^2 \right) \, dx + \epsilon \tag{2}\label{eq2} $$

where

$$ \epsilon = \int_a^b \left( \epsilon_1h^2 + \epsilon_2hh' + \epsilon_3h'^2 \right) \, dx $$

in Section 25. Step \ref{eq1} is just Taylor's theorem with remainder from David Widder's Advanced Calculus (Section 9.2, Eq. 3). Going from \ref{eq1} to \ref{eq2} however, Gelfand and Fomin state:

"If we replace $\bar{F}_{yy}$, $\bar{F}_{yy'}$, and $\bar{F}_{y'y'}$ by the derivatives $F_{yy}$, $F_{yy'}$, and $F_{y'y'}$ evaluated at the point $(x, y(x), y'(x))$, then [$\ref{eq1}$] becomes [$\ref{eq2}$]."

How do they do this? (Continuity or some application of mean value theorem perhaps?)

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Quantities like $\bar{F}_{yy}$ arise from an initial application of Taylor's theorem, that is, for some $\theta \in (0,1)$,

$$\bar{F}_{yy} = F_{yy}(x,y(x) +\theta h(x), y'(x) + \theta h'(x))$$

Thus,

$$\bar{F}_{yy} = F_{yy}(x,y(x), y'(x) ) + \underbrace{F_{yy}(x,y(x) +\theta h(x), y'(x) + \theta h'(x))- F_{yy}(x,y(x), y'(x) )}_{\epsilon_1(x)} $$

where for all $x \in [a,b]$ we have $\epsilon_1(x) \to 0$ uniformly as $\sup_{x\in[a,b]}h(x)\to 0$ and $\sup_{x\in[a,b]}h'(x),\to 0$ by continuity of $F_{yy}$.

The same applies to the other terms.

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  • $\begingroup$ Thank you for your reply! So how then is $\bar{F}_{yy}-F_{yy}=\bar{F}-F$? How did the derivatives vanish? I'm used to seeing Taylor's theorem as $f(x-a) = f(a) + f'(a)(x-a) + \frac{1}{2}f''(a)(x-a)^2 + ...$. $\endgroup$ – A. Hendry Apr 2 '19 at 13:30
  • $\begingroup$ That was a typo I just edited. I’m just working with the second partial derivative $F_{yy}$ here. It is assumed to be continuous so that $\epsilon_1$ vanishes when an appropriate norm of $h$ goes to $0$. $\endgroup$ – RRL Apr 2 '19 at 13:39
  • $\begingroup$ Is it understandable now? $\endgroup$ – RRL Apr 2 '19 at 13:56
  • $\begingroup$ Yes, this is more understandable now. Thank you. $\endgroup$ – A. Hendry Apr 3 '19 at 3:04

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