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## Source

IM Gelfand, SV Fomin. 1963. *Calculus of Variations*. Mineola, NY: Dover. Tr. RA Silverman. p. 18-19.

## Problem

In Sec. 4.2, Case 2, Gelfand and Fomin claim without further comment that we can move from

\( F_y – \frac{d}{dx} F_{y’} = F_{y} – F_{y’y}y’ – F_{y’y’}y^″ \)

to

\( F_y y’ – F_{y’y}y’^{2} – F_{y’y’}y’y^″ = \frac{d}{dx}(F – y’F_{y’}) \)

by multiplying the first equation by \(y’\). Verifying this turned out to be non-trivial for me, though (as is the way in math) transparent in retrospect, so I walk through it here.

## Walkthrough

We start with

\( F_y – \frac{d}{dx} F_{y’} = F_{y} – F_{y’y}y’ – F_{y’y’}y^″ \tag{1} \)

Multiply by \( y’ \) as instructed:

\( F_yy’ – y’ \frac{d}{dx} F_{y’} = F_{y}y’ – F_{y’y}y’^2 – F_{y’y’}y’y^″ \tag{2} \)

Reverse the order (you wouldn’t write this kind of thing out in a normal proof, of course, but this is a walkthrough):

\( F_{y}y’ – F_{y’y}y’^2 – F_{y’y’}y’y^″ = F_y y’ – y’ \frac{d}{dx} F_{y’} \tag{3} \)

You can see above that we already have the left-hand side of the target, so we start to work on the right-hand side. The key thing to do now is to insert \( F_{y’}y” – F_{y’}y” \) into the middle of the right-hand term.1 How on earth do we know to do this? I can’t speak for the Russians, whose mathematical powers are so far beyond mine as to seem magical to me, but I eventually figured it out by working backward. Before that, I went down innumerable blind alleys. Anyway, doing so gives us

\( F_{y}y’ – F_{y’y}y’^2 – F_{y’y’}y’y^″ = F_y y’ + \left( F_{y’}y^″ – F_{y’}y^″ \right) – y’ \frac{d}{dx}F_{y’} \tag{4} \)

I regroup the terms above, simply as a visual aid:

\( F_{y}y’ – F_{y’y}y’^2 – F_{y’y’}y’y^″ = \left( F_y y’ + F_{y’}y^″ \right) – \left( F_{y’}y^″ – y’ \frac{d}{dx}F_{y’} \right) \tag{5} \)

Rewrite the first group in 5 as \( \frac{dF}{dx} \). To see the equivalence, remember that in Case 2—the case Gelfand and Fomin are dealing with here—\( F \) is a function only of \(y\) and \(y’\). Hence, \( \frac{dF}{dx} = \frac{dF}{dy}\frac{dy}{dx} + \frac{dF}{dy’}\frac{dy’}{dx} \), which is just a different way of writing \( F_y y’ + F_{y’}y^″ \). So we have

\( F_{y}y’ – F_{y’y}y’^2 – F_{y’y’}y’y^″ = \frac{dF}{dx} – \left( F_{y’}y^″ – y’ \frac{d}{dx}F_{y’} \right) \tag{6} \)

Rewrite the second group as \(\frac{d}{dx}\left(y’F_{y’}\right)\). To see the equivalence, just apply the product rule to this new term, then do a tiny bit of rearranging, and you will get the second group in 5. We now have:

\( F_{y}y’ – F_{y’y}y’^2 – F_{y’y’}y’y^″ = \frac{dF}{dx} – \frac{d}{dx}\left(y’F_{y’}\right) \tag{7} \)

Finally, pull the \( \frac{d}{dx} \) out from the right-hand side of 7:

\( F_{y}y’ – F_{y’y}y’^2 – F_{y’y’}y’y^″ = \frac{d}{dx} \left( F – y’F_{y’} \right) \tag{8} \)

And there you go.

## Notes

1 We can do this because \( F_{y’}y^″ – F_{y’}y^″ = 0 \).