IM Gelfand and SV Fomin, Calculus of Variations
Sec. 4.2, Case 2: Walkthrough

Mark Vuletic

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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:

$$F_y - \frac{d}{dx} F_{y'} = F_{y} - F_{y'y}y' - F_{y'y'}y'' \tag{1}$$

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 $$F_{y'}y'' - F_{y'}y'' = 0$$.

Last updated: 25 Nov 2016

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