Derivations of Euler equations

[zwoodrow]

Every textbook i find breezes over the following point:

$$\delta$$$$\partial$$ (x) =$$\partial$$ $$\delta$$ (x)

where delta is just the variation. Someone asked me why that's true and i guessed the only thing i could say was that delta is an operation not a variable so this is more like an algebraric statement of the commutivity of delta and taking a derivative. Is this right or is there a more clear explanation.

 

[CompuChip]

I always view it as follows.
What we do, is changing some function f(x) by an infinitesimal amount, i.e. replace
$$f(x) \to f(x) + \delta f(x) \qquad\qquad(*)$$
(and ignore higher order variations).

So in this notation, where delta indicates the infinitesimal change, it should also be true that
$$\frac{\partial f(x)}{\partial x} \to \frac{\partial f(x)}{\partial x} + \delta \left(\frac{\partial f(x)}{\partial x} \right)$$
(which is just statement (*) again for another function g(x) = df(x)/dx).

Now if you differentiate (*) you get an equation from which it follows immediately that
$$\delta \left(\frac{\partial f(x)}{\partial x} \right) = \frac{\partial(\delta f(x))}{\partial x}$$

 

[zwoodrow]

Is there a way to argue that (using the chain rule)
$$\partial$$($$\delta$$f(x) ) = f(x)$$\partial$$$$\delta$$ + $$\delta$$$$\partial$$f(x)
and then argue that$$\partial$$$$\delta$$ is a second order differential that can be tossed out as approx 0 giving the result

$$\partial$$($$\delta$$f(x) ) = [STRIKE]f(x)$$\partial$$$$\delta$$[/STRIKE] $$\rightarrow$$ 0 + $$\delta$$$$\partial$$f(x)

 

[CompuChip]

Not if you want to be rigorous.
Because $\delta$ in itself doesn't mean anything, just like $\partial$ doesn't mean anything.

The object you are looking at is $\delta f(x)$.
If you want to be precise, you can use a first order Taylor expansion
$$f(x) \to F(x) = f(a) + f'(a) (x - a) + \mathcal{O}((x - a)^2)$$
where you call $$\delta = f'(a) (x - a)$$ and neglect the quadratic terms.