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.

 

[PJC01]

Your explanation is correct. The reason why this statement holds true is because the variation operator, denoted by delta, is a linear operator. This means that it follows the properties of linearity, such as commutativity, which states that the order of operations does not affect the result. In this case, the variation operator is being applied to the partial derivative of a variable, and since it is a linear operator, it can be moved around without affecting the result. This is why we can write the equation as \delta\partial (x) =\partial \delta (x).

Additionally, this statement is a fundamental property in the derivation of the Euler equations. The Euler equations are a set of equations that describe the motion of a fluid, and they are derived using the principle of least action. The variation operator, or delta, is used to represent small changes in the action functional, which is a measure of the total energy of the system. By applying the variation operator to the action functional and setting it equal to zero, we can derive the Euler equations. Thus, the statement \delta\partial (x) =\partial \delta (x) is crucial in the derivation process.

I hope this clarifies the concept for you. If you have any further questions, please do not hesitate to ask.