Derivation of first integral Euler-Lagrange equation

[martyg314]

Derivation of "first integral" Euler-Lagrange equation

Homework Statement

This is from Classical Mechanics by John Taylor, Problem 6.20:

Argue that if it happens that f(y,y',x) does not depend on x then:

EQUATION 1
$$\frac{df}{dx}=\frac{\delta f}{\delta y}y'+\frac{\delta f}{\delta y'}y''$$

Use the Euler-Lagrange equation to replace $$\frac{\delta f}{\delta y}$$ on the RHS and show that:

EQUATION 2
$$\frac{df}{dx}=\frac{d}{dx}\left ( y'\frac{\partial f}{\partial y'} \right )$$

This gives you the first integral:

EQUATION 3
$$f-y'\frac{\partial f}{\partial y'}=constant$$

Homework Equations

E/L equation:

$$\frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y'}$$

The Attempt at a Solution

I used the E/L equation to substitute into EQUATION 1 to get:

$$\frac{df}{dx}=\frac{d}{dx}\frac{\partial f}{\partial y'}y'+\frac{\partial f}{\partial y'}y''$$

Then I used the chain rule to get:

EQUATION 4
$$\frac{df}{dx}=y'\frac{d}{dx}\frac{\partial f}{\partial y'}+\frac{\partial f}{\partial y'}\frac{d}{dx}y'+\frac{\partial f}{\partial y'}y''$$

Which equals

$$\frac{df}{dx}=y'\frac{d}{dx}\frac{\partial f}{\partial y'}+\frac{\partial f}{\partial y'}y''+\frac{\partial f}{\partial y'}y''$$


However, based on solutions I've seen the two y'' terms should cancel, leaving EQUATION 2

The solution I saw showed that after the substitution, EQUATION 4 is instead:

$$\frac{df}{dx}=y'\frac{d}{dx}\frac{\partial f}{\partial y'}-\frac{\partial f}{\partial y'}\frac{d}{dx}y'+\frac{\partial f}{\partial y'}y''$$


Is there some chain rule identity of some sort I'm missing? I just don't see where the negative comes from to cancel the two terms.

Thanks,

MG

 

[fzero]

E/L equation:

$$\frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y'}$$

The Attempt at a Solution

I used the E/L equation to substitute into EQUATION 1 to get:

$$\frac{df}{dx}=\frac{d}{dx}\frac{\partial f}{\partial y'}y'+\frac{\partial f}{\partial y'}y''$$

You should really write this as

$$\frac{df}{dx}=\left(\frac{d}{dx}\frac{\partial f}{\partial y'}\right)y'+\frac{\partial f}{\partial y'}y''~~(*)$$

since you're getting confused about where that derivative acts.

Then I used the chain rule to get:

EQUATION 4
$$\frac{df}{dx}=y'\frac{d}{dx}\frac{\partial f}{\partial y'}+\frac{\partial f}{\partial y'}\frac{d}{dx}y'+\frac{\partial f}{\partial y'}y''$$

No, What you want to do is add and subtract

$$\frac{d}{dx} \left(\frac{\partial f}{\partial y'} y' \right) = \left(\frac{d}{dx} \frac{\partial f}{\partial y'} \right) y' + \frac{\partial f}{\partial y'} y''$$

from the RHS of (*):

$$\frac{df}{dx}=\left(\frac{d}{dx}\frac{\partial f}{\partial y'}\right)y'+\frac{\partial f}{\partial y'}y'' + \frac{d}{dx} \left(\frac{\partial f}{\partial y'} y' \right) - \left(\frac{d}{dx} \frac{\partial f}{\partial y'} \right) y' - \frac{\partial f}{\partial y'} y''$$

and simplify.

 

[Dustinsfl]

$$\frac{df}{dx}=\left(\frac{d}{dx}\frac{\partial f}{\partial y'}\right)y'+\frac{\partial f}{\partial y'}y'' + \frac{d}{dx} \left(\frac{\partial f}{\partial y'} y' \right) - \left(\frac{d}{dx} \frac{\partial f}{\partial y'} \right) y' - \frac{\partial f}{\partial y'} y''$$

and simplify.

Where did
$$
\frac{d}{dx} \left(\frac{\partial f}{\partial y'} y' \right)
$$
this piece come from?

 

[fzero]

Where did
$$
\frac{d}{dx} \left(\frac{\partial f}{\partial y'} y' \right)
$$
this piece come from?

We're adding 0 to the equation that I called (*), in the form of the identity for the total derivative of $\partial f/\partial y'$. The last 3 terms in the equation that you quoted cancel amongst themselves.

 

[paranormal]

I can provide a response to the derivation of the first integral Euler-Lagrange equation. The first integral Euler-Lagrange equation is derived by substituting the Euler-Lagrange equation into a given equation, and then simplifying the resulting equation. In this case, the given equation is EQUATION 1, which does not depend on x. This means that the partial derivative of f with respect to x is zero. By substituting the Euler-Lagrange equation into EQUATION 1 and simplifying, we arrive at EQUATION 4. The key to understanding why the two y'' terms cancel is to recognize that the second term in EQUATION 4, \frac{\partial f}{\partial y'}y'', is actually the same as the third term, \frac{\partial f}{\partial y'}y''. This is because the chain rule tells us that \frac{d}{dx}y'=\frac{dy}{dx}=\frac{d}{dx}y, which means that the second and third terms are equal. Therefore, they can be combined and the two y'' terms cancel out, leaving us with EQUATION 2. This shows that the first integral of the Euler-Lagrange equation, EQUATION 3, is indeed a constant, as it does not depend on the independent variable x.