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.