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
<br /> \frac{df}{dx}=\frac{\delta f}{\delta y}y&#039;+\frac{\delta f}{\delta y&#039;}y&#039;&#039;<br />

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

EQUATION 2
<br /> \frac{df}{dx}=\frac{d}{dx}\left ( y&#039;\frac{\partial f}{\partial y&#039;} \right )<br />

This gives you the first integral:

EQUATION 3
<br /> f-y&#039;\frac{\partial f}{\partial y&#039;}=constant<br />

Homework Equations

E/L equation:

<br /> \frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y&#039;}<br />

The Attempt at a Solution

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

<br /> \frac{df}{dx}=\frac{d}{dx}\frac{\partial f}{\partial y&#039;}y&#039;+\frac{\partial f}{\partial y&#039;}y&#039;&#039;<br />

Then I used the chain rule to get:

EQUATION 4
<br /> \frac{df}{dx}=y&#039;\frac{d}{dx}\frac{\partial f}{\partial y&#039;}+\frac{\partial f}{\partial y&#039;}\frac{d}{dx}y&#039;+\frac{\partial f}{\partial y&#039;}y&#039;&#039;<br />

Which equals

<br /> \frac{df}{dx}=y&#039;\frac{d}{dx}\frac{\partial f}{\partial y&#039;}+\frac{\partial f}{\partial y&#039;}y&#039;&#039;+\frac{\partial f}{\partial y&#039;}y&#039;&#039;<br />


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:

<br /> \frac{df}{dx}=y&#039;\frac{d}{dx}\frac{\partial f}{\partial y&#039;}-\frac{\partial f}{\partial y&#039;}\frac{d}{dx}y&#039;+\frac{\partial f}{\partial y&#039;}y&#039;&#039;<br />


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:

<br /> \frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y&#039;}<br />

The Attempt at a Solution

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

<br /> \frac{df}{dx}=\frac{d}{dx}\frac{\partial f}{\partial y&#039;}y&#039;+\frac{\partial f}{\partial y&#039;}y&#039;&#039;<br />

You should really write this as

<br /> \frac{df}{dx}=\left(\frac{d}{dx}\frac{\partial f}{\partial y&#039;}\right)y&#039;+\frac{\partial f}{\partial y&#039;}y&#039;&#039;~~(*)<br />

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

Then I used the chain rule to get:

EQUATION 4
<br /> \frac{df}{dx}=y&#039;\frac{d}{dx}\frac{\partial f}{\partial y&#039;}+\frac{\partial f}{\partial y&#039;}\frac{d}{dx}y&#039;+\frac{\partial f}{\partial y&#039;}y&#039;&#039;<br />

No, What you want to do is add and subtract

\frac{d}{dx} \left(\frac{\partial f}{\partial y&#039;} y&#039; \right) = \left(\frac{d}{dx} \frac{\partial f}{\partial y&#039;} \right) y&#039; + \frac{\partial f}{\partial y&#039;} y&#039;&#039;

from the RHS of (*):

<br /> \frac{df}{dx}=\left(\frac{d}{dx}\frac{\partial f}{\partial y&#039;}\right)y&#039;+\frac{\partial f}{\partial y&#039;}y&#039;&#039; + \frac{d}{dx} \left(\frac{\partial f}{\partial y&#039;} y&#039; \right) - \left(\frac{d}{dx} \frac{\partial f}{\partial y&#039;} \right) y&#039; - \frac{\partial f}{\partial y&#039;} y&#039;&#039;<br />

and simplify.

 

[Dustinsfl]

<br /> \frac{df}{dx}=\left(\frac{d}{dx}\frac{\partial f}{\partial y&#039;}\right)y&#039;+\frac{\partial f}{\partial y&#039;}y&#039;&#039; + \frac{d}{dx} \left(\frac{\partial f}{\partial y&#039;} y&#039; \right) - \left(\frac{d}{dx} \frac{\partial f}{\partial y&#039;} \right) y&#039; - \frac{\partial f}{\partial y&#039;} y&#039;&#039;<br />

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.