- #1
martyg314
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Derivation of "first integral" Euler-Lagrange equation
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
[tex]
\frac{df}{dx}=\frac{\delta f}{\delta y}y'+\frac{\delta f}{\delta y'}y''
[/tex]
Use the Euler-Lagrange equation to replace [tex]\frac{\delta f}{\delta y}[/tex] on the RHS and show that:
EQUATION 2
[tex]
\frac{df}{dx}=\frac{d}{dx}\left ( y'\frac{\partial f}{\partial y'} \right )
[/tex]
This gives you the first integral:
EQUATION 3
[tex]
f-y'\frac{\partial f}{\partial y'}=constant
[/tex]
E/L equation:
[tex]
\frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y'}
[/tex]
I used the E/L equation to substitute into EQUATION 1 to get:
[tex]
\frac{df}{dx}=\frac{d}{dx}\frac{\partial f}{\partial y'}y'+\frac{\partial f}{\partial y'}y''
[/tex]
Then I used the chain rule to get:
EQUATION 4
[tex]
\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''
[/tex]
Which equals
[tex]
\frac{df}{dx}=y'\frac{d}{dx}\frac{\partial f}{\partial y'}+\frac{\partial f}{\partial y'}y''+\frac{\partial f}{\partial y'}y''
[/tex]
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:
[tex]
\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''
[/tex]
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
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
[tex]
\frac{df}{dx}=\frac{\delta f}{\delta y}y'+\frac{\delta f}{\delta y'}y''
[/tex]
Use the Euler-Lagrange equation to replace [tex]\frac{\delta f}{\delta y}[/tex] on the RHS and show that:
EQUATION 2
[tex]
\frac{df}{dx}=\frac{d}{dx}\left ( y'\frac{\partial f}{\partial y'} \right )
[/tex]
This gives you the first integral:
EQUATION 3
[tex]
f-y'\frac{\partial f}{\partial y'}=constant
[/tex]
Homework Equations
E/L equation:
[tex]
\frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y'}
[/tex]
The Attempt at a Solution
I used the E/L equation to substitute into EQUATION 1 to get:
[tex]
\frac{df}{dx}=\frac{d}{dx}\frac{\partial f}{\partial y'}y'+\frac{\partial f}{\partial y'}y''
[/tex]
Then I used the chain rule to get:
EQUATION 4
[tex]
\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''
[/tex]
Which equals
[tex]
\frac{df}{dx}=y'\frac{d}{dx}\frac{\partial f}{\partial y'}+\frac{\partial f}{\partial y'}y''+\frac{\partial f}{\partial y'}y''
[/tex]
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:
[tex]
\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''
[/tex]
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