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Blue_Jaunte
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In my Classical Dynamics text (Thornton & Marion), there's one step in the derivation of Euler's equation that I don't understand. I think if I understood it, I'd be able to derive the equation on my own. I wrote out the entire derivation up to the point I don't get, just so you guys would understand the notation. You can skip to the section in the dotted lines for the specific step (it's about partial derivatives).
So the derivation starts with a quantity, J, that you want to minimize (or maximize). J is a function of the dependent variables y(x), y'(x), and the independent variable x.
[tex]\ J=\int{f(y(x),y'(x);x)dx}[/tex]
y(x) is varied until a minimum of J is found. This is accomplished by adding to y(x) a "neighboring function" such that y can be given the parametric representation:
[tex]y(\alpha, x) = y(0,x) + \alpha\eta(x)[/tex]
where α is a parameter and η(x) is some function of x that vanishes at the endpoints of integration.
Now, J can be written as a functional of α:
[tex]\ J(\alpha)=\int{f(y(\alpha,x),y'(\alpha,x);x)dx}[/tex]
The extreme value can be found, as in basic calculus, by taking the derivative of J with respect to α and setting it equal to zero:
[tex]\frac{\delta J}{\delta\alpha}=\frac{\delta}{\delta\alpha}\int{f(y,y';x)dx}[/tex]
And here's the part I don't understand. It moves the differential operator inside the integral and takes the derivative of the integrand.
[tex]\frac{\delta J}{\delta\alpha}=\int{\left(\frac{\delta f}{\delta y}\frac{\delta y}{\delta\alpha} + \frac{\delta f}{\delta y'}\frac{\delta y'}{\delta\alpha}\right)dx}[/tex]
---------------------------------------
[tex]f = f(y(\alpha,x), y'(\alpha,x); x) [/tex]
[tex]\frac{\delta f}{\delta\alpha}=\frac{\delta f}{\delta y}\frac{\delta y}{\delta\alpha}+\frac{\delta f}{\delta y'}\frac{\delta y'}{\delta\alpha}[/tex]
---------------------------------------
In case you're interested, here's the rest of the derivation:
y:
[tex]y(\alpha,x) = y(x) + \alpha\eta(x) [/tex]
[tex]\frac{\delta y}{\delta\alpha}= \eta(x) [/tex]
y':
[tex]y'(\alpha,x) = \frac{dy}{dx} = y'(x) + \alpha\frac{d\eta}{dx} [/tex]
[tex]\frac{\delta y'}{\delta\alpha} = \frac{d\eta}{dx} [/tex]
And so...
[tex]\frac{\delta J}{\delta\alpha}=\int{\left(\frac{\delta f}{\delta y}\frac{\delta y}{\delta\alpha} + \frac{\delta f}{\delta y'}\frac{\delta y'}{\delta\alpha}\right)dx} = \int{\left(\frac{\delta f}{\delta y}\eta(x) + \frac{\delta f}{\delta y'}\frac{d\eta}{dx}\right)dx}[/tex]
Integrating the second integrand term by parts gives:
[tex]\int{\frac{\delta f}{\delta y'}\frac{d\eta}{dx}dx} = \frac{\delta}{\delta y'}\eta(x) - \int{\frac{d}{dx}\left(\frac{\delta f}{\delta y'}\right)\eta(x)dx} [/tex]
But η=0 at the limits of integration, and so...
[tex]\frac{\delta J}{\delta\alpha} = \int{\left(\frac{\delta f}{\delta y}\eta(x) - \frac{d}{dx}\left(\frac{\delta f}{\delta y'}\right)\eta(x)\right)dx} [/tex]
[tex]=\int{\left(\frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'}\right)\eta(x)dx} = 0 [/tex]
Because η(x) is an arbitrary function (i.e. it can be anything, as long as it's differentiable and vanishes at the endpoints), the quantity in parentheses be equal to 0:
[tex] \frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'} = 0 [/tex]
This is Euler's equation.
Thanks
So the derivation starts with a quantity, J, that you want to minimize (or maximize). J is a function of the dependent variables y(x), y'(x), and the independent variable x.
[tex]\ J=\int{f(y(x),y'(x);x)dx}[/tex]
y(x) is varied until a minimum of J is found. This is accomplished by adding to y(x) a "neighboring function" such that y can be given the parametric representation:
[tex]y(\alpha, x) = y(0,x) + \alpha\eta(x)[/tex]
where α is a parameter and η(x) is some function of x that vanishes at the endpoints of integration.
Now, J can be written as a functional of α:
[tex]\ J(\alpha)=\int{f(y(\alpha,x),y'(\alpha,x);x)dx}[/tex]
The extreme value can be found, as in basic calculus, by taking the derivative of J with respect to α and setting it equal to zero:
[tex]\frac{\delta J}{\delta\alpha}=\frac{\delta}{\delta\alpha}\int{f(y,y';x)dx}[/tex]
And here's the part I don't understand. It moves the differential operator inside the integral and takes the derivative of the integrand.
[tex]\frac{\delta J}{\delta\alpha}=\int{\left(\frac{\delta f}{\delta y}\frac{\delta y}{\delta\alpha} + \frac{\delta f}{\delta y'}\frac{\delta y'}{\delta\alpha}\right)dx}[/tex]
---------------------------------------
[tex]f = f(y(\alpha,x), y'(\alpha,x); x) [/tex]
[tex]\frac{\delta f}{\delta\alpha}=\frac{\delta f}{\delta y}\frac{\delta y}{\delta\alpha}+\frac{\delta f}{\delta y'}\frac{\delta y'}{\delta\alpha}[/tex]
---------------------------------------
In case you're interested, here's the rest of the derivation:
y:
[tex]y(\alpha,x) = y(x) + \alpha\eta(x) [/tex]
[tex]\frac{\delta y}{\delta\alpha}= \eta(x) [/tex]
y':
[tex]y'(\alpha,x) = \frac{dy}{dx} = y'(x) + \alpha\frac{d\eta}{dx} [/tex]
[tex]\frac{\delta y'}{\delta\alpha} = \frac{d\eta}{dx} [/tex]
And so...
[tex]\frac{\delta J}{\delta\alpha}=\int{\left(\frac{\delta f}{\delta y}\frac{\delta y}{\delta\alpha} + \frac{\delta f}{\delta y'}\frac{\delta y'}{\delta\alpha}\right)dx} = \int{\left(\frac{\delta f}{\delta y}\eta(x) + \frac{\delta f}{\delta y'}\frac{d\eta}{dx}\right)dx}[/tex]
Integrating the second integrand term by parts gives:
[tex]\int{\frac{\delta f}{\delta y'}\frac{d\eta}{dx}dx} = \frac{\delta}{\delta y'}\eta(x) - \int{\frac{d}{dx}\left(\frac{\delta f}{\delta y'}\right)\eta(x)dx} [/tex]
But η=0 at the limits of integration, and so...
[tex]\frac{\delta J}{\delta\alpha} = \int{\left(\frac{\delta f}{\delta y}\eta(x) - \frac{d}{dx}\left(\frac{\delta f}{\delta y'}\right)\eta(x)\right)dx} [/tex]
[tex]=\int{\left(\frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'}\right)\eta(x)dx} = 0 [/tex]
Because η(x) is an arbitrary function (i.e. it can be anything, as long as it's differentiable and vanishes at the endpoints), the quantity in parentheses be equal to 0:
[tex] \frac{\delta f}{\delta y} - \frac{d}{dx}\frac{\delta f}{\delta y'} = 0 [/tex]
This is Euler's equation.
Thanks
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