Is My Calculus of Variations Approach Correct?

[erobz]

I would like to use the Calculus of Variations to show the minimum path connecting two points is a straight line, but I wish to do it from scratch without using the pre-packaged general result, because I'm having some trouble following it.

Points are ##(x_1,y_1),(x_2,y_2)##.

And we are to minimize this integral, whre ##y(x)## is the minimum path:

$$ \int_{x_1}^{x_2} ds = \int_{x_1}^{x_2} \sqrt{ 1 + y'(x)^2} dx $$

Then you make a new curve:

$$ Y(x) = y(x) + \beta \eta (x) $$

Differentiate ##Y(x)##:

$$ Y'(x) = y'(x) + \beta \eta '(x) $$

Sub into the integral and expand (dropping the function notation):

$$ I = \int_{x_1}^{x_2} \sqrt{ 1 + Y'(x)^2} dx = \int_{x_1}^{x_2} \sqrt{ 1 + y'^2 + 2 \beta y' \eta '+ \beta^2 \eta ' ^2 } dx $$

This you are supposed to take the derivative w.r.t. ##\beta##

$$ \frac{dI}{d \beta} = \int_{x_1}^{x_2} \frac{d}{d \beta}\left( \sqrt{ 1 + y'^2 + 2 \beta y' \eta '+ \beta^2 \eta ' ^2 } \right) dx $$

Am I doing this correctly to this point?

 

[wrobel]

Do not forget that ##\eta(x_1)=\eta(x_2)=0##.
It seems that deducing the Lagrange equations in general form takes less writing than this masochism

 

[erobz]

Do not forget that ##\eta(x_1)=\eta(x_2)=0##.
It seems that deducing the Lagrange equations in general form takes less writing than this masochism

Agreed, and that's probably why I don't find example worked this way. But Im trying to get to the bottom of something.

My problem is that In Classical Mechanics by Taylor there is a point in the derivation of the Euler-Lagrange Equation that he says:

$$ \frac{\partial f ( y+\alpha \eta, y' + \alpha \eta ', x) }{\partial \alpha} = \eta \frac{\partial f }{ \partial y} + \eta ' \frac{\partial f }{ \partial y'} $$

I can't figure it out! ##y## doesn't depend on ##\alpha##?

 

[wrobel]

y does not depend on ##\alpha## yes

 

[erobz]

y does not depend on ##\alpha## yes

Then that result is a mistake with notation?

 

[wrobel]

what "that"?
y is the fixed trajectory and ##y+\alpha \eta## its perturbation

 

[erobz]

what "that"?

post 3 is taken from Taylor.

 

[wrobel]

I do not see a problem with the formula from #3

 

[erobz]

I do not see a problem with the formula from #3

we are to be differentiating ##f## with respect to ##\alpha##. ##y## does not depend on ##\alpha##.

 

[wrobel]

we are to be differentiating f with respect to α. y does not depend on α.

exactly!

 

[erobz]

exactly!

Then #3 cannot be correct...

 

[pasmith]

we are to be differentiating ##f## with respect to ##\alpha##. ##y## does not depend on ##\alpha##.

In this case, [itex]\frac{\partial f}{\partial y}[/itex] means differentiation of [itex]f[/itex] with respect to its first argument, and [itex]\frac{\partial f}{\partial y'}[/itex] means differentiation of [itex]f[/itex] with respect to its second argument, in both cases with the other argument held constant.

You are differentiating not [itex]f[/itex], but the composite function [tex]g(y,y',\alpha) = f(y + \alpha \eta, y' + \alpha \eta')[/tex] with respect to [itex]\alpha[/itex]; by the chain rule this is then [tex]\frac{\partial g}{\partial \alpha} =
\frac{\partial f}{\partial y} \frac{\partial (y + \alpha \eta)}{\partial \alpha} + \frac{\partial f}{\partial y'} \frac{\partial (y' + \alpha \eta')}{\partial \alpha}.[/tex]

 

[erobz]

In this case, [itex]\frac{\partial f}{\partial y}[/itex] means differentiation of [itex]f[/itex] with respect to its first argument, and [itex]\frac{\partial f}{\partial y'}[/itex] means differentiation of [itex]f[/itex] with respect to its second argument, in both cases with the other argument held constant.

You are differntiating the composite function [itex]f(y + \alpha \eta, y' + \alpha \eta')[/itex] with respect to [itex]\alpha[/itex]; by the chain rule this is then [tex]
\frac{\partial f}{\partial y} \frac{\partial (y + \alpha \eta)}{\partial \alpha} + \frac{\partial f}{\partial y'} \frac{\partial (y' + \alpha \eta')}{\partial \alpha}.[/tex]

The argument being ##y + \alpha \eta ##.

So it is a typo. He defines ## Y(x) = y(x) + \alpha \eta ## earlier, and what they really meant to say is:

$$ \frac{ \partial f ( Y,Y',x) }{\partial \alpha} = \eta \frac{\partial f }{ \partial Y} + \eta ' \frac{\partial f }{ \partial Y'}$$

?