Undergrad 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, \frac{\partial f}{\partial y} means differentiation of f with respect to its first argument, and \frac{\partial f}{\partial y'} means differentiation of f with respect to its second argument, in both cases with the other argument held constant.

You are differentiating not f, but the composite function g(y,y&#039;,\alpha) = f(y + \alpha \eta, y&#039; + \alpha \eta&#039;) with respect to \alpha; by the chain rule this is then \frac{\partial g}{\partial \alpha} = <br /> \frac{\partial f}{\partial y} \frac{\partial (y + \alpha \eta)}{\partial \alpha} + \frac{\partial f}{\partial y&#039;} \frac{\partial (y&#039; + \alpha \eta&#039;)}{\partial \alpha}.

 

[erobz]

In this case, \frac{\partial f}{\partial y} means differentiation of f with respect to its first argument, and \frac{\partial f}{\partial y&#039;} means differentiation of f with respect to its second argument, in both cases with the other argument held constant.

You are differntiating the composite function f(y + \alpha \eta, y&#039; + \alpha \eta&#039;) with respect to \alpha; by the chain rule this is then <br /> \frac{\partial f}{\partial y} \frac{\partial (y + \alpha \eta)}{\partial \alpha} + \frac{\partial f}{\partial y&#039;} \frac{\partial (y&#039; + \alpha \eta&#039;)}{\partial \alpha}.

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:

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

?