Integration by parts, can you do this?

[earlofwessex]

I've seen this formula stated and used, ( in a stanford university video lecture)

\int \frac{dA}{dt}B\ dt = - \int \frac{dB}{dt}A\ dt

with the condition that you don't vary the end points.

but i don't understand how you can just remove the AB term from the right hand side, and I've not been able to find this written anywhere? i know the normal parts rule.

can anyone explain?

thanks

 

[LCKurtz]

I'll take a guess at it, not knowing the context. If you have

\int_a^b f'g\ dt = -\int_a^b fg'\ dt

then

0 = \int_a^b f'g + fg'\ dt= \int_a^b (fg)'\ dt = f(b)g(b)-f(a)g(a)

So if the product fg has the same values at a and b, it all works. Is that what you mean by "not varying the endpoints"?

 

[earlofwessex]

thanks LCKurtz,

hmm, that's true for this, though it seems a bit circular. the context is:
consider a trajectory between two points in space-time, where X is some function of t.

the action is described by A = \int L(x,x') dt where L is the Lagrangian, which depends on both position and velocity.
(is this an integral along the line or something else?)

in order to find the path of least action, we vary x(t) -> x(t) + \varepsilon f(t) and require
\delta A = \delta \int L(x,x') dt = 0

since \delta x(t) = \varepsilon f(t), [\tex]and \delta x&amp;#039;(t) = \varepsilon f&amp;#039;(t)<br /> <br /> we can write <br /> <br /> \delta \int L(x,x&amp;#039;) dt = \varepsilon \int \frac{dL}{dx}f(t) \ + \frac{dL}{dx&amp;#039;}f&amp;#039;(t)\ dt<br /> <br /> he then states the above rule and re-writes this as <br /> \varepsilon \int \frac{dL}{dx}f(t) \ - \frac{d}{dt} \frac{dL}{dx&amp;#039;}f(t)\ dt<br /> <br /> which would impy that the product of \frac{dL}{dx&amp;#039;}f(t) is the same for any t. um, which is true since he goes on to show that f(t) = 0 for any t, but he uses that final expression to show it. <br /> <br /> besides, he states the rule for a general case, not specific to this situation.<br /> <br /> sorry if that&#039;s a bit long winded, i&#039;d appreciate your thoughts

 

[ExtravagantDreams]

You derivation is not quite correct. I would advise you to see Goldstein (Classical Mechanics). It has a thorough explanation of Lagrangian mechanics starting with the variational principle.

You will be making variations with respect to \varepsilon, using x(t, \varepsilon ) = x(t, 0) + \varepsilon f(t), where x(t, 0) is the true solution and f(t) can be any function that vanishes at the end points.

When you take the derivative of the action w.r.t \varepsilon, using integration by parts, the second term will give you;
\int dt \frac{dL}{d\dot{x}} \frac{d\dot{x}}{d\varepsilon} = \frac{dL}{d\dot{x}} \frac{dx}{d\varepsilon} | - \int dt \frac{d}{dt} (\frac{dL}{d\dot{x}}) \frac{dx}{d\varepsilon}

The first of the terms on the right size is evaluated at the end points, and since \frac{dx}{d\varepsilon} = f(t) vanishes there, this term is zero.

thanks LCKurtz,

hmm, that's true for this, though it seems a bit circular. the context is:
consider a trajectory between two points in space-time, where X is some function of t.

the action is described by A = \int L(x,x&#039;) dt where L is the Lagrangian, which depends on both position and velocity.
(is this an integral along the line or something else?)

in order to find the path of least action, we vary x(t) -&gt; x(t) + \varepsilon f(t) and require
\delta A = \delta \int L(x,x&#039;) dt = 0

since \delta x(t) = \varepsilon f(t), [\tex]and \delta x&amp;#039;(t) = \varepsilon f&amp;#039;(t)<br /> <br /> we can write <br /> <br /> \delta \int L(x,x&amp;#039;) dt = \varepsilon \int \frac{dL}{dx}f(t) \ + \frac{dL}{dx&amp;#039;}f&amp;#039;(t)\ dt<br /> <br /> he then states the above rule and re-writes this as <br /> \varepsilon \int \frac{dL}{dx}f(t) \ - \frac{d}{dt} \frac{dL}{dx&amp;#039;}f(t)\ dt<br /> <br /> which would impy that the product of \frac{dL}{dx&amp;#039;}f(t) is the same for any t. um, which is true since he goes on to show that f(t) = 0 for any t, but he uses that final expression to show it. <br /> <br /> besides, he states the rule for a general case, not specific to this situation.<br /> <br /> sorry if that&#039;s a bit long winded, i&#039;d appreciate your thoughts

 

[ExtravagantDreams]

So, to generalize it, \int \frac{dA}{dt}B\ dt = - \int \frac{dB}{dt}A\ dt is true if one of the functions A or B vanishes at both endpoints, which is what LCKurtz showed. You just have to remember in the variational principle A and B will be derivatives.

I've seen this formula stated and used, ( in a stanford university video lecture)

\int \frac{dA}{dt}B\ dt = - \int \frac{dB}{dt}A\ dt

with the condition that you don't vary the end points.

but i don't understand how you can just remove the AB term from the right hand side, and I've not been able to find this written anywhere? i know the normal parts rule.

can anyone explain?

thanks

 

[earlofwessex]

ah ok, thanks that makes a lot of sense.

Goldstein (Classical Mechanics) is a textbook right?
what level of calculus do i need to follow it? i think i was confused above because I'm not familiar with the "types" of integral, open surface, closed loop and so on, just with everyday definite and indefinite area under a 2d curve. I'm definitely nowhere near vector or field calculus.

thanks

 

[ExtravagantDreams]

Yes, it's a book. Usually for graduate level, but the section on calculus of variations is just an ellaboration of what you've probably already learned.

This method uses integrating along a parametized line, which is something you learn early on in vector calculus. I think what throws many people for the first time is that it's a curve through phase-space, so it's difficult to visualize.