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 [tex]\delta x(t) = \varepsilon f(t), [\tex]and $$\delta x'(t) = \varepsilon f'(t)$$<br /> <br /> we can write <br /> <br /> $$\delta \int L(x,x') dt = \varepsilon \int \frac{dL}{dx}f(t) \ + \frac{dL}{dx'}f'(t)\ dt$$<br /> <br /> he then states the above rule and re-writes this as <br /> $$\varepsilon \int \frac{dL}{dx}f(t) \ - \frac{d}{dt} \frac{dL}{dx'}f(t)\ dt$$<br /> <br /> which would impy that the product of $$\frac{dL}{dx'}f(t)$$ 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's a bit long winded, i'd appreciate your thoughts[/tex]

 

[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 $$A = \int L(x,x') dt$$ 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 [tex]\delta x(t) = \varepsilon f(t), [\tex]and $$\delta x'(t) = \varepsilon f'(t)$$<br /> <br /> we can write <br /> <br /> $$\delta \int L(x,x') dt = \varepsilon \int \frac{dL}{dx}f(t) \ + \frac{dL}{dx'}f'(t)\ dt$$<br /> <br /> he then states the above rule and re-writes this as <br /> $$\varepsilon \int \frac{dL}{dx}f(t) \ - \frac{d}{dt} \frac{dL}{dx'}f(t)\ dt$$<br /> <br /> which would impy that the product of $$\frac{dL}{dx'}f(t)$$ 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's a bit long winded, i'd appreciate your thoughts[/tex]

[tex][/tex]

 

[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.