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Integration by parts, can you do this?

  1. Jan 28, 2010 #1
    I've seen this formula stated and used, ( in a stanford university video lecture)

    [tex] \int \frac{dA}{dt}B\ dt = - \int \frac{dB}{dt}A\ dt [/tex]

    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
     
  2. jcsd
  3. Jan 28, 2010 #2

    LCKurtz

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

    [tex]\int_a^b f'g\ dt = -\int_a^b fg'\ dt[/tex]

    then

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

    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"?
     
  4. Jan 29, 2010 #3
    thanks LCKurtz,

    hmm, thats 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 [tex] A = \int L(x,x') dt [/tex] 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 [tex] x(t) -> x(t) + \varepsilon f(t) [/tex] and require
    [tex] \delta A = \delta \int L(x,x') dt = 0 [/tex]

    since [tex] \delta x(t) = \varepsilon f(t), [\tex]and [tex] \delta x'(t) = \varepsilon f'(t)[/tex]

    we can write

    [tex] \delta \int L(x,x') dt = \varepsilon \int \frac{dL}{dx}f(t) \ + \frac{dL}{dx'}f'(t)\ dt [/tex]

    he then states the above rule and re-writes this as
    [tex]\varepsilon \int \frac{dL}{dx}f(t) \ - \frac{d}{dt} \frac{dL}{dx'}f(t)\ dt[/tex]

    which would impy that the product of [tex]\frac{dL}{dx'}f(t) [/tex] 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.

    besides, he states the rule for a general case, not specific to this situation.

    sorry if thats a bit long winded, i'd appreciate your thoughts
     
  5. Feb 1, 2010 #4
    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 [tex]\varepsilon [/tex], using [tex] x(t, \varepsilon ) = x(t, 0) + \varepsilon f(t) [/tex], 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 [tex]\varepsilon [/tex], using integration by parts, the second term will give you;
    [tex] \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}[/tex]

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


     
  6. Feb 1, 2010 #5
    So, to generalize it, [tex] \int \frac{dA}{dt}B\ dt = - \int \frac{dB}{dt}A\ dt [/tex] 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.


     
  7. Feb 2, 2010 #6
    ah ok, thanks that makes a lot of sense.

    Goldstein (Classical Mechanics) is a text book 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
     
  8. Feb 4, 2010 #7
    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.
     
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