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Find the equation

  1. Feb 27, 2006 #1
    Find the (generalized Euler) equation which a function y(x) must satisify in order that the action
    [tex] A = \int_{x_{1}}^{x_{2}} L(y(x),y'(x,),y''(x),x) dx [/tex] [/tex]
    be stationary that is [itex] \delta A = 0 [/itex] for arbitrary variations [itex] \delta y(x) [/itex] such taht
    [tex] \delta y(x) = \frac{d}{dx} \delta y(x)=0 \mbox{when} \ x=x_{1},x_{2} [/tex]
    Hint: [tex] \delta A = \int_{x_{1}}^{x_{2}} \left(\frac{\partial L}{\partial y} \delta y + \frac{\partial L}{\partial y'} \delta y'+ \frac{\partial L}{\partial y''} \delta y'' \right) dx, \mbox{where} \ \delta y' = \frac{d}{dx} \delta y \mbox{ and } \delta y'' = \frac{d}{dx} \delta y' = \frac{d^2}{dx^2} \delta y [/tex]


    by generalized Euler equation does the question ask the Euler Langrange equation?
    well in the hint the first term is zero...
    so we have
    [tex] \delta A = \int_{x_{1}}^{x_{2}} \left(\frac{\partial L}{\partial y} \delta y + \frac{\partial L}{\partial y'} \delta y'+ \frac{\partial L}{\partial y''} \delta y'' \right) dx = \int_{x_{1}}^{x_{2}} \left(\frac{\partial L}{\partial y} \frac{d}{dx} \delta y + \frac{\partial L}{\partial y''} \frac{d}{dx} \delta y'\right) dx = 0 [/tex]
    integration by parts so
    [tex] \left[ {\frac{\partial L}{\partial y} \delta y} \right]_{x_{1}}^{x_{2}} - \left[\frac{\partial^2 L}{\partial y \partial x} \delta y} \right]_{x_{1}}^{x_{2}} + \left[ \frac{\partial L}{\partial y''} \delta y'} \right]_{x_{1}}^{x_{2}} - \left[ \frac{\partial^2 L}{\partial y'' \partial x} \delta y'} \right]_{x_{1}}^{x_{2}} = 0 [/tex]
    not too sure about whrer this is going...
    Will continued working of thie problem yield the EUler Lagrange equations?
    Please help!
    Thank you
     
    Last edited: Feb 27, 2006
  2. jcsd
  3. Feb 27, 2006 #2

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    No it isn't.

    You want to integrate by parts to get rid of the [itex]\delta y'[/itex] and [itex]\delta y''[/itex]. The reason is, roughly speaking, we aren't free to vary these. However, we are free to vary [itex]\delta y[/itex]. You want to integrate by parts until you have some boundary terms, which should vanish since the endpoints are fixed, and an integral of some expression times [itex]\delta y[/itex]. Then you can use the arbitrariness of [itex]\delta y[/itex] to argue that for this integral to vanish, the expression multiplying [itex]\delta y[/itex] must be zero for all x.

    This is just a simple extension of the usual case, so I'd suggest going back over the derivation of the original Euler-Lagrange equation in your book.
     
    Last edited: Feb 27, 2006
  4. Feb 28, 2006 #3
    but isnt delta y (x) = 0 in the question? Doestnt the fndamental lemma follow?

    well ok integration by parts
    [tex] \delta A = \int_{x_{1}}^{x_{2}} \frac{\partial L}{\partial y} \delta y dx + \left[\frac{\partial L}{\partial y'} \delta y dx - \frac{\partial^2 L}{\partial y' \partial x} \delta y\right]_{x_{1}}^{x_{2}} + \left[\frac{\partial L}{\partial y''} \frac{d}{dx}\delta y - \frac{\partial^2 L}{\partial y'' \partial x} \frac{d}{dx} \delta y dx \right]_{x_{1}}^{x_{2}} [/tex]
    ok the first term is not zero... i m not sure why not delta y(x) = 0 isnt it?

    now i guess i could make the second and third term into the Euler Lagrange equation by saying that the second bracketed term is dM/dq and then the third bracketed term is d/dx (dM/dq) and that yield the euler lagrange... is that going in teh right direction?
    doesht that become zero afterward? So we are left iwth only
    [tex] \delta A = \int_{x_{1}}^{x_{2}} \frac{\partial L}{\partial y} \delta y dx [/tex]
    is that hte kind of expression we want in the first place?
     
  5. Feb 28, 2006 #4
    isnt delta y (x) = 0 at x1 and x2?
    SO doest the first term equal zero??
     
  6. Feb 28, 2006 #5

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    [itex]\delta y [/itex] is zero at the endpoints, yes, but not in between, so you can't assume the integral vanishes.

    You're on the right track with that equation you wrote in your last post, although there are a few mistakes. Remember from the derivation of the original Euler-Lagrange equations that:

    [tex]\begin{align} \int_{x_1}^{x_2} \frac{\partial L}{\partial y'} \delta y' dx

    &= \int_{x_1}^{x_2} \left( \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \delta y \right) - \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) \delta y \right) dx \\

    &=\left[ \frac{\partial L}{\partial y'} \delta y \right]_{x_1}^{x_2} - \int_{x_1}^{x_2} \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) \delta y dx \end{align} [/tex]

    You can argue that the boundary term vanishes. You will have to do something like this for the [itex] \frac{\partial L}{\partial y''}[/itex] term, although there will be an extra step.
     
    Last edited: Feb 28, 2006
  7. Feb 28, 2006 #6
    but waht about the Fundamental Lemma of the calculus of variations?
    doesnt it say just that... taht if delta y =0 then the integral is zero?
     
  8. Feb 28, 2006 #7
    well anyway... to correct what i posted in number 3

    [tex] \delta A = \int_{x_{1}}^{x_{2}} \frac{\partial L}{\partial y} \delta y dx + \left[\frac{\partial L}{\partial y'} \delta y dx \right]_{x_{1}}^{x_{2}} - \int_{x_{1}^{x_{2}} \frac{d}{dx} \frac{\partial L}{\partial y'} \frac{d}{dx
    \delta y dx + \left[\frac{\partial L}{\partial y''} \frac{d}{dx}\delta y \right]_{x_{1}}^{x_{2}} - \int_{x_{1}}^{x_{2}} \frac{d}{dx} \frac{\partial L}{\partial y''} \frac{d^2}{dx^2} \delta y dx [/tex]
    does this address the problems?

    in what you posted... whaat is the boundary term? is it
    [tex] \left[\frac{\partial L}{\partial y'} \delta y\right]_{x_{1}}^{x_{2}} [/tex]??
     
    Last edited: Feb 28, 2006
  9. Feb 28, 2006 #8

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    No, look carefully at what I did. You're evaluating the integrand at the endpoints where you should be integrating it. And the fundamental lemma states that if:

    [tex]\int_{x_1}^{x_2} f(y,...,x)\delta y dx = 0 [/tex]

    where [itex]\delta y[/itex] is arbitrary, then f(y,...,x)=0. You don't have something in this form yet, but that's what you need to get. And yes, that is what I was referring to as the boundary term. You can see it vanishes precisely because [itex]\delta y[/itex]=0 at the endpoints.
     
  10. Feb 28, 2006 #9
    ah i see
    i got the answer
    it is a second order ODE wrt dl/dq where q is the coordinate

    thanks for your help!
    if you can help with the particle in a field question... thats all the help i would need for tonight!

    thank you
     
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