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Hey guys, take a look at my work for a stationary integral?

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data

    Path between O=(0,0) and P=(0,4) for which the integral

    int. O to P (x (1-y'^2))^(1/2) dx

    becomes stationary.


    2. Relevant equations

    http://en.wikipedia.org/wiki/Calculus_of_variations#Example

    3. The attempt at a solution

    Okay guys,

    so mimicking the example from wiki but with L= (x)^1/2 (1-y'^2)^1/2

    dL/dy=0 and dL/dy'= -(x)^1/2 y' / (1-y'^2)^1/2 and then d/dx dL/df' = - y'/[(2x^1/2)(1-y'^2)^1/2]

    and since d/dx dL/df' =0 this implies y'=0 and we integrate to get y=c, adhering to boundary conditions y must equal 0... was that the correct way sorry for not writing in latex if I must I will redo the post
     
  2. jcsd
  3. Nov 13, 2012 #2

    haruspex

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    That last expression looks too simple. There should be a y" in there, and two different powers of x. Remember it's not a partial derivative this time.
     
  4. Nov 13, 2012 #3
    ahhhh you are correct! y is a function of x..
     
  5. Nov 13, 2012 #4
    well.. I used the product rule and the chain rule and now I am looking at an unbelievable

    2 x^1/2 y'^2 y'' - 1/2 x^-1/2 y' - x^1/2 y'' + 1/2 x^-1/2 y'^3 =0 .

    I believe I took the derivative correctly this time but now I am lost as what to do with this..
     
  6. Nov 13, 2012 #5
    well.. I used the product rule and the chain rule and now I am looking at an unbelievable

    2 x^1/2 y'^2 y'' - 1/2 x^-1/2 y' - x^1/2 y'' + 1/2 x^-1/2 y'^3 =0 .

    I believe I took the derivative correctly this time but now I am lost as what to do with this.
     
  7. Nov 13, 2012 #6

    haruspex

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    Allow me to put that in LaTex:

    [tex]2 x^{1/2} y'^2 y'' - (1/2)x^{-1/2} y' - x^{1/2} y'' + (1/2) x^{-1/2} y'^3 =0
    [/tex]
    Simplifying:
    [tex]4 x y'^2 y'' - y' - 2x y'' + y'^3 =0 [/tex]
    But that's not quite what I get. Writing K = (1-y'2)1/2:
    L = x1/2K
    ∂K/∂y' = -y'/K
    ∂L/∂y' = -x1/2y'/K
    [tex]0 = -\frac{d(∂L/∂y')}{dx} = \frac{x^{-1/2}y'}{2K} + \frac{x^{1/2}y''}{K} - \frac{x^{1/2}y'\frac{∂K}{∂y'}y''}{K^2}[/tex]
    [tex]0 = y'+ 2xy'' - 2\frac{xy'\frac{∂K}{∂y'}y''}{K}[/tex]
    [tex]0 = y'+ 2xy'' + 2\frac{xy'^2y''}{K^2}[/tex]
    [tex]0 = (y'+ 2xy'')(1-y'^2) + 2xy'^2y''[/tex]
    [tex]0 = y'+ 2xy'' - y'^3[/tex]
    Is that right or did I get sign wrong somewhere?
     
  8. Nov 13, 2012 #7
    hey thanks, jumped out of bed to respond.. I'm sure you have the math correct but I can't see how this helps us get a solution the the original problem. At this step we have a nonlinear ode so I'm a little frightened we are on the wrong track? Thank you for helping thus far
     
  9. Nov 13, 2012 #8

    haruspex

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    It's not hard from there. There are no y terms, so it's first order, and you can get it into the form f(x) = g(t)t' where t = y', so then it's just an integration to get an equation relating x to y'.
     
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