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Fermat's Principle

  1. Jun 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Need to minimize [tex]\int_{(x_1,y_1)}^{(x_2,y_2)} n(x,y)~ds[/tex] where [tex]n(x,y)=e^y[/tex] and [tex](x_1,y_1)=(-1,1)[/tex], [tex](x_2,y_2)=(1,1)[/tex].

    2. Relevant equations

    Euler-Lagrange equation

    3. The attempt at a solution

    [tex]\frac{d}{dx}\frac{dF}{dy'} - \frac{dF}{dy}=0[/tex]

    [tex]0 - e^y y' = 0[/tex]

    y' = 0 so y = constant or y = 1 considering the initial conditions?
     
    Last edited: Jun 6, 2009
  2. jcsd
  3. Jun 6, 2009 #2

    Dick

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    The independent variable here is time, s is arclength and ds=sqrt(x'^2+y'^2)*dt. So your F is not independent of y'=dy/dt.
     
  4. Jun 7, 2009 #3
    Oh ok, thanks. So,

    [tex]\frac{dF}{dy} = e^y \dot{y} (\dot{x}^2+\dot{y}^2)^{1/2}[/tex]

    [tex]\frac{dF}{d\dot{y}} = e^y \dot{y} (\dot{x}^2+\dot{y}^2)^{-1/2}[/tex]

    But then is [tex]\frac{d}{dx}\frac{dF}{d\dot{y}} = 0[/tex]?
     
  5. Jun 7, 2009 #4

    Dick

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    Thinking about it, I think what they actually want you to do is take x to be the parameter instead of t. ds=sqrt((dx/dt)^2+(dy/dt)^2)*dt=sqrt(1+(dy/dx)^2)*(dx/dt)*dt=sqrt(1+(dy/dx)^2)*dx. Doing it that way is much easier. So F=e^y*sqrt(1+y'^2) where y'=dy/dx. Try finding dF/dy, dF/dy' from that. And why would you think d/dx(dF/dy')=0? It's not a partial derivative.
     
  6. Jun 7, 2009 #5
    [tex]\frac{\partial F}{\partial y'}=e^y y'(1+y'^2)^{-1/2}[/tex]

    [tex]\frac{\partial F}{\partial y}=e^y (1+y'^2)^{1/2}[/tex]

    [tex]\frac{d}{dx}\frac{\partial F}{\partial y'} = e^y y'^2 (1+y'^2)^{-1/2} + e^y y'' (1+y'^2)^{-1/2} - e^y y'^2 y''(1+y'^2)^{-3/2}[/tex]

    [tex]\frac{d}{dx}\frac{\partial F}{\partial y'} - \frac{\partial F}{\partial y} = y'^2 (1+y'^2)^{-1/2} + y'' (1+y'^2)^{-1/2} - y'^2 y''(1+y'^2)^{-3/2} - (1+y'^2)^{1/2} = 0[/tex]
     
    Last edited: Jun 7, 2009
  7. Jun 7, 2009 #6

    Dick

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    dF/dy and dF/dy' look fine. d/dx(dF/dy') takes a bit of work...
     
  8. Jun 7, 2009 #7
    I'm wondering how I'm going to solve this ODE
     
  9. Jun 7, 2009 #8

    Dick

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    I got the same thing you did. Luckily, it simplifies. A lot. Set up dF/dy=d/dx(dF/dy') and multiply both sides by (1+y'^2)^(3/2), divide by e^y, etc. etc.
     
  10. Jun 7, 2009 #9
    [tex]y''-y'^2-1=0[/tex]?
     
  11. Jun 7, 2009 #10

    Dick

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    That's it.
     
  12. Jun 7, 2009 #11
    Thanks for your help.
     
  13. Jun 9, 2009 #12
    Is there a way to not explicitly compute the derivative [tex]\frac{d}{dx}\frac{\partial F}{\partial dy'}[/tex] to make the ODE easier to solve? What's the method for this non-linear second-order?
     
  14. Jun 9, 2009 #13

    Dick

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    That's not so hard so solve. First substitute y'=u and solve for u.
     
  15. Jun 9, 2009 #14
    For some reason I was thinking reduction of order could only be done on linear ODEs. This is nonlinear, true?
     
  16. Jun 9, 2009 #15

    Dick

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    It's nonlinear, true. But there's no reason you can't use a substitution on it! Solve for u and then integrate u to get y.
     
    Last edited: Jun 9, 2009
  17. Jun 9, 2009 #16
    Thanks. I have the full answer now.
     
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