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Simple existence/uniqueness proof for Newton's 2nd?

  1. Jun 6, 2010 #1

    Fredrik

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    Is there a simple proof of the existence and uniqueness of a solution of [itex]x''(t)=f(x'(t),x(t),t)[/itex], where f is an appropriately "nice" function? (Given an initial condition [itex]x(t_0)=x_0,\ x'(t_0)=v_0[/itex], or a boundary condition [itex]x(t_0)=x_0,\ x(t_1)=x_1[/itex]).
     
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  3. Jun 6, 2010 #2

    Office_Shredder

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    It's called Picard's theorem. It relates to first order differential equations, but through using vectors can solve a system of them. Any second order differential equation can be turned into a system of first order differential equations by the following (using your differential equation as the starting point)

    [tex] z_1(t)=x(t)[/tex] and [tex]z_2(t)=x'(t)[/tex] we get

    [tex]z_1'(t)=z_2(t)[/tex]

    and

    [tex] z_2'(t)=f(z_1(t), z_2(t),t)[/tex]

    Here's a wikipedia link to the theorem
    http://en.wikipedia.org/wiki/Picard–Lindelöf_theorem
     
  4. Jun 6, 2010 #3

    Fredrik

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    Thanks. I've seen references to Picard's theorem and I own a book that proves it, but I thought it was more general than that. I was hoping that there would exist a theorem with a particularly simple proof that covers the cases that are interesting for physics, but I guess not.

    What theorem do we use when we've been given boundary conditions instead of an initial condition? There's still a unique solution, right?
     
  5. Jun 6, 2010 #4

    Office_Shredder

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    When you're given boundary conditions it's a lot tougher. I forgot to mention a general technique for solving odes:
    http://en.wikipedia.org/wiki/Method_of_variation_of_parameters

    But you can have problems. Let's take the easy equation [tex]y''+y=0[/tex]. So [tex]y=Acos(x)+Bsin(x)[/tex]. Let's furthermore take the boundary conditions [tex]y(0)=0[/tex] and [tex]y(\pi)=0[/tex]. Then all we get is that the solution is something of the form [tex]y=Asin(x)[/tex]

    If instead we had [tex]y(0)=0[/tex] and [tex]y(\pi)=1[/tex] there would be no solution.
     
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