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How to generalize Newtonian and Lagrangian mechanics

  1. Sep 10, 2014 #1
    If I stated a problem that you have to find the solution

    [tex]
    [0,\infty[\;\to\mathbb{R},\quad t\mapsto x(t)
    [/tex]

    to the problem

    [tex]
    x(0) = x_0 < R
    [/tex]
    [tex]
    \dot{x}(0) = v_0 > 0
    [/tex]
    [tex]
    m\ddot{x}(t) = -\partial_x U\big(x(t)\big),\quad\quad m>0
    [/tex]

    where [itex]R, v_0, m[/itex] are some constants, and the function [itex]U[/itex] has been defined as

    [tex]
    U(x) = \left\{\begin{array}{ll}
    0, & x\leq R \\
    \Omega, & x > R \\
    \end{array}\right.
    [/tex]

    most people recognizing this as related to physics would probably state the solution as follows:

    If [itex]\frac{1}{2}mv_0^2 <\Omega[/itex], then

    [tex]
    x(t) = \left\{\begin{array}{ll}
    x_0 + v_0 t,\quad &t\leq \frac{R-x_0}{v_0} \\
    R - v_0\big(t - \frac{R-x_0}{v_0}\big),\quad &t > \frac{R-x_0}{v_0} \\
    \end{array}\right.
    [/tex]

    and if [itex]\frac{1}{2}mv_0^2 > \Omega[/itex], then

    [tex]
    x(t) = \left\{\begin{array}{ll}
    x_0 + v_0t,\quad & t\leq\frac{R-x_0}{v_0} \\
    R + \sqrt{v_0^2 - \frac{2\Omega}{m}}\big(t - \frac{R-x_0}{v_0}\big),\quad & t>\frac{R-x_0}{v_0} \\
    \end{array}\right.
    [/tex]

    and if [itex]\frac{1}{2}mv_0^2 = \Omega[/itex], then I guess we don't know which one of these to choose. Perhaps they are both ok.

    However, if we are speaking about differential equations like in mathematics, this would not be the correct answer. The correct answer is that the solution exists in the form

    [tex]
    x:\big[0,\frac{R-x_0}{v_0}\big[\;\to\mathbb{R}
    [/tex]

    and the domain cannot be extended from this, because the path hits a point where the function [itex]U[/itex] is not differentiable.

    My question is that is it possible formulate a well defined mathematical problem, into which the solution would be the previous solution which we recognize as the physical one?

    I'm asking this in the Topology and Analysis section of PF because I have a feeling that this could be related to distributions and related things. For example, in a sense the derivative of [itex]U[/itex] is infinite at the point [itex]x=R[/itex], and if the path [itex]x(t)[/itex] bounces back from it, in a sense the acceleration [itex]\ddot{x}[/itex] is momentarily infinite too, so it seems that the equation [itex]m\ddot{x}=-\partial_x U[/itex] could be making sense in some sense.

    Of course I know how to make the equation

    [tex]
    \partial_x \theta(x-x_0) = \delta(x-x_0)
    [/tex]

    rigorous by using test functions to reformulate the equation, but I still don't know the answer to above described problem. How precisely would you reformulate the differential equation for the time evolution using some test functions?

    If we approached this from the action point of view, what kind of function spaces would we need for the action? Would the test function spaces be related to the allowed variations?
     
  2. jcsd
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