I want that [itex][0,\infty[\to\mathbb{R}[/itex], [itex]t\mapsto x(t)[/itex] satisfies(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\ddot{x}(t) = -\partial_x U(x)

[/tex]

where [itex]U:\mathbb{R}\to\mathbb{R}[/itex] is some potential function. Then I set the initial conditions [itex]x(0) < 0[/itex], [itex]\dot{x}(0)>0[/itex], and define

[tex]

U(x) = \left\{\begin{array}{ll}

0,&\quad x < 0\\

\textrm{Cantor steps},&\quad 0\leq x\leq 1\\

1,&\quad x > 1\\

\end{array}\right.

[/tex]

What's going to happen? How should the Newton's law be interpreted?

I've got a feeling that this has something to do with weak solutions, but it doesn't seem clear to me. For example

[tex]

0 = \int\Big(\ddot{\rho}(t) x(t) + \rho(t) \partial_xU(x(t))\Big) dt

[/tex]

with some smooth test function [itex]\rho[/itex] doesn't solve the problem, because you cannot get rid of [itex]\partial_x U[/itex] with integration by parts.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Newton's law with Cantor potential!

**Physics Forums | Science Articles, Homework Help, Discussion**