- #1
jostpuur
- 2,116
- 19
I want that [itex][0,\infty[\to\mathbb{R}[/itex], [itex]t\mapsto x(t)[/itex] satisfies
[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.
[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.