- #1

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## Main Question or Discussion Point

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.