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Homework Help: Autonomous ODE: non-uniqueness of solutions

  1. Jun 20, 2012 #1
    1. The problem statement, all variables and given/known data
    We have the autonomous ODE:
    [itex]\dot{x} = f(x), x \in \mathbb{R}[/itex]
    first we define the following sets:
    [itex]E := \{f(x) = 0\}[/itex]
    [itex]E^+ := \{f(x) > 0\}[/itex]
    [itex]f(x)[/itex] is continuous so [itex]E[/itex] is a closed set and [itex]E^+[/itex] is an open set.
    [itex] x_0 \in (x_-,x_+)[/itex] where [itex](x_-,x_+)[/itex] is the connected component of [itex] E^+ [/itex] that contains [itex]x_0[/itex].
    In addition we define the function:
    [itex] \Psi : (x_-,x_+) \rightarrow \mathbb{R} [/itex]
    [itex] \Psi(z) := \int_{x_0}^z \frac{dy}{f(y)} [/itex]
    Now the texbook says that if:
    [itex]lim_{z\to x_{\pm}} \Psi(z) = l^{\pm}[/itex] is finite, we lose the uniqueness of the solutions and this is quite easy to understand, but what comes after is not :(
    It says that we have at least 2 solutions of the equation that at the istant [itex] t = l^{\pm}[/itex] coincide at [itex]x^+[/itex]; what are these? are there others?
    3. The attempt at a solution
    I struggled for hours and i think one of the solutions maybe the function:

    [itex] x(t) = \begin{cases} x_- & t = l^- \\ \Psi^{-1}(t-t_0) & l^- < t < l^+ \\ x^+ & t = l^+ \end{cases} [/itex]

    i.e. i extented the solution x(t) = [itex] \Psi^{-1}(t-t_0) [/itex] at the extremes of [itex] (x_-,x_+) [/itex]
    but i can't think of other solutions
  2. jcsd
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