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How to prove: Uniqueness of solution to first order autonomous ODE

  1. Jul 7, 2011 #1

    I would like to prove the following statement: Assume [itex]f\in C^{1}(\mathbb{R})[/itex]. Then the initial value problem [itex]\dot{x} = f(x),\quad x(0) = x_{0}[/itex] has a unique solution, on any interval on which a solution may be defined.

    I haven't been able to come up with a proof myself, but would really like to see a direct proof, not using too serious tools from a sophisticated theory of ODE's. I would very much appreciate it if someone could help me out.

    Thanks in advance!
  2. jcsd
  3. Jul 7, 2011 #2


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    Hello Jösus! :smile:
    So you can assume that there is a solution g, and you have to prove that there can't be two solutions, g and h.

    So suppose dg/dx = dh/dx = f.

    Then … ? :wink:
  4. Jul 8, 2011 #3
    Correct me if I'm wrong, but shouldn't the equation read [itex]dg/dt = f(g(t)), \quad dh/dt = f(h(t))[/itex], and thus there would be no apparent reason for these derivatives to be equal?

    I have thought about it some more, and found that if [itex]f(x_{0}) \neq 0[/itex] then there is an interval containing [itex]x_{0}[/itex] on which a unique solution exists (the equation is separable, so a simple integration trick works). When f reaches a zero in finite time, so that the inteval on which this solution is defined is cut off there may be extensions to the solutions beyond the problematic points. If, say, [itex]f(a) = 0[/itex] then setting [itex]x(t) = a[/itex] for t larger than (or smaller than if a is on the right of our starting point) will, I believe, do the trick. It is solutions of this type that aren't always unique, but with the requirement [itex]f \in C^{1}(\mathbb{R})[/itex] it should work. Any new ideas?
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