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Consider the following initial value problem (IVP) to the first order ODE:

$$\tag{1} \dot x = f(t, x), \ \ \ x(t_0) = x_0.$$

The Existence and Uniqueness Theorem describes when one has exactly one solutions. This is true (e.g.) when $f$ is continuously differentiable.

There are examples where uniqueness fails and existence fails (here or here):

In all of the examples where I am aware of, whenever one has more than one solutions, one actually has infinitely many. Hence my question:

Can an IVP has more than one solutions, but only finitely many solutions?

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  • $\begingroup$ I am new to this topic , so i will add more details of what i try if i got some ideas $\endgroup$
    – Cloud JR K
    Dec 5, 2018 at 16:28
  • $\begingroup$ How about $(y')^2=1$ with $y(0)=0$? The two solutions are $y=x$ and $y=-x$. $\endgroup$
    – Barry Cipra
    Dec 5, 2018 at 16:31
  • $\begingroup$ @BarryCipra thanks, but what about first order differential equaltion? $\endgroup$
    – Cloud JR K
    Dec 5, 2018 at 16:35
  • $\begingroup$ I don't why there is a negative vote! $\endgroup$
    – Cloud JR K
    Dec 5, 2018 at 16:36
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    $\begingroup$ @gt6989b, OK, I was going by a restricted definition of "first-order." It does seem clear, from the comments, that the OP is interested in the explicit variety, so the exercise suggested in Artem's answer would seem to be of interest. $\endgroup$
    – Barry Cipra
    Dec 5, 2018 at 16:51

1 Answer 1

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The answer is positive: in general (apart of some artificial examples) if an IVP for an ODE $$ \dot x=f(t,x),\quad x(t)\in \mathbb R^n $$ have non unique solution it implies that there are uncountably many solutions. In these general setting this is a very nontrivial theorem which can be found in Hartman's book (Kneser's theorem). If, however, you are dealing with an ODE $$ \dot x=f(t,x),\quad x(0)=x_0, $$ where $x(t)$ is one-dimensional, it is a good (and simple) exercise to prove that if there are two solutions to this problem then there are infinitely (uncountably) many solutions.

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  • $\begingroup$ Well, i will try your exercise! $\endgroup$
    – Cloud JR K
    Dec 5, 2018 at 16:48
  • $\begingroup$ @CloudJR Assume that there are two solutions. Draw a picture. Conclude. $\endgroup$
    – Artem
    Dec 5, 2018 at 17:04
  • $\begingroup$ I tried but its not that easy, i'm still stuck...lol $\endgroup$
    – Cloud JR K
    Dec 5, 2018 at 17:44
  • $\begingroup$ Could you point me to those artificial examples? I'd love to see exactly how one can break this theorem $\endgroup$
    – Isaac Browne
    Jun 19, 2019 at 20:40

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