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It is an old result of Kneser that if $f$ is a continuous function, and there are two solutions to the IVP $$y'=f(x,y), \quad y(x_0)=y_0,$$ then there are uncountably many solutions. I am interested in the general case, where $f$ is not continuous.

Is there an IVP with exactly two solutions?

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    $\begingroup$ So at some point $x_1$ you have two solution values $y_l(x_1)$ and $y_u(x_1)$. Now consider the initial value problem with $y(x_1)=(1-c)y_l(x_1)+cy_u(x_1)$, $0<c<1$, and trace its solution(s) back to $x_0$. What value $y(x_0)$ can you hope to achieve? Can you avoid $y_0$, and by what mechanism? $\endgroup$ Apr 2, 2020 at 12:18
  • $\begingroup$ @LutzLehmann Why is there a solution to the IVP with initial condition $y(x_1)=(1-c)y_l + cy_u$? $\endgroup$
    – Pachirisu
    Apr 3, 2020 at 2:02
  • $\begingroup$ Because on the segment between $x_0$ and $x_1$ it is bounded by $y_l(x)\le y(x)\le y_u(x)$ unless it crosses one of these solutions. As the right side is continuous everywhere that means that a local solution can always be continued until it leaves that region. But at the crossing point one can then continue with the bounding solution to still get a full solution that ends in $y_0$. $\endgroup$ Apr 3, 2020 at 5:06
  • $\begingroup$ @Lutz The right side is not continuous. $\endgroup$
    – Emolga
    Apr 7, 2020 at 11:35
  • $\begingroup$ I lack the imagination of how that could happen. You still need to account for what these inner solutions do if you trace them towards $x_0$. Their domain might end before that, which is quite typical of discontinuous right side if you do not generalize to sliding mode solutions. If that does not happen, they have to cross one of the original solution, and you need something that would prevent the inner solution to continue along the other one. I do not think that that is possible, but I also have no formal proof of it. $\endgroup$ Apr 7, 2020 at 11:49

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