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I have a misunderstanding regarding a very common reasoning:

Let's i.e. look at the IVP $\dot x=f(x), x(t_0)=x_0$ with $f(x)=(x-1)(x-2)$. Now, for $x_0\in ]1,2[$ there can be made an argument that the solution always stays in between $]1,2[$.

The usual way to reason this is since otherwise there would be a point, let's say $t_1$, such that $\lambda(t_1)=1$ or $\lambda(t_1)=2$ and that conflicts with the uniqueness of the solution since it would have an intersecting point with one of the constant solutions.

But I don't understand this: The constant solutions don't even solve the IVP if $x_0\in ]1,2[$, so why is it a problem that those solutions intersect?

Is the point here that a solution $\lambda$ for the IVP $x(t_0)=x_0$ would then also solve the IVP with $x(t_1)=1$ which is soled by the constant solution and therefor they must coincide?

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  • $\begingroup$ Yes, you get an IVP at $t_1$ that has by assumption two different solutions, contradicting uniqueness. $\endgroup$ – LutzL Dec 19 '18 at 16:18
  • $\begingroup$ Since we can solve IVP backwards, $x(t_1)=1$ determines past values of $x$ as $x(t)=1, t<t_1.$ $\endgroup$ – Song Dec 19 '18 at 16:36
  • $\begingroup$ Okay, thanks a lot! $\endgroup$ – RedLantern Dec 19 '18 at 16:43
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Firstly, the IVP $$ \dot{x}=f(x),x(t_0)=x_0.\tag{1} $$ has a unique solution for $x_0\in]1,2[$. Clearly $x=\underline{x}(t)\equiv1$ is a subsolution of (1) and $x=\bar{x}(t)\equiv2$ is a supersolution of (1). Since $\underline{x}(t)\le \bar{x}(t)$. Therefore there is a unique solution $x=x(t)$ such that $$ \underline{x}(t)\le x(t)\le \bar{x}(t). $$ Namely the solution always stays between $]1,2[$ if $x_0\in]1,2[$.

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