# Uniqueness of solutions of an IVP

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?

• Yes, you get an IVP at $t_1$ that has by assumption two different solutions, contradicting uniqueness. – Lutz Lehmann Dec 19 '18 at 16:18
• Since we can solve IVP backwards, $x(t_1)=1$ determines past values of $x$ as $x(t)=1, t<t_1.$ – Song Dec 19 '18 at 16:36
• Okay, thanks a lot! – RedLantern Dec 19 '18 at 16:43

## 1 Answer

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 $$x_0\in]1,2[$$.