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Let $f$ be lipschitz and consider the IVP $\dot x=f(x),\ x(t_0)=x_0$ with a solution $\varphi$. If $f(x_0)>0$, then $\varphi$ is monotonically increasing.

I have a question about the following reasoning for this fact:

Let's assume $\dot\varphi (t_1)=0$ for some $t_1\in\mathbb R$. Set $y:=\varphi(t_1)$, then we have $f(y)=0$ and thus $x_1(t)\equiv y$ solves the DE $\dot x=f(x)$. Because $x_1$ and $\varphi$ intersect in $y$, they must be equal, $\varphi =x_1$, which is a contradiction to $\dot\varphi(x_0)=f(x_0)\neq 0$.

Why must $\varphi$ and $x_1$ be equal? Picard-Lindelöf guarantees a unique solution since $f$ is lipschitz, but this is regarding the IVP. And, as I see it, $x_1 (t)\equiv y$ is a solution for the IVP if and only if $y=x_0$, but that would be a contradiction since $f(x_0)>0$, so we don't have the trivial solution $x(t)\equiv x_0$. Or, in other words, the constructed $x_1$ in the reasoning doesn't solve the IVP. Why should it be equal then with a solution that does(!) solve it?

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I'm very glad I ran across this statement, since I've been unfamiliar with it!

Keep in mind though that here, Picard‒Lindelöf is not applied to the original initial value problem, but to the IVP

$$ \begin{cases} x_1'(t) = f(x_1(t)) \\ x_1(t_1) = y; \end{cases} $$

the endpoint in the argument above for equality is $t_1$, not $t_0$.

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    $\begingroup$ Do you maybe mean the IVP with $x_1(t_1)=y$? For yours, $\varphi$ isn't a solution. $\endgroup$ – Buh Jul 16 '18 at 19:54
  • $\begingroup$ I do indeed @Buh $\endgroup$ – AlgebraicsAnonymous Jul 17 '18 at 6:08

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