# Prove that solutions to $\dot x=f(x)$ are monotonic

I have the following problem: let $$f(x)$$ be a continuous function with real argument. Consider the following equation:

$$\dot x=f(x) \tag{1}$$

Prove that all of its solutions are monotonic.

Here is the solution I came up with: let $$U=\{x:f(x)=0\}$$. If $$x_0\in U$$, then the function $$x\equiv x_0$$ satisfies the equation and it is monotonic.

Now let $$x_1\notin U$$. Let

$$m_1=\sup\{u\in U:u

By the theorem of existence and uniqueness, there exists a unique solution for any starting condition of kind $$x(t_0)=x_0$$ where $$t\in\mathbb R,x_0\in U$$ and we found such a solution above: $$x\equiv x_0$$. Therefore solutions to problems of kind $$x(t_0)=x_1$$ where $$t_0\in\mathbb R,x_0\in U$$ may not cross the line $$x=x_0$$ for any $$x_0\in U$$, therefore $$x(t)\in \langle m_1,M_1\rangle\;\;\forall t\in\mathbb R$$ where the interval is open on the corresponding end if $$m_1\in U$$ or $$M_1\in U$$ and closed otherwise.

Since $$f$$ is continuous, it has constant sign on this interval $$\langle m_1,M_1\rangle$$, because it never reaches zero on this interval, therefore $$\dot x=f(x)>0\;\;\forall t\in\mathbb R$$ or $$\dot x=f(x)<0\;\;\forall t\in\mathbb R$$, which means $$x$$ is strictly monotonic. Obviously, every solution can be expressed as a solution to a problem with starting conditions $$x(t_0)=x_2$$ where $$t_0\in\mathbb R$$ and $$x_2\in U$$ or $$x_2\notin U$$, therefore every solution is either constant or strictly monotonic.

However someone pointed out to me that the problem didn't state that $$f$$ is everywhere differentiable, which means that the theorem of existence and uniqueness doesn't quite hold, like for $$f(x)=x^{2/3}$$ solutions $$x(t)=0$$ and $$x(t)=t^3/27$$ intersect, yet they are both monotonic. And I am not sure what to do with that, because the statement of the problem is still supposed to be true. How can it be proved without using differentiability of $$f$$?

• Perhaps I am misunderstanding your attempted proof, but I do not understand why the differentiability of $f$ is relevant. If I understand correctly, you never actually used the differentiability of $f,$ only its continuity, which is explicitly given to you by the problem. Commented Dec 22, 2021 at 15:38
• @Angel, I use differentiability when assuming that solution $x$ for starting condition $x(t_0)=x_1,\;x_1\notin U$ never intersects with lines of kind $x=x_0,\;x_0\in U$. I make this conclusion based on the fact that solutions for the starting conditions $x(t_0)=x_0,\;x_0\in U$ are unique for any $t_0$ and we explicitly found them earlier, so there cannot be another one crossing these points $(t_0,x_0)$. But if $f$ is not differentiable, this is not true, there may be other solution than $x\equiv x_0$ satisfying $x(t_0)=x_0$ Commented Dec 22, 2021 at 17:19
• Does this answer your question? Solutions of autonomous ODEs are monotonic Commented May 3, 2022 at 21:01

I think the result is correct:

Assume that $$x:J \to \mathbb{R}$$ is a solution of $$x'(t)=f(x(t))$$ which is neither decreasing nor increasing on the interval $$J$$ (note that $$x \in C^1(J)$$). Then, there exist $$a, b \in J$$ such that $$x'(a) > 0$$ and $$x'(b) < 0$$. Without loss of generality consider the case $$a < b$$ and $$x(a) \le x(b)$$. Let $$\tau \in (a,b)$$ be such that $$x (\tau) = \max\{x(t):t \in [a,b]\}$$ (note that $$x(\tau)>x(b)$$), and let $$\sigma := \max \{t \in [a,\tau]:x(t) = x(b)\}$$. We have $$\sigma < \tau$$ and $$x(t) > x(\sigma)$$ $$(t \in (\sigma,\tau] )$$. Hence $$x'(\sigma) \ge 0$$. Since $$x(\sigma) = x(b)$$ we get $$0 > x'(b) = f(x(b)) = f(x(\sigma)) = x'(\sigma) \ge 0,$$ a contradiction.

• It took me quite a while to get, that this reasoning doesn't work to the example from another answer, because it has a wrong statement 😅. Anyway, it seems legitimate Commented Dec 22, 2021 at 17:38
• As a historical note: To my knowledge this result is due to Sylvan Wallach: Amer. J. Math. 70 (1948), 345-350. I gave an elaboration of the proof there.
– Gerd
Commented Dec 22, 2021 at 18:11
• @Gerd: Nice proof (+1). One question: How did you conclude that $x'(\sigma) \geq 0$? Commented May 26, 2022 at 22:59
• @Leonidas For $t \in (\sigma,\tau]$ we have $(x(t)-x(\sigma))/(t-\sigma) > 0$. Thus the limit for $t \to \sigma+$ is $\ge 0$ and this limit is $x'(\sigma)$.
– Gerd
Commented May 27, 2022 at 9:05
• @Gerd: Got it, thanks for the clarification! Commented May 27, 2022 at 18:57