Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $x'=f(t,x)$ be a differential equation with $f$ in the hypothesis of Picard's theorem. Let $\varphi$ be a solution such that its interval of definition contains $(t_0,+\infty)$ for some fixed $t_0\in \mathbb{R}$. Suppose $\lim_{t\to+\infty} \varphi(t)=a \in \mathbb{R}$. Must the constant function $t\mapsto a$ be a solution of the equation?

(This is a question I've asked myself while studying the logistic model $x'=ax(1-x)$ and wondering how to justify the solutions look the way they do.)

EDIT: By "the hypothesis of Picard's theorem", I mean $f$ continuous and locally Lipschitz with respect to $x$.

I'm interested also in particular cases (e.g. autonomous system) where this holds.

share|cite|improve this question
Yes. I'm sure there must be some straightforward way to see this, but my intuition is coming from integral curves on a manifold. All you have to do is prove that $x'=0$ at $x=a$. If it weren't, then any integral curve (solution) that approached $a$ would have to flow past that point and couldn't limit to it (the only rigorous proof I know of this fact involves choosing coordinates in a neighborhood so that the vector field is just $d/dx$ or in your case this just means you can rescale so that $x'=1$ on some open interval containing $a$). – Matt Oct 23 '11 at 3:02
It is true for autonomous equations $x'=f(x)$, but not for time dependent equations, as the example in Leslie Townes answer shows. – Julián Aguirre Oct 23 '11 at 8:32
@JuliánAguirre Would you please elaborate? I'm very interested in the autonomous case. – Bruno Stonek Oct 23 '11 at 12:50
up vote 2 down vote accepted

You have not been explicit about "the hypotheses of Picard's theorem" (and yes, different sources state these differently) but in any case I believe that the answer is no unless you are far more restrictive on what you require of $f$.

Consider $f$ defined on $(1, \infty) \times \mathbb{R}$ by $$ f(t,x) = \frac{\cos(t) - x}{t}, \qquad t > 1, \quad x \in \mathbb{R}. $$ Whatever your hypotheses are I expect they are satisfied by this $f$ on $(1, \infty) \times \mathbb{R}$.

Consider $y: (1, \infty) \to \mathbb{R}$ defined by $$ y(t) = \frac{\sin(t)}{t}, \qquad t > 1. $$ A short calculation (just the quotient rule) shows that $$ y'(t) = \frac{t \cos(t) - \sin(t)}{t^2} = \frac{\cos(t) - \frac{\sin(t)}{t}}{t} = \frac{\cos(t) - y(t)}{t} = f(t,y(t)), \qquad t > 1, $$ and clearly $$ \lim_{t \to \infty} y(t) = 0. $$ Yet $f(t,0) = \frac{\cos t}{t}$ is not identically zero, so the constant function $0$ is not a solution to $x' = f(t,x)$.

share|cite|improve this answer

This is an elaboration of my previuous comment, written at Bruno's request.

Consider the equation $x'=f(x)$ with $f\colon\mathbb{R}\to\mathbb{R}$ continuous and let $x\colon(t_0,\infty)\to\mathbb{R}$ be a solution such that $\lim_{t\to\infty}x(t)=a$. I claim that $f(a)=0$, and hence $z(t)\equiv a$ is a constant solution. The proof is by contradiction.

Suppose $f(a)\ne0$. Without loss of generality we may assume that $f(a)>0$. We have $$ \lim_{t\to\infty}x'(t)=\lim_{t\to\infty}f(x(t))=f(a). $$ Let $\delta=f(a)/2$. There exists $t_1\ge t_0$ such that $x'(t)\ge\delta$ if $t\ge t_1$. In particular, $$ x(t)=x(t_1)+\int_{t_1}^tx'(t)\,dt\ge x(t_1)+(t-t_1)\delta\quad\forall t\ge t_1, $$ which contradicts the fact that $\lim_{t\to\infty}x(t)=a$.

The proof carries over to autonomous systems of equations.

share|cite|improve this answer
Thank you very much, it's crystal clear. In fact, after the first simple manipulation, the proof is of the general fact that a function with a constant limit in infinity must have vanishing derivative in infinity, which is geometrically obvious. I will accept, though, the other answer as it adresses the original question, I hope you will understand. – Bruno Stonek Oct 26 '11 at 2:24
No problem. That's why I wrote a comment instead of an answer. As for the first part of your comment, it is not correct. Think of $f(t)=\sin(t^2)/t$; $\lim_{t\to\infty}f(t)=0$, but $f'(t)$ does not converge as $t\to\infty$. – Julián Aguirre Oct 26 '11 at 13:36
Oh, indeed. I believe this is the correct formulation of my comment: "If a function with constant limit in infinity has a derivative with existing (& finite) limit in infinity, then this limit must be zero." – Bruno Stonek Oct 26 '11 at 16:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.