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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.

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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
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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 $x' = f(t,x)$.

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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.

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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
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