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Let $f \in C^1(\mathbb R, \mathbb R)$ and suppose $$ \tag{H} \vert f(x) \vert\le \frac{1}{2}\vert x \vert + 3, \quad \forall x \in \mathbb R. $$

Then, every solution of $$ x'(t)+x(t)+f(x(t))=0, \quad t \in \mathbb R $$ is bounded on $[0,+\infty]$.

First of all, I would like to say that the text has been copied correctly: I mean, it's really $[0,+\infty]$ so I suppose the author wants me to prove $\displaystyle \lim_{t \to +\infty} x(t) <\infty$. Indeed, boundness on $[0,+\infty)$ is quite obvious, because we have global existence (its enough to write the equation as $x'=-x-f(x)$ and to observe that the RHS is sublinear thanks to $(H)$).

So, we have to prove $\displaystyle \lim_{t \to +\infty} x(t) <\infty$. How can we do? I've got some ideas but I can't conclude. First, I observe that the problem is autonomous: this implies that solutions are either constant either monotonic.

First idea: I've fixed $x_0 \in \mathbb R$ and I've written the equivalent integral equation: $$ x(t) = x_0 - \int_0^t [x(s)+f(x(s))]ds $$

Taking the absolute value and making some rough estimates, we get $$ \vert x(t) \vert \le \vert x_0 \vert + \left\vert \int_0^t [x(s)+f(x(s))]ds \right\vert \le \vert x_0 \vert + \int_0^t \frac{3}{2}\left\vert x(s) \right\vert +3 ds $$ but now I don't know how to conclude. Gronwall's lemma? But how can I use it?

Second idea: if $x_0 \in \mathbb R$ is s.t. $x_0 +f(x_0) \neq 0$, the solution of the Cauchy problem is not constant. I can divide both members of equation and I obtain (integrating on $[0,t]$) $$ -t=\int_0^t \frac{dx}{x+f(x)} $$ Now I let $t \to +\infty$ but... what can I conclude?

Thanks in advance for your help.

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In regards to your first paragraph - does global existence imply boundedness by definition? Otherwise, how? What does sublinearity mean (this?) and how does a derivative's sublinearity imply boundedness? –  anon Jul 17 '12 at 0:42
    
@anon I use a global existence theorem: you can find the statement here. As you can see, if the RHS has a growth which is at most linear we have global existence. In my opinion, this implies that the solutions are bounded on $[0,+\infty)$ (no vertical asymptotes). The only point where they could become unbounded is therefore $+\infty$: we have to prove that $\lim_{t \to + \infty} x(t) < \infty$. Do you agree? –  Romeo Jul 17 '12 at 8:32
    
"Bounded" does not mean "no vertical asymptotes." Bounded means the range is contained in an interval, i.e. the function has constant upper and lower bounds. In the DE $x'=x$ the RHS is at most linear, but $x=e^t$ is certainly not bounded. –  anon Jul 17 '12 at 12:29
    
I do agree with you about the definition of "bounded". Anyway, since we have global existence, the function $x(\cdot)$ exists and is of course continuous on $[0,+\infty)$: continuity implies that $x(\cdot)$ is bounded on every compact subset $[a,b] \subset [0,+\infty)$: therefore, the only problems of boundness can occur when $t \to +\infty$. Hope now it's clear what I meant. –  Romeo Jul 17 '12 at 15:03
    
A note, then: bounded functions need not have limits at infinity. (Consider $\sin$.) –  anon Jul 17 '12 at 15:12
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2 Answers

up vote 3 down vote accepted

By assumptions $$ x'(t) = -x(t) -f(x(t)) \leq -x(t) + \frac 1 2 \lvert x(t) \rvert + 3 $$ Let's suppose there is a point $t_0$ such that $$ x(t_0) > \max\{x(0), 6\} $$

Since $x(t_0)$ is positive the following relation is satisfied $$ x'(t_0) \leq -x(t_0) + \frac 1 2 \lvert x(t_0) \rvert + 3 = -\frac 1 2 x(t_0) + 3 < 0 $$

So the $\max$ of $x(t)$ on the interval $[0, t_0]$ must occur at $t_1\in (0, t_0)$, but that leads to contradiction because: $$ 0 = x'(t_1) \leq - \frac 1 2 x(t_1) + 3 < 0 $$ where the equal sign is there because $t_1$ is an extremum point of a differentiable function on an open interval.

A similar reasoning shows that $x(t)$ is bounded from below.

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Thank you very much, great proof. –  Romeo Jul 17 '12 at 17:42
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The goal is to show that the solution is bounded. (I have no idea what the $+\infty$ bit is about, it adds nothing as far as I can tell.)

We first need to show that the solution from any initial condition $x_0$ (at $t_0$) is defined on $[t_0,\infty)$. It is not immediately clear to me that 'we have global existence'. After that, it is fairly straightforward to show boundedness using a 'Lyapunov' function of sorts.

Let $\phi(x) =-x -f(x)$. Since $f$ is $C^1$, we have that $\phi$ is $C^1$, hence Lipschitz on any bounded set. From the usual considerations of ODEs, it is straightforward to establish that, for any $B>0$, if $|x_0| < B$, then there exists a $\delta>0$ (which depends only on $B$ and $f$) such that the solution to $\dot{x} = \phi(x)$ with initial condition $x(t_0) = x_0$ is defined on $[t_0,t_0+\delta]$. If we can establish that $|x(t_0+\delta)| < B$, then it is clear that the solution is defined on $[t_0,\infty)$. Furthermore, the solution is $C^1$.

Using the bound provided for $f$, we have:

If $x\geq 8$, $\phi(x) \leq -x -f(x) \leq -x +\frac{1}{2} |x| +3 = -\frac{1}{2} x+3 \leq -1$.

If $x\leq -8$, $\phi(x) \geq -x -f(x) \geq-x -\frac{1}{2} |x| -3 = -\frac{1}{2} x-3 \geq +1$.

Now let $x$ be as a solution to $\dot{x} = \phi(x)$, and define the 'Lyapunov' function $V(t) = \frac{1}{2} x(t)^2$. Then it is easy to see that $V$ is $C^1$ and $\dot{V}(t) = x(t) \phi(t)$. Consequently, if $|x(t)| \geq 8$, then $\dot{V}(t) \leq -8$.

Thus, if $|x_0| \geq 8$, then the solution will satisfy $|x(t)| < 8$ in finite time.

Now suppose $|x_0| < 8$, and that at some $t \in [t_0,t_0+\delta]$, we have $|x(t)| \geq 8$. Let $t_1$ be the first time at which $|x(t_1)| = 8$ ($x$ is continuous, so this is well defined). Since $V$ is $C^1$, we have, for some suitably small $\epsilon>0$, $\dot{V}(t) \leq -7$, for $t \in [t_1-\epsilon, t_1]$. Since $|x(t_1-\epsilon)| < 8$, this contradicts $|x(t_1)| = 8$, since $V(t_1) = V(t_1-\epsilon)+\int_{t_1-\epsilon}^{t_1} \dot{V}(\tau) d \tau$.

Hence $|x(t)| < 8$ for all $t \in [t_0,t_0+\delta]$, and from the considerations above, we see that $x$ is defined on $[t_0, \infty)$, and $|x(t)| < 8$, for all $t\geq t_0$.

If $\phi(x(t')) = 0$ for some $t'$, then it is clear that $x(t) = x(t')$, for all $t \geq t'$ (by uniqueness of solution). If $\phi(x(t')) \neq 0$ for all $t$, then it is clear that $\phi(x(t))$ has the same sign as $\phi(x(t'))$ for all $t\geq t'$. Hence either $x(t)$ is increasing and bounded above, or decreasing and bounded below. Hence $\lim_{t \to \infty} x(t)$ exists (and is in $[-8,8]$, of course).

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