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Consider a first order equation in $\mathbb{R}$ with $f(t,x)$ defined on $\mathbb{R}\times \mathbb{R}$. Assume the equation $x'=f(t,x)$. Suppose $xf(t,x)<0$ for $|x|>R$ where $R$ is a fixed positive constant number. I have to show that all solutions exist for all $t>0$.

The way I look at it is that $x x'=xf(t,x)<0$ thus $x'<0$ for $x> R$ and $x'>0$ for $x<-R$, therefore for solutions starting at $x>R$ the solution cannot pass $x=-R$ line since it would need to go back up and for solutions starting at $x<-R$ the solution starts growing but it cannot go beyond $x=R$ since it needs to start decreasing again.

I need help understanding the answer below.Why is $|x(\tau)|=2R$ below?

Thank you, Klara

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I changed $RxR$ to $R\times R$. –  Michael Hardy Nov 2 '12 at 3:54
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What is the $R$ in $|x| \gt R$? I suspect it is not the $\mathbb R$ that all the other $R$'s look like. –  Ross Millikan Nov 2 '12 at 4:18
    
I assume you mean the equation $\dot{x} = f(t,x)$? And do you have any kind of regularity assumptions on $f$, like Lipschitz-continuity in $x$ or something? –  Lukas Geyer Nov 2 '12 at 5:01
    
@ Ross I found R –  Klara Nov 2 '12 at 12:19
    
@ Lucas you assumed well. This is all I have no other assumptions. –  Klara Nov 2 '12 at 12:20
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1 Answer

up vote 2 down vote accepted

We have the following:

Proposition Let $x: [0,T) \to \mathbb{R}$ be continuous and differentiable and satisfies $x' = f(t,x)$, then for any increasing sequence of times $T_n \nearrow T$ we have that $\limsup_{n} |x(T_n) | < \infty$.

The above proposition implies that a solution cannot "blow up in finite future time", which I think is the proper interpretation of the question asked.

The proof of the proposition is simple: intuitively using that $(x^2)' = 2xx' = 2xf$ we have that by the condition given in the hypothesis, once $|x|$ reaches a value that is larger than $R$, it can only decrease (and not increase). This forces the value of $x$ to remain bounded and hence cannot blow up in finite time. To formalise the proof we write as follows.

Proof: By way of contradiction, we assume that there exists a increasing sequence of times $T_n$ such that $|x(T_n)| > n$. Then for $n > 2R$ we have that $|x(T_{n+1})| > n+1 > 2R$. By continuity we can find the value $$ \tau = \sup \{ t\in [T_n,T_{n+1}] | |x(t)| \leq 2R\} $$ and we know that $\tau < T_{n+1}$. Furthermore by continuity we know that $|x(\tau)| = 2R$. Apply now the mean value theorem to $|x|^2$ on the interval $[\tau,T_{n+1}]$, we have that there exists $\tau'\in [\tau,T_{n+1}]$ such that $$2x x' |_{t = \tau'} = (|x|^2)' = \frac{|x(T_{n+1}|^2 - 4R^2}{T_{n+1} - \tau} > 0 $$ On the other hand by the definition of $\tau$ we have $|x(\tau')| \geq 2R$ as $\tau' \geq \tau$. Hence we also have $$ xx'|_{t = \tau'} = xf < 0 $$ by the hypothesis of the problem. And we arrived at the contradiction.


A comment about existence: solutions to your differential equation do not need to exist if $f$ is sufficiently bad. It is well known that a derivative of a everywhere differentiable, continuous function cannot be too discontinuous. In particular, let $\chi(x)$ denote the function that equals $1$ when $x < 0$, and $-1$ when $x \geq 0$. Now let $\phi(t)$ be a nowhere continuous function such as the indicator of the rational numbers. Then for $f(x,t) = \phi(t) \chi(x)$ it is impossible for any continuous and differentiable function $x = x(t)$ to satisfy $x' = f(t,x)$.

So strictly speaking under the conditions given it is false that every solution exists for all positive time.

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You might want to add how you use $\phi$ and $\chi$ in the example, probably as $f(t,x)=\phi(t)\chi(x)$. –  Did Nov 2 '12 at 16:30
    
@ Willie Thank you so much! –  Klara Nov 2 '12 at 16:35
    
@did: oops! I can't believe I forgot that important piece of info! Thanks for the comment. –  Willie Wong Nov 5 '12 at 9:48
    
These things just happen... :-) –  Did Nov 5 '12 at 12:00
    
@Klara: I just saw your suggested edit; the other users rightly rejected the suggestions. Firstly, $x$ being a solution of an ordinary differential equation depends only on the independent variable $t$. As to the intuition behind the argument: the idea is that were $x$ to start bigger than $R$, then $x$ can only decrease until it becomes less than or equal to $R$, after which anything goes. Were $x$ to start smaller than $R$, it can do whatever it wants as long as it never gets bigger than $R$. –  Willie Wong Nov 20 '12 at 10:29
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