Sign up ×

Picard's existence and uniqueness theorem states that if a function satisfies some conditions on a particular interval then it has a solution in that interval. The function being f in the IVP:

$$ \begin{cases} x'=f(x(t),t) \\ x(t_0)=x_0 \end{cases}$$

But then there exists a theorem that states that these solutions can be continued further by repeating the argument with the old final time as the new initial time. I am looking for a detailed explanation and formulation of the proof for this theorem. Could somebody help me out? as I've been searching for quite a while now and can't seem to find anything.

share|cite|improve this question
I know one theorem which states that if the graph of the solution is contained in a compact set, then this solution can be extended outside that compact set. Is this what are you talking about? In general, there can be proved that there are maximal solutions, i.e. solutions that have maximal domain. Not every solution can be extended. – Beni Bogosel Jun 22 '11 at 10:44
One place would be pages 1-4 of Lars Hormander's Lectures on Nonlinear Hyperbolic Equations. (The relevant material is in the "preface" on ordinary differential equations.) – Willie Wong Jun 22 '11 at 11:59
Yes I think it is. I am trying to read "Coddington E. , Levinson N. - Theory of ordinary differential equations" and there the theorem is pretty much called continuation of solutions. It explains it by saying that if the solution is bounded, then it is somehow continuous at the boundary of the region and therefore somehow can be extended, I just need a better explanation or just an explanation from a different source. EDIT: Thank you Willie Wong, I'll check it out. – Al Sum Jun 22 '11 at 12:05

1 Answer 1

I'm expanding Beni Bogosel's comment. We have the IVP $$x'=f(t,x)\ ,\qquad x(t_0)=x_0\ ,$$ where $f$ is continuous on an open set $\Omega$ of the $(t,x)$-plane and locally Lipschitz-continuous with respect to $x$ on $\Omega$. This IVP has a ${\it maximal\ solution}$ $\ t\mapsto \phi(t) \ (\alpha < t < \beta)$ with the following properties:

(i) For any solution $\psi$ on some $t$-interval $(a,b)$ one has ${\rm graph}(\psi)\subset{\rm graph}(\phi)$.

(ii) For any compact set $K\subset\Omega$ with $(t_0,x_0)\in K$ the graph of $\phi$ will ultimately leave $K$.

I'll only sketch a proof of (ii). Assume that there is a sequence $t_n \to \beta\!-\ $ with $(t_n,\phi(t_n))\in K$. Taking a subsequence we may assume that $(t_n,\phi(t_n))\to (\tau,\xi)\in K\subset\Omega$. By assumption there are a neighbourhood $U \subset \Omega$ of $(\tau,\xi)$ and constants $L>0$, $M>0$ with $$|f(t,x)|\leq M\ , \qquad |f(t,x)-f(t,x')|\leq L|x-x'|\qquad\forall\ (t,x), (t,x')\in U\ .$$ There is a $\rho>0$ with $L\rho<1$ and $Q:=[\tau-\rho,\tau+\rho]\times[\xi-M\rho,\xi+M\rho]\subset U$.

Put $\rho':={3\over4}\rho$ and chose an $n$ with $t_n > \tau - {\rho\over4}$, $|\phi(t_n)-\xi|\leq {M\rho\over4}$. Then $$Q':=[t_n-\rho',t_n+\rho']\times[\phi(t_n)-M\rho',\phi(t_n)+M\rho']\subset Q\subset U\ .$$ The proof of the uniqueness theorem then shows that the IVP with initial point $(t_n, \phi(t_n))$ has a solution at least on the interval $[t_n-\rho',t_n+\rho']$, which in turn implies that $\phi$ can be extended to the right at least to $t_n+\rho'\geq \beta+\rho/2$, contradicting the definition of $\beta$.

share|cite|improve this answer

Your Answer


