I want to prove that every solution of $x'=f(t,x)$ can be enlarged in a maximal solution.

In order to prove that , I use the Zorn lemma defining $X=\{(\varphi,I);\varphi:I\to \mathbb R^n$ is solution of $x'=f(t,x);x(t_0)=x_0\}$. Now let's define a partial order $X$: $(\varphi_1,I_1)\leq(\varphi_2,I_2)\leftrightarrow I_1\subset I_2$ and $\varphi_2|I_1=\varphi_1$, if I take a chain in this set, I couldn't find the upper bound of this chain.

I need help


  • $\begingroup$ Did you mean to say that $\varphi_2|I_1=\varphi$? $\endgroup$ – Asaf Karagila Jul 26 '13 at 9:47
  • $\begingroup$ Please specify what kind of sets the domains $I$ are. Also, I would be curious to see why Zorn's lemma is needed for this. $\endgroup$ – Bob Terrell Jul 26 '13 at 12:25
  • $\begingroup$ @AsafKaragila I will edit it $\endgroup$ – user42912 Jul 27 '13 at 6:07
  • $\begingroup$ @BobTerrell $I_1$ and $I_2$ are intervals $\endgroup$ – user42912 Jul 27 '13 at 6:10
  • 1
    $\begingroup$ I think you have to say that the $I$ are open intervals containing $t_0$. $\endgroup$ – Bob Terrell Jul 27 '13 at 12:37

Suppose $\{(\varphi_{\alpha}, I_{\alpha}) \ | \ \alpha \in A \}$ is a chain. Take the function $\varphi$ which has domain $\bigcup_{\alpha \in A}I_{\alpha}$ and $\varphi(x) = \varphi_{\alpha}(x)$ for any $\alpha$ such that $x \in I_\alpha$ . Now show this is an upper bound and well defined.


If you think about functions are sets of ordered pairs, then the requirement that $\varphi_2|I_1=\varphi$ is essentially saying $\varphi\subseteq\varphi_2$.

It isn't very hard to show that the increasing union of functions is a function, whose domain is the increasing union of the domains. This should be enough in order to show that every chain has an upper bound.


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