Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let be given a map $F:(x,y)\in\mathbb{R}\times\mathbb{R}^n\to F(x,y)\in\mathbb{R}^n$.
Let us denote by $\mathcal{P}$ the set whose elements are the solutions of the ode $y'=F(x,y)$, i.e. the differentiable maps $u:J\to\mathbb{R}^n$, where $J\ $ is some open interval in $\mathbb{R}\ $, s.t. $u'(t)=F(t,u(t))$ for all $t\in J$.
Let $\mathcal{P}$ be endowed with the ordering by extension.

In order to prove that any element of $\mathcal{P}$ is extendable to a (not unique) maximal element, without particular hypothesis on $F$, I was wondering if the Zorn lemma can be used.

share|improve this question
    
I vaguely recall there might be something in Coddington & Levinson about this, in a section about existence theorems. –  Michael Hardy Oct 18 '11 at 21:15
    
You'll probably want to restrict your attention explicitly to functions defined on connected subsets of $\mathbb R$. Otherwise if you start with, say, a solution defined on the open unit interval, the "maximal extension" you get could be the original solution extended periodically to $\mathbb R\setminus\mathbb N$ with jump discontinuities at each integer. And that isn't very interesting... –  Henning Makholm Oct 18 '11 at 21:38
    
Dear Henning Makholm, I have edited the question after your comment. –  Giuseppe Tortorella Oct 18 '11 at 21:49
    
You need the existence theorem itself to guarantee $\mathcal{P}$ is not empty. The next issue, to apply Zorn's lemma you need upper bounds for each chain -- showing this strikes me as a big problem. –  Bill Cook Oct 18 '11 at 23:19
1  
@Bill, existence of a partial solution is assumed -- it is implicit in stating the proposition as "any element of $P$ ...". As for upper bounds, what could go wrong? Under the common definition of a function as the set of pairs $\langle x,f(x)\rangle$, the union of all functions in a chain will -- as far as I can see -- be a function with the required property. –  Henning Makholm Oct 20 '11 at 19:28
show 3 more comments

1 Answer

up vote 2 down vote accepted

Yes, Zorn's Lemma should be all you need. Take the set of partial solutions that extend your initial solution, and order them by the subset relation under the common definition of a function as the set of pairs $\langle x, f(x)\rangle$. Then the union of all functions in a chain will be another partial solution, so Zorn's Lemma applies.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.