Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

When teaching ODE's earlier this semester, one of my students asked the following question for which I didn't know the answer (and none of the textbooks I consulted seem to discuss it). It is standard that if $f(x,y)$ is Lipschitz, then the ODE $y'=f(x,y)$ can be solved uniquely for any initial condition (at least locally). The student asked for an example of a (non-Lipcshitz) $f(x,y)$ such that there is no solution. Of course, one can give stupid examples (eg functions $f(x,y)$ that are so non-continuous that they cannot be the derivatives of anything). However, I was unable to find a counterexample with $f(x,y)$ continuous. Can someone provide one?

share|improve this question
    
As the 3 simultaneous answers below point out, existence follows from continuity. I've deleted my answer as Pietro gives a better explanation. –  Vaughn Climenhaga Dec 5 '12 at 19:01
    
Ha. I was teeing the answer while my son watches Dumbo, and almost simultaneously was the comment from Will, as well as the answers from Michael, Vaughn and Pietro. The former gave my reply - Lipschitz is about uniqueness, the latter three contributed on your question, existence. –  Daniel Spector Dec 5 '12 at 19:03
    
Will, I think you mean $y=x^2/4$. Note that there is a continuum of solutions: $(x-a) _ + ^2/4$ . –  Pietro Majer Dec 5 '12 at 19:05
    
@Pietro, yes. There is a similar example as an answer. –  Will Jagy Dec 5 '12 at 19:54
add comment

4 Answers 4

up vote 14 down vote accepted

Indeed you can't, at least for ODE in $\mathbb{R^n}$: Peano's theorem asserts that any Cauchy problem for the ODE $\dot u(t)=f(t,u)$ with continuous non-linearity $f$, admits local solutions on some interval $I=[a,b]$, which are a non empty connected compact set in $C^0(I,\mathbb{R^n})$ (sometimes referred to as "Peano's phenomenon"). Note that in that generality ($f$ continuous) it is sufficient to consider autonomous equations. One can see this theorem as an instance of Schauder's fixed point theorem, or also prove it via approximation by Lipschitz problems with $f_k\to f$, using the Ascoli-Arzelà theorem to obtain a convergent subsequence of the solutions of the approximated problems.

The analogous equation $\dot u(t)=f(u)$ with continuous $f$ in a Banach space, e.g. $\ell_2$, may have no solution at all. The idea for a counterexample is coupling countably many scalar equations $\dot u_n(t)=f_n(u)$ with blow-ups at $T_n\to 0$ in such a way that the resulting$f$ be continuous. Check e.g. Dieudonné's Foundations of modern Analysis for such a counterexample. On the contrary, Lipschitz hypotheses ensure existence even for ODE in Banach spaces. The reason is, of course, that Peano's result relies on compactness, while Cauchy-Lipschitz on completeness.

[edit.] Just to chatter, the reason of the above mentioned Peano's phenomenon is the very same reason for which the limit points of the function $\sin(1/x)$, mentioned in a comment, as $x\to0$, are a compact interval, $[-1,+1]$. To fix things, let $X$ be the Banach space $C^0 _ b([0,1]\times\mathbb{R^n},\mathbb{R^n})$; let $D\subset X$ be the dense linear subspace of all $f$ that satisfy a Lipschitz condition on the second variable, and let $Y$ be the Banach space $C^0([0,1],\mathbb{R^n})$. By the Cauchy-Lipschitz existence theorem, there is a map $S$ taking $f\in D$ to the unique solution $u\in Y$ of $\dot u=f(t,u)$ with I.C. $u(0)=0$. This map $S:D\to Y$ is continuous at any $f\in D$, by the Gronwall lemma; moreover, it is a compact map just by the Ascoli-Arzelà theorem. Given $f\in X$ we may consider the limit set of $S$ as $g\to f$ : $${\mathrm{Lim} \atop{{g\to f}\atop g\in D } } {S(g)\atop } $$ defined as the set of all $ u\in Y$ such that there is a $g_k\to f$ such that $ S(g _k)\to u$ (in other words, the section at $f$ of the closure of the graph of $S$ in $X\times Y$). If $B(f,r)$ denotes the ball around $f$ in $X$, we may also express it as

$$ \bigcap_{ r >0 }\overline{S(B(f,r)\cap D)}\, , $$ a nested intersection of nonempty connected compact sets as a consequence of what we said. Therefore, it is a non-empty, connected compact set by a well known topological fact (a Kuratowski's lemma). Finally, it is clear that any element of the limit set is a solution of $\dot u=f(t,u)$ with $u(0)=0$ (by the theorem of "limit under the sign of derivative"; essentially, the fact that the derivative is a closed operator). A bit less obvious fact, yet not too hard to show, any solution $u$ of the above Cauchy problem may be obtained as a limit of a sequence $u _ k$ of solutions of approximating problems, $\dot u _ k =g _ k(t,u _ k )$ with $u _ k(0)=0$, that is $S(g _ k)\to u$, for a suitable sequence $g _ k $ in $D$, converging to $ f$, so in conclusion the set of solutions of $\dot u=f(t,u)$ with $u(0)=0$ is exactly the limit set of $S$ at $f$, which is a non-empty connected compact set for quite a general topological reason. This is how I like to see Peano's phenomenon.

share|improve this answer
    
Thanks, this is a very helpful answer (I'd vote it up, but I don't yet have the reputation to do so). –  Arthur Dec 5 '12 at 19:00
2  
It might be useful to mention that the ODE $y'=-\sqrt{y_+}$, for which non uniqueness holds, actually models a real physical system. $y(t)$ represents the water height in a pierced cylindrical bucket. And the non uniqueness of the Cauchy problem with initial condition $y(0)=0$ is natural in this model: if you know the bucket is empty at time $0$, it is hard to tell if there was water in it before and when it got empty. –  Thomas Richard Sep 5 '13 at 2:27
add comment

The reason for a Lipschitz condition is to guarantee uniqueness, as a standard example of non-uniqueness when $f$ is not Lipschitz is

$f(x,y)=y^{1/3}$

with the initial condition $y(0)=0$.

Then $y=0$ and $y=cx^\frac{3}{2}$ can be checked as two solutions satisfying the initial condition for the appropriate choice of $c$.

share|improve this answer
add comment

I think it is worth mentioning that although it is true that Lipschitz continuity guarantees unique solutions, there is a weaker regularity condition for $f$ which also gives uniqueness. The buzzword is "Osgood condition" and the theorem is given in Andrey Rekalo's answer to this MO question on "Existence/Uniqueness of solutions to quasi-Lipschitz ODEs".

share|improve this answer
add comment

It is well known that existence holds if f is continuous (Peano's existence theorem). This is documented in lots of textbooks.

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.