Question

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?

Answer 1

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.

Answer 2

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".

Answer 3

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 $c$.

Answer 4

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