Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider the following initial value problem: $$ \begin{cases} y^\prime = f(x,y)\\ y(0)=0 \end{cases} $$ where $f$ is the function $$ f(x,y) =\begin{cases} y\sin(1/y) & \text{if}\; y\neq 0\\ 0 & \text{if}\; y=0. \end{cases} $$ It is clear that $f$ is continuous, but not Lipschitz in a neighbourhood of $(0,0)$. Nevertheless, the IVP has one (and only one) solution in that neighbourhood. Which is the solution?

share|cite|improve this question
up vote 1 down vote accepted

"Which is the solution?" is not such an interesting question: of course $y\equiv 0$ is an equilibrium of this autonomous equation. The interesting question is why the solution is unique. Let's get rid of the (unused) dependency on $x$ and focus on $y'=f(y)$.

Claim. Suppose that $f:\mathbb R\to\mathbb R$ satisfies $f(0)=0$ and $\limsup_{y\to 0}|f(y)/y|<\infty$. Then the IVP $y'=f(y)$, $y(0)=0$ has a unique solution, namely $y\equiv 0$.

Proof. Suppose that $y$ solves the IVP but is not identically zero. The set $\{x:y(x)\ne 0\}$ is open and does not coincide with $\mathbb R$. Let $(a,b)$ be a connected component of this set. Consider the function $u(x)=\log|y(x)|$. By the chain rule, $$u'(x)=\frac{y'(x)}{y(x)}=\frac{f(y(x))}{y(x)}$$ Since $(a,b)\ne \mathbb R$, either $a$ or $b$ is a finite point. Suppose $a$ is finite; the other case is similar. Since $y(a)=0$, we have $\limsup_{x\to a+} |u'(x)|<\infty$. It follows that $u'$ is bounded on an interval of the form $(a,a+\epsilon)$. However, $u(x)\to-\infty$ as $x\to a+$. This is a contradiction. $\Box$

share|cite|improve this answer

Your Answer


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.