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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?

Flast9
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1 Answers1

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"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. $\Box$