Non-Lipschitz $f$ with $y' = f(y)$, $y(0) = 0$ has unique solution What is an example of an $f$ such that $f$ is not Lipschitz in any neighbourhood of 0 and $f(0) = 0$, but $y' = f(y)$ with $y(0) = 0$ has a unique solution?
 A: Consider
$$ \frac{dy}{dt} = f(y) = \cases{ y \log\left(2 + \frac1{|y|}\right) & if $ y \ne 0$ \cr 0 & if $y = 0$.}$$
I will show that the only solution with $y(0) = 0$ is $y(t) = 0$.  I will do this by showing that if $y(t) \ne 0$ for some $t \in \mathbb R$, then $y(t) \ne 0$ for all $t \in \mathbb R$.
Suppose there exists a $t^*$ such that $y(t^*) \ne 0$.  By replacing $y$ with $-y$ if necessary, we may suppose without loss of generality that $y(t^*) > 0$.  Let $(a,b)$ be the largest open interval on which $y$ is positive - remember that $y$ is continuous.
Suppose first that $b<\infty$, so $y(b) = 0$.  But then
$$ y(b) = y(t^*) + \int_{t^*}^b f(y(t)) \, dt \ge y(t^*) > 0,$$
a contradiction.
Next, suppose that $a > -\infty$, so $y(a) = 0$.
Let
$$ g(y) = \int_1^y \frac{dz}{z \log\left(2 + \frac1{|z|}\right)}, \qquad y>0.$$
Then the differential equation can be rewritten as
$$ \frac d{dt} [g(y(t))] = 1,  \qquad t \in (a,b) .$$
Integrate both sides with respect to $t$ from $a$ to $t^*$.  Then we get
$$ g(y(t^*)) - \lim_{t \searrow a} g(y(t)) = t^* - a .$$
But $y(a) = 0$, and $\lim_{y \searrow 0} g(y) = -\infty$, so we obtain a contradiction.
Hence $(a,b) = \mathbb R$.
Finally, it is easy to see that $f$ is both continuous, and non-Lipschitz, in any neighborhood of $0$.
A: Hint.
Consider the differential equation $y^\prime =\sqrt{y}+1$ with $y(0)=0$. Prove uniqueness of the solution considering $z(x)=(\sqrt{y_1(x)}-\sqrt{y_2(x}))^2$ where $y_1,y_2$ are solutions.
