On page 18 of the book Ordinary Differential Equations by Wolfgang Walter the following example is presented:
$$y'=-x(sgn( y))\sqrt{|y|}= \begin{cases} -x\sqrt{y} & \text{for } y\geq 0 \\ x\sqrt{-y} & \text{for } y< 0 \end{cases},$$
whose solution is
$$y(x;C)=\frac{1}{16} (C-x^2)^2 \text{ in } (-\sqrt{C}, \sqrt{C}) \text{ } (C>0)$$
It is said that this function can be extended by setting $y(x;C)=0$ for $|x|\ge \sqrt{C}$, in which case the solution will be defined in $\mathbb{R}$. Then it is also said that there are no other solutions, since $g(y)=\sqrt{|y|}$ vanishes only for $y=0$. Thus each IVP with $y(\xi)=\eta\ne 0$ is locally uniquely solvable.
I didn't understand from the above how it follows that $y(\xi)\ne 0$ is locally uniquely solvable.
Also, it is said that for the IVP with $y(\xi)=0$ there are infinitely many solutions in the case where $\xi \ne 0$, but only one solution in the case where $\xi=0$.
I also didn't get how the above part follows for $\xi=0$ and $\xi \ne 0$.
Could someone please clarify the above two points?
Note that $g(y)=\text{sgn}(y)\sqrt{|y|}=0$ if and only if $y=0$. For the IVP with $y(\xi)=\eta\neq0$, there is $\delta>0$ such that $|y(x)|>\frac{|\eta|}2$ for $\forall x\in[\xi-\delta,\xi+\delta]$. Let $f(x,y)=xg(y)$ and then satisfies the Lipchitz condition about $y$. In fact, for $y_1,y_2$ with $|y_1|>\frac{|\eta|}2$ $|y_2|>\frac{|\eta|}2$ \begin{eqnarray*} &&|f(x,y_1)-f(x,y_2)|\\ &=&|x||\sqrt{|y_1|}-\sqrt{|y_2|}|=|x|\frac{||y_1|-|y_2||}{\sqrt{|y_1|}+\sqrt{|y_2|}}\\ &\le&(|\xi|+\delta)\frac{|y_1-y_2|}{2\sqrt{\frac{\eta}{2}}}\\ &=&C|y_1-y_2| \end{eqnarray*} where $$ C= \frac{(|\xi|+\delta)}{2\sqrt{\frac{\eta}{2}}}. $$ Then by the existence and uniqueness theorem, the IVP is solvable in $[\xi-\delta,\xi+\delta]$. This is what it means that the IVP with $y(\xi)=\eta\neq0$ is locally solvable.