Consider the IVP $$y’=f(x,y), y(x_0)=y_0, x\in\Bbb R$$ where $f$ is continuous on an open connected set $D\subseteq\Bbb R^2$ and begin Lipschitz on any rectangle inside $D$. Then the above IVP has a unique solution (say) $ y$ in a certain neighbourhood $(x_0-h,x_0+h)$of $x_0$ if $(x_0,y_0)\in D$, $h $ is as given in Picard uniqueness theorem. Now my question is that can $y$ be continuously extend over whole domain $D$ in a unique way i.e. will be unique until it remains inside $D?$.
According to me it’s true as if $(x_0+h, f(x_0+h)\in D$ then again using uniqueness theorem on IVP $$y’=f(x,y), y(x_0+h)=f(x_0+h)$$ we can uniquely extend $y$ at the right end point of $(x_0-h,x_0+h)$ and similarly at left end point . Do the same process unitill $y$ remains inside $D$ and hence we have unique extension of $y$ over $D$ . Please suggest am i right ? If wrong please provide counter-example. Thank you .