I'm investigating where the IVP $$y'=-y, \enspace y(0)=3$$
has an unique solution. The right hand side satisfies a Lipschitz-condition for the constant $K$ with respect to variable $y$, and clearly the function $-y$ is continuous. This implies there is an interval $I:=(-\delta, \delta$) where the solution is unique. According to the proof, we have that $\delta = \min\{a,\frac{b}{2M}\} $, where $M = \sup f(y)_{y\in R}$. Define $R$ as the rectangle $$R : \vert x \vert \leq a, \enspace \enspace \vert y-3\vert\leq b$$
which means $M=(b+3)$, the maximal value of $f(x,y)=-y$ in R. The interval $I$ is maximal, when $\frac{b}{2M}$ is maximal, for we can then set $a$ as its equal. However, as the function $$q(p)=\frac{b}{2(b+3)}$$
has no extrema and no maximal points, this implies that $I$ can be made as large as we wish. Is it now correct to say that the unique solution exists globally?
Best Answer
Your equation is linear, with continuous coefficients on $\mathbb{R}$. Therefore, your solutions can be extended to all of $\mathbb{R}$ and there is a unique solution through every point.