Consider the following first order ODE:
$y' = f(t,y)$ subject to $y(t_{0}) = y_{0}$.
I would like to show that there exists a unique function $y(\cdot)$ that passes through ($t_{0}, y_{0}$)
The thing is $f$ is neither globally Lipschitz nor continuous. Therefore, I cannot apply standard theorems to prove the global existence. Instead, $f$ is locally Lipschitz, which implies the local existence at each point: For each point ($t_{0}, y_{0}$), there exists a unique function $y(\cdot)$ that satisfies $y' = f(t,y)$ for $t \in [t_{0} – \epsilon, t_{0}+\epsilon$] and the initial condition. My question is by pasting the local solutions, can we get a global solution? What are the necessary and sufficient conditions that make pasting a legitimate action for obtaining the global solution? I would appreciate your comments about this.
Best Answer
$y'=y^2$ is a pleasant ODE, but $$ y = \frac{1}{1-t} $$ is a solution with $y(0)=1$
Similar, $y'=1+y^2, \; \; y(0) = 0$ gives $y = \tan t.$