Consider a first order linear differential equation $y'+Py=Q,\,y(a)=y_0$ .Here continuity of $P$ and $Q$ ensure that the the ODE has an unique solution.But in case of non-linear initial value problem i.e $y'=f(x,y),\,y(x_0)=y_0$,continuity of $f$ does not ensure the unique solution.So we need to address the following question:
(1)Under what condition on $f$ the problem $y'=f(x,y),\,y(x_0)=y_0$ has a solution$?$
(2)If solution exists,whether it is unique or not$?$
$\to$First question is answered by the Peano existence theorem which states that "let $f$ be a continuous function in an interval $I$ containing the points $(x_0,y_0)$,then the the problem $y'=f(x,y),\,y(x_0)=y_0$ has a solution".
$\to$Second question is answered by Picard's uniqueness theorem which states that "let $f$ and $\frac{\delta f}{\delta y}$ are continuous in aregion R containing the initial points $(x_0,y_0)$ then the the problem $y'=f(x,y),\,y(x_0)=y_0$ has an unique solution"
Picard method for interval of definition:let $f$ and $\frac{\delta f}{\delta y}$ are continuous in a closed rectangle $$R=\{(x,y):|x-x_0|\leq a,|y-y_0|\leq b\}$$.Then the IVP $y'=f(x,y),\,y(x_0)=y_0$ has an unique solution in the interval $|x-x_0|\leq h=min{(a,\frac{b}{l})}$ where $l=MAX_{(x,y)\in R}|f(x,y)|$
Hope this will help you!!!
Best Answer
You need a global Lipschitz condition with the same constant to apply a global version of the theorem.
Your example $\dot x = x^3$ does not possess a global Lipschitz constant since its derivative $3x^2$ is unbounded. And indeed, the solution $x^{-2}=C-2t$ has a couple limitations on its domain.
However, there is a theorem of Cauchy that tells you that if the ODE function is continuously differentiable (and thus locally Lipschitz), then any solution can be uniquely continued to the boundary of its domain, which may be the $x$ infinity at finite time.