Consider the initial value problem:
$\frac{dy}{dt} = (t^2 + y^4)^{\frac{1}{2}}, \; y(t_0) = y_0$
Discuss existence, uniqueness, and continuation of solutions through each point $(t_0, y_0) ∈ \mathbb{R}^{2}.$
I know what that I need to show $f$ is locally Lipschitz w.r.t. $y$, in order to apply Fundamental Existence and Uniqueness Theorem however I am stuck on technicalities of my analysis. This is the definition of locally Lipschitz I am using:
${\it Def.}\;$ Let $f: D \subseteq (\mathbb{R} \;{\rm{x}}\; X) \rightarrow Y$. Then $f(t,x)$ is locally Lipschitz on $D$ w.r.t $x$, if for every $(t,x)\in D$, there exists an open set $U\in D$ with $(t,x)\in U$ s.t. $f$ is Lipschitz on $U$ w.r.t $x$ on D.
So far, I have that $f(t,y)=(t^2 + y^4)^{\frac{1}{2}}$ is continuous on its domain $\{ t,y \in \mathbb{R}\;| \;t^2+y^4\geq 0\}.$ Now I need to show $f$ is Lipschitz on an open set $D\subset \mathbb{R} \;{\rm{x}}\; \mathbb{R}^2$ that contains $(t_0,y_0)$, as this is the assumption of the Existance Uniqueness Theorem, but I am not sure if this set $D$ is my domain or if this is an open subset of my domain. Any help would be appreciated!