# Is this piecewise defined function in two variables lipschitz continuous?

According to Picard–Lindelöf theorem, IVP $$\begin{cases}y^\prime(t) = f(t,y(t))\\y(x_0)=y_0 \end{cases}$$ has a unique solution if $$f$$ is lipschitz continuous. What if my ODE contain piecewise defined function $$g:\mathbb{R}^2\rightarrow\mathbb{R}$$ $$\begin{equation} g(x,y) = \begin{cases} x^2\tanh(y)\text{ for } x>0\\ 0 \text{ for } x\leq 0 \end{cases} \end{equation}$$ Is $$g(x,y)$$ lipschitz continuous in $$\mathbb{R}^2$$ (what is the $$K$$ then?) and if not, does it automatically mean, that ODE with such function has no solution?

Edit. I am curious about the situation like $$\mathbf{x}^\prime(t) = \mathbf A^{n\times n}\mathbf{x}(t) + \begin{bmatrix} x_1(t)\\ x_2(t)\\ \vdots\\ g(x_1,x_2)\\ x_n(t)\\ \end{bmatrix}$$

• You don't need Lipschitz continuity in $\Bbb R^2$ but Lipschitz continuity in the second variable ($y$). Aug 27, 2021 at 20:29
• Maybe I messed up my original question. Please see edit Aug 27, 2021 at 20:52

No, $$g$$ is not Lipschitz, as $$x^2$$ is not Lipschitz. But $$g$$ is Lipschitz on $$(-\infty, r)\times\mathbb R$$ for all $$r$$.
Thus the ODE has a solution on each such set, so it also has a solution on $$\mathbb R$$. Picard Lindelöf only needs local lipschitz, and that only in the second variable.