# Proving $f(x,y)$ is not Lipschitz but still has a unique solution to initial value problem

Consider $g:\mathbb R\to\mathbb R$ defined by $g(y)=1+\sqrt |y|$. Let $h:\mathbb R\to \mathbb R$ be a continous function such that $h(x)\neq 0$ for all $x\neq 0$. I'm asked to prove the following:

• Prove the function $f(x,y)=\frac {h(x)}{g(y)}$ is not Lipschitz respect to $y$ around the origin.
• Prove the initial value problem associated with $(0,0)$ has a unique solution. Find the solution in the case $h(x)=1$.

The first point is clear since $\partial f(x,y)/\partial y \to \infty$ as $y\to 0$. My problem lies on the second point. My only tool for proving uniqueness is the Picard theorem but $f(x,y)$ is not Lipschitz so I don't know how to tackle this problem. I tried using the method of separating variables but don't really know if it ensures uniqueness. Any help is appreciated.

You can use the method of separating the variables. Namely, let $$G(y) := \int_0^y g(\xi)\, d\xi,\quad y\in\mathbb{R}, \qquad H(x) := \int_0^x h(s)\, ds, \quad x\in\mathbb{R}.$$ Since $G'(y) = g(y) \geq 1$ for every $y$, the function $G$ is strictly monotone increasing and $\text{Im}(G) = G(\mathbb{R}) = \mathbb{R}$, hence $G$ is bijective with inverse $G^{-1} \colon \mathbb{R}\to \mathbb{R}$.
Moreover, any solution $y(x)$ of the Cauchy problem satisfies the implicit relation $G(y(x)) = H(x)$, at least for those values of $x$ satisfying $H(x) \in \text{Im} G$.
Since $\text{Im} G = \mathbb{R}$, the implicit relation gives a unique solution $y(x) = G^{-1}(H(x))$ defined for all $x\in\mathbb{R}$.