# Cauchy problem with a parameter on the initial data

Consider

$$\begin{cases} y' = y^{\frac{1}{3}}\\ y(0)=k \in \mathbb{R} \end{cases}$$

• For which values of $$k$$ do the problem have a unique local solution?
• Show that for the other values of $$k$$ the problem has more than one solution

i) $$f(t,y)=y^{\frac{1}{3}}$$ is a continuous function over $$\mathbb{R}^2$$, while $$f_y=-\frac{2}{3 y^{2/3}}$$ which is discontinuous at $$0$$. Therefore, in any neigbourhood of $$(0,k)$$ with $$k\ne0$$, $$f_y$$ is continuous, and hence I have local existence and uniqueness of the solution.

ii) First I note that $$f(t,y)$$ is not Lipschitz, therefore I don't expect uniqueness. Indeed, for $$k=0$$, $$y(t)=0$$ is a solution, and,by integration I found also $$y(t)=\sqrt{\Bigl( \frac{3t}{2} \Bigr)^3}$$

**Is everything correct? **

$$y(t)=\sqrt{\big(\frac{3t}{2}}\big)^3=\Big(\frac{3t}{2}\Big)^\frac{3}{2}$$ then $$y'(t)=\frac{3}{2}\times\Big(\frac{3t}{2}\Big)^\frac{1}{2}\times\frac{3}{2}=\frac{3^2}{2^2}\times y^\frac{1}{3}\neq1\times y^\frac{1}{3}$$.
Consider now $$y(t)=\sqrt{\big(\frac{2t}{3}}\big)^3$$ and check that $$y'=y^\frac{1}{3}$$.