Cauchy problem with a parameter on the initial data

Question

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

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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? **

Answer 1

The first part is correct and the second part is not. But you had a good idea.

$y(t)=\sqrt{\big(\frac{3t}{2}}\big)^3=\Big(\frac{3t}{2}\Big)^\frac{3}{2}$ 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 $y'=y^\frac{1}{3}$.