1
$\begingroup$

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? **

$\endgroup$
1
$\begingroup$

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 $y'=y^\frac{1}{3}$.

$\endgroup$
1
  • 1
    $\begingroup$ Ups, thanks for the check. I need to check my computations... $\endgroup$ – andereBen Aug 20 '20 at 16:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.