Does there exist a unique solution for $y' = -y^{1/3}, y(0) = 0$?

Question

The given problem is to determine if the following initial value problems have a unique solution :

$y' = y^{\frac13}, y(0) = 0,$
$y' = -y^{\frac13}, y(0) = 0,$

We can not apply Picard's theorem here. For the first one, the answer in this post tells that solutions are infinite.

For the second one, I tried to construct a similar function given in the answer but was not successful. Since only the sign is different in the second one than the first one, I feel that the solutions are infinite but unable to prove or disprove them. Any help is higly appreciated.

@Surb Can you explain how $y(x) = x^{3/4}$ is a solution? I see that $y' = \frac34 x^{-1/4} \ne -y^{1/3}$ — VvCh, Oct 01 '22 at 15:50
@K.defaoite Can you again check if $y = (4/3)x^{3/4}$ satisfies the equation? I dont get it. — VvCh, Oct 01 '22 at 15:57

score 4 · Accepted Answer · answered Oct 01 '22 at 15:58

4

The second equation also has infinitely many solutions, in a very similar form as the infinite family for the first. Let $a<0$, then $$y(x)=\begin{cases}(2/3)^{3/2}(\color{blue}{a-x})^{3/2}&x\le a\\0&x>a\end{cases}$$ is the desired infinite family.

answered Oct 01 '22 at 15:58

Parcly Taxel

105,904

I also thought the same function initially but this does not satisfy $y(0) = 0 $ right? – VvCh Oct 01 '22 at 16:00
@VvCh It does, since $a$ is now negative! – Parcly Taxel Oct 01 '22 at 16:01
Ok, got it, I was taking $a$ as positive. Thank you. – VvCh Oct 01 '22 at 16:02
@VvCh don't forget to accept my answer too :) – Parcly Taxel Oct 01 '22 at 16:04

Does there exist a unique solution for $y' = -y^{1/3}, y(0) = 0$?

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