consider the initial value problem $ y'=y^{1/3}, y(0)=0$

Question

consider the initial value problem $y'=y^{\frac{1}{3}} , y(0)=0 $ This is already asked Here

I have a doubt in those answer, as that question is 6 years old, i ask this seperately.

Its is written in comment that the function is ill-defined in negative axis. What does it means? Please explain this intuitively

Also in that accepted answer how answerer got that idea of that function. It's not trivial

Federico · Answer 1 · 2018-12-05T17:07:21.330

The only reasonable definition of $y^{1/3}$ for $y<0$ in this context is $$ y^{1/3} = -|y|^{1/3} \qquad \forall y<0. $$ Whoever claims differently is just confusing people.

The non uniqueness of solutions is related to the non-Lipschitzianity of the function; it has nothing to do with it being "ill-defined in negative axis", whatever that means. You would have the same problem with the ODE $y'=|y|^{1/3}$. What happens on the negative $y$ axis does not matter at all in this exercise.

The example of a parametrized family of solutions starting at $y(0)=0$ is a standard exercise. The key insights are that:

the ODE is autonomous, so you can translate solutions,
$y=0$ is a solution,
you can "join" the $y=0$ solution with any one departing from it.

In this case we can say that the derivative is not continuous at 0, hence the uniqueness is not necessarily true, and so we must look for other possibilities. If I remember correctly Lipschitz condition is weaker than continuity and differentiability, so your answer is better, I am just confirming. — rhombicosicodecahedron, Feb 05 '19 at 19:17

mathcounterexamples.net · Accepted Answer · 2018-12-05T17:01:42.860

2

In general, $y \mapsto y^\beta$ with $\beta > 0$ real is only defined for $x\ge0$. How would you define $(-2)^\pi$? This is why the function is said to be ill-defined for $x<0$.
To find the solutions of $y^\prime=y^{\frac{1}{3}} , y(0)=0$, you can notice that functions always vanishing on an interval are solutions. And for $y(t) \neq 0$, you can rewrite the equation as $$\frac{y^\prime(t)}{(y(t))^{1/3}} = 1$$ and integrate it on $(\alpha,x)$. You get the solution $y(x) = (2/3)^{3/2} (x-\alpha)^{3/2}$. You then see that the functions

$$y(x)=\begin{cases} 0 & \text{ for } x < \alpha \\ (2/3)^{3/2} (x-\alpha)^{3/2} & \text{ for } x \ge \alpha \end{cases}$$

are also solutions.

edited Dec 05 '18 at 17:01

answered Dec 05 '18 at 16:40

mathcounterexamples.net

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But $\frac13\neq\pi$, and for $\frac13$ there is a preferred representative of the root also for negative arguments – Federico Dec 05 '18 at 16:58
you can notice that functions always vanishing on an interval are solutions---- please explain this a little – Cloud JR K Dec 05 '18 at 16:58
@CloudJR Let $y(t)=0$ on an interval. What is $y'$? Well, of course $y'(t)=0$, which happens to coincide with $y(t)^{1/3}=0$. So they are solutions. – Federico Dec 05 '18 at 16:59
If $y(t) = 0$ for $t \in (a,b)$ then $y^\prime(t) = 0$ also. So $y$ is a solution. – mathcounterexamples.net Dec 05 '18 at 17:00

consider the initial value problem $ y'=y^{1/3}, y(0)=0$

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