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I am just analyzing the IVP $x'=x^{\frac{2}{3}}$ with initial condition $x(0)=0$. It is obvious that there is a solution which is not unique, since it is not Lipschitz (not bounded near 0). My book says it has a solution that satifies $x(t)=0$ iff $t\in [t_1,t_2]$ for eached fixed $t_1<0<t_2$.

When I divide both sides by $x^{2/3}$ tae integrals etc. I get $x(t)=(\frac{t}{3})^3$. I do not see the the above equivalance.

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  • $\begingroup$ There seem to be $\geq1$ typos in your question. $\endgroup$ Oct 13, 2013 at 15:39
  • $\begingroup$ it has a solution that satifies $x(t)=0$ iff $t\in [t_1,t_2]$ for eached fixed $t_1<0<t_2$ - can you please put some quantifiers on this? I can't understand at all what you're asking. $\endgroup$
    – Git Gud
    Oct 13, 2013 at 15:46
  • $\begingroup$ How does this specific solution looks like? I want to show that the IVP has a solution that satifies $x(t)=0$ iff $t\in[t_1,t_2]$ for eached fix $t_1<0<t_2$, but I do not know how $\endgroup$
    – Alkibiades
    Oct 13, 2013 at 15:48
  • $\begingroup$ Ok, you have a if and only statement. On the left of $\iff$ you have $x(t)=0$. You need to quantify this. Is it $\forall t(x(t)=0)$? Maybe $\exists t(x(t)=0$? And where does $t$ range? $\endgroup$
    – Git Gud
    Oct 13, 2013 at 15:51
  • $\begingroup$ Well I guess it should mean $\exists t(x(t)=0)$. If it would be for all $t$ than the solution would be simply 0. It might be possible to conclude from the fact that there is such a $t$ that this $t$ must be an element of an interval $[t_1,t_2]$ $\endgroup$
    – Alkibiades
    Oct 13, 2013 at 15:55

1 Answer 1

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$$x(t) = \begin{cases} \left(\frac{t-t_1}{3}\right)^3 & \text{if } t<t_1 \\ 0 & \text{if } t_1\leq t\leq t_2 \\ \left(\frac{t-t_2}{3}\right)^3 & \text{if } t>t_2 \end{cases}$$

works.

You divide by $x^{2/3}$, but you cannot divide by $0$, so the validity of that division only holds when you know $x\neq 0$. The above shows all possible solutions; $x$ can only be zero on an interval, because once positive it stays positive to the right, and once negative it stays negative to the left. The initial condition just tells you that $t_1\leq 0\leq t_2$. ($t_1$ could be $-\infty$ and $t_2$ could be $+\infty$.)

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  • $\begingroup$ What is $x^{2/3}$ when $x\lt0$? $\endgroup$
    – Did
    Oct 13, 2013 at 16:10
  • $\begingroup$ @Did: It is either the real cube root of the square of $x$, or the square of the real cube root of $x$, which are equal. $\endgroup$ Oct 13, 2013 at 16:12
  • $\begingroup$ Right, I understand this unorthodox convention is needed to make the exercise true. $\endgroup$
    – Did
    Oct 13, 2013 at 16:17

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