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Given initial condition $y(0)=0$, solve the initial value problem $$y'(x)=xy^{1/3}.$$

I know that it is not Lipschitz at $y=0$ and by solving I arrived at the solution $$y=({x^3}/{3\sqrt{3}}).$$ Now how can I conclude that it has more than one solution? Please explain.

Kushal Bhuyan
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    How about the solution $y\equiv 0$? – frog Oct 01 '15 at 14:12
  • I disagree with the other solution – Empy2 Oct 01 '15 at 14:24
  • You can check this – Empty Oct 01 '15 at 14:40
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    For every $x_0\geqslant0$, a solution is $$y(x)=\left{\begin{array}{ccc} 0&\text{if}&0\leqslant x\leqslant x_0\ \left(\frac13(x^2-x_0^2)\right)^{3/2}&\text{if}&x\geqslant x_0\end{array}\right.$$ Extreme cases are $$y(x)=\frac{x^3}{3\sqrt{3}}$$ for every $x\geqslant0$, when $x_0=0$, and $$y(x)=0$$ for every $x\geqslant0$, when $x_0\to\infty$. Note that the function in the question is not a solution. – Did Oct 01 '15 at 14:44
  • When can we say that an IVP has a unique solution? I mean if it satisfies the Picard's theorem right? – Kushal Bhuyan Oct 01 '15 at 15:19

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For sure you have the solution $y(x)=0$, as frog suggested. Moreover, by separating variables, it seems to me the solution you get is \begin{equation} y(x)=\frac{1}{3\sqrt3}x^3, \end{equation} which again satisfies your initial datum.

popoolmica
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