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Why does the Cauchy problem $$ \begin{cases} y'= y^{3/4},\\ y(0)=0 \end{cases} $$

not admit a unique solution? And please how we justify that $(y^{3/4})'$ is not continuous?

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  • $\begingroup$ Probably because $y(x)=0$ is a trivial solution. $\endgroup$ – Alexis Olson May 23 '18 at 18:58
  • $\begingroup$ have a look at my thorough study here math.stackexchange.com/questions/2173984/… $\endgroup$ – zwim May 23 '18 at 19:46
  • $\begingroup$ a unique solution exists on an interval if both $y^{3/4}$ and $(y^{3/4})'$ are continuous on the interval. The problem is that neither are continuous at $0$ $\endgroup$ – Vasya May 23 '18 at 19:59
  • $\begingroup$ please how we justify that $(y^{3/4})'$ is not continuous? $\endgroup$ – rosy May 23 '18 at 20:21
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Because $y(x) \equiv 0$ and $y(x) = \frac{x^4}{4^4}$ are solutions.

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  • $\begingroup$ So we have to calculate solutions before deduce the non unicity of solution? $\endgroup$ – rosy May 23 '18 at 19:33
  • $\begingroup$ @rosy How else would you figure it out? $\endgroup$ – ericw31415 May 23 '18 at 20:12
  • $\begingroup$ please how we justify that $(y^{3/4})'$ is not continuous? $\endgroup$ – rosy May 23 '18 at 20:22
  • $\begingroup$ @rosy : Do you think that it is not continuous? $\endgroup$ – DisintegratingByParts May 23 '18 at 21:16
  • $\begingroup$ Vadya tells in an comment that $(y^{3/4})'$ is not continuous. Furetheremore, if there is no unicity, then $(y^{3/4})'$ is not continuous. No? $\endgroup$ – rosy May 23 '18 at 21:41

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