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Discuss the existence and uniqueness of solutions of the equation $X' = X^{a}$ where $a > 0$ and $x(0) = 0.$

Let $f(x) = X'$ I, then, take the derivative of $f(x)$ which gives me $f'(x) = ax^{a-1}$. Since a-1 on the exponential of X is negative. So this is not differentiable. Based on this, can I say this solution is not unique. If not, what can I say?

If a = 1, $f'(x) = 1$. So every number equals to 1. Then, this is not unique. Is it right?

I separate into three cases. If x > 1, $f'(x) = ax^{a-1}$ . This is a unique solution.

Am my reasoning right for all three cases? What are the key things that I need to test when considering existence and uniqueness?

Thank you

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    $\begingroup$ Why are you differentiating $x^a$? Your work makes no sense to me at all. $\endgroup$ – anon Feb 8 '14 at 7:58
  • $\begingroup$ What you need to do is to check wether the function $f(x)=x^a$ is Lipschitz or not. $\endgroup$ – PepeToro Feb 8 '14 at 11:41
  • $\begingroup$ @anon: the reason I differentiate $x^a$ with respect to x is to check whether it is differentiable since if it is not differentiable, then it is not continuous $\endgroup$ – afsdf dfsaf Feb 8 '14 at 16:39
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I'm 11 months late, so I don't know whether or not you care about this anymore, but your argument for the first and the last case seems legit.

Only for the second case where $a=1$, note that the nonlinear equation $x'=x^{a}$ reduces to the simple linear equation $x'=x$, and as I'm sure you know, $e^{t}$ is the unique solution to this equation.(Note that to any linear equation there's a unique solution.)

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