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

  • 1
    $\begingroup$ 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

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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.