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Let 0<ε<1 for fixed ε and the following initial value problem :

   { y'(t)=abs(y(t))^(1-ε) & y(0)=0    for 0<=t<=b 

show that the problem does not have a unique solution for any space [0,b].

By assuming y>0, I solved the above problem and found y(t)=(εt)^(1/ε). I know that in order to prove there's no unique solution y(t) in [o,b] for any b, I have to show that y doesn't satisfy the Lipschitz condition( nor local nor overall).

At this point I've been stuck...How do i prove that this function isn't Lipschitz for any 0<ε<1?

So I would appreciate any help... Thanks in advance!!

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You're misunderstanding. To show non-uniqueness, you need to exhibit two separate solutions. The first is the one you found, the second is $y(t) = 0$.

The theorem says that if $f(y)$ is Lipschitz, then $y' = f(y)$ has a unique solution. It does NOT say that if $f(y)$ is not Lipschitz, then $y' = f(y)$ does not have a unique solution.

Let $f(y) = y^{1-\epsilon}$ for $\epsilon \in (0,1)$. We see that $f$ is not Lipschitz because its derivative gets arbitrarily large. In detail: we see that $$f'(y) = \frac{1-\epsilon}{y^\epsilon}, y \in (0,b].$$ Since $f'(y) \to +\infty$ as $y \to 0^+$, For any $K > 0$, we can find $\delta \in (0,b)$, such that $$f'(y) > K, \,\,\,\,\,\, \text{ for all } y \in (0,\delta).$$ For $0 <y_1 < y_2 < \delta$, by the mean value theorem, there is $c \in (y_1, y_2)$ such that $$\lvert f(y_1) - f(y_2) \rvert = \lvert f'(c) \rvert \lvert y_1 - y_2 \rvert > K\lvert y_1 - y_2 \lvert.$$ Thus $K$ is not a Lipschitz constant for $f$. Since $K > 0$ was arbitrary, $f$ has no Lipschitz constant and is thus not Lipschitz.

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  • $\begingroup$ Yes you're right. I assumed the opposite of the theorem which is wrong! So I find these two separate solutions and the problem solved? But later the ex. asks to explain why Picard-Lindelof theorem of uniqueness does not apply to this problem ... why is that true? $\endgroup$ – kaithkolesidou Mar 24 '16 at 22:58
  • $\begingroup$ You were right about the latter question: the Picard-Lindelof theorem does not apply since $f(y) = y^{1-\epsilon}$ is not a Lipschitz map on $[0,b]$ for any $b >0 $ and $\epsilon \in (0,1)$. The exercise it pointing out to you what can happen when $f(y)$ is not Lipschitz. It is certainly not guaranteed to happen. $\endgroup$ – User8128 Mar 24 '16 at 23:04
  • $\begingroup$ great but could you help me show that f(y) is not Lipschitz? How do i prove this? $\endgroup$ – kaithkolesidou Mar 24 '16 at 23:08
  • $\begingroup$ Sure. It'll take just a sec; I'll append my above answer. $\endgroup$ – User8128 Mar 24 '16 at 23:09
  • $\begingroup$ ok.... thank you very very much for your time!!!! You helped me a lot $\endgroup$ – kaithkolesidou Mar 24 '16 at 23:12

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