# show that the Picard-Lindelöf Theorem does not imply uniqueness

Show that the initial value problem $\frac{dx}{dt} = \sqrt{x}$ for $x \ge 0$, $x(0) = 0$ has more than one solution $x(t), t \ge 0$.

Why does the Picard-Lindelöf Theorem not imply uniqueness?

for $\zeta \gt 0$
using separation of variables $\int_ {x_0} ^ x \zeta^{-\frac{1}{2}}\; d\zeta\; = \int_ {t_0} ^ t \; dy$

$2\zeta^{\frac{1}{2}}\; = t + t_0$

$\sqrt{x} - \sqrt{x_0} = \frac{t+t_0}{2}$

$x(t) = (\frac{t+t_0}{2} + \sqrt{x_0} )^2$

This function only being a solution if $\frac{t+t_0}{2} + \sqrt{x_0} x \ge 0$

Thus the solution is only valid for $t \ge -2\sqrt{x_0} - t_0$

Hence, $x(t) = (\frac{t+t_0}{2} + \sqrt{x_0} )^2\; , t\ge -2\sqrt{x_0} - t_0$

$x(t) = 0 \; , t\ge -2\sqrt{x_0} - t_0$

Checking C' at $-2\sqrt{x_0} - t_0.$

First showing it is differentiable by

$lim_{h\to0} \frac{x(-2\sqrt{x_0} - t_0 + h)-x(-2\sqrt{x_0} - t_0)}{h}$

The limit $h\to0^+$ is $0$ and the limit $h\to0^-$ is

$lim_{h\to0,h\lt0} \frac{x(-2\sqrt{x_0} - t_0 + h)-x(-2\sqrt{x_0} - t_0)}{h}$

$= lim_{h\to0,h\lt0} \frac{h^2}{h} = 0$

The derivative $x'(t) = \frac{t+t_0}{2} + \sqrt{x_0} \; , t\ge -2\sqrt{x_0} - t_0$

$x'(t) = 0 \;, t\le -2\sqrt{x_0} - t_0$

is continuous.

Hence the initial value problem with $x(0) = 0$ has solutions

$x(t) = (\frac{t+t_0}{2} + \sqrt{x_0} )^2\; , t\ge -2\sqrt{x_0} - t_0$

$x(t) = 0 \; , t\ge -2\sqrt{x_0} - t_0$

for any $x_0 \ge0$ and $t_0 \lt -2\sqrt{x_0}.$

$x(t) = 0$ for all $t \in \mathcal R$ is also a solution.

So i can get this far but i really am not sure how to show that the Picard-Lindelöf Theorem does not imply uniqueness...

Any help would be great.

Thanks

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Does your equation satisfy all the conditions in Picard-Lindelöf? –  mrf Feb 24 '13 at 21:47
No. Surely if it did the Picard-Lindelof Theorem would imply that it had a unique solution. –  user62909 Feb 24 '13 at 22:22
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If you want attention, see How to grab users' attention on an old question?. Don't jump up and down and shout, hey everyone look at my question right now. That's very annoying. –  Rahul Feb 24 '13 at 23:54
The Picard-Lindelof Theorem implies uniqueness under certain assumptions. What you want to show is that at least one of those assumptions does not hold for your equation. Which should lead you to examine the assumptions one by one and try them on your equation. –  user53153 Feb 25 '13 at 0:12
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## 2 Answers

For Picard-Lindelöf you need, that your right hand side is local lipschitz continuous. But as the deriviative of the root tends to infinity as $x\to 0$ it isn't local lipschitz continous around $0$.

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The two functions $$x_1(t):=\cases{0&(t\leq0)\cr {\textstyle{1\over 4}} t^2\qquad&(t\geq0)\cr}$$ and $$x_2(t):=0\qquad(-\infty<t<\infty)$$ both satisfy the initial value problem $$x'(t)=\sqrt{x(t)}\ ,\quad x(0)=0\ ;\tag{1}$$ the reason being that $(1)$ violates an essential assumption of the Picard-Lindelöf theorem, as explained in other answers and comments.

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