Prove that $f(t,y(t))=t \sqrt{1-y^2(t)}$ is not Lipschitz on $y$ I have the differential equation with initial condition,
$$\frac{dy}{dt} = t \sqrt{1-y^2(t)}\, , \, \, \, y(0)=1.$$
Now, this equation has 2 different solutions, $y(t) \equiv 1$ and $y(t)= \cos \big(\frac{t^2}{2} \big).$ The question is:

Why doesn't this contradict the existence and uniqueness theorem?

My attempt: My conjecture is that $f(t,y)=t \sqrt{1-y^2}$ is not Lipschitz on $y$, but I'm not very familiar with the Lipschitz condition, so I'm not sure my procedure is well justified.
If I take $t=1$ fixed (can I do this?), then we have that $f(1,y) = \sqrt{1-y^2}$. Now, if we assume that this function is Lipschitz, there is an $M>0$ such that for any $h \neq 0 $,
$$\frac{|f(1,y+h)- f(1,y)|}{|h|}\leq M,$$
but, this means that 
$$f'(1,y) = \lim_{h\to0} \frac{|f(1,y+h)- f(1,y)|}{|h|} \leq \lim_{h \to 0} M = M,$$
meaning that $f'$ is bounded by $M$, but 
$$f'(1,y)= -\frac{y}{\sqrt{1-y^2}}  \to -\infty \text{ when} \, y \to 1 \text{ and} \, f \to \infty, \text{ when} \,  y\to -1$$
so $f'(1,y)$ cannot be bounded, therefore $f$ is not Lipschitz on $y$. 

Is this reasoning correct? Have I use the Lipschitz condition correctly?

Thanks for any help you can give me.  
 A: First I should point out that your equation needs to be $dy/dt = - t \sqrt{ 1 - y^2}$ in order to get the solution $y(t) = \cos \left( \tfrac {t^2} 2 \right)$ for $t \in [0,\sqrt{2\pi})$.
Anyway, since your initial condition is $y(0) = 1$, you should test the Lipschitz condition on a neighbourhood of $t = 0$ and $ y = 1$.
In your example, the function is $f(t,y) = - t \sqrt{ 1 - y^2}$. You need to test if this $f(t,y) $ is uniformly Lipschitz contracting in the $y$-variable for $t \in [0, a] $ and $y \in [1-b, 1]$ for some $a > 0$ and $b > 0$.
(Wikipedia asks for the Lipschitz condition to be checked on symmetric intervals of the form $[-a, a]$ and $[1-b, 1+b]$, but since $f(t,y) \leq 0$ when $t \geq 0$, I see no reason why the proof in Wikipedia shouldn't work if we restrict to $[0,a]$ and $[1-b, 1]$. )
Anyway, let's go ahead and test this Lipschitz condition. Let's ask ourselves: Does there exist a $k \geq 0$ and $a, b > 0$ such that
$$ | f(t,y_1) - f(t, y_2) | \leq k | y_1 - y_2 |$$
for all $t \in [0, a]$ and all $y_1, y_2 \in [1 - b, 1]$?
In other words, does there exist a $k \geq 0$ and $a, b > 0$ such that
$$ \left| t \sqrt{1 - y_1^2} - t \sqrt{1 - y_2^2} \right| \leq k | y_1 - y_2 |$$
for all $t \in [0, a]$ and $y_1, y_2 \in [1 - b, 1]$?
The answer is clearly "no". For example, if you fix $t = a$ and $y_1 = 1$, then you would need to find a $k \geq 0 $ such that
$$ a\left|\sqrt{1 - y_2^2} \right| \leq k |1 - y_2 |$$
for all $y_2 \in [1-b,1]$, i.e. you would need to find a $k \geq 0$ such that
$$ \frac a k \leq \sqrt{\frac{1-y_2}{1+y_2}}$$
for all $y_2 \in [1-b,1]$. As the right-hand side tends to zero as $y_2$ tends to $ 1$ from above, there clearly does not exist such a $k$, no matter what you pick for $a$ and $b$.
So the Lipschitz condition is not satisfied, and the uniqueness theorem does not hold.
