is this a well posed problem ????

Is the following initial value problem well-posed? (i.e. does it satisfy the assumptions of the existence and uniqueness theorem) Why or why not?

y'=(1-y^2)^(1/2), with y(0)= 0

here is more detail about the existence and uniqueness theorem: Existence and Uniqueness

The derivative with respect to y of $\displaystyle (1- y^2)^{1/2}$ is $\displaystyle (1/2)(1- y^2)^{-1/2}(-2y)$$\displaystyle = \frac{-y}{\sqrt{1- y^2}}$. That is a perfectly well behaved problem in a small neighborhood of (0,0). What does the theorem you cite tell you about that?

(The answer would be very different if the initial value condition were y(0)= 1. Do you see why?)

For me, I think that this problem is also well posed.

The existence and uniqueness theorem states that a well posed problem will have a solution existence, uniqueness and continuous dependence on the data (Initial condition).

For y'=(1-y^2)^(1/2) with an initial condition y(t=0)=0
I know that y'=f(t,y) and f(t,y)=(1-y^2)^(1/2). After doing the integral, i have y(t)=sin(t+c), with the IC, I have y(t)=sin(t).

We know that the interval for sin is from 0 to pi( i am not sure I have this correct, is sine a infinite function? but i remember that when i was in high school the teacher said something about the domain for tan is from -pi/2 to pi/2, please clear this concept for me) , and the solution y is from -1 to 1 in this case.

There is an existence of a solution y.
1) there is a unique solution because any c, the solution y is differentiable in the interval from 0 to pi.
2) the solution y depends continuously on the IC. with y(0)=1, y then will become t(t)=sin(t+1).

any comments?

$\displaystyle y' = \sqrt {1 - {y^2}} \Rightarrow \frac{{dy}}{{dx}} = \sqrt {1 - {y^2}} \Rightarrow \int {\frac{{dy}}{{\sqrt {1 - {y^2}} }}} = \int {dx} \Rightarrow$

$\displaystyle \Rightarrow \arcsin y = x + C \Rightarrow y = \sin \left( {x + C} \right).$

$\displaystyle y\left( 0 \right) = \sin \left( C \right) = 0 \Rightarrow C = \pi k{\text{ where }}k \in \mathbb{Z}.$

Finally $\displaystyle y = \sin \left( {x + \pi k} \right)$.