# Thread: Existence and uniqueness theorem

1. ## Existence and uniqueness theorem

Hi,
I have the following equation $\frac{dy}{dt}=t\sqrt{1-y^2}$
I found solutions $*S:=\{1 , cos(t^2)\}$.
I don't really get why it would not violate the existence and uniqueness theorem.
I would have $R = \{(t,y), -\infty < t < \infty, -1 < y < 1 \}$.
Isn't f and f' continuous on R? If yes it should thus admit one solution.

2. Originally Posted by vincisonfire
Hi,
I have the following equation $\frac{dy}{dt}=t\sqrt{1-y^2}$
I found solutions $*S:=\{1 , cos(t^2)\}$.
I don't really get why it would not violate the existence and uniqueness theorem.
I would have $R = \{(t,y), -\infty < t < \infty, -1 < y < 1 \}$.
Isn't f and f' continuous on R? If yes it should thus admit one solution.
1. you forgot that the existence and uniqueness theorem is for differential equations with initial condition! i guess in your problem it's $y(0)=1.$

2. the function $y=\cos (t^2)$ is not a solution to your differential equation. [solve your equation again!]

3. one reason that the solution is not unique is that $\frac{\partial f}{\partial y}=\frac{-ty}{\sqrt{1-y^2}}$ is not continuous at $(0,1),$ and so it won't be continuous in any region containing that point.

4. basically there is no open rectangle containing $(0,1)$ in which $f$ is defined everywhere.

3. 1. Yes sorry about that $

y(0)=1.
$
is the initial condition.
2. $\int \frac{dy}{\sqrt{1-y^2}} = arcsin(y) = \frac{t^2}{2} + A = \int t dt$
$y(t) = sin(\frac{t^2}{2} + A)$
$y(0) = 1 = sin( A) \implies A = \frac{(4n+1)\pi}{2} n\in \mathbb Z$
$y(t) = sin(\frac{t^2}{2} + \frac{\pi}{2}) = cos(\frac{t^2}{2})$
But this does not satisfy the equation. Where am I mistaken?
3. & 4. Got it the largest "rectangle" is the point (0,1) where of course only one solution exists.