Results 1 to 3 of 3

Thread: Existence and uniqueness theorem

  1. #1
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    468
    Thanks
    2
    Awards
    1

    Existence and uniqueness theorem

    Hi,
    I have the following equation $\displaystyle \frac{dy}{dt}=t\sqrt{1-y^2} $
    I found solutions $\displaystyle *S:=\{1 , cos(t^2)\} $.
    I don't really get why it would not violate the existence and uniqueness theorem.
    I would have $\displaystyle 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by vincisonfire View Post
    Hi,
    I have the following equation $\displaystyle \frac{dy}{dt}=t\sqrt{1-y^2} $
    I found solutions $\displaystyle *S:=\{1 , cos(t^2)\} $.
    I don't really get why it would not violate the existence and uniqueness theorem.
    I would have $\displaystyle 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 $\displaystyle y(0)=1.$

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

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

    4. basically there is no open rectangle containing $\displaystyle (0,1)$ in which $\displaystyle f$ is defined everywhere.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member vincisonfire's Avatar
    Joined
    Oct 2008
    From
    Sainte-Flavie
    Posts
    468
    Thanks
    2
    Awards
    1
    1. Yes sorry about that $\displaystyle

    y(0)=1.
    $ is the initial condition.
    2. $\displaystyle \int \frac{dy}{\sqrt{1-y^2}} = arcsin(y) = \frac{t^2}{2} + A = \int t dt$
    $\displaystyle y(t) = sin(\frac{t^2}{2} + A) $
    $\displaystyle y(0) = 1 = sin( A) \implies A = \frac{(4n+1)\pi}{2} n\in \mathbb Z $
    $\displaystyle 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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Existence and Uniqueness Theorem
    Posted in the Differential Equations Forum
    Replies: 13
    Last Post: Oct 20th 2011, 05:49 AM
  2. Existence and Uniqueness Theorem
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: Sep 3rd 2011, 11:49 AM
  3. Existence and Uniqueness Theorem (Help)
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: Nov 12th 2010, 06:00 AM
  4. Existence and Uniqueness Theorem
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: Oct 15th 2010, 07:28 AM
  5. Existence Uniqueness Theorem
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: Feb 9th 2010, 06:18 PM

Search Tags


/mathhelpforum @mathhelpforum