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Math Help - Existence and uniqueness theorem

  1. #1
    Senior Member vincisonfire's Avatar
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    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.
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  2. #2
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    Quote Originally Posted by vincisonfire View Post
    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.
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    Senior Member vincisonfire's Avatar
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    1. Yes sorry about that <br /> <br />
y(0)=1.<br />
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.
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