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Math Help - is this a well posed problem ????

  1. #1
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    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
    Last edited by oxxiissiixxo; January 10th 2009 at 07:40 PM.
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  2. #2
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    The derivative with respect to y of (1- y^2)^{1/2} is (1/2)(1- y^2)^{-1/2}(-2y) = \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?)
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  3. #3
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    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).
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  4. #4
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    any comments?
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  5. #5
    Senior Member DeMath's Avatar
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    y' = \sqrt {1 - {y^2}}  \Rightarrow \frac{{dy}}{{dx}} = \sqrt {1 - {y^2}}  \Rightarrow \int {\frac{{dy}}{{\sqrt {1 - {y^2}} }}}  = \int {dx}  \Rightarrow

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

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

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