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Math Help - Uniqueness / Lipschitz condition

  1. #1
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    Uniqueness / Lipschitz condition

    Hi,

    I'm have some trouble understanding what my lecturer did.

    The question is:

    For the IVP, y' = \frac{x+y}{2y}, 0 \le x \le 1, y(0)=1,

    find an appropriate Lipschitz constant L in the domain \{(x,y) : 0\le x \le 1, y \ge 1\}. Hence determine whether or not the soln y(x) is unique.

    We've found L = 1/2 (by finding f_y and maximizing in our domain, but doesn't this function have a singularity at y=0? Which means the uniqueness thm doesn't apply?

    Can someone explain?

    EDIT: Also, he did things like:

    y(0)=1 => (x+y)/2y > 0 => y' > 0
    => y increases
    => y >= 1

    Which I don't understand.
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  2. #2
    MHF Contributor chisigma's Avatar
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    near Piacenza (Italy)
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    The DE can be written as...

    y^{'} = \frac{1}{2} + \frac{x}{2y}, x \in [0,1], y(0)=1 (1)

    This type of equation has been solved in eighteenth century by the Italian mathematician Gabriele Manfredi with the substitution...

    \frac{y}{x} = t \rightarrow dy= t\cdot dx + x\cdot dt (2)

    Applying (2) the (1) becomes...

    x\cdot \frac{dt}{dx} = \frac {1+t-2t^{2}}{2t} \rightarrow \frac{t\cdot dt}{(t-1)\cdot (t+\frac{1}{2})} = - \frac{dx}{x} (3)

    In (3) ther variables are separated, so that we can integrate both terms and obtain the general solution of (1)...

    \{(y-x)^{2}\cdot (y+\frac{x}{2}\}^{\frac{1}{3}} = c (4)

    Imposing the condition  y(0)=1 we obtain finally the solution in implicit form...

    (y-x)^{2} \cdot (y + \frac{x}{2}) -1 =0 (5)

    ... that is represented here...



    The Cauchy-Lipschitz condition in [0,1] are satisfied and the (1) has only one solution for y(0)=1. However in the case y(0)=0 that isn't and [easy to verify...] we have two different solution of (1)...

    y=x

    y=-\frac{x}{2} (6)

    Kind regards

    \chi \sigma
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