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Lipschitz Condition, Uniqueness and Existence of ODE

  1. Feb 7, 2013 #1
    1. The problem statement, all variables and given/known data

    Find a solution of the IVP

    [itex]\frac{dy}{dt}[/itex] = t(1-y2)[itex]\frac{1}{2}[/itex] and y(0)=0 (*)

    other than y(t) = 1. Does this violate the uniqueness part of the Existence/Uniqueness Theorem. Explain.

    2. Relevant equations

    Initial Value Problem [itex]\frac{dy}{dt}[/itex]=f(t,y) y(t0)=y0 has a solution if f is continuous on B = [t0,t0 + a] x [y0-b,y0+b]


    3. The attempt at a solution

    So when I solved * by separation of variables, my solution

    y(t)= sin(t2+[itex]\frac{πk}{2}[/itex]) where k = ±1,±2,...

    f(t,y) = t(1-y2)[itex]\frac{1}{2}[/itex]
    So [itex]\frac{∂f}{∂y}[/itex] = ty(1-y2)[itex]\frac{-1}{2}[/itex]

    I want to say that that as long as |y|≤ 1, f(t,y) is continuous and because the set B is closed, there exists a Maximum value M = max(t,y)[itex]\inB[/itex]|[itex]\frac{∂f}{∂y}[/itex]|, this is the Lipschitz condition and f satisfies this condition.
     
  2. jcsd
  3. Feb 7, 2013 #2

    HallsofIvy

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    No, that's not true. The partial derivative is what you say (you mean that last "-1/2" to be an exponent) but that goes to infinity at y= 1 so there is no such maximum value.

    f is continuous, and so has a maximum on a closed and bounded set, but [itex]\partial f/\partial y[/itex] is not.
     
  4. Feb 7, 2013 #3
    yes sorry that should be raised to the -1/2. I just saw that. So because ∂f/∂y goes to infinity at y=±1, does this mean that the Lipschitz condition failed and that means and that the existence and uniqueness theorem is violated?
     
    Last edited: Feb 7, 2013
  5. Feb 7, 2013 #4

    HallsofIvy

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    Yes, the Lipschitz condition fails so the existance and uniqueness theorem does NOT APPLY. (I would not say it is "violated". That would imply a situation where the hypotheses are true and the conclusion is false- a counterexample.)
     
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