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Homework Help: Problem for Theorem of Uniqueness

  1. Aug 15, 2011 #1
    1. The problem statement, all variables and given/known data

    Check if the given initial value problem has a unique solution

    2. Relevant equations

    y'=y^(1/2), y(4)=0

    3. The attempt at a solution

    f=y^(1/2) and its partial derivative 1/2(root of y) are continuous except where y<=0. We can take any rectangle R containing the initial value point (4,0). Then the hypothesis of theorem of uniqueness is satisfied.

    I want to make sure if this way is correct. Some help please.

  2. jcsd
  3. Aug 15, 2011 #2


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    Homework Helper

    The usual way you tackle this kind of thing is proof by contradiction and examine the integral:
    \int \left| \frac{dw}{dx}\right|^{2}dx
    where [itex]w=u-v[/itex] and u and v solve the original equation.

    I think in this case though, you can solve the general equation (without using the iniitial/boundary condition) and show that it's solution depends upon a single parameter. That parameter is determined with the condition you were given.
  4. Aug 15, 2011 #3
    Did you mean that my way and answer is wrong? This problem should be verified by using the theorem of uniqueness.

    I got y(t)=(4x^2)-64. Then can i say that this is a unique solution?
  5. Aug 15, 2011 #4


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    Homework Helper

    As I don't know the theorem you're using I can't say either way.
  6. Aug 15, 2011 #5
    sorry, you're right. The theorem I was trying to use is that if f and its partial derivative is continuous on the rectangle containing the given initial value problem, we can say that it has a unique theorem.
  7. Aug 15, 2011 #6


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    Science Advisor

    It would help if you actually stated the theorem you are using. I suspect you are using
    "The initial value problem y'= f(x,y), [itex]y(x_0)= y_0[/itex], has a unique solution in some neighborhood of the point [itex](x_0, y_0)[/itex] if both f(x,y) and [itex]f_y(x,y)[/itex] are continuous in some neighborhood of that point."
    No, you cannot "take any rectangle R containing the initial value point (4, 0)" because that will necessarily include points where y<= 0, where the functions are NOT continuous.

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