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Math Help - [SOLVED] Contraction

  1. #1
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    [SOLVED] Contraction

    Fix \alpha a positive real number, let X=[0,1] \subseteq R and define  f:X \rightarrow R by

     f(x)= \alpha x ( 1- x) for  x \in X

    a. For what values of  \alpha does the mapping  f:X \rightarrow R have the property that  f(X) \subseteq X ?

    b. For what values of  \alpha does the mapping  f:X \rightarrow R have the property that  f(X) \subseteq X and is a contraction?

    =========================================
    a. Since  f(x)= \alpha x ( 1- x) = \alpha x - \alpha x^2 , then the derivative is  f'(x) = \alpha (1- 2x)

    Since  f(X) \subseteq X = [0,1] , I need to find the maximum value of the function to be 1, since  f(x) \subseteq [0,1] to be a subset.

    Therefore,

     f'(x) = \alpha (1- 2x) = 0 \implies x = \frac{1}{2}

    Plug x = 1/2 back into, f(x) = \alpha x (1-x) \implies \alpha =4

    Thus, since we are given prior that alpha is a positive real number then  \boxed{0 < \alpha \leq 4} where  f(X) \subseteq X=[0,1]

    b. To be a contraction, then its Lipschitz constant C for the mapping is 0 \leq c < 1

    Then | f'(x) | \leq c

    So | \alpha ( 1- 2x) | < c

    I am having troubles with this inequality. Since  x \in [0,1] , then 0 \leq x \leq 1 and 0 \leq c < 1, how would I solve for the value of \alpha in  0 \leq \alpha \leq 4 ?

    Thank you for your time.
    Last edited by Paperwings; August 2nd 2008 at 11:31 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Paperwings View Post

    b. To be a contraction, then its Lipschitz constant C for the mapping is 0 \leq c < 1

    Then | f'(x) | \leq c

    So | \alpha ( 1- 2x) | < c
    We are looking for an \alpha such that:

    | \alpha ( 1- 2x) | < 1, \ \ \ \forall x \in [0,1]

    Now (1-2x) is a linear function so takes its extreme values on [0,1] at the end points of the interval so its maximum is 1 and its minimum is -1 . Hence, as \alpha>0:

    | \alpha ( 1- 2x) | \le \alpha |1-2x| \le \alpha, \ \ \ \forall x \in [0,1]

    So if \alpha<1, f is a contaraction mapping on [0,1] , and the required constant is \alpha (if \alpha \ge 1 then there are points in the interval where | \alpha ( 1- 2x) | \ge 1 and so f fails to be a contraction mapping)

    RonL
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  3. #3
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    Thank you very much CaptainBlack
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