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Thread: [SOLVED] Contraction

  1. #1
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    [SOLVED] Contraction

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

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

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

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

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

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

    Therefore,

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

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

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

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

    Then $\displaystyle | f'(x) | \leq c$

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

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

    Thank you for your time.
    Last edited by Paperwings; Aug 2nd 2008 at 10:31 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 $\displaystyle 0 \leq c < 1$

    Then $\displaystyle | f'(x) | \leq c$

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

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

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

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

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