[SOLVED] Contraction

**Paperwings** · August 2nd 2008, 09:40 AM

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?

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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.

**CaptainBlack** · August 2nd 2008, 10:07 AM

Originally Posted by Paperwings

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

**Paperwings** · August 2nd 2008, 10:36 AM

Thank you very much CaptainBlack

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