Determine $ \ a_{\max} \ $ and $ \ a_{\min} \ $ so that the above difference equation is well-defined.

Question

Consider the Logistic equation $ \ x_{n+1}=a x_n (1-x_n ) \ $ and constrain $ \ x_i \ $ in $ \ [0,1] \ $.

Determine $ \ a_{\max} \ $ and $ \ a_{\min} \ $ so that the above difference equation is well-defined.

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[My thoughts.]

If $a \in [0,4]$ , then the above logistic map will be well-defined. So $ \ a_{\max}=4 \ $ and $ \ a_{\min}=0 \ $.

But how to determine that $ \ a \in [0,4] \ $ is true?

Answer 1

We want $$0 \leq ax_n(1-x_n) \leq 1$$ for all $x_n \in [0,1]$.

We know that $0 \leq x_n(1-x_n) \leq \frac14$. Note that the range can be found easily since it is just a concave quadratic formula that attains the maximum at $\frac12$ and have end point values evaluated to $0$.

We can't pick $a$ to be negative as if we let $x_n = \frac12$, $ax_n(1-x_n)= \frac{a}4 <0$ which is beyond our desired range.

Similarly, if $a > 4$, then $\frac{a}4 > 1$.

If we choose $a \in [0,4]$, then $0 \leq ax_n(1-x_n) \leq \frac{a}{4} \leq 1$