= Solution
For the <cubic population iteration>, put $g(u)=\lambda u(1-u^2)$. Its <fixed points> solve
$$
u=g(u)\quad\Longleftrightarrow\quad u[(\lambda-1)-\lambda u^2]=0.
$$
Thus $u_*=0$ always exists. For $\lambda\ne0$ there are also
$$
\boxed{u_*=\pm\sqrt{1-\frac1\lambda}}
$$
whenever the expression under the square root is positive, namely $\lambda<0$ or $\lambda>1$. At $\lambda=1$ these coincide with zero; at $\lambda=0$ only zero exists.
The criterion for <fixed point stability for an iteration> is $|g'(u_*)|<1$. To justify it, <continuity> of $g'$ gives a small interval around the <fixed point> where $|g'|\leq q<1$. The <mean value theorem> then gives $|g(u)-u_*|\leq q|u-u_*|$, so this interval is invariant and the error decreases geometrically. This proves local <asymptotic stability> rather than merely linear boundedness.
Here $g'(u)=\lambda(1-3u^2)$. At zero its value is $\lambda$, so \b[zero is locally <asymptotically stable> for $-1<\lambda<1$]. At a nonzero <fixed point> it is
$$
g'(u_*)=\lambda\left[1-3\left(1-\frac1\lambda\right)\right]=3-2\lambda.
$$
The strict inequality $|3-2\lambda|<1$ is equivalent to $1<\lambda<2$. Both nonzero real <fixed points> are therefore locally <asymptotically stable> on that interval. This proves the two requested existence ranges. The strict <derivative> criterion alone makes no conclusion at the endpoint multipliers of modulus one.
Back to article page
