Past exam of the topics in applied mathematics course of the University of Oxford / 2007 / A / 1 / ii / Solution

A positive fixed point requires $\beta +u_{\ast }^{\gamma }=\alpha$, so it exists exactly when $\boxed{\alpha >\beta }$ and is $u_{\ast }=(\alpha -\beta {)}^{1/\gamma }$. Differentiating the map at it givesFor $1<\gamma \le 2$ and $0<\beta /\alpha <1$, this fixed-point multiplier is strictly between $-1$ and $1$. Hence the positive fixed point is linearly asymptotically stable, including $\gamma =2$; positivity of $\beta$ makes the lower bound strict.