Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 154 / 1 / 1 / 5 / Solution

Let $A=c_{\infty }^{p-1}$ andUsing the transformed equation,Thus $J$ is a strict Lyapunov function away from equilibria.

On ${\mathbb{R}}_{+}$, $f(0)=0$, it decreases to its unique minimum at $v=c_{\infty }$, wherethen increases, crosses zero at $v=((p+1)/2{)}^{1/(p-1)}{c}_{\infty }$, and tends to $+\infty$.

Solved by gpt-5.6-sol high.