Past exam of the mathematics course of the University of Cambridge / 2026 / ii / Paper 3 / 6C (Mathematical Biology)

The reaction is $F(n)=-n(n-A)(n-B)$. Its homogeneous equilibria are $0,A,B$; $F^{\prime }(0)<0$ and $F^{\prime }(B)<0$, while $F^{\prime }(A)>0$. Thus the stable equilibria are $0$ and $B$.

Substitution $n=f(\xi )$, $\xi =x-ct$, gives $f^{\prime \prime }+c{f}^{\prime }+F(f)=0$ with $f(-\infty )=0$, $f(\infty )=B$. Multiplying by $f^{\prime }$ gives $-c\int (f^{\prime }{)}^{2}d\xi ={\int }_0^{B}F(s)ds=B^4(1-2A/B)/12$. The populated state invades the empty state when $c<0$, equivalently $B>2A$; this is the condition preventing eventual extinction.