samples/texts/2274792/page_17.md

A.5 Stability of $A_3$ equilibrium with culture lost

To determine the stability of this equilibrium, I calculated the characteristic equation for the system given by Equations 2 and 3 evaluated where $x_{f30} = x_{m30} = 1$.

$$\frac{{\lambda }^9(1-\lambda {)}^3(2\lambda -(1+b-t))(2\lambda -(1+b-2t))}{2}=0\backslash frac\{\backslash lambda^9(1-\backslash lambda)^3(2\backslash lambda-(1+b-t))(2\backslash lambda-(1+b-2t))\}\{2\}\; =\; 0$$

Since three of the resulting eigenvalues are equal to one, the equilibrium is unstable or neutrally stable for all parameters. This makes intuitive sense since the fitnesses of all four alleles in both sexes are equivalent in the absence of the cultural trait.

A.6 Stability of $A_4$ equilibrium with culture lost

To determine the stability of this equilibrium, I calculated the characteristic equation for the system given by Equations 2 and 3 evaluated where $x_{f40} = x_{m40} = 1$.

$$\frac{{\lambda }^9(1-\lambda {)}^3(2\lambda -(1+b-t))(2\lambda -(1+b-2t))}{2}=0\backslash frac\{\backslash lambda^9(1-\backslash lambda)^3(2\backslash lambda-(1+b-t))(2\backslash lambda-(1+b-2t))\}\{2\}\; =\; 0$$

Since three of the resulting eigenvalues are equal to one, the equilibrium is unstable or neutrally stable for all parameters. This makes intuitive sense since the fitnesses of all four alleles in both sexes are equivalent in the absence of the cultural trait.