Study Questions for Final

n₁[t+1] = a₁ n₁[t] / (1 + b₁ n₁[t] + c₁ n₂[t])

n₂[t+1] = a₂ n₂[t] / (1 + b₂ n₂[t] + c₂ n₁[t])

(a) The equilibria must solve both: n₁ = a₁ n₁ / (1 + b₁ n₁ + c₁ n₂)

and n₂ = a₂ n₂ / (1 + b₂ n₂ + c₂ n₁)

One equilibrium is clearly n₁ = 0 and n₂ = 0.

Another occurs when n₁ = 0 and n₂ = (a₂-1)/b₂.

Another occurs when n₁ = (a₁-1)/b₁ and n₂ = 0.

Finally, the only equilibrium with both species present occurs when n₂ = (b₂a₁ - c₁a₂ + c₁ - b₂)/( b₁b₂ - c₁c₂) and n₂ = (b₁a₂ - c₂a₁ + c₂ - b₁)/( b₁b₂ - c₁c₂).

(b) The local stability matrices must be determined by finding the partial derivatives and plugging in the equilibria found in the previous equation.

For n₁ = 0 and n₂ = 0, this gives:

{{a₁,0},{0,a₂}}

Therefore if either a₁ or a₂ is greater than one then this equilibrium will be unstable. This corresponds to the condition that one or both species have a positive growth rate (r_i>0).

For n₁ = 0 and n₂ = (a₂-1)/b₂, the local stability matrix is:

{{a₁b₂/(b₂ - c₁ + a₂c₁) , 0}, {(1-a₂)c₂/(a₂b₂), 1/a₂}}

Since this is a lower triangular matrix, the eigenvalues are given by the diagonal elements. From the first eigenvalue, we know that the equilibrium will be unstable if the rare species 1 has a high growth rate (a₁ high), the common species 2 has strong intraspecific competition (b₂ high), and the common species 2 has little effect on the rare species 1 (c₁ low). From the second eigenvalue, we know this equilibrium will be unstable if a₂ is less than one (since then species 2 does not have a positive growth rate).

For n₁ = (a₁-1)/b₁ and n₂ = 0, the local stability matrix is:

{{1/a₁), (1-a₁)c₁/(a₁b₁)}, {0, a₂b₁/(b₁ - c₂ + a₁c₂}}

Since this is an upper triangular matrix, the eigenvalues are given by the diagonal elements. From the first eigenvalue, we know this equilibrium will be unstable if a₁ is less than one (since then species 1 does not have a positive growth rate). From the second eigenvalue, we know that the equilibrium will be unstable if the rare species 2 has a high growth rate (a₂ high), the common species 1 experiences strong intraspecific competition (b₁ high), and if species 1 has little effect on species 2 (c₂ low).

(c) Since a₁ and a₂ are both greater than one, the equilibrium with n₁ = 0 and n₂ = 0 is unstable. The eigenvalues for the case with n₁ = 0 and n₂ = (a₂-1)/b₂ are 0.94 and 0.96 so this equilibrium is stable. The eigenvalues for the case with n₁ = (a₁-1)/b₁ and n₂ = 0 are 0.76 and 1.06 so this equilibrium is unstable.

Species 1 (S. cerevisiae) is unable to invade the equilibrium with n₁ = 0 and n₂ = (a₂-1)/b₂ primarily because the value of c₁ is so high. This means that there is strong interspecific competition with S. pombe. S. pombe strongly suppresses the growth of S. cerevisiae.

Interestingly, the growth rate of S. cerevisiae is higher and so in Gause's experiments he observed a much more rapid increase in S. cerevisiae. The theory predicts, however, that had these experiments been run for longer periods of time, S. pombe would have displaced S. cerevisiae once it became established.