 Solution

ID: past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2007/a/4/ii/solution

The reaction Jacobian is

J = (- b^{2} b^{2} - 2 ab - 1 + 2 ab)_{(3/4, 1)} = (- 1 1 - 3/2 1/2) .

Its trace is

- 1/2

1

, so both eigenvalues have negative real parts. Explicitly they are

(- 1 \pm i 15) /4

. Therefore the homogeneous reaction equilibrium is a stable focus in the absence of diffusion.

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