The stability test for the core-periphery equilibrium case is slightly different since the core-periphery outcome entails 0K=1> q=landq*<l. The procedure, therefore, is to find the range of ф where q*<l, when 0K=1 • Since this is exactly the procedure used in Section 2 to determine the range of the core-periphery, we see that the core-periphery steady state is stable w’herever it exists. To examine the stability of the interior-non-symmetric steady state, we adopt the same procedure.
Namely, we study dq/50K evaluated at 0K given by. Given the complex nature of dq/50K and, we cannot sign the derivative analytically. However, for reasonable values of the parameters the derivative is negative. This finding is robust to sensitivity analysis on parameter values.
Finally, we turn to showing that there is no overlap in the zones of stability. With some difficulty, it is possible to show that фср>фса! and that the endpoints of the interior-non-symmetric steady state are exactly equal to фСЕ” and фС1>. Using these results. Figure 4 summarizes the model’s stability properties in a diagram with ф and 0K on the axes. More precisely, as already argued, ‘stable’ corresponds to a saddle while.‘unstable’ correponds to an unstable node/focus. The fact that interior-non-symmetric steady states must correspond to saddles follows from continuity arguments as the dynamical system undergoes a supercritical pitchfork bifurcation when ф crosses фса1: the symmetric steady state loses its stability to the two new neighboring steady states.