The appendix shows that in the neighbourhood of the symmetric equilibrium, the linearized system has two positive and one negative real roots when ф is less than a critical value. For this range of ф’з the system is saddle path stable, since only 0K is a nonjumper.
For ф beyond the critical value, the linearized system as three positive eigenvalues, so the symmetric equilibrium is unstable. As it turns out, however, an informal approach to stability provides the same answer with greater intuition.
Specifically, to study the symmetric equilibrium’s stability, we exogenously increase 0K by a small amount and check the impact of this perturbation on the regional q’s, allow ing expenditure shares to adjust according to. In particular, using,,, and, the steady-state q can be expressed as a function 0K and L,. Holding L, constant for the moment, the partial derivative of interest is <3q/d0K from ^=[0к,Ьк,0Е[0к];ф].
The symmetric equilibrium is stable, if and only if dq/30K is negative for a simple reason. If a unit of capital ’accidentally’ disturbed symmetry, the ‘accident’ lowers Tobin’s q in the north and raises it in the south (by symmetry <3q*/d0K and dq/30K have opposite signs). Moreover, these incipient changes in Tobin’s q are sufficient statistics for changes in regional investment levels (see Baldwin and Forslid 1997a Proposition 1 for details). Thus when dq/30K<O, the perturbation generates self-correcting forces in the sense that L, falls and L,* rises. If the derivative is positive, by contrast, the ‘accident’ boosts L, and lowers L,*, thus amplifying the initial shock to 0K. Plainly the symmetric equilibrium is unstable in this case. We turn now to signing 5q/50K.
Differentiating the definition of q with respect to 0K, we have:
Using to find d0E/d0K=2pA/[L(l+A)+p](l+A), we see that the system is unstable for sufficiently low trade costs (i.e. ф = 1). And under weak regularity conditions, the system is stable, i.e. is negative, for sufficiently high trade costs (ф =0).