5.5 EXAMPLE: A SIMPLE MIGRATION MODEL Assume that the population of a country is divided into two distinct segments: rural and urban. The natural yearly growth factors, due to procreation, in both segments are assumed to be identical and equal to o (that is, the population at year k +1 would be a times the population at year k). The population distribution, however, is modified by migration between the rural and urban segments. The rate of this migration is influenced by the need for a base of rural activity that is adequate to support the total population of the country- the optimal rural base being a given fraction y of the total population. The yearly level of migration itself, from rural to urban areas, is proportional to the excess of rural population over the optimal rural base. If the rural and urban populations at year k are denoted by rk) and u(k), respectively, then the total population is r(k)+ ulk), the optimal rural base is y[r(k)+ u(k)], and thus, the excess rural population is {r(k)yr(k)+ u(k)]]. A simple dynamic model of the migration process, based on the above assump- tions, is then r(k+ 1) = ark) - B{r(k)-[r(k) + ulk)]} u(k+1)=au(k)+ Br(k) - Yr(k) + (k)}} In this model, the growth factor is positive (and usually greater than unity). The migration factor B is positive, and is assumed to be less than a. The parametery is the ideal fraction of the total population that would be rural in order to support the total population. This parameter is a measure of rural productivity. Each of these parameters might normally change with time, but they are assumed constant for purposes of this example. The model can be easily put in the state vector form x(k+ 1) = Ax(k) where a-B(1-y) By 1 B(-y) - By and x(k)=["k)] k} = Lucky] At this point one might normally proceed by writing the characteristic polyno- mial as a first step toward finding the eigenvalues and eigenvectors of the matrix A, but in this case at least some of that information can be deduced by simple reasoning. Because the natural growth rates of both regions are identi- cal, it is clear that the total population grows at the common rate. Migration simply redistributes the population, it does not influence the overall growth. Therefore, we expect that the growth factor a is one eigenvalue of A and that the row vector f =[11] is a left eigenvector of A, because I x(k)=r(k)+ ulk) is the total population. Indeed, checking this mathematically, we find [11] a-B(1-y) By B(1-4) = a[11] which verifies our conjecture. This left eigenvector tells us what variable within the system (total population in this case) always grows by the factor o. The corresponding right eigenvector of A defines the distribution of population that is necessary in order for all variables to grow by the factor c. Again, this vector can be deduced by inspection. In order for both rural and urban population to grow by the factor a it is necessary that there be no net migration. Therefore, it is necessary that r(k)= [r(k) + u(k)]; or, equivalently, that rural and urban population be in the balanced proportions Y:(1Y) Therefore, we expect that is a right eigenvector of A corresponding to the eigenvalue a. Again, a simple check verifies this conjecture. By exploiting the simple observation about total population growth, we have deduced one eigenvalue together with its corresponding left and right eigenvectors. We can find the remaining eigenvalue and its corresponding left and right eigenvectors by use of the biorthogonality relations that hold among right and left eigenvectors. The second right eigenvector must be orthogonal to the first left eigenvec- tor. Thus. e-[-] Multiplying this by A verifies that it is an eigenvector and shows that the corresponding eigenvalue is a-. It represents the population distribution corresponding to zero total population. This eigenvector has both positive and negative components, and hence, is not a physically realizable state since population must be nonnegative. However, this eigenvector (or a multiple of it) will in general be a component of the actual population vector. To determine what variable in the system grows by the factor a-Bit is necessary to find the second left eigenvector. It will be orthogonal to the first right eigenvector and hence, 11 [1-y-y] This left eigenvector corresponds to the net rural population imbalance, since fx(k)=r(k) - Y[r(k)+u(k)). The various eigenvectors associated with this system together with their interpretations and interrelations are summarized in Table 5.1. A simple diagram in diagonal form summarizing the analysis is shown in Fig. 5.4. The above analysis can be translated into a fairly complete verbal descrip- tion of the general behavior of the migration model. Overall population grows by the factor ex each year. If there is an initial imbalance of rural versus urban population, with say more than a fraction y of the population in the rural Table 5.1. Eigenvectors for Migration Example Left Eigenvector Right Eigenvector corresponds to total population corresponds to the condi- tion of zero rural im- balance -1-y-y] corresponds to net rural imbalance corresponds to the condi- tion of zero population sector, then there is migration, and it grows by the factor a- B. The rural imbalance also grows by this factor. Since >B>0, the growth factor of rural imbalance is always less than that of population, so eventually the relative imbalance that is, the ratio of imbalance to total population) tends to disap- pear. If >a, then -BB>0, the growth factor of rural imbalance is always less than that of population, so eventually the relative imbalance that is, the ratio of imbalance to total population) tends to disap- pear. If >a, then -B