As this equation is algebraically very complicated, we use the graph of [alpha](r) in order to find out whether there exist any real

positive roots. From Figure 5 one can see that, for different values of the test particle's charge, a(r) given by (40) does not intersect the r-axis.

One can observe that if [b.sub.i] [greater than or equal to] 0, (i = 1, 2, 3, 4), then (42) has no

positive roots. Therefore (37) does not have any purely imaginary roots for all [[tau].sub.1] [member of] (0, [[tau].sup.*]) so that all roots of the characteristic equation (37) have negative real parts and the endemic equilibrium [E.sup.*] of (1) is stable for all [[tau].sub.1] [member of] (0, [[tau].sup.*]).

From Table 1, it is clear that the dispersion relation (2.12) has two non-zero real

positive roots which ensures the remark reported in Section 2.

If q has diagram [MATHEMATICAL EXPRESSION NOT REPRODUCIBLE IN ASCII] the set of

positive roots are, respectively,

Namely, the difference between the two consecutive

positive roots of [J.sub.[alpha]](t) = 0 is extremely close to n, where the bias is smaller than [10.sup.-3].

Without losing generality, we suppose that (22) has three

positive roots, which are defined by [z.sub.1], [z.sub.2], and [z.sub.3].

Suppose that (15) has

positive roots; without loss of generality, we assume that it has eight

positive roots, denoted by [z.sub.1], [z.sub.2], [z.sub.3], [z.sub.4], [z.sub.5], [z.sub.6], [z.sub.7], [z.sub.8], respectively.

(3) Assume one of the following conditions is satisfied, then (5) has one

positive root:

So it has m + 1

positive roots, if it also has a negative root [beta] then the negative root is unique as it has at most m + 2 real roots.

There exists at least one

positive root of g([rho]) = 0.

Then we will analyze the existence of

positive roots of F([N.sub.h]) = 0.

Then [PHI] = [[PHI].sup.+] [??] - [[PHI].sup.+] is the disjoint union of the set of

positive roots [[PHI].sup.+] and the set of negative roots -[[PHI].sup.+].