## Exercise 42 E

For more practice determining when a steady state is stable, do the computation for the discrete Ricker map

and show that the condition is |1 — r| < 1,or0 < r < 2.

But we have a two dimensional dynamical system. Since what follows is going to be a lot of work, we will do the analysis for the more general host-parasitoid dynamics. Basically, we do for the steady state of a two dimensional discrete dynamical system the same kind of analysis that we did for the two dimensional system of ordinary differential equations in Chapter 2. Because the procedure is similar, I will move along slightly faster (that is, skip a few more steps) than we did in Chapter 2. Our starting point is

H (t + 1)=RH (t) f (H (t), P(t)) P(t + 1) = H (t)(1 — f (H (t), P(t)))

which we assume has a steady state (H, P). We now assume that H(t) =H + h(t) and P(t) =P + p(t), substitute back into Eq. (4.6), Taylor expand keeping only linear terms and use o(h(t), p(t)) to represent terms that are higher order in h(t), p(t), or their product to obtain

H + h(t + 1)= R(H + h(t))[ f (H, P)+fHh(t)+fPp(t)\ + o(h(t), p(t))

P + p(t + 1) = (H + h(t))[1 — f (H, P)—fHh(t)—fPp(t)\+ o(h(t), p(t))

where fH = (d/dH)f(H,P)I(fif) and fP is defined analogously. Now, from the definition of the steady states we know that H = RHf (H, P), which also means that Rf (H, P) = 1, and that P = H(1 — f (H, P)). We now use these last observations concerning the steady state as we multiply through, collect terms, and simplify to obtain h(t + 1)= h(t)(1 + RHfn )+RHfpP{t) + o(h(t), p(t))

p(t + 1) = h(t)(1 -(1/R)-HfH) — HfPp(t) + o(h(t),p(t))

Unless you are really smart (probably too smart to find this book of any use to you), these equations should not be immediately obvious. On the other hand, you should be able to derive them from Eqs. (4.7), with the intermediate clues about properties of the steady states in about 3-4 lines of analysis for each line in Eqs. (4.8). If we ignore all but the linear terms in Eqs. (4.8) we have the linear system h(t + 1)= ah(t)+bp(t) p(t + 1) = ch(t)+dp(t)

with the coefficients a, b, c, and d suitably defined; as before, we can show that this is the same as the single equation h(t + 2) = (a + d)h(t + 1) + (bc - ad)h(t) (4.10)

by writing h(t + 2) = ah(t + 1) + bp(t + 1), p(t + 1) = ch(t) + dp(t) = ch(t) + (d/b)(h(t + 1) — ah(t)) and simplifying. (Once again you should not necessarily see how to do this in your head, but writing it out should make things obvious quickly.) If we now assume that h(t) ~ l (there is actually a constant in front of the right hand side, as in Chapter 2, but also as before it cancels), we obtain a quadratic equation for 1:

which I am going to write as I2 — pi + 7 = 0 with the obvious identification of the coefficients. Also as before, Eq. (4.11 ) will have two roots, which we will denote by X\ = (/3/2) + (\/p2 — 47/2) and 12 = (P/2) — (\/p2 — 47/2). The steady state will be stable if perturbations from the steady state become smaller in time, this requires that |11,2| < 1. We will now find conditions on the coefficients that makes this true. The analysis which we do follows Edelstein-Keshet (1988), who attributes it to May (1974). We will do the analysis for the case in which the eigenvalues are real (i.e. for which p > 47); this is our first condition. Figure 4.5 will be helpful in this analysis. The parabola I2 — pi + 7 has a minimum at p/2, and because we require — 1 < 12 < p/2 < 11 < 1 we know that one condition for stability is that |p/2| < 1, so that |p| < 2. The parabola is symmetric around the minimum. Now, if the roots lie between -1 and 1, the distance between the minimum and either root, which I have called D1, must be smaller than the distance between the minimum and -1 or 1, depending upon

Figure 4.5. The construction needed to determined when the solutions of the equation I2 - pi + 7 = 0 have absolute values less than 1, so that the linearized system in Eq. (4.9) has a stable steady state.

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