STABILITY AND PERSISTENCE IN ODE MODELS FOR POPULATIONS WITH MANY STAGES

. A model of ordinary diﬀerential equations is formulated for populations which are structured by many stages. The model is motivated by ticks which are vectors of infectious diseases, but is general enough to apply to many other species. Our analysis identiﬁes a basic reproduction number that acts as a threshold between population extinction and persistence. We establish con-ditions for the existence and uniqueness of nonzero equilibria and show that their local stability cannot be expected in general. Boundedness of solutions remains an open problem though we give some suﬃcient conditions.


1.
Introduction. Individual characteristics affect the development of populations and, to take this into account, structured population models are a golden middle ground between unstructured models of total densities and very detailed individual based models (see [6,22,24,39], e.g.).
In many mammalian and other species, development occurs in a rather continuous way and can best be traced along age or size structure leading to systems of η j (x) = γ j (x) + µ j (x), j = 1, . . . , n − 1 η n (x) = µ n (x) x ∈ R n + . (2.2) The system (2.1) then takes the form, We have more flexibility if we formulate our overall assumptions in terms of system (2.3).
Overall Assumptions. All functions g, γ j , η j : R n + → R + are Lipschitz continuous on bounded subsets of R n + , η j (x) ≥ γ j (x) ≥ 0, j = 1, . . . , n − 1, x ∈ R n + . (2.4) The population birth rate g : R n + → R + is differentiable at 0. The partial derivatives at 0, ∂ j g(0) =: β j , j = 1, . . . , n, (2.5) can be interpreted as the per capita birth rates of individuals in the j th stage if there is no competition. The numbers p j (x) = γ j (x) η j (x) ∈ [0, 1], j = 1, . . . , n − 1, (2.6) provide the probabilities of getting through the j th stage alive if the stage distribution of the population is x. Indeed, 1 ηj (x) is the mean sojourn time in the j th stage (counting death) and γ j (x) is the per capita rate of leaving the stage alive. See [36,Sec.13.6] for a more systematic exposition.

Cyclic stage models.
A special case is the cyclic stage model in which only the last stage produces offspring, While we keep the assumptions γ j (x) ≤ η j (x) for all j = 1, . . . , n − 1, no such assumption is made for γ n because γ n is the per capita reproduction rate and not a transition rate. The models considered in [21] and [43] fit into this framework by assuming γ n (x) =γ n (x n−2 ), γ j (x) =γ j , j = 1, . . . , n − 1, η j (x) =γ j +μ j (x j ), j = 1, . . . , n, (2.8) with some of theμ j being constant. Here x n is interpreted as the (number of) egg-laying adult ticks, x n−1 engorged adult ticks, and x n−2 feeding adult ticks. There is good reason to let γ j (x) depend on x because hosts can develop immunity or resistance to ticks that do not only lead to weight reduction in engorged ticks and to reduced tick fecundity but also to prolonged feeding times [42]. It is also suggestive to consider a more general dependence of µ j on x than x j : If host immunity or resistance which are triggered by feeding ticks increase the per capita mortality rate of feeding ticks, then they may also increase the per capita mortality rate of engorged ticks, i.e., µ n−1 could depend on x n−2 .

3.
Solutions. Let f : R n + → R n be the vector field given by the right hand side of (2.3) and let f j denote the components of f , j = 1, . . . , n. The assumptions mentioned before imply that the vector field is Lipschitz continuous on bounded subsets of R n + . By standard ODE theory, solutions to initial values in R n + exist on open intervals containing 0 and are uniquely determined by and continuously depend on those initial values.
To show that solutions that start in R n + stay in R n + in forward time as long as they exist, we use [33,Prop.B.7] (see also [36,Prop.A.1]). We only need to check the following: If x ∈ R n + and x 1 = 0, then f 1 (x) = g(x) ≥ 0. If j = 2, . . . , n, x ∈ R n + and x j = 0, then f j (x) = γ j−1 (x)x j−1 ≥ 0. To show that no solution blows up in finite forward time, we consider the total population size We add the equations in (2.3), We regroup and change the index of summation in one sum, Since the solution is non-negative, by (2.4), We make the following assumption.
Assumption 3.1. There exists some c > 0 such that g(x) ≤ c x for all x ∈ R n + . Here · is your favorite norm on R n . Then there exists somec such that y ≤cy and y(t) ≤ y(0)ec t for all t ≥ 0 for which the solution exists.
Theorem 3.2. Let Assumption 3.1 be satisfied. Then, for each x • ∈ R n + , there exists a unique solution x : 4. Stability of the extinction equilibrium. Since we have assumed that g(0) = 0, the origin is an equilibrium of (2.3) that represents a steady state at which the population is extinct (extinction equilibrium). We will analyze the stability of the origin in a way that does not require writing down huge matrices and right away leads to an expression for the reproduction number R 0 that is readily interpretable.
Since g is also assumed to be differentiable at 0, the vector field f is differentiable at 0 and We make another assumption.
Assumption 4.1. γ j (0) > 0 for all j = 1, . . . , n − 1 and η j (x) > 0 for all j = 1, . . . , n and x ∈ R n + . These assumptions are reasonable because, without competition, transition to the next stage should always be possible and exit from a stage should occur under all stage distributions at a positive rate by either transition or death. Since the system is large, we do our stability analysis ab ovo (from scratch) starting from eigenvalues and eigenvectors (cf. [36,Sec.23.3] and [43]).
Let λ ∈ C be an eigenvalue of f (0) and x ∈ C n , x = 0, an associated eigenvector. Then . . , n. This gives us a recursive relation for the coordinates of x, We solve, We notice that x = 0 implies x j = 0 for all j = 1, . . . , n. We substitute these expression into the first equation, notice that g (0)x = n k=1 β k x k with the per capita reproduction rates β k from (2.5) and divide by x 1 = 0. After some regrouping, we obtain the characteristic equation Here 0 j=1 := 1. Notice that has the interpretation of a basic reproduction number. Since p j is the probability of surviving the j th stage, q k is the probability of making it to the k th stage, while 1/η k (0) is the expected length of the k th stage and β k is the per capita reproduction rate during the k th stage (all without competition). So χ(0) is the expected amount of offspring an average newborn (or newly laid egg) can produce during its lifetime if there is no competition. Assume that R 0 = χ(0) > 1. Since χ is strictly decreasing in λ ≥ 0 and χ(λ) → 0 as λ → ∞ and χ is continuous, by the intermediate value theorem there exists some λ > 0 such that χ(λ) = 1. Working backwards and defining x by (4.2) with x 1 = 1, we see that this λ is an eigenvalue of f (0) associated with an eigenvector x ∈ (0, ∞) n . This implies that the origin is unstable. Now assume that R 0 = χ(0) < 1. Then χ(λ) < 1 for all λ ≥ 0 and there are no eigenvalues λ of f (0) with λ ≥ 0. Suppose there is an eigenvalue λ ∈ C with λ ≥ 0. We apply absolute values to the characteristic equation, use the triangle inequality and the multiplicity of the absolute value, a contradiction. This shows that all eigenvalues have negative real parts if R 0 < 1. We summarize.
Theorem 4.2. Let the Assumptions 4.1 be satisfied. Then the origin is locally asymptotically stable if R 0 < 1 and unstable if R 0 > 1.
In order to explore the possible global stability of the origin if R 0 < 1, we introduce a Lyapunov function to-be, with α j > 0, α 1 = 1. Let x be a solution of (2.3). Then We regroup, change the summation index in one sum and regroup again, We assume that g(x) ≤ n j=1 β j x j for all x ∈ R n + and η n (x) ≥ η n (0). Then We choose the coefficients α j recursively as . (4.6) The finite recursion (4.5) is solved by .
In order to enforce that (d/dt)V (x) ≤ 0, we make the following assumptions. Recall (2.6), the probabilities of surviving the j th stage at stage distribution x, Assumption (a) is very plausible stating that stage survival is larger without than with intraspecific competition. Assumption (b) may seem counterintuitive at the first glance. Notice, however, that it is trivially satisfied, if reproduction only occurs in the last stage or if the transition rates are constant. It can also be plausible if there are several reproductive stages and the later stages are less reproductive than the earlier ones. Then intraspecific competition would move reproductive individuals faster into the less reproductive stages.   Proof. By Theorem 4.4, every solution starting in R + is bounded in forward time and has a nonempty ω-limit set ω.
Since ω is invariant, there exists a solution y : R → ω with y(0) = x. By (4.11) and Assumption 4.3, (4.12) Since ω is invariant, there again exists a solution y : R → ω with y(0) = x. By our previous consideration, y n (t) = 0 for all t ∈ R and, from the differential equation for the last stage, Thus, again, x n−1 = 0. So, for all x ∈ ω(x), x n−1 = 0.
Continuing this way, we obtain that x = 0. So ω = {0}. Since the ω-limit set attracts the solution, all solutions converge to the origin as time tends to infinity. Theorem 4.6. Let the Assumptions 4.1 and 4.3 be satisfied and R 0 = 1. Further assume that there is some k ∈ {1, . . . , n − 1} with the following properties: + and x k > 0, • η j is bounded away from 0 for j = k + 1, . . . , n.
Then the origin is globally asymptotically stable.
Proof. Let ω be the ω-limit set of a solution. Let x ∈ ω. Then there exists a solution y : Let k be as specified in the assumption. By (4.12), x k = 0. So x k = 0 for all x ∈ ω.
As in the proof of Theorem 4.5, we obtain x j = 0 for j = 1, . . . , k and all x ∈ ω. Let x ∈ ω again. As before, there exists a solution y : R → ω with y(0) = x. By our previous results, y k (t) = 0 for all t ∈ R. So If y k+1 (r) > 0 for some r ∈ R, then y k+1 (t) → ∞ as t → −∞ because η k+1 is bounded away from 0. This contradicts the boundedness of y. So y k+1 (t) = 0 for all t ∈ R. Continuing this argument, we successively obtain that y j (t) = 0 for all j = k, . . . , n, t ∈ R. Thus x = y(0) = 0 and ω = {0}. Since a solution is attracted by its ω-limit set, all solutions converges to 0 as time tends to infinity.

5.
Persistence. We strive for persistence results that do not require any dissipativity of the model system. Even boundedness of solutions will not be required because it would need more restrictive assumptions than we would like to make (Section 7). The Assumptions 3.1 and 4.1 are supposed to hold throughout this section. We make another assumption.
Assumption 5.1. The per capita transition rates γ j are bounded and the exit rates η j are bounded away from 0. Further, for any δ ∈ (0, 1), there exists some > 0 such that where the b j : R n + → R + are Lipschitz continuous on bounded subsets on R n + . b j (x) can be interpreted as the per capita birth (or egg-laying) rate during stage j at stage distribution x. Then g satisfies the Assumptions 5.1. Further g is differentiable at 0 and ∂ j g(0) = b j (0) = β j .
In the following let · denote the sum-norm. Suppose the claim is not true. Let δ > 0 be arbitrary. Since η j and γ j are continuous and η j (x) > 0, there exists some > 0 such that Further, by assumption, Since we have assumed that the claim is false, for any˜ > 0 there exists a solution x (which depends on˜ ) such that x(0) ∈ R n + , x 1 (0) > 0, and lim sup t→∞ x 1 (t) <˜ . This implies that x 2 is bounded. By the fluctuation lemma [19] [36,Prop.A.22], there exists a sequence t k → ∞ such that From the differential equation for x 2 , We apply the fluctuation lemma multiple times and find some c ≥ 1 which only depends on the η j and γ j such that lim sup t→∞ x(t) < c˜ . Further x 1 (t) > 0 for all t ≥ 0. After a forward shift in time, we can assume that x(t) < c˜ and x 1 (t) > 0 for all t ≥ 0, j = 1, . . . , n. Choose˜ = /c. Then We reorganizex We solve these recursive inequalities, We substitute these formulas into the one forx 1 (λ) and divide byx 1 (λ), which is positive, and reorganize, Notice that this inequality no longer contains any terms belonging to the solution x (which depends on δ) and thus holds for any δ ∈ (0, 1) and λ > 0. So we can take the limit as δ, λ → 0 and obtain 1 ≥ R 0 (recall (4.3)), a contradiction.
Unfortunately, it seems to be difficult to turn the uniform weak persistence result in Theorem 5.3 into a uniform persistence result. So we try the next approach as well.
The linear map f (0) on R n can be written as with B = (B 1 , . . . , B n ) and C = (C 1 , . . . , C n ) and B j , C j : R n → R, and The eigenvalues of B are −η 1 (0), . . . , −η n (0), and so B has a negative spectral bound. Further B is quasipositive and C is positive. So the spectral bound of f (0) has the same sign as r(C(−B) −1 ) − 1 with r denoting the spectral radius. See [10], [37,Thm.3.5], [40]; the map C(−B) −1 can be interpreted as a next generation map (or matrix). Let λ be an eigenvalue of C(−B) −1 and y be an associated eigenvector. By the form of C, we have y j = 0 for j = 2, . . . , n and, without loss of generality, y 1 = 1. Then x = (−B) −1 y can be found by solving the system for x ∈ R n . This leads to the recursion This recursion is solved by (recall (4.3)) By (5.4), λ = n j=1 β j x j = R 0 and R 0 is the spectral radius of C(−B) −1 . We can now prove the following result.
Theorem 5.4. Let R 0 > 1 and β n = ∂ n g(0) > 0. Then there exists some > 0 such that The assumptions of this theorem and those of Theorem 5.3 are not comparable, but if it comes to applications the assumption β n = ∂ n g(0) > 0 will be more restrictive than those of Theorem 5.3.
Proof. By the preceding consideration, the spectral bound of f (0) is positive. Since β n > 0 and γ j (0) > 0 for j = 1, . . . , n − 1, f (0) is irreducible. By consequences of the Perron-Frobenius theory (see [36,Thm.A.45], e.g.), there exist λ > 0 and v ∈ (0, ∞) n such that v is a left eigenvector of the matrix associated with with , denoting the Euclidean inner product on R n . Since v ∈ (0, ∞) n , is a norm on R n which is equivalent to the sum-norm (and any other norm on R n ). We define We claim that the semiflow induced by (2.3) is uniformly weakly ρ-persistent, i.e., there exists some > 0 such that lim sup t→∞ ρ(x(t)) ≥ for all solutions with ρ(x(0)) > 0.
Remark 5.5. We could have used the relation between R 0 and the spectral bound of f (0) to prove a global stability result for the origin as in [21,Thm.2 This would require f (x) ≤ f (0)x for all x ∈ R n + which would in turn require γ j (x) ≤ γ j (0) and η j (x) ≥ η j (0). While the first assumption is suggestive and has been made in Assumption 4.3 (though only for those j for which β j > 0), the second assumption is less suggestive. Recall that η j = γ j (x) +µ j (x). Again, it is suggestive to assume that µ j (x) ≥ µ j (0), but this does not imply η j (x) ≥ η j (0). However, as assumed in Assumption 4.1, does follow. Further, we would not obtain a globally asymptotic stability result for R 0 = 1 (Theorem 4.6) this way.
6. Persistence equilibria. A vector x ∈ R n + is an equilibrium of (2.3) if and only if it satisfies the fixed point system We choose this reformulation because of the wealth of fixed point theorems available in the literature. Since 0 is a fixed point of F = (F 1 , . . . , F n ) : R n + → R n + , fixed point theorems in conical shells are particularly useful [8,Sec.20.2].
In this section, we assume that Here b j (x) is the per capita birth (or egg-laying) rate in stage j at stage distribution x. Analogously to (4.3), we define the reproduction number at stage distribution x by Solving the recursive equations in (6.1) and substituting the solutions into the first equation shows that any nonzero equilibrium x satisfies R(x) = 1.
We deal with the existence and the uniqueness of persistence equilibria separately because they hold under quite different assumptions.
Theorem 6.2. Assume that the functions b j are bounded on R n + and the functions η j are bounded away from 0. Assume that R 0 = R(0) > 1 and there exists some c > 0 such that Then there exists an equilibrium in (0, ∞) n .
Proof. We choose · as the sum norm on R n . Suppose that assumption (a) in Proposition 6.1 is not satisfied for any r 2 > 1. Then there exist sequences (x k ) in R n + and (λ k ) in (1, ∞) such that x k → ∞ and F (x k ) = λ k x k . By the definition of F in (6.1), By iteration, We conclude that x k 1 = 0. We substitute these equations into the equation for x k 1 and divide by x k 1 , Our assumptions imply that R(x k ) < 1 for sufficiently large k contradicting (6.5).
So assumption (a) of Proposition 6.1 is satisfied with some r 2 > 1. Now suppose that assumption (b) of Proposition 6.1 is not satisfied for any r 1 ∈ (0, 1) with u = (1, . . . , 1). Then there exists a sequence (x k ) in R n and some sequence (α k ) in (0, ∞) such that 0 = x k → 0 and Since F is differentiable at 0, After choosing a subsequence, we can assume that y k → y ∈ R n + with y = 1 and Then y − F (0)y ∈ R n + and A similar consideration as before implies that R 0 ≤ 1, a contradiction. So assumption (b) of Proposition 6.1 is satisfied and there exists a nonzero fixed point x of F in R n + . Then x ∈ (0, ∞) n . Theorem 6.2 does not cover the models in [21,43] because not all exit rates are density-dependent. Therefore, we establish an existence result that is tailored to the cyclic model (2.7). For this model, the reproduction rate at stage distribution x is So x → ∞ could mean that x 1 → ∞ and x j = 0 for j = 2, . . . , n. Since γ 1 /η 1 is assumed constant in [21,43], R(x) = R(0) as x → ∞ is possible. Theorem 6.3. Assume that the functions γ j are bounded and the functions η j are bounded away from 0 on R n + , j = 1, . . . , n. Assume that R 0 = R(0) > 1 and that there exists some c > 0 such that Then there exists an equilibrium of (2.7) in (0, ∞) n .
Proof. Suppose that assumption (a) in Proposition 6.1 is not satisfied for any r 2 > 1.
We conclude that x k 1 = 0. We substitute these equations into the equation for x k 1 and divide by x k 1 , This implies that the sequence (λ k ) is bounded. Assume that the sequence (x k n ) is bounded. Then, by (6.7), since the γ j are bounded and the η j are bounded away from 0, all sequences (x k j ) k∈N are bounded contradicting that x k → ∞. So (x k n ) is unbounded and converges to ∞ after choosing subsequences. Again by (6.7), all sequences (x k j ) k∈N converge to ∞. Our assumptions imply that R(x k ) < 1 for sufficiently large k contradicting (6.8). So assumption (a) of Proposition 6.1 is satisfied with some r 2 > 1. Assumption (b) of Proposition 6.1 follows in the same way as in the proof of Theorem 6.2.

(6.9)
Let ≤ denote the componentwise ordering in R n , x ≤x (equivalentlyx ≥ x) if and only if x j ≤x j for j = 1, . . . , n. We write x <x if x ≤x and x =x, and x x if x j <x j for j = 1, . . . , n. Assumption 6.4. The rate functions γ j and η j and the reproduction number function R have the following monotonicity properties: . . , n}, and x <x for = j, . . . , n, then x.
Theorem 6.6. If Assumption 6.4 holds, there is at most one nonzero equilibrium.
Proof. Let x,x ∈ R n + be two equilibria. Without loss of generality we can assume that x n ≤x n . By Assumption 6.4 (b), η n (x)x n ≤ η n (x)x with the inequality being strict if and only if x n <x n . Since η n (x)x n = γ n−1 (x)x n−1 and the same equation holds withx replacing x, it follows from Assumption 6.4 (a) that x n−1 ≤x n−1 with the inequality being strict if and only if x n <x n . Assumption 6.4 (b) now implies that η n−1 (x)x n−1 ≤ η n−1 (x)x n−1 with strict inequality if and only if x n <x n . Since η n−1 (x)x n−1 = γ n−2 (x)x n−2 and the same equation holds withx replacing x, it follows from Assumption 6.4 (a) that x n−2 ≤x n−2 with the inequality being strict if and only if x n <x n .
Proceeding this way, we find that x j ≤x j for j = 1, . . . , n, with equality holding for all j if x n =x n , and strict inequality holding for all j if x n <x n . Suppose that x n <x n . Then x x and R(x) > R(x), a contradiction because R(x) = 1 = R(x). So x n =x n and x =x. 7. Boundedness and dissipativity for cyclic stage structure. We consider the system (2.7).
Assumption 7.1. The functions γ j are bounded, and the functions η j are bounded away from 0.
We define These properties of ν j follow directly from the definition and from Assumption 7.1.
Further ν j is increasing (not necessarily strictly) on R + .
We set Γ j = sup x∈R n + γ j (x). From (2.7), (7.1) and our assumptions, we have the following system of differential inequalities, Let j = 1, . . . , n, and T > 0. We define Let j ≥ 2 and assume x T j > x j (0). Then x T j is taken at some t ∈ (0, T ], By the differential inequality for x j , Without assuming x T j > x j (0), x T j−1 , j = 2, . . . , n.
Similarly, we derive that x T n . (7.6) Theorem 7.3. Let Assumption 7.1 be satisfied and Γ j := sup x∈R n + γ j (x). Assume that there exists some c > 0 and ξ ∈ (0, 1) such that Then all solutions of (2.3) are bounded. Further the system is dissipative: There exists somec > 0 such that x ∞ j := lim sup t→∞ x j (t) ≤c for all solutions of (2.3). Finally min n j=1 x ∞ j ≤ c. Proof. Let c > 0 be as in the assumptions of the theorem. Since the ν j are increasing, Suppose that x n is not bounded. Then x T n → ∞ as T → ∞. It follows from (7.5) that x T j → ∞ as T → ∞. So, for sufficiently large T , (7.4) holds for j = 2, . . . , n. We solve (7.4) recursively, substitute the result into (7.6) and divide by x T 1 to obtain .
We take the limit as T → ∞ and obtain a contradiction to (7.7). Thus x n is bounded. By (7.6), x 1 is bounded and then the x j are bounded for j = 1, . . . , n.

GUIHONG FAN, YIJUN LOU, HORST R. THIEME AND JIANHONG WU
There exists a sequence ( and so Similarly We solve the first set of inequalities recursively and substitute it into the last, Without loss of generality we can assume that x ∞ 1 > 0 and divide by it. Since the ν j are increasing, and, by ( we find ac > 0 that only depends on Γ j and ν j (0) such that x ∞ j ≤c. 7.1. Back to persistence. In dynamical systems language, the result of Theorem 7.3 means that the semiflow induced by system (2.7) is point-dissipative. Since our state space is R n + , the semiflow is asymptotically smooth [ [45]).
Then there exists a bounded set that attracts all solutions. If R 0 > 1, the population is uniformly persistent.
8. Local stability studies in a special case. The purpose of this section is to illustrate that it may be very difficult to show general stability results for the persistence equilibrium in the models in [21,43].
We consider the case that only one particular stage exerts a negative feedback on procreating adults. Let n ≥ 3, ∈ {1, . . . , n}, and Here η j ≥ γ j > 0, j = 1, . . . , n − 1, η n > 0. Further h : R + → R + is decreasing and differentiable, h(0) > 0 ≥ h (0). For = n − 2, this is a special case of the ODE version of [26] investigated in [21,43]. Notice that, for < n, Theorem 7.3 does not provide any useful condition for the boundedness of solutions because its assumptions would imply that R 0 ≤ 1. Notice that γ n (x) = h(x ). So boundedness of solutions is an open problem for this system. Here we focus on the stability of persistence equilibria.
After scaling time, we can assume that If h (0) < 0, we can also assume that h (0) = −1 by scaling the dependent variables.
In this special case, we obtain from Section 6 that and, if R 0 > 1, there is a unique persistence equilibrium x ∈ (0, ∞) n determined by We will also need the relation (8.5) The variational system (linearization about the equilibrium) is Eigenvectors v of the right hand side with associated eigenvalues satisfy This is rewritten as . . , n.
We substitute this into the first equation and divide by v 1 , .
Assume that h (x ) < 0. We rewrite this equation as Then this equation, whatever , has no solutions λ ≥ 0. We explore several cases for .
8.1. Negative feedback from the first stage to itself. Let = 1. Then we can reorganize equation (8.10) as We observe that, if the real part of λ is nonnegative, the absolute value of the left hand side is greater than 1, and the absolute value of the right hand side is at most 1. So there are no roots with nonnegative real parts. Alternatively we can argue that the linearized system is quasipositive and the eigenvalue with leading real part is actually real. So the persistence equilibrium is locally asymptotically stable.

8.2.
Negative feedback from the last stage. Let = n. Then equation (8.10) becomes We take absolute values and find that there are no roots with nonnegative real part if ψ(x ) ≤ 2. In fact, one can show that the persistence equilibrium is globally asymptotically stable if h is strictly decreasing and s 2 h(s) is an increasing function of s. This is done by deriving a single integral equation for x n and combining Theorem B.40 and Corollary 9.9 in [36] (see also [13]).
By our previous consideration, this equation has a solution λ with positive real part if ψ(x ) > 1 and is large enough. By Rouché's theorem [32, A.3], the original equation has a solution λ with positive real part if ψ(x ) > 1 and is large enough and η n is small enough. We need to make sure that our choices are feasible. Recall η n /q = h(x ), q independent of η n . If h is strictly decreasing with limit 0 at infinity, then x → ∞ as η n → 0. For our purposes so far, we needed to have that ψ(x ) > 1 which we can achieve if ψ(∞) = lim s→∞ ψ(s) > 1 and η n > 0 is chosen sufficiently small. (a) h(y) = e −y . Then ψ(y) = y → ∞ as y → ∞.

9.
Epilog: A modified model. We developed a model for populations with many discrete stages. We obtained satisfactory results concerning the persistence and extinction of the population and the existence and uniqueness of positive equilibria. Some special cases of these models, however, present substantial difficulties for proving the boundedness of solutions and the stability of positive equilibria. In particular, we are not able to show boundedness of solutions for the system x j =γ j−1 x j−1 − η j x j , j = 2, . . . , n, (9.1) with n ≥ 3 and 1 < < n. Here η j ≥ γ j > 0, j = 1, . . . , n − 1, η n > 0 and h : R + → R + is strictly decreasing. (9.1) is a special case of the models in [21,43] with = n − 2 = 10, but we are not even able to prove boundedness of solutions if n = 3 and = 2. This is quite disturbing because one feels that a strong immune response or resistance of the hosts which is triggered when the adult ticks feed on them (represented by x ) and reduces the fertility of the ticks when they lay their eggs later (represented by x n ) should be able to keep tick numbers bounded even if there are no other overcrowding effects present. Sometimes mathematical difficulties can be a sign that something is wrong with the model [35], in our case with modeling the negative feedback of feeding adults on the fertility of egg-laying adults. A first observation is that, if the negative feedback is due to an immune reaction caused by feeding adults which only plays out during egg-laying, then the fertility reduction of egg-laying adults is caused by the ticks that were feeding at the same time as the egg-laying adults were, i.e., at some moment in the past. In a model of ordinary differential equations, where stage lengths are exponentially distributed, it is difficult to model the appropriate delay in a consistent way. In a model with discrete delays where stage lengths are the same for all individuals in the stage, this is possible in a more satisfactory way though it does not remove the mathematical difficulties [13].
Global stability of the persistence equilibrium can be enforced by assuming that the function yθ(y)τ n−2 (y) is increasing in y ≥ 0 and θ(y)τ n−2 (y) is strictly decreasing in y > 0 and that the functions yτ j (y) are increasing in y ≥ 0 while τ j is decreasing and θ strictly decreasing. Further µ j (x j ) should be increasing in x j . Then the system induces a monotone semiflow [31] with at most one nonzero equilibrium (Section 6.2). This is a bit of a balancing act because yθ(y)τ n−2 (y) should be both increasing and bounded as a function of y. Hopefully, some of these assumptions can be weakened.