Modeling the Argasid Tick (Ornithodoros moubata) Life Cycle

Abstract

The first mathematical models for an argasid tick are developed to explore the dynamics and identify knowledge gaps of these poorly studied ticks. These models focus on Ornithodoros moubata, an important tick species throughout Africa and Europe. Ornithodoros moubata is a known vector for African swine fever (ASF), a catastrophically fatal disease for domesticated pigs in Africa and Europe. In the absence of any previous models for soft-bodied ticks, we propose two mathematical models of the life cycle of O. moubata. One is a continuous-time differential equation model that simplifies the tick life cycle to two stages, and the second is a discrete-time difference equation model that uses four stages. Both models use two host types: small hosts and large hosts, and both models find that either host type alone could support the tick population and that the final tick density is a function of host density. While both models predict similar tick equilibrium values, we observe significant differences in the time to equilibrium. The results demonstrate the likely establishment of these ticks if introduced into a new area even if there is only one type of host. These models provide the basis for developing future models that include disease states to explore infection dynamics and possible management of ASF.

Appendix

1.1 Model Analysis of Continuous-Time Model

First, we prove that the solutions of the continuous-time model (4.3) and (4.4) are well-posed.

Theorem 1

For any given N(0), A(0) ≥ 0, where both cannot be zero, the solutions of (4.3) and (4.4) are positive and bounded.

Proof

We first rewrite the model (4.3) and (4.4) as $\dot N=F_1(N,\,A)$ and $\dot A=F_2(N,\,A)$, and find that if $(N,\,A) \in \mathcal {R}^2_{+} \,\cup \, (0,\,0)$, then F ₁(0, A) ≥ 0 and F ₂(N, 0) ≥ 0. Then, we apply Theorem A.4 in [34], and prove that solutions of (4.3) and (4.4) are non-negative if the initial values are non-negative.

Next, for the boundedness, we let $(N,\,A) \in \mathcal {R}^2_{+}$, adding up (4.3) and (4.4) yields

$$\displaystyle \begin{aligned} \frac{\mathrm{d} \left(N+A\right)}{\mathrm{d} t} = (A_1 -A_8) A -A_4 N- A_2 N^2 - A_6 A^2 -(A_3+A_7) N A. \end{aligned} $$

(1)

It yields $\frac {\mathrm {d} \left (N+A\right )}{\mathrm {d} t}<0$ with large positive values of N and A. Therefore, the value of $\left (N+A\right )$ is bounded. For notational simplicity, we denote parameters in model (4.3) and (4.4) as follows:

$$\displaystyle \begin{aligned} \begin{array}{ll} &A_1=b_S\,S+b_L\,L, \quad A_2=\frac{c_N}{M_S S+M_L L}, \quad A_3 =\alpha_N A_2, \quad A_4 = d_N, \\ &A_5 =\gamma_S S+\gamma_L L, \quad A_6 = \frac{c_A}{M_S S+M_L L}, \quad A_7 = A_6 \alpha_A, \quad A_8 = d_A. \end{array} {} \end{aligned} $$

(2)

□

Therefore, we focus on the system (4.3) and (4.4), which yields one tick-free equilibrium E ₀ = (0, 0), and positive equilibrium: $E_1=(\bar N,\,\bar A)$. Then, for E ₁, we have

$$\displaystyle \begin{aligned} \bar N = \frac{\bar N (A_6 \bar A+A_8)}{A_5-A_7 \bar A}, {} \end{aligned} $$

(3)

where $\bar A$ is determined by the following cubic equation:

$$\displaystyle \begin{aligned} F_1(\bar A) = C_0 \bar A^3 + C_1 \bar A^2 + C_2 \bar A + C_3. {} \end{aligned} $$

(4)

Here, C _i are in terms of A _i parameters in (2), as follows:

$$\displaystyle \begin{aligned} \begin{array}{ll} C_0 &=A_6^2 A_2 (\alpha_N \alpha_A-1),\\ C_1 &=\displaystyle A_6\left[ (A_1 \alpha_A +A_4+A_5) A_6 \alpha_2 -A_2 (A_5 \alpha_N+A_8) +A_8 \frac{C_0}{A_6^2}\right],\\ C_2 &= \displaystyle-\frac{C_3 A_6 \alpha_A}{A_5} -A_2 A_8 (A_5 \alpha_N+A_8) -(A_1 \alpha_A + A_5+A_4) A_5 A_6, \\ C_3 &= A_5 (A_1 A_5-A_4 A_8-A_5 A_8). \end{array} {} \end{aligned} $$

(5)

The local stability of the equilibrium solutions are determined by their corresponding eigenvalues solved from the corresponding characteristic polynomial as follows:

$$\displaystyle \begin{aligned} \begin{array}{ll} P(\lambda) &= \lambda^2 + B_1 \lambda + B_2, \\ B_1 &= (2 A_2+A_7)\bar N+(A_3+2 A_6 )\bar A+A_4+A_5+A_8,\\ B_2 &= 4 A_2 A_6 \bar N \bar A +2 A_2 A_7 \bar N^2 +2 A_3 A_6 \bar A^2 +(A_1 A_7+A_3 A_8) \bar A\\ & \quad {+}(2 A_2 A_8+A_3 A_5)\bar N {+}(A_4+A_5) (A_7\bar N{+}2 A_6 \bar A) {-}A_1 A_5{+}A_4 A_8{+}A_5 A_8: \end{array} {} \end{aligned} $$

(6)

Evaluating P(λ) at E ₀ yields

$$\displaystyle \begin{aligned} \begin{array}{l} P_0(\lambda) = \lambda^2 + B_{10} \lambda + B_{20}, \qquad \mathrm{where},\\ B_{10} = A_4+A_5+A_8, \qquad B_{20} = -A_1 A_5+A_4 A_8+A_5 A_8. \end{array} {} \end{aligned} $$

(7)

Theorem 2

In original parameter values, we define a threshold as

$$\displaystyle \begin{aligned} \begin{array}{ll} B_{20} = -(L b_L+S b_S) (L \gamma_L+S \gamma_S) +\left(d_N+L\gamma_L+S\gamma_S \right) d_A \qquad \mathrm{or}\\ R_0=\displaystyle\frac{L b_L+S b_S}{d_A} \frac{1}{\frac{d_N}{L \gamma_L+S \gamma_S}+1} \end{array} \end{aligned} $$

(8)

• when B ₂₀ > 0 or R ₀ < 1, the tick-free equilibrium E ₀ is locally asymptotically stable,

• when B ₂₀ < 0 or R ₀ > 1, E ₀ becomes unstable, while the positive equilibrium E ₁ emerges,

• when B ₂₀ = 0 or R ₀ = 1, a transcritical bifurcation occurs; moreover E ₀ and E ₁ intersect and exchange stability.

Proof

With all positive parameter values, the stability of E ₀ is easily derived from P ₀(λ) = 0 in (7). Since C ₃ = −A ₅ B ₂₀, we have C ₃ > 0 and C ₂ < 0, when B ₂₀ < 0. Therefore, the cubic equation (4) has at least one positive solution, denoted by E ₁. Moreover, evaluating P ₀(λ) = 0 at E ₁ yields $B_2|{ }_{E_1}=-B_{20} f_{E1}$, where f _E1 is in terms of A _i parameters. Therefore, E ₁ has one zero eigenvalue when B ₂₀ = 0. This proves the occurrence of the transcritical bifurcation. □

Theorem 3

With all positive parameter values and positive equilibrium solutions, no Hopf bifurcation occurs.

Proof

In (6), B ₁ is always positive with positive solutions (N, A) and positive parameter values. Therefore, B ₁ = 0 does not occur; thus, the necessary condition for a Hopf bifurcation is never satisfied for model (4.3)–(4.4). □

1.2 Model Analysis of Discrete-Time Model

In this section we examine the dynamics of the discrete-time model (5.3) assuming constant host densities. Model (5.3) can be represented by the matrix equation

$$\displaystyle \begin{aligned}\mathbf{T}(t+1) = P(S(t), L(t), \mathbf{T}(t)) \mathbf{T}(t), {}\end{aligned}$$

where the projection matrix P(S, L, T) is given by

$$\displaystyle \begin{aligned} { \left(\begin{array}{cccc} \sigma_{s_1}(S,L, \mathbf{T}) & 0\ & 0 & \beta(S, L, \mathbf{T})\\ \sigma_{g_1} \gamma_1(S,L, \mathbf{T}) & \sigma_{s_2}(S,L, \mathbf{T}) & 0 & 0\\ 0 & \sigma_{g_2} \gamma_{2,S} f_{2, S}(S,L, \mathbf{T}) & \sigma_{s_X}(S,L, \mathbf{T}) & 0\\ 0 & \sigma_{g_2}\gamma_{2,L} f_{2, L}(S,L, \mathbf{T}) & \sigma_{g_X} \gamma_X(S,L, \mathbf{T}) & \sigma_{s_A}(S,L, \mathbf{T}) \end{array}\right).} {} \end{aligned} $$

(9)

By the linearization principle [12], the extinction equilibrium T = 0 is stable when the dominant eigenvalue of the inherent projection matrix P(S, L, 0) is less than 1 and unstable when it is greater than 1. Since the dominant eigenvalue and the inherent net reproductive number R ₀ are on the same side of 1 [10], the same is true in terms of R ₀.

The inherent net reproductive number of model (5.3) is given by

$$\displaystyle \begin{aligned}R_0(S, L) = \frac{\beta\gamma_1\sigma_{g_1}\sigma_{g_2}\left(\gamma_{2,L}f_{2,L}(1-\sigma_{s_X})+\gamma_{2,S}f_{2,S}\sigma_{g_X}\gamma_X\right)}{(1-\sigma_{s_1})(1-\sigma_{s_2})(1-\sigma_{s_X})(1-\sigma_{s_A})} ,\end{aligned}$$

where the functional dependencies have been dropped to simplify notation and all functions are evaluated at (S, L, 0). This value is defined to be the dominant eigenvalue of the matrix F(I − U)⁻¹, where F and U are obtained by decomposing the projection matrix (9) into a fertility matrix F and a transition matrix U, so that P = F + U [10]. Theorem 4 establishes the condition for tick persistence and characterizes the behavior of model (5.3) in a neighborhood of R ₀ ≈ 1.

Theorem 4

Assume S and L are constant.

• (a) The extinction equilibrium T = 0 is globally asymptotically stable for R ₀(S, L) < 1.

• (b) For R ₀(S, L) > 1, the extinction equilibrium is unstable and system (5.3) is permanent; that is, there exists a positive constant δ > 0 such that

  $$\displaystyle \begin{aligned}\delta\leq \lim _{t\rightarrow\infty} \inf |\mathbf{T}(t)|\leq \lim _{t\rightarrow\infty} \sup |\mathbf{T}(t)|\leq \frac{1}{\delta}\end{aligned}$$

  for all solutions T(t) satisfying $\mathbf {T}(0)\in R_+^4$ and |T(0)| > 0.

• (c) For R ₀(S, L) > 1, a branch of positive equilibria bifurcates from the extinction equilibria. The positive equilibria are locally asymptotically stable in the neighborhood of $R_0(S,L)\gtrapprox 1$.

Proof

• (a) By Theorem 1.1.3 of [10], R ₀ and the dominant eigenvalue of P(S, L, 0) are on the same side of 1. Since the feeding functions (5.4) are decreasing functions of tick density, P(S, L, T) ≤ P(S, L, 0) holds for all $\mathbf {T}\in R_+^4$. Therefore, by Theorem 1.2.1 of [10], the extinction equilibrium is globally asymptotically stable for R ₀(S, L) < 1.

• (b) Assume R ₀(S, L) > 1, then by Theorem B2 of [21], system (5.3) is permanent if it is dissipative. Since all nonzero entries of projection matrix P are decreasing functions of tick density, there exists a K > 0 such that for |T| > K, the sum of each row of P is less than 1, that is, ∑_i p _ij(S, L, T) < 1. By Theorem B1 of [21], system (5.3) is dissipative.

• (c) By Theorem 1.2.5 of [10], a branch of positive equilibria bifurcates from the extinction equilibrium at R ₀(S, L) = 1. Since projection matrix P contains only negative tick density effects, the bifurcation is forward; that is, the positive equilibria exist for $R_0(S,L)\gtrapprox 1$. Since the bifurcation is forward, by Theorem 1.2.6 of [10], the equilibria are stable in a neighborhood of $R_0(S, L) \gtrapprox 1$. □