The effect of landscape heterogeneity and host movement on a tick-borne pathogen

Abstract

Landscape heterogeneity can be instrumental in determining local disease risk, pathogen persistence and spread. This is because different landscape features such as habitat type determine the abundance and spatial distributions of hosts and pathogen vectors. Therefore, disease prevalence and distribution are intrinsically linked to the hosts and vectors that utilise the different habitats. Here, we develop a simplified reaction diffusion model of the louping-ill virus and red grouse (Lagopus lagopus scoticus) system to investigate the occurrence of a tick-borne pathogen and the effect of host movement and landscape structure. Ticks (Ixodes ricinus), the virus-vector, are dispersed by a virally incompetent tick host, red deer (Cervus elephus), between different habitats, whilst the virus infects only red grouse. We investigated how deer movement between different habitats (forest and moorland) affected tick distribution and hence prevalence of infected ticks and grouse and hence, the effect of habitat size ratio and fragmentation on infection. When habitat type has a role in the survival of the pathogen vector, we demonstrated that habitat fragmentation can have a considerable effect on infection. These results highlight the importance of landscape heterogeneity and the proximity and size of adjacent habitats when predicting disease risk in a particular location. In addition, this model could be useful for other pathogen systems with generalist vectors and may inform policy on possible disease management strategies that incorporate host movements.

Appendix: Derivation of equations

We will now describe how we attained the system equations.

Grouse and corresponding (attached) tick Eqs. 15–21

Let $G_\sigma^j\left( {x,t} \right)$ be the density of grouse with j ticks, where σ ∊ {S, I} and $0 \leqslant j \leqslant N$. Let $G_R^{j,i}\left( {x,t} \right)$ be the density of recovered grouse with j susceptible and i infected ticks, respectively. We define the following

$$ {G_s} = \sum\limits_{j = 0}^N {G_s^j,{G_I} = \sum\limits_{j = 0}^N {G_I^j,{G_R} = } } \sum\limits_{k = 0}^N {\sum\limits_{j = 0}^N {G_R^{j,i}\quad and\quad G = {G_s} + {G_I} + {G_R}.} } $$

Equations describing the evolution of $G_s^0\left( {x,t} \right)$ and $G_I^0\left( {x,t} \right)$ over time can be written as

$$ \frac{{\partial G_s^0}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_s^0}}{{\partial {x^2}}} + \left( {{a_g} - {s_g}G} \right)G - {b_g}G_s^0 - {\beta_g}G_s^0\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right) + {\mu_g}G_s^1, $$

(15)

$$ \frac{{\partial G_I^0}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_I^0}}{{\partial {x^2}}} - \left( {{b_g} + {\alpha_g} + {\gamma_g}} \right)G_I^0 - {\beta_g}G_I^0\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right) + {\mu_g}G_s^1, $$

(16)

where we assume that all grouse can reproduce and each newborn is susceptible and has no ticks. Now, for $N - 1 \geqslant j \geqslant 1$, we have

$$ \begin{array}{*{20}{c}} {\frac{{\partial G_s^j}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_s^j}}{{\partial {x^2}}} - {b_g}G_s^j + {\beta_g}G_s^{j - 1}\left( {T_{\rm{off}}^s + \left( {1 - \kappa } \right)T_{\rm{off}}^I} \right) - {\beta_g}G_s^jT_{\rm{off}}^s} \\{ - {\mu_g}G_s^j + {\mu_g}G_s^{j + 1} - {\beta_g}G_s^jT_{\rm{off}}^I,} \\\end{array} $$

(17)

$$ \begin{array}{*{20}{c}} {\frac{{\partial G_I^j}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_I^j}}{{\partial {x^2}}} - \left( {{b_g} + {\alpha_g} + {\gamma_g}} \right)G_I^j + {\beta_g}G_I^{j - 1}\left( {T_{\rm{off}}^s + \kappa T_{\rm{off}}^I} \right) - {\beta_g}G_I^j\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right)} \\{ - {\mu_g}G_I^j + {\mu_g}G_I^{j + 1} + {\beta_g}G_s^{j - 1}T_{\rm{off}}^I,} \\\end{array} $$

(18)

and for j = N

$$ \frac{{\partial G_s^N}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_s^N}}{{\partial {x^2}}} - {b_g}G_s^N + {\beta_g}G_s^{N - 1}\left( {T_{\rm{off}}^s + \left( {1 - \kappa } \right)T_{\rm{off}}^I} \right) - {\mu_g}NG_s^N, $$

(19)

$$ \frac{{\partial G_I^N}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_I^N}}{{\partial {x^2}}} - \left( {{b_g} + {\alpha_g} + {\gamma_g}} \right)G_I^N + {\beta_g}G_I^{N - 1}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right) - {\mu_g}NG_I^N + \kappa {\beta_g}G_s^{N - 1}T_{\rm{off}}^I. $$

(20)

Summing over 0 ≤ j ≤ N then gives the following

$$ \frac{{\partial {G_s}}}{{\partial t}} = {M_G}\frac{{{\partial^2}{G_s}}}{{\partial {x^2}}} + \left( {{a_g} - {s_g}G} \right)G - {b_g}{G_s} - \kappa T_{\rm{off}}^i{\beta_g}{G_s}, $$

(21)

$$ \frac{{\partial {G_I}}}{{\partial t}} = {M_G}\frac{{{\partial^2}{G_I}}}{{\partial {x^2}}} - \left( {{b_g} + {\alpha_g} + {\gamma_g}} \right){G_I} + \kappa T_{\rm{off}}^i{\beta_g}{G_s}. $$

(22)

Note that for the derivation of (22), we have used the following:

$$ - {\beta_g}\sum\limits_{j = 0}^{N - 1} {G_s^j = - {\beta_g}{G_s} + {\beta_g}G_s^N \approx - {\beta_g}{G_s}.} $$

That is, we have assumed that N is sufficiently large that the corresponding density $G_s^N$ is small and can, therefore, be neglected to leading order. We define the following quantities

$$ {T_{{\rm{on}},s}} = \frac{{\sum\limits_{j = 1}^N {jG_s^j} }}{{\sum\limits_{j = 0}^N {G_s^j} }}\quad {\hbox{and}}\quad {T_{{\rm{on}},I}} = \frac{{\sum\limits_{j = 1}^N {jG_I^j} }}{{\sum\limits_{j = 0}^N {G_I^j} }} $$

to be the average number of (attached) ticks per susceptible and infected grouse, respectively. Rearranging gives

$$ {T_{{\rm{on}},s}}{G_s} = \sum\limits_{j = 1}^N {jG_s^j} \quad {\hbox{and}}\quad {T_{{\rm{on}},I}}{G_I} = \sum\limits_{j = 1}^N {jG_I^j.} $$

(23)

Differentiating with respect to time then gives

$$ \frac{{\partial {T_{{\rm{on}},\sigma }}}}{{\partial t}}{G_\sigma } + \frac{{\partial {G_\sigma }}}{{\partial t}}{T_{{\rm{on}},\sigma }} = \sum\limits_{j = 1}^N {j\frac{{\partial G_\sigma^j}}{{\partial t}},\,\sigma \in \left\{ {s,I} \right\}}, $$

which, upon substituting (21) and (15), (17), (19), yields

$$ \begin{array}{*{20}{c}} {\frac{{\partial {T_{{\rm{on}},s}}}}{{\partial t}}{G_s} + {T_{{\rm{on}},s}}\left[ {\left( {{a_g} - {s_g}G} \right)G - {b_g}{G_s} + {M_G}\frac{{{\partial^2}{G_s}}}{{\partial {x^2}}} - \kappa T_{\rm{off}}^i{\beta_g}{G_s}} \right]} \hfill \\{ = - {b_g}\sum\limits_{j = 1}^N {jG_s^j + {M_G}\sum\limits_{j = 1}^N {j\frac{{{\partial^2}G_s^j}}{{\partial {x^2}}} + {\beta_g}T_{\rm{off}}^s\sum\limits_{j = 0}^{N - 1} {G_s^j} } } } \hfill \\{ - \kappa {\beta_g}T_{\rm{off}}^I\sum\limits_{j = 1}^N {jG_s^j - {\mu_g}\sum\limits_{j = 1}^N {jG_s^j,} } } \hfill \\\end{array} $$

(24)

and, using (22) and (16), (18), (20),

$$ \begin{array}{*{20}{c}} {\frac{{\partial {T_{{\rm{on}},I}}}}{{\partial t}}{G_I} + {T_{{\rm{on}},I}}\left[ { - \left( {{b_g} + {\alpha_g} + {\gamma_g}} \right){G_I} + {M_G}\frac{{{\partial^2}{G_I}}}{{\partial {x^2}}} + \kappa T_{\rm{off}}^i{\beta_g}{G_s}} \right]} \hfill \\{ = - \left( {{b_g} + {\alpha_g} + {\gamma_g}} \right)\sum\limits_{j = 1}^N {jG_I^j + {M_G}\sum\limits_{j = 1}^N {j\frac{{{\partial^2}G_I^j}}{{\partial {x^2}}} + {\beta_g}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right)\sum\limits_{j = 0}^{N - 1} {G_I^j} } } } \hfill \\{ + \kappa {\beta_g}T_{\rm{off}}^I\sum\limits_{j = 1}^N {jG_s^{j - 1} - {\mu_g}\sum\limits_{j = 1}^N {jG_I^j} .} } \hfill \\\end{array} $$

(25)

Now differentiating the expressions in (23) twice with respect to x gives

$$ \frac{{{\partial^2}{T_{{\rm{on}},\sigma }}}}{{\partial {x^2}}}{G_\sigma } + 2\frac{{\partial {G_\sigma }}}{{\partial x}}\frac{{\partial {T_{{\rm{on}},\sigma }}}}{{\partial x}} + \frac{{{\partial^2}{G_\sigma }}}{{\partial {x^2}}}{T_{{\rm{on}},\sigma }} = \sum\limits_{j = 1}^N {j\frac{{{\partial^2}G_\sigma^j}}{{\partial {x^2}}}\,\sigma \in \left\{ {s,I} \right\}.} $$

Substitution into (24) and using (23) then simplifies (24), after a little re-arranging, to

$$ \frac{{\partial {T_{{\rm{on}},s}}}}{{\partial t}} = {M_G}\left( {\frac{{{\partial^2}{T_{{\rm{on}},s}}}}{{\partial {x^2}}} + \frac{2}{{{G_s}}}\frac{{\partial {G_s}}}{{\partial x}}\frac{{\partial {T_{{\rm{on}},s}}}}{{\partial x}}} \right) + T_{\rm{off}}^s{\beta_g} - {\mu_g}{T_{{\rm{on}},s}} - \frac{{{T_{{\rm{on}},s}}G}}{{{G_s}}}\left( {{a_g} - {s_g}G} \right). $$

(26)

The rationale behind the unattachment of ticks is that each individual tick has a mean feeding time of 1/μg and the rate of detachment will be the inverse of this value. Therefore, if there are j ticks on the host, then total rate of detachment will be j times this value, namely, jμg.

Similarly, (25) becomes

$$ \begin{array}{*{20}{c}} {\frac{{\partial {T_{{\rm{on}},I}}}}{{\partial t}} = {M_G}\left( {\frac{{{\partial^2}{T_{{\rm{on}},I}}}}{{\partial {x^2}}} + \frac{2}{{{G_I}}}\frac{{\partial {G_I}}}{{\partial x}}\frac{{\partial {T_{{\rm{on}},I}}}}{{\partial x}}} \right) + \left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right){\beta_g}} \hfill \\{\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad - {\mu_g}{T_{{\rm{on}},I}} + \frac{{\kappa {G_s}{\beta_g}T_{\rm{off}}^i}}{{{G_I}}}\left[ {\left( {{T_{{\rm{on}},s}} + 1} \right) - {T_{{\rm{on}},I}}} \right].} \hfill \\\end{array} $$

(27)

Note that we have again used the assumption that $G_s^N$ is small, so that $\sum\nolimits_{j = 0}^{N - 1} {G_s^j}$ approximates to G _s .

The equation for G _R can be derived in a similar way. Namely, we have:

$$ \frac{{\partial G_R^{0,0}}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_R^{0,0}}}{{\partial {x^2}}} - {b_g}G_R^{0,0} + {\gamma_g}G_I^0 - {\beta_g}G_R^{0,0}\left( {T_{off}^s + T_{off}^I} \right) + {\mu_g}\left( {G_R^{1,0} + G_R^{0,1}} \right); $$

(28)

for 1 ≤ i ≤ N−1,

$$ \begin{array}{*{20}{c}} {\frac{{\partial G_R^{0,i}}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_R^{0,i}}}{{\partial {x^2}}} - {b_g}G_R^{0,i} + {\gamma_g}G_I^i - {\beta_g}G_R^{0,i}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right) + {\beta_g}G_R^{0,i - 1}T_{\rm{off}}^I} \hfill \\{\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad - i{\mu_g}G_R^{0,i} + {\mu_g}G_R^{1,i} + \left( {i + 1} \right){\mu_g}G_R^{0,i + 1};} \hfill \\\end{array} $$

(29)

for 1 ≤ j ≤ N−1,

$$ \begin{array}{*{20}{c}} {\frac{{\partial G_R^{j,0}}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_R^{j,0}}}{{\partial {x^2}}} - {b_g}G_R^{j,0} - {\beta_g}G_R^{j,0}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right) + {\beta_g}G_R^{j - 1,0}T_{\rm{off}}^s} \hfill \\{\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad - j{\mu_g}G_R^{j,0} + {\mu_g}G_R^{j,1} + \left( {j + 1} \right){\mu_g}G_R^{j + 1,0};} \hfill \\\end{array} $$

(30)

$$ \frac{{\partial G_R^{0,N}}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_R^{0,N}}}{{\partial {x^2}}} - {b_g}G_R^{0,N} + {\gamma_g}G_I^N - {\beta_g}G_R^{0,N}T_{\rm{off}}^s + {\beta_g}G_R^{0,N - 1}T_{\rm{off}}^I + {\mu_g}G_R^{1,N} - {\mu_g}NG_R^{0,N}; $$

(31)

$$ \frac{{\partial G_R^{N,0}}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_R^{N,0}}}{{\partial {x^2}}} - {b_g}G_R^{N,0} - {\beta_g}G_R^{N,0}T_{\rm{off}}^I + {\beta_g}G_R^{N - 1,0}T_{\rm{off}}^s + {\mu_g}G_R^{N,1} - {\mu_g}NG_R^{N,0}; $$

(32)

for 1 ≤ i ≤ N-1,

$$ \begin{array}{*{20}{c}} {\frac{{\partial G_R^{N,i}}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_R^{N,i}}}{{\partial {x^2}}} - {b_g}G_R^{N,i} + {\beta_g}G_R^{N - 1,i}T_{\rm{off}}^s + {\beta_g}G_R^{N,i - 1}T_{\rm{off}}^I - {\beta_g}G_R^{N,i}T_{\rm{off}}^I} \hfill \\{ - N{\mu_g}G_R^{N,i} - i{\mu_g}G_R^{N,i} + \left( {i + 1} \right){\mu_g}G_R^{N,i + 1};} \hfill \\\end{array} $$

(33)

for 1 ≤ j ≤ N−1,

$$ \begin{array}{*{20}{c}} {\frac{{\partial G_R^{j,N}}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_R^{j,N}}}{{\partial {x^2}}} - {b_g}G_R^{j,N} + {\beta_g}G_R^{j - 1,N}T_{\rm{off}}^s + {\beta_g}G_R^{j,N - 1}T_{\rm{off}}^I - {\beta_g}G_R^{j,N}T_{\rm{off}}^s} \hfill \\{ - j{\mu_g}G_R^{j,N} - N{\mu_g}G_R^{j,N} + \left( {j + 1} \right){\mu_g}G_R^{j + 1,N};} \hfill \\\end{array} $$

(34)

$$ \frac{{\partial G_R^{N,N}}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_R^{N,N}}}{{\partial {x^2}}} - {b_g}G_R^{N,N} + {\beta_g}G_R^{N - 1,N}T_{\rm{off}}^s + {\beta_g}G_R^{N,N - 1}T_{\rm{off}}^I - 2N{\mu_g}G_R^{N,N}; $$

(35)

for 1 ≤ j,i ≤N −1,

$$ \begin{array}{*{20}{c}} {\frac{{\partial G_R^{j,i}}}{{\partial t}} = {M_G}\frac{{{\partial^2}G_R^{j,i}}}{{\partial {x^2}}} - {b_g}G_R^{j,i} + {\beta_g}G_R^{j - 1,i}T_{\rm{off}}^s + {\beta_g}G_R^{j,i - 1}T_{\rm{off}}^I - {\beta_g}G_R^{j,i}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right)} \hfill \\{ - \left( {j + i} \right){\mu_g}G_R^{j,i} - \left( {j + 1} \right){\mu_g}G_R^{j + 1,i} + \left( {i + 1} \right){\mu_g}G_R^{j,i + 1}.} \hfill \\\end{array} $$

(36)

Summing over 0 ≤ j,i ≤ N then gives

$$ \frac{{\partial {G_R}}}{{\partial t}} = {M_G}\frac{{{\partial^2}{G_R}}}{{\partial {x^2}}} - {b_g}{G_R} + {\gamma_g}{G_I}. $$

(37)

We define the following quantities

$$ T_{{\rm{on}},R}^s = \frac{{\sum\limits_{j = 1}^N {j\sum\limits_{i = 0}^N {G_R^{j,i}} } }}{{\sum\limits_{j,i = 0}^N {G_R^{j,i}} }}\quad {\hbox{and}}\quad T_{{\rm{on}},R}^I = \frac{{\sum\limits_{i = 1}^N i \sum\limits_{j = 0}^N {G_R^{j,i}} }}{{\sum\limits_{j,i = 0}^N {G_R^{j,i}} }} $$

to be the average number of (attached) susceptible and infected ticks per recovered grouse, respectively. Rearranging gives

$$ T_{{\rm{on}},R}^s{G_R} = \sum\limits_{j = 1}^N {j\sum\limits_{i = 0}^N {G_R^{j,i}} } \quad {\hbox{and}}\quad T_{{\rm{on}},R}^I{G_R} = \sum\limits_{i = 1}^N {j\sum\limits_{j = 0}^N {G_R^{j,i}.} } $$

(38)

Differentiating these expressions give

$$ \frac{{\partial T_{{\rm{on}},R}^s}}{{\partial t}}{G_R} + \frac{{\partial {G_R}}}{{\partial t}}T_{{\rm{on}},R}^s = \sum\limits_{j = 1}^N {\sum\limits_{i = 0}^N {j\frac{{\partial G_R^{j,i}}}{{\partial t}},} } $$

(39)

$$ \frac{{\partial T_{{\rm{on}},R}^I}}{{\partial t}}{G_R} + \frac{{\partial {G_R}}}{{\partial t}}T_{{\rm{on}},R}^I = \sum\limits_{i = 1}^N {\sum\limits_{j = 0}^N {i\frac{{\partial G_R^{j,i}}}{{\partial t}},} } $$

(40)

$$ \frac{{{\partial^2}T_{{\rm{on}},R}^s}}{{\partial {x^2}}}{G_R} + 2\frac{{\partial {G_R}}}{{\partial x}}\frac{{\partial T_{{\rm{on}},R}^s}}{{\partial x}} + \frac{{{\partial^2}{G_R}}}{{\partial {x^2}}}T_{{\rm{on}},R}^s = \sum\limits_{j = 1}^N {\sum\limits_{i = 0}^N {j\frac{{{\partial^2}G_R^{j,i}}}{{\partial {x^2}}},} } $$

(41)

$$ \frac{{{\partial^2}T_{{\rm{on}},R}^I}}{{\partial {x^2}}}{G_R} + 2\frac{{\partial {G_R}}}{{\partial x}}\frac{{\partial T_{{\rm{on}},R}^I}}{{\partial x}} + \frac{{{\partial^2}{G_R}}}{{\partial {x^2}}}T_{{\rm{on}},R}^I = \sum\limits_{i = 1}^N {\sum\limits_{j = 0}^N {i\frac{{{\partial^2}G_R^{j,i}}}{{\partial {x^2}}}.} } $$

(42)

Substituting (37) and (28) to (36) into (39) then yields, after a little re-arranging

$$ \begin{array}{*{20}{c}} {\frac{{\partial T_{{\rm{on}},R}^s}}{{\partial t}}{G_R} + T_{{\rm{on}},R}^s\left[ { - {b_g}{G_R} + {\gamma_g}{G_I} + {M_G}\frac{{{\partial^2}{G_R}}}{{\partial {x^2}}}} \right]} \hfill \\{ = - {b_g}\sum\limits_{j = 1}^N {\sum\limits_{i = 0}^N {jG_R^{j,i} + {M_G}\sum\limits_{j = 1}^N {\sum\limits_{i = 0}^N {j\frac{{{\partial^2}G_R^{j,i}}}{{\partial {x^2}}}} } } } } \hfill \\{ + {\beta_g}T_{\rm{off}}^s\sum\limits_{j = 0}^{N - 1} {\sum\limits_{i = 0}^N {G_R^{j,i} - {\mu_g}\sum\limits_{j = 1}^N {\sum\limits_{i = 0}^N {jG_R^{j,i}.} } } } } \hfill \\\end{array} $$

(43)

Substituting (41), we then obtain

$$ \frac{{\partial T_{{\rm{on}},R}^s}}{{\partial t}} = {M_G}\left( {\frac{{{\partial^2}T_{{\rm{on}},R}^s}}{{\partial {x^2}}} + \frac{2}{{{G_R}}}\frac{{\partial {G_R}}}{{\partial x}}\frac{{\partial T_{{\rm{on}},R}^s}}{{\partial x}}} \right) + T_{\rm{off}}^s{\beta_g} - {\mu_g}T_{{\rm{on}},R}^s - \frac{{{\gamma_g}T_{{\rm{on}},R}^s{G_I}}}{{{G_R}}}, $$

(44)

assuming, similar to above, that N is sufficiently large so that $\sum\nolimits_{i = 0}^N {\sum\nolimits_{j = 0}^{N - 1} {G_R^{j,i} \approx {G_R}} }$. Similarly, substituting (37) and (28) to (36) into (40) and using (42), we get

$$ \begin{array}{*{20}{c}} {\frac{{\partial T_{{\rm{on}},R}^I}}{{\partial t}}{G_R} + T_{{\rm{on}},R}^I\left[ { - {b_g}{G_R} + {\gamma_g}{G_I} + {M_G}\frac{{{\partial^2}{G_R}}}{{\partial {x^2}}}} \right]} \hfill \\{ = - {b_g}\sum\limits_{i = 1}^N {\sum\limits_{j = 0}^N {iG_R^{j,i} + {M_G}\sum\limits_{i = 0}^N {\sum\limits_{j = 0}^N {i\frac{{{\partial^2}G_R^{j,i}}}{{\partial {x^2}}} + {\gamma_g}\sum\limits_{i = 1}^N {iG_I^i} } } } } } \hfill \\{ + {\beta_g}T_{\rm{off}}^I\sum\limits_{i = 0}^{N - 1} {\sum\limits_{j = 0}^N {G_R^{j,i} - {\mu_g}\sum\limits_{i = 1}^N {\sum\limits_{j = 0}^N {iG_R^{j,i},} } } } } \hfill \\\end{array} $$

(45)

and

$$ \frac{{\partial T_{{\rm{on}},R}^I}}{{\partial t}} = {M_G}\left( {\frac{{{\partial^2}T_{{\rm{on}},R}^I}}{{\partial {x^2}}} + \frac{2}{{{G_R}}}\frac{{\partial {G_R}}}{{\partial x}}\frac{{\partial T_{{\rm{on}},R}^I}}{{\partial x}}} \right) + T_{\rm{off}}^I{\beta_g} - {\mu_g}T_{{\rm{on}},R}^I + \frac{{{\gamma_g}{G_I}}}{{{G_R}}}\left( {{T_{{\rm{on}},I}} - T_{{\rm{on}},R}^I} \right), $$

(46)

assuming that $\sum\nolimits_{j = 0}^N {\sum\nolimits_{i = 0}^{N - 1} {G_R^{j,i} \approx {G_R}} }$.

Deer Eqs. 22–23

Let D ^j (x, t) be the density of deer with j ticks, 0 ≤ j ≤ N. Note that attached ticks can be either susceptible or infected. Note also that, in this case, we also assume density dependence on tick uptake. Specifically, we assume that the uptake rate, β _D , is given by

$$ {\beta_D}(j) = {\beta_D}\left( {1 - \frac{j}{N}} \right). $$

That is, uptake rate decreases monotonically as the number of already attached ticks increases, representing a limit of tick occupancy space on the deer. We take ${\beta_D}\left( {j > N} \right) = 0$.

An equation describing the evolution of ${D^0}\left( {x,t} \right)$ over time can be written as

$$ \frac{{\partial {D^0}}}{{\partial t}} = {M_D}\frac{{{\partial^2}{D^0}}}{{\partial {x^2}}} + \left( {{a_D} - {s_D}D} \right)D - {b_D}{D^0} - {\beta_D}{D^0}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right) + {\mu_g}{D^1}, $$

(47)

where $D = \sum\nolimits_{j = 0}^N {{D^j}}$ and we assume, as for grouse, a logistic growth rate for deer and with each newborn having no ticks. For $N - 1 \geqslant j \geqslant 1$, we have

$$ \begin{array}{*{20}{c}} {\frac{{\partial {D^j}}}{{\partial t}} = {M_D}\frac{{{\partial^2}{D^j}}}{{\partial {x^2}}} - {b_D}{D^j} + {\beta_D}\left( {1 - \frac{{\left( {j - 1} \right)}}{N}} \right){D^{j - 1}}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right) - {\beta_D}\left( {1 - \frac{j}{N}} \right){D^j}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right)} \hfill \\{\quad \quad \quad \quad \quad \quad \quad \;\,16ex - {\mu_D}j{D^j} + {\mu_D}\left( {j + 1} \right){D^{j + 1}},} \hfill \\\end{array} $$

(48)

and for j=N

$$ \frac{{\partial {D^N}}}{{\partial t}} = {M_D}\frac{{{\partial^2}{D^N}}}{{\partial {x^2}}} - {b_D}{D^N} + {\beta_D}{D^{N - 1}}\left( {1 - \frac{{\left( {N - 1} \right)}}{N}} \right)\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right) - {\mu_g}N{D^N}. $$

(49)

Summing over $0 \leqslant j \leqslant N$ gives the following

$$ \frac{{\partial D}}{{\partial t}} = {M_D}\frac{{{\partial^2}D}}{{\partial {x^2}}} + \left( {{a_D} - {s_D}D} \right)D - {b_g}D. $$

(50)

Similar to before, we define

$$ {T_{{\rm{on}},D}} = \frac{{\sum\limits_{j = 1}^N {j{D^j}} }}{{\sum\limits_{j = 0}^N {{D^j}} }} $$

to be the average number of (attached) ticks per deer, so that

$$ {T_{{\rm{on}},D}}D = \sum\limits_{j = 1}^N {j{D^j}.} $$

(51)

Differentiating then gives

$$ \frac{{\partial {T_{{\rm{on}},D}}}}{{\partial t}}D + \frac{{\partial D}}{{\partial t}}{T_{{\rm{on}},D}} = \sum\limits_{j = 1}^N {j\frac{{\partial {D^j}}}{{\partial t}},} $$

and

$$ \frac{{{\partial^2}{T_{{\rm{on}},D}}}}{{\partial {x^2}}}D + 2\frac{{\partial D}}{{\partial x}}\frac{{\partial {T_{{\rm{on}},D}}}}{{\partial x}} + \frac{{{\partial^2}D}}{{\partial {x^2}}}{T_{{\rm{on}},D}} = \sum\limits_{j = 1}^N {j\frac{{{\partial^2}{D^j}}}{{\partial {x^2}}}}, $$

which, upon substituting (50) and (47) to (51), yields

$$ \begin{array}{*{20}{c}} {\frac{{\partial {T_{{\rm{on}},D}}}}{{\partial t}}D + {T_{{\rm{on}},D}}\left[ {\left( {{a_D} - {s_D}D} \right)D - {b_D}D + {M_D}\frac{{{\partial^2}D}}{{\partial {x^2}}}} \right]} \hfill \\{ = - {b_D}\sum\limits_{j = 1}^N {j{D^j} + {M_D}\sum\limits_{j = 1}^N {j\frac{{{\partial^2}{D^j}}}{{\partial {x^2}}}} } } \hfill \\{ + {\beta_D}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right)\sum\limits_{j = 0}^{N - 1} {\left( {1 - \frac{j}{N}} \right){D^j} - {\mu_D}\sum\limits_{j = 1}^N {j{D^j},} } } \hfill \\\end{array} $$

(52)

which simplifies to

$$ \frac{{\partial {T_{on,D}}}}{{\partial t}} = {M_D}\left( {\frac{{{\partial^2}{T_{\rm{on}}}}}{{\partial {x^2}}} + \frac{2}{D}\frac{{\partial D}}{{\partial x}}\frac{{\partial {T_{{\rm{on}},D}}}}{{\partial x}}} \right) + {\beta_D}\left( {T_{\rm{off}}^s + T_{\rm{off}}^I} \right)\left( {1 - \frac{{{T_{{\rm{on}},D}}}}{N}} \right) - {\mu_D}{T_{{\rm{on}},D}} - {T_{{\rm{on}},D}}\left( {{a_D} - {s_D}D} \right). $$

(53)

Questing tick Eqs. 24–25

Summing the uptake and drop-off terms from above gives the following

$$ \begin{array}{*{20}{c}} {\frac{{\partial T_{\rm{off}}^s}}{{\partial t}} = - T_{\rm{off}}^s\sum\limits_{j = 0}^{N - 1} {\left( {{\beta_g}\left[ {G_s^j + G_I^j + \sum\limits_{i = 0}^N {G_R^{j,i}} } \right] + {\beta_D}\left( {1 - \frac{j}{N}} \right){D^j}} \right)} } \hfill \\{\quad \quad \quad + {\mu_g}\sum\limits_{j = 1}^N {j\left( {G_s^j + \sum\limits_{i = 0}^N {G_R^{j,i}} } \right) + {\rho_t}{\mu_D}\sum\limits_{j = 1}^N {j{D^j} - {b_T}T_{\rm{off}}^s.} } } \hfill \\\end{array} $$

(54)

Note that we have additionally assumed a linear death term for ticks, with constant rate b _T and a proliferative rate ρ _t as the ticks drop-off deer (taking the newborn ticks to be susceptible).

From (54), again taking for simplicity $\sum\nolimits_{j = 0}^{N - 1} {G_s^j \approx {G_s}}$, $\sum\nolimits_{j = 0}^{N - 1} {G_I^j \approx {G_I}}$, $\sum\nolimits_{j = 0}^{N - 1} {\sum\nolimits_{i = 0}^N {G_R^{j,i} \approx {G_R}} }$, we obtain the leading order expression

$$ \begin{array}{*{20}{c}} {\frac{{\partial T_{\rm{off}}^s}}{{\partial t}} = - T_{\rm{off}}^s\left( {{\beta_g}G + {\beta_D}D\left( {1 - \frac{{{T_{{\rm{on}},D}}}}{N}} \right)} \right) + {\mu_g}\left( {{T_{{\rm{on}},s}}{G_s} + T_{{\rm{on}},R}^s{G_R}} \right)} \hfill \\{\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + {\rho_t}{\mu_D}{T_{{\rm{on}},D}}D - {b_T}T_{\rm{off}}^s.} \hfill \\\end{array} $$

(55)

Similarly, we can write

$$ \begin{array}{*{20}{c}} {\frac{{\partial T_{\rm{off}}^I}}{{\partial t}} = - T_{\rm{off}}^I\sum\limits_{i = 0}^{N - 1} {\left( {{\beta_g}\left[ {G_s^i + G_I^i + \sum\limits_{j = 0}^N {G_R^{j,i}} } \right] + {\beta_D}\left( {1 - \frac{i}{N}} \right){D^i}} \right)} } \hfill \\{\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + {\mu_g}\sum\limits_{i = 1}^N i \left( {G_I^i + \sum\limits_{j = 0}^N {G_R^{j,i}} } \right) - \left( {{b_T} + {\alpha_T}} \right)T_{\rm{off}}^I,} \hfill \\\end{array} $$

(56)

where α _T denotes the rate of tick virulence due to infection. Simplifying, gives to leading order

$$ \begin{array}{*{20}{c}} {\frac{{\partial T_{\rm{off}}^I}}{{\partial t}} = - T_{\rm{off}}^I\left( {{\beta_g}G + {\beta_D}D\left( {1 - \frac{{{T_{{\rm{on}},D}}}}{N}} \right)} \right)} \hfill \\{\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad + {\mu_g}\left( {{T_{{\rm{on}},I}}{G_I} + T_{{\rm{on}},R}^I{G_R}} \right) - \left( {{b_T} + {\alpha_T}} \right)T_{\rm{off}}^I} \hfill \\\end{array} $$

(57)