Models of Disease Vector Control: When Can Aggressive Initial Intervention Lower Long-Term Cost?

Abstract

Insecticide spraying of housing units is an important control measure for vector-borne infections such as Chagas disease. As vectors may invade both from other infested houses and sylvatic areas and as the effectiveness of insecticide wears off over time, the dynamics of (re)infestations can be approximated by ${ SIRS}$-type models with a reservoir, where housing units are treated as hosts, and insecticide spraying corresponds to removal of hosts. Here, we investigate three ODE-based models of this type. We describe a dual-rate effect where an initially very high spraying rate can push the system into a region of the state space with low endemic levels of infestation that can be maintained in the long run at relatively moderate cost, while in the absence of an aggressive initial intervention the same average cost would only allow a much less significant reduction in long-term infestation levels. We determine some sufficient and some necessary conditions under which this effect occurs and show that it is robust in models that incorporate some heterogeneity in the relevant properties of housing units.

Appendix

1.1 Calculating $I^*$ for the Basic Model with $c > 0$

At an equilibrium,

$$\begin{aligned} \begin{aligned} -\beta I^*S^*-cS^*+w(m-S^*-I^*)&=0 \\ \beta I^*S^* +cS^*-rI^*&=0.\\ \end{aligned} \end{aligned}$$

(28)

From the second line, we have

$$\begin{aligned} S^*=\frac{rI^*}{\beta I^*+c}. \end{aligned}$$

(29)

Substituting (29) into the first line of (28) and simplifying give a quadratic equation:

$$\begin{aligned} \beta (w+r)\left( I^*\right) ^{2}-(\beta wm-rw-c(w+r))I^*-wcm=0. \end{aligned}$$

(30)

The only solution of (30) with $I^* \ge 0$ is:

$$\begin{aligned} I^{*}=\frac{1}{2}\left( \left[ \frac{wm}{(w+r)}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }\right] + \sqrt{\left( \frac{wm}{(w+r)}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }\right) ^{2}+\frac{4wcm}{\beta (w+r)}}\right) . \end{aligned}$$

1.2 Proof of Inequality (22)

For $r_1=\frac{w(\beta m-c)}{w+c}$, we have $\frac{wm}{w+r_1}-\frac{r_1w}{\beta (w+r_1)}-\frac{c}{\beta } = 0$.

Recall that

$$\begin{aligned} I^*(r) =0.5\left( \frac{wm}{w+r}-\frac{r_1w}{\beta (w+r)}-\frac{c}{\beta } +\sqrt{\left( \frac{wm}{w+r}-\frac{r_1w}{\beta (w+r)}-\frac{c}{\beta }\right) ^2+\frac{4wcm}{\beta (w+r)}}\right) . \end{aligned}$$

Let $T(r)=\frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }$ and $P(r)=\frac{wcm}{\beta (w+r)}$. Then,

$$\begin{aligned} I^*(r) = 0.5T(r)+0.5\sqrt{(T(r))^2+4P(r)}. \end{aligned}$$

(31)

Now,

$$\begin{aligned} I^*(r_1)=0.5\left( 0\right) +0.5\sqrt{\left( 0\right) ^2+\frac{4wcm}{\beta (w+r_1)}}=\frac{1}{2}\left( \sqrt{\frac{4wcm}{\beta (w+r_1)}}\right) =\sqrt{\frac{wcm}{\beta (w+r_1)}}. \end{aligned}$$

(32)

We have

$$\begin{aligned} T^{'}(r) =-\frac{w(\beta m+w)}{\beta (w+r)^2} \quad \text{ and } \quad P^{'}(r) =\frac{-wcm}{\beta (w+r)^2}. \end{aligned}$$

From (31), we obtain

$$\begin{aligned} \left( I^*\right) '(r)=0.5T^{'}(r)+0.5\frac{T(r)T^{'}(r)+2P^{'}(r)}{\sqrt{T^2(r)+4P(r)}}, \end{aligned}$$

so that

$$\begin{aligned} \begin{aligned} \left( I^*\right) '(r_1)&= -\frac{wm}{2(w+r_1)^2}-\frac{w^2}{2\beta (w+r_1)^2}-\frac{wcm}{2\beta (w+r_1)^2\sqrt{\frac{wcm}{\beta (w+r_1)}}}\\ r_1\left( I^*\right) '(r_1)&=-\frac{wmr_1}{2(w+r_1)^2}-\frac{w^2r_1}{2\beta (w+r_1)^2}-\frac{wcmr_1}{2\beta (w+r_1)^2\sqrt{\frac{wcm}{\beta (w+r_1)}}}.\\ \end{aligned} \end{aligned}$$

Therefore, in view of (32),

$$\begin{aligned} \begin{aligned} C^{'}(r_1)&=r_1\left( I^*\right) '(r_1)+I^*(r_1)\\&=-\frac{wmr_1}{2(w+r_1)^2}-\frac{w^2r_1}{2\beta (w+r_1)^2}-\frac{wcmr_1}{2\beta (w+r_1)^2\sqrt{\frac{wcm}{\beta (w+r_1)}}}+\sqrt{\frac{wcm}{\beta (w+r_1)}}\\&=\frac{wm}{(w+r_1)^2\sqrt{\frac{wcm}{\beta (w+r_1)}}}\left( \frac{cw}{\beta }-\left[ \sqrt{\frac{wcm}{\beta (w+r_1)}}-\frac{c}{\beta }\right] \frac{r_1}{2}\right) -\frac{w^2r_1}{2\beta (w+r_1)^2}.\\ \end{aligned} \end{aligned}$$

Replace $r_1$ by $\frac{w(\beta m-c)}{w+c}$ and let

$$\begin{aligned} A=\sqrt{\frac{wcm}{\beta (w+r_1)}}-\frac{c}{\beta }=\sqrt{\frac{cm(w+c)}{\beta (w+\beta m)}}-\frac{c}{\beta }>0. \end{aligned}$$

Notice that for positive $\beta , c, m, w$ the following inequalities are all equivalent:

$$\begin{aligned} \begin{aligned} \frac{cm(w+c)}{\beta (w+\beta m)}&>\frac{c^2}{\beta ^2}\\ \frac{m(w+c)}{w+\beta m}&>\frac{c}{\beta }\\ \beta m(w+c) = \beta mw+\beta mc&> cw+\beta mc = c(w+\beta m)\\ \beta mw&> cw\\ m&> \frac{c}{\beta }. \end{aligned} \end{aligned}$$

Thus, $A > 0$ if, and only if, $m > \frac{c}{\beta }$.

Also,

$$\begin{aligned} w+r_1=w+\frac{w(\beta m-c)}{w+c}=\frac{w(w+c)+w(\beta m-c)}{w+c}=\frac{w(w+\beta m)}{w+c}. \end{aligned}$$

It follows that

$$\begin{aligned} C^{'}(r_1)&=\frac{m(w+c)^2}{(w+\beta m)^2\sqrt{\frac{cm(w+c)}{\beta (w+\beta m)}}}\left( \frac{c}{\beta }-A \frac{\beta m-c}{2(w+c)}\right) -\frac{w(\beta m-c)(w+c)}{2\beta (w+\beta m)^2}. \end{aligned}$$

Now, it suffices to show that when m is sufficiently large relative to the ratio $\frac{c}{\beta }$, the inequality

$$\begin{aligned} \frac{c}{\beta }-A \frac{\beta m-c}{2(w+c)}<0 \end{aligned}$$

(33)

holds so that

$$\begin{aligned} C^{'}(r_1)<0. \end{aligned}$$

In order to prove (33), it suffices to find some constant $B> 0$ such that $A \ge B$ for all sufficiently large m.

Here, we will show that

$$\begin{aligned} 0 <\sqrt{\frac{c(c+\beta )\left( w+c\right) }{\beta ^{2}\left( w+c+\beta \right) }}-\frac{c}{\beta }\le \sqrt{\frac{cm\left( w+c\right) }{\beta \left( w+\beta m\right) }}-\frac{c}{\beta } \end{aligned}$$

(34)

for all $m\ge \frac{c}{\beta }+1=\frac{c+\beta }{\beta }$.

For the first inequality in (34), notice that the following are all equivalent for any $\beta , w, c\ge 0$:

$$\begin{aligned} \begin{aligned} \beta w&> 0\\ \left( c+\beta \right) \left( w+c\right) = cw+c^2+\beta c+ \beta w&> cw+c^2+\beta c = c\left( w+c+\beta \right) \\ \frac{\left( c+\beta \right) \left( w+c\right) }{\left( w+c+\beta \right) }&\ge c\\ \frac{c\left( c+\beta \right) \left( w+c\right) }{\beta ^2\left( w+c+\beta \right) }&\ge \frac{c^2}{\beta ^2}\\ \sqrt{\frac{c\left( c+\beta \right) \left( w+c\right) }{\beta ^2\left( w+c+\beta \right) }}-\frac{c}{\beta }&> 0. \end{aligned} \end{aligned}$$

For the second inequality in (34), assume $\frac{c}{\beta }+1\le m$. Then,

$$\begin{aligned} \begin{aligned} cw+\beta w&\le \beta wm\\ \left( c+\beta \right) \left( w+\beta m\right) = cw+\beta w+c\beta m+\beta ^2m&\le \beta wm+c\beta m+\beta ^2m = \beta m\left( w+c+\beta \right) \\ \frac{c+\beta }{\beta \left( w+c+\beta \right) }&\le \frac{m}{w+\beta m}\\ \frac{c\left( c+\beta \right) \left( w+c\right) }{\beta ^2 \left( w+c+\beta \right) }&\le \frac{cm\left( w+c\right) }{\beta \left( w+\beta m\right) }\, , \end{aligned} \end{aligned}$$

and the result follows by taking square roots of both sides and subtracting $\frac{c}{\beta }$.

1.3 Proof of Theorem 3.6

Assume $\gamma =\frac{c}{\beta }$. Then,

$$\begin{aligned} \begin{aligned} I^{*}(r)&=\frac{1}{2}\left( \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\gamma +\sqrt{\left( \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\gamma \right) ^{2}+\frac{4\gamma wm}{w+r}}\right) \\&=\frac{1}{2}\left[ \frac{-\frac{4wm}{w+r}}{\left[ \frac{wm}{\gamma (w+r)}-\frac{rw}{c(w+r)}-1-\sqrt{\left( \frac{wm}{\gamma (w+r)}-\frac{rw}{c(w+r)}-1\right) ^{2}+\frac{4wm}{\gamma (w+r)}}\right] }\right] \\&=\frac{2wm}{\frac{-wm}{\gamma }+\frac{rw}{c}+w+r+\sqrt{\left( \frac{-wm}{\gamma }+\frac{rw}{c}+w+r\right) ^{2}+\frac{4wm(w+r)}{\gamma }}}. \end{aligned} \end{aligned}$$

(35)

We show that for $\gamma > m$ and any $r_2 > r_1 \ge 0$:

$$\begin{aligned} C(r_2)-C(r_1) =r_2I^{*}(r_2)-r_1I^{*}(r_1) \ge 0. \end{aligned}$$

(36)

By substituting the last line of (35) into the right-hand side of (36), canceling the positive factor 2mw and cross multiplying with the positive denominators, we see that it suffices to show

$$\begin{aligned} \begin{aligned}&r_2\left[ \frac{-wm}{\gamma }+\frac{r_1w}{c}+w+r_1+\sqrt{\left( \frac{-wm}{\gamma }+\frac{r_1w}{c}+w+r_1\right) ^{2}+\frac{4wm(w+r_1)}{\gamma }}\right] \\&\quad \ge r_1\left[ \frac{-wm}{\gamma }+\frac{r_2w}{c}+w+r_2+\sqrt{\left( \frac{-wm}{\gamma }+\frac{r_2w}{c}+w+r_2\right) ^{2}+\frac{4wm(w+r_2)}{\gamma }}\right] \end{aligned} \end{aligned}$$

Canceling some common terms gives

$$\begin{aligned} \begin{aligned}&r_2\left[ \frac{-wm}{\gamma }+w+\sqrt{\left( \frac{-wm}{\gamma }+\frac{r_1w}{c}+w+r_1\right) ^{2}+\frac{4wm(w+r_1)}{\gamma }}\right] \\&\quad \ge r_1\left[ \frac{-wm}{\gamma }+w+\sqrt{\left( \frac{-wm}{\gamma }+\frac{r_2w}{c}+w+r_2\right) ^{2}+\frac{4wm(w+r_2)}{\gamma }}\right] . \end{aligned} \end{aligned}$$

After rearranging terms, we obtain

$$\begin{aligned} \begin{aligned}&\left( r_2-r_1\right) \left[ w\left( 1-\frac{m}{\gamma }\right) +\sqrt{\left( \frac{-wmr_2}{\gamma }+\frac{r_1r_2w}{c}+wr_2+r_1r_2\right) ^{2}+\frac{4wmr^2_2(w+r_1)}{\gamma }}\right. \\&\quad \left. -\sqrt{\left( \frac{-wmr_1}{\gamma }+\frac{r_1r_2w}{c}+wr_1+r_1r_2\right) ^{2}+\frac{4wmr^2_1(w+r_2)}{\gamma }}\right] \ge 0. \end{aligned} \end{aligned}$$

If $\gamma > m$, the first term in the square brackets is nonnegative. Next, we will show that

$$\begin{aligned} \begin{aligned}&\sqrt{\left( \frac{-wmr_2}{\gamma }+\frac{r_1r_2w}{c}+wr_2+r_1r_2\right) ^{2}+\frac{4wmr^2_2(w+r_1)}{\gamma }}\\&\quad \ge \sqrt{\left( \frac{-wmr_1}{\gamma }+\frac{r_1r_2w}{c}+wr_1+r_1r_2\right) ^{2}+\frac{4wmr^2_1(w+r_2)}{\gamma }}. \end{aligned} \end{aligned}$$

(37)

Consider the following expression, which is always positive for $r_2>r_1$:

$$\begin{aligned} \begin{aligned}&\left[ (r_2+r_1)\left( w-\frac{wm}{\gamma }\right) +\frac{2r_1r_2w}{c}+2r_1r_2\right] \left[ \left( r_2-r_1\right) \left( w-\frac{wm}{\gamma }\right) \right] \\&\quad +\frac{4w^2m}{\gamma }\left( r^2_2-r^2_1\right) +\frac{4wmr_1r_2}{\gamma }\left( r_2-r_1\right) \ge 0. \end{aligned} \end{aligned}$$

(38)

By expanding the products inside each of the main factors in the first line of (38) and rearranging terms, we obtain:

$$\begin{aligned} \begin{aligned}&\left( \frac{-wmr_2}{\gamma }+\frac{r_1r_2w}{c}+wr_2+r_1r_2-\frac{wmr_1}{\gamma }+\frac{r_1r_2w}{c}+wr_1+r_1r_2\right) \\&\quad \left( \frac{-wmr_2}{\gamma }+\frac{r_1r_2w}{c}+wr_2+r_1r_2+\frac{wmr_1}{\gamma }-\frac{r_1r_2w}{c}-wr_1-r_1r_2\right) \\&\quad +\frac{4w^2mr^2_2}{\gamma }+\frac{4wmr_1r^2_2}{\gamma }-\frac{4w^2mr^2_1}{\gamma }-\frac{4wmr^2_1r_2}{\gamma }\ge 0. \end{aligned} \end{aligned}$$

(39)

By rewriting the first two lines of (39) as a difference of two squares and factoring out some terms in the third, we obtain

$$\begin{aligned} \begin{aligned}&\left( \frac{-wmr_2}{\gamma }+\frac{r_1r_2w}{c}+wr_2+r_1r_2\right) ^2-\left( \frac{-wmr_1}{\gamma }+\frac{r_1r_2w}{c}+wr_1+r_1r_2\right) \\&\quad +\frac{4wmr^2_2}{\gamma }\left( w+r_1\right) ^2-\frac{4wmr^2_1}{\gamma }\left( w+r_2\right) \ge 0. \end{aligned} \end{aligned}$$

(40)

Now, (37) follows from (40) by moving the negative terms to the right and taking square roots.

1.4 The Cost Function Increases When $\beta = 0$

When we set $\beta = 0$ in the basic model, we obtain a system of linear DEs

$$\begin{aligned} \begin{aligned} \frac{\hbox {d}I}{\hbox {d}t}&=c(m-I-R)-rI\\ \frac{\hbox {d}R}{\hbox {d}t}&=rI-wR.\\ \end{aligned} \end{aligned}$$

At equilibrium,

$$\begin{aligned} \begin{aligned} c(m-I^{*}-R^{*})-rI^{*}&=0\\ rI^{*}-wR^{*}&=0.\\ \end{aligned} \end{aligned}$$

(41)

Substitute $R^{*}=\frac{rI^{*}}{w}$ into the first line of (41) and solve for $I^{*}$ to get

$$\begin{aligned} \begin{aligned} I^{*}&=\frac{cwm}{wc+r(w+c)}\\ C(r)&=\frac{rcwm}{wc+r(w+c)}. \end{aligned} \end{aligned}$$

Thus, we can estimate the derivative of the cost function as follows:

$$\begin{aligned} \begin{aligned} \frac{\hbox {d}C}{\hbox {d}r}&=\frac{\left[ wc+r(w+c)\right] cwm-rcwm\left[ w+c\right] }{\left( wc+r(w+c)\right) ^2}\\&=\frac{w^2c^2m+rcw^2m+rc^2wm-rcw^2m-rc^2wm}{\left( wc+r(w+c)\right) ^2}\\&=\frac{w^2c^2m}{\left( wc+r(w+c)\right) ^2}>0.\\ \end{aligned} \end{aligned}$$

1.5 Proof of Proposition 3.8(a)

We consider any function of the form

$$\begin{aligned} f(h) = c - h + \sqrt{(c-h)^2 + dh}, \end{aligned}$$

(42)

where $c, d > 0$ are constants such that

$$\begin{aligned} d > \max \{4c, 0\}. \end{aligned}$$

(43)

The first derivative of f wrt h behaves as follows:

$$\begin{aligned} \begin{aligned} f'(h)&= -1 + \frac{-(c-h) + d/2}{\sqrt{(c-h)^2 + dh}},\\ f'(0)&= -1 + \frac{-c + d/2}{\sqrt{c^2 + 0}} = -1 -\frac{c}{|c|} + \frac{d}{2|c|} > 0,\\ \lim _{h\rightarrow \infty } f'(h)&= \lim _{h\rightarrow \infty } - 1 + \frac{-(c-h) + d/2}{\sqrt{(c-h)^2 + dh}}\\&= \lim _{h\rightarrow \infty } -1 +\frac{h-c}{\sqrt{(c-h)^2 + dh}} +\frac{ d}{2\sqrt{(c-h)^2 + dh}} = 0. \end{aligned} \end{aligned}$$

(44)

Let us calculate the second derivative of f wrt h.

$$\begin{aligned} \begin{aligned} f''(h)&= \frac{\sqrt{(c-h)^2 + dh} - \frac{((c-h) - d/2)^2}{\sqrt{(c-h)^2 + dh}}}{(c-h)^2 + dh} = \frac{(c-h)^2 + dh - ((c-h) - d/2)^2}{((c-h)^2 + dh)\sqrt{(c-h)^2 + dh}}\\&=\frac{(c-h)^2 + dh - (c-h)^2 - d^2/4 - dh + dc}{((c-h)^2 + dh)\sqrt{(c-h)^2 + dh}}\\&= \frac{ - d^2/4 + dc}{((c-h)^2 + dh)\sqrt{(c-h)^2 + dh}} < 0. \end{aligned} \end{aligned}$$

The last inequality follows from (43).

Thus, $f'(0) > 0$ and $f'$ decreases throughout $[0, \infty )$. Now, it follows from the last part of (44) that $f'(h) > 0$ throughout $[0, \infty )$. Proposition 3.8(a) follows by observing that the function $\frac{I^*(r_i, c_i)}{2\beta _{ii}}$ is of the form (42) with

$$\begin{aligned} \begin{aligned} h&= c_i\\ c&= \frac{\beta _{ii}w_im_i-w_ir_i}{w_i+r_i}\\ d&= \frac{4\beta _{ii}w_im_i}{w_i+r_i}, \end{aligned} \end{aligned}$$

so that (43) holds.

1.6 Proof of Proposition 3.8(c)

$$\begin{aligned}&\lim _{r\rightarrow \infty }C(r)= \lim _{r\rightarrow \infty } rI^*(r) \\&\quad = \frac{1}{2}\lim _{r \rightarrow \infty }r\left( \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }+\sqrt{\left( \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }\right) ^{2}+\frac{4wcm}{\beta (w+r)}}\right) \\&\quad =\frac{1}{2}\lim _{r \rightarrow \infty }r\left[ \frac{\left[ \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }\right] ^2-\left[ \left( \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }\right) ^{2}+\frac{4wcm}{\beta (w+r)}\right] }{\left[ \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }\right] -\sqrt{\left( \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }\right) ^{2}+\frac{4wcm}{\beta (w+r)}}}\right] \\&\quad =\frac{1}{2}\lim _{r \rightarrow \infty }\left[ \frac{-\frac{4rwcm}{\beta (w+r)}}{\left[ \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }\right] -\sqrt{\left( \frac{wm}{w+r}-\frac{rw}{\beta (w+r)}-\frac{c}{\beta }\right) ^{2}+\frac{4wcm}{\beta (w+r)}}}\right] \\&\quad =\frac{1}{2}\left[ \frac{-\frac{4wmc}{\beta }}{-\frac{w}{\beta }-\frac{c}{\beta }-\sqrt{\left( \frac{w}{\beta }+\frac{c}{\beta }\right) ^{2}}}\right] =\frac{1}{2}\left[ \frac{-\frac{4wmc}{\beta }}{-2\frac{(w+c)}{\beta }}\right] =\frac{wcm}{w+c}. \end{aligned}$$