Trade, Transboundary Pollution, and Foreign Lobbying

Abstract

In this paper, we explore the use of trade policy in addressing transboundary stock pollution problems such as acid rain and water pollution. We show that a tariff determined by the current level of accumulated pollution can induce the time path of emissions optimal for the downstream (polluted) country. But if the upstream (polluting) country can lobby the downstream government to impose lower tariffs, distortions brought by corruption and foreign lobbying lead to a rise in the upstream country’s social welfare, and to a decrease in social welfare in the downstream country. Thus, the usefulness of trade policy as a tool for encouraging cooperation and internalizing transboundary externalities depends critically on the degree of governments’ susceptibility to foreign political influence.

Appendices

Appendix 1: Details of Linear Quadratic Model

In this appendix we provide calculations leading to various equations in the text. We start with the description of payoffs. With linear demand the amount purchased when the price in D is p and the tariff is ${\uptau }$ equals $Q_i^d = a_i - bP$, where $P = p$ in D and $P = p - {\uptau }$ in U. At each point in time, market clearing requires that combined quantity demanded, $Q_u + Q_d$, equals quantity supplied, $Q = c(p-{\uptau })$. The equilibrium quantity is then $Q = Q_u + Q_d = a_u + a_d - 2bp +b{\uptau }$. It is easy to see that the resultant equilibrium price is a linear function of the tariff:

$$\begin{aligned} p = \hat{p} + (1-A) {\uptau }, \end{aligned}$$

(21)

where $\hat{p} = \frac{a_d + a_u}{2b+c}$ and $1-A = \frac{b+c}{2b+c}$.

Consumer surplus in country $i, \textit{CS}_i$, can be written as

$$\begin{aligned} \textit{CS}_i = \frac{(Q_i)^2}{2b} = \frac{(a_i - bP)^2}{2b}. \end{aligned}$$

Because the supply relation in U is $Q = cP$, marginal costs in U are Q / c and so total costs are $\frac{Q^2}{2c} = \frac{c}{2}(p-{\uptau })^2$. Thus, profits earned by sellers in U are $\frac{c}{2}(p-{\uptau })^2$. We note that this characterization implicitly embodies tariff payments, via the impact of ${\uptau }$ upon the net price received by sellers in U. Accordingly,$W_u$—the sum of consumer and profits in U, less tariff payments—equals

$$\begin{aligned} W_u = \frac{\left( a_u - b(p-{\uptau })\right) ^2}{2b} + \left( \frac{c}{2} \right) (p-{\uptau })^2. \end{aligned}$$

Straightforward algebraic manipulation of this expression yields Eq. (3) in the main body of the text. Tariff payments collected by D are ${\uptau }Q_d = {\uptau }(a_d - bp)$, so that the sum of consumer surplus and tariff payments are

$$\begin{aligned} \frac{\left( a_d - bp\right) ^2}{2b} + {\uptau }(a_d - bp). \end{aligned}$$

The optimal static tariff ${\uptau }^0$ maximizes $W_d$; making use of Eq. (21) we have

$$\begin{aligned} {\uptau }^0 = \frac{AY_0}{b(1-A^2)}, \end{aligned}$$

(22)

where $Y_0 = a_d - b \hat{ p}$, the exports to D in the absence of a tariff. The tariff ${\uptau }^0$ induces a price $p^0$, and an associated downstream demand $Q_d^0$; making use of Eq. (21), we have

$$\begin{aligned} p = p^0 + (1-A) ({\uptau }- {\uptau }^0). \end{aligned}$$

(23)

Let $V_d^0$ denote the value of gross payoffs to D based on the tariff ${\uptau }^0$, which we can write in terms of the quantity demanded $Q_d^0$ (which results from the tariff ${\uptau }^0$):

$$\begin{aligned} V_d^0 = \frac{(Q_d^0)^2}{2b} + {\uptau }Q_d^0. \end{aligned}$$

Then we can express gross payoffs at any tariff ${\uptau }$ as:

$$\begin{aligned} V_d&= \frac{(Q_d)^2}{2b} + {\uptau }Q_d \\&= V_d^0 + \frac{Q_d^2 - (Q_d^0)^2}{2b} + ({\uptau }- {\uptau }^0) Q_d +{\uptau }^0(Q_d - Q_d^0) \\&= V_d^0 + \frac{(Q_d - Q_d^0)^2}{2b} + \frac{Q_d^0(Q_d - Q_d^0)}{b} + {\uptau }^0(Q_d - Q_d^0) + ({\uptau }- {\uptau }^0) Q_d \end{aligned}$$

It is straightforward to show that the static optimal tariff satisfies $Q_d^0 + b{\uptau }^0 =\frac{Q_d^0}{1-A}$; making use of this fact, and noting also that $Q_d - Q_d^0 = -b(1-A)({\uptau }- {\uptau }^0)$, the expression for $V_d$ can be simplified to

$$\begin{aligned} V_d&= V_d^0 + \frac{b(1-A)^2}{b}\left( {\uptau }- {\uptau }^0\right) ^2 - b(1-A)({\uptau }- {\uptau }^0)^2 \\&= V_d^0 - \frac{b}{2}(1-A^2)({\uptau }- {\uptau }^0)^2. \end{aligned}$$

Subtracting damages $sZ^2/2$ then yields Downstream flow payoffs $W_d$, as given by Eq. (2) in the main body of the text.

Appendix 2: Deriving the Coefficients in the Markov Strategies

We obtain the Markov-perfect equilibrium strategies by solving each country’s optimization problem. To solve U’s dynamic optimization problem, we time-differentiate the first-order condition (8), assuming $I \ge 0$, and D plays the linear strategy given in Eq. (14); we then use Eq. (8) to substitute for ${{\upxi }}_P$ and Eq. (10) to substitute for ${\dot{{\upxi }}}_P$. This approach yields

$$\begin{aligned} \dot{I}=\left( r+{\updelta }-{\uplambda }{{\upalpha }}_3\right) I-{{\upalpha }}_3(Y_0-b(1-A){\uptau }). \end{aligned}$$

Then, using Eqs. (42) and (43), we express $\dot{I}$ in terms of Z and P:

$$\begin{aligned} \dot{I}&= \left( r+{\updelta }-{\uplambda }{\upalpha }_3\right) ({\upgamma }_1 + {\upgamma }_2Z + {\upgamma }_3P) - {\upalpha }_3[Y_0-b(1-A)({\upalpha }_1 + {\upalpha }_2 Z + {\upalpha }_3P)] \nonumber \\&=\left( r+{\updelta }-{\uplambda }{\upalpha }_3\right) {\upgamma }_1 - {\upalpha }_3[Y_0-b(1-A){\upalpha }_1] + ({\upgamma }_2 + b(1-A){\upalpha }_2 {\upalpha }_3)Z \nonumber \\&\quad + ({\upgamma }_3 + b(1-A){\upalpha }_3^2) P. \end{aligned}$$

(24)

The expression for $\dot{I}$ also satisfies a second equation, which can be obtained by time-differentiating Eq. (15) and then using Eqs. (17)–(18) to substitute for $\dot{Z}$ and $\dot{P}$:

$$\begin{aligned} \dot{I} = {\upgamma }_2 (E^0 - {\uplambda }{\upalpha }_1) + {\upgamma }_1 {\upgamma }_3 + {\upgamma }_2 \bigl ( {\upgamma }_3 - ({\uplambda }{\upalpha }_2 + k) \bigr )Z + \bigl ( {\uplambda }{\upalpha }_3 {\upgamma }_2 + {\upgamma }_3 ({\upgamma }_3 - {\updelta }) \bigr ) P. \end{aligned}$$

(25)

Comparing Eqs. (24) and (25), one see there are three restrictions corresponding to equating the intercepts, the slopes on Z and the slopes on P:

$$\begin{aligned} {\upgamma }_1&= \frac{(E^0 - {\uplambda }{\upalpha }_1){\upgamma }_2}{r+{\updelta }- {\upgamma }_3 -{\uplambda }{\upalpha }_3} + \frac{ {\upalpha }_3[Y_0-b(1-A){\upalpha }_1]}{r+{\updelta }- {\upgamma }_3 -{\uplambda }{\upalpha }_3} ; \end{aligned}$$

(26)

$$\begin{aligned} {\upgamma }_2&= -\frac{(b(1-A){\upalpha }_2 {\upalpha }_3}{1 + {\uplambda }{\upalpha }_2 + k -{\upgamma }_3}; \end{aligned}$$

(27)

$$\begin{aligned} {\upgamma }_3^3&- (2 + {\uplambda }{\upalpha }_2 + k + {\updelta }) {\upgamma }_3^2 + \bigl ( ({\updelta }+ 1) (1+ {\uplambda }{\upalpha }_2 + k) - b(1-A) {\upalpha }_3^2 \bigr ) {\upgamma }_3 \nonumber \\&\qquad + b(1-A){\upalpha }_3^2 (1 + 2{\uplambda }{\upalpha }_2 + k) = 0. \end{aligned}$$

(28)

For a given vector of parameters $({\upalpha }_1, {\upalpha }_2, {\upalpha }_3)$ describing D’s strategy, one can solve for the vector of parameters describing U’s strategy.²⁷

To solve D’s dynamic optimization problem, we time-differentiate the first-order condition (11), assuming U plays the linear strategy given in Eq. (15); we then use Eq. (12) to replace $\dot{{\upeta }}_Z$ and Eq. (11) to replace ${\upeta }_Z$:

$$\begin{aligned} \dot{{\uptau }}&= -\frac{{\uplambda }}{b\left( 1-A^2\right) }\dot{{\upeta }}_Z \nonumber \\&= -\frac{{\uplambda }}{b\left( 1-A^2\right) } \bigl \{(r+k) {\upeta }_Z + sZ - [{\upbeta }+ {\upeta }_P] {\upgamma }_2 \bigr \} \nonumber \\&= -(r+k){\uptau }_0 + \frac{{\uplambda }{\upbeta }{\upgamma }_2}{b\left( 1-A^2\right) } + (r+k) {\uptau }- \frac{{\uplambda }s}{b\left( 1-A^2\right) }Z + \frac{{\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } {\upeta }_P. \end{aligned}$$

(29)

The next step is to eliminate ${\upeta }_P$ from this expression. To that end, we first time-differentiate Eq. (29) and use Eq. (13) to substitute for $\dot{{\upeta }}_P$:

$$\begin{aligned} \ddot{{\uptau }}&= (r+k) \dot{{\uptau }}-\frac{{\uplambda }s}{b\left( 1-A^2\right) } \dot{Z} + \frac{{\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } \dot{{\upeta }}_P \\&= (r+k) \dot{{\uptau }}-\frac{{\uplambda }s}{b\left( 1-A^2\right) } \dot{Z} + \frac{{\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } \bigl ( (r+{\updelta }){\upeta }_P - f + P - [{\upbeta }+{\upeta }_P] {\upgamma }_3 \bigr ). \end{aligned}$$

Then, we use Eq. (29) to substitute for ${\upeta }_P$:

$$\begin{aligned} \ddot{{\uptau }}&= (r+k) \dot{{\uptau }}-\frac{{\uplambda }s}{b\left( 1-A^2\right) } \dot{Z} - \frac{{\uplambda }(f + {\upbeta }{\upgamma }_3) {\upgamma }_2}{b\left( 1-A^2\right) } + \frac{{\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } P \\&\quad + (r + {\updelta }- {\upgamma }_3) \bigl [ \dot{{\uptau }} + (r+k) ({\uptau }_0 - {\uptau }) - \frac{{\uplambda }{\upbeta }{\upgamma }_2}{b\left( 1-A^2\right) } + \frac{{\uplambda }s}{b\left( 1-A^2\right) } Z \bigr ] . \end{aligned}$$

Next, we use the linear Markov strategies, in Eqs. (14)–(15) to rewrite I and ${\uptau }$ in terms of Z and P, and $\dot{I}$ and $\dot{{\uptau }}$ in terms of $\dot{Z}$ and $\dot{P}$; then recalling the equations of motion for Z and P, we obtain an expression for $\ddot{{\uptau }}$ in terms of Z and P:

$$\begin{aligned} \ddot{{\uptau }}&= \left[ {\upalpha }_2 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }s}{b\left( 1-A^2\right) } \right] (E^0 - {\uplambda }{\upalpha }_1) + {\upalpha }_3 {\upgamma }_1 (2r + {\updelta }+ k - {\upgamma }_3) \\&\quad - \frac{{\uplambda }{\upgamma }_2}{b (1-A^2 )} \bigl (f + {\upbeta }[1+{\upgamma }_3] \bigr )+ (r+k)(r+{\updelta }- {\upgamma }_3)({\uptau }_0 - {\upalpha }_1) \\&\quad + \left\{ \frac{{\uplambda }s(1 + k + {\uplambda }{\upalpha }_2)}{b\left( 1-A^2\right) } - {\upalpha }_2 \bigr [2r + {\updelta }+ k - {\upgamma }_3 + (r+k)(r+{\updelta }- {\upgamma }_3) \bigr ] \right. \\&\quad +\left. {\upalpha }_3 {\upgamma }_2 (2r + {\updelta }+ k - {\upgamma }_3) \right\} Z \\&\quad + \left\{ \frac{ {\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } - {\upalpha }_3 \left( {\uplambda }\left[ {\upalpha }_2 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }s}{b\left( 1-A^2\right) } \right] \right. \right. \\&\quad \left. \left. + ({\upgamma }_3 - {\updelta })(2r + {\updelta }+ k - {\upgamma }_3) - (r+k)(r+{\updelta }- {\upgamma }_3) \right) \right\} P. \end{aligned}$$

To complete the solution, we obtain a second expression for $\ddot{{\uptau }}$ in terms of Z and I. Starting from Eq. (14) and time-differentiating, substituting for $\dot{Z}$ and $\dot{P}$, and then time-differentiating again and substituting for $\dot{Z}$ and $\dot{P}$ again yields:

$$\begin{aligned} \ddot{{\uptau }}&= {\upalpha }_3 \bigl [ {\upgamma }_2(E^0 - {\uplambda }{\upalpha }_1 - k{\upgamma }_1) - {\upgamma }_1( {\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ] -{\upalpha }_2( {\uplambda }{\upalpha }_2 + k)(E^0 - {\uplambda }{\upalpha }_1 - k{\upgamma }_1) \\&\quad - \bigl ( ({\uplambda }{\upalpha }_2 + k)({\upalpha }_3 {\upgamma }_2 - {\upalpha }_2({\uplambda }{\upalpha }_2 + k)) + {\upalpha }_2 {\upalpha }_3 ({\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ) Z \\&\quad - {\upalpha }_3 \bigl ({\uplambda }[ {\upalpha }_3 {\upgamma }_2 - {\upalpha }_2({\uplambda }{\upalpha }_2 + k)] + ({\upalpha }_3 - {\updelta })( {\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ) P. \end{aligned}$$

The two expressions for $\ddot{{\uptau }}$ must both apply for all values of Z and P; accordingly, one obtains three restrictions corresponding to equating the intercepts, the slopes on Z and the slopes on P):

$$\begin{aligned}&\left[ {\upalpha }_2 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }s}{b\left( 1-A^2\right) } \right] (E^0 - {\uplambda }{\upalpha }_1) + (r+k)(r+{\updelta }- {\upgamma }_3)({\uptau }_0 - {\upalpha }_1) \\&\qquad + {\upalpha }_3 {\upgamma }_1 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }{\upgamma }_2}{b (1-A^2 )} \bigl (f + {\upbeta }[1+{\upgamma }_3] \bigr ) \\&\quad ={\upalpha }_3 \bigl [ {\upgamma }_2(E^0 - {\uplambda }{\upalpha }_1 - k{\upgamma }_1) - {\upgamma }_1( {\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ] -{\upalpha }_2( {\uplambda }{\upalpha }_2 + k)(E^0 - {\uplambda }{\upalpha }_1 - k{\upgamma }_1); \\ \nonumber \\&\quad \frac{{\uplambda }s(1 + k + {\uplambda }{\upalpha }_2)}{b\left( 1-A^2\right) } - {\upalpha }_2 \bigr [2r + {\updelta }+ k - {\upgamma }_3 + (r+k)(r+{\updelta }- {\upgamma }_3) \bigl ] \,+\, {\upalpha }_3 {\upgamma }_2 (2r + {\updelta }+ k - {\upgamma }_3)\\&\quad = -(({\uplambda }{\upalpha }_2+k)({\upalpha }_3{\upgamma }_2-{\upalpha }_2({\uplambda }{\upalpha }_2+k))+{\upalpha }_2{\upalpha }_3({\uplambda }{\upalpha }_2+{\updelta }-{\upgamma }_3))\\&\quad {\upalpha }_3 \left( {\uplambda }\left[ {\upalpha }_2 (2r + {\updelta }+ k - {\upgamma }_3) - \frac{{\uplambda }s}{b\left( 1-A^2\right) } \right] \right. \nonumber \\&\quad +\left. ({\upgamma }_3 - {\updelta })(2r + {\updelta }+ k - {\upgamma }_3) + (r+k)(r+{\updelta }- {\upgamma }_3) \right) \nonumber \\&\quad - \frac{ {\uplambda }{\upgamma }_2}{b\left( 1-A^2\right) } = {\upalpha }_3 \bigl ({\uplambda }\left[ {\upalpha }_3 {\upgamma }_2 - {\upalpha }_2({\uplambda }{\upalpha }_2 + k)\right] + ({\upalpha }_3 - {\updelta })( {\uplambda }{\upalpha }_2 + {\updelta }- {\upgamma }_3) \bigr ). \end{aligned}$$

These three conditions can be manipulated to isolate expressions for the intercept (${\upalpha }_1$), slope on Z (${\upalpha }_2$) and the slope on P (${\upalpha }_3$) in D’s Markov strategy:

$$\begin{aligned} {\upalpha }_1&= \frac{ {\upalpha }_2 \bigl [2(r+k) + {\updelta }- {\upgamma }_3 + {\uplambda }{\upalpha }_2 \bigr ] E^0 + (r+k)(r+ {\updelta }- {\upgamma }_3) {\uptau }_0}{ {\uplambda }\bigl [2(r+k) + {\updelta }- {\upgamma }_3 + {\uplambda }{\upalpha }_2 \bigr ] {\upalpha }_2 - \frac{{\uplambda }s}{b\left( 1-A^2\right) } - {\upalpha }_3{\upgamma }_2 + (r+k)(r+ {\updelta }- {\upgamma }_3) } \nonumber \\&\quad - \frac{{\uplambda }\bigl (f {+} {\upbeta }(1{+}{\upgamma }_3) \bigr ) {\upgamma }_2}{b(1\!-A^2) \bigl ( {\uplambda }\bigl [2(r\!+k) {+} {\updelta }{-} {\upgamma }_3 {+} {\uplambda }{\upalpha }_2 \bigr ] {\upalpha }_2 \!- \frac{{\uplambda }s}{b\left( 1-A^2\right) } - {\upalpha }_3{\upgamma }_2 + (r+k)(r+ {\updelta }- {\upgamma }_3) \bigr )} \nonumber \\&\quad + \frac{{\upalpha }_3 {\upgamma }_1 [2(r+{\updelta }- {\upgamma }_3) + k(1+{\upgamma }_2) ] - {\upalpha }_2 {\upgamma }_1 k( {\uplambda }{\upalpha }_2 + k)}{ {\uplambda }\bigl [2(r+k) + {\updelta }- {\upgamma }_3 + {\uplambda }{\upalpha }_2 \bigr ] {\upalpha }_2 - \frac{{\uplambda }s}{b\left( 1-A^2\right) } - {\upalpha }_3{\upgamma }_2 + (r+k)(r+ {\updelta }- {\upgamma }_3) } ; \end{aligned}$$

(30)

$$\begin{aligned} {\upalpha }_2^2&- \bigl ({\upalpha }_3 + {\upgamma }_3 - 2(r + k + {\updelta }) \bigr ) {\upalpha }_2 + \frac{{\uplambda }s - [b (1-A^2){\upalpha }_3^2 + 1] {\upgamma }_2}{{\uplambda }b {\upalpha }_3(1-A^2)} \nonumber \\&\qquad \qquad + \frac{({\upgamma }_3 - {\updelta })(r + {\upalpha }_3 - {\upgamma }_3) + r(r+k)}{{\uplambda }^2} = 0; \end{aligned}$$

(31)

$$\begin{aligned} {\upalpha }_3&= \frac{ {\uplambda }s(1 + k + {\uplambda }{\upalpha }_2) }{b(1-A^2)\bigl (( {\upalpha }_2 + {\upgamma }_2) ({\upgamma }_3 - ({\uplambda }{\upalpha }_2 - {\updelta }) - 2(r+k) {\upgamma }_2\bigr )} \nonumber \\&\quad - \left( \frac{(r+k)(r+{\updelta }- b(1-A^2){\upgamma }_3) + ({\uplambda }{\upalpha }_2 + k)^2 + b(1-A^2)(2r + {\updelta }+ k )}{ b(1-A^2)( {\upalpha }_2 + {\upgamma }_2) ({\upgamma }_3 - ({\uplambda }{\upalpha }_2 - {\updelta }) - 2(r+k) {\upgamma }_2} \right) {\upalpha }_2. \end{aligned}$$

(32)

As for was the case when solving for the parameters in U’s strategy, once the rival’s strategy is specified it is straightforward to solve for the parameters in D’s strategy.

The equilibrium Markov strategies for U and D are described by the six coefficients ${\upalpha }_1, {\upalpha }_2, {\upalpha }_3, {\upgamma }_1, {\upgamma }_2$ and ${\upgamma }_3$, which are determined by the six equations (26)–(28) and (30)–(32).

As the system governing the state variables, (17)–(18), is a pair of linear first-order differential equations, its solution is given by the sum of a particular solution, ($Z^{\textit{MP}}, P^{\textit{MP}}$)—which correspond to the steady values—and the general solution to the system of homogenous differential equations

$$\begin{aligned} \dot{Z}&= -({\uplambda }{\upalpha }_2 + k) Z - {\uplambda }{\upalpha }_3P; \end{aligned}$$

(33)

$$\begin{aligned} \dot{P}&= {\upgamma }_2Z + ({\upgamma }_3 - {\updelta })P. \end{aligned}$$

(34)

The solution to the system (33)–(34) is of the form (Z, P) = (${\upnu }e^{{\upsigma }t}, {\upmu }e^{{\upsigma }t}$). Inserting these functions into Eqs. (33) and (34) leads to the characteristic equation

$$\begin{aligned} Q({\upsigma }) \equiv ({\upsigma }+{\uplambda }{{\upalpha }}_2+k)\left( {\upsigma }+{\updelta }-{{\upgamma }}_3\right) + {\uplambda }{{\upalpha }}_3{{\upgamma }}_2=0, \end{aligned}$$

(35)

which the parameter ${\upsigma }$ must satisfy. Denote the roots of this equation by ${\upsigma }_1$ and ${\upsigma }_2$, with ${\upsigma }_1$ the smaller root. It is straightforward to see that ${\upsigma }_1<$ min {$-({\uplambda }{\upalpha }_2 + k), {\upgamma }_3 - {\updelta }$} and ${\upsigma }_2>$ max{$-({\uplambda }{\upalpha }_2 + k), {\upgamma }_3 - {\updelta }$}; accordingly, ${\upsigma }_1$ is negative while ${\upsigma }_2$ could be either negative or positive. If ${\upsigma }_2 < 0$, the solution is a stable node, while if ${\upsigma }_2 > 0$ the solution is saddlepoint stable.²⁸ The state variables are subject to transversality conditions that ensure they converge to a steady state; it follows that the solution is ($Z_g, P_g$) = (${\upnu }_1 e^{{\upsigma }_1t}, {\upmu }_1 e^{{\upsigma }_1t}$) if ${\upsigma }_2> 0 > {\upsigma }_1$, and ($Z_{g}, P_g$) = (${\upnu }_1 e^{{\upsigma }_1t} + {\upnu }_2 e^{{\upsigma }_2t}, {\upmu }_1 e^{{\upsigma }_1t} + {\upmu }_2 e^{{\upsigma }_2t}$) if $0> {\upsigma }_{2} > {\upsigma }_{1}$.

The particular solution is found by setting $\dot{Z}=0=\dot{P}$ in Eqs. (17) and (18), and then solving the resultant system of two equations. By construction, then, the particular solution corresponds to the pair of steady state values for the two state variables:

$$\begin{aligned} Z^{\textit{MP}}&= [({\updelta }- {\upgamma }_3)(E^0 - {\uplambda }{\upalpha }_1) - {\uplambda }{\upalpha }_3 {\upgamma }_1] /\left[ ({\uplambda }{\upalpha }_2 + k)({\updelta }- {\upgamma }_3) + {\uplambda }{\upalpha }_3 {\upgamma }_2 \right] ; \end{aligned}$$

(36)

$$\begin{aligned} P^{\textit{MP}}&= [{\upgamma }_2 (E^0 - {\uplambda }{\upalpha }_1) + {\upgamma }_1 ({\uplambda }{\upalpha }_2 + k )/\left[ ({\uplambda }{\upalpha }_2 + k)({\updelta }- {\upgamma }_3) + {\uplambda }{\upalpha }_3 {\upgamma }_2 \right] . \end{aligned}$$

(37)

Altogether, the complete solution is

$$\begin{aligned} Z(t)&= {\upnu }_1 e^{{\upsigma }_1t} + {\upnu }_2 e^{{\upsigma }_2t} + Z^{\textit{MP}}, \end{aligned}$$

(38)

$$\begin{aligned} P(t)&= {\upmu }_1 e^{{\upsigma }_1t} + {\upmu }_2 e^{{\upsigma }_2t} + P^{\textit{MP}}, \end{aligned}$$

(39)

where ${\upnu }_2 = 0 = {\upmu }_2$ if ${\upsigma }_2 > 0$. Upon time-differentiating the right-hand side of Eqs. (38)–(39), and combining with Eqs. (33)–(34), one may derive

$$\begin{aligned} h_1\equiv \frac{{\upnu }_1}{{\upmu }_1} = \frac{{\upgamma }_2}{{\upsigma }_1 + {\updelta }- {\upgamma }_3}; \end{aligned}$$

(40)

Because ${\upsigma }_1<$ min{$-({\uplambda }{\upalpha }_2 + k$), ${\upgamma }_3 - {\updelta }$}, it follow that the sign of $h_1 < 0$ is opposite that of the sign of ${\upgamma }_2$. If in addition ${\upsigma }_2 < 0$, we have

$$\begin{aligned} h_2 \equiv \frac{{\upnu }_2}{{\upmu }_2} = \frac{{\upgamma }_2}{{\upsigma }_2 + {\upgamma }_3 + {\updelta }}. \end{aligned}$$

(41)

As ${\upsigma }_2>$ max{$-({\uplambda }{\upalpha }_2 + k$), $-({\upgamma }_3 + {\updelta }$)}, the sign of $h_2 > 0$ is the same as the sign of ${\upgamma }_2$.

Using these proportionality coefficients, the formulae for the state variables may be rewritten as

$$\begin{aligned} Z(t)&= {\upnu }_1 e^{{\upsigma }_1t} + {\upnu }_2 e^{{\upsigma }_2t} + Z^{\textit{MP}}, \end{aligned}$$

(42)

$$\begin{aligned} P(t)&= h_1 {\upnu }_1 e^{{\upsigma }_1t} + h_2 {\upnu }_2 e^{{\upsigma }_2t} + P^{\textit{MP}}. \end{aligned}$$

(43)

These expressions must hold at all points in time, in particular at t = 0. Noting that $Z(0) = Z_0$ and $P(0) = 0,$ it follows that:

$$\begin{aligned} {\upnu }_1&= [h_2\left( Z_0-Z^{\textit{MP}}\right) +P^{\textit{MP}}]/(h_2 - h_1); \end{aligned}$$

(44)

$$\begin{aligned} {\upnu }_2&= -[h_1\left( Z_0-Z^{\textit{MP}}\right) +P^{\textit{MP}}]/(h_2 - h_1). \end{aligned}$$

(45)

Combining Eqs. (40)–(45) then completes the solution for (Z, P).