Dynamic Fishing with Endogenous Habitat Damage

Abstract

The nature of fishing activities is such that marine habitats can be deteriorated when employing destructive fishing gear. This makes even more complex the determination of sustainable fishing policies and has led some authors to propose dynamic models which take into account this habitat degradation. In this work, we analyze in detail one of these models, an extension of the single-species Gordon–Schaefer model to two state interrelated variables: stock of fish and habitat. The model assumes that stock and carrying capacity are positively linked, and that the fishing activity has a direct and negative impact on the carrying capacity. We extend and characterize Clark’s most rapid approach optimal solution to this case.

Appendices

A Proofs of Propositions 2 and 3

It may be convenient in calculations to manipulate the following functions:

$$\begin{aligned} \Theta _X(t) = X(t) \lambda _X(t) \qquad \Theta _K(t) = K(t) \lambda _K(t) ~, \end{aligned}$$

with which $\chi$ can be written as:

$$\begin{aligned} \chi (t) ~ = ~ \Theta _X(t) + \beta \Theta _K(t) - P( X(t) ). \end{aligned}$$

Using (6) and (7), the first-order derivatives are given by:

$$\begin{aligned} \frac{\text{ d }\Theta _X}{\text{ d }t}&= \dot{X} \lambda _X + X \dot{\lambda }_X \nonumber \\&= \delta \Theta _X + ( F(X,K) - X F_X(X,K) ) \lambda _X - E X P'(X) \end{aligned}$$

(35)

$$\begin{aligned} \frac{\text{ d }\Theta _K}{\text{ d }t}&= \dot{K} \lambda _K + K \dot{\lambda }_K \nonumber \\&= \delta \Theta _K + ( G(K) - K G'(K) ) \lambda _K - K F_K(X,K) \lambda _X . \end{aligned}$$

(36)

Using (36) and (35) for the derivative of $\chi$, one obtains:

$$\begin{aligned} \dot{\chi }(t)&= \dot{\Theta }_X + \beta \dot{\Theta }_K - \dot{X} P'(X) \nonumber \\&= \delta ( \Theta _X + \beta \Theta _K ) + ( F - X F_X )\lambda _X - E X P'(X) \nonumber \\&+ \beta ( G - K G' )\lambda _K - \beta K F_K \lambda _X \nonumber \\&- ( F - EX ) P'(X) \nonumber \\ \dot{\chi }&= \delta ( \chi + P(X) ) + \lambda _X \left( F - X F_X - \beta K F_K \right) + \beta \lambda _K \left( G - K G' \right) - F P'(X). \end{aligned}$$

(37)

This derivative does not depend, formally, on the control function E(t). Of course, the trajectories of X, K, $\lambda _X$, and $\lambda _K$ do depend on the control.

1.1 A.1 Proof of Proposition 2

If the trajectory is constrained to stay on the curve defined by $\chi = 0$, then we must have, for all t, $\chi (t) = \dot{\chi }(t) = 0$. With (37), this leads to the system of equations:

$$\begin{aligned} X \lambda _X + \beta K \lambda _K&= P \\ \lambda _X \left( F - X F_X - \beta K F_K \right) + \beta \lambda _K \left( G - K G' \right)&= F P' - \delta P, \end{aligned}$$

which is linear in the variables $\lambda _X$ and $\beta \lambda _K$. Solving this linear system, we obtain

$$\begin{aligned} \lambda _X&= \frac{1}{\Delta } \left( ( G - K G' ) P + K ( \delta P(X) - F P') \right) \end{aligned}$$

(38)

$$\begin{aligned} \beta \lambda _K&= - \frac{1}{\Delta } \left( X ( \delta P - F P' ) + ( F - X F_X - \beta K F_K ) P ) \right) \\ \Delta&= X ( G - K G' ) - K ( F - X F_X - \beta K F_K ) . \nonumber \end{aligned}$$

(39)

For statement (b), we start from (37) but introduce the new functions $F^{(1)}(X,K) = F(X,K) - X F_X(X,K) - \beta K F_K(X,K)$, $G^{(1)}(K) = G(K) - K G'(K)$ and $P^{(1)}(X,K) = F(X,K) P'(X)$. Accordingly:

$$\begin{aligned} \dot{\chi }&= \delta ( \chi + P ) + \lambda _X F^{(1)} + \beta \lambda _K G^{(1)} - P^{(1)} \\ \ddot{\chi }&= \delta ( \dot{\chi } + \dot{X} P' ) + \dot{\lambda }_X F^{(1)} + \lambda _X ( \dot{X} F^{(1)}_X + \dot{K} F^{(1)}_K ) + \beta \dot{\lambda }_K G^{(1)} + \beta \lambda _K \dot{K} G^{(1)\prime }\\&- ( \dot{X} P^{(1)}_X + \dot{K} P^{(1)}_K ) . \end{aligned}$$

Then, applying (2), (3), (6) and (7), separating the terms with E from the rest, we have:

$$\begin{aligned} \ddot{\chi }&= \delta \dot{\chi } + \bigg \{ \delta F P' + \lambda _X ( \delta - F_X ) F^{(1)} + \lambda _X ( F F^{(1)}_X + G F^{(1)}_K ) \\&+ \beta \lambda _K (\delta - G') G^{(1)} - \beta \lambda _X F_K G^{(1)} + \beta \lambda _K G G^{(1)\prime } - ( F P^{(1)}_X + G P^{(1)}_K ) \bigg \} \nonumber \\&- E \bigg [ \delta X P' + ( P' - \lambda _X ) F^{(1)} + \lambda _X( X F^{(1)}_X + \beta K F^{(1)}_K ) - \beta ^2\lambda _K G^{(1)} + \beta ^2 K \lambda _K G^{(1)\prime } \nonumber \\&- ( X P^{(1)}_X + \beta K P^{(1)}_K ) \bigg ]. \nonumber \end{aligned}$$

(40)

If the trajectory is constrained to stay on the singular curve, then $\ddot{\chi }(t) = 0$. We can also eliminate $\lambda _X$ and $\lambda _K$ using (38) and (39). The function $\varphi (K,X)$ is defined as the term within braces, and the function $\psi (K,X)$ is defined as the term within square brackets, after this substitution. Then, solving $\ddot{\chi } = 0$, taking into account $\dot{\chi } = 0$, implies $0 = \varphi (K,X) - E \psi (K,X)$. This proves Proposition 2.

1.2 A. 2 Proof of Proposition 3

When $\beta$ is set to 0 in (37), the expression of $\dot{\chi }$ does not depend on $\lambda _K$ anymore. Also, $\chi = 0$ is equivalent to (9). Then, eliminating $\lambda _X$ from (37) gives (10). This proves statement (a) of Proposition 3.

For the proof of statement (b), we first expose preliminary considerations. Assume we want the state to stay on a particular curve, given in general by the equation $f(X,K) = 0$. Differentiating with respect to t, we have the identities:

$$\begin{aligned} 0&= \frac{\text {d}}{\text {d}t} f( X(t), K(t) )\nonumber \\&= \dot{X} \frac{\partial f}{\partial X} + \dot{K} \frac{\partial f}{\partial K} = \left( F(X,K) - E X \right) f_X + \left( G(K) - \beta E K \right) f_K \end{aligned}$$

(41)

$$\begin{aligned} E \left( X f_X + \beta K f_K \right)&= F(X,K) f_X + G(K) f_K \nonumber \\ E&= \frac{F(X,K) f_X + G(K) f_K}{ X f_X + \beta K f_K} . \end{aligned}$$

(42)

The expression (42) gives the control as a feedback of the state. If one plugs (42) in the dynamics of the state, one gets the differential system:

$$\begin{aligned} \dot{X}&= f_K ~ \frac{\beta K F(X,K) - X G(K)}{X f_X + \beta K f_K} \\ \dot{K}&= f_X ~ \frac{X G(X) - \beta K F(X,K)}{X f_X + \beta K f_K} \end{aligned}$$

and it is checked that this dynamics satisfies (41).

When $\beta =0$, these expressions are simplified as:

$$\begin{aligned} E(X,K)&= \frac{F(X,K)}{X} + \frac{G(K)}{X} \; \frac{f_K}{f_X} \\ \dot{X}&= - G(K) \frac{f_K}{f_X} \qquad \dot{K} = G(K) , \nonumber \end{aligned}$$

(43)

which proves statement (b) of Proposition 3. In (43), the term F(X, K)/X is the extraction that would make the stock X constant. It is corrected with a term, proportional to G(K), taking into account the growth of the carrying capacity K.

B Logistic Growth Functions

In this section, we list the formulas obtained by specifying the general results of Sect. 2 to the logistic growth functions.

1.1 B.1 Dynamics and the Singular Control

Assume the control has the objective of keeping the function $\chi$ equal to 0. Then, we have the following identities, in which we use the notation introduced in Section A.1.

$$\begin{aligned} \begin{aligned} G^{(1)}&:= G(K) - K G'(K) = \frac{\rho K^2}{\overline{K}}\\ F^{(1)}&:= F(X,K) - X F_X(X,K) - \beta K F_K(X,K) = r (1-\beta ) \frac{X^2}{K}. \end{aligned} \end{aligned}$$

(44)

1.1.1 B.1.1 Adjoint Variables

When using the logistic form of the growth functions, we obtain for the expressions (6) and (7):

$$\begin{aligned} \dot{\lambda }_X&= \delta \lambda _X - {p} E - \lambda _X \left( r - \frac{2rX}{K} - E \right) \end{aligned}$$

(45)

$$\begin{aligned} \dot{\lambda }_K&= \delta \lambda _K - r \lambda _X \frac{X^2}{K^2} - \lambda _K \left( \rho - \frac{2\rho K}{\overline{K}} - \beta E \right) . \end{aligned}$$

(46)

1.1.2 B.1.2 The Function $\chi (t)$

The formulas for the first derivatives of $\chi (t)$ are obtained from the generic formulas (37) and (40). Using the identities (44), we obtain:

$$\begin{aligned} \chi&= X \lambda _X + \beta K \lambda _K - {pX + c} ~ = ~ \Theta _X + \beta \Theta _K - {pX + c} \end{aligned}$$

(47)

$$\begin{aligned} \dot{\chi }&= \delta \left( \chi + {pX -c} \right) + \Theta _X (1-\beta ) \frac{rX}{K} + \beta \Theta _K \frac{\rho K}{\overline{K}} - {p} r X ( 1 - \frac{X}{K} ) \end{aligned}$$

(48)

$$\begin{aligned} \ddot{\chi }&= \delta \left[ \dot{\chi } + {p} r X ( 1 - \frac{X}{K} ) \right] \nonumber \\&+ r (1-\beta ) \frac{X}{K} \Theta _X \left[ \delta + r - \rho ( 1 -\frac{K}{\overline{K}} ) \right] + \rho \beta \frac{K}{\overline{K}} \left[ \Theta _K \left( \delta + \rho \right) - r \Theta _X \frac{X}{K} \right] \nonumber \\&- r X {p} \left( r ( 1 - \frac{X}{K} ) ( 1 - \frac{2X}{K} ) + \rho \frac{X}{K} ( 1 -\frac{K}{\overline{K}} ) \right) \nonumber \\&- E \bigg [ r ( 1 - \beta )^2 \frac{X}{K} \Theta _X + \rho \beta ^2 \frac{K}{\overline{K}} \Theta _K + \delta X {p} + r X {p} \left( ( 3 - 2\beta ) \frac{X}{K} - 1 \right) \bigg ] . \end{aligned}$$

(49)

1.1.3 B.1.3 Staying on the Curve $\chi = 0$

If the trajectory follows the curve $\chi (t) \equiv 0$, then $\chi = \dot{\chi } = 0$. Using (38) and (39) (or directly solving for $\Theta _X$ and $\Theta _K$ in (47) and (48)), we obtain the following values:

$$\begin{aligned} \Theta _X&= \frac{1}{\Delta } \left( {(pX -c)} ( \delta + \rho \frac{K}{\overline{K}} ) - r X {p} (1 - \frac{X}{K} ) \right) \end{aligned}$$

(50)

$$\begin{aligned} \beta \Theta _K&= \frac{1}{\Delta } \left( - {(pX -c)} ( \delta + (1-\beta ) r \frac{X}{K} ) + r X {p} (1 - \frac{X}{K} )\right) \end{aligned}$$

(51)

$$\begin{aligned} \Delta&= \rho \frac{K}{\overline{K}} - (1-\beta ) r \frac{X}{K} . \end{aligned}$$

(52)

1.2 B. 2 Proof of Lemma 1

The proof consists in enumerating the solutions of the system of four Eqs. (15)–(18) with four unknowns. We exclude $K = 0$ as a solution of (16) since in that case the dynamics (12) are not well defined. Then, the solution of (16) is

$$\begin{aligned} \qquad \qquad \qquad K^{\infty }&= \overline{K}\left( 1 - \frac{\beta E}{\rho } \right) . \quad \quad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad ~~~~~~~~~~~~~~~~~~~~~(19) \end{aligned}$$

Next, $X = 0$ is clearly a solution of (15). Replacing in (17) and (18), we find:

$$\begin{aligned} {p} E^\infty&= \lambda _X \left( \delta - r + E^\infty \right) \\ 0&= \lambda _K \left( \delta - \rho + E^\infty \right) , \end{aligned}$$

leading to $\lambda _K = 0$ and the value of $\lambda _X$ in (20). Assuming now that $K \not = 0$ and $X \not =0$, we have successively:

$$\begin{aligned} X^{\infty }&= K^{\infty }\left( 1 - \frac{E^\infty }{r} \right) ~ = ~ \overline{K}\left( 1 - \frac{\beta E^\infty }{\rho } \right) \left( 1 - \frac{E^\infty }{r} \right) \quad \quad \qquad \qquad \quad (21) \\ {p} E^\infty&= \lambda _X^{\infty }\left( \delta - r + 2r \left[ 1 - \frac{E^\infty }{r} \right] + E^\infty \right) \\ r \lambda _X^{\infty }\left( 1 - \frac{E^\infty }{r} \right) ^2&= \lambda _K^{\infty }\left( \delta - \rho + 2 \rho \left[ 1 - \frac{\beta E}{\rho } \right] + \beta E^\infty \right) . \end{aligned}$$

The two last equations are equivalent to (22) and (23) respectively. This concludes the proof of Lemma 1.

1.3 B. 3 Trajectories for E Constant

We investigate in this section the solutions to the differential systems when E is constant. To that end, it is convenient to perform the change of functions $Z = K^{-1}$ and $Y = X^{-1}$. This leads to the differential system:

$$\begin{aligned} \dot{Y}&= (E-r)Y + rZ \end{aligned}$$

(53)

$$\begin{aligned} \dot{Z}&= (\beta E - \rho )Z + \frac{\rho }{\overline{K}} ~. \end{aligned}$$

(54)

Trajectories of the state. The solution of (13) is, classically, (26). This is obtained, for instance, using the function $Z = K^{-1}$ and the linear differential equation (54), whose solution is:

$$\begin{aligned} Z(t) ~ = ~ \frac{1}{K(t)} ~ = ~ \frac{1}{K_0} ~ e^{(\beta E - \rho )t} + \frac{\rho }{\overline{K}} ~ \frac{e^{(\beta E - \rho )t}-1}{\beta E-\rho } ~. \end{aligned}$$

(55)

The solution of (12) can be deduced as follows. Using the function $Y = X^{-1}$ and (53), we have:

$$\begin{aligned} \dot{Y}&= (E - r) Y + r e^{(\beta E-\rho )t} \left( \frac{1}{K_0} - \frac{\rho }{\rho -\beta E} \frac{1}{\overline{K}} \right) + \frac{r\rho }{(\rho -\beta E)\overline{K}} ~. \end{aligned}$$

(56)

Consequently, the solution will be:

$$\begin{aligned} Y(t)&= Y(0) e^{(E-r)t} \\&~+~ \int _0^t \left[ r e^{(\beta E-\rho )u} \left( \frac{1}{K_0} - \frac{\rho }{\rho -\beta E} \frac{1}{\overline{K}} \right) + \frac{r\rho }{(\rho -\beta E)\overline{K}} \right] e^{(E-r)(t-u)} \text{ d }u\\&= Y(0) e^{(E-r)t}\\&~+~ r \left( \frac{1}{K_0} - \frac{\rho }{\rho -\beta E} \frac{1}{\overline{K}} \right) \frac{e^{(\beta E-\rho )t}-e^{(E-r)t}}{(\beta -1)E+r-\rho } + \frac{r\rho }{(\rho -\beta E)\overline{K}} \frac{1 - e^{(E-r)t}}{r-E} ~. \end{aligned}$$

This equation directly leads to (27).

Alternative representations. Using the asymptotic quantities when $t\rightarrow \infty$, derived in Sect. 3.1, the solution for E constant may be rewritten as:

$$\begin{aligned} Z(t) ~ = ~ \frac{1}{K(t)}&= \left( \frac{1}{K_0} - \frac{1}{K^{\infty }} \right) ~ e^{(\beta E - \rho )t} + \frac{1}{K^{\infty }} \end{aligned}$$

(57)

$$\begin{aligned} Y(t) ~ = ~ \frac{1}{X(t)}&= \left( \frac{1}{X_0} - \frac{1}{X^{\infty }} \right) e^{(E-r)t} ~+~ r \left( \frac{1}{K_0} - \frac{1}{K^{\infty }} \right) \frac{e^{(\beta E-\rho )t}-e^{(E-r)t}}{(\beta -1)E+r-\rho } + \frac{1}{X^{\infty }} ~. \end{aligned}$$

(58)

Equations in the (K, X)  plane. Provided $K_0 \not = K^{\infty }$, the variable t can be eliminated from (57):

$$\begin{aligned} t ~ = ~ \frac{1}{\rho - \beta E} ~ \log \left( \frac{\frac{1}{K_0}-\frac{1}{K^{\infty }}}{\frac{1}{K}-\frac{1}{K^{\infty }}} \right) ~ = ~ \frac{1}{\rho - \beta E} ~ \log \left( \frac{ K^{\infty }- K_0 }{ K^{\infty }- K } ~ \frac{K}{K_0} \right) ~. \end{aligned}$$

This is the amount of time needed to go from state $K_0$ to state K. Then, replacing in (58), we have:

$$\begin{aligned} \frac{1}{X} - \frac{1}{X^{\infty }}&= e^{(E-r)t} \left( \frac{1}{X_0} - \frac{1}{X^{\infty }} - \left( \frac{1}{K_0} - \frac{1}{K^{\infty }} \right) \frac{r}{(\beta -1)E+r-\rho } \right) \nonumber \\&~+~ e^{(\beta E-\rho )t} \left( \frac{1}{K_0} - \frac{1}{K^{\infty }} \right) \frac{r}{(\beta -1)E+r-\rho } \nonumber \\&= \left( \frac{ K^{\infty }- K_0 }{ K^{\infty }- K } ~ \frac{K}{K_0} \right) ^{\frac{E-r}{\rho -\beta E}} \left( \frac{1}{X_0} - \frac{1}{X^{\infty }} - \left( \frac{1}{K_0} - \frac{1}{K^{\infty }} \right) \frac{r}{(\beta -1)E+r-\rho } \right) \nonumber \\&~+~ \left( \frac{1}{K} - \frac{1}{K^{\infty }} \right) \frac{r}{(\beta -1)E+r-\rho } ~. \end{aligned}$$

(59)

This is the equation of the trajectory in the (K, X) plane, parameterized by any point $(K_0,X_0)$ on the curve.

If $K_0 = K^{\infty }$, the trajectory is included in the line $\{ K = K^{\infty }\}$. If needed, the time variable may be eliminated from (58):

$$\begin{aligned} t ~ = ~ \frac{1}{r - E} ~ \log \left( \frac{\frac{1}{X_0}-\frac{1}{X^{\infty }}}{\frac{1}{X}-\frac{1}{X^{\infty }}} \right) ~ = ~ \frac{1}{r - E} ~ \log \left( \frac{ X^{\infty }- X_0 }{ X^{\infty }- X } ~ \frac{X}{X_0} \right) . \end{aligned}$$

This is, in particular, the case in the standard model with K constant.

We now investigate the behaviors of trajectories in the (K, X) plane. From (59), and since with $E=0$ we have $X^{\infty }=K^{\infty }=\overline{K}$, these are the curves of equation

$$\begin{aligned} \frac{1}{X} ~ = ~ \frac{1}{\overline{K}} + \left( \frac{ \overline{K}- K_0 }{ \overline{K}- K } ~ \frac{K}{K_0} \right) ^{-r/\rho } \left( \frac{1}{X_0} - \frac{1}{\overline{K}} - \left( \frac{1}{K_0} - \frac{1}{\overline{K}} \right) \frac{r}{r-\rho } \right) ~+~ \left( \frac{1}{K} - \frac{1}{\overline{K}} \right) \frac{r}{r-\rho } \end{aligned}$$

parameterized by any point $(K_0,X_0)$ on the curve.⁴ These curves have the following property.

Lemma 3

Let f(K) be one of these curves, parameterized by $(K_0,X_0)$. There exists some $K_1\in (0,\overline{K})$ such that $\lim _{K\rightarrow K_1^+} f(K) = +\infty$ if:

• (a) $\rho > r$, or

• (b) when and only when

  $$\begin{aligned} X_0 ~ > ~ \frac{K_0 \overline{K}(r - \rho )}{r\overline{K}- \rho K_0} \end{aligned}$$

  (60)

  if $\rho < r$.

The proof consists in analyzing the behavior of the curves when $K\rightarrow 0$. If $\rho > r$, we find that $f(K) \sim K (r-\rho )/r < 0$: Then necessarily $f(\cdot )$ changes sign between $K=0$ and $K=\overline{K}$, which can occur only by diverging to $\pm \infty$. If $\rho < r$, we find that $f(K) \sim C(K_0,X_0) K^{r/\rho }$ for some value $C(K_0,X_0)$. If this value is negative, which is equivalent to condition (60), this is the same situation as when $\rho > r$ and there is a point $K_1$ where $f(\cdot )$ diverges. If this value is 0, the curve f has actually the equation

$$\begin{aligned} f_c(K) = \frac{K \overline{K}(r-\rho )}{r\overline{K}- \rho K_0} \end{aligned}$$

, and it does not diverge. If $C(K_0,X_0) > 0$, the trajectory lies below the curve $f_c(K)$ (solutions to the differential system cannot cross each other) so that it cannot diverge.

Adjoint variable trajectories with $E=0$. In the specific case where $E=0$, it is possible to obtain a formula for the adjoint variable $\lambda _X$. Start with the differential equation for $\Theta _X = X \lambda _X$

$$\begin{aligned} \dot{\Theta }_X&= \dot{X} \lambda _X + X \dot{\lambda }_X ~ = ~ \Theta _X \left( \delta + r \frac{X}{K} \right) \\ \frac{\dot{\Theta }_X}{\Theta _X}&= \delta + r - \frac{\dot{X}}{X}\\ \frac{\Theta _X(t)}{\Theta _X(0)}&= e^{(\delta +r)t} ~ \frac{X(0)}{X(t)}\\ \Theta _X(t)&= \Theta _X(0) X(0) \frac{e^{(\delta +r)t}}{X(t)} \end{aligned}$$

and this last equation yields (28).

C Proof of Lemma 2

Proof

(a) Consider the polynomial in (24), that is,

$$\begin{aligned} Pol(E)&= -c (E \beta - \delta - \rho ) r (E - \delta - r) \rho \\&+ p \overline{K}(E - r) (\beta E - \rho ) \left( 3 E^2 \beta - (\beta \delta + 2 \beta r + 2 \delta + 2 \rho ) E + (\delta + r)(\delta + \rho ) \right) . \end{aligned}$$

Recall that we are looking for admissible roots of Pol(E), that is, such that $0 \le E \le \min ( r, \rho /\beta )$. This polynomial has the following properties.

$$\begin{aligned} Pol(0)= & {} r \rho (\delta + \rho ) (\delta + r) (\overline{K}p - c), \\ Pol(r)= & {} r c \delta \rho (\beta r - \delta - \rho ), \quad Pol \left( \frac{\rho }{\beta } \right) = -\frac{\rho c \delta r (\beta \delta + \beta r - \rho )}{\beta }. \end{aligned}$$

The existence part is proved by showing that $Pol(\cdot )$ changes sign over the interval. The uniqueness follows from convexity.

Under the condition $\overline{K}p> c > 0$, $Pol(0) > 0$. Consider first the case $r \le {\rho }/{\beta }$. This implies $Pol(r) < 0$, and hence, $Pol(E)=0$ for at least one $E \in (0,r)$. Through tedious yet mechanical computations, using the assumption that $\overline{K}p - c > 0$, one shows that $Pol''(E) >0 \; \text{ if } \; E \in [0, r]$. Then, it is not possible to have three roots of $Pol(\cdot )$ in the interval. When $r > {\rho }/{\beta }$, then $Pol({\rho }/{\beta }) < 0$, and the reasoning is similar. From this analysis, we conclude not only that $Pol(\cdot )$ has a single root, but also that $Pol'(E) <0$ in the region where E is admissible.

Moreover from (24), and for admissible E,

$$\begin{aligned} \frac{\partial Pol}{\partial c}(E) ~ = ~ - r \rho (E \beta - \delta - \rho ) (E - \delta - r) ~ < ~ 0. \end{aligned}$$

Then using total differentiation for the root $E^\infty (c)$, we have:

$$\begin{aligned} { \frac{\partial E^\infty }{\partial c}(c) = - \frac{1}{Pol'(E^\infty (c))} \; \frac{\partial Pol}{\partial c}(E^\infty (c)) < 0. } \end{aligned}$$

(61)

(b) Now, we consider $c=0$. In this case, with $Pol_0(E) \equiv Pol(E)\vert _{c=0}$,

$$\begin{aligned} Pol_0(E) = p \overline{K}(E-r)(\beta E - \rho ) \left( 3 E^2 \beta - (\beta \delta + 2 \beta r + 2 \delta + 2 \rho ) E + (\delta + \rho ) (\delta + r) \right) . \end{aligned}$$

Let Q(E) denote the last polynomial of degree 2 in E. The discriminant of Q(E) is:

$$\begin{aligned} \Delta _\beta := (\delta + 2 r)^2 \beta ^2 - 4 (\delta + \rho ) (2 \delta + r) \beta + 4 (\delta + \rho )^2. \end{aligned}$$

This quadratic polynomial of $\beta$ has in turn as discriminant: $48 (\delta + \rho )^2(\delta - r) (\delta + r)$. Then, if $\delta < r$, this discriminant is negative and $\Delta _\beta > 0$ for all values of $\beta$. As a consequence, Q(E) has two real and positive roots $E_1 < E_2$, where $E_1$ is given explicitly by (25). We proceed with locating $E_1$ with respect to the other roots of $Pol_0(E)$: r and $\rho /\beta$. We have:

$$\begin{aligned} Q(0)&> 0,\\ Q\left( \frac{\delta +r}{2}\right)&= \frac{ \beta (\delta - r) (\delta + r)}{4}. \end{aligned}$$

If $\delta < r$, this implies that $0< E_1 < (\delta + r)/2$. If furthermore $r \le \rho /\beta$, then $(\delta + r)/2 < \min ( r, \rho /\beta )$ and therefore $E_1 = E_0^\infty$.

If $\delta < r$ but $r > \rho /\beta$, we just have $E_0^\infty = \min ( E_1, \rho /\beta )$. $\square$