Volatile Chemical Emission as a Weapon of Rearguard Action: A Game-Theoretic Model of Contest Behavior

Abstract

We use a game-theoretic model to explore whether volatile chemical (spiroacetal) emissions can serve as a weapon of rearguard action. Our basic model explores whether such emissions serve as a means of temporary withdrawal, preventing the winner of the current round of a contest from translating its victory into permanent possession of a contested resource. A variant of this model explores an alternative possibility, namely, that such emissions serve as a means of permanent retreat, attempting to prevent a winner from inflicting costs on a fleeing loser. Our results confirm that the underlying logic of either interpretation of weapons of rearguard action is sound; however, empirical observations on parasitoid wasp contests suggest that the more likely function of chemical weapons is to serve as a means of temporary withdrawal. While our work is centered around the particular biology of contest behavior in parasitoid wasps, it also provides the first contest model to explicitly consider self-inflicted damage costs and thus responds to a recent call by empiricists for theory in this area.

Appendices

Appendix 1: The Probability of Victory

Let us first suppose that $\mu = 0$. Then the probability of victory depends only on RHP difference and is the same in either role, that is, we can set $p_{\text {o}}(\varDelta ) = p_{\text {i}}(\varDelta ) = p(\varDelta )$ for all $\varDelta \in [-1,1]$ where $p^\prime (\varDelta ) > 0$ with $p(-1) = 0$, $p(0) = \frac{1}{2}$ and $p(1) = 1$ by (3)–(6). There only two ways to satisfy all of these constraints. The first is for p to be linear with slope $\frac{1}{2}$, specifically,

$$\begin{aligned} p(\varDelta ) = \tfrac{1}{2}(1+\varDelta ). \end{aligned}$$

(30)

The second way is for p to be sigmoidal with an inflection point where $\varDelta = 0$. Specifically, if $0< p^\prime (0) < \frac{1}{2}$ then $p^{\prime \prime }(\varDelta ) < 0$ for $\varDelta < 0$ and $p^{\prime \prime }(\varDelta ) > 0$ for $\varDelta > 0$, whereas if $\frac{1}{2}< p^\prime (0) < \infty$, then $p^{\prime \prime }(\varDelta ) > 0$ for $\varDelta < 0$ and $p^{\prime \prime }(\varDelta ) < 0$ for $\varDelta > 0$. In principle, a great many functions are of this type. In practice, however, what we need for modeling purposes is a known and well studied function that captures with a single parameter how the shape of the sigmoid changes from very flat as $p^\prime (0) \rightarrow 0$ to very steep as $p^\prime (0) \rightarrow \infty$, reflecting how the reliability of RHP difference as a predictor of fight outcome increases from very poor to almost perfect. Thus in practice there are relatively few sensible choices. All things considered, in our view the best function for our purposes is

(31)

where B denotes the incomplete Beta function, i.e., $B(w,p_1,p_2) = \int _0^w \xi ^{p_1- 1} \, (1-\xi )^{p_2 - 1}\,d\xi$, and r is the parameter. Note that $p^\prime (0)$ increases monotonically with r in such a way that $p^\prime (0) \rightarrow 0$, $0< p^\prime (0) < \frac{1}{2}$, $\frac{1}{2}< p^\prime (0) < \infty$ and $p^\prime (0) \rightarrow \infty$ correspond to $r \rightarrow 0$, $0< r < 1$, $1< r < \infty$ and $r \rightarrow \infty$, respectively, and in particular, (31) reduces to (30) for $r = 1$.

The effect of ownership advantage on the probability of victory at any RHP difference $\varDelta$ can now be most readily incorporated by writing

$$\begin{aligned} p_{\text {o}}(\varDelta ) = \bigl \{p(\varDelta )\bigr \}^z \end{aligned}$$

(32)

where $\bigl \{p(0)\bigr \}^z = \frac{1}{2}(1+\mu )$ to satisfy (6). Hence $\bigl \{\frac{1}{2}\bigr \}^z = \frac{1}{2}(1+\mu )$ or $z = 1- \ln (1+\mu )/\ln (2) > 0$. Note that $z \rightarrow 0$ as $\mu \rightarrow 1$ implying $p(\varDelta ) \rightarrow 1$ for any value of r—if the owner is guaranteed to win, then the reliability of RHP difference has no effect on the outcome. Substituting z back into (32) now yields (14).

Fig. 9

The sample space of pairs of strengths. For Model A, four possible cases are defined by the four rectangles. In the following, $i = 1$, $j = 2$ or $i = 2$, $j = 1$ according to whether the focal individual is the owner or the intruder. Case I $X< u_i, Y < v_j$ Neither contestant will release the chemical if it loses the first round. Case II $X > u_i, Y < v_j$ Only the u-strategist (focal individual or Player 1) will release the chemical if it loses the first round. Case III $X < u_i, Y > v_j$ Only the v-strategist (non-focal individual or Player 2) will release the chemical if it loses the first round. Case IV $X> u_i, Y > v_j$ Either contestant will release the chemical if it loses the first round

Appendix 2: Calculation of the Reward Function for Model A

Regardless of whether Player 1’s role is that of owner or intruder, the acceptance thresholds partition the sample space into four rectangular regions, which we denote by I, II, III and IV, as indicated in Fig. 9. Having already scaled fighting cost with respect to value, for consistency we must also scale fitness with respect to value. Accordingly, for $K = \text { I,}\ldots , \text {IV}$, let denote the payoff to the focal individual if $(X,Y) \in K$ when Player 1 is the owner, and let denote the corresponding payoff when Player 1 is the intruder.

Let us first suppose that Player 1 is the owner while Player 2 is the intruder. Region I is where either animal would accept defeat after losing Round 1, and so the payoff to the focal individual is $V - Vc(X)$ with probability $p_{\text {o}}(X-Y)$ and $0 - Vc(X)$ with probability $p_{\text {i}}(Y-X)$, or $\{V - Vc(X)\}p_{\text {o}}(X-Y) - Vc(X)p_{\text {i}}(Y-X) = V\{p_{\text {o}}(X-Y) - c(X)\}$ by (4), so that

$$\begin{aligned} f^{\text {o}}_{\text {I}}(X,Y) = p_{\text {o}}(X-Y) - c(X). \end{aligned}$$

(33)

Region II is where Player 1 instigates a second round after losing the first one, whereas Player 2 does not. So the payoff to the focal individual remains $V\{1 - c(X)\}$ if Player 1 wins the first round, that is, with probability $p_{\text {o}}(X-Y)$. If, however, Player 2 wins the first round, which happens with probability $p_{\text {i}}(Y-X)$, then the payoff to Player 1 is $V - Vc(X) - Vc(X_l)$ with probability $p_L(X_l-Y_w)$ and $0 - Vc(X) - Vc(X_l)$ with probability $p_w(Y_w-X_l)$, or $V\{p_L(X_l-Y_w) - c(X) - c(X_l)\}$, on setting $\varDelta = X_l-Y_w$ in (9), where $X_l$ and $Y_w$ are defined by (2). Multiplying the above conditional payoff by $p_{\text {i}}(Y-X)$, multiplying $V\{1 - c(X)\}$ by $p_{\text {o}}(X-Y)$, adding and using (4) with $\varDelta = X-Y$, we obtain

$$\begin{aligned} f^{\text {o}}_{\text {II}}(X,Y) = p_{\text {o}}(X-Y) + \{p_L(X_l-Y_w)-c(X_l)\}p_{\text {i}}(Y-X) - c(X). \end{aligned}$$

(34)

Correspondingly, Region III is where Player 2 instigates a second round after losing the first one, whereas Player 1 does not. So the payoff to the focal individual is $0 - Vc(X)$ if Player 2 wins the first round, that is, with probability $p_{\text {i}}(Y-X)$. If, however, Player 1 wins the first round, which happens with probability $p_{\text {o}}(X-Y)$, then the payoff to Player 1 is $V - Vc(X) - Vc(X_w)$ with probability $p_W(X_w-Y_l)$ and $0 - Vc(X) - Vc(X_w)$ with probability $p_l(Y_l-X_w)$, or $V\{p_W(X_w-Y_l)-c(X) - c(X_w)\}$ by (9) with $\varDelta = X_w-Y_l$, where $X_w$ and $Y_l$ are defined by (2). Multiplying the above conditional payoff by $p_{\text {o}}(X-Y)$, $0 - Vc(X)$ by $p_{\text {i}}(Y-X)$, adding and again using (4), we obtain, in lieu of (34),

$$\begin{aligned} f^{\text {o}}_{\text {III}}(X,Y) = \{p_W(X_w-Y_l)-c(X_w)\}p_{\text {o}}(X-Y) - c(X). \end{aligned}$$

(35)

Finally, Region IV is where either animal instigates a second round after losing the first one. So the payoff to the focal individual is $V\{p_W(X_w-Y_l)-c(X) - c(X_w)\}$ with probability $p_{\text {o}}(X-Y)$ or $V\{p_L(X_l-Y_w) - c(X) - c(X_l)\}$ with probability $p_{\text {i}}(Y-X)$. That is,

$$\begin{aligned} f^{\text {o}}_{\text {IV}}(X,Y)&= \{p_W(X_w-Y_l)-c(X_w)\}p_{\text {o}}(X-Y) + \{p_L(X_l-Y_w)\nonumber \\ {}&\quad -c(X_l)\}p_{\text {i}}(Y-X) - c(X). \end{aligned}$$

(36)

Let $f^{\text {o}}(u_1,v_2)$ denote the reward to a u-strategist in the role of owner against a v-strategist in the role of intruder, scaled with respect to value. Then

$$\begin{aligned} Vf^{\text {o}}(u_1,v_2) = \mathop {\sum }\limits _{K = \text {I}}^{\text {IV}}\;\iint \limits _K Vf^{\text {o}}_K(x,y)g(x)g(y)\,\mathrm{d}x\,\mathrm{d}y \end{aligned}$$

(37)

where g is the probability density function, implying

$$\begin{aligned}&f^{\text {o}}(u_1,v_2) = \int \limits _0^{u_1}\int \limits _0^{v_2} f^{\text {o}}_{\text {I}}(x,y)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x + \int \limits _{u_1}^1\int \limits _0^{v_2} f^{\text {o}}_{\text {II}}(x,y)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x \nonumber \\&\quad + \int \limits _0^{u_1}\int \limits _{v_2}^1 f^{\text {o}}_{\text {III}}(x,y)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x + \int \limits _{u_1}^1\int \limits _{v_2}^1 f^{\text {o}}_{\text {IV}}(x,y)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x. \end{aligned}$$

(38)

Considering cases when Player 1 is the intruder while Player 2 is the owner, the payoff to the focal individual in Region I becomes $\alpha V - Vc(X)$ with probability $p_{\text {i}}(X-Y)$ and $0 - Vc(X)$ with probability $p_{\text {o}}(Y-X)$, so that, by (4),

$$\begin{aligned} f^{\text {i}}_{\text {I}}(X,Y) = \alpha p_{\text {i}}(X-Y) - c(X). \end{aligned}$$

(39)

Continuing in this manner, (34)–(36) and (38) become modified to

$$\begin{aligned} f^{\text {i}}_{\text {II}}(X,Y) =&\alpha p_{\text {i}}(X-Y) + \{\alpha p_l(X_l-Y_w))-c(X_l)\}p_{\text {o}}(Y-X) - c(X), \end{aligned}$$

(40)

$$\begin{aligned} f^{\text {i}}_{\text {III}}(X,Y) =&\{\alpha p_w(X_w-Y_l)-c(X_w)\}p_{\text {i}}(X-Y) - c(X), \end{aligned}$$

(41)

$$\begin{aligned} f^{\text {i}}_{\text {IV}}(X,Y) =&\{\alpha p_w(X_w-Y_l)-c(X_w)\}p_{\text {i}}(X-Y) \nonumber \\&+ \{\alpha p_l(X_l-Y_w)-c(X_l)\}p_{\text {o}}(Y-X) - c(X) \end{aligned}$$

(42)

and

$$\begin{aligned}&f^{\text {i}}(u_2,v_1) = \int \limits _0^{u_2}\int \limits _0^{v_1} f^{\text {i}}_{\text {I}}(x,y)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x + \int \limits _{u_2}^1\int \limits _0^{v_1} f^{\text {i}}_{\text {II}}(x,y)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x\nonumber \\&\quad + \int \limits _0^{u_2}\int \limits _{v_1}^1 f^{\text {i}}_{\text {III}}(x,y)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x + \int \limits _{u_2}^1\int \limits _{v_1}^1 f^{\text {i}}_{\text {IV}}(x,y)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x \end{aligned}$$

(43)

where $f^{\text {i}}(u_2,v_1)$ denotes the reward to a u-strategist in the role of intruder against a v-strategist in the role of owner, scaled with respect to value. Let f(u, v) denote the unconditional reward to a u-strategist against a v-strategist, scaled with respect to value. Then assuming the roles of prior owner and intruder to be equally likely, we obtain

$$\begin{aligned} f(u,v) = \tfrac{1}{2}f^{\text {o}}(u_1,v_2) + \tfrac{1}{2} f^{\text {i}}(u_2,v_1). \end{aligned}$$

(44)

Straightforward partial differentiation with respect to $u_1$ shows that

$$\begin{aligned} \frac{\partial f}{\partial u_1}= & {} \tfrac{1}{2}\int \limits _0^{v_2} \{f^{\text {o}}_{\text {I}}(u_1,y) - f^{\text {o}}_{\text {II}}(u_1,y)\}g(u_1)g(y)\,\mathrm{d}y \nonumber \\&+ \tfrac{1}{2}\int \limits _{v_2}^1 \{f^{\text {o}}_{\text {III}}(u_1,y) - f^{\text {o}}_{\text {IV}}(u_1,y)\}g(u_1)g(y)\,\mathrm{d}y. \end{aligned}$$

(45a)

From (33)–(36), however, we obtain

$$\begin{aligned} f^{\text {o}}_{\text {I}}(X,Y) - f^{\text {o}}_{\text {II}}(X,Y) = f^{\text {o}}_{\text {III}}(X,Y) - f^{\text {o}}_{\text {IV}}(X,Y) = - \{p_L(X_l-Y_w)-c(X_l)\}p_{\text {i}}(Y-X) \end{aligned}$$

where $X_l = (1-\theta _l)X$ and $Y_w = (1-\theta _w)Y$ by (2). Setting $X = u_1$ and $Y = y$, substitution into (45a) now reduces it to

$$\begin{aligned} \frac{\partial f}{\partial u_1} = \tfrac{1}{2}\,g(u_1)\int \limits _0^1 \bigl \{c(\{1-\theta _l\}u_1)-p_L(\{1-\theta _l\}u_1-\{1-\theta _w\}y)\bigr \}p_{\text {i}}(y-u_1)g(y)\,\mathrm{d}y, \end{aligned}$$

(45b)

and further straightforward partial differentiation (with use of the product rule) yields

$$\begin{aligned} \frac{\partial ^2 f}{\partial {u_1}^2}&= \tfrac{1}{2}\{1-\theta _l\}\int \limits _0^1 \bigl \{c^{\prime }(\{1-\theta _l\}u_1)-p_L^{\prime } (\{1-\theta _l\}u_1-\{1-\theta _w\}y)\bigr \}\nonumber \\&\quad \times p_{\text {i}}(y-u_1)g(u_1)g(y)\,\mathrm{d}y\nonumber \\&\quad + \tfrac{1}{2}\int \limits _0^1 \bigl \{c(\{1-\theta _l\}u_1)-p_L(\{1-\theta _l\}u_1-\{1-\theta _w\}y)\bigr \}\nonumber \\&\quad \times \bigl \{p_{\text {i}}(y-u_1)g^{\prime }(u_1)-p_{\text {i}}^{\prime }(y-u_1)g(u_1)\bigr \}g(y)\,\mathrm{d}y. \end{aligned}$$

(46)

Likewise, differentiation with respect to $u_2$ instead yields

$$\begin{aligned} \frac{\partial f}{\partial u_2}&= \tfrac{1}{2}\int \limits _0^{v_1} \{f^{\text {i}}_{\text {I}}(u_2,y) - f^{\text {i}}_{\text {II}} (u_2,y)\}g(u_2)g(y)\,\mathrm{d}y \nonumber \\&\quad + \tfrac{1}{2}\int \limits _{v_1}^1 \{f^{\text {i}}_{\text {III}}(u_2,y) - f^{\text {i}}_{\text {IV}}(u_2,y)\}g(u_2)g(y)\,\mathrm{d}y \end{aligned}$$

(47a)

$$\begin{aligned}&= \tfrac{1}{2}\,g(u_2)\int \limits _0^1 \bigl \{c(\{1-\theta _l\}u_2)-\alpha p_l(\{1-\theta _l\}u_2-\{1-\theta _w\}y)\bigr \}p_{\text {o}}(y-u_2)g(y)\,\mathrm{d}y \end{aligned}$$

(47b)

with

$$\begin{aligned} \frac{\partial ^2 f}{\partial {u_2}^2}= & {} \tfrac{1}{2}\{1-\theta _l\}\int \limits _0^1 \bigl \{c^{\prime }(\{1-\theta _l\}u_2)-\alpha p_l^{\prime }(\{1-\theta _l\}u_2-\{1-\theta _w\}y)\bigr \}\nonumber \\&\times p_{\text {o}}(y-u_2)g(u_2)g(y)\,\mathrm{d}y\nonumber \\&+ \tfrac{1}{2}\int \limits _0^1 \bigl \{c(\{1-\theta _l\}u_2)-\alpha p_l(\{1-\theta _l\}u_2-\{1-\theta _w\}y)\bigr \}\nonumber \\&\times \bigl \{p_{\text {o}}(y-u_2)g^{\prime }(u_2)-p_{\text {o}}^{\prime }(y-u_2)g(u_2)\bigr \}g(y)\,\mathrm{d}y. \end{aligned}$$

(48)

Note that the first terms of (46) and (48) are invariably negative, by (3) and (7).

Appendix 3: Calculation of the ESS for Model A

A population strategy $v = (v_1,v_2)$ is a strong evolutionarily stable strategy or ESS sensu Maynard Smith (1982) when it is uniquely the best reply to itself, that is when $f(v,v) > f(u,v)$ for any potential mutant strategy $u \ne v$, which for $j = 1$ or $j = 2$ requires

$$\begin{aligned} \frac{\partial f}{\partial u_j}\Bigr |_{u = v} = 0 \end{aligned}$$

(49)

with

$$\begin{aligned} \frac{\partial ^2 f}{\partial {u_j}^2}\Bigr |_{u = v} < 0 \end{aligned}$$

(50)

for $0< v_j < 1$ at the ESS, but instead

$$\begin{aligned} \frac{\partial f}{\partial u_j}\Bigr |_{u = v} < 0 \end{aligned}$$

(51)

for $v_j = 0$ at the ESS and

$$\begin{aligned} \frac{\partial f}{\partial u_j}\Bigr |_{u = v} > 0 \end{aligned}$$

(52)

for $v_j = 1$ at the ESS (see, e.g., Broom and Rychtář 2013). Although (49)–(52) guarantee that v is a (strong) local ESS, to show that v is also a global ESS we must establish, either analytically or computationally, that no $u \in [0,1]\times [0,1]$—as opposed to no u in the vicinity of v—yields a fitness against v that is as high or higher. Various means can be used. In particular, for an interior ESS, it suffices (but is not necessary) to show that $\partial ^2 f/\partial {u_j}^2$ is negative throughout [0, 1], as opposed to only at $u_j = v_j$, and for $v_j = 0$ at the ESS, it suffices to show that $\partial f/\partial {u_j}$ is negative throughout the same interval. However, sufficiency for an interior ESS can instead be established analytically by showing that $\partial f/\partial {u_j}$ increases monotonically with a sign change at $u_j = v_j$, as illustrated in Appendix 5.1; or computationally by plotting $\{f(v,v)-f(u,v)\}\bigr |_{u_j = v_j}$ against $u_i$ for $i = 1, j = 2$ and $i = 2, j = 1$, as illustrated in, e.g., Appendix B of Mesterton-Gibbons and Heap (2014). All of these approaches have been used to confirm the ESSes in this paper.

We present the calculation of the ESS for Model A only for differential costs, because for constant costs the calculation is virtually unaltered. Setting $u_1 = 0$ or $u_1 = 1$ in (45b), we obtain

$$\begin{aligned}&\frac{\partial f}{\partial u_1}\Bigr |_{u_1 = 0} = \tfrac{1}{2}\,\int \limits _0^1 \bigl \{\gamma -p_L(-\{1-\theta _w\}y)\bigr \}p_{\text {i}}(y)\,\mathrm{d}y \end{aligned}$$

(53a)

$$\begin{aligned}&\frac{\partial f}{\partial u_1}\Bigr |_{u_1 = 1} = \tfrac{1}{2}\,\int \limits _0^1 \bigl \{\gamma (1 - \{1-\theta _l\}^k)-p_L(1-\theta _l-\{1-\theta _w\}y)\bigr \}p_{\text {i}}(y-1)\,\mathrm{d}y\nonumber \\ \end{aligned}$$

(53b)

because $g(y) = 1$ (implying in particular that $g(0) = g(1) = 1$) by (11), whereas $c(0) = \gamma$ and $c(1-\theta _l) = \gamma (1 - \{1-\theta _l\}^k)$ by (13). From (51) and (52), $v_1 = 0$ at the ESS if (53a) is negative and $v_1 = 1$ at the ESS if (53b) is positive. Hence $v_1 = 0$ at the ESS for $\gamma < \underline{\gamma }^{\text {o}}_{\text {c}}$ and $v_1 = 1$ at the ESS for $\gamma > \overline{\gamma }^{\text {o}}_{\text {c}}$, where

$$\begin{aligned} \underline{\gamma }^{\text {o}}_{\text {c}} = \frac{\int _0^1 p_L(-\{1-\theta _w\}y)p_{\text {i}}(y)\,\mathrm{d}y}{\int _0^1 p_{\text {i}}(y)\,\mathrm{d}y} \end{aligned}$$

(54a)

and

$$\begin{aligned} \overline{\gamma }^{\text {o}}_{\text {c}} = \frac{\int _0^1 p_L(1-\theta _l-\{1-\theta _w\}y)p_{\text {i}}(y-1)\,\mathrm{d}y}{(1-\{1-\theta _l\}^k)\int _0^1 p_{\text {i}}(y-1)\,\mathrm{d}y}. \end{aligned}$$

(54b)

It likewise follows from (7), (47b), (11), (13), (51) and (52) that $v_2 = 0$ at the ESS for $\gamma < \underline{\gamma }^{\text {i}}_{\text {c}}$ and $v_2 = 1$ at the ESS for $\gamma > \overline{\gamma }^{\text {i}}_{\text {c}}$, where

$$\begin{aligned} \underline{\gamma }_{\text {c}}^{{\text {i}}} = \frac{\alpha \int _0^1 p_l(-\{1-\theta _w\}y)p_{\text {o}}(y)\,\mathrm{d}y}{\int _0^1 p_{\text {o}}(y)\,\mathrm{d}y}, \end{aligned}$$

(55a)

which (10) reduces to the expression given in (16), and

$$\begin{aligned} \overline{\gamma }^{\text {i}}_{\text {c}} = \frac{\alpha \int _0^1 p_{\text {i}}(1-\theta _l-\{1-\theta _w\}y)p_{\text {o}}(y-1)\,\mathrm{d}y}{(1-\{1-\theta _l\}^k)\int _0^1 p_{\text {o}}(y-1)\,\mathrm{d}y}. \end{aligned}$$

(55b)

Note that $\underline{\gamma }^{\text {o}}_{\text {c}}$ and $\underline{\gamma }^{\text {i}}_{\text {c}}$ are independent of both k and $\theta _l$, and that $\underline{\gamma }^{\text {o}}_{\text {c}} < \frac{1}{2}$ and $\underline{\gamma }^{\text {i}}_{\text {c}} < \frac{1}{2}\alpha$ (although these values are approached in the limit as $\theta _W \rightarrow 1$). Note also that if $\theta _l = 0$, implying $c(1-\theta _l) = c(1) = \gamma (1-1^k) = 0$ by (13), then $v_j = 1$ can hold at the ESS neither for $j = 1$ nor for $j = 2$, because (45b) or (47b) implies $\frac{\partial f}{\partial u_j}\bigr |_{u_j = v_j = 1} < 0$, contradicting (52). Thus, because the RHP threshold for emission will always be less than its maximum value at the ESS when $\theta _l = 0$, it will always pay the strongest first-round losers to emit the chemical when the chemical is not toxic to the releaser. The very same result follows by noting that $\overline{\gamma }^{\text {o}}_{\text {c}} \rightarrow \infty$ as $\theta _l \rightarrow 0$ in (54b) and $\overline{\gamma }^{\text {i}}_{\text {c}} \rightarrow \infty$ as $\theta _l \rightarrow 0$ in (55b). When $\theta _l > 0$, on the other hand, both $\overline{\gamma }^{\text {o}}_{\text {c}}$ and $\overline{\gamma }^{\text {i}}_{\text {c}}$ are finite. If $\theta _l > 0$ is so large that $\overline{\gamma }^{\text {o}}_{\text {c}}$ and $\overline{\gamma }^{\text {i}}_{\text {c}}$ are not only finite but also less than 1, then it follows from (8) that there exists a range of values of the maximum cost $\gamma$ so close to its upper limit of 1 that even the strongest first-round losers would refrain from chemical emission; hence no first-round loser would emit the chemical, which is contrary to empirical evidence (Goubault et al. 2006, 2008). It therefore seems likely that self-toxicity $\theta _l$ is low enough to ensure $\overline{\gamma }^{\text {o}}_{\text {c}}$, $\overline{\gamma }^{\text {i}}_{\text {c}} > 1$ or $\gamma$ is never so close to its maximum, or both.

For values of $\gamma$ between the lower and upper critical values, it follows from (7), (45b), (47b), (11) and (49) that $v_1 = v_1^*$ at the ESS for $\underline{\gamma }^{\text {o}}_{\text {c}}< \gamma < \overline{\gamma }^{\text {o}}_{\text {c}}$ and $v_2 = v_2^*$ at the ESS for $\underline{\gamma }^{\text {i}}_{\text {c}}< \gamma < \overline{\gamma }^{\text {i}}_{\text {c}}$, where $v_1 = v_1^*$ and $v_2 = v_2^*$ are the only roots of the equations

$$\begin{aligned} \int \limits _0^1 \bigl \{c(\{1-\theta _l\}v_1)-p_L(\{1-\theta _l\}v_1-\{1-\theta _w\}y)\bigr \}p_{\text {i}}(y-v_1)\,\mathrm{d}y = 0 \end{aligned}$$

(56a)

and

$$\begin{aligned} \int \limits _0^1 \bigl \{c(\{1-\theta _l\}v_2)-\alpha p_l(\{1-\theta _l\}v_2-\{1-\theta _w\}y)\bigr \}p_{\text {o}}(y-v_2)\,\mathrm{d}y = 0, \end{aligned}$$

(56b)

respectively. Note from (10a) and (10b) that (54a), (54b) and (56a) differ between contests with final ($p_L = p_{\text {o}}$, prior owner retains owner advantage for the second round) and with intermediate ($p_L = p_{\text {i}}$, prior owner loses owner advantage for the second round) possession, whereas (55a), (55b) and (56b) are unaffected ($p_l = p_{\text {i}}$ in either case, a losing intruder lacks owner advantage).

We find that inequality (50) is invariably satisfied with $j = 1$ for $v_1 = v_1^*$ defined by (56a) and with $j = 2$ for $v_2 = v_2^*$ defined by (56b), thus confirming that $v^* = (v_1^*, v_2^*)$ is indeed an ESS. In some special cases we can verify these results analytically (see Appendix 5, in particular 79), although in general we can verify them only numerically. Nevertheless, inspection shows that (50) holds for $j = 1$ because (46) and (11) imply

$$\begin{aligned} \frac{\partial ^2 f}{\partial {u_1}^2}\Bigr |_{u = v}= & {} \tfrac{1}{2}\{1-\theta _l\}\int \limits _0^1 \bigl \{c^{\prime }(\{1-\theta _l\}v_1)-p_L^{\prime }(\{1-\theta _l\}v_1-\{1-\theta _w\}y)\bigr \}p_{\text {i}}(y-v_1)\,\mathrm{d}y\nonumber \\&- \tfrac{1}{2}\int \limits _0^1 \bigl \{c(\{1-\theta _l\}v_1)-p_L(\{1-\theta _l\}v_1-\{1-\theta _w\}y)\bigr \}p_{\text {i}}^{\prime }(y-v_1)\,\mathrm{d}y \end{aligned}$$

(57)

The first term must be negative by (3) and (7), and (3) implies that the second integral could be sufficiently negative to make (57) positive only if $c(\{1-\theta _l\}v_1)-p_L(\{1-\theta _l\}v_1-\{1-\theta _w\}y)$ were substantially negative on [0, 1], in which case (56a) could not hold. A very similar argument shows why (50) holds for $j = 2$.

The resultant ESS $v^* = (v_1^*,v_2^*)$ in general depends on seven parameters, namely, $\alpha$, $\gamma$, k, r, $\mu$, $\theta _l$ and $\theta _w$. At this ESS, Player 1 wins the contest after losing the first round if $(X,Y) \in \text {II}\cup \text {IV}$ in Fig. 9 and Player 2 wins the first round but Player 1 wins the second. With $\varDelta = X - Y$ and $\varDelta _{lw} = X_l - Y_w$ (defined by (2) with $S = X$ or Y), this event arises with probability $p_{\text {i}}(-\varDelta )p_L(\varDelta _{lw})$ if Player 1 is the prior owner or $p_{\text {o}}(-\varDelta )p_l(\varDelta _{lw})$ if Player 2 is the prior owner, for $(X,Y) \in \text {II}\cup \text {IV}$. So, the overall probability that Player 1 wins the contest after losing the first round is

$$\begin{aligned}&\tfrac{1}{2}\int \limits _{v_1^*}^1\int \limits _0^1 p_{\text {i}}(y-x)p_L(x_l - y_w)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x \nonumber \\&\qquad +\tfrac{1}{2} \int \limits _{v_2^*}^1\int \limits _0^1 p_{\text {o}}(y-x)p_l(x_l - y_w)g(y)g(x)\,\mathrm{d}y\,\mathrm{d}x \nonumber \\&\quad = \tfrac{1}{2}\int \limits _{v_1^*}^1\int \limits _0^1 p_{\text {i}}(y-x)p_L(x_l - y_w)\,\mathrm{d}y\,\mathrm{d}x + \tfrac{1}{2}\int \limits _{v_2^*}^1\int \limits _0^1 p_{\text {o}}(y-x)p_{\text {i}}(x_l - y_w)\,\mathrm{d}y\,\mathrm{d}x\nonumber \\ \end{aligned}$$

(58)

by (10) and (11), where $x_l = (1-\theta _l)x$ and $y_w = (1-\theta _w)y$; the first integral is the probability of winning after losing as owner and the second integral is the probability of winning after losing as intruder. The corresponding probability that Player 2 wins the contest after losing the first round, obtained by substituting $\text {III}$ for $\text {II}$ and interchanging x and y, equals the expression in (58). Hence, the probability $p_{\text {LW}}$ that the contest is won by a first-round loser is just twice the above expression, yielding (19). Finally, the overall probability that the prior owner wins the contest is

$$\begin{aligned} P_{\text {o}}= & {} \iint \limits _{\text {I}} p_{\text {o}}(x-y)g(x)g(y)\,\mathrm{d}x\,\mathrm{d}y \nonumber \\&+\iint \limits _{\text {II}}\{p_{\text {o}}(x-y)+ p_{\text {i}}(y-x)p_L(x_l - y_w)\}g(x)g(y)\,\mathrm{d}x\,\mathrm{d}y\nonumber \\&+\iint \limits _{\text {III}} p_{\text {o}}(x-y)p_W(x_w - y_l)g(x)g(y)\,\mathrm{d}x\,\mathrm{d}y \nonumber \\&+\iint \limits _{\text {IV}}\{p_{\text {o}}(x-y)p_W(x_w - y_l)+ p_{\text {i}}(y-x)p_L(x_l - y_w)\}g(x)g(y)\,\mathrm{d}x\,\mathrm{d}y,\nonumber \\ \end{aligned}$$

(59)

which (10) and (11) reduce to (21).

Appendix 4: Calculation of the ESS for Model B

For Model B, several of the distinctions observed in Sect. 2 and Appendix 2 disappear because there is no longer a second round of fighting, and the analysis must be modified accordingly. First, there is no longer a distinction between intermediate and final possession. Second, there is no longer a distinction between Regions I and III in Fig. 9. In either case, the focal individual emits the chemical and pays the discharge cost upon withdrawing. Its payoff is therefore $V - Vc(X)$ with probability $p_{\text {o}}(X-Y)$ and $-Vc(X) - V\delta$ with probability $p_{\text {i}}(Y-X)$ if it is the owner or $\alpha V - Vc(X)$ with probability $p_{\text {i}}(X-Y)$ and $-Vc(X) - V\delta$ with probability $p_{\text {o}}(Y-X)$ if it is the intruder, so that (33), (35), (39) and (41) reduce to

$$\begin{aligned}&f^{\text {o}}_{\text {I}}(X,Y) = f^{\text {o}}_{\text {III}}(X,Y) = p_{\text {o}}(X-Y) - \delta p_{\text {i}}(Y-X) - c(X) \end{aligned}$$

(60)

$$\begin{aligned}&f^{\text {i}}_{\text {I}}(X,Y) = f^{\text {i}}_{\text {III}}(X,Y) = \alpha p_{\text {i}}(X-Y) - \delta p_{\text {o}}(Y-X) - c(X). \end{aligned}$$

(61)

Third, there is likewise no longer a distinction between Regions II and IV in Fig. 9. In either case, the payoff to the focal individual is $V - Vc(X)$ with probability $p_{\text {o}}(X-Y)$ and $-Vc(X) - V\omega (X,Y)$ with probability $p_{\text {i}}(Y-X)$ for an owner and $\alpha V - Vc(X)$ with probability $p_{\text {i}}(X-Y)$ and $-Vc(X) - V\omega (X,Y)$ with probability $p_{\text {o}}(Y-X)$ for an intruder, so that (34), (36), (40) and (42) reduce to

$$\begin{aligned}&f^{\text {o}}_{\text {II}}(X,Y) = f^{\text {o}}_{\text {IV}}(X,Y) = p_{\text {o}}(X-Y) - \omega (X,Y)p_{\text {i}}(Y-X) - c(X). \end{aligned}$$

(62)

$$\begin{aligned}&f^{\text {i}}_{\text {II}}(X,Y) = f^{\text {i}}_{\text {IV}}(X,Y) = \alpha p_{\text {i}}(X-Y) - \omega (X,Y)p_{\text {o}}(Y-X) - c(X). \end{aligned}$$

(63)

Expressions (38), (43), (44), (45a) and (47a) all remain valid, reducing (45b) and (47b) to

$$\begin{aligned} \frac{\partial f}{\partial u_1} = \tfrac{1}{2}\,g(u_1)\int \limits _0^1 \bigl \{\omega (u_1,y)-\delta \bigr \}p_{\text {i}}(y-u_1)g(y)\,\mathrm{d}y \end{aligned}$$

(64)

and

$$\begin{aligned} \frac{\partial f}{\partial u_2} = \tfrac{1}{2}\,g(u_2)\int \limits _0^1 \bigl \{\omega (u_2,y)-\delta \bigr \}p_{\text {o}}(y-u_2)g(y)\,\mathrm{d}y \end{aligned}$$

(65)

respectively. Because $\omega (1,y) = 0$ by (24) and $\delta > 0$, it follows from (64) and (65) that $\frac{\partial f}{\partial u_j}$ is negative for $u_j = 1$ for both $j = 1$ and $j = 2$, and hence from (52) that we cannot have either $v_1 = 1$ or $v_2 = 1$ at the ESS. Thus the strongest losers should always refrain from releasing the chemical to guarantee their escape, regardless of whether they are owners or intruders.

From (11), (49), (51), (24) and (64), it now follows that $v_1 = 0$ at the ESS for $\delta > \delta ^{\text {o}}_c$ and that $v_1 = \psi _1(\delta )$ at the ESS for $\delta < \delta ^{\text {o}}_c$, where

$$\begin{aligned} \delta ^{\text {o}}_c = \frac{\int _0^1 \omega (0,y)p_{\text {i}}(y)\,\mathrm{d}y}{\int _0^1 p_{\text {i}}(y)\,\mathrm{d}y}, \end{aligned}$$

(66)

which reduces to the expression given in (26) on substitution from (24), and $v_1 = \psi _1(\delta )$ is the only root of (27). Note that $\delta ^{\text {o}}_c$ decreases with $\beta$, with $\delta ^{\text {o}}_c \rightarrow \eta$ as $\beta \rightarrow 0$ and $\delta ^{\text {o}}_c \rightarrow 0$ as $\beta \rightarrow 1$ as illustrated in Fig. 8. Likewise, $v_2 = 0$ at the ESS for $\delta > \delta ^{\text {i}}_c$ and $v_2 = \psi _2(\delta )$ at the ESS for $\delta < \delta ^{\text {i}}_c$, where

$$\begin{aligned} \delta ^{\text {i}}_c = \frac{\int _0^1 \omega (0,y)p_{\text {o}}(y)\,\mathrm{d}y}{\int _0^1 p_{\text {o}}(y)\,\mathrm{d}y}, \end{aligned}$$

(67)

which reduces to the expression given in (26) on substitution from (24), and $v_2 = \psi _2(\delta )$ is the only root of (28). It is found that (50) always holds for either $j = 1$ or $j = 2$, for reasons analogous to those identified in the discussion immediately below (57) in Appendix 3. Note that the ESS for Model B depends on neither the cost of fighting nor the intruder premium. On the contrary, it merely reflects a balance between the relative costs of emission and withdrawal.

Under this alternative scenario, the contest cannot be won by a first-round loser: $p_{\text {LW}} = 0$ in place of (19). In place of (20), the probability that the volatile chemical is released, by either contestant, becomes

$$\begin{aligned} p_{\text {VC}} = \int \limits _0^{v_1^*}\int \limits _0^1 p_{\text {i}}(y-x)\,\mathrm{d}y\,\mathrm{d}x + \int \limits _0^{v_2^*}\int \limits _0^1 p_{\text {o}}(y-x)\,\mathrm{d}y\,\mathrm{d}x \end{aligned}$$

(68)

and, in place of (21), the probability $P_{\text {o}}$ that the prior owner wins the contest is simply the integral of $p_{\text {o}}(x-y)$ over the whole of the sample space in Fig. 9.

Appendix 5: Some Analytical Results for Model A

In two special cases we have found purely analytical expressions for the ESS. Both obtain when $\mu = 0$ (no owner advantage, hence no distinction between intermediate and final possession), $r = 1$ (moderate reliability of RHP) and $\alpha = 1$ (no intruder premium). Because of the symmetry implied by $\mu = 0$ with $\alpha = 1$, we use (u, v) in place of both $(u_1,v_2)$ and $(u_2,v_1)$ in (44).

1.1 Appendix 5.1: Analytical Results for Constant Costs

In the first of the two special cases, costs are constant, so that substitution from (12) and (30) into (33)–(43) reduces (44) to

$$\begin{aligned} f(u,v)&\, = \tfrac{1}{12}\{1-\theta _l\}u^3 - \tfrac{1}{16}\{2(1 + 2\gamma ) - 3\theta _l - \theta _w\}u^2 - \tfrac{1}{24}\{2(2 - 9\gamma ) + 5\theta _w\}u\nonumber \\&+ \tfrac{1}{48}\{24(1 - 4\gamma ) + 2(2\{2 + 9\gamma \} + 5\theta _w)v \nonumber \\&+ 3(2\{1 - 2\gamma \} - 3\theta _l - \theta _w)v^2 - 4(1-\theta _l)v^3\} \end{aligned}$$

(69)

with $\gamma \le 1$ and $\theta _l \le \theta _w < 1$, so that

$$\begin{aligned} \tfrac{\partial f}{\partial u} = \tfrac{1}{4}\{1-\theta _l\}u^2 - \tfrac{1}{8}\{2(1 + 2\gamma ) - 3\theta _l - \theta _w\}u - \tfrac{1}{24}\{2(2 - 9\gamma ) + 5\theta _w\}. \end{aligned}$$

(70)

Let us define $\underline{\gamma }_{\text { c}}$ and $\overline{\gamma }_{\text { c}}$ by

$$\begin{aligned} \underline{\gamma }_{\text { c}} = \tfrac{2}{9} + \tfrac{5}{18}\theta _w, \quad \overline{\gamma }_{\text { c}} = \tfrac{2}{3} - \tfrac{1}{2}\theta _l + \tfrac{1}{3}\theta _w. \end{aligned}$$

(71)

Then it is readily shown that (70) is negative on [0, 1] for $\gamma < \underline{\gamma }_{\text { c}}$ and positive on [0, 1] for $\gamma > \overline{\gamma }_{\text { c}}$; whereas, for $\underline{\gamma }_{\text { c}}< \gamma < \overline{\gamma }_{\text { c}}$, (70) is positive at $u = 0$, negative at $u = 1$ and approaches $\infty$ as $u \rightarrow \infty$. The partial derivative therefore changes sign from positive to negative precisely once on $(\underline{\gamma }_{\text { c}}, \overline{\gamma }_{\text { c}})$ where $\tfrac{\partial f}{\partial u} = 0$, that is, at $u = \phi (\gamma ),$ where we define

$$\begin{aligned} \phi (\gamma ) = \frac{12(\gamma -\underline{\gamma }_{\text { c}})}{4\gamma - 3\theta _l - \theta _w + 2 + \sqrt{(4\gamma - 3\theta _l - \theta _w + 2)^2 + 48(1-\theta _l )(\underline{\gamma }_{\text { c}}-\gamma )}}. \end{aligned}$$

(72)

Thus f has a unique maximum on [0, 1] at $u = v^*$, where

$$\begin{aligned} v^* = {\left\{ \begin{array}{ll} 0&{}\text {if}\; 0< \gamma \le \underline{\gamma }_{\text { c}}\\ \phi (\gamma )&{}\text {if}\; \underline{\gamma }_{\text { c}}< \gamma \le \overline{\gamma }_{\text { c}}\\ 1&{}\text {if}\; \overline{\gamma }_{\text { c}} < \gamma \le 1. \end{array}\right. } \end{aligned}$$

(73)

Because $v^*$ is the best reply to any v, it must also be the best reply to itself. So $v^*$ is the unique strong ESS, and it is plotted in Fig. 7a, c for specific values of $\theta _l$ and $\theta _w$. We note in passing that the above results agree with the more general analysis in Sect. 3: by (12) and (14) with $\mu = 0$ and $r = 1 = \alpha$, (54) and (55) reduce to (71), while (23) reduces to $6(1-\theta _l)v^2 - 3(4\gamma - 3\theta _l - \theta _w + 2)v - 4 - 5\theta _w + 18\gamma = 0$ (with $v_1 = v_2 = v$), whose only solution between 0 and 1 is $v = \phi (\gamma )$. At this ESS, the overall probability that the contest is won by a first-round loser is

$$\begin{aligned} \begin{aligned} p_{LW}&= 2\int \limits _{v^*(\gamma )}^1\int \limits _0^1 p(y-x)p(x_l - y_w)\,\mathrm{d}y\,\mathrm{d}x\\ {}&= {\left\{ \begin{array}{ll}\frac{5}{12} + \frac{1}{24}(7\theta _w - 5\theta _l)&{}\text {if}\; 0< \gamma \le \underline{\gamma }_{\text { c}}\\ \frac{1}{24}\{1-\phi (\gamma )\}\{(10 - 5\theta _l + 7\theta _w +(2 - 5\theta _l - 3\theta _w)\phi (\gamma )-4(1-\theta _l)\phi (\gamma )^2\}&{}\text {if}\; \underline{\gamma }_{\text { c}}< \gamma \le \overline{\gamma }_{\text { c}}\\ 0&{}\text {if} \overline{\gamma }_{\text { c}} < \gamma \le 1 \end{array}\right. } \end{aligned} \end{aligned}$$

(74a)

and the probability that the volatile chemical is released is

$$\begin{aligned} \begin{aligned} p_{\text {VC}}&= 2\int \limits _{v^*(\gamma )}^1\int \limits _0^1 p(y-x)\,\mathrm{d}y\,\mathrm{d}x\\ {}&= {\left\{ \begin{array}{ll}1&{}\text {if}\; 0< \gamma \le \underline{\gamma }_{\text { c}}\\ \frac{1}{2}\{1-\phi (\gamma )\}\{2 -\phi (\gamma )\}&{}\text {if}\; \underline{\gamma }_{\text { c}}< \gamma \le \overline{\gamma }_{\text { c}}\\ 0&{}\text {if}\; \overline{\gamma }_{\text { c}} < \gamma \le 1 \end{array}\right. } \end{aligned}\end{aligned}$$

(74b)

after substitution from (30) and (73) into (19) and (20), respectively, with $p_L = p_{\text {o}} = p_{\text {i}} = p$. Both probabilities are plotted in Fig. 7b, d for specific values of $\theta _l$ and $\theta _w$.

1.2 Appendix 5.2: Analytical Results for Linear Differential Costs

In the second of the two special cases, costs are differential but linear, that is, $k = 1$ in (13). In place of (69) we obtain

$$\begin{aligned} f(u,v)&\, = \tfrac{1}{12}\{1-\theta _l\}(1+2\gamma )u^3 - \tfrac{1}{16}\{2 - 3\theta _l - \theta _w + 2 (5 - 3\theta _l)\gamma \}u^2 \nonumber \\&- \tfrac{1}{24}\{2(2 - 9\gamma ) + 5\theta _w\}u\nonumber \\&+ \tfrac{1}{48}\{24(1 - 2\gamma ) - 2(5\theta _l + 7\theta _w)\gamma + 2(4 + 5\theta _w)(1+2\gamma )v\nonumber \\&+ 3(2 - 3\theta _l - \theta _w - 2\{1 + \theta _w\}\gamma )v^2 - 4(1-\theta _l)v^3\} \end{aligned}$$

(75)

(again with $\gamma \le 1$ and $\theta _l \le \theta _w < 1$) after substitution from (13) and (30) into (33)–(44), so that

$$\begin{aligned} \tfrac{\partial f}{\partial u} = \tfrac{1}{4}\{1-\theta _l\}(1+2\gamma )u^2 - \tfrac{1}{8}\{2 - 3\theta _l - \theta _w + 2 (5 - 3\theta _l)\gamma \}u - \tfrac{1}{24}\{2(2 - 9\gamma ) + 5\theta _w\} \end{aligned}$$

(76)

in place of (70). We retain the definition of $\underline{\gamma }_{\text { c}}$ in (71), but redefine $\overline{\gamma }_{\text { c}}$ as

$$\begin{aligned} \overline{\gamma }_{\text {c}} = \frac{2(2+\theta _w)-3\theta _l}{6\theta _l}. \end{aligned}$$

(77)

Then, proceeding as in Appendix 5.1, we find that the ESS is still given by (73) and that (74) still holds, but with

$$\begin{aligned} \phi (\gamma ) = \frac{12(\gamma -\underline{\gamma }_{\text { c}})}{2(5-3\theta _l)\gamma - 3\theta _l - \theta _w + 2 + \sqrt{\{2(5-3\theta _l)\gamma - 3\theta _l - \theta _w + 2\}^2 + 48(1-\theta _l)(1+2\gamma )(\underline{\gamma }_{\text { c}}-\gamma )}}. \end{aligned}$$

(78)

in place of (72). Moreover, after differentiation of (76) and substitution from (78),

$$\begin{aligned} \frac{\partial ^2 f}{\partial u^2}\Bigr |_{u = v = \phi (\gamma )} = -\tfrac{1}{8}\sqrt{\{2(5-3\theta _l)\gamma - 3\theta _l - \theta _w + 2\}^2 + 48(1-\theta _l)(1+2\gamma )(\underline{\gamma }_{\text { c}}-\gamma )} \end{aligned}$$

(79)

confirms that (50) holds. Note, however, that $\overline{\gamma }_{\text { c}} < 1$ will be satisfied only if $\theta _l$ exceeds $\frac{2}{9}(1+\theta _w)$ and hence in particular exceeds $\frac{4}{9}$, an improbably large value. Thus we expect that the chemical will invariably be released by the strongest losers. In the case where there is no toxicity ($\theta _l = 0 = \theta _w$), the ESS is plotted in Fig. 3a as the lowest curve, and the corresponding probability $p_{LW}$ that the contest is won by a first-round loser and probability $p_{VC}$ that the chemical is released are plotted in Fig. 3b as the uppermost solid and dashed curves, respectively. This diagram illustrates that when costs are differential (as opposed to constant) and there is no toxicity, the probability that a first-round loser wins the contest never falls to zero, because the RHP threshold for release of the volatile chemical is exceeded by the RHPs of the strongest first-round losers.

Appendix 6: Some Analytical Results for Model B

When $r = 1$ and $\mu = 0$ as in Fig. 8, so that $p_{\text {o}} = p_{\text {i}} = p$ and $v_1^* = v_2^* = v^*$, (30) reduces (26) to $\delta ^{\text {o}}_c = \delta ^{\text {i}}_c = \delta _c$ where

$$\begin{aligned} \delta _c = \frac{2}{3}\Bigl (\frac{1}{1+\beta } + \frac{1}{2+\beta }\Bigr )\eta , \end{aligned}$$

(80)

while (30) reduces (27) and (28) to $2\eta (1-v)\{(2+\beta )(1-v) + 1 + \beta \} = \delta (1+\beta )(2+\beta )(3-2v)$ or $2(1-v)\{(2+\beta )(1-v) + 1 + \beta \} = \hat{\delta }(1+\beta )(2+\beta )(3-2v)$, where $\hat{\delta } = \delta /\eta$. The solution of this equation is $v = \hat{\psi }(\hat{\delta })$, where

$$\begin{aligned} \hat{\psi }(\hat{\delta }) = \frac{3(1+\beta )(2+\beta )(\hat{\delta }_c - \hat{\delta })}{5+3\beta - (1+\beta )(2+\beta )\hat{\delta } + \sqrt{(1+\beta )\{(5+\beta )\beta ^2\hat{\delta }^2 + (1+2\hat{\delta })^2 + \beta (1+2\hat{\delta } + 8\hat{\delta }^2)\}}}, \end{aligned}$$

(81)

so that the ESS becomes

$$\begin{aligned} v^* = {\left\{ \begin{array}{ll} \hat{\psi }(\hat{\delta })&{}\text {if}\; 0 \le \hat{\delta } \le \hat{\delta }_c\\ 0&{}\text {if}\; \hat{\delta }_c < \hat{\delta } \le 1. \end{array}\right. } \end{aligned}$$

(82)

It is plotted in Fig. 8a for specific values of $\beta$. At this ESS, the probability that the volatile chemical is released is

$$\begin{aligned} \begin{aligned} p_{\text {VC}}&= 2\int \limits _0^{v^*(\hat{\delta })}\int \limits _0^1 p(y-x)\,\mathrm{d}y\,\mathrm{d}x\\ {}&= {\left\{ \begin{array}{ll} \frac{1}{2}\hat{\psi }(\hat{\delta })\{3 -\hat{\psi }(\hat{\delta })\}&{}\text {if}\; 0 \le \hat{\delta } \le \hat{\delta }_c\\ 0&{}\text {if}\; \hat{\delta }_c < \hat{\delta } \le 1 \end{array}\right. } \end{aligned} \end{aligned}$$

(83)

by (68), and is plotted in Fig. 8b for specific values of $\beta$.