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Models for alarm call behaviour

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Abstract

The evolution of alarm call behaviour under individual selection is studied. Four mathematical models of increasing complexity are proposed and analysed. Theoretical conditions for the evolution of “selfish”, “mutualistic”, “altruistic” or “spiteful” alarm calls are established. The models indicate that the hypotheses of benefits of retaining group members or avoiding group detection are not sufficient to explain the evolution of alarm call behaviour, but serve as a complementary factor to facilitate its evolution in most cases. It is hypothesized that the evolution of alarm calls between non-kin should evolve probably when calls are mutualistic, mildly altruistic and there are beneficial group size effects against predation.

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Acknowledgements

Luciana Mafalda Elias de Assis acknowledges the support of UNEMAT. Raul Abreu de Assis acknowledges the support of CNPQ, grant 203654/2014-5 and the support of FAPEMAT, Edital Universal 005-2015, process number, 204271/2015. Moiseis Cecconello acknowledges the support of CNPQ, grant 200446/2015-0 and Ezio Venturino acknowledges the partial support of the program “Metodi numerici nelle scienze applicate” of the Dipartimento di Matematica “Giuseppe Peano”.

The authors thank the referees for their contributions for the improvement of this paper.

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Appendix A: Proofs of propositions

Appendix A: Proofs of propositions

In this appendix, we present proofs of propositions presented in the paper.

Proof

(Proposition 1): For the first case, it turns out that

$$\frac{dx}{dt}=\frac{x}{x+y}\left[(1-a)x+(1-b)y\right]<(1-p)x<0$$

in which p = min{a,b} > 1. On the otherhand, if both parameters are greater than 1, then we have then

$$\frac{dx}{dt}=\frac{x}{x+y}\left[(1-a)x+(1-b)y\right]>(1-p)x>0$$

in which p = max{a,b} < 1. □

Proof

(Proposition 2): In fact, since z(t) ∈ [0, 1]for all t ≥ 0,thefollowing inequality must be true

$$0<\frac{b-d}{c-a+b-d}<1.$$

Thus, clearly,we must have bd > 0and ca + bd > 0or bd < 0and ca + bd < 0. In the first case, itturns out that bd < ca + bd whichimplies that c > a. In thesecond, we have that bd > ca + bd.Hence a > c and the conclusion follows directly from this inequality.

The eigenvalues for this point are given by

$$\lambda_{\bar{z}_{3}} =\frac{\left( c - a\right)\, \left( b - d\right)}{c - a + b-d}. $$

Thus, in the first casethe, the numerator of \(\lambda _{\bar {z}_{3}}\)is positive whereas the denominator is negative. Therefore, the equilibrium point\(\bar {z}_{3}\)is feasible and stable onlywhen condition (i) is satisfied. □

Proof

(Proposition 3): As we saw previously, b > d implies that \(\bar {z}_{1}\)is an asymptoticallystable equilibrium point. Thus, if a > c then \(\bar {z}_{3}\)is no longerfeasible and z(t)decreases to zero as t for all z o ∈ (0, 1). On theother hand, when a < c then \(\bar {z}_{3}\)is unstableand z(t)decreasesto zero as t for all \(z_{o}\in (0,\bar {z}_{3})\).Writing x (t)and y (t)interms of z(t)we have:

$$\frac{dx}{dt}=x\left[1-az(t)-b(1-z(t))\right]$$
$$\frac{dy}{dt}=y\left[1-cz(t)-d(1-z(t))\right].$$

Since 1 − z(t) < 1and z(t)is amonotonic function,

$$\frac{dx}{dt}>x\left[1-az(T)-b\right]\qquad\frac{dy}{dt}>y\left[1-cz(T)-d\right] $$

for everyfixed T > 0. If d < b < 1thenthere is T > 0such that [1 − a z(T) − b] = α 1 > 0and [1 − c z(T) − d] = α 2 > 0forall t > T andtherefore x(t),y(t) → as t.

On the other hand,

$$\frac{dx}{dt}<x\left[1-b(1-z(T))\right]\qquad \frac{dy}{dt}<y\left[1-(1-d)z(T)\right]$$

for everyfixed T > 0. If 1 < d < b thenthere is T > 0such that [1 − a z(T) − b] = α 1 < 0and [1 − c z(T) − d] = α 2 < 0forall t > T andtherefore x(t),y(t) → 0as t.

Finally, when b > 1 > d,using theprevious inequalities for x (t)and y (t),thereare α 1 < 0and α 2 > 0such that x (t) < α 1 x and y (t) > α 2 y for all t > T for some T > 0. Thus, x(t) → 0and x(t) → as t and the propositionis proved. □

Proposition 4 can be proved using an analogous reasoning.

Proof

(Proposition 5): First, since a > c and d > b,\(\bar {z}_{3}\)is asymptoticallystable and \(z(t)\to \bar {z}_{3}\)as t for all z o ∈ (0, 1). As aconsequence, x(t) → if and only if y(t) → as t. Further, x(t) → 0if andonly if y(t) → 0as t. As we statedbefore, the line y = α x, α = (ac)/(db), defines an invariant set S and an initial condition (x o ,y o )belongs to S if and only if \(z_{o}=x_{o}/(x_{o}+y_{o})=\bar {z}_{3}\).In order to prove the statement, it is enough to analyse the signal of x (t)for initial conditions onthe set S. Now, for every (x,y) ∈ S,we have

$$\frac{dx}{dt}=\frac{1}{1+\alpha}\frac{a-c+b-d+bc-ad}{d-b}. $$

Therefore, since db > 0,it turns out that x (t) > 0if and only if ac + bd > a db c and the statementis proved. □

Proof

(Proposition 6): By Eqs. 12and 13, \(\bar {y}=(a-c)/(d-b)\bar {x}\).Assuming P 4feasiblewe must have (ac)/(db) > 0.That is, if a > c then d > d and when a < c we must have d < b. This proves the first part in both items. At P 4, the eigenvalues of the Jacobian matrix are givenby

$$\lambda_{1}=-\frac{\left( a - c\right) \left( d-b\right)}{a - c+d - b} \quad \lambda_{2}=a -1- \frac{\left( a - b\right)\, \left( a - c\right)}{a - c+d - b}.$$

Further, we havethat

$$\lambda_{1}\lambda_{2}=\frac{\left( a - c\right) \left( d - b\right)\left( a - c+ d- b - a d + b c\right)}{{\left( a - c + d- b \right)}^{2}}.$$

Since P 4is feasible we have that λ 1is always negative.Furthermore, when a > c and d > b Eq. 12and Eq. 13imply that ac + dba d + b c must be positive. Therefore,in this case, we have that λ 1 λ 2 > 0which implies that λ 2is also negative and this proves the first item of the statement. In a similar way, we can also prove the second claim. □

Proof

(Proposition 7): According to Proposition 6, P 4is not feasible for this configuration of parameters. The stability of P 2follows directly from the expression of the eigenvalues of the Jacobian matrix at P 2,which are λ 1 = db and λ 2 = d − 1. □

Proposition 8 is proved in an analogous way.

Proof

(Proposition 9): Since f (x) = −1 + (a/g) e x/gand f (x) = −(a/g 2) e x/git turns out that f(x)has a maximum at x m = g ln(a/g).Now, assuming a < 1,since f is concave, f(0) > 0and f(1) < 0,there is only one \(\bar {x}\)such that \(f(\bar {x})=0\).Hence, in this case, there is only an equilibrium point on the x-axis. It is not difficult to check that\(\bar {x}>x_{m}\).

Now, if c > a then Eqs. 15and 16imply that λ 1 < 1 − xa e x/g = 0.On the other hand, we have that λ 2 = x f (x)and since \(f^{\prime }(\bar {x})<f^{\prime }(x_{m})=0\)it follows that λ 2is also negative at the equilibrium point. Therefore, P 1is stable. In case of c < a,we have that λ 1 > 0which means that P 1is unstable. □

Proof

(Proposition 10): As before, the maximum value of f(x)is given by f(g ln(a/g)) = 1 − g ln(a/g) − g.Thus, since f is concave, f(0) < 0and f(1) < 0,the conclusion about the existence of equilibrium points depends on whether the maximum value of f(x)is negative, zero or positive.

Let us assume that P 1 = (x 1, 0)and P 2 = (x 2, 0)are feasible. By Eq. 16, λ 1is positive when a > c which implies that both equilibrium points are unstable. Assuming c > a we have that λ 1 < 0.Now, we can easily check that x 1 < x m < x 2in which x m is the maximum point of f(x).Since f (x)is a decreasing function it turns out that f (x 2) < 0 < f (x 1).Thus, λ 2 = x f (x)is negative at x 2and positive at x 1.Therefore, if c > a,then P 1is unstable and P 2is stable. □

Propositions 11 and 12 are also proved in a similar fashion.

Proof

(Proposition 13): Defining the function f(x)by

$$\begin{array}{@{}rcl@{}} f(x)=1-(1+\alpha)x-\beta e^{-(1+\alpha)x/g} \end{array} $$
(26)

it turns out that f (x) = −(1 + α) + (a + α b)g −1 e −(1 + α)x/gand f (x) = −(1 + α) (a + α b)g −1 e −(1 + α)x/gand thus f(x)is a concavefunction. Assuming a − 1 < α(1 − b),we have that f(0) > 0and f(1) < 0. Thus, since f(x)is concave, thereis only one point \(\bar {x}\)such that \(f(\bar {x})=0\)and,therefore \(P=(\bar {x},\alpha \bar {x})\)is an equilibrium point for Eq. 8.

Now, comparing Eqs. 18and 21, we can conclude that λ 1 < 0if and only if d > b. On the other hand,using Eq. 18again, we have that \(\lambda _{2}=\bar {x}f^{\prime }(\bar {x})\).Since \(\bar {x}\)must be greaterthan the point x m where f(x)reaches its maximumvalue, we have that \(f^{\prime }(\bar {x})<0\)and so λ 2 < 0. Now, by thefeasibility condition, \(P=(\bar {x},\bar {y})\)is stable when a > c and d > b andunstable when a < c and d < b. □

Proof

(Proposition 14): Defining f(x)as in Eq. 26, the function f reaches its maximum value at

$$\begin{array}{@{}rcl@{}} x_{m}=\frac{g}{1+\alpha}\ln\left( \beta g^{-1}\right) \end{array} $$
(27)

which is f(x m ) = 1 − g ln (β g −1) − g.Since f(0) < 0, f(1) < 0and f(x)is concave, the existence ofequilibrium points relies on having f(x m ) < 0, f(x m ) = 0or f(x m ) > 0. Thus, when f(x m ) > 0there arepoints \(\bar {x}_{1}<x_{m}<\bar {x}_{2}\)suchthat \(f(\bar {x}_{1})=f(\bar {x}_{2})=0\)and,therefore, \(P_{1}=(\bar {x}_{1},\alpha \bar {x}_{1})\)and \(P_{2}=(\bar {x}_{2},\alpha \bar {x}_{2})\)areequilibria.

By the concavity of f(x),we have \(f^{\prime }(\bar {x}_{1})>0\)and\(f^{\prime }(\bar {x}_{2})<0\). Using the same argument as inthe last proposition, we have that \(\lambda _{2}=\bar {x}_{1}f^{\prime }(\bar {x}_{1})>0\)and \(\lambda _{2}=\bar {x}_{2}f^{\prime }(\bar {x}_{2})<0\). This implies P 1is unstable. On the other hand,at both equilibria we have λ 1 < 0when d > b and λ 1 > 0when d < b. Now, using the feasibilityconditions for points on the line y = α x we can conclude that P 2is stable when a > c and d > b andunstable when a < c and d < b. □

Proof

(Proposition 15): As a,b > 1, a + α b > 1 + α which in turn implies that β > 1.The expression g[ln(p g −1) + 1] , p = max{a,d,β},converges monotonically to zero as g → 0.We also have that g[ln(p g −1) + 1] > 1when g = p.So, there is a \(\bar {g}\)such that g[ln(p g −1) + 1] < 1for each \(g<\bar {g}\).Therefore, for a g satisfying \(g<\bar {g},\)the hypotheses of the third item in Propositions 10, 12 and 14 are satisfied and Eq. 8has seven equilibrium points: P 1 = (0, 0); P 2 = (x 1, 0)and P 3 = (x 2, 0), x 1 < x 2; P 4 = (0,y 1)and P 5 = (0,y 2), y 1 < y 2; P 6 = (w 1,α w 1)and P 7 = (w 2,α w 2), 0 < w 1 < w 2.By the hypotheses, P 1and P 7are asymptotically stable.

We are going to prove that there is a heteroclinic orbit connecting P 2and P 6.In order to prove it, let (x o ,y o )be a initial condition on the (1-dimensional) stable manifold of P 6,below the line y = α x.Since it is bounded, the solution generated by this initial condition must converge to the equilibrium point as t →−.In this case, this solution cannot converge neither to P 1nor to P 7since these equilibrium points have empty unstable manifolds. Thus, the solution must converge either to P 2or P 3as t →−.

Suppose that it converges to P 3,that is, there is an heteroclinic orbit connecting P 6to P 3.Now, consider the set U enclosed by the line y = α x, xw 1,the x-axis from x ≥and the heteroclinic orbit connecting P 2to P 6.All solutions with initial condition on the interior of U must converge to P 7as t because the stable manifolds of P 6and P 3are not in the interior of U. The unstable manifold of P 6is the line segment connecting P 6to P 7and, therefore, it does not belong to the interior of U. Thus, solutions starting on U cannot converge to P 6as t →−.In this way, since all solutions are bounded, they would have to converge to P 3as t →− and, therefore, the unstable manifold of P 3would have dimension 2. But this is a contradiction that comes from considering an heteroclinic orbit connecting P 3to P 6.Therefore, the solution through a initial condition (x o ,y o )on the stable manifold of P 6must converge to P 2as t →− which implies that there is a heteroclinic orbit connecting P 2to P 6.Similarly, we can conclude that there is a heteroclinic orbit connecting P 4to P 6.

Now, consider the set Q bounded by the line segments connecting P 1to P 2and P 1to P 4as well as the heteroclinic orbits connecting P 2to P 6and P 4to P 6.Of course, this set is invariant and contains only the stable equilibrium P 1.Thus, all solutions starting on Q converge to P 1as t.Likewise, all solutions starting on the interior of the complement of A converge to P 7as t.Finally, to prove the statement, it is enough to define n 1and n 2as the minimum and maximum value of x + y,respectively, on the orbits connecting P 2to P 6and P 4to P 6. □

Proposition 16 is proved in a way analogous to Proposition 14.

Proof

(Proposition 17): The existence of nonzero equilibria relies on the existence of solutions for Eq. 24. Now, defining u = x + y and v = xy we can rewrite it as

$$\begin{array}{@{}rcl@{}} v=\frac{2u}{\gamma}\left( 1-\frac{\gamma}{2}-u\right), \qquad v=u\left( \frac{e^{-u/g}-\alpha}{e^{-u/g}+\alpha}\right) \end{array} $$
(28)

where γ = a + α b.Since we are looking for positive solutions, comparing these two equations it turns out that u must satisfy:

$$\begin{array}{@{}rcl@{}} s(u)=\gamma - (1-u)-\alpha(1-u)e^{u/g}=0 \end{array} $$
(29)

and by hypothesis we have that s(0) = γ − (1 + α) > 0and s(1) = γ > 0. Now,define f(u) = γ − 1 + u and h(u) = α(1 − u)e u/gso that s(u) = f(u) − h(u). Thus, the equilibriumsolution is given by f(u) = h(u).Taking g ≥ 1we can easilycheck that s (u) > 0for all u ∈ [0, 1]so that Eq. 29has no solution.On the other hand, we have that h (u) = (α/g)(1 − g + u)e u/gwhich implies that h(u)is an increasingfunction on the interval (0, 1 − g)anddecreasing on the interval (1 − g, 1).Furthermore, we can verify that \(h(\sqrt {g})\to \infty \)as g → 0so that\(s(\sqrt {g})<0\)for a fixed gg 1, with g 1arbitrarily small.Since s(u)is continuousthere must exist a \(u_{1}\in (0,\sqrt {g}),\)depending on g,such that s(u 1) = 0.Since s(u) > s(u 1)for 0 < u < u 1and s(u) < s(u 1)for u ∈ (u 1,g), it turnsout that s (u 1) < 0.Similarly, h(1 − g) → as g → 0and so, fora fixed gg 2, with g 2small, theremust be a u 2 ∈ (1 − g, 1),depending on g,such that s(u 2) = 0andat this point s (u 2) > 0.This proves the existence of equilibrium points as stated in the proposition.

Now, let F(x,y)and G(x,y)be the right-handsides of Eq. 9. Since x = 0.5(u + v)and y = 0.5(uv),usingthe first equality of Eq. 28, the partial derivatives of F and G can be written in terms of u. At the equilibrium points u i , we have\(e^{u_{i}/g}= (ad - bc - (b - d)(u_{i} - 1))/((a - c)(1 - u_{i}))=S\). Using these variables, the determinantof the Jacobian matrix J at P i = (x i ,y i )is given by

$$\begin{array}{@{}rcl@{}} \det(J_{P_{i}})(u_{i})=\frac{(a - c)u_{i}(u_{i} - 1)^{2}s^{\prime}(u_{i})}{g(ad - bc)^{2}} \end{array} $$
(30)

whereas the trace of J at P i = (x i ,y i )is given by

$$\begin{array}{@{}rcl@{}} {\text{tr}}(J_{P_{i}})(u_{i})=\frac{-g(ad - bc)p(u_{i})-(a - c)u_{i}(u_{i} - 1)q(u_{i})}{g(ad-bc)^{2}} \end{array} $$
(31)

with

$$p(u)=(a-c)(d-b)+ u\left[b(a-c)+c(d-b)\right], $$
$$q(u)=(u-1)(b-d)^{2} + d(ad - bc). $$

Assuming a > c,and thus d > b, we find\(\det (J_{P_{1}})<0\)which implies that P 1is an unstable equilibriumpoint. In this case \(\det (J_{P_{2}})>0\), sothat for the stability of P 2weconsider the value of \({\text {tr}}(J_{P_{2}})\).As we have shown, u 2 > 1 − g,i.e. u 2 − 1 > −g.Since a db c and p(u)arepositive, we have

$${\text{tr}}(J_{P_{2}})(u_{2})<\frac{(u_{2}-1)(ad - bc)p(u_{2})-(a - c)u_{2}(u_{2} - 1)q(u_{2})}{g(ad-bc)^{2}}. $$

Rewriting theright-hand side of last inequality, we come up with

$${\text{tr}}(J_{P_{2}})(u_{2})<\frac{(1-u_{2})(d-b)Q(u_{2})}{g(ad-bc)^{2}} $$

inwhich

$$Q(u)=\eta u^{2}+\left[(ad-bc)(a-2c)-\eta\right]u-(a-c)(ad-bc)$$

and η = (ac)(db) > 0. Thus,the function Q(u)isconvex, Q(0) = −(ac)(a db c) < 0and Q(1) = −c(a db c) < 0andso Q(u) < 0for all u ∈ [0, 1]. Therefore,\({\text {tr}}(J_{P_{2}})(u_{2})<0\)which impliesthat P 2isstable.

Now, assuming a < c, andthus d < b,it follows that\(\det (J_{P_{2}})<0\)which implies that P 2is unstable. In thiscase, \(\det (J_{P_{1}})>0\)and we want toprove the instability of P 1.Since \(u_{1}<\sqrt {g}\)it turns out that\(1-u_{1}>1-\sqrt {g}\). For this parameterconfiguration \(-q(u_{1})>-q(\sqrt {q})>0\)and therefore \(-q(u_{1})(c - a)u_{1}(1-u_{1})>-q(\sqrt {g})(c - a)u_{1}(1-\sqrt {g})\)which in turns implies

$${\text{tr}}(J_{P_{1}})(u_{1})>\frac{g(bc - ad)p(u_{1})-q(\sqrt{g})(c - a)(1-\sqrt{g})u_{1}}{g(ad-bc)^{2}}. $$

The right-hand side of lastinequality is a linear function of u,say l(u) = m u + n,whose coefficients are given by

$$m(g)=\frac{(1-\sqrt{g})\left[(1-\sqrt{g}){\eta_{2}^{2}} + d(bc - ad)\right]\eta_{1}}{g(bc-ad)^{2}}-\frac{b\eta_{1} + c\eta_{2}}{bc-ad}$$
$$n=\frac{\eta_{1}\eta_{2}}{bc-ad}$$

in which η 1 = ca > 0 and η 2 = bd > 0. Since l(0) > 0 and m(g) → as g → 0it turns out that there exists a g 3 > 0 such that l(u) ≥ 0 for all \(u\in [0,\sqrt {g}])\)and therefore \({\text {tr}}(J_{P_{1}})(u_{1})>0\)forall gg 3. Thus,taking \(\bar {g}=1\)and\(\hat {g}=\min \{g_{1},g_{2},g_{3}\}\), the statementis proved. □

Proposition 18 can be obtained as a consequence of Proposition 17.

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de Assis, L.M.E., de Assis, R.A., Cecconello, M. et al. Models for alarm call behaviour. Theor Ecol 11, 1–18 (2018). https://doi.org/10.1007/s12080-017-0345-0

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