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01 Jan 2016
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# On the interaction of species capable of explosive growth

###### Accepted: 05 Nov 2020
Journal Details
Format
Journal
eISSN
2444-8656
First Published
01 Jan 2016
Publication timeframe
2 times per year
Languages
English

In the classical Lotka-Volterra population models, the interacting species affect each other's growth rate. We propose an alternative model, in which the species affect each other through the limitation coefficients, rather then through the growth rates. This appears to be more realistic: the presence of foxes is not likely to diminish the fertility of rabbits, but will contribute to limiting rabbit's population. Both the cases of predation and of competition are considered, as well as competition in case of periodic coefficients. Our model becomes linear when one switches to the reciprocals of the variables. In another direction we use a similar idea to derive a multiplicity result for a class of periodic equations.

#### MSC 2010

Introduction

One way of solving the logistic population equation (here x = x(t)) $x′=ax−bx2$ x' = ax - b{x^2} is to divide this equation by x2, and obtain a linear equation for $u=1x$ u = {1 \over x} . Here a > 0 is the growth rate, and b > 0 is the limitation (or self-limitation) coefficient, both given numbers. We wish to explore the interactions of two species with populations x = x(t) and y = y(t) for which the substitution $u=1x$ u = {1 \over x} and $v=1y$ v = {1 \over y} leads to a linear system. The model we consider is $x′=ax+x2(by+e)y′=dy+y2(cx+f) ,$ \matrix{ {x' = ax + {x^2}\left( {{b \over y} + e} \right)} \cr {y' = dy + {y^2}\left( {{c \over x} + f} \right){\kern 1pt} ,} \cr } with constants a,b,c,d,e and f. Dividing the first equation by x2, the second one by y2, and setting $u=1x$ u = {1 \over x} and $v=1y$ v = {1 \over y} , gives a linear system $−u′=au+bv+e−v′=cu+dv+f .$ \matrix{ { - u' = au + bv + e} \cr { - v' = cu + dv + f{\kern 1pt} .} \cr } The signs of the coefficients determine the type of interaction, which will include both predator-prey and competing species cases.

Let us compare (1.2) with the classical Lotka-Volterra predator-prey model $x′=x(a−b y)y′=y(−c+d x) ,$ \matrix{ {x' = x\left( {a - b{\kern 1pt} y} \right)} \cr {y' = y\left( { - c + d{\kern 1pt} x} \right){\kern 1pt} ,} \cr } where the constants a,b,c,d are positive. In (1.4) the species affect each other through the growth rate: the prey, with the number given by x(t), improves the growth rate of the predator, with the number y(t), while the predator decreases the growth rate of the prey. In the model (1.2) the species affect each other through their limitation coefficients. This appears to be more realistic: the presence of foxes is not likely to decrease the fertility of rabbits (new rabbits will be born at the same rate), but will place a limitation on the growth of rabbit population.

Similarly to the Lotka-Volterra model, the proposed model (1.2) predicts oscillatory behavior for predator-prey interaction, and either stable coexistence or competitive exclusion for competing species. Unlike the Lotka-Volterra model, it is possible that the population number of one of the species goes to infinity in finite time, while the number of the other species remains finite and positive. Explosive growth of populations occurs often in nature, in connection with various “outbreaks”, see e.g., D. Ludwig, D.D. Jones and C.S. Holling [4] for a model of spruce budworm. Notice that our analysis leads to some non-standard questions about linear systems. For example, if a solution of (1.3) starts in the first quadrant of the xy-plane, will it stay in the first quadrant for all t?

Using the Floquet theory, we analyze a case of predator-prey interaction with periodic coefficients, and give a condition for the existence of a limit cycle.

In another direction we derive a multiplicity result on the periodic solutions for a class of periodic equations $x′(t)=f(t,x(t)) , with f(t+p,x)=f(t,x) .$ x'(t) = f(t,x(t)){\kern 1pt} ,\;\;\;\;{\kern 1pt} {\rm with}\;f(t + p,x) = f(t,x){\kern 1pt} {\kern 1pt} . The connection to the rest of this paper is both by the method (it uses the same transformation $u=1x$ u = {1 \over x} ), and by ecological applications, particularly to models of harvesting, see e.g., S. Ahmad and A.C. Lazer [1].

Explosive predator-prey model

Consider the model $x′=x2(by−1)y′=−y2(dx−1) .$ \matrix{ {x' = {x^2}\left( {{b \over y} - 1} \right)} \cr {y' = - {y^2}\left( {{d \over x} - 1} \right){\kern 1pt} .} \cr } Here x(t) gives the number of prey, and y(t) the number of predator. If y(t) is small, the prey grows explosively (with x′ behaving like αx2, α > 0). If the number of predators y(t) is large, then x′(t) < 0 and x(t) decreases. The number of predators y(t) decreases when x(t) small, and grows explosively for x(t) large. This model corresponds to (1.2), with a = d = 0. The coefficients e and f have been scaled out.

The system (2.1) has a rest point (d,b). Letting $X=d/bx$ X = {{\sqrt {d/b} } \over x} and $Y=1y$ Y = {1 \over y} transforms (2.1) into a perturbed harmonic oscillator $X′=−bd Y+d/bY′=bd X−1 .$ \matrix{ {X' = - \sqrt {bd} {\kern 1pt} Y + \sqrt {d/b} } \cr {Y' = \sqrt {bd} {\kern 1pt} X - 1{\kern 1pt} .} \cr } Setting $X(t)=ξ(t)+1bd$ X(t) = \xi (t) + {1 \over {\sqrt {bd} }} , $Y(t)=η(t)+1b$ Y(t) = \eta (t) + {1 \over b} leads to a harmonic oscillator $ξ′=−bd ηη′=bd ξ ,$ \matrix{ {\xi ' = - \sqrt {bd} {\kern 1pt} \eta } \cr {\eta ' = \sqrt {bd} {\kern 1pt} \xi {\kern 1pt} ,} \cr } so that the solution of (2.2) is $X(t)=1bd+c1cosbd t−c2sinbd tY(t)=1b+c1sinbd t+c2cosbd t ,$ \matrix{ {X(t) = {1 \over {\sqrt {bd} }} + {c_1}\cos \sqrt {bd} {\kern 1pt} t - {c_2}\sin \sqrt {bd} {\kern 1pt} t} \cr {Y(t) = {1 \over b} + {c_1}\sin \sqrt {bd} {\kern 1pt} t + {c_2}\cos \sqrt {bd} {\kern 1pt} t{\kern 1pt} ,} \cr } which is just a rotation of the point (X(0),Y (0)) around the point $(1bd,1b)$ ({1 \over {\sqrt {bd} }},{1 \over b}) , the rest point of (2.2), on the circle of radius $c12+c22$ \sqrt {c_1^2 + c_2^2} , which depends on the initial conditions. The solution of (2.1) is then $x(t)=d/b1bd+c1cosbd t−c2sinbd ty(t)=11b+c1sinbd t+c2cosbd t .$ \matrix{ {x(t) = {{\sqrt {d/b} } \over {{1 \over {\sqrt {bd} }} + {c_1}\cos \sqrt {bd} {\kern 1pt} t - {c_2}\sin \sqrt {bd} {\kern 1pt} t}}} \cr {y(t) = {1 \over {{1 \over b} + {c_1}\sin \sqrt {bd} {\kern 1pt} t + {c_2}\cos \sqrt {bd} {\kern 1pt} t}}{\kern 1pt} .} \cr } The constants c1 and c2 are determined from the initial values (x(0),y(0)): $c1=d/bx(0)−1bd , and c2=1y(0)−1b .$ {c_1} = {{\sqrt {d/b} } \over {x(0)}} - {1 \over {\sqrt {bd} }}{\kern 1pt} ,\;\;{\kern 1pt} \;{\rm and}{\kern 1pt} \;\;{c_2} = {1 \over {y(0)}} - {1 \over b}{\kern 1pt} . It is now clear that the rest point (d, b) is a center for (2.1), and we can give a complete description of the behavior of positive solutions.

Theorem 2.1

Given the initial point (x(0),y(0)), calculate c1 and c2 by (2.6), and $R=c12+c22$ R = \sqrt {c_1^2 + c_2^2} . If the circle C of radius R around the point ( $(1bd,1b)$ ({1 \over {\sqrt {bd} }},{1 \over b}) ) lies completely inside the first quadrant of the (X,Y) plane, then the corresponding solution (x(t),y(t)) of (2.1) is a closed curve around the rest point (d,b), given by (2.5). Moreover, the period of all these closed curves is the same, and $x(t)>d2$ x(t) > {d \over 2} , $y(t)>b2$ y(t) > {b \over 2} for all t. Assume now that this circle C, traveled counterclockwise beginning with the point $(X(0),Y(0))=(d/bx(0),1y(0))$ (X(0),Y(0)) = ({{\sqrt {d/b} } \over {x(0)}},{1 \over {y(0)}}) , hits one of the axes of the (X,Y) plane. If it hits the Y-axis first, then there is a time T > 0 so that limtT x(t) = ∞, while limtT y(t) is finite and positive. If C hits the X-axis first, then there is a time T > 0 so that limtT y(t) = ∞, while limtT x(t) is finite and positive.

Proof

The change of variables $X=d/bx$ X = {{\sqrt {d/b} } \over x} , $Y=1y$ Y = {1 \over y} , followed by $X(t)=ξ(t)+1bd$ X(t) = \xi (t) + {1 \over {\sqrt {bd} }} , $Y(t)=η(t)+1b$ Y(t) = \eta (t) + {1 \over b} leads to a harmonic oscillator (2.3). It follows that the solutions of the original system (2.1) are given by (2.4), while in the XY-plane solutions move on circles around the rest point $(1bd,1b)$ ({1 \over {\sqrt {bd} }},{1 \over b}) , starting with the point (X(0),Y (0)). If this circle lies completely in the first quadrant of the XY-plane, the corresponding solution of (2.1) is defined for all t > 0, and is periodic. If this circle touches first the Y-axis at some time t = T so that X(T) = 0, then limtT x(t) = ∞, while limtT y(t) is finite and positive. If the X-axis is touched first at some time t = T so that Y (T) = 0, then limtT y(t) = ∞, while limtT x(t) is finite and positive.

It remains to prove the lower bounds for the periodic solutions in the first part of the theorem. From (2.4) one sees that the positivity of X(t) and Y (t) implies that $X(t)<2bd$ X(t) < {2 \over {\sqrt {bd} }} and $Y<2b$ Y < {2 \over b} , from which one gets the lower bounds on x(t) and y(t).

Example

Using Mathematica, we computed four periodic solutions for the system (2.1), with b = 3 and d = 2, surrounding the rest point at (2,3), see Figure 1.

Explosive competing species model

Consider the model $x′=a x+x2(by−1), x(0)>0y′=d y+y2(cx−1), y(0)>0 ,$ \matrix{ {x' = a{\kern 1pt} x + {x^2}\left( {{b \over y} - 1} \right),\;\;x(0) > 0} \cr {y' = d{\kern 1pt} y + {y^2}\left( {{c \over x} - 1} \right),\;\;y(0) > 0{\kern 1pt} ,} \cr } with positive constants a,b,c and d. Each species grows explosively, if the number of the other one is small, while if the competitor's number is large, the growth is logistic-like. Clearly, the interaction is competitive in nature.

We begin with a simple observation: if x(0) > 0 and y(0) > 0, then x(t) > 0 and y(t) > 0 for all t > 0. Indeed, writing the first equation in the form x′ = A(t)x, with $A(t)≡a+x(t)(by(t)−1)$ A(t) \equiv a + x(t)\left( {{b \over {y(t)}} - 1} \right) , and integrating, obtain $x(t)=x(0)e∫0tA(s) ds>0$ x(t) = x(0){e^{\int_0^t A(s){\kern 1pt} ds}} > 0 . Similarly, y(t) > 0 for all t > 0. Hence, we can limit our study of (3.1) to the first quadrant of the (x,y) plane.

Setting $X=1x$ X = {1 \over x} and $Y=1y$ Y = {1 \over y} produces a linear system $X′=−aX−bY+1 , X(0)=1x(0)>0Y′=−cX−dY+1 , Y(0)=1y(0)>0 ,$ \matrix{ {X' = - aX - bY + 1{\kern 1pt} ,\;\;X(0) = {1 \over {x(0)}} > 0} \cr {Y' = - cX - dY + 1{\kern 1pt} ,\;\;Y(0) = {1 \over {y(0)}} > 0{\kern 1pt} ,} \cr } with a unique rest point (X0,Y0) given by $X0=d−bad−bc , Y0=a−cad−bc .$ {X_0} = {{d - b} \over {ad - bc}}{\kern 1pt} ,\;\;{Y_0} = {{a - c} \over {ad - bc}}{\kern 1pt} . Since x(t) > 0 and y(t) > 0 for all t > 0, we may restrict the system (3.2) to the first quadrant of the (X,Y) plane. The rest point (X0,Y0) lies in the first quadrant if either $d>b and a>c (and then ad>bc) ,$ d > b\;{\kern 1pt} \;{\rm and}{\kern 1pt} \;a > c\;{\kern 1pt} \;({\rm and}\;{\rm then}\;ad > bc){\kern 1pt} {\kern 1pt} , or $d d < b\;{\kern 1pt} \;{\rm and}{\kern 1pt} \;a < c\;{\kern 1pt} \;({\rm and}\;{\rm then}\;ad < bc){\kern 1pt} {\kern 1pt} . Letting ξ = XX0 and η = YY0, we translate the rest point to the origin, obtaining the system $ξ′=−aξ−bηη′=−cξ−dη ,$ \matrix{ {\xi ' = - a\xi - b\eta } \cr {\eta ' = - c\xi - d\eta {\kern 1pt} ,} \cr } with the matrix $A=[−a−b−c−d]$ A = \left[ {\matrix{ { - a} & { - b} \cr { - c} & { - d} \cr } } \right] . The eigenvalues of A are $λ1,2=12(−a−d±a2−2ad+4bc+d2) .$ {\lambda _{1,2}} = {1 \over 2}\left( { - a - d \pm \sqrt {{a^2} - 2ad + 4bc + {d^2}} } \right){\kern 1pt} . The corresponding (column) eigenvectors are $ξ1,2=(−−a+d±a2−2ad+4bc+d22c,1)T .$ {\xi _{1,2}} = {\left( { - {{ - a + d \pm \sqrt {{a^2} - 2ad + 4bc + {d^2}} } \over {2c}},1} \right)^T}{\kern 1pt} . In case (3.4) holds, both eigenvalues are negative, and the rest point (X0,Y0) is a stable node, while in case (3.5) holds, one eigenvalue is negative and the other one is positive, so that (X0,Y0) is a saddle.

Theorem 3.1

(i) Assume that the condition (3.5) holds. Then one of the species (depending on the initial conditions) grows explosively. Namely, for any solution of (3.1) there is a time T > 0 so that limtT x(t) = ∞, while limtT y(t) is finite and positive, or the other way around.

(ii) Assume that the condition (3.4) holds. If x(t) and y(t) remain finite for all t > 0 then $limt→∞x(t)=1X0$ \mathop {\lim }\nolimits_{t \to \infty } x(t) = {1 \over {{X_0}}} and $limt→∞y(t)=1Y0$ \mathop {\lim }\nolimits_{t \to \infty } y(t) = {1 \over {{Y_0}}} .

Proof

The change of variables $X=1x$ X = {1 \over x} , $Y=1y$ Y = {1 \over y} transforms (3.1) into the linear system (3.2). The matrix of (3.2) is $A=[−a−b−c−d]$ A = \left[ {\matrix{ { - a} & { - b} \cr { - c} & { - d} \cr } } \right] . The general solution of (3.2) is $(X(t),Y(t))T=(X0,Y0)T+c1eλ1tξ1+c2eλ2tξ2 ,$ {\left( {X(t),Y(t)} \right)^T} = {\left( {{X_0},{Y_0}} \right)^T} + {c_1}{e^{{\lambda _1}t}}{\xi _1} + {c_2}{e^{{\lambda _2}t}}{\xi _2}{\kern 1pt} , where λ1 and λ2 are the eigenvalues of A, given by (3.7), and the corresponding eigenvectors ξ1 and ξ2 are given by (3.8), while X0 and Y0 are given by (3.3).

(i) In case (3.5) holds, the eigenvalues of A are of opposite sign say λ1 < 0 < λ2. The term c1eλ1tξ1 is negligible in the long run. The eigenvector ξ2 corresponding to the positive eigenvalue (“plus” in front of the square root) has one component positive, and the other one is negative. It follows that all of the solutions of (3.2) eventually move either northwest or southeast of the rest point (X0,Y0) intersecting either the X or the Y axis.

(ii) In case (3.4) holds, the general solution of (3.2) is given by (3.9), with negative λ1 and λ2. It follows that the point (X(t),Y (t)) tends to the point (X0 > 0,Y0 > 0) as t → ∞. If the point (X(t),Y (t)) stays in the first quadrant, then x(t) and y(t) are defined for all t, otherwise one of the species becomes infinite in finite time.

Remark

In case (3.4) holds, the solution of (3.2) connects the points (X0,Y0) and (X(0),Y (0)) in the first quadrant. While it is rare for the solution (X(t),Y (t)) to exit the first quadrant, this may indeed happen if the points (X0,Y0) and (X(0),Y (0)) lie near one of the axes. We used Mathematica to solve (3.2) with a = 4, b = 1, c = 1, d = 5, X(0) = 5, Y (0) = 0.1. Here $X0=419>0$ {X_0} = {4 \over {19}} > 0 and $Y0=319>0$ {Y_0} = {3 \over {19}} > 0 . The graph of the solution in Figure 2 shows that Y (t) becomes zero at some T, which corresponds to limtT y(t) = ∞.

Explosive predator-prey model with periodic coefficients

We now consider a periodic perturbation of the explosive predator-prey model $x′=x2(b+β(t)y−1)y′=−y2(d+δ(t)x−1) ,$ \matrix{ {x' = {x^2}\left( {{{b + \beta (t)} \over y} - 1} \right)} \cr {y' = - {y^2}\left( {{{d + \delta (t)} \over x} - 1} \right){\kern 1pt} ,} \cr } with small continuous functions β (t) and δ (t) of period p, so that β (t + p) = β (t) and δ (t + p) = δ (t) for all t. (We make no assumptions on the sign of β (t) and δ (t).) The linear system for $X=1x$ X = {1 \over x} and $Y=1y$ Y = {1 \over y} $X′=−(b+β(t))Y+1Y′=(d+δ(t))X−1$ \matrix{ {X' = - \left( {b + \beta (t)} \right)Y + 1} \cr {Y' = \left( {d + \delta (t)} \right)X - 1} \cr } has p-periodic coefficients. Let F(t) be the normalized fundamental solution matrix (with F(0) = I, the identity matrix) of the corresponding homogeneous system $X′=−(b+β(t))YY′=(d+δ(t))X .$ \matrix{ {X' = - \left( {b + \beta (t)} \right)Y} \cr {Y' = \left( {d + \delta (t)} \right)X{\kern 1pt} .} \cr } For small β (t) and δ (t), F(t) is close for t ∈ [0, p] to the normalized fundamental solution matrix $F0(t)=[cosbd t−bdsinbd tdbsinbd tcosbd t]$ {F_0}(t) = \left[ {\matrix{\hfill {\cos \sqrt {bd} {\kern 1pt} t} & \hfill { - \sqrt {{b \over d}} \sin \sqrt {bd} {\kern 1pt} t} \cr \hfill {\sqrt {{d \over b}} \sin \sqrt {bd} {\kern 1pt} t} & \hfill {\cos \sqrt {bd} {\kern 1pt} t} \cr } } \right] of the unperturbed system $X′=−bYY′=dX .$ \matrix{ {X' = - bY} \cr {Y' = dX{\kern 1pt} .} \cr } By the continuous dependence of eigenvalues on the coefficients of the matrix, the Floquet multipliers of (4.3), i.e., the eigenvalues of F(p) are close to the eigenvalues ρ1 and ρ2 of F0(p). Clearly, $ρ1ρ2=1=detF0(p)ρ1+ρ2=2cosbd p=trace F0(p) .$ \matrix{ {{\rho _1}{\rho _2} = 1 = \det {F_0}(p)} \cr {{\rho _1} + {\rho _2} = 2\cos \sqrt {bd} {\kern 1pt} p = {\rm{trace}}{\kern 1pt} {F_0}(p){\kern 1pt} .} \cr }

Theorem 4.1

Assume that $bd p≠2πm$ \sqrt {bd} {\kern 1pt} p \ne 2\pi m , for any integer m. Then the system (4.1) has a unique positive p-periodic solution (xp(t),yp(t)) for sufficiently small β (t) and δ (t).

Proof

Observe that the Floquet multipliers of (4.4) satisfy ρi ≠ 1, for i = 1,2. Indeed, if ρ1 = 1, then from the first line in (4.5) ρ2 = 1, giving a contradiction in the second line in (4.5), because $cosbd p≠1$ \cos \sqrt {bd} {\kern 1pt} p \ne 1 . Since β (t) and δ (t) are small, the Floquet multipliers of the homogeneous problem (4.3) are different from one, so that (4.3) has no p-periodic solution, and then by a standard result the non-homogeneous system (4.2) (and hence the original system (4.1)) has a unique p-periodic solution (Xp(t),Yp(t)). It remains to show that Xp(t) > 0 and Yp(t) > 0 for all t.

We derive next an a priori bound on Xp(t) and Yp(t), uniform in β (t) and δ (t), provided that |β (t)|+|δ (t)| ≤ c0, for some constant c0. Indeed, integrating both equations in (4.2) over (0,t), with t ∈ (0, p), taking absolute values and then adding the corresponding inequalities, obtain $|Xp(t)|+|Yp(t)|≤a1∫0t(|Xp(s)|+|Yp(s)|) ds+a2 ,$ |{X_p}(t)| + |{Y_p}(t)| \le {a_1}\int_0^t \left( {|{X_p}(s)| + |{Y_p}(s)|} \right){\kern 1pt} ds + {a_2}{\kern 1pt} , for some positive constants a1 and a2. The desired bound over (0, p) follows by the Bellman-Gronwall lemma, see e.g., [2].

We claim that Xp(t) > 0 and Yp(t) > 0 for all t. Setting $Xp(t)=ξ(t)+1d$ {X_p}(t) = \xi (t) + {1 \over d} and $Yp(t)=η(t)+1b$ {Y_p}(t) = \eta (t) + {1 \over b} in (4.2) obtain $ξ′=−b η−β(t)Yp(t)η′=dξ+δ(t)Xp(t) .$ \matrix{ {\xi ' = - b{\kern 1pt} \eta - \beta (t){Y_p}(t)} \cr {\eta ' = d\xi + \delta (t){X_p}(t){\kern 1pt} .} \cr } Express $[ξ(t)η(t)]=F0(t)[c1c2]+F0(t)∫0tF0−1(s)f(s) ds ,$ \left[ {\matrix{ {\xi (t)} \cr {\eta (t)} \cr } } \right] = {F_0}(t)\left[ {\matrix{ {{c_1}} \cr {{c_2}} \cr } } \right] + {F_0}(t)\int_0^t F_0^{ - 1}(s)f(s){\kern 1pt} ds{\kern 1pt} , with some constants c1 and c2, and $f(t)=[−β(t)Yp(t)δ(t)Xp(t)]$ f(t) = \left[ {\matrix{ { - \beta (t){Y_p}(t)} \cr {\delta (t){X_p}(t)} \cr } } \right] . Since the vector $[ξ(t)η(t)]$ \left[ {\matrix{ {\xi (t)} \cr {\eta (t)} \cr } } \right] has period p, and the fundamental solution matrix F0(t) has period $2πmbd≠p$ {{2\pi m} \over {\sqrt {bd} }} \ne p , it follows that c1 = c2 = 0. The vector f (t) is small by our assumptions, and the a priori estimate (4.6). Both matrices F0(t) and $F0−1(s)$ F_0^{ - 1}(s) have bounded entries. Then the vector $[ξ(t)η(t)]$ \left[ {\matrix{ {\xi (t)} \cr {\eta (t)} \cr } } \right] is small, so that the trajectory (Xp(t),Yp(t)) remains near the point $(1d,1b)$ \left( {{1 \over d},{1 \over b}} \right) , and hence it stays in the first quadrant for all t.

Multiplicity of solutions for a class of periodic equations

The transformation $u(t)=1x(t)$ u(t) = {1 \over {x(t)}} of the preceding sections turns out to be useful for a class of first order equations with periodic coefficients. V.A. Pliss [6] considered the Abel equation: $x′(t)=a0(t)x3+a1(t)x2+a2(t)x+a3(t) .$ x'(t) = {a_0}(t){x^3} + {a_1}(t){x^2} + {a_2}(t)x + {a_3}(t){\kern 1pt} . Assuming that the given functions ai(t), 0 ≤ i ≤ 3, are of period p, and a0(t) is either positive or negative for all t, he proved that the equation (5.1) has at most three p-periodic solutions. The proof involved a clever combination of the equations that the inverses of solutions satisfy.

What if one changes the a0(t)x3 term to a0(t)x2n+1? In case it is a0(t)x5, the method of V.A. Pliss [6] still gives the same result with a little extra effort. For higher powers things get more involved, and in fact existence of at most three p-periodic solutions was proved by another elegant method in A.A. Panov [5]. It turns out that the following more general result was already known.

Theorem 5.1

For the equation $x′(t)=f(t,x)$ x'(t) = f(t,x) assume that the function f (x,t) is continuous and has three continuous derivatives in x, and also for some p > 0 and all real t and x one has $f(t+p,x)=f(t,x) ,$ f(t + p,x) = f(t,x){\kern 1pt} , $fxxx(t,x)>0 (or the opposite inequality holds) .$ {f_{xxx}}(t,x) > 0\;\;\;\;{\kern 1pt} (or\;the\;opposite\;inequality\;holds){\kern 1pt} {\kern 1pt} . Then the equation (5.2) has at most three p-periodic solutions.

This theorem follows from a more general result of A. Sandqvist and K.M. Andersen [7]. They considered the equation (5.2) on the interval (0, p) and called a solution to be closed if x(p) = x(0). Assuming the condition (5.4) holds, they showed that the problem (5.2) has at most three closed solutions, which implies the Theorem 5.1.

A simpler proof of the Theorem 5.1 was found in P. Korman and T. Ouyang [3]. We now simplify the presentation in that paper. The proof will follow from the following three simple lemmas.

Lemma 5.1

Assume the condition (5.4) holds and f (t,0) = 0 for all tR. Then for all tR and x > 0 one has $Q(t,x)≡2f(t,x)−2xfx(t,x)+x2fxx(t,x)>0 .$ Q(t,x) \equiv 2f(t,x) - 2x{f_x}(t,x) + {x^2}{f_{xx}}(t,x) > 0{\kern 1pt} .

Proof

Calculate Q(t,0) = 0 and Qx(t,x) = x2 fxxx(t,x) > 0.

Lemma 5.2

For the problem $y′(t)=g(t,y)$ y'(t) = g(t,y) assume that for some p > 0 and all tR and y > 0 one has $g(t+p,y)=g(t,y) ,gyy(t,y)>0 (or the opposite inequality holds) .$ \matrix{ {g(t + p,y) = g(t,y){\kern 1pt} ,} \cr {{g_{yy}}(t,y) > 0\;\;\;\;{\kern 1pt} (or\;the\;opposite\;inequality\;holds){\kern 1pt} {\kern 1pt} .} \cr }

Then the equation (5.5) has at most two positive p-periodic solutions.

The proof is standard, and it can be found in e.g., P. Korman [2], p. 245. The next lemma is crucial.

Lemma 5.3

For the problem (5.2) assume that f (t,0) = 0 for all tR, and the conditions (5.3), (5.4) hold for all tR and x > 0. Then the equation (5.2) has at most two positive p-periodic solutions.

Proof

Set $x=1y$ x = {1 \over y} in (5.2). Then $−y′=y2f(t,1y)≡g(t,y).$ - y' = {y^2}f(t,{1 \over y}) \equiv g(t,y). By Lemma 5.1 for any y > 0 $gyy=2f(t,1y)−2yfx(t,1y)+1y2fxx(t,1y)=2f(t,x)−2xfx(t,x)+x2fxx(t,x)>0 .$ {g_{yy}} = 2f(t,{1 \over y}) - {2 \over y}{f_x}(t,{1 \over y}) + {1 \over {{y^2}}}{f_{xx}}(t,{1 \over y}) = 2f(t,x) - 2x{f_x}(t,x) + {x^2}{f_{xx}}(t,x) > 0{\kern 1pt} . By Lemma 5.2 the equation (5.6) has at most two positive p-periodic solutions, and the same is true for (5.2).

Turning to the proof of the Theorem 5.1, observe that different solutions of (5.2) do not intersect by the uniqueness theorem. If the equation (5.2) has four p-periodic solutions, let ξ (t) be the smallest one. Then z(t) = x(t) − ξ (t) satisfies $z′=f(t,z+ξ)−f(t,ξ)≡g(t,z) ,$ z' = f(t,z + \xi ) - f(t,\xi ) \equiv g(t,z){\kern 1pt} , and the equation (5.7) has three positive p-periodic solutions. However, g(t,0) = 0 and gzzz(t,z) > 0 for z > 0, contradicting the Lemma 5.3.

Equations of the type (5.2) occur often in ecological problems, see e.g., S. Ahmad and A.C. Lazer [1], or P. Korman [2].

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