STUDY OF ECOLOGICAL COMPETITION AMONG FOUR SPECIES. E. Marchi and R. Velasco. IMA Preprint Series #2469. (March 2016)

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1 STUDY OF ECOLOGICAL COMPETITION AMONG FOUR SPECIES By E. Marchi and R. Velasco IMA Preprint Series #2469 (March 2016) INSTITUTE FOR MATHEMATICS AND ITS APPLICATIONS UNIVERSITY OF MINNESOTA 400 Lind Hall 207 Church Street S.E. Minneapolis, Minnesota Phone: Fax: URL:

2 1 STUDY OF ECOLOGICAL COMPETITION AMONG FOUR SPECIES E. Marchi *) and R. Velasco **) *) Founder and First Director IMASL, UNSL CONICET Emeritus Professor UNSL (Ex) Superior Researcher CONICET Corresponding author: emarchi1940 gmail.com **) Consejo Nacional de Investigaciones Científicas y Técnicas. Universidad Nacional de San Luis. Argentina. This paper has been partially supported by SECYT at UNSL. Argentina.

3 2 1- INTRODUCTION Following a previous paper (1), on the Lotka- Volterra equations for the interactive competition of three species in an habitat, we wish to analyze in the competition between four species, specially in relation with the important feature of the existence of cycles. In the literature, there are important studies, which attack different problems related to the Lotka- Volterra equations. For example Bojadziev and Chan (2), consider oscillations governed by the generalized Volterra- Gause- Witt equation for two species to include retardation effect in population dynamics. They use the Krylov- Bogolinbov- Mitropelskii perturbation method with an extension for differential equations with retarded argument. They also discuss models with small time delay. On the other hand in an interesting paper, De Angelis (3) applies perturbation methods to the differential equation of the predator- prey model, to find approximate amplitudes and periods of limit cycles solutions. Other important discussion of oscillatory models for two interacting species is found in Van de Vaart (4), where a complete enumeration is given of the condition that the parameters of these models must satisfy in order that a part of the phase space be filled with a family of closed curves. Recently there have been studies about the structural perturbations of the Lotka- Volterra equations. Let us mention the work by Ikeda and Siljak. Finally we note the paper by Bell (5) on predator prey type equations for simulating an immune response between antigen and antibody production. However there is not a detailed analysis for more than three species. In this paper we wish to develop a general- method for four species, which will give rise to an important step in the general study of the existence of cycles for an arbitrary number of species, a problem to be study a next paper.

4 3 2- FOUR SPECIES MODEL We now consider the Lotka- Volterra systems for four species, whose equations take the form: dx dt = x (ε (1) 1 a 1 2 y + a 1 3 z + a 1 4 u) dy dt = y ( ε 2 + a 2 1 x + a 2 3 z + a 2 4 u) dz dt = z (ε 3 + a 3 1 x + a 3 2 y + a 3 3 z + a 3 4 u) du dt = u (ε 4 + a 4 1 x + a 4 2 y + a 4 3 z + a 4 4 u) where x, y, z and u represent the population densities of the four species under consideration. The parameters have the standard meaning in this type of problem. We consider that ε 1, ε 2, a 12 and a 2 1 are positive, while the other parameters can take real value. Our task in this paper is to find the general condition for the existence of cyclic variation of the population densities. Therefore, we assume that the variable u, is related to the variables x, y and z through a function f 1 : (2) u = f 1 (x, y, z) Derivation with respect to time yields: du dt = f 1 x dx dt + f 1 y dy dt + f 1 z dz dt (3) Now, using system (1) in equation (3), we obtain: (4) u (ε 4 + a 4 1 x + a 4 2 y + a 4 3 z + a 4 4 u) = f 1 x x (ε 1 a 1 2 y + a 1 3 z + a 1 4 u) + f 1 y y ( ε 2 + a 2 1 x + a 2 3 z + a 2 4 u) + f 1 z z (ε 3 + a 3 1 x + a 3 2 y + a 3 3 z + a 3 4 u)

5 4 To solve this order partial differential equation, we first try to solve the following separate parts: (5) a 4 4 u = f 1 x a 1 4 x + f 1 y a 2 4 y + f 1 z a 3 4 z u (ε 4 + a 4 1 x + a 4 2 y + a 4 3 z ) = f 1 x x(ε 1 a 1 2 y + a 1 3 z ) + f 1 y( ε y 2 + a 2 1 x + a 2 3 z) + f 1 z z (ε 3 + a 3 1 x + a 3 2 y + a 3 3 z) (6) We remark that if we solve equation (5) in such a way that the arbitrary function which appears in the solution of equation (4). By standard methods, as it was obtained in the previous paper (1), the solution of equation (5) is: u = z a 44 F ( y1 a 24 x 1 a 14, z1 a34 y 1 ) a 24 (7) where F is an arbitrary function. Now, calling: ω 1 = y1 a 24 x 1 a 14, ω 2 = z1 y 1 a 24 and replacing the expression (7) for u in equation (6), we obtain the first order partial differential equation: (8) F(ε 4 + a 4 1 x + a 4 2 y + a 4 3 z ) = ω 1 a 14 + ( ω 1 a 24 F ω 1 (ε 1 a 1 2 y + a 1 3 z ) F ω 1 ω 2 a 24 F ω 2 ) ( ε 2 + a 2 1 x + a 2 3 z) + ( a 44 F + ω 2 F ω 2 ) (ε 3 + a 3 1 x + a 3 2 y + a 3 3 z)

6 5 In order to solve this last partial differential equation, we try the following system of first order partial differential equations: F = α 1 ω 1 F ω 1 + α 2 ω 2 F ω 2 (9) where the constant are: α 1 = ( ε 1 ε 2 a 44 ) (ε a 14 a 4 ε 3 ) ; α 24 a 2 = ( ε 3 + ε 2 ) (ε 34 a 4 a 44 ) 24 and F(a 41 x + a 42 y + a 43 z) (10) = ω 1 a 14 + ( ω 1 a 24 F ω 1 ( a 1 2 y + a 1 3 z ) F ω 1 ω 2 a 24 F ω 2 ) (a 2 1 x + a 2 3 z) + ( a 44 F + ω 2 F ω 2 ) (a 3 1 x + a 3 2 y + a 3 3 z) The general solution of equation (9) is given by: (1 α F = ω 1 ) (1 α 1 K(ω 1 1 ) (1 α ω 2 ) 2 ) (11) where K is an arbitrary function. Calling: (1 α ξ = ω 1 1 ) ω (1 α 2 2 ) and replacing the expression (11) in the partial differential equation (10), it turns out that:

7 6 (12) K(ξ)(a 4 1 x + a 4 2 y + a 4 3 z) = ( a 12 α 1 a 14 y a 13 α 1 a 14 z) (K(ξ) + ξ d dξ K(ξ)) + ( a 21 α 1 a 24 x + a 23 α 1 a 24 z) (K(ξ) + ξ d dξ K(ξ)) + ( a 21 α 1 a 24 x + a 23 α 1 a 24 z) (ξ d dξ K(ξ)) + ( a 31a 44 x + a 32a 44 y + a 33a 44 z) K(ξ) + ( a 31 α 2 x + a 32 α 2 y + a 33 α 2 z) ξ d dξ K(ξ) or, equivalently: K K = 1 ξ ( 1 x + 2 y + 3 z δ 1 x + δ 2 y + δ 3 z ) (12 ) where the constant appearing in the last equation take the form: 1 = a 41 a 21 α 1 a 24 a 31a 44 ; 2 = a 42 a 12 α 1 a 14 a 32a 44 3 = a 43 + a 13 α 1 a 14 a 23 α 1 a 24 a 33a 44 ; δ 1 = a 21 α 1 a 24 + a 21 α 2 a 24 a 31 α 2 δ 2 = a 12 α 1 a 14 a 32 α 2 ; δ 3 = a 13 α 1 a 14 + a 23 α 1 a 24 + a 23 α 2 a 24 a 33 α 2 To continue in our analysis, we impose the following restriction: 1 3 = δ 1 δ 3 and 2 3 = δ 2 δ 3

8 7 Under these condition we have that equation (12 ) becomes: The solution of this last equation is: K = 3 δ 3 1 ξ K (13) K(ξ) = C 1 ξ ( 3 δ 3 ) where C 1 is a constant to be determined. From here we obtain the following functional for u: with: u(x, y, z) = C 1 x r 1 y r 2 z r 3 r 1 = ( 1 α 1 a α 1 δ 3 a 14 ) r 2 = 1 α 1 a δ 3 a 24 ( 1 α α 2 ) (14) r 3 = a 44 3 α 2 δ 3 We note that the value of C 1 is determined with the initial conditions of the problem. namely: In this way, we have reduced the number of initial equations from four to three, dx dt = x (ε 1 a 12 y + a 13 z + a 14 C 1 x r 1 y r 2 z r 3) (15) dy dt = y ( ε 2 + a 21 x + a 23 y + a 24 C 1 x r 1 y r 2 z r 3) dz dt = z (ε 3 + a 31 x + a 32 y + a 33 z + C 1 x r 1 y r 2 z r 3) Next we are going to relate the variable z to the first two variables. The procedure is correct when a cyclic solution exists. Letting: z = f 2 (x, y) (16)

9 8 Again, to determine the function f 2, derivate z with respect to time: dz dt = f 2 x dx dt + f 2 dy y dt (17) and replace the values of the derivatives by what we have in system (15), obtaining the partial differential equation: z(ε 3 + a 31 x + a 32 y + a 33 z + C 1 x r 1 y r 2 z r 3) (17 ) = f 2 x x(ε 1 a 12 y + a 13 z + a 14 C 1 x r 1 y r 2 z r 3) + f 2 y y( ε 2 + a 21 x + a 23 z + a 24 C 1 x r 1 y r 2 z r 3) We solve this equation in steps. The first step is to consider the equation: whose solution is given by: where Q is an arbitrary function and: z = a 14 x f 2 x + a 24 y f 2 y z = y ( a 24 ) Q( ) = y(1 a 24) x (1 a 14) Replacing z by the expression (19), equation (17 ) turns out to be: (ε 3 + a 31 x + a 32 y + a 33 y ( a 24 ) Q)Q (18) (19) (20) or equivalently: With = ( 1 a 14 )Q ( )(ε 1 a 12 y + a 13 y ( a 24 ) Q) + ( a 24 ) + 1 a 24 Q ( )( ε 2 + a 21 x + a 23 y ( a 24 ) Q) Q Q = b(1 ) b = a 14(a 23 a 33 a 24 ) (a 13 a 24 a 23 a 14 ) (21) Under the conditions: ε 3 = ε 1 a 14 ε 2 a 24 (1 + )

10 9 a 31 = a 21 a 24 (1 + ) and a 32 = a 12 a 14 The solution of (21) is: Q( ) = C 2 b Where the arbitrary constant C 2 is determined by the initial condition of the problem. Thus the variable z becomes: z(x, y) = C 2 x S 1y S 2 (22) where: s 1 = b a 14 and s 2 = 1 a 24 (b + ) Replacing the value of z in the system of differential equations (15), it yields the reduced system: dx dt = x(ε 1 a 12 y + a 13 x s 1 y s 2 + a 14 x (r 1+s 1 r 3 ) y (r 2+s 2 r 3 ) ) (23) dy dt = y( ε 2 + a 21 x + a 23 x s 1 y s 2 + a 24 x (r 1+s 1 r 3 ) y (r 2+s 2 r 3 ) ) where: a 13 = C 2 a 13 ; a 23 = C 2 a 23 a 14 = C 1 C 2 r 3 a 14 ; a 24 = C 1 C 2 r 3 a SOLUTION OF THE REDUCED EQUATIONS Our task in the paragraph is to solve the systems (23), which is the reduced form of the system of four equation (1). We perform a substitution of variables similar to that used by Goel et al. in (6). Calling: a 12 = a β 1 ; a 21 = a β 2 q 1 = ε 2 β 2 a ; q 2 = ε 1 β 1 a with the change of variables: v 1 = log(x q 1 ) ; v 2 = log(y q 2 )

11 10 the system (23) becomes: dv 1 dt = ε 1 + a β 1 q 2 exp (v 2 ) + a 13 q 1 s 1 q 2 s 2 exp(s1 v 1 + s 2 v 2 ) + (a 14 q 1 (r 1 +s 1 r 3 ) q2 (r 2 +s 2 r 3 ) ) exp((r1 + s 1 r 3 )v 1 + (r 2 + s 2 r 3 )v 2 ) which with the constants: H 1 = β 1a 13 q 2 (s 2 1) a q 1 s 1 (r H 2 = (β 1 a) a 14 q 1 +s 1 r 3 ) (r 2 +s 2 r 3 1) 1 q2 becomes: dv 1 dt = a q 2{exp(v 2 ) 1 + H 1 exp(s 1 v 1 + s 2 v 2 ) + H 2 exp((r 1 + s 1 r 3 )v 1 + (r 2 + s 2 r 3 )v 2 )} (24) similarly, for the second equation: dv 2 dt = aq 1{exp(v 1 ) 1 + H 3 exp(s 1 v 1 + s 2 v 2 ) + H 4 exp((r 1 + s 1 r 3 )v 1 + (r 2 + s 2 r 3 )v 2 )} (25) where the coefficient are now: H 3 = (s β 1 a 13 q 2 ) 2 ( s a q 1 +1) ; H (r 4 = (β 2 a) a 24 q 1 +s 1 r 3 1) (r 2 +s 2 r 3 ) 1 q2 1 Cross- multiplication equation (24) and (25) yields after some manipulations: d dt {β 1q 1 (exp(v 1 ) v 1 )} (26) With the new constants: = d dt {β 2q 2 (exp(v 2 ) v 2 )} + exp(s 1 v 1 + s 2 v 2 ) (β 2 q 2 H 1 dv 2 dt β 1q 1 H 3 dv 1 dt ) + exp((r 1 + s 1 r 3 )v 1 + (r 2 + s 2 r 3 )v 2 ) (β 2 q 2 H 2 dv 2 dt β 1q 1 H 4 dv 1 dt ) β 2 q 2 H 1 = A 1 s 2 ; β 1 q 1 H 3 = A 2 s 1 β 2 q 2 H 2 = A 3 (r 2 + s 2 r 3 ) ; β 1 q 1 H 4 = A 4 (r 1 + s 1 r 3 )

12 11 And if: A = A 1 = A 2 ; B = A 3 = A 4 equation (26) becomes: d dt {β 1q 1 (exp(v 1 ) v 1 )} = d dt {β 2q 2 (exp(v 2 ) v 2 )} (27) dv 1 + A exp(s 1 v 1 + s 2 v 2 ) (s 1 dt + s dv 2 2 dt ) + Bexp((r 1 + s 1 r 3 )v 1 + (r 2 + s 2 r 3 )v 2 ) ((r 2 + s 2 r 3 ) dv 2 dt + (r 2 + s 2 r 3 ) dv 1 dt ) Now using the equality: d dt {exp( v 1 ± nv 2 )} = exp( v 1 ± nv 2 ) (( dv 1 dt ± n dv 2 dt )) equation (27) turns out to be: d dt {β 1q 1 (exp(v 1 ) v 1 )} (28) = d dt {β 2q 2 (exp(v 2 ) v 2 )} + d dt {A exp(s 1v 1 + s 2 v 2 )} + d dt {B exp((r 1 + s 1 r 3 )v 1 + (r 2 + s 2 r 3 )v 2 )} Integrating, we obtain: (29) β 1 q 1 (exp(v 1 ) v 1 ) A exp(s 1 v 1 + s 2 v 2 ) = C 3 + β 2 q 2 (exp(v 2 ) v 2 ) + B exp((r 1 + s 1 r 3 )v 1 + (r 2 + s 2 r 3 )v 2 ) where C 3 is an arbitrary constant to be determined with the initial values of the problem. Now going back to the old variables, we obtain for the solution the functional relation: 1 ( x log x ) + P ε 1 q 1 q 1 (y a 1 x b 1) = C ( y log y ) P 1 ε 2 q 2 q 2 (y a 2 x b 2) 2 (30) where: a 1 = s 2 ; b 1 = s 1 ; a 2 = r 2 + s 2 r 3 ; b 2 = s 1 r 3 r 1 C 3 = C 3 a q 1 q 2 ; P 1 = A a q 2 (s 2 +1) q1 (s 1 +1)

13 12 and (r P 2 = B (aq 1 +s 1 r 3 +1) (r 2 +s 2 r 3 +1) 1 q2 ) This equation can take the form: { (x q 1 ε1 1) exp(x q 1 ) } = K { (y q 1 ε2 2) exp(y q 2 ) } exp( P 1 y a 1 x b 1 P 2 y a 2 x b 2) (31) with the constant: K = exp (C 3 ) It is worth comparing equation (31) with the integral obtained by Volterra in these prey- predator systems, we note that equation (31) has an additional term, the last one of the right side. This term takes into consideration the competition of the third and fourth species between them and among with the first two species. 4- CYCLE EXISTENCE Having the integral or relationship given by equation (31) between the two primitive variables of the problem x and y, in this paragraph we are going to show the existence of a cyclic behavior in the population densities x and y. We study only a general situation, other cases might also be worked out, but for simplicity in the presentation, we do not include them here. As before, let us call 1 the function described in the left side of the equation (31), we call 2 the right side of the same equation. with: When: the last function becomes: a 1 = 1 and a 2 = 1 P 1 = P 1 < 0 and P 2 = P 2 < 0 2 (x, y) = K(1 q 2 ) ( 1 ε 2 ) y ( 1 ε 2 ) exp(f(x) y) (32) where: F(x) = (1 q 2 ε 2 ) + P 1(1 x b 1) + P 2(1 x b 2)

14 13 Now we follow the procedure out lined in reference (1), that is, we look for the dependence of the minimum of the function 2 with respect to y, parametrically in x. Hence we partially derive 2 and impose the minimum condition: 2 y = K(1 q 2) ( 1 ε 2 ) { 1 y 1 ε 2 1 ε 2 exp(f(x)y) + y 1 ε 2 The ordinate y r of such a minimum is related with the variable x by: or equivalently: (1 ε 2 )(1 y r ) + F(x) = 0 F(x) exp(f(x)y) } = 0 y r = (1 ε 2 F(x) ) (33) In order to obtain the parametric curves of the minima in terms of x, we replace equation (33) into equation (32), obtaining: 2 (x, y r ) = K (F(x)) 1 ε 2 (34) where: K = K(q 2 ε 2 ) 1 ε 2 exp (1 ε 2 ) Hence we now have the following qualitative curves for b 1, b 2 > 0: 1 F(x) 2 (x, y r ) 1 (x) 1 q 2 ε 2 x x 1 x 2 x Figure 1 At this point, by a analysis similar to the previous paper (1) we obtain the existence of the cycle under very general condition. (See the behavior of the functions 1 (x), 2 (x, y r )).

15 14 5- COMPUTATION OF THE PERIOD In the previous paragraphs we have proven the existence and computed the cycle for the competition among the four species under consideration. Here we are going to calculate the period of the general cycle. Let us consider equation (31), from which we have for a new variable: Ψ = { (x q 1 ε1 1) exp (x q 1 ) } = K { (1 q 1 ε2 2) exp(1 q 2 ) } exp( P 1 y a 1 x b 1 P 2 y a 2 x b 2) Now taking logarithm and derivating we get: (35) (36) 1 dψ Ψ dt = (1 ε 1) ( ε 2 2 a 2 21 x ε 2 a 21 x ) dx dt Now, replacing the derivate of x with respect to time in equation (23), we obtain: and therefore: T 2 1 dψ Ψ dt = 1 Ψ (ε 2 2 a 2 21 x ε 2 a 21 x ) (ε 1 a 12 y + a 13 y a 1x b 1 + a 14 y a 2x b 2) Ψ 1 2 x P = dt = ε 1 {( ε 2 2 a 21 ) (ε a 21 ε 1 a 12 y + a 13 y a 1x b 1 + a 14 y a 2x b 2)} 2 T 1 Ψ 1 1 dψ Ψ (37) (39) where T 1 is the time corresponding to Ψ 1, for the point with the minimum y, while T 2 is the time when the solution passes next through such a point Ψ 1 = Ψ 1. Thus the exact computation of the period P can be obtained. We now wish to estimate the period P. For this find a suitable relationship between Ψ an x, in order to replace it in the integral. From the first part of equation (35) we have: Ψ ε 1 = (a 21 ε 2 )x exp( a 21 ε 2 x) (40) Now taking: x = x 0 + w 1 (41) where x 0 is the equilibrium value of the number of individuals of the first species, an w 1 is the perturbation from this average value of the population. Replacing (41) in (40) it yields: ε 2 a 21 Ψ ε 1 exp(a 21 ε 2 x) = (x 0 + w 1 )exp( a 21 ε 2 w 1 ) (42) = (x 0 + w 1 ) (1 (a 21 ε 2 ) w (a 2 21 ε 2 ) 2 w 1 ) x 0 + w 1 (a 21 ε 2 )x 0 w 1

16 15 where in the last part we have neglected terms of higher order in w 1 since it is considered that the perturbation is small. where: consider: From equation (42) we get: w 1 = H 1Ψ ε 1 x 0 H 2 H 1 = (ε 2 a 21 )exp(a 21 ε 2 x 0 ) and H 2 = (1 (a 21 ε 2 )x 0 ) y = y 0 + w 2 (43) (44) (45) where y 0 is the equilibrium value of the number of individual in the second species. The value w 2 is a small perturbation. Now taking the right side of equation (35), we arrive to: (ε 1 a 12 )(Ψ K) ε 2 = (y 0 + w 2 ) exp{ ε 2 a 12 ε 1 (y 0 + w 2 ) + ε 2 P 1 (y 0 + v 2 ) a 1(x 0 + w 1 ) b 1 + ε 2 P 2 (y 0 + w 2 ) a 2(x 0 + w 1 ) b 2} (46) Expanding the binomial and exponential functions we arrive at: Ψ ε 2H 3 exp(h 4 w 1 ) y 0 = (H 5 1)w 2 (47) where: H 3 = K ε 2 ε 1 a 12 exp ((ε 2 a 12 a ε 1 )y 0 ) ε 2 P 1 y 1 b 0 x 1 a 0 ε 2 P 2 y 2 b 0 x 2 0 a H 4 = ε 2 P 1 b 1 y 1 (b 0 x 1 +1) a 0 + ε 2 P 2 b 2 y 2 (b 0 x 2 +1) 0 (48) a H 5 = ε 2 P 1 a 1 y 1 0 x b 1 a + ε 2 P 2 a 2 y 2 0 x b 2 (ε 2 a 12 ε 1 )y 0 From equation (47) we get: with: w 2 = H 7ex p( H 6 Ψ ε 1) Ψ ε 2 y 0 H 5 1 H 6 = H 4 H 1 H 2 and H 7 = H 3 exp(h 4 x 0 H 2 ) Now replacing the equation (41), (43), (45), and (49) in the integral (39) we arrive at the following expression: Ψ 1 Ψ 1 P = ε 1 ε 2 a 21 {(ε 2 2 a 2 21 (x 0 + w 1 )) (ε 1 a 12 (y 0 + w 2 ) + a 13 (y 0 + (49) (50) (51) w 2 ) a 1(x 0 + w 1 ) b 1) + a 14 (y 0 + w 2 ) a 2(x 0 + w 1 ) b 2} 1 dψ Ψ

17 16 get: Expanding the binomials and keeping only first order terms in the perturbations, we Ψ 1 ε 1 ε 2 a 21 dψ P = Ψ(ε 2 2 a 2 21 x 0 a 2 12 w 1 )(H 8 + H 9 w 1 + H 10 w 2 ) Ψ 1 (52) where: H 8 = ε 1 a 12 y + a 13 y 0 a 1 x 0 b 1 + a 14 y 0 a 2 x 0 b 2 (53) H 9 = a 13 b 1 y 0 a 1 x 0 b 1 1 a 14 b 2 y 0 a 2 x 0 b 2 1 H 10 = a 12 + a 13 a 1 y 0 a 1 1 x 0 b 1 + a 14 a 2 y 0 a 2 1 x 0 b 2 After some manipulation, we obtain the following integral: Ψ 1 H 11 dψ P = (H 16 + H 17 Ψ ε 1 + H18 Ψ ε 2 exp( H6 Ψ ε 1 )) Ψ 1 (54) with: (55) H 11 = ε 1 ε 2 a 21 ; H 12 = a 2 2 a 21 2 x 0 H 13 = H 12 H 8 ; H 14 = H 12 H 9 a 21 2 H 8 H 15 = H 12 H 1o ; H 16 = H 13 + H 14 x 0 H 2 H 15 y 0 (H 5 1) H 17 = H 14 H 1 H 2 ; H 18 = H 15 H 7 (H 5 1) 6- A PARTICULAR CASE In this paragraph we present a particular case with all the important features related with the cycle. We choose the particular set of parameters: ε 1 = 0,1 a 12 = 0,4 a 13 = 0,06 a 14 = 0,3 ε 2 = 0,4 a 21 = 0,3 a 23 = 0,2 a 24 = 0,4 ε 3 = 0,466 a 31 = 0,6 a 32 = 1,333 a 33 = 0,08 ε 4 = 0,3 a 41 = 0,337 a 42 = 0,6 a 43 = 0,12 = 0,2 a 44 = 0

18 17 values: From the general equations we obtain the curve show in Fig. 1 which with initial x(0) = 0,48 ; y(0) = 0,064 ; z(0) = 0,044 ; u(0) = 0,027 takes the shape shown in Fig. 2. In order to get the cycle, we perform the proyection on three planes of the phase space. In this way Figures 3, 4, 5 are obtained. Or the other hand, in order to obtain a sequence of the variables x, y, z and u it is convenient to study first the curve of the velocity dx dt as a function of x in the cycle. This can be done employing systems (23) together with the proyection of the cycle in the face (x, y). Thus we get the Figure 6. From here it is possible to derive the real function of the time as indicated in Figure 7. The period can be obtained as: P = 1,161 hare we have used the approximate formula (54) which have been evaluated by Simpson formula. The equilibrium values are: x e = 1,066 y e = 0,099 and it was integrated in four parts: (Ψ 1, Ψ 2 ); (Ψ 2, Ψ 3 ); (Ψ 3, Ψ 4 ) and (Ψ 4, Ψ 1 ) where: Ψ 1 = 0,011 Ψ 2 = 0,403 Ψ 3 = 0,977 Ψ 4 = 2,27 which correspond to the vertex points of the cycle in the phase space.

19 Figure 2 18 y 2 (x, y r ) 1 (x) x 1 x 2 x y 0,68 Figure 3 0,011 y 0,68 0,4 2,27 x Figure 4 0,011 0,3 1,93 z

20 0,76 u Figure ,014 0,4 2,27 x dx dt Figure 6 0,2 0,1-0,1 0,2 0,6 0,8 1 1,2 1,4 1,6 1,8 2 2,2 2,4 2,6-0,2-0,3-0,4-0,5-0,6 x Figure 7 z u y

21 20 7- REFERENCES [1] Marchi, E., Sales J. and Velasco R.: More General Ecological Competition among Three Species. UNSL. [2] Bojadziev G. and Chan S.: Asymtotiv Solutions of Differential Equations with Time Delay in Population Dynamics. Bull. of Math. Biol. Vol. 41. p.p (1979) [3] De Angelis D.: Estimates of Predator- Prey Limit Cycles. Bul of Math. Biol. Vol. 37. (1975) [4] Van der Vaart R.: Conditions for Periodic Solutions of Volterra differential systems. Bull. of Math. Biol. Vol. 40 (1978) [5] Bell G.: Predator- Prey Equations Simulating an Immune Response. Math. Bios. 16, (1978) [6] Goel N., Maitra S. and Montroll W.: On the Volterra and other Nonlinear Models of Interacting Populations. Rev. Mod. Phys (1971)