ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS. Sze-Bi Hsu. Christopher A. Klausmeier. Chiu-Ju Lin

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1 DISCRETE AND CONTINUOUS doi: /dcdsb DYNAMICAL SYSTEMS SERIES B Volume 12, Number 2, September 2009 pp ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS Sze-Bi Hsu Department of Mathematics National Tsing-Hua University Hsinchu, Taiwan 300 Christopher A. Klausmeier Kellogg Biological Station Michigan State University Hickory Corners MI 49060, USA Chiu-Ju Lin Department of Mathematics National Tsing-Hua University Hsinchu, Taiwan 300 Abstract. In this paper we study a mathematical model of two parallel food chains in a chemostat. Each food chain consists of a prey species x and a predator species y. Two food chains are symmetric in the sense that the prey species are identical and so are the specialized predator species. We assume that both of the prey species in the parallel food chains share the same nutrient R. In this paper we show that as the input concentration R (0) of the nutrient varies, there are several possible outcomes: (1) all species go extinct; (2) only the two prey species survive; (3) all species coexist at equilibrium; (4) all species coexist in the form of oscillations. We analyze cases (1) (3) rigorously; for case (4) we do extensive numerical studies to present all possible phenomena, which include limit cycles, heteroclinic cycles, and chaos. 1. Introduction and the model. Populations often vary in nature. While they are subject to external forcing, communities are also capable of generating sustained oscillations through interspecific interactions such as predator-prey [8, 15, 10, 13] and intransitive competition of three or more species [9, 7]. Key questions are: what other food web configurations are capable of producing internally-generated nonequilibrium dynamics, under what environmental conditions do we expect oscillations, and what are the characteristics of these oscillations? Here we investigate the dynamics of another simple food web module, that of two parallel food chains coupled by a shared basal resource. In the following we describe a mathematical model of symmetrical food chains in a chemostat. Let R(t) denote the concentration of nutrient (or resource) at time t. Let x 1 (t) and x 2 (t) be the population density of two identical prey at time t on the first and second food chain, respectively; y 1 (t) and y 2 (t) be the population density of two identical 2000 Mathematics Subject Classification. 34D05, 34D20, 92D25. Key words and phrases. food chains, limit cycles, heteroclinic orbits, chemostat, chaos. Research partially supported by National Council of Science, Republic of China. 337

2 338 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN predators respectively. Let R (0) be the input concentration and D be the dilution rate. The governing equations are R (t) = (R (0) R(t))D 1 γ x f(r(t))(x 1 (t)+x 2 (t)), x 1(t) = (f(r(t)) D)x 1 (t) 1 γ y g(x 1 (t))y 1 (t), x 2 (t) = (f(r(t)) D)x 2(t) 1 γ y g(x 2 (t))y 2 (t), y 1(t) = (g(x 1 (t)) D)y 1 (t), y 2(t) = (g(x 2 (t)) D)y 2 (t), f(r(t)) = mr(t) a+r(t), g(x i(t)) = μx i(t),i = 1,2, K +x i (t) R(0) 0, x 1 (0) > 0, x 2 (0) > 0, y 1 (0) > 0, y 2 (0) > 0, (1) where f(r) and g(x) are the growth rate of prey species x and predator y, respectively. They take the forms of the Michaelis-Menten formulation. m is the maximum growth rate, a is the half-saturation constant for the prey species x. μ is the maximal growth rate and K is the half-saturation constant for the predator species y. γ x and γ y are the yield constants for the prey species x and the predator species y respectively. By rescaling x and y, we may assume γ x = γ y = 1. Let Σ = R (0) (R+x 1 +x 2 +y 1 +y 2 ). Adding the equations in (1) yields Σ (t) = DΣ(t). It follows that Σ(t) = Σ(0)e Dt 0 as t. Since lim t (R(t)+x 1 (t)+x 2 (t)+ y 1 (t)+y 2 (t)) = R (0), we conclude that the omega limit set of the system (1) lies in the set Ω = {(R,x 1,x 2,y 1,y 2 ) : R+x 1 +x 2 +y 1 +y 2 = R (0) }. Consider the limiting system of (1) on Ω x 1(t) = (f(r(t)) D)x 1 (t) g(x 1 (t))y 1 (t), x 2(t) = (f(r(t)) D)x 2 (t) g(x 2 (t))y 2 (t), y 1 (t) = (g(x 1(t)) D)y 1 (t), y 2 (t) = (g(x 2(t)) D)y 2 (t), R(t) = R (0) (x 1 (t)+x 2 (t)) (y 1 (t)+y 2 (t)), x i (0) > 0, y i (0) > 0, i = 1,2, 0 < x 1 (0)+x 2 (0)+y 1 (0)+y 2 (0) < R (0). (2)

3 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 339 In the rest of section we shall consider the following model (3), which is a generalization of (2). x 1 (t) = (f(r(t)) d x)x 1 (t) g(x 1 (t))y 1 (t), x 2(t) = (f(r(t)) d x )x 2 (t) g(x 2 (t))y 2 (t), y 1(t) = (g(x 1 (t)) d y )y 1 (t), y 2(t) = (g(x 2 (t)) d y )y 2 (t), R(t) = R (0) (x 1 (t)+x 2 (t)) (y 1 (t)+y 2 (t)), x i (0) > 0,y i (0) > 0, i = 1,2, 0 < x 1 (0)+x 2 (0)+y 1 (0)+y 2 (0) < R (0), where d x and d y are the death rates of prey and predator, respectively. For convenience, we denote the break-even resource concentration λ x and λ y for prey species x and predator species y, respectively, λ x = f 1 a (d x ) = ( m d x ) 1, λ y = g 1 (d y ) = K ( μ d y ) 1. If m d x then x i (t) 0 as t, i = 1,2. Similarly if μ d y then y i (t) 0 as t, i = 1,2. Hence we assume λ x > 0, λ y > 0. The rest of the paper is organized as follows. In section 2, we state some preliminary results about the single food chain model. In section 3, we consider the two parallel food chains with either no predators or with only one predator. In section 4, we analyze and classify all cases on two parallel food chains sharing one nutrient. In section 5, we present our numerical studies and discuss their biological meanings. 2. Preliminary results for the single food chain. In this section we review some preliminary results about the single food chain model. Using the same notations in previous section, we consider the following system of the single food chain: x (t) = (f(r(t)) d x )x(t) g(x(t))y(t), y (t) = (g(x(t)) d y )y(t), R(t) = R (0) (x(t)+y(t)), x(0) > 0, y(0) > 0, 0 < x(0)+y(0) < R (0). Then from [12], we have the following results. Theorem 2.1. (a) If 0 < R (0) < λ x then x(t) 0, y(t) 0 as t. (b) If λ x < R (0) < λ x + λ y then x(t) x > 0, y(t) 0 as t where x = R (0) λ x. (c) If R (0) > λ x + λ y then there exists a unique equilibrium E c = (x c,y c ) of the system (4), where x c = λ y and y c satisfies the equation f(r (0) y x c y) d x = d y x c. Furthermore (x c,y c ) is locally stable if μ (K +x c ) 2y c < (3) (4) ma (a+r (0) x c y c ) 2. (*)

4 340 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN (d) ( ) is equivalent to R (0) < ˆR for some ˆR > 0. (e) If λ x +λ y < R (0) < ˆR then (x(t),y(t)) E c as t. (f) If R (0) > ˆR then there exists a limit cycle Γ. Remark 1. In (f), we conjecture that (x(t), y(t)) approaches a unique limit cycle Γ as t provided (x(0),y(0)) = (x c,y c ). 3. Two parallel food chains with either no predators or with only one predator. In this section we first consider two parallel food chains without predators. x 1(t) = (f(r(t)) d x )x 1 (t), x 2(t) = (f(r(t)) d x )x 2 (t), R(t) = R (0) (x 1 (t)+x 2 (t)), x 1 (0) > 0, x 2 (0) > 0, 0 < x 1 (0)+x 2 (0) < R (0). Theorem 3.1. If R (0) > λ x, then the solution (x 1 (t),x 2 (t)) (x 1,x 2 ) as t where the limit (x 1,x 2), x 1 0, x 2 0 depends on the initial condition (x 1 (0),x 2 (0)) satisfying x 1 +x 2 = R(0) λ x, x 1 = x1(0) x 2(0) x 2. Proof. Adding the two differential equations in (5) yields (x 1 +x 2 ) (t) = (f(r(t)) d x )(x 1 +x 2 )(t) Then x 1 (t)+x 2 (t) R (0) λ x as t. = (f(r (0) (x 1 +x 2 )(t)) d x )(x 1 +x 2 )(t) > 0 if (x 1 +x 2 )(t) < R (0) λ x, = 0 if (x 1 +x 2 )(t) = R (0) λ x, < 0 if (x 1 +x 2 )(t) > R (0) λ x. ( x1 ) From (5) x 1 x 1 = x 2 x x 2 = f(r(t)) d x, x 2 = 1 x 2 x1 x 2 x 2 x 2 = x 1 x 2 x1 x 1 x 2 x 1 = 0. Hence x 1(t) x1(0) x 2(t) = c, t 0 where c = x. Let x 2(0) 1(t) = cx 2 (t). From x 1 (t)+x 2 (t) R (0) λ x as t, we get (1 + c)x 2 (t) R (0) λ x, and x 2 (t) R(0) λ x = x 2, x 1(t) c(r (0) λ x) 1+c = x 1. Thus (x 1(t),x 2 (t)) (x 1,x 2 ) as t. Remark 2. Every point in the set {(x 1,x 2 )) : x 1 +x 2 = R (0) λ x } is an equilibrium for the system (5). It is easy to verify that each of them is stable, but not asymptotically stable. Consider two parallel food chains with only one predator. The equations take the following form: x 1(t) = (f(r(t)) d x )x 1 (t) g(x 1 (t))y 1 (t), x 2 (t) = (f(r(t)) d x)x 2 (t), y 1 (t) = (g(x 1(t)) d y )y 1 (t), R(t) = R (0) (x 1 (t)+x 2 (t)+y 1 (t)), x 1 (0) > 0,x 2 (0) > 0,y 1 (0) > 0, 0 < x 1 (0)+x 2 (0)+y 1 (0) < R (0). 1+c (5) (6)

5 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 341 In this case we obtain an interesting result that says that a single predator in the parallel food chains cannot survive. The following is a useful lemma in the proof of Theorem 3.3. Lemma 3.2. ([1]) If lim t f(t) exists, and f (t) is bounded, then lim t f (t) = 0. Theorem 3.3. If R (0) > λ x, then the solution (x 1 (t),x 2 (t),y 1 (t)) of (6) satisfies y 1 (t) 0 as t and (x 1 (t),x 2 (t),y 1 (t)) (x 1,x 2,0), for some x 1,x 2 0 satisfying x 1 +x 2 = R(0) λ x. Proof. First, we add the three differential equations in (6) and get (x 1 +x 2 +y 1 ) (t) = [f(r(t)) d x ](x 1 +x 2 )(t) d y y 1 (t). (7) From (7), if f(r(t)) d x 0 then the quantity (x 1 +x 2 +y 1 ) is decreasing. It is noted that f(r(t)) d x 0 R λ x R (0) (x 1 +x 2 +y 1 ) λ x R (0) λ x (x 1 +x 2 +y 1 ). So we divide the positive octant of R 3 into two regions. Let P = {(x 1,x 2,y 1 ) R 3 + x 1 +x 2 +y 1 = R (0) λ x }, Ω 1 = {(x 1,x 2,y 1 ) R 3 + R (0) > x 1 +x 2 +y 1 > R (0) λ x }, Ω 2 = {(x 1,x 2,y 1 ) R 3 + x 1 +x 2 +y 1 < R (0) λ x }. Suppose the trajectory (x 1 (t),x 2 (t),y 1 (t)) stays in the region Ω 1 for all t 0, then (x 1 + x 2 + y 1 )(t) is strictly decreasing and converges to a constant. It is easy to verify that x 1 (t) + x 2 (t) + y 1 (t) is bounded. Then from Lemma 3.2 we have (x 1 +x 2 +y 1 ) (t) 0 as t. Since f(r) d x < 0 in Ω 1, then from (7) we have that f(r) d x 0 and y 1 (t) 0 as t. Hence x 1 (t)+x 2 (t) converges to the constant R (0) λ x as t. If the trajectory (x 1 (t),x 2 (t),y 1 (t)) passes the plane P, then from (7) it enters the region Ω 2 and stays there. Now, consider the trajectory in the region Ω 2. Since R = R (0) (x 1 +x 2 +y 1 ), from (6) x 2(t) = (f(r(t)) d x )x 2 (t). x 2 (t) is strictly increasing and bounded above by R (0) λ x, then x 2 (t) converges, say x 2 (t) x 2 > 0 as t. Since x 2(t) converges and x 2 (t) is bounded, from Lemma 3.2 we obtain that x 2 (t) approaches zero as t. From the second equation of the system (6), it follows that f(r) d x 0 as t, R(t) λ x as t, R (0) (x 1 +x 2 +y 1 )(t) λ x as t, (x 1 +y 1 )(t) (R (0) λ x x 2) as t.

6 342 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN Similarly (x 1 +y 1 ) (t) is bounded, from Lemma 3.2 we have (x 1 +y 1 ) (t) 0 as t. Since f(r) d x 0 as t and x 1 (t) is bounded, it follows that (x 1 +y 1 ) (t) =[f(r(t)) d x ]x 1 (t) d y y 1 (t) d y y 1 ( ) as t. Thus y 1 (t) 0 and we have (x 1 (t),x 2 (t),y 1 (t)) (x 1,x 2,0) as t for some x 1,x 2 0, x 1 +x 2 = R(0) λ x. 4. Mathematical analysis of two parallel food chains system. In this section we focus on the limiting system (3). This is a system of four differential equations, and we proceed in the standard way: identify the rest points, determine their local stability, and discuss its global behavior Rest points and their local stability. The system has the following seven types of rest points: E 0 =(0,0,0,0),E 10 = ( x 1,0,0,0),E 20 = (0, x 2,0,0),E x = (x 1,x 2,0,0), E 1 = (x 1,0,y 1,0),E 2 = (0,x 2,0,y 2 ),E c = (x 1c,x 2c,y 1c,y 2c ). The rest point E 0 always exists; E 10 and E 20 exist if R (0) > λ x where x 1 = x 2 = R (0) λ x. The E x exists if R (0) > λ x where x 1 + x 2 = R (0) λ x. From (3), E 1 exists if R (0) > λ x +λ y, where x 1 = λ y and y 1 satisfies f(r (0) x 1 y 1 ) d x = d y y 1 x 1, (8) y 1 is uniquely determined from (8). Similarly E 2 has the same properties as E 1, so we just consider one of them. From (3) E c exists if and only if R (0) > λ x + 2λ y. It is easy to verify that x 1c = x 2c = λ y and y 1c = y 2c. We denote x c = x 1c = x 2c, y c = y 1c = y 2c. Then y c satisfies f(r (0) 2x c 2y c ) d x = d y y c x c. (9) Note that (9) has a unique positive solution if and only if R (0) > λ x +2λ y. Now we establish the asymptotic stability of the rest points by showing the real parts of the eigenvalues of the variational matrix around the equilibria are negative. The variational matrix J about equilibrium E i, which takes the form At E 0, J(E 0 ) = J = m 11 m 12 m 13 m 14 m 21 m 22 m 23 m 24 m 31 0 m m 42 0 m 44. f(r (0) ) d x f(r (0) ) d x d y d y The eigenvalues are the diagonals. Hence if R (0) < λ x then E 0 is asymptotically stable. If R (0) > λ x then E 0 is a saddle point with two dimensional stable manifold..

7 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 343 At E 10, J(E 10 ) = m 11 m 12 m 13 m m m 44 where m 11 = f ( R) x 1, m 12 = f ( R) x 1, m 13 = f ( R) x 1 g( x 1 ), m 14 = f ( R) x 1, m 33 = g( x 1 ) d y, m 44 = d y, R = R (0) x 1 = λ x. The eigenvalues are 0, m 11, m 33, m 44 with m 11 < 0 and m 44 < 0. We note that m 33 < 0 g( x 1 ) d y < 0 g(r (0) λ x ) < d y R (0) < λ x +λ y. If R (0) > λ x + λ y, then E 10 is a saddle. Similarly results hold for E 20 : if R (0) < λ x +λ y then J(E 20 ) has three negative eigenvalues and one zero eigenvalue, and E 20 is a saddle if R (0) > λ x +λ x. At E x, J(E x ) = m 11 m 12 m 13 m 14 m 21 m 22 m 23 m m m 44 where m 11 = m 12 = m 14 = f (R )x 1, m 13 = f (R )x 1 g(x 1 ), m 21 = m 22 = m 23 = f (R )x 2, m 24 = f (R )x 2 g(x 2), m 33 = g(x 1) d y, m 44 = g(x 2) d y, R = R (0) (x 1 +x 2 ) = λ x. It is easy to see that m 33 and m 44 are two eigenvalues. The remaining two eigenvalues are the eigenvalues of the matrix [ ] m11 m polynomial of 12 is computed as m 21 m 22,, [ ] m11 m 12. The characteristic m 21 m 22 [λ+f (λ x )x 1 ][λ+f (λ x )x 2 ] [f (λ x )] 2 x 1 x 2 = 0 λ 2 +[f (λ x )(R (0) λ x )]λ = 0. Clearly, the other two eigenvalues of J(E x ) are 0 and f (λ x )(R (0) λ x ). We note that m 33 < 0 iff x 1 < λ y and m 44 < 0 iff x 2 < λ y. When R (0) λ x < λ y we get x 1 < λ y, x 2 < λ y. Case 1: If λ x +λ y < R (0) < λ x +2λ y, then the line x 1 +x 2 = R (0) λ x is divided into three parts (See Fig. 1): 1. If(x 1,x 2) AD, then J(E x ) hasonezeroeigenvalue, twonegativeeigenvalues and one positive eigenvalue. The unstable manifold of E x points into the positive y 1 direction. 2. If (x 1,x 2 ) CD, then J(E x) has one zero eigenvalue, three negative eigenvalues. 3. If(x 1,x 2) BC, then J(E x ) hasonezeroeigenvalue, twonegativeeigenvalues and one positive eigenvalue. The unstable manifold of E x points into the positive y 2 direction. Case 2: If R (0) > λ x + 2λ y, then the line x 1 + x 2 = R (0) λ x is divided into three parts (See Fig. 2):

8 344 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN x 2 R (0) λ x λ y R (0) λ x 2 B 0 + C 0 D 0 + A x 1 R (0) λ x 2 λ y R (0) λ x Figure 1. Case 1. λ x + λ y < R (0) < λ x + 2λ y, AB is the line of equilibria (E x ). The plus and minus signs are the signs of the eigenvalues of J(E x ). x 2 R (0) λ x B 0 + C 0 ++ R (0) λ x 2 λ y D 0 + A x 1 λ y R (0) λ x R (0) λ x Figure 2. Case 2. R (0) > λ x +2λ y (x 1,x 2) AD, the eigenvalues of J(E x ) have same properties as in Case (x 1,x 2 ) CD, then J(E x) has one zero eigenvalue, one negative eigenvalue, two positive eigenvalues. 3. (x 1,x 2 ) BC, the eigenvalues of J(E x) have same properties as in Case 1. Case 3: If R (0) = λ x +2λ y, then the line x 1 +x 2 = R (0) λ x is divided into two parts (See Fig. 3): 1. If (x 1,x 2 ) AC, then J(E x) has one zeroeigenvalue, two negativeeigenvalues and one positive eigenvalue. 2. If(x 1,x 2 ) BC, then J(E x) hasonezeroeigenvalue, twonegativeeigenvalues and one positive eigenvalue. At E 1, J(E 1 ) = m 11 m 12 m 13 m 14 0 m m m 44,

9 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 345 x 2 R (0) λ x B 0 + C 0 + A x 1 λ y = R(0) λ x 2 R (0) λ x Figure 3. Case 3. R (0) = λ x +2λ y. λ y = R(0) λ x 2 where m 11 = (f(r) d x ) f (R)x 1 g (x 1 )y 1, m 12 = m 14 = f (R)x 1, m 13 = f (R)x 1 g(x 1 ), m 22 = f(r) d x, m 31 = g (x 1 )y 1, m 44 = d y, R = R (0) (x 1 + y 1 ). It is easy to see that one eigenvalue is m 44 = d y. The rest of three eigenvalues are the eigenvalues of m 11 m 12 m 13 0 m m The eigenvalues λ satisfy λ(m 11 λ)(m 22 λ) (m 22 λ)m 13 m 31 = 0 (m 22 λ)(λ 2 m 11 λ m 13 m 31 ) = 0. One of eigenvalues is m 22 = f(r (0) y x 1 y 1 ) d x = d 1 y x 1 > 0 by (8). The other two eigenvalues satisfy λ 2 m 11 λ m 13 m 31 = 0. Since m 13 m 31 < 0, it follows that the real part of the other two eigenvalues is negative if and only if m 11 < 0, i.e. y d 1 y x 1 f (R (0) x 1 y 1 )x 1 g (x 1 )y 1 < 0. We conclude that if m 11 < 0 then E 1 is a saddle point with three dimensional stable manifold, otherwise if m 11 > 0 then E 1 is saddle with one dimensional stable manifold. Similarly for E 2, if m 22 < 0 then E 2 is a saddle point with three dimensional stable manifold, otherwise if m 22 > 0 then E 2 is saddle with one dimensional stable manifold. At E c, where J(E c ) = m 11 m 12 m 13 m 14 m 21 m 22 m 23 m 24 m m = p q s q q p q s r r 0 0 p =(f(r c ) d x ) f (R c )x c g (x c )y c, q = f (R c )x c < 0, r =g (x c )y c > 0, s = f (R c )x c g(x c ) < 0, R c =R (0) 2(x c +y c ). The stability analysis around E c is presented in the following Lemma 4.1, whose proof is deferred to the appendix., (10)

10 346 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN Lemma 4.1. One of the following three cases holds: (a) E c is a repeller (each eigenvalue of the variational matrix J(E c ) has positive real part) or (b) E c is a saddle point with two dimensional stable manifold, or (c) J(E c ) has two eigenvalues that are purely imaginary, two eigenvalues with positive real part. The local behavior of the rest points are summarized in Table 1. Point Existence Stability E 0 = (0,0,0,0) Always Asymptotically stable if R (0) < λ x. Saddle with 2-D stable manifold if R (0) > λ x. E 10 = ( x 1,0,0,0) x 1 = R (0) λ x R (0) > λ x If R (0) < λ x + λ y, then J(E 10 ) has three negative eigenvalues and one zero eigenvalue. Saddle if R (0) > λ x +λ y. E 20 = (0, x 2,0,0) x 2 = R (0) λ x R (0) > λ x If R (0) < λ x + λ y, then J(E 20 ) has three negative eigenvalues and one zero eigenvalue. Saddle if R (0) > λ x +λ y. E x = (x 1,x 2,0,0) x 1 +x 2 = R(0) λ x R (0) > λ x If R (0) < λ x + λ y, then J(E x ) has three negative eigenvalues and one zero eigenvalue. If R (0) > λ x +λ y, there are Case 1, 2, 3 discsused above. E 1 = (x 1,0,y 1,0) x 1 = λ y, y 1 satisfies f(r (0) x 1 y 1 ) d x = d y y 1 x 1 E 2 = (0,x 2,0,y 2 ) x 2 = λ y, y 2 satisfies f(r (0) x 2 y 2 ) d x = d y y 2 x 2 E c = (x c,x c,y c,y c ) x c = λ y, y c satisfies f(r (0) 2x c 2y c ) d x = d y y c x c R (0) > λ x +λ y R (0) > λ x +λ y R (0) > λ x +2λ y Table 1 Saddle with 3-D stable manifold if y d 1 y x 1 f (R (0) (x 1 + y 1 )) g (x 1 )y 1 < 0. Saddle with 1-D stable manifold if y d 1 y x 1 f (R (0) (x 1 + y 1 )) g (x 1 )y 1 > 0. Saddle with 3-D stable manifold if y d 2 y x 2 f (R (0) (x 2 + y 2 )) g (x 2 )y 2 < 0. Saddle with 1-D stable manifold if y d 2 y x 2 f (R (0) (x 2 + y 2 )) g (x 2 )y 2 > 0. Saddle with 2-D stable manifold or a repeller or E c satisfies (3) of Lemma Global analysis of the two parallel food chains. We have established the existence and local stability of the rest points. As the parameter R (0) varies, the system (3) has different behavior. In the following, we study the global asymptotic behavior of the solutions of (3).

11 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 347 Theorem 4.2. (a) If 0 < R (0) < λ x then (x 1 (t),x 2 (t),y 1 (t),y 2 (t)) E 0 as t. (b) If λ x < R (0) < λ x +λ y then (x 1 (t),x 2 (t),y 1 (t),y 2 (t)) E x = (x 1,x 2,0,0) for some x 1 > 0, x 2 > 0, x 1 +x 2 = R(0) λ x as t. Proof. (a) If 0 < R (0) < λ x then f(r (0) ) d x < 0. From (3) we have x i(t) (f(r(t)) d x )x i (t), (f(r (0) ) d x )x i (t), = δx i (t),i = 1,2, where δ = d x f(r (0) ) > 0. It follows that lim t x i (t) = 0, i = 1,2. From (3) it is obvious that x i (t) 0 t, implies y i (t) 0 as t. (b) If λ x < R (0) < λ x +λ y then g(r (0) λ x ) < g(λ y ) = d y. From (3) we have (x 1 +x 2 ) (t) (f(r(t)) d x )(x 1 +x 2 )(t), (f(r (0) (x 1 +x 2 )(t)) d x )(x 1 +x 2 )(t). From the above differential inequality we get (x 1 +x 2 )(t) R (0) λ x +ε for ε > 0 small such that R (0) λ x +ε < λ y, and t sufficiently large. From (3) it follows that y i (t) y i (t) = g(x i(t)) d y g((x 1 +x 2 )(t)) d y g(r (0) λ x +ε) d y < 0,i = 1,2 if ε > 0 is small enough. Hence lim t y i (t) = 0, i = 1,2. Then we have (x 1 +x 2 ) (t) = (f(r(t)) d x )(x 1 +x 2 )(t) g(x 1 )y 1 g(x 2 )y 2 = (f(r (0) (x 1 +x 2 )(t)) d x )(x 1 +x 2 )(t)+o(1) as t. Thus (x 1 +x 2 )(t) R (0) λ x as t. Further analysis of the system (3) with larger R (0) is very technical. Next we will prove the extinction of top predators. The method is similar to those in the papers of Hsu, Hwang, and Kuang [5, 6], and Hsu [4]. Theorem 4.3. If λ x +λ y < R (0) < λ x +2λ y then lim t y 1 (t)y 2 (t) = 0. Proof. Without loss of generality we can assume that y 1 (0) > 0, y 2 (0) > 0. Let P be the hyperplane P = {(x 1,x 2,y 1,y 2 ) R 4 + x 1 +x 2 +y 1 +y 2 = R (0) λ x }, and the regions Ω 1 and Ω 2 be Ω 1 = {(x 1,x 2,y 1,y 2 ) R 4 + R0 > x 1 +x 2 +y 1 +y 2 > R (0) λ x }, Ω 2 = {(x 1,x 2,y 1,y 2 ) R 4 + x 1 +x 2 +y 1 +y 2 < R (0) λ x }. Adding all differential equations in (3) yields (x 1 +x 2 +y 1 +y 2 ) (t) = [f(r(t)) d x ](x 1 +x 2 )(t) d y (y 1 +y 2 )(t). (11)

12 348 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN If the trajectory (x 1 (t),x 2 (t),y 1 (t),y 2 (t)) stays in Ω 1 for t T where T is large, then (x 1 +x 2 +y 1 +y 2 )(t) is decreasing, since R (0) > x 1 +x 2 +y 1 +y 2 > R (0) λ x R (0) (x 1 +x 2 +y 1 +y 2 ) < R (0) (R (0) λ x ) = λ x f(r) < d x. From (11) and Lemma 3.2, we get (x 1 + x 2 + y 1 + y 2 ) 0 as t, and it follows that (y 1 +y 2 )(t) 0 as t. If the trajectory (x 1 (t),x 2 (t),y 1 (t),y 2 (t)) passes through hyperplane P, then obviously it enters the region Ω 2 and stays there for rest of time. Now we focus on the behavior of the trajectory in Ω 2. In this region we have Note that where x 1 +x 2 +y 1 +y 2 < R (0) λ x < (λ x +2λ y ) λ x = 2λ y. (12) y 1 y 1 + y 2 y 2 = g(x 1 )+g(x 2 ) 2d y = μx 1 K +x 1 + μx 2 K +x 2 2d y = μx 1(K +x 2 )+μx 2 (K +x 1 ) 2d y (K +x 1 )(K +x 2 ) P(x 1,x 2 ) = (K +x 1 )(K +x 2 ). P(x 1,x 2 ) =μx 1 (K +x 2 )+μx 2 (K +x 1 ) 2d y (K +x 1 )(K +x 2 ) =μx 1 (K +x 2 )+μx 2 (K +x 1 ) 2 μλ y K +λ y (K 2 +(x 1 +x 2 )K +x 1 x 2 ) = 1 K +λ y [μk((x 1 +x 2 )(K λ y )+2(x 1 x 2 Kλ y ))]. Thus from (12) P(x 1,x 2 ) μk [(x 1 +x 2 )(K λ y )+ (x 1 +x 2 ) 2 ] 2Kλ y K +λ y 2 < μk [(x 1 +x 2 )(K λ y )+λ y (x 1 +x 2 ) 2Kλ y ] K +λ y = μk K +λ y [(x 1 +x 2 )K 2Kλ y ] < μk K +λ y [2λ y K 2Kλ y ] =0. From above, we conclude that y 1 (t) y 1(t) + y 2 (t) y 2(t) < δ 1 < 0 for some δ 1 > 0 for all t 0. Then y 1 (t)y 2 (t) 0 exponentially as t. Conjecture 1. If λ x +λ y < R (0) < λ x +2λ y, the trajectory (x 1 (t),x 2 (t),y 1 (t),y 2 (t)) satisfies that lim t y i (t) = 0 for i = 1,2.

13 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 349 For the case R (0) > λ x + 2λ y, the global asymptotic behavior of the solution of (3) is more complicated. In this case the interior equilibrium E c exists and it is either a repeller or a saddle point with two dimensional stable manifold or it satisfies Lemma 4.1 (c). The following theorem describes what the stable manifold is. Theorem 4.4. Let R (0) > λ x +2λ y. Assume x 1 (0) = x 2 (0) > 0, y 1 (0) = y 2 (0) > 0. (a) If R (0) < R then (x 1 (t),x 2 (t),y 1 (t),y 2 (t)) E c as t, (b) If R (0) > R then there exists a limit cycle Γ, where R is the unique root of the equation of R (0), μ (K +x ) 2y = 2ma (a+r (0) 2(x +y )) 2, where x = λ y and y = y (R (0) ) satisfies f(r (0) 2x 2y) d x = d y y x. Remark 3. We conjecture (x 1 (t),x 2 (t),y 1 (t),y 2 (t)) approaches the unique limit cycle Γ as t. Proof. Let (x (t),y (t)) be the solution of the initial value problem x (t) =(f(r (0) 2(x+y)(t)) d x )x(t) g(x(t))y(t), y (t) =(g(x(t)) d y )y(t), x(0) = x 1 (0) = x 2 (0) > 0,y(0) = y 1 (0) = y 2 (0) > 0. Then (x (t),x (t),y (t),y (t)) satisfies (3). By uniqueness of the solution of an ordinary differential equation, it follows that x 1 (t) x 2 (t) x (t), y 1 (t) y 2 (t) y (t). Note that One makes the following changes: f(r (0) 2(x+y)) = m(r(0) 2(x+y)) a+(r (0) 2(x+y)) ( ) R 2m (0) 2 (x+y) = ( ). 2 a 2 +2 R (0) 2 (x+y) R (0) R(0) 2, a a 2, λ x λ x 2. Substituting these into above differential equations and dropping the bars yields x (t) =(f(r (0) (x+y)(t)) d x )x(t) g(x(t))y(t), y (t) =(g(x(t)) d y )y(t), x(0) > 0, y(0) > 0. It is the same as the system (4). Thus we complete the proof. Theorem 4.4 shows that when R (0) < R the two dimensional stable manifold of the saddle point E c contains {(x 1,x 2,y 1,y 2 ) R 4 + : x 1 = x 2,y 1 = y 2 }. In the following Theorem 4.5, we show that for the system (3) if the Hopf bifurcation occurs, it must occur at R (0) = R. The proof is deferred to appendix 6. Theorem 4.5. For the system (3) Hopf bifurcation occurs only at R (0) = R, where R is defined in Theorem 4.4.

14 350 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN 5. Discussion and numerical study. Our analytic study on the mathematical model of the two parallel food chains (3) reveals some interesting outcomes as the input concentration of the nutrient R (0) varies. If 0 < R (0) < λ x then both of the prey species and both of the predator species go extinct (Theorem 4.2 (a)). If λ x < R (0) < λ x + λ y then the prey species survive and the predator species go extinct with limiting value E x = (x 1,x 2,0,0) for some x 1,x 2 > 0 satisfying x 1 +x 2 = R(0) λ x (Theorem 4.2 (b)). If λ x +λ y < R (0) < λ x +2λ y then from Theorem 4.3 the predator species y 1 and y 2 satisfy lim t y 1 (t)y 2 (t) = 0. For the case R (0) > λ x + 2λ y, if the initial populations satisfy x 1 (0) = x 2 (0) > 0, y 1 (0) = y 2 (0) > 0 then the trajectory approaches E c = (x c,x c,y c,y c ) as t provided λ x +2λ y < R (0) < R (Theorem 4.4 (a)) or the trajectory approaches a unique limit cycle provided R (0) > R (Theorem 4.4 (b)). In the case λ x + λ y < R (0) < λ x + 2λ y, we know that y 1 and y 2 satisfy lim t y 1 (t)y 2 (t) = 0. Furthermore, by extensive numerical simulations, we conjecture that y 1 and y 2 approach zero as time goes to infinity (See Fig 4). For the case R > R (0) > λ x + 2λ y, we present some numerical simulations with varying initial conditions (Fig 5). The interior equilibrium E c is a saddle point with two-dimensional stable manifold. Each figure in (a) (e) shows the same behavior for the solutions of the system (3). The prey populations x 1 (t) and x 2 (t) alternatively exchange between the maximum x h and minimum x l, x h +x l = R (0) λ x. When the prey species x 1 (t) (x 2 (t)) decreases rapidly from the maximum x h to the minimum x l, the prey species x 2 (t) (x 1 (t)) increases rapidly from the minimum x l to the maximum x h and predator species y 1 (t) (y 2 (t)) behaves like a pulse when x 1 (t) (x 2 (t)) exchanges from x h to x l. Moreover, the prey population x 1 (t), x 2 (t) stays at the maximum x h and the minimum x l longer and longer as time becomes large. We may explain this phenomena by Fig 6. From Fig 2, for each equilibrium P on the segment AD the one-dimensional unstable manifold W (P) points into positive y 1 direction with zero y 2 -component. W (P) will approach the line x 1 + x 2 = R (0) λ x in x 1 -x 2 plane (Theorem 3.3). Similarly for each equilibrium Q on the segment BC, the one-dimensional unstable manifold W (Q) points into positive y 2 direction with zero y 1 -component. We conjecture that there exists a unique heteroclinic orbit Γ 1 from P = (x h,x l,0,0) to Q = (x l,x h,0,0) and a unique heteroclinic orbit Γ 2 from Q = (x l,x h,0,0) to P = (x h,x l,0,0) (See Fig 6 (a), (b)). Each time that the orbit γ(t) = (x 1 (t),x 2 (t),y 1 (t),y 2 (t)) approaches the equilibria (x h,x l,0,0) and (x l,x h,0,0), it stays there successively longer and longer. We note that in Fig 5 (f), the initial conditions satisfies x 1 (0) = x 2 (0), y 1 (0) = y 2 (0), i.e. γ(0) lies on the stable manifold of the interior equilibrium E c, the trajectory γ(t) approaches E c as t. In Figure 7 the input concentration R (0) satisfies R (0) > R and the interior equilibrium E c is a repeller (i.e. each eigenvalue of the variational matrix J(E c ) has positive real parts). In this case the behavior of the orbit γ(t) is similar to above case except that x h R (0) λ x and x l 0. Figure 7 (a) (e) show the prey population x 1 (t), x 2 (t) alternatively exchange between x h and x l and the predatorpopulation y 1 (t), y 2 (t) increaseand decreaserapidly when prey population exchange their values. We note that Figure 7 (f) shows that x(t) = x 1 (t) = x 2 (t) and y(t) = y 1 (t) = y 2 (t) oscillate periodically when x 1 (0) = x 2 (0), y 1 (0) = y 2 (0).

15 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 351 Figure 4. R (0) = 1, m = 2.2, a = 0.3, μ = 1.8, K = 0.4, d x = d y = 1, then λ x = 0.25, λ y = 0.5, 0.75 = λ x + λ y < 1 = R (0) < λ x +2λ y = The initial conditions are : x 1 (0) = 0.1, x 2 (0) = 0.2, y 1 (0) = 0.2, y 2 (0) = 0.1. Figure 5. R (0) = 2.8, m = 1.5, a = 0.3, μ = 1.5, K = 0.5, d x = d y = 1. Then λ x = 0.6, λ y = 1, R = > R (0) = 2.8 > λ x + 2λ y = 2.6, E c = (1,1,0.05,0.05). The initial conditions (x 1 (0),x 2 (0),y 1 (0),y 2 (0)) are (a) (1.1,1.05,0.01,0.3), (b) (0.06,0.12,1.2,1.03), (c) (0.05,1.1,1.32,0.2), (d) (0.75,0.7,0.05,0.05), (e)(0.7,0.7,0.05,0.1), (f) (0.7,0.7,0.05,0.05).

16 352 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN In [9, 11, 2] the authors studied the competition of three species with same intrinsic growth rate x 1 = x 1 (1 x 1 α 1 x 1 β 1 x 3 ), x 2 = x 2 (1 β 2 x 1 x 2 α 2 x 3 ), x 3 = x 3(1 α 3 x 1 β 3 x 2 x 3 ), x 1 (0) > 0, x 2 (0) > 0, x 3 (0) > 0, (13) where the parameters α i, β i, satisfy 0 < α i < 1 < β i, i = 1, 2, 3. (14) The condition (14) implies that there exists a heteroclinic orbit O 3 on the x 1 -x 2 plane from e 2 = (0,1,0) to e 1 = (1,0,0), a heteroclinic orbit O 2 on the x 1 -x 3 plane from e 1 to e 3 = (0,0,1) and a heteroclinic orbit O 1 on the x 2 -x 3 plane from e 3 to e 2. The system (13) is referred to as the asymmetric May-Leonard model or the rockpaper-scissors game. When α 1 = α 2 = α 3, β 1 = β 2 = β 3, (13) is referred to as the symmetricmay-leonardmodel. LetA i = 1 α i, B i = β i 1. IfA 1 A 2 A 3 < B 1 B 2 B 3 then it was shown [2, 11], the heteroclinic cycle O = 3 i=1 O i is an attractor. The system (3) exhibits similar behavior as (13) for the case R (0) > λ x + 2λ y with some parameters (m = 1.5, a = 0.3, μ = 1.5, K = 0.5, d x = d y = 1). For the case R (0) > λ x +2λ y, if the initial populations for either prey or predator are not identical, then from our extensive numerical simulations there are many different kinds of behaviors for the trajectory of the system (3). The trajectory may approach a unique heteroclinic orbit in some parameter range as we discuss above (See Fig 5 and Fig 7). The trajectory may approach a limit cycle (See Fig 8) in some parameter range. It is possible that the trajectory is chaotic in some parameter range (See Fig 9). For fixed R (0) we denoted y j 1 as the j-th local maximum of y 1 (t) for t [1000,2000]. In Fig 10, we plot the orbit diagram as R (0) varies. We note that in [14] Vandermeer studied a two-prey and two-predator system with switch predation mechanism. The behavior of the system also exhibits periodic oscillation and chaotic behavior as our system (3) does. There are still many open problems left for future investigation. We list as followings: (Q1): Under the assumption λ x +λ y < R (0) < λ x +2λ y, is it true that y 1 (t) 0 and y 2 (t) 0 as t. (Q2): Existence and stability of heteroclinic orbit. (Q3): When does the period doubling routes to chaos occur? (Q4): How does the asymptotic behavior change when the two food chains are not perfectly symmetrical? 6. Appendix. Proof of Lemma 4.1. Byroutinecomputation,thecharacteristicpolynomialofJ(E c ) is λ 4 +A 1 λ 3 +A 2 λ 2 +A 3 λ+a 4 = 0, (15)

17 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 353 where A 1 = 2p, The equation (15) can be rewritten as A 2 =p 2 q 2 2rs, A 3 =2r(ps q 2 ), A 4 =r 2 (s 2 q 2 ). Then eigenvalues of J(E c ) are the roots of (16) (λ 2 +a 1 λ+b 1 )(λ 2 +a 2 λ+b 2 ) = 0. (17) λ 2 +a 1 λ+b 1 = 0, λ 2 +a 2 λ+b 2 = 0. Let the roots of first and second equations be z 1, z 2 and z 3, z 4 respectively. Comparing (15) and (17) yields A 1 =a 1 +a 2 = 2p, A 2 =a 1 a 2 +b 1 +b 2 = p 2 q 2 2rs, A 3 =a 1 b 2 +b 1 a 2 = 2r(ps q 2 ), A 4 =b 1 b 2 = r 2 (s 2 q 2 ). From (11), it is easy to show that b 1 b 2 = r 2 (s 2 q 2 ) > 0. Thus 0 is not an eigenvalue. If Re(z i ) = 0 for all i = 1,2,3,4 then p = 0 and there are two pair of pure imaginary eigenvalues, therefore (17) becomes (λ 2 + b 1 )(λ 2 + b 2 ) = 0 and A 3 = 0. However, A 3 = 2rq 2 = 2g (λ y )y c [f (R c )λ y ] 2 < 0, a contradiction. From above observation, we know that there are at most one pair of pure imaginary eigenvalues. On the other hand, b 1 b 2 > 0 implies that both of b 1 and b 2 are either positive or negative. If b 1 < 0, b 2 < 0 then we have four real roots with two positive and two negative. Hence from now on, we only consider the case b 1 > 0, b 2 > 0. The following lemma is useful to the proof of Lemma 4.1. Lemma 6.1. p as a function of R (0), p(r (0) ) is increasing in R (0) and there exists a R (λ x +2λ y, ) such that p(r ) = 0. Proof. From (11) we let p(r (0) ) =f(r (0) 2λ y 2y c (R (0) )) d x f (R (0) 2λ y 2y c (R (0) ))λ y g (λ y )y c (R (0) ). Differentiating p(r (0) ) with respect to R (0) yields dp(r (0) ) dr (0) =f (R (0) 2λ y 2y c (R (0) )) (1 2y c (R(0) )) From (9) y c satisfies f (R (0) 2λ y 2y c (R (0) ))λ y (1 2y c(r (0) )) g (λ y )y c(r (0) ). (18) f(r (0) 2λ y 2y c (R (0) )) d x = d y λ y y c (R (0) ). (19)

18 354 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN Thus we have f (R (0) 2λ y 2y c (R (0) )) (1 2y c(r (0) )) = d y λ y y c(r (0) ), (20) y c f (R (0) 2λ y 2y x (R (0) )) (R(0) ) = d y > 0. (21) λ y +2f (R (0) 2λ y 2y x (R (0) )) From (20) we have f (R (0) 2λ y 2y c (R (0) )) (1 2y c(r (0) )) g (λ y )y c(r (0) ) = d y y λ c(r (0) ) g (λ y )y c(r (0) ) y [ ] =y c g(λy ) (R(0) ) g (λ y ) λ y =y c(r (0) μ λ y ) > 0. K +λ y K +λ y From above and f < 0 we have dp(r(0) ) > 0 i.e. p(r (0) ) is strictly increasing dr (0) in R (0). Next, we show that there is a unique root of p(r (0) ) = 0. When R (0) = λ x +2λ y, (19) implies y c (R (0) ) = 0. Then p(r (0) ) = f (λ x )λ y < 0. Consider the caser (0). From(19)and(21)itfollowsthat dy λ y y c (R (0) ) f(r (0) 2λ y ) d x < m d x, and lim R (0) y c (R (0) ) = y c ( ) = λy d y (m d x ). We note that Thus p(r (0) ) =f(r (0) 2λ y 2y c (R (0) )) d x f (R (0) 2λ y 2y c (R (0) ))λ y g (λ y )y c (R (0) ) = m(r(0) 2λ y 2y c (R (0) )) a+r (0) 2λ y 2y c (R (0) ) d x ma μk (a+r (0) 2λ y 2y c (R (0) )) 2λ y (K +λ y ) 2y c(r (0) ) = 2λy+2yc(R m(1 (0) ) ) R (0) a +1 2λy+2yc(R(0) ) R (0) R (0) ( m d x ma (R (0) ) 2 d x a +1 2λy+2yc(R(0) ) R (0) R (0) lim R (0) ) 2 λ y μk (K +λ y ) 2y c( ) as R (0). p(r (0) ) =p( ) = μk (K +λ y ) 2y c(r (0) ) [ ] dy μk λ y (K +λ y ) 2 y c ( ) = μλ y (K +λ y ) 2y c( ) > 0. μ Hence p( ) = (K+λ y) λ 2 y y c ( ) > 0. Therefore there exists a unique root R > λ x +2λ y s.t. p(r ) = 0.

19 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 355 In the following we discuss the properties of eigenvalues of J(E c ) as p varies. Case 1: If p = 0 then 4 i=1 Re(z i) = 0. From above, we note that it is impossible that Re(z i ) = 0 for all i = 1,2,3,4. If there exist Re(z i ) = 0 for some i, then there is exactly a pair of pure imaginary eigenvalues. Assume z i = ±iβ and z 3, z 4 are real. Then z 3 +z 4 = 0, b 2 = z 3 z 4 < 0, a contradiction to b 2 > 0. Thus Re(z i ) = 0 for all i. From p = 0 we have a 1 = a 2 and two of real part of z i are positive, two are negative. Case 2: If p > 0, then a 1 +a 2 = 2p < 0, a 1 b 2 +b 1 a 2 = 2r(ps q 2 ) < 0. W.L.O.G let a 1 a 2, there are three subcases. (i) a 1 a 2 < 0 a 1 < 0, a 2 > 0. (ii) a 1 a 2 > 0 a 1 < 0, a 2 < 0. (iii) a 1 a 2 = 0 a 1 < 0, a 2 = 0. For subcase (i) a 1 < 0, a 2 > 0, the eigenvalues z i, i = 1,2,3,4 satisfy { z1 z 2 = b 1 > 0, z 1 +z 2 = a 1 > 0. { z3 z 4 = b 2 > 0, z 3 +z 4 = a 2 < 0. From above we have that z 1, z 2 have positive real parts and z 3, z 4 have negative real parts. For subcase (ii) a 1 < 0, a 2 < 0, all 4 roots have positive real part. For subcase (iii) a 1 a 2 = 0 a 1 < 0, a 2 = 0, then there is a pair of pure imaginary eigenvalues z 3, z 4. Other eigenvalues z 1, z 2 have positive real parts. Case 3: If p < 0, then a 1 +a 2 = 2p > 0. Similarly, let a 1 a 2 and consider the following subcases. (i) a 1 a 2 < 0 a 1 > 0, a 2 < 0. (ii) a 1 a 2 > 0 a 1 > 0, a 2 > 0. (iii) a 1 a 2 = 0 a 1 > 0, a 2 = 0. Similar to the discussion of Case 2, we havethe followingresults. For the subcase (i) a 1 > 0, a 2 < 0, there aretworootswith positive realpartsandtwowith negative real parts. For subcase (ii) a 1 > 0, a 2 > 0 the real parts of all roots are negative. From Routh-Hurwitz criterion, we have A 1 > 0, A 2 > 0, A 3 > 0 and A 4 > 0 also A 3 (A 1 A 2 A 3 ) > A 2 1 A 4 ([3] p.55), or equivalently 2p > 0, p 2 q 2 2rs > 0, 2r(ps q 2 ) > 0, r 2 (s 2 q 2 ) > 0, (2r(ps q 2 ))(( 2p)(p 2 q 2 2rs) (2r(ps q 2 ))) > ( 2p) 2 (r 2 (s 2 q 2 )). (22) Assume (22) holds then from (11) and (18) 0 < a 1 +a 2 = 2p, 0 < a 1 a 2 +b 1 +b 2 = p 2 q 2 2rs q 2 p 2 < 2rs, 0 < a 1 b 2 +b 1 a 2 = 2r(ps q 2 ) q 2 ps < 0.

20 356 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN Note that Thus ( 2p)(p 2 q 2 2rs) (2r(ps q 2 )) =2[p(q 2 p 2 )+prs+q 2 r] <2[p( 2rs)+prs+q 2 r] =2[r(q 2 ps)] <0. (2r(ps q 2 ))(( 2p)(p 2 q 2 2rs) (2r(qs q 2 ))) < 0 < ( 2p) 2 (r 2 (s 2 q 2 )), which contradicts (22). Thus subcase (ii) cannot hold. (iii) a 1 > 0, a 2 = 0, thena 1 b 2 +b 1 a 2 = a 1 b 2 > 0. Ontheotherhanda 1 b 2 +b 1 a 2 = 2r(ps q 2 ) 0, a contradiction. From the above discussion, we complete the proof of Lemma 4.1. Proof of Theorem 4.5. From above discussion, Hopf bifurcation doesn t occur at p 0 i.e. R (0) R. Ifwereversethetimet tinthesystem(3), thenthecharacteristicpolynomial of J(E c ) becomes λ 4 A 1 λ 3 +A 2 λ 2 A 3 λ+a 4 = 0. (23) From (17) and Routh-Hurwitz criterion, Re(λ) < 0 if and only if B 1 = A 1 = 2p > 0, B 2 =A 2 = p 2 q 2 2rs > 0, B 3 = A 3 = 2r(ps q 2 ) > 0, B 4 =A 4 = r 2 (s 2 q 2 ) > 0, Δ =B 3 (B 1 B 2 B 3 ) B 2 1B 4 =[ 2r(ps q 2 )][2p(p 2 q 2 2rs)+2r(ps q 2 )] (2p) 2 r 2 (s 2 q 2 ) > 0. B 1 > 0, B 3 > 0 and B 4 > 0 hold spontaneously. If Δ > 0, then B 2 > 0. Thus Δ > 0 if and only if all eigenvalues λ have strictly negative real parts. We claim that Δ = 0 if and only if there are two purely imaginary eigenvalues and two eigenvalues with negative real parts. Thus, if Δ < 0 then there are two eigenvalues with positive real parts and two eigenvalues with negative real parts. Therefore bifurcation occurs as Δ = 0. To prove the claim, we use the similar discussion in the proof of Lemma 4.1, (23) can be rewritten as the form (17). Δ = 0 B 3 (B 1 B 2 B 3 ) B 2 1 B 4 = 0 (a 1 b 2 +a 2 b 1 )[(a 1 +a 2 )(a 1 a 2 +b 1 +b 2 ) (a 1 b 2 +a 2 b 1 )] (a 1 +a 2 ) 2 (b 1 b 2 ) = 0 a 1 a 2 [(b 1 b 2 ) 2 +(a 1 +a 2 )(a 1 b 2 +a 2 b 1 )] = 0. It is easy to get the sufficient direction. Conversely, we consider Δ =[ 2r(ps q 2 )][2p(p 2 q 2 2rs)+2r(ps q 2 )] (2p) 2 r 2 (s 2 q 2 ) =4r[(p q)(p+q)(pq 2 sp 2 +rq 2 )].

21 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 357 If Δ = 0, then p = q and it implies a 1 b 2 +a 2 b 1 = B 3 = 2rq(s +q) > 0. Hence [(b 1 b 2 ) 2 +(a 1 +a 2 )(a 1 b 2 +a 2 b 1 )] > 0 and a 1 a 2 = 0. The claim holds. From above Δ > 0 if and only if p + q > 0, i.e. (f(r c ) d x ) 2f (R c )λ y g (λ y )y c > 0. (f(r c ) d x ) 2f (R c )λ y g (λ y )y c = 1 ma μk d y y c 2 λ y (a+r c ) 2λ y (K +λ y ) 2y c [ ] μ ma =λ y (K +λ y ) 2y c 2 (a+r c ) 2. Thus(f(R c ) d x ) 2f (R c )λ y g μ ma (λ y )y c > 0ifandonlyif (K+λ y) y 2 c 2(a+R c) > 0. 2 The latter is the criterion in Theorem 4.4 such that R (0) > R. Therefore Δ > 0 if and only if R (0) > R. Thus if Hopf bifurcation occurs, it must occur at R (0) = R. Acknowledgments. We would like to thank Professor X. Q. Zhao for the stimulating discussion. Also we thank the referees for the improvement of the paper. REFERENCES [1] W. A. Coppell, Stability and Asymptotic Behavior of Solutions of Differential Equations, Heath, Boston, [2] C. W. Chi, S. B. Hsu and L. I. Wu, On the asymmetric May-Leonard model of three competing species, SIAM J. Appl. Math., 58 (1998), [3] S. B. Hsu, Ordinary Differential Equations and Applications, World Scientific Press, [4] S. B. Hsu, Predator-mediated coexistence and extinction, Math. Bios., 54 (1981), [5] S. B. Hsu, T. W. Hwang and Y. Kuang, A ratio-dependent food chain model and its applications to biological control, Math. Bios., 181 (2003), [6] S. B. Hsu, T. W. Hwang and Y. Kuang, Rich dynamics of a ratio-depend one-prey-twopredators model, J. Math. Biol., 43 (2001), [7] J. Huisman and F. J. Weissing, Biodiversity of plankton by species oscillations and chaos, Nature, 402 (1999), [8] A. J. Lotka, Elements of Physical Biology, Williams and Wilkins, [9] R. M. May and W. J. Leonard, Nonlinear aspects of competition between three species, SIAM J. Appl. Math., 29 (1975), [10] W. W. Murdoch, C. J. Briggs and R. M. Nisbet, Consumer-Resource Dynamics, Princeton University Press, [11] P. Schuster, K. Sigmund, and R. Wolf, On ω-limit for competition between three species, SIAM J. Appl. Math., 37 (1979), [12] H. L. Smith and P. Waltman, Theory of Chemostat, Cambridge University Press, [13] P. Turchin, Complex Population Dynamics: A Theoretical/Empirical Synthesis, Princeton University Press, [14] J. Vandermeer, Coupled oscillations in food webs: balancing competition and mutualism in simple ecological models, American Naturalist, 163 (2004), [15] V. Volterra, Variazioni e fluttuazioni del numero d individui in specie animali conviventi, Mem. R. Accad. Naz. dei Lincei, 2 (1926), Received November address: sbhsu@math.nthu.edu.tw address: klausme1@msu.edu address: d @oz.nthu.edu.tw

22 358 SZE-BI HSU, CHRISTOPHER A. KLAUSMEIER AND CHIU-JU LIN (a) y 1 Γ 1 (b) y 2 Γ 2 x 1 B x 2 B (x l,x h,0) (x l,x h,0) C C A D A D (x h,x l,0) (x h,x l,0) x 1 Figure 6 x 2 Figure 7. R (0) = 4.8, m = 1.5, a = 0.3, μ = 1.5, K = 0.5, d x = d y = 1, E c = (1,1,0.3175,0.3175). The initial conditions (x 1 (0),x 2 (0),y 1 (0),y 2 (0)) are (a) (0.2,0.6,0.05,1.3), (b) (2.3,1.5,0.05,0.3), (c) (0.05,0.1,0.08,0.1), (d) (0.6,0.6,1.1,1.05), (e) (0.65,0.6,1.1,1.1), (f) (0.6,0.6,1.1,1.1).

23 ANALYSIS OF A MODEL OF TWO PARALLEL FOOD CHAINS 359 Figure 8. R (0) = 1.05, m = 10, a = 0.3, μ = 10, K = 0.5, d x = d y = 1, then λ x = , λ y = The initial conditions (x 1 (0),x 2 (0),y 1 (0),y 2 (0)) are (0.1,0.2,0.2,0.1). Figure 9. R (0) = 1.3, m = 10, a = 0.3, μ = 10, K = 0.5, d x = d y = 1, then λ x = , λ y = The initial conditions (x 1 (0),x 2 (0),y 1 (0),y 2 (0)) are (0.1,0.2,0.2,0.1). Figure 10. Fixed m = 10, a = 0.3, μ = 10, K = 0.5, d x = d y = 1, then λ x = ,λ y = The range of R (0) is from 0.6 to 3.2.