AN EXTENDED ROSENZWEIG-MACARTHUR MODEL OF A TRITROPHIC FOOD CHAIN. Nicole Rocco

Transcription

1 AN EXTENDED ROSENZWEIG-MACARTHUR MODEL OF A TRITROPHIC FOOD CHAIN Nicole Rocco A Thesis Submitted to the University of North Carolina Wilmington in Partial Fulfillment of the Requirements for the Degree of Master of Science Department of Mathematics and Statistics University of North Carolina Wilmington 2011 Approved by Advisory Committee Michael Freeze Xin Lu Wei Feng Chair Accepted by Dean, Graduate School

2 This thesis has been prepared in the style and format consistent with the journal American Mathematical Monthly. ii

3 TABLE OF CONTENTS ABSTRACT iv ACKNOWLEDGMENTS v 1 HISTORY INTRODUCTION PRELIMINARY BOUNDS AND CONVERGENCE Ultimate Bounds for the Prey Population Ultimate Bounds for the Predator Population Ultimate Bounds for the Super-Predator Population STABILITY Jacobian Stability for the trivial equilibrium solution Stability with prey and no predator nor super-predator Stability with prey, predator and no super-predator Stability with prey, super-predator and no predator NUMERICAL SIMULATIONS Numerical Simulations for Equilibrium E Numerical Simulations for Equilibrium E Numerical Simulations for Equilibrium E COEXISTENCE, OSCILLATION AND COMPLEXITY OF THE MODEL Stability for Equilibrium E Oscillation for Equilibrium E CONCLUSIONS REFERENCES iii

4 ABSTRACT In this paper, we modify an extension of the Rosenzweig-MacArthur ditrophic food chain model to the tritrophic case. We study the dynamics of the three species food chain in which the top predator not only consumes the predator under it but also consumes the prey as well, a more realistic model. We obtain the ultimate bounds and exponential convergence for these populations, find all possible equilibrium solutions and investigate the stability and instability of equilibrium solutions in relation with all the ecological parameters. The population dynamics are then graphically represented by numerical simulations. iv

5 ACKNOWLEDGMENTS I would like to thank those who have helped me with this thesis and who have helped me throughout my college career. To Dr. Feng: Thank you for helping me complete this thesis. Even during the times when I got frustrated or got stuck on something, you pushed me to continue and believed that I could accomplish the task at hand. Thank you for all your support and encouragement. I would also like to thank my committee members Dr. Lu and Dr. Freeze. Thank you for your help and your suggestions with this thesis. To Dr. Freeze: Over the years, I have enjoyed taking your classes and working with you on my Honors project last year. Thank you for never just giving me the answers. You always provided guidance but believed that I could do something even when I didn t believe I could. To Dr. Karlof: Thank you for encouraging me to do the 5-year program. It was one of the best decisions I have made. To Brevin Rock: Thank you for your help with getting this thesis started by teaching me Maple. Without your help in explaining functional response and Maple, I would not have gained the insight I have today. v

6 1 HISTORY One of the first differential equation models that incorporated the interactions between two species is the Lotka-Volterra model. This model, proposed by Alfred Lotka and Vito Volterra in 1925 [10, 11], is a predator-prey model and is represented by, dx = x (α βy) dt dy = y (γ δx), (1.1) dt where y is the predator and x is the prey. The functions dy dt dx and dt are the growth rates of the populations over time, t. The parameters α, β, γ and δ represent the interaction between the two species. Prior to the Lotka-Volterra model, the models only considered single species. The Malthusian model, P (t) = P 0 e rt, (1.2) where P 0 is the initial population and r is the growth rate, is one of the simplest growth models with exponential growth. After this was the Logistic Equation, ( dp dt = rp 1 P ), (1.3) K where r is the birth rate and K the carrying capacity, once again representing a single species. In 1959, Holling suggested a function response model often referred to as the disk equation. The equation was developed using blindfolded human subjects trying to find and pickup small discs of sandpaper on a flat surface. It describes the relationship between search time, handling time, and consumption rate and is a Type II

7 response. Holling studied predation of small mammals on sawflies and found that the predation rates increased with increasing prey density [2, 6, 7]. The model assumes that the predator takes all the prey it encounters, searches randomly, and that the predators instantaneous rate of discovery, and the handling time required to kill and consume a prey item are constant and independent of prey density or feeding rate. Type II functional responses must ultimately reach an upper limit. Predator density increases with increasing prey density. Holling considered the responses of the prey and predator as functional response and numerical response. Later, we will discuss functional response pertaining to our model. Holling s three types of functional response: 1. Holling Type I Functional Response: This is a linear response in which the attack rate of the individual consumer increases linearly with prey density but then suddenly reaches a constant value when the consumer is full. It is found in passive predators like spiders. The number of flies caught is proportional to fly density. The mortality of the prey from predation is constant. Figure 1: Holling Type I 2. Holling Type II Functional Response: This response is cyrtoid, or having the curvature or shape of a hump, and is 2

8 the response where the attack rate increases at a decreasing rate with prey density until it becomes constant at saturation. It is the most typical function response. This is a typical response of predators that specialize on one or a few prey. The prey mortality declines with prey density and predators of this type cause maximum mortality at low prey density. An example of such would be small mammals that destroy most of gypsy moth pupae in sparse populations of gypsy moth. However in populations which deprive the trees and forests of leaves, these small mammals kill a minimal proportion of pupae. Figure 2: Holling Type II 3. Holling Type III Functional Response: This response is sigmoid, or logistic, in which the attack rate accelerates at first and then decelerates towards satiation. It is typical of natural enemies that easily switch from one food species to another and focus their consumption in areas where specific resources are the most abundant. This occurs with predators which increase their search activity with increasing prey density. Mortality first increases with prey increasing density, and then declines. For example, many vertebrate predators are able to switch to the most abundant species of prey by learning to visually recognize it and distinguish it from others. 3

9 Figure 3: Holling Type III The Lotka-Volterra models introduced the interactions of multiple species while Holling s functional responses extended them. The systems of equations is an example of a model that can illustrate the dynamics of biological systems with predatorprey interactions and competition. The Lotka-Volterra and Holling responses helped pave the way for other predator-prey equation models that considered multiple factors. Location, specific food consumption, limits on food or limits on the growth of the population are just a few factors that could now be incorporated into the models. In addition to considering various factors, these models next led to the research of three species models. 4

10 2 INTRODUCTION As with most models, the initial proposed model can always be improved. The Lotka- Volterra model makes two unrealistic assumptions. First, it assumes that in the absence of predators, the prey population will grow exponentially. Second, it implies that individual predators never get full. The Rosenzweig-MacArthur ditrophic food chain model (1963) corrects these assumptions: dx ( 1 dt = rx 1 1 x ) 1 x 1 a 2 x 2 K b 1 + x 1 dx 2 dt = c x 1 2a 2 x 2 d 2 x 2, (2.1) b 1 + x 1 where x 1 is the prey population, x 2 is the predator population, r is the intrinsic growth rate, K is the carrying capacity of the prey population and a 2 is the maximum predation rate for the predator. The parameter b 1 is the half saturation constant and c 2 is the predator efficiency. Parameter d 2 stands for the death rate of the predator. This thesis is influenced by an article published by Matteo Candaten and Sergio Rinaldi in 2003 [1]. The article discusses an extension of the Rosenzweig-MacArthur ditrophic food chain model, shown below: dx ( 1 dt = rx 1 1 x ) 1 x 1 a 2 x 2 K b 1 + x 1 dx 2 dt = c x 1 x 2 2a 2 x 2 d 2 x 2 a 3 x 3 b 1 + x 1 b 2 + x 2 dx 3 dt = c x 2 3a 3 x 3 d 3 x 3 (2.2) b 2 + x 2 This model has a prey, a predator and a super-predator. The predator consumes the prey and the super-predator consumes the predator.

11 The authors then present a model that includes a limiting nutrient for the prey and provides numerical simulations for this tritrophic food chain extension. The authors then go on to describe the food chain dynamics for various models [1, 5, 4]. This thesis is based on a further extension of the model presented in the article [1], which had added to the Rosenzweig-MacArthur model. We modify an extension of the Rosenzweig-MacArthur food chain (2.2) to fit a more realistic situation: dx ( 1 dt = rx 1 1 x ) 1 x 1 x 1 a 2 x 2 a 3 x 3 K b 1 + x 1 b 1 + x 1 dx 2 dt = c x 1 x 2 2a 2 x 2 d 2 x 2 a 3 x 3 b 1 + x 1 b 2 + x 2 dx 3 dt = c x 2 x 1 3a 3 x 3 d 3 x 3 + c 3 a 3 x 3 (2.3) b 2 + x 2 b 1 + x 1 This model has a prey (x 1 ), a predator(x 2 ) that consumes the prey and a superpredator (x 3 ) that has the ability to consume both the predator and the prey. Therefore, not only is the prey being attacked by the predator but it is also being attacked by the super-predator. In the model (2.3), r is the intrinsic growth rate, K is the carrying capacity of the prey population, a 2 is the maximum predation rate for the predator and a 3 is the maximum predation rate for the super-predator. There are also b 1 and b 2 which are the half saturation constants, c 2 which is the predator efficiency and c 3 which is the super-predator efficiency. Lastly, there are d 2 and d 3 which stand for the death rates. We also will consider the types of functional response in our new model. The numerical response is the intake rate of a species as a function of food density. The term c 2a 2 x 1 x 2 b 1 +x 1 in the equation calculates the number of prey killed by the predator. The numerator represents the potential kills and the number of possible pair-wise 6

12 interactions between the prey and predator. The denominator represents the relationship between the prey population and the number of prey killed and shows that there is an upper limit for the number of predators regardless of the amount of prey available. As the prey increases for any given number of predators, the number of prey killed per predator increases at some decreasing rate to a limiting upper value. c 2 a 2 x 1 x 2 In this case it is lim = c 2 a 2 x 2. Similar to the way c 2a 2 x 1 x 2 x 1 b b 1 + x 1 +x 1 describes the 1 relationship between prey and predator, c 3a 3 x 2 x 3 b 2 +x 2 describes the relationship between predator and super-predator using the same idea. Also, c 3a 3 x 1 x 3 b 1 +x 1 describes the relationship between the prey and super-predator using the same concepts. In the following descriptions of functional responses, we focus on c 2a 2 x 1 x 2 b 1 +x 1. We relate the types of Holling functional response to our model. 1. Holling Type I Functional Response: To achieve this behavior in our model, we would change the functional response term from c 2a 2 x 1 x 2 b 1 +x 1 to c 2 a 2 x 1 x 2. Notice that this change makes it a linear function of x 1. This means that as the prey population increases, the predator population increases its consumption. Figure 4: Holling Type I for the Model 2. Holling Type II Functional Response: This is the response applied in this paper. 7

13 Figure 5: Holling Type II for the Model 3. Holling Type III Functional Response: To obtain this behavior in our model we change the functional response to c 2 a 2 x 2 1+e b 1 x 1. Figure 6: Holling Type III for the Model In this paper, we seek to analyze the dynamics of the populations by: 1. determining the upper and lower bounds for the species, 2. examining the stability and instability of the equilibrium solutions, 3. providing numerical simulations for the dynamics found in the model. We begin in the next section by finding the equilibrium solutions and then proceeding to analyze the local stability of these points. 8

14 3 PRELIMINARY An important method that is employed in analyzing the dynamics of a system is the method of linearization. For more on the theorems presented here, see [10]. Suppose we have the system dx dt dy dt dz dt = f (x, y, z) = g (x, y, z) = h (x, y, z) (3.1) where we assume that f, g and h have continous partial derivatives. Then the solution of this initial value problem with x (t 0 ) = x 0, y (t 0 ) = y 0, z (t 0 ) = z 0 is unique. Now we let (x 0, y 0, z 0 ) be a critical point, also known as an equilibrium solution. That is, we assume f (x 0, y 0, z 0 ) = g (x 0, y 0, z 0 ) = h (x 0, y 0, z 0 ) = 0 and f x (x 0, y 0, z 0 ) g y (x 0, y 0, z 0 ) h z (x 0, y 0, z 0 ) f y (x 0, y 0, z 0 ) g z (x 0, y 0, z 0 ) h x (x 0, y 0, z 0 ) f z (x 0, y 0, z 0 ) g x (x 0, y 0, z 0 ) h y (x 0, y 0, z 0 ) 0. Then we may write (3.1) as dx dt = f x (x 0, y 0, z 0 ) (x x 0 ) + f y (x 0, y 0, z 0 ) (y y 0 ) + f z (x 0, y 0, z 0 ) (z z 0 ) + ɛ 1 (x x 0, y y 0, z z 0 ) dy dt = g x (x 0, y 0, z 0 ) (x x 0 ) + g y (x 0, y 0, z 0 ) (y y 0 ) + g z (x 0, y 0, z 0 ) (z z 0 ) + ɛ 2 (x x 0, y y 0, z z 0 ) dz dt = h x (x 0, y 0, z 0 ) (x x 0 ) + h y (x 0, y 0, z 0 ) (y y 0 ) + h z (x 0, y 0, z 0 ) (z z 0 ) + ɛ 3 (x x 0, y y 0, z z 0 ) (3.2) where ɛ 1, ɛ 2, ɛ 3 are higher order terms.

15 Now if (x 0, y 0, z 0 ) is translated to the origin by v 1 = x x 0, v 2 = y y 0, v 3 = z z 0 and if a = f x (x 0, y 0, z 0 ), b = f y (x 0, y 0, z 0 ), c = f z (x 0, y 0, z 0 ), d = g x (x 0, y 0, z 0 ), j = g y (x 0, y 0, z 0 ), k = g z (x 0, y 0, z 0 ), l = h x (x 0, y 0, z 0 ), m = h y (x 0, y 0, z 0 ), n = h z (x 0, y 0, z 0 ), then (3.2) can be written as dv 1 dt = av 1 + bv 2 + cv 3 + ɛ 1 (v 1, v 2, v 3 ) dv 2 dt = dv 1 + jv 2 + kv 3 + ɛ 1 (v 1, v 2, v 3 ) dv 3 dt = lv 1 + mv 2 + nv 3 + ɛ 1 (v 1, v 2, v 3 ). (3.3) The linear part of (3.3) is the same as dy 1 dt = f 1 (t, y 1, y 2, y 3 ) dy 2 dt = f 2 (t, y 1, y 2, y 3 ) dy 3 dt = f 3 (t, y 1, y 2, y 3 ) (3.4) which is a system of ordinary differential equations. If two systems have the same linear part, the real parts of the eigenvalues are not zero, then they have the same stability. If we have a nonlinear equation and a linear equation and we know the behavior of the linear system, then we can use this to analyze the behavior of the nonlinear system. For example, the linear part of (3.3), dξ 1 dt = aξ 1 + bξ 2 + cξ 3 dξ 2 dt = dξ 1 + jξ 2 + kξ 3 10

16 dξ 3 dt = lξ 1 + mξ 2 + nξ 3 is called the linearization of (3.1) about the equilibrium, critical point, (x 0, y 0, z 0 ). The stability of (3.2) near the critical point is determined by the stability of its linearization. In the small region, or neighborhood, around the equilibrium, solutions of (3.4) and its linearization may be mapped onto each other. Therefore, we are able to conclude that the stability of a system around the equilibrium can be found by analyzing the stability of the system s linearization. Since we will be analyzing the stability of the system, it is important to understand the different types of stability and what they mean. Let us consider the linear system dx dt = AX (3.5) where X is a vector and A is an n n matrix. The trivial solution is X = 0. Lemma 1 If the eigenvalues with zero real parts are simple and the remaining eigenvalues have negative real parts, then the trivial solution will be stable. This means that for every ɛ > 0 there exists δ > 0, where if x (t) is a solution of (3.5), with x(0) < δ, then x(t) < ɛ for t > 0. The notation, denotes how close a particular solution is to the trivial solution. The trivial solution is the present state of this system and x (t) is the solution that is a deviation from the present state of the system. If we have a stable trivial solution then x (t) will be close to the present state in the future as long as x (0) remains close to zero. 11

17 Figure 7: Stable Node Lemma 2 The trivial solution of differential equations is asymptotically stable if and only if all the eigenvalues have negative real parts. To be asymptotically stable means that solutions that begin sufficiently close to the trivial solution tend towards it in the future. A trivial solution is asymptotically stable if it is first and foremost stable and if there is ɛ > 0, where if x(0) < ɛ, we have lim t x(t) = 0. Figure 8: Asymptotically Stable Lemma 3 If at least one eigenvalue has a positive real part, then the trivial solution is unstable. 12

18 If a system is unstable it means that a small disturbance, from the initial condition, yields a major change of state in the future. Figure 9: An Unstable Node Now that we have a good understanding of stability, we need to recall some lemmas that will aid us in finding the bounds of the species in the model [3, 9, 8]. Lemma 4 Positivity Lemma. Let W (t) be a smooth function in [0, T ]. If W (t) satisfies W (t) + M (t) W (t) 0 in (0, T ] and W (0) 0, where function M (t) is bounded in [0, T ], then W (t) 0 in [0, T ]. Proof: We will prove this lemma by contradiction. Assume that W (t) 0 in [0, T ] is false. Then there exists a point t 0 [0, T ], where W (t 0 ) is a negative minimum of W on [0, T ]. Now since W (0) 0, then t 0 (0, T ]. This means that W (t 0 ) + M (t 0 ) W (t 0 ) 0. Since W (t) reaches its minimum value at t 0, then we have W (t 0 ) = 0 if t 0 T and W (t 0 ) 0 if t 0 = T. This guarantees that M (t 0 ) W (t 0 ) 0 which contradicts the assumption that W (t 0 ) < 0 when M (t 0 ) > 0. Now for the case when M (t 0 ) 0, let V (t) = e βt W (t) for some constant β with β > M (t) in (0, T ]. Then V satisfies V (t) + (β + M (t)) V (t) 0 in (0, T ] and 13

19 V (0) 0, where β + M (t) > 0 for all t (0, T ]. From the arguments above, we find V (t) 0 in [0, T ]. From W (t) = e βt V (t) it follows that W (t) 0 on [0, T ]. Lemma 5 The Comparison Argument. Let u 1 and u 2 be solutions to the initial value problem u i = f i (t, u i ) in (0, T ], where u i (0) = u i,0 and i = 1, 2. The functions f 1 and f 2 are continuous in [0, T ] R. Assume that f 1 and f 2 u u are both continuous on [0, T ] R. If f 1 (t, u) f 2 (t, u) in (0, T ] R and also u 1,0 u 2,0, then solutions u 1 and u 2 satisfy u 1 (t) u 2 (t) on [0, T ]. Proof: Let W (t) = u 2 (t) u 1 (t), and let M (t) be any bounded function in [0, T ]. Then by the previous lemma, W (t) satisfies W (t) + M (t) W (t) = M (t) [u 2 (t) u 1 (t)] + f 2 (t, u 2 (t)) f 1 (t, u 1 (t)) in (0, T ] W (0) = u 2,0 u 1,0 0 Since f 1 u is continous in u, by the mean value theorem, f 2 (t, u 2 ) f 1 (t, u 1 ) = [f 2 (t, u 2 ) f 1 (t, u 2 )] + [f 1 (t, u 2 ) f 1 (t, u 1 )] f 1 u (t, η) (u 2 u 1 ) where η = η (t) is an intermediate value between u 1 and u 2. So for the bounded function M (t) = f 1 u (t, η), W (t) satisfies W (t) + M (t) W (t) 0 in (0, T ]. We know from the previous lemma that W (t) 0, so u 2 (t) u 1 (t) on [0, T ]. In the next section, we find bounds for each of the populations and develop criteria to determine extinction. 14

20 4 BOUNDS AND CONVERGENCE In this section we find the ultimate bounds for the super-predator, predator and prey populations. By finding these bounds, we will be able to understand the overall behavior of the populations and determine whether they will co-exist or reach extinction. 4.1 Ultimate Bounds for the Prey Population We now find the upper and lower bounds for prey in this population. Theorem 1 The ultimate bounds for the prey are 0 x 1 (t) K. Proof: Recall that the original prey population is dx ( 1 dt = rx 1 1 x ) 1 K a 2 x 1 b 1 + x 1 x 2 a 3 x 1 b 1 + x 1 x 3. The non-negativity of the density functions allows us to discard some terms. This leaves us with: dx ( 1 dt rx 1 1 x 1 K ). We can see that lim x 1 (t) K. t If this case occurs where the prey is at its upper bound, then we have the situation where the predator and super-predator populations go to 0, resulting in their extinction.

21 4.2 Ultimate Bounds for the Predator Population Theorem 2 The upper bound for the predator population is ( ) c2 a 2 K b 0 x 2 (t) x 2 (0) e 1 +K d 2 t Proof: Our predator population is represented by dx 2 dt = c 2a 2 x 1 b 1 + x 1 x 2 d 2 x 2 a 3 x 2 b 2 + x 2 x 3. Since then we have 0 x 1 b 1 + x 1 dx 2 dt c 2a 2 K = K b 1 + K < 1, b 1 + K x 2 d 2 x 2 ( ) c2 a 2 K b 1 + K d 2 x 2 After solving the differential equation we obtain ( ) c2 a 2 K b 0 x 2 (t) x 2 (0) e 1 +K d 2 t. This leads us to the corollary regarding the extinction of the predator population. Corollary 1 If c 2a 2 K b 1 +K exponentially. < d 2, then the predator population, x 2 (t), is converging to zero This means that if we have the condition where c 2a 2 K b 1 +K population will go to extinction, that is, lim x 2 (t) = 0. If c 2a 2 K t b 1 +K < d 2, then the predator > d 2, then the upper bound for the predator population will go towards infinity. It can be seen 16

22 that predator population s survival depends upon the death rate, functional response and the carrying capacity of the prey, since the predator does not have a carrying capacity of its own. 4.3 Ultimate Bounds for the Super-Predator Population Theorem 3 The upper bound for the super-predator population is ( 0 x 3 (t) x 3 (0) e c 3 a 3 d 3 + c 3 a 3 K b 1 +K ) t Proof: Recall that dx 3 dt = c 3a 3 x 2 b 2 + x 2 x 3 d 3 x 3 + c 3 a 3 x 1 b 1 + x 1 x 3 Due to the non-negativity of the density functions, we can estimate 0 x 2 b 2 +x 2 1, leaving, dx 3 dt c K 3a 3 x 3 d 3 x 3 + c 3 a 3 b 1 + K x 3 ( ) K = c 3 a 3 d 3 + c 3 a 3 x 3 b 1 + K We can now solve this separable differential equation and obtain ( 0 x 3 (t) x 3 (0) e c 3 a 3 d 3 + c 3 a 3 K b 1 +K ) t This leads us to the next corollary about the extinction of the super-predator population. Corollary 2 If c 3 a 3 b 1 +2K b 1 +K < d 3, then the super-predator population, x 3 (t), is converging to zero exponentially. 17

23 This means that if we have the condition where c 3 a 3 b 1 +2K b 1 +K < d 3, then the superpredator population will go to extinction, that is, lim t x 3 (t) = 0. If c 3 a 3 b 1 +2K b 1 +K > d 3, then the upper bound for the super-predator population will go towards infinity. The survival of the super-predator is dependent on the death rate, functional response, carrying capacity of the prey and the abundance of predators. 18

24 5 STABILITY In the previous section we discovered the bounds for the prey, predator and superpredator populations. Now we will examine the stability of the equilibrium solutions. To first determine the stability, we must find the Jacobian of the model. The Jacobian of a model is a matrix of all first-order partial derivatives for a vector-valued function. In the Preliminary section, we introduced the method of linearization. We discussed the lemmas which help us analyze the eigenvalues to determine whether an equilibrium is stable, asymptotically stable or unstable. It was seen that those lemmas work when we are using a linear system. However, we stated that in order to analyze a nonlinear system we must first linearize it and then determine the stability. We do this because stability near a critical point is determined by the stability of its linearization. If we have a linear and a non-linear equation and we know the behavior of the linear system, then we can use this to analyze the nonlinear system. The system for our model is nonlinear which is why we must use the Jacobian matrix. The Jacobian matrix allows us to be able to linearize the system and determine the characteristic polynomial. We do so by calculating det J λi, where J is the Jacobian matrix and λi is the identity matrix. We then find the roots of the characteristic polynomial which allows us to determine the stability of the equilibrium solution. If the roots are all negative, then the equilibrium solution is locally asymptotically stable. If one or more of the roots is positive, then the equilibrium solution is unstable. We discuss more about the necessary criteria for stability later on in this section.

25 5.1 Jacobian Suppose we have G : R n R m is a function from Euclidean n-space to Euclidean m-space. The function is given by m real-valued component functions in (3.4). If the partial derivatives of these functions exist, they can be placed in a m-by-n matrix, called the Jacobian matrix J of G, shown below: y 1 x 1 J=..... y m x 1 Since we have three differential equations, we must use the Jacobian method. The Jacobian for our model is: x J 11 a 1 x 2 b 1 +x 1 a 1 3 ( ) J= 1 c 2 a 2 x 2 (b x 1 x 1 +x 1 J ) (b 1 +x 1 ) 2 22 a 2 3 ( ( 1 1 c 3 a 3 x 3 c 3 a 3 x 3 where, ( J 11 = r 2rx 1 a K 2x 2 ) x 1 (b 1 +x 1 ) (b 1 +x 1 ) 2 ( x J 22 = c 2 a 1 2 b 1 +x 1 d 2 a 3 x 3 ) 1 x 1 (b 1 +x 1 ) (b 1 +x 1 ) 2 J 33 = c 3 a 3 x 2 b 2 +x 2 d 3 + c 3 a 3 x 1 b 1 +x 1. y 1 x n y m x n a 3 x 3 ( ) 1 x 2 (b 2 +x 2 ) (b 2 +x 2 ) 2 ) x 2 (b 2 +x 2 ) (b 2 +x 2 ) 2 ) 1 x 1 (b 1 +x 1 ) (b 1 +x 1 ) 2, b 1 +x 1 b 2 +x 2 J 33 In order to begin finding the equilibrium solutions, we use the model and systematically begin setting some of the populations equal to zero and finding the representation for the other populations. To find the trivial equilibrium, we set x 1 = x 2 = x 3 = 0. To find E 2 we set x 2 = x 3 = 0 and substituting back into the model we solve and obtain a representation for x 1. Next, we set x 1 = 0 and solve for x 2 and x 3 ; however, this provides negative values for the predator and super-predator populations. Therefore, since this equilibrium is negative, we do not work with this critical point. To find E 3 we set x 3 = 0 and solve for x 1 and x 2. Similarly, to find 20

26 E 4, we solve for x 1 and x 3 after setting x 2 = 0. Finally, the coexistence equilibrium is one where the prey, predator and super-predator all survive and we will discuss this in greater detail later. The equilibriums are: 1. E 1 = (0, 0, 0) 2. E 2 = (K, 0, 0) d 2 b 1 3. E 3 = (, rb 1c 2 (Kc 2 a 2 b 1 d 2 Kd 2 ) c 2 a 2 d 2 K (c 2 a 2 d 2 ) 2, 0) To ensure that this equilibrium is non-negative, we must have c 2 a 2 > d 2 and K > b 1d 2 c 2 a 2 d 2. ( d3 b 1 4. E 4 =, 0, c ) 3b 1 r (c 3 a 3 K b 1 d 3 d 3 K) c 3 a 3 d 3 K (c 3 a 3 d 3 ) 2 To ensure that this equilibrium is non-negative, we must have c 3 a 3 > d 3 and K > b 1d 3 c 3 a 3 d Stability for the trivial equilibrium solution We now want to observe the stability of the trivial solution E 1 = (0, 0, 0). To do this, we must substitute the solution into the Jacobian to obtain the characteristic polynomial. We initially expect this solution to be unstable because otherwise it means that we have no surviving prey, predators or superpredators. This means our local solutions are being pushed away. If the local solutions were approaching E 1 this would mean that all our populations tend to extinction. Below is a figure of an unstable solution, a saddle point. 21

27 Figure 10: Instability at (0,0,0)- Saddle Point Theorem 4 One equilibrium solution, E 1, for this model is (0,0,0). This critical point is unstable. Proof: After plugging in the trivial equilibrium solution we find that the Jacobian matrix for E 1 is: r d d 3 We then take the determinant of the matrix and subtract the identity matrix λi to obtain J(0, 0, 0) λi = r λ d 2 λ d 3 λ The three eigenvalues are λ 1 = r, λ 2 = d 2 and λ 3 = d 3. Since r, d 2 and d 3 are positive, λ 2 and λ 3 are negative and λ 1 is positive. Therefore, since we have at least one positive, the critical point of (0,0,0) is unstable for this system. Since equilibrium solution E 1 describes an event where no prey, predator or superpredator exists, we prefer this outcome of instability. This means that our populations do not tend to extinction. If we had reached a conclusion of stability at 22

28 equilibrium E 1, this would mean that our species will all die off, which is clearly not a desired outcome. 5.3 Stability with prey and no predator nor super-predator Theorem 5 The second equilibrium solution for this model is E 2 =(K,0,0). This solution is: 1. asymptotically stable when c 2a 2 K b 1 +K < d 2 and c 3a 3 K b 1 +K < d 3, 2. unstable when c 2a 2 K b 1 +K > d 2 or c 3a 3 K b 1 +K > d 3. Proof: The Jacobian matrix for E 2 is: K K r a 2 b 1 a +K 3 b 1 +K K 0 c 2 a 2 b d 1 +K 2 0 K 0 0 d 3 + c 3 a 3 b 1 +K The determinant is found by J(K, 0, 0) λi = r λ a 2 K b 1 +K a 3 K b 1 +K 0 c 2 a 2 K b 1 +K d 2 λ d 3 + c 3 a 3 K b 1 +K λ The three eigenvalues are λ 1 = r, λ 2 = c 2a 2 K b 1 + K d 2 and λ 3 = d 3 + c 3a 3 K b 1 + K. The values for r, d 2 and d 3 are positive, so λ 1 is negative. Now, if c 2a 2 K b 1 + K < d 2 and c 3a 3 K b 1 + K < d 3, then λ 2 and λ 3 are both negative. Therefore, the eigenvalues are all negative so the equilibrium is asymptotically stable. This proves (1). If c 2a 2 K b 1 + K > d 2 or c 3a 3 K b 1 + K > d 3, then at least one of λ 2 and λ 3 is positive. Since at least one is positive, then the equilibrium becomes unstable. This proves (2). 23

29 This is the case described in Theorem 1, where the prey goes to its upper bound, K, and the predator and super-predator become extinct. 5.4 Stability with prey, predator and no super-predator Theorem 6 The third equilibrium solution for this model is d 2 b 1 E 3 = (, rb 1c 2 (Kc 2 a 2 b 1 d 2 Kd 2 ) c 2 a 2 d 2 K (c 2 a 2 d 2 ) 2, 0). In order to ensure that E 3 remains non-negative, we must have c 2 a 2 > d 2 and K > b 1d 2 c 2 a 2 d 2. This solution is: 1. unstable if d 2 > c 2a 2 d 3 c 3 a 3, 2. asymptotically stable if d 2 < c 2a 2 d 3 c 3 a 3. Proof: The Jacobian matrix here is d 2 b 1 J(, rb 1c 2 (Kc 2 a 2 b 1 d 2 Kd 2 ) c 2 a 2 d 2 K (c 2 a 2 d 2 ) 2, 0)= where α d 2 a 3 d 2 c 2 c 2 a 2 β 0 ξ 0 0 δ α = rkc 2a 2 (c 2 a 2 d 2 ) 2c 2 a 2 d 2 b 1 r r(c 2 a 2 d 2 )(Kc 2 a 2 b 1 d 2 Kd 2 ) c 2 a 2 K(c 2 a 2 d 2 ) β = r(kc 2a 2 d 2 b 1 Kd 2 ) Ka 2 ξ = δ = a 3 rb 1 c 2 (Kc 2 a 2 b 1 d 2 Kd 2 ) b 2 K(c 2 a 2 d 2 ) 2 +rb 1 c 2 (Kc 2 a 2 b 1 d 2 Kd 2 ) c 3 a 3 rb 1 c 2 (Kc 2 a 2 b 1 d 2 Kd 2 ) d b 2 K(c 2 a 2 d 2 ) 2 +rb 1 c 2 (Kc 2 a 2 b 1 d 2 Kd 2 ) 3 + c 3a 3 d 2 c 2 a 2 d 2 b 1 Then we have J(, rb 1c 2 (Kc 2 a 2 b 1 d 2 Kd 2 ) c 2 a 2 d 2 K (c 2 a 2 d 2 ) 2, 0) λi = α λ d 2 c 2 a 3 d 2 c 2 a 2 β 0 λ ξ 0 0 δ λ 24

30 Upon taking the determinant of this matrix and performing computations, we obtain the following: ( c 2 2 a 2c 3 a 3 rb 1 (Kc 2 a 2 b 1 d 2 Kd 2 )+( d 3 c 2 a 2 +c 3 a 3 d 2 )(b 2 K(c 2 a 2 d 2 ) 2 +rb 1 c 2 (Kc 2 a 2 b 1 d 2 Kd 2 )) ( λ 2 λ c 2 a 2[b 2 K(c 2 a 2 d 2 ) 2 +rb 1 c 2 (Kc 2 a 2 b 1 d 2 Kd 2 )] ( rkc2 a 2 (c 2 a 2 d 2 ) 2c 2 a 2 d 2 b 1 r r(c 2 a 2 d 2 )(Kc 2 a 2 b 1 d 2 Kd 2 ) c 2 a 2 K(c 2 a 2 d 2 ) ) ) λ + d 2r(Kc 2 a 2 b 1 d 2 Kd 2 ) c 2 a 2 K ). From this we can see right away that we have ( ) c λ 3 = 2 2 a 2c 3 a 3 rb 1 (Kc 2 a 2 b 1 d 2 Kd 2 )+( d 3 c 2 a 2 +c 3 a 3 d 2 )(b 2 K(c 2 a 2 d 2 ) 2 +rb 1 c 2 (Kc 2 a 2 b 1 d 2 Kd 2 )). c 2 a 2[b 2 K(c 2 a 2 d 2 ) 2 +rb 1 c 2 (Kc 2 a 2 b 1 d 2 Kd 2 )] We will first analyze this eigenvalue, λ 3. Since, we have conditions to ensure nonnegativity of the equilibrium, we will make use of these inequalities in our analysis. After applying these necessary conditions to λ 3, we see that this eigenvalue will be positive if d 2 > c 2a 2 d 3 c 3 a 3. Since this condition makes λ 3 positive then E 3 will be unstable and we do not need to analyze the characteristic polynomial. This proves (1). However, if d 2 < c 2a 2 d 3 c 3 a 3, then λ 3 becomes negative and we must then analyze the quadratic. To do so, we must present some well known facts here. Lemma 6 Let λ 1 and λ 2 be the roots of the quadratic equation λ 2 + bλ + c. (5.1) We can express this equation as λ 2 (λ 1 + λ 2 ) λ + λ 1 λ 2 (5.2) (a) When c < 0, both λ 1 and λ 2 are real and are opposite signs. Therefore, since 25

31 one of the λ s will be postive, the equilibrium solution will be unstable. (b) When c > 0, λ 1 and λ 2 are the same sign. To determine if they are negative or positive we must check b. (i) When b > 0, λ 1 and λ 2 are both negative. Therefore, the equilibrium is asypmtotically stable. (ii) When b < 0, λ 1 and λ 2 are both positive, in which case, the equilibrium solution is unstable. In order to effectively analyze the characteristic polynomial, we must use inequalities and the fact that every parameter is positive. The other factor of the characteristic polynomial is: ( λ 2 λ ( rkc2 a 2 (c 2 a 2 d 2 ) 2c 2 a 2 d 2 b 1 r r(c 2 a 2 d 2 )(Kc 2 a 2 b 1 d 2 Kd 2 ) c 2 a 2 K(c 2 a 2 d 2 ) ) ) + d 2r(Kc 2 a 2 b 1 d 2 Kd 2 ) c 2 a 2. K Again, we apply the necessary conditions c 2 a 2 > d 2 and K > b 1d 2 c 2 a 2 d 2. After analyzing c, we see that c > 0, which means we now must check b. ( ) Here b = rkc2 a 2 (c 2 a 2 d 2 ) 2c 2 a 2 d 2 b 1 r r(c 2 a 2 d 2 )(Kc 2 a 2 b 1 d 2 Kd 2 ) and after applying c 2 a 2 K(c 2 a 2 d 2 ) our conditions again, we see that b > 0. Therefore, two other roots of the characteristic polynomial will be negative. This ensures that we have asymptotic stability and proves (2). Therefore, we can conclude that along with the necessary conditions of c 2 a 2 > d 2 and K > b 1d 2 c 2 a 2 d 2, if d 2 > c 2a 2 d 3 c 3 a 3 then E 3 is unstable. If d 2 < c 2a 2 d 3 c 3 a 3 then λ 3 is negative and after analyzing the quadratic we see that the final two eigenvalues are also negative. Therefore, in this case E 3 is asymptotically stable. The equilibrium E 3 is referenced in Theorem 3, where the super-predator becomes extinct and the other species survive. 26

32 5.5 Stability with prey, super-predator and no predator Theorem 7 The fourth equilibrium for this model is ( ) d E 4 = 3 b 1 d 3 +c 3 a 3, 0, c 3b 1 r(c 3 a 3 K b 1 d 3 d 3 K). In order to ensure that E K( d 3 +c 3 a 3 ) 2 4 remains nonnegative, we must have c 3 a 3 > d 3 and K > b 1d 3 c 3 a 3 d 3. This solution is: 1. asymptotically stable when K < b 1(c 3 a 3 +d 3 ) c 3 a 3 d 3, 2. unstable when K > b 1(c 3 a 3 +d 3 ) c 3 a 3 d 3. Proof: The Jacobian matrix for E 4 is: where a 2 d 3 d 3 η c 3 a 3 c 3 0 τ 0 c 2 ς 3a 3 b 1 r (c 3 a 3 K b 1 d 3 d 3 K) b 2 K ( d 3 + c 3 a 3 ) 2 0 η = rd 3(c 3 a 3 K b 1 c 3 a 3 d 3 b 1 d 3 K) Ka 3 c 3 ( d 3 +c 3 a 3 ) τ = c 2a 2 d 3 b 2 K(c 3 a 3 d 3 ) 2 d 2 c 3 a 3 b 2 K(c 3 a 3 d 3 ) 2 a 2 3 c2 3 b 1r(c 3 a 3 K b 1 d 3 d 3 K) Kc 3 a 3 b 2 (c 3 a 3 d 3 ) 2 ς = r(c 3a 3 K b 1 d 3 d 3 K) Ka 3 ( We find the determinant by J ) d 3 b 1 d 3 +c 3 a 3, 0, c 3b 1 r(c 3 a 3 K b 1 d 3 d 3 K) K( d 3 +c 3 a 3 ) 2 λi = η λ a 2 d 3 c 3 a 3 d 3 c 3 0 τ λ 0 ς c 2 3a 3 b 1 r (c 3 a 3 K b 1 d 3 d 3 K) b 2 K ( d 3 + c 3 a 3 ) 2 0 λ After taking the determinant of this matrix and performing computations, we obtain the following: ( (c2 a 2 d 3 b 2 K(c 3 a 3 d 3 ) 2 d 2 c 3 a 3 b 2 K(c 3 a 3 d 3 ) 2 a 2 3 c2 3 b 1r(c 3 a 3 K b 1 d 3 d 3 K) c 3 a 3 Kb 2 (c 3 a 3 d 3 ) 2 ) λ 27

33 ( ( ) ) λ 2 λ rd3 (c 3 a 3 K b 1 c 3 a 3 d 3 b 1 d 3 K) Ka 3 c 3 ( d 3 +c 3 a 3 + rd 3(c 3 a 3 K b 1 d 3 d 3 K) ) Kc 3 a 3 Similar to the analysis of the previous equilibrium, we will again use inequalities to help determine the stability of this solution. We will make use of the conditions that ensure non-negativity of the equilibrium solution in analyzing the stability. We begin by analyzing: λ 3 = ( c2 a 2 d 3 b 2 K (c 3 a 3 d 3 ) 2 d 2 c 3 a 3 b 2 K (c 3 a 3 d 3 ) 2 a 2 3c 2 3b 1 r (c 3 a 3 K b 1 d 3 d 3 K) c 3 a 3 Kb 2 (c 3 a 3 d 3 ) 2. We can see that this eigenvalue is negative by the non-negativity conditions of c 3 a 3 > d 3 and K > b 1d 3 c 3 a 3 d 3, so we must now check the polynomial. Again, as we did in the last equilibrium, we begin by finding the value of c. Upon using the necessary inequalities, it can be seen easily that c > 0. Next, the sign of b must be found. Here we find that if K < b 1(c 3 a 3 +d 3 ) c 3 a 3 d 3, then b > 0. When this occurs, all three eigenvalues have negative real parts and so the equilibrium is asymptotically stable. This proves (1). If K > b 1(c 3 a 3 +d 3 ) c 3 a 3 d 3, then b < 0 and we will have positive real parts for both eigenvalues, λ 1 and λ 2, making the equilibrium unstable. This proves (2). Therefore, we clearly must have the necessary inequalities of c 3 a 3 > d 3 and K > b 1 d 3 c 3 a 3 d 3 hold. In addition, when K < b 1(c 3 a 3 +d 3 ) c 3 a 3 d 3, E 4 is asymptotically stable. Conversely, if K > b 1(c 3 a 3 +d 3 ) c 3 a 3 d 3, then E 4 is unstable. 28

34 This equilibrium is also described in Theorem 2, where the predator tends towards extinction and the prey and super-predator survive. In the next section, we will graphically view the results of this section. We will use our findings to demonstrate the stability and instability of the equilibrium solutions. 29

35 6 NUMERICAL SIMULATIONS In this section, we provide graphical representations of the population dynamics for this system. To do so, we make use of the conditions we specified in the previous section that determined the stability and instability of each equilibrium solution. The purpose of these numerical simulations is to visually see and obtain a better understanding of the overall behavior of the populations within the system over time. Since the functional response applied in this model is the Holling Type II as we described earlier, we will also increase functional response at times within this section. 6.1 Numerical Simulations for Equilibrium E 2 Figure 11: Stability of E 2 The above figure shows the stability of equilibrium E 2, with c 2a 2 K b 1 +K < d 2 and c 3 a 3 K b 1 +K < d 3. By choosing parameters r =.65, K = 1.9, b 1 =.65, b 2 =.65, d 2 =.25, d 3 =.15, a 2 =.70, a 3 =.25, c 2 =.35 and c 3 =.45 it is seen that the prey population will go to K = 1.9 and the predator and super-predator populations go to extinction. The death rate value is dependent on the functional response and the death rates are high for the predator and super-predator while they are also not consuming enough prey to survive.

36 In the next figure we show the instability of E 2, with c 2a 2 K b 1 +K > d 2. By satisfying this inequality, we have that at least one eigenvalue is positive for E 2 therefore the equilibrium is unstable. By choosing parameters r =.65, K = 1.9, b 1 =.75, b 2 =.65, Figure 12: Instability of E 2 d 2 =.10, d 3 =.15, a 2 =.70, a 3 =.25, c 2 =.35 and c 3 =.45 it is seen that the populations do not stabilize. The long term behavior of the simulation shows that the populations do not approach the desired equilibrium, therefore the equilibrium is unstable. 6.2 Numerical Simulations for Equilibrium E 3 Figure 13: Stability of E 3 The above figure shows the stability of equilibrium E 3, with d 2 < c 2a 2 d 3 c 3 a 3 K > b 1d 2 c 2 a 2 d 2. By choosing parameters r =.65, K = 1.9, b 1 =.65, b 2 =.65, d 2 =.15, d 3 =.40, a 2 =.60, a 3 =.25, c 2 =.40 and c 3 =.15 it is seen that the prey and 31 and

37 predator survive while the super-predator goes to extinction. By Corollary 2, we determined that when c 3 a 3 b 1 +2K b 1 +K and we have global stability for equilibrium solution E 3. < d 3, the super-predator population becomes extinct In the next figure, we show instability of E 3, with K > b 1d 2 c 2 a 2 d 2, the necessary condition for E 3 to be an equilibrium, and d 2 > c 2a 2 d 3 c 3 a 3. For this figure the parameters Figure 14: Instability of E 3 r =.65, K = 1.9, b 1 =.35, b 2 =.45, d 2 =.25, d 3 =.2, a 2 =.55, a 3 =.70, c 2 =.60 and c 3 =.38 were used. Here we see the predator and super-predator populations grow exponentially making this equilibrium case unstable. The populations do not stabilize over time. This can be verified by Corollary 2 regarding the ultimate bound on the super-predator population. When c 3 a 3 b 1 +2K b 1 +K > d 3, the super-predator does not stabilize and rather tends toward infinity. This case is verified in our simulation above. 6.3 Numerical Simulations for Equilibrium E 4 In this figure, we show the stability of equilibrium E 4, with K < b 1(c 3 a 3 +d 3 ) c 3 a 3 d 3 K > b 1d 3 c 3 a 3 d 3. Here the parameters r =.50, K = 1.9, b 1 =.40, b 2 =.75, d 2 =.50, d 3 =.15, a 2 =.25, a 3 =.55, c 2 =.25 and c 3 =.40 were chosen. It is seen that the prey and super-predator survive and their populations stabilize while the predator and 32

38 Figure 15: Stability of E 4 population becomes extinct. By Corollary 1, we determined that when c 2a 2 K b 1 +K < d 2, the predator population becomes extinct and we have global stability for equilibrium solution E 4. In the next figure, we increase the functional response value and the predation and consumption rates for the predator population (x 2 ). Figure 16: Instability of E 4 This figure shows instability of E 4, with K > b 1(c 3 a 3 +d 3 ) c 3 a 3 d 3. By choosing parameters r =.50, K = 1.9, b 1 =.45, b 2 =.75, d 2 =.40, d 3 =.10, a 2 =.40, a 3 =.50, c 2 =.30 and c 3 =.35 it is seen that the predator dies but the prey and super-predator remain unstable. They continue to oscillate as t approaches infinity. These populations do not stabilize and therefore do not reach our equilibrium. 33

39 7 COEXISTENCE, OSCILLATION AND COMPLEXITY OF THE MODEL The final equilibrium solution is E 5, the coexistence state. This equilibrium solution is one in which the prey, predator and super-predator populations all survive and are able to live amongst one another. While the representation for this final equilibrium was able to be found, it is too large to present within these margins. Since the representation proved to be too unwieldy to effectively present and analyze here, we show in this section that under certain parameters there is a stable solution for E 5. While we cannot provide criteria for the general case of E 5, we show that not only does it exist for specific parameter values, it is also complex and at times its stability coincides with other stable equilibriums. We will show this complexity as we examine several cases of stability for this coexistence state. 7.1 Stability for Equilibrium E 5 Figure 17: Stability of E 5 =( , , ) In the figure above, we choose parameters r =.65, K = 1.9, b 1 =.70, b 2 =.67, d 2 =.17, d 3 =.18, a 2 =.40, a 3 =.35, c 2 =.35 and c 3 =.45. For this figure, the instability conditions for E 3 and E 4 are satisfied, therefore for this solution, E 3 and E 4 are unstable. Additionally, while E 5 is stable, E 2 is stable as well and we will discuss more on this after verifying the stability of this E 5.

40 We use the parameters for this particular coexistence state along with the values x 1, x 2 and x 3 for this solution and plug into the Jacobian matrix. Then we find the determinant det J λi = λ λ λ Upon taking the determinant we find: ( λ) (( λ) ( λ) ) λ. After setting this equal to zero and solving for the eigenvalues, we see that: λ 1 = i, λ 2 = and λ 3 = i. By Lemma 2, we know that when we have all negative real parts of eigenvalues, the equilibrium is locally asymptotically stable. We cannot say that this solution is globally asymptotically stable because we do not have special criteria for this coexistence state as we do for the previous equilibriums. Since this is only locally asymptotically stable, we see that when we start with our initial conditions farther away from this E 5 solution, the numerical simulation will show the figure tending towards another equilibrium, namely E 2 which we stated earlier was stable when this E 5 is stable. We verify this graphically. In the figure below, x 1 [0] = 1.3, x 2 [0] =.45 and x 3 [0] =.3, which is near E 2 = (K, 0, 0) where K = 1.9 in these simulations. We are able to start this far away 35

41 from E 2 and see that this solution still tends towards (K, 0, 0). However, once we start farther away than x 1 [0] = 1.3, x 2 [0] =.45 and x 3 [0] =.3 from (1.9, 0, 0), we no longer tend towards E 2 and instead go to E 5. It is seen that when the ini- Figure 18: Effect on E 5 = ( , , ) when initial solutions begin close to E 2 tial conditions are changed, this E 5 tends towards a different equilibrium, further proving that we do not have global stability and only have asymptotic stability for E 5 = ( , , ). Below we show another coexistence stability case. Figure 19: Stability of E 5 =( , , ) In the figure above, we choose parameters r =.60, K = 1.9, b 1 =.60, b 2 =.65, d 2 =.23, d 3 =.215, a 2 =.55, a 3 =.31, c 2 =.30 and c 3 =.64. This figure uses 36

42 the instability conditions for the previous E 3 and E 4. This means that for this solution, E 3 and E 4 are unstable because their instability conditions have been satisfied. However, when this particular equilibrium is stable, E 2 is stable as well. This will be discussed after the stability for this case is verified. We can see that this figure represents a particular case where we have stability for the coexistence state. Using the parameters and the values for x 1, x 2 and x 3 which are the values for this solution, we plug into the Jacobian matrix. We then calculate the determinant det J λi = λ λ λ Taking the determinant of this matrix yields: ( λ) (( λ) ( λ) ) λ. Upon setting this expression equal to zero and solving for λ we obtain: λ 1 = i, λ 2 = and λ 3 = i. By Lemma 2, we have that this particular equilibrium is asymptotically stable since all the real parts of the three eigenvalues are negative. Therefore, we are able to numerically verify what we have observed in the simulation. However, we are only able to show that this solution is locally asymptotically stable. We cannot ensure global stablity for E 5 since we do not have any special criteria required for this condition. 37

43 Since this is only locally asymptotically stable, we are able to see that should we start with our initial conditions farther away from this E 5 solution, the numerical simulation will show the figure tending towards another equilibrium. This brings us back to the discussion earlier where it was stated that this E 5 is stable along with E 2. This will be verified graphically. In the figure below, x 1 [0] = 1.8, x 2 [0] =.55 and x 3 [0] =.35, which is near E 2 = (K, 0, 0) where K = 1.9 in these simulations. Figure 20: Effect on E 5 = ( , , ) when initial solutions begin close to E 2 It is seen that when the initial conditions are changed, this E 5 tends towards a different equilibrium, further proving that we only have asymptotic stability for E 5 = ( , , ) and not global stability. In this next figure, we see another case with a stable coexistence state. In the figure above, we choose parameters r =.5, K = 1.9, b 1 =.55, b 2 =.85, d 2 =.24, d 3 =.15, a 2 =.50, a 3 =.20, c 2 =.65 and c 3 =.70. This figure uses the instability conditions for the previous E 2 and E 4. This means that for this solution, E 2 and E 4 are unstable because their instability conditions have been satisfied. However, when this particular solution is stable, E 3 is stable as well. This will be discussed after 38

44 Figure 21: Stability of E 5 = ( , , ) the stability for this case is verified. Again, it is seen that this figure represents a particular case of stability for the coexistence state. Using the parameters and the values for x 1, x 2 and x 3 which are the values for this solution, we plug into the Jacobian matrix. We then calculate the determinant det J λi = λ λ λ When we take the determinant of this matrix, we obtain: ( λ) (( λ) ( λ) ) λ. After setting the expression equal to zero and solving for λ, we find that λ 1 = i, λ 2 = and λ 3 = i. As before, we know by Lemma 2 that when all the real parts of the eigenvalues are 39

45 negative, the equilibrium is asymptotically stable. As before, we are only able to ensure locally asymptotically stability, not global stability. Due to this, as we begin with our initial solutions farther away from the x 1, x 2 and x 3 for E 5, we expect the simulation to tend towards another equilibrium. In this case, we expect it to tend towards E 3, since when E 5 is stable for this case, so is E 3. Using these parameters and plugging into the representation for E 3, we see that E 3 = (1.55, 0.38, 0). Using our inital values for x 1, x 2 and x 3 near this E 3, we find that the simulation tends towards E 3. In the figure below we use x 1 [0] = 1.5, x 2 [0] = 0.3 and x 3 [0] = We see that when the initial conditions are changed, Figure 22: Effect on E 5 = ( , , ) when initial solutions begin close to E 3 this E 5 tends towards a different equilibrium, proving we only have asymptotic stability for E 5 = ( , , ). Below is a brief table of some other stable E 5 equilibriums that were found. The table lists the solution and parameters used. The equilibriums shown below are also stable with E 2. 40