Question

Exercise 3. (22 pts.) (Rabbits vs. Sheep) a)For each of the following ”rabbits vs. sheep” problems with x, y ≥ 0. Find the fixed points, classify them (type and stability), and find the eigenvalues and eigenvectors. Then sketch (by hand) a plausible phase portrait indicating nullclines, all relevant trajectories, and indicate all the different basins of attraction. Finally interpret the behavior of the population of the rabbits and sheep for the different systems (compare the three systems)

• x˙ = x(3 − 2x − 2y), ˙y = y(2 − x − y)

Answer 1

Given system. iu (3-2x-2y) e 30-24²-2my. = 9(2-*-y). 2y-ay-y for the fixed pts. x = 0 = ý. ..(3-2x-2y) > 0. y(2-x-y). When 12o. & 7 y(2-x-4) 20 - y(2-4)=0. 2) 920,2. When you &u(3-2 x 2) = 0 >> € 3.20) -- -> 0120,3 estarem 3-22-24 20 -2-n-y not possible, 40 gi - a c hap 1 So, the fixed pts. are (0,0), (122, 0), (0, 2) . (3-4u2y. Aa Jacobian of the system -2x .. ( - y 2-2-24) Consider the fixed pt copo). A 13 o eigenvaluen are 3,2. lo 2) so (oo) is an unstable node. eigenverters are (0) & ( respectively. (3,0). Now, wnsider a 1-3 A -3 eigenvalees are -3, & so, (3/10) is a sadale print which is unstable eigenverters are bored by resperti

Now, consider the fixed pt. (0,2). A2 / - 01 2 eigenvalues are at ,-2. 1-2 -2%. (0, 2) is a stable saadados a node eigenvertar are (+1) , ( i respectively,

cedex 5 al x+y=2. x+y = 3 n. 20 Y=0 The nullclinex are x+y a 2/3, n+y=2. Phase portrait Sheep Yabbit- The chase portrait interprets will be extinct as t o . & sheep will be only exists. that the poz rabbit population after the s'infinite.

Griven system. ni = x(3-28-4) y = y(2-x-4). for the fixed pts. ģ= 0 = ý. 2(3-2x-4) = 0 = 9 (2-3-4). When n2o. & 7 y(2-x-4) 20 - y(2-4) 20 > y=0,2. When you & M (3 - 2x-4) 20. > *€3-2) =0 -> 220,} When 3) 3-2*-y zo =2-3-4. 2 + y = 2. - 2x+y =3. -. - > x-2x = 2-3 » xal. ☺ => y = 2-x21. A So, the fixed pts. are (0,0), (2, 0), (0, 2), (1,1). A. Jacobian of the system = (3 1 I -4u-y. - y -R. 2-x-2y Consider the fixed pt Co,o). 13 ol eigenvakan are 3,2. A l o 2 so, (oo) is an unstal eigenverters are (0) & (1) respectively. Now, wnsider (3,0). A = (-3 2.) eigenvalees are -3, / so, (3/210) is an saddle point. which is unstable eigenverters are (6) & (3) respectively

Now, consider the fixed pt. (0,2). A o -2 1-2 eigenvalues 10.2) is a . are 1-2 unstable saddle point eigenvertar are (-2), (i) respectively. Consider the fixed pt (1,1). A2 (^2 -1) for eigenvalues. a +32 +120 2 -3 11 9 -4 -3 ± √5 eigenvalues are -3 tus -3-55 frace A -3 , det A=1. (Grace A) - 4 hora detta I-4=570. & (1.1) is a stable node. 2 + sk - For nullcliner i=0 & Ý 2o. (3-2x-4) 20 - 4(2-x-w). Nullcliner are. no, you, zuty=3 Anty=2 r. 2x+y (3 & nty(2, nyo, y7o. x70 k ý you D a +472, 2x+y <3 x70, 470 x>0, ý co. I xty22. & 2uty>3, 470, 47o. nico, ý 70. : +472, 2x+y)3 220 ; Lou

Phase portrait So any open neighbourhood of (1,1) will be the basin of attraction i.e. the positive quadraint of the IRL is the basin of attraction. The phase portraits interpret that no one specier cannot drives other to extinction as all the trajactaries converger to (i). it says that rabbit & sheep are will be in same no.in amount after infinite time Now, we consider the model. xam (3--2y) y ? 912--) Phase portrait shup stable manifold 10,2) (0.1 (30) rabbit This phase portrait interprets that one species driver other to extinction. If rajectorier strating below the stable manifold driver. Sheep to eat ination. Par