1. A simple prey-predator model can be written in normalized form as

(&N_1^' (t)=N_1 (t)(1-(N_1 (t))/K-c_12 N_2 (t))@&N_2^' (t)=N_2 (t)(-r+c_21 N_4 (t));)(1),

where N_1 (t) and N_2 (t) are the sizes of the prey and predator populations, respectively, at time t. The constant K is the carrying capacity of the prey and r is the normalized death rate of the predator, assuming the prey's growth rate is b=1. The term c_21 N_1 N_2 represents the gain in the growth rate of the predator due to its interaction with the prey, and c_12 N_1 N_2 is the corresponding loss in the growth rate of the prey. The constants c_12 and c_21 take into account the difference in the mean weight between the two species.

(a) Write system (1) in the form:

(&N_1^'=N_1-(N_1 )^2/-c_12 N_1 N_2F(N_1+N_2 )@&N_2^'=-rN_2+c_21 N_1 N_2G(N_1,N_2 )(2)

and compute the Jacobian matrix J(N_1,N_2 )=((F/(N_1 )&F/(N_2 )@G/(N_1 )&G/(N_2 ))) .

(b) Show that the equilibria are: (0,0),(K,0)",and " P=(r/c_21, (Kc_21-r)/(Kc_12 c_21 )) That is: Use the Jacobian found in Problem 1a to show that (0,0) is always a saddle point, and that if K>r/c_21 then (K,0) is also a saddle point. What type of equilibrium is (K,0), if K<r/c_21 . Note that in this case, the point K is no longer in the positive quadrant.

(c) if K>r/c_21 , show that the interior critical point P is asymptotically stable. For what value of r does it bifurcate between a spiral sink and a sink?

(d) Use the Euler's method to solve the system (1) for the following parameter values and initial conditions: r=0.2,K=10,c_12=0.45 and C_21=0.2, and initial conditions N_1 (0)=2 and N_2 (0)=1 for 0t100,t=0.05.

(e) What is the implication of the result of the numerical solution? Do the solutions converge to one of the equilibria solutions? Discuss it briefly.