1. (a) For Lotka-Volterra (predator prey) system the non-trivial (non-zero) steady state is (2, 2), i.e., ū = 2 and v = 2.

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1 M445: Dynamics; HW #6. OLUTON 1. (a) For Lotka-Volterra (predator prey) system the non-trivial (non-zero) steady state is (2, 2), i.e., ū = 2 and v = 2. (b) For the same system with constant effort harvesting coefficient h that appears in harvesting terms shows the fraction of current population taken per unit time as a result of harvesting. Note that for non-dimensionalized systems, due to various possible choices of characteristic time that may be used to scale time variable, this coefficient may be greater than one. The meaning of the fact that h is the same in both equations: the same effort is exerted to catch both species, or similarly, the same fraction of both species is taken per unit time. (c) Non-trivial (non-zero) steady state of the system with harvesting: ū = 2 + h, v = 2 h/2. The biologically meaningful non-trivial steady state exists for < h < 4 (for harvesting we must have h > ). The type of this non-trivial steady state (for h < 4) does not change compared to the original system since for the corresponding Jacobian matrix ( (4 + 2h) 2 h/2 we have tr and det (4 + 2h)(2 h/2) >, and thus, the steady state for linearized system is a center. We note that for < h < 4 the structure of the system does not change, so the steady state for the original system is a center as well. The values of h > 4 will lead to eventual extinction of both species: First, we note that for h > 4 the right hand side of the first equation becomes negative, so du/ <, and in the long run the prey species decay to monotonously. The right hand side of the second equation will also be zero for u(t) < 2 + h < 6, and this inequality will be eventually satisfied due to the fact that u(t) is a monotonously decaying function. o, we will have dv/ <, and in the long run the predator species also decay to. Harvesting is more beneficial for prey species since the steady state value of prey species population is higher with harvesting compared to the case without harvesting: for < h < 4, ū = 2 + h > 2, while v = 2 h/2 < 2. ) 1

2 2. For competition system du dv = 4u 2u 2 2uv = 4u(1 u/2 v/2), = 2v 2v 2 2uv = 2v(1 v/1 u/1), we introduce notation: M u = 2, M v = 1, N u = 1, N v = 2 (compare with notes!). (a) Following the analysis presented in the notes (with numbers substituted for parameters M u, M v, N u, and N v, we have that the system has three steady states (ū, v). teady state (, ) is an unstable node, (2,) is a stable node, and (,1) is a saddle. pecies v will eventually go extinct since M u = 2 > 1 = N u and M v = 1 < 2 = N v (see notes!). For the same system with constant effort harvesting for one of the species: du dv = 4u 2u 2 2uv hu = (4 h)u(1 2u/(4 h) 2v/(4 h)), = 2v 2v 2 2uv = 2v(1 v/1 u/1), we introduce notation: M u = (4 h)/2, M v = 1, N u = 1, N v = (4 h)/2 (compare with notes!). (b) Coefficient h that appears in a harvesting term shows the fraction of current population of u species taken per unit time as a result of harvesting. (c) n the presence of harvesting u species will go extinct instead of v species if h satisfies inequality: M u = (4 h)/2 < 1 = N u, or h > 2 (similarly, M v = 1 > (4 h)/2 = N v, or h > 2). 3. The model of interaction between deer population in a forest and hunters [under the assumptions (1) deer population grows logistically in the absence of hunters; (2) the presence of hunters depresses deer population at a rate jointly proportional to current deer population and hunters population; (3) the hunters are attracted to the forest at a rate directly proportional to current deer population; (4) The hunters are discouraged from hunting at a rate directly proportional to the number of hunters that are already in the forest] has the form du dv = ru(1 u/m) huv, = +au bv. Here u is deer population, v is the number of hunters, r, M, h, a, and b are positive parameters. The steady states are solutions of = ru(1 u/m) huv = u(r ru/m hv), = +au bv. 2

3 There are two possible steady states (ū, v): (,) and (r/(r/m + ha/b), a/b r/(r/m + ha/b)). The Jacobian matrix is ( ) r 2rū/M h v hū. a b For steady state (,) we have ( r a b ), with det A <, so this steady state is a saddle. For the non-trivial steady state (ū, v) = (r/(r/m + ha/b), a/b r/(r/m + ha/b)) we have ( ) rū/m hū. a b ince ū > (and v > ), we have that tr rū/m b < and det rbū/m + hūa >, so this steady state is TABLE (node or focus). t follows from the above analysis that under assumptions of the model, the number of deer and hunters in the forest will eventually converge to a non-trivial steady state. 4. Below, the figures produced by MATLAB code for the R model problem are presented. R model: case #1 Populations R Populations 4 2 R model: case # Time R 3

4 25 Phase plane: vector field of R model Phase plane: direction field (normalized vectors) Phase plane: vector field and phase trajectories

5 function []=R() % We solve R model % d/ = -alpha**+gamma*(n--) % d/ = alpha**-beta* % R(t) = N - (t) - (t) % % () = N - - R % () = % % Total population: N=5; % nitial conditions #1: u1= 5-1; v1= 1; y1= [u1;v1]; % nitial conditions #2: u2= 5-21; v2= 1; y2= [u2;v2]; % etup the ODE solver, these are default tolerances: options=odeset('reltol',1e-3,'abstol',1e-6); % Time span tspan=[ 2]; %Parameters: parameters.alpha =.1; parameters.beta = 1/1; parameters.gamma = 1/1; parameters.n = 5; % olve the equations. Chose the appropriate ODE solver: solver=@ode45; %olution for case #1: [t1 yout1]=solver(@sirs1,tspan,y1,options,parameters); %olution for case #2: [t2 yout2]=solver(@sirs1,tspan,y2,options,parameters); % Analyze the output: Convenient to map variables to something simpler: %olution for case #1 u1=yout1(:,1); v1=yout1(:,2); %olution for case #2 u2=yout2(:,1); v2=yout2(:,2); %Plotting results: %%%%%%%% Time series: model solutions %%%%%%%%%%%%%%%%%%%% figure(1) subplot(2,1,1),plot(t1,u1,'bo-',t1,v1,'r*-',t1,n-u1-v1,'g.-'); legend('','','r',-1); title('r model: case #1'); ylabel('populations'); axis([ 2-5 5]) subplot(2,1,2),plot(t2,u2,'bo-',t2,v2,'r*-',t2,n-u2-v2,'g.-'); legend('','','r',-1); title('r model: case #1'); xlabel('time'); ylabel('populations'); axis([ 2-5 5]) %%%%%%%%% Vector filed %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Let us plot the vector field: (du/,dv/), % which is the same as (f(u,v),g(u,v)) figure(2) % Get subdivision for u and v: number=15; % Number of subdivisions for the plot % Lenghts of intervals in u and v that are being subdivided: umax=n; vmax=n/2; ur=linspace(,umax,number); vr=linspace(,vmax,number); % Coordinates of sample points on the phase plane where the vectors

6 % (f(u,v),g(u,v)) will be constructed: [um vm]=meshgrid(ur,vr); % Have to call the sirs1 function many times to estimate % f(u,v) and g(u,v) values at sample points: for i=1:number for j=1:number sirs1righthandide=sirs1(,[um(i,j) vm(i,j)],parameters); fm(i,j)=sirs1righthandide(1); gm(i,j)=sirs1righthandide(2); end end % Plot velocity vector field: quiver(um,vm,fm,gm); axis([,umax,,vmax]); title('phase plane: vector field of R model') xlabel('') ylabel('') %%%%%%%%%%%%%% Direction field %%%%%%%%%%%%%%%%%%% figure(3) % Computing vector lengths: norm=sqrt(fm.^2+gm.^2); % Plot direction filed (normalized vectors): quiver(um,vm,fm./norm,gm./norm,'k'); axis([,umax,,vmax]); title('phase plane: direction field (normalized vectors)'); xlabel('') ylabel('') %%%%%%%%%%%%%% Phase trajectory %%%%%%%%%%%%%%%%%%% figure(4) % Here we superimpose the velocity vector field and the phase trajectory: quiver(um,vm,fm,gm); hold on axis([,umax,,vmax]); title('phase plane: vector field and phase trajectories') xlabel('') ylabel('') plot(u1,v1,'r-') plot(u2,v2,'m-') hold off % % ODE system to be integrated: % d/ = -alpha**+gamma*(n--) % d/ = alpha**-beta* function [ydot]=sirs1(t,y,parameters) % Parameters: alpha=parameters.alpha; beta=parameters.beta; gamma=parameters.gamma; N=parameters.N; % Here are the right hand sides of differential equations: ydot=[-alpha*y(1)*y(2)+gamma*(n-y(1)-y(2)); alpha*y(1)*y(2)-beta*y(2)];