dt (0) = C 1 + 3C 2 = 3. x(t) = exp(3t).

Transcription

1 M445: Dynamics; HW #5. SOLUTIONS so 1. a Characteristic polynomial is Corresponding general solution is λ 2 2λ 3 = λ 12 = 1 ± = 1 ± 2 = { 1 3 xt = C 1 exp t + C 2 exp3t. is The derivative is dt t = C 1 exp t + 3C 2 exp3t. x = C 1 + C 2 = 1 dt = C 1 + 3C 2 = 1. From the above C 2 = C 1 = 1. So the solution satisfying given initial conditions xt = exp t. Solution decays to zero as t. b Same as 1a. The general solution is so xt = C 1 exp t + C 2 exp3t. x = C 1 + C 2 = 1 dt = C 1 + 3C 2 = 3. Thus C 2 = 1 C 1 =. So the solution satisfying given initial conditions is Solution increases as t. c Characteristic polynomial is Corresponding general solution is xt = exp3t. λ 2 + 2λ + 1 = λ 12 = 1 ± 1 1 = 1 ± = 1. xt = C 1 exp t + C 2 t exp t. 1

2 so The derivative is dt t = C 1 exp t + C 2 exp t t exp t. x = C 1 = 1 dt = C 1 + C 2 = 1. Thus C 1 = 1 C 2 = 2 and the solution satisfying given initial conditions is Solution decreases as t. d Characteristic polynomial is xt = exp t + 2t exp t. λ 2 + 2λ + 2 = λ 12 = 1 ± 1 1 = 1 ± i = Corresponding general solution is The derivative is { 1 + i 1 i xt = C 1 exp t sint + C 2 exp t cost. dt t = C 1 exp t sint+exp t cost+c 2 exp t cost exp t sint. is x = C 2 = 1 dt = C 1 C 2 = 1. From the above C 2 = 1 C 1 = 2. So the solution satisfying given initial conditions xt = 2 exp t sint + exp t cost. Solution oscillates with the amplitude decreasing to zero as t. 2.a A = so tra = 2 > deta = 3 < and the steady state is a saddle. b 1 A = 1 1 2

3 so tra = 2 < deta = 1 > deta = tra/2 2 = 1 and the steady state is a stable node. c 1 1 A = 1 1 so tra = 2 < deta = 2 > deta = 2 > tra/2 2 = 1 and the steady state is a stable focus. 3. a Steady states satisfy satisfy the system: = x 2 y = x + y 2. The steady states are x 1 ȳ 1 = and x 2 ȳ 2 = 1 1. The Jacobian matrix is 2x 1 Jx y = 1 2y For x 1 ȳ 1 = J = A = 1 1 so tra = deta = 1 < and the steady state is a saddle. For x 2 ȳ 2 = J1 1 = A = 1 2 so tra = 4 > deta = 3 > deta = 3 < tra/2 2 = 4 and the steady state is an unstable node. bsteady states satisfy satisfy the system: = x + y 2 = x 2 y. The steady states are x 1 ȳ 1 = and x 2 ȳ 2 = 1 1. The Jacobian matrix is 1 2y Jx y = 2x 1 For x 1 ȳ 1 = J = A = 1 1 so tra = 2 deta = 1 > deta = tra/2 2 = 1 and the steady state is a stable node. For x 2 ȳ 2 = J1 1 = A = 2 1 3

4 so tra = 2 < deta = 3 < and the steady state is a saddle. csteady states satisfy satisfy the system: = x 2 y = x + y 2. The steady states are x 1 ȳ 1 = and x 2 ȳ 2 = 1 1. The Jacobian matrix is 2x 1 Jx y = 1 2y For x 1 ȳ 1 = J = A = 1 1 so tra = deta = 1 > and the steady state is a center. For x 2 ȳ 2 = 1 1 J 1 1 = A = so tra = deta = 3 < and the steady state is a saddle. 4. Below the figures produced by MATLAB code for the competition model problem are presented. Solution of Lotka Volterra model: case #1 Population 1 5 Prey Preditor Solution of Lotka Volterra model: case #2 Population 1 5 Prey Preditor Time 4

5 1 Phase plane: vector field 8 6 v u 1 Phase plane: direction field normalized vectors v u 1 Phase plane: vector field and phase trajectories v u 5

6 function []=competition % Solve competition model % du/dt = a1*u*1-b1*u-c1*v % dv/dt = a2*u*1-b2*u-c2*v % Initial conditions #1: u1= 2; v1= 1; y1= [u1;v1]; % Initial conditions #2: u2= 15; v2= 25; y2= [u2;v2]; % Setup the ODE solver these are default tolerances: options=odeset'reltol'1e-3'abstol'1e-6; % Time span tspan=[ 2]; % Solve the equations. Chose the appropriate ODE solver: solver=@ode45; %Solution for case #1: [t1 yout1]=solver@comp1tspany1options; %Solution for case #2: [t2 yout2]=solver@comp1tspany2options; % Analyze the output % Convenient to map variables to something simpler: %Solution for case #1 u1=yout1:1; v1=yout1:2; %Solution for case #2 u2=yout2:1; v2=yout2:2; %Plotting results. %%%%%%%% Time series: model solutions %%%%%%%%%%%%%%%%%%%% figure1 subplot211plott1u1'bo-'t1v1'r*-'; %plotting model solutions legend'prey''preditor'; title'solution of Lotka-Volterra model: case #1'; ylabel'population'; axis[ *max[u1;v1]] subplot212plott2u2'bo-'t2v2'r*-'; %plotting model solutions legend'prey''preditor'; title'solution of Lotka-Volterra model: case #2'; xlabel'time'; ylabel'population'; axis[ *max[u2;v2]] %%%%%%%%% Vector filed %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Let us plot the vector field: du/dtdv/dt % which is the same as fuvguv

7 figure2 % 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=1.1*max[u1;u2]; vmax=1.1*max[v1;v2]; ur=linspaceumaxnumber; vr=linspacevmaxnumber; % Coordinates of sample points on the phase plane where the vectors % fuvguv will be constructed: [um vm]=meshgridurvr; % Have to call the lv function many times to estimate % fuv and guv values at sample points: for i=1:number for j=1:number comp1righthandside=comp1[umij vmij]; fmij=comp1righthandside1; gmij=comp1righthandside2; end end % Plot velocity vector field: quiverumvmfmgm; axis[umaxvmax]; title'phase plane: vector field' xlabel'u' ylabel'v' %%%%%%%%%%%%%% Direction field %%%%%%%%%%%%%%%%%%% figure3 % Computing vector lengths: norm=sqrtfm.^2+gm.^2; % Plot direction filed normalized vectors: quiverumvmfm./normgm./norm'k'; axis[umaxvmax]; title'phase plane: direction field normalized vectors'; xlabel'u' ylabel'v' %%%%%%%%%%%%%% Phase trajectory %%%%%%%%%%%%%%%%%%% figure4 % Here we superimpose the velocity vector field and the phase trajectory: quiverumvmfmgm; hold on axis[umaxvmax]; title'phase plane: vector field and phase trajectories' xlabel'u' ylabel'v' plotu1v1'r-' plotu2v2'm-' hold off % % ODE function to be integrated function [ydot]=comp1ty % Here are the differential equations ydot=[1*y1*1-.1*y1-.2*y2; 2*y2*1-.1*y2-.2*y1]; %right hand side 1st eqn %right hand side 2nd eqn