Question

I have to finish the code given by my diff eq professor to analyze the lorenz function at different initial conditions and different values for p. I am not sure how to input the lorenz function in the way the code is set up.

Here is the code provided for matlab:

function lorenz

s = 10;
b = 8/3;
p = 0.15;

y = @(t,x) [ ; ; ]; % you are responsible for entering the lorenz system here

T = [0 200]; % sets the time interval; no need to edit
x0 = [1 1 1]; % initial conditions; this can and should be editied

[t,x] = ode45(y,T,x0); % ode solver; do not edit

plot3(x(:,1),x(:,2),x(:,3),'r') % plots 3d solution for entered conditions

grid on

Answer 1

%%Matlab code for Lorenz equation solution
clear all
close all

%value for alpha beta and eta
s=10; p=0.15; b=(8/3);
%and initial condition of y
y10=1; y20=1; y30=1;
%initial value for t
t0=0;
%t end values
tend=200;

y0=[y10;y20;y30];
%minimum and maximum time span
        tspan=[t0 tend];
        %Solution of ODEs using ode45 matlab function
        sol= ode45(@(t,y) odefcn1(t,y,s,p,b), tspan, y0);
        t1 = linspace(tspan(1),tspan(end),100001);
        %yy is the corresponding x y v and z
        yy1 = deval(sol,t1);
figure(1)     
plot3(yy1(1,:),yy1(2,:),yy1(3,:))
xlabel('y1');ylabel('y2');zlabel('y3')
title('3d line plot for Lorenz attractor initial cond [1 1 1] and p=0.15')

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%value for alpha beta and eta
s=10; p=10; b=(8/3);
%and initial condition of y
y10=1; y20=1; y30=1;
%initial value for t
t0=0;
%t end values
tend=200;

y0=[y10;y20;y30];
%minimum and maximum time span
        tspan=[t0 tend];
        %Solution of ODEs using ode45 matlab function
        sol= ode45(@(t,y) odefcn1(t,y,s,p,b), tspan, y0);
        t1 = linspace(tspan(1),tspan(end),100001);
        %yy is the corresponding x y v and z
        yy1 = deval(sol,t1);
figure(2)     
plot3(yy1(1,:),yy1(2,:),yy1(3,:))
xlabel('y1');ylabel('y2');zlabel('y3')
title('3d line plot for Lorenz attractor initial cond [1 1 1] and p=10')

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%value for alpha beta and eta
s=10; p=20; b=(8/3);
%and initial condition of y
y10=1; y20=2; y30=1;
%initial value for t
t0=0;
%t end values
tend=200;

y0=[y10;y20;y30];
%minimum and maximum time span
        tspan=[t0 tend];
        %Solution of ODEs using ode45 matlab function
        sol= ode45(@(t,y) odefcn1(t,y,s,p,b), tspan, y0);
        t1 = linspace(tspan(1),tspan(end),100001);
        %yy is the corresponding x y v and z
        yy1 = deval(sol,t1);
figure(3)     
plot3(yy1(1,:),yy1(2,:),yy1(3,:))
xlabel('y1');ylabel('y2');zlabel('y3')
title('3d line plot for Lorenz attractor initial cond [1 2 1] and p=20')

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%value for alpha beta and eta
s=10; p=20; b=(8/3);
%and initial condition of y
y10=1; y20=5; y30=1;
%initial value for t
t0=0;
%t end values
tend=200;

y0=[y10;y20;y30];
%minimum and maximum time span
        tspan=[t0 tend];
        %Solution of ODEs using ode45 matlab function
        sol= ode45(@(t,y) odefcn1(t,y,s,p,b), tspan, y0);
        t1 = linspace(tspan(1),tspan(end),100001);
        %yy is the corresponding x y v and z
        yy1 = deval(sol,t1);
figure(4)     
plot3(yy1(1,:),yy1(2,:),yy1(3,:))
xlabel('y1');ylabel('y2');zlabel('y3')
title('3d line plot for Lorenz attractor initial cond [1 5 1] and p=20')

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

%value for alpha beta and eta
s=10; p=28; b=(8/3);
%and initial condition of y
y10=1; y20=1; y30=1;
%initial value for t
t0=0;
%t end values
tend=200;

y0=[y10;y20;y30];
%minimum and maximum time span
        tspan=[t0 tend];
        %Solution of ODEs using ode45 matlab function
        sol= ode45(@(t,y) odefcn1(t,y,s,p,b), tspan, y0);
        t1 = linspace(tspan(1),tspan(end),100001);
        %yy is the corresponding x y v and z
        yy1 = deval(sol,t1);
figure(5)     
plot3(yy1(1,:),yy1(2,:),yy1(3,:))
xlabel('y1');ylabel('y2');zlabel('y3')
title('3d line plot for Lorenz attractor initial cond [1 1 1] and p=28')

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%value for alpha beta and eta
s=10; p=28; b=(8/3);
%and initial condition of y
y10=1; y20=5; y30=1;
%initial value for t
t0=0;
%t end values
tend=200;

y0=[y10;y20;y30];
%minimum and maximum time span
        tspan=[t0 tend];
        %Solution of ODEs using ode45 matlab function
        sol= ode45(@(t,y) odefcn1(t,y,s,p,b), tspan, y0);
        t1 = linspace(tspan(1),tspan(end),100001);
        %yy is the corresponding x y v and z
        yy1 = deval(sol,t1);
figure(6)     
plot3(yy1(1,:),yy1(2,:),yy1(3,:))
xlabel('y1');ylabel('y2');zlabel('y3')
title('3d line plot for Lorenz attractor initial cond [1 5 1] and p=28')


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%Function for evaluating the ODE
function dydt = odefcn1(t,y,s,p,b)

    eq1 = s*(y(2)-y(1));
    eq2 = y(1)*(p-y(3))-y(2);
    eq3 = y(1)*y(2)-b*y(3);

    %Evaluate the ODE for our present problem
    dydt = [eq1;eq2;eq3];
end

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%