Plotting phase space for 2nd order nonlinear ODE

Hello, I'm trying to figure out how to plot the phase space for a 2nd order nonlinear ODE.

Problem Statement

Given the 2nd order ODE, (d^2x/dt^2) - u*(1-x^2)*(dx/dt)+x=0, initial conditions are x(0) = 2 and x'(0) = 0. Find the solution if u = 1 and t is from 0-30 seconds. Plot time vs x and the phase space where x vs x'

MATLAB Code (what I have so far)

[t, y] = ode45(@ODE45SolvMain,[0 30],[2 0])

figure(1)

plot(t,y(:,1),t,y(:,2))

xlabel('Time t')

ylabel('Displacement')

title('Displacement vs Time')

legend('x_1','x_2')

---------------------------------------------------------

% Function Definition

function [x_primes] = ODE45SolvMain(t,x)

u = 1;

x_primes = [x(2); u*((1-x(1)^2))*x(2)-x(1)]

My solution so far

This figure is for the time vs x plot. I am unsure how to begin and plot the phase space for the this ODE.

Answer 1

[Star Strider]

I am not certain what you want to plot for the phase plot.

This calculates and plots as a funciton of and also calculates the derivatives (as the ‘derivs’ matrix) in the event those are also to be plotted —

[t, y] = ode45(@ODE45SolvMain,[0 30],[2 0]);
figure(1)
plot(t,y(:,1),t,y(:,2)) 
xlabel('Time t') 
ylabel('Displacement') 
title('Displacement vs Time') 
legend('x_1','x_2') 

figure(2)
plot(y(:,1),y(:,2)) 
xlabel('x_1')
% xlabel(']Time t') 
ylabel('x_2')
% ylabel('Displacement') 
title('Phase') 

% legend('x_1','x_2') 

% Derivative Calculations
for k = 1:numel(t)
    derivs(k,:) = ODE45SolvMain(t(k),y(k,:));
end
Derivs = table(t,derivs(:,1),derivs(:,2), 'VariableNames',{'t','dy1dt','dy2dt'})

Derivs = 341×3 table
        t            dy1dt        dy2dt 
    __________    ___________    _______

             0              0         -2
    2.5119e-05    -5.0236e-05    -1.9998
    5.0238e-05    -0.00010047    -1.9997
    7.5357e-05     -0.0001507    -1.9995
    0.00010048    -0.00020092    -1.9994
    0.00022607    -0.00045199    -1.9986
    0.00035166    -0.00070296    -1.9979
    0.00047726    -0.00095383    -1.9971
    0.00060285     -0.0012046    -1.9964
     0.0012308     -0.0024571    -1.9926
     0.0018588     -0.0037072    -1.9889
     0.0024868      -0.004955    -1.9851
     0.0031147     -0.0062005    -1.9814
     0.0062546      -0.012392    -1.9628
     0.0093945      -0.018526    -1.9443
      0.012534      -0.024603    -1.9261

% Function Definition
function [x_primes] = ODE45SolvMain(t,x)
u = 1;
x_primes = [x(2); u*((1-x(1)^2))*x(2)-x(1)];
end

.

Answer 2

[Steven Lord]

See this Answers post for an example that sounds like what you want to do.