MATLAB: 3D Phase portrait for a set of differential equations

3d phase portraits3d plotsMATLABplot3

I have a set of three differential equations and I want to make a phase portrait of them. I have some idea of using quiver or plot3 to get a phase portrait of a set of 3 differential equations. I am unable to do for this case.

Here is my attempt:

timerange= 0:0.5:350;
IC= [0.1,0,0];
IC1= [0.1,0.2,0];
[t,y] =ode45(@(t,y) fn(t,y),timerange, IC);
[t1,y1] =ode45(@(t,y) fn(t,y),timerange, IC1);
plot3(y(:,1),y(:,2),y(:,3)); 
hold on
plot3(y1(:,1),y1(:,2),y1(:,3)); 
grid 
xlabel('T')
ylabel('A')
zlabel('I')
function rk1 =fn(t,y)
r=0.00173;K=0.03166;A0=0.4;gammaA=0.04;eps = 0.00055;rho= 0.025;alpha1= 1.30187;c1=0.63433;
I0= 0.3;gammaI=0.0208;
alpha= 0.0002802;
n= y(1);
A= y(2);
I= y(3);
    
rk1(1)= r*n*(1- n/K)+ alpha*A*n -eps*n*I;
rk1(2) = A0 - gammaA*A;
rk1(3) = I0 + (rho*I*n)/(alpha1+n) -c1*I*n - gammaI*I;
rk1=rk1(:);
endI

What I am doing is merely changing the initial conditions The actual result should be something like this.

Any help would be greately helpful. This is what I am getting from above code.

Thank You

Best Answer

I tried to plot T A and I separately. X Y Z axis represents T A and I respectively. Color represents derivative of T A or I

I used surface to plot color lines (plotted curves as surfaces, made them nx2 size)

derivative

just difference

Related Solutions

MATLAB: Solving a system of differential equations using ODE45 by switching between two functions depending upon the solution obtained in last iteration.

According to the problem you defined in your comment, this is the correct way to implement it using ode45

timerange = 0:0.5:200;
IC= [0.1,0.1];%initial conditions
threshold=0.5;
[t,y] =ode45(@(t,y) fn_1(t,y), timerange, IC);
plot(t,y(:,1),'r');
hold on
plot(t,y(:,2),'b');
xlabel('n')
ylabel('I')
grid on
function rk1 =fn_1(t, y)
n = y(1);
I = y(2);
if n < 0.1
    r      = 0.5; K = 0.5;
    eps    = 0.001;  rho= 0.02;
    alpha1 = 2.15; c1= 0.025;
    gammaI = 0.02; I0= 0.01;
else
    r      = 0.5;   K= 0.5;
    eps    = 0.001; rho= 0.02;
    alpha1 = 2.15;  c1 = 0.025;
    I0     = 0.5;     gammaI = 0.2;
end
rk1(1) = r*n*(1- n/K)-eps*n*I;
rk1(2) = I0 + (rho*I*n)/(alpha1+n) -c1*I*n - gammaI*I;
rk1 = rk1(:);
end

However, even the 2nd set of parameters does not decrease the value of n.

MATLAB: Solving differential equation with ode45

func=@(t,y)(y-1)^2;
tspan = [0 10];
for x = 0.01:0.05:1
    y0 = x;
    [t,y] = ode45(func,tspan,y0);
    plot(t, y(:,1));
    hold on
end

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