Dear bosses, I want to draw the Hopf bifurcation image. Please help me, thank you. There are specific instructions below.

Dear bosses, in the ordinary differential equations, how to draw a parameter change to cause the value change, that is, Hopf bifurcation problem about ODE. For example, in the figure below, A = 0.5, \mu= 0.2, v = 0.3. \ beta is a variable. Now I want to draw the images of \beta and S and \ beta and R. Please help me. Please give me the MATLAB code. Thank you!

[darova]

Did you try ode45? DO you have any attempts? What problems are you facing? Can you elaborate?

[jianzhong gao]

Thank you for your reply. I did it carefully. I also used ode45, but there was some errors. Please help me to modify the code. Thank you! Here is the matlab code I wrote

function     bifurcation
global \beta
i=1;
for j=0.01:0.02:4
\beta=j;
sol=ode45(@bifupic_eq,@bifurpichist,[0,400],\beta);
p=sol.y(1,:);q=sol.y(2,:);t1=sol.x;
L=length(t1);
           for k=L-200:L
           x(i)=p(k);y(i)=q(k);t(i)=\beta;
            i=i+1;
             end
end
figure
plot(t,y,'.');
%plot(t,x,'.');
xlabel('\beta');
ylabel('Density of juvenile prey, x_0(t)');
%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function s=bifurpichist(t)  %%%%  initial value
s=[ 0.2*sin(t)+0.4
    0.1*cos(t)+0.5];
%%%%%%%%%%%%%%%%%%%%%%%%
function dydt=bifupic_eq(t,y,\beta);  %%%  differential equation
global \beta
dydt=[0.5-\beta*y(1)*y(2)-0.2*y(1)
    \beta*y(1)*y(2)-0.3*y(2)];

[darova]

try this

[jianzhong gao]

Thank you for your reply. According to your instructions, I modified the code, and the result was wrong. The result of running showed that the index must be a positive integer or a logical value. I think in ode45, the range of abscissa is [0.01, 4], not [0,400]. Is there a problem with the use rules of ode45? Please give me suggestions.Thank you！I look forward to your reply.

%%%a command file %%%
clear
i=1;
for j=0.01:0.02:4
\beta=j;
sol=ode45(@bifupic_eq,[0,400],[0.4 0.5],[],\beta);
p=sol.y(1,:);q=sol.y(2,:);t1=sol.x;
L=length(t1);
           for k=L-200:L
           x(i)=p(k);y(i)=q(k);t(i)=\beta;
            i=i+1;
             end
end
figure
plot(t,y,'.');
%plot(t,x,'.');
xlabel('\beta');
ylabel('Density of juvenile prey, x_0(t)');
%%%%%%%%%%%create a function file named bifupic_eq. m.%%%%%%
function dydt=bifupic_eq(t,y,\beta);  %%%  differential equation
dydt=[0.5-\beta*y(1)*y(2)-0.2*y(1)
    \beta*y(1)*y(2)-0.3*y(2)];

Answer 1

[darova]

What about this solution?

clc,clear
cla
for j = 0:4
    beta = j;
    F=@(t,y) [0.5-beta*y(1)*y(2)-0.2*y(1)
                beta*y(1)*y(2)-0.3*y(2)];
    [t,y] = ode45(F,[0 3],[0.4 0.5]);
    line(t,y,'color',rand(1,3))
end
xlabel('beta');
ylabel('Density of juvenile prey, x_0(t)');

[jianzhong gao]

Thank you for your reply. I want to draw the Hopf branch of \ beta and S or I. Can you draw a picture similar to the one below?

[darova]

Look here: LINK

[jianzhong gao]

Thank you!