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

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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!
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jianzhong gao
jianzhong gao 2020-5-21
编辑:jianzhong gao 2020-5-21
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)];

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回答(1 个)

darova
darova 2020-5-21
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)');
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