How to plot bifurcation with Delay Differential equations?

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I want to draw the bifurcation diagram for the model.
All parameters are positve constant.
The value of parameters are as:
A1 = 0.8463, A2 = 0.6891, K = 1.2708, beta1 = 0.4110, beta2 = 0.1421,
The diagram are vary tau from 68 to 72 in steps of 0.001. For inital conditions X(0) = 0.26 and Y(0) = 0.58.
Please ansers me for Matlab code to plot the bifurcation diagrams.
  7 个评论
kaushik dehingia
kaushik dehingia 2021-2-11
移动:Dyuman Joshi 2024-3-15
Can anyone share the Bifurcation diagram code for a delayed system? I t will be very helpful for me.

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采纳的回答

Alan Stevens
Alan Stevens 2020-11-17
How about the following for your loop (it assumed you have defined tau earlier in the file):
for j=1:N+1
if t(j)<=tau
xd = x(1);
yd = y(1);
else
d = ceil((t(j)-tau)/h);
xd = x(d);
yd = y(d);
end
t(j+1)=t(j)+h;
%tempx(j+1)=tempx(j)+h;
%tempy(j+1)=tempy(j)+h;
k1x=fx(t(j), x(j), y(j));
k1y=fy(t(j), xd, yd, y(j));
%k1y=fy(t(j), x(j), y(j), tempx(j), tempy(j));
k2x=fx(t(j)+h/2, x(j)+h/2*k1x, y(j)+h/2*k1y);
k2y=fy(t(j)+h/2, xd+h/2*k1x, yd+h/2*k1y, y(j)+h/2*k1y);
%k2y=fy(t(j)+h/2, x(j)+h/2*k1x, y(j)+h/2*k1y, tempx(j)+h/2*k1x, tempy(j)+h/2*k1y);
k3x=fx(t(j)+h/2, x(j)+h/2*k2x, y(j)+h/2*k2y);
k3y=fy(t(j)+h/2, xd+h/2*k2x, yd+h/2*k2y, y(j)+h/2*k2y);
%k3y=fy(t(j)+h/2, x(j)+h/2*k2x, y(j)+h/2*k2y, tempx(j)+h/2*k2x, tempy(j)+h/2*k2y);
k4x=fx(t(j)+h, x(j)+h*k3x, y(j)+h*k3y);
k4y=fy(t(j)+h, xd+h*k3x, yd+h*k3y, y(j)+h*k3y);
%k4y=fy(t(j)+h, x(j)+h*k3x, y(j)+h*k3y, tempx(j)+h*k3x, tempy(j)+h*k3y);
x(j+1)=x(j)+h/6*(k1x+2*k2x+2*k3x+k4x);
y(j+1)=y(j)+h/6*(k1y+2*k2x+2*k3y+k4y);
end
  20 个评论
ibtissam benamara
ibtissam benamara 2021-6-20
this is not my question, i undersand why you use (1,end-50:end);
the question why you will take x(i,:)=y(i,:), consequetly, it will give the same figure for the two species, this not true;
Akanksha Rajpal
Akanksha Rajpal 2022-1-30
Your code really helped, but I was wondering if we can use similar coding if we want to extend the work to two delays? I tried that but it was showing error.
If you could help me regarding this and provide a code for this example only where the delay residing in X and Y are tau1 and tau2 respectively.

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更多回答(1 个)

Priya Verma
Priya Verma 2024-3-15
In question, the denominator term is define in first delay variable term. Why are you all this term is defining in second delay term.
i. e. fy =@(t,x,y) A2*x*y/(1+y)-b2*y; in this denominator term is (1+y) .....?
A2*xd*yd/(1+yd)-b2*y; in this denominator term is (1+yd) .....?
please, explain...!
  21 个评论
Priya Verma
Priya Verma 2024-3-26
Is there any code, package, etc to fit the parameter values of dde?

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