Trouble with time delay differential equation

%% ddex1_func

function dYdt = ddex1_func(t,Y,Z)

a=8.5; b=-9.5; c=-1; d=4.55;

y1=Y(1);y2=Y(2);

y1_tau1 = Z(1,4.9);

y2_tau1 = Z(2,4.9);

dy1dt = a*y1*(1-y1) + b*y1*y2;

dy2dt = c*y2 + d*y1_tau1*y2_tau1;

dYdt = [dy1dt; dy2dt];

%% ddex1

clc;close all;clear all;

lags = [4.9];

tspan = [0 200];

sol = dde23(@ddex1_func,lags,@history,tspan)

y1 = sol.y(1,:); % note the y-solution is a row-wise matrix.

y2 = sol.y(2,:);

plot(y1,y2)

xlabel('y_1')

ylabel('y_2')

%% history

function y = history(t)

y = [0.5;0.4];

end

>> The error that shows is: Attempted to access Z(1,4.9); index must be a positive integer or logical.

>>Error in ddex1_func (line 5)

>>y1_tau1 = Z(1,4.9);

Can anyone help me rectify my mistake? The moment I replace 4.9 with 1 (wherever required i.e Z(:,4.9), lags=[4.9]), it provides a periodic graph, whereas 4.9 (which should have given aperiodic graph) yields error. I couldn't understand the logic behind the error.

Answer 1

[David Wilson]

I've only made one small change to your function. You have tried to index a non-integer number back, but the "Z" variable holds the values.

function dYdt = ddex1_func(t,Y,Z)
a=8.5; b=-9.5; c=-1; d=4.55;
y1=Y(1);y2=Y(2); 

%y1_tau1 = Z(1,4.9);
%y2_tau1 = Z(2,4.9);

y1_tau1 = Z(1,1); 
y2_tau1 = Z(2,1); 

dy1dt = a*y1*(1-y1) + b*y1*y2;
dy2dt = c*y2 + d*y1_tau1*y2_tau1;

dYdt = [dy1dt; dy2dt];
return 

Does this look like something you are expecting?

[Debanita Das]

Thank you so much for your help.

Yes the graph should look like this.

Query: The equation demands a lag of 4.9. Shouldn't I place it in the code?

[Torsten]

Query: The equation demands a lag of 4.9. Shouldn't I place it in the code?

You implemented it in your code by setting

lags = [4.9];

and calling dde23 as

sol = dde23(@ddex1_func,lags,@history,tspan)