I dont understand the error
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I would like it to stop calculating the ode when both functions meet the requirement that dx=0 , and I did:
alpha=0.5;
beta=0.5;
r1=2;
r2=3;
s1=1;
s2=1;
t0 = 0;
tfinal = 100;
y0 = [1;1];
AnonFun = @(t,y)diag([2+0.5*y(2)-1*y(1),3+0.5*y(1)-1*y(2)])*y;
if (alpha>0)&&(beta>0)
Opt=odeset('Events',@(t,y)myEvent1(t,y,AnonFun));
Other
Opt=odeset('Events',@(t,y)myEvent2(t,y,AnonFun));
end
[t,y,te,ye,ie] = ode23(AnonFun,[t0 tfinal],y0,Opt);
plot (t,y)
function [value, isterminal, direction] = myEvent1(t,y,AnonFun)
value = AnonFun(t,y)-1.0e-3;
isterminal = 1; % Stop the integration
direction = -1;
end
function [value, isterminal, direction] = myEvent2(t,y,AnonFun)
value=abs(AnonFun(t,y))-0.001;
isterminal = 1; % Stop the integration
direction = -1;
end
But when I change the vector y0 to [1;5] for example, I got this message:
Index exceeds the number of array elements. Index must not exceed 1.
Error in odezero (line 142) if any(isterminal(indzc))
Error in ode23 (line 335) odezero(@ntrp23,eventFcn,eventArgs,valt,t,y,tnew,ynew,t0,h,f,idxNonNegative);
Error in LogisticGrowthForTwoSpecies (line 17)
[t,y,te,ye,ie] = ode23(AnonFun,[t0 tfinal],y0,Opt);
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采纳的回答
Sam Chak
2022-8-24
编辑:Sam Chak
2022-8-27
Edit: After understanding what you really want in your latest clarification. There is a simpler and intuitive way to code the program such that the simulation only run for approximately 3 or 4 times the Settling Time, .
This allows or of the plot window to show the transient trajectories of the states from the initial values to the steady-state values. This method is only meaningful for systems that have stable equilibrium points.
Analysis shows that your system has a stable equilibrium point at and and three other unstable equilibrium points at , , and .
Note: I changed the initial values for and because and , and to show you the difference between the settling time approach and the event function approach.
alpha = 0.5;
beta = 0.5;
r1 = 2;
r2 = 3;
s1 = 1;
s2 = 1;
t0 = 0;
tfinal = 5; % Adjust this parameter roughly 4 times the Settling Time, Ts
y0 = [6 6];
AnonFun = @(t,y) diag([2 + 0.5*y(2) - 1*y(1), 3 + 0.5*y(1) - 1*y(2)])*y;
% AnonFun = @(t,y) [(2 + 0.5*y(2) - 1*y(1))*y(1);
% (3 + 0.5*y(1) - 1*y(2))*y(2)];
[t, y] = ode23(AnonFun, [t0 tfinal], y0);
plot(t, y), grid on
y1 = y(:,1);
y1e = 14/3;
idx = find(t > 1 & y1/y1e > 0.98 & y1/y1e < 1.02); % applying 2% criterion after 1 sec
Ts = t(idx(1))
14 个评论
Sam Chak
2022-8-27
I have updated my Answer to show you an alternative approach. But the following uses the Event function approach to force stop the ode45, so that you can see which one suits your needs.
alpha = 0.5;
beta = 0.5;
r1 = 2;
r2 = 3;
s1 = 1;
s2 = 1;
t0 = 0;
tfinal = 10;
y0 = [6 6];
AnonFun = @(t,y) diag([2 + 0.5*y(2) - 1*y(1), 3 + 0.5*y(1) - 1*y(2)])*y;
% AnonFun = @(t,y) [(2 + 0.5*y(2) - 1*y(1))*y(1);
% (3 + 0.5*y(1) - 1*y(2))*y(2)];
if (alpha > 0) && (beta > 0)
Opt = odeset('Events', @(t, y) myEvent1(t, y, AnonFun));
else
Opt = odeset('Events', @(t, y) myEvent2(t, y, AnonFun));
end
[t, y, te, ye, ie] = ode23(AnonFun, [t0 tfinal], y0, Opt);
plot(t, y), grid on
function [value, isterminal, direction] = myEvent1(t, y, AnonFun)
value = norm(AnonFun(t,y)) - 1e-2;
isterminal = 1; % Stop the integration
direction = -1;
end
function [value, isterminal, direction] = myEvent2(t, y, AnonFun)
value = norm(AnonFun(t,y)) - 1e-2;
isterminal = 1; % Stop the integration
direction = -1;
end
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