Event function and ODE.

2024 11 20
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更新时间：2024 11 22
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I was solving an one dimensional ODE using ODE45.
I have two events in event function, where the execution of second event is dependent on the time instant of the first event.
How could I do that? Please help!
Elaborately...
1) Second event will comes into existence only after the execution of first event and is dependent on the time instant of first event.
for example..
Let first event instant be t=te. Then second instant will be at t=te+1.

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Steven Lord
Steven Lord 2024-11-20
Can you provide the ODE you're trying to solve, describe what it represents (a ball bouncing on the ground, a baton twirling in the air, a ballistics problem, etc.), and explain what the events represent in the physical system?
Arnab
Arnab 2024-11-21
Thank you sir.
I am attaching the code below...which i tried
In the "events1" function I want second event occurs after some time at which first event occurs. That is , if "te" is the time aat which first event occurs then second event will occurs at time t=te+0.01.
Also I need y(te+0.01) as per defined in the else part i.e., y0 =z-0.5*ye(end);
Can I do such stuff or i have to do something else, kindly help. Thank you
function mtry2
y0 =2;
yplot = []; % store the solution
T=1; % constant
tp=[]; %collection of step size of time
tstar = 0;
z=[];
tau = 0.01; %delay
te=0.2652;
options1 = odeset('Events',@events1 );
while 1
tspan = linspace(tstar, tstar + 0.05, 50);
[t, y, te, ye, ie] = ode45(@event1, tspan, y0, options1); % solving the ode
%ie = intersection count of occuring of the events
%te = store the collection of intersecting oints
event1 =find(ie == 1);
event2 =find(ie == 2);
if isempty(event1) && isempty(event2)
% if isempty(ie) % Extend the interval
tstar = t(end);
y0 = y(end);
yplot = [yplot; y(:, 1)];
tp = [tp; t];
elseif ~isempty(event1) && isempty(event2)
tstar = te(end);
y0 = ye(end);
yplot = [yplot; y(:, 1)];
tp = [tp; t];
z=ye(end);
% event2 =find(ie == 2, 1);
% elseif isempty(event2)
% tstar = t(end); % Introduce the delay by adding tau
%
% y0 = y(end, :);
% % [t, y] = ode45(@event1, [tstar te(end)+tau], y0);
% yplot = [yplot; y(:, 1)];
% tp = [tp; t];
else
tstar=te(end);
y0 =z-0.5*ye(end);
yplot = [yplot; y(:, 1)];
tp = [tp; t];
end
% end
if (tstar >= 1)
break;
end
end
plot(tp,yplot,'b','LineWidth',2);
% hold on
% plot(tp,2*exp(-log(40)*tp/T));
grid on
ylabel('y(t)','FontSize',20,...
'FontWeight','bold','Color','k');
xlabel('t','FontSize',20,...
'FontWeight','bold','Color','k');
set(gca,'FontSize',25,'LineWidth',2.5);
grid on;
end
%==================================================
function dy = event1(~,y) % intermediate dunction to solve the ode
dy1=((y(1))^1.5)*sign(y(1)); % given ode
dy=dy1;
end
function [value,isterminal,direction] = events1(t,y) % when to stop the integration
% y is symbolic variable
% t is a vector
M=2;
tau = 0.01; %delay
te=0.2652;
ep=0.1;
T=1;
V=abs(y(1));
u=V/(1+V); %black curve
B=-log(M*M/ep)/T;
value =[u-M*exp(B*t);t-te-tau]; %blue curve
isterminal = [1;1]; %flag to stop the integration
direction = [0;0]; %dummy variable
end
Steven Lord
Steven Lord 2024-11-21
Can you show the mathematical form of the ODE you're trying to solve, not the code? Show it and describe what it's computing in text, please.

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 采纳的回答

Torsten
Torsten 2024-11-20
移动：Torsten 2024-11-20

0 个投票

Don't use both events simultaneously in your event function.
Start the solver with the first event function and integrate until the first event happens.
Return control to the calling program.
Restart the solver with the value of the solution variable obtained so far and the second event function (or up to te+1).

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Arnab
Arnab 2024-11-20
Thank you but
More specifically..i have to solve the ode
dy/dt=y
First event function is..... event(1)=y-e^-t. Let 'te' be such an instant.
Then we will go and check for the second event, which is
event(2)=t-te-1.
and the process will continue for 10 seconds.
My question is can i somehow give "te" as an input to event function or something like that!?
Thaank you
Torsten
Torsten 2024-11-20
编辑：Torsten 2024-11-20
Reread my answer. It seems you didn't understand it. For the second event, simply restart the solver with the value of the solution variable obtained after the first event and set the end time of the integration to te+10.
Arnab
Arnab 2024-11-20
Yah, i got it but i was trying this way. And thus asked if it could be possible .
By the can you plese provide the code corresponding to your answer, that will be of lot help as i am new to matlab and trying to work around this.
Thank you
Torsten
Torsten 2024-11-20
编辑：Torsten 2024-11-20
Ran in:
Ran in:
I don't understand your first event.
But here is code for your 2-stage events:
fun = @(t,y) y;
tspan = [0 3];
y0 = 1;
options = odeset('Events',@EventFcn);
%Integrate up to the first event (here: up to t such that exp(t) = 6)
[T1,Y1] = ode45(fun,tspan,y0,options);
%Continue integration for 1 second
tspan = [T1(end) T1(end)+1];
y0 = Y1(end,1);
[T2,Y2] = ode45(fun,tspan,y0);
T = [T1;T2];
Y = [Y1;Y2];
plot(T,Y)
grid on
function [position,isterminal,direction] = EventFcn(t,y)
position = y(1) - 6; % The value that we want to be zero
isterminal = 1; % Halt integration
direction = 0; % The zero can be approached from either direction
end
Arnab
Arnab 2024-11-21
Thank you , but i need to continue the loop to certain value .
Can you please have a look at code attached above and help me out.
Torsten
Torsten 2024-11-21
编辑：Torsten 2024-11-22
Ran in:
Ran in:
I'm not sure if this is what you want.
tfinish = 1.0;
tstart = 0.0;
treached = tstart;
tend = tfinish;
tspan = [tstart, tend];
y0 = 2;
options = odeset('Events',@event);
T = [];
Y = [];
while 1
[t, y] = ode45(@fun, tspan, y0, options); % solving the ode
T = [T;t];
Y = [Y;y];
treached = t(end);
if treached >= tfinish
break
end
tstart = treached;
tend = min(tfinish,treached + 0.01);
tspan = [tstart, tend];
y0 = y(end,1);
[t, y] = ode45(@fun, tspan, y0); % solving the ode
T = [T;t];
Y = [Y;y];
treached = t(end);
if treached >= tfinish
break
end
tstart = treached;
tend = tfinish;
tspan = [tstart,tend];
y0 = y0-0.5*y(end,1);
end
plot(T,Y)
function [value,isterminal,direction] = event(t,y) % when to stop the integration
% y is symbolic variable
% t is a vector
M=2;
ep=0.1;
T=1;
V=abs(y(1));
u=V/(1+V); %black curve
B=-log(M*M/ep)/T;
value =u-M*exp(B*t); %blue curve
isterminal = 1; %flag to stop the integration
direction = 0; %dummy variable
end
function dy = fun(~,y) % intermediate dunction to solve the ode
dy=((y(1))^1.5)*sign(y(1)); % given ode
end
Arnab
Arnab 2024-11-22
yes, i wanted something like this.
Thank you.

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