Event function and ODE.

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.

[Steven Lord]

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]

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]

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.

Answer 1

[Torsten]

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).

[Arnab]

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]

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]

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]

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]

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]

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]

yes, i wanted something like this.

Thank you.