while in ode45

Let me solve an ODE .

dy/dt=y.

y(0)=1.

Now i want to just solve the ODE for 5 seconds. Let y(5) denote the value of state 'y' at time =5.

At t=5 I want to update the initial condition as: y(5)=0.5*y(5) and again solve the given ODE for next 10 seconds.

I want to use while statement for this work. Can I do that ? If yes, please provide some outline of the code.

Thanks

[James Tursa]

What have you done so far? What specific problems are you having with your code? Can you code up the first part, without any while loop, to just solve for the first 5 seconds?

[Arnab]

Actually I was solving solving some different problem where I need this concept. Can you please solve the above ODE that will help me out, u think.

Answer 1

[Cris LaPierre]

One of your inputs to ode45 is tspan. If I were going to do this, I would look at building is so that my loop calls ode45 with the desired inputs rather than trying to do it inside the odefcn.

Here is an outline (untested)

y0 = 0;
tspan = [0 5;5 15]

y_out = [];
for ts = 1:sizse(tspan,1));
    [t,y] = ode45(odefcn,tspan(ts,:),y0);
    
    y0 = o.5*y(end);
    
    y_out = [y_out;y];
end

[Arnab]

Well, at first thanks.

Sir, can you please provide the whole code to solve the above ODE? That will give me some more insight to the problem and implementing somewhere in my original work.

Thanks

[Torsten]

If you choose to integrate from 0 to 5 and afterwards from 5 to 15, the solution becomes that large that your plot doesn't reveal what's happening. So I chose to integrate from 0 to 1 and from 1 to 2 for clarity.

y0 = 1;
odefcn = @(t,y) y;
tspan = [0 1;1 2];

y_out = [];
t_out = [];
for ts = 1:size(tspan,1);
    [t,y] = ode45(odefcn,tspan(ts,:),y0);
    
    y0 = 0.5*y(end);
    
    t_out = [t_out;t];
    y_out = [y_out;y];
end
plot(t_out,y_out)

[Arnab]

Thank you very much @torsten. Can you please elaborate a bit how it's working, it will help me to learn the concept.

By the way I was trying something to apply this in my main code.

Should I provide the main code and ask the problem there only?

[Walter Roberson]

Any one call to ode45() expects the implementing equations to be consistent, and to have continuous second derivatives.

In order to implement a discontinuity, it is necessary to stop the ode45 call at the boundary of the discontinuity, and adjust the boundary conditions, and then to call ode45 again to keep going.

If your implementing code had something like

if t == 5
    y = y./2;
end

then that would be a point of discontinuity in the first and second derivatives. The ode45() code would, in such a case, fail with a message about being unable to meet boundary conditions if you were lucky. If you are unlucky it would not notice and would silently return the wrong answer.

Actually, considering that ode45() uses adaptive step sizes, it is unlikely to happen to probe at exactly t == 5 unless 5 is one of the two tspan boundaries. So nearly all of the time it would simply produce the wrong answer.

The simplest form of discontinuity is if the discontinuity happens to occur at a strict pre-determined time boundary. Which is the case here -- the discontinuity occurs at time 5, which is pre-determined. In this particular case, you can "simply" ask ode45() to process up to time 5 during one call, then change the boundary conditions outside of ode45(), and then call ode45 again with an updated tspan.

In more complicated cases, discontinuities occur at unpredictable times. An example is a bouncing ball: the exact time it is going to hit the floor or an outside wall is difficult to predict. In such cases, you have to specify "event" functions through the ode options structure. The events functions would detect the boundary and set its outputs to signal whether or not a boundary had been it, and what to do if it is hit; most often the "what to do" is to terminate ode45() execution. (Termination of execution is not immediate; instead ode45() probes the boundary conditions and integrates right up to the boundary.)

[Arnab]

Yah, thanks. I genraly use event function .

By the way I was trying to solve something lik following:

I have to solve ODE given by:

dy/dt= y- (|y|^0.5)*signum(y).

And T1=2; is some given constant.

Where event points are generated by

s_(k+1)={ t >=t and t<T1_k: y(t)=e^-(t_k)}

where t_k= s_k+ r_k, r_k is the delay term . And at t=t_k, y(t) should update the initial condition as

y(t_k)=y((t_k)^)+ 0.5*|y(s_k)|/(1+0.5*)).

And for t>=T1 we should just solve the given ODE and there is event condition for t>=T1.

And we should plot t vs y(t) for 4 seconds.

Can u please help me out?

Thanks

[Walter Roberson]

You already asked about that in https://www.mathworks.com/matlabcentral/answers/2169883-event-triggered-control-using-ode45?s_tid=srchtitle

[Arnab]

Yes, thanks.

[Torsten]

@Arnab

As you are a beginner in MATLAB, use MATLAB Onramp to learn the basics of the new language:

https://matlabacademy.mathworks.com/details/matlab-onramp/gettingstarted

Concerning your question, it will help to study the "ballode" example here:

https://uk.mathworks.com/help/matlab/math/ode-event-location.html

[Arnab]

Ok, thank u for the help.