how to stop ode45 when one of states reach certain value

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Mohamed Aburakhis 2016-5-4

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https://ww2.mathworks.cn/matlabcentral/answers/282449-how-to-stop-ode45-when-one-of-states-reach-certain-value

评论： Walter Roberson 2023-10-20

采纳的回答： Jan

[x_dot] =derivative(t, x)

x is states [x(1).....x(4)] I need to stop the integration when x(3) reaches 0.1 when the integration stop record (t)

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Jan 2016-5-4

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https://ww2.mathworks.cn/matlabcentral/answers/282449-how-to-stop-ode45-when-one-of-states-reach-certain-value#answer_220745

在 MATLAB Online 中打开

Opt    = odeset('Events', @myEvent);
[T, Y] = ode45(@YourFun, T, Y0, Opt);
function [value, isterminal, direction] = myEvent(T, Y)
value      = (Y(3) == 0.1);
isterminal = 1;   % Stop the integration
direction  = 0;

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James Gilliam 2023-10-20

What is T in this senario and how is it defined? Thanks

Walter Roberson 2023-10-20

在 MATLAB Online 中打开

@James Gilliam

In the call

[T, Y] = ode45(@YourFun, T, Y0, Opt);

the input T is the timespan to integrate over. It is a vector that must have at least two elements, but may have more. If it has two elements then ode45() will decide by itself what times to output information at; if it has more than two elements then ode45() will output information at the times given in the vector.

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Walter Roberson 2016-5-4

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编辑：Walter Roberson 2017-9-13

You need to add an event function. See http://www.mathworks.com/help/matlab/ref/odeset.html#input_argument_d0e709348

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Mohamed Aburakhis 2016-5-4

I found a lot of options , which one should I use?

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Gustavo Lunardon 2020-8-17

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https://ww2.mathworks.cn/matlabcentral/answers/282449-how-to-stop-ode45-when-one-of-states-reach-certain-value#answer_481269

This is confusing. In matlab help it says: An event occurs when value(i) is equal to zero. All answers in this post make it value = 1 for the event to happen. Are the answers outdated somehow?

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

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Walter Roberson 2020-8-19

This is a valid concern.

value = (X(8) == 0.05 ) and (Y(3) == 0.1) would happen rarely, when the values were bit-for-bit identical to the representation of 0.05 and 0.1 . One bit different in the representation and the condition will not fire. Better is to write x(8) - 0.05 or 0.05 - x(8), and Y(3)-0.1 or 0.1-Y(3) -- zero crossings can be detected for those.

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