Change input at each time step of the ODE solver 'ode45'

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KC
KC am 5 Dez. 2015
Kommentiert: Jan am 5 Mai 2017
I am not sure how to change an input parameter 'β' at each time step. My code is below - which gives me an error. Can anybody help please!
t = [7 14 21 28 35 42 49 56 63 70 77 84];
for i=1:12;
beta(i) = 0.43e-08 + (4.28e-08 - 0.43e-08)*exp(-0.20*t(i));
end
f = @(t,x) [3494-0.054*x(1)-beta*x(1)*x(3); beta*x(1)*x(3) - 0.41*x(2); ...
50000*x(2) - 23*x(3)];
[t,xa1] = ode45(f,t,[64700 0 0.0033],beta);
  1 Kommentar
Jan
Jan am 5 Dez. 2015
And the error message is:
Error using vertcat
Dimensions of matrices being concatenated are not consistent.
Error in @(t,x)[3494-0.054*x(1)-beta*x(1)*x(3);beta*x(1)*x(3)-0.41*x(2);50000*x(2)-23*x(3)]

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Akzeptierte Antwort

Jan
Jan am 6 Dez. 2015
Please consider, that Matlab's ODE integrators cannot handle non-smooth functions sufficiently. See http://www.mathworks.com/matlabcentral/answers/59582#answer_72047 .
The only reliable method to run the integration is a loop over the time intervals:
function yourIntegration
tResult = [];
xResult = [];
tStep = [7 14 21 28 35 42 49 56 63 70 77 84];
x0 = [64700 0 0.0033];
for index = 2:numel(tStep)
% Integrate:
beta = 0.43e-08 + (4.28e-08 - 0.43e-08) * exp(-0.20*t(index - 1))
af = @(t,x) f(t, x, beta);
t = tStep(index-1:index);
[t, x] = ode45(af, t, x0);
% Collect the results:
tResult = cat(1, tResult, t);
xResult = cat(1, xResult, x);
% Final value of x is initial value for next step:
x0 = x(end, :);
end
function dx = f(t,x, beta)
dx = [3494-0.054*x(1)-beta*x(1)*x(3); ...
beta*x(1)*x(3) - 0.41*x(2); ...
50000*x(2) - 23*x(3)];
  7 Kommentare
Saiprasad Gore
Saiprasad Gore am 5 Mai 2017
Thanks a lot, I had a similar problem. I wanted to switch the eqn depending on condition after every step. I hope this will work in my case too. Can you tell me how to give ode45 just 1 step without intermediate adaptive steps?
Jan
Jan am 5 Mai 2017
@Saiprasad Gore: This is not possible with ode45.

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Weitere Antworten (1)

Walter Roberson
Walter Roberson am 6 Dez. 2015
f = @(T,x) [3494-0.054*x(1)-interp1(t,beta,T,'linear','extrap')*x(1)*x(3); interp1(t,beta,T,'linear','extrap')*x(1)*x(3) - 0.41*x(2); ...
50000*x(2) - 23*x(3)];
  2 Kommentare
KC
KC am 12 Dez. 2015
Thanks Walter!
sam
sam am 15 Jun. 2016
Bearbeitet: sam am 16 Jun. 2016
@Walter Roberson
Hi Walter,
Why do we have to do interpolation if we already know the exact expression of the variables? Couldnt we just input the exact expression of the variables into the Matlab ode45 solver? If we could, could you kindly tell me how to do this? Thanks.

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