Using loop for calculating ode with different parameter

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Martin Pecha
Martin Pecha on 9 Nov 2017
Commented: Steven Lord on 10 Nov 2017
How can I create a cycle, where I would use ode45 and be able to get a solution for 3 different values of a constant k so I don't have to change it manually.
example k=1 --> solution, then k=10 --> solution etc.
At the end, I would need to put all solutions into the same plot to compare.

Answers (1)

Eric
Eric on 9 Nov 2017
Use a for loop. If k is non-linear, just put your values of interest into a vector.
loop_i = 0;
for k = [1 10 ....Rest of your numbers];
loop_i = loop_i+1;
[t{loop_i}, y{loop_i}] = ode45(...Whatever involving k);
end
I have to admit I don't know what size outputs the function ode45() will return, so sticking them in cells was simply a safe bet for me. If you always expect the same size vector to be returned, you could use y(:,loop_i) instead, or y(:,:,loop_i), etc. If y(:,loop_i) works for you, plotting will be as easy as plot(t,y).
  5 Comments
Steven Lord
Steven Lord on 10 Nov 2017
Pass k into your ODE function (the first input to the ODE solver) as an additional parameter. The description of the odefun input argument in the documentation for ode45 contains a link "Parameterizing Functions" that describes various techniques you can use to do that.

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