How to load intermediate variable into workspace while using ode45 ?

function dx=system (t,x)
dx=-x(1);
a=1;
end
%%
function main
tspan=1:100;
x0=0;
[t,x]=ode45(@system, tspan,x0)
end
how to load variable "a" into the workspace ?

Réponses (2)

Stephen23
Stephen23 le 17 Juin 2022
Modifié(e) : Stephen23 le 17 Juin 2022
Here is the neat, easy, robust approach which returns exactly the a values at the exact t and x output values:
tspan = 1:100;
x0 = 0;
[t,x] = ode45(@system, tspan,x0);
[~,a] = arrayfun(@system,t,x)
a = 100×1
1 1 1 1 1 1 1 1 1 1
function [dx,a] = system(t,x)
dx = -x(1);
a = 1;
end
Note that trying to obtain the value generated during the ODE-solver routine is much more complex (although this is often what beginners think of, it is practically unsolvable because you have no easy way to distinguish between points which are kept vs. discarded, e.g. when the solver goes backwards a few steps. Not every call of the objective function corresponds to output values, and you have no way to know which... consider the implications of that).

2 commentaires

Pritesh Patel
Pritesh Patel le 23 Juin 2022
Modifié(e) : Torsten le 23 Juin 2022
function [dx,ss]=system(t,x)
dx=-x;
ts=40;
ssActive = [1,2,3,4,1,3,4,2,4,3,3,4,3,3,1,2,3,4,1,3,4,2];
k=floor(t/ts);
ss = ssActive(k+1);
end
tspan = 1:200;
x0 = zeros(38,1);
[t,x] = ode45(@system, tspan,x0);
[~,ss] = arrayfun(@system,t,x)
I want to plot ss. dimension of the state vector is 38*1
Code suggested by you is giving some dimensional error.
Ah, so in your actual problem x0 is not scalar.
tspan = 1:200;
x0 = zeros(38,1);
[t,x] = ode45(@system, tspan,x0);
[~,ss] = cellfun(@system,num2cell(t),num2cell(x,2),'uni',0);
ss = [ss{:}];
plot(t,ss)
function [dx,ss]=system(t,x)
dx=-x;
ts=40;
ssActive = [1,2,3,4,1,3,4,2,4,3,3,4,3,3,1,2,3,4,1,3,4,2];
k=floor(t/ts);
ss = ssActive(k+1);
end

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tspan = 1:200;
x0 = zeros(38,1);
[t,x] = ode45(@system, tspan,x0);
ts=40;
ssActive = [1,2,3,4,1,3,4,2,4,3,3,4,3,3,1,2,3,4,1,3,4,2];
k=floor(t/ts);
ss = ssActive(k+1);
plot(t,ss)
function [dx]=system(t,x)
dx=-x;
end

4 commentaires

That was nice....
But I am using value of 'ss' inside the system function. Now I have to plot that intermediate variable.
You can use them inside the "system" function as well as plot them after integration has finished.
In "system", you only have a single value for t - no chance for plotting there.
Pritesh Patel
Pritesh Patel le 24 Juin 2022
Modifié(e) : Pritesh Patel le 24 Juin 2022
ts=40;
ssActive = [1,2,3,4,1,3,4,2,4,3,3,4,3,3,1,2,3,4,1,3,4,2];
k=floor(t/ts);
ss = ssActive(k+1);
If I am using above equation inside the "system" function then after running the ode45 I am not able to store values in the workspace and hence not able to plot it.
Torsten
Torsten le 24 Juin 2022
Modifié(e) : Torsten le 24 Juin 2022
Do you need these 4 lines of code for the solution of your differential equation or only for postprocessing and plotting ?
If you need these 4 lines of code for the solution of your differential equation, please show how. In your code above, they are not used to calculate "dx".
If you need them for plotting, do it after ode45 has finished as I showed in the code I posted.

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Modifié(e) :

le 24 Juin 2022

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