How to load intermediate variable into workspace while using ode45 ?
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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 ?
0 个评论
回答(2 个)
Stephen23
2022-6-17
编辑:Stephen23
2022-6-17
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)
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 个评论
Stephen23
2022-6-23
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
Torsten
2022-6-23
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 个评论
Torsten
2022-6-24
编辑:Torsten
2022-6-24
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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