Efficient ODE with function handles
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Hi,
I am solving an ODE very often, so efficiency is important. The ODE I am solving calls other functions, because the ode is very general, but the details depend on the problem.
What I am doing now is this ( not working code, but the relevant snippets)
% just some parameters that are passed. This does not impact performance
params = {1,2,3};
% function handle of function that changes the ode in question
h1 = @f_function;
% function handle of the ode
%h_dydt = @(t,y)f_dydt(t,y,params); % aproach 1)
h_dydt = @(t,y)f_dydt(t,y,params,h1); % aproach 2)
solution = ode45(h_dydt,...)
Where the ode function is
%function [dydt] = f_dydt(t,y,params,h)% aproach 1)
function [dydt] = f_dydt(t,y,params,h) % aproach 2)
% a = f_function(y,params); % aproach 1)
a = h(y,params); % aproach 2)
%the rest is not important,but could be
dydt = y*t*a;
end
The problem is, that aproach 2) takes 60% more time than aproach 1), and I wonder if there is a way to have the same flexibility without the huge added cost. As it is now, I have to do aproach 1) and write a separate dydt function for each problem.
Thanks for looking into this!
采纳的回答
Bjorn Gustavsson
2021-9-3
编辑：Bjorn Gustavsson
2021-9-3
If the function-handle-argument is too costly and you still want "some general flexibility" you might try to simply extend the ODE-function with a switch-selecting between the different drivers? Something like this:
%function [dydt] = f_dydt(t,X,params,h)% aproach 1)
function [dydt] = f_dydt(t,X,params,n_RHS) % aproach 1.5)
% a = f_function(x,y,params); % aproach 1)
switch n_RHS
case 1
a = h1(X,params); % aproach 1.5)
case 2
a = h2(X,params); % aproach 1.5)
otherwise
a = 12;
end
%the rest is not important,but could be
dydt = y*t*a;
end
This should fall somewhere inbetween your two aproaches, and might be good enough as a solution to both issues. It will require some effort on writing the documentation of your function to provide "easy use".
HTH
更多回答（1 个）
Does f_function expect to be called with two input arguments or three?
x = [1; 2; 3];
y = [4; 5; 6];
value1 = min([x, y], 4) % This works
value2 = min(x, y, 4) % This does not work
How are the X input argument with which ode45 calls your function related to the two variables x and y whose definitions you've elided from your code?
I suspect you're comparing apples and oranges here, though with the limited information you've provided I can't be certain.
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