ODE45 with a vector input

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Daniel Grebler
Daniel Grebler 2018-6-28
评论： Torsten 2018-6-29
Hey, how can I solve an ode function with a changing input of my choosing: example:
I would like to solve the simple spring and defuse equation while inputting the external force at given time.
I would create a function for the differential equations:
if true
function dy = springdef(y,u)
k=1.5; %just values
m=3;
d=0.3;
% main differential equations
dy(1,1) = y(2);
dy(2,1) = -k/m*y(1)+u/m-d/m*y(2);
end
than I would calculate the result by using the ode45
if true
[t,y] = ode45(@(t,y)springdef(y,u),tspan,y0);
end
yet I don't find a way to choose u in advance (creating a vector for u for different values).
I can define a single input - u=0
or i can have a controller adjust the input - u= K*(y-[0,0]) but I don't find a way to preset the force u
such as u=[0,0,1,1,0,0] (of curse length(u) = length(tspan) )

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采纳的回答

Torsten
Torsten 2018-6-28
The example
"ODE with Time-Dependent Terms" under
https://de.mathworks.com/help/matlab/ref/ode45.html
should show you how to proceed.
Best wishes
Torsten.
  2 个评论
Daniel Grebler
Daniel Grebler 2018-6-28
Thank you Torsten, Allow me to ask another question:
How could I limit the max value of a solution (ode45 solution) without stopping the iterations
I can set a 'NoneNgeative' restrain I can think of maybe creating another variable such as
y(new)=y(restrained)-maxvalue
and have a NoneNegative on y(new) but how can I fuse it in the ode function since it has no derivative
my ode function is:
function dy = flywheeldef(t,y,u)
dy(1,1) = y(2);
dy(2,1) = (b*sin(y(1)))/a-(u/a);
dy(3,1) = y(4);
dy(4,1) = -b*sin(y(1))/a+u*(a+I2)/(I2*a);
--- how can I restrain y(4) to a max value? (and continue calculation)
Torsten
Torsten 2018-6-29
The ODEs uniquely determine the solution - you can't constrain a variable to a max value. If you want to do this, you will have to adapt your u dependent on time.
Best wishes
Torsten.

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