Ode45 initial conditions iteration

I am trying to solve the following 2nd order differential equation, which works fine but I need to iterate the initial conditions, so everytime the ode is solved, the last values for q(:,1) and q(:,2) should be the new value for X and Y respectivley.
%solve_2nd_order_diff
duf=@(t,q)[q(2);f*w^2*cos(w*t)-(Cmech/J)*q(2)-(K1/J)*q(1)-(K3/J)*q(1)^3];
for r=1:length(Rad_speed)%freq of harmonic fluctuations, rad/s
[t,q]=ode45(duf,Time,[X Y]);
Max_values=max(q);
plot(t,q(:,1),'k')
xlabel('TIme,s')
ylabel('Relative Displacement, Rad')
grid
end

 Respuesta aceptada

Alan Stevens
Alan Stevens el 19 de Nov. de 2020
Like so
...
[X Y] = ...
for r=1:length(Rad_speed)%freq of harmonic fluctuations, rad/s
[t,q]=ode45(duf,Time,[X Y]);
[X Y] = [q(end,1) q(end,2)];
...

5 comentarios

Adedayo Odeleye
Adedayo Odeleye el 19 de Nov. de 2020
Thanks for the swift reply.
it says "Too many output arguments." in regards to the [X Y] = [q(end,1) q(end,2)]; line
Alan Stevens
Alan Stevens el 19 de Nov. de 2020
Can you upload your full code?
Alan Stevens
Alan Stevens el 19 de Nov. de 2020
Editada: Alan Stevens el 19 de Nov. de 2020
Sorry, I should have written
X = q(end,1); Y = q(end,2);
rather than
[X Y] = [q(end,1) q(end,2)];
Incidentally, your plots will be overwritten unless you use the figure command or hold on (whichever is appropriate.
full code is below, it is working fine now.
J=3.37e-5; %inertia of rotor (kg/m2)
K1=0.7243; %Linear stiffness coef (N.m/rad)
K3=1409.7; %Nonlinear stiffness coef (N.m/rad3)
f=0.005;
Zeta=0.05;%damping ratio
Cmech=2*Zeta*(sqrt(K1/J))*J; %mechanical damping coef
ECF=0.2; %electromagnetic coupling factor
Rint=82;
Celec=ECF^2/2*Rint;%Electrical damping coef
L=160;%Inductance (mH)
X = 0;
Y = 0;
%shaft speed in RPM
Shaft_speed=2000;
%shaft speed in rad/s
Rad_speed=Shaft_speed*((2*pi)/60);
%freq of harmonic fluctuations
w=2*Rad_speed;
w_Hz=w/2*pi; %convert to Hertz
nt=1/w_Hz;
Time=0:nt/100:100*nt;
%solve_2nd_order_diff
duf=@(t,q)[q(2);f*w^2*cos(w*t)-(Cmech/J)*q(2)-(K1/J)*q(1)-(K3/J)*q(1)^3];
Iter=100;
for r=1:Iter
[t,q]=ode45(duf,Time,[X Y]);
X=q(end,1);
Y=q(end,2);
Max_values=max(q)
plot(t,q(:,1),'k')
xlabel('TIme,s')
ylabel('Relative Displacement, Rad')
grid
end
Alan Stevens
Alan Stevens el 19 de Nov. de 2020
Editada: Alan Stevens el 19 de Nov. de 2020
Excellent, but notice that your plot and your Max_values will only apply for the last iteration of the loop (they are overwritten every time through the loop). If that is what you want it is probably best to put that piece of coding outside the loop at the end. To store the Max_values (rather than just have them printed in the command window every iteration), you could use
Max_values(r) = max(q);

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el 19 de Nov. de 2020

Editada:

el 19 de Nov. de 2020

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