Solving differential equation with varying Constant

3 Ansichten (letzte 30 Tage)
P K
P K am 20 Dez. 2018
Kommentiert: Torsten am 20 Dez. 2018
Dimensions
G= 1x100, H=1X100;
I want to solve these 7 eqns. I am using ODE45. I am able to solve these equations for fixed T and Q. But i want to solve it for varying T and Q. Thats why i am using for loop. Can some one explain where am I wrong?. I have tried myself but unable to figure it out.
function [dUdt]=eqn(t,U)
dUdt=zeros(7,1);
K=0.003;
for i=1:100
T=G(1,i);
Q=H(1,i);
end
dUdt(1)=2*K*T(i)*(U(2)-U(1));
dUdt(2)=-2*K*U(2)*T(i);
dUdt(3)=K*T(i)*(2*Q(i)-U(3))-U(3)*K*(U(1)+2*U(2));
dUdt(4)=K*U(3)*T(i)-U(4)*K*(U(1)+2*U(2));
dUdt(5)=K*Q(i)*(U(1)+2*U(2))-2*K*U(5)*T(i);
dUdt(6)=K*U(3)*(U(1)+2*U(2))+2*K*U(5)*T(i)-K*U(6)*T(i);
dUdt(7)=K*U(6)*T(i)+U(4)*K*(U(1)+2*U(2));
end

Antworten (1)

Torsten
Torsten am 20 Dez. 2018
function [dUdt]=eqn(t,U)
dUdt=zeros(7,1);
K=0.003;
t_inter=0:99;
T_actual=interp1(t_inter,G,t);
Q_actual=interp1(t_inter,H,t);
dUdt(1)=2*K*T_actual*(U(2)-U(1));
dUdt(2)=-2*K*U(2)*T_actual;
dUdt(3)=K*T_actual*(2*Q_actual-U(3))-U(3)*K*(U(1)+2*U(2));
dUdt(4)=K*U(3)*T_actual-U(4)*K*(U(1)+2*U(2));
dUdt(5)=K*Q_actual*(U(1)+2*U(2))-2*K*U(5)*T_actual;
dUdt(6)=K*U(3)*(U(1)+2*U(2))+2*K*U(5)*T_actual-K*U(6)*T_actual;
dUdt(7)=K*U(6)*T_actual+U(4)*K*(U(1)+2*U(2));
end
  8 Kommentare
P K
P K am 20 Dez. 2018
Thank you @All . I tried with G=rand(1,10) and H=rand(1,10). It worked. I would find out why it is not working with MY G AND H.
Torsten
Torsten am 20 Dez. 2018
As a quick and dirty solution, add the line
global G H
in "eqn" as well as in the function of your program where you define G and H.

Melden Sie sich an, um zu kommentieren.

Kategorien

Mehr zu Loops and Conditional Statements finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by