Solving differential equation with varying Constant

3 ビュー (過去 30 日間)
P K
P K 2018 年 12 月 20 日
コメント済み: Torsten 2018 年 12 月 20 日
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

回答 (1 件)

Torsten
Torsten 2018 年 12 月 20 日
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 件のコメント
P K
P K 2018 年 12 月 20 日
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 2018 年 12 月 20 日
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.

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