problem in solving coupled non-linear Initial Value ODE using finite difference method
1 次查看(过去 30 天)
显示 更早的评论
Hi i wrote a code for 3 coupled non-linear Initial Value ODE ( u and w are second order and v is 1st order ODE) using finite difference method (forward scheme), but it doesn't work for me, the result for these ODE will be limit cycle oscillations but this generate wrong results. in need the result for t = 1:100 please help me in my code. Thanks.
Here is my code
t0=0;
Dt = 0.1;
tn = 6;
t = t0:Dt:tn;
n = (tn-t0)/Dt +1;%length(x);
u = zeros(1,n);
v = zeros(1,n);
w = zeros(1,n);
u0 = 0; %u(0)=0
u1 = 0.02; %u'(0)=0.02
v0 = 0; %v(0)=0
w0 = 0; %w(0)=0
w1 = 0; %w(0)=0
for j = 1:n-2
u(j+2) = 2*u(j+1)-u(j)+(Dt^2)*(-23.0665*(u(j)-v(j))+13.2617*((u(j)-v(j))^2)+5.3214*((u(j)-v(j))^3));
u(2)=u(1) - u1*Dt;
v(j+1) = v(j)+Dt*(0.2236*((u(j)-v(j))^3)-1.8198*((u(j)-v(j))^2)-0.0836*((u(j)-v(j)))+1.2263*(v(j)^2));
v(2)=v(1);
w(j+2) = 2*w(j+1)-w(j)+(Dt^2)*((-23.0665)*(u(j)-v(j))+0.0836*(0.02*Dt+(u(j)-v(j))+5.3214*((u(j)-v(j))^3)+13.2617*((u(j)-v(j))^2)+1.1872*(u(j)-v(j))*(0.02*Dt+u(j)-v(j))-0.6708*((u(j)-v(j))^2)*(0.02*Dt+u(j)-v(j))+2.4525*(u(j)-v(j))*(0.02*Dt+u(j))+2.4525*(0.02*Dt+u(j)-v(j))*u(j)-(-2.4525)*u(j)*(0.02*Dt+u(j))));
w(2)=w(1) - w1*Dt;
t(j) = t0 + (j-1)*Dt;
end
figure(1);
subplot(2,2,1);plot(t,u,'-');
subplot(2,2,2);plot(t,v,'-');
subplot(2,2,3);plot(t,w,'-');
0 个评论
回答(0 个)
另请参阅
类别
在 Help Center 和 File Exchange 中查找有关 Ordinary Differential Equations 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!