Runge-Kutta 3 variables, 3 equations

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jsparkes951 2015-5-4

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https://ww2.mathworks.cn/matlabcentral/answers/215025-runge-kutta-3-variables-3-equations

评论： SHIVANI TIWARI 2019-5-15

I am trying to use the 4th order Runge Kutta method to solve the Lorenz equations over a perios 0<=t<=250 seconds. I am able to solve when there are two equations involved but I do not know what do to for the third one. I have :

x(1)=1;
y(1)=2;
z(1)=0;
h=0.1;
sigma=10;
rho=28;
beta=(8/3);
f=sigma*(y-x);
g=rho*x-y-x*z;
w=x*y-beta*z;
for i=1:2500
    k1=h*f(x(i),y(i),z(i));
    l1=h*g(x(i),y(i),z(i));
    m1=h*w(x(i),y(i),z(i));

I do not know what to do when calculating k2 etc. I have:

k2=f(x(i)+h/2, y(i)+k1/2, z(i)+l1/2);

etc. but is there any change with m(i)? I only know how to solve this for two equations. Am I on the right track? I'm new to MATLAB and Runge-Kutta so any help would be greatly appreciated.

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Miguel Buitrago 2016-7-13

编辑：Miguel Buitrago 2016-7-13

在 MATLAB Online 中打开

I used the 4th order Runge Kutta method to solve the Lorenz equations:

   k1 = h*f(u1,u2);
   m1 = h*g(u1,u2,u3);
   n1 = h*e(u1,u2,u3);
   k2 = h*f(u1+k1/2, u2+m1/2);  
   m2 = h*g(u1+k1/2, u2+m1/2, u3+n1/2);  
   n2 = h*e(u1+k1/2, u2+m1/2, u3+n1/2);  
   k3 = h*f(u1+k2/2, u2+m2/2);
   m3 = h*g(u1+k2/2, u2+m2/2, u3+n2/2);  
   n3 = h*e(u1+k2/2, u2+m2/2, u3+n2/2);
   k4 = h*f(u1+k3, u2+m3);
   m4 = h*g(u1+k3, u2+m3, u3+n3);
   n4 = h*e(u1+k3, u2+m3, u3+n3);
   u1 = u1 + (k1+(k2*2)+(k3*2)+k4)/6;
   u2 = u2 + (m1+(m2*2)+(m3*2)+m4)/6;
   u3 = u3 + (n1+(n2*2)+(n3*2)+n4)/6;