part1.c

一类kdv方程数值解法,kdv方程的数值解法

  #include "stdlib.h"
  #include "math.h"
  void part1(t,y,n,h,k,z,n1,n2,n3,n4)
  int n,k;
  double t,h,y[],z[][],n1,n2,n3,n4;
  { int i,j;
    double f[50][314],q;
    double a,b,c,d,e,g,p,m,n;
    a=12*g/h;
    b=1-2*n3/(h*h);
    c=g/(2*h*h*h);
    d=g*(n3-1/h)/(h*h);
    e=g*(1/h+n3)/(h*h);
    p=-n2+n1*n1*n1;
    m=n3*n1*n1*n1+n4;
    n=12*n1;
    for(j=0;j<=n-1;j++)
    z[0][j]=y[j];
    for(i=0;i<=k-2;i++)
    { 
       for(j=0;j<=n-1;j++)
        {q=n1*h*j+n2*g*i;
         f[i][j]=p*sin(q)+m*cos(q)-n*pow(cos(q),3);
         
          }  
     }
    for(i=1;i<=k-1;i++)
       for(j=2;j<=n-3;j++)
        { 
          z[i][j]=a*pow(z[i-1][j],3)-a*pow(z[i-1][j],2)*z[i-1][j+1]+b*z[i-1][j]+g*f[i-1][j-2]-c*z[i-1][j+2]+c*z[i-1][j-2]+d*z[i-1][j-1]+e*z[i-1][j+1];
          j=j-2;
          n=n-2;
          }
   }