pbs1.c

计算方法中的常用程序

  #include "stdlib.h"
  #include "math.h"
  void gpbs1(t,h,n,y,eps,f)
  void  (*f)();
  int n;
  double t,h,eps,y[];
  { int i,j,k,m,nn,it;
    void rkt1();
    double x,hh,dd,q,p,g[10],*b,*d,*u,*v,*w,*e;
    b=malloc(10*n*sizeof(double));
    d=malloc(n*sizeof(double));
    u=malloc(n*sizeof(double));
    v=malloc(n*sizeof(double));
    w=malloc(n*sizeof(double));
    e=malloc(n*sizeof(double));
    for (j=0; j<=n-1; j++) v[j]=y[j];
    x=t; nn=1; hh=h; g[0]=hh;
    rkt1(x,hh,n,y,w,d,e,f);
    for (j=0; j<=n-1; j++)
      { b[j]=y[j]; u[j]=y[j];}
    k=1; it=1;
    while (it==1)
      { nn=nn+nn; hh=hh/2.0; it=0;
        g[k]=hh;
        for (j=0; j<=n-1; j++) y[j]=v[j];
        t=x;
        for (j=0; j<=nn-1; j++)
          { rkt1(t,hh,n,y,w,d,e,f);
            t=t+hh;
          }
        for (j=0; j<=n-1; j++)
          { dd=y[j]; m=0;
            for (i=0; i<=k-1; i++)
              if (m==0)
                { q=dd-b[i*n+j];
                  if (fabs(q)+1.0==1.0) m=1;
                  else dd=(g[k]-g[i])/q;
                }
            b[k*n+j]=dd;
            if (m!=0) b[k*n+j]=1.0e+35;
          }
        for (j=0; j<=n-1; j++)
          { dd=0.0;
            for (i=k-1; i>=0; i--)
              dd=-g[i]/(b[(i+1)*n+j]+dd);
            y[j]=dd+b[j];
          }
        p=0.0;
        for (j=0; j<=n-1; j++)
          { q=fabs(y[j]-u[j]);
            if (q>p) p=q;
          }
        if ((p>=eps)&&(k<7))
          { for (j=0; j<=n-1; j++) u[j]=y[j];
            k=k+1; it=1;
          }
      }
    free(b); free(d); free(u); free(v); free(w); free(e);
    return;
  }

  static void rkt1(t,h,n,y,b,d,e,f)
  void  (*f)();
  int n;
  double t,h,y[],b[],d[],e[];
  { int i,k;
    double a[4],tt;
    a[0]=h/2.0; a[1]=a[0]; a[2]=h; a[3]=h;
    (*f)(t,y,n,d);
    for (i=0; i<=n-1; i++) { b[i]=y[i]; e[i]=y[i];}
    for (k=0; k<=2; k++)
      { for (i=0; i<=n-1; i++)
          { y[i]=e[i]+a[k]*d[i];
            b[i]=b[i]+a[k+1]*d[i]/3.0;
          }
        tt=t+a[k];
        (*f)(tt,y,n,d);
      }
    for (i=0; i<=n-1; i++)
      y[i]=b[i]+h*d[i]/6.0;
    return;
  }