2d热方程：TDMA求解器

Question

我正在尝试使用2d TDMA求解器求解2d瞬态热方程。但是，我发现在定义温度场和实现求解器方面存在困难。 我使用了以下代码（LBL函数：逐行方法）但它没有用。

提前感谢您的帮助。

void LBL (int n, int m, int n1, int m1,int m2,
          double dt, double dx, double dy,
          vector <double> &x, vector<double> &y,
          vector< vector<double> > &aw,
          vector< vector<double> > &b,
          vector< vector<double> > &ap,
          vector< vector<double> > &ae,
          vector< vector<double> > &an,
          vector< vector<double> > &as,
          vector< vector<double> > &ap0,
          vector< vector<double> > &P,
          vector< vector<double> > &Q,
          vector< vector<double> > &Td,
          vector< vector<double> > &d,
          vector< vector<double> > &T0)
{
  double ERROR;
  double t=0;
  double r[n][m];

  for(int j=0; j<m; j++)
  {
    for(int i=0;i<n;i++)
    {
      d[i][j] =Tin;
      r[i][j]=Tin;
      T0[i][j]=Tin;
    }
  }

  int s=0;
  mi =100000;
  while (t<tm)
  {
    cout << " \n t=" << t<<endl;
    /****************calculation of dij****************/
    bi(t,n,m,dx,dy,dt,ap0,T0,b);
    do
    {
      for (int j=0; j<m; j++)
      {
        for (int i=0; i<n; i++)
        {
          d[i][j] = an[i][j] * r[i][j+1] + as[i][j] * r[i][j-1] + b[i][j];
        }
      }

      MAXI=  fabs(d[0][0]-r[0][0])/d[0][0];
      for ( int j =0; j < m; j++)
      {
        for(int i= 0; i<n; i++)
        {
          ERROR=fabs(d[i][j]-r[i][j])/d[i][j];
          if (MAXI < ERROR)     MAXI =ERROR;

        }
      }
      cout << " \n MAXI=" << MAXI<<endl;
      for ( int j =0; j < m; j++)
      {
        for(int i= 0; i<n; i++)
        {
          r[i][j]=d[i][j];
        }
      }

      s++;
    } while (MAXI>eps && s<mi );

    /***********calculation tdma******************/
    for (int j=0; j<m; j++)
    {
      P[0][j]=ae[0][j]/ap[0][j];
    }
    for (int j=0; j<m; j++)
    {
      Q[0][j]=d[0][j]/ap[0][j];
    }

    for (int j=0; j<m; j++)
    {
      for(int i=1; i<n; i++)
      {
        P[i][j]=an[i][j]/(ap[i][j]- (aw[i][j]*P[i-1][j]));
        Q[i][j]= (d[i][j] + (aw[i][j]*Q[i-1][j]) ) / (ap[i][j]-(aw[i][j]*P[i-1][j]));
      }
    }

    for(int j =0; j<m; j++)
    {
      Td[n-1][j]=Q[n-1][j];
    }

    for(int j =0;j<m;j++)
    {
      for (int i=n-2; i>=0; i--)
      {
        Td[i][j]=P[i][j]*Td[i+1][j] + Q[i][j];
      }
    }

    for (int j=0; j<m; j++)
    {
      for (int i =0; i<n; i++)
      {
        T0[i][j] =Td[i][j];
      }
    }

    t+=dt;
  }
}