implement crank-nicolson in c++

Question

i want to implement crank-nicolson method in c++ according to that :

−ru(i−1,j+1) + 2(1 + r)u(i,j+1) − ru(i+1,j+1) = 2(1 − r)u(i,j) + r (u(i−1,j)+ u(i+1,j) )

I use gauss-jordan method to solve the system but i can't figure how to implement the above formula.

const double pi=3.14159265;

double f (double x){
return sin(pi*x);

}

using namespace std;


//gauss-jordan
double* gauss(int n ,double **a){
    double factor;
    double *b,*x;
    x=new double[n];
    b=new double[n];


for (int k=1;k<=n;k++) {
  for (int i=1;i<=n;i++) {
    if (i!=k) {
     factor=a[i][k]/a[k][k];
     for (int j=1; j<=n; j++ ) {
       a[i][j]=a[k][j]*factor-a[i][j];

     }
b[i]=b[k]*factor -b[i];
    }

}
}
for (int i=1;i<=n;i++) {
 x[i]=b[i]/a[i][i];

}
return x;
}

int main()
{
    int i,j,n,m,xd,td;
    double h,k,r;
    double **t,**p;

    //----- initialize double pointer------------------
    double **u;
    u=new double *[m];
    for (int o=1;o<=m;o++){
        u[o]=new double [n];
    }
    //----- end of initialization-----------------------

   cout <<"\nEnter the value of x dimension : \n";
   cin >> xd;
   cout <<"\nEnter the step size h for x dimension : \n";
   cin >>h;
   cout <<"\nEnter the value of time dimension : \n";
   cin >>td;
   cout<<"\nEnter the step k of time dimension : \n";
   cin >>k;
   n=xd/h -1.0;
   m=td/k -1.0;
   cout <<"\n\nThe internal elements of x dimension are :"<<n;
   cout <<"\nThe internal elements of t dimension are :"<<m;
   r=k/(h*h);
   cout <<"\nThe r value is : "<<r;


   //initial conditions
    for (j=0;j<=m;j++){
    u[0][m]=0;
    u[10][m]=0;
    }

    //get the function
    for (i=1;i<n;i++){
    u[i][0]=f(i*h);
    }


    //apply crank-nicolson
    for (i=1;i<n;i++){
        for (j=1;j<n;j++){
              -r*u[i-1][j+1] +2.0*(1.0+r)*u[i][j+1] -r*u[i+1][j+1]=2.0*(1.0-r)*u[i][j] +r*(u[i-1][j]+u[i+1][j]);
    }  // here i can't figure the steps i must follow in order for this to work 



//-----delete double pointer-------------
    for(int o=1;o<m;o++){
    delete [] u[o];
    delete [] u;
    }
//---------------------------------------


    return 0;
}

Answer 1

I am assuming that the variable j represents the time steps. In order to implement Crank-Nicolson, you have to pose the problem as a system of linear equations and solve it. The matrix corresponding to the system will be of tridiagonal form, so it is better to use Thomas' algorithm rather than Gauss-Jordan.

The linear system will be of the form A x = b, with x being the vector (..., u(i-1, j+1), u(i, j+1), u(i+1, j+1), ...) and b being the vector (..., r u(i−1,j), 2(1 − r)u(i,j), r u(i+1,j), ...). The i-th row of the matrix A will be of the form (0, ..., 0, −r, 2(1 + r), −r, 0, ..., 0). You will have to be careful with the first and last rows, where you'll have to substitute the boundary conditions for your problem.

A good reference for finite difference methods and Crank-Nicolson in particular is the book by John Strikwerda.

Hope this helps.