Question

i want to solve a pde according to this matlab code (part of it):

.............
% Evaluate the initial conditions
x = xl : dx : xr;              % generate the grid point
% f(1:J+1) since array index starts from 1
f = sin(pi*x) + sin(2*pi*x);  

% store the solution at all grid points for all time steps
u = zeros(J+1,Nt);   

% Find the approximate solution at each time step
for n = 1:Nt
    t = n*dt;         % current time
    % boundary condition at left side
    gl = sin(pi*xl)*exp(-pi*pi*t)+sin(2*pi*xl)*exp(-4*pi*pi*t);   
    % boundary condition at right side
    gr = sin(pi*xr)*exp(-pi*pi*t)+sin(2*pi*xr)*exp(-4*pi*pi*t);  
    if n==1    % first time step
       for j=2:J    % interior nodes     
       u(j,n) = f(j) + mu*(f(j+1)-2*f(j)+f(j-1));
       end     
       u(1,n) = gl;   % the left-end point
       u(J+1,n) = gr; % the right-end point 
    else 
       for j=2:J    % interior nodes
         u(j,n)=u(j,n-1)+mu*(u(j+1,n-1)-2*u(j,n-1)+u(j-1,n-1));
       end
       u(1,n) = gl;   % the left-end point
       u(J+1,n) = gr; % the right-end point 
    end
 end

% Plot the results
tt = dt : dt : Nt*dt;
figure(1)
surf(x,tt, u');     % 3-D surface plot  
............ 

I did this in c++ :

double f(double x){

return sin(pi*x) + sin(2.0*pi*x);

}

double gl(double x,double t){

    return sin(pi*x)*exp(-pi*pi*t) +sin(2.0*pi*x)*exp(-4.0*pi*pi*t);
}

double gr(double x,double t){

    return sin(pi*x)*exp(-pi*pi*t) +sin(2.0*pi*x)*exp(-4.0*pi*pi*t);
}

using namespace std;

const int Nt=50;  // number of time steps
const int J=10; // number of division for x
double xl=0,xr=1.0; //left and right boundaries
double tf=0.1; //final simulation time
double dt,dx; //dx is the mesh size
double x;
double u[J][Nt];


int main()
{
dx=(xr-xl)/J;
dt=tf/Nt;
double m=dt/(dx*dx);


//Evaluate the initial conditions
//generate the grid point
for (x=xl;x<=xr;x+=dx) {
    f(x);
}

//store the solution at all grid points for all time steps
for (int i=0;i<J;i++){
    for (int j=0;j<Nt-1;j++){
        u[i][j]=0;
    }
}

//find the approximate solution at each time step
int n,t,j;
for (n=0;n<Nt;n++){
    t=(n+1)*dt; //current time
        // boundary condition at left side
        gl(xl,t);
        //boundary condition at right side
        gr(xr,t);

        if (n==0) {//first time step
            for (j=1;j<J-1;j++){
                u[j][n]=f(j)+m*(f(j+1)-2.0*f(j)+f(j-1));
                }
        u[0][n]=gl(xl,t);
        u[J-1][n]=gr(xr,t); }
        else  {
            for(j=1;j<J-1;j++){
                u[j][n]=u[j][n-1]+m*(u[j+1][n-1]-2.0*u[j][n-1]+u[j-1][n-1]);
                }
        u[0][n]=gl(xl,t);
        u[J-1][n]=gr(xr,t);
        }

    }

ofstream data;
data.open("Data.dat");


for (int tt=dt;tt<=Nt*dt;tt+=dt){
    data<< tt <<"\t"<<x<<"\t"<< u[J][Nt]<<endl;

}

cout <<"\nFiles written! "<<endl;

    return 0;
}

The problem is that i get a data file with only values 0 1.1 -3.58979e-09.
Maybe the problem is in the indexing?Or i messed up completely?

Answer 1

You have several serious problems here:

1. You have problems with all for lops. You always takes one value more than exist in arrays. Since you start your loops with zero you should use < instead of <=.

2. You use invalid indices in arrays: u[J][Nt-1] instead of u[i][j] in the initialization loop, and u[J][n] instead of u[J-1][n] in the main loop. For example the initialization loop should be:

   for (int i = 0; i < J; i++) {
       for (int j = 0; j < Nt-1; j++) {
           u[i][j] = 0;
       }
   }
   

   It’s strange that you don’t get access violation exception, since you are trying to access values outside the array.

3. You check n == 1 instead of n == 0 for the first iteration

4. For the calculation to be equal with Matlab you should calculate t = (n + 1) * dt;, since in the Matlab you n starts from 1.

5. Instead the loop for(j=0;j<=J-2;j++) there should be for(j=1; j < J-1; j++)

6. The f(x) in the first loop has no sense, just like the gl(xl, t) & gr(xr, t) in the beginning of the main loop.

By the way how have you defined these functions? There also could be problems.

Meanwhile you should fix these issues before you can continue testing your application.