lab5.cpp

The decision of a boundary problem for the differential equation of elliptic type of the second orde

#include <iostream.h>
#include <conio.h>
#include <math.h>

const double a1 = 0;
const double a2 = 1;
const double b1 = 0;
const double b2 = 1;
const double eps = 0.0001;

#define nX 5
#define nY 5
#define hX (a2 - a1)/nX
#define hY (b2 - b1)/nY

const double t = 0.5*(hY*hX*hY*hY)/(hX*hX + hY*hY);

int i, j;
double u[nX+1][nY+1], v[nX+1][nY+1], z[nX][nY];

double Function_F(double x, double y)
{
    return 0;
}

void main()
{
    double norm, s1, s2, s3, count=0;
    
    for(i = 0; i <= nX; i++)
        for(j = 0; j <= nY; j++)
            v[i][j] = 0;
    
    for(i = 1; i < nY; i++)
    {
        u[0][i] = 0;
        u[nX][i] = 0;
    }
    for(i = 1; i < nX; i++)
    {
        u[i][nY] = 10000;
    }
    for(i = 0; i <= nX; i++)
    {
        u[i][0] = 0;
    }
    u[nX][nY] = 5000;
    u[0][nY] = 5000;
    
    for(i = 1; i < nX; i++)
        for(j = 1; j < nY; j++)
            u[i][j] = 0;

    do
    {
        s2 = 0;

        for(i = 1; i < nX; i++)
        {
            for(j = 1; j < nY; j++)
            {
                z[i][j] = (u[i-1][j] - 2*u[i][j] + u[i+1][j])/(hX*hX) + 
                          (u[i][j-1] - 2*u[i][j] + u[i][j+1])/(hY*hY) - 
                          t*(u[i-1][j-1] - 2*u[i-1][j] + u[i-1][j+1] - 2*u[i][j-1] + 4*u[i][j] - 2*u[i][j+1] + u[i+1][j-1] - 2*u[i+1][j] + u[i+1][j+1])/(hX*hX*hY*hY);

                u[i][j] += t*z[i][j];
                
                s1 = fabs((u[i][j] - v[i][j])*(u[i][j] - v[i][j]));
                s2 += s1;
            }
        }
        norm = sqrt(s2*hX*hY);

        for(i = 1; i < nX; i++)
            for(j = 1; j < nY; j++)
                v[i][j] = u[i][j];

        s3 = 0;
        for(i = 1; i < nX; i++)
            for(j = 1; j < nY; j++)
                if(norm >= eps)
                    s3 = norm;
        count++;
    }while(s3 >= eps);

    cout<<"y/x";
    for(i = 0; i <= nX; i++)
        cout<<i*hX<<"    ";
    cout<<endl;
    for(j = 0; j <= nY; j++)
    {
        cout<<j*hY<<" ";
        for(i = 0; i <= nX; i++)
            cout<<u[i][j]<<" ";
        cout<<endl;
    }
}