# include # include # include # include # include # include # include using namespace std; int main ( ); int i4_modp ( int i, int j ); int i4_wrap ( int ival, int ilo, int ihi ); double *initial_condition ( int nx, double x[] ); double *r8vec_linspace_new ( int n, double a, double b ); void timestamp ( ); //****************************************************************************80 int main ( ) //****************************************************************************80 // // Purpose: // // FD1D_ADVECTION_LAX solves the advection equation using the Lax method. // // Discussion: // // The Lax method is stable for the advection problem, if the time step // satisifies the Courant-Friedrichs-Levy (CFL) condition: // // dt <= dx / c // // Licensing: // // This code is distributed under the MIT license. // // Modified: // // 27 January 2013 // // Author: // // John Burkardt // { double a; double b; double c; string command_filename = "advection_commands.txt"; ofstream command_unit; string data_filename = "advection_data.txt"; ofstream data_unit; double dt; double dx; int i; int j; int jm1; int jp1; int nx; int nt; int nt_step; double t; double *u; double *unew; double *x; timestamp ( ); cout << "\n"; cout << "FD1D_ADVECTION_LAX:\n"; cout << " C++ version\n"; cout << "\n"; cout << " Solve the constant-velocity advection equation in 1D,\n"; cout << " du/dt = - c du/dx\n"; cout << " over the interval:\n"; cout << " 0.0 <= x <= 1.0\n"; cout << " with periodic boundary conditions, and\n"; cout << " with a given initial condition\n"; cout << " u(0,x) = (10x-4)^2 (6-10x)^2 for 0.4 <= x <= 0.6\n"; cout << " = 0 elsewhere.\n"; cout << "\n"; cout << " We modify the FTCS method using the Lax method:\n"; cout << " du/dt = (u(t+dt,x)-0.5*u(t,x-dx)-0.5*u(t,x+dx))/dt\n"; cout << " du/dx = (u(t,x+dx)-u(t,x-dx))/2/dx\n"; nx = 101; dx = 1.0 / ( double ) ( nx - 1 ); a = 0.0; b = 1.0; x = r8vec_linspace_new ( nx, a, b ); nt = 1000; dt = 1.0 / ( double ) ( nt ); c = 1.0; u = initial_condition ( nx, x ); // // Open data file, and write solutions as they are computed. // data_unit.open ( data_filename.c_str ( ) ); t = 0.0; data_unit << " " << x[0] << " " << t << " " << u[0] << "\n"; for ( j = 0; j < nx; j++ ) { data_unit << " " << x[j] << " " << t << " " << u[j] << "\n"; } data_unit << "\n"; nt_step = 100; cout << "\n"; cout << " Number of nodes NX = " << nx << "\n"; cout << " Number of time steps NT = " << nt << "\n"; cout << " Constant velocity C = " << c << "\n"; cout << " CFL condition: dt (" << dt << ") <= dx / c (" << dx / c << ")\n"; unew = new double[nx]; for ( i = 0; i < nt; i++ ) { for ( j = 0; j < nx; j++ ) { jm1 = i4_wrap ( j - 1, 0, nx - 1 ); jp1 = i4_wrap ( j + 1, 0, nx - 1 ); unew[j] = 0.5 * u[jp1] + 0.5 * u[jm1] - c * dt / dx / 2.0 * ( u[jp1] - u[jm1] ); } for ( j = 0; j < nx; j++ ) { u[j] = unew[j]; } if ( i == nt_step - 1 ) { t = ( double ) ( i ) * dt; for ( j = 0; j < nx; j++ ) { data_unit << " " << x[j] << " " << t << " " << u[j] << "\n"; } data_unit << "\n"; nt_step = nt_step + 100; } } // // Close the data file once the computation is done. // data_unit.close ( ); cout << "\n"; cout << " Plot data written to the file \"" << data_filename << "\"\n"; // // Write gnuplot command file. // command_unit.open ( command_filename.c_str ( ) ); command_unit << "set term png\n"; command_unit << "set output 'advection_lax.png'\n"; command_unit << "set grid\n"; command_unit << "set style data lines\n"; command_unit << "unset key\n"; command_unit << "set xlabel '<---X--->'\n"; command_unit << "set ylabel '<---Time--->'\n"; command_unit << "splot '" << data_filename << "' using 1:2:3 with lines\n"; command_unit << "quit\n"; command_unit.close ( ); cout << " Gnuplot command data written to the file \"" << command_filename << "\"\n"; // // Free memory. // delete [] u; delete [] unew; delete [] x; // // Terminate. // cout << "\n"; cout << "FD1D_ADVECTION_LAX\n"; cout << " Normal end of execution.\n"; cout << "\n"; timestamp ( ); return 0; } //****************************************************************************80 int i4_modp ( int i, int j ) //****************************************************************************80 // // Purpose: // // I4_MODP returns the nonnegative remainder of I4 division. // // Discussion: // // If // NREM = I4_MODP ( I, J ) // NMULT = ( I - NREM ) / J // then // I = J * NMULT + NREM // where NREM is always nonnegative. // // The MOD function computes a result with the same sign as the // quantity being divided. Thus, suppose you had an angle A, // and you wanted to ensure that it was between 0 and 360. // Then mod(A,360) would do, if A was positive, but if A // was negative, your result would be between -360 and 0. // // On the other hand, I4_MODP(A,360) is between 0 and 360, always. // // I J MOD I4_MODP I4_MODP Factorization // // 107 50 7 7 107 = 2 * 50 + 7 // 107 -50 7 7 107 = -2 * -50 + 7 // -107 50 -7 43 -107 = -3 * 50 + 43 // -107 -50 -7 43 -107 = 3 * -50 + 43 // // Licensing: // // This code is distributed under the MIT license. // // Modified: // // 26 May 1999 // // Author: // // John Burkardt // // Parameters: // // Input, int I, the number to be divided. // // Input, int J, the number that divides I. // // Output, int I4_MODP, the nonnegative remainder when I is // divided by J. // { int value; if ( j == 0 ) { cerr << "\n"; cerr << "I4_MODP - Fatal error!\n"; cerr << " I4_MODP ( I, J ) called with J = " << j << "\n"; exit ( 1 ); } value = i % j; if ( value < 0 ) { value = value + abs ( j ); } return value; } //****************************************************************************80 int i4_wrap ( int ival, int ilo, int ihi ) //****************************************************************************80 // // Purpose: // // I4_WRAP forces an I4 to lie between given limits by wrapping. // // Example: // // ILO = 4, IHI = 8 // // I Value // // -2 8 // -1 4 // 0 5 // 1 6 // 2 7 // 3 8 // 4 4 // 5 5 // 6 6 // 7 7 // 8 8 // 9 4 // 10 5 // 11 6 // 12 7 // 13 8 // 14 4 // // Licensing: // // This code is distributed under the MIT license. // // Modified: // // 26 December 2012 // // Author: // // John Burkardt // // Parameters: // // Input, int IVAL, an integer value. // // Input, int ILO, IHI, the desired bounds for the integer value. // // Output, int I4_WRAP, a "wrapped" version of IVAL. // { int jhi; int jlo; int value; int wide; if ( ilo <= ihi ) { jlo = ilo; jhi = ihi; } else { jlo = ihi; jhi = ilo; } wide = jhi + 1 - jlo; if ( wide == 1 ) { value = jlo; } else { value = jlo + i4_modp ( ival - jlo, wide ); } return value; } //****************************************************************************80 double *initial_condition ( int nx, double x[] ) //****************************************************************************80 // // Purpose: // // INITIAL_CONDITION sets the initial condition. // // Licensing: // // This code is distributed under the MIT license. // // Modified: // // 26 December 2012 // // Author: // // John Burkardt // // Parameters: // // Input, int NX, the number of nodes. // // Input, double X[NX], the coordinates of the nodes. // // Output, double INITIAL_CONDITION[NX], the value of the initial condition. // { int i; double *u; u = new double[nx]; for ( i = 0; i < nx; i++ ) { if ( 0.4 <= x[i] && x[i] <= 0.6 ) { u[i] = pow ( 10.0 * x[i] - 4.0, 2 ) * pow ( 6.0 - 10.0 * x[i], 2 ); } else { u[i] = 0.0; } } return u; } //****************************************************************************80 double *r8vec_linspace_new ( int n, double a_first, double a_last ) //****************************************************************************80 // // Purpose: // // R8VEC_LINSPACE_NEW creates a vector of linearly spaced values. // // Discussion: // // An R8VEC is a vector of R8's. // // 4 points evenly spaced between 0 and 12 will yield 0, 4, 8, 12. // // In other words, the interval is divided into N-1 even subintervals, // and the endpoints of intervals are used as the points. // // Licensing: // // This code is distributed under the MIT license. // // Modified: // // 29 March 2011 // // Author: // // John Burkardt // // Parameters: // // Input, int N, the number of entries in the vector. // // Input, double A_FIRST, A_LAST, the first and last entries. // // Output, double R8VEC_LINSPACE_NEW[N], a vector of linearly spaced data. // { double *a; int i; a = new double[n]; if ( n == 1 ) { a[0] = ( a_first + a_last ) / 2.0; } else { for ( i = 0; i < n; i++ ) { a[i] = ( ( double ) ( n - 1 - i ) * a_first + ( double ) ( i ) * a_last ) / ( double ) ( n - 1 ); } } return a; } //****************************************************************************80 void timestamp ( ) //****************************************************************************80 // // Purpose: // // TIMESTAMP prints the current YMDHMS date as a time stamp. // // Example: // // 31 May 2001 09:45:54 AM // // Licensing: // // This code is distributed under the MIT license. // // Modified: // // 08 July 2009 // // Author: // // John Burkardt // // Parameters: // // None // { # define TIME_SIZE 40 static char time_buffer[TIME_SIZE]; const struct std::tm *tm_ptr; std::time_t now; now = std::time ( NULL ); tm_ptr = std::localtime ( &now ); std::strftime ( time_buffer, TIME_SIZE, "%d %B %Y %I:%M:%S %p", tm_ptr ); std::cout << time_buffer << "\n"; return; # undef TIME_SIZE }