(CFD) Problem with C code for 1D linear convection equation

[mcgrane5]

TL;DR

Hi im writing a c program on linear convection in 1D and im having problems with my final for loop. It returns bizarre results and i dont know why

I have the same problem coded in python and it works perfectly. Could anyone offer a fix to my code. Its only the last for loop where im having problems. I have attached my C code below as well as the working python script to help anyone debug

This is my c program below. Again everything works as expected up until the highlighted lines or the second for loop. When i run my program to
output the arrays contents i get bizarre numbers that are not correct.

[CODE lang="c" title="1D linear convection C code" highlight="63-68"]# include <math.h>
# include <stdlib.h>
# include <stdio.h>
# include <time.h>int main ( )
{
//eqyation to be solved 1D linear advection -- du/dt + c(du/dx) = 0
//initial conditions u(x,0) = u0(x)

//after discretisation using forward difference time and backward differemce space
//update equation becomes u = u0 - c * dt/dx * (u0 - u[i - 1]);
FILE *fpointer = NULL;
fpointer = fopen("bbb.txt", "w");

int nx = 41; //num of grid points

double dx = 2.0 / (nx - 1.0); //magnitude of the spacing between grid points

int nt = 25;//nt is the number of timesteps

double dt = 0.25; //the amount of time each timestep covers

int c = 1; //assume wavespeed
//double u0_array;

//set up our initial conditions. The initial velocity u_0
//is 2 across the interval of 0.5 <x < 1 and u_0 = 1 everywhere else.
//we will define an array of ones
double* u0_array = (double*)calloc(nx, sizeof(double));

for (int i = 0; i < nx-1; i++)
{
if (i * dx >= 0.5 && i * dx <= 1)
{
u0_array = 2;
}else
{
u0_array = 1;
}

}

//test code to make sure array is populated correctly
for (int i = 0; i < nx; i++)
{
//u0_array = 2;
//printf("%f, ", u0_array);

}

//make a temporary array that allows us to store the solution
double* usol_array = (double*)calloc(nx, sizeof(double));

//apply numerical scheme forward difference in
//time an backward difference in space
for (int i = 0; i < nt; i++)
{
usol_array = u0_array;

for (int j = 0; j < nx-1; j++)
{
//usol_array = u0_array;
u0_array[j] = usol_array[j] - c * dt/dx * (usol_array[j] - usol_array[j - 1]);
printf("%f\n", u0_array);
}
//first loop iterates through each time step
}


return EXIT_SUCCESS;
}

[/CODE]

BELOW IS THE WORKING PYTHON SCRIPT TO HELP ANYONE DEBUG MY C CODE[CODE lang="python" title="Working python script for same problem"]import numpy
import matplotlib.pyplot as plt
import time, sys

nx = 41 # try changing this number from 41 to 81 and Run All ... what happens?
dx = 2 / (nx-1)
nt = 25 #nt is the number of timesteps we want to calculate
dt = .025 #dt is the amount of time each timestep covers (delta t)
c = 1

u = numpy.ones(nx) #numpy function ones()
u[int(.5 / dx):int(1 / dx + 1)] = 2 #setting u = 2 between 0.5 and 1 as per our I.C.s
print(u)

un = numpy.ones(nx) #initialize a temporary array

for n in range(nt): #loop for values of n from 0 to nt, so it will run nt times
un = u.copy()
print(un)$copy the existing values of u into un for i in range(1, nx):$ you can try commenting this line and...
#for i in range(nx): ## ... uncommenting this line and see what happens!
u = un - c * dt / dx * (un - un[i-1])
#print(u)

plt.plot(numpy.linspace(0, 2, nx), u);
plt.show()[/CODE]

 

[bigfooted]

In your c file, manually compute what the value is that you get for u0_array when j=0 and you should see the problem.