(计算流体力学)数组问题。为什么我的C程序输出-1.#IND00

Question

我用C编写的程序有一个问题，它解决了一维线性对流方程。基本上，我已经初始化了两个数组。第一个数组(u0_array)是一个数组，数组元素在间隔上设置为两个或0.5

我遇到的问题是代码末尾的嵌套for循环。此循环将计算下一个点所需的更新方程应用于数组中的每个元素。当我运行我的脚本并试图打印结果时，我得到的输出只是-1.IND 00，用于循环的每一次迭代。(我是跟随Matlab风格的伪代码，我也附在下面)，我是非常新的C，所以我的经验显示。我不知道为什么会这样。如果有人能对此提出可能的解决办法，我将非常感激。到目前为止，我已经将我的代码附在下面，并附上了一些评论，这样您就可以遵循我的思维过程了。我还附加了Matlab风格的伪代码，我遵循。

# include <math.h>
# include <stdlib.h>
# include <stdio.h>
# include <time.h>


int main ( )
{
    //eqyation to be solved 1D linear convection -- 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[i] = u0[i] - c * dt/dx * (u0[i] - u[i - 1]);


    int nx = 41; //num of grid points

    int dx = 2 / (nx - 1); //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


    //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; i++)
    {
        u0_array[i] = 1;
    }

    // set u = 2 between 0.5 and 1 as per initial conditions
    //note 0.5/dx = 10, 1/dx+1 = 21
    for (int i = 10; i < 21; i++)
    {
        u0_array[i] = 2;
        //printf("%f, ", u0_array[i]);
    }

    //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++)
    {
        //first loop iterates through each time step
        usol_array[i] = u0_array[i];
        //printf("%f", usol_array[i]);
        

        //MY CODE WORKS FINE AS I WANT UP TO THIS LOOP
        //second array iterates through each grid point for that time step and applies
        //the update equation
        for (int j = 1; j < nx - 1; j++)
        {
            u0_array[j] = usol_array[j] - c * dt/dx * (usol_array[j] - usol_array[j - 1]);
            printf("%f, ", u0_array[j]);
        }
    }

    return EXIT_SUCCESS;
}

作为参考，我下面的伪代码附在一维线性对流伪码(Matlab风格)下面。

Answer 1

而不是整数除法，而是使用FP数学。

这避免了后面的除以0和-1.#IND00。

// int dx = 2 / (nx - 1);   quotient is 0.
double dx = 2.0 / (nx - 1);

OP的代码与注释不匹配

// u[i] = u0[i] - c * dt/dx * (u0[i] - u[i - 1]
u0_array[j] = usol_array[j] - c * dt/dx * (usol_array[j] - usol_array[j - 1]);

如果删除多余的_array，则更容易查看。

//                                     v correct?
// u[i] = u0[i] - c * dt/dx * (u0[i] - u[i - 1]
u0[j] = usol[j] - c * dt/dx * (usol[j] - usol[j - 1]);
//                                       ^^^^ Correct?

Answer 2

您可能想要的是matlab术语。

for i = 1:nt
    usol = u0
    u0(2:nx) = usol(2:nx) - c*dt/dx*(usol(2:nx)-usol(1:nx-1))
end%for

这意味着在空间维上有两个内部循环，两个向量操作中的每一个都有一个。这两个单独的循环在C代码中必须是显式的

    //apply numerical scheme forward difference in
    //time an backward difference in space
    for (int i = 0; i < nt; i++) {
        //first loop iterates through each time step
        for (int j = 0; j < nx; j++) {
            usol_array[j] = u0_array[j];
            //printf("%f", usol_array[i]);
        } 
        

        //second loop iterates through each grid point for that time step 
        //and applies the update equation
        for (int j = 1; j < nx; j++)
        {
            u0_array[j] = usol_array[j] - c * dt/dx * (usol_array[j] - usol_array[j - 1]);
            printf("%f, ", u0_array[j]);
        }
    }