计算流体动力学数组问题为什么我的 C 程序输出 -1.#IND00

编程问答 2022-05-26

问题描述

我在用 C 编写的用于求解一维线性对流方程的程序有问题。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，数组元素在区间内设置为等于 2 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尝试打印结果时，对于循环的每次迭代，我得到的输出只是 -1.IND00。 （我正在遵循我也在下面附上的 Matlab 风格的伪代码）我对 C 很陌生，所以我的经验不足。我不知道为什么会这样。如果有人可以提出可能的解决方案，我将不胜感激。到目前为止，我已经在下面附上了我的代码和一些评论，以便您可以遵循我的思考过程。我还附上了我遵循的 Matlab 风格的伪代码。

# 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;
}

下面附上我所遵循的伪代码以供参考 1D linear convection pseudo Code (Matlab Style)

解决方法

使用 FP 数学代替整数除法。
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]);

更容易查看是否删除了多余的 _array。

//                                     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?
,

你可能想要的是 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

这意味着您在空间维度上有 2 个内部循环，两个向量操作中的每一个都有一个。这两个单独的循环必须在 C 代码中显式

    //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]);
        }
    }
arraysarraysarrayscfluid-dynamicsnumerical-methods