为什么数值解与C语言中的解析解相同？

Question

我编写了一个1 Dimmension cfd问题,但我的数值解决方案与分析解决方案相同(最多6位小数).

我使用TDMA方法进行数值求解,对于解析解,我直接用函数T(x)中的x值代替.

分析解T(x)出来了 T(x) = -(x^2)/2 +11/21(x);

E. g.然后是4格点;

  x0  = 0.000000,    x1  = 0.333333 ,   x2  = 0.666666 ,   x3  = 0.999999 .
T(x0) = 0.000000 , T(x1) = 0.119048 , T(x2) = 0.126984 , T(x3) = 0.023810.

对于数值解决方案,我使用过TDMA技术,请参考下面的代码.

输入n = 4表示结果.

#include<stdio.h>

void temp_matrix(int n, double *a, double *b, double *c, double *d, double *T);

int main() {
  int Bi = 20.0;
  int n;
  printf("%s ", "Enter the Number of total Grid Points");
  scanf("%d", &n);
  float t = (n - 1);
  double dx = 1.0 / t;
  int i;
  printf("\n");

  double q; // analytical solution below
  double z[n];
  for (i = 0; i <= n - 1; i++) {
    q = (dx) * i;
    z[i] = -(q * q) / 2 + q * (11.0 / 21);
    printf("\nT analytical %lf ", z[i]);
  }

  double b[n - 1];
  b[n - 2] = -2.0 * Bi * dx - 2.0;
  for (i = 0; i <= n - 3; i++) {
    b[i] = -2.0;
  }

  double a[n - 1];
  a[n - 2] = 2.0;
  a[0] = 0;
  for (i = 1; i < n - 2; i++) {
    a[i] = 1.0;
  }

  double c[n - 1];
  for (i = 0; i <= n - 2; i++) {
    c[i] = 1.0;
  }

  double d[n - 1];
  for (i = 0; i <= n - 2; i++) {
    d[i] = -(dx * dx);
  }

  double T[n];
  temp_matrix(n, a, b, c, d, T);

  return 0;
}

void temp_matrix(int n, double *a, double *b, double *c, double *d, double *T) {
  int i;
  double beta[n - 1];
  double gama[n - 1];
  beta[0] = b[0];
  gama[0] = d[0] / beta[0];
  for (i = 1; i <= n - 2; i++) {
    beta[i] = b[i] - a[i] * (c[i - 1] / beta[i - 1]);
    gama[i] = (d[i] - a[i] * gama[i - 1]) / beta[i];
  }
  int loop;
  for (loop = 0; loop < n - 1; loop++)
    for (loop = 0; loop < n - 1; loop++)

      T[0] = 0;
  T[n - 1] = gama[n - 2];

  for (i = n - 2; i >= 1; i--) {
    T[i] = gama[i - 1] - (c[i - 1] * (T[i + 1])) / beta[i - 1];
  }
  printf("\n");
  for (i = 0; i < n; i++) {
    printf("\nT numerical %lf", T[i]);

  }
}

Answer 1

为什么数值解与C语言中的解析解相同？

它们有所不同,约为3位.

以足够的精度打印以查看差异.

使用下面的内容,我们看到x620与x619的显着性的最后一个十六进制数的差异T[3].这只是10 ¹⁵差异中的1个部分.

#include<float.h>
printf("T analytical %.*e\t%a\n", DBL_DECIMAL_DIG - 1, z[i], z[i]);
printf("T numerical  %.*e\t%a\n", DBL_DECIMAL_DIG - 1, T[i], T[i]);

C允许double数学在long double数学时执行,FLT_EVAL_METHOD == 2然后进行相同的分析/数值结果.您的结果可能与我的不同,因为它以及其他微妙的FP细微差别.

printf("FLT_EVAL_METHOD %d\n", FLT_EVAL_METHOD);

产量

T analytical 0.0000000000000000e+00 0x0p+0
T analytical 1.1904761904761907e-01 0x1.e79e79e79e7ap-4
T analytical 1.2698412698412700e-01 0x1.0410410410411p-3
T analytical 2.3809523809523836e-02 0x1.861861861862p-6

T numerical  0.0000000000000000e+00 0x0p+0
T numerical  1.1904761904761904e-01 0x1.e79e79e79e79ep-4
T numerical  1.2698412698412698e-01 0x1.041041041041p-3
T numerical  2.3809523809523812e-02 0x1.8618618618619p-6

FLT_EVAL_METHOD 0