双浮点精度

Question

I'm solving Laplace's Equation with Gauss-Seidel method, but in some regions, its showing a plateau-like aspect. Formally, ie, by numerical analysis, such regions should not exist, even if the gradient is almost zero.

I'm forced to believe that double precision isn't enough to perform the arithmetic and that a big number library need to be used (killing the performance, since now it will be done by software). Or, that I should do the operations in a different order, aiming to preserve some significance to the decimals.

Example

Cell (13, 14, 0) is being updated by 7-point mesh (in 3D), and its neighbours are:

(12,14,0)=  0.9999999999999936; // (x-)
(14,14,0)=  0.9999999999999969; // (x+)
(13,13,0)=  0.9999999999999938; // (y-)
(13,15,0)=  1.0000000000000000; // (y+)
(13,14,-1)= 1.0000000000000000; // (z-)
(13,14,1)=  0.9999999999999959; // (z+)

So, the new value of cell (13,14,0) would be evaluated as:

p_new = (0.9999999999999936 + 0.9999999999999969 + 0.9999999999999938 + 1.0000000000000000 + 1.0000000000000000 + 0.9999999999999959) / 6.0 ;

which leads to p_new being 1.0000000000000000, when it should be 0.9999999999999966.

Code

#include <stdio.h>

int main()
{
    double ad_neighboor[6] = {0.9999999999999936, 0.9999999999999969,
                              0.9999999999999938, 1.0000000000000000,
                              1.0000000000000000, 0.9999999999999959};

    double d_denom = 6.0;

    unsigned int i_xBackward=0;
    unsigned int i_xForward=1;

    unsigned int i_yBackward=2;
    unsigned int i_yForward=3;

    unsigned int i_zBackward=4;
    unsigned int i_zForward=5;

    double d_newPotential = (ad_neighboor[i_xForward] + ad_neighboor[i_xBackward] +
                             ad_neighboor[i_yForward] + ad_neighboor[i_yBackward] +
                             ad_neighboor[i_zForward] + ad_neighboor[i_zBackward] ) / d_denom;

    printf("%.16f\n", d_newPotential);
}

Answer 1

Since you are solving:

d²(phi)/dx² + d²(phi)/dy² = 0

Instead you can solve the equivalent problem:

d²(phi')/dx² + d²(phi')/dy² = 0

Where, phi' = phi - 1 .

Remember to apply the boundary conditions in terms of phi' .

Finally after the solution has converged, you can get the solution as phi = 1 + phi' .

I am assuming here that the boundary values are close to 1.

I haven't tried this, but I think that the numbers will be represented in their significant digits in a floating point notations, thus the truncation error will be reduced.

Answer 2

Your granularity is too fine for the double precision floating point type on your platform.

In the majority of cases, you'd address this by adjusting your granularity. If you need any convincing, 15 significant figures of granularity is enough to mesh the Solar system out to the orbit or Pluto in squares of 1cm length! For this method, I'd be inclined to reserve at least four orders of magnitude to obviate numerical noise.

Only in a very minimal number of cases ought you think in terms of switching to another data type, such as as long double (if different to double to your platform), or an arbitrary precision type.