Five-point stencil

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An illustration of the five-point stencil in one and two dimensions (top, and bottom, respectively).

In numerical analysis, given a square grid in one or two dimensions, the five-point stencil of a point in the grid is made up of the point itself together with its four "neighbors". It is used to write finite difference approximations to derivatives at grid points.

1 One dimension
- 1.1 First derivative
  - 1.1.1 Obtaining the formula
  - 1.1.2 Estimated error
- 1.2 Higher derivatives
  - 1.2.1 Estimated errors
2 Relationship to Lagrange interpolating polynomials
3 Two dimensions
4 Notes
5 References

[edit] One dimension

In one dimension, if the spacing between points in the grid is $h$ , then the five-point stencil of a point $x$ in the grid is

$\ \{x-2h, x-h, x, x+h, x+2h\}$ .

[edit] First derivative

The first derivative of a function $f$ of a real variable at a point $x$ can be approximated using a five-point stencil as

$f'(x) \approx \frac{-f(x+2 h)+8 f(x+h)-8 f(x-h)+f(x-2h)}{12 h}$

[edit] Obtaining the formula

This formula can be obtained by writing out the four Taylor series of $f(x \pm h)$ and $f(x \pm 2h)$ up to terms of $h 3$ (or up to terms of $h 5$ to get an error estimation as well) and solving this system of four equations to get $f'(x)$ . Actually, we have at points $(x + h)$ and $(x - h)$

$f(x \pm h) = f(x) \pm h f'(x) + \frac{h^2}{2}f''(x) \pm \frac{h^3}{6} f^{(3)}(x) + O_{1\pm}(h^4). \qquad (E_{1\pm})$

Evaluating $(E 1 +) - (E 1 -)$ gives us

$f(x+h) - f(x-h) = 2hf'(x) + \frac{h^3}{3}f^{(3)}(x) + O_1(h^4). \qquad (E_1)$

Note that the residual term $O 1 (h 4)$ should be of the order of $h 5$ instead of $h 4$ because if the terms of $h 4$ had been written out in $(E 1 +)$ and $(E 1 -)$ , it can be seen that they would have canceled each other out by $f (x + h) - f (x - h)$ . But for this calculation, it is left like that since the order of error estimation is not treated here (cf below).

Similarly, we have

$f(x \pm 2h) = f(x) \pm 2h f'(x) + 2h^2 f''(x) \pm \frac{4h^3}{3} f^{(3)}(x) + O_{2\pm}(h^4). \qquad (E_{2\pm})$

$(E 2 +) - (E 2 -)$ gives us

$f(x+2h) - f(x-2h) = 4hf'(x) + \frac{8h^3}{3}f^{(3)}(x) + O_2(h^4). \qquad (E_2)$

In order to eliminate the terms of $f (3) (x)$ , calculate $8 \times (E_1) - (E_2)$

$8f(x+h) - 8f(x-h) - f(x+2h) + f(x-2h) = 12h f'(x) + O(h^4) \,$

thus giving the formula as above.

[edit] Estimated error

The error in this approximation is of order $h 4$ . That can be seen from the expansion

$\frac{-f(x+2 h)+8 f(x+h)-8 f(x-h)+f(x-2h)}{12 h}=f'(x)-\frac{1}{30} f^{(5)}(x) h^4+O(h^5)$ ^[1]

which can be obtained by expanding the left-hand side in a Taylor series. Alternatively, apply Richardson extrapolation to the central difference approximation to $f'(x)$ on grids with spacing $2 h$ and $h$ .

[edit] Higher derivatives

The centered difference formulas for five-point stencils approximating second, third, and fourth derivatives are

$\begin{align} f''(x) &\approx \frac{-f(x+2 h)+16 f(x+h)-30 f(x) + 16 f(x-h) - f(x-2h)}{12 h^2}, \\ f^{(3)}(x) &\approx \frac{f(x+2 h)-2 f(x+h) + 2 f(x-h) - f(x-2h)}{2 h^3}, \\ f^{(4)}(x) &\approx \frac{f(x+2 h)-4 f(x+h)+6 f(x) - 4 f(x-h) + f(x-2h)}{h^4}. \end{align}$

[edit] Estimated errors

The errors in these approximations are $O (h 4)$ , $O (h 2)$ and $O (h 2)$ respectively.^[1]

[edit] Relationship to Lagrange interpolating polynomials

As an alternative to deriving the finite difference weights from the Taylor series, they may be obtained by differentiating the Lagrange polynomials

$\ell_j(\xi) = \prod_{i=0,\, i\neq j}^{k} \frac{\xi-x_i}{x_j-x_i},$

where the interpolation points are

$\begin{align} x_0=x-2h,\quad x_1=x-h,\quad x_2=x,\quad x_3=x+h,\quad x_4=x+2h. \end{align}$

Then, the quartic polynomial $p 4 (x)$ interpolating $f (x)$ at these five points is

$\begin{align} p_4(x) = \sum\limits_{j=0}^4 f(x_j) \ell_j(x) \end{align}$

and its derivative is

$\begin{align} p_4'(x) = \sum\limits_{j=0}^4 f(x_j) \ell'_j(x). \end{align}$

So, the finite difference approximation of $f'(x)$ at the middle point $x = x 2$ is

$\begin{align} f'(x_2) = \ell_0'(x_2) f(x_0) + \ell_1'(x_2) f(x_1) + \ell_2'(x_2) f(x_2) + \ell_3'(x_2) f(x_3) + \ell_4'(x_2) f(x_4) + O(h^4) \end{align}$

Evaluating the derivatives of the five Lagrange polynomials at $x = x 2$ gives the same weights as above. This method can be more flexible as the extension to a non-uniform grid is quite straightforward.