Round Two on Finite Difference Approximations of Derivatives |

There is another way to obtain the results stated in “Finite Difference Approximations of Derivatives“.

Let $y'_i, y''_i$ denotes $y'(x_i)$ and $y''(x_i)$ respectively and,

$h=x_{i+1}-x_i=x_i-x_{i-1} >0$ .

We define

$y'_i = \alpha_1 y_{i-1} + \alpha_2 y_{i+1}\quad\quad\quad(1)$ .

By Taylor’s expansion around $x_i$ ,

$y_{i-1} \approx y_i + y'_i(-h)+\frac{y''_i}{2!}(-h)^2\quad\quad\quad(1-1)$

$y_{i+1} \approx y_i + y'_ih+\frac{y''_i}{2!}h^2\quad\quad\quad(1-2)$

Substituting (1-1), (1-2) into (1),

$y'_i \approx \alpha_1 y_i - \alpha y'_ih +\alpha_1\frac{y''_i}{2}h^2 + \alpha_2 y_i +\alpha_2y'_i h +\alpha_2\frac{y''_i}{2}h^2$ .

That is,

$y'_i \approx (\alpha_1+\alpha_2) y_i +(\alpha_2-\alpha_1)h y'_i + (\alpha_1+\alpha_2)\frac{y''_i}{2}h^2.$

It follows that

$\begin{cases} \alpha_1+\alpha_2=0 \\ (\alpha_2-\alpha_1)h=1\\ \alpha_1+\alpha_2=0 \end{cases}\quad\quad\quad(1-3)$

Fig. 1

Solving (1-3) for $\alpha_1, \alpha_2$ (see Fig. 1) yields

$\alpha_1=-\frac{1}{2h}, \alpha_2=\frac{1}{2h}.$

Therefore,

$y'_i \approx -\frac{1}{2h} y_{i-1} + \frac{1}{2h} y_{i+1} = \frac{y_{i+1}-y_{i-1}}{2h}$

or,

$\boxed{y'_i \approx \frac{y_{i+1}-y_i}{2h}}$

Now, let

$y''_i = \alpha_1y_{i-1}+\alpha_2 y_i + \alpha_3 y_{i+1}$ .

From

$y_{i-1} \approx y_i - y'_i h + \frac{y''_i}{2!} h^2$

and

$y_{i+1} \approx y_i +y'_i h + \frac{y''_i}{2!}h^2,$

we have,

$y''_i \approx \alpha_1 y_i - \alpha_1 y'_i h + \frac{\alpha_1}{2}y''_i h^2 + \alpha_2 y_i + \alpha_3 y_i + \alpha_3 y'_i h + \frac{\alpha_3}{2}y''_i h^2$ .

It leads to

$\begin{cases} \alpha_1+\alpha_2+\alpha_3=0\\-\alpha_1+\alpha_3=0\\(\frac{\alpha_1}{2}+\frac{\alpha_3}{2})h^2=1 \end{cases}$

Fig. 2

whose solution (see Fig. 2) is

$\alpha_1 = \frac{1}{h^2}, \alpha_2 = -\frac{2}{h^2}, \alpha_3=\frac{1}{h^2}$ .

Hence,

$y''_i \approx \frac{1}{h^2}y_{i-1}-\frac{2}{h^2}y_i +\frac{1}{h^2}y_{i+1}=\frac{y_{i-1}-2y_i+y_{i+1}}{h^2}$

i.e.,

$\boxed{ y''_i \approx \frac{y_{i+1}-2y_i + y_{i-1}}{h^2}}$

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