4th order Padé scheme formula derivation

Question

I am trying to derive the formula of the 4th order Padé scheme that passes through the points $x_i$, $x_{i-1}$ and $x_{i+1}$

$$\Big(\frac{\partial\phi}{\partial x} \Big)_i = -\frac{1}{4}\Big(\frac{\partial\phi}{\partial x} \Big)_{i+1}-\frac{1}{4}\Big(\frac{\partial\phi}{\partial x} \Big)_{i-1} + \frac{3}{4}\frac{\phi_{i+1}-\phi_{i-1}}{\Delta x}$$

To achieve this I am thinking of solving it using linear algebra, Gauss elimination, to find the coefficients $a_i$

$$A=\pmatrix{ 1 & x_{i-1} & x_{i-1}^2 & x_{i-1}^3 &x_{i-1}^4 \\ 1 & x_{i} & x_{i}^2 & x_{i}^3 &x_{i}^4 \\ 1 & x_{i+1} & x_{i+1}^2 & x_{i+1}^3 &x_{i+1}^4 \\ } ~~~~~~ x=\pmatrix{ a_0\\a_1\\a_2\\a_3\\a_4 } ~~~~~~ y=\pmatrix{ \phi_{i-1} \\ \phi_{i} \\ \phi_{i+1} } $$

once I have the coefficients I can use them to find the derivatives in $x_i$, $x_{i-1}$ and $x_{i+1}$ and try to get the formula. I think this is a lengthy procedure, is this a good approach to derive the formula? Is there a better way (faster) to do it?

Answer 1

This is the procedure to derive formulations for Compact finite difference schemes.

First, assume the stencil and the general structure of the scheme. For the classical Padé scheme you have:

$$ \alpha f'_{n+1} + f'_{n} + \alpha f'_{n-1} = a \dfrac{f_{n+1} - f_{n-1}}{2 \Delta x} \qquad(1) $$

You write the Taylor series for the derivatives and the functions, e.g.,

$$ f'_{n+1}=f'_{n} + \Delta x f''_n + \dfrac{(\Delta x)^2}{2!} f^{(3)}_n + \cdots $$

You then replace the derivatives and the functions in Eq. (1) with their Taylor series. The system of equations to solve consists of the following equations:

• Set $f'_{n}$'s coefficients on both sides to be equal. Here we have, $$ 2\alpha + 1 = a $$

• $f_n$ coefficient could only appear on the right hand side and its coefficient must be set equal to zero. In this case, this does not give us any equations.

• Start with coefficients of the lowest powers of $\Delta x$ and set them to be equal, until you get enough equations, e.g., $$ a = 2 \dfrac{3!}{2!} \alpha $$

Answer 2

I would think a Taylor table would be the way to go. Moin's book "Fundamentals of Engineering Numerical Analysis" has good easy to follow examples.