What is the maximum attainable accuracy with a given set of $\alpha,\beta$?

Question

I am using LeVeque's book: https://faculty.washington.edu/rjl/fdmbook/

Suppose I want to compute $u''$ using FDM with $\alpha=\beta=2$ (centered) so the FDM is

$$ u''=\sum_{m = - \alpha}^\beta a_mu(x_i+mh) $$

Before acutally determining the coeficients ($a_m$'s) can we say anthying about how what the maximum attainable accuracy is?

Leveques does present methods the order of accuracy when the coefficients are already determined, but exactly how to determine the maximum order beforehand.

Answer 1

If you are looking for general accuracy of your stencil, it could be extracted by using Taylor expansion. Basically, if I want to write down your stencil explicitly, it contains terms of $u(x_{i}-2h)$, $u(x_{i}-h)$, $u(x_{i})$, $u(x_{i}+h)$, and $u(x_{i}+2h)$. Let's look at Taylor expansion of each of these terms:

$$u(x_{i}-2h) - u(x_{i}) = - u^{'}(x_{i})(2h) + \frac{1}{2} u^{''}(x_{i}) (4 h^{2}) - \frac{1}{6} u^{'''}(x_{i}) (8h^{3}) + \frac{1}{24} u^{''''}(x_{i}) (16h^{4}) + \mathcal{O}(h^{5})$$

$$u(x_{i}-h) - u(x_{i}) = - u^{'}(x_{i}) h + \frac{1}{2} u^{''}(x_{i}) h^{2} - \frac{1}{6} u^{'''}(x_{i})h^{3} + \frac{1}{24} u^{''''}(x_{i})h^{4} + \mathcal{O}(h^{5})$$

$$u(x_{i}+h) - u(x_{i}) = u^{'}(x_{i})h + \frac{1}{2} u^{''}(x_{i}) h^{2} + \frac{1}{6} u^{'''}(x_{i}) h^{3} + \frac{1}{24} u^{''''}(x_{i}) h^{4} + \mathcal{O}(h^{5})$$

$$u(x_{i}+2h) - u(x_{i}) = u^{'}(x_{i})(2h) + \frac{1}{2} u^{''}(x_{i})(4 h^{2}) + \frac{1}{6} u^{'''}(x_{i}) (8h^{3}) + \frac{1}{24} u^{''''}(x_{i}) (16h^{4}) + \mathcal{O}(h^{5})$$

You can write these equations in matrix form:

$$\begin{bmatrix} -2h & 2h^{2} & -\frac{4}{3} h^{3} & \frac{2}{3} h^{4} \\ -h & \frac{1}{2} h^{2} & -\frac{1}{6} h^{3} & \frac{1}{24} h^{4} \\ h & \frac{1}{2} h^{2} & \frac{1}{6} h^{3} & \frac{1}{24} h^{4} \\ 2h & 2h^{2} & \frac{4}{3} h^{3} & \frac{2}{3} h^{4} \end{bmatrix} \begin{bmatrix} u^{'}(x_{i}) \\ u^{''}(x_{i}) \\ u^{'''}(x_{i}) \\ u^{''''}(x_{i}) \end{bmatrix} = \begin{bmatrix} u(x_{i} - 2h) - u(x_{i}) \\ u(x_{i} - h) - u(x_{i}) \\ u(x_{i} + h) - u(x_{i}) \\ u(x_{i} + 2h) - u(x_{i}) \end{bmatrix}$$

When you solve this linear equation where you know $h$ and the values of $u(x_{i}-2h)$, $u(x_{i}-h)$, $u(x_{i})$, $u(x_{i}+h)$, and $u(x_{i}+2h)$, you would get unknowns of $u^{'}(x_{i})$, $u^{''}(x_{i})$, $u^{'''}(x_{i})$, and $u^{''''}(x_{i})$. You are interested particularly in $u^{''}(x_{i})$. The coefficients $a_{m}$ also will be determined here automatically.

In fact the inverse of above matrix could be extracted by sympy as:

$$\left[\begin{matrix}\frac{0.111111111111111}{0.444444444444444 h + 0.888888888888889} & - \frac{0.444444444444444 h + 0.444444444444444}{h \left(0.444444444444444 h + 0.888888888888889\right)} & \frac{0.444444444444444 \left(h + 1\right)}{h \left(0.444444444444444 h + 0.888888888888889\right)} & \frac{0.333333333333333}{- 1.33333333333333 h - 2.66666666666667}\\\frac{0.333333333333333}{h \left(- 1.33333333333333 h - 2.66666666666667\right)} & \frac{1.0 \left(0.740740740740741 h + 1.03703703703704\right)}{h^{2} \left(0.444444444444444 h + 0.888888888888889\right)} & \frac{1.0 \left(0.444444444444444 h + 1.33333333333333\right)}{h^{2} \left(0.444444444444444 h + 0.888888888888889\right)} & \frac{1.0 \left(0.111111111111111 h - 0.444444444444444\right)}{h^{2} \left(1.33333333333333 h + 2.66666666666667\right)}\\- \frac{0.666666666666667}{h^{2} \left(0.444444444444444 h + 0.888888888888889\right)} & \frac{1.33333333333333}{h^{2} \left(0.444444444444444 h + 0.888888888888889\right)} & - \frac{1.33333333333333}{h^{2} \left(0.444444444444444 h + 0.888888888888889\right)} & \frac{2.0}{h^{2} \left(1.33333333333333 h + 2.66666666666667\right)}\\\frac{4.0}{h^{3} \left(1.33333333333333 h + 2.66666666666667\right)} & \frac{10.6666666666667 h + 5.33333333333334}{h^{4} \left(- 1.33333333333333 h - 2.66666666666667\right)} & \frac{16.0}{h^{4} \left(- 1.33333333333333 h - 2.66666666666667\right)} & \frac{1.33333333333333 h - 5.33333333333333}{h^{4} \left(- 1.33333333333333 h - 2.66666666666667\right)}\end{matrix}\right] $$

So:

$$a_{-2} = \frac{0.333333333333333}{h \left(- 1.33333333333333 h - 2.66666666666667\right)}$$

$$a_{-1} = \frac{\left(0.740740740740741 h + 1.03703703703704\right)}{h^{2} \left(0.444444444444444 h + 0.888888888888889\right)}$$

$$a_{1} = \frac{\left(0.444444444444444 h + 1.33333333333333\right)}{h^{2} \left(0.444444444444444 h + 0.888888888888889\right)}$$

$$a_{2} = \frac{\left(0.111111111111111 h - 0.444444444444444\right)}{h^{2} \left(1.33333333333333 h + 2.66666666666667\right)}$$

$$a_{0} = - \frac{2.5}{h^{2}}$$

This is the code to generate these coefficients:

import sympy as sp
sp.init_printing(use_latex='mathjax')

h = sp.symbols('h')
A = sp.Matrix( [[-2*h,2*h**2,-(4/3)*h**2,(2/3)*h**4],[-h,(1/2)*h**2,-(1/6)*h**3,(1/24)*h**4],[h,(1/2)*h**2,(1/6)*h**3,(1/24)*h**4],[2*h,2*h**2,(4/3)*h**3,(2/3)*h**4]]) # Creates a matrix.
A_inverse = A.inv()

print(sp.printing.latex(sp.simplify(A_inverse)))

print(sp.printing.latex(sp.simplify(A_inverse[1,0])))
print(sp.printing.latex(sp.simplify(A_inverse[1,1])))
print(sp.printing.latex(sp.simplify(A_inverse[1,2])))
print(sp.printing.latex(sp.simplify(A_inverse[1,3])))

print(sp.printing.latex(sp.simplify(-A_inverse[1,0]-A_inverse[1,1]-A_inverse[1,2]-A_inverse[1,3])))

So the explicit answer of your question is that: this stencil is fourth-order accurate and the error is in the order of $\mathcal{O}(h^{5})$.