Question 6.1: Derive a centered finite difference formula for the second d...

For each evaluation point, we need to use the Taylor series

y(x) = y(x_i)+y^′(x−x_i)+\frac{1}{2}y^{′′}(x−xi)^2 +\frac{1}{6}y^{′′′}(x−xi)^3+\frac{1}{24}y^{′′′′}(x−xi)^4+···     (6.2.14)

to produce the equations

y_{i+2}=y_i+y^\prime (2\Delta x)+\frac{1}{2} y^{\prime \prime }(2\Delta x)^2 +\frac{1}{6}y^{\prime \prime \prime }(2\Delta x)^3+\frac{1}{24} y^{\prime \prime \prime \prime } (2\Delta x)^4+···     (6.2.15)

y_{i+1}=y_i+y^\prime (\Delta x)+\frac{1}{2} y^{\prime \prime }(\Delta x)^2 +\frac{1}{6}y^{\prime \prime \prime }(\Delta x)^3+\frac{1}{24} y^{\prime \prime \prime \prime } (\Delta x)^4+···     (6.2.16)

y_{i-1}=y_i+y^\prime (-\Delta x)+\frac{1}{2} y^{\prime \prime }(-\Delta x)^2 +\frac{1}{6}y^{\prime \prime \prime }(-\Delta x)^3+\frac{1}{24} y^{\prime \prime \prime \prime } (-\Delta x)^4+···     (6.2.17)

y_{i-2}=y_i+y^\prime (-2\Delta x)+\frac{1}{2} y^{\prime \prime }(-2\Delta x)^2 +\frac{1}{6}y^{\prime \prime \prime }(-2\Delta x)^3+\frac{1}{24} y^{\prime \prime \prime \prime } (-2\Delta x)^4+···     (6.2.18)

The idea is to multiply each equation by some weight c_i and add them together to remove all the derivatives except for y^{′′}. This leads to the system of equations

\begin{bmatrix} 2& 1& −1& −2\\ 4 &1 &1 &4 \\ 8& 1& −1& −8\\ 16& 1& 1& 16 \end{bmatrix} \begin{bmatrix} c_{i+2} \\c_{i+1} \\c_{i−1} \\c_{i−2} \end{bmatrix} =\begin{bmatrix} 0 \\ 2 \\ 0 \\0 \end{bmatrix}                  (6.2.19)
where we chose the value of 2 on the right-hand side to cancel the factor of 1/2 in the second derivative. We can solve by Gauss elimination to get

\begin{bmatrix} c_{i+2} \\c_{i+1} \\c_{i−1} \\c_{i−2} \end{bmatrix}=\frac{1}{12}\begin{bmatrix} −1 \\ 16 \\ 16 \\ −1 \end{bmatrix}                   (6.2.20)

Adding up the equations with these coefficients gives

\frac{−y_{i+2} + 16y_{i+1} + 16y_{i−1} − y_{i−2}}{12}=\frac{30}{12}y_i+y^{\prime \prime } \Delta x^2                  (6.2.21)

Solving for the derivative gives

y^{\prime \prime}= \frac{−y_{i+2} + 16y_{i+1} -30y_i+ 16y_{i−1} − y_{i−2}}{12\Delta x^2}                  (6.2.22)

What is the advantage of the higher-order formula? Let’s look at the error. If we continued our expansion out to the fifth derivatives, the weights and coefficients give

\frac{-(2\Delta x)^5+16(\Delta x)^5+16(-\Delta x)^5-(-2\Delta x)^5}{12}=0                  (6.2.23)

so the terms cancel out. When we take the expansion out to sixth derivatives we get a non-zero result,

\frac{-(2\Delta x)^6+16(\Delta x)^6+16(-\Delta x)^6-(-2\Delta x)^6}{12}=-8\Delta x^6                  (6.2.24)

Since we divided by Δx^2,  the  error  is  O( Δx^4).
This is one possible approach to derive higher-order finite difference approximations. We will discuss an alternate approach in the context of interpolation in Section 8.2.4.