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Truncation error

In order to investigate the truncation error associated with the finite difference scheme we consider the equation

$$-\frac{dp}{dz} = \Psi(p,\mbox{$<\!{\alpha}\!>$},\mbox{$<\!{... ...d \Upsilon(p,\mbox{$<\!{\alpha}\!>$},\mbox{$<\!{x}\!>$})}{dz}$$ (7.13)

which resembles both ordinary differential equations in our problem. As stated earlier we will discretize (7.14) about point $z_{i+\frac{1}{2}}$ which implies that we have to use the following Taylor series. Taylor series of $\Psi(z)$ about point z_i and z_i+1 (assuming $\Psi \in C^4 \mbox{ on } [z_i;z_{i+1}]$) respectively read

$$\Psi_{i+\frac{1}{2}} = \Psi_i + \frac{1}{2}k_i \left.\frac... ...^3 \left.\frac{d^3\Psi}{dz^3}\right\vert _i + {\cal O}(k_i^4)$$ (7.14)

$$\Psi_{i+\frac{1}{2}} = \Psi_{i+1} - \frac{1}{2}k_i \left.\... ...left.\frac{d^3\Psi}{dz^3}\right\vert _{i+1} + {\cal O}(k_i^4)$$ (7.15)

In addition we need Taylor series about $z_{i+\frac{1}{2}}$

$$\Psi_i = \Psi_{i+\frac{1}{2}} - \frac{1}{2}k_i \left.\frac... ...c{d^3\Psi}{dz^3}\right\vert _{i+\frac{1}{2}} + {\cal O}(k_i^4)$$ (7.16)

$$\Psi_{i+1} = \Psi_{i+\frac{1}{2}} + \frac{1}{2}k_i \left.\... ...c{d^3\Psi}{dz^3}\right\vert _{i+\frac{1}{2}} + {\cal O}(k_i^4)$$ (7.17)

Using the two last series (7.17) and (7.18) reveals that the difference

$$\frac{p_{i+1}-p_i}{k_i} = \left.\frac{dp}{dz}\right\vert _{... ...c{d^3\Psi}{dz^3}\right\vert _{i+\frac{1}{2}} + {\cal O}(k_i^3)$$ (7.18)

equals the derivative at point $z_{i+\frac{1}{2}}$ plus a local truncation error of size ${\cal O}(k_i^2)$.

It is a bit more involved to calculate the local truncation error which accompanies the difference for $\Psi_{z+\frac{1}{2}}$. From the Taylor series (7.15) and (7.16) we have

$$\frac{\Psi_i+\Psi_{i+1}}{2} = \Psi_{i+\frac{1}{2}} - \frac{... ...ac{d^2\Psi}{dz^2}\right\vert _{i+1} \right] + {\cal O}(k_i^3)$$ (7.19)

which at first glance indicates that we have a local truncation of ${\cal O}(k_i)$ but a Taylor expansion of $(d\Psi/dz)\vert _{i+1}$ about z_i yields

$$\left.\frac{d\Psi}{dz}\right\vert _{i+1} = \left.\frac{d\Ps... ..._i^2\left.\frac{d^3\Psi}{dz^3}\right\vert _i + {\cal O}(k_i^3)$$ (7.20)

which implies that the first square bracket in (7.20) can be rearranged in the following way

$$\left[ \left.\frac{d\Psi}{dz}\right\vert _i - \left.\frac{... ..._i^2\left.\frac{d^3\Psi}{dz^3}\right\vert _i + {\cal O}(k_i^3)$$ (7.21)

The final expression which includes a Taylor expansion of the last square bracket in (7.20) therefore reads

$$\begin{eqnarray*} \frac{\Psi_i+\Psi_{i+1}}{2} &=& \Psi_{i+\frac{1}{2}} - \frac... ...dz}\right\vert _i + {\cal O}(k_i^2) \right] + {\cal O}(k_i^3) \end{eqnarray*}$$

$\parbox{1.50cm}{\begin{eqnarray} \end{eqnarray}}$

which indicates a local truncation of order ${\cal O}(k_i^2)$.

We can conclude without actually writing out the full expression that the finite difference scheme (7.8)-(7.11) has a local truncation error of size

$$LTE_i = {\cal O}(k_i^2)$$ (7.22)

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