3.2 Handling differential term
Let \(\ \delta_{x}\ \)and\(\ {\delta\ }_{x}^{2}\) denote the first and
second order central difference approximations with respect to\(\ x\),
respectively, then
\begin{equation}
{\ \delta}_{x}V_{m}=\frac{V_{m+1}-V_{m-1}}{2h}\nonumber \\
\end{equation}and
\begin{equation}
{\ \delta}_{x}^{2}V_{m}=\frac{V_{m+1}-2V_{m}+V_{m-1}}{h^{2}}.\nonumber \\
\end{equation}At the internal node of region \(\Omega\) we obtain
\(V_{\tau}=\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)\delta_{x}V_{m}+\frac{1}{2}\sigma^{2}\delta_{x}^{2}V_{m}-rV_{m}+\lambda\omega_{m}+\varepsilon_{m},\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (3.2)\)
where\(\omega_{m}\mathcal{=g}\left(V_{m}\right)\mathcal{+l}\left(V_{m}\right).\)
Local truncation error
\begin{equation}
\varepsilon_{m}=\frac{h^{2}}{12}\left(2\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)\left(V_{\text{xxx}}\right)_{m}+\frac{1}{2}\sigma^{2}\left(V_{\text{xxxx}}\right)_{m}\right)\mathcal{+O}\left(h^{4}\right).\nonumber \\
\end{equation}Fourth-order discretization can be obtained by substituting higher
derivatives in terms of \(V_{x}\) and \(V_{\text{xx}}\) and using
central difference approximation. Referring to Thakoor et al [11],
there is
\begin{equation}
\frac{1}{2}\sigma^{2}V_{\text{xxx}}=V_{\text{xτ}}-\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)V_{\text{xx}}+rV_{x}-\lambda\omega_{x},\nonumber \\
\end{equation}and
\begin{equation}
\left(\frac{1}{2}\sigma^{2}\right)^{2}V_{\text{xxxx}}=\frac{1}{2}\sigma^{2}V_{\text{xxτ}}-\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)V_{\text{xτ}}+\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)^{2}V_{\text{xx}}-r\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)V_{x}\nonumber \\
\end{equation}\(+\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)\omega_{x}+\frac{r}{2}\sigma^{2}V_{\text{xx}}-\frac{1}{2}\sigma^{2}\omega_{\text{xx}}\).
Substituting \(V_{\text{xxx}}\) and \(V_{\text{xxxx}}\) into (3.2), we
obtain the numerical discretization form
\begin{equation}
\left(\frac{1}{2}\sigma^{2}+\frac{h^{2}}{12}\left(\frac{1}{2}\sigma^{2}\delta_{x}^{2}+\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)\delta_{x}\right)\right)\left(V_{\tau}\right)_{m}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left(\left(\frac{1}{2}\sigma^{2}\right)^{2}+\frac{h^{2}}{12}\left(\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)^{2}-\frac{r}{2}\sigma^{2}\right)\right)\delta_{x}^{2}V_{m}+\left(\frac{1}{2}\sigma^{2}\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)-\frac{h^{2}}{12}r\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)\right)\delta_{x}V_{m}-\frac{r}{2}\sigma^{2}V_{m}+\left(\frac{1}{2}\sigma^{2}+\frac{h^{2}}{12}\left(\frac{1}{2}\sigma^{2}\delta_{x}^{2}+\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)\delta_{x}\right)\right)\omega_{m}.\nonumber \\
\end{equation}It is noted that the finite difference operators \(\delta_{x}\) and\(\delta_{x}^{2}\) are\(\ \delta_{x}=\frac{1}{2h}\par
\begin{bmatrix}-1&0&1\\
\end{bmatrix}\),\(\delta_{x}^{2}=\frac{1}{h^{2}}\par
\begin{bmatrix}1&-2&1\\
\end{bmatrix}\), respectively, then
for\(\ \text{mϵ}\left[1,M-1\right]\),the operators
satisfy the following equation
\(\alpha\left(V_{\tau}\right)_{m-1}+\eta\left(V_{\tau}\right)_{m}+\gamma\left(V_{\tau}\right)_{m+1}=\left(\delta-\upsilon\right)V_{m-1}+\xi V_{m}+\left(\delta+\upsilon\right)V_{m+1}+\alpha\omega_{m-1}+\eta\omega_{m}+\gamma\omega_{m+1},\)
where
\begin{equation}
\omega_{m}\mathcal{=g}\left(x_{m}\right)\mathcal{+l}\left(x_{m}\right),\alpha=\frac{\sigma^{2}}{24}-\frac{\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)h}{24},\eta=\frac{5\sigma^{2}}{12},\gamma=\frac{\sigma^{2}}{24}+\frac{\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)h}{24},\nonumber \\
\end{equation}\begin{equation}
\delta=\frac{1}{12}\left(\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)^{2}-\frac{r}{2}\sigma^{2}\right)+\frac{\left(\sigma^{2}\right)^{2}}{{4h}^{2}},v=\frac{\sigma^{2}\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)}{4h}-\frac{r\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)h}{24},\nonumber \\
\end{equation}\(\xi=-\frac{1}{6}\left(\left(r-\text{βλ}-\frac{1}{2}\sigma^{2}\right)^{2}+\frac{5\sigma^{2}}{2}\right)\).