Finite difference : relationship involving gamma

Question

Given the following PDE,

$$ \frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2S^2\frac{\partial^2V}{\partial S^2}=0 $$ and its finite difference approximation, $$ \frac{V_n^{m+1}-V_n^m}{\Delta t} + \frac{1}{2}\sigma^2S_n^2\Gamma_n^m=0 \qquad (*) $$ where $$ \Gamma_n^m=\frac{V_{n+1}^{m}-2V_n^m + V_{n-1}^m}{\Delta S^2} $$

How do i show by considering suitable linear combination of (*) evaluated at mesh points $S_{n-1}$, $S_n$, $S_{n+1}$ that,

$$ \frac{\Gamma_n^{m+1}-\Gamma_n^m}{\Delta t} + \frac{1}{2}\sigma^2S_n^2\frac{\Gamma_{n+1}^m-2\Gamma_n^m+\Gamma_{n-1}^m}{\Delta S^2}+2\sigma^2S_n\frac{\Gamma_{n+1}^m-\Gamma_{n-1}^m}{2\Delta S}+\sigma^2\frac{\Gamma_{n+1}^m+\Gamma_{n-1}^m}{2}=0 $$

Answer 1

Actually I've managed to figure out that we just need linear combinations of 1, -2, 1 for (*) evaluated at $S_{n-1}$, $S_n$, $S_{n+1}$ respectively and we will get the required expression.

Answer 2

Basically we need to show that $$ D (S_n^2 \Gamma_n) = S_n^2 D \Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} + \Gamma_{n+1} + \Gamma_{n-1} $$ where $D$ is second order finite difference derivative operator $$ D u_n \equiv \frac{u_{n+1} -2 u_n+ u_{n-1}}{\Delta S^2} $$ Note, that $D$ can be written as $$ D = D_+ D_- = D_- D_+ $$ where $$ D_+ u_n = \frac{u_{n+1} - u_n}{\Delta S}\\ D_- u_n = \frac{u_{n} - u_{n-1}}{\Delta S}\\ $$ are one-sided finite difference first order derivative operators. Really, $$ D_+D_- u_n = D_+\left( \frac{u_{n} - u_{n-1}}{\Delta S}\right) = \frac{u_{n+1} - u_{n}}{\Delta S^2} - \frac{u_{n} - u_{n-1}}{\Delta S^2} = D u_n. $$ Same for the inverse order. Next, $$ D_{\pm} (a_n b_n) = \pm\frac{a_{n\pm 1}b_{n \pm 1} -a_nb_n}{\Delta S} = \pm\frac{a_{n\pm 1}b_{n \pm 1} -a_{n\pm 1}b_n+a_{n\pm 1}b_n-a_nb_n}{\Delta S} = a_{n \pm 1} D_\pm b_n + b_n D_\pm a_n. $$ Thus $$D_\pm S_n^2 = S_{n \pm 1} D_\pm S_n + S_n D_\pm S_n = S_n + S_{n \pm 1}\\ D_\pm \Gamma_n S_n^2 = \Gamma_{n \pm 1} (S_n + S_{n\pm 1}) + S_n^2 D_\pm \Gamma_n. $$ This far we found that $$ D_+ \Gamma_n S_n^2 = \Gamma_{n + 1} (S_n + S_{n+1}) + S_n^2 D_+ \Gamma_n. $$ Let's apply $D_-$ to that $$ D_- D_+ \Gamma_n S_n^2 = D_-\Gamma_{n + 1} (S_n + S_{n+1}) + D_- S_n^2 D_+ \Gamma_n. $$ Starting with the $D_- S_n^2 D_+ \Gamma_n$ we have $$ D_- S_n^2 D_+ \Gamma_n = S_{n-1}^2 D_-D_+ \Gamma_n + (S_n + S_{n-1})D_+\Gamma_n = S_{n-1}^2 D \Gamma_n + (S_n + S_{n-1})D_+ \Gamma_n. $$ Next $$ D_-\Gamma_{n + 1} (S_n + S_{n+1}) = \Gamma_n D_-(S_n+S_{n+1}) + (S_n+S_{n+2}) D_-\Gamma_{n+1} = 2\Gamma_n + (S_n + S_{n+1}) D_+\Gamma_n. $$ So far we have $$ D S_n^2 \Gamma_n = 2\Gamma_n + (S_{n-1} + 2S_n + S_{n+1}) D_+ \Gamma_n + S_{n-1}^2 D\Gamma_n. $$ Computing the same expression by reversing the $D_+D_-$ order, we have $$ D S_n^2 \Gamma_n = 2\Gamma_n + (S_{n-1} + 2S_n + S_{n+1}) D_- \Gamma_n + S_{n+1}^2 D\Gamma_n. $$ Also, since $S_{n\pm 1} = S_{n}\pm \Delta S$, $$ D S_n^2 \Gamma_n = 2\Gamma_n + 4S_n D_+ \Gamma_n + (S_{n}-\Delta S)^2 D\Gamma_n\\ D S_n^2 \Gamma_n = 2\Gamma_n + 4S_n D_- \Gamma_n + (S_{n}+\Delta S)^2 D\Gamma_n. $$ Halfsumming the equations finally we have $$ D S_n^2 \Gamma_n = 2\Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} + (S_n^2 + \Delta S^2)D\Gamma_n = \\ = S_n^2 D\Gamma_n + 4S_n \frac{\Gamma_{n+1} - \Gamma_{n-1}}{2\Delta S} + 2 \Gamma_n + \Delta S^2\frac{\Gamma_{n-1} - 2\Gamma_n + \Gamma_{n+1}}{\Delta S^2}. $$

Rest should be obvious now.