# Finite Difference implicit scheme

I'm trying to solve the following PDE numerically using an implicit FD scheme:

$$\frac{\sigma_s^2}{2}\frac{\partial^2 V}{\partial S^2} + \rho \sigma_S \sigma_\alpha\frac{\partial^2 V}{\partial S \partial \alpha} + \frac{\sigma_\alpha^2}{2}\frac{\partial^2 V}{\partial \alpha^2} + \mu_s \frac{\partial V}{\partial S} + \mu_\alpha \frac{\partial V}{\partial \alpha} + \frac{\partial V}{\partial t} - rV$$

This raises the following two questions I have not been able to find out yet:

• When substituting the derivatives with FD approximations, is the part $rV$ replaced by $rV_{i,j,k}$ or $rV_{i,j,k+1}$?

• When rewriting FD formula in the form of $V_{k}=AV_{k+1} - C$, are the boundary values needed to calculate $C$ taken from $V_{k+1}$ or $V_k$?

• Can you clarify your time notation: is $k+1$ before or after $k$? – Daneel Olivaw Jun 8 '18 at 13:55
• In my setup $k+1$ is before $k$. More precisely, I express the time dimension in terms of time to maturity $\tau$. Hence, in my code $k=0$ corresponds to $\tau_0=0$ (which corresponds to $t=T$, where $T$ denotes expiration). Similarly, $\tau_{max}=T$ (corresponding to $t=0$). – Pim Jun 8 '18 at 14:21
• Then you need to replace it by $rV_{i,j,k+1}$: if you consider two instants $t$ and $s$, $s>t$, the position $V_t$ earns the rate $r$ from $t$ to $s$ hence it is earned on the value $V_t$. – Daneel Olivaw Jun 8 '18 at 14:26
• alternatively you can get rid of the $rV$ term by solving the PDE for $U = e^{-rt}V$ – Antoine Conze Jun 8 '18 at 14:49
• Improved accuracy you don't add the extra error that comes from the scheme approximating $e^{-rdt}$ with $1-rdt$ or $1/(1+rdt)$ – Antoine Conze Jun 8 '18 at 18:25

When using the (Euler) Implicit scheme, the only thing that's taken at the previous time level (the one for which you have the solution already), is the $V_{i,j,k}$ that comes from the time derivative. Everything else in the discretized equation is taken at the next time level. So, for both your questions, it's k+1.