# Boundary conditions: Dirichlet vs Neumann

I'm thinking about the interplay of Dirichlet and Neumann BCs in a FDM scheme.

Let's assume a simple Black-Scholes call option problem, with BS PDE with constant coefficients, i.e. instead of $S$, in terms of $x=\ln(S)$.

In that case, the Dirichlet BC's are: \begin{equation} \begin{array}{l} V(t,{x_ + }) = \exp ({x_ + }) - K{e^{ - r(T - t)}}\\ V(t,{x_ - }) = 0 \end{array}. \end{equation} This is normally sufficient for solving the PDE. However, if I consider that \begin{equation} \frac{{\partial V(t,{x_ + })}}{{\partial x}} = \frac{{{\partial ^2}V(t,{x_ + })}}{{\partial {x^2}}}, \end{equation} this is also a valid BC for the upper boundary because $V(t,x) \propto {e^x}$ at the boundary.

1. can I drop the Dirichlet BC, if for that boundary I have also Neumann BC? I know the result won't be the same but is that approach correct?
2. can I use both types of BC at once? Would this yield a better approximation?
3. Is the role of the Neumann BCs important rather in the case of, say, defining the option's behaviour at the boundaries of the variance grid ($v_-,v_+$), e.g. in the Heston model, where no Dirichlet boundaries for $v$ exist?

You can certainly mix Dirichlet and Neumann boundary conditions, though the mixture has to be consistent. For example it is fine to use Neumann as $x \rightarrow \infty$ and Dirichlet as $x \rightarrow 0$. When pricing options on an $S$ grid rather than an $x$ grid this can make a lot of sense, because then you can put your bottom node right at zero.