# Finite Difference Method for Black-Scholes-Formula

Using finite difference method for the Black-Scholes-Partial Differential Equation one need to impose some boundary conditions on the edge of the grid, i.e for a Grid on $D=[a,b]\times R^+$ one need to impose the boundary condition on $V(a,t)$ and $V(b,t)$, where $V(x,t)$ is described by Black-Scholes-Partial Differential Equation with $x$ and $t$ representing respectively the underlying price and the time-to-maturity of the derivative $V$.

Question: By imposing those boundary condition, does it has impact on the well-posedness of the problem? From non-stochastic partial differential equation we know that adding boundary condition could make the initial value problem ill-posed. Is there any "rule" of choosing boundary conditions ensuring the problem to be well posed?

It depends on the type of payoff you want to price. If it is a call option, you know that $V(0,t) = 0$ and $V(x,t) \approx x$ when $x \rightarrow +\infty$ so you can use a dirichlet condition $V(a,t) = 0$ and $V(b,t) = b$. Alternatively you can use a linear condition $\frac{\partial^2V}{\partial x^2} = 0$ which in practice works fine for a variety of payoffs. The only case where you have to be carefull is when you price barrier options, for instance an up and out option, in which case $b$ will be set to the barrier and you have to use the dirichlet condition $V(b,t) =$ payoff on barrier.
Note that to improve numerical convergence of the scheme it is better to have constant coefficients if front of the $\frac{\partial^2V}{\partial x^2}$ term in the PDE when you use a uniform grid, so in the case of the Black & Scholes PDE you should work in $y = \log(x)$ space.
• Sorry I didn't see the question about the well-posedness of the problem. In pratical finance applications as long as you've done the appropriate change of variable and the volatility is bounded you don't really care about the boundary conditions as long as $a$ and $b$ are away enough from $x_0$, because the probability of reaching $a$ or $b$ is very small. This is why a linear condition, which is simple to implement and is not payoff dependent, is appropriate. – Antoine Conze Apr 28 '17 at 9:46