# PDE boundary condition question regarding limits

Just as a bit of background, I'm working with the Black-Scholes PDE and I'm testing some things out by taking an initial condition for it as $\sin(S/50)$, where $S$ is the spot price (but that's irrelevant to the question).

My issue comes with attempting to find the boundary condition as $S \rightarrow \infty$. Usually the limit of the initial condition is used to find the boundary condition as $S \rightarrow \infty$. In the case with $\sin(S/50)$, what would be an appropriate boundary condition to take? I've already tried taking boundary conditions such as $0$, $\sin(S_{T}/50)$ (where $S_{T}$ is the maximum spot price), and $\sin(S_{t}/50)$ (where $S_{t}$ is the current spot price), but none of these match with the Monte Carlo plot.

I'm fairly certain the Monte Carlo simulation is accurate since it is extremely easy to implement for this function, so I'm pretty sure the issue comes with the PDE simulation. If anyone can help me out with this I would greatly appreciate it, thanks in advance.

EDIT: Perhaps I should note that for the Monte Carlo simulation, I'm basically just taking $\sin(S_{t}/50) / (iterations)$ to find the expectation, where $iterations$ are the number of iterations performed and $S_{t} = \exp((r-0.5\sigma^{2})T + \sigma T N(0,1)$ where $\sigma$ is the volatility, $T$ is the maturity, $r$ is the interest rate, and $N(0,1)$ is the Normal distribution. Could it perhaps be an issue with the Monte Carlo code?

Since $\sin$ is $2\pi$-periodic, the only sensible boundary is to take $S_T = 2\pi \cdot 50$. This artificially restricts your domain to a compact interval of spot prices, but we lose no generality since any higher spot price can be simulated by using the corresponding one in the range from $[0,S_T]$.
• So as $S_{t} \rightarrow \infty$ I should take the boundary condition $S_{T} = 2\pi*50$? I tried that just now and the plot kinda... blew up compared to the Monte Carlo plot :/ Commented Jul 16, 2016 at 2:49
• No you should take $S \to 2 \pi \cdot 50$, not to $\infty$. Commented Jul 16, 2016 at 2:59