# About the boundary conditions of the Black-Scholes-Merton PDE

I have a question about the solution of the Black-Scholes PDE for the European call option when I read the book Stochastic Calculus for Finance II of Steven E.Shreve.

Let $c(t,x)$ be the value of the European call option at time $t$ if the stock price at that time is $S(t)=x$. Then, $c(t,x)$ satisfies the following equation: $$c_t(t,x) + r x c_x(t,x)+ \frac{1}{2} \sigma^2 x^2 c_{xx}(t,x)=rc(t,x) \text{ for all t\in [0, T), x\geq 0},$$ and $c(T, x)=(x-K)^{+}.$

To resolve the above equation, one needs boundary conditions at $x=0$ and $x= +\infty$. For $x=0$. It's easy to derive that $c(t,0)=0$ for all $t \in [0,T]$.

However, for $x=+\infty$, I do not know understand how the author finds out (w/o a detailed explanation)(c.f. page 158 in that book) that
$$\lim_{x \rightarrow +\infty} c(t,x)- (x- e^{-r (T-t)}K) =0 \text{ for all t \in [0, T]}?$$

• Informally: the only reason the call price is more than the discounted value of a forward purchase at the strike is that there is some positive probability that the asset price will drop below the strike and you'll be spared that loss. As the asset price goes to $\infty$, however, that probability clearly goes to $0$.
– lulu
Commented Jul 28, 2015 at 13:43
• Note: In your question, the "K" in the final equation should not be in the exponent. the exponent is just a discount factor and it should multiply K.
– lulu
Commented Jul 28, 2015 at 13:47
• Thanks a lot for your answer. Commented Jul 29, 2015 at 14:31

lulu's comment sums it up nicely. Shreve mentions this is a backward parabolic PDE, and for it to be well-defined we must specify boundary conditions. From a purely mathematical viewpoint, those can be almost anything we want. The condition $$\lim_{x \rightarrow +\infty} c(t,x)- (x- e^{-r (T-t) K}) =0 \text{ for all t \in [0, T]}$$ is purely an economic argument, as pointed out by lulu.