# 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 $t \in [0,T]$.

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

• The explanation is on the top of Page 159. Jul 28, 2015 at 14:02
• As Shreve has said: In this case, the price of the call is almost as much as the price of the forward contract.The value of a forward contract is given by $$f(t,S_t)=S_t-e^{-r(T-t)}K$$
– user16891
Jul 28, 2015 at 18:56