# 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. – Gordon Jul 28 '15 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 '15 at 18:56

In words the equation says that when the stock price is very high, the value of the call is (approximately) equal to the stock price minus the PV of the exercise. It is fairly intuitive: if K=100 and x is 1000, the stock is so much above K that exercise is for all practical purposes certain; the call today is worth x minus PV(k), since you can set aside PV(K) today today to exercise at maturity. So it is an arbitrage argument.

This equation is sometimes called the deep-in-the-money bound for the value of a European call.