
# 6.5: Black-Scholes Equation

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Solutions of the Black-Scholes equation define the value of a derivative, for example of a call or put option, which is based on an asset. An asset can be a stock or a derivative of it, for instance. In principle, there are infinitely many such products, for example n-th derivatives. The Black-Scholes equation for the value $$V(S,t)$$ of a derivative is

\label{BS1}
V_t+\frac{1}{2}\sigma^2 S^2V_{SS}+rSV_S-rV=0\ \ \mbox{in}\ \Omega,

where for a fixed $$T$$, $$0<T<\infty$$,

$$\Omega=\{(S,t)\in\mathbb{R}^2:\ 0<S<\infty,\ 0<t<T\},$$

and $$\sigma$$, $$r$$ are positive constants. More precisely, $$\sigma$$ is the volatility of the underlying asset $$S$$, $$r$$ is the guaranteed interest rate of a risk-free investment.

If $$S(t)$$ is the value of an asset at time $$t$$, then $$V(S(t),t)$$ is the value of the derivative at time $$t$$, where $$V(S,t)$$ is the solution of an appropriate initial-boundary value problem for the Black-Scholes equation, see below.

The Black-Scholes equation follows from Ito's Lemma under some assumptions on the random function associated to $$S(t)$$, see [26], for instance.

## Call option

Here is $$V(S,t):=C(S,t)$$, where $$C(S,t)$$ is the value of the (European) call option. In this case we have following side conditions to (\ref{BS1}):

\begin{eqnarray}
\label{BSC1}
C(S,T)&=&\max\{S-E,0\}\\
\label{BSC2}
C(0,t)&=&0\\
\label{BSC3}
C(S,t)&=&S+o(S)\ \mbox{as}\ S\to\infty, \ \mbox{uniformly in}\ t,
\end{eqnarray}

where $$E$$ and $$T$$ are positive constants, $$E$$ is the exercise price and $$T$$ the expiry.

Side condition (\ref{BSC1}) means that the value of the option has no value at time $$T$$ if $$S(T)\le E$$, condition (\ref{BSC2}) says that it makes no sense to buy assets if the value of the asset is zero, condition (\ref{BSC3}) means that we buy assets if its value becomes large, see Figure 6.5.1, where the side conditions are indicated.

Figure 6.5.1: Side conditions for a call option

Theorem 6.4 (Black-Scholes formula for European call options).
The solution $$C(S,t)$$, $$0\le S<\infty$$, $$0\le t\le T$$, of the initial-boundary value problem (\ref{BS1})-(\ref{BSC3}) is explicitly known and is given by

$$C(S,t)=SN(d_1)-Ee^{-r(T-t)}N(d_2),$$

where
\begin{eqnarray*}
N(x)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^x\ e^{-y^2/2}\ dy,\\
d_1&=&\frac{\ln(S/E)+(r+\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}},\\
d_2&=&\frac{\ln(S/E)+(r-\sigma^2/2)(T-t)}{\sigma\sqrt{T-t}}.
\end{eqnarray*}

Proof. Substitutions

$$S=Ee^x,\ \ t=T-\frac{\tau}{\sigma^2/2},\ \ C=Ev(x,\tau)$$

change equation (\ref{BS1}) to

\label{BS2}
v_\tau=v_{xx}+(k-1)v_x-kv,

where

$$k=\frac{r}{\sigma^2/2}.$$

Initial condition (\ref{BS2}) implies

\label{BS3}
v(x,0)=\max\{e^x-1,0\}.

For a solution of (\ref{BS2}) we make the ansatz

$$v=e^{\alpha x+\beta \tau}u(x,\tau),$$

where $$\alpha$$ and $$\beta$$ are constants which we will determine as follows. Inserting the ansatz into differential equation (\ref{BS2}), we get

$$\beta u+u_\tau=\alpha^2u+2\alpha u_x+u_{xx}+(k-1)(\alpha u+u_x)-ku.$$

Set $$\beta=\alpha^2+(k-1)\alpha-k$$ and choose $$\alpha$$ such that $$0=2\alpha+(k-1)$$, then $$u_\tau=u_{xx}$$. Thus

\label{BS4}
v(x,\tau)=e^{-(k-1)x/2-(k+1)^2\tau/4}u(x,\tau),

where $$u(x,\tau)$$ is a solution of the initial value problem

\begin{eqnarray*}
u_\tau&=&u_{xx},\ \ -\infty<x<\infty,\ \tau>0\\
u(x,0)&=&u_0(x),
\end{eqnarray*}

with

$$u_0(x)=\max\left\{e^{(k+1)x/2}-e^{(k-1)x/2},0\right\}.$$

A solution of this initial value problem is given by Poisson's formula

$$u(x,\tau)=\frac{1}{2\sqrt{\pi \tau}}\int_{-\infty}^{+\infty}\ u_0(s)e^{-(x-s)^2/(4\tau)}\ ds.$$

Changing variable by $$q=(s-x)/(\sqrt{2\tau})$$, we get
\begin{eqnarray*}
u(x,\tau)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}\ u_0(q\sqrt{2\tau}+x)e^{-q^2/2}\ dq\\
&=&I_1-I_2,
\end{eqnarray*}
where
\begin{eqnarray*}
I_1&=&\frac{1}{\sqrt{2\pi}}\int_{-x/(\sqrt{2\tau)}}^{\infty}\ e^{(k+1)(x+q\sqrt{2\tau})}e^{-q^2/2}\ dq\\
I_2&=&\frac{1}{\sqrt{2\pi}}\int_{-x/(\sqrt{2\tau)}}^{\infty}\ e^{(k-1)(x+q\sqrt{2\tau})}e^{-q^2/2}\ dq.
\end{eqnarray*}
An elementary calculation shows that
\begin{eqnarray*}
I_1&=&e^{(k+1)x/2+(k+1)^2\tau/4}N(d_1)\\
I_2&=&e^{(k-1)x/2+(k-1)^2\tau/4}N(d_2),
\end{eqnarray*}
where
\begin{eqnarray*}
d_1&=&\frac{x}{\sqrt{2\tau}}+\frac{1}{2}(k+1)\sqrt{2\tau}\\
d_2&=&\frac{x}{\sqrt{2\tau}}+\frac{1}{2}(k-1)\sqrt{2\tau}\\
N(d_i)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{d_i}\ e^{-s^2/2}\ ds,\ \ i=1,\ 2.
\end{eqnarray*}

Combining the formula for $$u(x,\tau)$$, definition (\ref{BS4}) of $$v(x,\tau)$$ and the previous settings $$x=\ln(S/E)$$, $$\tau=\sigma^2(T-t)/2$$ and $$C=Ev(x,\tau)$$, we get finally the formula of Theorem 6.4.

In general, the solution $$u$$ of the initial value problem for the heat equation is not uniquely defined, see for example [10], pp. 206.

Uniqueness. The uniqueness follows from the growth assumption (\ref{BSC3}). Assume there are two solutions of (\ref{BS1}), (\ref{BSC1})-(\ref{BSC3}), then the difference $$W(S,t)$$ satisfies the differential equation (\ref{BS1}) and the side conditions

$$W(S,T)=0,\ W(0,t)=0,\ W(S,t)=O(S)\ \mbox{as}\ S\to\infty$$

uniformly in $$0\le t\le T$$.

From a maximum principle consideration, see an exercise, it follows that $$|W(S,t)|\le cS$$ on $$S\ge 0$$, $$0\le t\le T$$. The constant $$c$$ is independent of $$S$$ and $$t$$. From the definition of $$u$$ we see that
$$u(x,\tau)=\frac{1}{E}e^{-\alpha x-\beta\tau} W(S,t),$$
where $$S=Ee^x$$, $$t=T-2\tau/(\sigma^2)$$. Thus we have the growth property

\label{wachs1}
|u(x,\tau)|\le Me^{a|x|},\ \ x\in\mathbb{R}^1,

with positive constants $$M$$ and $$a$$. Then the solution of $$u_\tau=u_{xx}$$, in $$-\infty<x<\infty$$,
$$0\le\tau\le\sigma^2 T/2$$, with the initial condition $$u(x,0)=0$$ is uniquely defined in the class of functions satisfying the growth condition (\ref{wachs1}), see Proposition 6.2 of this chapter.
That is, $$u(x,\tau)\equiv 0$$.

$$\Box$$

## Put option

Here is $$V(S,t):=P(S,t)$$, where $$P(S,t)$$ is the value of the (European) put option. In this case we have following side conditions to (\ref{BS1}):
\begin{eqnarray}
\label{BSP1}
P(S,T)&=&\max\{E-S,0\}\\
\label{BSP2}
P(0,t)&=&Ee^{-r(T-t)}\\
\label{BSP3}
P(S,t)&=&o(S)\ \mbox{as}\ S\to\infty,\ \mbox{uniformly in}\ 0\le t\le T.
\end{eqnarray}
Here $$E$$ is the exercise price and $$T$$ the expiry.

Side condition (\ref{BSP1}) means that the value of the option has no value at time $$T$$ if $$S(T)\ge E$$, condition (\ref{BSP2}) says that it makes no sense to sell assets if the value of the asset is zero, condition (\ref{BSP3}) means that it makes no sense to sell assets if its value becomes large.

Theorem 6.5 (Black-Scholes formula for European put options).

The solution $$P(S,t)$$, $$0<S<\infty$$, $$t<T$$ of the initial-boundary value problem (\ref{BS1}), (\ref{BSP1})-(\ref{BSP3}) is explicitly known and is given by

$$P(S,t)=Ee^{-r(T-t)}N(-d_2)-SN(-d_1)$$
where $$N(x)$$, $$d_1$$, $$d_2$$ are the same as in Theorem 6.4.

Proof. The formula for the put option follows by the same calculations as in the case of a call option or from the put-call parity

$$C(S,t)-P(S,t)=S-Ee^{-r(T-t)}$$

and from

$$N(x)+N(-x)=1.$$

Concerning the put-call parity see an exercise. See also [26], pp. 40, for a heuristic argument which leads to the formula for the put-call parity.

$$\Box$$

## Contributors

• Integrated by Justin Marshall.