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\section*{Key Concepts}

\begin{enumerate}
  \item The Black-Scholes equation for the value of a European
	    call option on a security can be solved by judicious
	    changes of variables that reduce the equation to the heat
	    equation, for which we can write down the solution.
  \item An example of how to choose and execute changes of
	    variables to solve a partial differential equation.
\end{enumerate}

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\section*{Vocabulary}
\begin{enumerate}
  \item A differential equation with auxiliary
	initial conditions and boundary conditions, that is an initial
	value problem, is said to be
	\textbf{well-posed} if the solution exists, is unique, and small
	changes in the equation parameters, the initial conditions or
	the boundary conditions produce only small changes in the
	solutions.
\end{enumerate}

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\section*{Mathematical Ideas}

\subsection*{Solution of the Black-Scholes Equation }

This discussion is drawn from Section 4.2, pages 59-63; Section
4.3, pages 66-69; Section 5.3, pages 75-76; and Section 5.4, pages 77
- 81 of \emph{The Mathematics of Financial Derivatives: A Student
Introduction} by P. Wilmott, S. Howison, J. Dewynne, Cambridge
University Press, Cambridge, 1995.
Some ideas are also taken from Chapter 11 of \emph{Stochastic
           Calculus and Financial Applications}  by
           J. Michael Steele,  Springer, New York, 2001.


We have to start somewhere, and to avoid the problem of deriving
everything back to calculus, we will assert that the
\emph{initial value problem for the heat equation on the infinite
interval is well-posed}, that is, 
the solution to the partial differential equation
$$
	\frac{\partial u}{\partial \tau} = \frac{\partial^2 u}{\partial
	x^2} \qquad -\infty < x < \infty, \quad \tau > 0
$$
with
$$
	u(x,0) = u_0(x)
$$
where the initial condition and the solution satisfy the following technical
requirements: 
\begin{enumerate}
   \item $u_0(x)$ has no more than a finite number of discontinuities
   of the jump kind,
   \item $\lim_{|x| \to \infty} u_0(x)  e^{-a x^2}  = 0$ for any $a >0$,
   \item $\lim_{|x| \to \infty} u(x, \tau)  e^{-a x^2}  = 0$ for any $a >0$
\end{enumerate}
exists for all time, is unique, and most importantly, can be
represented as
$$
	u(x, \tau) = \frac{1}{2\sqrt{\pi \tau}} \int_{-\infty}^{\infty}
	u_0(s) e^{-(x-s)^2/4\tau} \; ds
$$

\emph{Remark:}  This solution can derived in several different ways,
I think the easiest way is to use Fourier transforms.  The derivation
of this solution representation is standard in any course or book on
partial differential equations.

\emph{Remark:} Mathematically, the
 conditions above are unnecessarily restrictive, and can be
 considerably weakened, however, they will be more than sufficient for
 all practical situations we encounter in mathematical finance.

\emph{Remark:} The use of $\tau$ for the time variable (instead of the
 more natural $t$) is to avoid a
 conflict of notation in the several changes of variables we will soon
 have to make.

The Black-Scholes terminal value problem for the value $V(S,t)$ of a 
European call option on a security with price $S$ at time $t$ is
$$
	\frac{\partial V}{\partial t} + \frac{1}{2}\sigma^2 S^2
	\frac{\partial^2 V}{\partial S^2} + r S \frac{\partial
	V}{\partial S} - r V = 0
$$
with $V(0,t) = 0$, $V(S,t) \sim S$ as $S \to \infty$ and 
$$
	V(S,T) = \max( S- K, 0).
$$
Note that this looks a little like the heat equation on the infinite
interval in that it has a first derivative of the unknown with respect
to time and the second derivative of the unknown with respect to the
other (space) variable, but notice:
\begin{enumerate}
\item Each time the unknown is differentiated with respect to $S$, it
also multiplied by the independent variable 
$S$, so the equation is not a constant coefficient equation.
\item There is a first derivative of $V$ with respect to $S$ in the equation.
\item There is a zero-th order term $V$ in the equation.
\item The sign on the second derivative is the opposite of the heat
equation form, so the equation is of backward parabolic form.
\item The data of the problem is given at the final time $T$ instead
of the initial time $0$, consistent with the backward parabolic form of
the equation.
\item There is a \emph{boundary condition} $V(0,t) = 0$ specifying the
value of the solution at one sensible boundary of the problem.  The
boundary is sensible since security values must only be zero or positive.  This
boundary condition says that any time the security value is 0, then the
call value (with strike price $K$) is also worth 0.  
\item There is another boundary condition $V(S,t) \sim S$, as $S \to
\infty$, but although this is financially 
sensible, (it says that for very large security prices, the call value
with strike price $K$ is approximately $S$) it is more in the nature of a
technical condition, and we will ignore it without consequence.
\end{enumerate}

We will eliminate each and every one of these objections with a
suitable change of variables.  The plan is to change variables to
reduce the Black-Scholes terminal value problem to the heat equation,
and then to use the known solution of the heat equation to represent
the solution, and change variables back.  This is a standard technique
of solution in partial differential equations, and none of the
transformations we are making are strange, unmotivated, or unknown.

First we will take $t = T - \frac{\tau}{ (1/2)\sigma^2}$ and $S = K e^x$,
and we will set
$$
	V(S,t) = K v(x,\tau).
$$
Remember, $\sigma$ is the volatility, $r$ is the interest rate on a
risk-free bond, and $K$ is the strike price.  In the changes of
variables above, the choice for $t$ reverses the sense of time,
changing the problem from backward parabolic to forward parabolic.
The choice for $S$ is a well-known transformation based on experience
with the Euler equidimensional equation in differential equations.  
In addition, the variables
have been carefully scaled so as to make the transformed equation
expressed in dimensionless quantities.  All of these techniques are
standard and are covered in most courses and books on partial
differential equations and applied mathematics.

Some extremely wise advice adapted from
page 186 of \emph{Stochastic Calculus and Financial Applications} by
J. Michael Steele.  Springer, New York, 2001 is appropriate here.
\begin{quotation}
``There is nothing particularly difficult about changing variables and
transforming one equation to another, but there is an element of
tedium and complexity that slows us down.  There is no universal
remedy for this molasses effect, but the calculations do seem to go
more quickly if one follows a well-defined plan.  If we know that
$V(S,t)$ satisfies an equation (like the Black-Scholes equation) we
are guaranteed that we can make good use of the equation in the
derivation of the equation for a new function $v(x, \tau)$ defined in
terms of the old if we write the old $V$ as a function of the new $v$
and write the new $\tau$ and $x$ as functions of the old $t$ and $S$.
This order of things puts everything in the direct line of fire of the
chain rule; the partial derivatives $V_t$, $V_S$ and $V_{SS}$ are easy
to compute and at the end, the original equation stands ready for
immediate use.''
\end{quotation}

Following our advice, we first have to write
$$
	\tau = (1/2)\sigma^2( T - t)
$$
and
$$
	x = \log(S/K)
$$
so
$$
	\frac{\partial V}{\partial t} = K \frac{\partial v}{\partial
	\tau} \cdot \frac{ d\tau}{dt} = K \frac{\partial v}{\partial
	\tau} \cdot \frac{-\sigma^2}{2}
$$
and
$$
	\frac{\partial V}{\partial S} = K \frac{\partial v}{\partial
	x} \cdot \frac{dx}{dS} 
        =  K \frac{\partial v}{\partial x}\cdot\frac{1}{S}
$$
and
\begin{eqnarray*}
  \frac{\partial^2 V}{\partial S^2} 
  &=& \frac{\partial}{\partial S} \left( \frac{\partial V}{\partial S}   \right) \\
  &=& \frac{\partial}{\partial S}  \left( K \frac{\partial v}{\partial x} \frac{1}{S}
  \right) \\
  &=& K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} 
      + K \frac{\partial}{\partial S}  \left( \frac{\partial v}{\partial x} \right)
  \cdot \frac{1}{S} \\
  &=& K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} 
      +  K \frac{\partial}{\partial x}  \left( \frac{\partial v}{\partial x} \right)
                \cdot \frac{dx}{dS} \cdot \frac{1}{S} \\
  &=&  K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} + K \frac{\partial^2 v}{\partial x^2}
  \cdot \frac{1}{S^2}   
\end{eqnarray*}
and the terminal condition
$$
	 V(S,T) = \max(S-K, 0) = \max(K e^x - K, 0) 
$$
but $V(S,T) = K v(x,0)$ so $v(x,0) = \max( e^x - 1, 0)$.

Now substitute all of the derivatives into the Black-Scholes equation to obtain:
$$
 K \frac{\partial v}{\partial
	\tau} \cdot \frac{-\sigma^2}{2} + \frac{\sigma^2}{2} S^2 \left( 
      K \frac{\partial v}{\partial x} \cdot \frac{-1}{S^2} + K \frac{\partial^2 v}{\partial x^2}
  \cdot \frac{1}{S^2} \right) + r S \left(  K \frac{\partial
 v}{\partial x} \cdot \frac{1}{S} \right) - r K v
	= 0
$$
Now the simplification begins:
\begin{enumerate}
  \item Factor the common factor $K$ and cancel.
  \item Transpose the $\tau$-derivative to the other side, and divide through by
  $\sigma^2/2$
  \item Rename the remaining constant $r/(\sigma^2/2)$ as $k$. 
  $k$ measures the ratio
  between the risk-free interest rate and the volatility.
  \item Cancel the $S^2$ terms in the second derivative.
  \item Cancel the $S$ terms in the first derivative.
  \item Gather up like order terms.
\end{enumerate}

What remains is the rescaled, constant coefficient equation:
$$
	\frac{\partial v}{\partial \tau} = \frac{\partial^2 v}{\partial x^2} 
       + (k-1) \frac{\partial v}{\partial x} - k v.
$$
We have made considerable progress, because
\begin{enumerate}
   \item Now there is only one dimensionless parameter $k$  measuring the risk-free
   interest rate as a multiple of the volatility  and another (more or less
   hidden) rescaled time to expiry $(1/2) \sigma^2 T$ , not the original $4$ dimensioned quantities, namely $K$,
   $T$, $\sigma^2$ and $r$.
   \item The equation is defined on the interval $-\infty < x < \infty$, since this
   $x$-interval defines $ 0 < S < \infty$ through the change of variables $S = K e^x$.
   \item The equation is now constant coefficient.  In principle, we could solve the
   equation directly now,
\end{enumerate}

Instead, we will change the dependent variable scale yet again, by
$$
	v = e^{\alpha x + \beta \tau} u(x,\tau).
$$
where $\alpha$ and $\beta$ are yet to be determined.  Now using the product rule:
$$
	v_\tau = \beta  e^{\alpha x + \beta \tau} u(x,\tau) +  e^{\alpha x + \beta \tau}
	u_\tau
$$
and 
$$
	v_x = \alpha  e^{\alpha x + \beta \tau} u(x,\tau) +  e^{\alpha x + \beta \tau}
	u_x
$$
and
$$
	v_{xx} = \alpha^2  e^{\alpha x + \beta \tau} u(x,\tau) + 2\alpha e^{\alpha x + \beta \tau}
	u_x +  e^{\alpha x + \beta \tau} u_{xx}
$$
Put these into our constant coefficient partial differential equation, cancel the common
factor of $e^{\alpha x + \beta \tau}$ throughout and obtain:
$$
	\beta u + u_\tau = \alpha^2 u + 2\alpha u_x + u_{xx} + (k - 1) (\alpha u + u_x) - k
	u
$$
Now gather like terms:
$$
	u_\tau = u_{xx} + [ 2\alpha + (k-1)] u_x + [ \alpha^2 + (k-1) \alpha - k - \beta] u.
$$
Then choose $\alpha = -(k-1)/2$ so that the $u_x$ coefficient is $0$, and then choose
$\beta =   \alpha^2 + (k-1) \alpha - k = -(k+1)^2/4$ so the $u$ coefficient is likewise
$0$.  With this choice, the equation is reduced to
$$
	u_\tau = u_{xx}.
$$
We also need to transform the initial condition too.  This would be
\begin{eqnarray*}
    u(x,0) &=& e^{-(-(k-1)/2)x - (-(k+1)^2/4)0} v(x,0) \\
           &=& e^{((k-1)/2)x} \max( e^x - 1, 0) \\
           &=&  \max( e^{((k+1)/2)x} - e^{((k-1)/2)x}, 0 ).
\end{eqnarray*}
For future reference, we notice that this function is strictly positive when 
the argument $x$ is strictly positive, that is $u_0(x) > 0$ when $x > 0$, otherwise,
$u_0(x) = 0$ for $x \le 0$.

Now we are in the final stage, since we are ready to apply the solution representation
formula:
$$
	u(x, \tau) = \frac{1}{2\sqrt{\pi \tau}} \int_{-\infty}^{\infty}
	u_0(s) e^{-(x-s)^2/4\tau} \; ds.
$$
However, first we want to make a change of variable in the integration, by taking $z =
(s-x)/\sqrt{2\tau}$, (and thereby $dz = (-1/\sqrt{2\tau}) \; dx$)  so that the integration becomes:
$$
       u(x,\tau) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} u_0(z \sqrt{2\tau} + x)
       e^{-z^2/2} \; dz
$$
We may as well only integrate over the domain where $u_0 > 0$, that is for $z >
-x/\sqrt{2\tau}$.  On that domain, $u_0 = e^{((k+1)/2)(x + z\sqrt{2\tau})} -
e^{((k-1)/2)(x + z\sqrt{2\tau}) }$ so we are down to:
$$
\frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^{\infty}  e^{\frac{k+1}{2}(x + z\sqrt{2\tau})}
       e^{-z^2/2} \; dz - \frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^{\infty}  e^{\frac{k-1}{2}(x + z\sqrt{2\tau})}
       e^{-z^2/2} \; dz 
$$
Call the two integrals $I_1$ and $I_2$ respectively.

We will evaluate $I_1$ ( the one with the $k+1$ term) 
first.  This is easy, just completing the
square in the exponent yields a standard, tabulated integral.  The
exponent is
\begin{eqnarray*}
	((k+1)/2)(x + z\sqrt{2\tau}) - z^2/2 &=& 
            (-1/2)( z^2 -\sqrt{2\tau}(k+1)z) + ((k+1)/2) x \\
        & = & (-1/2)(z^2 - \sqrt{2\tau}(k+1)z + \tau(k+1)^2/2) + ((k+1)/2)x + \tau(k+1)^2/4 \\
        & = & (-1/2)(z - \sqrt{\tau/2}(k+1))^2 + (k+1)x/2 +\tau (k+1)^2/4
\end{eqnarray*}
Therefore
\[
    \frac{1}{\sqrt{2\pi}} \int_{-x/\sqrt{2\tau}}^{\infty}
    e^{\frac{k+1}{2} (x + z\sqrt{2\tau})}  e^{-z^2/2} \; dz =
    \frac{ e^{(k+1)x/2 +\tau (k+1)^2/4}}{\sqrt{2\pi}} 
    \int_{-x/\sqrt{2\tau}}^\infty e^{\frac{-1}{2} (z - \sqrt{\tau/2}(k+1))^2}
    \; dz
\]
Now, change variables again on the integral, choosing $y = z -
\sqrt{\tau/2}(k+1)$ so $dy = dz$, and all we need to change are the
limits of integration:
\[
	\frac{ e^{  (k+1)x/2 +\tau (k+1)^2/4} }{ \sqrt{2\pi}} 
    \int_{-x/\sqrt{2\tau} - \sqrt{\tau/2}(k+1)}^\infty e^{(-1/2)y^2}
    \; dz
\]
The integral can be represented in terms of the cumulative
distribution function of a  normal random variable, typically denoted
$\Phi$.  That is, $\Phi(d) = (1/\sqrt{2\pi})\int_{-\infty}^{d}
e^{-y^2/2} \; dy$ so
\[
	I_1 =  e^{  (k+1)x/2 +\tau (k+1)^2/4} \Phi(d_1)
\]
where $d_1 = x/\sqrt{2\tau} + \sqrt{\tau/2}(k+1)$ (note the use of the
symmetry of the integral!) The calculation of $I_2$ is identical,
except that $(k+1)$ is replaced by $(k-1)$ throughout.  

That, is the solution of the heat equation is 
\[
	u(x,\tau) =  e^{  (k+1)x/2 +\tau (k+1)^2/4} \Phi(d_1) + 
                       e^{  (k-1)x/2 +\tau (k-1)^2/4} \Phi(d_2)
\]

Now, we must systematically unwind each of the changes of variables,
from $u$, first $v(x,\tau) = e^{(-1/2)(k-1)x - (1/4)(k+1)^2 \tau}
u(x,\tau)$.  (Notice how the exponentials neatly combine and cancel!) 
Then put $x = \log(S/K)$, $\tau = (1/2)\sigma^2(T-t)$ and $V(S,t) =
Kv(x,\tau)$.  

The final solution is the Black-Scholes formula for the value of a
European call option at time $T$ with strike price $K$, if the current
time is $t$ and the underlying security price is $S$, the risk-free
interest rate is $r$ and the volatility is $\sigma^2$:
\[
	V(S,t) = S \Phi\left( \frac{\log(S/K) + ( r +
	\sigma^2/2)(T-t)}{\sigma \sqrt{T-t}} \right) 
         - K e^{-r(T-t)} \Phi\left( \frac{\log(S/K) + ( r -
	\sigma^2/2)(T-t)}{\sigma \sqrt{T-t}} \right)
\]

Note, usually one doesn't see the solution as this full closed form
solution.  Instead, most versions of the solution write intermediate
steps in small pieces, and then present the solution as an algorithm
putting the pieces together to obtain the final answer.

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\section*{Problems to Work for Understanding}
\begin{enumerate}
  \item  Explicitly evaluate the integral $I_2$ in terms of the
  c.d.f. $\Phi$ and other elementary functions as was done for the
  integral $I_1$.

  \HCode{<a href="problem1.shtml">} Solution \HCode{</a>}

  \item  What is the price of a European call option on a
  non-dividend-paying stock when the stock price is \$52, the strike
  price is \$50, the risk-free interest rate is 12\% per annum
  (compounded continuously), the volatility is 30\% per annum, and the
  time to maturity is 3 months?
  %% adapted from Hull, Problem 10.21, page 241.

  \HCode{<a href="problem2.shtml">} Solution \HCode{</a>}

  \item  What is the price of a European call option on a non-dividend
  paying stock when the stock price is \$30, the exercise price is
  \$29, the risk-free interest rate is 5\%, the volatility is 25\% per
  annum, and the time to maturity is 4 months?
  % adapted from Hull, Problem 10.23 (a) page 241.

  \HCode{<a href="problem3.shtml">} Solution \HCode{</a>}

  \item Show that the Black-Scholes formula for the price of a call
  option tends to $\max( S-K, 0)$ as $ t \to T$.

  \HCode{<a href="problem4.shtml">} Solution \HCode{</a>}

\end{enumerate}

\HCode{<hr>}

\section*{Reading Suggestion:}
  \begin{enumerate}
    \item Section 4.2, pages 59-63; Section
	4.3, pages 66-69; Section 5.3, pages 75-76; and Section 5.4, pages 77
	- 81 of \emph{ The Mathematics of Financial Derivatives: A Student
	Introduction }  by P. Wilmott, S. Howison, J. Dewynne, Cambridge
	University Press, Cambridge, 1995.
    \item 	 Chapter 11 of \emph{ Stochastic
           Calculus and Financial Applications }  by
           J. Michael Steele.  Springer, New York, 2001.
  \end{enumerate}

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\section*{Outside Readings and Links:}
\begin{enumerate}
  \item \Link{http://www.cs.cornell.edu/Info/Courses/Spring-98/CS522/content/lecture2.math.pdf}{}
      Cornell University , Department of Computer Science, 
      Prof. T. Coleman Rhodes and Prof. R. Jarrow,
 \EndLink Numerical Solution of Black-Scholes Equation, Submitted by Chun Fan, 
Nov. 12, 2002.
  \item  \Link{http://http://www.maths.monash.edu.au/mth3251/Lectures/NumericalSol.pdf}{}   Monash University , Department of Mathematical Science, 
      Eric. W. Chu \EndLink
 Applied mathematicians transform the Black-Scholes Equation into a 
simpler forward form and solve it.It gives some examples and maple commands,
      Submitted by Chun Fan, Nov. 12, 2002.
  \item  \Link{http://www.blobek.com/black-scholes.html}{}\EndLink
      </a>  An applet for calculating the option value based on Black-Scholes 
model. Also contains tips on options, business news and literature on options.
 Submitted by Yogesh Makkar, November 19, 2003.

  \item \Link{www.xleverywhere.com/samples/bs/bs.htm}{}
      ExcelEverywhere, a commercial application for spreadsheets on
      the Web \EndLink
      A sample spreadsheet based calculator for calculating the option values, 
based on Black-Scholes model.
 Submitted by Yogesh Makkar, November 19,2003
\end{enumerate}

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