\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small \emph{
Electronic Journal of Differential Equations}, 
Vol. 2007(2007), No. 153, pp. 1--14.\newline 
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu
or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2007 Texas State University - San Marcos.}
\vspace{8mm}}

\begin{document}
\title[\hfilneg EJDE-2007/153\hfil Construction of Green's functions]
{Construction of Green's functions for the Black-Scholes equation}

\author[M. Y. Melnikov, Y. A. Melnikov \hfil EJDE-2007/153\hfilneg]
{Max Y. Melnikov, Yuri A. Melnikov}

\address{Max Y. Melnikov \newline
Labry School of Business and Economics, 
Cumberland University,
Lebanon, TN 37087, USA}
\email{mmelnikov@cumberland.edu, Phone 615-547-1260}

\address{Yuri A. Melnikov \newline
Department of Mathematical Sciences, 
Middle Tennessee State University,
Murfreesboro, TN 37132, USA}
\email{ymelniko@mtsu.edu, Phone 615-898-2844, Fax 615-898-5422}

\thanks{Submitted August 28, 2007. Published November 14, 2007.}
\subjclass[2000]{35K20, 58J35}
\keywords{Black-Scholes equation; Green's function}

\begin{abstract}
A technique is proposed for the construction of Green's functions for
terminal-boundary value problems of the Black-Scholes equation. The
technique permits an application to a variety of problems that vary by
boundary conditions imposed. This is possible by extension of an approach
that was earlier developed for partial differential equations in applied
mechanics. The technique is based on the method of integral Laplace
transform and the method of variation of parameters. It provides closed form
analytic representations for the constructed Green's functions.
\end{abstract}

\maketitle

\numberwithin{equation}{section}

\section{Introduction}

The well-known function, in financial mathematics \cite{n1,s1,w1}, 
\begin{equation}
G(S,t;\widetilde{S})=\frac{\exp (-r(T-t))}{\widetilde{S}[2\pi \sigma
^{2}(T-t)]^{1/2}}\exp \Big(-\frac{[\ln (S/\widetilde{S})+(r-\sigma
^{2}/2)(T-t)]^{2}}{2\sigma ^{2}(T-t)}\Big)  \label{e1}
\end{equation}%
is referred to as the Green's function of the backward in time parabolic
partial differential equation 
\begin{equation}
\frac{\partial v(S,t)}{\partial t}+\frac{\sigma ^{2}S^{2}}{2}\frac{\partial
^{2}v(S,t)}{\partial S^{2}}+rS\frac{\partial v(S,t)}{\partial S}-rv(S,t)=0
\label{e2}
\end{equation}
which is called the Black-Scholes equation \cite{b1}. To be more specific
mathematically, we note that \eqref{e1} represents the Green's function for
the homogeneous terminal-boundary value problem corresponding to 
\begin{gather}
v(S,T)=f(S)  \label{e3} \\
|v(0,t)|<\infty \quad \text{and}\quad |v(\infty ,t)|<\infty .  \label{e4}
\end{gather}
This problem was posed for the Black-Scholes equation in the quarter-plane 
$\Omega =(0<S<\infty )\times (T>t>-\infty )$ of the $S,t$-plane.

In the above setting, $v=v(S,t)$ is the price of the derivative product, 
$f(S)$ is the pay-off function of a given derivative problem at the
expiration time $T$, with $S$ and $t$ being the share price of the
underlying asset and time, respectively. The parameters $\sigma $ and $r>0$
represent the volatility of the underlying asset and the risk-free interest
rate, respectively. The variable $\widetilde{S}\in (0,\infty )$ in \eqref{e1}
plays the role of a \textit{source point}.

A special comment is required as to the symbolism used in specifying the 
\textit{boundary} conditions in \eqref{e4}. Both the end-points of the
domain for the independent variable $S$ represent the so-called 
\textit{singular points} \cite{s2} to the Black-Scholes equation, in which case the
corresponding boundary conditions cannot formally assign certain values to
the solution of the governing differential equation. Instead, the conditions
in \eqref{e4} imply that the solution that we are looking for has to be
bounded as the variable $S$ approaches both zero and infinity.

The function in \eqref{e1} represents the only Green's function for %
\eqref{e2} that is available in financial mathematics for decades. This
study proposes a new approach that enables one to construct Green's
functions to the Black-Scholes equation not only for the boundary conditions
in \eqref{e4} but also for a variety of others. The approach flows out from
a technique proposed earlier \cite{m1} for boundary value problems in
applied mechanics. It is not based on the classical formalism for the
diffusion equation as in \cite{n1,s1}. Instead, the emphasis is made on the
parabolic single-parameter partial differential equation forward in time 
\begin{equation}  \label{e5}
\frac{\partial u(x,\tau )}{\partial \tau }=\frac{\partial ^{2}u(x,\tau ) }
{\partial x^{2}}+(c-1)\frac{\partial u(x,\tau )}{\partial x} -cu(x,\tau )
\end{equation}
which is traditionally obtained \cite{n1,w1} from \eqref{e2} by introducing
new independent variables 
\begin{equation}  \label{e6}
x=\ln S\quad\text{and}\quad \tau =\frac{\sigma ^{2}}{2}(T-t)
\end{equation}
and setting $u(x,\tau )=v(S,t)$.

The parameter $c$ in \eqref{e5} is defined in terms of $r$ and $\sigma ^{2}$
of the Black-Scholes equation as $c=2r/\sigma ^{2}$.

To illustrate the effectiveness of our approach, a validation example is
considered in the next section where we derive the Green's function of 
\eqref{e1}. After the approach is validated, it is used, in the following
sections, to tackle some other terminal-boundary value problems for the
Black-Scholes equation. New Green's functions are obtained none of which
have earlier been presented in literature.

\section{A validation example}

By introducing new variables $x$ and $\tau $ in compliance with the
relations in \eqref{e6}, the terminal-boundary value problem of 
\eqref{e2}-\eqref{e4} transforms to the following initial-boundary value problem 
\begin{gather}
u(x,0)=f(\exp x)  \label{e8} \\
|u(-\infty ,\tau )|<\infty ,\quad |u(\infty ,\tau )|<\infty \label{e9}
\end{gather}
for \eqref{e5} on the half-plane $(-\infty <x<\infty )\times (0<\tau <\infty
)$. Applying the Laplace transform 
\begin{equation*}
U(x;s)=L\{u(x,\tau )\}=\int_{0}^{\infty }\exp (-s\tau )u(x,\tau )d\tau
\end{equation*}
to the problem in \eqref{e5}, \eqref{e8} and \eqref{e9}, one arrives at the
boundary value problem 
\begin{gather}
\frac{d^{2}U(x;s)}{dx^{2}}+(c-1)\frac{dU(x;s)}{dx}-(s+c)U(x;s)=-f(\exp x),
\label{e10} \\
|U(-\infty ;s)|<\infty ,\quad |U(\infty ;s)|<\infty \label{e10b}
\end{gather}
for the Laplace transform $U(x;s)$ of $u(x,\tau )$. Note that \eqref{e10} is a
linear nonhomogeneous ordinary differential equation with constant
coefficients since $s$ is just a parameter and $U(x;s)$ is treated as a
single-variable function of $x$.

To find a fundamental set of solutions to the homogeneous equation
corresponding to \eqref{e10}, consider its characteristic equation 
\begin{equation*}
k^{2}+(c-1)k-(s+c)=0
\end{equation*}
whose roots are 
\begin{equation*}
k_{1}=\alpha +\omega ,\quad k_{2}=\alpha -\omega
\end{equation*}
where $\omega =(s+\beta )^{1/2}$, while the parameters $\alpha $ and $\beta $
are defined in terms of $c$ as 
\begin{equation}
\alpha =\frac{1-c}{2},\quad \beta =(\frac{1+c}{2})^{2}  \label{e11}
\end{equation}
This yields two linearly independent particular solutions to the homogeneous
equation corresponding to \eqref{e10} as 
\begin{equation*}
U_{1}(x;s)=\exp (\alpha +\omega )x,\quad U_{2}(x;s)=\exp (\alpha -\omega )x
\end{equation*}
with their linear combination 
\begin{equation}
U(x;s)=A(x;s)\exp (\alpha +\omega )x+B(x;s)\exp (\alpha -\omega )x
\label{e12}
\end{equation}
representing, according to the method of variation of parameters, the
general solution to \eqref{e10}. Following the procedure of this method, one
arrives at the well-posed system 
\begin{equation*}
\begin{pmatrix}
\exp (\alpha +\omega )x & \exp (\alpha -\omega )x \\ 
(\alpha +\omega )\exp (\alpha +\omega )x & (\alpha -\omega )\exp (\alpha
-\omega )x
\end{pmatrix}
\begin{pmatrix}
A^{\prime }(x;s) \\ 
B^{\prime }(x;s)
\end{pmatrix}
=\begin{pmatrix}
0 \\ 
-f(\exp x)
\end{pmatrix}
\end{equation*}
of linear algebraic equations in the derivatives with respect to $x$ of the
coefficients $A(x;s)$ and $B(x;s)$ of the linear combination in \eqref{e12}.
The solution of the above system is obtained as 
\begin{equation*}
A^{\prime }(x;s)=-\frac{\exp (-(\alpha +\omega )x)}{2\omega }f(\exp x),\quad
B^{\prime }(x;s)=\frac{\exp (-(\alpha -\omega )x)}{2\omega }f(\exp x)
\end{equation*}
Upon integration, the coefficients $A(x;s)$ and $B(x;s)$ are found in the
form 
\begin{gather*}
A(x;s)=-\frac{1}{2\omega }\int_{-\infty }^{x}\exp (-(\alpha +\omega )\xi
)f(\exp \xi )d\xi +M(s), \\
B(x;s)=\frac{1}{2\omega }\int_{-\infty }^{x}\exp (-(\alpha -\omega )\xi
)f(\exp \xi )d\xi +N(s)
\end{gather*}
Substitution of these in \eqref{e12} yields the general solution to \eqref{e10} in
the form 
\begin{equation}
\begin{aligned} U(x;s)&=\frac{1}{2\omega }\int_{-\infty }^{x}\exp \alpha
(x-\xi )[\exp \omega (\xi -x)-\exp \omega (x-\xi )]f(\exp \xi )d\xi \\
&\quad +M(s)\exp (\alpha +\omega )x+N(s)\exp (\alpha -\omega )x \end{aligned}
\label{e13}
\end{equation}%
The `constants of integration' $M(s)$ and $N(s)$ can be obtained upon
satisfying the boundary conditions of \eqref{e10b}. Omitting details, we have 
\begin{equation*}
N(s)=0,\quad M(s)=\frac{1}{2\omega }\int_{-\infty }^{\infty }\exp (-(\alpha
+\omega )\xi )f(\exp \xi )d\xi
\end{equation*}%
Upon substituting these in \eqref{e13}, one obtains the solution to the
boundary value problem in \eqref{e10} and \eqref{e10b} in the form 
\begin{align*}
U(x;s)& =\int_{-\infty }^{x}\frac{\exp \alpha (x-\xi )}{2\omega }[\exp
\omega (\xi -x)-\exp \omega (x-\xi )]f(\exp \xi )d\xi \\
& \quad +\int_{-\infty }^{\infty }\frac{\exp \alpha (x-\xi )}{2\omega }\exp
\omega (x-\xi )f(\exp \xi )d\xi
\end{align*}
which can be rewritten in a compact \textit{single-integral} form as 
\begin{equation}
U(x;s)=\int_{-\infty }^{\infty }\frac{\exp \alpha (x-\xi )}{2\omega }\exp
(-\omega |x-\xi |)f(\exp \xi )d\xi  \label{e14}
\end{equation}
The solution $u(x,\tau )$ to the initial-boundary value problem stated by 
\eqref{e5}, \eqref{e8} and \eqref{e9} can be obtained from $U(x;s)$ with the aid
of the inverse Laplace transform. In doing so, we keep in mind that the
parameter $\omega $ has earlier been introduced in terms of the parameter $s$
of the Laplace transform as $\omega =(s+\beta )^{1/2}$. This yields 
\begin{equation}
\begin{aligned} u(x,\tau )&=L^{-1}\{U(x;s)\} \\ 
&=\int_{-\infty }^{\infty
}\exp \alpha (x-\xi )L^{-1}\{ \frac{ \exp (-(s+\beta )^{1/2}|x-\xi
|)}{2(s+\beta )^{1/2}}\} f(\exp \xi )d\xi \\ 
&=\int_{-\infty }^{\infty
}\frac{\exp \alpha (x-\xi )\exp (-\beta \tau )}{2(\pi \tau )^{1/2}}\exp
(-\frac{(x-\xi )^{2}}{4\tau }) f(\exp \xi )d\xi \end{aligned}  \label{e15}
\end{equation}%
To obtain the solution $v(S,t)$ to the setting in \eqref{e2}-\eqref{e4}, we
make the backward substitutions in compliance with the relations of 
\eqref{e6}. This implies that the variables $x$, $\tau $ and $\xi $ ought to
be replaced with $S$, $t$ and $\widetilde{S}$ , respectively as 
\begin{equation*}
x=\ln S,\quad \tau =\frac{\sigma ^{2}}{2}(T-t),\quad \xi =\ln \widetilde{S}
\end{equation*}
The differential of the variable of integration $\xi $ in \eqref{e15}
converts to the form 
\begin{equation*}
d\xi =\frac{1}{\widetilde{S}}\,d\widetilde{S}\ 
\end{equation*}
while the interval of integration $(-\infty ,\infty )$ in \eqref{e15}
transforms, according to the relation $\xi =\ln \widetilde{S}$, to the
interval $[0,\infty )$ with respect to $\widetilde{S}$. With all this in
mind, one arrives at the solution to the terminal-boundary value problem in 
\eqref{e2}-\eqref{e4} as 
\begin{equation}
v(S,t)=\int_{0}^{\infty }\frac{1}{\sigma \widetilde{S}[2\pi (T-t)]^{1/2}}
\exp \Big(\alpha \ln (S/\widetilde{S})-\beta \frac{\sigma ^{2}}{2}(T-t)-
\frac{[\ln (S/\widetilde{S})]^{2}}{2\sigma ^{2}(T-t)}\Big)f(\widetilde{S})d
\widetilde{S}  \label{e16}
\end{equation}
revealing the Green's function to the problem in \eqref{e2}-\eqref{e4} in
the form 
\begin{equation}
G(S,t;\widetilde{S})=\frac{1}{\widetilde{S}[2\pi \sigma ^{2}(T-t)]^{1/2}}
\exp \Big(\alpha \ln (S/\widetilde{S})-\beta \frac{\sigma ^{2}}{2}(T-t)-
\frac{[\ln (S/\widetilde{S})]^{2}}{2\sigma ^{2}(T-t)}\Big)  \label{e17}
\end{equation}

It is not evident that the above representation for $G(S,t; \widetilde{S})$ 
and the one in \eqref{e1} are identical. To verify their identity, we
express $\alpha $ and $\beta $ in \eqref{e17} in terms of the original
parameters $\sigma ^{2}$ and $r$ of the Black-Scholes equation as 
\begin{equation*}
\alpha =\frac{\sigma ^{2}/2-r}{\sigma ^{2}},\quad \beta =(\frac{r+\sigma
^{2}/2}{\sigma ^{2}}) ^{2}
\end{equation*}
and then rewrite \eqref{e17} as 
\begin{equation}  \label{e18}
\begin{aligned} G(S,t;\widetilde{S}) &=\frac{1}{\widetilde{S}[2\pi \sigma
^{2}(T-t)]^{1/2}} \exp \Big(\frac{\sigma ^{2}/2-r}{\sigma ^{2}}\ln
(S/\widetilde{S})\\ &\quad -\frac{(r+\sigma ^{2}/2)^{2}}{2\sigma ^{2}}
(T-t)-\frac{[\ln (S/\widetilde{S})]^{2}}{2\sigma ^{2}(T-t)}\Big)
\end{aligned}
\end{equation}
Multiplying the above expression by the product of the two factors 
\begin{equation*}
\exp (-r(T-t))\exp r(T-t)
\end{equation*}
which is identically equal to one, we leave the first of these factors (the
negative exponent) in its current form, combine the second factor with the
existing exponential term in \eqref{e18} and rewrite subsequently the latter
as 
\begin{align*}
G(S,t;\widetilde{S}) &=\frac{\exp (-r(T-t))}{\widetilde{S}[2\pi \sigma
^{2}(T-t)]^{1/2}} \exp \Big(-\frac{r-\sigma ^{2}/2}{\sigma ^{2}}\ln (S/ 
\widetilde{S}) \\
&\quad +r(T-t)-\frac{(r+\sigma ^{2}/2)^{2}}{ 2\sigma ^{2}}(T-t)-\frac{[\ln
(S/\widetilde{S})]^{2}}{2\sigma ^{2}(T-t)}\Big)
\end{align*}
Combining the second and the third additive terms in the argument of the
extended exponential function, we reduce the latter to the form 
\begin{equation*}
\exp \Big(-\frac{r-\sigma ^{2}/2}{\sigma ^{2}}\ln (S/\widetilde{S })
-\frac{(r-\sigma ^{2}/2)^{2}}{2\sigma ^{2}}(T-t)
-\frac{ [\ln (S/\widetilde{S})]^{2}}{2\sigma ^{2}(T-t)}\Big)
\end{equation*}
which can immediately be transformed into 
\begin{equation*}
\exp \Big(-\frac{[\ln (S/\widetilde{S})]^{2}+2(r-\sigma ^{2}/2)(T-t)
\ln (S/\widetilde{S})+(r-\sigma ^{2}/2)^{2}(T-t)^{2}}{2\sigma ^{2}(T-t)}\Big)
\end{equation*}
It is evident that the numerator in the argument of this exponential
function represents a complete square, reducing the above to 
\begin{equation*}
\exp \Big(-\frac{[\ln (S/\widetilde{S}\ )+(r-\sigma ^{2}/2)(T-t)]^{2}}{
2\sigma ^{2}(T-t)}\Big)
\end{equation*}
Thus, the representation in \eqref{e17} is, indeed, identical to that of 
\eqref{e1}. This implies that the function that we came up with in 
\eqref{e17} does really represent the Green's function to the
terminal-boundary value problem in \eqref{e2}-\eqref{e4}. In other words,
our approach is proven productive, and in the next section we bring a
convincing justification of its successful applicability to other
terminal-boundary value problems for the Black-Scholes equation.

\section{Other Green's functions}

Two particular terminal-boundary value problems with different boundary
conditions imposed are considered as an illustration to the assertion made
in the previous section.

\subsection{Dirichlet boundary conditions}

As the first example, consider a terminal-boundary value problem stated for 
\eqref{e2} in the semi-infinite strip $\Omega =(S_{1}<S<S_{2})\times
(T>t>-\infty )$ of the $S,t$ -plane. Let the terminal condition be given by 
\eqref{e3}, while the Dirichlet boundary conditions 
\begin{equation}  \label{e19}
v(S_{1},t)=0, \quad v(S_{2},t)=0
\end{equation}
are imposed on the edges $S=S_{1}$ and $S=S_{2}$ of $\Omega $.

Note that the above setting for the Black-Scholes equation sounds quite
practical for the financial engineering, whereas its Green's function is not
yet available in literature.

By the transformations of \eqref{e6}, the setting in \eqref{e2}, \eqref{e3}
and \eqref{e19} converts to the following initial-boundary value problem 
\begin{gather}
u(x,0)=f(\exp x),  \label{e21} \\
u(a,\tau )=0,\quad u(b,\tau )=0 \label{e22} 
\end{gather}
for \eqref{e5} on the semi-infinite strip $(a<x<b)\times (0<\tau <\infty )$
in the $x,\tau $-plane, on which the region $\Omega $ maps by the change of
variables introduced in \eqref{e6}. The parameters $a$ and $b$ are
determined in terms of $S_{1}$ and $S_{2}$ as 
\begin{equation*}
a=\ln S_{1},\quad b=\ln S_{2}
\end{equation*}
The Laplace transform applied to the setting in \eqref{e5}, \eqref{e21} and
\eqref{e22} converts the latter into the boundary value problem 
\begin{gather}
\frac{d^{2}U(x;s)}{dx^{2}}+(c-1)\frac{dU(x;s)}{dx}-(s+c)U(x;s)=-f(\exp x),
\label{e23} \\
U(a;s)=0,\quad U(b;s)=0
\end{gather}
for the Laplace transform $U(x;s)$ of $u(x,\tau )$.

In compliance with the method of variation of parameters, the general
solution to \eqref{e23} is found, in this case, as 
\begin{equation}
\begin{aligned}
 U(x,s)&=\int_{a}^{x}\frac{\exp \alpha (x-\xi )}{2\omega
}[\exp \omega (\xi -x)-\exp \omega (x-\xi )]f(\exp \xi )d\xi \\ &\quad
+M(s)\exp (\alpha +\omega )x+N(s)\exp (\alpha -\omega )x \end{aligned}
\label{e24}
\end{equation}
where the parameter $\omega $ is defined as $\omega =(s+\beta )^{1/2}$.

Satisfying the boundary conditions of (3.5) yields the system of linear
algebraic equations 
\begin{equation*}
\begin{pmatrix}
\exp (\alpha +\omega )a & \exp (\alpha -\omega )a \\ 
\exp (\alpha +\omega )b & \exp (\alpha -\omega )b
\end{pmatrix}
\begin{pmatrix}
M(s) \\ 
N(s)
\end{pmatrix}
=\begin{pmatrix}
0 \\ 
\Psi (s)
\end{pmatrix}
\end{equation*}
in $M(s)$ and $N(s)$. Here 
\begin{equation*}
\Psi (s)=-\int_{a}^{b}\frac{1}{2\omega }[\exp (\alpha -\omega )(b-\xi )-\exp
(\alpha +\omega )(b-\xi )]f(\exp \xi )d\xi
\end{equation*}
Solving the above system, we obtain 
\begin{align*}
M(s)& =\int_{a}^{b}\frac{\exp (\alpha -\omega )a
\exp \alpha (b-\xi )}{2\omega [ \exp \omega (a-b)-\exp \omega (b-a)]} \\
& \quad \times [ \exp \omega (\xi -b)-\exp \omega (b-\xi )]f(\exp \xi
)d\xi
\end{align*}
and 
\begin{align*}
N(s)& =-\int_{a}^{b}\frac{\exp (\alpha +\omega )a\exp \alpha (b-\xi )}{%
2\omega [ \exp \omega (a-b)-\exp \omega (b-a)]} \\
& \quad \times [ \exp \omega (\xi -b)-\exp \omega (b-\xi )]f(\exp \xi
)d\xi
\end{align*}
Upon substituting these in \eqref{e24}, the latter reads 
\begin{align*}
U(x,s)& =\int_{a}^{x}\frac{\exp \alpha (x-\xi )}{2\omega }[\exp \omega (\xi
-x)-\exp \omega (x-\xi )]f(\exp \xi )d\xi \\
& \quad +\int_{a}^{b}\frac{\exp \alpha (x-\xi )[\exp \omega (x-a)-\exp
\omega (a-x)]}{2\omega [ \exp \omega (a-b)-\exp \omega (b-a)]} \\
& \quad \times [ \exp \omega (\xi -b)-\exp \omega (b-\xi )]f(\exp \xi
)d\xi
\end{align*}
which can be expressed in a \textit{single-integral} form as 
\begin{align*}
U(x;s)& =\int_{a}^{b}\frac{\exp \alpha (x-\xi )}{2\omega [ \exp \omega
(a-b)-\exp \omega (b-a)]} \\
& \quad \times \Big\{\exp \omega [ (x+\xi )-(a+b)]+\exp \omega [
(a+b)-(x+\xi )] \\
& \quad -\exp \omega [ |x-\xi |+(a-b)]-\exp \omega [
(b-a)-|x-\xi |]\Big\}f(\exp \xi )d\xi
\end{align*}
Transforming the bracket factor in the denominator as 
\begin{equation*}
\exp \omega (a-b)-\exp \omega (b-a)=-\exp \omega (b-a)[1-\exp 2\omega (a-b)]
\end{equation*}
we rewrite the above representation for $U(x;s)$ as 
\begin{equation}
\begin{aligned} U(x;s)&=-\int_{a}^{b}\frac{\exp \alpha (x-\xi )}{2\omega
\exp \omega (b-a)[ 1-\exp 2\omega (a-b)] } \\ &\quad \times \{ \exp \omega [
(x+\xi )-(a+b)]+\exp \omega [ (a+b)-(x+\xi )] \\ &\quad -\exp \omega [
|x-\xi |+(a-b)]-\exp \omega [ (b-a)-|x-\xi |]\} f(\exp \xi )d\xi
\end{aligned}  \label{e25}
\end{equation}
The inverse Laplace transform of $U(x;s)$ is problematic if the latter is
kept in its current form. Therefore, we adjust it first by representing the
factor 
\begin{equation*}
[ 1-\exp 2\omega (a-b)]^{-1}
\end{equation*}
in the integrand of \eqref{e25} as a geometric series, 
\begin{equation*}
\frac{1}{1-\exp 2\omega (a-b)}=\sum_{n=0}^{\infty }\exp 2n\omega (a-b)
\end{equation*}
whose common ratio $\exp 2\omega (a-b)$ represents a negative exponential
function ($a<b$) and is, therefore, less than one. This transforms %
\eqref{e25} to 
\begin{align*}
U(x;s)& =\int_{a}^{b}\frac{\exp \alpha (x-\xi )}{2\omega }\sum_{n=0}^{\infty
}\{\exp \omega [ |x-\xi |-2(n+1)(b-a)] \\
& \quad +\exp \omega [ 2(n+1)(a-b)-|x-\xi |]-\exp \omega [
2n(a-b)-2b+(x+\xi )] \\
& \quad -\exp \omega [ 2n(a-b)+2a-(x+\xi )]\}f(\exp \xi )d\xi
\end{align*}
and the inverse Laplace transform of the above can be accomplished in the
term-by-term manner. This yields the solution $u(x,\tau )$ to the
initial-boundary value problem in \eqref{e5}, \eqref{e21} and \eqref{e22} in the
form 
\begin{align*}
u(x,\tau )& =L^{-1}\{U(x,s)\} \\
& =\int_{a}^{b}\frac{\exp \alpha (x-\xi )\exp (-\beta \tau )}{2(\pi \tau
)^{1/2}}\sum_{n=0}^{\infty }
\Big\{\exp \Big(-\frac{[|x-\xi |+2(n+1)(a-b)]^{2}}{4\tau }\Big) \\
& \quad +\exp \Big(-\frac{[|x-\xi |-2n(a-b)]^{2}}{4\tau }\Big)-\exp 
\Big(-\frac{[2b-(x+\xi )-2n(a-b)]^{2}}{4\tau }\Big) \\
& \quad -\exp \Big(-\frac{[(x+\xi )-2a-2n(a-b)]^{2}}{4\tau }\Big)\Big\}
f(\exp \xi )d\xi
\end{align*}
which converts to a more compact form by rearranging the summation in the
above series. This yields 
\begin{align*}
u(x,\tau )& =\int_{a}^{b}\frac{\exp \alpha (x-\xi )\exp (-\beta \tau )}{
2(\pi \tau )^{1/2}}\sum_{m=-\infty }^{\infty }\Big\{\exp \Big(-\frac{[|x-\xi
|+2m(a-b)]^{2}}{4\tau }\Big) \\
& \quad -\exp \Big(-\frac{[2b-(x+\xi )-2m(a-b)]^{2}}{4\tau }\Big)\Big\}
f(\exp \xi )d\xi
\end{align*}

In compliance with the relations in \eqref{e6}, the solution $v(S,t)$ to the
setting in \eqref{e2}, \eqref{e3} and \eqref{e19} can be attained by the
backward replacement of the variables $x$, $\tau $ and $\xi $ with $S$, $t$
and $\widetilde{S}$ , respectively. Similarly to the analogous replacement
that has been performed in Section 2 (with $\alpha $ and $\beta $ replaced
with the original parameters $r$ and $\sigma ^{2}$ of the Black-Scholes
equation), we obtain $v(S,t)$ in the form 
\begin{align*}
v(S,t)& =\int_{S_{1}}^{S_{2}}\frac{\exp \big(-\frac{r-\sigma ^{2}/2}{\sigma
^{2}}\ln (S/\widetilde{S})-\frac{(r+\sigma ^{2}/2)^{2}}{2\sigma ^{2}}(T-t)
\big)}{\widetilde{S}[2\pi \sigma ^{2}(T-t)]^{1/2}} \\
& \quad \times \sum_{m=-\infty }^{\infty }\Big\{\exp \Big(-\frac{[\ln (S/
\widetilde{S})+2m\ln (S_{1}/S_{2})]^{2}}{2\sigma ^{2}(T-t)}\Big) \\
& \quad -\exp \Big(-\frac{[\ln (S_{2}^{2}/S\widetilde{S})-2m\ln
(S_{1}/S_{2})]^{2}}{2\sigma ^{2}(T-t)}\Big)\Big\}f(\widetilde{S})d
\widetilde{S}
\end{align*}
which can be transformed, by combining the logarithmic components in the
series factor. This yields 
\begin{equation}
\begin{aligned} 
v(S,t)&=\int_{S_{1}}^{S_{2}}\frac{\exp \big(-\frac{r-\sigma
^{2}/2}{\sigma ^{2}}\ln (S/\widetilde{S})-\frac{(r+\sigma
^{2}/2)^{2}}{2\sigma ^{2}}(T-t)\big) }{\widetilde{S}[2\pi \sigma
^{2}(T-t)]^{1/2}} \\ 
&\quad \times \sum_{m=-\infty }^{\infty }\Big\{ \exp
\Big(-\frac{ [\ln (SS_{1}^{2m}/\widetilde{S}S_{2}^{2m})]^{2}}{2\sigma
^{2}(T-t)} \Big) \\ &\quad -\exp \Big(-\frac{[\ln
(S_{2}^{2(m+1)}/S\widetilde{S} S_{1}^{2m})]^{2}}{2\sigma ^{2}(T-t)}\Big)
\Big\} f(\widetilde{S})d \widetilde{S} \end{aligned}  \label{e26}
\end{equation}%
Thus, the kernel in the above integral, 
\begin{equation}
\begin{aligned} 
G(S,t;\widetilde{S})&=\frac{\exp \big(-\frac{r-\sigma
^{2}/2}{\sigma ^{2}}\ln (S/\widetilde{S})-\frac{(r+\sigma
^{2}/2)^{2}}{2\sigma ^{2}}(T-t)\big) }{\widetilde{S}[2\pi \sigma
^{2}(T-t)]^{1/2}} \\ &\quad \times \sum_{m=-\infty }^{\infty }\Big\{ \exp
\Big(-\frac{ [\ln (SS_{1}^{2m}/\widetilde{S}S_{2}^{2m})]^{2}}{2\sigma
^{2}(T-t)} \Big) \\ &\quad -\exp \Big(-\frac{[\ln
(S_{2}^{2(m+1)}/S\widetilde{S} S_{1}^{2m})]^{2}}{2\sigma ^{2}(T-t)}\Big)
\Big\}, \end{aligned}  \label{e27}
\end{equation}
represents the Green's function to the setting in \eqref{e2}, \eqref{e3} and 
\eqref{e19}. The series in this representation converges at a \textit{high}
rate unless the term $(T-t)$ is very small. This implies that, in computing
values of $G(S,t;\widetilde{S})$, an accuracy level required for
applications can, in most cases, be attained by appropriately truncating the
series in (3.9) to a partial sum.

\subsection{Mixed boundary conditions}

Note that the qualitative theory of partial differential equations \cite{s2}
claims that if $G(S,t;\widetilde{S})$ is the Green's function to the setting
in, say, \eqref{e2}, \eqref{e3} and \eqref{e19}, then the solution to this
problem can be written in the integral form 
\begin{equation}  \label{e28}
v(S,t)=\int_{S_{1}}^{S_{2}}G(S,t;\widetilde{S})\ f(\widetilde{S})\, d 
\widetilde{S}
\end{equation}

This observation determined our strategy in the development of Sections 2
and 3.1 The strategy can also be applied while obtaining a Green's function
to the setting in \eqref{e2} and \eqref{e3}, with the following boundary
conditions 
\begin{equation}
|v(0,t)|<\infty ,\quad \frac{\partial v(D,t)}{\partial S}+\varrho
v(D,t)=0,\quad \varrho \geq 0  \label{e29}
\end{equation}
imposed on the boundary fragments $S=0$ and $S=D$ of the semi-infinite strip 
$\Omega =(0<S<D)\times (T>t>-\infty )$. Indeed, if we manage to find the
solution to the problem in \eqref{e2}, \eqref{e3} and \eqref{e29} in an 
\textit{integral form} like that in (3.10), then the kernel of the integral
represents the Green's function that we are looking for.

The second condition in \eqref{e29} is referred to, in mathematical physics,
as either \textit{mixed} or \textit{Robin }type. To our best knowledge,
mixed boundary conditions have never been considered yet in association with
the Black-Scholes equation. It is even unclear if such problem settings are
timely for financial engineering. But from mathematics stand-point, they do
not look unfeasible and could possibly find realistic applications in the
field of finance in years to come.

Upon introducing new variables $x$ and $\tau $ as suggested in \eqref{e6},
one converts the setting in \eqref{e2}, \eqref{e3} and \eqref{e29} to the
initial-boundary value problem 
\begin{gather}
u(x,0)=f(\exp x)  \label{e31} \\
|u(-\infty ,\tau )|<\infty ,\quad \frac{\partial u(b,\tau )}{\partial x}+
\overline{\varrho }u(b,\tau )=0
\end{gather}%
for \eqref{e5} on the quarter-plane $(-\infty <x<b)\times (0<\tau <\infty )$. 
The parameters $b$ and $\overline{\varrho }$ in \eqref{e31} are defined in
terms of the initial data in the original problem as 
\begin{equation*}
b=\ln D,\quad \overline{\varrho }=D\varrho
\end{equation*}
Applying the Laplace transform to the problem in \eqref{e5}, \eqref{e31} and
(3.13), one arrives at the boundary value problem 
\begin{equation}
|U(-\infty ;s)|<\infty ,\quad \frac{dU(b;s)}{dx}+\overline{\varrho }U(b;s)=0
\label{e32}
\end{equation}
for the equation in \eqref{e23}.

Aiming at the solution to the problem in \eqref{e2}, \eqref{e3} and 
\eqref{e29} in an integral form, we apply the method of variation of
parameters to the problem in \eqref{e23} and \eqref{e32}. This gives the general
solution of \eqref{e23} in the form 
\begin{equation}
\begin{aligned} U(x;s)&=\frac{1}{2\omega }\int_{-\infty }^{x}\exp \alpha
(x-\xi )[ \exp \omega (\xi -x)-\exp \omega (x-\xi )] f(\exp \xi )d\xi \\
&\quad +M(s)\exp (\alpha +\omega )x+N(s)\exp (\alpha -\omega )x \end{aligned}
\label{e33}
\end{equation}
To determine the functions $M(s)$ and $N(s)$, we take advantage of the
boundary conditions of \eqref{e32}. When $x$ approaches negative infinity,
the integral component in \eqref{e33} vanishes, while the $M(s)$-containing
component approaches zero. Hence, for the first condition in \eqref{e32} to
hold, $N(s)$ ought to be zero 
\begin{equation}
N(s)=0  \label{e34}
\end{equation}
because the exponential factor in the $N(s)$-containing component in 
\eqref{e33} is unbounded as $x$ approaches negative infinity.

In light of \eqref{e34}, the derivative of $U(x;s)$ reads as 
\begin{align*}
\frac{dU(x;s)}{dx}& =\frac{1}{2\omega }\int_{-\infty }^{x}[(\alpha -\omega
)\exp \omega (\xi -x)-(\alpha +\omega )\exp \omega (x-\xi )] \\
& \quad \times \exp \alpha (x-\xi )f(\exp \xi )d\xi +M(s)(\alpha +\omega
)\exp (\alpha +\omega )x
\end{align*}%
So, the second condition in \eqref{e32} yields the following equation in 
$M(s)$, 
\begin{align*}
& \frac{1}{2\omega }\int_{-\infty }^{b}[(\alpha -\omega )\exp \omega (\xi
-b)-(\alpha +\omega )\exp \omega (b-\xi )] \\
& \times \exp \alpha (b-\xi )f(\exp \xi )d\xi +M(s)(\alpha +\omega )\exp
(\alpha +\omega )b \\
& +\frac{\overline{\varrho }}{2\omega }\int_{-\infty }^{b}\exp \alpha (b-\xi
)[\exp \omega (\xi -b)-\exp \omega (b-\xi )]f(\exp \xi )d\xi \\
& +\overline{\varrho }M(s)\exp (\alpha +\omega )b=0
\end{align*}
from which $M(s)$ is found as 
\begin{equation*}
M(s)=-\frac{1}{2\omega }\int_{-\infty }^{b}[\frac{(\overline{\varrho }
+\alpha )-\omega }{(\overline{\varrho }+\alpha )+\omega }\exp \omega (\xi
-b)-\exp \omega (b-\xi )]\exp (-\alpha \xi -\omega b)f(\exp \xi )d\xi
\end{equation*}
Substituting now the above expression for $M(s)$ in \eqref{e33} and taking
into account \eqref{e34}, one obtains the solution to the boundary value
problem in \eqref{e23} and \eqref{e32} as 
\begin{align*}
U(x;s)& =\frac{1}{2\omega }\int_{-\infty }^{x}\exp \alpha (x-\xi )[\exp
\omega (\xi -x)-\exp \omega (x-\xi )]f(\exp \xi )d\xi \\
& \quad -\frac{1}{2\omega }\int_{-\infty }^{b}[\frac{(\overline{\varrho }
+\alpha )-\omega }{(\overline{\varrho }+\alpha )+\omega }\exp \omega (\xi
-b)-\exp \omega (b-\xi )] \\
& \quad \times \exp \alpha (x-\xi )\exp \omega (x-b)f(\exp \xi )d\xi
\end{align*}
To obtain the inverse Laplace transform of $U(x;s)$, $u(x,\tau
)=L^{-1}\{U(x;s)\}$ which represents the solution to the initial-boundary
value problem in \eqref{e5}, \eqref{e31} and (3.13), we simplify the above
expression for $U(x;s)$. Proceeding through a tedious but quite
straightforward algebra, one obtains a more compact form for $U(x;s)$ as 
\begin{equation}
\begin{aligned} U(x;s) &=\int_{-\infty }^{b}[ \exp (-\omega |x-\xi |)-
\frac{(\overline{\varrho }+\alpha )-\omega }{(\overline{\varrho } +\alpha
)+\omega }\exp \omega (x+\xi -2b)]\\ &\quad \times \frac{\exp \alpha (x-\xi
)}{2\omega }f(\exp \xi )d\xi \end{aligned}  \label{e35}
\end{equation}
which is not, unfortunately, convenient yet for the immediate inverse
Laplace transform. To facilitate the latter, we rewrite $U(x;s)$ in the
equivalent form 
\begin{align*}
U(x;s)& =\int_{-\infty }^{b}\Big\{\exp (-\omega |x-\xi |)-[\frac{2(\overline{%
\varrho }+\alpha )}{(\overline{\varrho }+\alpha )+\omega }-1]\exp \omega
(x+\xi -2b)\Big\} \\
& \quad \times \frac{\exp \alpha (x-\xi )}{2\omega }f(\exp \xi )d\xi
\end{align*}
or, recalling the expression for $\omega $ in terms of the parameter $s$ of
the Laplace transform, the above reads as 
\begin{equation}
\begin{aligned} U(x;s)&=\int_{-\infty }^{b}\frac{\exp \alpha (x-\xi
)}{2}\Big\{ \frac{\exp (-(s+\beta )^{1/2}|x-\xi |)}{(s+\beta )^{1/2}} \\
&\quad - [ \frac{2\Phi }{(\Phi +(s+\beta )^{1/2}) } -1] \frac{\exp ((s+\beta
)^{1/2}(x+\xi -2b)) }{ (s+\beta )^{1/2}}\Big\} f(\exp \xi )d\xi \end{aligned}
\label{e36}
\end{equation}
where we introduced, for compactness, $\Phi =\overline{\varrho }+\alpha $.

The inverse Laplace transform of $U(x;s)$ from \eqref{e36} represents the
solution $u(x,\tau )$ to the initial-boundary value problem in \eqref{e5}, 
\eqref{e31} and (3.13). It is found in the form 
\begin{equation}
\begin{aligned} u(x,\tau )
&=\int_{-\infty }^{b}\Big\{ \frac{1}{2(\pi \tau
)^{1/2}}\Big[ \exp \big(-\frac{(x-\xi )^{2}}{4\tau }\big) +\exp
\big(-\frac{(x+\xi -2b)^{2}}{4\tau }\big) \Big] \\ 
&\quad - \Phi \exp (\Phi
^{2}\tau -\Phi (x+\xi -2b))\mathop{\rm erfc}\Big( \Phi \tau
^{1/2}-\frac{x+\xi -2b}{2\tau ^{1/2}}\Big)\Big\} \\ &\quad \times \exp
\alpha (x-\xi )\exp (-\beta \tau )f(\exp \xi )d\xi \end{aligned}  \label{e37}
\end{equation}
where the $\mathop{\rm erfc}(\cdot )$ represents the \textit{complementary
error function} 
\begin{equation*}
\mathop{\rm erfc}(\varphi )=\frac{2}{\pi ^{1/2}}\int_{\varphi }^{\infty
}e^{-x^{2}}dx\,.
\end{equation*}
The solution $v(S,t)$ to the terminal-boundary value problem in \eqref{e2}, 
\eqref{e3} and \eqref{e29} can be obtained from \eqref{e37} by making the
backward substitution of the variables in compliance with the relations of 
\eqref{e6}. This implies 
\begin{align*}
v(S,t)& =\int_{0}^{D}\frac{1}{\widetilde{S}}\exp \Big(\alpha
 \ln (\frac{S}{\widetilde{S}})-\beta \frac{\sigma ^{2}}{2}(T-t)\Big) \\
& \quad \times \Big\{\frac{1}{[2\pi \sigma ^{2}(T-t)]^{1/2}}
\Big[\exp \Big(-\frac{[\ln (S/\widetilde{S})]^{2}}{2\sigma ^{2}(T-t)}\Big)
+\exp \Big(-\frac{[\ln (S\widetilde{S}/D^{2})]^{2}}{2\sigma ^{2}(T-t)}\Big)\Big] \\
& \quad -\Phi \exp (\Phi ^{2}\sigma ^{2}(T-t)/2-\Phi \ln (S\widetilde{S}%
/D^{2})) \\
& \quad \times \mathop{\rm erfc}(\frac{\Phi }{2}[2\sigma ^{2}(T-t)]^{1/2}-
\frac{\ln (S\widetilde{S}/D^{2})}{[2\sigma ^{2}(T-t)]^{1/2}})\}
f(\widetilde{S})d\widetilde{S}
\end{align*}
 From this, one arrives at a conclusion that the kernel $G(S,t;\widetilde{S})$
of the above integral represents the Green's function to the problem in 
\eqref{e2}, \eqref{e3} and \eqref{e29}. After a trivial algebra, 
$G(S,t;\widetilde{S})$ can be presented in the form 
\begin{equation}
\begin{aligned} G(S,t;\widetilde{S})&=\frac{1}{\widetilde{S}}
\big(\frac{S}{\widetilde{S} }\big) ^{\alpha }\exp \big(-\beta \frac{\sigma
^{2}}{2} (T-t)\big) \\ 
&\quad \times \Big\{ \frac{1}{[2\pi \sigma
^{2}(T-t)]^{1/2}} \Big[ \exp\Big(-\frac{[ \ln (S/\widetilde{S})]
^{2}}{2\sigma ^{2}(T-t)}\Big) +\exp \Big(-\frac{[ \ln (S
\widetilde{S}/D^{2})] ^{2}}{2\sigma ^{2}(T-t)}\Big) \Big] \\ 
&\quad -\Phi
\big(\frac{S\widetilde{S}}{D^{2}}\big) ^{-\Phi }\exp (\Phi ^{2}\sigma
^{2}(T-t)/2) \\ &\quad \times \mathop{\rm erfc}\Big(\frac{\Phi }{2}[2\sigma
^{2}(T-t)]^{1/2}-\frac{\ln (S\widetilde{S}/D^{2})}{[2\sigma
^{2}(T-t)]^{1/2}}\Big) \Big\} \end{aligned}  \label{e38}
\end{equation}%
Summarizing all the notations introduced at various stages of the present
development, we can express the parameters $\alpha $, $\beta $ and $\Phi $
in \eqref{e38} in terms of the original parameters $\sigma ^{2}$ and $r$ of
the Black-Scholes equation and the parameters $D$ and $\varrho $ as 
\begin{equation}
\alpha =\frac{\sigma ^{2}/2-r}{\sigma ^{2}},\quad \beta =\Big(\frac{r+\sigma
^{2}/2}{\sigma ^{2}}\Big)^{2},\quad \Phi =D\varrho +\alpha  \label{e39}
\end{equation}

\subsection{Particular cases}

Note that the problem statement in \eqref{e2}, \eqref{e3} and \eqref{e29}
allows two particular cases that might be of interest in option pricing
valuations. One of such cases occurs when the parameter $\varrho $ is set to
equal zero transforming the boundary conditions in \eqref{e29} into 
\begin{equation}  \label{e40}
|v(0,t)|<\infty ,\quad \frac{\partial v(D,t)}{\partial S}=0
\end{equation}
The Green's function for the problem in \eqref{e2}, \eqref{e3} and 
\eqref{e40}, 
\begin{equation}  \label{e41}
\begin{aligned}
G(S,t;\widetilde{S})&=\frac{1}{\widetilde{S}}\big(\frac{S}{\widetilde{S}
}\big) ^{\alpha }\exp \big(-\beta \frac{\sigma ^{2}}{2} (T-t)\big) \\ 
&\quad
\times \Big\{ \frac{1}{[2\pi \sigma ^{2}(T-t)]^{1/2}}\Big[ \exp
\Big(-\frac{[ \ln (S/\widetilde{S})] ^{2}}{2\sigma ^{2}(T-t)}\Big) +\exp
\Big(-\frac{[ \ln (S \widetilde{S}/D^{2})] ^{2}}{2\sigma ^{2}(T-t)}\Big)
\Big] \\ 
&\quad -\alpha (\frac{S\widetilde{S}}{D^{2}}) ^{-\alpha }\exp
(\alpha ^{2}\sigma ^{2}(T-t)/2) \\ &\quad \times \mathop{\rm
erfc}\Big(\frac{\alpha }{2}[2\sigma ^{2}(T-t)]^{1/2}-\frac{\ln
(S\widetilde{S}/D^{2})}{[2\sigma ^{2}(T-t)]^{1/2}}\Big) \Big\} 
\end{aligned}
\end{equation}
immediately arises from that of \eqref{e38} when the parameter $\Phi $ is
replaced with $\alpha $. Indeed, as it follows from \eqref{e39}, if
 $\varrho =0$, then $\Phi =\alpha $.

The second particular case of the problem statement in \eqref{e2}, \eqref{e3}
and \eqref{e29} occurs when the parameter $\varrho $ approaches infinity,
which transforms the boundary conditions in \eqref{e29} into 
\begin{equation}  \label{e42}
|v(0,t)|<\infty ,\quad v(D,t)=0\,.
\end{equation}

It is difficult to directly obtain Green's function to the terminal-boundary
value problem in \eqref{e2}, \eqref{e3} and \eqref{e42} from that of 
\eqref{e38}. The point is that taking a limit in the latter as $\varrho $
approaches infinity is not a trivial exercise. That is why an alternative
route is suggested. We revisit \eqref{e35} for $U(x;s)$ in the development
of the previous section and, observing that 
\begin{equation*}
\lim_{\varrho \to \infty }\frac{(\overline{\varrho }+\alpha )-\omega 
}{(\overline{\varrho }+\alpha )+\omega }=\lim_{\varrho \to \infty }
\frac{(D\varrho +\alpha )-\omega }{(D\varrho +\alpha )+\omega }=1
\end{equation*}
we rewrite \eqref{e35} for the setting in \eqref{e2}, \eqref{e3} and 
\eqref{e42} as 
\begin{align*}
U(x;s)&=\int_{-\infty }^{b}\frac{\exp \alpha (x-\xi )}{2\omega } [ \exp
(-\omega |x-\xi |-\exp \omega (x+\xi -2b)] f(\exp \xi )d\xi \\
&=\int_{-\infty }^{b}\frac{\exp \alpha (x-\xi )}{2}[ \frac{\exp (-(s+\beta
)^{1/2}|x-\xi |)}{(s+\beta )^{1/2}} \\
&\quad -\ \frac{\exp ((s+\beta )^{1/2}(x+\xi -2b))}{(s+\beta )^{1/2}} ]
f(\exp \xi )d\xi
\end{align*}
Taking the inverse Laplace transform of the above expression, we obtain 
\begin{align*}
u(x,\tau )&=\int_{-\infty }^{b}[ \exp (-\frac{ (x-\xi )^{2}}{4\tau }) -\exp
(-\frac{(x+\xi -2b)^{2}}{4\tau }) ] \\
&\quad \times \frac{\exp \alpha (x-\xi )\exp (-\beta \tau )}{2(\pi \tau
)^{1/2}} f(\exp \xi )d\xi
\end{align*}
allowing the solution to the problem in \eqref{e2}, \eqref{e3} and 
\eqref{e42} as 
\begin{align*}
v(S,t)&=\int_{0}^{D}\frac{1}{\widetilde{S}\big[2\pi \sigma ^{2}(T-t)\big]
^{1/2}}\exp \Big(\alpha \ln (\frac{S}{\widetilde{S}} ) -\beta \frac{\sigma
^{2}}{2}(T-t)\Big) \\
&\quad \times \Big[ \exp \Big(-\frac{[ \ln (S/\widetilde{S})] ^{2} }{2\sigma
^{2}(T-t)}\Big) -\exp \Big(-\frac{[ \ln (S \widetilde{S}/D^{2})] ^{2}}{
2\sigma ^{2}(T-t)}\Big) \Big] f(\widetilde{S})d\widetilde{S}
\end{align*}
that can easily be simplified to 
\begin{align*}
v(S,t)&=\int_{0}^{D}\frac{1}{\widetilde{S}}(\frac{S}{ \widetilde{S}})
^{\alpha }\frac{\exp (-\beta \sigma ^{2}(T-t)/2) }{[2\pi \sigma
^{2}(T-t)]^{1/2}} \\
&\quad \times \Big[ \exp \Big(-\frac{[ \ln (S/\widetilde{S})] ^{2} }{2\sigma
^{2}(T-t)}\Big) -\exp \Big(-\frac{[ \ln (S \widetilde{S}/D^{2})] ^{2}}{
2\sigma ^{2}(T-t)}\Big) \Big] f(\widetilde{S})d\widetilde{S}\ 
\end{align*}
from which it follows that 
\begin{align*}
G(S,t;\widetilde{S}\ )&=\frac{1}{\widetilde{S}}(\frac{S}{ \widetilde{S}})
^{\alpha }\frac{\exp \big(-\beta \sigma ^{2}(T-t)/2\big) }{[2\pi \sigma
^{2}(T-t)]^{1/2}} \\
&\times \Big[ \exp \Big(-\frac{[ \ln (S/\widetilde{S})] ^{2} }{2\sigma
^{2}(T-t)}\Big) -\exp \Big(-\frac{[ \ln (S\widetilde{S}/D^{2})] ^{2}}{
2\sigma ^{2}(T-t)}\Big) \Big]
\end{align*}
represents the closed form of the Green's function to the Black-Scholes
equation satisfying the boundary conditions in \eqref{e42}. Note that the
relations in \eqref{e39} bring the expressions of the parameters $\alpha $
and $\beta $ in terms of $\sigma ^{2}$and $r$.

\subsection*{Conclusion}

Compact analytic representations are derived for Green's functions for the
Black-Scholes equation. These and other Green's functions, whose compact
forms can be obtained by the approach suggested in the present study, are
easily accessible for both theoretical analysis and numerical work in the
field. They can readily be used in solving a variety of practical problem
settings in financial engineering.

\begin{thebibliography}{9}
\bibitem{b1} F. Black, M. S. Scholes; The pricing of options and corporate
liabilities. \textit{Journal of Political Economics,} \textbf{81,} (1973)
637-54.

\bibitem{m1} Y. A. Melnikov, \textit{Influence Functions and Matrices} (New
York--Basel: Marcel Dekker), 1998,

\bibitem{n1} S. N. Neftci; \textit{An Introduction to the Mathematics of
Financial Derivatives} (New York: Academic Press), 2000.

\bibitem{s1} D. Silverman; Solution of the Black-Scholes equation using the
Green's function of the diffusion equation. Manuscript. Department of
Physics and Astronomy. University of California. Irvine, 1999,

\bibitem{s2} V. I. Smirnov; \textit{A Course of Higher Mathematics }
(Oxford--New York: Pergamon Press), 1964.

\bibitem{w1} P. Wilmott, S. Howison, J. Dewynne; \textit{The Mathematics of
Financial Derivatives: A Student Introduction} (Cambridge: Cambridge
University Press), 1995.
\end{thebibliography}

\end{document}