\documentclass[twoside]{article}
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\markboth{\hfil Noncommutative operational calculus \hfil}%
{\hfil Henry E. Heatherly \&  Jason P. Huffman \hfil}
\begin{document} \setcounter{page}{11}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc 15th Annual Conference of Applied Mathematics, Univ. of Central Oklahoma},
\newline
Electronic Journal of Differential Equations, Conference~02, 1999, pp. 11--18. 
\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp  ejde.math.swt.edu (login: ftp)}
 \vspace{\bigskipamount} \\
  Noncommutative operational calculus 
\thanks{ {\em 1991 Mathematics Subject Classifications:} 
44A40, 45D05, 34A12, 16S60.
\hfil\break\indent
{\em Key words and phrases:} convolution, Mikusi\'{n}ski, Volterra integral equations, 
\hfil\break\indent
operational calculus, linear operators.
\hfil\break\indent
\copyright 1999 Southwest Texas State University  and University of
North Texas. \hfil\break\indent
Published November 24, 1999.} }

\date{}
\author{Henry E. Heatherly \&  Jason P. Huffman}
\maketitle

\begin{abstract} 
Oliver Heaviside's operational calculus was placed on a rigorous mathematical
basis by Jan Mikusi\'{n}ski, who constructed an algebraic setting for the
operational methods.  In this paper, we generalize Mikusi\'{n}ski's methods to
solve linear ordinary differential equations in which the unknown is a matrix-
or linear operator-valued function.  Because these functions can be
zero-divisors and do not necessarily commute, Mikusi\'{n}ski's one-dimensional
calculus cannot be used.  The noncommuative operational calculus developed here,
however, is used to solve a wide class of such equations.  In addition, we
provide new proofs of existence and uniqueness theorems for certain matrix- and
operator valued Volterra integral and integro-differential equations.  Several
examples are given which demonstrate these new methods.
\end{abstract}

\newtheorem{example}{Example}[section]
\newtheorem{proposition}[example]{Proposition}
\newtheorem{lemma}[example]{Lemma}

\section{Introduction}
Let $\mathfrak{M}$ be the linear space of all continuous, complex-valued 
functions defined on $[0, \infty)$. Taken with the Duhamel convolution 
operation, $f*g(t)=\int_0^t f(t-\tau) g(\tau) d\tau$, $\mathfrak{M}$ is a 
commutative, associative algebra
over $\mathbb{C}$. (We use $\mathbb{C}$ for the field of complex numbers and
$\mathbb{N}$ for the set of natural numbers.)
To put Heaviside's operational calculus on a rigorous mathematical
basis, Mikusi\'{n}ski considered the quotient field of $\mathfrak{M}$ 
consisting of all the (equivalence classes of) 
fractions $\frac{f}{g}$, where $f,g \in \mathfrak{M}$ and $g \neq 0$,
and denoted here by $Q(\mathfrak{M})$.
These fractions, which can be thought of as generalized functions, 
are called Mikusi\'{n}ski operators and are the basis for the operational calculus.
In the field $Q(\mathfrak{M})$ there exists an integral operator, i.e. the Heaviside
unit function $H(t) \equiv 1$ for all $t$, and a differential operator, $s=\frac{\delta}
{H}$, where $\delta$ is the unity in $Q(\mathfrak{M})$. Thus, for a function $f \in
\mathfrak{M}$, $H*f=\int_0^tf(\tau)d\tau$, and if $f$ 
is continuously differentiable, then $s*f=f'+f(0)$. (Note that  
$s*a$ is well-defined for all $a \in Q(\mathfrak{M})$, but the resulting
product may not be a continuous function.) 
Using the last equation above, Mikusi\'{n}ski developed algebraic expressions
for the $n$-th derivatives of a function, which allowed the transformation
of certain differential equations into algebraic equations. In this paper,
we expand on Mikusi\'{n}ski's methods to solve linear ordinary differential
equations in which the coefficients and the unknown function are 
matrix- or operator-valued.

We use $M_n[X]$ to denote the $n$-by-$n$ matrices over a set $X$.
Consider a matrix-valued function $F:[0, \infty) \rightarrow M_n[\mathbb{C}]$, 
continuous in each entry. It is easy
to see that we may identify $F$ with a matrix of complex-valued functions, so that
$F=[f_{ij}]$, where $f_{ij} \in \mathfrak{M}$ for all $i,j$, and then
$F(t)=[f_{ij}(t)]$, for all $t$. Thus, we consider the linear space
of all such functions (for a fixed $n$), denoted $M_n[\mathfrak{M}]$,
and define the convolution of two matrix-valued functions as follows:
\[
F*G(t)=\int_0^t F(t-\tau)G(\tau)d\tau=[f_{ij}][g_{ij}]
\]
where $F=[f_{ij}],G=[g_{ij}]$ and 
the right hand side denotes matrix multiplication with juxtaposition
in each entry taken as the Duhamel convolution. Thus, $M_n[\mathfrak{M}]$ is an 
associative $\mathbb{C}$-algebra. Two difficulties arise in $M_n[\mathfrak{M}]$
which are not present in $\mathfrak{M}$: the functions in $M_n[\mathfrak{M}]$
do not necessarily commute with each other, and there exist nonzero 
\textit{zero-divisors}, i.e. nonzero elements whose product is zero.
However, we are able to overcome these difficulties to develop a 
noncommutative operational calculus which
generalizes Mikusi\'{n}ski's methods and is used to solve a 
broad class of equations.


\section{A Matrix Operational Calculus}

It is easy to see that the algebra $M_n[\mathfrak{M}]$ embeds as a sub-algebra of
the $n$-by-$n$ matrices over the Mikusi\'{n}ski operators, $M_n[Q(\mathfrak{M})]$.
Because of the two limitations mentioned above, there is no field of fractions for
the algebra $M_n[Q(\mathfrak{M})]$.
However, a well-behaved subset of $M_n[Q(\mathfrak{M})]$ can be used to
construct a ring of fractions in a similar way. Let $\Delta_n$ be the set of all
matrices of operators whose entries along the main diagonal are all nonzero and 
all other entries are zero. We use the notation 
$\operatorname{Diag}(a_1, a_2, \dots , a_n)$ for such
a matrix. Let $R=M_n[Q(\mathfrak{M})]$. Then because $\Delta_n$ forms a
\textit{denominator set} for $R$, we form, in the standard way, the ring of quotients
$R \Delta_n^{-1}$ which consists of all fractions $\frac{a}{b}$ where $a \in R, b \in
\Delta_n$ and $b \neq 0$. (For more on general rings of quotients, see \cite[pp.50-61]{ste}.)

In this quotient ring $R \Delta_n^{-1}$ there exists an integral operator for 
matrix-valued functions, i.e. $H_n=\operatorname{Diag}(H,H,\dots,H)$, and a 
differential operator $S_n=\operatorname{Diag}
\left( s,s,...,s \right)$. (Recall that $s=\frac{\delta}{H}$.)
Thus, for $F \in
M_n[\mathfrak{M}]$, $H_n*F=\int_0^tF(\tau) d\tau$, and if $F$
is continuously differentiable, then
\[
S_n*F=F'+F(0).
\]
Using this last equation repeatedly, we develop algebraic expressions for any $n$-th
derivative of a function in $M_n[\mathfrak{M}]$ (assuming the derivative exists).
This enables us to solve differential
equations in an algebraic setting, as the following examples illustrate. 
(Note: we use $\alpha$ to denote the scalar multiple $\alpha \cdot \delta_n$
of the unity element $\delta_n \in R \Delta_n^{-1}$.) 

\begin{example}     \label{mat}
Let $\alpha =\left[ 
\begin{array}{ll}
\alpha _1 & \alpha _2 \\ 
\alpha _3 & \alpha _4
\end{array}
\right] $ and $\beta =\left[ 
\begin{array}{ll}
\beta _1 & \beta _2 \\ 
\beta _3 & \beta _4
\end{array}
\right] ,$ and $A=\left[ 
\begin{array}{ll}
a_1 & 0 \\ 
0 & a_2
\end{array}
\right] ,$ where $\alpha _i,\beta _i,a_i\in \mathbb{C}.$ We solve the initial
value problem $X^{\prime \prime }+AX=0;$ $X(0)=\alpha ,$ $X^{\prime
}(0)=\beta ,$ where $X\in M_2[\mathfrak{M}].$
\end{example}
\noindent {\bf Solution:} Using the operational calculus of this section, we
make the substitution $X^{\prime \prime }=S_2^2X-S_2X(0)-X^{\prime
}(0)=S_2^2X-S_2\alpha -\beta $. We
rewrite the equation as $(S_2^2+A)X=S_2\alpha +\beta $, or $%
X=(S_2^2+A)^{-1}(S_2\alpha +\beta ).$ Now, 
\begin{eqnarray*}
X &=&\left[ 
\begin{array}{ll}
\frac{\delta }{H^2}+a_1 & 0 \\ 
0 & \frac{\delta }{H^2}+a_2
\end{array}
\right] ^{-1}\left[ 
\begin{array}{ll}
\frac{\alpha _1}H+\beta _1 & \frac{\alpha _2}H+\beta _2 \\ 
\frac{\alpha _3}H+\beta _3 & \frac{\alpha _4}H+\beta _4
\end{array}
\right] \\
&=&\left[ 
\begin{array}{ll}
\frac{H^2}{\delta +a_1H^2} & 0 \\ 
0 & \frac{H^2}{\delta +a_2H^2}
\end{array}
\right] \left[ 
\begin{array}{ll}
\frac{\alpha _1}H+\beta _1 & \frac{\alpha _2}H+\beta _2 \\ 
\frac{\alpha _3}H+\beta _3 & \frac{\alpha _4}H+\beta _4
\end{array}
\right] \\
&=&\left[ 
\begin{array}{ll}
\frac{s\alpha _1+\beta _1}{s^2+a_1} & \frac{s\alpha _2+\beta _2}{s^2+a_1} \\ 
\frac{s\alpha _3+\beta _3}{s^2+a_2} & \frac{s\alpha _4+\beta _4}{s^2+a_2}
\end{array}
\right] \\
&=&\left[ 
\begin{array}{ll}
\alpha _1\cos \sqrt{a_1}t+\beta _1\sin \sqrt{a_1}t & \alpha _2\cos \sqrt{a_1}%
t+\beta _2\sin \sqrt{a_1}t \\ 
\alpha _3\cos \sqrt{a_2}t+\beta _3\sin \sqrt{a_2}t & \alpha _4\cos \sqrt{a_2}%
t+\beta _4\sin \sqrt{a_2}t
\end{array}
\right] .
\end{eqnarray*}
$\square$
 
This last step is obtained by identifying the rational expressions of $s$ in terms
of continuous functions of $t$. This process is similar to that used by
Mikusi\'{n}ski in the one-dimensional operational calculus. For more
on this, see \cite[pp.30-40]{mik}).

Generalizing this method from the two-dimensional case to the 
$n$-dimensional case, this matrix operational
calculus is well-suited to solve linear, matrix-valued ordinary differential
equations with coefficients which are diagonal scalar matrices. To expand
this method further, we broaden the class of coefficient matrices. Recall
that a matrix $A$ is said to be {\it diagonalizable} (by a similarity
transformation) if there exists an invertible matrix $P$ such that $P^{-1}AP$
is a diagonal matrix. 
\begin{example}     \label{diag}
Let $\alpha =\left[ 
\begin{array}{ll}
\alpha _1 & \alpha _2 \\ 
\alpha _3 & \alpha _4
\end{array}
\right] $ and $\beta =\left[ 
\begin{array}{ll}
\beta _1 & \beta _2 \\ 
\beta _3 & \beta _4
\end{array}
\right] ,$ and $A=\left[ 
\begin{array}{ll}
a_1 & a_2 \\ 
a_3 & a_4
\end{array}
\right] ,$ where $\alpha _i,\beta _i,a_i\in \mathbb{C}$, and $A$ is
diagonalizable. We solve the initial value problem $X^{\prime \prime }+AX=F;$
$X(0)=\alpha ,$ $X^{\prime }(0)=\beta ,$ where $X,F\in M_2[\mathfrak{M}].$
\end{example}
\noindent {\bf Solution:} There is an invertible matrix $P$ such that $%
P^{-1}AP=D,$ where $D \in \Delta_n$. Then, letting $Y=P^{-1}XP,$ we
rewrite the equation as follows: 
\[
(PYP^{-1})^{\prime \prime }+A(PYP^{-1})=F. 
\]
Bringing the derivatives inside the coefficient matrices and
substituting for $A,$ we have $PY^{\prime \prime }P^{-1}+PDP^{-1}PYP^{-1}=F.$
Multiply appropriately to get $Y^{\prime \prime }+DY=P^{-1}FP$.  We 
solve this initial value problem (with the appropriately modified
initial conditions) by following Example \ref{mat}. Thus, the unique
solution is simply $X=PYP^{-1},$ where $Y$ is the solution to the latter
initial value problem. \hfill $\square$ \medskip


Again, generalizing this method to $n$ dimensions,  
the matrix operational calculus is suitable for linear matrix-valued O.D.E.'s in which 
the coefficients are diagonalizable matrices. 
The Mikusi\'{n}ski one-dimensional operational
calculus is not particularly well-suited to handle linear O.D.E.'s with
variable coefficients. One reason for this ``defect'' is that the Duhamel
convolution operation does not agree well with pointwise multiplication of
functions. Because the matrix operational calculus presented here is a
direct generalization of Mikusi\'{n}ski's methods, a similar limitation
occurs.


\section{Volterra Integral and Integro-Differential Equations}

In this section, we give new
proofs of existence and uniqueness theorems for matrix-valued linear
Volterra integral and integro-differential equations. 
These existence and uniquess theorems are known, e.g. see
\cite[p.42]{gri}. However, the proofs provided here
offer the results immediately and easily in the algebraic setting, and
we avoid using any iterative methods.

Some brief comments on the background algebraic ideas helpful at this
point. Let $A$ be a linear associative algebra over $\mathbb{C}$.
An element $y \in A$ is said to be \textit{quasi-regular}
if there exists $\hat{y} \in A$ such that $y+\hat{y}+y\hat{y}=0$. 
The element $\hat{y}$ is uniquely determined by $y$ 
and is called the \textit{quasi-inverse} of $y$. If every element in $A$
is quasi-regular we write $\mathfrak{J}(A)=A$ and call $A$ a 
\textit{Jacobson radical algebra}.
It is well known that if $\mathfrak{J}(A)=A$, then 
$\mathfrak{J}(M_n[A])=M_n[\mathfrak{J}(A)]$, \cite[p.140]{sza}.
Highly pertinent to the development herein is that
$\mathfrak{M}$ is a Jacobson radical algebra, (for a proof, 
see \cite[p.195]{huf}).
Consequently, $\mathfrak{J}(M_n[ \mathfrak{M} ])= M_n[ \mathfrak{M} ]$. 


\begin{proposition}    \label{volt}
Let $K,F\in M_n[\mathfrak{M}].$ Then the matrix-valued integral equation 
\[
X+\int_0^tK(t-\tau )X(\tau )d\tau =F 
\]
has a unique, continuous solution $X \in M_n[\mathfrak{M}]$.
\end{proposition}
 
\begin{proof}
Since $M_n[ \mathfrak{M} ]$ is a Jacobson radical algebra, there
is a unique $\hat{K} \in M_n[ \mathfrak{M}]$ such that $K+\hat{K}=
-K*\hat{K}$. Then a routine calculation shows that $X+K*X=F$, when
$X=F+F*\hat{K}$, yielding the desired solution to the integral equation.
The uniqueness of $\hat{K}$ guarantees the uniqueness of this solution.
\end{proof}

It is worth noting that in the above proof $\hat{K}$ plays the role of the
\textit{resolvent} function in the theory of integral equations, 
\cite[p.44]{gri}.
\begin{proposition}
Let $K,F\in M_n[\mathfrak{M}]$ and let $A\in M_n[\mathbb{C}]$.
Then the matrix-valued integro-differential equation 
\begin{eqnarray*}
&X^{\prime }+AX+\int_0^tK(t-\tau )X(\tau )d\tau =F& \\
&X(0) =X_0&
\end{eqnarray*}
has a unique, continuous solution $X \in M_n[ \mathfrak{M} ]$.
\end{proposition}

\begin{proof}
Using $X^{\prime }=S_n*X-X(0)$, and working  
in the quotient ring $Q(M_n[ \mathfrak{M} ])$, we can rewrite the 
integro-differential equation as 
\[
S_n*X-X_0+AX+K*X=F. 
\]
Multiplying both sides of the equation by $H_n$ we have
\[
X-H_nX_0+H_n*AX+H_n*K*X=H_n*F. 
\]
Since $H_n$ is in the center of $M_n[\mathfrak{M}]$,
then $H_nX_0=X_0H_n$ and $%
H_n*AX=AH_n*X.$ Hence, the equation becomes $X+AH_n*X+H_n*K*X=H_n*F+X_0H_n$,
and then 
\[
X+[AH_n+H_n*K]*X=[H_n*F+X_0H_n]. 
\]
By Proposition \ref{volt} this last equation has a unique solution in 
$M_n[ \mathfrak{M} ],$ which is given by 
\[
X=P+P*\hat{Q},
\]
where $P=[H_n*F+X_0H_n]$ and $Q=[AH_n+H_n*K]$. 
\end{proof}


\section{Equations with Operator-Valued Functions}

In this section we consider equations whose coefficients and unknowns
can be bounded linear operators on a
separable Hilbert space.  We will make use of the well known
fact that if $\Omega$ is a separable Hilbert space with an orthonormal basis 
$\{e_k\}_{k=1}^\infty$, then a linear operator $A$, defined everywhere on 
$\Omega$, is bounded if and only if there exists a (unique) representation of $A$
as an infinite matrix $[\alpha _{ij}]_{i,j=1}^\infty $ with respect to the
basis $\{e_k\}.$ (For a proof of this, see \cite[p.49]{akh}.)
Thus, for any such bounded linear operator, we have the representation 
\[
A=\left[ 
\begin{array}{lll}
a_{11} & a_{21} & \cdots \\ 
a_{21} & a_{22} &  \\ 
\vdots &  & \ddots
\end{array}
\right] =[a_{ij}]. 
\]
Since $\sum_{i=1}^\infty \left|
a_{ik}\right| ^2<\infty ,$ for all $k\in \mathbb{N}$, matrix
multiplication is well defined, (i.e. the pertinent series all will
converge). 
With matrix multiplication, pointwise addition, and multiplication by a
complex scalar, the collection of all such bounded linear operators, 
here denoted $B_\infty$, is a $\mathbb{C}$-algebra.  To develop an 
operational calculus, we again identify the elements of our space with
equivalent elements of a more amenable space, (as with the matrix-valued 
functions in Section 2).  Here, we
identify a bounded linear operator $A=[\alpha _{ij}]$ with
the infinite matrix $[f_{ij}]$ of one-dimensional functions, $f_{ij} \in \mathfrak{M}$ 
and $f_{ij}(t)=\alpha _{ij}$. 
The collection of all such (countably) infinite matrices of functions, 
which is strictly larger than $B_\infty$, will be denoted $M_\infty [\mathfrak{M}]$.
Next, we embed the set $M_\infty [%
\mathfrak{M}]$ into the set of all infinite matrices of
Mikusi\'{n}ski operators, i.e. $M_\infty [Q(\mathfrak{M})]$.
It is clear that, with pointwise addition, $M_\infty
[\mathfrak{M}]$ embeds into $M_\infty [Q( \mathfrak{M})] $ as an 
abelian group. A fundamental
difficulty here is that, unlike $B_\infty$,
the sets $M_\infty [\mathfrak{M}]$ and $M_\infty [Q( \mathfrak{M})] $ are not
$\mathbb{C}$-algebras using matrix multiplication. This is because each
entry in a product matrix $C=AB$ is now an infinite sum of continuous
functions or Mikusi\'{n}ski operators. 
The convergence of these sums is necessary for a well-defined 
multiplication, and it is not difficult to find examples for which a sum of
functions or operators does not converge, 
\cite[p.372]{mik}.
We circumvent this difficulty by considering $M_\infty [Q( \mathfrak{M})] $ not as
an algebra over the complex numbers, but rather as a module over a 
well-behaved subset of $M_\infty [Q( \mathfrak{M})] $. In this case, the 
ring of scalars will be the
subset of diagonal matrices of operators, i.e.
$\Delta_\infty =\{\operatorname{Diag}(\frac{f_1}{g_1}, \frac{f_2}{g_2}, \frac{f_3}{g_3},...) :f_{i},g_{i}\in 
\mathfrak{M}$, $f_{i}\neq 0,g_{i}\neq 0\}\subset M_\infty [Q( \mathfrak{M})]$. 
It is clear that matrix multiplication is well defined in 
$\Delta_\infty $, for if $A,B\in \Delta_\infty ,$ then $AB=\operatorname{Diag}( a_1b_1, a_2b_2, a_3b_3,...)$.
Thus $(\Delta_\infty ,+,\cdot )$ is a commutative ring. Observe that the
mapping from $\Delta_\infty \times M_\infty [Q( \mathfrak{M})] \rightarrow 
M_\infty [Q( \mathfrak{M})] $ defined via 
\[
\operatorname{Diag}(\alpha_1, \alpha_2, \alpha_3,...)\cdot[f_{ii}]=\left[ 
\begin{array}{lll}
\alpha _{1}f_{11} & \alpha _{1}f_{12} & \cdots \\ 
\alpha _{2}f_{21} & \alpha _{2}f_{22} &  \\ 
\vdots &  & \ddots
\end{array}
\right]  
\]
is a well defined scalar multiplication, considering $M_\infty [Q( \mathfrak{M})] $
as a (left) $\Delta_\infty $-module. 

Note that infinite-dimensional integral and differential operators exist in 
$M_\infty [Q( \mathfrak{M})]$ and are denoted $H_\infty=\operatorname{Diag}(H,H,H,...)$ and 
$S_\infty=\operatorname{Diag}(s,s,s,...)$, respectively. 
Next, we give an example.
\begin{example}
Let $A$ and $B$ be bounded linear operators on a separable Hilbert space $\Omega$
such that $A\in \Delta_\infty .$ Then, with respect to an orthonormal basis $%
\{e_k\},$ we have the matrix representations $A=\operatorname{Diag}(a_1, a_2, a_3,...),B=(b_{ij}).$ We
solve the initial value problem $X^{\prime \prime }+AX=B;$ $X(0)=\alpha
,X^{\prime }(0)=\beta ,$ where $X$ is a bounded linear operator on $\Omega$ and 
$\alpha $ and $\beta $ are (infinite) coefficient matrices.
\end{example}
\noindent {\bf Solution:} Using the operational calculus, we make the
substitution $X^{\prime \prime }=S_\infty ^2X-S_\infty X(0)-X^{\prime
}(0)=S_\infty ^2X-S_\infty \alpha -\beta .$ Then we rewrite the equation as $%
(S_\infty ^2+A)X=S_\infty \alpha +\beta ,$ or $X=(S_\infty
^2+A)^{-1}(S_\infty \alpha +\beta ).$ We easily calculate $(S_\infty
^2+A)^{-1}$ in $M_\infty [Q( \mathfrak{M})] $ via 
\begin{eqnarray*}
(S_\infty ^2+A)^{-1}&=&\operatorname{Diag}(s^2+a_1, s^2+a_2, s^2+a_3,...)^{-1}\\
&=& \operatorname{Diag}\left( \frac \delta {s^2+a_1}, \frac \delta {s^2+a_2},  
\frac \delta {s^2+a_3},...  \right). 
\end{eqnarray*}
This gives the explicit solution 
\[
X=(S_\infty ^2+A)^{-1}(S_\infty \alpha +\beta )=\left[ 
\begin{array}{lll}
\frac{s\alpha _{11}+\beta _{11}}{s^2+a_{1}} & \frac{s\alpha _{12}+\beta _{12}%
}{s^2+a_{1}} & \cdots \\ 
\frac{s\alpha _{21}+\beta _{21}}{s^2+a_{2}} & \frac{s\alpha _{22}+\beta _{22}%
}{s^2+a_{2}} &  \\ 
\vdots &  & \ddots
\end{array}
\right] . 
\]
Finally, in a manner similar to the finite case, we identify 
this solution back in $M_\infty [\mathfrak{M}]$, where the entries
are all continuous functions. \hfill$\square$ \medskip

We solve similarly for linear ordinary differential equations in which the 
coefficient matrices are elements of $\Delta_\infty$.
Next, following the general methods of Section 3, we
give existence and uniqueness theorems for linear operator-valued
Volterra integral and integro-differential equations.
\begin{proposition}      \label{voltop}
Let $\Omega$ be a separable Hilbert space with orthonormal basis 
$\{e_k\}.$ If $A$ and $B$ are bounded linear operators defined everywhere on 
$\Omega$ such that $A \in \Delta_\infty$, then the Volterra integral
equation 
\[
X+\int_0^tA(t-\tau )X(\tau )d\tau =B 
\]
has a unique, bounded solution.
\end{proposition}

\begin{proof}
Observe that the bounded linear operator $X=B+B*\hat{A}$, where $\hat{A}$
denotes the quasi-inverse (i.e. the resolvent) of $A$, is the solution to the 
integral equation. We demonstrate that 
this operator $\hat{A}$ exists in $M_\infty [Q( \mathfrak{M})]$. Consider 
$A=\operatorname{Diag}(a_1,a_2,a_3,...) \in \Delta_\infty$. Routine calculation shows that the operator 
$\operatorname{Diag}( \hat{a_1}, \hat{a_2}, \hat{a_3},...)$, 
i.e. the infinite diagonal matrix each of whose entries is a
quasi-inverse of the appropriate entry in $A$, is the quasi-inverse
of $A.$ Thus we have that $\hat{A}= \operatorname{Diag}(\hat{a_1}, \hat{a_2}, \hat{a_3},...) 
\in M_\infty [Q( \mathfrak{M})]$. Hence, $B+B*\hat{A}=X\in M_\infty 
[Q( \mathfrak{M})]$. That the solution $X$ is unique follows from the
uniqueness of quasi-inverses.
\end{proof}
\begin{proposition}
Let $\Omega$ be a separable Hilbert space with orthonormal basis $\{e_k\}.$ If $A,B$, 
and $C$ are bounded linear operators defined everywhere on $\Omega$ such
that $A \in \Delta_\infty$, then the Volterra
integro-differential equation 
\begin{eqnarray*}
&X^{\prime }+BX+\int_0^tA(t-\tau )X(\tau )d\tau =C &\\
&X(0) = X_0&
\end{eqnarray*}
has a unique, bounded solution.
\end{proposition}

\begin{proof}
Using the operational calculus, we rewrite the equation as $S_\infty
X-X_0+BX+A*X=C.$ Multiplying both sides by the infinite-dimensional integral
operator $H_\infty$ yields $X-H_\infty X_0+H_\infty *BX+H_\infty
*A*X=H_\infty *C.$ Since $A$ and $H_\infty$ are in $\Delta_\infty $, 
we have $H_\infty X_0=X_0H_\infty $ and 
$H_\infty *BX=BH_\infty *X$. Thus, we rewrite the equation as $X+[BH_\infty
+H_\infty *A]*X=[H_\infty *C+X_0H_\infty ].$ Therefore, by Proposition \ref
{voltop}, there is a unique, bounded solution to the integro-differential 
equation. Letting $F=[H_\infty *C+X_0H_\infty ]$ and $K=
[BH_\infty+H_\infty *A]$, this solution is given by 
\[
X=F+F*\hat{K}.
\]
\end{proof}


\begin{thebibliography}{9}

\bibitem[Akh]{akh}
Akheizer, N.I., and I.M. Glazman, \textit{Theory of
Linear Operators in Hilbert Space}, Frederick Ungar, (New York, 1961).

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Gripenberg, G., S.O. Londen, and O. Staffans, 
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\end{thebibliography}  \bigskip

\noindent{\sc Henry E. Heatherly} \\
Department of Mathematics \\
University of  Louisiana, Lafayette \\
Lafayette, LA  70504, USA \\
e-mail: heh5820@usl.edu 
\medskip 

\noindent{\sc Jason P. Huffman} \\
Department of Mathematical, Computing, and Information Sciences \\
Jacksonville State University \\
Jacksonville, AL  36265, USA \\ 
e-mail: jhuffman@jsucc.jsu.edu 
\end{document}